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In the algebraic group $G=\operatorname {PCSp}(2^{r},K)$ where $K$ is an algebraically closed field of an odd characteristic, there is a conjugacy class of involutions with representative: $$ e=\left[ \begin{pmatrix} -wI_{2^{r-1}} & 0 \\ 0 & wI_{2^{r-1}} \\ \end{pmatrix} \right]$$ where $w$ is of order 4 in $K$.

The centralizer in $G$ is $C_G(e) =A_{2^{r-1}-1}T_1.2$. The "$2$" acts on $A_{2^{r-1}-1}$ as a graph automorphism and inverts the $T_1$ (one-dimensional torus).

Is there an explicit matrix form of the generator of the "$2$" group?

$\operatorname {PCSp}$ is the group which fixes a non-degenerate alternating bilinear form up to a scalar. And my bilinear form is:

$$ \Delta \left[ \begin{pmatrix} 0 & 1 \\ -1 & 0 \\ \end{pmatrix} \right]_m$$ where $\Delta$ means $m$-copies embedded diagonally.

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  • $\begingroup$ Your element doesn't lie in your generalised symplectic group when $m$ is odd and $m > 1$. Do you mean $e = \operatorname{diag}(-w, w, -w, w, \dotsc, -w, w)$? $\endgroup$
    – LSpice
    Commented Feb 4 at 23:40
  • $\begingroup$ I'll restrict myself. Thank you. $\endgroup$
    – scsnm
    Commented Feb 4 at 23:57
  • $\begingroup$ Re, but now your centraliser is wrong; the derived subgroup of the identity component of the centraliser is isogenous to $\operatorname{Sp}_m \times \operatorname{Sp}_m$ (where $m$ is as in your original notation, so $2m = 2^r$). $\endgroup$
    – LSpice
    Commented Feb 5 at 0:06

1 Answer 1

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I think that you must mean $e$ to be the image of $\operatorname{diag}(-w, w, -w, w, \dotsc, -w, w)$, since otherwise $e$ does not lie in $\operatorname{GSp}_{2m}(K)$ when $m$ is odd and greater than $1$.

I am more comfortable dealing with the symplectic form with matrix $\begin{pmatrix} & w_0 \\ -w_0 \end{pmatrix}$, where $w_0 = \operatorname{antidiag}(1, \dotsc, 1)$. This is conjugate to your form by the permutation matrix corresponding to the element of $\operatorname{Sym}_{2m}$ given by $i \mapsto 2i - 1$ and $2m - i + 1 \mapsto 2i$ for $i \le m$. In particular, in these coördinates, your element $e$ (revised as I propose) is identified to the element $e = \operatorname{diag}(-w, \dotsc, -w, w, \dotsc, w)$ that you originally proposed, and the answer is as in your other question, with an obvious sign change: $\operatorname{GL}_m$ embeds in $\operatorname{Sp}_{2m}$ by $g \mapsto \begin{pmatrix} g \\ & \operatorname{Int}(w_0)g^{-\mathsf T} \end{pmatrix}$, and we may take the "2" to be represented by the image in $\operatorname{PGSp}_{2m}$ of $\begin{pmatrix} & I_m \\ -I_m \end{pmatrix}$.

Conjugating this back to your original set-up gives the representative $\operatorname{antidiag}(1, -1, 1, -1, \dotsc, 1, -1)$ (or possibly its opposite—I didn't check carefully, but it doesn't matter, since we're projecting to $\operatorname{PGSp}_{2m}$).

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  • $\begingroup$ No worries that I restricted myself. Thank you again for the input. $\endgroup$
    – scsnm
    Commented Feb 5 at 0:00
  • $\begingroup$ Sorry, what "Int" means? $\endgroup$
    – scsnm
    Commented Feb 5 at 0:08
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    $\begingroup$ @scsnm, re, the inner-automorphism map $G \to \operatorname{Aut}(G)^\circ$, so that $\operatorname{Int}(w_0) : g \mapsto w_0 g w_0^{-1}$. $\endgroup$
    – LSpice
    Commented Feb 5 at 0:12

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