The Riemann Mapping Theorem and the Uniformization Theorem

One can ask when two domains (connected, open subsets of the complex plane) are conformally equivalent; that is, when there exists a holomorphic homeomorphism between them. The case of simply connected domains is answered by Riemann Mapping Theorem.

Theorem 2.1   If $\Omega$ is a simply connected subset of the complex plane, not equal to $\mathbb{C}$, then there exists a holomorphic homeomorphism from the unit disc onto $\Omega$.

Thus we have that there are two classes of simply connected domains in the complex plane, namely the complex plane and the unit disc. In particular the upper half plane is conformally equivalent to the unit disc; we will use this fact to make most of our statements on $\mathbb{H}$ rather than on $\mathbb{D}$.

We now consider the problem of classifying circular annuli, that is regions on the complex plane determined by two concentric circles. Set $A(z_0;r_1,r_2) = \{z\in {\mathbb{C}}; r_1 < \vert z - z_0\vert < r_2\}$, for $z_0 \in \mathbb{C}$ and $0 < r_1 < r_2$. We have that $A(z_0,r_1,r_2)$ is equivalent to $A(0;1,r_2/r_1)$. So we need to study only annuli of this form, which we denote by $A_R$ ( $R = r_2/r_1 > 1$). An analytic approach to this problem is given by the following result [42].

Theorem 2.2   The annuli $A_1 = A_{R_1}$ and $A_2 = A_{R_2}$ are conformally equivalent if and only if $R_1 = R_2$.

Proof. Let $f:A_1 \to A_2$ be a biholomorphic function. Let $C$ be the circle of centre the origin and radius $\sqrt{R_2}$. Since $f^{-1}(C)$ is compact we have that there exists an $\epsilon > 0$, such that $A(1,1+\epsilon)$ and $f^{-1}(C)$ have empty intersection. The set $V = f(A(1,1+\epsilon))$ is connected and does not intersect $C$, so $V \subset A(1,\sqrt{R_2})$ or $V \subset A(\sqrt{R_2},R_2)$. In the second case we can consider the function $R_2/f$ instead of $f$, so we can assume $V \subset A(1,r)$.

If $\{z_n\}$ is a sequence of complex numbers with $1 < \vert z_n\vert < 1 + \epsilon$ and $\vert z_n\vert \to 1$, then $\{f(z_n)\}$ is a sequence in $V$ without limit point in $A_2$, so $\vert f(z_n)\vert \to 1$. Similarly $\vert f(z_n)\vert \to R_2$ when $z_n \to R_1$.

Let $\alpha = \log R_2 / \log R_1$. Set $u(z) = 2 \log \vert f(z)\vert - 2 \alpha \log \vert z\vert$ on $A_1$. Since $\partial (2 \log\vert f\vert) = \partial (\log f\overline{f}) = f'/f$ we have $\partial u = f'/f - \alpha / z$, so $u$ is harmonic on $A_1$, and by the above remarks, $u$ extends to a continuous function on $\overline{A_1}$; then we get $u = 0$ and therefore $f'(z) / f(z) = \alpha / z$.

Let $\gamma(t) = \sqrt{R_1} e^{it}$, for $-\pi \leq t \leq \pi$ and $\Gamma = f \circ \gamma$. Then we have

$$\alpha = \frac{1}{2\pi i} \int_\gamma \frac{f'(z)}{f(z)} dz = Ind_\Gamma (0);$$

where $Ind_\Gamma(0)$ denotes the index of 0 with respect to $\Gamma$. So $\alpha$ is a positive integer, and $\partial(z^{-\alpha}f(z)) = 0$ on $A_1$, which implies that $f(z) = c z^\alpha$ for some constant $c$. Since $f$ is one-to-one we get $\alpha = 1$ and so $R_1 = R_2$. $\qedsymbol$

The classification of simply connected Riemann surfaces is given by the Uniformization theorem.

Theorem 2.3 (Poincaré-Koebe Uniformization Theorem)   Any simply connected Riemann sphere is biholomorphically equivalent to the Riemann sphere, the complex plane or the unit disc. These three surfaces are not conformally equivalent.

Using the Uniformization theorem one can approach the question of classification of annuli from a different point of view. First of all, it is easy to see that the universal covering of an annulus $A_R$ is given by $\pi_R:{\mathbb{H}} \to A_R$, where $\pi_R(z) = \exp (-2\pi i \log(z) / \log(\lambda))$; here $\lambda$ is a real number greater than $1$ satisfying $\exp (2\pi/\log(\lambda)) = R$. The covering group, $G_\lambda$, is generated by the transformation $g_\lambda(z) = \lambda z$. We will have that the annuli $G_{R_1}$ and $G_{R_2}$ are conformally equivalent if the corresponding groups, say $G_\lambda$ and $G_\mu$, are conjugate. This implies that there exists a Möbius transformation $A$ satisfying $g_\lambda = A g_\mu A^{-1}$. It is easy to see that then $g_\lambda = g_\mu$; that is, $R_1 = R_2$, which is the result proved above.