Welcome to MathOverflow

Question

5

3

Suppose you have a matrix $M$ in $SL(n, \mathbb{Z}).$ Question: is there a necessary and sufficient condition for $M$ to be conjugate to $N \in Sp(n, \mathbb{Z}).$ It is clearly necessary that the characteristic polynomial of $M$ be palindromic, but I would assume that this is not sufficient.

edited Jun 2 2012 at 9:25

Goldstern
5,310●1●12●27

asked May 23 2012 at 17:50

Igor Rivin
31.2k●3●37●71

Mark Sapir · Answer 1 · 2012-05-24 00:01:11Z UTC

4

You forgot the condition that $n$ is even. I do not think there is a better criterion than the tautology.

The unipotent matrix

$$\left[ \begin {array}{cccc} 1&1&2&3\\ 0&1&4&5\\ 0&0&1&1\\0&0&0&1\end{array} \right] $$ is not conjugate to a symplectic matrix in $SL(4,\mathbb{Z})$ (although it is conjugate to a symplectic matrix in $SL(4,\mathbb{C})$.

On the other hand the matrix

$$\left[ \begin {array}{cccc} 1&0&0&1\\ 0&1&0&0\\ 0&0&1&0\\0&0&0&1\end{array} \right] $$

is conjugate to a symplectic matrix.

Both facts can be easily verified by using Maple.

answered May 24 2012 at 0:01

Mark Sapir
27.9k●2●57●120

	Why is that matrix not conjugate to a symplectic matrix? Is there an obvious obstruction? – Igor Rivin May 24 2012 at 0:03
	The problem reduces to a system of polynomial equations. That system can be simplified and gives a quadratic equation whose discriminant is negative. You can do it yourself with Maple or any other CAS. – Mark Sapir May 24 2012 at 0:28
	Ah, so you are saying that the obstruction is getting it over the reals. After that we are down to a Diophantine problem, which is kind of what I was more interested in, but it's a start... – Igor Rivin May 24 2012 at 0:50
	Yes, there are no solutions over the reals (if I did not make errors). The sequence of polynomials corresponding to the first matrix has a Groebner basis (the one found by Maple had 56 polynomials, I am not sure it is optimal, perhaps I should have chosen a different order of variables). Then the standard procedure solves it. – Mark Sapir May 24 2012 at 0:56

determining symplecticity (if that’s a word)

Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

1 Answer

You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.