The Coxeter complex of $A_3$: a triangulation of the sphere and the residue of a proper parabolic

By [F2] and [F3] the group $W$ is finite, of order $24$; hence by [F7] the form $B$ is positive definite.

The subgroup orders of [F4] and [F10] give the coset counts $\mathrm{\mid }W\mathrm{\mid }\mathrm{/}\mathrm{\mid }{W}_I\mathrm{\mid }$ for the proper subsets $I$: $\mathrm{\mid }W\mathrm{\mid }\mathrm{/}\mathrm{\mid }{W}_{\{s_i\}}\mathrm{\mid }=12$ for the three one-element $I$, and for $\mathrm{\mid }I\mathrm{\mid }=2$ the values $24\mathrm{/}6=4$, $24\mathrm{/}4=6$ and $24\mathrm{/}6=4$ for $I=\{s_2,s_3\}$, $\{s_1,s_3\}$ and $\{s_1,s_2\}$.

The faces containing $w\overline{C_I}$ are exactly $u\overline{C_K}$ with $u\in w{W}_I$ and $K\subseteq I$. If $u=wx$, $x\in W_I$, then $u\overline{C_I}=w\overline{C_I}$ because $x$ fixes this face pointwise by [F5]; and $K\subseteq I$ gives $\overline{C_I}\subseteq \overline{C_K}$. Conversely, if $w\overline{C_I}\subseteq u\overline{C_K}$, their intersection is $w\overline{C_I}$, so [F5] and the dual-basis formula give $I\cup K\cup S(u^{-1}w)=I$. Hence $K\subseteq I$ and $u^{-1}w\in W_I$ by [F4], equivalently $u\in w{W}_I$.

Let ${\sigma }_0\in S_4$ be the reversal permutation $j\mapsto 3-j$ $(j\in \{0,1,2,3\})$; it satisfies $\mathrm{inv}({\sigma }_0)=6$ because every one of the six pairs $i<j$ is inverted, and $\mathrm{inv}(\tau )\le 6$ for every $\tau \in S_4$ because an inversion set consists of pairs. Hence $\mathrm{\ell }$ attains its maximum $6$ at ${\phi }^{-1}({\sigma }_0)$, and by [F6] the longest element $w_0$ is the unique element of maximal length; thus $w_0={\phi }^{-1}({\sigma }_0)$ corresponds to the reversal permutation, and $\mathrm{\ell }(w_0)=6=\mathrm{\mid }{\mathrm{\Phi }}_{+}\mathrm{\mid }=\mathrm{\mid }T\mathrm{\mid }$ and $w_0^2=1$ by [F6].

By [F5] the faces of $\mathrm{\Sigma }$ are the cosets $w{W}_I$ with $I⊊S$; a face of type $I$ is a spherical simplex on the $\mathrm{\mid }S\setminus I\mathrm{\mid }$ vertices $\rho (w)v_s\mathrm{/}\parallel v_s{\parallel }_B$ ($s\mathrm{\notin }I$). Hence the chambers (type $I=\mathrm{\varnothing }$) number $\mathrm{\mid }W\mathrm{\mid }\mathrm{/}\mathrm{\mid }{W}_{\mathrm{\varnothing }}\mathrm{\mid }=24$, the edges (type $\mathrm{\mid }I\mathrm{\mid }=1$) number $3\cdot 12=36$, and the vertices (type $\mathrm{\mid }I\mathrm{\mid }=2$, where the simplex is a single point) number $4+6+4=14$ by step 1.2; the alternating face count of this triangulation is $14-36+24=2$. Since $\mathrm{\mid }S\mathrm{\mid }=3$, [F5] realizes $\mathrm{\Sigma }$ as a triangulation of $S^2$. This proves the combinatorial part of (i).

Let $I=\{s,t\}$ with $\mathrm{\mid }I\mathrm{\mid }=2$ and let $\sigma :=w\overline{C_I}$ be a vertex. By step 1.3 the chambers containing $\sigma$ are the chambers $uC$ with $u\in w{W}_I$, so there are $\mathrm{\mid }{W}_I\mathrm{\mid }$ of them; the edges containing $\sigma$ are the faces $u\overline{C_{\{r\}}}$ with $u\in w{W}_I$, $r\in I$ (step 1.3 with $K=\{r\}$), and two such faces coincide exactly when the cosets $u{W}_{\{r\}}$ and $u^{\mathrm{\prime }}{W}_{\{r^{\mathrm{\prime }}\}}$ coincide, by the dictionary $u\overline{C_{\{r\}}}=u^{\mathrm{\prime }}\stackrel{‾}{C_{\{r^{\mathrm{\prime }}\}}}⟺u{W}_{\{r\}}=u^{\mathrm{\prime }}{W}_{\{r^{\mathrm{\prime }}\}}$ of [F5]. Each chamber $uC$ through $\sigma$ contains exactly the two edges $u\overline{C_{\{s\}}}$ and $u\overline{C_{\{t\}}}$ through $\sigma$; and each edge $u\overline{C_{\{r\}}}$ through $\sigma$ lies, by step 1.3 with $I=\{r\}$, in exactly the two chambers $uC$ and $urC$ through $\sigma$. Counting the incidences between the chambers and the edges through $\sigma$ by chambers gives twice $\mathrm{\mid }{W}_I\mathrm{\mid }$, and by edges gives twice the number $e$ of edges; hence $e=\mathrm{\mid }{W}_I\mathrm{\mid }$. The chambers through $\sigma$ form a connected graph under the relation of sharing an edge, because $W_I=⟨s,t⟩$ is generated by its two elements, so the consecutive chambers $uC$, $usC$ share $u\overline{C_{\{s\}}}$ and $uC$, $utC$ share $u\overline{C_{\{t\}}}$. A connected graph in which every vertex has degree $2$ is a cycle; hence the link of $\sigma$ is a cycle with $\mathrm{\mid }{W}_I\mathrm{\mid }$ edges and $\mathrm{\mid }{W}_I\mathrm{\mid }$ vertices, namely with $2\cdot m(s,t)$ of each. Its chamber edges are indexed by the coset $w{W}_I$ and its vertex edges by the cosets $u{W}_{\{r\}}$ ($u\in w{W}_I$, $r\in I$), which is the incidence structure of the Coxeter complex of the parabolic subsystem $(W_I,I)$: for $I=\{s_2,s_3\}$ (respectively $\{s_1,s_2\}$) this is the hexagon of $I_2(3)$ with $2\cdot 3=6$ edges, and for $I=\{s_1,s_3\}$ the $4$-cycle of $I_2(2)$. The same count with $\mathrm{\mid }I\mathrm{\mid }=1$ shows that an edge lies in exactly $\mathrm{\mid }{W}_I\mathrm{\mid }=2$ chambers, so the link of an edge is two points, and the link of a chamber is empty. This proves (ii).

To compute the round area, use orthonormal coordinates for $B$ (successively subtract projections from the three basis vectors and divide by their positive norms). Put $\mathrm{\Delta }=\{(u,t):u,t\ge 0,u+t\le 1\}$, $p(u,t)=(1-u-t)v_{s_1}+u{v}_{s_2}+t{v}_{s_3}$ and $\psi =p\mathrm{/}\parallel p{\parallel }_B$. The coefficient-sum functional $L({\sum }_i{a}_i{v}_{s_i})={\sum }_i{a}_i$ satisfies $L(p)=1$, so $p\mathrm{\ne }0$ on a neighbourhood of $\mathrm{\Delta }$; moreover radial projection is injective on the affine plane $L=1$, with inverse $x\mapsto x\mathrm{/}L(x)$ where $L(x)>0$. Its derivative on that plane is injective: $D(p\mathrm{/}\parallel p{\parallel }_B)h=0$ forces $h$ parallel to $p$, whereas $L(h)=0$ and $L(p)=1$. Thus $\psi$ is a regular patch for $C\cap S^2$. For every $w$, ${\psi }_w=\rho (w)\psi$ is a regular patch for its chamber, and $B$-invariance gives identical Gram matrices and therefore identical densities $J_{{\psi }_w}=J_{\psi }$. Let $a={\int }_{\mathrm{\Delta }}{J}_{\psi }$; each chamber has area $a$.

The three vertices of the spherical triangle $C\cap S^2$ lie each on a pair of the walls $H_{e_{s_1}},H_{e_{s_2}},H_{e_{s_3}}$, so by the dihedral-angle clause of [F5] its interior angles are $\pi \mathrm{/}m(s_1,s_2)=\pi \mathrm{/}3$, $\pi \mathrm{/}m(s_2,s_3)=\pi \mathrm{/}3$ and $\pi \mathrm{/}m(s_1,s_3)=\pi \mathrm{/}2$, with sum $7\pi \mathrm{/}6$; every chamber is $wC$ for a unique $w\in W$, and $\rho (w)$ preserves $B$ by [F1], hence is an isometry of $(V,B)$ carrying $C\cap S^2$ onto $wC\cap S^2$; so all $24$ chambers are congruent spherical triangles.

Here is the finite chart comparison needed to sum these areas; no independence of an arbitrary surface presentation is assumed. In orthonormal coordinates let $\eta (\varphi ,\theta )=(\sin \varphi \cos \theta ,\sin \varphi \sin \theta ,\cos \varphi )$ and $R_{\epsilon }=[\epsilon ,\pi -\epsilon ]\times [\epsilon ,2\pi -\epsilon ]$, $0<\epsilon <\pi \mathrm{/}2$. This is an injective regular chart, with Gram matrix $\mathrm{diag}(1,{\sin }^2\varphi )$, so $J_{\eta }=\sin \varphi$. Cut $R_{\epsilon }$ by the chamber walls and let $E_{w,\epsilon }={\eta }^{-1}({\psi }_w(\mathrm{\Delta }))\cap R_{\epsilon }$, $D_{w,\epsilon }={\psi }_w^{-1}(\eta (R_{\epsilon }))\cap \mathrm{\Delta }$. These compact sets are Jordan: away from the latitude-chart seam and poles, great circles have nonzero tangent and their chart preimages are locally smooth arcs; in the radial charts the chamber edges are straight segments, and latitude and longitude boundaries have smooth arc preimages. A finite cover of each compact arc by regular curve pieces suffices. A $C^1$ curve on a neighbourhood of a compact interval has bounded derivative by [F12], and hence a Lipschitz bound $M$ there by [F13]; it has content zero in the plane: divide the interval into $n$ equal parts and cover each image by a square of side at most $2M$ times the part length (enlarging by $1\mathrm{/}{n}^2$ if necessary); the total square area tends to zero. Finite unions and points have the same property, so the boundary criterion in [F8] applies, and the overlaps between different $E_{w,\epsilon }$ have content zero by [F5]. On a neighbourhood of $D_{w,\epsilon }$ the transition $h={\eta }^{-1}\circ {\psi }_w$ is an injective $C^1$ coordinate change with invertible derivative: radial projection has the explicit inverse in step 2.3 and the latitude chart has a $C^1$ inverse away from its seam and poles: at each point choose two ambient coordinate components on which its derivative has nonzero determinant and apply [F11]. The remaining sphere coordinate is locally the fixed-sign function $\pm \sqrt{1-x_i^2-x_j^2}$, since it is nonzero there; thus the local inverse also applies to ${\psi }_w$, and the inverses agree on overlaps by injectivity of $\eta$. The chain rule [F11] yields $G_{{\psi }_w}=D{h}^{\mathsf{T}}(G_{\eta }\circ h)Dh$, hence $J_{{\psi }_w}=(J_{\eta }\circ h)\mathrm{\mid }\det Dh\mathrm{\mid }$. Change of variables and finite additivity in [F8] now give ${\sum }_w{\int }_{D_{w,\epsilon }}{J}_{\psi }={\sum }_w{\int }_{E_{w,\epsilon }}\sin \varphi ={\int }_{R_{\epsilon }}\sin \varphi$.

The omitted radial parameter sets approach the preimage of the single longitude seam and the two poles. That compact preimage has content zero: the seam lies on a great circle, whose radial preimage lies on a straight line, and each pole has at most one preimage. For any finite open rectangle cover of that set, compactness places all omitted points in the cover once $\epsilon$ is sufficiently small: otherwise a compact subset outside the cover would meet arbitrarily narrow seam or pole strips despite its image avoiding the seam and poles. Since $J_{\psi }$ is continuous and bounded on $\mathrm{\Delta }$, the omitted integral is bounded by its bound times the total area of such a cover, and so tends to zero. There are only $24$ patches, hence the left side in step 3.2 tends to $24a$. By [F9] the right side is $(2\pi -2\epsilon ){\int }_{\epsilon }^{\pi -\epsilon }\sin \varphi d\varphi =(2\pi -2\epsilon )2\cos \epsilon$, tending to $4\pi$; the full latitude patch on $[0,\pi ]\times [0,2\pi ]$ itself has area ${\int }_0^{2\pi }{\int }_0^{\pi }\sin \varphi d\varphi d\theta =4\pi$, with its only seam identifications and rank failures on the parameter boundary. Thus the actual finite chamber sum equals the round sphere area, $24a=4\pi$, and $a=\pi \mathrm{/}6$. Together with step 3.1 this proves (iii). All covers and coordinate choices are finite; no choice principle is used.

Conjugating the adjacent transposition $(i-1i)$ by the reversal ${\sigma }_0$ gives $({\sigma }_0(i-1){\sigma }_0(i))=(4-i3-i)$, which is the adjacent transposition $(3-i4-i)$, that is $s_{4-i}$ under the identification of [F2]; hence $w_0{s}_i{w}_0=s_{4-i}$ for $i=1,2,3$. By [F6] the permutation $\sigma$ with $\rho (w_0)e_{s_i}=-e_{\sigma (s_i)}$ satisfies $\sigma (s_i)=s_{4-i}$, so $\rho (w_0)e_{s_i}=-e_{s_{4-i}}$. This proves (iv).