Lemma 11

Let $\mu\in Q$. If $\mu \notin R\cup \{0\}$, then $\mathfrak{g}^\mu= 0$. If $\mu\in R$, then dim$\mathfrak{g}^\mu= 1$.

1) If $\mu$ is not a multiple of an element of R, there exists $w\in W$ such that $w\mu \notin Q_+\cup Q_-$ (Lemma 10), so $\mathfrak{a}^{w\mu}= 0,\mathfrak{g}^{w\mu}= 0$, and hence $\mathfrak{g}^\mu= 0$ (Lemma 9).

2) Let $\alpha \in B$ and let $m$ be an integer. Since $\mathfrak{a}_+$ is a free Lie algebra with basic family $(x_{\alpha})_{\alpha\in B}$, we have dim$\mathfrak{a}^{\alpha}= 1$ and $\mathfrak{a}^{m\alpha}= 0$ for $m >1$ (Chap. II, §2, no. 6, Prop. 4). Hence dim$\mathfrak{g}^{\alpha}\leq 1$ and $\mathfrak{g}^{m\alpha}= 0$ for $m >1$. We cannot have $\mathfrak{g}^{\alpha}= 0$, as this would imply that $x_{\alpha}\in \mathfrak{n}+\theta \mathfrak{n}$, and hence that $\mathfrak{n}+\theta \mathfrak{n}$ contains $h_{\alpha}=-[x_{\alpha}, x_{-\alpha}]$, whereas $\mathfrak{a}^0\cap (\mathfrak{n}+\theta \mathfrak{n}) = 0$. Consequently, dim$\mathfrak{g}^{\alpha}= 1$.

3) If $\mu\in R$, there exists $w\in W(R)$ such that $w(\mu)\in B$ (Chap. VI, §1, no. 5, Prop. 15), so dim $\mathfrak{g}^\mu=$ dim $\mathfrak{g}^{w\mu}= 1$. Moreover, if $n$ is an integer $>1$ then $\mathfrak{g}^{nw(\mu)}= 0$ and so $\mathfrak{g}^{n\mu}= 0$.

Proof of Theorem 1.

Since dim$\mathfrak{g}^0=$ Card B, it follows from Lemma 11 that $\mathfrak{g}$ is of finite dimension equal to Card B + Card R. We show that $\mathfrak{g}$ is semi-simple. Let $\mathfrak{k}$ be a commutative ideal of $\mathfrak{g}$. Since $\mathfrak{k}$ is stable under ad($\mathfrak{h}$)$,\mathfrak{k}= (\mathfrak{k}\cap \mathfrak{h}) +\sum_{\mu\in R}(\mathfrak{k}\cap \mathfrak{g}^\mu)$.

It is clear that, for all $\alpha \in B,\mathfrak{g}^{\alpha}+\mathfrak{g}^{-\alpha}+kH_{\alpha}$ is isomorphic to $\mathfrak{s}\mathfrak{l}(2, k)$. In view of Lemma 9, for all $\mu\in R,\mathfrak{g}^\mu$ is contained in a subalgebra of $\mathfrak{g}$ isomorphic to $\mathfrak{s}\mathfrak{l}(2, k)$; consequently, $\mathfrak{k}\cap \mathfrak{g}^\mu= 0$, so $\mathfrak{k}\subset \mathfrak{h}$; hence

so $\mu_B(\mathfrak{k}) = 0$ for all $\mu\in R$. It follows that $\mathfrak{k}= 0$, which proves that $\mathfrak{g}$ is semi-simple.

Let $\mu\in R$. There exists $\alpha \in B$ such that $\langle \mu, \alpha^{\vee}\rangle \not= 0$, and (ad $H_{\alpha}$)$|\mathfrak{g}^\mu$ is then a non-zero homothety. Consequently, $\mathfrak{h}$ is equal to its own normalizer in $\mathfrak{g}$, and hence is a Cartan subalgebra of $\mathfrak{g}$. For all $u\in \mathfrak{h}$, ad $u$ is diagonalizable, so $(\mathfrak{g},\mathfrak{h})$ is a split semi-simple Lie algebra.

For all $\mu\in R$, it is clear that $\mu_B$ is a root of $(\mathfrak{g},\mathfrak{h})$ and that $\mathfrak{g}^\mu$ is the corresponding eigenspace. The number of roots of $(\mathfrak{g},\mathfrak{h})$ is dim$\mathfrak{g}-$ dim $\mathfrak{h}=$ Card R. Hence, the map $\mu \rightarrow \mu_B$ from V to $\mathfrak{h}^*$ maps R to the root system of $(\mathfrak{g},\mathfrak{h})$.