Uniformization theorem

Every simply connected Riemann surface is biholomorphic to the Riemann sphere, the complex plane, or the unit disc.

Every nonempty simply connected proper open subset of the complex plane is conformally equivalent to the unit disc.

For a bounded simply connected domain $D$ and a conformal bijection $f:\mathbb{D}\to D$, $f$ extends continuously to the closed disc if and only if $\partial D$ is a locally connected space. A Jordan boundary gives a homeomorphism of closed discs, but local connectedness alone need not give injectivity on the boundary: a slit has two boundary approaches. The same statement applies on the sphere to unbounded domains after a suitable change of coordinates. This boundary theorem is different from the Caratheodory extension theorem for measures.

The Koebe distortion theorem gives universal upper and lower bounds on a univalent function and its derivative in terms of the distance to the boundary. In particular, derivatives at two points whose hyperbolic distance is bounded differ by at most a universal multiplicative factor.

If $f$ is a univalent function on the unit disc, then $B(f(0),|f^{\prime }(0)|/4)\subseteq f(\mathbb{D})$. The constant $1/4$ is sharp, as shown by the Koebe function. This gives useful comparisons between the derivative of a conformal map, its conformal radius, and its distance to the domain boundary.

For a conformal bijection $f:D\to \tilde{D}$ between proper planar domains, put $d=\mathrm{dist}(z,\partial D)$ and $\tilde{d}=\mathrm{dist}(f(z),\partial \tilde{D})$. Apply the Koebe quarter theorem to $(f(z+dw)-f(z))/(d{f}^{\prime }(z))$ on the unit disc to obtain $d|f^{\prime }(z)|/4\le \tilde{d}$. Apply it again to $f^{-1}$ to get the opposite bound. Thus $\boxed{\tilde{d}/(4d)\le |f^{\prime }(z)|\le 4\tilde{d}/d}$. The domains need not be simply connected domains. For the whole plane, the boundary distances are infinite and these ratios are undefined.

The sphere has the Möbius group $PG{L}_2(\mathbb{C})$, the plane has the affine maps $z\mapsto az+b$ with $a\ne 0$, and the disc has the maps $e^{i\theta }(z-a)/(1-\bar{a}z)$ with $|a|<1$.

The quotients of $\mathbb{C}$ by free properly discontinuous conformal actions are $\mathbb{C}$, ${\mathbb{C}}^{\ast }$, and the complex tori $\mathbb{C}/\Lambda$. The translation group has respectively rank zero, one, or two.

Every plane domain whose complement contains at least two points is uniformized by the unit disc. A plane universal cover would give a nonconstant entire function omitting two values, contrary to the Little Picard theorem.

Let $R$ be a compact Riemann surface of genus $g$, and remove $n$ distinct points. The punctured surface is uniformized by the unit disc exactly when $2g-2+n>0$. It is uniformized by the complex plane exactly when $2g-2+n$ is $-1$ or $0$, corresponding respectively to $\mathbb{C}$, and to ${\mathbb{C}}^{\ast }$ or a complex torus. The sole spherical case is $g=n=0$.

If ${\mathbb{C}}^{\ast }$ embeds holomorphically in a compact Riemann surface $R$, the embedding extends across zero and infinity to a degree-one holomorphic map from the Riemann sphere to $R$. Hence $R$ is conformally the sphere.