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Recall that our goal is to determine, for a given map of graphs of spaces such as the one shown below, whether the map \Phi can be extended to a covering map \widehat{\Phi}.

Figure 1

Figure 1

Let \mathcal{X}, \mathcal{X}' be graphs of spaces equipped with maps \Xi' \to \Xi, \phi_{v'} : X_{v'} \to X_v and \phi_{e'} : X_{e'} \to X_e as before.  Recall that in Stallings’ proof of Hall’s Theorem, we completed an immersion to a covering map by gluing in edges.  We will aim to do the same thing with graphs of spaces.

Definition: Let \mathcal{X} be a graph of spaces, and let \eta : X_{\mathcal{X}} \to \Xi be the map to the underlying graph.  If \Delta \subseteq \Xi is a subgraph, then \eta^{-1}(\Delta) \subseteq X_{\mathcal{X}} has a graph-of-spaces structure \mathcal{Y} with underlying graph \Delta.  Call \mathcal{Y} a subgraph of spaces of \mathcal{X}.

We’re seeking a condition on \mathcal{X}' such that \mathcal{X}' is realized as a subgraph of spaces of some \widehat{\mathcal{X}} with a covering map \widehat{\Phi} : X_{\widehat{\mathcal{X}}} \to X_{\mathcal{X}} such that the following diagram commutes:

diagram1Definition: For each edge map \partial_e^{\pm} : X_e \to X_v of \mathcal{X}, and each v' \mapsto v a vertex of \mathcal{X}', let

\mathcal{E}^{\pm}(e) = \bigcup_{v' \mapsto v} \mathcal{E}^{\pm}(e, v').

For each possible degree \mathcal{D}, let \mathcal{E}_{\mathcal{D}}^{\pm}(e) \subseteq \mathcal{E}^{\pm}(e) be the set of elevations of degree \mathcal{D}.  We will say \mathcal{X}' satisfies Stallings’ condition if and only if the following two things hold:

(a) Every edge map of \mathcal{X}' is an elevation of the appropriate edge map of \mathcal{X}.
(b) For each e \in E(\Xi) and \mathcal{D}, there is a bijection \mathcal{E}_{\mathcal{D}}^+(e) \leftrightarrow \mathcal{E}_{\mathcal{D}}^-(e).

So in Figure 1, the graph of spaces \mathcal{X}' is something you might be able to turn into a covering.  In the picture, \mathcal{E}_{\mathcal{D}}^+(e) is represented by the blue circles, and \mathcal{E}_{\mathcal{D}}^-(e) is represented by the green circles.  Observe that the blue circles are in bijection with the green circles.

Corollary: \mathcal{X}' satisfies Stallings’ condition if and only if \mathcal{X}' can be realized as a subgraph of spaces of some \widehat{\mathcal{X}} such that

(a) V(\Xi ') = V(\widehat{\Xi}), and
(b) there is a covering map \widehat{\Phi} : X_{\widehat{\mathcal{X}}} \to X_{\mathcal{X}} such that the following diagram commutes:

diagram1

Proof of Corollary. First we’ll show that if \Phi can be extended to a covering map as described above, then \mathcal{X}' satisfies Stallings’ condition.  By Theorem 17, every edge map of \widehat{\mathcal{X}} is an elevation.  So there are inclusions

diagram2

Furthermore, these maps are surjective, and clearly degree-preserving.

Now assume that \mathcal{X}' satisfies Stallings’ condition.  Then we build \widehat{\mathcal{X}} as follows.  Let V(\widehat{\Xi}) = V(\Xi ').  As above, we have degree-preserving inclusions (this time, not surjections)

diagram3

Extend these inclusions to bijections \mathcal{E}_{\mathcal{D}}^+(e) \leftrightarrow \mathcal{E}_{\mathcal{D}}^-(e).  Now we set E(\widehat{\Xi}) = \bigcup_{e \in E(\Xi)} \mathcal{E}^+(e).  Each of these \widehat{e} is an elevation

diagram4

This defines an edge space X_{\widehat{e}} and an edge map \partial_{\widehat{e}}^+.  Consider the corresponding elevation in \bigcup_{e \in E(\Xi)} \mathcal{E}^-(e):

diagram5

Because \partial_{\widehat{e}}^+ and \partial_{\widehat{e}}^- are of the same degree, we have a covering transformation X_{\widehat{e}} ' \to X_{\widehat{e}}.  So we can identify them, and use \partial_e^- as the other edge map.  By construction, \widehat{\mathcal{X}} satisfies the conditions of Theorem 17, so there is a suitable covering map \widehat{\Phi} : X_{\widehat{\mathcal{X}}} \to X_{\mathcal{X}}.

Exercise 25: (This will be easier later, but we have the tools necessary to do this now.)  Prove that if G_1 and G_2 are LERF groups, then so is G_1 * G_2.

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.

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