Indecomposable continua and the Julia sets of polynomials, II

Abstract. Makienko's conjecture, a proposed addition to Sullivan's dictionary, can be stated as follows: The Julia set of a rational function R : C ∞ → C ∞ has buried points if and only if no component of the Fatou set is completely invariant under the second iterate of R. We prove Makienko's conjecture for rational functions with Julia sets that are decomposable continua. This is a very broad collection of Julia sets; it is not known if there exists a rational functions whose Julia set is an indecomposable continuum.

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“…We first state a useful theorem from [CMR06]. See also [R98], where a similar theorem first appeared, and [CMTT06] for variants.…”

Section: Lemma 2 There Exists a Fatou Componentmentioning

confidence: 99%

Any counterexample to Makienko’s conjecture is an indecomposable continuum

Curry

Mayer

Meddaugh

et al. 2009

Ergod. Th. Dynam. Sys.

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“…Without going into detail (but see [4]), the impression of a prime end of U is the intersection of the shadows of a sequence (A n ) ∞ n=1 of crosscuts of U having the property that for each n, (A m ) m>n is a pairwise closure disjoint null sequence contained in one of the crosscut neighborhoods of A n .…”

Section: Definition 13 a Sequence (U N )mentioning

confidence: 99%

“…In [4,5], dynamical considerations rule out that the Julia set of a polynomial can be the union of two proper indecomposable subcontinua. However, this is under the hypothesis that the Julia set is the boundary of one of its complementary domains.…”

Section: Corollary 24 (Burgess [2 Corollary To Theorem 9]) If Thementioning

confidence: 99%

“…A long-standing question in complex analytic dynamics asks: Can the Julia set of a rational function be an indecomposable continuum? Several authors have attacked this question, among them [14,4] for polynomials, [22] for bicritical rational maps (rational maps with exactly two critical points), and [5] for Julia sets of a class of rational functions with no buried points. In this situation, it is much easier to analyze the complement of the Julia set, called the Fatou set ; this motivates our interest in studying indecomposability from the point of view of a continuum's complement.…”

Section: Introductionmentioning

confidence: 99%

“…The second-named author, with various co-authors [14,4,5], investigated the recognition of indecomposable continua from their complements in the case that ∂U = X for some complementary domain U of X. The tool used was prime end theory, and a characterization was obtained in the context of X being a Julia set, making use of the dynamics.…”

Section: Introductionmentioning

confidence: 99%

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Characterizing indecomposable plane continua from their complements

Curry

Mayer

Tymchatyn

2008

Proc. Amer. Math. Soc.

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Abstract. We show that a plane continuum X is indecomposable iff X has a sequence (U n ) ∞ n=1 of not necessarily distinct complementary domains satisfying the double-pass condition: for any sequence (A n ) ∞ n=1 of open arcs, with A n ⊂ U n and A n \ A n ⊂ ∂U n , there is a sequence of shadows (S n ) ∞ n=1 , where each S n is a shadow of A n , such that lim S n = X. Such an open arc divides U n into disjoint subdomains V n,1 and V n,2 , and a shadow (of A n ) is one of the sets ∂V n,i ∩ ∂U.

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The Julia sets of quadratic Cremer polynomials

Blokh

Oversteegen

2006

Topology and its Applications

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Indecomposable continua and the Julia sets of polynomials, II

Cited by 9 publications

References 14 publications

Any counterexample to Makienko’s conjecture is an indecomposable continuum

Any counterexample to Makienko’s conjecture is an indecomposable continuum

Characterizing indecomposable plane continua from their complements

The Julia sets of quadratic Cremer polynomials

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