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RIEMANN SURFACES 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. Automorphisms of Riemann Surfaces of genus 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(C 0 )= 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

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RIEMANN SURFACES

AARON LANDESMAN

CONTENTS

1. Introduction 22. Maps of Riemann Surfaces 42.1. Defining the maps 42.2. The multiplicity of a map 42.3. Ramification Loci of maps 62.4. Applications 63. Properness 93.1. Definition of properness 93.2. Basic properties of proper morphisms 93.3. Constancy of degree of a map 104. Examples of Proper Maps of Riemann Surfaces 135. Riemann-Hurwitz 155.1. Statement of Riemann-Hurwitz 155.2. Applications 156. Automorphisms of Riemann Surfaces of genus ≥ 2 186.1. Statement of the bound 186.2. Proving the bound 186.3. We rule out g(Y) > 1 206.4. We rule out g(Y) = 1 206.5. We rule out g(Y) = 0, n ≥ 5 206.6. We rule out g(Y) = 0, n = 4 206.7. We rule out g(C′) = 0, n = 3 206.8. 217. Automorphisms in low genus 0 and 1 227.1. Genus 0 227.2. Genus 1 227.3. Example in Genus 3 23Appendix A. Proof of Riemann Hurwitz 25Appendix B. Quotients of Riemann surfaces by

automorphisms 29References 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 manyareas of math. In this way they connect seemingly disjoint fields,and also allow one to use tools from different areas of math to studythem. They are crucial objects of interest in algebraic geometry, num-ber theory, symplectic geometry, dynamics, and complex analysis,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 surface 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 sphere, C

(4) C/Λ for some lattice Λ(5) The g-holed torus(6) H := {x ∈ C : im x > 0} , the upper half plane(7) H/Γ where H is the upper half plane, and Γ is some torsion

free discrete group in SL2(R)

Remark 1.2. It is a nontrivial theorem (the uniformization theorem)that the above Riemann surfaces are an exhaustive list (though theabove list does contain some repetitions).

We’re ready to define Riemann surfaces. If you know about man-ifolds, a Riemann surface is just a 1-dimensional complex manifoldwith complex holomorphic transition functions.

Remark 1.3. One way to define a (compact) Riemann surface is as aconnected subset X ⊂ Cn such that for every point x ∈ X there isa biholomorphism (a holomorphic map with holomorphic inverse)between X and an open subset of C. However, that definition is notso great, because it requires that we specify an embedding X → Cn

for some n. It would be better to have an intrinsic definition, notdepending on an embedding.

Recall that a topological space X is Hausdorff if for any two pointsx, y ∈ X there are open sets U, V with x ∈ U, y ∈ V so that U ∩V =∅. A topological space is second countable if there is a countablecollection 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 open set is open. A homeomorphism is a continuous mapbetween two topological spaces 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 ⊂ Xand maps fi : Ui → C so that

(1) fi is a homeomorphism onto an open subset of C

(2) The transition functions f j ◦ f−1i | fi(Ui∩Uj)

: fi(Ui ∩ Uj) →f j(Ui ∩Uj) are complex analytic (i.e., holomorphic).

A collection of (Ui, fi)i∈I as above is called an atlas for X. A particu-lar (Ui, fi) is called a chart.

Remark 1.5. A Riemann surface is the datum of the topological spaceX together with the atlas (Ui, fi)i∈I . In particular, it is possible tohave two riemann surfaces with the same underlying topologicalspaces but different atlases.

Exercise 1.6. Show that C, the Riemann sphere (the one point com-pactification of C, whose underlying set is C∪ {∞} and whose opensets are either opens in C or sets whose complements are compactsets in C) is a Riemann surface. Hint: Cover it by two charts C− {0}and C−∞, with transition function given by z 7→ 1

z .

Exercise 1.7. Let E = C/Λ for Λ ⊂ C a lattice of rank 2, meaningΛ = Z⊕ aZ for a /∈ R. Show that E is a Riemann surface. Hint: Pickcharts for E as charts coming from open subsets U ⊂ C mappinghomeomorphically to E under the projection C → E. (Why is C

itself not a chart for E?)

4 AARON LANDESMAN

2. MAPS OF 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 saywhat 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 anatlas consisting of the single chart f1 : U1 → V1 and U2 has an atlasconsisting of the single chart f2 : U2 → V2, with Vi ⊂ C open. Thenwe say a map g : U1 → U2 is holomorphic if the map f2 ◦ g ◦ f−1

1 :V1 → V2

(2.1)U1 U2

V1 V2

g

f2

f2◦g◦ f−11

f−11

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 Xand (Vj, hj) are an atlas for Y, then the resulting restriction of f

f−1(Vj) ∩Uif−→ Vj

is holomorphic for all i, j.

Loosely speaking, a map of riemann surfaces is just a map whichis holomorphic when restricted to each of the open sets.

2.2. The multiplicity of a map. We next come to the notion of themultiplicity of a map of riemann surfaces. Roughly speaking, this isjust the number of preimages of “most points.” The following lemmamakes this precise, and is essential in what follows.

Lemma 2.4 (Key Lemma). For any map of Riemann surfaces g : X → Y,with f (p) = q there exists open sets U 3 p, V 3 q and charts f1 : U′ →U, f2 : V′ → V with h := f−1

2 ◦ g ◦ f1 : U′ → V′ given by h(z) = 0 orh(z) = zn for some n.

Proof. To start, choose charts f ′1 : U′′ → U and f2 : V′ → V withU′′ ⊂ C and V′ ⊂ C open sets containing 0 ∈ C with f ′1(0) = pand f2(0) = q. (Primes have nothing to do with derivatives here.)Define h′ := f−1

2 ◦ g ◦ f ′1. By assumption, h′ is holomorphic, so after

RIEMANN SURFACES 5

(2.2)

X Y

U V

U′′ V′

U′

g

g|U

α

f ′1h′

f2

f1

h

FIGURE 1. Diagram illustrating the proof of Lemma 2.4.

shrinking U′′, U, V′, V we can assume that h′ agrees with its Taylorexpansion about 0. If h′ is constant, we can take h = h′ and we aredone.

So, we assume h′ is not constant. Then, we can write h′(z) = znt(z)with t(0) 6= 0. After possibly shrinking our open sets further, wemay assume the power series t(z) has an nth root, say r(z)n = t(z),so h′(z) = (z · r(z))n.

Define a map α : U′′ → C sending z 7→ z · r(z) and let U′ ⊂ C

denote the image 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 forα, and hence, after further shrinking the open sets U, U′, U′′, V, V′,we may assume that α : U′′ → U′ is a biholomorphism.

Then, define f1 := α−1 ◦ f1 and h := h′ ◦ α−1. Since (h ◦ α)(z) =h′(z) = (z · r(z))n, and α(z) = z · r(z), we have h(z) = zn. �

Definition 2.5. For f : X → Y define the multiplicity of f at x to be

multx( f ) =

{d if f locally at x looks like z 7→ zd via Lemma 2.4∞ if f is constant in some neighborhood of x

Lemma 2.6. If f : X → Y is a map of Riemann surfaces, either f isconstant or multx( f ) < ∞ for all x ∈ X.

Proof. Note that the set of points where multx( f ) < ∞ is open andthe set where multx( f ) = ∞ is open, using Lemma 2.4. Since X isconnected, 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 ofbranch loci and ramification loci. These are just the places at whichthe 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 ford > 1. The branch locus, notated B( f ), is f (R( f )), the image of theramification 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 mappingtheorem, Louiville’s theorem, and the fundamental theorem of alge-bra.

Proposition 2.10 (Open Mapping Theorem). Any map f : X → Y isopen (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 setsU and V with x ∈ U ⊂ X, V ⊂ Y and f (U) ⊂ V, so that the mapf |U : U → V is open.

By Exercise 2.11, we can replace X and Y by sufficiently small opensubsets around each point. Hence, by Lemma 2.4, it suffices to checkfor 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 themap is of the form z 7→ zd, and the map is discrete when restrictedto such an open set. �

Exercise 2.14. For f : X → Y a map of Riemann surfaces, show thatR( 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 hasa finite subcover. If you haven’t seen compactness before, now is agood time to do the following exercises:

Exercise 2.15. Let X be a compact topological space. Show that anyclosed subspace Y ⊂ X is also compact.

Exercise 2.16. Suppose X is a compact Hausdorff space. Show thatif Y ⊂ X is compact (with the induced topology), it is also closed.Hint: Show that around each point x /∈ Y there is some open setUx 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 thatVxy ∩Uxy = ∅ with x ∈ Uxy, y ∈ Vxy. For a fixed x, use compactnessto produce a finite collection of such Vxy covering Y, and take Ux tobe 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 Riemannsurface 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 bothopen 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. Forthe 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 hencecompact. �

Exercise 2.19 (Generalization of Liouville’s Theorem). Prove the fol-lowing generalization of Liouville’s Theorem: Any analytic functionf : X → C for X compact is constant. In particular, taking X = C,we obtain Liouville’s theorem, that any bounded entire function isconstant. Hint: Use Lemma 2.18.

Corollary 2.20 (Fundamental theorem of algebra). Every nonconstantpolynomial over the complex numbers has a root.

Proof. Consider a polynomial as a function C → C (by extendingnonconstant polynomials 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 isa root of f . �

RIEMANN SURFACES 9

3. PROPERNESS

We next introduce the notion of properness. We have seen thatmaps of Riemann surfaces are always open in Proposition 2.10. Morallyspeaking properness means that the map is also closed. Technicallyspeaking, being proper is a bit stronger than being closed, but weshall see that proper maps are closed in Lemma 3.4.

3.1. Definition of properness. Here is the definition of what it meansfor a map to be proper.

Definition 3.1. A map f : X → Y is proper if for all compact K ⊂ Ywe 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 derivea 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 itis closed by Exercise 2.15 and Exercise 2.16. Then, the preimage of aclosed set is closed by continuity of f . �

For the next result, we recall a standard result from topology thatclosed bounded sets in Rn are compact. In particular, closed ballsare 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 isclosed.

Proof. If E is a closed set, then we wish to show f (E) is closed. Takey /∈ f (E). We wish to produce an open set around y not intersectingf (E). Take some small closed ball y ∈ V.

(3.1)E X f−1(V)

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 compactsubset of a compact set, hence closed. Since f (E ∩ f−1(V)) does notcontain y, we have int(V) − f (E ∩ f−1(V)) is open, which is ourdesired open neighborhood of y not intersecting f (E). �

10 AARON LANDESMAN

Corollary 3.5. If f : X → Y is a proper (nonconstant) map of Riemannsurfaces, it is surjective.

Proof. f (X) is open by the open mapping theorem, and closed byLemma 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 adiscrete 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 ) arediscrete.

3.3. Constancy of degree of a map. The next step to understandingmaps of Riemann surfaces is to define the notion of the degree of amap. 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 end of this section. There are two mainideas involved in the proof. First, we need to show the preimage ofany point only contains finitely many points, which we easily do inLemma 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 showingthat f is then a covering map away from the ramification points.Once we have both of these, we can analyze the map locally andconclude 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 discreteby Lemma 2.4.

Exercise 3.11. Conclude the proof by showing a compact discrete setis finite.

RIEMANN SURFACES 11

Our next goal is to show proper maps are covering maps, awayfrom their ramification points. To this end, we now define coveringmaps.

Definition 3.12. A map of topological spaces X → Y is a coveringmap if for each y ∈ Y we can find y ∈ V with f−1(V) homeomorphicto a disjoint union of open sets, each homeomorphic to V.

Definition 3.13. A map f : X → Y is a local homeomorphism if forevery x ∈ X there are some open sets x ∈ U ⊂ X and V ⊂ Y withf |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 inProposition 3.16.)

Exercise 3.15. Show that a holomorphic homeomorphisms betweenriemann surfaces is a biholomorphism.

Proposition 3.16. If f : X → Y is a proper local homeomorphism it is acovering map.

Proof. Choose y ∈ Y. By Lemma 3.10, f−1(y) is finite. Let x1, . . . , xnbe the preimages. Since f is a local homeomorphism, we may takeopen sets xi ∈ Ui so that f : Ui → Vi is a homeomorphism, whereVi = f (Ui). By shrinking the Ui, we may also assume their pairwiseintersections are empty. Then, define V = ∩iVi. Note that f−1(V) ⊂∪iUi, and f−1(V)∩Ui → V is a homeomorphism, as it is a restrictionof the homeomorphism Ui → Vi. So, f is a covering map. �

We can now prove the main result on constancy of the degree of aproper morphism.

Proof of Proposition 3.8. It suffices to show d(q) is locally constant asX is connected. Given a fixed q, we can find open sets Ui with xi ∈ Uiwhere {x1, . . . , xk} = f−1(q) and so that R( f ) ∩Ui ⊂ {xi} (i.e., thereare no ramification points in Ui, other than possibly xi). If thereare no branch points among the xi, the result follows from Propo-sition 3.16. Using this, it suffices to show degree is locally constanton each Ui. That is, we have reduced to the case f : Ui → Y is givenby z 7→ zd. In this case, we have d(q) = d at every point q ∈ Ui. �

We conclude with one additional exercise, good for practice withthe above concepts:

12 AARON LANDESMAN

Exercise 3.17. Let f : X → Y be a proper map of Riemann surfacesand q ∈ Y. Show that for U any neighborhood of f−1(q), there existssome neighborhood q ∈ V with f−1(V) ⊂ U.

RIEMANN SURFACES 13

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

4. EXAMPLES OF PROPER MAPS OF 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, directlyfrom 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 flocally 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 : C → C isproper. 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: Countthe number of preimages of any given point.

Then, R( f ) = {0, ∞} , B( f ) = {0, ∞} and away from 0 and ∞, themap is a covering map, by Lemma 4.2.

Example 4.6. The map f : C → C given by z 7→ z + 1z 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 branch point at x, then the polynomialz + 1

z = x has only a single solution in z. Multiplying by z, we seez2 − 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 casex = 2a = ±2.

14 AARON LANDESMAN

Exercise 4.7. Show that the rational map f : C → C given by z 7→zn + 1

zn also has branch locus B( f ) = {2,−2} and is a map of degree2n.

Exercise 4.8. In this exercise, we aim to show that if a compact Rie-mann surface X has a rational function with a single pole, then X 'C.

(1) Show all degree one covering maps are isomorphisms. Hint:Show that a map is an isomorphism 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 mapX → C (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 iswell defined map of Riemann surfaces En → E (given by sending apoint 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 andhas no ramification points.

RIEMANN SURFACES 15

5. RIEMANN-HURWITZ

In this section, we first state the Riemann Hurwitz theorem, thengive 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 topologicalspace, X is homeomorphic to either C or a g-holed torus for some g ≥ 1.

This theorem is not all too difficult to prove, but it lies orthogonalto the purposes of this course, so we do not provide one. UsingTheorem 5.1, we can define the genus of a compact Riemann surface.

Definition 5.2. Let X be a compact Riemann surface. Define thegenus of X to be g if X is a g-holed torus, and 0 if X ' C. Thisaccounts for every compact Riemann surface by Theorem 5.1. Wedenote 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 betweenthe 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 compactRiemann surfaces of degree d (see Definition 3.9). Let g(X) denote thegenus of X and g(Y) denote the genus of Y. Then,

2g(X)− 2 = d · (2g(Y)− 2) + ∑x∈X

(multx( f )− 1).

Remark 5.4. The term 2g− 2 on both sides may seem strange, but itwill appear naturally in the proof as the “Euler characteristic,” to bedefined in the course of the proof.

5.2. Applications. Before proving Riemann-Hurwitz, let’s see someapplications.

Example 5.5. Consider a map f : C → C of degree d (given by adegree d rational function). Suppose further that all branching issimple, meaning that for all x ∈ C, multx( f ) ≤ 2. We claim that fhas 2d− 2 ramification points. Indeed, by Riemann-Hurwitz Theo-rem 5.3, we see

2g(C)− 2 = d(2g(C)− 2) + ∑x∈X

(multx( f )− 1).

16 AARON LANDESMAN

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

−2 = d(−2) + ∑x∈X

(multx( f )− 1).

Hence,

2d− 2 = ∑x∈R( f )

(multx( f )− 1).

So, if multx( f ) ≤ 2, we obtain that

∑x∈X

(multx( f )− 1) = ∑x∈R( f )

(multx( f )− 1)

= ∑x∈R( f )

(2− 1)

= ∑x∈R( f )

1

= #R( f ),

so f has 2d− 2 ramification points.

Example 5.6. Suppose f : X → Y is a degree 2 map of compactRiemann surfaces where X has genus 1 and Y has genus 0. We willshow that X → Y has 4 branch points. To start, because the map hasdegree 2, the multiplicity of any point is at most 2. Therefore, anypoint 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) + ∑x∈X

(multx( f )− 1)

= −4 + ∑x∈R( f )

(mult( f , x)− 1)

= −4 + ∑x∈R( f )

(2− 1)

= −4 + ∑x∈R( f )

1

= −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 overdistinct points of Y, and so the map also has precisely four branchpoints.

RIEMANN SURFACES 17

Exercise 5.7. Suppose f : X → Y is a degree 2 map of compactriemann surfaces where Y has genus 0 and X has genus g. Computethe number of branch points of f and the number of ramificationpoints of f .

Exercise 5.8. Suppose f : X → Y is a map between two curves ofgenus 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 whatis meant by “counted with multiplicity.” The calculation isnearly completely carried out in Example 5.5.

Exercise 5.10 (Tricky exercise). Let E denote the quotient of the torusC/Λ with Λ the lattice Z⊕ i ·Z. Since multiplication by i (rotationby 90 degrees under the identification C ' R2) sends Λ 7→ Λ, wecan 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 ramificationpoints of multiplicity 4 and two ramification points of multiplicity 2.Thinking of E as a quotient C → C/Λ ' E, what are the preimagesof these ramification points in C?

18 AARON LANDESMAN

6. AUTOMORPHISMS OF RIEMANN SURFACES OF GENUS ≥ 2

In this section, we will begin by stating a bound on the number ofautomorphisms of a Riemann surface of genus at least 2. Then, wewill prove the claimed bound.

6.1. Statement of the bound.

Definition 6.1. An automorphism of a Riemann surface X is a mapof Riemann surfaces f : X → X.

Exercise 6.2. Verify that the collection of automorphisms of X forma group under composition, with the identity automorphism as theidentity element of the group. The collection of automorphisms withthis 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, theautomorphism group of X has order at most 84(g− 1).

This will be proved later starting in subsection 6.2 (the final proofoccurring in subsection 6.8).

Remark 6.4. Note that Theorem 5.3 emphatically does not hold inthe case g = 0, 1. In fact, curves of genus 0 or 1 always have infinitelymany automorphisms, as we show in section 7.

6.2. Proving the bound. We now are ready to prove the bound. Wetake as input the following surprisingly tricky theorem:

Theorem 6.5. Let X be a Riemann surface of genus g ≥ 2. Then X hasonly finitely many automorphisms.

As input we will also need that the quotient of a Riemann surfaceby 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, anduse this to deduce the desired bound.

Remark 6.6. The finiteness of automorphisms takes some more workto develop: The standard proof uses algebraic geometry and the no-tion of the “automorphisms schemes.” The vague idea is to con-struct a parameter space of the automorphisms (“the Aut scheme”)which is itself a geometric object. One can then use something called“deformation theory” to show that this parameter space has a 0-dimensional tangent space, so it is, in some sense discrete. If weknew it were compact and discrete, it would be finite.

RIEMANN SURFACES 19

For the reader familiar with algebraic geometry, one can showcompactness using the “canonical embedding” the automorphismsare a closed subscheme of PGLg(C), which is itself a finite type groupscheme, and hence the automorphism scheme is also finite type, and0-dimensional, hence a finite collection of points.

Alternatively, one can argue this compactness by appealing to anontrivial input from the theory of “Hilbert schemes” (with the Autscheme being an example of a Hilbert scheme).

The following lemma is crucial to our bound on the number ofautomorphisms.

Lemma 6.7. Let X be a Riemann surface and G be the subgroup of theautomorphism group of X generated by a single automorphism of X. LetY be the quotient of X by G and let f : X → Y be the quotient map (seeProposition B.1). Then, if f (x) = f (x′) we have multx( f ) = multx′( 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

2g(X)− 2 = |G|((2g(Y)− 2) +

n

∑i=1

ri − 1ri

).(6.1)

Proof. To show multx( f ) = multx′( f ) simply note that there is anautomorphism h : X → X with h(x) = x′.

Exercise 6.8. Show that multx( f ) = multx′( f ) Hint: Use the defini-tion of multiplicity as the power n such that the map looks locallylike z 7→ zn. Show that this n is preserved under automorphisms.

To complete the proof, it suffices to prove Equation 6.1. By theabove, the total degree of ramification above a branch point b isdeg π − |π−1(b)| = ∑q∈π−1(b)(ri − 1) = (ri − 1) · |π−1(b)|. Since,

deg π = |G|, |G| = ri · |π−1(b)|. Therefore, |π−1(b)| = |G|/ri.If we let B( f ) denote the set of branch points with multiplicity,

the ramification points summed with multiplicity is ∑b∈B|G|rb(rb− 1).

Then, applying Riemann Hurwitz, and replacing the sum over b ∈ Bby a sum from 1 to n, we get

2g− 2 = |G|(2g(Y)− 2) +n

∑i=1

|G|(ri − 1)ri

= |G|(

2g(Y)− 2 +n

∑i=1

ri − 1ri

).

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

2g(Y)− 2 +n

∑i=1

ri − 1ri

.(6.2)

First, note that the value 1/42 of the above expression is achievedwhen g(Y) = 0, n = 3 and r1 = 2, r2 = 3, r3 = 7. We now rule outthis value being less than 1/42 via the following casework:

6.3. We rule out g(Y) > 1. If g(Y) > 1, then since ∑iri−1

riis positive,

if g(Y) > 1 then Equation 6.2 is at least 2g(Y)− 2 ≥ 2.

6.4. We rule out g(Y) = 1. If g(Y) = 1, the result is then ∑ni=1

ri−1ri

isat least 1

2 assuming it is positive.For the rest of the problem, we are attempting to minimize ∑n

i=1ri−1

ri−

2 subject to the condition it is positive. We assume g(Y) = 0. Weknow 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−1ri

> 12 as

ri > 1, we obtain that ∑ni=1

ri−1ri

> 52 , and so such a sum cannot be

less than 2 + 142 .

6.6. We rule out g(Y) = 0, n = 4. Let us now rule out n = 4. Ofcourse, we cannot have ri =

12 for all i as then the sum ∑i

ri−1ri− 2 =

0. If three of the ri =12 and one of them is 2

3 then ∑iri−1

ri− 2 = 1

6 ,and all other values of the ri will yield strictly greater results, so wecannot make a sum less than 1

42 with n = 4.

6.7. We rule out g(C′) = 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. Wenow divide this into further subcases:

6.7.1. We rule out g(C′) = 0, n = 3 no ri = 2. Suppose none of theri are 2. Note that we cannot have r1 = r2 = r3 = 3 since then1/r1 + 1/r2 + 1/r3 = 1. Therefore, not all of them are 1

3 , so maximumpossible value of 1/r1 + 1/r2 + 1/r3 is attained when r1 = 3, r2 =3, r3 = 4, in which case 1/r1 + 1/r2 + 1/r3 = 11

12 < 4142 . This rules

out the case that no ri = 2. For the rest of the proof, we may assumer1 = 2.

RIEMANN SURFACES 21

6.7.2. We rule out g(C′) = 0, n = 3, r1 = 2, and r2, r3 > 3. In thecase 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 whenr2 = 4, r3 = 5. In this case, 1/r1 + 1/r2 + 1/r3 = 19

20 < 4142 .

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 testingvalues for r3 up to 7, we see that if 1/r1 + 1/r2 + 1/r3 < 1, we musthave r3 ≥ 7. The value maximizing the sum is then r3 = 7.

6.8. Finally we can put the above together to complete the proof ofTheorem 6.5:

Proof of Theorem 6.3. Let α denote the expression in Equation 6.2. Then,note that α|G| = 2g− 2 and α ≤ 1

42 by the above casework. Hence,|G| ≤ 42(2g− 2) = 84(g− 1). �

22 AARON LANDESMAN

7. AUTOMORPHISMS IN LOW GENUS 0 AND 1

In this subsection, we classify all possible automorphisms of genus0 and genus 1 curves, and give an example of a genus 3 curve with alarge, but finite, automorphism group. Many of the exercises assumebackground beyond the prerequisites for this course, so feel free toskip them if you are not do not have such background.

7.1. Genus 0.

Example 7.1 (Genus 0 automorphisms). Consider the rational func-tion f : C → C given by z 7→ az+b

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 becauseif you multiply a, b, c, d by a given scalar, the resulting map is thesame).

Exercise 7.2 (Tricky exercise, assuming knowledge of removable sin-gularities). Show that the automorphisms of C given in Example 7.1are all of the automorphisms of C, in the following steps:

(1) Show that the only automorphisms f : C→ C are of the formx 7→ ax + b for a, b ∈ C with a 6= 0. Hint: After a translation,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 thatthere is no z with g(z) = 0. Conclude that g is constant byCorollary 3.5.

(2) Show that the only automorphisms f : C→ C are those givenin Example 7.1. Hint: If you have any automorphism f , showthat you can compose it with one of Example 7.1 to assumef (∞) = ∞. Then use the first part.

7.2. Genus 1.

Example 7.3. Let E = C/Λ be a genus 1 Riemann surface. Showthat for any two points x, y ∈ E, there is an automorphism f : E→ Ewith f (x) = y. Conclude that E has infinitely many automorphisms.Hint: Consider the maps on E induced from the maps C → C givenby 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 automorphismsof E fixing e.

(1) Show that any automorphism of E fixing e is induced by aunique automorphism of C sending 0 7→ 0. Hint: Use thatC is the universal cover of E and path lifting lemmas (this iswhere 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→ Λand further sending 0 7→ 0.

(3) Show that if Λ, up to scaling, is equal to Z⊕ iZ ⊂ C thenΛ has automorphism group Z/4 generated by z 7→ iz. Hint:Use that the automorphisms of C fixing 0 are precisely thoseof 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 aprimitive cube root of unity then Λ has automorphism groupZ/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 arez 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 aroot of unity satisfies a degree 2 polynomial if and only if it isa 2nd root of unity, a 4th root of unity, or a 6th root of unity.(For this you may need some knowledge of algebraic numbertheory. 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 degreeof the cyclotomic polynomial is φ(n) for φ the Euler totientfunction.)

(6) Conclude that the automorphism group of E which fixes eis either Z/4, Z/6, or Z/2, depending on whether Λ, up toscaling, is equal to Z⊕ iZ, Z⊕ωZ, or something else.

(7) Combine the previous parts of this exercise with Example 7.3to deduce a complete description of all automorphisms of E.

7.3. Example in Genus 3.

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

24 AARON LANDESMAN

(1) Show that C has at least 168 automorphisms. Hint: Considerautomorphisms given by permuting the coordinates and thosemultiplying coordinates by seventh roots of unity. Find an au-tomorphism of order 7, and automorphism of order 3. Showthat 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: Writedown explicit charts via algebraic equations and concludethey are holomorphic.

(4) Construct a map f : C → C 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 andcompute its ramification points. Hint: The ramification pointswill exactly be the fixed points of the automorphisms foundin the first part.

(6) Use Riemann-Hurwitz to show g(C) = 3.

RIEMANN SURFACES 25

APPENDIX A. PROOF OF RIEMANN HURWITZ

In this appendix, we now provide a proof of the Riemann-Hurwitztheorem, Theorem 5.3. The key to the proof will be the notion of theEuler 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 fromtriangles 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 triangulation of a compact Riemannsurface has only finitely many triangles. Hint: Take an open coverby 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 comprisingT, fT be the number of faces, and vT be the number of vertices (so that iftwo triangles share an edge or a vertex, that edge or vertex is only countedonce). 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, letus 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) + ∑x∈X

(mult( f , x)− 1).

Instead, we will show its negation

2− 2g(X) = d · (2− 2g(Y))− ∑x∈X

(mult( f , x)− 1).

Recall that χ(X) = 2− 2g(X), by definition. Therefore, to concludethe 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 T′ of Y so that every branch point is the imageof a vertex of a triangle from the triangulation. By Lemma 4.2, weknow that away from the branch points, the map X → Y is a cov-ering map. Therefore, by taking a sufficiently fine refinement of T′,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-angle. This yields a corresponding triangulation of X, call it T. Wenow compute the Euler characteristic of X in terms of T. First, byconstruction eT = d · eT′ and fT = d · fT′ . It only remains to analyzewhat happens at any given vertex. If there is no ramification what-soever, we would then have vT = d · vT′ as well. However, if themultx( f ) = n, there will then only be a single preimage of f (x) atx, instead of the n preimages we “expected.” Summing this over allramification points, we have that the number of vertices in T is

vT = d · vT′ − ∑x∈R( f )

(multx( f )− 1).

This is exactly the “correction term” we were looking for from thestatement of Riemann-Hurwitz.

The rest is now formal. We have

χ(X) = fT − eT + vT

= d fT′ − deT′ + dvT′ − ∑x∈R( f )

(multx( f )− 1)

= d( fT′ − eT′ + vT′)− ∑x∈R( f )

(multx( f )− 1)

= d(χ(Y))− ∑x∈R( f )

(multx( f )− 1).

as we wanted to show. �

So, to conclude the proof of Theorem 5.3, we only need to proveProposition 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 proveProposition A.4, we will use the following lemma:

Lemma A.5. If T and T′ are two triangulations of a compact Riemannsurface X, then χT(X) = χT′(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 twotriangulations T and T′ have the same value χT(X). So, to show

RIEMANN SURFACES 27

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

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

To conclude our proof of Theorem 5.3, we only need to proveLemma 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 triangulationof 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 T′ be two separate triangulations of X.

Exercise A.8. Verify that one can construct a triangulation T′′ of X sothat the image of every triangle t′′i : T′′i → X in T′′ is contained bothin the image of some triangle t′i of T′ and the image of some triangleti of T. Such a T′′ is called a simultaneous refinement of T and T′.Hint: Intersect the images of T and T′ and triangulate the resultingintersections.

Let T′′ be a simultaneous refinement of T and T′. If we showχT(X) = χT′′(X) and χT′′(X) = χT(X), we will be done. Therefore,by relabeling T′′ by T′, we can reduce to the case that T′ is a refine-ment of T. That is, it suffices to prove Lemma A.5 in the special casethat T′ 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 (T′)j denote thetriangulation of tj(Tj) composed of those triangles t′i whose image iscontained in tj(Tj). Use Lemma A.7 to show that χT(X) = χT′(X)in the case T′ is a refinement of T by writing χT′(X) in terms of theχ(T′)j(ti(Ti)).

28 AARON LANDESMAN

Finally, to conclude our proof of Theorem 5.3, it suffices to proveLemma 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 eachstep, we either subdivide an edge into two edges by placing a vertexin the middle, or subdivide a face into two faces, by placing an edgein the middle. We claim that in the course of this process, there Eu-ler characteristic remains fixed. Indeed, when we add a vertex, thenumber of edges goes up by 1, but so does the number of vertices, sof − 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 soagain 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. QUOTIENTS OF RIEMANN SURFACES BYAUTOMORPHISMS

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, thereis a Riemann surface Y and a map of Riemann surfaces f : X → Y so thatfor any x, x′ ∈ X with f (x) = f (x′) there is a unique m with hmx = x′.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 thequotient of X by the equivalence relation x ∼ x′ if x = hmx′ for somem. Said yet another way, the group G acts simply transitively on thefibers of f . (If you haven’t heard these terms before, just ignore thisremark it; this remark is an immediate rephrasing of the propositionstatement if you are familiar with the words).

Corollary B.3. Let G be a finite group of automorphisms of some Riemannsurface X. Then, there is a Riemann surface Y and a map f : X → Y sothat for any x, x′ ∈ X with f (x) = f (x′) there is a unique h ∈ G withhx = x′. 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 automorphismh ∈ 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 subgroupgenerated by that element, and then repeating.

For the purposes of the proof, we will need discreteness of thefixed points, established in the next exercise.

Exercise B.5. Let h : X → X be an automorphism of order n. Showthat 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. Toconstruct Y as a set, simply define the set of points as points of Xmodulo the relation ∼ that x ∼ x′ if hkx = x′ for some k. Note thereis 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 thequotient topology meaning that the open sets U ⊂ Y are preciselythe sets such that f−1(U) is open in X.

To conclude, we only need to give Y the structure of a Riemannsurface. Indeed, this is the trickiest part, and will require some care.We will do this by defining the structure of a Riemann surface awayfrom 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 opencontaining 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, afterpossibly shrinking U, we may assume it is biholomorphic to an opensubset V ⊂ C. So, the map V ' U ' f (U) defines a chart aroundf (x).

Exercise B.7. Verify that the above charts defined above give Y∗ thestructure of a Riemann surface. Hint: You have to check that the“transition maps” are biholomorphic. To check this, use that thesemaps 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 manypoint Y−Y∗. Since these points are discrete, it suffices to choose onepoint y ∈ Y− Y∗. Take x ∈ X with f (x) = y. Suppose the subgroupof G which stabilizes x has size m (i.e., the powers of h fixing x areprecisely 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, themap hn/m 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 subsetof C, let Uy ⊂ C denote the open subset of C which is the image ofUx under the map C→ C given by z 7→ zm.

RIEMANN SURFACES 31

Exercise B.9 (Tricky exercise). Show that Uy is defines a chart for yand gives Y the structure of a riemann surface.

Exercise B.10. Show that the ramification points of the resulting mapf 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 thisstructure in the same way over the finitely many remaining pointsconcludes the proof. �

REFERENCES

[McM] C McMullen. Complex analysis on riemann surfaces. Harvard UniversityLecture Notes. http://www.math.harvard.edu/~ctm/home/text/class/

harvard/213b/13/html/home/course/course.pdf.