Infinite dihedral type: the chamber system is a line, not a sphere; the contractible model is deferred

In the coordinates $(y_s,y_t)$ the generators act by the displayed formulas [F2], and both fix $\mathrm{\Delta }$: $(-y_s)+(2y_s+y_t)=y_s+y_t$ and $(y_s+2y_t)+(-y_t)=y_s+y_t$. Since $S$ generates $W$ and the dual action is a left action, $\mathrm{\Delta }(w\cdot f)=\mathrm{\Delta }(f)$ for every $w\in W$.

The element $st$ satisfies $\rho (st)=r_s{r}_t$ and by [F1] its matrix on $P$ is $A=I+N$ with $N\mathrm{\ne }0$ and $N^2=0$; hence $\rho ((st{)}^k)=A^k=I+kN\mathrm{\ne }I$ for every $k\mathrm{\ne }0$ (for negative $k$ use $(I+N{)}^{-1}=I-N$, which holds because $N^2=0$). Therefore $(st{)}^k\mathrm{\ne }1$ in $W$ for every $k\mathrm{\ne }0$, and $W$ is infinite.

For $f$ with $\mathrm{\Delta }(f)=1$ write $a:=y_s$, so $y_t=1-a$; then [F2] gives $s\cdot f=(-a,2a+1-a)=(-a,1+a)$, that is, $s$ acts on the line $\{\mathrm{\Delta }=1\}$ by $a\mapsto -a$, and $t\cdot f=(a+2(1-a),-(1-a))=(2-a,a-1)$, that is, $a\mapsto 2-a$; since the action is a left action, $(st{)}^k$ acts by $a\mapsto a-2k$ and $s(st{)}^k$ by $a\mapsto -a+2k$, so the images of $C\cap \{\mathrm{\Delta }=1\}=\{0\le a\le 1\}=[0,1]$ under $W$ include every interval $[2k,2k+1]$ and $[2k-1,2k]$, that is, every unit interval $[k,k+1]$ with $k\in \mathbb{Z}$. Conversely $s$ sends a unit interval $[k,k+1]$ to $[-k-1,-k]$ and $t$ to $[1-k,2-k]$, both unit intervals, so by induction on length every $w$ sends $[0,1]$ to a unit interval. Since the $w$ act linearly and $C=\{y_s\ge 0,y_t\ge 0\}$ is the cone over $[0,1]$, each $wC$ is the cone over $w\cdot [0,1]$. Hence the chambers are exactly the cones over the unit intervals $[k,k+1]$.

By step 1.2 the group $W$ is infinite, so [F4] gives $0\mathrm{\notin }{U}^{\circ }$.

By [F1], $\rho ((st{)}^k)=I+kN$ on $P$ gives $\rho ((st{)}^k)e_s=e_s+k(2e_s+2e_t)=(1+2k)e_s+2k{e}_t$, and these are pairwise distinct roots for $k\in \mathbb{Z}$ because their $e_s$-coefficients $1+2k$ are distinct; hence $\mathrm{\Phi }$ is infinite. If $\mathrm{\ell }$ were bounded by some natural number $N$, then every element of $W$ would be the value of one of the finitely many words in $S$ of length at most $N$ (using $s^{-1}=s$), so $W$ would be finite, contradicting step 1.2. Thus $\mathrm{\ell }$ is unbounded on $W$ and $W$ has no longest element.

Every chamber is the cone over a unit interval of $\{\mathrm{\Delta }=1\}$ by step 2.1, and every point of such a cone is $\lambda g$ with $\lambda \ge 0$ and $\mathrm{\Delta }(g)=1$, so $\mathrm{\Delta }$ is nonnegative on $U$ and positive on $U\setminus \{0\}$. Conversely, if $\mathrm{\Delta }(f)>0$ then $a:=y_s\mathrm{/}\mathrm{\Delta }(f)$ is a real number, hence lies in some unit interval $[k,k+1]$, and then $f=\mathrm{\Delta }(f)(a,1-a)$ lies in the cone over $[k,k+1]$, which is a chamber; and $0\in C\subseteq U$. Therefore $U=\{f:\mathrm{\Delta }(f)>0\}\cup \{0\}$, so the nonzero points of $\{\mathrm{\Delta }=0\}$ lie outside $U$ and $U\mathrm{\ne }{V}^{\ast }$.

The relative interior of the chamber which is the cone over $[k,k+1]$ is the open cone over $(k,k+1)$, which is the open chamber $w{C}^{\circ }$ for the corresponding $w$; by [F3] these are pairwise disjoint.

If $wC=-C$ for some $w$, then $-C\subseteq \{\mathrm{\Delta }\ge 0\}$ because every chamber is contained in $\{\mathrm{\Delta }\ge 0\}$ by step 2.1; but $-C$ contains $-(v)$ for $v\in C^{\circ }$, and $\mathrm{\Delta }(-v)=-\mathrm{\Delta }(v)<0$ since $\mathrm{\Delta }(v)>0$ for $v\in C^{\circ }$ ($C^{\circ }$ is the open quadrant and $v\mathrm{\ne }0$). This contradiction shows that no chamber equals $-C$.

The map $f\mapsto f\mathrm{/}\mathrm{\Delta }(f)$ is well defined on $U\setminus \{0\}$ by step 3.1, has values in $\{\mathrm{\Delta }=1\}$, is invariant under positive scaling and separates distinct positive rays: if $f\mathrm{/}\mathrm{\Delta }(f)=g\mathrm{/}\mathrm{\Delta }(g)$ then $f=(\mathrm{\Delta }(f)\mathrm{/}\mathrm{\Delta }(g))g$ with positive factor, and conversely positive multiples have the same image; it is surjective onto $\{\mathrm{\Delta }=1\}$ because $\mathrm{\Delta }(g)=1$ gives $g=g\mathrm{/}\mathrm{\Delta }(g)$ with $g\in U$ by step 3.1. By step 2.1 it carries the chambers onto the unit intervals $[k,k+1]$. The subdivision is infinite because the intervals are pairwise distinct, and no finite union of chambers covers $U$: a finite union of cones over $[k_1,k_1+1],\dots ,[k_n,k_n+1]$ contains no point $f$ with $\mathrm{\Delta }(f)=1$ and $y_s$-coordinate $a>{\max }_i{k}_i+1$, while such points of $\{\mathrm{\Delta }=1\}\subseteq U$ exist. Hence this chamber system is not a finite sphere; the contractible comparison complex (the Davis complex) belongs to the later page named in the statement and is not used here.