Cohomology of the adjoint representation of $\mathrm{SL}_2(k)$

Question

Let $k$ be a finite field. Do we always have $H^1(\operatorname{PSL}_2(k), k^3) = 0$, where $\operatorname{PSL}_2(k)$ acts on $k^3$ via the adjoint representation (= conjugation action on trace zero matrices)? If not, is there a simple description of the $k$ for which this does or does not happen?

(For $k$ the prime field $\mathbf{F}_p$, with $p \ge 7$, the vanishing of this cohomology is shown in an old paper of mine with Sarah Zerbes, adapting an earlier argument of Flach for $\operatorname{GL}_2(k)$. But this argument fails for small $p$, and it relies on the $p$-Sylow of $\operatorname{SL}_2(\mathbf{F}_p)$ being cyclic, which is false for non-prime fields. So surely there must be a better way.)

Answer 1

$\DeclareMathOperator\PSL{PSL}\newcommand\Ad{\mathrm{Ad}}\newcommand\triv{\mathrm{triv}}$I believe that if the characteristic of $k$ is 2, then $H^1(\PSL_2(k),k^3)$ is nonzero, while if the characteristic of $k$ is at least 7, then $H^1(\PSL_2(k),k^3)=0$.

I will now write $\Ad$ for the adjoint representation and $\triv$ for the trivial representation. Note that $H^1(PSL_2(k),\Ad) \cong \operatorname{Ext}^1 (\triv,\Ad)$. So we need to know if there is a non-split short exact sequence

$$0\to \Ad \to X \to \triv\to 0.$$

If $k$ has characteristic 2, then we can take the tensor square of the defining representation $X=(k^2)^{\otimes 2}$.

If $k$ has characteristic at least 7, let me sketch my solution. Start with $X$ as above. Let $N$ be the subgroup which is the normaliser of a split torus. Since $\lvert N\rvert$ is coprime to the characteristic of $k$, we can, and do, split this short exact sequence in the category of $N$-modules.

Write $e,h,f$ for the usual basis of $\Ad$ and $v$ for a basis element of $\triv$. I claim that for every $x\in k$, there exist $\alpha,\beta\in k$ such that $$\left(\begin{smallmatrix} 1 & x \\ 0 & 1 \end{smallmatrix}\right) \cdot v=v+\alpha e,\qquad \left(\begin{smallmatrix} 1 & 0 \\ x & 1 \end{smallmatrix}\right) \cdot v=v+\beta f. $$ To prove this, I pick $t\in \mathbb{F}_p$ with $t^5\neq t$ (hence the assumption on the characteristic) and apply the identity $$ \left(\begin{smallmatrix} t & 0 \\ 0 & t^{-1} \end{smallmatrix}\right) \left(\begin{smallmatrix} 1 & x \\ 0 & 1 \end{smallmatrix}\right) = \left(\begin{smallmatrix} 1 & x \\ 0 & 1 \end{smallmatrix}\right)^{t^2} \left(\begin{smallmatrix} 1 & t \\ 0 & t^{-1} \end{smallmatrix}\right) $$ to $v$.

Now the product $\left(\begin{smallmatrix} 1 & x \\ 0 & 1 \end{smallmatrix}\right)\left(\begin{smallmatrix} 1 & 0 \\ -x^{-1} & 1 \end{smallmatrix}\right) \left(\begin{smallmatrix} 1 & x \\ 0 & 1 \end{smallmatrix}\right)$ lies in $N$ so must act trivially on $v$. Substituting in the equations above we end up with $\alpha=\beta=0$, which is enough to conclude as these elements generate $\PSL_2(k)$.

Answer 2

(Too long for a comment.) A Magma computation shows that for $k=\mathbf{F}_p$ with $p$ prime the group $H^1(\operatorname{PSL}_2(k);k^3)$ equals $0$ for $p=3$ and $7\le p\le 17$ while the cohomology group is $1$-dimensional for $p=2$ and $p=5$. I include the code and output below. If you are interested it is simple to convince Magma to get a generating cocycle for $p=5$. Also in this case note that $\operatorname{PSL}_2(5)=\operatorname{Alt}_5$ and the module is the unique simple $3$-dimensional module.

Program:

for p in [2,3,5,7,11,13,17] do  
  G:=AdjointChevalleyGroup("A",1,p);  
  M:=GModule(G);  
  X:=CohomologyModule(G,M);  
  print p,Dimension(CohomologyGroup(X,1));  
end for;  

Output:

2 1  
3 0  
5 1  
7 0  
11 0  
13 0  
17 0