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I have a couple questions regarding symplectic matrices, specifically those with integer entries.

1) Suppose we are trying to find symplectic matrices of dimension $4$ with integer entries. We know that eigenvalues come in reciprocal pairs; we are assuming that the symplectic matrix $A$ has irreducible characteristic polynomial so the eigenvalues are distinct. Also we want the eigenvalues real. Thus, it is necessary for the characteristic polynomial of $A$ to be of the form

$$f = x^4 - nx^3 + (2+m)x^2 -nx+1$$

where $n,m \in \mathbb{Z}$. The question is, if we have an irreducible $f$ of this form, with roots $\lambda, \lambda ^{-1}, \mu, \mu^{-1}$, is it possible to construct an integer matrix which is symplectic with this characteristic polynomial?

2) Suppose the construction in 1) is possible, and we have matrix A. Are there any properties on matrices in the conjugacy class of A over $\mathbb{Q}$ which ensure that those matrices are symplectic? That is, is there a way to find matrices conjugate to a symplectic matrix A which are also symplectic?

Thanks for any help or any suggestions on literature to read.

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The answer to the first question is "yes", see p 21 of

On that same page, if you look at conditions (6-8), that should answer your second question with a little work...

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Thank you for the reference; that is very helpful. – CCALNS Jul 23 '11 at 18:38
Thanks for the help - I still had a few questions, see below. – CCALNS Jul 24 '11 at 21:18
For some value of "below". – Gerry Myerson Jul 25 '11 at 5:49

I'm having some trouble constructing a 4 by 4 symplectic matrix with characteristic polynomial $f = x^4 - nx^3 + (2+m)x^2 -nx+1$ and I haven't been able to find where I'm going wrong. We have a matrix

$$\left(\begin{array}{cccc} 0 & 0 & 1 & \mathrm{b2}\newline 0 & 0 & \mathrm{b2} & \mathrm{b3}\newline -\frac{\mathrm{b3}}{\mathrm{b3} - {\mathrm{b2}}^2} & \frac{\mathrm{b2}}{\mathrm{b3} - {\mathrm{b2}}^2} & -1 & \mathrm{f1} + 1\newline \frac{\mathrm{b2}}{\mathrm{b3} - {\mathrm{b2}}^2} & -\frac{1}{\mathrm{b3} - {\mathrm{b2}}^2} & 1 & \mathrm{f2} \end{array}\right) $$

as constructed in the paper. We can compute the characteristic polynomial and obtain the conditions on $f1, f2$ that $f1 = -n-m-2$, $f2 = n+1$. Then by requiring that $M^TJM = J$, where $J = \left(\begin{array}{cccc} 0 & 0 & 1 & 0\newline 0 & 0 & 0 & 1\newline -1 & 0 & 0 & 0\newline 0 & -1 & 0 & 0 \end{array}\right)$, we have that $-1 - m - n + b2\cdot (n+2) = b3$. Then plugging in for $b3$ in $M$ and using the fact that $\det B = 1$ we find $b2^2 - b2\cdot (2 + n) + m + n =0$. We want $b2$ to be an integer but I can't figure out how to get this to always be an integer for any arbitrary $n,m$.

Also, I was wondering for the second part of the question, the above comment suggested that we look at the block form of the matrices. However, when we conjugate, we would need to get the inverse of a matrix in block form, so that looking at the matrix in block form does not seem to give an additional advantage. Are there any other conditions for matrices conjugate to a given symplectic matrix to be symplectic?

For example, given a symplectic matrix $A$ where $B$ and $C$ are symplectic matrices conjugate to $A$ over $\mathbb{Z}$ (or $\mathbb{Q}$) thus $B = S^{-1} C S$, must $S^{-1} A S$ be symplectic? That is, if $S$ conjugates one symplectic matrix to another symplectic matrix in the conjugacy class of $A$ (over $\mathbb{Z}$ or $\mathbb{Q}$), will $S$ also conjugate $A$ to a symplectic matrix?

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