The Coxeter complex of $I_2(5)$: the circle triangulated by the ten chambers

The number $c=\cos (\pi \mathrm{/}5)$ satisfies $4c^2-2c-1=0$: with $\theta :=\pi \mathrm{/}5$ one has $\cos 2\theta =2c^2-1$ and $\cos 3\theta =4c^3-3c$ by the addition formulas, while $3\theta =\pi -2\theta$ gives $\cos 3\theta =-\cos 2\theta =-(2c^2-1)$ by the shift formula $\cos (\pi -x)=-\cos x$; hence $4c^3-3c=-2c^2+1$, that is $(c+1)(4c^2-2c-1)=0$, and $c>0>-1$ forces $4c^2-2c-1=0$, equivalently $4c^2-1=2c$ and $4c^2-2c=1$.

$\rho (s{)}^2=\rho (s^2)=I$ and $[\rho (s)]$ is triangular with diagonal entries $-1,1$, so $\det \rho (s)=-1$; with $\det A=1$ and $\det A^k=(\det A{)}^k$ this gives $\det A^k=1$ for every $k\in \mathbb{Z}$, so no power of $A$ equals $\rho (s)$.

Put $z:=st$. Then $t=sz$, $szs=ts=z^{-1}$, and therefore $s{z}^k=z^{-k}s$ for every integer $k$. Substitute $t=sz$ into a word in $s,t$ and move every $s$ to the right by this identity, cancelling $s^2=1$; the value is $z^k$ or $z^{k}s$. Since $z^5=1$, reduce $k$ modulo $5$. Thus $W=\{(st{)}^k,(st{)}^{k}s:0\le k<5\}$ has at most ten elements and is finite.

By the positive-definiteness clause of [F1], $1-c^2={\sin }^2(\pi \mathrm{/}5)>0$. The dual vectors are $v_s=(e_s+c{e}_t)\mathrm{/}(1-c^2)$ and $v_t=(c{e}_s+e_t)\mathrm{/}(1-c^2)$: with $v_s=\alpha e_s+\beta e_t$ the conditions $B(v_s,e_s)=1$ and $B(v_s,e_t)=0$ read $\alpha -c\beta =1$ and $\beta -c\alpha =0$, that is $\beta =c\alpha$ and $\alpha (1-c^2)=1$; the computation for $v_t$ is symmetric. Then $B(v_s,v_s)=\alpha B(v_s,e_s)+\beta B(v_s,e_t)=\alpha =1\mathrm{/}(1-c^2)$ and $B(v_s,v_t)=B(v_s,c{e}_s+e_t)\mathrm{/}(1-c^2)=c\mathrm{/}(1-c^2)$, and $B(v_t,v_t)=1\mathrm{/}(1-c^2)$ similarly.

The faces containing a face $w\overline{C_I}$ are exactly the faces $u\overline{C_K}$ with $u\in w{W}_I$ and $K\subseteq I$: if $K\subseteq I$ and $u=wx$ with $x\in W_I$ then $u\overline{C_K}\supseteq u\overline{C_I}=w\overline{C_I}$ because $x$ fixes $\overline{C_I}$ pointwise and $\overline{C_I}\subseteq \overline{C_K}$; conversely $w\overline{C_I}\subseteq u^{\mathrm{\prime }}\overline{C_K}$ makes their intersection equal to $w\overline{C_I}$. By [F5] that intersection is $w\stackrel{‾}{C_{I\cup K\cup S(u^{\mathrm{\prime }-1}w)}}$; the dual-basis formula of [F5] distinguishes the face types, so $I\cup K\cup S(u^{\mathrm{\prime }-1}w)=I$. Hence $K\subseteq I$ and $S(u^{\mathrm{\prime }-1}w)\subseteq I$, giving $u^{\mathrm{\prime }-1}w\in W_I$ by [F6], equivalently $u^{\mathrm{\prime }}\in w{W}_I$. Here $\overline{C_{\mathrm{\varnothing }}}=C$.

$\rho (s)A\rho (s{)}^{-1}=\rho (s)\rho (s)\rho (t)\rho (s{)}^{-1}=\rho (t)\rho (s)=A^{-1}$, using $\rho (s{)}^{-1}=\rho (s)$ and $A^{-1}=\rho (t{)}^{-1}\rho (s{)}^{-1}=\rho (t)\rho (s)$; consequently $\rho (t)=\rho (s)A$ and every element of the subgroup generated by $\rho (s)$ and $\rho (t)$ has the form $A^k$ or $A^k\rho (s)$ with $k\in \mathbb{Z}$.

The root set is $\mathrm{\Phi }=\{\pm e_s,\pm e_t,\pm 2c(e_s+e_t),\pm (2c{e}_s+e_t),\pm (e_s+2c{e}_t)\}$: in the basis $(e_s,e_t)$ the matrices of $A^k$ applied to $e_s$ give $A{e}_s=2c(e_s+e_t)$, $A^2{e}_s=e_t$, $A^3{e}_s=-(2c{e}_s+e_t)$ and $A^4{e}_s=-(e_s+2c{e}_t)$ (using $4c^2-1=2c$ and $4c^2-2c=1$), and applying $\rho (s)$, whose matrix sends $e_s\mapsto -e_s$ and $e_t\mapsto 2c{e}_s+e_t$, to those five vectors gives $-e_s$, $e_s+2c{e}_t$, $2c{e}_s+e_t$, $-e_t$ and $-2c(e_s+e_t)$; since by step 1.3 every element of $W$ is $(st{)}^k$ or $(st{)}^{k}s$, the roots $\rho (w)e_s$ ($w\in W$) are exactly those ten vectors. The roots $\rho (w)e_t$ are the values $\rho ((st{)}^k)A^2{e}_s=A^{k+2}{e}_s$ and $\rho ((st{)}^{k}s)e_t=A^k(2c{e}_s+e_t)$; the first five are $e_t$, $-(2c{e}_s+e_t)$, $-(e_s+2c{e}_t)$, $e_s$ and $2c(e_s+e_t)$, and the second five are $2c{e}_s+e_t$, $e_s+2c{e}_t$, $-e_s$, $-2c(e_s+e_t)$ and $-e_t$, so both families lie in the displayed set and $\mathrm{\Phi }$ equals it. The five vectors $e_s$, $e_t$, $2c(e_s+e_t)$, $2c{e}_s+e_t$ and $e_s+2c{e}_t$ have pairwise different coordinate pairs in the basis $(e_s,e_t)$ (because $2c\mathrm{\ne }1$: $4c^2-2c-1$ vanishes at $c$ but not at $c=\frac{1}{2}$) and lie in $V_{+}\setminus \{0\}$, while their negatives lie in $-V_{+}\setminus \{0\}$; hence the five are pairwise distinct, and $V_{+}\cap (-V_{+})=\{0\}$ shows that none of the ten is a negative of another of the five, so the displayed set has ten elements. By [F4] each root lies in $V_{+}\setminus \{0\}\cup (-V_{+}\setminus \{0\})$, so ${\mathrm{\Phi }}_{+}=\mathrm{\Phi }\cap V_{+}$ consists exactly of those five vectors: $\mathrm{\mid }\mathrm{\Phi }\mathrm{\mid }=10$ and $\mathrm{\mid }{\mathrm{\Phi }}_{+}\mathrm{\mid }=5$.

The vertex $\overline{C_{\{t\}}}={\mathbb{R}}_{\ge 0}{v}_s$ lies in exactly two chambers, namely $C$ and $tC$: by step 1.5 with $I=\{t\}$ the chambers containing it are the $uC$ with $u\in W_{\{t\}}=\{1,t\}$, and these are distinct. Symmetrically the vertex $\overline{C_{\{s\}}}={\mathbb{R}}_{\ge 0}{v}_t$ lies in exactly the chambers $C$ and $sC$; every vertex of $\mathrm{\Sigma }$ is of the form $w\overline{C_{\{t\}}}$ or $w\overline{C_{\{s\}}}$ and hence, by the $W$-action, lies in exactly two chambers. Each chamber $wC$ has exactly the two vertices $w\overline{C_{\{s\}}}$ and $w\overline{C_{\{t\}}}$, since these are exactly the two one-dimensional faces of $C$ by the dual-basis formula of [F5]. The residue of the vertex $w{W}_{\{t\}}$ is therefore the two-point complex $\{wC,wtC\}$, which is the Coxeter complex of the rank-one system $(W_{\{t\}},\{t\})$, and symmetrically for $w{W}_{\{s\}}$. This proves (iv).

The ten elements $A^k$ and $A^k\rho (s)$ ($0\le k<5$) are pairwise distinct: the $A^k$ are distinct by $A^5=I$ and $A^k\mathrm{\ne }I$ for $0<k<5$, the $A^k\rho (s)$ are distinct for the same reason, and $A^k=A^j\rho (s)$ is impossible because the left side has determinant $1$ while $\det \rho (s)=-1$. Hence the image group $\rho (W)=⟨\rho (s),\rho (t)⟩$ has at least ten elements, so $\mathrm{\mid }W\mathrm{\mid }\ge 10$; combined with step 1.3 this gives $\mathrm{\mid }W\mathrm{\mid }=10$.

By steps 1.3 and 2.2 the group $W$ is finite, so the finiteness criterion applies; by clause (1) of Finiteness criterion: W is finite exactly when the Coxeter form is positive definite the form $B$ is positive definite. The map $\alpha \mapsto t_{\alpha }$ is a bijection ${\mathrm{\Phi }}_{+}\to T$ by [F4], so $\mathrm{\mid }T\mathrm{\mid }=5$; and $H_{-\alpha }=H_{\alpha }$ because $B(v,-\alpha )=-B(v,\alpha )$, so $\mathcal{A}=\{H_{\alpha }:\alpha \in {\mathrm{\Phi }}_{+}\}$. The five lines are pairwise distinct: the five positive roots correspond in the coordinates $(y_s,y_t)$ to the directions $(1,0)$, $(0,1)$, $(2c,2c)$, $(2c,1)$ and $(1,2c)$; two of these are proportional only if the corresponding pairs differ by a common nonzero factor, which fails for $(1,0)$ and $(0,1)$ against all others (a zero coordinate stays zero) and for the remaining three, as is checked by comparing ratios: $(2c,2c)$ versus $(1,2c)$ would force $2c=1$, $(2c,2c)$ versus $(2c,1)$ likewise, and $(2c,1)$ versus $(1,2c)$ would force $4c^2=1$, hence $2c+1=1$ and $c=0$, contrary to $4c^2-2c-1=-1\mathrm{\ne }0$. This proves (i).

Put $w_1:=(st{)}^{2}s$; then $\rho (w_1)=A^2\rho (s)$ has matrix $\left(\begin{array}{cc}0& -1\\ 1& -2c\end{array}\right)\left(\begin{array}{cc}-1& 2c\\ 0& 1\end{array}\right)=\left(\begin{array}{cc}0& -1\\ -1& 0\end{array}\right)$, so $\rho (w_1)e_s=-e_t$ and $\rho (w_1)e_t=-e_s$; applied to the five positive roots in the order of step 2.2 this gives $-e_t$, $-e_s$, $-2c(e_s+e_t)$, $-(e_s+2c{e}_t)$ and $-(2c{e}_s+e_t)$, so $\rho (w_1){\mathrm{\Phi }}_{+}={\mathrm{\Phi }}_{-}$ and hence $N(w_1)={\mathrm{\Phi }}_{+}$.

By [F5] the chamber $C$ equals $\{\lambda v_s+\mu v_t:\lambda ,\mu \ge 0\}$, the cone over the segment $[v_s,v_t]$, and $0\mathrm{\notin }[v_s,v_t]$ because $v_s,v_t$ are linearly independent; the radial projection $x\mapsto x\mathrm{/}\parallel x{\parallel }_B$ of that segment is therefore a continuous injective map of a connected set, so $C\cap S^1$ is an arc with endpoints $u_s$ and $u_t$. Its images $\rho (w)(C\cap S^1)=wC\cap S^1$ under the ten elements of $W$ are the ten spherical chambers; by [F5] the closed chambers cover $S^1$ and distinct closed chambers have disjoint interiors, so these ten closed arcs cover the circle and meet only in their endpoints, and the five root lines meet $S^1$ in exactly the ten endpoints; hence the five lines cut $S^1$ into the ten arcs $wC\cap S^1$. The triangulation $\mathrm{\Sigma }$ has one edge per chamber and one vertex per coset $w{W}_{\{s\}}$, $w{W}_{\{t\}}$; by [F6] and [F11] these cosets number $\mathrm{\mid }W\mathrm{\mid }\mathrm{/}\mathrm{\mid }{W}_{\{t\}}\mathrm{\mid }=10\mathrm{/}2=5$ and $\mathrm{\mid }W\mathrm{\mid }\mathrm{/}\mathrm{\mid }{W}_{\{s\}}\mathrm{\mid }=5$, so $\mathrm{\Sigma }$ is a circle with ten edges and ten vertices and Euler characteristic $10-10=0$.

Consequently $B(u_s,u_t)=B(v_s,v_t)\mathrm{/}(\parallel v_s{\parallel }_B\parallel v_t{\parallel }_B)=(c\mathrm{/}(1-c^2))\mathrm{/}(1\mathrm{/}(1-c^2))=c$, so the endpoints of the fundamental arc subtend the principal angle $\arccos c=\arccos (\cos (\pi \mathrm{/}5))=\pi \mathrm{/}5$. For each $w\in W$ the map $\rho (w)$ preserves $B$ by [F1] and therefore preserves $\parallel \cdot {\parallel }_B$, so $B(\rho (w)u_s,\rho (w)u_t)=B(u_s,u_t)=c$; hence every spherical chamber subtends the same angle $\pi \mathrm{/}5$, and the ten arcs account for the full turn $10\cdot (\pi \mathrm{/}5)=2\pi$.

By [F7] the element with $N(w_0)={\mathrm{\Phi }}_{+}$ is unique, so $w_1=w_0$; consequently $\mathrm{\ell }(w_0)=\mathrm{\mid }N(w_0)\mathrm{\mid }=\mathrm{\mid }{\mathrm{\Phi }}_{+}\mathrm{\mid }=5=\mathrm{\mid }T\mathrm{\mid }$, $w_0^2=1$, and the permutation $\sigma$ of [F7] satisfies $\rho (w_0)e_s=-e_{\sigma (s)}$; since $\rho (w_0)e_s=-e_t$ and $\rho (w_0)e_t=-e_s$, $\sigma$ interchanges $s$ and $t$. This proves (v) and completes all clauses.