Proposition 20

Let $P$ be a subset of $R$. The following conditions are equivalent:

(i) $P$ is parabolic; (ii) $P$ is closed and there exists a chamber $C$ of $R$ such that $P \supseteq R_+(C)$; (iii) there exist a chamber $C$ of $R$ and a subset $\Sigma$ of $B(C)$ such that $P$ is the union of $R_+(C)$ and the set $Q$ of roots that are linear combinations of elements of $\Sigma$ with non-positive integer coefficients.

(ii) $\Longrightarrow$ (iii): this follows from Lemma 3.

(iii) $\Longrightarrow$ (i): we adopt the assumptions and notation of (iii). It is clear that $P \cup (-P) = R$. We show that, if $\alpha, \beta \in P$ are such that $\alpha + \beta \in R$, then $\alpha + \beta \in P$. This is obvious if the root $\alpha + \beta$ is positive. Assume that $\alpha + \beta$ is negative. Then $\alpha + \beta = \sum_{\gamma \in B(C)} n_\gamma \gamma$, with $n_\gamma \leq 0$. But the coefficient of every element $\gamma$ of $B(C) - \Sigma$ in $\alpha$ or $\beta$ is $\geq 0$; hence $n_\gamma = 0$ if $\gamma \in B(C) - \Sigma$, so $\alpha + \beta \in Q \subseteq P$.

(i) $\Longrightarrow$ (ii): assume that $P$ is parabolic. Let $C$ be a chamber such that $\mathrm{Card}(P \cap R_+(C))$ is as large as possible. Let $\alpha \in B(C)$ and assume that $\alpha \in P$, so that $-\alpha \in P$. For all $\beta \in P \cap R_+(C)$, $\beta$ is not proportional to $\alpha$ (for the hypothesis $\beta = 2\alpha$ would imply that $\alpha = 2\alpha + (-\alpha) \in P$ since $P$ is closed). Thus $s_\alpha(\beta) \in R_+(C)$ (no. 6, Cor. 1 of Prop. 17). If we put $C' = s_\alpha(C)$, then $\beta = s_\alpha(s_\alpha(\beta)) \in s_\alpha(R_+(C)) = R_+(C)$, so $-\alpha \in P \cap R_+(C')$ and hence $\mathrm{Card}(P \cap R_+(C')) > \mathrm{Card}(P \cap R_+(C))$. This is absurd, since $\alpha \in P$. Thus $B(C) \subseteq P$, and consequently $R_+(C) \subseteq P$ by Prop. 19 and the fact that $P$ is closed.