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arXiv:1302.1637v1 [math.DS] 7 Feb 2013
Center foliation: absolute continuity,
disintegration and rigidity
R´egis Var˜ao ∗
February 8, 2013
Abstract
In this paper we address the issues of absolute continuity for the
center foliation (as well as the disintegration on the non-absolute con-
tinuous case) and rigidity of volume preserving partially hyperbolic
diffeomorphisms isotopic to a linear Anosov on T3. It is shown that
the disintegration of volume on center leaves may be neither atomic
nor Lebesgue, in contrast to the dichotomy (Lebesgue or atomic) ob-
tained by Avila, Viana, Wilkinson [1]. It is also obtained results con-
cerning the atomic disintegration. Moreover, the absolute continuity
of the center foliation does not imply smooth conjugacy with its lin-
earization. Imposing stronger conditions besides absolute continuity
on the center foliation, smooth conjugacy is obtained.
1 Introduction and statements
We study the measure-theoretical properties of the center foliation of par-
tially hyperbolic diffeomorphisms for which the center leaves are non-compact.
Two main issues are:
•absolute continuity: when is the center foliation absolutely continuous?
What can be said otherwise (disintegration)?
•rigidity: does absolute continuity imply greater regularity?
These issues are fairly well understood for certain volume preserving par-
tially hyperbolic diffeomorphism (perturbations of certain skew-products or
of time-one maps of Anosov flows) studied by Avila, Viana and Wilkinson
[1]. We state their dichotomy:
∗regisvarao@icmc.usp.br
1
•Atomic disintegration: If the center foliation is non-absolutely contin-
uous, then there exists k∈Nand a full volume subset that intersects
each center leaf on exactly kpoints/orbits.
•Rigidity: If the center foliation is absolutely continuous then the diffeo-
morphism is smoothly conjugate to a rigid model (a rotation extension
of an Anosov diffeomorphism or the time-one map of an Anosov flow);
On three-dimensional manifolds, the only known examples of partially
hyperbolic diffeomorphism are of skew-product type, perturbation of time-
one of Anosov flows and diffeomorphisms derives from a linear Anosov. In
fact it is conjecture by E. Pujals that these are all the possibilities (see
[4] for precise statements). Avila, Viana and Wilkinson [1] have treated
diffeomorphisms on the first two cases and we treat in this work the third
case.
We deal with derived from Anosov (DA) diffeomorphisms, that is, fis
a DA diffeomorphism if it partially hyperbolic and lies in the isotopy class
of some hyperbolic linear automorphism A. We refer to Aas the the lin-
earization of f. Every partially hyperbolic diffeomorphism has the following
splitting on the tangent space T M =Es⊕Ec⊕Eu(see §2 for definitions),
where Esis a contracting direction, Euis an expanding direction and the
center direction Echas an intermediate behavior. We consider Anosov diffeo-
morphisms (see §2 for definition) which are partially hyperbolic. That means
that they have the splitting T M =Es⊕Ec⊕Euand the center direction
is uniformly contracting or expanding. It then makes sense to talk about
the center foliation of an Anosov (partially hyperbolic) diffeomorphism, as
we do on Theorem 1.1 and Theorem 1.2. A center foliation (see §2) is an
invariant foliation by ftangent to the Ecdirection. We also mention that all
diffeomorphisms treated on this work are assumed to be at least C1+α. This
implies, in particular, that volume preserving Anosov on T3are ergodic.
We state now our results. For non-absolutely continuous center foliation,
we show that it is possible to have a disintegration (§2.2 for definition) which
is non-Lebesgue and non-atomic. Our conclusion is different from [1] and
indeed this is the first example of this kind:
Theorem 1.1. For partially hyperbolic Anosov diffeomorphisms on T3, vol-
ume preserving, for which the center foliation is non-absolutely continuous,
we have that
i) there exists fAnosov for which the disintegration of volume on the
center leaves are neither Lebesgue, nor atomic.
2
In fact, such diffeomorphisms fill a dense subset of an infinite-dimensional
manifold in the neighborhood of any hyperbolic linear automorphisms
in the space of volume preserving maps;
ii) the conditional measures are singular measures with respect to the vol-
ume on the center leaf;
iii) if the decomposition is atomic, then there is exactly one atom per leaf.
That is, there exists a set of full volume that intersects each center leaf
in one point;
iv) the disintegration of volume on the center leaves is atomic if and only
if the partition by center leaves is a measurable partition.
For item iv) above, we don’t need to suppose that we are on the non-
absolute continuous case. The next result shows, in contrast to the dichotomy
[1], that absolute continuity has no rigidity implications in our case:
Theorem 1.2. There exist volume preserving Anosov diffeomorphisms fon
T3for which the center foliation is absolutely continuous but fis not C1-
conjugate to its linearization.
In fact, such diffeomorphisms fill a dense subset of an infinite-dimensional
manifold in the neighborhood of any hyperbolic linear automorphism in the
space of volume preserving maps.
As we shall see, Theorem 1.2 will be just a corollary of the following
result, which is important on its own:
Lemma 1.1. Let fbe a volume preserving partially hyperbolic Anosov diffeo-
morphism on T3. Then, for any periodic points p, q the Lyapunov exponents
on each of the directions (stable, center, unstable) are the same if and only
if fis C1conjugate to its linearization.
Note that Theorem 1.2 implies that to obtain a rigidity result we must
impose some stronger conditions on the center foliation besides absolute con-
tinuity. And we do so to obtain the following rigidity result.
Theorem 1.3. Let fbe a volume preserving DA diffeomorphism on T3, with
the linearization A. If the center foliation is a C1foliation and the center
holonomies inside the center-unstable, Fcu
f, and center-stable, Fcs
f, leaves are
uniformly bounded, then fis C1conjugate to its linearization and, hence, is
an Anosov diffeomorphism.
3
Organization of the paper. In §2 we give some basic definitions such as
what we mean by to disintegrate a measure, absolute continuity, etc. In §3
we study the behavior of non-absolute continuous center foliation, where we
prove Theorem 1.1. We begin §4 understanding how Lyapunov exponents
vary with respect to their linearization, we then prove Lemma 1.1 and The-
orem 1.2. In §5 we construct some conditional measures (not probabilities)
on each center leaf with some dynamical meaning. We use these measures to
prove Theorem 1.3.
2 Preliminaries
2.1 Partially Hyperbolic Diffeomorphism.
A diffeomorphism fof a compact Riemannian manifold Mis called partially
hyperbolic if there are constants λ < ˆγ < 1< γ < µ and C > 1 and a Df
-invariant splitting of T M =Eu(x)⊕Ec(x)⊕Es(x) where
1
Cµn||v|| <||Df nv||, v ∈Eu
x− {0};
1
Cˆγn||v|| <||Df nv|| < Cγn||v||, v ∈Ec
x− {0};
||Df nv|| < Cλn||v||, v ∈Es
x− {0}.
We say that a partially hyperbolic diffeomorphism is dynamically coherent
if the subbundles Es⊕Ecand Ec⊕Euintegrate into invariant foliations,
Fcs,Fcu respectively. This implies in particular that there is a center foliation
Fc, which is obtained by an intersection of the other two: Fc=Fcs ∩ Fcu .
It was proved by Brin, Buragov, Ivanov [6] that
Theorem 2.1. Every partially hyperbolic diffeomorphism on T3is dynami-
cally coherent.
2.2 Decomposition of measure
Let (M, µ, B) be a probability space, where Mis a compact metric space,
µa probability and Bthe borelian σ-algebra. Given a partition Pof Mby
measurable sets, we associate the following probability space (P,eµ, e
B), where
eµ:= π∗µ,e
B:= π∗B. and π:M→ P is the canonical projection associate to
a point of Mthe partition element that contains it.
For a given a partition P, a family {µP}p∈P is a system of conditional
measures for µ(with respect to P) if
4
i) given φ∈C0(M), then P7→ RφµPis measurable;
ii) µP(P) = 1 eµ-a.e.;
iii) if φ∈C0(M), then ZM
φdµ =ZPZP
φdµPdeµ.
We call Pameasurable partition (w.r.t. µ) if there exist a family {Ai}i∈N
of borelian sets and a set Fof full µ-measure such that for every P∈ P there
exists a sequence {Bi}i∈N, where Bi∈ {Ai, Ac
i}such that P∩F=∩i∈NBi∩F.
The following result is also known as Rokhlin’s disintegration Theorem.
Theorem 2.2. Let Pbe a measurable partition of a compact metric space M
and µa borelian probability. Then there exists a disintegration by conditional
measures for µ.
Remark. On Theorem 1.1 the meaning of “disintegration of volume on
the center leaves are neither Lebesgue, nor atomic” means that on a foliated
box, since the center foliation form a measurable partition we can apply on
this foliated box the Rokhlin’s disintegration Theorem and the conditional
measures are neither Lebesgue, nor atomic. This is independent of the fo-
liated box (see Lemma 5.1) and that is why we don’t say instead that the
disintegration locally is neither Lebesgue nor atomic.
2.2.1 Absolute continuity
Let Fbe a foliation and disintegrate the volume inside a foliated box. If the
conditional measure mLon the leave satisfies that mL<< LebLfor almost
every leaf, then Fis said to be an absolutely continuous foliation, where LebL
is the Lebesgue measure on the leaf L.
We state a result due to Gogolev [7] which shall be our starting point to
understand absolute continuity for partially hyperbolic diffeomorphism with
non-compact center leaves.
Theorem 2.3. Let f:T3→T3be an Anosov diffeomorphism with splitting
of the form Es⊕Ewu ⊕Euu , then Fc
fis absolutely continuous if and only
λuu(p) = λuu (q)for all periodic points pand q.
Where λuu is the Lyapunov exponent on the Euu direction.
2.3 Geometric property
By a Derived from Anosov (DA) diffeomorphism f:T3→T3we mean a
partially hyperbolic homotopic to a linear Anosov diffeomorphism A. We
5
call this linear Anosov as the linearization of f. In fact, fis semi-conjugated
to its linearization. The itens from the Theorem below, which proof can
be found on Sambarino [10], show that the semi-conjugacy has in fact good
properties.
Theorem 2.4. Let B:R3→R3be a linear hyperbolic isomorphism. Then,
there exists C > 0such that if G:R3→R3is a homeomorphism such that
sup{||G(x)−Bx|| | x∈R3}=K < ∞then there exists H:R3→R3
continuous and surjective such that:
•B◦H=H◦G;
• ||H(x)−x|| ≤ CK for all x∈R3;
•H(x)is characterized as the unique point ysuch that
||Bn(y)−Gm(x)|| ≤ CK, ∀n∈Z;
•H(x) = H(y)if and only if ||Gn(x)−G(y)|| ≤ 2C K,∀n∈Z, and if
and only if supn∈Z{||Gn(x)−Gn(y)||} <∞;
•if B∈SL(3,Z)and Gis the lift of g:T3→T3then Hinduces h:
T3→T3continuous and onto such that B◦h=h◦gand distC0(h, id)≤
CdistC0(B, g).
The geometrical property we shall need later is given by Hammerlindl [9]:
Proposition 2.1. Let fbe a partially hyperbolic and Abe its linearization.
Denote by ˜
fand ˜
Athe lift to Rnof fand Arespectively. Then for each
k∈Zand C > 1there is M > 0and a linear map π:Rn→Rnsuch that
for all x, y ∈Rn
||x−y|| > M ⇒1
C<||π(˜
fk(x)−˜
fk(y))||
||π(˜
Ak(x)−˜
Ak(y))|| < C.
3 Non-absolute continuity
We dedicate this section for the proof of Theorem 1.1.
6
3.1 Proof of item i)
Consider a linear volume preserving Anosov with the following split T M =
Ess ⊕Ews ⊕Eu. Let φbe a volume preserving diffeomorphism which preserves
the Eudirection. By Baraviera, Bonnatti [2] Rλws
AdV ol 6=Rλws
A◦φdV ol. Let
hbe the conjugacy between Aand f,f◦h=h◦A. We claim that his volume
preserving and sends center leaves to center leaves. To see that his volume
preserving note that fand Ahave the same topological entropy λu
A. Hence,
h∗V ol is a measure of maximal entropy. Observe that the perturbation A◦ψ
of Ais such that it preserves the Euexponent, which means that by the
equilibrium state theory (see Bowen [5]) the potentials 0 and −log||Df|Eu||
are cohomological and therefore give the same equilibrium states. That is,
h∗V ol =V ol. And the fact that h(Fc) = Fccomes from Lemma 2 of [8].
Claim: Fc
A◦ψis not absolutely continuous.
Suppose, by contradiction, that it is absolutely continuous, then Theorem
2.3 implies λss
f(p) = cte for all periodic point p. By contruction we have
λu
f(p) = λu
A. Since we are on the volume preserving case, λws(p) is also
constant on periodic points. Therefore, by Lemma 1.1 fis C1-conjugate
to A, but this would imply Rλws
fdV ol =Rλws
AdV ol. Which is absurd by
Proposition 0.3 of Baravieira, Bonnatti [2].
Claim: The disintegration of volume on center leaves of A◦φis neihter
Lebesgue nor atomic.
It is not Lebesgue because it is not absolutely continuous. And to see that
it is not atomic, note that since his volume preserving and sends center leaves
onto center leaves we can induce (by push forward) the disintegration on the
center leaves of Ato the center leaves of A◦φ. And since the disintegration
for Ais Lebesgue, this means that the disintegration for A◦φis not atomic.
3.2 Proof of item ii)
By ergodicity we know that the Birkhoff set
B={x∈T3|1/n
n−1
X
i=0
δfi(x)→V ol as n→ ∞}
has full measure.
Claim. If there is a center leaf such that Fc∩Bhas positive Lebesgue
measure, then the center foliation is absolutely continuous.
Proof of the Claim. Let Dbe any disc on the central foliation and consider
7
the following construction
µn=1
n
n−1
X
j=0
fj
∗mD
mD(D),
where mDmeans the Lebesgue measure on the central leaf. It turns out that
these measures converge to a measure µsuch that the disintegration of µ
on the center leaves are absolutely continuous with respect to the Lebesgue
measure. This is a well-known construction of measures, studied by Pesin,
Sinai in the eighties. For more references see [3] Chapter 11 and the references
therein. Although Pesin, Sinai studied these measures for the case of the
disc Din the unstable foliation, for the center foliation, in our case, this
construction is the same. Gogolev, Guysinsky [8] have worked explicitly on
this case and the reader may check at [8] the construction.
We make a slightly different construction, instead of the disc D, as above,
we take D∩Bfor which it has positive Lebesgue measure on the center leaf.
By hypothesis there exists such a disc. It turns out that these measures still
converge to a measure with conditional measures absolutely continuous to
the Lebesgue measure on the center leaf (Lemma 11.12 [3]). Since the points
on Bhave the property 1/n Pn−1
i=0 δfi(x)→V ol, it turns out that the sequence
µnconverges to the volume. Hence, volume has Lebesgue disintegration on
the center leaves. Which proves the claim.
From the claim, since we are in the case where the center foliation is
non-absolutely continuous, we must have that the center foliation intersects
Bon a set of zero Lebesgue measure. But the conditional measures give full
measure to B, since Bhas full measure. Therefore the conditional measures
are singular with respect to the Lebesgue measure. And item ii) is proved.
3.3 Proof of item iii)
On what follows Riwill denote a rectangle of a fixed finite Markov partition.
The proof of item iii) will be a consequence of the following lemmas.
Lemma 3.1. All the atoms have the same weight when considering the dis-
integration of volume on the center leaves of Ri.
Proof. On each Markov rectangle we may apply Rokhlin’s disintegration the-
orem on center leaves. Therefore, when writing mxwe mean the conditional
measure for the disintegration on Markov rectangle that contains x. Consider
the set Aδ={x∈A|mx(x)≤δ}. Since f(Fc
R(x)(x)) ⊃ Fc
R(f(x))(f(x)), we
8
have that f∗mx(I)≤mf(x)(I) where Iis inside the connected component of
Fc
f(x)∩R(f(x)) that contains fn(x). If f(x)∈Aδ, then
mx(x) = f∗mx(f(x)) ≤mx(f(x)) ≤δ.
Hence, f−1(Aδ)⊂Aδ.
By ergodicity, since our Anosov is volume preserving on T3,Aδhas full
measure or zero measure. Let δ0be the discontinuity point of the function
δ∈[0,1] 7→ V ol(Aδ). This implies that almost every atom has weight δ0.
Lemma 3.2. On every Markov partition Rithe conditional measures have
the same number of atoms, with the same weight.
Proof. This is a direct consequence from the above lemma. Since all the
atoms have the same weight δ0the conditional measures must have 1/δ0
number of atoms.
Lemma 3.3. There is a set of full volume B1, of atoms, such that if x∈B1,
then B1∩ Fc
xis contained in the connected component of Ri(x)∩ Fc
xthat
contains x.
Proof. Let Abe the set of atoms and Tbe the set of transitive points. Both
sets have full volume measure by ergodicity. Suppose, by contradiction, that
there is a subset A1⊂Aof positive volume measure such that ∀x∈A1we
get A∩Rc
i(x)6=∅, where Rc
i(x)is the complement of the Markov partition
that contains x, note that V ol(A1∩T)>0. Define the following map
h:A1∩T→R
x7→ h(x) = dFc
x(Ri(x), R′
i(x)),
where dFc
x(Ri(x), R′
i(x)) means the distance inside the center leaf of the Markov
rectangle Ri(x)to the closest Markov rectangle that has an atom which we
call R′
i(x).
Since his a measurable map, there exists K1⊂A1∩T, with V ol(K1)>0
for which his a continuous map when restricted to K1. And since volume is
a regular measure, there is compact set K2⊂K1, also with positive volume
measure.
Let α=Maxx∈K2h(x). Fix z0∈Ri(z0), and consider a ball small enough
such that B(z0, r)⊂intRi(z0). Hence, ∀y∈K2, let ny∈Nbe an integer
big enough so that, since fis uniformly expanding in the center direction,
f−ny(Fc(y, α)) ⊂B(z0, r)⊂intRi(z0).
It means that we have at least doubled the number of atoms inside Ri(z0),
which is an absurd since we have already shown that the number of atoms
are constant on each Markov partition.
9
Lemma 3.4. There is a set of full volume B1, of atoms, such that if x∈B1,
then B1∩ Fc
xis contained in the connected component of Ri(x)∩ Fc
xthat
contains x.
Proof. Let Abe the set of atoms and Tbe the set of transitive points. Both
sets have full volume measure by ergodicity. Suppose, by contradiction, that
there is a subset A1⊂Aof positive volume measure such that ∀x∈A1we
get A∩Rc
i(x)6=∅, where Rc
i(x)is the complement of the Markov partition
that contains x, note that V ol(A1∩T)>0. Define the following map
h:A1∩T→R
x7→ h(x) = dFc
x(Ri(x), R′
i(x)),
where dFc
x(Ri(x), R′
i(x)) means the distance inside the center leaf of the Markov
rectangle Ri(x)to the closest Markov rectangle that has an atom which we
call R′
i(x).
Since his a measurable map, there exists K1⊂A1∩T, with V ol(K1)>0
for which his a continuous map when restricted to K1. And since volume is
a regular measure, there is compact set K2⊂K1, also with positive volume
measure.
Let α=Maxx∈K2h(x). Fix z0∈Ri(z0), and consider a ball small enough
such that B(z0, r)⊂intRi(z0). Hence, ∀y∈K2, let ny∈Nbe an integer
big enough so that, since fis uniformly expanding in the center direction,
f−ny(Fc(y, α)) ⊂B(z0, r)⊂intRi(z0).
It means that we have at least doubled the number of atoms inside Ri(z0),
which is an absurd since we have already shown that the number of atoms
are constant on each Markov partition.
Lemma 3.5. There is a set of full volume B2⊂B1such that the center
foliation intersects B2at most on one point.
Proof. By contradiction suppose that the number of atoms on all Markov
partition are greater than one. Let A2be a set with full volume measure
inside the union of the Markov rectangle such that if x∈A2, then A2∩ F c
x,loc
has the same number of points, in this case greater than one. Where Fc
x,loc is
the connected set of the center foliation restricted to the Markov rectangle
that intersects x. We define the map
h:A2→R
x7→ h(x)
10
where h(x) is the smallest distance between the atoms of Fc
x,loc. By Lusin’s
theorem there is a set K1⊂A2of positive measure for which his continuous.
Since volume is regular, there is a compact subset K2of K1with positive
measure. Let α= min
x∈K2
h(x).
Let β > 0 be an inferior bound for the length of Fc
loc. Let n0∈Nbig
enough so that any segment of a center leaf with length greater than or
equal to αhas the length of its n0th iterate greater than β. This means that
fn0(K2), which has positive measure, have all the atoms separated from each
other with respect to the Markov partition. Since we have a finite number of
Markov partition, one of them must have a set with positive measure such
that its leaves have only one atom. Hence all Markov partition must have
one atom, absurd.
3.4 Proof of item iv)
Suppose {F c
x}x∈Mis a measurable partition, then we can apply Rokhlin’s
theorem and we decompose the volume on probabilities mxon center leaves.
Let
AL={x∈M|mx(Fc
L(x)) ≥0.6},
where Fc
L(x) is the segment of Fc(x) of length Lon the induced metric and
centered at x.
Note that there is L∈Rsuch that vol(AL)>0. Let us suppose that f
contracts the center leaf, then f−1(Fc
L(f(x))) ⊃ Fc
L(x). Since f∗mx=mf(x),
for x∈AL,
mf(x)(Fc
L(f(x))) = mx(f−1(Fc
L(f(x)))) ≥mx(Fc
L(x)) ≥0.6.
So f(x)∈AL, by ergodicity f(AL)⊂ALimplies V ol(AL) = 1.
Claim: diamcAL∩ Fc
x≤2L, where diamcmeans the diameter of the set
inside the center leaf.
Suppose there exist y1, y2∈AL∩ F c
xwith dc(y1, y2)>2L. Then
Fc
L(y1)∩ Fc
L(y2) = ∅and mx(Fc
L(yi)) ≥0.6, i = 1,2.
Then
1≥mx(Fc
L(y1)∪ Fc
L(y2)) = mx(Fc
L(y1)) + mx(Fc
L(y2)) ≥0.6 + 0.6 = 1.2.
This absurd concludes the proof of the claim.
Claim: The decomposition has atom.
Define
L0=inf{L∈[0,∞)|V ol(AL) = 1}.
11
Note that V ol(AL0) = 1, to see that take a sequence Ln→L0and observe
that AL0=∩iALn. Let λ=inf ||Df −1|Ec||, let ε < 1 be such that ελ > 1.
For x∈AλL0
mf(x)(Fc
εL0(f(x))) = mx(f−1(Fc
εL0(f(x)))) ≥mx(Fc
L0(x)) ≥0.6.
Therefore f(x)∈AεL0. By ergodicity we may suppose AL0f-invariant,
hence V ol(AεL0) = 1. Absurd since εL0< L0. This means that L0= 0,
which implies atom.
Let us prove the converse. Suppose we have atomic decomposition, we
want to see that the partition through center leaves is a measurable partition.
Lift fto R3, by Hammerlindl [9] we may find a disk ˜
D2transverse to the
center foliation, by quasi-isometry of the center foliation we may take this
disk as big as we want. So take a disk such that its projection D2=π(˜
D2)
has the property:
Fc
x∩D26=∅,∀x∈T3.
Since the decomposition is atomic, we already know that it has one atom
per leaf. Let us define the following set of full measure:
ˆ
M=[
p∈A
Fc
loc(p),
where Ais the set of atoms, Fc
loc(p) is the segment of center leaf such that the
right extreme point is pand the left extreme point is on D2and #Fc
loc(p)∩
D2= 1.
Since D2is a separable metric space, {Fc
loc(p)}p∈Ais a measurable parti-
tion for ˆ
M. Therefore we have a family of subsets {Ai}i∈Nof ˆ
Msuch for all
p∈A
Fc
loc(p) = \
i∈N
Bi,where Bi∈ {Ai, Ac
i}.
4 Conjugacy
We begin by understanding how Lyapunov exponents vary with respect to
their linearization.
Proposition 4.1. Let f:T3→T3be a partially hyperbolic, not neces-
sarily ergodic nor volume preserving, and let Abe its linearization. Then
Rλu(f)dV ol ≤λu
A.
12
Proof. Suppose that Rλu
f(x)dV ol(x)> λu
A, then there exists a set Bof pos-
itive volume and a constant αsuch that λu
f(x)> α > λu
f∗,∀x∈B. Define
BN={x∈B| ||Df n|Eu
x|| ≥ enα;∀n≥N}.
Note that
B=
∞
[
N=1
BN,
this means that there is N0such that V ol(BN0)>0. Since Fu
fis absolutely
continuous then there is x∈Bsuch that Fu
f(x)∩BN0has positive volume
on the unstable leaf.
Let I⊂ Fu
f(x) be a compact segment with V olc(I∩BN0)>0 and
length(I) =: l(I)> M . Then
l(fn(I)) = Zfn(I)
dV olu=ZI
(fn)∗dV olu≥ZI∩AN0
(fn)∗dV olu
≥ZI∩BN0
||Df n|Eu
x||dV olu(x)≥enαV olc(I∩AN0).
Consider x, y the extremes of I= [x, y]. Then du(fn(x), f n(y)) = l(fn(I)).
Using quasi-isometry on the first inequality below we get
d(fn(x), fn(y))
d(An(x), An(y)) ≥ctedu(fn(x), f n(y))
d(An(x), An(y))
≥cte enα
enλu
A
V ol(I∩BN0)
d(x, y)
−→ ∞ as n→ ∞.
By Proposition 2.1 this ratio should be bounded. Absurd.
The same type of argument above give us:
Corollary 4.1. Zλs(f)≥λs(A).
We consider the following for the case of Anosov systems, for it will be
used later.
Corollary 4.2. Let fbe an Anosov diffeomorphism with the following split
on the tangent space T M =Ess ⊕Ews ⊕Euand Fws absolutely continuous.
Then λws
f≥λws
A.
13
Proof. The prove goes as before, with a minor change. We proceed, as pre-
viously, applying Proposition 2.1 with the following linear map π:Rn→Rn
which is the projection onto a center foliation of the linearization. The pro-
jection is with respect to the system of coordinate given by the foliations of
the linearization (xss, xws, xu)∈Rn.
4.1 Proof of Lemma 1.1
We only have to prove the implication, as the converse is a direct consequence
of the C1-conjugacy.
Let us suppose that fis partially hyperbolic with the following split of
the tangent space: T M =Ess ⊕Ews ⊕Eu. The next three lemmas concern
this case, the other case is reduced to this one by applying the inverse.
Lemma 4.1.
λu
f(m) = λu
f(p),∀p∈P er(f).
Proof. By ergodicity the set of transitive points Thas total volume. We
may assume that all points of Thave well defined Lyapunov exponents. For
x∈ T ; given ε > 0 let δ > 0 be such that by uniform continuity
|log||Df |Eu
y1|| − log||Df |Eu
y2|| | < ε, if d(y1, y2)< δ.
From the Shadowing lemma there is αsuch that for every α-pseudo orbit
is δshadowed by a real orbit. Given N0∈Nthere is n0∈Nand n0> N0such
that {...,fn0−1(x), x, f (x),...,fn0−1(x),...}is an α-pseudo orbit. Since it
is a pseudo-periodic orbit it is δshadowed by a periodic point with period
n0, call this point q. Using that Euis one dimensional, then
1
n0
log||Df n0|Eu
y1|| − 1
n0
log||Df n0|Eu
y2|| < ε.
Since we already know that λu
f(x) exists, this implies that λu
f(x) = λu
f(q),
hence λu
f(m) = λu
f(p) as we wanted.
Lemma 4.2.
λu
f(m) = λu
A.
Proof. We know that the topological entropy of Ais λu
A, the conjugacy gives
htop(f) = htop (A). From the theory of equilibrium states ([5]) the measure of
maximal entropy is given by the potential ψ= 0 and the equilibrium state
for the potential ψ=−logλugives the SRB measure, which is min our case.
And to see that both equilibrium states are the same we just need to see that
14
both potential are cohomologous ([5]). It means that both measures coincide
if, and only if,
1
n
n
X
i=1
(−log||Dffi(x)|Eu||) = cte, ∀xsuch that fn(x) = x.
Which is true by hypothesis.
Finally Pesin’s formula gives that hf(m) = Rλu
fdm =λu
f. Let us put all
this equalities below.
λu
A=htop(A) = htop (f) = hf(m) = Zλu
fdm =λu
f(p).
The lemma is then proved.
Lemma 4.3.
λws
f(p) = λws
A
Proof. By the above lemma we already know that λu
f(p) = λu
A; and λss
f(p)≥
λss
Aby Corollary 4.1. Hence, since we are on the volume preserving case
λss
f+λws
f+λu
f=λss
A+λws
A+λu
A, therefore we just need to see that λws
f(p)≥λws
A
which is the Corollary 4.2.
The above lemmas imply,
λ∗
f(p) = λ∗
A(h(p)),∀p∈P er(f).
The above equality gives what is known as periodic data, hence by Gogolev,
Guysinsky [8] fis C1conjugate to the linear one. Box
4.2 Proof of Theorem 1.2
We start from a linear Anosov with splitting T M =Ess ⊕Ews ⊕Eu. Let
φbe a volume preserving diffeomorphism which preserves the Ess direction.
This means it is absolutely continuous by Gogolev [7] and by Lemma 1.1 it
is not C1conjugate as we have changed the exponents.
5 Rigidity
The goal of this subsection is to prove Theorem 1.3. But first we construct
some Conditional measures with dynamical meaning. We shall associate to
15
each center leaf a class of measures differing from each other by a multipli-
cation of a positive real number in such a way that on each foliated box the
normalized element of this class will give the Rokhlin disintegration of the
measure. When the foliation satisfies the hypothesis on Theorem 1.3 we shall
be able to pick measurably on each leaf a representative with some dynam-
ical meaning, it will then help us to obtain some information on the center
Lyapunov exponent of f.
Lemma 5.1 (Avila, Viana, Wilkinson [1]).For any foliation boxes B,B′and
m-almost every x∈ B ∩ B′the restriction of mB
xand mB′
xto B ∩ B′coincide
up to a constant factor.
Proof. Let µBbe the measure on Σ obtained as the projection of m|B along
local leaves. Consider any C ⊂ B and let µCbe the projection of m|C on Σ,
dµC
dµB
∈(0,1], νCalmost every point.
For any measurable set E⊂ C
m(E) = ZΣ
mB
ξ(E)dµB(ξ) = ZΣ
mB
ξ(E)dµB
dµC
(ξ)dµC(ξ).
By essential uniqueness, this proves that the disintegration of m|C is given
by
mC
ξ=dµB
dµC
(ξ)mB
ξ;µC(ξ) almost every point.
Take C=B ∩ B′. Therefore dµB
dµC(ξ)mB
ξ|C =mC
ξ=dµB′
dµ′
C(ξ)mB′
ξ|C. Where µ′
C
is the projection of measure µon the transversal Σ′relative to the B′box.
Hence
mB
ξ|C =a(ξ)mB′
ξ|C,
where a(ξ) = dµB′
dµ′
C(ξ)(dµB
dµC(ξ))−1.
The above lemma implies the existence of a family {[mx]|x∈M}of
measures defined up to scaling and satisfying mx(M\Fx) = 0. The map
x7→ [mx] is constant on leaves of Fand the conditional probabilities mB
x
coincide almost everywhere with the normalized restrictions of [mx].
We observe that disintegration of a measure is an almost everywhere con-
cept, but in our case, since we shall be considering a C1center foliation,
we look to the conditional measures, of volume, defined everywhere. And,
16
more important, the number a(ξ) = dµB′
dµ′
C(ξ)(dµB
dµC(ξ))−1is indeed defined ev-
erywhere.
From now on we work on the lift. Let B:= Wsu (0) which is the saturation
by unstable leaves of the stable manifold of 0 ∈R3. By the semi-conjugacy
we know that every segment of center leaf which has size large enough keep
increasing by forward iteration. Let γ0be a length with this property. Let
B0be the two-dimensional topological surface such that each center leaf
intersects Band B0on two points, that are on the same center leaf and at
a distance γ0inside the center leaf. Let Bk:= fk(B0). Therefore, for each
point ξ∈Bthere is a unique point qk(ξ)∈Bkthat is on the same center
leaf as ξ. Since it will be clear to which point ξ qk(ξ) is associate, we use qk
instead to simplify notation.
Define the measure mξ,k by
mξ,k ([0, qk]) = λk,
where λis the center eigenvalue of the linearization, [0, qk] means the segment
[ξ, qk(ξ)] inside the center leaf of ξ.
Lemma 5.2.
f∗mx,k =λ−1mf(x),k+1.
Proof. Just see that
f∗mx,k([0, qk+1]) = λ−1mf(x),k+1 ([0, qk+1]).
Therefore if the sequence mx,k converges we would get
f∗mx=λ−1mf(x).
In general, by Lemma 5.1, for two foliated boxes Band B′we have
mB
x
dνB
dνC
=mB′
x
dνB′
dν′
C
.
We apply this formula to the following boxes: Band Bk, where Bcompre-
hend the segment of center leaves between Band B0, similarly Bkis formed
by the segment of center leaves bounded by Band Bk. Then
mB
x.1 = dµBk
dµB
mBk
x=dµBk
dµB
λ−kmx,k.
Note that λkmx,k =mBk
xby the definition of the disintegration. The
above proves
17
Lemma 5.3. On B:
mx,k = (dµBk
dµB
)−1λkmB
x.
To establish the convergence of the measures we shall need
Lemma 5.4. If Fcsatisfies the hypothesis of Theorem 1.3 then, there is a
uniform constant αsuch that
1
α
l(Fc
x∩ Bk)
l(Fc
x∩ B)≤dµBk
dµB
(x)≤αl(Fc
x∩ Bk)
l(Fc
x∩ B).
Proof. To calculate l(Fc
x∩Bk)
l(Fc
x∩B)we need to estimate the volume of a rectangular
box. The center holonomy on the center unstable and center stable folia-
tion are bounded by hypothesis. Therefore the volume can be calculated
(estimated) by height times base.
Hence,
dµBk
dµB
(x) = αx,k
l(Fc
x∩ Bk)
l(Fc
x∩ B),
where αx,k ∈[1/α, α], for all x∈R3and k∈N.
Therefore using Lemma 5.3 we get on B
mx,k =αx,k
l(Fc
x∩ Bk)
l(Fc
x∩ B)−1
λkmB
x.
For each xthere is a subsequence αx,ki(x)that converges to some ˜αxas
i(x)→ ∞.
Lemma 5.5. There is β > 0such that λk/l(Fc
x∩ Bk)∈[1/β, β ]for all x.
Proof. We need to estimate the fraction
||fn(H(x)) −fn(H(y))||
||An(x)−An(y)|| =||H◦An(x)−H◦An(x)||
||An(x)−An(y)|| .
By the triangular inequality:
||H◦An(x)−H◦An(y)||
||An(x)−An(y)|| ≤||H(An(x)) −An(x)||
||An(x)−An(y)|| +||An(x)−An(y)||
||An(x)−An(y)||
+||H(An(y)) −An(y)||
||An(x)−An(y)|| ,
18
and
||H◦An(x)−H◦An(y)||
||An(x)−An(y)|| ≥ −||H(An(x)) −An(x)||
||An(x)−An(y)|| +||An(x)−An(y)||
||An(x)−An(y)||
−||H(An(y)) −An(y)||
||An(x)−An(y)|| .
We know that His at a bounded distance of the identity and ||An(x)−An(y)||
is big.
By the above lemma we may assume that λk/l(Fc
x∩ Bk) goes to one as k
increases, otherwise incorporate it to the constant αx,k . Then sending ki(x)
to infinity
mx:= lim
ki(x)→∞ mx,ki(x)= (l(Fc
x∩ B)/˜αx)mB
x.
By going to a subsequence we obtained a convergent measure, but we
want it to have a specific property. Therefore we have to be more careful on
how to define them. We have seen above that f∗mx,k =λ−1mf(x),k+1, hence
for fixed xthere is ki(x)defined as above, but if we define ki(f(x)) =ki(x)+ 1
we obtain the limit satisfying f∗mx=λ−1mf(x). This means that for fixed x
we can define on the orbit of xmeasures satisfying the mentioned dynamical
property.
The measures are in fact indexed on a two dimensional plane manifold
Wsu. Hence, to define properly on the whole space, consider the rectangle
Asuch that the intersection of Ato the stable manifold of the origin is a
fundamental domain. And the sides formed by stable and unstable leaves.
Hence defining the measures as we mentioned above on Aand on its iterates
we get measures with dynamical properties.
From the above we conclude that we did get measures on each center leaf
with the property that f∗mx=λ−1mf(x). The construction of such measures
will help us to get information of the center Lyapunov exponent, since we
may recover λby the equality
df∗mx
dmf(x)
=λ−1.
Let us explore more deeply the above relation.
Lemma 5.6. By the above notation, the center Lyapunov exponent of f
exists everywhere and it is equal to λ.
19
Proof. Note that dfn
∗mx
dmfn(x)
(fn(x)) = λ−n.
Let us calculate the Radon-Nikodym derivative by another way. Let In
δ⊂
Fc
fn(x)be a segment of length δaround fn(x). Then
dfn
∗mx
dmfn(x)
(fn(x)) = lim
δ→0
fn
∗mx(In
δ)
mfn(x)(In
δ).
And
dfn
∗mx
dmfn(x)
(fn(x)) = lim
δ→0
mx(f−n(In
δ))
mfn(x)(In
δ)= lim
δ→0Rf−n(In
δ)ρxdλx
RIn
δρfn(x)dλf(x)
≈ρx(x)
ρfn(x)(fn(x)) lim
δ→0Rf−n(In
δ)dλx
RIn
δdλf(x)
≈lim
δ→0
ρx(x)
ρfn(x)RIn
δ||Df −n||dλx
RIn
δdλf(x)
≈ρx(x)
ρfn(x)(fn(x))||Df−n(x)||.
We then have
lim
δ→0
dfn
∗mx
dmfn(x)
(In
δ) = ρx(x)
ρfn(x)(fn(x))||Df−n(x)||.
From the other equalities we have
ρx(x)
ρfn(x)(fn(x))||Df−n(x)|| =λ−n.
By applying ”limn→∞1/n log” to the above equality we get
λc(x) = λ,
since the densities of mxare uniformly limited.
We are now ready for the
Proof of Theorem 1.3: First, let us prove that fis an Anosov diffeomor-
phism. We just need to analyze the behavior of Df on the center direction.
Let ε > 0 be such that λε:= λ−ε > 0. Since the center exponent exists for
every xthen, given x∈T3, there are nx∈Nand a neighborhood Uxof x
such that ∀x∈ Ux|Dfnx|Ec| ≥ enxλε. Since T3is a compact manifold take
a finite cover Ux1...Uxl. Let Ci<1 small enough so that for x∈ Uxithen
20
|Df n(x)|Ec| ≥ Cxienλεfor all n∈ {0,1,...,nxi}. Let C:= miniCxi, we
then have that |Dfn(x)|Ec| ≥ Cenλεfor all x∈T3and n∈N.
Since, in particular, the center foliation is absolutely continuous, from
Gogolev [7], one of the extremal exponents is constant on periodic points.
On the other hand the above theorem gives that in particular on the periodic
points the central exponent is also constant. Since we are on the conservative
case all Lyapunov exponents are constant on periodic points. Then Lemma
1.1 gives that fis C1-conjugate to its linearization.
Acknowledgements This article grew out of my PhD thesis, which was de-
fended at IMPA. I, therefore, would like to thank my former advisor Prof. M.
Viana to all the usefull conversations. This work was also greatly influenced
from the one month research period I spent at ICMC-USP on September
2011 working with Prof. A. Tahzibi. I leave here all my gratitute to them.
This work was partially supported by CNPq and FAPERJ. During the writ-
ing of this work the author counted with the support of FAPESP (process #
2011/21214-3).
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