Diagonalization of a quadratic form in integers

2012-08-09T10:30:24Z

Hello,

Recently I've been studying the problem of quadratic form diagonalization. Suppose that we have a form $F(x,y,z)$ with corresponding symmetric matrix $M$. This form is equivalent to another form $F'(x',y',z')$ which is in Smith normal form. The matrix that corresponds to $F'$ is $M'$ and it is diagonal. Note that both $F$ and $F'$ have integral coefficients. It is known that there exist two matrices $A$ and $B$ (which coefficients are integers as well) such that $M' = AMB$.

Now, do I understand correctly that $F$ and $F'$ represent exactly the same set of integers? If so, am I right that there must exist a change of variables of the form

$$ x = a_{11}x' + a_{12}y' + a_{13} z' $$ $$ y = a_{21}x' + a_{22}y' + a_{23} z' $$ $$ z = a_{31}x' + a_{32}y' + a_{33} z' $$

that transforms $F$ into $F'$? Finally, is it possible to deduce the matrix $(a_{ij})$ given $M'$, $M$, $A$ and $B$?

P.S. Here's an example. I have the following matrix $M$ of determinant 1:

$$ 5, 13, 1 $$

$$ 13, 34, 0 $$

$$ 1, 0, 35 $$

It corresponds to $M' = I$ where $I$ is an identity matrix. The Sage command M.smith_form() produced the following $A$:

$$ 0, 0, 1 $$ $$ 0, 1, 0 $$

$$ 1, 0, 0 $$

and $B$:

$$ -34, -455, 1190 $$ $$ 13, 174, -455 $$ $$ 1, 13, -34 $$

Now I need to find $(a_{ij})$.

Answer by Will Jagy for Diagonalization of a quadratic form in integers

2012-08-10T02:38:41Z

The actual correct manipulation is $$
\left( \begin{array}{rrr} 3 & -1 & 0 \\ -5 & 2 & 0 \\ -34 & 13 & 1
\end{array} \right) \cdot \left( \begin{array}{rrr} 5 & 13 & 1 \\ 13 & 34 & 0 \\ 1 & 0 & 35
\end{array} \right) \cdot \left( \begin{array}{rrr} 3 & -5 & -34 \\ -1 & 2 & 13 \\ 0 & 0 & 1
\end{array} \right) = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1
\end{array} \right) $$

Taking the sign for transpose as an apostrophe, we evaluate a quadratic form with symmetric matrix $A$ at a column vector $x$ as $$ x' A x. $$ We change from one symmetric matrix to another by taking a matrix $P$ of determinant $1$ and finding $P' A P,$ as I do above. This is called an equivalence. Typically, for dimension $3$ or higher, most authors allow $\det P = \pm 1.$

There is no guarantee that a for diagonalizes over the integers. There is also no guarantee of diagonalization over $\mathbb Q,$ as there may be the necessity for a few terms of type $xy$ or $x^2 + xy+y^2$ in $\mathbb Z_2,$ the $2$-adic integers.

I'm not entirely sure what to recommend, but Rational Quadratic Forms by Cassels is an inexpensive Dover reprint.

Diagonalization of a quadratic form in integers - MathOverflow

Diagonalization of a quadratic form in integers

Answer by Will Jagy for Diagonalization of a quadratic form in integers