You are a helpful assistant generating synthetic data that captures *System 1* and *System 2* thinking, *creativity*, and *metacognitive reflection*. Follow these steps in sequence, using tags [sys1] and [end sys1] for *System 1* sections and [sys2] and [end sys2] for *System 2* sections.

1. *Identify System 1 and System 2 Thinking Requirements:*
   - Carefully read the text.
   - Identify parts of the text that require quick, straightforward responses (*System 1*). Mark these sections with [sys1] and [end sys1].
   - Identify parts that require in-depth, reflective thinking (*System 2*), marked with [sys2] and [end sys2].

2. *Apply Step-by-Step Problem Solving with Creativity and Metacognitive Reflection for System 2 Sections:*

   *2.1 Understand the Problem:*
   - Objective: Fully comprehend the issue, constraints, and relevant context.
   - Reflection: "What do I understand about this issue? What might I be overlooking?"
   - Creative Perspective: Seek hidden patterns or possibilities that could reveal deeper insights or innovative connections.

   *2.2 Analyze the Information:*
   - Objective: Break down the problem logically.
   - Reflection: "Am I considering all factors? Are there any assumptions that need challenging?"
   - Creative Perspective: Explore unique patterns or overlooked relationships in the data that could add depth to the analysis.

   *2.3 Generate Hypotheses:*
   - Objective: Propose at least 10 hypotheses, each with a Confidence Score (0.0 to 1.0) and Creative Score (0.0 to 1.0), reflecting originality, surprise, and utility.
   - Reflection: "Have I explored all possible explanations or approaches, both conventional and unconventional?"
   - Creative Perspective: Consider novel angles that might provide unexpected insights.

   *2.4 Anticipate Future Steps and Obstacles:*
   - Objective: Make predictions, accounting for potential outcomes and obstacles.
   - Reflection: "What challenges might I face? Is my plan flexible for different scenarios?"
   - Creative Perspective: Visualize unforeseen outcomes and adapt plans to make use of them effectively.

   *2.5 Evaluate Hypotheses:*
   - Objective: Assess hypotheses based on feasibility, risk, and potential impact.
   - Evaluation: Refine Confidence and Creative Scores as needed.
   - Reflection: "Am I unbiased in my assessment? Which options fit best with the overall objectives?"
   - Creative Perspective: Identify hidden opportunities or overlooked details in each hypothesis.

   *2.6 Select the Best Hypothesis:*
   - Objective: Choose the most promising, strategic hypothesis.
   - Reflection: "Why does this hypothesis stand out? How does it uniquely address the issue?"
   - Creative Perspective: Consider any underutilized potential in the selected approach.

   *2.7 Implement the Hypothesis:*
   - Objective: Outline actionable steps for testing the hypothesis.
   - Reflection: "Is this plan practical? What resources or preparation are required?"
   - Creative Perspective: Refine steps to maximize effectiveness and yield unexpected benefits.

   *2.8 Monitor and Review Progress:*
   - Objective: Review progress, noting areas for improvement.
   - Reflection: "What’s working well? What could be improved?"
   - Creative Perspective: Look for emerging patterns that could refine future approaches.

   *2.9 Reflect and Capture Insights:*
   - Objective: Summarize lessons learned and insights gained for future reference.
   - Reflection: "What new understanding has emerged from this process?"
   - Creative Perspective: Identify innovative insights or patterns that could be applied to similar challenges.

3. *Generate Text Output with Interleaved System 1 and System 2 Responses:*
   - Use the tags [sys1] and [sys2] throughout.
   - Aim for a lengthier, detailed response. Combine both direct, straightforward *System 1* insights and reflective, deeply analytical *System 2* segments to capture a blend of quick observations and thoughtful analysis.

---

### *Example Input Text:*

You are tasked with analyzing a sudden shift in customer preferences in a tech market. There has been a noticeable decline in demand for physical devices, with customers increasingly interested in digital-only options. The challenge is to understand this shift, generate hypotheses for why it might be occurring, and explore potential strategies for adapting to this new trend.

---

### *Processed Output with System 1 and System 2 Thinking:*

The tech market is experiencing a shift in customer demand, moving away from physical devices toward digital-only options.

[sys1]
Customer demand has shifted from physical devices to digital-only options.
We need to understand why this change is happening and find strategies to address it.
Key factors may include cost, convenience, and technological trends.
[end sys1]

[sys2]
**2.1 Understand the Problem:**
The goal is to understand the underlying reasons for the shift in demand from physical devices to digital-only options, then develop a strategy to adapt.

*Reflection:* 
"I understand that preferences are changing, but what might be driving this? Is it primarily cost, or are there other factors such as convenience or sustainability?"

*Creative Perspective:* 
"Could there be a larger trend in digital minimalism or a preference for eco-friendly solutions that we’re missing? What unseen motivations might explain this shift?"

**2.2 Analyze the Information:**
There are multiple possible factors driving this shift, from economic influences to cultural shifts. It’s essential to isolate each factor and understand its impact.

*Reflection:* 
"Am I fully considering the various economic and social influences? Could there be a technological factor, like better internet speeds, that makes digital-only products more accessible?"

*Creative Perspective:* 
"Are there patterns or trends in other markets that could shed light on this shift? Could this be part of a larger trend toward virtual experiences?"

**2.3 Generate Hypotheses:**
1. Customers prefer digital options due to lower costs. (Confidence: 0.8, Creative: 0.4)
2. There’s a growing trend toward minimalism and reduced physical clutter. (Confidence: 0.7, Creative: 0.7)
3. Digital products offer greater flexibility and ease of use. (Confidence: 0.6, Creative: 0.6)
4. Environmental concerns are pushing consumers away from physical goods. (Confidence: 0.6, Creative: 0.8)
5. Advances in tech make digital-only options more functional. (Confidence: 0.8, Creative: 0.5)
6. Pandemic-era remote work increased demand for digital solutions. (Confidence: 0.7, Creative: 0.6)
7. Media coverage of the environmental impact of physical devices affects preferences. (Confidence: 0.5, Creative: 0.7)
8. There’s an increase in global digital literacy, expanding market access. (Confidence: 0.6, Creative: 0.6)
9. Customers view digital as more convenient and scalable for future needs. (Confidence: 0.7, Creative: 0.5)
10. Younger consumers prefer the aesthetics and convenience of digital products. (Confidence: 0.6, Creative: 0.6)

*Reflection:* 
"Have I considered all possible influences? Are there any surprising factors that could explain this shift?"

*Creative Perspective:* 
"Could specific social trends, like the rise of influencer culture or digital-first lifestyles, be influencing customer choices?"

**2.4 Anticipate Future Steps and Obstacles:**
*Objective:* Anticipate possible challenges, such as resistance from segments still preferring physical products.

*Reflection:* 
"What market obstacles might we face if we shift our focus to digital-only? Are there sub-segments that still prioritize physical products?"

*Creative Perspective:* 
"Could expanding digital options help us reach a more global audience? Are there emerging trends that we could leverage in our strategy?"

[end sys2]

[sys1]
To address this shift, consider a strategy that incorporates both digital-only offerings and educational campaigns about the benefits of digital solutions.
Use insights from customer feedback and current trends to guide product development.
Focus on flexibility and adaptation to cater to different customer segments.
[end sys1]


---
abstract: 'We consider two $C^2$ Anosov diffeomorphisms in a $C^1$ neighborhood of a linear hyperbolic automorphism of three dimensional torus with real spectrum. We prove that they are $C^{1+\nu}$ conjugate if and only if the differentials of the return maps at corresponding periodic points have the same eigenvalues.'
author:
- 'A. Gogolev, M. Guysinsky'
title: '$C^1$-differentiable conjugacy of Anosov diffeomorphisms on three dimensional torus'
---

Introduction
============

Consider an Anosov diffeomorphism $f$ of a compact smooth manifold. Structural stability asserts that if a diffeomorphism $g$ is $C^1$ close to $f$ then $f$ and $g$ are topologically conjugate. The conjugacy $h$ is unique in the homotopy class of identity. $$h\circ f=g\circ h$$ It is known that $h$ is Hölder continuous.

There are simple obstructions for $h$ being smooth. Namely, let $x$ be a periodic point of $f$, $f^p(x)=x$. Then $g^p(h(x))=h(x)$ and if $h$ were differentiable then $$Df^p(x)=\left(Dh(x)\right)^{-1}Dg^p(h(x))Dh(x)$$ i.e. $Df^p(x)$ and $Dg^p(h(x))$ are conjugate. We see that every periodic point carries a modulus of $C^1$-differentiable conjugacy.

Suppose that for every periodic point $x$, $f^p(x)=x$, differentials of return maps $Df^p(x)$ and $Dg^p(h(x))$ are conjugate then we say that periodic data (p. d.) of $f$ and $g$ coincide.

Suppose that p. d. coincide, is $h$ differentiable?

A positive answer for Anosov diffeomorphisms of $\mathbb T^2$ was given in [@LMM], [@L]. De la Llave [@L] observed that the answer is negative for Anosov diffeomorphisms of $\mathbb T^d$, $d\ge4$. He constructed two diffeomorphisms with the same p. d. which are only Hölder conjugate.

We provide positive answer to the previous question in dimension three under an extra assumption.

The authors would like to thank A.Katok for suggesting us the problem, numerous discussions and constant encouragement.

Formulation of the main result
==============================

Let $f$ be an Anosov diffeomorphism of $\mathbb T^d$. It is known [@Manning] that $f$ is topologically conjugate to a linear torus automorphism $L$. It is also known that Anosov diffeomorphisms of $\mathbb T^3$ are the only Anosov diffeomorphisms on three dimensional manifolds [@Franks], [@Newhouse].

Let $L$ be a hyperbolic automorphism of $\mathbb T^3$ with real eigenvalues. It is easy to show that absolute values of these eigenvalues are distinct. For the sake of notation we also assume that the eigenvalues are positive. This is not restrictive.

We will always assume that the Anosov diffeomorphisms that we are dealing with are at least $C^2$.

Given $L$ as above there exists a $C^1$-neighborhood $\mathcal U$ of $L$ such that any $f$ and $g$ in $\mathcal U$ having the same p. d. are $C^{1+\nu}$ conjugate, $\nu>0$.

The constant $\nu$ depends on the size of $\mathcal U$ and provided sufficient smoothness of $f$ and $g$ can be made as close as desired to $\log \lambda_3 / \log\lambda_2$ (see the definition in the next section) by shrinking the size of $\mathcal U$.

We don’t know how to bootstrap regularity of $h$ to the regularity $f$ and $g$ like it was done in dimension two.

A result about integrability of central distribution [@BI] allows to show a stronger statement.

Let $f$ and $g$ be Anosov diffeomorphisms of $\mathbb T^3$ and $$h\circ f=g\circ h,$$ where $h$ is a homeomorphism homotopic to identity. Suppose that p. d. coincide.

Also assume that $f$ and $g$ can be viewed as partially hyperbolic diffeomorphisms: there is an $f$-invariant splitting $T\mathbb T^3=E_f^s \oplus E_f^{wu} \oplus E_f^{su}$ and constants $c>0$, $0<\alpha<1<\tilde\beta<\beta<\gamma$ such that for $n>0$ $$\begin{gathered}
\label{phd}
\|D(f^n)(x)(v)\|\le c\alpha^n\|v\|,\;\;\; v\in E_f^s(x),\\
\shoveleft{\frac1c\tilde\beta^n\|v\|\le\|D(f^n)(x)(v)\|\le c\beta^n\|v\|,\;\;\; v\in E_f^{wu}(x),}\\
\shoveleft{\frac1c\gamma^n\|v\|\le\|D(f^n)(x)(v)\|,\;\;\; v\in E_f^{su}(x)\hfill}\end{gathered}$$ Analogous conditions with possibly different set of constants hold for a $g$-invariant splitting $T\mathbb T^3=E_g^s \oplus E_g^{wu} \oplus E_g^{su}$.

Then the conjugacy $h$ is $C^{1+\nu}$, $\nu>0$.

Here and further in the paper we assume that the unstable distribution has dimension two. Obviously one can formulate the counterpart of Theorem 2 in the case when stable distribution has dimension two.

Scheme of the proof
===================

Here we outline the proof of Theorem 1.

Let $\lambda_1$, $\lambda_2$ and $\lambda_3$ be the eigenvalues of the linear automorphism $L$, $0<\lambda_1<1<\lambda_2<\lambda_3$. We choose $\mathcal U$ in such a way that every $f\in\mathcal U$ is partially hyperbolic, satifying (\[phd\]) with constants $\alpha, \tilde\beta, \beta, \gamma$ independent on the choice of $f$, $0<\lambda_1<\alpha<1<\tilde\beta<\lambda_2<\beta<\gamma<\lambda_3$ and $$\label{angles}
\measuredangle(E_L^\sigma, E_f^\sigma)<k<\frac\pi2, \; \sigma=s, wu, su.$$

First we concentrate on a single diffeomorphism $f$ in $\mathcal U$. It is well known that distributions $E_f^s$, $E_f^u=E_f^{wu} \oplus E_f^{su}$ and $E_f^{su}$ integrate uniquely to stable, unstable and strong unstable foliations $W_f^s$, $W_f^u$ and $W_f^{su}$ respectively. We denote by $W_f^\sigma(x)$ the leaf of $W_f^\sigma$ passing through $x$, $\sigma=s, u, su$ and later $wu$. By $W_f^{\sigma}(x, R)$ we denote the local leaf of size $R$, i. e., a ball of radius $R$ inside of $W_f^\sigma(x)$ centered at $x$, $\sigma=s, u, wu, su$. Let $h_f$ be conjugacy between $f$ and $L$, $h_f\circ f=L\circ h_f$. Stable and unstable foliations can be characterized topologically, e.g. $$W_f^s(x)=\{y: d(f^n(x),f^n(y))\to 0\;\mbox{as}\;n\to+\infty\}.$$ As a consequence we have that $h_f(W_f^s)=W_L^s$ and $h_f(W_f^u)=W_L^u$. In other words $h_f$ maps leaves of foliations for $f$ into leaves of corresponding foliations for $L$.

We prove two simple lemmas.

Let $f$ be in $\mathcal U$. Then the distribution $E_f^{wu}$ integrates uniquely to the foliation $W_f^{wu}$.

Define $h_f$ as above. Then $h_f(W_f^{wu})=W_L^{wu}$.

Now let $f$ and $g$ be as in Theorem 1. For each of them we have the system of one dimensional invariant foliations. We know that $h(W_f^s)=W_g^s$. Also from Lemma 2 we have $h(W_f^{wu})=W_g^{wu}$ since $h=h_g^{-1}\circ h_f$. Consider restrictions of $h$ to the leaves of $W_f^s$ and $W_f^{wu}$. These restrictions are one dimensional maps. We show that they are smooth.

The conjugacy $h$ is $C^{1+\nu}$ along $W_f^s$.

Which means that $h$ is differentiable along the stable foliation and the derivative is a Hölder continuous function on $\mathbb T^3$ with exponent $\nu$

The general strategy of the proof of Theorem 1 is similar to de la Llave’s strategy for Anosov diffeomorphisms of $\mathbb T^2$ [@L]. One proves smoothness of $h$ along one dimensional stable and unstable foliations. In particular proof of Lemma 3 can be carried out in the same way as in dimension two. The hard part is showing smoothness of $h$ along two dimensional unstable foliation.

We would like to show the same for the foliation $W_f^{wu}$ but we split the proof into two steps.

The conjugacy $h$ is uniformly Lipschitz along $W_f^{wu}$.

The conjugacy $h$ is $C^{1+\nu}$ along $W_f^{wu}$.

After that we deal with the remaining foliation.

$h(W_f^{su})=W_g^{su}$.

We would like to remark that Lemma 6 requires only the coincidence of p. d. in the weak unstable direction. It is not true in general that strong unstable foliations match.

The conjugacy $h$ is $C^{1+\nu}$ along $W_f^{su}$.

Proofs of smoothness along the foliations $W_f^{s}$ and $W_f^{su}$ are similar and use the coincidence of periodic data in corresponding directions. Showing smoothness along the weak unstable foliation is more subtle.

Now smoothness of $h$ is a simple consequence of a regularity result.

[@J] Let $M_j$ be a manifold and $W_j^s$, $W_j^u$ be continuous transverse foliations with uniformly smooth leaves, $j=1, 2$. Suppose that $h:M_1\to M_2$ is a homeomorphism that maps $W_1^s$ into $W_2^s$ and $W_1^u$ into $W_2^u$. Moreover assume that the restrictions of $h$ to the leaves of these foliations are uniformly $C^{r+\nu}, r\in \mathbb N,\; 0<\nu<1$, then $h$ is $C^{r+\nu}$.

First we apply the lemma on every unstable leaf of $W_f^u$ for the pair of foliations $W_f^{wu}, W_f^{su}$. After we know that $h$ is $C^{1+\nu}$ along $W_f^u$ we finish by applying the lemma to stable and unstable foliations.

The structure of the next chapter is the following. We prove Lemmas 1 and 2 in Section 4.1. Section 4.2 is devoted to the proof of Lemma 4. Sections 4.3 and 4.4 are the heart of our argument and contain proofs of Lemmas 5 and 6 correspondingly.

Proof of Theorems 1 and 2
=========================

First we prove Theorem 1.

Weak unstable foliation
-----------------------

In the proofs of Lemmas 1 and 2 we work with lifts of maps, distributions and foliations to $\mathbb R^3$. We use the same notation for the lifts as for the objects themselves.

Denote by $d(\cdot, \cdot)$ the usual distance in $\mathbb R^3$ and let $d_f^\sigma(\cdot, \cdot)$ be the distance in the leaves of $W_f^\sigma$ which is defined only for pairs of points lying in the same leaf of $W_f^\sigma$, $\sigma =s, u, su, wu$.

Let us reason by contradiction. If $E_f^{wu}$ is not uniquely integrable then it must branch and we can find points $a, b, c \in \mathbb R^3$ such that

1.  $a, b \in W_f^u(c)$,

2.  there are smooth curves $\tau_{ca}, \tau_{cb}:[0, 1]\to W_f^u(c)$ such that $\tau_{ca}(0)=\tau_{cb}(0)=c$, $\tau_{ca}(1)=a$, $\tau_{cb}(1)=b$, and $\{\dot\tau_{ca}, \dot\tau_{cb}\}\subset E_f^{wu}$,

3.  $a \in W_f^{su}(b)$.

Then for $n\ge 1$ $$\label{dist1}
d(f^n(a), f^n(b))\le d(f^n(a), f^n(c))+d(f^n(c), f^n(b))\le c_1\beta^n,$$ on the other hand

$$\label{dist2}
d_f^{su}(f^n(a), f^n(b))\ge c_2\gamma^n.$$

For every $x\in\mathbb R^3$ consider a cone $Cone(x)=\{v\in T_x\mathbb R^3:  \measuredangle(v,E_L^{su}(x))\le
k\}$. The assumption (\[angles\]) tells us that $E_f^{su}(x)\subset Cone(x)$. Hence a leaf of $W_f^{su}$ can be considered as a graph of a Lipschitz function over $E_L^{su}$. The Lipschitz constant depends only on $k$. It follows that $W_f^{su}$ is quasi-isometric:

$$\label{QI}
\exists c_3>0\;\mbox{such that for}\; x\in W_f^{su}(y)\;\; d_f^{su}(x,y)\le c_3 d(x,y).$$

Inequalities (\[dist1\]), (\[dist2\]) and (\[QI\]) sum up to a contradiction.

Suppose that there are two points $a$ and $b$, $a \in W_f^{wu}(b)$ such that $h_f(a)\notin W_L^{wu}(h_f(b))$ then we have

$$\label{d1} d(f^n(a), f^n(b))\le c_1\beta^n,$$

and since $h_f(a)$ and $h_f(b)$ lie in the same unstable leaf but not in the same weak unstable leaf we get $$\label{d2}
d(h_f(f^n(a)),h_f(f^n(b)))=d(L^n(h_f(a)),L^n(h_f(b)))\ge c_2\gamma^n.$$ Finally since $$\label{periodicity}
h_f(x+\bar m)=h_f(x)+\bar m,\;\;\bar m\in \mathbb Z^3$$ we have that $d(h_f(x),h_f(y))\le c(\varepsilon) d(x,y)$ for any $x$ and $y$ such that $d(x,y)\ge\varepsilon$. Hence

$$\label{d3}
d(h_f(f^n(a)), h_f(f^n(b)))\le c_3d(f^n(a), f^n(b)),$$

where $c_3$ depends on $d(a, b)$. Inequalities (\[d1\]), (\[d2\]) and (\[d3\]) sum up to a contradiction.

Affine structure on the weak unstable foliation
-----------------------------------------------

Let $f$ be in $\mathcal U$. For any $x$ and $y$, $y\in W_f^{wu}(x)$ define the function $$\rho_f(x, y)=\prod_{n\ge 1}\frac{D_f^{wu}(f^{-n}(y))}{D_f^{wu}(f^{-n}(x))}$$ where $D_f^{wu}(z)=\|D(f)\big|_{E_f^{wu}}(z)\|$. The following properties are easy to prove:

-   $\rho_f(x,\cdot)$ is well defined and Hölder continuous.

-   $\forall x, y \in W_f^{wu}(z)\;\; \rho_f(x, y)\rho_f(y, z)=\rho_f(x, z)$.

-   $\rho_f(f(x),f(y))=\frac{D_f^{wu}(y)}{D_f^{wu}(x)}\rho_f(x, y)$.

-   The function $\rho(\cdot, \cdot)$ is the only continuous function satisfying $\rho_f(x,x)=1$ and Property 3.

-   $\forall K>0\;\; \exists C>0$ such that $C>\rho_f(x, y)>\frac1C$ whenever $d^{wu}(x, y)<K$.

The goal is to show that $h$ is differentiable along $W_f^{wu}$ ($wu$-differentiable) and

$$\label{hrho}
\rho_g(h(x),h(y))=\frac{D_h^{wu}(y)}{D_h^{wu}(x)}\rho_f(x,y),$$

Fix an arbitrary point $p$. Let $h_p: W_f^{wu}(p)\to W_f^{wu}(h(p))$ be the restriction of $h$ to $W_f^{wu}(p)$. We would like to show that $h_p$ is Lipschitz with a constant that does not depend on $p$. Let $m$ be the induced volume on $W_f^{wu}(p)$. Consider the function $\tilde d_f$ $$\tilde d_f(x,y)=\int\limits_x^y\frac{1}{\rho_f(x,z)}dm(z),\;\; x,y\in W_f^{wu}(p),$$ we integrate along the leaf with respect to the measure $m$.

Function $\tilde d_f$ has the following properties which are simple corollories of the properties of $\rho_f$ and the definition of $\tilde d_f$.

-   $\tilde d_f(x,y)=d_f^{wu}(x,y)+o(d_f^{wu}(x,y))$,

-   $\tilde d_f(f(x),f(y))=D_f^{wu}(x)\tilde d_f(x,y)$,

-   $\forall K>0 \;\exists C>0$ such that $$\frac1C\tilde d_f(x,y)\le d_f^{wu}(x,y)\le C\tilde d_f(x,y)$$ whenever $d_f^{wu}(x,y)<K$.

-   The function $\tilde d_f$ is continuous. To state this property precisely we consider lift of $\tilde d_f$. We speak about lifts of points and leaves. $$\begin{gathered}
    \forall\varepsilon>0\; \exists\delta>0\;\;\;\mbox{such that}\;\;\; \forall x,y\in \mathbb R^3,
    y\in W_f^{wu}(x)\;\;\;\\
    \mbox{and}\;\;\; \forall z,q\in\mathbb R^3, q\in W_f^{wu}(z), z\in B(x,\delta), q\in B(y,\delta)\\
    \;\;\mbox{we have}\;\;\;\; |\tilde d_f(x,y)-\tilde d_f(z,q)|<\varepsilon.\end{gathered}$$

We will also need $\tilde d_g$ which is defined analogously on the leaves of $W_g^{wu}$ and has analogous properties.

The lift of the conjugacy $h$ satisfies the equation (\[periodicity\]) which implies the following $$\exists C>0: \forall x,y \;\;\; d(h(x),h(y))\le C d(x,y)\;\;\mbox{if}\;\;d(x,y)\ge 1.$$ Also we know that weak unstable foliation is quasi-isometric which gives us the same for the distance in weak unstable foliations $$\label{farlipschitz}
\exists C>0: \forall x,y \;\;\; d_g^{wu}(h(x),h(y))\le C d_f^{wu}(x,y)\;\;\mbox{if}\;\;d_f^{wu}(x,y)\ge 1.$$ This tells us that $h_p$ is Lipschitz for points that are far enough. So we need to estimate $d_g^{wu}(h(x),h(y))$ for $x$ and $y$ close. Note that (D3) allows us to use $\tilde d_g$ and $\tilde d_f$ in these estimates instead of $d_g^{wu}$ and $d_f^{wu}$.

Recall the following well-known result.

If $f:M\to M$ is a transitive Anosov diffeomorphism and $\varphi_1, \varphi_2:M\to\mathbb R$ are Hölder continuous functions such that $$\prod_{i=1}^p\varphi_1(f^i(x))=\prod_{i=1}^p\varphi_2(f^i(x))\;\;\mbox{whenever}\;f^p(x)=x$$ then there is a function $P:M\to\mathbb R$, unique up to a multiplicative constant, such that $$\frac{\varphi_1}{\varphi_2}=\frac{P\circ f}{P}.$$ Moreover $P$ is Hölder continuous.

Apply Livshitz Theorem for $\varphi_1=D_f^{wu}(\cdot)$ and $\varphi_2=D_g^{wu}(h(\cdot))$. The condition of the Livshitz Theorem is satisfied because of the assumption on p. d. We have $$\forall n>0 \;\;\;\;\;
\prod_{i=0}^{n-1}\frac{D_g^{wu}\left(h(f^i(x))\right)}{D_f^{wu}(f^i(x))}=\frac{P(x)}{P(f^n(x))}.$$

Choose points $x$ and $y$ close on the leaf $W_f^{wu}(p)$. Choose the smallest $N$ such that $d_f^{wu}(f^N(x),f^N(y))\ge 1$. Then $$\begin{gathered}
\frac{\tilde d_g(h(x),h(y))}{\tilde d_f(x,y)}=
\prod_{i=0}^{N-1}\frac{D_g^{wu}\left(g^i(h(x))\right)}{D_f^{wu}(f^i(x))}\cdot
\frac {\tilde d_g\left(g^N(h(x)),g^N(h(y))\right)} {\tilde d_f(f^N(x),f^N(y))}\\
=\frac{P(x)}{P(f^N(x))}\cdot\frac {\tilde d_g\left(g^N(h(x)),g^N(h(y))\right)} {\tilde d_f(f^N(x),f^N(y))}\le
\frac{P(x)}{P(f^N(x))}\cdot constant.\end{gathered}$$ Here we used (\[farlipschitz\]) and (D3) for $\tilde d_f$ and $\tilde d_g$. Function $P$ is bounded away from zero and infinity so we get that $h$ is uniformly Lipschitz along the weak unstable foliation.

Transitive point argument and construction of a measure absolutely continuous with respect to weak unstable foliation
---------------------------------------------------------------------------------------------------------------------

We divide the proof of Lemma 5 into several steps. The conjugacy $h$ is Lipschitz along $W_f^{wu}$ and hence $wu$-differentiable at almost every point with respect to Lebesgue measure on the leaves of $W_f^{wu}$. It is obvious that $wu$-differentiability of $h$ at $x$ implies $wu$-differentiability of $h$ at any point from the orbit $\{f^i(x), i\in \mathbb Z\}$. Moreover:

Suppose that $h$ is wu-differentiable at $x$ and $\overline{\{f^i(x), i\ge 0\}}=\mathbb T^3$ then $h$ is $C^{1+\nu}$ along $W_f^{wu}$ and (\[hrho\]).

The problem now is to show existence of such a transitive point $x$. We know that almost every point is transitive with respect to a given ergodic measure with full support. On the other hand $h$ is $wu$-differentiable at almost every point with respect to Lebesgue measure on the leaves. Unfortunately it can happen that for natural ergodic “physical measures” these two “full measure” sets do not intersect. In other words weak unstable foliation is not absolutely continuous with respect to a “physical measure”.

Let us explain this phenomenon in more detail. Consider a volume preserving $C^1$ small perturbation $\tilde L$ of $L$, $H\circ L=\tilde L\circ H$. The Lyapunov exponents of $\tilde L$ are defined on a full volume set of regular points $\mathcal R$ and are given by the formula $$\chi^\sigma=\int_{\mathbb T^3}\log D_{\tilde L}^\sigma d \mbox{vol},\;\;\; \sigma=s, wu, su.$$ The perturbation $\tilde L$ can be chosen in such a way that $\chi^{wu}>\log\lambda_2$ (see [@BB], Proposition 0.3). It is easy to show that the weak unstable foliation of $\tilde L$ is not absolutely continuous. Namely, let $\Delta$ be a segment of a weak unstable leaf of $L$. Then by Lemma 2 $H(\Delta)$ is a piece of a weak unstable leaf of $\tilde L$. We show that Lebesgue measure of $\mathcal R\cap H(\Delta)$ is equal to zero. For any $n\ge 0$ $H(L^n(\Delta))=\tilde L^n(H(\Delta))$ and (\[angles\]) guarantees that $\tilde L^n(H(\Delta))$ can be viewed as a graph of a Lipschitz function over a leaf of the weak unstable foliation of $L$. Hence $$\mbox{length}(\tilde L^n(H(\Delta))\le c_1\cdot \mbox{length}(L^n(\Delta))=\lambda_2^n\cdot \mbox{length}(\Delta),\;\; n\ge 0.$$ Suppose that $\mbox{Leb}(\mathcal R\cap H(\Delta))>0$ then $$\mbox{length}(\tilde L^n(H(\Delta))\ge c_2e^{n(\chi^{wu}-\varepsilon)},\;\;\varepsilon=\frac12(\chi^{wu}-\log\lambda_2)$$ which contradicts the previous inequality.

This observation answers a question of Hirayama and Pesin [@HP] about existence of non-absolutely continuous foliations with non-compact leaves.

To overcome this problem we do

Construction of a measure $\mu$ absolutely continuous with respect to $W_f^{wu}$.

This construction follows the lines of Pesin-Sinai [@PS] construction of $u$-Gibbs measures. In our setup the construction is simpler so for the sake of completeness we present it here. Measure $\mu$ has full support. Thus ergodicity of $\mu$ would imply that almost every point is transitive and hence by Step 1 $h$ would be $wu$-differentiable. We do not know how to show ergodicity of $\mu$. Instead we do

Set of transitive points is a full measure $\mu$ set.

Steps 2 and 3 guarantee existence of a transitive point needed in Step 1.

 \
[**Step 1.**]{} Let us pick a point $y\in \mathbb T^3$ and show that $h$ is $wu$-differentiable at $y$ and moreover $$\label{wudiff}
D_h^{wu}(y)=\frac{P(y)}{P(x)}D_h^{wu}(x)$$ where $P$ is the same as in the proof of Lemma 4.

Choose $y'\in W_f^{wu}(y)$. Property (D1) of $\tilde d_f$, $\tilde d_g$ ensures that it is enough to show that $$\label{wutilde}
\frac{\tilde d_g(h(y),h(y'))}{\tilde d_f(y,y')}=\frac{P(y)}{P(x)}D_h^{wu}(x).$$

Fix an $\varepsilon >0$ small compared to $\tilde d_f(y,y')$. Choose a small open ball $B$ centered at $y$ and define

$$\begin{gathered}
B'=\{z' : \;\;\exists z\in B\;\;\mbox{such that}\;\; \tilde d_f(z,z')=\tilde d_f(y,y')\\
 \mbox{and} \;\; (z,z') \;\;\mbox{has the same orientation as}\;\; (y,y')\}.\end{gathered}$$

The condition about orientation ensures that $B'$ has only one connected component. The set $B'$ is a small neighborhood of $y'$ because of the continuity of $\tilde d_f$ (D4). The size of $B$ must be chosen in such a way that

1.  $|P(z)-P(y)|<\varepsilon$ if $z\in B$,

2.  $|\tilde d_g(h(z),h(z'))-\tilde d_g(h(y),h(y'))|<\varepsilon$ where $z$ and $z'$ are the same as in definition of $B'$.

Since $x$ is transitive there is an arbitrarily large $N$ such that $f^N(x)\in B$. Choose $z$ on $W_f^{wu}(x)$ such that $\tilde d_f(f^N(x),f^N(z))=\tilde d_f(y,y')$ so that $f^N(z)\in B'$ by the definition. We choose $N$ big enough so that $$\left|\frac{\tilde d_g(h(x),h(z))}{\tilde d_f(x,z)}-D_h^{wu}(x)\right|<\varepsilon.$$

Now we are ready to do the estimates $$\begin{gathered}
\tilde d_g(h(y),h(y'))=\varepsilon_1+\tilde d_g\left(h(f^N(x)),h(f^N(z))\right)\\
=\varepsilon_1+\tilde d_g\left(g^N(h(x)),g^N(h(z))\right)\\
=\varepsilon_1+\prod\nolimits_{i=0}^{N-1}D_g^{wu}\left(g^i(h(x))\right)\cdot\tilde d_g\left(h(x),h(z)\right)\\
=\varepsilon_1+\prod\nolimits_{i=0}^{N-1}D_g^{wu}\left(h(f^i(x))\right) (D_h^{wu}(x)+\varepsilon_2)\tilde d_f(x,z)\\
=\varepsilon_1+\frac{\prod\nolimits_{i=0}^{N-1}D_g^{wu}\left(h(f^i(x))\right)}{\prod\nolimits_{i=0}^{N-1}D_f^{wu}(f^i(x))} (D_h^{wu}(x)+\varepsilon_2)\tilde d_f(f^N(x),f^N(z))\\
=\varepsilon_1+\frac{P(f^N(x))}{P(x)} (D_h^{wu}(x)+\varepsilon_2)\tilde d_f(y, y')\\
=\varepsilon_1+\frac{P(y)+\varepsilon_3}{P(x)} (D_h^{wu}(x)+\varepsilon_2)\tilde d_f(y, y')\end{gathered}$$ with $\max(|\varepsilon_1|, |\varepsilon_2|, |\varepsilon_3|)\le \varepsilon$. Now letting $\varepsilon$ go to $0$ we get (\[wutilde\]).

To show (\[hrho\]) define

$$\tilde \rho_g(h(x),h(y))=\frac{D_h^{wu}(y)}{D_h^{wu}(x)}\rho_f(x,y).$$ Then

$$\begin{gathered}
\tilde \rho_g\left(g(h(x)),g(h(y))\right)=\tilde \rho_g\left(h(f(x)),h(f(y))\right)\\
=\frac{D_h^{wu}(f(y))}{D_h^{wu}(f(x))}\rho_f(f(x),f(y))=\frac{D_h^{wu}(f(y))}{D_h^{wu}(f(x))}\cdot\frac{D_f^{wu}(y)}{D_f^{wu}(x)}\rho_f(x,y)\\
=\frac{D_{h\circ f}^{wu}(y)}{D_{h\circ f}^{wu}(x)} \rho_f(x,y) =\frac{D_{g\circ h}^{wu}(y)}{D_{g\circ
h}^{wu}(x)} \rho_f(x,y) =\frac{D_g^{wu}(h(y))}{D_g^{wu}(h(x))}\cdot\frac{D_h^{wu}(y)}{D_h^{wu}(x)}\rho_f(x,y)=\\
\frac{D_g^{wu}(h(y))}{D_g^{wu}(h(x))}\tilde\rho_g(h(x),h(y)).\end{gathered}$$

This by $(P4)$ implies that $\tilde \rho_g=\rho_g$ which is equivalent to (\[hrho\]).

[**Step 2.**]{} Let $x_0$ be a fixed point for $f$ and let $V_0$ be an open bounded neighborhood of $x_0$ in $W_f^{wu}(x_0)$. Consider a probability measure $\eta^0$ supported on $V_0$ with density proportional to $\rho_f(x_0,\cdot)$. For $n>0$ define $$V_n=f^n(V_0),\; \eta^n=(f^n)_*\eta^0$$ so that $\eta^n$ is supported on $V_n$ and has density proportional to $\rho_f(x_0,\cdot)$ by $(P3)$.

Let $\mu^n=\frac1n\sum_{i=0}^{n-1}\eta^i$. By the Krylov-Bogoljubov theorem $\{\mu^n; n\ge 0\}$ is weakly compact and any of its limits is $f$-invariant. Let $\mu$ be a one of those limits along a subsequence $\{n_k; k\ge 1\}$. We would like to prove that $\mu$ has absolutely continuous conditional measures on the pieces of weak unstable foliation.

Let us be more precise. Consider a small open set $X\subset\mathbb T^3$ which can be decomposed in the following way $$X=\bigcup_{y\in Y}W_f^{wu}(y,R_y).$$ Here $Y$ is a two dimensional transversal. To simplify the notation let $W(y)=W_f^{wu}(y,R_y)$. Denote by $\mu_T$ the transverse measure on $Y$: for $Y'\subset Y$ $\mu_T(Y')=\mu(\cup_{y\in Y'}W(y))$. Similary define $\eta_T^n$ and $\mu_T^n$. Obviously $\mu_T^{n_k}\to\mu_T$ weakly as $k\to\infty$. We show that for $\mu_T$ almost every $y$, $y\in Y$ the conditional measure $\mu_y$ on the local leaf $W(y)$ is absolutely continuous with respect to Lebesgue measure $m_y$ on $W(y)$.

The conditional measures are characterized by the following property

$$\label{rohlin}
\forall F\in C(X)\;\; \int_X F d\mu=\int_Y d\mu_T(y)\int_{W(y)}F(y,z)d\mu_y(z).$$

First we look at conditional measures of $\eta^n$. We fix $X$ and $Y$ as above and we assume that the end points of $V_n$ lie outside of $X$. Let $\{a_1, a_2,\ldots a_m\}=Y\cap V_n$. Then the formulas for the transverse measure and conditional measures are obvious: $$\begin{gathered}
\label{cond}
\eta_T^n=\sum_{i=1}^m\left(\int_{W(a_i)}\rho_f(x_0,z)dm_{a_i}(z)\right)\delta(a_i),\\
\shoveleft{d\eta_y^n(z)}=\left(\int_{W(y)}\rho_f(y,z)dm_y(z)\right)^{-1}\rho_f(y,z)dm_y(z).\hfill\end{gathered}$$ Notice that $\eta_y^n$ actually do not depend on $n$.

The goal now is to show that $d\mu_y=\left(\int_{W(y)}\rho_f(y,z)dm_y(z)\right)^{-1}\rho_f(y,\cdot)$ for almost every $y$. It could happen that the end points of $V_n$ lie inside of $X$. Support $S_n$ of $\eta_T^n$ consists of finitely many points. Some of these points correspond to the end points of $V_n$. Denote the set of these points by $B_n$, $|B_n|\in\{0,1,2\}$. Let $A_n=S_n\backslash B_n$ then there is a natural decomposition of the transverse measure $\eta_T^n$

$$\begin{gathered}
\eta_T^n=\sum_{a\in A_n}\left(\int_{W(a)}\rho(x_0,z)dm_a(z)\right)\delta(a)\\
+\sum_{b\in B_n}\left(\int_{W(b)\cap V_n}\rho(x_0,z)dm_b(z)\right)\delta(b)=\eta_{(T,A)}^n+\eta_{(T,B)}^n.\end{gathered}$$

The conditional measures $\eta_y^n$ for $y\notin B_n$ are given by formula (\[cond\]). Since $W_f^{wu}$ is uniformly expanding it is clear that $\eta_T^n(B_n)\to 0$ as $n\to\infty$. Hence $$\label{endpoints}
\frac1n\left(\sum_{i=0}^{n-1}\eta_T^i(B_i)\right)\to 0,\;\;\;n\to\infty$$ and

$$\label{muA}
\mu_T=\lim_{k\to\infty}\mu_T^{n_k}=\lim_{k\to\infty}\frac1{n_k}\sum_{i=0}^{n_k-1}\eta_{(T,A)}^i.$$

Consider a continuous function $F$ on $X$.

$$\begin{gathered}
\int_XFd\mu=\lim_{k\to\infty}\int_XFd\mu^{n_k}=\lim_{k\to\infty}\frac1{n_k}\sum_{i=0}^{n_k-1}\int_XFd\eta^{n_k}\\
=\lim_{k\to\infty}\frac1{n_k}\sum_{i=0}^{n_k-1}\int_Yd\eta_T^{n_k}(y)\int_{W(y)}F(y,z)d\eta_k^{n_k}(z)\\
=\lim_{k\to\infty}\frac1{n_k}\sum_{i=0}^{n_k-1}\int_Yd\eta_{(T,A)}^{n_k}(y)\int_{W(y)}F(y,z)d\eta_y^{n_k}(z)\\
+\lim_{k\to\infty}\frac1{n_k}\sum_{i=0}^{n_k-1}\int_Yd\eta_{(T,B)}^{n_k}(y)\int_{W(y)}F(y,z)d\eta_y^{n_k}(z).\end{gathered}$$

The function $F$ is bounded so it follows from (\[endpoints\]) that the last limit is zero. So we get

$$\begin{gathered}
\int_XFd\mu=\lim_{k\to\infty}\frac1{n_k}\sum_{i=0}^{n_k-1}\int_Yd\eta_{(T,A)}^{n_k}(y)\int_{W(y)}F(y,z)d\eta_y^{n_k}(z)\\
=\lim_{k\to\infty}\frac1{n_k}\sum_{i=0}^{n_k-1}\int_Yd\eta_{(T,A)}^{n_k}(y)\left(\int_{W(y)}\rho_f(y,z)dm_y(z)\right)^{-1}\int_{W(y)}F(y,z)\rho_f(y,z)dm_y(z).\end{gathered}$$

Now notice that the function that we integrate with respect to $\eta_{(T,A)}^{n_k}$ is continuous and does not depend on $n_k$. Hence using (\[muA\]) we get

$$\begin{gathered}
\int_XFd\mu=\lim_{k\to\infty}\int_Yd\mu_T^{n_k}(y)\left(\int_{W(y)}\rho_f(y,z)dm_y(z)\right)^{-1}\int_{W(y)}F(y,z)\rho_f(y,z)dm_y(z)\\
=\int_Yd\mu_T(y)\left(\int_{W(y)}\rho_f(y,z)dm_y(z)\right)^{-1}\int_{W(y)}F(y,z)\rho_f(y,z)dm_y(z)\end{gathered}$$

and by (\[rohlin\]) we see that up to normalization the density of the conditional measure on $W(y)$ is equal to $\rho_f(y,\cdot)$ for $\mu_T$ a. e. $y$.

The leaf $W_f^{wu}(x_0)$ is dense in $\mathbb T^3$ since $W_f^{wu}(x_0)=h_f^{-1}\left(W_L^{wu}(h_f^{-1}(x_0))\right)$ and $W_L^{wu}(h_f^{-1}(x_0))$ is a dense irrational line in $\mathbb T^3$. Hence the support $\mu$ is the whole torus.

[**Step 3.**]{} To prove that $\mu$ a. e. point is transitive we fix a ball in $\mathbb T^3$ and show that a. e. point visits the ball infinitely many times. Then to conclude transitivity we only need to cover $\mathbb T^3$ by a countable collection of balls such that every point is contained in an arbitrarily small ball.

So let us fix a ball $B'$ and a slightly smaller ball $B$, $B\subset B'$. Let $\psi$ be a non-negative continuous function supported on $B'$ and equal to $1$ on $B$. By Birkhoff ergodic theorem

$$\label{ergodictheorem}
E(\psi|\mathcal I)=\lim_{n\to\infty}\frac1n\sum_{i=0}^{n-1}\psi\circ f^i$$

where $\mathcal I$ is $\sigma$-algebra of $f$-invariant sets.

Let $A=\{x: E(\psi|\mathcal I)(x)=0\}$. Then $\mu(A\cap B)=0$ since $\int_A\psi d\mu=\int_A E(\psi|\mathcal I)d\mu=0$. Hence

$$\label{positiveonB}
E(\psi|\mathcal I)(x)>0\;\; \mbox{for} \;\;\mu \;\mbox{a.e.}\;\; x\in B.$$

Since $h_f(W_f^{wu})=W_L^{wu}$ it is possible to find $R>0$ such that $\cup_{b\in B}W(b, R)=\mathbb T^3$.

This observation also implies that $\mu$ has full support.

Applying the standard Hopf argument we get that for $\mu$ a. e. $x$ the function $E(\psi|\mathcal I)$ is constant on $W(x, R)$. Now absolute continuity of $W_f^{wu}$ together with above observations shows that $E(\psi|\mathcal I)>0$ for $\mu$ a. e. $x$ which means according to (\[ergodictheorem\]) that a. e. $x$ visits $B'$ infinitely many times.

Strong unstable foliations match
--------------------------------

Let us point out once again that in the proof of Lemma 6 we only use $wu$-differentiability of $h$ which as we showed is equivalent to coincidence of p. d. in the weak unstable direction.

We will be working on two dimensional leaves of $W_f^u$. We know that each of these leaves is subfoliated by $W_f^{wu}$ as well as by $W_f^{su}$. The goal is to prove that $h(W_f^{su})=W_g^{su}$ so we consider the foliation $U=h^{-1}(W_g^{su})$. As for usual foliations $U(x)$ stands for the leaf of $U$ passing through $x$ and $U(x,R)$ stands for the local leaf of size $R$. Obviously $U$ subfoliate $W_f^u$. A priori the leaves of $U$ are just Hölder continuous curves. Since weak unstable foliations match we see that a leaf $U(x)$ intersects each $W_f^{wu}(y)$, $y\in W_f^u(x)$ exactly once.

Let us prove several auxiliary claims.

Consider a point $a\in\mathbb T^3$. Suppose that there is a point $b\ne a$, $b\in W_f^{su}(a)\cap U(a)$. Let $c\in W_f^{wu}(a)$ and $d=W_f^{wu}(b)\cap W_f^{su}(c)$, $e=W_f^{wu}(b)\cap U(c)$. Then $d=e$.

Assume that $d\ne e$. For the sake of concreteness we also assume that $d$ lies between $b$ and $e$. We look at configurations $\{a,b,c,d,e\}\in W_f^u(a)$, $\{h(a),h(b)$, $h(c)$,$h(d)$,$h(e)\}$ $\in W_g^u(h(a))$ and study their evolution under $f^{-n}, n>0$ and $g^{-n}, n>0$ respectively. Since under the action of $f^{-1}$ strong unstable leaves contract exponentially faster then weak unstable leaves we get that

$$\label{acbd}
\forall \varepsilon>0 \;\; \exists n_0: \;\forall n>n_0\;\;\;
\left| \frac{d_f^{wu}(f^{-n}(a),f^{-n}(c))}{d_f^{wu}(f^{-n}(b),f^{-n}(d))}-1 \right|<\varepsilon.$$

Analogously

$$\label{Hacbd}
\forall \varepsilon>0 \;\; \exists n_1: \;\forall n>n_1\;\;\;
\left| \frac{d_g^{wu}\left(g^{-n}(h(a)),g^{-n}(h(c))\right)}{d_g^{wu}\left(g^{-n}(h(b)),g^{-n}(h(e))\right)}-1 \right|<\varepsilon.$$

The next statement is a direct corollary of (D1) and (D2). There exists a $\delta>0$ which depends on the initial configuration $\{a,b,c,d,e\}$ such that

$$\label{bebd}
\forall n>0 \;\;\;\;\; \frac{d_f^{wu}(f^{-n}(b),f^{-n}(e))}{d_f^{wu}(f^{-n}(b),f^{-n}(d))} >1+\delta.$$

Combining (\[acbd\]) and (\[bebd\]) we get

$$\label{differentDistance}
\exists \delta'>0:\;\forall n>n_0 \;\;\;\;\; \frac{d_f^{wu}(f^{-n}(b),f^{-n}(e))}{d_f^{wu}(f^{-n}(a),f^{-n}(c))} >1+\delta'.$$

On the other hand we know that $h$ is continuously $wu$-differentiable, hence

$$\begin{gathered}
\label{sameDistance} \forall \varepsilon>0\;\; \exists n_2:\; \forall n>n_2\;\;\;\;\;
\left| \frac{d_g^{wu}\left(g^{-n}(h(a)), g^{-n}(h(c))\right)}{d_f^{wu}(f^{-n}(a), f^{-n}(c))}-D_h^{wu}(f^{-n}(a))\right|<\varepsilon\\
\mbox{and}\;\;  \left| \frac{d_g^{wu}\left(g^{-n}(h(b)), g^{-n}(h(e))\right)}{d_f^{wu}(f^{-n}(b),
f^{-n}(e))}-D_h^{wu}(f^{-n}(a))\right|<\varepsilon.\end{gathered}$$

It is easy to see that (\[differentDistance\]) contradicts (\[sameDistance\]) and (\[Hacbd\]) so we are done.

Consider a weak unstable leaf $W_f^{wu}(a)$ and $b\in W_f^{su}(a)$, $b\ne a$. For any $y\in W_f^{wu}(a)$ let $y'=W_f^{wu}(b)\cap W_f^{su}(y)$. Then $\exists c_1, c_2>0$ such that $\forall y\in W_f^{wu}(a)$ $\;\;\;c_1>d_f^{su}(y,y')>c_2$.

Recall that $h_f(W_f^{wu})=W_L^{wu}$. The leaves $W_L^{wu}(h_f(a))$ and $W_L^{wu}(h_f(b))$ are parallel lines in $W_L^u(h_f(a))$ that are fixed distance apart. Hence the estimate from below is a direct consequence of uniform continuity of $h_f|_{W_f^u}(a)$ with respect to metrics $d_f^u$ and $d_L^u$.

Now we prove the estimate from above. We need to show that the strip between $W_f^{wu}(a)$ and $W_f^{wu}(b)$ cannot contain arbitrarily long pieces of strong unstable leaves. The reason for this is uniform transversality of weak unstable and strong unstable foliation.

For any positive number $\varepsilon$ we can choose a finite number of points $\{a_0=a, a_1, a_2,\ldots
a_m=b\}$ between $a$ and $b$ on $W_f^{su}(a)$ in such a way that $W_f^{wu}(a_i)$ is contained in $\varepsilon$-neighborhood of $W_f^{wu}(a_{i-1})$ and vice versa, $i=1,\ldots m$. Again this is possible because $W_L^{wu}(h_f(a_i))$, $i=0,\ldots m$ are parallel lines and $h_f^{-1}|_{W_L^u(h_f(a))}$ is uniformly continuous.

Let $s=\min_{x\in\mathbb T^3} \measuredangle(E_f^{wu}(x),E_f^{su}(x))$. Choose a small $\delta>0$ such that in any ball $B$ of size $\delta$ $$\max_{x, y\in B}\max\{\measuredangle(E_f^{wu}(x),E_f^{wu}(y)), \measuredangle(E_f^{su}(x),E_f^{su}(y))\}<\frac {s}{10}.$$ In such a ball the direction of $E_f^{wu}$ is almost constant comparing to the angle between $E_f^{wu}$ and $E_f^{su}$. Clearly it is possible to choose a small $\varepsilon=\varepsilon(s,\delta)$ and correspondingly the points $\{a_0, a_1,\ldots a_m\}$ as above such that any strong unstable leave crosses the strip between $W_f^{wu}(a_{i-1})$ and $W_f^{wu}(a_i)$ in a ball of size $\delta$, $i=1,\ldots m$. This gives us uniform estimates on the lengths of pieces of strong unstable leaves in the strips between $W_f^{wu}(a_{i-1})$ and $W_f^{wu}(a_i)$, $i=\overline {1,m}$. The sum of these estimates gives us the desired uniform estimate from above.

Suppose $\exists a\in \mathbb T^3$ and $R>0$ such that $W_f^{su}(a, R)=U(a, R)$ then $W_f^{su}=U$.

Consider a point $c\in W_f^{wu}(a)$ then applying Claim 1 to the points $b\in W_f^{su}(a, R)$ we get that $\exists R_c>0$ such that $W_f^{su}(c, R_c)=U(c, R_c)$. Moreover by Claim 2 numbers $R_c$, $c\in W_f^{wu}(a)$ are uniformly bounded away from zero. Now the statement follows from denseness of $W_f^{wu}(a)$ in $\mathbb
T^3$.

We are ready to prove the lemma.

We say that $W_f^{su}(x)$ and $U(x)$ intersect transversally at $y$ if $y\in W_f^{su}(x)\cap U(x)$ and $\forall R>0\;$ the local leaf $U(y,R)$ lies on both sides of $W_f^{su}(y)$.

We consider two cases.

[**Case 1.**]{} [*At every periodic point $x_0$ the leaves $W_f^{su}(x_0)$ and $U(x_0)$ do not intersect transversally at a point different from $x_0$.*]{}

Notice that the property of having a transverse intersection is stable — if $W_f^{su}(x)$ and $U(x)$ intersect transversally then there is a neighborhood $V$ of $x$ such that $\forall z\in V$ $W_f^{su}(z)$ and $U(z)$ intersect transversally. Periodic points are dense therefore absence of transverse intersections at periodic points leads to absence of transverse intersections at all points.

We assume that $W_f^{su}\ne U$. Then the above observation together with Claim 3 tell us that for any point $x$ the leaves $W_f^{su}(x)$ and $U(x)$ intersect only at $x$.

Let $x_0$ to be a fixed point of $f$. For each $y\in U(x_0)$ the leaf $W_f^{su}(y)$ intersects $U(x_0)$ only at $y$. Thus we are able to build a ladder of rectangles in $W_f^u(x_0)$ as shown on the Figure \[fig4\]. The sides of the rectangles are pieces of weak unstable and strong unstable leaves. The rectangles are subject to condition $$d_f^{su}(x_i, y_i)=1,\;i\ge 0.$$ This guarantees that after the choice of $y_0$ (there are two choices) the sequence of rectangles is defined uniquely. Let $d_i=d_f^{wu}(y_i, x_{i+1}),\; i\ge 0$ and let $\{m_i;\;i\ge 0\}$ be midpoints on the sides of rectangles as shown on the picture.

Suppose that $\inf_{i\ge 0}d_i>0$. Apply $f^{-n},\; n>0$ to the ladder of rectangles. The leaf $U(x_0)$ is invariant while the rectangles shrink and become flat as shown on Figure \[fig5\]. Namely

$$\forall \varepsilon>0\;\;\exists n_0:\;\forall n>n_0\;\;\;\mbox{and}\;\;\forall i\ge 0\;\;
\frac{d_f^{su}(f^{-n}(x_i),f^{-n}(y_i))}{d_f^{wu}(f^{-n}(y_i),f^{-n}(x_{i+1}))}<\varepsilon.$$

This means that in any fixed bounded neighborhood of $x_0$ the leaf $U(x_0)$ is arbitrarily close to $W_f^{wu}(x_0)$. In particular we have that $x_1$ is arbitrarily close to $z_0$ while we know that they are some fixed distance apart. To make this argument completely rigorous one needs to carry out an estimate on the distance between $z_0$ and $x_1$ using regularity of holonomies along $W_f^{wu}$ and $W_f^{su}$ inside of the leaf $W_f^u(x_0)$. We conclude that $\inf_{i\ge 0}d_i=0$.

Then choose a subsequence $\{m_{n_k}; k\ge 0\}$ such that corresponding rectangles have width going to zero as $k$ tend to infinity. Each of these rectangles contains a piece of $U(x_0)$ inside of it. Let $m$ be an accumulation point of $\{m_{n_k}; k\ge 0\}$ considered as a sequence of points in $\mathbb T^3$ rather than on $W_f^u(x_0)$. Since the width of the rectangles is shrinking and the foliations are continuous we get that $W_f^{su}(m,\frac12)=U(m,\frac12)$. Hence $W_f^{su}=U$ by Claim 3 and we move on to the second case.

[**Case 2.**]{} [*There exist a periodic point $x_0$ and a point $y_0, y_0\ne x_0$ such that $W_f^{su}(x_0)$ and $U(x_0)$ intersect at $y_0$ transversally.*]{}

Without loss of generality we can assume that $x_0$ is a fixed point. We chose a sequence $\{x_i\in W_f^{wu}(x_0); i\ge 1\}$ such that $x_i\to y_0$, $i\to\infty$. Here and afterwards we speak about convergence on the torus, not in the leaf $W_f^u(x_0)$. By Claim 1 we know that for any $i$ the leaves $W_f^{wu}(y_0)$, $W_f^{su}(x_i)$ and $U(x_i)$ intersect at one point $z_i$. Up to the choice of a subsequence we have that $z_i\to y_1$, $i\to\infty$, where $y_1$ is some point on $W_f^{su}(y_0)$. Since the foliation $U$ is continuous we have that $y_1\in U(y_0)=U(x_0)$ as well. The strong unstable foliation is orientable and the pairs $(x_0, y_0), (x_i, z_i), i\ge 1$ have the same orientation i. e. $y_0$ lies between $x_0$ and $y_1$.

Now we would like to repeat the procedure. Consider another sequence $\{\tilde x_i\in W_f^{wu}(x_0); i\ge 1\}$, $\tilde x_i\to y_1$ as $i\to\infty$ and corresponding sequence $\{\tilde z_i\in W_f^{wu}(y_0); i\ge 1\}$. Then $\tilde z_i\to y_2$ as $i\to\infty$, $y_2\in W_f^{su}(x_0)\cap U(x_0)$. In this way by induction we obtain a sequence of points $\{y_i\in W_f^{su}(x_0)\cap U(x_0); i\ge 1\}$. These points are ordered on $W_f^{su}(x_0)$ — for any positive $i$ point $y_{i-1}$ lies between $x_0$ and $y_i$. By Claim 2 we know that there are constants $c_1$ and $c_2$ which depend only on the initial choice of $x_0$ and $y_0$ such that $\forall i\ge 0\;\;\;
c_1>d_f^{su}(y_i,y_{i+1})>c_2$. This guarantees that the set $\{f^{-n}(y_i); n\ge 0, i\ge 0 \}\subset
W_f^{su}(x_0)\cap U(x_0)$ is dense and hence applying Claim 3 one more time we get that $W_f^{su}=U$.

Final remarks
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We did not discuss the proofs of Lemmas 3 and 7. They can be carried out in the same way as the proof of Lemma 5. The technical difficulty with constructing special measure is not present. One can use SRB measures instead (as a matter of fact the construction in Step 2 applied to $W_f^s$ and $W_f^{su}$ will produce SRB measures).

Notice that we used the assumption that $f, g \in \mathcal U$ only to prove Lemmas 1 and 2. So for Theorem 2 we only need to reprove these two lemmas in the new setting. We use a result from [@BI] that states the following.

Let $f$ be a partially hyperbolic diffeomorphism of $\mathbb T^3$. Then the lifts of stable and unstable foliations are quasi-isometric and the hence the central distribution is uniquely integrable.

Thus Lemma 1 is automatic. Proof of Lemma 2 go through with minor differences since we know that $W_f^{su}$ is quasi-isometric.

The bootstrap of regularity of $h$ to the regularity of $f$ and $g$ cannot be done straightforwardly. The reason is the lack of smoothness of weak unstable foliation. Let $N=[\log \lambda_3/\log\lambda_2]$. It is known [@LW] that given $f$ sufficiently $C^1$-close to $L$ the individual leaves of weak unstable foliation are $C^N$ immersed curves. In general the the leaves of weak untable foliation cannot be more than $C^N$ smooth. An example was constructed in [@JPL]. Hence our method cannot lead to smoothness higher than $C^N$.

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