# Matrix version of number theoretic integral lattice claim

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 $$Q_1 \; = \; \left( \begin{array}{cccc} -3 & -2 & 6 & 0 \\\ 3 & -3 & -1 & 0 \\\ -1 & -4 & -3 & 0 \\\ 0 & 1 & -1 & -5 \end{array} \right).$$

• Here is an idea where I haven't filled in the details: Take a look at mathoverflow.net/questions/97151/… . You want to know when a matrix $P$ will have the property that all $(n-2) \times (n-2)$ minors are divisible by $r^{n-2}$. I point out that one good explanation is if $P \mod r$ has rank $1$. That's true in your example; is it always true? – David E Speyer May 30 '12 at 20:55
• @David, thanks, I will look. I think it is all $(n-1) \times (n-1)$ minors. The rank 1 thing may be true, I need to think. I have little objection to $r$ being prime for the moment, it may ease things. – Will Jagy May 30 '12 at 21:04
• Whoops, $(n-1) \times (n-1)$ minors is right, thanks. – David E Speyer May 30 '12 at 22:32
• @David, rank 1 is too strong a statement, I edited in Schiemann's quaternary forms. We still get the nice condition on the adjoint, though. This is all Watson's sort of thinking, I cannot seem to find quite this. And, it may have been known to Minkowski. I wasn't there at the time. – Will Jagy May 30 '12 at 23:44
• Gotcha. I'll think a bit about this. – David E Speyer May 31 '12 at 2:00

## 2 Answers

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 $Q_1$.

• Printed out to study with my little brain. You know, if nobody has told you, you are really quite good at this, you ought to consider a career in mathematics. – Will Jagy May 31 '12 at 3:30
• Please excuse Will, sometimes he is slow on the uptake. Only the other day I was reading a post of his about an idea on propulsion he was considering: the selfmobile. Gerhard "Not Always Faster Than Will" Paseman, 2012.05.30 – Gerhard Paseman May 31 '12 at 4:07
• @Gerhard, now I must leave it in place. I know where David works. Right, I get a little typing difference with inverses or not for $S,T$ but no matter. That is a new one on me, if we have a square matrix over a field with a rectangular $a \times b$ block of nothing but $0'$s, and $a + b \geq n+1,$ the the matrix is singular. Very nice. – Will Jagy May 31 '12 at 4:17
• @Gerhard, I still think the selfmobile has a future, I just need some venture capital. – Will Jagy May 31 '12 at 4:22
• Will, the observation two comments up can be understood in terms of Hall's Marriage Theorem. – Gerry Myerson May 31 '12 at 7:05

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.

• Thanks again, David. I think I am getting somewhere. If the rank of $P \pmod r$ is $w,$ then i get that $[ L : L \cap M] = r^{n-w}$ as long as $r$ is prime, anyway. – Will Jagy May 31 '12 at 16:51