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.
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asked May 23, 2012 at 17:50
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1 Answer
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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.
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$\begingroup$ Why is that matrix not conjugate to a symplectic matrix? Is there an obvious obstruction? $\endgroup$ Commented May 24, 2012 at 0:03
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$\begingroup$ 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. $\endgroup$– user6976Commented May 24, 2012 at 0:28
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$\begingroup$ 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... $\endgroup$ Commented May 24, 2012 at 0:50
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$\begingroup$ 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. $\endgroup$– user6976Commented May 24, 2012 at 0:56