Carathéodory topology and the space of Riemann mappings

Riemann mapping theorem states that every simply connected open proper subset $U$ of the complex plane $\mathbb{C}$ is conformally isomorphic to the unit disk $\mathbb{D}$. In other words, there exists a conformal isomorphism $f: \mathbb{D} \to U$. What I would like to discuss is a way to topologise the set of simply connected open proper subsets $U$ of $\mathbb{C}$ and to naturally normalise the corresponding conformal isomorphisms $f$ in order to establish a natural topological correspondence $U \mapsto (f: \mathbb{D} \to U)$.

Limits of Sets

Let $X$ be a set. Any infinite sequence $A_n$ of subsets admits two limits:

The inner limit $\liminf A_n$ is the set of points in $X$ which are in all but finitely many $A_n$’s, whereas the outer limit $\limsup A_n$ is the set of points in $X$ which are contained in infinitely many $A_n$’s. As the names suggest, the former is a subset of the latter.

In set and measure theories, it is common to say that $A_n$ has a limit $A$ when both limits coincide and are equal to $A$. When $X$ is a topological space, this notion of convergence may not respect the topological properties of these subsets. In $\mathbb{R}$, open intervals can converge to non-open subsets: $(-\frac{1}{n}, \frac{1}{n}) \to \{0\}$.

Hausdorff Metric Topology

Now, assume that $(X,d)$ is a compact metric space and let’s have a look at the set $C(X)$ of all non-empty closed subsets of $X$. We can fix the earlier problem by endowing $C(X)$ with the topology where closed subsets $A_n$ converge to $A$ if and only if the closure of the two limits coincide with $A$:

In fact, $C(X)$ is metrisable. Define $D: C(X) \times C(X) \to [0,\infty)$ by

where $ A_{\epsilon} := \{ x \in X : d(x,A) \leq \epsilon \}$ denotes the closed $\epsilon$-neighbourhood of the closed subset $A$. I’ll leave it to you to check that $D$ really is a metric on $C(X)$. $D$ is often called the Hausdorff metric³ and it measures the maximum distance a point in $A$ has to travel to go to some point in $B$, and vice versa. For example, if $X = [0,1]$ is the Euclidean unit interval, $D(\{0,1\}, [0,0.4]) = 0.6$.

The set $O(X)$ of all proper open subsets of $X$ can also be topologised by saying that $U_n \to U$ if and only if

It can also be endowed with a similar metric which we shall again call the Hausdorff metric denoted by $D$:

where $U^{\epsilon} := \{ x \in U : d(x, X \backslash U) > \epsilon \}$ denotes the set of points in $U$ which is at least $\epsilon$ distance away from its complement. This definition coincides with our previous one by the relation $D(U,V) = D(X\backslash U, X \backslash V)$.

Riemann mappings

Let’s look at the one-point compactification $\mathbb{P}^1 = \mathbb{C} \cup \{ \infty \}$ of the complex plane $\mathbb{C}$. As a Riemann surface, the Riemann sphere $\mathbb{P}^1$ can be endowed with the spherical metric $d$.

Let’s call $U$ a topological disk if it is a simply connected domain (connected open subset) in $\mathbb{P}^1$ whose complement consists of at least two distinct points. The set $\mathcal{D}$ of all topological disks can be viewed as a subspace of the Hausdorff metric space $( O(\mathbb{P}^1), D)$.

The uniformisation theorem asserts that every topological disk $U$ admits a conformal isomorphism $f: \mathbb{D} \to U$. By studying the conformal automorphisms of $\mathbb{D}$, we can also say that for every basepoint $u \in U$ there is a unique $f$ such that $f(0)=u$ and $f’(0) \in \mathbb{R}_+$. (See Ahlfors¹ Chapter 6.1) We shall call such function the Riemann mapping of the pointed topological disk $(U,u)$. Riemann mappings provide a bijective correspondence $\Phi: (U,u) \mapsto f$ between the set $\mathcal{E}$ of all pointed topological disks and the set $\mathcal{F}$ of all conformal injections from $\mathbb{D}$ to $\mathbb{P}^1$ with positive derivative at $0$.

We can view $\mathcal{E}$ as both a subspace of the product $\mathcal{D} \times \mathbb{P}^1$ and a fiber bundle of $\mathcal{D}$ over $\mathbb{D}$. In particular, we can endow $\mathcal{E}$ with the metric $d_\mathcal{E}$ where

The resulting topological structure is often called the Carathéodory topology. Meanwhile, the set $\mathcal{F}$ can be endowed with the compact-open topology, that is, $f_n \to f$ if and only if $f_n$ converges to $f$ uniformly on every compact subset of $\mathbb{D}$. The following theorem by Carathéodory² states that $\Phi$ is much more than just a bijection. With respect to the topologies stated before, pointed disks and their respective Riemann mappings are topologically equivalent!

Theorem: The map $\Phi: \mathcal{E} \to \mathcal{F}$ is a homeomorphism.

To prove the theorem, it is sufficient to prove that the map $\Psi : \mathcal{D}_0 \to \mathcal{G}$ where $\mathcal{D}_0 \subset \mathcal{D}$ is the space of all topological disks containing $0$, $\mathcal{G} \subset \mathcal{F}$ is the space of all conformal injections fixing $0$ and $\Psi(U) = \Phi((U,0))$. Indeed, suppose $(U_n, u_n)$, $n \geq 1$, and $(U,u)$ are pointed disks with Riemann mappings $f_n$ and $f$. If $u \neq \infty$, then we may assume all $u_n$’s are finite too. Then, we can post-compose $f_n$ with the translation $z \mapsto z - u_n$ and $f$ with $z \mapsto z-u$. By applying triangle inequality and the fact that $\Psi$ is a homeomorphism, we can show the following:

When $u = \infty$, then we may assume that all $u_n$’s are non-zero and additionally post-compose $f_n$’s and $f$ with the inversion map $z \mapsto \frac{1}{z}$ in order to make the basepoints finite.

Now, let’s have a look at a disk $U_0$ and a sequence of disks $U_n$ in $\mathcal{D}_0$ as well as their respective Riemann mappings $f_0 = \Psi(U_0)$ and $f_n = \Psi(U_n)$ for $n\geq 1$.

Suppose $f_n \to f_0$ in $\mathcal{G}$. Let $A_0$ be a continuum inside the disk $U_0$. We wish to show that $A_0$ is contained in $U_n$ for sufficiently high $n$. Since its preimage $A = f_0^{-1}(A_0)$ is compact in $\mathbb{D}$, $A$ must be contained in an open disk $\mathbb{D}_r$ of some radius $r < 1$ centered at $0$. Suppose for a contradiction that for some subsequence $f_{n_k}$, $A_0$ is not contained in $f_{n_k} (\mathbb{D}_r)$ for all $k$. Then, there must be some sequence of points $w_{n_k} \in A_0 \cap f_{n_k}(\mathbb{D}_r)$. The sequence of preimages $z_{n_k} = f_{n_k}^{-1} (w_{n_k})$ lies in the compact set $A$ and, by passing to some further subsequence, converges to some $z_0 \in \partial \mathbb{D}_r$. Therefore,

This contradicts the fact that $A$ is disjoint from the circle $\partial \mathbb{D}_r$. As $A_0$ is arbitrary, we have shown that $U_0 \subset \text{int} \left( \liminf U_n \right) $.

Let’s pick an arbitrary non-zero point $ w \in \text{int} \left( \limsup U_n \right) $. There is a open neighbourhood $B$ of both $0$ and $w$ contained in some infinite subsequence $U_{n_l}$. Obviously, the family $\{ f_{n_l}^{-1} : B \to \mathbb{D} \}$ forms a normal family as it is uniformly bounded. Passing to a subsequence, $f_{n_l}^{-1}$ converges to some map $g$. Let $z_l = f_{n_l}^{-1}(w)$ and $z = g(w)$. Since

then $w \in U_0$. Therefore, $\text{int} \left( \limsup U_n \right) \subset U_0$, and we have then shown that $U_n \to U_0$.

Suppose instead that $U_n \to U$. Since $U_n \to U$, the sequence $\{f_n \mid_{\mathbb{D}_r } \}$ is uniformly bounded for each $r <1$ and consequently, $\{f_n \}$ forms a normal family. Every limit in this family will have image $U_0$, so then by uniqueness of Riemann mappings, this limit has to be $f_0$. Thus, $f_n \to f_0$, and we have finally proven the theorem.

References

1: L. Ahlfors. Complex Analysis. McGraw-Hill, 1979.
2: C. Carathéodory. Conformal Representation. Cambridge University Press, 1952.
3: F. Hausdorff. Set Theory. Chelsea Publishing Co., 1957.