Davis CAT(0) Geometry and Finite Subgroup Fixed Points — Examples

By [F2,F3], $st$ has infinite order, so $W_{\{s,t\}}=W$ is infinite, whereas $W_{\mathrm{\varnothing }}$, $W_{\{s\}}$ and $W_{\{t\}}$ are finite. Thus the spherical subsets are exactly $\mathrm{\varnothing },\{s\},\{t\}$, and the Davis cells are vertices and edges only. By [F6,F7] the $1$-skeleton is $G=\mathrm{Cay}(W,\{s,t\})$.

Put $p:=st$ and, for $n\in \mathbb{Z}$, set $v_{2n}:=p^n$ and $v_{2n+1}:=p^{n}s$. The reduced-word normal form shows these are all distinct and exhaust $W$: the empty word and the even-length reduced words are $p^n$ for unique $n\in \mathbb{Z}$, while every odd-length reduced word is $p^{n}s$ for a unique $n\in \mathbb{Z}$. Consecutive vertices $v_{2n},v_{2n+1}$ and $v_{2n+1},v_{2n+2}$ differ by right multiplication by $s$ and $t$, respectively. Since $s$ and $t$ are distinct reduced one-syllable words, each vertex has exactly these two distinct neighbors, and this enumeration identifies $G$ with the bi-infinite line.

Each edge has length $2d_s=2d_t=1$ by [F8, F9]. Send $v_k$ to $k$ and extend linearly over each edge. Every cell is a point or one of these intervals by [F5], so this map is isometric on each cell. For any chain from $x$ to $y$, the sum of its cellwise lengths is at least the absolute difference of the endpoint coordinates; conversely, the finite line segment between them is a chain with exactly that length. Thus by [F10] the chain metric candidate is the usual real-line metric under this map, so it is a metric and gives an isometry $\mathrm{\Sigma }\to \mathbb{R}$. It follows from [F12, F13] that $\mathrm{\Sigma }$ is complete and CAT(0). Left multiplication by any $g\in W$ sends each vertex $w$ to $gw$ and each edge $\{w,ws\}$ or $\{w,wt\}$ to the corresponding edge at $gw$; its length-preserving extension is a bijective isometry for this metric by [F14].

Let $Y\subseteq \mathrm{\Sigma }\cong \mathbb{R}$ be nonempty and finite. By [F11], the set of line coordinates of $Y$ has a minimum $a$ and maximum $b$. Put $D:=b-a$; since all coordinates lie between $a$ and $b$ and both endpoints belong to $Y$, this is the diameter of $Y$. Let $m$ be the point with coordinate $(a+b)\mathrm{/}2$. Every $y\in Y$ has coordinate between $a$ and $b$, so $d(y,m)\le D\mathrm{/}2$; the endpoints $a,b$ each lie at distance $D\mathrm{/}2$, hence $r_Y(m)=D\mathrm{/}2$.

For any $x\in \mathrm{\Sigma }\cong \mathbb{R}$, $r_Y(x)\ge \max \{\mathrm{\mid }x-a\mathrm{\mid },\mathrm{\mid }x-b\mathrm{\mid }\}$, so the inequalities $\mathrm{\mid }x-a\mathrm{\mid }\le r_Y(x)$ and $\mathrm{\mid }x-b\mathrm{\mid }\le r_Y(x)$ imply $2r_Y(x)\ge \mathrm{\mid }x-a\mathrm{\mid }+\mathrm{\mid }x-b\mathrm{\mid }\ge b-a=D$. If $r_Y(x)=D\mathrm{/}2$, then both $\mathrm{\mid }x-a\mathrm{\mid }$ and $\mathrm{\mid }x-b\mathrm{\mid }$ are at most $D\mathrm{/}2$, whose two closed intervals intersect only at $m=(a+b)\mathrm{/}2$ (also when $D=0$). Hence $m$ is the unique minimizer and the unique center of $Y$.

Write $H\le W$ for a finite subgroup. The normal-form indexing in step 1.2 says every element is either $p^n$ or $p^{n}s$. If $n\mathrm{\ne }0$, $p^n$ has infinite order; each $p^{n}s$ is an involution because $s{p}^{n}s=p^{-n}$. Two distinct involutions $p^{m}s$ and $p^{n}s$ have product $p^{m-n}$ with $m\mathrm{\ne }n$, which has infinite order. Consequently a finite subgroup is either $\{1\}$ or $\{1,r\}$ for one reflection $r=p^{n}s$. If $H=\{1\}$, the orbit $H{x}_0=\{x_0\}$ has center $x_0$. If $H=\{1,r\}$ and $r{x}_0=x_0$, its orbit again has center $x_0$; otherwise the orbit is $\{x_0,r{x}_0\}$ and step 2.1 gives its unique center as their midpoint. Since $r$ is an isometry interchanging these endpoints, it fixes that midpoint. Let $f$ be an isometry of $\mathbb{R}$, put $c=f(0)$ and write $f(1)=c+ϵ$ with $ϵ\in \{1,-1\}$. For $y=f(x)-c$, the two distance equalities give $\mathrm{\mid }y\mathrm{\mid }=\mathrm{\mid }x\mathrm{\mid }$ and $\mathrm{\mid }y-ϵ\mathrm{\mid }=\mathrm{\mid }x-1\mathrm{\mid }$; subtracting their squares gives $y=x$ if $ϵ=1$ and $y=-x$ if $ϵ=-1$. Thus every isometry is a translation $x\mapsto x+c$ or a reflection $x\mapsto -x+c$. An involutive translation is the identity, while a reflection has the unique fixed point $c\mathrm{/}2$; the left action is faithful on vertices, so nonidentity $r$ is not the identity isometry and hence has a unique fixed point. Thus in every case the orbit center is fixed by $H$.

Under AC [F15], [F16] applies to $X=\mathrm{\Sigma }$ and $Y=H{x}_0$: step 1.3 gives completeness, CAT(0), and an isometric action; the orbit is nonempty and finite, hence bounded. The general theorem's center is the unique minimizer of the same radius function used in steps 1.4 and 2.1, so it equals the explicitly computed orbit center. This is precisely the companion A-page center map restricted to this line. The local orbit classification and fixed-point calculation in step 3.1 do not use AC; AC enters here only through the general theorem's minimizing-sequence argument.

Clause (i). Under AC [F14], invoke clauses (2)-(4) of the A-page link lemma [F3,F15,F16] for the vertex links and their local metrics. For $S=\{s,t\}$ with $m(s,t)=3$, [F4] gives $B_S=\left(\begin{array}{cc}1& -1\mathrm{/}2\\ -1\mathrm{/}2& 1\end{array}\right)$, whose eigenvalues are $1\mathrm{/}2$ and $3\mathrm{/}2$, so it is positive definite. The assignments $s\mapsto (12)$ and $t\mapsto (23)$ satisfy the presentation relators, so [F1] gives a homomorphism $W_S\to S_3$; it is onto. Put $a=st$. The presentation gives $a^3=1$, $t=sa$, and $sas=a^{-1}$, so every word reduces to $a^i$ or $a^{i}s$ for $i=0,1,2$. Hence $\mathrm{\mid }{W}_S\mathrm{\mid }\le 6$; the surjection gives $\mathrm{\mid }{W}_S\mathrm{\mid }\ge 6$, so $W_S\cong S_3$, the dihedral group of order six. Every subset is spherical [F2]; $L$ is one edge and [F3,F15] give edge length $2\pi \mathrm{/}3$ with cosine matrix $B_S$. The empty cosine matrix is positive definite by [F17], and every nonempty subset has a positive-definite cosine matrix as a principal submatrix of $B_S$; since every subset is spherical, the metric-flag equivalence holds [F10]. When $d_s=d_t$, [F8] gives the regular hexagonal cell.

Clause (ii). Let $S=\{s,t,u\}$ and $m(s,s)=1$ and $m(s,t)=3$ for distinct generators. By [F4], $B_S=\frac{3}{2}I-\frac{1}{2}J$. For $x=(x_1,x_2,x_3)$, $2x^T{B}_{S}x=3{\sum }_i{x}_i^2-({\sum }_i{x}_i{)}^2=(x_1-x_2{)}^2+(x_2-x_3{)}^2+(x_3-x_1{)}^2,$ so $B_S$ is positive semidefinite with kernel $\mathbb{R}(1,1,1)$ and is not positive definite. Hence $W_S$ is infinite by [F6]. Each pair has matrix $\left(\begin{array}{cc}1& -1\mathrm{/}2\\ -1\mathrm{/}2& 1\end{array}\right)$ with eigenvalues $1\mathrm{/}2$ and $3\mathrm{/}2$, so it is positive definite by [F5] and its parabolic is finite by [F6,F13]. The nerve therefore has all three edges but no 2-simplex [F2]. By [F3,F15], $X$ is the cycle of three edges of length $2\pi \mathrm{/}3$, hence the round circle of circumference $2\pi$. Its pairwise adjacent triple has cosine matrix $B_S$, which is not positive definite, so the metric flag test leaves it unfilled [F10]. The empty matrix is positive definite by [F17], each singleton has matrix $[1]$, and every pair is an edge with positive-definite matrix by the preceding calculation. Thus the only pairwise adjacent set failing to span a simplex is the triple, whose matrix is not positive definite, so both directions of the metric-flag condition hold. The three incident rank-two cells contribute local link angles $2\pi \mathrm{/}3$ each by [F15], for total angle $2\pi$; when $d_s=d_t=d_u$, they are regular hexagons by [F8].

Clause (iii). Let every distinct pair in the finite set $S$ have label $\mathrm{\infty }$. For each distinct $s,t$, [F4,F5] give the pair matrix $\left(\begin{array}{cc}1& -1\\ -1& 1\end{array}\right)$, which is positive semidefinite with radical $\mathbb{R}(e_s+e_t)$ and is not positive definite; hence $W_{\{s,t\}}$ is infinite by [F6,F13]. No pair is spherical, so [F2] makes $L$ discrete; [F7] gives Davis cells of dimension at most one and [F3] identifies the link with the points of $S$ at pairwise truncated distance $\pi$. By [F9], points of radii $r,q$ on the same ray have distance $\mathrm{\mid }r-q\mathrm{\mid }$, and points on distinct rays have distance $r+q$; if either radius is zero, the cone-apex formula gives the same result. Thus the cone is the metric star of $\mathrm{\mid }S\mathrm{\mid }$ rays: a point for $S=\mathrm{\varnothing }$, a ray for $\mathrm{\mid }S\mathrm{\mid }=1$, and for $\mathrm{\mid }S\mathrm{\mid }=2$ an isometric copy of $\mathbb{R}$ by sending the two rays to opposite half-lines. The almost-negative matrix has off-diagonal entries $-1$; for $\mathrm{\mid }S\mathrm{\mid }=2$ its quadratic form is $(x_1-x_2{)}^2$ and its kernel is $\mathbb{R}(1,1)$. The metric flag test holds: a pairwise adjacent set has at most one vertex, the empty matrix is positive definite by [F17], and a singleton has matrix $[1]$ [F10].

Clause (i), CAT(1). Step 1.1 identifies the link with an interval of length $L=2\pi \mathrm{/}3$. For any three points ordered along it, the side lengths are $a,b,a+b$ with $a+b\le L$, so the perimeter is $2(a+b)\le 4\pi \mathrm{/}3<2\pi$. The spherical comparison triangle is the same degenerate great-circle segment, since the longest side is the sum of the other two and is less than $\pi$; corresponding side-point distances therefore agree. Thus the interval satisfies the CAT(1) comparison. Its closed midpoint ball of radius $\pi \mathrm{/}3$ is the whole interval and is convex.

Clause (ii), CAT(1). Step 1.2 identifies the link with the round circle of circumference $2\pi$. By [F12], this is the equality case of the criterion that $S_{\mathrm{\ell }}^1$ is CAT(1) exactly when $\mathrm{\ell }\ge 2\pi$. The three incident rank-two cells contribute the local angles $2\pi \mathrm{/}3$ each by [F15], for total angle $2\pi$; when $d_s=d_t=d_u$, they are regular hexagons by [F8].

Clause (iii), CAT(1). Step 2.1 gives a discrete link with distinct points at distance $\pi$. Every pair at distance $<\pi$ is identical and has the constant geodesic; a triangle with any two distinct vertices has perimeter at least $2\pi$, so the only tested triangles are constant and satisfy comparison with equality.

Clause (iv) and conclusion. By [F15], each edge has length $\pi -\pi \mathrm{/}m(s,t)$ and cosine $B(e_s,e_t)$, while each non-edge has $B(e_s,e_t)=-1$. Thus the edge length is $\pi -\pi \mathrm{/}m(s,t)$; the two formulas coincide for $m(s,t)=2$ and differ for $m(s,t)>2$. In the affine case the pairwise adjacent triple is unfilled precisely because its cosine matrix is semidefinite, not positive definite [F10, step 1.2]. The A-page link description records $X$ as finite large metric flag with associated matrix $B$ [F16]; the three metric-flag tests here are checked directly in steps 1.1-2.1. The local group, matrix, cone and CAT(1) calculations use no choice; AC is used only for the A-page link-lemma invocation in step 1.1 [F14]. No general 3-circuit classification or CAT(1) theorem for all large metric flag complexes is asserted.

Clause (i). By [F1] every element has a unique reduced syllable word. A word of length $1$ is a generator and is an involution. If an odd reduced word $w=g_1\cdots g_k$ has $k\ge 3$, then $g_k=g_1$ and $g_{1}w{g}_1=g_2\cdots g_{k-1}$ is a shorter odd reduced word. Induction shows that $w$ is conjugate to $s$ or $t$, so has order $2$. If $k=0$, $w=1$. If $k=2j>0$, the word is $(st{)}^j$ or $(ts{)}^j=t(st{)}^{j}t$; [F1] says $st$ has infinite order, so every such word has infinite order. Thus every element is the identity, a reflection, or an infinite-order translation. If ${\rho }_1\mathrm{\ne }{\rho }_2$ are reflections, then ${\rho }_1{\rho }_2\mathrm{\ne }1$ and its reduced syllable word is obtained from a concatenation of two odd-length words by cancelling equal adjacent pairs. The resulting word is nonempty and has positive even length, so it is a translation of infinite order by the preceding classification. A finite subgroup therefore contains no translation and at most one reflection, so it is trivial or has order two generated by a reflection.

The line structure. By [F2] the only spherical subsets are $\mathrm{\varnothing },\{s\},\{t\}$, so there are only vertices and edges, and by [F3] the $1$-skeleton is $G=\mathrm{Cay}(W,\{s,t\})$. Let $p:=st$ and, for $n\in \mathbb{Z}$, set $v_{2n}:=p^n$ and $v_{2n+1}:=p^{n}s$. The normal forms of [F1] show these vertices are distinct and exhaust $W$: even reduced words are $p^n$; odd words starting with $s$ are $p^{n}s$ for $n\ge 0$, and odd words starting with $t$ are $p^{n}s=(ts{)}^{-n-1}t$ for $n<0$. Consecutive vertices differ by right multiplication by $s$ and $t$, respectively, so this indexes the Cayley graph as a bi-infinite line. Send $v_k$ to $k$ and extend linearly over each edge. Each edge has length $1$ by [F3]. For any cell chain, the sum of its cellwise lengths is at least the absolute difference of the endpoint coordinates by the triangle inequality on $\mathbb{R}$; conversely, the finite line segment between two points is a finite chain of edge subsegments with length equal to that coordinate difference. Thus [F8] gives the chain metric as the usual metric on $\mathbb{R}$, and $\mathrm{\Sigma }$ is a metric line. Every isometry of this line is either a translation or a reflection: after writing $f(0)=c$, one has $f(1)=c+1$ or $c-1$, and the distances to these two points determine $f(x)$ uniquely as $c+x$ or $c-x$. Therefore a nonidentity involution is a reflection with exactly one fixed point. The action on vertices is free by [F4].

Clause (ii). Let $w\in W$. By [F2] the vertex $w$ is the $0$-cell $w{W}_{\mathrm{\varnothing }}$, whose relative interior is $\{w\}$; by [F4] applied with $q=w{W}_{\mathrm{\varnothing }}$ and $T=\mathrm{\varnothing }$, ${\mathrm{Stab}}_W(w)=\{1\}$. A finite subgroup fixing a vertex $v$ lies in ${\mathrm{Stab}}_W(v)=\{1\}$, so it is trivial.

Clause (iv). Assume now that $(W,S)$ is finite with $S=\{s,t\}$ and $m(s,t)=3$. Then $B$ is positive definite by [F6], $S$ is spherical and $W=W_S$, so every spherical coset is contained in $W_S=W$ and indexes a face of the single top cell $C_S$ by The cellulation of the Davis complex: incidence, stabilizers and the model U(W,K) (1),(2). Hence $\mathrm{\Sigma }=C_S$; the universal-system description in [F2] is not used in this finite case. The point $0$ lies in the interior of $C_S$ by [F6], so in the top-cell chart it has $w_0=1$ and $I=S$; the point-stabilizer formula of [F4] gives ${\mathrm{Stab}}_W(0)=W_S=W$. Here $W$ is dihedral of order $2m=6$ by [F6], and $W=W_S$ is a spherical parabolic of full rank.

Clause (iii). Let $w$ be a nonidentity reflection. By step 1.1 it is an involution; by step 1.2 its action on the line is a reflection with a unique fixed point $m$. By step 1.3 no nontrivial element fixes a vertex, so $m$ lies in the interior of an edge. The edge enumeration in step 1.2 writes its carrier as $q=p^k{W}_r$ for some $k\in \mathbb{Z}$, $r\in \{s,t\}$, with endpoints $a=p^k$ and $b=p^{k}r$. The reflection swaps these endpoints, so $wa=b$ and $w=b{a}^{-1}=p^{k}r{p}^{-k}$. In the $\dot{q}$-chart the midpoint has coordinate $0\in C_{\{r\}}$, in the chamber face $w_0=1$, $I=\{r\}$. The formula of [F4] gives ${\mathrm{Stab}}_W(m)=(\dot{q})W_r(\dot{q}{)}^{-1}$. Since $\dot{q}\in p^k{W}_r$, this is $p^k{W}_r{p}^{-k}=\{1,w\}=⟨w⟩$; it is the setwise stabilizer of the carrier edge and a conjugate of the rank-one spherical parabolic $W_r$ by [F5]. Under AC, this is the equality case of the general point-stabilizer formula [F7]; the explicit computation here does not rely on that theorem.

Clause (v) and the Choice bookkeeping. Steps 1.1-1.4 and 2.1 compute the two examples: the finite subgroups of the universal rank-two group are trivial or generated by one reflection, each nontrivial finite subgroup fixing the midpoint of a $1$-cell with stabilizer the conjugate spherical parabolic $p^k{W}_r{p}^{-k}$; in the finite rank-two case the whole group fixes the interior point $0$ of the top cell, with stabilizer $W_S=W$. These are the rank-one and top-rank extremes of the containment statement in [F7]. The normal-form, line, midpoint and cell calculations are choice-free. AC is assumed only for the comparison with [F7], and no conclusion of that general theorem is used in the explicit proof. No assertion is made about infinite subgroups, higher-rank trees or the number of conjugacy classes of finite subgroups in general.