3Actions on S^{1}

IV Bounded Cohomology

3.1 The bounded Euler class

We are now going going to apply the machinery of bounded cohomology to

understand actions on S

1

. Recall that the central extension

0 Z Homeo

+

Z

(R) Homeo

+

(S

1

) 0

defines the Euler class

e ∈ H

2

(

Homeo

+

(

S

1

)

, Z

). We have also shown that there

is a representative cocycle c(f, g) taking the values in {0, 1}, defined by

f ◦ g ◦ T

c(f,g)

=

¯

f ◦ ¯g,

where for any f, the map

¯

f is the unique lift to R such that

¯

f(0) ∈ [0, 1).

Since

c

takes values in

{

0

,

1

}

, in particular, it is a bounded cocycle. So we

can use it to define

Definition (Bounded Euler class). The bounded Euler class

e

b

∈ H

2

b

(Homeo

+

(S

1

), Z)

is the bounded cohomology class represented by the cocycle c.

By construction, e

b

is sent to e via the comparison map

c

2

: H

2

b

(Homeo

+

(S

1

), Z) H

2

(Homeo

+

(S

1

), Z) .

In fact, the comparison map is injective. So this

e

b

is the unique element that is

sent to

e

, and doesn’t depend on us arbitrarily choosing

c

as the representative.

Definition

(Bounded Euler class of action)

.

The bounded Euler class of an

action h: Γ → Homeo

+

(S

1

) is h

∗

(e

b

) ∈ H

2

b

(Γ, Z).

By naturality (proof as exercise),

h

∗

(

e

b

) maps to

h

∗

(

e

) under the comparison

map. The bounded Euler class is actually a rather concrete and computable

object. Note that if we have an element

ϕ ∈ Homeo

+

(

S

1

), then we obtain a

group homomorphism

Z → Homeo

+

(

S

1

) that sends 1 to

ϕ

, and vice versa. In

other words, we can identify elements of

Homeo

+

(

S

1

) with homomorphisms

h

:

Z → Homeo

+

(

S

1

). Any such homomorphism will give a bounded Euler class

h

∗

(e

b

) ∈ H

2

b

(Z, Z)

∼

=

R/Z.

Exercise.

If

h: Z → Homeo

+

(

S

1

) and

ϕ

=

h

(1), then under the isomorphism

H

2

b

(Z, Z)

∼

=

R/Z, we have h

∗

(e

b

) = Rot(ϕ), the Poincar´e rotation number of ϕ.

Thus, one way to think about the bounded Euler class is as a generalization

of the Poincar´e rotation number.

Exercise.

Assume

h:

Γ

→ Homeo

+

(

S

1

) takes values in the rotations

Rot

. Let

χ:

Γ

→ R/Z

the corresponding homomorphism. Then under the connecting

homomorphism

Hom(Γ, R/Z) H

2

b

(Γ, Z)

δ

,

we have δ(χ) = h

∗

(e

b

).

Exercise.

If

h

1

and

h

2

are conjugate in

Homeo

+

(

S

1

), i.e. there exists a

ϕ ∈

Homeo

+

(S

1

) such that h

1

(γ) = ϕh

2

(γ)ϕ

−1

, then

h

∗

1

(e) = h

∗

2

(e), h

∗

1

(e

b

) = h

∗

2

(e

b

).

The proof involves writing out a lot of terms explicitly.

How powerful is this bounded Euler class in distinguishing actions? We just

saw that conjugate actions have the same bounded Euler class. The converse

does not hold. For example, one can show that any action with a global fixed

point has trivial bounded Euler class, and there are certainly non-conjugate

actions that both have global fixed points (e.g. take one of them to be the trivial

action).

It turns out there is a way to extend the notion of conjugacy so that the

bounded Euler class becomes a complete invariant.

Definition

(Increasing map of degree 1)

.

A map

ϕ: S

1

→ S

1

is increasing of

degree 1 if there is some

˜ϕ: R → R

lifting

ϕ

such that

˜ϕ

is is monotonically

increasing and

˜ϕ(x + 1) = ˜ϕ(x) + 1

for all x ∈ R.

Note that there is no continuity assumption made on

ϕ

. On the other hand,

it is an easy exercise to see that any monotonic map

R → R

has a countable set

of discontinuities. This is also not necessarily injective.

Example.

The constant map

S

1

→ S

1

sending

x 7→

0 is increasing of degree 1,

as it has a lift ˜ϕ(x) = [x].

Equivalently, such a map is one that sends a positive 4-tuple to a weakly

positive 4-tuple (exercise!).

Definition

(Semiconjugate action)

.

Two actions

h

1

, h

2

:

Γ

→ Homeo

+

(

S

1

) are

semi-conjugate if there are increasing maps of degree 1

ϕ

1

, ϕ

2

: S

1

→ S

1

such

that

(i) h

1

(γ)ϕ

1

= ϕ

1

h

2

(γ) for all γ ∈ Γ;

(ii) h

2

(γ)ϕ

2

= ϕ

2

h

1

(γ) for all γ ∈ Γ.

One can check that the identity action is semiconjugate to any action with a

global fixed point.

Recall the following definition:

Definition

(Minimal action)

.

An action on

S

1

is minimal if every orbit is dense.

Lemma.

If

h

1

and

h

2

are minimal actions that are semiconjugate via

ϕ

1

and

ϕ

2

, then ϕ

1

and ϕ

2

are homeomorphisms and are inverses of each other.

Proof. The condition (i) tells us that

h

1

(γ)(ϕ

1

(x)) = ϕ

1

(h

2

(γ)(x)).

for all

x ∈ S

1

and

γ ∈

Γ. This means

im ϕ

1

is

h

1

(Γ)-invariant, hence dense in

S

1

. Thus, we know that

im ˜ϕ

1

is dense in

R

. But

˜ϕ

is increasing. So

˜ϕ

1

must

be continuous. Indeed, we can look at the two limits

lim

x%y

˜ϕ

1

(x) ≤ lim

x&y

˜ϕ

1

(x).

But since

˜ϕ

1

is increasing, if

˜ϕ

1

were discontinuous at

y ∈ R

, then the inequality

would be strict, and hence the image misses a non-trivial interval. So

˜ϕ

1

is

continuous.

We next claim that

˜ϕ

1

is injective. Suppose not. Say

ϕ

(

x

1

) =

ϕ

(

x

2

). Then

by looking at the lift, we deduce that

ϕ

((

x

1

, x

2

)) =

{x}

for some

x

. Then by

minimality, it follows that

ϕ

is locally constant, hence constant, which is absurd.

We can continue on and then decide that ϕ

1

, ϕ

2

are homeomorphisms.

Theorem

(F. Ghys, 1984)

.

Two actions

h

1

and

h

2

are semiconjugate iff

h

∗

1

(

e

b

) =

h

∗

2

(e

b

).

Thus, in the case of minimal actions, the bounded Euler class is a complete

invariant of actions up to conjugacy.

Proof.

We shall only prove one direction, that if the bounded Euler classes agree,

then the actions are semi-conjugate.

Let

h

1

, h

2

:

Γ

→ Homeo

+

(

S

1

). Recall that

c

(

f, g

)

∈ {

0

,

1

}

refers to the

(normalized) cocycle defining the bounded Euler class. Therefore

c

1

(γ, η) = c(h

1

(γ), h

1

(η))

c

2

(γ, η) = c(h

2

(γ), h

2

(η)).

are representative cocycles of h

∗

1

(e

b

), h

∗

2

(e

b

) ∈ H

2

b

(Γ, Z).

By the hypothesis, there exists u : Γ → Z bounded such that

c

2

(γ, η) = c

1

(γ, η) + u(γ) − u(γη) + u(η)

for all γ, η ∈ Γ.

Let

¯

Γ = Γ ×

c

1

Z be constructed with c

1

, with group law

(γ, n)(η, m) = (γη, c

1

(γ, η) + n + m)

We have a section

s

1

: Γ →

¯

Γ

γ 7→ (γ, 0).

We also write

δ

= (

e,

1)

∈

¯

Γ

, which generates the copy of

Z

in

¯

Γ

. Then we have

s

1

(γη)δ

c

1

(γ,η)

= s

1

(γ)s

2

(η).

Likewise, we can define a section by

s

2

(γ) = s

1

(γ)δ

u(γ)

.

Then we have

s

2

(γη) = s

1

(γη)δ

u(γη)

= δ

−c

1

(γ,η)

s

1

(γ)s

1

(η)δ

u(γη)

= δ

−c

1

(γ,η)

δ

−u(γ)

s

2

(γ)δ

−u(η)

s

2

(η)δ

u(γη)

= δ

−c

1

(γ,η)−u(γ)+u(γη)−u(η)

s

2

(γ)s

2

(η)

= δ

−c

2

(γ,η)

s

2

(γ)s

2

(η).

Now every element in

¯

Γ

can be uniuely written as a product

s

1

(

γ

)

δ

n

, and the

same holds for s

2

(γ)δ

m

.

Recall that for

f ∈ Homeo

+

(

S

1

), we write

¯

f

for the unique lift with

¯

f

(0)

∈

[0, 1). We define

Φ

i

(s

i

(γ)δ

n

) = h

i

(γ) · T

n

.

We claim that this is a homomorphism! We simply compute

Φ

i

(s

i

(γ)δ

n

s

i

(η)δ

m

) = Φ

i

(s

i

(γ)s

i

(η)δ

n+m

)

= Φ

i

(s

i

(γη)δ

c

i

(γ,η)+n+m

)

= h

i

(γη)T

c

i

(γ,η)

T

n+m

= h

i

(γ)h

i

(η)T

n+m

= h

i

(γ)T

n

h

i

(η)T

m

= Φ

i

(s

i

(γ)δ

n

)Φ

i

(s

i

(η)δ

m

).

So we get group homomorphisms Φ

i

:

¯

Γ → Homeo

+

Z

(R).

Claim. For any x ∈ R, the map

¯

Γ → R

g 7→ Φ

1

(g)

−1

Φ

2

(g)(x)

is bounded.

Proof. We define

v(g, x) = Φ

1

(g)

−1

Φ(g)x.

We notice that

v(gδ

m

, x) = Φ

1

(gδ

m

)

−1

Φ

2

(gδ

m

)(x)

= Φ

1

(g)

−1

T

−m

T

m

Φ

2

(g)

= v(g, x).

Also, for all g, the map x 7→ v(g, x) is in Homeo

+

Z

(R).

Hence it is sufficient to show that

γ 7→ v(s

2

(γ), 0)

is bounded. Indeed, we just have

v(s

2

(γ), 0) = Φ

1

(s

2

(γ)

−1

Φ

2

(s

2

(γ))(0)

= Φ

1

(s

1

(γ)δ

u(γ)

)

−1

Φ

2

(s

2

(γ))(0)

= δ

−u(γ)

h

1

(γ)

−1

h

2

(γ)(0)

= −u(γ) + h

1

(γ)

−1

(h

2

(γ)(0)).

But u is bounded, and also

h

1

(γ)

−1

(h

2

(γ)(0)) ∈ (−1, 1).

So we are done.

Finally, we can write down our two quasi-conjugations. We define

˜ϕ(x) = sup

g∈

¯

Γ

v(g, x).

Then we verify that

˜ϕ(Φ

2

(h)x) = Φ

1

(h)(ϕ(x)).

Reducing everything modulo Z, we find that

ϕh

2

(γ) = h

1

(γ)ϕ.

The other direction is symmetric.