Conjugacy classes of symplectic group $\mathrm{Sp}(4,q)$

Question

$\DeclareMathOperator\Sp{Sp}$I was reading the famous paper of Bhama Srinivasan "The characters of the finite symplectic group $\Sp(4,q)$". An AMS link for the paper is here. $\Sp(4,q)$ is the group of all invertible $4\times 4$ matrices $X$ over $F_q$ satisfying $XAX^t=A,$ where

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

On pages 489-491, the author mentions conjugacy classes of the group $\Sp(4,q).$ In the second last row on page 490, the author mentions the representative of a conjugacy class denoted by $B_9(i)$ as the following matrix

$B_9(i)=\left( \begin{array}{cccc} \gamma^i & 0 & 1 & 0 \\ 0 & \gamma^{-i} & 0 & 1 \\ 0 & 0 & \gamma^i & 0\\ 0& 0& 0& \gamma^{-i} \end{array} \right)$

where $i\in \{1,2,\dots,\frac{1}{2}(q-3)\} $ and $\gamma \in F_q.$

Now if I compute $B_9(1)AB_9(1)^t,$ I get

$ \left( \begin{array}{cccc} 0 & 2& 0 & \gamma^{-1} \\ -2 & 0& -\gamma & 0 \\ 0 & \gamma & 0 & 1\\ -\gamma^{-1}&0 &-1 &0 \end{array} \right) \neq A$

As per definition of $\Sp(4,q), B_9(1)AB_9(1)^t$ should be equal to $A.$ I am not getting what I am doing wrong. I will be grateful for any help in understanding my mistake or any other good reference for conjugacy classes of $\Sp(4,q)$

Answer 1

Keep in mind that it's tricky to assemble a character table for all finite groups of Lie type in a family, and in particular the thesis work by Srinivasan at Manchester (supervised by J.A. Green) excluded the prime $p=2$ (later filled in by H. Enomoto here. Moreover, tables often have errors; here there are small errors found by A. Pryzgocki here. And as the comment by Bullet51 shows, not everything is straightforward: at the time there was mainly a paper by G.E. Wall on the conjugacy classes of finire classical groups. Her thesis work was influential in the later development of sophisticated methods by Deligne and Lusztig (1976). But it is still difficult to summarize character tables for an entire Lie family.

Concerning your attempt to work out the classes in a concrete way, I'm not sure exactly what goes wrong. But this does illustrate the perils of trying to be too computational: such an approach will be troublesome for larger matrices.

Answer 2

So, we know of a conjugacy class in $Sp_4(F_q)$ which has $(x-\gamma^i)^2(x-\gamma^{-i})^2$ as its minimal polynomial. In Bhama Srinivasan's paper, the matrix she's mentioned is a Jordan form conjugated by some elementary matrix. And the product $B_9(i)AB_9(i)^t$ mentioned in the question is rightly calculated, and thus the representative for $B_9(i)$ isn't the one suited to be in $Sp_4(F_q)$. So instead, we can take this one.

$$\begin{pmatrix} \gamma^i & 1 & 0 & 0\\ 0&\gamma^i &0&0\\ 0&0&\gamma^{-i}&0\\ 0&0&-\gamma^{-2i} & \gamma^{-i}\end{pmatrix}$$

This is with respect to the bilinear form $\begin{pmatrix} \mathbf{0} & I_2\\ -I_2 &\mathbf{0}\end{pmatrix}$.