RATIONAL MAPS WITH DISCONNECTED JULIA SET by Kevin …tanlei/papers/disconn.pdf · RATIONAL MAPS...

RATIONAL MAPS WITH DISCONNECTED JULIA SET

by

Kevin PILGRIM & TAN Lei

Abstract. — We show that if f is a hyperbolic rational map with disconnected Julia

set J , then with the possible exception of finitely many periodic components of Jand their countable collection of preimages, every connected component of J is apoint or a Jordan curve. As a corollary, every component of J is locally connected.We also discuss when a Jordan curve Julia component is a quasicircle and give an

explicit example of a hyperbolic rational map with a Jordan curve Julia componentwhich is not a quasicircle.

Resume. — Soit f une fraction rationnelle hyperbolique. On suppose que son en-semble de Julia J n’est pas connexe. Nous allons montrer que, a l’exception d’unnombre fini de composantes periodiques de J , et la collection denombrable de leurs

composantes preimages, toute composante de J est soit un point soit une courbede Jordan. Par consequent, toute composante de J est localement connexe. Nousdiscutons egalement quand une telle courbe de Jordan est aussi un quasi-cercle. Nousdonnerons un exemple explicit d’une fraction rationnelle ayant une composante de

Julia qui est une courbe de Jordan mais pas un quasi-cercle.

1. Introduction

For a rational map f of the Riemann sphere C to itself with disconnected Juliaset J , we investigate the topological and geometric possibilities for a connected com-ponent of J . If J is disconnected, then f maps components of J onto componentsof J , and there are uncountably many such components (cf. [Mi] and [Be]). Thepostcritical set of f

P :=⋃n>0

f ′(c)=0

f◦n(c)

2000 Mathematics Subject Classification. — Primary 58F23, Secondary 30D05.Key words and phrases. — iteration of rational maps, Julia set, connected components, Jordancurve.

2 KEVIN PILGRIM & TAN LEI

plays a crucial role in our study. We say that f is hyperbolic if P∩J = ∅, geometricallyfinite if P ∩ J is finite, and nice if P ∩ J is contained in finitely many connectedcomponents of J .

Theorem 1.1. — Let f be a polynomial with disconnected filled Julia set K. As-sume that only finitely many connected components of K intersect P. Then, with thepossible exception of finitely many periodic components and their countable collectionof preimages, every connected component of K is a point.

Using a variant of this theorem, we establish

Theorem 1.2. — Let f be a hyperbolic rational map with disconnected Julia setJ . Then, with the possible exception of finitely many periodic components and theircountable collection of preimages, every connected component of J is either a pointor a Jordan curve.

The same result for geometrically finite maps can be proved by similar methods aswell. We will only sketch the necessary modifications at the end of the paper.

We establish also weaker results for nice maps. The precise statements are givenin Propositions [Case 2], [Case 3], [Case 4] and Theorem 9.2.

It is known that J can be a Cantor set ([Be], §1.8), or homeomorphic to theproduct of a Cantor set with a quasicircle, where each component is a K-quasicirclefor some fixed K independent of the component ([Mc1]). In §8 we give an explicitexample of a hyperbolic rational map which has a Jordan curve Julia componentwhich is not a quasicircle.

Results from plane topology imply that at most countably many Julia componentscontain an embedding of the letter “Y ”. Our theorems make precise which ones theyare. It is also interesting to see that at most countably many Julia components canbe a segment, which a priori is not a restriction from plane topology alone.

By a theorem of McMullen ([Mc1], Corollary 3.5), there are at most countablymany periodic components of J . Since periodic points of f are dense in J , if Jis disconnected there must be exactly countably many periodic Julia components.Since the degree of f is finite there must be exactly countably many preperiodic Juliacomponents. Hence there are uncountably many wandering Julia components. Ourtheorems show that under the stated assumptions, no wandering Julia component canbe a segment or contain an embedding of the letter “Y ”, since they must either bepoints or Jordan curves.

Combining the above theorem with a result of Tan-Yin ([TY]), which shows thatevery preperiodic Julia component for a hyperbolic rational map is locally connected,we get:

Corollary 1.3. — For a hyperbolic rational map, each Julia component is locallyconnected.

This corollary completes another entry in the growing dictionary between the the-ories of rational maps and Kleinian groups. The analogs of a hyperbolic rational mapf and its Julia set J are a convex compact (or expanding) Kleinian group Γ andits limit set Λ. It is known that each component of Λ is locally connected; the proof

RATIONAL MAPS WITH DISCONNECTED JULIA SET 3

depends on the fact that a “wandering” component of Λ, i.e. a component with trivialstabilizer, is necessarily a point. See [AM] and [Mc3] (Theorem 4.18).

The main ideas in our proof are a canonical decomposition C = E t U , whereE is a finite collection of Julia components such that f(E) ⊂ E, and the fact thata hyperbolic rational map f is uniformly expanding on a neighborhood of its Juliaset with respect to the Poincare metric on C − P. Our goal is to show that anyJulia component which does not land in E is either a point or a Jordan curve. Tothis end, we further decompose the sphere into several canonical pieces and measurethe itinerary of the orbit of a Julia component J0 under f with respect to thesepieces. We use a combinatorial analysis combined with the lemma below to showthat components with certain kinds of itineraries are points. A separate argumenttreats the case of Jordan curves.

We say that K ⊂ C is a full continuum if K is compact, connected, and C − Kis connected. The following Lemma was essentially known to Fatou (see [Br], Thm.6.2)

Lemma 1.4. — (Fatou) Let f be a rational map, Q =⊔k

i=1 Qi be the union offinitely many disjoint full continua, such that Q ∩ P = ∅. Then any connected setJ ⊂ J satisfying fn(J) ⊂ Q for infinitely many n is a point.

Contents. In §2 we give some motivating examples, define the above mentioneddecomposition and state four basic lemmas for nice maps, and give a more precisestatement (Theorem 1.2’) of Theorem 1.2. We then reduce the proof of Theorem 1.2’to three cases, Cases 2, 3 and 4. Related results for nice maps are stated as well. In§3 we analyze the topology and dynamics of the decomposition and prove the fourbasic lemmas. §4 contains analytic preliminaries for use in §§5 and 6. In §§5, 6, and7 we prove the Propositions in Cases 2, 3, and 4, respectively; §7 contains also theproof of Theorem 1.1. §8 lists related results and discusses when a Jordan curve Juliacomponent is a quasicircle. §9 contains sketches of proofs–a generalization our resultsto the geometrically finite case and some further results for nice maps. §10 is anappendix of technical topological results used in our proofs.Acknowledgments. Recently G. Cui, Y. Jiang, and D. Sullivan [CJS] have alsoproven Theorem 1.2 for geometrically finite maps in a different context. Their meth-ods are in some respects similar, but they do not make use of a canonical decomposi-tion. The authors would like to thank Cui for providing a copy of their manuscript,A. Douady, D. Epstein, M. Lyubich, C. McMullen, B. Sevennec and M. Shishikurafor many useful discussions, and MSRI for financial support.

2. The decomposition and the reduction to three cases

Let f be a rational map with Julia set J = J (f) and postcritical set P = P(f).For J ′ a continuum (i.e. a compact, connected set) in C, and P a compact set

disjoint from J ′, we say that J ′ separates P if either J ′ ∩ P 6= ∅, or J ′ ∩ P = ∅ andthere are at least two components of C− J ′ intersecting P . We say that J ′ separates


Figure 1. Julia set of 1z◦ (z2 − 1) ◦ 1

z+ 10−11z−3

P into exactly q parts if J ′ ∩P = ∅ and C−J ′ has exactly q components intersectingP .

Let J ′ be a component of J . We say that J ′ is critically separating if J ′ separatesP. We say that two distinct components J ′ and J ′′ of J are parallel, if they are bothcritically separating and the unique annulus component in C− (J ′ ∪J ′′) (see Lemma3.1) does not intersect P.Definition (decomposition C = E t U , first step). Let E be the union of Juliacomponents J ′ such that either

1. J ′ ∩ P 6= ∅, or2. J ′ ∩ P = ∅ and J ′ separates P into three or more parts, or3. J ′ separates P into exactly two parts and J ′ separates no two Julia components

which are parallel to J ′, i.e. all Julia components J ′′ parallel to J ′ are containedin the same component of C − J ′.

We think of J ′ as an extremal Julia component. Set U = C − E. The set E may beempty, e.g. if J is a Cantor set.Examples. Let f0(z) = z2 + 10−9z−3 (McMullen) and f1(z) = 1

z ◦ (z2 − 1) ◦ 1z +

10−11z−3. The Julia set of f1, in log(z)-coordinates, is shown in Figure 1. The Juliaset of f0 is homeomorphic to product of a Cantor set with a quasicircle ([Mc1]).

For f1, the point at infinity and −1 form the unique attracting cycle. There arefive critical points in the annular Fatou component near the center of the picture,which maps to the Fatou component containing zero (at left) by degree five. P iscontained in the union of the disc Fatou component containing zero (at left) and theimmediate basins of infinity (at far right) and -1 (the prominent disc at right) . Theset E consists of a homeomorphic copy J+ of the Julia set of z2 − 1, at right, and itspreimage J− (at left) which is a threefold cover of J+. J+ and J− are parallel, J−

maps to J+, and J+ is fixed.

Lemma 2.1. — If f is nice, the set E consists of at most finitely many Julia com-ponents, f(E) ⊂ E, f−1U ⊂ U , and every component of E is either periodic orpreperiodic.

This lemma will be proved in the next section. One can also show that E is closedand forward-invariant for any rational map f ; we omit the proof.


Definition (decomposition of U , second step) DefineA =

⋃{ annular components of U disjoint from P };

D =⋃

{ disc components of U disjoint from P };L =

⋃{components of U not in A or D }

=⋃

{ non-disc non-annular components of U , or components of U intersecting P }.D′ =

⋃{ disc components of U disjoint from P but intersect f−1E};

D′′ =⋃

{ disc components of U disjoint from P ∪ f−1E }.Note that U = A tD t L = A tD′ t D′′ t L.

Example. Let f = f1. Then U = C − E is decomposed into:– A = a single annulus, bounded by J− and J+ (A is not a closed annulus).– L = three disc components, each intersecting P. Two are disc components of

U containing the attractor at infinity and −1 with boundaries contained in J+,and the other is the component of U containing zero with boundary containedin J−.

– D = D′′ = the countable set of remaining components of U , all discs withboundaries in E.

– D′ = ∅.The set f−1E consists of E plus two other components contained in A and parallelto components in E.Definition (decomposition of f−1A, third and final step) We denote by

– AS , the union of components A′ of f−1A such that A′ ⊂ A and A′ ↪→ A is nothomotopic to a constant map, and

– AO, the union of components A′ of f−1A such that A′ ⊂ A and A′ ↪→ A ishomotopic to a constant map.

For f1, the set AS consists of two essential subannuli of A, and the set AO is empty.Here is a more precise statement of Theorem 1.2.

Theorem 1.2’ Let f be a hyperbolic rational map and C = EtU be the decompositionabove. Let J0 be a Julia component. Set Jn = fnJ0. Then exactly one of the followingoccurs:

1. Jn ⊂ E for n ≥ n0, in which case J0 is preperiodic, or2. Jn ⊂ AS for n ≥ n0, in which case J0 is a Jordan curve, or3. there is a sequence nk → ∞ such that Jnk

⊂ U − AS, in which case J0 is apoint.

If E = ∅, the Julia set J is totally disconnected.This decomposition is canonical and natural with respect to conjugation by Mobius

transformations. While the first decomposition C = U tE is the same for any iterateof f , the further decomposition can change.

The following lemmas will be proved in the next section:

Lemma 2.2. — If f is nice, the sets L, A, D′, AS and AO all have finitely manycomponents.

Lemma 2.3. — Let f be a nice map. Then each component L of L contains a uniqueFatou component W such that ∂W ⊃ ∂L and W ∩ P = L ∩ P.


Lemma 2.4. — Let f be a nice map. Then every Julia component J0 is in one ofthe following four cases: let Jn = fn(J0),Case 1. There is n0 such that Jn ⊂ E for n ≥ n0.Case 2. There is n0 such that Jn ⊂ AS for n ≥ n0.Case 3. Jn ⊂ AO ∪ D′ for infinitely many n.Case 4. Jn ⊂ L for infinitely many n.

While these cases cover all possibilities, the last two are not mutually disjoint.

This result tegether with Lemma 2.2 means that some finite part of the decom-position encodes a significant portion of the orbit of each Julia components. We aregoing to prove:Proposition [Case 2]. In Case 2, J0 is a Jordan curve if f is hyperbolic, or C−J0

has exactly two components if f is nice.Proposition [Case 3]. In Case 3, J0 is a point if f is hyperbolic, or C − J0 isconnected if f is nice.Proposition [Case 4]. In Case 4, J0 is a point if f is nice (in particular if f ishyperbolic).

Our cases are also distinguished by our methods of proof. In Case 2, we extracta dynamical system consisting of a finite collection of annuli and covering maps andanalyze this restricted system. Case 3 is similar to Case 2. Case 4 is more delicate.We actually prove a stronger result, Theorem 7.1, from which both Proposition [Case4] and Theorem 1.1 follow as corollaries.

Theorem 1.2 and Theorem 1.2’ are direct consequences of the above Propositions.

3. Topology and dynamics of the decomposition

Throughout this section, f denotes a nice rational map. We first prove Lemmas2.1 and 2.2, and then analyze the topological and dynamical possibilities for the setsin our decomposition in order to prove Lemmas 2.3 and 2.4. We will frequently usethe following result from plane topology (see [Ne] for a proof):

Lemma 3.1. — For a nonempty set J of disjoint continua J ′ in S2, every compo-nent of S2 − J ′ is a disc (simply connected). Given J ′ and J ′′ two disjoint continua,the set S2 − (J ′ ∪ J ′′) has a unique annulus component A(J ′, J ′′), the component J ′′

is contained in a component U ′ of S2 − J ′, and

U ′ = A(J ′, J ′′) t J ′′ t⋃

{V | V is a component of S2 − J ′′ and V ∩ J ′ = ∅ } .

If J0 is a continuum disjoint from J ′∪J ′′ but separating J ′ and J ′′, then A(J ′, J ′′) =

A(J ′, J0)tA(J0, J ′′)tJ0t⋃

{V | V is a component of S2−J0 and V ∩(J ′∪J ′′) = ∅ } .

Proof of Lemma 2.1. — Since f is nice, there exists a compact set B ⊂ C such that1. B has finitely many connected components,2. B ⊃ P, and each connected component of B intersects P,3. B contains every Julia component intersecting P and no other Julia components.


B may be taken to be the union of Julia components intersecting P together withthe suitable preimages of the following: closed, forward-invariant neighborhoods ofattracting and superattracting basins; closed, forward-invariant attracting parabolicpetals, invariant closed sub-discs of Siegel discs containing points of P, and invariantsub-rings of Herman rings containing points of P. Then a Julia component J iscritically separating if and only if it is either contained in B, or is disjoint from Pand separates components of B.

An easy induction argument shows that the number of Julia components J ′ whichseparate B into three or more pieces is finite and bounded by k − 2 if B has kcomponents. Such Julia components are in E by definition.

Now we deal with the Julia components in E that separate B into exactly twoparts. By a method similar to the above, one can prove that if each continuum of theset is disjoint from B and separates B into exactly two parts, and no two continua areparallel (relative to B), then this set of continua is finite. Now assume J0 ⊂ E andJ0 separates P into two parts. We will see that among the Julia components parallelto J0 at most one of them is contained in E. Let J1, J2 be two distinct parallels ofJ0. Since J0 does not separates its parallels, J1 and J2 are contained in the samecomponent of C − J0. There are two possible cases:

a. The component J1, say, separates J0 from J2. Then, according to Lemma 3.1,J1 is contained in A(J0, J2) and J1 ↪→ A(J0, J2) is homotopically non trivial. SinceA(J0, J2) ∩ P = ∅, the component J1 is also parallel to J2. So J1 separates P intoexactly two parts, and separates parallels of J1. Thus J1 is not contained in E.

b. We have J1 ⊂ A(J0, J2) but J1 does not separate J0 ∪ J2 (therefore J2 ⊂A(J0, J1) but J2 does not separate J0 ∪ J1). This is impossible for the followingreason: The annulus A(J0, J2) contains J1 and all but one (disc)-components ofC − J1. Since J1 is critically separating, at least one of these components containspoints of P. So A(J0, J2)∩P 6= ∅. This is a contradiction to the assumption that J0

and J2 are parallel.So E contains at most finitely many components that separate B into two or three

parts, hence E has at most finitely many components.The following lemma implies that f(E) ⊂ E; since U = C − E, f−1U ⊂ U .

Therefore each component of E is either periodic or preperiodic. If E is not empty,it consists of finitely many periodic cycles of Julia components, and some (finitelymany) of their preimages.

Proof of Lemma 2.2. — If E = ∅, we have L = C and A = D′ = ∅. There is nothingto prove.

Assume now E 6= ∅. Since f−1E has finitely many components, so is D′. Itremains to prove that A ∪ L has finitely many components. Using the notation inthe proof of Lemma 2.1, the set P is contained in B, which consists of finitely manyconnected components and contains finitely many Julia components. Let U be adisc-component of L intersecting P. Then either U contains a component of B, orU contains a preimage of a closed parabolic attracting petal containing points of Pused in the construction of B. Since the number of components of B and the numberof such petals is finite, the number of disc-components of L intersecting P is finite.


The other components of A∪L are precisely the non simply connected components ofC−E. Since E consists of finitely many components, only finitely many componentsof C − E can be non simply connected.

Lemma 3.2. — If a Julia component J ′ is critically separating then f(J ′) is alsocritically separating. If f(J ′′) of a Julia component J ′′ separates P into two parts andseparates the parallels of f(J ′′), then either J ′′ does not separate P or J ′′ separatesthe parallels of J ′′ and separates P into two parts.

Proof. — We prove that if f(J ′) is not critically separating then neither is J ′. IfJ ′ ∩ P 6= ∅ then f(J ′) ∩ P 6= ∅. Hence we may assume that J ′ and f(J ′) are disjointfrom P. Assume that P is contained in one (disc)-component of C − f(J ′). Thereis a Jordan curve γ separating f(J ′) and P. Let C be the disc-component of C − γcontaining f(J ′). Since C ∩ P = ∅, every component of f−1(C) is again a disc, andagain disjoint from P since f−1P ⊃ P. One of them, say bounded by γ′, contains J ′.Thus J ′ does not separate P.

Now we prove the second statement of the lemma. By assumption C − f(J ′′) hasexactly two components U1 and U2 intersecting P, and there are Julia componentsJ1 ⊂ U1 and J2 ⊂ U2 parallel to f(J ′′). Then A(J1, J2) ∩ P = ∅, since, accordingto Lemma 3.1, A(J1, J2) is the union of A(J1, f(J ′′)), A(f(J ′′), J2), f(J ′′) and thecomponents of C− f(J ′′) distinct from U1 and U2, and none of these sets intersectsP.

Now each component of f−1(A(J1, J2)) is again an annulus, and again disjointfrom P. One of them, say A′′, contains J ′′, and the components of ∂A′′ are containedin two distinct Julia components J ′

1 and J ′2 which are separated by J ′′.

Since all but two components of C− J ′′ are contained in A′′, and A′′ ∩P = ∅, thecomponent J ′′ separates P into at most two parts. If it does separate P, so do J ′

1

and J ′2. Hence J ′

1 and J ′2 are parallel to J ′′.

Lemma 3.3. — Suppose f : S2 → S2 is a branched covering, U, V ⊂ S2 are finitelyconnected open subsets with f(U) = V and f |U : U → V proper. Then f(∂U) = ∂Vand f maps connected components of ∂U onto connected components of ∂V .

Remark. A subtlety is that the map f : ∂U → ∂V need not be open in the subspacetopology, in other words, a component of ∂U may be a proper subset of a componentof f−1(∂V ).

Proof. — That f(∂U) ⊂ ∂V follows by properness. Since f(U) = V, f(∂U) ⊂ ∂V ,and f(U) is a closed subset containing V , we have f(∂U) = ∂V . Finally, let K ′ bea boundary component of U and let K be the component of ∂V containing f(K ′).Since f is a branched covering and V has finitely many boundary components, thereare open annuli A′ ⊂ U,A ⊂ V such that K ′,K are boundary components of A′, Arespectively and f : A′ → A is a covering map. Then f |A′ : A′ → A satisfies thehypotheses of the first conclusion, so f(∂A′) = ∂A. It follows easily that f(K ′) =K.


Lemma 3.4. — Given any integer n and any component V of f−nU , any Juliacomponent is either contained in V or disjoint from V . The boundary ∂V has finitelymany components, and each component is contained in a different Julia componentwhich is disjoint from V .

Proof. — fn(V ) is a component of U and the mapping fn : V → fn(V ) is properand finite-to-one. Hence V has finitely many boundary components since fnV ∈ Uhas finitely many boundary components. Since fn preserves the Julia components, aJulia component intersecting both V and ∂V would be mapped to a Julia componentintersecting both fn(V ) and ∂fn(V ), which is impossible.

Lemma 3.5. — Any Julia component J0 not in E separating E ∪P is disjoint fromP and critically separating.

Proof. — Since J0 is not in E, if it separates E it must separate components of E.Assume that J1 and J2 are two Julia components in E, separated by J0. Then thereis a component U i of C−J0 containing J i, i = 1, 2, and U1 ∩U2 = ∅. By Lemma 3.1the set U1 contains all but one component of C−J1. Since J1 is critically separating,U1 ∩ P 6= ∅. Similarly U2 ∩ P 6= ∅. Thus J0 separates P.

Assume now J1 is a Julia component in E and x ∈ P such that J1 and x areseparated by J0. One can show similarly that J0 is also critically separating.

Lemma 3.6. — Critically separating Julia components are contained in EtAEtAS.

Proof. — Here we denote by UV the set of z ∈ U for which f(z) ∈ V .We show at first that those Julia components are contained in EtA. If J ′∩P 6= ∅

then J ′ ∈ E by definition. Thus we may assume that J ′ ∩ P = ∅ and that J ′ iscritically separating and is not in E. Then there are exactly two components U1 andU2 of C − J ′ meeting P, and each U i contains parallels of J ′. Set P i = U i ∩ P. Wewill construct a component of A containing J ′.

For i = 1 and 2, let Wi =

{ U | U is a component of the complement of some Julia component and U∩P = P i }.

Set W i =⋃

U∈WiU .

Here we apply the topological result Lemma A.1 to conclude that W i is an opendisc which is also an element of Wi and is in fact the unique maximal element. Becauseeach U i contains parallels of J ′, we have ∂W 1 ∩ ∂W 2 = ∅ and W 1 ∪ W 2 = C. ThusW 1 ∩W 2 is an annulus. To show that this annulus is a component of A, we just needto show ∂(W 1 ∩ W 2) ⊂ E and W 1 ∩ W 2 ∩ E = ∅.

Denote by J i the Julia component containing ∂W i, i = 1, 2.Assume by contradiction that J1, say, is not contained in E. Since W 1, as a

component of C − J1, meets only part of P, the Julia component J1 is criticallyseparating. By definition of E, the only possibility for J1 not being in E is that thereis another component W of C − J1 such that P 2 = P − P 1 is contained in W , andthere is a Julia component J0 ⊂ W parallel to J1. Thus C − J0 has a componentU containing J1 ∪ W 1. Furthermore U ∩ P = W 1 ∩ P = P 1 (Lemma 3.1). In other


words, U is also an element of W1. This contradicts the fact that W 1 is the maximalelement of W1.

Thus ∂(W 1∩W 2) ⊂ E. Now any critically separating Julia component in W 1∩W 2

would also separate J1 and J2, therefore separate P into two parts and be parallelto both J1 and J2. So W 1 ∩ W 2 contains no component of E. As a consequence,W 1∩W 2 is an annulus component of C−E disjoint from P. By definition, W 1∩W 2

is a component of A.So J ′ ⊂ A. Thus every critically separating Julia component is contained in EtA.Now A is decomposed into AE t AS t AO t AL t AD. For any Julia component

J ′ in AL tAD, we have f(J ′) ⊂ L t D. Hence f(J ′) is not critically separating. ByLemma 3.2, the component J ′ is not critically separating either. Since the inclusionmap of each component of AO into A is homotopic to a constant map, and A isdisjoint from P, no continuum in AO can be critically separating.

Thus all critically separating Julia components are contained in E tAE tAS .

Lemma 3.7. — For U a component of L t D, there is a unique component UR off−1U which we call a reduced component with the following properties:

1. UR ⊂ U , ∂U ⊂ ∂UR, and each component of ∂U is a component of ∂UR;2. If U ∩ f−1E = ∅ then UR = U . Otherwise UR is the complement in U of the

union of finitely many disjoint full continua, each of which is contained in U ;3. (U − UR) ∩ P = ∅;4. If U is a component of L then f(UR) is also a component of L. In particular,

f(∂L) ⊂ ∂L.5. f(UR) is a component V of U , f(∂UR) = ∂V , and f maps connected components

of ∂UR onto connected components of ∂V .6. There are finitely many components U in D such that U ∩ f−1(P ∪E) 6= ∅. For

any such U , f(UR) is a component of L ∪ A.

Proof. — 1 and 2. Note that f−1U = C − f−1E. If U ∩ f−1E = ∅, the set U isalso a component of C − f−1E (since f−1E ⊃ E), and we set UR = U . Otherwise,let C1, . . . , Ck denote the components of f−1E which are contained in U . Lemmas3.5 and 3.6 imply that no Ci separates components of ∂U . Hence for each i, there isa unique component Vi of C − Ci containing ∂U . Let Ki = C − Vi and K = ∪iKi.Lemma 3.1 implies that either Ki ∩ Kj = ∅ or Ki ⊂ Kj or Kj ⊂ Ki. Each Ki is fullsince Vi is connected. Then UR := U − K has the first two properties in the lemma.3. Now we show that no component of ∂UR separates (P ∩ U) ∪ ∂U . This is trivialfor components of ∂UR which are also components of ∂U . For the other componentsof ∂UR, if this does not hold, there would be a Julia component J ′ in U separatingP ∪E, and thus J ′ would be critically separating (Lemma 3.5). This is impossible byLemma 3.6. Therefore UR ∩ P = U ∩ P and (U − UR) ∩ P = ∅.4. Assume now that U is a component of L. f(UR) is a component of U . Bydefinition of L, either U ∩ P 6= ∅ or U is neither a disc nor an annulus. In the firstcase f(UR) ∩ P ⊃ f(UR ∩ P) = f(U ∩ P) 6= ∅, so f(UR) is again a component of L.In the second case, either UR ∩ f−1P 6= ∅ (in which case f(UR) ∩ P 6= ∅ and hencef(UR) is in L), or f : UR → f(UR) is an unbranched covering. Since U is not a disc


or annulus neither is UR. Any covering over a disc (resp. an annulus) is again a disc(resp. an annulus), so f(UR) is neither a disc nor an annulus and hence f(UR) is acomponent of L.5. This follows immediately from the properness of f : UR → V and Lemma 3.3.6. Suppose U ∩ f−1(P ∪ E) 6= ∅. If U ∩ f−1E 6= ∅, then U is a component ofD′. Otherwise U contains a Fatou component intersecting P. By the No WanderingDomains theorem, the number of such Fatou components is finite, hence the numberof components U of the latter type is finite. Combining with the fact that D′ hasonly finitely many components (Lemma 2.2), we get the finiteness. Now if f(UR) wasan element of D, then f(UR) would be an open disc disjoint from P and E, henceU would be an open disc disjoint from f−1(P ∪ E). Hence f(UR) is a component ofA ∪ L.

Lemma 3.8. — Let A be a component of A and δ+ be a component of ∂A. Thenthere is a unique component A+ of f−1U with the following properties:

1. A+ ⊂ A and δ+ ⊂ ∂A+;2. Either A+ is a component of AS or A+ is a component of AL, i.e. f(A+) is a

component of A or L;3. f(∂A+) = ∂f(A+) and f maps connected components of ∂A+ onto connected

components of ∂f(A+). Thus f maps boundary components of components ofAS onto boundary components of components of A or L.

The proof is similar to the one above.

Lemma 3.9. — Assume E 6= ∅. For U a component of U , the set f(U) is again acomponent of U if and only if U ∩ f−1E = ∅. Either there is a minimal integer k ≥ 0such that fkU ∩ f−1E 6= ∅, or some iterate V of U is a periodic Fatou component.In the latter case, V is finitely connected, is itself a component of U , and either

1. V ∩ P 6= ∅, and V is a component of L which is either a simply-connectedattracting or parabolic basin, or a Siegel disc or Herman ring intersecting P, or

2. V ∩ P = ∅, and V is either a Siegel disc and a component of D, or a Hermanring and a component of A.

Proof. — If U ∩ f−1E 6= ∅ then f(U) ∩E 6= ∅, and so f(U) can not be a componentof U = C − E. Otherwise U ∩ f−1E = ∅ and so U is a component of f−1U whichmaps properly under f onto a component of U .

Assume now that for every n ≥ 0, fn(U) ∩ f−1E = ∅. Then fnU is a componentof U for every n ≥ 0. By Montel’s theorem the family {fn|U}n is then normal (sinceE is uncountable if it is nonempty), so U coincides with a Fatou component (since∂U ⊂ J ). By the No Wandering Domains Theorem, some iterate V of U is a periodicFatou component, and so V is a component of L. Since components of U are finitelyconnected, V is finitely connected. The Lemma then follows from the classification ofperiodic Fatou components and the fact that attracting or parabolic basins are eithersimply connected or infinitely connected ([Be], §7.5).

The following lemma is a more precise version of Lemma 2.3:


Lemma 3.10. — Assume E 6= ∅. Let L0, L1, · · · , Lm be the (finitely many) compo-nents of L. Then each Li contains a unique Fatou component Wi such that ∂Wi ⊃∂Li. Moreover the components of ∂Li are precisely the components of ∂Wi separatingP.

We say f∗(Li) = Lj if f(LRi ) = Lj. In this case every Li is preperiodic under

f∗ and f(Wi) = Wj. Furthermore, if {L0, · · · , Lp−1} is a periodic cycle of f∗, theneither

1. LRi = Li for all 0 ≤ i ≤ p − 1, in which case Wi = Li and either(a) Wi ∩ P 6= ∅ for all i, in which case Wi is a simply connected attracting

or parabolic basin, a Siegel disc or a Herman ring intersecting P, or(b) Wi ∩ P = ∅ for all i, in which case Wi is a Siegel disc or Herman ring

disjoint from P; or2. LR

i 6= Li for some i, in which case Wi is an infinitely connected attracting orparabolic basin for each 0 ≤ i ≤ p − 1.

Proof. — By Lemma 2.2 the set L has only finitely many components. By Lemma 3.6no Julia component separates ∂Li (resp. ∂LR

i ) or is critically separating. CorollaryA.5 implies that there is a unique Fatou component Wi (resp. WR

i ) such that Wi ⊂ Li

and ∂Wi ⊃ ∂Li (resp. WRi ⊂ LR

i and ∂WRi ⊃ ∂LR

i ).By Lemma 3.7, we have ∂LR

i ⊃ ∂Li, thus by uniqueness with respect to theproperty of containing ∂Li, we have WR

i = Wi.Note that every connected component of ∂Wi is either contained in ∂Li or is

contained in Li. Since no Julia component in Li is critically separating, and everycomponent of ∂Li is critically separating, the components of ∂Wi separating P areprecisely those in ∂Li.

If f∗(Li) = Lj , by uniqueness of Wj , we have f(Wi) = f(WRi ) = Wj .

By Lemma 3.7, for any i, f(LRi ) is again a component of L, thus coincides with

some Lj . So f∗(Li) is well defined for each i. Since there are only finitely manycomponents in L, each of them is eventually periodic under f∗.

Let {L0, · · · , Lp−1} be a periodic cycle of f∗. Then {W0, · · · , Wp−1} forms a peri-odic cycle of Fatou components.

If LRi = Li, 0 ≤ i ≤ p−1, then the conclusion (1) follows by Lemma 3.9. Otherwise,

LRi has at least two boundary components, hence Wi has at least two boundary

components. Wi cannot be a Siegel disc or Herman ring. For in these cases, ∂Wi∩P 6=∅ and so Li = Wi is itself a component of U , contradicting LR

i 6= Li for some i. HenceWi is either an attracting or parabolic basin with at least two boundary components,hence is infinitely connected.

For our example f = f1 above, L has a periodic cycle of period 2 formed by theFatou components containing infinity and -1.

Proof of Lemma 2.4. — Since f(E) ⊂ E, if Jn0 ⊂ E for some n0, then Jn ⊂ E forall n ≥ n0. This is our case 1.

Assume now Jn ⊂ U for all n. Assume furthermore that J0 is not in Case 2, thatis, Jn ∩ AS = ∅ for infinitely many n. We are going to show Jn ⊂ AO ∪ D′ ∪ L forinfinitely many n (so J0 is in Case 3 or 4 or both).


We show at first that Jn ⊂ AO ∪D ∪L = AO ∪D′ ∪D′′ ∪L for infinitely many n.Denote by AB the set of z ∈ A for which f(z) ∈ B. We have Jn ∩ AE = ∅ for all n.If Jn ⊂ AD ∪AL for some n, then Jn+1 ⊂ D ∪L. Since U = AE ∪AS ∪AO ∪AD ∪AL ∪ D ∪ L, we are done.

We now show that if Jn ⊂ D′′ for infinitely many n, then Jn ⊂ L∪D′ for infinitelymany n. Assume Jn1 ⊂ D for D a component of D′′. By definition, D ∩ f−1E = ∅.By Lemma 3.7, either f(D) is a component of D′ ∪ D′′ (this corresponds to the caseD ∩ f−1P = ∅), or f(D) is a component of L. As a consequence of Sullivan’s non-wandering domain theorem, there is an integer 0 < k < ∞ such that D, f(D), · · · ,fk−1(D) are components of D′′ and fk(D) is a component of D′ ∪ L. ThereforeJn1+k ⊂ D′ ∪ L.

4. Analytic preliminaries

We now restrict to the case when f is hyperbolic. The results generalize to geomet-rically finite maps; the Poincare metric ρ is replaced by a more complicated metricfor which the map is still expanding (cf. [TY] and §9).

Recall that f is hyperbolic if and only if J ∩P = ∅. If |P| = 2 then f is conjugateto zn and J is connected. Moreover, C − P is connected. Let ρ|dz| denote thePoincare Riemannian metric on C−P, dρ(x, y) the corresponding distance, and lρ(γ)the length of a curve with respect to ρ. Then f : C− f−1(P) → C−P is expandingwith respect to ρ. If B is the subset given in the proof of Lemma 2.1, then since f ishyperbolic we have P ⊂ B ⊂ L ∩ (C − J ), and f : C − f−1(int(B)) → C − int(B)expands ρ uniformly by some definite factor λ > 1. The inverse of f is then uniformlycontracting, in the following sense: if γ : [0, 1] → C− int(B), then lρ(γ) < (1/λn)lρ(γ)for any lift γ of γ under fn. This observation will be the main tool in our proofs ofPropositions [Case 2] and [Case 3].

If U is a path-connected subset of C − P, we define the path metric dpathU (x, y)on U with respect to ρ by

dpathU (x, y) = infγ{lρ(γ)|ρ : [0, 1] → U, γ(0) = x, γ(1) = y}.

Lemma 4.1. — Let J ′ be a periodic or preperiodic Julia component and U a com-ponent of C − J ′. Then there is a C1 Jordan curve γ : S1 → U − B, a continuoussurjective map h : S1 → ∂U and a constant L depending on U , such that for eacht ∈ S1, one can find a path ηt : [0, 1] → U with ρ-length at most L, such thatηt(]0, 1]) ⊂ U , ηt(0) = h(t) and ηt(1) = γ(t).

Recall that we have a partition E t U of C, and A t D is the union of disc andannulus components of U disjoint from P. Moreover ∂(A t D) ⊂ E, and E consistsof finitely many preperiodic Julia components.

Corollary 4.2. — Each component U of A,D,AO, and AS has finite path-diameterrelative to ρ, and each has locally connected boundary.

Proof. — We first assume that J ′ is fixed and that the ideal boundary (cf. [Mc1])of U is also fixed by f . In other words, there is a component U ′ of f−1U such


that U ′ ⊂ U , but ∂U ⊂ ∂U ′ and f(∂U) = ∂U . We have f(U ′) = U . Chooseγ = γ0 : S1 → U such that the annulus A between ∂U and γ(S1) contains no pointsof f−1B. Denote by A′ the component of f−1A such that A′ ⊂ U ′ and ∂U ⊂ ∂A′.We may adjust γ so that A′ ⊂ A, see [Mc1].

Choose x′ ∈ A′ such that f(x′) = γ(0). The degree d = deg(f : A′ → A) is apositive integer. We define γ1(t) so that γ1(0) = x′ and f(γ1(t)) = γ(d · t). LetH0 : [0, 1] × S1 → A be a C1 map such that H0(0, ·) = γ1 and H0(1, ·) = γ.

Since f : A′ → A is a covering, one can lift H0 to get H1 : [0, 1] × S1 → A′ suchthat H1(1, ·) = H0(0, ·). Define γ2(·) = H1(0, ·). One can then define Hn−1 and γn

by induction.To control the convergence, we proceed as follows. For each t0 ∈ S1, the ρ-length

of the curve {H0(s, t0), s ∈ [0, 1]} is finite, depending continuously on t0 ∈ S1. Soit has a finite maximum, say C ′. For t ∈ S1, dρ(γn(t), γn−1(t)) is smaller than orequal to the length of the curve {Hn(s, t), s ∈ [0, 1]} which is smaller than or equalto C ′/λn−1. So {γn} forms a Cauchy sequence.

Therefore γn(t) converges uniformly to a limit map, h, and the path distancebetween γ(t) and γn(t) is uniformly bounded by C ′λ/(λ − 1).

To define ηt for each t ∈ S1, one note that the set h(t)∪ (⋃

n≥0

⋃s∈[0,1] Hn(s, t)) is

an embedded closed arc with finite length. Reparametrizing it we get η(t).For periodic J ′ or periodic ideal boundary, we consider an iterate of f . For prepe-

riodic cases, we pull back the curves given by the result for the periodic cases.

5. Proof of Proposition [Case 2]

This is the case where a Julia component J0 satisfies Jn = fn(J0) ⊂ AS for n ≥ n0.We may assume n0 = 0.Part I, f is hyperbolic. We will show that J0 is a Jordan curve.

Recall that AS consists of components of f−1A parallel to some components of A.Denote by A1, A2, · · · , Ap the components of AS , and by Aij the set of points z

such that z ∈ Ai, f(z) ∈ Aj .If Aij 6= ∅, it is an essential subannulus of Ai, and f : Aij → Aj is a covering. For

f : Ai → f(Ai) is a covering, f(Ai) is a component of A and Aj is a subannulus off(Ai) parallel to f(Ai).

For each Aj , choose γj : S1 → Aj an injective homotopically non-trivial C1 curve.For each i such that Aij 6= ∅, choose xij ∈ Aij such that f(xij) = γj(0). Then thehomotopy classes of γi and γj determine uniquely generators for π1(Aij) and π1(Aj).The degree dij = deg(f : Aij → Aj) is then a positive or negative integer. Define a liftγij(t) of γj so that γij(0) = xij and f(γij(t)) = γj(dij · t). Let Hij : [0, 1] × S1 → Ai

be a C1 map such that Hij(0, ·) = γi and Hij(1, ·) = γij .Let α = (a0a1 · · · ) be any infinite sequence such that for all n ≥ 0 we have

1 ≤ an ≤ p and Aanan+1 6= ∅ and call such a sequence admissible.For n ≥ 0, denote by Tn = Tn(α) the set of points z such that fk(z) ∈ Aak

for0 ≤ k ≤ n. Then Tn+1 is an essential subannulus of Tn for n ≥ 0 and fn : Tn → Aan is


a covering. The curve γa0 determines a generator for π1(Tn) and we let dn = deg(fn :Tn → Aan); it can be a positive or negative integer.

Set J ′ = J ′(α) =⋂

n Tn. Since Tn forms a nested sequence of compact connectedsets which are critically separating, J ′ is also compact connected and critically sepa-rating. We will show that either J ′ ⊂ ∂TN for some N , or J ′ is a Jordan curve, anda Julia component. The proof is split into several lemmas.Lemma 0. For each n ≥ 0, there is a (parametrized) C1 curve ζn(t), and for n ≥ 1, ahomotopy Gn : [0, 1] × S1 → Tn−1 such that ζn(S1) ⊂ Tn ⊂ Aa0 , Gn(0, t) = ζn−1(t),Gn(1, t) = ζn(t). Moreover fn(ζn(t)) = γan(dn · t).Proof. Set ζ0 = γa0 . Assume we have constructed ζn−1 and Gn−1. Since the mapfn−1 : Tn−1 → Aan−1 is a covering, mapping Tn onto Aan−1,an , one can lift thehomotopy Han−1,an to a map Gn : [0, 1] × S1 → Tn−1 with Gn(0, t) = ζn−1(t). Setζn(t) = Gn(1, t). These are the maps required by the lemma. Finally, by our choiceof generators of fundamental groups, we have fn(ζn(t)) = γan(dn · t).

Next, by our choice of B (in particular B ∩ J = ∅) we have⋃

Ai ⊂ C − B andAij ⊂ C − f−1(B) for all possible pairs (i, j). By Corollary 4.2, there is a positivenumber M such that for i = 1, · · · , p, the path diameter of Ai ∈ AS with respect to ρ

is at most M . Moreover, on AS ⊂ C− f−1(int(B)), f expands ρ by a definite factorλ > 1.Lemma 1. The curves ζn(t) defined in Lemma 0 converge uniformly to a continuousmap ζ : S1 → Aa0 .Proof. For each possible pair (i, j), the ρ-length of the curve {Hij(s, t0), s ∈ [0, 1]}is finite, depending continuously on t0 ∈ S1. So it has a finite maximum. Let C ′ bethe maximum of the ρ-length of the curves among all possible couples (i, j) and allt0 ∈ S1; it is again finite. For t ∈ S1, dρ(ζn(t), ζn−1(t)) is less than or equal to thelength of the curve {Gn(s, t), s ∈ [0, 1]} which is smaller than or equal to C ′/λn−1.So {ζn} forms a Cauchy sequence.

Choose a base point x± in each component of ∂Aa0 . Denote by δ+n (resp. δ−n ) the

component of ∂Tn which either contains x+ (resp. x−) or which separates x+ (resp.x−) and Tn. Denote by DH the Hausdorff distance on compact subsets of C− int(B)with respect to the metric dρ. By definition

DH(F,G) = max ( maxx∈F

miny∈G

dρ(x, y) , maxy∈G

minx∈F

dρ(x, y) ) .

Lemma 2. For every ε > 0, there exists an N independent of α such that for everyn ≥ N , DH(J ′, δ+

n ) < ε, DH(J ′, δ−n ) < ε and DH(J ′, Tn) < ε.Lemma 3. DH(J ′, ζn(S1)) → 0.Proof of Lemmas 2 and 3. Fix y ∈ δ+

n . We first show

minx∈J ′

dρ(x, y) < M/λn.

Let yn = fn(y), n ≥ 0, and for each n choose xn ∈ fn(J ′). Then for each n, thereis a path ηn : [0, 1] → Aan such that ηn(0) = yn, ηn(1) = xn, ηn(]0, 1[) ⊂ Aan , andlρ(ηn) ≤ M . For any n, f−n(fn(J ′)) ∩ Tn = J ′, since fn(

⋂k T k) =

⋂k fnT k. Hence

there is a lift ηn : [0, 1] → Tn of ηn under fn joining y to some point x′n ∈ J ′. Hence


by expansionminx∈J ′

dρ(x, y) ≤ dρ(x′n, y) ≤ lρ(ηn) ≤ M/λn.

Hencemaxy∈δ+

n

minx∈J ′

dρ(x, y) ≤ M/λn.

A similar argument bounds maxx∈J′ miny∈δ+n

dρ(x, y) by the same quantity. Theremainder of the two lemmas are proved similarly.

Lemmas 1 and 3 imply that ζ(S1) = J ′. As a consequence, J ′ is locally connected.Lemma 4. Either J ′ coincides with one boundary component of TN for some N , orJ ′ ⊂ Tn for all n. In the second case, J ′ is a Jordan curve, and a Julia component.

Proof. We first show J ′ ⊂ J . First, ∂J ′ ⊂ J since ∂Tn ⊂ J for all n and J isclosed. Second, J ′ = ∂J ′. For otherwise there is a nonempty component W of int(J ′).Then W ⊂ Tn for all n, and as a consequence fn(W ) ⊂ A for all n. Thus {fn} isnormal on W . So W coincides with a Fatou component. But for a hyperbolic mapevery Fatou component is eventually attracting, and meets eventually P, contradictingA ∩ P = ∅.

There are thus two possibilities: either J ′ ⊂ ∂TN for some N , or J ′ ⊂ Tn for alln. In the first case J ′ coincides with one boundary component of Tk for all k ≥ N(Lemma 2). In the second case, J ′ must be a Julia component. For otherwise, J ′ isa nonempty proper closed subset of some Julia component J ′′; if x ∈ J ′′ − J ′ thenDH(x, J ′) > 0, and hence Lemma 2 implies that for some n, ∂Tn either separatesx and J ′ or ∂Tn contains x. But this implies that J ′′ intersects ∂Tn, hence J ′ iscontained in a boundary component of Tn, violating our assumption.

Moreover, Lemma 2 implies that C − J ′ has exactly two components U1, U2, and∂U1 = ∂U2 = J ′. The lemma below (pointed out to us by M. Lyubich) allows us toconclude that J ′ is a Jordan curve.

Consider now our Julia component J0 such that Jn ⊂ AS for all n. It determines anadmissible sequence α = (a0a1 · · · ) by setting an = m if Jn ⊂ Am. Then J0 = J ′(α),and it is a Jordan curve.(Remark. The proof actually shows much more; see §8. )Part II, f is nice. Let J0 be a Julia component such that Jn ⊂ AS for all n. DefineTn to be the component of f−nU containing J0 (it is in fact the same Tn as in PartI). Then each Tn is an open annulus, contained essentially in Tn−1. With the help ofSullivan’s non-wandering domain theorem, one can show easily that J0 =

⋂n Tn. On

the other hand, since J0 is disjoint from ∂Tn ⊂ f−n(E) for all n, there is a sequencenk → ∞ such that Tnk

⊂ Tnk−1 . Therefore C − J0 has exactly two components.

Lemma 5.1. — Assume that K is a closed subset of C satisfying either conditionsa) and b) or condition c):

a) K is the common boundary of two disjoint open connected sets U1 and U2.b) K is locally connected.c) C−K has exactly two components V1 and V2 and each point of K is accessible

from both V1 and V2.Then K is a Jordan curve. There are counter examples if one of the above condi-

tions is not satisfied.


Proof. — Condition a) shows that U1 and U2 are simply connected and K is compactconnected. By b) and Caratheodory’s theorem, a Riemann map φ : ∆ → U1 extendscontinuously to the boundary, and the extension is locally non constant. But theextension is also injective, for otherwise the image by φ of a pair of distinct radialsegments of ∆ would form a Jordan curve ν and each component of C − ν wouldintersect U2. The contradicts the fact that ν ∩ U2 = ∅ and U2 is connected.

For a proof in condition c) case, see Newman ([Ne]), Theorem 16.1.

6. Proof of Proposition [Case 3]

This is the case where a Julia component J0 satisfies Jn = fn(J0) ⊂ AO ∪ D′ forinfinitely many n.Part I. f is hyperbolic. We will need the following variant of Lemma 1.4; thedifference is that we do not assume Qi is full.

Lemma 6.1. — Let f be a hyperbolic rational map and Q = tki=1Qi be a finite family

of disjoint path-connected subsets of C such that

1. Qi ∩ P = ∅ for each i,2. the inclusion maps ιi : Qi → C − P are homotopic to constant maps, and3. the path-diameter of each Qi with respect to dpathQi

is finite.

Then any connected set J satisfying fn(J) ⊂ Q for infinitely many n is a point.

Proof. — We may assume Qi ∩ B = ∅, where B is a closed neighborhood of Pconstructed previously. Then since f is hyperbolic, f expands ρ uniformly by somefactor λ > 1 on C − f−1(int(B)). Let Jn = fn(J) and let Qn denote the componentof Q containing Jn. Choose x0, y0 ∈ J0 and let xn = fn(x0), yn = fn(y0). Let M bean upper bound on the path-diameters of the Qi with respect to ρ. Then for each n,there is a path ηn : [0, 1] → Qn, ηn(0) = xn, ηn(1) = yn for which lρ(ηn) ≤ M . SinceQn ↪→ C−P is homotopic to a constant map, for each n there is a lift ηn of ηn underfn joining x0 to y0. By expansion,

dρ(x0, y0) ≤ lρ(ηn) ≤ lρ(ηn)/λn ≤ M/λn → 0.

Hence x0 = y0 and J is a point.

Now assume that J0 is a Julia component for f , and Jn ⊂ AO ∪ D′ for infinitelymany n. By Lemma 2.2 the set AO ∪D′ has finitely many components, each of finitepath diameter by Corollary 4.2. The above Lemma then applies and hence J0 is apoint.Part II. f is nice. For each component U of AO ⊂ A, take a simple closed curve γwhich is a generator of the fundamental group of U . Then C− γ has exactly one disccomponent V contained in A. The set U = U ∪ V is an open disc contained in A (sois disjoint from P. Therefore the enlarged set AO ∪D′ consists of finitely many opendiscs disjoint from P. The rest of the proof is very similar to Part II in the proof ofProposition [Case 2]. We omit the details here.


7. Proof of Proposition [Case 4] and Theorem 1.1

In this section we derive Theorem 1.1 and Proposition [Case 4] from

Theorem 7.1. — Let f be a rational map (not necessarily hyperbolic or nice). LetW be the union of finitely many Fatou components W0, · · · ,Wm such that

1. f(W) ⊂ W;2. each Wi eventually lands on an attracting or parabolic periodic basin under

iteration of f , and Wi ∩ P 6= ∅ for all i;3. for each Wi, only finitely many components K1,i, · · · , Kmi,i of C−Wi intersect

P.

Then there is a finite union Q of disjoint full continua in C−P such that any Juliacomponent J0 satisfying fn(J0) ⊂

⋃i

(C −

⋃mi

j=1 Kj,i

)for infinitely many n passes

infinitely often through Q, and is a point.

Proof of Theorem 1.1. — Let W0 be the basin of infinity of the polynomial f . Itis a fixed attracting component, and infinitely connected. Since only finitely manycomponents of K = C − W0 intersect P, we may apply the above theorem to provethat every Julia component of f passing infinitely many times through U0 is a point,where

U0 = W0

⋃{K | K is a K-component disjoint from P} .

On the other hand, every Julia component is the boundary of a K-component. Soevery K-component passing infinitely many times through U0 is a point. But in thisparticular case, we know also that the orbit of a K-component either stays entirely inU0 or lands eventually on a K-component intersecting P, which is preperiodic, sincethere are only finitely many of such K-components and the union of them is forwardinvariant.

Proof of Proposition [Case 4]. — Let f be a nice map and J0 be a Julia component.Assume that Jn = fn(J0) ⊂ L for infinitely many n. For each component Li of L,Lemma 3.10 provides a unique Fatou component Wi such that Wi ⊂ Li and ∂Li ⊂∂Wi. The union of these Wi’s satisfies the conditions of Theorem 7.1. Moreover,in the notation of the statement and the proof of the theorem, we have Li = Ui =C−

⋃mi

j=1 Kj,i. Therefore we can apply Theorem 7.1 to conclude that J0 is a point.

We now start the proof of Theorem 7.1. — We will use the following notation. LetS be a closed subset of S2 and W be an open connected subset of S2. Define

U(W,S) = W ∪⋃

{K | K a component of S2 − W such that K ∩ S = ∅} .

Then U(W,S) is an open set.For i = 0, · · · ,m, set Ui = U(Wi,P) = C−

⋃mi

j=1 Kj,i. Our aim is to find a compactset Q satisfying the properties require in the theorem. We then apply Lemma 1.4 toconclude.

The proposition below is proved in the Appendix.


Proposition 7.2. — Let W be a Fatou component of a rational map f . Supposethat P ∩ (C−W ) is contained in the union K1 t · · · tKr of finitely many connectedcomponents of S2 − W . Let W ′ be a connected component of f−1(W ) and let U =U(W,P) and U ′ = U(W ′, f−1(P)). Then

1. U and U ′ are connected open subsets with finitely many boundary components;2. f : U ′ → U is proper, surjective and a branched covering;3. f : ∂U ′ → ∂U is surjective, and given any connected component K ′

i of S2 −U ′,f maps ∂K ′

i surjectively onto ∂Kj for some unique component Kj of S2 − U .

Since each component Wi contains points of P, Ui ∩ Uj = ∅, i 6= j. For each i, setU ′

i = U(Wi, f−1P). Then U ′

i ⊂ Ui since f−1(P) ⊃ P. By Proposition 7.2, U ′i has

finitely many boundary components, hence Qi := Ui − U ′i consists of finitely many

full continua disjoint from f−1(P), hence disjoint from P. This will be one piece ofour set Q.

Proposition 7.2 also implies that f : U ′i → Uj is proper if f(Wi) = Wj . We

analyze the dynamical system f : tiU′i → tiUi. Define f∗(Ui) = Uj if f(U ′

i) = Uj

(equivalently, if f(Wi) = Wj).Assume that a Julia component J0 satisfies that fn(J0) ⊂

⋃i Ui for infinitely many

n. Then either fn(J0) ⊂⋃

i Qi for infinitely many n or fn(J0) ⊂⋃

i U ′i for all n ≥ n1.

In the former case fn(J0) ⊂ Q for infinitely many n too since Q is going to be definedas a set containing

⋃i Qi. In the latter case the orbit of J0 lands eventually into a

periodic cycle of f∗. We now analyze this second case.Let U0, · · · , Up−1 be a periodic cycle of f∗. Set g = fp. Then J (g) = J (f) and

Ui = U(Wi,P(f)) = U(Wi,P(g)).For any Julia component J0 such that J0 ⊂ U ′

0 and fn(J0) ⊂⋃

i U ′i for all n ≥ 0,

we have f(J0) ⊂ U ′1, · · · , fp−1(J0) ⊂ U ′

p−1 and fp(J0) ⊂ U ′0 and so on. Therefore

gn(J0) = fnp(J0) ⊂ U ′0 ⊂ U0 for all n.

Set W = W0 and U = U0 = U(W,P). Then W is a fixed attracting or parabolicFatou component of g.

We are going to find a compact set Q′0 which is the union of finitely many disjoint

full continua such that, if the g-orbit of a Julia component J0 passes infinitely manytimes through U , then it passes infinitely many times through Q′

0 as well.Denote by X the empty set in case W is an attracting basin, or the set of one

single element which is the fixed parabolic point of W , in case W is a parabolic basin.Set Xn =

⋃0≤j≤n g−jX.

One can find a disc V ⊂ W with Jordan curve boundary such that V ⊂ W ∪ Xand g(V ) ⊂ V ∪X . We may choose V such that (∂V −X)∩P = ∅. For any n, let Vn

be the unique component of g−n(V ) containing V0 = V . Then V n ⊂ Vn+1 ∪ Xn forany n. There is an integer N guaranteed by Lemma 7.3, such that every componentof C − VN contains at most one of the finitely many components of C − W whichintersect P, and P ∩ W ⊂ VN .

The set C−VN is a disjoint union of finitely many full continua with J ⊂ int(C−VN ) ∪XN . Denote the components of C− VN by D1, · · · , Dl, · · · , Dn such that, forj > l, Dj ∩ P = ∅; and for 1 ≤ j ≤ l, Dj contains a unique component Kj of C − Wsuch that Kj ∩ P 6= ∅.


For 1 ≤ j ≤ l, set Aj = int(Dj) − Kj . Then Aj is an open annulus with possiblyfinitely many pinched points at points in XN . Moreover Aj contains no critical value(by the choice of N).

Now we look at the level N +1. For U = U(W,P) and U ′ = U(W, g−1P), the mapg : U ′ → U is a branched covering (Proposition 7.2). Since VN∪(

⋃lj=1 Aj∪

⋃j>l Dj) =

U , we conclude that U ′ − g−1(⋃l

j=1 Aj ∪⋃

j>l Dj) is connected, and coincides withg−1VN ∩ U ′. So g−1VN has a unique component in U ′, which is VN+1.

Denote the components of C − VN+1 by D′1, · · · , D′

l, · · · , D′k such that Kj ⊂

D′j ⊂ Dj for 1 ≤ j ≤ l, and D′

j ∩ P = ∅ for j > l.We set g∗(Ki) = Kj if g(∂Ki) = ∂Kj , which is well-defined by Proposition 7.2.Set A′

j = int(D′j) − Kj , j = 1, · · · , l. If g∗(Ki) = Kj , then g(A′

i) = Aj andg : A′

i → Aj is a covering map.Note that Aj −A′

j = int(Dj)− int(D′j). Moreover J ∩ (int(Dj)−D′

j) is containedin

⋃s>l D

′s.

We claim then every Julia point x in⋃l

j=0 A′j must have some iterate in

⋃lj=0(Aj−

A′j). If not, there is n0 > 0 such that for all n > n0, gn(x) ∈

⋃{A′

j | Kj is periodicfor g∗ }. On the other hand, for K0, · · · ,Kq−1 a periodic cycle of g∗, and for anyy ∈ A′

0, there is a minimal integer s > 0 such that gs(y) /∈⋃q−1

j=0 A′j , for Lemma 7.3

below implies that each Kj is the nested intersection of sets of the form ∩nBn, whereBn is a component of C − Vn, n ≥ 0.

Therefore if a Julia component J0 satisfies that gn(J0) ⊂⋃l

j=0 A′j ∪

⋃s>l D

′s for

infinitely many n, then gn(J0) ⊂⋃

s>l D′s for infinitely many n.

On the other hand, for our set U = U(W,P), we have J ∩U ⊂⋃l

j=0 A′j∪

⋃s>l D

′s∪

XN+1. Thus saying gn(J0) ⊂ U for infinitely many n is the same as saying gn(J0) ⊂⋃lj=0 A′

j ∪⋃

s>l D′s for infinitely many n.

Return to our original map f now. The components of Q0 = U0 − U ′0 are disjoint

full continua contained in⋃l

j=0 A′j ∪

⋃s>l D

′s. Therefore Q′

0 = Q0 ∪⋃

s>l D′s is again

a union of finitely many disjoint full continua.For each periodic cycle of f∗, choose one representative Ui in the cycle, and de-

fine Q′i in the same way, for the other i in the cycle, define Q′

i = Qi. We write{0, 1, . . . ,m} = I ′ t I, where j ∈ I ′ if Uj is f∗-periodic and j ∈ I otherwise.

Set Q =⋃

i∈I′ Q′i ∪

⋃i∈I Qi. This is again a union of finitely many disjoint full

continua, disjoint from P. Now let J0 be a Julia component passing infinitely manytimes through

⋃i Ui. Then either it passes infinitely many times through

⋃i Qi ⊂ Q,

or there is some i ∈ I ′, such that J0 passes infinitely many times through Q′i ⊂ Q. In

both cases J0 must pass infinitely many times through Q.Finally we apply Lemma 1.4 to conclude that such Julia components are points.We mention here two particular cases.1. The component W is simply connected. In this case U = W and no Julia

component passes through U .2. The set U(W,P) coincides with C, that is P ⊂ W . In this case l = 0 and each

Julia component is a point. That is J is totally disconnected.


A variant of the following lemma can be found in [St], page 63 and 117.

Lemma 7.3. — Let W be an fixed attracting or parabolic basin of a rational mapf . Let V be a either a disc neighborhood of the attracting fixed point with Jordancurve boundary or a Fatou petal in W of the parabolic fixed point. Then for Vn thecomponent of f−n(V ) containing V , we have W =

⋃n Vn, and each component of

C− Vn is closed disc with possibly finitely many pinching points. Furthermore, givenany two components K1 and K2 of C − W , there is an integer N such that VN

separates K1 and K2.

8. Further results

8.1. Diameter of Julia components. — Let f be a hyperbolic rational map.

Corollary 8.1. — A Julia component J0 of J is not a point if and only if there isan integer N such that Jn = fn(J0) ⊂ AS t f−1E for any n ≥ N .

If J0 is a point, we may regard N as +∞.One can show that the set f−1(AS t f−1E)− (AS t f−1E) is contained in finitely

many discs. Thus the same technique as in the above sections can prove also thefollowing results: to each integer N , there is a positive number ε(N), with ε(N) → 0as N → ∞, such that for J0 a Julia component, and N the minimal integer such thatfn(J0) ⊂ AS t f−1E for n ≥ N , the spherical diameter of J0 is at most ε(N).

8.2. Symbolic dynamics of Julia components in AS. — Let f be a hyperbolicrational map. The arguments given in §5 actually show much more. The space Σof admissible sequences {a0a1 . . .} equipped with the product topology and the one-sided shift map σ is a subshift of finite type. Let Cl(C) denote the space of all closedsubsets of C in the Hausdorff topology. The map f induces a map from this spaceto itself which we again denote by f . Given α ∈ Σ, we have defined in §5 a nestedsequence of annuli Tn(α) and a continuum J ′(α) = ∩nTn(α). The set J ′(α) may ormay not be a Julia component.

Theorem 8.2. — The map

Φ : α 7→ J ′(α); Σ → Cl(C)

defines a uniformly continuous injective map conjugating σ to f |Φ(Σ). Moreover,1. there is a (at most) countable subset Σe and a finite subset Σe,0 ⊂ Σ such that

(a) α ∈ Σe,0 if and only if J ′(α) is a boundary component of Aa0 for someAa0 ∈ AS, i.e. is a boundary component of T0(α).

(b) α ∈ Σe if and only if J ′(α) is a boundary component of Tn(α) for somen.

(c) σ(Σe,0) ⊂ Σe,0.(d) Σe = ∪n≥0σ

−n(Σe,0).2. α /∈ Σe if and only if J ′(α) is a Jordan curve Julia component satisfying

fn(J ′(α)) ⊂ As for all n ≥ 0.


3. Let Je,0 = Φ(Σe,0). Then as a subset of C,⋃α∈Σ

Φ(α) = {z|fn(z) ∈ As t (∪δ∈Je,0δ)∀n ≥ 0}.

The elements of the set of continua Je := Φ(Σe) may be thought of as “exposed”boundary components, in the sense that they are boundary components of Tn.

Corollary 8.3. — The following are equivalent:1. Σ is uncountable.2. There is a wandering component of the Julia set which is a Jordan curve.3. There are uncountably many wandering components of the Julia set which are

Jordan curves.4. There are infinitely many periodic Jordan curve Julia components.5. There exists a component C ′ of A, disjoint essential subannuli A,B ⊂ C ′, and

integers m,n > 0 such that fm : A → C ′, fn : B → C ′ are covering maps.

Proof. — For any subshift of finite type (Σ, σ), the following are equivalent: 1. thespace of admissible sequences Σ is uncountable; 2. there is a wandering sequence;3. there are uncountably many wandering sequences; 4. there are infinitely manyperiodic sequences; 5. there are two finite-length sequences α = (a0, . . . , am), β =(b0, . . . , bn) with c := a0 = am = b0 = bn and ai 6= bi for some 0 ≤ i ≤ minm, n.

By the above Theorem and the preceding facts, the first three conditions are thusequivalent, and (1) implies (4). The set E contains finitely many Julia components,and each periodic Jordan curve Julia component which is not in E must be containedin AS . Each such component is J ′(α) for some periodic α, by the above Theorem,hence Σ has infinitely many periodic sequences and so (4) implies (1). We now showthe equivalence of (4) and (5). First, note that (5) is equivalent, by pulling back, tothe same condition with C ′, a component of A, replaced by C, a component of AS .The equivalence of (4) and (5) then follows immediately from the above Theorem andthe preceding paragraph.

Proof of Theorem 8.2. — The continuity of Φ follows immediately from Lemma 2 of§5. That Φ is a semiconjugacy follows from the fact that f : Tn(α) → Tn−1(σ(α))is a covering map, hence f : Tn(α) → Tn−1(σ(α)) is surjective, and so f(Φ(α)) =f(J ′(α)) = f(∩nTn(α)) = ∩nTn−1(σ(α)) = J ′(σ(α)) = Φ(σ(α)).

To see that Φ is injective, first note that given a boundary component δ of A ∈ A, noother component B ∈ A has δ as a boundary component. For δ is locally connected(e.g. by Corollary 4.2) and hence δ is homeomorphic to S1 if it is the commonboundary of two disjoint open annuli, by Lemma 5.1. But if this occurs then δ is acomponent of the Julia set separating its parallels, violating the construction of A.More generally, since f sends boundary components of Tn(α) to boundary componentsof Tn−1(σ(α)), if δ is the common boundary of Tn(α) and Tn(β), then Tn(α) = Tn(β),i.e. if α = {a0a1 . . .}, β = {b0b1 . . .} then ai = bi, 0 ≤ i ≤ n. If J ′(α) = J ′(β), theneither J ′(α) = J ′(β) is a boundary component of Tn(α), some n, or J ′(α) ⊂ Tn(α)for all n. In the former case we have Tn(α) = Tn(β) for all n by the above observation


while in the latter we must have Tn(α) = Tn(β) for all n since if α 6= β then for somen, Tn(α) ∩ Tn(β) = ∅. Hence α = β.(1). Define Σe,0 as in 1(a). We first prove that Je,0 = Φ(Σe,0) is forward-invariantunder f . Let J0 ∈ Je,0 and let J1 = f(J0). Then J0 is a boundary component ofT0(α) for some α. The proof of Lemma 4 of §5 shows that then J0 is a boundarycomponent of Tk(α) for all k ≥ 0, hence J0 is a boundary component of T1(α). Sincef : T1(α) → T0(σ(α)) is a covering, f(J0) = J1 is a boundary component of T0(σ(α))and so J1 ∈ Je,0. Now 1(a) holds by definition, 1(c) follows from the above result andthe fact that Φ is a semiconjugacy. Define Σe by 1(d). 1(b) then follows immediatelyfrom the fact that Φ is a semiconjugacy and the fact that fn : Tn(α) → T0(σn(α)).(2). Note that α /∈ Σe if and only if J ′(α) ⊂ Tn(α) for all n, by 1b). The result thenfollows by Lemma 4 of §5.(3). First, let J0 = J ′(α). If α /∈ Σe, then J0 ⊂ Tn(α) for all n ≥ 0, hencefn(J0) ⊂ T0(σn(α)) ⊂ AS for all n ≥ 0. Otherwise, there is a minimal N ≥ 0 suchthat for 0 ≤ i < N , J0 ⊂ Ti(α) and for i = N , J0 is a boundary component of Ti(α).Hence for 0 ≤ i < N , f i(J0) ⊂ T0(σi(α)) ⊂ AS , and for i = N , f i(J0) is a boundarycomponent of T0(σi(α)) ∈ Je,0. Since Je,0 is forward-invariant under f the inclusionin this direction is proved.

To prove the other direction, given z an element of the right-hand side we mustproduce α so that z ∈ J ′(α). Let zn = fn(z). If zn ∈ Abn

∈ AS for all 0 ≤ n < Nwe set an = bn. Hence we may assume zn ∈ ∂AS for all n ≥ 0. Choose a componentAa0 so that z0 ∈ J0, a boundary component of Aa0 (there may be more than onesuch component, since the annuli in AS may have closures which intersect). ThenJ0 ∈ Je,0. Since Je,0 is forward-invariant, Jn = fn(J0) is a boundary component of aunique component Aan of AS . We now claim that α = (a0a1 . . .) chosen in this fashionis admissible, i.e. that Aaiai+1 6= ∅ for all i. For on the one hand, f(Aai) = A ∈ A,hence Ji+1 = f(Ji) is a boundary component of A. On the other hand, Ji+1 is aboundary component of Aai+1 ⊂ B ∈ A. If A 6= B then we must have that Ji+1 is thecommon boundary of A and B, which we have previously noted is impossible. HenceA = B and so Aaiai+1 6= ∅. Hence J0 = J ′(α) contains z.

8.3. Moduli restrictions and description of the shift. — Let f be a nicerational map. Let {C1, . . . , Ck} be the set of annuli in A. Denote by mj the modulusof Cj . Note that 0 < mj < ∞. For each j choose a Jordan curve γj ⊂ Cj which is ahomotopically non trivial in Cj . Set Γ = {γ1, · · · , γk}. Let Λ(i, j) denote the set ofcomponents Ci,j,λ of f−1(Cj) homotopic to Ci and let di,j,λ be the positive degree off : Ci,j,λ → Cj .

Define two linear maps

fΓ, fΓ,# : RΓ → RΓ

by

fΓ(γj) =∑γi∈Γ

(∑

λ∈Λ(i,j)

1di,j,λ

)γi


andfΓ,#(γj) =

∑γi∈Γ

(∑

λ∈Λ(i,j)

1)γi.

If Λ(i, j) = ∅ we take the coefficient of γi to be zero. The map fΓ is the Thurstonlinear transformation defined by the multicurve Γ. The Grotzsch inequality impliesthat if ~m = (mj) ∈ RΓ is the vector of moduli mj , then

(fΓ(~m))j < mj , 1 ≤ j ≤ k.

As a consequence, the leading eigenvalue of fΓ is smaller than 1 (consider the productvfΓm with v a leading eigenvector of (fΓ)t).

The map fΓ,# is almost the transition matrix for the subshift (Σ, σ). The compo-nents of AS are in one-to-one correspondence with the collection of elements {λ|λ ∈Λ(i, j)}. If Aλ, Aµ ∈ AS where λ ∈ Λ(i, j) 6= ∅ and µ ∈ Λ(k, l) 6= ∅ then Aλµ ={z|z ∈ Aλ, f(z) ∈ Aµ} 6= ∅ if and only if j = k. An admissible sequence α ∈ Σ is thusgiven by an infinite sequence λn where successive terms satisfy the above condition.Hence by removing any basis elements γj for which the jth row of fΓ,# consists onlyof zeros we obtain a new matrix. Iterating this removal process we obtain a matrixwhich gives exactly the subshift (Σ, σ).

8.4. Quasicircles and non-quasicircles. — A Jordan curve J in the sphere is aK-quasicircle if it is the image of a round circle under a K-quasiconformal map, andit is said to have K ′-bounded turning if the ratio

diam(L)/d(x, y) ≤ K ′

for all x, y ∈ J , where L is the component of J − {x, y} with smallest diameter.Here distance is measured with respect to the spherical metric d. J is said to be aquasicircle if it is a K-quasicircle for some K. It is known that J is a quasicircle ifand only if it has bounded turning ([LV] II, §8).

If a Jordan curve component J of J is preperiodic and f is hyperbolic, it is a qua-sicircle by the surgery argument of McMullen [Mc1]. However, a wandering Jordancurve component J need not be a quasicircle. We first show

Theorem 8.4. — Let f be a hyperbolic rational map. If the orbit of α under σ doesnot accumulate on Σe,0 then Φ(α) is a quasicircle.

(Here we use the same notation as in §8.2). The proof also shows the followingcorollary. However, we have no example where the hypotheses are satisfied.

Corollary 8.5. — If Σe,0 is empty (i.e. if every boundary component of a componentof AS maps to a boundary component of L which is not also a boundary componentof any component of AS), then there exists a K such that for every α ∈ Σ, Φ(α) is aJulia component which is a K−quasicircle.

Proof. — Let J0 = Φ(α). If α does not accumulate on Σe,0 then there is a positiveinteger N0 such that

T ′0 = {TN0(σ

i(α)), i ≥ 0}


is a disjoint union of open annuli which is compactly contained in AS , and whichcontains the entire forward orbit of J0. Since the orbit of J0 does not accumulate ona boundary component of AS , it does not accumulate on boundary components ofT ′

0 . Hence there is an integer N1 > N0 such that

T ′1 = {TN1(σ

i(α)), i ≥ 0}

is a disjoint union of open annuli which is compactly contained in T ′0 and which

contains the entire forward orbit of J0. Then fN1−N0(T ′1 ) = T ′

0 .Enlarge T ′

0 slightly to a disjoint union T0 of open annuli whose boundaries arereal-analytic Jordan curves and which is compactly contained in AS . Let T1 be theunion of preimages of components of T0 under fN1−N0 containing components of T ′

1 .Then T1 is compactly contained in T0, and

g = fN1−N0 : T1 → T0

is an expanding conformal dynamical system.On the other hand, we may build another model for this dynamical system of the

formh : R0 → R1

where Ri consist of round annuli and the map h on each component is z 7→ zd, somed. We may also find smooth maps φi : Ri → Ti such that φ0 ◦ h = g ◦ φ1 on R1 andφ0 ' φ1 rel ∂R0. That is, the pair (φ0, φ1) gives a combinatorial equivalence betweenthe two dynamical systems. By Theorem A.1 of [Mc2], φi are isotopic rel ∂R0 toa quasiconformal conjugacy ψ. Since the Julia set of h is the product of a Cantorset with a round circle, J0 is a component of the Julia set of g and hence J0 is aquasicircle.

If Σe,0 is empty, then f−1(AS) ∩ AS is compactly contained in AS . We may thenapply the argument above with T ′

0 = AS and T ′1 = f−1(AS) ∩ AS to prove the

Corollary.

Example. We illustrate this by our example f = f1. In this case AS has twocomponents A0 and A1. We choose A0 to be the outermost one, i.e. the one witha fixed boundary (the component J+). There are no components of J which arepoints, and indeed J = {z|fn(z) ∈ AS∀n}. Moreover, the boundary components ofcomponents of AS are entire Julia components, and A0 ∩ A1 = ∅. It follows that theconnected components of J are precisely the sets of the form J ′(α), α ∈ Σ. Henceby Theorem 8.2, the dynamics on the space of connected components of J in theHausdorff topology is conjugate to the one-sided shift on two symbols 0 and 1. Notethat then Σe,0 = {(000 . . .)} and hence that Σe = {(a0a1 . . .)|an = 0∀n ≥ n0}.

The dynamical system (Σ, σ) is conjugate via φ to (C, g), where C is the invariantCantor set for the interval map g : [0, 1/3] t [2/3, 1] → [0, 1] given by g(c) = 3c for0 ≤ c ≤ 1/3 and g(c) = 3(1 − c) for 2/3 ≤ c ≤ 1. The conjugacy φ is defined byφ(c) = (a0a1a2 . . .) where an = 0 if gn(c) ∈ [0, 1/3] and an = 1 otherwise. Note thatthe point 0 ∈ C corresponds to (0000 . . .) and the point 1 ∈ C to (1000 . . .). Orderthe components of J so that J > J ′ if J ′ separates J and the point at infinity. Then


the composition Φ ◦ φ is order-preserving with respect to the usual ordering on theinterval and >.

Say c ∈ C is exposed if c is a boundary point of a component of the complement ofC and is buried otherwise. Similarly, we say a Julia component is exposed if it meetsthe boundary of a Fatou component of f and is buried otherwise. Then c is exposedif and only if φ(c) ∈ Σe if and only if Φ ◦ φ(c) is exposed if and only if Φ ◦ φ(c) is acovering of J+ if and only if Φ ◦ φ(c) is not a Jordan curve.

Finally, we note that Σe is dense in Σ.We now show

Theorem 8.6. — Φ ◦ φ(c) is a quasicircle if and only if c does not accumulate at0 under g, i.e. if and only if α = φ(c) does not contain arbitrarily long strings ofconsecutive zeros.

Thus quasi-circle components form a dense subset in the space of Julia components.On the other hand, the non quasi-circle components form a residual set in Baire’scategory.

Proof. — Let J0 = Φ(c0) and Jn = fn(J0) = Φ(gn(c0)), n ≥ 0. By Theorem 8.4 itsuffices to show that if zero is a limit point of the orbit of c0 then J0 is not a quasicircle.On a compact neighborhood of AS avoiding the point at infinity the spherical andplanar metrics are equivalent, hence J0 is a quasicircle if and only if it has boundedturning with respect to the planar metric. For simplicity we conjugate the map by1/z so that J+ is near zero and the Jn separate J+ from infinity.

By Theorem 8.2, if gn(c0) has zero as limit point, then there is a subsequenceJnk

→ J+ in the Hausdorff topology. In J+ we may find a cut point p and small opendiscs W ′ and W with W ⊂ W ′ and such that W ′ ∩ P = ∅, p ∈ W , and W containsa connected component L of J+ − {p}. Since p is a cut point p is accessible from thecomponent of C − J+ containing infinity (recall we conjugated our map by 1/z) viatwo distinct accesses ηx, ηy. Since Jnk

→ J+ in the Hausdorff topology and the Jnk

are critically separating, for k sufficiently large, there are points xk ∈ ηx ∩ Jnk∩ W

and yk ∈ ηy ∩ Jnk∩ W such that xk, yk → p and for which the component Lk of

Jnk−{xk, yk} of smallest diameter is also contained in W and has diameter bounded

below by D = diam(L).Since W ′ ∩ P = ∅, there is a univalent branch hk of (fnk)−1 on W ′ sending Lk

to a subarc of J0. Let x0k = hk(xk), y0k = hk(yk), and L0k = hk(Lk). Then theKoebe distortion theorem implies that since W is compactly contained in W ′, thereis a constant C ≥ 1 independent of k such that hk distorts ratios of planar distancesbetween points of W by at most a factor of C. Hence for all k

C−1 diam(Lk)|xk − yk|

≤ diam(L0k)|x0k − y0k|

≤ Cdiam(Lk)|xk − yk|

.

But diam(Lk) is bounded from below by D while |xk − yk| → 0. Hence J0 hasunbounded turning and so is not a quasicircle.


8.5. Constructing other examples. — The map f1 was initially found by usingquasiconformal surgery to glue z2 − 1 with 1/z3 in the appropriate fashion. Thisconstruction and its generalizations are the subject of work in progress. For example,one may apply the same construction by replacing z2−1 with any polynomial p whosefilled Julia set has interior, and glue it to 1/z3.Question If p is a quadratic polynomial with non-locally connected Julia set and aSiegel disc, can one obtain, by this gluing, a map with wandering Julia componentswhich are critically separating but not Jordan curves?

8.6. How to find the set E. — There are finitely many Fatou componentsW0, · · · ,Wm either intersecting P or separating P into at least three parts (Lemma2.2). For each Wi take the finitely many boundary components which are criticallyseparating. Saturated them into Julia components. Call the union of them E′. It isfinite and forward invariant. Fill in the disc components of C − E′ disjoint from P.We get a fattened E′′ of E′. We distinguish three types of components of C−E′′: typeI, components containing exactly one Wi (these are also the components intersecting⋃

Wi); type II, annulus components disjoint from⋃

Wi; and type III, non-annuluscomponents disjoint from

⋃Wi. In each of type III component, there are finitely

many Julia components separating P into at least three parts. Adding them to E′,we get E. Note that the set L is precisely the union of components of type I (see alsoLemma 3.10).

Let W be a fixed attracting basin for a rational map f . How do we know thatonly finitely many components of C − W intersect P? And if so how can we findVN in Lemma 7.3 and in the proof of Theorem 7.1? Here is a constructive answer.Take V0 a open disc with Jordan curve boundary in W such that f(V0) ⊂ V0. Let Vn

be the component of f−n(V0) containing V0. Then only finitely many components ofC−W intersect P if and only if there is a minimal integer N such that each boundarycurve of VN+1 is either not critically separating, or is parallel to a boundary curve ofVN . And for this VN , each component of C− VN contains at most one component ofC − W intersecting P.

9. Generalizations

9.1. Geometrically finite maps. — Our techniques allow a generalization of The-orem 1.2 to the case of geometrically finite maps f . In [TY] (§1, Step 4 and Prop.1.3)) (cf. also [DH2]) a Riemannian metric is constructed for which f is uniformlyexpanding on a neighborhood of its Julia set. The arguments given above for Propo-sitions [Case 2] and [Case 3] then apply in this more general setting. Proposition[Case 4] is proved for nice maps, therefore applies automatically to geometricallyfinite maps.

We may also recover a variant of Theorem 8.2 for geometrically finite maps as well,though we may lose the injectivity of Φ–it may be possible to construct a map fwith a periodic Jordan curve Julia component J0 which intersects P and which is thecommon boundary of two components of AS . Then J0 may be J ′(α) for two distinct


admissible sequences α. At worst, however, Φ is two-to-one. We define Σe,0 and Σe

the same as for hyperbolic maps.Theorem 8.4 holds for geometrically finite maps as well, and we may recover a

result of Cui et. al. ([CJS], Prop. 6.2):

Proposition 9.1. — If f is geometrically finite, then there are at most finitely manyperiodic Jordan curve Julia components J0 which are not quasicircles.

Proof. — Let Jn = fn(J0). Then either Jn ⊂ As for all n, or Jn is a boundarycomponent of a component of AS for all n, since J0 is periodic. The latter set of suchJ0 is finite, while in the former case J0 = J ′(α) for some unique α ∈ Σ. The sequenceα cannot accumulate on Σe,0 under σ since otherwise J0 accumulates on a boundarycomponent of AS , by Theorem 8.2. Hence J0 is a quasicircle by Theorem 8.4.

9.2. A further result for nice maps. —

Theorem 9.2. — If f is nice and every component of AS is also a component of A,then every Julia component J0 not eventually landing on a component of E is a point.

Proof. — The hypotheses imply that every component of AS is a Herman ring or apreimage of a Herman ring, therefore contains no Julia components. So Case 2 ofLemma 2.4 does not occur.

By Proposition [Case 4], if J0 is a Julia component such that Jn = fn(J0) ⊂ L forinfinitely many n, then J0 is a point.

Assume now J0 is a Julia component that is not in Cases 1, 2 and 4 of Lemma 2.4.Replacing J0 by a forward iterate of it if necessary, we may assume that Jn ∩ (E ∪AS ∪ L) = ∅ for all n ≥ 0. Furthermore, by Lemma 2.4, Jn ⊂ AO ∪ D′ for infinitelymany n. Therefore

Jn ⊂ Q = (AO −AOAS ∪ AOAL) ∪ (D′ −D′AS ∪ D′L ∪ D′AL)

for infinitely many n, where the notation ABC means {z|z ∈ A, f(z) ∈ B, f2(z) ∈ C}.We claim that Q is contained in the disjoint union of finitely many full continua

disjoint from P. We can then apply Lemma 1.4 to conclude that J0 is a point.Let A′ be a component of AO. Either it coincides with a component of AOAS ,

or A = f(A′) is a component of A − AS . Assume we are in the latter case. ByLemma 3.8, for δ± the boundary components of A, there are components V +, V − ofAL (which may coincide) such that δ+ ⊂ ∂V + and δ− ⊂ ∂V −. Thus A − (V + ∪ V −)is disjoint from ∂A. Hence A′ − A′AL is contained in A′. Since the inclusion mapfrom A′ into C−P is homotopic to a constant map, A′ − A′AL is contained in a fullcontinuum disjoint from P.

The case of a component of D′ is similar, using Lemma 3.7.

ATechnical results about plane topology

In this appendix we collect technical results used in the course of our proofs.The following lemma was used in the proof of Lemma 3.6.


Lemma A.1. — Let P 1 and P 2 be two disjoint non empty closed sets of S2. SetW = {U | U is a complement component of some component of J , P 1 ⊂ U andP 2 ∩ U = ∅}. Then W is either empty or totally ordered with respect to inclusion,and W =

⋃U∈W U as the unique maximal element, and ∂W is connected.

Proof. — That W is totally ordered with respect to inclusion if it is nonempty followsfrom Lemma 3.1. Since each U ∈ W is a disc and the U ’s are nested, we have thatW is a disc. Hence ∂W is connected.

We now show that W is also an element of W. Let x ∈ ∂W . For any integer n,there is a point x′ in W such that the distance between x′ and x is less than 1/n, andthere is U ∈ W such that x′ ∈ U . The segment [x, x′] intersects ∂U , which is a subsetof J . We conclude then either x ∈ ∂U or there are points of J arbitrarily close to x.In both cases x ∈ J .

Since ∂W is a connected subset of J , it is contained in some component S of J .Now W must be a component of C−S. This is because S ∩U = ∅ for any U ∈ W,

therefore S ∩ W = ∅.Moreover W ⊃ P 1, but W ∩ P 2 = ∅. So W is indeed the maximal element of

W.

The remaining results are essentially ingredients in the proof of Theorem 7.1. Thelogical dependencies are:

Lemma A.2 → Corollary A.3 → Corollary A.4 → Corollary A.5 → Lemma 3.10 →Theorem 7.1.

We will need the following fact from general topology.Let X be a compact Hausdorff space. Then every component Y of X coincides

with the intersection of open and closed subsets of X containing Y . Moreover, forany open neighborhood U of Y , there is a closed and open subset V of X such thatY ⊂ V ⊂ U . The component Y is always closed. In case Y is not open in X, there isa sequence of closed and open set Vn such that Y ⊂6= Vn ⊂6= Vn−1, and

⋂n Vn = Y .

If X is the Julia set of some map, and is disconnected, no component of X is openin X (see [Be]).

In the next three results, let J t F be a decomposition of S2 with J closed anddisconnected, let d denote spherical distance and dH the Hausdorff distance on closedsets with respect to d.

The following Lemma is known as Zoretti’s Theorem ([Wh], p. 109).

Lemma A.2. — Given J ′ and J ′′ two distinct components of J , and ε > 0, there isa Jordan curve γ in F separating J ′ and J ′′ such that supx∈γ d(x, J ′) < ε.

The next two results are easy consequences of Lemma A.2.

Corollary A.3. — 1. Let U be a connected component of S2 − J ′ for some con-nected component of J . Then there is a sequence of closed discs Dn bounded byJordan curves γn such that γn ⊂ U ∩ F , Dn ⊂ int(Dn−1), ∂U ⊂ int(Dn), andDH(∂U, γn) → 0 (where DH denotes the Hausdorff distance of compact sets).

2. Given W a component of F and K a component of S2 −W , there is a sequenceof closed disks Dn such that Dn ⊂ int(Dn−1), ∂Dn ⊂ W and

⋂n Dn = K.


Proof. — Let {Km}∞m=0 be a sequence of closed discs such that Km−1 ⊂ int(Km),∪mKm = U , and dH(∂U, ∂Km) → 0. Let J0 = J ∪ K0, F0 = S2 − J0. Then J0 isclosed and disconnected and Lemma A.2 implies that there is a Jordan curve γ0 ⊂F0 ⊂ F separating ∂U from ∂K0. Let D0 be the disc bounded by γ0 and containing∂U . Inductively define γn as follows. There is an m(n) such that γn−1 ⊂ int(Km(n)).Let Jn = J ∪ Km(n), Fn = S2 − Jn. Lemma A.2 implies that there is γn ⊂ Fn ⊂ Fseparating ∂U from Km(n) and bounding a closed disc Dn containing ∂U . This showsthe first part; the second is similar.

Corollary A.4. — The following conditions are equivalent:1. there is a component W of F contained in U such that ∂W ∩ ∂U 6= ∅;2. there is a component W of F contained in U such that ∂U is a component of

∂W ;3. there is an n, such that no component of J separates ∂U and γn, where γn is a

sequence of Jordan curves as in Corollary A.3, Part 1.

Proof. — That 2 =⇒ 1 is obvious; 2 =⇒ 3 follows directly from Part 2 of CorollaryA.3 and the hypothesis that ∂U is a component of ∂W . To see 3 =⇒ 2, note thatF ∩ (U −Dn) must be connected, where the Dn are as in Part 1 above. Let X = ∂W .Then ∂U is contained in a nested intersection of sets which are open and closed in X,hence ∂U is a connected component of ∂W . Finally, that 1 =⇒ 2 follows easily fromthe observation that if C0 is the connected component of ∂W which intersects ∂U ,then C0 is contained in some component J0 of J , hence C0 cannot separate points ofU .

Corollary A.5. — Let E be a nonempty set of finitely many disjoint components ofJ and let U be a connected component of S2 − E.

1. If ∂U is connected, let W0 be a component of F contained in U , and assume nocomponent of J separates W0 from ∂U .

Then ∂U is a connected component of ∂W0, and is the unique component ofF contained in U whose boundary intersects ∂U .

2. If ∂U is not connected, assume no component of J separates components of E.Then there exists a component W0 of F contained in U such that every con-

nected component of ∂U is a connected component of ∂W0, and W0 is the uniquecomponent of F contained in U whose boundary intersects each component of∂U .

Proof. — Assume first that ∂U is connected. By Corollary A.3, Part 1, there is asequence γn → ∂U of Jordan curves in F ∩ U ; we may take these curves to separatesome point w0 of W0 from ∂U . By hypothesis and Corollary A.4, Part 3, there existsa component W of F contained in U such that ∂U is a connected component of ∂W .If W 6= W0, then since W0,W ⊂ U we must have that W and W0 are separated by acomponent J ′ of J which is contained in U . Then W ∪ ∂U is a connected set whichis separated from W0 by J ′, hence ∂U is separated from W0 by J ′, a contradiction.The uniqueness assertion follows from the equivalence of the first two parts in theprevious lemma, and the proof of the second case is similar.


BProof of Proposition 7.2

A subtlety to prove Proposition 7.2 is that the map f : ∂W ′ → ∂W need not beopen in the subspace topology. We first establishClaim Let f be a rational map and let W ′,W be two Fatou components of f suchthat f(W ′) = W . Let K (resp. K ′) be a component of C − W (resp. C − W ′) suchthat f(∂K ′) ∩ ∂K 6= ∅. Then

(1) f(∂K ′) = ∂K .(2) f(K ′) ⊃ K.(3) K ∩ P = ∅ if and only if K ′ ∩ f−1P = ∅. In this case f(K ′) = K.(4) For any set S ⊃ P , K ′ ∩ f−1S 6= ∅ if and only if K ∩ S 6= ∅.

Proof of Proposition 7.2. — 1. By (1) above, if K ′ is a connected component ofC − W ′, then f(∂K ′) = ∂K where K is a connected component of C − W . By (1)and (4), there are finitely many such K ′ for which K ′ ∩ f−1(P) 6= ∅. 2. It sufficesto show f(U ′) ⊂ U and f(∂U ′) ⊂ ∂U . Since U ′ is the union of W ′ with componentsof C − W ′ disjoint from f−1P, this follows from (1) and (3). 3. Again this followsdirectly from the finiteness of the number of boundary components and (1).

Proof of Claim. — (1) Denote by J the Julia component containing ∂K.Denote by V the component of C − J containing W and by V ′ the component of

C − f−1J containing W ′ . Then f : V ′ → V is proper and f maps each boundarycomponent of V ′ onto the boundary of V , which is ∂K ′ (see Lemma 3.3).

Clearly f(∂K ′) ⊂ ∂K. So ∂K ′ ∩ V ′ = ∅. Since W ′ ⊂ V ′ and ∂K ′ ⊂ ∂W ′, wehave ∂K ′ ⊂ ∂V ′. So ∂K ′ is contained in a component S′ of ∂V ′. But S′ is in theJulia set, and the Julia component of ∂K ′ is contained in K ′, so S′ ⊂ K ′. Since ∂K ′

separates W ′ from int(K ′), no point of int(K ′) can be in V ′. Therefore S′ ⊂ ∂K ′.Thus S′ = ∂K ′ and f(∂K ′) = ∂K.(2) A rational map f maps connected components of preimages of a compact con-nected set surjectively onto its image (see [Be], Ch. 5). Let L′ be the connected com-ponent of the preimage of K intersecting ∂K ′. Then f(L′) = K and ∂K ′ ⊂ L′ ⊂ K ′.Hence f(K ′) ⊃ K.(3) Assume at first that K ∩ P = ∅. Since K is full the component L′ in point (2)is also full. A simple topological argument then shows K ′ = L′. So f(K ′) = K andK ′ ∩ f−1P = ∅.

Assume now that K ′ ∩ f−1P = ∅. If K ∩ P 6= ∅, since f(K ′) ⊃ K we would haveK ′ ∩ f−1P 6= ∅ which is impossible. So K ∩ P = ∅, K ′ = L′ and f(K ′) = K.(4) Assume at first that K ∩ P = ∅. Since f(K ′) = K, we get 4). Assume nowK ∩ P 6= ∅. Then K ∩ S 6= ∅ and K ′ ∩ f−1S 6= ∅ (for otherwise K ′ ∩ f−1P = ∅ andK ∩ P = ∅).

References

[AM] J. Anderson and B. Maskit, On the local connectivity of limit sets of Kleiniangroups, Complex Variables, 31 (1996), 177-183.


[Be] A. Beardon, Iteration of Rational Functions. Springer Verlag Graduate Texts inMathematics 132, 1991.

[Br] H. Brolin, Invariant sets under iteration of rational functions. Arkiv for Matematic6(6)(1965), 103-144.

[CJS] G. Cui, Y. Jiang, and D. Sullivan, Dynamics of geometrically finite rational maps.Manuscript 1996.

[DH1] A. Douady and J.H. Hubbard, On the dynamics of polynomial-like mappings.

Ann. Sci. Ec. Norm. Sup.18 (1985), 287-344.

[DH2] A. Douady and J.H. Hubbard, Etude dynamique des polynomes complexes, II.Pub. Math. d’Orsay, 1984.

[LV] O. Lehto and K.J. Virtanen, Quasiconformal mappings in the plane. Berlin-Heidelberg-New York: Springer, 1973.

[Mc1] C. McMullen, Automorphisms of rational maps. In Holomorphic Functions andModuli I, pages 31-60. Springer Verlag, 1988.

[Mc2] C. McMullen, Self-similarity of Siegel disks and Hausdorff dimension of Julia sets.Preprint 1995.

[Mc3] C. McMullen, The classification of conformal dynamical systems, Current Devel-opments in Mathematics 1995, International Press, 1995, pp. 323-360.

[Mi] J. Milnor, Dynamics in One Complex Variable: Introductory Lectures, preprint1990/5 of SUNY at Stony Brook.

[Ne] M.H.A. Newman, Elements of the topology of planesets of points. Cambridge Uni-versity Press, 1961.

[St] N. Steinmetz, Rational Iteration: Complex Analytic Dynamical Systems. DeGruyter Studies in Mathematics 16, 1993.

[TY] Tan Lei and Yin Yongcheng, Local connectivity of the Julia set for geometricallyfinite rational maps. Science in China, Series A (1) 39(1996), pages 39-47.

[Wh] G. T. Whyburn, Analytic Topology. AMS Colloquium Publications vol. 28 (1942).

Kevin PILGRIM, Department of Mathematics and Statistics, University of Missouri, Rolla

Building, 1870 Minor Circle, Rolla, MO 65409-0020, U.S.A. • E-mail : [email protected]

Url : http://www.umr.edu/~mathstatTAN Lei, Department of Mathematics, University of Warwick, Coventry CV4 7AL, U.K.

E-mail : [email protected] • Url : http://www.maths.warwick.ac.uk

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