Proposition 25

Assume that $R$ is irreducible. Let $C$ be a chamber of $R$, and let $B(C) = \{\alpha_1, \ldots, \alpha_l\}$ be the corresponding basis.

(i) There exists a root $\tilde{\alpha} = \sum_{i=1}^{l} n_i \alpha_i$ such that, for every root $\sum_{i=1}^{l} p_i \alpha_i$, we have $n_1 \geq p_1, n_2 \geq p_2, \ldots, n_l \geq p_l$. In other words, $R$ has a largest element for the ordering defined by $C$. (ii) We have $\tilde{\alpha} \in \overline{C}$. (iii) We have $(\tilde{\alpha}|\tilde{\alpha}) \geq (\alpha|\alpha)$ for every root $\alpha$. (iv) For every positive root $\alpha'$ not proportional to $\tilde{\alpha}$, we have $n(\alpha', \tilde{\alpha}) = 0$ or $1$.

1) Let $\alpha = \sum_{i=1}^{l} n_i \alpha_i$, $\beta = \sum_{i=1}^{l} p_i \alpha_i$ be two maximal roots for the ordering defined by $C$. We shall prove that $\alpha = \beta$, which will establish (i).

2) If $(\alpha|\alpha_i) < 0$ for some index $i$, it follows that either $\alpha + \alpha_i \in R$ or $\alpha = -\alpha_i$ (Cor. of Th. 1, no. 3), and both possibilities are absurd by the maximality of $\alpha$. Thus $(\alpha|\alpha_i) \geq 0$ for all $i$.

3) If $\alpha < 0$, then $\alpha < -\alpha$, which is absurd. Thus $n_i \geq 0$ for all $i$. Let $J$ be the set of $i$ such that $n_i > 0$, and $J'$ the complement of $J$ in $\{1, 2, \ldots, l\}$. Then $J \neq \varnothing$. If $J'$ were non-empty, there would exist an $i \in J$ and an $i' \in J'$ such that $(\alpha_i|\alpha_{i'}) < 0$ (Cor. 5 of Prop. 20, no. 7); we would then have

since $(\alpha_j | \alpha_k) \leq 0$ whenever $j$ and $k$ are distinct, which would contradict 2). Thus $J' = \varnothing$ and $n_i > 0$ for all $i$.

4) We have $(\beta | \alpha_i) \geq 0$ for all $i$ by 2). We cannot have $(\beta | \alpha_i) = 0$ for all $i$ since $\beta \neq 0$. We deduce from 3) that

If $\gamma = \alpha - \beta \in \mathbf{R}$, either $\alpha > \beta$ or $\beta > \alpha$ (Th. 3, no. 6), which contradicts the maximality of $\alpha$ and $\beta$. Thus $\alpha = \beta$ (Cor. of Th. 1, no. 3).

5) By 2), $\tilde{\alpha} \in \overline{C}$. We shall prove that $(\alpha' | \alpha') \leq (\tilde{\alpha} | \tilde{\alpha})$ for every $\alpha' \in \mathbf{R}$. Since $\overline{C}$ is a fundamental domain for $W(\mathbf{R})$, we can assume that $\alpha' \in \overline{C}$. We have $\tilde{\alpha} - \alpha' \geq 0$, so $(\tilde{\alpha} - \alpha' | x) \geq 0$ for all $x \in \overline{C}$. In particular $(\tilde{\alpha} - \alpha' | \tilde{\alpha}) \geq 0$ and $(\tilde{\alpha} - \alpha' | \alpha') \geq 0$, hence $(\tilde{\alpha} | \tilde{\alpha}) \geq (\alpha' | \tilde{\alpha}) \geq (\alpha' | \alpha')$. Thus $n(\alpha', \tilde{\alpha})$ must be equal to 0, 1 or -1 if $\alpha'$ is not proportional to $\tilde{\alpha}$. If $\alpha' \geq 0$, then $(\tilde{\alpha} | \alpha') \geq 0$ by 2), so $n(\alpha', \tilde{\alpha}) \geq 0$ and $n(\alpha', \tilde{\alpha})$ must be either 0 or 1. Q.E.D.