Proof. Obviously, the point x is contained in $\Gamma (f)$ due to the compactness of $\Gamma (f)$ (see Proposition 2.5). Suppose that $E^c_+$ is not uniquely integrable at x. Take a point $z\in \omega (x)$ , then the iterations of z under f are all distinct points in M since f has no periodic point. By the compactness of M, there is a small open set U of diameter $\delta /2$ , for a small $\delta $ chosen as in Lemma 4.1, containing three iterates of z, denoted by $z_1, z_2,z_3$ . Notice that each pair of iterations of z cannot belong to the same $su$ -plaque, otherwise, we have a periodic point by the Anosov closing lemma (Lemma 4.3). Let $U'$ be a $\delta $ -neighborhood ofU. As $E^c$ is one-dimensional and oriented, the $su$ -plaques of $\Gamma (f)$ can be ordered in $U'$ . Here, we allow all $su$ -plaques of $\Gamma (f)$ to project to a disconnected set in an open interval when $U'$ contains some points in the complement of $\Gamma (f)$ . For this order, we assume that $z_1<z_3<z_2$ .

As $z_i\in \omega (x)$ for $i=1,2, 3$ , there are $n_1, n_2\in \mathbb {N}$ such that $d(f^{n_1}(x), z_1)\leq \delta /8$ and $d(f^{n_2}(x), z_2)\leq \delta /8$ . Notice that $E^c_+$ is neither uniquely integrable at $f^{n_1}(x)$ or $f^{n_2}(x)$ . Let $c_i\in \mathcal {C}^+(f^{n_i}(x))$ be a center curve through $f^{n_i}(x)$ of length less than $\delta /8$ for each $i=1, 2$ . By assumption, the curve $c_i$ intersects $\Gamma (f)$ in infinitely many points. Up to taking a sub-curve, we assume that both endpoints of $c_i$ are contained in $\Gamma (f)$ for $i=1, 2$ . By Lemma 4.2, there is an integer $N_i\in \mathbb {N}$ such that the length of $f^n(c_i)$ is greater than $\delta $ for any $n\geq N_i$ , $i=1,2$ . Pick a large number $n_3\in \mathbb {N}$ such that $n_3-n_i>N_i$ and $d(f^{n_3}(x), z_3)\leq \delta /8$ for $i=1,2$ . Projecting along $su$ -plaques, for either $i=1$ or $2$ , we obtain that the map $f^{n_i-n_3}$ acts contractively on an interval I that has one endpoint corresponding to the $su$ -plaque through $f^{n_3}(x)$ . By the f-invariance of $\Gamma (f)$ , both endpoints of I correspond to $su$ -plaques in $\Gamma (f)$ . Then, there is a fixed point of $f^{n_i-n_3}$ in I, which corresponds to an $su$ -plaque in $\Gamma (f)$ since $\Gamma (f)$ is compact. This $su$ -plaque contains a point p and its iteration $f^{n_i-n_3}(p)$ . Thus, by Lemma 4.3, this yields a periodic point of f, which is absurd.

Proof. Under the existence of a compact leaf of $\Gamma (f)$ , the set $\Lambda $ of all compact leaves forms a compact invariant lamination by Theorem 2.7. Suppose that there are finitely many compact leaves in $\Lambda $ . Then, we can find a periodic one of some period $K\in \mathbb {N}$ . The restriction of $f^K$ on this periodic leaf is an Anosov diffeomorphism. It turns out the existence of periodic points of f and, thus, contradicts the assumption. Then, there are infinitely many compact leaves in $\Lambda $ and no one is periodic.

Suppose that $E^c_+$ is not uniquely integrable at some point x. Then, $\mathcal {C}^+(x)$ consists of distinct center curves through x as the only common point. Let $\delta>0$ and $\rho>0$ be two sufficiently small constants given in Lemmas 4.1 and 4.2. By Lemma 4.2, there is $N\in \mathbb {N}$ such that the length of $f^n(c)$ is larger than $\delta $ for any center curve $c\in \mathcal {C}^+(x)$ and any $n\geq N$ or $n\leq -N$ . We assume the case $n\geq N$ and an analogous argument applies directly for the other. Suppose that there exists a curve $c\in \mathcal {C}^+(x)$ entirely contained in a connected compact region $R\subset \Lambda $ . Notice that $\Lambda $ is a compact sublamination. Denote by $\hat {R}\subset \Lambda $ the maximal connected region containing R, which has several compact boundary leaves. As no leaf in $\Lambda $ is periodic, the regions $f^{i}(\hat {R})$ and $f^{j}(\hat {R})$ are disjoint for any $i\neq j\in \mathbb {Z}$ . However, for each $n\geq N$ , $f^n(\hat {R})$ contains a center curve $f^n(c)$ of length larger than $\delta $ . Then, by Lemma 4.1, $f^n(\hat {R})$ contains a ball of radius bounded from below by $\rho $ for every $n\geq N$ . This contradicts the compactness of the manifold M.

Suppose that there exists a center curve $c\in \mathcal {C}^+(x)$ entirely contained in the complement of $\Lambda $ or intersecting $\Lambda $ in the unique point x (which happens if $x\in \Lambda $ ). Denote by V the maximal connected component of the complement $M\setminus \Lambda $ containing c and $L\in \Lambda $ one of its boundary leaves. As $\Lambda $ is compact, it contains a compact leaf $L_0\in \Lambda $ accumulated by $f^{n_i}(L)$ . There is an integer $m\in \mathbb {N}$ such that $f^{n_i}(L)$ is contained in the $\rho $ -neighborhood $U_{\rho }$ of $L_0$ for each $i\geq m$ . It implies that $f^{n_i}(V)$ belongs to $U_\rho $ for each $i\geq m$ . Using Lemmas 4.1 and 4.2, there are integers $j\neq i\geq m$ with $n_j\neq n_i\geq N$ such that $f^{n_i}(c)$ intersects both $f^{n_i}(L)$ and $f^{n_j}(L)$ , which are two distinct leaves in $\Lambda $ . By the f-invariance of $\Lambda $ , we obtain that c intersects $\Lambda $ in two distinct points, arriving at a contradiction.

Now, the only remaining case is that x is accumulated by a sequence of leaves $L_i\in \Lambda $ , and any curve in $\mathcal {C}^+(x)$ intersects both the sequence $L_i$ and $M\setminus \Lambda $ . In particular, one can deduce that $x\in \Lambda $ by the compactness of $\Lambda $ . Using Lemma 4.4, we obtain the unique integrability of $E^c_+$ at x. The same argument applies analogously for the unique integrability of $E^c_-$ at x. Therefore, we obtain that $E^c$ is uniquely integrable at any point x and, thus, finish the proof.

Proof. Choose interstitial regions sufficiently thin such that any center curve within them connects two boundary leaves, and every center segment between boundary leaves has length at most $\delta /2$ , where $\delta> 0$ is the constant from Lemma 4.2. This is achievable because $E^c$ is transverse to $E^s \oplus E^u$ and interstitial regions can be made arbitrarily thin.

Suppose $E^s\oplus E^c$ is not uniquely integrable at some point x and, for every integer ${n \in \mathbb {N}}$ , there exists $k> n$ such that $f^k(x)$ lies in an interstitial region. Fix a center curve $\gamma (x)$ through x. By Lemma 4.2, there exists $N> 0$ such that the length of $f^k(\gamma (x))$ is greater than $\delta $ for all $k> N$ . Without loss of generality, assume $f^i(x)$ resides in an interstitial region for all $i> N$ .

For each $i> N$ , let $\alpha (f^i(x))$ be a connected subsegment of $f^i(\gamma (x))$ containing $f^i(x)$ and entirely contained within its interstitial region. Since non-unique integrability persists at $f^i(x)$ , Lemma 4.2 yields $N_i'> 0$ such that the length of $f^{k'}(\alpha (f^i(x)))$ is greater than $\delta $ for all $k'> N_i'$ . Select $j> N_i'$ such that $f^{i+j}(x)$ also lies in an interstitial region. By f-invariance of the lamination decomposition, $f^j(\alpha (f^i(x)))$ remains entirely within an interstitial region. However, this curve simultaneously satisfies that the length is greater than $\delta $ yet must be at most $\delta /2$ by our thinness choice, which is a contradiction.

Thus, we conclude that $E^s\oplus E^c$ is uniquely integrable at x. The same argument for $f^{-1}$ implies the unique integrability of $E^c\oplus E^u$ at x. Hence, $E^c$ is uniquely integrable at x.

Proof. Consider first a point x in the boundary of a gut. Its orbit contains at most finitely many points within that gut boundary. Suppose otherwise that infinitely many orbit points lie in the gut boundary. By Lemma 4.7, these occupy finitely many $su$ -plaques. The pigeonhole principle implies one $su$ -plaque contains two distinct orbit points. The Anosov closing lemma (Lemma 4.3) then produces a periodic point, which contradicts our assumption.

Now, let x be in the boundary of an interstitial region. The previous argument shows finitely many orbit points lie in any single gut boundary. Compactness of M ensures finitely many guts, so altogether, only finitely many orbit points occupy gut boundaries. Since boundary leaves partition into gut-boundary and interstitial-boundary components, infinitely many forward-orbit and backward-orbit points must lie in interstitial boundaries. Lemma 4.6 then establishes unique integrability in the complementary region direction.

Proof. It is immediate for x in an interstitial region boundary. Since interstitial regions can be made arbitrarily thin, every center curve through x terminates at the opposite boundary leaf.

Consider a point x in a gut boundary. The proof of Proposition 4.8 implies only finitely many points in the orbit of x lie in gut boundaries. Choose $N> 0$ such that $f^n(x)$ lies in an interstitial boundary for all $|n| \geq N$ . By thinness of interstitial regions, every center curve through $f^n(x)$ connects two boundary leaves of a complementary region. Applying $f^{-n}$ preserves this connectivity (as the set of boundary leaves is f-invariant), yielding a center curve through x connecting two boundary leaves.

Proof. Let $U\subset V$ be a path-connected complementary region of $\Lambda $ . Fix a boundary leaf $L_1$ . By Lemma 4.9, every $x \in L_1$ has a center curve to another boundary leaf of U. Suppose U has at least three boundary leaves. If all center curves from $L_1$ terminate at a single other boundary leaf $L'$ , replace $L_1$ with $L'$ and repeat. Since U has finitely many boundary leaves, we eventually find a boundary leaf (still called $L_1$ ) containing points $x,y$ whose center curves reach two distinct boundary leaves $L_2$ and $L_3$ , respectively.

Let $\{L_i\}_{i=1}^k$ be all boundary leaves of U. Define $A_i \subset L_1$ ( $i=2,\ldots ,k$ ) as the set of points with center curves connecting to $L_i$ . Sets $A_2$ and $A_3$ are non-empty by construction and pairwise disjoint by boundary leaf distinctness. Lemma 4.9 implies $\bigcup _{i=2}^k A_i = L_1$ . This yields a contradiction: $L_1$ is connected but partitioned into finitely many disjoint non-empty open subsets, which is impossible for a leaf in a lamination.

Proof. One can see that A is an open set by the continuity of center curves. We claim that the length of center curves in A is uniformly bounded from above. Indeed, it is obvious for points lying in the interstitial regions. We can define a function $\tau :A\rightarrow \mathbb {R}^+$ such that $\tau (x)$ represents the length of the center curve through x connecting two boundary leaves of U. This function is clearly continuous. As the gut of U is compact, continuity of $\tau $ gives a uniform upper bound.

If A does not coincide with U, then there exists $y\in \overline {A}\setminus A$ accumulated by $x_n\in A$ as $n\rightarrow \infty $ . Since $\tau (x_n)$ is uniformly bounded, the center segments of $x_n$ joining the two boundary leaves of U converge to a center curve through z joining the same boundary leaves with a bounded length. It implies $z\in A$ and, thus, $A=U$ .

Proof. First, we prove unique integrability of $E^c$ in $A\subset U$ for any complementary region $U\subset V$ . Take any point $y \in A$ . By definition of A, there exists a center curve $\gamma $ through y connecting both boundary leaves. The argument in Proposition 4.8 implies that after finitely many iterates, $\gamma $ lies entirely within interstitial regions. Lemma 4.6 then forces unique integrability at y. Lemma 4.11 establishes unique integrability throughout U. The connectivity via center curves of uniformly bounded lengths implies the I-bundle of U, which completes the proof.

Proof. For each point $x\in \Gamma (f)$ , it suffices to show that both $E^c_+(x)$ and $E^c_-(x)$ are uniquely integrable. Recall that we denote by $\mathcal {C}^+(x)$ and $\mathcal {C}^-(x)$ the collection of integral curves of $E^c_+(x)$ and $E^c_-(x)$ , respectively, with the common endpoint x. If x is an interior point of $\Gamma (f)$ , then both $\mathcal {C}^+(x)$ and $\mathcal {C}^-(x)$ contain small curves intersecting infinitely many plaques of $\Gamma (f)$ . Lemma 4.4 then implies unique integrability of $E^c$ at x.

If x is a boundary point of $\Gamma (f)$ , then there exists a small curve in either $\mathcal {C}^+(x)$ or $\mathcal {C}^-(x)$ intersecting infinitely many plaques of $\Gamma (f)$ . We assume this for $\mathcal {C}^+(x)$ . Then, by Lemma 4.4, $E^c_+(x)$ is uniquely integrable. As x is a boundary point of $\Gamma (f)$ , $\mathcal {C}^-(x)$ contains a small curve lying in a complementary region. Proposition 4.8 ensures unique integrability for $E^c_-(x)$ . Thus, $E^c(x)$ is uniquely integrable at all points of $\Gamma (f)$ .

Proof of Theorem 1.5.

As explained in the beginning of the section, we can assume that the manifold M and the center bundle $E^c$ are both oriented with orientations preserved by f. Since f is not accessible, the non-open accessibility classes constitute a non-empty codimension-one compact set $\Gamma (f)\neq \emptyset $ by Proposition 2.5.

The center bundle $E^c$ is uniquely integrable if $\Gamma (f)=M$ by Proposition 2.6. It suffices to consider $\emptyset \neq \Gamma (f)\neq M$ as a proper lamination. We obtain the unique integrability in the case where $\Gamma (f)$ contains a compact leaf in Proposition 4.5. In the absence of compact leaves, we obtain the central unique integrability in $\Gamma (f)$ (Proposition 4.13) and all its complementary regions (Proposition 4.12). Thus, we complete the proof.

Proof. If $\Gamma (f)$ contains a compact $su$ -leaf, then there must be infinitely many compact leaves since f has no periodic points. Assume that $\Gamma (f)$ contains no compact leaves. By Proposition 4.12, if $\Gamma (f)$ is a proper lamination, then it trivially extends to a foliation, denoted by $\mathcal {F}$ , without compact leaves. When $\Gamma (f)$ is a foliation, we denote $\mathcal {F}=\Gamma (f)$ . By Theorem 4.14, the foliation $\mathcal {F}$ has finitely many minimal sets; hence, so does $\Gamma (f)$ . The union of all minimal sets in $\Gamma (f)$ is f-invariant by the invariance of $\Gamma (f)$ . Suppose that there are two distinct minimal sets $\Lambda \neq \Lambda '\subset \Gamma (f)$ . Then, up to a finite iterate, we can assume that both $\Lambda $ and $\Lambda '$ are f-invariant.

By Proposition 4.12, each closed complementary region of $\Lambda $ forms an I-bundle with exactly two boundary leaves. Moreover, the fibers of each I-bundle can be taken as center segments. If there is a leaf $L\in \Lambda '$ intersecting an interstitial region of any complementary region of $\Lambda $ , then it must intersect all I-fibers in interstitial regions by transversality. As any interstitial region becomes arbitrarily thin away from the gut, the leaf L accumulates on some boundary leaf of $\Lambda $ . However, then $\overline {L}\subset \Lambda '$ has non-empty intersection with $\Lambda $ . This is impossible since $\Lambda $ and $\Lambda '$ are disjoint.

It turns out that $\Lambda '$ lies entirely within a gut piece $D\subset G$ of some complementary region of $\Lambda $ . Denote by $\hat {U}$ the closed complementary region containing D. As $\Lambda $ and $\Lambda '$ are both compact, they are uniformly separated; that is, their distance is bounded from below. Then, each boundary leaf of $\Lambda '$ is entirely contained in the gut piece D. Note that the union $\Lambda \cup \Lambda '$ is also an f-invariant compact lamination. By Proposition 4.12, each closed complementary region of $\Lambda \cup \Lambda '$ is an I-bundle and has exactly two boundary leaves. Since $\Lambda $ and $\Lambda '$ are disjoint, the boundary of $\Lambda \cup \Lambda '$ consists of the boundary leaves of $\Lambda $ and $\Lambda '$ . Now, the closed complementary region of $\Lambda \cup \Lambda '$ in $\hat {U}$ has at least three boundary leaves, including two boundary leaves of $\Lambda $ and at least one boundary leaf of $\Lambda '$ . Thus, we arrive at a contradiction, which implies that $\Gamma (f)$ has at most one minimal set.

Since $\Gamma (f)$ is a compact lamination without compact leaves, it admits at least one minimal set. It implies that $\Gamma (f)$ admits exactly one minimal set, which could be $\Gamma (f)$ or a proper subset. Hence, we finish the proof.

Proof of Theorem 1.6.

Note that it suffices to show the result up to a finite lift and a finite iterate. Indeed, any finite iterate of f admits the same accessibility classes as f. Any accessibility class in a finite cover projects by the covering map to a subset in an accessibility class of f in M. Each minimal set in a finite cover also projects to a minimal set in M. Moreover, the absence of periodic points persists under finite iterates and finite covers. Thus, by taking a finite lift and an iterate if necessary, we can assume that M and $E^c$ are both oriented and f preserves their orientations.

Assume that f is not accessible, which means that the set $\Gamma (f)$ is not empty. If $\Gamma (f)$ contains finitely many compact leaves, then each one of them is f-periodic. Up to a finite iterate, the restriction of f to any such compact leaf is uniformly hyperbolic, implying the existence of periodic points by the Anosov closing lemma (see also Lemma 4.3). Thus, $\Gamma (f)$ contains infinitely many compact accessibility classes whenever it contains at least one compact leaf.

Assume that $\Gamma (f)$ has no compact leaves. By Proposition 4.15, there exists a unique minimal set $\Lambda \subset \Gamma (f)$ , necessarily f-invariant. The case where $\Lambda =M$ forces $\Gamma (f)$ to be a minimal foliation without compact leaves. If $\Lambda $ is a proper lamination, then by Proposition 4.12, the closed complementary regions of $\Lambda $ are all I-bundles by center segments. Moreover, $\Lambda $ trivially extends to a foliation without compact leaves.

If we further assume $NW(f)=M$ , then $\Lambda $ must be a minimal foliation and ${\Lambda =\Gamma (f)=M}$ . Otherwise, if $\Lambda $ is a proper subset, then it has finitely many complementary regions with gut pieces. Moreover, each boundary leaf of such a complementary region is f-periodic. As shown by [Reference Hertz, Hertz and UresHHU08a, Proposition A.5], such boundary leaves of $\Lambda $ contain dense periodic points, which contradicts the absence of periodic points.

Hence, we finish the proof.

Proof. If y is contained in $\Gamma (f)$ , then the limit set $\omega (y)$ belongs to $\Gamma (f)$ by the invariance and compactness of $\Gamma (f)$ . It suffices to consider the case where $\Gamma (f)$ is a proper lamination and y lies in the complement of $\Gamma (f)$ . We will discuss separately the situations whether $\Gamma (f)$ contains a compact leaf or not.

Let $\delta>0$ be a small number such that for any point $x\in M$ , the $\delta $ -neighborhood of the leaf $W^c(x)$ in $\mathcal {F}^{cs}(x)$ is contained in $W^{s}(W^c(x))$ . Such a $\delta $ exists by transversality of the center and stable foliations; see Lemma 4.1. Since the center curve $\gamma $ can be taken as short as desired, we assume that the length of $\gamma $ is less than ${\delta }/{16}$ . Note that $f^i(y)$ is at least $\delta $ -away from $W^c(f^i(x))$ for any $i\in \mathbb {Z}$ by the choice of $\delta $ . Otherwise, $f^i(y)$ would be contained in $W^s(W^c(f^i(x)))$ and then y would be contained in $W^s(W^c(x))$ . Moreover, as $\gamma (0) \in W^s(W^c(x))$ , there is $N\in \mathbb {N}$ such that for any $m\geq N$ , the distance from $f^m(\gamma (0))$ to $W^c(f^m(x))$ in the leaf $\mathcal {F}^{cs}(f^m(x))$ is smaller than ${\delta }/{2}$ . This implies that the length of $f^m(\gamma )$ is greater than ${\delta }/{2}$ for any $m\geq N$ .

When $\Gamma (f)$ contains a compact leaf, it contains infinitely many compact leaves by Theorem 1.6. Denote by $\mathcal {K}\subset \Gamma (f)$ the set of all compact leaves, forming a compact invariant lamination by Theorem 2.7. Denote by R the maximal connected complementary region of $\mathcal {K}$ containing y, which is non-empty since y lies in the complement of $\Gamma (f)$ . By taking a smaller $\delta>0$ if necessary, we assume the center curve $\gamma $ through y lies entirely within R of length less than ${\delta }/{16}$ . The argument above establishes that the length of $f^m(\gamma )\subset f^m(R)$ is greater than ${\delta }/{2}$ for any $m\geq N$ . Lemma 4.1 then ensures a constant $\rho>0$ dominating the radius of $f^m(R)$ for each $m\geq N$ . This contradicts with the compactness of M.

Thus, we conclude $y\in \mathcal {K}$ whenever there exists a compact accessibility class and, thus, $\omega (y)\subset \mathcal {K}\subset \Gamma (f)$ . Now, let us consider the case without compact leaves.

By the absence of compact accessibility class, the complementary regions of the lamination $\Gamma (f)$ consist of finitely many gut pieces and some interstitial regions. Assume that y is contained in a complementary region U of $\Gamma (f)$ . By Proposition 4.12, all complementary regions of $\Gamma (f)$ are I-bundles by center segments. The point y admits a center segment c whose two endpoints are contained in two boundary leaves of $\hat {U}$ and the interior points are contained in U. Let $y^+$ be one of the endpoints of c. By Lemma 4.7, there are finitely many $su$ -plaques in the boundary of each gut. Lemma 4.3 ensures that the iterations of $y^+$ are all contained in distinct $su$ -plaques. Note that there are finitely many guts due to the compactness of the manifold. Then, there is $k\in \mathbb {N}$ such that for any $n\in \mathbb {Z}$ with $|n|\geq k$ , the point $f^n(y^+)$ is contained in the boundary of the interstitial regions. As $\Gamma (f)$ and its complement are f-invariant, the center segment $f^n(c)$ is contained in the interstitial regions for $|n|\geq k$ except its two endpoints. In particular, each point $f^n(y)$ is contained in the interstitial regions.

Suppose that there is a point $p\in \omega (y)$ lying in the complement of $\Gamma (f)$ . Then, it must be contained in an interstitial region, denoted by O, since $f^n(y)$ cannot intersect any gut for every $|n|\geq k$ . The center curve through p intersects the boundary of O in a point, denoted by $p^+\in \Gamma (f)$ . Pick a small $\epsilon _0>0$ as in Lemma 4.3. Recall that $E^c$ is uniquely integrable by Theorem 1.5. By Lemma 4.1 and openness of O, there exists $\tau>0$ such that $B_{\tau }(p)\subset O$ and any point $z\in B_{\tau }(p)$ admits a center curve intersecting $W^s_{\epsilon _0}(W^u_{\epsilon _0}(p^+))$ in a unique point. Let $\{n_i\}_{i\in \mathbb {N}}$ be a sequence of integers such that $f^{n_i}(y)$ converges to p as $i\rightarrow \infty $ . There exists $N\in \mathbb {N}$ such that for any $i\geq N$ , the point $f^{n_i}(y)$ is contained in $B_{\tau }(p)$ . Thus, for any $i\geq N$ , $f^{n_i}(y)$ admits a center curve intersecting $W^s_{\epsilon _0}(W^u_{\epsilon _0}(p^+))$ in a unique point. Taking $i> j\geq N$ , we denote by $y_i, y_j\in W^s_{\epsilon _0}(W^u_{\epsilon _0}(p^+))$ two points in center curves through $f^{n_i}(y), f^{n_j}(y)$ , respectively. Since the boundary of $\Gamma (f)$ is f-invariant, we deduce $f^{n_i-n_j}(y_j)=y_i$ . Then, Lemma 4.3 produces a periodic point in $W^s_{\epsilon _0}(W^u_{\epsilon _0}(p))$ , which is absurd.

Therefore, the limit set $\omega (y)$ must be a subset of $\Gamma (f)$ and we finish the proof.

Proof. Since f is not accessible, the set $\Gamma (f)$ is either a foliation or a proper lamination. We only present our proof in the case where $\Gamma (f)$ is a lamination since the foliation case follows in a simpler way. Suppose to the contrary that $\mathcal {F}^{cs}$ is not complete, implying $\mathcal {F}^{cs}(x) \neq W^{s}(W^c(x))$ for some $x\in M$ . Let $y\in \mathcal {F}^{cs}(x)$ be an accessible boundary point of $W^{s}(W^c(x))$ , and $\gamma : [0, 1]\rightarrow M$ be a center curve satisfying that $ \gamma (1)= y$ and $\gamma ([0, 1))\subset W^s(W^c(x))$ .

Let $\delta>0$ be given as in the proof of Lemma 5.1. Replacing $\gamma $ by a shorter curve if necessary, the argument in Lemma 5.1 shows the existence of $N\in \mathbb {N}$ such that the length of $f^m(\gamma )$ is greater than ${\delta }/{2}$ for each $m\geq N$ .

As the manifold is compact and $E^c$ is one-dimensional, by taking a smaller $\delta $ if necessary, we can order all $su$ -plaques of $\Gamma (f)$ in each coordinate cube of radius at most $\delta $ . Take a point $z\in \omega (y)$ . Lemma 5.1 implies $z\in \Gamma (f)$ . Since the manifold is compact and f has no periodic points, there are three iterations of z in an open set with order $\omega _1<\omega _3<\omega _2$ such that $dist_c(\omega _3, \omega _i)<{\delta }/{8}$ , $i=1,2$ , where $dist_c$ is the infimum of the length of center curves joining corresponding $su$ -plaques. Take two integers $n_1, n_2\in \mathbb {N}$ such that $dist_c(f^{n_i}(y), \omega _i)<{\delta }/{16}$ , $i=1, 2$ . Note that $f^{n_i}(y)$ , $i=1,2$ , are also accessible boundary points (see Figure 1).

Figure 1

The $su$ -plaques through $\omega _i$ and center curves through $f^{n_i}(y)$ in a coordinate cube.

Suppose that y is not contained in $\Gamma (f)$ . Then, any iteration of y cannot be contained in $\Gamma (f)$ by the invariance of $\Gamma (f)$ . We can take two center curves $c_1$ and $c_2$ starting at $f^{n_1}(y)$ and $f^{n_2}(y)$ , respectively, as given in the definition of accessible boundary points such that they are disjoint with $\Gamma (f)$ . So, there exists $N_i\in \mathbb {N}$ such that for any $n\geq N_i$ , the length of $f^n(c_i)$ is greater than ${\delta }/{2}$ for $i=1,2$ . Pick an integer $n_3\in \mathbb {N}$ such that $n_3-n_i\geq N_i$ and $dist_c(f^{n_3}(y), \omega _3)< {\delta }/{16}$ for $i=1,2$ . Then, for either $i=1$ or $i=2$ , the center curve $f^{n_3-n_i}(c_i)$ intersects the $su$ -plaque through the point $\omega _i$ . By the invariance of $\Gamma (f)$ and our choice of $c_i$ , the center curve $f^{n_3-n_i}(c_i)$ is disjoint with $\Gamma (f)$ . This is a contradiction as $\omega _i\in \omega (y)$ is contained in $\Gamma (f)$ . Thus, we have $y\in \Gamma (f)$ .

Now, we take a center curve $c_i$ of length less than ${\delta }/{16}$ satisfying the definition of accessible boundary point such that it is bounded by $f^{n_i}(y)$ and some other point in $\Gamma (f)$ for $i=1,2$ . This is achievable for large $n_i$ when $c_i$ intersects the complement of $\Gamma (f)$ as the interstitial regions can be made arbitrarily thin. However, the interior of the center curve $c_i$ must intersect infinitely many $su$ -plaques of $\Gamma (f)$ . Otherwise, one can take $c_i$ with interior lying in the complement of $\Gamma (f)$ and deduce a contradiction by applying the argument above. For each $i=1,2$ , there is an integer $N_i\in \mathbb {N}$ so that for any $n\geq N_i$ , the length of $f^{n}(c_i)$ is greater than ${\delta }/{2}$ . Take $n_3\in \mathbb {N}$ such that $n_3-n_i\geq N_i$ and ${dist_c(f^{n_3}(y), \omega _3)<{\delta }/{16}}$ . Then, the center curve $f^{n_3-n_i}(c_i)$ has length greater than ${\delta }/{2}$ . Under the distance $dist_c$ , the center curve $c_i$ is entirely contained in the ${\delta }/{2}$ -neighborhood of $f^{n_3}(y)$ for $i=1,2$ . Then, for either $i=1$ or $i=2$ , projecting along $su$ -plaques, we obtain a map $f^{n_i-n_3}$ acting contractively in an interval I whose two endpoints correspond to two $su$ -plaques in $\Gamma (f)$ . By the invariance and compactness of $\Gamma (f)$ , there is an $su$ -plaque in $\Gamma (f)$ containing some point p and its iteration $f^{n_i-n_3}(p)$ . Lemma 4.3 implies the existence of a periodic point of f, which contradicts with the assumption.

Hence, we complete the proof.

Proof of Corollary 1.3.

Accessibility clearly implies the absence of $su$ -tori. Assume that $f: M\to M$ is a non-accessible partially hyperbolic diffeomorphism of a closed 3-manifold without periodic points. By Theorem 1.2, after passing to a finite cover, the fundamental group $\pi _1(M)$ is solvable. According to [Reference Hertz, Hertz and UresHHU08b] and [Reference HammerlindlHam17, Theorem 2.5], either there exists a 2-torus tangent to $E^s \oplus E^u$ or f lies in the homotopy class of an Anosov automorphism on $\mathbb {T}^3$ . In the latter case, f must possess periodic points, which contradicts our assumption. Therefore, an $su$ -torus must exist.

Assuming the existence of $su$ -tori, let $\Lambda $ denote the collection of all such tori. Theorem 2.7 establishes that $\Lambda $ forms a compact f-invariant lamination. No leaf in $\Lambda $ can be periodic under f, because otherwise, an iterate of f restricted to a periodic $su$ -torus would be Anosov and yield periodic points. Theorem 1.5 ensures unique integrability of $E^c$ , implying f is dynamically coherent. Since $\pi _1(M)$ is solvable, Theorem 2.4 guarantees an f-periodic compact center curve $\gamma $ . Since f is $C^2$ and $\gamma $ is 2-normally hyperbolic, $\gamma $ forms a $C^2$ circle. After adjusting by the period, we may assume $\gamma $ is f-invariant. The restriction $f|_{\gamma }$ has no periodic points, resulting in an irrational rotation number. As $f|_{\gamma }$ is $C^2$ , Denjoy’s theorem ensures every orbit is dense. Note that $\Lambda \cap \gamma $ is $f|_{\gamma }$ -invariant. Consequently, $\Lambda $ must have trivial complement and therefore constitutes a foliation of M.

Thus, we complete the proof.

Proof of Theorem 1.7.

Let $f:M\rightarrow M$ be a partially hyperbolic diffeomorphism of a closed 3-manifold without periodic points, and assume f is non-accessible. By Theorem 1.5, the center bundle $E^c$ is uniquely integrable, which implies f is dynamically coherent. As shown in Theorem 1.2, the ambient manifold M has virtually solvable fundamental group. Then, up to a finite lift and iterate, f is leaf conjugate to one of three diffeomorphisms listed in Theorem 2.4. The last two cases are desired, so we need only eliminate the first case. In this excluded scenario, the manifold M is the 3-torus and f is homotopic to an Anosov automorphism. However, this would force periodic points, which contradicts our assumption. Therefore, we finish the proof of Theorem 1.7.

Proof. Let $f: M\rightarrow M$ be a $C^1$ partially hyperbolic diffeomorphism of a closed 3-manifold without periodic points and assume f is non-accessible. Theorem 1.5 establishes unique integrability of the center bundle, yielding a unique center foliation $\mathcal {F}^c$ . By Theorem 1.7, up to a finite cover and iterate, f becomes leaf-conjugate to either a skew product over a linear Anosov automorphism of $\mathbb {T}^2$ or the time-one map of a suspension Anosov flow.

In both cases, f possesses a quasi-isometric action along the center foliation $\mathcal {F}^c$ (see [Reference FengFen26, Reference MartinchichMar23]). This property persists prior to taking covers or iterates. Consequently, f exhibits plaque expansiveness on its unique center foliation $\mathcal {F}^c$ . According to [Reference Hirsch, Pugh and ShubHPS77, Theorem 7.4], there exists a $C^1$ -neighborhood $U_1$ of f where every $g \in U_1$ admits a unique plaque expansive center foliation and is leaf-conjugate to f.

When f is accessible, [Reference DidierDid03] guarantees a $C^1$ -neighborhood $U_2$ of f such that every $g \in U_2$ remains accessible. Hence, we complete the proof by taking $U = U_1 \cap U_2$ .

Proof. This is a direct consequence of Theorems 1.5, 1.6, Proposition 5.2, [Reference CarrascoCar15, Theorem 1.2], and [Reference CarrascoCar11, Theorem 4.6].