# bounds on dimension of Julia sets inside Mandelbrot set

$\dim_H J(f_c) \ge 1$ for $c \in M$ by connectedness and uncountability of $J(f_c)$. For which points is there equality? $c=0$ and $c=-2$ for sure, but is this an exhaustive list?

Notation: $\dim_H$ is Hausdorff dimension, $J$ is Julia set, $f_c(z) = z^2 + c$, $M = \{ c : J(f_c) \text{ is connected}\} \subset \mathbb{C}$ is the Mandelbrot set.

• Presumably $M$ is the Mandelbrot set, and $J(f_c)$ is the Julia set with parameter $c$? Mar 5, 2018 at 3:11
• @XanderHenderson yes - I added some notes on notation to the question Mar 5, 2018 at 9:10
• By the observations in section 11.1 of Beardon's Iteration of Rational Functions, for any polynomial, the only smooth Julia sets are circles arising from $f(z)=a z^n$ and line segments arising from the Chebyshev polynomials. Since the Mandelbrot set parametrizes all the quadratics uniquely up to conjugacy, it follows that $c=-2$ and $c=0$ are the only values of $c$ such that $J(f_c)$ is a smooth manifold. Now that's not quite what your asking, since a non-smooth manifold can have dimension 1, but perhaps it's a step. Mar 6, 2018 at 10:54
• @MarkMcClure in fact this question was inspired by fractalforums.org/fractal-mathematics-and-new-theories/28/… which talks about non-smooth shapes with dimension 1 Mar 7, 2018 at 19:08
• section 6.5 in Potential theory in the complex plane of Thomas Ransford can be usedfull with the upperbound of $dim(J_f)$. But for your question about another point of lower bound you can take a look at this preprint arxiv.org/pdf/1712.03102.pdf May 25, 2018 at 8:16

This theorem states that if $f$ is a rational map, and $m$ is its measure of maximal entropy, then the dimension of the Julia set is strictly larger than the dimension of $m$ unless $f$ has a parabolic orbifold.
For a polynomial with connected Julia set (in particular for a quadratic polynomial in the Mandebrot set), it is known that the dimension of $m$ is 1 (this is due to Makarov).
It is too complicated to explain here what it means for $f$ to have a parabolic orbifold, but suffice it to say that the only quadratic polynomials with that property are (up to affine conjugacy) $z^2$ and $z^2-2$.