Proposition 2

Let $\mathfrak{a}$ be a subalgebra of $\mathfrak{g}$ stable under ad$\mathfrak{h}$, and let $\mathfrak{h}'\subset \mathfrak{h}$, $P\subset R$ be such that $\mathfrak{a}=\mathfrak{h}'+\mathfrak{g}^P$.

(i) Let $\mathfrak{k}$ be the set of $x\in \mathfrak{h}'$ such that $\alpha (x) = 0$ for all $\alpha \in P\cap (-P)$. The radical of $\mathfrak{a}$ is $\mathfrak{k}+\mathfrak{g}^Q$, where Q is the set of $\alpha \in P$ such that $-\alpha \notin P$. Moreover, $\mathfrak{g}^Q$ is a nilpotent ideal of $\mathfrak{a}$.

(ii) $\mathfrak{a}$ is semi-simple if and only if $P =-P$ and $\mathfrak{h}'=\mathfrak{h}_P$.

(iii) $\mathfrak{a}$ is solvable if and only if $P\cap (-P) =\emptyset$. In that case $[\mathfrak{a},\mathfrak{a}] =\mathfrak{g}^S$, where

(iv) $\mathfrak{a}$ is reductive in $\mathfrak{g}$ if and only if $P =-P$.

(v) $\mathfrak{a}$ consists of nilpotent elements if and only if $\mathfrak{h}'= 0$. Then $P\cap (-P) =$ $\emptyset$, and $\mathfrak{a}$ is nilpotent.

We prove (v). If $\mathfrak{a}$ consists of nilpotent elements, $\mathfrak{a}$ is clearly nilpotent, and $\mathfrak{h}'= 0$ since the elements of $\mathfrak{h}$ are semi-simple. Assume that $\mathfrak{h}'= 0$. By Prop. 1 (i), $P\cap (-P) =\emptyset$. By Chap. VI, §1, no. 7, Prop. 22, there exists a chamber C of R such that $P\subset R_+(C)$. Hence, there exists an integer $n >0$ with the following properties: if $\alpha_1, . . . , \alpha_n\in P$ and $\beta \in R\cup \{0\}$, then

This implies that every element of $\mathfrak{g}^P$ is nilpotent, hence (v).

We prove (iii). If $P\cap (-P) =\emptyset ,\mathfrak{g}^P$ is a subalgebra of $\mathfrak{g}$ (Prop. 1 (i)), and is nilpotent by (v). Now

so $\mathfrak{a}$ is solvable and $[\mathfrak{a},\mathfrak{a}]$ is given by the formula in the proposition. If $P\cap (-P)\not=\emptyset$, let $\alpha \in P$ be such that $-\alpha \in P$. Then $\mathfrak{h}_{\alpha}+\mathfrak{g}^{\alpha}+\mathfrak{g}^{-\alpha}$ is a simple subalgebra of $\mathfrak{a}$ so $\mathfrak{a}$ is not solvable.

We prove (i). Since P is closed, $(P + Q)\cap R\subset P$. If $\alpha \in P, \beta \in Q$ and $\alpha +\beta \in R$, we cannot have $\alpha +\beta \in -P$, for, P being closed, this would imply that $-\beta =-(\alpha +\beta ) +\alpha \in P$ whereas $\beta \in Q$; thus, $(P + Q)\cap R\subset Q$. This proves that $\mathfrak{g}^Q$ is an ideal of $\mathfrak{a}$, nilpotent by (v). We have $P\cap (-Q) =\emptyset$, and $P\cap (-P) = P\cap \complement Q$, so $\mathfrak{h}_{P\cap(-Q)}\subset \mathfrak{k}\subset \bigcap_{\alpha\in P,\alpha \notin Q}$ Ker$\alpha$. By Prop. 1 (ii),

$\mathfrak{k}+\mathfrak{g}^Q$ is an ideal of $\mathfrak{a}$. Since $Q\cap (-Q) =\emptyset$, this ideal is solvable by (iii). It is therefore contained in the radical $\mathfrak{r}$ of $\mathfrak{a}$. Since $\mathfrak{r}$ is stable under every derivation of $\mathfrak{a},\mathfrak{r}$ is stable under ad $\mathfrak{h}$. Hence there exists a subset S of P such that $\mathfrak{r}= (\mathfrak{r}\cap \mathfrak{h}) +\mathfrak{g}^S$. Suppose that $\alpha \in S$ and that $-\alpha \in P$. Then $\mathfrak{h}_{\alpha}= [\mathfrak{g}^{\alpha},\mathfrak{g}^{-\alpha}]\subset \mathfrak{r}$, so $\mathfrak{g}^{-\alpha}= [\mathfrak{h}_{\alpha},\mathfrak{g}^{-\alpha}]\subset \mathfrak{r}= 0$, so that $-\alpha \in S$; by (iii), this contradicts the fact that $\mathfrak{r}$ is solvable. Consequently, $S\subset Q$. Finally, if $x\in \mathfrak{r}\cap \mathfrak{h}$ and if $\alpha \in P\cap (-P)$, then $[x,\mathfrak{g}^{\alpha}]\subset \mathfrak{g}^{\alpha}\cap \mathfrak{r}= 0$, so $\alpha (x) = 0$; this shows that $x\in \mathfrak{k}$. Hence $\mathfrak{r}\subset \mathfrak{k}+\mathfrak{g}^Q$ and the proof of (i) is complete.

We prove (iv). By (i), the adjoint representation of $\mathfrak{a}$ on $\mathfrak{g}$ is semi-simple if and only if ad$_{\mathfrak{g}}x$ is semi-simple for all $x\in \mathfrak{k}+\mathfrak{g}^Q$ (Chap. I, §6, no. 5, Th. 4); by (v), this is the case if and only if $Q =\emptyset$, in other words $P =-P$.

We prove (ii). If $\mathfrak{a}$ is semi-simple, $P =-P$ by (i), so $\mathfrak{h}_P\subset \mathfrak{h}'$; further, $\mathfrak{a}= [\mathfrak{a},\mathfrak{a}]\subset \mathfrak{h}_P+\mathfrak{g}^P$ and consequently $\mathfrak{h}'=\mathfrak{h}_P$. If $P =-P$ and $\mathfrak{h}'=\mathfrak{h}_P,\mathfrak{a}$ is reductive by (iv), and $\mathfrak{a}=\sum_{\alpha\in P}\mathfrak{s}_{\alpha}$, so $\mathfrak{a}= [\mathfrak{a},\mathfrak{a}]$ and $\mathfrak{a}$ is semi-simple.