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$\DeclareMathOperator\Sp{Sp}$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 $\Sp(4,q)$

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    $\begingroup$ “We remark that in the first column we give as class representatives not necessarily elements of $G$, but their canonical forms in an extension field of $F$”, from page 492 $\endgroup$ Feb 6, 2019 at 13:13
  • $\begingroup$ I understand this. The matrix I am talking about is defined over $F_q.$ As per notations in paper, $\gamma$ lies in the field $F_q.$ $\endgroup$ Feb 8, 2019 at 4:20
  • $\begingroup$ Perhaps what was meant in the quotation cited by Bullet51 was: canonical forms in $GL_4(F^*)$ for some extension field $F^*$ of $G$, not in $Sp_4(F^*)$. $\endgroup$ Feb 9, 2019 at 20:00

2 Answers 2

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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.

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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 $\begin{pmatrix} \mathbf{0} & I_2\\ -I_2 &\mathbf{0}\end{pmatrix}$.

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  • $\begingroup$ How does this answer the question? $\endgroup$
    – Derek Holt
    Feb 3, 2021 at 11:24

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