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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

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

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 $J = \left(\begin{array}{cccc} 0 & 0 & 1 & 0\ 0\newline 0 & 0 & 0 & 1\ 1\newline -1 & 0 & 0 & 0\ 0\newline 0 & -1 & 0 & 0 \end{array}\right)$, 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?

1

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

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

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 $J = \left(\begin{array}{cccc} 0 & 0 & 1 & 0\ 0 & 0 & 0 & 1\ -1 & 0 & 0 & 0\ 0 & -1 & 0 & 0 \end{array}\right)$, 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?