Expansive and fixed point free homeomorphisms of the plane

The aim of this work is to describe the set of fixed point free homeomorphisms of the plane under certain expansive conditions.


Introduction.
In [1], necessary and sufficient conditions for a homeomorphism of the plane with one fixed point to be topologically conjugate to a linear hyperbolic automorphism was proved. The discovery of a hypothesis about the behavior of Lyapunov functions at infinity was essential for this purpose. In this work we will describe the set of fixed point free homeomorphisms of the plane which admit a Lyapunov metric function with the same condition imposed in the previously cited article. These homeomorphisms are expansive respect to the Lyapunov metric U, meaning U : IR 2 × IR 2 → IR continuous and positive (i.e. it is equal to zero only on the diagonal) and W = ∆(∆U) positive with ∆U(x, y) = U(f (x), f (y)) − U(x, y). As it is well known, expansive homeomorphisms on compact surfaces were classified by Lewowicz in [2] and Hiraide in [7]. 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). These arguments will allow us to construct singular transverse foliations F s and F u . The main result of this article, theorem 4.0.3, states: Let f be a homeomorphism of IR 2 which is fixed point free and admits a Lyapunov function U : IR 2 × IR 2 → IR such that the following properties hold: U is a metric in IR 2 and induces the same topology in the plane as the usual metric, For each point x ∈ IR 2 and for each k > 0 there exist points y and z on the boundary of B k (x) such that V (x, y) = U(f (x), f (y)) − U(x, y) > 0 and V (x, z) = U(f (x), f (z)) − U(x, z) < 0, respectively, f does not admit singularities, f is topologically conjugate to a translation of the plane if and only if U admits condition HP which states that given any compact set C ⊂ IR 2 the following property holds: uniformly with y, z in C. Let´s take as an example a translation of the plane and show that admits a Lyapunov metric function with the condition HP. For this aim we will adapt some arguments used by White in [6]. Let T be a translation of the plane defined by T (x, y) = (x + 1, y). Fix λ > 1 and consider the Riemannian structure defined by where e s = (1, 0) and e u = (0, 1). The metric induced on the plane by this Riemannian metric is defined by where γ is an arc joining points p and q. The topology induced by d σ on the plane is the same that is induced by the usual metric. Let γ = (γ 1 , γ 2 ) : [0, 1] → IR 2 be a curve joining points p and q of IR 2 . We can consider two pseudo metrics D s and D u defined by By a simple computation we can prove that and D u (T (p), T (q)) = λD u (p, q).
Consider the Lyapunov metric D of IR 2 defined by D = D s + D u . Let´s test some conditions required for D: For every point x ∈ IR 2 and for every k > 0, there are points y in the boundary of B k (x) such that D u (x, y) = 0 (this is true because the stable set separates the plane). Therefore, ∆D(x, y) < 0 as we wanted. A similar argument lets us find points z ∈ IR 2 such that ∆D(x, z) > 0.
(II) Property HP. Let V = ∆D and W = ∆ 2 D. We´ll prove that given any compact set C ⊂ IR 2 , the following property holds we can conclude that ∆ 2 D(x, y) tends to infinity when x tends to infinity. Now, Then |∆D(x, y) − ∆D(x, z)| is uniformly bounded when points y and z lie on a compact set. Then property HP holds.

Constructing foliations.
Let f : IR 2 → IR 2 be a homeomorphism of the plane that admits a Lyapunov metric function U, meaning U : IR 2 × IR 2 → IR continuous and positive (i.e. it is equal to zero only on the diagonal) and W = ∆(∆U) positive with ∆U(x, y) = U(f (x), f (y)) − U(x, y). In this section we will resume some results of Lewowicz in [2], and Groisman in [1]. We will work with the topology induced by a Lyapunov function U and define the k-stable set in the following way: Similar definition for the k-unstable set. Let f be a homeomorphism of the plane that admits a Lyapunov function U : IR 2 × IR 2 → IR such that the following properties hold: (1) U is a metric in I R 2 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.
(2) Existence of both signs for the first difference of U. For each point x ∈ IR 2 and for each k > 0 there exist points y and z on the boundary of B k (x) such that Remark 2.0.1 A homeomorphism f that admits a Lyapunov function U defined at IR 2 × IR 2 is U-expansive. This means that given two different points of the plane x, y and given any k > 0, there exists n ∈ Z such that U(f n (x), f n (y)) > k.
Proof: Let x and y be two different points of the plane such that V (x, y) = U(f (x), f (y))− U(x, y) > 0. Since ∆V > 0, then V (f n (x), f n (y)) > V (x, y) holds for n > 0. This means that U(f n (x), f n (y)) grows to infinity, since Thus, given k > 0 there exists n ∈ IN such that By using similar arguments we can prove the case when ) > 0 and this is precisely our first case.

Remark 2.0.2 Property (2) for U implies the non-existence of stable (unstable) points.
Proof: Given the existence of both signs for V (x, y) = U(f (x), f (y)) − U(x, y) in any neighborhood of x, we can state that for each k > 0, there exists a point y in B k (x) such that V (x, y) > 0. Since ∆V > 0, we can state that V (f n (x), f n (y)) > V (x, y) for n > 0, so U(f n (x), f n (y)) grows to infinity. Thus, there are no stable points. We can use similar arguments for the unstable case.
There exists a compact connected set C with x ∈ C ⊂ A, C ∩ ∂(A) = ∅ such that, for all y ∈ C and n ≥ 0, U(f n (x), f n (y)) ≤ k holds.
Proof: See section 3 [1] For x ∈ IR 2 and k > 0, let S k (x) be the k−stable set for x, defined by Proof: See section 3 [1] Let C k (x) be the connected component of S k (x) that contains x.
The following lemmas refer to these stable and unstable curves. Proof: If they intersect each other more than once, we would contradict U-expansiveness: if two different points x and y belong to the intersection of a stable and an unstable curve, then there exists k 0 > 0 such that U(f n (x), f n (y) < k 0 for all n ∈ Z.
Lemma 2.0.5 Every stable (unstable) curve separates the plane.
Proof: See section 3 [1] Finally, we have conditions to state the following theorem: Theorem 2.0.1 Let f be a homeomorphism of the plane that admits a Lyapunov function U : IR 2 × IR 2 → IR such that the following properties hold: U is a metric in IR 2 and induces the same topology in the plane as the usual metric.
For each point x ∈ IR 2 and for each k > 0 there exist points y and z on the Then, f admits transverse singular foliations F s and F u . Leaves of F s (F u ) are the stable (unstable) curves constructed in this section.

Foliation description.
In this section we will describe those foliations introduced in the previous section with one additional condition for the Lyapunov function U: Property HP. Let V = ∆U and W = ∆ 2 U. Given any compact set C ⊂ IR 2 , the following property holds uniformly with y, z in C.
In [1] these foliations were characterized in the case when the homeomorphism f does not admit singularities and it has a fixed point. Now we will generalize these results for the case when f is fixed point free. for each point x ∈ IR 2 and any k > 0, there exist points y and z in the boundary of Property HP. Given any compact set C ⊂ IR 2 the following property holds: uniformly with y, z in C.
Then, there exist transverse foliations F s and F u such that every leaf of F s intersects every leaf of F u .
Proof: Let F s and F u be the stable and unstable foliations constructed in the last section. Let x an arbitrarily point of IR 2 and denote by W s (x) and W u (x) the leaves of F s and F u through x. We will divide this proof in two steps: If not let z be the first point of W u (y) such that W s (z) ∩ W u (x) = ∅ (see fig. 1).
Let (z n ) be a sequence of W u (y) such that z n converges to z and W s (z n )∩W u (x) = ∅, for all n. As n grows, the behavior of W s (z n ) has two possibilities as shown in fig 2. In the first situation we could find points w n = W s (z n ) ∩ W u (x) arbitrarily close to infinity. Since for each n we have that V (w n , z n ) < 0 (because they belong to the same stable leaf) and V (w n , x) > 0 (because they belong to the same unstable leaf), there exists a point q n , belong to the segment line z n x, such that V (w n , q n ) = 0.

Figura 2:
Since z n and q n belong to a compact set for all n, we can apply condition HP and find n ∈ IN such that |V (w n , z n ) − V (w n , q n )| W (w n , z n ) < 1, which implies that W (w n , z n ) + V (w n , z n ) > 0, and then V (f (w n ), f (z n )) > 0.
This yields a contradiction since points f (w n ), f (z n ) are in the same stable leaf. In the second situation (see fig 3), the set {w n : n ∈ IN} is bounded. Let us consider, as figure 3 shows, the segment line w n z n and let h n be the point of the arc belong to W s (z n ) determined by the points w n and z n , which is farthest from the point x.
W u (h n ) must intersect segment w n z n . Otherwise, it would cut W s (z n ) more than once. Let r n be that intersection point. We want to apply our condition HP. Observe that points r n , z n would be in a compact set for all n ∈ IN and h n tends to infinity when z n tends to z. V (h n , r n ) > 0 because they belong to the same unstable leaf, and V (h n , z n ) < 0 because they belong to the same stable leaf. So, there exists a point q n that belongs to segment line z n r n such that V (h n , q n ) = 0. Then Therefore, we can choose h n such that This contradicts the fact that points f (h n ), f (z n ) are in the same stable leaf.
W u (y)∩W s (x) = ∅ and W s (y)∩W u (x) = ∅, for every point y in IR 2 . Let us consider the set A consisting of the points whose stable (unstable) leaf intersects the unstable (stable) leaf of point x. It is clear that A is open. Let us prove that it is also closed. Let (q n ) be a sequence of A, convergent to some point q (see figure 4).

Figura 4: Coordinates
Let V (q) be a neighborhood of q with local product structure. Let us consider q n 0 ∈ V (q). So, we have that W s (q n 0 ) ∩ W u (q) = α n 0 as a consequence of the local product structure and W s (q n 0 ) ∩ W u (p) = ∅ since q n 0 ∈ A. But then α n 0 is a point in W s (q n 0 ) that cuts the unstable leaf of point x, and then, applying the previous step we have that W u (q) = W u (α n 0 ) must cut the stable leaf of point x. A similar argument lets us prove that the stable leaf of q must cut the unstable leaf of point x. Therefore q belongs to the set A and consequently A is closed. Then A is the whole plane.
Since x is an arbitrary point, this proof is finished.

Main section.
Let f be a fixed point free homeomorphism of IR 2 . Brouwer's translation theorem (see [3], [4]) asserts that if f preserves orientation, then every x 0 ∈ IR 2 is contained in a domain of translation for f , i.e. an open connected subset of IR 2 whose boundary is L ∪ f (L) where L is the image of a proper embedding of IR in IR 2 , such that L separates f (L) and f −1 (L). The purpose of this section is to find some domain of translation of a fixed point free homeomorphism of IR 2 which admits a Lyapunov metric function U such that properties presented in the last section hold. This situation will allow us to prove that the homeomorphisms discussed in theorem 3.0.2 are topologically conjugate to a translation of the plane. U is a metric in IR 2 and induces the same topology in the plane as the usual metric; for each point x ∈ IR 2 and for each k > 0 there exist points y and z on the boun- Proof: If U = IR 2 then there exists a point p and a sequence (x n ) such that x n ∈ f n (W s ) n > 0(n < 0) and lím x n = p. Let W s (p) be the leaf of F s through p. Using continuity of F s respect to the initial point we could state that every point of W s (p) is the limit of a sequence (y n ) such that y n ∈ f n (W s ) n > 0(n < 0). Thus, W s (p) is included in the boundary of U. If W s (p) is not invariant then all the iterates of W s (p) are also included in the boundary of U.
Lemma 4.0.7 Let f be a homeomorphism of IR 2 which is fixed point free and admits a Lyapunov function U : IR 2 × IR 2 → IR such that the following properties hold: U is a metric in IR 2 and induces the same topology in the plane as the usual metric, for each point x ∈ IR 2 and for each k > 0 there exist points y and z on the boun- f has no singularities.

Figura 5:
Proof: Let W s be an arbitrary non invariant leaf of F s and consider the leaves f (W s ) and f −1 (W s ). Let's suppose that none of these three leaves separates the others (see fig.  5). Take any point x in W s . Applying theorem 3.0.2 we have that W u (x) must intersect transversally f (W s ) and f −1 (W s ). Also, recall that in lemma 2.0.4 we proved that stable and unstable leaves intersect each other at most once. Then W u (x) goes from one component to the other determined by W s only once (remember that W s separates the plane). Since we are assuming that W s does not separate f (W s ) and f −1 (W s ), then these two leaves are in the same component determined by W s . So, if W u (x) intersects f (W s ) then it can not intersect f −1 (W s ) (because if it does, we would have either more than one intersection between W u (x) and f (W s ) or an auto intersection of W u (x)). That yields a contradiction. Then, one of the three stable leaves separates the other two. Then W s must separate f 2 (W s ) and f −2 (W s ).
Proof: Let's suppose that f (W s ) separates W s and f −1 (W s ). As fig. 6 shows let us consider an unstable arc a joining a point x of f (W s ) with a point y of W s . Since f n (W s ) must separate f n−1 (W s ) and f n−2 (W s ) we conclude that f n (W s ) must intersect the compact arc a in a point x n . Using lemma 4.0.7, we can state that sequence x 2n is monotone and bounded. Let's denote by z its limit. Since W s (z) separates f −2 (W s ) and f 2 (W s ) we conclude that W s (z) is f 2 -invariant. The other case is analogous. U is a metric in IR 2 and induces the same topology in the plane as the usual metric, for each point x ∈ IR 2 and for each k > 0 there exist points y and z on the boun- Reciprocally, let´s consider the translation of the plane T defined by T (x, y) = (x + 1, y).
It was proved at the introduction that T admits a Lyapunov metric function D with the conditions required in this theorem. Now, let´s see the case when f is conjugated to a translation T . Let us define a Lyapunov function for f such as L(p 1 , p 2 ) = D(H(p 1 ), H(p 2 )), where D is the previous defined Lyapunov metric function for T and H is a homeomorphism from IR 2 over IR 2 . It follows easily that L is a Lyapunov metric function for f such that property HP holds. This conclude the prove of the theorem.
Using the results of [1] and the last theorem we can state the following general characterization theorem: