Bourbaki

Corollary 2

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

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

Let φ\varphi be a map from RR to a abelian group Γ\Gamma having the following properties:

1) φ(−α)=−φ(α)\varphi(-\alpha) = -\varphi(\alpha) for α∈R\alpha \in R; 2) if α∈R,β∈R\alpha \in R, \beta \in R are such that α+β∈R\alpha + \beta \in R, then φ(α+β)=φ(α)+φ(β)\varphi(\alpha + \beta) = \varphi(\alpha) + \varphi(\beta).

Let QQ be the subgroup of VV generated by RR. Then φ\varphi extends to a homomorphism from QQ to Γ\Gamma.

Let BB be a basis of RR. Let ψ\psi be the unique homomorphism from QQ to Γ\Gamma that coincides with φ\varphi on BB. It suffices to show that ψ(α)=φ(α)\psi(\alpha) = \varphi(\alpha) when α\alpha is a positive root relative to BB. We have α=β1+⋯+βm\alpha = \beta_1 + \cdots + \beta_m with βi∈B\beta_i \in B for all ii, and β1+⋯+βh∈R\beta_1 + \cdots + \beta_h \in R for all hh (Prop. 19). We show that ψ(α)=φ(α)\psi(\alpha) = \varphi(\alpha) by induction on mm. This is clear if m=1m = 1. The induction hypothesis gives

ψ(β1+⋯+βm−1)=φ(β1+⋯+βm−1), \psi(\beta_1 + \cdots + \beta_{m-1}) = \varphi(\beta_1 + \cdots + \beta_{m-1}),

and we have ψ(βm)=φ(βm)\psi(\beta_m) = \varphi(\beta_m), hence ψ(α)=φ(α)\psi(\alpha) = \varphi(\alpha), which proves the corollary.

For any root α=∑β∈B(C)nββ\alpha = \sum_{\beta \in B(C)} n_\beta \beta in RR, denote by Y(α)Y(\alpha) the set of β∈B(C)\beta \in B(C) such that nβ≠0n_\beta \neq 0. Moreover, observe that B(C)B(C) can be identified with the set of vertices of the graph of the Coxeter system formed by W(R)W(R) and the sαis_{\alpha_i} (cf. Chap. IV, § 1, no. 9 and Chap. V, § 3, no. 2).

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.