Root-space decomposition

By [L1] the subalgebra $\mathfrak{h}$ is abelian and ${\mathrm{ad}}_H$ is semisimple for every $H\in \mathfrak{h}$; the family $\{{\mathrm{ad}}_H{\}}_{H\in \mathfrak{h}}$ is therefore pairwise commuting and [L2] makes it simultaneously diagonalisable. Hence $\mathfrak{g}={⨁}_{\alpha \in {\mathfrak{h}}^{\ast }}{\mathfrak{g}}_{\alpha }$ with the ${\mathfrak{g}}_{\alpha }$ of [L3].

The zero weight space is ${\mathfrak{g}}_0=\{x:[H,x]=0\text{ for all }H\in \mathfrak{h}\}=C_{\mathfrak{g}}(\mathfrak{h})$. Since $\mathfrak{h}$ is abelian we have $\mathfrak{h}\subseteq C_{\mathfrak{g}}(\mathfrak{h})$, and $C_{\mathfrak{g}}(\mathfrak{h})\subseteq N_{\mathfrak{g}}(\mathfrak{h})=\mathfrak{h}$ by [L1]; hence ${\mathfrak{g}}_0=\mathfrak{h}$.

Consequently $\mathfrak{g}=\mathfrak{h}\oplus {⨁}_{\alpha \mathrm{\ne }0}{\mathfrak{g}}_{\alpha }$ where the sum runs over all nonzero functionals, and deleting the zero summands leaves precisely the sum over the roots; the decomposition is direct because it is a subsum of a direct sum. Only finitely many root spaces are nonzero, because $\mathfrak{g}$ is finite-dimensional and the summands are linearly independent nonzero subspaces, so $\mathrm{\Phi }$ is finite. The Axiom of Choice was inherited from [L1].