Exercise 9, § 1

Let $R$ be a root system, $C$ a chamber. Put $B(C) = \{\alpha_1, \ldots, \alpha_l\}$,

(the $\alpha_i$ being pairwise distinct), and $\alpha = \alpha_1 + \cdots + \alpha_s$.

a) For $i = 1, 2, \ldots, s$, put $\varepsilon_i = \pm 1$. Let $\alpha^* = \varepsilon_1 \alpha_1 + \cdots + \varepsilon_s \alpha_s$. If $(\alpha^*|\alpha_i) > 0$ for $i = 1, \ldots, l$, then $\alpha^* = \alpha$ and $\varepsilon_i = 1$ for all $i$. (Let $\gamma$ (resp. $\delta$) be the sum of the $\alpha_i$ for which $\varepsilon_i = 1$ (resp. $-1$). Then, $(\gamma|\alpha_i) - (\delta|\alpha_i) > 0$, $(\gamma|\alpha_i) + (\delta|\alpha_i) = (\alpha_i|\alpha_i)$, so

since $2(\gamma|\alpha_i)/(\alpha_i|\alpha_i) \in \mathbf{Z}$, $(\gamma|\alpha_i) \geq (\alpha_i|\alpha_i) = (\alpha|\alpha_i)$, so $\gamma - \alpha \in \overline{C}$, $\gamma \geq \alpha$, $\gamma = \alpha$ and $\delta = 0$.

b) For $i = 1, 2, \ldots, s$, let $\varepsilon_i = \pm 1$. The set of $\varepsilon_i \alpha_i$ is the set of roots $> 0$ relative to a chamber if and only if $\alpha^* = \sum_i \varepsilon_i \alpha_i$ belongs to a chamber. (To show that the condition is sufficient, let $\mu_i = \pm 1$ be such that $(\alpha^*|\mu_i \alpha_i) > 0$. There exists an element $w \in W(R)$ that transforms the set of $\mu_i \alpha_i$ into the set of $\alpha_i$, and hence $\alpha^*$ into $\alpha^{**} = \sum_i \varepsilon_i \mu_i \alpha_{\sigma(i)}$, where $\sigma \in S_s$. Applying a), show that $\alpha^{**} = \alpha$, so $\mu_i \varepsilon_i = 1$.)