the 290 theorem

Quadratic forms and the 290 theorem Quadratic forms and Lattices A sketch proof of the 290 theorem

The 290 Theorem

Dan Fretwell

SoMaS postgrad seminar26th October 2014

Outline of talk

1 Quadratic forms and the 290 theorem

2 Quadratic forms and Lattices

3 A sketch proof of the 290 theorem

Lagrange’s four square TheoremEvery positive integer is a sum of four square numbers.

7 = 12 + 12 + 12 + 22

1831 = 22 + 32 + 272 + 332

231287 = 12 + 32 + 3062 + 3712

A quadratic form over a ring R is a polynomialQ(x1, x2, ..., xn) ∈ R[x1, x2, ..., xn] of the form:

Q(x1, x2, ..., xn) =∑

1≤i,j≤n

ai,jxixj .

If R ⊆ R we say that Q is positive definite ifQ(α1, α2, ..., αn) > 0 whenever α1, α2, ..., αn ∈ R\{0}.

For example:x2 + y2 + z2 + w2

x2 − 4xy + 6y2 = (x − 2y)2 + 2y2

are positive definite quadratic forms over Z.

A quadratic form over a ring R is a polynomialQ(x1, x2, ..., xn) ∈ R[x1, x2, ..., xn] of the form:

Q(x1, x2, ..., xn) =∑

1≤i,j≤n

ai,jxixj .

If R ⊆ R we say that Q is positive definite ifQ(α1, α2, ..., αn) > 0 whenever α1, α2, ..., αn ∈ R\{0}.

For example:x2 + y2 + z2 + w2

x2 − 4xy + 6y2 = (x − 2y)2 + 2y2

are positive definite quadratic forms over Z.

Say that a quadratic form Q over Z represents m ∈ N if thereexists a1,a2, ...,an ∈ Z such that Q(a1,a2, ...,an) = m.

Question: Is there a simple way to tell if a quadratic formrepresents all positive integers?

We call such a quadratic form universal. We have already seenthat universal forms exist, for example x2 + y2 + z2 + w2 isuniversal by Lagrange’s four square theorem.

The 290 theoremLet Q be a positive definite quadratic form over Z. Then Q isuniversal if and only if it represents the 29 integers1,2,3,5,6,7,10,13,14,15,17,19,21,22,23,26,29,30,31,34,35,37,42,58,93,110,145,203,290.

The 290 theorem was first proved by Conway andSchneeberger in 1993 but their proof was complicated and sowas never published. It wasn’t until 2000 that a simpler proofwas found by Bhargava, one of the recipients of this years fieldsmedals.

The 290 theoremLet Q be a positive definite quadratic form over Z. Then Q isuniversal if and only if it represents the 29 integers1,2,3,5,6,7,10,13,14,15,17,19,21,22,23,26,29,30,31,34,35,37,42,58,93,110,145,203,290.

The 290 theorem was first proved by Conway andSchneeberger in 1993 but their proof was complicated and sowas never published. It wasn’t until 2000 that a simpler proofwas found by Bhargava, one of the recipients of this years fieldsmedals.

Outline of talk

Let v1,v2, ...,vn be a basis of Rn. Then a “generic” elementv = x1v1 + x2v2 + ...+ xnvn ∈ Rn satisfies:

v · v =∑

1≤i,j≤n

(vi · vj)xixj .

This is a positive definite quadratic form over R!

For example the standard basisv1 = (1,0, ...,0),v2 = (0,1, ...,0), ...,vn = (0, ...,0,1) gives usthe quadratic form x2

1 + x22 + ...+ x2

Let v1,v2, ...,vn be a basis of Rn. Then a “generic” elementv = x1v1 + x2v2 + ...+ xnvn ∈ Rn satisfies:

v · v =∑

1≤i,j≤n

(vi · vj)xixj .

This is a positive definite quadratic form over R!

For example the standard basisv1 = (1,0, ...,0),v2 = (0,1, ...,0), ...,vn = (0, ...,0,1) gives usthe quadratic form x2

1 + x22 + ...+ x2

The following guarantees that we get quadratic forms over Z.

An integral basis is one such that vi · vj ∈ Z for each i , j .

We also want the variables in our forms to take integer valuesonly. We can make this happen by considering only the integerlinear combinations of basis vectors.

The following guarantees that we get quadratic forms over Z.

An integral basis is one such that vi · vj ∈ Z for each i , j .

We also want the variables in our forms to take integer valuesonly. We can make this happen by considering only the integerlinear combinations of basis vectors.

DefinitionLet Λ ⊆ Rn. Then Λ is a lattice if there exists a basisv1,v2, ...,vn of Rn such that:

Λ = Zv1 ⊕ Zv2 ⊕ ...⊕ Zvn

An integral lattice is one corresponding to an integral basis.

We then have the following correspondence:

TheoremThere is a one-to-one correspondence between integral latticesin Rn and positive definite quadratic forms in n variables over Z(up to equivalence of lattices and forms).

DefinitionLet Λ ⊆ Rn. Then Λ is a lattice if there exists a basisv1,v2, ...,vn of Rn such that:

Λ = Zv1 ⊕ Zv2 ⊕ ...⊕ Zvn

An integral lattice is one corresponding to an integral basis.

We then have the following correspondence:

TheoremThere is a one-to-one correspondence between integral latticesin Rn and positive definite quadratic forms in n variables over Z(up to equivalence of lattices and forms).

The Gram matrix of a lattice Λ = Zv1 ⊕ Zv2 ⊕ ...⊕ Zvn is thematrix MΛ = (ai,j), where:

ai,j =

{vi · vj if i = j

vi ·vj2 otherwise

If Λ is integral then the entries of this matrix lie in 12Z.

The Gram matrix encodes within it the quadratic form attachedto Λ. In fact the quadratic form is QΛ = xMΛxT , wherex = (x1, x2, ..., xn).

The Gram matrix of a lattice Λ = Zv1 ⊕ Zv2 ⊕ ...⊕ Zvn is thematrix MΛ = (ai,j), where:

ai,j =

{vi · vj if i = j

vi ·vj2 otherwise

If Λ is integral then the entries of this matrix lie in 12Z.

The Gram matrix encodes within it the quadratic form attachedto Λ. In fact the quadratic form is QΛ = xMΛxT , wherex = (x1, x2, ..., xn).

Outline of talk

We will not be able to see a full proof of the 290 theorem but wewill be able to see a part of Bhargava’s elegant argument.

Let Q be a non-universal positive definite quadratic form overZ. The truant of Q is defined to be the smallest positive integernot represented by Q.

Let Λ be an integral lattice with Q as its quadratic form. Thenan escalation of Λ is an integral lattice Λ′ generated by Λ and avector v such that v · v is the truant of Q.

We will not be able to see a full proof of the 290 theorem but wewill be able to see a part of Bhargava’s elegant argument.

Let Q be a non-universal positive definite quadratic form overZ. The truant of Q is defined to be the smallest positive integernot represented by Q.

Let Λ be an integral lattice with Q as its quadratic form. Thenan escalation of Λ is an integral lattice Λ′ generated by Λ and avector v such that v · v is the truant of Q.

Bhargava’s idea is to consider chains of escalators that comefrom the zero dimensional lattice {0}.

Clearly the truant of {0} is 1 so we may escalate by creatingthe lattice Λ1 = Z. This is the only possible choice.

Now QΛ1(x) = x2 and so the truant is 2. Thus we may escalateby creating a vector v ∈ R2 such that v · v = 2 and settingΛ2 = Z(1,0)⊕ Zv.However there are genuinely different lattices that can arise!

Consider the Gram matrix of such an escalation:

MΛ2 =

(1 aa 2

)We know that a = (1,0)·v

2 ∈ 12Z and also by the Cauchy-Schwarz

inequality a2 ≤ 2 and so a = 0,±12 ,±1.

Up to equivalence we get the following three possibilities (inreduced form):(

1 00 2

(1 00 1

These give quadratic forms x2 + 2y2, x2 + xy + 2y2 and x2 + y2

respectively. The truants are then 5, 3 and 3 respectively.

MΛ2 =

(1 aa 2

1 00 2

(1 00 1

MΛ2 =

(1 aa 2

1 00 2

(1 00 1

We can play the same game with each Λ2 (since each has atruant), generating escalator lattices Λ3 ⊆ R3. We find this timethat there are 34 possibilities (up to equivalence).

Each of the 34 forms has a truant and so we continue to playthe game.

A crazy thing happens once we reach four dimensions. We nowexpect some of the quadratic forms to be universal (we knowthat x2 + y2 + z2 + w2 is by Lagrange).

Indeed we get a collection of 6560 possibilities for escalators Λ4of Λ3 (up to equivalence).

Luckily 6402 of these are universal! (Non-trivial, needsmodular forms etc).153 happen to represent all but 3 or less positive integersand so become universal after at most 3 escalations!The other 5 forms are problematic but all have truant 14and came from an escalation of a Λ3 that has truant 10.Swapping the order of these truants turns out to giveescalations that fall into the second case above.

ResultThe zero dimensional lattice can be escalated at most seventimes (producing collections of lattices Λi for i = 1,2, ...,7).

So how does this prove the 290 theorem?

Facts:Every integral lattice Λ must contain one of the escalatorsΛi for some i .If Λ is non-universal then its truant is the truant of one ofthe non-universal Λi .

Since we know all truants for the Λi we know that exactly thesenumbers must be represented for Λ to be universal. This list ofnumbers matches the one in the 290 theorem.

Thanks for listening!

the 290 theorem

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