7
$\begingroup$

This question is related to a classification of rational maps in terms of properties of their Julia set.

Let $f= z^2 + c$, with $c\in \mathbb{C}$ be a quadratic polynomial such that its Julia set $J(f)$ is connected.

  • Q1: If there exists a relatively open set of $J(f)$ that is (the support of) a smooth curve. Is $f$ conjugate (resp equal) to a Tchebychev polynomial or power map $z^2$?
  • Q2: Is the answer to Q1 yes under the additional assumption that $J(f)$ is also locally connected?
  • Q3: If the answer to Q1 is no, can one describe the set of such counterexamples in terms of the parameter $c$?

Thanks a lot.

Improve this question
$\endgroup$

2 Answers 2

Reset to default
7
$\begingroup$

The answer to a) is yes, and this was proved by Fatou in 1919. Sur les équations fonctionnelles Bulletin de la S. M. F., tome 48 (1920), p. 208-314. There are many generalizations of this fact. For one generalization, and further references you may look to Meromorphic functions with linearly distributed values and Julia sets of rational functions, Proc. AMS. 137 (2009), 2329-2333.

Improve this answer
$\endgroup$
3
  • $\begingroup$ Thanks a lot! I haven't gone through Fatou's argument yet but I am more convinced now that it is in fact true. $\endgroup$
    – Gari
    Commented Jun 22, 2019 at 11:05
  • $\begingroup$ I don't see yet why the Julia set being equal to a circle or an arc of a circle implies that the answer to Q1 is yes but I will try. $\endgroup$
    – Gari
    Commented Jun 22, 2019 at 11:06
  • 2
    $\begingroup$ @Gari: This is explained in Fatou, archive.numdam.org/article/BSMF_1920__48__208_1.pdf $\endgroup$
    – Alexandre Eremenko
    Commented Jun 22, 2019 at 16:22
5
$\begingroup$

From this paper of Bedford and Kim (arxiv link):

Fatou showed that if the Julia set $J$ is a smooth curve, then either $J$ is the unit circle, or $J$ is a real interval. If $J$ is the circle, then $f$ is equivalent to $z → z^d$ , where $d$ is an integer with $|d| ≥ 2$; if $J$ is the interval, then $f$ is equivalent to a Chebyshev polynomial.

Improve this answer
$\endgroup$
2
  • 2
    $\begingroup$ They take $f$ arbitrary holomorphic. What's meant by "is the unit circle"? If one conjugates $z\mapsto z^2$ by an affine map, one can get any circle. $\endgroup$
    – YCor
    Commented Jun 21, 2019 at 17:41
  • $\begingroup$ Thanks for the answer. However I wasn't able to track down the statement using their reference. Maybe it is somewhere in Milnor and I have overlooked it. $\endgroup$
    – Gari
    Commented Jun 22, 2019 at 10:53

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .