Bourbaki

Theorem 1

Tag 00V8, permanent. Cite it as [Bourbaki, Tag 00V8].

Lie Groups and Lie Algebras, chapter VII, § 3. Conjugacy theorems

Assume that kk is algebraically closed. Let g\mathfrak{g} be a Lie algebra. The group E is normal in Aut(g\mathfrak{g}) and operates transitively on the set of Cartan subalgebras of g\mathfrak{g}.

Let h\mathfrak{h} be a Cartan subalgebra of g\mathfrak{g}, and v∈v\in Aut(g\mathfrak{g}). Then

vE(h)v−1=E(v(h))=E(h) vE(\mathfrak{h})v^{-1}= E(v(\mathfrak{h})) = E(\mathfrak{h})

so E(h)=EE(\mathfrak{h}) = E is normal in Aut(g\mathfrak{g}). If h′\mathfrak{h}' is another Cartan subalgebra of g\mathfrak{g}, then, in the notation of Prop. 2,u′−1u(h)=h′2,u^{'-1}u(\mathfrak{h}) =\mathfrak{h}', and u′−1u∈Eu^{'-1}u\in E.

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.

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Cites

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Transcribed from the printed volume. Not translated.