7
$\begingroup$

I am looking to find an exact solution to the Hausdorff dimension of a Julia set $J(f)$ for a polynomial $f: z \mapsto z^2 +c$ given an arbitrary $c$.

I know this question is known for a number of special cases. For example, if the $c$ is on the boundary of the Mandelbrot set, it has dimension 2. The dimension for $c=0$ is obvious as well. Are there any other cases known exactly? If so, how are they found? I'd imagine there are a number of measurements using box-counting methods to approximate the dimension for various cases.

Also, have there been efforts to calculate the dimension of a Julia set for any polynomial or rational function $p : \Bbb{C} \to \Bbb{C}$?

Any knowledge of work done in this area or a place to start would be awesome.

EDIT: Googling the question led to a number of papers. These are the results I have found:

This paper gives the dimension of some set of points, although googling the word "biaccesible" only brings up that paper and references to it.

This one shows that Julia sets for $c$ arbitrarily close to the boundary of the Mandelbrot set have dimensions arbitrarily close to 2.

This one [pdf] gives a number of results:

  • The dimension of $J(f)$ is less than 2 if $f$ has no non-periodic recurrent critical points
  • The Julia set of rational $f$ is hyperbolic $\implies$ The Hausdorff dimension as a function of $c$ is continuous
  • Some other results that seem to require another paper

The introduction of this paper Says that for a rational $f$ we have yet to find a Julia set with positive area and doesn't contain the whole Riemann sphere. (Doesn't $z \mapsto z^2$ provide a counterexample? I'm not sure I understand this one) I also proves a result about the Julia set of a $\sin $ function.

A paper from Harvard [pdf] gives ways of calculating the dimension numerically, and proves that the Hausdorff dimension is continuous from the Feigenbaum point (is this the same one from bifurcation diagrams?) to 1/4.

I'm going to read through more and these ones more carefully. In the meantime any guidance would help.

$\endgroup$
6
  • $\begingroup$ A quick google search brought me a bunch of papers... I have not yet read through them, but wouldnt that be a start? $\endgroup$
    – CBenni
    Commented Jan 27, 2013 at 20:04
  • 1
    $\begingroup$ I just found some too. I'll read through them and compile what I find into my question, but I'm still looking for someone who knows the question far better than me to sum up present knowledge. $\endgroup$ Commented Jan 27, 2013 at 20:07
  • 1
    $\begingroup$ I wonder what the Hausdorff dimension for C=0.25, the parabolic boundary case, is $\endgroup$
    – Sheldon L
    Commented Jan 28, 2013 at 20:08
  • $\begingroup$ @SheldonL It is approximately 1.0812 according to the last harvard paper. It's on the first table in Appendix A. $\endgroup$ Commented Jan 28, 2013 at 20:45
  • $\begingroup$ Regarding z↦z^2 , the Julia set is the unit circle, which has area zero (and Hausdorff dimension 1). So it is not a counterexample. $\endgroup$ Commented Sep 26, 2019 at 16:00

2 Answers 2

3
$\begingroup$

There is NO exact solution for the Hausdorff dimension of Julia sets for $z^2+c$ for general $c$. There are numerical ways to compute approximately (which I'm not vary familiar with), but the exact value is known only for very exceptional case.

Also, it is true that the Hausdorff dimension is equal to two for GENERIC c in the boundary of the Mandelbrot set: see Mitsuhiro Shishikura, "The Hausdorff Dimension of the Boundary of the Mandelbrot Set and Julia Sets".
Moreover, it is known that the area of the Julia set can be positive for some $c$: see Xavier Buff and Arnaud Chéritat, "Quadratic Julia sets with positive area" or by Artur Avila and Mikhail Lyubich, "Lebesgue measure of Feigenbaum Julia sets" (Note that the McMullen's paper on positive area is not for rational maps, but for sine family $f(z) = \lambda \sin z$ (and similar ones), which is a family of transcendental entire maps).

However, at the same time, there are many cases that the Hausdorff dimension is strictly smaller than two. For example, as already mentioned, if the critical points are non-recurrent, the Hausdorff dimension is strictly less than two: see Mariusz Urbański, "Rational functions with no recurrent critical points". And there are many $c$ in the boundary of the Mandelbrot set satisfying this property.

$\endgroup$
1
$\begingroup$

See also

  • Saupe, Dietmar, Efficient computation of Julia sets and their fractal dimension. Phys. D 28 (1987), no. 3, 358–370. DOI: 10.1016/0167-2789(87)90024-8

  • Jenkinson, Oliver; Pollicott, Mark, Calculating Hausdorff dimensions of Julia sets and Kleinian limit sets. Amer. J. Math. 124 (2002), no. 3, 495–545. DOI 10.1353/ajm.2002.0015

$\endgroup$
2

You must log in to answer this question.

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