Bourbaki

Corollary

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

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

*Let V1V_1 be a vector subspace of VV, and let V2V_2 be the vector subspace generated by R∩V1R \cap V_1. Then R∩V1R \cap V_1 is a root system in V2V_2.

This follows from (ii) applied to X=R∩V1X = R \cap V_1.

For α,β∈R\alpha, \beta \in R, put

⟨α,β−⟩=n(α,β). \langle \alpha, \beta^- \rangle = n(\alpha, \beta).
(1)

Then

n(α,α)=2 n(\alpha, \alpha) = 2
(2)
n(−α,β)=n(α,−β)=−n(α,β). n(-\alpha, \beta) = n(\alpha, -\beta) = -n(\alpha, \beta).
(3)

By (RSIII),

n(α,β)∈Z. n(\alpha, \beta) \in \mathbf{Z}.
(4)

By the definition of n(α,β)n(\alpha, \beta),

sβ(α)=α−n(α,β)β. s_\beta(\alpha) = \alpha - n(\alpha, \beta)\beta.
(5)

Formula (1) and Prop. 2 imply that

n(α,β)=n(β−,α−). n(\alpha, \beta) = n(\beta^-, \alpha^-).
(6)

Let (x∣y)(x|y) be a symmetric bilinear form on VV, non-degenerate and invariant under W(R)W(R) (Prop. 3). By Lemma 2,

n(α,β)=2(α∣β)(β∣β). n(\alpha, \beta) = \frac{2(\alpha|\beta)}{(\beta|\beta)}.
(7)

It follows that

n(α,β)=0  ⟺  n(β,α)=0  ⟺  (α∣β)=0  ⟺  sα and sβ commute. n(\alpha, \beta) = 0 \iff n(\beta, \alpha) = 0 \iff (\alpha|\beta) = 0 \iff s_\alpha \text{ and } s_\beta \text{ commute.}
(8)

If (α∣β)≠0(\alpha|\beta) \neq 0, then

n(β,α)n(α,β)=(β∣β)(α∣α). \frac{n(\beta, \alpha)}{n(\alpha, \beta)} = \frac{(\beta|\beta)}{(\alpha|\alpha)}.
(9)

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.