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
).