Canonical form of symmetric integer matrix M

Question

Let $M$, $N$ be a symmetric matrix over a ring $R$. $M$ and $N$ are said to be equivalent if there exist an invertible matrix $U$ (over the same ring $R$) such that $N=U M U^T$ ($U^T$ is the transpose of $U$). A question is that what is the simple canonical form of $M$ under such an equivalent relation.

We know that when $R$ is the ring of real numbers, every real symmetric matrix is equivalent to an diagonal matrix with diagonal entries being 1, -1, or 0.

When $R$ is the ring of integers, do we have a similar result?

If there is no nice results, we may assume $M$ to satisfy additional conditions:

(a) $|\det(M)|=1$

(b) There exist a $J$ such that $J^2=1$ and $JMJ^T=-M$.

Thanks!

Edit: I am also interested in finding the simple canonical form of integer symmetric matrices $M$, that satisfy

(a) $|\det(M)|=1$

(b) There exist a $J$ such that $J^2=-1$ and $JMJ^T=-M$.

Answer 1

It's all in the correct reference. Cassels, Rational Quadratic Forms, chapter 9 "Integral Forms over the Rational Integers," pages 163-164, Examples 9-11. Example 11(i) says that, for "odd" matrices, we can cut down the dimension by 2 and write $y_1^2 - y_2^2 + g(z_1, \ldots , z_{n-2}).$ The determinant of $g$ is still $\pm 1,$ so the only problem is that $g$ may be "even."

Next, if $f$ is "even" the quadratic form can, in fact, be written $ 2y_1 y_2 + g(z_1, \ldots , z_{n-2}).$

So, all we really need is to show, as in Sylvester's Law of Inertia, that the resulting form $g$ continues to be indefinite. Presumably your condition with $J M J^T = -M$ can do this.

Otherwise, without your $J$ condition, Example 11(vi) says that either $f$ or $-f,$ if "even," is equivalent to a sum of some $2x_j y_j$ terms along with a single $\mathbb E_8$ lattice. CASSELS

I was uneasy about the possible need to mix 2 by 2 blocks of both types, despite Hahn's statement, but $$ \left( \begin{array}{cccc} 3 & 4 & 2 & 2 \\\ 2 & 3 & 1 & 2 \\\ 0 & 1 & 1 & 1 \\\ 2 & 3 & 2 & 1 \end{array} \right) \left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\\ 0 & -1 & 0 & 0 \\\ 0 & 0 & 0 & 1 \\\ 0 & 0 & 1 & 0 \end{array} \right) \left( \begin{array}{cccc} 3 & 2 & 0 & 2 \\\ 4 & 3 & 1 & 3 \\\ 2 & 1 & 1 & 2 \\\ 2 & 2 & 1 & 1 \end{array} \right) = \left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\\ 0 & -1 & 0 & 0 \\\ 0 & 0 & 1 & 0 \\\ 0 & 0 & 0 & -1 \end{array} \right) $$

Answer 2

For symplectic unimodular symmetric (or skew) matrices, such a result is shown in Zarrow, Robert A canonical form for symmetric and skew-symmetric extended symplectic modular matrices with applications to Riemann surface theory. Trans. Amer. Math. Soc. 204 (1975), 207–227.

You might be able to extend it to the nonsymplectic case (though I am a bit skeptical).

Answer 3

I don't see how to diagonalize the quadratic form $2xy.$ As far as your conditions, we have $$ M = \left( \begin{array}{cc} 0 & 1 \\\ 1 & 0 \end{array} \right) $$ and $$ J = \left( \begin{array}{cc} 1 & 0 \\\ 0 & -1 \end{array} \right), $$ with $$ \left( \begin{array}{cc} 1 & 0 \\\ 0 & -1 \end{array} \right) \left( \begin{array}{cc} 0 & 1 \\\ 1 & 0 \end{array} \right) \left( \begin{array}{cc} 1 & 0 \\\ 0 & -1 \end{array} \right) = \left( \begin{array}{cc} 0 & -1 \\\ -1 & 0 \end{array} \right) $$ However, the only diagonal matrices with determinant $-1$ are $\pm J,$ which does not work as $x^2 - y^2$ does not represent any numbers congruent to $2 \pmod 4.$

Let's see, Conway and Sloane refer to Watson for his 2-adic canonical form for their work on the Mass Formula, so I can recommend the book Integral Quadratic Forms by George Leo Watson. More recent, SPLAG, which is Sphere Packings, Lattices, and Groups by Conway and Sloane.