Matrix version of number theoretic integral lattice claim

Question

I know this not, everybody else seems to. This is from page 215 of Cassels, Rational Quadratic Forms, formula 4.1, or SPLAG, page 389 formula (36). quote:

If $f$ and $g$ are forms of determinant $d$ in the same genus, then they are rationally equivalent by some transformation whose denominator is prime to $2d.$ Hence we can find corresponding lattices $L,M$ for which $$ [ L : L \cap M ] = [ M : L \cap M] > = r, $$ say, for some number $r$ which is prime to $2d.$

We are talking about Siegel's definition of forms being in the same genus if they are rationally equivalent without essential denominator.

So, here is an example in matrix slang. Given quadratic forms with symmetric matrices $$ F \; = \; \left( \begin{array}{cccc} 2 & 1 & 0 & 1 \\\ 1 & 2 & 0 & 0 \\\ 0 & 0 & 2 & 1 \\\ 1 & 0 & 1 & 6 \end{array} \right) $$ and $$ G \; = \; \left( \begin{array}{cccc} 2 & 1 & 1 & 0 \\\ 1 & 2 & 0 & 1 \\\ 1 & 0 & 2 & 0 \\\ 0 & 1 & 0 & 8 \end{array} \right) $$ Next, I take $r=3,$ because the discriminant is $29,$ and find $$ P \; = \; \left( \begin{array}{cccc} 1 & 1 & 1 & 7 \\\ 1 & -2 & 1 & -5 \\\ 1 & 1 & -2 & 1 \\\ 1 & 1 & 1 & -2 \end{array} \right) $$ that satisfies $$ P^T \; F \; P \; = \; 9 \; G = r^2 \; G. $$

Now, here is the trick. To get back from $G$ to $F,$ it appears that one needs to use $Q = \; \mbox{adj} \; P$ which has a much larger determinant, so things look asymmetric. Indeed, $$ Q \; = \; \; \mbox{adj} \; P \; = \; \left( \begin{array}{cccc} -18 & -27 & -27 & -9 \\\ 9 & 27 & 0 & -36 \\\ -9 & 0 & 27 & -18 \\\ -9 & 0 & 0 & 9 \end{array} \right) $$ However, a miracle! The GCD of these entries is 9, and we get the improved $$ Q_1 \; = \; \left( \begin{array}{cccc} -2 & -3 & -3 & -1 \\\ 1 & 3 & 0 & -4 \\\ -1 & 0 & 3 & -2 \\\ -1 & 0 & 0 & 1 \end{array} \right). $$ The we really do get what we wanted, $$ P Q_1 = -9 I = \pm r^2 I $$ and $$ Q_1^T \; G \; Q_1 \; = \; 9 \; F = \; r^2 F. $$

Alright, so here is the question, with the dimension $n$ thrown in: Suppose $F,G$ are symmetric positive definite matrices of integers with the same determinant $d.$ Suppose we have an integer $r$ with $\gcd (r, 2 d) = 1.$ Suppose that we have a matrix $P$ of integers, with $\det P = \pm r^n,$ such that $ P^T \; F \; P \; = \; r^2 \; G. $ Take $Q = \; \mbox{adj} \; P,$ so that $\det Q = \pm r^{n^2 - n}$ and $PQ = QP = (\det P) I = \pm r^n I.$ Is it always the case that $$ \gcd Q = r^{n-2} ?$$

I think this is progress. Since 1994, it is only 18 years that I have been completely confused on this point and unaware that I was confused.

EDIT: The matrix $P$ is not necessarily rank $1 \pmod r.$ Here are the two forms by Schiemann of discriminant 1729, not equivalent but in the same genus and, wait for it, the same theta series.

$$ F \; = \; \left( \begin{array}{cccc} 4 & 1 & 0 & 1 \\\ 1 & 8 & 1 & 3 \\\ 0 & 1 & 8 & 4 \\\ 1 & 3 & 4 & 10 \end{array} \right) $$ and $$ G \; = \; \left( \begin{array}{cccc} 4 & 2 & 1 & 1 \\\ 2 & 8 & -2 & 1 \\\ 1 & -2 & 10 & 5 \\\ 1 & 1 & 5 & 10 \end{array} \right) $$ Next, I take $r=5,$ and find $$ P \; = \; \left( \begin{array}{cccc} -1 & 6 & -4 & 0 \\\ -2 & -3 & -3 & 0 \\\ 3 & 2 & -3 & 0 \\\ -1 & -1 & 0 & -5 \end{array} \right) $$ that satisfies $$ P^T \; F \; P \; = \; 25 \; G = r^2 \; G. $$ We still wind up, in the same manner, with a very pleasant $$ Q_1 \; = \; \left( \begin{array}{cccc} -3 & -2 & 6 & 0 \\\ 3 & -3 & -1 & 0 \\\ -1 & -4 & -3 & 0 \\\ 0 & 1 & -1 & -5 \end{array} \right). $$

Answer 1

Write $P$ in Smith normal form as $SDT$, where $S$ and $T$ have determinant $\pm 1$ and $D$ is diagonal with diagonal entries $d_1 | d_2 | d_3 | \cdots | d_n$. We may replace $(F, G, P)$ by $(S^{-T} F S^{-1}, T^{T} G T, D)$ and therefore assume $P$ is diagonal with entries $d_i$ as stated. Note that this replacement preserves all the hypotheses and the desired conclusion.

Claim 1: We have $r^2 | d_k d_{n+1-k}$ for all $1 \leq k \leq n$.

Proof: Suppose, to the contrary, that $p$ is a prime dividing $r^2$ more times than it divides $d_k d_{n+1-k}$. The condition $D^T F D \equiv 0 \mod r^2$ means $d_i d_j F_{ij} \equiv 0 \mod r^2$ for all $(i,j)$. So, for all $i \leq k$ and $j \leq n+1-k$, we have $F_{ij} \equiv 0 \mod p$. So the first $k$ rows of $F$ span a $\leq k-1$ dimensional space modulo $p$, contradicting that $p$ does not divide $\det F$. $\square$

Claim 1 is used only to prove the stronger:

Claim 2: For all $1 \leq k \leq n$, we have $d_k d_{n+1-k} = r^2$.

Proof: Since $D^T F D = r^2 G$ and $\det F= \det G$, we deduce that $\det D = r^n$. So $\prod_{k=1}^n d_k = r^n$ and $\prod (d_k d_{n+1-k}/ r^2) = 1$. By the previous claim, all of the factors in this product are positive integers, so they must all be $1$. $\square$

In particular, every $d_k | r^2$. This means that every principal minor $\det D/d_k$ must be divisible by $r^n/r^2 = r^{n-2}$, and the nonprincipal minors are zero, giving the desired conclusion. More conceptually, if we define $D'$ to be $D$ with the elements in the opposite order, then $D D' = r^2$. Conjugating back to the original basis, this $D'$ will turn into the matrix you called $Q_1$.

Answer 2

A much shorter proof, though one that doesn't explain as much to me. From $P^T F P = r^2 G$, we deduce $P^{-1} = r^{-2} G^{-1} P^T F$ and thus $P^{-1} \in \frac{1}{r^2 \det G} \mathrm{Mat}(\mathbb{Z})$. But also $P^{-1} = \frac{1}{\det P} \mathrm{Adj}(P)$ so $P^{-1} \in \frac{1}{r^n} \mathrm{Mat}(\mathbb{Z})$. Since $GCD(r^2 \det G, r^n) = r^2$, we have $P^{-1} \in \frac{1}{r^2} \mathrm{Mat}(\mathbb{Z})$, as desired.