Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 157 / 2

Let $z_0$ be attracting and let $U$ be its immediate basin. If $z_0$ itself is critical there is nothing to prove. Otherwise suppose $U$ contains no critical point. The restriction $f:U\to U$ is then an unbranched covering. Since the complement of $U$ contains the Julia set and hence at least three points, $U$ is hyperbolic. Lift the covering to the universal cover $\mathbb{D}\to U$, choosing a lift that fixes a point above $z_0$. Because both maps are universal coverings, the lift is an automorphism of $\mathbb{D}$. A disc automorphism fixing an interior point has derivative of hyperbolic norm one there, whereas the multiplier at $z_0$ has modulus strictly below one. This contradiction proves that $U$ contains a critical point, whose orbit converges to $z_0$. Thus every attracting fixed point attracts a critical point.

The fixed Fatou component classification has four cases:

Choose a $d$th root of unity $r\ne 1$ far enough from one that $d|1-r|>1$, put $a=1-r$, and defineThis map has degree $d$ and only two critical points, $0$ and infinity, each of multiplicity $d-1$. Their orbits areThe multiplier at $r$ iswhose modulus exceeds one. Thus every critical orbit lands at a repelling fixed point.

An attracting or parabolic periodic Fatou component would capture a critical orbit, contrary to the displayed dynamics. A Siegel disc or Herman ring would have boundary in the closure of the postcritical set, but that set is finite and contained in the repelling grand orbit, whereas a rotation-domain boundary is infinite. By the Sullivan no-wandering-domain theorem, every Fatou component is eventually periodic, so the classification leaves no Fatou component. Therefore this rational map with Julia set equal to the Riemann sphere satisfies

Take the degree-$d$ Chebyshev polynomial $T_d$, characterized byIts Julia set is the interval $[-1,1]$. The complement $\hat{\mathbb{C}}\setminus [-1,1]$ is connected, and it is exactly the basin of infinity. Hence $T_d$ has exactly one Fatou component. This is the Chebyshev polynomial Julia set example.