genus 2 Siegel theta series of 3-dimensional lattices

Question

Let $(V,f)$ be a $3$-dimensional positive definite quadratic space over $\mathbf Q$.

Let $G(V)$ be a set of representatives of the isometry classes of maximal integral lattices on $V$.

To an element $L$ of $G(V)$, one associates the usual Theta series $$\Theta_L:=\sum_{x\in L} q^{2f(x)} \in \mathbf Q[[q]]\ \ \ .$$

Let $T(V)$ be the image of $L\mapsto \Theta_L$ and let $\mathbf T(V)$ be the subspace of $\mathbf Q[[q]]$ generated by $T(V)$.

One has the inequalities $$\dim(\mathbf T(V))\leq \text{card}(T(V))\leq \text{card}(G(V))\ \ \ .$$

The first inequality is often strict (maybe it can be an equality in only a finite number of cases ?). The second inequality is always an equality (Schiemann).

Let $\mathbf G(V)$ be the $\mathbf Q$-vector space generated by $G(V)$. Let $\mathbf G_0(V)$ be the kernel of the linear extension of $L\mapsto L_f$.

Let $\mathcal P$ be the set of isometry classes of positive definite quadratic $\mathbf Z$-planes. For $L$ in $G(V)$ and $P\in\mathcal P$, let us call $a(L,P)$ the number of submodules of $L$ that are isometric to $P$. Let $\mathbf P$ be the $\mathbf Q$-vector space of functions $\mathcal P\to \mathbf Q$ and let us write $$\theta^{(2)}_L:=\left[P\mapsto a(L,P)\right] \in\mathbf P\ \ \ .$$

(This is an avatar of Siegel genus 2 Theta.)

Is it possible for $L\mapsto\theta_L^{(2)}$ to have a non-trivial kernel on $\mathbf G_0(V)$ ?

(In higher dimensions, this can happen, but in all the examples I know, $\theta^{i}$ has no kernel on $\mathbf G(V)$ when $i> \dim(V)/2$.)

Answer 1

It is known, by Kitaoka's theory of characteristic sublattices, that if two lattices of rank $n$ with the same discriminant representing the same collection of lattices of rank $n - 1$, then the two lattices must be isometric. See, for example, Kitaoka's book "Arithmetic of quadratic forms", Chapter 6, Section 4.

Answer 2

The modern way to attack this problem would be through the theory of the Weil representation, as developed by Waldspurger, Howe, Rallis, etc. This would tell you that the kind of result you expect involves conditions on the existence of particular kinds of models locally, and a possible global condition.

The result for your problem would be a Siegel weight $3/2$ form, which lives on the metaplectic group. It's possible to write down formulas for the Petersson product of these forms, and evaluate them on the $O(3)$ side, but I haven't carried out this calculation.

For $O(4)$ the answer is no, as described by Ralf Schmidt in "The Saito-Kurokawa Lifting and Functoriality".

Answer 3

Edit : @WKC furnished the valuable reference Yoshiyuki Kitaoka, Arithmetic of quadratic forms, where it is proved that two $n$-dim. positive definite lattices that have the same determinant and represent the same $(n-1)$-dim lattices are in fact isometric. Kitaoka's proof is more technical that the one given below, but doesn't need such a massive hammer as the one I use at the end.

Original :

I think I finally managed to find a proof by myself. Here is how it goes :

Let $V$ be a positive definite quadratic space over $\mathbf Q$ of dimension $n$. Let $\tilde G$ be the set of maximal integral lattices on $V$. The determinant of an element of $\tilde G$ is denoted by $\Delta$.

Definition : Two elements of $\tilde G$ are said to be $k$-equivalent if they represent the same $k$-dimensional quadratic spaces.

Theorem : If two elements of $\tilde G$ are $(n-1)$-equivalent, then they are isometric.

Corollary : The map $L\mapsto\theta^{(n-1)}_L$ assumes linearly independant values on $G(V)=\tilde G/\mathbf{O}(V)$, equivalently $\theta^{(n-1)}$ is an injection on $\mathbf G(V)$.

Note : for $n=2$, the result is very well known and follows for example from the reduction theory of positive definite quadratic forms on $\mathbf R^2$. Thus we will assume $n\geq 3$.

The proof of the theorem uses the following elementary result.

Lemma : If two elements $L$ and $L'$ of $\tilde G$ are $(n-1)$-equivalent and if one of them represents a prime $p$ not dividing $2\Delta$, then they are isometric.

Proof of the lemma : if $L$ represents $p$ at a vector $v$, then we have an inclusion $<v>^\perp \perp <v> \subset L$, and we have the identity $[L:<v>^\perp \perp <v>]=p^2$ (because $L\otimes\mathbf Z_p$ is unimodular). The $(n-1)$-dimensional $K:=<v>^\perp$ has determinant $p\Delta$. If $L$ and $L'$ are $(n-1)$-equivalent, then there exists a sublattice $K'$ of $L'$ that is isomorphic to $K$. The orthogonal of $K'$ in $L'$ is a one dimensional lattice of determinant $p$, hence is isomorphic to $[p]$. Thus, without loss of generality, we can assume that $L$ and $L'$ both contain $M:=K\perp <v>$ as a submodule of index $p^2$. One then notices that $M$ is contained in exactly two elements of $\tilde G$, and that these lattices are exchanged by the automorphism of $M$ that is the identity on $K$ and sends $v$ to $-v$. $\square$

Proof of the theorem : it suffices to prove that any element $L$ of $\tilde G$ represents a prime $p> 2\Delta$. This is well known, but seems to be quite a hard theorem. The only proof I can see at the moment (but maybe there's a much simpler one):

• there exists a bound $M$ such that all integers $d\geq M$ that are representable (i.e. for whom there are no local obstruction) are represented (and this asks for a more or less weakened version of the Ramanujan conjecture - see e.g. Schulze-Pillot : Thetareihen positiv definiter quadratischer Formen, Invent. Math. 75 (1984) for the most difficult case $n=3$),

• there are infinitely many primes that are representable (and this is a consequence of Dirichlet theorem on primes in arithmetic progressions).

$\square$