I am reading the research article *"[The Hausdorff dimension of the boundary of Mandelbrot set and Julia sets](https://doi.org/10.2307/121009)"* 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 $f(z) = z^2$.



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


