Let $g \geq 2$ be an integer and consider the symmetric group $S\_n$ where $n = 2g+1$ or $n = 2g+2$ as a subgroup of the symplectic group $\mathrm{Sp}\_{2g}(\mathbb{F}\_2)$ via the standard representation(s). (See [this question](https://mathoverflow.net/questions/290225/low-dimensional-irreducible-2-modular-representations-of-the-symmetric-group) for a description of standard representations of symmetric groups.) This induces the adjoint action of $S\_n$ on the Lie algebra $\mathfrak{sp}\_{2g}(\mathbb{F}\_2)$ (which is an $\mathbb{F}\_2$-vector space of dimension $2g^2 + g$). Can anyone tell me the dimension of the first group cohomology $H^1(S\_n, \mathfrak{sp}\_{2g}(\mathbb{F}\_2))$? In particular, I want to know if this first cohomology vanishes (I expect the answer might be different for $n = 2g+1$ and $n = 2g+2$).


EDIT: Actually, I see that what I really need to understand for my purposes is the first group cohomology of $S\_{2g+1}$ or $S\_{2g+2}$ with coefficients in $\mathbb{F}\_2^{2g}$ via the standard representation(s) mentioned above. I'm not sure if this is a simpler or more difficult thing to solve, but for the moment I don't know how to solve it except by attempting brute force. Of course I'm still curious about my initial question.


I see that some similar questions have been asked here which sort of skirt around this, but most of the literature referenced in the answers seems to address either much more complicated questions or simply cohomology with coefficients in $\mathbb{F}\_2$. It would take me a while to extract what I need from it as my primary background is not in Lie theory.



first group cohomology for the standard representation of $S_n$ over $\mathbb{F}_2$
You are right that the answer differs for $2g+1$ and $2g+2$. The cohomology with coefficients in $F\_2^{2g}$ is trivial for $S\_{2g+1}$, but nontrivial for $S\_{2g+2}$. Here is an elementary sketch why. 


In general if $M$ is an $F\_2[G]$-module, $H^1(G,M)$ vanishes if and only if every exact sequence $0\to M\to E\to F\_2\to 0$ splits. (The sequence splits if and only if for any $e\in E-M$, there is a complement of the form $F\_2(e+m)$, $m\in M$, and such a complement exists iff the cocycle $g\mapsto e-ge$ is a coboundary.) 


Let $F=F\_2$, let $F\_0^k$ be the set of $k$-tuples of sum $0$, and let $I\_k\subseteq F^k$ the set of constant $k$-tuples. If $k=2g+2$, then your "standard rep" is the module $M=F\_0^k/I\_k$, and $0\to M\to F^k/I\_k\to (F^k/F\_0^k\cong F)\to 0$ does not split. On the other hand if $k=2g+1$, your "standard rep" is the module $M=F\_0^k$. Consider an exact sequence $0\to M\to E\to F\to 0$ of $S\_{2g+1}$-modules. Note that $M$ is self-dual via the standard inner product on $F^{2g+1}$. So it is enough to show that the dual sequence $0\to F\to E^\*\to M\to 0$ splits. As a module for the alternating group $A\_{2g}$, $M$ has a unique $1$-dimensional trivial submodule $M\_1$. The preimage $E\_1^\*$ of $M\_1$ in $E^\*$ is also a trivial ($2$-dimensional) $A\_{2g}$-submodule since there are no nontrivial homomorphisms $A\_{2g}\to F^+$. Let $e\in E\_1^\*-F$. Then the set $\{ge\,|\,g\in A\_{2g+1}\}$ of images of $e$ under $A\_{2g+1}$ has cardinality $2g+1$. So one checks that the $A\_{2g+1}$-submodule of $E^\*$ spanned by $e$ is a complement to $F$ and the sequence splits as a sequence of $A\_{2g+1}$-modules. Therefore the original sequence $0\to M\to E\to F\to 0$ splits as a sequence of $A\_{2g+1}$-modules. Write $E=M\oplus N$ with $N\cong F$ as $A\_{2g+1}$-module. Then $N$ is the full set of fixed points of $A\_{2g+1}$ on $E$, so $N$ is invariant under the action of $S\_{2g+1}$, i.e., $N$ is the desired complementary $S\_{2g+1}$-submodule. 


