Proposition 6

For all $i \in J$, the element $\gamma_i = t(\overline{\omega}_i)w_iw_0$ belongs to $\Gamma_C$ and the map $i \mapsto \gamma_i$ is a bijection from J to $\Gamma_C - \{1\}$.

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.