Proposition 17

Assume that $R$ is reduced and let $w \in W$ and $\alpha \in R$.

(i) We have $\psi(w(\alpha)) = U_w(\psi(\alpha))$. (ii) Assume that $\alpha$ is positive. The root $w(\alpha)$ is negative if and only if

in other words if $s_{\alpha} \in T_{w^{-1}}$.

(iii) We have $\eta(w, s_\alpha) = -1$ if and only if the chambers C and $w(C)$ are on opposite sides of the hyperplane $L_\alpha$. In other words, the set $T_w$ consists of the reflections with respect to the walls separating C and $w(C)$.

Let $\beta \in B(C)$ and put $s = s_\beta$. Clearly $T_s = \{ s \}$ and consequently

On the other hand, let $\rho = \sum_{\gamma \in B(C)} n_\gamma(\rho) \gamma$ be a positive root. Put

If $\rho \neq \beta$, there exists an element $\gamma \in B(C)$ with $\gamma \neq \beta$, such that $n_\gamma(\rho) > 0$, and we have $n_\gamma(s(\rho)) = n_\gamma(\rho) > 0$ (no. 1, formula (5)). Hence $s(\rho)$ is positive. We deduce immediately that

Comparison of (12) and (13) now shows that $U_s(\psi(\gamma)) = \psi(s(\gamma))$ for all roots $\gamma$ and all $s \in S$. Since S generates W, (i) follows.

On the other hand, saying that $w(\alpha)$ is negative is equivalent to saying that

or, by (i), that $U_w(\psi(\alpha)) = (-1, ws_\alpha w^{-1})$. If in addition $\alpha$ is positive, then $\psi(\alpha) = (+1, s_\alpha)$ and $U_w(\psi(\alpha)) = (\eta(w^{-1}, s_\alpha), ws_\alpha w^{-1})$, hence (ii).

Finally, by (ii), $\eta(w, s_\alpha) = -1$ if and only if one of the roots $\alpha$ and $w^{-1}(\alpha)$ is positive and the other negative. This is equivalent to saying that $(\alpha|x).(w^{-1}(\alpha)|x) = (\alpha|x).(w|x) < 0$ for all $x \in C$, hence the first assertion in (iii). The second assertion in (iii) follows immediately.