Proposition 7

Let $\mathfrak{b}=\mathfrak{h}+\mathfrak{g}^P$ be a subalgebra of $\mathfrak{g}$ containing $\mathfrak{h}$. The following conditions are equivalent:

(i) $\mathfrak{b}$ is a maximal solvable subalgebra of $\mathfrak{g}$;

(ii) there exists a chamber C of R such that $P = R_+(C)$;

(iii) $P\cap (-P) =\emptyset$ and $P\cup (-P) = R$.

(i) $=\Rightarrow$ (ii): If $\mathfrak{b}$ is solvable, $P\cap (-P) =\emptyset$. Then there exists a chamber C of R such that $P\subset R_+(C)$ (Chap. VI, §1, no. 7, Prop. 22). Then $\mathfrak{h}+\mathfrak{g}^{R_+(C)}$ is a solvable subalgebra of $\mathfrak{g}$ containing $\mathfrak{b}$, hence equal to $\mathfrak{b}$ if $\mathfrak{b}$ is maximal.

(ii) $=\Rightarrow$ (iii): This is clear.

(iii) $=\Rightarrow$ (i): Assume that $P\cap (-P) =\emptyset$ and that $P\cup (-P) = R$. Then $\mathfrak{b}$ is solvable. Let $\mathfrak{b}'$ be a solvable subalgebra of $\mathfrak{g}$ containing $\mathfrak{b}$. There exists a subset Q of R such that $\mathfrak{b}'=\mathfrak{h}+\mathfrak{g}^Q$. Then $Q\cap (-Q) =\emptyset$ and $Q\supset P$, so Q = P and $\mathfrak{b}'=\mathfrak{b}$.