Wandering continua for rational maps

We prove that a Latt' es map admits an eventually simply-connected wandering continuum precisely when it is flexible. The simply-connected wandering continuum is a line segment in a bi-infinite geodesic under the flat metric.


Introduction
Let f be a rational map of the Riemann sphere C with deg f ≥ 2. Denote by J f the Julia set of f and F f the Fatou set of f . One may refer to [Mi1] for their definitions and basic properties. By a wandering continuum we mean a non-degenerate continuum K ⊂ J f (i.e. K is a connected compact set consisting more than one point) such that f n (K) ∩ f m (K) = ∅ for any n > m ≥ 0.
The existence of wandering continua for polynomials has been studied by many authors. For a polynomial with disconnected Julia set, it is proved that all wandering components of its Julia set are points [BH, KS, QY]. For a polynomial with connected Julia set, it is proved that a polynomial without irrational indifferent periodic cycles has no wandering continuum if and only if the Julia set is locally connected [K].
The situation for non-polynomial rational maps is different. There is a hyperbolic rational map with disconnected Julia set such that it has non-degenerate wandering components of its Julia set. The first example was given by McMullen, where the wandering Julia components are Jordan curves ([Mc2]). In fact, it is proved that for a geometrically finite rational map, a wandering component of its Julia set is either a Jordan curve or a single point [PT].
In this work we study wandering continua for rational maps with connected Julia sets. Let E ⊂ C be continuum. E is called simply-connected if C\E is connected, or 2connected if C\E contains exactly two components. The following results were proved in [CPT].

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Theorem A. Let f be a post-critically finite rational map. Let E ⊂ J f be a wandering continuum. Then E is simply-connected or 2-connected. Moreover, either (1) f n (E) is simply-connected for all n ≥ 0, or (2) There exists an integer N ≥ 0 such that f n (E) is 2-connected for n ≥ N .
Theorem B. Let f be a post-critically finite rational map. Then f admits a 2-connected wandering continuum if and only if f has a Cantor multicurve. In this case the 2-connected wandering continuum must be a Jordan curve.
Refer to [CPT] for the definition of Cantor multicurves. As a result, we are interested to the following problem: Problem: Under what condition does a post-critically finite rational map f admit an eventually simply-connected wandering continuum?
In this paper, we solve this problem for Lattès maps (refer to §2 for its definition).
Here is the main theorem: Theorem 1.1. A Lattès map f admits an eventually simply-connected wandering continuum if and only if it is flexible. In this case the wandering continuum is a line segment in an infinite geodesic under the flat metrics.

Lattès maps
This section is a brief review about Lattès maps. Refer to [Mc1,Mi1] or [Mi2] for details. Let f : C → C be a rational map with degree deg f ≥ 2. Denote by deg z f the local degree of f at z, the critical set of f and Let f be a post-critically finite rational map. Define ν f (z) for each point z ∈ C to be the least common multiple of the local degrees deg y f n for all n > 0 and y ∈ C such that f n (y) = z. By convention ν f (z) = ∞ if these local degrees are unbounded, equivalently the point z is in a super-attracting cycle. The orbiford of f is defined by O f = ( C, ν f ).
Note that ν f (z) > 1 if and only if z ∈ P f . The signature of the orbifold O f is the list of the values of ν on P f . The Euler Characteristic of O f is given by .
1. Power maps. Suppose that the signature of O f is (∞, ∞). Then P f is exactly the set of exceptional points. So the map f is holomorphically conjugate to a power map z → z d with |d| ≥ 2.
2. Chebyshev maps. Suppose that the signature of O f is (2, 2, ∞). Then the point z 0 ∈ C with ν f (z 0 ) = ∞ is an exceptional point of the map f . Let O be the orbifold with underlying space C and the multiplicity map ν(z) = 2 on {2, −2} and ν(z) = 1 otherwise. Then there exists a bi-holomorphic map γ : O → O f \{z 0 }. Denote by τ (z) = z + 1/z. Then there exist a covering P : C\{0} → C\{0} and a finite covering Φ : O → O such that the following diagram commutes: Note that P fixes the infinity and maps 3. Lattès maps. A post-critically finite rational map f with parabolic orbifold is called a Lattès map if ν f (z) = ∞ for any point z ∈ C. Thus its Julia set is the whole sphere. Denote by ν(O f ) = max{ν f (z), z ∈ C}. Refer to [Mi1] Theorem 3.1 for the next theorem.
Theorem 2.1. Let f be a Lattès map. Then there exist a lattice Λ = {n + mω, n, m ∈ Z} (Im ω > 0), a finite holomorphic cover Θ : C/Λ → O f , a finite cyclic group G of order ν(O f ) generated by a conformal self-map ρ of C/Λ with fixed points, and an affine map for some n ∈ Z, and the following diagram commutes: Suppose now that f is a Lattès map. Then C/Λ is a torus. Let ρ : C → C be a lift of the generator ρ of G under the natural projection π : C → C/Λ. Let z 0 ∈ C be its fixed point. Then When #P f = 3, the complex structure of O f is uniquely determined. Therefore if the map f is topologically conjugate to another rational map, then they are holomorphically conjugate.
Now assume that #P f = 4, i.e., the signature of O f is (2, 2, 2, 2). Denote by Q ⊂ C/Λ the set of fixed points of the generator ρ of G. Then #Q = 4 and Θ(Q) = P f . Let ρ : C → C be a lift of the generator ρ of G. Let z 0 be the fixed point of ρ. Then ρ(z) = 2z 0 − z. Therefore (1)

Consider the affine map
Equivalently, there exist integers (p, q, r, s) such that a = p + qω, and aω = r + sω. ( It follows that If a is a real number, then q = r = 0 and a = p = s. Thus a must be an integer and the equation (2) holds for any complex number ω. This shows that one may make a quasi-conformal deformation for the map f to get another rational map such that they are not holomorphically conjugate.
If a is not real, then q = 0 and thus the complex number ω with Im ω > 0 is uniquely determined by the integers (p, q, r, s) from equation (3). This shows that if the map f is topologically conjugate to another rational map, then they are holomorphically conjugate.
A Lattès map f is called flexible if it admits a non-trivial quasi-conformal deformation, or equivalently, if its orbifold O f has siganature (2, 2, 2, 2) and the affine map A(z) = az + b (mod Λ) : C/Λ → C/Λ in Theorem 2.1 has integer derivative A = a ∈ Z.

Wandering continua of torus coverings
Let Λ = {n + mω : n, m ∈ Z} with Im ω > 0. A continuum E ⊂ C/Λ is simplyconnected if there exists a simply-connected domain U ⊂ C/Λ such that E ⊂ U and U \E is connected. Let π : C → C/Λ be the natural projection. If E ⊂ C/Λ is a simplyconnected continuum, then so is each component of π −1 (E). In this section, we will prove the next theorem which implies Theorem 1.1 by Theorem 2.1.
Then the map A admits an eventually simply-connected wandering continuum E if and only if its derivative a is an integer. In this case, the wandering continuum E is a line segment in an infinite geodesic under the flat metric.
The proof of this theorem is based on the following lemmas.
Lemma 3.2. Let E ⊂ C/Λ be a simply-connected continuum. Then for any line L ⊂ C and any connected component B of π −1 (E), if I is a bounded component of L\B, then π is injective on I.
Proof. Assume by contradiction that there exists a bounded component I = (x 1 , y 1 ) of L\B and two distinct points x, y ∈ I such that π(x) = π(y). Then there are exactly two components U, V of C\(L ∪ B) such that their boundaries contain the interval I. We claim that at least one of them, denoted by U , is bounded. Otherwise one can find a Jordan curve γ in U ∪ V ∪ I such that γ separates the point x 1 from the point y 1 . Since γ is disjoint from B and both x 1 and y 1 are contained in B, this contradicts that fact that B is connected.  Set T (z) = z + (x − y) and H(G 0 ) = [x 2 , y 2 ]. Then there exists an integer n such that T n (x 2 ) ∈ [x, y] and hence T n (y 2 ) / ∈ I. Let H be the component of C\L that contains U , then T n (G 0 ) is a continuum in H ∪ L connecting T n (x 2 ) and T n (y 2 ), whereas G 0 is a continuum in H ∪ L connecting x 2 and y 2 . Thus G 0 must intersect T n (G 0 ).

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On the other hand, since π(x) = π(y), we have (x − y) ∈ Λ. Thus T n (B) is another component of π −1 (E) and hence is disjoint from B. This contradicts the facts that G 0 ⊂ B and G 0 intersects T n (G 0 ). Lemma 3.3. Let A(z) = az + b (mod Λ) : C/Λ → C/Λ be a covering with deg A ≥ 2 and E ⊂ C/Λ be an eventually simply-connected wandering continuum. Then E must be a line segment.
Proof. Let B be a component of π −1 (E). Assume by contradiction that B is not a line segment. We claim that there exists a line L ⊂ C such that L\B has a bounded component I. Otherwise, each line segment connecting two points in B must be contained in B. Thus B is convex and hence has positive measure since it is not a line segment. This is impossible since A is expanding and E is wandering.
As in the proof of Lemma 3.2, there exists a bounded component U of C\(L ∪ B) such that I ⊂ ∂U . Denote by z 0 ∈ C the fixed point of the map α(z) = az + b : C → C. Denote by Γ 0 = {n + mω + z 0 : n, m ∈ Z} and Γ n = α −n (Γ 0 ). Then there exists two distinct points x n , y n ∈ U ∩ Γ n for some integer n ≥ 0 such that for the line L n that passes through the points x n , y n , the set L n ∩ U has a component I n which contains both x n and y n , and both ends of I n are contained in B. Now consider the simply-connected continuum α n (B) and the line α n (L n ). The set α n (L n )\α n (B) has a component α n (I n ), which contains x = α n (x n ) and y = α n (y n ). Since x n , y n ∈ Γ n , we have x, y ∈ Γ 0 and hence π(x) = π(y). This contradicts Lemma 3.2.
Lemma 3.4. Let A(z) = az + b (mod Λ) : C/Λ → C/Λ be a torus covering with degree d ≥ 2. If a is not real, then any line segment in C/Λ is not wandering. Proof. Let E ⊂ C/Λ be a line segment. We want to show that there exists an integer n > 0 such that A n (E) intersects A n+1 (E).
Let R be the parallelogram with vertices 0, 1, ω and 1 + ω. Then R is a fundamental domain of the group Λ. Thus for any n ≥ 0, the set π −1 (A n (E)) has a component B n such that the midpoint of the line segment B n is contained in the closure of R.
Denote by a = |a| exp(iθ). Since a is not real, we have 0 < |θ| < π and hence | cos θ| < 1. Thus the angle formed by the two lines containing B n and B n+1 respectively, is |θ|. Note that the diameter of R is less than 1 + |ω|. If B n is disjoint from B n+1 , we have: where |B n | is the length of B n . But |B n | = |a| n |B 0 | → ∞ as n → ∞. This is impossible. Now suppose that A(z) = az + b (mod Λ) : C/Λ → C/Λ is a covering with deg A ≥ 2 and a is an integer. Let L ⊂ C be a line. Then either π is injective on L and π(L) is an infinite geodesic under the flat metric, or π(L) is a Jordan curve on C/Λ. In the former case, either π(L) is eventually periodic under A or wandering under A. The next Lemma is easy to check and we omit its proof. Proof of Theorem 1.1. Let f be a Lattès map. By Theorem 2.1, there exist a lattice Λ = {n + mω, n, m ∈ Z} with Im ω > 0, a finite holomorphic cover Θ : C/Λ → O f , a finite cyclic group G of order ν(O f ) generated by a conformal self-map ρ of C/Λ with fixed points, and an affine map A(z) = az + b (mod Λ) : C/Λ → C/Λ, such that Θ(z 1 ) = Θ(z 2 ) ⇔ z 1 = ρ n (z 2 ) for some n ∈ Z, and the following diagram commutes: is a simply-connected wandering continuum of the map A. Therefore the constant a is an integer and K is a line segment in an infinite geodesic under the flat metric by Theorem 3.1.
Let E n be a component of Θ −1 (f n (K)). Then ρ(E n ) is also a component of Θ −1 (f n (K)), where ρ is the generator of the group G. Let R be the parallelogram with vertices 0, 1, ω and 1 + ω. Then R is a fundamental domain of the group Λ. Thus there are components I n , J n of π −1 (E n ) and π −1 (ρ(E n )), respectively such that the midpoints of I n and J n are contained in R.
Assume that #P f = 3. Let ρ be a lift of the map ρ under the projection π. Then ρ is a rotation around its fixed point with angle (2π)/ν, where ν = 3, 4 or 6. Thus the angle formed by the two lines containing I n and J n respectively, is (2π)/ν. As in the proof of Lemma 3.4, we have: where |I n | is the length of I n . This leads to a contradiction since |I n | = |a| n |I 0 | → ∞ as n → ∞. Therefore #P f = 4 and f is a flexible Lattès map.
Conversely, suppose that the map f is flexible. Denote by Q the set of fixed points of ρ.
If π(L) ⊂ C/Λ is a wandering line, then A n (π(L)) is disjoint from Q for all n ≥ 0. Thus Θ is injective on each line A n (π(L)). On the other hand, if Θ(A n (π(L))) intersects Θ(A n+p (π(L))) for some integer p > 0, then they coincide since the map ρ in G preserves the slopes of the lines. Thus A n+p (π(L))) = ρ(A n (π(L)). Therefore A n+2p (π(L)) = ρ(A n+p (π(L)) since A • ρ = ρ • A. But ρ 2 is the identity. So A n+2p (π(L)) = A n+p (π(L)). This is a contradiction. Therefore Θ(A n (π(L))) is disjoint pairwise. Thus for any line segment E ⊂ π(L), the set Θ(E) is a simply-connected wandering continuum of f . Now suppose that π(L) ⊂ C/Λ is an eventually periodic line with period p ≥ 1. Then Θ(A n (π(L))) are either bi-infinite geodesics or rays depending on whether A n (π(L)) passes through a point in Q. Since ρ 2 is the identity, either they are disjoint and have the same period, or two of them coincide and the period is p/2. Let E ⊂ π(L) be a wandering line segment. In the former case, Θ(E) is a simply-connected wandering continuum of f . In the latter case, there exists a line segment E 0 ⊂ E such that Θ(E 0 ) is a simply-connected wandering continuum of f .