1Quasihomomorphisms
IV Bounded Cohomology
1.1 Quasihomomorphisms
In this chapter,
A
will denote
Z
or
R
. Let
G
be a group. The usual definition of
a group homomorphism f : G → A requires that for all x, y ∈ G, we have
f(xy) = f(x) + f(y).
In a quasihomomorphism, we replace the equality with a weaker notion, and
allow for some “errors”.
Definition
(Quasihomomorphism)
.
Let
G
be a group. A function
f : G → A
is a quasihomomorphism if the function
df : G × G → A
(x, y) 7→ f(x) + f(y) − f(xy)
is bounded. We define the defect of f to be
D(f) = sup
x,y∈G
df(x, y).
We write QH(G, A) for the Amodule of quasihomomorphisms.
Example.
Every homomorphism is a quasihomomorphism with
D
(
f
) = 0.
Conversely, a quasihomomorphism with D(f) = 0 is a homomorphism.
We can obtain some “trivial” quasihomomorphisms as follows — we take
any homomorphism, and then edit finitely many values of the homomorphism.
Then this is a quasihomomorphism. More generally, we can add any bounded
function to a quasihomomorphism and still get a quasihomomorphism.
Notation. We write
`
∞
(G, A) = {f : G → A : f is bounded}.
Thus, we are largely interested in the quasihomomorphisms modulo
`
∞
(
G, A
).
Often, we also want to quotient out by the genuine homomorphisms, and obtain
QH(G, A)
`
∞
(G, A) + Hom(G, A)
.
This contains subtle algebraic and geometric information about G, and we will
later see this is related to the second bounded cohomology H
2
b
(G, A).
We first prove a few elementary facts about quasihomomorphisms. The
first task is to find canonical representatives of the classes in the quotient
QH(G, R)/`
∞
(G, R).
Definition
(Homogeneous function)
.
A function
f : G → R
is homogeneous if
for all n ∈ Z and g ∈ G, we have f(g
n
) = nf(g).
Lemma. Let f ∈ QH(G, A). Then for every g ∈ G, the limit
Hf(g) = lim
n→∞
f(g
n
)
n
exists in R. Moreover,
(i) Hf : G → R is a homogeneous quasihomomorphism.
(ii) f − Hf ∈ `
∞
(G, R).
Proof. We iterate the quasihomomorphism property
f(ab) − f (a) − f(b) ≤ D(f).
Then, viewing g
mn
= g
m
· · · g
m
, we obtain
f(g
mn
) − nf (g
m
) ≤ (n − 1)D(f).
Similarly, we also have
f(g
mn
) − mf (g
n
) ≤ (m − 1)D(f).
Thus, dividing by nm, we find
f(g
mn
)
nm
−
f(g
m
)
m
≤
1
m
D(f)
f(g
mn
)
nm
−
f(g
n
)
n
≤
1
n
D(f).
So we find that
f(g
n
)
n
−
f(g
m
)
m
≤
1
m
+
1
n
D(f). (∗)
Hence the sequence
f(g
n
)
n
is Cauchy, and the limit exists.
The fact that
Hf
is a quasihomomorphism follows from the second assertion.
To prove the second assertion, we can just take
n
= 1 in (
∗
) and take
m → ∞
.
Then we find
f(g) − Hf(g) ≤ D(f).
So this shows that f − Hf is bounded, hence Hf is a quasihomomorphism.
The homogeneity is left as an easy exercise.
Notation.
We write
QH
h
(
G, R
) for the vector space of homogeneous quasi
homomorphisms G → R.
Then the above theorem gives
Corollary. We have
QH(G, R) = QH
h
(G, R) ⊕ `
∞
(G, R)
Proof.
Indeed, observe that a bounded homogeneous quasihomomorphism must
be identically zero.
Thus, if we want to study
QH
(
G, R
), it suffices to just look at the ho
mogeneous quasihomomorphisms. It turns out these have some very nice
perhapsunexpected properties.
Lemma. Let f : G → R be a homogeneous quasihomomorphism.
(i) We have f(xyx
−1
) = f(y) for all x, y ∈ G.
(ii) If G is abelian, then f is in fact a homomorphism. Thus
QH
h
(G, R) = Hom(G, R).
Thus, quasihomomorphisms are only interesting for nonabelian groups.
Proof.
(i) Note that for any x, the function
y 7→ f(xyx
−1
)
is a homogeneous quasihomomorphism. It suffices to show that the
function
y 7→ f(xyx
−1
) − f (y)
is a bounded homogeneous quasihomomorphism, since all such functions
must be zero. Homogeneity is clear, and the quasihomomorphism property
follows from the computation
f(xyx
−1
) − f (y) ≤ f(x) + f(y) + f(x
−1
) − f (y) + 2D(f) = 2D(f ),
using the fact that f (x
−1
) = −f(x) by homogeneity.
(ii)
If
x
and
y
commute, then (
xy
)
n
=
x
n
y
n
. So we can use homogeneity to
write
f(xy) − f(x) − f(y) =
1
n
f((xy)
n
) − f (x
n
) − f (y
n
)
=
1
n
f(x
n
y
n
) − f (x
n
) − f (y
n
)
≤
1
n
D(f).
Since n is arbitrary, the difference must vanish.
The case of
QH
(
G, Z
)
/`
∞
(
G, Z
) is more complicated. For example, we have
the following nice result:
Example. Given α ∈ R, define the map g
α
: Z → Z by
g
α
(m) = [mα].
Then this is a homomorphism, and one can check that the map
R −→
QH(Z, Z)
`
∞
(Z, Z)
α 7−→ g
α
is an isomorphism. This gives a further isomorphism
R/Z
∼
=
QH(Z, Z)
`
∞
(Z, Z) + Hom(Z, Z)
.
We next turn to the case
G
=
F
2
, the free group on two generators
a, b
. We
will try to work out explicitly a lot of nontrivial elements of
QH
(
F
2
, R
). In
general, when we try to construct quasihomomorphisms, what we manage to
get are not homogeneous. So when we construct several quasihomomorphisms,
it takes some nontrivial work to show that they are distinct. Our construction
will be one such that this is relatively easy to see.
Consider the vector space:
`
∞
odd
(Z) = {α: Z → R : α bounded and α(−n) = −α(n)}.
Note that in particular, we have α(0) = 0.
Given
α, β ∈ `
∞
odd
(
Z
), we define a quasihomomorphisms
f
α,β
: F
2
→ R
as
follows — given a reduced word w = a
n
1
b
m
1
· · · a
n
k
b
m
k
, we let
f
α,β
(w) =
k
X
i=1
α(n
i
) +
k
X
j=1
β(m
j
).
Allowing for
n
1
= 0 or
m
k
= 0, this gives a welldefined function
f
α,β
defined on
all of F
2
.
Let’s see what this does on some special sequences.
Example. We have
f
α,β
(a
n
) = α(n), f
α,β
(b
m
) = β(m),
and these are bounded functions of n, m.
So we see that f
α,β
is never homogeneous unless α = β = 0.
Example. Pick k
1
, k
2
, n 6= 0, and set
w = a
nk
1
b
nk
2
(b
k
2
a
k
1
)
−n
= a
nk
1
b
nk
2
a
−k
1
b
−k
2
· · · a
−k
1
b
−k
2

{z }
n times
.
This is now in reduced form. So we have
f
α,β
(w) = α(nk
1
) + β(nk
2
) − nα(k
1
) − nβ(k
2
).
This example is important. If
α
(
k
1
) +
β
(
k
2
)
6
= 0, then this is an unbounded
function as
n → ∞
. However, we know any genuine homomorphisms
f : F
2
→ R
must factor through the abelianization, and
w
vanishes in the abelianization. So
this suggests our
f
α,β
is in some sense very far away from being a homomorphism.
Theorem
(P. Rolli, 2009)
.
The function
f
α,β
is a quasihomomorphism, and
the map
`
∞
odd
(Z) ⊕ `
∞
odd
(Z) →
QH(F
2
, R)
`
∞
(F
2
, R) + Hom(F
2
, R)
is injective.
This tells us there are a lot of nontrivial elements in QH(F
2
, R).
The advantage of this construction is that the map above is a linear map.
So to see it is injective, it suffices to see that it has trivial kernel.
Proof.
Let
α, β ∈ `
∞
odd
(
Z, R
), and define
f
α,β
as before. By staring at it long
enough, we find that
f(xy) − f(x) − f(y) ≤ 3 max(kαk
∞
, kβk
∞
),
and so it is a quasihomomorphism. The main idea is that
f(b
n
) + f (b
−n
) = f(a
n
) + f (a
−n
) = 0
by oddness of
α
and
β
. So when we do the word reduction in the product, the
amount of error we can introduce is at most 3 max(kαk
∞
, kβk
∞
).
To show that the map is injective, suppose
f
α,β
= ϕ + h,
where
ϕ: F
2
→ R
is bounded and
h: F
2
→ R
is a homomorphism. Then we
must have
h(a
`
) = f(a
`
) − ϕ(a
`
) = α(`) − ψ(a
`
),
which is bounded. So the map
` 7→ h
(
a
`
) =
`h
(
a
) is bounded, and so
h
(
a
) = 0.
Similarly, h(b) = 0. So h ≡ 0. In other words, f
α,β
is bounded.
Finally,
f((a
`
1
b
`
2
)
k
) = k(α(`
1
) + β(`
2
)) = 0.
Since this is bounded, we must have
α
(
`
1
) +
β
(
`
2
) = 0 for all
`
1
, `
2
6
= 0. Using
the fact that
α
and
β
are odd, this easily implies that
α
(
`
1
) =
β
(
`
2
) = 0 for all
`
1
and `
2
.
More generally, we have the following theorem, which we shall not prove, or
even explain what the words mean:
Theorem (Hull–Osin 2013). The space
QH(G, R)
`
∞
(G, R) + Hom(G, R)
is infinitedimensional if G is acylindrically hyperbolic.