Spherical Parabolic Cosets and the Davis Complex — Examples

By [F9], the relations are $s^2=t^2=1$ and $(st{)}^3=1$. Put $r=st$; then $srs=r^{-1}$, so every word reduces to $r^k$ or $s{r}^k$ with $0\le k<3$. The map $s\mapsto (12)$, $t\mapsto (23)$ is onto $S_3$, so the six normal forms are distinct and $\mathrm{\mid }W\mathrm{\mid }=6$, in agreement with [F10]. Thus $W_{\{s\}}=\{1,s\}$ and $W_{\{t\}}=\{1,t\}$ each have index $3$, while $W_S=W$. The cosets are $W_{\{s\}},t{W}_{\{s\}},st{W}_{\{s\}};W_{\{t\}},s{W}_{\{t\}},ts{W}_{\{t\}};W,$ and the six singleton cosets $w{W}_{\mathrm{\varnothing }}$. The equality criterion [F2] shows that these are distinct within each type, and different types give different cells. Since every subset of $S$ is spherical, these are all cells; their dimensions are $0,1,2$ by [F3],[F5]. This gives $6+3+3+1=13$ cells, agreeing with the coarser counts of [F13].

In the rank-two reflection plane, the reflecting lines bound a sector of angle $\pi \mathrm{/}3$: by [F7] their unit normals have inner product $-\cos (\pi \mathrm{/}3)=-1\mathrm{/}2$, so the normals meet at angle $2\pi \mathrm{/}3$ and their perpendicular lines meet at angle $\pi \mathrm{/}3$. By [F8], the action defining $C_S$ sends $s,t$ to these two reflections. The distances of $x_S$ from the two walls are $B(x_S,e_s)=d_s$ and $B(x_S,e_t)=d_t$ by [F3]. If $d_s=d_t$, then $x_S$ lies on the sector's internal angle bisector. Choose angular coordinates with the walls at $0$ and $\pi \mathrm{/}3$; then $x_S$ has angle $\theta =\pi \mathrm{/}6$. The reflections preserve $B$, and their product has determinant $1$ and trace $2\cos (2\pi \mathrm{/}3)=-1$ by [F7], so in this Euclidean plane it is a rotation by $2\pi \mathrm{/}3$ or $-2\pi \mathrm{/}3$. The orbit angles are therefore $\theta +2k\pi \mathrm{/}3$ and $-\theta +2k\pi \mathrm{/}3$ for $k=0,1,2$. These are the six equally spaced angles $\theta +j\pi \mathrm{/}3$, $0\le j<6$. Their convex hull is a regular hexagon, as also recorded in [F12].

The proper nonempty faces of $C_S$ are its six vertices and six edge cosets by [F3] and the lists in step 1.1. The boundary walk alternates the two labels and has vertices $1,s,st,sts,ts,t,1$: they are distinct up to the repeated endpoint by the six normal forms in step 1.1. Hence the proper cells form a $6$-cycle, which is the rank-two Coxeter complex described by [F11]. Since $S$ itself is spherical, [F1] makes $L$ the single $1$-simplex on $s,t$. At each hexagon vertex there are exactly two incident edges, one of each label, and the unique $2$-cell joins their directions; its local link is therefore one closed interval, the same $1$-simplex $L$.

The order complex of the four-element poset $\mathbb{S}$ has the two triangles $\mathrm{\varnothing }<\{s\}<S$ and $\mathrm{\varnothing }<\{t\}<S$, glued along $\mathrm{\varnothing }<S$, so its realization is the stated square chamber $K$. By [F4], the subposet below the maximal coset $W$ (which is all of $WS$) realizes the barycentric subdivision of $C_S$; consequently $\mathrm{\Sigma }$ is homeomorphic to the closed disk $C_S$. Its coarse cellulation has the six vertices, six edges, and one face from step 1.1, giving Euler characteristic $1$, in agreement with [F13]. Since $C_S$ is convex and contains $0$ by [F3], the homotopy $H(u,z)=(1-u)z$ contracts it to $0$. The quotient statement follows from [F6]. All group and coset lists are finite and explicit, so no Choice is used.

The diagram $\mathrm{\Gamma }$ has no edges by [F1], so its components are the singletons $\{a\},\{b\},\{c\}$. For each singleton $s$, [F2] gives the restricted presentation with only the relator $s^2=1$: every word reduces to $1$ or $s$, while the map sending $s$ to the nonidentity element of $\mathbb{Z}\mathrm{/}2$ respects the relator and separates them. Thus $W_{\{s\}}=\{1,s\}$ has order $2$. By [F3], multiplication is an isomorphism $W_{\{a\}}\times W_{\{b\}}\times W_{\{c\}}\to W$, so $W\cong (\mathbb{Z}\mathrm{/}2{)}^3$ and $\mathrm{\mid }W\mathrm{\mid }=8$. For $T=\mathrm{\varnothing }$, $W_T=\{1\}$; for $\mathrm{\mid }T\mathrm{\mid }=1$ the order-two calculation applies; and for $\mathrm{\mid }T\mathrm{\mid }\ge 2$ the restricted diagram has singleton components, so [F3] gives $W_T\cong (\mathbb{Z}\mathrm{/}2{)}^T$ and $\mathrm{\mid }{W}_T\mathrm{\mid }=2^{\mathrm{\mid }T\mathrm{\mid }}$. Hence every subset of $S$ is spherical and $\mathbb{S}$ is the full power set.

For $T=\mathrm{\varnothing }$, step 1.1 and the convention $⟨\mathrm{\varnothing }⟩=\{1\}$ give $W_T=\{1\}$, while $V_T=\{0\}$, $x_T=0$, and $C_T=\{0\}$, the product over the empty set. For every nonempty $T$, step 1.1 gives sphericality; fix such a $T$ and let $x_T={\sum }_{s\in T}{d}_s{v}_s^{(T)}$ be as in [F4]. By [F15] applied to $T$, the sum $V_T={⨁}_{s\in T}\mathbb{R}{e}_s$ is $B$-orthogonal, and by [F18] $B(e_s,e_s)=1$, so $B(v_s^{(T)},e_t)={\delta }_{st}$ forces $v_s^{(T)}=e_s$ and $x_T={\sum }_{s\in T}{d}_s{e}_s$. For $s\in T$ the reflection $\rho (s)=r_{e_s}$ fixes $e_t$ pointwise for $t\in T\setminus \{s\}$, because $B(e_t,e_s)=0$ and the reflection formula of [F5] reduces to $r_{e_s}(v)=v$ when $B(v,e_s)=0$, while $r_{e_s}(e_s)=-e_s$ since $B(e_s,e_s)=1$; hence $\rho (W_T)x_T$ is exactly the set of points ${\sum }_{s\in T}{\epsilon }_s{d}_s{e}_s$ with ${\epsilon }_s\in \{+1,-1\}$, the vertex set of the box of (i).

By step 1.1 every $T\subseteq S$ is spherical. The cells of the cellulation are the cosets $w{W}_T$, one for each element of $WS$, and have dimension $\mathrm{\mid }T\mathrm{\mid }$ by [F7,F8]. The cosets of each $W_T$ partition $W$ by [F6], and Lagrange's formula [F16] gives their number as $\mathrm{\mid }W\mathrm{\mid }\mathrm{/}\mathrm{\mid }{W}_T\mathrm{\mid }=8\mathrm{/}{2}^{\mathrm{\mid }T\mathrm{\mid }}$: there are $8$ cosets of $W_{\mathrm{\varnothing }}$ (the vertices), $3\cdot 4=12$ of the rank-one parabolics (the edges), $3\cdot 2=6$ of the rank-two parabolics (the rectangular $2$-cells, combinatorial squares) and one coset $W_S$ (the top $3$-cell), so there are $27$ cells in all. Since $W_S=W$ is the maximum coset, every cell is a face of the top cell, and $\mathrm{\Sigma }=\mathrm{\mid }WS\mathrm{\mid }=\mathrm{\mid }W{S}_{\le W_S}\mathrm{\mid }$ is the barycentric subdivision of $C_S$ by [F7]; the contraction $H(t,x)=(1-t)x$ of the convex box $C_S$ to $0$ transfers through this homeomorphism to a contraction of $\mathrm{\Sigma }$, while the alternating count of [F13] gives $\chi =8-12+6-1=1$.

Write $I_s:=[-d_s{e}_s,d_s{e}_s]$ and $A_s:=\{-d_s{e}_s,d_s{e}_s\}$. The convex hull of a product of finite sets is the product of the convex hulls: a convex combination ${\sum }_i{\lambda }_i(a_i^1,\dots ,a_i^k)$ of product points has coordinates ${\sum }_i{\lambda }_i{a}_i^s\in \mathrm{conv}(A_s)$, and conversely a tuple of convex combinations with coefficient vectors ${\lambda }^1,\dots ,{\lambda }^k$ is the convex combination of product points with weights ${\lambda }_{i_1}^1\cdots {\lambda }_{i_k}^k$; therefore $C_T=\mathrm{conv}(\rho (W_T)x_T)={\prod }_{s\in T}{I}_s$ by [step 2.1], a box of dimension $\mathrm{\mid }T\mathrm{\mid }$, agreeing with the product-of-intervals description [F14]. Its nonempty faces are products with a set $U\subseteq T$ of free coordinates and signs fixed on $T\setminus U$; the face indexed by $u{W}_U$ has the signs of $u$ on $T\setminus U$. For $u,v\in W_T$ these faces satisfy $F(u,U)\subseteq F(v,V)$ exactly when $U\subseteq V$ and $u,v$ have the same signs outside $V$, which, by [F17], is equivalent to $u{W}_U\subseteq v{W}_V$. Thus every nonempty box face is indexed by exactly one parabolic coset, and face-containment and coset-containment agree (so the reverse-inclusion nonempty face and coset posets agree as well); the additional empty face has no coset index.

By step 1.1 all proper subsets are spherical, and they index exactly the proper nonempty faces of the top cell $C_S$ by [F4]; by [F7] those faces are the boundary cells of $\mathrm{\Sigma }$, so they are the $8$ vertices, $12$ edges and $6$ rectangular $2$-faces of [step 2.2]. By step 3.1, $C_S={\prod }_{s\in S}[-d_s{e}_s,d_s{e}_s]$ in the orthonormal coordinates $e_a,e_b,e_c$; radial projection $x\mapsto x\mathrm{/}\mathrm{\parallel }x\mathrm{\parallel }$ sends $\mathrm{\partial }{C}_S$ to $S^2$ and has continuous inverse $u\mapsto u\mathrm{/}{\max }_{s\in S}(\mathrm{\mid }{u}_s\mathrm{\mid }\mathrm{/}{d}_s)$, so this boundary is a $2$-sphere; the alternating count gives $8-12+6=2$ by [F13].

The chamber $K=\mathrm{\mid }\mathbb{S}\mathrm{\mid }$ is the order complex of the full power set of $S$ by [F10] and [step 1.1]. Every chain extends to a maximal chain. Map each vertex $T$ to its characteristic vector ${\mathbf{1}}_T\in [0,1{]}^S$. For a permutation $(s_1,s_2,s_3)$, the chain $\mathrm{\varnothing }\subset \{s_1\}\subset \{s_1,s_2\}\subset S$ gives a tetrahedron with vertices $0,e_{s_1},e_{s_1}+e_{s_2},{\mathbf{1}}_S$. Every point $y\in [0,1{]}^S$ lies in one of these tetrahedra: choose an ordering $y_{s_1}\ge y_{s_2}\ge y_{s_3}$ and write it as the convex combination with coefficients $1-y_{s_1}$, $y_{s_1}-y_{s_2}$, $y_{s_2}-y_{s_3}$, and $y_{s_3}$ on those four vertices. The coefficients are nonnegative and sum to one; strict coordinate orders give disjoint tetrahedron interiors, while ties make the corresponding coefficient differences zero and put the point in a shared face. Hence these six characteristic-vector tetrahedra triangulate the cube and give a homeomorphism $K\cong [0,1{]}^S$; $K$ has six top simplices. By [F9] the quotient $W\mathrm{\backslash }\mathrm{\Sigma }$ is homeomorphic to $K$, and by [F11] the chambers are the eight distinct translates $wK$ [step 1.1]. By step 3.1, $C_S$ is a box; its barycentric subdivision has $8\cdot 3\cdot 2=48$ top simplices, by choosing a box vertex, an incident edge and an incident $2$-face; this agrees with the $8\cdot 6=48$ tetrahedra in the chamber translates.

Write an element of $W$ as its sign vector $({\epsilon }_a,{\epsilon }_b,{\epsilon }_c)\in \{\pm 1{\}}^3$ under the direct-product isomorphism of step 1.1. A Coxeter-complex vertex of type $S\setminus \{s\}$ is the coset $w{W}_{S\setminus \{s\}}$; it is determined exactly by the $s$-coordinate ${\epsilon }_s$, since that parabolic changes the other two coordinates freely. Thus the six Coxeter-complex vertices are the three opposite pairs $(s,+1),(s,-1)$ for $s\in S$. The chamber indexed by $w$ is the triangle with vertices $(a,{\epsilon }_a),(b,{\epsilon }_b),(c,{\epsilon }_c)$, so the eight chambers are precisely all choices of one vertex from each opposite pair. This is the octahedral triangulation: its vertices are the six signed coordinate directions and its triangles choose one from each opposite pair. By step 3.1 these labels are the coordinate faces of the box; a rectangular $2$-face with fixed $s$-coordinate ${\epsilon }_s$ corresponds to the Coxeter vertex $(s,{\epsilon }_s)$; a box vertex $({\epsilon }_a,{\epsilon }_b,{\epsilon }_c)$ corresponds to the Coxeter triangle with those three vertices; a box edge with two fixed coordinates corresponds to the Coxeter edge joining the two matching signed vertices; and their incidences are reversed. This proves that the cubical boundary and the Coxeter complex are dual cellulations of the same $2$-sphere, not the same cellulation.

The clauses are proved: (i) is [step 2.1] with [step 3.1], (ii) is [step 2.2], (iii) is [step 4.1] with [step 5.1], (iv) is [step 4.2], and (v) restates (i). No Choice is used: all groups, hulls and cell families here are finite, and the only identifications are the explicit ones of the cited clauses.

By [F6], $W=⟨s,t\mid s^2=t^2=(st{)}^4=1⟩$. Put $r=st$; then $r^4=1$, $srs=r^{-1}$, and $t=sr$, so $sr=r^{-1}s$. Moving every $s$ to the right and reducing powers of $r$ shows that every element is one of $r^k$ or $r^{k}s$ for $k=0,1,2,3$, hence $\mathrm{\mid }W\mathrm{\mid }\le 8$. To separate these eight forms, let $s(x,y)=(x,-y)$ and $t(x,y)=(y,x)$ on ${\mathbb{R}}^2$; both are reflections, $st$ is a quarter-turn, and the eight maps $(st{)}^k$ and $(st{)}^{k}s$ are distinct. By [F6] these assignments define a homomorphism from $W$ onto the eight-element symmetry group of the square, so $\mathrm{\mid }W\mathrm{\mid }=8$ and the forms are distinct. The images also show $s,t\mathrm{\ne }1$, hence $W_{\{s\}}=\{1,s\}$ and $W_{\{t\}}=\{1,t\}$; all four subsets are spherical by [F1]. By [F7], each singleton parabolic has four left cosets in $W$. There are eight singleton cosets and one top coset $W_S=W$, so [F1]–[F3] give $8+4+4+1=17$ cells.

By [F2] and [step 1.1], the top cell is a two-dimensional convex polytope with eight vertices and eight edges, so it is an octagon; [F4] makes it regular when $d_s=d_t$. At each vertex $\{w\}$, the incident rank-one cosets are exactly $w{W}_{\{s\}}$ and $w{W}_{\{t\}}$: both contain $w$, and any coset of either type containing $w$ equals the corresponding one by [F1]. Thus the edge labels alternate. All cosets lie below $W_S=W$, so [F3] identifies $\mathrm{\Sigma }$ with its barycentric subdivision, and [F4] gives a closed disk. The boundary consists of its eight vertices and eight edges. These proper cosets label the rank-two Coxeter complex by [F5]; both graphs are cycles with alternating singleton types, and exchanging their vertex and edge labels gives the dual circle identification.

The spherical-subset poset has exactly two maximal chains $\mathrm{\varnothing }<\{s\}<S$ and $\mathrm{\varnothing }<\{t\}<S$; their triangles meet in the diagonal $\mathrm{\varnothing }<S$, producing $K$. By [F3] it is the compact quotient. Every maximal coset chain chooses one of eight vertices and one of its two incident edges before the top cell, giving $16$ triangles. Each vertex $w$ gives the two chains $\{w\}<w{W}_{\{s\}}<W$ and $\{w\}<w{W}_{\{t\}}<W$, precisely the translate $wK$.

Each edge of label $s$ or $t$ is isometric to its rank-one cell by [F2], and has length $2d_s$ or $2d_t$ by [F4]. Labels alternate around the octagon, so unequal distances give unequal side lengths and exclude regularity. For every positive distance family [F2] gives the same coset faces and [F4] gives the disk topology. All calculations and constructions are finite, so no Choice is used.

Let $R$ be the finite words in $S$ with no equal adjacent letters, including the empty word. Define ${\lambda }_s$ on $R$ by deleting the initial $s$ if present and otherwise prefixing $s$. This is an involution, so [F2] gives an action of $W$ on $R$. Cancellation of adjacent equal letters reduces every word to a member of $R$. With composition right-to-left, a reduced word $s_1\cdots s_k$ sends the empty word to exactly that word; distinct reduced words therefore represent distinct elements of $W$. In particular each $s$ has order two, and alternating words in distinct $s,t$ give infinitely many elements of $⟨s,t⟩$. Any type of size at least two is consequently infinite, while the empty and singleton types are finite, proving (i) by [F1]. For factor homomorphisms into an arbitrary group, the images of the generators satisfy precisely $s^2=1$, so [F2] extends them uniquely to $W$. This is the universal property [F5], proving the free-product assertion.

By (i) and [F3], only the Cayley graph occurs. It is connected because $S$ generates $W$. A closed nonbacktracking edge walk would give a nonempty reduced word representing $1$, contradicting step 1.1. Thus it is a tree. The neighbors $ws$ are distinct for distinct $s$ by the same normal form, so its valence is $\mathrm{\mid }S\mathrm{\mid }$. For two generators the unique reduced words alternate and give the bi-infinite line; three give valence three. The zero-generator group is trivial and gives a point; one generator gives two vertices and one interval. Lengths follow from [F3]. Each vertex has one incident edge of each label and no higher cells, giving the discrete link $S$.

The poset of spherical types has a least element $\mathrm{\varnothing }$ and the incomparable singleton types. Its realization $K$ is therefore the asserted fan, and [F4] supplies its compact quotient. The barycentric subdivision of the Cayley tree has vertices $w$ and midpoint vertices $w⟨s⟩$, and one half-edge $(w,s)$ joining each such pair. Left multiplication acts on all labels. A central vertex has trivial stabilizer. A half-edge has one central and one midpoint endpoint, so an element preserving it must fix its central endpoint and is therefore trivial; a midpoint vertex has stabilizer $w⟨s⟩w^{-1}$ by coset equality. For the graph of groups on this fan, put the trivial group at the center and every edge, and $⟨s⟩$ at leaf $s$, with the unique injections from the trivial edge groups. The fan itself is a maximal tree; [F6] kills all edge symbols, and trivial edge groups add no conjugation relations, leaving exactly the presentation in [F2]. The Bass–Serre coset incidence of [F6] now joins $w$ to $w⟨s⟩$, exactly the subdivision just described. This proves the full graph-of-groups claim without an ordinary covering assertion.

Root the metric tree at $1$. Each point $x$ has a unique finite arc $[1,x]$: connectivity supplies a finite edge route to a cell containing $x$, and the absence of cycles makes its reduced route unique. Define $H(t,x)$ to be the point at distance $(1-t)d(1,x)$ along that arc. To check continuity, for $x,y$ let their root arcs share the initial segment of length $c$, and write $a=d(1,x)$, $b=d(1,y)$. If both contracted points lie beyond that common segment, their distance is $(1-t)(a+b)-2c\le d(x,y)$; if both are on the common segment it is $(1-t)\mathrm{\mid }a-b\mathrm{\mid }\le d(x,y)$; if just one is beyond it, it is $(1-t)\mathrm{\mid }a-b\mathrm{\mid }\le d(x,y)$. Thus $d(H(t,x),H(t,y))\le d(x,y)$, and $d(H(t,y),H(u,y))=\mathrm{\mid }t-u\mathrm{\mid }d(1,y)$. The triangle inequality gives joint continuity, including at the root. By [F4] this is continuity in the Davis topology. Since $H(0,x)=x$, $H(1,x)=1$, and $H(t,1)=1$, the tree is contractible. All normal forms and routes are explicit finite constructions, so no Choice is used.