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Let $f\colon \widehat{\mathbb{C}}\to \widehat{\mathbb{C}}$ be a rational function of degree two or greater whose Julia set $J_f$ is connected. If $S\subseteq J_f$ is a finite set of periodic points, is it possible that the complement $J_f\setminus S$ has infinitely many connected components? I am particularly interested in the case where $f$ is hyperbolic.

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For a polynomial, this is equivalent to asking whether there can be infinitely many external rays landing at a single point. This could happen only if the function has a Cremer point (i.e., a non-linearisable irrationally indifferent periodic point). If the Cremer point is accessible from the complement, then any external ray landing at it would have to be non-periodic by the so-called "snail lemma", which means that then there would indeed be infinitely many rays landing at the same point. However, I believe that it is still an open question whether or not this actually happens.

Since hyperbolic maps have no Cremer points, what you ask about is impossible in the hyperbolic case. I think that, similarly, for hyperbolic rational maps it is impossible, and probably more generally whenever the Julia set is locally connected.

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  • $\begingroup$ Thanks for the answer. Do you have an idea for how the argument would go in the rational hyperbolic case? There aren't external rays to work with, and it's hard to see how to prove anything about connectivity when you don't have the Carathéodory loop. $\endgroup$ – Jim Belk Jan 15 at 18:14
  • $\begingroup$ @JimBelk If you have infinitely many accesses from the same invariant component to a periodic point, then you should basically be in the same situation as in the polynomial case. On the other hand, you should not be able to have a point accessible from infinitely many different Fatou components. $\endgroup$ – Lasse Rempe Jan 15 at 21:03
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Well, $f(z)=z^2$ has degree 2, and $S=\{1,-1\}$ is a set of periodic points. Then $J_f \setminus S$ consists of two components.

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    $\begingroup$ Thanks, but my question is whether the number of components can be infinite. I've edited the title to make this more clear. $\endgroup$ – Jim Belk Jan 15 at 17:29

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