3Actions on S1
IV Bounded Cohomology
3.2 The real bounded Euler class
The next thing we do might be a bit unexpected. We are going to forget that
the cocycle
c
takes values in
Z
, and view it as an element in the real bounded
cohomology group.
Definition
(Real bounded Euler class)
.
The real bounded Euler class is the
class e
b
R
∈ H
2
b
(Homeo
+
(S
1
), R) obtained by change of coefficients from Z → R.
The real bounded Euler class of an action
h:
Γ
→ Homeo
+
(
S
1
) is the pullback
h
∗
(e
b
R
) ∈ H
2
b
(Γ, R).
A priori, this class contains less information that the original Euler class.
However, it turns out the real bounded Euler class can distinguish between
very different dynamical properties. Recall that we had the Gersten long exact
sequence
0 Hom(Γ, R/Z) H
2
b
(Γ, Z) → H
2
b
(Γ, R)
δ
.
By exactness, the real bounded Euler class vanishes if and only if
e
b
is in
the image of
δ
. But we can characterize the image of
δ
rather easily. Each
homomorphism
χ
: Γ
→ R/Z
gives an action by rotation, and in a previous
exercise, we saw the bounded Euler class of this action is
δ
(
χ
). So the image of
δ
is exactly the bounded Euler classes of actions by rotations. On the other hand,
we know that the bounded Euler class classifies the action up to semi-conjugacy.
So we know that
Corollary.
An action
h
is semi-conjugate to an action by rotations iff
h
∗
(
e
b
R
) = 0.
We want to use the real bounded Euler class to classify different kinds of
actions. Before we do that, we first classify actions without using the real
bounded Euler class, and then later see how this classification is related to the
real bounded Euler class.
Theorem.
Let
h:
Γ
→ Homeo
+
(
S
1
) be an action. Then one of the following
holds:
(i) There is a finite orbit, and all finite orbits have the same cardinality.
(ii) The action is minimal.
(iii)
There is a closed, minimal, invariant, infinite, proper subset
K ( S
1
such
that any x ∈ S
1
, the closure of the orbit h(Γ)x contains K.
We will provide a proof sketch. More details can be found in Hector–Hirsch’s
Introduction to the geometry of foliations.
Proof sketch.
By compactness and Zorn’s lemma, we can find a minimal, non-
empty, closed, invariant subset
K ⊆ S
1
. Let
∂K
=
K \
˚
K
, and let
K
0
be the
set of all accumulation points of
K
(i.e. the set of all points
x
such that every
neighbourhood of
x
contains infinitely many points of
K
). Clearly
K
0
and
∂K
are closed and invariant as well, and are contained in
K
. By minimality, they
must be K or empty.
(i)
If
K
0
=
∅
, then
K
is finite. It is an exercise to show that all orbits have
the same size.
(ii)
If
K
0
=
K
, and
∂K
=
∅
, then
K
=
˚
K
, and hence is open. Since
S
1
is
connected, K = S
1
, and the action is minimal.
(iii)
If
K
0
=
K
=
∂K
, then
K
is perfect, i.e. every point is an accumulation
point, and
K
is totally disconnected. We definitely have
K 6
=
S
1
and
K
is
infinite. It is also minimal and invariant.
Let
x ∈ S
1
. We want to show that the closure of its orbit contains
K
.
Since
K
is minimal, it suffices to show that
h(Γ)x
contains a point in
K
.
If
x ∈ K
, then we are done. Otherwise, the complement of
K
is open,
hence a disjoint union of open intervals.
For the sake of explicitness, we define an interval of a circle as follows — if
a, b ∈ S
1
and a 6= b, then
(a, b) = {z ∈ S
1
: (a, z, b) is positively oriented}.
Now let (
a, b
) be the connected component of
S
1
\ K
containing
x
. Then
we know a ∈ K.
We observe that
S
1
\ K
has to be the union of countably many intervals,
and moreover
h
(Γ)
a
consists of end points of these open intervals. So
h
(Γ)
a
is a countable set. On the other hand, since
K
is perfect, we know
K
is uncountable. The point is that this allows us to pick some element
y ∈ K \ h(Γ)a.
Since
a ∈ K
, minimality tells us there exists a sequence (
γ
n
)
n≥1
such that
h
(
γ
n
)
a → y
. But since
y 6∈ h
(Γ)
a
, we may wlog assume that all the points
{h
(
γ
n
)
a
:
n ≥
1
}
are distinct. Hence
{h
(
γ
n
)(
a, b
)
}
n≥1
is a collection of
disjoint intervals in
S
1
. This forces their lengths tend to 0. We are now
done, because then h(γ
n
)x gets arbitrarily close to h(γ
n
)a as well.
We shall try to rephrase this result in terms of the real bounded Euler class.
It will take some work, but we shall state the result as follows:
Corollary. Let h: Γ → S
1
be an action. Then one of the following is true:
(i) h
∗
(e
b
R
) = 0 and h is semi-conjugate to an action by rotations.
(ii) h
∗
(
e
b
R
)
6
= 0, and then
h
is semi-conjugate to a minimal unbounded action,
i.e. {h(γ) : γ ∈ Γ} is not equicontinuous.
Observe that if Λ
⊆ Homeo
+
(
S
1
) is equicontinuous, then by Arzela–Ascoli,
its closure
¯
Λ is compact.
To prove this, we first need the following lemma:
Lemma.
A minimal compact subgroup
U ⊆ Homeo
+
(
S
1
) is conjugate to a
subgroup of Rot.
Proof.
By Kakutani fixed point theorem, we can pick an
U
-invariant probability
measure on S
1
, say µ, such that µ(S
1
) = 2π.
We parametrize the circle by
p:
[0
,
2
π
)
→ S
1
. We define
ϕ ∈ Homeo
+
(
S
1
)
by
ϕ(p(t)) = p(s),
where s ∈ [0, 2π) is unique with the property that
µ(p([0, s)) = t.
One then verifies that ϕ is a homeomorphism, and ϕU ϕ
−1
⊆ Rot.
Proof of corollary.
Suppose
h
∗
(
e
b
R
)
6
= 0. Thus we are in case (ii) or (iii) of the
previous trichotomy.
We first show how to reduce (iii) to (ii). Let
K ( S
1
be the minimal
h
(Γ)-
invariant closed set given by the trichotomy theorem. The idea is that this
K
misses a lot of open intervals, and we want to collapse those intervals.
We define the equivalence relation on
S
1
by
x ∼ y
if
{x, y} ⊆
¯
I
for some
connected component
I
of
S
1
\ K
. Then
∼
is an equivalence relation that is
h
(Γ)-invariant, and the quotient map is homeomorphic to
S
1
(exercise!). Write
i: S
1
/ ∼→ S
1
for the isomorphism.
In this way, we obtain an action of
ρ:
Γ
→ Homeo
+
(
S
1
) which is minimal,
and the map
ϕ: S
1
S
1
/ ∼ S
1
pr
i
intertwines the two actions, i.e.
ϕh(γ) = ρ(γ)ϕ.
Then one shows that
ϕ
is increasing of degree 1. Then we would need to find
ψ : S
1
→ S
1
which is increasing of degree 1 with
ψρ(γ) = h(γ)ψ.
But
ϕ
is surjective, and picking an appropriate section of this would give the
ψ
desired.
So h is semi-conjugate to ρ, and 0 6= h
∗
(e
b
R
) = ρ
∗
(e
b
R
).
Thus we are left with
ρ
minimal, with
ρ
∗
(
e
b
R
)
6
= 0. We have to show that
ρ
is
not equicontinuous. But if it were, then
ρ
(Γ) would be contained in a compact
subgroup of
Homeo
+
(
S
1
), and hence by the previous lemma, would be conjugate
to an action by rotation.
The following theorem gives us a glimpse of what unbounded actions look
like:
Theorem
(Ghys, Margulis)
.
If
ρ:
Γ
→ Homeo
+
(
S
1
) is an action which is
minimal and unbounded. Then the centralizer
C
Homeo
+
(S
1
)
(
ρ
(Γ)) is finite cyclic,
say
hϕi
, and the factor action
ρ
0
on
S
1
/hϕi
∼
=
S
1
is minimal and strongly
proximal. We call this action the strongly proximal quotient of ρ.
Definition
(Strongly proximal action)
.
A Γ-action by homeomorphisms on a
compact metrizable space
X
is strongly proximal if for all probability measures
µ on X, the weak-∗ closure Γ
∗
µ contains a Dirac mass.
For a minimal action on
X
=
S
1
, the property is equivalent to the following:
–
Every proper closed interval can be contracted. In other words, for every
interval
J ⊆ S
1
, there exists a sequence (
γ
n
)
n≥1
such that
diam
(
ρ
(
γ
n
)
J
)
→
0 as n → ∞.
Proof of theorem.
Let
ψ
commute with all
ρ
(
γ
) for all
γ ∈
Γ, and assume
ψ 6
=
id
.
Claim. ψ has no fixed points.
Proof. Otherwise, if ψ(p) = p, then
ψ(ρ(γ)p) = ρ(γ)ψ(p) = ρ(γ)(p).
Then by density of {ρ(γ)p : γ ∈ Γ}, we have ψ = id.
Hence we can find
ε >
0 such that
length
([
x, ψ
(
x
)])
≥ ε
for all
x
by compact-
ness. Observe that
ρ(γ)[x, ψ(x)] = [ρ(γ)x, ρ(γ)ψ(x)] = [ρ(γ)x, ψ(ρ(γ)x)].
This is just an element of the above kind. So length(ρ(γ)[x, ψ(x)]) ≥ ε.
Now assume ρ(Γ) is minimal and not equicontinuous.
Claim. Every point x ∈ S
1
has a neighbourhood that can be contracted.
Proof.
Indeed, since
ρ
(Γ) is not equicontinuous, there exists
ε >
0, a sequence
(γ
n
)
n≥1
and intervals I
k
such that length(I
k
) & 0 and length(ρ(γ
n
)I
n
) ≥ ε.
Since we are on a compact space, after passing to a subsequence, we may
assume that for
n
large enough, we can find some interval
J
such that
length
(
J
)
≥
ε
2
and J ⊆ ρ(γ
n
)I
n
.
But this means
ρ(γ
n
)
−1
J ⊆ I
n
.
So J can be contracted. Since the action is minimal,
[
γ∈Γ
ρ(γ)J = S
1
.
So every point in S
1
is contained in some interval that can be contracted.
We shall now write down what the homeomorphism that generates the
centralizer. Fix x ∈ S
1
. Then the set
C
x
= {[x, y) ∈ S
1
: [x, y) can be contracted}
is totally ordered ordered by inclusion. Define
ϕ(x) sup C
x
.
Then
[x, ϕ(x)) =
[
C
x
.
This gives a well-defined map
ϕ
that commutes with the action of
γ
. It is then
an interesting exercise to verify all the desired properties.
–
To show
ϕ
is homeomorphism, we show
ϕ
is increasing of degree 1, and
since it commutes with a minimal action, it is a homeomorphism.
–
If
ϕ
is not periodic, then there is some
n
such that
ϕ
n
(
x
) is between
x
and
ϕ
(
x
). But since
ϕ
commutes with the action of Γ, this implies [
x, ϕ
n
(
x
)]
cannot be contracted, which is a contradiction.
Exercise. We have
ρ
∗
(e
b
) = kρ
∗
0
(e
b
),
where k is the cardinality of the centralizer.
Example. We can decompose PSL(2, R) = PSO(2)AN , where
A =
λ 0
0 λ
−1
: λ > 0
, N =
1 x
0 1
.
More precisely,
SO
(2)
× A × N → SL
(2
, R
) is a diffeomorphism and induces
on
PSO
(2)
× A × N → PSL
(2
, R
). In particular, the inclusion
i: PSO
(2)
→
PSL(2, R) induces an isomorphism on the level of π
1
∼
=
Z.
We can consider the subgroup
kZ ⊆ Z
. which gives us a covering of
PSO
(2)
and PSL(2, R) that fits in the diagram
PSO(2)
k
PSL(2, R)
PSO(2) PSL(2, R)
i
k
p p
i
.
On the other hand, if we put
B
=
A · N
, which is a contractible subgroup, we
obtain a homomorphism s : B → PSL(2, R)
k
, and we find that
PSL(2, R)
k
∼
=
PSO(2)
k
· s(B).
So we have
PSL(2, R)
k
s(B)
∼
=
PSO(2)
k
.
So
PSL
(2
, R
)
k
/s
(
B
) is homeomorphic to a circle. So we obtain an action of
PSL(2, R)
k
on the circle.
Now we can think of Γ
∼
=
F
r
as a lattice in
PSL
(2
, R
). Take any section
σ :
Γ
→ PSL
(2
, R
)
k
. This way, we obtain an unbounded minimal action with
centralizer isomorphic to Z/kZ.
Definition
(Lattice)
.
A lattice in a locally compact group
G
is a discrete
subgroup Γ such that on Γ\G, there is a G-invariant probability measure.
Example.
Let
O
be the ring of integers of a finite extension
k/Q
. Then
SL
(
n, O
) is a lattice in an appropriate Lie group. To construct this, we write
[
k
:
Q
] =
r
+ 2
s
, where
r
and 2
s
are the number of real and complex field
embeddings of k. Using these field embeddings, we obtain an injection
SL(n, O) → SL(n, R)
r
× SL(n, C)
s
,
and the image is a lattice.
Example.
If
X
is a complete proper CAT(0) space, then
Isom
(
X
) is locally
compact, and in many cases conatins lattices.
Theorem
(Burger, 2007)
.
Let
G
be a second-countable locally compact group,
and Γ
< G
be a lattice, and
ρ:
Γ
→ Homeo
+
(
S
1
) a minimal unbounded action.
Then the following are equivalent:
– ρ
∗
(e
b
R
) is in the image of the restriction map H
2
bc
(G, R) → H
2
b
(Γ, R)
–
The strongly proximal quotient
ρ
ss
:
Γ
→ Homeo
+
(
S
1
) extends continu-
ously to G.
Theorem
(Burger–Monod, 2002)
.
The restriction map
H
2
bc
(
G
)
→ H
2
b
(Γ
, R
) is
an isomorphism in the following cases:
(i) G
=
G
1
× · · · × G
n
is a cartesian product of locally compact groups and Γ
has dense projections on each individual factor.
(ii) G
is a connected semisimple Lie group with finite center and rank
G ≥
2,
and Γ is irreducible.
Example.
Let
k/Q
be a finite extension that is not an imaginary quadratic
extension. Then we have an inclusion
SL(2, O) → SL(2, R)
r
× SL(2, C)
s
and is a product of more than one thing. One can actually explicitly compute
the continuous bounded cohomology group of the right hand side.
Exercise.
Let Γ
< SL
(3
, R
) be any lattice. Are there any actions by oriented
homeomorphisms on S
1
?
Let’s discuss according to ρ
∗
(e
b
R
).
–
If
ρ
∗
(
e
b
R
) = 0, then there is a finite orbit. Then we are stuck, and don’t
know what to say.
–
If
ρ
∗
(
e
b
R
)
6
= 0, then we have an unbounded minimal action. This leads
to a strongly proximal action
ρ
ss
:
Γ
→ Homeo
+
(
S
1
). But by the above
results, this implies the action extends continuously to an action of
SL
(3
, R
)
on
S
1
. But
SL
(3
, R
) contains
SO
(3), which is a compact group. But we
know what compact subgroups of
Homeo
+
(
S
1
) look like, and it eventually
follows that the action is trivial. So this case is not possible.