Proof. From (2.1), we have $f\circ \phi _t\circ f^{-1}(x)=\phi _{\unicode{x3bb} t}(x)$ . By taking derivatives on both sides with respect to t, we have

$$ \begin{align*}Df|_{\mathcal X(f^{-1}x)}\cdot \mathcal X(f^{-1}x)=\unicode{x3bb} \mathcal X(x).\end{align*} $$

Hence,

$$ \begin{align*}Df^n|_{\mathcal X(x)}\cdot \mathcal X(x)=\unicode{x3bb}^n \mathcal X(f^n(x));\end{align*} $$

therefore,

$$ \begin{align*}\|Df^n|_{\mathcal X(x)}\|= \unicode{x3bb}^n\frac{\|\mathcal X(f^n(x))\|}{\|\mathcal X(x)\|}.\end{align*} $$

The proof is complete by setting $C= {\max _x\{\|\mathcal X(x)\|\}}/{\min _x\{\|\mathcal X(x)\|\}}$ .

Proof. Since f is uniformly expanding along $\mathcal F^u$ , $f^{-1}$ uniformly contracts leaves of $\mathcal F^u$ . This implies that $\mathcal F^u$ has no closed leaves. Otherwise, assume $\gamma ^u\in \mathcal F^u$ is a closed leaf. Then, the length of $f^{-n}(\gamma ^u)$ tends to zero as $n\to +\infty $ . By taking a subsequence if necessary, $f^{-k_n}(\gamma ^u)\to z\in \mathbb T^2$ , which implies that the leaf $\mathcal F^u(z)=\{z\}$ and contradicts with the condition on the foliation $\mathcal F^u$ .

Since $\mathcal F^u$ has no closed leaves, [Reference Hector and Hirsch27, Theorem 4.3.3] shows that $\mathcal F^u$ is a suspension foliation with an irrational rotation number, that is, there exists a simple closed circle $\gamma :\mathbb {S}^1\to \mathbb T^2$ which transversally intersects every leaf of $\mathcal F^u$ , and the holonomy map induced by $\mathcal F^u$ on $\gamma $ has an irrational rotation number. In particular, if $\mathcal F^u$ is $C^2$ -smooth, then Denjoy’s theorem shows that $\mathcal F^u$ is minimal. This proves items (1) and (2).

Finally, we show that the lifting foliation $\tilde {\mathcal F}^u$ is quasi-isometric. Since $\gamma $ is a simple closed curve that intersects every leaf of ${\mathcal F}^u$ , there exists $a_1>0$ such that, for every $x\in \gamma $ and $y\in {\mathcal F}^u(x)\cap \gamma $ which contains no point in $\gamma $ between x and y in ${\mathcal F}^u(x)$ , we have

$$ \begin{align*} d_{\mathcal F^u}(x,y)<a_1. \end{align*} $$

Since $\gamma $ is a simple closed curve in $\mathbb T^2$ and transverse to the one-dimensional foliation ${\mathcal F}^u$ , we have that:

• • $\gamma $ is homotopically non-trivial (otherwise it bounds a disk and Poincare–Hopf theorem implies ${\mathcal F}^u$ has singularities);

• • there exists $(k,l)\in {\mathbb Z}^2$ with $k,l$ coprime, such that for every lift $\tilde \gamma \subset \mathbb R^2$ of $\gamma $ ,

  $$ \begin{align*} \tilde\gamma=\tilde\gamma+(k,l); \end{align*} $$

• • there exists $b_1>0$ such that, for every $x\in \tilde \gamma $ , the line $L(x)=\{x+t\cdot (k,l):t\in {\mathbb R}\}$ , satisfies that the Hausdorff distance

  $$ \begin{align*} d_H (\tilde\gamma,L(x))= \max\Big\{\max_{z\in\tilde\gamma}\{d_{\mathbb R^2}(z,L(x))\}, \max_{z\in L(x)}\{d_{\mathbb R^2}(z,\tilde\gamma)\}\Big\} < b_1. \end{align*} $$

Taking $N_1\in \mathbb N$ large enough such that $3b_1\ll N_1\cdot \|(l,-k)\|$ , then

$$ \begin{align*} &\inf\{d_{\mathbb R^2}(z,y): z\in\tilde\gamma, y\in(\tilde\gamma+N_1(l,-k))\}\\&\quad \geq \inf \{d_{\mathbb R^2}(z,y): z\in L(x), y\in L(x+N_1(l,-k))\} \\&\qquad - d_H(\tilde\gamma,L(x)) - d_H(\tilde\gamma+N_1(l,-k),L(x+N_1(l,-k)))\\&\quad \geq \ N_1\cdot\|(l,-k)\|-b_1-b_1 \\&\quad \geq \ b_1. \end{align*} $$

There exists $N_2>0$ (which depends on $\gamma $ only), such that, for every $x\in \tilde \gamma $ and $y\in \tilde {\mathcal F}^u(x)\cap (\tilde \gamma +N_1(l,-k))$ , the segment $[x,y]^u\subset \mathcal F^u(x)$ with endpoints $x,y$ intersects $\tilde \gamma +{\mathbb Z}^2$ (all lifts of $\gamma $ ) with $(N_2+1)$ -points. Thus, we have

$$ \begin{align*} d_{\mathbb R^2}(x,y)\geq b_1\quad \text{and}\quad d_{\tilde{\mathcal F}^u}(x,y)\leq N_2\cdot a_1. \end{align*} $$

For every $x\in \mathbb R^2$ and $y\in \tilde {\mathcal F}^u(x)$ with $d_{\tilde {\mathcal F}^u}(x,y)$ large enough, there exist:

• • $x_1\in \tilde {\mathcal F}^u(x)\cap \tilde \gamma _1$ with $d_{\tilde {\mathcal F}^u}(x,x_1)<a_1$ , where $\tilde \gamma _1$ is a lift of $\gamma $ ;

• • $m\in {\mathbb Z}$ with $|m|$ large, and $y_1\in \tilde {\mathcal F}^u(x)\cap (\tilde \gamma _1+mN_1(l,-k))$ such that $d_{\tilde {\mathcal F}^u}(y,y_1)<N_2\cdot a_1$ .

Then, we have

$$ \begin{align*} d_{\mathbb R^2}(x,y)\geq |m|\cdot b_1-(N_2+1)a_1 \quad \text{and} \quad d_{\tilde{\mathcal F}^u}(x,y)\leq |m|\cdot N_2\cdot a_1+(N_2+1)a_1, \end{align*} $$

which implies

$$ \begin{align*} d_{\tilde{\mathcal F}^u}(x,y) \leq \bigg[\frac{N_2\cdot a_1}{b_1}\bigg]\cdot d_{\mathbb R^2}(x,y)+ \bigg[\frac{N_2(N_2+1)a_1^2}{b_1}+(N_2+1)a_1\bigg]. \end{align*} $$

This proves that $\tilde {\mathcal F}^u$ is quasi-isometric.

Proof. Assume A is not hyperbolic. Let $F:\mathbb R^2\to \mathbb R^2$ be a lift of f. Then $F(x)=Ax+G(x)$ for every $x\in \mathbb R^2$ , where $G:\mathbb R^2\to \mathbb R^2$ is a ${\mathbb Z}^2$ -periodic continuous function:

$$ \begin{align*} G(x+m)=G(x)\quad\text{for all } x\in{\mathbb R}^2, \text{ for all } m\in{\mathbb Z}^2. \end{align*} $$

In particular, there exists $C_0>0$ such that $\|G(x)\|\leq C_0$ for every $x\in \mathbb R^2$ .

For any bounded set $\gamma \subset \mathbb R^2$ , denote $\|\gamma \|=\sup _{x\in \gamma }\{\|x\|\}$ . Then we inductively have

$$ \begin{align*} \|F(\gamma)\|&=\|A(\gamma)+G(\gamma)\| \leq\|A\|\cdot\|\gamma\|+C_0 \\ \|F^2(\gamma)\|&=\|F(A(\gamma)+G(\gamma))\| \leq\|A^2(\gamma)+A\circ G(\gamma)\|+C_0 \\ &\leq\|A^2\|\cdot\|\gamma\|+C_0\|A\|+C_0 \\ &\cdots \cdots \\ \|F^k(\gamma)\|&\leq\|A^k\|\cdot\|\gamma\|+ C_0\cdot\bigg(\sum_{i=0}^{k-1}\|A^i\|\bigg). \end{align*} $$

Since A is not hyperbolic, $\|F^k(\gamma )\|$ has at most polynomial growth rate in $\mathbb R^2$ with respect to k.

However, if we take a segment $\gamma ^u\subset \tilde {\mathcal F}^u(x)$ for any $x\in \mathbb R^2$ with two endpoints $x,y\in \gamma ^u$ , then

$$ \begin{align*} d_{\tilde{\mathcal F}^u}(F^n(x),F^n(y))\geq C^{-1}\unicode{x3bb}^n\cdot d_{\tilde{\mathcal F}^u}(x,y). \end{align*} $$

The third item of Lemma 2.2 shows that $\tilde {\mathcal F}^u$ is quasi-isometric, and thus we have

$$ \begin{align*} d_{\mathbb R^2}(F^n(x),F^n(y)) \geq \frac{1}{a}\cdot (d_{\tilde{\mathcal F}^u}(F^n(x), F^n(y))-b) \geq \frac{1}{a}\cdot (C^{-1}\unicode{x3bb}^n\cdot d_{\tilde{\mathcal F}^u}(x,y)-b). \end{align*} $$

This implies

$$ \begin{align*} \|F^n(\gamma^u)\| \geq \frac{1}{2}\cdot d_{\mathbb R^2}(F^n(x), F^n(y))\geq \frac{1}{2a}\cdot (C^{-1}\unicode{x3bb}^n\cdot d_{\tilde{\mathcal F}^u}(x,y)-b), \end{align*} $$

which has exponential growth rate. This is a contradiction, so $f_*=A\in \mathrm {GL}(2,{\mathbb Z})$ is hyperbolic.

Proof of Proposition 3.1

Since $H:\mathbb R^2\to \mathbb R^2$ is ${\mathbb Z}^2$ -periodic which induces a continuous surjective map $h:\mathbb T^2\to \mathbb T^2$ , we only need to show that H is injective, which will guarantee that both H and h are homeomorphisms.

We have the following claim which implies that $h({\mathcal F}^u)=L^u$ .

Claim 3.3. Let $\tilde L^u$ be the expanding line foliation of A on $\mathbb R^2$ . For any $x\in \mathbb R^2$ , the map H satisfies

$$ \begin{align*} H(\tilde{\mathcal F}^u(x))=\tilde L^u(H(x))\end{align*} $$

and it is a homeomorphism.

Proof of the claim

First, we have $H(\tilde {\mathcal F}^u(x))\subset \tilde L^u(H(x))$ . For any $y\in \tilde {\mathcal F}^u(x)$ , by (2.1),

$$ \begin{align*} d_{\mathbb R^2} (F^{-n}(x), F^{-n}(y))\to 0\quad \text{as } n \to +\infty. \end{align*} $$

Together with $\|H-\mathrm {Id}\|_{C^0}<K$ , this implies that there exists $C>0$ such that

$$ \begin{align*} d_{\mathbb R^2} (H\circ F^{-n}(x), H\circ F^{-n}(y))<C\quad \text{for all } n\geq0. \end{align*} $$

Then, by the semi-conjugacy $H\circ F(x)=A\circ H(x)$ , we have

$$ \begin{align*} d_{\mathbb R^2} ( A^{-n}(H(x)), A^{-n}(H(y)))<C. \end{align*} $$

Hence, $H(y)\in \tilde L^u(H(x))$ .

Moreover, $H:\tilde {\mathcal F}^u(x)\to \tilde L^u(H(x))$ is injective. Actually, for any $y,z\in \tilde {\mathcal F}^u(x)$ , since $\tilde {\mathcal F}^u$ is quasi-isometric,

$$ \begin{align*} d_{\mathbb R^2} (F^n(y), F^n(z))\geq \frac{1}{a} (d_{\tilde{\mathcal F}^u}(F^n(y),F^n(z))-b)\to\infty\quad\text{as } n\to+\infty. \end{align*} $$

If $H(y)=H(z)$ , then $H\circ F^n(y)=A^n\circ H(y)=A^n\circ H(z)=H\circ F^n(z)$ and hence

$$ \begin{align*} d_{\mathbb R^2} (F^n(y), F^n(z))\leq d_{\mathbb R^2}(F^n(y), H\circ F^n(y))+d_{\mathbb R^2}(H\circ F^n(z), F^n(z))\leq 2K\\\quad \text{for all } n \geq 0. \end{align*} $$

This is a contradiction, and thus $H(y)\neq H(z)$ and $H:\tilde {\mathcal F}^u(x)\to \tilde L^u(H(x))$ is injective.

Finally, since H is continuous and $\tilde {\mathcal F}^u(x)$ is simply connected, it follows that

$$ \begin{align*} H(\tilde {\mathcal F}^u(x))\subset \tilde L^u(H(x)) \end{align*} $$

is also simply connected. This together with $\|H-\mathrm {Id}\|_{C^0}<K$ implies that

$$ \begin{align*} H(\tilde{\mathcal F}^u(x))= \tilde L^u(H(x)), \end{align*} $$

proving that H is surjective.

Therefore, $H:\tilde {\mathcal F}^u(x)\to \tilde L^u(H(x))$ is a homeomorphism for every $x\in \mathbb R^2$ as claimed.

To complete the proof, we consider the corresponding quotient maps. Recall that $F:\mathbb R^2\to \mathbb R^2$ preserving the foliation $\tilde {\mathcal F}^u$ and $A:\mathbb R^2\to \mathbb R^2$ preserving $\tilde L^u$ . Since H maps the foliation $\tilde {\mathcal F}^u(x)$ onto the foliation $\tilde L^u(H(x))$ , namely

$$ \begin{align*} H(\tilde{\mathcal F}^u(x))=\tilde L^u(H(x)) \end{align*} $$

for every $x\in \mathbb R^2$ , the commutative diagram $H\circ F=A\circ H$ reduces to a diagram of the corresponding quotient maps on quotient spaces $\mathbb R^2/\tilde {\mathcal F}^u$ and $\mathbb R^2/\tilde L^u$ . Namely, we have the following diagram:

(3.1)

where $\hat F$ (respectively $\hat A, \hat H$ ) is induced by F (respectively $A, H$ ) on the quotient space. Since by Lemma 2.2 both ${\mathcal F}^u$ and $L^u=\pi (\tilde L^u)$ are irrational minimal foliations on $\mathbb T^2$ , both quotient spaces $\mathbb R^2/\tilde {\mathcal F}^u$ and $\mathbb R^2/\tilde L^u$ are necessarily isomorphic to $\mathbb R$ . We denote $\mathbb R_{\tilde {\mathcal F}^u}=\mathbb R^2/\tilde {\mathcal F}^u$ . Notice that the quotient space $\mathbb R^2/\tilde L^u$ is equal to the stable leaf $\tilde L^s(0)$ of A at $0\in \mathbb R^2$ : $\tilde L^s(0)=\mathbb R^2/\tilde L^u$ , and the quotient map $\hat A=A$ on $\tilde L^s(0)$ . Thus, the diagram in (3.1) induces a new diagram

(3.2)

where $\hat A=A:\tilde L^s(0)\to \tilde L^s(0)$ is the linear contracting map $A(w)=\unicode{x3bb} ^{-1}w$ for every $w\in \tilde L^s(0)\subset \mathbb R^2$ .

We have the following claim.

Proof of the claim

Since $\|H-\mathrm {Id}\|_{C^0}<K$ , the orientation of $\mathbb R_{\tilde {\mathcal F}^u}$ induces an orientation on $\tilde L^s(0)$ by H globally on $\mathbb R^2$ . Since $F:\mathbb R^2\to \mathbb R^2$ is a diffeomorphism, the quotient map $\hat F:\mathbb R_{\tilde {\mathcal F}^u}\to \mathbb R_{\tilde {\mathcal F}^u}$ is a homeomorphism. Since iterating $\hat F$ is necessary, we can assume that $\hat F:\mathbb R_{\tilde {\mathcal F}^u}\to \mathbb R_{\tilde {\mathcal F}^u}$ preserves the orientation and so does $A:\tilde L^s(0)\to \tilde L^s(0)$ .

For claim (a1), assume otherwise that two points $\hat x, \hat y\in \mathbb R_{\tilde {\mathcal F}^u}$ satisfy $\hat x<\hat y$ and $\hat H(\hat x)>\hat H(\hat y)$ in $\tilde L^s(0)$ . Since $\hat F$ preserves the orientation, for $\tilde {\mathcal F}^u(x)=\hat x$ and $\tilde {\mathcal F}^u(y)=\hat y$ , we have

$$ \begin{align*} \hat F^{-n}(\hat x)<\hat F^{-n}(\hat y)\quad \text{and} \quad F^{-n}(\tilde{\mathcal F}^u(x))<F^{-n}(\tilde{\mathcal F}^u(y)) \quad\text{for all } n>0. \end{align*} $$

For $\hat H(\hat x)>\hat H(\hat y)$ and $\hat H(\hat x)=\hat L^u(H(x))>\hat H(\hat y)=\hat L^u(H(y))$ , since $A^{-1}$ is uniformly expanding along $\tilde {L}^s(0)$ , we have

$$ \begin{align*} \hat A^{-n}(\hat H(\hat x))-\hat A^{-n}(\hat H(\hat y))\to +\infty\quad \text{as } n\to+\infty. \end{align*} $$

This is equivalent to that the Hausdorff distance between $A^{-n}(\tilde L^u(H(x)))$ and $A^{-n}(\tilde L^u(H(y)))$ tends to infinity as $n\to +\infty $ and $A^{-n}(\tilde L^u(H(x)))>A^{-n}(\tilde L^u(H(y)))$ .

However, since $F^{-n}(\tilde {\mathcal F}^u(x))<F^{-n}(\tilde {\mathcal F}^u(y))$ for every $n>0$ , the semi-conjugation

$$ \begin{align*} A^{-n}(\tilde L^u(H(x)))=H\circ F^{-n}(\tilde{\mathcal F}^u(x)), A^{-n}(\tilde L^u(H(y)))=H\circ F^{-n}(\tilde{\mathcal F}^u(y)), \end{align*} $$

and $\|H-\mathrm {Id}\|_{C_0}<K$ shows that $A^{-n}(\tilde L^u(H(x))$ has $2K$ -bounded distance with the negative component of $\mathbb R^2\setminus A^{-n}(\tilde L^u(H(y)))$ for every $n>0$ . This contradicts the fact that the Hausdorff distance between $A^{-n}(\tilde L^u(H(x)))$ and $A^{-n}(\tilde L^u(H(y)))$ tends to infinity as $n\to +\infty $ and

$$ \begin{align*} A^{-n}(\tilde L^u(H(x)))>A^{-n}(\tilde L^u(H(y))). \end{align*} $$

This proves claim (a1).

For claim (a2), the surjective part of $\hat H$ comes from the fact that $H:\mathbb R^2\to \mathbb R^2$ is surjective. We only need to show that $\hat H$ is injective. Otherwise, $\hat H(\hat x)=\hat H(\hat y)$ , meaning that both $\tilde {\mathcal F}^u(x)$ and $\tilde {\mathcal F}^u(y)$ are mapped to a single line $\tilde L^u(H(x))=\tilde L^u(H(y))$ . Since $\hat H$ is orientation-preserving and increasing, it follows that H maps the region $R_{x,y}\subset \mathbb R^2$ bounded by $\tilde {\mathcal F}^u(x)$ and $\tilde {\mathcal F}^u(y)$ in $\mathbb R^2$ to $\tilde L^u(H(x))$ . However, since $\tilde {\mathcal F}^u$ is an irrational minimal foliation, we have

$$ \begin{align*} \mathbb T^2=\pi (R_{x,y}) \quad\text{and}\quad \mathbb T^2=h(\mathbb T^2)=h\circ\pi (R_{x,y}))= \pi (H(R_{x,y}))=\pi (\tilde L^u(H(x))). \end{align*} $$

This is a contradiction since $\pi (\tilde L^u(H(x)))$ is a single one-dimensional leaf in $\mathbb T^2$ . This proves the claim.

Finally, the injectivity of H follows from the injectivity of both $H|_{\tilde {\mathcal F}^u(x)}$ for every $x\in \mathbb R^2$ and $\hat H$ . Thus, $H:\mathbb R^2\to \mathbb R^2$ is a homeomorphism.

Proof. By Lemma 2.1, for every periodic point p of f, $f$ has one positive Lyapunov exponent along $\mathcal F^u$ which is $\log \unicode{x3bb} $ . We denote it as $\unicode{x3bb} ^u(p)=\log \unicode{x3bb} $ . By Proposition 3.1, f is topologically conjugate to A by h and $h(\mathcal F^u)=L^u$ . Since A is uniformly contracting along the transversal direction of $L^u$ , f is topologically contracting in the transversal direction of $\mathcal F^u$ . Thus, p has another Lyapunov exponent $\unicode{x3bb} ^{cs}(p)\leq 0$ .

Moreover, the periodic measures of A are dense in the space of ergodic measures of A. By the topological conjugacy, the periodic measures of f are also dense in the space of ergodic measures of f. Thus, for every ergodic measure $\mu $ of f, $f$ has two Lyapunov exponents

$$ \begin{align*} \unicode{x3bb}^{cs}(\mu)\leq 0<\unicode{x3bb}^u(\mu)=\log\unicode{x3bb}. \end{align*} $$

Now, we only need to show that f admits a dominated splitting, which implies that f is partially hyperbolic. That is the following claim.

Claim 4.2. There exists a $Df$ -dominated splitting $T\mathbb T^2=E^{cs}\oplus E^u$ with $E^u=T\mathcal F^u$ , that is, the splitting is continuous, $Df$ -invariant, and there exist two constants $0<\eta <1, C>1$ , such that

$$ \begin{align*} \frac{\|Df^n|_{E^{cs}(x)}\|}{\|Df^n|_{E^u(x)}\|} \leq C\cdot\eta^n\quad \text{for all } x\in\mathbb T^2, n\geq0. \end{align*} $$

Proof of the claim

The proof follows exactly the same form as in [Reference Gogolev and Shi25, Proposition 5.9]. Since $\mathcal F^u$ is a $C^2$ -foliation, the $Df$ -invariant $E^u=T\mathcal F^u$ is a $C^1$ -bundle. We have a $C^1$ -smooth splitting $T\mathbb T^2=E^u\oplus E^{\perp }$ where $E^{\perp }$ is perpendicular to $E^u$ . We take continuous families of unit vectors in $\{e^u(x),e^{\perp }(x)\}_{x\in \mathbb T^2}$ in $E^u,E^{\perp }$ , respectively, which form a $C^1$ base on $T\mathbb T^2$ .

Since F is $C^2$ -smooth, there exist three families of $C^1$ -functions $\{A(x)\}_{x\in \mathbb T^2}$ , $\{B(x)\}_{x\in \mathbb T^2}$ , and $\{C(x)\}_{x\in \mathbb T^2}$ , such that in the base $\{e^u(x),e^{\perp }(x)\}_{x\in \mathbb T^2}$ ,

$$ \begin{align*} Df(x) = \left( \begin{array}{@{}cc@{}} A(x) & B(x) \\ 0 & C(x) \\ \end{array} \right) \quad \text{for all } x\in\mathbb T^2. \end{align*} $$

Then, we have

$$ \begin{align*} Df(e^u(x))=A(x)e^u(fx), \quad \text{and} \quad \textit{proj}^{\perp}\circ Df(e^{\perp}(x))=C(x)e^{\perp}(fx), \end{align*} $$

where $\textit {proj}^{\perp }:T\mathbb T^2\to E^{\perp }$ is the projection through $E^u$ .

For every $x\in \mathbb T^2$ and $n\geq 1$ , we introduce the following notation for the cocycles:

$$ \begin{align*} A^n(x)=\prod_{i=0}^{n-1}A(f^i(x))\quad \text{and}\quad C^n(x)=\prod_{i=0}^{n-1}C(f^i(x)). \end{align*} $$

Lemma 2.1 shows that there exists $C>1$ , such that $|A^n(x)|\geq C^{-1}\unicode{x3bb} ^n$ for every $x\in \mathbb T^2$ and $n\in \mathbb N$ . Moreover, for every $\epsilon>0$ and every periodic point p of f, since the other Lyapunov exponent of p is non-positive, we have

$$ \begin{align*} \lim_{n\to+\infty}\frac{1}{n}\log|C^n(p)|\leq\epsilon. \end{align*} $$

Since periodic measures are dense in all invariant measures, we have

$$ \begin{align*} \lim_{n\to+\infty}\frac{1}{n}\log|C^n(x)|\leq\epsilon\quad \text{for all } x\in\mathbb T^2. \end{align*} $$

We fix $0<\epsilon \ll \log \unicode{x3bb} $ , and [Reference Kalinin31, Theorem 1.3] shows that there exists some $N=N(\epsilon )$ , such that

$$ \begin{align*} |C^n(x)|\leq\exp(n\epsilon)\quad \text{for all } n\geq N, \text{ for all } x\in\mathbb T^2. \end{align*} $$

Finally, since $B(x)$ varies $C^1$ -smoothly with respect to $x\in \mathbb T^2$ and is uniformly bounded, there exists a continuous cone field $\{\mathcal C(x)\}_{x\in \mathbb T^2}$ containing $E^u$ , such that

$$ \begin{align*} Df (\overline{\mathcal C(x)})\subset {\mathcal C}(f(x))\quad \text{for all } x\in\mathbb T^2. \end{align*} $$

Therefore, by the cone-field criterion [Reference Crovisier and Potrie9, Theorem 2.6], there exists a dominated splitting

$$ \begin{align*} T\mathbb T^2=E^{cs}\oplus E^u \quad\text{with } T{\mathcal F}^u=E^u. \end{align*} $$

This proves the claim.

Finally, [Reference Potrie37, Proposition 4.A.7] shows that a partially hyperbolic diffeomorphism on $\mathbb T^2$ is dynamically coherent, that is, there exists an f-invariant foliation $\mathcal F^{cs}$ tangent to $E^{cs}$ . Moreover, by the topological conjugacy $h\circ f=A\circ h$ , the foliation $h(\mathcal F^{cs})$ is A-invariant and transverse to $L^u=h(\mathcal F^u)$ , which is unique and $L^s=h(\mathcal F^{cs})$ .

Proof. First of all, let $\mu _{\mathrm {\max }}$ be the measure with maximal entropy of f, which is also the measure with maximal entropy of $f^{-1}$ . Since f is topologically conjugate to A, the measure entropy of $\mu _{\mathrm {\max }}$ associated to $f^{-1}$ is equal to the topological entropy of A which is $\log \unicode{x3bb} $ . From Ruelle’s inequality, the largest Lyapunov exponent of $f^{-1}$ in $\mu _{\mathrm {\max }}$ satisfies

$$ \begin{align*} \log\unicode{x3bb} \leq \unicode{x3bb}^+(\mu_{\mathrm{\max}},f^{-1}) = -\unicode{x3bb}^{cs}(\mu_{\max},f). \end{align*} $$

From the density of periodic measures, there exists a sequence of periodic points $p_n$ whose periodic measures converge to $\mu _{\mathrm {\max }}$ . Then, we have

(4.1) $$ \begin{align} \lim_{n\to\infty}\unicode{x3bb}^{cs}(p_n) = \unicode{x3bb}^{cs}(\mu_{\mathrm{\max}},f) \leq -\log\unicode{x3bb}. \end{align} $$

In particular, $\unicode{x3bb} ^{cs}(p_n)<0$ and $p_n$ is hyperbolic for n large enough.

Proof of the claim

Since $\phi _t$ is $C^2$ , when restricted to a smooth transversal cross section which is diffeomorphic to a unit circle $\mathbb S^1$ , the induced map $\hat \phi $ is a $C^2 $ irrational rotation, whose rotation number is a degree-2 algebraic number (that is, the largest eigenvalue $\unicode{x3bb} $ of $f_*$ ). By Herman [Reference Herman28], see also [Reference Katznelson and Ornstein33, Reference Khanin and Teplinsky34], $\hat \phi $ is bi-Lipschitz conjugate to $R_\unicode{x3bb} $ . As a result, there exists a constant $C_2>0$ such that, for any small segment I that is contained in a leaf of $\mathcal F^{cs}(x)$ ,

(4.2) $$ \begin{align} \frac{1}{C_2}\le\frac{|I|}{|\mathrm{Hol}_t(I)|}\le C_2. \end{align} $$

Here, $|\cdot |$ is the length function and $\mathrm {Hol}_t:\mathcal F^{cs}(x)\to \mathcal F^{cs}(\phi _t(x))$ is the holonomy map induced by $\phi _t$ satisfying

$$ \begin{align*} \mathrm{Hol}_t(x)=\phi_t(x)\quad \text{and}\quad \mathrm{Hol}_t(I)\subset\mathcal F^{cs}(\phi_t(x)) \end{align*} $$

for any segment $I\subset \mathcal F^{cs}(x)$ of length small enough. We remark that the constant $C_2$ is independent of I and t.

Now, fix two distinct hyperbolic periodic points $p,q$ . Then the Lyapunov exponents of $p,q$ along $E^{cs}$ satisfy $\unicode{x3bb} ^{cs}(p),\unicode{x3bb} ^{cs}(q)<0$ . In particular, we have that both $\mathcal F^{cs}(p)$ and $\mathcal F^{cs}(q)$ are contained in the stable manifolds of p and q, respectively. Denote $\pi>0$ to be the common period of p, and q: $f^\pi (p)=p$ and $f^\pi (q)=q$ .

Let x be an intersecting point of $\mathcal F^u(q)$ with the local stable manifold $\mathcal F^{cs}_{\mathrm {loc}}(p)$ . Then there exists $t\in \mathbb R$ such that $x=\phi _t(q)\in \mathcal F^{cs}(p)$ . Moreover, we can define the holonomy map

$$ \begin{align*} \mathrm{Hol}_t:\mathcal F^{cs}(q)\to\mathcal F^{cs}(x)=\mathcal F^{cs}(p)\quad \text{with } \mathrm{Hol}_t(q)=x. \end{align*} $$

Take another point $y\in \mathcal F^{cs}_{\mathrm {loc}}(q)$ and a segment $J\subset \mathcal F^{cs}(q)$ with two endpoints $q,y$ . Then there exists a unique point $z=\mathrm { Hol}_t(y)\in \mathcal F^{cs}(x)=\mathcal F^{cs}(p)$ . Since we can take y close to q and $|J|$ is small, (4.2) implies $|\mathrm {Hol}_t(J)|$ is small, and $\mathrm {Hol}_t(J)\subset \mathcal F^{cs}(p)$ with endpoints $x=\mathrm {Hol}_t(q)$ and $z=\mathrm {Hol}_t(y)$ .

Since f is $C^r$ -smooth with $r\geq 2$ , $E^{cs}$ is Hölder continuous. This implies both $\mathcal F^{cs}_{\mathrm {loc}}(p)$ and $\mathcal F^{cs}_{\mathrm {loc}}(q)$ are $C^{1+\text {H\"older}}$ -smooth sub-manifolds. Since f is uniformly contracting in $\mathcal F^{cs}_{\mathrm {loc}}(p)$ and $\mathcal F^{cs}_{\mathrm {loc}}(q)$ , the distortion control argument shows that there exists $K>0$ such that, for every $n\geq 0$ ,

$$ \begin{align*} \frac{1}{K}\leq\frac{|f^{\pi n}(J)|}{\exp(\unicode{x3bb}^{cs}(q)\pi n)}\leq K \quad \text{and} \quad \frac{1}{K}\leq\frac{|f^{\pi n}\circ\mathrm{Hol}_t(J)|}{\exp(\unicode{x3bb}^{cs}(p)\pi n)} \leq K. \end{align*} $$

However, since both $\mathcal F^{cs}$ and $\mathcal F^u$ are f-invariant, the holonomy map $\mathrm {Hol}_t$ is commuting with f, and thus, for every $n>0$ , there exists $t_n\in \mathbb R$ such that

$$ \begin{align*} f^{\pi n}{\kern-1pt}\circ{\kern-1pt}\mathrm{Hol}_t(J){\kern-1pt}={\kern-1pt}\mathrm{Hol}_{t_n}{\kern-1pt}\circ{\kern-1pt} f^{\pi n}(J) \quad\text{and}\quad \frac{1}{C_2}{\kern-1pt}\leq{\kern-1pt}\frac{|f^{\pi n}(J)|}{|\mathrm{Hol}_{t_n}\circ f^{\pi n}(J)|}{\kern-1pt} ={\kern-1pt}\frac{|f^{\pi n}(J)|}{|f^{\pi n}\circ\mathrm{ Hol}_{t}(J)|}{\kern-1pt}\leq{\kern-1pt} C_2. \end{align*} $$

This implies

$$ \begin{align*} \frac{1}{KC_2}\leq\frac{\exp(\unicode{x3bb}^{cs}(p)\pi n)}{\exp(\unicode{x3bb}^{cs}(q)\pi n)} \leq KC_2\quad \text{for all } n>0. \end{align*} $$

Thus, we must have $\unicode{x3bb} ^{cs}(p)=\unicode{x3bb} ^{cs}(q)$ for every pair of hyperbolic periodic points p and q.

From this claim and (4.1), we know that

(4.3) $$ \begin{align} \unicode{x3bb}^{cs}(p)\leq-\log\unicode{x3bb}<0 \end{align} $$

for every hyperbolic periodic point p of f.

Since $f:\mathbb T^2\to \mathbb T^2$ is topologically conjugate to $A:\mathbb T^2\to \mathbb T^2$ , it also satisfies the specification property in [Reference Sigmund40]. If there exists some periodic point $p\in \mathrm { Per}(f)$ satisfying $\unicode{x3bb} ^{cs}(p)=0$ , then by the specification property, there exist hyperbolic periodic points of f with Lyapunov exponents arbitrarily close to zero along $E^{cs}$ . This is absurd since $\unicode{x3bb} ^{cs}(p)\leq -\log \unicode{x3bb} $ for every hyperbolic periodic point p. Thus, every periodic point p of f is hyperbolic with $\unicode{x3bb} ^{cs}(p)\leq -\log \unicode{x3bb} $ .

This implies f is Anosov and $\unicode{x3bb} ^{cs}(\mu )\leq -\log \unicode{x3bb} $ for every ergodic measure $\mu $ of f. If we consider the Sinai–Ruelle–Bowen measure $\mu ^-$ of $f^{-1}$ , its Lyapunov exponent along $E^{cs}$ is equal to its measure entropy $h(\mu ^-,f^{-1})$ , which is smaller than $\log \unicode{x3bb} $ . So we have

$$ \begin{align*} -\unicode{x3bb}^{cs}(\mu^-)=\unicode{x3bb}^{u}(\mu^-,f^{-1})=h(\mu^-,f^{-1})\leq h_{\mathrm{top}}(f^{-1})=\log\unicode{x3bb}. \end{align*} $$

This implies $\unicode{x3bb} ^{cs}(p)=\unicode{x3bb} ^{cs}(\mu ^-)\geq -\log \unicode{x3bb} $ . Combined with (4.3) and Lemma 2.1, we have

$$ \begin{align*} \unicode{x3bb}^{cs}(p)=-\log\unicode{x3bb} \quad \text{and} \quad \unicode{x3bb}^u(p)=\log\unicode{x3bb} \quad \text{for all } p\in\mathrm{Per}(f). \end{align*} $$

Finally, the work of de la Llave [Reference de la Llave13, Reference de la Llave14] (e.g. [Reference de la Llave14, Theorem 1.1]) indicates that the conjugacy h is $C^{r-\epsilon }$ -smooth when all periodic points of f have the same Lyapunov exponents to A.

Proof of Theorem 1.1

We have shown that by the $C^{r-\epsilon }$ -smooth conjugacy,

$$ \begin{align*} A=h\circ f\circ h^{-1}=h\circ\rho(1,0)\circ h^{-1}. \end{align*} $$

Now, we can define a $C^{r-\epsilon }$ -smooth flow on $\mathbb T^2$ :

$$ \begin{align*} \psi_t=h\circ\phi_t\circ h^{-1}=h\circ\rho(0,t)\circ h^{-1}\quad\text{satisfying } A\circ\psi_t=\psi_{\unicode{x3bb} t}\circ A. \end{align*} $$

Moreover, we have shown that $h(\mathcal F^u)=L^u$ which maps the orbit of $\phi _t$ to the linear unstable foliation of A. Thus, the orbit of $\psi _t$ is $L^u$ . To prove Theorem 1.1, we only need to show that $\psi _t$ has constant velocity.

Denote

$$ \begin{align*} \mathcal Z(x)=\frac{d}{dt}|_{t=0}\psi_t(x). \end{align*} $$

Let p be a fixed point of A and $x\in L^u(p)$ with $\psi _t(p)=x$ for some $t\in \mathbb R$ . Then, we have

$$ \begin{align*} A^n\circ\psi_t(p)=\psi_{\unicode{x3bb}^nt}\circ A(p)\quad \text{and} \quad DA^n\circ D\psi_t(\mathcal Z(p))=D\psi_{\unicode{x3bb}^nt}\circ DA^n(\mathcal Z(p)). \end{align*} $$

This implies that $DA^n(\mathcal Z(x))=D\psi _{\unicode{x3bb} ^nt}(\unicode{x3bb} ^n\cdot \mathcal Z(p))$ . By taking the norm, we have

$$ \begin{align*} \unicode{x3bb}^n\cdot\|\mathcal Z(x)\|=\unicode{x3bb}^n\cdot\|\mathcal Z(\psi_{\unicode{x3bb}^n t}(p))\| \quad \text{for all } n\in{\mathbb Z}. \end{align*} $$

Letting $n\to -\infty $ , we have

$$ \begin{align*} \psi_{\unicode{x3bb}^n t}(p)\to p\quad\text{and}\quad \|\mathcal Z(\psi_{\unicode{x3bb}^n t}(p))\|\to\|\mathcal Z(p)\|. \end{align*} $$

This implies that $\|\mathcal Z(x)\|=\|\mathcal Z(p)\|$ for every $x\in L^u(p)$ .

Since $L^u(p)$ is dense in $\mathbb T^2$ , we have $\|\mathcal Z(x)\|=\|\mathcal Z(p)\|\triangleq a$ for every $x\in \mathbb T^2$ . Thus, $\psi _t$ is the linear flow with constant velocity. This proves Theorem 1.1, that $\rho $ is $C^{r-\epsilon }$ conjugate to the affine action $\{A,v_{at}\}$ .