The Universal Covering Space of the Twice Punctured Complex Plane

Gustav Jonzon U.U.D.M. Project Report 2007:12

Examensarbete i matematik, 20 poäng Handledare och examinator: Karl-Heinz Fieseler

Mars 2007

Department of Mathematics Uppsala University

Abstract In this paper we present the basics of holomorphic covering maps between Riemann surfaces and we discuss and study the structure of the holomorphic covering map given by the modular function λ onto the base space C r {0, 1} from its universal covering space H, where the plane regions C r {0, 1} and H are considered as Riemann surfaces. We develop the basics of Riemann surface theory, tie it together with complex analysis and topological covering space theory as well as the study of transformation groups on H and elliptic functions. We then utilize these tools in our analysis of the universal covering map λ : H → C r {0, 1}. Preface

First and foremost, I would like to express my gratitude to the supervisor of this paper, Karl-Heinz Fieseler, for suggesting the project and for his kind help along the way. I would also like to thank Sandra for helping me out when I got stuck with TEX and Daja for mathematical feedback. Contents

1 Introduction and Background 1

2 Preliminaries 2 2.1 A Quick Review of Topological Covering Spaces ...... 2 2.1.1 Covering Maps and Their Basic Properties ...... 2 2.2 An Introduction to Riemann Surfaces ...... 6 2.2.1 The Basic Deﬁnitions ...... 6 2.2.2 Elementary Properties of Holomorphic Mappings Between Riemann Surfaces ...... 7 2.2.3 Holomorphic Covering Maps ...... 12 2.3 Properly Discontinuous Group Actions and Fundamental Regions 18 2.3.1 The Upper Half-Plane and its Automorphisms ...... 20 2.3.2 The Modular Group Γ and Modular Functions on H . . . . 21

3 The Universal Covering Space of C r {0, 1} 22 3.1 The Subgroup G of Γ ...... 22 3.2 The Fundamental Region Ω ⊂ H for G ...... 26 3.3 The G - Modular Function λ : H → C r {0, 1} ...... 36 3.3.1 An Application of λ ...... 45 λ 3.4 The covering group Deck (H −−→ C r {0, 1}) ...... 47 3.5 A Relation Between λ and J ...... 47 3.5.1 A Quick Review of Elliptic Functions ...... 47 3.5.2 A Branched Holomorphic Covering Map f Factoring J Through λ ...... 49

4 Bibliography 53 1 INTRODUCTION AND BACKGROUND 1

1 Introduction and Background

In the theory of functions of one complex variable we study holomorphic or meromorphic functions defined on subregions of the complex plane. However, the naïve approach to this study involves vague notions of multiple valued functions. Consider, for example, the complex logarithm which cannot be globally defined on C without discontinuities. Such phenomena can however be dealt with by considering a more natural domain of definition than a subregion of the plane. This gives rise to the theory of Riemann surfaces on which multi-valued holomorphic functions become single valued. By abstracting the notion of Riemann surfaces corresponding to functions and defining them as surfaces in their own right with a complex structure we bring the study into the setting of manifold theory. Each Riemann surface corresponding to a function turns out to be a Riemann surface in the abstract sense as well. A powerful tool for understanding topological spaces, and Riemann surfaces in particular, is given by covering spaces and fundamental groups. Covering spaces of a given base space mimic the base space to a large extent but are often more susceptible to analysis. This applies in particular if the covering space is simply connected, i.e., if it is the universal covering of the given space. In analyzing Riemann surfaces, it is convenient to require that the covering maps are holomorphic. This actually turns out to be automatic given the proper definition of complex structure. Basic topology assures us that every manifold in fact has a universal covering space. The uniformization theorem1(see, e.g., [6]) then guarantees that the universal covering space of any region in C is conformally equivalent either to the complex plane C, the open unit disc D or the Riemann sphere P1. The most interesting cases yield D as its universal covering space. However, since D is conformally equivalent to the upper half plane H we might as well study the latter space. In fact H turns out to be a more convenient space to work with. It is thus useful to study spaces which have H as its universal covering space. A practical device for constructing such spaces is by considering a discrete group of automorphisms of H and its corresponding natural action, and taking the quotient of H with respect to this action. The resulting quotient space is seen to have a natural complex structure turning the quotient projection into a holomorphic covering map, possibly having branch points in its domain. This construction is aided by considering a fundamental region for the given discrete group of automorphisms and functions that are invariant under the action of this group. In this master thesis we consider in detail a construction of the universal covering space of the Riemann surface Cr{0, 1} and its affiliated covering map. We will

1This is basically a generalization of the Riemann mapping theorem to general Riemann surfaces. 2 PRELIMINARIES 2 begin by recalling several deﬁnitions and theorems of topological covering space theory. Secondly, we will bring covering space theory into the setting of complex analysis and Riemann surfaces. Only the most pertinent notions and results will be reviewed. For a more thorough account of these topics the reader is referred to [6]. We will continue by introducing the concept of discrete group actions and automorphisms. Our construction now begins by giving a discrete subgroup G of the group PSL2(R) of automorphisms of H. The corresponding fundamental region Ω is described in detail, as are results concerning the group G and its relation to Ω. We proceed to construct a covering map λ : H → C r {0, 1}. The relevant properties of this map are then deduced. In particular, λ is constructed to be invariant under the action of G, which is also shown to be the group of Deck transformations of the covering. Finally, we produce a mapping relating λ to the classical modular function J . The reader is presumed to have working knowledge of topology up to and including an exposure to manifolds and the basic theory of fundamental groups and covering spaces. Some good sources for this are given by [13], [16] and [14]. In addition, a solid background in complex analysis including the Riemann mapping theorem and the Carathéodory theorem is assumed. Some nice references here are [1], [4] and [9]. In the last section, we also presume some acquaintance with the very basics of elliptic function theory. This is treated in [1], [7] and [3].

2 Preliminaries

2.1 A Quick Review of Topological Covering Spaces We will here very brieﬂy recall the main deﬁnitions and results of topological covering space theory that we will need. The reader is referred to [14] or [13] for a thorough account. A review of the bare essentials needed for Riemann surface theory can be also be found in the initial chapters of [6].

2.1.1 Covering Maps and Their Basic Properties Let X and Y be topological spaces, and let p : Y → X be a continuous mapping. An open set U ⊆ X is said to be evenly covered by p if p−1(U) is a disjoint union of open sets, each of which is mapped homeomorphically onto U by p, i.e., F −1 ∃ {Uα}α∈I ⊂ ℘(X): p (U) = α∈I Uα and p |Uα : Uα −→ U ∀α ∈ I. The mapping p : Y → X is said to be a covering map if every point x in X has an open neighborhood Ux which is evenly covered by p. If p : Y → X is a covering map, then Y is said to be a covering space of X. Clearly, covering maps are local homeomorphisms. 2 PRELIMINARIES 3

We now state the results on covering maps that we are going to need below. The theorems stated below are not in their most general forms, and we refer to any of the texts [13], [16] or [14] on topology found in the bibliography for proofs and more generality.

Lifting Properties: Let X and Y be topological spaces and let p : Y → X be a continuous mapping. Let Z be a topological space and let f : Z → X be a continuous mapping. Then a continuous mapping g : Z → Y is said to be a lifting of f w.r.t. p if we have f = p ◦ g, i.e., if the following diagram commutes.

Y ? g p / Z f X

Theorem 2.1.1 (Uniqueness of Liftings) Suppose that X and Y are Hausdorﬀ topological spaces and that p : Y → X is a local homeomorphism. Suppose further that Z is a connected topological space and that f : Z → X is a continuous mapping. If g1 : Z → Y and g2 : Z → Y both are liftings of f w.r.t. p and if ∃ z0 ∈ Z : g1(z0) = g2(z0), then we have g1 = g2.

Theorem 2.1.2 (Unique Curve Lifting Property) Suppose that X and Y are topological spaces and that p : Y → X is a covering map. Let γ : [0, 1] → X be a curve in X and let y0 ∈ Y be a point in Y such that p(y0) = γ(0). Then there exists a unique lifting γ˜ : [0, 1] → X of γ w.r.t. p such that we have γ˜(0) = y0.

< Y zz γ˜ zz zz p zz / [0, 1] γ X

Theorem 2.1.3 Suppose that X is a Hausdorﬀ space, that Y is a path-connected Hausdorﬀ space and that p : Y → X is a covering map. Then, ∀x1, x2 ∈ X, we −1 −1 have |p ({x1})| = |p ({x2})|. In particular, if Y , ∅, then p is surjective.

Given x ∈ X, the cardinality |p−1({x})| ∈ N ∪ {∞} of p−1({x}) is called the number of sheets of the covering. The above theorem assures us that this notion is well- deﬁned. 2 PRELIMINARIES 4

Theorem 2.1.4 Suppose that X and Y are Hausdorﬀ spaces and that p : Y → X is a covering map. Let Z be a simply connected, pathwise connected and locally pathwise connected topological space and let f : Z → X be a continuous mapping. Let z0 ∈ Z, y0 ∈ Y : f (z0) = p(y0). Then there exists a unique lifting fˆ : Z → Y of f w.r.t. p such that we have fˆ(z0) = y0.

Y ? f˜ p / Z f X

The Universal Covering Space: Suppose that X, Y1 and Y2 are topological spaces and that p1 : Y1 → X and p2 : Y2 → X are covering maps. Then the covering spaces Y1 and Y2 of X are said to be isomorphic if there exists a ﬁber-preserving homeomorphism between Y1 and Y2.

/ Y1 @ Y2 @@ ~~ @@ ~ p @ ~~p 1 @ ~~ 2 X

Suppose that X and Y are connected topological spaces and that p : Y → X is a covering map. Suppose that if Z is a connected topological space and q : Z → X is a covering map and z0 ∈ Z, y0 ∈ Y : f (z0) = p(y0) then there exists a unique ﬁber-preserving mapping f : Y → Z such that we have f (y0) = z0. Then the covering (Y, p) is called universal.

Y f Z p ?? ?? ?? q ? X

Given a connected topological space X, its universal covering space is unique up to an isomorphism of covering spaces.

Theorem 2.1.5 Suppose that M and N are connected topological manifolds and p : N → M is a covering map. If N is simply connected then (N, p) is the universal covering space of M. 2 PRELIMINARIES 5

Theorem 2.1.6 Let M be a connected topological manifold. Then M has a universal covering manifold, i.e., there exists a connected and simply connected manifold M˜ and a covering map p : M˜ → M.

Deck Transformations: Suppose that X and Y are topological spaces, and p : Y → X is a covering map. Then a ﬁber-preserving homeomorphism φ : Y → Y is called a deck transformation of the covering. They clearly form a group, which we denote p by Deck(Y −−→ X).

φ Y / Y ?? ?? ?? p ? p X Suppose that X and Y are connected Hausdorﬀ spaces and p : Y → X is a covering map. Then the covering is said to be Galois if, for all y1, y2 ∈ Y such that p(y1) = p(y2), there exists a deck transformation φ : Y → Y such that we have f (y1) = y2. Such a deck transformation must then be unique.

Theorem 2.1.7 Let M be a connected manifold and let (M˜ , p) be its universal p covering manifold. Then p is a Galois covering and we have Deck(Y −−→ X) π1(M).

Suppose that X and Y are topological spaces and p : Y → X is a covering map. p Let G ≤ Deck(Y −−→ X) and let y1, y2 ∈ Y. Then y1 and y2 are said to be equivalent modulo G, y1 ∼G y2, if ∃ φ ∈ G : φ(y1) = y2.

Theorem 2.1.8 Suppose that M and N are connected topological manifolds and q : N → M is a covering map. Let (M˜ , p) be the universal covering space of X. Let f : M˜ → N be a ﬁber-preserving continuous mapping. Then f is a (universal) p covering map. Moreover, there exists a subgroup G π1(N) of Deck(M˜ −−→ M) such that, ∀x1, x2 ∈ X, we have f (x1) = f (x2) ⇔ x1 ∼G x2.

M˜ ~ f ~~ ~~ ~~ N p AA AA AA q A M This pretty much sums up the topological results needed below. 2 PRELIMINARIES 6

2.2 An Introduction to Riemann Surfaces We will begin by presenting the foundations of Riemann surface theory. Our approach will be abstract, as opposed to considering Riemann surfaces corresponding to given holomorphic functions. For the latter, the reader is referred to chapter IX of [4], or any one of the texts [11] or [6].

2.2.1 The Basic Deﬁnitions Recall that an n-dimensional (real) manifold is a 2nd-countable Hausdorﬀ topological space M such that each point in M has an open neighborhood which is homeomorphic to an open subset of Rn. Let X be a 2-dimensional manifold. Then a complex chart on X is a homeomorphism φ : U → V of an open subset U ⊂ X onto an open subset V ⊂ C. Two complex charts φ1 : U1 → V1 and φ2 : U2 → V2 on X are said to be holomorphically compatible if the mapping

−1 φ2 ◦ φ1 : φ1(U1 ∩ U2) → φ2(U1 ∩ U2)

2 is biholomorphic .A complex atlas for X is a collectionS {φi : Ui → Vi}i∈I of pairwise holomorphically compatible charts such that i∈I Ui = X. Two complex atlases A and B for X are said to be holomorphically equivalent if each complex chart in A is holomorphically compatible with each complex chart in B. This is easily seen to be an equivalence relation on the set of all complex atlases on X, and a complex structure on X is defined to be an equivalence class of complex atlases on X under this relation. We are now ready to define our main object of study. A Riemann surface is a pair X = (X, [A]), where X is a 2-dimensional manifold and [A] is a complex structure on X. In general, an n-(complex-)dimensional complex (analytic) manifold is defined just as (real) differential manifolds, but with Rn replaced by Cn and the requirement of real differentiability of the transition functions replaced by the condition of holomorphicity. In this setting, a Riemann surface is a connected 1-(complex-) dimensional complex manifold. We take note of the fact that an n-dimensional complex manifold has more structure than a 2n-dimensional smooth manifold, since holomorphy is a stronger requirement than smoothness. As is conventional, we endow Cˆ with a topology by declaring a set to be open if and only if it is an open subset of C or it is (C r K) ∪ {∞} with a compact subset K ⊂ C, whereupon Cˆ becomes homeomorphic to the 2-sphere S2. The mappings ˆ ˆ 1 id : C r {∞} → C and C r {0} 3 z 7→ z ∈ C comprise an atlas giving a complex

2That is, holomorphic and bijective with a holomorphic inverse mapping. 2 PRELIMINARIES 7 structure on S2, making it into a Riemann surface, the Riemann sphere, sometimes denoted by P1. Of course, any region U in C is a Riemann surface with the identity mapping idU as a globally defined chart giving an atlas on U. More generally, any connected open subset of a Riemann surface is again a Riemann surface. Let X be a Riemann surface, and let A ⊂ X be an open subset. Then a function f : A → C is said to be holomorphic if the function f ◦ ψ−1 : ψ(U ∩ A) → C is holomorphic (in the usual sense) for each complex chart ψ : U → V on X. A function f : A → C is said to be meromorphic if there exists an open subset 0 0 0 A ⊂ A such that f |A0 ∈ O(A ) and A r A only contains isolated points, and such 0 0 that, ∀p ∈ A r A , limx→p | f (x)| = ∞, the points in A r A then being called the poles of f . The sets of holomorphic and meromorphic functions on a given Riemann surface X are denoted by O(X) and M (X) respectively, and both have natural structures of C-algebras. The meromorphic functions on a Riemann surface X can also be identified with holomorphic mappings of X into the Riemann sphere, in the sense to be defined now. Given two Riemann surfaces X and Y, a continuous mapping f : X → Y is said to be holomorphic if, for each complex chart ψ1 : U1 → V1 on X and for each complex chart ψ2 : U2 → V2 on Y with f (U1) ⊂ U2, the mapping −1 ψ2 ◦ f ◦ ψ1 : V1 → V2 is holomorphic (in the usual sense). Notice that compositions of holomorphic mappings are again holomorphic, and that the holomorphic mappings from X to C are precisely the holomorphic functions on X. If f : X → Y is holomorphic and bijective with a holomorphic inverse mapping, then f is called biholomorphic. Just as in classical complex analysis, we will interchangeably use the synonyms "biholomorphic", "conformally equivalent" and "analytically isomorphic". Also, a biholomorphism f : X → X is called an automorphism of X. It turns out that for each proper non-constant holomorphic mapping f : X → Y between two Riemann surfaces X and Y there exists a closed and discrete subset A ⊂ X such that X r f −1( f (A)) and Y r f (A) are Riemann surfaces and f : X r f −1( f (A)) → Y r f (A) is a (holomorphic) covering map. Such a mapping will be called a "branched" covering map (see subsection 2.2.3 below).

2.2.2 Elementary Properties of Holomorphic Mappings Between Riemann Surfaces The properties of holomorphic functions and conformal mappings on open subsets of the complex plane transfer through charts to Riemann surfaces. Here follow a 2 PRELIMINARIES 8 few useful examples. For more information, see [6] or [12].

Theorem 2.2.1 (The Riemann Removable Singularities Theorem) Let X be a Rie- mann surface, let U ⊆ X be a non-empty open subset of X, let ζ0 ∈ U and let f ∈ O(U r {ζ0}) be a function which is bounded in some punctured open neighborhood of ζ0 in U r {ζ0} ( X. Then f can be uniquely extended to holomorphic function fˆ ∈ O(U) on U.

Proof Let (V, φ) be a complex coordinate chart on X around the point ζ0 ∈ X such that we have V ⊆ U. Then (V r {ζ0}, φ|Vr{ζ0}) is a again a complex chart of 3 X and we have V r {ζ0} ⊆ U r {ζ0}. Thus, since we have f ∈ O(U r {ζ0}), and φ|Vr{ζ0}(V r {ζ0}) = φ(V r {ζ0}) = φ(V) r {φ(ζ0)}, we know that the function −1 −1 f ◦ (φ|Vr{ζ0}) : φ(V) r {φ(ζ0)} → C is holomorphic, i.e., we have f ◦ (φ|Vr{ζ0}) ∈ −1 O(φ(V) r {φ(ζ0)}). Since f is bounded near ζ0 ∈ X, we see that f ◦ (φ|Vr{ζ0}) is bounded near φ(ζ0) ∈ φ(V) ⊂ C. Hence, by Riemann´s removable singularities 4 −1 theorem , we know that f ◦ (φ|Vr{ζ0}) can be uniquely extended to a holomorphic −1 function f ◦ (φ|Vr{ζ0}) ∈ O(φ(V)). ˆ ˆ −1 We extend f to f on U by deﬁning f (ζ0):= f ◦ (φ|Vr{ζ0}) (φ(ζ0)) ∈ C. Then ˆ −1 ˆ we have f |V = f ◦ (φ|Vr{ζ0}) ◦ φ, so f is holomorphic on U, as desired.

Theorem 2.2.2 (The Identity Theorem) Let X and Y be Riemann surfaces and let f : X → Y and g : X → Y be holomorphic mappings. Let A := {z ∈ X : f (z) = g(z)} ⊆ X. Suppose that A has a limit point in X. Then we have f = g.

Proof Let B := {z ∈ X : f |W = g|W f or some open neighborhood W ⊆ X o f zinX} ⊆ X. Let us show that we have B = X. Since X is connected, it suﬃces to show that B ⊆ X is open, closed and non-empty. B is open: Let z ∈ B. Then there exists an open neighborhood W ⊆ X of z in X such that we have f |W = g|W . Let ζ ∈ W. Then W ⊆ X is an open neighborhood of ζ in X such that f |W = g|W . Hence ζ ∈ B. Therefore, we have W ⊆ B, so that B is open. B is closed: Let z ∈ ∂B. Then, since we have ( f − g)(ζ) = 0 ∀ζ ∈ B, and f − g is continuous, we must have ( f − g)(z) = 0 too, i.e., we have f (z) = g(z). Let (U, φ) and (V, ψ) be charts around z and f (z) = g(z) respectively such that U is connected, f (U), g(U) ⊂ V and ψ ◦ f ◦ φ−1, ψ ◦ g ◦ φ−1 : φ(U) → ψ(V) are holomorphic. Since z ∈ U ∩ ∂B, we have U ∩ B , ∅. Thus, φ(U ∩ B) = {ζ ∈ φ(U): ψ ◦ f ◦ φ−1(ζ) = ψ ◦ g ◦ φ−1(ζ)} is a non-empty open subset of φ(U) ⊂ C. Hence by the identity theorem5 and the fact that φ(U) is connected, we see that we

3belonging to the same complex structure on X as the previous one 4for functions deﬁned and holomorphic on open subsets of the complex plane 5for holomorphic functions on regions in C 2 PRELIMINARIES 9

−1 −1 have ψ ◦ f ◦ φ = ψ ◦ g ◦ φ . It follows that f |U = g|U . Therefore, z ∈ B, so B is closed. B is non-empty: Let α ∈ X be a limit point of A. Then, just as in the preceding continuity argument, we ﬁnd that f (α) = g(α), i.e., α ∈ A (so A is closed). Let (V, ψ) be a chart around f (α) = g(α) in Y. Then there is a chart (U, φ) around α in X such that f (U), g(U) ⊂ V and ψ ◦ f ◦ φ−1, ψ ◦ g ◦ φ−1 : φ(U) → ψ(V) are holomorphic. But since α ∈ U ∩ A, we also have φ(α) ∈ φ(U ∩ A) = {z ∈ φ(U): ψ ◦ f ◦ φ−1(z) = ψ ◦ g ◦ φ−1(z)}. Again, by the identity theorem in the −1 −1 plane, we get ψ ◦ f ◦ φ |W = ψ ◦ g ◦ φ |W for some open disc W ⊂ φ(U) around φ(α). Consequently, f |U∩φ−1(W) = g|U∩φ−1(W), so that α ∈ B. In particular, we get B , ∅. Since B is open, closed and non-empty, and X is connected, we have B = X, as claimed.

Theorem 2.2.3 (The Open Mapping Theorem) Let X and Y be Riemann surfaces and let f : X → Y be a non-constant holomorphic mapping. Then f is an open mapping.

Proof Let U ⊂ X be an open subset of X. Let ζ ∈ f (U). Then ∃ z ∈ U : f (z) = ζ. Let (U1, φ1) and (U2, φ2) be charts around z and ζ respectively, such that f (U1) ⊆ −1 0 U2 and φ2◦ f ◦φ1 : φ1(U1) → φ2(U2) is holomorphic. Let U := U∩U1 ⊂ X. Then 0 0 6 0 U is an open subset of U1, so (U , φ1|U0 ) is a chart on X , and φ1(U ) must be an −1 open subset of φ1(U1) ⊂ C. Since f is not constant, neither is φ2◦ f ◦φ1 (If it were, then f |U1 would be too. Hence, by the identity theorem 2.2.2, f would be constant, contradicting our assumption.). Therefore, the open mapping theorem7, assures us −1 that the mapping φ2 ◦ f ◦ φ1 : φ1(U1) → φ2(U2) is open. Consequently, since 0 0 −1 0 φ1(U ) is open in φ1(U1), φ2( f (U )) = φ2 ◦ f ◦ φ1 (φ1(U )) is open in φ2(U2) ⊂ C, 0 −1 0 0 which implies that f (U ) = φ2 (φ2( f (U ))) is open in U2 ⊂ Y. But z ∈ f (U ) ⊂ f (U), i.e., f (U0) is an open neighborhood of z in Y which is contained in f (U). Hence, f (U) is open in Y, as claimed.

Remark Notice that the open mapping theorem together with the fact that continuous mappings take connected sets to connected sets, imply that images of Riemann surfaces under non-constant holomorphic mappings are again Riemann surfaces.

Theorem 2.2.4 (The Maximum Modulus Theorem) Let X be a Riemann surface and let f ∈ O(X) be a non-constant holomorphic function on X. Then the real- valued function | f | : X → R≥0 on X does not attain its supremum in X.

6 belonging to the same equivalence class of atlases as (U1, φ1) 7for holomorphic functions on open subsets of C 2 PRELIMINARIES 10

Proof Suppose, to get a contradiction, that ∃ ζ ∈ X : | f (ζ)| = sup{| f (z)| : z ∈ X}. Then we have | f (ζ)| > 0 (since f is non-constant), and f (X) ⊂ B(0; | f (ζ)|). But by the open mapping theorem, f (X) is open in Y. Hence, we in fact have f (X) ⊂ B(0; | f (ζ)|), contradicting the fact that ζ ∈ X.

Theorem 2.2.5 Let X and Y be Riemann surfaces and let f : X → Y be a non- constant holomorphic mapping. Suppose that X is compact. Then Y is compact and f is surjective.

Proof By the open mapping theorem, f (X) is open in Y. Since X is compact, f (X) is compact, and thus closed, in (the Hausdorﬀ space) Y. Hence, since Y is connected, we must have f (X) = Y. Consequently, f is surjective and Y is compact, as claimed.

An immediate consequence is the classic result that C is an algebraically closed ﬁeld:

Corollary 2.2.1 (The Fundamental Theorem of Algebra) Every non-constant polynomial in C[X] has a zero in C.

Proof Any non-constant polynomial p(X) ∈ C[X] may be considered as a non- constant holomorphic mapping p : Cˆ → Cˆ , where p(∞) = ∞ (and f (C) ⊂ C). Then, by the above theorem, p is surjective (since Cˆ is compact), i.e., p(Cˆ ) = Cˆ . Thus, since p(∞) = ∞, we get p(C) = C. In particular, we have 0 ∈ f (C). Therefore, ∃ ζ ∈ C : p(ζ) = 0, i.e., ζ is a zero of p in C.

Theorem 2.2.6 (The Liouville Theorem) Let X be a compact Riemann surface and let f ∈ O(X). Then f is a constant function.

Proof This follows immediately from the previous theorem, since C is a non- compact Riemann surface.

Remark Notice that the above theorem implies that if X is a compact Riemann surface, then O(X) C as C-algebras.

We notice that the above theorem is a generalization of the following result from the theory of functions of one complex variable.

Corollary 2.2.2 (The Classical Liouville´s Theorem) Every bounded entire function is constant. 2 PRELIMINARIES 11

Proof Let f ∈ O(C), and suppose that f is bounded. Since f : C → C is a holomorphic function, we know that, by Riemann´s removable singularities theorem 2.2.1, f can be analytically continued to a holomorphic mapping fˆ : Cˆ → C, which must then be constant, in view of the above theorem (since Cˆ is compact). This implies that f is constant too, as claimed.

Another direct consequence is the following:

Theorem 2.2.7 Each f ∈ M (Cˆ ) is a rational function.

Proof Clearly, f : Cˆ → C can only have ﬁnitely many poles8. We may assume, 1 without loss of generality, that ∞ is not a pole of f (Otherwise, we consider f instead.). Let ζ1, ..., ζn ∈ C be the poles of f , and let, for eachP i ∈ {1, ..., n}, ˆ −1 j hi : C → C be the principal part of f at the pole ζi, so hi(z):= j=−ki ci, j(z − ζi) + for some positiveP integer ki ∈ Z and some complex constants ci,−ki , ..., ci,−1 ∈ C. n ˆ ˆ Let g := f − i=1 hi : C → C. Then the function Pg : C → C is holomorphic, i.e., n g ∈ O(X). Hence, by the above theorem, g = f − i=1 hi is a constant function. It follows directly that f is rational, as claimed.

The following theorem is a very useful biholomorphy criterion.

Theorem 2.2.8 Let X and Y be Riemann surfaces, and let f : X → Y be an injective holomorphic mapping. Then f is a biholomorphic mapping of X onto its image f (X) in Y under f .

Proof By the open mapping theorem 2.2.3 and the remark following it, f (X) is an open Riemann sub-surface of Y. Let g : f (X) → X denote the inverse mapping of the holomorphic bijection f : X → f (X). We want to show that g is holomorphic too. Let (V, ψ) and (U, φ) be charts on f (X) and X respectively, such that g(V) ⊂ U (Notice that (Vψ) is also a chart on Y.). Then (g(V), φ|g(V)) is a chart on X and f (g(V)) = V. Thus, since the mapping f : X → Y is holomorphic, we know that −1 the transition function ψ ◦ f ◦ φ|g(V) : φ(g(V)) → ψ(V) is holomorphic. Moreover, since φ|g(V) : g(V) → φ(g(V)) and ψ : V → ψ(V) are biholomorphisms, and f is a −1 holomorphic injection, ψ ◦ f ◦ φ|g(V) : φ(g(V)) → ψ(V) is a holomorphic injection as well. Hence, by the corresponding biholomorphy criterion for holomorphic −1 injective functions on subregions of C, we know that ψ ◦ f ◦ φ|g(V) maps φ(g(V)) −1 biholomorphically onto ψ(V) with the inverse of ψ ◦ f ◦ φ|g(V) : φ(g(V)) → ψ(V) given by φ ◦ g ◦ ψ−1 : ψ(V) → φ(g(V)). In particular, φ ◦ g ◦ ψ−1 : ψ(V) → φ(U) is a holomorphic function. Therefore, g is a holomorphic mapping, as claimed.

8Otherwise, since Cˆ is compact, the set of poles of f would have a limit point in Cˆ , and hence, the identity theorem 2.2.2 would yield f ≡ ∞, which is forbidden for meromorphic functions. 2 PRELIMINARIES 12

2.2.3 Holomorphic Covering Maps A mapping f : X → Y between topological spaces X and Y is said to be a discrete mapping if f −1({x}) is a discrete subset of Y for each x ∈ X.

Theorem 2.2.9 Let X and Y be Riemann surfaces and let p : Y → X be a non- constant holomorphic mapping. Then p is discrete.

−1 Proof Suppose that p is not a discrete mapping. Then ∃ x0 ∈ X : p ({x0}) has a limit− point in Y. Thus, by the identity theorem 2.2.2, we have f (x) = f (x0) ∀x ∈ X. In particular, p is constant, contradicting our assumption. Hence p is discrete, as desired.

Remark Let X and Y be Riemann surfaces and suppose that p : Y → X is a non-constant holomorphic mapping. Each holomorphic function f ∈ O(Y) on Y may also be considered as a multiple-valued holomorphic function f˜ ∈ O(X) on X in the following way. Let, −1 ∀x ∈ X, p ({x}) = {yi}i∈I ⊂ Y. Then, ∀x ∈ X, C ⊃ { f (yi)}i∈I =: f˜(x).

f / {yi}i∈I ⊂ Y C ⊃ { f (yi)}i∈I m6 mmm p mmm mmm mmm f˜ x ∈ X In an analogous way, each meromorphic function f ∈ M (Y) on Y may also be considered as a multiple-valued meromorphic function f˜ ∈ M (X) on X.

Suppose that X and Y are Riemann surfaces and that p : Y → X is a non-constant holomorphic mapping. A point y ∈ Y of Y is said to be a branch point of p if there exists no open neighborhood V ⊆ Y of y in Y such that p|V is injective, and p is said to be unbranched if it has no branch points.

Theorem 2.2.10 Let X and Y be Riemann surfaces and suppose that p : Y → X is a non-constant holomorphic mapping. Then the following statements are equivalent:

(i) p is unbranched

(ii) p is a local homeomorphism

(iii) p is a local biholomorphism 2 PRELIMINARIES 13

Proof Suppose ﬁrst that p is unbranched. Let y ∈ Y. Then y has an open neighborhood V in Y such that p|V is an injection. Hence, by the open mapping theorem, p|V is a topological embedding. Thus p is a local homeomorphism. Conversely, suppose now that p is a local homeomorphism. Let y ∈ Y. Then there exist open neighborhoods V of y in Y and U of p(y) in X such that p maps V homeomorphically onto U. Thus, in particular, p|V is injective. Therefore, p is unbranched. This establishes the equivalence of (i) and (ii). The equivalence of (ii) and (iii) follows directly from theorem 2.2.8.

Recall the following results from topology:

Lemma 2.2.1 Suppose that X and Y are topological spaces and that p : Y → X is a (continuous) local homeomorphism.

• Suppose that X is locally Euclidean of dimension n ∈ N. Then so is Y.

• Assume that Y is Hausdorﬀ. Suppose that X is a (real) topological manifold of dimension n ∈ N. Then so is Y.

Proof The 1st statement follows by locally composing p with charts on X. The 2nd follows directly from the 1st.

Lemma 2.2.2 Suppose that X and Y are topological spaces and that p : Y → X is a (topological) covering map.

• Suppose that X is Hausdorﬀ. Then so is Y.

• Suppose that X is a (real) topological manifold of dimension n ∈ N. Then so is Y.

Proof Let y1, y2 ∈ Y : y1 , y2. Suppose first that we have p(y1) = p(y2). Let U be an open neighborhood of p(y1) = p(y2) which is evenly covered by p. Then p−1(U) is a disjoint union of open subsets of Y, each of which is mapped homeomorphically onto U by p. Let V1 and V2 be the (unique) ones containing y1 and y2 respectively. Then, in particular, V1 and V2 are disjoint open neighborhoods of y1 and y2 in Y. Suppose now that we have p(y1) , p(y2). Then, since X is Haus- dorff, there exist open neighborhoods U1 of p(y1) and U2 of p(y2) in X such that −1 −1 U1 ∩ U2 , ∅. Let V1 := p (U1) and V2 := p (U2). Then V1 and V2 are disjoint open neighborhoods of y1 and y2 in Y. Hence Y is Hausdorff, as desired. The latter statement follows immediately from the former statement together with the preceding lemma. 2 PRELIMINARIES 14

In fact, in the setting of Riemann surface theory, we have the following result:

Theorem 2.2.11 Let X be a Riemann surface and let Y be a Hausdorﬀ topological space. Suppose that p : Y → X is a local homeomorphism. Then there exists a unique complex structure on Y 9 such that p is a holomorphic mapping. With this complex structure on Y, p is even locally biholomorphic.

Proof Let AX be the maximal atlas of the complex structure on X. Let, ∀y ∈ Y, Vy ⊆ Y be an open neighborhood of y in Y such that p maps Vy homeomorphically onto an open neighborhood Uy of p(y) in X and such that there exists a topological embedding φy : Uy → C such that (Uy, φy) ∈ AX. Then , ∀y ∈ Y, φy ◦ p : Vy → C is a topological embedding. Let AY := {(Vy, φy ◦ p): y ∈ Y}. Then AY is a complex atlas on Y. We deﬁne Y to have the corresponding complex structure [AY]. Thus, (Y, [AY]) is a Riemann surface. We notice that, ∀y ∈ Y, p maps Vy biholomorphically onto Uy. Thus, p is a locally biholomorphic mapping. Hence, in particular, p is holomorphic. 0 Suppose that AY is an atlas in another complex structure on Y also making p into a holomorphic mapping. Since p is a holomorphic local homeomorphism, it is by theorem 2.2.8 in fact also a local biholomorphism. Thus, the identity 0 mapping idY :(Y, [AY]) → (Y, [AY]) is a locally biholomorphic bijection and hence a biholomorphic mapping.

id / 0 (Y, [AY]) (Y, [AY]) LL LL rrr LLL rr p LL rr p L& rx rr (X, [AX])

0 Consequently, the Riemann surfaces (Y, [AX]) and (Y, [AX]) are conformally equivalent.

The following theorem guarantees that we can treat analytic covering spaces just as topological ones.

Corollary 2.2.3 Let X be a Riemann surface and let Y be a topological space. Suppose that p : Y → X is a (topological) covering map. Then there exists a unique complex structure on Y (thus naturally making it into a Riemann surface) such that p is a holomorphic (and locally biholomorphic) mapping.

Proof This follows directly from lemma 2.2.2 and theorem 2.2.11.

Remark We take notice of the following consequences.

9thus naturally making it into a Riemann surface 2 PRELIMINARIES 15

• If U ⊆ X is an open subset of X which is evenly covered by p so that p−1(U) is a disjoint union of open subsets {Vi}i∈I of Y where p maps Vi homeomorphically onto U ∀i ∈ I then p actually maps Vi biholomorphically onto U ∀i ∈ I . p p • We have Deck(Y −−→ X) ≤ Aut(Y). Proof: Recall that, a priori, Deck(Y −−→ X) is just a group of (p-ﬁber-preserving) homeomorphisms of Y. That is, if φ : Y → Y is a deck transformation of the covering so that φ is a homeomorphism of Y onto itself and preserves the ﬁbers of p then φ is actually an automorphism of Y. This follows from the fact that if (V1, ψ1) and (V2, ψ2) are complex charts on Y belonging to the complex atlas AY on Y 10 given in the proof of theorem 2.2.11 above such that we have φ(V1) ⊆ V2 11 then there exist complex charts (U1, ϕ1) and (U2, ϕ2) on X such that p maps V1 homeomorphically onto U1 and V2 homeomorphically onto U2 and such that we have ψ1 = ϕ1 ◦ p and ψ2 = ϕ2 ◦ p. It follows that we have ψ1(V1) = ϕ1(U1) and ψ2(V2) = ϕ2(U2) and that the transition function −1 −1 −1 ψ2◦φ◦ψ1 : ψ1(V1) → ψ2(V2) is given by ψ2◦φ◦ψ1 = (ϕ2◦p)◦φ◦(ϕ1◦p) = −1 −1 −1 −1 −1 ϕ2 ◦ (p ◦ φ) ◦ p ◦ ϕ1 = ϕ2 ◦ p ◦ p ◦ ϕ1 = ϕ2 ◦ ϕ1 : ϕ1(U1) → ϕ2(U2), which is holomorphic.

If X and Y are Riemann surfaces, and p : X → Y is a holomorphic covering map, then the corresponding (topological) covering space Y is referred to as analytic. The following theorem guarantees that liftings of holomorphic mappings to analytic covering spaces are again holomorphic.

Theorem 2.2.12 Let X and Y be Riemann surfaces and suppose that p : Y → X is an unbranched non-constant holomorphic mapping. Let Z be a Riemann surface and let f : Z → X be a holomorphic mapping. Suppose that g : Z → Y is a lifting of f w.r.t p. Then g is a holomorphic mapping.

Y Ð@ g Ð ÐÐ p ÐÐ ÐÐ Z / f X

Proof Let z ∈ Z, y := g(z) ∈ Y and x := p(y) = f (z) ∈ X. By theorem 2.2.10, p is a locally biholomorphic mapping. Thus, there exist open neighborhoods V ⊆ Y of y in Y and U ⊆ X of x in X such that p maps V biholomorphically onto U. Let ξ : U → V be the inverse mapping. Since g is a continuous mapping, there exists

10 AY is by deﬁnition in the complex structure of Y 11 (U1, ϕ1) and (U2, ϕ2) are assumed to belong to a complex atlas in the complex structure of X 2 PRELIMINARIES 16 a connected open neighborhood W ⊆ Z of z in Z such that g(W) ⊆ V. Since g is a lifting of f w.r.t. p, we have f = p◦g. Hence, we have g|W = ξ ◦( f |W ): W → V ⊂ Y. Thus, in particular, the mapping g|W : W → Y is holomorphic. Consequently, g is holomorphic, as desired.

Corollary 2.2.4 Suppose that X, Y and Z are Riemann surfaces and that p : Y → X and q : Z → X are unbranched non-constant holomorphic mappings. Then every ﬁber-preserving continuous mapping f : Y → Z is holomorphic.

f / Y > Z >> ÐÐ > ÐÐ p >> Ð q > Ð Ð X

Proof This follows immediately from the preceding theorem, since f is merely a lifting of p w.r.t. q.

p Remark The above corollary again implies that we have Deck(Y −−→ X) ≤ Aut(Y) as remarked above. It follows also that isomorphisms of analytic covering spaces (i.e. ﬁber-preserving homeomorphisms) are in fact also biholomorphisms.

As mentioned earlier, it turns out that proper non-constant holomorphic mappings between Riemann surfaces behave almost like (holomorphic) covering maps. In fact, unbranched proper non-constant holomorphic mappings between Riemann surfaces are (holomorphic) covering maps, whereas the images of the branch points of a branched (proper) non-constant holomorphic mapping between two Riemann surfaces under the branched (proper) non-constant holomorphic mapping have no (open) neighborhoods which are evenly covered by the branched (proper) non-constant holomorphic mapping. Recall, by the way, that a proper mapping p : Y → X between topological spaces Y and X is a mapping (continuous or not) such that whenever K ⊆ Y is a compact set then so is f −1(K) ⊆ X. Proper continuous mappings have nice properties in locally compact Hausdorﬀ spaces, e.g., they are closed mappings. Being manifolds, Riemann surfaces are of course locally compact.

Lemma 2.2.3 Let X and Y be locally compact Hausdorﬀ spaces, and suppose that p : Y → X is a proper and discrete continuous mapping. Let x ∈ X. Then we have |p−1({x})| < ∞ and for each open neighborhood V of p−1({x}) in Y there exists an open neighborhood U of x in X such that p−1(U) ⊆ V. 2 PRELIMINARIES 17

Proof Since p−1({x}) ⊂ Y is a compact and discrete subset of Y, it must be ﬁnite. Let V ⊆ Y be an open neighborhood of p−1({x}) in Y. Then Y r V ⊆ Y is a closed subset of Y. Since p is a closed mapping, it follows that p(Y r V) ⊆ X is a closed subset of X. Thus, we see that X r p(Y rV) ⊆ X is an open subset of X. Moreover, we have x ∈ X r p(Y r V). Hence, X r p(Y r V) is an open neighborhood of x in X. It is easily seen that we have p−1(X r p(Y r V)) ⊆ V.

Theorem 2.2.13 Let X and Y be locally compact Hausdorﬀ spaces and suppose that p : Y → X is a proper local homeomorphism. Then p is a covering map with ﬁnitely many sheets.

Proof We first notice that p must be a discrete mapping. If not, then ∃ x0 ∈ X : −1 0 −1 p ({x0}) has a limitpoint y in Y. Since p ({x0}) ⊂ Y is compact and Y is Haus- −1 0 −1 dorff, we know that p ({x0}) is closed in Y. Thus, we must have y ∈ p ({x0}). But then y0 has no open neighborhood in which p is injective, contradicting the fact that p is a locally homeomorphic mapping. Let x ∈ X. Then, by the preceding lemma and the fact that p is also discrete, we get |p−1({x})| < ∞. Suppose −1 that we have p ({x}) = {y1, ..., yn}, where the elements yi are pairwise distinct. Since p is a local homeomorphism, we know that, ∀i ∈ {1, ..., n}, there exists open neighborhoods Wi ⊂ Y of yi ∈ Y in Y and Ui ⊂ X of x ∈ X in X such that p maps Wi homeomorphically onto Ui. WeF may assume, w.l.o.g., that the sets n n {Wi}i=1 are pairwise disjoint. Then the set i=1 Wi ⊂ Y is an open neighborhood −1 of the set p ({x}) ⊂ Y in TY. Thus, again by the preceding lemma, there exists an n open neighborhoodF U ⊂ i=1 Ui ⊂ X of the point x ∈ X in X such that we have −1 n −1 n p (U) ⊆ i=1 Wi (⊂ Y). Let, ∀i ∈ {1, ..., n}, Vi := Wi ∩ p (U) ⊂ Y. ThenF {Vi}i=1 −1 n is a disjoint collection of open subsets of Y such that we have p (U) = i=1 Vi (⊂ Y), and such that, ∀i ∈ {1, ..., n}, p maps the open set Vi ⊂ Y homeomorphically onto the open set U ⊂ X. Hence, the open subset U ⊂ X of X is an open neighborhood of the point p ∈ X in X which is evenly covered by the continuous mapping p. It follows that p is a covering map, as desired. Since p is proper, it has only finitely many sheets.

Remark • If X and Y are Riemann surfaces and f : X → Y is an unbranched proper non-constant holomorphic mapping, then f is a finitely- sheeted covering map. This follows from theorem 2.2.10 and theorem 2.2.13. In fact, more generally, the following statement holds. If X is any Hausdorff space and Y is any non-empty path-wise connected Hausdorff space and f : X → Y is any proper local homeomorphism, then f is a covering map with a (well-defined) finite number n ∈ N r {0} of sheets 12. 12Conversely, all ∞ > n-sheeted (holomorphic) covering maps are proper mappings. 2 PRELIMINARIES 18

• Let us suppose that X and Y are Riemann surfaces, and that f : X → Y is a proper non-constant holomorphic mapping. We let A f ⊂ X be the set of branch-points of f . Then A f ⊂ X is a closed and discrete subset of X. Thus, since f is a proper mapping, f (A f ) ⊂ Y is a closed and discrete subset of −1 Y. We take notice of the fact that X r f ( f (A f )) ⊂ X r A f ⊂ X and Y r f (A f ) ⊂ Y are open Riemann sub-surfaces of X and Y respectively and that −1 −1 f |Xr f ( f (A f )) : X r f ( f (Ap)) → Y r f (A f ) is an un-branched proper non- −1 −1 constant holomorphic mapping. Thus, f |Xr f ( f (A f )) : X r f ( f (A f )) → Y r f (A f ) is a ﬁnite-sheeted holomorphic covering map. The original mapping f : X → Y is referred to as a branched (holomorphic) covering map. That is, by deﬁnition, given two Riemann surfaces X and Y, a mapping f : X → Y is a branched (holomorphic) covering map if and only if it is a proper non-constant holomorphic mapping . A branched (holomorphic) covering map is a covering map if and only if it is un-branched.

2.3 Properly Discontinuous Group Actions and Fundamental Regions Let X be a topological space, let G be a group and suppose that

α : G × X → X, (g, x) 7→ gx, is a left group action of G on X. The action α is said to be

• continuous if it is continuous when G is endowed with the discrete topol- ogy13

• faithful if, ∀g ∈ G, we have

gx = x ∀x ∈ X ⇒ g = e

• free if, ∀g ∈ G, ∀x ∈ X, we have

gx = x ⇒ g = e

• properly discontinuous if it is continuous (sic) and each point x ∈ X of X has an open neighborhood U ⊂ X in X such that

|{g ∈ G : gU ∩ U , ∅}| < ∞

13thus making it into a topological group 2 PRELIMINARIES 19

Remark • The group G acts continuously on the topological space X ⇔ the mapping X 3 x 7→ gx ∈ X is continuous ∀g ∈ G ⇔ the mapping X 3 x 7→ gx ∈ X is a homeomorphism ∀g ∈ G. • Suppose that G acts properly discontinuously on X. Then, ∀x ∈ X, the orbit Gx ⊂ X of x under G is a closed and discrete subset of X. a b Let R be a commutative ring with unity, let GL2(R):= { c d : a, b, c, d ∈ ∗ R & ad − bc ∈ R }, notice that GL 2(R) is a group under matrix multiplication ∗ a b and that detR : GL2(R) → R , c d 7→ ad − bc, is a surjective group homomorphism, and let SL (R):= ker(det ) E GL (R). We let I := 1R 0R be the identity. 2 R 2 R 0R 1R We notice that if S is a subring of R then GL2(S ) ≤ GL2(R) and SL2(S ) ≤ SL2(R) are subgroups. Let us deﬁne a left group action of the group GL2(C) on the space Cˆ := C ∪ {∞} by a b az + b • z := ∀z ∈ Cˆ , ∀ a b ∈ GL (C). c d cz + d c d 2 A straightforward check veriﬁes that this is indeed a left group action. Remark The complex projective line CP1 is the set of all one-(complex-)dimensional (complex-)linear subspaces of the C-linear space C2, i.e.,

1 2 2 CP := {U ⊂ C : U is a C − linear subspace o f C and we have dimC(U) = 1}. Thus, CP1 is the quotient of C2 r{0} modulo the equivalence relation ∼ on C2 r{0} defined by z ∼ w :⇔ ∃ λ ∈ C r {0} : w = λz. Given z ∈ C2 r{0}, we denote its equivalence class under ∼ by [z]. We endow CP1 with the quotient topology induced by the natural projection π : C2 r {0} → CP1. Let U1 := {[(z1, z2)] : z1 , 0} and U2 := {[(z1, z2)] : z2 , 0} and let φ1 : U1 → C z2 and φ2 : U2 → C be given respectively by φ1([(z1, z2)]) := and φ2([(z1, z2)]) := z1 z1 1 . Then {(U1, φ1), (U2, φ2)} is a complex atlas on CP , turning it into a Riemann z2 1 surface. We see that U1 = CP r {[(1, 0)]} and that φ1 identifies each U ∈ U1 with 2 1 its intersection with the affine subspace C × {1} of C . Extending φ1 to all of CP by defining φ1([(1, 0)]) := ∞, we obtain a biholomorphic mapping identifying CP1 with the extended complex plane Cˆ . It is now evident that the action defined 2 above is natural, since by the natural action of GL2(C) on C , we have a b z az + b az+b = ∼ cz+d , c d 1 cz + d 1 a b ˆ so that by the identification above, the action of c d ∈ GL2(C) on z ∈ C is given az+b by cz+d . 2 PRELIMINARIES 20

Let, for each A ∈ GL2(C), the mapping φA : Cˆ → Cˆ be the mapping deﬁned by φA(z):= A • z ∀z ∈ Cˆ .

Notice that we may assume, w.l.o.g., that we have A ∈ SL2(C). Let

M := {φA : A ∈ GL2(C)} = {φA : A ∈ SL2(C)}, i.e., M is just the set of all Möbius transformations on Cˆ . Clearly, M is a group under composition. Let R ⊆ C be a subring of C and let ΦR : GL2(R) → M be given by ΦR(A):= φA ∀A ∈ GL2(R). ∗ Notice that ΦR is a group homomorphism, with kernel given by ker(ΦR) = R IR = ∗ {λIR : λ ∈ R } E GL2(R) and that ΦR |SL2(R): SL2(R) → M is a group homomorphism with kernel ker(ΦR |SL2(R)) = {±IR} E SL2(R). Let PGL2(R):= st GL2(R)/ker(ΦR) and PSL2(R):= SL2(R)/ker(ΦR |SL2(R)). By the 1 isomorphism theorem for groups, we have im(ΦR) GL2(R)/ker(ΦR) and im(ΦR)

SL2(R)/ker(ΦR). Notice that ΦC : GL2(C) → M and ΦC |SL2(C): SL2(C) → M both are surjective, i.e., ΦC(GL2(C)) = ΦC |SL2(C) (SL2(C)) = M. Hence, we have

M PGL2(C) PSL2(C). Moreover, we see that

i f H ≤ GL2(C) then ΦC(H) ≤ M.

2.3.1 The Upper Half-Plane and its Automorphisms Recall that the upper half-plane H in the complex plane C is the proper unbounded region in C given by H := {z ∈ C : =(z) > 0} .

Recall the following result, or refer to [11], pages 200-202, for a proof.

Theorem 2.3.1 • Aut(Cˆ ) = M PGL2(C) PSL2(C) • Aut(C) = {z 7→ az + b : a ∈ C∗, b ∈ C}

• Aut(H) = {φA : A ∈ SL2(R)} PSL2(R) We state here for future reference the following useful formula. a b az+b Lemma 2.3.1 Let ϕ ∈ Aut(H), i.e., ∃ c d ∈ SL2(R): ϕ(z) = cz+d ∀z ∈ H. Then we have =(z) =(ϕ(z)) = ∀z ∈ H. |cz + d|2 2 PRELIMINARIES 21

az+b Proof This follows directly from the fact that if f ∈ M is given by f (z) = cz+d then we have ac|z|2 + (ad + bc)<(z) + bd (ad − bc)=(z) f (z) = + i ∀z ∈ H. |cz + d|2 |cz + d|2

2.3.2 The Modular Group Γ and Modular Functions on H

A discrete subgroup of Aut(H) PSL2(R) is called a Fuchsian group. All Fuch- sian groups act properly discontinuously on H. A subgroup G of Aut(H) is a Fuchsian group if and only if , ∀z ∈ H, the G-orbit of z, Gz ⊂ H, is a discrete subset of H. Holomorphic functions on H which are invariant under the action of a Fuchsian group are called automorphic. The subgroup Γ := {φA : A ∈ SL2(Z)} PSL2(Z) < PSL2(R) of Aut(H) is called the modular group, and will be of signiﬁcant interest to us. Clearly, Γ and all its subgroups are Fuchsian groups and thus act properly discontinuously and faithfully on H. Sometimes, any subgroup of Γ is referred to as a modular group.

Let G < Aut(H). We deﬁne an equivalence relation ∼G on H by

z1 ∼G z2 :⇔ ∃ g ∈ G : gz1 = z2.

We let H/G be the set of equivalence classes under ∼G, and we equip H/G with the corresponding quotient topology. Suppose that G acts properly discontinuously and freely on H 14. Then, in a natural way, H/G is a Riemann surface. Let z˜ ∈ H/G. Let z ∈ π−1(˜z). Since G acts freely and properly discontinuously on H, z ∈ H has an open neighborhood U ⊂ H such that gU ∩ U = ∅ ∀g ∈ G r {idH}. Thus, π : U → π(U) is a homeomorphism. Let F ⊂ H be a subregion of H. Then F is said to be a fundamental region for G if ∀z ∈ H ∃ w ∈ F : z ∼G w and z /G w ∀z, w ∈ F .

Remark If F ⊂ H is a fundamental region for G, then we have [ H = gF . g∈G

14 This implies that |G| ≤ ℵ0. 3 THE UNIVERSAL COVERING SPACE OF C r {0, 1} 22

For example, a fundamental region for the full modular group Γ is given by R˚Γ ( H, where 1 1 R := {z ∈ H : |z| ≥ 1, − ≤ <(z) ≤ 0} t {z ∈ H : |z| > 1, 0 < <(z) < }, Γ 2 2 as is shown in, e.g., [3], pages 31-33, and we have G H = ϕ(RΓ). ϕ∈Γ

RΓ

u ui e ρ ρ + 1

-1 -0.5 0 0.5 1

2πi Figure 1 : ρ = e 3 A meromorphic function f ∈ M (H) on H is said to be a modular function if there exists a subgroup G ≤ Γ of the modular group Γ such that f is invariant under the natural action of G on H, i.e., such that we have f ◦ φ = f ∀φ ∈ G, and the Fourier expansion of f has the form X∞ 2πinz f (z) = ane n=−m for some integer n ∈ Z. If we need to be speciﬁc, we will call f a G-modular function. For example, the Klein function J : H → C is Γ-modular (see [3] for information about J ).

3 The Universal Covering Space of C r {0, 1}

3.1 The Subgroup G of Γ

+ Let, for each n ∈ Z , Γ(n):= ker(SL2(Z) → SL2(Zn)) E SL2(Z) be the kernel of the element-wise reduction modulo n homomorphism mapping, i.e., Γ(n) = 3 THE UNIVERSAL COVERING SPACE OF C r {0, 1} 23

{A ∈ SL2(Z): A ≡ ± I (mod n)}. Notice that we have Γ(1) = SL2(Z), and that, ∀n ∈ Z+ r {1}, we have Γ(n) / Γ(1). Given n ∈ Z+, the subgroup Γ(n) of SL2(Z) is called the principal congruence subgroup mod n (of Γ), or the principal congruence subgroup of level n (of Γ). We will primarily be interested in the case where we have n = 2. Let G < Aut(Cˆ ) be the group of automorphisms of Cˆ corresponding to the action of the (discrete) principal congruence subgroup Γ(2) mod 2 on the Riemann sphere Cˆ , i.e., we let

G := {φA : A ∈ Γ(2)}.

Let Zo := {2n + 1 : n ∈ Z} be the set of all odd integers, and let Ze := {2n : n ∈ Z} be the set of all even integers. Let Λ := {(a, b, c, d) ∈ Zo × Ze × Ze × Zo : ad − bc = 1}. Then we have az + b G = {z 7→ :(a, b, c, d) ∈ Λ}. cz + d Let τ ∈ Γ be given by τ(z):= z + 2, and let σ ∈ Γ be given by z σ(z):= . 2z + 1 Notice that we actually have τ, σ ∈ G. In fact, we will show soon that the mappings τ and σ together generate all of G.

Remark Since τ is a non-trivial translation, it has the single ﬁxed point ∞. The curves in Cˆ left invariant by τ are the extended straight lines in C parallel to R, i.e., ˆ ˆ 1 the generalized circles in C tangent to R at ∞. Let ι ∈ M be deﬁned by ι(z):= z . Then a straightforward calculation reveals that

σ = ι ◦ τ ◦ ι−1. (1)

Notice, moreover, that we have ι−1 = ι, ι(0) = ∞ and ι(∞) = 0. The origin 0 is the sole ﬁxed point of σ. Since ι is a Möbius transformation, it takes generalized circles to generalized circles, and being an automorphism of Cˆ , it preserves an- gles. Consequently, since Rˆ clearly is invariant under ι, ι takes generalized circles tangent to Rˆ at ∞ to generalized circles tangent to Rˆ at 0. Hence, by equation (1), the curves in Cˆ invariant under σ are precisely Rˆ together with the circles in C tangent to R at 0. 3 THE UNIVERSAL COVERING SPACE OF C r {0, 1} 24

'$- - - - - - - - - - m 0 r - &%'$ - m - 0 - - - - - - &%-

Figure 2: The τ-invariant curves. Figure 3: The σ-invariant curves.

Lemma 3.1.1 : We shall make use of the following formulae below. Let n ∈ Z. Then a straightforward calculation shows that (i) τn ∈ G is given by τn(z) = z + 2n,

n n (4n+1)z+2n (ii) τ σ ∈ G is given by τ σ(z) = 2z+1 ,

−n n −1 −1 −n z−2n (iii) στ = (τ σ ) ∈ G is given by στ (z) = 2z+1−4n ,

−1 −n n −1 −1 −n z−2n (iv) σ τ = (τ σ) ∈ G is given by σ τ (z) = −2z+4n+1 . az+b ˆ Proposition 3.1.1 : Let φ ∈ G, i.e. ∃ (a, b, c, d) ∈ Λ : φ(z) = cz+d ∀z ∈ C. Then the following statements hold. (i) We have c = 0 ⇔ ∃ n ∈ Z : φ = τn. (ii) We have c = 2d , 0 ⇔ ∃ n ∈ Z : φ = τnσ. Proof : (i) Suppose that ∃ n ∈ Z : φ = τn. Then , clearly, we have (a, b, c, d) = ± (1, 2n, 0, 1). Thus, in particular, we get c = 0. Suppose, conversely, that we have c = 0. Then, since ad − bc = 1, we have ad = 1. Thus, since a, d ∈ Z, we get a = d = ±1. Hence, we have φ(z) = z±b ∀z ∈ Cˆ . Therefore, n since b ∈ Ze, ∃ n ∈ Z : φ(z) = z + 2n ∀z ∈ Cˆ . That is, ∃ n ∈ Z : φ = τ . (ii) Suppose that ∃ n ∈ Z : φ = τnσ. Then, by lemma 3.1.1, we have (a, b, c, d) = ± (4n + 1, 2n, 2, 1). Thus, in particular, it holds that c = 2d. Suppose, conversely, that we have c = 2d. Then, since ad − bc = 1, we have d(a − 2b) = 1. Thus, since d, a − 2b ∈ Z, we get d = a − 2b = ±1. Hence, we z ˆ see that φ(z) = 2z+1 ± b ∀z ∈ C. Therefore, since b ∈ Ze, we know that z ˆ n ∃ n ∈ Z : φ(z) = 2z+1 + 2n ∀z ∈ C. That is, ∃ n ∈ Z : φ = τ σ. 3 THE UNIVERSAL COVERING SPACE OF C r {0, 1} 25

Theorem 3.1.1 We have G = hτ, σi.

Proof Notice that ∀φ ∈ G ∃!(a, b, c, d) ∈ Λ :(c > 0 ∨ c = 0 ∧ d > 0) & az + b φ(z) = ∀z ∈ Cˆ . cz + d

We proceed by induction on c ∈ Ze ∩ N. Suppose that we have c = 0. Then, by proposition 3.1.1(i), ∃ n ∈ Z :

φ = τn.

Thus, in particular, we see that

φ ∈ hτ, σi, as required. Let c0 ∈ 2Z+ and suppose inductively that we have

φ ∈ hτ, σi f or all φ ∈ G such that c < c0.

0 0 We want to show that φ ∈ hτ, σi also holds for φ ∈ G with c = c . Since c ∈ Zer{0} d d and d ∈ Zo, we have c0 ∈ Q r Z. Thus, in particular, we have c0 < Zo. Hence, we see that d d Z ∩ (−1 − , 1 − ) , ∅. e c0 c0 d d d d Let n ∈ Z : 2n ∈ Ze ∩ (−1 − c0 , 1 − c0 ). Then, since −1 − c0 < 2n < 1 − c0 and d 2n , − c0 , we get 0 < |2nc0 + d| < c0. 0 0 0 n Let b := 2na + b ∈ Ze and d := 2nc + d ∈ Zo. Then φ ◦ τ ∈ G is given by az + b0 φ ◦ τn(z) = ∀z ∈ Cˆ . c0z + d0

d Let m ∈ Z be an integer closest to the quotient c0 . Then, we have |c0 + 2md0| ≤ |d0|.

0 0 00 0 0 n m Let a := a + 2mb ∈ Zo and c := c + 2md ∈ Ze. Then φ ◦ τ ◦ σ ∈ G is given by a0z + b0 φ ◦ τn ◦ σm(z) = ∀z ∈ Cˆ . c00z + d0 3 THE UNIVERSAL COVERING SPACE OF C r {0, 1} 26

But by construction we get |c00| ≤ |d0| < c0. Hence, by the induction hypothesis, we conclude that φ ◦ τn ◦ σm ∈ hτ, σi. Therefore, since φ = (φ ◦ τn ◦ σm) ◦ σ−m ◦ τ−n, we have φ ∈ hτ, σi, as required. The proposition now follows by induction. In fact we can say more. Recall that a group G is said to be free on a subset A of G if there exists a bijection between G and the set of reduced words consisting of symbols in A. We will show shortly that G is in fact free on {τ, σ}. Lemma 3.1.2 Let (a, b, c, d) ∈ Z4 : ad − bc = 1. Then

d 1 1 1 c = 2d ⇔ 0 , c = 2d ⇔ (c, d) = ± (2, 1) ⇔ (− , ) = (− , ). c |c| 2 2 Proof If c = 2d, then, since ad − bc = 1, we have (a − 2b)d = 1. Thus, since a − 2b, d ∈ Z, we get a − 2b = d = ±1. Hence, in particular, (c, d) = ± (2, 1) d 1 1 1 holds. Suppose that we have c , 0 and (− c , |c| ) = (− 2 , 2 ). Then we get 2d = c and 2 = |c|. Thus, in particular, we see that c = 2d. The remaining implications are even more trivial.

3.2 The Fundamental Region Ω ⊂ H for G We deﬁne the set Ω (H(C by Ω := {z ∈ H : −1 ≤ <(z) < 1, |2z + 1| ≥ 1, |2z − 1| > 1} =

= {z ∈ C : =(z) > 0, −1 ≤ <(z) < 1, | 2z + 1 |≥ 1, | 2z − 1 |> 1}.

Ω

ee e -1 0 1 Figure 4 3 THE UNIVERSAL COVERING SPACE OF C r {0, 1} 27

We shall make use of the following notations. Let L− := {z ∈ C : <(z) = −1}, 1 1 L0 := {z ∈ C : <(z) = 0}, L+ := {z ∈ C : <(z) = 1}, C− := S (− 2 ; 2 ) = {z ∈ C : 1 1 H |2z + 1| = 1} and C+ := S ( 2 ; 2 ) = {z ∈ C : |2z − 1| = 1}. Also, let L− := L− ∩ H, H H H H L0 := L0 ∩ H, L+ := L+ ∩ H, C− := C− ∩ H and C− := C− ∩ H. Clearly, we have H H H H ∂(Ω) = L− t {−1} t C− t {0} t C+ t {1} t L+ . We notice here, for future reference, the following fact.

+ 1 1 Lemma 3.2.1 Let (z, r) ∈ R × R :(z, r) , (− 2 , 2 ). Suppose that we have

B(z; r) ∩ Ω , ∅. Then we have

B(z; r) ∩ {−1, 0, 1} , ∅.

Proof This clearly holds by the deﬁnition and geometry of Ω.

Proposition 3.2.1 : Let (a, b, c, d) ∈ Λ and let φ ∈ G be given by az + b φ(z) = ∀z ∈ Cˆ . cz + d (i) Then we have =(φ(z)) ≤ =(z) ∀z ∈ Ω.

(ii) Suppose further that we have 0 , c , 2d. Then we even have

=(φ(z)) < =(z) ∀z ∈ Ω.

Proof :

(i) By (ii), which will be veriﬁed independently, it suﬃces for now to assume that we have c = 2d. Then, by lemma 3.1.2, we have (c, d) = ± (2, 1). Hence, ∀z ∈ Cˆ , we see that |cz + d| = |2z + 1|. But, by lemma 2.3.1, we have =(z) ˆ =(φ(z)) = |cz+d|2 ∀z ∈ C. Hence, it holds that

=(z) =(φ(z)) = ∀z ∈ Cˆ . (2) |2z + 1|2 Let z ∈ Ω. Then, by the deﬁnition of Ω, we have |2z + 1| ≥ 1. Thus, by (2), we get =(φ(z)) ≤ =(z). Hence, we have =(φ(z)) ≤ =(z) ∀z ∈ Ω, as claimed. 3 THE UNIVERSAL COVERING SPACE OF C r {0, 1} 28

(ii) Suppose that we have 0 , c , 2d. We will show that, ∀z ∈ Ω, we have |cz + d| > 1. Suppose, to get a contradiction, that ∃ z ∈ Ω : |cz + d| ≤ 1. d 1 d 1 Then we must have z ∈ B(− c , |c| ). Thus, we see that B(− c , |c| ) ∩ Ω , ∅. d 1 Hence, by lemma 3.2.1, we must have B(− c , |c| ) ∩ {−1, 0, 1} , ∅, i.e., ∃ k ∈ d 1 B(− c , |c| ) ∩ {−1, 0, 1}. Thus, we conclude that k ∈ {−1, 0, 1}, (3)

and |ck + d| < 1. (4)

In particular, by (3), we have k ∈ Z. Hence, since c ∈ Ze and d ∈ Zo, we must have cz + d ∈ Zo, contradicting (4). Consequently, it holds that

|cz + d| > 1 ∀z ∈ Ω. (5)

=(z) ˆ But, by lemma 2.3.1, we have =(φ(z)) = |cz+d|2 ∀z ∈ C. Therefore, by (5), it follows that =(φ(z)) < =(z) ∀z ∈ Ω, as desired.

Lemma 3.2.2 We have

σ(Ω) = {z ∈ H : |2z − 1| ≤ 1, |4z − 3| ≥ 1, |6z − 1| > 1, |12z − 5| > 1}.

σ(Ω)

1 1 0 3 2 1

Figure 5

Proof Recall that the boundary of Ω in C is given by

H H H H ∂(Ω) = L− t {−1} t C− t {0} t C+ t {1} t L+ . Notice that we have L−, L+, C−, C+ ⊥ R. 3 THE UNIVERSAL COVERING SPACE OF C r {0, 1} 29

Thus, since σ is conformal, we must also have

σ(L−), σ(L+), σ(C−), σ(C+) ⊥ σ(R) = R.

Recall that Möbius transformations map generalized circles to generalized circles. Thus, since σ in particular is a Möbius transformation, this applies here. Notice that 1 1 σ(−1) = 1, σ(0) = 0, σ(1) = , & σ(∞) = . 3 2 Thus, we have 3 1 1 1 1 1 5 1 σ(L ) = S ( ; ), σ(C ) = S ( ; ), σ(C ) = S ( ; ), σ(L ) = S ( ; ). − 4 4 − 2 2 + 6 6 + 12 12 Hence, by the orientation principle, we have σ(Ω) =

= {z ∈ C : =(z) > 0, |2z − 1| ≤ 1, |4z − 3| ≥ 1, |6z − 1| > 1, |12z − 5| > 1} =

= {z ∈ H : |2z − 1| ≤ 1, |4z − 3| ≥ 1, |6z − 1| > 1, |12z − 5| > 1}, as desired.

Notice, in particular, that we have σ(Ω) ∩ Ω = ∅.

Lemma 3.2.3 :

(i) Let n ∈ Z r {0}. Then we have

τn(Ω) ∩ Ω = ∅.

(ii) Let n ∈ Z. Then we have τnσ(Ω) ∩ Ω = ∅.

Proof :

(i) This clearly holds by the deﬁnition of Ω.

(ii) This clearly holds by lemma 3.2.2 and the deﬁnition of Ω.

Proposition 3.2.2 Let φ ∈ G r {idCˆ }. Then we have φ(Ω) ∩ Ω = ∅. 3 THE UNIVERSAL COVERING SPACE OF C r {0, 1} 30

Proof Let φ ∈ G : φ , idCˆ . Then, by the deﬁnition of G, az + b ∃ (a, b, c, d) ∈ Λ r {± (1, 0, 1, 0)} : φ(z) = ∀z ∈ Cˆ . cz + d Case 1: Suppose that we have c = 0. Then, by proposition 3.1.1(i),

∃ n ∈ Z r {0} : φ = τn.

Thus, by lemma 3.2.3(i), we have φ(Ω) ∩ Ω = ∅. Case 2 : Suppose that we have 0 , c = 2d. Then, by proposition 3.1.1(ii),

∃ n ∈ Z : φ = τnσ.

Thus, by lemma 3.2.3(ii), we have φ(Ω) ∩ Ω = ∅. Case 3 : Suppose that we have 0 , c , 2d. Suppose, to get a contradiction, that we have φ(Ω) ∩ Ω , ∅. Let z ∈ φ(Ω) ∩ Ω. Then ∃ ζ ∈ Ω : φ(ζ) = z (∈ Ω). Thus, by proposition 3.2.1,

=(z) = =(φ(ζ)) < =(ζ), and =(ζ) = =(φ−1(z)) ≤ =(z), yielding the contradiction =(z) < =(z). Hence, in this case too, we get φ(Ω)∩Ω = ∅.

Corollary 3.2.1 Let φ, ψ ∈ G : φ , ψ. Then we have

φ(Ω) ∩ ψ(Ω) = ∅.

Proof Let φ, ψ ∈ G : φ , ψ. Then it follows that

−1 −1 ψ φ ∈ G : ψ φ , idCˆ .

Thus, by proposition 3.2.2, we get

ψ−1φ(Ω) ∩ Ω = ∅.

Hence, we see that φ(Ω) ∩ ψ(Ω) = ∅, as claimed.

Lemma 3.2.4 : 3 THE UNIVERSAL COVERING SPACE OF C r {0, 1} 31

(i) Let ζ ∈ H. Then ∃ (c0, d0) ∈ Ze × Zo : |c0ζ + d0| ≤ |cζ + d| ∀ (c, d) ∈ Ze × Zo.

(ii) Let (c, d) ∈ Ze × Zo. Then ∃ a, b ∈ Z :(a, b, c, d) ∈ Λ.

Proof (i) Let S := {cζ + d :(c, d) ∈ Ze × Zo} ⊂ C. Notice that S is a discrete subset of C. Thus |S ∩ B(0; r)| < ∞ ∀r ∈ R+. The conclusion now follows immediately.

(ii) If (c, d) ∈ Ze × Zo, then, in particular, gcd(c, d) = 1. Thus ∃ (α, β) ∈ Z × Z : αd − βc = 1. But we also have

(α + kc)d − (β + kd)c = 1 ∀k ∈ Z.

Since d ∈ Zo, ∃ κ ∈ Z : β + κd ∈ Ze. Then, since c ∈ Ze and (α + κc)d − (β + κd)c = 1, we have α + κc < Ze. Hence, if we let a := α + κc and c := β + κd, then (a, b) ∈ Zo × Ze : ad − bc = 1. Therefore, (a, b, c, d) ∈ Λ.

Proposition 3.2.3 : Let ζ ∈ H. Then ∃ φ0 ∈ G : =(φ0(ζ)) ≥ =(φ(ζ)) ∀ φ ∈ G.

Proof Let ζ ∈ H. Then, by lemma 3.2.4(i), ∃ (c0, d0) ∈ Ze × Zo :

|c0ζ + d0| ≤ |cζ + d| ∀ (c, d) ∈ Ze × Zo.

Then, by lemma 3.2.4(ii), ∃ (a0, b0) ∈ Zo × Ze : a0d0 − b0c0 = 1. Notice that we have (a0, b0, c0, d0) ∈ Λ.

Let φ0 : Cˆ → Cˆ be given by

a0z + b0 φ0(z):= . c0z + d0

Then we have φ0 ∈ G and

=(φ0(ζ)) ≥ =(φ(ζ)) ∀ φ ∈ G, as desired.

SLemma 3.2.5 Let V := {z ∈ H : |2z − k| ≥ 1 ∀k ∈ Zo}. Then we have V ⊂ φ∈G φ(Ω). 3 THE UNIVERSAL COVERING SPACE OF C r {0, 1} 32

S n n Figure 6: V ⊂ n∈Z τ (Ω) ∪ τ σ(Ω)

ProofS Let z ∈ V. Suppose ﬁrst that we have |2z − k| > 1 ∀k ∈ Zo. Then we have n z ∈ n∈Z τ (Ω). SupposeS now that ∃k ∈ Zo : |2z−k| = 1. If k ∈ {4n−1S : n ∈ Z} then n n we again have z ∈ n∈Z τ (Ω). If k ∈ {4n+1 : n ∈ Z} then we have z ∈ n∈Z τ σ(Ω). Thus, we conclude that [ [ [ V ⊂ τn(Ω) ∪ τnσ(Ω) ⊂ φ(Ω). n∈Z n∈Z φ∈G S Proposition 3.2.4 We have φ∈G φ(Ω) = H. S Proof Let U := φ∈G φ(Ω). We want to show that U = H holds. Recall that we have G < Γ < {φA : A ∈ SL2(R)} = Aut(H) and Ω ⊂ H. Thus, we see that φ(Ω) ⊂ H ∀φ ∈ G, i.e., that {φ(Ω): φ ∈ G} ⊂ ℘(H). Hence, we conclude that U ⊆ H. (6)

Let ζ ∈ H. Then, by proposition 3.2.3, ∃ φ0 ∈ G :

=(φ(ζ)) ≤ =(φ0(ζ)) ∀φ ∈ G. (7)

−1 Let z := φ0(ζ) ∈ H. Then we have ζ = φ0 (z), and, ∀φ ∈ G, we have

−1 −1 φ(ζ) = φ(φ0 (z)) = φφ0 (z). Thus, by (7), we have −1 =(φφ0 (z)) ≤ =(z) ∀φ ∈ G. (8) −1 But, since G is a group and φ0 ∈ G, we know that, ∀φ ∈ G,

−1 φ ∈ G ⇔ φφ0 ∈ G, 3 THE UNIVERSAL COVERING SPACE OF C r {0, 1} 33

−1 i.e., as φ ranges over G, so does φφ0 , and vice versa. Thus, by (8), we get =(φ(z)) ≤ =(z) ∀φ ∈ G. Thus, in particular, we have =(στn(z)) ≤ =(z) ∀n ∈ Z, (9) and =(σ−1τn(z)) ≤ =(z) ∀n ∈ Z. (10) But, by lemma 3.1.1 and lemma 2.3.1, we must have z + 2n =(z) στn(z) = =( ) = ∀n ∈ Z, 2z + 1 − 4n |2z + 1 − 4n|2 and z − 2n =(z) σ−1τn(z) = =( ) = ∀n ∈ Z. −2z + 4n + 1 | − 2z + 4n + 1|2 Thus, by (4) and (5), we get =(z) ≤ =(z) ∀n ∈ Z, (11) |2z + 1 − 4n|2 and =(z) ≤ =(z) ∀n ∈ Z. (12) | − 2z + 4n + 1|2 But we have z ∈ H. Thus =(z) > 0 holds. Hence, by (6) and (7), we see that 1 ≤ |2z + 1 − 4n| ∀n ∈ Z, and 1 ≤ | − 2z + 4n + 1| ∀n ∈ Z. That is, we get |2z − (4n − 1)| ≥ 1 , |2z − (4n + 1)| ≥ 1 ∀n ∈ Z. (13) But we clearly have {4n − 1, 4n + 1 : n ∈ Z} = Zo. Thus, by (13), it must hold that k 1 |z − | ≥ ∀k ∈ Z . 2 2 o −1 Consequently, byS lemma 3.2.5,S we have z ∈ U. Hence,S we also haveS ζ = φ0 (z) ∈ −1 −1 −1 −1 φ0 (U)) = φ0 ( φ∈G φ(Ω)) = φ∈G φ0 (φ(Ω)) = φ∈G φ0 φ(Ω) = φ∈G φ(Ω) = U. Therefore, we must have H ⊆ U. (14) Finally, by (6) and (14),we get U = H, as desired. 3 THE UNIVERSAL COVERING SPACE OF C r {0, 1} 34

Theorem 3.2.1 The collection {φ(Ω): φ ∈ G} ⊂ ℘(H) is a partition of H, i.e., G H = φ(Ω). φ∈G

Proof This follows immediately from proposition 3.2.2 and proposition 3.2.4.

Corollary 3.2.2 The region Ω˚ ⊂ H is a fundamental domain for the group G < Γ.

Proof This is simply a reformulation of the previous theorem.

We now show that G is free, as previously promised.

Theorem 3.2.2 The group G is free on the two generators τ and σ.

Proof By the preceding proposition, it suﬃces to show that any reduced expression in terms of τ and σ is unique. Let ζ ∈ Ω˚ . Then we have

φ(ζ) ∈ φ(Ω˚ ) ∀φ ∈ G.

Let G = (V, E) be the inﬁnite planar graph embedded in H with vertex set

V := {φ(ζ): φ ∈ G} ⊂ H and edge set E deﬁned by connecting two vertices φ(ζ) and ψ(ζ) if and only if ∂φ(Ω˚ ) and ∂ψ(Ω˚ ) share a common boundary curve segment

H H H H C ∈ {φ(L− ), φ(C− ), φ(C+ ), φ(L+ ): φ ∈ G}, in which case we connect φ(ζ) to ψ(ζ) with an edge e embedded in H in such a way that e crosses C exactly once and lies entirely in the region

φ(Ω˚ ) ∪ C ∪ ψ(Ω˚ ) ⊂ H.

Then, since H r Ω has the four connected components A, B, C and D as shown in the picture below, we see that G must be a tree 15, i.e.,

f or each φ ∈ G, there exists exactly one path in G f rom ζ to φ(ζ).

15It is evident that once a reduced/simple path has joined the vertices φ(ζ) and ψ(ζ) of two neighboring regions φ(Ω˚ ) and ψ(Ω˚ ) via an edge e, the only way to return is by crossing C := φ(Ω) ∩ ψ(Ω) ∩ H, and thus by construction necessarily utilizing the very same edge e, yielding a contradiction. 3 THE UNIVERSAL COVERING SPACE OF C r {0, 1} 35

A Ω˚ D

BC -1 0 1 Figure 7 : H r Ω has 4 connected components A, B, C and D, H H which are regions in C with boundaries in H given by L− , C− , H H C+ and L+ respectively.

The ﬁgure above makes it evident that once an edge in H joins a point in the region Ω˚ ⊂ H to a point in the open set H r Ω = A t B t C t D ⊂ H it must traverse one H H H H of the boundary curves L− , C− , C+ , L+ ⊂ H, whereupon it can never reach any of the remaining component regions Ω˚ , A, B, C, D of H r ∂Ω without traversing the very same boundary curve. However, paths from η to φ(ζ), φ ∈ G, can in a natural way be encoded by the expression of φ in terms of τ and σ assured by the previous proposition; one merely moves across boundaries according to the maps τ and σ occurring in the expression (reading from right to left, of course). For this correspondence to hold, it is vital, of course, that these expressions are reduced as words over {τ, σ}, since we forbid the edges of G to cross a boundary curve segment of any region φ(Ω˚ ), φ ∈ G, more than once. Suppose that there are two distinct reduced expansions of some element φ ∈ G in terms of of the generators τ and σ. Then they encode two distinct paths in G from ζ to φ(ζ). Hence, tracing one of these from ζ to φ(ζ) and then the other one backwards returning to ζ, we have detected a cycle in G, contradicting the fact that G i a tree. The theorem now follows.

Remark Consider the Poincaré upper half-plane model of hyperbolic geometry, |dz| which is given by introducing the metric ds = y on H (where z = x + iy and s = |z|). Then ∂Ω is a hyperbolic geodesic quadrangle, and {φ(Ω): φ ∈ G} is a hyperbolic tessellation of H by geodesic quadrangles. In fact, a proof of Corollary 3.2.2 based on hyperbolic geometry can be found in [10]. Notice also that RΓ is a hyperbolic geodesic triangle and that {ϕ(RΓ): ϕ ∈ Γ} is a tessellation by triangles (In fact, Aut(H)(⊃ Γ ⊃ G) is the group of direct isometries with respect |dz| to ds = y ,i.e., the group of orientation-preserving hyperbolic motions of H.). For additional information, consult [2]. 3 THE UNIVERSAL COVERING SPACE OF C r {0, 1} 36

3.3 The G - Modular Function λ : H → C r {0, 1} Recall the following theorems of complex analysis.

Theorem 3.3.1 (The Riemann Mapping theorem): Let U (C be a simply connected proper region in C. Let a ∈ U . Then there exists a unique biholomorphic mapping f : U → D := B(0; 1) such that we have f (a) = 0 and f 0(a) > 0.

Proof See any thorough introduction to complex variables, e.g., [4], pages 160- 163.

Corollary 3.3.1 Let U be a simply connected proper region in C. Then there exists a biholomorphic mapping f : U → H.

Proof This follows immediately by composition with the Cayley mapping z + 1 D 3 z 7→ −i ∈ H, z − 1 which is a conformal equivalence of D with H.

Given a set U ⊆ C, let ∂∞(U) denote its boundary in Cˆ , so that ∂∞(U) = ∂U if U is bounded and ∂∞(U) = ∂U ∪ {∞} if U is unbounded. Likewise, let cl∞(U) be the closure of U in Cˆ . Thus, for instance, we have ∂∞(H) = ∂H ∪ {∞} = R ∪ {∞} and 1 cl∞(Ω) = Ω ∪ {∞}. Recall that a Jordan curve is a homeomorphic image of S .

Theorem 3.3.2 (The Osgood-Carathéodory Theorem): Let Ω1 and Ω2 be regions in C such that ∂∞(Ω1) and ∂∞(Ω1) each consist of a single Jordan curve in Cˆ . Suppose that f : Ω1 → Ω2 is a biholomorphism. Then f extends to a homeomorphism

f˜ : cl∞(Ω1) → cl∞(Ω2).

Proof See [5] (page 51 for statement, pages 48-51 for proof).

Given a set U ⊂ C, we let U ∗ := {z : z ∈ U }, i.e., U ∗ is the reﬂection of the set U in the line {z ∈ C : =(z) = 0} = R.

Theorem 3.3.3 (The Schwarz Reﬂection Principle): Let U be a region such that U ∗ = U and U ∩ R is a non-empty open interval. Let f : U ∩ (H t R) → C be a function such that 3 THE UNIVERSAL COVERING SPACE OF C r {0, 1} 37

• f is continuous,

• f |U ∩H is holomorphic,

• f |U ∩R is real-valued. Let F : U → C be the function given by ¨ f (z) if z ∈ U ∩ (H t R) F(z):= f (z) if z ∈ U ∩ (C r H)

Then F is holomorphic.

Proof The proof consists of a straightforward application of Morera´s theorem together with a continuity argument, and the details can be found in any good introductory text on complex variables, e.g., [4].

Let us denote by H the right half-plane {z ∈ C : <(z) > 0} and, given a ? set U ⊂ C, let us denote by U the reﬂection of the set U in the line L0, i.e., U ? := {−z : z ∈ U }. We will make use of the following direct consequence of the reﬂection principle.

? Corollary 3.3.2 Let U be a region such that U = U and U ∩L0 is non-empty, open and connected. Let f : U ∩ (H t L0) → C be a continuous function such that f |U ∩H is holomorphic and f |U ∩L0 is real-valued. Let F : U → C be the function given by ¨ f (z) if z ∈ U ∩ (H t L ) F(z):= 0 f (−z) if z ∈ U ∩ (C r H)

Then F is holomorphic.

Proof Let ρ : Cˆ → Cˆ be given by

i π ρ(z):= e 2 z = iz ∀z ∈ Cˆ .

Then ρ is an automorphism of Cˆ , and since ρ(∞) = ∞, ρ | C → C is an automorphism of C. Moreover, ρ(H) = H and ρ(L0) = R and, for each U ∈ ℘(C), ρ(U ?) = ρ(U )∗. −1 Let g := f ◦ ( ρ |ρ(U )): ρ(U ) → C. Notice that g : ρ(U ) → C is holomorphic, ∗ ? ρ(U ) = ρ(U ) = ρ(U ) and g(ρ(U ) ∩ R) = f (U ∩ L0) ⊂ R. Let G : ρ(U ) → C be deﬁned by ¨ g(z) if z ∈ ρ(U ) ∩ (H t R) = ρ(U ∩ (H t L )) G(z):= 0 g(−z) if z ∈ ρ(U ) ∩ (C r H) = ρ(U ∩ (C r H)) 3 THE UNIVERSAL COVERING SPACE OF C r {0, 1} 38

Then, by the Schwarz reﬂection principle, G is holomorphic. −1 Let F := G ◦ (ρ |U ): U → C. Then F is holomorphic and, since f (ρ (ρ(z))) = f (−i · iz) = f (−z) for each z ∈ U , F : U → C is given by ¨ f (z) if z ∈ U ∩ (H t L ) F(z) = 0 f (−z) if z ∈ U ∩ (C r H)

We are now ready to begin our construction of λ. Let Ω0 be the region in C deﬁned by

Ω0 := {z ∈ Ω : <(z) > 0} = Ω ∩ H.

˚ H ? Then we have Ω0 ⊂ H ∩ H and Ω = Ω0 t L0 t Ω0 .

Ω0

0 1 Figure 8

Theorem 3.3.4 There exists a continuous mapping

λ0 : cl∞(Ω) → Cˆ

The ﬁgure below illustrates the mapping properties of λ0. 3 THE UNIVERSAL COVERING SPACE OF C r {0, 1} 39

H L0

λ ? Ω0 Ω0 R H H L− L+ λ(Ω0) = H

H H C− C+ λ(L H) λ(C H) λ(L H) ee e 0 e+ e + H H -1 0 1 λ(C− ) λ(L− ) ? ? λ(Ω0 ) = H

Figure 9: Mapping properties of λ : H → C r {0, 1}

Proof We notice that Ω0 is a simply connected proper region in C. Thus, by corollary 3.3.1 of the Riemann mapping theorem, there exists a biholomorphic mapping f : Ω0 → H. H H H Notice that ∂∞(Ω0) = L0 t {0} t C+ t {1} t L+ t {∞} and ∂∞(H) = R t {∞}. Thus, in particular, ∂∞(Ω0) and ∂∞(H) are both Jordan curves in Cˆ . Hence, by the Osgood-Carathéodory theorem, f extends to a homeomorphism

f˜ : cl∞(Ω0) → cl∞(H).

It follows of course that f˜|∂∞(Ω0) → ∂∞(H) is a homeomorphism too. Therefore, in particular, we have f˜(∂∞(Ω0)) = ∂∞(H) = R t {∞}. Since 0, 1 and ∞ are distinct points of ∂∞(Ω0) and f˜ is injective, f˜(0), f˜(1) and f˜(∞) are distinct points in R t {∞} = ∂∞(H) ⊂ cl∞(H). Consequently, since the natural action of M on Cˆ is triply transitive, ∃! µ ∈ M = Aut(Cˆ ):

µ( f˜(0)) = 0 & µ( f˜(1)) = 1 & µ( f˜(∞)) = ∞.

It must therefore also satisfy µ(R)) = R and µ(H)) = H. Hence,

µ | H → H is a biholomorphism and µ | cl∞(H) → cl∞(H) is a homeomorphism. 3 THE UNIVERSAL COVERING SPACE OF C r {0, 1} 40

Let ξ := µ ◦ f˜ : cl∞(Ω0) → cl∞(H). Then

ξ : cl∞(Ω0) → cl∞(H) is a homeomorphism such that

ξ(0) = 0, ξ(1) = 1, ξ(∞) = ∞, and ξ | Ω0 → H is a biholomorphism. It follows that ξ | ∂∞(Ω0) → ∂∞(H), ξ | Ω0 → H and ξ | ∂Ω0 → ∂H are also homeomorphisms. Thus, in particular, we have ξ(∂Ω0) = R. In fact, since ξ, by virtue of continuity, must preserve connectedness, we must H H H ˚ ? ˚ have ξ(L0 ) = (−∞, 0), ξ(C+ ) = (0, 1) and ξ(L+ ) = (1, +∞). Clearly, Ω = Ω. ˚ H ˚ Moreover, we note that Ω ∩ L0 = L0 , so ξ(Ω ∩ L0) = (−∞, 0) ⊂ R. Let λ0 : cl∞(Ω) → Cˆ be deﬁned by ¨ ξ(z) if z ∈ cl∞(Ω0) λ0(z):= ? ξ(−z) if z ∈ cl∞(Ω0 )

? H Then λ0 is well-deﬁned, since cl∞(Ω0) ∩ cl∞(Ω0 ) = {0} t L0 t {∞} ⊂ L0 t {∞} H and ξ({0} t L0 t {∞}) = {0} t (−∞, 0) t {∞} ⊂ R t {∞} so ξ(−z) = ξ(z) for ? all z ∈ cl∞(Ω0) ∩ cl∞(Ω0 ). Clearly, λ0 is continuous and, by corollary 3.3.2 of the Schwarz reﬂection principle, λ0 |Ω˚ is holomorphic. Additionally, λ0(Ω0) = H ? ∗ h h and λ0(Ω0 ) = H . Finally, we note that λ0 | C± → (0, 1), λ0 | L± → (1, ∞) and λ | h h all are homeomorphisms. This completes the proof. 0 cl∞(Ω)r(C± tL± )

Remark Notice that the mapping λ0 : cl∞(Ω) → Cˆ of the previous theorem is not uniquely determined since it depends on the choice of the Riemann mapping f : Ω0 → H in the above proof.

Remark Notice that by the deﬁnition of λ0, τ and σ, as well as the fact that λ0(∂Ω) = R, we have

λ0(−1 + iy) = λ0(1 + iy) = λ0(1 + iy) = λ0(τ(−1 + iy)) (15) for each y ∈ R+ and 1 1 iθ 1 1 i(π−θ) λ0 − + e = λ0 + e 2 2 2 2 (16) 1 1 1 1 = λ + ei(π−θ) = λ σ − + eiθ 0 2 2 0 2 2 3 THE UNIVERSAL COVERING SPACE OF C r {0, 1} 41 for each θ ∈ (0, π). That is,

H λ0(z) = λ0(τ(z)) , ∀z ∈ L− , (17) and H λ0(z) = λ0(σ(z)) , ∀z ∈ C− . (18)

It follows from theorem 3.3.4 above especially that the mapping

λ0 | Ω → C r {0, 1} is a univalent (i.e. holomorphic injection) bijection. Recall that {φ(Ω): φ ∈ G} is a partition of H. Let us deﬁne the mapping

λ : H → C r {0, 1} by −1 λ |φ(Ω):= λ0 ◦ (φ |φ(Ω)): φ(Ω) → C r {0, 1} , ∀ φ ∈ G. Then λ is at least well-deﬁned. Moreover, we have

λ ◦ φ = λ , ∀ φ ∈ G.

Furthermore, we have λ(H) = λ(Ω) = C r {0, 1}, so λ is surjective. To show that λ is holomorphic we shall make use of the following result.

Theorem 3.3.5 (Analytic Continuation by Continuity): Let A and B be two disjoint simply connected regions in C whose boundaries intersect in a simple (non- self-intersecting) smooth(C1) curve γ, and let f : A → C and g : B → C be holomorphic. Let C := A ∪ γ((0, 1)) ∪ B, and notice that C is again a simply connected region in C. Suppose that

• each point in γ((0, 1)) has an open neighborhood in C,

• f and g are continuous on A ∪ γ([0, 1]), and

• for each ζ ∈ γ([0, 1]), we have limA3z→ζ f (z) = limB3z→ζ g(z). Then there is a unique holomorphic extension of f and g to C.

Proof Like the proof of the Schwarz reﬂection principle, the proof of this theorem is based on Moreras theorem and can be found, e.g., in [15], page ??.

Theorem 3.3.6 The mapping λ : H → C r {0, 1} deﬁned above is holomorphic. 3 THE UNIVERSAL COVERING SPACE OF C r {0, 1} 42

Proof Let Υ := Ω ∪ τ−1(Ω) ∪ σ−1(Ω) ⊃ Ω. It follows from (15) and (16) that λ is continuous on Υ. Recall λ is holomorphic on int(Ω), int(τ−1(Ω)) and int(σ−1(Ω)). −1 H −1 H Notice also that ∂Ω∩∂τ (Ω) = L− ∪{−1} and ∂Ω∩∂σ (Ω) = C− ∪{−1}, both of H H which are simple smooth curves in C. Finally, we note that ∀ζ ∈ L− ∪ C− ∃ r > 0 : B(ζ; r) ⊂ Υ. Hence, by theorem 3.3.5, we see that λ is holomorphic on Υ˚ . But clearly φ(Υ˚ ) is an open neighborhood of the set φ(Ω˚ ) for each φ ∈ G. Consequently, λ is holomorphic on all of H, as claimed.

Theorem 3.3.7 The mapping λ : H → Cr{0, 1} deﬁned above is a covering map.

Proof We have already shown that λ : H → C r {0, 1} is continuous (in fact, holomorphic) and surjective. It remains only to show that each point in C r {0, 1} has an open neighborhood which is evenly covered by λ. Let ζ ∈ C r {0, 1}. It suﬃces to check that ζ has an evenly covered open neighborhood. Suppose ﬁrst that we have ζ ∈ C r [0, ∞). Then, since C r [0, ∞) is open,

∃ δ ∈ R+ : B(ζ; δ) ⊂ C r [0, ∞).

Let −1 ˚ U0 := λ0 (B(ζ; δ)) ⊂ Ω.

Then, clearly, G −1 λ (B(ζ; δ)) = φ(U0), φ∈G −1 and {φ(U0): φ ∈ G} is the set of connected components of λ (B(ζ; δ)). By construction, these are all mapped homeomorphically onto B(ζ; δ)) by λ. Therefore, B(ζ; δ) is an evenly covered set. Suppose now that we have ζ ∈ (0, 1). Then, since (0, 1) ⊂ C r {0, 1} and C r {0, 1} is open, ∃δ ∈ R+ : B(ζ; δ) ⊂ C r {0, 1}. Then we have −1 H λ0 (ζ) = {z−, z+}, where z± ∈ C± ⊂ ∂Ω, and −1 λ (B(ζ; δ)) = U− t U+, where U− and U+ are connected sets containing z− and z+, respectively. We have

λ0(U+) = B(ζ; δ) ∩ H, and ∗ λ0(U−) = B(ζ; δ) ∩ H . 3 THE UNIVERSAL COVERING SPACE OF C r {0, 1} 43

H Recall (from the proof of theorem 3.3.4) that σ maps C− (homeomorphically) H onto C+ . Thus, since ζ = λ0(z±) = λ(z±) = λ(σ(z±)), we must have σ(z−) = z+. Let U0 := U+ ∪ σ(U−) = U˚+ t σ(U−).

Then U0 is an open neighborhood of z+ and

λ(U0) = λ(U˚+ t σ(U−)) = λ(U˚+) t λ(σ(U−)) = λ0(U˚+) t λ0(U−) = B(ζ; δ). −1 Hence, the open sets {φ(U0): φ ∈ G} are the connected components of λ (B(ζ; δ)) and each of these are mapped homeomorphically onto B(ζ; δ) by λ. Consequently, B(ζ; δ) is an evenly covered open neighborhood of ζ in C r {0, 1}. Suppose ﬁnally that we have ζ ∈ (1, ∞). Then, since (1, ∞) ⊂ C r {0, 1} and C r {0, 1} is open, ∃δ ∈ R+ : B(ζ; δ) ⊂ C r {0, 1}. Then we have −1 H λ0 (ζ) = {z−, z+}, where z± ∈ L± ⊂ ∂Ω, and −1 λ (B(ζ; δ)) = U− t U+, where U− and U+ are connected sets containing z− and z+, respectively. We have

λ0(U+) = B(ζ; δ) ∩ H, and ∗ λ0(U−) = B(ζ; δ) ∩ H . H H Obviously, τ maps L− (homeomorphically) onto L+ . Thus, since ζ = λ0(z±) = λ(z±) = λ(τ(z±)), we must have τ(z−) = z+. Let

U0 := U+ ∪ τ(U−) = U˚+ t τ(U−).

Then U0 is an open neighborhood of z+ and

λ(U0) = λ(U˚+ t τ(U−)) = λ(U˚+) t λ(τ(U−)) = λ0(U˚+) t λ0(U−) = B(ζ; δ). −1 Hence, the open sets {φ(U0): φ ∈ G} are the connected components of λ (B(ζ; δ)) and each of these are mapped homeomorphically onto B(ζ; δ) by λ. Consequently, B(ζ; δ) is an evenly covered open neighborhood of ζ in C r {0, 1}. This concludes the proof.

Hence λ : H → C r {0, 1} is a (ℵ0-sheeted) holomorphic covering map, and, since H is simply connected it must be the universal covering space of C r {0, 1}. We proceed to show that H is the natural domain for λ, in the sense that λ cannot be extended to any holomorphic function on any region properly containing H. First we prove a lemma. 3 THE UNIVERSAL COVERING SPACE OF C r {0, 1} 44

Lemma 3.3.1 The set 16 G0 = {φ(0) : φ ∈ G} (R is dense in R, i.e., we have

{φ(0) : φ ∈ G} = R.

Proof Let us first show that we have b {φ(0) : φ ∈ G} = { : b ∈ Z , d ∈ Z } ((Q). (19) d e o b Clearly, ∀φ ∈ G ∃ b ∈ Ze, d ∈ Zo : gcd(b, d) = 1 & φ(0) = d . Conversely, let b ∈ Ze, d ∈ Zo : gcd(b, d) = 1. We want to show that ∃ φ ∈ G : b φ(0) = d . It suffices to show that ∃ a ∈ Zo, c ∈ Ze : ad − bc = 1. We reformulate the problem as follows. Let m, n ∈ Z : gcd(2m + 1, 2n) = 1. We want to show that ∃p, q ∈ Z : (2p + 1)(2m + 1) − (2n)(2q) = 1. But, ∀p, q ∈ Z, we have (2p + 1)(2m + 1) − (2n)(2q) = 4pm + 2p + 2m + 1 − 4nq, so that we have (2p + 1)(2m + 1) − (2n)(2q) = 1 ⇔ 2pm + p + m − 2nq = 0 ⇔ −m = p(2m + 1) − (2n)q. We conclude that it is sufficient to show that ∃ p, q ∈ Z : −m = p(2m + 1) − (2n)q, i.e., it is enough to prove that we have −m ∈ {p(2m + 1) − (2n)q : p, q ∈ Z}. We notice that {p(2m + 1) − (2n)q : p, q ∈ Z} ⊆ Z is the ideal in Z generated by the two integers 2m + 1 ∈ Z and 2n ∈ Z. But we have gcd(2m + 1, 2n) = 1. Hence, we must have {p(2m + 1) − (2n)q : p, q ∈ Z} = Z. Thus, in particular, we have −m ∈ {p(2m + 1) − (2n)q : p, q ∈ Z}, as desired. Consequently, equation ( 19 ) holds true.

b It remains only to show that the set G0 = { d : b ∈ Ze, d ∈ Zo} (Q is dense in R, i.e., that it satisﬁes G0 = R. However, since Q is dense in R, it suﬃces to show that G0((Q) is dense in Q((R), i.e., that we have G0 ⊇ Q. Let q ∈ Q. If we have q ∈ G0 then we are done. Otherwise, ∃ i, j, k ∈ Z : 2i+1 2k2l 2i+1 2k2l (2i+1)2l q = 2k(2 j+1) . However, ∀l ∈ Z, we have q 2k2l+1 = 2k(2 j+1) 2k2l+1 = (2 j+1)(2k2l+1) = 2l(2i+1) 2kl 2( j2k+1l+ j+2kl)+1 ∈ G0 and liml→∞ q 2kl+1 = q. Hence, we have G0 = R, as desired. Theorem 3.3.8 The holomorphic function λ ∈ O(H) cannot be analytically continued to any region in C which properly contains H.

Proof Suppose, to get a contradiction, that λ can be analytically continued to a region U ⊆ C which properly contains H. That is, we suppose that there exists a region U in C such that we have H( U and there exists a holomorphic function λˆ ∈ O(U) deﬁned and holomorphic on U, such that we have λˆ|H = λ. Since U ⊆ C is an open set strictly containing H, we see that U must contain a non-trivial open sub-interval of R, i.e., ∃ a, b ∈ R : a < b &(a, b) ⊂ U.

16Recall that the natural action of G on H extends naturally to H as well. 3 THE UNIVERSAL COVERING SPACE OF C r {0, 1} 45

Let c ∈ (a, b). Then, by lemma 3.3.1, we know that there exists a sequence {φn}n∈N ⊂ G such that we have φn(0) → c as n → ∞. Let N ∈ N : ∀n ∈ N, n ≥ N implies φn(0) ∈ (a, b). Let n ∈ N : n ≥ N. Then we have λˆ(φn(0)) = λˆ(φn(lim0

3.3.1 An Application of λ We end this chapter with a classic application of the modular function λ. Theorem 3.3.9 ( The Little Picard Theorem ) Suppose that f : C → C is an entire function that omits two distinct values in C. Then f is a constant function. Proof We let f : C → C be an entire function, and we suppose that ∃ a, b ∈ C : a , b & a, b < f (C), that is, we have a, b ∈ C : a , b & f (C) ⊆ C r {a, b}. We let µ ∈ M = Aut(Cˆ ): µ(0) = a & µ(1) = b & µ(∞) = ∞. Then, we have µ ∈ Aut(C): µ(C r {a, b}) = C r {0, 1}, µ ◦ f : C → C is still an entire function and (µ ◦ f )(C) = µ( f (C)) ⊆ µ(C r {a, b}) = C r {0, 1}. Thus, we see that µ ◦ f : C → C r {0, 1} is a holomorphic mapping. Hence, by the fact that C is simply connected and µ ◦ f is continuous, there exists a (unique after the choice of one image point) continuous mapping µß◦ f : C → H such that we have λ ◦ µß◦ f = µ ◦ f , i.e., such that the diagram

u: H Ý uu µ◦ f uu uu λ uu uu / C r {0, 1} C µ◦ f

17 By the construction of the mapping λ0 : cl∞(Ω) → Cˆ given in the proof of theorem 3.3.4 it is clear that we have λ0(0) = 0. 3 THE UNIVERSAL COVERING SPACE OF C r {0, 1} 46 is commutative , and by theorem 2.2.12, µß◦ f is in fact holomorphic . z−i Let ξ : H → D, z 7→ z+i , be the biholomorphic Cayley mapping. Then, we see that ξ ◦ µß◦ f : C → D is a holomorphic mapping. That is, ξ◦µß◦ f is an entire function, and ξ◦µß◦ f (C) ⊆ D, i.e., |ξ ◦ µß◦ f (z)| < 1 ∀z ∈ C, so that ξ ◦ µß◦ f is also bounded. Consequently, by Liouville´s theorem, ξ ◦ µß◦ f : C → D is a constant function, i.e., ∃ c ∈ D ⊂ C :

ξ ◦ µß◦ f (z) = c ∀z ∈ C.

Therefore, we have µß◦ f (z) = ξ−1(c) ∀z ∈ C, so that we have µ ◦ f (z) = λ ◦ µß◦ f (z) = λ(ξ−1(c)) ∀z ∈ C, which in turn implies that we have

f (z) = µ−1(λ(ξ−1(c))) ∀z ∈ C.

Hence, in particular, we see that the function f : C → C is constant, as claimed.

The diagram shown below illustrates the proof.

ξ ◦ µß◦ f

R 1 HD- ß ξ µ ◦ f λ ? - - C f C r {a, b} µ C r {0, 1}

µß◦ f

Figure 10 : Picard´s Little Theorem

Remark We take note of the fact that the above result is "best possible" in the sense that there exist entire functions that omit a single value in the complex plane. Consider, e.g., the complex exponential function exp : C → C, z 7→ ez, which is entire and satisﬁes exp(C) = C r {0}. 3 THE UNIVERSAL COVERING SPACE OF C r {0, 1} 47

λ 3.4 The covering group Deck (H −−→ C r {0, 1}) Theorem 3.4.1 We have λ Deck(H −−→ C r {0, 1}) = G.

Proof Recall that

G = {φA : A ∈ Γ(2)} < {φA : A ∈ SL2(Z)} ( Aut(H). Moreover, since λ ◦ φ = λ ∀φ ∈ G, each mapping φ ∈ G is ﬁber-preserving (with respect to λ). Hence, we certainly have λ G ≤ Deck(H −−→ C r {0, 1}).

However, since λ|Ω : Ω → C r {0, 1} is a bijection, we must in fact have

λ G = Deck(H −−→ C r {0, 1}), as claimed.

3.5 A Relation Between λ and J 3.5.1 A Quick Review of Elliptic Functions We begin by briefly recalling the definition of Klein´s modular function J . This function originally grew out of a wish to classify all complex tori up to biholomorphism. The quotient space C/Λ of C under the equivalence relation z ∼ w :⇔ z − w ∈ Λ on C, where Λ is a lattice in C, turns out to be a Riemann surface homeomorphic to the torus S1 × S1. Moreover, each complex structure on S1 × S1 can be constructed in this way, and two lattices Λ and Λ0 in C give rise to equivalent complex structures on the torus, if and only if, there exists an a ∈ C∗ such that Λ0 = aΛ, in which case the lattices are said to be equivalent18 and we write Λ ' Λ0. That is, two complex tori are conformally equivalent if and only if their underlying lattices are equivalent, and the study of complex tori is essentially that of equivalence classes of lattices in the plane. Let ω1, ω2 ∈ C be linearly indepen- dent over R, and consider the lattice Λ := Zω1 + Zω2 that they generate. We see −1 by multiplying by ω1 that ∃ τ ∈ C : Λ ' Z1 + Zτ, where we may assume that =(τ) > 0. It is easily verified that if τ0 ∈ H then aτ + b Z + Zτ0 ' Z + Zτ ⇔ ∃ a b ∈ SL (Z): τ0 = . c d 2 cτ + d 18This clearly defines an equivalence relation on the set of lattices in the complex plane. 3 THE UNIVERSAL COVERING SPACE OF C r {0, 1} 48

Thus, the lattices in the complex plane C are parameterized by the upper half- plane H modulo the action of the group Γ PSL2(Z). That is, there is a one- to-one correspondence between H/Γ and the set of isomorphy-classes of complex tori. We now turn to meromorphic functions invariant under a given lattice. Fixing a lattice Λ = Zω1 + Zω2 ' Z1 + Zτ, we define a meromorphic function f ∈ M (C) on C to be Λ − elliptic if f (z + ω) = f (z), ∀ω ∈ Λ (i.e., it is doubly periodic w.r.t. ω1 and ω2). Note that non constant elliptic functions necessarily have poles (by the maximum modulus principle and Liouville´s theorem). An example of a Λ-elliptic function is the Weierstraß ℘ − f unction, defined on C by 1 X 1 1 ℘(z):= 2 + 2 − 2 . z ω∈Λ (z + ω) ω This series is locally uniformly convergent, so ℘ is well-defined. It is holomorphic on CrΛ, has poles of order 2 at points of Λ and is Λ-periodic. We can differentiate ℘ term wise, and it is not hard to show that it satisfies the non-linear differential equation 0 2 3 (℘ ) = 4℘ − g2℘ − g3, P P 1 1 where g2 = 60 ω∈Λ ω4 and g3 = 140 ω∈Λ ω6 . Denoting by e1,e2 and e3 the values of ℘ at the half-periods, i.e., ω ω ω + ω e := ℘( 1 ), e := ℘( 2 ), e := ℘( 1 2 ), (20) 1 2 2 2 3 2 we can show that

3 4℘ − g2℘ − g3 = 4(℘(z) − e1)(℘(z) − e2)(℘(z) − e3), and that the roots e1,e2 and e3 are distinct, so that the discriminant

3 2 4 := g2 − 27g3 is non-zero. Finally, J is deﬁned by g3 J := 2 4 3 and depends only on the equivalence class of Λ, g2 and 4 being homogeneous of the same degree −12. Since Λ can be parameterized by τ, as explained above, we may consider J to be deﬁned on the upper half-plane through J (τ):= J (Z + Zτ), and it can be shown that the functions g2(τ), g3(τ), 4(τ) and J (τ) all are holomorphic on H, and that J is invariant under unimodular transformations, i.e., J ◦ φ = J ∀φ ∈ Γ. 3 THE UNIVERSAL COVERING SPACE OF C r {0, 1} 49

2πi Furthermore, J is surjective, satisﬁes J (e 3 ) = 0, J (i) = 1 and J (i∞) = ∞. Also, J has a 1st order pole at i∞ (i.e. the Fourier series of J has the form P ∞ 2πint rd 2πi nd n=−1 ane where a−1 , 0 ), a 3 order zero at e 3 and J (τ)−1 has a 2 order zero at τ = i. Moreover, J takes every value exactly once in the closure of the fundamental region RΓ of Γ (see ﬁgure 1 above), hence H/Γ = C.

3.5.2 A Branched Holomorphic Covering Map f Factoring J Through λ

Given a subgroup G ≤ Γ of the modular group Γ = {φA : A ∈ SL2(Z)} < {φA : A ∈ SL2(R)} = Aut(H), we let H/G denote the corresponding orbit space. Let π : H → H/G be the natural quotient projection H 3 z 7→ Gz ∈ H/G, and endow the orbit space with the quotient topology with respect to π. In fact, H/G even has a natural complex structure, as is the content of the following theorem.

Theorem 3.5.1 The space H/G is a (connected) Riemann surface, and the quotient projection π : H → H/G, z 7→ Gz, is a branched holomorphic covering map. Moreover, π is unbranched (i.e., is an ordinary holomorphic covering map) if and only if the action of G on H is free (i.e., its isotropy subgroups are all trivial).

Proof See, e.g., [11], pages 248-250.

For the subgroup G of Γ, this quotient map is simply given by the covering map λ and we have H/G C r {0, 1}. However, in general we cannot expect π : H → H/G to be a covering map in the strict sense. When we take the quotient of H with the full modular group Γ, the corresponding quotient mapping is given by the modular function J : H → C. This is a branched covering map (with branch points of orders 2 and 3 on the −1 −1 2πi orbits Γi = J (1) and Γρ = J (0) containing i and ρ := e 3 respectively; see [11], page 290-291). Clearly, it is a modular function, since it satisﬁes

J ◦ φ = J ∀φ ∈ Γ.

In this subsection we will construct a mapping f : C r {0, 1} → C such that J = f ◦ λ, i.e., such that the following diagram commutes.

m H M mmm MMM λ mmm MMJ mmm MMM mmm MM vm MM& C r {0, 1} / f C 3 THE UNIVERSAL COVERING SPACE OF C r {0, 1} 50

Since J is invariant under the action on H of the full modular group Γ we necessarily have f ◦ λ ◦ φ = f ◦ λ ∀φ ∈ Γ. (21) Before going further, we recall that Γ(2) is the kernel of the surjective group homomorphism SL2(Z) → SL2(Z2) given by reduction (mod 2) of the integer matrix entries. Thus, by the 1st group isomorphism theorem and the fact that SL2(Z2) is isomorphic to S 3, we see that

Γ/G SL2(Z)/Γ(2) SL2(Z2) S 3. In particular, G is a subgroup of index 3! = 6 in Γ. Furthermore, there is a natural induced action of Γ/G on H/G. Since |Γ/G| < ∞, Γ/G acts properly discontinuously and faithfully on H/G. Thus, in particular, (H/G)/(Γ/G) is a Riemann surface. We know that Γ/G is a subgroup of Aut(H/G) = Aut(C r {0, 1}) of order |Γ/G| = 6. However, Aut(C r {0, 1}) = Aut(Cˆ r {0, 1, ∞}) = {φ ∈ Aut(Cˆ ): φ({0, 1, ∞}) = {0, 1, ∞} = {φ ∈ M : φ({0, 1, ∞}) = {0, 1, ∞}}. That is, the group of automorphisms of Cr{0, 1} consists precisely of the Möbius transformations that map the set {0, 1, ∞} onto itself, i.e., the Möbius transformations that permute the points 0, 1, ∞ ∈ Cˆ . In particular, we see that we must have Γ/G = Aut(C r {0, 1}). Let the mappings

ψ1, ψ2, ψ3, ψ4, ψ5, ψ6 : Cˆ → Cˆ be given by 1 ψ = id , ψ (z):= 1 − z, ψ (z):= , 1 Cˆ 2 3 z 1 z − 1 z ψ (z):= , ψ (z):= , ψ (z):= 4 1 − z 5 z 6 z − 1 respectively. Then the automorphisms of Cr{0, 1} are the restrictions to Cr{0, 1} of these mappings, i.e., we have 1 1 z − 1 z Aut(Cr{0, 1}) = {z 7→ z, z 7→ 1−z, z 7→ , z 7→ , z 7→ , z 7→ } S . z 1 − z z z − 1 3

In fact, we see that Aut(C r {0, 1}) S 3 is generated by ψ2 and ψ3. Defining f : C r {0, 1} → C to be the canonical quotient projection H/G → (H/G)/(Γ/G) H/Γ, we are guaranteed a holomorphic mapping satisfying f ◦ ψ = f ∀ψ ∈ Γ/G = Aut(C r {0, 1}), (22) which is equivalent to the condition (21). But as explained above, we have Aut(Cr {0, 1}) = hψ2, ψ3i. Hence it suffices to require that f satisfies

f ◦ ψi = f f or i ∈ {2, 3}, 3 THE UNIVERSAL COVERING SPACE OF C r {0, 1} 51 i.e., it suﬃces to demand that f satisﬁes

f (1 − z) = f (z) ∀z ∈ H (23) and 1 f ( ) = f (z) ∀z ∈ H. (24) z Each point in Cˆ is taken by f with total multiplicity 6. More precisely, the cardinality of the fiber f −1(z) above each point z ∈ Cˆ r {0, 1, ∞} is precisely |Aut(C r {0, 1})| = 6, whereas we have f −1(∞) = {0, 1, ∞}, so that 0, 1 and ∞ are poles of f of order 2. Therefore, f must have a partial fraction expansion of the form a b c d f (z) = + + + + e + f z + gz2, z z2 z − 1 (z − 1)2 where a, b, c, d, e, f, g ∈ C. Imposing the restriction (23) on f , we find that c = −a and d = b and f = −g and then, by imposing (24), we find that b = −a and g = a. Hence the partial fraction expansion of f is of the form a −a −a −a f (z) = + + + + e + az + −az2, (25) z z2 z − 1 (z − 1)2 for some complex constants a, e ∈ C. Recall that the modular function λ constructed in the previous chapter was not uniquely determined by our construction. It is, however, not difficult to see that λ can be normalized to fulfill

2πi 2πi λ(i) = −1 and λ(e 3 + 1) = e 3 + 1 (26)

2πi (in which case it is unique). Then, since we have J (e 3 ) = 0, J (i) = 1, J (i∞) = ∞ and J (z) = J (−z), ∀z ∈ H (see [3], page 39-42), we get

2πi 2πi 2πi 0 = J (e 3 ) = J (−e 3 ) = J (e 3 + 1) = 2πi 2πi = f (λ(e 3 + 1)) = f (e 3 + 1) = 3a + e and 15 1 = J (i) = f (λ(i)) = f (−1) = e − a, 4 4 4 which yields a = − 27 and e = 9 . Consequently, by factorizing the expression (25), we see that f : C r {0, 1} → C is given by the formula

4 (z2 − z + 1)3 f (z) = . 27 z2(z − 1)2 3 THE UNIVERSAL COVERING SPACE OF C r {0, 1} 52

Remark When λ is normalized according to (26), we actually have e (τ) − e (τ) λ(τ) = 3 2 ∀τ ∈ H, (27) e2(τ) − e1(τ) where e1, e2, e3 : H → C are given by (20). In fact, (27) is often taken as the deﬁnition of λ, a route which is more direct but where the mapping properties of λ are not as transparent as in the existence proof given previously. 4 BIBLIOGRAPHY 53

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