**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).

$\square$