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.

  • 2
    $\begingroup$ 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. $\endgroup$
    – YCor
    Dec 24, 2021 at 19:30

1 Answer 1


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.

  • $\begingroup$ Here by "parameter", do you mean the constants for the quasi-isometry? $\endgroup$
    – Muduri
    Dec 27, 2021 at 10:48
  • $\begingroup$ And is it $\partial Y=\partial X$ instead of $\overline{Y}=\overline{X}$ that you wanted to say in the second last paragraph? $\endgroup$
    – Muduri
    Dec 27, 2021 at 10:50
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    $\begingroup$ 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. $\endgroup$
    – AGenevois
    Dec 28, 2021 at 12:31

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