The disconnected universal-Coxeter nerve and the angular truncation convention

The nerve and its path distance. Fix distinct $s,t\in S$. Send $s$ to the involution $a(u)=-u$, $t$ to $b(u)=2-u$, and every other generator to the identity in $\mathrm{Sym}(\mathbb{R})$. By [F9] this extends to a homomorphism from $W$. The product $ab$ is the translation $u\mapsto u-2$, whose $k$-th power sends $0$ to $-2k$; these images are distinct for distinct nonnegative integers $k$. Hence $W_{\{s,t\}}$ is infinite. Every $W_T$ with $s,t\in T$ contains it, and is infinite; a singleton generates at most the two elements $1,s$, so the nerve consists exactly of the empty simplex and the $n$ singletons. Every cell $\mathrm{\Sigma }(C_i)$ of its spherical realization $L$ is a point by [F1]. Any step of a chain in $L$ therefore has equal endpoints, and no chain joins distinct points. Thus [F2] gives $d_{\mathrm{p}\mathrm{a}\mathrm{t}\mathrm{h}}(x,x)=0$ and $d_{\mathrm{p}\mathrm{a}\mathrm{t}\mathrm{h}}(x,y)=+\mathrm{\infty }$ for $x\mathrm{\ne }y$.

The truncated metric. By [F3] and the convention $\min \{\pi ,\mathrm{\infty }\}=\pi$, one has $d_{\pi }(x,y)=\min \{\pi ,d_{\mathrm{p}\mathrm{a}\mathrm{t}\mathrm{h}}(x,y)\}=\pi$ for $x\mathrm{\ne }y$, while $d_{\pi }(x,x)=0$ by step 1.1. By [F4] the function $d_{\pi }$ is a metric of diameter at most $\pi$, and since $n\ge 2$ there is a pair $x\mathrm{\ne }y$ with $d_{\pi }(x,y)=\pi$, so the diameter is exactly $\pi$. Comparing values, $d_{\pi }$ agrees with $d_{\mathrm{p}\mathrm{a}\mathrm{t}\mathrm{h}}$ exactly for $x=y$ and nowhere off the diagonal, and by [F5] a metric takes real values while $+\mathrm{\infty }$ is not a real number, so $d_{\mathrm{p}\mathrm{a}\mathrm{t}\mathrm{h}}$ itself is not a metric on $L$ and the truncation is what produces one.

The cone distances. For $r,s>0$ the cone formula of [F3] and step 2.1 give, for $x=y$, $d_C((r,x),(s,x){)}^2=r^2+s^2-2rs\cos 0=(r-s{)}^2$, so that $d_C=\mid r-s\mid$, and, for $x\mathrm{\ne }y$, $d_C((r,x),(s,y){)}^2=r^2+s^2-2rs\cos \pi =(r+s{)}^2$ with $\cos \pi =-1$ by [F7], so that $d_C=r+s$; also $d_C(o,(r,x))=r$. Restricting to one ray, the map $t\mapsto (t,x)$ with $0\mapsto o$ therefore satisfies $d_C=\mid t-t^{\mathrm{\prime }}\mid$ on $[0,\mathrm{\infty })$, an isometry onto the ray $R_x:=\{o\}\cup \{(t,x):t>0\}$, and for $x\mathrm{\ne }y$ all pairs on different rays satisfy $d_C=t+t^{\mathrm{\prime }}$.

The rays meet only at the apex. Let $x\mathrm{\ne }y$, $r,s>0$ and let $\gamma \text{\hspace{0.17em}⁣}:[0,1]\to C(L)$ be a path with $\gamma (0)=(r,x)$ and $\gamma (1)=(s,y)$. By step 3.1 the ball of radius $r$ about $(r,x)$ consists of the points $(t,x)$ with $\mid t-r\mid <r$, since a point $(t,z)$ with $z\mathrm{\ne }x$ has distance $r+t\ge r$; so the open ray $R_x\setminus \{o\}$ is open in $C(L)$, and likewise every $R_z\setminus \{o\}$. These open rays are pairwise disjoint with union $C(L)\setminus \{o\}$, and the projection $\pi$ sending $R_z\setminus \{o\}$ to $z$ is continuous for the discrete topology on $L$, its fibres being open. If $\gamma$ avoided $o$, then $\pi \circ \gamma$ would be a continuous map $[0,1]\to L$: by [F8] the interval $[0,1]$ is connected, so its image is connected, and a connected subset of the discrete space $L$ is a singleton, contradicting $\pi (\gamma (0))=x\mathrm{\ne }y=\pi (\gamma (1))$. Hence every path from $(r,x)$ to $(s,y)$ passes through the apex $o$.

The through-apex geodesic and geodesic space. For $x\mathrm{\ne }y$ and $r,s>0$, step 2.1 gives $d_{\pi }(x,y)=\pi$, so [F4] shows that the two-segment path through $o$ has length $r+s=d_C((r,x),(s,y))$ and is minimizing; in particular it is a geodesic segment by [F6]. For the $D_{\pi }$-condition, a pair $(z,w)$ with $d_{\pi }(z,w)<\pi$ has $z=w$ by step 2.1 and is joined by the constant segment $\gamma \text{\hspace{0.17em}⁣}:[0,0]\to L$ with $\gamma (0)=z$, so the hypothesis of [F4] holds vacuously and $C(L)$ is a geodesic metric space. Finally, if $\gamma \text{\hspace{0.17em}⁣}:[0,r+s]\to C(L)$ is a geodesic from $(r,x)$ to $(s,y)$ with $x\mathrm{\ne }y$, then $\gamma$ is a path, so by step 4.1 there is $t_0$ with $\gamma (t_0)=o$, and since $\gamma$ is distance-preserving of length $r+s=d_C((r,x),(s,y))$ while $d_C((r,x),o)=r$ and $d_C(o,(s,y))=s$, one gets $t_0=r$; the restriction of $\gamma$ to $[0,r)$ avoids $o$, hence lies in the single ray $R_x$ by the argument of step 4.1, and for $u\in [0,r)$ the distance to $\gamma (r)=o$ gives $\gamma (u)=(r-u,x)$; symmetrically $\gamma (r+u^{\mathrm{\prime }})=(u^{\mathrm{\prime }},y)$ for $u^{\mathrm{\prime }}\in (0,s]$, with $\gamma (r)=o$. So the geodesic is the two-segment path through the apex.

The truncation value is forced. Let $x\mathrm{\ne }y$ and $r,s>0$. The value $\theta =0$ is excluded at once: it would give $d_C((r,x),(r,y))=0$ for distinct points and violate separation, so let $\theta \in (0,\pi )$. If the cone formula received the value $\theta$ in place of $\pi$, it would assign the points $(r,x)$ and $(s,y)$ the distance $L:=\sqrt{r^2+s^2-2rs\cos \theta }$, which is $<r+s$ because cosine is strictly decreasing on $[0,\pi ]$ by [F7]; and the open rays would still be open for the distance function so defined, since a point of another branch is at distance at least $t_0\sin \theta >0$ from $(t_0,x)$ for every $t_0>0$ (the quadratic $t_0^2+t^2-2t_{0}t\cos \theta$ in $t\ge 0$ is minimised at $t=t_0\cos \theta$ when this is nonnegative and at $t=0$ otherwise). Hence the argument of step 4.1 applies verbatim: a geodesic $\gamma \text{\hspace{0.17em}⁣}:[0,L^{\mathrm{\prime }}]\to C(L)$ from $(r,x)$ to $(s,y)$ would be a path, hence would pass through the apex at some time $t_0$; distance-preservation would then give $t_0=d_C((r,x),o)=r$ and $L^{\mathrm{\prime }}-t_0=d_C(o,(s,y))=s$, so $L^{\mathrm{\prime }}=r+s$, contradicting $L^{\mathrm{\prime }}=d_C((r,x),(s,y))=L<r+s$. Hence no geodesic joins the two branches once a value $\theta <\pi$ is used, whereas with the value $\pi$ the through-apex path is a geodesic by step 5.1: among the values in $[0,\pi ]$, only $\pi$ makes $C(L)$ the metric star with its through-apex geodesics. Since a metric value must be a real number by [F5], the auxiliary value $+\mathrm{\infty }$ cannot be passed either; so the formula receives $\pi$, the unique $\theta \in [0,\pi ]$ with $\cos \theta =-1$ by [F7].