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Home , Canonical bundle, Riemann sphere, Riemann surface, Serre duality

Universit´edu Qu´ebec `aMontr´eal D´epartement de math´ematiques Marco A. P´erezB. [email protected]

THE -ROCH THEOREM

January 2012

Contents

Introduction i

1 Algebraic and some properties of its 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 expressions in some nonconstant ...... 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 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 , for the computation of the of the space of meromorphic functions with prescribed zeroes and allowed poles. Roughly speaking, it links the of a compact Riemann with its topological . Initially proved as Riemann’s inequality by in 1857,

Theorem 1. Let X be a 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 , one of Riemann’s students, in 1865 providing the error term:

Theorem 2. Let X be a Riemann surface of genus g. Then for any divisor D and any canonical 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 a purely 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’s principle. 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 appropriate notion of divisor, or . The general formulation of the theorem depends on splitting it into two parts. One, which would now be called , interprets the L(K − D) term as a dimension of a first cohomology ; with L(D) the dimension of a zeroth cohomology group, or space of sections, the left-hand side of the theorem becomes an , and the right-hand side a computation of it as a degree corrected according to the of the Riemann surface. An n-dimensional generalization, the Hirzebruch-Riemann-Roch Theorem, was found and proved by Friedrich Hirzebruch, as an application of characteristic classes in .

i At about the same time Jean-Pierre Serre was giving the general form of Serre Duality. proved a far-reaching generalization in 1957, now known as the Grothendieck-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 of . Then we give the concept of an algebraic and study some of its properties, such as the Laurent Series Approximation Theorem. In the last section of this chapter, we study the field of meromorphic functions on an X, M(X), as a field extension of certain subfield. Such a subfield is the field of rational expressions in a nonconstant meromorphic function on X. The degree of this extension shall be an important tool to prove the Riemann-Roch Theorem. In the second chapter we prove the Riemann-Roch Theorem for algebraic curves, by studying the 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 Duality Theorem, 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-Roch Theorem for holomorphic line bundles, the Hirzebruch-Riemann-Roch Theorem, and the Grothendick-Riemann-Roch Theorem.

ii Chapter 1

Algebraic curves and some properties of its function field

This chapter is devoted to the study of several properties of a certain type of Riemann surfaces called algebraic curves. In the first section we shall recall some notions of complex geometry, such as meromorphic functions, divisors, and meromorphic 1-forms. We shall also give 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 of points in X and tails of Laurent series such that at each of these points the Laurent series of the function starts with one of these tails. Another property of the algebraic curves is that one can always find a nonconstant meromorphic function on it. So given a nonconstant meromorphic function f on al algebraic curve X, one can consider the field C(f) of rational expressions of f with coefficients in C, which is a subfield of the field M(X) of meromorphic functions on X. The Laurent Series Approximation Theorem shall help us to compute the degree 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= 0 is called the support of D.A divisor on X is a function D : X −→ Z whose support is a discrete subset of X. We shall denote a divisor D by using the following summation notation: X D = D(p) · p. p∈X

1 The divisor of a meromorphic function f : X −→ C, denoted by (f), is the divisor defined by the order function: X (f) = ordp(f) · p. p∈X

The divisor of poles of f, denoted by (f)∞, is the divisor X (f)∞ = (−ordp(f)) · p.

p with ordp(f)<0

The space of meromorphic functions with poles bounded by D, denoted L(D), is the set 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. Then the 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 V ⊆ C is an expression of the form ω = f(z)dz, where f is a meromorphic function on V . Let ω1 = f(z)dz and ω2 = g(w)dw be two meromorphic differentials defined on V1 and V2. Let T be a from V2 to V1. We say that ω1 transforms to ω2 under T if

g(w) = f(T (w))T 0(w).

If X is a Riemann surface, a meromorphic 1-form on X is a collection of meromorphic differentials (ωϕ), one for each chart ϕ : U −→ V and defined on V , such that if two charts

ϕ1 : U1 −→ V1 and ϕ2 : U2 −→ V2 have overlapping domains, then ωϕ1 transforms to ωϕ2 −1 under the transition function ϕ1 ◦ ϕ2 . The notion of holomorphic 1-form is defined in a similar 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 the Laurent series of f in the coordinate zp:

∞ ! X n ω = f(zp)dzp = cnzp dzp n=−M

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

Theorem 1.1.1 (The Residue Theorem). Let ω be a meromorphic 1-form on a compact Riemann surface X. Then X Resp(ω) = 0. p∈X

The divisor of ω, denoted by (ω), is the divisor defined by the order function X (ω) = ordp(ω) · p. 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 nonconstant meromorphic 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 by D, 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 of holomorphic 1-forms. The following result gives a way to relate the spaces L(−) and L(1)(−). You can check the proof 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. Then the 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 ∈ X if either f is holomorphic at p ∈ X and ordp(f −f(p)) = 1, or f has a simple pole at p. Recall that for every meromorphic function f : X −→ C, there corresponds a unique holomorphic map F : X −→ C ∪ {∞} which is not identically ∞. Such a map is given by  f(x) ∈ if x is not a pole of f, F (x) = C ∞ 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.

3 (1) We say that S separates points in X if for every pair of distinct points p and q in X 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 meromorphic functions on X separates points and tangents of X. Note that (1) is easy to understand. But (2) is not so clear. Suppose that S separates tangents. Then there is a meromorphic function f : X −→ C having multiplicity one at each p ∈ X. Then the associated holomorphic map F : X −→ C ∪ {∞} has multiplicity one at 0 each p ∈ X. So ordp(F − F (p)) = 1, i.e. (F − F (p)) (p) 6= 0. Hence the derivative map 0 F : TpX −→ TF (p)C ∪ {∞} is one-to-one. Note also that if ordp(f − f(p)) = 1, then f is a local coordinate chart, since it is locally a biholomorphic map from an open neighbourhood U ⊆ X to an open set V ⊆ C. Therefore, S separates tangents if and only if at every point p ∈ X, there is a local coordinate which extends to a meromorphic function on all X. It follows that if X is an algebraic curve, then for every p ∈ X there exists a meromorphic function g on X such that ordp(g) = 1. For take a function f separating tangents at p, if f is holomorphic at p, set g = f − f(p), and if f has a simple pole at p set g = 1/f. S Example 1.1.1. The Riemann C {∞} and the complex 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 in the 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 there is a global meromorphic function f on X with ordp(f) = N.

Pm i A Laurent r(z) = i=n ciz 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 coordinate z centred at p. Fix any Laurent polynomial r(z) in z. Then there exists a meromorphic function f on X whose Laurent series at p has r(z) as a Laurent tail.

4 Pm i Proof: Let r(z) = i=n ciz 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 of r(z). Suppose m − n + 1 = 1. Then z(z) = czm. By the previous lemma, there exists a meromorphic function f on X with ordp(f) = m. Then its Laurent series in the coordinate z has the form P∞ a zi, where a 6= 0. Then c f is the desired meromorphic function. i=m i m am Now assume that m − n + 1 ≥ 2. We know that we can find a meromorphic function h n having cnz as a Laurent tail. Consider the meromorphic function h − r. Let s(z) be the Laurent polynomial which is the tail of the Laurent series of h − r at p, up through the zm term. Note that h(z) − r(z) starts at n + 1. It is clear that s(z) has fewer terms than r(z). By induction, there is a meromorphic function g on X whose Laurent series at p has 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, . . . , qn in 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 meromorphic function on X such that g(p) 6= g(q). Suppose that p is not a pole of g. We may assume that g(p) = 0. There is nothing to show if q is a pole of g. If not, then consider f = g/(g(q) − g). Then f(p) = 0 and f has a pole at q. Now if p is a pole of g, repeat the previous argument with 1/g. Suppose n ≥ 2 and let g be a meromorphic function with a zero at p and a pole at q1, . . . , qn−1. On the other hand, there is a meromorphic function h with a zero at p and a 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 a m m pole at q1. For m sufficiently large, g + h has a pole at q1 since the order of h at q1 is greater than the order of g at the same point. Increasing m if necessary, we may assume m that g + h has a pole at each qi with i ≤ n − 1. We know that g has a pole at qn, but it may 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, . . . , qn in X, and any N ≥ 1, there is a global meromorphic function f on X with ordp(f − 1) ≥ N and ordqi (f) ≥ N for each i.

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

that ordqi (f) ≥ N.

Theorem 1.2.1 (Laurent Series Approximation). Suppose X is an algebraic curve. Fix a finite number of points p1, . . . , pn in X, choose a local coordinate zi at each , and finally choose Laurent ri(zi) for each i. Then there is a meromorphic function f on X such 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 as a Laurent tail at pi. Let

M = min{ordpi (ri)/ i = 1, . . . , n} = min{ordpi (gi)/ i = 1, . . . , n}.

Let N be a fixed integer larger than every exponent of every term of every ri. By Lemma i 1.2.4, there are meromorphic function hi on X such that for each i, ordpi (h −1) ≥ N −M Pn and ordpj (hi) ≥ N − M for j 6= i. The function f = i=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 in X, and a finite number of integers mi. Then there exists a meromorphic function f on X such that ordpi (f) = mi for each i.

1.3 The function field M(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 meromorphic function on an algebraic curve X. We have the inclusions of fields C ⊆ C(f) ⊆ M(X). Then M(X) is a field extension of C(f), and hence M(X) can be considered as a vector space over C(f). Let [M(X): C(f)] denote the dimension of M(X) over C(f), which is called the degree of the extension M(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 this result will only help to show the inequality [M(X): C(f)] ≥ deg((f)∞). We shall need other results to prove the other inequality.

6 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 a meromorphic function g on X such that A − (g) ≤ mD. Moreover, g can be taken to be a polynomial 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 zero A(pi) at pi or order at least 1, and having the same poles as f. It follows that (f − f(pi)) is a meromorphic function with a zero at pi of order at least A(pi), and whose poles are Qk A(pi) the poles of f. Let g = i=1(f − f(pi)) . Then g is a meromorphic function such that A − (g) is positive only at the poles of f. Now suppose that p is a pole of f, we have

k X A(p) − (g)(p) = A(p) − ordp(g) = A(p) − A(pi)ordp(f − f(pi)) i=1 k k ! X X = A(p) − A(pi)ordp(f) = A(p) − A(pi) · ordp(f) i=1 i=1 k ! X = A(p) + A(pi) · (f)∞(p) i=1

≤ 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 nonconstant meromorphic functions on X. Then there is a polynomial r(t) ∈ C[t] such that the function r(f)h has no poles outside of the poles of f. In this case there is an integer m such that r(f)h ∈ L(mD), where D = (f)∞.

Proof: Put A = −(h) in the previous lemma. There exists a meromorphic function g on X such that −(h) − (g) ≤ m(f)∞, for some integer m > 0, where such a g can be chosen 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 of r(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 let D = (f)∞. Suppose that [M(X): C(f)] ≥ k. Then there is an integer m0 such that for all m ≥ m0, dim(L(mD)) ≥ (m − m0 + 1)k.

Proof: Since [M(X): C(f)] ≥ k, we can choose linearly independent meromorphic functions g1, . . . , gk over C(f). By the previous corollary, for each i there exists a nonzero rational 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 an integer mi such that hi ∈ L(miD). Let p ∈ X, we have (hi)(p) ≥ −miD(p). Let m0 = max{mi : 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.

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

i i i (f hj)(p) = ordp(f hj) = ordp(f ) + 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).

i i Then f hj ∈ L(mD). Note that the f hj’s are linearly independent over C(f). Then i the C(f)-vector space generated by the f hj’s has dimension (m − m0 + 1)k. It follows dim(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)∞.

8 Proof: We first show that [M(X): C(f)] ≤ deg(D). Suppose [M(X): C(f)] is at least deg(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. P 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 meromorphic function gij with a pole at pi or order j, and no zero or pole at any of the other pk’s. The set {gij / 1 ≤ j ≤ ni} is linearly independent over C(f), for each i. Then we have X X [M(X): C(f)] ≥ card({gij / 1 ≤ j ≤ ni}) = ni = deg(D). i i

9 10 Chapter 2

The Grothendieck-Riemann-Roch Theorem

In this chapter we shall prove the Riemann-Roch Theorem for algebraic curves, and comment two generalizations known as the Hirzebruch-Riemann-Roch Theorem and the Grothendieck- Riemann-Roch Theorem. In the first section we study certain sums called Laurent tail divisors. We can relate Laurent tail divisors and ordinary divisors via an operation called truncation. This allows us to define a group T [D](X), where X is an algebraic curve, formed by Laurent tail divisors that are bounded, in some sense, by a divisor D. We can also relate meromorphic functions to Laurent tail divisors. In fact, for every meromorphic function there corresponds an element in T [D] via a group homomorphism αD : M(X) −→ T [D](X). This map is of vital importance to prove the Riemann-Roch Theorem. Recall that this theorem gives us a formula to compute the dimension of the space L(D), which turns out to be the kernel of the homomorphism αD. A first form of this theorem involves the cokernel of this homomorphism. At this point, the problem is that we are dealing with two spaces, namely L(D) and the cokernel of αD. We shall prove and use a result called Serre Duality to give an easy way two compute the dimension of the cokernel of αD, and so we shall obtain a refined version of the Riemann-Roch Theorem that allows us to compute the dimension of L(D) only in 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 zp centred at p.A Laurent tail divisor on X is a finite formal sum of the form X rp · p, p

11 where rp(zp) is a Laurent polynomial in the coordinate zp. The set of Laurent tail divisors forms a group under formal , denoted T (X). We shall focus on special of P T (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 that T [D](X) is a of T (X). From this point, we shall construct group homomorphisms concerning the previous groups. P Consider a Laurent tail divisor p rp · p and a divisor D. At each p, we have

mp X p i rp(zp) = ai zp. i=np

Let iD be the smallest integer between np and mp such that iD + 1 ≥ −D(p). We can define a truncation of rp(zp) as mp iD X p i X p i ai zp 7→ ai zp. i=np i=np This mapping defines a group homomorphism

tD : T (X) −→ T [D](X) P P that sends each p rp · p to p rˆp · p, wherer ˆp denotes the truncation of rp whose top term is the largest integer between np and mp strictly smaller than −D(p), in other words, tD is defined 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) for P every p ∈ X. Then −D2(p) ≤ −D1(p) for every p ∈ X. Let p rp · p ∈ T [D1](X). For each Pmp p i p with rp 6= 0, consider the Laurent polynomial rp(zp) = i=np ai zp, where mp < −D1(p). Suppose there exists an integer between np and mp greater or equal than −D2(p), and let iD2 + 1 be the smallest of such integers. Then we can truncate rp(zp):

mp iD2 X p i X p i ai zp 7→ ai zp. i=np i=np

Pmp p i If such an iD2 does not exist, then just map i=np ai zp to 0. This truncation defines a group homomorphism tD1 : T [D ](X) −→ T [D ](X) D2 1 2 defined by removing from each rp(zp) those terms of degree greater or equal than −D2(p). We shall call the maps tD1 truncation maps. D2 P Consider a meromorphic function f and a divisor D. Let p rp · p be a Laurent tail in P∞ i T [D](X). Let i=n aizp be the Laurent series of f in the coordinate zp, and let rp(zp) = p     Pk b zj. We take the product P∞ a zi · Pk b zj = P∞ a b zi+j, and j=mp j p i=np i p j=mp j p i=np, j=mp i j p

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

D µf : T [D](X) −→ T [D − (f)](X) P P mapping each p rp · p to p(frp) · p, where f · rp is the Laurent polynomial of the above truncation of the series P∞ a b zi+j. It is straightforward to check that µD has an i=np, j=mp i j p f D−(f) inverse µ1/f . Now consider again the Laurent series of f: P∞ a zi . Given a divisor D, there exists a i=np i p smallest integer mp such that mp + 1 ≥ −D(p). Then we can truncate the previous series Pmp i and get a Laurent polynomial rp(zp) = i=np aizp. 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 then the following triangle commutes:

αD1 M(X) T [D1](X)

α tD1 D D2 2

T [D2](X)

(2) αD is compatible with the operators: If f and g are meromorphic func- tions on X, then D µf (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 P∞ a zi be the Laurent series of f in the coordinate z , and let m be the i=np i p p p largest integer greater or equal than np such that mp < −D1(p). Then αD(f) = P Pmp i p rp · p, where rp(zp) = i=np aizp. 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 that kp < mp. Then kp X tD1 ◦ α (f) = a zi = α (f). D2 D1 i p D2 i=np

(2) Consider the Laurent series f(z ) = P∞ a zp and g(z ) = P∞ b zj of f and g p i=np i i p j=mp j p in the coordinate z . We have (f · g)(z ) = P∞ a b zi+j. On the one hand, p p i=np, j=mp i j p Pk j αD(g) = j=m bjzp, where k ≥ mp is the largest integer satisfying k < −D(p). p     Map α (g) to the series P∞ a zi · Pk b zj = P∞ Pk a b zi+j. On D i=np i p j=mp j p i=np j=mp i j p

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

k < −D(p) + (f)(p) = −D(p) + ordp(f).

Then µD(α (g)) = Pk a b zi+j = α (f · g). f D i=np, j=mp i j p D−(f) (3) Let f ∈ L(D). Then (f) ≥ −D. Consider the Laurent series f(z ) = P∞ a zi . p i=np i p Since np = ordp(f) ≥ −D(p), the previous series is mapped to 0 by αD. Now let f ∈ 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 about CoKer(αD)?. Define

1 H (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 the map 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. 1 Also, note that Im(αD) = Im(αD) = Ker(T [D](X) −→ H (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 map tD1 : T [D ](X) −→ T [D ](X). Also, L(D ) ⊆ L(D ). For each D there is a short exact D2 1 2 1 2 i sequence 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.

1 1 (ii) Now define a map G : H (D1) −→ H (D2) by

G(Z + Im(α )) = tD1 (Z) + Im(α ). D1 D2 D2

0 0 We check G is well defined. Suppose Z − Z ∈ Im(αD1 ). Then Z − Z = αD1 (f) for some meromorphic function f. By the first part of the previous proposition, we have

tD1 (Z) − tD1 (Z0) = tD1 ◦ α (f) = α (f) ∈ Im(α ) D2 D2 D2 D1 D2 D2 It is clear that G is a group homomorphism.

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

α γD D1 1 1 0 M(X)/L(D1) T [D1](X) H (D1) 0

tD1 F D2 G

1 0 M(X)/L(D2) T [D2](X) H (D2) 0 γD αD2 2

where γD1 and γD2 are the maps.

Proof: On the one hand

tD1 ◦ α (f + L(D )) = tD1 (α (f)) = α (f), by (1) of the previous proposition D2 D1 1 D2 D1 D2

= αD2 (f + L(D2))

= αD2 ◦ F (f + L(D1)).

15 So the left square commutes. On the other hand,

G ◦ γ (Z) = G(Z + Im(α )) = tD1 (Z) + Im(α ) = γ (tD1 (Z)). D1 D1 D2 D2 D2 D2 Hence the right square commutes.

1 1 Let H (D1/D2) := Ker(G). We use this diagram two show that H (D1/D2) is finite dimen- sional. This fact shall help us to prove the finite dimensionality of H1(D). Using the diagram above and the Snake Lemma, we obtain the following exact sequence

0 −→ Ker(F ) −→ Ker(tD1 ) −→ Ker(G) −→ CoKer(F ) −→ CoKer(tD1 ) −→ CoKer(G) −→ 0. D2 D2

Note that F is surjective. We shall see that so are tD1 and G. Let Z = P r · p ∈ T [D ](X), D2 p p 2 Pmp p i where rp(zp) = i=np ai zp and mp ≥ np is the largest integer with mp < −D2(p) ≤ −D1(p). So Z ∈ T [D ](X) and Z = tD1 (Z). Hence tD1 is surjective, and this implies that so is G. 1 D2 D2

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 r · p ∈ Ker(tD1 ), where r (z ) = Pmp apzi . Since P apzi is mapped to p p D2 p p i=np i p p i p 0, we have n ≥ −D (p). Hence Ker(tD1 ) is the space of all P r · p such that the top term p 2 D2 p p of rp has order less than −D1(p) and the bottom term has order at least −D2(p). At each i p, we have D2(p) − D1(p) possible monomials zp, −D2(p) ≤ i ≤ −D1(p), that are linearly independent. Hence X X X dim(Ker(tD1 )) = (D (p) − D (p)) = D (p) − D (p) = deg(D ) − deg(D ) D2 2 1 2 1 2 1 p p p

Recall that we have a short exact sequence

0 −→ L(D )/L(D ) −→ Ker(tD1 ) −→ H1(D /D ) −→ 0 2 1 D2 1 2

1 Since H (D1/D2) is a free C-module, the previous sequence splits, so M Ker(tD1 ) ∼ (L(D )/L(D )) H1(D /D ) D2 = 2 1 1 2

It follows

dim(Ker(tD1 )) = dim(L(D )/L(D )) + dim(H1(D /D )) D2 2 1 1 2 1 dim(H (D1/D2)) = deg(D2) − deg(D1) + dim(L(D1)) − dim(L(D2)).

1 We have proven that H (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 1 dim(H (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 dimensional vector 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, and 1 let D = (f)∞. Then for any large m, the dimension of H (0/mD) is constant, independent of 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 on ∼ X 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 is nondecreasing. For suppose 0 < m1 < m2. Then 0 < m1D < m2D. We have surjective 1 G1,0 1 1 G2,1 1 maps H (0) −→ H (m1D) and H (m1D) −→ H (m2D). Note that

1 1 G2,1 ◦ G1,0 = G2,0 : H (0) −→ H (m2D).

1 1 It is easy to see that Ker(G1,0) ⊆ Ker(G2,0). Then H (0/m1D) ⊆ H (0/m2D) and hence

1 1 dim(H (0/m1D)) ≤ dim(H (0/m2D)).

17 We have that dim(H1(0/mD)) is a nondecreasing bounded function of m. Therefore, for m 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 integer M 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 meromorphic function g on X and an integer m such that B = A − (g) ≤ mD. We shall see A and B have the same degree. Since g is a nonconstant meromorphic function on a compact Riemann surface X, we have X X deg((g)) = (g)(p) = ordp(g) = 0. p p

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 of 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 difference deg(A0) − dim(L(A0)) is maximal.

1 Lemma 2.2.4. H (A0) = 0.

1 Proof: Suppose that H (A0) = T [A0](X)/Im(αA0 ) 6= 0. Then there exists a Laurent

tail divisor Z such that Z 6∈ Im(αA0 ). Let B be a divisor with B ≥ A0. Recall that P A0 Z = p rp · p ∈ Ker(tB ) 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 that A0 A0 Z ∈ Ker(tB ). Hence, tB (Z) + Im(αB) = 0 + Im(αB). So

G 1 B,A0 1 1 Z + Im(αA0 ) ∈ Ker(H (A0) −→ H (B)) = H (A0/B).

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

1 1 ≤ dim(H (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 map 1 1 1 1 H (A0) −→ H (A0 + P ), where H (A0) = 0. It follows H (A0 + P ) = 0. Since A0 + P − N ≤ A0 + P , we have a short exact sequence

1 1 1 0 −→ H (A0 + P − N/A0 + P ) −→ H (A0 + P − N) −→ H (A0 + P ) −→ 0,

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

1 1 ∼ 1 H (D) = H (A0 + P − N) = H (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 short exact sequence 1 1 1 0 −→ H (D1/D2) −→ H (D1) −→ H (D2) −→ 0 1 ∼ 1 L 1 of finite dimensional vector spaces, where H (D1) = H (D1/D2) H (D2) since this se- quence splits. It follows

1 1 1 dim(H (D1/D2)) = dim(H (D1)) − dim(H (D2)).

By Lemma 2.2.1, we also have

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

So we get

1 1 deg(D2) − dim(L(D2)) − deg(D1) + dim(L(D1)) = dim(H (D1)) − dim(H (D2)) 1 1 dim(L(D1)) − deg(D1) − dim(H (D1)) = dim(L(D2)) − deg(D2) − dim(H (D2)).

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

1 1 dim(L(D1)) − deg(D1) − dim(H (D1)) = dim(L(D)) − deg(D) − dim(H (D)) 1 = dim(L(D2)) − deg(D2) − dim(H (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, we need to consider three spaces: L(D), H1(D) and H1(0). Our next goal is to find more suitable expressions for dim(H1(D)) and dim(H1(0)). Such expressions will be consequences of a result called Serre Duality. Roughly speaking, Serre duality states that the space of meromorphic 1-forms with poles bounded by −D, L(1)(−D), and the to H1(D) are isomorphic. We first construct a linear map L(1)(−D) −→ H1(D)∗, which we shall call the Residue map. Suppose D is a divisor on X and ω a meromorphic 1-form on X in the (1) space L (−D), i.e. ordp(ω) ≥ D(p) for all p ∈ X. It follows we can write

 ∞  X n ω =  cnzp  dzp n=D(p) in a local coordinate zp about p, for each p. We define the following linear map:

Resω : T [D](X) −→ C X X rp · p 7→ Resp(rp · ω). p p

P k Suppose f is a meromorphic function on X. Write f = k akzp in the coordinate zp. Near p, we have !  ∞  X k X n fω = akzp ·  cnzp  dzp. k n=D(p) P∞ The coefficient of 1/zp is given by n=D(p) cna−n−1. So

∞ X Resp(fω) = cna−n−1. n=D(p) P The expression Resp(fω) depends only on the Laurent tail divisor αD(f) = p rp · p, since P∞ n=D(p) cna−n−1 depends on the coefficients ai for f with i < −D(p). Then we have X Resω(αD(f)) = Resp(fω). p

21 By the Residue Theorem, we get Resω(αD(f)) = 0. This means that the previous map Resω descends to a map 1 Resω : H (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 D+(f) meromorphic function on X, then φ ◦ µf : 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

D+(f) Resω ◦ µf = Resfω as functionals on T [D + (f)](X)

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

D+(f) φ(µf (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 1 Lemma 2.3.1. Suppose that φ1 and φ2 are two linear functionals on H (A), for some divisor A. Then there is a positive divisor C and nonzero meromorphic functions f1 and f2 in L(C) such that A−C−(f1) A−C−(f2) φ1 ◦ tA ◦ µf1 = φ2 ◦ tA ◦ µf2 as functionals on H1(A − C). In other words, the two maps on T [A − C](X) in the diagram

A−C−(f1) tA T [A − C − (f1)](X) T [A](X) µf1 φ1 T [A − C](X) C µ f2 φ2 T [A − C − (f2)](X) T [A](X) A−C−(f2) tA

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)

A−C−(f1) A−C−(f2) (f1, f2) 7→ φ1 ◦ tA ◦ µf1 − φ2 ◦ tA ◦ 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 limit deg(C) −→ ∞, getting a contradiction.

(1) Lemma 2.3.2. Suppose that D1 is a divisor on X with ω ∈ L (−D1), so that Resω : T [D1](X) −→ C is well defined. Suppose that D2 ≥ D1 and that Resω vanishes on the kernel of tD1 : T [D ](X) −→ T [D ](X). Then ω ∈ L(1)(−D ). D2 1 2 2

(1) Proof: Suppose ω 6∈ L (−D2). Then there is a point p ∈ X with k = ordp(ω) < D2(p). −k−1 Consider Z = zp ·p, which is a Laurent tail divisor in T [D1](X) since −k−1 < −D1(p). D1 On the other hand, −k − 1 ≥ −D2(p), and so Z ∈ Ker(tD ).   2 Write ω = P∞ c zn dz . Then n=D1(p) n p p

 ∞  −k−1 X n−k−1 zp · ω =  cnzp  dzp.

n=D1(p)

−k−1 Since k = ordp(ω), we have Resω(Z) = Resp(zp · 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} such 1 P P that Res(ω) ≡ 0 on H (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) since (1) −k−1 ω ∈ L (−D). We have −k − 1 < D(p) and so zp · p ∈ T [D](X). Write

∞ ! X n ω = cnzp dzp n=k

where ck 6= 0. We have

∞ ! −k−1 −k−1 X n Resω(zp · p) = Resp zp · cnzp = ck 6= 0, n=k

contradicting the fact that Resω ≡ 0.

(2) Res is surjective: Let φ : H1(D) −→ C be a functional in H1(D)∗, which can be considered as a functional T [D](X) −→ C vanishing on αD(M(X)). Let ω 6= 0 be a meromorphic 1-form, and consider the canonical divisor K = (ω). We can find a divisor A such that A ≤ D and A ≤ K. Then ω ∈ L(1)(−A). Consider the A tD φ composite map φA : T [A](X) −→ T [D](X) −→ C. We have two linear functionals on T [A](X), namely φA and Resω. Note that Resω vanishes on αA(M(X)) since ω ∈ L−1(−A). On the other hand,

A φA(αA(f)) = φ ◦ tD ◦ 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 positive divisor C and meromorphic functions f1, f2 ∈ L(C) such that

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

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

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

= φ ◦ αD(f1 · g) = 0.

A−C−(f2) A similar argument shows that Resω ◦tA ◦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

µA−C f2 T [A − C](X) T [A − C − (f2)](X)

tA−C A−C−(f2) A+(f2) tA

T [A + (f2)](X) T [A](X) µA+(f2) f2

Then we have

Res ◦ tA−C−(f2) ◦ µA−C = Res ◦ µA+(f2) ◦ tA−C ω A f2 ω f2 A+(f2) φ ◦ tA−C−(f1) ◦ µA−C = Res ◦ µA+(f2) ◦ tA−C = Res ◦ tA−C A A f1 ω f2 A+(f2) f2ω A+(f2) φ ◦ tA−C−(f1) = Res ◦ tA−C ◦ µA−C−(f1). A A f2ω A+(f2) 1/f1 Similarly, we have another commutative square

µA−C−(f1) 1/f1 T [A − C − (f1)](X) T [A − C](X)

tA−C−(f1) tA−C A+(f2/f1) A+(f2)

T [A + (f2/f1)](X) T [A + (f2)](X) µA+(f2/f1) 1/f1

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

φ ◦ tA−C−(f1) = Res ◦ tA−C ◦ µA−C−(f1) A A f2ω A+(f2) 1/f1 = Res ◦ µA+(f2/f1) ◦ tA−C−(f1) f2ω 1/f1 A+(f2/f1) = Res ◦ tA−C−(f1) (f2/f1)ω A+(f2/f1) Note that

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

(1) So (f2/f1)ω ∈ L (A − C − (f1)). Then Res(f2/f1)ω : T [A + (f2/f1)](X) −→ C can be considered as a map Res(f2/f1)ω : T [A − C − (f1)](X) −→ C. Now if A−C−(f1) Z ∈ Ker(tA ), then Res(f2/f1)ω(Z) = 0. By Lemma 2.3.2, we have (f2/f1)ω belongs to L(1)(−A).

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

tA−C−(f1) A+(f2/f1) T [A − C − (f1)](X) T [A + (f2/f1)](X)

A−C−(f1) Res tA (f2/f1)ω

T [A](X) C Res(f2/f1)ω

A−C−(f1) A−C−(f1) A−C−(f1) Then we have φA ◦ tA = Res(f2/f1)ω ◦ tA . Since tA is surjective, we get A φ ◦ tD = φA = Res(f2/f1)ω. (1) Similarly, one can show that (f2/f1)ω ∈ L (−D) using Lemma 2.3.2. So A Res(f2/f1)ω = Res(f2/f1)ω ◦ tD, and we get

A A φ ◦ tD = Res(f2/f!)ω ◦ tD.

A Hence φ = Res((f2/f1)ω) since tD 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 K0 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(K0 − 0)) = dim(L(K0)), dim(H1(K0)) = dim(L(K0 − K0)) = dim(L(0)) = 1.

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

dim(L(K0)) − dim(H1(K0)) = deg(K0) + 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:

27 Theorem 2.3.2 (Riemann-Roch). Let X be an algebraic curve of genus g. Then for any divisor 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 refer the 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 for holomorphic line bundles. Let L be a holomorphic line bundle on a compact Riemann surface X of genus g and let Γ(X,L) denote the space of holomorphic sections of L. This space is finite-dimensional. Let K denote the on X, i.e. K = Λn(T ∨) where T ∨ is the and n = dim(X). Now consider a general section σ : X −→ L. We can produce such a section by giving it locally and then glueing it together using a partition of unity. Locally, the line bundle L is trivial so it looks like C × ∆ −→ ∆ where ∆ is the open unit . In this local picture, σ is just a map ∆ −→ C. Locally, the inverse image of 0 ∈ C under the map σ : ∆ −→ C has a finite number of points. Each point p ∈ X where σ intersects the zero section is called a zero 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. Let sgn(p) denote the sign of the of this matrix. The degree of L is defined by P deg(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 on X, 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-Roch problem for complex algebraic varieties of all . The theorem applies to any holo- morphic E on a compact complex X, to calculate the holomorphic Euler characteristic of E in , 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 bundle of X. These all lie in the cohomology ring of X.

28 Theorem 2.4.2. Let E be a holomorphic vector bundle on a compact X. Then n X Tj χ(X,E) = ch (E) n−j j! j=0 using the Chern character ch(E) in cohomology.

Finally, we comment the Grothendieck-Riemann-Roch Theorem. It is a generalization of the Hirzebruch-Riemann-Roch Theorem, about complex . Let X be a smooth quasi- projective over a field. The Grothendieck group K0(X) of bounded complexes of coherent sheaves is canonically isomorphic to the Grothendieck group of bounded complexes of finite-rank vector bundles. Using this isomorphism, consider the Chern character as a functorial 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 . Now consider a f : X −→ Y between 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 the i pushforward f∗ by R f∗. The Grothendieck-Riemann-Roch Theorem goes as follows:

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

X i i f! : (−1) R f∗K0(X) −→ K0(Y ) and f∗ : A(X) −→ A(Y ).

29 30 Bibliography

[1] Harder, G¨unter. Lectures on II. Vieweg+Teubner. Heidelberg, Ger- many (2011).

[2] Javanpeykat, Ariyan. The Grothendieck-Riemann-Roch Theorem with an Application to 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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