Lemma 2

Let C be a chamber. (i) For any x ∈ E, there exists an element w ∈ W such that w(x) ∈ $\overline{C}$. (ii) For any chamber C', there is an element w ∈ W such that w(C') = C. (iii) The group W is generated by the set of orthogonal reflections with respect to the walls of C.

Let M be the set of walls of C and let W_M be the subgroup of W generated by the reflections with respect to the walls of C.

(i) Let x ∈ E and let J be the orbit of x under the group W_M. It suffices to prove that J meets $\overline{C}$.

Let a be a point of C; there is a closed ball B with centre a meeting J; since B is compact, property (D'2) of no. 1 shows that B ∩ J is finite. Hence, there exists a point y of J such that

We shall prove that y ∈ $\overline{C}$. For this, it suffices to show that if H is a wall of C then y ∈ D_H(C) (cf. $§ 1$, no. 4, Prop. 9). Since s_H ∈ W_M, we have s_H(y) ∈ J and so (fig. 1)

by (1). There exist b ∈ H and two vectors t and u such that a = b + t and y = b + u, the vector u being orthogonal to H; then s_H(y) = b − u, and (2) is equivalent to $(t-u|t-u) \leq (t+u|t+u)$, or to $(t|u) \geq 0$. This inequality implies that $y \in D_H(C)$.


(ii) Let $C'$ be a chamber and $a' \in C'$. By what we have proved, there exists $w \in W_{\mathfrak{m}}$ such that $w^{-1}(a') \in \overline{C}$; hence, the chamber $C'$ meets $w(C)$; since $w(C)$ is the union of $w(C)$ and facets with empty interior (§ 1, no. 2, Prop. 3), we have $C' = w(C)$.

(iii) We have to prove that $W = W_{\mathfrak{m}}$ and for this it suffices to prove that $s_{H'} \in W_{\mathfrak{m}}$ for all $H' \in \mathfrak{H}$. Now $H'$ is a wall of at least one chamber $C'$ (§ 1, no. 4, Prop. 8); we have seen that there exists $w \in W_{\mathfrak{m}}$ such that $C' = w(C)$; consequently, there exists a wall $H$ of $C$ such that $H' = w(H)$, hence $s_{H'} = w.s_H.w^{-1} \in W_{\mathfrak{m}}$.