You are currently browsing the monthly archive for February 2009.

Our goal is to understand the abelian subgroups of a hyperbolic group; e.g., can a hyperbolic group contain as a subgroup? We already know that it cannot be a quasiconvex subgroup, but it may be possible for to be “twisted” in some manner.

**Theorem 10.** The intersection of two quasiconvex subgroups is quasiconvex.

**Pf.** As usual, we work in a -hyperbolic . Let be quasiconvex subgroups each with the same corresponding constant . Let , and let . Our goal is therefore to show that is in a bounded neighborhood of . Let be the (or more precisely, a particular) closest element of to , and suppose that . Let be such that and let be such that ; such elements exist since and are quasiconvex. Let be such that . We sketch the situation below.

Consider the geodesic triangle with vertices , , and . Because this triangle is -slim, for each there exists some such that .

Likewise, for each there exists such that . Next, let and . Then .

Suppose . Then by the Pigeonhole Principle there are integers and with such that and . Now, we can use this information to find a closer element of .

Consider . By the triangle inequality,

.

All that remains is to prove that . But since ,

.

Playing this same game with , we get , and hence we have found our contradiction.

For a group , recall that is the **center** of .

**Corollary.** For any (where is -hyperbolic) of infinite order, the subgroup generated by is quasiconvex. Equivalently, the map sending is a quasigeodesic (which is a sensible statement to make given that is quasi-isometric to ).

**Pf.** By Theorem 9, we deduce that is quasiconvex, and so is finitely generated. Let be a finite generating set for . Notice

where is the centralizer in of , so is quasiconvex by Theorem 10 and thus is a finitely generated abelian group containing . By Exercise 16 (below), we deduce that is quasi-isometrically embedded in , and hence is quasiconvex in .

**Exercise 16.** Cyclic subgroups of finitely generated abelian groups are quasi-isometrically embedded.

**Lemma 10.** Suppose the order of is infinite. If is conjugate to , then .

**Pf.** Suppose . An easy induction on shows . Therefore, applying the triangle inequality,

.

But is a -quasi-isometric embedding, so

.

This is impossible unless (eventually the exponential growth dominates). Similarly, reversing the roles of and in the above argument implies that , so .

**Theorem 8: **Let be a -hyperbolic group with respect to . If are conjugate then there exists such that

where depends only on .

**Proof:** We work in . Let be such that . Let be such that . We want to find a bound on .

Let . By Lemma 9,

Also

So . Thus . Suppose that . By the Pigeonhole Principle there exist integers such that . It follows that one can find a shorter conjugating element by cutting out the section of between and .

Recall, for , is the *centralizer* of .

**Theorem 9:** If is -hyperbolic with respect to and , then is quasi-convex in .

**Proof: **Again we work in . Let , . We need to prove that is in a bounded neighborhood .

Just as in the proof of Theorem 8,

Well, and are conjugate. By Theorem 8 there exists such that

But so that and .

**Exercise 15: **Prove that

is not hyperbolic for any Anosov .

We will see two examples of non-quasiconvex subgroups in this section. The first one is NOT a hyperbolic group, while the second one is.

**Example:** For the first example, let

,

with one eigenvalue (the larger one) . Notice that does not fix any non-zero vectors in (such a map is called Anosov).

Now let . This is a group. The group law works like this: for any , . Pick , . The map is, by the following analysis, NOT a quasi-embedding:

Choose such that . All norms on are bilipschitz, so there exists such that . Therefore, for sufficiently large , , and so . On the other side, we have . It follows that is not a quasi-embedding.

**Example: **For the second example, let be a hyperbolic surface. An automorphism of is called pseudo-Anosov if for any smooth closed curve on and any , is not homotopic to . Let be the mapping torus of , i.e., , with the relation generated by .

Under these assumptions, we are able to use a theorem of Thurston asserting that, must be a hyperbolic 3-manifold. (W. Thurston, “On the geometry and dynamics of diffeomorphisms of surfaces,” Bull. Amer. Math. Soc. vol 19 (1988), 417-431)

Hence, if is closed, then is also closed. So acts nicely on (actually ), and so is word-hyperbolic by the *Švarc*–*Milnor* Lemma. Then a similar argument to the previous shows the natural map is NOT a quasi-embedding.

For concrete examples, see A. Casson & S. Bleiler, “Automorphisms of Surfaces After Nielsen and Thurston”.

After the two examples, let us switch to a property for all hyperbolic groups:

**Theorem 7:**** **Hyperbolic groups are finitely presented.

In order to prove this theorem, we need the following lemma:

**Lemma 9: **Let be two geodesics in a -hyperbolic metric space , . (If is longer than , say, then extend by the constant map). Then for any , .

**Proof: **Case 1: there is such that . Without loss of generality, assume , then . So, .

Case 2: there is no such that . Then must be within distance of . Apply a similar argument to the previous, we see .

**Proof of Theorem 7: **Let be -hyperbolic, with the generating set . Let be any relation, which corresponds to a loop in the Cayley graph . We can always take with and geodesics in and , by “triangulating”.

Write , . Denote , , . An easy induction shows that

.

But Lemma 9 implies that for all , so we have written the loop as a product of conjugates of loops of length at most . Therefore, the set of all loops of length at most is a finite set of relations for .

## Recent Comments