\delta-Hyperbolicity is Quasi-Isometric Invariant

On Wednesday, when we introduced delta-hyperbolicity, we used the usually Cayley graph \Gamma of \mathbb{Z}_3*\mathbb{Z}_4 (triangles glued to rectangles) as an example. Specifically, triangles in \Gamma are 2-slim so that \Gamma is 2-hyperbolic. On the other hand, we note that \delta-hyperbolicity is a generalization of trees because all triangles in trees are 0-slim. Indeed, by using cosets of \mathbb{Z}_3 and \mathbb{Z}_4 as vertices, \mathbb{Z}_3*\mathbb{Z}_4 also acts on the Bass-Serre Tree, which we call T.

By observing that T and \Gamma are quasi-isometric (the idea is to inject T onto \Gamma so that a net can be formed and R, in this case, is 2), we ask if we can deduce that \Gamma is \delta-hyperbolic from the fact that T and \Gamma are quasi-isometric? In other words, is \delta-hyperbolicity a quasi-isometric invariant?

The title and the class on Friday spoil the answer, so it is a firm YES! In this post, I will walk you through proving the following Theorem.

Theorem. If X is \delta-hyperbolic, and X and Y are quasi-isometric through f:X\to Y. Then, there exists \delta' such that Y is \delta'-hyperbolic.

To prove the Theorem, we first need to define quasi-geodesics.

Definition. A quasi-geodesic is the image of an interval [0, L] (L can also be \infty) under a quasi-isometric map.

Note that the above definition is generalized from the fact that a geodesic is the image of the interval [0, L] under an isometric map. For the reasoning, we defined geodesics as paths with the shortest length, and we know that the shortest distance on \mathbb{R} with the Euclidean metric is realized by intervals. Thus, since isometry preserves geodesics (a good exercise for you! Hint: use the inverse isometry), we see that a geodesic is the image of the interval [0, L] under the isometric map.

While geodesics are rigid and friendly, quasi-geodesics are, in general sneaky! Since quasi-isometric maps can stretch and translate spaces, quasi-geodesics can wind around the space for a while before it hits the final destination. I provide two examples.

Example (1) Logarithmic Spiral. Let g:[1,\infty) \to \mathbb{E}^2 be defined by x\to (x\cos \ln x, x \sin \ln x). Then, the image of g, for example, restricted to [1,,10] is a quasi-geodesic (we will check this claim in homework!).

Logarithmic Spiral. (From Wikipedia)

Note that in this example, our quasi-geodesics are very quasi! They take too many unnecessary detours. Indeed, let’s compare quasi-geodesics and geodesics with endpoints on the positive x-axis so that geodesics are just straight lines. There is no common constant C such that all quasi-geodesics are within C-neighborhood of geodesics since the spiral’s radius is strictly increasing.

(2) I give another example of a three-dimensional Swirl. Let g:[0,\infty) \to \mathbb{E}^3 be defined by x\to (x\cos x, x \sin x, x).

Then, we can also check that g is a quasi-isometry. Similarly, this curve always takes too many details rather than directly traveling in a straight line. Also, we cannot find a common constant C such that all quasi-geodesics following g are in C-neighborhood of geodesics with the same endpoints.

After those two examples, we understand that quasi-geodesics are generally hard to control. However, as a recurring theme that quasi-ness works well in hyperbolic geometry, we state the following lemma needed to prove the Theorem. Essentially, it says that we can find a constant C in \delta-hyperbolic spaces in the context of the previous two examples.

Lemma (Morse Lemma). If \gamma is a quasi-geodesics in \delta-hyperbolic space X, then there exists a constant C, depending only on \delta,K,\epsilon such that there exists a geodesic \gamma' so that \gamma \subset N_C(\gamma') and \gamma' \subset N_C(\gamma). In particular, if \gamma has endpoints, we may assume \gamma' has the same endpoints.

We do not give proof of this lemma. But the rough idea is to use the lemma proved last time about the “closeness” of projection points and the “thinness” of rectangles to restrict quasi-geodesics close to geodesics by triangle inequalities.

Next, we are ready to prove the Theorem!

Proof of the Theorem. Let a,b,c form an arbitrary geodesic triangle T in Y. To show that Y is \deta'-hyperbolic, it suffices to show that this arbitrary geodesic triangle is \delta'-slim.

For the first step, we note that by the existence of quasi-isometric inverse g to f, we name x=g(a), y=g(b), z=g(c) such that d(f(x), a), d(f(y),b), d(f(z),c)\leq R. Then, we claim that quasi-geodesics are quasi-isometrically mapped to quasi-geodesics. To prove the claim, note that quasi-isometries are transitive. Therefore, we compose the quasi-isometry from the quasi-geodesic with the given quasi-isometry to obtain a quasi-geodesic in the codomain.

With the above claim, we may choose an appropriate g so that the geodesic triangle T is mapped to a quasi-geometric triangle S in X (isometry is trivially quasi-isometric). Then, we connect x,y,z in X with geodesics into a triangle T' and use Morse Lemma to conclude that there exists constant C so that T' and S are close.

We choose an arbitrary point r on T. Note that g(r) is on S by construction. Therefore, we may find a point m on T' so that d(g(r),m)<C. Using that T' is \delta-thin, we know that there exists n on some other side of the triangle so that d(m,n)<\delta. Finally, there exists a point w back to T' so that d(n,w)<C. All together, we note the following:

    \[d(g(r),w)\leq d(g(r),m)+ d(m,n) + d(n,w)<2C+\delta.\]

Now, we have that f(w) is at most R distance away from a point a point s on T at a side other than the one r sits on. Thus, we derive d(r,s)<2C+\delta+R. We complete the proof by letting \delta'=2C+\delta+R.

To close out the discussion, I give two following corollaries.

Corollary 1. If space X is \delta-hyperbolic and Y is not \delta-hyperbolic, then, there exists no quasi-isometry (with the net condition) between them.

From the above Corollary, we immediately know that there exists no quasi-isometry between the hyperbolic plane and \mathbb{E}^2.

Corollary 1. If X is quasi-isometric to a tree, then it is \delta-hyperbolic.

Note that this corollary implies precisely that all Cayley graphs of free products are \delta-hyperbolic.

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5 Responses to \delta-Hyperbolicity is Quasi-Isometric Invariant

  1. Akash Ganguly says:

    Nice post Shuhangster! I really appreciated the diagrams. I’m yet to take an analysis class yet, but getting a taste of the more analytic side of things is really cool! Some parts feel much more concrete – like the concrete bounds on some things, and moreover it’s very interesting to me to see all the different algebraic properties we can get from these ideas. To give an example stolen from Office hours with a Geometric Group theorist: If G is quasi-isometric to Zn then G has a finite index subgroup
    isomorphic to Zn.

  2. Horace Fusco says:

    I really like the diagrams!! It’s interesting to see what happens when you weaken some notion of “sameness” in math and then all of a sudden find that two things which don’t at all look the same are actually equivalent under your new definition.

  3. Osip Surdutovich says:

    Great post Shuhang and really nice pictures provided. I’m a bit confused whether the three dimensional swirl is quasi-isometric? I assume it is not since the 2D version is not.

    • Shuhang Xue says:

      I agree with what you said. Indeed, I had this post before the homework problem. It turns out that the spiral’s logarithmic nature is crucial. Let’s say we have a spiral of equal distance from each other. Then, one cannot give both the upper bound and lower bound at the same time. Nice correction.

  4. Michaela Polley says:

    Nice post, Shuhang! I feel like a lot of upper-level math is about breaking the rules slightly and seeing where they take you. I find it interesting that the logarithmic spiral, which feels very not straight is a quasi-geodesic. It really shows us that definitions can be very helpful in one situation and not as helpful in another situation and we need to be very precise about what we mean.

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