Proposition 8

Let F be a real Hilbert space of finite dimension l. Let $\mathfrak{H}$ be the set of affine hyperplanes of F and G the group generated by the orthogonal reflections $s_H$ with respect to the hyperplanes $H \in \mathfrak{H}$. Assume that the conditions of Chap. V, § 3 are satisfied (i.e. that $g(H) \in \mathfrak{H}$ for all $H \in \mathfrak{H}$ and $g \in G$, and that G acts properly on F). Assume also that 0 is a special point for G and that the group T of translations belonging to G is of rank l. Then there exists a unique reduced root system R in $V = F^*$ such that the canonical isomorphism from F to $V^*$ transforms G to the affine Weyl group $W_a$ of R.

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.