Let $z = x e_s + y e_{s'}$, with $x, y \in \mathbf{R}$, be an element of $E_{s,s'}$. We have

which shows that $B_M$ is positive on $E_{s,s'}$, and that it is non-degenerate there if and only if $\sin \frac{\pi}{m} \neq 0$. The proposition follows.

The reflections $\sigma_s$ and $\sigma_{s'}$ leave $E_{s,s'}$ stable. We are going to determine the order of the restriction of $\sigma_s \sigma_{s'}$ to $E_{s,s'}$. We distinguish two cases:

$a) \ m = +\infty$.

Let $u = e_s + e_{s'}$. We have $B_M(u, e_s) = B_M(u, e_{s'}) = 0$, so $u$ is invariant under $\sigma_s$ and $\sigma_{s'}$. Moreover,

hence It follows that the restriction of $\sigma_s \sigma_{s'}$ to $E_{s,s'}$ is of infinite order.

$b) \ m$ is finite.

The form $B_H$ provides $E_{s,s'}$ with the structure of a euclidean plane. Since the scalar product of $e_s$ and $e_{s'}$ is equal to $-\cos \frac{\pi}{m} = \cos (\pi - \frac{\pi}{m})$, we can orient $E_{s,s'}$ so that the angle between the half-lines $\mathbf{R}_+ e_s$ and $\mathbf{R}_+ e_{s'}$ is equal to $\pi - \frac{\pi}{m}$. If D and D' denote the lines orthogonal to $e_s$ and $e_{s'}$,

Now, the restrictions $\overline{\sigma}_s$ and $\overline{\sigma}_{s'}$ of $\sigma_s$ and $\sigma_{s'}$ to $E_{s,s'}$ are the orthogonal symmetries with respect to D and D'. By the Cor. to Prop. 6 of § 2, no. 5, it follows that $\overline{\sigma}_s \overline{\sigma}_{s'}$ is the rotation with angle $\frac{2\pi}{m}$. In particular, its order is $m$.

We now return to E:

Since $\sigma_s$ and $\sigma_{s'}$ are of order 2, and are distinct, it is enough to show that their product $\sigma_s \sigma_{s'}$ is of order $m(s, s')$. When $m(s, s')$ is infinite, this follows from $a$ above. When $m(s, s')$ is finite, it follows from Prop. 1 that E is the direct sum of $E_{s,s'}$ and its orthogonal complement $V_{s,s'}$; since $\sigma_s$ and $\sigma_{s'}$ are the identity on $V_{s,s'}$, and since the restriction of $\sigma_s \sigma_{s'}$ to $E_{s,s'}$ is of order $m(s, s')$ by $b$, the order of $\sigma_s \sigma_{s'}$ is indeed equal to $m(s, s')$.

such that $\sigma(g_s) = \sigma_s$ for all $s \in S$. The elements of $\sigma(W)$ preserve the bilinear form $B_M$.

To prove the existence and uniqueness of $\sigma$, it is enough to show that $(\sigma_s \sigma_{s'})^{m(s, s')} = 1$ if $m(s, s') \neq +\infty$. Now, if $s = s'$, this follows from the fact that $\sigma_s$ is of order 2; if $s \neq s'$, it follows from what we proved in no. 2. Finally, since the reflections $\sigma_s$ preserve $B_M$, so do the elements of $\sigma(W)$.

b) Each of the $g_s$ is of order 2.

c) If $s, s' \in S$, $g_s g_{s'}$ is of order $m(s, s')$.

Assertion a) follows from the fact that the composite map

from $S$ to $\mathbf{GL}(E)$ is injective.

For b) (resp. c)), we remark that the order of $g_s$ (resp. the order of $g_s g_{s'}$) is at most 2 (resp. at most $m(s, s')$). Since we have seen in no. 2 that the order of $\sigma_s$ (resp. of $\sigma_s \sigma_{s'}$) is 2 (resp. $m(s, s')$), we must have equality.

In view of a), $S$ can be identified with a subset of $W$ by means of the map $s \mapsto g_s$.

¹ This means that, if $L_S$ denotes the free group on $S$, $W$ is the quotient of $L_S$ by the smallest normal subgroup of $L_S$ containing the elements $(ss')^{m(s, s')}$, for $m(s, s') \neq +\infty$.

This is simply the content of properties b) and c), together with the definition of W.

We indicate immediately several consequences of this theorem:

This is clear.

Indeed, if σ*(w) = 1, then w(C) = C, so w = 1 by the theorem. The injectivity of σ follows from that of σ*.

Let x* ∈ C. The set U of g ∈ \mathbf{GL}(E^*) such that g(x*) ∈ C is a neighbourhood of the identity element in \mathbf{GL}(E^*); by the theorem,

thus, σ*(W) is a discrete subgroup of \mathbf{GL}(E^*). By transport of structure, it follows that σ(W) is discrete in \mathbf{GL}(E).

Proof of Theorem 1.

If $w \in W$, denote by $l(w)$ the length of $w$ with respect to $S$ (Chap. IV, § 1, no. 1).

We are going to prove the following assertions, where $n$ denotes an integer $\geqslant 0$:

(P_n) Let $w \in W$, with $l(w) = n$, and $s \in S$. Then: either $w(C) \subset A_s$; or $w(C) \subset s(A_s)$ and $l(sw) = l(w) - 1$.

(Q_n) Let $w \in W$, with $l(w) = n$, and $s, s' \in S, s \neq s'$. Let $W_{s,s'}$ be the subgroup of $W$ generated by $s$ and $s'$. There exists $u \in W_{s,s'}$ such that

These assertions are trivial for $n = 0$. We prove them by induction on $n$, according to the scheme

Proof that $((P_n) \text{ and } (Q_n)) \Longrightarrow (P_{n+1})$.

Let $w \in W$, with $l(w) = n + 1$, and $s \in S$. We can write $w$ in the form $w = s'w'$ with $s' \in S$ and $l(w') = n$. If $s' = s$, $(P_n)$ applied to $w'$ shows that $w'(C) \subset A_s$, hence $w(C) \subset s(A_s)$ and $l(sw) = l(w') = l(w) - 1$. If $s' \neq s$, $(Q_n)$ applied to $w'$ shows that there exists $u \in W_{s,s'}$ such that

We have $w(C) = s'w'(C) \subset s'u(A_s \cap A_{s'})$.

The proof will be given in no. 5.

We apply the lemma to the element $v = s'u$. There are two possibilities: either

or Moreover, in the second case, $l(ss'u) = l(s'u) - 1$. Hence and we know that this implies that $l(sw) = l(w) - 1$.

Proof that ((P_{n+1}) and (Q_n)) \Longrightarrow (Q_{n+1}).

Let w \in W, with l(w) = n + 1, and s, s' \in S, s \neq s'. If w(C) is contained in A_s \cap A_{s'}, condition (Q_{n+1}) is satisfied with u = 1. For if not, suppose for example that w(C) is not contained in A_s. By (P_{n+1}), w(C) \subset s(A_s) and l(sw) = n. By (Q_n), applied to sw, there exists v \in W_{s,s'} such that

sw(C) \subset v(A_s \cap A_{s'}) \quad \text{and} \quad l(sw) = l(v) + l(v^{-1}sw).

Then

w(C) \subset sv(A_s \cap A_{s'})

and

l(w) = 1 + l(sw) = 1 + l(v) + l(v^{-1}sw) \geq l(sv) + l((sv)^{-1}w) \geq l(w),

so the inequalities above must be equalities. It follows that (Q_{n+1}) is satisfied with u = sv.

Proof of the theorem.

Let w \in W, with w \neq 1. We can write w in the form sw' with s \in S and l(w') = l(w) - 1. By (P_n), applied to w' and n = l(w'), we have w'(C) \subset A_s, since the case w'(C) \subset s(A_s) is excluded because l(sw') = l(w) = l(w') + 1. So w(C) = sw'(C) \subset s(A_s), and since A_s and s(A_s) are disjoint, we have C \cap w(C) = \varnothing. Q.E.D.

We are reduced immediately to the case $w' = 1$. The proof is by induction on the length $n$ of $w$. If $n = 0$, the assertion is clear. If $l(w) > 0$, there exists $s \in S$ such that $l(sw) = l(w) - 1$ and then (cf. the end of no. 4) $w(C) \subset s(A_s)$, hence $w(\overline{C}) \subset s(\overline{A}_s)$. Since $\overline{C} \subset \overline{A}_s$, it follows that

Then $s(x^*) = x^*$ for all $x^* \in \overline{C} \cap w(\overline{C})$, and a fortiori for all

Consequently, the relation $C_{X'} \cap w(C_X) \neq \varnothing$ implies on the one hand that $C_{X'} \cap H_s \neq \varnothing$, and hence that $s \in X'$, and on the other hand that $C_{X'} \cap sw(C_X) \neq \varnothing$. The induction hypothesis then implies that $X = X'$ and $swW_X = W_{X'} = W_X$, so $sw \in W_X$ and $w \in W_X$ since $s \in W_X$. It follows that $wW_X = W_{X'}$ and that $w(C_X) = C_X = C_{X'}$.

Now let $U$ be the union of the $w(\overline{C})$ for $w \in W$, and let $\mathscr{F}$ be the set of subsets of $U$ of the form $w(C_X)$, with $X \subset S$ and $w \in W$. By the above, $\mathscr{F}$ is a partition of $U$.

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

If $x \in E^0$, then $\sigma_s(x) = x - 2B_M(e_s, x)e_s = x$ for all $s \in S$. Since W is generated by S, it follows that W acts trivially on $E^0$.

Let $E'$ be a subspace of E stable under W. Let $s, s' \in S$ be two elements that are linked in the graph $\Gamma$ of $(W, S)$ (Chap. IV, §1, no. 9); recall that this means that $m(s, s') \geq 3$. Suppose that $e_s \in E'$. Then $\sigma_{s'}(e_s) \in E'$ and since the coefficient of $e_{s'}$ in $\sigma_{s'}(e_s)$ is non-zero, we have $e_{s'} \in E'$. Since $\Gamma$ is connected, it follows that, if $E'$ contains one of the $e_s$ it contains them all and coincides with E. Except in this case, it follows from Prop. 3 of §2, no. 2 that, for all $s \in S$, $E'$ is contained in the hyperplane $H_s$ orthogonal to $e_s$. Since the intersection of the $H_s$ is equal to $E^0$, this proves the proposition.

In case a), Prop. 7 shows that $E$ is simple, hence also absolutely simple ($§ 2$, no. 1, Prop. 1).

In case b), $E^0 \neq 0$, $E \neq E^0$ (since $B_M \neq 0$), and Prop. 7 shows that $E^0$ has no complement stable under $W$; thus, the W-module $E$ is not semi-simple.

(1) $\Longrightarrow$ (2). Let $S = \bigcup_i S_i$ be the decomposition of $S$ into connected components (Chap. IV, $§ 1$, no. 9), and let $W = \prod_i W_i$ be the corresponding decomposition of $W$. The space $E$ can be identified with the direct sum of the spaces $E_i = \mathbf{R}^{S_i}$, and $B_M$ can be identified with the direct sum of the corresponding forms $B_{M_i}$. We are thus reduced to the case when $(W, S)$ is irreducible. Since $W$ is assumed to be finite, $E$ is a semi-simple W-module (Appendix, Prop. 2). By the Cor. to Prop. 5, it follows that $E$ is absolutely simple. Let $B'$ be a positive non-degenerate form on $E$, and let $B''$ be the sum of its transforms under $W$. Since $B''$ is invariant under $W$, it is proportional to $B_M$ ($§ 2$, no. 1, Prop. 1); since $B_M(e_s, c_s) = 1$ for all $s \in S$, the coefficient of proportionality is $> 0$, and since $B''$ is positive so is $B_M$, which proves (2).

(2) $\Longrightarrow$ (1). If $B_M$ is positive non-degenerate, the orthogonal group $O(B_M)$ is compact (Integration, Chap. VII, $§ 3$, no. 1). Since $\sigma(W)$ is a discrete subgroup of $O(B_M)$ (Cor. 3 of Th. 1), it follows that $\sigma(W)$ is finite, hence so is $W$. Q.E.D.

The following result has been established in the course of the proof:

The finiteness criterion provided by Th. 2 permits the classification of all finite Coxeter groups (cf. Chap. VI, $§ 4$). We restrict ourselves here to the following preliminary result:

If not, this graph would contain a circuit $(s_1, \ldots, s_n)$, $n \geq 3$. Putting $m_i = m(s_1, s_{i+1})$, $1 \leq i < n$, and $m_n = m(s_n, s_1)$, this means that $m_i \geq 3$ for all $i$. Let

Then $B_M(x, x) = n + 2 \sum_{i<j} B_M(e_{s_i}, e_{s_j})$. Now$^2$ and similarly for $B_M(e_{s_n}, e_{s_1})$. Since the other terms in the sum are $\leq 0$, we obtain contrary to the fact that $B_M$ is positive non-degenerate.

First of all let $(W, S)$ be a finite Coxeter group. Denote by $(x|y)$ the form $B_M(x, y)$; by Th. 2, this is a scalar product on $E$. For all $s \in S$, let $H_s$ be the hyperplane associated to the orthogonal reflection $\sigma_s$, and let $\mathfrak{H}$ be the family of hyperplanes $w(H_s)$, for $s \in S$, $w \in W$. Let $C_0$ be the set of $x \in E$ such that $(x|e_s) > 0$ for all $s \in S$. Finally, identify $W$ (by means of $\sigma$) with a subgroup of the orthogonal group $\mathbf{O}(E)$ of the space $E$.