Bourbaki

Lemma 1

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

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

Let V be a vector subspace of g\mathfrak{g} and R(V) the set of α∈R\alpha \in R such that gα⊂V\mathfrak{g}^{\alpha}\subset V. Then, (V∩h)+∑α∈R(V)gα(V\cap \mathfrak{h}) +\sum_{\alpha\in R(V)}\mathfrak{g}^{\alpha} is the largest vector subspace of V stable under ad h\mathfrak{h}.

A vector subspace W of V is stable under ad h\mathfrak{h} if and only if

W=(W∩h)+∑α∈R(W∩gα) W = (W\cap \mathfrak{h}) +\sum_{\alpha\in R}(W\cap \mathfrak{g}^{\alpha})

(Algebra, Chap. VII, §2, no. 2, Cor. 1 of Th. 1). The largest vector subspace of V stable under ad h\mathfrak{h} is thus (V∩h)+∑α∈R(V∩gα)(V\cap \mathfrak{h}) +\sum_{\alpha\in R}(V\cap \mathfrak{g}^{\alpha}). But V∩gα=gαV\cap \mathfrak{g}^{\alpha}=\mathfrak{g}^{\alpha} for α∈R(V)\alpha \in R(V), and V∩gα=0V\cap \mathfrak{g}^{\alpha}= 0 for α∉R(V)\alpha \notin R(V) since dim gα=1\mathfrak{g}^{\alpha}= 1. Q.E.D.

For any subset P of R, put

gP=∑α∈PgαhP=∑α∈Phα \mathfrak{g}^P=\sum_{\alpha\in P}\mathfrak{g}^{\alpha}\mathfrak{h}_P=\sum_{\alpha\in P}\mathfrak{h}_{\alpha}

If P⊂RP\subset R and Q⊂RQ\subset R, we clearly have

[h,gP]⊂gP(1) [\mathfrak{h},\mathfrak{g}^P]\subset \mathfrak{g}^P \tag{1}
[gP,gQ]=g(P+Q)∩R+hP∩(−Q)(2) [\mathfrak{g}^P,\mathfrak{g}^Q] =\mathfrak{g}^{(P+Q)\cap R}+\mathfrak{h}_{P\cap(-Q)} \tag{2}

Recall (Chap. VI, §1, no. 7, Def. 4) that a subset P of R is said to be closed if the conditions α∈P,β∈P,α+β∈R\alpha \in P, \beta \in P, \alpha +\beta \in R imply α+β∈P\alpha +\beta \in P, in other words if (P+P)∩R⊂P(P + P)\cap R\subset P.

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

  • Algebra, Chap. VII, §2, no. 2, Cor. 1 of Th. 1 not in this corpus
  • Chap. VI, §1, no. 7, Def. 4 not in this corpus
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