1Quasi-homomorphisms

IV Bounded Cohomology

1.1 Quasi-homomorphisms
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 quasi-homomorphism, we replace the equality with a weaker notion, and
allow for some “errors”.
Definition
(Quasi-homomorphism)
.
Let
G
be a group. A function
f : G A
is a quasi-homomorphism 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,yG
|df(x, y)|.
We write QH(G, A) for the A-module of quasi-homomorphisms.
Example.
Every homomorphism is a quasi-homomorphism with
D
(
f
) = 0.
Conversely, a quasi-homomorphism with D(f) = 0 is a homomorphism.
We can obtain some “trivial” quasi-homomorphisms as follows we take
any homomorphism, and then edit finitely many values of the homomorphism.
Then this is a quasi-homomorphism. More generally, we can add any bounded
function to a quasi-homomorphism and still get a quasi-homomorphism.
Notation. We write
`
(G, A) = {f : G A : f is bounded}.
Thus, we are largely interested in the quasi-homomorphisms 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 quasi-homomorphisms. 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 quasi-homomorphism.
(ii) f Hf `
(G, R).
Proof. We iterate the quasi-homomorphism 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 quasi-homomorphism 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 quasi-homomorphism.
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 quasi-homomorphism must
be identically zero.
Thus, if we want to study
QH
(
G, R
), it suffices to just look at the ho-
mogeneous quasi-homomorphisms. It turns out these have some very nice
perhaps-unexpected properties.
Lemma. Let f : G R be a homogeneous quasi-homomorphism.
(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, quasi-homomorphisms are only interesting for non-abelian groups.
Proof.
(i) Note that for any x, the function
y 7→ f(xyx
1
)
is a homogeneous quasi-homomorphism. It suffices to show that the
function
y 7→ f(xyx
1
) f (y)
is a bounded homogeneous quasi-homomorphism, since all such functions
must be zero. Homogeneity is clear, and the quasi-homomorphism 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) = [].
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 non-trivial elements of
QH
(
F
2
, R
). In
general, when we try to construct quasi-homomorphisms, what we manage to
get are not homogeneous. So when we construct several quasi-homomorphisms,
it takes some non-trivial 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 quasi-homomorphisms
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 well-defined 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
) (k
1
) (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 quasi-homomorphism, 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 non-trivial 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 quasi-homomorphism. 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 infinite-dimensional if G is acylindrically hyperbolic.