Bourbaki

Corollary

Tag 0104, permanent. Cite it as [Bourbaki, Tag 0104].

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

The group Aute(g)_e(\mathfrak{g}) operates transitively on the set of pairs (t,b)(\mathfrak{t},\mathfrak{b}) where t\mathfrak{t} is a splitting Cartan subalgebra of g\mathfrak{g} and b\mathfrak{b} is a Borel subalgebra of (g,t)(\mathfrak{g},\mathfrak{t}).

Let (t1,b1)(\mathfrak{t}_1,\mathfrak{b}_1) and (t2,b2)(\mathfrak{t}_2,\mathfrak{b}_2) be two such pairs. There exists a splitting Cartan subalgebra t\mathfrak{t} of g\mathfrak{g} contained in b1∩b2\mathfrak{b}_1\cap \mathfrak{b}_2 (Prop. 10). By the Cor. of Prop. 9, we are reduced to the case in which t1=t2=t\mathfrak{t}_1=\mathfrak{t}_2=\mathfrak{t}. Let S be the root system of (g,t)(\mathfrak{g},\mathfrak{t}). There exists bases B1,B2B_1,B_2 of S such that bi\mathfrak{b}_i is associated to Bi(i=1,2)B_i(i= 1,2), and there exists s∈W(S)s\in W(S) which transforms B1B_1 into B2B_2. Finally, there exists a∈a\in Aute(g)_e(\mathfrak{g}) such that a∣t=s(a|\mathfrak{t}=s(§2, no. 2, Cor. of Th. 2). Then a(t)=ta(\mathfrak{t}) =\mathfrak{t} and a(b1)=b2a(\mathfrak{b}_1) =\mathfrak{b}_2.

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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Transcribed from the printed volume. Not translated.