Bourbaki

Corollary 3

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

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

a) Let α∈R\alpha \in \mathbf{R}. Then Y(α)Y(\alpha) is a connected subset of B(C)B(C) (Chap. IV, Appendix).

b) Let YY be a non-empty connected subset of B(C)B(C). Then ∑β∈Yβ\sum_{\beta \in Y} \beta belongs to R\mathbf{R}.

To prove a), we can assume that α\alpha is positive. We argue by induction on Card(Y(α))\mathrm{Card}(Y(\alpha)), the assertion being trivial if Card(Y(α))=1\mathrm{Card}(Y(\alpha)) = 1. By Prop. 19, there exists β∈B(C)\beta \in B(C) such that α−β∈R\alpha - \beta \in \mathbf{R}. Let pp be the largest integer ⩾0\geqslant 0 such that γ=α−pβ∈R\gamma = \alpha - p\beta \in \mathbf{R}. Since γ−β∉R\gamma - \beta \notin \mathbf{R} and γ+pβ∈R\gamma + p\beta \in \mathbf{R}, (γ∣β)≠0(\gamma|\beta) \neq 0 (Prop. 9); thus β\beta is linked to at least one element of Y(γ)Y(\gamma). But Y(α)=Y(γ)∪{β}Y(\alpha) = Y(\gamma) \cup \{\beta\}, and Y(γ)Y(\gamma) is connected by the induction hypothesis. Thus Y(α)Y(\alpha) is connected, which proves a).

Now let YY be a non-empty connected subset of B(C)B(C); we show by induction on Card(Y)\mathrm{Card}(Y) that ∑β∈Yβ\sum_{\beta \in Y} \beta is a root. The case in which Card(Y)⩽1\mathrm{Card}(Y) \leqslant 1 is trivial. Assume that Card(Y)⩾2\mathrm{Card}(Y) \geqslant 2. Since XX is a forest (Chap. V, § 4, no. 8, Prop. 8), YY is a tree and has a terminal vertex β\beta (Chap. IV, Appendix). The set Y−{β}Y - \{\beta\} is connected, and one of its element is linked to β\beta. By the induction hypothesis, α=∑γ∈Y−{β}γ∈R\alpha = \sum_{\gamma \in Y - \{\beta\}} \gamma \in \mathbf{R}, and since (α∣β)<0(\alpha|\beta) < 0, it follows that α+β∈R\alpha + \beta \in \mathbf{R} (Th. 1). Q.E.D.

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

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