How to find the hyperbolic dimension of map $f(z) = z^2$ of $\overline{\mathbb{C}}$ onto itself?

Question

I am reading the research article "The Hausdorff dimension of the boundary of Mandelbrot set and Julia sets" by Shishikura. In his article, he defined hyperbolic sets and hyperbolic dimensions for any rational map of $\overline{\mathbb{C}}$ onto itself, where $\overline{\mathbb{C}}$ is Riemann sphere.

Let $f$ be rational map on $\overline{\mathbb{C}}$. A closed subset $E$ of $\overline{\mathbb{C}}$ is called a hyperbolic subset for $f$ if

1. $f(E) \subset E$ and
2. there exist a positive constant $c$ and $\kappa$ $> 1$ such that $\lVert (f^n)' \rVert \geq c \kappa^{n}$ on $E$ for $n \geq 0$. Here $\rVert \cdot \rVert$ denotes the norm of derivative with respect to the spherical metric on $\overline{\mathbb{C}}$.

The hyperbolic dimension of $f$ is defined as \begin{align} \operatorname{hyp-dim}(f):= \sup\{\operatorname{H-dim}(E) : E \; \text{is hyperbolic set of $f$} \;\} \end{align} where $\operatorname{H-dim}(E)$ is Hausdorff dimension of $E$.

Since he didn't give any example of this. Therefore I want to work with some examples for these definitions. Suppose I want to find the hyperbolic dimension of one of the simplest nonlinear rational map $f(z) = z^2$. How to proceed? I believe that then I need to tackle all the hyperbolic sets for $f(z) = z^2$. But this does not seem trivial. I am looking forward for the help regarding finding the hyperbolic dimension of $f(z) = z^2$.

Answer 1

The hyperbolic dimension of $f$ is 1 and its maximal hyperbolic set is the unit circle $\mathbb{S}^1$.

First we show that a hyperbolic set for $f$ must be contained in the unit circle $\mathbb{S}^1$. If $E$ is a closed, $f$-invariant set that is not contained in $\mathbb{S}^1$ then it must contain $0$ or $\infty$. The derivative of $f$ vanishing at $0$ and $\infty$, the set $E$ cannot be hyperbolic.

Then it is easily seen that $\mathbb{S}^1$ is hyperbolic, by computing the derivative of $f^n(z) = z^{2^n}$ on $\mathbb{S}^1$. The restriction of the spherical metric to $\mathbb{S}^1$ is equivalent to the constant one, so it enough to observe that $ (f^n)^{'}(z)=2^n z^{2^n - 1}$ and $ \lvert (f^n)^{'}(z) \rvert =2^n $ for $ z \in \mathbb{S}^1 $. The Hausdorff dimension of $\mathbb{S}^1$ is 1.

As any hyperbolic set is contained in $\mathbb{S}^1$, and the Hausdorff dimension is non-decreasing with respect to inclusion, we conclude that the hyperbolic dimension of $f$ is 1.

Edit: the length element of the spherical metric is $ds^2 = \frac{dz d\bar{z}}{(1 + \lvert z \rvert^2)^2}$. This implies that the norm of the derivative in this metric is $ \lVert f' \rVert_z = \frac{1 + \lvert z \rvert^2}{1 + \lvert f(z) \rvert^2} \lvert f'(z) \rvert$ .