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Home , Covering space, Monodromy, Simply connected space

15 Universal Covering Spaces

Uniformization theorem Let us assert the following theorem:

Theorem 15.1 () Every connected is biholomor- phic to one of the following:

∗ CP1, C , C , or C/L , ∆(1)/Γ where Γ ⊂ P SU(1, 1) is a discrete of automorphisms acting freely and properly dis- continuously.

The uniformization theorem breaks up into two parts.

Theorem 15.2 (hard) Every simply connected Riemann surface is biholomorphic to CP1, C or ∆(1).

Theorem 15.3 (fairly easy) Every connected Riemann surface X biholomorphic to a quo- e e e tient X/Γ of a simply connected Riemann surface X by a group Γ ⊂ Aut(X) of automor- e phisms acting freely and properly discontinuously. Moreover, X and Γ are unique up to biholomorphism.

To prove the hard theorem, the difficulty is to construct a biholomorphic map from the simply connected Riemann surface to CP1, C or ∆(1). You may recall that the proof of is quite non-trivial. To prove the easy theorem, the proof turns out to be purely topological, which involves “” and “” and are discussed in this section. Covering space and covering map We first introduce a definition from . Let X,Y be topological spaces and f : Y → X is called a covering map if and only if every −1 x ∈ X has a neighborhood U in X so that f (U)= ∪jVj is a disjoint union with so that for each Vj, the restriction map f : Vj → U is a . Such Y is called a covering space of X. Similarly, we can define holomorphic covering space and holomorphic covering map (for one dimension one or higher dimension). Here is the definition for one dimension. Let X,Y be Riemann surfaces and a holomorphic map f : Y → X is called a holomorphic covering

93 −1 map if and only if every x ∈ X has a neighborhood U in X so that f (U) = ∪jVj is a disjoint union with so that for each Vj, the restriction map f : Vj → U is a biholomorphic map. Here are some examples:

1. Let f(z)= ez : C → C∗ := C −{0}. It is a holomorphic covering map.

2. Let g(z)= zn : C∗ → C∗. It is also a holomorphic covering map. However, g(z)= zn : C → C is not a covering map when n> 1.

3. In general, a quotient space M/G where M is a complex and G is a of Aut(M) may not be a complex manifold. Some conditions are required.

For a covering map π : X → Y , in addition, if X is simply connected, such covering map π : X → Y is called a universal covering map, and the X is called a universal covering space. Similar definitions for holomorphic case. Lifting properties of covering spaces in topology Let us assume existence of covering space and discuss its some basic properties. Our basic model is a complex X = C/Γ e where the natural quotient map π : X = C → X, z 7→ [z], is a universal covering map.

e e Proposition 15.4 Let f : X → X be a covering spaces where X and X are topological spaces. e 1. ( lifting property) For any p ∈ X, pe ∈ X with f(pe)= p, and any continuous curve e e φ : [0, 1] → X with φ(0) = p, there is a unique continuous curve φ : [0, 1] → X such e that f ◦ φ = φ. e 2. (lifting of simply connected spaces) For any w0 ∈ W , p ∈ X and pe ∈ X with f(pe)= p, and a continuous map g : W → X from a path connected, simply W to X, with g(w0) = p where w0 ∈ W and p ∈ X. Then there is a unique lifting e ge : W → X satisfying f ◦ ge = g with ge(x0)= pe. e 3. (Uniqueness of simply connected covering spaces) Assume that X is connected and X is simply connected, then f is a homeomorphism.

Proof: 1. By the property in the definition of covering spaces. 2. By the part 1. and the same argument on theorem.

94 − e 3. As the part 2, we construct the inverse map f 1 from X o X.  e Quotient space and covering spaces Let f : X → X be a covering map. Let a group e e Γ act on X by . The action is properly discontinuous if every xe ∈ X has e e some neighbourhood U whose translates γU, as γ ranges over Γ, are mutually disjoint. The between quotient space and covering space can be seen in the following result, which is immediate from the definition.

e e Proposition 15.5 If Γ acts properly discontinuously on X, then the quotient map X → e X/Γ is a covering space.

Moreover we have e e Lemma 15.6 Let f : X → X be a covering map where X is simply connected. Let Γ be e the group of self-homeomorphisms of X which commutes with f (i.e., f ◦ φ = f for any e e e e φ : X → X in Γ). Then Γ acts X properly discontinuously and X ≃ X/Γ.

Proof: For any p ∈ X and any a, b ∈ f −1(p), by the lifting property of covering maps, e e ′ e e there exists a unique γ : X → X such that γ(a)= b, and another unique γ : X → X such that γ′(b) = a. Then γ′ ◦ γ fixes a and γ ◦ γ′ fixes b, so that they are the identity maps e of X, by the uniqueness of lifting. So γ is a homeomorphism commuting with f. We just established that Γ acts simply transitively on f −1(p). As p is arbitrary, this implies that e − X = X/Γ. Choosing a small connected neighbourhood U of p for which f 1(U) is a disjoint union of copies of U, it follows that Γ acts simply transitively on these copies, hence it acts e properly discontinuously on X.  What is application for “covering space?” or “lifting ?” As an application, we’ll prove by assuming the exitence of holomorphic covering space for every Riemann surface.

Theorem 15.7 (Picard Little Theorem) If an entire f ∈ Hol(C) misses two values, 36 then f = constant.

36An entire function f : C → C missing one point is not necessary to be a constant. For example, f(z)= ez misses only one value 0.

95 To prove this theorem, we consider

f : C → C −{0, 1} and want to show: f = constant. We claim that the universal covering space of C = {0, 1, } = ∆(1), which will be proved in the next section. Then we apply Proposition 22.4 to left f to fˆ:

∆(1) fˆ ր ↓ π f C −→ C −{0, 1} and shall prove that fˆ is holomorphic. By Louville’s theorem, from fˆ : C → ∆(1), it implies that f = constant. It remains to prove existence of the universal covering space for C −{0, 1}. To prove this, we need Schwarz Reflection principle and the following result.

Theorem 15.8 Let D ⊂ Cˆ be a simply connected domain such that D =6 Cˆ. Assume that the boundary ∂D of D is given by a closed Jordan curve. Then every biholomorphic map f : ∆(1) → D extends to a homeomorphism f : ∆(1) → D. In particular, the restriction

f|∂∆(1) : ∂∆(1) → ∂D is homeomorphic.

96