You are currently browsing mathjoker’s articles.

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

Let be graphs of spaces equipped with maps , and 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 be a graph of spaces, and let be the map to the underlying graph. If is a subgraph, then has a graph-of-spaces structure with underlying graph . Call a *subgraph of spaces of *.

We’re seeking a condition on such that is realized as a subgraph of spaces of some with a covering map such that the following diagram commutes:

**Definition: **For each edge map of , and each a vertex of , let

.

For each possible degree , let be the set of elevations of degree . We will say satisfies *Stallings’ condition* if and only if the following two things hold:

**(a)** Every edge map of is an elevation of the appropriate edge map of .

**(b)** For each and , there is a bijection .

So in Figure 1, the graph of spaces is something you might be able to turn into a covering. In the picture, is represented by the blue circles, and is represented by the green circles. Observe that the blue circles are in bijection with the green circles.

**Corollary:** satisfies Stallings’ condition if and only if can be realized as a subgraph of spaces of some such that

**(a)** , and

**(b)** there is a covering map such that the following diagram commutes:

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

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

Now assume that satisfies Stallings’ condition. Then we build as follows. Let . As above, we have degree-preserving inclusions (this time, not surjections)

Extend these inclusions to bijections . Now we set . Each of these is an elevation

This defines an edge space and an edge map . Consider the corresponding elevation in :

Because and are of the same degree, we have a covering transformation . So we can identify them, and use as the other edge map. By construction, satisfies the conditions of Theorem 17, so there is a suitable covering map .

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

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 .

**(ii)** .

**(iii) ** for all and in , where and .

**(iv) **.

**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.

## Recent Comments