Bourbaki

Lemma 1

Tag 011S, permanent. Cite it as [Bourbaki, Tag 011S].

Lie Groups and Lie Algebras, chapter VIII, § 5. Automorphisms of a semi-simple Lie algebra

Let α∈R\alpha \in R and t∈k∗t\in k^*. Let φ\varphi be the homomorphism λ→tλ(Hα)\lambda \rightarrow t^{\lambda(H_{\alpha})} from Q(R) to k∗k^*. Then f(φ)=θα(t)θα(−1)f(\varphi ) =\theta_{\alpha}(t)\theta_{\alpha}(-1).

Let ρ\rho be the representation of sl(2,k)\mathfrak{s}\mathfrak{l}(2, k) on g\mathfrak{g} associated to XαX_{\alpha}. Let π\pi be the representation of SL(2,k)\mathbf{S}\mathbf{L}(2, k) compatible with ρ\rho. Introduce the notations θ(t),h(t)\theta (t), h(t) of §1, no. 5. Since ρ(H)=\rho (H) = ad HαH_{\alpha}, the elements of gλ\mathfrak{g}^{\lambda} are of weight λ(Hα)\lambda (H_{\alpha}) for ρ\rho. By §2, no. 2,θα(t)θα(−1)=π(θ(t)θ(−1))=π(h(t))2,\theta_{\alpha}(t)\theta_{\alpha}(-1) =\pi (\theta (t)\theta (-1)) =\pi (h(t)). Hence the restriction of θα(t)θα(−1)\theta_{\alpha}(t)\theta_{\alpha}(-1) to gλ\mathfrak{g}^{\lambda} is the homothety of ratio tλ(Hα)(t^{\lambda(H_{\alpha})}(§1, no. 5, Prop. 6), hence the lemma.

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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Transcribed from the printed volume. Not translated.