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 \(\delta\).

Trees are $$0$$-hyperbolic

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.