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, X is always a geodesic metric space.)

Definition: If A, B \subseteq X are compact, then

d_{Haus}(A, B) = \max \lbrace \sup_{a\in A} \ \inf_{b \in B} \ d(a, b), \sup_{b \in B} \ \inf_{a \in A} \ d(a, b) \rbrace

is the Hausdorff distance between A and B.

Recall: A quasi-geodesic is a quasi-isometrically embedded interval.

Theorem 6: For all \delta \geq 0, \lambda \geq 1, \epsilon \geq 0, there exists R = R(\delta, \lambda, \epsilon) with the following property: If X is a \delta-hyperbolic metric space, c : [a, b] \to X is a (\lambda, \epsilon)-quasi-geodesic, and \ [c(a), c(b)] is any geodesic from c(a) to c(b), then d_{Haus} (im \ c, [c(a), c(b)]) \leq R(\delta, \lambda, \epsilon).

Corollary: A geodesic metric space X is hyperbolic if and only if for every \lambda \geq 1, \epsilon \geq 0, there exists an M such that every (\lambda, \epsilon)-quasi-geodesic triangle is M-slim.

Corollary: If X is \delta-hyperbolic, Y is geodesic, and f : Y \to X is a quasi-isometric embedding, then Y 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.

rectifyDefinition: A continuous path c : [a, b] \to X has length

l(c) = \sup_{a = t_0 < t_1 < \cdots < t_n = b} \sum_{i=1}^n d(c(t_{i-1}), c(t_i)).

If l(c) < \infty, then we say that c 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 X be \delta-hyperbolic. Let c be a continuous, rectifiable path from p to q. Then for any x \in [p, q],

d(x, im \ c) \leq \delta | \log_2 l(c) | + 1.

quasigeodesic1Proof. Without loss of generality, we may assume that c : [0, 1] \to X is parametrized proportionally to length. Let N \in \mathbb{N} such that

\frac{l(c)}{2^N} < 1 \leq \frac{l(c)}{2^{N-1}}.

It’s enough to prove that d(x, im \ c) \leq \delta N + 1. The proof is by induction on N. If N = 0, then l(c) < 1, so a point on \ [p, q] can’t be more than one unit away from the image of c. So in this case the inequality follows immediately.

For the inductive step, consider the triangle \Delta = \Delta(p, c(\frac{1}{2}), q):

quasigeodesic2Now d(x, x') \leq \delta for some x' on one of the edges of the triangle other than \ [p, q]; without loss of generality we’ll say \ [p, c(\frac{1}{2})]. By induction, we have d(x', im \ c) \leq \delta(N - 1) + 1. So d(x, im \ c) \leq d(x, x') + d(x', im \ c) \leq \delta + \delta(N - 1) + 1 = \delta N + 1, 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 X be a geodesic metric space. Given any (\lambda, \epsilon)-quasi-geodesic c : [a, b] \to X, there exists a continuous (\lambda, \epsilon')-quasi-geodesic c' : [a, b] \to X such that

(i) c'(a) = c(a) and c'(b) = c(b).
(ii) \epsilon' = 2(\lambda + \epsilon).
(iii) l(c'|_{[s, t]}) \leq k_1 d(c'(s), c'(t)) + k_2 for all s and t in \ [a, b], where k_1 = \lambda(\lambda + \epsilon) and k_2 = (\lambda \epsilon' + 3)(\lambda + \epsilon).
(iv) d_{Haus} (im \ c, im \ c') \leq \lambda + \epsilon.

Proof. First we’ll choose some points where c' will coincide with c. These will be

\Sigma = \{ a, b \} \cup ([a, b] \cap \mathbb{Z}).

So we set c'(t) = c(t) for all t \in \Sigma (and thus (i) immediately follows). Choose geodesic segments joining these points, and parametrize c' linearly along these segments.

So each segment is of length at most \lambda + \epsilon, since c is a (\lambda, \epsilon)-quasi-geodesic. Every point of im \ c \cup im \ c' is at most \frac{\lambda + \epsilon}{2} from c(\Sigma) = c'(\Sigma), and so (iv) follows.

Now we prove (ii). For t \in [a, b], let \ [t] be a choice of nearest element of \Sigma. Then for s, t \in [a, b], we have

d(c'(s), c'(t)) \leq d(c([s]), c([t])) + (\lambda + \epsilon)
\quad \leq \lambda | [s] - [t] | + \epsilon + (\lambda + \epsilon)
\quad \leq \lambda (|s - t| + 1) + (\lambda + 2\epsilon)
\quad = \lambda |s - t| + 2(\lambda + \epsilon).

The other inequality is similar, and (ii) follows.

Finally, to prove (iii), we’ll start by looking at integer subintervals. For all integers n, m \in [a, b] with n < m, we have

l(c'|_{[n, m]}) = \sum_{i=n}^{m-1} d(c(i), c(i+1))
\quad \leq (\lambda + \epsilon) |m - n|.

Similarly, l(c'|_{[a, m]}) \leq (\lambda + \epsilon)(m - a + 1) and l(c'|_{[n, b]}) \leq (\lambda + \epsilon)(b - n + 1). Therefore, for all s, t \in [a, b] we have

l(c'|_{[s, t]}) \leq (\lambda + \epsilon)(|[s] - [t]| + 2).

Combine this with

d(c'(s), c'(t)) \geq \frac{1}{\lambda} |s - t| - \epsilon' \geq \frac{1}{\lambda}(|[s] - [t]| - 1) - \epsilon'.

Then (iii) follows.