how can I show $H^1(g , Hom_C(g,M))=0$?

Question

For a simple Lie algebra $g$ and a finite dimensional vector space $M$ with a trivial $g-$action, how can I show $H^1(g , Hom_C(g,M))=0$?

Answer 1

Given two $\mathfrak{g}$-modules $A$ and $B$, $M=Hom(A,B)$ is a $\mathfrak{g}$-module via $$ (x\cdot \phi)(a)=x.\phi(a)-\phi(x\cdot a), $$ for $a\in A$, $x\in \mathfrak{g}$ and $\phi\in M$. By Whitehead's first lemma, $H^1(\mathfrak{g},M)=0$ for semisimple Lie algebras $\mathfrak{g}$ and for all finite-dimensional $\mathfrak{g}$-modules, in characteristic zero. In the above case, $A=\mathfrak{g}$ is the adjoint module. Note that $$ H^1(\mathfrak{g},Hom(A,B))=Ext(A,B), $$ the space of equivalence classes of module extensions of $A$ by $B$.