Classical free action and covering exact sequence

The coordinate description in [F1] makes $\bar{C}$ a closed convex polyhedral cone with nonempty interior $C$. For each $s\in S$, the dual action of $s$ has order two and fixes pointwise the wall $H_{e_s}$: indeed $(sf)(v)=f(v)-2B(v,e_s)f(e_s)$. Its fixed space is exactly $H_{e_s}$, since evaluation at $e_s$ changes sign. These walls are precisely the coordinate walls of $\bar{C}$. The faithful dual action identifies $W$ with the group generated by these wall reflections, and [F2] supplies the required disjointness of $C$ and its nonidentity translates. Thus this is a Vinberg system in the sense of [F3].

Let $r\in W$ act on $\mathrm{\Theta }$ as any linear reflection. Choose $u$ of minimum word length among all conjugates of $r$, possible by [F6]. Its dual action is again a linear reflection. By faithfulness, $u^2=1$ and $u\mathrm{\ne }1$. Put $A=\rho (u)$. Then $A^2=I$, and the dual action has matrix $(A^{-1}{)}^{\mathsf{T}}=A^{\mathsf{T}}$ in dual bases. Transposition preserves rank, so the hyperplane fixed space of the dual action gives $\mathrm{rank}(A-I)=1$. Also, reversing a word in the involutions $S$ represents its inverse with the same length; doing this in both directions gives $\mathrm{\ell }(v^{-1})=\mathrm{\ell }(v)$ for every $v\in W$.

Let $s$ be the last letter of a reduced expression of $u$. Deleting it gives $\mathrm{\ell }(us)=\mathrm{\ell }(u)-1$: the remaining word gives the upper bound, and a shorter word for $us$, followed by $s$, would contradict the minimal length of $u$. Since $u=u^{-1}$, inversion invariance gives $\mathrm{\ell }(su)=\mathrm{\ell }(us)=\mathrm{\ell }(u)-1$. By [F2], $A{e}_s\in {\mathrm{\Phi }}_{-}$. If $A{e}_s\mathrm{\ne }-e_s$, [F5] gives $\rho (su)e_s=s(A{e}_s)\in {\mathrm{\Phi }}_{-}$. Applying [F2] to $su$ then gives $\mathrm{\ell }(sus)<\mathrm{\ell }(su)<\mathrm{\ell }(u)$, contradicting the choice of $u$ among conjugates of $r$. Therefore $A{e}_s=-e_s$.

Since $(A-I)e_s=-2e_s$ and $A-I$ has rank one, every $v\in V$ satisfies $Av=v+\lambda (v)e_s$ for some scalar $\lambda (v)$. Form preservation and $A{e}_s=-e_s$ give $B(v,e_s)=B(Av,A{e}_s)=-B(v,e_s)-\lambda (v)$, because $B(e_s,e_s)=1$. Thus $Av=v-2B(v,e_s)e_s=\rho (s)v$ for every $v$. Faithfulness gives $u=s$. Since $u$ is conjugate to $r$, every linear reflection in the dual action is conjugate to a simple reflection. This argument does not assume nondegeneracy of $B$.

By step 3.1, every reflection used in [F3] is $ws{w}^{-1}$ for some $w\in W$ and $s\in S$. Its fixed hyperplane is $w{H}_{e_s}=H_{w{e}_s}$ by the contragredient action. Conversely, each root is $w{e}_s$, and $ws{w}^{-1}$ is an order-two linear map whose fixed space is $H_{w{e}_s}$. Thus the fixed hyperplanes of all linear reflections in [F3] are exactly the members of $\mathcal{H}$. The open cone in [F3] is $\mathrm{int}({\bigcup }_{w}w\bar{C})={\mathrm{C}\mathrm{o}\mathrm{n}\mathrm{e}}^{\circ }$ by [F1]. Consequently its product complement is exactly ${\mathcal{X}}^{\mathbb{C}}$, with the same topology and diagonal action by [F4].

Apply [F3] to the Vinberg system identified in step 1.1 and the complement identified in step 4.1. It gives freeness, proper discontinuity and the disjoint-neighbourhood condition, as well as the regular covering with deck group $W$. Its endpoint-translate sequence at $Z$ is the exact sequence displayed in the Statement; regular and normal have the stated meaning here. This proves all assertions.