Bourbaki

Theorem 1

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

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

Let α,β\alpha, \beta be two roots.

(i) If n(α,β)>0n(\alpha, \beta) > 0, α−β\alpha - \beta is a root unless α=β\alpha = \beta. (ii) If n(α,β)<0n(\alpha, \beta) < 0, α+β\alpha + \beta is a root unless α=−β\alpha = -\beta.

If n(α,β)>0n(\alpha, \beta) > 0, the possibilities, by the list above, are the following:

1) n(α,β)=1n(\alpha, \beta) = 1; then α−β−sβ(α)∈R\alpha - \beta - s_{\beta}(\alpha) \in \mathbf{R}; 2) n(β,α)=1n(\beta, \alpha) = 1; then β−α=sα(β)∈R\beta - \alpha = s_{\alpha}(\beta) \in \mathbf{R}, so α−β∈R\alpha - \beta \in \mathbf{R}; 3) β=α\beta = \alpha.

This proves (i), and (ii) follows by changing β\beta to −β-\beta.

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

Cited by

Cites

Nothing.

Transcribed from the printed volume. Not translated.