The Maximal Entropy Measure of Fatou Boundaries
aa r X i v : . [ m a t h . D S ] A ug The Maximal Entropy Measure of FatouBoundaries ∗ Jane HawkinsDepartment of Mathematics, University of North CarolinaChapel Hill, NC 27599, USAMichael TaylorDepartment of Mathematics, University of North CarolinaChapel Hill, NC 27599, USA
Abstract
We look at the maximal entropy (MME) measure of the boundariesof connected components of the Fatou set of a rational map of degree ≥
2. We show that if there are infinitely many Fatou components,and if either the Julia set is disconnected or the map is hyperbolic,then there can be at most one Fatou component whose boundary haspositive MME measure. We also replace hyperbolicity by the moregeneral hypothesis of geometric finiteness.
Let R : b C → b C be a rational map of degree d ≥
2, defined on the Riemannsphere b C , F its Fatou set, J its Julia set, and λ the unique maximal entropymeasure (MME) on J (i.e., h λ ( R ) = log d ). Recall that F and J areinvariant under R , R preserves λ and acts ergodically on ( J , B , λ ) , (1.1)where B consists of the Borel subsets of J , andsupp λ = J . (1.2) ∗ The second author was partially supported by NSF grant DMS-1500817
Key words. complex dynamics, Julia sets, Fatou sets, measure § λ ( ∂ O ), when O is a connected component of F , and in particular to understand when wecan say λ ( ∂ O ) = 0.The results in this paper hold for any Borel probability measure µ satis-fying (1.1) and (1.2) It is well known that for many rational maps there aremeasures in addition to the MME measure λ that have these properties. Wefocus on λ to complement an earlier result by the authors on the construc-tion of the MME. In the jointly written appendix of [5], the authors showedthat the backward random iteration method works for drawing the Julia setof any rational map. They proved the result by showing the associated deltamass measures, averaged along almost any randomly chosen backward path,converge to λ . It is observed that certain details of Julia sets have a fuzzyappearance for many maps when this algorithm is used, and the results ofthis paper provide an explanation of this observation.To get started, given a rational map R and a component O ⊂ F , Sul-livan’s Non-wandering Theorem implies there exist m ∈ N and a collection O , . . . , O k of mutually disjoint connected components of F such that R : O j −→ O j +1 , j, j + 1 ∈ Z / ( k ) , (1.3)and R m ( O ) ⊂ O , hence R m ( ∂ O ) ⊂ ∂ O . (1.4)Our first goal is to establish the following. Theorem 1.1
Let O , . . . , O k be mutually disjoint connected components of F such that (1.3) holds. Then either λ ( ∂ O j ) = 0 , ∀ j ∈ Z / ( k ) , (1.5) or ∂ O = · · · = ∂ O k = J . (1.6) If (1.5) holds, then λ ( ∂ O ) = 0 whenever O is a component of F such that O ⊂ R − M ( O j ) , M ∈ N , and any j . The following general result will be useful in the proof of Theorem 1.1.
Lemma 1.2
Let ( J , B , λ ) be a probability space, and assume R : J → J is B -measurable and λ is ergodic and invariant under R . Then, for E ∈ B , R − ( E ) ⊃ E = ⇒ λ ( E ) = 0 or . (1.7)2 roof. Let F j = R − j ( E ). Then E = F ⊂ F j ր F ∈ B . The hypothesesalso imply λ ( F ) = λ ( E ) and R − ( F ) = F . Ergodicity implies λ ( F ) = 0 or1, so we have (1.7). (cid:3) Proof of Theorem 1.1.
Lemma 1.2 applies to E = ∂ O ∪ · · · ∪ ∂ O k , and wehave (1.7). Also R − ( ∂ O j +1 ) ⊃ ∂ O j , so λ ( ∂ O j +1 ) ≥ λ ( ∂ O j ) , ∀ j ∈ Z / ( k ) . (1.8)This implies λ ( ∂ O ) = · · · = λ ( ∂ O k ) . (1.9)Applying (1.7) leads to two cases: λ ( ∂ O ∪ · · · ∪ ∂ O k ) = 0 , or λ ( ∂ O ∪ · · · ∪ ∂ O k ) = 1 . (1.10)In the first case, (1.5) holds. Furthermore, λ ( ∂ O ) = 0 for each O ⊂ R − M ( O j ), M ∈ N , since R M ( O ) = O j ⇒ R M ( O ) = O j and R M ( ∂ O ) ⊂ ∂ O j ⇒ ∂ O ⊂ R − M ( ∂ O j ) . (1.11)In the second case, since ∂ O ∪ · · · ∪ ∂ O k is compact and (1.2) holds, wehave ∂ O ∪ · · · ∪ ∂ O k = J . (1.12)Going further, let us note that R and R k have the same Julia set andthe same maximal entropy measure. Given (1.3), we have R k : O j → O j foreach j ∈ Z / ( k ), hence R k : ∂ O j → ∂ O j . Hence, by Lemma 1.2 (applied to R k ), for each such j , λ ( ∂ O j ) = 0 or 1 . (1.13)Then λ ( ∂ O j ) = 1 = ⇒ ∂ O j = J . (1.14)We still have (1.9), so λ ( ∂ O j ) = 1 = ⇒ ∂ O = · · · = ∂ O k = J . (1.15)This proves Theorem 1.1. (cid:3) The following result of [3] is a significant consequence of Theorem 1.1.3 heorem 1.3
Either λ ( ∂ O ) = 0 for each connected component O of F , (1.16) or there is a connected component O of F such that ∂ O = J . (1.17) More precisely, O can be taken as in (1.3) – (1.4) , and then (1.6) holds. We can rephrase Theorem 1.3 using the notion of residual Julia set ,defined by J = J \ [ j ∂ O j , (1.18)where {O j } consists of all the connected components of F . Note that(1.16) ⇒ J = ∅ , and (1.17) ⇒ J = ∅ . (1.19)As noted in [3], it follows from Theorem 1.3 that J = ∅ = ⇒ λ ( J ) = 1 , (1.20)and we have the following basic result of [9]. Corollary 1.4
Either J = ∅ or there is a Fatou component O such that ∂ O = J . If J = ∅ , then J is a dense, G δ subset of J . To establish the results in the last sentence of Corollary 1.4, we notethat denseness of J in J follows from (1.20) and (1.2). The fact that J is obtained from J by successively removing ∂ O j , j ∈ N , implies J is a G δ subset of J .Our goal is the study of λ ( ∂ O ) for various Fatou components in caseswhere this measure is not identically zero on these boundaries. An im-portant class of rational maps with empty residual Julia set is the class ofpolynomials of degree d ≥
2, which we tackle in §
2. In this case, the Fatoucomponent O ∞ containing ∞ satisfies ∂ O ∞ = J . (1.21)We give conditions under which we can show that, if O is another componentof F , i.e., a bounded component of F , then λ ( ∂ O ) = 0 . (1.22)4e demonstrate this for a class of polynomials whose Fatou sets have aninfinite number of components. One important property of O ∞ used in theanalysis is its complete invariance: R ( O ∞ ) ⊂ O ∞ and R − ( O ∞ ) ⊂ O ∞ .In § R for which there is a completely invariant Fatou component O , so R ( O ) ⊂ O and R − ( O ) ⊂ O . (1.23)We extend the basic results of § § §§ Theorem 1.5
Let R be a rational map of degree ≥ , and assume F hasinfinitely many connected components. Assume there is a Fatou component O such that λ ( ∂ O ) = 0 . Then λ ( ∂ O ) = 0 for each Fatou component O 6 = O , under either of the following conditions: J is disconnected, (A) or J is connected and R is hyperbolic. (B) Moreover, if (A) or (B) hold, then ∂ O = J , and O is completely invariant. Theorem 1.5 follows from Propositions 3.3 and 4.1. We mention thatProposition 3.2 has the same conclusion, with hypothesis (B) replaced by J is connected and locally connected, and F has a completely invariant component. (B1)The special case when R is a polynomial and J is connected and locallyconnected is Proposition 2.3. As we show in §
4, using (B1), we can alsoreplace (B) by J is connected and R is geometrically finite. (B2)We end with an appendix, giving examples to illustrate our results. Here we explore consequences of Theorem 1.1 for the class of polynomialsof degree d ≥
2, e.g., R ( z ) = z d + a d − z d − + · · · + a , d ≥ . (2.1)5If the leading term were a d z d , we could conjugate by z cz with c d − = a d ,to obtain the form (2.1).) In such a case, F has a connected component O ∞ ,containing ∞ , and (cf. [1], Theorem 5.2.1) J = ∂ O ∞ . (2.2)It clearly follows from (2.2) that for polynomials, J = ∅ and λ ( ∂ O ∞ ) = 1 . The set K = b C \ O ∞ is called the filled Julia set of R , and one has J = ∂ K .In some cases, F = O ∞ . If F has exactly 2 connected components, say O ∞ and O , then it is also the case that ∂ O = J . Otherwise, F must havean infinite number of connected components (cf. [1], Theorem 5.6.2). Wecall the connected components of F other than O ∞ , i.e., those contained in K , bounded Fatou components . The component O ∞ is completely invariant.Hence, if O is a bounded Fatou component, so is R ( O ) and so is eachconnected component of R − ( O ). We now look into when a bounded Fatoucomponent O can be shown to satisfy λ ( ∂ O ) = 0. Proposition 2.1
Let R ( z ) have the form (2.1) , and filled Julia set K . As-sume there is a point z ∈ K such that K \ { z } is not connected. Then λ ( ∂ O ) = 0 for each bounded Fatou component O of R .Proof. Take a bounded Fatou component
O ⊂ K . The non-wandering the-orem gives m ∈ N and Fatou components O , . . . , O k (necessarily boundedFatou components) such that (1.3)–(1.4) hold. Say K and K are disjointconnected components of K \ { z } . We can assume O ⊂ K . Then O doesnot contain ∂ K , so (1.6) fails. By Theorem 1.1, this forces λ ( ∂ O ) = 0,and hence λ ( ∂ O ) = 0. (cid:3) Of course, Proposition 2.1 applies if K is not connected. We will nowassume K is connected. Hence b C \ K is simply connected. By the Riemannmapping theorem, there is a unique biholomorphic map ϕ : b C \ D −→ b C \ K , (2.3)such that ϕ ( ∞ ) = ∞ , Dϕ ( ∞ ) = α > . (2.4)Here D is the disk { z ∈ C : | z | ≤ } . We use this to recall from Chapter6 of [7] conditions under which the hypothesis of Proposition 2.1 holds. Itinvolves the notion of external rays, ψ θ ( r ) = ϕ ( re iθ ) , r ∈ (1 , ∞ ) , θ ∈ T = R / (2 π Z ) . (2.5)6iven θ ∈ T , we say ψ θ lands at a point z ∈ ∂ K providedlim r ց ψ θ ( r ) = z. (2.6)We mention two classical results, given as Theorems 6.1 and 6.2 of [7]. Thefirst is that the limit in (2.6) exists for almost all θ ∈ T . The second is thefollowing:If E ⊂ T has positive measure, then there exist θ , θ ∈ E such thatlim r ց ψ θ ( r ) = z , lim r ց ψ θ ( r ) = z , and z = z . (2.7)These results lead to the following, which is part of Corollary 6.7 of [7]. Proposition 2.2
Let R have the form (2.1) and assume K is connected.Take ϕ and ψ θ as in (2.3) – (2.5) . Assume there exist ξ = ξ ∈ T such that lim r ց ψ ξ ( r ) = lim r ց ψ ξ ( r ) = z . (2.8) Then
K \ { z } is not connected.Proof. The union γ of the two rays { ψ ξ ( r ) : 1 < r < ∞} and { ψ ξ ( r ) :1 < r < ∞} , together with their endpoints z and ∞ , forms a simple closedcurve in b C . The Jordan curve theorem implies that b C \ γ has exactly twoconnected components. Our hypotheses imply K ∩ γ = { z } . (2.9)It remains to note that each connected component of b C \ γ contains a pointof K , and this follows from (2.7), first taking E to be the open arc from ξ to ξ in T , then taking E to be the complementary open arc in T . (cid:3) Regarding the question of when Proposition 2.2 applies, we note that adefinite answer can be given under the additional hypothesis that J is connected and locally connected, (2.10)or equivalently, that K is connected and locally connected. In that case,a classical result of Caratheodory (cf. [8], Theorem 17.14) implies that ϕ in (2.3), mapping D ∞ = b C \ D to O ∞ = b C \ K , has a unique continuousextension to ϕ : D ∞ −→ O ∞ . (2.11)7he image of ϕ in (2.11) is both compact and dense in O ∞ , so ϕ in (2.11)is surjective. It restricts to a continuous map ϕ : ∂ D −→ J , (2.12)also surjective. In view of (2.5), we have ϕ ( e iθ ) = lim r ց ψ θ ( r ) . (2.13)Now if ϕ in (2.12) is also one-to-one, this makes J ⊂ b C a simple closedcurve, so F = b C \ J would have just two connected components. We havethe following conclusion. Proposition 2.3
Let R have the form (2.1) and assume J is connectedand locally connected. If F has infinitely many connected components, then λ ( ∂ O ) = 0 for each bounded component O of F . (2.14) Proof.
As we have just seen, the hypotheses imply that ϕ in (2.12) is notone-to-one. Hence there exist ξ = ξ ∈ T such that ϕ ( e iξ ) = ϕ ( e iξ )in (2.13). Thus, with z = ϕ ( e iξ ) = ϕ ( e iξ ), Proposition 3.2 implies that K \ { z } is not connected, so (2.14) follows from Proposition 2.1. (cid:3) Extending our scope a bit, let us now assume that R (of degree d ≥
2) hasthe property that F has a completely invariant connected component O ,i.e., R ( O ) ⊂ O and R − ( O ) ⊂ O . (3.1)If there is a fixed point p ∈ O of R , then, conjugating by a linear fractionaltransformation, we can take p = ∞ . If R − ( p ) = p as well, then R is apolynomial (which guarantees that p must be a superattracting fixed point).There are many examples of rational maps with completely invariant Fatoudomains and attracting fixed points that are not polynomials. We discusssome in Appendix A.Generally when (3.1) holds, we have ∂ O = J . (3.2)8his follows from Theorem 5.2.1 of [1], which also contains the results that O is either simply connected or infinitely connected; all the other compo-nents of F are simply connected; and O is simply connected if and onlyif J is connected. Using these facts, we can set things up to obtain resultsparallel to Propositions 2.1–2.3. To formulate these results, let us arrangethat ∞ ∈ O . Then O is the unbounded component of F , and we callthe other components of F bounded Fatou components. As in the case ofpolynomials, we see that if O is a bounded Fatou component, so is R ( O )and so is each connected component of R − ( O ). We set K = b C \ O , andcall this the filled Julia set of R . The following result has the same proof asProposition 2.1. Proposition 3.1
Let R be a rational map of degree ≥ for which a Fatoucomponent O satisfies (3.1) , and let K = b C \ O be its filled Julia set.Assume there exists z ∈ K such that K \ { z } is not connected. Then λ ( ∂ O ) = 0 for each Fatou component O 6 = O . As before, Proposition 3.1 certainly applies if K is not connected. If K is connected, then O is simply connected, and we again have the set-up(2.3)–(2.7), and Proposition 2.2 extends to this setting, as does the analysisleading to the following extension of Proposition 2.3. Proposition 3.2
Let R be a rational map of degree ≥ for which there is aFatou component O satisfying (3.1) , and assume J is connected and locallyconnected. If F has infinitely many connected components, then λ ( ∂ O ) = 0 for each Fatou component O 6 = O . In counterpoint, we have the following result when J is not connected. Proposition 3.3
Let R be a rational map of degree ≥ , and assume J isdisconnected. If there is a Fatou component O such that λ ( ∂ O ) = 0 , then λ ( ∂ O ) = 0 for each Fatou component O 6 = O .Proof. The hypotheses imply that the residual set J is empty. Hence,given given J disconnected, by Theorem 4.4.19 of [10], F has a completelyinvariant component O . By Proposition 3.1, λ ( ∂ O ) = 0 for each Fatoucomponent O 6 = O . This proves the proposition (and forces O = O ). (cid:3) Remark.
As seen in §
1, the hypothesis that λ ( ∂ O ) = 0 for some Fatoucomponent O is equivalent to the hypothesis that J = ∅ . The Makienko onjecture can be stated as saying that, if J = ∅ , then there is a Fatoucomponent that is completely invariant under R . A discussion of conditionsunder which Makienko’s conjecture has been proved can be found in [2]. Ofparticular use here is the following result of [11]:The Makienko conjecture holds provided J is locally connected. (3.3)We make use of this in the following section. A rational map R : b C → b C , of degree d ≥
2, with Julia set J , is said to be hyperbolic provided J ∩ C + ( R ) = ∅ , where C + ( R ) = [ k ≥ R k ( C R ) , (4.1)where C R = { z ∈ b C : DR ( z ) = 0 } is the set of critical points of R . Anequivalent condition for hyperbolicity (cf. [10], Theorem 4.4.2) is that eachcritical point of R is in F and each forward orbit of a critical point convergesto an attracting cycle. Hyperbolic rational maps are relatively “tame,” froma topological point of view. For example (cf. [10], Theorem 4.4.5), if R ishyperbolic, J connected = ⇒ J locally connected. (4.2)There is also a partial converse to the implication (3.1) ⇒ (3.2) for hyperbolicmaps (cf. [10], Theorem 4.4.16):If R is hyperbolic and J is connected, and O is a component of F , then R ( O ) ⊂ O , ∂ O = J = ⇒ O is completely invariant. (4.3)We therefore have the following. Proposition 4.1
Assume that R is hyperbolic and J is connected, that F has infinitely many connected components, and that there is a Fatou compo-nent O such that λ ( ∂ O ) = 0 . Then λ ( ∂ O ) = 0 for each Fatou component O 6 = O .Proof. By Theorem 1.3, there must be a component O of F such that ∂ O = J . Furthermore, O is invariant under R k , which is also hyperbolic.By (4.3), such O must be completely invariant (under R k ). Thanks to104.2), Proposition 3.2 applies (to R k ), and it implies that λ ( ∂ O ) = 0 foreach component O 6 = O .Incidentally, this forces O = O . Furthermore, taking into account(1.6), we see that k must be 1. Hence O must be completely invariantunder R . (cid:3) For further results, we bring in the following two generalizations of hy-perbolicity, taking definitions from [10], p. 153.
Definition.
A rational map R : b C → b C of degree ≥ subhy-perbolic provided the following two conditions hold:the forward orbit of each critical point in F converges to an attracting cycle , (4.4)and the forward orbit of each critical point in J is eventually periodic. (4.5)If we merely assume (4.5) holds, we say R is geometrically finite .Clearly R hyperbolic ⇒ R subhyperbolic ⇒ R geometrically finite. Theusefulness of geometric finiteness for the work here stems from the followinggeneralization of (4.2), established in [12], Theorem A:If R is geometrically finite and J is connected,then J is locally connected. (4.6)In concert with (3.3), this yields the following: Lemma 4.2
Assume that R is geometrically finite and J is connected.Then either J = ∅ or F has a completely invariant component for R . We now establish the following extension of Proposition 4.1.
Proposition 4.3
Assume R is geometrically finite, J is connected, and F has infinitely many connected components. Assume further that there is aFatou component O such that λ ( ∂ O ) = 0 . Then λ ( ∂ O ) = 0 for each Fatoucomponent O 6 = O .Proof. The hypotheses imply J = ∅ , by (1.20). Hence, by Lemma 4.2,there is a component O of F that is completely invariant under R . As in113.2), this implies ∂ O = J . We are assuming J is connected, and by (4.6) J is also locally connected. Thus Proposition 3.2 applies (to R ), giving λ ( ∂ O ) = 0 for each Fatou component O 6 = O .As before, this forces O = O . It also implies that O is completelyinvariant under R . (cid:3) We turn to a proof of Theorem 1.5.
Proof of Theorem 1.5 . If (A) holds, the conclusion that λ ( ∂ O ) = 0 for eachFatou component O 6 = O is a direct consequence of Proposition 3.3. Theproof of Proposition 3.3 also ends with the comment that O must coincidewith a completely invariant component O . Similarly, if (B) holds, the con-clusion of the theorem follows from Proposition 4.1, the complete invarianceof O again established in the course of the proof of Proposition 4.1. Validityof Theorem 1.5 with (B) replaced by (B1) follows from Proposition 3.2, andits validity with (B) replaced by (B2) follows from Proposition 4.3. (cid:3) A Examples
We include a few basic examples illustrating the results of §§ J = ∅ . Before getting to them, we make a few useful generalcomments whose proofs are found in the literature in the bibliography. If R is a rational map of degree d ≥ k -periodic point z whose multipliersatisfies: DR k ( z ) = exp(2 πi/m ), m ∈ N , then the k -cycle is called parabolic ; k always refers to the minimum period. Remarks A.
1. If a rational map R has an attracting k -periodic point in F , then thereare at least k Fatou components in F .2. If R has a parabolic k -periodic point z with multiplier DR k ( z ) =exp(2 πi/m ), then there are at least mk Fatou components in F .3. These lead to the following lemma. Lemma A.1
Let R : b C → b C be a rational map of degree ≥ . Assumeits Fatou set F has a completely invariant component and that R hasan attracting or parabolic k -periodic point with k ≥ . Then F hasinfinitely many components. roof. The completely invariant component cannot be one of the com-ponents in the n -cycle, n ≥
2, of components in F induced by thenon-repelling k -cycle for R so F has more than two components. (cid:3)
4. Whenever R : b C → b C is a rational map of degree 2, the Julia set J iseither connected or totally disconnected.5. If every critical point in J has a finite forward orbit (i.e., R is geomet-rically finite), then all non-repelling periodic points of R are attractingor parabolic. If R has degree d ≥ d − F , then R is geometrically finite. Example 1.
As an example of a non-polynomial map with a completelyinvariant Fatou component, neither hyperbolic nor subhyperbolic, but geo-metrically finite, we set: R ( z ) = z + 1 z + 32 . (A.1) R has critical points c = 1 and c = − ∞ is a fixed point with multiplier 1 so the map isparabolic; this implies that ∞ , ∈ J and there is an attracting petal in F which sits in a component O ⊂ F containing a critical point. We have: −
7→ − /
7→ − F , O and O = R ( O ). Therefore c ∈ O . Since there are no other critical points, R isgeometrically finite by Remark A.5.There are infinitely many components O ⊂ F ; in fact all
O 6 = O mapto O ∪ O ; the first statement follows from Lemma A.1, and the secondsince we have exhausted the critical points so no other Fatou componentsare possible. Every component of F is simply connected by Remark A.4.Since c ∈ O and R ( O ) = O , O is a degree 2 branched cover ofitself, so O is completely invariant. The map R is geometrically finite, and J is connected so it is locally connected using (4.6). Then λ ( ∂ O ) = 1 and λ ( ∂ O ) = λ ( ∂ O ) = 0 by Proposition 3.2.Because of the parabolic point at ∞ , R is neither hyperbolic nor subhy-perbolic, since lim n →∞ R n ( c ) ∈ J causing (4.1) and (4.4) to fail.13 xample 2. Here is another example of a geometrically finite but nonhyperbolic rational map. R ( z ) = z − z z + 1 . (A.2)We list a few easily verifiable properties of R : • There are 3 fixed points at − , , and ∞ , with respective multipliers1 / , −
1, and 2. • Since − ∈ J has twoattracting petals, F has infinitely many components by Lemma A.1. • Since R is quadratic, and J is not a Cantor set, J is connected usingA.4; therefore each component O ⊂ F is simply connected. • Let O denote the component of F containing the attracting point at −
2. Then since O contains a critical point we see that R : O → O is a 2-fold branched covering of itself, and since R is degree 2, thereare no other components of R − O so it is completely invariant. • The geometric finiteness comes from the fixed point at − F , and similarly the fixed point at 0 attracts the othercritical point, which is also in F .This shows the hypotheses of Theorem 1.5 (B2) are satisfied; therefore λ ( ∂ O ) = 1, and for any component O mapping onto the components of F containing the petals, we have λ ( ∂O ) = 0. Remark.
An interesting contrast between Examples 1 and 2 lies in thedifferent natures of their respective completely invariant Fatou components.
Example 3.
We turn to a hyperbolic polynomial example. The map P ( z ) = z −