Proposition 5

Tag 00ZU, permanent. Cite it as [Bourbaki, Tag 00ZU].

Lie Groups and Lie Algebras, chapter VIII, § 3. Subalgebras of split semi-simple Lie algebras

Let P be a closed subset of R.

(i) $\mathfrak{h}+\mathfrak{g}^P$ is solvable if and only if $P\cap (-P) =\emptyset$ . In that case, $[\mathfrak{h}+\mathfrak{g}^P,\mathfrak{h}+\mathfrak{g}^P] =\mathfrak{g}^P$ .

(ii) $\mathfrak{h}+\mathfrak{g}^P$ is reductive if and only if $P =-P$ .

Assertion (i) follows from Prop. 2 (iii). If $P =-P,\mathfrak{h}+\mathfrak{g}^P$ is reductive (Prop. 2 (iv)). Assume that $\mathfrak{a}=\mathfrak{h}+\mathfrak{g}^P$ is reductive. Then

\mathfrak{g}^P= [\mathfrak{h},\mathfrak{g}^P]\subset [\mathfrak{a},\mathfrak{a}]\subset \mathfrak{h}+\mathfrak{g}^P

so $[\mathfrak{a},\mathfrak{a}]$ is of the form $\mathfrak{h}'+\mathfrak{g}^P$ with $\mathfrak{h}'\subset \mathfrak{h}$ ; since $[\mathfrak{a},\mathfrak{a}]$ is semi-simple, $P =-P$ (Prop. 2 (ii)).

Bourbaki sets the proof directly after the statement, and the corpus does not separate them, so what is above is the statement and its proof together.

Cited by

Proposition 9, § 3

Cites

Prop. 2