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.
Take the 2minute tour
×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.
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. 

