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Let $f$ be a function, holomorphic in $\mathbb{C}$, and $K(f)$ its non-escaping set : $$K(f) = \{ z \in \mathbb{C} : f^{(k)}(z) \nrightarrow_{k \to \infty} \infty \} $$

Question : If $K(f)$ is connected, is it also contractible ?

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    $\begingroup$ It is not even known if $K(f)$ must be path connected. $\endgroup$ Nov 16, 2013 at 22:44
  • $\begingroup$ @AdamEpstein : thank you, for your comment. So it's an hard open problem... Do you know if "$K(f)$ path-connected $\Rightarrow$ contractible" is also open ? $\endgroup$ Nov 16, 2013 at 22:55
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    $\begingroup$ Offhand I don't know. But any locally connected example will be a dendrite, hence contractible. $\endgroup$ Nov 16, 2013 at 22:57
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    $\begingroup$ What do you really mean by "holomorphic function"? A polynomial? Or your question applies to transcendental entire functons (for which $K(f)$ is NOT called the filled in Julia set). $\endgroup$ Nov 16, 2013 at 23:31
  • $\begingroup$ @AlexandreEremenko : My question applies to all the holomorphic functions. What's the name of $K(f)$ in the general case ? Also, is it known to be true in the polynomial case ? $\endgroup$ Nov 16, 2013 at 23:42

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The answer to your question is negative.

EDIT I have added some additional details and made some corrections.

In the entire case, it is possible to construct an entire function with the following properties: (a) The Fatou set consists of a single conneted attracting basin; (b) If $C$ is a component of the Julia set, then the set of non-escaping points in $C$ is totally disconnected (in fact has Hausdorff dimension zero), (c) There is a component of $J(f)$ that contains a non-escaping point, but no point that is accessible from $F(f)$.

Now, by (a), the nonescaping set is connected (since it contains a dense connected subset of the plane). On the other hand, it can be shown that there is no curve connecting the non-escaping points in (c) to a point in the Fatou set without intersecting the escaping set.

Hence the non-escaping set is not path-wise connected, and hence not contractible. (The construction is contained in an upcoming article of mine, dealing more generally with the topology of transcendental Julia sets.)

For quadratic Cremer polynomials, the key point is that the Cremer point $z_0$ is accumulated on by small cycles by work of Yoccoz. Now, if the Julia set is path-connected (otherwise, there is nothing to prove), then there is a unique arc connecting each of these periodic points to $z_0$.

Now, for any cycle, it follows from the work of Perez-Marco that at least one of the corresponding arcs has diameter at least $\delta$, for $\delta$ independent of the cycle. Indeed, I believe it follows that at least one of them must contain the critical point.

From this, one can deduce (although I haven't made sure to check all the details) that the Julia set is not contractible.

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  • $\begingroup$ Thank you for this outline ! I'm not sure to well understand about the quadratic Cremer polynomials: do they disprove the weaker property "$0$-connected $\Rightarrow$ $1$-connected"? (If no, is this property still open in the entire case?) $\endgroup$ Nov 17, 2013 at 21:10
  • $\begingroup$ In the entire transcendental case, are you sure that the non-escaping set is connected? (I did not understand your "hence", because the non-escaping set contains much more than the Fatou component). $\endgroup$ Nov 17, 2013 at 21:35
  • $\begingroup$ @Alex - I will clarify. Sebastien - I don't understand. I believe contractibility is a stronger property than being 1-connected. In any case, I am not sure that it is known whether Cremer Julia sets can be pathwise connected. $\endgroup$ Nov 18, 2013 at 5:48
  • $\begingroup$ @LasseRempe-Gillen : I thought you prove that it's not $1$-connected (because $1$-connected is weaker than contractible). But, in fact, you prove directly that it's not contractible. Thank you ! If $0$-connected Cremer filled Julia sets exist, do you know if they could be not $1$-connected ? $\endgroup$ Nov 18, 2013 at 11:17
  • $\begingroup$ @Sebastien A Cremer Julia set has no interior and can never disconnect the plane, so - if I have my definitions right - every such Julia set is 1-connected. $\endgroup$ Nov 18, 2013 at 17:37

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