transcendental number theory: schanuel's conjecture
TRANSCRIPT
Heidelberg Colloquium organized by the Institute for Mathematics
httpwwwmathiuni-heidelbergdevortraghtmls=1 January 22 2009
Transcendental Number TheorySchanuelrsquos Conjecture
Michel Waldschmidt
Institut de Mathematiques de Jussieu amp Paris VIhttpwwwmathjussieufrsimmiw
1 59
Abstract
One of the main open problems in transcendental numbertheory is Schanuelrsquos Conjecture which was stated in the1960rsquos
If x1 xn are Qndashlinearly independent complexnumbers then among the 2n numbers x1 xnex1 exn at least n are algebraically independent
We first give a list of consequences of this statement next wedescribe the state of the art by giving special cases of theconjecture which have been proved and finally we introduce apromising approach which has been initiated in 1999 byD Roy
2 59
Algebraic and transcendental numbersAlgebraic numbers Roots of polynomials with rationalcoefficients Algebraic closure of Q in C
Q sub Q sub C
Transcendental numbers Complex numbers that are notalgebraic
The set of algebraic numbers behaves well with respect toaddition multiplication division It is a fieldThe set of transcendental numbers is the complement in thefield C of the field Q The sum of transcendental numbersmay be rational algebraic or transcendental The same for theproductHowever the sum of a transcendental number and an algebraicnumber is transcendental and the product of a transcendentalnumber and a nonndashzero algebraic number is transcendental
3 59
The exponential function
For z isin C
ez = exp(z) = 1 + z +z2
2+z3
6+ middot middot middot =
sumnge0
zn
nmiddot
Addition Theorem ez1+z2 = ez1ez2 e2iπ = 1
exp C minusrarr Ctimes ker(exp) = 2iπZ C2iπZ Ctimes
Differential equation
d
dzez = ez
4 59
First examples of transcendental numbers
sumnge1 2minusn Liouville 1844
e Hermite 1872
π Lindemann 1881
5 59
Ch Hermite ndash F Lindemann ndash C Weierstraszlig
log 2 eradic
2 Theorem of HermitendashLindemann
eradic
2 +radic
3eradic
6 LindemannndashWeierstraszlig 1888
6 59
Question of Euler and Hilbert
Leonhard Euler (1707 ndash 1783)Introductio in analysin infinitorum (1737)
Transcendence of 2radic
2
David Hilbert (1862 - 1943)ICM 1900 7ndashth Problem
Transcendence of log a log 2
7 59
AO Gelrsquofond ndash Th Schneider ndash A Baker
eπ Gelrsquofond 1929
2radic
2 log 2 log 3 Gelrsquofond and Schneider 1934
log 2 +radic
3 log 3 eradic
22radic
35radic
7 Baker 1968
8 59
Some open problems
For each of the following numbers it is expected that it istranscendental but it is not even known whether it is rationalor not
e+ π eπ πe ee ee2
eee
ππ ππ2
πππ
log π log(log 2) π log 2 (log 2)(log 3) 2log 2 (log 2)log 3
In other words we do not know whether a degree 1 polynomialcould vanish at the corresponding point but we expect thatno nonndashzero polynomial of any degree vanishes at this point
9 59
Algebraic independence
Algebraicity and transcendence deal with a single complexnumber and one variable polynomialsAlgebraic dependence or independence is the same but fortuples and multivariate polynomials
Let K sup k be an extension of fields and (θ1 θm) be amndashtuple of elements in K We say that θ1 θm arealgebraically dependent over k if there exists a nonndashzeropolynomial f isin k[X1 Xm] which vanishes at the point(θ1 θm) isin KmOtherwise we say that θ1 θm are algebraically independentover k
10 59
Special cases
In the case m = 1 to say that θ1 is algebraically independentover k just means that it is transcendental over k
Dealing with complex numbers K = C the words algebraictranscendental algebraically dependent algebraicallyindependent refer to the case k = Q
11 59
Examples
The numbersradic
2 and π are algebraically dependent The polynomial X2 minus 2 isin Z[X Y ] vanishes at (
radic2 π)
The numbers π andradicπ2 + 1 are algebraically dependent (and
both are transcendental numbers) The polynomial Y minusX2 minus 1 vanishes at (π
radicπ2 + 1)
The two numbers e and eradic
2 are algebraically independentwhich means that for any nonndashzero polynomial f isin Z[X Y ]
with rational integer coefficients the number f(e eradic
2) is notzeroSpecial case of the Lindemann-Weierstraszlig Theorem
12 59
Transcendence degree
Let Kk be a field extension The maximal number ofelements in K which are algebraically independent over k iscalled the transcendence degree of K over k and denotedtr degkK
Let t = tr degkK A subset θ1 θt of K with t elementswhich are algebraically independent is called a transcendencebasis of K over k It is the same as a maximal subset ofkndashalgebraically independent elements in KHence K is an algebraic extension of k(θ1 θt)
13 59
Transcendence degree of extensions
Assumek sub K sub L
The union of a transcendence basis of K over k and of atranscendence basis of L over K produces a transcendencebasis of L over kHence
tr degkL = tr degkK + tr degKL
An algebraic extension Kk is an extension of transcendencedegree 0 This means that there is no transcendental elementin K over k (any element in K is algebraic over k)
14 59
Algebraic independence of complex numbers
For complex numbers algebraic independence over Q or overQ is the same In particular if θ1 θm are algebraicallyindependent complex numbers then for any non-constantpolynomial f with algebraic coefficients the numberf(θ1 θm) is transcendental
For instance the two numbers e and eradic
2 are algebraicallyindependent As a consequence for any nonndashconstantpolynomial f isin Q[X Y ] with algebraic coefficients the
number f(e eradic
2) is transcendental
15 59
Schanuelrsquos Conjecture
Let x1 xn be Q-linearly independent complex numbersThen at least n of the 2n numbers x1 xn e
x1 exn arealgebraically independent
Since there are 2n numbers only the transcendence degreeover Q of the field
Q(x1 xn e
x1 exn)
is at most 2n The conjecture is that this transcendencedegree is always ge n
n le tr degQQ(x1 xn e
x1 exn)le 2n
16 59
Origin of Schanuelrsquos ConjectureCourse given by Serge Lang (1927ndash2005) at Yale in the 60rsquos
S Lang ndash Introduction to transcendental numbersAddison-Wesley 1966
also attended by M Nagata (1927ndash2008)(14th Problem of Hilbert)
Nagatarsquos Conjecture solved by E Bombieri
17 59
Schanuelrsquos Conjecture
Let x1 xn be Q-linearly independent complex numbersThen
tr degQQ(x1 xn e
x1 exn)ge n
Remark For almost all tuples (with respect to the Lebesguemeasure) the transcendence degree is 2n
18 59
AO Gelrsquofond CRAS 1934
19 59
Statement by Gelrsquofond (1934)
Let β1 βn be Q-linearly independent algebraic numbersand let logα1 logαm be Q-linearly independentlogarithms of algebraic numbers Then the numbers
eβ1 eβn logα1 logαm
are algebraically independent over Q
20 59
Further statement by Gelrsquofond
Let β1 βn be algebraic numbers with β1 6= 0 and letα1 αm be algebraic numbers with α1 6= 0 1 α2 6= 0 1αi 6= 0 Then the numbers
eβ1eβ2e βnminus1e
βn
and αα αm
21
are transcendental and there is no nontrivial algebraic relationbetween such numbers
Remark The condition on α2 should be that it is irrational
21 59
Easy consequence of Schanuelrsquos Conjecture
According to Schanuelrsquos Conjecture the following numbers arealgebraically independent
e+ π eπ πe ee ee2
eee
ππ ππ2
πππ
log π log(log 2) π log 2 (log 2)(log 3) 2log 2 (log 2)log 3
Proof This is an easy exercise
22 59
Langrsquos exercise
Define E0 = Q Inductivelyfor n ge 1 define En as thealgebraic closure of the fieldgenerated over Enminus1 by thenumbers exp(x) = ex wherex ranges over Enminus1 Let E bethe union of En n ge 0Then Schanuelrsquos Conjectureimplies that the number πdoes not belong to E
More precisely Schanuelrsquos Conjecture implies that thenumbers π log π log log π log log log π are algebraicallyindependent over E
23 59
A variant of Langrsquos exercise
Define L0 = Q Inductively for n ge 1 define Ln as thealgebraic closure of the field generated over Lnminus1 by thenumbers y where y ranges over the set of complex numberssuch that ey isin Lnminus1 Let L be the union of Ln n ge 0Then Schanuelrsquos Conjecture implies that the number e doesnot belong to L
More precisely Schanuelrsquos Conjecture implies that the
numbers e ee eee ee
ee
are algebraically independent over L
24 59
Arizona Winter School AWS2008 Tucson
Theorem [Jonathan Bober Chuangxun Cheng Brian DietelMathilde Herblot Jingjing Huang Holly Krieger DiegoMarques Jonathan Mason Martin Mereb and Robert Wilson]Schanuelrsquos Conjecture implies that the fields E and L arelinearly disjoint over Q
Definition Given a field extension FK and twosubextensions F1 F2 sube F we say F1 F2 are linearly disjointover K when the following holds Any set x1 xn sube F1
of Kndash linearly independent elements is linearly independentover F2
Reference arXiv08043550 [mathNT] 2008
25 59
Formal analogs
WD Brownawell(was a student of Schanuel)
J Axrsquos Theorem (1968) Version of SchanuelrsquosConjecture for power seriesover C(and R Coleman for powerseries over Q)
Work by WD Brownawelland K Kubota on the ellipticanalog of Axrsquos Theorem
26 59
Conjectures by A Grothendieck and Y Andre
Generalized Conjecture onPeriods by Grothendieck Dimension of theMumfordndashTate group of asmooth projective variety
Generalization by Y Andre tomotives
Case of 1ndashmotives Elliptico-Toric Conjecture ofC Bertolin
27 59
Ubiquity of Schanuelrsquos Conjecture
Other contexts pndashadic numbers Leopoldtrsquos Conjecture onthe pndashadic rank of the units of an algebraic number fieldNon-vanishing of RegulatorsNonndashdegenerescence of heightsConjecture of B Mazur on rational pointsDiophantine approximation on tori
Dipendra Prasad Gopal Prasad
28 59
Methods from logicEhud Hrushovski Boris Zilber Jonathan Kirby
Calculus of rdquopredimension functionsrdquo (E Hrushovski)
Zilberrsquos construction of a rdquopseudoexponentiationrdquo
Also A Macintyre DE Marker G Terzo AJ WilkieD Bertrand
29 59
Methods from logic Model theory
Exponential algebraicity in exponential fieldsby
Jonathan Kirby
The dimension of the exponential algebraic closure operator inan exponential field satisfies a weak Schanuel property
A corollary is that there are at most countably many essentialcounterexamples to Schanuelrsquos conjecture
arXiv 08104285v2
30 59
Known
LindemannndashWeierstraszlig Theorem = case where x1 xn arealgebraic
Let β1 βn be algebraic numbers which are linearlyindependent over Q Then the numbers eβ1 eβn arealgebraically independent over Q
31 59
Problem of Gelrsquofond and SchneiderRaised by AO Gelrsquofond in 1948 and Th Schneider in 1952
Conjecture If α is an algebraic number α 6= 0 α 6= 1 and ifβ is an irrational algebraic number of degree d then the dminus 1numbers
αβ αβ2
αβdminus1
are algebraically independent
Special case of Schanuelrsquos Conjecture Take xi = βiminus1 logαn = d so that
x1 xn e
x1 exn
islogα β logα βdminus1 logα α αβ αβ
dminus1
The conclusion of Schanuelrsquos Conjecture is
tr degQQ(logα αβ αβ
2
αβdminus1)
= d
32 59
Algebraic independence method of Gelrsquofond
AO Gelrsquofond (1948)
The two numbers 23radic2 and
23radic4 are algebraically
independentMore generally if α is analgebraic number α 6= 0α 6= 1 and if β is a algebraicnumber of degree d ge 3 thentwo at least of the numbers
αβ αβ2
αβdminus1
are algebraically independent
33 59
Tools
Transcendence criterion Replaces Liouvillersquos inequality intranscendence proofsLiouville A nonndashzero rational integer n isin Z satisfies |n| ge 1Gelrsquofond Needs to give a lower bound for |P (θ)| withP isin Z[X] 0 when θ is transcendental
Zero estimate for exponential polynomials C Hermite P Turan K Mahler R Tijdeman
Small transcendence degree AO Gelrsquofond AA Smelev R Tijdeman WD Brownawell
34 59
Large transcendence degree
GV Chudnovsky (1976)Among the numbers
αβ αβ2
αβdminus1
at least [log2 d] arealgebraically independent
GV Chudnovsky ndash On the path to Schanuelrsquosconjecture Algebraic curves close to a pointI General theory of colored sequencesII Fields of finite transcendence type and coloredsequences ResultantsStudia Sci Math Hungar 12 (1977) 125ndash157 (1980)
35 59
Partial result on the problem of Gelrsquofond and
Schneider
AO Gelrsquofond GV Chudnovskii P Philippon YuV Nesterenko
G Diaz If α is an algebraic number α 6= 0 α 6= 1 and if β isan irrational algebraic number of degree d then
tr degQQ(αβ αβ
2
αβdminus1) ge [d+ 1
2
]
36 59
Conjecture of algebraic independence of logarithms
of algebraic numbers
Denote by L the set of complex numbers λ for which eλ isalgebraic
L = logα α isin Qtimes
Hence L is a Q-vector subspace of CThe most important special case of Schanuelrsquos Conjecture is
Conjecture Let λ1 λn be Q-linearly independent elementsin L Then the numbers λ1 λn are algebraicallyindependent over Q
Not yet known that the transcendence degree is ge 2 Open problem Among all logarithms of algebraic numbersone at least is transcendental over Q(π)
37 59
Structural rank of a matrix
Let K be a field k a subfield and M a matrix with entries in KConsider the k-vector subspace E of K spanned by the entriesof M Choose an injective morphism ϕ of E into a k-vectorspace kX1 + middot middot middot+ kXn The image ϕ(M) of M is a matrixwhose entries are in the field k(X1 Xn) of rationalfractions Its rank does not depend on the choice of ϕ
This is the structural rank of M with respect to k
38 59
Example
LetM =
(bij + λij
)1leiled1lejle`
be a matrix with coefficients in Q + L Consider a basis of theQndashvector spanned by the entries and replace the elements inthis basis by unknowns This gives a new matrix M withcoefficients in a field of rational fractions the rank of which isthe structural rank of M (with respect to Q)
As a consequence of the conjecture of algebraic independenceof logarithms of algebraic numbers the rank of M should bethe same as the rank of its specialization M
39 59
Equivalence between the two conjectures
Following D Roy theconjecture on algebraicindependence of logarithms ofalgebraic numbers isequivalent to
Conjecture Any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L has a rank equal to its structuralrank
40 59
Any Polynomial is the Determinant of a Matrix
The proof of the equivalence uses the nice auxiliary result
For any P isin k[X1 Xn] there exists a square matrix withentries in the k-vector space k + kX1 + middot middot middot+ kXn whosedeterminant is P
41 59
Half of the Conjecture is solved
From a certain point of view half of the conjecture of algebraicindependence of logarithms of algebraic numbers is solved
Theorem [D Roy] The rank of any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L is at least half its structural rank
42 59
Reformulation by D Roy
Instead of taking logarithmsof algebraic numbers andlooking for the algebraicindependence relationsD Roy fixes a polynomial andlooks at the points withcoordinates logarithms ofalgebraic numbers on thecorresponding hypersurface
43 59
Reformulation by D Roy
Royrsquos reformulation of the conjecture of algebraicindependence of logarithms is
Conjecture For any algebraic subvariety V of Cn definedover the field Q of algebraic numbers the set V cap Ln is theunion of the sets E cap Ln where E ranges over the set ofvector subspaces of Cn which are defined over Q andcontained in V
Trivial Any element in E cap Ln where E is a vector subspaceof Cn defined over Q and contained in V belongs to V cap Ln
44 59
Example The Four Exponentials ConjectureTake for V the hypersurface of C4 defined by the equation
z1z4 = z2z3
The maximal Cndashvector subspaces of C4 defined over Q andcontained in V are the planes
az1 = bz2 bz4 = az3
and the planesaz1 = bz3 bz4 = az2
with (a b) isin Q2 (0 0)Hence Schanuelrsquos Conjecture implies that if λ1 λ2 λ3 λ4 arelogarithms of algebraic numbers satisfying
λ1λ4 = λ2λ3
then either λ1 λ2 are linearly dependent over Q or else λ1 λ3
are linearly dependent over Q45 59
Six Exponentials Theorem
and Four Exponentials Conjecture
A Selberg CL Siegel S Lang K Ramachandra
46 59
The Four Exponentials Conjecture
Conjecture (Th Schneider S Lang K Ramachandra) Ifx1 x2 are Qndashlinearly independent complex numbers and y1 y2
are Qndashlinearly independent complex numbers then one atleast of the four numbers
ex1y1 ex1y2 ex2y1 ex2y2
is transcendentalEquivalent statement Any 2times 2 matrix(
λ1 λ2
λ3 λ4
)with entries in L and with Qndashlinearly independent rows andQndashlinearly independent columns has maximal rank 2
47 59
How could we attack Schanuelrsquos Conjecture
Let x1 xn be Qndashlinearly independent complex numbersFollowing the transcendence methods of Hermite GelrsquofondSchneider one may start by introducing an auxiliaryfunction
F (z) = P (z ez)
where P isin Z[X0 X1] is a nonndashzero polynomial One considersthe derivatives (
d
dz
)kF = (DkP )(z ez)
at the pointsm1x1 + middot middot middot+mnxn
for various values of (m1 mn) isin Zn
48 59
Auxiliary functionLet D denote the derivation
D =part
partX0
+X1part
partX1
over the ring C[X0 X1] Recall that x1 xn are Qndashlinearlyindependent complex numbers Let α1 αn be nonndashzerocomplex numbersThe transcendence machinery produces a sequence (PN)Nge0
of polynomials with integer coefficients satisfying∣∣∣∣∣(DkPN)(nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1
49 59
Royrsquos approach to Schanuelrsquos Conjecture (1999)
If the number of equations we produce is too small such a setof relations does not contain any information The existenceof a sequence of nonndashtrivial polynomials (PN)Nge0 follows fromlinear algebra
On the other hand following D Roy one may expect that theexistence of such a sequence (PN)Nge0 producing sufficientlymany such equations will yield the desired conclusion
tr degQQ(x1 xn α1 αn
)ge n
50 59
Royrsquos Conjecture (1999)Let s0 s1 t0 t1 u positive real numbers satisfying
max1 t0 2t1 lt mins0 2s1and
maxs0 s1 + t1 lt u lt1
2(1 + t0 + t1)
Assume that for any sufficiently large positive integer N there exists a nonndashzero polynomial PN isin Z[X0 X1] withpartial degree le N t0 in X0 partial degree le N t1 in X1 andheight le eN which satisfies∣∣∣∣∣(DkPN)(
nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1 Then
tr degQQ(x1 xn α1 αn) ge n51 59
Royrsquos Theorem (1999)
Royrsquos conjecture is equivalent to Schanuelrsquos Conjecture
52 59
Equivalence between Schanuel and Roy
Let (x α) isin CtimesCtimes and let s0 s1 t0 t1 u be positive realnumbers satisfying the inequalities of Royrsquos Conjecture Thenthe following conditions are equivalent (a) The number αeminusx is a root of unity
(b) For any sufficiently large positive integer N there exists anonndashzero polynomial QN isin Z[X0 X1] with partial degreele N t0 in X0 partial degree le N t1 in X1 and heightH(QN) le eN such that∣∣(DkQN)(mx αm)
∣∣ le exp(minusNu)
for any km isin N with k le N s0 and m le N s1
53 59
Gelrsquofondrsquos transcendence criterion
Simple form Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le n and height le en such that
|Pn(ϑ)| le eminus7n2
for all n ge 1 then ϑ is algebraic and Pn(ϑ) = 0 for all n ge 1
First extension Replace the bound for the degree by le dnthe bound for the height by ebn and the bound for |Pn(ϑ)| byeminuscdnbn with some constant c gt 0 independent of n
Some mild conditions are required on the sequences (dn)nge1 and(bn)nge1
54 59
Transcendence criterion with multiplicitiesWith derivatives Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le dn and height le ebn such that∣∣P (j)
n (ϑ)∣∣ le eminuscdnbntn
for all j in the range 0 le j lt tn and all n ge 1 then ϑ isalgebraicDue to M Laurent and D Roy applications to algebraicindependence
55 59
Criterion with several points
Goal Given a sequence of complex numbers (ϑi)ige1 if thereexists a sequence (Pn)nge1 of nonndashzero polynomials in Z[X]with Pn of degree le dn and height le ebn such that∣∣P (j)
n (ϑi)∣∣ le eminuscdnbntnsn
for 0 le j lt tn 1 le i le sn and all n ge 1 then the numbers ϑiare algebraic
D Roy Not true in general but true in some special caseswith a structure on the sequence (ϑi)ige1Combines the elimination arguments used for criteria ofalgebraic independence and for zero estimates
56 59
Small value estimates for the additive group
D Roy Small value estimates for the additive group Intern JNumber Theory to appear
Let ξ be a transcendental complex number let β σ τ and νbe non-negative real numbers let n0 be a positive integer andlet (Pn)ngen0 be a sequence of nonndashzero polynomials in Z[T ]satisfying deg(Pn) le n and H(Pn) le exp(nβ) for eachn ge n0 Suppose that β gt 1 (34)σ + τ lt 1 andν gt 1 + β minus (34)σ minus τ Then for infinitely many n we have
max|P [j]n (iξ)| 0 le i le nσ 0 le j le nτ
gt exp(minusnν)
57 59
Small value estimates for the multiplicative group
D Roy Small value estimates for the multiplicative groupActa Arith to appear
Let ξ1 ξm be multiplicatively independent complexnumbers in a field of transcendence degree 1 Under suitableassumptions on the parameters β σ τ ν for infinitely manypositive integers n there exists no nonndashzero polynomialP isin Z[T ] satisfying deg(P ) le n H(P ) le exp(nβ) and
max|P [j](ξi11 middot middot middot ξimm )| 0 le i1 im le nσ 0 le j le nτ
gt exp(minusnν)
58 59
Heidelberg Colloquium organized by the Institute for Mathematics
httpwwwmathiuni-heidelbergdevortraghtmls=1 January 22 2009
Transcendental Number TheorySchanuelrsquos Conjecture
Michel Waldschmidt
Institut de Mathematiques de Jussieu amp Paris VIhttpwwwmathjussieufrsimmiw
59 59
Abstract
One of the main open problems in transcendental numbertheory is Schanuelrsquos Conjecture which was stated in the1960rsquos
If x1 xn are Qndashlinearly independent complexnumbers then among the 2n numbers x1 xnex1 exn at least n are algebraically independent
We first give a list of consequences of this statement next wedescribe the state of the art by giving special cases of theconjecture which have been proved and finally we introduce apromising approach which has been initiated in 1999 byD Roy
2 59
Algebraic and transcendental numbersAlgebraic numbers Roots of polynomials with rationalcoefficients Algebraic closure of Q in C
Q sub Q sub C
Transcendental numbers Complex numbers that are notalgebraic
The set of algebraic numbers behaves well with respect toaddition multiplication division It is a fieldThe set of transcendental numbers is the complement in thefield C of the field Q The sum of transcendental numbersmay be rational algebraic or transcendental The same for theproductHowever the sum of a transcendental number and an algebraicnumber is transcendental and the product of a transcendentalnumber and a nonndashzero algebraic number is transcendental
3 59
The exponential function
For z isin C
ez = exp(z) = 1 + z +z2
2+z3
6+ middot middot middot =
sumnge0
zn
nmiddot
Addition Theorem ez1+z2 = ez1ez2 e2iπ = 1
exp C minusrarr Ctimes ker(exp) = 2iπZ C2iπZ Ctimes
Differential equation
d
dzez = ez
4 59
First examples of transcendental numbers
sumnge1 2minusn Liouville 1844
e Hermite 1872
π Lindemann 1881
5 59
Ch Hermite ndash F Lindemann ndash C Weierstraszlig
log 2 eradic
2 Theorem of HermitendashLindemann
eradic
2 +radic
3eradic
6 LindemannndashWeierstraszlig 1888
6 59
Question of Euler and Hilbert
Leonhard Euler (1707 ndash 1783)Introductio in analysin infinitorum (1737)
Transcendence of 2radic
2
David Hilbert (1862 - 1943)ICM 1900 7ndashth Problem
Transcendence of log a log 2
7 59
AO Gelrsquofond ndash Th Schneider ndash A Baker
eπ Gelrsquofond 1929
2radic
2 log 2 log 3 Gelrsquofond and Schneider 1934
log 2 +radic
3 log 3 eradic
22radic
35radic
7 Baker 1968
8 59
Some open problems
For each of the following numbers it is expected that it istranscendental but it is not even known whether it is rationalor not
e+ π eπ πe ee ee2
eee
ππ ππ2
πππ
log π log(log 2) π log 2 (log 2)(log 3) 2log 2 (log 2)log 3
In other words we do not know whether a degree 1 polynomialcould vanish at the corresponding point but we expect thatno nonndashzero polynomial of any degree vanishes at this point
9 59
Algebraic independence
Algebraicity and transcendence deal with a single complexnumber and one variable polynomialsAlgebraic dependence or independence is the same but fortuples and multivariate polynomials
Let K sup k be an extension of fields and (θ1 θm) be amndashtuple of elements in K We say that θ1 θm arealgebraically dependent over k if there exists a nonndashzeropolynomial f isin k[X1 Xm] which vanishes at the point(θ1 θm) isin KmOtherwise we say that θ1 θm are algebraically independentover k
10 59
Special cases
In the case m = 1 to say that θ1 is algebraically independentover k just means that it is transcendental over k
Dealing with complex numbers K = C the words algebraictranscendental algebraically dependent algebraicallyindependent refer to the case k = Q
11 59
Examples
The numbersradic
2 and π are algebraically dependent The polynomial X2 minus 2 isin Z[X Y ] vanishes at (
radic2 π)
The numbers π andradicπ2 + 1 are algebraically dependent (and
both are transcendental numbers) The polynomial Y minusX2 minus 1 vanishes at (π
radicπ2 + 1)
The two numbers e and eradic
2 are algebraically independentwhich means that for any nonndashzero polynomial f isin Z[X Y ]
with rational integer coefficients the number f(e eradic
2) is notzeroSpecial case of the Lindemann-Weierstraszlig Theorem
12 59
Transcendence degree
Let Kk be a field extension The maximal number ofelements in K which are algebraically independent over k iscalled the transcendence degree of K over k and denotedtr degkK
Let t = tr degkK A subset θ1 θt of K with t elementswhich are algebraically independent is called a transcendencebasis of K over k It is the same as a maximal subset ofkndashalgebraically independent elements in KHence K is an algebraic extension of k(θ1 θt)
13 59
Transcendence degree of extensions
Assumek sub K sub L
The union of a transcendence basis of K over k and of atranscendence basis of L over K produces a transcendencebasis of L over kHence
tr degkL = tr degkK + tr degKL
An algebraic extension Kk is an extension of transcendencedegree 0 This means that there is no transcendental elementin K over k (any element in K is algebraic over k)
14 59
Algebraic independence of complex numbers
For complex numbers algebraic independence over Q or overQ is the same In particular if θ1 θm are algebraicallyindependent complex numbers then for any non-constantpolynomial f with algebraic coefficients the numberf(θ1 θm) is transcendental
For instance the two numbers e and eradic
2 are algebraicallyindependent As a consequence for any nonndashconstantpolynomial f isin Q[X Y ] with algebraic coefficients the
number f(e eradic
2) is transcendental
15 59
Schanuelrsquos Conjecture
Let x1 xn be Q-linearly independent complex numbersThen at least n of the 2n numbers x1 xn e
x1 exn arealgebraically independent
Since there are 2n numbers only the transcendence degreeover Q of the field
Q(x1 xn e
x1 exn)
is at most 2n The conjecture is that this transcendencedegree is always ge n
n le tr degQQ(x1 xn e
x1 exn)le 2n
16 59
Origin of Schanuelrsquos ConjectureCourse given by Serge Lang (1927ndash2005) at Yale in the 60rsquos
S Lang ndash Introduction to transcendental numbersAddison-Wesley 1966
also attended by M Nagata (1927ndash2008)(14th Problem of Hilbert)
Nagatarsquos Conjecture solved by E Bombieri
17 59
Schanuelrsquos Conjecture
Let x1 xn be Q-linearly independent complex numbersThen
tr degQQ(x1 xn e
x1 exn)ge n
Remark For almost all tuples (with respect to the Lebesguemeasure) the transcendence degree is 2n
18 59
AO Gelrsquofond CRAS 1934
19 59
Statement by Gelrsquofond (1934)
Let β1 βn be Q-linearly independent algebraic numbersand let logα1 logαm be Q-linearly independentlogarithms of algebraic numbers Then the numbers
eβ1 eβn logα1 logαm
are algebraically independent over Q
20 59
Further statement by Gelrsquofond
Let β1 βn be algebraic numbers with β1 6= 0 and letα1 αm be algebraic numbers with α1 6= 0 1 α2 6= 0 1αi 6= 0 Then the numbers
eβ1eβ2e βnminus1e
βn
and αα αm
21
are transcendental and there is no nontrivial algebraic relationbetween such numbers
Remark The condition on α2 should be that it is irrational
21 59
Easy consequence of Schanuelrsquos Conjecture
According to Schanuelrsquos Conjecture the following numbers arealgebraically independent
e+ π eπ πe ee ee2
eee
ππ ππ2
πππ
log π log(log 2) π log 2 (log 2)(log 3) 2log 2 (log 2)log 3
Proof This is an easy exercise
22 59
Langrsquos exercise
Define E0 = Q Inductivelyfor n ge 1 define En as thealgebraic closure of the fieldgenerated over Enminus1 by thenumbers exp(x) = ex wherex ranges over Enminus1 Let E bethe union of En n ge 0Then Schanuelrsquos Conjectureimplies that the number πdoes not belong to E
More precisely Schanuelrsquos Conjecture implies that thenumbers π log π log log π log log log π are algebraicallyindependent over E
23 59
A variant of Langrsquos exercise
Define L0 = Q Inductively for n ge 1 define Ln as thealgebraic closure of the field generated over Lnminus1 by thenumbers y where y ranges over the set of complex numberssuch that ey isin Lnminus1 Let L be the union of Ln n ge 0Then Schanuelrsquos Conjecture implies that the number e doesnot belong to L
More precisely Schanuelrsquos Conjecture implies that the
numbers e ee eee ee
ee
are algebraically independent over L
24 59
Arizona Winter School AWS2008 Tucson
Theorem [Jonathan Bober Chuangxun Cheng Brian DietelMathilde Herblot Jingjing Huang Holly Krieger DiegoMarques Jonathan Mason Martin Mereb and Robert Wilson]Schanuelrsquos Conjecture implies that the fields E and L arelinearly disjoint over Q
Definition Given a field extension FK and twosubextensions F1 F2 sube F we say F1 F2 are linearly disjointover K when the following holds Any set x1 xn sube F1
of Kndash linearly independent elements is linearly independentover F2
Reference arXiv08043550 [mathNT] 2008
25 59
Formal analogs
WD Brownawell(was a student of Schanuel)
J Axrsquos Theorem (1968) Version of SchanuelrsquosConjecture for power seriesover C(and R Coleman for powerseries over Q)
Work by WD Brownawelland K Kubota on the ellipticanalog of Axrsquos Theorem
26 59
Conjectures by A Grothendieck and Y Andre
Generalized Conjecture onPeriods by Grothendieck Dimension of theMumfordndashTate group of asmooth projective variety
Generalization by Y Andre tomotives
Case of 1ndashmotives Elliptico-Toric Conjecture ofC Bertolin
27 59
Ubiquity of Schanuelrsquos Conjecture
Other contexts pndashadic numbers Leopoldtrsquos Conjecture onthe pndashadic rank of the units of an algebraic number fieldNon-vanishing of RegulatorsNonndashdegenerescence of heightsConjecture of B Mazur on rational pointsDiophantine approximation on tori
Dipendra Prasad Gopal Prasad
28 59
Methods from logicEhud Hrushovski Boris Zilber Jonathan Kirby
Calculus of rdquopredimension functionsrdquo (E Hrushovski)
Zilberrsquos construction of a rdquopseudoexponentiationrdquo
Also A Macintyre DE Marker G Terzo AJ WilkieD Bertrand
29 59
Methods from logic Model theory
Exponential algebraicity in exponential fieldsby
Jonathan Kirby
The dimension of the exponential algebraic closure operator inan exponential field satisfies a weak Schanuel property
A corollary is that there are at most countably many essentialcounterexamples to Schanuelrsquos conjecture
arXiv 08104285v2
30 59
Known
LindemannndashWeierstraszlig Theorem = case where x1 xn arealgebraic
Let β1 βn be algebraic numbers which are linearlyindependent over Q Then the numbers eβ1 eβn arealgebraically independent over Q
31 59
Problem of Gelrsquofond and SchneiderRaised by AO Gelrsquofond in 1948 and Th Schneider in 1952
Conjecture If α is an algebraic number α 6= 0 α 6= 1 and ifβ is an irrational algebraic number of degree d then the dminus 1numbers
αβ αβ2
αβdminus1
are algebraically independent
Special case of Schanuelrsquos Conjecture Take xi = βiminus1 logαn = d so that
x1 xn e
x1 exn
islogα β logα βdminus1 logα α αβ αβ
dminus1
The conclusion of Schanuelrsquos Conjecture is
tr degQQ(logα αβ αβ
2
αβdminus1)
= d
32 59
Algebraic independence method of Gelrsquofond
AO Gelrsquofond (1948)
The two numbers 23radic2 and
23radic4 are algebraically
independentMore generally if α is analgebraic number α 6= 0α 6= 1 and if β is a algebraicnumber of degree d ge 3 thentwo at least of the numbers
αβ αβ2
αβdminus1
are algebraically independent
33 59
Tools
Transcendence criterion Replaces Liouvillersquos inequality intranscendence proofsLiouville A nonndashzero rational integer n isin Z satisfies |n| ge 1Gelrsquofond Needs to give a lower bound for |P (θ)| withP isin Z[X] 0 when θ is transcendental
Zero estimate for exponential polynomials C Hermite P Turan K Mahler R Tijdeman
Small transcendence degree AO Gelrsquofond AA Smelev R Tijdeman WD Brownawell
34 59
Large transcendence degree
GV Chudnovsky (1976)Among the numbers
αβ αβ2
αβdminus1
at least [log2 d] arealgebraically independent
GV Chudnovsky ndash On the path to Schanuelrsquosconjecture Algebraic curves close to a pointI General theory of colored sequencesII Fields of finite transcendence type and coloredsequences ResultantsStudia Sci Math Hungar 12 (1977) 125ndash157 (1980)
35 59
Partial result on the problem of Gelrsquofond and
Schneider
AO Gelrsquofond GV Chudnovskii P Philippon YuV Nesterenko
G Diaz If α is an algebraic number α 6= 0 α 6= 1 and if β isan irrational algebraic number of degree d then
tr degQQ(αβ αβ
2
αβdminus1) ge [d+ 1
2
]
36 59
Conjecture of algebraic independence of logarithms
of algebraic numbers
Denote by L the set of complex numbers λ for which eλ isalgebraic
L = logα α isin Qtimes
Hence L is a Q-vector subspace of CThe most important special case of Schanuelrsquos Conjecture is
Conjecture Let λ1 λn be Q-linearly independent elementsin L Then the numbers λ1 λn are algebraicallyindependent over Q
Not yet known that the transcendence degree is ge 2 Open problem Among all logarithms of algebraic numbersone at least is transcendental over Q(π)
37 59
Structural rank of a matrix
Let K be a field k a subfield and M a matrix with entries in KConsider the k-vector subspace E of K spanned by the entriesof M Choose an injective morphism ϕ of E into a k-vectorspace kX1 + middot middot middot+ kXn The image ϕ(M) of M is a matrixwhose entries are in the field k(X1 Xn) of rationalfractions Its rank does not depend on the choice of ϕ
This is the structural rank of M with respect to k
38 59
Example
LetM =
(bij + λij
)1leiled1lejle`
be a matrix with coefficients in Q + L Consider a basis of theQndashvector spanned by the entries and replace the elements inthis basis by unknowns This gives a new matrix M withcoefficients in a field of rational fractions the rank of which isthe structural rank of M (with respect to Q)
As a consequence of the conjecture of algebraic independenceof logarithms of algebraic numbers the rank of M should bethe same as the rank of its specialization M
39 59
Equivalence between the two conjectures
Following D Roy theconjecture on algebraicindependence of logarithms ofalgebraic numbers isequivalent to
Conjecture Any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L has a rank equal to its structuralrank
40 59
Any Polynomial is the Determinant of a Matrix
The proof of the equivalence uses the nice auxiliary result
For any P isin k[X1 Xn] there exists a square matrix withentries in the k-vector space k + kX1 + middot middot middot+ kXn whosedeterminant is P
41 59
Half of the Conjecture is solved
From a certain point of view half of the conjecture of algebraicindependence of logarithms of algebraic numbers is solved
Theorem [D Roy] The rank of any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L is at least half its structural rank
42 59
Reformulation by D Roy
Instead of taking logarithmsof algebraic numbers andlooking for the algebraicindependence relationsD Roy fixes a polynomial andlooks at the points withcoordinates logarithms ofalgebraic numbers on thecorresponding hypersurface
43 59
Reformulation by D Roy
Royrsquos reformulation of the conjecture of algebraicindependence of logarithms is
Conjecture For any algebraic subvariety V of Cn definedover the field Q of algebraic numbers the set V cap Ln is theunion of the sets E cap Ln where E ranges over the set ofvector subspaces of Cn which are defined over Q andcontained in V
Trivial Any element in E cap Ln where E is a vector subspaceof Cn defined over Q and contained in V belongs to V cap Ln
44 59
Example The Four Exponentials ConjectureTake for V the hypersurface of C4 defined by the equation
z1z4 = z2z3
The maximal Cndashvector subspaces of C4 defined over Q andcontained in V are the planes
az1 = bz2 bz4 = az3
and the planesaz1 = bz3 bz4 = az2
with (a b) isin Q2 (0 0)Hence Schanuelrsquos Conjecture implies that if λ1 λ2 λ3 λ4 arelogarithms of algebraic numbers satisfying
λ1λ4 = λ2λ3
then either λ1 λ2 are linearly dependent over Q or else λ1 λ3
are linearly dependent over Q45 59
Six Exponentials Theorem
and Four Exponentials Conjecture
A Selberg CL Siegel S Lang K Ramachandra
46 59
The Four Exponentials Conjecture
Conjecture (Th Schneider S Lang K Ramachandra) Ifx1 x2 are Qndashlinearly independent complex numbers and y1 y2
are Qndashlinearly independent complex numbers then one atleast of the four numbers
ex1y1 ex1y2 ex2y1 ex2y2
is transcendentalEquivalent statement Any 2times 2 matrix(
λ1 λ2
λ3 λ4
)with entries in L and with Qndashlinearly independent rows andQndashlinearly independent columns has maximal rank 2
47 59
How could we attack Schanuelrsquos Conjecture
Let x1 xn be Qndashlinearly independent complex numbersFollowing the transcendence methods of Hermite GelrsquofondSchneider one may start by introducing an auxiliaryfunction
F (z) = P (z ez)
where P isin Z[X0 X1] is a nonndashzero polynomial One considersthe derivatives (
d
dz
)kF = (DkP )(z ez)
at the pointsm1x1 + middot middot middot+mnxn
for various values of (m1 mn) isin Zn
48 59
Auxiliary functionLet D denote the derivation
D =part
partX0
+X1part
partX1
over the ring C[X0 X1] Recall that x1 xn are Qndashlinearlyindependent complex numbers Let α1 αn be nonndashzerocomplex numbersThe transcendence machinery produces a sequence (PN)Nge0
of polynomials with integer coefficients satisfying∣∣∣∣∣(DkPN)(nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1
49 59
Royrsquos approach to Schanuelrsquos Conjecture (1999)
If the number of equations we produce is too small such a setof relations does not contain any information The existenceof a sequence of nonndashtrivial polynomials (PN)Nge0 follows fromlinear algebra
On the other hand following D Roy one may expect that theexistence of such a sequence (PN)Nge0 producing sufficientlymany such equations will yield the desired conclusion
tr degQQ(x1 xn α1 αn
)ge n
50 59
Royrsquos Conjecture (1999)Let s0 s1 t0 t1 u positive real numbers satisfying
max1 t0 2t1 lt mins0 2s1and
maxs0 s1 + t1 lt u lt1
2(1 + t0 + t1)
Assume that for any sufficiently large positive integer N there exists a nonndashzero polynomial PN isin Z[X0 X1] withpartial degree le N t0 in X0 partial degree le N t1 in X1 andheight le eN which satisfies∣∣∣∣∣(DkPN)(
nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1 Then
tr degQQ(x1 xn α1 αn) ge n51 59
Royrsquos Theorem (1999)
Royrsquos conjecture is equivalent to Schanuelrsquos Conjecture
52 59
Equivalence between Schanuel and Roy
Let (x α) isin CtimesCtimes and let s0 s1 t0 t1 u be positive realnumbers satisfying the inequalities of Royrsquos Conjecture Thenthe following conditions are equivalent (a) The number αeminusx is a root of unity
(b) For any sufficiently large positive integer N there exists anonndashzero polynomial QN isin Z[X0 X1] with partial degreele N t0 in X0 partial degree le N t1 in X1 and heightH(QN) le eN such that∣∣(DkQN)(mx αm)
∣∣ le exp(minusNu)
for any km isin N with k le N s0 and m le N s1
53 59
Gelrsquofondrsquos transcendence criterion
Simple form Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le n and height le en such that
|Pn(ϑ)| le eminus7n2
for all n ge 1 then ϑ is algebraic and Pn(ϑ) = 0 for all n ge 1
First extension Replace the bound for the degree by le dnthe bound for the height by ebn and the bound for |Pn(ϑ)| byeminuscdnbn with some constant c gt 0 independent of n
Some mild conditions are required on the sequences (dn)nge1 and(bn)nge1
54 59
Transcendence criterion with multiplicitiesWith derivatives Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le dn and height le ebn such that∣∣P (j)
n (ϑ)∣∣ le eminuscdnbntn
for all j in the range 0 le j lt tn and all n ge 1 then ϑ isalgebraicDue to M Laurent and D Roy applications to algebraicindependence
55 59
Criterion with several points
Goal Given a sequence of complex numbers (ϑi)ige1 if thereexists a sequence (Pn)nge1 of nonndashzero polynomials in Z[X]with Pn of degree le dn and height le ebn such that∣∣P (j)
n (ϑi)∣∣ le eminuscdnbntnsn
for 0 le j lt tn 1 le i le sn and all n ge 1 then the numbers ϑiare algebraic
D Roy Not true in general but true in some special caseswith a structure on the sequence (ϑi)ige1Combines the elimination arguments used for criteria ofalgebraic independence and for zero estimates
56 59
Small value estimates for the additive group
D Roy Small value estimates for the additive group Intern JNumber Theory to appear
Let ξ be a transcendental complex number let β σ τ and νbe non-negative real numbers let n0 be a positive integer andlet (Pn)ngen0 be a sequence of nonndashzero polynomials in Z[T ]satisfying deg(Pn) le n and H(Pn) le exp(nβ) for eachn ge n0 Suppose that β gt 1 (34)σ + τ lt 1 andν gt 1 + β minus (34)σ minus τ Then for infinitely many n we have
max|P [j]n (iξ)| 0 le i le nσ 0 le j le nτ
gt exp(minusnν)
57 59
Small value estimates for the multiplicative group
D Roy Small value estimates for the multiplicative groupActa Arith to appear
Let ξ1 ξm be multiplicatively independent complexnumbers in a field of transcendence degree 1 Under suitableassumptions on the parameters β σ τ ν for infinitely manypositive integers n there exists no nonndashzero polynomialP isin Z[T ] satisfying deg(P ) le n H(P ) le exp(nβ) and
max|P [j](ξi11 middot middot middot ξimm )| 0 le i1 im le nσ 0 le j le nτ
gt exp(minusnν)
58 59
Heidelberg Colloquium organized by the Institute for Mathematics
httpwwwmathiuni-heidelbergdevortraghtmls=1 January 22 2009
Transcendental Number TheorySchanuelrsquos Conjecture
Michel Waldschmidt
Institut de Mathematiques de Jussieu amp Paris VIhttpwwwmathjussieufrsimmiw
59 59
Algebraic and transcendental numbersAlgebraic numbers Roots of polynomials with rationalcoefficients Algebraic closure of Q in C
Q sub Q sub C
Transcendental numbers Complex numbers that are notalgebraic
The set of algebraic numbers behaves well with respect toaddition multiplication division It is a fieldThe set of transcendental numbers is the complement in thefield C of the field Q The sum of transcendental numbersmay be rational algebraic or transcendental The same for theproductHowever the sum of a transcendental number and an algebraicnumber is transcendental and the product of a transcendentalnumber and a nonndashzero algebraic number is transcendental
3 59
The exponential function
For z isin C
ez = exp(z) = 1 + z +z2
2+z3
6+ middot middot middot =
sumnge0
zn
nmiddot
Addition Theorem ez1+z2 = ez1ez2 e2iπ = 1
exp C minusrarr Ctimes ker(exp) = 2iπZ C2iπZ Ctimes
Differential equation
d
dzez = ez
4 59
First examples of transcendental numbers
sumnge1 2minusn Liouville 1844
e Hermite 1872
π Lindemann 1881
5 59
Ch Hermite ndash F Lindemann ndash C Weierstraszlig
log 2 eradic
2 Theorem of HermitendashLindemann
eradic
2 +radic
3eradic
6 LindemannndashWeierstraszlig 1888
6 59
Question of Euler and Hilbert
Leonhard Euler (1707 ndash 1783)Introductio in analysin infinitorum (1737)
Transcendence of 2radic
2
David Hilbert (1862 - 1943)ICM 1900 7ndashth Problem
Transcendence of log a log 2
7 59
AO Gelrsquofond ndash Th Schneider ndash A Baker
eπ Gelrsquofond 1929
2radic
2 log 2 log 3 Gelrsquofond and Schneider 1934
log 2 +radic
3 log 3 eradic
22radic
35radic
7 Baker 1968
8 59
Some open problems
For each of the following numbers it is expected that it istranscendental but it is not even known whether it is rationalor not
e+ π eπ πe ee ee2
eee
ππ ππ2
πππ
log π log(log 2) π log 2 (log 2)(log 3) 2log 2 (log 2)log 3
In other words we do not know whether a degree 1 polynomialcould vanish at the corresponding point but we expect thatno nonndashzero polynomial of any degree vanishes at this point
9 59
Algebraic independence
Algebraicity and transcendence deal with a single complexnumber and one variable polynomialsAlgebraic dependence or independence is the same but fortuples and multivariate polynomials
Let K sup k be an extension of fields and (θ1 θm) be amndashtuple of elements in K We say that θ1 θm arealgebraically dependent over k if there exists a nonndashzeropolynomial f isin k[X1 Xm] which vanishes at the point(θ1 θm) isin KmOtherwise we say that θ1 θm are algebraically independentover k
10 59
Special cases
In the case m = 1 to say that θ1 is algebraically independentover k just means that it is transcendental over k
Dealing with complex numbers K = C the words algebraictranscendental algebraically dependent algebraicallyindependent refer to the case k = Q
11 59
Examples
The numbersradic
2 and π are algebraically dependent The polynomial X2 minus 2 isin Z[X Y ] vanishes at (
radic2 π)
The numbers π andradicπ2 + 1 are algebraically dependent (and
both are transcendental numbers) The polynomial Y minusX2 minus 1 vanishes at (π
radicπ2 + 1)
The two numbers e and eradic
2 are algebraically independentwhich means that for any nonndashzero polynomial f isin Z[X Y ]
with rational integer coefficients the number f(e eradic
2) is notzeroSpecial case of the Lindemann-Weierstraszlig Theorem
12 59
Transcendence degree
Let Kk be a field extension The maximal number ofelements in K which are algebraically independent over k iscalled the transcendence degree of K over k and denotedtr degkK
Let t = tr degkK A subset θ1 θt of K with t elementswhich are algebraically independent is called a transcendencebasis of K over k It is the same as a maximal subset ofkndashalgebraically independent elements in KHence K is an algebraic extension of k(θ1 θt)
13 59
Transcendence degree of extensions
Assumek sub K sub L
The union of a transcendence basis of K over k and of atranscendence basis of L over K produces a transcendencebasis of L over kHence
tr degkL = tr degkK + tr degKL
An algebraic extension Kk is an extension of transcendencedegree 0 This means that there is no transcendental elementin K over k (any element in K is algebraic over k)
14 59
Algebraic independence of complex numbers
For complex numbers algebraic independence over Q or overQ is the same In particular if θ1 θm are algebraicallyindependent complex numbers then for any non-constantpolynomial f with algebraic coefficients the numberf(θ1 θm) is transcendental
For instance the two numbers e and eradic
2 are algebraicallyindependent As a consequence for any nonndashconstantpolynomial f isin Q[X Y ] with algebraic coefficients the
number f(e eradic
2) is transcendental
15 59
Schanuelrsquos Conjecture
Let x1 xn be Q-linearly independent complex numbersThen at least n of the 2n numbers x1 xn e
x1 exn arealgebraically independent
Since there are 2n numbers only the transcendence degreeover Q of the field
Q(x1 xn e
x1 exn)
is at most 2n The conjecture is that this transcendencedegree is always ge n
n le tr degQQ(x1 xn e
x1 exn)le 2n
16 59
Origin of Schanuelrsquos ConjectureCourse given by Serge Lang (1927ndash2005) at Yale in the 60rsquos
S Lang ndash Introduction to transcendental numbersAddison-Wesley 1966
also attended by M Nagata (1927ndash2008)(14th Problem of Hilbert)
Nagatarsquos Conjecture solved by E Bombieri
17 59
Schanuelrsquos Conjecture
Let x1 xn be Q-linearly independent complex numbersThen
tr degQQ(x1 xn e
x1 exn)ge n
Remark For almost all tuples (with respect to the Lebesguemeasure) the transcendence degree is 2n
18 59
AO Gelrsquofond CRAS 1934
19 59
Statement by Gelrsquofond (1934)
Let β1 βn be Q-linearly independent algebraic numbersand let logα1 logαm be Q-linearly independentlogarithms of algebraic numbers Then the numbers
eβ1 eβn logα1 logαm
are algebraically independent over Q
20 59
Further statement by Gelrsquofond
Let β1 βn be algebraic numbers with β1 6= 0 and letα1 αm be algebraic numbers with α1 6= 0 1 α2 6= 0 1αi 6= 0 Then the numbers
eβ1eβ2e βnminus1e
βn
and αα αm
21
are transcendental and there is no nontrivial algebraic relationbetween such numbers
Remark The condition on α2 should be that it is irrational
21 59
Easy consequence of Schanuelrsquos Conjecture
According to Schanuelrsquos Conjecture the following numbers arealgebraically independent
e+ π eπ πe ee ee2
eee
ππ ππ2
πππ
log π log(log 2) π log 2 (log 2)(log 3) 2log 2 (log 2)log 3
Proof This is an easy exercise
22 59
Langrsquos exercise
Define E0 = Q Inductivelyfor n ge 1 define En as thealgebraic closure of the fieldgenerated over Enminus1 by thenumbers exp(x) = ex wherex ranges over Enminus1 Let E bethe union of En n ge 0Then Schanuelrsquos Conjectureimplies that the number πdoes not belong to E
More precisely Schanuelrsquos Conjecture implies that thenumbers π log π log log π log log log π are algebraicallyindependent over E
23 59
A variant of Langrsquos exercise
Define L0 = Q Inductively for n ge 1 define Ln as thealgebraic closure of the field generated over Lnminus1 by thenumbers y where y ranges over the set of complex numberssuch that ey isin Lnminus1 Let L be the union of Ln n ge 0Then Schanuelrsquos Conjecture implies that the number e doesnot belong to L
More precisely Schanuelrsquos Conjecture implies that the
numbers e ee eee ee
ee
are algebraically independent over L
24 59
Arizona Winter School AWS2008 Tucson
Theorem [Jonathan Bober Chuangxun Cheng Brian DietelMathilde Herblot Jingjing Huang Holly Krieger DiegoMarques Jonathan Mason Martin Mereb and Robert Wilson]Schanuelrsquos Conjecture implies that the fields E and L arelinearly disjoint over Q
Definition Given a field extension FK and twosubextensions F1 F2 sube F we say F1 F2 are linearly disjointover K when the following holds Any set x1 xn sube F1
of Kndash linearly independent elements is linearly independentover F2
Reference arXiv08043550 [mathNT] 2008
25 59
Formal analogs
WD Brownawell(was a student of Schanuel)
J Axrsquos Theorem (1968) Version of SchanuelrsquosConjecture for power seriesover C(and R Coleman for powerseries over Q)
Work by WD Brownawelland K Kubota on the ellipticanalog of Axrsquos Theorem
26 59
Conjectures by A Grothendieck and Y Andre
Generalized Conjecture onPeriods by Grothendieck Dimension of theMumfordndashTate group of asmooth projective variety
Generalization by Y Andre tomotives
Case of 1ndashmotives Elliptico-Toric Conjecture ofC Bertolin
27 59
Ubiquity of Schanuelrsquos Conjecture
Other contexts pndashadic numbers Leopoldtrsquos Conjecture onthe pndashadic rank of the units of an algebraic number fieldNon-vanishing of RegulatorsNonndashdegenerescence of heightsConjecture of B Mazur on rational pointsDiophantine approximation on tori
Dipendra Prasad Gopal Prasad
28 59
Methods from logicEhud Hrushovski Boris Zilber Jonathan Kirby
Calculus of rdquopredimension functionsrdquo (E Hrushovski)
Zilberrsquos construction of a rdquopseudoexponentiationrdquo
Also A Macintyre DE Marker G Terzo AJ WilkieD Bertrand
29 59
Methods from logic Model theory
Exponential algebraicity in exponential fieldsby
Jonathan Kirby
The dimension of the exponential algebraic closure operator inan exponential field satisfies a weak Schanuel property
A corollary is that there are at most countably many essentialcounterexamples to Schanuelrsquos conjecture
arXiv 08104285v2
30 59
Known
LindemannndashWeierstraszlig Theorem = case where x1 xn arealgebraic
Let β1 βn be algebraic numbers which are linearlyindependent over Q Then the numbers eβ1 eβn arealgebraically independent over Q
31 59
Problem of Gelrsquofond and SchneiderRaised by AO Gelrsquofond in 1948 and Th Schneider in 1952
Conjecture If α is an algebraic number α 6= 0 α 6= 1 and ifβ is an irrational algebraic number of degree d then the dminus 1numbers
αβ αβ2
αβdminus1
are algebraically independent
Special case of Schanuelrsquos Conjecture Take xi = βiminus1 logαn = d so that
x1 xn e
x1 exn
islogα β logα βdminus1 logα α αβ αβ
dminus1
The conclusion of Schanuelrsquos Conjecture is
tr degQQ(logα αβ αβ
2
αβdminus1)
= d
32 59
Algebraic independence method of Gelrsquofond
AO Gelrsquofond (1948)
The two numbers 23radic2 and
23radic4 are algebraically
independentMore generally if α is analgebraic number α 6= 0α 6= 1 and if β is a algebraicnumber of degree d ge 3 thentwo at least of the numbers
αβ αβ2
αβdminus1
are algebraically independent
33 59
Tools
Transcendence criterion Replaces Liouvillersquos inequality intranscendence proofsLiouville A nonndashzero rational integer n isin Z satisfies |n| ge 1Gelrsquofond Needs to give a lower bound for |P (θ)| withP isin Z[X] 0 when θ is transcendental
Zero estimate for exponential polynomials C Hermite P Turan K Mahler R Tijdeman
Small transcendence degree AO Gelrsquofond AA Smelev R Tijdeman WD Brownawell
34 59
Large transcendence degree
GV Chudnovsky (1976)Among the numbers
αβ αβ2
αβdminus1
at least [log2 d] arealgebraically independent
GV Chudnovsky ndash On the path to Schanuelrsquosconjecture Algebraic curves close to a pointI General theory of colored sequencesII Fields of finite transcendence type and coloredsequences ResultantsStudia Sci Math Hungar 12 (1977) 125ndash157 (1980)
35 59
Partial result on the problem of Gelrsquofond and
Schneider
AO Gelrsquofond GV Chudnovskii P Philippon YuV Nesterenko
G Diaz If α is an algebraic number α 6= 0 α 6= 1 and if β isan irrational algebraic number of degree d then
tr degQQ(αβ αβ
2
αβdminus1) ge [d+ 1
2
]
36 59
Conjecture of algebraic independence of logarithms
of algebraic numbers
Denote by L the set of complex numbers λ for which eλ isalgebraic
L = logα α isin Qtimes
Hence L is a Q-vector subspace of CThe most important special case of Schanuelrsquos Conjecture is
Conjecture Let λ1 λn be Q-linearly independent elementsin L Then the numbers λ1 λn are algebraicallyindependent over Q
Not yet known that the transcendence degree is ge 2 Open problem Among all logarithms of algebraic numbersone at least is transcendental over Q(π)
37 59
Structural rank of a matrix
Let K be a field k a subfield and M a matrix with entries in KConsider the k-vector subspace E of K spanned by the entriesof M Choose an injective morphism ϕ of E into a k-vectorspace kX1 + middot middot middot+ kXn The image ϕ(M) of M is a matrixwhose entries are in the field k(X1 Xn) of rationalfractions Its rank does not depend on the choice of ϕ
This is the structural rank of M with respect to k
38 59
Example
LetM =
(bij + λij
)1leiled1lejle`
be a matrix with coefficients in Q + L Consider a basis of theQndashvector spanned by the entries and replace the elements inthis basis by unknowns This gives a new matrix M withcoefficients in a field of rational fractions the rank of which isthe structural rank of M (with respect to Q)
As a consequence of the conjecture of algebraic independenceof logarithms of algebraic numbers the rank of M should bethe same as the rank of its specialization M
39 59
Equivalence between the two conjectures
Following D Roy theconjecture on algebraicindependence of logarithms ofalgebraic numbers isequivalent to
Conjecture Any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L has a rank equal to its structuralrank
40 59
Any Polynomial is the Determinant of a Matrix
The proof of the equivalence uses the nice auxiliary result
For any P isin k[X1 Xn] there exists a square matrix withentries in the k-vector space k + kX1 + middot middot middot+ kXn whosedeterminant is P
41 59
Half of the Conjecture is solved
From a certain point of view half of the conjecture of algebraicindependence of logarithms of algebraic numbers is solved
Theorem [D Roy] The rank of any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L is at least half its structural rank
42 59
Reformulation by D Roy
Instead of taking logarithmsof algebraic numbers andlooking for the algebraicindependence relationsD Roy fixes a polynomial andlooks at the points withcoordinates logarithms ofalgebraic numbers on thecorresponding hypersurface
43 59
Reformulation by D Roy
Royrsquos reformulation of the conjecture of algebraicindependence of logarithms is
Conjecture For any algebraic subvariety V of Cn definedover the field Q of algebraic numbers the set V cap Ln is theunion of the sets E cap Ln where E ranges over the set ofvector subspaces of Cn which are defined over Q andcontained in V
Trivial Any element in E cap Ln where E is a vector subspaceof Cn defined over Q and contained in V belongs to V cap Ln
44 59
Example The Four Exponentials ConjectureTake for V the hypersurface of C4 defined by the equation
z1z4 = z2z3
The maximal Cndashvector subspaces of C4 defined over Q andcontained in V are the planes
az1 = bz2 bz4 = az3
and the planesaz1 = bz3 bz4 = az2
with (a b) isin Q2 (0 0)Hence Schanuelrsquos Conjecture implies that if λ1 λ2 λ3 λ4 arelogarithms of algebraic numbers satisfying
λ1λ4 = λ2λ3
then either λ1 λ2 are linearly dependent over Q or else λ1 λ3
are linearly dependent over Q45 59
Six Exponentials Theorem
and Four Exponentials Conjecture
A Selberg CL Siegel S Lang K Ramachandra
46 59
The Four Exponentials Conjecture
Conjecture (Th Schneider S Lang K Ramachandra) Ifx1 x2 are Qndashlinearly independent complex numbers and y1 y2
are Qndashlinearly independent complex numbers then one atleast of the four numbers
ex1y1 ex1y2 ex2y1 ex2y2
is transcendentalEquivalent statement Any 2times 2 matrix(
λ1 λ2
λ3 λ4
)with entries in L and with Qndashlinearly independent rows andQndashlinearly independent columns has maximal rank 2
47 59
How could we attack Schanuelrsquos Conjecture
Let x1 xn be Qndashlinearly independent complex numbersFollowing the transcendence methods of Hermite GelrsquofondSchneider one may start by introducing an auxiliaryfunction
F (z) = P (z ez)
where P isin Z[X0 X1] is a nonndashzero polynomial One considersthe derivatives (
d
dz
)kF = (DkP )(z ez)
at the pointsm1x1 + middot middot middot+mnxn
for various values of (m1 mn) isin Zn
48 59
Auxiliary functionLet D denote the derivation
D =part
partX0
+X1part
partX1
over the ring C[X0 X1] Recall that x1 xn are Qndashlinearlyindependent complex numbers Let α1 αn be nonndashzerocomplex numbersThe transcendence machinery produces a sequence (PN)Nge0
of polynomials with integer coefficients satisfying∣∣∣∣∣(DkPN)(nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1
49 59
Royrsquos approach to Schanuelrsquos Conjecture (1999)
If the number of equations we produce is too small such a setof relations does not contain any information The existenceof a sequence of nonndashtrivial polynomials (PN)Nge0 follows fromlinear algebra
On the other hand following D Roy one may expect that theexistence of such a sequence (PN)Nge0 producing sufficientlymany such equations will yield the desired conclusion
tr degQQ(x1 xn α1 αn
)ge n
50 59
Royrsquos Conjecture (1999)Let s0 s1 t0 t1 u positive real numbers satisfying
max1 t0 2t1 lt mins0 2s1and
maxs0 s1 + t1 lt u lt1
2(1 + t0 + t1)
Assume that for any sufficiently large positive integer N there exists a nonndashzero polynomial PN isin Z[X0 X1] withpartial degree le N t0 in X0 partial degree le N t1 in X1 andheight le eN which satisfies∣∣∣∣∣(DkPN)(
nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1 Then
tr degQQ(x1 xn α1 αn) ge n51 59
Royrsquos Theorem (1999)
Royrsquos conjecture is equivalent to Schanuelrsquos Conjecture
52 59
Equivalence between Schanuel and Roy
Let (x α) isin CtimesCtimes and let s0 s1 t0 t1 u be positive realnumbers satisfying the inequalities of Royrsquos Conjecture Thenthe following conditions are equivalent (a) The number αeminusx is a root of unity
(b) For any sufficiently large positive integer N there exists anonndashzero polynomial QN isin Z[X0 X1] with partial degreele N t0 in X0 partial degree le N t1 in X1 and heightH(QN) le eN such that∣∣(DkQN)(mx αm)
∣∣ le exp(minusNu)
for any km isin N with k le N s0 and m le N s1
53 59
Gelrsquofondrsquos transcendence criterion
Simple form Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le n and height le en such that
|Pn(ϑ)| le eminus7n2
for all n ge 1 then ϑ is algebraic and Pn(ϑ) = 0 for all n ge 1
First extension Replace the bound for the degree by le dnthe bound for the height by ebn and the bound for |Pn(ϑ)| byeminuscdnbn with some constant c gt 0 independent of n
Some mild conditions are required on the sequences (dn)nge1 and(bn)nge1
54 59
Transcendence criterion with multiplicitiesWith derivatives Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le dn and height le ebn such that∣∣P (j)
n (ϑ)∣∣ le eminuscdnbntn
for all j in the range 0 le j lt tn and all n ge 1 then ϑ isalgebraicDue to M Laurent and D Roy applications to algebraicindependence
55 59
Criterion with several points
Goal Given a sequence of complex numbers (ϑi)ige1 if thereexists a sequence (Pn)nge1 of nonndashzero polynomials in Z[X]with Pn of degree le dn and height le ebn such that∣∣P (j)
n (ϑi)∣∣ le eminuscdnbntnsn
for 0 le j lt tn 1 le i le sn and all n ge 1 then the numbers ϑiare algebraic
D Roy Not true in general but true in some special caseswith a structure on the sequence (ϑi)ige1Combines the elimination arguments used for criteria ofalgebraic independence and for zero estimates
56 59
Small value estimates for the additive group
D Roy Small value estimates for the additive group Intern JNumber Theory to appear
Let ξ be a transcendental complex number let β σ τ and νbe non-negative real numbers let n0 be a positive integer andlet (Pn)ngen0 be a sequence of nonndashzero polynomials in Z[T ]satisfying deg(Pn) le n and H(Pn) le exp(nβ) for eachn ge n0 Suppose that β gt 1 (34)σ + τ lt 1 andν gt 1 + β minus (34)σ minus τ Then for infinitely many n we have
max|P [j]n (iξ)| 0 le i le nσ 0 le j le nτ
gt exp(minusnν)
57 59
Small value estimates for the multiplicative group
D Roy Small value estimates for the multiplicative groupActa Arith to appear
Let ξ1 ξm be multiplicatively independent complexnumbers in a field of transcendence degree 1 Under suitableassumptions on the parameters β σ τ ν for infinitely manypositive integers n there exists no nonndashzero polynomialP isin Z[T ] satisfying deg(P ) le n H(P ) le exp(nβ) and
max|P [j](ξi11 middot middot middot ξimm )| 0 le i1 im le nσ 0 le j le nτ
gt exp(minusnν)
58 59
Heidelberg Colloquium organized by the Institute for Mathematics
httpwwwmathiuni-heidelbergdevortraghtmls=1 January 22 2009
Transcendental Number TheorySchanuelrsquos Conjecture
Michel Waldschmidt
Institut de Mathematiques de Jussieu amp Paris VIhttpwwwmathjussieufrsimmiw
59 59
The exponential function
For z isin C
ez = exp(z) = 1 + z +z2
2+z3
6+ middot middot middot =
sumnge0
zn
nmiddot
Addition Theorem ez1+z2 = ez1ez2 e2iπ = 1
exp C minusrarr Ctimes ker(exp) = 2iπZ C2iπZ Ctimes
Differential equation
d
dzez = ez
4 59
First examples of transcendental numbers
sumnge1 2minusn Liouville 1844
e Hermite 1872
π Lindemann 1881
5 59
Ch Hermite ndash F Lindemann ndash C Weierstraszlig
log 2 eradic
2 Theorem of HermitendashLindemann
eradic
2 +radic
3eradic
6 LindemannndashWeierstraszlig 1888
6 59
Question of Euler and Hilbert
Leonhard Euler (1707 ndash 1783)Introductio in analysin infinitorum (1737)
Transcendence of 2radic
2
David Hilbert (1862 - 1943)ICM 1900 7ndashth Problem
Transcendence of log a log 2
7 59
AO Gelrsquofond ndash Th Schneider ndash A Baker
eπ Gelrsquofond 1929
2radic
2 log 2 log 3 Gelrsquofond and Schneider 1934
log 2 +radic
3 log 3 eradic
22radic
35radic
7 Baker 1968
8 59
Some open problems
For each of the following numbers it is expected that it istranscendental but it is not even known whether it is rationalor not
e+ π eπ πe ee ee2
eee
ππ ππ2
πππ
log π log(log 2) π log 2 (log 2)(log 3) 2log 2 (log 2)log 3
In other words we do not know whether a degree 1 polynomialcould vanish at the corresponding point but we expect thatno nonndashzero polynomial of any degree vanishes at this point
9 59
Algebraic independence
Algebraicity and transcendence deal with a single complexnumber and one variable polynomialsAlgebraic dependence or independence is the same but fortuples and multivariate polynomials
Let K sup k be an extension of fields and (θ1 θm) be amndashtuple of elements in K We say that θ1 θm arealgebraically dependent over k if there exists a nonndashzeropolynomial f isin k[X1 Xm] which vanishes at the point(θ1 θm) isin KmOtherwise we say that θ1 θm are algebraically independentover k
10 59
Special cases
In the case m = 1 to say that θ1 is algebraically independentover k just means that it is transcendental over k
Dealing with complex numbers K = C the words algebraictranscendental algebraically dependent algebraicallyindependent refer to the case k = Q
11 59
Examples
The numbersradic
2 and π are algebraically dependent The polynomial X2 minus 2 isin Z[X Y ] vanishes at (
radic2 π)
The numbers π andradicπ2 + 1 are algebraically dependent (and
both are transcendental numbers) The polynomial Y minusX2 minus 1 vanishes at (π
radicπ2 + 1)
The two numbers e and eradic
2 are algebraically independentwhich means that for any nonndashzero polynomial f isin Z[X Y ]
with rational integer coefficients the number f(e eradic
2) is notzeroSpecial case of the Lindemann-Weierstraszlig Theorem
12 59
Transcendence degree
Let Kk be a field extension The maximal number ofelements in K which are algebraically independent over k iscalled the transcendence degree of K over k and denotedtr degkK
Let t = tr degkK A subset θ1 θt of K with t elementswhich are algebraically independent is called a transcendencebasis of K over k It is the same as a maximal subset ofkndashalgebraically independent elements in KHence K is an algebraic extension of k(θ1 θt)
13 59
Transcendence degree of extensions
Assumek sub K sub L
The union of a transcendence basis of K over k and of atranscendence basis of L over K produces a transcendencebasis of L over kHence
tr degkL = tr degkK + tr degKL
An algebraic extension Kk is an extension of transcendencedegree 0 This means that there is no transcendental elementin K over k (any element in K is algebraic over k)
14 59
Algebraic independence of complex numbers
For complex numbers algebraic independence over Q or overQ is the same In particular if θ1 θm are algebraicallyindependent complex numbers then for any non-constantpolynomial f with algebraic coefficients the numberf(θ1 θm) is transcendental
For instance the two numbers e and eradic
2 are algebraicallyindependent As a consequence for any nonndashconstantpolynomial f isin Q[X Y ] with algebraic coefficients the
number f(e eradic
2) is transcendental
15 59
Schanuelrsquos Conjecture
Let x1 xn be Q-linearly independent complex numbersThen at least n of the 2n numbers x1 xn e
x1 exn arealgebraically independent
Since there are 2n numbers only the transcendence degreeover Q of the field
Q(x1 xn e
x1 exn)
is at most 2n The conjecture is that this transcendencedegree is always ge n
n le tr degQQ(x1 xn e
x1 exn)le 2n
16 59
Origin of Schanuelrsquos ConjectureCourse given by Serge Lang (1927ndash2005) at Yale in the 60rsquos
S Lang ndash Introduction to transcendental numbersAddison-Wesley 1966
also attended by M Nagata (1927ndash2008)(14th Problem of Hilbert)
Nagatarsquos Conjecture solved by E Bombieri
17 59
Schanuelrsquos Conjecture
Let x1 xn be Q-linearly independent complex numbersThen
tr degQQ(x1 xn e
x1 exn)ge n
Remark For almost all tuples (with respect to the Lebesguemeasure) the transcendence degree is 2n
18 59
AO Gelrsquofond CRAS 1934
19 59
Statement by Gelrsquofond (1934)
Let β1 βn be Q-linearly independent algebraic numbersand let logα1 logαm be Q-linearly independentlogarithms of algebraic numbers Then the numbers
eβ1 eβn logα1 logαm
are algebraically independent over Q
20 59
Further statement by Gelrsquofond
Let β1 βn be algebraic numbers with β1 6= 0 and letα1 αm be algebraic numbers with α1 6= 0 1 α2 6= 0 1αi 6= 0 Then the numbers
eβ1eβ2e βnminus1e
βn
and αα αm
21
are transcendental and there is no nontrivial algebraic relationbetween such numbers
Remark The condition on α2 should be that it is irrational
21 59
Easy consequence of Schanuelrsquos Conjecture
According to Schanuelrsquos Conjecture the following numbers arealgebraically independent
e+ π eπ πe ee ee2
eee
ππ ππ2
πππ
log π log(log 2) π log 2 (log 2)(log 3) 2log 2 (log 2)log 3
Proof This is an easy exercise
22 59
Langrsquos exercise
Define E0 = Q Inductivelyfor n ge 1 define En as thealgebraic closure of the fieldgenerated over Enminus1 by thenumbers exp(x) = ex wherex ranges over Enminus1 Let E bethe union of En n ge 0Then Schanuelrsquos Conjectureimplies that the number πdoes not belong to E
More precisely Schanuelrsquos Conjecture implies that thenumbers π log π log log π log log log π are algebraicallyindependent over E
23 59
A variant of Langrsquos exercise
Define L0 = Q Inductively for n ge 1 define Ln as thealgebraic closure of the field generated over Lnminus1 by thenumbers y where y ranges over the set of complex numberssuch that ey isin Lnminus1 Let L be the union of Ln n ge 0Then Schanuelrsquos Conjecture implies that the number e doesnot belong to L
More precisely Schanuelrsquos Conjecture implies that the
numbers e ee eee ee
ee
are algebraically independent over L
24 59
Arizona Winter School AWS2008 Tucson
Theorem [Jonathan Bober Chuangxun Cheng Brian DietelMathilde Herblot Jingjing Huang Holly Krieger DiegoMarques Jonathan Mason Martin Mereb and Robert Wilson]Schanuelrsquos Conjecture implies that the fields E and L arelinearly disjoint over Q
Definition Given a field extension FK and twosubextensions F1 F2 sube F we say F1 F2 are linearly disjointover K when the following holds Any set x1 xn sube F1
of Kndash linearly independent elements is linearly independentover F2
Reference arXiv08043550 [mathNT] 2008
25 59
Formal analogs
WD Brownawell(was a student of Schanuel)
J Axrsquos Theorem (1968) Version of SchanuelrsquosConjecture for power seriesover C(and R Coleman for powerseries over Q)
Work by WD Brownawelland K Kubota on the ellipticanalog of Axrsquos Theorem
26 59
Conjectures by A Grothendieck and Y Andre
Generalized Conjecture onPeriods by Grothendieck Dimension of theMumfordndashTate group of asmooth projective variety
Generalization by Y Andre tomotives
Case of 1ndashmotives Elliptico-Toric Conjecture ofC Bertolin
27 59
Ubiquity of Schanuelrsquos Conjecture
Other contexts pndashadic numbers Leopoldtrsquos Conjecture onthe pndashadic rank of the units of an algebraic number fieldNon-vanishing of RegulatorsNonndashdegenerescence of heightsConjecture of B Mazur on rational pointsDiophantine approximation on tori
Dipendra Prasad Gopal Prasad
28 59
Methods from logicEhud Hrushovski Boris Zilber Jonathan Kirby
Calculus of rdquopredimension functionsrdquo (E Hrushovski)
Zilberrsquos construction of a rdquopseudoexponentiationrdquo
Also A Macintyre DE Marker G Terzo AJ WilkieD Bertrand
29 59
Methods from logic Model theory
Exponential algebraicity in exponential fieldsby
Jonathan Kirby
The dimension of the exponential algebraic closure operator inan exponential field satisfies a weak Schanuel property
A corollary is that there are at most countably many essentialcounterexamples to Schanuelrsquos conjecture
arXiv 08104285v2
30 59
Known
LindemannndashWeierstraszlig Theorem = case where x1 xn arealgebraic
Let β1 βn be algebraic numbers which are linearlyindependent over Q Then the numbers eβ1 eβn arealgebraically independent over Q
31 59
Problem of Gelrsquofond and SchneiderRaised by AO Gelrsquofond in 1948 and Th Schneider in 1952
Conjecture If α is an algebraic number α 6= 0 α 6= 1 and ifβ is an irrational algebraic number of degree d then the dminus 1numbers
αβ αβ2
αβdminus1
are algebraically independent
Special case of Schanuelrsquos Conjecture Take xi = βiminus1 logαn = d so that
x1 xn e
x1 exn
islogα β logα βdminus1 logα α αβ αβ
dminus1
The conclusion of Schanuelrsquos Conjecture is
tr degQQ(logα αβ αβ
2
αβdminus1)
= d
32 59
Algebraic independence method of Gelrsquofond
AO Gelrsquofond (1948)
The two numbers 23radic2 and
23radic4 are algebraically
independentMore generally if α is analgebraic number α 6= 0α 6= 1 and if β is a algebraicnumber of degree d ge 3 thentwo at least of the numbers
αβ αβ2
αβdminus1
are algebraically independent
33 59
Tools
Transcendence criterion Replaces Liouvillersquos inequality intranscendence proofsLiouville A nonndashzero rational integer n isin Z satisfies |n| ge 1Gelrsquofond Needs to give a lower bound for |P (θ)| withP isin Z[X] 0 when θ is transcendental
Zero estimate for exponential polynomials C Hermite P Turan K Mahler R Tijdeman
Small transcendence degree AO Gelrsquofond AA Smelev R Tijdeman WD Brownawell
34 59
Large transcendence degree
GV Chudnovsky (1976)Among the numbers
αβ αβ2
αβdminus1
at least [log2 d] arealgebraically independent
GV Chudnovsky ndash On the path to Schanuelrsquosconjecture Algebraic curves close to a pointI General theory of colored sequencesII Fields of finite transcendence type and coloredsequences ResultantsStudia Sci Math Hungar 12 (1977) 125ndash157 (1980)
35 59
Partial result on the problem of Gelrsquofond and
Schneider
AO Gelrsquofond GV Chudnovskii P Philippon YuV Nesterenko
G Diaz If α is an algebraic number α 6= 0 α 6= 1 and if β isan irrational algebraic number of degree d then
tr degQQ(αβ αβ
2
αβdminus1) ge [d+ 1
2
]
36 59
Conjecture of algebraic independence of logarithms
of algebraic numbers
Denote by L the set of complex numbers λ for which eλ isalgebraic
L = logα α isin Qtimes
Hence L is a Q-vector subspace of CThe most important special case of Schanuelrsquos Conjecture is
Conjecture Let λ1 λn be Q-linearly independent elementsin L Then the numbers λ1 λn are algebraicallyindependent over Q
Not yet known that the transcendence degree is ge 2 Open problem Among all logarithms of algebraic numbersone at least is transcendental over Q(π)
37 59
Structural rank of a matrix
Let K be a field k a subfield and M a matrix with entries in KConsider the k-vector subspace E of K spanned by the entriesof M Choose an injective morphism ϕ of E into a k-vectorspace kX1 + middot middot middot+ kXn The image ϕ(M) of M is a matrixwhose entries are in the field k(X1 Xn) of rationalfractions Its rank does not depend on the choice of ϕ
This is the structural rank of M with respect to k
38 59
Example
LetM =
(bij + λij
)1leiled1lejle`
be a matrix with coefficients in Q + L Consider a basis of theQndashvector spanned by the entries and replace the elements inthis basis by unknowns This gives a new matrix M withcoefficients in a field of rational fractions the rank of which isthe structural rank of M (with respect to Q)
As a consequence of the conjecture of algebraic independenceof logarithms of algebraic numbers the rank of M should bethe same as the rank of its specialization M
39 59
Equivalence between the two conjectures
Following D Roy theconjecture on algebraicindependence of logarithms ofalgebraic numbers isequivalent to
Conjecture Any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L has a rank equal to its structuralrank
40 59
Any Polynomial is the Determinant of a Matrix
The proof of the equivalence uses the nice auxiliary result
For any P isin k[X1 Xn] there exists a square matrix withentries in the k-vector space k + kX1 + middot middot middot+ kXn whosedeterminant is P
41 59
Half of the Conjecture is solved
From a certain point of view half of the conjecture of algebraicindependence of logarithms of algebraic numbers is solved
Theorem [D Roy] The rank of any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L is at least half its structural rank
42 59
Reformulation by D Roy
Instead of taking logarithmsof algebraic numbers andlooking for the algebraicindependence relationsD Roy fixes a polynomial andlooks at the points withcoordinates logarithms ofalgebraic numbers on thecorresponding hypersurface
43 59
Reformulation by D Roy
Royrsquos reformulation of the conjecture of algebraicindependence of logarithms is
Conjecture For any algebraic subvariety V of Cn definedover the field Q of algebraic numbers the set V cap Ln is theunion of the sets E cap Ln where E ranges over the set ofvector subspaces of Cn which are defined over Q andcontained in V
Trivial Any element in E cap Ln where E is a vector subspaceof Cn defined over Q and contained in V belongs to V cap Ln
44 59
Example The Four Exponentials ConjectureTake for V the hypersurface of C4 defined by the equation
z1z4 = z2z3
The maximal Cndashvector subspaces of C4 defined over Q andcontained in V are the planes
az1 = bz2 bz4 = az3
and the planesaz1 = bz3 bz4 = az2
with (a b) isin Q2 (0 0)Hence Schanuelrsquos Conjecture implies that if λ1 λ2 λ3 λ4 arelogarithms of algebraic numbers satisfying
λ1λ4 = λ2λ3
then either λ1 λ2 are linearly dependent over Q or else λ1 λ3
are linearly dependent over Q45 59
Six Exponentials Theorem
and Four Exponentials Conjecture
A Selberg CL Siegel S Lang K Ramachandra
46 59
The Four Exponentials Conjecture
Conjecture (Th Schneider S Lang K Ramachandra) Ifx1 x2 are Qndashlinearly independent complex numbers and y1 y2
are Qndashlinearly independent complex numbers then one atleast of the four numbers
ex1y1 ex1y2 ex2y1 ex2y2
is transcendentalEquivalent statement Any 2times 2 matrix(
λ1 λ2
λ3 λ4
)with entries in L and with Qndashlinearly independent rows andQndashlinearly independent columns has maximal rank 2
47 59
How could we attack Schanuelrsquos Conjecture
Let x1 xn be Qndashlinearly independent complex numbersFollowing the transcendence methods of Hermite GelrsquofondSchneider one may start by introducing an auxiliaryfunction
F (z) = P (z ez)
where P isin Z[X0 X1] is a nonndashzero polynomial One considersthe derivatives (
d
dz
)kF = (DkP )(z ez)
at the pointsm1x1 + middot middot middot+mnxn
for various values of (m1 mn) isin Zn
48 59
Auxiliary functionLet D denote the derivation
D =part
partX0
+X1part
partX1
over the ring C[X0 X1] Recall that x1 xn are Qndashlinearlyindependent complex numbers Let α1 αn be nonndashzerocomplex numbersThe transcendence machinery produces a sequence (PN)Nge0
of polynomials with integer coefficients satisfying∣∣∣∣∣(DkPN)(nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1
49 59
Royrsquos approach to Schanuelrsquos Conjecture (1999)
If the number of equations we produce is too small such a setof relations does not contain any information The existenceof a sequence of nonndashtrivial polynomials (PN)Nge0 follows fromlinear algebra
On the other hand following D Roy one may expect that theexistence of such a sequence (PN)Nge0 producing sufficientlymany such equations will yield the desired conclusion
tr degQQ(x1 xn α1 αn
)ge n
50 59
Royrsquos Conjecture (1999)Let s0 s1 t0 t1 u positive real numbers satisfying
max1 t0 2t1 lt mins0 2s1and
maxs0 s1 + t1 lt u lt1
2(1 + t0 + t1)
Assume that for any sufficiently large positive integer N there exists a nonndashzero polynomial PN isin Z[X0 X1] withpartial degree le N t0 in X0 partial degree le N t1 in X1 andheight le eN which satisfies∣∣∣∣∣(DkPN)(
nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1 Then
tr degQQ(x1 xn α1 αn) ge n51 59
Royrsquos Theorem (1999)
Royrsquos conjecture is equivalent to Schanuelrsquos Conjecture
52 59
Equivalence between Schanuel and Roy
Let (x α) isin CtimesCtimes and let s0 s1 t0 t1 u be positive realnumbers satisfying the inequalities of Royrsquos Conjecture Thenthe following conditions are equivalent (a) The number αeminusx is a root of unity
(b) For any sufficiently large positive integer N there exists anonndashzero polynomial QN isin Z[X0 X1] with partial degreele N t0 in X0 partial degree le N t1 in X1 and heightH(QN) le eN such that∣∣(DkQN)(mx αm)
∣∣ le exp(minusNu)
for any km isin N with k le N s0 and m le N s1
53 59
Gelrsquofondrsquos transcendence criterion
Simple form Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le n and height le en such that
|Pn(ϑ)| le eminus7n2
for all n ge 1 then ϑ is algebraic and Pn(ϑ) = 0 for all n ge 1
First extension Replace the bound for the degree by le dnthe bound for the height by ebn and the bound for |Pn(ϑ)| byeminuscdnbn with some constant c gt 0 independent of n
Some mild conditions are required on the sequences (dn)nge1 and(bn)nge1
54 59
Transcendence criterion with multiplicitiesWith derivatives Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le dn and height le ebn such that∣∣P (j)
n (ϑ)∣∣ le eminuscdnbntn
for all j in the range 0 le j lt tn and all n ge 1 then ϑ isalgebraicDue to M Laurent and D Roy applications to algebraicindependence
55 59
Criterion with several points
Goal Given a sequence of complex numbers (ϑi)ige1 if thereexists a sequence (Pn)nge1 of nonndashzero polynomials in Z[X]with Pn of degree le dn and height le ebn such that∣∣P (j)
n (ϑi)∣∣ le eminuscdnbntnsn
for 0 le j lt tn 1 le i le sn and all n ge 1 then the numbers ϑiare algebraic
D Roy Not true in general but true in some special caseswith a structure on the sequence (ϑi)ige1Combines the elimination arguments used for criteria ofalgebraic independence and for zero estimates
56 59
Small value estimates for the additive group
D Roy Small value estimates for the additive group Intern JNumber Theory to appear
Let ξ be a transcendental complex number let β σ τ and νbe non-negative real numbers let n0 be a positive integer andlet (Pn)ngen0 be a sequence of nonndashzero polynomials in Z[T ]satisfying deg(Pn) le n and H(Pn) le exp(nβ) for eachn ge n0 Suppose that β gt 1 (34)σ + τ lt 1 andν gt 1 + β minus (34)σ minus τ Then for infinitely many n we have
max|P [j]n (iξ)| 0 le i le nσ 0 le j le nτ
gt exp(minusnν)
57 59
Small value estimates for the multiplicative group
D Roy Small value estimates for the multiplicative groupActa Arith to appear
Let ξ1 ξm be multiplicatively independent complexnumbers in a field of transcendence degree 1 Under suitableassumptions on the parameters β σ τ ν for infinitely manypositive integers n there exists no nonndashzero polynomialP isin Z[T ] satisfying deg(P ) le n H(P ) le exp(nβ) and
max|P [j](ξi11 middot middot middot ξimm )| 0 le i1 im le nσ 0 le j le nτ
gt exp(minusnν)
58 59
Heidelberg Colloquium organized by the Institute for Mathematics
httpwwwmathiuni-heidelbergdevortraghtmls=1 January 22 2009
Transcendental Number TheorySchanuelrsquos Conjecture
Michel Waldschmidt
Institut de Mathematiques de Jussieu amp Paris VIhttpwwwmathjussieufrsimmiw
59 59
First examples of transcendental numbers
sumnge1 2minusn Liouville 1844
e Hermite 1872
π Lindemann 1881
5 59
Ch Hermite ndash F Lindemann ndash C Weierstraszlig
log 2 eradic
2 Theorem of HermitendashLindemann
eradic
2 +radic
3eradic
6 LindemannndashWeierstraszlig 1888
6 59
Question of Euler and Hilbert
Leonhard Euler (1707 ndash 1783)Introductio in analysin infinitorum (1737)
Transcendence of 2radic
2
David Hilbert (1862 - 1943)ICM 1900 7ndashth Problem
Transcendence of log a log 2
7 59
AO Gelrsquofond ndash Th Schneider ndash A Baker
eπ Gelrsquofond 1929
2radic
2 log 2 log 3 Gelrsquofond and Schneider 1934
log 2 +radic
3 log 3 eradic
22radic
35radic
7 Baker 1968
8 59
Some open problems
For each of the following numbers it is expected that it istranscendental but it is not even known whether it is rationalor not
e+ π eπ πe ee ee2
eee
ππ ππ2
πππ
log π log(log 2) π log 2 (log 2)(log 3) 2log 2 (log 2)log 3
In other words we do not know whether a degree 1 polynomialcould vanish at the corresponding point but we expect thatno nonndashzero polynomial of any degree vanishes at this point
9 59
Algebraic independence
Algebraicity and transcendence deal with a single complexnumber and one variable polynomialsAlgebraic dependence or independence is the same but fortuples and multivariate polynomials
Let K sup k be an extension of fields and (θ1 θm) be amndashtuple of elements in K We say that θ1 θm arealgebraically dependent over k if there exists a nonndashzeropolynomial f isin k[X1 Xm] which vanishes at the point(θ1 θm) isin KmOtherwise we say that θ1 θm are algebraically independentover k
10 59
Special cases
In the case m = 1 to say that θ1 is algebraically independentover k just means that it is transcendental over k
Dealing with complex numbers K = C the words algebraictranscendental algebraically dependent algebraicallyindependent refer to the case k = Q
11 59
Examples
The numbersradic
2 and π are algebraically dependent The polynomial X2 minus 2 isin Z[X Y ] vanishes at (
radic2 π)
The numbers π andradicπ2 + 1 are algebraically dependent (and
both are transcendental numbers) The polynomial Y minusX2 minus 1 vanishes at (π
radicπ2 + 1)
The two numbers e and eradic
2 are algebraically independentwhich means that for any nonndashzero polynomial f isin Z[X Y ]
with rational integer coefficients the number f(e eradic
2) is notzeroSpecial case of the Lindemann-Weierstraszlig Theorem
12 59
Transcendence degree
Let Kk be a field extension The maximal number ofelements in K which are algebraically independent over k iscalled the transcendence degree of K over k and denotedtr degkK
Let t = tr degkK A subset θ1 θt of K with t elementswhich are algebraically independent is called a transcendencebasis of K over k It is the same as a maximal subset ofkndashalgebraically independent elements in KHence K is an algebraic extension of k(θ1 θt)
13 59
Transcendence degree of extensions
Assumek sub K sub L
The union of a transcendence basis of K over k and of atranscendence basis of L over K produces a transcendencebasis of L over kHence
tr degkL = tr degkK + tr degKL
An algebraic extension Kk is an extension of transcendencedegree 0 This means that there is no transcendental elementin K over k (any element in K is algebraic over k)
14 59
Algebraic independence of complex numbers
For complex numbers algebraic independence over Q or overQ is the same In particular if θ1 θm are algebraicallyindependent complex numbers then for any non-constantpolynomial f with algebraic coefficients the numberf(θ1 θm) is transcendental
For instance the two numbers e and eradic
2 are algebraicallyindependent As a consequence for any nonndashconstantpolynomial f isin Q[X Y ] with algebraic coefficients the
number f(e eradic
2) is transcendental
15 59
Schanuelrsquos Conjecture
Let x1 xn be Q-linearly independent complex numbersThen at least n of the 2n numbers x1 xn e
x1 exn arealgebraically independent
Since there are 2n numbers only the transcendence degreeover Q of the field
Q(x1 xn e
x1 exn)
is at most 2n The conjecture is that this transcendencedegree is always ge n
n le tr degQQ(x1 xn e
x1 exn)le 2n
16 59
Origin of Schanuelrsquos ConjectureCourse given by Serge Lang (1927ndash2005) at Yale in the 60rsquos
S Lang ndash Introduction to transcendental numbersAddison-Wesley 1966
also attended by M Nagata (1927ndash2008)(14th Problem of Hilbert)
Nagatarsquos Conjecture solved by E Bombieri
17 59
Schanuelrsquos Conjecture
Let x1 xn be Q-linearly independent complex numbersThen
tr degQQ(x1 xn e
x1 exn)ge n
Remark For almost all tuples (with respect to the Lebesguemeasure) the transcendence degree is 2n
18 59
AO Gelrsquofond CRAS 1934
19 59
Statement by Gelrsquofond (1934)
Let β1 βn be Q-linearly independent algebraic numbersand let logα1 logαm be Q-linearly independentlogarithms of algebraic numbers Then the numbers
eβ1 eβn logα1 logαm
are algebraically independent over Q
20 59
Further statement by Gelrsquofond
Let β1 βn be algebraic numbers with β1 6= 0 and letα1 αm be algebraic numbers with α1 6= 0 1 α2 6= 0 1αi 6= 0 Then the numbers
eβ1eβ2e βnminus1e
βn
and αα αm
21
are transcendental and there is no nontrivial algebraic relationbetween such numbers
Remark The condition on α2 should be that it is irrational
21 59
Easy consequence of Schanuelrsquos Conjecture
According to Schanuelrsquos Conjecture the following numbers arealgebraically independent
e+ π eπ πe ee ee2
eee
ππ ππ2
πππ
log π log(log 2) π log 2 (log 2)(log 3) 2log 2 (log 2)log 3
Proof This is an easy exercise
22 59
Langrsquos exercise
Define E0 = Q Inductivelyfor n ge 1 define En as thealgebraic closure of the fieldgenerated over Enminus1 by thenumbers exp(x) = ex wherex ranges over Enminus1 Let E bethe union of En n ge 0Then Schanuelrsquos Conjectureimplies that the number πdoes not belong to E
More precisely Schanuelrsquos Conjecture implies that thenumbers π log π log log π log log log π are algebraicallyindependent over E
23 59
A variant of Langrsquos exercise
Define L0 = Q Inductively for n ge 1 define Ln as thealgebraic closure of the field generated over Lnminus1 by thenumbers y where y ranges over the set of complex numberssuch that ey isin Lnminus1 Let L be the union of Ln n ge 0Then Schanuelrsquos Conjecture implies that the number e doesnot belong to L
More precisely Schanuelrsquos Conjecture implies that the
numbers e ee eee ee
ee
are algebraically independent over L
24 59
Arizona Winter School AWS2008 Tucson
Theorem [Jonathan Bober Chuangxun Cheng Brian DietelMathilde Herblot Jingjing Huang Holly Krieger DiegoMarques Jonathan Mason Martin Mereb and Robert Wilson]Schanuelrsquos Conjecture implies that the fields E and L arelinearly disjoint over Q
Definition Given a field extension FK and twosubextensions F1 F2 sube F we say F1 F2 are linearly disjointover K when the following holds Any set x1 xn sube F1
of Kndash linearly independent elements is linearly independentover F2
Reference arXiv08043550 [mathNT] 2008
25 59
Formal analogs
WD Brownawell(was a student of Schanuel)
J Axrsquos Theorem (1968) Version of SchanuelrsquosConjecture for power seriesover C(and R Coleman for powerseries over Q)
Work by WD Brownawelland K Kubota on the ellipticanalog of Axrsquos Theorem
26 59
Conjectures by A Grothendieck and Y Andre
Generalized Conjecture onPeriods by Grothendieck Dimension of theMumfordndashTate group of asmooth projective variety
Generalization by Y Andre tomotives
Case of 1ndashmotives Elliptico-Toric Conjecture ofC Bertolin
27 59
Ubiquity of Schanuelrsquos Conjecture
Other contexts pndashadic numbers Leopoldtrsquos Conjecture onthe pndashadic rank of the units of an algebraic number fieldNon-vanishing of RegulatorsNonndashdegenerescence of heightsConjecture of B Mazur on rational pointsDiophantine approximation on tori
Dipendra Prasad Gopal Prasad
28 59
Methods from logicEhud Hrushovski Boris Zilber Jonathan Kirby
Calculus of rdquopredimension functionsrdquo (E Hrushovski)
Zilberrsquos construction of a rdquopseudoexponentiationrdquo
Also A Macintyre DE Marker G Terzo AJ WilkieD Bertrand
29 59
Methods from logic Model theory
Exponential algebraicity in exponential fieldsby
Jonathan Kirby
The dimension of the exponential algebraic closure operator inan exponential field satisfies a weak Schanuel property
A corollary is that there are at most countably many essentialcounterexamples to Schanuelrsquos conjecture
arXiv 08104285v2
30 59
Known
LindemannndashWeierstraszlig Theorem = case where x1 xn arealgebraic
Let β1 βn be algebraic numbers which are linearlyindependent over Q Then the numbers eβ1 eβn arealgebraically independent over Q
31 59
Problem of Gelrsquofond and SchneiderRaised by AO Gelrsquofond in 1948 and Th Schneider in 1952
Conjecture If α is an algebraic number α 6= 0 α 6= 1 and ifβ is an irrational algebraic number of degree d then the dminus 1numbers
αβ αβ2
αβdminus1
are algebraically independent
Special case of Schanuelrsquos Conjecture Take xi = βiminus1 logαn = d so that
x1 xn e
x1 exn
islogα β logα βdminus1 logα α αβ αβ
dminus1
The conclusion of Schanuelrsquos Conjecture is
tr degQQ(logα αβ αβ
2
αβdminus1)
= d
32 59
Algebraic independence method of Gelrsquofond
AO Gelrsquofond (1948)
The two numbers 23radic2 and
23radic4 are algebraically
independentMore generally if α is analgebraic number α 6= 0α 6= 1 and if β is a algebraicnumber of degree d ge 3 thentwo at least of the numbers
αβ αβ2
αβdminus1
are algebraically independent
33 59
Tools
Transcendence criterion Replaces Liouvillersquos inequality intranscendence proofsLiouville A nonndashzero rational integer n isin Z satisfies |n| ge 1Gelrsquofond Needs to give a lower bound for |P (θ)| withP isin Z[X] 0 when θ is transcendental
Zero estimate for exponential polynomials C Hermite P Turan K Mahler R Tijdeman
Small transcendence degree AO Gelrsquofond AA Smelev R Tijdeman WD Brownawell
34 59
Large transcendence degree
GV Chudnovsky (1976)Among the numbers
αβ αβ2
αβdminus1
at least [log2 d] arealgebraically independent
GV Chudnovsky ndash On the path to Schanuelrsquosconjecture Algebraic curves close to a pointI General theory of colored sequencesII Fields of finite transcendence type and coloredsequences ResultantsStudia Sci Math Hungar 12 (1977) 125ndash157 (1980)
35 59
Partial result on the problem of Gelrsquofond and
Schneider
AO Gelrsquofond GV Chudnovskii P Philippon YuV Nesterenko
G Diaz If α is an algebraic number α 6= 0 α 6= 1 and if β isan irrational algebraic number of degree d then
tr degQQ(αβ αβ
2
αβdminus1) ge [d+ 1
2
]
36 59
Conjecture of algebraic independence of logarithms
of algebraic numbers
Denote by L the set of complex numbers λ for which eλ isalgebraic
L = logα α isin Qtimes
Hence L is a Q-vector subspace of CThe most important special case of Schanuelrsquos Conjecture is
Conjecture Let λ1 λn be Q-linearly independent elementsin L Then the numbers λ1 λn are algebraicallyindependent over Q
Not yet known that the transcendence degree is ge 2 Open problem Among all logarithms of algebraic numbersone at least is transcendental over Q(π)
37 59
Structural rank of a matrix
Let K be a field k a subfield and M a matrix with entries in KConsider the k-vector subspace E of K spanned by the entriesof M Choose an injective morphism ϕ of E into a k-vectorspace kX1 + middot middot middot+ kXn The image ϕ(M) of M is a matrixwhose entries are in the field k(X1 Xn) of rationalfractions Its rank does not depend on the choice of ϕ
This is the structural rank of M with respect to k
38 59
Example
LetM =
(bij + λij
)1leiled1lejle`
be a matrix with coefficients in Q + L Consider a basis of theQndashvector spanned by the entries and replace the elements inthis basis by unknowns This gives a new matrix M withcoefficients in a field of rational fractions the rank of which isthe structural rank of M (with respect to Q)
As a consequence of the conjecture of algebraic independenceof logarithms of algebraic numbers the rank of M should bethe same as the rank of its specialization M
39 59
Equivalence between the two conjectures
Following D Roy theconjecture on algebraicindependence of logarithms ofalgebraic numbers isequivalent to
Conjecture Any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L has a rank equal to its structuralrank
40 59
Any Polynomial is the Determinant of a Matrix
The proof of the equivalence uses the nice auxiliary result
For any P isin k[X1 Xn] there exists a square matrix withentries in the k-vector space k + kX1 + middot middot middot+ kXn whosedeterminant is P
41 59
Half of the Conjecture is solved
From a certain point of view half of the conjecture of algebraicindependence of logarithms of algebraic numbers is solved
Theorem [D Roy] The rank of any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L is at least half its structural rank
42 59
Reformulation by D Roy
Instead of taking logarithmsof algebraic numbers andlooking for the algebraicindependence relationsD Roy fixes a polynomial andlooks at the points withcoordinates logarithms ofalgebraic numbers on thecorresponding hypersurface
43 59
Reformulation by D Roy
Royrsquos reformulation of the conjecture of algebraicindependence of logarithms is
Conjecture For any algebraic subvariety V of Cn definedover the field Q of algebraic numbers the set V cap Ln is theunion of the sets E cap Ln where E ranges over the set ofvector subspaces of Cn which are defined over Q andcontained in V
Trivial Any element in E cap Ln where E is a vector subspaceof Cn defined over Q and contained in V belongs to V cap Ln
44 59
Example The Four Exponentials ConjectureTake for V the hypersurface of C4 defined by the equation
z1z4 = z2z3
The maximal Cndashvector subspaces of C4 defined over Q andcontained in V are the planes
az1 = bz2 bz4 = az3
and the planesaz1 = bz3 bz4 = az2
with (a b) isin Q2 (0 0)Hence Schanuelrsquos Conjecture implies that if λ1 λ2 λ3 λ4 arelogarithms of algebraic numbers satisfying
λ1λ4 = λ2λ3
then either λ1 λ2 are linearly dependent over Q or else λ1 λ3
are linearly dependent over Q45 59
Six Exponentials Theorem
and Four Exponentials Conjecture
A Selberg CL Siegel S Lang K Ramachandra
46 59
The Four Exponentials Conjecture
Conjecture (Th Schneider S Lang K Ramachandra) Ifx1 x2 are Qndashlinearly independent complex numbers and y1 y2
are Qndashlinearly independent complex numbers then one atleast of the four numbers
ex1y1 ex1y2 ex2y1 ex2y2
is transcendentalEquivalent statement Any 2times 2 matrix(
λ1 λ2
λ3 λ4
)with entries in L and with Qndashlinearly independent rows andQndashlinearly independent columns has maximal rank 2
47 59
How could we attack Schanuelrsquos Conjecture
Let x1 xn be Qndashlinearly independent complex numbersFollowing the transcendence methods of Hermite GelrsquofondSchneider one may start by introducing an auxiliaryfunction
F (z) = P (z ez)
where P isin Z[X0 X1] is a nonndashzero polynomial One considersthe derivatives (
d
dz
)kF = (DkP )(z ez)
at the pointsm1x1 + middot middot middot+mnxn
for various values of (m1 mn) isin Zn
48 59
Auxiliary functionLet D denote the derivation
D =part
partX0
+X1part
partX1
over the ring C[X0 X1] Recall that x1 xn are Qndashlinearlyindependent complex numbers Let α1 αn be nonndashzerocomplex numbersThe transcendence machinery produces a sequence (PN)Nge0
of polynomials with integer coefficients satisfying∣∣∣∣∣(DkPN)(nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1
49 59
Royrsquos approach to Schanuelrsquos Conjecture (1999)
If the number of equations we produce is too small such a setof relations does not contain any information The existenceof a sequence of nonndashtrivial polynomials (PN)Nge0 follows fromlinear algebra
On the other hand following D Roy one may expect that theexistence of such a sequence (PN)Nge0 producing sufficientlymany such equations will yield the desired conclusion
tr degQQ(x1 xn α1 αn
)ge n
50 59
Royrsquos Conjecture (1999)Let s0 s1 t0 t1 u positive real numbers satisfying
max1 t0 2t1 lt mins0 2s1and
maxs0 s1 + t1 lt u lt1
2(1 + t0 + t1)
Assume that for any sufficiently large positive integer N there exists a nonndashzero polynomial PN isin Z[X0 X1] withpartial degree le N t0 in X0 partial degree le N t1 in X1 andheight le eN which satisfies∣∣∣∣∣(DkPN)(
nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1 Then
tr degQQ(x1 xn α1 αn) ge n51 59
Royrsquos Theorem (1999)
Royrsquos conjecture is equivalent to Schanuelrsquos Conjecture
52 59
Equivalence between Schanuel and Roy
Let (x α) isin CtimesCtimes and let s0 s1 t0 t1 u be positive realnumbers satisfying the inequalities of Royrsquos Conjecture Thenthe following conditions are equivalent (a) The number αeminusx is a root of unity
(b) For any sufficiently large positive integer N there exists anonndashzero polynomial QN isin Z[X0 X1] with partial degreele N t0 in X0 partial degree le N t1 in X1 and heightH(QN) le eN such that∣∣(DkQN)(mx αm)
∣∣ le exp(minusNu)
for any km isin N with k le N s0 and m le N s1
53 59
Gelrsquofondrsquos transcendence criterion
Simple form Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le n and height le en such that
|Pn(ϑ)| le eminus7n2
for all n ge 1 then ϑ is algebraic and Pn(ϑ) = 0 for all n ge 1
First extension Replace the bound for the degree by le dnthe bound for the height by ebn and the bound for |Pn(ϑ)| byeminuscdnbn with some constant c gt 0 independent of n
Some mild conditions are required on the sequences (dn)nge1 and(bn)nge1
54 59
Transcendence criterion with multiplicitiesWith derivatives Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le dn and height le ebn such that∣∣P (j)
n (ϑ)∣∣ le eminuscdnbntn
for all j in the range 0 le j lt tn and all n ge 1 then ϑ isalgebraicDue to M Laurent and D Roy applications to algebraicindependence
55 59
Criterion with several points
Goal Given a sequence of complex numbers (ϑi)ige1 if thereexists a sequence (Pn)nge1 of nonndashzero polynomials in Z[X]with Pn of degree le dn and height le ebn such that∣∣P (j)
n (ϑi)∣∣ le eminuscdnbntnsn
for 0 le j lt tn 1 le i le sn and all n ge 1 then the numbers ϑiare algebraic
D Roy Not true in general but true in some special caseswith a structure on the sequence (ϑi)ige1Combines the elimination arguments used for criteria ofalgebraic independence and for zero estimates
56 59
Small value estimates for the additive group
D Roy Small value estimates for the additive group Intern JNumber Theory to appear
Let ξ be a transcendental complex number let β σ τ and νbe non-negative real numbers let n0 be a positive integer andlet (Pn)ngen0 be a sequence of nonndashzero polynomials in Z[T ]satisfying deg(Pn) le n and H(Pn) le exp(nβ) for eachn ge n0 Suppose that β gt 1 (34)σ + τ lt 1 andν gt 1 + β minus (34)σ minus τ Then for infinitely many n we have
max|P [j]n (iξ)| 0 le i le nσ 0 le j le nτ
gt exp(minusnν)
57 59
Small value estimates for the multiplicative group
D Roy Small value estimates for the multiplicative groupActa Arith to appear
Let ξ1 ξm be multiplicatively independent complexnumbers in a field of transcendence degree 1 Under suitableassumptions on the parameters β σ τ ν for infinitely manypositive integers n there exists no nonndashzero polynomialP isin Z[T ] satisfying deg(P ) le n H(P ) le exp(nβ) and
max|P [j](ξi11 middot middot middot ξimm )| 0 le i1 im le nσ 0 le j le nτ
gt exp(minusnν)
58 59
Heidelberg Colloquium organized by the Institute for Mathematics
httpwwwmathiuni-heidelbergdevortraghtmls=1 January 22 2009
Transcendental Number TheorySchanuelrsquos Conjecture
Michel Waldschmidt
Institut de Mathematiques de Jussieu amp Paris VIhttpwwwmathjussieufrsimmiw
59 59
Ch Hermite ndash F Lindemann ndash C Weierstraszlig
log 2 eradic
2 Theorem of HermitendashLindemann
eradic
2 +radic
3eradic
6 LindemannndashWeierstraszlig 1888
6 59
Question of Euler and Hilbert
Leonhard Euler (1707 ndash 1783)Introductio in analysin infinitorum (1737)
Transcendence of 2radic
2
David Hilbert (1862 - 1943)ICM 1900 7ndashth Problem
Transcendence of log a log 2
7 59
AO Gelrsquofond ndash Th Schneider ndash A Baker
eπ Gelrsquofond 1929
2radic
2 log 2 log 3 Gelrsquofond and Schneider 1934
log 2 +radic
3 log 3 eradic
22radic
35radic
7 Baker 1968
8 59
Some open problems
For each of the following numbers it is expected that it istranscendental but it is not even known whether it is rationalor not
e+ π eπ πe ee ee2
eee
ππ ππ2
πππ
log π log(log 2) π log 2 (log 2)(log 3) 2log 2 (log 2)log 3
In other words we do not know whether a degree 1 polynomialcould vanish at the corresponding point but we expect thatno nonndashzero polynomial of any degree vanishes at this point
9 59
Algebraic independence
Algebraicity and transcendence deal with a single complexnumber and one variable polynomialsAlgebraic dependence or independence is the same but fortuples and multivariate polynomials
Let K sup k be an extension of fields and (θ1 θm) be amndashtuple of elements in K We say that θ1 θm arealgebraically dependent over k if there exists a nonndashzeropolynomial f isin k[X1 Xm] which vanishes at the point(θ1 θm) isin KmOtherwise we say that θ1 θm are algebraically independentover k
10 59
Special cases
In the case m = 1 to say that θ1 is algebraically independentover k just means that it is transcendental over k
Dealing with complex numbers K = C the words algebraictranscendental algebraically dependent algebraicallyindependent refer to the case k = Q
11 59
Examples
The numbersradic
2 and π are algebraically dependent The polynomial X2 minus 2 isin Z[X Y ] vanishes at (
radic2 π)
The numbers π andradicπ2 + 1 are algebraically dependent (and
both are transcendental numbers) The polynomial Y minusX2 minus 1 vanishes at (π
radicπ2 + 1)
The two numbers e and eradic
2 are algebraically independentwhich means that for any nonndashzero polynomial f isin Z[X Y ]
with rational integer coefficients the number f(e eradic
2) is notzeroSpecial case of the Lindemann-Weierstraszlig Theorem
12 59
Transcendence degree
Let Kk be a field extension The maximal number ofelements in K which are algebraically independent over k iscalled the transcendence degree of K over k and denotedtr degkK
Let t = tr degkK A subset θ1 θt of K with t elementswhich are algebraically independent is called a transcendencebasis of K over k It is the same as a maximal subset ofkndashalgebraically independent elements in KHence K is an algebraic extension of k(θ1 θt)
13 59
Transcendence degree of extensions
Assumek sub K sub L
The union of a transcendence basis of K over k and of atranscendence basis of L over K produces a transcendencebasis of L over kHence
tr degkL = tr degkK + tr degKL
An algebraic extension Kk is an extension of transcendencedegree 0 This means that there is no transcendental elementin K over k (any element in K is algebraic over k)
14 59
Algebraic independence of complex numbers
For complex numbers algebraic independence over Q or overQ is the same In particular if θ1 θm are algebraicallyindependent complex numbers then for any non-constantpolynomial f with algebraic coefficients the numberf(θ1 θm) is transcendental
For instance the two numbers e and eradic
2 are algebraicallyindependent As a consequence for any nonndashconstantpolynomial f isin Q[X Y ] with algebraic coefficients the
number f(e eradic
2) is transcendental
15 59
Schanuelrsquos Conjecture
Let x1 xn be Q-linearly independent complex numbersThen at least n of the 2n numbers x1 xn e
x1 exn arealgebraically independent
Since there are 2n numbers only the transcendence degreeover Q of the field
Q(x1 xn e
x1 exn)
is at most 2n The conjecture is that this transcendencedegree is always ge n
n le tr degQQ(x1 xn e
x1 exn)le 2n
16 59
Origin of Schanuelrsquos ConjectureCourse given by Serge Lang (1927ndash2005) at Yale in the 60rsquos
S Lang ndash Introduction to transcendental numbersAddison-Wesley 1966
also attended by M Nagata (1927ndash2008)(14th Problem of Hilbert)
Nagatarsquos Conjecture solved by E Bombieri
17 59
Schanuelrsquos Conjecture
Let x1 xn be Q-linearly independent complex numbersThen
tr degQQ(x1 xn e
x1 exn)ge n
Remark For almost all tuples (with respect to the Lebesguemeasure) the transcendence degree is 2n
18 59
AO Gelrsquofond CRAS 1934
19 59
Statement by Gelrsquofond (1934)
Let β1 βn be Q-linearly independent algebraic numbersand let logα1 logαm be Q-linearly independentlogarithms of algebraic numbers Then the numbers
eβ1 eβn logα1 logαm
are algebraically independent over Q
20 59
Further statement by Gelrsquofond
Let β1 βn be algebraic numbers with β1 6= 0 and letα1 αm be algebraic numbers with α1 6= 0 1 α2 6= 0 1αi 6= 0 Then the numbers
eβ1eβ2e βnminus1e
βn
and αα αm
21
are transcendental and there is no nontrivial algebraic relationbetween such numbers
Remark The condition on α2 should be that it is irrational
21 59
Easy consequence of Schanuelrsquos Conjecture
According to Schanuelrsquos Conjecture the following numbers arealgebraically independent
e+ π eπ πe ee ee2
eee
ππ ππ2
πππ
log π log(log 2) π log 2 (log 2)(log 3) 2log 2 (log 2)log 3
Proof This is an easy exercise
22 59
Langrsquos exercise
Define E0 = Q Inductivelyfor n ge 1 define En as thealgebraic closure of the fieldgenerated over Enminus1 by thenumbers exp(x) = ex wherex ranges over Enminus1 Let E bethe union of En n ge 0Then Schanuelrsquos Conjectureimplies that the number πdoes not belong to E
More precisely Schanuelrsquos Conjecture implies that thenumbers π log π log log π log log log π are algebraicallyindependent over E
23 59
A variant of Langrsquos exercise
Define L0 = Q Inductively for n ge 1 define Ln as thealgebraic closure of the field generated over Lnminus1 by thenumbers y where y ranges over the set of complex numberssuch that ey isin Lnminus1 Let L be the union of Ln n ge 0Then Schanuelrsquos Conjecture implies that the number e doesnot belong to L
More precisely Schanuelrsquos Conjecture implies that the
numbers e ee eee ee
ee
are algebraically independent over L
24 59
Arizona Winter School AWS2008 Tucson
Theorem [Jonathan Bober Chuangxun Cheng Brian DietelMathilde Herblot Jingjing Huang Holly Krieger DiegoMarques Jonathan Mason Martin Mereb and Robert Wilson]Schanuelrsquos Conjecture implies that the fields E and L arelinearly disjoint over Q
Definition Given a field extension FK and twosubextensions F1 F2 sube F we say F1 F2 are linearly disjointover K when the following holds Any set x1 xn sube F1
of Kndash linearly independent elements is linearly independentover F2
Reference arXiv08043550 [mathNT] 2008
25 59
Formal analogs
WD Brownawell(was a student of Schanuel)
J Axrsquos Theorem (1968) Version of SchanuelrsquosConjecture for power seriesover C(and R Coleman for powerseries over Q)
Work by WD Brownawelland K Kubota on the ellipticanalog of Axrsquos Theorem
26 59
Conjectures by A Grothendieck and Y Andre
Generalized Conjecture onPeriods by Grothendieck Dimension of theMumfordndashTate group of asmooth projective variety
Generalization by Y Andre tomotives
Case of 1ndashmotives Elliptico-Toric Conjecture ofC Bertolin
27 59
Ubiquity of Schanuelrsquos Conjecture
Other contexts pndashadic numbers Leopoldtrsquos Conjecture onthe pndashadic rank of the units of an algebraic number fieldNon-vanishing of RegulatorsNonndashdegenerescence of heightsConjecture of B Mazur on rational pointsDiophantine approximation on tori
Dipendra Prasad Gopal Prasad
28 59
Methods from logicEhud Hrushovski Boris Zilber Jonathan Kirby
Calculus of rdquopredimension functionsrdquo (E Hrushovski)
Zilberrsquos construction of a rdquopseudoexponentiationrdquo
Also A Macintyre DE Marker G Terzo AJ WilkieD Bertrand
29 59
Methods from logic Model theory
Exponential algebraicity in exponential fieldsby
Jonathan Kirby
The dimension of the exponential algebraic closure operator inan exponential field satisfies a weak Schanuel property
A corollary is that there are at most countably many essentialcounterexamples to Schanuelrsquos conjecture
arXiv 08104285v2
30 59
Known
LindemannndashWeierstraszlig Theorem = case where x1 xn arealgebraic
Let β1 βn be algebraic numbers which are linearlyindependent over Q Then the numbers eβ1 eβn arealgebraically independent over Q
31 59
Problem of Gelrsquofond and SchneiderRaised by AO Gelrsquofond in 1948 and Th Schneider in 1952
Conjecture If α is an algebraic number α 6= 0 α 6= 1 and ifβ is an irrational algebraic number of degree d then the dminus 1numbers
αβ αβ2
αβdminus1
are algebraically independent
Special case of Schanuelrsquos Conjecture Take xi = βiminus1 logαn = d so that
x1 xn e
x1 exn
islogα β logα βdminus1 logα α αβ αβ
dminus1
The conclusion of Schanuelrsquos Conjecture is
tr degQQ(logα αβ αβ
2
αβdminus1)
= d
32 59
Algebraic independence method of Gelrsquofond
AO Gelrsquofond (1948)
The two numbers 23radic2 and
23radic4 are algebraically
independentMore generally if α is analgebraic number α 6= 0α 6= 1 and if β is a algebraicnumber of degree d ge 3 thentwo at least of the numbers
αβ αβ2
αβdminus1
are algebraically independent
33 59
Tools
Transcendence criterion Replaces Liouvillersquos inequality intranscendence proofsLiouville A nonndashzero rational integer n isin Z satisfies |n| ge 1Gelrsquofond Needs to give a lower bound for |P (θ)| withP isin Z[X] 0 when θ is transcendental
Zero estimate for exponential polynomials C Hermite P Turan K Mahler R Tijdeman
Small transcendence degree AO Gelrsquofond AA Smelev R Tijdeman WD Brownawell
34 59
Large transcendence degree
GV Chudnovsky (1976)Among the numbers
αβ αβ2
αβdminus1
at least [log2 d] arealgebraically independent
GV Chudnovsky ndash On the path to Schanuelrsquosconjecture Algebraic curves close to a pointI General theory of colored sequencesII Fields of finite transcendence type and coloredsequences ResultantsStudia Sci Math Hungar 12 (1977) 125ndash157 (1980)
35 59
Partial result on the problem of Gelrsquofond and
Schneider
AO Gelrsquofond GV Chudnovskii P Philippon YuV Nesterenko
G Diaz If α is an algebraic number α 6= 0 α 6= 1 and if β isan irrational algebraic number of degree d then
tr degQQ(αβ αβ
2
αβdminus1) ge [d+ 1
2
]
36 59
Conjecture of algebraic independence of logarithms
of algebraic numbers
Denote by L the set of complex numbers λ for which eλ isalgebraic
L = logα α isin Qtimes
Hence L is a Q-vector subspace of CThe most important special case of Schanuelrsquos Conjecture is
Conjecture Let λ1 λn be Q-linearly independent elementsin L Then the numbers λ1 λn are algebraicallyindependent over Q
Not yet known that the transcendence degree is ge 2 Open problem Among all logarithms of algebraic numbersone at least is transcendental over Q(π)
37 59
Structural rank of a matrix
Let K be a field k a subfield and M a matrix with entries in KConsider the k-vector subspace E of K spanned by the entriesof M Choose an injective morphism ϕ of E into a k-vectorspace kX1 + middot middot middot+ kXn The image ϕ(M) of M is a matrixwhose entries are in the field k(X1 Xn) of rationalfractions Its rank does not depend on the choice of ϕ
This is the structural rank of M with respect to k
38 59
Example
LetM =
(bij + λij
)1leiled1lejle`
be a matrix with coefficients in Q + L Consider a basis of theQndashvector spanned by the entries and replace the elements inthis basis by unknowns This gives a new matrix M withcoefficients in a field of rational fractions the rank of which isthe structural rank of M (with respect to Q)
As a consequence of the conjecture of algebraic independenceof logarithms of algebraic numbers the rank of M should bethe same as the rank of its specialization M
39 59
Equivalence between the two conjectures
Following D Roy theconjecture on algebraicindependence of logarithms ofalgebraic numbers isequivalent to
Conjecture Any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L has a rank equal to its structuralrank
40 59
Any Polynomial is the Determinant of a Matrix
The proof of the equivalence uses the nice auxiliary result
For any P isin k[X1 Xn] there exists a square matrix withentries in the k-vector space k + kX1 + middot middot middot+ kXn whosedeterminant is P
41 59
Half of the Conjecture is solved
From a certain point of view half of the conjecture of algebraicindependence of logarithms of algebraic numbers is solved
Theorem [D Roy] The rank of any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L is at least half its structural rank
42 59
Reformulation by D Roy
Instead of taking logarithmsof algebraic numbers andlooking for the algebraicindependence relationsD Roy fixes a polynomial andlooks at the points withcoordinates logarithms ofalgebraic numbers on thecorresponding hypersurface
43 59
Reformulation by D Roy
Royrsquos reformulation of the conjecture of algebraicindependence of logarithms is
Conjecture For any algebraic subvariety V of Cn definedover the field Q of algebraic numbers the set V cap Ln is theunion of the sets E cap Ln where E ranges over the set ofvector subspaces of Cn which are defined over Q andcontained in V
Trivial Any element in E cap Ln where E is a vector subspaceof Cn defined over Q and contained in V belongs to V cap Ln
44 59
Example The Four Exponentials ConjectureTake for V the hypersurface of C4 defined by the equation
z1z4 = z2z3
The maximal Cndashvector subspaces of C4 defined over Q andcontained in V are the planes
az1 = bz2 bz4 = az3
and the planesaz1 = bz3 bz4 = az2
with (a b) isin Q2 (0 0)Hence Schanuelrsquos Conjecture implies that if λ1 λ2 λ3 λ4 arelogarithms of algebraic numbers satisfying
λ1λ4 = λ2λ3
then either λ1 λ2 are linearly dependent over Q or else λ1 λ3
are linearly dependent over Q45 59
Six Exponentials Theorem
and Four Exponentials Conjecture
A Selberg CL Siegel S Lang K Ramachandra
46 59
The Four Exponentials Conjecture
Conjecture (Th Schneider S Lang K Ramachandra) Ifx1 x2 are Qndashlinearly independent complex numbers and y1 y2
are Qndashlinearly independent complex numbers then one atleast of the four numbers
ex1y1 ex1y2 ex2y1 ex2y2
is transcendentalEquivalent statement Any 2times 2 matrix(
λ1 λ2
λ3 λ4
)with entries in L and with Qndashlinearly independent rows andQndashlinearly independent columns has maximal rank 2
47 59
How could we attack Schanuelrsquos Conjecture
Let x1 xn be Qndashlinearly independent complex numbersFollowing the transcendence methods of Hermite GelrsquofondSchneider one may start by introducing an auxiliaryfunction
F (z) = P (z ez)
where P isin Z[X0 X1] is a nonndashzero polynomial One considersthe derivatives (
d
dz
)kF = (DkP )(z ez)
at the pointsm1x1 + middot middot middot+mnxn
for various values of (m1 mn) isin Zn
48 59
Auxiliary functionLet D denote the derivation
D =part
partX0
+X1part
partX1
over the ring C[X0 X1] Recall that x1 xn are Qndashlinearlyindependent complex numbers Let α1 αn be nonndashzerocomplex numbersThe transcendence machinery produces a sequence (PN)Nge0
of polynomials with integer coefficients satisfying∣∣∣∣∣(DkPN)(nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1
49 59
Royrsquos approach to Schanuelrsquos Conjecture (1999)
If the number of equations we produce is too small such a setof relations does not contain any information The existenceof a sequence of nonndashtrivial polynomials (PN)Nge0 follows fromlinear algebra
On the other hand following D Roy one may expect that theexistence of such a sequence (PN)Nge0 producing sufficientlymany such equations will yield the desired conclusion
tr degQQ(x1 xn α1 αn
)ge n
50 59
Royrsquos Conjecture (1999)Let s0 s1 t0 t1 u positive real numbers satisfying
max1 t0 2t1 lt mins0 2s1and
maxs0 s1 + t1 lt u lt1
2(1 + t0 + t1)
Assume that for any sufficiently large positive integer N there exists a nonndashzero polynomial PN isin Z[X0 X1] withpartial degree le N t0 in X0 partial degree le N t1 in X1 andheight le eN which satisfies∣∣∣∣∣(DkPN)(
nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1 Then
tr degQQ(x1 xn α1 αn) ge n51 59
Royrsquos Theorem (1999)
Royrsquos conjecture is equivalent to Schanuelrsquos Conjecture
52 59
Equivalence between Schanuel and Roy
Let (x α) isin CtimesCtimes and let s0 s1 t0 t1 u be positive realnumbers satisfying the inequalities of Royrsquos Conjecture Thenthe following conditions are equivalent (a) The number αeminusx is a root of unity
(b) For any sufficiently large positive integer N there exists anonndashzero polynomial QN isin Z[X0 X1] with partial degreele N t0 in X0 partial degree le N t1 in X1 and heightH(QN) le eN such that∣∣(DkQN)(mx αm)
∣∣ le exp(minusNu)
for any km isin N with k le N s0 and m le N s1
53 59
Gelrsquofondrsquos transcendence criterion
Simple form Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le n and height le en such that
|Pn(ϑ)| le eminus7n2
for all n ge 1 then ϑ is algebraic and Pn(ϑ) = 0 for all n ge 1
First extension Replace the bound for the degree by le dnthe bound for the height by ebn and the bound for |Pn(ϑ)| byeminuscdnbn with some constant c gt 0 independent of n
Some mild conditions are required on the sequences (dn)nge1 and(bn)nge1
54 59
Transcendence criterion with multiplicitiesWith derivatives Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le dn and height le ebn such that∣∣P (j)
n (ϑ)∣∣ le eminuscdnbntn
for all j in the range 0 le j lt tn and all n ge 1 then ϑ isalgebraicDue to M Laurent and D Roy applications to algebraicindependence
55 59
Criterion with several points
Goal Given a sequence of complex numbers (ϑi)ige1 if thereexists a sequence (Pn)nge1 of nonndashzero polynomials in Z[X]with Pn of degree le dn and height le ebn such that∣∣P (j)
n (ϑi)∣∣ le eminuscdnbntnsn
for 0 le j lt tn 1 le i le sn and all n ge 1 then the numbers ϑiare algebraic
D Roy Not true in general but true in some special caseswith a structure on the sequence (ϑi)ige1Combines the elimination arguments used for criteria ofalgebraic independence and for zero estimates
56 59
Small value estimates for the additive group
D Roy Small value estimates for the additive group Intern JNumber Theory to appear
Let ξ be a transcendental complex number let β σ τ and νbe non-negative real numbers let n0 be a positive integer andlet (Pn)ngen0 be a sequence of nonndashzero polynomials in Z[T ]satisfying deg(Pn) le n and H(Pn) le exp(nβ) for eachn ge n0 Suppose that β gt 1 (34)σ + τ lt 1 andν gt 1 + β minus (34)σ minus τ Then for infinitely many n we have
max|P [j]n (iξ)| 0 le i le nσ 0 le j le nτ
gt exp(minusnν)
57 59
Small value estimates for the multiplicative group
D Roy Small value estimates for the multiplicative groupActa Arith to appear
Let ξ1 ξm be multiplicatively independent complexnumbers in a field of transcendence degree 1 Under suitableassumptions on the parameters β σ τ ν for infinitely manypositive integers n there exists no nonndashzero polynomialP isin Z[T ] satisfying deg(P ) le n H(P ) le exp(nβ) and
max|P [j](ξi11 middot middot middot ξimm )| 0 le i1 im le nσ 0 le j le nτ
gt exp(minusnν)
58 59
Heidelberg Colloquium organized by the Institute for Mathematics
httpwwwmathiuni-heidelbergdevortraghtmls=1 January 22 2009
Transcendental Number TheorySchanuelrsquos Conjecture
Michel Waldschmidt
Institut de Mathematiques de Jussieu amp Paris VIhttpwwwmathjussieufrsimmiw
59 59
Question of Euler and Hilbert
Leonhard Euler (1707 ndash 1783)Introductio in analysin infinitorum (1737)
Transcendence of 2radic
2
David Hilbert (1862 - 1943)ICM 1900 7ndashth Problem
Transcendence of log a log 2
7 59
AO Gelrsquofond ndash Th Schneider ndash A Baker
eπ Gelrsquofond 1929
2radic
2 log 2 log 3 Gelrsquofond and Schneider 1934
log 2 +radic
3 log 3 eradic
22radic
35radic
7 Baker 1968
8 59
Some open problems
For each of the following numbers it is expected that it istranscendental but it is not even known whether it is rationalor not
e+ π eπ πe ee ee2
eee
ππ ππ2
πππ
log π log(log 2) π log 2 (log 2)(log 3) 2log 2 (log 2)log 3
In other words we do not know whether a degree 1 polynomialcould vanish at the corresponding point but we expect thatno nonndashzero polynomial of any degree vanishes at this point
9 59
Algebraic independence
Algebraicity and transcendence deal with a single complexnumber and one variable polynomialsAlgebraic dependence or independence is the same but fortuples and multivariate polynomials
Let K sup k be an extension of fields and (θ1 θm) be amndashtuple of elements in K We say that θ1 θm arealgebraically dependent over k if there exists a nonndashzeropolynomial f isin k[X1 Xm] which vanishes at the point(θ1 θm) isin KmOtherwise we say that θ1 θm are algebraically independentover k
10 59
Special cases
In the case m = 1 to say that θ1 is algebraically independentover k just means that it is transcendental over k
Dealing with complex numbers K = C the words algebraictranscendental algebraically dependent algebraicallyindependent refer to the case k = Q
11 59
Examples
The numbersradic
2 and π are algebraically dependent The polynomial X2 minus 2 isin Z[X Y ] vanishes at (
radic2 π)
The numbers π andradicπ2 + 1 are algebraically dependent (and
both are transcendental numbers) The polynomial Y minusX2 minus 1 vanishes at (π
radicπ2 + 1)
The two numbers e and eradic
2 are algebraically independentwhich means that for any nonndashzero polynomial f isin Z[X Y ]
with rational integer coefficients the number f(e eradic
2) is notzeroSpecial case of the Lindemann-Weierstraszlig Theorem
12 59
Transcendence degree
Let Kk be a field extension The maximal number ofelements in K which are algebraically independent over k iscalled the transcendence degree of K over k and denotedtr degkK
Let t = tr degkK A subset θ1 θt of K with t elementswhich are algebraically independent is called a transcendencebasis of K over k It is the same as a maximal subset ofkndashalgebraically independent elements in KHence K is an algebraic extension of k(θ1 θt)
13 59
Transcendence degree of extensions
Assumek sub K sub L
The union of a transcendence basis of K over k and of atranscendence basis of L over K produces a transcendencebasis of L over kHence
tr degkL = tr degkK + tr degKL
An algebraic extension Kk is an extension of transcendencedegree 0 This means that there is no transcendental elementin K over k (any element in K is algebraic over k)
14 59
Algebraic independence of complex numbers
For complex numbers algebraic independence over Q or overQ is the same In particular if θ1 θm are algebraicallyindependent complex numbers then for any non-constantpolynomial f with algebraic coefficients the numberf(θ1 θm) is transcendental
For instance the two numbers e and eradic
2 are algebraicallyindependent As a consequence for any nonndashconstantpolynomial f isin Q[X Y ] with algebraic coefficients the
number f(e eradic
2) is transcendental
15 59
Schanuelrsquos Conjecture
Let x1 xn be Q-linearly independent complex numbersThen at least n of the 2n numbers x1 xn e
x1 exn arealgebraically independent
Since there are 2n numbers only the transcendence degreeover Q of the field
Q(x1 xn e
x1 exn)
is at most 2n The conjecture is that this transcendencedegree is always ge n
n le tr degQQ(x1 xn e
x1 exn)le 2n
16 59
Origin of Schanuelrsquos ConjectureCourse given by Serge Lang (1927ndash2005) at Yale in the 60rsquos
S Lang ndash Introduction to transcendental numbersAddison-Wesley 1966
also attended by M Nagata (1927ndash2008)(14th Problem of Hilbert)
Nagatarsquos Conjecture solved by E Bombieri
17 59
Schanuelrsquos Conjecture
Let x1 xn be Q-linearly independent complex numbersThen
tr degQQ(x1 xn e
x1 exn)ge n
Remark For almost all tuples (with respect to the Lebesguemeasure) the transcendence degree is 2n
18 59
AO Gelrsquofond CRAS 1934
19 59
Statement by Gelrsquofond (1934)
Let β1 βn be Q-linearly independent algebraic numbersand let logα1 logαm be Q-linearly independentlogarithms of algebraic numbers Then the numbers
eβ1 eβn logα1 logαm
are algebraically independent over Q
20 59
Further statement by Gelrsquofond
Let β1 βn be algebraic numbers with β1 6= 0 and letα1 αm be algebraic numbers with α1 6= 0 1 α2 6= 0 1αi 6= 0 Then the numbers
eβ1eβ2e βnminus1e
βn
and αα αm
21
are transcendental and there is no nontrivial algebraic relationbetween such numbers
Remark The condition on α2 should be that it is irrational
21 59
Easy consequence of Schanuelrsquos Conjecture
According to Schanuelrsquos Conjecture the following numbers arealgebraically independent
e+ π eπ πe ee ee2
eee
ππ ππ2
πππ
log π log(log 2) π log 2 (log 2)(log 3) 2log 2 (log 2)log 3
Proof This is an easy exercise
22 59
Langrsquos exercise
Define E0 = Q Inductivelyfor n ge 1 define En as thealgebraic closure of the fieldgenerated over Enminus1 by thenumbers exp(x) = ex wherex ranges over Enminus1 Let E bethe union of En n ge 0Then Schanuelrsquos Conjectureimplies that the number πdoes not belong to E
More precisely Schanuelrsquos Conjecture implies that thenumbers π log π log log π log log log π are algebraicallyindependent over E
23 59
A variant of Langrsquos exercise
Define L0 = Q Inductively for n ge 1 define Ln as thealgebraic closure of the field generated over Lnminus1 by thenumbers y where y ranges over the set of complex numberssuch that ey isin Lnminus1 Let L be the union of Ln n ge 0Then Schanuelrsquos Conjecture implies that the number e doesnot belong to L
More precisely Schanuelrsquos Conjecture implies that the
numbers e ee eee ee
ee
are algebraically independent over L
24 59
Arizona Winter School AWS2008 Tucson
Theorem [Jonathan Bober Chuangxun Cheng Brian DietelMathilde Herblot Jingjing Huang Holly Krieger DiegoMarques Jonathan Mason Martin Mereb and Robert Wilson]Schanuelrsquos Conjecture implies that the fields E and L arelinearly disjoint over Q
Definition Given a field extension FK and twosubextensions F1 F2 sube F we say F1 F2 are linearly disjointover K when the following holds Any set x1 xn sube F1
of Kndash linearly independent elements is linearly independentover F2
Reference arXiv08043550 [mathNT] 2008
25 59
Formal analogs
WD Brownawell(was a student of Schanuel)
J Axrsquos Theorem (1968) Version of SchanuelrsquosConjecture for power seriesover C(and R Coleman for powerseries over Q)
Work by WD Brownawelland K Kubota on the ellipticanalog of Axrsquos Theorem
26 59
Conjectures by A Grothendieck and Y Andre
Generalized Conjecture onPeriods by Grothendieck Dimension of theMumfordndashTate group of asmooth projective variety
Generalization by Y Andre tomotives
Case of 1ndashmotives Elliptico-Toric Conjecture ofC Bertolin
27 59
Ubiquity of Schanuelrsquos Conjecture
Other contexts pndashadic numbers Leopoldtrsquos Conjecture onthe pndashadic rank of the units of an algebraic number fieldNon-vanishing of RegulatorsNonndashdegenerescence of heightsConjecture of B Mazur on rational pointsDiophantine approximation on tori
Dipendra Prasad Gopal Prasad
28 59
Methods from logicEhud Hrushovski Boris Zilber Jonathan Kirby
Calculus of rdquopredimension functionsrdquo (E Hrushovski)
Zilberrsquos construction of a rdquopseudoexponentiationrdquo
Also A Macintyre DE Marker G Terzo AJ WilkieD Bertrand
29 59
Methods from logic Model theory
Exponential algebraicity in exponential fieldsby
Jonathan Kirby
The dimension of the exponential algebraic closure operator inan exponential field satisfies a weak Schanuel property
A corollary is that there are at most countably many essentialcounterexamples to Schanuelrsquos conjecture
arXiv 08104285v2
30 59
Known
LindemannndashWeierstraszlig Theorem = case where x1 xn arealgebraic
Let β1 βn be algebraic numbers which are linearlyindependent over Q Then the numbers eβ1 eβn arealgebraically independent over Q
31 59
Problem of Gelrsquofond and SchneiderRaised by AO Gelrsquofond in 1948 and Th Schneider in 1952
Conjecture If α is an algebraic number α 6= 0 α 6= 1 and ifβ is an irrational algebraic number of degree d then the dminus 1numbers
αβ αβ2
αβdminus1
are algebraically independent
Special case of Schanuelrsquos Conjecture Take xi = βiminus1 logαn = d so that
x1 xn e
x1 exn
islogα β logα βdminus1 logα α αβ αβ
dminus1
The conclusion of Schanuelrsquos Conjecture is
tr degQQ(logα αβ αβ
2
αβdminus1)
= d
32 59
Algebraic independence method of Gelrsquofond
AO Gelrsquofond (1948)
The two numbers 23radic2 and
23radic4 are algebraically
independentMore generally if α is analgebraic number α 6= 0α 6= 1 and if β is a algebraicnumber of degree d ge 3 thentwo at least of the numbers
αβ αβ2
αβdminus1
are algebraically independent
33 59
Tools
Transcendence criterion Replaces Liouvillersquos inequality intranscendence proofsLiouville A nonndashzero rational integer n isin Z satisfies |n| ge 1Gelrsquofond Needs to give a lower bound for |P (θ)| withP isin Z[X] 0 when θ is transcendental
Zero estimate for exponential polynomials C Hermite P Turan K Mahler R Tijdeman
Small transcendence degree AO Gelrsquofond AA Smelev R Tijdeman WD Brownawell
34 59
Large transcendence degree
GV Chudnovsky (1976)Among the numbers
αβ αβ2
αβdminus1
at least [log2 d] arealgebraically independent
GV Chudnovsky ndash On the path to Schanuelrsquosconjecture Algebraic curves close to a pointI General theory of colored sequencesII Fields of finite transcendence type and coloredsequences ResultantsStudia Sci Math Hungar 12 (1977) 125ndash157 (1980)
35 59
Partial result on the problem of Gelrsquofond and
Schneider
AO Gelrsquofond GV Chudnovskii P Philippon YuV Nesterenko
G Diaz If α is an algebraic number α 6= 0 α 6= 1 and if β isan irrational algebraic number of degree d then
tr degQQ(αβ αβ
2
αβdminus1) ge [d+ 1
2
]
36 59
Conjecture of algebraic independence of logarithms
of algebraic numbers
Denote by L the set of complex numbers λ for which eλ isalgebraic
L = logα α isin Qtimes
Hence L is a Q-vector subspace of CThe most important special case of Schanuelrsquos Conjecture is
Conjecture Let λ1 λn be Q-linearly independent elementsin L Then the numbers λ1 λn are algebraicallyindependent over Q
Not yet known that the transcendence degree is ge 2 Open problem Among all logarithms of algebraic numbersone at least is transcendental over Q(π)
37 59
Structural rank of a matrix
Let K be a field k a subfield and M a matrix with entries in KConsider the k-vector subspace E of K spanned by the entriesof M Choose an injective morphism ϕ of E into a k-vectorspace kX1 + middot middot middot+ kXn The image ϕ(M) of M is a matrixwhose entries are in the field k(X1 Xn) of rationalfractions Its rank does not depend on the choice of ϕ
This is the structural rank of M with respect to k
38 59
Example
LetM =
(bij + λij
)1leiled1lejle`
be a matrix with coefficients in Q + L Consider a basis of theQndashvector spanned by the entries and replace the elements inthis basis by unknowns This gives a new matrix M withcoefficients in a field of rational fractions the rank of which isthe structural rank of M (with respect to Q)
As a consequence of the conjecture of algebraic independenceof logarithms of algebraic numbers the rank of M should bethe same as the rank of its specialization M
39 59
Equivalence between the two conjectures
Following D Roy theconjecture on algebraicindependence of logarithms ofalgebraic numbers isequivalent to
Conjecture Any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L has a rank equal to its structuralrank
40 59
Any Polynomial is the Determinant of a Matrix
The proof of the equivalence uses the nice auxiliary result
For any P isin k[X1 Xn] there exists a square matrix withentries in the k-vector space k + kX1 + middot middot middot+ kXn whosedeterminant is P
41 59
Half of the Conjecture is solved
From a certain point of view half of the conjecture of algebraicindependence of logarithms of algebraic numbers is solved
Theorem [D Roy] The rank of any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L is at least half its structural rank
42 59
Reformulation by D Roy
Instead of taking logarithmsof algebraic numbers andlooking for the algebraicindependence relationsD Roy fixes a polynomial andlooks at the points withcoordinates logarithms ofalgebraic numbers on thecorresponding hypersurface
43 59
Reformulation by D Roy
Royrsquos reformulation of the conjecture of algebraicindependence of logarithms is
Conjecture For any algebraic subvariety V of Cn definedover the field Q of algebraic numbers the set V cap Ln is theunion of the sets E cap Ln where E ranges over the set ofvector subspaces of Cn which are defined over Q andcontained in V
Trivial Any element in E cap Ln where E is a vector subspaceof Cn defined over Q and contained in V belongs to V cap Ln
44 59
Example The Four Exponentials ConjectureTake for V the hypersurface of C4 defined by the equation
z1z4 = z2z3
The maximal Cndashvector subspaces of C4 defined over Q andcontained in V are the planes
az1 = bz2 bz4 = az3
and the planesaz1 = bz3 bz4 = az2
with (a b) isin Q2 (0 0)Hence Schanuelrsquos Conjecture implies that if λ1 λ2 λ3 λ4 arelogarithms of algebraic numbers satisfying
λ1λ4 = λ2λ3
then either λ1 λ2 are linearly dependent over Q or else λ1 λ3
are linearly dependent over Q45 59
Six Exponentials Theorem
and Four Exponentials Conjecture
A Selberg CL Siegel S Lang K Ramachandra
46 59
The Four Exponentials Conjecture
Conjecture (Th Schneider S Lang K Ramachandra) Ifx1 x2 are Qndashlinearly independent complex numbers and y1 y2
are Qndashlinearly independent complex numbers then one atleast of the four numbers
ex1y1 ex1y2 ex2y1 ex2y2
is transcendentalEquivalent statement Any 2times 2 matrix(
λ1 λ2
λ3 λ4
)with entries in L and with Qndashlinearly independent rows andQndashlinearly independent columns has maximal rank 2
47 59
How could we attack Schanuelrsquos Conjecture
Let x1 xn be Qndashlinearly independent complex numbersFollowing the transcendence methods of Hermite GelrsquofondSchneider one may start by introducing an auxiliaryfunction
F (z) = P (z ez)
where P isin Z[X0 X1] is a nonndashzero polynomial One considersthe derivatives (
d
dz
)kF = (DkP )(z ez)
at the pointsm1x1 + middot middot middot+mnxn
for various values of (m1 mn) isin Zn
48 59
Auxiliary functionLet D denote the derivation
D =part
partX0
+X1part
partX1
over the ring C[X0 X1] Recall that x1 xn are Qndashlinearlyindependent complex numbers Let α1 αn be nonndashzerocomplex numbersThe transcendence machinery produces a sequence (PN)Nge0
of polynomials with integer coefficients satisfying∣∣∣∣∣(DkPN)(nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1
49 59
Royrsquos approach to Schanuelrsquos Conjecture (1999)
If the number of equations we produce is too small such a setof relations does not contain any information The existenceof a sequence of nonndashtrivial polynomials (PN)Nge0 follows fromlinear algebra
On the other hand following D Roy one may expect that theexistence of such a sequence (PN)Nge0 producing sufficientlymany such equations will yield the desired conclusion
tr degQQ(x1 xn α1 αn
)ge n
50 59
Royrsquos Conjecture (1999)Let s0 s1 t0 t1 u positive real numbers satisfying
max1 t0 2t1 lt mins0 2s1and
maxs0 s1 + t1 lt u lt1
2(1 + t0 + t1)
Assume that for any sufficiently large positive integer N there exists a nonndashzero polynomial PN isin Z[X0 X1] withpartial degree le N t0 in X0 partial degree le N t1 in X1 andheight le eN which satisfies∣∣∣∣∣(DkPN)(
nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1 Then
tr degQQ(x1 xn α1 αn) ge n51 59
Royrsquos Theorem (1999)
Royrsquos conjecture is equivalent to Schanuelrsquos Conjecture
52 59
Equivalence between Schanuel and Roy
Let (x α) isin CtimesCtimes and let s0 s1 t0 t1 u be positive realnumbers satisfying the inequalities of Royrsquos Conjecture Thenthe following conditions are equivalent (a) The number αeminusx is a root of unity
(b) For any sufficiently large positive integer N there exists anonndashzero polynomial QN isin Z[X0 X1] with partial degreele N t0 in X0 partial degree le N t1 in X1 and heightH(QN) le eN such that∣∣(DkQN)(mx αm)
∣∣ le exp(minusNu)
for any km isin N with k le N s0 and m le N s1
53 59
Gelrsquofondrsquos transcendence criterion
Simple form Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le n and height le en such that
|Pn(ϑ)| le eminus7n2
for all n ge 1 then ϑ is algebraic and Pn(ϑ) = 0 for all n ge 1
First extension Replace the bound for the degree by le dnthe bound for the height by ebn and the bound for |Pn(ϑ)| byeminuscdnbn with some constant c gt 0 independent of n
Some mild conditions are required on the sequences (dn)nge1 and(bn)nge1
54 59
Transcendence criterion with multiplicitiesWith derivatives Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le dn and height le ebn such that∣∣P (j)
n (ϑ)∣∣ le eminuscdnbntn
for all j in the range 0 le j lt tn and all n ge 1 then ϑ isalgebraicDue to M Laurent and D Roy applications to algebraicindependence
55 59
Criterion with several points
Goal Given a sequence of complex numbers (ϑi)ige1 if thereexists a sequence (Pn)nge1 of nonndashzero polynomials in Z[X]with Pn of degree le dn and height le ebn such that∣∣P (j)
n (ϑi)∣∣ le eminuscdnbntnsn
for 0 le j lt tn 1 le i le sn and all n ge 1 then the numbers ϑiare algebraic
D Roy Not true in general but true in some special caseswith a structure on the sequence (ϑi)ige1Combines the elimination arguments used for criteria ofalgebraic independence and for zero estimates
56 59
Small value estimates for the additive group
D Roy Small value estimates for the additive group Intern JNumber Theory to appear
Let ξ be a transcendental complex number let β σ τ and νbe non-negative real numbers let n0 be a positive integer andlet (Pn)ngen0 be a sequence of nonndashzero polynomials in Z[T ]satisfying deg(Pn) le n and H(Pn) le exp(nβ) for eachn ge n0 Suppose that β gt 1 (34)σ + τ lt 1 andν gt 1 + β minus (34)σ minus τ Then for infinitely many n we have
max|P [j]n (iξ)| 0 le i le nσ 0 le j le nτ
gt exp(minusnν)
57 59
Small value estimates for the multiplicative group
D Roy Small value estimates for the multiplicative groupActa Arith to appear
Let ξ1 ξm be multiplicatively independent complexnumbers in a field of transcendence degree 1 Under suitableassumptions on the parameters β σ τ ν for infinitely manypositive integers n there exists no nonndashzero polynomialP isin Z[T ] satisfying deg(P ) le n H(P ) le exp(nβ) and
max|P [j](ξi11 middot middot middot ξimm )| 0 le i1 im le nσ 0 le j le nτ
gt exp(minusnν)
58 59
Heidelberg Colloquium organized by the Institute for Mathematics
httpwwwmathiuni-heidelbergdevortraghtmls=1 January 22 2009
Transcendental Number TheorySchanuelrsquos Conjecture
Michel Waldschmidt
Institut de Mathematiques de Jussieu amp Paris VIhttpwwwmathjussieufrsimmiw
59 59
AO Gelrsquofond ndash Th Schneider ndash A Baker
eπ Gelrsquofond 1929
2radic
2 log 2 log 3 Gelrsquofond and Schneider 1934
log 2 +radic
3 log 3 eradic
22radic
35radic
7 Baker 1968
8 59
Some open problems
For each of the following numbers it is expected that it istranscendental but it is not even known whether it is rationalor not
e+ π eπ πe ee ee2
eee
ππ ππ2
πππ
log π log(log 2) π log 2 (log 2)(log 3) 2log 2 (log 2)log 3
In other words we do not know whether a degree 1 polynomialcould vanish at the corresponding point but we expect thatno nonndashzero polynomial of any degree vanishes at this point
9 59
Algebraic independence
Algebraicity and transcendence deal with a single complexnumber and one variable polynomialsAlgebraic dependence or independence is the same but fortuples and multivariate polynomials
Let K sup k be an extension of fields and (θ1 θm) be amndashtuple of elements in K We say that θ1 θm arealgebraically dependent over k if there exists a nonndashzeropolynomial f isin k[X1 Xm] which vanishes at the point(θ1 θm) isin KmOtherwise we say that θ1 θm are algebraically independentover k
10 59
Special cases
In the case m = 1 to say that θ1 is algebraically independentover k just means that it is transcendental over k
Dealing with complex numbers K = C the words algebraictranscendental algebraically dependent algebraicallyindependent refer to the case k = Q
11 59
Examples
The numbersradic
2 and π are algebraically dependent The polynomial X2 minus 2 isin Z[X Y ] vanishes at (
radic2 π)
The numbers π andradicπ2 + 1 are algebraically dependent (and
both are transcendental numbers) The polynomial Y minusX2 minus 1 vanishes at (π
radicπ2 + 1)
The two numbers e and eradic
2 are algebraically independentwhich means that for any nonndashzero polynomial f isin Z[X Y ]
with rational integer coefficients the number f(e eradic
2) is notzeroSpecial case of the Lindemann-Weierstraszlig Theorem
12 59
Transcendence degree
Let Kk be a field extension The maximal number ofelements in K which are algebraically independent over k iscalled the transcendence degree of K over k and denotedtr degkK
Let t = tr degkK A subset θ1 θt of K with t elementswhich are algebraically independent is called a transcendencebasis of K over k It is the same as a maximal subset ofkndashalgebraically independent elements in KHence K is an algebraic extension of k(θ1 θt)
13 59
Transcendence degree of extensions
Assumek sub K sub L
The union of a transcendence basis of K over k and of atranscendence basis of L over K produces a transcendencebasis of L over kHence
tr degkL = tr degkK + tr degKL
An algebraic extension Kk is an extension of transcendencedegree 0 This means that there is no transcendental elementin K over k (any element in K is algebraic over k)
14 59
Algebraic independence of complex numbers
For complex numbers algebraic independence over Q or overQ is the same In particular if θ1 θm are algebraicallyindependent complex numbers then for any non-constantpolynomial f with algebraic coefficients the numberf(θ1 θm) is transcendental
For instance the two numbers e and eradic
2 are algebraicallyindependent As a consequence for any nonndashconstantpolynomial f isin Q[X Y ] with algebraic coefficients the
number f(e eradic
2) is transcendental
15 59
Schanuelrsquos Conjecture
Let x1 xn be Q-linearly independent complex numbersThen at least n of the 2n numbers x1 xn e
x1 exn arealgebraically independent
Since there are 2n numbers only the transcendence degreeover Q of the field
Q(x1 xn e
x1 exn)
is at most 2n The conjecture is that this transcendencedegree is always ge n
n le tr degQQ(x1 xn e
x1 exn)le 2n
16 59
Origin of Schanuelrsquos ConjectureCourse given by Serge Lang (1927ndash2005) at Yale in the 60rsquos
S Lang ndash Introduction to transcendental numbersAddison-Wesley 1966
also attended by M Nagata (1927ndash2008)(14th Problem of Hilbert)
Nagatarsquos Conjecture solved by E Bombieri
17 59
Schanuelrsquos Conjecture
Let x1 xn be Q-linearly independent complex numbersThen
tr degQQ(x1 xn e
x1 exn)ge n
Remark For almost all tuples (with respect to the Lebesguemeasure) the transcendence degree is 2n
18 59
AO Gelrsquofond CRAS 1934
19 59
Statement by Gelrsquofond (1934)
Let β1 βn be Q-linearly independent algebraic numbersand let logα1 logαm be Q-linearly independentlogarithms of algebraic numbers Then the numbers
eβ1 eβn logα1 logαm
are algebraically independent over Q
20 59
Further statement by Gelrsquofond
Let β1 βn be algebraic numbers with β1 6= 0 and letα1 αm be algebraic numbers with α1 6= 0 1 α2 6= 0 1αi 6= 0 Then the numbers
eβ1eβ2e βnminus1e
βn
and αα αm
21
are transcendental and there is no nontrivial algebraic relationbetween such numbers
Remark The condition on α2 should be that it is irrational
21 59
Easy consequence of Schanuelrsquos Conjecture
According to Schanuelrsquos Conjecture the following numbers arealgebraically independent
e+ π eπ πe ee ee2
eee
ππ ππ2
πππ
log π log(log 2) π log 2 (log 2)(log 3) 2log 2 (log 2)log 3
Proof This is an easy exercise
22 59
Langrsquos exercise
Define E0 = Q Inductivelyfor n ge 1 define En as thealgebraic closure of the fieldgenerated over Enminus1 by thenumbers exp(x) = ex wherex ranges over Enminus1 Let E bethe union of En n ge 0Then Schanuelrsquos Conjectureimplies that the number πdoes not belong to E
More precisely Schanuelrsquos Conjecture implies that thenumbers π log π log log π log log log π are algebraicallyindependent over E
23 59
A variant of Langrsquos exercise
Define L0 = Q Inductively for n ge 1 define Ln as thealgebraic closure of the field generated over Lnminus1 by thenumbers y where y ranges over the set of complex numberssuch that ey isin Lnminus1 Let L be the union of Ln n ge 0Then Schanuelrsquos Conjecture implies that the number e doesnot belong to L
More precisely Schanuelrsquos Conjecture implies that the
numbers e ee eee ee
ee
are algebraically independent over L
24 59
Arizona Winter School AWS2008 Tucson
Theorem [Jonathan Bober Chuangxun Cheng Brian DietelMathilde Herblot Jingjing Huang Holly Krieger DiegoMarques Jonathan Mason Martin Mereb and Robert Wilson]Schanuelrsquos Conjecture implies that the fields E and L arelinearly disjoint over Q
Definition Given a field extension FK and twosubextensions F1 F2 sube F we say F1 F2 are linearly disjointover K when the following holds Any set x1 xn sube F1
of Kndash linearly independent elements is linearly independentover F2
Reference arXiv08043550 [mathNT] 2008
25 59
Formal analogs
WD Brownawell(was a student of Schanuel)
J Axrsquos Theorem (1968) Version of SchanuelrsquosConjecture for power seriesover C(and R Coleman for powerseries over Q)
Work by WD Brownawelland K Kubota on the ellipticanalog of Axrsquos Theorem
26 59
Conjectures by A Grothendieck and Y Andre
Generalized Conjecture onPeriods by Grothendieck Dimension of theMumfordndashTate group of asmooth projective variety
Generalization by Y Andre tomotives
Case of 1ndashmotives Elliptico-Toric Conjecture ofC Bertolin
27 59
Ubiquity of Schanuelrsquos Conjecture
Other contexts pndashadic numbers Leopoldtrsquos Conjecture onthe pndashadic rank of the units of an algebraic number fieldNon-vanishing of RegulatorsNonndashdegenerescence of heightsConjecture of B Mazur on rational pointsDiophantine approximation on tori
Dipendra Prasad Gopal Prasad
28 59
Methods from logicEhud Hrushovski Boris Zilber Jonathan Kirby
Calculus of rdquopredimension functionsrdquo (E Hrushovski)
Zilberrsquos construction of a rdquopseudoexponentiationrdquo
Also A Macintyre DE Marker G Terzo AJ WilkieD Bertrand
29 59
Methods from logic Model theory
Exponential algebraicity in exponential fieldsby
Jonathan Kirby
The dimension of the exponential algebraic closure operator inan exponential field satisfies a weak Schanuel property
A corollary is that there are at most countably many essentialcounterexamples to Schanuelrsquos conjecture
arXiv 08104285v2
30 59
Known
LindemannndashWeierstraszlig Theorem = case where x1 xn arealgebraic
Let β1 βn be algebraic numbers which are linearlyindependent over Q Then the numbers eβ1 eβn arealgebraically independent over Q
31 59
Problem of Gelrsquofond and SchneiderRaised by AO Gelrsquofond in 1948 and Th Schneider in 1952
Conjecture If α is an algebraic number α 6= 0 α 6= 1 and ifβ is an irrational algebraic number of degree d then the dminus 1numbers
αβ αβ2
αβdminus1
are algebraically independent
Special case of Schanuelrsquos Conjecture Take xi = βiminus1 logαn = d so that
x1 xn e
x1 exn
islogα β logα βdminus1 logα α αβ αβ
dminus1
The conclusion of Schanuelrsquos Conjecture is
tr degQQ(logα αβ αβ
2
αβdminus1)
= d
32 59
Algebraic independence method of Gelrsquofond
AO Gelrsquofond (1948)
The two numbers 23radic2 and
23radic4 are algebraically
independentMore generally if α is analgebraic number α 6= 0α 6= 1 and if β is a algebraicnumber of degree d ge 3 thentwo at least of the numbers
αβ αβ2
αβdminus1
are algebraically independent
33 59
Tools
Transcendence criterion Replaces Liouvillersquos inequality intranscendence proofsLiouville A nonndashzero rational integer n isin Z satisfies |n| ge 1Gelrsquofond Needs to give a lower bound for |P (θ)| withP isin Z[X] 0 when θ is transcendental
Zero estimate for exponential polynomials C Hermite P Turan K Mahler R Tijdeman
Small transcendence degree AO Gelrsquofond AA Smelev R Tijdeman WD Brownawell
34 59
Large transcendence degree
GV Chudnovsky (1976)Among the numbers
αβ αβ2
αβdminus1
at least [log2 d] arealgebraically independent
GV Chudnovsky ndash On the path to Schanuelrsquosconjecture Algebraic curves close to a pointI General theory of colored sequencesII Fields of finite transcendence type and coloredsequences ResultantsStudia Sci Math Hungar 12 (1977) 125ndash157 (1980)
35 59
Partial result on the problem of Gelrsquofond and
Schneider
AO Gelrsquofond GV Chudnovskii P Philippon YuV Nesterenko
G Diaz If α is an algebraic number α 6= 0 α 6= 1 and if β isan irrational algebraic number of degree d then
tr degQQ(αβ αβ
2
αβdminus1) ge [d+ 1
2
]
36 59
Conjecture of algebraic independence of logarithms
of algebraic numbers
Denote by L the set of complex numbers λ for which eλ isalgebraic
L = logα α isin Qtimes
Hence L is a Q-vector subspace of CThe most important special case of Schanuelrsquos Conjecture is
Conjecture Let λ1 λn be Q-linearly independent elementsin L Then the numbers λ1 λn are algebraicallyindependent over Q
Not yet known that the transcendence degree is ge 2 Open problem Among all logarithms of algebraic numbersone at least is transcendental over Q(π)
37 59
Structural rank of a matrix
Let K be a field k a subfield and M a matrix with entries in KConsider the k-vector subspace E of K spanned by the entriesof M Choose an injective morphism ϕ of E into a k-vectorspace kX1 + middot middot middot+ kXn The image ϕ(M) of M is a matrixwhose entries are in the field k(X1 Xn) of rationalfractions Its rank does not depend on the choice of ϕ
This is the structural rank of M with respect to k
38 59
Example
LetM =
(bij + λij
)1leiled1lejle`
be a matrix with coefficients in Q + L Consider a basis of theQndashvector spanned by the entries and replace the elements inthis basis by unknowns This gives a new matrix M withcoefficients in a field of rational fractions the rank of which isthe structural rank of M (with respect to Q)
As a consequence of the conjecture of algebraic independenceof logarithms of algebraic numbers the rank of M should bethe same as the rank of its specialization M
39 59
Equivalence between the two conjectures
Following D Roy theconjecture on algebraicindependence of logarithms ofalgebraic numbers isequivalent to
Conjecture Any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L has a rank equal to its structuralrank
40 59
Any Polynomial is the Determinant of a Matrix
The proof of the equivalence uses the nice auxiliary result
For any P isin k[X1 Xn] there exists a square matrix withentries in the k-vector space k + kX1 + middot middot middot+ kXn whosedeterminant is P
41 59
Half of the Conjecture is solved
From a certain point of view half of the conjecture of algebraicindependence of logarithms of algebraic numbers is solved
Theorem [D Roy] The rank of any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L is at least half its structural rank
42 59
Reformulation by D Roy
Instead of taking logarithmsof algebraic numbers andlooking for the algebraicindependence relationsD Roy fixes a polynomial andlooks at the points withcoordinates logarithms ofalgebraic numbers on thecorresponding hypersurface
43 59
Reformulation by D Roy
Royrsquos reformulation of the conjecture of algebraicindependence of logarithms is
Conjecture For any algebraic subvariety V of Cn definedover the field Q of algebraic numbers the set V cap Ln is theunion of the sets E cap Ln where E ranges over the set ofvector subspaces of Cn which are defined over Q andcontained in V
Trivial Any element in E cap Ln where E is a vector subspaceof Cn defined over Q and contained in V belongs to V cap Ln
44 59
Example The Four Exponentials ConjectureTake for V the hypersurface of C4 defined by the equation
z1z4 = z2z3
The maximal Cndashvector subspaces of C4 defined over Q andcontained in V are the planes
az1 = bz2 bz4 = az3
and the planesaz1 = bz3 bz4 = az2
with (a b) isin Q2 (0 0)Hence Schanuelrsquos Conjecture implies that if λ1 λ2 λ3 λ4 arelogarithms of algebraic numbers satisfying
λ1λ4 = λ2λ3
then either λ1 λ2 are linearly dependent over Q or else λ1 λ3
are linearly dependent over Q45 59
Six Exponentials Theorem
and Four Exponentials Conjecture
A Selberg CL Siegel S Lang K Ramachandra
46 59
The Four Exponentials Conjecture
Conjecture (Th Schneider S Lang K Ramachandra) Ifx1 x2 are Qndashlinearly independent complex numbers and y1 y2
are Qndashlinearly independent complex numbers then one atleast of the four numbers
ex1y1 ex1y2 ex2y1 ex2y2
is transcendentalEquivalent statement Any 2times 2 matrix(
λ1 λ2
λ3 λ4
)with entries in L and with Qndashlinearly independent rows andQndashlinearly independent columns has maximal rank 2
47 59
How could we attack Schanuelrsquos Conjecture
Let x1 xn be Qndashlinearly independent complex numbersFollowing the transcendence methods of Hermite GelrsquofondSchneider one may start by introducing an auxiliaryfunction
F (z) = P (z ez)
where P isin Z[X0 X1] is a nonndashzero polynomial One considersthe derivatives (
d
dz
)kF = (DkP )(z ez)
at the pointsm1x1 + middot middot middot+mnxn
for various values of (m1 mn) isin Zn
48 59
Auxiliary functionLet D denote the derivation
D =part
partX0
+X1part
partX1
over the ring C[X0 X1] Recall that x1 xn are Qndashlinearlyindependent complex numbers Let α1 αn be nonndashzerocomplex numbersThe transcendence machinery produces a sequence (PN)Nge0
of polynomials with integer coefficients satisfying∣∣∣∣∣(DkPN)(nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1
49 59
Royrsquos approach to Schanuelrsquos Conjecture (1999)
If the number of equations we produce is too small such a setof relations does not contain any information The existenceof a sequence of nonndashtrivial polynomials (PN)Nge0 follows fromlinear algebra
On the other hand following D Roy one may expect that theexistence of such a sequence (PN)Nge0 producing sufficientlymany such equations will yield the desired conclusion
tr degQQ(x1 xn α1 αn
)ge n
50 59
Royrsquos Conjecture (1999)Let s0 s1 t0 t1 u positive real numbers satisfying
max1 t0 2t1 lt mins0 2s1and
maxs0 s1 + t1 lt u lt1
2(1 + t0 + t1)
Assume that for any sufficiently large positive integer N there exists a nonndashzero polynomial PN isin Z[X0 X1] withpartial degree le N t0 in X0 partial degree le N t1 in X1 andheight le eN which satisfies∣∣∣∣∣(DkPN)(
nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1 Then
tr degQQ(x1 xn α1 αn) ge n51 59
Royrsquos Theorem (1999)
Royrsquos conjecture is equivalent to Schanuelrsquos Conjecture
52 59
Equivalence between Schanuel and Roy
Let (x α) isin CtimesCtimes and let s0 s1 t0 t1 u be positive realnumbers satisfying the inequalities of Royrsquos Conjecture Thenthe following conditions are equivalent (a) The number αeminusx is a root of unity
(b) For any sufficiently large positive integer N there exists anonndashzero polynomial QN isin Z[X0 X1] with partial degreele N t0 in X0 partial degree le N t1 in X1 and heightH(QN) le eN such that∣∣(DkQN)(mx αm)
∣∣ le exp(minusNu)
for any km isin N with k le N s0 and m le N s1
53 59
Gelrsquofondrsquos transcendence criterion
Simple form Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le n and height le en such that
|Pn(ϑ)| le eminus7n2
for all n ge 1 then ϑ is algebraic and Pn(ϑ) = 0 for all n ge 1
First extension Replace the bound for the degree by le dnthe bound for the height by ebn and the bound for |Pn(ϑ)| byeminuscdnbn with some constant c gt 0 independent of n
Some mild conditions are required on the sequences (dn)nge1 and(bn)nge1
54 59
Transcendence criterion with multiplicitiesWith derivatives Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le dn and height le ebn such that∣∣P (j)
n (ϑ)∣∣ le eminuscdnbntn
for all j in the range 0 le j lt tn and all n ge 1 then ϑ isalgebraicDue to M Laurent and D Roy applications to algebraicindependence
55 59
Criterion with several points
Goal Given a sequence of complex numbers (ϑi)ige1 if thereexists a sequence (Pn)nge1 of nonndashzero polynomials in Z[X]with Pn of degree le dn and height le ebn such that∣∣P (j)
n (ϑi)∣∣ le eminuscdnbntnsn
for 0 le j lt tn 1 le i le sn and all n ge 1 then the numbers ϑiare algebraic
D Roy Not true in general but true in some special caseswith a structure on the sequence (ϑi)ige1Combines the elimination arguments used for criteria ofalgebraic independence and for zero estimates
56 59
Small value estimates for the additive group
D Roy Small value estimates for the additive group Intern JNumber Theory to appear
Let ξ be a transcendental complex number let β σ τ and νbe non-negative real numbers let n0 be a positive integer andlet (Pn)ngen0 be a sequence of nonndashzero polynomials in Z[T ]satisfying deg(Pn) le n and H(Pn) le exp(nβ) for eachn ge n0 Suppose that β gt 1 (34)σ + τ lt 1 andν gt 1 + β minus (34)σ minus τ Then for infinitely many n we have
max|P [j]n (iξ)| 0 le i le nσ 0 le j le nτ
gt exp(minusnν)
57 59
Small value estimates for the multiplicative group
D Roy Small value estimates for the multiplicative groupActa Arith to appear
Let ξ1 ξm be multiplicatively independent complexnumbers in a field of transcendence degree 1 Under suitableassumptions on the parameters β σ τ ν for infinitely manypositive integers n there exists no nonndashzero polynomialP isin Z[T ] satisfying deg(P ) le n H(P ) le exp(nβ) and
max|P [j](ξi11 middot middot middot ξimm )| 0 le i1 im le nσ 0 le j le nτ
gt exp(minusnν)
58 59
Heidelberg Colloquium organized by the Institute for Mathematics
httpwwwmathiuni-heidelbergdevortraghtmls=1 January 22 2009
Transcendental Number TheorySchanuelrsquos Conjecture
Michel Waldschmidt
Institut de Mathematiques de Jussieu amp Paris VIhttpwwwmathjussieufrsimmiw
59 59
Some open problems
For each of the following numbers it is expected that it istranscendental but it is not even known whether it is rationalor not
e+ π eπ πe ee ee2
eee
ππ ππ2
πππ
log π log(log 2) π log 2 (log 2)(log 3) 2log 2 (log 2)log 3
In other words we do not know whether a degree 1 polynomialcould vanish at the corresponding point but we expect thatno nonndashzero polynomial of any degree vanishes at this point
9 59
Algebraic independence
Algebraicity and transcendence deal with a single complexnumber and one variable polynomialsAlgebraic dependence or independence is the same but fortuples and multivariate polynomials
Let K sup k be an extension of fields and (θ1 θm) be amndashtuple of elements in K We say that θ1 θm arealgebraically dependent over k if there exists a nonndashzeropolynomial f isin k[X1 Xm] which vanishes at the point(θ1 θm) isin KmOtherwise we say that θ1 θm are algebraically independentover k
10 59
Special cases
In the case m = 1 to say that θ1 is algebraically independentover k just means that it is transcendental over k
Dealing with complex numbers K = C the words algebraictranscendental algebraically dependent algebraicallyindependent refer to the case k = Q
11 59
Examples
The numbersradic
2 and π are algebraically dependent The polynomial X2 minus 2 isin Z[X Y ] vanishes at (
radic2 π)
The numbers π andradicπ2 + 1 are algebraically dependent (and
both are transcendental numbers) The polynomial Y minusX2 minus 1 vanishes at (π
radicπ2 + 1)
The two numbers e and eradic
2 are algebraically independentwhich means that for any nonndashzero polynomial f isin Z[X Y ]
with rational integer coefficients the number f(e eradic
2) is notzeroSpecial case of the Lindemann-Weierstraszlig Theorem
12 59
Transcendence degree
Let Kk be a field extension The maximal number ofelements in K which are algebraically independent over k iscalled the transcendence degree of K over k and denotedtr degkK
Let t = tr degkK A subset θ1 θt of K with t elementswhich are algebraically independent is called a transcendencebasis of K over k It is the same as a maximal subset ofkndashalgebraically independent elements in KHence K is an algebraic extension of k(θ1 θt)
13 59
Transcendence degree of extensions
Assumek sub K sub L
The union of a transcendence basis of K over k and of atranscendence basis of L over K produces a transcendencebasis of L over kHence
tr degkL = tr degkK + tr degKL
An algebraic extension Kk is an extension of transcendencedegree 0 This means that there is no transcendental elementin K over k (any element in K is algebraic over k)
14 59
Algebraic independence of complex numbers
For complex numbers algebraic independence over Q or overQ is the same In particular if θ1 θm are algebraicallyindependent complex numbers then for any non-constantpolynomial f with algebraic coefficients the numberf(θ1 θm) is transcendental
For instance the two numbers e and eradic
2 are algebraicallyindependent As a consequence for any nonndashconstantpolynomial f isin Q[X Y ] with algebraic coefficients the
number f(e eradic
2) is transcendental
15 59
Schanuelrsquos Conjecture
Let x1 xn be Q-linearly independent complex numbersThen at least n of the 2n numbers x1 xn e
x1 exn arealgebraically independent
Since there are 2n numbers only the transcendence degreeover Q of the field
Q(x1 xn e
x1 exn)
is at most 2n The conjecture is that this transcendencedegree is always ge n
n le tr degQQ(x1 xn e
x1 exn)le 2n
16 59
Origin of Schanuelrsquos ConjectureCourse given by Serge Lang (1927ndash2005) at Yale in the 60rsquos
S Lang ndash Introduction to transcendental numbersAddison-Wesley 1966
also attended by M Nagata (1927ndash2008)(14th Problem of Hilbert)
Nagatarsquos Conjecture solved by E Bombieri
17 59
Schanuelrsquos Conjecture
Let x1 xn be Q-linearly independent complex numbersThen
tr degQQ(x1 xn e
x1 exn)ge n
Remark For almost all tuples (with respect to the Lebesguemeasure) the transcendence degree is 2n
18 59
AO Gelrsquofond CRAS 1934
19 59
Statement by Gelrsquofond (1934)
Let β1 βn be Q-linearly independent algebraic numbersand let logα1 logαm be Q-linearly independentlogarithms of algebraic numbers Then the numbers
eβ1 eβn logα1 logαm
are algebraically independent over Q
20 59
Further statement by Gelrsquofond
Let β1 βn be algebraic numbers with β1 6= 0 and letα1 αm be algebraic numbers with α1 6= 0 1 α2 6= 0 1αi 6= 0 Then the numbers
eβ1eβ2e βnminus1e
βn
and αα αm
21
are transcendental and there is no nontrivial algebraic relationbetween such numbers
Remark The condition on α2 should be that it is irrational
21 59
Easy consequence of Schanuelrsquos Conjecture
According to Schanuelrsquos Conjecture the following numbers arealgebraically independent
e+ π eπ πe ee ee2
eee
ππ ππ2
πππ
log π log(log 2) π log 2 (log 2)(log 3) 2log 2 (log 2)log 3
Proof This is an easy exercise
22 59
Langrsquos exercise
Define E0 = Q Inductivelyfor n ge 1 define En as thealgebraic closure of the fieldgenerated over Enminus1 by thenumbers exp(x) = ex wherex ranges over Enminus1 Let E bethe union of En n ge 0Then Schanuelrsquos Conjectureimplies that the number πdoes not belong to E
More precisely Schanuelrsquos Conjecture implies that thenumbers π log π log log π log log log π are algebraicallyindependent over E
23 59
A variant of Langrsquos exercise
Define L0 = Q Inductively for n ge 1 define Ln as thealgebraic closure of the field generated over Lnminus1 by thenumbers y where y ranges over the set of complex numberssuch that ey isin Lnminus1 Let L be the union of Ln n ge 0Then Schanuelrsquos Conjecture implies that the number e doesnot belong to L
More precisely Schanuelrsquos Conjecture implies that the
numbers e ee eee ee
ee
are algebraically independent over L
24 59
Arizona Winter School AWS2008 Tucson
Theorem [Jonathan Bober Chuangxun Cheng Brian DietelMathilde Herblot Jingjing Huang Holly Krieger DiegoMarques Jonathan Mason Martin Mereb and Robert Wilson]Schanuelrsquos Conjecture implies that the fields E and L arelinearly disjoint over Q
Definition Given a field extension FK and twosubextensions F1 F2 sube F we say F1 F2 are linearly disjointover K when the following holds Any set x1 xn sube F1
of Kndash linearly independent elements is linearly independentover F2
Reference arXiv08043550 [mathNT] 2008
25 59
Formal analogs
WD Brownawell(was a student of Schanuel)
J Axrsquos Theorem (1968) Version of SchanuelrsquosConjecture for power seriesover C(and R Coleman for powerseries over Q)
Work by WD Brownawelland K Kubota on the ellipticanalog of Axrsquos Theorem
26 59
Conjectures by A Grothendieck and Y Andre
Generalized Conjecture onPeriods by Grothendieck Dimension of theMumfordndashTate group of asmooth projective variety
Generalization by Y Andre tomotives
Case of 1ndashmotives Elliptico-Toric Conjecture ofC Bertolin
27 59
Ubiquity of Schanuelrsquos Conjecture
Other contexts pndashadic numbers Leopoldtrsquos Conjecture onthe pndashadic rank of the units of an algebraic number fieldNon-vanishing of RegulatorsNonndashdegenerescence of heightsConjecture of B Mazur on rational pointsDiophantine approximation on tori
Dipendra Prasad Gopal Prasad
28 59
Methods from logicEhud Hrushovski Boris Zilber Jonathan Kirby
Calculus of rdquopredimension functionsrdquo (E Hrushovski)
Zilberrsquos construction of a rdquopseudoexponentiationrdquo
Also A Macintyre DE Marker G Terzo AJ WilkieD Bertrand
29 59
Methods from logic Model theory
Exponential algebraicity in exponential fieldsby
Jonathan Kirby
The dimension of the exponential algebraic closure operator inan exponential field satisfies a weak Schanuel property
A corollary is that there are at most countably many essentialcounterexamples to Schanuelrsquos conjecture
arXiv 08104285v2
30 59
Known
LindemannndashWeierstraszlig Theorem = case where x1 xn arealgebraic
Let β1 βn be algebraic numbers which are linearlyindependent over Q Then the numbers eβ1 eβn arealgebraically independent over Q
31 59
Problem of Gelrsquofond and SchneiderRaised by AO Gelrsquofond in 1948 and Th Schneider in 1952
Conjecture If α is an algebraic number α 6= 0 α 6= 1 and ifβ is an irrational algebraic number of degree d then the dminus 1numbers
αβ αβ2
αβdminus1
are algebraically independent
Special case of Schanuelrsquos Conjecture Take xi = βiminus1 logαn = d so that
x1 xn e
x1 exn
islogα β logα βdminus1 logα α αβ αβ
dminus1
The conclusion of Schanuelrsquos Conjecture is
tr degQQ(logα αβ αβ
2
αβdminus1)
= d
32 59
Algebraic independence method of Gelrsquofond
AO Gelrsquofond (1948)
The two numbers 23radic2 and
23radic4 are algebraically
independentMore generally if α is analgebraic number α 6= 0α 6= 1 and if β is a algebraicnumber of degree d ge 3 thentwo at least of the numbers
αβ αβ2
αβdminus1
are algebraically independent
33 59
Tools
Transcendence criterion Replaces Liouvillersquos inequality intranscendence proofsLiouville A nonndashzero rational integer n isin Z satisfies |n| ge 1Gelrsquofond Needs to give a lower bound for |P (θ)| withP isin Z[X] 0 when θ is transcendental
Zero estimate for exponential polynomials C Hermite P Turan K Mahler R Tijdeman
Small transcendence degree AO Gelrsquofond AA Smelev R Tijdeman WD Brownawell
34 59
Large transcendence degree
GV Chudnovsky (1976)Among the numbers
αβ αβ2
αβdminus1
at least [log2 d] arealgebraically independent
GV Chudnovsky ndash On the path to Schanuelrsquosconjecture Algebraic curves close to a pointI General theory of colored sequencesII Fields of finite transcendence type and coloredsequences ResultantsStudia Sci Math Hungar 12 (1977) 125ndash157 (1980)
35 59
Partial result on the problem of Gelrsquofond and
Schneider
AO Gelrsquofond GV Chudnovskii P Philippon YuV Nesterenko
G Diaz If α is an algebraic number α 6= 0 α 6= 1 and if β isan irrational algebraic number of degree d then
tr degQQ(αβ αβ
2
αβdminus1) ge [d+ 1
2
]
36 59
Conjecture of algebraic independence of logarithms
of algebraic numbers
Denote by L the set of complex numbers λ for which eλ isalgebraic
L = logα α isin Qtimes
Hence L is a Q-vector subspace of CThe most important special case of Schanuelrsquos Conjecture is
Conjecture Let λ1 λn be Q-linearly independent elementsin L Then the numbers λ1 λn are algebraicallyindependent over Q
Not yet known that the transcendence degree is ge 2 Open problem Among all logarithms of algebraic numbersone at least is transcendental over Q(π)
37 59
Structural rank of a matrix
Let K be a field k a subfield and M a matrix with entries in KConsider the k-vector subspace E of K spanned by the entriesof M Choose an injective morphism ϕ of E into a k-vectorspace kX1 + middot middot middot+ kXn The image ϕ(M) of M is a matrixwhose entries are in the field k(X1 Xn) of rationalfractions Its rank does not depend on the choice of ϕ
This is the structural rank of M with respect to k
38 59
Example
LetM =
(bij + λij
)1leiled1lejle`
be a matrix with coefficients in Q + L Consider a basis of theQndashvector spanned by the entries and replace the elements inthis basis by unknowns This gives a new matrix M withcoefficients in a field of rational fractions the rank of which isthe structural rank of M (with respect to Q)
As a consequence of the conjecture of algebraic independenceof logarithms of algebraic numbers the rank of M should bethe same as the rank of its specialization M
39 59
Equivalence between the two conjectures
Following D Roy theconjecture on algebraicindependence of logarithms ofalgebraic numbers isequivalent to
Conjecture Any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L has a rank equal to its structuralrank
40 59
Any Polynomial is the Determinant of a Matrix
The proof of the equivalence uses the nice auxiliary result
For any P isin k[X1 Xn] there exists a square matrix withentries in the k-vector space k + kX1 + middot middot middot+ kXn whosedeterminant is P
41 59
Half of the Conjecture is solved
From a certain point of view half of the conjecture of algebraicindependence of logarithms of algebraic numbers is solved
Theorem [D Roy] The rank of any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L is at least half its structural rank
42 59
Reformulation by D Roy
Instead of taking logarithmsof algebraic numbers andlooking for the algebraicindependence relationsD Roy fixes a polynomial andlooks at the points withcoordinates logarithms ofalgebraic numbers on thecorresponding hypersurface
43 59
Reformulation by D Roy
Royrsquos reformulation of the conjecture of algebraicindependence of logarithms is
Conjecture For any algebraic subvariety V of Cn definedover the field Q of algebraic numbers the set V cap Ln is theunion of the sets E cap Ln where E ranges over the set ofvector subspaces of Cn which are defined over Q andcontained in V
Trivial Any element in E cap Ln where E is a vector subspaceof Cn defined over Q and contained in V belongs to V cap Ln
44 59
Example The Four Exponentials ConjectureTake for V the hypersurface of C4 defined by the equation
z1z4 = z2z3
The maximal Cndashvector subspaces of C4 defined over Q andcontained in V are the planes
az1 = bz2 bz4 = az3
and the planesaz1 = bz3 bz4 = az2
with (a b) isin Q2 (0 0)Hence Schanuelrsquos Conjecture implies that if λ1 λ2 λ3 λ4 arelogarithms of algebraic numbers satisfying
λ1λ4 = λ2λ3
then either λ1 λ2 are linearly dependent over Q or else λ1 λ3
are linearly dependent over Q45 59
Six Exponentials Theorem
and Four Exponentials Conjecture
A Selberg CL Siegel S Lang K Ramachandra
46 59
The Four Exponentials Conjecture
Conjecture (Th Schneider S Lang K Ramachandra) Ifx1 x2 are Qndashlinearly independent complex numbers and y1 y2
are Qndashlinearly independent complex numbers then one atleast of the four numbers
ex1y1 ex1y2 ex2y1 ex2y2
is transcendentalEquivalent statement Any 2times 2 matrix(
λ1 λ2
λ3 λ4
)with entries in L and with Qndashlinearly independent rows andQndashlinearly independent columns has maximal rank 2
47 59
How could we attack Schanuelrsquos Conjecture
Let x1 xn be Qndashlinearly independent complex numbersFollowing the transcendence methods of Hermite GelrsquofondSchneider one may start by introducing an auxiliaryfunction
F (z) = P (z ez)
where P isin Z[X0 X1] is a nonndashzero polynomial One considersthe derivatives (
d
dz
)kF = (DkP )(z ez)
at the pointsm1x1 + middot middot middot+mnxn
for various values of (m1 mn) isin Zn
48 59
Auxiliary functionLet D denote the derivation
D =part
partX0
+X1part
partX1
over the ring C[X0 X1] Recall that x1 xn are Qndashlinearlyindependent complex numbers Let α1 αn be nonndashzerocomplex numbersThe transcendence machinery produces a sequence (PN)Nge0
of polynomials with integer coefficients satisfying∣∣∣∣∣(DkPN)(nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1
49 59
Royrsquos approach to Schanuelrsquos Conjecture (1999)
If the number of equations we produce is too small such a setof relations does not contain any information The existenceof a sequence of nonndashtrivial polynomials (PN)Nge0 follows fromlinear algebra
On the other hand following D Roy one may expect that theexistence of such a sequence (PN)Nge0 producing sufficientlymany such equations will yield the desired conclusion
tr degQQ(x1 xn α1 αn
)ge n
50 59
Royrsquos Conjecture (1999)Let s0 s1 t0 t1 u positive real numbers satisfying
max1 t0 2t1 lt mins0 2s1and
maxs0 s1 + t1 lt u lt1
2(1 + t0 + t1)
Assume that for any sufficiently large positive integer N there exists a nonndashzero polynomial PN isin Z[X0 X1] withpartial degree le N t0 in X0 partial degree le N t1 in X1 andheight le eN which satisfies∣∣∣∣∣(DkPN)(
nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1 Then
tr degQQ(x1 xn α1 αn) ge n51 59
Royrsquos Theorem (1999)
Royrsquos conjecture is equivalent to Schanuelrsquos Conjecture
52 59
Equivalence between Schanuel and Roy
Let (x α) isin CtimesCtimes and let s0 s1 t0 t1 u be positive realnumbers satisfying the inequalities of Royrsquos Conjecture Thenthe following conditions are equivalent (a) The number αeminusx is a root of unity
(b) For any sufficiently large positive integer N there exists anonndashzero polynomial QN isin Z[X0 X1] with partial degreele N t0 in X0 partial degree le N t1 in X1 and heightH(QN) le eN such that∣∣(DkQN)(mx αm)
∣∣ le exp(minusNu)
for any km isin N with k le N s0 and m le N s1
53 59
Gelrsquofondrsquos transcendence criterion
Simple form Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le n and height le en such that
|Pn(ϑ)| le eminus7n2
for all n ge 1 then ϑ is algebraic and Pn(ϑ) = 0 for all n ge 1
First extension Replace the bound for the degree by le dnthe bound for the height by ebn and the bound for |Pn(ϑ)| byeminuscdnbn with some constant c gt 0 independent of n
Some mild conditions are required on the sequences (dn)nge1 and(bn)nge1
54 59
Transcendence criterion with multiplicitiesWith derivatives Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le dn and height le ebn such that∣∣P (j)
n (ϑ)∣∣ le eminuscdnbntn
for all j in the range 0 le j lt tn and all n ge 1 then ϑ isalgebraicDue to M Laurent and D Roy applications to algebraicindependence
55 59
Criterion with several points
Goal Given a sequence of complex numbers (ϑi)ige1 if thereexists a sequence (Pn)nge1 of nonndashzero polynomials in Z[X]with Pn of degree le dn and height le ebn such that∣∣P (j)
n (ϑi)∣∣ le eminuscdnbntnsn
for 0 le j lt tn 1 le i le sn and all n ge 1 then the numbers ϑiare algebraic
D Roy Not true in general but true in some special caseswith a structure on the sequence (ϑi)ige1Combines the elimination arguments used for criteria ofalgebraic independence and for zero estimates
56 59
Small value estimates for the additive group
D Roy Small value estimates for the additive group Intern JNumber Theory to appear
Let ξ be a transcendental complex number let β σ τ and νbe non-negative real numbers let n0 be a positive integer andlet (Pn)ngen0 be a sequence of nonndashzero polynomials in Z[T ]satisfying deg(Pn) le n and H(Pn) le exp(nβ) for eachn ge n0 Suppose that β gt 1 (34)σ + τ lt 1 andν gt 1 + β minus (34)σ minus τ Then for infinitely many n we have
max|P [j]n (iξ)| 0 le i le nσ 0 le j le nτ
gt exp(minusnν)
57 59
Small value estimates for the multiplicative group
D Roy Small value estimates for the multiplicative groupActa Arith to appear
Let ξ1 ξm be multiplicatively independent complexnumbers in a field of transcendence degree 1 Under suitableassumptions on the parameters β σ τ ν for infinitely manypositive integers n there exists no nonndashzero polynomialP isin Z[T ] satisfying deg(P ) le n H(P ) le exp(nβ) and
max|P [j](ξi11 middot middot middot ξimm )| 0 le i1 im le nσ 0 le j le nτ
gt exp(minusnν)
58 59
Heidelberg Colloquium organized by the Institute for Mathematics
httpwwwmathiuni-heidelbergdevortraghtmls=1 January 22 2009
Transcendental Number TheorySchanuelrsquos Conjecture
Michel Waldschmidt
Institut de Mathematiques de Jussieu amp Paris VIhttpwwwmathjussieufrsimmiw
59 59
Algebraic independence
Algebraicity and transcendence deal with a single complexnumber and one variable polynomialsAlgebraic dependence or independence is the same but fortuples and multivariate polynomials
Let K sup k be an extension of fields and (θ1 θm) be amndashtuple of elements in K We say that θ1 θm arealgebraically dependent over k if there exists a nonndashzeropolynomial f isin k[X1 Xm] which vanishes at the point(θ1 θm) isin KmOtherwise we say that θ1 θm are algebraically independentover k
10 59
Special cases
In the case m = 1 to say that θ1 is algebraically independentover k just means that it is transcendental over k
Dealing with complex numbers K = C the words algebraictranscendental algebraically dependent algebraicallyindependent refer to the case k = Q
11 59
Examples
The numbersradic
2 and π are algebraically dependent The polynomial X2 minus 2 isin Z[X Y ] vanishes at (
radic2 π)
The numbers π andradicπ2 + 1 are algebraically dependent (and
both are transcendental numbers) The polynomial Y minusX2 minus 1 vanishes at (π
radicπ2 + 1)
The two numbers e and eradic
2 are algebraically independentwhich means that for any nonndashzero polynomial f isin Z[X Y ]
with rational integer coefficients the number f(e eradic
2) is notzeroSpecial case of the Lindemann-Weierstraszlig Theorem
12 59
Transcendence degree
Let Kk be a field extension The maximal number ofelements in K which are algebraically independent over k iscalled the transcendence degree of K over k and denotedtr degkK
Let t = tr degkK A subset θ1 θt of K with t elementswhich are algebraically independent is called a transcendencebasis of K over k It is the same as a maximal subset ofkndashalgebraically independent elements in KHence K is an algebraic extension of k(θ1 θt)
13 59
Transcendence degree of extensions
Assumek sub K sub L
The union of a transcendence basis of K over k and of atranscendence basis of L over K produces a transcendencebasis of L over kHence
tr degkL = tr degkK + tr degKL
An algebraic extension Kk is an extension of transcendencedegree 0 This means that there is no transcendental elementin K over k (any element in K is algebraic over k)
14 59
Algebraic independence of complex numbers
For complex numbers algebraic independence over Q or overQ is the same In particular if θ1 θm are algebraicallyindependent complex numbers then for any non-constantpolynomial f with algebraic coefficients the numberf(θ1 θm) is transcendental
For instance the two numbers e and eradic
2 are algebraicallyindependent As a consequence for any nonndashconstantpolynomial f isin Q[X Y ] with algebraic coefficients the
number f(e eradic
2) is transcendental
15 59
Schanuelrsquos Conjecture
Let x1 xn be Q-linearly independent complex numbersThen at least n of the 2n numbers x1 xn e
x1 exn arealgebraically independent
Since there are 2n numbers only the transcendence degreeover Q of the field
Q(x1 xn e
x1 exn)
is at most 2n The conjecture is that this transcendencedegree is always ge n
n le tr degQQ(x1 xn e
x1 exn)le 2n
16 59
Origin of Schanuelrsquos ConjectureCourse given by Serge Lang (1927ndash2005) at Yale in the 60rsquos
S Lang ndash Introduction to transcendental numbersAddison-Wesley 1966
also attended by M Nagata (1927ndash2008)(14th Problem of Hilbert)
Nagatarsquos Conjecture solved by E Bombieri
17 59
Schanuelrsquos Conjecture
Let x1 xn be Q-linearly independent complex numbersThen
tr degQQ(x1 xn e
x1 exn)ge n
Remark For almost all tuples (with respect to the Lebesguemeasure) the transcendence degree is 2n
18 59
AO Gelrsquofond CRAS 1934
19 59
Statement by Gelrsquofond (1934)
Let β1 βn be Q-linearly independent algebraic numbersand let logα1 logαm be Q-linearly independentlogarithms of algebraic numbers Then the numbers
eβ1 eβn logα1 logαm
are algebraically independent over Q
20 59
Further statement by Gelrsquofond
Let β1 βn be algebraic numbers with β1 6= 0 and letα1 αm be algebraic numbers with α1 6= 0 1 α2 6= 0 1αi 6= 0 Then the numbers
eβ1eβ2e βnminus1e
βn
and αα αm
21
are transcendental and there is no nontrivial algebraic relationbetween such numbers
Remark The condition on α2 should be that it is irrational
21 59
Easy consequence of Schanuelrsquos Conjecture
According to Schanuelrsquos Conjecture the following numbers arealgebraically independent
e+ π eπ πe ee ee2
eee
ππ ππ2
πππ
log π log(log 2) π log 2 (log 2)(log 3) 2log 2 (log 2)log 3
Proof This is an easy exercise
22 59
Langrsquos exercise
Define E0 = Q Inductivelyfor n ge 1 define En as thealgebraic closure of the fieldgenerated over Enminus1 by thenumbers exp(x) = ex wherex ranges over Enminus1 Let E bethe union of En n ge 0Then Schanuelrsquos Conjectureimplies that the number πdoes not belong to E
More precisely Schanuelrsquos Conjecture implies that thenumbers π log π log log π log log log π are algebraicallyindependent over E
23 59
A variant of Langrsquos exercise
Define L0 = Q Inductively for n ge 1 define Ln as thealgebraic closure of the field generated over Lnminus1 by thenumbers y where y ranges over the set of complex numberssuch that ey isin Lnminus1 Let L be the union of Ln n ge 0Then Schanuelrsquos Conjecture implies that the number e doesnot belong to L
More precisely Schanuelrsquos Conjecture implies that the
numbers e ee eee ee
ee
are algebraically independent over L
24 59
Arizona Winter School AWS2008 Tucson
Theorem [Jonathan Bober Chuangxun Cheng Brian DietelMathilde Herblot Jingjing Huang Holly Krieger DiegoMarques Jonathan Mason Martin Mereb and Robert Wilson]Schanuelrsquos Conjecture implies that the fields E and L arelinearly disjoint over Q
Definition Given a field extension FK and twosubextensions F1 F2 sube F we say F1 F2 are linearly disjointover K when the following holds Any set x1 xn sube F1
of Kndash linearly independent elements is linearly independentover F2
Reference arXiv08043550 [mathNT] 2008
25 59
Formal analogs
WD Brownawell(was a student of Schanuel)
J Axrsquos Theorem (1968) Version of SchanuelrsquosConjecture for power seriesover C(and R Coleman for powerseries over Q)
Work by WD Brownawelland K Kubota on the ellipticanalog of Axrsquos Theorem
26 59
Conjectures by A Grothendieck and Y Andre
Generalized Conjecture onPeriods by Grothendieck Dimension of theMumfordndashTate group of asmooth projective variety
Generalization by Y Andre tomotives
Case of 1ndashmotives Elliptico-Toric Conjecture ofC Bertolin
27 59
Ubiquity of Schanuelrsquos Conjecture
Other contexts pndashadic numbers Leopoldtrsquos Conjecture onthe pndashadic rank of the units of an algebraic number fieldNon-vanishing of RegulatorsNonndashdegenerescence of heightsConjecture of B Mazur on rational pointsDiophantine approximation on tori
Dipendra Prasad Gopal Prasad
28 59
Methods from logicEhud Hrushovski Boris Zilber Jonathan Kirby
Calculus of rdquopredimension functionsrdquo (E Hrushovski)
Zilberrsquos construction of a rdquopseudoexponentiationrdquo
Also A Macintyre DE Marker G Terzo AJ WilkieD Bertrand
29 59
Methods from logic Model theory
Exponential algebraicity in exponential fieldsby
Jonathan Kirby
The dimension of the exponential algebraic closure operator inan exponential field satisfies a weak Schanuel property
A corollary is that there are at most countably many essentialcounterexamples to Schanuelrsquos conjecture
arXiv 08104285v2
30 59
Known
LindemannndashWeierstraszlig Theorem = case where x1 xn arealgebraic
Let β1 βn be algebraic numbers which are linearlyindependent over Q Then the numbers eβ1 eβn arealgebraically independent over Q
31 59
Problem of Gelrsquofond and SchneiderRaised by AO Gelrsquofond in 1948 and Th Schneider in 1952
Conjecture If α is an algebraic number α 6= 0 α 6= 1 and ifβ is an irrational algebraic number of degree d then the dminus 1numbers
αβ αβ2
αβdminus1
are algebraically independent
Special case of Schanuelrsquos Conjecture Take xi = βiminus1 logαn = d so that
x1 xn e
x1 exn
islogα β logα βdminus1 logα α αβ αβ
dminus1
The conclusion of Schanuelrsquos Conjecture is
tr degQQ(logα αβ αβ
2
αβdminus1)
= d
32 59
Algebraic independence method of Gelrsquofond
AO Gelrsquofond (1948)
The two numbers 23radic2 and
23radic4 are algebraically
independentMore generally if α is analgebraic number α 6= 0α 6= 1 and if β is a algebraicnumber of degree d ge 3 thentwo at least of the numbers
αβ αβ2
αβdminus1
are algebraically independent
33 59
Tools
Transcendence criterion Replaces Liouvillersquos inequality intranscendence proofsLiouville A nonndashzero rational integer n isin Z satisfies |n| ge 1Gelrsquofond Needs to give a lower bound for |P (θ)| withP isin Z[X] 0 when θ is transcendental
Zero estimate for exponential polynomials C Hermite P Turan K Mahler R Tijdeman
Small transcendence degree AO Gelrsquofond AA Smelev R Tijdeman WD Brownawell
34 59
Large transcendence degree
GV Chudnovsky (1976)Among the numbers
αβ αβ2
αβdminus1
at least [log2 d] arealgebraically independent
GV Chudnovsky ndash On the path to Schanuelrsquosconjecture Algebraic curves close to a pointI General theory of colored sequencesII Fields of finite transcendence type and coloredsequences ResultantsStudia Sci Math Hungar 12 (1977) 125ndash157 (1980)
35 59
Partial result on the problem of Gelrsquofond and
Schneider
AO Gelrsquofond GV Chudnovskii P Philippon YuV Nesterenko
G Diaz If α is an algebraic number α 6= 0 α 6= 1 and if β isan irrational algebraic number of degree d then
tr degQQ(αβ αβ
2
αβdminus1) ge [d+ 1
2
]
36 59
Conjecture of algebraic independence of logarithms
of algebraic numbers
Denote by L the set of complex numbers λ for which eλ isalgebraic
L = logα α isin Qtimes
Hence L is a Q-vector subspace of CThe most important special case of Schanuelrsquos Conjecture is
Conjecture Let λ1 λn be Q-linearly independent elementsin L Then the numbers λ1 λn are algebraicallyindependent over Q
Not yet known that the transcendence degree is ge 2 Open problem Among all logarithms of algebraic numbersone at least is transcendental over Q(π)
37 59
Structural rank of a matrix
Let K be a field k a subfield and M a matrix with entries in KConsider the k-vector subspace E of K spanned by the entriesof M Choose an injective morphism ϕ of E into a k-vectorspace kX1 + middot middot middot+ kXn The image ϕ(M) of M is a matrixwhose entries are in the field k(X1 Xn) of rationalfractions Its rank does not depend on the choice of ϕ
This is the structural rank of M with respect to k
38 59
Example
LetM =
(bij + λij
)1leiled1lejle`
be a matrix with coefficients in Q + L Consider a basis of theQndashvector spanned by the entries and replace the elements inthis basis by unknowns This gives a new matrix M withcoefficients in a field of rational fractions the rank of which isthe structural rank of M (with respect to Q)
As a consequence of the conjecture of algebraic independenceof logarithms of algebraic numbers the rank of M should bethe same as the rank of its specialization M
39 59
Equivalence between the two conjectures
Following D Roy theconjecture on algebraicindependence of logarithms ofalgebraic numbers isequivalent to
Conjecture Any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L has a rank equal to its structuralrank
40 59
Any Polynomial is the Determinant of a Matrix
The proof of the equivalence uses the nice auxiliary result
For any P isin k[X1 Xn] there exists a square matrix withentries in the k-vector space k + kX1 + middot middot middot+ kXn whosedeterminant is P
41 59
Half of the Conjecture is solved
From a certain point of view half of the conjecture of algebraicindependence of logarithms of algebraic numbers is solved
Theorem [D Roy] The rank of any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L is at least half its structural rank
42 59
Reformulation by D Roy
Instead of taking logarithmsof algebraic numbers andlooking for the algebraicindependence relationsD Roy fixes a polynomial andlooks at the points withcoordinates logarithms ofalgebraic numbers on thecorresponding hypersurface
43 59
Reformulation by D Roy
Royrsquos reformulation of the conjecture of algebraicindependence of logarithms is
Conjecture For any algebraic subvariety V of Cn definedover the field Q of algebraic numbers the set V cap Ln is theunion of the sets E cap Ln where E ranges over the set ofvector subspaces of Cn which are defined over Q andcontained in V
Trivial Any element in E cap Ln where E is a vector subspaceof Cn defined over Q and contained in V belongs to V cap Ln
44 59
Example The Four Exponentials ConjectureTake for V the hypersurface of C4 defined by the equation
z1z4 = z2z3
The maximal Cndashvector subspaces of C4 defined over Q andcontained in V are the planes
az1 = bz2 bz4 = az3
and the planesaz1 = bz3 bz4 = az2
with (a b) isin Q2 (0 0)Hence Schanuelrsquos Conjecture implies that if λ1 λ2 λ3 λ4 arelogarithms of algebraic numbers satisfying
λ1λ4 = λ2λ3
then either λ1 λ2 are linearly dependent over Q or else λ1 λ3
are linearly dependent over Q45 59
Six Exponentials Theorem
and Four Exponentials Conjecture
A Selberg CL Siegel S Lang K Ramachandra
46 59
The Four Exponentials Conjecture
Conjecture (Th Schneider S Lang K Ramachandra) Ifx1 x2 are Qndashlinearly independent complex numbers and y1 y2
are Qndashlinearly independent complex numbers then one atleast of the four numbers
ex1y1 ex1y2 ex2y1 ex2y2
is transcendentalEquivalent statement Any 2times 2 matrix(
λ1 λ2
λ3 λ4
)with entries in L and with Qndashlinearly independent rows andQndashlinearly independent columns has maximal rank 2
47 59
How could we attack Schanuelrsquos Conjecture
Let x1 xn be Qndashlinearly independent complex numbersFollowing the transcendence methods of Hermite GelrsquofondSchneider one may start by introducing an auxiliaryfunction
F (z) = P (z ez)
where P isin Z[X0 X1] is a nonndashzero polynomial One considersthe derivatives (
d
dz
)kF = (DkP )(z ez)
at the pointsm1x1 + middot middot middot+mnxn
for various values of (m1 mn) isin Zn
48 59
Auxiliary functionLet D denote the derivation
D =part
partX0
+X1part
partX1
over the ring C[X0 X1] Recall that x1 xn are Qndashlinearlyindependent complex numbers Let α1 αn be nonndashzerocomplex numbersThe transcendence machinery produces a sequence (PN)Nge0
of polynomials with integer coefficients satisfying∣∣∣∣∣(DkPN)(nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1
49 59
Royrsquos approach to Schanuelrsquos Conjecture (1999)
If the number of equations we produce is too small such a setof relations does not contain any information The existenceof a sequence of nonndashtrivial polynomials (PN)Nge0 follows fromlinear algebra
On the other hand following D Roy one may expect that theexistence of such a sequence (PN)Nge0 producing sufficientlymany such equations will yield the desired conclusion
tr degQQ(x1 xn α1 αn
)ge n
50 59
Royrsquos Conjecture (1999)Let s0 s1 t0 t1 u positive real numbers satisfying
max1 t0 2t1 lt mins0 2s1and
maxs0 s1 + t1 lt u lt1
2(1 + t0 + t1)
Assume that for any sufficiently large positive integer N there exists a nonndashzero polynomial PN isin Z[X0 X1] withpartial degree le N t0 in X0 partial degree le N t1 in X1 andheight le eN which satisfies∣∣∣∣∣(DkPN)(
nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1 Then
tr degQQ(x1 xn α1 αn) ge n51 59
Royrsquos Theorem (1999)
Royrsquos conjecture is equivalent to Schanuelrsquos Conjecture
52 59
Equivalence between Schanuel and Roy
Let (x α) isin CtimesCtimes and let s0 s1 t0 t1 u be positive realnumbers satisfying the inequalities of Royrsquos Conjecture Thenthe following conditions are equivalent (a) The number αeminusx is a root of unity
(b) For any sufficiently large positive integer N there exists anonndashzero polynomial QN isin Z[X0 X1] with partial degreele N t0 in X0 partial degree le N t1 in X1 and heightH(QN) le eN such that∣∣(DkQN)(mx αm)
∣∣ le exp(minusNu)
for any km isin N with k le N s0 and m le N s1
53 59
Gelrsquofondrsquos transcendence criterion
Simple form Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le n and height le en such that
|Pn(ϑ)| le eminus7n2
for all n ge 1 then ϑ is algebraic and Pn(ϑ) = 0 for all n ge 1
First extension Replace the bound for the degree by le dnthe bound for the height by ebn and the bound for |Pn(ϑ)| byeminuscdnbn with some constant c gt 0 independent of n
Some mild conditions are required on the sequences (dn)nge1 and(bn)nge1
54 59
Transcendence criterion with multiplicitiesWith derivatives Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le dn and height le ebn such that∣∣P (j)
n (ϑ)∣∣ le eminuscdnbntn
for all j in the range 0 le j lt tn and all n ge 1 then ϑ isalgebraicDue to M Laurent and D Roy applications to algebraicindependence
55 59
Criterion with several points
Goal Given a sequence of complex numbers (ϑi)ige1 if thereexists a sequence (Pn)nge1 of nonndashzero polynomials in Z[X]with Pn of degree le dn and height le ebn such that∣∣P (j)
n (ϑi)∣∣ le eminuscdnbntnsn
for 0 le j lt tn 1 le i le sn and all n ge 1 then the numbers ϑiare algebraic
D Roy Not true in general but true in some special caseswith a structure on the sequence (ϑi)ige1Combines the elimination arguments used for criteria ofalgebraic independence and for zero estimates
56 59
Small value estimates for the additive group
D Roy Small value estimates for the additive group Intern JNumber Theory to appear
Let ξ be a transcendental complex number let β σ τ and νbe non-negative real numbers let n0 be a positive integer andlet (Pn)ngen0 be a sequence of nonndashzero polynomials in Z[T ]satisfying deg(Pn) le n and H(Pn) le exp(nβ) for eachn ge n0 Suppose that β gt 1 (34)σ + τ lt 1 andν gt 1 + β minus (34)σ minus τ Then for infinitely many n we have
max|P [j]n (iξ)| 0 le i le nσ 0 le j le nτ
gt exp(minusnν)
57 59
Small value estimates for the multiplicative group
D Roy Small value estimates for the multiplicative groupActa Arith to appear
Let ξ1 ξm be multiplicatively independent complexnumbers in a field of transcendence degree 1 Under suitableassumptions on the parameters β σ τ ν for infinitely manypositive integers n there exists no nonndashzero polynomialP isin Z[T ] satisfying deg(P ) le n H(P ) le exp(nβ) and
max|P [j](ξi11 middot middot middot ξimm )| 0 le i1 im le nσ 0 le j le nτ
gt exp(minusnν)
58 59
Heidelberg Colloquium organized by the Institute for Mathematics
httpwwwmathiuni-heidelbergdevortraghtmls=1 January 22 2009
Transcendental Number TheorySchanuelrsquos Conjecture
Michel Waldschmidt
Institut de Mathematiques de Jussieu amp Paris VIhttpwwwmathjussieufrsimmiw
59 59
Special cases
In the case m = 1 to say that θ1 is algebraically independentover k just means that it is transcendental over k
Dealing with complex numbers K = C the words algebraictranscendental algebraically dependent algebraicallyindependent refer to the case k = Q
11 59
Examples
The numbersradic
2 and π are algebraically dependent The polynomial X2 minus 2 isin Z[X Y ] vanishes at (
radic2 π)
The numbers π andradicπ2 + 1 are algebraically dependent (and
both are transcendental numbers) The polynomial Y minusX2 minus 1 vanishes at (π
radicπ2 + 1)
The two numbers e and eradic
2 are algebraically independentwhich means that for any nonndashzero polynomial f isin Z[X Y ]
with rational integer coefficients the number f(e eradic
2) is notzeroSpecial case of the Lindemann-Weierstraszlig Theorem
12 59
Transcendence degree
Let Kk be a field extension The maximal number ofelements in K which are algebraically independent over k iscalled the transcendence degree of K over k and denotedtr degkK
Let t = tr degkK A subset θ1 θt of K with t elementswhich are algebraically independent is called a transcendencebasis of K over k It is the same as a maximal subset ofkndashalgebraically independent elements in KHence K is an algebraic extension of k(θ1 θt)
13 59
Transcendence degree of extensions
Assumek sub K sub L
The union of a transcendence basis of K over k and of atranscendence basis of L over K produces a transcendencebasis of L over kHence
tr degkL = tr degkK + tr degKL
An algebraic extension Kk is an extension of transcendencedegree 0 This means that there is no transcendental elementin K over k (any element in K is algebraic over k)
14 59
Algebraic independence of complex numbers
For complex numbers algebraic independence over Q or overQ is the same In particular if θ1 θm are algebraicallyindependent complex numbers then for any non-constantpolynomial f with algebraic coefficients the numberf(θ1 θm) is transcendental
For instance the two numbers e and eradic
2 are algebraicallyindependent As a consequence for any nonndashconstantpolynomial f isin Q[X Y ] with algebraic coefficients the
number f(e eradic
2) is transcendental
15 59
Schanuelrsquos Conjecture
Let x1 xn be Q-linearly independent complex numbersThen at least n of the 2n numbers x1 xn e
x1 exn arealgebraically independent
Since there are 2n numbers only the transcendence degreeover Q of the field
Q(x1 xn e
x1 exn)
is at most 2n The conjecture is that this transcendencedegree is always ge n
n le tr degQQ(x1 xn e
x1 exn)le 2n
16 59
Origin of Schanuelrsquos ConjectureCourse given by Serge Lang (1927ndash2005) at Yale in the 60rsquos
S Lang ndash Introduction to transcendental numbersAddison-Wesley 1966
also attended by M Nagata (1927ndash2008)(14th Problem of Hilbert)
Nagatarsquos Conjecture solved by E Bombieri
17 59
Schanuelrsquos Conjecture
Let x1 xn be Q-linearly independent complex numbersThen
tr degQQ(x1 xn e
x1 exn)ge n
Remark For almost all tuples (with respect to the Lebesguemeasure) the transcendence degree is 2n
18 59
AO Gelrsquofond CRAS 1934
19 59
Statement by Gelrsquofond (1934)
Let β1 βn be Q-linearly independent algebraic numbersand let logα1 logαm be Q-linearly independentlogarithms of algebraic numbers Then the numbers
eβ1 eβn logα1 logαm
are algebraically independent over Q
20 59
Further statement by Gelrsquofond
Let β1 βn be algebraic numbers with β1 6= 0 and letα1 αm be algebraic numbers with α1 6= 0 1 α2 6= 0 1αi 6= 0 Then the numbers
eβ1eβ2e βnminus1e
βn
and αα αm
21
are transcendental and there is no nontrivial algebraic relationbetween such numbers
Remark The condition on α2 should be that it is irrational
21 59
Easy consequence of Schanuelrsquos Conjecture
According to Schanuelrsquos Conjecture the following numbers arealgebraically independent
e+ π eπ πe ee ee2
eee
ππ ππ2
πππ
log π log(log 2) π log 2 (log 2)(log 3) 2log 2 (log 2)log 3
Proof This is an easy exercise
22 59
Langrsquos exercise
Define E0 = Q Inductivelyfor n ge 1 define En as thealgebraic closure of the fieldgenerated over Enminus1 by thenumbers exp(x) = ex wherex ranges over Enminus1 Let E bethe union of En n ge 0Then Schanuelrsquos Conjectureimplies that the number πdoes not belong to E
More precisely Schanuelrsquos Conjecture implies that thenumbers π log π log log π log log log π are algebraicallyindependent over E
23 59
A variant of Langrsquos exercise
Define L0 = Q Inductively for n ge 1 define Ln as thealgebraic closure of the field generated over Lnminus1 by thenumbers y where y ranges over the set of complex numberssuch that ey isin Lnminus1 Let L be the union of Ln n ge 0Then Schanuelrsquos Conjecture implies that the number e doesnot belong to L
More precisely Schanuelrsquos Conjecture implies that the
numbers e ee eee ee
ee
are algebraically independent over L
24 59
Arizona Winter School AWS2008 Tucson
Theorem [Jonathan Bober Chuangxun Cheng Brian DietelMathilde Herblot Jingjing Huang Holly Krieger DiegoMarques Jonathan Mason Martin Mereb and Robert Wilson]Schanuelrsquos Conjecture implies that the fields E and L arelinearly disjoint over Q
Definition Given a field extension FK and twosubextensions F1 F2 sube F we say F1 F2 are linearly disjointover K when the following holds Any set x1 xn sube F1
of Kndash linearly independent elements is linearly independentover F2
Reference arXiv08043550 [mathNT] 2008
25 59
Formal analogs
WD Brownawell(was a student of Schanuel)
J Axrsquos Theorem (1968) Version of SchanuelrsquosConjecture for power seriesover C(and R Coleman for powerseries over Q)
Work by WD Brownawelland K Kubota on the ellipticanalog of Axrsquos Theorem
26 59
Conjectures by A Grothendieck and Y Andre
Generalized Conjecture onPeriods by Grothendieck Dimension of theMumfordndashTate group of asmooth projective variety
Generalization by Y Andre tomotives
Case of 1ndashmotives Elliptico-Toric Conjecture ofC Bertolin
27 59
Ubiquity of Schanuelrsquos Conjecture
Other contexts pndashadic numbers Leopoldtrsquos Conjecture onthe pndashadic rank of the units of an algebraic number fieldNon-vanishing of RegulatorsNonndashdegenerescence of heightsConjecture of B Mazur on rational pointsDiophantine approximation on tori
Dipendra Prasad Gopal Prasad
28 59
Methods from logicEhud Hrushovski Boris Zilber Jonathan Kirby
Calculus of rdquopredimension functionsrdquo (E Hrushovski)
Zilberrsquos construction of a rdquopseudoexponentiationrdquo
Also A Macintyre DE Marker G Terzo AJ WilkieD Bertrand
29 59
Methods from logic Model theory
Exponential algebraicity in exponential fieldsby
Jonathan Kirby
The dimension of the exponential algebraic closure operator inan exponential field satisfies a weak Schanuel property
A corollary is that there are at most countably many essentialcounterexamples to Schanuelrsquos conjecture
arXiv 08104285v2
30 59
Known
LindemannndashWeierstraszlig Theorem = case where x1 xn arealgebraic
Let β1 βn be algebraic numbers which are linearlyindependent over Q Then the numbers eβ1 eβn arealgebraically independent over Q
31 59
Problem of Gelrsquofond and SchneiderRaised by AO Gelrsquofond in 1948 and Th Schneider in 1952
Conjecture If α is an algebraic number α 6= 0 α 6= 1 and ifβ is an irrational algebraic number of degree d then the dminus 1numbers
αβ αβ2
αβdminus1
are algebraically independent
Special case of Schanuelrsquos Conjecture Take xi = βiminus1 logαn = d so that
x1 xn e
x1 exn
islogα β logα βdminus1 logα α αβ αβ
dminus1
The conclusion of Schanuelrsquos Conjecture is
tr degQQ(logα αβ αβ
2
αβdminus1)
= d
32 59
Algebraic independence method of Gelrsquofond
AO Gelrsquofond (1948)
The two numbers 23radic2 and
23radic4 are algebraically
independentMore generally if α is analgebraic number α 6= 0α 6= 1 and if β is a algebraicnumber of degree d ge 3 thentwo at least of the numbers
αβ αβ2
αβdminus1
are algebraically independent
33 59
Tools
Transcendence criterion Replaces Liouvillersquos inequality intranscendence proofsLiouville A nonndashzero rational integer n isin Z satisfies |n| ge 1Gelrsquofond Needs to give a lower bound for |P (θ)| withP isin Z[X] 0 when θ is transcendental
Zero estimate for exponential polynomials C Hermite P Turan K Mahler R Tijdeman
Small transcendence degree AO Gelrsquofond AA Smelev R Tijdeman WD Brownawell
34 59
Large transcendence degree
GV Chudnovsky (1976)Among the numbers
αβ αβ2
αβdminus1
at least [log2 d] arealgebraically independent
GV Chudnovsky ndash On the path to Schanuelrsquosconjecture Algebraic curves close to a pointI General theory of colored sequencesII Fields of finite transcendence type and coloredsequences ResultantsStudia Sci Math Hungar 12 (1977) 125ndash157 (1980)
35 59
Partial result on the problem of Gelrsquofond and
Schneider
AO Gelrsquofond GV Chudnovskii P Philippon YuV Nesterenko
G Diaz If α is an algebraic number α 6= 0 α 6= 1 and if β isan irrational algebraic number of degree d then
tr degQQ(αβ αβ
2
αβdminus1) ge [d+ 1
2
]
36 59
Conjecture of algebraic independence of logarithms
of algebraic numbers
Denote by L the set of complex numbers λ for which eλ isalgebraic
L = logα α isin Qtimes
Hence L is a Q-vector subspace of CThe most important special case of Schanuelrsquos Conjecture is
Conjecture Let λ1 λn be Q-linearly independent elementsin L Then the numbers λ1 λn are algebraicallyindependent over Q
Not yet known that the transcendence degree is ge 2 Open problem Among all logarithms of algebraic numbersone at least is transcendental over Q(π)
37 59
Structural rank of a matrix
Let K be a field k a subfield and M a matrix with entries in KConsider the k-vector subspace E of K spanned by the entriesof M Choose an injective morphism ϕ of E into a k-vectorspace kX1 + middot middot middot+ kXn The image ϕ(M) of M is a matrixwhose entries are in the field k(X1 Xn) of rationalfractions Its rank does not depend on the choice of ϕ
This is the structural rank of M with respect to k
38 59
Example
LetM =
(bij + λij
)1leiled1lejle`
be a matrix with coefficients in Q + L Consider a basis of theQndashvector spanned by the entries and replace the elements inthis basis by unknowns This gives a new matrix M withcoefficients in a field of rational fractions the rank of which isthe structural rank of M (with respect to Q)
As a consequence of the conjecture of algebraic independenceof logarithms of algebraic numbers the rank of M should bethe same as the rank of its specialization M
39 59
Equivalence between the two conjectures
Following D Roy theconjecture on algebraicindependence of logarithms ofalgebraic numbers isequivalent to
Conjecture Any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L has a rank equal to its structuralrank
40 59
Any Polynomial is the Determinant of a Matrix
The proof of the equivalence uses the nice auxiliary result
For any P isin k[X1 Xn] there exists a square matrix withentries in the k-vector space k + kX1 + middot middot middot+ kXn whosedeterminant is P
41 59
Half of the Conjecture is solved
From a certain point of view half of the conjecture of algebraicindependence of logarithms of algebraic numbers is solved
Theorem [D Roy] The rank of any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L is at least half its structural rank
42 59
Reformulation by D Roy
Instead of taking logarithmsof algebraic numbers andlooking for the algebraicindependence relationsD Roy fixes a polynomial andlooks at the points withcoordinates logarithms ofalgebraic numbers on thecorresponding hypersurface
43 59
Reformulation by D Roy
Royrsquos reformulation of the conjecture of algebraicindependence of logarithms is
Conjecture For any algebraic subvariety V of Cn definedover the field Q of algebraic numbers the set V cap Ln is theunion of the sets E cap Ln where E ranges over the set ofvector subspaces of Cn which are defined over Q andcontained in V
Trivial Any element in E cap Ln where E is a vector subspaceof Cn defined over Q and contained in V belongs to V cap Ln
44 59
Example The Four Exponentials ConjectureTake for V the hypersurface of C4 defined by the equation
z1z4 = z2z3
The maximal Cndashvector subspaces of C4 defined over Q andcontained in V are the planes
az1 = bz2 bz4 = az3
and the planesaz1 = bz3 bz4 = az2
with (a b) isin Q2 (0 0)Hence Schanuelrsquos Conjecture implies that if λ1 λ2 λ3 λ4 arelogarithms of algebraic numbers satisfying
λ1λ4 = λ2λ3
then either λ1 λ2 are linearly dependent over Q or else λ1 λ3
are linearly dependent over Q45 59
Six Exponentials Theorem
and Four Exponentials Conjecture
A Selberg CL Siegel S Lang K Ramachandra
46 59
The Four Exponentials Conjecture
Conjecture (Th Schneider S Lang K Ramachandra) Ifx1 x2 are Qndashlinearly independent complex numbers and y1 y2
are Qndashlinearly independent complex numbers then one atleast of the four numbers
ex1y1 ex1y2 ex2y1 ex2y2
is transcendentalEquivalent statement Any 2times 2 matrix(
λ1 λ2
λ3 λ4
)with entries in L and with Qndashlinearly independent rows andQndashlinearly independent columns has maximal rank 2
47 59
How could we attack Schanuelrsquos Conjecture
Let x1 xn be Qndashlinearly independent complex numbersFollowing the transcendence methods of Hermite GelrsquofondSchneider one may start by introducing an auxiliaryfunction
F (z) = P (z ez)
where P isin Z[X0 X1] is a nonndashzero polynomial One considersthe derivatives (
d
dz
)kF = (DkP )(z ez)
at the pointsm1x1 + middot middot middot+mnxn
for various values of (m1 mn) isin Zn
48 59
Auxiliary functionLet D denote the derivation
D =part
partX0
+X1part
partX1
over the ring C[X0 X1] Recall that x1 xn are Qndashlinearlyindependent complex numbers Let α1 αn be nonndashzerocomplex numbersThe transcendence machinery produces a sequence (PN)Nge0
of polynomials with integer coefficients satisfying∣∣∣∣∣(DkPN)(nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1
49 59
Royrsquos approach to Schanuelrsquos Conjecture (1999)
If the number of equations we produce is too small such a setof relations does not contain any information The existenceof a sequence of nonndashtrivial polynomials (PN)Nge0 follows fromlinear algebra
On the other hand following D Roy one may expect that theexistence of such a sequence (PN)Nge0 producing sufficientlymany such equations will yield the desired conclusion
tr degQQ(x1 xn α1 αn
)ge n
50 59
Royrsquos Conjecture (1999)Let s0 s1 t0 t1 u positive real numbers satisfying
max1 t0 2t1 lt mins0 2s1and
maxs0 s1 + t1 lt u lt1
2(1 + t0 + t1)
Assume that for any sufficiently large positive integer N there exists a nonndashzero polynomial PN isin Z[X0 X1] withpartial degree le N t0 in X0 partial degree le N t1 in X1 andheight le eN which satisfies∣∣∣∣∣(DkPN)(
nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1 Then
tr degQQ(x1 xn α1 αn) ge n51 59
Royrsquos Theorem (1999)
Royrsquos conjecture is equivalent to Schanuelrsquos Conjecture
52 59
Equivalence between Schanuel and Roy
Let (x α) isin CtimesCtimes and let s0 s1 t0 t1 u be positive realnumbers satisfying the inequalities of Royrsquos Conjecture Thenthe following conditions are equivalent (a) The number αeminusx is a root of unity
(b) For any sufficiently large positive integer N there exists anonndashzero polynomial QN isin Z[X0 X1] with partial degreele N t0 in X0 partial degree le N t1 in X1 and heightH(QN) le eN such that∣∣(DkQN)(mx αm)
∣∣ le exp(minusNu)
for any km isin N with k le N s0 and m le N s1
53 59
Gelrsquofondrsquos transcendence criterion
Simple form Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le n and height le en such that
|Pn(ϑ)| le eminus7n2
for all n ge 1 then ϑ is algebraic and Pn(ϑ) = 0 for all n ge 1
First extension Replace the bound for the degree by le dnthe bound for the height by ebn and the bound for |Pn(ϑ)| byeminuscdnbn with some constant c gt 0 independent of n
Some mild conditions are required on the sequences (dn)nge1 and(bn)nge1
54 59
Transcendence criterion with multiplicitiesWith derivatives Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le dn and height le ebn such that∣∣P (j)
n (ϑ)∣∣ le eminuscdnbntn
for all j in the range 0 le j lt tn and all n ge 1 then ϑ isalgebraicDue to M Laurent and D Roy applications to algebraicindependence
55 59
Criterion with several points
Goal Given a sequence of complex numbers (ϑi)ige1 if thereexists a sequence (Pn)nge1 of nonndashzero polynomials in Z[X]with Pn of degree le dn and height le ebn such that∣∣P (j)
n (ϑi)∣∣ le eminuscdnbntnsn
for 0 le j lt tn 1 le i le sn and all n ge 1 then the numbers ϑiare algebraic
D Roy Not true in general but true in some special caseswith a structure on the sequence (ϑi)ige1Combines the elimination arguments used for criteria ofalgebraic independence and for zero estimates
56 59
Small value estimates for the additive group
D Roy Small value estimates for the additive group Intern JNumber Theory to appear
Let ξ be a transcendental complex number let β σ τ and νbe non-negative real numbers let n0 be a positive integer andlet (Pn)ngen0 be a sequence of nonndashzero polynomials in Z[T ]satisfying deg(Pn) le n and H(Pn) le exp(nβ) for eachn ge n0 Suppose that β gt 1 (34)σ + τ lt 1 andν gt 1 + β minus (34)σ minus τ Then for infinitely many n we have
max|P [j]n (iξ)| 0 le i le nσ 0 le j le nτ
gt exp(minusnν)
57 59
Small value estimates for the multiplicative group
D Roy Small value estimates for the multiplicative groupActa Arith to appear
Let ξ1 ξm be multiplicatively independent complexnumbers in a field of transcendence degree 1 Under suitableassumptions on the parameters β σ τ ν for infinitely manypositive integers n there exists no nonndashzero polynomialP isin Z[T ] satisfying deg(P ) le n H(P ) le exp(nβ) and
max|P [j](ξi11 middot middot middot ξimm )| 0 le i1 im le nσ 0 le j le nτ
gt exp(minusnν)
58 59
Heidelberg Colloquium organized by the Institute for Mathematics
httpwwwmathiuni-heidelbergdevortraghtmls=1 January 22 2009
Transcendental Number TheorySchanuelrsquos Conjecture
Michel Waldschmidt
Institut de Mathematiques de Jussieu amp Paris VIhttpwwwmathjussieufrsimmiw
59 59
Examples
The numbersradic
2 and π are algebraically dependent The polynomial X2 minus 2 isin Z[X Y ] vanishes at (
radic2 π)
The numbers π andradicπ2 + 1 are algebraically dependent (and
both are transcendental numbers) The polynomial Y minusX2 minus 1 vanishes at (π
radicπ2 + 1)
The two numbers e and eradic
2 are algebraically independentwhich means that for any nonndashzero polynomial f isin Z[X Y ]
with rational integer coefficients the number f(e eradic
2) is notzeroSpecial case of the Lindemann-Weierstraszlig Theorem
12 59
Transcendence degree
Let Kk be a field extension The maximal number ofelements in K which are algebraically independent over k iscalled the transcendence degree of K over k and denotedtr degkK
Let t = tr degkK A subset θ1 θt of K with t elementswhich are algebraically independent is called a transcendencebasis of K over k It is the same as a maximal subset ofkndashalgebraically independent elements in KHence K is an algebraic extension of k(θ1 θt)
13 59
Transcendence degree of extensions
Assumek sub K sub L
The union of a transcendence basis of K over k and of atranscendence basis of L over K produces a transcendencebasis of L over kHence
tr degkL = tr degkK + tr degKL
An algebraic extension Kk is an extension of transcendencedegree 0 This means that there is no transcendental elementin K over k (any element in K is algebraic over k)
14 59
Algebraic independence of complex numbers
For complex numbers algebraic independence over Q or overQ is the same In particular if θ1 θm are algebraicallyindependent complex numbers then for any non-constantpolynomial f with algebraic coefficients the numberf(θ1 θm) is transcendental
For instance the two numbers e and eradic
2 are algebraicallyindependent As a consequence for any nonndashconstantpolynomial f isin Q[X Y ] with algebraic coefficients the
number f(e eradic
2) is transcendental
15 59
Schanuelrsquos Conjecture
Let x1 xn be Q-linearly independent complex numbersThen at least n of the 2n numbers x1 xn e
x1 exn arealgebraically independent
Since there are 2n numbers only the transcendence degreeover Q of the field
Q(x1 xn e
x1 exn)
is at most 2n The conjecture is that this transcendencedegree is always ge n
n le tr degQQ(x1 xn e
x1 exn)le 2n
16 59
Origin of Schanuelrsquos ConjectureCourse given by Serge Lang (1927ndash2005) at Yale in the 60rsquos
S Lang ndash Introduction to transcendental numbersAddison-Wesley 1966
also attended by M Nagata (1927ndash2008)(14th Problem of Hilbert)
Nagatarsquos Conjecture solved by E Bombieri
17 59
Schanuelrsquos Conjecture
Let x1 xn be Q-linearly independent complex numbersThen
tr degQQ(x1 xn e
x1 exn)ge n
Remark For almost all tuples (with respect to the Lebesguemeasure) the transcendence degree is 2n
18 59
AO Gelrsquofond CRAS 1934
19 59
Statement by Gelrsquofond (1934)
Let β1 βn be Q-linearly independent algebraic numbersand let logα1 logαm be Q-linearly independentlogarithms of algebraic numbers Then the numbers
eβ1 eβn logα1 logαm
are algebraically independent over Q
20 59
Further statement by Gelrsquofond
Let β1 βn be algebraic numbers with β1 6= 0 and letα1 αm be algebraic numbers with α1 6= 0 1 α2 6= 0 1αi 6= 0 Then the numbers
eβ1eβ2e βnminus1e
βn
and αα αm
21
are transcendental and there is no nontrivial algebraic relationbetween such numbers
Remark The condition on α2 should be that it is irrational
21 59
Easy consequence of Schanuelrsquos Conjecture
According to Schanuelrsquos Conjecture the following numbers arealgebraically independent
e+ π eπ πe ee ee2
eee
ππ ππ2
πππ
log π log(log 2) π log 2 (log 2)(log 3) 2log 2 (log 2)log 3
Proof This is an easy exercise
22 59
Langrsquos exercise
Define E0 = Q Inductivelyfor n ge 1 define En as thealgebraic closure of the fieldgenerated over Enminus1 by thenumbers exp(x) = ex wherex ranges over Enminus1 Let E bethe union of En n ge 0Then Schanuelrsquos Conjectureimplies that the number πdoes not belong to E
More precisely Schanuelrsquos Conjecture implies that thenumbers π log π log log π log log log π are algebraicallyindependent over E
23 59
A variant of Langrsquos exercise
Define L0 = Q Inductively for n ge 1 define Ln as thealgebraic closure of the field generated over Lnminus1 by thenumbers y where y ranges over the set of complex numberssuch that ey isin Lnminus1 Let L be the union of Ln n ge 0Then Schanuelrsquos Conjecture implies that the number e doesnot belong to L
More precisely Schanuelrsquos Conjecture implies that the
numbers e ee eee ee
ee
are algebraically independent over L
24 59
Arizona Winter School AWS2008 Tucson
Theorem [Jonathan Bober Chuangxun Cheng Brian DietelMathilde Herblot Jingjing Huang Holly Krieger DiegoMarques Jonathan Mason Martin Mereb and Robert Wilson]Schanuelrsquos Conjecture implies that the fields E and L arelinearly disjoint over Q
Definition Given a field extension FK and twosubextensions F1 F2 sube F we say F1 F2 are linearly disjointover K when the following holds Any set x1 xn sube F1
of Kndash linearly independent elements is linearly independentover F2
Reference arXiv08043550 [mathNT] 2008
25 59
Formal analogs
WD Brownawell(was a student of Schanuel)
J Axrsquos Theorem (1968) Version of SchanuelrsquosConjecture for power seriesover C(and R Coleman for powerseries over Q)
Work by WD Brownawelland K Kubota on the ellipticanalog of Axrsquos Theorem
26 59
Conjectures by A Grothendieck and Y Andre
Generalized Conjecture onPeriods by Grothendieck Dimension of theMumfordndashTate group of asmooth projective variety
Generalization by Y Andre tomotives
Case of 1ndashmotives Elliptico-Toric Conjecture ofC Bertolin
27 59
Ubiquity of Schanuelrsquos Conjecture
Other contexts pndashadic numbers Leopoldtrsquos Conjecture onthe pndashadic rank of the units of an algebraic number fieldNon-vanishing of RegulatorsNonndashdegenerescence of heightsConjecture of B Mazur on rational pointsDiophantine approximation on tori
Dipendra Prasad Gopal Prasad
28 59
Methods from logicEhud Hrushovski Boris Zilber Jonathan Kirby
Calculus of rdquopredimension functionsrdquo (E Hrushovski)
Zilberrsquos construction of a rdquopseudoexponentiationrdquo
Also A Macintyre DE Marker G Terzo AJ WilkieD Bertrand
29 59
Methods from logic Model theory
Exponential algebraicity in exponential fieldsby
Jonathan Kirby
The dimension of the exponential algebraic closure operator inan exponential field satisfies a weak Schanuel property
A corollary is that there are at most countably many essentialcounterexamples to Schanuelrsquos conjecture
arXiv 08104285v2
30 59
Known
LindemannndashWeierstraszlig Theorem = case where x1 xn arealgebraic
Let β1 βn be algebraic numbers which are linearlyindependent over Q Then the numbers eβ1 eβn arealgebraically independent over Q
31 59
Problem of Gelrsquofond and SchneiderRaised by AO Gelrsquofond in 1948 and Th Schneider in 1952
Conjecture If α is an algebraic number α 6= 0 α 6= 1 and ifβ is an irrational algebraic number of degree d then the dminus 1numbers
αβ αβ2
αβdminus1
are algebraically independent
Special case of Schanuelrsquos Conjecture Take xi = βiminus1 logαn = d so that
x1 xn e
x1 exn
islogα β logα βdminus1 logα α αβ αβ
dminus1
The conclusion of Schanuelrsquos Conjecture is
tr degQQ(logα αβ αβ
2
αβdminus1)
= d
32 59
Algebraic independence method of Gelrsquofond
AO Gelrsquofond (1948)
The two numbers 23radic2 and
23radic4 are algebraically
independentMore generally if α is analgebraic number α 6= 0α 6= 1 and if β is a algebraicnumber of degree d ge 3 thentwo at least of the numbers
αβ αβ2
αβdminus1
are algebraically independent
33 59
Tools
Transcendence criterion Replaces Liouvillersquos inequality intranscendence proofsLiouville A nonndashzero rational integer n isin Z satisfies |n| ge 1Gelrsquofond Needs to give a lower bound for |P (θ)| withP isin Z[X] 0 when θ is transcendental
Zero estimate for exponential polynomials C Hermite P Turan K Mahler R Tijdeman
Small transcendence degree AO Gelrsquofond AA Smelev R Tijdeman WD Brownawell
34 59
Large transcendence degree
GV Chudnovsky (1976)Among the numbers
αβ αβ2
αβdminus1
at least [log2 d] arealgebraically independent
GV Chudnovsky ndash On the path to Schanuelrsquosconjecture Algebraic curves close to a pointI General theory of colored sequencesII Fields of finite transcendence type and coloredsequences ResultantsStudia Sci Math Hungar 12 (1977) 125ndash157 (1980)
35 59
Partial result on the problem of Gelrsquofond and
Schneider
AO Gelrsquofond GV Chudnovskii P Philippon YuV Nesterenko
G Diaz If α is an algebraic number α 6= 0 α 6= 1 and if β isan irrational algebraic number of degree d then
tr degQQ(αβ αβ
2
αβdminus1) ge [d+ 1
2
]
36 59
Conjecture of algebraic independence of logarithms
of algebraic numbers
Denote by L the set of complex numbers λ for which eλ isalgebraic
L = logα α isin Qtimes
Hence L is a Q-vector subspace of CThe most important special case of Schanuelrsquos Conjecture is
Conjecture Let λ1 λn be Q-linearly independent elementsin L Then the numbers λ1 λn are algebraicallyindependent over Q
Not yet known that the transcendence degree is ge 2 Open problem Among all logarithms of algebraic numbersone at least is transcendental over Q(π)
37 59
Structural rank of a matrix
Let K be a field k a subfield and M a matrix with entries in KConsider the k-vector subspace E of K spanned by the entriesof M Choose an injective morphism ϕ of E into a k-vectorspace kX1 + middot middot middot+ kXn The image ϕ(M) of M is a matrixwhose entries are in the field k(X1 Xn) of rationalfractions Its rank does not depend on the choice of ϕ
This is the structural rank of M with respect to k
38 59
Example
LetM =
(bij + λij
)1leiled1lejle`
be a matrix with coefficients in Q + L Consider a basis of theQndashvector spanned by the entries and replace the elements inthis basis by unknowns This gives a new matrix M withcoefficients in a field of rational fractions the rank of which isthe structural rank of M (with respect to Q)
As a consequence of the conjecture of algebraic independenceof logarithms of algebraic numbers the rank of M should bethe same as the rank of its specialization M
39 59
Equivalence between the two conjectures
Following D Roy theconjecture on algebraicindependence of logarithms ofalgebraic numbers isequivalent to
Conjecture Any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L has a rank equal to its structuralrank
40 59
Any Polynomial is the Determinant of a Matrix
The proof of the equivalence uses the nice auxiliary result
For any P isin k[X1 Xn] there exists a square matrix withentries in the k-vector space k + kX1 + middot middot middot+ kXn whosedeterminant is P
41 59
Half of the Conjecture is solved
From a certain point of view half of the conjecture of algebraicindependence of logarithms of algebraic numbers is solved
Theorem [D Roy] The rank of any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L is at least half its structural rank
42 59
Reformulation by D Roy
Instead of taking logarithmsof algebraic numbers andlooking for the algebraicindependence relationsD Roy fixes a polynomial andlooks at the points withcoordinates logarithms ofalgebraic numbers on thecorresponding hypersurface
43 59
Reformulation by D Roy
Royrsquos reformulation of the conjecture of algebraicindependence of logarithms is
Conjecture For any algebraic subvariety V of Cn definedover the field Q of algebraic numbers the set V cap Ln is theunion of the sets E cap Ln where E ranges over the set ofvector subspaces of Cn which are defined over Q andcontained in V
Trivial Any element in E cap Ln where E is a vector subspaceof Cn defined over Q and contained in V belongs to V cap Ln
44 59
Example The Four Exponentials ConjectureTake for V the hypersurface of C4 defined by the equation
z1z4 = z2z3
The maximal Cndashvector subspaces of C4 defined over Q andcontained in V are the planes
az1 = bz2 bz4 = az3
and the planesaz1 = bz3 bz4 = az2
with (a b) isin Q2 (0 0)Hence Schanuelrsquos Conjecture implies that if λ1 λ2 λ3 λ4 arelogarithms of algebraic numbers satisfying
λ1λ4 = λ2λ3
then either λ1 λ2 are linearly dependent over Q or else λ1 λ3
are linearly dependent over Q45 59
Six Exponentials Theorem
and Four Exponentials Conjecture
A Selberg CL Siegel S Lang K Ramachandra
46 59
The Four Exponentials Conjecture
Conjecture (Th Schneider S Lang K Ramachandra) Ifx1 x2 are Qndashlinearly independent complex numbers and y1 y2
are Qndashlinearly independent complex numbers then one atleast of the four numbers
ex1y1 ex1y2 ex2y1 ex2y2
is transcendentalEquivalent statement Any 2times 2 matrix(
λ1 λ2
λ3 λ4
)with entries in L and with Qndashlinearly independent rows andQndashlinearly independent columns has maximal rank 2
47 59
How could we attack Schanuelrsquos Conjecture
Let x1 xn be Qndashlinearly independent complex numbersFollowing the transcendence methods of Hermite GelrsquofondSchneider one may start by introducing an auxiliaryfunction
F (z) = P (z ez)
where P isin Z[X0 X1] is a nonndashzero polynomial One considersthe derivatives (
d
dz
)kF = (DkP )(z ez)
at the pointsm1x1 + middot middot middot+mnxn
for various values of (m1 mn) isin Zn
48 59
Auxiliary functionLet D denote the derivation
D =part
partX0
+X1part
partX1
over the ring C[X0 X1] Recall that x1 xn are Qndashlinearlyindependent complex numbers Let α1 αn be nonndashzerocomplex numbersThe transcendence machinery produces a sequence (PN)Nge0
of polynomials with integer coefficients satisfying∣∣∣∣∣(DkPN)(nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1
49 59
Royrsquos approach to Schanuelrsquos Conjecture (1999)
If the number of equations we produce is too small such a setof relations does not contain any information The existenceof a sequence of nonndashtrivial polynomials (PN)Nge0 follows fromlinear algebra
On the other hand following D Roy one may expect that theexistence of such a sequence (PN)Nge0 producing sufficientlymany such equations will yield the desired conclusion
tr degQQ(x1 xn α1 αn
)ge n
50 59
Royrsquos Conjecture (1999)Let s0 s1 t0 t1 u positive real numbers satisfying
max1 t0 2t1 lt mins0 2s1and
maxs0 s1 + t1 lt u lt1
2(1 + t0 + t1)
Assume that for any sufficiently large positive integer N there exists a nonndashzero polynomial PN isin Z[X0 X1] withpartial degree le N t0 in X0 partial degree le N t1 in X1 andheight le eN which satisfies∣∣∣∣∣(DkPN)(
nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1 Then
tr degQQ(x1 xn α1 αn) ge n51 59
Royrsquos Theorem (1999)
Royrsquos conjecture is equivalent to Schanuelrsquos Conjecture
52 59
Equivalence between Schanuel and Roy
Let (x α) isin CtimesCtimes and let s0 s1 t0 t1 u be positive realnumbers satisfying the inequalities of Royrsquos Conjecture Thenthe following conditions are equivalent (a) The number αeminusx is a root of unity
(b) For any sufficiently large positive integer N there exists anonndashzero polynomial QN isin Z[X0 X1] with partial degreele N t0 in X0 partial degree le N t1 in X1 and heightH(QN) le eN such that∣∣(DkQN)(mx αm)
∣∣ le exp(minusNu)
for any km isin N with k le N s0 and m le N s1
53 59
Gelrsquofondrsquos transcendence criterion
Simple form Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le n and height le en such that
|Pn(ϑ)| le eminus7n2
for all n ge 1 then ϑ is algebraic and Pn(ϑ) = 0 for all n ge 1
First extension Replace the bound for the degree by le dnthe bound for the height by ebn and the bound for |Pn(ϑ)| byeminuscdnbn with some constant c gt 0 independent of n
Some mild conditions are required on the sequences (dn)nge1 and(bn)nge1
54 59
Transcendence criterion with multiplicitiesWith derivatives Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le dn and height le ebn such that∣∣P (j)
n (ϑ)∣∣ le eminuscdnbntn
for all j in the range 0 le j lt tn and all n ge 1 then ϑ isalgebraicDue to M Laurent and D Roy applications to algebraicindependence
55 59
Criterion with several points
Goal Given a sequence of complex numbers (ϑi)ige1 if thereexists a sequence (Pn)nge1 of nonndashzero polynomials in Z[X]with Pn of degree le dn and height le ebn such that∣∣P (j)
n (ϑi)∣∣ le eminuscdnbntnsn
for 0 le j lt tn 1 le i le sn and all n ge 1 then the numbers ϑiare algebraic
D Roy Not true in general but true in some special caseswith a structure on the sequence (ϑi)ige1Combines the elimination arguments used for criteria ofalgebraic independence and for zero estimates
56 59
Small value estimates for the additive group
D Roy Small value estimates for the additive group Intern JNumber Theory to appear
Let ξ be a transcendental complex number let β σ τ and νbe non-negative real numbers let n0 be a positive integer andlet (Pn)ngen0 be a sequence of nonndashzero polynomials in Z[T ]satisfying deg(Pn) le n and H(Pn) le exp(nβ) for eachn ge n0 Suppose that β gt 1 (34)σ + τ lt 1 andν gt 1 + β minus (34)σ minus τ Then for infinitely many n we have
max|P [j]n (iξ)| 0 le i le nσ 0 le j le nτ
gt exp(minusnν)
57 59
Small value estimates for the multiplicative group
D Roy Small value estimates for the multiplicative groupActa Arith to appear
Let ξ1 ξm be multiplicatively independent complexnumbers in a field of transcendence degree 1 Under suitableassumptions on the parameters β σ τ ν for infinitely manypositive integers n there exists no nonndashzero polynomialP isin Z[T ] satisfying deg(P ) le n H(P ) le exp(nβ) and
max|P [j](ξi11 middot middot middot ξimm )| 0 le i1 im le nσ 0 le j le nτ
gt exp(minusnν)
58 59
Heidelberg Colloquium organized by the Institute for Mathematics
httpwwwmathiuni-heidelbergdevortraghtmls=1 January 22 2009
Transcendental Number TheorySchanuelrsquos Conjecture
Michel Waldschmidt
Institut de Mathematiques de Jussieu amp Paris VIhttpwwwmathjussieufrsimmiw
59 59
Transcendence degree
Let Kk be a field extension The maximal number ofelements in K which are algebraically independent over k iscalled the transcendence degree of K over k and denotedtr degkK
Let t = tr degkK A subset θ1 θt of K with t elementswhich are algebraically independent is called a transcendencebasis of K over k It is the same as a maximal subset ofkndashalgebraically independent elements in KHence K is an algebraic extension of k(θ1 θt)
13 59
Transcendence degree of extensions
Assumek sub K sub L
The union of a transcendence basis of K over k and of atranscendence basis of L over K produces a transcendencebasis of L over kHence
tr degkL = tr degkK + tr degKL
An algebraic extension Kk is an extension of transcendencedegree 0 This means that there is no transcendental elementin K over k (any element in K is algebraic over k)
14 59
Algebraic independence of complex numbers
For complex numbers algebraic independence over Q or overQ is the same In particular if θ1 θm are algebraicallyindependent complex numbers then for any non-constantpolynomial f with algebraic coefficients the numberf(θ1 θm) is transcendental
For instance the two numbers e and eradic
2 are algebraicallyindependent As a consequence for any nonndashconstantpolynomial f isin Q[X Y ] with algebraic coefficients the
number f(e eradic
2) is transcendental
15 59
Schanuelrsquos Conjecture
Let x1 xn be Q-linearly independent complex numbersThen at least n of the 2n numbers x1 xn e
x1 exn arealgebraically independent
Since there are 2n numbers only the transcendence degreeover Q of the field
Q(x1 xn e
x1 exn)
is at most 2n The conjecture is that this transcendencedegree is always ge n
n le tr degQQ(x1 xn e
x1 exn)le 2n
16 59
Origin of Schanuelrsquos ConjectureCourse given by Serge Lang (1927ndash2005) at Yale in the 60rsquos
S Lang ndash Introduction to transcendental numbersAddison-Wesley 1966
also attended by M Nagata (1927ndash2008)(14th Problem of Hilbert)
Nagatarsquos Conjecture solved by E Bombieri
17 59
Schanuelrsquos Conjecture
Let x1 xn be Q-linearly independent complex numbersThen
tr degQQ(x1 xn e
x1 exn)ge n
Remark For almost all tuples (with respect to the Lebesguemeasure) the transcendence degree is 2n
18 59
AO Gelrsquofond CRAS 1934
19 59
Statement by Gelrsquofond (1934)
Let β1 βn be Q-linearly independent algebraic numbersand let logα1 logαm be Q-linearly independentlogarithms of algebraic numbers Then the numbers
eβ1 eβn logα1 logαm
are algebraically independent over Q
20 59
Further statement by Gelrsquofond
Let β1 βn be algebraic numbers with β1 6= 0 and letα1 αm be algebraic numbers with α1 6= 0 1 α2 6= 0 1αi 6= 0 Then the numbers
eβ1eβ2e βnminus1e
βn
and αα αm
21
are transcendental and there is no nontrivial algebraic relationbetween such numbers
Remark The condition on α2 should be that it is irrational
21 59
Easy consequence of Schanuelrsquos Conjecture
According to Schanuelrsquos Conjecture the following numbers arealgebraically independent
e+ π eπ πe ee ee2
eee
ππ ππ2
πππ
log π log(log 2) π log 2 (log 2)(log 3) 2log 2 (log 2)log 3
Proof This is an easy exercise
22 59
Langrsquos exercise
Define E0 = Q Inductivelyfor n ge 1 define En as thealgebraic closure of the fieldgenerated over Enminus1 by thenumbers exp(x) = ex wherex ranges over Enminus1 Let E bethe union of En n ge 0Then Schanuelrsquos Conjectureimplies that the number πdoes not belong to E
More precisely Schanuelrsquos Conjecture implies that thenumbers π log π log log π log log log π are algebraicallyindependent over E
23 59
A variant of Langrsquos exercise
Define L0 = Q Inductively for n ge 1 define Ln as thealgebraic closure of the field generated over Lnminus1 by thenumbers y where y ranges over the set of complex numberssuch that ey isin Lnminus1 Let L be the union of Ln n ge 0Then Schanuelrsquos Conjecture implies that the number e doesnot belong to L
More precisely Schanuelrsquos Conjecture implies that the
numbers e ee eee ee
ee
are algebraically independent over L
24 59
Arizona Winter School AWS2008 Tucson
Theorem [Jonathan Bober Chuangxun Cheng Brian DietelMathilde Herblot Jingjing Huang Holly Krieger DiegoMarques Jonathan Mason Martin Mereb and Robert Wilson]Schanuelrsquos Conjecture implies that the fields E and L arelinearly disjoint over Q
Definition Given a field extension FK and twosubextensions F1 F2 sube F we say F1 F2 are linearly disjointover K when the following holds Any set x1 xn sube F1
of Kndash linearly independent elements is linearly independentover F2
Reference arXiv08043550 [mathNT] 2008
25 59
Formal analogs
WD Brownawell(was a student of Schanuel)
J Axrsquos Theorem (1968) Version of SchanuelrsquosConjecture for power seriesover C(and R Coleman for powerseries over Q)
Work by WD Brownawelland K Kubota on the ellipticanalog of Axrsquos Theorem
26 59
Conjectures by A Grothendieck and Y Andre
Generalized Conjecture onPeriods by Grothendieck Dimension of theMumfordndashTate group of asmooth projective variety
Generalization by Y Andre tomotives
Case of 1ndashmotives Elliptico-Toric Conjecture ofC Bertolin
27 59
Ubiquity of Schanuelrsquos Conjecture
Other contexts pndashadic numbers Leopoldtrsquos Conjecture onthe pndashadic rank of the units of an algebraic number fieldNon-vanishing of RegulatorsNonndashdegenerescence of heightsConjecture of B Mazur on rational pointsDiophantine approximation on tori
Dipendra Prasad Gopal Prasad
28 59
Methods from logicEhud Hrushovski Boris Zilber Jonathan Kirby
Calculus of rdquopredimension functionsrdquo (E Hrushovski)
Zilberrsquos construction of a rdquopseudoexponentiationrdquo
Also A Macintyre DE Marker G Terzo AJ WilkieD Bertrand
29 59
Methods from logic Model theory
Exponential algebraicity in exponential fieldsby
Jonathan Kirby
The dimension of the exponential algebraic closure operator inan exponential field satisfies a weak Schanuel property
A corollary is that there are at most countably many essentialcounterexamples to Schanuelrsquos conjecture
arXiv 08104285v2
30 59
Known
LindemannndashWeierstraszlig Theorem = case where x1 xn arealgebraic
Let β1 βn be algebraic numbers which are linearlyindependent over Q Then the numbers eβ1 eβn arealgebraically independent over Q
31 59
Problem of Gelrsquofond and SchneiderRaised by AO Gelrsquofond in 1948 and Th Schneider in 1952
Conjecture If α is an algebraic number α 6= 0 α 6= 1 and ifβ is an irrational algebraic number of degree d then the dminus 1numbers
αβ αβ2
αβdminus1
are algebraically independent
Special case of Schanuelrsquos Conjecture Take xi = βiminus1 logαn = d so that
x1 xn e
x1 exn
islogα β logα βdminus1 logα α αβ αβ
dminus1
The conclusion of Schanuelrsquos Conjecture is
tr degQQ(logα αβ αβ
2
αβdminus1)
= d
32 59
Algebraic independence method of Gelrsquofond
AO Gelrsquofond (1948)
The two numbers 23radic2 and
23radic4 are algebraically
independentMore generally if α is analgebraic number α 6= 0α 6= 1 and if β is a algebraicnumber of degree d ge 3 thentwo at least of the numbers
αβ αβ2
αβdminus1
are algebraically independent
33 59
Tools
Transcendence criterion Replaces Liouvillersquos inequality intranscendence proofsLiouville A nonndashzero rational integer n isin Z satisfies |n| ge 1Gelrsquofond Needs to give a lower bound for |P (θ)| withP isin Z[X] 0 when θ is transcendental
Zero estimate for exponential polynomials C Hermite P Turan K Mahler R Tijdeman
Small transcendence degree AO Gelrsquofond AA Smelev R Tijdeman WD Brownawell
34 59
Large transcendence degree
GV Chudnovsky (1976)Among the numbers
αβ αβ2
αβdminus1
at least [log2 d] arealgebraically independent
GV Chudnovsky ndash On the path to Schanuelrsquosconjecture Algebraic curves close to a pointI General theory of colored sequencesII Fields of finite transcendence type and coloredsequences ResultantsStudia Sci Math Hungar 12 (1977) 125ndash157 (1980)
35 59
Partial result on the problem of Gelrsquofond and
Schneider
AO Gelrsquofond GV Chudnovskii P Philippon YuV Nesterenko
G Diaz If α is an algebraic number α 6= 0 α 6= 1 and if β isan irrational algebraic number of degree d then
tr degQQ(αβ αβ
2
αβdminus1) ge [d+ 1
2
]
36 59
Conjecture of algebraic independence of logarithms
of algebraic numbers
Denote by L the set of complex numbers λ for which eλ isalgebraic
L = logα α isin Qtimes
Hence L is a Q-vector subspace of CThe most important special case of Schanuelrsquos Conjecture is
Conjecture Let λ1 λn be Q-linearly independent elementsin L Then the numbers λ1 λn are algebraicallyindependent over Q
Not yet known that the transcendence degree is ge 2 Open problem Among all logarithms of algebraic numbersone at least is transcendental over Q(π)
37 59
Structural rank of a matrix
Let K be a field k a subfield and M a matrix with entries in KConsider the k-vector subspace E of K spanned by the entriesof M Choose an injective morphism ϕ of E into a k-vectorspace kX1 + middot middot middot+ kXn The image ϕ(M) of M is a matrixwhose entries are in the field k(X1 Xn) of rationalfractions Its rank does not depend on the choice of ϕ
This is the structural rank of M with respect to k
38 59
Example
LetM =
(bij + λij
)1leiled1lejle`
be a matrix with coefficients in Q + L Consider a basis of theQndashvector spanned by the entries and replace the elements inthis basis by unknowns This gives a new matrix M withcoefficients in a field of rational fractions the rank of which isthe structural rank of M (with respect to Q)
As a consequence of the conjecture of algebraic independenceof logarithms of algebraic numbers the rank of M should bethe same as the rank of its specialization M
39 59
Equivalence between the two conjectures
Following D Roy theconjecture on algebraicindependence of logarithms ofalgebraic numbers isequivalent to
Conjecture Any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L has a rank equal to its structuralrank
40 59
Any Polynomial is the Determinant of a Matrix
The proof of the equivalence uses the nice auxiliary result
For any P isin k[X1 Xn] there exists a square matrix withentries in the k-vector space k + kX1 + middot middot middot+ kXn whosedeterminant is P
41 59
Half of the Conjecture is solved
From a certain point of view half of the conjecture of algebraicindependence of logarithms of algebraic numbers is solved
Theorem [D Roy] The rank of any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L is at least half its structural rank
42 59
Reformulation by D Roy
Instead of taking logarithmsof algebraic numbers andlooking for the algebraicindependence relationsD Roy fixes a polynomial andlooks at the points withcoordinates logarithms ofalgebraic numbers on thecorresponding hypersurface
43 59
Reformulation by D Roy
Royrsquos reformulation of the conjecture of algebraicindependence of logarithms is
Conjecture For any algebraic subvariety V of Cn definedover the field Q of algebraic numbers the set V cap Ln is theunion of the sets E cap Ln where E ranges over the set ofvector subspaces of Cn which are defined over Q andcontained in V
Trivial Any element in E cap Ln where E is a vector subspaceof Cn defined over Q and contained in V belongs to V cap Ln
44 59
Example The Four Exponentials ConjectureTake for V the hypersurface of C4 defined by the equation
z1z4 = z2z3
The maximal Cndashvector subspaces of C4 defined over Q andcontained in V are the planes
az1 = bz2 bz4 = az3
and the planesaz1 = bz3 bz4 = az2
with (a b) isin Q2 (0 0)Hence Schanuelrsquos Conjecture implies that if λ1 λ2 λ3 λ4 arelogarithms of algebraic numbers satisfying
λ1λ4 = λ2λ3
then either λ1 λ2 are linearly dependent over Q or else λ1 λ3
are linearly dependent over Q45 59
Six Exponentials Theorem
and Four Exponentials Conjecture
A Selberg CL Siegel S Lang K Ramachandra
46 59
The Four Exponentials Conjecture
Conjecture (Th Schneider S Lang K Ramachandra) Ifx1 x2 are Qndashlinearly independent complex numbers and y1 y2
are Qndashlinearly independent complex numbers then one atleast of the four numbers
ex1y1 ex1y2 ex2y1 ex2y2
is transcendentalEquivalent statement Any 2times 2 matrix(
λ1 λ2
λ3 λ4
)with entries in L and with Qndashlinearly independent rows andQndashlinearly independent columns has maximal rank 2
47 59
How could we attack Schanuelrsquos Conjecture
Let x1 xn be Qndashlinearly independent complex numbersFollowing the transcendence methods of Hermite GelrsquofondSchneider one may start by introducing an auxiliaryfunction
F (z) = P (z ez)
where P isin Z[X0 X1] is a nonndashzero polynomial One considersthe derivatives (
d
dz
)kF = (DkP )(z ez)
at the pointsm1x1 + middot middot middot+mnxn
for various values of (m1 mn) isin Zn
48 59
Auxiliary functionLet D denote the derivation
D =part
partX0
+X1part
partX1
over the ring C[X0 X1] Recall that x1 xn are Qndashlinearlyindependent complex numbers Let α1 αn be nonndashzerocomplex numbersThe transcendence machinery produces a sequence (PN)Nge0
of polynomials with integer coefficients satisfying∣∣∣∣∣(DkPN)(nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1
49 59
Royrsquos approach to Schanuelrsquos Conjecture (1999)
If the number of equations we produce is too small such a setof relations does not contain any information The existenceof a sequence of nonndashtrivial polynomials (PN)Nge0 follows fromlinear algebra
On the other hand following D Roy one may expect that theexistence of such a sequence (PN)Nge0 producing sufficientlymany such equations will yield the desired conclusion
tr degQQ(x1 xn α1 αn
)ge n
50 59
Royrsquos Conjecture (1999)Let s0 s1 t0 t1 u positive real numbers satisfying
max1 t0 2t1 lt mins0 2s1and
maxs0 s1 + t1 lt u lt1
2(1 + t0 + t1)
Assume that for any sufficiently large positive integer N there exists a nonndashzero polynomial PN isin Z[X0 X1] withpartial degree le N t0 in X0 partial degree le N t1 in X1 andheight le eN which satisfies∣∣∣∣∣(DkPN)(
nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1 Then
tr degQQ(x1 xn α1 αn) ge n51 59
Royrsquos Theorem (1999)
Royrsquos conjecture is equivalent to Schanuelrsquos Conjecture
52 59
Equivalence between Schanuel and Roy
Let (x α) isin CtimesCtimes and let s0 s1 t0 t1 u be positive realnumbers satisfying the inequalities of Royrsquos Conjecture Thenthe following conditions are equivalent (a) The number αeminusx is a root of unity
(b) For any sufficiently large positive integer N there exists anonndashzero polynomial QN isin Z[X0 X1] with partial degreele N t0 in X0 partial degree le N t1 in X1 and heightH(QN) le eN such that∣∣(DkQN)(mx αm)
∣∣ le exp(minusNu)
for any km isin N with k le N s0 and m le N s1
53 59
Gelrsquofondrsquos transcendence criterion
Simple form Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le n and height le en such that
|Pn(ϑ)| le eminus7n2
for all n ge 1 then ϑ is algebraic and Pn(ϑ) = 0 for all n ge 1
First extension Replace the bound for the degree by le dnthe bound for the height by ebn and the bound for |Pn(ϑ)| byeminuscdnbn with some constant c gt 0 independent of n
Some mild conditions are required on the sequences (dn)nge1 and(bn)nge1
54 59
Transcendence criterion with multiplicitiesWith derivatives Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le dn and height le ebn such that∣∣P (j)
n (ϑ)∣∣ le eminuscdnbntn
for all j in the range 0 le j lt tn and all n ge 1 then ϑ isalgebraicDue to M Laurent and D Roy applications to algebraicindependence
55 59
Criterion with several points
Goal Given a sequence of complex numbers (ϑi)ige1 if thereexists a sequence (Pn)nge1 of nonndashzero polynomials in Z[X]with Pn of degree le dn and height le ebn such that∣∣P (j)
n (ϑi)∣∣ le eminuscdnbntnsn
for 0 le j lt tn 1 le i le sn and all n ge 1 then the numbers ϑiare algebraic
D Roy Not true in general but true in some special caseswith a structure on the sequence (ϑi)ige1Combines the elimination arguments used for criteria ofalgebraic independence and for zero estimates
56 59
Small value estimates for the additive group
D Roy Small value estimates for the additive group Intern JNumber Theory to appear
Let ξ be a transcendental complex number let β σ τ and νbe non-negative real numbers let n0 be a positive integer andlet (Pn)ngen0 be a sequence of nonndashzero polynomials in Z[T ]satisfying deg(Pn) le n and H(Pn) le exp(nβ) for eachn ge n0 Suppose that β gt 1 (34)σ + τ lt 1 andν gt 1 + β minus (34)σ minus τ Then for infinitely many n we have
max|P [j]n (iξ)| 0 le i le nσ 0 le j le nτ
gt exp(minusnν)
57 59
Small value estimates for the multiplicative group
D Roy Small value estimates for the multiplicative groupActa Arith to appear
Let ξ1 ξm be multiplicatively independent complexnumbers in a field of transcendence degree 1 Under suitableassumptions on the parameters β σ τ ν for infinitely manypositive integers n there exists no nonndashzero polynomialP isin Z[T ] satisfying deg(P ) le n H(P ) le exp(nβ) and
max|P [j](ξi11 middot middot middot ξimm )| 0 le i1 im le nσ 0 le j le nτ
gt exp(minusnν)
58 59
Heidelberg Colloquium organized by the Institute for Mathematics
httpwwwmathiuni-heidelbergdevortraghtmls=1 January 22 2009
Transcendental Number TheorySchanuelrsquos Conjecture
Michel Waldschmidt
Institut de Mathematiques de Jussieu amp Paris VIhttpwwwmathjussieufrsimmiw
59 59
Transcendence degree of extensions
Assumek sub K sub L
The union of a transcendence basis of K over k and of atranscendence basis of L over K produces a transcendencebasis of L over kHence
tr degkL = tr degkK + tr degKL
An algebraic extension Kk is an extension of transcendencedegree 0 This means that there is no transcendental elementin K over k (any element in K is algebraic over k)
14 59
Algebraic independence of complex numbers
For complex numbers algebraic independence over Q or overQ is the same In particular if θ1 θm are algebraicallyindependent complex numbers then for any non-constantpolynomial f with algebraic coefficients the numberf(θ1 θm) is transcendental
For instance the two numbers e and eradic
2 are algebraicallyindependent As a consequence for any nonndashconstantpolynomial f isin Q[X Y ] with algebraic coefficients the
number f(e eradic
2) is transcendental
15 59
Schanuelrsquos Conjecture
Let x1 xn be Q-linearly independent complex numbersThen at least n of the 2n numbers x1 xn e
x1 exn arealgebraically independent
Since there are 2n numbers only the transcendence degreeover Q of the field
Q(x1 xn e
x1 exn)
is at most 2n The conjecture is that this transcendencedegree is always ge n
n le tr degQQ(x1 xn e
x1 exn)le 2n
16 59
Origin of Schanuelrsquos ConjectureCourse given by Serge Lang (1927ndash2005) at Yale in the 60rsquos
S Lang ndash Introduction to transcendental numbersAddison-Wesley 1966
also attended by M Nagata (1927ndash2008)(14th Problem of Hilbert)
Nagatarsquos Conjecture solved by E Bombieri
17 59
Schanuelrsquos Conjecture
Let x1 xn be Q-linearly independent complex numbersThen
tr degQQ(x1 xn e
x1 exn)ge n
Remark For almost all tuples (with respect to the Lebesguemeasure) the transcendence degree is 2n
18 59
AO Gelrsquofond CRAS 1934
19 59
Statement by Gelrsquofond (1934)
Let β1 βn be Q-linearly independent algebraic numbersand let logα1 logαm be Q-linearly independentlogarithms of algebraic numbers Then the numbers
eβ1 eβn logα1 logαm
are algebraically independent over Q
20 59
Further statement by Gelrsquofond
Let β1 βn be algebraic numbers with β1 6= 0 and letα1 αm be algebraic numbers with α1 6= 0 1 α2 6= 0 1αi 6= 0 Then the numbers
eβ1eβ2e βnminus1e
βn
and αα αm
21
are transcendental and there is no nontrivial algebraic relationbetween such numbers
Remark The condition on α2 should be that it is irrational
21 59
Easy consequence of Schanuelrsquos Conjecture
According to Schanuelrsquos Conjecture the following numbers arealgebraically independent
e+ π eπ πe ee ee2
eee
ππ ππ2
πππ
log π log(log 2) π log 2 (log 2)(log 3) 2log 2 (log 2)log 3
Proof This is an easy exercise
22 59
Langrsquos exercise
Define E0 = Q Inductivelyfor n ge 1 define En as thealgebraic closure of the fieldgenerated over Enminus1 by thenumbers exp(x) = ex wherex ranges over Enminus1 Let E bethe union of En n ge 0Then Schanuelrsquos Conjectureimplies that the number πdoes not belong to E
More precisely Schanuelrsquos Conjecture implies that thenumbers π log π log log π log log log π are algebraicallyindependent over E
23 59
A variant of Langrsquos exercise
Define L0 = Q Inductively for n ge 1 define Ln as thealgebraic closure of the field generated over Lnminus1 by thenumbers y where y ranges over the set of complex numberssuch that ey isin Lnminus1 Let L be the union of Ln n ge 0Then Schanuelrsquos Conjecture implies that the number e doesnot belong to L
More precisely Schanuelrsquos Conjecture implies that the
numbers e ee eee ee
ee
are algebraically independent over L
24 59
Arizona Winter School AWS2008 Tucson
Theorem [Jonathan Bober Chuangxun Cheng Brian DietelMathilde Herblot Jingjing Huang Holly Krieger DiegoMarques Jonathan Mason Martin Mereb and Robert Wilson]Schanuelrsquos Conjecture implies that the fields E and L arelinearly disjoint over Q
Definition Given a field extension FK and twosubextensions F1 F2 sube F we say F1 F2 are linearly disjointover K when the following holds Any set x1 xn sube F1
of Kndash linearly independent elements is linearly independentover F2
Reference arXiv08043550 [mathNT] 2008
25 59
Formal analogs
WD Brownawell(was a student of Schanuel)
J Axrsquos Theorem (1968) Version of SchanuelrsquosConjecture for power seriesover C(and R Coleman for powerseries over Q)
Work by WD Brownawelland K Kubota on the ellipticanalog of Axrsquos Theorem
26 59
Conjectures by A Grothendieck and Y Andre
Generalized Conjecture onPeriods by Grothendieck Dimension of theMumfordndashTate group of asmooth projective variety
Generalization by Y Andre tomotives
Case of 1ndashmotives Elliptico-Toric Conjecture ofC Bertolin
27 59
Ubiquity of Schanuelrsquos Conjecture
Other contexts pndashadic numbers Leopoldtrsquos Conjecture onthe pndashadic rank of the units of an algebraic number fieldNon-vanishing of RegulatorsNonndashdegenerescence of heightsConjecture of B Mazur on rational pointsDiophantine approximation on tori
Dipendra Prasad Gopal Prasad
28 59
Methods from logicEhud Hrushovski Boris Zilber Jonathan Kirby
Calculus of rdquopredimension functionsrdquo (E Hrushovski)
Zilberrsquos construction of a rdquopseudoexponentiationrdquo
Also A Macintyre DE Marker G Terzo AJ WilkieD Bertrand
29 59
Methods from logic Model theory
Exponential algebraicity in exponential fieldsby
Jonathan Kirby
The dimension of the exponential algebraic closure operator inan exponential field satisfies a weak Schanuel property
A corollary is that there are at most countably many essentialcounterexamples to Schanuelrsquos conjecture
arXiv 08104285v2
30 59
Known
LindemannndashWeierstraszlig Theorem = case where x1 xn arealgebraic
Let β1 βn be algebraic numbers which are linearlyindependent over Q Then the numbers eβ1 eβn arealgebraically independent over Q
31 59
Problem of Gelrsquofond and SchneiderRaised by AO Gelrsquofond in 1948 and Th Schneider in 1952
Conjecture If α is an algebraic number α 6= 0 α 6= 1 and ifβ is an irrational algebraic number of degree d then the dminus 1numbers
αβ αβ2
αβdminus1
are algebraically independent
Special case of Schanuelrsquos Conjecture Take xi = βiminus1 logαn = d so that
x1 xn e
x1 exn
islogα β logα βdminus1 logα α αβ αβ
dminus1
The conclusion of Schanuelrsquos Conjecture is
tr degQQ(logα αβ αβ
2
αβdminus1)
= d
32 59
Algebraic independence method of Gelrsquofond
AO Gelrsquofond (1948)
The two numbers 23radic2 and
23radic4 are algebraically
independentMore generally if α is analgebraic number α 6= 0α 6= 1 and if β is a algebraicnumber of degree d ge 3 thentwo at least of the numbers
αβ αβ2
αβdminus1
are algebraically independent
33 59
Tools
Transcendence criterion Replaces Liouvillersquos inequality intranscendence proofsLiouville A nonndashzero rational integer n isin Z satisfies |n| ge 1Gelrsquofond Needs to give a lower bound for |P (θ)| withP isin Z[X] 0 when θ is transcendental
Zero estimate for exponential polynomials C Hermite P Turan K Mahler R Tijdeman
Small transcendence degree AO Gelrsquofond AA Smelev R Tijdeman WD Brownawell
34 59
Large transcendence degree
GV Chudnovsky (1976)Among the numbers
αβ αβ2
αβdminus1
at least [log2 d] arealgebraically independent
GV Chudnovsky ndash On the path to Schanuelrsquosconjecture Algebraic curves close to a pointI General theory of colored sequencesII Fields of finite transcendence type and coloredsequences ResultantsStudia Sci Math Hungar 12 (1977) 125ndash157 (1980)
35 59
Partial result on the problem of Gelrsquofond and
Schneider
AO Gelrsquofond GV Chudnovskii P Philippon YuV Nesterenko
G Diaz If α is an algebraic number α 6= 0 α 6= 1 and if β isan irrational algebraic number of degree d then
tr degQQ(αβ αβ
2
αβdminus1) ge [d+ 1
2
]
36 59
Conjecture of algebraic independence of logarithms
of algebraic numbers
Denote by L the set of complex numbers λ for which eλ isalgebraic
L = logα α isin Qtimes
Hence L is a Q-vector subspace of CThe most important special case of Schanuelrsquos Conjecture is
Conjecture Let λ1 λn be Q-linearly independent elementsin L Then the numbers λ1 λn are algebraicallyindependent over Q
Not yet known that the transcendence degree is ge 2 Open problem Among all logarithms of algebraic numbersone at least is transcendental over Q(π)
37 59
Structural rank of a matrix
Let K be a field k a subfield and M a matrix with entries in KConsider the k-vector subspace E of K spanned by the entriesof M Choose an injective morphism ϕ of E into a k-vectorspace kX1 + middot middot middot+ kXn The image ϕ(M) of M is a matrixwhose entries are in the field k(X1 Xn) of rationalfractions Its rank does not depend on the choice of ϕ
This is the structural rank of M with respect to k
38 59
Example
LetM =
(bij + λij
)1leiled1lejle`
be a matrix with coefficients in Q + L Consider a basis of theQndashvector spanned by the entries and replace the elements inthis basis by unknowns This gives a new matrix M withcoefficients in a field of rational fractions the rank of which isthe structural rank of M (with respect to Q)
As a consequence of the conjecture of algebraic independenceof logarithms of algebraic numbers the rank of M should bethe same as the rank of its specialization M
39 59
Equivalence between the two conjectures
Following D Roy theconjecture on algebraicindependence of logarithms ofalgebraic numbers isequivalent to
Conjecture Any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L has a rank equal to its structuralrank
40 59
Any Polynomial is the Determinant of a Matrix
The proof of the equivalence uses the nice auxiliary result
For any P isin k[X1 Xn] there exists a square matrix withentries in the k-vector space k + kX1 + middot middot middot+ kXn whosedeterminant is P
41 59
Half of the Conjecture is solved
From a certain point of view half of the conjecture of algebraicindependence of logarithms of algebraic numbers is solved
Theorem [D Roy] The rank of any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L is at least half its structural rank
42 59
Reformulation by D Roy
Instead of taking logarithmsof algebraic numbers andlooking for the algebraicindependence relationsD Roy fixes a polynomial andlooks at the points withcoordinates logarithms ofalgebraic numbers on thecorresponding hypersurface
43 59
Reformulation by D Roy
Royrsquos reformulation of the conjecture of algebraicindependence of logarithms is
Conjecture For any algebraic subvariety V of Cn definedover the field Q of algebraic numbers the set V cap Ln is theunion of the sets E cap Ln where E ranges over the set ofvector subspaces of Cn which are defined over Q andcontained in V
Trivial Any element in E cap Ln where E is a vector subspaceof Cn defined over Q and contained in V belongs to V cap Ln
44 59
Example The Four Exponentials ConjectureTake for V the hypersurface of C4 defined by the equation
z1z4 = z2z3
The maximal Cndashvector subspaces of C4 defined over Q andcontained in V are the planes
az1 = bz2 bz4 = az3
and the planesaz1 = bz3 bz4 = az2
with (a b) isin Q2 (0 0)Hence Schanuelrsquos Conjecture implies that if λ1 λ2 λ3 λ4 arelogarithms of algebraic numbers satisfying
λ1λ4 = λ2λ3
then either λ1 λ2 are linearly dependent over Q or else λ1 λ3
are linearly dependent over Q45 59
Six Exponentials Theorem
and Four Exponentials Conjecture
A Selberg CL Siegel S Lang K Ramachandra
46 59
The Four Exponentials Conjecture
Conjecture (Th Schneider S Lang K Ramachandra) Ifx1 x2 are Qndashlinearly independent complex numbers and y1 y2
are Qndashlinearly independent complex numbers then one atleast of the four numbers
ex1y1 ex1y2 ex2y1 ex2y2
is transcendentalEquivalent statement Any 2times 2 matrix(
λ1 λ2
λ3 λ4
)with entries in L and with Qndashlinearly independent rows andQndashlinearly independent columns has maximal rank 2
47 59
How could we attack Schanuelrsquos Conjecture
Let x1 xn be Qndashlinearly independent complex numbersFollowing the transcendence methods of Hermite GelrsquofondSchneider one may start by introducing an auxiliaryfunction
F (z) = P (z ez)
where P isin Z[X0 X1] is a nonndashzero polynomial One considersthe derivatives (
d
dz
)kF = (DkP )(z ez)
at the pointsm1x1 + middot middot middot+mnxn
for various values of (m1 mn) isin Zn
48 59
Auxiliary functionLet D denote the derivation
D =part
partX0
+X1part
partX1
over the ring C[X0 X1] Recall that x1 xn are Qndashlinearlyindependent complex numbers Let α1 αn be nonndashzerocomplex numbersThe transcendence machinery produces a sequence (PN)Nge0
of polynomials with integer coefficients satisfying∣∣∣∣∣(DkPN)(nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1
49 59
Royrsquos approach to Schanuelrsquos Conjecture (1999)
If the number of equations we produce is too small such a setof relations does not contain any information The existenceof a sequence of nonndashtrivial polynomials (PN)Nge0 follows fromlinear algebra
On the other hand following D Roy one may expect that theexistence of such a sequence (PN)Nge0 producing sufficientlymany such equations will yield the desired conclusion
tr degQQ(x1 xn α1 αn
)ge n
50 59
Royrsquos Conjecture (1999)Let s0 s1 t0 t1 u positive real numbers satisfying
max1 t0 2t1 lt mins0 2s1and
maxs0 s1 + t1 lt u lt1
2(1 + t0 + t1)
Assume that for any sufficiently large positive integer N there exists a nonndashzero polynomial PN isin Z[X0 X1] withpartial degree le N t0 in X0 partial degree le N t1 in X1 andheight le eN which satisfies∣∣∣∣∣(DkPN)(
nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1 Then
tr degQQ(x1 xn α1 αn) ge n51 59
Royrsquos Theorem (1999)
Royrsquos conjecture is equivalent to Schanuelrsquos Conjecture
52 59
Equivalence between Schanuel and Roy
Let (x α) isin CtimesCtimes and let s0 s1 t0 t1 u be positive realnumbers satisfying the inequalities of Royrsquos Conjecture Thenthe following conditions are equivalent (a) The number αeminusx is a root of unity
(b) For any sufficiently large positive integer N there exists anonndashzero polynomial QN isin Z[X0 X1] with partial degreele N t0 in X0 partial degree le N t1 in X1 and heightH(QN) le eN such that∣∣(DkQN)(mx αm)
∣∣ le exp(minusNu)
for any km isin N with k le N s0 and m le N s1
53 59
Gelrsquofondrsquos transcendence criterion
Simple form Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le n and height le en such that
|Pn(ϑ)| le eminus7n2
for all n ge 1 then ϑ is algebraic and Pn(ϑ) = 0 for all n ge 1
First extension Replace the bound for the degree by le dnthe bound for the height by ebn and the bound for |Pn(ϑ)| byeminuscdnbn with some constant c gt 0 independent of n
Some mild conditions are required on the sequences (dn)nge1 and(bn)nge1
54 59
Transcendence criterion with multiplicitiesWith derivatives Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le dn and height le ebn such that∣∣P (j)
n (ϑ)∣∣ le eminuscdnbntn
for all j in the range 0 le j lt tn and all n ge 1 then ϑ isalgebraicDue to M Laurent and D Roy applications to algebraicindependence
55 59
Criterion with several points
Goal Given a sequence of complex numbers (ϑi)ige1 if thereexists a sequence (Pn)nge1 of nonndashzero polynomials in Z[X]with Pn of degree le dn and height le ebn such that∣∣P (j)
n (ϑi)∣∣ le eminuscdnbntnsn
for 0 le j lt tn 1 le i le sn and all n ge 1 then the numbers ϑiare algebraic
D Roy Not true in general but true in some special caseswith a structure on the sequence (ϑi)ige1Combines the elimination arguments used for criteria ofalgebraic independence and for zero estimates
56 59
Small value estimates for the additive group
D Roy Small value estimates for the additive group Intern JNumber Theory to appear
Let ξ be a transcendental complex number let β σ τ and νbe non-negative real numbers let n0 be a positive integer andlet (Pn)ngen0 be a sequence of nonndashzero polynomials in Z[T ]satisfying deg(Pn) le n and H(Pn) le exp(nβ) for eachn ge n0 Suppose that β gt 1 (34)σ + τ lt 1 andν gt 1 + β minus (34)σ minus τ Then for infinitely many n we have
max|P [j]n (iξ)| 0 le i le nσ 0 le j le nτ
gt exp(minusnν)
57 59
Small value estimates for the multiplicative group
D Roy Small value estimates for the multiplicative groupActa Arith to appear
Let ξ1 ξm be multiplicatively independent complexnumbers in a field of transcendence degree 1 Under suitableassumptions on the parameters β σ τ ν for infinitely manypositive integers n there exists no nonndashzero polynomialP isin Z[T ] satisfying deg(P ) le n H(P ) le exp(nβ) and
max|P [j](ξi11 middot middot middot ξimm )| 0 le i1 im le nσ 0 le j le nτ
gt exp(minusnν)
58 59
Heidelberg Colloquium organized by the Institute for Mathematics
httpwwwmathiuni-heidelbergdevortraghtmls=1 January 22 2009
Transcendental Number TheorySchanuelrsquos Conjecture
Michel Waldschmidt
Institut de Mathematiques de Jussieu amp Paris VIhttpwwwmathjussieufrsimmiw
59 59
Algebraic independence of complex numbers
For complex numbers algebraic independence over Q or overQ is the same In particular if θ1 θm are algebraicallyindependent complex numbers then for any non-constantpolynomial f with algebraic coefficients the numberf(θ1 θm) is transcendental
For instance the two numbers e and eradic
2 are algebraicallyindependent As a consequence for any nonndashconstantpolynomial f isin Q[X Y ] with algebraic coefficients the
number f(e eradic
2) is transcendental
15 59
Schanuelrsquos Conjecture
Let x1 xn be Q-linearly independent complex numbersThen at least n of the 2n numbers x1 xn e
x1 exn arealgebraically independent
Since there are 2n numbers only the transcendence degreeover Q of the field
Q(x1 xn e
x1 exn)
is at most 2n The conjecture is that this transcendencedegree is always ge n
n le tr degQQ(x1 xn e
x1 exn)le 2n
16 59
Origin of Schanuelrsquos ConjectureCourse given by Serge Lang (1927ndash2005) at Yale in the 60rsquos
S Lang ndash Introduction to transcendental numbersAddison-Wesley 1966
also attended by M Nagata (1927ndash2008)(14th Problem of Hilbert)
Nagatarsquos Conjecture solved by E Bombieri
17 59
Schanuelrsquos Conjecture
Let x1 xn be Q-linearly independent complex numbersThen
tr degQQ(x1 xn e
x1 exn)ge n
Remark For almost all tuples (with respect to the Lebesguemeasure) the transcendence degree is 2n
18 59
AO Gelrsquofond CRAS 1934
19 59
Statement by Gelrsquofond (1934)
Let β1 βn be Q-linearly independent algebraic numbersand let logα1 logαm be Q-linearly independentlogarithms of algebraic numbers Then the numbers
eβ1 eβn logα1 logαm
are algebraically independent over Q
20 59
Further statement by Gelrsquofond
Let β1 βn be algebraic numbers with β1 6= 0 and letα1 αm be algebraic numbers with α1 6= 0 1 α2 6= 0 1αi 6= 0 Then the numbers
eβ1eβ2e βnminus1e
βn
and αα αm
21
are transcendental and there is no nontrivial algebraic relationbetween such numbers
Remark The condition on α2 should be that it is irrational
21 59
Easy consequence of Schanuelrsquos Conjecture
According to Schanuelrsquos Conjecture the following numbers arealgebraically independent
e+ π eπ πe ee ee2
eee
ππ ππ2
πππ
log π log(log 2) π log 2 (log 2)(log 3) 2log 2 (log 2)log 3
Proof This is an easy exercise
22 59
Langrsquos exercise
Define E0 = Q Inductivelyfor n ge 1 define En as thealgebraic closure of the fieldgenerated over Enminus1 by thenumbers exp(x) = ex wherex ranges over Enminus1 Let E bethe union of En n ge 0Then Schanuelrsquos Conjectureimplies that the number πdoes not belong to E
More precisely Schanuelrsquos Conjecture implies that thenumbers π log π log log π log log log π are algebraicallyindependent over E
23 59
A variant of Langrsquos exercise
Define L0 = Q Inductively for n ge 1 define Ln as thealgebraic closure of the field generated over Lnminus1 by thenumbers y where y ranges over the set of complex numberssuch that ey isin Lnminus1 Let L be the union of Ln n ge 0Then Schanuelrsquos Conjecture implies that the number e doesnot belong to L
More precisely Schanuelrsquos Conjecture implies that the
numbers e ee eee ee
ee
are algebraically independent over L
24 59
Arizona Winter School AWS2008 Tucson
Theorem [Jonathan Bober Chuangxun Cheng Brian DietelMathilde Herblot Jingjing Huang Holly Krieger DiegoMarques Jonathan Mason Martin Mereb and Robert Wilson]Schanuelrsquos Conjecture implies that the fields E and L arelinearly disjoint over Q
Definition Given a field extension FK and twosubextensions F1 F2 sube F we say F1 F2 are linearly disjointover K when the following holds Any set x1 xn sube F1
of Kndash linearly independent elements is linearly independentover F2
Reference arXiv08043550 [mathNT] 2008
25 59
Formal analogs
WD Brownawell(was a student of Schanuel)
J Axrsquos Theorem (1968) Version of SchanuelrsquosConjecture for power seriesover C(and R Coleman for powerseries over Q)
Work by WD Brownawelland K Kubota on the ellipticanalog of Axrsquos Theorem
26 59
Conjectures by A Grothendieck and Y Andre
Generalized Conjecture onPeriods by Grothendieck Dimension of theMumfordndashTate group of asmooth projective variety
Generalization by Y Andre tomotives
Case of 1ndashmotives Elliptico-Toric Conjecture ofC Bertolin
27 59
Ubiquity of Schanuelrsquos Conjecture
Other contexts pndashadic numbers Leopoldtrsquos Conjecture onthe pndashadic rank of the units of an algebraic number fieldNon-vanishing of RegulatorsNonndashdegenerescence of heightsConjecture of B Mazur on rational pointsDiophantine approximation on tori
Dipendra Prasad Gopal Prasad
28 59
Methods from logicEhud Hrushovski Boris Zilber Jonathan Kirby
Calculus of rdquopredimension functionsrdquo (E Hrushovski)
Zilberrsquos construction of a rdquopseudoexponentiationrdquo
Also A Macintyre DE Marker G Terzo AJ WilkieD Bertrand
29 59
Methods from logic Model theory
Exponential algebraicity in exponential fieldsby
Jonathan Kirby
The dimension of the exponential algebraic closure operator inan exponential field satisfies a weak Schanuel property
A corollary is that there are at most countably many essentialcounterexamples to Schanuelrsquos conjecture
arXiv 08104285v2
30 59
Known
LindemannndashWeierstraszlig Theorem = case where x1 xn arealgebraic
Let β1 βn be algebraic numbers which are linearlyindependent over Q Then the numbers eβ1 eβn arealgebraically independent over Q
31 59
Problem of Gelrsquofond and SchneiderRaised by AO Gelrsquofond in 1948 and Th Schneider in 1952
Conjecture If α is an algebraic number α 6= 0 α 6= 1 and ifβ is an irrational algebraic number of degree d then the dminus 1numbers
αβ αβ2
αβdminus1
are algebraically independent
Special case of Schanuelrsquos Conjecture Take xi = βiminus1 logαn = d so that
x1 xn e
x1 exn
islogα β logα βdminus1 logα α αβ αβ
dminus1
The conclusion of Schanuelrsquos Conjecture is
tr degQQ(logα αβ αβ
2
αβdminus1)
= d
32 59
Algebraic independence method of Gelrsquofond
AO Gelrsquofond (1948)
The two numbers 23radic2 and
23radic4 are algebraically
independentMore generally if α is analgebraic number α 6= 0α 6= 1 and if β is a algebraicnumber of degree d ge 3 thentwo at least of the numbers
αβ αβ2
αβdminus1
are algebraically independent
33 59
Tools
Transcendence criterion Replaces Liouvillersquos inequality intranscendence proofsLiouville A nonndashzero rational integer n isin Z satisfies |n| ge 1Gelrsquofond Needs to give a lower bound for |P (θ)| withP isin Z[X] 0 when θ is transcendental
Zero estimate for exponential polynomials C Hermite P Turan K Mahler R Tijdeman
Small transcendence degree AO Gelrsquofond AA Smelev R Tijdeman WD Brownawell
34 59
Large transcendence degree
GV Chudnovsky (1976)Among the numbers
αβ αβ2
αβdminus1
at least [log2 d] arealgebraically independent
GV Chudnovsky ndash On the path to Schanuelrsquosconjecture Algebraic curves close to a pointI General theory of colored sequencesII Fields of finite transcendence type and coloredsequences ResultantsStudia Sci Math Hungar 12 (1977) 125ndash157 (1980)
35 59
Partial result on the problem of Gelrsquofond and
Schneider
AO Gelrsquofond GV Chudnovskii P Philippon YuV Nesterenko
G Diaz If α is an algebraic number α 6= 0 α 6= 1 and if β isan irrational algebraic number of degree d then
tr degQQ(αβ αβ
2
αβdminus1) ge [d+ 1
2
]
36 59
Conjecture of algebraic independence of logarithms
of algebraic numbers
Denote by L the set of complex numbers λ for which eλ isalgebraic
L = logα α isin Qtimes
Hence L is a Q-vector subspace of CThe most important special case of Schanuelrsquos Conjecture is
Conjecture Let λ1 λn be Q-linearly independent elementsin L Then the numbers λ1 λn are algebraicallyindependent over Q
Not yet known that the transcendence degree is ge 2 Open problem Among all logarithms of algebraic numbersone at least is transcendental over Q(π)
37 59
Structural rank of a matrix
Let K be a field k a subfield and M a matrix with entries in KConsider the k-vector subspace E of K spanned by the entriesof M Choose an injective morphism ϕ of E into a k-vectorspace kX1 + middot middot middot+ kXn The image ϕ(M) of M is a matrixwhose entries are in the field k(X1 Xn) of rationalfractions Its rank does not depend on the choice of ϕ
This is the structural rank of M with respect to k
38 59
Example
LetM =
(bij + λij
)1leiled1lejle`
be a matrix with coefficients in Q + L Consider a basis of theQndashvector spanned by the entries and replace the elements inthis basis by unknowns This gives a new matrix M withcoefficients in a field of rational fractions the rank of which isthe structural rank of M (with respect to Q)
As a consequence of the conjecture of algebraic independenceof logarithms of algebraic numbers the rank of M should bethe same as the rank of its specialization M
39 59
Equivalence between the two conjectures
Following D Roy theconjecture on algebraicindependence of logarithms ofalgebraic numbers isequivalent to
Conjecture Any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L has a rank equal to its structuralrank
40 59
Any Polynomial is the Determinant of a Matrix
The proof of the equivalence uses the nice auxiliary result
For any P isin k[X1 Xn] there exists a square matrix withentries in the k-vector space k + kX1 + middot middot middot+ kXn whosedeterminant is P
41 59
Half of the Conjecture is solved
From a certain point of view half of the conjecture of algebraicindependence of logarithms of algebraic numbers is solved
Theorem [D Roy] The rank of any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L is at least half its structural rank
42 59
Reformulation by D Roy
Instead of taking logarithmsof algebraic numbers andlooking for the algebraicindependence relationsD Roy fixes a polynomial andlooks at the points withcoordinates logarithms ofalgebraic numbers on thecorresponding hypersurface
43 59
Reformulation by D Roy
Royrsquos reformulation of the conjecture of algebraicindependence of logarithms is
Conjecture For any algebraic subvariety V of Cn definedover the field Q of algebraic numbers the set V cap Ln is theunion of the sets E cap Ln where E ranges over the set ofvector subspaces of Cn which are defined over Q andcontained in V
Trivial Any element in E cap Ln where E is a vector subspaceof Cn defined over Q and contained in V belongs to V cap Ln
44 59
Example The Four Exponentials ConjectureTake for V the hypersurface of C4 defined by the equation
z1z4 = z2z3
The maximal Cndashvector subspaces of C4 defined over Q andcontained in V are the planes
az1 = bz2 bz4 = az3
and the planesaz1 = bz3 bz4 = az2
with (a b) isin Q2 (0 0)Hence Schanuelrsquos Conjecture implies that if λ1 λ2 λ3 λ4 arelogarithms of algebraic numbers satisfying
λ1λ4 = λ2λ3
then either λ1 λ2 are linearly dependent over Q or else λ1 λ3
are linearly dependent over Q45 59
Six Exponentials Theorem
and Four Exponentials Conjecture
A Selberg CL Siegel S Lang K Ramachandra
46 59
The Four Exponentials Conjecture
Conjecture (Th Schneider S Lang K Ramachandra) Ifx1 x2 are Qndashlinearly independent complex numbers and y1 y2
are Qndashlinearly independent complex numbers then one atleast of the four numbers
ex1y1 ex1y2 ex2y1 ex2y2
is transcendentalEquivalent statement Any 2times 2 matrix(
λ1 λ2
λ3 λ4
)with entries in L and with Qndashlinearly independent rows andQndashlinearly independent columns has maximal rank 2
47 59
How could we attack Schanuelrsquos Conjecture
Let x1 xn be Qndashlinearly independent complex numbersFollowing the transcendence methods of Hermite GelrsquofondSchneider one may start by introducing an auxiliaryfunction
F (z) = P (z ez)
where P isin Z[X0 X1] is a nonndashzero polynomial One considersthe derivatives (
d
dz
)kF = (DkP )(z ez)
at the pointsm1x1 + middot middot middot+mnxn
for various values of (m1 mn) isin Zn
48 59
Auxiliary functionLet D denote the derivation
D =part
partX0
+X1part
partX1
over the ring C[X0 X1] Recall that x1 xn are Qndashlinearlyindependent complex numbers Let α1 αn be nonndashzerocomplex numbersThe transcendence machinery produces a sequence (PN)Nge0
of polynomials with integer coefficients satisfying∣∣∣∣∣(DkPN)(nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1
49 59
Royrsquos approach to Schanuelrsquos Conjecture (1999)
If the number of equations we produce is too small such a setof relations does not contain any information The existenceof a sequence of nonndashtrivial polynomials (PN)Nge0 follows fromlinear algebra
On the other hand following D Roy one may expect that theexistence of such a sequence (PN)Nge0 producing sufficientlymany such equations will yield the desired conclusion
tr degQQ(x1 xn α1 αn
)ge n
50 59
Royrsquos Conjecture (1999)Let s0 s1 t0 t1 u positive real numbers satisfying
max1 t0 2t1 lt mins0 2s1and
maxs0 s1 + t1 lt u lt1
2(1 + t0 + t1)
Assume that for any sufficiently large positive integer N there exists a nonndashzero polynomial PN isin Z[X0 X1] withpartial degree le N t0 in X0 partial degree le N t1 in X1 andheight le eN which satisfies∣∣∣∣∣(DkPN)(
nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1 Then
tr degQQ(x1 xn α1 αn) ge n51 59
Royrsquos Theorem (1999)
Royrsquos conjecture is equivalent to Schanuelrsquos Conjecture
52 59
Equivalence between Schanuel and Roy
Let (x α) isin CtimesCtimes and let s0 s1 t0 t1 u be positive realnumbers satisfying the inequalities of Royrsquos Conjecture Thenthe following conditions are equivalent (a) The number αeminusx is a root of unity
(b) For any sufficiently large positive integer N there exists anonndashzero polynomial QN isin Z[X0 X1] with partial degreele N t0 in X0 partial degree le N t1 in X1 and heightH(QN) le eN such that∣∣(DkQN)(mx αm)
∣∣ le exp(minusNu)
for any km isin N with k le N s0 and m le N s1
53 59
Gelrsquofondrsquos transcendence criterion
Simple form Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le n and height le en such that
|Pn(ϑ)| le eminus7n2
for all n ge 1 then ϑ is algebraic and Pn(ϑ) = 0 for all n ge 1
First extension Replace the bound for the degree by le dnthe bound for the height by ebn and the bound for |Pn(ϑ)| byeminuscdnbn with some constant c gt 0 independent of n
Some mild conditions are required on the sequences (dn)nge1 and(bn)nge1
54 59
Transcendence criterion with multiplicitiesWith derivatives Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le dn and height le ebn such that∣∣P (j)
n (ϑ)∣∣ le eminuscdnbntn
for all j in the range 0 le j lt tn and all n ge 1 then ϑ isalgebraicDue to M Laurent and D Roy applications to algebraicindependence
55 59
Criterion with several points
Goal Given a sequence of complex numbers (ϑi)ige1 if thereexists a sequence (Pn)nge1 of nonndashzero polynomials in Z[X]with Pn of degree le dn and height le ebn such that∣∣P (j)
n (ϑi)∣∣ le eminuscdnbntnsn
for 0 le j lt tn 1 le i le sn and all n ge 1 then the numbers ϑiare algebraic
D Roy Not true in general but true in some special caseswith a structure on the sequence (ϑi)ige1Combines the elimination arguments used for criteria ofalgebraic independence and for zero estimates
56 59
Small value estimates for the additive group
D Roy Small value estimates for the additive group Intern JNumber Theory to appear
Let ξ be a transcendental complex number let β σ τ and νbe non-negative real numbers let n0 be a positive integer andlet (Pn)ngen0 be a sequence of nonndashzero polynomials in Z[T ]satisfying deg(Pn) le n and H(Pn) le exp(nβ) for eachn ge n0 Suppose that β gt 1 (34)σ + τ lt 1 andν gt 1 + β minus (34)σ minus τ Then for infinitely many n we have
max|P [j]n (iξ)| 0 le i le nσ 0 le j le nτ
gt exp(minusnν)
57 59
Small value estimates for the multiplicative group
D Roy Small value estimates for the multiplicative groupActa Arith to appear
Let ξ1 ξm be multiplicatively independent complexnumbers in a field of transcendence degree 1 Under suitableassumptions on the parameters β σ τ ν for infinitely manypositive integers n there exists no nonndashzero polynomialP isin Z[T ] satisfying deg(P ) le n H(P ) le exp(nβ) and
max|P [j](ξi11 middot middot middot ξimm )| 0 le i1 im le nσ 0 le j le nτ
gt exp(minusnν)
58 59
Heidelberg Colloquium organized by the Institute for Mathematics
httpwwwmathiuni-heidelbergdevortraghtmls=1 January 22 2009
Transcendental Number TheorySchanuelrsquos Conjecture
Michel Waldschmidt
Institut de Mathematiques de Jussieu amp Paris VIhttpwwwmathjussieufrsimmiw
59 59
Schanuelrsquos Conjecture
Let x1 xn be Q-linearly independent complex numbersThen at least n of the 2n numbers x1 xn e
x1 exn arealgebraically independent
Since there are 2n numbers only the transcendence degreeover Q of the field
Q(x1 xn e
x1 exn)
is at most 2n The conjecture is that this transcendencedegree is always ge n
n le tr degQQ(x1 xn e
x1 exn)le 2n
16 59
Origin of Schanuelrsquos ConjectureCourse given by Serge Lang (1927ndash2005) at Yale in the 60rsquos
S Lang ndash Introduction to transcendental numbersAddison-Wesley 1966
also attended by M Nagata (1927ndash2008)(14th Problem of Hilbert)
Nagatarsquos Conjecture solved by E Bombieri
17 59
Schanuelrsquos Conjecture
Let x1 xn be Q-linearly independent complex numbersThen
tr degQQ(x1 xn e
x1 exn)ge n
Remark For almost all tuples (with respect to the Lebesguemeasure) the transcendence degree is 2n
18 59
AO Gelrsquofond CRAS 1934
19 59
Statement by Gelrsquofond (1934)
Let β1 βn be Q-linearly independent algebraic numbersand let logα1 logαm be Q-linearly independentlogarithms of algebraic numbers Then the numbers
eβ1 eβn logα1 logαm
are algebraically independent over Q
20 59
Further statement by Gelrsquofond
Let β1 βn be algebraic numbers with β1 6= 0 and letα1 αm be algebraic numbers with α1 6= 0 1 α2 6= 0 1αi 6= 0 Then the numbers
eβ1eβ2e βnminus1e
βn
and αα αm
21
are transcendental and there is no nontrivial algebraic relationbetween such numbers
Remark The condition on α2 should be that it is irrational
21 59
Easy consequence of Schanuelrsquos Conjecture
According to Schanuelrsquos Conjecture the following numbers arealgebraically independent
e+ π eπ πe ee ee2
eee
ππ ππ2
πππ
log π log(log 2) π log 2 (log 2)(log 3) 2log 2 (log 2)log 3
Proof This is an easy exercise
22 59
Langrsquos exercise
Define E0 = Q Inductivelyfor n ge 1 define En as thealgebraic closure of the fieldgenerated over Enminus1 by thenumbers exp(x) = ex wherex ranges over Enminus1 Let E bethe union of En n ge 0Then Schanuelrsquos Conjectureimplies that the number πdoes not belong to E
More precisely Schanuelrsquos Conjecture implies that thenumbers π log π log log π log log log π are algebraicallyindependent over E
23 59
A variant of Langrsquos exercise
Define L0 = Q Inductively for n ge 1 define Ln as thealgebraic closure of the field generated over Lnminus1 by thenumbers y where y ranges over the set of complex numberssuch that ey isin Lnminus1 Let L be the union of Ln n ge 0Then Schanuelrsquos Conjecture implies that the number e doesnot belong to L
More precisely Schanuelrsquos Conjecture implies that the
numbers e ee eee ee
ee
are algebraically independent over L
24 59
Arizona Winter School AWS2008 Tucson
Theorem [Jonathan Bober Chuangxun Cheng Brian DietelMathilde Herblot Jingjing Huang Holly Krieger DiegoMarques Jonathan Mason Martin Mereb and Robert Wilson]Schanuelrsquos Conjecture implies that the fields E and L arelinearly disjoint over Q
Definition Given a field extension FK and twosubextensions F1 F2 sube F we say F1 F2 are linearly disjointover K when the following holds Any set x1 xn sube F1
of Kndash linearly independent elements is linearly independentover F2
Reference arXiv08043550 [mathNT] 2008
25 59
Formal analogs
WD Brownawell(was a student of Schanuel)
J Axrsquos Theorem (1968) Version of SchanuelrsquosConjecture for power seriesover C(and R Coleman for powerseries over Q)
Work by WD Brownawelland K Kubota on the ellipticanalog of Axrsquos Theorem
26 59
Conjectures by A Grothendieck and Y Andre
Generalized Conjecture onPeriods by Grothendieck Dimension of theMumfordndashTate group of asmooth projective variety
Generalization by Y Andre tomotives
Case of 1ndashmotives Elliptico-Toric Conjecture ofC Bertolin
27 59
Ubiquity of Schanuelrsquos Conjecture
Other contexts pndashadic numbers Leopoldtrsquos Conjecture onthe pndashadic rank of the units of an algebraic number fieldNon-vanishing of RegulatorsNonndashdegenerescence of heightsConjecture of B Mazur on rational pointsDiophantine approximation on tori
Dipendra Prasad Gopal Prasad
28 59
Methods from logicEhud Hrushovski Boris Zilber Jonathan Kirby
Calculus of rdquopredimension functionsrdquo (E Hrushovski)
Zilberrsquos construction of a rdquopseudoexponentiationrdquo
Also A Macintyre DE Marker G Terzo AJ WilkieD Bertrand
29 59
Methods from logic Model theory
Exponential algebraicity in exponential fieldsby
Jonathan Kirby
The dimension of the exponential algebraic closure operator inan exponential field satisfies a weak Schanuel property
A corollary is that there are at most countably many essentialcounterexamples to Schanuelrsquos conjecture
arXiv 08104285v2
30 59
Known
LindemannndashWeierstraszlig Theorem = case where x1 xn arealgebraic
Let β1 βn be algebraic numbers which are linearlyindependent over Q Then the numbers eβ1 eβn arealgebraically independent over Q
31 59
Problem of Gelrsquofond and SchneiderRaised by AO Gelrsquofond in 1948 and Th Schneider in 1952
Conjecture If α is an algebraic number α 6= 0 α 6= 1 and ifβ is an irrational algebraic number of degree d then the dminus 1numbers
αβ αβ2
αβdminus1
are algebraically independent
Special case of Schanuelrsquos Conjecture Take xi = βiminus1 logαn = d so that
x1 xn e
x1 exn
islogα β logα βdminus1 logα α αβ αβ
dminus1
The conclusion of Schanuelrsquos Conjecture is
tr degQQ(logα αβ αβ
2
αβdminus1)
= d
32 59
Algebraic independence method of Gelrsquofond
AO Gelrsquofond (1948)
The two numbers 23radic2 and
23radic4 are algebraically
independentMore generally if α is analgebraic number α 6= 0α 6= 1 and if β is a algebraicnumber of degree d ge 3 thentwo at least of the numbers
αβ αβ2
αβdminus1
are algebraically independent
33 59
Tools
Transcendence criterion Replaces Liouvillersquos inequality intranscendence proofsLiouville A nonndashzero rational integer n isin Z satisfies |n| ge 1Gelrsquofond Needs to give a lower bound for |P (θ)| withP isin Z[X] 0 when θ is transcendental
Zero estimate for exponential polynomials C Hermite P Turan K Mahler R Tijdeman
Small transcendence degree AO Gelrsquofond AA Smelev R Tijdeman WD Brownawell
34 59
Large transcendence degree
GV Chudnovsky (1976)Among the numbers
αβ αβ2
αβdminus1
at least [log2 d] arealgebraically independent
GV Chudnovsky ndash On the path to Schanuelrsquosconjecture Algebraic curves close to a pointI General theory of colored sequencesII Fields of finite transcendence type and coloredsequences ResultantsStudia Sci Math Hungar 12 (1977) 125ndash157 (1980)
35 59
Partial result on the problem of Gelrsquofond and
Schneider
AO Gelrsquofond GV Chudnovskii P Philippon YuV Nesterenko
G Diaz If α is an algebraic number α 6= 0 α 6= 1 and if β isan irrational algebraic number of degree d then
tr degQQ(αβ αβ
2
αβdminus1) ge [d+ 1
2
]
36 59
Conjecture of algebraic independence of logarithms
of algebraic numbers
Denote by L the set of complex numbers λ for which eλ isalgebraic
L = logα α isin Qtimes
Hence L is a Q-vector subspace of CThe most important special case of Schanuelrsquos Conjecture is
Conjecture Let λ1 λn be Q-linearly independent elementsin L Then the numbers λ1 λn are algebraicallyindependent over Q
Not yet known that the transcendence degree is ge 2 Open problem Among all logarithms of algebraic numbersone at least is transcendental over Q(π)
37 59
Structural rank of a matrix
Let K be a field k a subfield and M a matrix with entries in KConsider the k-vector subspace E of K spanned by the entriesof M Choose an injective morphism ϕ of E into a k-vectorspace kX1 + middot middot middot+ kXn The image ϕ(M) of M is a matrixwhose entries are in the field k(X1 Xn) of rationalfractions Its rank does not depend on the choice of ϕ
This is the structural rank of M with respect to k
38 59
Example
LetM =
(bij + λij
)1leiled1lejle`
be a matrix with coefficients in Q + L Consider a basis of theQndashvector spanned by the entries and replace the elements inthis basis by unknowns This gives a new matrix M withcoefficients in a field of rational fractions the rank of which isthe structural rank of M (with respect to Q)
As a consequence of the conjecture of algebraic independenceof logarithms of algebraic numbers the rank of M should bethe same as the rank of its specialization M
39 59
Equivalence between the two conjectures
Following D Roy theconjecture on algebraicindependence of logarithms ofalgebraic numbers isequivalent to
Conjecture Any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L has a rank equal to its structuralrank
40 59
Any Polynomial is the Determinant of a Matrix
The proof of the equivalence uses the nice auxiliary result
For any P isin k[X1 Xn] there exists a square matrix withentries in the k-vector space k + kX1 + middot middot middot+ kXn whosedeterminant is P
41 59
Half of the Conjecture is solved
From a certain point of view half of the conjecture of algebraicindependence of logarithms of algebraic numbers is solved
Theorem [D Roy] The rank of any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L is at least half its structural rank
42 59
Reformulation by D Roy
Instead of taking logarithmsof algebraic numbers andlooking for the algebraicindependence relationsD Roy fixes a polynomial andlooks at the points withcoordinates logarithms ofalgebraic numbers on thecorresponding hypersurface
43 59
Reformulation by D Roy
Royrsquos reformulation of the conjecture of algebraicindependence of logarithms is
Conjecture For any algebraic subvariety V of Cn definedover the field Q of algebraic numbers the set V cap Ln is theunion of the sets E cap Ln where E ranges over the set ofvector subspaces of Cn which are defined over Q andcontained in V
Trivial Any element in E cap Ln where E is a vector subspaceof Cn defined over Q and contained in V belongs to V cap Ln
44 59
Example The Four Exponentials ConjectureTake for V the hypersurface of C4 defined by the equation
z1z4 = z2z3
The maximal Cndashvector subspaces of C4 defined over Q andcontained in V are the planes
az1 = bz2 bz4 = az3
and the planesaz1 = bz3 bz4 = az2
with (a b) isin Q2 (0 0)Hence Schanuelrsquos Conjecture implies that if λ1 λ2 λ3 λ4 arelogarithms of algebraic numbers satisfying
λ1λ4 = λ2λ3
then either λ1 λ2 are linearly dependent over Q or else λ1 λ3
are linearly dependent over Q45 59
Six Exponentials Theorem
and Four Exponentials Conjecture
A Selberg CL Siegel S Lang K Ramachandra
46 59
The Four Exponentials Conjecture
Conjecture (Th Schneider S Lang K Ramachandra) Ifx1 x2 are Qndashlinearly independent complex numbers and y1 y2
are Qndashlinearly independent complex numbers then one atleast of the four numbers
ex1y1 ex1y2 ex2y1 ex2y2
is transcendentalEquivalent statement Any 2times 2 matrix(
λ1 λ2
λ3 λ4
)with entries in L and with Qndashlinearly independent rows andQndashlinearly independent columns has maximal rank 2
47 59
How could we attack Schanuelrsquos Conjecture
Let x1 xn be Qndashlinearly independent complex numbersFollowing the transcendence methods of Hermite GelrsquofondSchneider one may start by introducing an auxiliaryfunction
F (z) = P (z ez)
where P isin Z[X0 X1] is a nonndashzero polynomial One considersthe derivatives (
d
dz
)kF = (DkP )(z ez)
at the pointsm1x1 + middot middot middot+mnxn
for various values of (m1 mn) isin Zn
48 59
Auxiliary functionLet D denote the derivation
D =part
partX0
+X1part
partX1
over the ring C[X0 X1] Recall that x1 xn are Qndashlinearlyindependent complex numbers Let α1 αn be nonndashzerocomplex numbersThe transcendence machinery produces a sequence (PN)Nge0
of polynomials with integer coefficients satisfying∣∣∣∣∣(DkPN)(nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1
49 59
Royrsquos approach to Schanuelrsquos Conjecture (1999)
If the number of equations we produce is too small such a setof relations does not contain any information The existenceof a sequence of nonndashtrivial polynomials (PN)Nge0 follows fromlinear algebra
On the other hand following D Roy one may expect that theexistence of such a sequence (PN)Nge0 producing sufficientlymany such equations will yield the desired conclusion
tr degQQ(x1 xn α1 αn
)ge n
50 59
Royrsquos Conjecture (1999)Let s0 s1 t0 t1 u positive real numbers satisfying
max1 t0 2t1 lt mins0 2s1and
maxs0 s1 + t1 lt u lt1
2(1 + t0 + t1)
Assume that for any sufficiently large positive integer N there exists a nonndashzero polynomial PN isin Z[X0 X1] withpartial degree le N t0 in X0 partial degree le N t1 in X1 andheight le eN which satisfies∣∣∣∣∣(DkPN)(
nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1 Then
tr degQQ(x1 xn α1 αn) ge n51 59
Royrsquos Theorem (1999)
Royrsquos conjecture is equivalent to Schanuelrsquos Conjecture
52 59
Equivalence between Schanuel and Roy
Let (x α) isin CtimesCtimes and let s0 s1 t0 t1 u be positive realnumbers satisfying the inequalities of Royrsquos Conjecture Thenthe following conditions are equivalent (a) The number αeminusx is a root of unity
(b) For any sufficiently large positive integer N there exists anonndashzero polynomial QN isin Z[X0 X1] with partial degreele N t0 in X0 partial degree le N t1 in X1 and heightH(QN) le eN such that∣∣(DkQN)(mx αm)
∣∣ le exp(minusNu)
for any km isin N with k le N s0 and m le N s1
53 59
Gelrsquofondrsquos transcendence criterion
Simple form Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le n and height le en such that
|Pn(ϑ)| le eminus7n2
for all n ge 1 then ϑ is algebraic and Pn(ϑ) = 0 for all n ge 1
First extension Replace the bound for the degree by le dnthe bound for the height by ebn and the bound for |Pn(ϑ)| byeminuscdnbn with some constant c gt 0 independent of n
Some mild conditions are required on the sequences (dn)nge1 and(bn)nge1
54 59
Transcendence criterion with multiplicitiesWith derivatives Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le dn and height le ebn such that∣∣P (j)
n (ϑ)∣∣ le eminuscdnbntn
for all j in the range 0 le j lt tn and all n ge 1 then ϑ isalgebraicDue to M Laurent and D Roy applications to algebraicindependence
55 59
Criterion with several points
Goal Given a sequence of complex numbers (ϑi)ige1 if thereexists a sequence (Pn)nge1 of nonndashzero polynomials in Z[X]with Pn of degree le dn and height le ebn such that∣∣P (j)
n (ϑi)∣∣ le eminuscdnbntnsn
for 0 le j lt tn 1 le i le sn and all n ge 1 then the numbers ϑiare algebraic
D Roy Not true in general but true in some special caseswith a structure on the sequence (ϑi)ige1Combines the elimination arguments used for criteria ofalgebraic independence and for zero estimates
56 59
Small value estimates for the additive group
D Roy Small value estimates for the additive group Intern JNumber Theory to appear
Let ξ be a transcendental complex number let β σ τ and νbe non-negative real numbers let n0 be a positive integer andlet (Pn)ngen0 be a sequence of nonndashzero polynomials in Z[T ]satisfying deg(Pn) le n and H(Pn) le exp(nβ) for eachn ge n0 Suppose that β gt 1 (34)σ + τ lt 1 andν gt 1 + β minus (34)σ minus τ Then for infinitely many n we have
max|P [j]n (iξ)| 0 le i le nσ 0 le j le nτ
gt exp(minusnν)
57 59
Small value estimates for the multiplicative group
D Roy Small value estimates for the multiplicative groupActa Arith to appear
Let ξ1 ξm be multiplicatively independent complexnumbers in a field of transcendence degree 1 Under suitableassumptions on the parameters β σ τ ν for infinitely manypositive integers n there exists no nonndashzero polynomialP isin Z[T ] satisfying deg(P ) le n H(P ) le exp(nβ) and
max|P [j](ξi11 middot middot middot ξimm )| 0 le i1 im le nσ 0 le j le nτ
gt exp(minusnν)
58 59
Heidelberg Colloquium organized by the Institute for Mathematics
httpwwwmathiuni-heidelbergdevortraghtmls=1 January 22 2009
Transcendental Number TheorySchanuelrsquos Conjecture
Michel Waldschmidt
Institut de Mathematiques de Jussieu amp Paris VIhttpwwwmathjussieufrsimmiw
59 59
Origin of Schanuelrsquos ConjectureCourse given by Serge Lang (1927ndash2005) at Yale in the 60rsquos
S Lang ndash Introduction to transcendental numbersAddison-Wesley 1966
also attended by M Nagata (1927ndash2008)(14th Problem of Hilbert)
Nagatarsquos Conjecture solved by E Bombieri
17 59
Schanuelrsquos Conjecture
Let x1 xn be Q-linearly independent complex numbersThen
tr degQQ(x1 xn e
x1 exn)ge n
Remark For almost all tuples (with respect to the Lebesguemeasure) the transcendence degree is 2n
18 59
AO Gelrsquofond CRAS 1934
19 59
Statement by Gelrsquofond (1934)
Let β1 βn be Q-linearly independent algebraic numbersand let logα1 logαm be Q-linearly independentlogarithms of algebraic numbers Then the numbers
eβ1 eβn logα1 logαm
are algebraically independent over Q
20 59
Further statement by Gelrsquofond
Let β1 βn be algebraic numbers with β1 6= 0 and letα1 αm be algebraic numbers with α1 6= 0 1 α2 6= 0 1αi 6= 0 Then the numbers
eβ1eβ2e βnminus1e
βn
and αα αm
21
are transcendental and there is no nontrivial algebraic relationbetween such numbers
Remark The condition on α2 should be that it is irrational
21 59
Easy consequence of Schanuelrsquos Conjecture
According to Schanuelrsquos Conjecture the following numbers arealgebraically independent
e+ π eπ πe ee ee2
eee
ππ ππ2
πππ
log π log(log 2) π log 2 (log 2)(log 3) 2log 2 (log 2)log 3
Proof This is an easy exercise
22 59
Langrsquos exercise
Define E0 = Q Inductivelyfor n ge 1 define En as thealgebraic closure of the fieldgenerated over Enminus1 by thenumbers exp(x) = ex wherex ranges over Enminus1 Let E bethe union of En n ge 0Then Schanuelrsquos Conjectureimplies that the number πdoes not belong to E
More precisely Schanuelrsquos Conjecture implies that thenumbers π log π log log π log log log π are algebraicallyindependent over E
23 59
A variant of Langrsquos exercise
Define L0 = Q Inductively for n ge 1 define Ln as thealgebraic closure of the field generated over Lnminus1 by thenumbers y where y ranges over the set of complex numberssuch that ey isin Lnminus1 Let L be the union of Ln n ge 0Then Schanuelrsquos Conjecture implies that the number e doesnot belong to L
More precisely Schanuelrsquos Conjecture implies that the
numbers e ee eee ee
ee
are algebraically independent over L
24 59
Arizona Winter School AWS2008 Tucson
Theorem [Jonathan Bober Chuangxun Cheng Brian DietelMathilde Herblot Jingjing Huang Holly Krieger DiegoMarques Jonathan Mason Martin Mereb and Robert Wilson]Schanuelrsquos Conjecture implies that the fields E and L arelinearly disjoint over Q
Definition Given a field extension FK and twosubextensions F1 F2 sube F we say F1 F2 are linearly disjointover K when the following holds Any set x1 xn sube F1
of Kndash linearly independent elements is linearly independentover F2
Reference arXiv08043550 [mathNT] 2008
25 59
Formal analogs
WD Brownawell(was a student of Schanuel)
J Axrsquos Theorem (1968) Version of SchanuelrsquosConjecture for power seriesover C(and R Coleman for powerseries over Q)
Work by WD Brownawelland K Kubota on the ellipticanalog of Axrsquos Theorem
26 59
Conjectures by A Grothendieck and Y Andre
Generalized Conjecture onPeriods by Grothendieck Dimension of theMumfordndashTate group of asmooth projective variety
Generalization by Y Andre tomotives
Case of 1ndashmotives Elliptico-Toric Conjecture ofC Bertolin
27 59
Ubiquity of Schanuelrsquos Conjecture
Other contexts pndashadic numbers Leopoldtrsquos Conjecture onthe pndashadic rank of the units of an algebraic number fieldNon-vanishing of RegulatorsNonndashdegenerescence of heightsConjecture of B Mazur on rational pointsDiophantine approximation on tori
Dipendra Prasad Gopal Prasad
28 59
Methods from logicEhud Hrushovski Boris Zilber Jonathan Kirby
Calculus of rdquopredimension functionsrdquo (E Hrushovski)
Zilberrsquos construction of a rdquopseudoexponentiationrdquo
Also A Macintyre DE Marker G Terzo AJ WilkieD Bertrand
29 59
Methods from logic Model theory
Exponential algebraicity in exponential fieldsby
Jonathan Kirby
The dimension of the exponential algebraic closure operator inan exponential field satisfies a weak Schanuel property
A corollary is that there are at most countably many essentialcounterexamples to Schanuelrsquos conjecture
arXiv 08104285v2
30 59
Known
LindemannndashWeierstraszlig Theorem = case where x1 xn arealgebraic
Let β1 βn be algebraic numbers which are linearlyindependent over Q Then the numbers eβ1 eβn arealgebraically independent over Q
31 59
Problem of Gelrsquofond and SchneiderRaised by AO Gelrsquofond in 1948 and Th Schneider in 1952
Conjecture If α is an algebraic number α 6= 0 α 6= 1 and ifβ is an irrational algebraic number of degree d then the dminus 1numbers
αβ αβ2
αβdminus1
are algebraically independent
Special case of Schanuelrsquos Conjecture Take xi = βiminus1 logαn = d so that
x1 xn e
x1 exn
islogα β logα βdminus1 logα α αβ αβ
dminus1
The conclusion of Schanuelrsquos Conjecture is
tr degQQ(logα αβ αβ
2
αβdminus1)
= d
32 59
Algebraic independence method of Gelrsquofond
AO Gelrsquofond (1948)
The two numbers 23radic2 and
23radic4 are algebraically
independentMore generally if α is analgebraic number α 6= 0α 6= 1 and if β is a algebraicnumber of degree d ge 3 thentwo at least of the numbers
αβ αβ2
αβdminus1
are algebraically independent
33 59
Tools
Transcendence criterion Replaces Liouvillersquos inequality intranscendence proofsLiouville A nonndashzero rational integer n isin Z satisfies |n| ge 1Gelrsquofond Needs to give a lower bound for |P (θ)| withP isin Z[X] 0 when θ is transcendental
Zero estimate for exponential polynomials C Hermite P Turan K Mahler R Tijdeman
Small transcendence degree AO Gelrsquofond AA Smelev R Tijdeman WD Brownawell
34 59
Large transcendence degree
GV Chudnovsky (1976)Among the numbers
αβ αβ2
αβdminus1
at least [log2 d] arealgebraically independent
GV Chudnovsky ndash On the path to Schanuelrsquosconjecture Algebraic curves close to a pointI General theory of colored sequencesII Fields of finite transcendence type and coloredsequences ResultantsStudia Sci Math Hungar 12 (1977) 125ndash157 (1980)
35 59
Partial result on the problem of Gelrsquofond and
Schneider
AO Gelrsquofond GV Chudnovskii P Philippon YuV Nesterenko
G Diaz If α is an algebraic number α 6= 0 α 6= 1 and if β isan irrational algebraic number of degree d then
tr degQQ(αβ αβ
2
αβdminus1) ge [d+ 1
2
]
36 59
Conjecture of algebraic independence of logarithms
of algebraic numbers
Denote by L the set of complex numbers λ for which eλ isalgebraic
L = logα α isin Qtimes
Hence L is a Q-vector subspace of CThe most important special case of Schanuelrsquos Conjecture is
Conjecture Let λ1 λn be Q-linearly independent elementsin L Then the numbers λ1 λn are algebraicallyindependent over Q
Not yet known that the transcendence degree is ge 2 Open problem Among all logarithms of algebraic numbersone at least is transcendental over Q(π)
37 59
Structural rank of a matrix
Let K be a field k a subfield and M a matrix with entries in KConsider the k-vector subspace E of K spanned by the entriesof M Choose an injective morphism ϕ of E into a k-vectorspace kX1 + middot middot middot+ kXn The image ϕ(M) of M is a matrixwhose entries are in the field k(X1 Xn) of rationalfractions Its rank does not depend on the choice of ϕ
This is the structural rank of M with respect to k
38 59
Example
LetM =
(bij + λij
)1leiled1lejle`
be a matrix with coefficients in Q + L Consider a basis of theQndashvector spanned by the entries and replace the elements inthis basis by unknowns This gives a new matrix M withcoefficients in a field of rational fractions the rank of which isthe structural rank of M (with respect to Q)
As a consequence of the conjecture of algebraic independenceof logarithms of algebraic numbers the rank of M should bethe same as the rank of its specialization M
39 59
Equivalence between the two conjectures
Following D Roy theconjecture on algebraicindependence of logarithms ofalgebraic numbers isequivalent to
Conjecture Any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L has a rank equal to its structuralrank
40 59
Any Polynomial is the Determinant of a Matrix
The proof of the equivalence uses the nice auxiliary result
For any P isin k[X1 Xn] there exists a square matrix withentries in the k-vector space k + kX1 + middot middot middot+ kXn whosedeterminant is P
41 59
Half of the Conjecture is solved
From a certain point of view half of the conjecture of algebraicindependence of logarithms of algebraic numbers is solved
Theorem [D Roy] The rank of any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L is at least half its structural rank
42 59
Reformulation by D Roy
Instead of taking logarithmsof algebraic numbers andlooking for the algebraicindependence relationsD Roy fixes a polynomial andlooks at the points withcoordinates logarithms ofalgebraic numbers on thecorresponding hypersurface
43 59
Reformulation by D Roy
Royrsquos reformulation of the conjecture of algebraicindependence of logarithms is
Conjecture For any algebraic subvariety V of Cn definedover the field Q of algebraic numbers the set V cap Ln is theunion of the sets E cap Ln where E ranges over the set ofvector subspaces of Cn which are defined over Q andcontained in V
Trivial Any element in E cap Ln where E is a vector subspaceof Cn defined over Q and contained in V belongs to V cap Ln
44 59
Example The Four Exponentials ConjectureTake for V the hypersurface of C4 defined by the equation
z1z4 = z2z3
The maximal Cndashvector subspaces of C4 defined over Q andcontained in V are the planes
az1 = bz2 bz4 = az3
and the planesaz1 = bz3 bz4 = az2
with (a b) isin Q2 (0 0)Hence Schanuelrsquos Conjecture implies that if λ1 λ2 λ3 λ4 arelogarithms of algebraic numbers satisfying
λ1λ4 = λ2λ3
then either λ1 λ2 are linearly dependent over Q or else λ1 λ3
are linearly dependent over Q45 59
Six Exponentials Theorem
and Four Exponentials Conjecture
A Selberg CL Siegel S Lang K Ramachandra
46 59
The Four Exponentials Conjecture
Conjecture (Th Schneider S Lang K Ramachandra) Ifx1 x2 are Qndashlinearly independent complex numbers and y1 y2
are Qndashlinearly independent complex numbers then one atleast of the four numbers
ex1y1 ex1y2 ex2y1 ex2y2
is transcendentalEquivalent statement Any 2times 2 matrix(
λ1 λ2
λ3 λ4
)with entries in L and with Qndashlinearly independent rows andQndashlinearly independent columns has maximal rank 2
47 59
How could we attack Schanuelrsquos Conjecture
Let x1 xn be Qndashlinearly independent complex numbersFollowing the transcendence methods of Hermite GelrsquofondSchneider one may start by introducing an auxiliaryfunction
F (z) = P (z ez)
where P isin Z[X0 X1] is a nonndashzero polynomial One considersthe derivatives (
d
dz
)kF = (DkP )(z ez)
at the pointsm1x1 + middot middot middot+mnxn
for various values of (m1 mn) isin Zn
48 59
Auxiliary functionLet D denote the derivation
D =part
partX0
+X1part
partX1
over the ring C[X0 X1] Recall that x1 xn are Qndashlinearlyindependent complex numbers Let α1 αn be nonndashzerocomplex numbersThe transcendence machinery produces a sequence (PN)Nge0
of polynomials with integer coefficients satisfying∣∣∣∣∣(DkPN)(nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1
49 59
Royrsquos approach to Schanuelrsquos Conjecture (1999)
If the number of equations we produce is too small such a setof relations does not contain any information The existenceof a sequence of nonndashtrivial polynomials (PN)Nge0 follows fromlinear algebra
On the other hand following D Roy one may expect that theexistence of such a sequence (PN)Nge0 producing sufficientlymany such equations will yield the desired conclusion
tr degQQ(x1 xn α1 αn
)ge n
50 59
Royrsquos Conjecture (1999)Let s0 s1 t0 t1 u positive real numbers satisfying
max1 t0 2t1 lt mins0 2s1and
maxs0 s1 + t1 lt u lt1
2(1 + t0 + t1)
Assume that for any sufficiently large positive integer N there exists a nonndashzero polynomial PN isin Z[X0 X1] withpartial degree le N t0 in X0 partial degree le N t1 in X1 andheight le eN which satisfies∣∣∣∣∣(DkPN)(
nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1 Then
tr degQQ(x1 xn α1 αn) ge n51 59
Royrsquos Theorem (1999)
Royrsquos conjecture is equivalent to Schanuelrsquos Conjecture
52 59
Equivalence between Schanuel and Roy
Let (x α) isin CtimesCtimes and let s0 s1 t0 t1 u be positive realnumbers satisfying the inequalities of Royrsquos Conjecture Thenthe following conditions are equivalent (a) The number αeminusx is a root of unity
(b) For any sufficiently large positive integer N there exists anonndashzero polynomial QN isin Z[X0 X1] with partial degreele N t0 in X0 partial degree le N t1 in X1 and heightH(QN) le eN such that∣∣(DkQN)(mx αm)
∣∣ le exp(minusNu)
for any km isin N with k le N s0 and m le N s1
53 59
Gelrsquofondrsquos transcendence criterion
Simple form Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le n and height le en such that
|Pn(ϑ)| le eminus7n2
for all n ge 1 then ϑ is algebraic and Pn(ϑ) = 0 for all n ge 1
First extension Replace the bound for the degree by le dnthe bound for the height by ebn and the bound for |Pn(ϑ)| byeminuscdnbn with some constant c gt 0 independent of n
Some mild conditions are required on the sequences (dn)nge1 and(bn)nge1
54 59
Transcendence criterion with multiplicitiesWith derivatives Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le dn and height le ebn such that∣∣P (j)
n (ϑ)∣∣ le eminuscdnbntn
for all j in the range 0 le j lt tn and all n ge 1 then ϑ isalgebraicDue to M Laurent and D Roy applications to algebraicindependence
55 59
Criterion with several points
Goal Given a sequence of complex numbers (ϑi)ige1 if thereexists a sequence (Pn)nge1 of nonndashzero polynomials in Z[X]with Pn of degree le dn and height le ebn such that∣∣P (j)
n (ϑi)∣∣ le eminuscdnbntnsn
for 0 le j lt tn 1 le i le sn and all n ge 1 then the numbers ϑiare algebraic
D Roy Not true in general but true in some special caseswith a structure on the sequence (ϑi)ige1Combines the elimination arguments used for criteria ofalgebraic independence and for zero estimates
56 59
Small value estimates for the additive group
D Roy Small value estimates for the additive group Intern JNumber Theory to appear
Let ξ be a transcendental complex number let β σ τ and νbe non-negative real numbers let n0 be a positive integer andlet (Pn)ngen0 be a sequence of nonndashzero polynomials in Z[T ]satisfying deg(Pn) le n and H(Pn) le exp(nβ) for eachn ge n0 Suppose that β gt 1 (34)σ + τ lt 1 andν gt 1 + β minus (34)σ minus τ Then for infinitely many n we have
max|P [j]n (iξ)| 0 le i le nσ 0 le j le nτ
gt exp(minusnν)
57 59
Small value estimates for the multiplicative group
D Roy Small value estimates for the multiplicative groupActa Arith to appear
Let ξ1 ξm be multiplicatively independent complexnumbers in a field of transcendence degree 1 Under suitableassumptions on the parameters β σ τ ν for infinitely manypositive integers n there exists no nonndashzero polynomialP isin Z[T ] satisfying deg(P ) le n H(P ) le exp(nβ) and
max|P [j](ξi11 middot middot middot ξimm )| 0 le i1 im le nσ 0 le j le nτ
gt exp(minusnν)
58 59
Heidelberg Colloquium organized by the Institute for Mathematics
httpwwwmathiuni-heidelbergdevortraghtmls=1 January 22 2009
Transcendental Number TheorySchanuelrsquos Conjecture
Michel Waldschmidt
Institut de Mathematiques de Jussieu amp Paris VIhttpwwwmathjussieufrsimmiw
59 59
Schanuelrsquos Conjecture
Let x1 xn be Q-linearly independent complex numbersThen
tr degQQ(x1 xn e
x1 exn)ge n
Remark For almost all tuples (with respect to the Lebesguemeasure) the transcendence degree is 2n
18 59
AO Gelrsquofond CRAS 1934
19 59
Statement by Gelrsquofond (1934)
Let β1 βn be Q-linearly independent algebraic numbersand let logα1 logαm be Q-linearly independentlogarithms of algebraic numbers Then the numbers
eβ1 eβn logα1 logαm
are algebraically independent over Q
20 59
Further statement by Gelrsquofond
Let β1 βn be algebraic numbers with β1 6= 0 and letα1 αm be algebraic numbers with α1 6= 0 1 α2 6= 0 1αi 6= 0 Then the numbers
eβ1eβ2e βnminus1e
βn
and αα αm
21
are transcendental and there is no nontrivial algebraic relationbetween such numbers
Remark The condition on α2 should be that it is irrational
21 59
Easy consequence of Schanuelrsquos Conjecture
According to Schanuelrsquos Conjecture the following numbers arealgebraically independent
e+ π eπ πe ee ee2
eee
ππ ππ2
πππ
log π log(log 2) π log 2 (log 2)(log 3) 2log 2 (log 2)log 3
Proof This is an easy exercise
22 59
Langrsquos exercise
Define E0 = Q Inductivelyfor n ge 1 define En as thealgebraic closure of the fieldgenerated over Enminus1 by thenumbers exp(x) = ex wherex ranges over Enminus1 Let E bethe union of En n ge 0Then Schanuelrsquos Conjectureimplies that the number πdoes not belong to E
More precisely Schanuelrsquos Conjecture implies that thenumbers π log π log log π log log log π are algebraicallyindependent over E
23 59
A variant of Langrsquos exercise
Define L0 = Q Inductively for n ge 1 define Ln as thealgebraic closure of the field generated over Lnminus1 by thenumbers y where y ranges over the set of complex numberssuch that ey isin Lnminus1 Let L be the union of Ln n ge 0Then Schanuelrsquos Conjecture implies that the number e doesnot belong to L
More precisely Schanuelrsquos Conjecture implies that the
numbers e ee eee ee
ee
are algebraically independent over L
24 59
Arizona Winter School AWS2008 Tucson
Theorem [Jonathan Bober Chuangxun Cheng Brian DietelMathilde Herblot Jingjing Huang Holly Krieger DiegoMarques Jonathan Mason Martin Mereb and Robert Wilson]Schanuelrsquos Conjecture implies that the fields E and L arelinearly disjoint over Q
Definition Given a field extension FK and twosubextensions F1 F2 sube F we say F1 F2 are linearly disjointover K when the following holds Any set x1 xn sube F1
of Kndash linearly independent elements is linearly independentover F2
Reference arXiv08043550 [mathNT] 2008
25 59
Formal analogs
WD Brownawell(was a student of Schanuel)
J Axrsquos Theorem (1968) Version of SchanuelrsquosConjecture for power seriesover C(and R Coleman for powerseries over Q)
Work by WD Brownawelland K Kubota on the ellipticanalog of Axrsquos Theorem
26 59
Conjectures by A Grothendieck and Y Andre
Generalized Conjecture onPeriods by Grothendieck Dimension of theMumfordndashTate group of asmooth projective variety
Generalization by Y Andre tomotives
Case of 1ndashmotives Elliptico-Toric Conjecture ofC Bertolin
27 59
Ubiquity of Schanuelrsquos Conjecture
Other contexts pndashadic numbers Leopoldtrsquos Conjecture onthe pndashadic rank of the units of an algebraic number fieldNon-vanishing of RegulatorsNonndashdegenerescence of heightsConjecture of B Mazur on rational pointsDiophantine approximation on tori
Dipendra Prasad Gopal Prasad
28 59
Methods from logicEhud Hrushovski Boris Zilber Jonathan Kirby
Calculus of rdquopredimension functionsrdquo (E Hrushovski)
Zilberrsquos construction of a rdquopseudoexponentiationrdquo
Also A Macintyre DE Marker G Terzo AJ WilkieD Bertrand
29 59
Methods from logic Model theory
Exponential algebraicity in exponential fieldsby
Jonathan Kirby
The dimension of the exponential algebraic closure operator inan exponential field satisfies a weak Schanuel property
A corollary is that there are at most countably many essentialcounterexamples to Schanuelrsquos conjecture
arXiv 08104285v2
30 59
Known
LindemannndashWeierstraszlig Theorem = case where x1 xn arealgebraic
Let β1 βn be algebraic numbers which are linearlyindependent over Q Then the numbers eβ1 eβn arealgebraically independent over Q
31 59
Problem of Gelrsquofond and SchneiderRaised by AO Gelrsquofond in 1948 and Th Schneider in 1952
Conjecture If α is an algebraic number α 6= 0 α 6= 1 and ifβ is an irrational algebraic number of degree d then the dminus 1numbers
αβ αβ2
αβdminus1
are algebraically independent
Special case of Schanuelrsquos Conjecture Take xi = βiminus1 logαn = d so that
x1 xn e
x1 exn
islogα β logα βdminus1 logα α αβ αβ
dminus1
The conclusion of Schanuelrsquos Conjecture is
tr degQQ(logα αβ αβ
2
αβdminus1)
= d
32 59
Algebraic independence method of Gelrsquofond
AO Gelrsquofond (1948)
The two numbers 23radic2 and
23radic4 are algebraically
independentMore generally if α is analgebraic number α 6= 0α 6= 1 and if β is a algebraicnumber of degree d ge 3 thentwo at least of the numbers
αβ αβ2
αβdminus1
are algebraically independent
33 59
Tools
Transcendence criterion Replaces Liouvillersquos inequality intranscendence proofsLiouville A nonndashzero rational integer n isin Z satisfies |n| ge 1Gelrsquofond Needs to give a lower bound for |P (θ)| withP isin Z[X] 0 when θ is transcendental
Zero estimate for exponential polynomials C Hermite P Turan K Mahler R Tijdeman
Small transcendence degree AO Gelrsquofond AA Smelev R Tijdeman WD Brownawell
34 59
Large transcendence degree
GV Chudnovsky (1976)Among the numbers
αβ αβ2
αβdminus1
at least [log2 d] arealgebraically independent
GV Chudnovsky ndash On the path to Schanuelrsquosconjecture Algebraic curves close to a pointI General theory of colored sequencesII Fields of finite transcendence type and coloredsequences ResultantsStudia Sci Math Hungar 12 (1977) 125ndash157 (1980)
35 59
Partial result on the problem of Gelrsquofond and
Schneider
AO Gelrsquofond GV Chudnovskii P Philippon YuV Nesterenko
G Diaz If α is an algebraic number α 6= 0 α 6= 1 and if β isan irrational algebraic number of degree d then
tr degQQ(αβ αβ
2
αβdminus1) ge [d+ 1
2
]
36 59
Conjecture of algebraic independence of logarithms
of algebraic numbers
Denote by L the set of complex numbers λ for which eλ isalgebraic
L = logα α isin Qtimes
Hence L is a Q-vector subspace of CThe most important special case of Schanuelrsquos Conjecture is
Conjecture Let λ1 λn be Q-linearly independent elementsin L Then the numbers λ1 λn are algebraicallyindependent over Q
Not yet known that the transcendence degree is ge 2 Open problem Among all logarithms of algebraic numbersone at least is transcendental over Q(π)
37 59
Structural rank of a matrix
Let K be a field k a subfield and M a matrix with entries in KConsider the k-vector subspace E of K spanned by the entriesof M Choose an injective morphism ϕ of E into a k-vectorspace kX1 + middot middot middot+ kXn The image ϕ(M) of M is a matrixwhose entries are in the field k(X1 Xn) of rationalfractions Its rank does not depend on the choice of ϕ
This is the structural rank of M with respect to k
38 59
Example
LetM =
(bij + λij
)1leiled1lejle`
be a matrix with coefficients in Q + L Consider a basis of theQndashvector spanned by the entries and replace the elements inthis basis by unknowns This gives a new matrix M withcoefficients in a field of rational fractions the rank of which isthe structural rank of M (with respect to Q)
As a consequence of the conjecture of algebraic independenceof logarithms of algebraic numbers the rank of M should bethe same as the rank of its specialization M
39 59
Equivalence between the two conjectures
Following D Roy theconjecture on algebraicindependence of logarithms ofalgebraic numbers isequivalent to
Conjecture Any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L has a rank equal to its structuralrank
40 59
Any Polynomial is the Determinant of a Matrix
The proof of the equivalence uses the nice auxiliary result
For any P isin k[X1 Xn] there exists a square matrix withentries in the k-vector space k + kX1 + middot middot middot+ kXn whosedeterminant is P
41 59
Half of the Conjecture is solved
From a certain point of view half of the conjecture of algebraicindependence of logarithms of algebraic numbers is solved
Theorem [D Roy] The rank of any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L is at least half its structural rank
42 59
Reformulation by D Roy
Instead of taking logarithmsof algebraic numbers andlooking for the algebraicindependence relationsD Roy fixes a polynomial andlooks at the points withcoordinates logarithms ofalgebraic numbers on thecorresponding hypersurface
43 59
Reformulation by D Roy
Royrsquos reformulation of the conjecture of algebraicindependence of logarithms is
Conjecture For any algebraic subvariety V of Cn definedover the field Q of algebraic numbers the set V cap Ln is theunion of the sets E cap Ln where E ranges over the set ofvector subspaces of Cn which are defined over Q andcontained in V
Trivial Any element in E cap Ln where E is a vector subspaceof Cn defined over Q and contained in V belongs to V cap Ln
44 59
Example The Four Exponentials ConjectureTake for V the hypersurface of C4 defined by the equation
z1z4 = z2z3
The maximal Cndashvector subspaces of C4 defined over Q andcontained in V are the planes
az1 = bz2 bz4 = az3
and the planesaz1 = bz3 bz4 = az2
with (a b) isin Q2 (0 0)Hence Schanuelrsquos Conjecture implies that if λ1 λ2 λ3 λ4 arelogarithms of algebraic numbers satisfying
λ1λ4 = λ2λ3
then either λ1 λ2 are linearly dependent over Q or else λ1 λ3
are linearly dependent over Q45 59
Six Exponentials Theorem
and Four Exponentials Conjecture
A Selberg CL Siegel S Lang K Ramachandra
46 59
The Four Exponentials Conjecture
Conjecture (Th Schneider S Lang K Ramachandra) Ifx1 x2 are Qndashlinearly independent complex numbers and y1 y2
are Qndashlinearly independent complex numbers then one atleast of the four numbers
ex1y1 ex1y2 ex2y1 ex2y2
is transcendentalEquivalent statement Any 2times 2 matrix(
λ1 λ2
λ3 λ4
)with entries in L and with Qndashlinearly independent rows andQndashlinearly independent columns has maximal rank 2
47 59
How could we attack Schanuelrsquos Conjecture
Let x1 xn be Qndashlinearly independent complex numbersFollowing the transcendence methods of Hermite GelrsquofondSchneider one may start by introducing an auxiliaryfunction
F (z) = P (z ez)
where P isin Z[X0 X1] is a nonndashzero polynomial One considersthe derivatives (
d
dz
)kF = (DkP )(z ez)
at the pointsm1x1 + middot middot middot+mnxn
for various values of (m1 mn) isin Zn
48 59
Auxiliary functionLet D denote the derivation
D =part
partX0
+X1part
partX1
over the ring C[X0 X1] Recall that x1 xn are Qndashlinearlyindependent complex numbers Let α1 αn be nonndashzerocomplex numbersThe transcendence machinery produces a sequence (PN)Nge0
of polynomials with integer coefficients satisfying∣∣∣∣∣(DkPN)(nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1
49 59
Royrsquos approach to Schanuelrsquos Conjecture (1999)
If the number of equations we produce is too small such a setof relations does not contain any information The existenceof a sequence of nonndashtrivial polynomials (PN)Nge0 follows fromlinear algebra
On the other hand following D Roy one may expect that theexistence of such a sequence (PN)Nge0 producing sufficientlymany such equations will yield the desired conclusion
tr degQQ(x1 xn α1 αn
)ge n
50 59
Royrsquos Conjecture (1999)Let s0 s1 t0 t1 u positive real numbers satisfying
max1 t0 2t1 lt mins0 2s1and
maxs0 s1 + t1 lt u lt1
2(1 + t0 + t1)
Assume that for any sufficiently large positive integer N there exists a nonndashzero polynomial PN isin Z[X0 X1] withpartial degree le N t0 in X0 partial degree le N t1 in X1 andheight le eN which satisfies∣∣∣∣∣(DkPN)(
nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1 Then
tr degQQ(x1 xn α1 αn) ge n51 59
Royrsquos Theorem (1999)
Royrsquos conjecture is equivalent to Schanuelrsquos Conjecture
52 59
Equivalence between Schanuel and Roy
Let (x α) isin CtimesCtimes and let s0 s1 t0 t1 u be positive realnumbers satisfying the inequalities of Royrsquos Conjecture Thenthe following conditions are equivalent (a) The number αeminusx is a root of unity
(b) For any sufficiently large positive integer N there exists anonndashzero polynomial QN isin Z[X0 X1] with partial degreele N t0 in X0 partial degree le N t1 in X1 and heightH(QN) le eN such that∣∣(DkQN)(mx αm)
∣∣ le exp(minusNu)
for any km isin N with k le N s0 and m le N s1
53 59
Gelrsquofondrsquos transcendence criterion
Simple form Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le n and height le en such that
|Pn(ϑ)| le eminus7n2
for all n ge 1 then ϑ is algebraic and Pn(ϑ) = 0 for all n ge 1
First extension Replace the bound for the degree by le dnthe bound for the height by ebn and the bound for |Pn(ϑ)| byeminuscdnbn with some constant c gt 0 independent of n
Some mild conditions are required on the sequences (dn)nge1 and(bn)nge1
54 59
Transcendence criterion with multiplicitiesWith derivatives Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le dn and height le ebn such that∣∣P (j)
n (ϑ)∣∣ le eminuscdnbntn
for all j in the range 0 le j lt tn and all n ge 1 then ϑ isalgebraicDue to M Laurent and D Roy applications to algebraicindependence
55 59
Criterion with several points
Goal Given a sequence of complex numbers (ϑi)ige1 if thereexists a sequence (Pn)nge1 of nonndashzero polynomials in Z[X]with Pn of degree le dn and height le ebn such that∣∣P (j)
n (ϑi)∣∣ le eminuscdnbntnsn
for 0 le j lt tn 1 le i le sn and all n ge 1 then the numbers ϑiare algebraic
D Roy Not true in general but true in some special caseswith a structure on the sequence (ϑi)ige1Combines the elimination arguments used for criteria ofalgebraic independence and for zero estimates
56 59
Small value estimates for the additive group
D Roy Small value estimates for the additive group Intern JNumber Theory to appear
Let ξ be a transcendental complex number let β σ τ and νbe non-negative real numbers let n0 be a positive integer andlet (Pn)ngen0 be a sequence of nonndashzero polynomials in Z[T ]satisfying deg(Pn) le n and H(Pn) le exp(nβ) for eachn ge n0 Suppose that β gt 1 (34)σ + τ lt 1 andν gt 1 + β minus (34)σ minus τ Then for infinitely many n we have
max|P [j]n (iξ)| 0 le i le nσ 0 le j le nτ
gt exp(minusnν)
57 59
Small value estimates for the multiplicative group
D Roy Small value estimates for the multiplicative groupActa Arith to appear
Let ξ1 ξm be multiplicatively independent complexnumbers in a field of transcendence degree 1 Under suitableassumptions on the parameters β σ τ ν for infinitely manypositive integers n there exists no nonndashzero polynomialP isin Z[T ] satisfying deg(P ) le n H(P ) le exp(nβ) and
max|P [j](ξi11 middot middot middot ξimm )| 0 le i1 im le nσ 0 le j le nτ
gt exp(minusnν)
58 59
Heidelberg Colloquium organized by the Institute for Mathematics
httpwwwmathiuni-heidelbergdevortraghtmls=1 January 22 2009
Transcendental Number TheorySchanuelrsquos Conjecture
Michel Waldschmidt
Institut de Mathematiques de Jussieu amp Paris VIhttpwwwmathjussieufrsimmiw
59 59
AO Gelrsquofond CRAS 1934
19 59
Statement by Gelrsquofond (1934)
Let β1 βn be Q-linearly independent algebraic numbersand let logα1 logαm be Q-linearly independentlogarithms of algebraic numbers Then the numbers
eβ1 eβn logα1 logαm
are algebraically independent over Q
20 59
Further statement by Gelrsquofond
Let β1 βn be algebraic numbers with β1 6= 0 and letα1 αm be algebraic numbers with α1 6= 0 1 α2 6= 0 1αi 6= 0 Then the numbers
eβ1eβ2e βnminus1e
βn
and αα αm
21
are transcendental and there is no nontrivial algebraic relationbetween such numbers
Remark The condition on α2 should be that it is irrational
21 59
Easy consequence of Schanuelrsquos Conjecture
According to Schanuelrsquos Conjecture the following numbers arealgebraically independent
e+ π eπ πe ee ee2
eee
ππ ππ2
πππ
log π log(log 2) π log 2 (log 2)(log 3) 2log 2 (log 2)log 3
Proof This is an easy exercise
22 59
Langrsquos exercise
Define E0 = Q Inductivelyfor n ge 1 define En as thealgebraic closure of the fieldgenerated over Enminus1 by thenumbers exp(x) = ex wherex ranges over Enminus1 Let E bethe union of En n ge 0Then Schanuelrsquos Conjectureimplies that the number πdoes not belong to E
More precisely Schanuelrsquos Conjecture implies that thenumbers π log π log log π log log log π are algebraicallyindependent over E
23 59
A variant of Langrsquos exercise
Define L0 = Q Inductively for n ge 1 define Ln as thealgebraic closure of the field generated over Lnminus1 by thenumbers y where y ranges over the set of complex numberssuch that ey isin Lnminus1 Let L be the union of Ln n ge 0Then Schanuelrsquos Conjecture implies that the number e doesnot belong to L
More precisely Schanuelrsquos Conjecture implies that the
numbers e ee eee ee
ee
are algebraically independent over L
24 59
Arizona Winter School AWS2008 Tucson
Theorem [Jonathan Bober Chuangxun Cheng Brian DietelMathilde Herblot Jingjing Huang Holly Krieger DiegoMarques Jonathan Mason Martin Mereb and Robert Wilson]Schanuelrsquos Conjecture implies that the fields E and L arelinearly disjoint over Q
Definition Given a field extension FK and twosubextensions F1 F2 sube F we say F1 F2 are linearly disjointover K when the following holds Any set x1 xn sube F1
of Kndash linearly independent elements is linearly independentover F2
Reference arXiv08043550 [mathNT] 2008
25 59
Formal analogs
WD Brownawell(was a student of Schanuel)
J Axrsquos Theorem (1968) Version of SchanuelrsquosConjecture for power seriesover C(and R Coleman for powerseries over Q)
Work by WD Brownawelland K Kubota on the ellipticanalog of Axrsquos Theorem
26 59
Conjectures by A Grothendieck and Y Andre
Generalized Conjecture onPeriods by Grothendieck Dimension of theMumfordndashTate group of asmooth projective variety
Generalization by Y Andre tomotives
Case of 1ndashmotives Elliptico-Toric Conjecture ofC Bertolin
27 59
Ubiquity of Schanuelrsquos Conjecture
Other contexts pndashadic numbers Leopoldtrsquos Conjecture onthe pndashadic rank of the units of an algebraic number fieldNon-vanishing of RegulatorsNonndashdegenerescence of heightsConjecture of B Mazur on rational pointsDiophantine approximation on tori
Dipendra Prasad Gopal Prasad
28 59
Methods from logicEhud Hrushovski Boris Zilber Jonathan Kirby
Calculus of rdquopredimension functionsrdquo (E Hrushovski)
Zilberrsquos construction of a rdquopseudoexponentiationrdquo
Also A Macintyre DE Marker G Terzo AJ WilkieD Bertrand
29 59
Methods from logic Model theory
Exponential algebraicity in exponential fieldsby
Jonathan Kirby
The dimension of the exponential algebraic closure operator inan exponential field satisfies a weak Schanuel property
A corollary is that there are at most countably many essentialcounterexamples to Schanuelrsquos conjecture
arXiv 08104285v2
30 59
Known
LindemannndashWeierstraszlig Theorem = case where x1 xn arealgebraic
Let β1 βn be algebraic numbers which are linearlyindependent over Q Then the numbers eβ1 eβn arealgebraically independent over Q
31 59
Problem of Gelrsquofond and SchneiderRaised by AO Gelrsquofond in 1948 and Th Schneider in 1952
Conjecture If α is an algebraic number α 6= 0 α 6= 1 and ifβ is an irrational algebraic number of degree d then the dminus 1numbers
αβ αβ2
αβdminus1
are algebraically independent
Special case of Schanuelrsquos Conjecture Take xi = βiminus1 logαn = d so that
x1 xn e
x1 exn
islogα β logα βdminus1 logα α αβ αβ
dminus1
The conclusion of Schanuelrsquos Conjecture is
tr degQQ(logα αβ αβ
2
αβdminus1)
= d
32 59
Algebraic independence method of Gelrsquofond
AO Gelrsquofond (1948)
The two numbers 23radic2 and
23radic4 are algebraically
independentMore generally if α is analgebraic number α 6= 0α 6= 1 and if β is a algebraicnumber of degree d ge 3 thentwo at least of the numbers
αβ αβ2
αβdminus1
are algebraically independent
33 59
Tools
Transcendence criterion Replaces Liouvillersquos inequality intranscendence proofsLiouville A nonndashzero rational integer n isin Z satisfies |n| ge 1Gelrsquofond Needs to give a lower bound for |P (θ)| withP isin Z[X] 0 when θ is transcendental
Zero estimate for exponential polynomials C Hermite P Turan K Mahler R Tijdeman
Small transcendence degree AO Gelrsquofond AA Smelev R Tijdeman WD Brownawell
34 59
Large transcendence degree
GV Chudnovsky (1976)Among the numbers
αβ αβ2
αβdminus1
at least [log2 d] arealgebraically independent
GV Chudnovsky ndash On the path to Schanuelrsquosconjecture Algebraic curves close to a pointI General theory of colored sequencesII Fields of finite transcendence type and coloredsequences ResultantsStudia Sci Math Hungar 12 (1977) 125ndash157 (1980)
35 59
Partial result on the problem of Gelrsquofond and
Schneider
AO Gelrsquofond GV Chudnovskii P Philippon YuV Nesterenko
G Diaz If α is an algebraic number α 6= 0 α 6= 1 and if β isan irrational algebraic number of degree d then
tr degQQ(αβ αβ
2
αβdminus1) ge [d+ 1
2
]
36 59
Conjecture of algebraic independence of logarithms
of algebraic numbers
Denote by L the set of complex numbers λ for which eλ isalgebraic
L = logα α isin Qtimes
Hence L is a Q-vector subspace of CThe most important special case of Schanuelrsquos Conjecture is
Conjecture Let λ1 λn be Q-linearly independent elementsin L Then the numbers λ1 λn are algebraicallyindependent over Q
Not yet known that the transcendence degree is ge 2 Open problem Among all logarithms of algebraic numbersone at least is transcendental over Q(π)
37 59
Structural rank of a matrix
Let K be a field k a subfield and M a matrix with entries in KConsider the k-vector subspace E of K spanned by the entriesof M Choose an injective morphism ϕ of E into a k-vectorspace kX1 + middot middot middot+ kXn The image ϕ(M) of M is a matrixwhose entries are in the field k(X1 Xn) of rationalfractions Its rank does not depend on the choice of ϕ
This is the structural rank of M with respect to k
38 59
Example
LetM =
(bij + λij
)1leiled1lejle`
be a matrix with coefficients in Q + L Consider a basis of theQndashvector spanned by the entries and replace the elements inthis basis by unknowns This gives a new matrix M withcoefficients in a field of rational fractions the rank of which isthe structural rank of M (with respect to Q)
As a consequence of the conjecture of algebraic independenceof logarithms of algebraic numbers the rank of M should bethe same as the rank of its specialization M
39 59
Equivalence between the two conjectures
Following D Roy theconjecture on algebraicindependence of logarithms ofalgebraic numbers isequivalent to
Conjecture Any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L has a rank equal to its structuralrank
40 59
Any Polynomial is the Determinant of a Matrix
The proof of the equivalence uses the nice auxiliary result
For any P isin k[X1 Xn] there exists a square matrix withentries in the k-vector space k + kX1 + middot middot middot+ kXn whosedeterminant is P
41 59
Half of the Conjecture is solved
From a certain point of view half of the conjecture of algebraicindependence of logarithms of algebraic numbers is solved
Theorem [D Roy] The rank of any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L is at least half its structural rank
42 59
Reformulation by D Roy
Instead of taking logarithmsof algebraic numbers andlooking for the algebraicindependence relationsD Roy fixes a polynomial andlooks at the points withcoordinates logarithms ofalgebraic numbers on thecorresponding hypersurface
43 59
Reformulation by D Roy
Royrsquos reformulation of the conjecture of algebraicindependence of logarithms is
Conjecture For any algebraic subvariety V of Cn definedover the field Q of algebraic numbers the set V cap Ln is theunion of the sets E cap Ln where E ranges over the set ofvector subspaces of Cn which are defined over Q andcontained in V
Trivial Any element in E cap Ln where E is a vector subspaceof Cn defined over Q and contained in V belongs to V cap Ln
44 59
Example The Four Exponentials ConjectureTake for V the hypersurface of C4 defined by the equation
z1z4 = z2z3
The maximal Cndashvector subspaces of C4 defined over Q andcontained in V are the planes
az1 = bz2 bz4 = az3
and the planesaz1 = bz3 bz4 = az2
with (a b) isin Q2 (0 0)Hence Schanuelrsquos Conjecture implies that if λ1 λ2 λ3 λ4 arelogarithms of algebraic numbers satisfying
λ1λ4 = λ2λ3
then either λ1 λ2 are linearly dependent over Q or else λ1 λ3
are linearly dependent over Q45 59
Six Exponentials Theorem
and Four Exponentials Conjecture
A Selberg CL Siegel S Lang K Ramachandra
46 59
The Four Exponentials Conjecture
Conjecture (Th Schneider S Lang K Ramachandra) Ifx1 x2 are Qndashlinearly independent complex numbers and y1 y2
are Qndashlinearly independent complex numbers then one atleast of the four numbers
ex1y1 ex1y2 ex2y1 ex2y2
is transcendentalEquivalent statement Any 2times 2 matrix(
λ1 λ2
λ3 λ4
)with entries in L and with Qndashlinearly independent rows andQndashlinearly independent columns has maximal rank 2
47 59
How could we attack Schanuelrsquos Conjecture
Let x1 xn be Qndashlinearly independent complex numbersFollowing the transcendence methods of Hermite GelrsquofondSchneider one may start by introducing an auxiliaryfunction
F (z) = P (z ez)
where P isin Z[X0 X1] is a nonndashzero polynomial One considersthe derivatives (
d
dz
)kF = (DkP )(z ez)
at the pointsm1x1 + middot middot middot+mnxn
for various values of (m1 mn) isin Zn
48 59
Auxiliary functionLet D denote the derivation
D =part
partX0
+X1part
partX1
over the ring C[X0 X1] Recall that x1 xn are Qndashlinearlyindependent complex numbers Let α1 αn be nonndashzerocomplex numbersThe transcendence machinery produces a sequence (PN)Nge0
of polynomials with integer coefficients satisfying∣∣∣∣∣(DkPN)(nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1
49 59
Royrsquos approach to Schanuelrsquos Conjecture (1999)
If the number of equations we produce is too small such a setof relations does not contain any information The existenceof a sequence of nonndashtrivial polynomials (PN)Nge0 follows fromlinear algebra
On the other hand following D Roy one may expect that theexistence of such a sequence (PN)Nge0 producing sufficientlymany such equations will yield the desired conclusion
tr degQQ(x1 xn α1 αn
)ge n
50 59
Royrsquos Conjecture (1999)Let s0 s1 t0 t1 u positive real numbers satisfying
max1 t0 2t1 lt mins0 2s1and
maxs0 s1 + t1 lt u lt1
2(1 + t0 + t1)
Assume that for any sufficiently large positive integer N there exists a nonndashzero polynomial PN isin Z[X0 X1] withpartial degree le N t0 in X0 partial degree le N t1 in X1 andheight le eN which satisfies∣∣∣∣∣(DkPN)(
nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1 Then
tr degQQ(x1 xn α1 αn) ge n51 59
Royrsquos Theorem (1999)
Royrsquos conjecture is equivalent to Schanuelrsquos Conjecture
52 59
Equivalence between Schanuel and Roy
Let (x α) isin CtimesCtimes and let s0 s1 t0 t1 u be positive realnumbers satisfying the inequalities of Royrsquos Conjecture Thenthe following conditions are equivalent (a) The number αeminusx is a root of unity
(b) For any sufficiently large positive integer N there exists anonndashzero polynomial QN isin Z[X0 X1] with partial degreele N t0 in X0 partial degree le N t1 in X1 and heightH(QN) le eN such that∣∣(DkQN)(mx αm)
∣∣ le exp(minusNu)
for any km isin N with k le N s0 and m le N s1
53 59
Gelrsquofondrsquos transcendence criterion
Simple form Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le n and height le en such that
|Pn(ϑ)| le eminus7n2
for all n ge 1 then ϑ is algebraic and Pn(ϑ) = 0 for all n ge 1
First extension Replace the bound for the degree by le dnthe bound for the height by ebn and the bound for |Pn(ϑ)| byeminuscdnbn with some constant c gt 0 independent of n
Some mild conditions are required on the sequences (dn)nge1 and(bn)nge1
54 59
Transcendence criterion with multiplicitiesWith derivatives Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le dn and height le ebn such that∣∣P (j)
n (ϑ)∣∣ le eminuscdnbntn
for all j in the range 0 le j lt tn and all n ge 1 then ϑ isalgebraicDue to M Laurent and D Roy applications to algebraicindependence
55 59
Criterion with several points
Goal Given a sequence of complex numbers (ϑi)ige1 if thereexists a sequence (Pn)nge1 of nonndashzero polynomials in Z[X]with Pn of degree le dn and height le ebn such that∣∣P (j)
n (ϑi)∣∣ le eminuscdnbntnsn
for 0 le j lt tn 1 le i le sn and all n ge 1 then the numbers ϑiare algebraic
D Roy Not true in general but true in some special caseswith a structure on the sequence (ϑi)ige1Combines the elimination arguments used for criteria ofalgebraic independence and for zero estimates
56 59
Small value estimates for the additive group
D Roy Small value estimates for the additive group Intern JNumber Theory to appear
Let ξ be a transcendental complex number let β σ τ and νbe non-negative real numbers let n0 be a positive integer andlet (Pn)ngen0 be a sequence of nonndashzero polynomials in Z[T ]satisfying deg(Pn) le n and H(Pn) le exp(nβ) for eachn ge n0 Suppose that β gt 1 (34)σ + τ lt 1 andν gt 1 + β minus (34)σ minus τ Then for infinitely many n we have
max|P [j]n (iξ)| 0 le i le nσ 0 le j le nτ
gt exp(minusnν)
57 59
Small value estimates for the multiplicative group
D Roy Small value estimates for the multiplicative groupActa Arith to appear
Let ξ1 ξm be multiplicatively independent complexnumbers in a field of transcendence degree 1 Under suitableassumptions on the parameters β σ τ ν for infinitely manypositive integers n there exists no nonndashzero polynomialP isin Z[T ] satisfying deg(P ) le n H(P ) le exp(nβ) and
max|P [j](ξi11 middot middot middot ξimm )| 0 le i1 im le nσ 0 le j le nτ
gt exp(minusnν)
58 59
Heidelberg Colloquium organized by the Institute for Mathematics
httpwwwmathiuni-heidelbergdevortraghtmls=1 January 22 2009
Transcendental Number TheorySchanuelrsquos Conjecture
Michel Waldschmidt
Institut de Mathematiques de Jussieu amp Paris VIhttpwwwmathjussieufrsimmiw
59 59
Statement by Gelrsquofond (1934)
Let β1 βn be Q-linearly independent algebraic numbersand let logα1 logαm be Q-linearly independentlogarithms of algebraic numbers Then the numbers
eβ1 eβn logα1 logαm
are algebraically independent over Q
20 59
Further statement by Gelrsquofond
Let β1 βn be algebraic numbers with β1 6= 0 and letα1 αm be algebraic numbers with α1 6= 0 1 α2 6= 0 1αi 6= 0 Then the numbers
eβ1eβ2e βnminus1e
βn
and αα αm
21
are transcendental and there is no nontrivial algebraic relationbetween such numbers
Remark The condition on α2 should be that it is irrational
21 59
Easy consequence of Schanuelrsquos Conjecture
According to Schanuelrsquos Conjecture the following numbers arealgebraically independent
e+ π eπ πe ee ee2
eee
ππ ππ2
πππ
log π log(log 2) π log 2 (log 2)(log 3) 2log 2 (log 2)log 3
Proof This is an easy exercise
22 59
Langrsquos exercise
Define E0 = Q Inductivelyfor n ge 1 define En as thealgebraic closure of the fieldgenerated over Enminus1 by thenumbers exp(x) = ex wherex ranges over Enminus1 Let E bethe union of En n ge 0Then Schanuelrsquos Conjectureimplies that the number πdoes not belong to E
More precisely Schanuelrsquos Conjecture implies that thenumbers π log π log log π log log log π are algebraicallyindependent over E
23 59
A variant of Langrsquos exercise
Define L0 = Q Inductively for n ge 1 define Ln as thealgebraic closure of the field generated over Lnminus1 by thenumbers y where y ranges over the set of complex numberssuch that ey isin Lnminus1 Let L be the union of Ln n ge 0Then Schanuelrsquos Conjecture implies that the number e doesnot belong to L
More precisely Schanuelrsquos Conjecture implies that the
numbers e ee eee ee
ee
are algebraically independent over L
24 59
Arizona Winter School AWS2008 Tucson
Theorem [Jonathan Bober Chuangxun Cheng Brian DietelMathilde Herblot Jingjing Huang Holly Krieger DiegoMarques Jonathan Mason Martin Mereb and Robert Wilson]Schanuelrsquos Conjecture implies that the fields E and L arelinearly disjoint over Q
Definition Given a field extension FK and twosubextensions F1 F2 sube F we say F1 F2 are linearly disjointover K when the following holds Any set x1 xn sube F1
of Kndash linearly independent elements is linearly independentover F2
Reference arXiv08043550 [mathNT] 2008
25 59
Formal analogs
WD Brownawell(was a student of Schanuel)
J Axrsquos Theorem (1968) Version of SchanuelrsquosConjecture for power seriesover C(and R Coleman for powerseries over Q)
Work by WD Brownawelland K Kubota on the ellipticanalog of Axrsquos Theorem
26 59
Conjectures by A Grothendieck and Y Andre
Generalized Conjecture onPeriods by Grothendieck Dimension of theMumfordndashTate group of asmooth projective variety
Generalization by Y Andre tomotives
Case of 1ndashmotives Elliptico-Toric Conjecture ofC Bertolin
27 59
Ubiquity of Schanuelrsquos Conjecture
Other contexts pndashadic numbers Leopoldtrsquos Conjecture onthe pndashadic rank of the units of an algebraic number fieldNon-vanishing of RegulatorsNonndashdegenerescence of heightsConjecture of B Mazur on rational pointsDiophantine approximation on tori
Dipendra Prasad Gopal Prasad
28 59
Methods from logicEhud Hrushovski Boris Zilber Jonathan Kirby
Calculus of rdquopredimension functionsrdquo (E Hrushovski)
Zilberrsquos construction of a rdquopseudoexponentiationrdquo
Also A Macintyre DE Marker G Terzo AJ WilkieD Bertrand
29 59
Methods from logic Model theory
Exponential algebraicity in exponential fieldsby
Jonathan Kirby
The dimension of the exponential algebraic closure operator inan exponential field satisfies a weak Schanuel property
A corollary is that there are at most countably many essentialcounterexamples to Schanuelrsquos conjecture
arXiv 08104285v2
30 59
Known
LindemannndashWeierstraszlig Theorem = case where x1 xn arealgebraic
Let β1 βn be algebraic numbers which are linearlyindependent over Q Then the numbers eβ1 eβn arealgebraically independent over Q
31 59
Problem of Gelrsquofond and SchneiderRaised by AO Gelrsquofond in 1948 and Th Schneider in 1952
Conjecture If α is an algebraic number α 6= 0 α 6= 1 and ifβ is an irrational algebraic number of degree d then the dminus 1numbers
αβ αβ2
αβdminus1
are algebraically independent
Special case of Schanuelrsquos Conjecture Take xi = βiminus1 logαn = d so that
x1 xn e
x1 exn
islogα β logα βdminus1 logα α αβ αβ
dminus1
The conclusion of Schanuelrsquos Conjecture is
tr degQQ(logα αβ αβ
2
αβdminus1)
= d
32 59
Algebraic independence method of Gelrsquofond
AO Gelrsquofond (1948)
The two numbers 23radic2 and
23radic4 are algebraically
independentMore generally if α is analgebraic number α 6= 0α 6= 1 and if β is a algebraicnumber of degree d ge 3 thentwo at least of the numbers
αβ αβ2
αβdminus1
are algebraically independent
33 59
Tools
Transcendence criterion Replaces Liouvillersquos inequality intranscendence proofsLiouville A nonndashzero rational integer n isin Z satisfies |n| ge 1Gelrsquofond Needs to give a lower bound for |P (θ)| withP isin Z[X] 0 when θ is transcendental
Zero estimate for exponential polynomials C Hermite P Turan K Mahler R Tijdeman
Small transcendence degree AO Gelrsquofond AA Smelev R Tijdeman WD Brownawell
34 59
Large transcendence degree
GV Chudnovsky (1976)Among the numbers
αβ αβ2
αβdminus1
at least [log2 d] arealgebraically independent
GV Chudnovsky ndash On the path to Schanuelrsquosconjecture Algebraic curves close to a pointI General theory of colored sequencesII Fields of finite transcendence type and coloredsequences ResultantsStudia Sci Math Hungar 12 (1977) 125ndash157 (1980)
35 59
Partial result on the problem of Gelrsquofond and
Schneider
AO Gelrsquofond GV Chudnovskii P Philippon YuV Nesterenko
G Diaz If α is an algebraic number α 6= 0 α 6= 1 and if β isan irrational algebraic number of degree d then
tr degQQ(αβ αβ
2
αβdminus1) ge [d+ 1
2
]
36 59
Conjecture of algebraic independence of logarithms
of algebraic numbers
Denote by L the set of complex numbers λ for which eλ isalgebraic
L = logα α isin Qtimes
Hence L is a Q-vector subspace of CThe most important special case of Schanuelrsquos Conjecture is
Conjecture Let λ1 λn be Q-linearly independent elementsin L Then the numbers λ1 λn are algebraicallyindependent over Q
Not yet known that the transcendence degree is ge 2 Open problem Among all logarithms of algebraic numbersone at least is transcendental over Q(π)
37 59
Structural rank of a matrix
Let K be a field k a subfield and M a matrix with entries in KConsider the k-vector subspace E of K spanned by the entriesof M Choose an injective morphism ϕ of E into a k-vectorspace kX1 + middot middot middot+ kXn The image ϕ(M) of M is a matrixwhose entries are in the field k(X1 Xn) of rationalfractions Its rank does not depend on the choice of ϕ
This is the structural rank of M with respect to k
38 59
Example
LetM =
(bij + λij
)1leiled1lejle`
be a matrix with coefficients in Q + L Consider a basis of theQndashvector spanned by the entries and replace the elements inthis basis by unknowns This gives a new matrix M withcoefficients in a field of rational fractions the rank of which isthe structural rank of M (with respect to Q)
As a consequence of the conjecture of algebraic independenceof logarithms of algebraic numbers the rank of M should bethe same as the rank of its specialization M
39 59
Equivalence between the two conjectures
Following D Roy theconjecture on algebraicindependence of logarithms ofalgebraic numbers isequivalent to
Conjecture Any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L has a rank equal to its structuralrank
40 59
Any Polynomial is the Determinant of a Matrix
The proof of the equivalence uses the nice auxiliary result
For any P isin k[X1 Xn] there exists a square matrix withentries in the k-vector space k + kX1 + middot middot middot+ kXn whosedeterminant is P
41 59
Half of the Conjecture is solved
From a certain point of view half of the conjecture of algebraicindependence of logarithms of algebraic numbers is solved
Theorem [D Roy] The rank of any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L is at least half its structural rank
42 59
Reformulation by D Roy
Instead of taking logarithmsof algebraic numbers andlooking for the algebraicindependence relationsD Roy fixes a polynomial andlooks at the points withcoordinates logarithms ofalgebraic numbers on thecorresponding hypersurface
43 59
Reformulation by D Roy
Royrsquos reformulation of the conjecture of algebraicindependence of logarithms is
Conjecture For any algebraic subvariety V of Cn definedover the field Q of algebraic numbers the set V cap Ln is theunion of the sets E cap Ln where E ranges over the set ofvector subspaces of Cn which are defined over Q andcontained in V
Trivial Any element in E cap Ln where E is a vector subspaceof Cn defined over Q and contained in V belongs to V cap Ln
44 59
Example The Four Exponentials ConjectureTake for V the hypersurface of C4 defined by the equation
z1z4 = z2z3
The maximal Cndashvector subspaces of C4 defined over Q andcontained in V are the planes
az1 = bz2 bz4 = az3
and the planesaz1 = bz3 bz4 = az2
with (a b) isin Q2 (0 0)Hence Schanuelrsquos Conjecture implies that if λ1 λ2 λ3 λ4 arelogarithms of algebraic numbers satisfying
λ1λ4 = λ2λ3
then either λ1 λ2 are linearly dependent over Q or else λ1 λ3
are linearly dependent over Q45 59
Six Exponentials Theorem
and Four Exponentials Conjecture
A Selberg CL Siegel S Lang K Ramachandra
46 59
The Four Exponentials Conjecture
Conjecture (Th Schneider S Lang K Ramachandra) Ifx1 x2 are Qndashlinearly independent complex numbers and y1 y2
are Qndashlinearly independent complex numbers then one atleast of the four numbers
ex1y1 ex1y2 ex2y1 ex2y2
is transcendentalEquivalent statement Any 2times 2 matrix(
λ1 λ2
λ3 λ4
)with entries in L and with Qndashlinearly independent rows andQndashlinearly independent columns has maximal rank 2
47 59
How could we attack Schanuelrsquos Conjecture
Let x1 xn be Qndashlinearly independent complex numbersFollowing the transcendence methods of Hermite GelrsquofondSchneider one may start by introducing an auxiliaryfunction
F (z) = P (z ez)
where P isin Z[X0 X1] is a nonndashzero polynomial One considersthe derivatives (
d
dz
)kF = (DkP )(z ez)
at the pointsm1x1 + middot middot middot+mnxn
for various values of (m1 mn) isin Zn
48 59
Auxiliary functionLet D denote the derivation
D =part
partX0
+X1part
partX1
over the ring C[X0 X1] Recall that x1 xn are Qndashlinearlyindependent complex numbers Let α1 αn be nonndashzerocomplex numbersThe transcendence machinery produces a sequence (PN)Nge0
of polynomials with integer coefficients satisfying∣∣∣∣∣(DkPN)(nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1
49 59
Royrsquos approach to Schanuelrsquos Conjecture (1999)
If the number of equations we produce is too small such a setof relations does not contain any information The existenceof a sequence of nonndashtrivial polynomials (PN)Nge0 follows fromlinear algebra
On the other hand following D Roy one may expect that theexistence of such a sequence (PN)Nge0 producing sufficientlymany such equations will yield the desired conclusion
tr degQQ(x1 xn α1 αn
)ge n
50 59
Royrsquos Conjecture (1999)Let s0 s1 t0 t1 u positive real numbers satisfying
max1 t0 2t1 lt mins0 2s1and
maxs0 s1 + t1 lt u lt1
2(1 + t0 + t1)
Assume that for any sufficiently large positive integer N there exists a nonndashzero polynomial PN isin Z[X0 X1] withpartial degree le N t0 in X0 partial degree le N t1 in X1 andheight le eN which satisfies∣∣∣∣∣(DkPN)(
nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1 Then
tr degQQ(x1 xn α1 αn) ge n51 59
Royrsquos Theorem (1999)
Royrsquos conjecture is equivalent to Schanuelrsquos Conjecture
52 59
Equivalence between Schanuel and Roy
Let (x α) isin CtimesCtimes and let s0 s1 t0 t1 u be positive realnumbers satisfying the inequalities of Royrsquos Conjecture Thenthe following conditions are equivalent (a) The number αeminusx is a root of unity
(b) For any sufficiently large positive integer N there exists anonndashzero polynomial QN isin Z[X0 X1] with partial degreele N t0 in X0 partial degree le N t1 in X1 and heightH(QN) le eN such that∣∣(DkQN)(mx αm)
∣∣ le exp(minusNu)
for any km isin N with k le N s0 and m le N s1
53 59
Gelrsquofondrsquos transcendence criterion
Simple form Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le n and height le en such that
|Pn(ϑ)| le eminus7n2
for all n ge 1 then ϑ is algebraic and Pn(ϑ) = 0 for all n ge 1
First extension Replace the bound for the degree by le dnthe bound for the height by ebn and the bound for |Pn(ϑ)| byeminuscdnbn with some constant c gt 0 independent of n
Some mild conditions are required on the sequences (dn)nge1 and(bn)nge1
54 59
Transcendence criterion with multiplicitiesWith derivatives Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le dn and height le ebn such that∣∣P (j)
n (ϑ)∣∣ le eminuscdnbntn
for all j in the range 0 le j lt tn and all n ge 1 then ϑ isalgebraicDue to M Laurent and D Roy applications to algebraicindependence
55 59
Criterion with several points
Goal Given a sequence of complex numbers (ϑi)ige1 if thereexists a sequence (Pn)nge1 of nonndashzero polynomials in Z[X]with Pn of degree le dn and height le ebn such that∣∣P (j)
n (ϑi)∣∣ le eminuscdnbntnsn
for 0 le j lt tn 1 le i le sn and all n ge 1 then the numbers ϑiare algebraic
D Roy Not true in general but true in some special caseswith a structure on the sequence (ϑi)ige1Combines the elimination arguments used for criteria ofalgebraic independence and for zero estimates
56 59
Small value estimates for the additive group
D Roy Small value estimates for the additive group Intern JNumber Theory to appear
Let ξ be a transcendental complex number let β σ τ and νbe non-negative real numbers let n0 be a positive integer andlet (Pn)ngen0 be a sequence of nonndashzero polynomials in Z[T ]satisfying deg(Pn) le n and H(Pn) le exp(nβ) for eachn ge n0 Suppose that β gt 1 (34)σ + τ lt 1 andν gt 1 + β minus (34)σ minus τ Then for infinitely many n we have
max|P [j]n (iξ)| 0 le i le nσ 0 le j le nτ
gt exp(minusnν)
57 59
Small value estimates for the multiplicative group
D Roy Small value estimates for the multiplicative groupActa Arith to appear
Let ξ1 ξm be multiplicatively independent complexnumbers in a field of transcendence degree 1 Under suitableassumptions on the parameters β σ τ ν for infinitely manypositive integers n there exists no nonndashzero polynomialP isin Z[T ] satisfying deg(P ) le n H(P ) le exp(nβ) and
max|P [j](ξi11 middot middot middot ξimm )| 0 le i1 im le nσ 0 le j le nτ
gt exp(minusnν)
58 59
Heidelberg Colloquium organized by the Institute for Mathematics
httpwwwmathiuni-heidelbergdevortraghtmls=1 January 22 2009
Transcendental Number TheorySchanuelrsquos Conjecture
Michel Waldschmidt
Institut de Mathematiques de Jussieu amp Paris VIhttpwwwmathjussieufrsimmiw
59 59
Further statement by Gelrsquofond
Let β1 βn be algebraic numbers with β1 6= 0 and letα1 αm be algebraic numbers with α1 6= 0 1 α2 6= 0 1αi 6= 0 Then the numbers
eβ1eβ2e βnminus1e
βn
and αα αm
21
are transcendental and there is no nontrivial algebraic relationbetween such numbers
Remark The condition on α2 should be that it is irrational
21 59
Easy consequence of Schanuelrsquos Conjecture
According to Schanuelrsquos Conjecture the following numbers arealgebraically independent
e+ π eπ πe ee ee2
eee
ππ ππ2
πππ
log π log(log 2) π log 2 (log 2)(log 3) 2log 2 (log 2)log 3
Proof This is an easy exercise
22 59
Langrsquos exercise
Define E0 = Q Inductivelyfor n ge 1 define En as thealgebraic closure of the fieldgenerated over Enminus1 by thenumbers exp(x) = ex wherex ranges over Enminus1 Let E bethe union of En n ge 0Then Schanuelrsquos Conjectureimplies that the number πdoes not belong to E
More precisely Schanuelrsquos Conjecture implies that thenumbers π log π log log π log log log π are algebraicallyindependent over E
23 59
A variant of Langrsquos exercise
Define L0 = Q Inductively for n ge 1 define Ln as thealgebraic closure of the field generated over Lnminus1 by thenumbers y where y ranges over the set of complex numberssuch that ey isin Lnminus1 Let L be the union of Ln n ge 0Then Schanuelrsquos Conjecture implies that the number e doesnot belong to L
More precisely Schanuelrsquos Conjecture implies that the
numbers e ee eee ee
ee
are algebraically independent over L
24 59
Arizona Winter School AWS2008 Tucson
Theorem [Jonathan Bober Chuangxun Cheng Brian DietelMathilde Herblot Jingjing Huang Holly Krieger DiegoMarques Jonathan Mason Martin Mereb and Robert Wilson]Schanuelrsquos Conjecture implies that the fields E and L arelinearly disjoint over Q
Definition Given a field extension FK and twosubextensions F1 F2 sube F we say F1 F2 are linearly disjointover K when the following holds Any set x1 xn sube F1
of Kndash linearly independent elements is linearly independentover F2
Reference arXiv08043550 [mathNT] 2008
25 59
Formal analogs
WD Brownawell(was a student of Schanuel)
J Axrsquos Theorem (1968) Version of SchanuelrsquosConjecture for power seriesover C(and R Coleman for powerseries over Q)
Work by WD Brownawelland K Kubota on the ellipticanalog of Axrsquos Theorem
26 59
Conjectures by A Grothendieck and Y Andre
Generalized Conjecture onPeriods by Grothendieck Dimension of theMumfordndashTate group of asmooth projective variety
Generalization by Y Andre tomotives
Case of 1ndashmotives Elliptico-Toric Conjecture ofC Bertolin
27 59
Ubiquity of Schanuelrsquos Conjecture
Other contexts pndashadic numbers Leopoldtrsquos Conjecture onthe pndashadic rank of the units of an algebraic number fieldNon-vanishing of RegulatorsNonndashdegenerescence of heightsConjecture of B Mazur on rational pointsDiophantine approximation on tori
Dipendra Prasad Gopal Prasad
28 59
Methods from logicEhud Hrushovski Boris Zilber Jonathan Kirby
Calculus of rdquopredimension functionsrdquo (E Hrushovski)
Zilberrsquos construction of a rdquopseudoexponentiationrdquo
Also A Macintyre DE Marker G Terzo AJ WilkieD Bertrand
29 59
Methods from logic Model theory
Exponential algebraicity in exponential fieldsby
Jonathan Kirby
The dimension of the exponential algebraic closure operator inan exponential field satisfies a weak Schanuel property
A corollary is that there are at most countably many essentialcounterexamples to Schanuelrsquos conjecture
arXiv 08104285v2
30 59
Known
LindemannndashWeierstraszlig Theorem = case where x1 xn arealgebraic
Let β1 βn be algebraic numbers which are linearlyindependent over Q Then the numbers eβ1 eβn arealgebraically independent over Q
31 59
Problem of Gelrsquofond and SchneiderRaised by AO Gelrsquofond in 1948 and Th Schneider in 1952
Conjecture If α is an algebraic number α 6= 0 α 6= 1 and ifβ is an irrational algebraic number of degree d then the dminus 1numbers
αβ αβ2
αβdminus1
are algebraically independent
Special case of Schanuelrsquos Conjecture Take xi = βiminus1 logαn = d so that
x1 xn e
x1 exn
islogα β logα βdminus1 logα α αβ αβ
dminus1
The conclusion of Schanuelrsquos Conjecture is
tr degQQ(logα αβ αβ
2
αβdminus1)
= d
32 59
Algebraic independence method of Gelrsquofond
AO Gelrsquofond (1948)
The two numbers 23radic2 and
23radic4 are algebraically
independentMore generally if α is analgebraic number α 6= 0α 6= 1 and if β is a algebraicnumber of degree d ge 3 thentwo at least of the numbers
αβ αβ2
αβdminus1
are algebraically independent
33 59
Tools
Transcendence criterion Replaces Liouvillersquos inequality intranscendence proofsLiouville A nonndashzero rational integer n isin Z satisfies |n| ge 1Gelrsquofond Needs to give a lower bound for |P (θ)| withP isin Z[X] 0 when θ is transcendental
Zero estimate for exponential polynomials C Hermite P Turan K Mahler R Tijdeman
Small transcendence degree AO Gelrsquofond AA Smelev R Tijdeman WD Brownawell
34 59
Large transcendence degree
GV Chudnovsky (1976)Among the numbers
αβ αβ2
αβdminus1
at least [log2 d] arealgebraically independent
GV Chudnovsky ndash On the path to Schanuelrsquosconjecture Algebraic curves close to a pointI General theory of colored sequencesII Fields of finite transcendence type and coloredsequences ResultantsStudia Sci Math Hungar 12 (1977) 125ndash157 (1980)
35 59
Partial result on the problem of Gelrsquofond and
Schneider
AO Gelrsquofond GV Chudnovskii P Philippon YuV Nesterenko
G Diaz If α is an algebraic number α 6= 0 α 6= 1 and if β isan irrational algebraic number of degree d then
tr degQQ(αβ αβ
2
αβdminus1) ge [d+ 1
2
]
36 59
Conjecture of algebraic independence of logarithms
of algebraic numbers
Denote by L the set of complex numbers λ for which eλ isalgebraic
L = logα α isin Qtimes
Hence L is a Q-vector subspace of CThe most important special case of Schanuelrsquos Conjecture is
Conjecture Let λ1 λn be Q-linearly independent elementsin L Then the numbers λ1 λn are algebraicallyindependent over Q
Not yet known that the transcendence degree is ge 2 Open problem Among all logarithms of algebraic numbersone at least is transcendental over Q(π)
37 59
Structural rank of a matrix
Let K be a field k a subfield and M a matrix with entries in KConsider the k-vector subspace E of K spanned by the entriesof M Choose an injective morphism ϕ of E into a k-vectorspace kX1 + middot middot middot+ kXn The image ϕ(M) of M is a matrixwhose entries are in the field k(X1 Xn) of rationalfractions Its rank does not depend on the choice of ϕ
This is the structural rank of M with respect to k
38 59
Example
LetM =
(bij + λij
)1leiled1lejle`
be a matrix with coefficients in Q + L Consider a basis of theQndashvector spanned by the entries and replace the elements inthis basis by unknowns This gives a new matrix M withcoefficients in a field of rational fractions the rank of which isthe structural rank of M (with respect to Q)
As a consequence of the conjecture of algebraic independenceof logarithms of algebraic numbers the rank of M should bethe same as the rank of its specialization M
39 59
Equivalence between the two conjectures
Following D Roy theconjecture on algebraicindependence of logarithms ofalgebraic numbers isequivalent to
Conjecture Any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L has a rank equal to its structuralrank
40 59
Any Polynomial is the Determinant of a Matrix
The proof of the equivalence uses the nice auxiliary result
For any P isin k[X1 Xn] there exists a square matrix withentries in the k-vector space k + kX1 + middot middot middot+ kXn whosedeterminant is P
41 59
Half of the Conjecture is solved
From a certain point of view half of the conjecture of algebraicindependence of logarithms of algebraic numbers is solved
Theorem [D Roy] The rank of any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L is at least half its structural rank
42 59
Reformulation by D Roy
Instead of taking logarithmsof algebraic numbers andlooking for the algebraicindependence relationsD Roy fixes a polynomial andlooks at the points withcoordinates logarithms ofalgebraic numbers on thecorresponding hypersurface
43 59
Reformulation by D Roy
Royrsquos reformulation of the conjecture of algebraicindependence of logarithms is
Conjecture For any algebraic subvariety V of Cn definedover the field Q of algebraic numbers the set V cap Ln is theunion of the sets E cap Ln where E ranges over the set ofvector subspaces of Cn which are defined over Q andcontained in V
Trivial Any element in E cap Ln where E is a vector subspaceof Cn defined over Q and contained in V belongs to V cap Ln
44 59
Example The Four Exponentials ConjectureTake for V the hypersurface of C4 defined by the equation
z1z4 = z2z3
The maximal Cndashvector subspaces of C4 defined over Q andcontained in V are the planes
az1 = bz2 bz4 = az3
and the planesaz1 = bz3 bz4 = az2
with (a b) isin Q2 (0 0)Hence Schanuelrsquos Conjecture implies that if λ1 λ2 λ3 λ4 arelogarithms of algebraic numbers satisfying
λ1λ4 = λ2λ3
then either λ1 λ2 are linearly dependent over Q or else λ1 λ3
are linearly dependent over Q45 59
Six Exponentials Theorem
and Four Exponentials Conjecture
A Selberg CL Siegel S Lang K Ramachandra
46 59
The Four Exponentials Conjecture
Conjecture (Th Schneider S Lang K Ramachandra) Ifx1 x2 are Qndashlinearly independent complex numbers and y1 y2
are Qndashlinearly independent complex numbers then one atleast of the four numbers
ex1y1 ex1y2 ex2y1 ex2y2
is transcendentalEquivalent statement Any 2times 2 matrix(
λ1 λ2
λ3 λ4
)with entries in L and with Qndashlinearly independent rows andQndashlinearly independent columns has maximal rank 2
47 59
How could we attack Schanuelrsquos Conjecture
Let x1 xn be Qndashlinearly independent complex numbersFollowing the transcendence methods of Hermite GelrsquofondSchneider one may start by introducing an auxiliaryfunction
F (z) = P (z ez)
where P isin Z[X0 X1] is a nonndashzero polynomial One considersthe derivatives (
d
dz
)kF = (DkP )(z ez)
at the pointsm1x1 + middot middot middot+mnxn
for various values of (m1 mn) isin Zn
48 59
Auxiliary functionLet D denote the derivation
D =part
partX0
+X1part
partX1
over the ring C[X0 X1] Recall that x1 xn are Qndashlinearlyindependent complex numbers Let α1 αn be nonndashzerocomplex numbersThe transcendence machinery produces a sequence (PN)Nge0
of polynomials with integer coefficients satisfying∣∣∣∣∣(DkPN)(nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1
49 59
Royrsquos approach to Schanuelrsquos Conjecture (1999)
If the number of equations we produce is too small such a setof relations does not contain any information The existenceof a sequence of nonndashtrivial polynomials (PN)Nge0 follows fromlinear algebra
On the other hand following D Roy one may expect that theexistence of such a sequence (PN)Nge0 producing sufficientlymany such equations will yield the desired conclusion
tr degQQ(x1 xn α1 αn
)ge n
50 59
Royrsquos Conjecture (1999)Let s0 s1 t0 t1 u positive real numbers satisfying
max1 t0 2t1 lt mins0 2s1and
maxs0 s1 + t1 lt u lt1
2(1 + t0 + t1)
Assume that for any sufficiently large positive integer N there exists a nonndashzero polynomial PN isin Z[X0 X1] withpartial degree le N t0 in X0 partial degree le N t1 in X1 andheight le eN which satisfies∣∣∣∣∣(DkPN)(
nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1 Then
tr degQQ(x1 xn α1 αn) ge n51 59
Royrsquos Theorem (1999)
Royrsquos conjecture is equivalent to Schanuelrsquos Conjecture
52 59
Equivalence between Schanuel and Roy
Let (x α) isin CtimesCtimes and let s0 s1 t0 t1 u be positive realnumbers satisfying the inequalities of Royrsquos Conjecture Thenthe following conditions are equivalent (a) The number αeminusx is a root of unity
(b) For any sufficiently large positive integer N there exists anonndashzero polynomial QN isin Z[X0 X1] with partial degreele N t0 in X0 partial degree le N t1 in X1 and heightH(QN) le eN such that∣∣(DkQN)(mx αm)
∣∣ le exp(minusNu)
for any km isin N with k le N s0 and m le N s1
53 59
Gelrsquofondrsquos transcendence criterion
Simple form Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le n and height le en such that
|Pn(ϑ)| le eminus7n2
for all n ge 1 then ϑ is algebraic and Pn(ϑ) = 0 for all n ge 1
First extension Replace the bound for the degree by le dnthe bound for the height by ebn and the bound for |Pn(ϑ)| byeminuscdnbn with some constant c gt 0 independent of n
Some mild conditions are required on the sequences (dn)nge1 and(bn)nge1
54 59
Transcendence criterion with multiplicitiesWith derivatives Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le dn and height le ebn such that∣∣P (j)
n (ϑ)∣∣ le eminuscdnbntn
for all j in the range 0 le j lt tn and all n ge 1 then ϑ isalgebraicDue to M Laurent and D Roy applications to algebraicindependence
55 59
Criterion with several points
Goal Given a sequence of complex numbers (ϑi)ige1 if thereexists a sequence (Pn)nge1 of nonndashzero polynomials in Z[X]with Pn of degree le dn and height le ebn such that∣∣P (j)
n (ϑi)∣∣ le eminuscdnbntnsn
for 0 le j lt tn 1 le i le sn and all n ge 1 then the numbers ϑiare algebraic
D Roy Not true in general but true in some special caseswith a structure on the sequence (ϑi)ige1Combines the elimination arguments used for criteria ofalgebraic independence and for zero estimates
56 59
Small value estimates for the additive group
D Roy Small value estimates for the additive group Intern JNumber Theory to appear
Let ξ be a transcendental complex number let β σ τ and νbe non-negative real numbers let n0 be a positive integer andlet (Pn)ngen0 be a sequence of nonndashzero polynomials in Z[T ]satisfying deg(Pn) le n and H(Pn) le exp(nβ) for eachn ge n0 Suppose that β gt 1 (34)σ + τ lt 1 andν gt 1 + β minus (34)σ minus τ Then for infinitely many n we have
max|P [j]n (iξ)| 0 le i le nσ 0 le j le nτ
gt exp(minusnν)
57 59
Small value estimates for the multiplicative group
D Roy Small value estimates for the multiplicative groupActa Arith to appear
Let ξ1 ξm be multiplicatively independent complexnumbers in a field of transcendence degree 1 Under suitableassumptions on the parameters β σ τ ν for infinitely manypositive integers n there exists no nonndashzero polynomialP isin Z[T ] satisfying deg(P ) le n H(P ) le exp(nβ) and
max|P [j](ξi11 middot middot middot ξimm )| 0 le i1 im le nσ 0 le j le nτ
gt exp(minusnν)
58 59
Heidelberg Colloquium organized by the Institute for Mathematics
httpwwwmathiuni-heidelbergdevortraghtmls=1 January 22 2009
Transcendental Number TheorySchanuelrsquos Conjecture
Michel Waldschmidt
Institut de Mathematiques de Jussieu amp Paris VIhttpwwwmathjussieufrsimmiw
59 59
Easy consequence of Schanuelrsquos Conjecture
According to Schanuelrsquos Conjecture the following numbers arealgebraically independent
e+ π eπ πe ee ee2
eee
ππ ππ2
πππ
log π log(log 2) π log 2 (log 2)(log 3) 2log 2 (log 2)log 3
Proof This is an easy exercise
22 59
Langrsquos exercise
Define E0 = Q Inductivelyfor n ge 1 define En as thealgebraic closure of the fieldgenerated over Enminus1 by thenumbers exp(x) = ex wherex ranges over Enminus1 Let E bethe union of En n ge 0Then Schanuelrsquos Conjectureimplies that the number πdoes not belong to E
More precisely Schanuelrsquos Conjecture implies that thenumbers π log π log log π log log log π are algebraicallyindependent over E
23 59
A variant of Langrsquos exercise
Define L0 = Q Inductively for n ge 1 define Ln as thealgebraic closure of the field generated over Lnminus1 by thenumbers y where y ranges over the set of complex numberssuch that ey isin Lnminus1 Let L be the union of Ln n ge 0Then Schanuelrsquos Conjecture implies that the number e doesnot belong to L
More precisely Schanuelrsquos Conjecture implies that the
numbers e ee eee ee
ee
are algebraically independent over L
24 59
Arizona Winter School AWS2008 Tucson
Theorem [Jonathan Bober Chuangxun Cheng Brian DietelMathilde Herblot Jingjing Huang Holly Krieger DiegoMarques Jonathan Mason Martin Mereb and Robert Wilson]Schanuelrsquos Conjecture implies that the fields E and L arelinearly disjoint over Q
Definition Given a field extension FK and twosubextensions F1 F2 sube F we say F1 F2 are linearly disjointover K when the following holds Any set x1 xn sube F1
of Kndash linearly independent elements is linearly independentover F2
Reference arXiv08043550 [mathNT] 2008
25 59
Formal analogs
WD Brownawell(was a student of Schanuel)
J Axrsquos Theorem (1968) Version of SchanuelrsquosConjecture for power seriesover C(and R Coleman for powerseries over Q)
Work by WD Brownawelland K Kubota on the ellipticanalog of Axrsquos Theorem
26 59
Conjectures by A Grothendieck and Y Andre
Generalized Conjecture onPeriods by Grothendieck Dimension of theMumfordndashTate group of asmooth projective variety
Generalization by Y Andre tomotives
Case of 1ndashmotives Elliptico-Toric Conjecture ofC Bertolin
27 59
Ubiquity of Schanuelrsquos Conjecture
Other contexts pndashadic numbers Leopoldtrsquos Conjecture onthe pndashadic rank of the units of an algebraic number fieldNon-vanishing of RegulatorsNonndashdegenerescence of heightsConjecture of B Mazur on rational pointsDiophantine approximation on tori
Dipendra Prasad Gopal Prasad
28 59
Methods from logicEhud Hrushovski Boris Zilber Jonathan Kirby
Calculus of rdquopredimension functionsrdquo (E Hrushovski)
Zilberrsquos construction of a rdquopseudoexponentiationrdquo
Also A Macintyre DE Marker G Terzo AJ WilkieD Bertrand
29 59
Methods from logic Model theory
Exponential algebraicity in exponential fieldsby
Jonathan Kirby
The dimension of the exponential algebraic closure operator inan exponential field satisfies a weak Schanuel property
A corollary is that there are at most countably many essentialcounterexamples to Schanuelrsquos conjecture
arXiv 08104285v2
30 59
Known
LindemannndashWeierstraszlig Theorem = case where x1 xn arealgebraic
Let β1 βn be algebraic numbers which are linearlyindependent over Q Then the numbers eβ1 eβn arealgebraically independent over Q
31 59
Problem of Gelrsquofond and SchneiderRaised by AO Gelrsquofond in 1948 and Th Schneider in 1952
Conjecture If α is an algebraic number α 6= 0 α 6= 1 and ifβ is an irrational algebraic number of degree d then the dminus 1numbers
αβ αβ2
αβdminus1
are algebraically independent
Special case of Schanuelrsquos Conjecture Take xi = βiminus1 logαn = d so that
x1 xn e
x1 exn
islogα β logα βdminus1 logα α αβ αβ
dminus1
The conclusion of Schanuelrsquos Conjecture is
tr degQQ(logα αβ αβ
2
αβdminus1)
= d
32 59
Algebraic independence method of Gelrsquofond
AO Gelrsquofond (1948)
The two numbers 23radic2 and
23radic4 are algebraically
independentMore generally if α is analgebraic number α 6= 0α 6= 1 and if β is a algebraicnumber of degree d ge 3 thentwo at least of the numbers
αβ αβ2
αβdminus1
are algebraically independent
33 59
Tools
Transcendence criterion Replaces Liouvillersquos inequality intranscendence proofsLiouville A nonndashzero rational integer n isin Z satisfies |n| ge 1Gelrsquofond Needs to give a lower bound for |P (θ)| withP isin Z[X] 0 when θ is transcendental
Zero estimate for exponential polynomials C Hermite P Turan K Mahler R Tijdeman
Small transcendence degree AO Gelrsquofond AA Smelev R Tijdeman WD Brownawell
34 59
Large transcendence degree
GV Chudnovsky (1976)Among the numbers
αβ αβ2
αβdminus1
at least [log2 d] arealgebraically independent
GV Chudnovsky ndash On the path to Schanuelrsquosconjecture Algebraic curves close to a pointI General theory of colored sequencesII Fields of finite transcendence type and coloredsequences ResultantsStudia Sci Math Hungar 12 (1977) 125ndash157 (1980)
35 59
Partial result on the problem of Gelrsquofond and
Schneider
AO Gelrsquofond GV Chudnovskii P Philippon YuV Nesterenko
G Diaz If α is an algebraic number α 6= 0 α 6= 1 and if β isan irrational algebraic number of degree d then
tr degQQ(αβ αβ
2
αβdminus1) ge [d+ 1
2
]
36 59
Conjecture of algebraic independence of logarithms
of algebraic numbers
Denote by L the set of complex numbers λ for which eλ isalgebraic
L = logα α isin Qtimes
Hence L is a Q-vector subspace of CThe most important special case of Schanuelrsquos Conjecture is
Conjecture Let λ1 λn be Q-linearly independent elementsin L Then the numbers λ1 λn are algebraicallyindependent over Q
Not yet known that the transcendence degree is ge 2 Open problem Among all logarithms of algebraic numbersone at least is transcendental over Q(π)
37 59
Structural rank of a matrix
Let K be a field k a subfield and M a matrix with entries in KConsider the k-vector subspace E of K spanned by the entriesof M Choose an injective morphism ϕ of E into a k-vectorspace kX1 + middot middot middot+ kXn The image ϕ(M) of M is a matrixwhose entries are in the field k(X1 Xn) of rationalfractions Its rank does not depend on the choice of ϕ
This is the structural rank of M with respect to k
38 59
Example
LetM =
(bij + λij
)1leiled1lejle`
be a matrix with coefficients in Q + L Consider a basis of theQndashvector spanned by the entries and replace the elements inthis basis by unknowns This gives a new matrix M withcoefficients in a field of rational fractions the rank of which isthe structural rank of M (with respect to Q)
As a consequence of the conjecture of algebraic independenceof logarithms of algebraic numbers the rank of M should bethe same as the rank of its specialization M
39 59
Equivalence between the two conjectures
Following D Roy theconjecture on algebraicindependence of logarithms ofalgebraic numbers isequivalent to
Conjecture Any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L has a rank equal to its structuralrank
40 59
Any Polynomial is the Determinant of a Matrix
The proof of the equivalence uses the nice auxiliary result
For any P isin k[X1 Xn] there exists a square matrix withentries in the k-vector space k + kX1 + middot middot middot+ kXn whosedeterminant is P
41 59
Half of the Conjecture is solved
From a certain point of view half of the conjecture of algebraicindependence of logarithms of algebraic numbers is solved
Theorem [D Roy] The rank of any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L is at least half its structural rank
42 59
Reformulation by D Roy
Instead of taking logarithmsof algebraic numbers andlooking for the algebraicindependence relationsD Roy fixes a polynomial andlooks at the points withcoordinates logarithms ofalgebraic numbers on thecorresponding hypersurface
43 59
Reformulation by D Roy
Royrsquos reformulation of the conjecture of algebraicindependence of logarithms is
Conjecture For any algebraic subvariety V of Cn definedover the field Q of algebraic numbers the set V cap Ln is theunion of the sets E cap Ln where E ranges over the set ofvector subspaces of Cn which are defined over Q andcontained in V
Trivial Any element in E cap Ln where E is a vector subspaceof Cn defined over Q and contained in V belongs to V cap Ln
44 59
Example The Four Exponentials ConjectureTake for V the hypersurface of C4 defined by the equation
z1z4 = z2z3
The maximal Cndashvector subspaces of C4 defined over Q andcontained in V are the planes
az1 = bz2 bz4 = az3
and the planesaz1 = bz3 bz4 = az2
with (a b) isin Q2 (0 0)Hence Schanuelrsquos Conjecture implies that if λ1 λ2 λ3 λ4 arelogarithms of algebraic numbers satisfying
λ1λ4 = λ2λ3
then either λ1 λ2 are linearly dependent over Q or else λ1 λ3
are linearly dependent over Q45 59
Six Exponentials Theorem
and Four Exponentials Conjecture
A Selberg CL Siegel S Lang K Ramachandra
46 59
The Four Exponentials Conjecture
Conjecture (Th Schneider S Lang K Ramachandra) Ifx1 x2 are Qndashlinearly independent complex numbers and y1 y2
are Qndashlinearly independent complex numbers then one atleast of the four numbers
ex1y1 ex1y2 ex2y1 ex2y2
is transcendentalEquivalent statement Any 2times 2 matrix(
λ1 λ2
λ3 λ4
)with entries in L and with Qndashlinearly independent rows andQndashlinearly independent columns has maximal rank 2
47 59
How could we attack Schanuelrsquos Conjecture
Let x1 xn be Qndashlinearly independent complex numbersFollowing the transcendence methods of Hermite GelrsquofondSchneider one may start by introducing an auxiliaryfunction
F (z) = P (z ez)
where P isin Z[X0 X1] is a nonndashzero polynomial One considersthe derivatives (
d
dz
)kF = (DkP )(z ez)
at the pointsm1x1 + middot middot middot+mnxn
for various values of (m1 mn) isin Zn
48 59
Auxiliary functionLet D denote the derivation
D =part
partX0
+X1part
partX1
over the ring C[X0 X1] Recall that x1 xn are Qndashlinearlyindependent complex numbers Let α1 αn be nonndashzerocomplex numbersThe transcendence machinery produces a sequence (PN)Nge0
of polynomials with integer coefficients satisfying∣∣∣∣∣(DkPN)(nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1
49 59
Royrsquos approach to Schanuelrsquos Conjecture (1999)
If the number of equations we produce is too small such a setof relations does not contain any information The existenceof a sequence of nonndashtrivial polynomials (PN)Nge0 follows fromlinear algebra
On the other hand following D Roy one may expect that theexistence of such a sequence (PN)Nge0 producing sufficientlymany such equations will yield the desired conclusion
tr degQQ(x1 xn α1 αn
)ge n
50 59
Royrsquos Conjecture (1999)Let s0 s1 t0 t1 u positive real numbers satisfying
max1 t0 2t1 lt mins0 2s1and
maxs0 s1 + t1 lt u lt1
2(1 + t0 + t1)
Assume that for any sufficiently large positive integer N there exists a nonndashzero polynomial PN isin Z[X0 X1] withpartial degree le N t0 in X0 partial degree le N t1 in X1 andheight le eN which satisfies∣∣∣∣∣(DkPN)(
nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1 Then
tr degQQ(x1 xn α1 αn) ge n51 59
Royrsquos Theorem (1999)
Royrsquos conjecture is equivalent to Schanuelrsquos Conjecture
52 59
Equivalence between Schanuel and Roy
Let (x α) isin CtimesCtimes and let s0 s1 t0 t1 u be positive realnumbers satisfying the inequalities of Royrsquos Conjecture Thenthe following conditions are equivalent (a) The number αeminusx is a root of unity
(b) For any sufficiently large positive integer N there exists anonndashzero polynomial QN isin Z[X0 X1] with partial degreele N t0 in X0 partial degree le N t1 in X1 and heightH(QN) le eN such that∣∣(DkQN)(mx αm)
∣∣ le exp(minusNu)
for any km isin N with k le N s0 and m le N s1
53 59
Gelrsquofondrsquos transcendence criterion
Simple form Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le n and height le en such that
|Pn(ϑ)| le eminus7n2
for all n ge 1 then ϑ is algebraic and Pn(ϑ) = 0 for all n ge 1
First extension Replace the bound for the degree by le dnthe bound for the height by ebn and the bound for |Pn(ϑ)| byeminuscdnbn with some constant c gt 0 independent of n
Some mild conditions are required on the sequences (dn)nge1 and(bn)nge1
54 59
Transcendence criterion with multiplicitiesWith derivatives Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le dn and height le ebn such that∣∣P (j)
n (ϑ)∣∣ le eminuscdnbntn
for all j in the range 0 le j lt tn and all n ge 1 then ϑ isalgebraicDue to M Laurent and D Roy applications to algebraicindependence
55 59
Criterion with several points
Goal Given a sequence of complex numbers (ϑi)ige1 if thereexists a sequence (Pn)nge1 of nonndashzero polynomials in Z[X]with Pn of degree le dn and height le ebn such that∣∣P (j)
n (ϑi)∣∣ le eminuscdnbntnsn
for 0 le j lt tn 1 le i le sn and all n ge 1 then the numbers ϑiare algebraic
D Roy Not true in general but true in some special caseswith a structure on the sequence (ϑi)ige1Combines the elimination arguments used for criteria ofalgebraic independence and for zero estimates
56 59
Small value estimates for the additive group
D Roy Small value estimates for the additive group Intern JNumber Theory to appear
Let ξ be a transcendental complex number let β σ τ and νbe non-negative real numbers let n0 be a positive integer andlet (Pn)ngen0 be a sequence of nonndashzero polynomials in Z[T ]satisfying deg(Pn) le n and H(Pn) le exp(nβ) for eachn ge n0 Suppose that β gt 1 (34)σ + τ lt 1 andν gt 1 + β minus (34)σ minus τ Then for infinitely many n we have
max|P [j]n (iξ)| 0 le i le nσ 0 le j le nτ
gt exp(minusnν)
57 59
Small value estimates for the multiplicative group
D Roy Small value estimates for the multiplicative groupActa Arith to appear
Let ξ1 ξm be multiplicatively independent complexnumbers in a field of transcendence degree 1 Under suitableassumptions on the parameters β σ τ ν for infinitely manypositive integers n there exists no nonndashzero polynomialP isin Z[T ] satisfying deg(P ) le n H(P ) le exp(nβ) and
max|P [j](ξi11 middot middot middot ξimm )| 0 le i1 im le nσ 0 le j le nτ
gt exp(minusnν)
58 59
Heidelberg Colloquium organized by the Institute for Mathematics
httpwwwmathiuni-heidelbergdevortraghtmls=1 January 22 2009
Transcendental Number TheorySchanuelrsquos Conjecture
Michel Waldschmidt
Institut de Mathematiques de Jussieu amp Paris VIhttpwwwmathjussieufrsimmiw
59 59
Langrsquos exercise
Define E0 = Q Inductivelyfor n ge 1 define En as thealgebraic closure of the fieldgenerated over Enminus1 by thenumbers exp(x) = ex wherex ranges over Enminus1 Let E bethe union of En n ge 0Then Schanuelrsquos Conjectureimplies that the number πdoes not belong to E
More precisely Schanuelrsquos Conjecture implies that thenumbers π log π log log π log log log π are algebraicallyindependent over E
23 59
A variant of Langrsquos exercise
Define L0 = Q Inductively for n ge 1 define Ln as thealgebraic closure of the field generated over Lnminus1 by thenumbers y where y ranges over the set of complex numberssuch that ey isin Lnminus1 Let L be the union of Ln n ge 0Then Schanuelrsquos Conjecture implies that the number e doesnot belong to L
More precisely Schanuelrsquos Conjecture implies that the
numbers e ee eee ee
ee
are algebraically independent over L
24 59
Arizona Winter School AWS2008 Tucson
Theorem [Jonathan Bober Chuangxun Cheng Brian DietelMathilde Herblot Jingjing Huang Holly Krieger DiegoMarques Jonathan Mason Martin Mereb and Robert Wilson]Schanuelrsquos Conjecture implies that the fields E and L arelinearly disjoint over Q
Definition Given a field extension FK and twosubextensions F1 F2 sube F we say F1 F2 are linearly disjointover K when the following holds Any set x1 xn sube F1
of Kndash linearly independent elements is linearly independentover F2
Reference arXiv08043550 [mathNT] 2008
25 59
Formal analogs
WD Brownawell(was a student of Schanuel)
J Axrsquos Theorem (1968) Version of SchanuelrsquosConjecture for power seriesover C(and R Coleman for powerseries over Q)
Work by WD Brownawelland K Kubota on the ellipticanalog of Axrsquos Theorem
26 59
Conjectures by A Grothendieck and Y Andre
Generalized Conjecture onPeriods by Grothendieck Dimension of theMumfordndashTate group of asmooth projective variety
Generalization by Y Andre tomotives
Case of 1ndashmotives Elliptico-Toric Conjecture ofC Bertolin
27 59
Ubiquity of Schanuelrsquos Conjecture
Other contexts pndashadic numbers Leopoldtrsquos Conjecture onthe pndashadic rank of the units of an algebraic number fieldNon-vanishing of RegulatorsNonndashdegenerescence of heightsConjecture of B Mazur on rational pointsDiophantine approximation on tori
Dipendra Prasad Gopal Prasad
28 59
Methods from logicEhud Hrushovski Boris Zilber Jonathan Kirby
Calculus of rdquopredimension functionsrdquo (E Hrushovski)
Zilberrsquos construction of a rdquopseudoexponentiationrdquo
Also A Macintyre DE Marker G Terzo AJ WilkieD Bertrand
29 59
Methods from logic Model theory
Exponential algebraicity in exponential fieldsby
Jonathan Kirby
The dimension of the exponential algebraic closure operator inan exponential field satisfies a weak Schanuel property
A corollary is that there are at most countably many essentialcounterexamples to Schanuelrsquos conjecture
arXiv 08104285v2
30 59
Known
LindemannndashWeierstraszlig Theorem = case where x1 xn arealgebraic
Let β1 βn be algebraic numbers which are linearlyindependent over Q Then the numbers eβ1 eβn arealgebraically independent over Q
31 59
Problem of Gelrsquofond and SchneiderRaised by AO Gelrsquofond in 1948 and Th Schneider in 1952
Conjecture If α is an algebraic number α 6= 0 α 6= 1 and ifβ is an irrational algebraic number of degree d then the dminus 1numbers
αβ αβ2
αβdminus1
are algebraically independent
Special case of Schanuelrsquos Conjecture Take xi = βiminus1 logαn = d so that
x1 xn e
x1 exn
islogα β logα βdminus1 logα α αβ αβ
dminus1
The conclusion of Schanuelrsquos Conjecture is
tr degQQ(logα αβ αβ
2
αβdminus1)
= d
32 59
Algebraic independence method of Gelrsquofond
AO Gelrsquofond (1948)
The two numbers 23radic2 and
23radic4 are algebraically
independentMore generally if α is analgebraic number α 6= 0α 6= 1 and if β is a algebraicnumber of degree d ge 3 thentwo at least of the numbers
αβ αβ2
αβdminus1
are algebraically independent
33 59
Tools
Transcendence criterion Replaces Liouvillersquos inequality intranscendence proofsLiouville A nonndashzero rational integer n isin Z satisfies |n| ge 1Gelrsquofond Needs to give a lower bound for |P (θ)| withP isin Z[X] 0 when θ is transcendental
Zero estimate for exponential polynomials C Hermite P Turan K Mahler R Tijdeman
Small transcendence degree AO Gelrsquofond AA Smelev R Tijdeman WD Brownawell
34 59
Large transcendence degree
GV Chudnovsky (1976)Among the numbers
αβ αβ2
αβdminus1
at least [log2 d] arealgebraically independent
GV Chudnovsky ndash On the path to Schanuelrsquosconjecture Algebraic curves close to a pointI General theory of colored sequencesII Fields of finite transcendence type and coloredsequences ResultantsStudia Sci Math Hungar 12 (1977) 125ndash157 (1980)
35 59
Partial result on the problem of Gelrsquofond and
Schneider
AO Gelrsquofond GV Chudnovskii P Philippon YuV Nesterenko
G Diaz If α is an algebraic number α 6= 0 α 6= 1 and if β isan irrational algebraic number of degree d then
tr degQQ(αβ αβ
2
αβdminus1) ge [d+ 1
2
]
36 59
Conjecture of algebraic independence of logarithms
of algebraic numbers
Denote by L the set of complex numbers λ for which eλ isalgebraic
L = logα α isin Qtimes
Hence L is a Q-vector subspace of CThe most important special case of Schanuelrsquos Conjecture is
Conjecture Let λ1 λn be Q-linearly independent elementsin L Then the numbers λ1 λn are algebraicallyindependent over Q
Not yet known that the transcendence degree is ge 2 Open problem Among all logarithms of algebraic numbersone at least is transcendental over Q(π)
37 59
Structural rank of a matrix
Let K be a field k a subfield and M a matrix with entries in KConsider the k-vector subspace E of K spanned by the entriesof M Choose an injective morphism ϕ of E into a k-vectorspace kX1 + middot middot middot+ kXn The image ϕ(M) of M is a matrixwhose entries are in the field k(X1 Xn) of rationalfractions Its rank does not depend on the choice of ϕ
This is the structural rank of M with respect to k
38 59
Example
LetM =
(bij + λij
)1leiled1lejle`
be a matrix with coefficients in Q + L Consider a basis of theQndashvector spanned by the entries and replace the elements inthis basis by unknowns This gives a new matrix M withcoefficients in a field of rational fractions the rank of which isthe structural rank of M (with respect to Q)
As a consequence of the conjecture of algebraic independenceof logarithms of algebraic numbers the rank of M should bethe same as the rank of its specialization M
39 59
Equivalence between the two conjectures
Following D Roy theconjecture on algebraicindependence of logarithms ofalgebraic numbers isequivalent to
Conjecture Any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L has a rank equal to its structuralrank
40 59
Any Polynomial is the Determinant of a Matrix
The proof of the equivalence uses the nice auxiliary result
For any P isin k[X1 Xn] there exists a square matrix withentries in the k-vector space k + kX1 + middot middot middot+ kXn whosedeterminant is P
41 59
Half of the Conjecture is solved
From a certain point of view half of the conjecture of algebraicindependence of logarithms of algebraic numbers is solved
Theorem [D Roy] The rank of any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L is at least half its structural rank
42 59
Reformulation by D Roy
Instead of taking logarithmsof algebraic numbers andlooking for the algebraicindependence relationsD Roy fixes a polynomial andlooks at the points withcoordinates logarithms ofalgebraic numbers on thecorresponding hypersurface
43 59
Reformulation by D Roy
Royrsquos reformulation of the conjecture of algebraicindependence of logarithms is
Conjecture For any algebraic subvariety V of Cn definedover the field Q of algebraic numbers the set V cap Ln is theunion of the sets E cap Ln where E ranges over the set ofvector subspaces of Cn which are defined over Q andcontained in V
Trivial Any element in E cap Ln where E is a vector subspaceof Cn defined over Q and contained in V belongs to V cap Ln
44 59
Example The Four Exponentials ConjectureTake for V the hypersurface of C4 defined by the equation
z1z4 = z2z3
The maximal Cndashvector subspaces of C4 defined over Q andcontained in V are the planes
az1 = bz2 bz4 = az3
and the planesaz1 = bz3 bz4 = az2
with (a b) isin Q2 (0 0)Hence Schanuelrsquos Conjecture implies that if λ1 λ2 λ3 λ4 arelogarithms of algebraic numbers satisfying
λ1λ4 = λ2λ3
then either λ1 λ2 are linearly dependent over Q or else λ1 λ3
are linearly dependent over Q45 59
Six Exponentials Theorem
and Four Exponentials Conjecture
A Selberg CL Siegel S Lang K Ramachandra
46 59
The Four Exponentials Conjecture
Conjecture (Th Schneider S Lang K Ramachandra) Ifx1 x2 are Qndashlinearly independent complex numbers and y1 y2
are Qndashlinearly independent complex numbers then one atleast of the four numbers
ex1y1 ex1y2 ex2y1 ex2y2
is transcendentalEquivalent statement Any 2times 2 matrix(
λ1 λ2
λ3 λ4
)with entries in L and with Qndashlinearly independent rows andQndashlinearly independent columns has maximal rank 2
47 59
How could we attack Schanuelrsquos Conjecture
Let x1 xn be Qndashlinearly independent complex numbersFollowing the transcendence methods of Hermite GelrsquofondSchneider one may start by introducing an auxiliaryfunction
F (z) = P (z ez)
where P isin Z[X0 X1] is a nonndashzero polynomial One considersthe derivatives (
d
dz
)kF = (DkP )(z ez)
at the pointsm1x1 + middot middot middot+mnxn
for various values of (m1 mn) isin Zn
48 59
Auxiliary functionLet D denote the derivation
D =part
partX0
+X1part
partX1
over the ring C[X0 X1] Recall that x1 xn are Qndashlinearlyindependent complex numbers Let α1 αn be nonndashzerocomplex numbersThe transcendence machinery produces a sequence (PN)Nge0
of polynomials with integer coefficients satisfying∣∣∣∣∣(DkPN)(nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1
49 59
Royrsquos approach to Schanuelrsquos Conjecture (1999)
If the number of equations we produce is too small such a setof relations does not contain any information The existenceof a sequence of nonndashtrivial polynomials (PN)Nge0 follows fromlinear algebra
On the other hand following D Roy one may expect that theexistence of such a sequence (PN)Nge0 producing sufficientlymany such equations will yield the desired conclusion
tr degQQ(x1 xn α1 αn
)ge n
50 59
Royrsquos Conjecture (1999)Let s0 s1 t0 t1 u positive real numbers satisfying
max1 t0 2t1 lt mins0 2s1and
maxs0 s1 + t1 lt u lt1
2(1 + t0 + t1)
Assume that for any sufficiently large positive integer N there exists a nonndashzero polynomial PN isin Z[X0 X1] withpartial degree le N t0 in X0 partial degree le N t1 in X1 andheight le eN which satisfies∣∣∣∣∣(DkPN)(
nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1 Then
tr degQQ(x1 xn α1 αn) ge n51 59
Royrsquos Theorem (1999)
Royrsquos conjecture is equivalent to Schanuelrsquos Conjecture
52 59
Equivalence between Schanuel and Roy
Let (x α) isin CtimesCtimes and let s0 s1 t0 t1 u be positive realnumbers satisfying the inequalities of Royrsquos Conjecture Thenthe following conditions are equivalent (a) The number αeminusx is a root of unity
(b) For any sufficiently large positive integer N there exists anonndashzero polynomial QN isin Z[X0 X1] with partial degreele N t0 in X0 partial degree le N t1 in X1 and heightH(QN) le eN such that∣∣(DkQN)(mx αm)
∣∣ le exp(minusNu)
for any km isin N with k le N s0 and m le N s1
53 59
Gelrsquofondrsquos transcendence criterion
Simple form Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le n and height le en such that
|Pn(ϑ)| le eminus7n2
for all n ge 1 then ϑ is algebraic and Pn(ϑ) = 0 for all n ge 1
First extension Replace the bound for the degree by le dnthe bound for the height by ebn and the bound for |Pn(ϑ)| byeminuscdnbn with some constant c gt 0 independent of n
Some mild conditions are required on the sequences (dn)nge1 and(bn)nge1
54 59
Transcendence criterion with multiplicitiesWith derivatives Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le dn and height le ebn such that∣∣P (j)
n (ϑ)∣∣ le eminuscdnbntn
for all j in the range 0 le j lt tn and all n ge 1 then ϑ isalgebraicDue to M Laurent and D Roy applications to algebraicindependence
55 59
Criterion with several points
Goal Given a sequence of complex numbers (ϑi)ige1 if thereexists a sequence (Pn)nge1 of nonndashzero polynomials in Z[X]with Pn of degree le dn and height le ebn such that∣∣P (j)
n (ϑi)∣∣ le eminuscdnbntnsn
for 0 le j lt tn 1 le i le sn and all n ge 1 then the numbers ϑiare algebraic
D Roy Not true in general but true in some special caseswith a structure on the sequence (ϑi)ige1Combines the elimination arguments used for criteria ofalgebraic independence and for zero estimates
56 59
Small value estimates for the additive group
D Roy Small value estimates for the additive group Intern JNumber Theory to appear
Let ξ be a transcendental complex number let β σ τ and νbe non-negative real numbers let n0 be a positive integer andlet (Pn)ngen0 be a sequence of nonndashzero polynomials in Z[T ]satisfying deg(Pn) le n and H(Pn) le exp(nβ) for eachn ge n0 Suppose that β gt 1 (34)σ + τ lt 1 andν gt 1 + β minus (34)σ minus τ Then for infinitely many n we have
max|P [j]n (iξ)| 0 le i le nσ 0 le j le nτ
gt exp(minusnν)
57 59
Small value estimates for the multiplicative group
D Roy Small value estimates for the multiplicative groupActa Arith to appear
Let ξ1 ξm be multiplicatively independent complexnumbers in a field of transcendence degree 1 Under suitableassumptions on the parameters β σ τ ν for infinitely manypositive integers n there exists no nonndashzero polynomialP isin Z[T ] satisfying deg(P ) le n H(P ) le exp(nβ) and
max|P [j](ξi11 middot middot middot ξimm )| 0 le i1 im le nσ 0 le j le nτ
gt exp(minusnν)
58 59
Heidelberg Colloquium organized by the Institute for Mathematics
httpwwwmathiuni-heidelbergdevortraghtmls=1 January 22 2009
Transcendental Number TheorySchanuelrsquos Conjecture
Michel Waldschmidt
Institut de Mathematiques de Jussieu amp Paris VIhttpwwwmathjussieufrsimmiw
59 59
A variant of Langrsquos exercise
Define L0 = Q Inductively for n ge 1 define Ln as thealgebraic closure of the field generated over Lnminus1 by thenumbers y where y ranges over the set of complex numberssuch that ey isin Lnminus1 Let L be the union of Ln n ge 0Then Schanuelrsquos Conjecture implies that the number e doesnot belong to L
More precisely Schanuelrsquos Conjecture implies that the
numbers e ee eee ee
ee
are algebraically independent over L
24 59
Arizona Winter School AWS2008 Tucson
Theorem [Jonathan Bober Chuangxun Cheng Brian DietelMathilde Herblot Jingjing Huang Holly Krieger DiegoMarques Jonathan Mason Martin Mereb and Robert Wilson]Schanuelrsquos Conjecture implies that the fields E and L arelinearly disjoint over Q
Definition Given a field extension FK and twosubextensions F1 F2 sube F we say F1 F2 are linearly disjointover K when the following holds Any set x1 xn sube F1
of Kndash linearly independent elements is linearly independentover F2
Reference arXiv08043550 [mathNT] 2008
25 59
Formal analogs
WD Brownawell(was a student of Schanuel)
J Axrsquos Theorem (1968) Version of SchanuelrsquosConjecture for power seriesover C(and R Coleman for powerseries over Q)
Work by WD Brownawelland K Kubota on the ellipticanalog of Axrsquos Theorem
26 59
Conjectures by A Grothendieck and Y Andre
Generalized Conjecture onPeriods by Grothendieck Dimension of theMumfordndashTate group of asmooth projective variety
Generalization by Y Andre tomotives
Case of 1ndashmotives Elliptico-Toric Conjecture ofC Bertolin
27 59
Ubiquity of Schanuelrsquos Conjecture
Other contexts pndashadic numbers Leopoldtrsquos Conjecture onthe pndashadic rank of the units of an algebraic number fieldNon-vanishing of RegulatorsNonndashdegenerescence of heightsConjecture of B Mazur on rational pointsDiophantine approximation on tori
Dipendra Prasad Gopal Prasad
28 59
Methods from logicEhud Hrushovski Boris Zilber Jonathan Kirby
Calculus of rdquopredimension functionsrdquo (E Hrushovski)
Zilberrsquos construction of a rdquopseudoexponentiationrdquo
Also A Macintyre DE Marker G Terzo AJ WilkieD Bertrand
29 59
Methods from logic Model theory
Exponential algebraicity in exponential fieldsby
Jonathan Kirby
The dimension of the exponential algebraic closure operator inan exponential field satisfies a weak Schanuel property
A corollary is that there are at most countably many essentialcounterexamples to Schanuelrsquos conjecture
arXiv 08104285v2
30 59
Known
LindemannndashWeierstraszlig Theorem = case where x1 xn arealgebraic
Let β1 βn be algebraic numbers which are linearlyindependent over Q Then the numbers eβ1 eβn arealgebraically independent over Q
31 59
Problem of Gelrsquofond and SchneiderRaised by AO Gelrsquofond in 1948 and Th Schneider in 1952
Conjecture If α is an algebraic number α 6= 0 α 6= 1 and ifβ is an irrational algebraic number of degree d then the dminus 1numbers
αβ αβ2
αβdminus1
are algebraically independent
Special case of Schanuelrsquos Conjecture Take xi = βiminus1 logαn = d so that
x1 xn e
x1 exn
islogα β logα βdminus1 logα α αβ αβ
dminus1
The conclusion of Schanuelrsquos Conjecture is
tr degQQ(logα αβ αβ
2
αβdminus1)
= d
32 59
Algebraic independence method of Gelrsquofond
AO Gelrsquofond (1948)
The two numbers 23radic2 and
23radic4 are algebraically
independentMore generally if α is analgebraic number α 6= 0α 6= 1 and if β is a algebraicnumber of degree d ge 3 thentwo at least of the numbers
αβ αβ2
αβdminus1
are algebraically independent
33 59
Tools
Transcendence criterion Replaces Liouvillersquos inequality intranscendence proofsLiouville A nonndashzero rational integer n isin Z satisfies |n| ge 1Gelrsquofond Needs to give a lower bound for |P (θ)| withP isin Z[X] 0 when θ is transcendental
Zero estimate for exponential polynomials C Hermite P Turan K Mahler R Tijdeman
Small transcendence degree AO Gelrsquofond AA Smelev R Tijdeman WD Brownawell
34 59
Large transcendence degree
GV Chudnovsky (1976)Among the numbers
αβ αβ2
αβdminus1
at least [log2 d] arealgebraically independent
GV Chudnovsky ndash On the path to Schanuelrsquosconjecture Algebraic curves close to a pointI General theory of colored sequencesII Fields of finite transcendence type and coloredsequences ResultantsStudia Sci Math Hungar 12 (1977) 125ndash157 (1980)
35 59
Partial result on the problem of Gelrsquofond and
Schneider
AO Gelrsquofond GV Chudnovskii P Philippon YuV Nesterenko
G Diaz If α is an algebraic number α 6= 0 α 6= 1 and if β isan irrational algebraic number of degree d then
tr degQQ(αβ αβ
2
αβdminus1) ge [d+ 1
2
]
36 59
Conjecture of algebraic independence of logarithms
of algebraic numbers
Denote by L the set of complex numbers λ for which eλ isalgebraic
L = logα α isin Qtimes
Hence L is a Q-vector subspace of CThe most important special case of Schanuelrsquos Conjecture is
Conjecture Let λ1 λn be Q-linearly independent elementsin L Then the numbers λ1 λn are algebraicallyindependent over Q
Not yet known that the transcendence degree is ge 2 Open problem Among all logarithms of algebraic numbersone at least is transcendental over Q(π)
37 59
Structural rank of a matrix
Let K be a field k a subfield and M a matrix with entries in KConsider the k-vector subspace E of K spanned by the entriesof M Choose an injective morphism ϕ of E into a k-vectorspace kX1 + middot middot middot+ kXn The image ϕ(M) of M is a matrixwhose entries are in the field k(X1 Xn) of rationalfractions Its rank does not depend on the choice of ϕ
This is the structural rank of M with respect to k
38 59
Example
LetM =
(bij + λij
)1leiled1lejle`
be a matrix with coefficients in Q + L Consider a basis of theQndashvector spanned by the entries and replace the elements inthis basis by unknowns This gives a new matrix M withcoefficients in a field of rational fractions the rank of which isthe structural rank of M (with respect to Q)
As a consequence of the conjecture of algebraic independenceof logarithms of algebraic numbers the rank of M should bethe same as the rank of its specialization M
39 59
Equivalence between the two conjectures
Following D Roy theconjecture on algebraicindependence of logarithms ofalgebraic numbers isequivalent to
Conjecture Any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L has a rank equal to its structuralrank
40 59
Any Polynomial is the Determinant of a Matrix
The proof of the equivalence uses the nice auxiliary result
For any P isin k[X1 Xn] there exists a square matrix withentries in the k-vector space k + kX1 + middot middot middot+ kXn whosedeterminant is P
41 59
Half of the Conjecture is solved
From a certain point of view half of the conjecture of algebraicindependence of logarithms of algebraic numbers is solved
Theorem [D Roy] The rank of any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L is at least half its structural rank
42 59
Reformulation by D Roy
Instead of taking logarithmsof algebraic numbers andlooking for the algebraicindependence relationsD Roy fixes a polynomial andlooks at the points withcoordinates logarithms ofalgebraic numbers on thecorresponding hypersurface
43 59
Reformulation by D Roy
Royrsquos reformulation of the conjecture of algebraicindependence of logarithms is
Conjecture For any algebraic subvariety V of Cn definedover the field Q of algebraic numbers the set V cap Ln is theunion of the sets E cap Ln where E ranges over the set ofvector subspaces of Cn which are defined over Q andcontained in V
Trivial Any element in E cap Ln where E is a vector subspaceof Cn defined over Q and contained in V belongs to V cap Ln
44 59
Example The Four Exponentials ConjectureTake for V the hypersurface of C4 defined by the equation
z1z4 = z2z3
The maximal Cndashvector subspaces of C4 defined over Q andcontained in V are the planes
az1 = bz2 bz4 = az3
and the planesaz1 = bz3 bz4 = az2
with (a b) isin Q2 (0 0)Hence Schanuelrsquos Conjecture implies that if λ1 λ2 λ3 λ4 arelogarithms of algebraic numbers satisfying
λ1λ4 = λ2λ3
then either λ1 λ2 are linearly dependent over Q or else λ1 λ3
are linearly dependent over Q45 59
Six Exponentials Theorem
and Four Exponentials Conjecture
A Selberg CL Siegel S Lang K Ramachandra
46 59
The Four Exponentials Conjecture
Conjecture (Th Schneider S Lang K Ramachandra) Ifx1 x2 are Qndashlinearly independent complex numbers and y1 y2
are Qndashlinearly independent complex numbers then one atleast of the four numbers
ex1y1 ex1y2 ex2y1 ex2y2
is transcendentalEquivalent statement Any 2times 2 matrix(
λ1 λ2
λ3 λ4
)with entries in L and with Qndashlinearly independent rows andQndashlinearly independent columns has maximal rank 2
47 59
How could we attack Schanuelrsquos Conjecture
Let x1 xn be Qndashlinearly independent complex numbersFollowing the transcendence methods of Hermite GelrsquofondSchneider one may start by introducing an auxiliaryfunction
F (z) = P (z ez)
where P isin Z[X0 X1] is a nonndashzero polynomial One considersthe derivatives (
d
dz
)kF = (DkP )(z ez)
at the pointsm1x1 + middot middot middot+mnxn
for various values of (m1 mn) isin Zn
48 59
Auxiliary functionLet D denote the derivation
D =part
partX0
+X1part
partX1
over the ring C[X0 X1] Recall that x1 xn are Qndashlinearlyindependent complex numbers Let α1 αn be nonndashzerocomplex numbersThe transcendence machinery produces a sequence (PN)Nge0
of polynomials with integer coefficients satisfying∣∣∣∣∣(DkPN)(nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1
49 59
Royrsquos approach to Schanuelrsquos Conjecture (1999)
If the number of equations we produce is too small such a setof relations does not contain any information The existenceof a sequence of nonndashtrivial polynomials (PN)Nge0 follows fromlinear algebra
On the other hand following D Roy one may expect that theexistence of such a sequence (PN)Nge0 producing sufficientlymany such equations will yield the desired conclusion
tr degQQ(x1 xn α1 αn
)ge n
50 59
Royrsquos Conjecture (1999)Let s0 s1 t0 t1 u positive real numbers satisfying
max1 t0 2t1 lt mins0 2s1and
maxs0 s1 + t1 lt u lt1
2(1 + t0 + t1)
Assume that for any sufficiently large positive integer N there exists a nonndashzero polynomial PN isin Z[X0 X1] withpartial degree le N t0 in X0 partial degree le N t1 in X1 andheight le eN which satisfies∣∣∣∣∣(DkPN)(
nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1 Then
tr degQQ(x1 xn α1 αn) ge n51 59
Royrsquos Theorem (1999)
Royrsquos conjecture is equivalent to Schanuelrsquos Conjecture
52 59
Equivalence between Schanuel and Roy
Let (x α) isin CtimesCtimes and let s0 s1 t0 t1 u be positive realnumbers satisfying the inequalities of Royrsquos Conjecture Thenthe following conditions are equivalent (a) The number αeminusx is a root of unity
(b) For any sufficiently large positive integer N there exists anonndashzero polynomial QN isin Z[X0 X1] with partial degreele N t0 in X0 partial degree le N t1 in X1 and heightH(QN) le eN such that∣∣(DkQN)(mx αm)
∣∣ le exp(minusNu)
for any km isin N with k le N s0 and m le N s1
53 59
Gelrsquofondrsquos transcendence criterion
Simple form Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le n and height le en such that
|Pn(ϑ)| le eminus7n2
for all n ge 1 then ϑ is algebraic and Pn(ϑ) = 0 for all n ge 1
First extension Replace the bound for the degree by le dnthe bound for the height by ebn and the bound for |Pn(ϑ)| byeminuscdnbn with some constant c gt 0 independent of n
Some mild conditions are required on the sequences (dn)nge1 and(bn)nge1
54 59
Transcendence criterion with multiplicitiesWith derivatives Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le dn and height le ebn such that∣∣P (j)
n (ϑ)∣∣ le eminuscdnbntn
for all j in the range 0 le j lt tn and all n ge 1 then ϑ isalgebraicDue to M Laurent and D Roy applications to algebraicindependence
55 59
Criterion with several points
Goal Given a sequence of complex numbers (ϑi)ige1 if thereexists a sequence (Pn)nge1 of nonndashzero polynomials in Z[X]with Pn of degree le dn and height le ebn such that∣∣P (j)
n (ϑi)∣∣ le eminuscdnbntnsn
for 0 le j lt tn 1 le i le sn and all n ge 1 then the numbers ϑiare algebraic
D Roy Not true in general but true in some special caseswith a structure on the sequence (ϑi)ige1Combines the elimination arguments used for criteria ofalgebraic independence and for zero estimates
56 59
Small value estimates for the additive group
D Roy Small value estimates for the additive group Intern JNumber Theory to appear
Let ξ be a transcendental complex number let β σ τ and νbe non-negative real numbers let n0 be a positive integer andlet (Pn)ngen0 be a sequence of nonndashzero polynomials in Z[T ]satisfying deg(Pn) le n and H(Pn) le exp(nβ) for eachn ge n0 Suppose that β gt 1 (34)σ + τ lt 1 andν gt 1 + β minus (34)σ minus τ Then for infinitely many n we have
max|P [j]n (iξ)| 0 le i le nσ 0 le j le nτ
gt exp(minusnν)
57 59
Small value estimates for the multiplicative group
D Roy Small value estimates for the multiplicative groupActa Arith to appear
Let ξ1 ξm be multiplicatively independent complexnumbers in a field of transcendence degree 1 Under suitableassumptions on the parameters β σ τ ν for infinitely manypositive integers n there exists no nonndashzero polynomialP isin Z[T ] satisfying deg(P ) le n H(P ) le exp(nβ) and
max|P [j](ξi11 middot middot middot ξimm )| 0 le i1 im le nσ 0 le j le nτ
gt exp(minusnν)
58 59
Heidelberg Colloquium organized by the Institute for Mathematics
httpwwwmathiuni-heidelbergdevortraghtmls=1 January 22 2009
Transcendental Number TheorySchanuelrsquos Conjecture
Michel Waldschmidt
Institut de Mathematiques de Jussieu amp Paris VIhttpwwwmathjussieufrsimmiw
59 59
Arizona Winter School AWS2008 Tucson
Theorem [Jonathan Bober Chuangxun Cheng Brian DietelMathilde Herblot Jingjing Huang Holly Krieger DiegoMarques Jonathan Mason Martin Mereb and Robert Wilson]Schanuelrsquos Conjecture implies that the fields E and L arelinearly disjoint over Q
Definition Given a field extension FK and twosubextensions F1 F2 sube F we say F1 F2 are linearly disjointover K when the following holds Any set x1 xn sube F1
of Kndash linearly independent elements is linearly independentover F2
Reference arXiv08043550 [mathNT] 2008
25 59
Formal analogs
WD Brownawell(was a student of Schanuel)
J Axrsquos Theorem (1968) Version of SchanuelrsquosConjecture for power seriesover C(and R Coleman for powerseries over Q)
Work by WD Brownawelland K Kubota on the ellipticanalog of Axrsquos Theorem
26 59
Conjectures by A Grothendieck and Y Andre
Generalized Conjecture onPeriods by Grothendieck Dimension of theMumfordndashTate group of asmooth projective variety
Generalization by Y Andre tomotives
Case of 1ndashmotives Elliptico-Toric Conjecture ofC Bertolin
27 59
Ubiquity of Schanuelrsquos Conjecture
Other contexts pndashadic numbers Leopoldtrsquos Conjecture onthe pndashadic rank of the units of an algebraic number fieldNon-vanishing of RegulatorsNonndashdegenerescence of heightsConjecture of B Mazur on rational pointsDiophantine approximation on tori
Dipendra Prasad Gopal Prasad
28 59
Methods from logicEhud Hrushovski Boris Zilber Jonathan Kirby
Calculus of rdquopredimension functionsrdquo (E Hrushovski)
Zilberrsquos construction of a rdquopseudoexponentiationrdquo
Also A Macintyre DE Marker G Terzo AJ WilkieD Bertrand
29 59
Methods from logic Model theory
Exponential algebraicity in exponential fieldsby
Jonathan Kirby
The dimension of the exponential algebraic closure operator inan exponential field satisfies a weak Schanuel property
A corollary is that there are at most countably many essentialcounterexamples to Schanuelrsquos conjecture
arXiv 08104285v2
30 59
Known
LindemannndashWeierstraszlig Theorem = case where x1 xn arealgebraic
Let β1 βn be algebraic numbers which are linearlyindependent over Q Then the numbers eβ1 eβn arealgebraically independent over Q
31 59
Problem of Gelrsquofond and SchneiderRaised by AO Gelrsquofond in 1948 and Th Schneider in 1952
Conjecture If α is an algebraic number α 6= 0 α 6= 1 and ifβ is an irrational algebraic number of degree d then the dminus 1numbers
αβ αβ2
αβdminus1
are algebraically independent
Special case of Schanuelrsquos Conjecture Take xi = βiminus1 logαn = d so that
x1 xn e
x1 exn
islogα β logα βdminus1 logα α αβ αβ
dminus1
The conclusion of Schanuelrsquos Conjecture is
tr degQQ(logα αβ αβ
2
αβdminus1)
= d
32 59
Algebraic independence method of Gelrsquofond
AO Gelrsquofond (1948)
The two numbers 23radic2 and
23radic4 are algebraically
independentMore generally if α is analgebraic number α 6= 0α 6= 1 and if β is a algebraicnumber of degree d ge 3 thentwo at least of the numbers
αβ αβ2
αβdminus1
are algebraically independent
33 59
Tools
Transcendence criterion Replaces Liouvillersquos inequality intranscendence proofsLiouville A nonndashzero rational integer n isin Z satisfies |n| ge 1Gelrsquofond Needs to give a lower bound for |P (θ)| withP isin Z[X] 0 when θ is transcendental
Zero estimate for exponential polynomials C Hermite P Turan K Mahler R Tijdeman
Small transcendence degree AO Gelrsquofond AA Smelev R Tijdeman WD Brownawell
34 59
Large transcendence degree
GV Chudnovsky (1976)Among the numbers
αβ αβ2
αβdminus1
at least [log2 d] arealgebraically independent
GV Chudnovsky ndash On the path to Schanuelrsquosconjecture Algebraic curves close to a pointI General theory of colored sequencesII Fields of finite transcendence type and coloredsequences ResultantsStudia Sci Math Hungar 12 (1977) 125ndash157 (1980)
35 59
Partial result on the problem of Gelrsquofond and
Schneider
AO Gelrsquofond GV Chudnovskii P Philippon YuV Nesterenko
G Diaz If α is an algebraic number α 6= 0 α 6= 1 and if β isan irrational algebraic number of degree d then
tr degQQ(αβ αβ
2
αβdminus1) ge [d+ 1
2
]
36 59
Conjecture of algebraic independence of logarithms
of algebraic numbers
Denote by L the set of complex numbers λ for which eλ isalgebraic
L = logα α isin Qtimes
Hence L is a Q-vector subspace of CThe most important special case of Schanuelrsquos Conjecture is
Conjecture Let λ1 λn be Q-linearly independent elementsin L Then the numbers λ1 λn are algebraicallyindependent over Q
Not yet known that the transcendence degree is ge 2 Open problem Among all logarithms of algebraic numbersone at least is transcendental over Q(π)
37 59
Structural rank of a matrix
Let K be a field k a subfield and M a matrix with entries in KConsider the k-vector subspace E of K spanned by the entriesof M Choose an injective morphism ϕ of E into a k-vectorspace kX1 + middot middot middot+ kXn The image ϕ(M) of M is a matrixwhose entries are in the field k(X1 Xn) of rationalfractions Its rank does not depend on the choice of ϕ
This is the structural rank of M with respect to k
38 59
Example
LetM =
(bij + λij
)1leiled1lejle`
be a matrix with coefficients in Q + L Consider a basis of theQndashvector spanned by the entries and replace the elements inthis basis by unknowns This gives a new matrix M withcoefficients in a field of rational fractions the rank of which isthe structural rank of M (with respect to Q)
As a consequence of the conjecture of algebraic independenceof logarithms of algebraic numbers the rank of M should bethe same as the rank of its specialization M
39 59
Equivalence between the two conjectures
Following D Roy theconjecture on algebraicindependence of logarithms ofalgebraic numbers isequivalent to
Conjecture Any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L has a rank equal to its structuralrank
40 59
Any Polynomial is the Determinant of a Matrix
The proof of the equivalence uses the nice auxiliary result
For any P isin k[X1 Xn] there exists a square matrix withentries in the k-vector space k + kX1 + middot middot middot+ kXn whosedeterminant is P
41 59
Half of the Conjecture is solved
From a certain point of view half of the conjecture of algebraicindependence of logarithms of algebraic numbers is solved
Theorem [D Roy] The rank of any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L is at least half its structural rank
42 59
Reformulation by D Roy
Instead of taking logarithmsof algebraic numbers andlooking for the algebraicindependence relationsD Roy fixes a polynomial andlooks at the points withcoordinates logarithms ofalgebraic numbers on thecorresponding hypersurface
43 59
Reformulation by D Roy
Royrsquos reformulation of the conjecture of algebraicindependence of logarithms is
Conjecture For any algebraic subvariety V of Cn definedover the field Q of algebraic numbers the set V cap Ln is theunion of the sets E cap Ln where E ranges over the set ofvector subspaces of Cn which are defined over Q andcontained in V
Trivial Any element in E cap Ln where E is a vector subspaceof Cn defined over Q and contained in V belongs to V cap Ln
44 59
Example The Four Exponentials ConjectureTake for V the hypersurface of C4 defined by the equation
z1z4 = z2z3
The maximal Cndashvector subspaces of C4 defined over Q andcontained in V are the planes
az1 = bz2 bz4 = az3
and the planesaz1 = bz3 bz4 = az2
with (a b) isin Q2 (0 0)Hence Schanuelrsquos Conjecture implies that if λ1 λ2 λ3 λ4 arelogarithms of algebraic numbers satisfying
λ1λ4 = λ2λ3
then either λ1 λ2 are linearly dependent over Q or else λ1 λ3
are linearly dependent over Q45 59
Six Exponentials Theorem
and Four Exponentials Conjecture
A Selberg CL Siegel S Lang K Ramachandra
46 59
The Four Exponentials Conjecture
Conjecture (Th Schneider S Lang K Ramachandra) Ifx1 x2 are Qndashlinearly independent complex numbers and y1 y2
are Qndashlinearly independent complex numbers then one atleast of the four numbers
ex1y1 ex1y2 ex2y1 ex2y2
is transcendentalEquivalent statement Any 2times 2 matrix(
λ1 λ2
λ3 λ4
)with entries in L and with Qndashlinearly independent rows andQndashlinearly independent columns has maximal rank 2
47 59
How could we attack Schanuelrsquos Conjecture
Let x1 xn be Qndashlinearly independent complex numbersFollowing the transcendence methods of Hermite GelrsquofondSchneider one may start by introducing an auxiliaryfunction
F (z) = P (z ez)
where P isin Z[X0 X1] is a nonndashzero polynomial One considersthe derivatives (
d
dz
)kF = (DkP )(z ez)
at the pointsm1x1 + middot middot middot+mnxn
for various values of (m1 mn) isin Zn
48 59
Auxiliary functionLet D denote the derivation
D =part
partX0
+X1part
partX1
over the ring C[X0 X1] Recall that x1 xn are Qndashlinearlyindependent complex numbers Let α1 αn be nonndashzerocomplex numbersThe transcendence machinery produces a sequence (PN)Nge0
of polynomials with integer coefficients satisfying∣∣∣∣∣(DkPN)(nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1
49 59
Royrsquos approach to Schanuelrsquos Conjecture (1999)
If the number of equations we produce is too small such a setof relations does not contain any information The existenceof a sequence of nonndashtrivial polynomials (PN)Nge0 follows fromlinear algebra
On the other hand following D Roy one may expect that theexistence of such a sequence (PN)Nge0 producing sufficientlymany such equations will yield the desired conclusion
tr degQQ(x1 xn α1 αn
)ge n
50 59
Royrsquos Conjecture (1999)Let s0 s1 t0 t1 u positive real numbers satisfying
max1 t0 2t1 lt mins0 2s1and
maxs0 s1 + t1 lt u lt1
2(1 + t0 + t1)
Assume that for any sufficiently large positive integer N there exists a nonndashzero polynomial PN isin Z[X0 X1] withpartial degree le N t0 in X0 partial degree le N t1 in X1 andheight le eN which satisfies∣∣∣∣∣(DkPN)(
nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1 Then
tr degQQ(x1 xn α1 αn) ge n51 59
Royrsquos Theorem (1999)
Royrsquos conjecture is equivalent to Schanuelrsquos Conjecture
52 59
Equivalence between Schanuel and Roy
Let (x α) isin CtimesCtimes and let s0 s1 t0 t1 u be positive realnumbers satisfying the inequalities of Royrsquos Conjecture Thenthe following conditions are equivalent (a) The number αeminusx is a root of unity
(b) For any sufficiently large positive integer N there exists anonndashzero polynomial QN isin Z[X0 X1] with partial degreele N t0 in X0 partial degree le N t1 in X1 and heightH(QN) le eN such that∣∣(DkQN)(mx αm)
∣∣ le exp(minusNu)
for any km isin N with k le N s0 and m le N s1
53 59
Gelrsquofondrsquos transcendence criterion
Simple form Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le n and height le en such that
|Pn(ϑ)| le eminus7n2
for all n ge 1 then ϑ is algebraic and Pn(ϑ) = 0 for all n ge 1
First extension Replace the bound for the degree by le dnthe bound for the height by ebn and the bound for |Pn(ϑ)| byeminuscdnbn with some constant c gt 0 independent of n
Some mild conditions are required on the sequences (dn)nge1 and(bn)nge1
54 59
Transcendence criterion with multiplicitiesWith derivatives Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le dn and height le ebn such that∣∣P (j)
n (ϑ)∣∣ le eminuscdnbntn
for all j in the range 0 le j lt tn and all n ge 1 then ϑ isalgebraicDue to M Laurent and D Roy applications to algebraicindependence
55 59
Criterion with several points
Goal Given a sequence of complex numbers (ϑi)ige1 if thereexists a sequence (Pn)nge1 of nonndashzero polynomials in Z[X]with Pn of degree le dn and height le ebn such that∣∣P (j)
n (ϑi)∣∣ le eminuscdnbntnsn
for 0 le j lt tn 1 le i le sn and all n ge 1 then the numbers ϑiare algebraic
D Roy Not true in general but true in some special caseswith a structure on the sequence (ϑi)ige1Combines the elimination arguments used for criteria ofalgebraic independence and for zero estimates
56 59
Small value estimates for the additive group
D Roy Small value estimates for the additive group Intern JNumber Theory to appear
Let ξ be a transcendental complex number let β σ τ and νbe non-negative real numbers let n0 be a positive integer andlet (Pn)ngen0 be a sequence of nonndashzero polynomials in Z[T ]satisfying deg(Pn) le n and H(Pn) le exp(nβ) for eachn ge n0 Suppose that β gt 1 (34)σ + τ lt 1 andν gt 1 + β minus (34)σ minus τ Then for infinitely many n we have
max|P [j]n (iξ)| 0 le i le nσ 0 le j le nτ
gt exp(minusnν)
57 59
Small value estimates for the multiplicative group
D Roy Small value estimates for the multiplicative groupActa Arith to appear
Let ξ1 ξm be multiplicatively independent complexnumbers in a field of transcendence degree 1 Under suitableassumptions on the parameters β σ τ ν for infinitely manypositive integers n there exists no nonndashzero polynomialP isin Z[T ] satisfying deg(P ) le n H(P ) le exp(nβ) and
max|P [j](ξi11 middot middot middot ξimm )| 0 le i1 im le nσ 0 le j le nτ
gt exp(minusnν)
58 59
Heidelberg Colloquium organized by the Institute for Mathematics
httpwwwmathiuni-heidelbergdevortraghtmls=1 January 22 2009
Transcendental Number TheorySchanuelrsquos Conjecture
Michel Waldschmidt
Institut de Mathematiques de Jussieu amp Paris VIhttpwwwmathjussieufrsimmiw
59 59
Formal analogs
WD Brownawell(was a student of Schanuel)
J Axrsquos Theorem (1968) Version of SchanuelrsquosConjecture for power seriesover C(and R Coleman for powerseries over Q)
Work by WD Brownawelland K Kubota on the ellipticanalog of Axrsquos Theorem
26 59
Conjectures by A Grothendieck and Y Andre
Generalized Conjecture onPeriods by Grothendieck Dimension of theMumfordndashTate group of asmooth projective variety
Generalization by Y Andre tomotives
Case of 1ndashmotives Elliptico-Toric Conjecture ofC Bertolin
27 59
Ubiquity of Schanuelrsquos Conjecture
Other contexts pndashadic numbers Leopoldtrsquos Conjecture onthe pndashadic rank of the units of an algebraic number fieldNon-vanishing of RegulatorsNonndashdegenerescence of heightsConjecture of B Mazur on rational pointsDiophantine approximation on tori
Dipendra Prasad Gopal Prasad
28 59
Methods from logicEhud Hrushovski Boris Zilber Jonathan Kirby
Calculus of rdquopredimension functionsrdquo (E Hrushovski)
Zilberrsquos construction of a rdquopseudoexponentiationrdquo
Also A Macintyre DE Marker G Terzo AJ WilkieD Bertrand
29 59
Methods from logic Model theory
Exponential algebraicity in exponential fieldsby
Jonathan Kirby
The dimension of the exponential algebraic closure operator inan exponential field satisfies a weak Schanuel property
A corollary is that there are at most countably many essentialcounterexamples to Schanuelrsquos conjecture
arXiv 08104285v2
30 59
Known
LindemannndashWeierstraszlig Theorem = case where x1 xn arealgebraic
Let β1 βn be algebraic numbers which are linearlyindependent over Q Then the numbers eβ1 eβn arealgebraically independent over Q
31 59
Problem of Gelrsquofond and SchneiderRaised by AO Gelrsquofond in 1948 and Th Schneider in 1952
Conjecture If α is an algebraic number α 6= 0 α 6= 1 and ifβ is an irrational algebraic number of degree d then the dminus 1numbers
αβ αβ2
αβdminus1
are algebraically independent
Special case of Schanuelrsquos Conjecture Take xi = βiminus1 logαn = d so that
x1 xn e
x1 exn
islogα β logα βdminus1 logα α αβ αβ
dminus1
The conclusion of Schanuelrsquos Conjecture is
tr degQQ(logα αβ αβ
2
αβdminus1)
= d
32 59
Algebraic independence method of Gelrsquofond
AO Gelrsquofond (1948)
The two numbers 23radic2 and
23radic4 are algebraically
independentMore generally if α is analgebraic number α 6= 0α 6= 1 and if β is a algebraicnumber of degree d ge 3 thentwo at least of the numbers
αβ αβ2
αβdminus1
are algebraically independent
33 59
Tools
Transcendence criterion Replaces Liouvillersquos inequality intranscendence proofsLiouville A nonndashzero rational integer n isin Z satisfies |n| ge 1Gelrsquofond Needs to give a lower bound for |P (θ)| withP isin Z[X] 0 when θ is transcendental
Zero estimate for exponential polynomials C Hermite P Turan K Mahler R Tijdeman
Small transcendence degree AO Gelrsquofond AA Smelev R Tijdeman WD Brownawell
34 59
Large transcendence degree
GV Chudnovsky (1976)Among the numbers
αβ αβ2
αβdminus1
at least [log2 d] arealgebraically independent
GV Chudnovsky ndash On the path to Schanuelrsquosconjecture Algebraic curves close to a pointI General theory of colored sequencesII Fields of finite transcendence type and coloredsequences ResultantsStudia Sci Math Hungar 12 (1977) 125ndash157 (1980)
35 59
Partial result on the problem of Gelrsquofond and
Schneider
AO Gelrsquofond GV Chudnovskii P Philippon YuV Nesterenko
G Diaz If α is an algebraic number α 6= 0 α 6= 1 and if β isan irrational algebraic number of degree d then
tr degQQ(αβ αβ
2
αβdminus1) ge [d+ 1
2
]
36 59
Conjecture of algebraic independence of logarithms
of algebraic numbers
Denote by L the set of complex numbers λ for which eλ isalgebraic
L = logα α isin Qtimes
Hence L is a Q-vector subspace of CThe most important special case of Schanuelrsquos Conjecture is
Conjecture Let λ1 λn be Q-linearly independent elementsin L Then the numbers λ1 λn are algebraicallyindependent over Q
Not yet known that the transcendence degree is ge 2 Open problem Among all logarithms of algebraic numbersone at least is transcendental over Q(π)
37 59
Structural rank of a matrix
Let K be a field k a subfield and M a matrix with entries in KConsider the k-vector subspace E of K spanned by the entriesof M Choose an injective morphism ϕ of E into a k-vectorspace kX1 + middot middot middot+ kXn The image ϕ(M) of M is a matrixwhose entries are in the field k(X1 Xn) of rationalfractions Its rank does not depend on the choice of ϕ
This is the structural rank of M with respect to k
38 59
Example
LetM =
(bij + λij
)1leiled1lejle`
be a matrix with coefficients in Q + L Consider a basis of theQndashvector spanned by the entries and replace the elements inthis basis by unknowns This gives a new matrix M withcoefficients in a field of rational fractions the rank of which isthe structural rank of M (with respect to Q)
As a consequence of the conjecture of algebraic independenceof logarithms of algebraic numbers the rank of M should bethe same as the rank of its specialization M
39 59
Equivalence between the two conjectures
Following D Roy theconjecture on algebraicindependence of logarithms ofalgebraic numbers isequivalent to
Conjecture Any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L has a rank equal to its structuralrank
40 59
Any Polynomial is the Determinant of a Matrix
The proof of the equivalence uses the nice auxiliary result
For any P isin k[X1 Xn] there exists a square matrix withentries in the k-vector space k + kX1 + middot middot middot+ kXn whosedeterminant is P
41 59
Half of the Conjecture is solved
From a certain point of view half of the conjecture of algebraicindependence of logarithms of algebraic numbers is solved
Theorem [D Roy] The rank of any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L is at least half its structural rank
42 59
Reformulation by D Roy
Instead of taking logarithmsof algebraic numbers andlooking for the algebraicindependence relationsD Roy fixes a polynomial andlooks at the points withcoordinates logarithms ofalgebraic numbers on thecorresponding hypersurface
43 59
Reformulation by D Roy
Royrsquos reformulation of the conjecture of algebraicindependence of logarithms is
Conjecture For any algebraic subvariety V of Cn definedover the field Q of algebraic numbers the set V cap Ln is theunion of the sets E cap Ln where E ranges over the set ofvector subspaces of Cn which are defined over Q andcontained in V
Trivial Any element in E cap Ln where E is a vector subspaceof Cn defined over Q and contained in V belongs to V cap Ln
44 59
Example The Four Exponentials ConjectureTake for V the hypersurface of C4 defined by the equation
z1z4 = z2z3
The maximal Cndashvector subspaces of C4 defined over Q andcontained in V are the planes
az1 = bz2 bz4 = az3
and the planesaz1 = bz3 bz4 = az2
with (a b) isin Q2 (0 0)Hence Schanuelrsquos Conjecture implies that if λ1 λ2 λ3 λ4 arelogarithms of algebraic numbers satisfying
λ1λ4 = λ2λ3
then either λ1 λ2 are linearly dependent over Q or else λ1 λ3
are linearly dependent over Q45 59
Six Exponentials Theorem
and Four Exponentials Conjecture
A Selberg CL Siegel S Lang K Ramachandra
46 59
The Four Exponentials Conjecture
Conjecture (Th Schneider S Lang K Ramachandra) Ifx1 x2 are Qndashlinearly independent complex numbers and y1 y2
are Qndashlinearly independent complex numbers then one atleast of the four numbers
ex1y1 ex1y2 ex2y1 ex2y2
is transcendentalEquivalent statement Any 2times 2 matrix(
λ1 λ2
λ3 λ4
)with entries in L and with Qndashlinearly independent rows andQndashlinearly independent columns has maximal rank 2
47 59
How could we attack Schanuelrsquos Conjecture
Let x1 xn be Qndashlinearly independent complex numbersFollowing the transcendence methods of Hermite GelrsquofondSchneider one may start by introducing an auxiliaryfunction
F (z) = P (z ez)
where P isin Z[X0 X1] is a nonndashzero polynomial One considersthe derivatives (
d
dz
)kF = (DkP )(z ez)
at the pointsm1x1 + middot middot middot+mnxn
for various values of (m1 mn) isin Zn
48 59
Auxiliary functionLet D denote the derivation
D =part
partX0
+X1part
partX1
over the ring C[X0 X1] Recall that x1 xn are Qndashlinearlyindependent complex numbers Let α1 αn be nonndashzerocomplex numbersThe transcendence machinery produces a sequence (PN)Nge0
of polynomials with integer coefficients satisfying∣∣∣∣∣(DkPN)(nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1
49 59
Royrsquos approach to Schanuelrsquos Conjecture (1999)
If the number of equations we produce is too small such a setof relations does not contain any information The existenceof a sequence of nonndashtrivial polynomials (PN)Nge0 follows fromlinear algebra
On the other hand following D Roy one may expect that theexistence of such a sequence (PN)Nge0 producing sufficientlymany such equations will yield the desired conclusion
tr degQQ(x1 xn α1 αn
)ge n
50 59
Royrsquos Conjecture (1999)Let s0 s1 t0 t1 u positive real numbers satisfying
max1 t0 2t1 lt mins0 2s1and
maxs0 s1 + t1 lt u lt1
2(1 + t0 + t1)
Assume that for any sufficiently large positive integer N there exists a nonndashzero polynomial PN isin Z[X0 X1] withpartial degree le N t0 in X0 partial degree le N t1 in X1 andheight le eN which satisfies∣∣∣∣∣(DkPN)(
nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1 Then
tr degQQ(x1 xn α1 αn) ge n51 59
Royrsquos Theorem (1999)
Royrsquos conjecture is equivalent to Schanuelrsquos Conjecture
52 59
Equivalence between Schanuel and Roy
Let (x α) isin CtimesCtimes and let s0 s1 t0 t1 u be positive realnumbers satisfying the inequalities of Royrsquos Conjecture Thenthe following conditions are equivalent (a) The number αeminusx is a root of unity
(b) For any sufficiently large positive integer N there exists anonndashzero polynomial QN isin Z[X0 X1] with partial degreele N t0 in X0 partial degree le N t1 in X1 and heightH(QN) le eN such that∣∣(DkQN)(mx αm)
∣∣ le exp(minusNu)
for any km isin N with k le N s0 and m le N s1
53 59
Gelrsquofondrsquos transcendence criterion
Simple form Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le n and height le en such that
|Pn(ϑ)| le eminus7n2
for all n ge 1 then ϑ is algebraic and Pn(ϑ) = 0 for all n ge 1
First extension Replace the bound for the degree by le dnthe bound for the height by ebn and the bound for |Pn(ϑ)| byeminuscdnbn with some constant c gt 0 independent of n
Some mild conditions are required on the sequences (dn)nge1 and(bn)nge1
54 59
Transcendence criterion with multiplicitiesWith derivatives Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le dn and height le ebn such that∣∣P (j)
n (ϑ)∣∣ le eminuscdnbntn
for all j in the range 0 le j lt tn and all n ge 1 then ϑ isalgebraicDue to M Laurent and D Roy applications to algebraicindependence
55 59
Criterion with several points
Goal Given a sequence of complex numbers (ϑi)ige1 if thereexists a sequence (Pn)nge1 of nonndashzero polynomials in Z[X]with Pn of degree le dn and height le ebn such that∣∣P (j)
n (ϑi)∣∣ le eminuscdnbntnsn
for 0 le j lt tn 1 le i le sn and all n ge 1 then the numbers ϑiare algebraic
D Roy Not true in general but true in some special caseswith a structure on the sequence (ϑi)ige1Combines the elimination arguments used for criteria ofalgebraic independence and for zero estimates
56 59
Small value estimates for the additive group
D Roy Small value estimates for the additive group Intern JNumber Theory to appear
Let ξ be a transcendental complex number let β σ τ and νbe non-negative real numbers let n0 be a positive integer andlet (Pn)ngen0 be a sequence of nonndashzero polynomials in Z[T ]satisfying deg(Pn) le n and H(Pn) le exp(nβ) for eachn ge n0 Suppose that β gt 1 (34)σ + τ lt 1 andν gt 1 + β minus (34)σ minus τ Then for infinitely many n we have
max|P [j]n (iξ)| 0 le i le nσ 0 le j le nτ
gt exp(minusnν)
57 59
Small value estimates for the multiplicative group
D Roy Small value estimates for the multiplicative groupActa Arith to appear
Let ξ1 ξm be multiplicatively independent complexnumbers in a field of transcendence degree 1 Under suitableassumptions on the parameters β σ τ ν for infinitely manypositive integers n there exists no nonndashzero polynomialP isin Z[T ] satisfying deg(P ) le n H(P ) le exp(nβ) and
max|P [j](ξi11 middot middot middot ξimm )| 0 le i1 im le nσ 0 le j le nτ
gt exp(minusnν)
58 59
Heidelberg Colloquium organized by the Institute for Mathematics
httpwwwmathiuni-heidelbergdevortraghtmls=1 January 22 2009
Transcendental Number TheorySchanuelrsquos Conjecture
Michel Waldschmidt
Institut de Mathematiques de Jussieu amp Paris VIhttpwwwmathjussieufrsimmiw
59 59
Conjectures by A Grothendieck and Y Andre
Generalized Conjecture onPeriods by Grothendieck Dimension of theMumfordndashTate group of asmooth projective variety
Generalization by Y Andre tomotives
Case of 1ndashmotives Elliptico-Toric Conjecture ofC Bertolin
27 59
Ubiquity of Schanuelrsquos Conjecture
Other contexts pndashadic numbers Leopoldtrsquos Conjecture onthe pndashadic rank of the units of an algebraic number fieldNon-vanishing of RegulatorsNonndashdegenerescence of heightsConjecture of B Mazur on rational pointsDiophantine approximation on tori
Dipendra Prasad Gopal Prasad
28 59
Methods from logicEhud Hrushovski Boris Zilber Jonathan Kirby
Calculus of rdquopredimension functionsrdquo (E Hrushovski)
Zilberrsquos construction of a rdquopseudoexponentiationrdquo
Also A Macintyre DE Marker G Terzo AJ WilkieD Bertrand
29 59
Methods from logic Model theory
Exponential algebraicity in exponential fieldsby
Jonathan Kirby
The dimension of the exponential algebraic closure operator inan exponential field satisfies a weak Schanuel property
A corollary is that there are at most countably many essentialcounterexamples to Schanuelrsquos conjecture
arXiv 08104285v2
30 59
Known
LindemannndashWeierstraszlig Theorem = case where x1 xn arealgebraic
Let β1 βn be algebraic numbers which are linearlyindependent over Q Then the numbers eβ1 eβn arealgebraically independent over Q
31 59
Problem of Gelrsquofond and SchneiderRaised by AO Gelrsquofond in 1948 and Th Schneider in 1952
Conjecture If α is an algebraic number α 6= 0 α 6= 1 and ifβ is an irrational algebraic number of degree d then the dminus 1numbers
αβ αβ2
αβdminus1
are algebraically independent
Special case of Schanuelrsquos Conjecture Take xi = βiminus1 logαn = d so that
x1 xn e
x1 exn
islogα β logα βdminus1 logα α αβ αβ
dminus1
The conclusion of Schanuelrsquos Conjecture is
tr degQQ(logα αβ αβ
2
αβdminus1)
= d
32 59
Algebraic independence method of Gelrsquofond
AO Gelrsquofond (1948)
The two numbers 23radic2 and
23radic4 are algebraically
independentMore generally if α is analgebraic number α 6= 0α 6= 1 and if β is a algebraicnumber of degree d ge 3 thentwo at least of the numbers
αβ αβ2
αβdminus1
are algebraically independent
33 59
Tools
Transcendence criterion Replaces Liouvillersquos inequality intranscendence proofsLiouville A nonndashzero rational integer n isin Z satisfies |n| ge 1Gelrsquofond Needs to give a lower bound for |P (θ)| withP isin Z[X] 0 when θ is transcendental
Zero estimate for exponential polynomials C Hermite P Turan K Mahler R Tijdeman
Small transcendence degree AO Gelrsquofond AA Smelev R Tijdeman WD Brownawell
34 59
Large transcendence degree
GV Chudnovsky (1976)Among the numbers
αβ αβ2
αβdminus1
at least [log2 d] arealgebraically independent
GV Chudnovsky ndash On the path to Schanuelrsquosconjecture Algebraic curves close to a pointI General theory of colored sequencesII Fields of finite transcendence type and coloredsequences ResultantsStudia Sci Math Hungar 12 (1977) 125ndash157 (1980)
35 59
Partial result on the problem of Gelrsquofond and
Schneider
AO Gelrsquofond GV Chudnovskii P Philippon YuV Nesterenko
G Diaz If α is an algebraic number α 6= 0 α 6= 1 and if β isan irrational algebraic number of degree d then
tr degQQ(αβ αβ
2
αβdminus1) ge [d+ 1
2
]
36 59
Conjecture of algebraic independence of logarithms
of algebraic numbers
Denote by L the set of complex numbers λ for which eλ isalgebraic
L = logα α isin Qtimes
Hence L is a Q-vector subspace of CThe most important special case of Schanuelrsquos Conjecture is
Conjecture Let λ1 λn be Q-linearly independent elementsin L Then the numbers λ1 λn are algebraicallyindependent over Q
Not yet known that the transcendence degree is ge 2 Open problem Among all logarithms of algebraic numbersone at least is transcendental over Q(π)
37 59
Structural rank of a matrix
Let K be a field k a subfield and M a matrix with entries in KConsider the k-vector subspace E of K spanned by the entriesof M Choose an injective morphism ϕ of E into a k-vectorspace kX1 + middot middot middot+ kXn The image ϕ(M) of M is a matrixwhose entries are in the field k(X1 Xn) of rationalfractions Its rank does not depend on the choice of ϕ
This is the structural rank of M with respect to k
38 59
Example
LetM =
(bij + λij
)1leiled1lejle`
be a matrix with coefficients in Q + L Consider a basis of theQndashvector spanned by the entries and replace the elements inthis basis by unknowns This gives a new matrix M withcoefficients in a field of rational fractions the rank of which isthe structural rank of M (with respect to Q)
As a consequence of the conjecture of algebraic independenceof logarithms of algebraic numbers the rank of M should bethe same as the rank of its specialization M
39 59
Equivalence between the two conjectures
Following D Roy theconjecture on algebraicindependence of logarithms ofalgebraic numbers isequivalent to
Conjecture Any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L has a rank equal to its structuralrank
40 59
Any Polynomial is the Determinant of a Matrix
The proof of the equivalence uses the nice auxiliary result
For any P isin k[X1 Xn] there exists a square matrix withentries in the k-vector space k + kX1 + middot middot middot+ kXn whosedeterminant is P
41 59
Half of the Conjecture is solved
From a certain point of view half of the conjecture of algebraicindependence of logarithms of algebraic numbers is solved
Theorem [D Roy] The rank of any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L is at least half its structural rank
42 59
Reformulation by D Roy
Instead of taking logarithmsof algebraic numbers andlooking for the algebraicindependence relationsD Roy fixes a polynomial andlooks at the points withcoordinates logarithms ofalgebraic numbers on thecorresponding hypersurface
43 59
Reformulation by D Roy
Royrsquos reformulation of the conjecture of algebraicindependence of logarithms is
Conjecture For any algebraic subvariety V of Cn definedover the field Q of algebraic numbers the set V cap Ln is theunion of the sets E cap Ln where E ranges over the set ofvector subspaces of Cn which are defined over Q andcontained in V
Trivial Any element in E cap Ln where E is a vector subspaceof Cn defined over Q and contained in V belongs to V cap Ln
44 59
Example The Four Exponentials ConjectureTake for V the hypersurface of C4 defined by the equation
z1z4 = z2z3
The maximal Cndashvector subspaces of C4 defined over Q andcontained in V are the planes
az1 = bz2 bz4 = az3
and the planesaz1 = bz3 bz4 = az2
with (a b) isin Q2 (0 0)Hence Schanuelrsquos Conjecture implies that if λ1 λ2 λ3 λ4 arelogarithms of algebraic numbers satisfying
λ1λ4 = λ2λ3
then either λ1 λ2 are linearly dependent over Q or else λ1 λ3
are linearly dependent over Q45 59
Six Exponentials Theorem
and Four Exponentials Conjecture
A Selberg CL Siegel S Lang K Ramachandra
46 59
The Four Exponentials Conjecture
Conjecture (Th Schneider S Lang K Ramachandra) Ifx1 x2 are Qndashlinearly independent complex numbers and y1 y2
are Qndashlinearly independent complex numbers then one atleast of the four numbers
ex1y1 ex1y2 ex2y1 ex2y2
is transcendentalEquivalent statement Any 2times 2 matrix(
λ1 λ2
λ3 λ4
)with entries in L and with Qndashlinearly independent rows andQndashlinearly independent columns has maximal rank 2
47 59
How could we attack Schanuelrsquos Conjecture
Let x1 xn be Qndashlinearly independent complex numbersFollowing the transcendence methods of Hermite GelrsquofondSchneider one may start by introducing an auxiliaryfunction
F (z) = P (z ez)
where P isin Z[X0 X1] is a nonndashzero polynomial One considersthe derivatives (
d
dz
)kF = (DkP )(z ez)
at the pointsm1x1 + middot middot middot+mnxn
for various values of (m1 mn) isin Zn
48 59
Auxiliary functionLet D denote the derivation
D =part
partX0
+X1part
partX1
over the ring C[X0 X1] Recall that x1 xn are Qndashlinearlyindependent complex numbers Let α1 αn be nonndashzerocomplex numbersThe transcendence machinery produces a sequence (PN)Nge0
of polynomials with integer coefficients satisfying∣∣∣∣∣(DkPN)(nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1
49 59
Royrsquos approach to Schanuelrsquos Conjecture (1999)
If the number of equations we produce is too small such a setof relations does not contain any information The existenceof a sequence of nonndashtrivial polynomials (PN)Nge0 follows fromlinear algebra
On the other hand following D Roy one may expect that theexistence of such a sequence (PN)Nge0 producing sufficientlymany such equations will yield the desired conclusion
tr degQQ(x1 xn α1 αn
)ge n
50 59
Royrsquos Conjecture (1999)Let s0 s1 t0 t1 u positive real numbers satisfying
max1 t0 2t1 lt mins0 2s1and
maxs0 s1 + t1 lt u lt1
2(1 + t0 + t1)
Assume that for any sufficiently large positive integer N there exists a nonndashzero polynomial PN isin Z[X0 X1] withpartial degree le N t0 in X0 partial degree le N t1 in X1 andheight le eN which satisfies∣∣∣∣∣(DkPN)(
nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1 Then
tr degQQ(x1 xn α1 αn) ge n51 59
Royrsquos Theorem (1999)
Royrsquos conjecture is equivalent to Schanuelrsquos Conjecture
52 59
Equivalence between Schanuel and Roy
Let (x α) isin CtimesCtimes and let s0 s1 t0 t1 u be positive realnumbers satisfying the inequalities of Royrsquos Conjecture Thenthe following conditions are equivalent (a) The number αeminusx is a root of unity
(b) For any sufficiently large positive integer N there exists anonndashzero polynomial QN isin Z[X0 X1] with partial degreele N t0 in X0 partial degree le N t1 in X1 and heightH(QN) le eN such that∣∣(DkQN)(mx αm)
∣∣ le exp(minusNu)
for any km isin N with k le N s0 and m le N s1
53 59
Gelrsquofondrsquos transcendence criterion
Simple form Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le n and height le en such that
|Pn(ϑ)| le eminus7n2
for all n ge 1 then ϑ is algebraic and Pn(ϑ) = 0 for all n ge 1
First extension Replace the bound for the degree by le dnthe bound for the height by ebn and the bound for |Pn(ϑ)| byeminuscdnbn with some constant c gt 0 independent of n
Some mild conditions are required on the sequences (dn)nge1 and(bn)nge1
54 59
Transcendence criterion with multiplicitiesWith derivatives Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le dn and height le ebn such that∣∣P (j)
n (ϑ)∣∣ le eminuscdnbntn
for all j in the range 0 le j lt tn and all n ge 1 then ϑ isalgebraicDue to M Laurent and D Roy applications to algebraicindependence
55 59
Criterion with several points
Goal Given a sequence of complex numbers (ϑi)ige1 if thereexists a sequence (Pn)nge1 of nonndashzero polynomials in Z[X]with Pn of degree le dn and height le ebn such that∣∣P (j)
n (ϑi)∣∣ le eminuscdnbntnsn
for 0 le j lt tn 1 le i le sn and all n ge 1 then the numbers ϑiare algebraic
D Roy Not true in general but true in some special caseswith a structure on the sequence (ϑi)ige1Combines the elimination arguments used for criteria ofalgebraic independence and for zero estimates
56 59
Small value estimates for the additive group
D Roy Small value estimates for the additive group Intern JNumber Theory to appear
Let ξ be a transcendental complex number let β σ τ and νbe non-negative real numbers let n0 be a positive integer andlet (Pn)ngen0 be a sequence of nonndashzero polynomials in Z[T ]satisfying deg(Pn) le n and H(Pn) le exp(nβ) for eachn ge n0 Suppose that β gt 1 (34)σ + τ lt 1 andν gt 1 + β minus (34)σ minus τ Then for infinitely many n we have
max|P [j]n (iξ)| 0 le i le nσ 0 le j le nτ
gt exp(minusnν)
57 59
Small value estimates for the multiplicative group
D Roy Small value estimates for the multiplicative groupActa Arith to appear
Let ξ1 ξm be multiplicatively independent complexnumbers in a field of transcendence degree 1 Under suitableassumptions on the parameters β σ τ ν for infinitely manypositive integers n there exists no nonndashzero polynomialP isin Z[T ] satisfying deg(P ) le n H(P ) le exp(nβ) and
max|P [j](ξi11 middot middot middot ξimm )| 0 le i1 im le nσ 0 le j le nτ
gt exp(minusnν)
58 59
Heidelberg Colloquium organized by the Institute for Mathematics
httpwwwmathiuni-heidelbergdevortraghtmls=1 January 22 2009
Transcendental Number TheorySchanuelrsquos Conjecture
Michel Waldschmidt
Institut de Mathematiques de Jussieu amp Paris VIhttpwwwmathjussieufrsimmiw
59 59
Ubiquity of Schanuelrsquos Conjecture
Other contexts pndashadic numbers Leopoldtrsquos Conjecture onthe pndashadic rank of the units of an algebraic number fieldNon-vanishing of RegulatorsNonndashdegenerescence of heightsConjecture of B Mazur on rational pointsDiophantine approximation on tori
Dipendra Prasad Gopal Prasad
28 59
Methods from logicEhud Hrushovski Boris Zilber Jonathan Kirby
Calculus of rdquopredimension functionsrdquo (E Hrushovski)
Zilberrsquos construction of a rdquopseudoexponentiationrdquo
Also A Macintyre DE Marker G Terzo AJ WilkieD Bertrand
29 59
Methods from logic Model theory
Exponential algebraicity in exponential fieldsby
Jonathan Kirby
The dimension of the exponential algebraic closure operator inan exponential field satisfies a weak Schanuel property
A corollary is that there are at most countably many essentialcounterexamples to Schanuelrsquos conjecture
arXiv 08104285v2
30 59
Known
LindemannndashWeierstraszlig Theorem = case where x1 xn arealgebraic
Let β1 βn be algebraic numbers which are linearlyindependent over Q Then the numbers eβ1 eβn arealgebraically independent over Q
31 59
Problem of Gelrsquofond and SchneiderRaised by AO Gelrsquofond in 1948 and Th Schneider in 1952
Conjecture If α is an algebraic number α 6= 0 α 6= 1 and ifβ is an irrational algebraic number of degree d then the dminus 1numbers
αβ αβ2
αβdminus1
are algebraically independent
Special case of Schanuelrsquos Conjecture Take xi = βiminus1 logαn = d so that
x1 xn e
x1 exn
islogα β logα βdminus1 logα α αβ αβ
dminus1
The conclusion of Schanuelrsquos Conjecture is
tr degQQ(logα αβ αβ
2
αβdminus1)
= d
32 59
Algebraic independence method of Gelrsquofond
AO Gelrsquofond (1948)
The two numbers 23radic2 and
23radic4 are algebraically
independentMore generally if α is analgebraic number α 6= 0α 6= 1 and if β is a algebraicnumber of degree d ge 3 thentwo at least of the numbers
αβ αβ2
αβdminus1
are algebraically independent
33 59
Tools
Transcendence criterion Replaces Liouvillersquos inequality intranscendence proofsLiouville A nonndashzero rational integer n isin Z satisfies |n| ge 1Gelrsquofond Needs to give a lower bound for |P (θ)| withP isin Z[X] 0 when θ is transcendental
Zero estimate for exponential polynomials C Hermite P Turan K Mahler R Tijdeman
Small transcendence degree AO Gelrsquofond AA Smelev R Tijdeman WD Brownawell
34 59
Large transcendence degree
GV Chudnovsky (1976)Among the numbers
αβ αβ2
αβdminus1
at least [log2 d] arealgebraically independent
GV Chudnovsky ndash On the path to Schanuelrsquosconjecture Algebraic curves close to a pointI General theory of colored sequencesII Fields of finite transcendence type and coloredsequences ResultantsStudia Sci Math Hungar 12 (1977) 125ndash157 (1980)
35 59
Partial result on the problem of Gelrsquofond and
Schneider
AO Gelrsquofond GV Chudnovskii P Philippon YuV Nesterenko
G Diaz If α is an algebraic number α 6= 0 α 6= 1 and if β isan irrational algebraic number of degree d then
tr degQQ(αβ αβ
2
αβdminus1) ge [d+ 1
2
]
36 59
Conjecture of algebraic independence of logarithms
of algebraic numbers
Denote by L the set of complex numbers λ for which eλ isalgebraic
L = logα α isin Qtimes
Hence L is a Q-vector subspace of CThe most important special case of Schanuelrsquos Conjecture is
Conjecture Let λ1 λn be Q-linearly independent elementsin L Then the numbers λ1 λn are algebraicallyindependent over Q
Not yet known that the transcendence degree is ge 2 Open problem Among all logarithms of algebraic numbersone at least is transcendental over Q(π)
37 59
Structural rank of a matrix
Let K be a field k a subfield and M a matrix with entries in KConsider the k-vector subspace E of K spanned by the entriesof M Choose an injective morphism ϕ of E into a k-vectorspace kX1 + middot middot middot+ kXn The image ϕ(M) of M is a matrixwhose entries are in the field k(X1 Xn) of rationalfractions Its rank does not depend on the choice of ϕ
This is the structural rank of M with respect to k
38 59
Example
LetM =
(bij + λij
)1leiled1lejle`
be a matrix with coefficients in Q + L Consider a basis of theQndashvector spanned by the entries and replace the elements inthis basis by unknowns This gives a new matrix M withcoefficients in a field of rational fractions the rank of which isthe structural rank of M (with respect to Q)
As a consequence of the conjecture of algebraic independenceof logarithms of algebraic numbers the rank of M should bethe same as the rank of its specialization M
39 59
Equivalence between the two conjectures
Following D Roy theconjecture on algebraicindependence of logarithms ofalgebraic numbers isequivalent to
Conjecture Any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L has a rank equal to its structuralrank
40 59
Any Polynomial is the Determinant of a Matrix
The proof of the equivalence uses the nice auxiliary result
For any P isin k[X1 Xn] there exists a square matrix withentries in the k-vector space k + kX1 + middot middot middot+ kXn whosedeterminant is P
41 59
Half of the Conjecture is solved
From a certain point of view half of the conjecture of algebraicindependence of logarithms of algebraic numbers is solved
Theorem [D Roy] The rank of any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L is at least half its structural rank
42 59
Reformulation by D Roy
Instead of taking logarithmsof algebraic numbers andlooking for the algebraicindependence relationsD Roy fixes a polynomial andlooks at the points withcoordinates logarithms ofalgebraic numbers on thecorresponding hypersurface
43 59
Reformulation by D Roy
Royrsquos reformulation of the conjecture of algebraicindependence of logarithms is
Conjecture For any algebraic subvariety V of Cn definedover the field Q of algebraic numbers the set V cap Ln is theunion of the sets E cap Ln where E ranges over the set ofvector subspaces of Cn which are defined over Q andcontained in V
Trivial Any element in E cap Ln where E is a vector subspaceof Cn defined over Q and contained in V belongs to V cap Ln
44 59
Example The Four Exponentials ConjectureTake for V the hypersurface of C4 defined by the equation
z1z4 = z2z3
The maximal Cndashvector subspaces of C4 defined over Q andcontained in V are the planes
az1 = bz2 bz4 = az3
and the planesaz1 = bz3 bz4 = az2
with (a b) isin Q2 (0 0)Hence Schanuelrsquos Conjecture implies that if λ1 λ2 λ3 λ4 arelogarithms of algebraic numbers satisfying
λ1λ4 = λ2λ3
then either λ1 λ2 are linearly dependent over Q or else λ1 λ3
are linearly dependent over Q45 59
Six Exponentials Theorem
and Four Exponentials Conjecture
A Selberg CL Siegel S Lang K Ramachandra
46 59
The Four Exponentials Conjecture
Conjecture (Th Schneider S Lang K Ramachandra) Ifx1 x2 are Qndashlinearly independent complex numbers and y1 y2
are Qndashlinearly independent complex numbers then one atleast of the four numbers
ex1y1 ex1y2 ex2y1 ex2y2
is transcendentalEquivalent statement Any 2times 2 matrix(
λ1 λ2
λ3 λ4
)with entries in L and with Qndashlinearly independent rows andQndashlinearly independent columns has maximal rank 2
47 59
How could we attack Schanuelrsquos Conjecture
Let x1 xn be Qndashlinearly independent complex numbersFollowing the transcendence methods of Hermite GelrsquofondSchneider one may start by introducing an auxiliaryfunction
F (z) = P (z ez)
where P isin Z[X0 X1] is a nonndashzero polynomial One considersthe derivatives (
d
dz
)kF = (DkP )(z ez)
at the pointsm1x1 + middot middot middot+mnxn
for various values of (m1 mn) isin Zn
48 59
Auxiliary functionLet D denote the derivation
D =part
partX0
+X1part
partX1
over the ring C[X0 X1] Recall that x1 xn are Qndashlinearlyindependent complex numbers Let α1 αn be nonndashzerocomplex numbersThe transcendence machinery produces a sequence (PN)Nge0
of polynomials with integer coefficients satisfying∣∣∣∣∣(DkPN)(nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1
49 59
Royrsquos approach to Schanuelrsquos Conjecture (1999)
If the number of equations we produce is too small such a setof relations does not contain any information The existenceof a sequence of nonndashtrivial polynomials (PN)Nge0 follows fromlinear algebra
On the other hand following D Roy one may expect that theexistence of such a sequence (PN)Nge0 producing sufficientlymany such equations will yield the desired conclusion
tr degQQ(x1 xn α1 αn
)ge n
50 59
Royrsquos Conjecture (1999)Let s0 s1 t0 t1 u positive real numbers satisfying
max1 t0 2t1 lt mins0 2s1and
maxs0 s1 + t1 lt u lt1
2(1 + t0 + t1)
Assume that for any sufficiently large positive integer N there exists a nonndashzero polynomial PN isin Z[X0 X1] withpartial degree le N t0 in X0 partial degree le N t1 in X1 andheight le eN which satisfies∣∣∣∣∣(DkPN)(
nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1 Then
tr degQQ(x1 xn α1 αn) ge n51 59
Royrsquos Theorem (1999)
Royrsquos conjecture is equivalent to Schanuelrsquos Conjecture
52 59
Equivalence between Schanuel and Roy
Let (x α) isin CtimesCtimes and let s0 s1 t0 t1 u be positive realnumbers satisfying the inequalities of Royrsquos Conjecture Thenthe following conditions are equivalent (a) The number αeminusx is a root of unity
(b) For any sufficiently large positive integer N there exists anonndashzero polynomial QN isin Z[X0 X1] with partial degreele N t0 in X0 partial degree le N t1 in X1 and heightH(QN) le eN such that∣∣(DkQN)(mx αm)
∣∣ le exp(minusNu)
for any km isin N with k le N s0 and m le N s1
53 59
Gelrsquofondrsquos transcendence criterion
Simple form Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le n and height le en such that
|Pn(ϑ)| le eminus7n2
for all n ge 1 then ϑ is algebraic and Pn(ϑ) = 0 for all n ge 1
First extension Replace the bound for the degree by le dnthe bound for the height by ebn and the bound for |Pn(ϑ)| byeminuscdnbn with some constant c gt 0 independent of n
Some mild conditions are required on the sequences (dn)nge1 and(bn)nge1
54 59
Transcendence criterion with multiplicitiesWith derivatives Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le dn and height le ebn such that∣∣P (j)
n (ϑ)∣∣ le eminuscdnbntn
for all j in the range 0 le j lt tn and all n ge 1 then ϑ isalgebraicDue to M Laurent and D Roy applications to algebraicindependence
55 59
Criterion with several points
Goal Given a sequence of complex numbers (ϑi)ige1 if thereexists a sequence (Pn)nge1 of nonndashzero polynomials in Z[X]with Pn of degree le dn and height le ebn such that∣∣P (j)
n (ϑi)∣∣ le eminuscdnbntnsn
for 0 le j lt tn 1 le i le sn and all n ge 1 then the numbers ϑiare algebraic
D Roy Not true in general but true in some special caseswith a structure on the sequence (ϑi)ige1Combines the elimination arguments used for criteria ofalgebraic independence and for zero estimates
56 59
Small value estimates for the additive group
D Roy Small value estimates for the additive group Intern JNumber Theory to appear
Let ξ be a transcendental complex number let β σ τ and νbe non-negative real numbers let n0 be a positive integer andlet (Pn)ngen0 be a sequence of nonndashzero polynomials in Z[T ]satisfying deg(Pn) le n and H(Pn) le exp(nβ) for eachn ge n0 Suppose that β gt 1 (34)σ + τ lt 1 andν gt 1 + β minus (34)σ minus τ Then for infinitely many n we have
max|P [j]n (iξ)| 0 le i le nσ 0 le j le nτ
gt exp(minusnν)
57 59
Small value estimates for the multiplicative group
D Roy Small value estimates for the multiplicative groupActa Arith to appear
Let ξ1 ξm be multiplicatively independent complexnumbers in a field of transcendence degree 1 Under suitableassumptions on the parameters β σ τ ν for infinitely manypositive integers n there exists no nonndashzero polynomialP isin Z[T ] satisfying deg(P ) le n H(P ) le exp(nβ) and
max|P [j](ξi11 middot middot middot ξimm )| 0 le i1 im le nσ 0 le j le nτ
gt exp(minusnν)
58 59
Heidelberg Colloquium organized by the Institute for Mathematics
httpwwwmathiuni-heidelbergdevortraghtmls=1 January 22 2009
Transcendental Number TheorySchanuelrsquos Conjecture
Michel Waldschmidt
Institut de Mathematiques de Jussieu amp Paris VIhttpwwwmathjussieufrsimmiw
59 59
Methods from logicEhud Hrushovski Boris Zilber Jonathan Kirby
Calculus of rdquopredimension functionsrdquo (E Hrushovski)
Zilberrsquos construction of a rdquopseudoexponentiationrdquo
Also A Macintyre DE Marker G Terzo AJ WilkieD Bertrand
29 59
Methods from logic Model theory
Exponential algebraicity in exponential fieldsby
Jonathan Kirby
The dimension of the exponential algebraic closure operator inan exponential field satisfies a weak Schanuel property
A corollary is that there are at most countably many essentialcounterexamples to Schanuelrsquos conjecture
arXiv 08104285v2
30 59
Known
LindemannndashWeierstraszlig Theorem = case where x1 xn arealgebraic
Let β1 βn be algebraic numbers which are linearlyindependent over Q Then the numbers eβ1 eβn arealgebraically independent over Q
31 59
Problem of Gelrsquofond and SchneiderRaised by AO Gelrsquofond in 1948 and Th Schneider in 1952
Conjecture If α is an algebraic number α 6= 0 α 6= 1 and ifβ is an irrational algebraic number of degree d then the dminus 1numbers
αβ αβ2
αβdminus1
are algebraically independent
Special case of Schanuelrsquos Conjecture Take xi = βiminus1 logαn = d so that
x1 xn e
x1 exn
islogα β logα βdminus1 logα α αβ αβ
dminus1
The conclusion of Schanuelrsquos Conjecture is
tr degQQ(logα αβ αβ
2
αβdminus1)
= d
32 59
Algebraic independence method of Gelrsquofond
AO Gelrsquofond (1948)
The two numbers 23radic2 and
23radic4 are algebraically
independentMore generally if α is analgebraic number α 6= 0α 6= 1 and if β is a algebraicnumber of degree d ge 3 thentwo at least of the numbers
αβ αβ2
αβdminus1
are algebraically independent
33 59
Tools
Transcendence criterion Replaces Liouvillersquos inequality intranscendence proofsLiouville A nonndashzero rational integer n isin Z satisfies |n| ge 1Gelrsquofond Needs to give a lower bound for |P (θ)| withP isin Z[X] 0 when θ is transcendental
Zero estimate for exponential polynomials C Hermite P Turan K Mahler R Tijdeman
Small transcendence degree AO Gelrsquofond AA Smelev R Tijdeman WD Brownawell
34 59
Large transcendence degree
GV Chudnovsky (1976)Among the numbers
αβ αβ2
αβdminus1
at least [log2 d] arealgebraically independent
GV Chudnovsky ndash On the path to Schanuelrsquosconjecture Algebraic curves close to a pointI General theory of colored sequencesII Fields of finite transcendence type and coloredsequences ResultantsStudia Sci Math Hungar 12 (1977) 125ndash157 (1980)
35 59
Partial result on the problem of Gelrsquofond and
Schneider
AO Gelrsquofond GV Chudnovskii P Philippon YuV Nesterenko
G Diaz If α is an algebraic number α 6= 0 α 6= 1 and if β isan irrational algebraic number of degree d then
tr degQQ(αβ αβ
2
αβdminus1) ge [d+ 1
2
]
36 59
Conjecture of algebraic independence of logarithms
of algebraic numbers
Denote by L the set of complex numbers λ for which eλ isalgebraic
L = logα α isin Qtimes
Hence L is a Q-vector subspace of CThe most important special case of Schanuelrsquos Conjecture is
Conjecture Let λ1 λn be Q-linearly independent elementsin L Then the numbers λ1 λn are algebraicallyindependent over Q
Not yet known that the transcendence degree is ge 2 Open problem Among all logarithms of algebraic numbersone at least is transcendental over Q(π)
37 59
Structural rank of a matrix
Let K be a field k a subfield and M a matrix with entries in KConsider the k-vector subspace E of K spanned by the entriesof M Choose an injective morphism ϕ of E into a k-vectorspace kX1 + middot middot middot+ kXn The image ϕ(M) of M is a matrixwhose entries are in the field k(X1 Xn) of rationalfractions Its rank does not depend on the choice of ϕ
This is the structural rank of M with respect to k
38 59
Example
LetM =
(bij + λij
)1leiled1lejle`
be a matrix with coefficients in Q + L Consider a basis of theQndashvector spanned by the entries and replace the elements inthis basis by unknowns This gives a new matrix M withcoefficients in a field of rational fractions the rank of which isthe structural rank of M (with respect to Q)
As a consequence of the conjecture of algebraic independenceof logarithms of algebraic numbers the rank of M should bethe same as the rank of its specialization M
39 59
Equivalence between the two conjectures
Following D Roy theconjecture on algebraicindependence of logarithms ofalgebraic numbers isequivalent to
Conjecture Any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L has a rank equal to its structuralrank
40 59
Any Polynomial is the Determinant of a Matrix
The proof of the equivalence uses the nice auxiliary result
For any P isin k[X1 Xn] there exists a square matrix withentries in the k-vector space k + kX1 + middot middot middot+ kXn whosedeterminant is P
41 59
Half of the Conjecture is solved
From a certain point of view half of the conjecture of algebraicindependence of logarithms of algebraic numbers is solved
Theorem [D Roy] The rank of any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L is at least half its structural rank
42 59
Reformulation by D Roy
Instead of taking logarithmsof algebraic numbers andlooking for the algebraicindependence relationsD Roy fixes a polynomial andlooks at the points withcoordinates logarithms ofalgebraic numbers on thecorresponding hypersurface
43 59
Reformulation by D Roy
Royrsquos reformulation of the conjecture of algebraicindependence of logarithms is
Conjecture For any algebraic subvariety V of Cn definedover the field Q of algebraic numbers the set V cap Ln is theunion of the sets E cap Ln where E ranges over the set ofvector subspaces of Cn which are defined over Q andcontained in V
Trivial Any element in E cap Ln where E is a vector subspaceof Cn defined over Q and contained in V belongs to V cap Ln
44 59
Example The Four Exponentials ConjectureTake for V the hypersurface of C4 defined by the equation
z1z4 = z2z3
The maximal Cndashvector subspaces of C4 defined over Q andcontained in V are the planes
az1 = bz2 bz4 = az3
and the planesaz1 = bz3 bz4 = az2
with (a b) isin Q2 (0 0)Hence Schanuelrsquos Conjecture implies that if λ1 λ2 λ3 λ4 arelogarithms of algebraic numbers satisfying
λ1λ4 = λ2λ3
then either λ1 λ2 are linearly dependent over Q or else λ1 λ3
are linearly dependent over Q45 59
Six Exponentials Theorem
and Four Exponentials Conjecture
A Selberg CL Siegel S Lang K Ramachandra
46 59
The Four Exponentials Conjecture
Conjecture (Th Schneider S Lang K Ramachandra) Ifx1 x2 are Qndashlinearly independent complex numbers and y1 y2
are Qndashlinearly independent complex numbers then one atleast of the four numbers
ex1y1 ex1y2 ex2y1 ex2y2
is transcendentalEquivalent statement Any 2times 2 matrix(
λ1 λ2
λ3 λ4
)with entries in L and with Qndashlinearly independent rows andQndashlinearly independent columns has maximal rank 2
47 59
How could we attack Schanuelrsquos Conjecture
Let x1 xn be Qndashlinearly independent complex numbersFollowing the transcendence methods of Hermite GelrsquofondSchneider one may start by introducing an auxiliaryfunction
F (z) = P (z ez)
where P isin Z[X0 X1] is a nonndashzero polynomial One considersthe derivatives (
d
dz
)kF = (DkP )(z ez)
at the pointsm1x1 + middot middot middot+mnxn
for various values of (m1 mn) isin Zn
48 59
Auxiliary functionLet D denote the derivation
D =part
partX0
+X1part
partX1
over the ring C[X0 X1] Recall that x1 xn are Qndashlinearlyindependent complex numbers Let α1 αn be nonndashzerocomplex numbersThe transcendence machinery produces a sequence (PN)Nge0
of polynomials with integer coefficients satisfying∣∣∣∣∣(DkPN)(nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1
49 59
Royrsquos approach to Schanuelrsquos Conjecture (1999)
If the number of equations we produce is too small such a setof relations does not contain any information The existenceof a sequence of nonndashtrivial polynomials (PN)Nge0 follows fromlinear algebra
On the other hand following D Roy one may expect that theexistence of such a sequence (PN)Nge0 producing sufficientlymany such equations will yield the desired conclusion
tr degQQ(x1 xn α1 αn
)ge n
50 59
Royrsquos Conjecture (1999)Let s0 s1 t0 t1 u positive real numbers satisfying
max1 t0 2t1 lt mins0 2s1and
maxs0 s1 + t1 lt u lt1
2(1 + t0 + t1)
Assume that for any sufficiently large positive integer N there exists a nonndashzero polynomial PN isin Z[X0 X1] withpartial degree le N t0 in X0 partial degree le N t1 in X1 andheight le eN which satisfies∣∣∣∣∣(DkPN)(
nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1 Then
tr degQQ(x1 xn α1 αn) ge n51 59
Royrsquos Theorem (1999)
Royrsquos conjecture is equivalent to Schanuelrsquos Conjecture
52 59
Equivalence between Schanuel and Roy
Let (x α) isin CtimesCtimes and let s0 s1 t0 t1 u be positive realnumbers satisfying the inequalities of Royrsquos Conjecture Thenthe following conditions are equivalent (a) The number αeminusx is a root of unity
(b) For any sufficiently large positive integer N there exists anonndashzero polynomial QN isin Z[X0 X1] with partial degreele N t0 in X0 partial degree le N t1 in X1 and heightH(QN) le eN such that∣∣(DkQN)(mx αm)
∣∣ le exp(minusNu)
for any km isin N with k le N s0 and m le N s1
53 59
Gelrsquofondrsquos transcendence criterion
Simple form Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le n and height le en such that
|Pn(ϑ)| le eminus7n2
for all n ge 1 then ϑ is algebraic and Pn(ϑ) = 0 for all n ge 1
First extension Replace the bound for the degree by le dnthe bound for the height by ebn and the bound for |Pn(ϑ)| byeminuscdnbn with some constant c gt 0 independent of n
Some mild conditions are required on the sequences (dn)nge1 and(bn)nge1
54 59
Transcendence criterion with multiplicitiesWith derivatives Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le dn and height le ebn such that∣∣P (j)
n (ϑ)∣∣ le eminuscdnbntn
for all j in the range 0 le j lt tn and all n ge 1 then ϑ isalgebraicDue to M Laurent and D Roy applications to algebraicindependence
55 59
Criterion with several points
Goal Given a sequence of complex numbers (ϑi)ige1 if thereexists a sequence (Pn)nge1 of nonndashzero polynomials in Z[X]with Pn of degree le dn and height le ebn such that∣∣P (j)
n (ϑi)∣∣ le eminuscdnbntnsn
for 0 le j lt tn 1 le i le sn and all n ge 1 then the numbers ϑiare algebraic
D Roy Not true in general but true in some special caseswith a structure on the sequence (ϑi)ige1Combines the elimination arguments used for criteria ofalgebraic independence and for zero estimates
56 59
Small value estimates for the additive group
D Roy Small value estimates for the additive group Intern JNumber Theory to appear
Let ξ be a transcendental complex number let β σ τ and νbe non-negative real numbers let n0 be a positive integer andlet (Pn)ngen0 be a sequence of nonndashzero polynomials in Z[T ]satisfying deg(Pn) le n and H(Pn) le exp(nβ) for eachn ge n0 Suppose that β gt 1 (34)σ + τ lt 1 andν gt 1 + β minus (34)σ minus τ Then for infinitely many n we have
max|P [j]n (iξ)| 0 le i le nσ 0 le j le nτ
gt exp(minusnν)
57 59
Small value estimates for the multiplicative group
D Roy Small value estimates for the multiplicative groupActa Arith to appear
Let ξ1 ξm be multiplicatively independent complexnumbers in a field of transcendence degree 1 Under suitableassumptions on the parameters β σ τ ν for infinitely manypositive integers n there exists no nonndashzero polynomialP isin Z[T ] satisfying deg(P ) le n H(P ) le exp(nβ) and
max|P [j](ξi11 middot middot middot ξimm )| 0 le i1 im le nσ 0 le j le nτ
gt exp(minusnν)
58 59
Heidelberg Colloquium organized by the Institute for Mathematics
httpwwwmathiuni-heidelbergdevortraghtmls=1 January 22 2009
Transcendental Number TheorySchanuelrsquos Conjecture
Michel Waldschmidt
Institut de Mathematiques de Jussieu amp Paris VIhttpwwwmathjussieufrsimmiw
59 59
Methods from logic Model theory
Exponential algebraicity in exponential fieldsby
Jonathan Kirby
The dimension of the exponential algebraic closure operator inan exponential field satisfies a weak Schanuel property
A corollary is that there are at most countably many essentialcounterexamples to Schanuelrsquos conjecture
arXiv 08104285v2
30 59
Known
LindemannndashWeierstraszlig Theorem = case where x1 xn arealgebraic
Let β1 βn be algebraic numbers which are linearlyindependent over Q Then the numbers eβ1 eβn arealgebraically independent over Q
31 59
Problem of Gelrsquofond and SchneiderRaised by AO Gelrsquofond in 1948 and Th Schneider in 1952
Conjecture If α is an algebraic number α 6= 0 α 6= 1 and ifβ is an irrational algebraic number of degree d then the dminus 1numbers
αβ αβ2
αβdminus1
are algebraically independent
Special case of Schanuelrsquos Conjecture Take xi = βiminus1 logαn = d so that
x1 xn e
x1 exn
islogα β logα βdminus1 logα α αβ αβ
dminus1
The conclusion of Schanuelrsquos Conjecture is
tr degQQ(logα αβ αβ
2
αβdminus1)
= d
32 59
Algebraic independence method of Gelrsquofond
AO Gelrsquofond (1948)
The two numbers 23radic2 and
23radic4 are algebraically
independentMore generally if α is analgebraic number α 6= 0α 6= 1 and if β is a algebraicnumber of degree d ge 3 thentwo at least of the numbers
αβ αβ2
αβdminus1
are algebraically independent
33 59
Tools
Transcendence criterion Replaces Liouvillersquos inequality intranscendence proofsLiouville A nonndashzero rational integer n isin Z satisfies |n| ge 1Gelrsquofond Needs to give a lower bound for |P (θ)| withP isin Z[X] 0 when θ is transcendental
Zero estimate for exponential polynomials C Hermite P Turan K Mahler R Tijdeman
Small transcendence degree AO Gelrsquofond AA Smelev R Tijdeman WD Brownawell
34 59
Large transcendence degree
GV Chudnovsky (1976)Among the numbers
αβ αβ2
αβdminus1
at least [log2 d] arealgebraically independent
GV Chudnovsky ndash On the path to Schanuelrsquosconjecture Algebraic curves close to a pointI General theory of colored sequencesII Fields of finite transcendence type and coloredsequences ResultantsStudia Sci Math Hungar 12 (1977) 125ndash157 (1980)
35 59
Partial result on the problem of Gelrsquofond and
Schneider
AO Gelrsquofond GV Chudnovskii P Philippon YuV Nesterenko
G Diaz If α is an algebraic number α 6= 0 α 6= 1 and if β isan irrational algebraic number of degree d then
tr degQQ(αβ αβ
2
αβdminus1) ge [d+ 1
2
]
36 59
Conjecture of algebraic independence of logarithms
of algebraic numbers
Denote by L the set of complex numbers λ for which eλ isalgebraic
L = logα α isin Qtimes
Hence L is a Q-vector subspace of CThe most important special case of Schanuelrsquos Conjecture is
Conjecture Let λ1 λn be Q-linearly independent elementsin L Then the numbers λ1 λn are algebraicallyindependent over Q
Not yet known that the transcendence degree is ge 2 Open problem Among all logarithms of algebraic numbersone at least is transcendental over Q(π)
37 59
Structural rank of a matrix
Let K be a field k a subfield and M a matrix with entries in KConsider the k-vector subspace E of K spanned by the entriesof M Choose an injective morphism ϕ of E into a k-vectorspace kX1 + middot middot middot+ kXn The image ϕ(M) of M is a matrixwhose entries are in the field k(X1 Xn) of rationalfractions Its rank does not depend on the choice of ϕ
This is the structural rank of M with respect to k
38 59
Example
LetM =
(bij + λij
)1leiled1lejle`
be a matrix with coefficients in Q + L Consider a basis of theQndashvector spanned by the entries and replace the elements inthis basis by unknowns This gives a new matrix M withcoefficients in a field of rational fractions the rank of which isthe structural rank of M (with respect to Q)
As a consequence of the conjecture of algebraic independenceof logarithms of algebraic numbers the rank of M should bethe same as the rank of its specialization M
39 59
Equivalence between the two conjectures
Following D Roy theconjecture on algebraicindependence of logarithms ofalgebraic numbers isequivalent to
Conjecture Any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L has a rank equal to its structuralrank
40 59
Any Polynomial is the Determinant of a Matrix
The proof of the equivalence uses the nice auxiliary result
For any P isin k[X1 Xn] there exists a square matrix withentries in the k-vector space k + kX1 + middot middot middot+ kXn whosedeterminant is P
41 59
Half of the Conjecture is solved
From a certain point of view half of the conjecture of algebraicindependence of logarithms of algebraic numbers is solved
Theorem [D Roy] The rank of any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L is at least half its structural rank
42 59
Reformulation by D Roy
Instead of taking logarithmsof algebraic numbers andlooking for the algebraicindependence relationsD Roy fixes a polynomial andlooks at the points withcoordinates logarithms ofalgebraic numbers on thecorresponding hypersurface
43 59
Reformulation by D Roy
Royrsquos reformulation of the conjecture of algebraicindependence of logarithms is
Conjecture For any algebraic subvariety V of Cn definedover the field Q of algebraic numbers the set V cap Ln is theunion of the sets E cap Ln where E ranges over the set ofvector subspaces of Cn which are defined over Q andcontained in V
Trivial Any element in E cap Ln where E is a vector subspaceof Cn defined over Q and contained in V belongs to V cap Ln
44 59
Example The Four Exponentials ConjectureTake for V the hypersurface of C4 defined by the equation
z1z4 = z2z3
The maximal Cndashvector subspaces of C4 defined over Q andcontained in V are the planes
az1 = bz2 bz4 = az3
and the planesaz1 = bz3 bz4 = az2
with (a b) isin Q2 (0 0)Hence Schanuelrsquos Conjecture implies that if λ1 λ2 λ3 λ4 arelogarithms of algebraic numbers satisfying
λ1λ4 = λ2λ3
then either λ1 λ2 are linearly dependent over Q or else λ1 λ3
are linearly dependent over Q45 59
Six Exponentials Theorem
and Four Exponentials Conjecture
A Selberg CL Siegel S Lang K Ramachandra
46 59
The Four Exponentials Conjecture
Conjecture (Th Schneider S Lang K Ramachandra) Ifx1 x2 are Qndashlinearly independent complex numbers and y1 y2
are Qndashlinearly independent complex numbers then one atleast of the four numbers
ex1y1 ex1y2 ex2y1 ex2y2
is transcendentalEquivalent statement Any 2times 2 matrix(
λ1 λ2
λ3 λ4
)with entries in L and with Qndashlinearly independent rows andQndashlinearly independent columns has maximal rank 2
47 59
How could we attack Schanuelrsquos Conjecture
Let x1 xn be Qndashlinearly independent complex numbersFollowing the transcendence methods of Hermite GelrsquofondSchneider one may start by introducing an auxiliaryfunction
F (z) = P (z ez)
where P isin Z[X0 X1] is a nonndashzero polynomial One considersthe derivatives (
d
dz
)kF = (DkP )(z ez)
at the pointsm1x1 + middot middot middot+mnxn
for various values of (m1 mn) isin Zn
48 59
Auxiliary functionLet D denote the derivation
D =part
partX0
+X1part
partX1
over the ring C[X0 X1] Recall that x1 xn are Qndashlinearlyindependent complex numbers Let α1 αn be nonndashzerocomplex numbersThe transcendence machinery produces a sequence (PN)Nge0
of polynomials with integer coefficients satisfying∣∣∣∣∣(DkPN)(nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1
49 59
Royrsquos approach to Schanuelrsquos Conjecture (1999)
If the number of equations we produce is too small such a setof relations does not contain any information The existenceof a sequence of nonndashtrivial polynomials (PN)Nge0 follows fromlinear algebra
On the other hand following D Roy one may expect that theexistence of such a sequence (PN)Nge0 producing sufficientlymany such equations will yield the desired conclusion
tr degQQ(x1 xn α1 αn
)ge n
50 59
Royrsquos Conjecture (1999)Let s0 s1 t0 t1 u positive real numbers satisfying
max1 t0 2t1 lt mins0 2s1and
maxs0 s1 + t1 lt u lt1
2(1 + t0 + t1)
Assume that for any sufficiently large positive integer N there exists a nonndashzero polynomial PN isin Z[X0 X1] withpartial degree le N t0 in X0 partial degree le N t1 in X1 andheight le eN which satisfies∣∣∣∣∣(DkPN)(
nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1 Then
tr degQQ(x1 xn α1 αn) ge n51 59
Royrsquos Theorem (1999)
Royrsquos conjecture is equivalent to Schanuelrsquos Conjecture
52 59
Equivalence between Schanuel and Roy
Let (x α) isin CtimesCtimes and let s0 s1 t0 t1 u be positive realnumbers satisfying the inequalities of Royrsquos Conjecture Thenthe following conditions are equivalent (a) The number αeminusx is a root of unity
(b) For any sufficiently large positive integer N there exists anonndashzero polynomial QN isin Z[X0 X1] with partial degreele N t0 in X0 partial degree le N t1 in X1 and heightH(QN) le eN such that∣∣(DkQN)(mx αm)
∣∣ le exp(minusNu)
for any km isin N with k le N s0 and m le N s1
53 59
Gelrsquofondrsquos transcendence criterion
Simple form Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le n and height le en such that
|Pn(ϑ)| le eminus7n2
for all n ge 1 then ϑ is algebraic and Pn(ϑ) = 0 for all n ge 1
First extension Replace the bound for the degree by le dnthe bound for the height by ebn and the bound for |Pn(ϑ)| byeminuscdnbn with some constant c gt 0 independent of n
Some mild conditions are required on the sequences (dn)nge1 and(bn)nge1
54 59
Transcendence criterion with multiplicitiesWith derivatives Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le dn and height le ebn such that∣∣P (j)
n (ϑ)∣∣ le eminuscdnbntn
for all j in the range 0 le j lt tn and all n ge 1 then ϑ isalgebraicDue to M Laurent and D Roy applications to algebraicindependence
55 59
Criterion with several points
Goal Given a sequence of complex numbers (ϑi)ige1 if thereexists a sequence (Pn)nge1 of nonndashzero polynomials in Z[X]with Pn of degree le dn and height le ebn such that∣∣P (j)
n (ϑi)∣∣ le eminuscdnbntnsn
for 0 le j lt tn 1 le i le sn and all n ge 1 then the numbers ϑiare algebraic
D Roy Not true in general but true in some special caseswith a structure on the sequence (ϑi)ige1Combines the elimination arguments used for criteria ofalgebraic independence and for zero estimates
56 59
Small value estimates for the additive group
D Roy Small value estimates for the additive group Intern JNumber Theory to appear
Let ξ be a transcendental complex number let β σ τ and νbe non-negative real numbers let n0 be a positive integer andlet (Pn)ngen0 be a sequence of nonndashzero polynomials in Z[T ]satisfying deg(Pn) le n and H(Pn) le exp(nβ) for eachn ge n0 Suppose that β gt 1 (34)σ + τ lt 1 andν gt 1 + β minus (34)σ minus τ Then for infinitely many n we have
max|P [j]n (iξ)| 0 le i le nσ 0 le j le nτ
gt exp(minusnν)
57 59
Small value estimates for the multiplicative group
D Roy Small value estimates for the multiplicative groupActa Arith to appear
Let ξ1 ξm be multiplicatively independent complexnumbers in a field of transcendence degree 1 Under suitableassumptions on the parameters β σ τ ν for infinitely manypositive integers n there exists no nonndashzero polynomialP isin Z[T ] satisfying deg(P ) le n H(P ) le exp(nβ) and
max|P [j](ξi11 middot middot middot ξimm )| 0 le i1 im le nσ 0 le j le nτ
gt exp(minusnν)
58 59
Heidelberg Colloquium organized by the Institute for Mathematics
httpwwwmathiuni-heidelbergdevortraghtmls=1 January 22 2009
Transcendental Number TheorySchanuelrsquos Conjecture
Michel Waldschmidt
Institut de Mathematiques de Jussieu amp Paris VIhttpwwwmathjussieufrsimmiw
59 59
Known
LindemannndashWeierstraszlig Theorem = case where x1 xn arealgebraic
Let β1 βn be algebraic numbers which are linearlyindependent over Q Then the numbers eβ1 eβn arealgebraically independent over Q
31 59
Problem of Gelrsquofond and SchneiderRaised by AO Gelrsquofond in 1948 and Th Schneider in 1952
Conjecture If α is an algebraic number α 6= 0 α 6= 1 and ifβ is an irrational algebraic number of degree d then the dminus 1numbers
αβ αβ2
αβdminus1
are algebraically independent
Special case of Schanuelrsquos Conjecture Take xi = βiminus1 logαn = d so that
x1 xn e
x1 exn
islogα β logα βdminus1 logα α αβ αβ
dminus1
The conclusion of Schanuelrsquos Conjecture is
tr degQQ(logα αβ αβ
2
αβdminus1)
= d
32 59
Algebraic independence method of Gelrsquofond
AO Gelrsquofond (1948)
The two numbers 23radic2 and
23radic4 are algebraically
independentMore generally if α is analgebraic number α 6= 0α 6= 1 and if β is a algebraicnumber of degree d ge 3 thentwo at least of the numbers
αβ αβ2
αβdminus1
are algebraically independent
33 59
Tools
Transcendence criterion Replaces Liouvillersquos inequality intranscendence proofsLiouville A nonndashzero rational integer n isin Z satisfies |n| ge 1Gelrsquofond Needs to give a lower bound for |P (θ)| withP isin Z[X] 0 when θ is transcendental
Zero estimate for exponential polynomials C Hermite P Turan K Mahler R Tijdeman
Small transcendence degree AO Gelrsquofond AA Smelev R Tijdeman WD Brownawell
34 59
Large transcendence degree
GV Chudnovsky (1976)Among the numbers
αβ αβ2
αβdminus1
at least [log2 d] arealgebraically independent
GV Chudnovsky ndash On the path to Schanuelrsquosconjecture Algebraic curves close to a pointI General theory of colored sequencesII Fields of finite transcendence type and coloredsequences ResultantsStudia Sci Math Hungar 12 (1977) 125ndash157 (1980)
35 59
Partial result on the problem of Gelrsquofond and
Schneider
AO Gelrsquofond GV Chudnovskii P Philippon YuV Nesterenko
G Diaz If α is an algebraic number α 6= 0 α 6= 1 and if β isan irrational algebraic number of degree d then
tr degQQ(αβ αβ
2
αβdminus1) ge [d+ 1
2
]
36 59
Conjecture of algebraic independence of logarithms
of algebraic numbers
Denote by L the set of complex numbers λ for which eλ isalgebraic
L = logα α isin Qtimes
Hence L is a Q-vector subspace of CThe most important special case of Schanuelrsquos Conjecture is
Conjecture Let λ1 λn be Q-linearly independent elementsin L Then the numbers λ1 λn are algebraicallyindependent over Q
Not yet known that the transcendence degree is ge 2 Open problem Among all logarithms of algebraic numbersone at least is transcendental over Q(π)
37 59
Structural rank of a matrix
Let K be a field k a subfield and M a matrix with entries in KConsider the k-vector subspace E of K spanned by the entriesof M Choose an injective morphism ϕ of E into a k-vectorspace kX1 + middot middot middot+ kXn The image ϕ(M) of M is a matrixwhose entries are in the field k(X1 Xn) of rationalfractions Its rank does not depend on the choice of ϕ
This is the structural rank of M with respect to k
38 59
Example
LetM =
(bij + λij
)1leiled1lejle`
be a matrix with coefficients in Q + L Consider a basis of theQndashvector spanned by the entries and replace the elements inthis basis by unknowns This gives a new matrix M withcoefficients in a field of rational fractions the rank of which isthe structural rank of M (with respect to Q)
As a consequence of the conjecture of algebraic independenceof logarithms of algebraic numbers the rank of M should bethe same as the rank of its specialization M
39 59
Equivalence between the two conjectures
Following D Roy theconjecture on algebraicindependence of logarithms ofalgebraic numbers isequivalent to
Conjecture Any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L has a rank equal to its structuralrank
40 59
Any Polynomial is the Determinant of a Matrix
The proof of the equivalence uses the nice auxiliary result
For any P isin k[X1 Xn] there exists a square matrix withentries in the k-vector space k + kX1 + middot middot middot+ kXn whosedeterminant is P
41 59
Half of the Conjecture is solved
From a certain point of view half of the conjecture of algebraicindependence of logarithms of algebraic numbers is solved
Theorem [D Roy] The rank of any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L is at least half its structural rank
42 59
Reformulation by D Roy
Instead of taking logarithmsof algebraic numbers andlooking for the algebraicindependence relationsD Roy fixes a polynomial andlooks at the points withcoordinates logarithms ofalgebraic numbers on thecorresponding hypersurface
43 59
Reformulation by D Roy
Royrsquos reformulation of the conjecture of algebraicindependence of logarithms is
Conjecture For any algebraic subvariety V of Cn definedover the field Q of algebraic numbers the set V cap Ln is theunion of the sets E cap Ln where E ranges over the set ofvector subspaces of Cn which are defined over Q andcontained in V
Trivial Any element in E cap Ln where E is a vector subspaceof Cn defined over Q and contained in V belongs to V cap Ln
44 59
Example The Four Exponentials ConjectureTake for V the hypersurface of C4 defined by the equation
z1z4 = z2z3
The maximal Cndashvector subspaces of C4 defined over Q andcontained in V are the planes
az1 = bz2 bz4 = az3
and the planesaz1 = bz3 bz4 = az2
with (a b) isin Q2 (0 0)Hence Schanuelrsquos Conjecture implies that if λ1 λ2 λ3 λ4 arelogarithms of algebraic numbers satisfying
λ1λ4 = λ2λ3
then either λ1 λ2 are linearly dependent over Q or else λ1 λ3
are linearly dependent over Q45 59
Six Exponentials Theorem
and Four Exponentials Conjecture
A Selberg CL Siegel S Lang K Ramachandra
46 59
The Four Exponentials Conjecture
Conjecture (Th Schneider S Lang K Ramachandra) Ifx1 x2 are Qndashlinearly independent complex numbers and y1 y2
are Qndashlinearly independent complex numbers then one atleast of the four numbers
ex1y1 ex1y2 ex2y1 ex2y2
is transcendentalEquivalent statement Any 2times 2 matrix(
λ1 λ2
λ3 λ4
)with entries in L and with Qndashlinearly independent rows andQndashlinearly independent columns has maximal rank 2
47 59
How could we attack Schanuelrsquos Conjecture
Let x1 xn be Qndashlinearly independent complex numbersFollowing the transcendence methods of Hermite GelrsquofondSchneider one may start by introducing an auxiliaryfunction
F (z) = P (z ez)
where P isin Z[X0 X1] is a nonndashzero polynomial One considersthe derivatives (
d
dz
)kF = (DkP )(z ez)
at the pointsm1x1 + middot middot middot+mnxn
for various values of (m1 mn) isin Zn
48 59
Auxiliary functionLet D denote the derivation
D =part
partX0
+X1part
partX1
over the ring C[X0 X1] Recall that x1 xn are Qndashlinearlyindependent complex numbers Let α1 αn be nonndashzerocomplex numbersThe transcendence machinery produces a sequence (PN)Nge0
of polynomials with integer coefficients satisfying∣∣∣∣∣(DkPN)(nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1
49 59
Royrsquos approach to Schanuelrsquos Conjecture (1999)
If the number of equations we produce is too small such a setof relations does not contain any information The existenceof a sequence of nonndashtrivial polynomials (PN)Nge0 follows fromlinear algebra
On the other hand following D Roy one may expect that theexistence of such a sequence (PN)Nge0 producing sufficientlymany such equations will yield the desired conclusion
tr degQQ(x1 xn α1 αn
)ge n
50 59
Royrsquos Conjecture (1999)Let s0 s1 t0 t1 u positive real numbers satisfying
max1 t0 2t1 lt mins0 2s1and
maxs0 s1 + t1 lt u lt1
2(1 + t0 + t1)
Assume that for any sufficiently large positive integer N there exists a nonndashzero polynomial PN isin Z[X0 X1] withpartial degree le N t0 in X0 partial degree le N t1 in X1 andheight le eN which satisfies∣∣∣∣∣(DkPN)(
nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1 Then
tr degQQ(x1 xn α1 αn) ge n51 59
Royrsquos Theorem (1999)
Royrsquos conjecture is equivalent to Schanuelrsquos Conjecture
52 59
Equivalence between Schanuel and Roy
Let (x α) isin CtimesCtimes and let s0 s1 t0 t1 u be positive realnumbers satisfying the inequalities of Royrsquos Conjecture Thenthe following conditions are equivalent (a) The number αeminusx is a root of unity
(b) For any sufficiently large positive integer N there exists anonndashzero polynomial QN isin Z[X0 X1] with partial degreele N t0 in X0 partial degree le N t1 in X1 and heightH(QN) le eN such that∣∣(DkQN)(mx αm)
∣∣ le exp(minusNu)
for any km isin N with k le N s0 and m le N s1
53 59
Gelrsquofondrsquos transcendence criterion
Simple form Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le n and height le en such that
|Pn(ϑ)| le eminus7n2
for all n ge 1 then ϑ is algebraic and Pn(ϑ) = 0 for all n ge 1
First extension Replace the bound for the degree by le dnthe bound for the height by ebn and the bound for |Pn(ϑ)| byeminuscdnbn with some constant c gt 0 independent of n
Some mild conditions are required on the sequences (dn)nge1 and(bn)nge1
54 59
Transcendence criterion with multiplicitiesWith derivatives Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le dn and height le ebn such that∣∣P (j)
n (ϑ)∣∣ le eminuscdnbntn
for all j in the range 0 le j lt tn and all n ge 1 then ϑ isalgebraicDue to M Laurent and D Roy applications to algebraicindependence
55 59
Criterion with several points
Goal Given a sequence of complex numbers (ϑi)ige1 if thereexists a sequence (Pn)nge1 of nonndashzero polynomials in Z[X]with Pn of degree le dn and height le ebn such that∣∣P (j)
n (ϑi)∣∣ le eminuscdnbntnsn
for 0 le j lt tn 1 le i le sn and all n ge 1 then the numbers ϑiare algebraic
D Roy Not true in general but true in some special caseswith a structure on the sequence (ϑi)ige1Combines the elimination arguments used for criteria ofalgebraic independence and for zero estimates
56 59
Small value estimates for the additive group
D Roy Small value estimates for the additive group Intern JNumber Theory to appear
Let ξ be a transcendental complex number let β σ τ and νbe non-negative real numbers let n0 be a positive integer andlet (Pn)ngen0 be a sequence of nonndashzero polynomials in Z[T ]satisfying deg(Pn) le n and H(Pn) le exp(nβ) for eachn ge n0 Suppose that β gt 1 (34)σ + τ lt 1 andν gt 1 + β minus (34)σ minus τ Then for infinitely many n we have
max|P [j]n (iξ)| 0 le i le nσ 0 le j le nτ
gt exp(minusnν)
57 59
Small value estimates for the multiplicative group
D Roy Small value estimates for the multiplicative groupActa Arith to appear
Let ξ1 ξm be multiplicatively independent complexnumbers in a field of transcendence degree 1 Under suitableassumptions on the parameters β σ τ ν for infinitely manypositive integers n there exists no nonndashzero polynomialP isin Z[T ] satisfying deg(P ) le n H(P ) le exp(nβ) and
max|P [j](ξi11 middot middot middot ξimm )| 0 le i1 im le nσ 0 le j le nτ
gt exp(minusnν)
58 59
Heidelberg Colloquium organized by the Institute for Mathematics
httpwwwmathiuni-heidelbergdevortraghtmls=1 January 22 2009
Transcendental Number TheorySchanuelrsquos Conjecture
Michel Waldschmidt
Institut de Mathematiques de Jussieu amp Paris VIhttpwwwmathjussieufrsimmiw
59 59
Problem of Gelrsquofond and SchneiderRaised by AO Gelrsquofond in 1948 and Th Schneider in 1952
Conjecture If α is an algebraic number α 6= 0 α 6= 1 and ifβ is an irrational algebraic number of degree d then the dminus 1numbers
αβ αβ2
αβdminus1
are algebraically independent
Special case of Schanuelrsquos Conjecture Take xi = βiminus1 logαn = d so that
x1 xn e
x1 exn
islogα β logα βdminus1 logα α αβ αβ
dminus1
The conclusion of Schanuelrsquos Conjecture is
tr degQQ(logα αβ αβ
2
αβdminus1)
= d
32 59
Algebraic independence method of Gelrsquofond
AO Gelrsquofond (1948)
The two numbers 23radic2 and
23radic4 are algebraically
independentMore generally if α is analgebraic number α 6= 0α 6= 1 and if β is a algebraicnumber of degree d ge 3 thentwo at least of the numbers
αβ αβ2
αβdminus1
are algebraically independent
33 59
Tools
Transcendence criterion Replaces Liouvillersquos inequality intranscendence proofsLiouville A nonndashzero rational integer n isin Z satisfies |n| ge 1Gelrsquofond Needs to give a lower bound for |P (θ)| withP isin Z[X] 0 when θ is transcendental
Zero estimate for exponential polynomials C Hermite P Turan K Mahler R Tijdeman
Small transcendence degree AO Gelrsquofond AA Smelev R Tijdeman WD Brownawell
34 59
Large transcendence degree
GV Chudnovsky (1976)Among the numbers
αβ αβ2
αβdminus1
at least [log2 d] arealgebraically independent
GV Chudnovsky ndash On the path to Schanuelrsquosconjecture Algebraic curves close to a pointI General theory of colored sequencesII Fields of finite transcendence type and coloredsequences ResultantsStudia Sci Math Hungar 12 (1977) 125ndash157 (1980)
35 59
Partial result on the problem of Gelrsquofond and
Schneider
AO Gelrsquofond GV Chudnovskii P Philippon YuV Nesterenko
G Diaz If α is an algebraic number α 6= 0 α 6= 1 and if β isan irrational algebraic number of degree d then
tr degQQ(αβ αβ
2
αβdminus1) ge [d+ 1
2
]
36 59
Conjecture of algebraic independence of logarithms
of algebraic numbers
Denote by L the set of complex numbers λ for which eλ isalgebraic
L = logα α isin Qtimes
Hence L is a Q-vector subspace of CThe most important special case of Schanuelrsquos Conjecture is
Conjecture Let λ1 λn be Q-linearly independent elementsin L Then the numbers λ1 λn are algebraicallyindependent over Q
Not yet known that the transcendence degree is ge 2 Open problem Among all logarithms of algebraic numbersone at least is transcendental over Q(π)
37 59
Structural rank of a matrix
Let K be a field k a subfield and M a matrix with entries in KConsider the k-vector subspace E of K spanned by the entriesof M Choose an injective morphism ϕ of E into a k-vectorspace kX1 + middot middot middot+ kXn The image ϕ(M) of M is a matrixwhose entries are in the field k(X1 Xn) of rationalfractions Its rank does not depend on the choice of ϕ
This is the structural rank of M with respect to k
38 59
Example
LetM =
(bij + λij
)1leiled1lejle`
be a matrix with coefficients in Q + L Consider a basis of theQndashvector spanned by the entries and replace the elements inthis basis by unknowns This gives a new matrix M withcoefficients in a field of rational fractions the rank of which isthe structural rank of M (with respect to Q)
As a consequence of the conjecture of algebraic independenceof logarithms of algebraic numbers the rank of M should bethe same as the rank of its specialization M
39 59
Equivalence between the two conjectures
Following D Roy theconjecture on algebraicindependence of logarithms ofalgebraic numbers isequivalent to
Conjecture Any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L has a rank equal to its structuralrank
40 59
Any Polynomial is the Determinant of a Matrix
The proof of the equivalence uses the nice auxiliary result
For any P isin k[X1 Xn] there exists a square matrix withentries in the k-vector space k + kX1 + middot middot middot+ kXn whosedeterminant is P
41 59
Half of the Conjecture is solved
From a certain point of view half of the conjecture of algebraicindependence of logarithms of algebraic numbers is solved
Theorem [D Roy] The rank of any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L is at least half its structural rank
42 59
Reformulation by D Roy
Instead of taking logarithmsof algebraic numbers andlooking for the algebraicindependence relationsD Roy fixes a polynomial andlooks at the points withcoordinates logarithms ofalgebraic numbers on thecorresponding hypersurface
43 59
Reformulation by D Roy
Royrsquos reformulation of the conjecture of algebraicindependence of logarithms is
Conjecture For any algebraic subvariety V of Cn definedover the field Q of algebraic numbers the set V cap Ln is theunion of the sets E cap Ln where E ranges over the set ofvector subspaces of Cn which are defined over Q andcontained in V
Trivial Any element in E cap Ln where E is a vector subspaceof Cn defined over Q and contained in V belongs to V cap Ln
44 59
Example The Four Exponentials ConjectureTake for V the hypersurface of C4 defined by the equation
z1z4 = z2z3
The maximal Cndashvector subspaces of C4 defined over Q andcontained in V are the planes
az1 = bz2 bz4 = az3
and the planesaz1 = bz3 bz4 = az2
with (a b) isin Q2 (0 0)Hence Schanuelrsquos Conjecture implies that if λ1 λ2 λ3 λ4 arelogarithms of algebraic numbers satisfying
λ1λ4 = λ2λ3
then either λ1 λ2 are linearly dependent over Q or else λ1 λ3
are linearly dependent over Q45 59
Six Exponentials Theorem
and Four Exponentials Conjecture
A Selberg CL Siegel S Lang K Ramachandra
46 59
The Four Exponentials Conjecture
Conjecture (Th Schneider S Lang K Ramachandra) Ifx1 x2 are Qndashlinearly independent complex numbers and y1 y2
are Qndashlinearly independent complex numbers then one atleast of the four numbers
ex1y1 ex1y2 ex2y1 ex2y2
is transcendentalEquivalent statement Any 2times 2 matrix(
λ1 λ2
λ3 λ4
)with entries in L and with Qndashlinearly independent rows andQndashlinearly independent columns has maximal rank 2
47 59
How could we attack Schanuelrsquos Conjecture
Let x1 xn be Qndashlinearly independent complex numbersFollowing the transcendence methods of Hermite GelrsquofondSchneider one may start by introducing an auxiliaryfunction
F (z) = P (z ez)
where P isin Z[X0 X1] is a nonndashzero polynomial One considersthe derivatives (
d
dz
)kF = (DkP )(z ez)
at the pointsm1x1 + middot middot middot+mnxn
for various values of (m1 mn) isin Zn
48 59
Auxiliary functionLet D denote the derivation
D =part
partX0
+X1part
partX1
over the ring C[X0 X1] Recall that x1 xn are Qndashlinearlyindependent complex numbers Let α1 αn be nonndashzerocomplex numbersThe transcendence machinery produces a sequence (PN)Nge0
of polynomials with integer coefficients satisfying∣∣∣∣∣(DkPN)(nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1
49 59
Royrsquos approach to Schanuelrsquos Conjecture (1999)
If the number of equations we produce is too small such a setof relations does not contain any information The existenceof a sequence of nonndashtrivial polynomials (PN)Nge0 follows fromlinear algebra
On the other hand following D Roy one may expect that theexistence of such a sequence (PN)Nge0 producing sufficientlymany such equations will yield the desired conclusion
tr degQQ(x1 xn α1 αn
)ge n
50 59
Royrsquos Conjecture (1999)Let s0 s1 t0 t1 u positive real numbers satisfying
max1 t0 2t1 lt mins0 2s1and
maxs0 s1 + t1 lt u lt1
2(1 + t0 + t1)
Assume that for any sufficiently large positive integer N there exists a nonndashzero polynomial PN isin Z[X0 X1] withpartial degree le N t0 in X0 partial degree le N t1 in X1 andheight le eN which satisfies∣∣∣∣∣(DkPN)(
nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1 Then
tr degQQ(x1 xn α1 αn) ge n51 59
Royrsquos Theorem (1999)
Royrsquos conjecture is equivalent to Schanuelrsquos Conjecture
52 59
Equivalence between Schanuel and Roy
Let (x α) isin CtimesCtimes and let s0 s1 t0 t1 u be positive realnumbers satisfying the inequalities of Royrsquos Conjecture Thenthe following conditions are equivalent (a) The number αeminusx is a root of unity
(b) For any sufficiently large positive integer N there exists anonndashzero polynomial QN isin Z[X0 X1] with partial degreele N t0 in X0 partial degree le N t1 in X1 and heightH(QN) le eN such that∣∣(DkQN)(mx αm)
∣∣ le exp(minusNu)
for any km isin N with k le N s0 and m le N s1
53 59
Gelrsquofondrsquos transcendence criterion
Simple form Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le n and height le en such that
|Pn(ϑ)| le eminus7n2
for all n ge 1 then ϑ is algebraic and Pn(ϑ) = 0 for all n ge 1
First extension Replace the bound for the degree by le dnthe bound for the height by ebn and the bound for |Pn(ϑ)| byeminuscdnbn with some constant c gt 0 independent of n
Some mild conditions are required on the sequences (dn)nge1 and(bn)nge1
54 59
Transcendence criterion with multiplicitiesWith derivatives Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le dn and height le ebn such that∣∣P (j)
n (ϑ)∣∣ le eminuscdnbntn
for all j in the range 0 le j lt tn and all n ge 1 then ϑ isalgebraicDue to M Laurent and D Roy applications to algebraicindependence
55 59
Criterion with several points
Goal Given a sequence of complex numbers (ϑi)ige1 if thereexists a sequence (Pn)nge1 of nonndashzero polynomials in Z[X]with Pn of degree le dn and height le ebn such that∣∣P (j)
n (ϑi)∣∣ le eminuscdnbntnsn
for 0 le j lt tn 1 le i le sn and all n ge 1 then the numbers ϑiare algebraic
D Roy Not true in general but true in some special caseswith a structure on the sequence (ϑi)ige1Combines the elimination arguments used for criteria ofalgebraic independence and for zero estimates
56 59
Small value estimates for the additive group
D Roy Small value estimates for the additive group Intern JNumber Theory to appear
Let ξ be a transcendental complex number let β σ τ and νbe non-negative real numbers let n0 be a positive integer andlet (Pn)ngen0 be a sequence of nonndashzero polynomials in Z[T ]satisfying deg(Pn) le n and H(Pn) le exp(nβ) for eachn ge n0 Suppose that β gt 1 (34)σ + τ lt 1 andν gt 1 + β minus (34)σ minus τ Then for infinitely many n we have
max|P [j]n (iξ)| 0 le i le nσ 0 le j le nτ
gt exp(minusnν)
57 59
Small value estimates for the multiplicative group
D Roy Small value estimates for the multiplicative groupActa Arith to appear
Let ξ1 ξm be multiplicatively independent complexnumbers in a field of transcendence degree 1 Under suitableassumptions on the parameters β σ τ ν for infinitely manypositive integers n there exists no nonndashzero polynomialP isin Z[T ] satisfying deg(P ) le n H(P ) le exp(nβ) and
max|P [j](ξi11 middot middot middot ξimm )| 0 le i1 im le nσ 0 le j le nτ
gt exp(minusnν)
58 59
Heidelberg Colloquium organized by the Institute for Mathematics
httpwwwmathiuni-heidelbergdevortraghtmls=1 January 22 2009
Transcendental Number TheorySchanuelrsquos Conjecture
Michel Waldschmidt
Institut de Mathematiques de Jussieu amp Paris VIhttpwwwmathjussieufrsimmiw
59 59
Algebraic independence method of Gelrsquofond
AO Gelrsquofond (1948)
The two numbers 23radic2 and
23radic4 are algebraically
independentMore generally if α is analgebraic number α 6= 0α 6= 1 and if β is a algebraicnumber of degree d ge 3 thentwo at least of the numbers
αβ αβ2
αβdminus1
are algebraically independent
33 59
Tools
Transcendence criterion Replaces Liouvillersquos inequality intranscendence proofsLiouville A nonndashzero rational integer n isin Z satisfies |n| ge 1Gelrsquofond Needs to give a lower bound for |P (θ)| withP isin Z[X] 0 when θ is transcendental
Zero estimate for exponential polynomials C Hermite P Turan K Mahler R Tijdeman
Small transcendence degree AO Gelrsquofond AA Smelev R Tijdeman WD Brownawell
34 59
Large transcendence degree
GV Chudnovsky (1976)Among the numbers
αβ αβ2
αβdminus1
at least [log2 d] arealgebraically independent
GV Chudnovsky ndash On the path to Schanuelrsquosconjecture Algebraic curves close to a pointI General theory of colored sequencesII Fields of finite transcendence type and coloredsequences ResultantsStudia Sci Math Hungar 12 (1977) 125ndash157 (1980)
35 59
Partial result on the problem of Gelrsquofond and
Schneider
AO Gelrsquofond GV Chudnovskii P Philippon YuV Nesterenko
G Diaz If α is an algebraic number α 6= 0 α 6= 1 and if β isan irrational algebraic number of degree d then
tr degQQ(αβ αβ
2
αβdminus1) ge [d+ 1
2
]
36 59
Conjecture of algebraic independence of logarithms
of algebraic numbers
Denote by L the set of complex numbers λ for which eλ isalgebraic
L = logα α isin Qtimes
Hence L is a Q-vector subspace of CThe most important special case of Schanuelrsquos Conjecture is
Conjecture Let λ1 λn be Q-linearly independent elementsin L Then the numbers λ1 λn are algebraicallyindependent over Q
Not yet known that the transcendence degree is ge 2 Open problem Among all logarithms of algebraic numbersone at least is transcendental over Q(π)
37 59
Structural rank of a matrix
Let K be a field k a subfield and M a matrix with entries in KConsider the k-vector subspace E of K spanned by the entriesof M Choose an injective morphism ϕ of E into a k-vectorspace kX1 + middot middot middot+ kXn The image ϕ(M) of M is a matrixwhose entries are in the field k(X1 Xn) of rationalfractions Its rank does not depend on the choice of ϕ
This is the structural rank of M with respect to k
38 59
Example
LetM =
(bij + λij
)1leiled1lejle`
be a matrix with coefficients in Q + L Consider a basis of theQndashvector spanned by the entries and replace the elements inthis basis by unknowns This gives a new matrix M withcoefficients in a field of rational fractions the rank of which isthe structural rank of M (with respect to Q)
As a consequence of the conjecture of algebraic independenceof logarithms of algebraic numbers the rank of M should bethe same as the rank of its specialization M
39 59
Equivalence between the two conjectures
Following D Roy theconjecture on algebraicindependence of logarithms ofalgebraic numbers isequivalent to
Conjecture Any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L has a rank equal to its structuralrank
40 59
Any Polynomial is the Determinant of a Matrix
The proof of the equivalence uses the nice auxiliary result
For any P isin k[X1 Xn] there exists a square matrix withentries in the k-vector space k + kX1 + middot middot middot+ kXn whosedeterminant is P
41 59
Half of the Conjecture is solved
From a certain point of view half of the conjecture of algebraicindependence of logarithms of algebraic numbers is solved
Theorem [D Roy] The rank of any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L is at least half its structural rank
42 59
Reformulation by D Roy
Instead of taking logarithmsof algebraic numbers andlooking for the algebraicindependence relationsD Roy fixes a polynomial andlooks at the points withcoordinates logarithms ofalgebraic numbers on thecorresponding hypersurface
43 59
Reformulation by D Roy
Royrsquos reformulation of the conjecture of algebraicindependence of logarithms is
Conjecture For any algebraic subvariety V of Cn definedover the field Q of algebraic numbers the set V cap Ln is theunion of the sets E cap Ln where E ranges over the set ofvector subspaces of Cn which are defined over Q andcontained in V
Trivial Any element in E cap Ln where E is a vector subspaceof Cn defined over Q and contained in V belongs to V cap Ln
44 59
Example The Four Exponentials ConjectureTake for V the hypersurface of C4 defined by the equation
z1z4 = z2z3
The maximal Cndashvector subspaces of C4 defined over Q andcontained in V are the planes
az1 = bz2 bz4 = az3
and the planesaz1 = bz3 bz4 = az2
with (a b) isin Q2 (0 0)Hence Schanuelrsquos Conjecture implies that if λ1 λ2 λ3 λ4 arelogarithms of algebraic numbers satisfying
λ1λ4 = λ2λ3
then either λ1 λ2 are linearly dependent over Q or else λ1 λ3
are linearly dependent over Q45 59
Six Exponentials Theorem
and Four Exponentials Conjecture
A Selberg CL Siegel S Lang K Ramachandra
46 59
The Four Exponentials Conjecture
Conjecture (Th Schneider S Lang K Ramachandra) Ifx1 x2 are Qndashlinearly independent complex numbers and y1 y2
are Qndashlinearly independent complex numbers then one atleast of the four numbers
ex1y1 ex1y2 ex2y1 ex2y2
is transcendentalEquivalent statement Any 2times 2 matrix(
λ1 λ2
λ3 λ4
)with entries in L and with Qndashlinearly independent rows andQndashlinearly independent columns has maximal rank 2
47 59
How could we attack Schanuelrsquos Conjecture
Let x1 xn be Qndashlinearly independent complex numbersFollowing the transcendence methods of Hermite GelrsquofondSchneider one may start by introducing an auxiliaryfunction
F (z) = P (z ez)
where P isin Z[X0 X1] is a nonndashzero polynomial One considersthe derivatives (
d
dz
)kF = (DkP )(z ez)
at the pointsm1x1 + middot middot middot+mnxn
for various values of (m1 mn) isin Zn
48 59
Auxiliary functionLet D denote the derivation
D =part
partX0
+X1part
partX1
over the ring C[X0 X1] Recall that x1 xn are Qndashlinearlyindependent complex numbers Let α1 αn be nonndashzerocomplex numbersThe transcendence machinery produces a sequence (PN)Nge0
of polynomials with integer coefficients satisfying∣∣∣∣∣(DkPN)(nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1
49 59
Royrsquos approach to Schanuelrsquos Conjecture (1999)
If the number of equations we produce is too small such a setof relations does not contain any information The existenceof a sequence of nonndashtrivial polynomials (PN)Nge0 follows fromlinear algebra
On the other hand following D Roy one may expect that theexistence of such a sequence (PN)Nge0 producing sufficientlymany such equations will yield the desired conclusion
tr degQQ(x1 xn α1 αn
)ge n
50 59
Royrsquos Conjecture (1999)Let s0 s1 t0 t1 u positive real numbers satisfying
max1 t0 2t1 lt mins0 2s1and
maxs0 s1 + t1 lt u lt1
2(1 + t0 + t1)
Assume that for any sufficiently large positive integer N there exists a nonndashzero polynomial PN isin Z[X0 X1] withpartial degree le N t0 in X0 partial degree le N t1 in X1 andheight le eN which satisfies∣∣∣∣∣(DkPN)(
nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1 Then
tr degQQ(x1 xn α1 αn) ge n51 59
Royrsquos Theorem (1999)
Royrsquos conjecture is equivalent to Schanuelrsquos Conjecture
52 59
Equivalence between Schanuel and Roy
Let (x α) isin CtimesCtimes and let s0 s1 t0 t1 u be positive realnumbers satisfying the inequalities of Royrsquos Conjecture Thenthe following conditions are equivalent (a) The number αeminusx is a root of unity
(b) For any sufficiently large positive integer N there exists anonndashzero polynomial QN isin Z[X0 X1] with partial degreele N t0 in X0 partial degree le N t1 in X1 and heightH(QN) le eN such that∣∣(DkQN)(mx αm)
∣∣ le exp(minusNu)
for any km isin N with k le N s0 and m le N s1
53 59
Gelrsquofondrsquos transcendence criterion
Simple form Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le n and height le en such that
|Pn(ϑ)| le eminus7n2
for all n ge 1 then ϑ is algebraic and Pn(ϑ) = 0 for all n ge 1
First extension Replace the bound for the degree by le dnthe bound for the height by ebn and the bound for |Pn(ϑ)| byeminuscdnbn with some constant c gt 0 independent of n
Some mild conditions are required on the sequences (dn)nge1 and(bn)nge1
54 59
Transcendence criterion with multiplicitiesWith derivatives Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le dn and height le ebn such that∣∣P (j)
n (ϑ)∣∣ le eminuscdnbntn
for all j in the range 0 le j lt tn and all n ge 1 then ϑ isalgebraicDue to M Laurent and D Roy applications to algebraicindependence
55 59
Criterion with several points
Goal Given a sequence of complex numbers (ϑi)ige1 if thereexists a sequence (Pn)nge1 of nonndashzero polynomials in Z[X]with Pn of degree le dn and height le ebn such that∣∣P (j)
n (ϑi)∣∣ le eminuscdnbntnsn
for 0 le j lt tn 1 le i le sn and all n ge 1 then the numbers ϑiare algebraic
D Roy Not true in general but true in some special caseswith a structure on the sequence (ϑi)ige1Combines the elimination arguments used for criteria ofalgebraic independence and for zero estimates
56 59
Small value estimates for the additive group
D Roy Small value estimates for the additive group Intern JNumber Theory to appear
Let ξ be a transcendental complex number let β σ τ and νbe non-negative real numbers let n0 be a positive integer andlet (Pn)ngen0 be a sequence of nonndashzero polynomials in Z[T ]satisfying deg(Pn) le n and H(Pn) le exp(nβ) for eachn ge n0 Suppose that β gt 1 (34)σ + τ lt 1 andν gt 1 + β minus (34)σ minus τ Then for infinitely many n we have
max|P [j]n (iξ)| 0 le i le nσ 0 le j le nτ
gt exp(minusnν)
57 59
Small value estimates for the multiplicative group
D Roy Small value estimates for the multiplicative groupActa Arith to appear
Let ξ1 ξm be multiplicatively independent complexnumbers in a field of transcendence degree 1 Under suitableassumptions on the parameters β σ τ ν for infinitely manypositive integers n there exists no nonndashzero polynomialP isin Z[T ] satisfying deg(P ) le n H(P ) le exp(nβ) and
max|P [j](ξi11 middot middot middot ξimm )| 0 le i1 im le nσ 0 le j le nτ
gt exp(minusnν)
58 59
Heidelberg Colloquium organized by the Institute for Mathematics
httpwwwmathiuni-heidelbergdevortraghtmls=1 January 22 2009
Transcendental Number TheorySchanuelrsquos Conjecture
Michel Waldschmidt
Institut de Mathematiques de Jussieu amp Paris VIhttpwwwmathjussieufrsimmiw
59 59
Tools
Transcendence criterion Replaces Liouvillersquos inequality intranscendence proofsLiouville A nonndashzero rational integer n isin Z satisfies |n| ge 1Gelrsquofond Needs to give a lower bound for |P (θ)| withP isin Z[X] 0 when θ is transcendental
Zero estimate for exponential polynomials C Hermite P Turan K Mahler R Tijdeman
Small transcendence degree AO Gelrsquofond AA Smelev R Tijdeman WD Brownawell
34 59
Large transcendence degree
GV Chudnovsky (1976)Among the numbers
αβ αβ2
αβdminus1
at least [log2 d] arealgebraically independent
GV Chudnovsky ndash On the path to Schanuelrsquosconjecture Algebraic curves close to a pointI General theory of colored sequencesII Fields of finite transcendence type and coloredsequences ResultantsStudia Sci Math Hungar 12 (1977) 125ndash157 (1980)
35 59
Partial result on the problem of Gelrsquofond and
Schneider
AO Gelrsquofond GV Chudnovskii P Philippon YuV Nesterenko
G Diaz If α is an algebraic number α 6= 0 α 6= 1 and if β isan irrational algebraic number of degree d then
tr degQQ(αβ αβ
2
αβdminus1) ge [d+ 1
2
]
36 59
Conjecture of algebraic independence of logarithms
of algebraic numbers
Denote by L the set of complex numbers λ for which eλ isalgebraic
L = logα α isin Qtimes
Hence L is a Q-vector subspace of CThe most important special case of Schanuelrsquos Conjecture is
Conjecture Let λ1 λn be Q-linearly independent elementsin L Then the numbers λ1 λn are algebraicallyindependent over Q
Not yet known that the transcendence degree is ge 2 Open problem Among all logarithms of algebraic numbersone at least is transcendental over Q(π)
37 59
Structural rank of a matrix
Let K be a field k a subfield and M a matrix with entries in KConsider the k-vector subspace E of K spanned by the entriesof M Choose an injective morphism ϕ of E into a k-vectorspace kX1 + middot middot middot+ kXn The image ϕ(M) of M is a matrixwhose entries are in the field k(X1 Xn) of rationalfractions Its rank does not depend on the choice of ϕ
This is the structural rank of M with respect to k
38 59
Example
LetM =
(bij + λij
)1leiled1lejle`
be a matrix with coefficients in Q + L Consider a basis of theQndashvector spanned by the entries and replace the elements inthis basis by unknowns This gives a new matrix M withcoefficients in a field of rational fractions the rank of which isthe structural rank of M (with respect to Q)
As a consequence of the conjecture of algebraic independenceof logarithms of algebraic numbers the rank of M should bethe same as the rank of its specialization M
39 59
Equivalence between the two conjectures
Following D Roy theconjecture on algebraicindependence of logarithms ofalgebraic numbers isequivalent to
Conjecture Any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L has a rank equal to its structuralrank
40 59
Any Polynomial is the Determinant of a Matrix
The proof of the equivalence uses the nice auxiliary result
For any P isin k[X1 Xn] there exists a square matrix withentries in the k-vector space k + kX1 + middot middot middot+ kXn whosedeterminant is P
41 59
Half of the Conjecture is solved
From a certain point of view half of the conjecture of algebraicindependence of logarithms of algebraic numbers is solved
Theorem [D Roy] The rank of any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L is at least half its structural rank
42 59
Reformulation by D Roy
Instead of taking logarithmsof algebraic numbers andlooking for the algebraicindependence relationsD Roy fixes a polynomial andlooks at the points withcoordinates logarithms ofalgebraic numbers on thecorresponding hypersurface
43 59
Reformulation by D Roy
Royrsquos reformulation of the conjecture of algebraicindependence of logarithms is
Conjecture For any algebraic subvariety V of Cn definedover the field Q of algebraic numbers the set V cap Ln is theunion of the sets E cap Ln where E ranges over the set ofvector subspaces of Cn which are defined over Q andcontained in V
Trivial Any element in E cap Ln where E is a vector subspaceof Cn defined over Q and contained in V belongs to V cap Ln
44 59
Example The Four Exponentials ConjectureTake for V the hypersurface of C4 defined by the equation
z1z4 = z2z3
The maximal Cndashvector subspaces of C4 defined over Q andcontained in V are the planes
az1 = bz2 bz4 = az3
and the planesaz1 = bz3 bz4 = az2
with (a b) isin Q2 (0 0)Hence Schanuelrsquos Conjecture implies that if λ1 λ2 λ3 λ4 arelogarithms of algebraic numbers satisfying
λ1λ4 = λ2λ3
then either λ1 λ2 are linearly dependent over Q or else λ1 λ3
are linearly dependent over Q45 59
Six Exponentials Theorem
and Four Exponentials Conjecture
A Selberg CL Siegel S Lang K Ramachandra
46 59
The Four Exponentials Conjecture
Conjecture (Th Schneider S Lang K Ramachandra) Ifx1 x2 are Qndashlinearly independent complex numbers and y1 y2
are Qndashlinearly independent complex numbers then one atleast of the four numbers
ex1y1 ex1y2 ex2y1 ex2y2
is transcendentalEquivalent statement Any 2times 2 matrix(
λ1 λ2
λ3 λ4
)with entries in L and with Qndashlinearly independent rows andQndashlinearly independent columns has maximal rank 2
47 59
How could we attack Schanuelrsquos Conjecture
Let x1 xn be Qndashlinearly independent complex numbersFollowing the transcendence methods of Hermite GelrsquofondSchneider one may start by introducing an auxiliaryfunction
F (z) = P (z ez)
where P isin Z[X0 X1] is a nonndashzero polynomial One considersthe derivatives (
d
dz
)kF = (DkP )(z ez)
at the pointsm1x1 + middot middot middot+mnxn
for various values of (m1 mn) isin Zn
48 59
Auxiliary functionLet D denote the derivation
D =part
partX0
+X1part
partX1
over the ring C[X0 X1] Recall that x1 xn are Qndashlinearlyindependent complex numbers Let α1 αn be nonndashzerocomplex numbersThe transcendence machinery produces a sequence (PN)Nge0
of polynomials with integer coefficients satisfying∣∣∣∣∣(DkPN)(nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1
49 59
Royrsquos approach to Schanuelrsquos Conjecture (1999)
If the number of equations we produce is too small such a setof relations does not contain any information The existenceof a sequence of nonndashtrivial polynomials (PN)Nge0 follows fromlinear algebra
On the other hand following D Roy one may expect that theexistence of such a sequence (PN)Nge0 producing sufficientlymany such equations will yield the desired conclusion
tr degQQ(x1 xn α1 αn
)ge n
50 59
Royrsquos Conjecture (1999)Let s0 s1 t0 t1 u positive real numbers satisfying
max1 t0 2t1 lt mins0 2s1and
maxs0 s1 + t1 lt u lt1
2(1 + t0 + t1)
Assume that for any sufficiently large positive integer N there exists a nonndashzero polynomial PN isin Z[X0 X1] withpartial degree le N t0 in X0 partial degree le N t1 in X1 andheight le eN which satisfies∣∣∣∣∣(DkPN)(
nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1 Then
tr degQQ(x1 xn α1 αn) ge n51 59
Royrsquos Theorem (1999)
Royrsquos conjecture is equivalent to Schanuelrsquos Conjecture
52 59
Equivalence between Schanuel and Roy
Let (x α) isin CtimesCtimes and let s0 s1 t0 t1 u be positive realnumbers satisfying the inequalities of Royrsquos Conjecture Thenthe following conditions are equivalent (a) The number αeminusx is a root of unity
(b) For any sufficiently large positive integer N there exists anonndashzero polynomial QN isin Z[X0 X1] with partial degreele N t0 in X0 partial degree le N t1 in X1 and heightH(QN) le eN such that∣∣(DkQN)(mx αm)
∣∣ le exp(minusNu)
for any km isin N with k le N s0 and m le N s1
53 59
Gelrsquofondrsquos transcendence criterion
Simple form Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le n and height le en such that
|Pn(ϑ)| le eminus7n2
for all n ge 1 then ϑ is algebraic and Pn(ϑ) = 0 for all n ge 1
First extension Replace the bound for the degree by le dnthe bound for the height by ebn and the bound for |Pn(ϑ)| byeminuscdnbn with some constant c gt 0 independent of n
Some mild conditions are required on the sequences (dn)nge1 and(bn)nge1
54 59
Transcendence criterion with multiplicitiesWith derivatives Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le dn and height le ebn such that∣∣P (j)
n (ϑ)∣∣ le eminuscdnbntn
for all j in the range 0 le j lt tn and all n ge 1 then ϑ isalgebraicDue to M Laurent and D Roy applications to algebraicindependence
55 59
Criterion with several points
Goal Given a sequence of complex numbers (ϑi)ige1 if thereexists a sequence (Pn)nge1 of nonndashzero polynomials in Z[X]with Pn of degree le dn and height le ebn such that∣∣P (j)
n (ϑi)∣∣ le eminuscdnbntnsn
for 0 le j lt tn 1 le i le sn and all n ge 1 then the numbers ϑiare algebraic
D Roy Not true in general but true in some special caseswith a structure on the sequence (ϑi)ige1Combines the elimination arguments used for criteria ofalgebraic independence and for zero estimates
56 59
Small value estimates for the additive group
D Roy Small value estimates for the additive group Intern JNumber Theory to appear
Let ξ be a transcendental complex number let β σ τ and νbe non-negative real numbers let n0 be a positive integer andlet (Pn)ngen0 be a sequence of nonndashzero polynomials in Z[T ]satisfying deg(Pn) le n and H(Pn) le exp(nβ) for eachn ge n0 Suppose that β gt 1 (34)σ + τ lt 1 andν gt 1 + β minus (34)σ minus τ Then for infinitely many n we have
max|P [j]n (iξ)| 0 le i le nσ 0 le j le nτ
gt exp(minusnν)
57 59
Small value estimates for the multiplicative group
D Roy Small value estimates for the multiplicative groupActa Arith to appear
Let ξ1 ξm be multiplicatively independent complexnumbers in a field of transcendence degree 1 Under suitableassumptions on the parameters β σ τ ν for infinitely manypositive integers n there exists no nonndashzero polynomialP isin Z[T ] satisfying deg(P ) le n H(P ) le exp(nβ) and
max|P [j](ξi11 middot middot middot ξimm )| 0 le i1 im le nσ 0 le j le nτ
gt exp(minusnν)
58 59
Heidelberg Colloquium organized by the Institute for Mathematics
httpwwwmathiuni-heidelbergdevortraghtmls=1 January 22 2009
Transcendental Number TheorySchanuelrsquos Conjecture
Michel Waldschmidt
Institut de Mathematiques de Jussieu amp Paris VIhttpwwwmathjussieufrsimmiw
59 59
Large transcendence degree
GV Chudnovsky (1976)Among the numbers
αβ αβ2
αβdminus1
at least [log2 d] arealgebraically independent
GV Chudnovsky ndash On the path to Schanuelrsquosconjecture Algebraic curves close to a pointI General theory of colored sequencesII Fields of finite transcendence type and coloredsequences ResultantsStudia Sci Math Hungar 12 (1977) 125ndash157 (1980)
35 59
Partial result on the problem of Gelrsquofond and
Schneider
AO Gelrsquofond GV Chudnovskii P Philippon YuV Nesterenko
G Diaz If α is an algebraic number α 6= 0 α 6= 1 and if β isan irrational algebraic number of degree d then
tr degQQ(αβ αβ
2
αβdminus1) ge [d+ 1
2
]
36 59
Conjecture of algebraic independence of logarithms
of algebraic numbers
Denote by L the set of complex numbers λ for which eλ isalgebraic
L = logα α isin Qtimes
Hence L is a Q-vector subspace of CThe most important special case of Schanuelrsquos Conjecture is
Conjecture Let λ1 λn be Q-linearly independent elementsin L Then the numbers λ1 λn are algebraicallyindependent over Q
Not yet known that the transcendence degree is ge 2 Open problem Among all logarithms of algebraic numbersone at least is transcendental over Q(π)
37 59
Structural rank of a matrix
Let K be a field k a subfield and M a matrix with entries in KConsider the k-vector subspace E of K spanned by the entriesof M Choose an injective morphism ϕ of E into a k-vectorspace kX1 + middot middot middot+ kXn The image ϕ(M) of M is a matrixwhose entries are in the field k(X1 Xn) of rationalfractions Its rank does not depend on the choice of ϕ
This is the structural rank of M with respect to k
38 59
Example
LetM =
(bij + λij
)1leiled1lejle`
be a matrix with coefficients in Q + L Consider a basis of theQndashvector spanned by the entries and replace the elements inthis basis by unknowns This gives a new matrix M withcoefficients in a field of rational fractions the rank of which isthe structural rank of M (with respect to Q)
As a consequence of the conjecture of algebraic independenceof logarithms of algebraic numbers the rank of M should bethe same as the rank of its specialization M
39 59
Equivalence between the two conjectures
Following D Roy theconjecture on algebraicindependence of logarithms ofalgebraic numbers isequivalent to
Conjecture Any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L has a rank equal to its structuralrank
40 59
Any Polynomial is the Determinant of a Matrix
The proof of the equivalence uses the nice auxiliary result
For any P isin k[X1 Xn] there exists a square matrix withentries in the k-vector space k + kX1 + middot middot middot+ kXn whosedeterminant is P
41 59
Half of the Conjecture is solved
From a certain point of view half of the conjecture of algebraicindependence of logarithms of algebraic numbers is solved
Theorem [D Roy] The rank of any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L is at least half its structural rank
42 59
Reformulation by D Roy
Instead of taking logarithmsof algebraic numbers andlooking for the algebraicindependence relationsD Roy fixes a polynomial andlooks at the points withcoordinates logarithms ofalgebraic numbers on thecorresponding hypersurface
43 59
Reformulation by D Roy
Royrsquos reformulation of the conjecture of algebraicindependence of logarithms is
Conjecture For any algebraic subvariety V of Cn definedover the field Q of algebraic numbers the set V cap Ln is theunion of the sets E cap Ln where E ranges over the set ofvector subspaces of Cn which are defined over Q andcontained in V
Trivial Any element in E cap Ln where E is a vector subspaceof Cn defined over Q and contained in V belongs to V cap Ln
44 59
Example The Four Exponentials ConjectureTake for V the hypersurface of C4 defined by the equation
z1z4 = z2z3
The maximal Cndashvector subspaces of C4 defined over Q andcontained in V are the planes
az1 = bz2 bz4 = az3
and the planesaz1 = bz3 bz4 = az2
with (a b) isin Q2 (0 0)Hence Schanuelrsquos Conjecture implies that if λ1 λ2 λ3 λ4 arelogarithms of algebraic numbers satisfying
λ1λ4 = λ2λ3
then either λ1 λ2 are linearly dependent over Q or else λ1 λ3
are linearly dependent over Q45 59
Six Exponentials Theorem
and Four Exponentials Conjecture
A Selberg CL Siegel S Lang K Ramachandra
46 59
The Four Exponentials Conjecture
Conjecture (Th Schneider S Lang K Ramachandra) Ifx1 x2 are Qndashlinearly independent complex numbers and y1 y2
are Qndashlinearly independent complex numbers then one atleast of the four numbers
ex1y1 ex1y2 ex2y1 ex2y2
is transcendentalEquivalent statement Any 2times 2 matrix(
λ1 λ2
λ3 λ4
)with entries in L and with Qndashlinearly independent rows andQndashlinearly independent columns has maximal rank 2
47 59
How could we attack Schanuelrsquos Conjecture
Let x1 xn be Qndashlinearly independent complex numbersFollowing the transcendence methods of Hermite GelrsquofondSchneider one may start by introducing an auxiliaryfunction
F (z) = P (z ez)
where P isin Z[X0 X1] is a nonndashzero polynomial One considersthe derivatives (
d
dz
)kF = (DkP )(z ez)
at the pointsm1x1 + middot middot middot+mnxn
for various values of (m1 mn) isin Zn
48 59
Auxiliary functionLet D denote the derivation
D =part
partX0
+X1part
partX1
over the ring C[X0 X1] Recall that x1 xn are Qndashlinearlyindependent complex numbers Let α1 αn be nonndashzerocomplex numbersThe transcendence machinery produces a sequence (PN)Nge0
of polynomials with integer coefficients satisfying∣∣∣∣∣(DkPN)(nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1
49 59
Royrsquos approach to Schanuelrsquos Conjecture (1999)
If the number of equations we produce is too small such a setof relations does not contain any information The existenceof a sequence of nonndashtrivial polynomials (PN)Nge0 follows fromlinear algebra
On the other hand following D Roy one may expect that theexistence of such a sequence (PN)Nge0 producing sufficientlymany such equations will yield the desired conclusion
tr degQQ(x1 xn α1 αn
)ge n
50 59
Royrsquos Conjecture (1999)Let s0 s1 t0 t1 u positive real numbers satisfying
max1 t0 2t1 lt mins0 2s1and
maxs0 s1 + t1 lt u lt1
2(1 + t0 + t1)
Assume that for any sufficiently large positive integer N there exists a nonndashzero polynomial PN isin Z[X0 X1] withpartial degree le N t0 in X0 partial degree le N t1 in X1 andheight le eN which satisfies∣∣∣∣∣(DkPN)(
nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1 Then
tr degQQ(x1 xn α1 αn) ge n51 59
Royrsquos Theorem (1999)
Royrsquos conjecture is equivalent to Schanuelrsquos Conjecture
52 59
Equivalence between Schanuel and Roy
Let (x α) isin CtimesCtimes and let s0 s1 t0 t1 u be positive realnumbers satisfying the inequalities of Royrsquos Conjecture Thenthe following conditions are equivalent (a) The number αeminusx is a root of unity
(b) For any sufficiently large positive integer N there exists anonndashzero polynomial QN isin Z[X0 X1] with partial degreele N t0 in X0 partial degree le N t1 in X1 and heightH(QN) le eN such that∣∣(DkQN)(mx αm)
∣∣ le exp(minusNu)
for any km isin N with k le N s0 and m le N s1
53 59
Gelrsquofondrsquos transcendence criterion
Simple form Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le n and height le en such that
|Pn(ϑ)| le eminus7n2
for all n ge 1 then ϑ is algebraic and Pn(ϑ) = 0 for all n ge 1
First extension Replace the bound for the degree by le dnthe bound for the height by ebn and the bound for |Pn(ϑ)| byeminuscdnbn with some constant c gt 0 independent of n
Some mild conditions are required on the sequences (dn)nge1 and(bn)nge1
54 59
Transcendence criterion with multiplicitiesWith derivatives Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le dn and height le ebn such that∣∣P (j)
n (ϑ)∣∣ le eminuscdnbntn
for all j in the range 0 le j lt tn and all n ge 1 then ϑ isalgebraicDue to M Laurent and D Roy applications to algebraicindependence
55 59
Criterion with several points
Goal Given a sequence of complex numbers (ϑi)ige1 if thereexists a sequence (Pn)nge1 of nonndashzero polynomials in Z[X]with Pn of degree le dn and height le ebn such that∣∣P (j)
n (ϑi)∣∣ le eminuscdnbntnsn
for 0 le j lt tn 1 le i le sn and all n ge 1 then the numbers ϑiare algebraic
D Roy Not true in general but true in some special caseswith a structure on the sequence (ϑi)ige1Combines the elimination arguments used for criteria ofalgebraic independence and for zero estimates
56 59
Small value estimates for the additive group
D Roy Small value estimates for the additive group Intern JNumber Theory to appear
Let ξ be a transcendental complex number let β σ τ and νbe non-negative real numbers let n0 be a positive integer andlet (Pn)ngen0 be a sequence of nonndashzero polynomials in Z[T ]satisfying deg(Pn) le n and H(Pn) le exp(nβ) for eachn ge n0 Suppose that β gt 1 (34)σ + τ lt 1 andν gt 1 + β minus (34)σ minus τ Then for infinitely many n we have
max|P [j]n (iξ)| 0 le i le nσ 0 le j le nτ
gt exp(minusnν)
57 59
Small value estimates for the multiplicative group
D Roy Small value estimates for the multiplicative groupActa Arith to appear
Let ξ1 ξm be multiplicatively independent complexnumbers in a field of transcendence degree 1 Under suitableassumptions on the parameters β σ τ ν for infinitely manypositive integers n there exists no nonndashzero polynomialP isin Z[T ] satisfying deg(P ) le n H(P ) le exp(nβ) and
max|P [j](ξi11 middot middot middot ξimm )| 0 le i1 im le nσ 0 le j le nτ
gt exp(minusnν)
58 59
Heidelberg Colloquium organized by the Institute for Mathematics
httpwwwmathiuni-heidelbergdevortraghtmls=1 January 22 2009
Transcendental Number TheorySchanuelrsquos Conjecture
Michel Waldschmidt
Institut de Mathematiques de Jussieu amp Paris VIhttpwwwmathjussieufrsimmiw
59 59
Partial result on the problem of Gelrsquofond and
Schneider
AO Gelrsquofond GV Chudnovskii P Philippon YuV Nesterenko
G Diaz If α is an algebraic number α 6= 0 α 6= 1 and if β isan irrational algebraic number of degree d then
tr degQQ(αβ αβ
2
αβdminus1) ge [d+ 1
2
]
36 59
Conjecture of algebraic independence of logarithms
of algebraic numbers
Denote by L the set of complex numbers λ for which eλ isalgebraic
L = logα α isin Qtimes
Hence L is a Q-vector subspace of CThe most important special case of Schanuelrsquos Conjecture is
Conjecture Let λ1 λn be Q-linearly independent elementsin L Then the numbers λ1 λn are algebraicallyindependent over Q
Not yet known that the transcendence degree is ge 2 Open problem Among all logarithms of algebraic numbersone at least is transcendental over Q(π)
37 59
Structural rank of a matrix
Let K be a field k a subfield and M a matrix with entries in KConsider the k-vector subspace E of K spanned by the entriesof M Choose an injective morphism ϕ of E into a k-vectorspace kX1 + middot middot middot+ kXn The image ϕ(M) of M is a matrixwhose entries are in the field k(X1 Xn) of rationalfractions Its rank does not depend on the choice of ϕ
This is the structural rank of M with respect to k
38 59
Example
LetM =
(bij + λij
)1leiled1lejle`
be a matrix with coefficients in Q + L Consider a basis of theQndashvector spanned by the entries and replace the elements inthis basis by unknowns This gives a new matrix M withcoefficients in a field of rational fractions the rank of which isthe structural rank of M (with respect to Q)
As a consequence of the conjecture of algebraic independenceof logarithms of algebraic numbers the rank of M should bethe same as the rank of its specialization M
39 59
Equivalence between the two conjectures
Following D Roy theconjecture on algebraicindependence of logarithms ofalgebraic numbers isequivalent to
Conjecture Any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L has a rank equal to its structuralrank
40 59
Any Polynomial is the Determinant of a Matrix
The proof of the equivalence uses the nice auxiliary result
For any P isin k[X1 Xn] there exists a square matrix withentries in the k-vector space k + kX1 + middot middot middot+ kXn whosedeterminant is P
41 59
Half of the Conjecture is solved
From a certain point of view half of the conjecture of algebraicindependence of logarithms of algebraic numbers is solved
Theorem [D Roy] The rank of any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L is at least half its structural rank
42 59
Reformulation by D Roy
Instead of taking logarithmsof algebraic numbers andlooking for the algebraicindependence relationsD Roy fixes a polynomial andlooks at the points withcoordinates logarithms ofalgebraic numbers on thecorresponding hypersurface
43 59
Reformulation by D Roy
Royrsquos reformulation of the conjecture of algebraicindependence of logarithms is
Conjecture For any algebraic subvariety V of Cn definedover the field Q of algebraic numbers the set V cap Ln is theunion of the sets E cap Ln where E ranges over the set ofvector subspaces of Cn which are defined over Q andcontained in V
Trivial Any element in E cap Ln where E is a vector subspaceof Cn defined over Q and contained in V belongs to V cap Ln
44 59
Example The Four Exponentials ConjectureTake for V the hypersurface of C4 defined by the equation
z1z4 = z2z3
The maximal Cndashvector subspaces of C4 defined over Q andcontained in V are the planes
az1 = bz2 bz4 = az3
and the planesaz1 = bz3 bz4 = az2
with (a b) isin Q2 (0 0)Hence Schanuelrsquos Conjecture implies that if λ1 λ2 λ3 λ4 arelogarithms of algebraic numbers satisfying
λ1λ4 = λ2λ3
then either λ1 λ2 are linearly dependent over Q or else λ1 λ3
are linearly dependent over Q45 59
Six Exponentials Theorem
and Four Exponentials Conjecture
A Selberg CL Siegel S Lang K Ramachandra
46 59
The Four Exponentials Conjecture
Conjecture (Th Schneider S Lang K Ramachandra) Ifx1 x2 are Qndashlinearly independent complex numbers and y1 y2
are Qndashlinearly independent complex numbers then one atleast of the four numbers
ex1y1 ex1y2 ex2y1 ex2y2
is transcendentalEquivalent statement Any 2times 2 matrix(
λ1 λ2
λ3 λ4
)with entries in L and with Qndashlinearly independent rows andQndashlinearly independent columns has maximal rank 2
47 59
How could we attack Schanuelrsquos Conjecture
Let x1 xn be Qndashlinearly independent complex numbersFollowing the transcendence methods of Hermite GelrsquofondSchneider one may start by introducing an auxiliaryfunction
F (z) = P (z ez)
where P isin Z[X0 X1] is a nonndashzero polynomial One considersthe derivatives (
d
dz
)kF = (DkP )(z ez)
at the pointsm1x1 + middot middot middot+mnxn
for various values of (m1 mn) isin Zn
48 59
Auxiliary functionLet D denote the derivation
D =part
partX0
+X1part
partX1
over the ring C[X0 X1] Recall that x1 xn are Qndashlinearlyindependent complex numbers Let α1 αn be nonndashzerocomplex numbersThe transcendence machinery produces a sequence (PN)Nge0
of polynomials with integer coefficients satisfying∣∣∣∣∣(DkPN)(nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1
49 59
Royrsquos approach to Schanuelrsquos Conjecture (1999)
If the number of equations we produce is too small such a setof relations does not contain any information The existenceof a sequence of nonndashtrivial polynomials (PN)Nge0 follows fromlinear algebra
On the other hand following D Roy one may expect that theexistence of such a sequence (PN)Nge0 producing sufficientlymany such equations will yield the desired conclusion
tr degQQ(x1 xn α1 αn
)ge n
50 59
Royrsquos Conjecture (1999)Let s0 s1 t0 t1 u positive real numbers satisfying
max1 t0 2t1 lt mins0 2s1and
maxs0 s1 + t1 lt u lt1
2(1 + t0 + t1)
Assume that for any sufficiently large positive integer N there exists a nonndashzero polynomial PN isin Z[X0 X1] withpartial degree le N t0 in X0 partial degree le N t1 in X1 andheight le eN which satisfies∣∣∣∣∣(DkPN)(
nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1 Then
tr degQQ(x1 xn α1 αn) ge n51 59
Royrsquos Theorem (1999)
Royrsquos conjecture is equivalent to Schanuelrsquos Conjecture
52 59
Equivalence between Schanuel and Roy
Let (x α) isin CtimesCtimes and let s0 s1 t0 t1 u be positive realnumbers satisfying the inequalities of Royrsquos Conjecture Thenthe following conditions are equivalent (a) The number αeminusx is a root of unity
(b) For any sufficiently large positive integer N there exists anonndashzero polynomial QN isin Z[X0 X1] with partial degreele N t0 in X0 partial degree le N t1 in X1 and heightH(QN) le eN such that∣∣(DkQN)(mx αm)
∣∣ le exp(minusNu)
for any km isin N with k le N s0 and m le N s1
53 59
Gelrsquofondrsquos transcendence criterion
Simple form Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le n and height le en such that
|Pn(ϑ)| le eminus7n2
for all n ge 1 then ϑ is algebraic and Pn(ϑ) = 0 for all n ge 1
First extension Replace the bound for the degree by le dnthe bound for the height by ebn and the bound for |Pn(ϑ)| byeminuscdnbn with some constant c gt 0 independent of n
Some mild conditions are required on the sequences (dn)nge1 and(bn)nge1
54 59
Transcendence criterion with multiplicitiesWith derivatives Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le dn and height le ebn such that∣∣P (j)
n (ϑ)∣∣ le eminuscdnbntn
for all j in the range 0 le j lt tn and all n ge 1 then ϑ isalgebraicDue to M Laurent and D Roy applications to algebraicindependence
55 59
Criterion with several points
Goal Given a sequence of complex numbers (ϑi)ige1 if thereexists a sequence (Pn)nge1 of nonndashzero polynomials in Z[X]with Pn of degree le dn and height le ebn such that∣∣P (j)
n (ϑi)∣∣ le eminuscdnbntnsn
for 0 le j lt tn 1 le i le sn and all n ge 1 then the numbers ϑiare algebraic
D Roy Not true in general but true in some special caseswith a structure on the sequence (ϑi)ige1Combines the elimination arguments used for criteria ofalgebraic independence and for zero estimates
56 59
Small value estimates for the additive group
D Roy Small value estimates for the additive group Intern JNumber Theory to appear
Let ξ be a transcendental complex number let β σ τ and νbe non-negative real numbers let n0 be a positive integer andlet (Pn)ngen0 be a sequence of nonndashzero polynomials in Z[T ]satisfying deg(Pn) le n and H(Pn) le exp(nβ) for eachn ge n0 Suppose that β gt 1 (34)σ + τ lt 1 andν gt 1 + β minus (34)σ minus τ Then for infinitely many n we have
max|P [j]n (iξ)| 0 le i le nσ 0 le j le nτ
gt exp(minusnν)
57 59
Small value estimates for the multiplicative group
D Roy Small value estimates for the multiplicative groupActa Arith to appear
Let ξ1 ξm be multiplicatively independent complexnumbers in a field of transcendence degree 1 Under suitableassumptions on the parameters β σ τ ν for infinitely manypositive integers n there exists no nonndashzero polynomialP isin Z[T ] satisfying deg(P ) le n H(P ) le exp(nβ) and
max|P [j](ξi11 middot middot middot ξimm )| 0 le i1 im le nσ 0 le j le nτ
gt exp(minusnν)
58 59
Heidelberg Colloquium organized by the Institute for Mathematics
httpwwwmathiuni-heidelbergdevortraghtmls=1 January 22 2009
Transcendental Number TheorySchanuelrsquos Conjecture
Michel Waldschmidt
Institut de Mathematiques de Jussieu amp Paris VIhttpwwwmathjussieufrsimmiw
59 59
Conjecture of algebraic independence of logarithms
of algebraic numbers
Denote by L the set of complex numbers λ for which eλ isalgebraic
L = logα α isin Qtimes
Hence L is a Q-vector subspace of CThe most important special case of Schanuelrsquos Conjecture is
Conjecture Let λ1 λn be Q-linearly independent elementsin L Then the numbers λ1 λn are algebraicallyindependent over Q
Not yet known that the transcendence degree is ge 2 Open problem Among all logarithms of algebraic numbersone at least is transcendental over Q(π)
37 59
Structural rank of a matrix
Let K be a field k a subfield and M a matrix with entries in KConsider the k-vector subspace E of K spanned by the entriesof M Choose an injective morphism ϕ of E into a k-vectorspace kX1 + middot middot middot+ kXn The image ϕ(M) of M is a matrixwhose entries are in the field k(X1 Xn) of rationalfractions Its rank does not depend on the choice of ϕ
This is the structural rank of M with respect to k
38 59
Example
LetM =
(bij + λij
)1leiled1lejle`
be a matrix with coefficients in Q + L Consider a basis of theQndashvector spanned by the entries and replace the elements inthis basis by unknowns This gives a new matrix M withcoefficients in a field of rational fractions the rank of which isthe structural rank of M (with respect to Q)
As a consequence of the conjecture of algebraic independenceof logarithms of algebraic numbers the rank of M should bethe same as the rank of its specialization M
39 59
Equivalence between the two conjectures
Following D Roy theconjecture on algebraicindependence of logarithms ofalgebraic numbers isequivalent to
Conjecture Any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L has a rank equal to its structuralrank
40 59
Any Polynomial is the Determinant of a Matrix
The proof of the equivalence uses the nice auxiliary result
For any P isin k[X1 Xn] there exists a square matrix withentries in the k-vector space k + kX1 + middot middot middot+ kXn whosedeterminant is P
41 59
Half of the Conjecture is solved
From a certain point of view half of the conjecture of algebraicindependence of logarithms of algebraic numbers is solved
Theorem [D Roy] The rank of any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L is at least half its structural rank
42 59
Reformulation by D Roy
Instead of taking logarithmsof algebraic numbers andlooking for the algebraicindependence relationsD Roy fixes a polynomial andlooks at the points withcoordinates logarithms ofalgebraic numbers on thecorresponding hypersurface
43 59
Reformulation by D Roy
Royrsquos reformulation of the conjecture of algebraicindependence of logarithms is
Conjecture For any algebraic subvariety V of Cn definedover the field Q of algebraic numbers the set V cap Ln is theunion of the sets E cap Ln where E ranges over the set ofvector subspaces of Cn which are defined over Q andcontained in V
Trivial Any element in E cap Ln where E is a vector subspaceof Cn defined over Q and contained in V belongs to V cap Ln
44 59
Example The Four Exponentials ConjectureTake for V the hypersurface of C4 defined by the equation
z1z4 = z2z3
The maximal Cndashvector subspaces of C4 defined over Q andcontained in V are the planes
az1 = bz2 bz4 = az3
and the planesaz1 = bz3 bz4 = az2
with (a b) isin Q2 (0 0)Hence Schanuelrsquos Conjecture implies that if λ1 λ2 λ3 λ4 arelogarithms of algebraic numbers satisfying
λ1λ4 = λ2λ3
then either λ1 λ2 are linearly dependent over Q or else λ1 λ3
are linearly dependent over Q45 59
Six Exponentials Theorem
and Four Exponentials Conjecture
A Selberg CL Siegel S Lang K Ramachandra
46 59
The Four Exponentials Conjecture
Conjecture (Th Schneider S Lang K Ramachandra) Ifx1 x2 are Qndashlinearly independent complex numbers and y1 y2
are Qndashlinearly independent complex numbers then one atleast of the four numbers
ex1y1 ex1y2 ex2y1 ex2y2
is transcendentalEquivalent statement Any 2times 2 matrix(
λ1 λ2
λ3 λ4
)with entries in L and with Qndashlinearly independent rows andQndashlinearly independent columns has maximal rank 2
47 59
How could we attack Schanuelrsquos Conjecture
Let x1 xn be Qndashlinearly independent complex numbersFollowing the transcendence methods of Hermite GelrsquofondSchneider one may start by introducing an auxiliaryfunction
F (z) = P (z ez)
where P isin Z[X0 X1] is a nonndashzero polynomial One considersthe derivatives (
d
dz
)kF = (DkP )(z ez)
at the pointsm1x1 + middot middot middot+mnxn
for various values of (m1 mn) isin Zn
48 59
Auxiliary functionLet D denote the derivation
D =part
partX0
+X1part
partX1
over the ring C[X0 X1] Recall that x1 xn are Qndashlinearlyindependent complex numbers Let α1 αn be nonndashzerocomplex numbersThe transcendence machinery produces a sequence (PN)Nge0
of polynomials with integer coefficients satisfying∣∣∣∣∣(DkPN)(nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1
49 59
Royrsquos approach to Schanuelrsquos Conjecture (1999)
If the number of equations we produce is too small such a setof relations does not contain any information The existenceof a sequence of nonndashtrivial polynomials (PN)Nge0 follows fromlinear algebra
On the other hand following D Roy one may expect that theexistence of such a sequence (PN)Nge0 producing sufficientlymany such equations will yield the desired conclusion
tr degQQ(x1 xn α1 αn
)ge n
50 59
Royrsquos Conjecture (1999)Let s0 s1 t0 t1 u positive real numbers satisfying
max1 t0 2t1 lt mins0 2s1and
maxs0 s1 + t1 lt u lt1
2(1 + t0 + t1)
Assume that for any sufficiently large positive integer N there exists a nonndashzero polynomial PN isin Z[X0 X1] withpartial degree le N t0 in X0 partial degree le N t1 in X1 andheight le eN which satisfies∣∣∣∣∣(DkPN)(
nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1 Then
tr degQQ(x1 xn α1 αn) ge n51 59
Royrsquos Theorem (1999)
Royrsquos conjecture is equivalent to Schanuelrsquos Conjecture
52 59
Equivalence between Schanuel and Roy
Let (x α) isin CtimesCtimes and let s0 s1 t0 t1 u be positive realnumbers satisfying the inequalities of Royrsquos Conjecture Thenthe following conditions are equivalent (a) The number αeminusx is a root of unity
(b) For any sufficiently large positive integer N there exists anonndashzero polynomial QN isin Z[X0 X1] with partial degreele N t0 in X0 partial degree le N t1 in X1 and heightH(QN) le eN such that∣∣(DkQN)(mx αm)
∣∣ le exp(minusNu)
for any km isin N with k le N s0 and m le N s1
53 59
Gelrsquofondrsquos transcendence criterion
Simple form Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le n and height le en such that
|Pn(ϑ)| le eminus7n2
for all n ge 1 then ϑ is algebraic and Pn(ϑ) = 0 for all n ge 1
First extension Replace the bound for the degree by le dnthe bound for the height by ebn and the bound for |Pn(ϑ)| byeminuscdnbn with some constant c gt 0 independent of n
Some mild conditions are required on the sequences (dn)nge1 and(bn)nge1
54 59
Transcendence criterion with multiplicitiesWith derivatives Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le dn and height le ebn such that∣∣P (j)
n (ϑ)∣∣ le eminuscdnbntn
for all j in the range 0 le j lt tn and all n ge 1 then ϑ isalgebraicDue to M Laurent and D Roy applications to algebraicindependence
55 59
Criterion with several points
Goal Given a sequence of complex numbers (ϑi)ige1 if thereexists a sequence (Pn)nge1 of nonndashzero polynomials in Z[X]with Pn of degree le dn and height le ebn such that∣∣P (j)
n (ϑi)∣∣ le eminuscdnbntnsn
for 0 le j lt tn 1 le i le sn and all n ge 1 then the numbers ϑiare algebraic
D Roy Not true in general but true in some special caseswith a structure on the sequence (ϑi)ige1Combines the elimination arguments used for criteria ofalgebraic independence and for zero estimates
56 59
Small value estimates for the additive group
D Roy Small value estimates for the additive group Intern JNumber Theory to appear
Let ξ be a transcendental complex number let β σ τ and νbe non-negative real numbers let n0 be a positive integer andlet (Pn)ngen0 be a sequence of nonndashzero polynomials in Z[T ]satisfying deg(Pn) le n and H(Pn) le exp(nβ) for eachn ge n0 Suppose that β gt 1 (34)σ + τ lt 1 andν gt 1 + β minus (34)σ minus τ Then for infinitely many n we have
max|P [j]n (iξ)| 0 le i le nσ 0 le j le nτ
gt exp(minusnν)
57 59
Small value estimates for the multiplicative group
D Roy Small value estimates for the multiplicative groupActa Arith to appear
Let ξ1 ξm be multiplicatively independent complexnumbers in a field of transcendence degree 1 Under suitableassumptions on the parameters β σ τ ν for infinitely manypositive integers n there exists no nonndashzero polynomialP isin Z[T ] satisfying deg(P ) le n H(P ) le exp(nβ) and
max|P [j](ξi11 middot middot middot ξimm )| 0 le i1 im le nσ 0 le j le nτ
gt exp(minusnν)
58 59
Heidelberg Colloquium organized by the Institute for Mathematics
httpwwwmathiuni-heidelbergdevortraghtmls=1 January 22 2009
Transcendental Number TheorySchanuelrsquos Conjecture
Michel Waldschmidt
Institut de Mathematiques de Jussieu amp Paris VIhttpwwwmathjussieufrsimmiw
59 59
Structural rank of a matrix
Let K be a field k a subfield and M a matrix with entries in KConsider the k-vector subspace E of K spanned by the entriesof M Choose an injective morphism ϕ of E into a k-vectorspace kX1 + middot middot middot+ kXn The image ϕ(M) of M is a matrixwhose entries are in the field k(X1 Xn) of rationalfractions Its rank does not depend on the choice of ϕ
This is the structural rank of M with respect to k
38 59
Example
LetM =
(bij + λij
)1leiled1lejle`
be a matrix with coefficients in Q + L Consider a basis of theQndashvector spanned by the entries and replace the elements inthis basis by unknowns This gives a new matrix M withcoefficients in a field of rational fractions the rank of which isthe structural rank of M (with respect to Q)
As a consequence of the conjecture of algebraic independenceof logarithms of algebraic numbers the rank of M should bethe same as the rank of its specialization M
39 59
Equivalence between the two conjectures
Following D Roy theconjecture on algebraicindependence of logarithms ofalgebraic numbers isequivalent to
Conjecture Any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L has a rank equal to its structuralrank
40 59
Any Polynomial is the Determinant of a Matrix
The proof of the equivalence uses the nice auxiliary result
For any P isin k[X1 Xn] there exists a square matrix withentries in the k-vector space k + kX1 + middot middot middot+ kXn whosedeterminant is P
41 59
Half of the Conjecture is solved
From a certain point of view half of the conjecture of algebraicindependence of logarithms of algebraic numbers is solved
Theorem [D Roy] The rank of any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L is at least half its structural rank
42 59
Reformulation by D Roy
Instead of taking logarithmsof algebraic numbers andlooking for the algebraicindependence relationsD Roy fixes a polynomial andlooks at the points withcoordinates logarithms ofalgebraic numbers on thecorresponding hypersurface
43 59
Reformulation by D Roy
Royrsquos reformulation of the conjecture of algebraicindependence of logarithms is
Conjecture For any algebraic subvariety V of Cn definedover the field Q of algebraic numbers the set V cap Ln is theunion of the sets E cap Ln where E ranges over the set ofvector subspaces of Cn which are defined over Q andcontained in V
Trivial Any element in E cap Ln where E is a vector subspaceof Cn defined over Q and contained in V belongs to V cap Ln
44 59
Example The Four Exponentials ConjectureTake for V the hypersurface of C4 defined by the equation
z1z4 = z2z3
The maximal Cndashvector subspaces of C4 defined over Q andcontained in V are the planes
az1 = bz2 bz4 = az3
and the planesaz1 = bz3 bz4 = az2
with (a b) isin Q2 (0 0)Hence Schanuelrsquos Conjecture implies that if λ1 λ2 λ3 λ4 arelogarithms of algebraic numbers satisfying
λ1λ4 = λ2λ3
then either λ1 λ2 are linearly dependent over Q or else λ1 λ3
are linearly dependent over Q45 59
Six Exponentials Theorem
and Four Exponentials Conjecture
A Selberg CL Siegel S Lang K Ramachandra
46 59
The Four Exponentials Conjecture
Conjecture (Th Schneider S Lang K Ramachandra) Ifx1 x2 are Qndashlinearly independent complex numbers and y1 y2
are Qndashlinearly independent complex numbers then one atleast of the four numbers
ex1y1 ex1y2 ex2y1 ex2y2
is transcendentalEquivalent statement Any 2times 2 matrix(
λ1 λ2
λ3 λ4
)with entries in L and with Qndashlinearly independent rows andQndashlinearly independent columns has maximal rank 2
47 59
How could we attack Schanuelrsquos Conjecture
Let x1 xn be Qndashlinearly independent complex numbersFollowing the transcendence methods of Hermite GelrsquofondSchneider one may start by introducing an auxiliaryfunction
F (z) = P (z ez)
where P isin Z[X0 X1] is a nonndashzero polynomial One considersthe derivatives (
d
dz
)kF = (DkP )(z ez)
at the pointsm1x1 + middot middot middot+mnxn
for various values of (m1 mn) isin Zn
48 59
Auxiliary functionLet D denote the derivation
D =part
partX0
+X1part
partX1
over the ring C[X0 X1] Recall that x1 xn are Qndashlinearlyindependent complex numbers Let α1 αn be nonndashzerocomplex numbersThe transcendence machinery produces a sequence (PN)Nge0
of polynomials with integer coefficients satisfying∣∣∣∣∣(DkPN)(nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1
49 59
Royrsquos approach to Schanuelrsquos Conjecture (1999)
If the number of equations we produce is too small such a setof relations does not contain any information The existenceof a sequence of nonndashtrivial polynomials (PN)Nge0 follows fromlinear algebra
On the other hand following D Roy one may expect that theexistence of such a sequence (PN)Nge0 producing sufficientlymany such equations will yield the desired conclusion
tr degQQ(x1 xn α1 αn
)ge n
50 59
Royrsquos Conjecture (1999)Let s0 s1 t0 t1 u positive real numbers satisfying
max1 t0 2t1 lt mins0 2s1and
maxs0 s1 + t1 lt u lt1
2(1 + t0 + t1)
Assume that for any sufficiently large positive integer N there exists a nonndashzero polynomial PN isin Z[X0 X1] withpartial degree le N t0 in X0 partial degree le N t1 in X1 andheight le eN which satisfies∣∣∣∣∣(DkPN)(
nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1 Then
tr degQQ(x1 xn α1 αn) ge n51 59
Royrsquos Theorem (1999)
Royrsquos conjecture is equivalent to Schanuelrsquos Conjecture
52 59
Equivalence between Schanuel and Roy
Let (x α) isin CtimesCtimes and let s0 s1 t0 t1 u be positive realnumbers satisfying the inequalities of Royrsquos Conjecture Thenthe following conditions are equivalent (a) The number αeminusx is a root of unity
(b) For any sufficiently large positive integer N there exists anonndashzero polynomial QN isin Z[X0 X1] with partial degreele N t0 in X0 partial degree le N t1 in X1 and heightH(QN) le eN such that∣∣(DkQN)(mx αm)
∣∣ le exp(minusNu)
for any km isin N with k le N s0 and m le N s1
53 59
Gelrsquofondrsquos transcendence criterion
Simple form Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le n and height le en such that
|Pn(ϑ)| le eminus7n2
for all n ge 1 then ϑ is algebraic and Pn(ϑ) = 0 for all n ge 1
First extension Replace the bound for the degree by le dnthe bound for the height by ebn and the bound for |Pn(ϑ)| byeminuscdnbn with some constant c gt 0 independent of n
Some mild conditions are required on the sequences (dn)nge1 and(bn)nge1
54 59
Transcendence criterion with multiplicitiesWith derivatives Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le dn and height le ebn such that∣∣P (j)
n (ϑ)∣∣ le eminuscdnbntn
for all j in the range 0 le j lt tn and all n ge 1 then ϑ isalgebraicDue to M Laurent and D Roy applications to algebraicindependence
55 59
Criterion with several points
Goal Given a sequence of complex numbers (ϑi)ige1 if thereexists a sequence (Pn)nge1 of nonndashzero polynomials in Z[X]with Pn of degree le dn and height le ebn such that∣∣P (j)
n (ϑi)∣∣ le eminuscdnbntnsn
for 0 le j lt tn 1 le i le sn and all n ge 1 then the numbers ϑiare algebraic
D Roy Not true in general but true in some special caseswith a structure on the sequence (ϑi)ige1Combines the elimination arguments used for criteria ofalgebraic independence and for zero estimates
56 59
Small value estimates for the additive group
D Roy Small value estimates for the additive group Intern JNumber Theory to appear
Let ξ be a transcendental complex number let β σ τ and νbe non-negative real numbers let n0 be a positive integer andlet (Pn)ngen0 be a sequence of nonndashzero polynomials in Z[T ]satisfying deg(Pn) le n and H(Pn) le exp(nβ) for eachn ge n0 Suppose that β gt 1 (34)σ + τ lt 1 andν gt 1 + β minus (34)σ minus τ Then for infinitely many n we have
max|P [j]n (iξ)| 0 le i le nσ 0 le j le nτ
gt exp(minusnν)
57 59
Small value estimates for the multiplicative group
D Roy Small value estimates for the multiplicative groupActa Arith to appear
Let ξ1 ξm be multiplicatively independent complexnumbers in a field of transcendence degree 1 Under suitableassumptions on the parameters β σ τ ν for infinitely manypositive integers n there exists no nonndashzero polynomialP isin Z[T ] satisfying deg(P ) le n H(P ) le exp(nβ) and
max|P [j](ξi11 middot middot middot ξimm )| 0 le i1 im le nσ 0 le j le nτ
gt exp(minusnν)
58 59
Heidelberg Colloquium organized by the Institute for Mathematics
httpwwwmathiuni-heidelbergdevortraghtmls=1 January 22 2009
Transcendental Number TheorySchanuelrsquos Conjecture
Michel Waldschmidt
Institut de Mathematiques de Jussieu amp Paris VIhttpwwwmathjussieufrsimmiw
59 59
Example
LetM =
(bij + λij
)1leiled1lejle`
be a matrix with coefficients in Q + L Consider a basis of theQndashvector spanned by the entries and replace the elements inthis basis by unknowns This gives a new matrix M withcoefficients in a field of rational fractions the rank of which isthe structural rank of M (with respect to Q)
As a consequence of the conjecture of algebraic independenceof logarithms of algebraic numbers the rank of M should bethe same as the rank of its specialization M
39 59
Equivalence between the two conjectures
Following D Roy theconjecture on algebraicindependence of logarithms ofalgebraic numbers isequivalent to
Conjecture Any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L has a rank equal to its structuralrank
40 59
Any Polynomial is the Determinant of a Matrix
The proof of the equivalence uses the nice auxiliary result
For any P isin k[X1 Xn] there exists a square matrix withentries in the k-vector space k + kX1 + middot middot middot+ kXn whosedeterminant is P
41 59
Half of the Conjecture is solved
From a certain point of view half of the conjecture of algebraicindependence of logarithms of algebraic numbers is solved
Theorem [D Roy] The rank of any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L is at least half its structural rank
42 59
Reformulation by D Roy
Instead of taking logarithmsof algebraic numbers andlooking for the algebraicindependence relationsD Roy fixes a polynomial andlooks at the points withcoordinates logarithms ofalgebraic numbers on thecorresponding hypersurface
43 59
Reformulation by D Roy
Royrsquos reformulation of the conjecture of algebraicindependence of logarithms is
Conjecture For any algebraic subvariety V of Cn definedover the field Q of algebraic numbers the set V cap Ln is theunion of the sets E cap Ln where E ranges over the set ofvector subspaces of Cn which are defined over Q andcontained in V
Trivial Any element in E cap Ln where E is a vector subspaceof Cn defined over Q and contained in V belongs to V cap Ln
44 59
Example The Four Exponentials ConjectureTake for V the hypersurface of C4 defined by the equation
z1z4 = z2z3
The maximal Cndashvector subspaces of C4 defined over Q andcontained in V are the planes
az1 = bz2 bz4 = az3
and the planesaz1 = bz3 bz4 = az2
with (a b) isin Q2 (0 0)Hence Schanuelrsquos Conjecture implies that if λ1 λ2 λ3 λ4 arelogarithms of algebraic numbers satisfying
λ1λ4 = λ2λ3
then either λ1 λ2 are linearly dependent over Q or else λ1 λ3
are linearly dependent over Q45 59
Six Exponentials Theorem
and Four Exponentials Conjecture
A Selberg CL Siegel S Lang K Ramachandra
46 59
The Four Exponentials Conjecture
Conjecture (Th Schneider S Lang K Ramachandra) Ifx1 x2 are Qndashlinearly independent complex numbers and y1 y2
are Qndashlinearly independent complex numbers then one atleast of the four numbers
ex1y1 ex1y2 ex2y1 ex2y2
is transcendentalEquivalent statement Any 2times 2 matrix(
λ1 λ2
λ3 λ4
)with entries in L and with Qndashlinearly independent rows andQndashlinearly independent columns has maximal rank 2
47 59
How could we attack Schanuelrsquos Conjecture
Let x1 xn be Qndashlinearly independent complex numbersFollowing the transcendence methods of Hermite GelrsquofondSchneider one may start by introducing an auxiliaryfunction
F (z) = P (z ez)
where P isin Z[X0 X1] is a nonndashzero polynomial One considersthe derivatives (
d
dz
)kF = (DkP )(z ez)
at the pointsm1x1 + middot middot middot+mnxn
for various values of (m1 mn) isin Zn
48 59
Auxiliary functionLet D denote the derivation
D =part
partX0
+X1part
partX1
over the ring C[X0 X1] Recall that x1 xn are Qndashlinearlyindependent complex numbers Let α1 αn be nonndashzerocomplex numbersThe transcendence machinery produces a sequence (PN)Nge0
of polynomials with integer coefficients satisfying∣∣∣∣∣(DkPN)(nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1
49 59
Royrsquos approach to Schanuelrsquos Conjecture (1999)
If the number of equations we produce is too small such a setof relations does not contain any information The existenceof a sequence of nonndashtrivial polynomials (PN)Nge0 follows fromlinear algebra
On the other hand following D Roy one may expect that theexistence of such a sequence (PN)Nge0 producing sufficientlymany such equations will yield the desired conclusion
tr degQQ(x1 xn α1 αn
)ge n
50 59
Royrsquos Conjecture (1999)Let s0 s1 t0 t1 u positive real numbers satisfying
max1 t0 2t1 lt mins0 2s1and
maxs0 s1 + t1 lt u lt1
2(1 + t0 + t1)
Assume that for any sufficiently large positive integer N there exists a nonndashzero polynomial PN isin Z[X0 X1] withpartial degree le N t0 in X0 partial degree le N t1 in X1 andheight le eN which satisfies∣∣∣∣∣(DkPN)(
nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1 Then
tr degQQ(x1 xn α1 αn) ge n51 59
Royrsquos Theorem (1999)
Royrsquos conjecture is equivalent to Schanuelrsquos Conjecture
52 59
Equivalence between Schanuel and Roy
Let (x α) isin CtimesCtimes and let s0 s1 t0 t1 u be positive realnumbers satisfying the inequalities of Royrsquos Conjecture Thenthe following conditions are equivalent (a) The number αeminusx is a root of unity
(b) For any sufficiently large positive integer N there exists anonndashzero polynomial QN isin Z[X0 X1] with partial degreele N t0 in X0 partial degree le N t1 in X1 and heightH(QN) le eN such that∣∣(DkQN)(mx αm)
∣∣ le exp(minusNu)
for any km isin N with k le N s0 and m le N s1
53 59
Gelrsquofondrsquos transcendence criterion
Simple form Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le n and height le en such that
|Pn(ϑ)| le eminus7n2
for all n ge 1 then ϑ is algebraic and Pn(ϑ) = 0 for all n ge 1
First extension Replace the bound for the degree by le dnthe bound for the height by ebn and the bound for |Pn(ϑ)| byeminuscdnbn with some constant c gt 0 independent of n
Some mild conditions are required on the sequences (dn)nge1 and(bn)nge1
54 59
Transcendence criterion with multiplicitiesWith derivatives Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le dn and height le ebn such that∣∣P (j)
n (ϑ)∣∣ le eminuscdnbntn
for all j in the range 0 le j lt tn and all n ge 1 then ϑ isalgebraicDue to M Laurent and D Roy applications to algebraicindependence
55 59
Criterion with several points
Goal Given a sequence of complex numbers (ϑi)ige1 if thereexists a sequence (Pn)nge1 of nonndashzero polynomials in Z[X]with Pn of degree le dn and height le ebn such that∣∣P (j)
n (ϑi)∣∣ le eminuscdnbntnsn
for 0 le j lt tn 1 le i le sn and all n ge 1 then the numbers ϑiare algebraic
D Roy Not true in general but true in some special caseswith a structure on the sequence (ϑi)ige1Combines the elimination arguments used for criteria ofalgebraic independence and for zero estimates
56 59
Small value estimates for the additive group
D Roy Small value estimates for the additive group Intern JNumber Theory to appear
Let ξ be a transcendental complex number let β σ τ and νbe non-negative real numbers let n0 be a positive integer andlet (Pn)ngen0 be a sequence of nonndashzero polynomials in Z[T ]satisfying deg(Pn) le n and H(Pn) le exp(nβ) for eachn ge n0 Suppose that β gt 1 (34)σ + τ lt 1 andν gt 1 + β minus (34)σ minus τ Then for infinitely many n we have
max|P [j]n (iξ)| 0 le i le nσ 0 le j le nτ
gt exp(minusnν)
57 59
Small value estimates for the multiplicative group
D Roy Small value estimates for the multiplicative groupActa Arith to appear
Let ξ1 ξm be multiplicatively independent complexnumbers in a field of transcendence degree 1 Under suitableassumptions on the parameters β σ τ ν for infinitely manypositive integers n there exists no nonndashzero polynomialP isin Z[T ] satisfying deg(P ) le n H(P ) le exp(nβ) and
max|P [j](ξi11 middot middot middot ξimm )| 0 le i1 im le nσ 0 le j le nτ
gt exp(minusnν)
58 59
Heidelberg Colloquium organized by the Institute for Mathematics
httpwwwmathiuni-heidelbergdevortraghtmls=1 January 22 2009
Transcendental Number TheorySchanuelrsquos Conjecture
Michel Waldschmidt
Institut de Mathematiques de Jussieu amp Paris VIhttpwwwmathjussieufrsimmiw
59 59
Equivalence between the two conjectures
Following D Roy theconjecture on algebraicindependence of logarithms ofalgebraic numbers isequivalent to
Conjecture Any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L has a rank equal to its structuralrank
40 59
Any Polynomial is the Determinant of a Matrix
The proof of the equivalence uses the nice auxiliary result
For any P isin k[X1 Xn] there exists a square matrix withentries in the k-vector space k + kX1 + middot middot middot+ kXn whosedeterminant is P
41 59
Half of the Conjecture is solved
From a certain point of view half of the conjecture of algebraicindependence of logarithms of algebraic numbers is solved
Theorem [D Roy] The rank of any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L is at least half its structural rank
42 59
Reformulation by D Roy
Instead of taking logarithmsof algebraic numbers andlooking for the algebraicindependence relationsD Roy fixes a polynomial andlooks at the points withcoordinates logarithms ofalgebraic numbers on thecorresponding hypersurface
43 59
Reformulation by D Roy
Royrsquos reformulation of the conjecture of algebraicindependence of logarithms is
Conjecture For any algebraic subvariety V of Cn definedover the field Q of algebraic numbers the set V cap Ln is theunion of the sets E cap Ln where E ranges over the set ofvector subspaces of Cn which are defined over Q andcontained in V
Trivial Any element in E cap Ln where E is a vector subspaceof Cn defined over Q and contained in V belongs to V cap Ln
44 59
Example The Four Exponentials ConjectureTake for V the hypersurface of C4 defined by the equation
z1z4 = z2z3
The maximal Cndashvector subspaces of C4 defined over Q andcontained in V are the planes
az1 = bz2 bz4 = az3
and the planesaz1 = bz3 bz4 = az2
with (a b) isin Q2 (0 0)Hence Schanuelrsquos Conjecture implies that if λ1 λ2 λ3 λ4 arelogarithms of algebraic numbers satisfying
λ1λ4 = λ2λ3
then either λ1 λ2 are linearly dependent over Q or else λ1 λ3
are linearly dependent over Q45 59
Six Exponentials Theorem
and Four Exponentials Conjecture
A Selberg CL Siegel S Lang K Ramachandra
46 59
The Four Exponentials Conjecture
Conjecture (Th Schneider S Lang K Ramachandra) Ifx1 x2 are Qndashlinearly independent complex numbers and y1 y2
are Qndashlinearly independent complex numbers then one atleast of the four numbers
ex1y1 ex1y2 ex2y1 ex2y2
is transcendentalEquivalent statement Any 2times 2 matrix(
λ1 λ2
λ3 λ4
)with entries in L and with Qndashlinearly independent rows andQndashlinearly independent columns has maximal rank 2
47 59
How could we attack Schanuelrsquos Conjecture
Let x1 xn be Qndashlinearly independent complex numbersFollowing the transcendence methods of Hermite GelrsquofondSchneider one may start by introducing an auxiliaryfunction
F (z) = P (z ez)
where P isin Z[X0 X1] is a nonndashzero polynomial One considersthe derivatives (
d
dz
)kF = (DkP )(z ez)
at the pointsm1x1 + middot middot middot+mnxn
for various values of (m1 mn) isin Zn
48 59
Auxiliary functionLet D denote the derivation
D =part
partX0
+X1part
partX1
over the ring C[X0 X1] Recall that x1 xn are Qndashlinearlyindependent complex numbers Let α1 αn be nonndashzerocomplex numbersThe transcendence machinery produces a sequence (PN)Nge0
of polynomials with integer coefficients satisfying∣∣∣∣∣(DkPN)(nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1
49 59
Royrsquos approach to Schanuelrsquos Conjecture (1999)
If the number of equations we produce is too small such a setof relations does not contain any information The existenceof a sequence of nonndashtrivial polynomials (PN)Nge0 follows fromlinear algebra
On the other hand following D Roy one may expect that theexistence of such a sequence (PN)Nge0 producing sufficientlymany such equations will yield the desired conclusion
tr degQQ(x1 xn α1 αn
)ge n
50 59
Royrsquos Conjecture (1999)Let s0 s1 t0 t1 u positive real numbers satisfying
max1 t0 2t1 lt mins0 2s1and
maxs0 s1 + t1 lt u lt1
2(1 + t0 + t1)
Assume that for any sufficiently large positive integer N there exists a nonndashzero polynomial PN isin Z[X0 X1] withpartial degree le N t0 in X0 partial degree le N t1 in X1 andheight le eN which satisfies∣∣∣∣∣(DkPN)(
nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1 Then
tr degQQ(x1 xn α1 αn) ge n51 59
Royrsquos Theorem (1999)
Royrsquos conjecture is equivalent to Schanuelrsquos Conjecture
52 59
Equivalence between Schanuel and Roy
Let (x α) isin CtimesCtimes and let s0 s1 t0 t1 u be positive realnumbers satisfying the inequalities of Royrsquos Conjecture Thenthe following conditions are equivalent (a) The number αeminusx is a root of unity
(b) For any sufficiently large positive integer N there exists anonndashzero polynomial QN isin Z[X0 X1] with partial degreele N t0 in X0 partial degree le N t1 in X1 and heightH(QN) le eN such that∣∣(DkQN)(mx αm)
∣∣ le exp(minusNu)
for any km isin N with k le N s0 and m le N s1
53 59
Gelrsquofondrsquos transcendence criterion
Simple form Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le n and height le en such that
|Pn(ϑ)| le eminus7n2
for all n ge 1 then ϑ is algebraic and Pn(ϑ) = 0 for all n ge 1
First extension Replace the bound for the degree by le dnthe bound for the height by ebn and the bound for |Pn(ϑ)| byeminuscdnbn with some constant c gt 0 independent of n
Some mild conditions are required on the sequences (dn)nge1 and(bn)nge1
54 59
Transcendence criterion with multiplicitiesWith derivatives Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le dn and height le ebn such that∣∣P (j)
n (ϑ)∣∣ le eminuscdnbntn
for all j in the range 0 le j lt tn and all n ge 1 then ϑ isalgebraicDue to M Laurent and D Roy applications to algebraicindependence
55 59
Criterion with several points
Goal Given a sequence of complex numbers (ϑi)ige1 if thereexists a sequence (Pn)nge1 of nonndashzero polynomials in Z[X]with Pn of degree le dn and height le ebn such that∣∣P (j)
n (ϑi)∣∣ le eminuscdnbntnsn
for 0 le j lt tn 1 le i le sn and all n ge 1 then the numbers ϑiare algebraic
D Roy Not true in general but true in some special caseswith a structure on the sequence (ϑi)ige1Combines the elimination arguments used for criteria ofalgebraic independence and for zero estimates
56 59
Small value estimates for the additive group
D Roy Small value estimates for the additive group Intern JNumber Theory to appear
Let ξ be a transcendental complex number let β σ τ and νbe non-negative real numbers let n0 be a positive integer andlet (Pn)ngen0 be a sequence of nonndashzero polynomials in Z[T ]satisfying deg(Pn) le n and H(Pn) le exp(nβ) for eachn ge n0 Suppose that β gt 1 (34)σ + τ lt 1 andν gt 1 + β minus (34)σ minus τ Then for infinitely many n we have
max|P [j]n (iξ)| 0 le i le nσ 0 le j le nτ
gt exp(minusnν)
57 59
Small value estimates for the multiplicative group
D Roy Small value estimates for the multiplicative groupActa Arith to appear
Let ξ1 ξm be multiplicatively independent complexnumbers in a field of transcendence degree 1 Under suitableassumptions on the parameters β σ τ ν for infinitely manypositive integers n there exists no nonndashzero polynomialP isin Z[T ] satisfying deg(P ) le n H(P ) le exp(nβ) and
max|P [j](ξi11 middot middot middot ξimm )| 0 le i1 im le nσ 0 le j le nτ
gt exp(minusnν)
58 59
Heidelberg Colloquium organized by the Institute for Mathematics
httpwwwmathiuni-heidelbergdevortraghtmls=1 January 22 2009
Transcendental Number TheorySchanuelrsquos Conjecture
Michel Waldschmidt
Institut de Mathematiques de Jussieu amp Paris VIhttpwwwmathjussieufrsimmiw
59 59
Any Polynomial is the Determinant of a Matrix
The proof of the equivalence uses the nice auxiliary result
For any P isin k[X1 Xn] there exists a square matrix withentries in the k-vector space k + kX1 + middot middot middot+ kXn whosedeterminant is P
41 59
Half of the Conjecture is solved
From a certain point of view half of the conjecture of algebraicindependence of logarithms of algebraic numbers is solved
Theorem [D Roy] The rank of any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L is at least half its structural rank
42 59
Reformulation by D Roy
Instead of taking logarithmsof algebraic numbers andlooking for the algebraicindependence relationsD Roy fixes a polynomial andlooks at the points withcoordinates logarithms ofalgebraic numbers on thecorresponding hypersurface
43 59
Reformulation by D Roy
Royrsquos reformulation of the conjecture of algebraicindependence of logarithms is
Conjecture For any algebraic subvariety V of Cn definedover the field Q of algebraic numbers the set V cap Ln is theunion of the sets E cap Ln where E ranges over the set ofvector subspaces of Cn which are defined over Q andcontained in V
Trivial Any element in E cap Ln where E is a vector subspaceof Cn defined over Q and contained in V belongs to V cap Ln
44 59
Example The Four Exponentials ConjectureTake for V the hypersurface of C4 defined by the equation
z1z4 = z2z3
The maximal Cndashvector subspaces of C4 defined over Q andcontained in V are the planes
az1 = bz2 bz4 = az3
and the planesaz1 = bz3 bz4 = az2
with (a b) isin Q2 (0 0)Hence Schanuelrsquos Conjecture implies that if λ1 λ2 λ3 λ4 arelogarithms of algebraic numbers satisfying
λ1λ4 = λ2λ3
then either λ1 λ2 are linearly dependent over Q or else λ1 λ3
are linearly dependent over Q45 59
Six Exponentials Theorem
and Four Exponentials Conjecture
A Selberg CL Siegel S Lang K Ramachandra
46 59
The Four Exponentials Conjecture
Conjecture (Th Schneider S Lang K Ramachandra) Ifx1 x2 are Qndashlinearly independent complex numbers and y1 y2
are Qndashlinearly independent complex numbers then one atleast of the four numbers
ex1y1 ex1y2 ex2y1 ex2y2
is transcendentalEquivalent statement Any 2times 2 matrix(
λ1 λ2
λ3 λ4
)with entries in L and with Qndashlinearly independent rows andQndashlinearly independent columns has maximal rank 2
47 59
How could we attack Schanuelrsquos Conjecture
Let x1 xn be Qndashlinearly independent complex numbersFollowing the transcendence methods of Hermite GelrsquofondSchneider one may start by introducing an auxiliaryfunction
F (z) = P (z ez)
where P isin Z[X0 X1] is a nonndashzero polynomial One considersthe derivatives (
d
dz
)kF = (DkP )(z ez)
at the pointsm1x1 + middot middot middot+mnxn
for various values of (m1 mn) isin Zn
48 59
Auxiliary functionLet D denote the derivation
D =part
partX0
+X1part
partX1
over the ring C[X0 X1] Recall that x1 xn are Qndashlinearlyindependent complex numbers Let α1 αn be nonndashzerocomplex numbersThe transcendence machinery produces a sequence (PN)Nge0
of polynomials with integer coefficients satisfying∣∣∣∣∣(DkPN)(nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1
49 59
Royrsquos approach to Schanuelrsquos Conjecture (1999)
If the number of equations we produce is too small such a setof relations does not contain any information The existenceof a sequence of nonndashtrivial polynomials (PN)Nge0 follows fromlinear algebra
On the other hand following D Roy one may expect that theexistence of such a sequence (PN)Nge0 producing sufficientlymany such equations will yield the desired conclusion
tr degQQ(x1 xn α1 αn
)ge n
50 59
Royrsquos Conjecture (1999)Let s0 s1 t0 t1 u positive real numbers satisfying
max1 t0 2t1 lt mins0 2s1and
maxs0 s1 + t1 lt u lt1
2(1 + t0 + t1)
Assume that for any sufficiently large positive integer N there exists a nonndashzero polynomial PN isin Z[X0 X1] withpartial degree le N t0 in X0 partial degree le N t1 in X1 andheight le eN which satisfies∣∣∣∣∣(DkPN)(
nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1 Then
tr degQQ(x1 xn α1 αn) ge n51 59
Royrsquos Theorem (1999)
Royrsquos conjecture is equivalent to Schanuelrsquos Conjecture
52 59
Equivalence between Schanuel and Roy
Let (x α) isin CtimesCtimes and let s0 s1 t0 t1 u be positive realnumbers satisfying the inequalities of Royrsquos Conjecture Thenthe following conditions are equivalent (a) The number αeminusx is a root of unity
(b) For any sufficiently large positive integer N there exists anonndashzero polynomial QN isin Z[X0 X1] with partial degreele N t0 in X0 partial degree le N t1 in X1 and heightH(QN) le eN such that∣∣(DkQN)(mx αm)
∣∣ le exp(minusNu)
for any km isin N with k le N s0 and m le N s1
53 59
Gelrsquofondrsquos transcendence criterion
Simple form Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le n and height le en such that
|Pn(ϑ)| le eminus7n2
for all n ge 1 then ϑ is algebraic and Pn(ϑ) = 0 for all n ge 1
First extension Replace the bound for the degree by le dnthe bound for the height by ebn and the bound for |Pn(ϑ)| byeminuscdnbn with some constant c gt 0 independent of n
Some mild conditions are required on the sequences (dn)nge1 and(bn)nge1
54 59
Transcendence criterion with multiplicitiesWith derivatives Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le dn and height le ebn such that∣∣P (j)
n (ϑ)∣∣ le eminuscdnbntn
for all j in the range 0 le j lt tn and all n ge 1 then ϑ isalgebraicDue to M Laurent and D Roy applications to algebraicindependence
55 59
Criterion with several points
Goal Given a sequence of complex numbers (ϑi)ige1 if thereexists a sequence (Pn)nge1 of nonndashzero polynomials in Z[X]with Pn of degree le dn and height le ebn such that∣∣P (j)
n (ϑi)∣∣ le eminuscdnbntnsn
for 0 le j lt tn 1 le i le sn and all n ge 1 then the numbers ϑiare algebraic
D Roy Not true in general but true in some special caseswith a structure on the sequence (ϑi)ige1Combines the elimination arguments used for criteria ofalgebraic independence and for zero estimates
56 59
Small value estimates for the additive group
D Roy Small value estimates for the additive group Intern JNumber Theory to appear
Let ξ be a transcendental complex number let β σ τ and νbe non-negative real numbers let n0 be a positive integer andlet (Pn)ngen0 be a sequence of nonndashzero polynomials in Z[T ]satisfying deg(Pn) le n and H(Pn) le exp(nβ) for eachn ge n0 Suppose that β gt 1 (34)σ + τ lt 1 andν gt 1 + β minus (34)σ minus τ Then for infinitely many n we have
max|P [j]n (iξ)| 0 le i le nσ 0 le j le nτ
gt exp(minusnν)
57 59
Small value estimates for the multiplicative group
D Roy Small value estimates for the multiplicative groupActa Arith to appear
Let ξ1 ξm be multiplicatively independent complexnumbers in a field of transcendence degree 1 Under suitableassumptions on the parameters β σ τ ν for infinitely manypositive integers n there exists no nonndashzero polynomialP isin Z[T ] satisfying deg(P ) le n H(P ) le exp(nβ) and
max|P [j](ξi11 middot middot middot ξimm )| 0 le i1 im le nσ 0 le j le nτ
gt exp(minusnν)
58 59
Heidelberg Colloquium organized by the Institute for Mathematics
httpwwwmathiuni-heidelbergdevortraghtmls=1 January 22 2009
Transcendental Number TheorySchanuelrsquos Conjecture
Michel Waldschmidt
Institut de Mathematiques de Jussieu amp Paris VIhttpwwwmathjussieufrsimmiw
59 59
Half of the Conjecture is solved
From a certain point of view half of the conjecture of algebraicindependence of logarithms of algebraic numbers is solved
Theorem [D Roy] The rank of any matrix(bij + λij
)1leiled1lejle`
with bij isin Q and λij isin L is at least half its structural rank
42 59
Reformulation by D Roy
Instead of taking logarithmsof algebraic numbers andlooking for the algebraicindependence relationsD Roy fixes a polynomial andlooks at the points withcoordinates logarithms ofalgebraic numbers on thecorresponding hypersurface
43 59
Reformulation by D Roy
Royrsquos reformulation of the conjecture of algebraicindependence of logarithms is
Conjecture For any algebraic subvariety V of Cn definedover the field Q of algebraic numbers the set V cap Ln is theunion of the sets E cap Ln where E ranges over the set ofvector subspaces of Cn which are defined over Q andcontained in V
Trivial Any element in E cap Ln where E is a vector subspaceof Cn defined over Q and contained in V belongs to V cap Ln
44 59
Example The Four Exponentials ConjectureTake for V the hypersurface of C4 defined by the equation
z1z4 = z2z3
The maximal Cndashvector subspaces of C4 defined over Q andcontained in V are the planes
az1 = bz2 bz4 = az3
and the planesaz1 = bz3 bz4 = az2
with (a b) isin Q2 (0 0)Hence Schanuelrsquos Conjecture implies that if λ1 λ2 λ3 λ4 arelogarithms of algebraic numbers satisfying
λ1λ4 = λ2λ3
then either λ1 λ2 are linearly dependent over Q or else λ1 λ3
are linearly dependent over Q45 59
Six Exponentials Theorem
and Four Exponentials Conjecture
A Selberg CL Siegel S Lang K Ramachandra
46 59
The Four Exponentials Conjecture
Conjecture (Th Schneider S Lang K Ramachandra) Ifx1 x2 are Qndashlinearly independent complex numbers and y1 y2
are Qndashlinearly independent complex numbers then one atleast of the four numbers
ex1y1 ex1y2 ex2y1 ex2y2
is transcendentalEquivalent statement Any 2times 2 matrix(
λ1 λ2
λ3 λ4
)with entries in L and with Qndashlinearly independent rows andQndashlinearly independent columns has maximal rank 2
47 59
How could we attack Schanuelrsquos Conjecture
Let x1 xn be Qndashlinearly independent complex numbersFollowing the transcendence methods of Hermite GelrsquofondSchneider one may start by introducing an auxiliaryfunction
F (z) = P (z ez)
where P isin Z[X0 X1] is a nonndashzero polynomial One considersthe derivatives (
d
dz
)kF = (DkP )(z ez)
at the pointsm1x1 + middot middot middot+mnxn
for various values of (m1 mn) isin Zn
48 59
Auxiliary functionLet D denote the derivation
D =part
partX0
+X1part
partX1
over the ring C[X0 X1] Recall that x1 xn are Qndashlinearlyindependent complex numbers Let α1 αn be nonndashzerocomplex numbersThe transcendence machinery produces a sequence (PN)Nge0
of polynomials with integer coefficients satisfying∣∣∣∣∣(DkPN)(nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1
49 59
Royrsquos approach to Schanuelrsquos Conjecture (1999)
If the number of equations we produce is too small such a setof relations does not contain any information The existenceof a sequence of nonndashtrivial polynomials (PN)Nge0 follows fromlinear algebra
On the other hand following D Roy one may expect that theexistence of such a sequence (PN)Nge0 producing sufficientlymany such equations will yield the desired conclusion
tr degQQ(x1 xn α1 αn
)ge n
50 59
Royrsquos Conjecture (1999)Let s0 s1 t0 t1 u positive real numbers satisfying
max1 t0 2t1 lt mins0 2s1and
maxs0 s1 + t1 lt u lt1
2(1 + t0 + t1)
Assume that for any sufficiently large positive integer N there exists a nonndashzero polynomial PN isin Z[X0 X1] withpartial degree le N t0 in X0 partial degree le N t1 in X1 andheight le eN which satisfies∣∣∣∣∣(DkPN)(
nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1 Then
tr degQQ(x1 xn α1 αn) ge n51 59
Royrsquos Theorem (1999)
Royrsquos conjecture is equivalent to Schanuelrsquos Conjecture
52 59
Equivalence between Schanuel and Roy
Let (x α) isin CtimesCtimes and let s0 s1 t0 t1 u be positive realnumbers satisfying the inequalities of Royrsquos Conjecture Thenthe following conditions are equivalent (a) The number αeminusx is a root of unity
(b) For any sufficiently large positive integer N there exists anonndashzero polynomial QN isin Z[X0 X1] with partial degreele N t0 in X0 partial degree le N t1 in X1 and heightH(QN) le eN such that∣∣(DkQN)(mx αm)
∣∣ le exp(minusNu)
for any km isin N with k le N s0 and m le N s1
53 59
Gelrsquofondrsquos transcendence criterion
Simple form Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le n and height le en such that
|Pn(ϑ)| le eminus7n2
for all n ge 1 then ϑ is algebraic and Pn(ϑ) = 0 for all n ge 1
First extension Replace the bound for the degree by le dnthe bound for the height by ebn and the bound for |Pn(ϑ)| byeminuscdnbn with some constant c gt 0 independent of n
Some mild conditions are required on the sequences (dn)nge1 and(bn)nge1
54 59
Transcendence criterion with multiplicitiesWith derivatives Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le dn and height le ebn such that∣∣P (j)
n (ϑ)∣∣ le eminuscdnbntn
for all j in the range 0 le j lt tn and all n ge 1 then ϑ isalgebraicDue to M Laurent and D Roy applications to algebraicindependence
55 59
Criterion with several points
Goal Given a sequence of complex numbers (ϑi)ige1 if thereexists a sequence (Pn)nge1 of nonndashzero polynomials in Z[X]with Pn of degree le dn and height le ebn such that∣∣P (j)
n (ϑi)∣∣ le eminuscdnbntnsn
for 0 le j lt tn 1 le i le sn and all n ge 1 then the numbers ϑiare algebraic
D Roy Not true in general but true in some special caseswith a structure on the sequence (ϑi)ige1Combines the elimination arguments used for criteria ofalgebraic independence and for zero estimates
56 59
Small value estimates for the additive group
D Roy Small value estimates for the additive group Intern JNumber Theory to appear
Let ξ be a transcendental complex number let β σ τ and νbe non-negative real numbers let n0 be a positive integer andlet (Pn)ngen0 be a sequence of nonndashzero polynomials in Z[T ]satisfying deg(Pn) le n and H(Pn) le exp(nβ) for eachn ge n0 Suppose that β gt 1 (34)σ + τ lt 1 andν gt 1 + β minus (34)σ minus τ Then for infinitely many n we have
max|P [j]n (iξ)| 0 le i le nσ 0 le j le nτ
gt exp(minusnν)
57 59
Small value estimates for the multiplicative group
D Roy Small value estimates for the multiplicative groupActa Arith to appear
Let ξ1 ξm be multiplicatively independent complexnumbers in a field of transcendence degree 1 Under suitableassumptions on the parameters β σ τ ν for infinitely manypositive integers n there exists no nonndashzero polynomialP isin Z[T ] satisfying deg(P ) le n H(P ) le exp(nβ) and
max|P [j](ξi11 middot middot middot ξimm )| 0 le i1 im le nσ 0 le j le nτ
gt exp(minusnν)
58 59
Heidelberg Colloquium organized by the Institute for Mathematics
httpwwwmathiuni-heidelbergdevortraghtmls=1 January 22 2009
Transcendental Number TheorySchanuelrsquos Conjecture
Michel Waldschmidt
Institut de Mathematiques de Jussieu amp Paris VIhttpwwwmathjussieufrsimmiw
59 59
Reformulation by D Roy
Instead of taking logarithmsof algebraic numbers andlooking for the algebraicindependence relationsD Roy fixes a polynomial andlooks at the points withcoordinates logarithms ofalgebraic numbers on thecorresponding hypersurface
43 59
Reformulation by D Roy
Royrsquos reformulation of the conjecture of algebraicindependence of logarithms is
Conjecture For any algebraic subvariety V of Cn definedover the field Q of algebraic numbers the set V cap Ln is theunion of the sets E cap Ln where E ranges over the set ofvector subspaces of Cn which are defined over Q andcontained in V
Trivial Any element in E cap Ln where E is a vector subspaceof Cn defined over Q and contained in V belongs to V cap Ln
44 59
Example The Four Exponentials ConjectureTake for V the hypersurface of C4 defined by the equation
z1z4 = z2z3
The maximal Cndashvector subspaces of C4 defined over Q andcontained in V are the planes
az1 = bz2 bz4 = az3
and the planesaz1 = bz3 bz4 = az2
with (a b) isin Q2 (0 0)Hence Schanuelrsquos Conjecture implies that if λ1 λ2 λ3 λ4 arelogarithms of algebraic numbers satisfying
λ1λ4 = λ2λ3
then either λ1 λ2 are linearly dependent over Q or else λ1 λ3
are linearly dependent over Q45 59
Six Exponentials Theorem
and Four Exponentials Conjecture
A Selberg CL Siegel S Lang K Ramachandra
46 59
The Four Exponentials Conjecture
Conjecture (Th Schneider S Lang K Ramachandra) Ifx1 x2 are Qndashlinearly independent complex numbers and y1 y2
are Qndashlinearly independent complex numbers then one atleast of the four numbers
ex1y1 ex1y2 ex2y1 ex2y2
is transcendentalEquivalent statement Any 2times 2 matrix(
λ1 λ2
λ3 λ4
)with entries in L and with Qndashlinearly independent rows andQndashlinearly independent columns has maximal rank 2
47 59
How could we attack Schanuelrsquos Conjecture
Let x1 xn be Qndashlinearly independent complex numbersFollowing the transcendence methods of Hermite GelrsquofondSchneider one may start by introducing an auxiliaryfunction
F (z) = P (z ez)
where P isin Z[X0 X1] is a nonndashzero polynomial One considersthe derivatives (
d
dz
)kF = (DkP )(z ez)
at the pointsm1x1 + middot middot middot+mnxn
for various values of (m1 mn) isin Zn
48 59
Auxiliary functionLet D denote the derivation
D =part
partX0
+X1part
partX1
over the ring C[X0 X1] Recall that x1 xn are Qndashlinearlyindependent complex numbers Let α1 αn be nonndashzerocomplex numbersThe transcendence machinery produces a sequence (PN)Nge0
of polynomials with integer coefficients satisfying∣∣∣∣∣(DkPN)(nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1
49 59
Royrsquos approach to Schanuelrsquos Conjecture (1999)
If the number of equations we produce is too small such a setof relations does not contain any information The existenceof a sequence of nonndashtrivial polynomials (PN)Nge0 follows fromlinear algebra
On the other hand following D Roy one may expect that theexistence of such a sequence (PN)Nge0 producing sufficientlymany such equations will yield the desired conclusion
tr degQQ(x1 xn α1 αn
)ge n
50 59
Royrsquos Conjecture (1999)Let s0 s1 t0 t1 u positive real numbers satisfying
max1 t0 2t1 lt mins0 2s1and
maxs0 s1 + t1 lt u lt1
2(1 + t0 + t1)
Assume that for any sufficiently large positive integer N there exists a nonndashzero polynomial PN isin Z[X0 X1] withpartial degree le N t0 in X0 partial degree le N t1 in X1 andheight le eN which satisfies∣∣∣∣∣(DkPN)(
nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1 Then
tr degQQ(x1 xn α1 αn) ge n51 59
Royrsquos Theorem (1999)
Royrsquos conjecture is equivalent to Schanuelrsquos Conjecture
52 59
Equivalence between Schanuel and Roy
Let (x α) isin CtimesCtimes and let s0 s1 t0 t1 u be positive realnumbers satisfying the inequalities of Royrsquos Conjecture Thenthe following conditions are equivalent (a) The number αeminusx is a root of unity
(b) For any sufficiently large positive integer N there exists anonndashzero polynomial QN isin Z[X0 X1] with partial degreele N t0 in X0 partial degree le N t1 in X1 and heightH(QN) le eN such that∣∣(DkQN)(mx αm)
∣∣ le exp(minusNu)
for any km isin N with k le N s0 and m le N s1
53 59
Gelrsquofondrsquos transcendence criterion
Simple form Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le n and height le en such that
|Pn(ϑ)| le eminus7n2
for all n ge 1 then ϑ is algebraic and Pn(ϑ) = 0 for all n ge 1
First extension Replace the bound for the degree by le dnthe bound for the height by ebn and the bound for |Pn(ϑ)| byeminuscdnbn with some constant c gt 0 independent of n
Some mild conditions are required on the sequences (dn)nge1 and(bn)nge1
54 59
Transcendence criterion with multiplicitiesWith derivatives Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le dn and height le ebn such that∣∣P (j)
n (ϑ)∣∣ le eminuscdnbntn
for all j in the range 0 le j lt tn and all n ge 1 then ϑ isalgebraicDue to M Laurent and D Roy applications to algebraicindependence
55 59
Criterion with several points
Goal Given a sequence of complex numbers (ϑi)ige1 if thereexists a sequence (Pn)nge1 of nonndashzero polynomials in Z[X]with Pn of degree le dn and height le ebn such that∣∣P (j)
n (ϑi)∣∣ le eminuscdnbntnsn
for 0 le j lt tn 1 le i le sn and all n ge 1 then the numbers ϑiare algebraic
D Roy Not true in general but true in some special caseswith a structure on the sequence (ϑi)ige1Combines the elimination arguments used for criteria ofalgebraic independence and for zero estimates
56 59
Small value estimates for the additive group
D Roy Small value estimates for the additive group Intern JNumber Theory to appear
Let ξ be a transcendental complex number let β σ τ and νbe non-negative real numbers let n0 be a positive integer andlet (Pn)ngen0 be a sequence of nonndashzero polynomials in Z[T ]satisfying deg(Pn) le n and H(Pn) le exp(nβ) for eachn ge n0 Suppose that β gt 1 (34)σ + τ lt 1 andν gt 1 + β minus (34)σ minus τ Then for infinitely many n we have
max|P [j]n (iξ)| 0 le i le nσ 0 le j le nτ
gt exp(minusnν)
57 59
Small value estimates for the multiplicative group
D Roy Small value estimates for the multiplicative groupActa Arith to appear
Let ξ1 ξm be multiplicatively independent complexnumbers in a field of transcendence degree 1 Under suitableassumptions on the parameters β σ τ ν for infinitely manypositive integers n there exists no nonndashzero polynomialP isin Z[T ] satisfying deg(P ) le n H(P ) le exp(nβ) and
max|P [j](ξi11 middot middot middot ξimm )| 0 le i1 im le nσ 0 le j le nτ
gt exp(minusnν)
58 59
Heidelberg Colloquium organized by the Institute for Mathematics
httpwwwmathiuni-heidelbergdevortraghtmls=1 January 22 2009
Transcendental Number TheorySchanuelrsquos Conjecture
Michel Waldschmidt
Institut de Mathematiques de Jussieu amp Paris VIhttpwwwmathjussieufrsimmiw
59 59
Reformulation by D Roy
Royrsquos reformulation of the conjecture of algebraicindependence of logarithms is
Conjecture For any algebraic subvariety V of Cn definedover the field Q of algebraic numbers the set V cap Ln is theunion of the sets E cap Ln where E ranges over the set ofvector subspaces of Cn which are defined over Q andcontained in V
Trivial Any element in E cap Ln where E is a vector subspaceof Cn defined over Q and contained in V belongs to V cap Ln
44 59
Example The Four Exponentials ConjectureTake for V the hypersurface of C4 defined by the equation
z1z4 = z2z3
The maximal Cndashvector subspaces of C4 defined over Q andcontained in V are the planes
az1 = bz2 bz4 = az3
and the planesaz1 = bz3 bz4 = az2
with (a b) isin Q2 (0 0)Hence Schanuelrsquos Conjecture implies that if λ1 λ2 λ3 λ4 arelogarithms of algebraic numbers satisfying
λ1λ4 = λ2λ3
then either λ1 λ2 are linearly dependent over Q or else λ1 λ3
are linearly dependent over Q45 59
Six Exponentials Theorem
and Four Exponentials Conjecture
A Selberg CL Siegel S Lang K Ramachandra
46 59
The Four Exponentials Conjecture
Conjecture (Th Schneider S Lang K Ramachandra) Ifx1 x2 are Qndashlinearly independent complex numbers and y1 y2
are Qndashlinearly independent complex numbers then one atleast of the four numbers
ex1y1 ex1y2 ex2y1 ex2y2
is transcendentalEquivalent statement Any 2times 2 matrix(
λ1 λ2
λ3 λ4
)with entries in L and with Qndashlinearly independent rows andQndashlinearly independent columns has maximal rank 2
47 59
How could we attack Schanuelrsquos Conjecture
Let x1 xn be Qndashlinearly independent complex numbersFollowing the transcendence methods of Hermite GelrsquofondSchneider one may start by introducing an auxiliaryfunction
F (z) = P (z ez)
where P isin Z[X0 X1] is a nonndashzero polynomial One considersthe derivatives (
d
dz
)kF = (DkP )(z ez)
at the pointsm1x1 + middot middot middot+mnxn
for various values of (m1 mn) isin Zn
48 59
Auxiliary functionLet D denote the derivation
D =part
partX0
+X1part
partX1
over the ring C[X0 X1] Recall that x1 xn are Qndashlinearlyindependent complex numbers Let α1 αn be nonndashzerocomplex numbersThe transcendence machinery produces a sequence (PN)Nge0
of polynomials with integer coefficients satisfying∣∣∣∣∣(DkPN)(nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1
49 59
Royrsquos approach to Schanuelrsquos Conjecture (1999)
If the number of equations we produce is too small such a setof relations does not contain any information The existenceof a sequence of nonndashtrivial polynomials (PN)Nge0 follows fromlinear algebra
On the other hand following D Roy one may expect that theexistence of such a sequence (PN)Nge0 producing sufficientlymany such equations will yield the desired conclusion
tr degQQ(x1 xn α1 αn
)ge n
50 59
Royrsquos Conjecture (1999)Let s0 s1 t0 t1 u positive real numbers satisfying
max1 t0 2t1 lt mins0 2s1and
maxs0 s1 + t1 lt u lt1
2(1 + t0 + t1)
Assume that for any sufficiently large positive integer N there exists a nonndashzero polynomial PN isin Z[X0 X1] withpartial degree le N t0 in X0 partial degree le N t1 in X1 andheight le eN which satisfies∣∣∣∣∣(DkPN)(
nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1 Then
tr degQQ(x1 xn α1 αn) ge n51 59
Royrsquos Theorem (1999)
Royrsquos conjecture is equivalent to Schanuelrsquos Conjecture
52 59
Equivalence between Schanuel and Roy
Let (x α) isin CtimesCtimes and let s0 s1 t0 t1 u be positive realnumbers satisfying the inequalities of Royrsquos Conjecture Thenthe following conditions are equivalent (a) The number αeminusx is a root of unity
(b) For any sufficiently large positive integer N there exists anonndashzero polynomial QN isin Z[X0 X1] with partial degreele N t0 in X0 partial degree le N t1 in X1 and heightH(QN) le eN such that∣∣(DkQN)(mx αm)
∣∣ le exp(minusNu)
for any km isin N with k le N s0 and m le N s1
53 59
Gelrsquofondrsquos transcendence criterion
Simple form Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le n and height le en such that
|Pn(ϑ)| le eminus7n2
for all n ge 1 then ϑ is algebraic and Pn(ϑ) = 0 for all n ge 1
First extension Replace the bound for the degree by le dnthe bound for the height by ebn and the bound for |Pn(ϑ)| byeminuscdnbn with some constant c gt 0 independent of n
Some mild conditions are required on the sequences (dn)nge1 and(bn)nge1
54 59
Transcendence criterion with multiplicitiesWith derivatives Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le dn and height le ebn such that∣∣P (j)
n (ϑ)∣∣ le eminuscdnbntn
for all j in the range 0 le j lt tn and all n ge 1 then ϑ isalgebraicDue to M Laurent and D Roy applications to algebraicindependence
55 59
Criterion with several points
Goal Given a sequence of complex numbers (ϑi)ige1 if thereexists a sequence (Pn)nge1 of nonndashzero polynomials in Z[X]with Pn of degree le dn and height le ebn such that∣∣P (j)
n (ϑi)∣∣ le eminuscdnbntnsn
for 0 le j lt tn 1 le i le sn and all n ge 1 then the numbers ϑiare algebraic
D Roy Not true in general but true in some special caseswith a structure on the sequence (ϑi)ige1Combines the elimination arguments used for criteria ofalgebraic independence and for zero estimates
56 59
Small value estimates for the additive group
D Roy Small value estimates for the additive group Intern JNumber Theory to appear
Let ξ be a transcendental complex number let β σ τ and νbe non-negative real numbers let n0 be a positive integer andlet (Pn)ngen0 be a sequence of nonndashzero polynomials in Z[T ]satisfying deg(Pn) le n and H(Pn) le exp(nβ) for eachn ge n0 Suppose that β gt 1 (34)σ + τ lt 1 andν gt 1 + β minus (34)σ minus τ Then for infinitely many n we have
max|P [j]n (iξ)| 0 le i le nσ 0 le j le nτ
gt exp(minusnν)
57 59
Small value estimates for the multiplicative group
D Roy Small value estimates for the multiplicative groupActa Arith to appear
Let ξ1 ξm be multiplicatively independent complexnumbers in a field of transcendence degree 1 Under suitableassumptions on the parameters β σ τ ν for infinitely manypositive integers n there exists no nonndashzero polynomialP isin Z[T ] satisfying deg(P ) le n H(P ) le exp(nβ) and
max|P [j](ξi11 middot middot middot ξimm )| 0 le i1 im le nσ 0 le j le nτ
gt exp(minusnν)
58 59
Heidelberg Colloquium organized by the Institute for Mathematics
httpwwwmathiuni-heidelbergdevortraghtmls=1 January 22 2009
Transcendental Number TheorySchanuelrsquos Conjecture
Michel Waldschmidt
Institut de Mathematiques de Jussieu amp Paris VIhttpwwwmathjussieufrsimmiw
59 59
Example The Four Exponentials ConjectureTake for V the hypersurface of C4 defined by the equation
z1z4 = z2z3
The maximal Cndashvector subspaces of C4 defined over Q andcontained in V are the planes
az1 = bz2 bz4 = az3
and the planesaz1 = bz3 bz4 = az2
with (a b) isin Q2 (0 0)Hence Schanuelrsquos Conjecture implies that if λ1 λ2 λ3 λ4 arelogarithms of algebraic numbers satisfying
λ1λ4 = λ2λ3
then either λ1 λ2 are linearly dependent over Q or else λ1 λ3
are linearly dependent over Q45 59
Six Exponentials Theorem
and Four Exponentials Conjecture
A Selberg CL Siegel S Lang K Ramachandra
46 59
The Four Exponentials Conjecture
Conjecture (Th Schneider S Lang K Ramachandra) Ifx1 x2 are Qndashlinearly independent complex numbers and y1 y2
are Qndashlinearly independent complex numbers then one atleast of the four numbers
ex1y1 ex1y2 ex2y1 ex2y2
is transcendentalEquivalent statement Any 2times 2 matrix(
λ1 λ2
λ3 λ4
)with entries in L and with Qndashlinearly independent rows andQndashlinearly independent columns has maximal rank 2
47 59
How could we attack Schanuelrsquos Conjecture
Let x1 xn be Qndashlinearly independent complex numbersFollowing the transcendence methods of Hermite GelrsquofondSchneider one may start by introducing an auxiliaryfunction
F (z) = P (z ez)
where P isin Z[X0 X1] is a nonndashzero polynomial One considersthe derivatives (
d
dz
)kF = (DkP )(z ez)
at the pointsm1x1 + middot middot middot+mnxn
for various values of (m1 mn) isin Zn
48 59
Auxiliary functionLet D denote the derivation
D =part
partX0
+X1part
partX1
over the ring C[X0 X1] Recall that x1 xn are Qndashlinearlyindependent complex numbers Let α1 αn be nonndashzerocomplex numbersThe transcendence machinery produces a sequence (PN)Nge0
of polynomials with integer coefficients satisfying∣∣∣∣∣(DkPN)(nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1
49 59
Royrsquos approach to Schanuelrsquos Conjecture (1999)
If the number of equations we produce is too small such a setof relations does not contain any information The existenceof a sequence of nonndashtrivial polynomials (PN)Nge0 follows fromlinear algebra
On the other hand following D Roy one may expect that theexistence of such a sequence (PN)Nge0 producing sufficientlymany such equations will yield the desired conclusion
tr degQQ(x1 xn α1 αn
)ge n
50 59
Royrsquos Conjecture (1999)Let s0 s1 t0 t1 u positive real numbers satisfying
max1 t0 2t1 lt mins0 2s1and
maxs0 s1 + t1 lt u lt1
2(1 + t0 + t1)
Assume that for any sufficiently large positive integer N there exists a nonndashzero polynomial PN isin Z[X0 X1] withpartial degree le N t0 in X0 partial degree le N t1 in X1 andheight le eN which satisfies∣∣∣∣∣(DkPN)(
nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1 Then
tr degQQ(x1 xn α1 αn) ge n51 59
Royrsquos Theorem (1999)
Royrsquos conjecture is equivalent to Schanuelrsquos Conjecture
52 59
Equivalence between Schanuel and Roy
Let (x α) isin CtimesCtimes and let s0 s1 t0 t1 u be positive realnumbers satisfying the inequalities of Royrsquos Conjecture Thenthe following conditions are equivalent (a) The number αeminusx is a root of unity
(b) For any sufficiently large positive integer N there exists anonndashzero polynomial QN isin Z[X0 X1] with partial degreele N t0 in X0 partial degree le N t1 in X1 and heightH(QN) le eN such that∣∣(DkQN)(mx αm)
∣∣ le exp(minusNu)
for any km isin N with k le N s0 and m le N s1
53 59
Gelrsquofondrsquos transcendence criterion
Simple form Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le n and height le en such that
|Pn(ϑ)| le eminus7n2
for all n ge 1 then ϑ is algebraic and Pn(ϑ) = 0 for all n ge 1
First extension Replace the bound for the degree by le dnthe bound for the height by ebn and the bound for |Pn(ϑ)| byeminuscdnbn with some constant c gt 0 independent of n
Some mild conditions are required on the sequences (dn)nge1 and(bn)nge1
54 59
Transcendence criterion with multiplicitiesWith derivatives Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le dn and height le ebn such that∣∣P (j)
n (ϑ)∣∣ le eminuscdnbntn
for all j in the range 0 le j lt tn and all n ge 1 then ϑ isalgebraicDue to M Laurent and D Roy applications to algebraicindependence
55 59
Criterion with several points
Goal Given a sequence of complex numbers (ϑi)ige1 if thereexists a sequence (Pn)nge1 of nonndashzero polynomials in Z[X]with Pn of degree le dn and height le ebn such that∣∣P (j)
n (ϑi)∣∣ le eminuscdnbntnsn
for 0 le j lt tn 1 le i le sn and all n ge 1 then the numbers ϑiare algebraic
D Roy Not true in general but true in some special caseswith a structure on the sequence (ϑi)ige1Combines the elimination arguments used for criteria ofalgebraic independence and for zero estimates
56 59
Small value estimates for the additive group
D Roy Small value estimates for the additive group Intern JNumber Theory to appear
Let ξ be a transcendental complex number let β σ τ and νbe non-negative real numbers let n0 be a positive integer andlet (Pn)ngen0 be a sequence of nonndashzero polynomials in Z[T ]satisfying deg(Pn) le n and H(Pn) le exp(nβ) for eachn ge n0 Suppose that β gt 1 (34)σ + τ lt 1 andν gt 1 + β minus (34)σ minus τ Then for infinitely many n we have
max|P [j]n (iξ)| 0 le i le nσ 0 le j le nτ
gt exp(minusnν)
57 59
Small value estimates for the multiplicative group
D Roy Small value estimates for the multiplicative groupActa Arith to appear
Let ξ1 ξm be multiplicatively independent complexnumbers in a field of transcendence degree 1 Under suitableassumptions on the parameters β σ τ ν for infinitely manypositive integers n there exists no nonndashzero polynomialP isin Z[T ] satisfying deg(P ) le n H(P ) le exp(nβ) and
max|P [j](ξi11 middot middot middot ξimm )| 0 le i1 im le nσ 0 le j le nτ
gt exp(minusnν)
58 59
Heidelberg Colloquium organized by the Institute for Mathematics
httpwwwmathiuni-heidelbergdevortraghtmls=1 January 22 2009
Transcendental Number TheorySchanuelrsquos Conjecture
Michel Waldschmidt
Institut de Mathematiques de Jussieu amp Paris VIhttpwwwmathjussieufrsimmiw
59 59
Six Exponentials Theorem
and Four Exponentials Conjecture
A Selberg CL Siegel S Lang K Ramachandra
46 59
The Four Exponentials Conjecture
Conjecture (Th Schneider S Lang K Ramachandra) Ifx1 x2 are Qndashlinearly independent complex numbers and y1 y2
are Qndashlinearly independent complex numbers then one atleast of the four numbers
ex1y1 ex1y2 ex2y1 ex2y2
is transcendentalEquivalent statement Any 2times 2 matrix(
λ1 λ2
λ3 λ4
)with entries in L and with Qndashlinearly independent rows andQndashlinearly independent columns has maximal rank 2
47 59
How could we attack Schanuelrsquos Conjecture
Let x1 xn be Qndashlinearly independent complex numbersFollowing the transcendence methods of Hermite GelrsquofondSchneider one may start by introducing an auxiliaryfunction
F (z) = P (z ez)
where P isin Z[X0 X1] is a nonndashzero polynomial One considersthe derivatives (
d
dz
)kF = (DkP )(z ez)
at the pointsm1x1 + middot middot middot+mnxn
for various values of (m1 mn) isin Zn
48 59
Auxiliary functionLet D denote the derivation
D =part
partX0
+X1part
partX1
over the ring C[X0 X1] Recall that x1 xn are Qndashlinearlyindependent complex numbers Let α1 αn be nonndashzerocomplex numbersThe transcendence machinery produces a sequence (PN)Nge0
of polynomials with integer coefficients satisfying∣∣∣∣∣(DkPN)(nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1
49 59
Royrsquos approach to Schanuelrsquos Conjecture (1999)
If the number of equations we produce is too small such a setof relations does not contain any information The existenceof a sequence of nonndashtrivial polynomials (PN)Nge0 follows fromlinear algebra
On the other hand following D Roy one may expect that theexistence of such a sequence (PN)Nge0 producing sufficientlymany such equations will yield the desired conclusion
tr degQQ(x1 xn α1 αn
)ge n
50 59
Royrsquos Conjecture (1999)Let s0 s1 t0 t1 u positive real numbers satisfying
max1 t0 2t1 lt mins0 2s1and
maxs0 s1 + t1 lt u lt1
2(1 + t0 + t1)
Assume that for any sufficiently large positive integer N there exists a nonndashzero polynomial PN isin Z[X0 X1] withpartial degree le N t0 in X0 partial degree le N t1 in X1 andheight le eN which satisfies∣∣∣∣∣(DkPN)(
nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1 Then
tr degQQ(x1 xn α1 αn) ge n51 59
Royrsquos Theorem (1999)
Royrsquos conjecture is equivalent to Schanuelrsquos Conjecture
52 59
Equivalence between Schanuel and Roy
Let (x α) isin CtimesCtimes and let s0 s1 t0 t1 u be positive realnumbers satisfying the inequalities of Royrsquos Conjecture Thenthe following conditions are equivalent (a) The number αeminusx is a root of unity
(b) For any sufficiently large positive integer N there exists anonndashzero polynomial QN isin Z[X0 X1] with partial degreele N t0 in X0 partial degree le N t1 in X1 and heightH(QN) le eN such that∣∣(DkQN)(mx αm)
∣∣ le exp(minusNu)
for any km isin N with k le N s0 and m le N s1
53 59
Gelrsquofondrsquos transcendence criterion
Simple form Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le n and height le en such that
|Pn(ϑ)| le eminus7n2
for all n ge 1 then ϑ is algebraic and Pn(ϑ) = 0 for all n ge 1
First extension Replace the bound for the degree by le dnthe bound for the height by ebn and the bound for |Pn(ϑ)| byeminuscdnbn with some constant c gt 0 independent of n
Some mild conditions are required on the sequences (dn)nge1 and(bn)nge1
54 59
Transcendence criterion with multiplicitiesWith derivatives Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le dn and height le ebn such that∣∣P (j)
n (ϑ)∣∣ le eminuscdnbntn
for all j in the range 0 le j lt tn and all n ge 1 then ϑ isalgebraicDue to M Laurent and D Roy applications to algebraicindependence
55 59
Criterion with several points
Goal Given a sequence of complex numbers (ϑi)ige1 if thereexists a sequence (Pn)nge1 of nonndashzero polynomials in Z[X]with Pn of degree le dn and height le ebn such that∣∣P (j)
n (ϑi)∣∣ le eminuscdnbntnsn
for 0 le j lt tn 1 le i le sn and all n ge 1 then the numbers ϑiare algebraic
D Roy Not true in general but true in some special caseswith a structure on the sequence (ϑi)ige1Combines the elimination arguments used for criteria ofalgebraic independence and for zero estimates
56 59
Small value estimates for the additive group
D Roy Small value estimates for the additive group Intern JNumber Theory to appear
Let ξ be a transcendental complex number let β σ τ and νbe non-negative real numbers let n0 be a positive integer andlet (Pn)ngen0 be a sequence of nonndashzero polynomials in Z[T ]satisfying deg(Pn) le n and H(Pn) le exp(nβ) for eachn ge n0 Suppose that β gt 1 (34)σ + τ lt 1 andν gt 1 + β minus (34)σ minus τ Then for infinitely many n we have
max|P [j]n (iξ)| 0 le i le nσ 0 le j le nτ
gt exp(minusnν)
57 59
Small value estimates for the multiplicative group
D Roy Small value estimates for the multiplicative groupActa Arith to appear
Let ξ1 ξm be multiplicatively independent complexnumbers in a field of transcendence degree 1 Under suitableassumptions on the parameters β σ τ ν for infinitely manypositive integers n there exists no nonndashzero polynomialP isin Z[T ] satisfying deg(P ) le n H(P ) le exp(nβ) and
max|P [j](ξi11 middot middot middot ξimm )| 0 le i1 im le nσ 0 le j le nτ
gt exp(minusnν)
58 59
Heidelberg Colloquium organized by the Institute for Mathematics
httpwwwmathiuni-heidelbergdevortraghtmls=1 January 22 2009
Transcendental Number TheorySchanuelrsquos Conjecture
Michel Waldschmidt
Institut de Mathematiques de Jussieu amp Paris VIhttpwwwmathjussieufrsimmiw
59 59
The Four Exponentials Conjecture
Conjecture (Th Schneider S Lang K Ramachandra) Ifx1 x2 are Qndashlinearly independent complex numbers and y1 y2
are Qndashlinearly independent complex numbers then one atleast of the four numbers
ex1y1 ex1y2 ex2y1 ex2y2
is transcendentalEquivalent statement Any 2times 2 matrix(
λ1 λ2
λ3 λ4
)with entries in L and with Qndashlinearly independent rows andQndashlinearly independent columns has maximal rank 2
47 59
How could we attack Schanuelrsquos Conjecture
Let x1 xn be Qndashlinearly independent complex numbersFollowing the transcendence methods of Hermite GelrsquofondSchneider one may start by introducing an auxiliaryfunction
F (z) = P (z ez)
where P isin Z[X0 X1] is a nonndashzero polynomial One considersthe derivatives (
d
dz
)kF = (DkP )(z ez)
at the pointsm1x1 + middot middot middot+mnxn
for various values of (m1 mn) isin Zn
48 59
Auxiliary functionLet D denote the derivation
D =part
partX0
+X1part
partX1
over the ring C[X0 X1] Recall that x1 xn are Qndashlinearlyindependent complex numbers Let α1 αn be nonndashzerocomplex numbersThe transcendence machinery produces a sequence (PN)Nge0
of polynomials with integer coefficients satisfying∣∣∣∣∣(DkPN)(nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1
49 59
Royrsquos approach to Schanuelrsquos Conjecture (1999)
If the number of equations we produce is too small such a setof relations does not contain any information The existenceof a sequence of nonndashtrivial polynomials (PN)Nge0 follows fromlinear algebra
On the other hand following D Roy one may expect that theexistence of such a sequence (PN)Nge0 producing sufficientlymany such equations will yield the desired conclusion
tr degQQ(x1 xn α1 αn
)ge n
50 59
Royrsquos Conjecture (1999)Let s0 s1 t0 t1 u positive real numbers satisfying
max1 t0 2t1 lt mins0 2s1and
maxs0 s1 + t1 lt u lt1
2(1 + t0 + t1)
Assume that for any sufficiently large positive integer N there exists a nonndashzero polynomial PN isin Z[X0 X1] withpartial degree le N t0 in X0 partial degree le N t1 in X1 andheight le eN which satisfies∣∣∣∣∣(DkPN)(
nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1 Then
tr degQQ(x1 xn α1 αn) ge n51 59
Royrsquos Theorem (1999)
Royrsquos conjecture is equivalent to Schanuelrsquos Conjecture
52 59
Equivalence between Schanuel and Roy
Let (x α) isin CtimesCtimes and let s0 s1 t0 t1 u be positive realnumbers satisfying the inequalities of Royrsquos Conjecture Thenthe following conditions are equivalent (a) The number αeminusx is a root of unity
(b) For any sufficiently large positive integer N there exists anonndashzero polynomial QN isin Z[X0 X1] with partial degreele N t0 in X0 partial degree le N t1 in X1 and heightH(QN) le eN such that∣∣(DkQN)(mx αm)
∣∣ le exp(minusNu)
for any km isin N with k le N s0 and m le N s1
53 59
Gelrsquofondrsquos transcendence criterion
Simple form Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le n and height le en such that
|Pn(ϑ)| le eminus7n2
for all n ge 1 then ϑ is algebraic and Pn(ϑ) = 0 for all n ge 1
First extension Replace the bound for the degree by le dnthe bound for the height by ebn and the bound for |Pn(ϑ)| byeminuscdnbn with some constant c gt 0 independent of n
Some mild conditions are required on the sequences (dn)nge1 and(bn)nge1
54 59
Transcendence criterion with multiplicitiesWith derivatives Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le dn and height le ebn such that∣∣P (j)
n (ϑ)∣∣ le eminuscdnbntn
for all j in the range 0 le j lt tn and all n ge 1 then ϑ isalgebraicDue to M Laurent and D Roy applications to algebraicindependence
55 59
Criterion with several points
Goal Given a sequence of complex numbers (ϑi)ige1 if thereexists a sequence (Pn)nge1 of nonndashzero polynomials in Z[X]with Pn of degree le dn and height le ebn such that∣∣P (j)
n (ϑi)∣∣ le eminuscdnbntnsn
for 0 le j lt tn 1 le i le sn and all n ge 1 then the numbers ϑiare algebraic
D Roy Not true in general but true in some special caseswith a structure on the sequence (ϑi)ige1Combines the elimination arguments used for criteria ofalgebraic independence and for zero estimates
56 59
Small value estimates for the additive group
D Roy Small value estimates for the additive group Intern JNumber Theory to appear
Let ξ be a transcendental complex number let β σ τ and νbe non-negative real numbers let n0 be a positive integer andlet (Pn)ngen0 be a sequence of nonndashzero polynomials in Z[T ]satisfying deg(Pn) le n and H(Pn) le exp(nβ) for eachn ge n0 Suppose that β gt 1 (34)σ + τ lt 1 andν gt 1 + β minus (34)σ minus τ Then for infinitely many n we have
max|P [j]n (iξ)| 0 le i le nσ 0 le j le nτ
gt exp(minusnν)
57 59
Small value estimates for the multiplicative group
D Roy Small value estimates for the multiplicative groupActa Arith to appear
Let ξ1 ξm be multiplicatively independent complexnumbers in a field of transcendence degree 1 Under suitableassumptions on the parameters β σ τ ν for infinitely manypositive integers n there exists no nonndashzero polynomialP isin Z[T ] satisfying deg(P ) le n H(P ) le exp(nβ) and
max|P [j](ξi11 middot middot middot ξimm )| 0 le i1 im le nσ 0 le j le nτ
gt exp(minusnν)
58 59
Heidelberg Colloquium organized by the Institute for Mathematics
httpwwwmathiuni-heidelbergdevortraghtmls=1 January 22 2009
Transcendental Number TheorySchanuelrsquos Conjecture
Michel Waldschmidt
Institut de Mathematiques de Jussieu amp Paris VIhttpwwwmathjussieufrsimmiw
59 59
How could we attack Schanuelrsquos Conjecture
Let x1 xn be Qndashlinearly independent complex numbersFollowing the transcendence methods of Hermite GelrsquofondSchneider one may start by introducing an auxiliaryfunction
F (z) = P (z ez)
where P isin Z[X0 X1] is a nonndashzero polynomial One considersthe derivatives (
d
dz
)kF = (DkP )(z ez)
at the pointsm1x1 + middot middot middot+mnxn
for various values of (m1 mn) isin Zn
48 59
Auxiliary functionLet D denote the derivation
D =part
partX0
+X1part
partX1
over the ring C[X0 X1] Recall that x1 xn are Qndashlinearlyindependent complex numbers Let α1 αn be nonndashzerocomplex numbersThe transcendence machinery produces a sequence (PN)Nge0
of polynomials with integer coefficients satisfying∣∣∣∣∣(DkPN)(nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1
49 59
Royrsquos approach to Schanuelrsquos Conjecture (1999)
If the number of equations we produce is too small such a setof relations does not contain any information The existenceof a sequence of nonndashtrivial polynomials (PN)Nge0 follows fromlinear algebra
On the other hand following D Roy one may expect that theexistence of such a sequence (PN)Nge0 producing sufficientlymany such equations will yield the desired conclusion
tr degQQ(x1 xn α1 αn
)ge n
50 59
Royrsquos Conjecture (1999)Let s0 s1 t0 t1 u positive real numbers satisfying
max1 t0 2t1 lt mins0 2s1and
maxs0 s1 + t1 lt u lt1
2(1 + t0 + t1)
Assume that for any sufficiently large positive integer N there exists a nonndashzero polynomial PN isin Z[X0 X1] withpartial degree le N t0 in X0 partial degree le N t1 in X1 andheight le eN which satisfies∣∣∣∣∣(DkPN)(
nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1 Then
tr degQQ(x1 xn α1 αn) ge n51 59
Royrsquos Theorem (1999)
Royrsquos conjecture is equivalent to Schanuelrsquos Conjecture
52 59
Equivalence between Schanuel and Roy
Let (x α) isin CtimesCtimes and let s0 s1 t0 t1 u be positive realnumbers satisfying the inequalities of Royrsquos Conjecture Thenthe following conditions are equivalent (a) The number αeminusx is a root of unity
(b) For any sufficiently large positive integer N there exists anonndashzero polynomial QN isin Z[X0 X1] with partial degreele N t0 in X0 partial degree le N t1 in X1 and heightH(QN) le eN such that∣∣(DkQN)(mx αm)
∣∣ le exp(minusNu)
for any km isin N with k le N s0 and m le N s1
53 59
Gelrsquofondrsquos transcendence criterion
Simple form Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le n and height le en such that
|Pn(ϑ)| le eminus7n2
for all n ge 1 then ϑ is algebraic and Pn(ϑ) = 0 for all n ge 1
First extension Replace the bound for the degree by le dnthe bound for the height by ebn and the bound for |Pn(ϑ)| byeminuscdnbn with some constant c gt 0 independent of n
Some mild conditions are required on the sequences (dn)nge1 and(bn)nge1
54 59
Transcendence criterion with multiplicitiesWith derivatives Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le dn and height le ebn such that∣∣P (j)
n (ϑ)∣∣ le eminuscdnbntn
for all j in the range 0 le j lt tn and all n ge 1 then ϑ isalgebraicDue to M Laurent and D Roy applications to algebraicindependence
55 59
Criterion with several points
Goal Given a sequence of complex numbers (ϑi)ige1 if thereexists a sequence (Pn)nge1 of nonndashzero polynomials in Z[X]with Pn of degree le dn and height le ebn such that∣∣P (j)
n (ϑi)∣∣ le eminuscdnbntnsn
for 0 le j lt tn 1 le i le sn and all n ge 1 then the numbers ϑiare algebraic
D Roy Not true in general but true in some special caseswith a structure on the sequence (ϑi)ige1Combines the elimination arguments used for criteria ofalgebraic independence and for zero estimates
56 59
Small value estimates for the additive group
D Roy Small value estimates for the additive group Intern JNumber Theory to appear
Let ξ be a transcendental complex number let β σ τ and νbe non-negative real numbers let n0 be a positive integer andlet (Pn)ngen0 be a sequence of nonndashzero polynomials in Z[T ]satisfying deg(Pn) le n and H(Pn) le exp(nβ) for eachn ge n0 Suppose that β gt 1 (34)σ + τ lt 1 andν gt 1 + β minus (34)σ minus τ Then for infinitely many n we have
max|P [j]n (iξ)| 0 le i le nσ 0 le j le nτ
gt exp(minusnν)
57 59
Small value estimates for the multiplicative group
D Roy Small value estimates for the multiplicative groupActa Arith to appear
Let ξ1 ξm be multiplicatively independent complexnumbers in a field of transcendence degree 1 Under suitableassumptions on the parameters β σ τ ν for infinitely manypositive integers n there exists no nonndashzero polynomialP isin Z[T ] satisfying deg(P ) le n H(P ) le exp(nβ) and
max|P [j](ξi11 middot middot middot ξimm )| 0 le i1 im le nσ 0 le j le nτ
gt exp(minusnν)
58 59
Heidelberg Colloquium organized by the Institute for Mathematics
httpwwwmathiuni-heidelbergdevortraghtmls=1 January 22 2009
Transcendental Number TheorySchanuelrsquos Conjecture
Michel Waldschmidt
Institut de Mathematiques de Jussieu amp Paris VIhttpwwwmathjussieufrsimmiw
59 59
Auxiliary functionLet D denote the derivation
D =part
partX0
+X1part
partX1
over the ring C[X0 X1] Recall that x1 xn are Qndashlinearlyindependent complex numbers Let α1 αn be nonndashzerocomplex numbersThe transcendence machinery produces a sequence (PN)Nge0
of polynomials with integer coefficients satisfying∣∣∣∣∣(DkPN)(nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1
49 59
Royrsquos approach to Schanuelrsquos Conjecture (1999)
If the number of equations we produce is too small such a setof relations does not contain any information The existenceof a sequence of nonndashtrivial polynomials (PN)Nge0 follows fromlinear algebra
On the other hand following D Roy one may expect that theexistence of such a sequence (PN)Nge0 producing sufficientlymany such equations will yield the desired conclusion
tr degQQ(x1 xn α1 αn
)ge n
50 59
Royrsquos Conjecture (1999)Let s0 s1 t0 t1 u positive real numbers satisfying
max1 t0 2t1 lt mins0 2s1and
maxs0 s1 + t1 lt u lt1
2(1 + t0 + t1)
Assume that for any sufficiently large positive integer N there exists a nonndashzero polynomial PN isin Z[X0 X1] withpartial degree le N t0 in X0 partial degree le N t1 in X1 andheight le eN which satisfies∣∣∣∣∣(DkPN)(
nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1 Then
tr degQQ(x1 xn α1 αn) ge n51 59
Royrsquos Theorem (1999)
Royrsquos conjecture is equivalent to Schanuelrsquos Conjecture
52 59
Equivalence between Schanuel and Roy
Let (x α) isin CtimesCtimes and let s0 s1 t0 t1 u be positive realnumbers satisfying the inequalities of Royrsquos Conjecture Thenthe following conditions are equivalent (a) The number αeminusx is a root of unity
(b) For any sufficiently large positive integer N there exists anonndashzero polynomial QN isin Z[X0 X1] with partial degreele N t0 in X0 partial degree le N t1 in X1 and heightH(QN) le eN such that∣∣(DkQN)(mx αm)
∣∣ le exp(minusNu)
for any km isin N with k le N s0 and m le N s1
53 59
Gelrsquofondrsquos transcendence criterion
Simple form Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le n and height le en such that
|Pn(ϑ)| le eminus7n2
for all n ge 1 then ϑ is algebraic and Pn(ϑ) = 0 for all n ge 1
First extension Replace the bound for the degree by le dnthe bound for the height by ebn and the bound for |Pn(ϑ)| byeminuscdnbn with some constant c gt 0 independent of n
Some mild conditions are required on the sequences (dn)nge1 and(bn)nge1
54 59
Transcendence criterion with multiplicitiesWith derivatives Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le dn and height le ebn such that∣∣P (j)
n (ϑ)∣∣ le eminuscdnbntn
for all j in the range 0 le j lt tn and all n ge 1 then ϑ isalgebraicDue to M Laurent and D Roy applications to algebraicindependence
55 59
Criterion with several points
Goal Given a sequence of complex numbers (ϑi)ige1 if thereexists a sequence (Pn)nge1 of nonndashzero polynomials in Z[X]with Pn of degree le dn and height le ebn such that∣∣P (j)
n (ϑi)∣∣ le eminuscdnbntnsn
for 0 le j lt tn 1 le i le sn and all n ge 1 then the numbers ϑiare algebraic
D Roy Not true in general but true in some special caseswith a structure on the sequence (ϑi)ige1Combines the elimination arguments used for criteria ofalgebraic independence and for zero estimates
56 59
Small value estimates for the additive group
D Roy Small value estimates for the additive group Intern JNumber Theory to appear
Let ξ be a transcendental complex number let β σ τ and νbe non-negative real numbers let n0 be a positive integer andlet (Pn)ngen0 be a sequence of nonndashzero polynomials in Z[T ]satisfying deg(Pn) le n and H(Pn) le exp(nβ) for eachn ge n0 Suppose that β gt 1 (34)σ + τ lt 1 andν gt 1 + β minus (34)σ minus τ Then for infinitely many n we have
max|P [j]n (iξ)| 0 le i le nσ 0 le j le nτ
gt exp(minusnν)
57 59
Small value estimates for the multiplicative group
D Roy Small value estimates for the multiplicative groupActa Arith to appear
Let ξ1 ξm be multiplicatively independent complexnumbers in a field of transcendence degree 1 Under suitableassumptions on the parameters β σ τ ν for infinitely manypositive integers n there exists no nonndashzero polynomialP isin Z[T ] satisfying deg(P ) le n H(P ) le exp(nβ) and
max|P [j](ξi11 middot middot middot ξimm )| 0 le i1 im le nσ 0 le j le nτ
gt exp(minusnν)
58 59
Heidelberg Colloquium organized by the Institute for Mathematics
httpwwwmathiuni-heidelbergdevortraghtmls=1 January 22 2009
Transcendental Number TheorySchanuelrsquos Conjecture
Michel Waldschmidt
Institut de Mathematiques de Jussieu amp Paris VIhttpwwwmathjussieufrsimmiw
59 59
Royrsquos approach to Schanuelrsquos Conjecture (1999)
If the number of equations we produce is too small such a setof relations does not contain any information The existenceof a sequence of nonndashtrivial polynomials (PN)Nge0 follows fromlinear algebra
On the other hand following D Roy one may expect that theexistence of such a sequence (PN)Nge0 producing sufficientlymany such equations will yield the desired conclusion
tr degQQ(x1 xn α1 αn
)ge n
50 59
Royrsquos Conjecture (1999)Let s0 s1 t0 t1 u positive real numbers satisfying
max1 t0 2t1 lt mins0 2s1and
maxs0 s1 + t1 lt u lt1
2(1 + t0 + t1)
Assume that for any sufficiently large positive integer N there exists a nonndashzero polynomial PN isin Z[X0 X1] withpartial degree le N t0 in X0 partial degree le N t1 in X1 andheight le eN which satisfies∣∣∣∣∣(DkPN)(
nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1 Then
tr degQQ(x1 xn α1 αn) ge n51 59
Royrsquos Theorem (1999)
Royrsquos conjecture is equivalent to Schanuelrsquos Conjecture
52 59
Equivalence between Schanuel and Roy
Let (x α) isin CtimesCtimes and let s0 s1 t0 t1 u be positive realnumbers satisfying the inequalities of Royrsquos Conjecture Thenthe following conditions are equivalent (a) The number αeminusx is a root of unity
(b) For any sufficiently large positive integer N there exists anonndashzero polynomial QN isin Z[X0 X1] with partial degreele N t0 in X0 partial degree le N t1 in X1 and heightH(QN) le eN such that∣∣(DkQN)(mx αm)
∣∣ le exp(minusNu)
for any km isin N with k le N s0 and m le N s1
53 59
Gelrsquofondrsquos transcendence criterion
Simple form Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le n and height le en such that
|Pn(ϑ)| le eminus7n2
for all n ge 1 then ϑ is algebraic and Pn(ϑ) = 0 for all n ge 1
First extension Replace the bound for the degree by le dnthe bound for the height by ebn and the bound for |Pn(ϑ)| byeminuscdnbn with some constant c gt 0 independent of n
Some mild conditions are required on the sequences (dn)nge1 and(bn)nge1
54 59
Transcendence criterion with multiplicitiesWith derivatives Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le dn and height le ebn such that∣∣P (j)
n (ϑ)∣∣ le eminuscdnbntn
for all j in the range 0 le j lt tn and all n ge 1 then ϑ isalgebraicDue to M Laurent and D Roy applications to algebraicindependence
55 59
Criterion with several points
Goal Given a sequence of complex numbers (ϑi)ige1 if thereexists a sequence (Pn)nge1 of nonndashzero polynomials in Z[X]with Pn of degree le dn and height le ebn such that∣∣P (j)
n (ϑi)∣∣ le eminuscdnbntnsn
for 0 le j lt tn 1 le i le sn and all n ge 1 then the numbers ϑiare algebraic
D Roy Not true in general but true in some special caseswith a structure on the sequence (ϑi)ige1Combines the elimination arguments used for criteria ofalgebraic independence and for zero estimates
56 59
Small value estimates for the additive group
D Roy Small value estimates for the additive group Intern JNumber Theory to appear
Let ξ be a transcendental complex number let β σ τ and νbe non-negative real numbers let n0 be a positive integer andlet (Pn)ngen0 be a sequence of nonndashzero polynomials in Z[T ]satisfying deg(Pn) le n and H(Pn) le exp(nβ) for eachn ge n0 Suppose that β gt 1 (34)σ + τ lt 1 andν gt 1 + β minus (34)σ minus τ Then for infinitely many n we have
max|P [j]n (iξ)| 0 le i le nσ 0 le j le nτ
gt exp(minusnν)
57 59
Small value estimates for the multiplicative group
D Roy Small value estimates for the multiplicative groupActa Arith to appear
Let ξ1 ξm be multiplicatively independent complexnumbers in a field of transcendence degree 1 Under suitableassumptions on the parameters β σ τ ν for infinitely manypositive integers n there exists no nonndashzero polynomialP isin Z[T ] satisfying deg(P ) le n H(P ) le exp(nβ) and
max|P [j](ξi11 middot middot middot ξimm )| 0 le i1 im le nσ 0 le j le nτ
gt exp(minusnν)
58 59
Heidelberg Colloquium organized by the Institute for Mathematics
httpwwwmathiuni-heidelbergdevortraghtmls=1 January 22 2009
Transcendental Number TheorySchanuelrsquos Conjecture
Michel Waldschmidt
Institut de Mathematiques de Jussieu amp Paris VIhttpwwwmathjussieufrsimmiw
59 59
Royrsquos Conjecture (1999)Let s0 s1 t0 t1 u positive real numbers satisfying
max1 t0 2t1 lt mins0 2s1and
maxs0 s1 + t1 lt u lt1
2(1 + t0 + t1)
Assume that for any sufficiently large positive integer N there exists a nonndashzero polynomial PN isin Z[X0 X1] withpartial degree le N t0 in X0 partial degree le N t1 in X1 andheight le eN which satisfies∣∣∣∣∣(DkPN)(
nsumj=1
mjxjnprodj=1
αmjj
)∣∣∣∣∣ le exp(minusNu)
for any non-negative integers k m1 mn with k le N s0 andmaxm1 mn le N s1 Then
tr degQQ(x1 xn α1 αn) ge n51 59
Royrsquos Theorem (1999)
Royrsquos conjecture is equivalent to Schanuelrsquos Conjecture
52 59
Equivalence between Schanuel and Roy
Let (x α) isin CtimesCtimes and let s0 s1 t0 t1 u be positive realnumbers satisfying the inequalities of Royrsquos Conjecture Thenthe following conditions are equivalent (a) The number αeminusx is a root of unity
(b) For any sufficiently large positive integer N there exists anonndashzero polynomial QN isin Z[X0 X1] with partial degreele N t0 in X0 partial degree le N t1 in X1 and heightH(QN) le eN such that∣∣(DkQN)(mx αm)
∣∣ le exp(minusNu)
for any km isin N with k le N s0 and m le N s1
53 59
Gelrsquofondrsquos transcendence criterion
Simple form Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le n and height le en such that
|Pn(ϑ)| le eminus7n2
for all n ge 1 then ϑ is algebraic and Pn(ϑ) = 0 for all n ge 1
First extension Replace the bound for the degree by le dnthe bound for the height by ebn and the bound for |Pn(ϑ)| byeminuscdnbn with some constant c gt 0 independent of n
Some mild conditions are required on the sequences (dn)nge1 and(bn)nge1
54 59
Transcendence criterion with multiplicitiesWith derivatives Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le dn and height le ebn such that∣∣P (j)
n (ϑ)∣∣ le eminuscdnbntn
for all j in the range 0 le j lt tn and all n ge 1 then ϑ isalgebraicDue to M Laurent and D Roy applications to algebraicindependence
55 59
Criterion with several points
Goal Given a sequence of complex numbers (ϑi)ige1 if thereexists a sequence (Pn)nge1 of nonndashzero polynomials in Z[X]with Pn of degree le dn and height le ebn such that∣∣P (j)
n (ϑi)∣∣ le eminuscdnbntnsn
for 0 le j lt tn 1 le i le sn and all n ge 1 then the numbers ϑiare algebraic
D Roy Not true in general but true in some special caseswith a structure on the sequence (ϑi)ige1Combines the elimination arguments used for criteria ofalgebraic independence and for zero estimates
56 59
Small value estimates for the additive group
D Roy Small value estimates for the additive group Intern JNumber Theory to appear
Let ξ be a transcendental complex number let β σ τ and νbe non-negative real numbers let n0 be a positive integer andlet (Pn)ngen0 be a sequence of nonndashzero polynomials in Z[T ]satisfying deg(Pn) le n and H(Pn) le exp(nβ) for eachn ge n0 Suppose that β gt 1 (34)σ + τ lt 1 andν gt 1 + β minus (34)σ minus τ Then for infinitely many n we have
max|P [j]n (iξ)| 0 le i le nσ 0 le j le nτ
gt exp(minusnν)
57 59
Small value estimates for the multiplicative group
D Roy Small value estimates for the multiplicative groupActa Arith to appear
Let ξ1 ξm be multiplicatively independent complexnumbers in a field of transcendence degree 1 Under suitableassumptions on the parameters β σ τ ν for infinitely manypositive integers n there exists no nonndashzero polynomialP isin Z[T ] satisfying deg(P ) le n H(P ) le exp(nβ) and
max|P [j](ξi11 middot middot middot ξimm )| 0 le i1 im le nσ 0 le j le nτ
gt exp(minusnν)
58 59
Heidelberg Colloquium organized by the Institute for Mathematics
httpwwwmathiuni-heidelbergdevortraghtmls=1 January 22 2009
Transcendental Number TheorySchanuelrsquos Conjecture
Michel Waldschmidt
Institut de Mathematiques de Jussieu amp Paris VIhttpwwwmathjussieufrsimmiw
59 59
Royrsquos Theorem (1999)
Royrsquos conjecture is equivalent to Schanuelrsquos Conjecture
52 59
Equivalence between Schanuel and Roy
Let (x α) isin CtimesCtimes and let s0 s1 t0 t1 u be positive realnumbers satisfying the inequalities of Royrsquos Conjecture Thenthe following conditions are equivalent (a) The number αeminusx is a root of unity
(b) For any sufficiently large positive integer N there exists anonndashzero polynomial QN isin Z[X0 X1] with partial degreele N t0 in X0 partial degree le N t1 in X1 and heightH(QN) le eN such that∣∣(DkQN)(mx αm)
∣∣ le exp(minusNu)
for any km isin N with k le N s0 and m le N s1
53 59
Gelrsquofondrsquos transcendence criterion
Simple form Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le n and height le en such that
|Pn(ϑ)| le eminus7n2
for all n ge 1 then ϑ is algebraic and Pn(ϑ) = 0 for all n ge 1
First extension Replace the bound for the degree by le dnthe bound for the height by ebn and the bound for |Pn(ϑ)| byeminuscdnbn with some constant c gt 0 independent of n
Some mild conditions are required on the sequences (dn)nge1 and(bn)nge1
54 59
Transcendence criterion with multiplicitiesWith derivatives Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le dn and height le ebn such that∣∣P (j)
n (ϑ)∣∣ le eminuscdnbntn
for all j in the range 0 le j lt tn and all n ge 1 then ϑ isalgebraicDue to M Laurent and D Roy applications to algebraicindependence
55 59
Criterion with several points
Goal Given a sequence of complex numbers (ϑi)ige1 if thereexists a sequence (Pn)nge1 of nonndashzero polynomials in Z[X]with Pn of degree le dn and height le ebn such that∣∣P (j)
n (ϑi)∣∣ le eminuscdnbntnsn
for 0 le j lt tn 1 le i le sn and all n ge 1 then the numbers ϑiare algebraic
D Roy Not true in general but true in some special caseswith a structure on the sequence (ϑi)ige1Combines the elimination arguments used for criteria ofalgebraic independence and for zero estimates
56 59
Small value estimates for the additive group
D Roy Small value estimates for the additive group Intern JNumber Theory to appear
Let ξ be a transcendental complex number let β σ τ and νbe non-negative real numbers let n0 be a positive integer andlet (Pn)ngen0 be a sequence of nonndashzero polynomials in Z[T ]satisfying deg(Pn) le n and H(Pn) le exp(nβ) for eachn ge n0 Suppose that β gt 1 (34)σ + τ lt 1 andν gt 1 + β minus (34)σ minus τ Then for infinitely many n we have
max|P [j]n (iξ)| 0 le i le nσ 0 le j le nτ
gt exp(minusnν)
57 59
Small value estimates for the multiplicative group
D Roy Small value estimates for the multiplicative groupActa Arith to appear
Let ξ1 ξm be multiplicatively independent complexnumbers in a field of transcendence degree 1 Under suitableassumptions on the parameters β σ τ ν for infinitely manypositive integers n there exists no nonndashzero polynomialP isin Z[T ] satisfying deg(P ) le n H(P ) le exp(nβ) and
max|P [j](ξi11 middot middot middot ξimm )| 0 le i1 im le nσ 0 le j le nτ
gt exp(minusnν)
58 59
Heidelberg Colloquium organized by the Institute for Mathematics
httpwwwmathiuni-heidelbergdevortraghtmls=1 January 22 2009
Transcendental Number TheorySchanuelrsquos Conjecture
Michel Waldschmidt
Institut de Mathematiques de Jussieu amp Paris VIhttpwwwmathjussieufrsimmiw
59 59
Equivalence between Schanuel and Roy
Let (x α) isin CtimesCtimes and let s0 s1 t0 t1 u be positive realnumbers satisfying the inequalities of Royrsquos Conjecture Thenthe following conditions are equivalent (a) The number αeminusx is a root of unity
(b) For any sufficiently large positive integer N there exists anonndashzero polynomial QN isin Z[X0 X1] with partial degreele N t0 in X0 partial degree le N t1 in X1 and heightH(QN) le eN such that∣∣(DkQN)(mx αm)
∣∣ le exp(minusNu)
for any km isin N with k le N s0 and m le N s1
53 59
Gelrsquofondrsquos transcendence criterion
Simple form Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le n and height le en such that
|Pn(ϑ)| le eminus7n2
for all n ge 1 then ϑ is algebraic and Pn(ϑ) = 0 for all n ge 1
First extension Replace the bound for the degree by le dnthe bound for the height by ebn and the bound for |Pn(ϑ)| byeminuscdnbn with some constant c gt 0 independent of n
Some mild conditions are required on the sequences (dn)nge1 and(bn)nge1
54 59
Transcendence criterion with multiplicitiesWith derivatives Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le dn and height le ebn such that∣∣P (j)
n (ϑ)∣∣ le eminuscdnbntn
for all j in the range 0 le j lt tn and all n ge 1 then ϑ isalgebraicDue to M Laurent and D Roy applications to algebraicindependence
55 59
Criterion with several points
Goal Given a sequence of complex numbers (ϑi)ige1 if thereexists a sequence (Pn)nge1 of nonndashzero polynomials in Z[X]with Pn of degree le dn and height le ebn such that∣∣P (j)
n (ϑi)∣∣ le eminuscdnbntnsn
for 0 le j lt tn 1 le i le sn and all n ge 1 then the numbers ϑiare algebraic
D Roy Not true in general but true in some special caseswith a structure on the sequence (ϑi)ige1Combines the elimination arguments used for criteria ofalgebraic independence and for zero estimates
56 59
Small value estimates for the additive group
D Roy Small value estimates for the additive group Intern JNumber Theory to appear
Let ξ be a transcendental complex number let β σ τ and νbe non-negative real numbers let n0 be a positive integer andlet (Pn)ngen0 be a sequence of nonndashzero polynomials in Z[T ]satisfying deg(Pn) le n and H(Pn) le exp(nβ) for eachn ge n0 Suppose that β gt 1 (34)σ + τ lt 1 andν gt 1 + β minus (34)σ minus τ Then for infinitely many n we have
max|P [j]n (iξ)| 0 le i le nσ 0 le j le nτ
gt exp(minusnν)
57 59
Small value estimates for the multiplicative group
D Roy Small value estimates for the multiplicative groupActa Arith to appear
Let ξ1 ξm be multiplicatively independent complexnumbers in a field of transcendence degree 1 Under suitableassumptions on the parameters β σ τ ν for infinitely manypositive integers n there exists no nonndashzero polynomialP isin Z[T ] satisfying deg(P ) le n H(P ) le exp(nβ) and
max|P [j](ξi11 middot middot middot ξimm )| 0 le i1 im le nσ 0 le j le nτ
gt exp(minusnν)
58 59
Heidelberg Colloquium organized by the Institute for Mathematics
httpwwwmathiuni-heidelbergdevortraghtmls=1 January 22 2009
Transcendental Number TheorySchanuelrsquos Conjecture
Michel Waldschmidt
Institut de Mathematiques de Jussieu amp Paris VIhttpwwwmathjussieufrsimmiw
59 59
Gelrsquofondrsquos transcendence criterion
Simple form Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le n and height le en such that
|Pn(ϑ)| le eminus7n2
for all n ge 1 then ϑ is algebraic and Pn(ϑ) = 0 for all n ge 1
First extension Replace the bound for the degree by le dnthe bound for the height by ebn and the bound for |Pn(ϑ)| byeminuscdnbn with some constant c gt 0 independent of n
Some mild conditions are required on the sequences (dn)nge1 and(bn)nge1
54 59
Transcendence criterion with multiplicitiesWith derivatives Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le dn and height le ebn such that∣∣P (j)
n (ϑ)∣∣ le eminuscdnbntn
for all j in the range 0 le j lt tn and all n ge 1 then ϑ isalgebraicDue to M Laurent and D Roy applications to algebraicindependence
55 59
Criterion with several points
Goal Given a sequence of complex numbers (ϑi)ige1 if thereexists a sequence (Pn)nge1 of nonndashzero polynomials in Z[X]with Pn of degree le dn and height le ebn such that∣∣P (j)
n (ϑi)∣∣ le eminuscdnbntnsn
for 0 le j lt tn 1 le i le sn and all n ge 1 then the numbers ϑiare algebraic
D Roy Not true in general but true in some special caseswith a structure on the sequence (ϑi)ige1Combines the elimination arguments used for criteria ofalgebraic independence and for zero estimates
56 59
Small value estimates for the additive group
D Roy Small value estimates for the additive group Intern JNumber Theory to appear
Let ξ be a transcendental complex number let β σ τ and νbe non-negative real numbers let n0 be a positive integer andlet (Pn)ngen0 be a sequence of nonndashzero polynomials in Z[T ]satisfying deg(Pn) le n and H(Pn) le exp(nβ) for eachn ge n0 Suppose that β gt 1 (34)σ + τ lt 1 andν gt 1 + β minus (34)σ minus τ Then for infinitely many n we have
max|P [j]n (iξ)| 0 le i le nσ 0 le j le nτ
gt exp(minusnν)
57 59
Small value estimates for the multiplicative group
D Roy Small value estimates for the multiplicative groupActa Arith to appear
Let ξ1 ξm be multiplicatively independent complexnumbers in a field of transcendence degree 1 Under suitableassumptions on the parameters β σ τ ν for infinitely manypositive integers n there exists no nonndashzero polynomialP isin Z[T ] satisfying deg(P ) le n H(P ) le exp(nβ) and
max|P [j](ξi11 middot middot middot ξimm )| 0 le i1 im le nσ 0 le j le nτ
gt exp(minusnν)
58 59
Heidelberg Colloquium organized by the Institute for Mathematics
httpwwwmathiuni-heidelbergdevortraghtmls=1 January 22 2009
Transcendental Number TheorySchanuelrsquos Conjecture
Michel Waldschmidt
Institut de Mathematiques de Jussieu amp Paris VIhttpwwwmathjussieufrsimmiw
59 59
Transcendence criterion with multiplicitiesWith derivatives Given a complex number ϑ if there exists asequence (Pn)nge1 of nonndashzero polynomials in Z[X] with Pnof degree le dn and height le ebn such that∣∣P (j)
n (ϑ)∣∣ le eminuscdnbntn
for all j in the range 0 le j lt tn and all n ge 1 then ϑ isalgebraicDue to M Laurent and D Roy applications to algebraicindependence
55 59
Criterion with several points
Goal Given a sequence of complex numbers (ϑi)ige1 if thereexists a sequence (Pn)nge1 of nonndashzero polynomials in Z[X]with Pn of degree le dn and height le ebn such that∣∣P (j)
n (ϑi)∣∣ le eminuscdnbntnsn
for 0 le j lt tn 1 le i le sn and all n ge 1 then the numbers ϑiare algebraic
D Roy Not true in general but true in some special caseswith a structure on the sequence (ϑi)ige1Combines the elimination arguments used for criteria ofalgebraic independence and for zero estimates
56 59
Small value estimates for the additive group
D Roy Small value estimates for the additive group Intern JNumber Theory to appear
Let ξ be a transcendental complex number let β σ τ and νbe non-negative real numbers let n0 be a positive integer andlet (Pn)ngen0 be a sequence of nonndashzero polynomials in Z[T ]satisfying deg(Pn) le n and H(Pn) le exp(nβ) for eachn ge n0 Suppose that β gt 1 (34)σ + τ lt 1 andν gt 1 + β minus (34)σ minus τ Then for infinitely many n we have
max|P [j]n (iξ)| 0 le i le nσ 0 le j le nτ
gt exp(minusnν)
57 59
Small value estimates for the multiplicative group
D Roy Small value estimates for the multiplicative groupActa Arith to appear
Let ξ1 ξm be multiplicatively independent complexnumbers in a field of transcendence degree 1 Under suitableassumptions on the parameters β σ τ ν for infinitely manypositive integers n there exists no nonndashzero polynomialP isin Z[T ] satisfying deg(P ) le n H(P ) le exp(nβ) and
max|P [j](ξi11 middot middot middot ξimm )| 0 le i1 im le nσ 0 le j le nτ
gt exp(minusnν)
58 59
Heidelberg Colloquium organized by the Institute for Mathematics
httpwwwmathiuni-heidelbergdevortraghtmls=1 January 22 2009
Transcendental Number TheorySchanuelrsquos Conjecture
Michel Waldschmidt
Institut de Mathematiques de Jussieu amp Paris VIhttpwwwmathjussieufrsimmiw
59 59
Criterion with several points
Goal Given a sequence of complex numbers (ϑi)ige1 if thereexists a sequence (Pn)nge1 of nonndashzero polynomials in Z[X]with Pn of degree le dn and height le ebn such that∣∣P (j)
n (ϑi)∣∣ le eminuscdnbntnsn
for 0 le j lt tn 1 le i le sn and all n ge 1 then the numbers ϑiare algebraic
D Roy Not true in general but true in some special caseswith a structure on the sequence (ϑi)ige1Combines the elimination arguments used for criteria ofalgebraic independence and for zero estimates
56 59
Small value estimates for the additive group
D Roy Small value estimates for the additive group Intern JNumber Theory to appear
Let ξ be a transcendental complex number let β σ τ and νbe non-negative real numbers let n0 be a positive integer andlet (Pn)ngen0 be a sequence of nonndashzero polynomials in Z[T ]satisfying deg(Pn) le n and H(Pn) le exp(nβ) for eachn ge n0 Suppose that β gt 1 (34)σ + τ lt 1 andν gt 1 + β minus (34)σ minus τ Then for infinitely many n we have
max|P [j]n (iξ)| 0 le i le nσ 0 le j le nτ
gt exp(minusnν)
57 59
Small value estimates for the multiplicative group
D Roy Small value estimates for the multiplicative groupActa Arith to appear
Let ξ1 ξm be multiplicatively independent complexnumbers in a field of transcendence degree 1 Under suitableassumptions on the parameters β σ τ ν for infinitely manypositive integers n there exists no nonndashzero polynomialP isin Z[T ] satisfying deg(P ) le n H(P ) le exp(nβ) and
max|P [j](ξi11 middot middot middot ξimm )| 0 le i1 im le nσ 0 le j le nτ
gt exp(minusnν)
58 59
Heidelberg Colloquium organized by the Institute for Mathematics
httpwwwmathiuni-heidelbergdevortraghtmls=1 January 22 2009
Transcendental Number TheorySchanuelrsquos Conjecture
Michel Waldschmidt
Institut de Mathematiques de Jussieu amp Paris VIhttpwwwmathjussieufrsimmiw
59 59
Small value estimates for the additive group
D Roy Small value estimates for the additive group Intern JNumber Theory to appear
Let ξ be a transcendental complex number let β σ τ and νbe non-negative real numbers let n0 be a positive integer andlet (Pn)ngen0 be a sequence of nonndashzero polynomials in Z[T ]satisfying deg(Pn) le n and H(Pn) le exp(nβ) for eachn ge n0 Suppose that β gt 1 (34)σ + τ lt 1 andν gt 1 + β minus (34)σ minus τ Then for infinitely many n we have
max|P [j]n (iξ)| 0 le i le nσ 0 le j le nτ
gt exp(minusnν)
57 59
Small value estimates for the multiplicative group
D Roy Small value estimates for the multiplicative groupActa Arith to appear
Let ξ1 ξm be multiplicatively independent complexnumbers in a field of transcendence degree 1 Under suitableassumptions on the parameters β σ τ ν for infinitely manypositive integers n there exists no nonndashzero polynomialP isin Z[T ] satisfying deg(P ) le n H(P ) le exp(nβ) and
max|P [j](ξi11 middot middot middot ξimm )| 0 le i1 im le nσ 0 le j le nτ
gt exp(minusnν)
58 59
Heidelberg Colloquium organized by the Institute for Mathematics
httpwwwmathiuni-heidelbergdevortraghtmls=1 January 22 2009
Transcendental Number TheorySchanuelrsquos Conjecture
Michel Waldschmidt
Institut de Mathematiques de Jussieu amp Paris VIhttpwwwmathjussieufrsimmiw
59 59
Small value estimates for the multiplicative group
D Roy Small value estimates for the multiplicative groupActa Arith to appear
Let ξ1 ξm be multiplicatively independent complexnumbers in a field of transcendence degree 1 Under suitableassumptions on the parameters β σ τ ν for infinitely manypositive integers n there exists no nonndashzero polynomialP isin Z[T ] satisfying deg(P ) le n H(P ) le exp(nβ) and
max|P [j](ξi11 middot middot middot ξimm )| 0 le i1 im le nσ 0 le j le nτ
gt exp(minusnν)
58 59
Heidelberg Colloquium organized by the Institute for Mathematics
httpwwwmathiuni-heidelbergdevortraghtmls=1 January 22 2009
Transcendental Number TheorySchanuelrsquos Conjecture
Michel Waldschmidt
Institut de Mathematiques de Jussieu amp Paris VIhttpwwwmathjussieufrsimmiw
59 59
Heidelberg Colloquium organized by the Institute for Mathematics
httpwwwmathiuni-heidelbergdevortraghtmls=1 January 22 2009
Transcendental Number TheorySchanuelrsquos Conjecture
Michel Waldschmidt
Institut de Mathematiques de Jussieu amp Paris VIhttpwwwmathjussieufrsimmiw
59 59