Proof. Let $T^1{\mathbb T}^4$ be the unit tangent bundle of ${\mathbb T}^4$ (which can be identified with ${\mathbb T}^4\times \mathbb S^3$ ) with $Df_*$ being the $C^1$ diffeomorphism induced by f. We will consider the central unit tangent bundle $S_f:=\bigcup _{x\in {\mathbb T}^4}S(f,x)$ , where $S(f,x)=T_x^1{\mathcal F}^c_f(x)$ is the unit circle in $E^c_f(x)$ . Then, $S_f$ is a Hölder submanifold of $T^1{\mathbb T}^4$ invariant under $Df_*$ , which is also a Hölder bundle over ${\mathbb T}^4$ .

We claim that there exists a continuous unstable foliation on $S_f$ and that $Dh_{f,x,y*}^u$ is exactly the unstable holonomy for this foliation between the transversals $T^1{\mathcal F}^c_f(x)$ and $T^1{\mathcal F}^c_f(y)$ . We apply the standard construction of the local unstable leaves as the invariant section of a bundle contraction map (see [Reference Hirsch, Pugh and Shub13] for example), with a minor difficulty arising from the lack of smoothness.

For any $x,y\in {\mathbb T}^4$ , we define the $\pi _{f,y,x}:E^c_f(y)\rightarrow E^c_f(x)$ as the projection parallel to $E^s_f(x)\oplus E^u_f(x)$ . The maps $\pi _{f,y,x}$ depend continuously on $x,y\in {\mathbb T}^4$ and f (in the $C^1$ topology). For x close to y, this is invertible and close to the identity, and its projectivization $\pi _{f,x,y*}$ is bi-Lipschitz with Lipschitz constant close to 1.

For $\delta>0$ and $x\in {\mathbb T}^4$ , let $\alpha _{f,x}:[-\delta ,\delta ]\rightarrow {\mathcal F}^u_f(x)$ be the length parameterization of the local unstable manifold of f at x. Since the unstable foliation is orientable and depends continuously in the $C^1$ topology with respect to x and f, we have that $\alpha _{f,x}$ is continuous in x and f (in the $C^1$ topology).

For any $\delta>0$ , there exists $\epsilon _{\delta }>0$ such that for any $x,y$ such that $d(x,y)<\delta $ , we have:

• • $\|\pi _{f,y,x}^{\pm 1}-\mathrm {Id}\|<\epsilon _{\delta }$ ;

• • $\pi _{f,y,x*}^{\pm 1}$ is bi-Lipschitz with constant $(1+\epsilon _{\delta })$ ;

• • $(1+\epsilon _{\delta })^{-1}\unicode{x3bb} _c^-(x)<\unicode{x3bb} _c^-(y)\leq \unicode{x3bb} _c^+(y)<(1+\epsilon _{\delta })\unicode{x3bb} _c^+(x)$ .

• • If furthermore $y\in {\mathcal F}^u_{f,\delta }(x)$ , then $d_u(f(x),f(y))\geq (1+\epsilon _{\delta })^{-1}\unicode{x3bb} _u(x)d_u(x,y)$ , where $d_u$ is the distance along the unstable leaves.

We can choose $\epsilon _{\delta }$ independent of f in a $C^1$ neighborhood and $\lim _{\delta \rightarrow 0}\epsilon _{\delta }=0$ .

Now we will construct the bundle with the candidates for the local unstable manifolds in $S_f\:\!$ . Consider $\delta>0$ (small) to be specified later. Let

$$ \begin{align*} B=\bigg\{\sigma:[-\delta,\delta])\rightarrow \mathbb R:\ \sigma(0)=0,\ \bigg|\frac{\sigma(t)}{t}\bigg|<\infty\bigg\} \end{align*} $$

be the Banach space of functions $\sigma $ bounded for the norm

$$ \begin{align*} \|\sigma\|=\sup_{t\in [-\delta,\delta]}\bigg|\frac{\sigma(t)}{t}\bigg|. \end{align*} $$

Then,

$$ \begin{align*} V(f):=S_f\times B \end{align*} $$

is a continuous (in fact, Hölder) Banach bundle over $S_f$ .

Remark 2.4. The maps $\sigma $ are candidates for unstable manifolds in $S_f$ in the following sense. For any $(x,v)\in S_f$ and $\sigma \in B$ , we can define a section $\tilde \sigma :{\mathcal F}^u_{f,\delta }(x)\rightarrow S_f$ in the following way:

$$ \begin{align*} \tilde\sigma(y):=\pi_{f,y,x*}^{-1}(v+\sigma(\alpha_{f,x}^{-1}(y)))\in S(f,y). \end{align*} $$

The graph of this section $\tilde \sigma $ is a natural candidate for the local unstable manifold in ${(x,v)\in S_f}\:\!$ . We construct it as a fixed point of the natural graph transformation.

Let $T:V(f)\rightarrow V(f)$ be the bundle map which fibers over $Df_*$ on $S_f$ and is given by

$$ \begin{align*} T\sigma_{(x,v)}(t)&=(\pi_{f,f(y(t)),f(x)}\circ Df(y(t))\circ\pi_{f,y(t),x}^{-1})_*(v+\sigma(\alpha_{f,x}^{-1}(y(t))))\\ &\quad-Df(x)_*(v),\\ y(t)&=f^{-1}\circ\alpha_{f,f(x)}(t). \end{align*} $$

One can check that in fact T is defined in such a way so that we have $\tilde {T\sigma }=Df_*\tilde \sigma $ . Let us check that T is a continuous bundle map on $V(f)$ , which is also a fiber contraction.

Proof. Remember that $y(t)=f^{-1}\circ \alpha _{f,f(x)}(t)$ , and let us denote

$$ \begin{align*} G(t):=(\pi_{f,f(y(t)),f(x)}\circ Df(y(t))\circ\pi_{f,x,y}^{-1}). \end{align*} $$

Observe that $G(t)_*$ is Lipschitz with the Lipschitz constant

Also,

Then,

where in the last line, we used the inequality

$$ \begin{align*} \bigg\|\frac a{\|a\|}-\frac b{\|b\|}\bigg\|\leq\bigg\|\frac a{\|a\|}-\frac a{\|b\|}\bigg\|+\bigg\|\frac a{\|b\|}-\frac b{\|b\|}\bigg\|\leq\frac 2{\|b\|}\|a-b\|. \end{align*} $$

Let us remark that if $v\in E^c_f(x)$ , then $\pi _{f,f(y(t)),f(x)}\circ Df(x)\circ \pi _{f,y,x}^{-1}(v)=Df(x)(v)$ , because the partially hyperbolic splitting is invariant under $Df$ . Then,

Finally, we obtain the desired bound:

Proof. We have

Now all we have to do is to choose $\delta $ small enough so that $\epsilon _{\delta }$ is close enough to zero so that we have

Claims 1 and 2 show that we are in the conditions of the invariant section theorem from [Reference Hirsch, Pugh and Shub13], so there exists a unique bounded continuous invariant section.

From [Reference Pugh, Shub and Wilkinson24], we know that the unstable holonomy along center leaves is uniformly differentiable. The projectivization of the derivative of this local holonomy will then correspond to a bounded invariant section for the transfer operator T, so it has to coincide with the unique continuous invariant section constructed above. This concludes the proof of the continuity of $Dh^u_{f,x,y*}$ with respect to the points $x,y$ (we proved it for $d(x,y)<\delta $ , but this can be easily extended to larger distances).

If $f_n$ converges to f, then $S_{f_n}$ converges to $S_f$ (this can be made explicit by projecting $E_{f_n}^c$ to $E^c_f$ parallel to $E_f^s\oplus E_f^u$ for example). One can check that the corresponding transfer operators $T_{f_n}$ also converge to $T_f$ . Since the invariant section is continuous with respect to the bundle map, we obtain the continuity of $Dh^u_{f,x,y*}$ with respect to f.

Proof. This is explained in [Reference Ben Ovadia3, Theorem 2.4] and [Reference Buzzi, Crovisier and Sarig7, §1.6]. See also [Reference Ben Ovadia2] and [Reference Buzzi, Crovisier and Sarig7, Corollary 3.3].

Proof. Denote by ${\mathcal F}^c_f$ the center foliation of f. By Proposition 1.3, the projection map $\pi ^c_f$ along the center foliation induces a topological Anosov homeomorphism $\overline {f}$ on the quotient space $\overline {{\mathbb T}}^2_f={\mathbb T}^4/{\mathcal F}^c_f$ , which is topological conjugate to A, so we may in fact identify $\overline {{\mathbb T}}^2_f$ with ${\mathbb T}^2$ and $\overline {f}$ with A.

Denote by ${\mathcal F}^s_A$ (respectively ${\mathcal F}^u_A$ ) the stable (respectively unstable) foliation of A. The projection map $\pi ^c_f$ maps each center unstable leaf ${\mathcal F}^{cu}$ of f to an unstable leaf ${\mathcal F}^u_A$ of A. In particular, $\pi ^c_f$ maps every unstable leaf ${\mathcal F}^u$ of f to an unstable leaf ${\mathcal F}^u_A$ of A. Proposition 1.3 implies that all the hypotheses of Tahzibi and Yang [Reference Tahzibi and Yang28, Theorem A] are satisfied (see also [Reference Álvarez1, §2.5]), and this implies that $h_\mu (f,{\mathcal F}^u)\leq h_{{\mathrm {top}}}(A)=\log \unicode{x3bb} _A$ .

Proof. We will show that $\unicode{x3bb} ^c_1>0$ . To prove that $\unicode{x3bb} ^c_2<0$ , one only needs to consider the diffeomorphism $f^{-1}$ instead of diffeomorphism f.

Suppose by contradiction that $\unicode{x3bb} ^c_1\leq 0$ . The entropy formula of Ledrappier and Young (see [Reference Brown5, Reference Ledrappier and Strelcyn15]) implies that $h_\mu (f)=h_\mu (f,{\mathcal F}^u)$ .

Combining with the previous lemma, we obtain that $h_\mu (f)\leq \log \unicode{x3bb} _A$ , which is a contradiction with the hypothesis that $h_\mu (f)>\log \unicode{x3bb} _A$ . The proof is complete.

Proof. An immediate consequence of the compactness of ${\mathcal F}^c(p)$ and of the fact that the stable and unstable holonomies depend continuously in the $C^1$ topology with respect to the points (see Remark 1.7) is the following lemma.

However, the holonomies along the center leaves depend continuously in the $C^1$ topology with respect to the map f (in $C^2$ topology), so the images of $\tilde g_{q_1}\times \tilde g_{q_2}\times \cdots \times \tilde g_{q_k}$ and $\pm \tilde v^{s^k}$ depend continuously on the map f. Since these images are compact, this concludes the $C^2$ openness of $\mathcal U_s^r$ . The proof for $\mathcal U_u^r$ is similar, so $\mathcal U^r=\mathcal U_s^r\cap \mathcal U_u^r$ is $C^2$ open.

Proof. Part (1). Since ${\mathcal F}^c(p)$ is compact, it is enough to prove that $\tilde g_{q(f\circ \phi _T),f\circ \phi _T}(x)$ is $C^1$ in $(x,T)$ in a small neighborhood of every point $(x_0,0^n)\in {\mathcal F}^c(p)\times \mathbb R^n$ .

Let $x_0\in {\mathcal F}^c(p)$ and denote $y_0=h_{p,q}^u(x_0)\in {\mathcal F}^c(q)$ , $y_1=f(y_0)\in {\mathcal F}^c(f(q))$ , $z_1=h_{f(q),p}^s(y_1)\in {\mathcal F}^c(p)$ , and $z_0=f^{-1}(z_1)\in {\mathcal F}^c(p)$ . The f-invariance of the stable holonomy implies that $h_{q,p}^s(y_0)=z_0$ , or $\tilde h_q(z_0)=x_0$ .

Let $\psi _p:\mathbb T^2\rightarrow {\mathcal F}^c(p)$ be a $C^2$ embedding, $a_0=\psi _p^{-1}(x_0),\ b_0=\psi _p^{-1}(z_0)$ . Let $I_{\delta }=[-\delta ,\delta ]$ . There exist $C^2$ foliations charts of ${\mathcal F}^s$ (respectively ${\mathcal F}^u$ ) on a small neighborhood of $y_1$ (respectively $y_0$ ) inside ${\mathcal F}^{cs}(p)$ (respectively ${\mathcal F}^{cu}(p)$ ):

$$ \begin{align*} &\alpha^s:B^c_{y_1}\times I_{\delta}\rightarrow B^{cs}_{y_1}, \alpha^s(\cdot,0)=\mathrm{Id}_{{\mathcal F}^c_{\mathrm{loc}}(y_1)}, \alpha^s(\{y\}\times I_{\delta})={\mathcal F}^s_{\mathrm{loc}}(y)\quad \text{for all } y\in B^c_{y_1};\\ &\alpha ^u:B^c_{y_0}\times I_{\delta}\rightarrow B^{cu}_{y_0}, \alpha^u(\cdot,0)=\mathrm{Id}_{{\mathcal F}^c_{\mathrm{loc}}(y_0)}, \alpha^u(\{y\}\times I_{\delta})={\mathcal F}^u_{\mathrm{loc}}(y)\quad \text{for all } y\in B^c_{y_0}, \end{align*} $$

where $B_x^*$ denotes a small ball centered in x inside ${\mathcal F}^*(x)$ . Define the $C^2$ maps

$$ \begin{align*} &\beta^s:B_{b_0}\times I_{\delta}\rightarrow B^{cs}_{y_1},\ \beta^s(b,s)=\alpha^s(h^s_{p,f(q)}(f(\psi_p(b))),s);\\ &\beta ^u:B_{a_0}\times I_{\delta}\rightarrow B^{cu}_{y_0},\ \beta^u(a,r)=\alpha^u(h_{p,q}^u(\psi_p(a)),r), \end{align*} $$

where $B_x$ is a small ball centered at x in $\mathbb T^2$ .

We know that the support of $\phi _T$ does not intersect $f^k({\mathcal F}^{cs}_{\mathrm {loc}}(y_1)) \text { for all } k\geq 0$ and $f^l({\mathcal F}^{cu}_{\mathrm {loc}}(y_0)) \text { for all } l<0$ . This implies that ${\mathcal F}^{cs}_{\mathrm {loc}}(y_1)$ remains a local center-stable leaf for $f_T:=f\circ \phi _T$ for all T, $\beta ^s(a,\cdot )$ remain parameterizations of the strong stable manifolds inside ${\mathcal F}^{cs}_{\mathrm {loc}}(y_i)$ , ${\mathcal F}^{cs}_{\mathrm {loc}}(y_1)$ remains inside ${\mathcal F}^{cs}(p)$ and the stable holonomy between ${\mathcal F}^c(p)$ and ${\mathcal F}^{cs}_{\mathrm {loc}}(y_1)$ is unchanged. A similar statement holds for ${\mathcal F}^{cu}_{\mathrm {loc}}(y_0)$ .

The maps $f_T$ do change the center leaves ${\mathcal F}^c(q)$ , we have that ${\mathcal F}^c(q(f_T),f_T)=f^{-1}({\mathcal F}^s_{\mathrm {loc}}({\mathcal F}^c(y_1)))\cap {\mathcal F}^u_{\mathrm {loc}}({\mathcal F}^c(y_0))$ . We can in fact compute implicitly the homoclinic stable–unstable holonomy $\tilde h_{q(f_T), f_T}$ in a neighborhood of $z_0$ in the following way:

$$ \begin{align*} \psi_p(a)=\tilde h_{q(f_T),f_T}(\psi_p(b)) &\! \iff h^u_{p,q(f_T),f_T}(\psi_p(a))=h^s_{p,q(f_T),f_T}(\psi_p(b))\\ &\! \iff h^u_{p,q(f_T),f_T}(\psi_p(a))=f_T^{-1}(h^s_{p,q(f_T),f_T}(f(\psi_p(b)))\\ &\!\iff \beta^u(a,r)=f_T^{-1}(\beta^s(b,s)) \quad\mbox{for some } r,s\in I_{\delta}\\ &\!\iff \phi_T(\beta^u(a,r))=f^{-1}(\beta^s(b,s)) \quad\mbox{for some } r,s\in I_{\delta}. \end{align*} $$

In conclusion, denoting $h_T=\psi _p^{-1}\circ \tilde h_{q(f_T),f_T}\circ \psi _p$ (the map $\tilde h_{q(f_T),f_T}$ in the chart $\psi _p$ ), we have

(10) $$ \begin{align} a=h_T(b)\!\iff \phi_T(\beta^u(a,r))=f^{-1}(\beta^s(b,s)) \quad\mbox{for some } r,s\in I_{\delta}. \end{align} $$

We choose a $C^{\infty }$ chart $\psi _q:B_{y_0}\rightarrow \mathbb R^4$ (can also be volume preserving) such that:

• • $\psi _q(y_0)=0^4$ ;

• • $D\psi _q(y_0)(E^c(q_0))=\mathrm {span}\{e_1,e_2\}$ ;

• • $D\psi _q(y_0)(E^u(q_0))=\mathrm {span}\{e_3\}$ ;

• • $D\psi _q(y_0)(E^s(q_0))=\mathrm {span}\{e_4\}$ .

Let $E:B_{a_0}\times B_{b_0}\times I_{\delta }^{n+2}\rightarrow \mathbb R^4$ ,

(11) $$ \begin{align} E(a,b,r,s,T)=\psi_q(\phi_T(\beta^u(a,r)))-\psi_q(f^{-1}(\beta^s(b,s))). \end{align} $$

We have that E is $C^2$ and $E(a_0,b_0,0,0,0^n)=\psi _q(y_0)-\psi _q(f^{-1}(y_1))=0$ .

Now let us finish the proof of the first part of the lemma. The implicit function theorem gives us the existence of a $C^2$ function $H:B_{a_0}\times I_{\delta }^n\rightarrow B_{b_0}\times I_{\delta }^2$ , $H(a,T)=(h(a,T), r(a,T),s(a,T))$ such that $E(a,h(a,T),r(a,T),s(a,T),T)=0$ (eventually by making smaller the balls and the intervals). Then the map $h_T(a)=h(a,T)$ is $C^2$ in both variables, which means that $\tilde h_{q(f_T), f_T}(x)$ is $C^2$ in both variables, and then $\tilde g_{q(f\circ \phi _T),f\circ \phi _T}(x)$ is $C^1$ in both variables. This finishes the proof of the first part.

Part (2). We will use the same notation from part (1). Let $\rho :\mathbb R^4\rightarrow [0,\infty )$ be a smooth bump function supported on a small ball centered at the origin, and constantly equal to one near the origin. The family $\phi _t:\mathbb T^4\rightarrow \mathbb T^4$ is defined as

$$ \begin{align*} \phi_t:=\psi_{q}^{-1}\circ (R_{\rho t}\times \mathrm{Id}_{\mathbb R^2})\circ\psi_q, \end{align*} $$

where $R_t$ is the rotation of angle t in $\mathbb R^2$ . Assume that the support of $\rho $ is small enough so that the support of $\phi _t$ is inside $B(y_0,r_0)$ and disjoint of all the other iterates of $W^c(q)$ . From part (1), we have

(12) $$ \begin{align} E(a,b,r,s,t)=(R_{\rho t}\times \mathrm{Id}|{\mathbb R^2})(\psi_q(\beta^u(a,r)))-\psi_q(f^{-1}(\beta^s(b,s))). \end{align} $$

We will compute $DE(a_0,b_0,0,0,t)$ . Observe that $L_b:=DE(a_0,b_0,0,0,t)\mid _{T_{b_0}B_{b_0}}:T_{b_0}B_{B_0}\!\rightarrow \text {span}\{e_1,e_2\}$ and $L_s:=DE(a_0,b_0,0,0,t)\mid _{\text {span}\{{\partial }/{\partial s}\}}:\text {span}\{\partial /{\partial s}\}\!\rightarrow \text {span}\{e_4\}$ are isomorphisms independent of t. Since $D(R_{\rho t}\times \mathrm {Id}_{\mathbb R^2})$ keeps $e_3$ invariant, we have that also $L_r:=DE(a_0,b_0,0,0,t)\mid _{\text {span}\{{\partial }/{\partial r}\}}:\text {span}\{\partial /{\partial r}\}\rightarrow \text {span}\{e_3\}$ is also an isomorphism independent of t, and

$$ \begin{align*} \frac{\partial E}{\partial(b,r,s)}(a_0,b_0,0,0,t)=L_b\times L_r\times L_s. \end{align*} $$

From equation (12), we can compute

$$ \begin{align*} DE(a_0,b_0,0,0,t)\mid_{T_{a_0}B_{a_0}}=R_t\circ L_a:T_{a_0}B_{a_0}\rightarrow \text{span}\{e_1,e_2\}, \end{align*} $$

where $L_a:=DE(a_0,b_0,0,0,0)\mid _{T_{a_0}B_{a_0}}:T_{a_0}B_{a_0}\rightarrow \text {span}\{e_1,e_2\}$ is an isomorphism. From the implicit function theorem, we deduce that

(13) $$ \begin{align} Dh_t(a_0)=L_b^{-1}\circ R_t\circ L_a. \end{align} $$

Define $g:B_{a_0}\times I_{\delta }\rightarrow \mathbb T^1$ ,

$$ \begin{align*} g(a,t):=D\psi_p^{-1}(a)_*\tilde g_{q(f\circ\phi_t),f\circ\phi_t}(\psi(a))=\frac{Dh_t(a)(v^s(a))}{\|Dh_t(a)(v^s(a))\|}, \end{align*} $$

where $D\psi _p^{-1}(a)_*$ is the diffeomorphism induced by $D\psi _p^{-1}(a)$ on the unit tangent bundles and $v^s(a)=D\psi _p^{-1}(a)(\tilde v^s(\psi _p(a)))$ . In other words, $g(\cdot ,t)$ is in fact the map $\tilde g_{q(f\circ \phi _t),f\circ \phi _t}$ seen in the chart $\psi _p$ which identifies $W^c(p)$ with $\mathbb T^2$ and the unit tangent spaces to $W^c(p)$ with $\mathbb T^1$ . To prove equation (9), it is enough to show that

$$ \begin{align*} \frac{\partial}{\partial t}g(a,t)\mid_{(a,t)=(a_0,0)}\neq 0, \end{align*} $$

which in turns is equivalent to the fact that $Dh_0(a_0)(v^s(a_0))$ and ${\partial }/{\partial t}Dh_t(a_0)(v^s(a_0))|_{t=0}$ are not collinear. Using equation (13), we obtain $Dh_0(a_0)(v^s(a_0))=L_b^{-1}\circ L_a(v^s(a_0))$ and ${\partial }/{\partial t}Dh_t(a_0)(v^s(a_0))\mid _{t=0}=L_b^{-1}\circ R_{{\pi }/2}\circ L_a(v^s(a_0))$ , which are clearly non-collinear since $L_a$ and $L_b$ are isomorphisms while $v^s(a_0)$ is non-zero. This finishes the proof of part (2).

Proof of Proposition 3.4

Let $f\in \mathcal {SH}^r_b$ . We need to find maps in $\mathcal U^r_s$ arbitrarily $C^r$ close to f. Since the $C^{\infty }$ maps are dense in the $C^r$ maps in the $C^r$ topology (even inside the volume preserving class), we can assume that f is $C^{\infty }$ .

Choose $q_1,q_2,q_3$ homoclinic points of ${\mathcal F}^c(p)$ such that the orbits of the homoclinic leaves ${\mathcal F}^c(q_i)$ are mutually disjoint, $i\in \{1,2,3\}$ .

For any $x\in {\mathcal F}^c(p)$ and any $1\leq i\leq 3$ , there exists $r_{x,i}>0$ such that if $y_i:=h_{p,q_i}^u(x)\in {\mathcal F}^c(q_i)$ , then the ball $B(y_i,r_{x,i})$ is disjoint from ${\mathcal F}^c(p)$ , from all the iterates $f^k({\mathcal F}^c(q_i))\text { for all } k\neq 0$ , and from all the iterates of ${\mathcal F}^c(q_j)$ , $j\neq i$ . Applying Lemma 3.6 part (2), we obtain the family of perturbations $\phi _{t,x,i}$ such that the derivative of $\tilde g_{q_i,f\circ \phi _{t,x,i}}$ with respect to t in $(x,0)$ does not vanish. By the continuity of the derivative, there exists a neighborhood $U_{x,i}$ of x such that $({\partial }/{\partial t})\tilde g_{q_i,f\circ \phi _{t,x,i}}$ is non-zero on $\overline U_{x,i}\times \{0\}$ .

Let $U_x=\bigcap _{i=1}^3U_{x,i}$ . By compactness of ${\mathcal F}^c(p)$ , there exist finitely many $x^1,x^2,\ldots x^K\in {\mathcal F}^c(p)$ such that ${\mathcal F}^c(p)=\bigcup _{j=1}^KU_{x^j}$ .

Let us fix some notation. Denote

$$ \begin{align*} T=(t_i^j)_{1\leq i\leq 3, 1\leq j\leq K} = (T_i)_{1\leq i\leq 3}=(T^j)_{1\leq j\leq K}\in I^{3K}:=[-\delta,\delta]^{3K}, \end{align*} $$

with $T_i=(t_i^j)_{1\leq j\leq K}\in I^K$ , $i\in \{1,2,3\}$ and $T^j=(t_i^j)_{1\leq i\leq 3}\in I^3$ , $j\in \{1,2,\ldots K\}$ .

For every $1\leq i\leq 3$ , we let $\phi _i:\mathbb T^4\times I^K\rightarrow \mathbb T^4$ given by

(14) $$ \begin{align} \phi_i(\cdot, T_i)=\phi_{t_i^1,x^1,i}\circ\phi_{t_i^2,x^2,i}\circ\cdots\circ\phi_{t_i^K,x^K,i}\quad \text{for all } T_i\in I^K. \end{align} $$

We define $\phi ,\ F:\mathbb T^4\times I^{3K}\rightarrow \mathbb T^4$ ,

$$ \begin{align*} \phi(\cdot,T)=\phi_T(\cdot)&:= \phi_1(\cdot, T_1)\circ\phi_2(\cdot,T_2)\circ \phi_3(\cdot,T_3),\\ F(\cdot,T)=F_T(\cdot)&:= f\circ\phi_T. \end{align*} $$

The maps $\phi _i$ , $\phi $ , and F have the following properties:

1. (1) $\phi _i$ , $\phi $ , and F are of class $C^{\infty }$ on $(x,T)$ ;

2. (2) $\phi _i$ is a small perturbation of the identity on a small neighborhood of ${\mathcal F}^c(q_i)$ , in particular, it leaves the other homoclinic orbits of ${\mathcal F}^c(q_j)$ unchanged for $j\neq i$ ;

3. (3) $F_T$ is equal to f on a neighborhood of ${\mathcal F}^c(p)$ , so it does not change ${\mathcal F}^c(p)$ and the function $\tilde v^s$ .

Let $V_j=\psi _p^{-1}(U_{x^j})$ , where $\psi _p:\mathbb T^2\rightarrow {\mathcal F}^c(p)$ is the $C^2$ embedding. For every $1\leq i\leq~3$ , define $g_i:\mathbb T^2\times I^{3K}\rightarrow \mathbb T^1$ ,

$$ \begin{align*} g_i(x,T)=D\psi_p^{-1}(x)_*\tilde g_{q_i(f_T),f_T}(\psi_p(x)). \end{align*} $$

In other words, $g_i(\cdot ,T)$ is again the map $\tilde g_{q_i(f_T),f_T}$ seen in the chart $\psi _p$ which identifies ${\mathcal F}^c(p)$ with $\mathbb T^2$ and the unit tangent spaces $T^1{\mathcal F}^c(p)$ with $\mathbb T^1$ . Lemma 3.6 part (1) tells us that $g_i$ is $C^1$ with respect to $(x,T)\in \mathbb T^2\times I^{3K}$ (maybe for a smaller interval I). Furthermore,

(15) $$ \begin{align} \frac{\partial g_i}{\partial t_i^j}(x,T)\neq 0\quad \text{for all } (x,T)\in \overline V_j\times\{0\}^{3K}, \text{ for all } 1\leq j\leq K. \end{align} $$

However, because for $l\neq i$ , the perturbation $\phi _l$ does not touch the orbit of ${\mathcal F}^c(q_i)$ , we have

(16) $$ \begin{align} \frac{\partial g_i}{\partial t_l^j}(x,T)= 0\quad \text{for all } (x,T), \text{ for all } l\neq i, \text{ for all } 1\leq j\leq K. \end{align} $$

Define $G:\mathbb T^2\times I^{3K}\rightarrow \mathbb T^3$

(17) $$ \begin{align} G(x,T)=(g_1(x,T),g_2(x,T),g_3(x,T)). \end{align} $$

Again, G is $C^1$ in $(x,T)\in \mathbb T^2\times I^{3K}$ . Equations (15) and (16) tell us that for every $1\leq j\leq K$ , we have

$$ \begin{align*} \det\bigg(\frac{\partial G}{\partial T^j}(x,T)\bigg)=\det\bigg(\frac {\partial g_i}{\partial t_i^j}(x,T)\bigg)=\prod_{i=1}^3\frac {\partial g_i}{\partial t_i^j}(x,T)\neq 0\quad \text{for all } (x,T)\in \overline V_j\times\{0\}^{3K}. \end{align*} $$

From the compactness of $\overline V_j$ and the $C^1$ continuity of G with respect to T, there exists $J\subset I$ with $0\in J$ such that, for all $1\leq j\leq K$ , we have

(18) $$ \begin{align} \det\bigg(\frac{\partial G}{\partial T^j}(x,T)\bigg)\neq 0\quad \text{ for all } (x,T)\in V_j\times J^{3K}, \end{align} $$

and since every point from $\mathbb T^2$ is inside some $V_j$ , we conclude that G has maximal rank at every point in $\mathbb T^2\times J^{3K}$ .

Remember that $v^s:\mathbb T^2\rightarrow \mathbb T^1$ is the $C^1$ map given by $v^s(x)=D\psi _p^{-1}(x)_*\tilde v^s(\psi (x))$ . Let $A:=\{(x,T)\in \mathbb T^2\times J^{3K}:\ G(x,T)\in \{-v^s(x),v^s(x)\}^3\}$ and $B=\pi _2(A)$ , where $\pi _2$ is the projection from $\mathbb T^2\times J^{3K}$ on the T component in $J^{3K}$ .

A simple consequence of the above definitions is the fact that if $T\notin B$ , then $f_T\in \mathcal U_s^r$ . To finish the proof of the density of $\mathcal U_s^r$ , we have to find T arbitrarily close to $0^{NK}$ such that $T\notin B$ . We will prove in fact that B has Lebesgue measure zero in $J^{3K}$ .

It is enough to show this for $B_1=\pi _2(A_1)$ , where $A_1=\{(x,T)\in \mathbb T^2\times J^{3K}:\ G(x,T)=v^s(x)^3\}$ , the other combinations of $\pm v^s$ work similarly. Let $H(x, T)=G(x,T)-v^s(x)^3$ , this is a $C^1$ map from $\mathbb T^2\times J^{3K}$ to $\mathbb T^3$ . Equation (18) tells us that H has maximal rank equal to 3 at every point ( $v^s$ is independent of T), so $H^{-1}(0^3)$ is a $C^1$ submanifold of codimension 3 (or dimension $3K-1$ ) inside $\mathbb T^2\times J^{3K}$ . Since $\pi _2\mid _{H^{-1}(0^3)}:H^{-1}(0^3)\rightarrow J^{3K}$ is a $C^1$ map, Sard’s theorem tells us that the image $B_1$ has Lebesgue measure zero.

This implies that we can find arbitrarily small $T\notin B$ , which finishes the proof of the $C^r$ density of $\mathcal U_s^r$ .

Proof. We first remark that, because of the transitivity of the homoclinic relation, it is enough to show that every hyperbolic periodic point of index 2 of $f\in \mathcal U^r$ is homoclinically related to the fixed point p of the hyperbolic fixed leaf ${\mathcal F}^c(p)$ .

Let x be a hyperbolic point of $f\in \mathcal U^r$ of index 2. Let $\tilde v^u(x)$ be the unit tangent vector to the weak unstable direction inside $T_x{\mathcal F}^c(x)$ . The strong unstable manifold ${\mathcal F}^u(x)$ must accumulate on the fixed hyperbolic leaf ${\mathcal F}^c(p)$ , so there exists a sequence of homoclinic points $p_n\in {\mathcal F}^u(x)\cap {\mathcal F}^s_{\mathrm {loc}}({\mathcal F}^c(p))$ such that $\lim _{n\rightarrow \infty }p_n=p_0\in {\mathcal F}^c(p)$ . If for some $p_n$ we have that $Dh^u_{x,p_n*}(\tilde v^u(x))\neq \pm Dh^s_{p,p_n*}(\tilde v^s(h^{s^{-1}}_{p,p_n}(p_n)))$ , then the two-dimensional unstable manifold of x, $W^u(x)$ , intersects ${\mathcal F}^c(p_n)$ in a $C^1$ curve locally transverse to the weak stable foliation inside ${\mathcal F}^c(p_n)$ (which is then pushed forward by the stable holonomy of the weak stable foliation in ${\mathcal F}^c(p)$ ). Since the two-dimensional global stable manifold of p, $W^s(p)$ , is dense inside the weak stable foliation of ${\mathcal F}^c(p_n)$ , we obtain a transverse homoclinic intersection from x to p.

Suppose that $W^u(x)\cap W^s(p)=\emptyset $ . The above argument implies that

$$ \begin{align*} Dh^u_{x,p_n*}(\tilde v^u)=\pm Dh^s_{p,p_n*}(\tilde v^s(h^{s^{-1}}_{p,p_n}(p_n)))\quad \mbox{for all } n\in\mathbb N. \end{align*} $$

Since $f\in \mathcal U^r$ , there exists a homoclinic point q of ${\mathcal F}^c(p)$ such that $\tilde g_q(p_0)\neq \pm \tilde v^s(p_0)\in T^1{\mathcal F}^c(p)$ . Let $q_0:=h^u_{p,q}(p_0)$ , consider the strong unstable holonomy $h^u_{\mathrm {loc}}:{\mathcal F}^{cs}_{\mathrm {loc}}(p_0)\rightarrow {\mathcal F}^{cs}_{\mathrm {loc}}(q_0)$ , and let $q_n:=h^u_{\mathrm {loc}}(p_n)\in {\mathcal F}^{cs}_{\mathrm {loc}}(q_0)$ . Then, $q_n\rightarrow q_0$ . The lack of homoclinic relations between x and p implies that also

$$ \begin{align*} Dh^u_{x,q_n*}(\tilde v^u)=\pm Dh^s_{p,q_n*}(\tilde v^s(h^{s^{-1}}_{p,q_n}(q_n)))\quad \mbox{for all } n\in\mathbb N. \end{align*} $$

Since $h_{x,q_n}^u=h_{p_n,q_n}^u\circ h_{x,p_n}^u$ , we obtain that

$$ \begin{align*} Dh^s_{p,q_n*}(\tilde v^s(h^{s^{-1}}_{p,q_n}(q_n)))=\pm Dh^u_{p_n,q_n*}\circ Dh^s_{p,p_n*}(\tilde v^s(h^{s^{-1}}_{p,p_n}(p_n))). \end{align*} $$

Using the continuity of $\tilde v^s$ and of $Dh^{s,u}$ , we can pass to the limit and obtain that

$$ \begin{align*} Dh^s_{p,q*}(\tilde v^s(h^{s^{-1}}_{p,q}(q_0)))=\pm Dh^u_{p,q*}(\tilde v^s(p_0)), \end{align*} $$

or $\tilde g_q(p_0)=\pm \tilde v^s(p_0)$ , which is a contradiction.

The proof of the intersection of the global stable manifold of x with the global unstable manifold of p is similar. This concludes the proof.