Gromov’s original definition of Gromov hyperbolicity makes sense for arbitrary metric spaces. However, it is only a quasi-isometry invariant for geodesic metric spaces. I learned this from a paper of Väisälä. The purpose of this post is to understand the counterexample he gives. I also define Gromov hyperbolicity and quasi-isometry in this post, which might make it useful for future reference. The reader already familiar with Gromov hyperbolicity and quasi-isometries might wish to skip ahead to the heading below.

Let $(X,d)$
be a metric space.
We say that $X$
is a *length space* if the distance between any two points $x$
and $y$
in $X$
is equal to the infimum of the length of a path joining them.
A *geodesic* in $X$
is a path $\gamma\colon [a,b] \to X$
from a closed interval $[a,b] \subset \mathbb{R}$
such that $d(\gamma(s),\gamma(t)) = |s-t|$
for all $s$
and $t$
in the interval $[a,b]$
.
We say that $X$
is a *geodesic* metric space if any two points can be joined by a geodesic.
Geodesic metric spaces are length spaces but not necessarily conversely.

Gromov’s original definition of hyperbolicity of $X$
is the following.
Given points $x$
, $y$
and $z$
in $X$
, define the *Gromov product*
of $x$
and $y$
with respect to $z$
to be the quantity

Then we say that $X$
is *$\delta$
-hyperbolic* if there exists some $\delta \ge 0$
with the property that for any four points $x$
, $y$
, $z$
and $w$
in $X$
,
we have

Let’s try and make sense of this.
The intuition for the Gromov product is the following.
A *tripod* is a metric tree with four vertices, three of which are leaves and one of which has valence three.
Given any triple of points $x$
, $y$
and $z$
in a metric space,
you should convince yourself that we can always construct a tripod
with leaves $\hat x$
, $\hat y$
and $\hat z$
such that we have

In this situation, the Gromov product $(x,y)_z$ is equal to the distance from $\hat z$ to the center vertex of the tripod. To make sense of the hyperbolicity condition, let’s continue to think about trees, only in this case assume that $X$ itself is a metric tree. Then in this case it’s not too hard to argue that the hyperbolicity condition holds with $\delta = 0$ ; a picture illustrating the general case is below. Thus one way to think of $\delta$ -hyperbolicity is to think that from the point of view of any point $w \in X$ , triangles in $X$ look like triangles in trees, up to an error of $\delta$ .

If $X$ is a geodesic metric space, there are other conditions equivalent to this one that are more directly about the geometry of, e.g. triangles in $X$ . I’ve been collecting as many of these conditions as the community will give me in a MathOverflow post. Eventually I expect to make a blog post or several about the various definitions.

Given metric spaces $(X,d_X)$
and $(Y,d_Y)$
and constants $K \ge 1$
and $C \ge 0$
,
a *$(K,C)$
-quasi-isometric embedding* is a function $f\colon X \to Y$
(it need not be continuous)
such that for all points $x$
and $y$
in $X$
, the following inequality holds

So we see that $f$
is allowed to distort distances by a bounded multiplicative and additive amount.
A *$(K,C)$
-quasi-isometry* is a $(K,C)$
-quasi-isometric embedding with the additional property that
for every point $y \in Y$
there is a point $x \in X$
such that $d_Y(y,f(x)) \le C$
.

## The theorem and the counterexample

Väisälä proves the following theorem.

TheoremLet $X$ and $Y$ be length spaces and $f\colon X \to Y$ a quasi-isometric embedding. If $Y$ is $\delta$ -hyperbolic, then $X$ is $\delta'$ -hyperbolic for some $\delta'$ depending only on $\delta$ and the quasi-isometry constants for $f$ .

He then observes that the theorem is not true if one does not assume that $X$ is a length space. Here is the counterexample. The hyperbolic space $Y$ is the real line $\mathbb{R}$ with the standard metric. (Recall that $\mathbb{R}$ is a real tree, and thus $0$ -hyperbolic.) The space $X$ is the graph of the absolute value function $\{(x,y) \in \mathbb{R}^2 : y = |x| \}$ with the Euclidean metric. The claim is that the map $f\colon X \to Y$ defined as $f(x,y) = x$ is a quasi-isometric embedding, but that $X$ is not hyperbolic.

First we show that $f$ is a quasi-isometric embedding. Indeed, it is clear that given $\vec x$ and $\vec y$ in $X$ , we have $d_Y(f(\vec x),f(\vec y)) \le d_X(\vec x,\vec y)$ . On the other hand, note that $d_X(\vec x,\vec y)$ is furthest from $d_Y(f(\vec x),f(\vec y))$ when $\vec x$ and $\vec y$ are in the same quadrant, where we have $d_X(\vec x,\vec y) = \sqrt{2}d_Y(f(\vec x),f(\vec y))$ . Thus the map $f$ is a $(\sqrt 2,0)$ -quasi-isometric embedding, in fact a $(\sqrt 2,0)$ -quasi-isometry.

Finally we show that $X$ is not hyperbolic. Consider the four points $x = (0,0)$ , $y = (-2t,2t)$ , $z = (2t,2t)$ and $w = (t,t)$ . Then we have

$ (x,z)_w = \frac 12\left(\sqrt 2t + \sqrt 2t - 2\sqrt 2t\right) = 0,$ $ (x,y)_w = \frac 12\left(\sqrt 2t + 2\sqrt 5t - 2\sqrt 2t\right) \approx 1.53t$ $ (y,z)_w = \frac 12\left(2\sqrt 5t + \sqrt 2t - 4t\right) \approx 0.94t$But observe that as $t \to \infty$ , we have that $(x,y)_w$ and $(y,z)_w$ tend to $\infty$ , so no uniform choice of $\delta$ makes the inequality $(x,z)_w \ge \min\{(x,y)_w,(y,z)_w\} - \delta$ hold. The problem, roughly, is that the quasi-isometry condition cannot see the “kink” in the space $X$ , while the $\delta$ -hyperbolicity condition, at least for this family of triangles, can.