Is there an approach to understanding solution counts to quadratic forms that doesn't involve modular forms?

Question

Given a quadratic form Q in k variables, there is an associated theta series $$\theta_Q(z) = \sum_{x\in \mathbb{Z}^k}q^{Q(x)}$$ where $q = e^{2\pi i z}$ which is a modular form of weight $k/2$. Thus the coefficient on $q^n$ counts the number of solutions in $\mathbb{Z}^k$ to $Q(x) = n$; denote this quantity $r_Q(n)$. The level of these modular forms is related to the level of their corresponding quadratic forms. Since these modular forms live inside finite-dimensional vector spaces, one question to ask is whether there are any linear relationships among them. At least among the ternary forms for which I've looked at this computationally, there often is a great deal of linear dependence among forms of this type that live in the same space, and so there are a lot of linear relationships among the $r_Q(n)$ for various $Q$.

My question is if there are "other reasons" (i.e. not related to modular considerations) to expect that the $r_Q(n)$ for various $Q$ should be related in this fashion. Is there a reason a priori to expect that these theta series should be heavily linearly-dependent on one another? Is there a more combinatorial approach (or an alternative number-theoretic approach) to counting solutions to quadratic forms that could suggest the form that these relationships might take? I'm asking because in the course of my computations with ternary forms I observed that twisting forms by quadratic characters always (at least for all the examples I computed) gave a form that was linearly dependent on untwisted forms that lived in the same space. Is there a reason to suspect why this should always be the case?

Answer 1

Hi There, It would help if you gave me some examples of actual positive ternary forms with specific linear dependencies. The main source of dependencies is the Siegel representation formula, which calculates the weighted average of representation numbers in terms of a product of ``local densities.'' In practice, what this means is that by restricting the target number n to some appropriate arithmetic progression and relating representations by two genera one may get an explicit linear dependence among representation counts. Very much in this spirit is the viewpoint of Jones, given for example in a 1999 paper by Ono and Soundararajan called "Integers Represented by Ternary Quadratic Forms" where they point out that the number of essentially distinct representations of an (eligible) number $N$ by $ x^2 + y^2 + 10 z^2$ is just $ h( -40 N) / 4 $ when $N$ and 10 are coprime. However, the main thing that would surprise me is linear dependence for all n among primitive forms. For instance, Schiemann showed that no two positive ternary forms (inequivalent) have the same theta series.

I'm not sure this facility allows chats back and forth, if you want to try emailing me in person get my address from http://www.ams.org/cml, and for that matter google me as "Will Jagy" in double quotes.

William C. Jagy

JULY: follow-up email sent. I posted something here, JSE felt it might be too revealing. I did not think so, but there is little harm in deleting it and sending you email instead.

Answer 2

The best short reference for this is Lehman Math Comp 1992 PDF. A more elaborate discussion is in ME and ALEX which appeared in the Journal of Number Theory, January 2012, volume 132, number 1, pages 258-275. I attempted to include this material in a final section of the JNT paper, that did not work out, see META

A (ternary) positive quadratic form is indicated by $\langle a,b,c,r,s,t \rangle$ which refers to $$ f(x,y,z) = a x^2 + b y^2 + c z^2 + r y z + s z x + t x y. $$ The discriminant is $$ \Delta = 4 a b c + r s t - a r^2 - b s^2 - c t^2. $$ Forms are gathered together into genera when they are equivalent locally. A fundamental result of Siegel is that we may calculate the weighted average of representations, over a genus, of a given target number. Siegel's result relates quadratic forms and modular forms.

We have genera labelled $G_1, G_2, G_3.$ Given an odd prime $p,$ define useful integers $u,v$ such that $$ (-u | p) = -1, \; \; \; (-v | p) = +1. $$ The first one, $G_1,$ is the only genus of discriminant $p^2.$ Then we have two of the six genera of discriminant $4 p^2,$ these have level $4 p$ and are classically integral. Together $$ \begin{array}{lccccc} \mbox{Genus} & \Delta & \mbox{Level} & \mbox{2-adic} & \mbox{$p$-adic} & \mbox{Mass} \\\ G_1 & p^2 & 4 p & y z - x^2 & u x^2 + p(y^2 + u z^2) & (p-1)/48 \\\ G_2 & 4p^2 & 4 p & 2 y z - x^2 & u x^2 + p(y^2 + u z^2) & (p-1)/32 \\\ G_3 & 4p^2 & 4 p & x^2 + y^2 + z^2 & v x^2 + p(y^2 + v z^2) & (p+1)/96 \end{array} $$ Note that $G_1$ and $G_2$ represent exactly the same numbers, but with different representation measures. Furthermore, when $p \equiv 3 \pmod 4,$ then $h(x,y,z) = x^2 + p y^2 + p z^2 \in G_2,$ but when $p \equiv 1 \pmod 4,$ then $h(x,y,z) = x^2 + p y^2 + p z^2 \in G_3.$

Let $s(n)$ be the number of representations of $n$ as the sum of three squares. Then, taking one form $g$ per equivalence class in the specified genus, let $$ R_j(n) = \sum_{g \in G_j} \frac{r_g(n)}{|\mbox{Aut} g|} .$$

The two new identities are $$ s(p^2 n) \; - \; p s(n) \; = \; 96 \; R_1(n)\; - \; 96 \; R_2(n), $$

$$ (p+2) \; s( n) \; - \; s(p^2 n) \; = \; 96 \; R_3(n) .$$