We want to show that the notion of hyperbolicity is preserved under quasi-isometry. One problem is that geodesics are not preserved under quasi-isometry; one can show that there are quasi-geodesics in the plane that look nothing like geodesics.
(In the following, is always a geodesic metric space.)
Definition: If are compact, then
is the Hausdorff distance between and .
Recall: A quasi-geodesic is a quasi-isometrically embedded interval.
Theorem 6: For all , , , there exists with the following property: If is a -hyperbolic metric space, is a -quasi-geodesic, and is any geodesic from to , then .
Corollary: A geodesic metric space is hyperbolic if and only if for every , , there exists an such that every -quasi-geodesic triangle is -slim.
Corollary: If is -hyperbolic, is geodesic, and is a quasi-isometric embedding, then is hyperbolic.
Corollary: Hyperbolicity is a quasi-isometry invariant of geodesic metric spaces.
(Note: Gromov provides a definition of hyperbolicity that works for non-geodesic metric spaces, but this notion of hyperbolicity is not quasi-isometry invariant.)
To prove Theorem 6, we must first think about how to find the length of a curve in a metric space. The idea is to choose several points on the curve, and draw geodesic segments between pairs of consecutive points. The length of the curve should then be greater than or equal to the total of the lengths of these segments. We then define the length of the curve to be the supremum of this sum over all possible choices of points on the curve.
Definition: A continuous path has length
If , then we say that is rectifiable.
Now we’ll show that a path in a hyperbolic metric space can’t go very far from a geodesic between its endpoints unless the path is very long.
Lemma 7: Let be -hyperbolic. Let be a continuous, rectifiable path from to . Then for any ,
Proof. Without loss of generality, we may assume that is parametrized proportionally to length. Let such that
It’s enough to prove that . The proof is by induction on . If , then , so a point on can’t be more than one unit away from the image of . So in this case the inequality follows immediately.
For the inductive step, consider the triangle :
Now for some on one of the edges of the triangle other than ; without loss of generality we’ll say . By induction, we have . So , as desired.
In the next lemma, we show that given an arbitrary quasi-geodesic, we can find a “nicer” quasi-geodesic that is close to the given quasi-geodesic.
Lemma 8: Let be a geodesic metric space. Given any -quasi-geodesic , there exists a continuous -quasi-geodesic such that
(i) and .
(iii) for all and in , where and .
Proof. First we’ll choose some points where will coincide with . These will be
So we set for all (and thus (i) immediately follows). Choose geodesic segments joining these points, and parametrize linearly along these segments.
So each segment is of length at most , since is a -quasi-geodesic. Every point of is at most from , and so (iv) follows.
Now we prove (ii). For , let be a choice of nearest element of . Then for , we have
The other inequality is similar, and (ii) follows.
Finally, to prove (iii), we’ll start by looking at integer subintervals. For all integers with , we have
Similarly, and . Therefore, for all we have
Combine this with
Then (iii) follows.