Transitivity and rotation sets with nonempty interior for homeomorphisms of the 2-Torus

We show that, if $f$ is a homeomorphism of the 2--torus isotopic to the identity, and its lift $\widetilde f$ is transitive, or even if it is transitive outside of the lift of the elliptic islands, then $(0,0)$ is in the interior of the rotation set of $\widetilde f.$ This proves a particular case of Boyland's conjecture.


Introduction
In this work we study homeomorphisms of the 2 torus isotopic to the identity. Let f be such a homeomorphism, and let f be a fixed lift of f to the plane, a homeomorphism of R 2 that satisfies f (x + 1, y + 1) = f (x, y) + (1, 1), and π( f (x)) = f (π(x)), where π : R 2 → T 2 is the covering map.
Given f and f we define, following [9]and [6], the rotation set of f as the set of accumulation points of the subset of R 2 f n (x) − x n | x ∈ R 2 , n ∈ Z Z + , which is a compact convex set (see [9]). One important question is to characterize for which homeomorphisms of the torus the rotation set of its lifts has non-empty interior. There are several consequences associated with this property. In [8], it is shown that if a lift of f has a rotation set with non-empty interior, then f has positive topological entropy. In [9] it is shown that if a point ( p1 q , p2 q ) ∈ Q 2 is in the interior of ρ( f ), then there exists a point x in the plane such that f q (x) = x + (p 1 , p 2 ), while the results from [7] and [10] show that, if α is a point in the interior of ρ( f ), with both coordinates irrational, then there exists a minimal set K α such that the restriction of f to K α is regularly semi-conjugate to the translation by α.
While the dynamics tied to the interior of the rotation set is somewhat well understood, a description of the dynamics implied by the extremal points of the rotation set when it has interior is somewhat lacking, even for the cases where f is area preserving. One standing conjecture in this direction is the Boyland conjecture, explained below: Given an area preserving homeomorphism of the torus f and a fixed lift f , one can consider the rotation vector of the Lebesgue measure λ for f , 1] f (x) − xdλ.
An homeomorphisms f isotopic to the identity such that ρ λ ( f ) = (0, 0) is usually called in the literature an irrotational homeomorphism. Whenever we speak of an irrotational homeomorphisms, an specific lift for f is fixed.
Boyland's conjecture for the torus claims that, if f is an irrotational homeomorphisms and the interior of ρ( f ) is non-empty, then (0, 0) is in the interior of the rotation set.
Another important dynamical property of homeomorphisms which has received attention recently is transitivity. In [4] it was shown that C 1 generically, an area preserving diffeomorphism of the torus is transitive. In [3], the authors showed that, C 0 generically, if f is irrotational, then f is transitive in the whole plane. The work [4] suggests this is also true C 1 generically.
In [1] it is shown that, if g is a homeomorphism of the closed annulus whose lift g to the strip R × [0, 1] is transitive, and if there are no fixed point in the boundary of the annulus, then 0 is in the interior of the rotation interval of g. Some of the ideas presented in this note follow from that paper.
In [5] the authors study the relationship between transitivity, entropy and rotation set, showing that if f is C 1+ε , has topological entropy, and the lift of f to every finite covering of the torus is transitive, then the rotation set of f has non-empty interior.
Here we study homeomorphism of the 2-torus isotopic to the identity which have the following property: Definition 1. A lift f of a homeomorphism of the torus isotopic to the identity is said to have the Property T if there exists a point x T in R 2 such that the omegalimit of the orbit of x T by f , ω(x T ), contains all Z Z 2 translates of x T , that is, The above hypothesis is clearly weaker than requesting transitivity of f . In fact, it covers the following case, which is of relevance for measure-preserving homeomorphisms, a hypothesis we do not require. A common and stable trait of sufficiently differentiable area preserving diffeomorphisms of the torus is the existence of elliptic islands. Of course such diffeomorphisms cannot be transitive, nor can their lift, but it is still possible for such mappings to be transitive in the complement of the interior of these islands, which in many relevant cases is a torus with infinitely many holes. If a diffeomorphism f has a lift f which is transitive in the complement of the lift of the elliptic islands, then f obeys Property T.
Our main result is: Theorem 1. Let f be a homeomorphism of the 2-torus isotopic to the identity such that its lift f obeys Property T. Then (0, 0) is in the interior of ρ( f ).
In section 2 we introduce the sets B θ and ω(B θ ), and we study their structure. In section 3 we study the projection of the sets B θ and ω(B θ ) to the torus, and derive some properties of its closure. The final section is devoted to proving the main theorem.
Let z ∈ Ω C , and let Γ be the closed connected component of Ω C that contains z. Since Ω C is forward invariant, for every positive integer j we have f j (Γ) ∩ Ω = ∅ but since (V + θ ) C ⊂ Ω, Γ is a closed connected set whose forward orbit remains in V + θ . Furthermore, since ∂(Ω C ) ⊂ B θ , then Γ ∩ B θ = ∅. Let z ∈ Γ ∩ B θ and Θ be the connected component of B θ that contains z. Θ is unbounded and since Θ ⊂ Ω C and Θ∩Γ = ∅, then Θ ⊂ Γ. So Γ is unbounded, closed and its forward orbit remains in V + θ , therefore Γ ⊂ B θ . This shows that Ω C = B θ and implies property 4 Proof. We assume, without loss of generality, that 0 ≤ θ < π 2 . Since f satisfies Property T, there is y a translate of x T which lies in V + θ+π such that, for every positive integer m, there exists a positive integer n 0 (m) such that f n0(m) (y) ∈ V + θ + (m, 0). Let n = n(m) be the first time that f n (V + θ+π ) intersects Let m k be a sequence of integers such that lim k→∞ n(m k ) = ∞. For each k there is an unbounded closed curve γ k that starts at f n(m k ) (V θ ) and such that γ k ⊂ f n(m k ) (V + θ ) ∩ V + θ + (m, 0). It follows that the sets β k = f −n(m k ) (γ k ) all intersect V θ , and that for 0 ≤ i ≤ n(m k ), f i (β k ) ⊂ V + θ . Also, since for every point z in V θ there are integers p and q such that (p, q); e θ ≥ 0 and such that z + (p, q) is in [0, 1] × [0, 1], we have that there is a curve α k = β k +(p, q) that starts in [0, 1]×[0, 1] and such that, for 0 ≤ i ≤ n(m k ), f i (α k ) ⊂ V + θ . Now consider the closed sets α k in S 2 corresponding to the sets α k . α k is a sequence of compact connected sets, and as such there is a subsequence α kj that is convergent in the Hausdorff topology to a closed connected set Γ. It is clear that α k is in B θ , and that Γ intersects [0, 1] × [0, 1]. The subset Γ of R 2 corresponding to Γ shows the lemma is true Another interesting property of the set B θ is the following: Proof. Since f is biperiodic, there exists a positive real K such that, for all Let y ∈ B θ , and let (p, q) ∈ Z Z 2 be such that (p, q); e θ ≥ 2K. Let Γ be the connected component of y.
We will need to also consider the following sets: , where the last equality comes from the positive invariance of B θ , and the respective The following properties of ω(B θ ) and ω(B θ ) follow directly from the definition and proposition 1.

Proposition 2.
For every θ in [0, 2π), we have: θ , for all positive integers k, which shows that Γ ⊂ C θ . Since the choice of Γ was arbitrary, we have ω(B θ ) ⊂ C θ . Again, we use the invariance of ω(B θ ) and get The following lemma will be our main tool in establishing the existence of vectors in the rotation set of f .

Lemma 4.
If ω(B θ ) = ∅, then there exists a positive real number a such that: Proof. We again assume, without loss of generality, that θ is in [0, π/2), and we prove just the first item. We claim that there is a positive n such that f i (B θ ) is contained in V + θ + (1, 0), for all i > n. If this was not the case, then there would be a sequence i j → ∞ such that, for all j, there is a point 0). But this implies that there is a sequence of pair of integers p j , q j such that (p j , q j ); e θ ≥ 0 and such that f ij (x j ) + (p j , q j ) is in [1,2] 0). As the choice of Γ was arbitrary, we have A simple induction argument now shows that, for every positive integer k, which ends the proof The following is an immediate consequence of the previous lemma: We can now state the main lemma of this work: which will be proved in the last section: We will postpone the proof of this result to the last section, but let us now show how this lemma yields theorem 1.
Proof of theorem 1: We first note that, since f satisfies Property T, it has a recurrent point, and as so (0, 0) ∈ ρ( f ). Since ρ( f ) is convex, either (0, 0) is in the interior of ρ( f ), and we are done, or there exists a straight line passing through (0, 0) such that ρ( f ) is contained in one side of this line. Assume by contradiction that the latter happens. This means that there is θ 1 such that for all r ∈ ρ( f ), r; e θ1 ≥ 0. But from lemma 5 either ω(B θ 1 ) = ∅ or ω(B θ 1 +π ) = ∅, and in both cases this implies, by corollary 1, that there exists r ∈ ρ( f ) with r; e θ 1 < 0.
3. The sets π(B θ ) and π(ω(B θ )) Let us state this general result: Lemma 6. Let g be a homeomorphism of the torus isotopic to the identity and g be its lift to the plane. Let A ⊂ R 2 be such that if z ∈ A, then z + (i, 0) ∈ A for all integer i. Assume that there exists a point x in R 2 such that . Then, for every z ∈ A = π( A) and every ε > 0, there exists a curve γ ⊂ R 2 which is unbounded in the e 0 direction but bounded in the e θ/2 direction, and such that, for every point in y ∈ γ there exists a positive n(y) such that π( g n(y) (y)) ∈ B ε (z).
Also, since x 1 and y 1 belong to α 1 , and since y 1 = g k1 (x 1 ) belongs to g k1 (α 1 ), we have that g k1 (α 1 ) ∪ α 1 is a connected set that contains g k1 (x 1 ). A simple induction shows that is a connected arc joining g −k1k2 (x 1 ) and y 1 .
Consider the set which is also connected and bounded in the e π/2 direction, and which contains both z and z − (k 1 , 0). If γ = ∞ i=−∞ (γ 0 + (ik 1 , 0)), then γ is connected, unbounded in the e 0 direction and bounded in the e π/2 direction, and satisfies that, if y ∈ γ, then there is a n(y) > 0 such that π( g n(y) (y)) ∈ π(α i ) for some 0 ≤ i ≤ 4. Since π(α i ) ⊂ B ε (z), 0 ≤ i ≤ 4, we have the result The previous lemma yields this two simple corollaries, the first one which will be used in the remainder of the paper, the second interesting in itself.

Corollary 2. Let g be a homeomorphism of the torus and g be its lift to the plane.
Assume there is a set A ⊂ R 2 which is Z Z 2 invariant, i.e., for every y in A, and every pair of integers (p, q) ∈ Z Z2, y + (p, q) ∈ A. Assume further than there is a point x in A such that A ⊂ ω( x).
Let B ⊂ R 2 be a set such that g(B) ⊂ B and such that B has an unbounded connected component. Then π( A) ⊂ π(B).
Proof. Let z ∈ B be a point in an unbounded connected component of B. Suppose, by contradiction, that there exists a point y ∈ A and a positive real ε such that π(B) ∩ B ε (π( y)) = ∅.
The previous lemma implies that there are two connected sets, α and β, with the following property: • For all x in α ∪ β, there exists a positive integer n( x) such that π( g n ( x)) ∈ B ε (π( y)). • α is unbounded in the direction of e π/2 , bounded in the direction of e 0 and disconnects the plane.
• β is bounded in the direction of e π/2 , unbounded in the direction of e 0 and disconnects the plane. Now, by the contradiction hypothesis and the first item, By the second item, there is an integer k 1 such that z is to the left of α + (k 1 , 0), to the right of α − (k 1 , 0), below β + (0, k 1 ) and above β − (0, k 1 ). This shows that z is in a bounded connected component of the complement But, from (1), B ∩ D = ∅, which is absurd since the connected component of B that contains z must be unbounded by hypothesis and remain in the same connected component of D C that contains z.
We finish this section with 2 direct applications of this results: Corollary 3. Let g be a homeomorphism of the torus isotopic to the identity and g be its lift. Assume g is transitive in R 2 . Let B ⊂ R 2 be a set such that g(B) ⊂ B and such that B has an unbounded connected component. Then π(B) = T 2 .
Let y ∈ x T + Z Z 2 be such that 1/4 ≤ y; e θ ≤ 3/4. We will show that at least N of the points y, y + i(n 1 , m 1 ), y + j(n 2 , m 2 ), for 1 ≤ i ≤ N, 1 ≤ j ≤ N, are in distinct connected components of O 1 , which clearly proves the proposition.
But since ω(B θ )∩ω(B θ+π )+(p, q) is not empty, there exists x in the intersection. Let Γ be the connected component of ω(B θ ) that contains x, and let Θ be the connected component of ω(B θ+π ) + (p, q) that contains x. Then Γ 1 = Γ ∪ Θ is a connected component of ω(B θ ) ∩ ω(B θ+π ) + (p, q) such that sup w∈Γ1 w; e θ ) = +∞ and such that inf w∈Γ1 w; e θ = −∞. But this contradicts the fact that γ ∩ Γ 1 = ∅ and that γ is connected Since f satisfies Property T, the forward orbit of x T is not confined to any half-plane and, as such, x T / ∈ (ω(B θ ) ∪ ω(B θ+π ) + (p, q), so that x T ∈ O 1 . Now x T is a recurrent point, so there is an n such that f n (x T ) and x T are in the same connected component of O 1 . Therefore, if U is the connected component of O 1 that contains x T , then f n (U ) ∩ U is not empty, and since O 1 is an invariant set, this implies that f n (U ) = U. Thus, the orbit of x T only intersects n connected components of O 1 .
But lemma 7 tells us that the set x T + Z Z 2 both intersects infinitely many connected components of O 1 , and is contained in the closure of the forward orbit of x T which is absurd. Therefore Case 1 cannot be.
We begin with the following proposition, whose proof can be seen in [11] Proposition 4. Let K 1 and K 2 be compact connected sets of R 2 such that K 1 ∩K 2 is connected. If x and y are in the same connected component of K C 1 and are also in the same connected component of K C 2 , then x and y are in the same connected component of (K 1 ∪ K 2 ) C .
Proof. We already know that ω(B θ ) + (p 1 , q 1 ) ∩ ω(B θ+π ) + (p 2 , q 2 ) = ∅. We also know that, from lemma 1, that both the complements of ω(B θ ) + (p 1 , q 1 ) and of ω(B θ+π ) + (p 2 , q 2 ) have a single connected component. This means that, if we consider the corresponding sets in S 2 , ω(B θ ) + (p 1 , q 1 ) and ω(B θ+π ) + (p 2 , q 2 ), their intersection consists of the point ∞, and so we can apply the previous proposition, which gives us the result The following is result follows from a construction very similar to the one done in the proof of lemma 7.
Proof. The first claim follows from proposition 5.
Let p be the midpoint of the segment v, and let r = dist(p, Θ) = dist(p, Γ). The closed ball B r [p] is splitted in two by v. Of course, since (Γ ∪ Θ) C has a single connected component, (Γ ∪ Θ ∪ v) C has, at most, two components. Let p be a point in one of the connected components of B r [p] \ v and let q be a point in the second connected component. Suppose, by contradiction, that p and q are in the same connected component of (Γ ∪ Θ ∪ v) C . Then, there is an arc α connecting p and q lying entirely on this set. We can add to α the segment joining p and q to obtain a Jordan curve J, which does not intersect Γ ∪ Θ. But one endpoint of v must be in the int(J), and so one of Γ and Θ must be in the int(J), absurd since both are unbounded. Proof. Let y be a fixed point of f (Since x T is a recurrent point of f , f cannot be a Brouwer homeomorphism and so it has fixed points). By proposition 6, there is an arc γ which connects infinitely many translates of y and that separates ω(B θ ) and ω(B θ+π ) + (p, q). Since Γ ∪ Θ ∪ v is connected, γ ∩ v = ∅, and so there must be translates of y both in Ω 1 and in Ω 2 Lemma 9. Either f (Ω 1 ) ⊂ Ω 1 or f (Ω 2 ) ⊂ Ω 2 .
This means that either f (v) ⊂ Ω 1 or f (v) ⊂ Ω 2 . We will assume, without loss of generality, that the former is true.
But since f satisfies property T the future orbit of x T must visit any neighborhood of any Z Z 2 translate of x T infinitely many times, and so, from lemma 8, there must be integers n 1 < n 2 < n 3 such that But, from the previous lemma, if f (Ω 1 ) ⊂ Ω 1 , then for all n ≥ n 1 , f n (x T ) / ∈ Ω 2 , which violates (3) and if f (Ω 2 ) ⊂ Ω 2 , then for all n ≥ n 2 , f n (x T ) / ∈ Ω 1 , which again violates (3), and we are done.

aknowledgment
I'd like to thank Prof. Addas-Zanata for the insightful discussions on this work, and presenting ideas that helped to clarify some of the proofs presented here.