2Group cohomology and bounded cohomology

IV Bounded Cohomology

2.1 Group cohomology

We can start talking about cohomology. Before doing bounded cohomology, we

first try to understand usual group cohomology. In this section,

A

will be any

abelian group. Ultimately, we are interested in the case

A

=

Z

or

R

, but we can

develop the theory in this generality.

The general idea is that to a group Γ, we are going to associate a sequence

of abelian groups H

k

(Γ, A) that is

– covariant in A; and

– contravariant in Γ.

It is true, but we will not prove or use, that if

X

=

K

(Γ

,

1), i.e.

X

is a CW-

complex whose fundamental group is Γ and has a contractible universal cover,

then there is a natural isomorphism

H

k

(Γ, A)

∼

=

H

k

sing

(X, A).

There are several ways we can define group cohomology. A rather powerful

way of doing so is via the theory of derived functors. However, developing the

machinery requires considerable effort, and to avoid scaring people off, we will

use a more down-to-earth construction. We begin with the following definition:

Definition

(Homogeneous

k

-cochain)

.

A homogeneous

k

-cochain with values

in A is a function f : Γ

k+1

→ A. The set C(Γ

k+1

, A) is an abelian group and Γ

acts on it by automorphisms in the following way:

(γ

∗

f)(γ

0

, · · · , γ

m

) = f(γ

−1

γ

0

, · · · , γ

−1

γ

k

).

By convention, we set C(Γ

0

, A)

∼

=

A.

Definition

(Differential

d

(k)

)

.

We define the differential

d

(k)

: C

(Γ

k

, A

)

→

C(Γ

k+1

, A) by

(d

(k)

f)(γ

0

, · · · , γ

k

) =

k

X

j=0

(−1)

j

f(γ

0

, · · · , ˆγ

j

, · · · , γ

k

).

In particular, we set d

(0)

(a) to be the function that is constantly a.

Example. We have

d

(1)

f(γ

0

, γ

1

) = f(γ

1

) − f(γ

0

)

d

(2)

f(γ

0

, γ

1

, γ

2

) = f(γ

1

, γ

2

) − f(γ

0

, γ

2

) + f(γ

0

, γ

1

).

Thus, we obtain a complex of abelian groups

0 A C(Γ, A) C(Γ

2

, A) · · ·

d

(0)

d

(1)

d

(2)

.

The following are crucial properties of this complex.

Lemma.

(i) d

(k)

is a Γ-equivariant group homomorphism.

(ii) d

(k+1)

◦ d

(k)

= 0. So im d

(k)

⊆ ker d

(k+1)

.

(iii) In fact, we have im d

(k)

= ker d

(k+1)

.

Proof.

(i) This is clear.

(ii) You just expand it out and see it is zero.

(iii) If f ∈ ker d

(k)

, then setting γ

k

= e, we have

0 = d

(k)

f(γ

0

, · · · , γ

k−1

, e) = (−1)

k

f(γ

0

, · · · , γ

k−1

)

+

k−1

X

j=0

(−1)

j

f(γ

0

, · · · , ˆγ

j

, · · · , γ

k−1

, e).

Now define the following (k − 1)-cochain

h(γ

0

, · · · , γ

k−2

) = (−1)

k

f(γ

0

, · · · , γ

k−2

, e).

Then the above reads

f = d

(k−1)

h.

We make the following definitions:

Definition (k-cocycle and k-coboundaries).

– The k-cocycles are ker d

(k+1)

.

– The k-coboundaries are im d

(k)

.

So far, every cocycle is a coboundary, so nothing interesting is happening.

To obtain interesting things, we use the action of Γ on C(Γ

k

, A). We denote

C(Γ

k

, A)

Γ

= {f : Γ

k

→ A | f is Γ-invariant}.

Since the differentials

d

(k)

commute with the Γ-action, it restricts to a map

C(Γ

k

, A)

Γ

→ C(Γ

k+1

, A)

Γ

. We can arrange these into a new complex

0 A C(Γ, A)

Γ

C(Γ

2

, A)

Γ

· · ·

0 A C(Γ, A) C(Γ

2

, A) · · ·

d

(0)

d

(1)

d

(2)

d

(0)

d

(1)

d

(2)

.

We are now in a position to define group cohomology.

Definition

(Group cohomology

H

k

(Γ

, A

))

.

We define the

k

th cohomology group

to be

H

k

=

(ker d

(k+1)

)

Γ

d

(k)

(C(Γ

k

, A)

Γ

)

=

(d

(k)

(C(Γ

k

, A)))

Γ

d

(k)

(C(Γ

k

, A)

Γ

)

.

Before we do anything with group cohomology, we provide a slightly different

description of group cohomology, using inhomogeneous cochains. The idea is to

find a concrete description of all invariant cochains.

Observe that if we have a function

f :

Γ

k+1

→ A

that is invariant under the

action of Γ, then it is uniquely determined by the value on

{

(

e, γ

1

, · · · , γ

k

) :

γ

i

∈

Γ

}

. So we can identify invariant functions

f :

Γ

k+1

→ A

with arbitrary functions

Γ

k

→ A

. So we have one variable less to worry about, but on the other hand,

the coboundary maps are much more complicated.

More explicitly, we construct an isomorphism

C(Γ

k

, A)

Γ

C(Γ

k−1

, A)

ρ

(k−1)

τ

(k)

,

by setting

(ρ

(k−1)

f)(g

1

, · · · , g

k−1

) = f(e, g

1

, g

2

, · · · , g

1

· · · g

k−1

)

(τ

(k)

h)(g

1

, · · · , g

k

) = h(g

−1

1

g

2

, g

−1

2

g

3

, · · · , g

−1

k−1

g

k

).

These homomorphisms are inverses of each other. Then under this identifica-

tion, we obtain a new complex

C(Γ

k

, A)

Γ

C(Γ

k+1

, A)

Γ

C(Γ

k−1

, A) C(Γ

k

, A)

d

(k)

ρ

(k)

τ

(k)

d

k

where

d

k

= ρ

k

◦ d

(k)

◦ τ

k

.

A computation shows that

(d

k

f)(g

1

, · · · , g

k

) = f(g

2

, · · · , g

k

) +

k−1

X

j=1

(−1)

j

f(g

1

, · · · , g

j

g

j+1

, · · · , g

k

)

+(−1)

k

f(g

1

, · · · , g

k−1

).

It is customary to denote

Z

k

(Γ, A) = ker

d+1

⊆ C(Γ

k

, A)

B

k

(Γ, A) = im d

k

⊆ C(Γ

k

, A),

the inhomogeneous

k

-cocycles and inhomogeneous

k

-coboundaries. Then we

simply have

H

k

(Γ, A) =

Z

k

(Γ, A)

B

k

(Γ, A)

.

It is an exercise to prove the following:

Lemma.

A homomorphism

f

: Γ

→

Γ

0

of groups induces a natural map

f

∗

:

H

k

(Γ

0

, Z

)

→ H

k

(Γ

, Z

) for all

k

. Moreover, if

g

: Γ

0

→

Γ

00

is another group

homomorphism, then f

∗

◦ g

∗

= (gf)

∗

.

Computation in degrees k = 0, 1, 2

It is instructive to compute explicitly what these groups mean in low degrees.

We begin with the boring one:

Proposition. H

0

(Γ, A)

∼

=

A.

Proof. The relevant part of the cochain is

0 A C(Γ, A)

d

1

=0

.

The k = 1 case is not too much more interesting.

Proposition. H

1

(Γ, A) = Hom(Γ, A).

Proof. The relevant part of the complex is

A C(Γ, A) C(Γ

2

, A)

d

1

=0 d

2

,

and we have

(d

2

f)(γ

1

, γ

2

) = f(γ

1

) − f(γ

1

γ

2

) + f(γ

2

).

The k = 2 part is more interesting. The relevant part of the complex is

C(Γ, A) C(Γ

2

, A) C(Γ

3

, A)

d

2

d

3

.

Here d

3

is given by

d

3

α(g

1

, g

2

, g

3

) = α(g

2

, g

3

) − α(g

1

g

2

, g

3

) + α(g

1

, g

2

g

3

) − α(g

1

, g

2

).

Suppose that

d

3

α

(

g

1

, g

2

, g

3

) = 0, and, in addition, by some magic, we managed

to pick

α

such that

α

(

g

1

, e

) =

α

(

e, g

2

) = 0. This is known as a normalized

cocycle. We can now define the following operation on Γ × A:

(γ

1

, a

2

)(γ

2

, a

2

) = (γ

1

γ

2

, a

1

+ a

2

+ α(γ

1

, γ

2

)).

Then the property that

α

is a normalized cocycle is equivalent to the assertion

that this is an associative group law with identity element (

e,

0). We will write

this group as Γ ×

α

A.

We can think of this as a generalized version of the semi-direct product. This

group here has a special property. We can organize it into an exact sequence

0 A Γ ×

α

A Γ 0 .

Moreover, the image of

A

is in the center of Γ

×

α

A

. This is known as a central

extension.

Definition

(Central extension)

.

Let

A

be an abelian group, and Γ a group.

Then a central extension of Γ by A is an exact sequence

0 A

˜

Γ Γ 0

such that the image of A is contained in the center of

˜

Γ.

The claim is now that

Proposition. H

2

(Γ

, A

) parametrizes the set of isomorphism classes of central

extensions of Γ by A.

Proof sketch. Consider a central extension

0 A G Γ 0

i

p

.

Arbitrarily choose a section

s:

Γ

→ G

of

p

, as a function of sets. Then we know

there is a unique α(γ

1

, γ

2

) such that

s(γ

1

γ

2

)α(γ

1

, γ

2

) = s(γ

1

)s(γ

2

).

We then check that α is a (normalized) 2-cocycle, i.e. α(γ

1

, e) = γ(e, γ

2

) = 0.

One then verifies that different choices of

s

give cohomologous choices of

α

,

i.e. they represent the same class in H

2

(Γ, A).

Conversely, given a 2-cocycle

β

, we can show that it is cohomologous to a

normalized 2-cocycle

α

. This gives rise to a central extension

G

= Γ

×

α

A

as

constructed before (and also a canonical section s(γ) = (γ, 0)).

One then checks this is a bijection.

Exercise. H

2

(Γ

, A

) has a natural structure as an abelian group. Then by the

proposition, we should be able to “add” two central extensions. Figure out what

this means.

Example. As usual, write F

r

for the free group on r generators. Then

H

k

(F

r

, A) =

A k = 0

A

r

k = 1

0 k = 2

.

The fact that

H

2

(

F

r

, A

) vanishes is due to the fact that

F

r

is free, so every short

exact sequence splits.

Example. Consider Γ

g

= π

1

(S

g

) for g > 0. Explicitly, we can write

Γ

g

=

(

a

1

, b

1

, · · · , a

g

, b

g

:

g

Y

i=1

[a

i

, b

i

] = e

)

Then we have H

1

(Γ

g

, Z) = Z

2g

and H

2

(Γ

g

, Z)

∼

=

Z.

We can provide a very explicit isomorphism for H

2

(Γ

g

, Z). We let

0 Z G Γ 0

i

p

be a central extension. Observe that whenever

γ, η ∈

Γ

g

, and

˜γ, ˜η ∈ G

are lifts,

then [

˜γ, ˜η

] is a lift of [

γ, η

] and doesn’t depend on the choice of

˜γ

and

˜η

. Thus,

we can pick ˜a

1

,

˜

b

1

, · · · , ˜a

g

,

˜

b

g

. Then notice that

g

Y

i=1

[˜a

i

,

˜

b

i

]

is in the kernel of p, and is hence in Z.

Alternatively, we can compute the group cohomology using topology. We

notice that

R

2

is the universal cover of

S

g

, and it is contractible. So we know

S

g

=

K

(Γ

g

,

1). Hence, by the remark at the beginning of the section (which

we did not prove), it follows that

H

1

(Γ

g

, Z

)

∼

=

H

1

sing

(

S

g

;

Z

), and the latter is a

standard computation in algebraic topology.

Finally, we look at actions on a circle. Recall that we previously had the

central extension

0 Z Homeo

+

Z

(R) Homeo

+

(S

1

) 0

i

p

.

This corresponds to the Euler class e ∈ H

2

(Homeo

+

(S

1

), Z).

We can in fact construct a representative cocycle of

e

. To do so, we pick

a section

s: Homeo

+

(

S

1

)

→ Homeo

+

Z

(

R

) by sending

f ∈ Homeo

+

(

S

1

) to the

unique lift

¯

f : R → R such that

¯

f(0) ∈ [0, 1).

Then we find that

s(f

1

, f

2

)T

c(f

1

,f

2

)

= s(f

1

)s(f

2

)

for some c(f

1

, f

2

) ∈ Z.

Lemma. We have c(f

1

, f

2

) ∈ {0, 1}.

Proof. We have f

1

f

2

(0) ∈ [0, 1), while

¯

f

2

(0) ∈ [0, 1). So we find that

¯

f

1

(

¯

f

2

(0)) ∈ [

¯

f

1

(0),

¯

f

1

(1)) = [

¯

f

1

(0),

¯

f

1

(0) + 1) ⊆ [0, 2).

But we also know that c(f

1

, f

2

) is an integer. So c(f

1

, f

2

) ∈ {0, 1}.

Definition

(Euler class)

.

The Euler class of the Γ-action by orientation-

preserving homeomorphisms of S

1

is

h

∗

(e) ∈ H

2

(Γ, Z),

where h: Γ → Homeo

+

(S

1

) is the map defining the action.

For example, if Γ

g

is a surface group, then we obtain an invariant of actions

valued in Z.

There are some interesting theorems about this Euler class that we will not

prove.

Theorem (Milnor–Wood). If h : Γ

g

→ Homeo

+

(S

1

), then |h

∗

(e)| ≤ 2g − 2.

Theorem

(Gauss–Bonnet)

.

If

h:

Γ

g

→ PSL

(2

, R

)

⊆ Homeo

+

(

S

1

) is the holon-

omy representation of a hyperbolic structure, then

h

∗

(e) = ±(2g − 2).

Theorem

(Matsumoko, 1986)

.

If

h

defines a minimal action of Γ

g

on

S

1

and

|h

∗

(e)| = 2g − 2, then h is conjugate to a hyperbolization.