In this post I’ll discuss a cool lemma due to Brian Bowditch (from this paper) which is very useful when you want to show that a space is hyperbolic.

The most direct way of showing that is hyperbolic is trying to figure out how geodesics in look like, show that you guessed the geodesics correctly and then show that triangles are thin, right? Well, Bowditch tells you that you can skip the second step…

This might not seem much, but showing that a given path is (close to) a geodesic might be quite annoying… and instead you get this for free!

More formally, the statement is the following (in this form it’s due to Ursula Hamenstädt, see Proposition 3.5 here). We denote the -neighborhood by .

**Guessing Geodesics Lemma:** Suppose that we’re given, for each pair of points in the geodesic metric space , a path connecting them so that, for each ,

for some constant . Then is -hyperbolic and each is -Hausdorff-close to a geodesic, where .

(The statement is almost true: You also have to require two “coherence conditions”, i.e. that if and that for each the Hausdorff distance between and the subpath of between and is at most .)

This trick has been used many times in the literature: Bowditch used it to show that curve complexes are hyperbolic, Hamenstädt to study convex-cocompact subgroups of mapping class groups, Druţu and Sapir to show that asymptotic tree-graded is the same as relatively hyperbolic, Ilya Kapovich and Rafi to show that the free-factor complex is hyperbolic, etc.

The proof is quite neat. First, one shows a weaker estimate, i.e. that if is any path connecting, say, to , then the distance from any from is bounded, roughly, by .

To show this, you just split in two halves of equal length, draw the corresponding ‘s and notice that is close to one of them.

Then repeat the procedure logarithmically many times or, more formally, use induction (starting from when ).

Ok, now that we have the weaker estimate (for any rectifiable path) we have to improve it (for geodesics).

Let now be a *geodesic* from to , let be the furthest point from , and set . Pick before and after at distance from (let’s just ignore the case when or , I hope you trust it’s not hard to handle).

As is the worst point, we have , so that we can draw a red path of length at most like so:

*And now the magic happens*. In view of the weak estimate for the red path , we get

,

which gives a bound on . (You might have noticed that here we need the second “coherence condition”). That’s it!

Exercise: complete the proof showing that each is close to .

A relatively hyperbolic version of this trick will be available soon… 🙂