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Multidimensional continued fractions andDiophantine exponents of lattices

UDT 2018, CIRM, Luminy

OlegN.German

Moscow State University

October 02, 2018

OlegN.German Multidimensional continued fractions and Diophantine exponents of lattices 1 / 19

Irrationality exponent vs Growth of partial quotients

Irrationality exponent θ ∈ R\Q

µ(θ) = sup{γ ∈ R

∣∣∣ ∣∣θ − p/q∣∣ 6 |q|−γ admits ∞ solutions in (q, p) ∈ Z2}

Relation to the growth of partial quotientsIf

θ = [a0; a1, a2, . . .] = a0 +1

a1 +1

a2 + . . .

,pnqn

= [a0; a1, . . . , an],

thenµ(θ)− 2 = lim sup

n→+∞

ln an+1

ln qn


Geometry of continued fractions

y = θ1x

y = θ2x

K1

K2

µ(θ1)− 2 = lim supn→+∞

ln an+1

ln qnθ1 > 1, −1 < θ2 < 0

θ1 = [a0; a1, a2, . . .]

−1/θ2 = [a−1; a−2, a−3, . . .]

vn =

(qnpn

), vn = anvn−1 + vn−2



y = θ1x

y = θ2x

a0

a−2

a2a1

a−1

v0

v−2

v−1v−3

K1

K2


ln an+1

ln qnθ1 > 1, −1 < θ2 < 0

θ1 = [a0; a1, a2, . . .]

−1/θ2 = [a−1; a−2, a−3, . . .]

vn =

(qnpn

), vn = anvn−1 + vn−2



y = θ1x

y = θ2x

a0

a0

a−2

a2a1

a−1

v0

v−2

v−1v−3

K1

K2


ln an+1

ln qnθ1 > 1, −1 < θ2 < 0

θ1 = [a0; a1, a2, . . .]

−1/θ2 = [a−1; a−2, a−3, . . .]

vn =

(qnpn

), vn = anvn−1 + vn−2

an+1 = int.lenght of [vn−1,vn+1]

an+1 = int.angle α(vn) at vn



y = θ1x

y = θ2x

a0

a1

a−1

a−2

a2

a0

a1

a−1a−2

v−2

v−1v−3

K1

K2


ln an+1

ln qnθ1 > 1, −1 < θ2 < 0

θ1 = [a0; a1, a2, . . .]

−1/θ2 = [a−1; a−2, a−3, . . .]

vn =

(qnpn

), vn = anvn−1 + vn−2





y = θ1x

y = θ2x

a0

a1

a−1

a−2

a2

a0

a1

a−1a−2

v−2

v−1v−3

K1

K2


ln an+1

ln qnθ1 > 1, −1 < θ2 < 0

θ1 = [a0; a1, a2, . . .]

−1/θ2 = [a−1; a−2, a−3, . . .]

vn =

(qnpn

), vn = anvn−1 + vn−2



|qn| � |vn|



y = θ1x

y = θ2x

a0

a1

a−1

a−2

a2

a0

a1

a−1a−2

v−2

v−1v−3

K1

K2


ln an+1

ln qnθ1 > 1, −1 < θ2 < 0

θ1 = [a0; a1, a2, . . .]

−1/θ2 = [a−1; a−2, a−3, . . .]

vn =

(qnpn

), vn = anvn−1 + vn−2

an+1 = α(vn), |qn| � |vn|


Irrationality exponent in terms of lattice exponents

y = θ1x

y = θ2x

v0

v−2

v−1v−3

K1

K2


ln an+1

ln qnL1(q, p) = qθ1 − p, L2(q, p) = qθ2 − p

Λ ={(L1(z), L2(z)

) ∣∣∣ z ∈ Z2}



y = θ1x

y = θ2x

v0

v−2

v−1v−3

K1

K2


ln an+1


Λ ={(L1(z), L2(z)

) ∣∣∣ z ∈ Z2}

µ(θ1) = lim supq,p∈Z|q|→∞

log(|θ1−p/q|−1

)log |q| =

= 2 + lim supq,p∈Z|q|→∞

log(|q(qθ1−p)|−1

)log |q| =

= 2 + lim supn→+∞

log(|L1(vn)·L2(vn)|−1

)log |vn|



y = θ1x

y = θ2x

v0

v−2

v−1v−3

K1

K2


ln an+1


Λ ={(L1(z), L2(z)

) ∣∣∣ z ∈ Z2}

µ(θ2) = lim supq,p∈Z|q|→∞

log(|θ2−p/q|−1

)log |q| =

= 2 + lim supq,p∈Z|q|→∞

log(|q(qθ2−p)|−1

)log |q| =

= 2 + lim supn→−∞


)log |vn|



y = θ1x

y = θ2x

v0

v−2

v−1v−3

K1

K2


ln an+1


Λ ={(L1(z), L2(z)

) ∣∣∣ z ∈ Z2}

µ(θ1) = 2 + lim supn→+∞


)log |vn|

µ(θ2) = 2 + lim supn→−∞


)log |vn|



y = θ1x

y = θ2x

v0

v−2

v−1v−3

K1

K2


ln an+1


Λ ={(L1(z), L2(z)

) ∣∣∣ z ∈ Z2}

µ(θ1) = 2 + lim supn→+∞


)log |vn|

µ(θ2) = 2 + lim supn→−∞


)log |vn|

max(µ(θ1), µ(θ2)) = 2 + 2ω(Λ)



y = θ1x

y = θ2x

v0

v−2

v−1v−3

K1

K2


ln an+1


Λ ={(L1(z), L2(z)

) ∣∣∣ z ∈ Z2}

µ(θ1) = 2 + lim supn→+∞


)log |vn|

µ(θ2) = 2 + lim supn→−∞


)log |vn|

max(µ(θ1), µ(θ2)) = 2 + 2ω(Λ)

ω(Λ) = lim supx=(x1,x2)∈Λ|x|→∞

log(|x1x2|−1/2

)log |x|


Reformulation

y = θ1x

y = θ2x

v0

v−2

v−1v−3

K1

K2


ln an+1


Λ ={(L1(z), L2(z)

) ∣∣∣ z ∈ Z2}

ω(Λ) = lim supx=(x1,x2)∈Λ|x|→∞

log(|x1x2|−1/2

)log |x|

ω(Λ) = 12 lim sup

|v|→∞v∈V(K1)∪V(K2)

log(α(v))

log |v|


Higher dimensions: lattice exponents and Klein polyhedra

Λ a lattice in Rd, det Λ = 1

Π(x) = |x1 . . . xd|1/d for x = (x1, . . . , xd) ∈ Rd

Lattice exponent

ω(Λ) = lim supx∈Λ|x|→∞

log(Π(x)−1

)log |x|

= sup{γ ∈ R

∣∣∣Π(x) 6 |x|−γ for infinitely many x ∈ Λ}

O an orthant

Klein polyhedron

K = conv(O ∩ Λ\{0}

)




Π(x) = |x1 . . . xd|1/d for x = (x1, . . . , xd) ∈ Rd

Lattice exponent

ω(Λ) = lim supx∈Λ|x|→∞

log(Π(x)−1

)log |x|

= sup{γ ∈ R

∣∣∣Π(x) 6 |x|−γ for infinitely many x ∈ Λ}

O an orthant

Klein polyhedron

K = conv(O ∩ Λ\{0}

)OlegN.German Multidimensional continued fractions and Diophantine exponents of lattices 16 / 19



Klein polyhedron

K = conv(O ∩ Λ\{0}

)

Determinant of a face FLet v1, . . . ,vk be the vertices of F . Then

detF =∑

16i1<...<id6k

|det(vi1 , . . . ,vid)|

Determinant of an edge star StvLet r1, . . . , rk be the primitive vectors parallel to the edges incident to v. Then

detStv =∑

16i1<...<id6k

|det(ri1 , . . . , rid)|




Klein polyhedron

K = conv(O ∩ Λ\{0}

)Determinant of a face FLet v1, . . . ,vk be the vertices of F . Then

detF =∑

16i1<...<id6k

|det(vi1 , . . . ,vid)|

Determinant of an edge star StvLet r1, . . . , rk be the primitive vectors parallel to the edges incident to v. Then

detStv =∑

16i1<...<id6k

|det(ri1 , . . . , rid)|



Λ a lattice in Rd, K1, . . . ,K2d its Klein polyhedra, V(Ki) set of vertices

Two-dimensional statement

ω(Λ) = 12 lim sup

|v|→∞v∈V(K1)∪V(K2)

log(detStv)

log |v|

Question, arbitrary dimension

Is it true that ω(Λ) � lim sup|v|→∞

v∈⋃

i V(Ki)

log(detStv)

log |v|?

E.Bigushev, O.G., 2018

For d = 3 we have ω(Λ) 6 23 lim sup|v|→∞

v∈⋃

i V(Ki)

log(detStv)

log |v|


Thank you!


Download - Multidimensional continued fractions and Diophantine ... · Multidimensional continued fractions and Diophantine exponents of lattices UDT2018,CIRM,Luminy OlegN.German Moscow State

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