Dimension formula from roots

By [L1] the vector space $\mathfrak{g}$ is the direct sum of $\mathfrak{h}$ and one one-dimensional space for each of the $\mathrm{\mid }\mathrm{\Phi }\mathrm{\mid }$ roots; dimensions are additive over direct sums, so $\dim \mathfrak{g}=\dim \mathfrak{h}+\mathrm{\mid }\mathrm{\Phi }\mathrm{\mid }$.

Let ${\mathfrak{h}}^{\circ }$ be the complement in $\mathfrak{h}$ of the finitely many root hyperplanes $\ker \alpha$. A finite union of proper linear subspaces cannot cover a complex vector space, so ${\mathfrak{h}}^{\circ }$ is nonempty (and equals $\{0\}$ when $\mathfrak{g}=0$). For $H\in {\mathfrak{h}}^{\circ }$, [L1] gives ${\mathfrak{g}}^H=\ker ({\mathrm{ad}}_H)=\mathfrak{h}$, because ${\mathrm{ad}}_H$ acts by the nonzero scalar $\alpha (H)$ on every ${\mathfrak{g}}_{\alpha }$.

Let $G$ be the connected real Lie group supplied by [L3] for the underlying real Lie algebra of $\mathfrak{g}$, and define $F:G\times {\mathfrak{h}}^{\circ }⟶\mathfrak{g}$ by $F(g,H)={\mathrm{Ad}}_{g}H$. At $(e,H)$ its differential is $(Y,K)\mapsto [Y,H]+K$ by [L3]. The root decomposition [L1] and the inequalities $\alpha (H)\mathrm{\ne }0$ give $[\mathfrak{g},H]={⨁}_{\alpha \in \mathrm{\Phi }}{\mathfrak{g}}_{\alpha }$, so this differential is onto. Translation in $G$ and composition with ${\mathrm{Ad}}_g$ show the same at every $(g,H)$; hence $F$ is a submersion. By [L3] its image $U$ is a nonempty open subset of $\mathfrak{g}$, and every point of $U$ has centralizer dimension $\dim \mathfrak{h}$ by step 1.2 and conjugation invariance.

Put $r=\mathrm{rank}\mathfrak{g}$ and $n=\dim \mathfrak{g}$. In a fixed basis the entries of ${\mathrm{ad}}_x$ depend linearly on $x$. Choose $x_0$ with $\dim \ker ({\mathrm{ad}}_{x_0})=r$ and an $(n-r)\times (n-r)$ minor nonzero at $x_0$. The nonvanishing set of this minor is a nonempty dense open subset $R$ of the complex vector space $\mathfrak{g}$; at every point of $R$, the adjoint map has rank at least $n-r$, and maximality of $n-r$ forces kernel dimension exactly $r$. Thus $R$ consists of regular elements. Since $R$ is dense and $U$ from step 2.1 is nonempty open, choose $x\in R\cap U$. Then $r=\dim {\mathfrak{g}}^x=\dim \mathfrak{h}$.

By [L2], all Cartan subalgebras have this same dimension; step 3.1 identifies it with $\mathrm{rank}\mathfrak{g}$. Substituting in step 1.1 yields $\dim \mathfrak{g}=\mathrm{rank}\mathfrak{g}+\mathrm{\mid }\mathrm{\Phi }\mathrm{\mid }$, including the zero algebra.