Uniformization theorem

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

The level-two principal congruence subgroup $\Gamma (2)\subset {\mathrm{PSL}}_2(\mathbb{R})$ consists of integer determinant-one matrices congruent to the identity modulo two, modulo the scalar sign. It acts freely and properly discontinuously on the complex upper half-plane. An ideal quadrilateral with vertices $-1,0,1,\infty$ has its paired sides identified by $\tau \mapsto \tau +2$ and $\tau \mapsto \tau /(1-2\tau )$. The quotient has genus zero and three punctures, so it is conformally the sphere with three points removed. The invariant metric $|d\tau |/\mathrm{Im}\tau$ descends to its curvature-minus-one hyperbolic metric.

// Destination: analysis.bigb

Exhaust a simply connected hyperbolic Riemann surface in potential theory by compact bordered Jordan domains $R_n$. Their Green functions on a Riemann surface $G_n(\cdot ,p)$ have logarithmic poles and zero boundary values. Exponentiating their harmonic conjugates produces proper degree-one maps $f_n:R_n\to \mathbb{D}$ with $f_n(p)=0$. If $G_n=-\log |\zeta |+c_n+o(1)$, normalize $f_n^{\prime }(p)=e^{-c_n}>0$. Domain monotonicity and a global positive Green function bound $c_n$ increasingly from above. Thus Montel theorem gives a nonconstant limit $F$, which is injective by the locally uniform limit of univalent functions theorem. The Schwarz lemma makes $F^{\prime }(p)$ extremal among normalized disk-valued maps. If $F$ omitted $a\ne 0$, taking a global holomorphic square root of a disk automorphism applied to $F$, then renormalizing at $p$, would multiply its derivative by $(1+|a|)/(2\sqrt{|a|})>1$. Thus $F$ is onto and gives the desired biholomorphism.

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

For a univalent function $f:\Omega \to \mathbb{D}$ with $f(z_0)=0$, suppose $a\ne 0$ is omitted. Apply the disk automorphism $T_a$ so the resulting image omits zero, take a holomorphic square root, then apply the disk automorphism taking the value at $z_0$ to zero. If $\Omega$ is a simply connected domain, the root exists. The resulting normalized univalent map has derivative magnitude multiplied by $(1+|a|)/(2\sqrt{|a|})>1$, since $(1-\sqrt{|a|}{)}^2>0$. Thus a map maximizing the derivative cannot omit an interior disk point.

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.

For a normalized univalent function $f$ on the unit disc, put $r=|z|<1$. The displayed sharp two-sided bounds hold. Rotations $z/(1-e^{i\theta }z{)}^2$ of the Koebe function attain the bounds along suitable radial directions. In particular, bounding the modulus controls the radius of a disk containing each smaller image $f(r\mathbb{D})$.

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.