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Fall 2018 - Problem 11

Riemann mapping theorem

An analytic Jordan curve is a set of the form

Γ=f({∣z∣=1})\Gamma = f\p{\set{\abs{z} = 1}}

where ff is analytic and 1-1 on an annulus {r<∣z∣<1r}\set{r < \abs{z} < \frac{1}{r}}, 0<r<10 < r < 1.

Let C∗=C∪{∞}\C^* = \C \cup \set{\infty} be the Riemann sphere, let N<∞N < \infty, and let Ω⊆C∗\Omega \subseteq \C^* be a domain for which ∂Ω\partial\Omega has NN connected components, none of which are single points. Prove there is a conformal (i.e., one-to-one analytic) mapping from Ω\Omega onto a domain bounded by NN pairwise disjoint analytic Jordan curves.
Solution.

This problem is probably incorrect as stated. Consider this StackExchange post. The boundary of the curve in the question is self-intersecting, so it cannot be the image of any 1-1 function. Instead, the problem should say that C∗∖Ω\C^* \setminus \Omega has NN connected components, rather than the boundary itself having NN connected components.

Let NN be the number of connected components of Ω\Omega. We will show that Ω\Omega is conformally equivalent to D∖(E1∪⋯∪EN−1)\D \setminus \p{E_1 \cup \cdots \cup E_{N-1}} where E1,…,EN−1⊆DE_1, \ldots, E_{N-1} \subseteq \D are simply connected by induction on NN:

If N=1N = 1, then Ω\Omega is simply connected. If this were not the case, then there exists a simple closed curve γ⊆Ω\gamma \subseteq \Omega which is not contractible to a point. Hence, the interior of γ\gamma contains a point z0∈C∗∖Ωz_0 \in \C^* \setminus \Omega. This means that C∗∖Ω\C^* \setminus \Omega is contained in the interior of γ\gamma since it is connected, but this implies that z0z_0 is in a different component of the portion of C∗∖Ω\C^* \setminus \Omega which lies in the exterior of γ\gamma, a contradiction. Hence, Ω\Omega is simply connected with more than one point in its complement, so by the Riemann mapping theorem, Ω\Omega is conformally equivalent to D\D. Since ∂D\partial\D is the unit circle, it is trivially an analytic Jordan curve.

Now suppose that the claim is true when for domains with NN connected components in its complement, and assume C∗∖Ω\C^* \setminus \Omega has N+1N + 1 connected components, C1,…,CN+1C_1, \ldots, C_{N+1}. Let Ω′=Ω∪CN+1\Omega' = \Omega \cup C_{N+1}, which has C∗∖Ω′=C1∪⋯∪CN\C^* \setminus \Omega' = C_1 \cup \cdots \cup C_N. By the inductive hypothesis, we get a conformal map φ ⁣:Ω′→D∖(E1∪⋯∪EN−1)\func{\phi}{\Omega'}{\D \setminus \p{E_1 \cup \cdots \cup E_{N-1}}} with the EkE_k's simply connected. On the other hand, if we let U=C∗∖φ(CN+1)U = \C^* \setminus \phi\p{C_{N+1}}, then C∗∖U\C^* \setminus U has only one connected component, so by the argument above, we see that UU is simply connected. Thus, because CN+1C_{N+1} is not a singleton, the Riemann mapping theorem gives a conformal map ψ ⁣:U→D\func{\psi}{U}{\D}.

If EN+1=φ(CN+1)E_{N+1} = \phi\p{C_{N+1}}, then we get the conformal map

φ∣Ω ⁣:Ω→D∖(E1∪⋯∪EN+1)\func{\res{\phi}{\Omega}}{\Omega}{\D \setminus \p{E_1 \cup \cdots \cup E_{N+1}}}

with

ψ(∂EN+1)=ψ(∂(C∗∖U))=∂D,\psi\p{\partial E_{N+1}} = \psi\p{\partial\p{\C^* \setminus U}} = \partial\D,

i.e., ∂EN+1\partial E_{N+1} is conformally equivalent to S1S^1, which is an analytic Jordan curve, and this completes the proof.