Dynamics of homeomorphisms of the torus homotopic to Dehn twists

Dynamics of homeomorphisms of the torus homotopic to Dehn twists

Salvador Addas-Zanata, F´ abio A. Tal and Br´ aulio A. Garcia

Instituto de Matem´atica e Estat´ıstica Universidade de S˜ao Paulo

Rua do Mat˜ao 1010, Cidade Universit´aria, 05508-090 S˜ao Paulo, SP, Brazil

Abstract

In this paper we consider torus homeomorphismsfhomotopic to Dehn twists. We prove that if the vertical rotation set off is reduced to zero, then there exists a compact connected essential ”horizontal” set K, invari- ant underf. In other words, if we consider the liftfboff to the cylinder, which has zero vertical rotation number, then all points have uniformly bounded motion under iterates off. Also, we give a simple explicit condi-b tion which, when satisfied, implies that the vertical rotation set contains an interval and thus also implies positive topological entropy. As a corol- lary of the above results, we prove a version of Boyland’s conjecture to this setting: If f is area preserving and has a liftfbto the cylinder with zero Lebesgue measure vertical rotation number, then either the orbits of all points are uniformly bounded under f, or there are points in theb cylinder with positive vertical velocity and others with negative vertical velocity.

Key words: vertical rotation set, omega limits, brick decompositions

e-mail: [email protected], [email protected] and [email protected]

2000 Mathematics Subject Classification: 37C25, 37C50, 37E30, 37E45

The first two authors are partially supported by CNPq, grants: 304803/06-5 and 304360/05-8.

The third is supported by a FAPESP grant 2008/10363-5.

1 Introduction and main results

In this paper we study homeomorphisms f of the torus homotopic to Dehn twists. These homotopy classes are in some way simpler to analyze than the identity case. One of the reasons for this is the fact that there is no sense in defining a two dimensional rotation set for torus maps homotopic to Dehn twists, instead a vertical rotation set is defined, see expression (1).

Many important conjectures for homotopic to the identity maps have their analogs in this setting. For instance, how is the rotation interval of a min- imal Dehn twist homeomorphism? Does the set of minimal Dehn twist C^r- diffeomorphisms (r≥2) have no interior? Iff is a Dehn twist homeomorphism which preserves area and has zero Lebesgue measure vertical rotation number, is it true that eitherf is more or less like an annulus homeomorphism or the vertical rotation interval has no empty interior?

One of the main motivations for our work is a recent example of F. Tal and A. Koropecki, where they present an area preserving torus homeomorphism h homotopic to the identity, such that its rotation set is only (0,0),satisfying the following property:

• hhas a lift to the plane, denotedeh,such thatehhas fixed points and some points in the plane have unboundedeh-orbits in every direction

In other words, this example implies that the existence of sub-linear dis- placement does not imply linear displacement, at least in the homotopic to the identity class. In this work we show that maps homotopic to Dehn twists have a different behavior. Before presenting our results, we need some definitions.

Definitions:

1. Let T²= IR²/ZZ² be the flat torus and letp: IR² −→T² andπ: IR²−→

S¹×IR be the associated covering maps. Coordinates are denoted as (ex,y)e ∈IR²,(x,b by)∈S¹×IR and (x, y)∈T².

2. Let DT(T²) be the set of homeomorphisms of the torus homotopic to a Dehn twist (x, y)−→(x+kymod 1, ymod 1), for somek∈ZZ^∗, and let

DT(S¹×IR) andDT(IR²) be the sets of lifts of elements from DT(T²) to the cylinder and plane. Homeomorphisms fromDT(T²) are denotedf and their lifts to the vertical cylinder and plane are respectively denoted fbandf .e

3. Let p_1,2 : IR² −→ IR be the standard projections; p₁(˜x,y) = ˜˜ x and p2(˜x,y) = ˜˜ y. Projections on the cylinder are also denoted by p1 and p2.

4. Given f ∈ DT(T²) and a lift fb∈ DT(S¹ ×IR), the so called vertical rotation set can be defined as follows, see [12]:

ρV(fb) = \ i≥1

[ n≥i

(p2◦fbⁿ(bz)−p2(z)b

n :bz∈S¹×IR )

(1)

This set is a closed interval (maybe a single point, but never empty) and it was proved in [1] and [3] (and much earlier in [6], although the first author discovered this only recently) that all numbers in its interior are realized by compact f-invariant subsets of T², which are periodic orbits in the rational case. From its definition, it is easy to see that

ρ_V(fb^m+ (0, n)) =m.ρ_V(fb) +nfor any integersn, m.

5. Given f ∈ DT(T²) and a lift fb∈ DT(S¹×IR), let µ be a f-invariant Borel probability measure. We define the vertical rotation number ofµas follows:

ρV(µ) = Z

T²

φ(x, y)dµ,

where the vertical displacement functionφ: T²→IR is given byφ(x, y) = p₂◦fb(x,b y)b −y,b for any (x,b y)b ∈S¹×IR such thatπ⁻¹(x,b y)b ⊂p⁻¹(x, y).

So, givenf ∈DT(T²) andfb∈DT(S¹×IR),as we said above, one wants to know, under which conditions canf be minimal? It is not difficult to see that in this case the vertical rotation interval must be a single point, otherwise there would be infinitely many periodic orbits. But more can be said.

Theorem 1 : Given f ∈DT(T²) and a liftfb∈DT(S¹×IR),suppose that f is minimal. Then,ρV(fb) ={α} for some irrational numberα.

So, iff is aC^rdiffeomorphism, for somer≥2,is there a natural perturba- tion that destroys minimality? As the extreme points ofρ_V(fb) vary continuously withfb∈DT(S¹×IR) (see [7]), a way to attack this problem is by showing that irrational extremes are not stable under perturbations. This was done in [2] for twist mappings on the torus.

The main problem addressed in this paper is in a way, complementary to the above. Suppose for instance that ρV(fb) contains a single rational number p/q.What can we say about the dynamics off? And iff preserves area and the center of gravity, that is Lebesgue measure has zero vertical rotation number, what can we say about its vertical rotation interval? When it is not reduced to zero, is zero always an interior point? This is the so called Boyland’s Conjecture.

Below we state our main results:

Theorem 2 : Givenf ∈DT(T²)and a liftfb∈DT(S¹×IR),ifρV(fb) ={p/q}, for some rational p/q, then there exists a compact connected setK ⊂S¹×IR, invariant under fb^q −(0, p), which separates the ends of the cylinder. So, all points have uniformly bounded orbits under the action offb^q−(0, p).

Note that no area preservation hypothesis appear in our theorem. The fol- lowing corollary is almost immediate:

Corollary 1 : Given f ∈ DT(T²) and a lift fb∈ DT(S¹×IR), suppose that ρ_V(f) = [a, p/q],b for some rationalp/qand some realasmaller thanp/q. Then there existsM >0such that for all bz∈S¹×IR, p2◦fbⁿ(bz)−p2(bz)−np/q < M, for all integersn >0.

The next result gives an explicit criteria which implies non-degenerate ver- tical rotation sets and thus by a result analogous to the one in [11], implies positive topological entropy (see for instance [6] and [1]).

Theorem 3 : Given f ∈ DT(T²) and a lift fb∈ DT(S¹×IR), there exists M >0 (which can be explicitly computed) such that if for some points bz1,bz2∈

S¹×IR, we have p2◦fbⁿ¹(bz1)−p2(zb1)<−M and p2◦fbⁿ²(bz2)−p2(zb2)> M, for certain positive integersn1 andn2,then 0is an interior point of ρV(fb).

The next result gives a positive answer for Boyland’s conjecture in this set- ting:

Corollary 2 : Given an area-preserving f ∈DT(T²) and a lift fb∈DT(S¹× IR) with zero Lebesgue measure vertical rotation number, then either ρV(fb) is reduced to0,or0 is an interior point of ρ_V(fb).

This paper is organized as follows. In the second section we present some background results we use, with references and a few proofs and in the third section we prove our main results. From now on we assume, without loss of generality, that any f ∈ DT(T²) we consider, is homotopic to a Dehn twist (x, y)−→(x+kDehny mod 1, ymod 1) withkDehn>0.

2 Basic Tools

2.1 Brick Decompositions of the plane

We define a brick decomposition of the plane as follows:

IR²=^∞∪

i=0Di,

where each D_i ∈ Brick Decomposition is the closure of a connected simply connected open set, such that∂Diis a polygonal simple curve andinterior(Di)∩

interior(D_j) =∅,for i6=j. Moreover, the decomposition is locally finite, that is, ^∞∪

i=0∂Di is a graph whose vertices have three edges adjacent to them and the number of elements of the decomposition contained in any compact subset of the plane is finite.

Given an orientation preserving homeomorphism of the planeeh,we say that the brick decomposition is free, if all its bricks are free, that is,eh(Di)∩Di=∅, for alli∈IN.Given two bricks,DandE,we say that there is a chain connecting them, if there are bricks

D=D0, D1, D2, ..., D_n−1, Dn=E

such thateh(Di)∩Di+16=∅,fori= 0,1, ..., n−1.IfD=E,the chain is said to be closed.

In the following we will present a version of a theorem of J. Franks [8] due to Le Roux and Guillou, see [9], page 39:

Lemma 1 : The existence of a closed chain of free closed bricks implies that there exists a simple closed curveγ⊂IR², such that

index(γ,eh) =degree(γ, eh(z)−z

eh(z)−z

) = 1.

This result is a clever application of Brouwer’s lemma on translation arcs.

2.2 On the sets B

and B

Here we present a theory developed in [4] and extend some constructions to our new setting. For this, consider a homeomorphism f ∈DT(T²), a liftfb∈ DT(S¹×IR) and a lift of fbto the plane, denotedfe∈DT(IR²). Given a real numbera, let

Ha=S¹× {a},

H_a⁻=S¹×]− ∞, a] andH_a⁺=S¹×[a,+∞[.

We will also denote the sets H0, H₀⁻ and H₀⁺ simply by H, H⁻ and H⁺ respectively. If we consider the closed sets,

B⁻=\

n≤0

fbⁿ(H⁻) and B⁺=\

n≤0

fbⁿ(H⁺),

we get that they are both closed and positively f-invariant. For each of theseb sets, consider the following subsets: B_S⁻ ⊂B⁻ and B_N⁺ ⊂ B⁺, each of which consisting of exactly all unbounded connected components of respectively,B⁻ and B⁺. The sets B_S⁻ and B_N⁺ are always closed, but in some cases may be empty. The next lemma tells us that under certain conditions, they really exist.

Lemma 2 : Suppose0∈ρV(fb).ThenB⁺_N andB⁻_S are not empty.

Proof:

The proof of this result goes back to Le Calvez [10] and even Birkhoff [5].

First, suppose that ∪

n≥0 fbⁿ(H) is unbounded both from above and from below. In this case, considering the setB_S⁻, the only thing we have to prove is that, for alla≤ −1,there exists a first positive integern=n(a),such that

fb⁻ⁿ(Ha)∩H 6=∅andn(a)→ ∞asa→ −∞. (2) Our assumption on ∪

n≥0fbⁿ(H) implies thatfb^N(H)∩H_a⁻ 6=∅ for some integer N >0.If expression (2) does not hold forN,thenfb^−N(H_a)⊂H⁺⊂H_a⁺+(0,1), which would imply that 0 ∈/ ρV(fb), a contradiction. So expression (2) is true and the proof continues, for instance as in lemma 6 of [4]. A similar argument holds forB_N⁺ (in this casea≥1).

If for some integerM0>0,fbⁿ(S¹× {0})⊂S¹×[−M0,+∞[ for all integers n≥0,then clearlyfbⁿ(S¹×[M₀,+∞[)⊂S¹×[0,+∞[ for all integersn≥0,so B_N⁺⊃S¹×[M0,+∞[ and thus, it is not empty.

To prove thatB_S⁻ is also not empty, we have to work a little more.

LetO^∗= ∪

n≥0fbⁿ(S¹×]0,+∞[) and letObe the complement of the connected component of (O^∗)^cwhich contains the lower end of the cylinder. We claim that O^c is connected and the same holds for ∂O^def.= K. This follows if we consider the N orth−South compactification of the cylinder and remember that it is a classical result, in the plane or sphere, that the frontier of any connected component of the complement of a compact connected subset is also connected.

Clearly,O^∗⊂O(we just fill the holes),Ocontains the upper end of the cylinder andfb(O)⊂O.

If ∩

n≤0fbⁿ(O^c) =∅,then 0∈/ ρ_V(fb).So ∩

n≤0fbⁿ(O^c)6=∅and as each connected component of this closedfb-invariant set is bounded from above and unbounded, we get that for a sufficiently large integerj≥0, ∩

n≤0fbⁿ(O^c)−(0, j)⊂B_S⁻6=∅.

The remaining possibility can be treated in an analogous way. 2

2.3 The ω-limit sets of B

and B

In this subsection we examine some properties of the set ω(B_S⁻)^def.=

∞

\

n=0

∞

[

i=n

fbⁱ(B⁻_S). (3)

Due to the fact thatfb(B_S⁻)⊂B_S⁻=B_S⁻,we get that ω(B_S⁻) =

∞

\

n=0

fbⁿ(B_S⁻) =

∞

\

n=−∞

fbⁿ(B_S⁻). (4) Lemma 3 : ω(B⁻_S) is a closed, fb-invariant set, whose connected components are all unbounded.

Proof:

See lemma 7 of [4]. 2

So from (4),ω(B⁻_S)⊂B_S⁻ and it is still possible thatω(B_S⁻) =∅.The next lemma tells us that in this case, things are easier.

Lemma 4 : Suppose 0 ∈ ρV(fb), which implies that B_S⁻ is not empty. If ω(B⁻_S) =∅,thenρV(fb)⊃[−,0], for some >0.

Proof:

See the proof of lemma 10 of [4] and the paragraph below it. 2

Now, if we consider the set B_S⁻ for fb⁻¹, denoted B_S⁻(inv), we get the fol- lowing:

Lemma 5 : The setsω(B_S⁻)andω(B_S⁻(inv))are equal.

Proof:

Let Γ be a connected component ofω(B_S⁻).From the definition,fbⁿ(Γ)⊂H⁻ for all integersn.So Γ⊂B_S⁻(inv) and moreover, for each positive integern,as fbⁿ(Γ) is contained inH⁻, we get that Γ ⊂fb⁻ⁿ(B_S⁻(inv)), which means that Γ⊂ω(B_S⁻(inv)).Thus ω(B⁻_S)⊂ω(B_S⁻(inv)). The other inclusion is proved in an analogous way. 2

The following are important results on the structure of these sets.

Lemma 6 : Any connected component eΓ of π⁻¹(ω(B_S⁻)) is unbounded, not necessarily in they-direction.e

Proof:

Letd be the metric onS¹× IR and letdebe the lifted metric on the plane.

Consider a point Pe ∈ Γ and lete P = π(Pe). As P ∈ ω(B_S⁻), there exists a connected component Γ ofω(B_S⁻) that contains P.Since by lemma 3 Γ is un- bounded, for every sufficiently large integernthere exists a simple continuous arcγ_n ⊂S¹×IR such that:

• P is one endpoint ofγn;

• γnis contained inS¹×[−n,0] and it intersectsS¹× {−n}only at its other endpoint;

• γn is contained in a (1/n, d)-neighborhood of Γ;

Now let eγn be the connected component of π⁻¹(γn) that contains P .e This arcγe_nis contained in a (1/n,d)-neighborhood ofe π⁻¹(Γ)⊂π⁻¹(ω(B⁻_S)) because the covering map is locally an isometry.

Now, embed the plane in the sphereS²= IR²t {∞}equipped with a met- ricD topologically equivalent to the metricdeon the plane. Then there exists a subsequence eγn_i

i→∞→ Θ in the Hausdorff topology, for some compact con- nected set Θ⊂ S². Clearly, both ∞ and Pe belong to Θ. Furthermore, since π⁻¹(ω(B⁻_S))∪ {∞}is a closed set and

n→∞lim sup e^z∈e^γn

d(ez, πe ⁻¹(ω(B⁻_S)))

!

= 0,

we get thatπ⁻¹(ω(B_S⁻))∪ {∞}contains Θ and the proof is over. 2

Lemma 7 : For any connected componenteΓof π⁻¹(ω(B_S⁻)), eΓ^c is connected.

Proof:

Take a connected component eΓ ofπ⁻¹(ω(B_S⁻)).First note that eΓ^c has one connected component, denotedO⁺, which contains IR×]0,+∞[.So, if there is

another one, denotedO1,it must be contained in IR×]− ∞,0].In the following we will prove thatfeⁿ(O1)⊂IR×]− ∞,0] for all integersn.

By contradiction, suppose that

there is an integern0 such thatfeⁿ⁰(O1) is not contained in IR×]− ∞,0]. (5) There exists a numberm0>0 such that ify > me 0,then the pointfe⁻ⁿ⁰(x,e ey) has positive ey-coordinate, for all ex ∈ IR (see (6)). So our hypothesis in (5) implies thatfe⁻ⁿ⁰(IR×]0,∞[)∩∂O16=∅,which means thatfeⁿ⁰(∂O1) intersects IR×]0,∞[,a contradiction with the fact that

feⁿ⁰(∂O₁)⊂feⁿ⁰(eΓ)⊂π⁻¹(ω(B_S⁻))⊂IR×]− ∞,0].

So (5) does not hold. To conclude, let Γ be the connected component ofω(B⁻_S) that contains π(eΓ), which as we know by lemma 3 is unbounded. The set O₁∪eΓ is connected as well as π(O₁ ∪Γ)e ∪Γ = π(O₁)∪Γ and the later is contained inω(B_S⁻) becausefbⁿ(π(O1))⊂H⁻ for all integersn.It follows that π(O1)∪Γ = Γ ⊂ω(B_S⁻) and therefore O1∪eΓ is contained in π⁻¹(ω(B_S⁻)),a contradiction with the choice ofeΓ.2

Clearly, similar results hold forB_N⁺.

3 Proofs

3.1 Proof of theorem 1

Assume f ∈DT(T²) and its lift fb∈DT(S¹×IR) are such thatf is minimal andρV(fb) is rational. Without loss of generality we can assume thatρV(fb) = 0, because iff is minimal, the same happens for all its iterates. This follows from the fact that, if for some integerq >0, f^q is not minimal, then it has a compact invariant minimal setK ⊂T², which by minimality, has empty interior. But then,

K∪f(K)∪...∪f^q−1(K)

is invariant underf and asK^cis open and dense, Baire ’s property also implies that K∪f(K)∪...∪f^q−1(K) has empty interior, a contradiction with the minimality off.

As fbhas no fixed points, lemma 2 of [3] implies that there exists a homo- topically non trivial simple closed curveγin the cylinder such thatγ∩f(γ) =b ∅.

Without loss of generality, we can suppose thatfb(γ)⊂γ⁻,the connected com- ponent ofγ^cwhich is belowγ.Letk >0 be an integer such thatγ−(0, k)⊂γ⁻. If for somen >0,fbⁿ(γ)⊂(γ−(0, k))⁻,then 0 would not belong toρ_V(fb).So, for alln >0,there exists a pointbzn,abovefb(γ) and belowγ,such that

{bzn,fb(zbn),fb²(zbn), ...,fbⁿ(bzn)}is above γ−(0, k).

Taking a subsequence if necessary, we can assume thatbz_ni ^i→∞→ bz^∗, a point in the closure of the region between fb(γ) andγ. Clearly, the positive orbit of bz^∗ is bounded in the cylinder and so itsω-limit setω(bz^∗) is a compactf-invariantb subset of the cylinder. Moreover, as any integer vertical translate ofω(bz^∗) is alsofb-invariant, if we pick a minimal fb-invariant compact set K contained in ω(zb^∗),clearly, by minimality it satisfiesK∩K+ (0, n) =∅for alln6= 0.

As f is minimal, when K is projected to the torus is must be the whole torus, a contradiction. 2

3.2 Proof of theorem 2

Givenf ∈DT(T²) and a lift fb∈DT(S¹×IR), without any loss of generality we can assume thatρV(fb) = 0.

Lemma 2 implies that B_N⁺ 6=∅ andB_S⁻ 6=∅, and lemma 4 implies that the same holds for theirω-limits,ω(B_N⁺)6=∅ andω(B_S⁻)6=∅.

In the following we will present two technical results. For each xb ∈ S¹, consider the following functions, which as the next lemma shows, are well defined at allbx∈S¹:

µ(x) = max{b yb∈IR : (x,b y)b ∈ω(B_S⁻)}

ν(x) = min{b yb∈IR : (x,b y)b ∈ω(B_N⁺)}

Lemma 8 : There exists a constantM_f >0 such that sup

bx,

by∈S¹

|µ(bx)−µ(y)| ≤b Mf and sup b^x,b^y∈S

1

|ν(bx)−ν(y)| ≤b Mf. Proof:

The proof is analogous for both cases, so let us only consider the function µ.Asω(B_S⁻) is closed and bounded from above, choose somebx0∈S¹ such that {xb₀}×]− ∞,0]∩ω(B_S⁻)6=∅and for someyb₀≤0,(bx₀,by₀) belongs toω(B_S⁻) and has maximaly-coordinate. Thenb µ(xb0) =yb0is well defined.

Note that as f is homotopic to a Dehn twist, for all (ex,ey)∈IR² there are constantsAf >0 andBf>0 such that

p2◦fe(ex,y)e −ye

< Af and

p1◦fe(x,e ey)−ex−kDehnye

< Bf. (6) So for any compact setG⊂IR²with

|p2(G)|^def.= max(p2(G))−min(p2(G))≥Vf

def.= (3 + 2Bf) k_Dehn and

|p1(G)|^def.= max(p1(G))−min(p1(G))<1, we have:

p1(fe(G))

>2 andp2|

f(G)e >min(p2(G))−Af.

Consider the intersectionπ⁻¹(ω(B_S⁻))∩IR×[µ(xb0)−Vf, µ(xb0)].If all vertical segmentsSeg

ex ={ex} ×[µ(xb₀)−V_f, µ(bx₀)] intersect π⁻¹(ω(B⁻_S)), then for all bx∈S¹, µ(xb0)−Vf ≤µ(x)b ≤0 and the proof is over. So, suppose that there exists a real number xe^∗ such that Seg

e^x

∗ do not intersect π⁻¹(ω(B⁻_S)). This implies that for any integern, Seg

ex^∗+ (n,0) do not intersectπ⁻¹(ω(B_S⁻)).Let θ be the connected component of ω(B_S⁻) containing (bx0,by0) and let Θ be a component ofπ⁻¹(θ).The set Θ is also a connected component ofπ⁻¹(ω(B⁻_S)), so by lemma 6 it is unbounded. It is now clear that Θ intersects the two horizontal boundaries of [ex^∗+nΘ,ex^∗+nΘ+ 1]×[µ(bx0)−Vf, µ(bx0)] for some integern_Θ,because it can not meet the open half plane{y > µ(e xb₀)}.

Thus,

p1(fe(Θ))

>2 andp2|

f(Θ)e > µ(xb0)−Vf−Af.Asω(B⁻_S) is invariant, π

fe(Θ)

⊂ω(B_S⁻) and so for anyxb∈S¹, µ(bx0)−Vf−Af < µ(x)b ≤0.

The above argument implies that if we chooseM_f =V_f +A_f,then we are done. 2

Now, let us define the number

MDehn= 2 +Bf

k_Dehn >0. (7)

A simple computation shows that for all (x,e ey)∈IR² withy > Me _Dehn,we have

p1◦fe(x,e ey)>xe+ 2 and p1◦fe(ex,−y)e <xe−2.

The construction performed below is analogous for bothω(B_N⁺) andω(B_S⁻).

The details will be presented for ω(B⁻_S). First, note that for every xb ∈ S¹, µ(x)+b

−max bz∈S¹

µ(z) +b M_f

+M_Dehn≥M_Dehn.This means that if we define the following positive integer numberntrans

def.=

−max b^z∈S

1

µ(z) +b Mf+MDehn

+ 1 (bacis the integer part ofa), then the set

ω(B⁻_S)_trans^def.= ω(B_S⁻) + (0, n_trans) (8) has, for every xb ∈ S¹, a point of the form (x,b y),b with by > MDehn. In other words, the function µtrans associated with ω(B_S⁻)trans satisfies µtrans(bx) ^def.= µ(x) +b ntrans> MDehn,for allbx∈S¹.

Now, for a fixed xe∈IR, consider the semi-line {x} ×e [MDehn,+∞[. When we intersect it with

ω(Bg_S⁻)_trans^def.= π⁻¹ ω(B_S⁻)trans

we get that{ex}×]µtrans(π(x)),e +∞[∩ω(Bg⁻_S)_trans=∅ (note thatω(Bg⁻_S)_trans is also afe-invariant set).

Letv ={ex}×]µtrans(π(ex)),+∞[ and let Θ be the connected component of ω(Bg⁻_S)_trans that contains (ex, µtrans(π(x))).e

Lemma 9 : The following holds: Θ∪v is a closed connected set, (Θ∪v)^c has two open connected components, one of which is positively invariant and feⁿ(v)∩v=∅ for all integers n6= 0.

Proof:

The fact that Θ∪v is closed and connected is obvious. As Θ is a connected component of ω(Bg_S⁻)_trans, it is unbounded and limited from above in the y-e direction.

By the Jordan separation theorem, we get that (Θ∪v)^c has at least two connected components, O_L and O_R, defined as follows: For any point Pe ∈v, there exists δ>0 such thatBδ(Pe)∩Θ =∅.Moreover, Bδ(P)\ve has exactly 2 connected components, one to the left ofv,contained inOL and the other one to the right ofv,contained inO_R.So their closures,O_LandO_Rboth containv.

Now, suppose (Θ∪v)^c has another connected component, denotedO^∗.Clearly

∂O^∗ do not intersect vbecause all points sufficiently close to a point in v and, not inv,are contained in O_L∪O_R. So, ∂O^∗ ⊂Θ andO^∗ is then a connected component of Θ^c bounded from above in they-direction. And this contradictse lemma 7. So, (Θ∪v)^c=OL∪OR.

Note thatfe(v)∩v=fe(v)∩Θ =fe⁻¹(v)∩Θ =∅.The paragraph after defini- tion (7) implies thatf(v)e ⊂OR.In the following we will show thatfe(OR)⊂OR.

There are 2 possibilities:

1. ˜f(Θ) 6= Θ⇒ f˜(Θ)∩Θ =∅, because Θ is a connected component of an invariant set;

2. ˜f(Θ) = Θ;

Assume first that ˜f(Θ)∩Θ =∅. Then

fe(Θ∪v)∩(Θ∪v) =∅.

Since fe(v) ⊂ OR and fe(Θ∪v) is connected, we get that fe(Θ∪v) ⊂OR, so O_L ∪Θ∪v is contained either in fe(O_L) or fe(O_R). It can not be contained in fe(OR) because a point of the form (−a, a) for a sufficiently large a > 0 is contained in O_L and fe⁻¹(−a, a) is also contained in O_L, see (6). Thus, OL∪Θ∪v⊂f(Oe L), which implies that,f(Oe R)⊂OR.

Now suppose ˜f(Θ) = Θ. This implies that OL∪v ⊂ (fe(v∪Θ))^c because fe(v)⊂O_R and fe(Θ) = Θ.So, by connectedness,O_L∪v is contained either in

fe(OR) or infe(OL).As in the case ˜f(Θ)∩Θ =∅,one actually getsOL∪v⊂fe(OL) so

fe(O_R)⊂(fe(O_L))^c⊂(O_L∪v)^c=O_R∪Θ and sincefe(Θ) = Θ,we finally get thatfe(OR)⊂OR.

In order to finish the proof, note that, as fe(v)∩v = ∅, for any n ≥ 2, feⁿ(v) ⊂fe(OR), which do not intersectv. So feⁿ(v)∩v = ∅. This finishes the proof of our lemma. 2

Remarks:

• as µtrans(π(x))e < Mf+MDehn+ 2 for all ex∈IR, we get thatfeⁿ({ex} × [Mf+MDehn+ 2,+∞[)∩ {ex} ×[Mf+MDehn+ 2,+∞[=∅for all integers n >0.

• an analogous argument applied to ω(B_N⁺) implies that for any xe∈IR, if w={ex}×]− ∞, ν(π(x))e −

inf b^z∈S

1

ν(z) +b Mf+MDehn

−1[,thenfeⁿ(w)∩ w =∅ for all integersn > 0. So as in the above remark, νtrans(π(ex)) >

−2−Mf−M_Dehnfor allxe∈IR, which implies thatfeⁿ({x}×]e − ∞,−Mf− MDehn−2[)∩ {x}×]e − ∞,−Mf−MDehn−2[=∅for all integersn >0.

Summarizing, there exists a real number M⁰ > 0 such that for all xe ∈IR, feⁿ({x} ×e [M⁰,+∞[)∩ {x} ×e [M⁰,+∞[=∅ andfeⁿ({ex}×]− ∞,−M⁰])∩ {x}×]e −

∞,−M⁰] =∅ for all integersn >0 and

M^0def.= Mf+MDehn+ 2 = 5 + 3Bf

kDehn

+Af+ 2 (9)

Now let us suppose by contradiction that there exists a pointbzin the cylinder and an integern0>0 such that

p2(fbⁿ⁰(bz))−p2(z)b

>2M⁰+ 8.

Without loss of generality, we can assume thatp2(bz)<−M⁰−3 andp2(fbⁿ⁰(z))b >

M⁰+ 3.

Let us also consider the fixed point free mapping of the plane eg(•) =feⁿ⁰(•)−(0,1).

To see that it is actually fixed point free, note that ifeg has a fixed point, then 1/n0 ∈ ρV(fb), a contradiction. Now, note that for allxe ∈IR, eg({x} ×e [M⁰ + 2,+∞[)∩{x}×[Me ⁰+2,+∞[=∅andeg({x}×]−∞,e −M⁰−2])∩{x}×]−∞,e −M⁰− 2] =∅.Moreover, using the fact thategis also the lift of a torus homeomorphism homotopic to a Dehn twist and a compacity argument, one can prove that there exists an integerN >0,such that for all integersn,the sets

F_n⁻= [n/N,(n+ 1)/N]×]− ∞,−M⁰−2]

and

F_n⁺= [n/N,(n+ 1)/N]×[M⁰+ 2,∞[

(10)

are free under eg,that is, eg(F_n^+or−)∩F_n^+or− =∅,for all integers n.Moreover, the fact thatk_Dehn >0 (see the end of section 1) implies that there exists an integerKcrit>0,such that for all integersn

eg(F_n⁺)∩F_m⁺6=∅, for allm≥n+K_crit and

g(Fe _n⁻)∩F_m⁻6=∅, for allm≤n−Kcrit.

These will be important bricks in a special brick decomposition of the plane ineg-free sets we will construct, which will be invariant under integer horizontal translations (x,e y)e →(ex+ 1,ey).

Clearly, such a construction is possible, because as eg(ex+ 1,ey) =eg(ex,y) +e (1,0),we just have to decomposeS¹×[−M⁰−2, M⁰+ 2] into a union of bricks with sufficiently small diameter, so that their pre-images underπareeg-free.

To conclude our proof, we will show that this brick decomposition has a closed brick chain, a contradiction with the fact that eg is fixed point free, see lemma 1. This idea was already used in the proof of theorem 4 of [3].

Consider a point ez ∈ π⁻¹(z) and a brickb F_i⁻₀ that contains ez. From our choices,

g(Fe _i⁻

0)∩F_i⁺

1 6=∅, for some integeri₁.

AsρV(fb) ={0},let us choose a pointwb∈S¹×]M⁰+ 2,+∞[ such that p2(bgⁿ(w))b ^n→∞→ −∞,

where bg(•)^def.= fbⁿ⁰(•)−(0,1) (as ρ_V(bg) = {−1}, all points in S¹×IR satisfy

the above condition). So, we can choose a pointwe∈F_i⁺

2, for some integeri2, such that:

• i₂> i₁+K_crit,so eg(F_i⁺

1)∩F_i⁺

2 6=∅;

• egⁿ²(w)e ∈F_i⁻

3, for some integersn2>0 andi3> i0+Kcrit; As eg(F_i⁻

3)∩F_i⁻

0 6= ∅, we get there exists a closed brick chain starting at F_i⁻

0. As we said, this is a contradiction because eg is fixed point free. Thus fbⁿ(S¹× {0}) ⊂ S¹×[−8−2M⁰,2M⁰+ 8] for all integers n >0. In order to conclude the proof, letKbe the only connected component of the frontier of

n≥0∩ fbⁿ(closure( ∪

m≥0fb^m(S¹×]0,+∞[)))

which does not bound a disc. ThenKis a compact connected set that separates the ends of the cylinder,fb(K+(0, l)) =K+(0, l),for all integersland|p2(K)| ≤ 4M⁰+ 20.2

3.3 Proof of corollary 1

Without loss of generality, by considering fb^q −(0, p), we can suppose that ρV(f) = [a,b 0], for some a < 0. As in the proof of theorem 2, lemma 2 im- plies that B_N⁺ 6= ∅, B_S⁻ 6= ∅ and B_N⁺(inv) 6= ∅, B⁻_S(inv) 6= ∅. If for instance ω(B⁻_S) =∅,then lemma 5 implies thatω(B_S⁻(inv)) =∅and so lemma 4 implies that there exists >0 such thatρV(fb⁻¹)⊃[−,0],which givesρV(fb)⊃[0, ],a contradiction.So, we can assume thatω(B⁺_N)6=∅andω(B_S⁻)6=∅.If we suppose that for everyM > 0, there exists a point bz ∈ S¹×IR and an integer n > 0 such that

p₂(fbⁿ(bz))−p₂(z)b > M,

then following exactly the same ideas used in theorem 2, we arrive at a contra- diction which proves the corollary. 2

3.4 Proof of theorem 3

As in theorem 2, let us fix afe∈ DT(IR²), which is a lift off .b First, we will show that if

M ≥M₀^def.= (20 + 2B_f)/k_Dehn+ 10 (see (6)),

thenfbhas a fixed point. In case fbis fixed point free, lemma 2 of [3] tells us that there exists a homotopically non-trivial simple closed curveγ ⊂S¹×IR such thatfb(γ)∩γ=∅andγ⊂S¹×[−mD, m_D],wherem_D>0 is the smallest real number that satisfies

fe({ex} ×[mD,+∞[)⊂[ex+ 10,+∞[×IR and

fe({x} ×e [−∞,−mD])⊂]− ∞,xe−10]×IR,

(11)

for all xe ∈ IR. A simple computation shows that if we take mD equal (10 + B_f)/k_Dehn,then (11) is satisfied.

So, asM ≥2mD+10,the theorem hypotheses imply thatfbhas a fixed point.

Thus 0∈ρ_V(fb) and lemma 2 implies thatB_N⁺ 6=∅, B_S⁻6=∅and the same holds for the inverse off ,b namely,B_S⁻(inv)6=∅andB_N⁺(inv)6=∅.Ifω(B_N⁺) =∅,then lemma 4 implies that there exists δ > 0 such that ρV(fb) ⊃ [0, δ]. Also, from lemma 5 we get that ω(B_N⁺(inv)) = ∅ and so again by lemma 4, there exists >0 such that ρV(fb⁻¹)⊃[0, ], which givesρV(fb)⊃[−, δ] and the theorem is proved. So, again we can suppose thatω(B_S⁻)6=∅ andω(B_N⁺)6=∅.

Ifρ_V(fb) = [a,0] for somea≤0, then if M ≥M1

def.= 2M⁰+ 8 = 10 + 6Bf

kDehn

+ 2Af + 12,

by the same argument used to prove theorem 2, we arrive at a contradiction.

The same happens in the other possibility, that is, ifρ_V(fb) = [0, b], for some b >0.

So, it is enough to choose

M = max{M0, M₁} ≤ 20 + 6B_f kDehn

+ 2A_f+ 12 to finish the proof.2

3.5 Proof of Corollary 2

Let us start by showing that there are two possibilities:

1) ∪

n≥0fbⁿ(H) is bounded and this means thatρ_V(fb) ={0};

2) ∪

n≥0fbⁿ(H) is unbounded from above and from below;

In order to understand that the above are the only possible cases, suppose for instance that ∪

n≥0fbⁿ(H) is unbounded and contained in H_a⁺ for some real numbera <0.

As in lemma 2, letO^∗= ∪

n≥0fbⁿ(S¹×]0,+∞[) and letObe the complement of the connected component of (O^∗)^c which contains the lower end of the cylinder.

As in that lemma,∂O^def.= Kis a compact connected set that separates the ends of the cylinder. Clearly,O^∗ ⊂O (we just fill the holes), H₁⁺⊂O ⊂H_a⁺, O is an open set homeomorphic to the cylinder andfb(O)⊂O.

Let us state a simple result, but before we present a definition:

Definition : Ifγis a homotopically non trivial simple closed curve inS¹×IR, then γ^{c def.}= γ^−o∪γ^+o, whereγ^−o(+o) is the open connected component of γ^c which contains the lower (upper) end of the cylinder. We define γ^−def.= closure(γ^−o) =γ^−o∪γ and the same forγ⁺.

Proposition 1 : Given an area-preservingf ∈DT(T²)and a liftfb∈DT(S¹× IR) with zero Lebesgue measure vertical rotation number, for any b ∈ IR the following equality holds (in this casefbis said to be exact):

Leb(H_b⁺∩(fb(Hb))⁻) =Leb(H_b⁻∩(fb(Hb))⁺),

where for any measurable setD, Leb(D)^def.= Lebesgue measureof D.

Proof:

If we remember (6), we get that there exists an integerN >0 such that, for any givenb∈IR,fb(Hb)∩(Hb+N ∪H_b−N) =∅.So, consider the finite annulus Ω^def.=S¹×[b, b+N].As it is a finite union of fundamental domains of the torus, we get that

Z

Ω

h

p2◦fb(x,b by)−ybi

dbxdyb= 0 (this follows fromρV(Leb) = 0). (12)

Note that we can write

Ω =

fb(Ω)∩Ω

∪

H_b⁺∩(fb(H_b))^−o

∪

H_b⁻∩(fb(H_b))^+o+ (0, N)

and fb(Ω) =

fb(Ω)∩Ω

∪

H_b^+o∩(fb(Hb))⁻+ (0, N)

∪

H_b^−o∩(fb(Hb))⁺ , where the unions are disjoint. Expression (12) together with the preservation of area imply that they-coordinate of the geometric center of Ω and ofb fb(Ω) are equal. So, let us compute them (for a measurable set Π in the cylinder, we denote theby-coordinate of its geometric center byyb_G.C.(Π)):

yb_G.C.(Ω)=









yb_G.C.(

f(Ω)∩Ω)b .Leb(f(Ω)b ∩Ω)+

+yb

G.C.(H_b⁺∩(^f(Hb ^b))−)

.Leb(H_b⁺∩(fb(H_b))⁻)+

+ by

G.C.(H⁻_b∩(^f(Hb ^b))+) +N

.Leb(H_b⁻∩(fb(Hb))⁺)









/Leb(Ω)

by

G.C.(^f(Ω))b

=









yb_G.C.(

f(Ω)∩Ω)b .Leb(fb(Ω)∩Ω)+

+ yb

G.C.(H_b⁺∩(^f(Hb ^b))−) +N

.Leb(H_b⁺∩(f(Hb b))⁻)+

+by

G.C.(H⁻_b∩(^f(Hb ^b))+)

.Leb(H_b⁻∩(f(Hb _b))⁺)









/Leb(fb(Ω))

AsLeb(fb(Ω)) =Leb(Ω) andyb_G.C.(

f(Ω))b =ybG.C.(Ω), we get that N.Leb(H_b⁺∩(fb(Hb))⁻) =N.Leb(H_b⁻∩(fb(Hb))⁺),

which proves the proposition (note that we used the fact thatLeb(Hb) = 0). 2 Now let us choosec∈IR such that{K∪fb(K)} ⊂interior(H_c⁻∩(fb(Hc))⁻).

From the preservation of Lebesgue measure and the above proposition, we get that

Leb(O∩H_c⁻) =Leb(fb(O)∩(fb(Hc))⁻) =Leb(fb(O)∩H_c⁻).

The choice ofc,together with the fact thatfb(O)⊂O,implies thatclosure(O) = closure(fb(O)) = fb(closure(O)). So ∂(closure(O)) separates the ends of the cylinder and isfb-invariant. But this means that all orbits are uniformly bounded,

a contradiction with our hypothesis that ∪

n≥0 fbⁿ(H) is unbounded. So, either 1) or 2) from the beginning of the proof of the corollary can happen.

And in possibility 2) we can apply theorem 3 to conclude the proof. 2

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