Expansive homeomorphisms of the plane.

# Expansive homeomorphisms of the plane.

J. Groisman
April, 2008
###### Abstract

This article tackles the problem of the classification of expansive homeomorphisms of the plane. Necessary and sufficient conditions for a homeomorphism to be conjugate to a linear hyperbolic automorphism will be presented. The techniques involve topological and metric aspects of the plane. The use of a Lyapunov metric function which defines the same topology as the one induced by the usual metric but that, in general, is not equivalent to it is an example of such techniques. The discovery of a hypothesis about the behavior of Lyapunov functions at infinity allows us to generalize some results that are valid in the compact context. Additional local properties allow us to obtain another classification theorem.

(e-mail: jorgeg@fing.edu.uy)

## 1 Introduction

The aim of this work is to describe the set of expansive homeomorphisms of the plane with one fixed point under certain conditions. The original question that we asked ourselves was whether every expansive homeomorphism of the plane was a lift of an expansive homeomorphism on some compact surface. As it is well known, such expansive homeomorphisms were classified by Lewowicz in [7] and Hiraide in [5]. As a matter of fact, we began by studying whether some of the results obtained in the previously cited article could be adapted to our new context (i.e. without working in a compact environment but having the local compactness of the plane). In this work I study expansive homeomorphisms with one fixed point, singular or not, and without stable (unstable) points. The existence of a Lyapunov function that allows us, among other things, to generalize Lewowicz’s results on stable and unstable sets will be essential. In fact, it will also allow us to obtain a characterization of those homeomorphisms of the plane which are liftings of expansive homeomorphisms on . The result can be tested in any given homeomorphism provided with a suitable Lyapunov function. Although many of the techniques used in this work are valid for the case where there are many singularities, we leave the study of this situation for forthcoming papers.
Let be a homeomorphism of the plane that admits a Lyapunov metric function , meaning continuous and positive (i.e. it is equal to zero only on the diagonal) and positive with . We define as being -expansive if given two different points of the plane the following property holds: for every , there exists such that

 U(fn(x),fn(y))>k.

The main objective of this work is to describe every expansive homeomorphism with one fixed point where some Lyapunov metric function verifies certain conditions concerning . During this work we will require the existence of such a Lyapunov function , unlike in the compact case where expansiveness is a necessary and sufficient condition for the existence of a Lyapunov function (see [7] ). In the previous reference, Lewowicz classifies expansive homeomorphisms on compact surfaces. Our main results is (Theorem 4.2.1): A homeomorphism with a fixed point is conjugate to a linear hyperbolic automorphism if and only if it admits a Lyapunov metric function that satisfies condition HP and it has not singular points. Condition HP establishes that: given any compact set of , the following property holds

 limx→∞|V(x,y)−V(x,z)|W(x,y)=0,

uniformly with in and .
Without condition HP and demanding other kind of conditions for , different behaviors appear. These are described in Theorem 5.1.1: Let be a homeomorphism of the plane with a fixed point. admits a Lyapunov function that verifies hypothesis HL if and only if restricted to each quadrant determined by the stable and unstable curves of the fixed point is conjugated (such conjugations must preserve stable and unstable curves) either to a linear hyperbolic automorphism or to a restriction of a linear hyperbolic automorphism to certain invariant region. The most important part of condition HL establishes that:

• the first difference verifies the following property: given there exists such that if then , ;

• the second difference verifies the following property: given , there exists such that for every on the plane with .

The difference between the two cases (presented in Theorem 5.1.1) consists of the existence of stable and unstable curves that do not intersect each other. In 5.2 we will show examples about the case where there are stable and unstable curves that do not intersect each other.
We also conjecture that if is a preserving-orientation and fixed point free homeomorphism that admits a Lyapunov function satisfying condition HP, then it must be topologically conjugate to a translation of the plane. We believe that the proof of this assertion is a consequence of Brouwer’s translation theorem (see [1], [3]) and of some techniques used in this article. We leave the study of this situation for forthcoming works.
Regarding the structure of the paper, we begin section 2 by studying some properties that are verified by a Lyapunov function associated to a lift of an expansive homeomorphism in the compact case, as well as to homeomorphisms conjugated to it. In section 3 we describe stable and unstable sets, by adapting Lewowicz’s arguments used on [7]. In Section 4 we show our main result. In Section 5 we study an other context and some examples.

## 2 Preliminaries

During the course of this work we will consider homeomorphisms of the plane which admit a Lyapunov metric function with certain characteristics. These properties are natural since they are verified by a Lyapunov function of a lift of an expansive homeomorphism in the compact case. In this section we will verify some of these properties. In [2], [9] is proved that every lift of an expansive homeomorphism on compact surfaces satisfies the existence of pseudo-metrics , and such that for every the following holds:

 Ds(F−1(ξ),F−1(η))=λDs(ξ, η);
 Du(F(ξ),F(η))=λDu(ξ, η),

where is a Lyapunov metric in for .
We will test the following properties:

• Signs for . We shall use the notation for the connected component of set , which contains . For every point and for every , there are points in the border of such that and points in the border of such that . This property will be essential to describe stable and unstable sets.

• Property HP. Let and . Given any compact set , the following property holds

 lim∥x∥→∞|V(x,y)−V(x,z)|W(x,y)=0,

uniformly with in . This property will be essential to prove the main result on this work.

### 2.1 Lifted case.

Let be the Lyapunov metric function that we introduced at the beginning of this section.

• Signs for . Proof:

 ΔD(x,y)=D(f(x),f(y))−D(x,y)=
 Ds(f(x),f(y))−Ds(x,y)+Du(f(x),f(y))−Du(x,y)=
 (λ−1)Du(x,y)−(1−1/λ)Ds(x,y).

For every point and for every , there are points in the border of such that (this is true because the stable set separates the plane). Therefore, as we wanted. A similar argument lets us find points such that .

• Property HP. Proof: Since

 Δ2D(x,y)=ΔD(f(x),f(y))−ΔD(x,y)=
 (λ−1)2Du(x,y)+(1−1/λ)2Ds(x,y),

we can conclude that tends to infinity when tends to infinity. Now,

 |ΔD(x,y)−ΔD(x,z)|≤
 (λ−1)|Du(x,y)−Du(x,z)|+(1−1/λ)|Ds(x,y)−Ds(x,z)|≤
 (λ−1)Du(z,y)+(1−1/λ)Ds(z,y).

Then is uniformly bounded when points and lie on a compact set. Then property HP holds.

### 2.2 Lifted conjugated case.

Now, let us start with the case where is conjugated to a lift of an expansive homeomorphism on a compact surface. Let us define a Lyapunov function for such as

 L(p1,p2)=D(H(p1),H(p2)),

where is the previous defined Lyapunov metric function for and is a homeomorphism from over . It follows easily that is a Lyapunov function for and a metric in .

• Signs for . Proof: It is clear since

 ΔL(p1,p2)=ΔD(H(p1),H(p2)),

and is continuous at infinity.

• Property HP. Proof:

 |Δ(L)(p,q)−Δ(L)(p,r)|=
 |ΔD(H(p),H(q))−ΔD(H(p),H(r))|≤
 (λ−1)Du(H(q),H(r))+(1−1/λ)Ds(H(q),H(r))≤K,

since and are in a compact set and is a homeomorphism. Since

 Δ2(L)(p,q)=Δ2D(H(p),H(q))

and is continuous at infinity we conclude that tends to infinite when tends to infinity. Then, if is conjugated to a lift of an expansive homeomorphism on a compact surface, it admits a Lyapunov function such that condition (HP) holds.

## 3 Stable and unstable sets.

In this section we will stay close to the arguments used by Lewowicz in [7], Sambarino in [8] and Groisman in [4]. We have to adapt them for our non-compact context. We will work with the topology induced by a Lyapunov function and define the -stable set in the following way:

 Sk(x)={y∈IR2: U(fn(x),fn(y))≤k,∀n∈IN}.

Similar definition for the -unstable set . Let be a homeomorphism of the plane that admits a Lyapunov function such that the following properties hold:

• is a metric in and induces the same topology in the plane as the usual metric. Observe that given any Lyapunov function it is possible to obtain another Lyapunov function that verifies all the properties of a metric except, perhaps, for the triangular property.

• Existence of both signs for the first difference of . For each point and for each there exist points and on the border of such that and , respectively.

###### Remark 3.0.1

A homeomorphism that admits a Lyapunov function defined at is U-expansive. This means that given two different points of the plane and given any , there exists such that

 U(fn(x),fn(y))>k.

Proof: Let and be two different points of the plane such that . Since , then holds for . This means that grows to infinity, since

 U(fn(x),fn(y))=U(x,y)+n∑j=1V(fj(x),fj(y))>
 U(x,y)+nV(x,y).

Thus, given there exists such that

 U(fn(x),fn(y))>k.

By using similar arguments we can prove the case when . If , then and this is precisely our first case.

###### Definition 3.0.1

Let be a homeomorphism of the plane that admits a Lyapunov metric function . A point is a stable (unstable) point if given any there exists such that for every , it follows that for each ().

###### Remark 3.0.2

Property for implies the non-existence of stable (unstable) points.

Proof: Given the existence of both signs for in any neighborhood of , we can state that for each , there exists a point in such that . Since , we can state that for , so grows to infinity. Thus, there are no stable points. We can use similar arguments for the unstable case.

###### Remark 3.0.3

There does not exist and such that

 U(fn+1(y),fn+1(x))>U(fn(y),fn(x))

and

 U(fn+1(y),fn+1(x))>U(fn+2(y),fn+2(x)).

Proof: Suppose that there exist two different points and such that they do not verify the thesis. Since

 △(△U)(fn(x),fn(y))=
 U(fn+2(x),fn+2(y))−2U(fn+1(x),fn+1(y))+U(fn(x),fn(y)),

we have that

 △(△U)(fn(x),fn(y))
 U(fn+1(x),fn+1(y))=0,

which is not possible.

###### Lemma 3.0.1

Let be an open set of with . There exists a compact connected set with , such that, for all and , holds.

Proof: Suppose that there exists such that for each compact connected set that joins with the border of , there exists and with such that . Otherwise, for each we would find that joins with the border of such that for every , , . Then

 D∞=∞⋂n=0¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯(∞⋃j=nDj),

is a connected compact set that satisfies our assertion. Let us go back to the prior assumption. Consider a point in the border of that belongs to the region where , this means that . Then , because otherwise, we would contradict the previous remark. So, contains points such that does not belong to . Let us take and a point , like we did before. Let us base our reasoning in the connected component of which contains . Let be an arc such that , , and let be the supremum of such that, for all , , for all and . Since , for all and then belongs to the border of . Hence is a connected compact set that joins with the and remains inside for all , which contradicts our assumption. For and , let be the stable set for , defined by

 Sk(x)={y∈IR2: U(fn(x),fn(y))≤k, n≥0}.
###### Lemma 3.0.2

Let us consider . There exists such that if and , then .

Proof: Let us consider and such that and . Then we state that for each . If there exists such that , then for each since . But then, is unbounded, which contradicts the fact that . But if for each , we have that for all , and then this proves the lemma. Let () be the connected component of () that contains .

###### Lemma 3.0.3

is locally connected at .

Proof: (See Lemma , [7])

###### Corollary 3.0.1

For each in , is connected and locally connected.

Proof: (See [7])

###### Corollary 3.0.2

For any in and any pair of points and in , there exists an arc included in that joins and .

Proof: (See Topology, Kuratowski, [6], section )

###### Definition 3.0.2

We say that has local product structure if a map which is a homeomorphism over its image () exists and there exists such that for all it is verified that and .

###### Proposition 3.0.1

Except for a discrete set of points, that we shall call singular, every in has local product structure. The stable (unstable) sets of a singular point consists of the union of arcs, with that meet only at . The stable (unstable) arcs separate unstable (stable) sectors.

Proof: (See Section , [7])

###### Remark 3.0.4

The neighborhood’s size where there exists a local product structure may become arbitrarily small. However we are able to extend these stable and unstable arcs getting curves that we will denote as and , respectively. If two points and belong to (), then for some and for all ().

The following lemmas refer to these stable and unstable curves.

###### Lemma 3.0.4

Let be a homeomorphism of the plane which verifies the conditions of this section. Then stable and unstable curves intersect each other at most once.

Proof: If they intersect each other more than once, we would contradict expansiveness: if two different points and belong to the intersection of a stable and an unstable curve, then there exists such that for all .

###### Lemma 3.0.5

Every stable (unstable) curve separates the plane.

Proof: Let be a parametrization of a stable curve , such that . We will first prove that , i.e. given any closed neighborhood of there exists such that does not belong to for each . We will work with . Arguments for are identical. Let us assume that there exists such that for each there exists with . Let us consider the set . This set accumulates at a point of . Let us take a large enough such that is close enough to and such that intersects . But then intersects for each , which contradicts the previous lemma (see figure 1).

Then we proved that . This implies that has more than one connected component. Since a stable curve can not auto intersect (because there are no stable points), then there exist exactly two components in the complement of .

## 4 Main section.

In this section, we will prove the main result of this work. Let be a homeomorphism of the plane which verifies the following conditions:

• has a fixed point;

• the quadrants determined by the stable and the unstable curves of the fixed point are -invariant;

• admits a Lyapunov metric function . This metric induces in the plane the same topology as the usual distance;

• has no singularities;

• for each point and any , there exist points and in the border of such that and , where .

• Property HP. Given any compact set the following property holds:

 lim∥x∥→∞|V(x,y)−V(x,z)|W(x,y)=0,

uniformly with in .

### 4.1 Previous lemmas.

###### Lemma 4.1.1

Let be a homeomorphism of the plane under the hypothesis described at the beginning of this section. If the unstable (stable) curve of a point intersects the stable (unstable) curve of the fixed point, then the stable (unstable) curve of intersects the unstable (stable) curve of the fixed point.

Proof: Let us assume that there exists a point in the stable curve () of the fixed point, such that there exists a point that verifies . As there are no singular points, let be the first point in such that its stable curve does not intersect the unstable curve of the fixed point. (See figure 2 )

We will prove that is -invariant. Let us concentrate on one of the quadrants determined by the stable and the unstable curves of the fixed point. Since separates the first quadrant, denote by zone the region that does not include the fixed point in its border, and zone the other region. We will divide the proof in three steps:

• belongs to zone .
If is included in zone (see figure 2), then and the same happens to every point in a neighborhood of . But then is an open set that contains and has the property that the stable curves of their points do not cut the unstable curve of the fixed point (since if some point of verified that its stable curve cuts the unstable curve of the fixed point, then would have the same property, which is a contradiction). Then would not be the first point of such that its stable curve does not cut the unstable curve of the fixed point (see figure 2).

• belongs to zone and .
Consider the simple closed curve determined by the unstable arc , the stable arc , the unstable arc and the stable arc (see figure 3).

intersects the bounded component determined by and intersects in . We will prove that does not intersect in any other point, which be a contradiction since separates the plane. Indeed, can not cut the stable arcs of because they are different stable curves. can not cut the unstable arc in a point different from because we would have a stable curve and an unstable curve cutting each other in two different points, and we already have proved this is not possible. If the intersection point was , we would have a closed curve of a stable arc which would imply the existence of a stable point, and this is not possible. Finally, the intersection of with can not belong to the unstable arc since in this case would not be the first point of that arc such that does not intersect the unstable curve of the fixed point.

• belongs to zone and .
Let us consider a sequence in converging to such that the stable curve of each point cuts the unstable curve of the fixed point.

The behavior of the stable curves of points must be the one shown in the figure 4: as grows the period of time for which they remain close to grows arbitrarily. Denote by the intersections of with . divides the unstable curve in a bounded arc and an unbounded arc. We will prove that belongs to the compact arc for each . If some belongs to the unbounded arc, let us consider the closed curve determined by the stable arc , the stable arc , the unstable arc and the stable arc (see figure 5).

Similar arguments as those used in the previous case allow us to state that can not intersect twice, which is a contradiction. This proves that is in the bounded arc , for every . Let us consider, as figure 6 shows, the segment and let be the point of which is farthest from the fixed point.

must intersect segment . Otherwise, it would cut more than once. Let be that intersection point. We want to apply our condition HP. Observe that points would be in a compact set and tends to infinity when tends to . because they are in the same unstable curve, and because they are on the same stable curve. So, there exists a point that belongs to segment such that . Then

 limn→∞|V(wn,yn)−V(wn,qn)|W(wn,yn)=0.

Therefore, we can choose such that

 W(wn,yn)+V(wn,yn)>0,

which implies that

 V(f(wn),f(yn))>0.

This contradicts the fact that points are in the same stable curve.

Thus the existence of our invariant stable curve is proved. From now on, we will denote it by . Next, we will prove, using again condition HP, that the existence of this invariant stable curve is not possible, so we will end the proof. Let us take any point in the stable curve of the fixed point. We state that the unstable curve through intersects . Otherwise, there would exist a point in the stable curve of the fixed point such that its unstable curve, , is the first one that does not intersect . But then would intersect . This is a contradiction since it is one of the previous cases. Then, we have that every point of the stable curve of the fixed point has the property that its unstable curve intersects . Moreover, as point comes closer to the fixed point, the intersection, , gets closer to infinity, since separates the plane. We want to apply our hypothesis HP. Let us fix a point at the invariant stable curve and a point in the unstable curve of the fixed point (see figure 7).

Then, let us consider close enough to the fixed point, and let be the intersection of with segment . Let be the intersection of with . Reasoning in a similar way to other parts of this proof, we have that because they are in the same unstable curve, and because they are in the same stable curve. So, there exists a point in segment such that . If gets closer to , tends to infinity, and then we are able to choose such that

 |V(zt,y)−V(zt,qt)|W(zt,y)<1,

which implies that

 W(zt,y)+V(zt,y)>0,

and then

 V(f(zt),f(y))>0.

This yields a contradiction since points are in the same stable curve.

###### Lemma 4.1.2

Let be a homeomorphism of the plane that verifies all the conditions described at the beginning of this section. Then the stable (unstable) curve of every point intersects the unstable (stable) curve of the fixed point.

Proof: Let us consider set consisting of the points whose stable (unstable) curve intersects the (unstable) stable curve of the fixed point. It is clear that is open. Let us prove that it is also closed. Let be a sequence of , convergent to some point (see figure 8). Let be a neighborhood of with local product structure.

Let us consider . So, we have that as a consequence of the local product structure and since . But then is a point in that cuts the unstable curve of the fixed point, and then, applying lemma 4.1.1 we have that must cut the stable curve of the fixed point. A similar argument lets us prove that the stable curve of must cut the unstable curve of the fixed point. Therefore belongs to set and consequently is closed. Then is the whole plane.

###### Remark 4.1.1

At this point we can not ensure that every stable (unstable) curve cuts every unstable (stable) curve. Theorem 4.2.1, one of the main results in this work, shows that under our conditions, which means admitting the existence of a Lyapunov metric function with the required hypothesis, every stable (unstable) curve cuts every unstable (stable) curve.

### 4.2 Main result.

In this section we will prove one of the main results of this work. We obtain a characterization theorem (Theorem 4.2.1) of expansive homeomorphisms which verify the set of conditions exposed at the beginning.

###### Proposition 4.2.1

Let be a homeomorphism such that:

• has a fixed point;

• the quadrants determined by the stable and unstable curves of the fixed point are -invariant;

• admits a Lyapunov metric function . induces on the plane the same topology than the usual distance;

• does not have singularities;

• for each point and any there exist points and in the border of such that and , where .

Then, if admits the condition (HP), then is conjugated to a linear hyperbolic automorphism.

Proof: Let be a homeomorphism of the plane that admits a Lyapunov metric function with the given hypothesis. If we proved that every stable curve intersects every unstable curve, we would be able to define a conjugation between a linear hyperbolic automorphism and , in the following way: first, we define sending the stable (unstable) curve of the fixed point of to the stable (unstable) curve of the fixed point of and such that . Since every on the plane is determined by with belonging respectively to the unstable and stable curves of the fixed point, we define . If we prove that every stable curve intersects every unstable curve, we could say that is a homeomorphism of over . Let us prove that every stable curve intersects every unstable curve. Let us suppose that, as shown in figure 9, there exist and ( is the fixed point) such that

 Wu(p2)∩Ws(p1)=∅.

We are also under the assumption that is the first point in such that its stable curve does not intersect .

Thus

 Wu(p2)∩Ws(x)={qx},

were is a point in the unstable arc and near infinity as we want, when tends to . Let be the first difference of the Lyapunov function. As points are in a compact set, we are able to apply condition HP concerning the Lyapunov function in the following way: , since both points are in the same unstable curve and , since both points are in the same stable curve. So, there exists a point on segment such that . Then

 lim∥x∥→p1|V(qx,x)−V(qx,zx)|W(qx,x)=0.

Then, we can choose such that

 W(qx,x)+V(qx,x)>0,

which implies that

 V(f(qx),f(x))>0.

This contradicts the fact that points and are in the same stable curve.

###### Theorem 4.2.1

In the same conditions we had in the previous proposition, is conjugated to a linear hyperbolic automorphism if and only if it admits a Lyapunov metric function satisfying condition (HP).

Proof: It is a consequence of the last proposition and section 2.

###### Corollary 4.2.1

Under the same conditions of the previous theorem, admits a Lyapunov function satisfying condition (HP) if and only if it is conjugated to a lift of an expansive homeomorphism on .

The stable curve and the unstable curve which we built in this article, have the property that given two points of () there exists such that for ( for ). Denote by () the stable (unstable) curves in the usual metric sense, this means that verifies that given two points of () there exists such that for ( for ). Observe that the conjugations that appear in Proposition 4.2.1 send stable (unstable) curves () of the linear automorphism into stable (unstable) curves () of . The following proposition is a necessary and sufficient condition to preserve the stable curves and unstable curves in the sense of the usual metric.

###### Proposition 4.2.2

A necessary and sufficient condition for the conjugation of the theorem 4.2.1 to preserve stable and unstable curves (in the sense of the usual distance of the plane) is that the homeomorphism admits a Lyapunov function that verifies the following property: given any there exists such that implies , for each in .

Proof: Let us consider a function that admits a Lyapunov function with the property of the statement. Because of Theorem 4.2.1, is conjugated to a linear hyperbolic automorphism and the conjugacy maps linear stable (unstable) curves into stable (unstable) curves of in the sense of function . But, precisely this stable (unstable) curve satisfies for some and every (). Then, using the property, we have that there exists such that , which proves that preserves stable and unstable curves in the sense of the usual metric . Let us suppose now that preserves stable and unstable curves in the sense of the usual metric. Let us consider two points such that , where (). Let be the intersection of the unstable curve of with the stable curve of (see figure 10).

As implies , then

 Ds(H(x),H(q))=D(H(x),H(q))

and

 Du(H(y),H(q))=D(H(y),H(q))

Since preserves stable and unstable curves,

 d(x,q)

and

 d(y,q)

with a uniform . This implies that there exists uniform such that which concludes the proof.

###### Corollary 4.2.2

A homeomorphism of the plane with the conditions given in this section is the time of a flow.

Proof: We have that

 f=H−1φ1H,

where is a linear automorphism. Then we can consider the flow

 ψt=H−1φtH

and this implies that is .

## 5 Another context and some examples.

In this section we will show some generalizations about the hypothesis we asked for our homeomorphisms. The main difference with what we have exposed until now, is that we will work without condition HP. Instead of this we will ask for uniform local conditions concerning the first and the second difference ( and ) of the Lyapunov function . In this new context, we will get a new characterization result in Theorem 5.1.1, which shows two possible behaviors of our homeomorphisms. At