Covering Spaces, Uniformization and Picard Theorems
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Covering Spaces, Uniformization and Picard Theorems Complex Analysis - Spring 2017 - Dupuy Abstract The purpose of this note to develop enough covering space theory to allow us to talk about the covering space proof of the Little Picard Theorem. ♠♠♠ Taylor: [add references] 1 Fundamental Groups Definition 1.1. Let (X; x0) be a pointed topological space. Exercise 1. 1. Check that π1(X; x0) is a group. 2. Assume that X is a path connected topological space. Show that π1(X; x0) ∼= π1(X; x1) as groups. Given f :(X; x ) (Y; y ) we get an induced map of fundamental groups 0 ! 0 f∗ : π (X; x ) π (Y; y ) 1 0 ! 1 0 f∗([γ]) = [f γ] ◦ where [γ] denotes the homotopy class of the loop γ : [0; 1] X. ! Example 1.2. 1. Let D be the unit disc. We have π (D 0 ; 1=2) = Z. 1 n f g ∼ 2. Let S1 = z C : z = 1 = @D. We have π (S1; 1) = Z. f 2 j j 1 The above example shows homotopy invariance of π1. We have that D 0 S1. nf g ∼ Example 1.3. Let X be the figure eight. We have π1(X; x0) = Z Z = α; β = 1 ∼ ∗ h g F2 the free group on two generators. The figure eight is an example of a \smash product". The figure eight is the smash product of S1 with itself. A smash product is where you take two topological spaces and identify two points between them. 1 G1 ∗ G2 is the free product of two groups. It is just the word build out of elements from G1 and G2 subject to the obvious relations. 1 Exercise 2. Show that the fundamental group of a smash product is the free product of the fundamental groups. 1 Example 1.4. 1. π1(P ; x0) = 1 (it doesn't matter what the base point is). 2. π (P1 0 ; x ). Note that is doesn't matter which point we remove and 1 n f g 0 we have P1 a = P1 = C. n f g ∼ n f1g ∼ 3. π (P1 0; 1 ; x ) = π (P1 ; 0 ; x ) = π (C 0 ; x ) = Z. 1 n f g 0 1 n f1 g 0 1 n f g 0 ∼ 4. π (P1 0; 1; ; x ) = Z Z. 1 n f 1g 0 ∼ ∗ What about an elliptic curve? Example 1.5. Let Λ = !1Z !2Z. Recall the fundamental domain for C=Λ. Consider our basepoint C=Λ ⊕z = 0 ! ! ! + ! . Let α be the path 3 0 ≡ 1 ≡ 2 ≡ 1 2 from 0 to !1 and β be the path from 0 to !2 Any loop can be pushed to the boundary we see that any element is generated by α and β. We can also see from the fundamental domain that αβα−1β−1 = 1 and hence that αβ = βα. This show that the group is commutative. It remains to show that α and β are not homotopic. We can try to work this out but we won't. Instead we use that C=Λ = S1 S1 as topological spaces and then use ∼ × that π (S1 S1; (x ; x )) = π (S1; x ) π (S1; x ) = Z Z. 2 1 × 1 2 ∼ 1 1 × 1 2 ∼ × To completely justify this we need the following exercise. Exercise 3. 1. Let γ; γ0 : J X Y be a morphisms of topological spaces. Write γ(t) = (α(t); β(t)) and! γ×0(t) = (α0(t); β0(t)) where α; α0 : J X and β; β0 : J Y are the maps obtained by projecting γ; γ0 to X and! Y respectively.! Show that γ γ0 α α0 and β β0: ∼ () ∼ ∼ 2. Show that the map π (X Y; (x ; y )) π (X; x ) π (Y; y ) given by 1 × 0 0 ! 1 0 × 1 0 [(α; β)] ([α0; β0]) 7! is a well-defined, bijective group homomorphism. 2Warning: Although it is true that for every lattice Λ ⊂ C we have C=Λ =∼ S1 × S1 as topological spaces we do not have C=Λ1 =∼ C=Λ2 as Riemann Surfaces for Λ1 6= Λ2 as general 3 3 3 topological spaces. The j-invariant, j = 1728g2(Λ) =∆(Λ) where ∆(Λ) = g2(Λ) − 27g2(Λ) determines whether we have isomorphic Riemann surfaces. 2 2 Covering spaces In this section we work with Manifolds. Definition 2.1. A covering map of X is a surjective p : Y X such that for every x X there exist some U x open with ! 2 3 −1 a p (U) = Vi i2I with Vi Y open such that p V : Vi U is a homeomorphism. ⊂ j i ! Morphism of covering spaces: Let p1 : Y1 X and p1 : Y2 X be covering maps. A morphisms of coverings of X is a! morphisms of topological! spaces f : Y Y such that the following diagram commutes: 1 ! 2 f Y1 / Y2 : p1 p2 ~ X The collection of objects and morphisms allow us to define a category Cov(X) of coverings of X. We denote the category of coverings of X by Cov(X): We denote the category of connected coverings by Cov(X)0: Definition 2.2. For a pointed topological space we define Cov(X; x0) and 0 Cov(X; x0) in the same way only using morphisms of pointed topological spaces. Remark 2.3. The space Y in the definition above is a covering space and we will use the term covering space and covering map interchangably understanding that covering spaces come with covering maps. Definition 2.4. An automorphism of a covering is called a deck transforma- tion. For a cover X Y the group of deck transformations are denoted by G(Y=X). ! Definition 2.5. A universal covering space is an initial object of Cov(X)0 [unique up to isomorphism]. Spelled-out version of universal covering space: suppose that Xe is a universal cover of X. This means for every cover Y X there exist some map Xe Y such that ! ! Xe / Y : X 3 Definition 2.6. Let p : Y X be a covering of topological spaces (or pointed topological spaces). Consider! a map γ : Z X of topological spaces (I can be anything here). A lift of γ is a map γ such! that p γ = γ. e ◦ e Lemma 2.7 (Path Lifting). Let p :(Y; y ) (X; x ) be a covering space. 0 ! 0 For any γ : [0; 1] (X; x0) such that γ(0) = x0) there exists a unique lifting γ : [0; 1] Y such! that e ! 1. γe is a lift. 2. γe(0) = y0 Sn Proof. Since [0; 1] is compact we can assume that γ([0; 1]) i=1 Ui X with −1 ` ⊂ ⊂ p (Ui) = j2Ji Vi;j local trivializations. We perform induction on n. If n = 1 then let V1;j1 be the unique open set containing y0. The lifting • of the path follows from the isomorphism V U . 1;j1 ∼= 1 Suppose now the theorem holds for covers containing m < n open sets. • Break γ into parts γ = γ1 γ2 such that γ1([0; 1]) and γ2([0; 1]) are con- · 3 tained in a union of fewer than n elements of = Ui : 1 i n . By U f ≤ ≤ g inductive hypothesis, γ1 has a unique lift γe1 with γe1(0) = y0. By induc- tive hypothesis, γ2 has a unique lift γe2 with γe2(0) = γe1(1). The conjuction γ := γ γ works. e e1 · e2 Lemma 2.8. Let Z = [0; 1]. Let p : Y X be a covering. Let γ : Z X be a map with a lift γ : Z Y . Any homotopy! H : [0; 1] Z X lifts to! a unique e ! × ! homotopy He : [0; 1] Z Y . × ! Proof. First we prove the uniqueness statement in the theorem. Suppose • He1; He2 : [0; 1] Z Y are two lifts of H : [0; 1] Z X with He1(0; z) = × ! × ! He2(0; z) = γ(z). Then for each z0 Z, and each i 1; 2 , the function e 2 2 f g Hei(t; z0) is a path lifting of H(t; z0) with Hei(0; z0) = γe(z0) the lifted point. By uniqueness of path lifting we have He1(t; z0) = He2(t; z0). Since I Z is compact there exists some n and some = Ui : 1 i n • × U f Sn ≤ ≤ g a collection of trivializing open sets such that H(I Z) i=1 Ui X; −1 ` × ⊂ ⊂ where the Ui are trivializing: p (Ui) = Vi;j where Ji is some index ∼ j2Ji set. Let's suppose in addition that such a collection is minimal. We perform induction on n. Base case of induction: Suppose n = 1. Then U = V ;j for some unique • 1 ∼ 1 1 V1;j1 y0 (we are just using something here to pin down the sheet in the cover3 that we want to lift to). We use this isomorphism to transport the homotopy. 3 If no such break point exists then for every subinterval [a; b] we have γ([a; b]) touching every one of these Ui's. Letting a ! b we get a discontinuity which is a contradiction. 4 y0 Y y1 p x0 X Figure 1: Here is a cover of a surface with three holes to a surface with holes. The map here is degree two. Note that loops on the bottom do not necessarily lift to loops on the top. To perform the inductive step we must suppose there exists a partition • [ [0; 1] Z = [0; 1] [0; 1] = [ti; ti ] [zj; zj ] (2.9) × × +1 × +1 1≤i≤r;1≤j≤s such that H([ti; ti+1] [zj; zj+1])intersects a proper subset of the Ui : 1 i n .