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

The centralizer in $G$ is $C_G(e) =A_{m-1}T_1.2$. The "$2$" acts on $A_{m-1}$ 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?

$G=\operatorname {PCGO}(2m,K)$ is the group which fixes a non-degenerate quadratic from up to a scalar.

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  • $\begingroup$ In the coördinates used for your matrix realisation, what is your orthogonal form? The one with matrix $\operatorname{antidiag}(1, \dotsc, 1)$? \\ I am used to CO (for conformal orthogonal) and GO (for general orthogonal), but why CGO? $\endgroup$
    – LSpice
    Commented Feb 4 at 22:18

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It looks like you are working with respect to the orthogonal form with matrix $\begin{pmatrix} & w_0 \\ w_0 \end{pmatrix}$, where $w_0 = \operatorname{antidiag}(1, \dotsc, 1)$. That's the one that I'm used to, so it works for me.

$\operatorname{GL}_m$ embeds in $\operatorname{SO}_{2m}$ as $g \mapsto \begin{pmatrix} g \\ & \tau(g) \end{pmatrix}$, where $\tau : g \mapsto \operatorname{Int}(w_0)g^{-\mathsf T}$ is an outer involution of $\operatorname{GL}_m$. This embedding gives the isomorphism $\operatorname{GL}_m/\mu_2 \cong \operatorname C_{\operatorname{PGO}_{2m}}(e)^\circ$.

The non-identity component of $\operatorname C_{\operatorname{PGO}_{2m}}(e)$ contains the image in $\operatorname{PGO}_{2m}(e)$ of $\begin{pmatrix} & I_m \\ I_m \end{pmatrix}$, which may therefore be taken as an explicit matrix form of a generator of the "2" in which you are interested.

By the way, this is all probably easiest to picture when $m = 1$, when the images in $\operatorname{PGL}_2$ of $\begin{pmatrix} i \\ & -i \end{pmatrix}$ and $\begin{pmatrix} & 1 \\ 1 \end{pmatrix}$ are the classical example of commuting, semisimple elements that do not belong to a common torus.

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  • $\begingroup$ Thank you for the input. If my form is $$\Delta \left[ \begin{pmatrix} 0 & 1 \\ -1 & 0 \\ \end{pmatrix} \right]_m,$$ would the answer be$$\Delta \left[ \begin{pmatrix} 0 & 1 \\ 1 & 0 \\ \end{pmatrix} \right]_m$$ instead? $\endgroup$
    – scsnm
    Commented Feb 4 at 22:51
  • $\begingroup$ @scsnm, re, I don't understand your notation. What is $\Delta[\cdot]_m$? $\endgroup$
    – LSpice
    Commented Feb 4 at 22:57
  • $\begingroup$ Sorry about that. It means $m$-copies embedded diagonally. $\endgroup$
    – scsnm
    Commented Feb 4 at 22:59
  • $\begingroup$ @scsnm, re, then I don't understand what it means for the form to be $\Delta\left[\begin{pmatrix} & 1 \\ -1 \end{pmatrix}\right]_m$. An orthogonal form should have a symmetric matrix. $\endgroup$
    – LSpice
    Commented Feb 4 at 23:00
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    $\begingroup$ I apologize. I didn't realize I put the group $G$ mistakenly. It is supposed to be a symplectic group. I will write another question. So sorry. $\endgroup$
    – scsnm
    Commented Feb 4 at 23:03

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