The cohomology group $H^{1}(GL_{2}(\mathbb{F}_{p}), M_{2}(\mathbb{F}_{p}))$

Question

Let $M_{2}(\mathbb{F}_{p})$ be the vector space of 2$\times$2 matrices over the finite field $\mathbb{F}_{p}$ where $p$ is a prime number, and let $GL_{2}(\mathbb{F}_{p})$ be the group of invertible 2$\times$2 matrices over $\mathbb{F}_{p}$. We let $GL_{2}(\mathbb{F}_{p})$ act on $M_{2}(\mathbb{F}_{p})$ by conjugation. My question is how to compute the cohomology group $H^{1}(GL_{2}(\mathbb{F}_{p}), M_{2}(\mathbb{F}_{p}))$ ?

Answer 1

Here's a (sketch of a) proof that this cohomology group is always zero using the fact that $G=GL_2(\mathbb{F}_p)$ has a cyclic Sylow $p$-subgroup (and so it definitely doesn't generalize easily to $GL_n$ for larger $n$ or $\mathbb{F}_{p^k}$ for $k>1$).

First, if $V$ is the natural $2$-dimensional module for $G$, then $M_2(\mathbb{F}_p)\cong V\otimes V^*$ as a $G$-module, so $$H^1(G,M_2(\mathbb{F}_p)\cong \operatorname{Ext}^1_{\mathbb{F}_pG}(\mathbb{F}_p,V\otimes V^*)\cong\operatorname{Ext}^1_{\mathbb{F}_pG}(V,V).$$

Let $H$ be the subgroup of $G$ consisting of upper triangular matrices. Since the index of $H$ in $G$ is coprime to $p$, restriction to $H$ is injective on positive-degree cohomology, so it's enough to show $\operatorname{Ext}^1_{\mathbb{F}_pH}(V,V)=0$.

It's easy to check that (if $p>2$) $V$ is indecomposable, with two non-isomorphic composition factors, as an $\mathbb{F}_pH$-module. Also, $H$ has a normal cyclic Sylow $p$-subgroup, and the representation theory of such groups is very well understood: the group algebras are products of "Brauer star algebras", and it's an easy direct calculation for such algebras to check that $\operatorname{Ext}^1(V,V)=0$ for any indecomposable module $V$ with two non-isomorphic composition factors.

The case $p=2$ is easy, as then $H$ is cyclic of order two, and $V$ is a projective $\mathbb{F}_pH$-module.

I'm sure there must be a simpler proof for this particular case that avoids the theory of Brauer tree algebras, though.

Answer 2

Let $G=GL_2(\mathbb F_p)$, $M=M_2(\mathbb F_p)$, $M^0=$ matrices of trace 0 in $M$. If you consider first $H^1(G,M^0)$, then it is 0 as soon as $p \geq 7$ by CPS (Cline, Parshall, Scott, cohomology of finite groups of Lie type, Publ. IHES 45, (1975)) Theorem 4.2. (This is a result sometimes useful in Galois deformation theory, cf. Mazur, "Deforming Galois Representations", pages 401-402).

As for $H^1(G,M)$ you get that it has dimension 0 (edited after the comments) if $p\geq7$ since $M=M_0 \oplus \mathbb F_p$ as $G$-modules when $p$ is odd.

Answer 3

There are really two questions here: (1) What is the dimension of this cohomology? (2) How do I compute it? Though it may not be strictly necessary, it's probably best to treat the cases $p=2$ and $p$ odd separately. As Joel proposes, the answer to (1) should be 0 or "almost 0" though I'd have to spend more time thinking about it to be confident.

Concerning question (2), two different approaches are possible: either work directly in the finite group framework or else relate these particular finite groups to the ambient algebraic groups (as Cline-Parshall-Scott and others have done extensively). A complication is that much of the literature in both directions focuses on $\mathrm{SL}(2,p^n)$ rather than on reductive groups such as the general linear group. So one has to make some transition. For example, in Proc. London Math. Soc. 47 (1983), Jon Carlson lays out a program for computing cohomology of all degrees for these finite special linear groups and their irreducible modules. Though it gets quite tricky to do explicit computation in high degrees, he is for instance able to reproduce the Ext results in degree 1 published earlier that year in the same journal (volume 46) by Andersen-Jorgensen-Landrock based on the algebraic groups. (Keep in mind that the $H^1$ here is the Ext group classifying possibly nonsplit extensions of the trivial 1-dimensional module by the given module.)

In your situation, the essential action (leaving aside a pointwise fixed line in the 4-dimensional space of matrices) is the natural 3-dimensional representation of the rank 1 adjoint group $\mathrm{PGL}(2,p)$ (or of $\mathrm{SL}(2,p)$). For this set-up the algebraic group results expressed in terms of highest weights adapt nicely to the finite groups, as in CPS or AJL or other more recent papers. Again you probably want to treat separately the case $p=2$, but for all odd primes the analysis is uniform from a representation-theoretic viewpoint.

Answer 4

I would like to sketch a different argument which is maybe rather roundabout but whose method generalizes to linear groups beyond the finite field case. Unfortunately, I am lazy so I will exclude characteristic $2$.

The obvious first reduction step uses the inflation-restriction sequence $$ 0\to H^1(PGL_2(k),M_2(k))\to H^1(GL_2(k),M_2(k))\to H^1(k^\times,M_2(k))^{PGL_2(k)}\to $$ The action of the center $k^\times$ on $M_2(k)$ is trivial, hence the universal coefficient sequence shows $H^1(k^\times,M_2(k))^{PGL_2(k)}\cong Hom_k((k^\times\otimes_{\mathbb{Z}} k),k)$. The fact that $H^1(\mathbb{F}_q^\times,\mathbb{F}_q)=0$ implies $H^1(\mathbb{F}_q^\times,M_2(\mathbb{F}_q))=0$. Second, we split $M_2(k)\cong k\oplus\mathfrak{sl}_2(k)$, and we have $H^1(PGL_2(k),k)\cong H_1(PGL_2(k),k)^\vee\cong Hom(k^\times/(k^\times)^2,k)$. The latter is trivial for $k=\mathbb{F}_q$.

We are left with computing $H^1(PGL_2(k),\mathfrak{sl}_2(k))$, which can be done using a resolution of $\mathfrak{sl}_2(k)$ as $PGL_2(k)$-module in J.-L. Cathelineau: On the homology of $SL_2$, a complement. Math. Scand. 66 (1990), 17-20. Cathelineau proves that for any field, there is an exact sequence of $PGL_2(k)$-modules $$ \coprod_{(a,b)\in\mathbb{P}^1(k)^2,a\neq b}\mathfrak{b}_a\cap \mathfrak{b}_b\stackrel{\partial}{\longrightarrow} \coprod_{a\in\mathbb{P}^1(k)}\mathfrak{b}_a \stackrel{\omega}{\longrightarrow} \mathfrak{sl}_2(k)\to 0 $$ where $\mathfrak{b}_a$ is the Lie algebra of the Borel subgroup $B_a$ stabilizing $a\in\mathbb{P}^1$ in $PGL_2(k)$. The cohomology $H^1(PGL_2(k),\mathfrak{sl}_2(k))$ can then be computed using the long exact sequences associated to the short exact sequences from the resolution, following J.-L. Cathelineau: Sur l'homologie de $SL_2$ à coefficients dans l'action adjointe. Math. Scand. 63 (1988), 51-86. I'll spare you the details, the main point is that $PGL_2(k)$ acts transitively on $\ker\left(\coprod\mathfrak{b}_a\to\mathfrak{sl}_2(k)\right)$ which implies that there is an injection $H^1(PGL_2(k),\mathfrak{sl}_2(k))\hookrightarrow H^1(B,\mathfrak{b})$; but the latter is trivial over a finite field (again because $\mathbb{F}_q^\times\otimes\mathbb{F}_q=0$).

Moving to homology and away from finite fields: Conceptually (though this is not strictly true), I would interpret this as saying that $H^1(PGL_2(k),M_2(k))$ is computed by the dual of the tangent of the Bloch-Wigner-Dupont-Sah sequence (which computes $H_2$ and $H_3$ of $GL_2(k)$ with constant coefficients). The essential point that implies the vanishing of $H^1$ (for finite fields) is $\mathbb{F}_q^\times\otimes\mathbb{F}_q=0$. There is a bulk of further literature on $H_1(SL_2(k),\mathfrak{sl}_2(k))$ as related to scissors congruences and tangents to K-theory: see the papers of Dupont-Sah, Cathelineau and Elbaz-Vincent.