Bourbaki

Lemma 1

Tag 0AU4, permanent. Cite it as [Bourbaki, Tag 0AU4].

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

Let VV be a vector space over kk, RR a finite subset of VV generating VV. For any α∈R\alpha \in R such that α≠0\alpha \neq 0, there exists at most one reflection ss of VV such that s(α)=−αs(\alpha) = -\alpha and s(R)=Rs(R) = R.

Let GG be the group of automorphisms of VV leaving RR stable. Since RR generates VV, GG is isomorphic to a subgroup of the symmetric group of RR, and hence is finite. Let s,s′s, s' be reflections of VV such that s(α)=s′(α)=−αs(\alpha) = s'(\alpha) = -\alpha, s(R)=Rs(R) = R, s′(R)=Rs'(R) = R. Then t=ss′t = ss' belongs to GG, and hence is of finite order mm. On the other hand,

t(α)=αandt(x)≡xmod  .kαfor all x∈V. t(\alpha) = \alpha \quad \text{and} \quad t(x) \equiv x \mod. k\alpha \quad \text{for all } x \in V.

Hence, there exists a linear form ff on VV such that

t(x)=x+f(x)αfor all x∈V t(x) = x + f(x)\alpha \quad \text{for all } x \in V

and f(α)=0f(\alpha) = 0. By induction on nn, it follows that

tn(x)=x+nf(x)α for all x∈V. t^n(x) = x + nf(x)\alpha \text{ for all } x \in V.

Taking nn equal to mm, we see that mf(x)=0mf(x) = 0 for all x∈Vx \in V, so f=0f = 0, t=1t = 1 and s=s′s = s'.

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.