Perturbing homeomorphisms of the torus whose rotation sets have rationals in their boundaries

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1 Perturbing homeomorphisms of the torus whose rotation sets have rationals in their boundaries Patrice Le Calvez and Salvador Addas-Zanata Institut de Mathématiques de Jussieu Université Pierre et Marie Curie-Université Paris Diderot 4 place Jussieu, Case Paris Cedex 5, France Instituto de Matemática e Estatística Universidade de São Paulo Rua do Matão 1010, Cidade Universitária, São Paulo, SP, Brazil Abstract In this paper we consider homeomorphisms of the torus homotopic to the identity and their rotation sets. Let f : T 2 T 2 be such a homeomorphism, f : IR 2 IR 2 be a fixed lift and ρ( f) be its rotation set, which we assume to have interior. We also assume that some rational point ( p, r ) ρ( f) and we want to understand how stable this situation is. q q To be more precise, we want to know if it is possible to find two different homeomorphisms, which are arbitrarily small C 0 -perturbations of f, denoted f 1 and f 2, in a way that ( p, r ) does not belong to the rotation q q set of f 1 and ( p, r ) is contained in the interior of the rotation set of f2. q q We give two examples in this direction. The first is a C -diffeomorphism f dissip, such that (0,0) ρ( f dissip ), f dissip has only one fixed point with zero rotation vector and there are maps f 1 and f 2 satisfying the conditions above. The second is an area preserving version of the above, but in this conservative setting we obtain only a C 0 example. We also present two theorems in the opposite direction. The first says that if f is area preserving and analytic, then there can not be f 1 and f 2 as above. The second result, implies that for a generic (in the sense of Brunovsky) one parameter family f t : T 2 T 2 of C 1 -diffeomorphisms such that for some parameter t, ρ( f t ) has interior, ( p, r ) ρ( f q q t ) and ( p, r ) / ρ( f q q t) for t < t, then for all t > t sufficiently close to t, ( p q, r q ) / int(ρ( f t )). This kind of result gives some sort of local stability for the rotation set near rational points in its boundary. patrice.le-calvez@im-prg.fr and sazanata@ime.usp.br 2010 Mathematics Subect Classification: 37E30, 37E45, 37C25, 37C15, 37C29 The second author is partially supported by CNPq, grant: /2012-0

2 1 Introduction and main results The main motivation for this paper is to study how the rotation set of a homeomorphism of the two dimensional torus T 2 changes as the homeomorphism changes. For instance, suppose we consider a one parameter family f t : T 2 T 2 of such maps and want to study how the parametrized rotation set t ρ( f t ) IR 2 varies. The rotation set is a compact convex subset of the plane (see definition below), which varies continuously with the homeomorphism, at least in the special situation when it has interior (see [14]). In particular, we are interested in the following problem: Suppose f : T 2 T 2 is a homeomorphism homotopic to the identity and its rotation set, which is supposed to have interior, has a point ρ in its boundary with both coordinates rational. Is it possible to find two different arbitrarily small C 0 - perturbations of f, denoted f 1 and f 2 in a way that ρ does not belong to the rotation set of f 1 and ρ is contained in the interior of the rotation set of f 2? In other words we are asking if the rational mode locking found by A. de Carvalho, P. Boyland and T. Hall [5] in their particular family of homeomorphisms is, in a certain sense, a general fenomena or not. Our main theorems and examples will show that the answer to this question depends on the set of hypotheses we have. In general, we can find such maps f 1 and f 2 as described above, but if we assume certain hypotheses on f, then this sort of local mode locking happens. Even in the much simpler context of orientation preserving circle homeomorphisms, the only maps with rational rotation number which can be perturbed in an arbitrarily C 0 -small way in order to decrease or increase their rotation numbers (the analogous one dimensional version of our condition) are the ones conugate to rational rotations. In the context of degree one circle endomorphisms, P. Boyland has shown that if f : S 1 S 1 is such an endomorphism, f : IR IR is a fixed lift and ρ( f) is its rotation set (which in this case is a compact interval), if we assume that it is not degenerate to a point and has a rational extreme p/q, then it is not possible to find C 0 neighbors of f, one whose rotation set does not contain p/q and the other with p/q in the interior of its rotation set, see [4]. 1

3 In order to make things precise and to present our main results, a few definitions are necessary: Basic notation and some definitions: 1. Let T 2 = IR 2 /ZZ 2 be the flat torus and let p : IR 2 T 2 be the associated covering map. Coordinates are denoted as z IR 2 and z T Let Diff0 r(t2 ) be the set of C r (for r = 0,1,2,...,,ω) diffeomorphisms (homeomorphisms if r = 0) of the torus homotopic to the identity and let Diff0 r(ir2 ) be the set of lifts of elements from Diff0 r(t2 ) to the plane. Maps from Diff0(T r 2 ) are denoted f and their lifts to the plane are denoted f. 3. Let p 1,2 : IR 2 IR be the standard proections, respectively in the horizontal and vertical components; 4. Given f Diff0(T 0 2 ) and a lift f Diff0(IR 0 2 ), the so called rotation set of f, ρ( f), can be defined following Misiurewicz and Ziemian [13] as: ρ( f) = { } fn ( z) z : z IR n 2 (1) i 1 n i This set is a compact convex subset of IR 2 (see [13]), and it was proved in [10] and [13] that all points in its interior are realized by compact f-invariant subsets of T 2, which can be chosen as periodic orbits in the rationalcase. Bysayingthatsomevectorρ ρ( f)isrealizedbyacompact f-invariant set, we mean that there exists a compact f-invariant subset K T 2 such that for all z K and any z p 1 (z) lim n f n ( z) z n = ρ. (2) Moreover, the above limit, whenever it exists, is called the rotation vector of the point z, denoted ρ(z). As the rotation set is a compact convex subset of the plane, there are three possibilities for its shape: 2

4 1. it is a point; 2. it is a linear segment; 3. it has interior; In this paper we only consider the situation when the rotation set has interior. Our main results are: Theorem 1 : Let f Diff ω 0 (T2 ) be an analytic area preserving diffeomorphism such that ρ( f) has interior. Assume that some rational point in the plane ( p q, r q ) ρ( f), with (p,r,q) = 1 and there are arbitrarily small C 0 -perturbations of f which do not contain ( p q, r q ) in their rotation sets. Then for any homeomorphism sufficiently C 0 -close to f, ( p q, r q ) does not belong to the interior of its rotation set. It either belongs to the boundary or does not belong to the rotation set at all. The next result deals with the C 1 -generic situation. Theorem 2 : Let f Diff 1 0 (T2 ) be such that ρ( f) has interior. Assume that some rational point in the plane ( p q, r q ) ρ( f) and there are arbitrarily small C 0 -perturbations of f which do not contain ( p q, r q ) in their rotation sets. In particular, this implies that for f, either there is no q-periodic point with rotation vector equal to ( p q, r q ) or in case there are such points, they are topologically degenerate. So, assume also that these points appear in finite number and the determinant of Df q at each of them is different from 1 (these are generic conditions). Then, as in theorem 1, for all g Diff 1 0 (T2 ) sufficiently C 1 close to f, ( p q, r q ) does not belong to the interior of its rotation set. It either belongs to the boundary or does not belong to the rotation set at all. This is the theorem that implies the statement about generic families explained in the abstract. More precisely, for C 1 -generic one parameter families of diffeomorphisms, Brunovski has shown, see [7], that the only bifurcations that create periodic points are saddle-nodes or period doubling. And theorem 2 hypotheses imply that the creation of the periodic orbits with rotation vector 3

5 ( p q, r q ) had to be through a saddle-node type of bifurcation because the map f has neighbors without q-periodic points with rotation vector equal to ( p q, r q ). In the next results we present examples that do not satisfy the hypotheses of the above theorems, neither their conclusions. Theorem 3 : There exists a map f dissip Diff0 (T 2 ) such that it has only one index zero fixed point with zero rotation vector, its rotation set has interior and there are maps arbitrarily C 0 -close to f dissip, denoted g 1,g 2 Diff0 0(T2 ) such that (0,0) / ρ( g 1 ) and (0,0) int(ρ( g 2 )). The map f dissip does not preserve area. In the case of area preserving maps, the example we obtain is only C 0 : Theorem 4 : There exists an area preserving map f conserv Diff0(T 0 2 ) such that (0,0) ρ( f conserv ), its rotation set has interior and there are area preserving maps arbitrarily C 0 -close to f conserv, denoted f 1,f 2 Diff0(T 0 2 ) such that (0,0) / ρ( f 1 ) and (0,0) int(ρ( f 2 )). This paper is organized as follows. In the next section we present a brief summary on the local dynamics near fixed points of area preserving analytic diffeomorphisms of the plane and a sketch of the proof of theorem 1. In the third section we prove our results. 2 Some background results and a sketch of the proof of theorem On the dynamics near fixed points of analytic diffeomorphisms of the plane The dynamics near isolated singularities of analytic vector fields in the plane is very well understood, at least when the topological index of the singularity is not 1 (see for instance [8]). It can be proved that, if the singularity is not a focus or a center (which have topological index equal 1), then the dynamics near it can be obtained from a finite number of sectors, glued in an adequate way. Topologically, these sectors can be classified in 4 types: elliptic, hyperbolic, 4

6 expanding and attracting. Dumortier et al. studied this problem for planar diffeomorphisms near fixed points in [9]. The situation we want to understand in this sub-section is the following: Is there a topological picture of the dynamics near an index zero isolated fixed point of an analytic area-preserving diffeomorphism of the plane? It turns out that the area-preservation together with the zero index hypotheses imply that the eigenvalues of the derivative of the diffeomorphism at the fixed point are both equal to 1. The area-preservation implies that in a sufficiently small neighborhood of the fixed point, the diffeomorphism is the time one mapping of a formal vector field (defined by a formal series), which has zero divergent, see [16] and [15]. As we are assuming that the diffeomorphism is analytic, the singularity of this formal vector field, which coincides with the diffeomorphism s fixed point, is of Loasiewicz type because the vector field can not have a totally null et at the singularity. As the singularity has zero topological index, it must have a characteristic curve (see [9]), which is a simple arc with one end point at the singularity, either positively or negatively invariant under the formal flow, with a well defined tangent at the singularity (all these assertions are in a formal sense, in other words, the characteristic curve is given by a formal power series). So we are able to apply results from [9], which say that in this situation, at least in a topologicalsense, the dynamics near the fixed point of the diffeomorphism can be obtained as in the vector field setting, gluing a finite number of sectors. As we are supposing that area is preserved by the diffeomorphism, there can not be elliptic, expanding and attracting sectors. As the topological index of the fixed point is zero, there must be exactly 2 invariant hyperbolic sectors and the dynamics is topologically as in figure Sketch of the proof of theorem 1 The fundamental tool for this proof is the concept of free modifications developed by M. Brown, see [6]. The main steps in the proof are the following: 5

7 2.2.1 Step 1: Here we show that after some simple modifications and a linear coordinate change, we can assume that the rational in the boundary is (0,0) and the map f has two different periodic orbits whose rotation vectors are ( ρ 1, ρ 2 ) and (ρ 1, ρ 2 ), for some rational numbers ρ 1,ρ 2 > 0 and both these rotation vectors are contained in the interior of the rotation set of f. We also prove that under the hypothesis that it is possible to perturb f in order to destroy all its fixed points with zero rotation vector, these fixed points must be finite (here we use the analiticity hypothesis). And again, the fact that f is analytic implies that the dynamics in a neighborhood of each of these fixed points with zero rotation vector is as in figure 1 (this follows from the arguments in the previous subsection). In order to simplify things in this sketch, we assume that there is only one such fixed point, denoted P T 2, and V P is a neighborhood of P disoint from all the periodic orbits with rotation vectors equal to ( ρ 1, ρ 2 ) and (ρ 1, ρ 2 ) and prime periods. Moreover, suppose still that the dynamics inside V is as in figure Step 2: Now, assumeby contradictionthat therearehomeomorphismsf i arbitrarilyc 0 - close to f such that {(0,0),( ρ 1, ρ 2 ),(ρ 1, ρ 2 )} int.(ρ( f i )) and f i i f in the C 0 topology. So, for each integer i > 0 there exists a rational vector (0,ρ i ) int.(ρ( f i )), for some rational ρ i > 0. Related to this, it is easy to see that for any real number a > 0, (0,0) interior(convex Hull{( ρ 1, ρ 2 ),(ρ 1, ρ 2 ),(0,a)}). (3) Now we take an integer i 0 > 0 sufficiently large in a way that all the f i0 - periodic orbits with rotation vectors equal to ( ρ 1, ρ 2 ) and (ρ 1, ρ 2 ) (and prime periods) avoid V, and all the f i0 -fixed points with zero rotation vector are contained in B λ/10 (P) B λ (P) V, for some small λ > 0 and V V which satisfies f 1 (V) V f(v) V, see figure 3. 6

8 2.2.3 Step 3: The idea now is to pick a f i0 -periodic point Q i0 T 2 \V whose rotation vector is (0,ρ i0 ) int.(ρ( f i0 )) (for some rational ρ i0 > 0) and to deform f i0 inside a small neighborhood of f i0 1 (V) V f i0 (V) and outside B λ (P), in order to obtain another homeomorphism φ f i0, which coincides with f i0 outside the small neighborhood of f i0 1 (V) V f i0 (V) and inside B λ (P) and whose set of fixed points with zero rotation vector coincides with the one for f i0, and is thus contained in B λ/10 (P). Moreover, the mapping φ : T 2 T 2 is a homeomorphism supported in finitely many disoint closed disks contained in V\B λ (P), each of them free under f i0 (free means that each disk is disoint from its image under f i0 ) constructed in a way that Q i0 is still a periodic point for φ f i0 whose rotation vector is of the form (0,ρ i 0 ), for some rational ρ i 0 > 0 and its whole orbit avoids B λ (P). This means that we can deform φ f i0 Bλ (P) in order to get a mapping without fixed points with zero rotation vector, whose rotation set still contains ( ρ 1, ρ 2 ),(ρ 1, ρ 2 ) and (0,ρ i 0 ). And this is a contradiction with (3), which concludes the proof. Finally, we give a brief explanation why such a construction can be done. Suppose for the sake of simplicity, that the orbit of Q i0 under f i0 has only one point z inside B λ (P). So, there are points z,z + V, the first is a negative iterate of z and the second is a positive one, such that z belongs to a small neighborhood of the entrance of V under the action of the original map f, and z + belongs to a small neighborhood of the exit of V also under the action of f, see figure 4. We also fix a point w that belongs to the entrance of V, such that f nw i 0 (w) belongs to the exit of V for some integer n w > 0 and {w,f nw i 0 (w),...,f nw i 0 (w)} avoids B λ (P). Now we take closed disks D and D +, such that D contains z and w, D + contains z + and f nw i 0 (w), both free under f i0. If φ is a homeomorphism supported in D and D +, such that φ(z ) = w and φ(f nw i 0 (w)) = z +, then Q i0 is still periodic under φ f i0, its rotation vector is (0,ρ i 0 ) (for some ρ i 0 > 0). It is from the dynamics of f V (which is, except for the fixed point, given by a horizontal translation to the right for each point in V) and the fact that 7

9 f i0 is close to f, that we are able to construct free disks like D and D +, again see figure 4. 3 Proofs 3.1 Proof of theorem 1 The proof we present is by contradiction. So we assume that there exists a sequence of homeomorphisms f f in the C 0 topology, such that ( p q, r q ) int(ρ( f )). If we consider the maps ) ) F ( ) def. = R ( fq ( ) (p,r) R 1 def. and F( ) = R ( fq ( ) (p,r) R 1, (4) for some suitable 2 2 integer matrix R with determinant equal to 1, then we can assume that: (0,0) (ρ( F)) and (0,0) int(ρ( F )) for all positive integers ; there are rational points ( p1 ) and ( p1 ) contained in int(ρ( F)) for some sufficiently small rational numbers p1, p2 > 0 (g.c.d.(p 1,p 2, ) = 1); by continuity of the rotation set (see [14]), ( p1 ) and ( p1 ) are contained in int(ρ( F )) for all sufficiently large s. Neglecting the initial terms of the sequence F if necessary, we can suppose that ( p1 ) and ( p1 ) are contained in int(ρ( F )) for all s; The theorem hypotheses imply that F has arbitrarily C 0 -close neighbors without fixed points with zero rotation vector (because (0, 0) does not belong to the rotation sets of this neighbors). In the next lemma we will show that the analiticity of f implies that in this situation, F has finitely many fixed points with zero rotation vector. Lemma 5 : In the above hypotheses, F has finitely many fixed points with zero rotation vector. 8

10 Proof: For each z T 2, let us define a function M : T 2 IR given by M(z) = (p 1 F( z) p 1 ( z)) 2 +(p 2 F( z) p 2 ( z)) 2, for any z p 1 (z). The set of F-fixed points with zero rotation vector now denoted as Fix 0 (F) clearly coincides with M 1 (0). So if we show that #M 1 (0) <, then we are done. M is an analytic function, so if #Fix 0 (F) =, as Fix 0 (F) is compact (it is a closed subset of T 2 ), by some results in [12] and [1], there exists a simple closed curve γ T 2 such that M γ = 0 γ Fix 0 (F). As F(γ) = γ, the fact that ρ( F) has interior implies that γ must be homotopically trivial. Let γ be a lift of γ to the plane. As γ is a simple closed curve, F-invariant, it is the boundary of a closed disk D, which is also F-invariant. As F is analytic, it must have a non-fixed point inside D. So a sufficiently small ball centered at this point is disoint from its image under F. But the area preservation implies that a future iterate of this ball under F intersects the ball. And by a well known consequence of Brouwer theory (see [3]), this implies the existence of a simple closed curve contained in interior( D) with F-topological index equal 1. Something that contradicts the fact that there are homeomorphisms arbitrarily close to F without fixed points of zero rotation vector. So the lemma is proved. Let us denote the F-fixed points with zero rotation vector by {Q 1,Q 2,...,Q n }, where n 0 is the number of such points. If n = 0, then we are done because all mappings sufficiently close to F, by continuity, can not have fixed points with zero rotation vector, therefore (0, 0) can not belong to the interiorof their rotation sets. So n 1 and we can assume without loss of generality (applying a local coordinate change if necessary) that the dynamics of F near each of the Q i s is horizontal, from left to right, as in figure 5. This follows from the fact that F is analytic, preserves area and all the Q i s have zero topological indexes, as explained in subsection

11 The vectors (0,0),( p1 ) and ( p1 ) are contained in int(ρ( F )), as is a point of the form (0, L T ), with L T > 0 being a rational number. (5) Also,astheF- -periodicpointswithrotationvectorsequal( p1 )and( p1 ) are disoint from {Q 1,Q 2,...,Q n }, there exists ǫ 0 > 0 such that n i=1 B ǫ0 (Q i ) is a disoint union and the sets ( n B ǫ 0 (Q i ), i=1 F- -periodic points with rotation vector equal ( p1 ( and ) F- -periodic points with rotation vector equal ( p1 ) are two by two disoint (by (A) ǫ0 we mean the ǫ 0 -neighborhood of the set A). Moreover, the dynamics inside B ǫ0 (Q i ) consists of, apart from the fixed point Q i, horizontally moving points to the right (this is true for all sufficiently small ǫ 0 > 0). Our obective now is to show that we can choose closed rectangular neighborhoods V i of each Q i, of horizontal and vertical sides, contained in B ǫ0 (Q i ), satisfying several conditions listed below (for i = 1, 2..., n). Construction of the V i s : 1. F 1 (V i ) V i F(V i ) B ǫ0 (Q i ); 2. In each V i we denote by E i the subset of V i defined as follows. A point z E i if and only if, z V i and F 1 (z) / V i. And analogously, S i is the subset of V i such that z S i if and only if, z V i and F(z) / V i. Intuitively speaking, if a point is outside V i and its iterate belongs to V i, then the iterate belongs to E i (the entrance of V i ) and if a point is in V i and its iterate is not, then the point belongs to S i (the exit from V i ). Fixed the size of the horizontal sides of each V i, it is easy to see that if the size of the vertical sides is sufficiently small, then closure(e i ) closure(s i ) = and moreover we can take two vertical segments Σ Ei,Σ Si inside each V i, Σ Ei close to E i and Σ Si close to S i, both avoiding E i and S i, in a way ) ) ǫ 0 ǫ 0 10

12 that Q i belongs to the region between these vertical segments, see figure 6 ; 3. there exists a small closed ball θ i E i and an integer m i > 0 such that F mi (θ i ) S i. For instance, one can think that θ i is very close to the upper (or the lower) boundary of V i. Moreover, F(θ i ) θ i = and so F mi+1 (θ i ) F mi (θ i ) = ; 4. let B i be a sufficiently small ball centered at Q i in a way that closure(f 1 (B i ) B i F(B i )) interior(region beetwen the vertical segments Σ Ei and Σ Si ) and {θ i,f(θ i ),...,F mi (θ i )} is disoint from the connected component of F l (B i ) V i which contains Q i for all integers l; 5. there exists a subset of E i, called E Bi, such that every z B i that has a negative iterate outside V i, has a negative iterate F s E i (z) (z) in E Bi and F (z) V i for all s Ei(z) 0. Analogously, there exists a subset of S i, called S Bi, such that every z B i that has a positive iterate outside V i, has a positive iterate F s S i (z) (z) in S Bi and F (z) V i for all 0 s Si(z). From the choice of θ i and B i made above, we get that closure(θ i ) closure(e Bi ) = closure(f mi (θ i )) closure(s Bi ) = ; 6. θ i has such a small radius in a way that any linear segment λ i whose endpoints are, one in closure(e Bi ) and the other in θ i, do not intersect F 1 (θ i ) and F(θ i ). Analogously, any linear segment σ i whose endpoints are, one in closure(s Bi ) and the other in F mi (θ i ), do not intersect F mi 1 (θ i ) and F mi+1 (θ i ). This can be achieved if diameter(θ i ) is sufficiently small because as the vertical coordinate of a point inside V i is preserved under iterates of F, we get that a linear segment intersecting θ i and either F 1 (θ i ) or F(θ i ) is almost horizontal (the smaller diameter(θ i ) is, the closer to horizontal the segment becomes), the same for any linear segment intersecting F mi (θ i ) and either F mi 1 (θ i ) or F mi+1 (θ i ). And the anglesλ i and σ i makewith the horizontaldirectionareuniformly bounded away from 0 and π, no matter how small is the diameter(θ i ); 11

13 See figure 6 for a picture of a V i and all the sets constructed above. Now, let us choose 0 < ǫ 1 < ǫ 0 such that for all i {1,2,...,n} : 1. (F 1 (V i ) V i F(V i )) 2ǫ1 B ǫ0 (Q i ); 2. (F 1 (B i ) B i F(B i )) 2ǫ1 interior(regionbeetwentheverticalsegments Σ Ei and Σ Si ); 3. (E i ) 2ǫ1 (S i ) 2ǫ1 = ; 4. (θ i ) 2ǫ1 E i, (F mi (θ i )) 2ǫ1 S i and (θ i ) 2ǫ1 (E Bi ) 2ǫ1 =, (F mi (θ i )) 2ǫ1 (S Bi ) 2ǫ1 =. Moreover (θ i ) 2ǫ1 (F ±1 (θ i )) 2ǫ1 = and (F mi (θ i )) 2ǫ1 (F ±1 ((F mi (θ i )) 2ǫ1 )) 2ǫ1 = ; 5. Any linear segment λ i whose endpoints are, one in closure((e B i ) 2ǫ1 ) and the other in closure((θ i ) 2ǫ1 ), do not intersect closure((f 1 (θ i )) 2ǫ1 ) and closure((f(θ i )) 2ǫ1 ). Analogously, any linear segment σ i whose endpoints are, one in closure((s Bi ) 2ǫ1 ) and the other in closure((f mi (θ i )) 2ǫ1 ), do not intersect closure((f mi 1 (θ i )) 2ǫ1 ) and closure((f mi+1 (θ i )) 2ǫ1 ); For each i {1,2,...,n}, from the dynamics inside each V i, there exists an integer N i > 0 such that: For all z (E i ) 2ǫ1 \(E Bi ) 2ǫ1, sup{ IN :F s (z) V i for all 1 s } = N + i (z) is smaller then N i. Clearly, F N+ i (z)+1 (z) / V i ; For all z (S i ) 2ǫ1 \(S Bi ) 2ǫ1, sup{ IN :F s (z) V i for all s 1} = N i (z) is smaller then N i. Clearly, F N i (z) 1 (z) / V i ; Moreover, for all z (E i ) 2ǫ1 \(E Bi ) 2ǫ1, dist.(f s (z),b i ) 2ǫ 1, for all 0 s N + i (z) (6) and analogously, for all z (S i ) 2ǫ1 \(S Bi ) 2ǫ1, dist.(f s (z),b i ) 2ǫ 1, for all N i (z) s 0. (7) 12

14 Now we will choose a map denoted G, which will be equal to some F as in (4), sufficiently C 0 -close to F in a way that conditions (8), (9), (11) and (12) hold (remember that (0,0) int.(ρ( G))) : dist.(g s (z),f s (z)) < ǫ 1 /10, for all z T 2 and all max {N i} 10 s max {N i}+10; 0 i n 0 i n for all 0 i n and 0 m i, G (θ i ) (F (θ i )) ǫ1, which implies that G mi (θ i ) (S Bi ) 2ǫ1 = ; (8) any linear segment connecting a point in θ i to a point in (E Bi ) 2ǫ1 is disoint from its image under G; analogously, any linear segment connecting a point in (F mi (θ i )) 2ǫ1 to a point in (S Bi ) 2ǫ1 is disoint from its image under G; In order to show that all G sufficiently close to F satisfies the conditions in (9), proceed as follows. First, note that there exists a number c > 0 such that for any i {1,2,...,n} and any points z θ i, w closure((e Bi ) 2ǫ1 ), if zw is the linear segment connecting them, then (9) dist(zw,f(zw)) > c. (10) Thishappensbecauseforeveryi {1,2,...,n}andanyz θ i,w closure((e Bi ) 2ǫ1 ), from the choice of θ i and B i (see figure 6), the angle zw forms with the horizontal direction is uniformly bounded away from 0 and π. As the F-iterate of a point inside any V i \B i is in the same horizontal of the point, translated a little bit to the right, we get that (10) holds. Finally, if dist.(g(z),f(z)) < c/10, for all z T 2, we get that the part of (9) related θ i and (E Bi ) 2ǫ1 holds. The other assertion in (9) is proven in an analogous way. the G- -periodic points with rotation vector equal ( p1 ) are contained in ( ) F- -periodic points with rotation vector equal ( p1 ) the G- -periodic points with rotation vector equal ( p1 ) are contained in ( ) F- -periodic points with rotation vector equal ( p1 ) ǫ 0 ; ǫ 0 ; (11) 13

15 The conditions above are clearly satisfied if G is sufficiently C 0 -close to F. From the choice of ǫ 0 and ǫ 1, the G- -periodic points whose rotation vectors are equal to ( p1 ) and to ( p1 ) are disoint from all the B ǫ0 (Q i s). So if we perturb G inside some of the (F 1 (V i s) V i s F(V i s)) 2ǫ1 B ǫ0 (Q i s), we do not destroy orbits with rotation vector equal ( p1 ) and ( p1 ). The last condition on G we assume is the following: there exists an isotopy G t supported in n i=1 B i, from G to some map G 1 which has no fixed points with zero rotation vector; (12) The above condition clearly holds for all maps sufficiently close to F. The map G has a periodic orbit with rotation vector of the form (0, L T ) (for some rational L T > 0) and period T, see expression (5). Denote this orbit by χ = {W 1,W 2,...,W T }. If χ ( n B i ) =, then we are done, because we can i=1 perturb G inside n B i in order to get a homeomorphism G 1 without fixed i=1 points with zero rotation vector (see condition (12)) and such that ( p1 ), ( p1 ) and (0, L T ) belong to its rotation set. As the rotation set is convex, (0,0) belongs to the interior of the rotation set of G 1, a contradiction because G 1 does not have fixed points with zero rotation vector. And so this concludes the proof in this case. Now, assume without loss of generality that W 1 / n i=0 B i. For each i {1,2,...,n}, by choosing a smaller closed ball contained in the original θ i if necessary,andrenamingit asθ i again,wecansupposethatθ i χ = (remember that n is the number of F-fixed points with zero rotation vector). For each i {1,2,...,n}, let {W s i 1,W s i 2,...,W s } be certain points in χ which belong im(i) to B i, defined by the following conditions: 1. if W B i, then there exists an integer l such that G l (W ) {W s i 1,W s i 2,...,W s } and for all integers t between 0 and l, im(i) G t (W ) B i ; 2. the set {W s i 1,W s i 2,...,W s } has the smallest possible cardinality satis- im(i) fying condition above; 14

16 Intuitively, for each i {1,2,...,n}, {W s i 1,W s i 2,...,W s i } is a subset of χ M(i) with smallest possible cardinality such that every time a point in the future orbit of W 1 enters B i, before leaving B i it must coincide with some point in {W s i 1,W s i 2,...,W s i }. M(i) Moreover, from condition (8), for each W s i B i, there exist points W Es i χ (E Bi ) 2ǫ1 and W Ss i χ (S Bi ) 2ǫ1 such that W Ss i is in the future G- orbit of W Es i and the only point from {W s i 1,W s i 2,...,W s } which belongs to im(i) {W Es i,g(w Es i ),G 2 (W Es i ),...,W Ss i } is W s i. In order to see this, note that if a point z is outside some V i and G(z) is in V i, then G(z) belongs to (E i ) 2ǫ1 {ǫ 1 - neighborhood of the horizontal sides of V i }. So, if G l (z) belongs to B i for some l > 1 and {G(z),G 2 (z),...,g l (z)} V i, then G(z) (E Bi ) 2ǫ1 (see condition 5 on the choice of the V i s and expressions (6) and (7)). Analogously, if a point z is in some V i and G(z) is not, then z belongs to (S i ) 2ǫ1 {ǫ 1 -neighborhood of the horizontal sides of V i }. So, if G l (z) belongs to B i for some l < 0 and {G l (z),g l+1 (z),...,z} V i, then z (S Bi ) 2ǫ1. The obective now is to connect all the points in {W Es i 1,W Es i 2,...,W Es im(i) } to M(i) disoint chosen points {U i 1,Ui 2,...,Ui M(i) } interior(θ i) by disoint linear segments, which by the construction of the V i s, do not intersect the B i s, see figure 7. It is clearly possible to make such a choice of M(i) points in the interior of each θ i. Changing these linear segments locally if necessary (in an arbitrarily C 0 -small way), we can assume that apart from their endpoints in {W Es i 1,W Es i 2,...,W Es im(i) } {Ui 1,U2,...,U i M(i) i }, each of them avoids ( mi ) χ s=0 Gs ({U1,U i 2,...,U i M(i) i }).Let usdenoteeachoftheseperturbed (ornot) linear segments, connecting W Es i to U i by τei, {1,2,...,M(i)}. The fundamentalpropertytheyhaveisthefactthattheyareg-free, thatis,g(τ Ei ) τ Ei =, for all i {1,2,...,n} and {1,2,...,M(i)}. This follows from condition (9) on G. An analogousconstruction must be performed in (S Bi ) 2ǫ1, but we have to be more careful. This happens because we are not allowed to choose which points in interior(g mi (θ i )) we will connect to the {W Ss i 1,W Ss i 2,...,W Ss }. These im(i) pointsmustbe{g mi (U1 i (U i ),Gmi 2 (U i ),...,Gmi M(i) )} (θ interior(gmi i )). What 15

17 we do is the following: We connect each point W Ss i (0 M(i)) to some point P i Gmi (θ i ) by disoint linear segments τ Si. Again from condition (9) on G, these segments are G-free. Now, for each 0 M(i), we connect each P i to G mi (U i ) by disoint simple arcs τ Si G mi (θ i ). Clearly, these arcs are G-free because G mi (θ i ) is G-free. In the same way as we did for{w Es i 1,W Es i 2,...,W Es i } and {U1 i,ui 2,...,Ui M(i) M(i) }, changing the disoint simple arcs τ Si τ Si locally if necessary (in an arbitrarily C 0 -small way), we can assume that apart from their endpoints in {W Ss i 1,W Ss i 2,...,W Ss i } M(i) ( {G mi (U1 i (U i ),Gmi 2 (U i ),...,Gmi M(i) )},theyavoidχ mi ) s=0 Gs ({U1 i,ui 2,...,Ui M(i) }). Finally, we claim that each τ Si def. = τ Si τ Si is free under G. To see this, note that τ Si ) τ Si G(τ Si = (G(τ Si ) τ Si ) (G(τ Si ) (τ Si τ Si ) (G(τ Si ) = ) τ Si ) (G(τ Si ) τ Si ). As we already said, the first and the fourth terms in the union above are empty. If we prove that τ Si done, because G 1 (τ Si follows from the fact that the angle τ Si does not intersect G mi 1 (θ i ) and G mi+1 (θ i ) then we are ) G mi 1 (θ i ) and G(τ Si ) G mi+1 (θ i ). And this makes with the horizontal is uniformly bounded away from 0 and π (it depends only on (F mi (θ i )) ǫ1 and (S Bi ) 2ǫ1 ) and any linear segment which contains a point in G mi (θ i ) and another one either in G mi 1 (θ i ) or in G mi+1 (θ i ) must be closer to the horizontal then it is allowed to τ Si (see condition 7 on the construction of the V i s and condition 6 on how to choose ǫ 1 > 0). So for each τ Ei,τ Si, i {1,2,...,n} and,k {1,2,...,M(i)}, it is possible to choose open disks D Ei τ Ei and D Si τ Si such that: D Ei (V i ) ǫ1 and D Ei ( n l=1 B l ) = ; D Ei D Ei k =, if k; D Ei (χ D Ei is G-free; ( mi ) s=0 Gs ({U1 i,ui 2,...,Ui M(i) }) ) = {W Es i,u i}; D Si (V i ) ǫ1 and D Si ( n l=1 B l ) = ; 16

18 D Si D Si k =, if k; ( mi ) D Si (χ s=0 Gs ({U1 i,ui 2,...,Ui M(i) }) ) = {W Ss i,g mi (U i)}; D Si is G-free; In particular, the existence of the vertical segments Σ Ei,Σ Si inside each V i as in figure 7, implies that τ Ei τ Si k =. So, it is possible to choose the disks D Ei,DSi k in a way that n i=1 M(i) =1 (DEi D Si ) is a disoint union. See figure 7 for a picture of all the arcs above and the disks which contain them. So now we consider a homeomorphism φ: T 2 T 2, supported in n (D Ei M(i) M(i) i=1 =1 D Si ) (it is equal the identity on the boundary of each of the disoint disks in n i=1 =1 (DEi D Si )), such that φ(w Es i) = Ui and φ(gm(i) (U i)) = W Ss i, for all i {1,2,...,n} and {1,2,...,M(i)}. The homeomorphism φ G has periodic orbits with rotation vectors equal to ( p1 ), ( p1 ) and (0, L T ) (with p1, p2, L T > 0, for some T > 0) and these orbits avoid n B i by the i=1 choice of ǫ 0,ǫ 1,G and φ. So if we perturb φ G inside n B l (φ G is equal l=1 to G in this set) in order to destroy all its fixed points with zero rotation vector, as φ is supported on disoint G-free disks, we get that this perturbed homeomorphism does not have fixed points with zero rotation vector, although ( p1 ), ( p1 ) and (0, L T ) are contained in its rotation set. As (0,0) is contained in the interior of the convex hull of {( p1 ),( p1 ),(0, L T )}, this is the contradiction which proves the theorem. 3.2 Proof of theorem 2 With the same construction performed in the beginning of the proof of theorem 1, without loss of generality we can suppose that ( p q, r q ) = (0,0) and that there are rational points ( p1 ) and ( p1 ) contained in int(ρ( f)) for some sufficiently small rational numbers p1, p2 > 0 (g.c.d.(p 1,p 2, ) = 1). Now suppose by contradiction that there exists g Diff 1 0 (T2 ) arbitrarily C 1 -close to f such that (0,0) int(ρ( f)). As f has a finite number of fixed points {Q 1,Q 2,...,Q n } (n 1 otherwise the theorem is proved, as in theorem 1) with 17

19 zero rotation vector, such that det(df(p i )) 1, we get that there are disoint neighborhoods V i s of the Q i s such that the dynamics inside each V i is the one of a saddle-node or saddle-source, see figure 8. So if g is sufficiently C 1 -close to f, then all the fixed points with zero rotation vector g may have are contained in n i=1 V i and the g-dynamics inside each V i is for example like the one in figure 9. In particular, each V i can be divided in the following way: there exists an arc whose endpoints are in V i, this arc passes through a g-fixed point, it is either positively invariant or negatively invariant and it divides V i into two sets, V i and V i. Every point in V i converges under iterates of g to a fixed point of zero rotation vector inside V i, either for positive or negative iterates. So all periodic points do not enter V i. On the other hand, V i is always the union of two local saddle sectors. So, the way periodic orbits enter each V i is through a saddle sector, that is, they can only belong to V i, see figure 9. In this way, it is possible to deform g, the deformation supported in sub-neighborhoods of each of the V i s, containing all fixed points with zero rotation vector, in a way that this sub-neighborhood avoids three periodic orbits with rotation vectors equal to ( p1 ), ( p1 ) and (0, L T ), for some rational L T > 0, and the deformed homeomorphism does not have fixed points with zero rotation vector. As the previous g-periodic orbits with rotation vectors equal to ( p1 ), ( p1 ) and (0, L T ) are unaffected by the deformation (they do not intersect the support of the deformation), we get a contradiction because (0,0) is contained in the interior of the rotation set of the deformed homeomorphism and it does not have fixed points with zero rotation vector. 3.3 Proof of theorem 3 The first step in the proof is to construct a C diffeomorphism of the torus h : T 2 T 2 satisfying certain properties ( h : IR 2 IR 2 is a fixed lift of h). For that, in figure 10 we present a representation of the flat torus partitioned into three closed sets B,C and V. If we think of T 2 as a square, then rotating it by 90, V coincides with B and so B coincides with V (clearly, C falls on C). Below we list the properties: 18

20 h fixes every point z p 1 (B) and these are its only fixed points; for every z p 1 (V C), p 1 h( z) p 1 ( z) 0 and for every z p 1 (C), p 1 h( z) p 1 ( z) > 0 and p 2 h( z) = p 2 ( z); there exists a point z 0 {0} [0,1] such that h n ( z 0 ) {0} [0,1] for all integers n, p 2 h n+1 ( z 0 ) > p 2 h n ( z 0 ), p 2 h n ( z 0 ) n (0,0) and p 2 h n ( z 0 ) n (0,1); there exists another point z 1 V, p 2 (z 1 ) = 1/2 such that for any z 1 p 1 (z 1 ), h( z 1 ) = z 1 +(1,0); We will assume that such a diffeomorphism exists and show how the proof goes. After that, we construct it. We need two diffeomorphisms as above to conclude the proof, one denoted h Hor exactly as above and another one denoted h Ver as follows: h V er fixes every point z p 1 (V) and these are its only fixed points; forevery z p 1 (B C),p 2 h Ver ( z) p 2 ( z) 0andforevery z p 1 (C), p 2 h Ver ( z) p 2 ( z) < 0 and p 1 h ver ( z) = p 1 ( z); there exists a point z 0 [0,1] {0} such that h n V er ( z 0 ) [0,1] {0} for all integers n, p 1 h n+1 Ver ( z 0) < p 1 h n Ver ( z 0), p 1 h n Ver ( z 0) n (1,0) and p 1 h n Ver ( z 0 ) n (0,0); there exists another point z 1 B, p 1(z 1 ) = 1/2 such that for any z 1 p 1 (z 1), h V er ( z 1) = z 1 (0,1); Now consider the torus mapping h Ver h Hor : T 2 T 2 and its lift h V er hhor : IR 2 IR 2, where h Ver, h Hor are the lifts fixed above. The mapping f dissip def. = h Ver h Hor is homotopic to the identity, C, it has only one fixed point whose rotation vector is (0,0) (at P = (0,0)), its rotation set contains the triangle whose vertices are (0,0),(0, 1) and (1,0), and there are orbits for hver h Hor, one horizontal whose α limit set is ( 1,0) and ω limit set is (0,0) and another one, which is vertical and its α limit set is (0,0) and ω limit set is (0,1). 19

21 So, if we perturb f dissip (in a C 0 small way) in a neighborhood of P, we can turn the horizontal homoclinic orbit into a periodic orbit with rotation vector of the form ( p q,0) for some rational number p q > 0 and analogously, we can turn the vertical homoclinic orbit into a periodic orbit with rotation vector of the form (0, r s ) for some rational number r s > 0, without destroying the fixed points z 1 and z 1. Thus for this perturbed mapping, (0,0) belongs to the interior of its rotation set. And if we consider f dissip ( ) + (v 1,v 2 ) for any real numbers v 1 > 0 and v 2 < 0 we will show that (0,0) does not belong to its rotation set. This happens because for every z IR 2 we have: p 2 ( f dissip ( z)+(v 1,v 2 )) p 2 ( z) < v 2 < 0 or p 1 ( f dissip ( z)+(v 1,v 2 )) p 1 ( z) > v 1 > 0 So f dissip ( )+(v 1,v 2 ) has no fixed points. In this way, if (0,0) ρ( f dissip + (v 1,v 2 )), it must belong to the boundary of the rotation set and it can not be an extremal point (see [10] and [11]). So there are two extremal points ω 1,ω 2 ρ( f dissip + (v 1,v 2 )) such that (0,0) belongs to the linear segment connecting ω 1 to ω 2. Also note that the image under f dissip ( )+(v 1,v 2 ) of any horizontal straight line of the form IR {n} (for any integer n) is disoint from it and strictly below. And analogously, the image under f dissip ( ) + (v 1,v 2 ) of any vertical straight line of the form {n} IR (for any integer n) is also disoint from it and strictly to the right. Thus, for any ω ρ( f dissip +(v 1,v 2 )), p 1 (ω) 0 and p 2 (ω) 0, a contradiction with the existence of ω 1,ω 2 as above. This proves the theorem. In the following we construct the annular diffeomorphism described in the beginning of the proof. Our construction will give h : IR 2 IR 2 the lift of a C diffeomorphism of the torus which fixes pointwise the straight lines IR {n} for all integers n (this is why we call it annular diffeomorphism). First, we partition the plane into three closed equivariant sets Ṽ, C and B as in figure 11 and construct the map hǫ ( z) def. = z +ǫ.(l( z),m( z)), (13) 20

22 for certain real valued C 1-periodic functions l,m and a sufficiently small ǫ > 0. We will choose l,m satisfying certain properties and then taking ǫ > 0 suff. small, h ǫ will be the lift of a diffeomorphism of the torus. So choose any pair of C periodic functions l,m as below: 1. l,m B 0; 2. m C 0 and l {n} IR 0 for all integers n; 3. m int( Ṽ ) > 0; 4. l int( C) ( Ṽ\ n Z {n} IR)> 0 As the set of C diffeomorphisms of the torus is open, if ǫ > 0 is sufficiently small, then h ǫ ( z) as in (13) is the lift of a diffeomorphism of the torus which fixes pointwise the horizontals IR {n} B for all integers n and B = Fix( h ǫ ). The map h ǫ leaves {0} IR Ṽ invariant, fixes points of the form (0,n) for all integers n and moves the other points in {0} IR positively in the vertical direction. So we can take a point z 0 {0} S 1 T 2 whose orbit avoids S 1 { 1 2 δ, 1 2 +δ} (for a suff. small δ > 0) and construct a C diffeomorphism of T 2, denoted φ, such that φ (S 1 { 1 2 δ,1 2 +δ})c Id; p 2 φ(z) = p 2 (z) for all z T 2 ; for some z 1 V S 1 { 1 2 } (there are two such points in the previous intersection), φ( hǫ ( z 1 )) = z 1 + (1,0), where φ is the lift of φ which is the identity except at the horizontal strips of height 2.δ, and z 1 is any point in p 1 (z 1 ). The number δ > 0 is sufficiently small in a way that S 1 { 1 2 δ, 1 2 +δ} does not intersect the set B T2. It is not hard to see that the mapping we are looking for is h def. = φ h ǫ. 21

23 4 Proof of theorem 4 The main idea used to construct this example is analogous to the previous one, namely we compose two homeomorphisms of the torus, the first one leaves a horizontal annulus invariant and the second a vertical annulus, in order to produce the example. But things here are harder, so our construction is only C 0. Probably there are smooth examples, but as theorem 1 shows, no analytic example can be found. The first part of the proof is to construct an area preserving homeomorphism ofthetorush:t 2 T 2 homotopictothe identity, whichhasalift h : IR 2 IR 2 satisfying the following properties: 1. Fix( h) = n Z IR {n}; 2. there exists a point z 0 {0} [0,1] such that h n ( z 0 ) {0} [0,1] for all integers n, p 2 h n+1 ( z 0 ) > p 2 h n ( z 0 ), p 2 h n ( z 0 ) n (0,0) and p 2 h n ( z 0 ) n (0,1); 3. the rotation set of h, ρ( h) is equal to [0,a] {0} for some a > 0 and so there exists z 1 T 2 which is periodic and has a rotation vector of the form ( p q,0), for some rational p q > 0; 4. there existsaverticalannulus [ 1 2 c, 1 2 +c] S1 T 2 (for some0 < c < suff. small) which avoids the orbit of the point z 1 above and for every z p 1 ([c,d] ]0,1[), p 1 h( z) p 1 ( z) > 0 and p 2 h( z) = p 2 ( z); To construct such a mapping we start with h : IR 2 IR 2 given by h ( z) = h ( x,ỹ) = ( x+m(ỹ),ỹ), where m : IR IR is a 1-periodic continuous non-negative function such that m(ỹ) = 0 ỹ is an integer, it has a maximum at 1 2 (m(1 2 ) < ) and it is affine increasing between 0 and 1 2 and affine decreasing between 1 2 figure 12. Now let c n, n ZZ, be a sequence of real numbers such that: and 1, see 22

24 c n 1 2 and 0 < c n < c n+1 < 1, for all integers n; for all integers n, c n+1 c n < 1 100, c n n n 0 and c n 1; n 0 is the integer such that 1 2 [c n 0,c n0+1]; For each integer n, denote by θ n the linear segment connecting h (0,c n ) to (0,c n+1 ). Clearly, θ n θ m =, for all integers n m. From the choice of the function m we get that (h ) 1 (θ n ) θ n = for all integers n, diam.((h ) 1 (θ n )) n ± 0 and for n n 0, (h ) 1 (θ n ) is a linear segment. For n = n 0, (h ) 1 (θ n0 ) is piecewise linear and it proects inectively in the x and ỹ axis, see figure 13. Now we fix some point z 1 T 2 which is periodic under h, p 2 (z 1 ) c n forall integers n and it has a rotation vector of the form ( p q,0), for some rational p q > 0 (clearly a necessary and sufficient condition for this is that m(p 2 (z 1 )) = p q ). As allpointsinthe horizontalcirclecontainingz 1 areq-periodicforh withrotation vector equal ( p q,0), we can suppose that the orbit of z 1 does not intersect any of the (h ) 1 (θ n ) and avoids { 1 2 } S1. So, there exists 0 < c < the orbit of z 1 avoids [ 1 2 c, 1 2 +c] S1. such that The next step is to fix an area and orientation preserving coordinate change η : IR [0,1] Mod.Strip, which sends the horizontal strip IR [0,1] into the modified strip M od.strip, described in figure 14, which will be important later. It is possible to choose η in a way that: η( z +(1,0)) = η( z) for any z IR [0,1]; η maps verticals into verticals; ( h ) 1 ([ 1 2 c, 1 2 +c] [0,1]) [1 2 c, 1 2 +c] [0,1] h ([ 1 2 c, 1 2 +c] [0,1]) V 3, a set described in figure 14 and η V3 acts linearly, sending horizontal lines into horizontal lines and vertical lines into vertical lines. It expands the horizontal lines and contracts the vertical ones. Clearly, as η preserves area, the product of this expansion and contraction is equal to 1; 23

25 η maps the vertical rectangle [ 1 2 c, 1 2 +c] [0,1] IR [0,1] linearly into the square Q, whose side length is 2.c and Q is exactly in the middle of the fundamental domain of M od.strip it belongs, as described in figure 14. Also remember that [ 1 2 c, 1 2 +c] [0,1] is a connected component of the lift of the vertical annulus [ 1 2 c, 1 2 +c] S1 T 2 to IR [0,1]. This annulus avoids the orbit of the periodic point z 1 above and also avoids θ n (h ) 1 (θ n ); n Z the height of Mod.Strip is equal to the width of each of its fundamental domains (denoted L(η) > 0), which is a little less than 1 2.c ; the angle α described in figure 14 is less then 30 ; η V1 sends vertical lines into vertical lines, stretching their lengths by a factor greater or equal to 2. So the width of η(v 1 ) is less then half the width of V 1, which we denote by 2.K(η); The next step is to perturb h in order to create a heteroclinic orbit from (0,0) to (0,1). With this in mind, for all integers n, let us define a sequence of numbers δ n and ǫ n such that (for any set K T 2, as in the proof of theorem 1, we denote by (K) α the α-open neighborhood of K) 1) δ n,ǫ n < min{ cn+1 cn 100,K(η)}, (θ n ) ǫn interior(h (((h ) 1 (θ n )) δn )) and the orbit of z 1 avoids all ((h ) 1 (θ n )) δn (remember that 2.K(η) is the width of V 1 ); 2) ((h ) 1 (θ n )) δn ((h ) 1 (θ m )) δm =, for all integers n m; 3) ((h ) 1 (θ n )) δn h (((h ) 1 (θ n )) δn ) = for all integers n; The idea now is to modify h in the interior of the ((h ) 1 (θ n s)) δn in order to create the heteroclinic orbit from (0, 0) to (0, 1) without creating fixed points not in S 1 {0}. This is possible because each ((h ) 1 (θ n )) δn is free under h (as we already said, free means disoint from its image), so changing h inside n Z ((h ) 1 (θ n )) δn does not create fixed points (again, see [6]). Also, a deformation supported in this set will keep the orbit of z 1 unaltered. 24

26 As is explained in [2], it is not difficult to construct an area preserving homeomorphism φ : T 2 T 2 homotopic to the identity such that: φ ( n Z Vǫn(θn))c Id; φ(h (0,c n )) = (0,c n+1 ); So, h def. = φ h : T 2 T 2 is the desired homeomorphism. As the support of φ does not intersect the vertical annulus [ 1 2 c, 1 2 +c] S1, h restricted to it coincides with h. One of the crucial properties this map has is the following. Crucial Property 4: If h : IR 2 IR 2 is the lift of h that fixes n Z IR {n} pointwise, then it may have points z IR [0,1] such that p 1 ( h( z)) p 1 ( z) < 0. From the choices above, such a point belongs to V 1 +(n,0) for some integer n. But for these points, p 2 ( h( z)) p 2 ( z) > 5.(p 1 ( h( z)) p 1 ( z)), see figure 15. In this way, if we consider h restricted to IR [0,1], it induces a periodic homeomorphism in Mod.Strip, given by h Hor def. = η h η 1. And finally, we can extend h Hor to the strip IR [0,L(η)](remember that L(η) it the height and width of a fundamental domain of Mod.Strip) by imposing that h Hor restricted to the set B described in figure 14 is the identity. As h Hor IR {0} IR {L(η)} is the identity, we can extend it to the whole plane in a way that it is the lift of a homeomorphism of the torus homotopic to the identity. So h Hor : IR 2 IR 2 inducesahomeomorphismh Hor : T 2 L(η) T2 L(η), wheret2 L(η) = IR2 /(L(η).ZZ) 2, such that its rotation set is ρ( h Hor ) = [0,a] {0}, for some a > 0. Moreover, hhor fixes (0,0) and (0,L(η)) (these points correspond to the same point in the torus T 2 L(η) ) and h Hor has a heteroclinic orbit from (0,0) to (0,L(η)). Now we perform an analogous construction in the vertical direction in order toobtainahomeomorphism h Ver : IR 2 IR 2 whichinducesh V er : T 2 L(η) T2 L(η), such that its rotation set is ρ( h V er ) = {0} [ a,0]. Moreover, h V er leaves a 90 rotation of Mod.Strip invariant, fixes (0,0) and ( L(η),0) (these points correspond to the same point in the torus T 2 L(η) ) and it has a heteroclinic orbit 25