Bourbaki

Corollary 2

Tag 012Z, permanent. Cite it as [Bourbaki, Tag 012Z].

Lie Groups and Lie Algebras, chapter VIII, § 6. Modules over a split semi-simple Lie algebra

Let ρ=12∑α∈R+α\rho =\frac{1}{2}\sum_{\alpha\in R_+}\alpha, and λ,μ∈h∗\lambda , \mu\in \mathfrak{h}^*. Assume that λ+ρ\lambda +\rho is a dominant weight in R, and that there exists w∈Ww\in W with μ+ρ=w(λ+ρ)\mu+\rho =w(\lambda +\rho ). Then Z(μ)Z(\mu) is isomorphic to a submodule of Z(λ)Z(\lambda ).

The assertion is clear when w=1w= 1. Assume that it is established whenever ww is of length <q< q. If ww is of length qq, there exists α∈B\alpha \in B such that w=sαw′−1w=s_{\alpha}w^{'-1}, with l(w′)=q−1l(w') =q-1. We have w′(α)∈R+w'(\alpha )\in R_+ (Chap. VI, §1, no. 6, Cor. 2 of Prop. 17), and hence w′−1(λ+ρ)(Hα)=(λ+ρ)(Hw′α)w^{'-1}(\lambda +\rho )(H_{\alpha}) = (\lambda +\rho )(H_{w'\alpha}) is an integer ≥0\geq 0. Put

μ′=w′−1(λ+ρ)−ρ \mu'=w^{'-1}(\lambda +\rho )-\rho

By the induction hypothesis, Z(μ′)Z(\mu') is isomorphic to a submodule of Z(λ)Z(\lambda ). On the other hand, by Chap. VI, §1, no. 10, Prop. 29 (ii),

−α+sαμ′=−α+sαw′−1(λ+ρ)−sαρ=w(λ+ρ)−ρ=μ -\alpha +s_{\alpha}\mu'=-\alpha +s_{\alpha}w^{'-1}(\lambda +\rho )-s_{\alpha}\rho =w(\lambda +\rho )-\rho =\mu

Moreover, ρ(Hα)=1\rho (H_{\alpha}) = 1 (Chap. VI, §1, Prop. 29 (iii)), so μ′(Hα)+1∈N\mu'(H_{\alpha})+1\in \mathbf{N}. Cor. 1 now implies that Z(μ)Z(\mu) is isomorphic to a submodule of Z(μ′)Z(\mu'), and hence also to a submodule of Z(λ)Z(\lambda ).

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. VI, §1, no. 6, Cor. 2 of Prop. 17 not in this corpus
  • Chap. VI, §1, Prop. 29 not in this corpus
  • Chap. VI, §1, no. 10, Prop. 29 not in this corpus
  • Cor. 1
Transcribed from the printed volume. Not translated.