Bourbaki

Proposition 19

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

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

Let (βi)1≤i≤n(\beta_i)_{1 \leq i \leq n} be a sequence of positive roots for the chamber CC such that β1+β2+⋯+βn\beta_1 + \beta_2 + \cdots + \beta_n is a root. Then there exists a permutation π∈Sn\pi \in S_n such that, for all i∈{1,2,…,n}i \in \{1, 2, \ldots, n\}, βπ(1)+βπ(2)+⋯+βπ(i)\beta_{\pi(1)} + \beta_{\pi(2)} + \cdots + \beta_{\pi(i)} is a root.

We argue by induction on nn, the proposition being clear for n≤2n \leq 2. Put β=β1+⋯+βn\beta = \beta_1 + \cdots + \beta_n. Then ∑i=1n(β∣βi)=(β∣β)>0\sum_{i=1}^n (\beta|\beta_i) = (\beta|\beta) > 0, so there exists an index kk such that (β∣βk)>0(\beta|\beta_k) > 0. If β=βk\beta = \beta_k, then n=1n = 1 since βi>0\beta_i > 0 for all ii. Otherwise β−βk\beta - \beta_k is a root (no. 3, Cor. of Th. 1); it then suffices to apply the induction hypothesis to β−βk=∑i≠kβi\beta - \beta_k = \sum_{i \neq k} \beta_i.

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.