Bourbaki

Corollary 2

Tag 01J4, permanent. Cite it as [Bourbaki, Tag 01J4].

Lie Groups and Lie Algebras, chapter VIII, § 7. Finite dimensional modules over a split semi-simple Lie algebra

Let μ∈P++\mu\in P_{++} and ν∈Q+\nu \in Q_+. If μ+ν∈X\mu+\nu \in \mathscr{X}, then μ∈X\mu\in \mathscr{X}.

Write ν=∑α∈Bcα.α\nu =\sum_{\alpha\in B}c_{\alpha}.\alpha, where cα∈Nc_{\alpha}\in \mathbf{N} for all α∈B\alpha \in B. The corollary is clear

when ∑α∈Bcα=0\sum_{\alpha\in B}c_{\alpha}= 0; assume that ∑α∈Bcα>0\sum_{\alpha\in B}c_{\alpha}>0 and argue by induction on ∑α∈Bcα\sum_{\alpha\in B}c_{\alpha}. Let (⋅∣⋅)(\cdot | \cdot ) be a W-invariant non-degenerate positive symmetric bilinear form on hR∗\mathfrak{h}^*_{\mathbf{R}}. Then (ν∣∑α∈Bcα.α)>0(\nu |\sum_{\alpha\in B}c_{\alpha}.\alpha )>0, so there exists β∈B\beta \in B such that cβ≥1c_{\beta}\geq 1 and

(ν∣β)>0(\nu |\beta )>0, hence ν(Hβ)≥1\nu (H_{\beta})\geq 1. Since μ∈P++\mu\in P_{++}, it follows that (μ+ν)(Hβ)≥1(\mu+\nu )(H_{\beta})\geq 1. By Cor. 1,μ+(ν−β)∈X1,\mu+ (\nu -\beta )\in \mathscr{X}, and it suffices to apply the induction hypothesis.

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.