Convex geodesic bicombings and hyperbolicity

Abstract

A geodesic bicombing on a metric space selects for every pair of points a geodesic connecting them. We prove existence and uniqueness results for geodesic bicombings satisfying different convexity conditions. In combination with recent work by the second author on injective hulls, this shows that every word hyperbolic group acts geometrically on a proper, finite dimensional space $X$ with a unique (hence equivariant) convex geodesic bicombing of the strongest type. Furthermore, the Gromov boundary of $X$ is a $Z$-set in the closure of $X$, and the latter is a metrizable absolute retract, in analogy with the Bestvina–Mess theorem on the Rips complex.

Appendix: Proof of Theorem 4.1

We use the notation from Sect. 4.

First we show the “if” direction. Let $Y \subset X$ be a finite set. If $|Y| \le 2n + 1$, the dimension of $E(Y)$ is at most $n$. Now suppose that $|Y| \ge 2n + 2$, $f \in E(Y)$, and $Z \subset Y$ is a set with $|Z| = 2n + 2$ and with a fixed point free involution $i :Z \rightarrow Z$ such that $\{z,i(z)\} \in A(f)$ for every $z \in Z$. By assumption, there exists a fixed point free bijection $j :Z \rightarrow Z$ such that $j \ne i$ and

$$\begin{aligned} \sum _{z \in Z} d(z,i(z)) \le \sum _{z \in Z} d(z,j(z)). \end{aligned}$$

Since $f(z) + f(i(z)) = d(z,i(z))$ and $d(z,j(z)) \le f(z) + f(j(z))$, this gives

$$\begin{aligned} \sum _{z \in Z} f(z) + f(i(z)) \le \sum _{z \in Z} f(z) + f(j(z)). \end{aligned}$$

However, since both $i$ and $j$ are bijections, these two sums agree, so each of the inequalities $d(z,j(z)) \le f(z) + f(j(z))$ must in fact be an equality, that is, $\{z,j(z)\} \in A(f)$. There is at least one $z \in Z$ such that $j(z) \ne i(z)$, so the graph with vertex set $Z$ and edge set $\bigcup _{z \in Z}\{\{z,i(z)\},\{z,j(z)\}\}$ has at most $n$ connected components. As this holds for every $Z$ and $i$ as above, we conclude that also the graph $(Y,A(f))$ has no more than $n$ components. Since $f \in E(Y)$ was arbitrary, this shows that the dimension of $E(Y)$ is less than or equal to $n$.

Now we prove the other implication. Suppose that $Z \subset X$ is a set with $|Z| = 2n+2$, and $i :Z \rightarrow Z$ is a fixed point free involution. Let $Z_2$ denote the set of all subsets of cardinality two of $Z$. The involution $i$ selects a subset $Z_i {:=} \{\{z,i(z)\} : z \in Z\}$ of $Z_2$ of $n+1$ disjoint pairs. For every function $h :Z \rightarrow \mathbb {R}$ we consider the set $W(h)$ of all $w :Z_2 \rightarrow \mathbb {R}$ such that $w \le 0$ on $Z_i$, $w \ge 0$ on $Z_2 \setminus Z_i$, and

$$\begin{aligned} \sum _{z' \in Z \setminus \{z\}} w(\{z,z'\}) = h(z) \end{aligned}$$

for all $z \in Z$. First we observe that if $z \in Z$, and if $\{x,y\} \in Z_i$ is chosen such that $z \not \in \{x,y\}$, then the function $w_{z,\{x,y\}} {:=} (\delta _{\{x,z\}} + \delta _{\{y,z\}} - \delta _{\{x,y\}})/2$ on $Z_2$ belongs to $W(\delta _z)$, and $w_{i(z),\{x,y\}} - \delta _{\{z,i(z)\}}$ is in $W(-\delta _z)$. It follows that $W(h)$ is non-empty for every $h :Z \rightarrow \mathbb {R}$. Put

$$\begin{aligned} \mu (h) {:=} \sup \{ S(w) : w \in W(h) \}, \quad S(w) {:=} \sum _{\{x,y\} \in Z_2} w(\{x,y\}) d(x,y); \end{aligned}$$

note that $\mu (h) > -\infty$ but possibly $\mu (h) = \infty$. For $w_{z,\{x,y\}}$ as above we have $S(w_{z,\{x,y\}}) \ge 0$ by the triangle inequality, thus $\mu (\delta _z) \ge 0$ for all $z \in Z$. Furthermore, since $0 \in W(0)$, also $\mu (0) \ge 0$. In fact, if $w \in W(0)$, then $\lambda w \in W(0)$ for all $\lambda \ge 0$, so we have either $\mu (0) = 0$ or $\mu (0) = \infty$.

In case $\mu (0) = \infty$, choose $w \in W(0)$ with $S(w) > 0$ such that no $w' \in W(0)$ with $S(w') > 0$ has strictly smaller support. It follows that every nonzero $v \in W(0)$ with support $\mathrm{spt }(v) \subset \mathrm{spt }(w)$ satisfies $S(v) > 0$, for if $\lambda > 0$ is the maximal number with the property that $|\lambda v| \le |w|$, then $v' {:=} w - \lambda v$ belongs to $W(0)$ and $\mathrm{spt }(v')$ is a strict subset of $\mathrm{spt }(w)$, so $S(v') \le 0$ and $\lambda S(v) = S(w) - S(v') > 0$. (In fact $\lambda v = w$, because $v' \ne 0$ would likewise imply $S(v') > 0$.) Since for fixed $z$ the sum of the weights $w(\{z,z'\})$ is zero, it is not difficult to see that there exist pairwise distinct points $z_0,z_1,\dots ,z_l$, with $l \ge 1$, such that $w(\{z_k,i(z_k)\}) < 0 < w(\{i(z_k),z_{k+1}\})$ for $k = 0,\dots ,l$, where $z_{l+1} {:=} z_0$. Then the function

$$\begin{aligned} v {:=} \sum _{k = 0}^l - \delta _{\{z_k,i(z_k)\}} + \delta _{\{i(z_k),z_{k+1}\}} \end{aligned}$$

on $Z_2$ belongs to $W(0)$, and $\mathrm{spt }(v) \subset \mathrm{spt }(w)$. Hence $S(v) > 0$. This means that

$$\begin{aligned} \sum _{k = 0}^l d(z_k,i(z_k)) < \sum _{k = 0}^l d(i(z_k),z_{k+1}). \end{aligned}$$

The points $i(z_0),\dots ,i(z_l)$ are distinct, so there is a well-defined map $j :Z \rightarrow Z$ such that $j(i(z_k)) = z_{k+1}$ for $k = 0,\dots ,l$ and $j(z) = i(z)$ otherwise. Note that $j$ is fixed point free since $\{i(z_k),z_{k+1}\} \in Z_2$ and $i(z) \ne z$ for all $z \in Z$. Furthermore, $j$ is injective because $i$ is injective, $z_0,\dots ,z_l$ are distinct, and $j(z) = i(z) = z_k$ would imply $j(z) = j(i(z_k)) = z_{k+1} \ne z_k$. Now it is clear that (4.1) holds, even with strict inequality. Note that this part of the proof does not use the assumption $\dim _\mathrm{comb}(X) \le n$.

It remains to consider the case $\mu (0) = 0$. Let $h,h' \in \mathbb {R}^Z$. For all $w \in W(h)$ and $w' \in W(h')$ we have $w + w' \in W(h + h')$, therefore

$$\begin{aligned} \mu (h) + \mu (h') \le \mu (h + h'); \end{aligned}$$

in particular $\mu (h) \le \mu (0) - \mu (-h) = -\mu (-h) < \infty$. Put

$$\begin{aligned} \nu (h) {:=} \frac{\mu (h) - \mu (-h)}{2}. \end{aligned}$$

We have $\mu (h) - \mu (-h') \ge \mu (h+h')$ and $\mu (h') - \mu (-h) \ge \mu (h+h')$, so

$$\begin{aligned} \nu (h) + \nu (h') \ge \mu (h+h'). \end{aligned}$$

We have already observed that $\mu (\delta _z) \ge 0$ for all $z \in Z$, hence $-\mu (-\delta _z) \ge 0$ and $\nu (\delta _z) \ge 0$. If $\{x,y\} \in Z_2 \setminus Z_i$, then $\delta _{\{x,y\}} \in W(\delta _x + \delta _y)$, thus

$$\begin{aligned} \nu (\delta _x) + \nu (\delta _y) \ge \mu (\delta _x + \delta _y) \ge d(x,y). \end{aligned}$$

Similarly, if $\{x,y\} \in Z_i$, then $-\delta _{\{x,y\}} \in W(-\delta _x-\delta _y)$, hence

$$\begin{aligned} \nu (-\delta _x) + \nu (-\delta _y) \ge \mu (-\delta _x - \delta _y) \ge -d(x,y) \end{aligned}$$

and so $\nu (\delta _x) + \nu (\delta _y) = -\nu (-\delta _x) - \nu (-\delta _y) \le d(x,y)$. Now it is clear that there exists a function $f \in \mathbb {R}^Z$ such that $f(z) \ge \nu (\delta _z)$ for all $z \in Z$ and $f(x) + f(y) = d(x,y)$ for all $\{x,y\} \in Z_i$. Hence, $f \in \varDelta (Z)$ and $Z_i \subset A(f)$, so in fact $f \in E(Z)$. Since $\dim _\mathrm{comb}(X) \le n$ by assumption, we have $\mathrm{rk }(A(f)) \le n$, and there is no loss of generality in assuming that there is no $g \in E(Z)$ with $Z_i \subset A(g)$ and $\mathrm{rk }(A(g)) > \mathrm{rk }(A(f))$. Since $Z$ has at most $n$ even $A(f)$-components, some such component $Z'$ contains more than one of the $n+1$ edges in $Z_i$. Note that $i$ maps $Z'$ onto $Z'$. We claim that for every $z' \in Z'$ there is a $z'' \in Z' \setminus \{z',i(z')\}$ such that $\{z',z''\} \in A(f)$. If, to the contrary, there were a $z' \in Z'$ with no such $z''$, we could choose $g \in \mathbb {R}^Z$ such that $f(z') > g(z') > d(z',z) - f(z)$ and $g(z) = f(z)$ for all $z \in Z \setminus \{z',i(z')\}$, and $g(i(z')) = d(z',i(z')) - g(z')$. This function would satisfy $\mathrm{rk }(A(g)) > \mathrm{rk }(A(f))$ and $Z_i \subset A(g)$, in contradiction to the assumption on $f$. It follows easily that there exist $z_0,z_1,\dots ,z_l \in Z'$, with $l \ge 1$, such that $\{z_0,i(z_0)\},\dots ,\{z_l,i(z_l)\}$ are pairwise disjoint and $\{i(z_k),z_{k+1}\} \in A(f)$ for $k = 0,\dots ,l$, where $z_{l+1} = z_0$. Now

$$\begin{aligned} \sum _{k=0}^l d(z_k,i(z_k)) = \sum _{k=0}^l f(z_k) + f(i(z_k)) = \sum _{k=0}^l d(i(z_k),z_{k+1}). \end{aligned}$$

Putting $j(i(z_k)) = z_{k+1}$ and $j(z_{k+1}) = i(z_k)$ for $k = 0,\dots ,l$ and $j(z) = i(z)$ otherwise, we get a fixed point free involution $j :Z \rightarrow Z$ distinct from $i$ such that (4.1) holds with equality.