## Section3.4The critical orbit

Each function $f_c(z)=z^2+c$ has exactly one critical point at the origin. It turns out that the orbit of this critical point dominates the global dynamics of the iteration of $f_c\text{.}$

For the time being, we will content ourselves with a proof that the Julia set is totally disconnected when $|c| \gt 2$ and sketch of the rest. The general idea is based on a process called inverse iteration. Given and function $f:\mathbb C \to \mathbb C$ and a set $S\in\mathbb C$ let

\begin{equation*} f^{-1}(S) = \{z\in\mathbb C:f(z)\in S\}. \end{equation*}

The set $f^{-1}(S)$ is called the inverse image or pre-image of $S\text{.}$ We define an initial approximation $J_0$ and let $J_n=f_c^{-n}(J_0)\equiv f_c^{-1}(J_{n-1})\text{.}$ As we will see, if the initial approximation is chosen correctly, the sequence of iterates will collapse down to filled Julia set.

Suppose that $|c|\gt 2$ and consider the disk $D$ of radius $|c|$ centered at the origin. By the proof of theorem Theorem 3.2.1, every point on the boundary of this disk maps to a point whose absolute value is larger than $|c|\text{.}$ As a result, the image of the boundary is a loop that completely encircles $D\text{.}$ (In fact, it's a circle of radius $|c|^2$ centered at $c\text{,}$ as you'll show in exercise Exercise 3.7.1.) Thus, $f(D)\supset D\text{.}$ As a result, $f^{-1}(D)\subset D\text{.}$ We now perform inverse iteration from $D\text{.}$

To help us understand $f^{-1}(D)\text{,}$ note that $f_c(0)=c\text{,}$ which lies on the boundary of $D\text{.}$ In fact, $f_c(z) = z^2+c=c$ is equivalent to $z^2=0$ so that zero is the only point that maps to $c\text{.}$ Every other point on the boundary of $D$ has exactly two pre-images. Thus, the pre-image of the boundary of $D$ is a figure 8 with a cut point at the origin. The pre-image of $D$ itself consists of two lobes bounded by the two curves forming the figure 8.

We next consider $f_c^{-2}(D)\text{.}$ This set consists of two separate pieces, one in each of the lobes of $f_c^{-1}(D)\text{.}$ More generally, $f_c^{-n}(D)$ consists of $2^{n-1}$ obtained by cutting the pieces of $f_c^{-(n-1)}$ in half.

Finally, the filled Julia set itself is exactly

\begin{equation*} \bigcap_{n=1}^{\infty} f_c^{-n}(D). \end{equation*}

As the diameters of these sets decrease down to zero, the filled Julia set coincides exactly with the Julia set. The construction above yields an obvious correspondence between the points in the Julia set and the points in the Cantor set.

The construction just described in illustrated in figure Figure 3.4.2(a).

The situation just described is more complicated when $|c| \leq 2\text{.}$ If $|c| \leq 2$ and the critical orbit escapes, then there is some first value of $n$ with the property that $|f_c^{n}(0)| \gt 2\text{.}$ We can then apply the inverse iteration idea to the disk of radius $|f_c^{n}(0)|$ centered at the origin. The figure 8 configuration then appears after $n$ iterates and the ideas of the proof are then applicable. This situation is illustrated in figure Figure 3.4.2(b).

If the critical orbit does not escape, then we can attempt inverse iteration with a disk $D$ whose radius is larger than two. In this case, the figure 8 never forms so that $f^{-n}(D)$ is a connected set for all $n\text{.}$ As a result, the filled Julia set is a connected set since it's the intersection of a nested family of connected sets. This set is illustrated in Figure 3.4.2(c) and (d). In figure (c), $f_c$ has an attractive orbit. In figure (d), $f_c$ does not have an attractive orbit so that the sets $f^{-n}(D)$ collapse down to have area zero. Since any attractive behavior must attract the critical orbit, one easy way to ensure the lack of attractive behavior is to chose $c$ so that the orbit of zero lands exactly on a repelling orbit. In fact, one can prove that this always happens when zero is pre-periodic. In Figure 3.4.2(d), $c=i$ and zero eventually maps to an orbit of period 2.