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It is classically known that every positive integer is a sum of at most four squares of integers, i.e. every sum of squares of integers is a sum of four squares of integers. Now consider a symmetric $n\times n$ matrix $M$ with integer entries which can be written as $M= Q^{\rm T} Q$ for an $m \times n$ matrix $Q$ with integer entries. Can we bound $m$ in some way? I.e. is there a constant $c(n)\in \mathbb{Z}$ depending only on $n$ such that in this situation we can always find a decomposition $M= \widetilde{Q}^{\rm T} \widetilde{Q}$ for a $c(n) \times n$ matrix $\widetilde{Q}$ with integer entries? What is the smallest possible such constant (perhaps even linear growth?)?

If we look at the same situation over the polynomial ring $\mathbb{R}[T]$ in one variable, we have the following: Every sum of squares in $\mathbb{R}[T]$ is a sum of two squares and one can show that $c(n)$ grows linearly in $n$: $c(n)=2n$. Thus, I hope that a similar result might hold over the integers too. Perhaps $c(n)$ even grows linearly in $n$?

Is someone aware of something in that direction?

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  • $\begingroup$ As an alternative to my answer below, which doesn't seem very efficient, there might be some Siegel Theta argument, but I know almost nothing about that and my frenetic googling gave nothing. So consider Siegel's $\theta^{(m)}$ for $\mathrm{I}_n$. It is the sum of an Eisenstein (or linear comb. of such) series and a cuspidal one. If you have enough inequalities on their Fourier coeffs to ensure that, say, for $n>2m$, the Eisenstein coefs are very big while the cusp. coefs are relatively small, then you are done. Some MO experts might tell you if this is realistic or not. $\endgroup$ – few_reps Oct 13 '15 at 19:47
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Here is an answer (see the last point). It differs from what I had been claiming in my first post. There I was saying that any positive bilinear module $\Lambda$ over $\mathbf Z[\frac 12]$ was representable by some euclidean module $\mathrm{I}_n\otimes\mathbf Z[\frac 12]$. This is true (with $n\leq \text{rk}(\Lambda)+3$), but is of no use in our situation.

Nevertheless the same technics show that any positive bilinear $m$-dimensional module over $\mathbf Z$ is representable by somebody in the same genus than the euclidean module of rank $m+4$. This is proved here in the third point, since it might be of interest.

In terms of matrices, as in the OP, it says that for any symmetric positive definite matrix $M\in\mathrm{Sym}_m(\mathbf Z)$ there exists a symmetric positive definite matrix $G\in\mathrm{Sym}_{m+4}(\mathbf Z)$ with determinant $1$ and at least one odd diagonal entry, and a matrix $Q\in \mathrm{Mat}_{m+4,m}(\mathbf Z)$ such that the following holds : $$Q^t.G.Q=M\ \ \ .$$

Let $M$ be a positive definite symmetric $m\times m$ matrix. Let $\mathrm{M}$ be the bilinear module $(\mathbf{Z}^m,M)$. The question can be rephrased :

What is the smallest value $n$ such that the standard euclidean bilinear module $\mathrm{I}_n$ represents $\mathrm{M}$.

$\blacktriangleright$ First note that in general, such an $n$ doesn't exist. Indeed, a positive definite integral lattice decomposes as the orthogonal sum of indecomposable (for the orthogonal direct sum) lattices in a unique manner. In particular, the only unimodular (i.e. $\det(M)=1$) lattice $\mathrm{M}$ represented by $\mathrm{I}_n$ are isomorphic to $\mathrm{I}_m$.

$\blacktriangleright$ Over $\mathbf Q$, the space $\mathrm{M}\otimes\mathbf Q$ is represented by $\mathrm{I}_{n}\otimes\mathbf Q$ for some $n\leq m+3$. This follows from the Hasse principle (that implies that a rank $4$ positive definite space represents $1$), Witt cancellation, and the fact that, for example, $(\mathrm{M}\otimes\mathbf Q)^{\perp 4}$ is euclidean, as a computation of Hasse-Minkowski symbols shows.

$\blacktriangleright$ It follows that over $\mathbf Z$, any bilinear module $\mathrm{M}$ is represented by some bilinear module $\mathrm{N}$ lying in the genus of $\mathrm{I}_{n}$, for some $n\leq m+4$ (the addition of a one dimensional module $\mathrm{I}_1$ might be necessary when $m+3$ is a multiple of $8$, in order to ensure that $\mathrm{N}$ is odd).

$\blacktriangleright$ Finally, here is an answer to the question. Let $M$ be an $m$-dimensional submodule of $\mathrm{I}_n$. Let $P$ be its orthogonal. Let $\pi : \mathrm{I}_n\otimes Q\to M\otimes Q$ be the orthogonal projection. Then $\pi(\mathrm{I}_n)$ is contained in $M^\sharp$ (the dual lattice of $M$ in $M\otimes \mathbf Q$), and there exists a smallest $d$ such that $M^\sharp\subset d^{-1}M$. Let $S$ be the sphere of unitary vectors in $\mathrm{I}_n$. Then the vectors of $d.\pi(S)\subset M$ have squared length smaller than $d^2$. If the number of such vectors is smaller than $\mathrm{card}(S)=2^n$, this means that a vector of squared length $2$ in $I_n$ lies in $P$, so $\mathrm{M}$ is in fact represented by $\mathrm{I}_{n-2}\perp <2>$. If once again the same phenomenon occurs, then $\mathrm{M}$ is in fact represented by $\mathrm{I}_{n-4}\perp <2,2>$ which is represented by $\mathrm{I}_{n-2}$. So what we get is :

Let $d$ be the smallest integer such that $\mathrm{M}^\sharp\subset \frac 1d\mathrm{M}$. Let $B_d(\mathrm{M})$ denote the number of non-trivial vectors of $M$ in the ball of radius $d$. Let $n_0$ be the floor of $2+\log_2(B_d(\mathrm{M}))$. If $\mathrm{M}$ is represented by $\mathrm{I}_n$ for some $n$, then $\mathrm{M}$ is represented by $\mathrm{I}_{n_0}$.

I hope there are better bounds, but I cannot see how to get them.

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  • $\begingroup$ Thanks, that sounds great. Can you give me some references for the two statements that you use? $\endgroup$ – Hans Oct 13 '15 at 10:04
  • $\begingroup$ Everything is in O'Meara's Introduction to quadratic forms, but you might as well be content with Cassel's Rational quadratic forms. $\endgroup$ – few_reps Oct 13 '15 at 10:18
  • $\begingroup$ Do you mean that the margin is too narrow for your beautiful prof ? $\endgroup$ – Denis Serre Oct 13 '15 at 11:21
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    $\begingroup$ @DenisSerre He did promise to write up the proof tomorrow, so I don't think you should be snarky yet... $\endgroup$ – Igor Rivin Oct 13 '15 at 11:45
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    $\begingroup$ @Denis, lack of time, not space – Galois, not Fermat. $\endgroup$ – Gerry Myerson Oct 13 '15 at 12:04
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In Maria Icaza's Ohio State PhD thesis (1992?) she established the following result. For any positive integer $n$, let $S(n)$ be the set of $\mathbb Z$-lattices (or integral quadratic forms, if one prefers polynomials) of rank $n$ that can be represented by some sums of squares. Then there exists an integer $g(n)$ such that all lattices in $S(n)$ are represented by $g(n)$ number of squares. She also obtained an explicit upper bound on $g(n)$, which is certainly not optimal. All of these were published later in a couple of papers which you can find in MathSciNet.

Precise values of $g(n)$ are known up to $n = 6$. For $n \leq 5$, $g(n) = n + 3$. However, $g(6) = 10$ which was proved by M-H Kim and B-K Oh in the 90's (in Journal of Number Theory). Later they obtained much better upper bounds for $g(n)$ in a series of papers.

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  • $\begingroup$ In the first paragraph you mean $S(n)$ is the set of $n$-dimensional lattices that are represented by some euclidean $\mathbf Z^m$ (and the result is that there exists $g(n)$ such that any element of $S(n)$ is represented by the euclidean $\mathbf Z^{g(n)}$, right ? $\endgroup$ – few_reps Oct 14 '15 at 5:30
  • $\begingroup$ @few_reps Yes, you are right. Thanks. $\endgroup$ – WKC Oct 14 '15 at 5:33
  • $\begingroup$ After looking at mathscinet notices (not at the papers themselves), Icaza's result seems to be the only general theorem ... the bound she obtains has the form $g(n)\leq 8+\sum_{k=1}^n c_k$ where $c_k$ is the (highly explosive) number of isometry classes of $k$-dim. unimodular lattices. All other results seem to find better bounds for small values of $n$. $\endgroup$ – few_reps Oct 14 '15 at 5:43
  • $\begingroup$ Kim and Oh (2005?) give an explicit upper bound for $g(n)$ for all sufficiently large $n$, which is more or less exponential in $n$. $\endgroup$ – WKC Oct 14 '15 at 6:03
  • $\begingroup$ Right ! Strangely, the review of this paper doesn't speak about Icaza's result, hence did not appear when I tried with "Icaza" in the field "anywhere" ! Their result says that $g(n)=O(3^{\frac n2}\log(n))$ which is quite better than Icaza's bound. $\endgroup$ – few_reps Oct 14 '15 at 6:27

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