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Home , Branch point, Complex manifold, Genus (mathematics), Riemann sphere, Riemann surface, Surface (topology)

AARON LANDESMAN

CONTENTS 1. Introduction 2 2. Maps of Riemann Surfaces 4 2.1. Defining the maps 4 2.2. The multiplicity of a map 4 2.3. Ramification Loci of maps 6 2.4. Applications 6 3. Properness 9 3.1. Definition of properness 9 3.2. Basic properties of proper morphisms 9 3.3. Constancy of degree of a map 10 4. Examples of Proper Maps of Riemann Surfaces 13 5. Riemann-Hurwitz 15 5.1. Statement of Riemann-Hurwitz 15 5.2. Applications 15 6. of Riemann Surfaces of ≥ 2 18 6.1. Statement of the bound 18 6.2. Proving the bound 18 6.3. We rule out g(Y) > 1 20 6.4. We rule out g(Y) = 1 20 6.5. We rule out g(Y) = 0, n ≥ 5 20 6.6. We rule out g(Y) = 0, n = 4 20 6.7. We rule out g(C0) = 0, n = 3 20 6.8. 21 7. Automorphisms in low genus 0 and 1 22 7.1. Genus 0 22 7.2. Genus 1 22 7.3. Example in Genus 3 23 Appendix A. Proof of Riemann Hurwitz 25 Appendix B. Quotients of Riemann surfaces by automorphisms 29 References 31 1 2 AARON LANDESMAN

1. INTRODUCTION In this course, we’ll discuss the theory of Riemann surfaces. Rie- mann surfaces are a beautiful breeding ground for ideas from many of math. In this way they connect seemingly disjoint fields, and also allow one to use tools from different areas of math to study them. They are crucial objects of interest in algebraic , num- ber theory, , dynamics, and , just to name a few. One good, albeit advanced reference for this ma- terial is [McM], where I learned much of the material in these notes. To start, we begin by giving some examples of Riemann surfaces. Intuitively, a Riemann is just an object which looks like C when you zoom in. Example 1.1. Here are some examples of Riemann surfaces. (1) C (2) C× = C − {0} (3) The Riemann , Cˆ (4) C/Λ for some lattice Λ (5) The g-holed (6) H := {x ∈ C : im x > 0} , the upper half (7) H/Γ where H is the upper half plane, and Γ is some torsion free discrete in SL2(R) Remark 1.2. It is a nontrivial theorem (the ) that the above Riemann surfaces are an exhaustive list (though the above list does contain some repetitions). We’re ready to define Riemann surfaces. If you know about man- ifolds, a is just a 1-dimensional complex with complex holomorphic transition functions. Remark 1.3. One way to define a (compact) Riemann surface is as a connected subset X ⊂ Cn such that for every point x ∈ X there is a (a holomorphic map with holomorphic inverse) between X and an open subset of C. However, that definition is not so great, because it requires that we specify an X → Cn for some n. It would be better to have an intrinsic definition, not depending on an embedding. Recall that a topological X is Hausdorff if for any two points x, y ∈ X there are open sets U, V with x ∈ U, y ∈ V so that U ∩ V = ∅. A is second countable if there is a countable collection of open sets Ui so that every open is a union of such Ui. A topological space is connected if it cannot be written as a disjoint RIEMANN SURFACES 3 union of two nonempty open sets. A map is continuous if the preim- age of an is open. A is a continuous map between two topological that has a continuous inverse. Definition 1.4. A Riemann surface is a Hausdorff, second countable, connected topological space X with an open cover by sets Ui ⊂ X and maps fi : Ui → C so that (1) fi is a homeomorphism onto an open subset of C ◦ −1| ( ∩ ) → (2) The transition functions fj fi fi(Ui∩Uj) : fi Ui Uj fj(Ui ∩ Uj) are complex analytic (i.e., holomorphic). A collection of (Ui, fi)i∈I as above is called an for X. A particu- lar (Ui, fi) is called a chart. Remark 1.5. A Riemann surface is the datum of the topological space X together with the atlas (Ui, fi)i∈I. In particular, it is possible to have two riemann surfaces with the same underlying topological spaces but different atlases. Exercise 1.6. Show that Cˆ , the (the one point com- pactification of C, whose underlying set is C ∪ {∞} and whose open sets are either opens in C or sets whose complements are compact sets in C) is a Riemann surface. Hint: Cover it by two charts Cˆ − {0} ˆ 1 and C − ∞, with transition given by z 7→ z . Exercise 1.7. Let E = C/Λ for Λ ⊂ C a lattice of rank 2, meaning Λ = Z ⊕ aZ for a ∈/ R. Show that E is a Riemann surface. Hint: Pick charts for E as charts coming from open subsets U ⊂ C mapping homeomorphically to E under the C → E. (Why is C itself not a chart for E?) 4 AARON LANDESMAN

2. MAPSOF RIEMANN SURFACES 2.1. Defining the maps. As is ubiquitous in mathematics, after defin- ing the objects of our category (the Riemann surfaces) we should say what maps between them are. Definition 2.1. A biholomorphism between two open sets X, Y ⊂ C is a holomorphic map f : X → Y with holomorphic inverse.

Definition 2.2. Let U1, U2 be two Riemann surfaces where U1 has an atlas consisting of the single chart f1 : U1 → V1 and U2 has an atlas consisting of the single chart f2 : U2 → V2, with Vi ⊂ C open. Then −1 we say a map g : U1 → U2 is holomorphic if the map f2 ◦ g ◦ f1 : V1 → V2 g U1 U2 (2.1) −1 f f1 2 −1 f2◦g◦ f1 V1 V2 is holomorphic. Definition 2.3. Let X and Y be two Riemann surfaces. A map f : X → Y is a continuous map such that if (Ui, gi) are an atlas for X and (Vj, hj) are an atlas for Y, then the resulting restriction of f

−1 f f (Vj) ∩ Ui −→ Vj is holomorphic for all i, j. Loosely speaking, a map of riemann surfaces is just a map which is holomorphic when restricted to each of the open sets. 2.2. The multiplicity of a map. We next come to the notion of the multiplicity of a map of riemann surfaces. Roughly speaking, this is just the of preimages of “most points.” The following lemma makes this precise, and is essential in what follows. Lemma 2.4 (Key Lemma). For any map of Riemann surfaces g : X → Y, 0 with f (p) = q there exists open sets U 3 p, V 3 q and charts f1 : U → 0 −1 0 0 U, f2 : V → V with h := f2 ◦ g ◦ f1 : U → V given by h(z) = 0 or h(z) = zn for some n. 0 00 0 Proof. To start, choose charts f1 : U → U and f2 : V → V with 00 0 0 U ⊂ C and V ⊂ C open sets containing 0 ∈ C with f1(0) = p and f2(0) = q. (Primes have nothing to do with derivatives here.) 0 −1 0 0 Define h := f2 ◦ g ◦ f1. By assumption, h is holomorphic, so after RIEMANN SURFACES 5

g X Y

g| U U V 0 (2.2) f1 f2 00 h0 0 f1 U V α h U0

FIGURE 1. Diagram illustrating the proof of Lemma 2.4. shrinking U00, U, V0, V we can assume that h0 agrees with its Taylor expansion about 0. If h0 is constant, we can take h = h0 and we are done. So, we assume h0 is not constant. Then, we can write h0(z) = znt(z) with t(0) 6= 0. After possibly shrinking our open sets further, we may assume the power series t(z) has an nth root, say r(z)n = t(z), so h0(z) = (z · r(z))n. Define a map α : U00 → C sending z 7→ z · r(z) and let U0 ⊂ C denote the of α. Because t(0) 6= 0, we also have r(0) 6= 0, and so α has nonvanish- ing derivative at 0. We can then formally invert the power series for α, and hence, after further shrinking the open sets U, U0, U00, V, V0, we may assume that α : U00 → U0 is a biholomorphism. −1 0 −1 Then, define f1 := α ◦ f1 and h := h ◦ α . Since (h ◦ α)(z) = 0 n n h (z) = (z · r(z)) , and α(z) = z · r(z), we have h(z) = z .  Definition 2.5. For f : X → Y define the multiplicity of f at x to be ( d if f locally at x looks like z 7→ zd via Lemma 2.4 multx( f ) = ∞ if f is constant in some neighborhood of x Lemma 2.6. If f : X → Y is a map of Riemann surfaces, either f is constant or multx( f ) < ∞ for all x ∈ X.

Proof. Note that the set of points where multx( f ) < ∞ is open and the set where multx( f ) = ∞ is open, using Lemma 2.4. Since X is connected, only one of these two conditions may hold.  Remark 2.7. Unless otherwise stated, we will assume f is noncon- stant. 6 AARON LANDESMAN

2.3. Ramification Loci of maps. We next introduce the notions of branch loci and ramification loci. These are just the places at which the map does not look like z 7→ z. Definition 2.8. The ramification locus, notated R( f ), of a map f : X → Y is the set of points at which f is locally of the form z 7→ zd for d > 1. The branch locus, notated B( f ), is f (R( f )), the image of the ramification locus. Exercise 2.9. Directly from the definitions, show that for f : X → Y, g : Y → Z, we have B(g ◦ f ) = g(B( f )) ∪ B(g). Possible hint: Draw a picture! 2.4. Applications. We now use the notions of maps of riemann sur- faces just introduced to deduce some pleasant applications. In par- ticular, we will use them to give easy proofs of the open mapping theorem, Louiville’s theorem, and the fundamental theorem of alge- bra. Proposition 2.10 (Open Mapping Theorem). Any map f : X → Y is open (meaning that the image of any open set is again open). Proof. Exercise 2.11. Show that in order to verify a map f : X → Y is open, it suffices to check that for any point x ∈ X there are some open sets U and V with x ∈ U ⊂ X, V ⊂ Y and f (U) ⊂ V, so that the map f |U : U → V is open. By Exercise 2.11, we can replace X and Y by sufficiently small open subsets around each point. Hence, by Lemma 2.4, it suffices to check for a map f of the form z 7→ zn. Exercise 2.12. Verify such maps are open, completing the proof.  Lemma 2.13. Any map f : X → Y is discrete, meaning for q ∈ Y, f −1(q) is a discrete set. Proof. Let x ∈ f −1(q). We wish to show there is an open neighbor- hood around x on which where are no other points mapping to q. This follows as we can choose an open neighborhood on which the map is of the form z 7→ zd, and the map is discrete when restricted to such an open set.  Exercise 2.14. For f : X → Y a map of Riemann surfaces, show that R( f ) is discrete. Hint: Use the same idea as in Lemma 2.13. RIEMANN SURFACES 7

Recall that a topological space is compact if every open cover has a finite subcover. If you haven’t seen compactness before, now is a good time to do the following exercises: Exercise 2.15. Let X be a compact topological space. Show that any closed subspace Y ⊂ X is also compact. Exercise 2.16. Suppose X is a compact . Show that if Y ⊂ X is compact (with the induced topology), it is also closed. Hint: Show that around each point x ∈/ Y there is some open set Ux not meeting Y. To construct Ux, for each y ∈ Y, x ∈ X − Y, use that X is Hausdorff to construct open sets Vxy, Uxy with so that Vxy ∩ Uxy = ∅ with x ∈ Uxy, y ∈ Vxy. For a fixed x, use compactness to produce a finite collection of such Vxy covering Y, and take Ux to be the intersection of the resulting Uxy. Exercise 2.17. Show that the image of a compact set under a contin- uous map is compact. Lemma 2.18. If X is a compact Riemann surface and Y is a Riemann surface with a map f : X → Y (assumed to be nonconstant by Remark 2.7, then (1) f is surjective, (2) Y is compact, (3) the set f −1(q) is finite for each q ∈ Y, and (4) both R( f ), B( f ) are finite. Proof. To show f is surjective, note that f is open, by Proposition 2.10. The image of f is also compact, hence closed. Therefore, f (X) is both open and closed. Since Y is connected, f (X) = Y. Since f is surjec- tive, f (X) = Y is the image of a compact set, hence compact. For the last two parts, we have f −1(q), B( f ), R( f ) are all discrete sub- sets of the compact set X, by Lemma 2.13 and Exercise 2.14 hence compact.  Exercise 2.19 (Generalization of Liouville’s Theorem). Prove the fol- lowing generalization of Liouville’s Theorem: Any f : X → C for X compact is constant. In particular, taking X = Cˆ , we obtain Liouville’s theorem, that any bounded entire function is constant. Hint: Use Lemma 2.18. Corollary 2.20 (Fundamental theorem of algebra). Every nonconstant over the complex has a root. Proof. Consider a polynomial as a function Cˆ → Cˆ (by extending nonconstant to send ∞ 7→ ∞. They are surjective by 8 AARON LANDESMAN

Lemma 2.18. In particular, there is some x with f (x) = 0, and so x is a root of f .  RIEMANN SURFACES 9

3. PROPERNESS We next introduce the notion of properness. We have seen that maps of Riemann surfaces are always open in Proposition 2.10. Morally speaking properness means that the map is also closed. Technically speaking, being proper is a bit stronger than being closed, but we shall see that proper maps are closed in Lemma 3.4. 3.1. Definition of properness. Here is the definition of what it means for a map to be proper. Definition 3.1. A map f : X → Y is proper if for all compact K ⊂ Y we have f −1(K) is compact. Remark 3.2. Note that continuous maps send compact sets to com- pact sets, but proper maps mean that f −1 of a compact set is compact. 3.2. Basic properties of proper morphisms. We now quickly derive a number of pleasing properties of proper morphisms. Lemma 3.3. If f : X → Y is a map between compact Hausdorff spaces, then f is proper. Proof. In a compact Hausdorff space, a set is compact if and only if it is closed by Exercise 2.15 and Exercise 2.16. Then, the preimage of a closed set is closed by continuity of f .  For the next result, we recall a standard result from topology that closed bounded sets in Rn are compact. In particular, closed balls are compact. This is known as the Heine-Borel theorem. Lemma 3.4. If f : X → Y is a proper map of Riemann surfaces, then f is closed. Proof. If E is a closed set, then we wish to show f (E) is closed. Take y ∈/ f (E). We wish to produce an open set around y not intersecting f (E). Take some small closed ball y ∈ V.

E X f −1(V) (3.1)

f (E) Y V y

Since V is compact and f is proper, f −1(V) is compact. Therefore, E ∩ f −1(V) is compact. Hence, f (E ∩ f −1(V)) ⊂ V is a compact subset of a compact set, hence closed. Since f (E ∩ f −1(V)) does not contain y, we have int(V) − f (E ∩ f −1(V)) is open, which is our desired open neighborhood of y not intersecting f (E).  10 AARON LANDESMAN

Corollary 3.5. If f : X → Y is a proper (nonconstant) map of Riemann surfaces, it is surjective. Proof.f (X) is open by the open mapping theorem, and closed by Lemma 3.4. Hence, the f (X) = Y as Y is connected.  Exercise 3.6. For f : X → Y a proper map, if D ⊂ X is discrete, show that f (D) is also. Hint: Suppose f (D) were not discrete. Then, f (D) would have some limit point p. Take a closed ball V 3 p. Show f −1(V) is a compact set containing infinitely many points in a discrete set. Exercise 3.7. For f : X → Y a proper map of Riemann surfaces, show that the ramification locus R( f ) and the branch locus B( f ) are discrete. 3.3. Constancy of degree of a map. The next step to understanding maps of Riemann surfaces is to define the notion of the degree of a map. The main result to this effect is the following: Proposition 3.8. Let f : X → Y be a proper, nonconstant morphism. Then, the function d(q) := ∑ f (p)=q multp( f ) is finite and constant (inde- pendent of the point q). Assuming Proposition 3.8, we can make the following definition Definition 3.9. For f : X → Y a proper map of Riemann surfaces, the constant value d(q) of Proposition 3.8 is called the degree of f . We now build the necessary machinery to prove Proposition 3.8, which we will do at the of this section. There are two main ideas involved in the proof. First, we need to show the preimage of any point only contains finitely many points, which we easily do in Lemma 3.10. This allows us to make sense of the sum in the defini- tion of the degree. Following this, most of the work lies in showing that f is then a covering map away from the ramification points. Once we have both of these, we can analyze the map locally and conclude the result. Lemma 3.10. If f : X → Y is proper then f −1(q) is finite for all q ∈ Y. Proof. We have f −1(q) is compact by properness. It is also discrete by Lemma 2.4. Exercise 3.11. Conclude the proof by showing a compact discrete set is finite.  RIEMANN SURFACES 11

Our next goal is to show proper maps are covering maps, away from their ramification points. To this end, we now define covering maps. Definition 3.12. A map of topological spaces X → Y is a covering map if for each y ∈ Y we can find y ∈ V with f −1(V) homeomorphic to a disjoint union of open sets, each homeomorphic to V. Definition 3.13. A map f : X → Y is a local homeomorphism if for every x ∈ X there are some open sets x ∈ U ⊂ X and V ⊂ Y with f |U : U → V a homeomorphism. Exercise 3.14 (Easy exercise). Show covering maps are local homeo- morphisms. (The converse will be shown to hold for proper maps in Proposition 3.16.) Exercise 3.15. Show that a holomorphic between riemann surfaces is a biholomorphism. Proposition 3.16. If f : X → Y is a proper local homeomorphism it is a covering map.

−1 Proof. Choose y ∈ Y. By Lemma 3.10, f (y) is finite. Let x1,..., xn be the preimages. Since f is a local homeomorphism, we may take open sets xi ∈ Ui so that f : Ui → Vi is a homeomorphism, where Vi = f (Ui). By shrinking the Ui, we may also assume their pairwise −1 intersections are empty. Then, define V = ∩iVi. Note that f (V) ⊂ −1 ∪iUi, and f (V) ∩ Ui → V is a homeomorphism, as it is a restriction of the homeomorphism Ui → Vi. So, f is a covering map.  We can now prove the main result on constancy of the degree of a proper morphism. Proof of Proposition 3.8. It suffices to show d(q) is locally constant as X is connected. Given a fixed q, we can find open sets Ui with xi ∈ Ui −1 where {x1,..., xk} = f (q) and so that R( f ) ∩ Ui ⊂ {xi} (i.e., there are no ramification points in Ui, other than possibly xi). If there are no branch points among the xi, the result follows from Propo- sition 3.16. Using this, it suffices to show degree is locally constant on each Ui. That is, we have reduced to the case f : Ui → Y is given d by z 7→ z . In this case, we have d(q) = d at every point q ∈ Ui.  We conclude with one additional exercise, good for practice with the above concepts: 12 AARON LANDESMAN

Exercise 3.17. Let f : X → Y be a proper map of Riemann surfaces and q ∈ Y. Show that for U any neighborhood of f −1(q), there exists some neighborhood q ∈ V with f −1(V) ⊂ U. RIEMANN SURFACES 13

FIGURE 2. A picture of a proper map of Riemann surfaces

4. EXAMPLESOF PROPER MAPSOF RIEMANN SURFACES In this section, we’ll give some examples of proper maps of Rie- mann surfaces. To start, we show that away from the branch locus, a map of Rie- mann surfaces is a covering map. Let Y∗ := Y − B( f ) and X∗ = f −1(Y∗). Exercise 4.1 (Easy exercise). Verify that X∗ → Y∗ is proper, directly from the definition, using that X → Y is proper. Lemma 4.2. The map f : X∗ → Y∗ is a covering map. Proof. By removing all ramification points, we have ensured that f locally looks like z 7→ z, and hence is a proper local homeomor- phism. Therefore, by Proposition 3.16, it is a covering map.  Exercise 4.3. Show that a map of compact Riemann surfaces is proper. Hint: Use Lemma 3.3. Example 4.4. By the preceding exercise, every map f : Cˆ → Cˆ is proper. Consider the map f (z) = zd. Exercise 4.5. Show the map f (z) = zd has degree d and it has rami- fication points of order d at 0 and ∞, but nowhere else. Hint: Count the number of preimages of any given point. Then, R( f ) = {0, ∞} , B( f ) = {0, ∞} and away from 0 and ∞, the map is a covering map, by Lemma 4.2. ˆ ˆ 1 Example 4.6. The map f : C → C given by z 7→ z + z is a degree 2 proper map. The branch points of the map are {−2, 2}. To see why, note that if the map has a at x, then the polynomial 1 z + z = x has only a single solution in z. Multiplying by z, we see z2 − zx + 1 = 0 has a double root, which means it factors as (z − a)(z − a). Therefore, we must have a2 = 1, so a = ±1, in which case x = 2a = ±2. 14 AARON LANDESMAN

Exercise 4.7. Show that the rational map f : Cˆ → Cˆ given by z 7→ n 1 z + zn also has branch locus B( f ) = {2, −2} and is a map of degree 2n. Exercise 4.8. In this exercise, we aim to show that if a compact Rie- mann surface X has a with a single pole, then X ' Cˆ . (1) Show all degree one covering maps are . Hint: Show that a map is an if it defines an isomor- phism between a neighborhood of every point and its preim- age. (2) Show that any compact Riemann surface with a degree 1 map X → Cˆ (i.e., a rational function with a single pole) is an iso- morphism.

Exercise 4.9. Let E = C/Λ and let En = C/nΛ. Show that there is well defined map of Riemann surfaces En → E (given by sending a point x ∈ En thought of as an equivalence class of points x + (nΛ) ∈ C to the point x + Λ ∈ C/Λ = E. Show this map has degree n2 and has no ramification points. RIEMANN SURFACES 15

5. RIEMANN-HURWITZ In this section, we first state the Riemann Hurwitz theorem, then give some applications. A proof is given in the appendix, Appen- dix A. 5.1. Statement of Riemann-Hurwitz. To start, we state the surpris- ingly simple classification of compact Riemann surfaces: Theorem 5.1. Let X be a compact Riemann surface. Then, as a topological space, X is homeomorphic to either Cˆ or a g-holed torus for some g ≥ 1. This theorem is not all too difficult to prove, but it lies orthogonal to the purposes of this course, so we do not provide one. Using Theorem 5.1, we can define the genus of a compact Riemann surface.

Definition 5.2. Let X be a compact Riemann surface. Define the genus of X to be g if X is a g-holed torus, and 0 if X ' Cˆ . This accounts for every compact Riemann surface by Theorem 5.1. We denote the genus of X by g(X). We are now ready to state the Riemann-Hurwitz theorem, which, given a map between Riemann surfaces, yields a relation between the genera of the two surfaces and the ramification points of the map.

Theorem 5.3 (Riemann-Hurwitz). Let f : X → Y be a map of compact Riemann surfaces of degree d (see Definition 3.9). Let g(X) denote the genus of X and g(Y) denote the genus of Y. Then, 2g(X) − 2 = d · (2g(Y) − 2) + ∑ (multx( f ) − 1). x∈X Remark 5.4. The term 2g − 2 on both sides may seem strange, but it will appear naturally in the proof as the “,” to be defined in the course of the proof. 5.2. Applications. Before proving Riemann-Hurwitz, let’s see some applications. Example 5.5. Consider a map f : Cˆ → Cˆ of degree d (given by a degree d rational function). Suppose further that all branching is simple, meaning that for all x ∈ Cˆ , multx( f ) ≤ 2. We claim that f has 2d − 2 ramification points. Indeed, by Riemann-Hurwitz Theo- rem 5.3, we see ˆ ˆ 2g(C) − 2 = d(2g(C) − 2) + ∑ (multx( f ) − 1). x∈X 16 AARON LANDESMAN

Since g(Cˆ ) = 0, simplifying this, we obtain

−2 = d(−2) + ∑ (multx( f ) − 1). x∈X Hence,

2d − 2 = ∑ (multx( f ) − 1). x∈R( f )

So, if multx( f ) ≤ 2, we obtain that

∑ (multx( f ) − 1) = ∑ (multx( f ) − 1) x∈X x∈R( f ) = ∑ (2 − 1) x∈R( f ) = ∑ 1 x∈R( f ) = #R( f ), so f has 2d − 2 ramification points. Example 5.6. Suppose f : X → Y is a degree 2 map of compact Riemann surfaces where X has genus 1 and Y has genus 0. We will show that X → Y has 4 branch points. To start, because the map has degree 2, the multiplicity of any point is at most 2. Therefore, any point with multiplicity more than 1 will have multiplicity exactly 2. It follows from Theorem 5.3 that

0 = 2 · 1 − 2 = 2 · (2 · 0 − 2) + ∑ (multx( f ) − 1) x∈X = −4 + ∑ (mult( f , x) − 1) x∈R( f ) = −4 + ∑ (2 − 1) x∈R( f ) = −4 + ∑ 1 x∈R( f ) = −4 + #R( f )

So 4 = #R( f ), meaning f has precisely four ramification points. Since the map has degree 2, all the ramification points must lie over distinct points of Y, and so the map also has precisely four branch points. RIEMANN SURFACES 17

Exercise 5.7. Suppose f : X → Y is a degree 2 map of compact riemann surfaces where Y has genus 0 and X has genus g. Compute the number of branch points of f and the number of ramification points of f . Exercise 5.8. Suppose f : X → Y is a map between two of genus 1. Show that f has no ramification points. Exercise 5.9. Suppose f : X → X is a map of degree d > 1. (1) Show that either g(X) = 1 or g(X) = 0. (2) Show that if g(X) = 1 then R( f ) = ∅ and B( f ) = ∅. (3) If g(X) = 0, show that X has 2d − 2 ramification points, counted with multiplicity. Hint: Figure out rigorously what is meant by “counted with multiplicity.” The calculation is nearly completely carried out in Example 5.5. Exercise 5.10 (Tricky exercise). Let E denote the quotient of the torus C/Λ with Λ the lattice Z ⊕ i · Z. Since by i ( by 90 degrees under the identification C ' R2) sends Λ 7→ Λ, we can quotient C/Λ by the equivalence x ∼ ix and we obtain a map f : E → E/ ∼ (you may assume the latter exists as a Riemann surface, as is shown in Proposition B.1). Show that f has two ramification points of multiplicity 4 and two ramification points of multiplicity 2. Thinking of E as a quotient C → C/Λ ' E, what are the preimages of these ramification points in C? 18 AARON LANDESMAN

6. AUTOMORPHISMSOF RIEMANN SURFACES OF GENUS ≥ 2 In this section, we will begin by stating a bound on the number of automorphisms of a Riemann surface of genus at least 2. Then, we will prove the claimed bound. 6.1. Statement of the bound. Definition 6.1. An of a Riemann surface X is a map of Riemann surfaces f : X → X. Exercise 6.2. Verify that the collection of automorphisms of X form a group under composition, with the identity automorphism as the of the group. The collection of automorphisms with this group structure is called the automorphism group of X. The main result of this section is the following: Theorem 6.3. Let X be a Riemann surface of genus g ≥ 2. Then, the automorphism group of X has order at most 84(g − 1). This will be proved later starting in subsection 6.2 (the final proof occurring in subsection 6.8). Remark 6.4. Note that Theorem 5.3 emphatically does not hold in the case g = 0, 1. In fact, curves of genus 0 or 1 always have infinitely many automorphisms, as we show in section 7. 6.2. Proving the bound. We now are ready to prove the bound. We take as input the following surprisingly tricky theorem: Theorem 6.5. Let X be a Riemann surface of genus g ≥ 2. Then X has only finitely many automorphisms. As input we will also need that the quotient of a Riemann surface by a finite automorphism group is again a Riemann surface. Indeed, this is shown in Corollary B.3, in the appendix. So, we will assume X has only finitely many automorphisms, and use this to deduce the desired bound. Remark 6.6. The finiteness of automorphisms takes some more work to develop: The standard proof uses and the no- tion of the “automorphisms schemes.” The vague idea is to con- struct a parameter space of the automorphisms (“the Aut ”) which is itself a geometric object. One can then use something called “deformation theory” to show that this parameter space has a 0- dimensional space, so it is, in some sense discrete. If we knew it were compact and discrete, it would be finite. RIEMANN SURFACES 19

For the reader familiar with algebraic geometry, one can show compactness using the “canonical embedding” the automorphisms are a closed subscheme of PGLg(C), which is itself a finite type group scheme, and hence the automorphism scheme is also finite type, and 0-dimensional, hence a finite collection of points. Alternatively, one can argue this compactness by appealing to a nontrivial input from the theory of “Hilbert schemes” (with the Aut scheme being an example of a Hilbert scheme). The following lemma is crucial to our bound on the number of automorphisms. Lemma 6.7. Let X be a Riemann surface and G be the of the automorphism group of X generated by a single automorphism of X. Let Y be the quotient of X by G and let f : X → Y be the quotient map (see 0 Proposition B.1). Then, if f (x) = f (x ) we have multx( f ) = multx0 ( f ). Further, if y1,..., yn denote the branch points of f and ri denotes multx( f ) for f (x) = yi (independent of the choice of x by the above), we have ! n r − 1 2g(X) − 2 = |G| (2g(Y) − 2) + ∑ i .(6.1) i=1 ri

Proof. To show multx( f ) = multx0 ( f ) simply note that there is an automorphism h : X → X with h(x) = x0.

Exercise 6.8. Show that multx( f ) = multx0 ( f ) Hint: Use the defini- tion of multiplicity as the power n such that the map looks locally like z 7→ zn. Show that this n is preserved under automorphisms. To complete the proof, it suffices to prove 6.1. By the above, the total degree of ramification above a branch point b is − | −1( )| = ( − ) = ( − ) · | −1( )| deg π π b ∑q∈π−1(b) ri 1 ri 1 π b . Since, −1 −1 deg π = |G|, |G| = ri · |π (b)|. Therefore, |π (b)| = |G|/ri. If we let B( f ) denote the set of branch points with multiplicity, |G| the ramification points summed with multiplicity is ∑ (r − 1). b∈B rb b Then, applying Riemann Hurwitz, and replacing the sum over b ∈ B by a sum from 1 to n, we get ! n |G|(r − 1) n r − 1 2g − 2 = |G|(2g(Y) − 2) + ∑ i = |G| 2g(Y) − 2 + ∑ i . i=1 ri i=1 ri  We now continue with the proof of Theorem 6.3. Recall from The- orem 6.5 that for g(X) ≥ 2, the automorphism group G of X is finite. 20 AARON LANDESMAN

Hence, from Corollary B.3, the quotient of X by G is a Riemann sur- face. With notation as in the statement of Lemma 6.7, to maximize |G| we want to minimize n r − 1 2g(Y) − 2 + ∑ i .(6.2) i=1 ri First, note that the value 1/42 of the above expression is achieved when g(Y) = 0, n = 3 and r1 = 2, r2 = 3, r3 = 7. We now rule out this value being less than 1/42 via the following casework:

r − 6.3. We rule out g(Y) > 1. If g(Y) > 1, then since ∑ i 1 is positive, i ri if g(Y) > 1 then Equation 6.2 is at least 2g(Y) − 2 ≥ 2.

r − 6.4. We rule out g(Y) = 1. If g(Y) = 1, the result is then ∑n i 1 is i=1 ri 1 at least 2 assuming it is positive. r − For the rest of the problem, we are attempting to minimize ∑n i 1 − i=1 ri 2 subject to the condition it is positive. We assume g(Y) = 0. We know that n > 2, as if n = 2 then this quantity is negative.

6.5. We rule out g(Y) = 0, n ≥ 5. Suppose n ≥ 5. Since ri−1 > 1 as ri 2 r − r > 1, we obtain that ∑n i 1 > 5 , and so such a sum cannot be i i=1 ri 2 1 less than 2 + 42 . 6.6. We rule out g(Y) = 0, n = 4. Let us now rule out n = 4. Of r − course, we cannot have r = 1 for all i as then the sum ∑ i 1 − 2 = i 2 i ri r − 0. If three of the r = 1 and one of them is 2 then ∑ i 1 − 2 = 1 , i 2 3 i ri 6 and all other values of the ri will yield strictly greater results, so we 1 cannot make a sum less than 42 with n = 4. 6.7. We rule out g(C0) = 0, n = 3. Finally, we come to the tricki- est case of n = 3. The problem is equivalent to maximizing 1/r1 + 1/r2 + 1/r3 subject to the condition that this sum is less than 1. We now divide this into further subcases: 0 6.7.1. We rule out g(C ) = 0, n = 3 no ri = 2. Suppose none of the ri are 2. Note that we cannot have r1 = r2 = r3 = 3 since then 1 1/r1 + 1/r2 + 1/r3 = 1. Therefore, not all of them are 3 , so maximum possible value of 1/r1 + 1/r2 + 1/r3 is attained when r1 = 3, r2 = 11 41 3, r3 = 4, in which case 1/r1 + 1/r2 + 1/r3 = 12 < 42 . This rules out the case that no ri = 2. For the rest of the proof, we may assume r1 = 2. RIEMANN SURFACES 21

0 6.7.2. We rule out g(C ) = 0, n = 3, r1 = 2, and r2, r3 > 3. In the case r1 = 2 and r2, r3 > 3, we cannot have both r2 = r3 = 4. Hence, the maximum possible value of 1/r1 + 1/r2 + 1/r3 is attained when 19 41 r2 = 4, r3 = 5. In this case, 1/r1 + 1/r2 + 1/r3 = 20 < 42 . 0 6.7.3. We deal with the case g(C ) = 0, n = 3, r1 = 2, and r2 = 3. Hence, we may assume r1 = 2, r2 = 3. Then, by simply testing values for r3 up to 7, we see that if 1/r1 + 1/r2 + 1/r3 < 1, we must have r3 ≥ 7. The value maximizing the sum is then r3 = 7. 6.8. Finally we can put the above together to complete the proof of Theorem 6.5: Proof of Theorem 6.3. Let α denote the expression in Equation 6.2. Then, 1 note that α|G| = 2g − 2 and α ≤ 42 by the above casework. Hence, |G| ≤ 42(2g − 2) = 84(g − 1).  22 AARON LANDESMAN

7. AUTOMORPHISMSINLOWGENUS 0 AND 1 In this subsection, we classify all possible automorphisms of genus 0 and genus 1 curves, and give an example of a genus 3 with a large, but finite, automorphism group. Many of the exercises assume background beyond the prerequisites for this course, so feel free to skip them if you are not do not have such background. 7.1. Genus 0. Example 7.1 (Genus 0 automorphisms). Consider the rational func- ˆ ˆ az+b tion f : C → C given by z 7→ cz+d for a, b, c, d ∈ C with ad 6= bc. Then, f is a degree 1 map of Riemann surfaces so it is an iso- morphism by Exercise 4.8. Therefore, C has infinitely many auto- morphisms. In fact, in a sense that can be made precise, there is a “3-dimensional” space of automorphisms (it is 3 and not 4 because if you multiply a, b, c, d by a given scalar, the resulting map is the same). Exercise 7.2 (Tricky exercise, assuming knowledge of removable sin- gularities). Show that the automorphisms of Cˆ given in Example 7.1 are all of the automorphisms of Cˆ , in the following steps: (1) Show that the only automorphisms f : C → C are of the form x 7→ ax + b for a, b ∈ C with a 6= 0. Hint: After a , you can assume f (0) = 0. Consider the map g(z) := f (z)/z (defined away from z = 0). Show that 0 is a removable sin- gularity of g and so g(z) extends to a map C → C. Show that there is no z with g(z) = 0. Conclude that g is constant by Corollary 3.5. (2) Show that the only automorphisms f : Cˆ → Cˆ are those given in Example 7.1. Hint: If you have any automorphism f , show that you can compose it with one of Example 7.1 to assume f (∞) = ∞. Then use the first part. 7.2. Genus 1. Example 7.3. Let E = C/Λ be a genus 1 Riemann surface. Show that for any two points x, y ∈ E, there is an automorphism f : E → E with f (x) = y. Conclude that E has infinitely many automorphisms. Hint: Consider the maps on E induced from the maps C → C given by z 7→ z + a for a ∈ C. Exercise 7.4 (Tricky exercise, assuming an understanding of cover- ing spaces and algebraic number theory relating to roots of unity). Suppose E = C/Λ is a genus 1 riemann surface. Let e ∈ E denote RIEMANN SURFACES 23 the image of 0 ∈ C. In this exercise, we classify the automorphisms of E fixing e. (1) Show that any automorphism of E fixing e is induced by a unique automorphism of C sending 0 7→ 0. Hint: Use that C is the universal cover of E and path lifting lemmas (this is where the understanding of covering spaces is necessary). (2) Show that any automorphism E → E sending e 7→ e is in- duced by a unique automorphism C → C sending Λ 7→ Λ and further sending 0 7→ 0. (3) Show that if Λ, up to scaling, is equal to Z ⊕ iZ ⊂ C then Λ has automorphism group Z/4 generated by z 7→ iz. Hint: Use that the automorphisms of C fixing 0 are precisely those of the form z 7→ az for a ∈ C − {0}, by the first part of Exer- cise 7.2. (4) Show that if Λ, up to scaling, is equal to Z ⊕ ωZ ⊂ C, for ω a primitive of unity then Λ has automorphism group Z/6Z generated by z 7→ −ωz. (5) If Λ, up to scaling, is neither Z ⊕ iZ nor Z ⊕ ωZ for ω a prim- itive cube root of unity then the only automorphisms of Λ are z 7→ −z. Hint: Argue that the lattice is of the form Z ⊕ ζZ for ζ some root of unity. Show that ζ2 = aζ + b. Show that a root of unity satisfies a degree 2 polynomial if and only if it is a 2nd root of unity, a 4th root of unity, or a 6th root of unity. (For this you may need some knowledge of algebraic number theory. For example, if you are familiar with cyclotomic poly- nomials, the cyclotomic polynomial is the polynomial of min- imal degree which the root of unity satisfies, and the degree of the cyclotomic polynomial is φ(n) for φ the Euler totient function.) (6) Conclude that the automorphism group of E which fixes e is either Z/4, Z/6, or Z/2, depending on whether Λ, up to scaling, is equal to Z ⊕ iZ, Z ⊕ ωZ, or something else. (7) Combine the previous parts of this exercise with Example 7.3 to deduce a complete description of all automorphisms of E.

7.3. Example in Genus 3.

Example 7.5 (Strenuous exercise, assuming knowledge of ). This exercise assumes a familiarity with projective space. If you have not seen it, skip this exercise. In this exercise, we com- pute the automorphism group of the “,” C, which is the vanishing locus in P2 of the equation x3y + y3z + z3x = 0. 24 AARON LANDESMAN

(1) Show that C has at least 168 automorphisms. Hint: Consider automorphisms given by permuting the coordinates and those multiplying coordinates by seventh roots of unity. Find an au- tomorphism of order 7, and automorphism of order 3. Show that it has an subgroup of automorphisms of order 8. (2) Assuming C has genus 3, show that these are all the auto- morphisms using Theorem 6.3. The rest of the exercise is con- cerned with showing C has genus 3. (3) Show that the Klein quartic is a Riemann surface. Hint: Write down explicit charts via algebraic and conclude they are holomorphic. (4) Construct a map f : C → Cˆ given by sending [x : y : z] 7→ [x : y] when z 6= 0. Although this map is not defined when z = 0, show it extends over the set z = 0. (5) Show that the map f of the previous part has degree 3 and compute its ramification points. Hint: The ramification points will exactly be the fixed points of the automorphisms found in the first part. (6) Use Riemann-Hurwitz to show g(C) = 3. RIEMANN SURFACES 25

APPENDIX A. PROOFOF RIEMANN HURWITZ In this appendix, we now provide a proof of the Riemann-Hurwitz theorem, Theorem 5.3. The key to the proof will be the notion of the Euler characteristic of a surface, which is a topological notion. Definition A.1. Let X be a compact Riemann surface of genus g. Then, the Euler characteristic of X, denoted χ(X) is 2 − 2g. The key result to proving Riemann-Hurwitz is that the Euler char- acteristic can be computed in terms of triangulations of X: Definition A.2. Let X be a compact Riemann surface. A triangula- tion T of X is a collection of continuous maps, ti : Ti → X for Ti from triangles in C onto X so that (1) the ti are homeomorphisms onto their image (2) the union of the images of all ti cover X (3) the intersection of any two triangles is either an edge or a point. Exercise A.3. Show that any of a compact Riemann surface has only finitely many triangles. Hint: Take an open cover by slight enlargements of the triangles and use compactness. We will need the following. Proposition A.4. Let T be any triangulation of a compact Riemann sur- face X. Let eT be the number of edges in the image of the maps comprising T, fT be the number of faces, and vT be the number of vertices (so that if two triangles share an edge or a vertex, that edge or vertex is only counted once). Then, χ(X) = fT − eT + vT (which is finite by Exercise A.3). We will come back and prove Proposition A.4 soon, but first, let us use it to deduce Theorem 5.3. Proof of Theorem 5.3 assuming Proposition A.4. We need to show 2g(X) − 2 = d · (2g(Y) − 2) + ∑ (mult( f , x) − 1). x∈X Instead, we will show its negation 2 − 2g(X) = d · (2 − 2g(Y)) − ∑ (mult( f , x) − 1). x∈X Recall that χ(X) = 2 − 2g(X), by definition. Therefore, to conclude the proof, we only need to show d · (2 − 2g(Y)) − ∑x∈X(mult( f , x) − 1) = χ(X). To show this, by Proposition A.4, it suffices to con- struct a particular triangulation T of X so that fT − eT + vT = d · (2 − 2g(Y)) − ∑x∈X(mult( f , x) − 1). 26 AARON LANDESMAN

We will construct a suitable triangulation of X as follows: First, take a triangulation T0 of Y so that every branch point is the image of a vertex of a triangle from the triangulation. By Lemma 4.2, we know that away from the branch points, the map X → Y is a cov- ering map. Therefore, by taking a sufficiently fine refinement of T0, we may further assume that away from the ramification locus of X, the preimage of any triangle is a disjoint union of copies of that tri- . This yields a corresponding triangulation of X, call it T. We now compute the Euler characteristic of X in terms of T. First, by construction eT = d · eT0 and fT = d · fT0 . It only remains to analyze what happens at any given vertex. If there is no ramification what- soever, we would then have vT = d · vT0 as well. However, if the multx( f ) = n, there will then only be a single preimage of f (x) at x, instead of the n preimages we “expected.” Summing this over all ramification points, we have that the number of vertices in T is

vT = d · vT0 − ∑ (multx( f ) − 1). x∈R( f ) This is exactly the “correction term” we were looking for from the statement of Riemann-Hurwitz. The rest is now formal. We have

χ(X) = fT − eT + vT = d fT0 − deT0 + dvT0 − ∑ (multx( f ) − 1) x∈R( f )

= d( fT0 − eT0 + vT0 ) − ∑ (multx( f ) − 1) x∈R( f ) = d(χ(Y)) − ∑ (multx( f ) − 1). x∈R( f ) as we wanted to show.  So, to conclude the proof of Theorem 5.3, we only need to prove Proposition A.4. For notational purposes of proving Proposition A.4, let χT(X) be fT − eT + vT, for T a triangulation of X. In order to prove Proposition A.4, we will use the following lemma: Lemma A.5. If T and T0 are two triangulations of a compact Riemann surface X, then χT(X) = χT0 (X). Let’s now see why Proposition A.4 follows from Lemma A.5. Proof of Proposition A.4 assuming Lemma A.5. By Lemma A.5, any two 0 triangulations T and T have the same value χT(X). So, to show RIEMANN SURFACES 27

0 χT0 (X) = 2 − 2g for an arbitrary choice of T , it suffices to show χT(X) = 2 − 2g for a single choice of T. In this case, we may take any subdivision we like, and so the result follows from the next exercise.

Exercise A.6. Find a particular triangulation T of a genus g Riemann surface X with χT(X) = 2 − 2g. Hint: First do this explicitly for the sphere, then explicitly for the torus, and then explicitly for the torus with either 1 or 2 nonadjacent faces of triangles removed. Express the Euler characteristic of X in terms of the Euler characteristic of the above objects.

 To conclude our proof of Theorem 5.3, we only need to prove Lemma A.5. This in turn will follow from the following lemma: Lemma A.7. Let X ⊂ R2 be a single triangle and let T be a triangulation of X. Then, for any such T, χT(X) = 1. Again, we will now prove Lemma A.5 assuming Lemma A.7. Proof. Let T and T0 be two separate triangulations of X. Exercise A.8. Verify that one can construct a triangulation T00 of X so 00 00 00 that the image of every triangle ti : Ti → X in T is contained both 0 0 in the image of some triangle ti of T and the image of some triangle 00 0 ti of T. Such a T is called a simultaneous refinement of T and T . Hint: Intersect the images of T and T0 and triangulate the resulting intersections. Let T00 be a simultaneous refinement of T and T0. If we show χT(X) = χT00 (X) and χT00 (X) = χT(X), we will be done. Therefore, by relabeling T00 by T0, we can reduce to the case that T0 is a refine- ment of T. That is, it suffices to prove Lemma A.5 in the special case that T0 is a refinement of T.  ( ) Exercise A.9. Let tj Tj j∈I denote the collection of the images of the triangles of T in X. For a particular j ∈ I, let (T0)j denote the 0 triangulation of tj(Tj) composed of those triangles ti whose image is contained in tj(Tj). Use Lemma A.7 to show that χT(X) = χT0 (X) 0 in the case T is a refinement of T by writing χT0 (X) in terms of the χ(T0)j (ti(Ti)).  28 AARON LANDESMAN

Finally, to conclude our proof of Theorem 5.3, it suffices to prove Lemma A.7. Proof of Lemma A.7. Let X be a triangle. The construction of a trian- gulation of X can be viewed as a sequence of moves, where at each step, we either subdivide an edge into two edges by placing a vertex in the middle, or subdivide a face into two faces, by placing an edge in the middle. We claim that in the course of this process, there Eu- ler characteristic remains fixed. Indeed, when we add a vertex, the number of edges goes up by 1, but so does the number of vertices, so f − e + v stays constant. Similarly, when we add an edge, the num- ber of edges goes up by 1, but so does the number of faces, and so again f − e + v stays constant. Exercise A.10. Make the details of the above precise.  This completes our proof of Theorem 5.3. RIEMANN SURFACES 29

APPENDIX B. QUOTIENTSOF RIEMANN SURFACES BY AUTOMORPHISMS The main goal of this section is to prove the following: Proposition B.1. Let X be a Riemann surface and h : X → X an auto- morphism of finite order (meaning hn = id for some n ∈ Z). Then, there is a Riemann surface Y and a map of Riemann surfaces f : X → Y so that for any x, x0 ∈ X with f (x) = f (x0) there is a unique m with hmx = x0. The ramification points are precisely the points fixed by hk for some k < n. Remark B.2. Alternatively put, Proposition B.1 says that Y is the quotient of X by the equivalence relation x ∼ x0 if x = hmx0 for some m. Said yet another way, the group G acts simply transitively on the fibers of f . (If you haven’t heard these terms before, just ignore this remark it; this remark is an immediate rephrasing of the proposition statement if you are familiar with the words). Corollary B.3. Let G be a finite group of automorphisms of some Riemann surface X. Then, there is a Riemann surface Y and a map f : X → Y so that for any x, x0 ∈ X with f (x) = f (x0) there is a unique h ∈ G with hx = x0. This Y is called the quotient of X by G and is denoted X/G. The ramification points are precisely the points fixed by some automorphism h ∈ G with h 6= id. Proof. Exercise B.4. Deduce this from Proposition B.1 by choosing a non- identity element in G, constructing the quotient by the subgroup generated by that element, and then repeating.  For the purposes of the proof, we will need discreteness of the fixed points, established in the next exercise. Exercise B.5. Let h : X → X be an automorphism of order n. Show that the set of points x ∈ X with hkx = x for some k < n is discrete. We are now ready to prove Proposition B.1.

Proof of Proposition B.1. We construct Y in three steps. First as a set, then as a topological space, and finally as a Riemann surface. To construct Y as a set, simply define the set of points as points of X modulo the relation ∼ that x ∼ x0 if hkx = x0 for some k. Note there is a map of sets f : X → Y sending a point x to its equivalence class. 30 AARON LANDESMAN

Next, we give Y the structure of a topological space. We give Y the quotient topology meaning that the open sets U ⊂ Y are precisely the sets such that f −1(U) is open in X. To conclude, we only need to give Y the structure of a Riemann surface. Indeed, this is the trickiest part, and will require some care. We will do this by defining the structure of a Riemann surface away from finitely many points, and then extending it over those points. To this end, let D ⊂ X denote the finite set of points fixed by hk for k < n (discrete by Exercise B.5). Let X∗ := X − D. Let Y∗ := f (X − D). We will now define the structure of a Riemann surface on Y∗. Re- ∗ ∗ stricting f , we have a map f |X∗ : X → Y . Exercise B.6. For every point x ∈ X∗, show there is an open set U ⊂ X∗ with hkU ∩ U = ∅ for every k < n. Hint: Choose any open containing x, translate it, and remove the intersections. By Exercise B.6, for every x ∈ X∗ there is an open set U ⊂ X∗ not intersecting any translate of U. Therefore, f maps U homeomor- phically (hence biholomorphically) to its image under f . Then, after possibly shrinking U, we may assume it is biholomorphic to an open subset V ⊂ C. So, the map V ' U ' f (U) defines a chart around f (x). Exercise B.7. Verify that the above charts defined above give Y∗ the structure of a Riemann surface. Hint: You have to check that the “transition maps” are biholomorphic. To check this, use that these maps are biholomorphic on X∗. To conclude the proof, we have given Y∗ the structure of a Rie- mann surface, and we only need to extend this to the finitely many point Y − Y∗. Since these points are discrete, it suffices to choose one point y ∈ Y − Y∗. Take x ∈ X with f (x) = y. Suppose the subgroup of G which stabilizes x has size m (i.e., the powers of h fixing x are precisely those of the form ht·n/m for t ∈ Z).

Exercise B.8. Show that one can choose a suitably small Ux con- nected open neighborhood of x so that, after choosing charts, the n/m map h looks like z 7→ ζmz for ζm a primitive mth root of unity, and further that hn/m(U) = U.

Under the identification of Ux of Exercise B.8 with an open subset of C, let Uy ⊂ C denote the open subset of C which is the image of m Ux under the map C → C given by z 7→ z . RIEMANN SURFACES 31

Exercise B.9 (Tricky exercise). Show that Uy is defines a chart for y and gives Y the structure of a riemann surface. Exercise B.10. Show that the ramification points of the resulting map f are precisely those x ∈ X − X∗. We have now extended the structure from Y∗ to Y∗ ∪ {y}, show- ing that the preimages of y are branch points, and extending this structure in the same way over the finitely many remaining points concludes the proof. 

REFERENCES [McM] C McMullen. Complex analysis on riemann surfaces. Harvard University Lecture Notes. http://www.math.harvard.edu/~ctm/home/text/class/ harvard/213b/13/html/home/course/course.pdf.