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What I mean by that is, consider, say, the "Koch snowflake" curve. It is formed by repeatedly applying a substitution to the lines of a triangle to get the final curve in the limit.

What I am after is whether or not you can find a substitution and base figure, both of which can be described in "elementary" terms, such that the Julia sets, namely of the complex quadratic map $z \mapsto z^2 + c$, for at least some values of $c$, can be described in an analogous way. Is that possible for any value of $c$ for which the Julia set is both connected and fractal (i.e. has Hausdorff dimension > 1)? If so, what is an example, with attendant explicit description of substitution and base figure? If not, why is it not possible, and what is the proof?

The_Sympathizer
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  • The Julia set is the unit circle for $c-0$ and the interval $[-2,2]$ for $c=-2$. For other values, it is a fractal. – lhf Jun 17 '16 at 19:07
  • The definition of the Julia set is pretty elementary. Do you have some other notion of "elementary" in mind? – Lee Mosher Jun 18 '16 at 01:05
  • What I mean is, whether or not one can describe them as the result of repeated substitution of some explicitly-described geometric figure into a base geometric figure as in the Koch curve and snowflake. – The_Sympathizer Jun 18 '16 at 08:04

1 Answers1

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For the dynamical system,

$$f_c(z_n)=z_{n+1}=z_{n}^2-c$$

The Julia set for a value $c$ is the collection of points $z_0$ such that the iterates stay bounded. The Mandelbrot set is the collection of points such that $z_0$ with $c=z_0$ such that the iterates stay bounded.

Denote the Julia set by $J_c$. Then assuming the attractor, fractal, for the dynamical system exists, the attractor for the system is $J_c$. By definition, this means,

$$f_c(J_c)=J_c$$

However, that also means,

$$J_c=f^{-1}_c(J_c)=\pm \sqrt{z+c}$$

Since the inverse is multivalued and we need all the values, we need to apply both, which is to say we take their union. This yields,

$$J_c=(\sqrt{J_c+c}) \cup (-\sqrt{J_c+c})=w_1(J_c) \cup w_2(J_c)$$

So these Julia sets can cast into the form of a nonlinear IFS with transformations given by,

$$\ w_1(z)=\sqrt{z+c}, \ w_2(z)=-\sqrt{z+c}; \ z \in C$$

With the Julia set $J_c$ as the attractor.

Zach466920
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