Bourbaki

Lemma 2

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

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

Let h′\mathfrak{h}' be a vector subspace of h\mathfrak{h} and P a subset of R. Then h′+gP\mathfrak{h}'+\mathfrak{g}^P is a subalgebra of g\mathfrak{g} if and only if P is a closed subset of R and

h′⊃hP∩(−P) \mathfrak{h}'\supset \mathfrak{h}_{P\cap(-P)}

Indeed,

[h′+gP,h′+gP]=[h′,gP]+[gP,gP]=hP∩(−P)+[h′,gP]+g(P+P)∩R [\mathfrak{h}'+\mathfrak{g}^P,\mathfrak{h}'+\mathfrak{g}^P] = [\mathfrak{h}',\mathfrak{g}^P] + [\mathfrak{g}^P,\mathfrak{g}^P] =\mathfrak{h}_{P\cap(-P)}+ [\mathfrak{h}',\mathfrak{g}^P] +\mathfrak{g}^{(P+P)\cap R}

Hence h′+gP\mathfrak{h}'+\mathfrak{g}^P is a subalgebra of g\mathfrak{g} if and only if

hP∩(−P)⊂h′\mathfrak{h}_{P\cap(-P)}\subset \mathfrak{h}' and g(P+P)∩R⊂gP\mathfrak{g}^{(P+P)\cap R}\subset \mathfrak{g}^P

which proves the lemma.

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

Cites

Nothing.

Transcribed from the printed volume. Not translated.