Bourbaki

Lemma 4

Tag 00Z8, permanent. Cite it as [Bourbaki, Tag 00Z8].

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

Let α,β∈R\alpha , \beta \in R be such that α+β∈R\alpha +\beta \in R. Let pp (resp. qq) be the largest integer jj such that β+jα∈R\beta +j\alpha \in R (resp. β−jα∈R\beta -j\alpha \in R). Then,

Nα,βN−α,α+β=−p(q+1)(3) N_{\alpha ,\beta}N_{-\alpha ,\alpha+\beta}=-p(q+ 1) \tag{3}
N−α,α+β⟨Hβ,Hβ⟩=−N−α,−β⟨Hα+β,Hα+β⟩(4) N_{-\alpha ,\alpha+\beta}\langle H_{\beta}, H_{\beta}\rangle =-N_{-\alpha ,-\beta}\langle H_{\alpha+\beta}, H_{\alpha+\beta}\rangle \tag{4}
Nα,βN−α,−β=(q+1)2(5) N_{\alpha ,\beta}N_{-\alpha ,-\beta}= (q+ 1)^2 \tag{5}

Let ρ\rho be the representation of sl(2,k)\mathfrak{s}\mathfrak{l}(2, k) on g\mathfrak{g} defined by XαX_{\alpha}. The element e=Xβ+pαe=X_{\beta+p\alpha} is primitive of weight p+qp+q (Prop. 4 (i)). Put

en=(−n1)e_n=(-n1)! nρ(X−)ne^n\rho (X_-)^ne for n≥0n\geq 0.

By Prop. 1 of §1,

(ad XαX_{\alpha})ep=(q+1)ep−1e_p= (q+ 1)e_{p-1}

(ad X−αX_{-\alpha})(ad XαX_{\alpha})ep=−p(q+1)epe_p=-p(q+ 1)e_p.

This proves (3) since epe_p is a non-zero element of gβ\mathfrak{g}^{\beta}.

The form ⟨⋅,⋅⟩\langle \cdot ,\cdot \rangle being invariant, we have

⟨[X−α,Xα+β],X−β⟩=−⟨Xα+β,[X−α,X−β]⟩ \langle [X_{-\alpha}, X_{\alpha+\beta}], X_{-\beta}\rangle =-\langle X_{\alpha+\beta},[X_{-\alpha}, X_{-\beta}]\rangle

so

N−α,α+β⟨Xβ,X−β⟩=−N−α,−β⟨Xα+β,X−α−β⟩ N_{-\alpha ,\alpha+\beta}\langle X_{\beta}, X_{-\beta}\rangle =-N_{-\alpha ,-\beta}\langle X_{\alpha+\beta}, X_{-\alpha-\beta}\rangle

which, in view of Lemma 3, proves (4).

The restriction of ⟨⋅,⋅⟩\langle \cdot ,\cdot \rangle to h\mathfrak{h} is non-degenerate and invariant under the Weyl group (Prop. 5). Identify h\mathfrak{h} and h∗\mathfrak{h}^* by means of this restriction. If γ∈R,Hγ\gamma \in R,H_{\gamma} is identified with 2γ/⟨γ,γ⟩2\gamma /\langle \gamma , \gamma \rangle (Chap. VI, §1, no. 1, Lemma 2); hence, for all γ,δ∈R\gamma , \delta \in R,

⟨γ,γ⟩⟨Hδ,Hδ⟩ \langle \gamma , \gamma \rangle \langle H_{\delta}, H_{\delta}\rangle
⟨δ,δ⟩‾=⟨Hγ,Hγ⟩‾(6) \overline{\langle\delta , \delta\rangle}=\overline{\langle H_{\gamma}, H_{\gamma}\rangle} \tag{6}

Now, by Chap. VI, §1, no. 3, Prop. 10,

⟨α+β,α+β⟩⟨β,β⟩=q+1p(7) \frac{\langle\alpha + \beta , \alpha + \beta\rangle}{\langle\beta , \beta\rangle}=\frac{q + 1}{p} \tag{7}

so, by (3), (4), (6), (7),

Nα,βN−α,−β=−Nα,βN−α,α+β⟨Hβ,Hβ⟩⟨Hα+β,Hα+β⟩ N_{\alpha ,\beta}N_{-\alpha ,-\beta}=-N_{\alpha ,\beta}N_{-\alpha ,\alpha+\beta}\frac{\langle H_{\beta}, H_{\beta}\rangle}{\langle H_{\alpha+\beta}, H_{\alpha+\beta}\rangle}
=−Nα,βN−α,α+βq+1p=(q+1)2 =-N_{\alpha ,\beta}N_{-\alpha ,\alpha+\beta}\frac{q + 1}{p}= (q+ 1)^2

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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