Let $k$ be an algebraically closed field and $\Gamma$ a finite group. $\Gamma$ acts on itself via conjugation, and it is true that the induced action on the cohomology algebra $H^{*}(\Gamma,k)$ is trivial.

Now, let $p$ be a prime number, suppose that $\operatorname{char}k=p$, and let $G$ be an algebraic group over $k$. Denote the Lie algebra of $G$ by $\mathfrak{g}$. $G$ acts on $\mathfrak{g}$ via the adjoint action, $\operatorname{Ad}:G\to\operatorname{Aut}\mathfrak{g}$. In this setting, is the induced action on Lie algebra cohomology $H^*(\mathfrak{g},k)$ also trivial? Here I take $H^*(\mathfrak{g},k)$ to be the algebra of extensions of $k$ by $k$ in the category of $u(\mathfrak{g})$-modules, where $u(\mathfrak{g})$ is the restricted enveloping algebra of the restricted Lie algebra $\mathfrak{g}$.

  • $\begingroup$ $H^*(\mathfrak{g}, - )$ are the higher derived functors of $Hom_{\mathfrak{g}}(k , - )$ that is $\mathfrak{g}-$invariants. It is clear that the action is trivial here and hence action on the derived functors is also trivial. $\endgroup$ – random123 Dec 6 '17 at 11:05

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