Bourbaki

Lemma 3

Tag 0108, permanent. Cite it as [Bourbaki, Tag 0108].

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

Let V be a finite dimensional real vector space, S a root system in V∗,PV^*,\mathscr{P} the set of parabolic subsets of S; let H\mathscr{H} be the set of Ker α\alpha for α∈S\alpha \in S, and F\mathscr{F} the set of facets of V relative to H\mathscr{H} (Chap. V, §1, no. 2, Def. 1).

If P∈PP\in \mathscr{P}, let F(P) be the set of v∈Vv\in V such that α(v)≥0\alpha (v)\geq 0 for all α∈P\alpha \in P. If F∈FF\in \mathscr{F}, let P(F) be the set of α∈R\alpha \in R such that α(v)≥0\alpha (v)\geq 0 for all v∈Fv\in F.

Then F→P(F)F \rightarrow P(F) is a bijection from F\mathscr{F} to P\mathscr{P}; for all F∈FF\in \mathscr{F}, F(P(F)) is the closure of F.

a) Let P∈PP\in \mathscr{P}. There exists a chamber C of S and a subset Σ\Sigma of the basis B(C) such that P=S+(C)∪QP = S_+(C)\cup Q where Q is the set of linear combinations of elements of Σ\Sigma with non-positive integer coefficients (Chap. VI, §1, no. 7, Prop. 20). Put

B(C)={α1,...,αl},Σ={α1,...,αm} B(C) =\{\alpha_1, . . . , \alpha_l\}, \Sigma =\{\alpha_1, . . . , \alpha_m\}

If v∈Vv\in V, we have the following equivalences:

α(v)≥0\alpha (v)\geq 0 for all α∈P\alpha \in P

⇐⇒α1(v)≥0,...αl(v)≥0,α1(v)≤0,...,αm(v)≤0 \Leftarrow \Rightarrow \alpha_1(v)\geq 0, . . . \alpha_l(v)\geq 0, \alpha_1(v)\leq 0, . . . , \alpha_m(v)\leq 0
⇐⇒α1(v)=⋯=αm(v)=0,αm+1(v)≥0,...,αl(v)≥0 \Leftarrow \Rightarrow \alpha_1(v) =\cdots =\alpha_m(v) = 0, \alpha_{m+1}(v)\geq 0, . . . , \alpha_l(v)\geq 0

so F(P) is the closure of the set

{v∈V∣α1(v)=⋯=αm(v)=0,αm+1(v)>0,...,αl(v)>0} \{v\in V|\alpha_1(v) =\cdots =\alpha_m(v) = 0, \alpha_{m+1}(v)>0, . . . , \alpha_l(v)>0\}

a set which is a facet F relative to H\mathscr{H} since every element of S is a linear combination of α1,...,αl\alpha_1, . . . , \alpha_l in which the coefficients are either all ≥0\geq 0 or all ≤0\leq 0. Moreover, if β=u1α1+⋯+ulαl∈S\beta =u_1\alpha_1+\cdots +u_l\alpha_l\in S,

β∈P(F)⇐⇒um+1≥0,...,ul≥0 \beta \in P(F)\Leftarrow \Rightarrow u_{m+1}\geq 0, . . . , u_l\geq 0

⇐⇒β∈S+(C)\Leftarrow \Rightarrow \beta \in S_+(C) or (−β∈S+(C)(-\beta \in S_+(C) and um+1=...=ul=0)u_{m+1}=. . .=u_l= 0)

⇐⇒β∈S+(C)∪Q=P \Leftarrow \Rightarrow \beta \in S_+(C)\cup Q = P

so P(F) = P.

b) Let F∈FF\in \mathscr{F}. It is clear that P(F)∈PP(F)\in \mathscr{P}. On the other hand, F is contained in the closure of a chamber relative to H\mathscr{H} (Chap. V, §1, no. 3, formulas (6)), and so is a facet relative to the set of walls of this chamber (Chap. V, §1, no. 4, Prop. 9). Consequently, F‾\overline{F} is of the form {v∈V∣α(v)≥0\{v\in V|\alpha (v)\geq 0 for all α∈T}\alpha \in T\}, where T is a subset of S which we can clearly take to be equal to P(F). Thus, F‾=F(P(F))\overline{F}= F(P(F)). Q.E.D.

If P∈PP\in \mathscr{P}, the facet F such that P = P(F) is said to be associated to P; we denote it by F(P). We extend these conventions to the case in which (g,h)(\mathfrak{g},\mathfrak{h}) is split reductive.

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

  • Chap. V, §1, no. 3
  • Chap. V, §1, no. 2, Def. 1 not in this corpus
  • Chap. V, §1, no. 4, Prop. 9 not in this corpus
  • Chap. VI, §1, no. 7, Prop. 20 not in this corpus
Transcribed from the printed volume. Not translated.