Proposition 6

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

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

(i) The group $W_p$ is a group of displacements of the euclidean affine space $E_p$ ; it is generated by reflections; provided with the discrete topology, it acts properly on $E_p$ ; it is irreducible.

(ii) Let $\mathfrak{H}_p$ be the set of hyperplanes $H$ of $E_p$ such that $s_H \in W_p$ . The set $\mathfrak{H}$ consists of the hyperplanes of the form

H = E_0 \times E_1 \times \cdots \times E_{p-1} \times H_p \times E_{p+1} \times \cdots \times E_s,

with $p = 1, \ldots, s$ and $H_p \in \mathfrak{H}_p$ .

(iii) Every chamber $C$ is of the form $E_0 \times C_1 \times \cdots \times C_s$ , where for each $p$ the set $C_p$ is a chamber defined in $E_p$ by the set of hyperplanes $\mathfrak{H}_p$ ; moreover, the walls of $C_p$ are the hyperplanes $\varphi_p(H_i(C))$ for $i \in J_p$ .

Let $C$ be a chamber. Put $H_i = H_i(C)$ , $e_i = e_i(C)$ and $s_i = s_i(C)$ for $i \in I$ (in the notation of no. 4).

(i) Let i be in J_p; since e_i \in T_p and T is the direct sum of the mutually orthogonal subspaces T_0, T_1, \ldots, T_s, the hyperplane of T orthogonal to e_i is of the form L_i + T'_p, where L_i is the hyperplane of T_p orthogonal to e_i. The affine hyperplane H_i of E is of the form L_i + T'_p + x, with x \in E, and we have

H_i = E_0 \times E_1 \times \cdots \times E_{p-1} \times H'_i \times E_{p+1} \times \cdots \times E_s

with H'_i = L_i + \varphi_p(x) = \varphi_p(H_i). It is now immediate that s_i acts in E_p by the reflection associated to the hyperplane H'_i of E_p. Thus, the group W_p is a group of displacements generated by reflections in E_p; the verification of the properness criterion (D'2) is immediate. Finally, Prop. 5, (v) shows that W_p is irreducible. This proves (i).

(ii) By the Cor. of Th. 1, the set \mathscr{H}_p consists of the hyperplanes of the form w_p(H'_i) for i in J_p and w_p in W_p. Further, if w = w_1 \ldots w_s with w_p \in W_p for all p, formulas (5) and (6) imply that

w(H_i) = E_0 \times E_1 \times \cdots \times E_{p-1} \times w_p(H'_i) \times E_{p+1} \times \cdots \times E_s,

from which (ii) follows immediately.

(iii) Let i be in J_p; by formula (6), the open half-space D_i bounded by H_i and containing C is of the form

D_i = E_0 \times E_1 \times \cdots \times E_{p-1} \times D'_i \times E_{p+1} \times \cdots \times E_s,

where D'_i is an open half-space bounded by H'_i in E_p. Put C_p = \bigcap_{i \in J_p} D'_i; since C = \bigcap_{i \in I} D_i, it follows immediately that

C = E_0 \times C_1 \times \cdots \times C_s;

consequently, none of the sets C_p is empty, and since C does not meet any hyperplane belonging to \mathscr{H}, the set C_p does not meet any hyperplane belonging to \mathscr{H}_p. Prop. 5 of § 1, no. 3 now shows that C_p is one of the chambers defined by \mathscr{H}_p in E_p. By using Prop. 4 of § 1, no. 2, it is easy to see that the walls of C_p are the hyperplanes H'_i = \varphi_p(H_i) for i \in J_p.

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

Proposition 6

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