# Embedding coboundedly into ultracomplete hyperbolic space

A Gromov-hyperbolic space $$X$$ is called ultracomplete if any two distinct points $$x,y\in\overline{X}$$ in the Gromov bordification can be connected by a geodesic segment, ray or line (depending on whether the end points are on the boundary or not). A segment, ray or line is geodesic if it is isometric to a closed interval in $$\mathbb{R}$$.

In $$\S$$7.5 from Hyperbolic groups, Gromov states without proof that "by a trivial argument", one has the following embedding:

Every geodesic hyperbolic $$X$$ isometrically embeds into some ultracomplete space $$Y$$, such that $$\mathrm{dist}_Y(y,X)\leq C<\infty$$ for a constant $$C$$ and all $$y\in Y$$. In particular, $$\partial Y=\partial X$$. Furthermore, one can choose a $$Y$$, such that every isometry of $$X$$ extends to a unique isometry of $$Y$$.

I understand this statement by two parts: first part is that one can "add" something, not much, to $$X$$ to make it ultracomplete while preserving the hyperbolicity; the second part is that there is a universality in this extension.

I suppose that the embedding in the first part is "trivially" achieved via a $$(1,C)$$-quasi-isometry. But I cannot find any reference that gives details. So I came to MO to see if anyone can provide one.

• I guess the argument is as follows: first embed $X$ into its metric ultrapower $X^*$. This is an isometric embedding into a geodesic and ultracomplete space. But $X$ is not cobounded in $X^*$ in general. Then the idea would be to find a subset in between. I'm not sure exactly how. Maybe the union of geodesics in $X^*$ between boundary points of $X$, or something of this flavor.
– YCor
Dec 24, 2021 at 19:30

For all $$x,y \in \overline{X}$$, fix a quasi-geodesic $$\gamma_{x,y}$$ between $$x$$ and $$y$$ with fixed parameters $$(A,B)$$ depending only on the hyperbolicity constant of $$X$$. Because $$X$$ is geodesic, we can choose a geodesic for $$\gamma_{x,y}$$ if $$x,y \in X$$. Now fix two parameters $$C,D$$ that are sufficiently large compared to $$A,B$$ respectively. Construct a new length metric space $$Y$$ as follows:

• Start by rescalling the metric of $$X$$ by a factor $$C$$: $$(X,Cd)$$.
• For all $$x,y \in \overline{X}$$, glue a copy $$\hat{\gamma}_{x,y}$$ of $$\gamma_{x,y}$$ by identifying the endpoints of these two lines.
• For all $$x,y \in \overline{X}$$ and every integer $$0< k< d(x,y)$$, connect $$\hat{\gamma}_{x,y}(k)$$ to $$\gamma_{x,y}(k)$$ with a segment of length $$D$$.

There is an obvious embedding $$X \hookrightarrow Y$$, and every point of $$Y$$ lies at distance at most $$D+1$$ from a point in $$X$$. If $$C,D$$ are chosen large enough, then $$\hat{\gamma}_{x,y}$$ is a geodesic in $$Y$$ for all $$x,y \in \overline{X}$$. Consequently, $$X$$ is isometrically embedded in $$Y$$ and any two points in $$\overline{Y} \supset \overline{X}$$ are connected by a geodesic.

The argument is an adaptation of the proof of Proposition 6 (2) from Bestvina and Fujiwara's article Bounded cohomology of subgroups of mapping class groups.

• Here by "parameter", do you mean the constants for the quasi-isometry? Dec 27, 2021 at 10:48
• And is it $\partial Y=\partial X$ instead of $\overline{Y}=\overline{X}$ that you wanted to say in the second last paragraph? Dec 27, 2021 at 10:50
• Yes, by $(A,B)$ being the parameters of a quasi-geodesic $\gamma$, I mean that $\gamma$ is an $(A,B)$-quasi-isometry. For your second remark, I should have said $\overline{Y} \supset \overline{X}$. I will correct my answer accordingly. Dec 28, 2021 at 12:31