The adjoint representation of the symplectic group in characteristic 2

Question

For a prime $p$ and some $g \geq 2$, consider the adjoint representation $\mathfrak{sp}_{2g}(\mathbb{F}_p)$ of the symplectic group $\text{Sp}_{2g}(\mathbb{F}_p)$. For $p \geq 3$, it is not hard to show that this is an irreducible representation. However, it is reducible in characteristic $2$.

To explain this, we'll choose coordinates. Regard $\text{Sp}_{2g}(\mathbb{F}_p)$ as the set of $2g \times 2g$ block matrices $M$ such that $$M \left(\begin{array}{c|c} 0 & 1 \\ \hline -1 & 0 \end{array}\right) M^t = \left(\begin{array}{c|c} 0 & 1 \\ \hline -1 & 0 \end{array}\right).$$ With these coordinates, $$\mathfrak{sp}_{2g}(\mathbb{F}_p) = \{\text{$\left(\begin{array}{c|c} A & B \\ \hline C & -A^t \end{array}\right)$ $|$ $B^t=B$ and $C^t=C$}\}.$$ The group $\text{Sp}_{2g}(\mathbb{F}_p)$ acts on this by conjugation. Define $$V = \{\text{$\left(\begin{array}{c|c} A & B \\ \hline C & -A^t \end{array}\right) \in \mathfrak{sp}_{2g}(\mathbb{F}_2)$ $|$ the diagonal entries of $B$ and $C$ are $0$}\}.$$ One can then calculate that the subspace $V$ is preserved by $\text{Sp}_{2g}(\mathbb{F}_2)$ and that as a representation of $\text{Sp}_{2g}(\mathbb{F}_2)$ we have $$\mathfrak{sp}_{2g}(\mathbb{F}_2) / V \cong \mathbb{F}_2^{2g}$$ with the evident action of $\text{Sp}_{2g}(\mathbb{F}_2)$.

The quotient map $$\Psi\colon \mathfrak{sp}_{2g}(\mathbb{F}_2) \longrightarrow \mathbb{F}_2^{2g}$$ takes an element of $\mathfrak{sp}_{2g}(\mathbb{F}_2)$ to the vector whose entries are the diagonal entries of $B$ and $C$. This brings me to my question:

Question: Does anyone know a conceptual explanation for $\Psi$?

I learned about $\Psi$ from Igusa's classical paper "On the Graded Ring of Theta-Constants", where it is implicit in some of his calculations with the symplectic group. However, it basically just falls out of a bunch of matrix calculations, and I have no deep understanding of the reason it exists.

(By the way, my earlier question here was motivated by trying to understand Igusa's calculations, which play a role in a paper I am writing. The great answer I got inspired me to ask this followup!)

Answer 1

Let $W$ be a vector space over a field $K$ of characteristic two, let $\beta$ be a non-degenerate alternating bilinear form on $W$.

Let $$G = \operatorname{Sp}(V) = \{ g \in \operatorname{End}(W) : \beta(gv, gv') = \beta(v,v') \text{ for all } v,v' \in W \}.$$ Now the adjoint representation of $G$ that you consider is $\mathfrak{g} = \mathfrak{sp}(W)$, that is, $$\mathfrak{g} = \{X \in \operatorname{End}(W) : \beta(Xv,v') = -\beta(v, Xv') \text{ for all } v,v' \in W \}.$$

We know that $\operatorname{End}(W) \cong W \otimes W^*$ as $G$-modules. Also $W \cong W^*$, so $\operatorname{End}(W) \cong (W \otimes W)^*$ and you can identify $\mathfrak{g}$ as a particular submodule of $(W \otimes W)^*$. You can prove that this is exactly $S^2(W)^*$, that is, the dual of the symmetric square of $W$. Here the embedding of $S^2(W)^*$ is given by the dual of the usual quotient map $W \otimes W \rightarrow S^2(W)$.

There is a short exact sequence $$0 \rightarrow W^{[1]} \rightarrow S^2(W) \xrightarrow{\phi} \wedge^2(W) \rightarrow 0,$$ where $\phi(xy) = x \wedge y$ for all $x,y \in W$ and where $W^{[1]}$ is the submodule of $S^2(W)$ spanned by all $x^2$, $x \in W$.

The restriction map $S^2(W)^* \rightarrow (W^{[1]})^*$ corresponds to the map $\Psi$ in your question, and it has kernel isomorphic to $\wedge^2(W)^*$. Here $\wedge^2(W)^*$ corresponds to the $V$ in your question. Also $(W^{[1]})^* \cong W$ when $K = \mathbb{F}_2$, but this is no longer true over $K = \mathbb{F}_{4}$ for example.