# Gromov Hyperbolicity For Arbitrary Metric Spaces

01 Jul 2021

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

$(x,y)_z = \frac{1}{2}\left(d(x,z) + d(y,z) - d(x,y)\right).$

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

$(x,z)_w \ge \min\{(x,y)_w,(y,z)_w\} - \delta.$

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

$d(x,y) = d(\hat x,\hat y),\quad d(x,z) = d(\hat x,\hat z),\quad\text{and}\quad d(y,z) = d(\hat y,\hat z).$

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

$\frac 1Kd_X(x,y) - C \le d_Y(f(x),f(y)) \le Kd_X(x,y) + C.$

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.

Theorem Let $$X$$ 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.