Universite du Quebec a MontrealDepartement de mathematiquesMarco A. Perez B.marco.perez@cirget.ca

THE RIEMANN-ROCH THEOREM

January 2012

Contents

Introduction i

1 Algebraic curves and some properties of its function field 1

1.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.2 Laurent series approximation . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.3 The function field M(X) as an extension of the field of rational expressionsin some nonconstant meromorphic function . . . . . . . . . . . . . . . . . . . 6

2 The Grothendieck-Riemann-Roch Theorem 11

2.1 Laurent tail divisors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

2.2 The first form of the Riemann-Roch Theorem . . . . . . . . . . . . . . . . . 14

2.3 Serre Duality and the Riemann-Roch Theorem . . . . . . . . . . . . . . . . . 21

2.4 Generalizations of the Riemann-Roch Theorem . . . . . . . . . . . . . . . . . 28

Bibliography 31

Introduction

The Riemann-Roch theorem is an important result in complex geometry, for the computationof the dimension of the space of meromorphic functions with prescribed zeroes and allowedpoles. Roughly speaking, it links the complex analysis of a compact Riemann surface with itstopological genus. Initially proved as Riemann’s inequality by Bernhard Riemann in 1857,

Theorem 1. Let X be a Riemann surface of genus g, then

dim(L(D)) ≥ deg(D) + 1− g,

where L(D) is the space of meromorphic functions with poles bounded by a divisor D.

the theorem reached its definitive form thanks to the work of Gustav Roch, one of Riemann’sstudents, in 1865 providing the error term:

Theorem 2. Let X be a Riemann surface of genus g. Then for any divisor D and anycanonical divisor K, we have

dim(L(D))− dim(L(K −D)) = deg(D) + 1− g.

Originally developed in complex function theory, Riemann and Roch proved the theorem in apurely analytic context. The legitimacy of the proof was called into question by Weierstrass,who found a counterexample to a main tool in the proof that Riemann called Dirichlet’sprinciple. In spite of this challenge, the theorems ideas were too useful to do without.

The Riemann-Roch Theorem was proved for algebraic curves by F. K. Schmidt in 1929.It was later generalized to higher-dimensional varieties and beyond, using the appropriatenotion of divisor, or line bundle. The general formulation of the theorem depends on splittingit into two parts. One, which would now be called Serre Duality, interprets the L(K − D)term as a dimension of a first sheaf cohomology group; with L(D) the dimension of a zerothcohomology group, or space of sections, the left-hand side of the theorem becomes an Eulercharacteristic, and the right-hand side a computation of it as a degree corrected accordingto the topology of the Riemann surface.

An n-dimensional generalization, the Hirzebruch-Riemann-Roch Theorem, was found andproved by Friedrich Hirzebruch, as an application of characteristic classes in algebraic topology.

i

At about the same time Jean-Pierre Serre was giving the general form of Serre Duality.Alexander Grothendieck proved a far-reaching generalization in 1957, now known as theGrothendieck-Riemann-Roch Theorem. His work reinterprets Riemann-Roch not as a theo-rem about a variety, but about a morphism between two varieties.

These notes are divided into two chapters. In the first chapter we recall some notions ofcomplex geometry. Then we give the concept of an algebraic curve and study some of itsproperties, such as the Laurent Series Approximation Theorem. In the last section of thischapter, we study the field of meromorphic functions on an algebraic curve X, M(X), asa field extension of certain subfield. Such a subfield is the field of rational expressions in anonconstant meromorphic function on X. The degree of this extension shall be an importanttool to prove the Riemann-Roch Theorem.

In the second chapter we prove the Riemann-Roch Theorem for algebraic curves, by studyingthe spaces L(D) and L(1)(D) of meromorphic functions and meromorphic 1-forms, respec-tively, with poles bounded by a divisor D. We shall prove a result called the Serre DualityTheorem, which is the most important tool to prove the Riemann-Roch Theorem. Finally,we shall comment some of the generalizations of this theorem, such as the Riemann-RochTheorem for holomorphic line bundles, the Hirzebruch-Riemann-Roch Theorem, and theGrothendick-Riemann-Roch Theorem.

ii

Chapter 1

Algebraic curves and some propertiesof its function field

This chapter is devoted to the study of several properties of a certain type of Riemannsurfaces called algebraic curves. In the first section we shall recall some notions of complexgeometry, such as meromorphic functions, divisors, and meromorphic 1-forms. We shall alsogive the notion of an algebraic curve. These structures have very interesting properties.One of them is known as the Laurent Series Approximation Theorem. Roughly speaking,this theorem allows us to construct a meromorphic function on X from a finite number ofpoints in X and tails of Laurent series such that at each of these points the Laurent seriesof the function starts with one of these tails. Another property of the algebraic curves isthat one can always find a nonconstant meromorphic function on it. So given a nonconstantmeromorphic function f on al algebraic curve X, one can consider the field C(f) of rationalexpressions of f with coefficients in C, which is a subfield of the fieldM(X) of meromorphicfunctions on X. The Laurent Series Approximation Theorem shall help us to compute thedegree of the extension M(X)/C(f).

1.1 Preliminaries

Recall that a function f : X −→ C is meromorphic at p ∈ X if it is either holomorphic,has a removable singularity, or has a pole at p.

Given a function D : X −→ Z on a Riemann surface X, the set of all p such that D(p) 6= 0is called the support of D. A divisor on X is a function D : X −→ Z whose support is adiscrete subset of X. We shall denote a divisor D by using the following summation notation:

D =∑p∈X

D(p) · p.

1

The divisor of a meromorphic function f : X −→ C, denoted by (f), is the divisor definedby the order function:

(f) =∑p∈X

ordp(f) · p.

The divisor of poles of f , denoted by (f)∞, is the divisor

(f)∞ =∑

p with ordp(f)<0

(−ordp(f)) · p.

The space of meromorphic functions with poles bounded by D, denoted L(D), is theset of meromorphic functions

L(D) = f ∈M(X) / (f) ≥ −D,

where (f) ≥ −D means (f)(p) ≥ −D(p) for every p ∈ X.

Proposition 1.1.1. Let X be a compact Riemann surface, and let D be a divisor on X. Thenthe space L(D) is a finite dimensional complex vector space. Indeed, if we write D = P −N ,with P and N nonnegative divisors with disjoint supports, then dim(L(D)) ≤ 1 + deg(P ).In particular, if D is a nonnegative divisor, then

dim(L(D)) ≤ 1 + deg(D).

Recall that a meromorphic differential on an open set V ⊆ C is an expression of the formω = f(z)dz, where f is a meromorphic function on V . Let ω1 = f(z)dz and ω2 = g(w)dw betwo meromorphic differentials defined on V1 and V2. Let T be a holomorphic function fromV2 to V1. We say that ω1 transforms to ω2 under T if

g(w) = f(T (w))T ′(w).

If X is a Riemann surface, a meromorphic 1-form on X is a collection of meromorphicdifferentials (ωϕ), one for each chart ϕ : U −→ V and defined on V , such that if two chartsϕ1 : U1 −→ V1 and ϕ2 : U2 −→ V2 have overlapping domains, then ωϕ1 transforms to ωϕ2

under the transition function ϕ1 ϕ−12 . The notion of holomorphic 1-form is defined in asimilar way.

Given a meromorphic 1-form ω on X, for each p choose a local coordinate zp centred at p.We may write ω = f(zp)dzp, where f is a meromorphic function at z = 0. The order of p,denoted by ordp(ω), is the order of the function f at zp = 0. We can also write ω via theLaurent series of f in the coordinate zp:

ω = f(zp)dzp =

(∞∑

n=−M

cnznp

)dzp

2

where c−M 6= 0, so that ordp(ω) = −M . The residue of ω at p, denoted by Resp(ω), is thecoefficient c−1 in a Laurent series for ω at p. You can see the proof of the following theoremin [3, Theorem 3.17].

Theorem 1.1.1 (The Residue Theorem). Let ω be a meromorphic 1-form on a compactRiemann surface X. Then ∑

p∈X

Resp(ω) = 0.

The divisor of ω, denoted by (ω), is the divisor defined by the order function

(ω) =∑p

ordp(ω) · p.

Any divisor of this form is called a canonical divisor.

Proposition 1.1.2. If X is a compact Riemann surface of genus g which has a nonconstantmeromorphic function, then there is a canonical divisor on X of degree 2g − 2.

Given a divisor D on X, the space of meromorphic 1-forms with poles bounded byD, denoted L(1)(D), is the set of meromorphic 1-forms

L(1)(D) = ω / (ω) ≥ −D.

It is clear that L(1)(D) is a complex vector space. Note that L(1)(0) = Ω(X), the space ofholomorphic 1-forms.

The following result gives a way to relate the spaces L(−) and L(1)(−). You can check theproof in [3, Lemma 3.11].

Proposition 1.1.3. Let D be a divisor on X, and let K be a canonical divisor on X. Thenthe spaces L(1)(D) and L(D +K) are isomorphic.

A meromorphic function f on a Riemann surface X has multiplicity one at a point p ∈ Xif either f is holomorphic at p ∈ X and ordp(f−f(p)) = 1, or f has a simple pole at p. Recallthat for every meromorphic function f : X −→ C, there corresponds a unique holomorphicmap F : X −→ C ∪ ∞ which is not identically ∞. Such a map is given by

F (x) =

f(x) ∈ C if x is not a pole of f,∞ if x is a pole of f.

It is clear that f has multiplicity one at p if and only if F does.

Let S be a set of meromorphic functions on a compact Riemann surface X.

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(1) We say that S separates points in X if for every pair of distinct points p and q inX there is a meromorphic function f ∈ S such that f(p) 6= f(q).

(2) We say that S separates tangents of X if for every point p ∈ X there is a meromor-phic function f ∈ S which has multiplicity one at p.

A compact Riemann surface X is an algebraic curve if the field M(X) of meromorphicfunctions on X separates points and tangents of X.

Note that (1) is easy to understand. But (2) is not so clear. Suppose that S separatestangents. Then there is a meromorphic function f : X −→ C having multiplicity one at eachp ∈ X. Then the associated holomorphic map F : X −→ C ∪ ∞ has multiplicity one ateach p ∈ X. So ordp(F − F (p)) = 1, i.e. (F − F (p))′(p) 6= 0. Hence the derivative mapF ′ : TpX −→ TF (p)C ∪ ∞ is one-to-one. Note also that if ordp(f − f(p)) = 1, then f is alocal coordinate chart, since it is locally a biholomorphic map from an open neighbourhoodU ⊆ X to an open set V ⊆ C. Therefore, S separates tangents if and only if at every pointp ∈ X, there is a local coordinate which extends to a meromorphic function on all X. Itfollows that if X is an algebraic curve, then for every p ∈ X there exists a meromorphicfunction g on X such that ordp(g) = 1. For take a function f separating tangents at p, if fis holomorphic at p, set g = f − f(p), and if f has a simple pole at p set g = 1/f .

Example 1.1.1. The Riemann sphere C⋃∞ and the complex torus C/Γ are examples of

algebraic curves. It is known that every compact Riemann surface is an algebraic curve.

1.2 Laurent series approximation

In this section we prove the Laurent Series Approximation Theorem. By the comments inthe previous section, we can find a meromorphic function g on X such that ordp(g) = 1.Then if we set f = gN , we get the following result.

Lemma 1.2.1. Let X be an algebraic curve, and let p ∈ X. Then for any integer N thereis a global meromorphic function f on X with ordp(f) = N .

A Laurent polynomial r(z) =∑m

i=n cizi is called a Laurent tail of a Laurent series h(z) if

the Laurent series starts with r(z).

Lemma 1.2.2. Let X be an algebraic curve. Fix a point p ∈ X and a local coordinatez centred at p. Fix any Laurent polynomial r(z) in z. Then there exists a meromorphicfunction f on X whose Laurent series at p has r(z) as a Laurent tail.

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Proof: Let r(z) =∑m

i=n cizi be our Laurent polynomial, where cn, cm 6= 0. Note that

r(z) has m − n + 1 terms. The result follows by induction on the number of terms ofr(z). Suppose m − n + 1 = 1. Then z(z) = czm. By the previous lemma, there exists ameromorphic function f onX with ordp(f) = m. Then its Laurent series in the coordinatez has the form

∑∞i=m aiz

i, where am 6= 0. Then camf is the desired meromorphic function.

Now assume that m − n + 1 ≥ 2. We know that we can find a meromorphic function hhaving cnz

n as a Laurent tail. Consider the meromorphic function h− r. Let s(z) be theLaurent polynomial which is the tail of the Laurent series of h− r at p, up through thezm term. Note that h(z)− r(z) starts at n+ 1. It is clear that s(z) has fewer terms thanr(z). By induction, there is a meromorphic function g on X whose Laurent series at phas s as a tail. Now consider f = h− g. We have

f = (h− r)− g + r = · · ·+ s(z)− · · · − s(z) + r(z) = · · ·+ r(z),

and so f has r as a tail.

Lemma 1.2.3. Let X be an algebraic curve. Then for any finite number of points p, q1, . . . , qnin X, there is a meromorphic function f on X with a zero at p and a pole at each qi.

Proof: We proceed by induction on n. For the case n = 1, let g be a meromorphicfunction on X such that g(p) 6= g(q). Suppose that p is not a pole of g. We mayassume that g(p) = 0. There is nothing to show if q is a pole of g. If not, then considerf = g/(g(q) − g). Then f(p) = 0 and f has a pole at q. Now if p is a pole of g, repeatthe previous argument with 1/g.

Suppose n ≥ 2 and let g be a meromorphic function with a zero at p and a pole atq1, . . . , qn−1. On the other hand, there is a meromorphic function h with a zero at p anda pole at qn. Consider the point q1 and the function g + h. If h is holomorphic at q1,then g + h has a pole at q1. If h has a pole at q1, then it may occur that g + h has not apole at q1. For m sufficiently large, g + hm has a pole at q1 since the order of hm at q1 isgreater than the order of g at the same point. Increasing m if necessary, we may assumethat g+ hm has a pole at each qi with i ≤ n− 1. We know that g has a pole at qn, but itmay occur that g+ hm does not. In this case, increase the value of m. Thus, f = g+ hm

has a zero at p and a pole at q1, . . . , qn for a large m.

Lemma 1.2.4. Let X be an algebraic curve. Then for any finite number of points p, q1, . . . , qnin X, and any N ≥ 1, there is a global meromorphic function f on X with ordp(f − 1) ≥ Nand ordqi(f) ≥ N for each i.

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Proof: By the previous lemma, there exists a meromorphic function g on X with azero at p and a pole at each qi. Consider the function f = 1/(1 + gN). We havef(z)− 1 = −gN/(1 + gN). Then p is a zero of f − 1 of order at least N . It is easy to seethat ordqi(f) ≥ N .

Theorem 1.2.1 (Laurent Series Approximation). Suppose X is an algebraic curve. Fix afinite number of points p1, . . . , pn in X, choose a local coordinate zi at each pi, and finallychoose Laurent polynomials ri(zi) for each i. Then there is a meromorphic function f on Xsuch that for every i, f has ri as a Laurent tail at pi.

Proof: By Lemma 1.2.2 there are meromorphic functions gi on X such that gi has ri asa Laurent tail at pi. Let

M = minordpi(ri) / i = 1, . . . , n = minordpi(gi) / i = 1, . . . , n.

Let N be a fixed integer larger than every exponent of every term of every ri. By Lemma1.2.4, there are meromorphic function hi on X such that for each i, ordpi(h

i−1) ≥ N−Mand ordpj(hi) ≥ N −M for j 6= i. The function f =

∑ni=1 hi · gi has ri as a Laurent tail

at pi, for each pi.

Corollary 1.2.1. Let X be an algebraic curve. Fix a finite number of points p1, . . . , pn inX, and a finite number of integers mi. Then there exists a meromorphic function f on Xsuch that ordpi(f) = mi for each i.

1.3 The function fieldM(X) as an extension of the field

of rational expressions in some nonconstant mero-

morphic function

Let C(f) denote the field of rational expressions in f , where f is a nonconstant meromorphicfunction on an algebraic curve X. We have the inclusions of fields C ⊆ C(f) ⊆M(X). ThenM(X) is a field extension of C(f), and hence M(X) can be considered as a vector spaceover C(f). Let [M(X) : C(f)] denote the dimension of M(X) over C(f), which is calledthe degree of the extensionM(X)/C(f). In this section we prove that [M(X) : C(f)] =deg((f)∞). For this it is necessary the theorem of Laurent Series Approximation. But thisresult will only help to show the inequality [M(X) : C(f)] ≥ deg((f)∞). We shall need otherresults to prove the other inequality.

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Lemma 1.3.1. Let A be a divisor on a compact Riemann surface X, and let D = (f)∞for a nonconstant meromorphic function f on X. Then there is an integer m > 0 and ameromorphic function g on X such that A − (g) ≤ mD. Moreover, g can be taken to be apolynomial in f : g = r(f) for some polynomial r(t) ∈ C[t].

Proof: Let p1, . . . , pk be points in the support of A such that

(i) p1, . . . , pk are not poles of f .

(ii) A(pi) ≥ 1 for each 1 ≤ i ≤ k.

Then f(pi) ∈ C, and so we can consider f − f(pi) as a meromorphic function with a zeroat pi or order at least 1, and having the same poles as f . It follows that (f − f(pi))

A(pi)

is a meromorphic function with a zero at pi of order at least A(pi), and whose poles arethe poles of f . Let g =

∏ki=1(f −f(pi))

A(pi). Then g is a meromorphic function such thatA− (g) is positive only at the poles of f . Now suppose that p is a pole of f , we have

A(p)− (g)(p) = A(p)− ordp(g) = A(p)−k∑i=1

A(pi)ordp(f − f(pi))

= A(p)−k∑i=1

A(pi)ordp(f) = A(p)−

(k∑i=1

A(pi)

)· ordp(f)

= A(p) +

(k∑i=1

A(pi)

)· (f)∞(p)

≤ m(f)∞, for some m sufficiently large.

If p is not a pole of f , then we have that A(p)− (g)(p) < 0 < m(f)∞(p).

Corollary 1.3.1. Let X be a compact Riemann surface, and let f and h be nonconstantmeromorphic functions on X. Then there is a polynomial r(t) ∈ C[t] such that the functionr(f)h has no poles outside of the poles of f . In this case there is an integer m such thatr(f)h ∈ L(mD), where D = (f)∞.

Proof: Put A = −(h) in the previous lemma. There exists a meromorphic function gon X such that −(h) − (g) ≤ m(f)∞, for some integer m > 0, where such a g can bechosen as a polynomial function in f , say g = r(f). First, we show that r(f)h ∈ L(mD),i.e. (r(f)h) ≥ −mD. Let p ∈ X, we have (r(f)h)(p) = (g)(p) + (h)(p) ≥ −mD(p). Now

7

we show that r(f)h has no poles outside of the poles of f . Suppose p ∈ X is a pole ofr(f)h. Then ordp(r(f)h) < 0. We have 0 > ordp(r(f)h) ≥ −m(f)∞(p) = m · ordp(f).Since m > 0, we have ordp(f) < 0, i.e. p is a pole of f .

Lemma 1.3.2. Fix a meromorphic function f on a compact Riemann surface X, and letD = (f)∞. Suppose that [M(X) : C(f)] ≥ k. Then there is an integer m0 such that for allm ≥ m0,

dim(L(mD)) ≥ (m−m0 + 1)k.

Proof: Since [M(X) : C(f)] ≥ k, we can choose linearly independent meromorphicfunctions g1, . . . , gk over C(f). By the previous corollary, for each i there exists a nonzerorational expression ri(f) such that hi = ri(f)gi has no poles outside of the poles of f .It is clear that the h1, . . . , hk are linearly independent over C(f). For each i there is aninteger mi such that hi ∈ L(miD). Let p ∈ X, we have (hi)(p) ≥ −miD(p). Let m0 =maxmi : i = 1, . . . , k. Then (hi)(p) ≥ −m0D(p), for every i, that is hi ∈ L(m0D).Now we show that for any integer m ≥ m0, one has

dim(L(mD)) ≥ (m−m0 + 1)k.

First, note that f ihj ∈ L(mD) if i ≤ m−m0. For let p ∈ X, we have

(f ihj)(p) = ordp(fihj) = ordp(f

i) + ordp(hj) = i · ordp(f) + ordp(hj)

≥ i · ordp(f)−m0D(p) = i · ordp(f) +m0 · ordp(f)

= (i+m0) · ordp(f) = −(i+m0)(f)∞(p)

≥ −mD(p).

Then f ihj ∈ L(mD). Note that the f ihj’s are linearly independent over C(f). Thenthe C(f)-vector space generated by the f ihj’s has dimension (m −m0 + 1)k. It followsdim(L(mD)) ≥ (m−m0 + 1)k.

Now we are ready to compute [M(X) : C(f)].

Theorem 1.3.1. Let f be a nonconstant meromorphic function on an algebraic curve X.Then [M(X) : C(f)] = deg(D), where D = (f)∞.

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Proof: We first show that [M(X) : C(f)] ≤ deg(D). Suppose [M(X) : C(f)] is at leastdeg(D) + 1. By the previous lemma, there exists an integer m0 such that for all m ≥ m0,

dim(L(mD)) ≥ (m−m0 + 1)(1 + deg(D)).

On the other hand,

dim(L(mD)) ≤ 1 + deg(mD) = 1 +mdeg(D).

So we get

(m−m0 + 1)(1 + deg(D)) ≤ 1 +mdeg(D)

m−m0 −m0deg(D) + deg(D) ≤ 0

m+ deg(D) ≤ m0 +m0deg(D),

for every m, getting a contradiction.

To show the other inequality, write D =∑ni · pi, with each ni ≥ 1, since D is a

nonnegative divisor. By Corollary 1.2.1, for each pi and each j there is a meromorphicfunction gij with a pole at pi or order j, and no zero or pole at any of the other pk’s. Theset gij / 1 ≤ j ≤ ni is linearly independent over C(f), for each i. Then we have

[M(X) : C(f)] ≥∑i

card(gij / 1 ≤ j ≤ ni) =∑i

ni = deg(D).

9

10

Chapter 2

The Grothendieck-Riemann-RochTheorem

In this chapter we shall prove the Riemann-Roch Theorem for algebraic curves, and commenttwo generalizations known as the Hirzebruch-Riemann-Roch Theorem and the Grothendieck-Riemann-Roch Theorem. In the first section we study certain sums called Laurent taildivisors. We can relate Laurent tail divisors and ordinary divisors via an operation calledtruncation. This allows us to define a group T [D](X), where X is an algebraic curve, formedby Laurent tail divisors that are bounded, in some sense, by a divisor D. We can also relatemeromorphic functions to Laurent tail divisors. In fact, for every meromorphic function therecorresponds an element in T [D] via a group homomorphism αD :M(X) −→ T [D](X). Thismap is of vital importance to prove the Riemann-Roch Theorem. Recall that this theoremgives us a formula to compute the dimension of the space L(D), which turns out to be thekernel of the homomorphism αD. A first form of this theorem involves the cokernel of thishomomorphism. At this point, the problem is that we are dealing with two spaces, namelyL(D) and the cokernel of αD. We shall prove and use a result called Serre Duality to give aneasy way two compute the dimension of the cokernel of αD, and so we shall obtain a refinedversion of the Riemann-Roch Theorem that allows us to compute the dimension of L(D) onlyin terms of the degree of D and the genus of X.

2.1 Laurent tail divisors

Let X be a compact Riemann surface. For each point p ∈ X, choose a local coordinate zpcentred at p. A Laurent tail divisor on X is a finite formal sum of the form∑

p

rp · p,

11

where rp(zp) is a Laurent polynomial in the coordinate zp. The set of Laurent tail divisorsforms a group under formal addition, denoted T (X). We shall focus on special subgroups ofT (X): for any divisor D, define T [D](X) as the set of all finite formal sums

∑p rp · p such

that for all p with rp 6= 0, the top term of rp has degree strictly less that −D(p). Note thatT [D](X) is a subgroup of T (X).

From this point, we shall construct group homomorphisms concerning the previous groups.Consider a Laurent tail divisor

∑p rp · p and a divisor D. At each p, we have

rp(zp) =

mp∑i=np

api zip.

Let iD be the smallest integer between np and mp such that iD + 1 ≥ −D(p). We can definea truncation of rp(zp) as

mp∑i=np

api zip 7→

iD∑i=np

api zip.

This mapping defines a group homomorphism

tD : T (X) −→ T [D](X)

that sends each∑

p rp · p to∑

p rp · p, where rp denotes the truncation of rp whose top termis the largest integer between np and mp strictly smaller than −D(p), in other words, tD isdefined by removing from each rp(zp) those terms of degree greater or equal than −D(p).

Now suppose that we have two divisors D1 and D2 such that D1 ≤ D2, i.e. D1(p) ≤ D2(p) forevery p ∈ X. Then −D2(p) ≤ −D1(p) for every p ∈ X. Let

∑p rp · p ∈ T [D1](X). For each

p with rp 6= 0, consider the Laurent polynomial rp(zp) =∑mp

i=npapi z

ip, where mp < −D1(p).

Suppose there exists an integer between np and mp greater or equal than −D2(p), and letiD2 + 1 be the smallest of such integers. Then we can truncate rp(zp):

mp∑i=np

api zip 7→

iD2∑i=np

api zip.

If such an iD2 does not exist, then just map∑mp

i=npapi z

ip to 0. This truncation defines a group

homomorphismtD1D2

: T [D1](X) −→ T [D2](X)

defined by removing from each rp(zp) those terms of degree greater or equal than −D2(p).We shall call the maps tD1

D2truncation maps.

Consider a meromorphic function f and a divisor D. Let∑

p rp · p be a Laurent tail in

T [D](X). Let∑∞

i=npaiz

ip be the Laurent series of f in the coordinate zp, and let rp(zp) =∑k

j=mpbjz

jp. We take the product

(∑∞i=np

aizip

)·(∑k

j=mpbjz

jp

)=∑∞

i=np, j=mpaibjz

i+jp , and

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truncate it by removing those terms of degree greater or equal than −D(p) + (f)(p). Thisgives rise to a group homomorphism

µDf : T [D](X) −→ T [D − (f)](X)

mapping each∑

p rp · p to∑

p(frp) · p, where f · rp is the Laurent polynomial of the above

truncation of the series∑∞

i=np, j=mpaibjz

i+jp . It is straightforward to check that µDf has an

inverse µD−(f)1/f .

Now consider again the Laurent series of f :∑∞

i=npaiz

ip. Given a divisor D, there exists a

smallest integer mp such that mp + 1 ≥ −D(p). Then we can truncate the previous seriesand get a Laurent polynomial rp(zp) =

∑mp

i=npaiz

ip. This way we get a map

αD :M(X) −→ T [D](X)

which turns out to be a group homomorphism.

Proposition 2.1.1 (Properties of αD).

(1) αD commutes with the truncation maps: If D1 and D2 are divisors with D1 ≤ D2 thenthe following triangle commutes:

M(X) T [D1](X)

T [D2](X)

αD1

αD2

tD1D2

(2) αD is compatible with the multiplication operators: If f and g are meromorphic func-tions on X, then

µDf (αD(g)) = αD−(f)(f · g)

for any divisor D.

(3) L(D) = Ker(αD).

Proof: Let p ∈ X and let zp be a coordinate centred at p.

(1) Let∑∞

i=npaiz

ip be the Laurent series of f in the coordinate zp, and let mp be the

largest integer greater or equal than np such that mp < −D1(p). Then αD(f) =∑p rp · p, where rp(zp) =

∑mp

i=npaiz

ip. Notice that −D2(p) ≤ −D1(p). Let kp be

13

the largest integer greater or equal than np such that kp + 1 ≥ −D2(p). Note thatkp < mp. Then

tD1D2 αD1(f) =

kp∑i=np

aizip = αD2(f).

(2) Consider the Laurent series f(zp) =∑∞

i=npaiz

pi and g(zp) =

∑∞j=mp

bjzjp of f and g

in the coordinate zp. We have (f · g)(zp) =∑∞

i=np, j=mpaibjz

i+jp . On the one hand,

αD(g) =∑k

j=mpbjz

jp, where k ≥ mp is the largest integer satisfying k < −D(p).

Map αD(g) to the series(∑∞

i=npaiz

ip

)·(∑k

j=mpbjz

jp

)=∑∞

i=np

∑kj=mp

aibjzi+jp . On

the other hand, let k ≥ mp + np be the largest integer such that

k < −D(p) + (f)(p) = −D(p) + ordp(f).

Then µDf (αD(g)) =∑k

i=np, j=mpaibjz

i+jp = αD−(f)(f · g).

(3) Let f ∈ L(D). Then (f) ≥ −D. Consider the Laurent series f(zp) =∑∞

i=npaiz

ip.

Since np = ordp(f) ≥ −D(p), the previous series is mapped to 0 by αD. Now letf ∈ Ker(αD). Then each rp(zp) = 0 and hence np ≥ −D(p), i.e. (f)(p) ≥ −D(p).

2.2 The first form of the Riemann-Roch Theorem

In the previous section, we proved that Ker(αD) = L(D). What could we say aboutCoKer(αD)?. Define

H1(D) := CoKer(αD) = T [D](X)/Im(αD)

We shall prove that the space H1(D) is finite dimensional. By (3) of the previous proposition,we have a short exact sequence

0 −→ L(D) −→M(X)αD−→ T [D](X) −→ H1(D) −→ 0

Consider the quotient space M(X)/L(D). Since αD vanishes on L(D), we can consider themap

αD :M(X)/L(D) −→ T [D](X)

given by αD(f + L(D)) = αD(f). It is clear that it is a well defined group monomorphism.Also, note that Im(αD) = Im(αD) = Ker(T [D](X) −→ H1(D)). Then we get the following

14

short exact sequence:

0 −→M(X)/L(D) −→ T [D](X) −→ H1(D) −→ 0

Now let D1 and D2 be two divisors satisfying D1 ≤ D2. Then we have a truncation maptD1D2

: T [D1](X) −→ T [D2](X). Also, L(D1) ⊆ L(D2). For each Di there is a short exactsequence as above. We shall connect these sequences constructing two homomorphisms.

(i) Let F :M(X)/L(D1) −→M(X)/L(D2) be the map given by

F (f + L(D1)) = f + L(D2).

This map is well defined. For if f − g ∈ L(D1) then (f) − (g) ≥ −D1 ≥ −D2, i.e.f − g ∈ L(D2). Also, it is clear that F is a group homomorphism.

(ii) Now define a map G : H1(D1) −→ H1(D2) by

G(Z + Im(αD1)) = tD1D2

(Z) + Im(αD2).

We check G is well defined. Suppose Z − Z ′ ∈ Im(αD1). Then Z − Z ′ = αD1(f) forsome meromorphic function f . By the first part of the previous proposition, we have

tD1D2

(Z)− tD1D2

(Z ′) = tD1D2 αD1(f) = αD2(f) ∈ Im(αD2)

It is clear that G is a group homomorphism.

Proposition 2.2.1. The following diagram commutes and has exact rows:

0 M(X)/L(D1) T [D1](X) H1(D1) 0

0 M(X)/L(D2) T [D2](X) H1(D2) 0

αD1

F

γD1

tD1D2

G

αD2γD2

where γD1 and γD2 are the projection maps.

Proof: On the one hand

tD1D2 αD1(f + L(D1)) = tD1

D2(αD1(f)) = αD2(f), by (1) of the previous proposition

= αD2(f + L(D2))

= αD2 F (f + L(D1)).

15

So the left square commutes. On the other hand,

G γD1(Z) = G(Z + Im(αD1)) = tD1D2

(Z) + Im(αD2) = γD2(tD1D2

(Z)).

Hence the right square commutes.

Let H1(D1/D2) := Ker(G). We use this diagram two show that H1(D1/D2) is finite dimen-sional. This fact shall help us to prove the finite dimensionality of H1(D). Using the diagramabove and the Snake Lemma, we obtain the following exact sequence

0 −→ Ker(F ) −→ Ker(tD1D2

) −→ Ker(G) −→ CoKer(F ) −→ CoKer(tD1D2

) −→ CoKer(G) −→ 0.

Note that F is surjective. We shall see that so are tD1D2

and G. Let Z =∑

p rp ·p ∈ T [D2](X),

where rp(zp) =∑mp

i=npapi z

ip and mp ≥ np is the largest integer with mp < −D2(p) ≤ −D1(p).

So Z ∈ T [D1](X) and Z = tD1D2

(Z). Hence tD1D2

is surjective, and this implies that so is G.

On the other hand, note that Ker(F ) = L(D2)/L(D1). So we have

dim(Ker(F )) = dim(L(D2))− dim(L(D1))

Now consider∑

p rp · p ∈ Ker(tD1D2

), where rp(zp) =∑mp

i=npapi z

ip. Since

∑p a

pi zip is mapped to

0, we have np ≥ −D2(p). Hence Ker(tD1D2

) is the space of all∑

p rp · p such that the top termof rp has order less than −D1(p) and the bottom term has order at least −D2(p). At eachp, we have D2(p) − D1(p) possible monomials zip, −D2(p) ≤ i ≤ −D1(p), that are linearlyindependent. Hence

dim(Ker(tD1D2

)) =∑p

(D2(p)−D1(p)) =∑p

D2(p)−∑p

D1(p) = deg(D2)− deg(D1)

Recall that we have a short exact sequence

0 −→ L(D2)/L(D1) −→ Ker(tD1D2

) −→ H1(D1/D2) −→ 0

Since H1(D1/D2) is a free C-module, the previous sequence splits, so

Ker(tD1D2

) ∼= (L(D2)/L(D1))⊕

H1(D1/D2)

It follows

dim(Ker(tD1D2

)) = dim(L(D2)/L(D1)) + dim(H1(D1/D2))

dim(H1(D1/D2)) = deg(D2)− deg(D1) + dim(L(D1))− dim(L(D2)).

We have proven that H1(D1/D2) is finite dimensional. Summarizing, we have

16

Lemma 2.2.1. If D1 and D2 are divisors on a compact Riemann surface X, with D1 ≤ D2,then

dim(H1(D1/D2)) = [deg(D2)− dim(L(D2))]− [deg(D1)− dim(L(D1))]

As we commented above, we shall use the previous formula to prove the following result:

Proposition 2.2.2. For any divisor D on an algebraic curve X, H1(D) is a finite dimensionalvector space over C.

Before giving a proof, we need some lemmas.

Lemma 2.2.2. Let f be a nonconstant meromorphic function on an algebraic curve X, andlet D = (f)∞. Then for any large m, the dimension of H1(0/mD) is constant, independentof m.

Proof: First, apply the previous lemma putting D1 = 0 and D2 = mD:

dim(H1(0/mD)) = [deg(mD)− dim(L(mD))]− [deg(0)− dim(L(0))]

= deg(mD)− dim(L(mD)) + 1.

Note that we have used the equality dim(L(0)) = 1, where L(0) is the space of holomor-phic functions on X. Since X is compact, we have that every holomorphic function onX is constant and hence L(0) ∼= C has dimension one. Since [M(X) : C(f)] = deg(D)by Theorem 1.3.1, by Lemma 1.3.2 there is an integer m0 such that

dim(L(mD)) ≥ (m−m0 + 1)deg(D)

for all m ≥ m0. Then we have

dim(H1(0/mD)) = deg(mD)− dim(L(mD)) + 1

≤ deg(mD)− (m−m0 + 1)deg(D) + 1

= (m0 − 1)deg(D) + 1.

Considering dim(H1(0/mD)) as a function of m, we have it is bounded. We check it isnondecreasing. For suppose 0 < m1 < m2. Then 0 < m1D < m2D. We have surjective

maps H1(0)G1,0−→ H1(m1D) and H1(m1D)

G2,1−→ H1(m2D). Note that

G2,1 G1,0 = G2,0 : H1(0) −→ H1(m2D).

It is easy to see that Ker(G1,0) ⊆ Ker(G2,0). Then H1(0/m1D) ⊆ H1(0/m2D) and hence

dim(H1(0/m1D)) ≤ dim(H1(0/m2D)).

17

We have that dim(H1(0/mD)) is a nondecreasing bounded function of m. Therefore, form sufficiently large, we have H1(0/mD) is constant.

Lemma 2.2.3. For any algebraic curve X and for any divisor A on X, there is an integerM such that

deg(A)− dim(L(A)) ≤M.

Proof: Fix a meromorphic function f and let D = (f)∞. Note that

deg(mD)− dim(L(mD)) = dim(H1(0/mD))− 1 ≤M,

since dim(H1(0/mD)) is bounded. We know by Lemma 1.3.1 that there is a meromorphicfunction g on X and an integer m such that B = A − (g) ≤ mD. We shall see A andB have the same degree. Since g is a nonconstant meromorphic function on a compactRiemann surface X, we have

deg((g)) =∑p

(g)(p) =∑p

ordp(g) = 0.

Then we get

deg(B) = deg(A− (g)) = deg(A)− deg((g)) = deg(A).

Now we show that L(A) ∼= L(B). If f ∈ L(A) then (f) ≥ −A = −B − (g). We have

ordp(f) ≥ −B(p)− ordp(g)

ordp(f) + ordp(g) ≥ −B(p)

ordp(f · g) ≥ −B(p).

Then we can define a map L(A) −→ L(B) by f 7→ f · g, which is clearly an isomorphismof vector spaces, whose inverse is given by f 7→ f/g. It follows dim(L(A)) = dim(L(B)).On the other hand, by Lemma 2.2.1, B ≤ mD implies

dim(H1(B/mD)) = [deg(mD)− dim(L(mD))]− [deg(B)− dim(L(B))].

18

Hence, we have

deg(A)− dim(L(A)) = deg(B)− dim(L(B))

= deg(mD)− dim(L(mD))− dim(H1(B/mD))

≤ deg(mD)− dim(L(mD))

≤M.

Note the previous lemma implies that there exists a divisor A0 on X such that the differencedeg(A0)− dim(L(A0)) is maximal.

Lemma 2.2.4. H1(A0) = 0.

Proof: Suppose that H1(A0) = T [A0](X)/Im(αA0) 6= 0. Then there exists a Laurenttail divisor Z such that Z 6∈ Im(αA0). Let B be a divisor with B ≥ A0. Recall thatZ =

∑p rp · p ∈ Ker(tA0

B ) if and only if the top term of each rp has order smaller than−A0(p) and the bottom term has order at least −B(p). Choose a divisor B such thatZ ∈ Ker(tA0

B ). Hence, tA0B (Z) + Im(αB) = 0 + Im(αB). So

Z + Im(αA0) ∈ Ker(H1(A0)GB,A0−→ H1(B)) = H1(A0/B).

Hence H1(A0/B) 6= 0 and dim(H1(A0/B)) ≥ 1. By Lemma 2.2.1, we get

1 ≤ dim(H1(A0/B)) = [deg(B)− dim(L(B))]− [deg(A0)− dim(L(A0))].

On the other hand,

[deg(B)− dim(L(B))]− [deg(A0)− dim(L(A0))] < 0

since deg(A0)− dim(L(A0)) is maximal, getting a contradiction.

19

Proof of the Proposition 2.2.2: Write D − A0 = P − N , where P and N are non-negative divisors. We have that A0 ≤ A0 + P and so we have a surjective mapH1(A0) −→ H1(A0 + P ), where H1(A0) = 0. It follows H1(A0 + P ) = 0. SinceA0 + P −N ≤ A0 + P , we have a short exact sequence

0 −→ H1(A0 + P −N/A0 + P ) −→ H1(A0 + P −N) −→ H1(A0 + P ) −→ 0,

where H1(A0 + P ) = 0. By exactness, we get

H1(D) = H1(A0 + P −N) ∼= H1(A0 + P −N/A0 + P ),

where the last space is finite dimensional by Lemma 2.2.1.

So far we have proven that for any two divisors D1 and D2 with D1 ≤ D2, there is a shortexact sequence

0 −→ H1(D1/D2) −→ H1(D1) −→ H1(D2) −→ 0

of finite dimensional vector spaces, where H1(D1) ∼= H1(D1/D2)⊕

H1(D2) since this se-quence splits. It follows

dim(H1(D1/D2)) = dim(H1(D1))− dim(H1(D2)).

By Lemma 2.2.1, we also have

dim(H1(D1/D2)) = [deg(D2)− dim(L(D2))]− [deg(D1)− dim(L(D1))].

So we get

deg(D2)− dim(L(D2))− deg(D1) + dim(L(D1)) = dim(H1(D1))− dim(H1(D2))

dim(L(D1))− deg(D1)− dim(H1(D1)) = dim(L(D2))− deg(D2)− dim(H1(D2)).

Now let D1 and D2 be any divisors on X, and let D be a common maximum of D1 and D2.Then

dim(L(D1))− deg(D1)− dim(H1(D1)) = dim(L(D))− deg(D)− dim(H1(D))

= dim(L(D2))− deg(D2)− dim(H1(D2)).

It follows that the integer dim(L(D))− deg(D)− dim(H1(D)) is constant and it is equal to

dim(L(0))− deg(0)− dim(H1(0)) = 1− dim(H1(0)).

Summarizing, we have

20

Theorem 2.2.1 (Riemann-Roch: First Form). Let D be a divisor on an algebraic curve X.Then

dim(L(D))− dim(H1(D)) = deg(D) + 1− dim(H1(0)).

2.3 Serre Duality and the Riemann-Roch Theorem

The First Form of the Riemann-Roch Theorem have some problems. In its formula, weneed to consider three spaces: L(D), H1(D) and H1(0). Our next goal is to find moresuitable expressions for dim(H1(D)) and dim(H1(0)). Such expressions will be consequencesof a result called Serre Duality. Roughly speaking, Serre duality states that the space ofmeromorphic 1-forms with poles bounded by −D, L(1)(−D), and the dual space to H1(D)are isomorphic. We first construct a linear map L(1)(−D) −→ H1(D)∗, which we shall callthe Residue map. Suppose D is a divisor on X and ω a meromorphic 1-form on X in thespace L(1)(−D), i.e. ordp(ω) ≥ D(p) for all p ∈ X. It follows we can write

ω =

∞∑n=D(p)

cnznp

dzp

in a local coordinate zp about p, for each p. We define the following linear map:

Resω : T [D](X) −→ C∑p

rp · p 7→∑p

Resp(rp · ω).

Suppose f is a meromorphic function on X. Write f =∑

k akzkp in the coordinate zp. Near

p, we have

fω =

(∑k

akzkp

)·

∞∑n=D(p)

cnznp

dzp.

The coefficient of 1/zp is given by∑∞

n=D(p) cna−n−1. So

Resp(fω) =∞∑

n=D(p)

cna−n−1.

The expression Resp(fω) depends only on the Laurent tail divisor αD(f) =∑

p rp · p, since∑∞n=D(p) cna−n−1 depends on the coefficients ai for f with i < −D(p). Then we have

Resω(αD(f)) =∑p

Resp(fω).

21

By the Residue Theorem, we get Resω(αD(f)) = 0. This means that the previous map Resωdescends to a map

Resω : H1(D) −→ C

in H1(D)∗. Therefore, we obtain a linear map

Res : L(1)(−D) −→ H1(D)∗

ω 7→ Resω.

Theorem 2.3.1 (Serre Duality). For any divisor D on an algebraic curve X, the map

Res : L(1)(−D) −→ H1(D)∗

is an isomorphism of complex vector spaces.

Before proving this theorem, we need two lemmas and some properties.

Proposition 2.3.1.

(1) If φ : T [D](X) −→ C is a linear functional that vanishes on αD(M(X)), and f is any

meromorphic function on X, then φ µD+(f)f : T [D + (f)](X) −→ C is also a linear

map vanishing on αD+(f)(M(X)).

(2) Resω is compatible with the multiplication map µf : Suppose f is a meromorphic func-tion on X and ω ∈ L(1)(−D). Then fω ∈ L(1)(−D − (f)) and

Resω µD+(f)f = Resfω

as functionals on T [D + (f)](X)

Proof: By the first part of Proposition 2.1.1, we have

φ(µD+(f)f (αD+(f)(g))) = φ(αD+(f)−(f)(f · g)) = φ(αD(f · g)) = 0.

Hence (1) holds. Part (2) follows by noticing that

Resp((frp) · ω) = Resp(rp · (fω)).

22

Lemma 2.3.1. Suppose that φ1 and φ2 are two linear functionals on H1(A), for some divisorA. Then there is a positive divisor C and nonzero meromorphic functions f1 and f2 in L(C)such that

φ1 tA−C−(f1)A µf1 = φ2 tA−C−(f2)A µf2as functionals on H1(A−C). In other words, the two maps on T [A−C](X) in the diagram

T [A− C − (f1)](X) T [A](X)

T [A− C](X) C

T [A− C − (f2)](X) T [A](X)

µf1

µf2

tA−C−(f1)A

tA−C−(f2)A

φ1

φ2

are equal for some C and some f1, f2 ∈ L(C)\0.

Proof: First, note that φ1 and φ2 can be considered as linear functionals on T [A](X)that vanish on αA(M(X)). Suppose no such divisor C and functions f1 and f2 exist.Then for every positive divisor C, we have a linear map

L(C)× L(C) −→ H1(A− C)

(f1, f2) 7→ φ1 tA−C−(f1)A µf1 − φ2 tA−C−(f2)A µf2

which turns out to be injective. It follows

(1) dim(H1(A− C)) ≥ dim(L(C)× L(C)) = 2dim(L(C)).

By the First Form of the Riemann-Roch Theorem applied to A− C and C, we get

dim(L(A− C))− dim(H1(A− C)) = deg(A− C) + 1− dim(H1(0))

dim(H1(A− C)) = dim(L(A− C))− deg(A− C)− 1 + dim(H1(0))

(2) dim(H1(A− C)) ≤ dim(L(A))− deg(A)− 1 + dim(H1(0)) + deg(C),

dim(L(C))− deg(H1(C)) = deg(C) + 1− dim(H1(0))

dim(L(C)) = dim(H1(C)) + deg(C) + 1− dim(H1(0))

(3) dim(L(C)) ≥ deg(C) + 1− dim(H1(0)).

23

Connecting (1), (2) and (3), we have

2deg(C) + 2− 2dim(H1(0)) ≤ 2dim(L(C))

≤ dim(H1(A− C))

≤ dim(L(A))− deg(A)− 1 + dim(H1(0)) + deg(C)

2deg(C) + 2− 2dim(H1(0)) ≤ dim(L(A))− deg(A)− 1 + dim(H1(0)) + deg(C)

deg(C) + 3− 3dim(H1(0)) ≤ dim(L(A))− deg(A)

deg(C) ≤ dim(L(A))− deg(A)− 3 + 3dim(H1(0)).

We have that deg(C) is bounded. Since C is arbitrary and positive, we can take the limitdeg(C) −→∞, getting a contradiction.

Lemma 2.3.2. Suppose that D1 is a divisor on X with ω ∈ L(1)(−D1), so that Resω :T [D1](X) −→ C is well defined. Suppose that D2 ≥ D1 and that Resω vanishes on thekernel of tD1

D2: T [D1](X) −→ T [D2](X). Then ω ∈ L(1)(−D2).

Proof: Suppose ω 6∈ L(1)(−D2). Then there is a point p ∈ X with k = ordp(ω) < D2(p).Consider Z = z−k−1p ·p, which is a Laurent tail divisor in T [D1](X) since −k−1 < −D1(p).

On the other hand, −k − 1 ≥ −D2(p), and so Z ∈ Ker(tD1D2

).

Write ω =(∑∞

n=D1(p)cnz

np

)dzp. Then

z−k−1p · ω =

∞∑n=D1(p)

cnzn−k−1p

dzp.

Since k = ordp(ω), we have Resω(Z) = Resp(z−k−1p · ω) = ck 6= 0, getting a contradiction.

24

Proof of Serre Duality Theorem:

(1) Res is injective: Suppose the converse, i.e. that there exists ω ∈ L(1)(−D)\0 suchthat Res(ω) ≡ 0 on H1(D). Then

∑p Resp(rp ·ω) = 0 for every

∑rp ·p ∈ T [D](X).

Now fix p ∈ X and a coordinate zp about p. Note that k = ordp(ω) ≥ D(p) sinceω ∈ L(1)(−D). We have −k − 1 < D(p) and so z−k−1p · p ∈ T [D](X). Write

ω =

(∞∑n=k

cnznp

)dzp

where ck 6= 0. We have

Resω(z−k−1p · p) = Resp

(z−k−1p ·

∞∑n=k

cnznp

)= ck 6= 0,

contradicting the fact that Resω ≡ 0.

(2) Res is surjective: Let φ : H1(D) −→ C be a functional in H1(D)∗, which can beconsidered as a functional T [D](X) −→ C vanishing on αD(M(X)). Let ω 6= 0be a meromorphic 1-form, and consider the canonical divisor K = (ω). We canfind a divisor A such that A ≤ D and A ≤ K. Then ω ∈ L(1)(−A). Consider the

composite map φA : T [A](X)tAD−→ T [D](X)

φ−→ C. We have two linear functionalson T [A](X), namely φA and Resω. Note that Resω vanishes on αA(M(X)) sinceω ∈ L−1(−A). On the other hand,

φA(αA(f)) = φ tAD αA(f) = φ αD(f) = 0.

So φA also vanishes on αA(M(X)). Now we can apply Lemma 2.3.1 to get a positivedivisor C and meromorphic functions f1, f2 ∈ L(C) such that

φA tA−C−(f1)A µf1 = Resω tA−C−(f2)A µf2

where both sides can be considered as functionals on H1(A− C). For

φA tA−C−(f1)A µf1(αA−C(g)) = φA tA−C−(f1)A αA−C−(f1)(f1 · g)

= φ tAD αA(f1 · g)

= φ αD(f1 · g) = 0.

A similar argument shows that Resω tA−C−(f2)A µf2 also vanishes on αA−C(M(X)).We shall prove that φ = Res((f2/f1)ω). Since −C ≤ (f2), we have A−C ≤ A+(f2).

25

Then we can consider the following commutative diagram

T [A− C](X) T [A− C − (f2)](X)

T [A+ (f2)](X) T [A](X)

µA−Cf2

tA−CA+(f2) tA−C−(f2)A

µA+(f2)f2

Then we have

Resω tA−C−(f2)A µA−Cf2= Resω µA+(f2)

f2 tA−CA+(f2)

φA tA−C−(f1)A µA−Cf1= Resω µA+(f2)

f2 tA−CA+(f2)

= Resf2ω tA−CA+(f2)

φA tA−C−(f1)A = Resf2ω tA−CA+(f2) µA−C−(f1)1/f1

.

Similarly, we have another commutative square

T [A− C − (f1)](X) T [A− C](X)

T [A+ (f2/f1)](X) T [A+ (f2)](X)

µA−C−(f1)1/f1

tA−C−(f1)A+(f2/f1)

tA−CA+(f2)

µA+(f2/f1)1/f1

since A− C − (f1) ≤ A+ (f2/f1). Then we have

φA tA−C−(f1)A = Resf2ω tA−CA+(f2) µA−C−(f1)1/f1

= Resf2ω µA+(f2/f1)1/f1

tA−C−(f1)A+(f2/f1)

= Res(f2/f1)ω tA−C−(f1)A+(f2/f1)

Note that

((f2/f1)ω) = (f2/f1) + (ω) = (f2)− (f1) + (ω) ≥ A− C − (f1).

So (f2/f1)ω ∈ L(1)(A − C − (f1)). Then Res(f2/f1)ω : T [A + (f2/f1)](X) −→ Ccan be considered as a map Res(f2/f1)ω : T [A − C − (f1)](X) −→ C. Now if

Z ∈ Ker(tA−C−(f1)A ), then Res(f2/f1)ω(Z) = 0. By Lemma 2.3.2, we have (f2/f1)ω

belongs to L(1)(−A).

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It follows we can consider the map Res(f2/f1)ω defined on T [A](X). Note that thefollowing square commutes:

T [A− C − (f1)](X) T [A+ (f2/f1)](X)

T [A](X) C

tA−C−(f1)A+(f2/f1)

tA−C−(f1)A

Res(f2/f1)ω

Res(f2/f1)ω

Then we have φA tA−C−(f1)A = Res(f2/f1)ω tA−C−(f1)A . Since t

A−C−(f1)A is surjective,

we getφ tAD = φA = Res(f2/f1)ω.

Similarly, one can show that (f2/f1)ω ∈ L(1)(−D) using Lemma 2.3.2. SoRes(f2/f1)ω = Res(f2/f1)ω tAD, and we get

φ tAD = Res(f2/f!)ω tAD.

Hence φ = Res((f2/f1)ω) since tAD is surjective.

Fix a canonical divisor K = (ω) for some meromorphic 1-form ω on X. By Proposition 1.1.3,we know that L(1)(−D) and L(K −D) are isomorphic. Hence, by Serre Duality we get

dim(H1(D)) = dim(L(1)(−D)) = dim(L(K −D)).

Finally, we show that dim(H1(0)) = g. Let K ′ be a canonical divisor on X of degree 2g − 2,which exists by Proposition 1.1.2. By Serre Duality, we have

dim(H1(0)) = dim(L(K ′ − 0)) = dim(L(K ′)),

dim(H1(K ′)) = dim(L(K ′ −K ′)) = dim(L(0)) = 1.

By the First Form of the Riemann-Roch Theorem and the previous equality, we have

dim(L(K ′))− dim(H1(K ′)) = deg(K ′) + 1− dim(H1(0))

dim(H1(0))− 1 = 2g − 2 + 1− dim(H1(0))

dim(H1(0)) = g.

We have obtained a more refined form of Theorem 2.2.1:

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Theorem 2.3.2 (Riemann-Roch). Let X be an algebraic curve of genus g. Then for anydivisor D and any canonical divisor K, we have

dim(L(D))− dim(L(K −D)) = deg(D) + 1− g.

2.4 Generalizations of the Riemann-Roch Theorem

In this section we shall comment three of the generalizations of the Riemann-Roch Theorem.We shall neither prove these results nor explain in detail the constructions used. We referthe reader to [1] and [2] for a more detailed exposition of this matter.

The first generalization that we shall give of the Riemann-Roch Theorem is its version forholomorphic line bundles. Let L be a holomorphic line bundle on a compact Riemann surfaceX of genus g and let Γ(X,L) denote the space of holomorphic sections of L. This space isfinite-dimensional. Let K denote the canonical bundle on X, i.e. K = Λn(T∨) where T∨ isthe cotangent bundle and n = dim(X).

Now consider a general section σ : X −→ L. We can produce such a section by giving itlocally and then glueing it together using a partition of unity. Locally, the line bundle L istrivial so it looks like C ×∆ −→ ∆ where ∆ is the open unit disk. In this local picture, σis just a map ∆ −→ C. Locally, the inverse image of 0 ∈ C under the map σ : ∆ −→ C hasa finite number of points. Each point p ∈ X where σ intersects the zero section is called azero of σ. Around each such point p the section σ is a map σ : ∆ −→ C where p = 0 ∈ ∆and σ(0) = 0. The differential Tpσ : T0∆ −→ T0C is a nonsingular two-by-two matrix. Letsgn(p) denote the sign of the determinant of this matrix. The degree of L is defined bydeg(L) =

∑p∈X sgn(p) where the sum is over all points p where a section σ is zero.

The Riemann-Roch Theorem for holomorphic line bundles goes as follows:

Theorem 2.4.1. If L is a holomorphic line bundle on X and K is the canonical bundle onX, then

dim(Γ(X,L))− dim(Γ(X,L−1 ⊗K)) = deg(L) + 1− g.

The Hirzebruch-Riemann-Roch Theorem is a result that contributes to the Riemann-Rochproblem for complex algebraic varieties of all dimensions. The theorem applies to any holo-morphic vector bundle E on a compact complex manifold X, to calculate the holomorphicEuler characteristic of E in sheaf cohomology, namely the alternating sum

χ(X,E) = dim(H0(X,E))− dim(H1(X,E)) + dim(H2(X,E))− · · ·

of the dimensions as complex vector spaces.

Hirzebruch’s theorem states that χ(X,E) is computable in terms of the Chern classes Cj(E)of E, and the Todd polynomials Tj in the Chern classes of the holomorphic tangent bundleof X. These all lie in the cohomology ring of X.

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Theorem 2.4.2. Let E be a holomorphic vector bundle on a compact complex manifold X.Then

χ(X,E) =n∑j=0

chn−j(E)Tjj!

using the Chern character ch(E) in cohomology.

Finally, we comment the Grothendieck-Riemann-Roch Theorem. It is a generalization of theHirzebruch-Riemann-Roch Theorem, about complex manifolds. Let X be a smooth quasi-projective scheme over a field. The Grothendieck group K0(X) of bounded complexes ofcoherent sheaves is canonically isomorphic to the Grothendieck group of bounded complexesof finite-rank vector bundles. Using this isomorphism, consider the Chern character as afunctorial transformation

ch : K0(X) −→ Ad(X,Q)

where Ad(X,Q) is the Chow group of cycles on X of dimension d modulo rational equiva-lence, tensored with the rational numbers. Now consider a proper morphism f : X −→ Ybetween smooth quasi-projective schemes and a bounded complex of sheaves F•. Let td(X)be the Todd genus of the tangent bundle of X. We denote the i-right derived functor of thepushforward f∗ by Rif∗. The Grothendieck-Riemann-Roch Theorem goes as follows:

Theorem 2.4.3. ch(f!F•)td(Y ) = f∗(ch(F•)td(X)), where

f! :∑

(−1)iRif∗K0(X) −→ K0(Y ) and f∗ : A(X) −→ A(Y ).

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Bibliography

[1] Harder, Gunter. Lectures on Algebraic Geometry II. Vieweg+Teubner. Heidelberg, Ger-many (2011).

[2] Javanpeykat, Ariyan. The Grothendieck-Riemann-Roch Theorem with an Applicationto Covers of Varieties. Master’s Thesis. Mathematisch Instituut, Universiteit Leiden(2010).

[3] Miranda, Rick. Algebraic Curves and Riemann Surfaces. Graduate Studies in Mathe-matics. American Mathematical Society. Providence (1995).

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