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
k+1
, A) denoting the functions Γ
k+1
A. Likewise, we denote
C
b
k+1
, A) = {f C
k+1
, A) : f is bounded} C
k+1
, A).
We have
d
(k)
(
C
b
k
, A
))
C
b
k+1
, A
), and so as before, we obtain a chain
complexes
0 A C
b
, A)
Γ
C
b
2
, A)
Γ
· · ·
0 A C
b
, A) C
b
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
, A) =
ker(d
(k+1)
: C
b
k+1
, A)
Γ
C
b
k+2
, A)
Γ
)
d
(k)
(C
b
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
k+1
, A), we can define
kfk
= sup
xΓ
k+1
|f(x)|.
Then
k · k
makes
C
b
k+1
, A
) into a normed abelian group, and in the
case A = R, a Banach space.
Then for [f] H
k
b
, A), we define
k[f]k
= inf{kf + dgk
: g C
b
k
, A)
Γ
}.
This induces a semi-norm on
H
k
b
, A
). This is called the canonical semi-
norm.
(ii) We have a map of chain complexes
C
b
, A)
Γ
C
b
2
, A)
Γ
C
b
3
, A)
Γ
· · ·
C, A)
Γ
C
2
, A)
Γ
C
3
, A)
Γ
· · ·
Thus, this induces a natural map
c
k
: H
k
b
, A
)
H
k
, 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
, A) C
b
2
, A) · · ·
d
1
=0 d
2
d
3
In degree 0, the boundedness condition is useless, and we have
H
0
b
, A) = H
0
, A) = A.
For
k
= 1, we have
im d
1
= 0. So we just have to compute the cocycles. For
f C
b
, A
), we have
d
2
f
= 0 iff
f
(
g
1
)
f
(
g
1
g
2
) +
f
(
g
2
) = 0, iff
f Hom
, 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
, A) = 0.
If we allow non-trivial coefficients, then
H
1
b
, A
) may be always be zero. But
that’s another story.
The interesting part starts at
H
2
b
, A
). To understand this, We are going to
determine the kernel of the comparison map
c
2
: H
2
b
, A) H
2
, A).
We consider the relevant of the defining complexes, where we take inhomogeneous
cochains
C, A) C
2
, A) C
3
, A)
C
b
, A) C
b
2
, A) C
b
3
, A)
d
2
d
3
d
2
d
3
By definition, the kernel of
c
2
consists of the [
α
]
H
2
b
, A
) such that
α
=
d
2
f
for some
f C
, A
). But
d
2
f
=
α
being bounded tells us
f
is a quasi-
homomorphism! Thus, we have a map
¯
d
2
: QH, A) ker c
2
f [d
2
f].
Proposition. The map
¯
d
2
induces an isomorphism
QH, A)
`
, A) + Hom(Γ, A)
=
ker c
2
.
Proof.
We know that
¯
d
2
is surjective. So it suffices to show that the kernel is
`
, A) + Hom(Γ, A).
Suppose
f QH
, A
) is such that
¯
d
2
f H
2
b
, A
) = 0. Then there exists
some g C
b
, A) such that
d
2
f = d
2
g.
So it follows that
d
2
(
f g
) = 0. That is,
f g Hom
, A
). Hence it follows
that
ker
¯
d
2
`
, A) + Hom(Γ, A).
The other inclusion is clear.
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
, A
) =
`
, 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(kk
, kk
).
Returning to general theory, a natural question to ask ourselves is how the
groups
H
·
b
, Z
) and
H
·
b
, 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
k1
, B/A) H
k
, A) H
k
, B) H
k
, 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
·
, B) C
·
, A) C
·
, 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
·
, 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
·
, Z) C
b
·
, R) C
·
, R/Z) 0
is short exact. Thus, snake lemma tells us we have a long exact sequence
· · · H
k1
, R/Z) H
k
b
, Z) H
k
b
, R) H
k
, 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
, R) Hom(Γ, R/Z) H
2
b
, Z) H
2
b
, 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
, Z
)
H
2
, 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
, 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: `
, 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
, 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
k1
) = ϕ(γ
0
, · · · , γ
k1
).
Recall that for η Γ, we can define
h
η
(γ
0
, · · · , γ
k1
) = (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
k1
) = h
γ
1
η
(γ
0
, · · · , γ
k1
).
To fix this, let
m: `
(Γ)
R
be a left-invariant mean. We notice that the map
η 7→ h
η
(γ
0
, · · · , γ
k1
)
is bounded by kfk
. So we can define
ϕ(γ
0
, · · · , γ
k1
) = m
n
η 7→ h
η
(γ
0
, · · · , γ
k1
)
o
.
Then this is the ϕ we want. Indeed, we have
ϕ(γγ
0
, · · · , γγ
k1
) = m
n
η 7→ h
γ
1
η
(γ
0
, · · · , γ
k1
)
o
.
But this is just the mean of a left translation of the original function. So this is
just ϕ(γ
0
, · · · , γ
k1
). 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
).