## BI: Guessing geodesics in hyperbolic spaces

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 $X$ is hyperbolic is trying to figure out how geodesics in $X$ 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 $D$-neighborhood by $N_D$.

Guessing Geodesics Lemma: Suppose that we’re given, for each pair of points $x,y$ in the geodesic metric space $X$, a path $\eta(x,y)$ connecting them so that, for each $x,y,z$,

$\eta(x,y)\subseteq N_D(\eta(x,z)\cup \eta(z,y)),$

for some constant $D$. Then $X$ is $K$-hyperbolic and each $\eta(x,y)$ is $K$-Hausdorff-close to a geodesic, where $K=K(D)$.

(The statement is almost true: You also have to require two “coherence conditions”, i.e. that $diam(\eta(x,y))\leq D$ if $d(x,y)\leq 1$ and that for each $x',y'\in \eta(x,y)$ the Hausdorff distance between $\eta(x',y')$ and the subpath of $\eta(x,y)$ between $x'$ and $y'$ is at most $D$.)

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 $\beta$ is any path connecting, say, $x$ to $y$, then the distance from any $p\in\eta(x,y)$ from $\beta$ is bounded, roughly, by $\log(length(\beta))$.

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

Then repeat the procedure logarithmically many times or, more formally, use induction (starting from $diam(x,y)\leq D$ when $d(x,y)\leq 1$).

Ok, now that we have the weaker estimate (for any rectifiable path) we have to improve it (for geodesics).
Let now $\beta$ be a geodesic from $x$ to $y$, let $p\in\eta(x,y)$ be the furthest point from $\beta$, and set $d(p,\beta)=\xi$. Pick $x',y'\in\eta(x,y)$ before and after $p$ at distance $2\xi$ from $p$ (let’s just ignore the case when $d(p,x)<2\xi$ or $d(p,y)<2\xi$, I hope you trust it’s not hard to handle).

As $p$ is the worst point, we have $d(x',\beta),d(y',\beta)\leq \xi$, so that we can draw a red path of length at most $8\xi$ like so:

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

$\xi= d(p,\alpha)\leq O(\log(\xi))$,

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

Exercise: complete the proof showing that each $q\in \beta$ is close to $\eta(x,y)$.

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

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