Indeed, let K be a compact subset of E. If a hyperplane H ∈ $\mathscr{H}$ meets K, the set s_H(K) also meets K, since every point of K ∩ H is fixed by s_H. The set of H ∈ $\mathscr{H}$ meeting K is thus finite by (D'2).

We can thus apply to E and $\mathscr{H}$ the definitions and results of $§ 1$. We shall call the chambers, facets, walls, etc. defined in E by $\mathscr{H}$ simply the chambers, facets, walls, etc. relative to W. Any displacement w ∈ W permutes the chambers, facets, walls, etc.

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 $s_{H'} = w.s_H.w^{-1} \in W_{\mathfrak{m}}$.

(i) The pair $(W, S)$ is a Coxeter system.

(ii) Let $w \in W$ and let $H$ be a wall of $C$. The relation $l(s_H w) > l(w)$ implies that the chambers $C$ and $w(C)$ are on the same side of $H$.

(iii) For any chamber $C'$, there exists a unique element $w \in W$ such that $w(C) = C'$.

(iv) The set of hyperplanes $H$ such that $s_H \in W$ is equal to $\mathfrak{H}$.

Every element of $S$ is of order 2 and Lemma 2 shows that $S$ generates $W$. For any wall $H$ of $C$, denote by $P_H$ the set of elements $w \in W$ such that the chambers $C$ and $w(C)$ (which do not meet $H$) are on the same side of $H$. We shall verify conditions (A'), (B') and (C) of Chap. IV, § 1, no. 7.

(A') $1 \in P_H$: Trivial.

(B') $P_H$ and $s_H.P_H$ are disjoint: Indeed, $w(C)$ and $s_H w(C)$ are on opposite sides of $H$, so if $w(C)$ is on the same side of $H$ as $C$, it is not on the same side as $s_H w(C)$.

(C) Let $w \in P_H$ and let $H'$ be a wall of $C$ such that $w s_{H'} \notin P_H$; then $w s_{H'} = s_H w$: By assumption, $w(C)$ is on the same side of $H$ as $C$ and $w s_{H'}(C)$ is on the other side. Thus, $w s_{H'}(C)$ and $w(C)$ are on opposite sides of H; hence, the chambers $s_{H'}(C)$ and C are on opposite sides of the hyperplane $w^{-1}(H)$. Let $a$ be a point of the face of C with support $H'$. The point $a = s_{H'}(a)$ is in the closure of the two chambers C and $s_{H'}(C)$ that are contained respectively in the two open half-spaces bounded by $w^{-1}(H)$; thus, $a \in w^{-1}(H)$, so $H' = w^{-1}(H)$. From this we deduce that $s_{H'} = w^{-1}s_Hw$, hence $ws_{H'} = s_Hw$.

Assertions (i) and (ii) follow from this and Prop. 6 of Chap. IV, § 1, no. 7. Moreover, we have (loc. cit., condition (A))

Lemma 2 shows that W acts transitively on the set of chambers. Moreover, if $w \in W$ is such that $w(C) = C$, then $w \in P_H$ for every wall H of C, hence $w = 1$ by (3). This proves (iii).

Finally, let H be a hyperplane such that $s_H \in W$. If H did not belong to $\mathfrak{H}$, there would be at least one chamber $C'$ meeting H (§ 1, no. 3, Prop. 7). Every point of $H \cap C'$ is invariant under $s_H$, and thus belongs to the chambers $C'$ and $s_H(C')$; thus, $C' = s_H(C')$, which contradicts (iii) since $s_H \neq 1$.

Let $\mathfrak{H}'$ be the set of hyperplanes of the form $w(H)$ with $w \in W$ and $H \in \mathfrak{H}$ such that $s_H \in \Sigma$. Since W is generated by the family $(s_H)_{H \in \mathfrak{H}'}$ and since $\mathfrak{H}'$ is stable under W, we can apply all the results of this no. to $\mathfrak{H}'$ instead of $\mathfrak{H}$; but Th. 1, (iv) shows that every reflection in W is of the form $s_H$ with $H \in \mathfrak{H}'$, hence the corollary.

(iv) $w$ fixes at least one point of $F$; (v) $w$ fixes every point of $F$; (vi) $w$ fixes every point of $\overline{F}$; (vii) $w$ belongs to the subgroup of $W$ generated by the reflections with respect to the walls of $C$ containing $F$.

For every subset $A$ of $E$, denote by $W(A)$ the subgroup of $W$ consisting of the elements that fix every point of $A$.

We first prove the following assertion:

(I) Let $C$ be a chamber, let $x$ and $y$ be two points of $\overline{C}$ and let $w \in W$ be such that $w(x) = y$. Then $x = y$ and $w$ belongs to the subgroup $W_{\mathfrak{N}}$, where $\mathfrak{N}$ is the set of walls of $C$ containing $x$.

We argue by induction on the length $q$ of $w$ (relative to the set $S$ of reflections with respect to the walls of $C$), the case $q = 0$ being obvious. If $q \geq 1$, there exist a wall $H$ of $C$ and an element $w' \in W$ such that $w = s_H w'$ and $l(w') = q - 1$. Since $l(s_H w) < l(w)$, the chambers $C$ and $w(C)$ are on opposite sides of $H$ by Th. 1 of no. 2. Thus, $\overline{C} \cap w(\overline{C}) \subset H$, so $y \in H$. Thus $y = w'(x)$ and the induction hypothesis implies that $x = y$ and $w' \in W_{\mathfrak{N}}$. Since $y \in H$, it follows that $H \in \mathfrak{N}$, and hence that $w = s_H w' \in W_{\mathfrak{N}}$, completing the proof of (I).

Proof of Theorem 2: this follows from (I) and Lemma 2.

Proof of Proposition 1: we know that two distinct facets are disjoint and have distinct closures (§ 1, no. 2, Cor. of Prop. 3). The equivalence of (i), (ii) and (iii) follows. On the other hand, it is clear that

and assertion (I) shows that (i) $\Longrightarrow$ (vii).

Proof of Proposition 2: let $A''$ be the affine subspace of $E$ generated by $A$. Clearly, $W(A) = W(A'')$. By Prop. 7 of § 1, no. 3, there exists a point $x \in A''$ that does not belong to any hyperplane $H \in \mathfrak{H} - \mathfrak{H}_A$. Let $F_x$ be the facet containing $x$: it is open in $A''$ and Prop. 1 shows that

hence $W(A) = W(A') = W(F_x)$. Replacing $A$ by $F$, we also have

hence the proposition.

(i) If $H_i$ and $H_j$ are parallel, then $m_{ij} = \infty$ and $e_i = -e_j$.

(ii) If $H_i$ and $H_j$ are not parallel, then $m_{ij}$ is finite and

(iii) We have $(e_i|e_j) \leq 0$.

If $H_i$ and $H_j$ are parallel, $s_i s_j$ is a translation (§ 2, no. 4, Prop. 5), so $m_{ij} = \infty$. Moreover, either $e_i = e_j$ or $e_i = -e_j$. Now, there exists a point $a$ (resp. $a'$) in the closure of C that belongs to $H_i$ (resp. $H_j$) but not to $H_j$ (resp. $H_i$). Then $(a' - a|e_i) > 0$ and $(a - a'|e_j) > 0$, which excludes the case $e_i = e_j$ and proves (i).

Assume now that $H_i$ and $H_j$ are not parallel. Choose an origin $a \in H_i \cap H_j$ and identify T with E by the bijection $t \mapsto a + t$. Let V be the plane orthogonal to $H_i \cap H_j$ and passing through $a$. Put $\Gamma = V \cap D_{H_i}(C) \cap D_{H_j}(C)$ (where $D_H(C)$ denotes the open half-space bounded by H and containing C ($§ 1$, no. 1)) and let D (resp. $D'$) be the half-line in V contained in $H_i \cap V$ (resp. $H_j \cap V$) and in the closure of $\Gamma$. For a suitable orientation of V, the set $\Gamma$ is the union of the open half-lines $\Delta$ in V such that

Let $W'$ be the subgroup of W generated by $s_i$ and $s_j$. For any $w \in W'$, the hyperplanes $w(H_i)$ and $w(H_j)$ belong to $\mathscr{H}$, contain $H_i \cap H_j$ and do not meet C. It follows that they do not meet $\Gamma$ ($§ 1$, no. 5, Prop. 10). The Cor. of Prop. 7 of $§ 2$, no. 5 thus implies (ii).

Finally, assertion (iii) follows immediately from (i) and (ii), since $m_{ij} \geq 2$ for $i \neq j$.

(i) If $c_1, \ldots, c_n$ are real numbers such that $q \left( \sum_i c_i a_i \right) = 0$, then

(ii) If q is non-degenerate and if there exists a linear form f on V such that $f(a_i) > 0$ for all i, the vectors $a_1, \ldots, a_n$ are linearly independent.

The relation $B(a_i, a_j) \leq 0$ for $i \neq j$ immediately implies that

hence (i). If q is non-degenerate, the relation $\sum_i c_i a_i = 0$ thus implies that it follows that, for any linear form f on V, we have $\sum_i |c_i| f(a_i) = 0$, and hence $c_i = 0$ for all i if we also have $f(a_i) > 0$ for all i. This proves (ii).

Then: (i) The kernel N of q is of dimension 0 or 1. If $\dim N = 1$, N is generated by a vector all of whose coordinates are > 0. (ii) The smallest eigenvalue of Q is of multiplicity 1 and a corresponding eigenvector has all its coordinates > 0 or all its coordinates < 0.

Since q is a positive quadratic form, the kernel N of q is the set of isotropic vectors for q (Algebra, Chap. IX, § 7, no. 1, Cor. of Prop. 2). Let $a_1, \ldots, a_n$ be the canonical basis of $\mathbf{R}^n$. If $\sum_i c_i a_i \in N$, Lemma 3 shows that we also have $\sum_i |c_i| a_i \in N$, and hence $\sum_i q_{ji} |c_i| = 0$ for all j. Let I be the set of i such that $c_i \neq 0$. If $j \notin I$, then $q_{ji} |c_i| \leq 0$ for $i \in I$ and $q_{ji} |c_i| = 0$ for $i \notin I$, so $q_{ji} = 0$ for $j \notin I$ and $i \in I$. Assumption b) thus implies that either $I = \varnothing$ or $I = \{1, \ldots, n\}$. Consequently, every non-zero vector in N has all its coordinates $\neq 0$. If $\dim N \geq 2$, the intersection of N with the hyperplane with equation $x_i = 0$ would be of dimension $\geq 1$, contrary to what we have just shown. The preceding argument also shows that, if $\dim N = 1$, then N contains a vector all of whose coordinates are > 0. This proves (i).

On the other hand, we know that the eigenvalues of Q are real (Algebra, Chap. IX, § 7, no. 3, Prop. 5) and positive since q is positive. Let $\lambda$ be the smallest of them. The matrix $Q' = Q - \lambda I_n$ is then the matrix of a degenerate positive form $q'$ and the off-diagonal elements of $Q'$ are the same as those of Q. Consequently, $Q'$ satisfies conditions a), b) and c) of the statement of the lemma. Since the kernel $N'$ of $q'$ is the eigenspace of Q corresponding to the eigenvalue $\lambda$, assertion (ii) follows from (i).

Then there are two possibilities: 1) $(e_1, \ldots, e_n)$ is a basis of T; 2) $n = \dim T + 1$; there exists a family $(c_i)_{1 \leq i \leq n}$ of real numbers $> 0$ such that $\sum_i c_i e_i = 0$, and any family $(c'_i)_{1 \leq i \leq n}$ of real numbers such that $\sum_i c'_i e_i = 0$ is proportional to $(c_i)_{1 \leq i \leq n}$.

Put $q_{ij} = (e_i|e_j)$. The matrix $Q = (q_{ij})$ then satisfies the hypotheses of Lemma 4: conditions a) and b) of Lemma 4 are the same as conditions a) and b) above, and c) is satisfied since $\sum_{i,j} q_{ij} x_i x_j = \| \sum_i x_i e_i \|_2^2$. The kernel N of the quadratic form $q$ on $\mathbf{R}^n$, with matrix $Q$, is the set of $(c_1, \ldots, c_n) \in \mathbf{R}^n$ such that $\sum_i c_i e_i = 0$. If $N = \{0\}$, the $e_i$ are linearly dependent and we are in case 1). If $\dim N > 0$, Lemma 4 (i) shows that we are in case 2).

(i) If $x = \sum_i c_i e_i \in T$ is such that $(x|e_i) \geq 0$ for all $i$, then $c_i \geq 0$ for all $i$.

(ii) If $x$ and $y$ are two elements of $T$ such that $(x|e_i) \geq 0$ and $(y|e_i) \geq 0$ for all $i$, then $(x|y) \geq 0$. If $(x|e_i) > 0$ and $(y|e_i) > 0$ for all $i$, then $(x|y) > 0$.

Under the hypotheses of (i), assume that $c_i < 0$ for some $i$. Let $f$ be the linear form on $T$ defined by $f(e_i) = 1$ and

The vectors $-x, e_1, \ldots, e_n$ then satisfy the hypotheses of Lemma 3 (ii) (taking for $q$ the metric form on $T$). We conclude that they are linearly independent, which is absurd. Hence (i). Moreover, if $x = \sum_i c_i e_i \in T$ and if $y \in T$, then $(x|y) = \sum_i c_i (e_i|y)$, so (ii) follows immediately from (i).

For $a, a' \in A$ such that $a \neq a'$, we have

Now, the unit sphere $S$ of $T$ being compact, there exists a finite covering of $S$ by sets of diameter $< (2 - 2\lambda)^{1/2}$ and each of these sets contains at most one point of $A$, hence the lemma.

Denote by $U(w)$ the automorphism of $T$ associated to the affine map $w \in W$ from $E$ to itself. We have

This defines a homomorphism $U$ from the group $W$ to the orthogonal group of $T$; the kernel of $U$ is the set of translations belonging to $W$.

Assertion (i) follows immediately from Prop. 3, (iii) and Lemma 7.

We prove (ii). Let $C$ be a chamber and $M$ the set of its walls. The facets of $\overline{C}$ (relative to $\mathfrak{H}$) are the same as those relative to $M$ (§ 1, no. 4, Prop. 9).

Since $\mathfrak{M}$ is finite, they are finite in number. Since a facet meets only finitely many hyperplanes belonging to $\mathfrak{H}$, the set of hyperplanes belonging to $\mathfrak{H}$ and meeting $\overline{C}$ is finite, hence so is the set $A(C)$ of unit vectors in $T$ orthogonal to some hyperplane belonging to $\mathfrak{H}$ and meeting $\overline{C}$. Consequently, there exists a real number $\lambda < 1$ such that $(a|a') \leq \lambda$ for $a, a' \in A(C)$ and $a \neq a'$.

Let $A$ be the set of unit vectors in $T$ orthogonal to a hyperplane belonging to $\mathfrak{H}$. Let $a, a' \in A$ with $a \neq a'$. If $a$ and $a'$ are parallel, then $a = -a'$ and $(a|a') = -1$. Otherwise, let $H \in \mathfrak{H}$ (resp. $H' \in \mathfrak{H}$) be such that $a$ (resp. $a'$) is orthogonal to $H$ (resp. $H'$). We have $H \cap H' \neq \varnothing$, and if $x \in H \cap H'$ there exists an element $w \in W$ such that $x \in w(\overline{C})$. The vectors $U(w).a$ and $U(w).a'$ then belong to $A(C)$, we have

and the set $A$ is finite by Lemma 7. Hence (ii).

Now let $w \in W$ be such that $U(w).a = a$ for all $a \in A$. Then $U(w).t = t$ for all $t$ belonging to the subspace of $T$ generated by $A$. On the other hand, if $t \in T$ is orthogonal to $A$, we have $U(s_H).t = t$ for all $H \in \mathfrak{H}$, hence $U(w).t = t$ and finally $U(w) = 1$. Since $U(w)(A) = A$ for all $w \in W$, we deduce that $U(W)$ is isomorphic to a group of permutations of the finite set $A$, hence (iii).

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

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

Assume that there exists a point $a$ of $E$ such that $a \in H$ for all $H \in \mathfrak{N}$ and let $t \in T$ be such that $a + t \in C$. Since $(e_H|e_{H'}) \leq 0$ for $H, H' \in \mathfrak{N}$ such that $H \neq H'$ (Prop. 3), and since $(t|e_H) > 0$ for all $H \in \mathfrak{N}$, Lemma 3 (ii) implies that the $e_H$ for $H \in \mathfrak{N}$ are linearly independent. Consequently, (iii) implies (iv).

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

(i) The group W is the direct product of the groups $W_p$ ($1 \leq p \leq s$). (ii) The space T is the orthogonal direct sum of the subspaces $T_0, T_1, \ldots, T_s$, which are all stable under U(W). (iii) For all q such that $1 \leq q \leq s$, the subspace $T'_q$ of T consists of the vectors invariant under U(Wq); it is the direct sum of the $T_p$ for $1 \leq p \leq s$ and $p \neq q$.*

(iv) Let C be a chamber. The subspace T_p (for $1 \leq p \leq s$) is generated by the vectors $e_i(C)$ for $i \in J_p$ (in the notation of no. 4).

(v) The representations of W in the subspaces T_p ($1 \leq p \leq s$) are absolutely irreducible, non-trivial, and pairwise inequivalent.

Assertion (i) follows from Chap. IV, §1, no. 9. On the other hand, we have already seen that the subspaces T_p are invariant under U(W), and so is T_0. Let C be a chamber; since W_p is generated by the reflections $s_i(C)$ for $i \in J_p$, it is clear that $T'_p$ is the subspace orthogonal to the $e_i(C)$ for $i \in J_p$, hence (iv). Moreover, if $i \in J_p, j \in J_q$ with $p \neq q$, then $m_{ij} = 2$ since $\{i, j\}$ is not an edge of the Coxeter graph of W, so $(e_i|e_j) = 0$. Assertion (ii) is now immediate. And assertion (iii) follows, since $T'_q$ is the orthogonal complement of $T_q$.

Finally, let V be a subspace of T_p invariant under U(W_p). For all $i \in J_p$, either $e_i \in V$ or $e_i$ is orthogonal to V (§ 2, no. 2, Prop. 3). Let A (resp. B) be the set of $i \in J_p$ such that $e_i \in V$ (resp. $e_i$ is orthogonal to V). Clearly $(e_i|e_j) = 0$ for $i \in A$ and $j \in B$, and since $J_p$ is connected, it follows that either $A = \varnothing$ and $V = \{0\}$, or $A = J_p$ and $V = T_p$. Consequently, the representation of W_p on T_p is irreducible, hence absolutely irreducible (§ 2, no. 1, Prop. 1). It is non-trivial by the very definition of T_p. Finally, the last assertion in (v) follows immediately from (iii).

If the subspace T_0 of vectors in T invariant under U(W) reduces to $\{0\}$, then W is said to be essential; if the representation U of W on T is irreducible, then W is said to be irreducible.