Since W permutes the chambers determined by $\mathfrak{H}$ transitively (§ 3, no. 2, Th. 1), we are reduced to proving the following: let $(C, (H_i)_{1 \leq i \leq l})$ be an ordered chamber and $\pi \in S_l$; then $s_{H_1} s_{H_2} \cdots s_{H_l}$ and $s_{H_{\pi(1)}} s_{H_{\pi(2)}} \cdots s_{H_{\pi(l)}}$ are conjugate in W. Taking into account § 4, no. 8, Prop. 8, this will follow immediately from the following lemma:

1) We proceed by induction on $n = \mathrm{Card}\, X$. The case $n = 1$ is immediate, so assume that $n \geqslant 2$. There exists in $X$ a terminal vertex $a$ (Chap. IV, Appendix, no. 3, Prop. 2). Let $b \in X - \{a\}$ be a vertex linked to $a$ if one exists; if $a$ is not linked to any vertex in $X - \{a\}$, let $b$ in $X - \{a\}$ be arbitrary. In all cases, $g_a$ commutes with $g_x$ for $x \neq b$. Let $\eta \in \mathscr{T}$ be such that $a$ is the largest element of $X$ and $b$ the largest element of $X - \{a\}$; we let $\xi \in \mathscr{T}$ and prove that $p_{\xi}, p_{\eta}$ are conjugate.

2) Assume first that, for $\xi$, $a$ is the largest element of $X$ and $b$ the largest element of $X - \{a\}$. Let $X'$ be the full subgraph $X - \{a\}$, which is a forest. Define a map $x \mapsto g'_x$ from $X'$ to $\Gamma$ by putting $g'_x = g_x$ if $x \neq b$, $g'_b = g_bg_a$. Let $\xi', \eta'$ be the restrictions of $\xi, \eta$ to $X'$. The induction hypothesis applies, so $p_{\xi'}$ and $p_{\eta'}$ are conjugate. But it is clear that $p_{\xi'} = p_{\xi}, p_{\eta'} = p_{\eta}$, proving the lemma in this case.

3) Assume that $a$ is the largest element of $X$ for $\xi$. Let $X_1$ (resp. $X_2$) be the set of elements of $X - \{a\}$ strictly larger (resp. smaller) than $b$; let $\xi_i$ be the restriction of $\xi$ to $X_i$. Then

and this element is conjugate to $p_{\xi_2} p_{\xi_1} g_b g_a$. We are thus reduced to case 2).

4) In the general case, let $X_3$ (resp. $X_4$) be the set of elements of $X$ strictly larger (resp. smaller) than $a$; let $\xi_i$ be the restriction of $\xi$ to $X_i$. Then $p_{\xi} = p_{\xi_3} g_a p_{\xi_4}$, and this element is conjugate to $p_{\xi_4} p_{\xi_3} g_a$. We are thus reduced to case 3).

It follows from Prop. 1 that all the Coxeter transformations are of the same order $h = h(W)$. This number is called the Coxeter number of $W$.

Let $(e^1, \ldots, e^l)$ be the basis of V such that $(e^i|e_j) = \delta_{ij}$. Then C is the open simplicial cone determined by the $e^i$ (§ 3, no. 9, Prop. 7). It is clear that $V'$ is generated by $e^{r+1}, \ldots, e^l$ and $V''$ by $e^1, \ldots, e^r$. Let q be the endomorphism of V such that $q(e^1) = e_1, \ldots, q(e^l) = e_l$. Its matrix with respect to $(e^1, \ldots, e^l)$ is $Q = ((e_i|e_j))$. We have $(e_i|e_j) \leq 0$ for $i \neq j$ (§ 3, no. 4, Prop. 3). Since W is irreducible, there does not exist any partition $\{1, 2, \ldots, l\} = I_1 \cup I_2$ such that $(e_i|e_j) = 0$ for $i \in I_1$ and $j \in I_2$. Thus (§ 3, no. 5, Lemma 4) Q has an eigenvector $(a_1, \ldots, a_l)$ all of whose coordinates are > 0; let a be the corresponding eigenvalue. Put

and let $P$ be the plane generated by $z'$ and $z''$. Then $P \cap C$ is the set of linear combinations of $z'$ and $z''$ with coefficients $> 0$. The relation $q(z) = az$ gives $\sum_{j=1}^l a_j e_j = \sum_{j=1}^l aa_j e^j$; scalar multiplying by $e_k$ (where $k \leq r$) gives

Thus, $(a-1)z'' + z'$ is orthogonal to $e^1, \ldots, e^r$, that is, to $V''$. Hence, $s''$ leaves stable the plane generated by $z''$ and $(a-1)z'' + z'$, that is, $P$. Similarly, $s'$ leaves $P$ stable. Since $z' \in P \cap V'$ and $z'' \in P \cap V''$, $s'|P$ and $s''|P$ are the reflections with respect to $Rz'$ and $Rz''$.

We retain the preceding notation. The restriction of $c = s's''$ to $P$ is the rotation with angle $2(\widehat{z'', z'})$ (§ 2, no. 5, Cor. of Prop. 6). Since $c$ is of order $h$, the $h$ elements $1, c, \ldots, c^{h-1}$ of $W$ are pairwise distinct; the elements $s', s'c, \ldots, s'c^{h-1}$ are thus pairwise distinct, and are distinct from the preceding elements since $c^i|P$ is a rotation and $s'c^j|P$ is a reflection. The set

is the subgroup $W'$ of $W$ generated by $s'$ and $s''$, and induces on $P$ the group $W''$ generated by the orthogonal reflections with respect to $Rz', Rz''$. The transform of $C$ by an element of $W'$ is either disjoint from $-C$ or equal to $-C$. Thus, the transform of $P \cap C$ by an element of $W''$ is either disjoint from $-(P \cap C)$ or equal to $-(P \cap C)$. Hence, for a suitable orientation of P, there exists an integer $m > 0$ such that $(\overline{z''}, z') = \frac{\pi}{m}$ (§ 2, no. 5, Cor. of Prop. 7). Moreover, the sets $g'(C)$, for $g' \in W'$, are pairwise disjoint; the sets $g''(P \cap C)$, for $g'' \in W''$, are thus pairwise disjoint; so $W''$ is of order $2h$. Hence $m = h$. By definition, $c|P$ is a rotation with angle $\frac{2\pi}{m}$, and so has eigenvalues $\exp \frac{2i\pi}{h}$, $\exp \frac{2i\pi(h-1)}{h}$. This proves that $m_1 = 1, m_l = h - 1$.

The transforms of $Rz'$ and $Rz''$ by $W'$ are $h$ lines $D_1, \ldots, D_h$ of P, and the points of $P - (D_1 \cup \cdots \cup D_h)$ are transforms by the elements of $W'$ of points of $P \cap C$. Thus, a hyperplane of $\mathfrak{H}$ necessarily cuts P along one of the lines $D_i$, and consequently is a transform by an operation of $W'$ of a hyperplane of $\mathfrak{H}$ containing $Rz'$ or $Rz''$.

Now, any $H \in \mathfrak{H}$ that contains $Rz'$ is one of the hyperplanes $H_1, \ldots, H_r$. Indeed, let $e_H$ be the unit vector orthogonal to H and on the same side of H as C. Then $e_H = \lambda_1 e_1 + \cdots + \lambda_l e_l$ with the $\lambda_i$ all $\geq 0$ (§ 3, no. 5, Lemma 6, (i)). Now $0 = (e_H|z') = \lambda_{r+1} a_{r+1} + \cdots + \lambda_l a_l$, so

Suppose that two of the $\lambda_i$ were non-zero, for example $\lambda_1$ and $\lambda_2$; since $e_1, \ldots, e_r$ are pairwise orthogonal, we would have and the coordinates of $s_1(e_H)$ would not be of the same sign, which is absurd (loc. cit.). Thus, $e_H$ is proportional to one of the vectors $e_1, \ldots, e_r$, which proves our assertion. Similarly, any $H \in \mathfrak{H}$ that contains $Rz''$ is one of the hyperplanes $H_{r+1}, \ldots, H_l$.

The number of elements of $\mathfrak{H}$ containing $Rz'$ or $Rz''$ is therefore $l$. If $h$ is even, $\mathrm{Card}(\mathfrak{H})$ is thus equal to $\frac{h}{2} l$. If $h$ is odd, $\mathrm{Card}(\mathfrak{H})$ is equal to $\frac{h-1}{2} l + r$, and also to $\frac{h-1}{2} l + (l - r)$; hence $r = l - r$, so $r = \frac{l}{2}$, and $\mathrm{Card}(\mathfrak{H}) = \frac{h-1}{2} l + \frac{l}{2} = \frac{h}{2} l$.

for all $x \in V$.

Put $f(x) = \sum_{u \in R_0} (x|u)^2$. It is clear that $f$ is a positive quadratic form invariant under $W$, and non-degenerate since the $e_i$ form a basis of $V$. Since

W is irreducible, there exists a constant $\beta$ such that $f(x) = \beta(x|x)$ (§ 2, no. 1, Prop. 1). If $(x_i)_{1 \leq i \leq l}$ is an orthonormal basis of V for the scalar product $(x|y)$, then

Thus $\beta = h$, which proves (4).

We use the notation in the proof of Th. 1. Since $c|P$ is a rotation through an angle $\frac{2\pi}{h}$, $c^{h/2}$ transforms $z'$ to $-z'$, $z''$ to $-z''$, and hence $z' + z'' = z$ to $-z$. Now $z \in C$, so the chamber $c^{h/2}(C)$ is necessarily $-C$.

Put $V' = V \otimes_{\mathbf{R}} \mathbf{C}$, $S' = S(V') = S \otimes_{\mathbf{R}} \mathbf{C}$, and extend the scalar product on V to a hermitian form on $V'$. If $c$ is a Coxeter transformation of W, there exists an orthonormal basis $(X_i)_{1 \leq i \leq l}$ of $V'$ consisting of eigenvectors of $c \otimes 1$ (Algebra, Chap. IX, § 7, no. 3, Prop. 4); moreover, we can assume that, for $1 \leq j \leq l$, $X_j$ corresponds to the eigenvalue $\exp \frac{2i\pi m_j}{h}$ of $c \otimes 1$. It is clear that $S'$ can be identified with the algebra $\mathbf{C}[X_1, \ldots, X_l]$, and that we can write $u_j \otimes 1 = f_j(X_1, \ldots, X_l)$, where $f_j$ is a homogeneous polynomial of degree $p_j$ in $\mathbf{C}[X_1, \ldots, X_l]$. Put $D_j = \frac{\partial}{\partial X_j}$ and $J = \det(D_k f_j)$. Recall (§ 5, no. 4, Prop. 5) that $J(X_1, \ldots, X_l)$ is proportional to the product in $S'$ of $\mathrm{Card}(\mathfrak{H})$ vectors $y_k$ of V each of which is orthogonal to a hyperplane of $\mathfrak{H}$. Since we can assume that $X_1 \notin H \otimes \mathbf{C}$ for all $H \in \mathfrak{H}$ (Remark), the $X_1$ component of each of the vectors $y_k$ is non-zero, so $J(1, 0, \ldots, 0) \neq 0$. The rule for expanding determinants now proves the existence of a permutation $\sigma$ of $\{1, 2, \ldots, l\}$ such that $(D_{\sigma(j)} f_j)(1, 0, \ldots, 0) \neq 0$ for all $j$. Since $D_{\sigma(j)} f_j$ is homogeneous of degree $p_j - 1$, the coefficient of $X_1^{p_j - 1} X_{\sigma(j)}$ in $f_j(X_1, \ldots, X_l)$ is non-zero. Now $f_j(X_1, \ldots, X_l)$ is invariant under $c \otimes 1$, and

This proves that $p_j - 1 + m_{\sigma(j)} \equiv 0$ (mod. $h$). Now $h - m_{\sigma(j)}$ is an exponent (formula (2)). Permuting the $u_j$ if necessary, we can assume that $p_j - 1 \equiv m_j$ (mod. $h$) for all $j$. Since $p_j - 1 \geq 0$ and $m_j < h$, we have $p_j - 1 = m_j + \mu_j h$ with $\mu_j$ an integer $\geq 0$. By § 5, Prop. 3, we see that

Taking into account formula (3) and Th. 1 (ii), we obtain $h \sum_{j=1}^{l} \mu_j = 0$, so $\mu_j = 0$ for all $j$, and finally $p_j - 1 = m_j$ for all $j$.

This follows from the relations $m_j + 1 = p_j$ and § 5, Cor. of Th. 3.

are eigenvalues of c of multiplicity 1.

Otherwise, there would exist two non-proportional homogeneous invariants of degree 2 in S, and hence two non-proportional quadratic forms on V* invariant under W, contrary to § 2, no. 1, Prop. 1.

The first assertion follows from § 5, no. 3, Prop. 4. Assume that the exponents of W are odd. Then h is even by formula (2), and

thus $c^{h/2} = -1$ since c is a semi-simple automorphism of V (Algebra, Chap. IX, § 7, no. 3, Prop. 4).