Contractibility of connected holomorphic dynamics?

Question

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 ?

Answer 1

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.