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

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 $$f(z) = z^2$$.

• Does ℂ with an overline denote the Riemann sphere? Or something else? Commented Mar 30, 2023 at 1:30
• TeX note: ' in TeX already has a superscript built in (it essentially translates to ^{\prime}, so you almost never want ^{'}. Compare $f' f^\prime f^{'}$ f' f^\prime f^{'}. I edited accordingly. Commented Mar 30, 2023 at 2:46
• @DanielAsimov Yes $\overline{\mathbb{C}}$ is Riemann sphere. Commented Mar 30, 2023 at 8:47

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 $$\lVert f' \rVert_z = \frac{1 + \lvert z \rvert^2}{1 + \lvert f(z) \rvert^2} \lvert f'(z) \rvert$$ .

• It means the overall idea is that we need to try to show that Julia set is a hyperbolic set for $f$ because we know that every hyperbolic set is contained in Julia set. Due to this, we will get that Julia set is the maximal hyperbolic set. But what if there is a rational map for which the Julia set is not a hyperbolic set? In that case, how to find the hyperbolic dimension of it? Commented Mar 29, 2023 at 19:46
• @Nirmal Yes, there exists rational maps whose Julia set is not a hyperbolic set. I don't think we can hope for a general method to find the hyperbolic dimension of rational maps (there is a criterion for the equality between the Hausdorff dimension of the Julia set and the hyperbolic dimension in the Shishikura's paper you mention).
– FMB
Commented Mar 29, 2023 at 21:51
• @Nirmal for example, the Julia set of $f(z)=z^2-2$ is $\left[ -2, 2 \right]$ and it is not a hyperbolic set as it contains the critical point $0$ (the computation of this Julia set is done in Milnor's book).
– FMB
Commented Mar 29, 2023 at 22:10
• @Nirmal this is standard. The spherical metric in this case is also called the Fubini–Study metric (you can look that up on Wikipedia for example). The main point is that because the Riemann sphere is compact, you can choose any Riemannian metric in the definition of hyperbolicity (the constant will be differents but the exponential growth does not depend on the metric).
– FMB
Commented Mar 30, 2023 at 16:59
• I added an explanation. For the record, I think this kind of questions is best suited for math.stackexchange .
– FMB
Commented Mar 30, 2023 at 19:43