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We know that there exists a polynomial the Fatou set $F(P)$ is connected, which is just an attracting basin for infinity.

I have a question: Given a rational function $R$ such that $F(R)$ is connected, is this always true that $F(R)$ is just a attracting basin?

If the question is not, I wondered whether there exists a rational function $R$, such that $F(R)$ is connected and $F(R)$ is a completely invariant parabolic petal.

Further question: Is it possible to classify all rational maps $R$ with exactly one one Fatou component.

Any comments and remarks will be appreciated.

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2 Answers 2

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Clearly if $F(R)$ is connected, then $F(R)$ consists of a single attracting or parabolic basin.

Both cases can occur. Indeed, if you consider the slice $\operatorname{Per}_1(1)$ of quadratic rational maps having a fixed point with multiplier one, then this slice has an analog of the Mandelbrot set inside it, which is homemorphic to the Mandelbrot set, with corresponding maps being topologically conjugate on their Julia sets, by work of Petersen and Roesch. ("The parabolic Mandelbrot set", see the slides at http://www.math.univ-toulouse.fr/~roesch/Banff.pdf .) So there are many maps of this type, having a completely invariant parabolic basin. It won't be hard work to find an explicit example.

In general, it might be plausible that any example is topologically conjugate to a polynomial (on the corresponding Julia sets). But I am not aware that this has been proven.

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To complement the answer of Lasse Rempe-Gillen. If there is one component, then it is completely invariant. It is either an immediate domain of an attracting fixed point or of a petal. If it is the domain of an attracting fixed point, then (I believe) there is a polynomial, which is (topologicaly, quasiconformally) conjugate with our rational function in a neighborhood of the Julia set. I do not have a reference but this seems to be easy to prove. Then the problem of classification of such rational functions is reduced to classification of polynomials with only one Fatou component. But I am not sure in what sense a "classification" is required.

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  • $\begingroup$ Dear Alex, surely the "classification" of polynomials with only one Fatou components are those not having attracting, parabolic or Siegel periodic points. I agree that in the attracting case what you say is OK, but presumably in the parabolic case it would require a good bit more serious work (given that the work of Petersen and Roesch, which gives this for the slice Per_1(1), seems quite intricate). $\endgroup$
    – Lasse Rempe
    Commented Sep 10, 2015 at 11:39
  • $\begingroup$ @Lasse Rempe-Gillen: I agree with what you say about neutral case. But "classification" of polynomials with one Fatou component, on my point of view, includes MLC conjecture:-) $\endgroup$
    – Alexandre Eremenko
    Commented Sep 10, 2015 at 12:10
  • $\begingroup$ @AlexandreEremenko Thank you, professor. Thanks for your nice answer and comments. It is really a pity that we can only accept only one answer on MO. $\endgroup$
    – yaoxiao
    Commented Sep 14, 2015 at 13:57

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