Since W is generated by the reflections $s_\alpha$, it is contained in $W_a$. On the other hand, $t(\alpha^\sim) = s_{\alpha,1} \circ s_\alpha$ if $\alpha \in \mathbf{R}$, which shows that $Q \subseteq W_a$.

Since W leaves $Q(\mathbf{R}^\sim)$ stable (§ 1, no. 9), the group G of affine transformations generated by W and Q is the semi-direct product of W by Q. Now $G \subseteq W_a$ from above and $s_{\alpha,k} \in G$ for all $\alpha \in \mathbf{R}$ and $k \in \mathbf{Z}$ by (1). It follows that $W_a = G$.

Since $Q(\mathbf{R}^\sim)$ is a discrete subgroup of $V^*$, the group Q acts properly on E. Hence, so does $W_a = W.Q$, since W is finite. Moreover, for $\alpha, \beta \in \mathbf{R}$ and $k \in \mathbf{Z}$:

where $n(\alpha,\beta) = \langle \beta^\sim, \alpha \rangle$ is an integer, hence the second assertion.

We can thus apply the results of Chapter V, § 3 to $W_a$ acting on E. To avoid any confusion with the chambers of the Weyl group W in $V^*$, we shall call the chambers determined by the system of hyperplanes $L_{\alpha,k}$ (for $\alpha \in \mathbf{R}$ and $k \in \mathbf{Z}$) in E alcoves. The group $W_a$ thus acts simply transitively on the set of alcoves and the closure of an alcove is a fundamental domain for $W_a$ acting on E (Chap. V, § 3, no. 2, Th. 1 and no. 3, Th. 2). It is clear that the Weyl group W is identified with the canonical image $U(W_a)$ of $W_a$ in the orthogonal group of V* (cf. Chap. V, § 3, no. 6). It follows that W_a is essential (Chap. V, § 3, no. 7) and that W_a is irreducible if and only if the root system R is (§ 1, no. 2, Cor. of Prop. 5). If R is irreducible, every alcove is an open simplex (Chap. V, § 3, no. 9, Prop. 8). In the general case, the canonical product decomposition of the affine space E (Chap. V, § 3, no. 8) corresponds to the decomposition of R into irreducible components. In particular, the alcoves are products of open simplexes.

Note also that the Cor. of Th. 1 of Chap. V, § 3, no. 2 shows that the s_{\alpha,k} are the only reflections in W_a.

Let x_0 \in E and let \alpha \in R. The hyperplane L parallel to Ker \alpha and passing through x_0 has equation \langle \alpha, x \rangle = \langle \alpha, x_0 \rangle. To be equal to some L_{\beta,k}, it is necessary on the one hand that \alpha and \beta are proportional, and so, since R is reduced, that \beta = \pm \alpha, and on the other hand that \langle \alpha, x_0 \rangle is an integer. It follows immediately that x_0 is a special point of W_a if and only if \langle \alpha, x_0 \rangle \in \mathbf{Z} for all \alpha \in R, in other words, if and only if x_0 \in P(R') (§ 1, no. 9).

This follows from Prop. 3, in view of the Cor. of Prop. 11 of Chap. V, § 3, no. 10 and Prop. 12 of Chap. V, § 3, no. 10.

(i) There exists a unique alcove C contained in C' such that 0 \in \overline{C}. (ii) The union of the w(\overline{C}) for w \in W is a neighbourhood of 0 in E. (iii) Every wall of C' is a wall of C.

This follows from Prop. 11 of Chap. V, § 3, no. 10.

Assume now that R is irreducible. Let (\alpha_i)_{i \in I} be a basis of R (§ 1, no. 5, Def. 2), and let (\overline{\omega}_i)_{i \in I} be the dual basis. The \overline{\omega}_i are the fundamental weights of R' for the chamber D' of R' corresponding to the basis (\alpha_i). Let

be the highest root of R (§ 1, no. 8), and let J be the set of i \in I such that n_i = 1.

(i) $C$ is the set of $x \in E$ such that $\langle \alpha_i, x \rangle > 0$ for all $i \in I$ and $\langle \tilde{\alpha}, x \rangle < 1$.

(ii) The set $\overline{C} \cap P(R^-)$ consists of 0 and the $\overline{\omega}_i$ for $i \in J$.

Let $D$ be the set of $x \in E$ such that $\langle \tilde{\alpha}, x \rangle < 1$ and set $C_1 = C' \cap D$. Since $0 \in \overline{C}$, we have $C \subseteq D$ and hence $C \subseteq C_1$. We are going to show that, for all $\alpha \in R$ and all $k \in \mathbf{Z}$, the sets $C$ and $C_1$ are on the same side of the hyperplane $L_{\alpha,k}$. This will prove that $C_1 \subseteq C$ and so will establish assertion (i). If $k = 0$, the whole of the chamber $C'$ is on one side of $L_{\alpha,0}$, which establishes our assertion in this case. If $k \neq 0$, we may, by replacing $\alpha$ by $-\alpha$, assume that $k > 0$. Then $\langle \alpha, x \rangle < k$ on $C$, since $0 \in \overline{C}$. On the other hand, $\tilde{\alpha} - \alpha$ is positive on $C'$ (§ 1, no. 8, Prop. 25). Thus, for $y \in C_1$, we have $\langle \alpha, y \rangle \leq \langle \tilde{\alpha}, y \rangle < 1 \leq k$. Consequently, $C$ and $C_1$ are on the same side of $L_{\alpha,k}$.

Now let $\overline{\omega} \in P(R^-)$. Then $\overline{\omega} = \sum_i p_i \overline{\omega}_i$, with $p_i \in \mathbf{Z}$ (§ 1, no. 10), and $\overline{\omega} \in \overline{C'}$ if and only if the integers $p_i$ are positive. If $\overline{\omega} \in \overline{C'}$, then $\overline{\omega} \in \overline{C}$ if and only if $\langle \tilde{\alpha}, \overline{\omega} \rangle = \sum_i n_i p_i$ is $\leq 1$, hence (ii).

This follows from (i).

We remark first of all that the root $w_i(\tilde{\alpha})$ is of the form

and hence is positive.

We show that, if $i \in J$, then $\gamma_i \in \Gamma_C$. Indeed, let $a \in C$ and $b = \gamma_i(a)$. For $1 \leq j \leq l$ and $j \neq i$,

since $w_0(a) \in -C'$ and $w_i(\alpha_j)$ is negative. On the other hand, since $w_0(a) \in -C'$, $\tilde{\alpha} - w_i(\alpha_i)$ takes negative values on $-C'$, and $\langle w_0(a), \tilde{\alpha} \rangle > -1$. Finally, since $w_0(a) \in -C'$ and $w_i(\tilde{\alpha})$ is a positive root. The relations (2), (3) and (4) then imply that $b \in C$, and hence that $\gamma_i \in \Gamma_C$. It is clear that the map $i \mapsto \gamma_i$ is injective, since $\gamma_i(0) = \overline{\omega}_i$. Finally, let $\gamma \in \Gamma_C$ with $\gamma \neq 1$, and put $\gamma = tw$ with $t \in P$ and $w \in W$. Then $t \neq 1$ since $\Gamma_C \cap W = \{1\}$. On the other hand, $t(0) = \gamma(0) \in \overline{C} \cap P(R^-)$ and Prop. 5 implies that there exists $i \in J$ such that $t(0) = \overline{\omega}_i$. Then $\gamma_i^{-1}\gamma(0) = 0$, hence $\gamma = \gamma_i$ since $\Gamma_C \cap W = \{1\}$. This completes the proof.

Indeed, if we identify $\Gamma_c$ with $P(R^-)/Q(R^-)$, the element $\gamma_i$ is identified with the class of $\overline{\omega}_i$ mod. $Q(R^-)$.

Let $(s_\lambda)_{\lambda \in \Lambda}$ be a family of representatives of the right cosets of $G'$ in $G$. Let $U_1$ be the union of the $s_\lambda U$. Then the $s'U_1$, for $s' \in G'$, are pairwise disjoint and have union $M = \bigcup_{s \in G} sU$. Let $M' = \bigcup_{s' \in G'} s'U'$. The union of $U'$ (resp. $U_1$) and a suitable subset of $X - M'$ (resp. $X - M$) is a fundamental domain, evidently $\mu$-measurable, for $G'$. By Integration, Chap. VII, § 2, no. 10, Cor. of Th. 4, $\mu(U') = \mu(U_1)$. This proves that Card $\Lambda = (G : G')$ is finite, and that $\mu(U') = (\text{Card } \Lambda)\mu(U)$.

Let $(\overline{\omega}_1, \ldots, \overline{\omega}_l)$ be the basis of $P(R^-)$ dual to $B$. By the Cor. of Prop. 5, the open simplex $C$ with vertices $0, n_1^{-1} \overline{\omega}_1, \ldots, n_l^{-1} \overline{\omega}_l$ is an alcove of $E$. Choose a Haar measure $\mu$ on the additive group $V^*$. Let $A$ be the set of elements of $V^*$ of the form $\xi_1 \overline{\omega}_1 + \cdots + \xi_l \overline{\omega}_l$, with $0 < \xi_i < 1$ for $i = 1, \ldots, l$. By Cor. 2 of Prop. 15 of Integration, Chap. VII, § 1, no. 10,

On the other hand, let $A'$ be the set of elements of $V^*$ of the form with $0 < \xi_i < 1$ for $i = 1, \ldots, l$. Since $(\alpha_{\tilde{1}}, \ldots, \alpha_{\tilde{l}})$ is a basis of the $\mathbf{Z}$-module $Q(R^-)$, we can apply Lemma 1 with $X = V^*$, $G = W_a$, $G' = Q$, $U = C$ and $U' = A'$. This gives

Finally, Lemma 1 can be applied again, taking X = V*, G = P, G' = Q, U = A and U' = A'. This gives

The proposition now follows by comparing formulas (5), (6) and (7).

We remark first of all that the assumption on T implies that G is essential: otherwise, the affine space F would decompose into a product $F_0 \times F_1$, with $\dim F_1 < l$, the group G being identified with a group of displacements acting properly on $F_1$ (Chap. V, § 3, no. 8, Prop. 6), and T would not be of rank l.

Let $\mathfrak{H}_0$ be the set of $H \in \mathfrak{H}$ such that $0 \in H$. For $H \in \mathfrak{H}_0$, let $\mathfrak{H}_H$ be the set of elements of $\mathfrak{H}$ parallel to H. Since 0 is a special point, $\mathfrak{H}$ is the union of the $\mathfrak{H}_H$ for $H \in \mathfrak{H}_0$. Since T is of rank l, there exists a $v \in F$ such that the translation by the vector v belongs to T and $v \notin H$. The hyperplanes $H + kv$ for $k \in \mathbf{Z}$ are pairwise distinct and belong to $\mathfrak{H}_H$. Now let a be a unit vector of F orthogonal to H: then $H + (v|a)a \in \mathfrak{H}_H$ and since $\mathfrak{H}$ is locally finite (Chap. V, § 1, no. 1, Lemma 1), there is a smallest real number $\lambda > 0$ such that $H + \lambda a \in \mathfrak{H}$. We are going to show that $\mathfrak{H}_H$ is the set of hyperplanes $H + k\lambda a$ for $k \in \mathbf{Z}$. Indeed,

and the element $s_{H'} \circ s_H$ of G is the translation by the vector $2\lambda a$ (Chap. V, § 2, no. 4, Prop. 5). Consequently, $H + 2n\lambda a = (s_{H'} s_H)^n(H)$ and $H + (2n + 1)\lambda a = (s_{H'} s_H)^n(H')$ belongs to $\mathfrak{H}_H$. On the other hand, if $L \in \mathfrak{H}_H$, there exists $\xi \in \mathbf{R}$ such that $L = H + \xi \lambda a$ and there exists an integer n such that

either $2n < \xi \leq 2n + 1$, or $2n - 1 < \xi \leq 2n$.

In the first case, $(s_H s_{H'})^n(L) = H + (\xi - 2n)\lambda a$ with

and the definition of $\lambda$ implies that $\xi = 2n + 1$; in the second case,

and the definition of $\lambda$ implies that $\xi = 2n$.

It follows that if $\alpha_H$ is the linear form on $F$ such that

the set $\mathscr{H}_H$ is the set of hyperplanes $L_{\alpha_H, k} = \{ x \in F \mid \langle \alpha_H, x \rangle = k \}$ for $k \in \mathbf{Z}$, and $\alpha_H$ and $-\alpha_H$ are the only linear forms with this property.

Consequently, the proposition will be proved if we show that the set R of elements of V of the form $\pm \alpha_H$ is a reduced root system in V.

a) We prove condition (RS_I): it is clear that R is finite (since $\mathscr{H}_0$ is finite) and does not contain 0. Moreover, R generates V. Indeed, if $x \in F$ is orthogonal to R, then $x \in H$ for all $H \in \mathscr{H}_0$ and the translation by the vector $x$ commutes with every element of G. Since G is essential, this implies that $x = 0$.

b) We prove (RS_{II}). For $v \in V$ and $r \in \mathbf{R}$, put $L_{v, r} = \{ x \in F \mid \langle v, x \rangle = r \}$ as above; if $\alpha \in R$, put $H_\alpha = L_{\alpha, 0}$, and let $s_\alpha$ be the transpose of $s_{H_\alpha}$. There exists a unique element $\alpha^\sim \in F$ orthogonal to $H_\alpha$ and such that $\langle \alpha^\sim, \alpha \rangle = 2$. Then $s_{H_\alpha} = s_{\alpha^\sim, \alpha}$ and $s_\alpha = s_{\alpha, \alpha^\sim}$. For $\beta \in R$,

and there exist $\gamma \in R$ and $n \in \mathbf{N}^*$ such that $L_{s_\alpha(\beta), 1} = L_{\gamma, n}$. Then and so $1/n \in \mathbf{Z}$. Thus $n = 1$ and $s_\alpha(\beta) = \gamma \in R$. This proves (RS_{II}).

c) We prove (RS_{III}). Let $\alpha \in R$ and set $H'_\alpha = L_{\alpha, 1}$. Then

since the translation $t(\alpha^\sim)$ by the vector $\alpha^\sim$ is the product $s_{H'_\alpha} s_{H_\alpha}$ (Chap. V, § 2, no. 4, Prop. 5), it belongs to T and $\alpha^\sim = t(\alpha^\sim)(0)$ is a special point for G. Consequently, for all $\beta \in R$, there exists a hyperplane $L_{\beta, k}$ passing through $\alpha^\sim$, with $k$ an integer, which shows that $\langle \beta, \alpha^\sim \rangle \in \mathbf{Z}$, and proves (RS_{III}).

d) Finally, it is clear that R is reduced, for if $H, H' \in \mathscr{H}_0, H \neq H'$, the linear forms $\alpha_H$ and $\alpha_{H'}$ are not proportional.

(i) There exists a discrete subgroup of V of rank l that is stable under G. (ii) There exists a $\mathbf{Q}$-structure on V (Algebra, Chap. II, § 8, no. 1, Def. 1) invariant under G. (iii) There exists a root system in V whose Weyl group is G. (iv) There exists a discrete group $G'$ of displacements of V, acting properly on V, and generated by reflections, such that $G'$ is the semi-direct product of G and a group of translations of rank l.

(ii) $\Longrightarrow$ (i): let $V' \subseteq V$ be a $\mathbf{Q}$-structure on V invariant under G. Let A be a finite subset of $V'$ generating the $\mathbf{Q}$-vector space $V'$. Replacing A by $\bigcup_{s \in G} s(A)$, we can assume that A is stable under G. Let B be the subgroup of V generated by A. Then B is stable under G, of finite type and torsion-free, so has a basis over $\mathbf{Z}$ which is both a basis of $V'$ over $\mathbf{Q}$ and a basis of V over $\mathbf{R}$.

(iii) $\Longrightarrow$ (ii): this follows from Prop. 1 of § 1, no. 1, for example. (iv) $\Longrightarrow$ (iii): let $G'$ be a group satisfying condition (iv). The group of translations of $G'$ is of rank l, and 0 is a special point for $G'$ by Prop. 9 of Chap. V, § 3, no. 10. Prop. 6 shows that there exists a reduced root system $R_0$ in $V^*$ such that $G'$ is identified with $W_\alpha(R_0)$; the group G is then the Weyl group of the inverse root system of $R_0$.

(i) $\Longrightarrow$ (iv): assume that G leaves stable a discrete subgroup M of V, of rank l. For any reflection $s \in G$, $s(x) - x \in M$ for all $x \in M$, so the line $D_s$ orthogonal to $H_s$ meets M; let $\alpha_s, -\alpha_s$ be the generators of the cyclic group $D_s \cap M$; the set A of the $\alpha_s$ and $-\alpha_s$ is stable under G, hence generates a subgroup $M'$ of M stable under G; the discrete group $M'$ is of rank l because G is essential. Let $G'$ be the group of affine transformations of V that is the semi-direct product of G and the group of translations whose vectors belong to $M'$. Let $G'_1$ be the subgroup of $G'$ generated by the reflections of $G'$. We shall show that $G'_1 = G'$, which will complete the proof. First, $G'_1 \supseteq G$ since G is generated by reflections. On the other hand, for any reflection s of G, let $t_s$ be the translation with vector $\alpha_s$. The transformation $s \circ t_s$ is a reflection, and $s \circ t_s \in G'$; thus $t_s$ is a product of two reflections of $G'$; this being true for every reflection s of G, the translations whose vector belongs to $M'$ are all in $G'_1$.