Bourbaki

Proposition 1

Tag 0138, permanent. Cite it as [Bourbaki, Tag 0138].

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

Let V be a finite dimensional g\mathfrak{g}-module.

(i) All the weights of V belong to P.

(ii) V=⨁μ∈PVμV =\bigoplus_{\mu\in P}V^\mu.

(iii) For all μ∈h∗,Vμ\mu\in \mathfrak{h}^*,V^\mu is the set of x∈Vx\in V such that h.x=μ(h)xh.x=\mu(h)x for all h∈hh\in \mathfrak{h}.

For all α∈B\alpha \in B, there exists a homomorphism from sl(2,k)\mathfrak{s}\mathfrak{l}(2, k) to g\mathfrak{g} that takes HH to HαH_{\alpha}. Thus, by §1, no. 2, Cor. of Prop. 2,(Hα)V2, (H_{\alpha})_V is diagonalizable and its eigenvalues are integers. Hence, the set of (Hα)V(H_{\alpha})_V, for α∈B\alpha \in B, is diagonalizable (Algebra, Chap. VII, §5, no. 6, Prop. 13). Consequently, for all h∈h,hVh\in \mathfrak{h},h_V is diagonalizable. By Chap. VII, §1, no. 3, Prop. 9,V=⨁μ∈h∗Vμ9, V =\bigoplus_{\mu\in\mathfrak{h}^*}V^\mu. On the other hand, if Vμ≠0V^\mu\not= 0, the preceding shows that μ(Hα)∈Z\mu(H_{\alpha})\in \mathbf{Z} for all α∈B\alpha \in B, so μ∈P\mu\in P. This proves (i) and (ii). We see in the same way that hV\mathfrak{h}_V is diagonalizable, hence (iii).

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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