Bourbaki

Proposition 31

Tag 0AW8, permanent. Cite it as [Bourbaki, Tag 0AW8].

Lie Groups and Lie Algebras, chapter VI, § 1. Root systems

Assume that RR is irreducible and reduced, and let β=n1α1+⋯+nlαl\beta = n_1 \alpha_1 + \cdots + n_l \alpha_l be the highest root of RR (cf. no. 8). Then n1+⋯+nl=h−1n_1 + \cdots + n_l = h - 1.

Let R+R_+ be the set of positive roots relative to CC. Then (no. 10, Cor. of Prop. 29):

n1+⋯+nl=12∑α∈R+n(β,α) n_1 + \cdots + n_l = \frac{1}{2} \sum_{\alpha \in R_+} n(\beta, \alpha)
=1+12∑α∈R+,α≠βn(β,α)=1+∑α∈R+,α≠β(α∣β)(α∣α) = 1 + \frac{1}{2} \sum_{\alpha \in R_+, \alpha \neq \beta} n(\beta, \alpha) = 1 + \sum_{\alpha \in R_+, \alpha \neq \beta} \frac{(\alpha | \beta)}{(\alpha | \alpha)}

By no. 8, Prop. 25 (iv), for all α∈R+\alpha \in R_+ and α≠β\alpha \neq \beta, n(α,β)=0n(\alpha, \beta) = 0 or 1, so n(α,β)2=n(α,β)n(\alpha, \beta)^2 = n(\alpha, \beta), that is 4(α∣β)2(β∣β)2=2(α∣β)(β∣β)\frac{4(\alpha | \beta)^2}{(\beta | \beta)^2} = \frac{2(\alpha | \beta)}{(\beta | \beta)}. Hence:

n1+⋯+nl+1=2+2∑α∈R+,α≠β(α∣β)2(α∣α)(β∣β)=2∑α∈R+(α∣β)2(α∣α)(β∣β)=(β∣β)−1∑α∈R(α∥α∥∣β)2. n_1 + \cdots + n_l + 1 = 2 + 2 \sum_{\alpha \in R_+, \alpha \neq \beta} \frac{(\alpha | \beta)^2}{(\alpha | \alpha)(\beta | \beta)} = 2 \sum_{\alpha \in R_+} \frac{(\alpha | \beta)^2}{(\alpha | \alpha)(\beta | \beta)} = (\beta | \beta)^{-1} \sum_{\alpha \in R} \left( \frac{\alpha}{\| \alpha \|} | \beta \right)^2.

By Chap. V, § 6, no. 2, Cor. of Th. 1,

∑α∈R(α∥α∥∣β)2=h(β∣β) \sum_{\alpha \in R} \left( \frac{\alpha}{\| \alpha \|} | \beta \right)^2 = h(\beta | \beta)

so n1+⋯+nl+1=hn_1 + \cdots + n_l + 1 = h.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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Transcribed from the printed volume. Not translated.