Bourbaki

Lemma 1

Tag 00YV, permanent. Cite it as [Bourbaki, Tag 00YV].

Lie Groups and Lie Algebras, chapter VIII, § 2. Root system of a split semi-simple Lie algebra

(i) For all h∈h,θα(t).h=h−α(h)Hαh\in \mathfrak{h},\theta_{\alpha}(t).h=h-\alpha (h)H_{\alpha}.

(ii) For all β∈R,θα(t)(gβ)=gβ−β(Hα)α\beta \in R,\theta_{\alpha}(t)(\mathfrak{g}^{\beta}) =\mathfrak{g}^{\beta-\beta(H_{\alpha})\alpha}.

(iii) If α,β∈R,β−β(Hα)α∈R\alpha , \beta \in R,\beta -\beta (H_{\alpha})\alpha \in R.

Let h∈hh\in \mathfrak{h}. If α(h)=0,[Xα,h]=[X−α,h]=0\alpha (h) = 0, [X_{\alpha}, h] = [X_{-\alpha}, h] = 0, so θα(t).h=h\theta_{\alpha}(t).h=h. On the other hand, the formulas (5) of §1, no. 5 show that θα(t).Hα=−Hα\theta_{\alpha}(t).H_{\alpha}=-H_{\alpha}. This proves assertion (i). It follows that θα(t)2∣h=\theta_{\alpha}(t)^2|\mathfrak{h}= Id. If x∈gβx\in \mathfrak{g}^{\beta} and h∈hh\in \mathfrak{h},

[h,θα(t)x]=θα(t).[θα(t)h,x]−β(θα(t)h).θα(t)x [h, \theta_{\alpha}(t)x] =\theta_{\alpha}(t).[\theta_{\alpha}(t)h, x]-\beta (\theta_{\alpha}(t)h).\theta_{\alpha}(t)x
=(β(h)−α(h)β(Hα)).θα(t)x = (\beta (h)-\alpha (h)\beta (H_{\alpha})).\theta_{\alpha}(t)x
=(β−β(Hα)α)(h).θα(t)x = (\beta -\beta (H_{\alpha})\alpha )(h).\theta_{\alpha}(t)x

so θα(t)x∈gβ−β(Hα)α\theta_{\alpha}(t)x\in \mathfrak{g}^{\beta-\beta(H_{\alpha})\alpha}. This proves (ii). Assertion (iii) follows from (ii).

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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