Bourbaki

Lemma 2

Tag 0143, permanent. Cite it as [Bourbaki, Tag 0143].

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

Let h0=∑α∈R+Hαh^0=\sum_{\alpha\in R_+}H_{\alpha}. Then h0=∑α∈BaαHαh^0=\sum_{\alpha\in B}a_{\alpha}H_{\alpha}, where the aαa_{\alpha} are integers ≥1\geq 1. Let (bα)α∈B,(cα)α∈B(b_{\alpha})_{\alpha\in B},(c_{\alpha})_{\alpha\in B} be families of scalars such that bαcα=aαb_{\alpha}c_{\alpha}=a_{\alpha} for all α∈B\alpha \in B. Put x=∑α∈BbαXα,y=∑α∈BcαX−αx=\sum_{\alpha\in B}b_{\alpha}X_{\alpha}, y=\sum_{\alpha\in B}c_{\alpha}X_{-\alpha}. There exists a homomorphism φ\varphi

from sl(2,k)\mathfrak{s}\mathfrak{l}(2, k) to g\mathfrak{g} such that φ(H)=h0,φ(X+)=x,φ(X−)=y\varphi (H) =h^0, \varphi (X_+) =x, \varphi (X_-) =y.

The fact that the aαa_{\alpha} are integers ≥1\geq 1 follows from the fact that (Hα)α∈B(H_{\alpha})_{\alpha\in B} is a basis of the root system (Hα)α∈B(H_{\alpha})_{\alpha\in B} (cf. Chap. VI, §1, no. 5, Remark 5). We have:

α(h0)=2(1) \alpha (h^0) = 2 \tag{1}

for all α∈B\alpha \in B (Chap. VI, §1, no. 10, Cor. of Prop. 29), so

[h0,x]=∑α∈Bbαα(h0)Xα=2x(2) [h^0, x] =\sum_{\alpha\in B}b_{\alpha}\alpha (h^0)X_{\alpha}= 2x \tag{2}
[h0,y]=∑α∈Bcα(−α(h0))X−α=−2y(3) [h^0, y] =\sum_{\alpha\in B}c_{\alpha}(-\alpha (h^0))X_{-\alpha}=-2y \tag{3}

On the other hand,

[x,y]=∑α,β∈Bbαcβ[Xα,X−β]=∑α∈Bbαcα[Xα,X−α]=−∑α∈BaαHα=−h0(4) [x, y] =\sum_{\alpha ,\beta\in B}b_{\alpha}c_{\beta}[X_{\alpha}, X_{-\beta}] =\sum_{\alpha\in B}b_{\alpha}c_{\alpha}[X_{\alpha}, X_{-\alpha}] =-\sum_{\alpha\in B}a_{\alpha}H_{\alpha}=-h^0 \tag{4}

hence the existence of the homomorphism φ\varphi.

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. 10, Cor. of Prop. 29 not in this corpus
  • Chap. VI, §1, no. 5, Remark 5 not in this corpus
Transcribed from the printed volume. Not translated.