Clustering of periodic points for a polynomial iteration of $\mathbb{C}$

Question

Let $f : \mathbb{C} \to \mathbb{C}$ be a polynomial map of degree $q > 1$. Consider $E_n \subset \mathbb{C}$ the set of periodic points with period (dividing) $n$; generally, $|E_n| = q^n$. Since all the peridic points are limited to a bounded subset of $\mathbb{C}$, we know that there are always two among them within a distance of $O(q^{-n/2})$.

Question. (i) How close can two distinct elements of $E_n$ get? In the asymptotic where $n \to \infty$ along a sequence, can it happen that $E_n$ contains a pair of points within a distance of $o(1/q^n)$?

(ii) Given any point $z_0 \in \mathbb{C}$ and $C$, does the disk $|z-z_0| < C/q^n$ contain only $O_{C,z_0}(1)$ points from $E_n$?

Answer 1

Consider $f(x) = x^2 - 2$. This has the property that $f(2 \cos (x)) = 2 \cos(2x)$, so that $f_n(2 \cos (x)) = 2 \cos(2^n x)$ (where $f_n$ is $f$ iterated $n$ times). Thus $E_n = \{2 \cos (x): \cos(x) = \cos(2^n x)\}$, and the two greatest members of $E_n$ are both greater than $2 \cos(\pi 2^{1-n})\approx 2 - \pi^2 2^{2-2n}$

Answer 2

The answer to both of your questions is negative, even if you replace $q^{-n}$ by any other sequence $(a_n)$ of natural numbers.

The reason is, broadly speaking, that you may have of course have maps having multiple periodic points, and under small perturbations these will yield maps having a cycle of potentially large period nearby.

A little bit more precisely, let us consider the family of quadratic polynomials, $f_c(z) = z^2+c$. The following is well-known:

Proposition. Suppose that $f_c$ has a periodic point $z_0$ of period $n$, with corresponding multiplier $\mu = (f_c^n)'(z_0) = e^{2\pi i p/q}$, for some $p,q\in\mathbb{N}$.

Then, for any neighbourhood $U$ of $z_0$, there is a parameter $c'$ (arbitrarily close to $c$) such that $f_{c'}$ has a periodic point $z'$ of period $nq$ such that $f_{c'}^{kn}(z')\in U$ for $k\geq 0$, and such that $z'$ itself has multiplier $\mu' = (f_{c'}^{nq})'(z') = e^{2\pi i /q'}$ for some (large) integer $q'$.

(In addition, the map $f_{c'}$ also has a (repelling) periodic point of period $n$ in $U$.)

(What happens is that the parameter $c$ is on the boundary of two components of the interior of the Mandelbrot set, one consisting of points having an attracting orbit of period $n$ and one having an attracting orbit of period $nq$. The perturbation is obtained by passing a little bit along the boundary of the latter component.)

Now start e.g. with the map $f_{1/4}$ and $z=1/2$, and apply this proposition inductively. Take a limit of the resulting parameters (taking care to make each perturbation small enough as to not destroy the features already constructed). Clearly in this way we can find a quadratic polynomial having periodic points $(z_j)_{j\geq 0}$ of periods $(n_j)$, where $$ n_j = \prod_{k=1}^j q_j$$. (Here $q_j$ is some rapidly increasing sequence of positive integers obtained in the construction.)

Given any sequence $(\varepsilon_n)$, we can inductively carry out the construction so that $q_j$ points of the orbit of $z_j$ are within distance $\varepsilon_{n_j}$ of each other. Letting $\varepsilon_n$ tend to zero faster than $a_n$, the claim follows.

Answer 3

While Robert Israel's answer is correct, $x^2-2$ is a very exceptional case.

In a way, how close do they get is not the interesting question: rather, what gets strange is how far apart do the closest ones remain? We know the periodic points are dense in the Julia set, but in the case of weird ones (like the ones with Cremer points, or even some with Siegel disks where the disk itself is very 'deep' within the Julia set, as measured by the external rays), the periodic points tend to avoid certain parts of the Julia set as long as possible. This is what causes the 'inverse method' of rendering images of Julia sets to be so bad for those cases.