4Hyperbolic groups
IV Topics in Geometric Group Theory
4.2 Quasigeodesics and hyperbolicity
Definition
(Quasigeodesic)
.
A (
λ, ε
)quasigeodesic for
λ ≥
1
, ε ≥
0 is a
(λ, ε)quasiisometric embedding I → X, where I ⊆ R is a closed interval.
Note that our definition allows for
I
to be unbounded, i.e.
I
may be of the
forms [
a, b
], [0
, ∞
) or
R
respectively. We call these quasigeodesic intervals,
quasigeodesic rays and quasigeodesic lines respectively.
Example. The map [0, ∞) → R
2
given in polar coordinates by
t 7→ (t, log(1 + t))
is a quasigeodesic ray.
This should be alarming. The quasigeodesics in the Euclidean plane can
look unlike any genuine geodesic. However, it turns out things are wellbehaved
in hyperbolic spaces.
Theorem
(Morse lemma)
.
For all
δ ≥
0
, λ ≥
1 there is
R
(
δ, λ, ε
) such that the
following holds:
If
X
is a
δ
hyperbolic metric space, and
c
: [
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
d
Haus
([p, q], im(c)) ≤ R(δ, λ, ε),
where
d
Haus
(A, B) = inf{ε > 0  A ⊆ N
ε
(B) and B ⊆ N
ε
(A)}
is the Hausdorff distance.
This has nothing to do with the Morse lemma in differential topology.
Corollary.
There is an
M
(
δ, λ, ε
) such that a geodesic metric space
X
is
δ

hyperbolic iff any (λ, ε)quasigeodesic triangle is M slim.
Corollary.
Suppose
X, X
0
are geodesic metric spaces, and
f
:
X → X
00
is a
quasiisometric embedding. If X
0
is hyperbolic, then so is X.
In particular, hyperbolicity is a quasiisometrically 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 finitelygenerated 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
c
: [
a, b
]
→ X
be a continuous path. Let
a = t
0
< t
1
< · · · < t
n
= b be a dissection D of [a, b]. Define
`(c) = sup
D
n
X
i=1
d(c(t
i−1
), c(t
i
)).
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
X
be a geodesic space. For any (
λ, ε
)quasigeodesic
c
: [
a, b
]
→ X
,
there exists a continuous, rectifiable (
λ, ε
0
)quasigeodesic
c
0
: [
a, b
]
→ X
with
ε
0
= 2(λ + ε) such that
(i) c
0
(a) = c(a), c
0
(b) = c(b).
(ii) For all a ≤ t < t
0
≤ b, we have
`(c
0

[t,t
0
]
) ≤ k
1
d(c
0
(t), c
0
(t
0
)) + k
2
where k
1
= λ(λ + ε) and k
2
= (λε
0
+ 3)(λ + 3).
(iii) d
Haus
(im c, im c
0
) ≤ λ + ε.
Proof sketch.
Let Σ =
{a, b} ∪
((
a, b
)
∩ Z
). For
t ∈
Σ, we let
c
0
(
t
) =
c
(
t
), and
define
c
0
to be geodesic between the points of Σ. Then claims (i) and (iii) are clear,
and to prove quasigeodesicity and (ii), let
σ
: [
a, b
]
→
Σ be a choice of closest
point in Σ, and then estimate d(c
0
(t), c
0
(t
0
)) in terms of d(c(σ(t)), c(σ(t
0
))).
Lemma.
let
X
be
δ
hyperbolic, and
c
: [
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
2
`(c) + 1.
Proof.
We may assume
c
: [0
,
1]
→ X
is parametrized proportional to arc length.
Suppose
`(c)
2
N
< 1 ≤
`(c)
2
N−1
.
Let
x
0
=
x
. Pick a geodesic triangle between
p, q, c
(
1
2
). By
δ
hyperbolicity, there
exists a point
x
1
lying on the other two edges such that
d
(
x
0
, x
1
)
≤ δ
. We wlog
assume x
1
∈ [p, c(
1
2
)]. We can repeat the argument with c
[0,
1
2
]
.
p q
c(
1
2
)
x
0
x
1
Formally, we proceed by induction on
N
. If
N
= 0 so that
`
(
c
)
<
1, then we
are done by taking desired point on
im c
to be
p
(or
q
). Otherwise, there is some
x
1
∈ [p, c(
1
2
)] such that d(x
0
, x
1
) ≤ δ. Then
`(c
[0,
1
2
]
)
2
N−1
< 1 ≤
`(c
[0,
1
2
])
2
N−2
.
So by the induction hypothesis,
d(x
1
, im c) ≤ δ log
2
`(c
[0,
1
2
]
 + 1
= δ
1
2
log
2
`(c)
+ 1
= δ(log
2
`(c) − 1) + 1.
Note that we used the fact that `(c) > 1, so that log
2
`(c) > 0.
Then we are done since
d(x, im c) ≤ d(x, x
1
) + d(x
1
, im c).
Proof of Morse lemma.
By the first lemma, we may assume that
c
is continuous
and rectifiable, and satisfies the properties as in the lemma.
Let
p, q
be the end points of
c
, 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
x
0
∈
[
p, q
] where the supremum is attained.
Then by assumption,
im c
lies outside
˚
B
(
x
0
, D
). Choose
y, z ∈
[
p, q
] be such that
d
(
x
0
, y
) =
d
(
x
0
, z
) = 2
D
and
y, x
0
, z
appear on the geodesic in this order (take
y = p, or z = q if that is not possible).
Let
y
0
=
c
(
s
)
∈ im c
be such that
d
(
y
0
, y
)
≤ D
, and similarly let
z
0
=
c
(
t
)
∈
im c be such that d(z, z
0
) ≤ D.
p q
c
x
0
D
y z
2D 2D
y
0
z
0
≤ D ≤ D
Let γ = [y, y
0
] · c
[s,t]
· [z
0
, z]. Then
`(γ) = d(y, y
0
) + d(z, z
0
) + `(c
[s,t]
) ≤ D + D + k
1
d(y
0
, z
0
) + k
2
,
by assumption. Also, we know that d(y
0
, z
0
) ≤ 6D. So we have
`(γ) ≤ 6k
1
D + 2D + k
2
.
But we know that
d(x
0
, im γ) = D.
So the second lemma tells us
D ≤ δ log
2
(6k
1
D + 2D + k
2
) + 1.
The left hand side is linear in
D
, and the right hand side is a logarithm in
D
.
So it must be the case that
D
is bounded. Hence [
p, q
]
⊆ N
D
0
(
im c
), where
D
0
is some constant.
It remains to find a bound
M
such that
im c ⊆ N
M
([
p, q
]). Let [
a
0
, b
0
] be a
maximal subinterval of [
a, b
] such that
c
[
a
0
, b
0
] lies entirely outside
˚
N
D
0
([
p, q
]).
Since
¯
N
D
(
c
[
a, a
0
]) and
¯
N
D
(
c
[
b
0
, b
]) are both closed, and they collectively cover
the connected set [p, q], there exists
w ∈ [p, q] ∩
¯
N
D
0
(c[a, a
0
]) ∩
¯
N
D
0
(c[b
0
, b]).
Therefore there exists
t ∈
[
a, a
0
] and
t
0
∈
[
b
0
, b
] such that
d
(
w, c
(
t
))
≤ D
0
and
d(w, c(t
0
)) ≤ D
0
. In particular, d(c(t), c(t
0
)) ≤ 2D
0
.
By the first lemma, we know
`(c
[t,t
0
]
) ≤ 2k
1
D
0
+ k
2
.
So we know that for s ∈ [a
0
, b
0
], we have
d(c(s), [p, q]) ≤ d(c(s), w)
≤ d(c(s), c(t)) + d(c(t), w)
≤ `(c
[t,t
0
]
) + D
0
≤ D
0
+ 2k
1
D
0
+ k
2
.
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.
Finitelygenerated free groups are hyperbolic, since their Cayley
graphs are trees.
Example.
Surface groups Σ
g
for
g ≥
2 acts on the hyperbolic plane properly
discontinuously and cocompactly by isometries (since the universal cover of the
genus
g
surface is the hyperbolic plane by tessellating the hyperbolic plane with
4ggons, cf. IB Geometry).
The following notion of “virtually” would be helpful
Definition
(virtuallyP)
.
Let P be a property of groups. Then we say
G
is
virtuallyP if there is a finite index subgroup G
0
≤ G such that G
0
is P.
Example. Finite groups are virtually trivial!
Note that
G
0
≤ G
is finiteindex, then
G
0
acts on
Cay
S
(
G
) 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 finiteindex for) and hence
G
0
'
qi
G. In particular,
Example.
A virtually hyperbolic group is hyperbolic. For example, virtually
free groups such that (Z/2) ∗ (Z/3) = PSL
2
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
1
, . . . , a
m
 r
1
, . . . , r
n
i.
such that
r
i
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 groupvalued 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
quasiisomorphism, then we cannot go completely local, because locally, you are
allowed to be anything.
Definition
(
k
local geodesic)
.
Let
X
be a geodesic metric space, and
k >
0. A
path c : [a, b] → X is a klocal geodesic if
d(c(t), c(t
0
)) = t − t
0

for all t, t
0
∈ [a, b] with t − t
0
 ≤ 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
X
be
δ
hyperbolic and
c
: [
a, b
]
→ X
be a
k
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
X
be
δ
hyperbolic and
k >
8
δ
. If
c
: [
a, b
]
→ X
is a
k
local
geodesic, then im c is contained in the 2δneighbourhood of [c(a), c(b)].
Observe that by iterating the definition of hyperbolicity, on a
δ
hyperbolic
space, any point on an
n
gon is at most (
n −
2)
δ
away from a point on another
side.
Proof. Let x = c(t) maximize d(x, [c(a), c(b)]). Let
y = c
t −
k
2
, z = c
t −
k
2
.
If t −
k
2
< a, we set y = c(a) instead, and similarly for z.
Let
y
0
∈
[
c
(
a
)
, c
(
b
)] minimize
d
(
y, y
0
), and likewise let
z
0
∈
[
c
(
a
)
, c
(
b
)] minimize
d(z, z
0
).
x
y
0
z
0
y
z
w
Fix geodesics [
y, y
0
] and [
z, z
0
]. Then we have a geodesic rectangle with
vertices
y, y
0
, z, z
0
. By
δ
hyperbolicity, there exists
w
on the rectangle not on
im c, such that d(x, w) = 2δ.
If
w ∈
[
y
0
, z
0
], then we win. Otherwise, we may wlog assume
w ∈
[
y, y
0
].
Note that in the case
y
=
c
(
a
), we must have
y
=
y
0
, and so this would imply
w
=
y ∈
[
c
(
a
)
, c
(
b
)]. So we are only worried about the case
y
=
c
t −
k
2
.
So
d
(
y, x
) =
k
2
>
4
δ
. But then by the triangle inequality, we must have
d(y, w) > 2δ = d(x, w).
However,
d(x, y
0
) ≤ d(x, w) + d(w, y
0
) < d(y, w) + d(w, y
0
) = d(y, y
0
).
So it follows that
d(x, [c(a), c(b)]) < d(y, y
0
) = d(y, [c(a), c(b)]).
This contradicts our choice of x.
Proof of theorem.
Let
c
: [
a, b
]
→ X
be a
k
local geodesic, and
t ≤ t
0
∈
[
a, b
].
Choose
t
0
=
t < t
1
< · · · < t
n
< t
0
such that
t
i
=
t
i−1
+
k
for all
i
and
t
0
−t
n
< k
.
Then by definition, we have
d(c(t
i−1
), c(t
i
)) = k, d(c(t
n
), c(t
0
)) = t
n
− t
0
.
for all i. So by the triangle inequality, we have
d(c(t), c(t
0
)) ≤
n
X
i=1
d(c(t
i−1
), c(t
i
)) + d(t
n
, t
0
) = t − t
0
.
We now have to establish a coarse lower bound on d(c(t), c(t
0
)).
We may wlog assume t = a and t
0
= b. We need to show that
d(c(a), c(b)) ≥
1
λ
b − a − ε.
We divide
c
up into regular subintervals [
x
i
, x
i+1
], and choose
x
0
i
close to
x
i
. The
goal is then to prove that the x
0
i
appear in order along [c(a), c(b)].
Let
k
0
=
k
2
+ 2δ > 6δ.
Let
b − a
=
Mk
0
+
η
for 0
≤ η < k
0
and
M ∈ N
. Put
x
i
=
c
(
ik
0
) for
i
= 1
, . . . , M
,
and let
x
0
i
be a closest point on [
c
(
a
)
, c
(
b
)] to
x
i
. By the lemma, we know
d(x
i
, x
0
i
) ≤ 2δ.
Claim. x
0
1
, . . . , x
0
m
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
0
i
, x
0
i+1
) ≥ k
0
− 4δ > 2δ
because we know d(x
i
, x
i+1
) = 6δ, and also d(x
m
, c(b)) ≥ η − 2δ. Therefore
d(c(a), c(b)) =
M
X
i=1
d(x
i
, x
i−1
) + d(x
m
, c(b)) ≥ 2δM + η − 2δ ≥ 2δ(M − 1).
On the other hand, we have
M =
b − a − η
k
0
≥
b − a
k
0
− 1.
Thus, we find that
d(c(a), c(b)) ≥
2δ
k
0
b − a − 4δ.
To prove the claim, let x
i
= c(t
i
) for all i. We let
y = c(t
i−1
+ 2δ)
z = c(t
i+1
− 2δ).
Define
∆
−
= ∆(x
i−1
, x
0
i−1
, y)
∆
+
= ∆(x
i+1
, x
0
i+1
, z).
Both ∆
−
and ∆
+
are disjoint from
B
(
x
i
,
3
δ
). Indeed, if
w ∈
∆
−
with
d
(
x
i
, w
)
≤
3
δ
, then by
δ
slimness of ∆
−
, we know
d
(
w, x
i−1
)
≤
3
δ
, and so
d
(
x
i
, x
i−1
)
≤
6
δ
,
which is not possible.
Therefore, since the rectangle
y, z, x
0
i+1
, x
0
i−1
is 2
δ
slim, and
x
i
is more than
2
δ
away from the sides
yx
0
i−1
and
zx
0
i+1
. So there must be some
x
00
i
∈
[
x
0
i−1
, x
0
i+1
]
with d(x
i
, x
00
i
) ≤ 2δ.
Now consider ∆ = ∆(
x
i
, x
0
i
, x
00
i
). We know
x
i
x
0
i
and
x
i
x
00
i
are both of length
≤
2
δ
. Note that every point in this triangle is within 3
δ
of
x
i
by
δ
slimness. So
∆
⊆ B
(
x
i
,
3
δ
), and this implies ∆ is disjoint from
B
(
x
i−1
,
3
δ
) and
B
(
x
i+1
,
3
δ
)
as before.
But
x
0
i−1
∈ B
(
x
i−1
,
3
δ
) and
x
0
i+1
∈ B
(
x
i+1
,
3
δ
). Moreover, ∆ contains the
segment of [
c
(
a
)
, c
(
b
)] joining
x
0
i
and
x
00
i
. Therefore, it must be the case that
x
0
i
∈ [x
0
i−1
, x
0
i+1
].