Proposition 6

(i) The cone U is convex.

(ii) Every closed segment contained in U meets only finitely many elements of $\mathfrak{F}$.

(iii) The cone $\overline{C}$ is a fundamental domain for the action of W on U.

To prove (iii), it is enough to show that, if $x^*, y^* \in \overline{C}$ and $w \in W$ are such that $w(x^*) = y^*$, then $x^* = y^*$. Now there exist two subsets X and Y of S such that $x^* \in C_X$ and $y^* \in C_Y$; we have $w(C_X) \cap C_Y \neq \varnothing$, and Prop. 5 shows that $X = Y$ and $w \in W_X$, which implies that $x^* = y^*$.

Now let $x^*, y^* \in U$; we shall show that the segment $[x^*y^*]$ is covered by finitely many elements of $\mathfrak{F}$, which will establish both (i) and (ii). By transforming $x^*$ and $y^*$ by the same element of W, we can assume that $x^* \in \overline{C}$. Let $w \in W$ be such that $y^* \in w(\overline{C})$. We argue by induction on the length of $w$. For $s \in S$, the relation $w(\overline{C}) \not\subset \overline{A_s}$ is equivalent to $w(C) \not\subset A_s$ and hence to $l(sw) < l(w)$ (cf. no. 4). Prop. 7 of no. 8 of Chap. IV, §1 now implies that there exist only finitely many $s \in S$ such that $w(\overline{C}) \not\subset \overline{A_s}$. The set T of $s \in S$ such that $\langle y^*, e_s \rangle < 0$ is thus finite. On the other hand, the intersection $\overline{C} \cap [x^*y^*]$ is a closed segment $[x^*z^*]$. If $z^* = y^*$, that is if $y^* \in \overline{C}$, there exist subsets X and Y of S such that $x^* \in C_X$ and $y^* \in C_Y$. The open segment $]x^*y^*[$ is then contained in $C_{X \cap Y}$, so $[x^*y^*] \subset C_X \cup C_Y \cup C_{X \cap Y}$. If $z^* \neq y^*$, there exists an $s \in T$ such that $z^* \in H_s$. Then $w(C) \not\subset A_s$ and $l(sw) < l(w)$. The induction hypothesis thus implies that the segment $[z^*y^*] = s([z^*(s(y^*))])$ is covered by a finite number of elements of $\mathfrak{F}$, and hence so is

since $[x^*z^*] \subset \overline{C}$.