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 $i> \dim(V)/2$.)



genus 2 Siegel theta series of 3-dimensional lattices
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.


