Whitehead's lemmas

Remarks

For an irreducible module $V$ on which $\mathfrak{g}$ acts nontrivially, the Casimir element of $V$ acts on $V$ by a nonzero scalar. It also acts by zero on the cohomology, so all cohomology groups $H^n(\mathfrak{g}, V)$ vanish. For the trivial module $k$, one has $H^1(\mathfrak{g}, k) \cong (\mathfrak{g}/[\mathfrak{g}, \mathfrak{g}])^*$, which is zero because $\mathfrak{g} = [\mathfrak{g}, \mathfrak{g}]$. The vanishing of $H^2(\mathfrak{g}, k)$ uses the nondegeneracy of the Killing form. A long exact sequence argument then passes from irreducible modules to all finite-dimensional modules.

The hypothesis of semisimplicity is necessary. For the one-dimensional abelian Lie algebra $\mathfrak{g} = k$ and the trivial module $k$, every linear form on $\mathfrak{g}$ is a $1$-cocycle and the only coboundary is $0$, so $H^1(\mathfrak{g}, k) \cong k \neq 0$. The lemmas do not extend to $H^3$: for a simple Lie algebra $\mathfrak{g}$, the $3$-form $(x, y, z) \mapsto \kappa([x, y], z)$ built from the Killing form $\kappa$ gives a nonzero class in $H^3(\mathfrak{g}, k)$.

Going deeper

For a student

The first lemma gives a quick proof of Weyl's theorem on complete reducibility. A submodule $W \subseteq V$ gives an extension of $\mathfrak{g}$-modules, and the obstruction to a complement lies in $H^1(\mathfrak{g}, \Hom(V/W, W))$, which is zero. The second lemma is the key step in the proof of the Levi decomposition. An abelian extension of a semisimple Lie algebra is classified by $H^2$, so it splits, and induction on the radical gives a semisimple complement.