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 \{ } ∼ 2. Let S1 = z C : z = 1 = ∂D. We have π (S1, 1) = Z. { ∈ | | 1 The above example shows homotopy invariance of π1. We have that D 0 S1. \{ } ∼

Example 1.3. Let X be the figure eight. We have π1(X, x0) = Z Z = α, β = 1 ∼ ∗ h } 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 \{ } 0 we have P1 a = P1 = C. \{ } ∼ \ {∞} ∼ 3. π (P1 0, 1 , x ) = π (P1 , 0 , x ) = π (C 0 , x ) = Z. 1 \{ } 0 1 \ {∞ } 0 1 \{ } 0 ∼ 4. π (P1 0, 1, , x ) = Z Z. 1 \{ ∞} 0 ∼ ∗ What about an elliptic curve?

Example 1.5. Let Λ = ω1Z ω2Z. Recall the fundamental domain for C/Λ. Consider our basepoint C/Λ ⊕z = 0 ω ω ω + ω . Let α be the path 3 0 ≡ 1 ≡ 2 ≡ 1 2 from 0 to ω1 and β be the path from 0 to ω2 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 → ∈ 3 −1 a p (U) = Vi i∈I with Vi Y open such that p V : Vi U is a homeomorphism. ⊂ | 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 .

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) = j∈Ji 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 { ≤ ≤ } inductive hypothesis, γ1 has a unique lift γe1 with γe1(0) = y0. By inductive 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 ∈ ∈ { } 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 { Sn ≤ ≤ } 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 ∼ j∈Ji 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

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 . We will show× that such a partition exists in a subsequent{ bullet.≤ ≤ } Supposing the existence of the partition (2.9) exists, we perform the induc- • tive step: By inductive hypothesis there exist a unique Fe1,j : [0, t1] Zj × → X for j = 1, . . . , s that has Fe1,j(0, z) = γe(z) lifting γ(z). By the uniqueness lemma, Fe1,j(t1, z) = Fe1,j+1(t1, z),

and hence we get He1 : [0, t1] Y lifting H ,t ×Z : deﬁned by → |[0 1]  Fe1,1(t, z), z [z1, z2]  ∈ Fe1,2(t, z), z [z2, z3] He1(t, z) = ∈ .  .  Fe1,s−1(t, z), z [zs−1, zs]. ∈

5 This ﬁts into a diagram

Y . : He1

H [0, t ] Z / X 1 ×

Now we repeat this process applying the inductive hypothesis the next intervals [t1, z] Zj where we apply our lifting to “initial data” He1(t1, z) × (the lift we just constructed). This gives a lifts Fe2,j :[t1, t2] Zj Y × → lifting H [t1,t2]×Zj . As before these glue together to give some He2 :[t1, t2] Z Y lifting| H :[t , t ] Z X. × → |[t1,t2]×Z 1 2 × → Repeating this process we get a sequence Hej :[tj, tj+1] Z Y and we deﬁne × →  He1(t, z), t [t1, t2]  ∈ He2(t, z), t [t2, t3] He(t, z) := ∈ . .  .  Her−1(t, z), t [tr−1, tr]. ∈ By uniqueness of lifting, all of the components agree on their intersection and we have a lifting. Proof of existence of the partition (2.9): Suppose no such partition exists. • We derive a contradiction in the style of Goursat’s Theorem or Bolzano- Weierstrass: By subdividing I Z = [0, 1] [0, 1] into fourths there exists × × some B1 in that partition which intersects all of the Ui in our cover. By subdividing B into quarters again, there exists some B B one 1 2 ⊂ 1 quarter of the size whose image intersects all of the Ui again. Repeating this process gives us a decreasing sequence

B B B ... 1 ⊃ 2 ⊃ 3 ⊃ T with Bi = b∞ I Z (as the sets are converging we get a single point) such that{ for} ⊂ all j×,

H(Bj) Ui = for 1 i n . ∩ 6 ∅ ≤ ≤

If we pick bj Bj such that if j i mod n we have ∈ ≡ [ H(bj) Ui Uk ∈ \ k:k=6 i

then limj→∞ H(bj) doesn’t exist while (bj) b∞ I Z. Since H is continuous this is a contradiction. → ∈ ×

6 X x X \{ } X x0 x0

x x

not contractible contractible inclusion

Figure 2: A loop around x in X x is not contractible, but after adding the point x back it, it becomes contractible.\{ }

Lemma 2.10 (Coverings give Subgroups of Fundamental Groups). Let p : (Y, y ) (X, x ) be a covering map. The map p∗ :(Y, y ) (X, x ) is injective. 0 → 0 0 → 0

p∗π (Y, y ) = loops based at x with lifts that are loops based at y / 1 0 { 0 0 } ∼ Remark 2.11. Can we draw a picture of Lemma 2.10? Not really. Lemma 2.10 states essentially that is impossible to draw two loops in Figure 1 which are not homotopic in the bottom but homotopic in the top. Note that for non-covering maps this is a common thing. Consider for example (Z, z0) = (X x , x0) (X, x0) together with its inclusion map. Here loops around x in X\{ x} can⊂ be not homotopic to zero but become homotopic to zero one we consider\{ their} pushforward in X. See ﬁgure 2.4

Proof. By the Homotopy Lifting Lemma, the map is injective.

Remark 2.12. For covering map (Y, y0) (X, x0) it is useful to make the notational convention →

π (Y, y ) = p∗π (Y, y ) π (X, x ) 1 0 1 0 ⊂ 1 0 viewing fundamental groups of coverer as subgroups of the fundamental group of the coveree. 4 The kernel of the map π1(X \{x}, x0) → π1(X, x0) is called the inertia group and generated by loops around x. These actually correspond to stabilizers for actions of fundamental groups on universal covers. For ramiﬁed coverings of Riemann surface, if one deletes the branch points and branch values one gets a covering space. The inertia group of a ram- iﬁcation value is then a subgroup of the group of deck transformations corresponding to the kernel of this inclusion. ♠♠♠ Taylor: [write down exact sequence in later version]

7 Lemma 2.13 (Lifting Criterion). Let p :(Y, y0) (X, x0) be a covering map. Let f :(Z, z ) (X, x ) be any morphism. There→ exists some f 0 :(Z, z ) 0 → 0 0 → (Y, y0) such that the following diagram commutes

(Y, y0) o (Z, z0) f 0 p f z (X, x0) if and only if f∗π (Z, z ) π (Y, y ) π (X, x ). 1 0 ⊂ 1 0 ⊂ 1 0 Proof. We deﬁne the map Lemma 2.14 (Universal Covers = Simply Connected Covers). Consider the diagram

(Y, y0) covering map f (Z, z0) / (X, x0)

If Z is simply connected then there exists a unique fe :(Z, z0) (Y, y0) such that → (Y, y0) . : f e covering map f (Z, z0) / (X, x0)

In particular any covering map (Z, z0) (X, x0) with Z simply connected is a universal covering map. →

Proof. Taylor: [Add the proof from class. This is essentially from Munkres.] ♠♠♠

0 Theorem 2.15 (Galois Correspondence). Let Cov(X, x0) be the category of connected pointed covers.

π : Cov(X, x )0 subgroups of π (X, x ) . 1 0 → { 1 0 } Taylor: [Finish Adding proofs] Covers of covers are covers so higher covers♠♠♠ corrspond to smaller subgroups. The universal cover corresponds to the trivial subgroup.

Note that Nπ1(X,x0)(π1(Y, y0)) is a subcover. This is the Galois hull.

8 3 Aside: Seifert-Van Kampen

We can now have enough machinery to prove some things about fundamental groups. Theorem 3.1 (Seifert-Van Kampen). Let X be a manifold and assume X = U1 U2 with U1 an U2 non-empty open sets. Assume that U1 U2 is also non-empty.∪ Then for x U U ∩ 0 ∈ 1 ∩ 2 π (U U , x ) / π (U , x ) 1 1 ∩ 2 0 1 1 0

π2(U2, x0) / π1(X, x0) is a pushout. Proof. The idea of the proof is to use the Galois correspondence to construct a cover. Taylor: [Finish me] ♠♠♠ Taylor: [Applications] ♠♠♠ Taylor: [In the category of topological spaces the universal cover exists because♠♠♠ it is the inverse limit of topological spaces. ]

4 Galois Coverings

Deﬁnition 4.1. A covering p : Y X is called normal or Galois if and only if x X, y, y0 p−1(x), g G(Y/X→ ) such that ∀ ∈ ∀ ∈ ∃ ∈ g(y) = y0. Lemma 4.2. Let p :(Y, y ) (X, x ) be a covering map. 0 → 0 1. G(Y/X) N (π (Y, y ))/π (Y, y ).5 ∼= π1(X,x0) 1 0 1 0

2. The covering p is Galois if and only if π1(Y, y0) π1(X, x0) is normal. In this case we have ⊂

G(Y/X) = π1(X, x0)/π1(Y, y0).

Lemma 4.3. If p : Y X is Galois then X = Y/G where G = G(Y/X). → ∼ Remark 4.4. In the theory of Riemann surfaces (and more generally algebraic geometry) it is useful to consider so called “branched coverings”. There are maps p : Y X where outside some small set R X we have a genuine covering: → ⊂ Y p−1(R) X R. \ → \ We will not consider these here, but you should keep this in mind when reading things elsewhere.

5 Here Nπ1(X,x0)(π1(Y, y0)) denotes the normalizer.

9 5 Coverings of Riemann Surfaces

Lemma 5.1. Let p : Y X be a covering. If X is a Riemann surface then so is Y . → Proof Idea. Y is locally isomorphic to X. Theorem 5.2 (Uniformization Theorem). Every simply connected Riemann surface is isomorphic to one of the following:

P1 • C • H = D the upper-half plane or unit disc. • ∼ Lemma 5.3 (Classiﬁcation of Riemann Surfaces). Any Riemann surface is isomorphic to one of the following:

1. X = P1 2. X = C/Γ, Γ Aut(C) = AL (C) ⊂ 1 3. X = H/Γ, Γ Aut(H) = PSL (R). ⊂ 2 Proof. Let X be a Riemann surface. Let Xe be its universal cover.

By the characterization of universal covers (Lemma 2.14) we know that • Xe is simply connected. By the characterization of simply connected Riemann surfaces we know • 1 that Xe ∼= P , C or H.

Since π1(Xe) = 1 we have that Xe X is Galois and hence that • →

X ∼= X/Ge (X/Xe ).

Lemma 5.4. Consider X = P1 p , p , . . . , r if r 3 then the universal \{ 1 2 } ≥ cover Xe is D.

6 Little Picard Theorem

Theorem 6.1 (Little Picard). Let f : C C be an entire function. If the image of f omits more than two points then→f is constant. Proof. If suppose f omits more than two points. We can view this as a • 1 map f : C X = P p , p , . . . , pr with r 3. → \{ 1 2 } ≥

10 1 6 There is a cover of P p , p , . . . , pr by D = H = z : Im z > 0 . • \{ 1 2 } ∼ { } By the lifting lemma, since C is simply connected there exists some C D • such that → F X = C / Y = D f p ) 1 X = P p , p , . . . , pr \{ 1 2 } . By Liouville’s Theorem F is constant and hence so is f. •

Taylor: [ It suﬃces to prove this for two points missing since having more points♠♠♠ missing still deﬁnes a map f : C C 0, 1 .] → \{ } 7 Galois theory of functions

Theorem 7.1. The contravariant functor

Compact Riemann surfaces Algebraic extensions of C(z) { } → { } X Mer(X) 7→ which assigns a compact Riemann surface to its ﬁeld of meromorphic functions is an equivalence of categories.

First things ﬁrst, it is not clear that for a compact Riemann surface X that Mer(X)

6In the case of three points it is enough to show that there is a map H → C \{0, 1} and the map 3 j(z) = 1728g2 (z)/∆(z) 3 2 where ∆(z) = g2 (z) − 27g3 (z) for z ∈ H does the trick. If we let q = exp(2πiz) then j = 1/q + 744 + 196844q + 21493760q2 + 86429970q3 + ··· .