Center foliation: absolute continuity, disintegration and rigidity
aa r X i v : . [ m a t h . D S ] F e b Center foliation: absolute continuity,disintegration and rigidity
R´egis Var˜ao ∗ October 18, 2018
Abstract
In this paper we address the issues of absolute continuity for thecenter foliation (as well as the disintegration on the non-absolute con-tinuous case) and rigidity of volume preserving partially hyperbolicdiffeomorphisms isotopic to a linear Anosov on T . It is shown thatthe disintegration of volume on center leaves may be neither atomicnor 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 continuityof the center foliation does not imply smooth conjugacy with its lin-earization. Imposing stronger conditions besides absolute continuityon the center foliation, smooth conjugacy is obtained. 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 orof time-one maps of Anosov flows) studied by Avila, Viana and Wilkinson[1]. We state their dichotomy: ∗ [email protected] Atomic disintegration : If the center foliation is non-absolutely contin-uous, then there exists k ∈ N and a full volume subset that intersectseach center leaf on exactly k points/orbits. • Rigidity : If the center foliation is absolutely continuous then the diffeo-morphism is smoothly conjugate to a rigid model (a rotation extensionof an Anosov diffeomorphism or the time-one map of an Anosov flow);On three-dimensional manifolds, the only known examples of partiallyhyperbolic diffeomorphism are of skew-product type, perturbation of time-one of Anosov flows and diffeomorphisms derives from a linear Anosov. Infact it is conjecture by E. Pujals that these are all the possibilities (see[4] for precise statements). Avila, Viana and Wilkinson [1] have treateddiffeomorphisms on the first two cases and we treat in this work the thirdcase.We deal with derived from Anosov (DA) diffeomorphisms, that is, f isa DA diffeomorphism if it partially hyperbolic and lies in the isotopy classof some hyperbolic linear automorphism A . We refer to A as the the lin-earization of f . Every partially hyperbolic diffeomorphism has the followingsplitting on the tangent space T M = E s ⊕ E c ⊕ E u (see § E s is a contracting direction, E u is an expanding direction and thecenter direction E c has an intermediate behavior. We consider Anosov diffeo-morphisms (see § T M = E s ⊕ E c ⊕ E u and the center directionis uniformly contracting or expanding. It then makes sense to talk aboutthe center foliation of an Anosov (partially hyperbolic) diffeomorphism, aswe do on Theorem 1.1 and Theorem 1.2. A center foliation (see §
2) is aninvariant foliation by f tangent to the E c direction. We also mention that alldiffeomorphisms treated on this work are assumed to be at least C α . Thisimplies, in particular, that volume preserving Anosov on T are ergodic.We state now our results. For non-absolutely continuous center foliation,we show that it is possible to have a disintegration ( § Theorem 1.1.
For partially hyperbolic Anosov diffeomorphisms on T , vol-ume preserving, for which the center foliation is non-absolutely continuous,we have thati) there exists f Anosov for which the disintegration of volume on thecenter leaves are neither Lebesgue, nor atomic. n fact, such diffeomorphisms fill a dense subset of an infinite-dimensionalmanifold in the neighborhood of any hyperbolic linear automorphismsin 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 leafin one point;iv) the disintegration of volume on the center leaves is atomic if and onlyif 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 f on T for which the center foliation is absolutely continuous but f is not C -conjugate to its linearization.In fact, such diffeomorphisms fill a dense subset of an infinite-dimensionalmanifold in the neighborhood of any hyperbolic linear automorphism in thespace of volume preserving maps. As we shall see, Theorem 1.2 will be just a corollary of the followingresult, which is important on its own:
Lemma 1.1.
Let f be a volume preserving partially hyperbolic Anosov diffeo-morphism on T . Then, for any periodic points p, q the Lyapunov exponentson each of the directions (stable, center, unstable) are the same if and onlyif f is C conjugate to its linearization. Note that Theorem 1.2 implies that to obtain a rigidity result we mustimpose 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 f be a volume preserving DA diffeomorphism on T , withthe linearization A . If the center foliation is a C foliation and the centerholonomies inside the center-unstable, F cuf , and center-stable, F csf , leaves areuniformly bounded, then f is C conjugate to its linearization and, hence, isan Anosov diffeomorphism. rganization of the paper. In § § § § A diffeomorphism f of a compact Riemannian manifold M is called partiallyhyperbolic if there are constants λ < ˆ γ < < γ < µ and C > Df -invariant splitting of T M = E u ( x ) ⊕ E c ( x ) ⊕ E s ( x ) where1 C µ n || v || < || Df n v || , v ∈ E ux − { } ;1 C ˆ γ n || v || < || Df n v || < Cγ n || v || , v ∈ E cx − { } ; || Df n v || < Cλ n || v || , v ∈ E sx − { } . We say that a partially hyperbolic diffeomorphism is dynamically coherent if the subbundles E s ⊕ E c and E c ⊕ E u integrate into invariant foliations, F cs , F cu respectively. This implies in particular that there is a center foliation F c , which is obtained by an intersection of the other two: F c = F cs ∩ F cu .It was proved by Brin, Buragov, Ivanov [6] that Theorem 2.1.
Every partially hyperbolic diffeomorphism on T is dynami-cally coherent. Let (
M, µ, B ) be a probability space, where M is a compact metric space, µ a probability and B the borelian σ -algebra. Given a partition P of M bymeasurable sets, we associate the following probability space ( P , e µ, e B ), where e µ := π ∗ µ , e B := π ∗ B . and π : M → P is the canonical projection associate toa point of M the partition element that contains it.For a given a partition P , a family { µ P } p ∈P is a system of conditionalmeasures for µ (with respect to P ) if 4) given φ ∈ C ( M ), then P R φµ P is measurable;ii) µ P ( P ) = 1 e µ -a.e.;iii) if φ ∈ C ( M ), then Z M φdµ = Z P (cid:18)Z P φdµ P (cid:19) d e µ .We call P a measurable partition (w.r.t. µ ) if there exist a family { A i } i ∈ N of borelian sets and a set F of full µ -measure such that for every P ∈ P thereexists a sequence { B i } i ∈ N , where B i ∈ { A i , A ci } such that P ∩ F = ∩ i ∈ N B i ∩ F .The following result is also known as Rokhlin’s disintegration Theorem. Theorem 2.2.
Let P be a measurable partition of a compact metric space M and µ a borelian probability. Then there exists a disintegration by conditionalmeasures for µ .Remark. On Theorem 1.1 the meaning of “disintegration of volume onthe center leaves are neither Lebesgue, nor atomic” means that on a foliatedbox, since the center foliation form a measurable partition we can apply onthis foliated box the Rokhlin’s disintegration Theorem and the conditionalmeasures 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 thedisintegration locally is neither Lebesgue nor atomic.
Let F be a foliation and disintegrate the volume inside a foliated box. If theconditional measure m L on the leave satisfies that m L << Leb L for almostevery leaf, then F is said to be an absolutely continuous foliation , where Leb L is the Lebesgue measure on the leaf L .We state a result due to Gogolev [7] which shall be our starting point tounderstand absolute continuity for partially hyperbolic diffeomorphism withnon-compact center leaves. Theorem 2.3.
Let f : T → T be an Anosov diffeomorphism with splittingof the form E s ⊕ E wu ⊕ E uu , then F cf is absolutely continuous if and only λ uu ( p ) = λ uu ( q ) for all periodic points p and q . Where λ uu is the Lyapunov exponent on the E uu direction. By a Derived from Anosov (DA) diffeomorphism f : T → T we mean apartially hyperbolic homotopic to a linear Anosov diffeomorphism A . We5all this linear Anosov as the linearization of f . In fact, f is semi-conjugatedto its linearization. The itens from the Theorem below, which proof canbe found on Sambarino [10], show that the semi-conjugacy has in fact goodproperties. Theorem 2.4.
Let B : R → R be a linear hyperbolic isomorphism. Then,there exists C > such that if G : R → R is a homeomorphism such that sup {|| G ( x ) − Bx || | x ∈ R } = K < ∞ then there exists H : R → R continuous and surjective such that: • B ◦ H = H ◦ G ; • || H ( x ) − x || ≤ CK for all x ∈ R ; • H ( x ) is characterized as the unique point y such that || B n ( y ) − G m ( x ) || ≤ CK, ∀ n ∈ Z ; • H ( x ) = H ( y ) if and only if || G n ( x ) − G ( y ) || ≤ CK , ∀ n ∈ Z , and ifand only if sup n ∈ Z {|| G n ( x ) − G n ( y ) ||} < ∞ ; • if B ∈ SL (3 , Z ) and G is the lift of g : T → T then H induces h : T → T continuous and onto such that B ◦ h = h ◦ g and dist C ( h, id ) ≤ Cdist C ( B, g ) . The geometrical property we shall need later is given by Hammerlindl [9]:
Proposition 2.1.
Let f be a partially hyperbolic and A be its linearization.Denote by ˜ f and ˜ A the lift to R n of f and A respectively. Then for each k ∈ Z and C > there is M > and a linear map π : R n → R n such thatfor all x, y ∈ R n || x − y || > M ⇒ C < || π ( ˜ f k ( x ) − ˜ f k ( y )) |||| π ( ˜ A k ( x ) − ˜ A k ( y )) || < C. We dedicate this section for the proof of Theorem 1.1.6 .1 Proof of item i)
Consider a linear volume preserving Anosov with the following split
T M = E ss ⊕ E ws ⊕ E u . Let φ be a volume preserving diffeomorphism which preservesthe E u direction. By Baraviera, Bonnatti [2] R λ wsA dV ol = R λ wsA ◦ φ dV ol . Let h be the conjugacy between A and f , f ◦ h = h ◦ A . We claim that h is volumepreserving and sends center leaves to center leaves. To see that h is volumepreserving note that f and A have the same topological entropy λ uA . Hence, h ∗ V ol is a measure of maximal entropy. Observe that the perturbation A ◦ ψ of A is such that it preserves the E u exponent, which means that by theequilibrium state theory (see Bowen [5]) the potentials 0 and − log || Df | E u || are cohomological and therefore give the same equilibrium states. That is, h ∗ V ol = V ol . And the fact that h ( F c ) = F c comes from Lemma 2 of [8]. Claim: F cA ◦ ψ is not absolutely continuous.Suppose, by contradiction, that it is absolutely continuous, then Theorem2.3 implies λ ssf ( p ) = cte for all periodic point p . By contruction we have λ uf ( p ) = λ uA . Since we are on the volume preserving case, λ ws ( p ) is alsoconstant on periodic points. Therefore, by Lemma 1.1 f is C -conjugateto A , but this would imply R λ wsf dV ol = R λ wsA dV ol . Which is absurd byProposition 0.3 of Baravieira, Bonnatti [2]. Claim:
The disintegration of volume on center leaves of A ◦ φ is neihterLebesgue nor atomic.It is not Lebesgue because it is not absolutely continuous. And to see thatit is not atomic, note that since h is volume preserving and sends center leavesonto center leaves we can induce (by push forward) the disintegration on thecenter leaves of A to the center leaves of A ◦ φ . And since the disintegrationfor A is Lebesgue, this means that the disintegration for A ◦ φ is not atomic. (cid:3) By ergodicity we know that the Birkhoff set B = { x ∈ T | /n n − X i =0 δ f i ( x ) → V ol as n → ∞} has full measure. Claim.
If there is a center leaf such that F c ∩ B has positive Lebesguemeasure, then the center foliation is absolutely continuous. Proof of the Claim.
Let D be any disc on the central foliation and consider7he following construction µ n = 1 n n − X j =0 f j ∗ (cid:18) m D m D ( D ) (cid:19) , where m D means the Lebesgue measure on the central leaf. It turns out thatthese measures converge to a measure µ such that the disintegration of µ on the center leaves are absolutely continuous with respect to the Lebesguemeasure. This is a well-known construction of measures, studied by Pesin,Sinai in the eighties. For more references see [3] Chapter 11 and the referencestherein. Although Pesin, Sinai studied these measures for the case of thedisc D in the unstable foliation, for the center foliation, in our case, thisconstruction is the same. Gogolev, Guysinsky [8] have worked explicitly onthis 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 ∩ B for which it has positive Lebesgue measure on the center leaf.By hypothesis there exists such a disc. It turns out that these measures stillconverge to a measure with conditional measures absolutely continuous tothe Lebesgue measure on the center leaf (Lemma 11.12 [3]). Since the pointson B have the property 1 /n P n − i =0 δ f i ( x ) → V ol , it turns out that the sequence µ n converges to the volume. Hence, volume has Lebesgue disintegration onthe center leaves. Which proves the claim.From the claim, since we are in the case where the center foliation isnon-absolutely continuous, we must have that the center foliation intersects B on a set of zero Lebesgue measure. But the conditional measures give fullmeasure to B , since B has full measure. Therefore the conditional measuresare singular with respect to the Lebesgue measure. And item ii) is proved. (cid:3) On what follows R i will 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 R i .Proof. On each Markov rectangle we may apply Rokhlin’s disintegration the-orem on center leaves. Therefore, when writing m x we mean the conditionalmeasure for the disintegration on Markov rectangle that contains x . Considerthe set A δ = { x ∈ A | m x ( x ) ≤ δ } . Since f ( F cR ( x ) ( x )) ⊃ F cR ( f ( x )) ( f ( x )), we8ave that f ∗ m x ( I ) ≤ m f ( x ) ( I ) where I is inside the connected component of F cf ( x ) ∩ R ( f ( x )) that contains f n ( x ). If f ( x ) ∈ A δ , then m x ( x ) = f ∗ m x ( f ( x )) ≤ m x ( f ( x )) ≤ δ. Hence, f − ( A δ ) ⊂ A δ .By ergodicity, since our Anosov is volume preserving on T , A δ has fullmeasure or zero measure. Let δ be the discontinuity point of the function δ ∈ [0 , V ol ( A δ ). This implies that almost every atom has weight δ . Lemma 3.2.
On every Markov partition R i the conditional measures havethe same number of atoms, with the same weight.Proof. This is a direct consequence from the above lemma. Since all theatoms have the same weight δ the conditional measures must have 1 /δ number of atoms. Lemma 3.3.
There is a set of full volume B , of atoms, such that if x ∈ B ,then B ∩ F cx is contained in the connected component of R i ( x ) ∩ F cx thatcontains x .Proof. Let A be the set of atoms and T be the set of transitive points. Bothsets have full volume measure by ergodicity. Suppose, by contradiction, thatthere is a subset A ⊂ A of positive volume measure such that ∀ x ∈ A weget A ∩ R ci ( x ) = ∅ , where R ci ( x ) is the complement of the Markov partitionthat contains x , note that V ol ( A ∩ T ) >
0. Define the following map h : A ∩ T → R x h ( x ) = d F cx ( R i ( x ) , R ′ i ( x ) ) , where d F cx ( R i ( x ) , R ′ i ( x ) ) means the distance inside the center leaf of the Markovrectangle R i ( x ) to the closest Markov rectangle that has an atom which wecall R ′ i ( x ) .Since h is a measurable map, there exists K ⊂ A ∩ T , with V ol ( K ) > h is a continuous map when restricted to K . And since volume isa regular measure, there is compact set K ⊂ K , also with positive volumemeasure.Let α = M ax x ∈ K h ( x ). Fix z ∈ R i ( z ) , and consider a ball small enoughsuch that B ( z , r ) ⊂ intR i ( z ) . Hence, ∀ y ∈ K , let n y ∈ N be an integerbig enough so that, since f is uniformly expanding in the center direction, f − n y ( F c ( y, α )) ⊂ B ( z , r ) ⊂ intR i ( z ) .It means that we have at least doubled the number of atoms inside R i ( z ) ,which is an absurd since we have already shown that the number of atomsare constant on each Markov partition.9 emma 3.4. There is a set of full volume B , of atoms, such that if x ∈ B ,then B ∩ F cx is contained in the connected component of R i ( x ) ∩ F cx thatcontains x .Proof. Let A be the set of atoms and T be the set of transitive points. Bothsets have full volume measure by ergodicity. Suppose, by contradiction, thatthere is a subset A ⊂ A of positive volume measure such that ∀ x ∈ A weget A ∩ R ci ( x ) = ∅ , where R ci ( x ) is the complement of the Markov partitionthat contains x , note that V ol ( A ∩ T ) >
0. Define the following map h : A ∩ T → R x h ( x ) = d F cx ( R i ( x ) , R ′ i ( x ) ) , where d F cx ( R i ( x ) , R ′ i ( x ) ) means the distance inside the center leaf of the Markovrectangle R i ( x ) to the closest Markov rectangle that has an atom which wecall R ′ i ( x ) .Since h is a measurable map, there exists K ⊂ A ∩ T , with V ol ( K ) > h is a continuous map when restricted to K . And since volume isa regular measure, there is compact set K ⊂ K , also with positive volumemeasure.Let α = M ax x ∈ K h ( x ). Fix z ∈ R i ( z ) , and consider a ball small enoughsuch that B ( z , r ) ⊂ intR i ( z ) . Hence, ∀ y ∈ K , let n y ∈ N be an integerbig enough so that, since f is uniformly expanding in the center direction, f − n y ( F c ( y, α )) ⊂ B ( z , r ) ⊂ intR i ( z ) .It means that we have at least doubled the number of atoms inside R i ( z ) ,which is an absurd since we have already shown that the number of atomsare constant on each Markov partition. Lemma 3.5.
There is a set of full volume B ⊂ B such that the centerfoliation intersects B at most on one point.Proof. By contradiction suppose that the number of atoms on all Markovpartition are greater than one. Let A be a set with full volume measureinside the union of the Markov rectangle such that if x ∈ A , then A ∩ F cx,loc has the same number of points, in this case greater than one. Where F cx,loc isthe connected set of the center foliation restricted to the Markov rectanglethat intersects x . We define the map h : A → R x h ( x )10here h ( x ) is the smallest distance between the atoms of F cx,loc . By Lusin’stheorem there is a set K ⊂ A of positive measure for which h is continuous.Since volume is regular, there is a compact subset K of K with positivemeasure. Let α = min x ∈ K h ( x ).Let β > F cloc . Let n ∈ N bigenough so that any segment of a center leaf with length greater than orequal to α has the length of its n th iterate greater than β . This means that f n ( K ), which has positive measure, have all the atoms separated from eachother with respect to the Markov partition. Since we have a finite number ofMarkov partition, one of them must have a set with positive measure suchthat its leaves have only one atom. Hence all Markov partition must haveone atom, absurd. Suppose {F cx } x ∈ M is a measurable partition, then we can apply Rokhlin’stheorem and we decompose the volume on probabilities m x on center leaves.Let A L = { x ∈ M | m x ( F cL ( x )) ≥ . } , where F cL ( x ) is the segment of F c ( x ) of length L on the induced metric andcentered at x .Note that there is L ∈ R such that vol ( A L ) >
0. Let us suppose that f contracts the center leaf, then f − ( F cL ( f ( x ))) ⊃ F cL ( x ). Since f ∗ m x = m f ( x ) ,for x ∈ A L , m f ( x ) ( F cL ( f ( x ))) = m x ( f − ( F cL ( f ( x )))) ≥ m x ( F cL ( x )) ≥ . . So f ( x ) ∈ A L , by ergodicity f ( A L ) ⊂ A L implies V ol ( A L ) = 1. Claim: diam c A L ∩ F cx ≤ L , where diam c means the diameter of the setinside the center leaf.Suppose there exist y , y ∈ A L ∩ F cx with d c ( y , y ) > L . Then F cL ( y ) ∩ F cL ( y ) = ∅ and m x ( F cL ( y i )) ≥ . , i = 1 , . Then1 ≥ m x ( F cL ( y ) ∪ F cL ( y )) = m x ( F cL ( y )) + m x ( F cL ( y )) ≥ . . . . This absurd concludes the proof of the claim.
Claim:
The decomposition has atom.Define L = inf { L ∈ [0 , ∞ ) | V ol ( A L ) = 1 } . V ol ( A L ) = 1, to see that take a sequence L n → L and observethat A L = ∩ i A L n . Let λ = inf || Df − | E c || , let ε < ελ > x ∈ A λL m f ( x ) ( F cεL ( f ( x ))) = m x ( f − ( F cεL ( f ( x )))) ≥ m x ( F cL ( x )) ≥ . . Therefore f ( x ) ∈ A εL . By ergodicity we may suppose A L f -invariant,hence V ol ( A εL ) = 1. Absurd since εL < L . This means that L = 0,which implies atom.Let us prove the converse. Suppose we have atomic decomposition, wewant to see that the partition through center leaves is a measurable partition.Lift f to R , by Hammerlindl [9] we may find a disk ˜ D transverse to thecenter foliation, by quasi-isometry of the center foliation we may take thisdisk as big as we want. So take a disk such that its projection D = π ( ˜ D )has the property: F cx ∩ D = ∅ , ∀ x ∈ T . Since the decomposition is atomic, we already know that it has one atomper leaf. Let us define the following set of full measure:ˆ M = [ p ∈ A F cloc ( p ) , where A is the set of atoms, F cloc ( p ) is the segment of center leaf such that theright extreme point is p and the left extreme point is on D and F cloc ( p ) ∩ D = 1.Since D is a separable metric space, {F cloc ( p ) } p ∈ A is a measurable parti-tion for ˆ M . Therefore we have a family of subsets { A i } i ∈ N of ˆ M such for all p ∈ A F cloc ( p ) = \ i ∈ N B i , where B i ∈ { A i , A ci } . (cid:3) We begin by understanding how Lyapunov exponents vary with respect totheir linearization.
Proposition 4.1.
Let f : T → T be a partially hyperbolic, not neces-sarily ergodic nor volume preserving, and let A be its linearization. Then R λ u ( f ) dV ol ≤ λ uA . roof. Suppose that R λ uf ( x ) dV ol ( x ) > λ uA , then there exists a set B of pos-itive volume and a constant α such that λ uf ( x ) > α > λ uf ∗ , ∀ x ∈ B . Define B N = { x ∈ B | || Df n | E ux || ≥ e nα ; ∀ n ≥ N } . Note that B = ∞ [ N =1 B N , this means that there is N such that V ol ( B N ) >
0. Since F uf is absolutelycontinuous then there is x ∈ B such that F uf ( x ) ∩ B N has positive volumeon the unstable leaf.Let I ⊂ F uf ( x ) be a compact segment with V ol c ( I ∩ B N ) > length ( I ) =: l ( I ) > M . Then l ( f n ( I )) = Z f n ( I ) dV ol u = Z I ( f n ) ∗ dV ol u ≥ Z I ∩ A N ( f n ) ∗ dV ol u ≥ Z I ∩ B N || Df n | E ux || dV ol u ( x ) ≥ e nα V ol c ( I ∩ A N ) . Consider x, y the extremes of I = [ x, y ]. Then d u ( f n ( x ) , f n ( y )) = l ( f n ( I )).Using quasi-isometry on the first inequality below we get d ( f n ( x ) , f n ( y )) d ( A n ( x ) , A n ( y )) ≥ cte d u ( f n ( x ) , f n ( y )) d ( A n ( x ) , A n ( y )) ≥ cte e nα e nλ uA V ol ( I ∩ B N ) 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. Z λ s ( f ) ≥ λ s ( A ) . We consider the following for the case of Anosov systems, for it will beused later.
Corollary 4.2.
Let f be an Anosov diffeomorphism with the following spliton the tangent space T M = E ss ⊕ E ws ⊕ E u and F ws absolutely continuous.Then λ wsf ≥ λ wsA . roof. The prove goes as before, with a minor change. We proceed, as pre-viously, applying Proposition 2.1 with the following linear map π : R n → R n 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 ofthe linearization ( x ss , x ws , x u ) ∈ R n . We only have to prove the implication, as the converse is a direct consequenceof the C -conjugacy.Let us suppose that f is partially hyperbolic with the following split ofthe tangent space: T M = E ss ⊕ E ws ⊕ E u . The next three lemmas concernthis case, the other case is reduced to this one by applying the inverse. Lemma 4.1. λ uf ( m ) = λ uf ( p ) , ∀ p ∈ P er ( f ) . Proof.
By ergodicity the set of transitive points T has total volume. Wemay assume that all points of T have well defined Lyapunov exponents. For x ∈ T ; given ε > δ > | log || Df | E uy || − log || Df | E uy || | < ε, if d ( y , y ) < δ. From the Shadowing lemma there is α such that for every α -pseudo orbitis δ shadowed by a real orbit. Given N ∈ N there is n ∈ N and n > N suchthat { . . . , f n − ( x ) , x, f ( x ) , . . . , f n − ( x ) , . . . } is an α -pseudo orbit. Since itis a pseudo-periodic orbit it is δ shadowed by a periodic point with period n , call this point q . Using that E u is one dimensional, then (cid:12)(cid:12)(cid:12)(cid:12) n log || Df n | E uy || − n log || Df n | E uy || (cid:12)(cid:12)(cid:12)(cid:12) < ε. Since we already know that λ uf ( x ) exists, this implies that λ uf ( x ) = λ uf ( q ),hence λ uf ( m ) = λ uf ( p ) as we wanted. Lemma 4.2. λ uf ( m ) = λ uA . Proof.
We know that the topological entropy of A is λ uA , the conjugacy gives h top ( f ) = h top ( A ). From the theory of equilibrium states ([5]) the measure ofmaximal entropy is given by the potential ψ = 0 and the equilibrium statefor the potential ψ = − logλ u gives the SRB measure, which is m in our case.And to see that both equilibrium states are the same we just need to see that14oth potential are cohomologous ([5]). It means that both measures coincideif, and only if,1 n n X i =1 ( − log || Df f i ( x ) | E u || ) = cte, ∀ x such that f n ( x ) = x. Which is true by hypothesis.Finally Pesin’s formula gives that h f ( m ) = R λ uf dm = λ uf . Let us put allthis equalities below. λ uA = h top ( A ) = h top ( f ) = h f ( m ) = Z λ uf dm = λ uf ( p ) . The lemma is then proved.
Lemma 4.3. λ wsf ( p ) = λ wsA Proof.
By the above lemma we already know that λ uf ( p ) = λ uA ; and λ ssf ( p ) ≥ λ ssA by Corollary 4.1. Hence, since we are on the volume preserving case λ ssf + λ wsf + λ uf = λ ssA + λ wsA + λ uA , therefore we just need to see that λ wsf ( p ) ≥ λ wsA 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] f is C conjugate to the linear one. Box
We start from a linear Anosov with splitting
T M = E ss ⊕ E ws ⊕ E u . Let φ be a volume preserving diffeomorphism which preserves the E ss direction.This means it is absolutely continuous by Gogolev [7] and by Lemma 1.1 itis not C conjugate as we have changed the exponents. (cid:3) The goal of this subsection is to prove Theorem 1.3. But first we constructsome
Conditional measures with dynamical meaning . We shall associate to15ach 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 thenormalized element of this class will give the Rokhlin disintegration of themeasure. When the foliation satisfies the hypothesis on Theorem 1.3 we shallbe 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 centerLyapunov exponent of f . Lemma 5.1 (Avila, Viana, Wilkinson [1]) . For any foliation boxes B , B ′ andm-almost every x ∈ B ∩ B ′ the restriction of m B x and m B ′ x to B ∩ B ′ coincideup to a constant factor.Proof. Let µ B be the measure on Σ obtained as the projection of m |B alonglocal leaves. Consider any C ⊂ B and let µ C be the projection of m |C on Σ, dµ C dµ B ∈ (0 , , ν C almost every point . For any measurable set E ⊂ C m ( E ) = Z Σ m B ξ ( E ) dµ B ( ξ ) = Z Σ m B ξ ( E ) dµ B dµ C ( ξ ) dµ C ( ξ ) . By essential uniqueness, this proves that the disintegration of m |C is givenby m C ξ = dµ B dµ C ( ξ ) m B ξ ; µ C ( ξ ) almost every point . Take C = B ∩ B ′ . Therefore dµ B dµ C ( ξ ) m B ξ |C = m C ξ = dµ B′ dµ ′C ( ξ ) m B ′ ξ |C . Where µ ′C is the projection of measure µ on the transversal Σ ′ relative to the B ′ box.Hence m B ξ |C = a ( ξ ) m B ′ ξ |C , where a ( ξ ) = dµ B′ dµ ′C ( ξ )( dµ B dµ C ( ξ )) − . The above lemma implies the existence of a family { [ m x ] | x ∈ M } ofmeasures defined up to scaling and satisfying m x ( M \F x ) = 0. The map x [ m x ] is constant on leaves of F and the conditional probabilities m B x coincide almost everywhere with the normalized restrictions of [ m x ].We observe that disintegration of a measure is an almost everywhere con-cept, but in our case, since we shall be considering a C center foliation,we look to the conditional measures, of volume, defined everywhere. And,16ore important, the number a ( ξ ) = dµ B′ dµ ′C ( ξ )( dµ B dµ C ( ξ )) − is indeed defined ev-erywhere.From now on we work on the lift. Let B := W su (0) which is the saturationby unstable leaves of the stable manifold of 0 ∈ R . By the semi-conjugacywe know that every segment of center leaf which has size large enough keepincreasing by forward iteration. Let γ be a length with this property. Let B be the two-dimensional topological surface such that each center leafintersects B and B on two points, that are on the same center leaf and ata distance γ inside the center leaf. Let B k := f k ( B ). Therefore, for eachpoint ξ ∈ B there is a unique point q k ( ξ ) ∈ B k that is on the same centerleaf as ξ . Since it will be clear to which point ξ q k ( ξ ) is associate, we use q k instead to simplify notation.Define the measure m ξ,k by m ξ,k ([0 , q k ]) = λ k , where λ is the center eigenvalue of the linearization, [0 , q k ] means the segment[ ξ, q k ( ξ )] inside the center leaf of ξ . Lemma 5.2. f ∗ m x,k = λ − m f ( x ) ,k +1 . Proof.
Just see that f ∗ m x,k ([0 , q k +1 ]) = λ − m f ( x ) ,k +1 ([0 , q k +1 ]) . Therefore if the sequence m x,k converges we would get f ∗ m x = λ − m f ( x ) . In general, by Lemma 5.1, for two foliated boxes B and B ′ we have m B x dν B dν C = m B ′ x dν B ′ dν ′C . We apply this formula to the following boxes: B and B k , where B compre-hend the segment of center leaves between B and B , similarly B k is formedby the segment of center leaves bounded by B and B k . Then m B x . dµ B k dµ B m B k x = dµ B k dµ B λ − k m x,k . Note that λ k m x,k = m B k x by the definition of the disintegration. Theabove proves 17 emma 5.3. On B : m x,k = ( dµ B k dµ B ) − λ k m B x . To establish the convergence of the measures we shall need
Lemma 5.4. If F c satisfies the hypothesis of Theorem 1.3 then, there is auniform constant α such that α l ( F cx ∩ B k ) l ( F cx ∩ B ) ≤ dµ B k dµ B ( x ) ≤ α l ( F cx ∩ B k ) l ( F cx ∩ B ) . Proof.
To calculate l ( F cx ∩B k ) l ( F cx ∩B ) we need to estimate the volume of a rectangularbox. 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, dµ B k dµ B ( x ) = α x,k l ( F cx ∩ B k ) l ( F cx ∩ B ) , where α x,k ∈ [1 /α, α ], for all x ∈ R and k ∈ N .Therefore using Lemma 5.3 we get on B m x,k = (cid:18) α x,k l ( F cx ∩ B k ) l ( F cx ∩ B ) (cid:19) − λ k m B x . For each x there is a subsequence α x,k i ( x ) that converges to some ˜ α x as i ( x ) → ∞ . Lemma 5.5.
There is β > such that λ k /l ( F cx ∩ B k ) ∈ [1 /β, β ] for all x .Proof. We need to estimate the fraction || f n ( H ( x )) − f n ( H ( y )) |||| A n ( x ) − A n ( y ) || = || H ◦ A n ( x ) − H ◦ A n ( x ) |||| A n ( x ) − A n ( y ) || . By the triangular inequality: || H ◦ A n ( x ) − H ◦ A n ( y ) |||| A n ( x ) − A n ( y ) || ≤ || H ( A n ( x )) − A n ( x ) |||| A n ( x ) − A n ( y ) || + || A n ( x ) − A n ( y ) |||| A n ( x ) − A n ( y ) || + || H ( A n ( y )) − A n ( y ) |||| A n ( x ) − A n ( y ) || , || H ◦ A n ( x ) − H ◦ A n ( y ) |||| A n ( x ) − A n ( y ) || ≥ − || H ( A n ( x )) − A n ( x ) |||| A n ( x ) − A n ( y ) || + || A n ( x ) − A n ( y ) |||| A n ( x ) − A n ( y ) ||− || H ( A n ( y )) − A n ( y ) |||| A n ( x ) − A n ( y ) || . We know that H is at a bounded distance of the identity and || A n ( x ) − A n ( y ) || is big.By the above lemma we may assume that λ k /l ( F cx ∩ B k ) goes to one as k increases, otherwise incorporate it to the constant α x,k . Then sending k i ( x ) to infinity m x := lim k i ( x ) →∞ m x,k i ( x ) = ( l ( F cx ∩ B ) / ˜ α x ) m B x . By going to a subsequence we obtained a convergent measure, but wewant it to have a specific property. Therefore we have to be more careful onhow to define them. We have seen above that f ∗ m x,k = λ − m f ( x ) ,k +1 , hencefor fixed x there is k i ( x ) defined as above, but if we define k i ( f ( x )) = k i ( x ) + 1we obtain the limit satisfying f ∗ m x = λ − m f ( x ) . This means that for fixed x we can define on the orbit of x measures satisfying the mentioned dynamicalproperty.The measures are in fact indexed on a two dimensional plane manifold W su . Hence, to define properly on the whole space, consider the rectangle A such that the intersection of A to the stable manifold of the origin is afundamental domain. And the sides formed by stable and unstable leaves.Hence defining the measures as we mentioned above on A and on its iterateswe get measures with dynamical properties.From the above we conclude that we did get measures on each center leafwith the property that f ∗ m x = λ − m f ( x ) . The construction of such measureswill help us to get information of the center Lyapunov exponent, since wemay recover λ by the equality df ∗ m x dm f ( x ) = λ − . 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 . roof. Note that df n ∗ m x dm f n ( x ) ( f n ( x )) = λ − n . Let us calculate the Radon-Nikodym derivative by another way. Let I nδ ⊂F cf n ( x ) be a segment of length δ around f n ( x ). Then df n ∗ m x dm f n ( x ) ( f n ( x )) = lim δ → f n ∗ m x ( I nδ ) m f n ( x ) ( I nδ ) . And df n ∗ m x dm f n ( x ) ( f n ( x )) = lim δ → m x ( f − n ( I nδ )) m f n ( x ) ( I nδ ) = lim δ → R f − n ( I nδ ) ρ x dλ x R I nδ ρ f n ( x ) dλ f ( x ) ≈ ρ x ( x ) ρ f n ( x ) ( f n ( x )) lim δ → R f − n ( I nδ ) dλ x R I nδ dλ f ( x ) ≈ lim δ → ρ x ( x ) ρ f n ( x ) R I nδ || Df − n || dλ x R I nδ dλ f ( x ) ≈ ρ x ( x ) ρ f n ( x ) ( f n ( x )) || Df − n ( x ) || . We then havelim δ → df n ∗ m x dm f n ( x ) ( I nδ ) = ρ x ( x ) ρ f n ( x ) ( f n ( x )) || Df − n ( x ) || . From the other equalities we have ρ x ( x ) ρ f n ( x ) ( f n ( x )) || Df − n ( x ) || = λ − n . By applying ” lim n →∞ /n log ” to the above equality we get λ c ( x ) = λ, since the densities of m x are uniformly limited.We are now ready for the Proof of Theorem 1.3:
First, let us prove that f is an Anosov diffeomor-phism. We just need to analyze the behavior of Df on the center direction.Let ε > λ ε := λ − ε >
0. Since the center exponent exists forevery x then, given x ∈ T , there are n x ∈ N and a neighborhood U x of x such that ∀ x ∈ U x | Df n x | E c | ≥ e n x λ ε . Since T is a compact manifold takea finite cover U x . . . U x l . Let C i < x ∈ U x i then20 Df n ( x ) | E c | ≥ C x i e nλ ε for all n ∈ { , , . . . , n x i } . Let C := min i C x i , wethen have that | Df n ( x ) | E c | ≥ Ce nλ ε for all x ∈ T and n ∈ N .Since, in particular, the center foliation is absolutely continuous, fromGogolev [7], one of the extremal exponents is constant on periodic points.On the other hand the above theorem gives that in particular on the periodicpoints the central exponent is also constant. Since we are on the conservativecase all Lyapunov exponents are constant on periodic points. Then Lemma1.1 gives that f is C -conjugate to its linearization. (cid:3) 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 influencedfrom the one month research period I spent at ICMC-USP on September2011 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
References [1] A. Avila, M. Viana and A. Wilkinson.
Absolute continuity, Lyapunovexponents and rigidity I: Geodesic flows , Preprint, 2012.[2] A. Baraviera and C. Bonatti.
Removing zero Lyapunov exponents , ErgodicTheory Dynam. Systems, 2003.[3] C. Bonatti, L. Diaz and M. Viana.
Dynamics Beyond Uniform Hyperbol-icity , Springer-Verlag, 2004.[4] C. Bonatti and A. Wilkinson.
Transitive partially hyperbolic diffeomor-phisms on 3-manifolds , Topology, 2005.[5] Bowen, R.;
Equilibrium States and the Ergodic Theory of Anosov Diffeo-morphisms , Lecture Notes in Mathematics 470, 2008.[6] M. Brin, D. Burago and S. Ivanov.
Dynamical coherence of partially hy-perbolic diffeomorphisms of the 3-torus , J. Mod. Dyn. 3, 2009.[7] A. Gogolev.
How typical are pathological foliations in partially hyperbolicdynamics: an example , accepted to Israel Journal of Mathematics, 2012.[8] A. Gogolev and M. Guysinsky.
C1-differentiable conjugacy of Anosov dif-feomorphisms on three dimensional torus , Discrete and Continuous Dy-namical System-A, 22, 2008. 219] A. Hammerlindl.
Leaf conjugacies on the torus , Ph.D. Thesis, 2009.[10] M. Sambarino.