Second Whitehead lemma

Let $S$ be a simple module with nontrivial action, put $\mathfrak{k}=\ker \rho$, and use [L4] to choose a complementary ideal $\mathfrak{h}$ in $\mathfrak{g}$. The two ideals commute, $\rho {\mathrm{\mid }}_{\mathfrak{h}}$ is faithful, and $\mathfrak{h}$ is nonzero semisimple. The radical of its trace form on $S$ is an ideal by [L6]; its restricted trace form meets the hypothesis of [L5], so that radical is solvable and hence zero. Thus the trace form on $\mathfrak{h}$ is nondegenerate and defines the dual-basis Casimir operator $C$ of [L7]. By [L2], $C$ intertwines the simple module. Its trace is $\dim \mathfrak{h}\mathrm{\ne }0$, so $C$ is nonzero and therefore invertible by the kernel-image argument for an endomorphism of a simple module.

It remains to treat the trivial simple module $k$. By [L3], take a central extension $0\to k\to \mathfrak{e}\stackrel{\pi }{\to }\mathfrak{g}\to 0$. For $x\in \mathfrak{g}$, choose a lift $\tilde{x}$ and define $x\cdot y=[\tilde{x},y]$ on $\mathfrak{e}$. Centrality makes this independent of the lift, Jacobi makes it a representation, and $\pi$ is a $\mathfrak{g}$-map for the adjoint action on $\mathfrak{g}$. By [L1], the surjection has a module section $\sigma$. Taking $\tilde{x}=\sigma (x)$, equivariance gives $[\sigma (x),\sigma (y)]=\sigma ([x,y])$, so $\sigma$ is a Lie section. The extension splits and [L3] gives $H^2(\mathfrak{g},k)=0$.

For the trace-dual bases $(e_i)$ and $(e^i)$ from [L7] in the ideal $\mathfrak{h}$, define $H={\sum }_i\rho (e_i){\iota }_{e^i}$ on the CE cochains of $\mathfrak{g}$. The inverse tensor ${\sum }_i{e}_i\otimes e^i$ is invariant under $\mathfrak{h}$; it is also invariant under $\mathfrak{k}$ because the complementary ideals commute. Expanding the CE differential therefore gives $dH+Hd=C$: the value-action terms give ${\sum }_i\rho (e_i)\rho (e^i)$ and the argument-action terms cancel in pairs by this invariance. Hence $C$ acts null-homotopically in positive degrees. Since $C$ is invertible by step 1.1 and commutes with $d$, composing $H$ with $C^{-1}$ contracts every positive-degree cocycle. In particular $H^2(\mathfrak{g},S)=0$.

The CE complex commutes with finite direct sums in the coefficient module. Decompose $M$ by [L1]; steps 2.1 and 1.2 make the second cohomology of every simple summand zero, hence $H^2(\mathfrak{g},M)=0$. The zero module and zero algebra are included: for $\mathfrak{g}=0$, ${\mathrm{\Lambda }}^2\mathfrak{g}=0$. No choice principle is used beyond finite-dimensional basis choices.