IV Topics in Geometric Group Theory

4Hyperbolic groups

4.2 Quasi-geodesics and hyperbolicity

Definition

(Quasi-geodesic)

A (

λ, ε

)-quasi-geodesic for

λ ≥

, ε ≥

0 is a

(λ, ε)-quasi-isometric embedding I → X, where I ⊆ R is a closed interval.

Note that our definition allows for

to be unbounded, i.e.

may be of the

forms [

a, b

], [0

, ∞

) or

respectively. We call these quasi-geodesic intervals,

quasi-geodesic rays and quasi-geodesic lines respectively.

Example. The map [0, ∞) → R

given in polar coordinates by

t 7→ (t, log(1 + t))

is a quasigeodesic ray.

This should be alarming. The quasi-geodesics in the Euclidean plane can

look unlike any genuine geodesic. However, it turns out things are well-behaved

in hyperbolic spaces.

Theorem

(Morse lemma)

For all

δ ≥

, λ ≥

1 there is

(

δ, λ, ε

) such that the

following holds:

is a

-hyperbolic metric space, and

: [

a, b

]

→ X

is a (

λ, ε

)-quasigeodesic

from p to q, and [p, q] is a choice of geodesic from p to q, then

Haus

([p, q], im(c)) ≤ R(δ, λ, ε),

where

Haus

(A, B) = inf{ε > 0 | A ⊆ N

(B) and B ⊆ N

(A)}

is the Hausdorff distance.

This has nothing to do with the Morse lemma in differential topology.

Corollary.

There is an

(

δ, λ, ε

) such that a geodesic metric space

hyperbolic iff any (λ, ε)-quasigeodesic triangle is M -slim.

Corollary.

Suppose

X, X

are geodesic metric spaces, and

X → X

is a

quasi-isometric embedding. If X

is hyperbolic, then so is X.

In particular, hyperbolicity is a quasi-isometrically invariant property, when

restricted to geodesic metric spaces.

Thus, we can make the following definition:

Definition

(Hyperbolic group)

A group Γ is hyperbolic if it acts properly

discontinuously and cocompactly by isometries on a proper, geodesic hyperbolic

metric space. Equivalently, if it is finitely-generated and with a hyperbolic

Cayley graph.

Let’s prove the Morse lemma. To do so, we need the following definition:

Definition

(Length of path)

Let

: [

a, b

]

→ X

be a continuous path. Let

a = t

< t

< · · · < t

= b be a dissection D of [a, b]. Define

`(c) = sup

i=1

d(c(t

i−1

), c(t

)).

In general, this number is extremely hard to compute, and may even be

infinite.

Definition (Rectifiable path). We say a path c is rectifiable if `(c) < ∞.

Example. Piecewise geodesic paths are rectifiable.

Lemma.

Let

be a geodesic space. For any (

λ, ε

)-quasigeodesic

: [

a, b

]

→ X

there exists a continuous, rectifiable (

λ, ε

)-quasigeodesic

: [

a, b

]

→ X

with

= 2(λ + ε) such that

(i) c

(a) = c(a), c

(b) = c(b).

(ii) For all a ≤ t < t

≤ b, we have

`(c

[t,t

]

) ≤ k

d(c

(t), c

)) + k

where k

= λ(λ + ε) and k

= (λε

+ 3)(λ + 3).

(iii) d

Haus

(im c, im c

) ≤ λ + ε.

Proof sketch.

Let Σ =

{a, b} ∪

((

a, b

)

∩ Z

). For

t ∈

Σ, we let

(

) =

(

), and

define

to be geodesic between the points of Σ. Then claims (i) and (iii) are clear,

and to prove quasigeodesicity and (ii), let

: [

a, b

]

→

Σ be a choice of closest

point in Σ, and then estimate d(c

(t), c

)) in terms of d(c(σ(t)), c(σ(t

))).

Lemma.

let

-hyperbolic, and

: [

a, b

]

→ X

a continuous, rectifiable path

in X joining p to q. For [p, q] a geodesic, for any x ∈ [p, q], we have

d(x, im c) ≤ δ| log

`(c)| + 1.

Proof.

We may assume

: [0

→ X

is parametrized proportional to arc length.

Suppose

`(c)

< 1 ≤

`(c)

N−1

Let

. Pick a geodesic triangle between

p, q, c

(

). By

-hyperbolicity, there

exists a point

lying on the other two edges such that

(

, x

)

≤ δ

. We wlog

assume x

∈ [p, c(

)]. We can repeat the argument with c|

[0,

]

p q

)

Formally, we proceed by induction on

. If

= 0 so that

(

)

1, then we

are done by taking desired point on

im c

to be

(or

). Otherwise, there is some

∈ [p, c(

)] such that d(x

, x

) ≤ δ. Then

`(c|

[0,

]

)

N−1

< 1 ≤

`(c|

[0,

])

N−2

So by the induction hypothesis,

d(x

, im c) ≤ δ| log

`(c|

[0,

]

| + 1

= δ



log

`(c)



+ 1

= δ(|log

`(c)| − 1) + 1.

Note that we used the fact that `(c) > 1, so that log

`(c) > 0.

Then we are done since

d(x, im c) ≤ d(x, x

) + d(x

, im c).

Proof of Morse lemma.

By the first lemma, we may assume that

is continuous

and rectifiable, and satisfies the properties as in the lemma.

Let

p, q

be the end points of

, and [

p, q

] a geodesic. First we show that [

p, q

]

is contained in a bounded neighbourhood of im c. Let

D = sup

x∈[p,q]

d(x, im c).

By compactness of the interval, let

∈

[

p, q

] where the supremum is attained.

Then by assumption,

im c

lies outside

(

, D

). Choose

y, z ∈

[

p, q

] be such that

(

, y

) =

(

, z

) = 2

and

y, x

, z

appear on the geodesic in this order (take

y = p, or z = q if that is not possible).

Let

(

)

∈ im c

be such that

(

, y

)

≤ D

, and similarly let

(

)

∈

im c be such that d(z, z

) ≤ D.

p q

y z

2D 2D

≤ D ≤ D

Let γ = [y, y

] · c|

[s,t]

· [z

, z]. Then

`(γ) = d(y, y

) + d(z, z

) + `(c|

[s,t]

) ≤ D + D + k

d(y

, z

) + k

by assumption. Also, we know that d(y

, z

) ≤ 6D. So we have

`(γ) ≤ 6k

D + 2D + k

But we know that

d(x

, im γ) = D.

So the second lemma tells us

D ≤ δ| log

(6k

D + 2D + k

)| + 1.

The left hand side is linear in

, and the right hand side is a logarithm in

So it must be the case that

is bounded. Hence [

p, q

]

⊆ N

(

im c

), where

is some constant.

It remains to find a bound

such that

im c ⊆ N

([

p, q

]). Let [

, b

] be a

maximal subinterval of [

a, b

] such that

[

, b

] lies entirely outside

([

p, q

]).

Since

(

[

a, a

]) and

(

[

, b

]) are both closed, and they collectively cover

the connected set [p, q], there exists

w ∈ [p, q] ∩

(c[a, a

]) ∩

(c[b

, b]).

Therefore there exists

t ∈

[

a, a

] and

∈

[

, b

] such that

(

w, c

(

))

≤ D

and

d(w, c(t

)) ≤ D

. In particular, d(c(t), c(t

)) ≤ 2D

By the first lemma, we know

`(c|

[t,t

]

) ≤ 2k

+ k

So we know that for s ∈ [a

, b

], we have

d(c(s), [p, q]) ≤ d(c(s), w)

≤ d(c(s), c(t)) + d(c(t), w)

≤ `(c|

[t,t

]

) + D

≤ D

+ 2k

+ k

With the Morse lemma proven, we can sensibly talk about hyperbolic groups.

We have already seen many examples of hyperbolic spaces, such as trees and the

hyperbolic plane.

Example.

Any finite group is hyperbolic since they have bounded Cayley

graphs.

Example.

Finitely-generated free groups are hyperbolic, since their Cayley

graphs are trees.

Example.

Surface groups Σ

for

g ≥

2 acts on the hyperbolic plane properly

discontinuously and cocompactly by isometries (since the universal cover of the

genus

surface is the hyperbolic plane by tessellating the hyperbolic plane with

4g-gons, cf. IB Geometry).

The following notion of “virtually” would be helpful

Definition

(virtually-P)

Let P be a property of groups. Then we say

virtually-P if there is a finite index subgroup G

≤ G such that G

is P.

Example. Finite groups are virtually trivial!

Note that

≤ G

is finite-index, then

acts on

Cay

(

) in a way that

satisfies the properties of the Schwarz–Milnor lemma (the only property we may

worry about is cocompactness, which is what we need finite-index for) and hence

G. In particular,

Example.

A virtually hyperbolic group is hyperbolic. For example, virtually

free groups such that (Z/2) ∗ (Z/3) = PSL

Z are hyperbolic.

These are some nice classes of examples, but they will be dwarfed by our

next class of examples.

A “random group” is hyperbolic. More precisely, fix

m ≥

2 and

n ≥

1. We

temporarily fix ` ≥ 0. Consider groups of the form

Γ = ha

, . . . , a

| r

, . . . , r

such that

are all cyclicly reduced words of length

(i.e. reduced words such

that the first letter is not the inverse of the last). Put the uniform probability

distribution on the set of all such groups. This defines a group-valued random

variable Γ

. For a property P , we say that “a random group is P ” if

lim

`→∞

P(Γ

is hyperbolic) = 1

for all n, m.

Theorem (Gromov). A random group is infinite and hyperbolic.

There are, of course, other ways to define a “random group”. As long as we

control the number of relations well so that we don’t end up with finite groups

all the time, the “random group” should still be hyperbolic.

Recall that in differential geometry, geodesics are defined locally. However,

we defined our geodesics to be embedded interval, which is necessarily a global

notion. We want an analogous local version. However, if we want to work up to

quasi-isomorphism, then we cannot go completely local, because locally, you are

allowed to be anything.

Definition

(

-local geodesic)

Let

be a geodesic metric space, and

k >

0. A

path c : [a, b] → X is a k-local geodesic if

d(c(t), c(t

)) = |t − t

for all t, t

∈ [a, b] with |t − t

| ≤ k.

We know very well that on a general Riemannian manifold, local geodesics

need not be actual geodesics. In fact, they can be far from being an actual

geodesic, since, for example, we can wrap around the sphere many times. Again,

things are much better in the hyperbolic world.

Theorem.

Let

-hyperbolic and

: [

a, b

]

→ X

be a

-local geodesic where

k > 8δ. Then c is a (λ, ε)-quasigeodesic for some λ = λ(δ, k) and ε = ε(δ, k).

First, we prove

Lemma.

Let

-hyperbolic and

k >

. If

: [

a, b

]

→ X

is a

-local

geodesic, then im c is contained in the 2δ-neighbourhood of [c(a), c(b)].

Observe that by iterating the definition of hyperbolicity, on a

-hyperbolic

space, any point on an

-gon is at most (

n −

away from a point on another

side.

Proof. Let x = c(t) maximize d(x, [c(a), c(b)]). Let

y = c



t −



, z = c



t −



If t −

< a, we set y = c(a) instead, and similarly for z.

Let

∈

[

(

)

, c

(

)] minimize

(

y, y

), and likewise let

∈

[

(

)

, c

(

)] minimize

d(z, z

Fix geodesics [

y, y

] and [

z, z

]. Then we have a geodesic rectangle with

vertices

y, y

, z, z

. By

-hyperbolicity, there exists

on the rectangle not on

im c, such that d(x, w) = 2δ.

w ∈

[

, z

], then we win. Otherwise, we may wlog assume

w ∈

[

y, y

Note that in the case

(

), we must have

, and so this would imply

y ∈

[

(

)

, c

(

)]. So we are only worried about the case



t −



(

y, x

) =

. But then by the triangle inequality, we must have

d(y, w) > 2δ = d(x, w).

However,

d(x, y

) ≤ d(x, w) + d(w, y

) < d(y, w) + d(w, y

) = d(y, y

So it follows that

d(x, [c(a), c(b)]) < d(y, y

) = d(y, [c(a), c(b)]).

This contradicts our choice of x.

Proof of theorem.

Let

: [

a, b

]

→ X

be a

-local geodesic, and

t ≤ t

∈

[

a, b

Choose

t < t

< · · · < t

< t

such that

i−1

for all

and

−t

< k

Then by definition, we have

d(c(t

i−1

), c(t

)) = k, d(c(t

), c(t

)) = |t

− t

for all i. So by the triangle inequality, we have

d(c(t), c(t

)) ≤

i=1

d(c(t

i−1

), c(t

)) + d(t

, t

) = |t − t

We now have to establish a coarse lower bound on d(c(t), c(t

)).

We may wlog assume t = a and t

= b. We need to show that

d(c(a), c(b)) ≥

|b − a| − ε.

We divide

up into regular subintervals [

, x

i+1

], and choose

close to

. The

goal is then to prove that the x

appear in order along [c(a), c(b)].

Let

+ 2δ > 6δ.

Let

b − a

for 0

≤ η < k

and

M ∈ N

. Put

(

) for

= 1

, . . . , M

and let

be a closest point on [

(

)

, c

(

)] to

. By the lemma, we know

d(x

, x

) ≤ 2δ.

Claim. x

, . . . , x

appear in the correct order along [c(a), c(b)].

Let’s finish the proof assuming the claim. If this holds, then note that

d(x

, x

i+1

) ≥ k

− 4δ > 2δ

because we know d(x

, x

i+1

) = 6δ, and also d(x

, c(b)) ≥ η − 2δ. Therefore

d(c(a), c(b)) =

i=1

d(x

, x

i−1

) + d(x

, c(b)) ≥ 2δM + η − 2δ ≥ 2δ(M − 1).

On the other hand, we have

M =

|b − a| − η

≥

|b − a|

− 1.

Thus, we find that

d(c(a), c(b)) ≥

2δ

|b − a| − 4δ.

To prove the claim, let x

= c(t

) for all i. We let

y = c(t

i−1

+ 2δ)

z = c(t

i+1

− 2δ).

Define

∆

−

= ∆(x

i−1

, x

i−1

, y)

∆

= ∆(x

i+1

, x

i+1

, z).

Both ∆

−

and ∆

are disjoint from

(

). Indeed, if

w ∈

∆

−

with

(

, w

)

≤

, then by

-slimness of ∆

−

, we know

(

w, x

i−1

)

≤

, and so

(

, x

i−1

)

≤

which is not possible.

Therefore, since the rectangle

y, z, x

i+1

, x

i−1

is 2

-slim, and

is more than

away from the sides

i−1

and

i+1

. So there must be some

∈

[

i−1

, x

i+1

]

with d(x

, x

) ≤ 2δ.

Now consider ∆ = ∆(

, x

). We know

and

are both of length

≤

. Note that every point in this triangle is within 3

-slimness. So

∆

⊆ B

(

), and this implies ∆ is disjoint from

(

i−1

) and

(

i+1

)

as before.

But

i−1

∈ B

(

i−1

) and

i+1

∈ B

(

i+1

). Moreover, ∆ contains the

segment of [

(

)

, c

(

)] joining

and

. Therefore, it must be the case that

∈ [x

i−1

, x

i+1