Our main source for this post is Bridson and Haefliger’s book [BrH].
However, the material has origins in
- Alonso, T. Brady, Cooper, Ferlini, Lustig, Mihalik, Shapiro and Short (also ed.),
- Gersten and Short,
- Ghys and de al Harpe, Sur les groupes hyperboliques d’après Mikhael Gromov, Progr. Math., 83, Birkhauser, 1990, (partial English translation here),
- Gromov, Hyperbolic groups, Essays on group theory, MSRI series vol. 8, Springer–Verlag, 1987,
- Short, Regular subgroups of automatic groups, MSRI preprint no. 07723–89, 1989.
We’ve completed our overview of the landscape of non–positively curved groups and now turn to the primary subject of this blog — the subgroups. We’ve promised “Here there be dragons!” — but first we’ll cast an eye over the tamer inhabitants. In this post and the next we’ll describe contexts where the subgroup structure can be said to be well–behaved.
We begin here with hyperbolic groups, where the crucial notion in the study of “well–behaved” subgroups was identified by Gromov (Hyperbolic groups, Essays on group theory, MSRI series vol. 8, Springer–Verlag, 1987) as being quasi-convexity.
Here is how the the usual definition of convexity is “quasifyied”.
Definition. A subspace of a geodesic metric space is quasi–convex when there exists some such that every geodesic in that connects a pair of points in lies inside the –neighbourhood of .
Definition. Suppose is a group with finite generating set . A subgroup of a group is quasi–convex if it’s quasi-convex in the Cayley graph .
For hyperbolic groups, this does not depend on .
Lemma 1. Suppose is a group with finite generating set and that is a –quasi-convex subgroup of . Then is finitely generated and the inclusion is a quasi–isometric embedding.
Proof: Let be set of all elements of a distance at most from in . We claim that .
Suppose is a geodesic edge–path in from to an element of . Consider inserting detours into the path : on arriving at each successive vertex travel to a nearest element of and then back. By the quasi–convexity of , each detour has length at most . So our new path is a concatenation of at most paths each of which travels between elements of and has length at most — see below.
That is a quasi–isometric embedding follows from
Distortion and chacterizing quasi–convexity in hyperbolic groups
Definition. The distortion of a finitely generated subgroup in a finitely generated group , is defined by
Here and denote word metrics associated to some choices of finite generating sets. Those choices are not important: different choices lead to –equivalent distortion functions.
Definition. is undistorted in when there exists such that for all .
Lemma 2. For a finitely generated subgroup of a hyperbolic group , the following are equivalent.
- is quasi–convex,
- the inclusion is a quasi-isometric embedding,
- is undistorted in .
Proof: Lemma 1 gives 1 2. For 2 1, note that a geodesic in the Cayley graph of (with respect to some finite generating set) connecting a pair of elements maps to a quasi-geodesic connecting them in the Cayley graph of (with respect to some finite generating set) since is a quasi-isometric embedding. But, as is –hyperbolic, that quasi–geodesic is close to any geodesic connecting to (as we’ve previously discussed), and so any geodesic in the Cayley graph of between and is in some uniform neighbourhood of . For 2 3, unravel the definitions to see they are two ways of expressing the same condition.
Quasi–convex subgroups are hyperbolic
Proof: The work for this was done in Lemma 2. Suppose is a quasi–convex subgroup of a hyperbolic group . A geodesic triangle in (a Cayley graph of) becomes a quasi–geodesic triangle in (a Cayley graph of) , since is a quasi-isometric embedding. But in –hyperbolic spaces, quasi–geodesic are uniformly close to geodesics and geodesic triangles are –thin, so is uniformly thin in and hence also in .
The converse to Theorem 1 is false. As we’ll see in some future post (e.g. on Cannon–Thurston maps or on hydra groups), there are hyperbolic (finite rank free, indeed) subgroups of hyperbolic groups that are heavily distorted.
Applications of quasi–convexity
The applications we will draw are that is never a subgroup of a hyperbolic group and that subgroups are always undistorted.
First we need:
Lemma 3 (adapted from [BrH], pages 476–477). If and are –quasi-convex subgroups of a group with finite generating set , then their intersection is quasi–convex.
Proof: Let denote the word metric on with respect to . Let
We will show that is –quasi–convex. It is enough to prove that if is a geodesic word on representing an element of and is a prefix of , then represents an element of a distance at most from .
Let be the set of words such that and for every prefix of , the group element represented by is a distance at most from each of and . This set is non–empty as one could take to be the suffix of such that as words.
Let be a minimal length word in . We claim that , which will complete the proof of the lemma. Well, suppose and are prefixes of and , for , are group elements with such that represent elements of . Suppose — then, as words, and for some words and . If , then would be a shorter word than in . So, as there are at most possibilities for the pairs , the length of is at most .
Incidentally, our next lemma implies the Conjugacy Problem is decidable in hyperbolic groups. (In fact, it also works in semi–hyperbolic groups.)
Lemma 4. Suppose is a hyperbolic group and denotes the word metric on with respect to finite generating set . Then there is a constant such that if and are conjugate elements in and , then there is a word such that in and .
Proof: Let be a minimal length word such that in . Let denote the length– prefix of , and let be a minimal length word such that in . (See below.) The words and must be different for all between and , as otherwise we could find a word shorter than such that : if and as words, then take .
By the convexity of the metric, there is a constant , depending only on and such that for all . But there are no more than words of length at most , which proves the lemma.
Proposition 1. (adapted from [BrH], pages 477–478). The centralizer of any element in a hyperbolic group is quasi-convex.
Proof: Fix a finite generating set for . Suppose is represented by a geodesic word and is a prefix of . We aim to show that there is a constant , depending only on , and such that the group element represented by is within distance from .
Let . By applying the convexity of the metric to the geodesic rectangle displayed below, we learn there is some constant depending only on and such that .
By definition, and are conjugate. Let be a minimal length word such that . Note that because
Applying Lemma 4 together with the inequality , we get an upper bound,
which depends only on , and . Thus is distance at most from .
Another way of stating the conclusion of the next theorem is that the subgroups are bi–infinite quasi–geodesics.
Theorem 2 (adapted from [BrH], page 462). subgroups of a hyperbolc group are quasi-convex.
Proof: Suppose is an element of infinite order in a hyperbolic group . By Proposition 1, is quasi–convex, and so, by Theorem 1, is hyperbolic and, in particular, finitely generated. Its centre is the intersection of the centralizers of each of the elements of a finite generating set, so is itself quasi–convex, by Lemma 3. So is hyperbolic by Theorem 1. But is abelian and hyperbolic abelian groups and either finite or virtually . As contains , it must be virtually .
Now the inclusions
are all quasi-isometric embeddings: the first because is of finite index in , and the second and third by Lemma 2. So the composition is a quasi-isometric embedding, and therefore is quasi–convex by Lemma 2.
The theorem above means many non–abelian solvable groups cannot be subgroups of hyperbolic groups — they have distorted subgroups on account of calculations such as the following.
for all integers .
Proof: Induct on :
The following proposition appears in Gromov’s original article on hyperbolic groups, but his proof is very different: he defines and analyzes the boundary of .
Proposition 2 (adapted from [BrH], page 462). If is an infinite order element of a hyperbolic group , then is a finite index subgroup of its centralizer .
Proof: Fix a finite generating set for and a such that is –hyperbolic. Lemma 5 implies that the positive powers of are in distinct conjugacy classes since if for some and some integers and with , then
and would fail to be a quasi–isometric embedding. So, by replacing with a power if necessary, we may assume is not conjugate to an element of within a distance from .
Now suppose . We aim to show that where , which will suffice to prove the proposition. We will suppose otherwise and seek a contradiction.
Suppose . By replacing by (which is also in ) we can assume .
Let be a geodesic edge–path in from to , parametrized proportional to arc–length — so is a distance from . The translate is a geodesic edge–path from to . Consider a geodesic rectangle which has and as two of its sides:
The fact that geodesic triangles in are –thin implies that for all , the point is within a distance from one of the other three sides of the rectangle. The assumption that allows us to choose a for which that other side must be — that is, there is some such that . But and , so . And therefore, .
Let be a group element (i.e. a vertex) on closest to . Then and so , which is a contradiction.
As an immediate consequence of Proposition 2, we have:
Theorem 3. Abelian subgroups of hyperbolic groups are either finite or virtually . In particular, is never a subgroup of a hyperbolic group.