Finite Weyl closed chambers and stabilizers

Choose $\rho \in E$ with $(\rho ,{\alpha }_i)>0$ for every $i$, by taking the sum of the basis dual to the simple roots; in rank zero use zero. For $x\in E$, the finite orbit $Wx$ has a point $\eta$ maximizing $(\rho ,\eta )$. If $(\eta ,{\alpha }_i)<0$, then $(\rho ,s_i\eta -\eta )=-2(\eta ,{\alpha }_i)(\rho ,{\alpha }_i)\mathrm{/}({\alpha }_i,{\alpha }_i)>0$, a contradiction. Thus $\eta \in \bar{\mathcal{C}}$.

Suppose $\lambda ,\mu \in \bar{\mathcal{C}}$ and $\lambda =w\mu$. Choose among such $w$ one of least word length. If nonidentity, write a reduced expression $w=s_{i}v$. By F2, $w^{-1}{\alpha }_i<0$. Since $\mu$ pairs nonnegatively with every positive root by F1, $(\lambda ,{\alpha }_i)=(\mu ,w^{-1}{\alpha }_i)\le 0.$ Dominance of $\lambda$ forces equality. Hence $s_i\lambda =\lambda$ and $v\mu =s_{i}w\mu =\lambda$, contradicting minimal length. Thus $w=1$ and $\lambda =\mu$. Lattice preservation then gives the unique dominant representative of each orbit in $P$.

For $\eta \in \bar{\mathcal{C}}$ and any nonidentity $w$ fixing it, write a reduced expression $w=s_{i}v$. The same scalar-product argument as in step 1.2 gives $(\eta ,{\alpha }_i)=0$. Then $s_i$ fixes $\eta$ and $v=s_{i}w$ does too, with smaller length. Induction writes $w$ as a product of zero-label simple reflections. Conversely each of those reflections visibly fixes $\eta$, proving exactly the stabilizer assertion, including $\eta =0$.

For any regular point, all roots have fixed nonzero signs on its connected component of the hyperplane complement, since a continuous real linear functional cannot change sign without vanishing. Conversely a nonempty prescribed sign region is an intersection of open linear halfspaces, hence convex and connected. Thus the chambers are exactly those sign regions. F1 gives $\mathcal{C}=\{x:(x,\alpha )>0\text{ for all }\alpha \in {\mathrm{\Phi }}^{+}\}$. Step 1.1 sends every regular point into this region (its image remains regular), so $W$ acts transitively on chambers. If $w\mathcal{C}=\mathcal{C}$, then $w^{-1}$ preserves positive roots by testing their signs on this chamber. F2 gives length zero, so $w=1$. This proves simple transitivity without a chamber-faithfulness assumption in the proof of exchange.

The negative region $-\mathcal{C}$ is a chamber, so step 2.2 gives a unique $w_0$ sending $\mathcal{C}$ to it. Equivalently $w_0^{-1}{\mathrm{\Phi }}^{+}={\mathrm{\Phi }}^{-}$, hence also $w_0{\mathrm{\Phi }}^{+}={\mathrm{\Phi }}^{-}$. F2 gives length $\mathrm{\mid }{\mathrm{\Phi }}^{+}\mathrm{\mid }$, the largest possible inversion count. Any element of that length reverses all positive roots and hence has the same chamber image, so equals $w_0$. Moreover $w_0^2$ preserves the positive chamber and is therefore the identity. In rank zero the hyperplane complement and its unique chamber are the one-point space, all groups and stabilizer claims are trivial and $w_0=1$.