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Let $\mathfrak g$ be a semisimple Lie algebra and let $M$ be a nontrivial simple $\mathfrak g$-module. Then quadratic Casimir of $\mathfrak g$ acts on $M$ as multiplication by a nonzero scalar $c_M$. We also have representation $\widehat \rho$ of $\mathfrak g$ on cochain spaces $M^n= M \otimes \mathfrak g^{* \otimes n}$. It is defined as the tensor product of $\rho$ and the $n$-fold tensor product of the coadjoint representation. It is simple to show that $\widehat \rho(x)$ commutes with the codifferential for each $x \in \mathfrak g$, hence we have a representation of $\mathfrak g$ on the cohomology groups. It can be shown that each $\widehat \rho(x)$ is chain-homotopic to the zero morphism, so action on cohomology groups happens to be trivial. Some sources claim that the quadratic Casimir $C'$ of modules $(H^p(\mathfrak g,M),\widehat \rho)$ acts as multiplication by $c_M$. Since we have already seen that $\mathfrak g$-action is trivial, this implies that $H^p(\mathfrak g, M)=0$. I don't understand why $C'=c_M$.

Remark: Actually I'm interested in Lie superalgebras, but I will probably be able to work out this case once I learn the argument for ordinary Lie algebras.

Blazej
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    The Casimir element does not in general act as a scalar on $\frak g$-modules. This is true if $M$ is a simple module, but not say if $M$ is a direct sum of non-isomorphic simple modules. – Angina Seng Sep 01 '18 at 15:29
  • Thank you for this comment. Indeed, I assume that $M$ is simple and forgot to write this. – Blazej Sep 01 '18 at 15:33
  • Whitehead's first lemma states that $H^1(\mathfrak{g},M)=0$ when $\frak g$ is semisimple. Whitehead's second lemma states that $H^2(\mathfrak{g},M)=0$ when $\frak g$ is semisimple. Whitehead didn't have a third lemma: $H^3(\mathfrak{g},M)$ may be nonzero. – Angina Seng Sep 01 '18 at 15:37
  • Dear Lord Shark, I believe this is because theorem I am refering to works only for nontrivial modules (these with nonzero $\mathfrak g$-action). In particular cohomology valued in trivial module may be still nonzero. General modules may contain the trivial submodule, hence have nonzero cohomology. Two Whitehead's lemmas may be proved by using the fact I'm refering to, checking independently that $H^1(\mathfrak g, \mathbb C)=H^2(\mathfrak g, \mathbb C)=0$ and analyzing the long exact sequence in cohomology. – Blazej Sep 01 '18 at 15:43

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