Bourbaki

Proposition 4

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

Lie Groups and Lie Algebras, chapter V, § 3. Groups of displacements generated by reflections

Let CC be a chamber and N\mathfrak{N} a set of walls of CC. Let WNW_{\mathfrak{N}} be the subgroup of WW generated by the orthogonal reflections with respect to the elements of N\mathfrak{N}. For H∈NH \in \mathfrak{N}, denote by eHe_H the unit vector orthogonal to HH on the same side of HH as CC. The following conditions are equivalent: (i) The group WNW_{\mathfrak{N}} is finite. (ii) There exists a point of EE invariant under every element of WNW_{\mathfrak{N}}. (iii) The hyperplanes belonging to N\mathfrak{N} have non-empty intersection. (iv) The family of vectors (eH)H∈N(e_H)_{H \in \mathfrak{N}} is free in TT.

By property (D2) at the beginning of § 3, the stabiliser in WW of every point of EE is finite, so (ii) implies (i).

Since the group WNW_{\mathfrak{N}} is generated by the set of reflections with respect to the hyperplanes belonging to N\mathfrak{N}, the fixed points of WNW_{\mathfrak{N}} are the points of EE belonging to every hyperplane H∈NH \in \mathfrak{N}, hence the equivalence of (ii) and (iii).

Assume that there exists a point aa of EE such that a∈Ha \in H for all H∈NH \in \mathfrak{N} and let t∈Tt \in T be such that a+t∈Ca + t \in C. Since (eH∣eH′)≤0(e_H|e_{H'}) \leq 0 for H,H′∈NH, H' \in \mathfrak{N} such that H≠H′H \neq H' (Prop. 3), and since (t∣eH)>0(t|e_H) > 0 for all H∈NH \in \mathfrak{N}, Lemma 3 (ii) implies that the eHe_H for H∈NH \in \mathfrak{N} are linearly independent. Consequently, (iii) implies (iv).

Suppose finally that the family (eH)H∈N(e_H)_{H \in \mathfrak{N}} is free. Let aa be a point of EE. For any hyperplane H∈NH \in \mathfrak{N}, there exists a real number cHc_H such that HH consists of the points a+ta + t of EE with (t∣eH)=cH(t|e_H) = c_H. Since the family (eH)(e_H) is free, there exists t∈Tt \in T such that (t∣eH)=cH(t|e_H) = c_H for all H∈NH \in \mathfrak{N}, and the point a+ta + t of EE belongs to all the hyperplanes H∈NH \in \mathfrak{N}. Thus, (iv) implies (iii).

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.