I am interested in the finite unitary reflection group $G= G_{32}$, the group No. 32 in Table VII on page 301 of the paper: Shephard, G. C.; Todd, J., A. Finite unitary reflection groups. Canad. J. Math. 6 (1954), 274–304.

This is a group of order $2^7 3^5 5 = 155520$. Its commutator subgroup $H=(G,G)$ is of index 3, and a computer calculation shows that $H$ is isomorphic to ${\rm Sp}(4,3):={\rm Sp}_4({\Bbb F}_3)$, the symplectic group of of $4\times 4$ matrices over the finite field ${\Bbb F}_3$.

This group $G$ is given with a faithful 4-dimensional complex representation $$\rho: G\to {\rm GL}(4, {\Bbb C}).$$ Moreover, it is is stable under the standard complex conjugation in ${\Bbb C}^4$, and so we obtain an involutive automorphism (an automorphism of order 2) $\ \sigma\colon H\to H$.

I am trying to guess this involution $\sigma$ and to compute the first nonabelian cohomology set $H^1(\langle\sigma\rangle, H)$. A computer calculation shows that the $H^1$ is trivial, and I would like to understand this without computer.

Question 1. What are the nontrivial 4-dimensional complex representations of the finite group ${\rm Sp}(4,3)$?

Question 2. What are the involutive automorphisms of ${\rm Sp}(4,3)$ ? In particular, is it true that all nontrivial involutive automorphisms of ${\rm Sp}(4,3)$ come from elements of order 2 in the projective symplectic group ${\rm PSp}_4({\Bbb F}_3)$ ?

Question 3. Which of those involutive automorphisms of $H={\rm Sp}(4,3)$ can come from the complex conjugation in a 4-dimensional complex representation of $H$?

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    $\begingroup$ I imagine that the involutive automorphism is the outer automorphism induced within the conformal group ${\rm CSp}(4,3)$, which contains ${\rm Sp}(4,3)$ as a subgroup of index $2$ (some people denote that by ${\rm GSp}(4,3)$ ). I am not really sure what you are asking in Question 1 - there are two 4-dimensional complex representations, which are interchanged by complex conjugation, and also by the outer automorphism. (So the representations are not fixed by complex conjugation, but their images can be chosen so that they are fixed.) $\endgroup$ – Derek Holt May 17 '20 at 22:06

We can take $H={\rm Sp}(4,3)$ to be the group $\{ A \in {\rm GL}(4,3) \mid AFA^{\mathsf T} = F\}$, where $$F=\left(\begin{array}{rrrr}0&0&0&1\\0&0&1&0\\0&-1&0&0\\-1&0&0&0\end{array}\right)$$ is the matrix of the preserved symplectic form.

The the matrix $$C =\left(\begin{array}{rrrr}1&0&0&0\\0&1&0&0\\0&0&-1&0\\0&0&0&-1\end{array}\right),$$ satisfies $CFC^{\mathsf T}= -F$, it normalizes and induces an involutory outer automorphism of $H$, and $\langle H,C \rangle$ is the conformal symplectic group, which I prefer to denote by ${\rm CSp}(4,3)$ (although it is sometimes written as ${\rm GSp}(4,3)$).

There are two dual 4-dimensional complex representations of $H$, which are interchanged by the outer automorphism induced by $C$, so this appears to be the automorphism that you are looking for.

From your description, I think the only possible structure of the group $G$ is the direct product $H \times C_3$.

To answer you specific questions, I am not sure what you are looking for in Question 1.

For Question 2, the full automorphism group group of $H$ is the image of ${\rm CSp}(4,3)$ mod scalars, which we can denote by ${\rm PCSp}(4,3)$: it has order $2|{\rm PSp}(4,3)| = 51840$. The involutory automorphism in question is an outer automorphism, and is not induced by an element of ${\rm PSp}(4,3)$.

For Question 3, I am not completely sure. There are actually two conjugacy classes of involutory outer automorphisms of $H$, one of which is induced by the matrix $C$ above, and I am not sure whether both can be induced by complex conjugation or only one of them.

An example of an element of ${\rm CSp}(4,3)$ that induces an involutory automorhism from the other class is $$C' =\left(\begin{array}{rrrr}0&0&1&0\\1&0&0&-1\\-1&0&0&0\\0&1&1&0\end{array}\right).$$ This has order 4 in ${\rm CSp}(4,3)$, but its square is $-I$, so it induces an involutory automorphism. It is interesting that its centralizer in $H$ has order $720$, whereas the centralizer of $C$ has order $48$. That might be useful in deciding which automorphism is induced by complex conjugation.

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    $\begingroup$ @LSpice Thanks! "..e looking for". $\endgroup$ – Derek Holt May 18 '20 at 8:02
  • $\begingroup$ Many thanks! A stupid question: I know the definition of the symplectic group ${\rm Sp}(4,3):={\rm Sp}_4( {\Bbb F}_3)$ as the group $$G_+=\{ A \in {\rm GL}(4,3) \mid AFA^{\mathsf T} = F\}.$$ However, you take the group $$G_-=\{ A \in {\rm GL}(4,3) \mid AFA^{\mathsf T} = -F\}.$$ How can one define an isomorphism between $G_+$ and $G_-$? $\endgroup$ – Mikhail Borovoi May 18 '20 at 11:47
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    $\begingroup$ $G_-$ is not a group. It is a coset of $G_+ = {\rm Sp}(4,3)$ in the larger group ${\rm CSp}(4,3)$. $\endgroup$ – Derek Holt May 18 '20 at 12:18
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    $\begingroup$ Concerning Question 2: Of course you are right! I used bad notation. By ${\rm PSp}_4(\Bbb F_3)$ I meant ${\rm PCSp}(4,3)$. $\endgroup$ – Mikhail Borovoi May 18 '20 at 12:19
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    $\begingroup$ ``$G_-$ is not a group.'' Good! However, you write: " We can take $H={\rm Sp}(4,3)$ to be the group $\{ A \in {\rm GL}(4,3) \mid AFA^{\mathsf T} = -F\}$". I was a bit puzzled.... $\endgroup$ – Mikhail Borovoi May 18 '20 at 12:26

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