2Group cohomology and bounded cohomology

IV Bounded Cohomology

2.2 Bounded cohomology of groups

We now move on to bounded cohomology. We will take

A

=

Z

or

R

now. The

idea is to put the word “bounded” everywhere. For example, we previously had

C(Γ

k+1

, A) denoting the functions Γ

k+1

→ A. Likewise, we denote

C

b

(Γ

k+1

, A) = {f ∈ C(Γ

k+1

, A) : f is bounded} ⊆ C(Γ

k+1

, A).

We have

d

(k)

(

C

b

(Γ

k

, A

))

⊆ C

b

(Γ

k+1

, A

), and so as before, we obtain a chain

complexes

0 A C

b

(Γ, A)

Γ

C

b

(Γ

2

, A)

Γ

· · ·

0 A C

b

(Γ, A) C

b

(Γ

2

, A) · · ·

d

(0)

d

(1)

d

(2)

d

(0)

d

(1)

d

(2)

.

This allows us to define

Definition

(Bounded cohomology)

.

The

k

-th bounded cohomology group of Γ

with coefficients in A Is

H

k

b

(Γ, A) =

ker(d

(k+1)

: C

b

(Γ

k+1

, A)

Γ

→ C

b

(Γ

k+2

, A)

Γ

)

d

(k)

(C

b

(Γ

k

, A)

Γ

)

.

This comes with two additional features.

(i)

As one would expect, a bounded cochain is bounded. So given an element

f ∈ C

b

(Γ

k+1

, A), we can define

kfk

∞

= sup

x∈Γ

k+1

|f(x)|.

Then

k · k

∞

makes

C

b

(Γ

k+1

, A

) into a normed abelian group, and in the

case A = R, a Banach space.

Then for [f] ∈ H

k

b

(Γ, A), we define

k[f]k

∞

= inf{kf + dgk

∞

: g ∈ C

b

(Γ

k

, A)

Γ

}.

This induces a semi-norm on

H

k

b

(Γ

, A

). This is called the canonical semi-

norm.

(ii) We have a map of chain complexes

C

b

(Γ, A)

Γ

C

b

(Γ

2

, A)

Γ

C

b

(Γ

3

, A)

Γ

· · ·

C(Γ, A)

Γ

C(Γ

2

, A)

Γ

C(Γ

3

, A)

Γ

· · ·

Thus, this induces a natural map

c

k

: H

k

b

(Γ

, A

)

→ H

k

(Γ

, A

), known as the

comparison map. In general, c

k

need not be injective or surjective.

As before, we can instead use the complex of inhomogeneous cochains. Then

we have a complex that looks like

0 A C

b

(Γ, A) C

b

(Γ

2

, A) · · ·

d

1

=0 d

2

d

3

In degree 0, the boundedness condition is useless, and we have

H

0

b

(Γ, A) = H

0

(Γ, A) = A.

For

k

= 1, we have

im d

1

= 0. So we just have to compute the cocycles. For

f ∈ C

b

(Γ

, A

), we have

d

2

f

= 0 iff

f

(

g

1

)

− f

(

g

1

g

2

) +

f

(

g

2

) = 0, iff

f ∈ Hom

(Γ

, A

).

But we have the additional information that

f

is bounded, and there are no

non-zero bounded homomorphisms to Γ or A! So we have

H

1

b

(Γ, A) = 0.

If we allow non-trivial coefficients, then

H

1

b

(Γ

, A

) may be always be zero. But

that’s another story.

The interesting part starts at

H

2

b

(Γ

, A

). To understand this, We are going to

determine the kernel of the comparison map

c

2

: H

2

b

(Γ, A) → H

2

(Γ, A).

We consider the relevant of the defining complexes, where we take inhomogeneous

cochains

C(Γ, A) C(Γ

2

, A) C(Γ

3

, A)

C

b

(Γ, A) C

b

(Γ

2

, A) C

b

(Γ

3

, A)

d

2

d

3

d

2

d

3

By definition, the kernel of

c

2

consists of the [

α

]

∈ H

2

b

(Γ

, A

) such that

α

=

d

2

f

for some

f ∈ C

(Γ

, A

). But

d

2

f

=

α

being bounded tells us

f

is a quasi-

homomorphism! Thus, we have a map

¯

d

2

: QH(Γ, A) ker c

2

f [d

2

f].

Proposition. The map

¯

d

2

induces an isomorphism

QH(Γ, A)

`

∞

(Γ, A) + Hom(Γ, A)

∼

=

ker c

2

.

Proof.

We know that

¯

d

2

is surjective. So it suffices to show that the kernel is

`

∞

(Γ, A) + Hom(Γ, A).

Suppose

f ∈ QH

(Γ

, A

) is such that

¯

d

2

f ∈ H

2

b

(Γ

, A

) = 0. Then there exists

some g ∈ C

b

(Γ, A) such that

d

2

f = d

2

g.

So it follows that

d

2

(

f − g

) = 0. That is,

f − g ∈ Hom

(Γ

, A

). Hence it follows

that

ker

¯

d

2

⊆ `

∞

(Γ, A) + Hom(Γ, A).

The other inclusion is clear.

Since we already know about group cohomology, the determination of the

kernel can help us compute the bounded cohomology. In certain degenerate

cases, it can help us determine it completely.

Example.

For

G

abelian and

A

=

R

, we saw that

QH

(Γ

, A

) =

`

∞

(Γ

, A

) +

Hom(Γ, A). So it follows that c

2

is injective.

Example.

For

H

2

b

(

Z, Z

), we know

H

2

(

Z, Z

) = 0 since

Z

is a free group (hence,

e.g. every extension splits, and in particular all central extensions do). Then we

know

H

2

b

(Z, Z)

∼

=

QH(Z, Z)

`

∞

(Z, Z) + Hom(Z, Z)

∼

=

R/Z.

Example.

Consider

H

2

b

(

F

r

, R

). We know that

H

2

(

F

r

, R

) = 0. So again

H

2

b

(

F

r

, R

) is given by the quasi-homomorphisms. We previously found many

such quasi-homomorphisms — by Rollis’ theorem, we have an inclusion

`

∞

odd

(Z, R) ⊕ `

∞

odd

(Z, R) H

2

b

(F

r

, R)

(α, β) [d

2

f

α,β

]

Recall that

H

2

b

(

F

r

, R

) has the structure of a semi-normed space, which we called

the canonical norm. One can show that

k[d

2

f

α,β

]k = max(kdαk

∞

, kdβk

∞

).

Returning to general theory, a natural question to ask ourselves is how the

groups

H

·

b

(Γ

, Z

) and

H

·

b

(Γ

, R

) are related. For ordinary group cohomology, if

A ≤ B

is a subgroup (we are interested in

Z ≤ R

), then we have a long exact

sequence of the form

· · · H

k−1

(Γ, B/A) H

k

(Γ, A) H

k

(Γ, B) H

k

(Γ, B/A) · · ·

β

,

where

β

is known as the Bockstein homomorphism. This long exact sequence

comes from looking at the short exact sequence of chain complexes (of inhomo-

geneous cochains)

0 C(Γ

·

, B) C(Γ

·

, A) C(Γ

·

, B/A) 0 ,

and then applying the snake lemma.

If we want to perform the analogous construction for bounded cohomology,

we might worry that we don’t know what

C

b

(Γ

·

, R/Z

) means. However, if we

stare at it long enough, we realize that we don’t have to worry about that. It

turns out the sequence

0 C

b

(Γ

·

, Z) C

b

(Γ

·

, R) C(Γ

·

, R/Z) 0

is short exact. Thus, snake lemma tells us we have a long exact sequence

· · · H

k−1

(Γ, R/Z) H

k

b

(Γ, Z) H

k

b

(Γ, R) H

k

(Γ, R/Z) · · ·

δ

.

This is known as the Gersten long exact sequence

Example. We can look at the beginning of the sequence, with

0 = H

1

b

(Γ, R) Hom(Γ, R/Z) H

2

b

(Γ, Z) H

2

b

(Γ, R)

δ

.

In the case Γ =

Z

, from our first example, we know

c

2

:

H

2

b

(

Z, R

)

→ H

2

(

Z, R

) = 0

is an injective map. So we recover the isomorphism

R/Z = Hom(Z, R/Z)

∼

=

H

2

b

(Z, Z)

we found previously by direct computation.

We’ve been talking about the kernel of

c

2

so far. In Gersten’s Bounded

cocycles and combing of groups (1992) paper, it was shown that the image of

the comparison map

c

2

: H

2

b

(Γ

, Z

)

→ H

2

(Γ

, Z

) describes central extensions with

special metric features. We shall not pursue this too far, but the theorem is as

follows:

Theorem.

Assume Γ is finitely-generated. Let

G

α

be the central extension of

Γ by

Z

, defined by a class in

H

2

(Γ

, Z

) which admits a bounded representative.

Then with any word metric, Γ

α

is quasi-isometric to Γ

× Z

via the “identity

map”.

Before we end the chapter, we produce a large class of groups for which

bounded cohomology (with real coefficients) vanish, namely amenable groups.

Definition

(Amenable group)

.

A discrete group Γ is amenable if there is a

linear form m: `

∞

(Γ, R) → R such that

– m(f) ≥ 0 if f ≥ 0;

– m(1) = 1; and

– m is left-invariant, i.e. m(γ

∗

f) = m(f), where (γ

∗

f)(x) = f(γ

−1

x).

A linear form that satisfies the first two properties is known as a mean, and

we can think of this as a way of integrating functions. Then an amenable group

is a group with a left invariant mean. Note that the first two properties imply

|m(f)| ≤ kfk

∞

.

Example.

– Abelian groups are amenable, and finite groups are.

– Subgroups of amenable groups are amenable.

– If

0 Γ

1

Γ

2

Γ

3

0

is a short exact sequence, then Γ

2

is amenable iff Γ

1

and Γ

3

are amenable.

–

Let Γ =

hSi

for

S

a finite set. Given a finite set

A ⊆

Γ, we define

∂A

to

be the set of all edges with exactly one vertex in A.

For example, Z

2

with the canonical generators has Cayley graph

Then if

A

consists of the red points, then the boundary consists of the

orange edges.

It is a theorem that a group Γ is non-amenable iff there exists a constant

c = c(S, Γ) > 0 such that for all A ⊆ Γ, we have |∂A| ≥ c|A|.

– There exists infinite, finitely generated, simple, anemable groups.

–

If Γ

⊆ GL

(

n, C

), then Γ is amenable iff it contains a finite-index subgroup

which is solvable.

– F

2

is non-amenable.

– Any non-elementary word-hyperbolic group is non-amenable.

Proposition. Let Γ be an amenable group. Then H

k

b

(Γ, R) = 0 for k ≥ 1.

The proof requires absolutely no idea.

Proof.

Let

k ≥

1 and

f :

Γ

k+1

→ R

a Γ-invariant bounded cocycle. In other

words,

d

(k+1)

f = 0

f(γγ

0

, · · · , γγ

k

) = f(γ

0

, · · · , γ

k

).

We have to find ϕ : Γ

k

→ R bounded such that

d

(k)

ϕ = f

ϕ(γγ

0

, · · · , γγ

k−1

) = ϕ(γ

0

, · · · , γ

k−1

).

Recall that for η ∈ Γ, we can define

h

η

(γ

0

, · · · , γ

k−1

) = (−1)

k+1

f(γ

0

, · · · , γ

k+1

, η),

and then

d

(k+1)

f = 0 ⇐⇒ f = d

(k)

(h

η

).

However, h

η

need not be invariant. Instead, we have

h

η

(γγ

0

, · · · , γγ

k−1

) = h

γ

−1

η

(γ

0

, · · · , γ

k−1

).

To fix this, let

m: `

∞

(Γ)

→ R

be a left-invariant mean. We notice that the map

η 7→ h

η

(γ

0

, · · · , γ

k−1

)

is bounded by kfk

∞

. So we can define

ϕ(γ

0

, · · · , γ

k−1

) = m

n

η 7→ h

η

(γ

0

, · · · , γ

k−1

)

o

.

Then this is the ϕ we want. Indeed, we have

ϕ(γγ

0

, · · · , γγ

k−1

) = m

n

η 7→ h

γ

−1

η

(γ

0

, · · · , γ

k−1

)

o

.

But this is just the mean of a left translation of the original function. So this is

just ϕ(γ

0

, · · · , γ

k−1

). Also, by properties of the mean, we know kϕk

∞

≤ kfk

∞

.

Finally, by linearity, we have

d

(k)

ϕ(γ

0

, · · · , γ

k

) = m

n

η 7→ d

(k)

h

η

(γ

0

, · · · , γ

k

)

o

= m

n

f(γ

0

, · · · , γ

k

) · 1

Γ

o

= f(γ

0

, · · · , γ

k

)m(1

Γ

)

= f(γ

0

, · · · , γ

k

).