Theory of Algebraic Surfaces [1st ed.] 9789811573798, 9789811573804

Table of contents :
Front Matter ....Pages i-xiii
Fundamentals of Algebraic Surfaces (Kunihiko Kodaira)....Pages 1-45
Pluri-Canonical Systems on Algebraic Surfaces of General Type (Kunihiko Kodaira)....Pages 47-73
Back Matter ....Pages 75-75

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SPRINGER BRIEFS IN MATHEMATICS

Kunihiko Kodaira

Theory of Algebraic Surfaces

SpringerBriefs in Mathematics Series Editors Nicola Bellomo, Torino, Italy Michele Benzi, Pisa, Italy Palle Jorgensen, Iowa, USA Tatsien Li, Shanghai, China Roderick Melnik, Waterloo, Canada Otmar Scherzer, Linz, Austria Benjamin Steinberg, New York, USA Lothar Reichel, Kent, USA Yuri Tschinkel, New York, USA George Yin, Detroit, USA Ping Zhang, Kalamazoo, USA

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Kunihiko Kodaira

Theory of Algebraic Surfaces

Kunihiko Kodaira University of Tokyo Tokyo, Japan Translated by Kazuhiro Konno Takatsuki, Japan

ISSN 2191-8198 ISSN 2191-8201 (electronic) SpringerBriefs in Mathematics ISBN 978-981-15-7379-8 ISBN 978-981-15-7380-4 (eBook) https://doi.org/10.1007/978-981-15-7380-4 © The Author(s), under exclusive licence to Springer Nature Singapore Pte Ltd. 2020 This work is subject to copyright. All rights are solely and exclusively licensed by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors, and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, expressed or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. This Springer imprint is published by the registered company Springer Nature Singapore Pte Ltd. The registered company address is: 152 Beach Road, #21-01/04 Gateway East, Singapore 189721, Singapore

Notes taken by Shigeho Yamashima

Kunihiko Kodaira at his home in Tokyo, 1990 © Springer Japan

Foreword

In the academic year 1967, Kunihiko Kodaira gave a course of lectures at the University of Tokyo on the theory of complex algebraic surfaces. The lecture notes were published in 1968 as Volume 20 in the series of “Seminary Notes” by the University of Tokyo. That was the copy of the handwritten manuscript in Japanese by Shigeho Yamashima, based on his beautiful notes reflecting faithfully the atmosphere of Kodaira’s lectures. The present book is an English translation of that volume with slight modifications, correcting typos, etc. in the Japanese version. The readers are expected to have only the elementary prerequisites on complex manifolds as the background. The book consists of two parts: Chaps. 1 and 2. After stating the goal of the lecture in the Introduction, in Chap. 1, basic facts on algebraic surfaces are reviewed, touching upon divisors, linear systems, intersection theory, and the Riemann–Roch theorem. It provides an elegant introduction to the theory of algebraic surfaces covering some classical materials whose modern proofs first appeared in Kodaira’s papers. Among others, one can find a concise analytic proof of Gorenstein’s theorem for curves on a non-singular surface, which is a detailed explanation of the one given in Appendix I to “On Compact Complex Analytic Surfaces, I,” Ann. of Math. 71 (1960). Another highlight is the elementary proof of Noether’s formula for the arithmetic genus of an algebraic surface. Nowadays, the formula is known and treated as a special case of Hirzebruch’s Riemann–Roch theorem. Kodaira’s approach is based on the standard fact that, via generic projections, every algebraic surface can be obtained as the normalization of a surface with only ordinary singularities in the projective 3-space. However, unlike the other modern proofs, the argument does not rely on general facts, such as Porteus’ formula, which require a separate treatment. It is self-contained and follows a classical line, using Lefschetz pencils, much more in the style of Noether’s original proof. The second part, Chap. 2, discusses the behaviors of the Pluri-canonical maps of algebraic surfaces of general type, as an application of the general theory provided in Chap. 1. It gives a detailed account of “Pluri-canonical Systems on Algebraic Surfaces of General Type,” J. Math. Soc. Japan 20 (1968). The main tool is a vii

viii

Foreword

vanishing criterion for the first cohomology groups, which is known as Mumford’s vanishing theorem. It is shown by the analytic method in essentially the same way as in the proof of Kodaira’s vanishing theorem. It gives us, for example, the plurigenus formula for algebraic surfaces of general type. Then by means of various composition series for pluri-canonical divisors, one can discuss when the pluricanonical system is free from base points and when the pluri-canonical map is birational onto its image. Such excellent results have been extended by Kodaira himself, Enrico Bombieri, and others. So, we can now answer affirmatively to a question stated in Remark 9 in the Introduction. Takatsuki, Japan April 2020

Kazuhiro Konno

Preface

This grew out of a course of lectures given by Professor Kunihiko Kodaira at the Department of Mathematics, Faculty of Science, the University of Tokyo, two hours per week from September 1967 to February 1968. The notes were taken and carefully arranged by Mr. Shigeho Yamashima. I would like to express a deep gratitude to him. May 1968

Yukiyoshi Kawada

ix

Introduction (Purposes and Known Results)

Throughout, we let S denote a non-singular projective algebraic surface defined over the field C of complex numbers, that is, it is a two-dimensional compact complex manifold embedded into a complex projective space PN (C). In general, a two-dimensional complete algebraic variety without singular points is projective. Notation 1 K: the canonical line bundle on S. mK = K ⊗ · · · ⊗ K = K ⊗m : m-times tensor product of K. L (mK): the linear space consisting of (global) holomorphic sections of mK. Pm = dimC L (mK): the m-genus of S. {ϕ0 , ϕ1 , . . . , ϕn }: a basis for L (mK) over C. Pn

the rational map defined by mK.

∈

∈

mK : S −→

z  (ϕ0 (z), ϕ1 (z), . . . , ϕn (z)) c1 :

the first Chern class of S.

c12 can be considered as an integer via H 4 (S, Z) ∼ = Z. Assumption 2 S contains no exceptional curves of the first kind. (An exceptional curve of the first kind is a non-singular rational curve with self-intersection number −1.) Then, S is said to be relatively minimal. Assumption 3 P2 > 0 and c12 > 0. Definition 4 When Assumptions 2 and 3 are satisfied, S is called a minimal nonsingular algebraic surface of general type.

xi

xii

Introduction (Purposes and Known Results)

Remark 5 (i) If Pm > 0 for some m, then a relatively minimal model S is the minimal model. (ii) In Šafareviˇc [8], the term of fundamental type is used for of general type. We assume that S satisfies 2 and 3. Problem 6 Study mK . 7 Known results (a) D. Mumford [5]: There exists an integer m0 (S) such that mK is a birational holomorphic map for any m  m0 (S). (b) I.P. Šafareviˇc [8]: 9K is a birational map (not referred to whether it is holomorphic or not). The main purpose of the book is to prove the following: Theorem 8 (Theorem 8.17) mK is a birational holomorphic map when m  6. Remark 9 Can we relax the condition to m  5? There is an example for which 5K is a birational holomorphic map, while 4K isnot birational. In fact, (i) let W ⊂ P3 be the surface defined by 3i=0 zi5 = 0. For ε = exp( 2πi 5 ), put g : (z0 , z1 , z2 .z3 ) → (z0 , εz1 , ε−1 z2 , z3 ) and G = {g n | n = 0, 1, . . . , 4}. Then W/G has 5 singular points corresponding to (1, 0, 0, εν ) (0  ν  4). If S is its desingularization [1], then 4K is notbirational, but 5K is a birational holomorphic map. (ii) When W is defined by 3i=0 zi6 = 0, a similar construction gives us a surface S for which 4K is birational. Remark 10 If S satisfies Assumption 2 but not Assumption 3 (that is, either P2 = 0 or c12  0), then S is one of the following five types of surfaces: (a) (b) (c) (d) (e)

the projective plane P2 , a ruled surface, a K3 surface, an Abelian surface (two-dimensional complex torus), an elliptic surface.

Contents

1

Fundamentals of Algebraic Surfaces . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . 1 1.1 Exact Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . 1 1.2 Divisors and Linear Systems . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . 14 1.3 Intersection Multiplicities and the Adjunction Formula . . . . . . . . . . . . . . . 16 1.4 Riemann–Roch Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . 25

2 Pluri-Canonical Systems on Algebraic Surfaces of General Type . . . . . . . 2.5 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . 2.6 Vanishing Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . 2.7 Composition Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . 2.8 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . .

47 47 49 54 60

Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . 75

xiii

Chapter 1

Fundamentals of Algebraic Surfaces

Abstract Fundamental properties of algebraic surfaces are collected and reviewed, such as curves on a surface, divisors and linear systems and the Riemann–Roch theorem. An elementary proof of Noether’s formula is also given.

1.1 Exact Sequences Notation 1.1 S: C ⊂ S:  → C: μ:C O = OS : O(−C):

a non-singular algebraic surface. an algebraic curve on S (assumed to be irreducible).  is a non-singular model of C the desingularization of C, that is, C (unique up to isomorphisms) and μ is a birational holomorphic map. the sheaf of germs of holomorphic functions on S. the sheaf of germs of holomorphic functions on S vanishing on C (this is a subsheaf of O).

We put OC = O/O(−C). 1.2 We study the structure of OC . For x ∈ S, we denote by (w, z) a system of local coordinates of the center x on S. Then Ox = OS,x = C{w, z}, where C{w, z} denotes the ring of convergent power series in the indeterminates w, z with coefficients in C. In the following, we separately consider three cases (a), (b), and (c) divided according to the position of x for C: (a) The case where x ∈ C. Obviously, we have (OC )x = {0}. (b) The case where x is a simple point of C. As in the figure, we take a system of local coordinates (w, z) on S with the center x such that C is defined by w = 0 in a neighborhood of x. Then O(−C)x = wC{w, z} and, hence, (OC )x = Ox /O(−C)x = C{w, z}/wC{w, z} ∼ = C{z}. © The Author(s), under exclusive licence to Springer Nature Singapore Pte Ltd. 2020 K. Kodaira, Theory of Algebraic Surfaces, SpringerBriefs in Mathematics, https://doi.org/10.1007/978-981-15-7380-4_1

1

2

1 Fundamentals of Algebraic Surfaces

(c) The case where x is a singular point of C.

The inverse image μ−1 (x) consists of a finite number of points {p1 , . . . ,  We take a local parameter tλ around pλ on C  such that μ can pλ , . . . , pr } on C. be expressed as μ : tλ → (w, z) = (Pλ (tλ ), tλmλ ) in a neighborhood of pλ , where Pλ (tλ ) ∈ tλmλ C{tλ }. Let Cλ = {μ(tλ ) | |tλ | < α} be the irreducible analytic branch of C corresponding to pλ . We write μλ for μ as above to specify the index (that is, μλ is the restriction of μ to a neighborhood of pλ ). We define Rλ , R as follows: Rλ (w, z) =

m λ −1 

 w − Pλ (ελk z1/mλ ) = wmλ + Aλ,1 (z)wmλ −1 + · · · + Aλ,mλ (z),

k=0

R(w, z) =

r 

Rλ (w, z) = wm + A1 (z)wm−1 + · · · + Am (z),

λ=1

r where ελ = exp(2πi/mλ ), Ak (z) ∈ zk C{z} and m = λ=1 mλ . Then C : R(w, z) = 0 in a neighborhood of x. We have a homomorphism μ∗λ : O → C{tλ }(=: oλ ) defined by μ∗λ : (w, z) → (μ∗λ )(tλ ) = (Pλ (tλ ), tλmλ ).

1.1 Exact Sequences

3

Put o = ⊕rλ=1 oλ . Then, we have the homomorphism μ∗ : Ox → o defined by μ∗ :  → μ∗1  + · · · + μ∗λ  + · · · + μ∗r , where  = (w, z) ∈ Ox = C{w, z}. We obviously have O(−C)x = { ∈ Ox | μ∗  = 0} and, therefore, (OC )x = Ox /O(−C)x = μ∗ Ox ⊂ o. So, it suffices to study μ∗ Ox , which will be done later in 1.4. Example 1.2.1 Here is an example in which C has a unique irreducible analytic branch at x ∈ C. Put C = C1 in a neighborhood of x and μ = μ1 : t → (w, z) = (t q , t m ), where q > m > 0, (q, m) = 1.

Since we have (μ∗ )(t) = obtain



h,k0 ahk t

hq+mk

for  =

μ∗ Ox = {μ∗  |  ∈ Ox } = {ϕ(t) | ϕ(t) =



ahk wh zk ∈ Ox , we



bn t n }.

n = hq + km h, k  0 If we put m = 5, q = 7 for example, then we have {hq + km}h,k0 = {7h + 5k}h,k0 = {0, 5, 7, 10, 12, 14, 15, 17, 19, 20, 21, 22, 24, 25, 26, 27, . . .}. In this sequence, the number of elements from 0 to 22 is 12 = 12 (m − 1)(q − 1) and we have all integers greater than or equal to 24 = (m−1)(q −1). See Example 1.4.1. Definition 1.3 In the situation of 1.2(c), we put d(tλmλ ) σλ dtλ = Rw (Pλ (tλ ), tλmλ )

 ∂R(w, z) where Rw (w, z) = . ∂w

(1.3.1)

4

1 Fundamentals of Algebraic Surfaces

Then we can express σλ as σλ = tλ−cλ (aλ,0 + aλ,1 tλ + · · · )

(aλ,0 = 0).

Since x is a singular point, cλ is a positive integer. We put  o=

r

 tλcλ oλ

⊂o=

λ=1

r

oλ .

(1.3.2)

λ=1

Remark 1.3.3 Even when we are in the situation of 1.2(b), we can consider 1.3.1. Since then mλ = 1, R(w, z) = w, we get σλ = 1 and cλ = 0. (In fact, r = 1.) Theorem 1.4 (Gorenstein [3] and Rosenlicht [7]) (1)  o is a subring of μ∗ Ox , that is,  o ⊂ μ∗ Ox ⊂ o. r 1 ∗ ∗ (2) dim [o/μ Ox ] = dim [μ Ox /˜o] = cλ . 2 λ=1

Example 1.4.1 In the situation of Example 1.2.1, we have R(w, z) = wm − zq , and, hence, Rw = mwm−1 and σ dt = m = 5 and q = 7,

P (t) = t q

d(t m ) dt = (m−1)(q−1) . In particular, when mt q(m−1) t

⎧ ⎪ o ←→ Z0 = {0, 1, 2, 3, . . . } ⎪ ⎪ ⎨ μ∗ O ←→ {7h + 5k} (see 1.2.1) x h,k0 ⎪  o ←→ {24, 25, . . . } ⎪ ⎪ ⎩ c = c1 = (m − 1)(q − 1) = 4 × 6 = 24. In fact, we have o = C{t} and  o = t 24 C{t}. So, in this case Theorem 1.4 has been confirmed. Problem 1.4.2 Generalize Theorem 1.4 to higher-dimensional cases. 1.5 From here to the end of 1.6, we prepare things for the proof of Theorem 1.4 which will be given in 1.7.

1.1 Exact Sequences

5

Let the situation be as in 1.2(c). We define fλ , f, R as follows: fλ = the quotient field of oλ = {ϕ(tλ ) | ϕ(tλ ) =

+∞ 

an tλn } = C({tλ })

n=m

(hereafter, C({z}) = the quotient field of C{z}), f=

r

fλ ,

λ=1

 R=

    ∗ ,  ∈ Ox , μλ  = 0 for any λ .  

The homomorpism μ∗ : O → o can be extended to the homomorphism μ˜ ∗ : R → f, μ˜ ∗ |Ox = μ∗ . In fact, it suffices to put μ˜ ∗ :

 μ∗   λ → .  μ∗λ  r

λ=1

Lemma 1.5.1 μ˜ ∗ : R → f is surjective. In fact, any ξ ∈ f can be expressed in the form ξ = μ˜ ∗ F, F = F (w, z) = F0 (z)wm−1 + F1 (z)wm−2 + · · · + Fm−1 (z), where Fk (z) ∈ C({z}), m = determined.

r

λ=1 mλ .

Moreover, such an F is uniquely

Proof (a) Surjectivity of μ˜ ∗ : We put wλ∗ = Pλ (tλ ), zλ∗ = tλmλ and f∗λ = C({zλ∗ }) (⊂ fλ ). Then we obviously have [fλ : f∗λ ] = mλ . On the other hand, since Rλ (w, zλ∗ ) = wmλ + Aλ1(zλ∗ )wmλ −1 + · · · + Aλmλ (zλ∗ ) (Aλk (zλ∗ ) ∈ f∗λ ) is irreducible and Rλ (wλ∗ , zλ∗ ) = 0, we have [f∗λ (wλ∗ ) : f∗λ ] = mλ . Since we have fλ ⊆ f∗λ (wλ∗ ) and [fλ : f∗λ ] = mλ , we obtain fλ = f∗λ (wλ∗ ). Next, we put Sλ = Sλ (w, z) =



(*)

Rν (w, z) and sλ = μ∗λ Sλ (∈ oλ ). Express

ν=λ  the given ξ ∈ f as ξ = rλ=1 ξλ (ξλ ∈ fλ ). Noting that sλ = 0, we put ηλ = sλ−1 ξλ (∈ fλ ). Then by (∗) we can write ∗ ∗ ηλ = Fλ0 · (wλ∗ )mλ −1 + · · · + Fλm , λ −1

6

1 Fundamentals of Algebraic Surfaces ∗ = F (z∗ ) (∈ f∗ ) is a meromorphic function in z∗ . Noting that where Fλk λk λ λ λ Fλk (z) ∈ C({z}), if we put

Fλ (w, z) = Fλ0 (z)wmλ −1 + · · · + Fλmλ −1 (z), then ηλ = μ˜ ∗λ Fλ , where μ˜ ∗λ : R → fλ denotes the λ-th component of μ˜ ∗ . We r  Sλ (w, z)Fλ (w, z). Then it is of the form put F (w, z) = λ=1

F (w, z) = F0 (z)wm−1 + F1 (z)wm−2 + · · · + Fm−1 (z) as wished and we have μ˜ ∗ F =

r 

sλ ηλ =

λ=1

r 

ξλ = ξ .

λ=1

(b) The unicity of F : We shall show that we have F = 0 if either F = 0 or F is a polynomial in w of degree m − 1 satisfying μ˜ ∗ F = 0 (F ∈ R). Recall that Rλ (w, z) =

m λ −1

1/mλ

(w − Pλ (ελk zλ

))

k=0

= wmλ + Aλ1 (z)wmλ −1 + · · · + Aλmλ (z)

(ελ = exp(

2πi )) mλ

is irreducible and Rλ (Pλ (tλ ), tλmλ ) = 0. On the other hand, we have F (Pλ (tλ ), tλmλ ) = 0. Therefore, F (w, z) ≡ 0 (Rλ ) and, hence, F (w, z) ≡ r  0 (R). However, since R = Rλ = wm + A1 (z)wm−1 + · · · + Am (z) (Ak (z) λ=1

∈ zk C{z}) is a polynomial of degree m in w, we get F (w, z) = 0. Definition 1.5.2 For a given η =

r 

 

ηλ ∈ f, ηλ = ηλ (tλ ) ∈ fλ , we put

λ=1

1  2πi r

ρ(η) =

 ηλ (tλ )σλ (tλ ) dtλ ,

λ=1

 where means the integral along a sufficiently small circle “|tλ | = constant”, and σλ = σλ (tλ ) is as in 1.3.1. Definition 1.5.3 We put Bh (w, z) = wh + A1 (z)wh−1 + · · · + Ah (z) (recall: R(w, z) = wm + A1 (z)wm−1 + · · · + Am (z)).

1.1 Exact Sequences

7

Remark 1.5.4 When R(ζ, z) = 0, R(w, z) = wm−1 + B1 (ζ, z)wm−2 + B2 (ζ, z)wm−3 + · · · + Bm−1 (ζ, z). w−ζ Proof We let Q(w, z; ζ ) be the right hand side of the above equation. Then Q = wm−1 +(ζ +A1 )wm−2 +(ζ 2 +A1 ζ +A2 )wm−3 +· · ·+(ζ m−1 +A1 ζ m−2 +· · ·+Am−1 ). We have (w − ζ )Q = wm + (ζ + A1 )wm−1 + (ζ 2 + A1 ζ + A2 )wm−2 + · · · + (ζ m−1 + A1 ζ m−2 + · · · + Am−1 )w − ζ wm−1 − (ζ 2 + A1 ζ )wm−2 − · · · − (ζ m + A1 ζ m−1 + · · · + Am−1 ζ ) = wm + A1 wm−1 + · · · + Am (z) = R(w, z),

because −(ζ m + A1 ζ m−1 + · · · + Am−1 ζ ) = Am by R(ζ, z) = 0. Lemma 1.5.5 F as in 1.5.1 is given by the following formula: ⎧ m−1 + F1 (z)wm−1 + · · · + Fm−1 (z) ⎪ ⎪ F (w, z) = F0 (z)w ⎪ ⎪ ⎪ ⎪ F (z) =  F zn ⎪ ⎪ h hn ⎪ ⎪ ⎨ n ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

Fhn = ρ(ξ · μ˜ ∗ (z−n−1 Bh )) (cf. 1.5.2)  m 1  ξλ (tλ )Bh (Pλ (tλ ), tλ λ )mλ dtλ = mλ nmλ +1 2πi ∂ R(P (t ), t )t w λ λ λ λ λ

Proof Because R(w, z) =

r m λ −1 

1 m

(w − ζλk ), ζλk = Pλ (ελk zλ λ ), we have

λ=1 k=0

F (w, z) =

  R(w, z) F (ζλk , z) w − ζλk ∂w R(ζλk , z) λ

k

by Lagrange’s interpolarion formula. By 1.5.4, however, R(w, z) = wm−1 + B1 (ζλk , z)wm−2 + · · · + Bh (ζλk , z)wm−h−1 + · · · . w − ζλk

 

8

1 Fundamentals of Algebraic Surfaces



Hence, Fh (z) =

λ

Bh (ζλk , z)

k

F (ζλk , z) . On the other hand, since we have ∂w R(ζλk , z) 1

ξλ (tλ ) = μ∗λ F = F (Pλ (tλ ), tλmλ ), we get F (ζλk , z) = ξλ (ελk z mλ ). It follows that 1  ξλ (ελk z mλ ) Fh (z) = . Bh (ζλk , z) ∂w R(ζλk , z) λ

k

Now, if we put

+∞  Bh (Pλ (tλ ), tλmλ )ξλ (tλ ) = γν(λ) tλν , then we have ∂w R(Pλ (tλ ), tλmλ ) ν=ν 0

 k

1

m λ −1   ν ξλ (ελk z mλ ) (λ) = Bh (ζλk , z) γν(λ) ελkν z mλ = γnm m zn , λ λ ∂w R(ζλk , z) ν n k=0

because m λ −1

 ελkν

k=0

Hence, Fh (z) =



=

mλ , ν ≡ 0 (mλ ) 0, ν ≡  0 (mλ )

Fhn zn =



n

Fhn =

 λ

λ (λ) γnm m = λ λ

(where ελ = exp( 2πi mλ )).

(λ) γnm m zn and we have λ λ

n

 1  Bh (Pλ (tλ ), t mλ )ξλ (tλ )mλ λ dtλ . 2πi ∂w R(Pλ (tλ ), tλmλ )tλnmλ +1 λ  

Theorem 1.6 A necessary and sufficient condition for ξ ∈ f to be in μ∗ Ox is that ρ(ξ · ϕ) = 0 holds for all ϕ ∈ μ∗ Ox . Proof (i) Sufficiency: We take ξ ∈ f satisfying ρ(ξ · ϕ) = 0 for all ϕ ∈ μ∗ Ox . Since, by 1.5.5, it is given by ξ = μ˜ ∗ F , Fhn = ρ(ξ · μ˜ ∗ (z−n−1 Bh )) (F ∈ R), we only have to show that F ∈ Ox = C{w, z}. That is, it suffices to show that Fhn = 0 for n  −1. When n  −1, however, we have z−n−1 Bh (w, z) ∈ Ox and it follows that μ˜ ∗ (z−n−1 Bh ) ∈ μ˜ ∗ Ox ⊂ o. Hence Fhn = ρ(ξ · μ˜ ∗ (z−n−1 Bh )) = 0 by the hypothesis. (ii) Necessity: ξ can be written as ξ = μ˜ ∗ F with some F ∈ R as in 1.5.5. Now, we assume that ξ ∈ μ˜ ∗ Ox . Hence we can express ξ = μ˜ ∗  with some  ∈ Ox . Firstly, we shall show that F ∈ Ox . Recall that F = F0 (z)wm−1 + · · · + Fm−1 (z) (Fh (z) ∈ C({z})). We assume that F ∈ Ox and derive a contradiction.

1.1 Exact Sequences

9

Assume that Fh (z) ∈ C{z} for some h. If we write Fh (z) = Fh,−ah z−ah + · · · (Fh,−ah = 0) and put max{a0 , a1 , . . . , ah , . . . } = b, then 

zb F (w, z) = G(w, z) ∈ C{w, z}, G(w, 0) = 0.

Since μ˜ ∗  = μ˜ ∗ F , we have μ˜ ∗ (G − zb ) = μ˜ ∗ (zb F − zb ) = 0. It follows that G − zb  ≡ 0 (R), that is, we can write G(w, z) − zb (w, z) = Q(w, z)R(w, z) (Q(w, z) ∈ C{w, z}). Noting that R = wm + · · · + Ah (z)wm−h + · · · (Ah (z) ∈ zh C{z}), we get G(w, 0) = Q(w, 0)wm . Since, however, G(w, 0) is a polynomial in w of degree  m−1 and Q(w, 0) = 0, this is impossible. Therefore, F ∈ Ox . We infer from 1.5.5 that for n  −1 Fhn = ρ(ξ · z−n−1 Bh ) = 0. In particular, F0,−1 = ρ(ξ ) = 0. We have shown that ρ(η) = 0 holds for any η ∈ μ∗ Ox . We need to show that ρ(ξ · ϕ) = 0 for any ϕ ∈ μ∗ Ox . But, since we have assumed that ξ ∈ μ∗ Ox , we have ξ · ϕ ∈ μ∗ Ox . Therefore, ρ(ξ · ϕ) = 0.   1.7 Here we show Theorem 1.4. For any submodule m of f, we define a submodule m of f by m = {ξ ∈ f | ρ(ξ · η) = 0 for all η ∈ m}. Recall that o = (a)

o

r

r

oλ and σλ (tλ ) = tλ−cλ (aλ0 + aλ1tλ + · · · ) (aλ0 = 0).

λ=1

Hence o ⊂ o.  1  This is because ρ(ξ · η) = ξλ (tλ )ηλ (tλ )σλ (tλ ) dtλ and, therefore, 2πi

=

cλ λ=1 tλ oλ .

λ

 ρ(ξ · η) = 0 for ∀η ∈ o ⇔

ξλ (tλ )ηλ (tλ )σλ (tλ ) dtλ = 0 for ∀ηλ ∈ C{tλ }, ∀λ

⇔ ξλ (tλ )ηλ (tλ )σλ (tλ ) is holomorphic for ∀ηλ ∈ C{tλ }, ∀λ ⇔ ξλ (tλ ) ≡ 0 (tλcλ ), ∀λ.

10

1 Fundamentals of Algebraic Surfaces

If ξ, η ∈ o, then ρ(ξ · η) is determined modulo o and, hence, we regard ρ(ξ · η) as a (non-degenerate) bilinear function over o/o . Then, by duality, we have (b) o = o. In general, if o ⊂ m ⊂ o, then we infer from (b) that (c) o ⊂ m ⊂ o and dim(m/o ) = dim(o/m ). Now, by 1.6, we have (μ∗ Ox ) = μ∗ Ox . Therefore, since μ∗ Ox ⊂ o, we get o ⊂ (μ∗ Ox ) = μ∗ Ox . This shows 1.4(1). Applying (c) to o ⊂ μ∗ Ox ⊂ o, we get 1 dim [o/o ]. 2 r On the other hand, we infer from (a) that dim [o/o ] = λ=1 cλ . We have shown 1.4(2). dim [μ∗ Ox /o ] = dim [o/μ∗ Ox ] =

1.8 For any singular point x ∈ C, we write o = ox ,

o = ox

for the notation in 1.7. Recall that μ∗ Ox ∼ = (OC )x (cf. 1.2(c)). We denote this isomorphism by τx . If we apply the above notations to a simple point x ∈ C, we have ox = μ∗ Ox = ox = o and, of course, τx : μ∗ Ox ∼ = (OC )x . Definition 1.8.1 We denote by O  the subsheaf of OC whose stalk over x ∈ C is given by Ox = τx (ox ) (⊆ (OC )x ) (recall ox ⊂ μ∗ Ox ). We define M = O/O  . Then 0 → O  → OC → M → 0

(exact).

1.8.2 For x ∈ C, Mx ∼ = μ∗ Ox /ox is a finite-dimensional vector space. In fact, we have ⎧ ⎨ 0 · · · x : simple point, dim Mx = 1 ⎩ cx · · · x : singular point, 2 where we put cx =

 pλ ∈μ−1 (x)

cλ .

1.1 Exact Sequences

11

1.9 Let F be a complex line bundle over S.

Then F can be described as follows: (a) A surjective holomorphic map  : F →  S (called the canonical projection). (b) A sufficiently fine open covering S = j Uj such that  −1 (Uj ) = Uj × C. (c) Uj × C and Uk × C are glued by the rule: for z ∈ Uj ∩ Uk , (z, ζj ) ∈ Uj × C and (z, ζk ) ∈ Uk × C, (z, ζj ) = (z, ζk ) ⇐⇒ ζj = fj k (z)ζk , where fj k is a (non-vanishing) holomorphic function defined on Uj ∩ Uk = ∅ and we have fik = fij fj k on Ui ∩ Uj ∩ Uk . We call ζj the fibre coordinate of (z, ζj ) over Uj . 1.9.1 By O ∗ , we denote the multiplicative subsheaf of O such that (U, O ∗ ) consists of all invertible elements of (U, O). The 1-cocycle {fij } as above determines an element of H 1 (S, O ∗ ), the isomorphism class of F . It is obvious that H 1 (S, O ∗ ) becomes an abelian group with product F1 ⊗ F2 . We regard H 1 (S, O ∗ ) as an additive group and write F1 + F2 instead of F1 ⊗ F2 . 1.9.2 A holomorphic section ϕ of F over S is a holomorphic map from S to F such that the composite  ◦ ϕ is the identity map. So we can express it as: 

ϕ : z → ϕ(z) = (z, ϕj (z)) (over Uj ), ϕj (z) = fj k (z)ϕk (z) on Uj ∩ Uk .

That is, ϕ can be identified with a collection {ϕj (z)} of holomorphic functions ϕj (z) on Uj satisfying ϕj (z) = fj k (z)ϕk (z).

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1 Fundamentals of Algebraic Surfaces

Definition 1.9.3 A collection ϕ = {ϕj (z)} of meromorphic functions ϕj (z) on Uj is called a meromorphic section of F if ϕj (z) = fj k (z)ϕk (z) holds on Uj ∩ Uk . 1.10 Let F and C be a complex line bundle and a curve on S, respectively. We define sheaves as in the following (a), (b), and (c): (a) (b) (c)

O(F ) = the sheaf of germs of holomorphic sections of F (a sheaf over S). O(F − C) = {ϕ ∈ O(F ) | ϕ(z) = (z, 0) for z ∈ C}. O(F )C = O(F )/O(F − C): the restriction to C. Therefore, O(F )C = OC ⊗O O(F ) = OC ⊗ F . In fact: −1 (U )) (OC |Uj = τ j OC ⊗ F = j OC |Uj is patched up by the following rule:



For z ∈ Uj ∩ Uk , ϕj ∈ (OC |Uj )z and ϕk ∈ (OC |Uk )z , ϕj = ϕk ⇐⇒ ϕj = fj k ϕk .

1.11 O  (F ) = O  ⊗ F

(cf. 1.8.1).

Then we have the following exact sequences (a), (b): (a) 0 → O  (F ) → O(F )C → M → 0, (b) 0 → O(F − C) → O(F ) → O(F )C → 0. As to (a), we infer from 1.8.1 that the sequence 0 → O  ⊗ F → OC ⊗ F → M ⊗ F → 0 is exact and, since the sheaf M over C has non-zero stalks only over the singular points of C, we have M ⊗ F ∼ = M ; (b) is clear.

1.1 Exact Sequences

13

 → C ⊂ S denote the desingularization of a curve C and Definition 1.12 Let μ : C  let F be a complex line bundle over S. We denote by μ∗ F the line bundle over C  of germs of holomorphic induced from F by μ, and by O(μ∗ F ) the sheaf over C sections of μ∗ F .  is called the conductor on C: Definition 1.13 The following divisor c on C c=





cλ p λ

(see 1.3 for the symbols).

x∈Sing(C) pλ ∈μ−1 (x)

Then we put O(μ∗ F − c) = {η ∈ O(μ∗ F ) | η ≡ 0 (c)}. ∼ =

Lemma 1.13.1 μ∗ O(μ∗ F − c) ←− O  (F ). Proof The natural homomorphism τ : O  (F ) → O(μ∗ F − c) is given by τ : ϕ(w, z) → (μ∗ ϕ)(t). It suffices to show that it induces the isomorphism on each stalk over any point on C. It is clear at simple points on C. So, let us consider a singular point x ∈ C. We employ the notation in 1.2. If we put r

μ−1 (x) = {p1 , . . . , pλ , . . . , pr }, then μ∗ O(μ∗ F −c)x ∼ O(μ∗ F − c)pλ . On the = λ=1

other hand, we infer from 1.13, 1.7, and 1.8 that are canonical isomorphisms there c O(μ∗ F − c)pλ ∼ tλcλ oλ . Then, by the nature of the = tλλ oλ and O  (F )x ∼ = Ox ∼ = ∼ =

map μ, we get τx : O  (F )x → μ∗ (O(μ∗ F − c))x .

 

Theorem 1.14 The following sequences are exact: 0 → μ∗ O(μ∗ F − c) → O(F )C → M → 0. 0 → O(F − C) → O(F ) → O(F )C → 0. Proof The former follows from 1.11 and 1.13. The latter is obvious. We remark that O(FC ) = O(F )C .

 

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1 Fundamentals of Algebraic Surfaces

Theorem 1.15 The following sequences are exact: 0 → H 0 (S, O(F − C)) → H 0 (S, O(F )) → H 0 (C, O(F )C ) → H 1 (S, O(F − C)) → H 1 (S, O(F )) → H 1 (C, O(F )C ) → H 2 (S, O(F − C)) → H 2 (S, O(F )) → 0.  O(μ∗ F − c)) → H 0 (C, O(FC )) → H 0 (C, M ) 0 → H 0 (C, 1  O(μ∗ F − c)) → H 1 (C, O(FC )) → 0. → H (C, Moreover, dim H 0 (C, M ) =

1 2

 x∈Sing(C)

cx

(cx =



cλ ).

pλ ∈μ−1 (x)

 O(μ∗ F − c)) ∼ Proof This follows from 1.14. We also remark that H ν (C, = ν  H (C, O (F )) holds by 1.13.1. The last equality follows from 1.8.2.  

1.2 Divisors and Linear Systems 2.1 Let C be an irreducible curve on S. We shall define the line bundle [C] over S. For any point x ∈ C, if we take a sufficiently small neighborhood U of x ∈ S, then we can write C ∩ U = {R(w, z) = 0}. Here, (w, z) is a system of local coordinates of S with the center x, R(z) = R(z1 , z2 ) = z2m + A1 (z1 )z2m−1 + · · · + Am (z1 ) is a holomorphic function such that R(z) = 0 is the minimal local equation (1.2), that is, for y ∈ C ∩ U and  ∈ Oy , if (z) = 0 holds for any z ∈ C in a neighborhood of y, then  ≡ 0 (R) in Oy .  Let S = j Uj be an open covering such that the minimal local equation of C R (z)

on Uj is Rj (z) = 0. Then fj k = Rjk (z) is a non-vanishing holomorphic function on Uj ∩ Uk . We of course have fik = fij fj k on Ui ∩ Uj ∩ Uk and {fj k } defines a complex line bundle F . Definition 2.1.1 We write F = [C]. Since Rj = fj k Rk on Uj ∩ Uk , ϕ : z → ϕ(z) = (z, Rj (z)) is a holomorphic section of F , and C = {z ∈ S | ϕ(z) = (z, 0)}. Definition 2.1.2 We write C = (ϕ) and call it the divisor of ϕ.  2.2 We extend the above discussion to general divisors D = rν=1 mν Cν . (mν ∈ Z, Cν is an irreducible curve on S.)

1.2 Divisors and Linear Systems

15

 Definition 2.2.1 [D] = rν=1 mν [Cν ].  In fact, let S = j Uj be an open covering such that the minimal local equation r   Rνj mν . Then [D] is determined by of Cν on Uj is Rνj (z) = 0 and put fj k = Rνk ν=1    {fj k } and ϕ : z → ϕ(z) = z, rν=1 Rνj (z)mν is a meromorphic section of [D]; we write D = (ϕ) and call it the divisor of ϕ. Conversely, given a (not identically zero) meromorphic section ϕ of a line bundle F , it is immediate that we have a unique divisor D satisfying F = [D] and D = (ϕ). Definition 2.2.2 When [D] = 0, where 0 denotes the trivial line bundle, that is, one defined by fj k (z) = 1), we say that D is linearly equivalent to 0 and write D ≈ 0. (Two divisors D1 and D2 are linearly equivalent, D1 ≈ D2 , if D1 − D2 ≈ 0.) 2.2.3 The necessary and sufficient condition that D ≈ 0 is that there exists a meromorphic function f on S satisfying D = (f ). Proof We can write D = (ϕ) with a meromorphic section ϕ(z) = (z, ϕj ) of [D]. If D ≈ 0, then [D] = 0. Hence we have ϕj (z) = ϕk (z) on Uj ∩Uk , that is, there exists a meromorphic function f on S such that f (z) = ϕj (z) on Uj . Then D = (f ). The converse is clear.   2.3 Let F be a line bundle over S. Then H 0 (S, O(F )) is the space of holomorphic sections of F . Definition 2.3.1 We call |F | = {(ϕ) | ϕ ∈ H 0 (S, O(F )), ϕ = 0} the complete linear system determined by F . (Here, ϕ = 0 means that ϕ is not identically zero.) For any D ∈ |F |, we have D  0 (see Remark below for the meaning of ).  Remark 2.3.2 For D = rν=1 mν Cν , 

when mν > 0, D is said to be positive and we write D > 0, when mν  0, D is said to be non-negative (or effective) and we write D  0.

Definition 2.3.3 We put |X| = |[X]| for a divisor X. Proposition 2.3.4 |X| = {D | D ≈ X, D  0}. Proof Clear.

 

Definition 2.4 Let D be a divisor and F a line bundle on S. We define the sheaf O(F − D) over S as follows: (a) The case D > 0. We take a holomorphic section ϕ(z) = (z, ϕj (z)) of F satisfying D = (ϕ) and let O(F − D) be the subsheaf of O(F ) determined by the following

16

1 Fundamentals of Algebraic Surfaces

condition (obviously, it does not depend on the choice of ϕ): The stalk over x ∈ S is O(F − D)x = {ψ ∈ O(F )x | ψ/ϕj ∈ Ox } = {ψ ∈ O(F )x | ψ ≡ 0 (D)}. (b) General case. We take a meromorphic section ϕ(z) = (z, ϕj (z)) of F satisfying D = (ϕ). O(F − D) consists of meromorphic sections ψ(z) = (z, ψj (z)) of F satisfying ψj /ϕj ∈ Ox for x ∈ S. Remark In the above argument, the expressions e.g., a section ϕ(z) = (z, ϕj (z)),  are considered after taking a suitable open covering S = j Uj as in 2.1. We apply the same rule in what follows. Proposition 2.4.1 There is a natural isomorphism O(F − D) ∼ = O(F − [D]). Proof We employ the notation in 2.4(b). We shall see that there is a natural isomorphism between sections over any open subset of S. In the following, we consider over an arbitrary fixed open subset of S. For any section ψ : z → ψ(z, ψj (z)) of O(F − D), we put ηj = ψj /ϕj . Then η : z → (z, ηj (z)) is a section of O(F − [D]) = O(F ⊗ [D]−1 ), because ψj = fj k ψk , ψj /ϕj is holomorphic on Uj and ηj /ηk = (ψj /ψk )(ϕk /ϕj ) = fj k ·gj−1 k (=: hj k ) (where we put gj k = ϕj /ϕk ; [D] is defined by {gj k }). Therefore, {hj k } defines F ⊗ [D]−1 , since ηj = hj k ηk , we see that η is a holomorphic section of F ⊗ [D]−1 . It is obvious that the map ψ → η gives us an isomorphism.  

1.3 Intersection Multiplicities and the Adjunction Formula 3.1 Let (S, O) be a non-singular algebraic surface and O ∗ the multiplicative sheaf of germs of non-vanishing holomorphic functions (1.9.1). Then we have the following exact sequence: τ

0 → Z → O −→ O ∗ → 0 (where τ : f → e2πif ), and hence c

F

∈

∈

· · · → H 1 (S, O ∗ ) −→ H 2 (S, Z) → · · · 

c(F )

1.3 Intersection Multiplicities and the Adjunction Formula

17

In fact, for a line bundle F , c(F ) denotes the Chern class of F . We look at this correspondence concretely:  If F is given by the 1-cocycle {fj k } for a covering S = j Uj (where fj k is a non-vanishing holomorphic function on Uj ∩ Uk ) and H 2 (S, Z)  c(F ) = {cij k } (cij k ∈ Z), then it is easy to see that cij k =

1 {log fij (z) + log fj k (z) + log fki (z)}. 2πi

3.2 Let F, G(∈ H 1 (S, O ∗ )) be line bundles and C, D divisors on S. We denote the value (c(F )c(G))[S] ∈ Z of c(F ) · c(G) ∈ H 4 (S, Z) on the fundamental 4-cycle S also by the same symbol c(F ) · c(G). Definition 3.2.1 F · G = c(F ) · c(G), C · D = c([C]) · c([D]), F · D = c(F ) · c([D]). Proposition 3.3 Let (R, OR ) be a compact Riemann surface. Then the sheaf OR∗ can be defined as before and there is an exact sequence c

∈

∈

· · · → H 1 (R, OR∗ ) −→ H 2 (R, Z) →, 

f

c(f)

 where f denotes a line bundle over R. If d = ν mν pν (mν ∈ Z, pν ∈ R) is a divisor on R, then  c([d]) = mν (= deg d). ν

 Proof We take a suitable open covering R = j Uj and put c([d]) = {cij k } ∈ H 2 (R, Z) (an abbreviation of the representative {cij k } ∈ Z 2 (U , Z), U = (Uj )). Of course, we may think {cij k } ∈ H 2 (R, R). By de Rham’s theorem, there is a 2form ξ on R corresponding to {cij k }. In the following, we shall write ξ explicitly. Firstly, there exists a 1-cochain {μj k } (where μik is a C ∞ function on Ui ∩ Uk ) such that cij k = μij +μj k +μki and dμij +dμj k +dμki = 0. {dμj k } is the coboundary of a 0-cochain {σj } (where σi is a C ∞ 1-form on Ui ). That is, we have dμij = σj − σi . Therefore, dσi = dσj on Ui ∩ Uj . This implies that ξ = dσi = dσj = · · · is a 2-form on R. It is clear that c([d]) = {cij k } ← ξ (not necessarily unique, but the meaning may be clear). We also have the correspondence {cij k } ↔ c ∈ Z, because H 2 (R, Z) ∼ = Z. In order to get this c, we take (i) a simplicial decomposition of R and (ii) its dual cellular decomposition. (We enlarge the cells a bit to get the Uj ’s  and an open covering R = j Uj .)

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1 Fundamentals of Algebraic Surfaces

 Then we have c = sgn ij k · cij k (the sum is taken over all simplexes ij k ). Moreover, we have the following:  3.3.1 − ξ = c. R

Proof We have  ξ= R

 i

i

=

ξ=

=−

σi =

∂i

 i,j

i

i

 i

  i,j

(σi − σj )

(by Green’s formula)

ij

dμij = − ij

dσi

 [μij (zij k ) − μij (zij h )] i,j

 =− [μij (zij k ) + μj k (zij k ) + μki (zij k )] zijk

=−



(by μij + μj k + μki = cij k )

cij k

zijk

= −c,

where i ⊂ Ui denotes a cell. In order to prove c([d]) = deg d (deg d = c([p]) = 1 for any p ∈ R.



  mν ), it suffices to show that

1.3 Intersection Multiplicities and the Adjunction Formula

(a)

19

(b)

We take a sufficiently fine open covering R = Ui0 ∪ Uj ∪ Uk ∪ · · · (p ∈ Ui0 ) and let fj (z) = 0 be the equation in Uj defining d = p. That is, for j = i0 , fj (z) = 1 and fi0 (z) has a simple zero at p and fi0 (z) = 0 for z = p. If we put f fj k = fjk , then [p] is determined by {fj k }. We put c([p]) = {cij k }. Then, from the

1 arguments above, cij k = 2πi (log fij + log fj k + log fki ), there is a C ∞ function ∞ μik on Ui ∩ Uk and a C 1-form σi on Ui satisfying cij k = μij + μj k + μki (on Ui ∩ Uj ∩ Uk ), dμij = σj − σi (on Ui ∩ Uj ) and ξ = dσi (on Ui ). Recall  1 log fij , then γij + γj k + γki = 0 that c([p]) = − R ξ . If we put γij = μij − 2πi ∞ and there exists a C function τi on Ui satisfying γij = τj − τi on Ui ∩ Uj . 1 1 Hence, ω = σi + 2πi d log fi − dτi = σj + 2πi d log fj − dτj is a 1-form on R. ∞ It is clear that ω is C on R \ {p}. We have ξ =    dσi = dω. In view of Fig. b, we 1 have d log fi0 = −1. Hence ξ= ξ = lim dω = − lim →0 →0 2πi R R\W ∂W  R\{p}   c([p]) = − R ξ = 1.

Proposition 3.4 Let D be a divisor on S. Then c([D]) and D are dual to each other (note that a divisor D is a 2-cycle on S). In other words, for all a ∈ H 2 (S, Z), c([D]) · a = a(D). Proof By linearity, it suffices to show the assertion for an irreducible curve C instead of a given D.  We take a suitable open covering S = j Uj and let fj (z) = 0 be the minimal local equation of C in Uj . If we put fj k = fj /fk on Uj ∩ Uk , then [C] = {fj k }. 1 If c([C]) = {cij k }, then cij k = 2πi (log fij + log fj k + log fki ) and, by de Rham’s theorem, we can express cij k with a 2-form ξ . We go along the same line as in the proof of 3.3 employing similar notation. Putting ξ = dσi = dσj = · · · , ω = 1 σi + 2πi d log fi − dτi , we obtain ξ = dω, where σi is a C ∞ 1-form and τi is a C ∞ function on Ui . In this way, we get the correspondence c([D]) ← ξ (not unique but the class of ξ modulo exact forms is unique). Similarly, there is a 2-form η with dη = 0 suchthat a ← η. By deRham’s theorem, we have c([C]) · a ← ξ ∧ η and c([C])·a = S ξ ∧η, a(C)  = − C a. Therefore, in order to prove c([C])·a = a(C), it suffices to show that S ξ ∧ η = − C η.

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1 Fundamentals of Algebraic Surfaces

We can choose η in such a way that η vanishes at any singular points of C (in fact, take a sufficiently small neighborhood W of a singular point and subtract a suitable exact form with support in W from η). ω is C ∞ on S \ C. Hence







ξ ∧ η = lim S

→0 S\T

ξ ∧ η = lim



= − lim

→0 ∂T

→0 S\T

d(ω ∧ η)



(∗)

ω ∧ η = − lim

→0

1 dfi (∗∗) ∧η = − 2πi fi

 η C

Here the equality (∗) can be obtained by substituting the expression of ω given above. To see that the equality (∗∗) holds, we calculate as follows.

Since η = 0 at a singular point of C, we consider around a smooth point of C. We let (w, z) denote a system of local coordinates of S around that point. We can assume that w = fi , since C : fi (z) = 0. Writing η = Adw ∧ dz + Bdw¯ ∧ d¯z + Cdw ∧ d¯z + Ddz ∧ d¯z + Edw¯ ∧ dz + F dw ∧ dw, ¯ we get dw dw ∧η = ∧ Ddz ∧ d¯z + ()dw ∧ dw. ¯ w w

1.3 Intersection Multiplicities and the Adjunction Formula

21

However, fixing a small ε > 0 and putting w = εeiθ , we get dw = εieiθ dθ , dw¯ = −εie−iθ dθ and, hence, dw ∧ dw¯ = 0. So, we have    dfi ∧ η = 2πi Ddz ∧ d¯z = 2πi η. fi C C  

This completes the proof.

3.5 Let C and F be an irreducible curve and a complex line bundle over S,  → C ⊂ S be the desingularization of C. We let F  = μ∗ F respectively, and let μ : C  be the line bundle induced on C. For two line bundles F , G over S, we have c(F ) · c(G) = c(G) · c(F ). ). Proposition 3.5.1 F · C = c(F Proof We use 3.4. We have F · C = c(F ) · c([C]) = c([C]) · c(F ) = c(F )(C) = ). c(F   3.6 Let B, C be irreducible curves on S, B = C. Then C ∩ B consists of a finite number of points. We define the intersection multiplicity Ip (B, C) at p ∈ C ∩ B.

→ C We denote by R(w, z) = 0 the minimal local equation of B at p, μ : C the desingularization and μ−1 (p) = {p1 , . . . , pλ , . . . , pr }. In a neighborhood of pλ , we may assume μ : tλ → (w, z) = (Pλ (tλ ), tλmλ ) (cf. 1.2, take a suitable tλ ). Put R(Pλ (tλ ), tλmλ ) = tλnλ (aλ0 + aλ1 tλ + · · · ) (aλ0 = 0). Obviously, nλ is determined by pλ .  Definition 3.6.1 Ip (B, C) = nλ . pλ ∈μ−1 (p)

Here we remark that Ip (B, C) = Ip (C, B). Proposition 3.7 For irreducible curves B, C on S, B·C =



Ip (B, C).

p∈B∩C

 Proof Takinga suitable open covering S = Uj , we let F = [B] be defined   Rj  = λ in such a way that U by {fj k } = Rk . We take an open covering C

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1 Fundamentals of Algebraic Surfaces

λ ) ⊂ Uj = Uj (λ). Put R λ (t) = Rj (λ) (μ(t)) on U λ and f˜λν (t) = fj (λ)j (ν)(μ(t)) μ(U λ (t ) R ν . Hence, putting f˜λν (t) =  ˜  λ ∩ U on U ν (t ) , F is defined by {fλν }. ϕ˜ = {Rλ } is a R  and, putting d = (ϕ),  = [d]. Therefore, holomorphic section of F ˜ we have F ) = deg d. On the other from 3.5.1 and  3.3, we have B · C = F · C = c( F   hand, deg d = nλ = p∈B∩C Ip (B, C) (because, taking Uλ as a coordinate mλ nλ  neighborhood around pλ , we have Rλ (t) = R(Pλ (tλ ), tλ ) = tλ (aλ0 +aλ1 tλ +· · · ), aλ0 = 0).   3.8 We give the definition of the canonical bundle K over S. Let (zj1 , zj2 ) = (wj , zj )  be a system of local coordinates on a coordinate neighborhood Uj , S = Uj . 1 2 ϕ = {ϕj (z)dzj ∧ dzj } is said to be a meromorphic differential form if: (a) ϕj (z) is a meromorphic function on Uj , and (b) ϕj dzj1 ∧ dzj2 = ϕk dzk1 ∧ dzk2 . Definition 3.8.1 D = (ϕ) = (ϕj ) is called a canonical divisor. It is uniquely determined modulo linear equivalence. Definition 3.8.2 K = [D] is called the canonical bundle over S. If we put Jj k (z) = det

∂(zk1 ,zk2 ) , ∂(zj1 ,zj2 )

then ϕj = Jj k ϕk and we see that ϕ is a

meromorphic section of the line bundle defined by the 1-cocycle {Jj k }. Hence one may say that K is defined by {Jj k }. Definition 3.8.3 c1 = −c(K) is called the first Chern class of S.  is a Riemann surface. We let tλ be a local Example 3.8.4 In the situation of 3.5, C λ , C  = λ∈ U   uniformizing parameter on U   λ . The canonical bundle k over C is dtν ˜  by π(C), . If we denote the genus of C determined by the 1-cocyle {kλν } = dtλ

 − 2 from the Riemann–Roch theorem. then c(k) = 2π(C)  → C ⊂ S the Adjunction formula 3.9 Let C be an irreducible curve, μ : C  desingularization, K and k the canonical bundles over S and C, respectively, and let c denote the conductor of C. Then, k = μ∗ [C] + μ∗ K − [c].    = λU λ and Proof We take a sufficiently fine open covering S = j Uj , C λ , respectively, so that the following system of local coordinates (wj , zj ), tλ on Uj , U (a), (b), and (c) hold: λ ) ⊂ Uj (j = j (λ)). (a) μ(U λ ) ∩ s = ∅, then Uj (λ) ∩ s = ∅, where s = Sing(C). (b) If μ(U λ . Hence the minimal In this case, we may put μ : tλ → (wj , zj ) = (0, tλ ) on U local equation of C on Uj is Rj (wj , zj ) = wj = 0.

1.3 Intersection Multiplicities and the Adjunction Formula

23

λ )  p (p ∈ s), then pλ : tλ = 0, p = μ(0) = (0, 0) and μ can be (c) If μ(U λ as expressed on U μ : tλ → (wj , zj ) = (Pλ (tλ ), tλmλ ). We denote the minimal local equation of C on Uj by Rj (wj , zj ) = 0.

In the notation of 1.3, we have σλ dtλ = ∂wj Rj =

∂Rj ∂wj

) and σλ =

1 c (a0 tλ λ

d(t mλ ) (where ∂wj Rj (Pλ (tλ ), tλmλ )

+ a1 tλ + · · · ) (a0 = 0). In particular, on

λ ) ∩ s = ∅, we have ∂wj Rj = ∂wj = 1 (μ(U λ ) ⊂ Uj ). Hence, λ with μ(U U ∂wj  since c = ν cν pν , we see that σ −1= 0  is a local equation of c. It follows that [c] is determined by the 1-cocycle

σλ−1 σν−1

. In view of 3.7 and 3.8.1, we have

R [C] = {fj k }, K = {Jj k }, μ∗ [C] = {f˜λν } and μ∗ K = {Jλν }, where fj k = Rjk , f˜λν (t) = fj (λ)j (ν)(μ(t)) and Jλν = Jj (λ)j (ν)(μ(t)). We define gj k (z) by the following (d): (d) dRj ∧ dzj = gj k (z)dRk ∧ dzk . Then, since dRj = ∂wj Rj dwj + ∂zj Rj dzj , substituting it for (d), we get ∂wj Rj dwj ∧ dzj = gj k ∂wk Rk dwk ∧ dzk . Hence, (e) ∂wj Rj = gj k Jj k ∂wk Rk . On the other hand, since Rj = fj k Rk , we have dRj (z) = fj k (z)dRk (z) + Rk (z)dfj k . Hence, for z ∈ C, dRj (z) = fj k (z)dRk (z). This and (d) imply that fj k dzj = gj k dzk for z ∈ C. Applying μ∗ , we get (f) f˜λν (t)d(tλmλ ) = gj (λ)j (ν)(μ(t))d(tνmν ).

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1 Fundamentals of Algebraic Surfaces

By (e) and (f), f˜λν (t)

d(tλmλ ) 1 d(tνmν ) mλ =  m ∂wj (λ) Rj (λ) (Pλ (tλ ), tλ ) Jλν ∂wj (ν) Rj (ν) (Pν (tν ), tν ν )

and, hence, f˜λν σλ dtλ = J1 σν dtν . It follows that kλν = λν Considering the corresponding line bundles, we get

dtν dtλ

= f˜λν · Jλν ·



σλ σν

 .

k = μ∗ [C] + μ∗ K − [c]  

as wished. Corollary 3.10 Under the notation in 3.9,  = π(C)

1 1 2 (C + KC) + 1 − deg c. 2 2

In particular, if C has no singular points, then π(C) =

1 2 (C + KC) + 1. 2

Proof We have c(μ∗ F ) = F C, c(μ∗ [C]) = C · C = C 2 , c([d]) = deg d by  3.3, 3.5.1. By the Riemann–Roch theorem, c(k) = 2π(C)−2, while we have c(k) = 2 C + KC − deg c by 3.9.   Definition 3.11 For a divisor D and a line bundle F over S, 1 2 (D + KD) + 1, 2 1 π(F ) = (F 2 + KF ) + 1. 2

π(D) =

We call them the virtual genera of D, F , respectively. Proposition 3.12 For any irreducible curve C,  + π(C) = π(C)

1 deg c. 2

In particular, the following (a), (b) and (c) hold:  ⇐⇒ C = C.  (a) π(C) > π(C)  ⇐⇒ C = C.  (b) π(C) = π(C) (c) π(C) = 0 ⇐⇒ C : a non-singular rational curve.

1.4 Riemann–Roch Theorem

25

1.4 Riemann–Roch Theorem Notation 4.0 For an algebraic (analytic) sheaf , a line bundle F and a divisor D, we employ the following notation: χ(S, ) =



(−1)ν dim H ν (S, ),

ν

χ(S, F ) = χ(S, O(F )), χ(S, D) = χ(S, O([D])). Definition 4.1 (a) q = dim H 1 (S, O): the irregularity of S. It holds that q = Betti number of S). (b) pg = dim H 2 (S, O): the geometric genus of S. (c) pa = pg − q: the arithmetic genus of S.

1 2 b1

(b1 : the first

Remark 4.1.1 On the naming of “irregularity”(?): Let M be a surface defined by  its non-singular model. Then we have f (z0 , z1 , z2 , z3 ) = 0 in P3 , and S = M q(S) = pg − pa = 0 in almost all cases. In fact, for M : f (z0 , . . . , z3 ) = 0, if we consider the correspondence f ←→ (a1 , a2 , . . . , aN ) ∈ CN (the coefficients of f ), then the measure of {f | b1 (S) > 0} is 0. (Recall b1 = 2q.) Riemann–Roch Theorem 4.2 χ(S, D) =

1 2 (D − KD) + χ(S, O), 2

i.e., χ(S, D) = 12 (D 2 − KD) + pa + 1. (The latter follows from dim H 0 (S, O) = 1 and 4.1.) Proof Put ψ(F ) = χ(S, F ) − 12 (F 2 − KF ). If we have ψ(F − [C]) = ψ(F ) for any irreducible curve C, then, since we can express as D = mi Ci , we will have ψ([D]) = ψ(0) = χ(S, O) = pa + 1 and complete the proof. Therefore, it suffices to prove the following lemma.   Lemma 4.2.1 ψ(F − [C]) = ψ(F ). Proof We have the following exact sequences (1.14): (a) 0 → O(F − [C]) → O(F ) → O(F )C → 0, (b) 0 → μ∗ O(μ∗ F − [c]) → O(F )C → M → 0. From (a), we have χ(S, F ) − χ(S, F − [C]) = χ(C, O(F )C ). From (b), we have  μ∗ F − [c]) + dim H 0 (C, M ). We infer from 1.15 and the χ(C, O(F )C ) = χ(C,  μ∗ F − Riemann–Roch theorem for curves that dim H 0 (C, M ) = 12 deg c and χ(C, ∗   = F ·C −deg c−π(C)+1, respectively. Therefore, [c]) = c(μ F −[c])−π(C)+1

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1 Fundamentals of Algebraic Surfaces

using 3.12, we get χ(S, F ) − χ(S, F − [C]) = F · C − π(C) + 1. On the other hand, 1 1 1 2 2 2 2 (F −KF )− 2 ((F −C) −K(F −C)) = F ·C − 2 (C +K ·C) = F ·C −π(C)+1 (3.11). Hence, by the definition of ψ(·), we get ψ(F ) − ψ(F − [C]) = 0.   Riemann–Roch–Hirzebruch Formula 4.3 χ(S, D) =

1 2 1 (D − KD) + (c12 + c2 ). 2 12

(cν : the ν-th Chern class of S. c2 equals the Euler number of S.) By virtue of 4.2, it is sufficient for this to prove the following formula: Noether’s formula 4.4 pa + 1 =

1 2 12 (c1

+ c2 ).

Proof (Enriques [2]) It suffices to show Theorem 4.6 below. We will prepare things in 4.5 for that.  

4.5 On generic projections. (I) Projection of a non-singular curve C ⊂ P3 . Take linear subspaces P0 , P2 ⊂ P3 in general position. We let M be the image of C under the projection λ : P3 → P2 with center P0 . Then the singular points of M are at most nodes; we let d denote the number of nodes. We call the degree of the irreducible equation defining M the degree of M, and put it n. In fact, n equals the number of intersection points of M and a general line in P2 (that is, the number of generic hyperplane section of M). Then π(C) = 1 2 (n − 1)(n − 2) − d as is well-known. (II) Projection of a non-singular surface S → PN . We take linear subspaces P3 , PN−4 → PN in general position. We let M be the image of S under the projection λ : PN → P3 with center PN−4 . Then M is a model of S. We study  = Sing(M).

1.4 Riemann–Roch Theorem

27

If we take a system of local coordinates (x, y, z) of P3 around p ∈  suitably, then the equation of M in a neighborhood of p can take one of the following three forms: (a) yz = 0 (p : an ordinary double point), (b) xyz = 0 (p : a triple point), (c) xy 2 − z2 = 0 (p : a cuspidal (or pinch) point). In the following, we study these three cases separately. We employ notation  , λ−1 (p) = {p˜ 1 , . . . }. such as λ−1 () =  (a) Around a double point, taking a suitable system of local coordinates (u, v) on S, λ can be expressed as λ : (u, v) −→ (x, y, z) = (u, 0, v)

in a neighborhood of p˜ 1 ,

λ : (u, v) −→ (x, y, z) = (u, v, 0)

in a neighborhood of p˜ 2 .

(b) Around a triple point, similarly to (a), we have λ : (u, v) −→ (x, y, z) = (u, v, 0)

in a neighborhood of p˜ 1 ,

λ : (u, v) −→ (x, y, z) = (u, 0, v)

in a neighborhood of p˜ 2 ,

λ : (u, v) −→ (x, y, z) = (0, u, v)

in a neighborhood of p˜ 3 .

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1 Fundamentals of Algebraic Surfaces

(c) Around a cuspidal point, taking a suitable system of local coordinates (u, v) on S, λ can be expressed as λ : (u, v) −→ (x, y, z) = (u2 , v, uv) in a neighborhood of p˜ .

As is easily seen, the cuspidal point of M is a simple point of . We put: • n: the degree of M (that is, the degree of the irreducible equation defining M).

• m: the degree of  ( is the double curve). In fact, if we denote by  and L a general line and a general plane, respectively, then n = Card(M ∩ ),

m = Card( ∩ L).

• t: the number of triple points of M. (From the above observations (a), (b), and (c),any singular point of  is a triple point of M.) ρ •  = ν=1 ν : the decomposition of  into irreducible curves ν . We ˆ ν the non-singular model of ν . denote by 

1.4 Riemann–Roch Theorem

29

 ˆ ν ) − 1} + 1 + 2t. • Then π() = ν {π( (If O denotes the sheaf of germs of holomorphic functions on P3 and if O is the restriction of O to , then χ(, O ) = dim H 0 (, O ) − dim H 1 (, O ) = 1 − π().) Theorem 4.6 (Enriques [2]) Let the notation be as in 4.5, (II). Then, pa =

 n−1 − (n − 4)m + π() − 1, 3

c12 = n(n − 4)2 − (5n − 24)m + 4π() − 4 + t, c2 = n(n2 − 4n + 6) − (7n − 24)m + 8π() − 8 − t, where pa : the arithmetic genus of S, and c1 , c2 : Chern classes of S. Proof We will calculate pa in 4.12. We devote 4.8 and 4.9 to the preparations. The calculation of c12 and c2 will be done in 4.13.   Corollary 4.7 c12 + c2 = 12(pa + 1). λ

Adjunction formula 4.8 Let S −→ M → P3 be the generic projection. If K and K respectively denote the canonical bundles over S and P3 , then ], K = λ∗ K + λ∗ [M] − [  = λ−1 () ([]  : the conductor). where  = Sing(M) and  Proof Similar to the case of one dimension lower (3.9). If we put locally λ : (u, v) −→ (x, y, z) = (μ1 (u, v), μ2 (u, v), μ3 (u, v)), R(x, y, z) = 0 : the minimal local equation of M, σ =

dμ1 (u, v) ∧ dμ2 (u, v) , ∂R ∂z (μ1 (u, v), μ2 (u, v), μ3 (u, v))

] = −(σ ). For example, around a cuspidal point, then the conductor is given by [ we have R = xy 2 − z2 = 0, λ : (u, v) −→ (x, y, z) = (u2 , v, uv), 1 2udu ∧ dv 1 d(u2 ) ∧ dv =− = − du ∧ dv. −2z 2 uv v   ] = −(σ ) = (v) = − 1 . Therefore, [ v σ =

 

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From now on until 4.13, we let λ, S, M, , n, m, t, . . . , etc. be as in 4.5, (II). 4.9 Preparations for calculating pg and pa . Let L : a0 ζ0 + a1 ζ1 + a2 ζ2 + a3 ζ3 = 0 be a plane in P3 and put E = [L]. Then M ≈ nL (linearly equivalent) and [M] = nE. On the other hand, the canonical bundle over P3 is K = −4E (in fact, the canonical bundle over PN is OPN (−N −1)). Hence, it follows from 4.8 that ]. 4.9.1 K = (n − 4)λ∗ E − [ For any line bundle F over P3 , we can define OP3 (F), OP3 (F − ), OP3 (F − M) in a similar way to S (1.10). Moreover, we put OP3 (F) = OP3 (F)/OP3 (F − ), OP3 (F − )M = OP3 (F − )/OP3 (F − M). Hence we obtain the following exact sequences 4.9.2, 4.9.3: 4.9.2 0 → OP3 (F − M) → OP3 (F − ) → OP3 (F − )M → 0, 4.9.3 0 → OP3 (F − ) → OP3 (F) → OP3 (F) → 0. We have: !

 ←− OP3 (F − )M . 4.9.4 λ∗ (O(λ∗ F − ))  ) is given by Proof The natural λ homorphism ρ : OP3 (F − ) → O(λ∗ F −  ∗ ρ : ϕ(x, y, z) → (λ ϕ)(u, v). So, we need to show that it is an isomorphism on the stalk over any point on M. For this purpose, it suffices to consider p ∈ . Here, we consider the case that p is a cuspidal point (and the other cases are left to the readers). Then (λ∗ ϕ)(u, v) = ϕ(u2 , v, uv). (We follow Fig. c in 4.5(II).)

 )p˜ . Since (a) ρ is surjective: Suppose that we are given ψ(u, v) ∈ O(λ∗ F −   we have ψ(u, 0) = 0 (recall that in a neighborhood of ψ vanishes along ,  : v = 0). Hence we can write ψ(u, v) = ω(u, v)v with a holomorphic p˜ ,  function ω(u, v). Since λ : (u, v) → (x, y, z) = (u2 , v, uv), we can find some ϕ(x, y, z) ∈ OP3 (F)p such that ψ(u, v) = ϕ(u2 , v, uv) = (λ∗ ϕ)(u, v). Moreover, since ϕ(x, 0, 0) = ψ(u, 0) = 0 (recall that  : y = z = 0 in a neighborhood of p), we have ϕ(x, y, z) ∈ (OP3 (F − )M )p .

1.4 Riemann–Roch Theorem

31

(b) The injectivity of ρ clearly follows from the fact that ϕ modulo (xy 2 − z2 ) can be determined uniquely.   μ  −→ Remark 4.9.5 For C C → S, recall that we have an exact sequence κ

0 → μ∗ (O(μ∗ F − c)) → O(F )C −→ M → 0 (1.14) ∼ =

that is, an isomorphism μ∗ (O(μ∗ F − c)) ←− Ker(O(F )C → M ). 4.9.4 is an extension of it to the one dimension higher case. In fact, Ker(κ) = O(F − Sing(C))C .     f = f (ζ ) is a homogeneous polynomial Notation 4.10 Lk (−) = f  . of degree k satisfying f (ξ ) = 0 for ξ ∈  4.11 pg = dim Ln−4 (−). Proof Put F = (n − 4)E. Then we get pg = dim H 2 (S, O) = dim H 0 (S, O(K))

(by Serre duality)

 = dim H 0 (S, O(λ∗ F − []))

(by 4.9.1)

= dim H 0 (M, OP3 (F − )M )

(by 4.9.4).

On the other hand, since [M] = nE, we have OP3 (F − M) = OP3 (−4E). It follows from H ν (P3 , OP3 (kE)) = 0 (ν = 1, 2) and H 0 (P3 , OP3 (−4E)) = 0 ∼ =

that H 0 (P3 , OP3 (F − )) → H 0 (M, OP3 (F − )M ). Therefore, pg dim H 0 (P3 , OP3 (F − )) = dim Ln−4 (−), because F = (n − 4)[L].  n−1 4.12 pa = − (n − 4)m + π() − 1. 3

=  

Proof We put F = (n − 4)E. Then pa + 1 = χ(S, O) = χ(S, O(K))

(by Serre duality)

 = χ(S, O(λ∗ F − ))

(by 4.9.1)

= χ(M, OP3 (F − )M )

(by 4.9.4)

= χ(P3 , OP3 (F − )) − χ(P3 , OP3 (K))

(by 4.9.2).

(The last equality comes from OP3 (F − M) = OP3 (K).) By the Serre duality theorem, however, we have χ(P3 , OP3 (K)) = −χ(P3 , OP3 ) = −1. Hence, pa + 1 = χ(P3 , OP3 (F)) − χ(, OP3 (F) ) + 1 by 4.9.3. Since χ(P3 , OP3 (F)) =   dim H 0 (P3 , OP3 ((n − 4)E)) = dim Ln−4 = n−1 3 , we will complete the proof of 4.12 if we can show the following.  

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Lemma 4.12.1 χ(, OP3 (F) ) = (n − 4)m − π() + 1.  (If we denote the non-singular model of  = j j (irreducible decomposition)  ˆ j ) − 1) + 2t.) ˆ = j ˆ j , then π() − 1 = j (π( by  Proof We put: ⎧ ⎪ L : a general plane in P3 , ⎪ ⎪ ⎪ ⎪ ⎨ E = [L], F = (n − 4)E, Sing() = {t1 , . . . , tk , . . . , tt } (all of them are triple points of (4.5, II)), ⎪ ⎪ ˆ →  ⊂ M ⊂ P3 desingularization, ⎪ μ ˆ: ⎪ ⎪ ⎩ μˆ −1 (t ) = {ˆt , ˆt , ˆt }. k k1 k2 k3

Then the conductor of  is

3 t  

ˆtkν . Since we have the natural isomorphism

k=1 ν=1 3

Oˆ (−ˆtkν )tˆkν ∼ = O (tk )tk , we obtain:

ν=1

(a) μˆ ∗ (Oˆ (μˆ ∗ F −

t  3 

∼ ˆtkν )) ← OP3 (F −

t 

t k ) ,

 where F denotes the restriction of F to  and OP3 (F − k tk ) = {ϕ ∈ OP3 (F) | ϕ(tk ) = 0 (k = 1, 2, . . . )}. Then, we clearly have the exact sequence: t  tk ) → OP3 (F) → T → 0, (b) 0 → OP3 (F − k=1 ν=1

k=1

k=1

where T is a sheaf over  satisfying Supp(T ) = Sing() = ∪tk=1 tk , dim H 0 (, T ) = t and, hence, χ(, T ) = t. This yields: t  3  ˆ O ˆ (μˆ ∗ F − ˆtkν )) + t. (c) χ(, OP3 (F) ) = χ(,  k=1 ν=1

1.4 Riemann–Roch Theorem

33

If d = L ·  (hence deg d = m), then F = (n − 4)[d]. Similarly,  if we put dj = L · j , dˆ = μˆ ∗ d and dˆ j = μˆ ∗ dj , then obviously dˆ = j dˆ j . We    ˆ j and write j aj next distribute the conductor a = tk=1 3ν=1 ˆtkν to each  ˆ j ). It is clear that deg a = 3t. Noting ˆ j,  ˆ = j  (where aj is the divisor on   ∗ ˆ ˆ ˆ that  = j j and μˆ F = (n − 4)[d] = j (n − 4)dˆ , we get: t  3

 ˆtkν ) = (d) Oˆ (μˆ ∗ F − Oˆ j ((n − 4)dˆ − aj ). k=1 ν=1

(e)

j

By the Riemann–Roch theorem for curves, we have ˆ j , O ˆ ((n − 4)dˆ − aj )) = (n − 4) deg dˆ j − deg aj − π( ˆ j ) + 1. χ( j From (c), (d), and (e), χ(, OP3 (F) ) =



χ(j , Oˆ j ((n − 4)dˆ − aj )) + t

j

= (n − 4) deg dˆ − deg a −



ˆ j ) − 1) + t (π(

j

= (n − 4)m − π() + 1 which is what we want.

 

4.13 In order to compute the Chern numbers c12 , c2 , we prepare things in (I), (II), and (III) below. The actual calculations will be done in (IV). Let λ : S → M → P3 be the projection, let (w, x, y, z) denote a system of homogeneous coordinates of P3 (which is assumed to be taken sufficiently general with respect to M) and let f (w, x, y, z) = 0 be the irreducible equation defining M. The degree of f is n. (I) Hyperplane sections of M (finding the formula for c2 (S)). Put  (τ ∈ P1 ), Lτ : w − τ z = 0 L∞ : z = 0,  Cτ : Lτ ∩ M, : w = z = 0 (of course  ⊂ Lτ ).

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1 Fundamentals of Algebraic Surfaces

For a general τ , we have



Lτ ∩  = δτ,1 + δτ,2 + · · · + δτ,m (m : degree of ), (n : degree of M).  ∩ M = b1 + b2 + · · · + bn

τ be the proper inverse image of Cτ in S, that is, C τ = λ−1 [Cτ ]. Then Let C  Cτ is non-singular for a general τ . On the other hand, there are a finite number of points τ1 , . . . , τj , . . . , τc (∈ P1 ) such that Lτj contacts M at a point oj , that is, c is the number of tangent planes of M passing through . The number c is known as the class of M.

Since (w, x, y, z) is taken generally with respect to M, we can assume oj ∈ . τj . Then, o˜ j = λ−1 (oj ) is a node (ordinary double point) of C τ is non-singular. 4.13.1 If τ = τj (1  j  c), then C

1.4 Riemann–Roch Theorem

35

τ in a neighborhood of a point in λ−1 (q) when Lτ passes Proof We investigate C through a cuspidal point q of M. (In fact, since τ = τj , Sing(Cτ ) ⊂ Lτ ∩ . Hence τ ), it suffices to work in a neighborhood of λ−1 (Lτ ∩ ) in in order to study Sing(C τ .) If Lτ passes through a singular point p other than a cuspidal point, it is almost C τ is non-singular in a neighborhood of λ−1 (p).) clear that C

Letting (X, Y, Z) and (u, v) denote local coordinates around q and q˜ = λ−1 (q), respectively, we may write locally ⎧ 2 ⎪ ⎪ λ : (u, v)2 −→2(X, Y, Z) = (u , v, uv), ⎨ M : XY − Z = 0, ⎪ L : AX + BY + CZ = 0, ⎪ ⎩ τ τ : Au2 + Bv + Cuv = 0. C We may assume that q = (0, 0, 0), A = 0, B = 0, C = 0 and, furthermore, τ : (B+Cu)v+Au2 = 0, B+Cu = 0 in a neighborhood of q˜ = (0, 0). Then, since C  we see that q˜ = (0, 0) is a simple point of Cτ . τ is non-singular when τ = τj (1  j  c). Therefore, C   τ ). Definition 4.13.2 For a general τ , we put g = π(C Since Cτ is a plane curve of degree n having m nodes, we have the following: (a)

2g − 2 = n(n !− 3) − 2m. τ , Cτ ∩ Cλ = b1 + b2 + · · · + bn (=  ∩ M). Moreover, C Now, S = τ ∈P1

λ = b˜1 + b˜2 + · · · + b˜n and assume that C τ and C λ meet τ ∩ C we can write C normally at b˜ k . τ }τ ∈P1 , in order Since {b˜k }1kn is the set of all base points of the pencil {C to have a fibration from it, we perform a quadratic transformation Qb˜k with center b˜k to S (1  k  n) to get " S, that is, " S = Q ˜ Q ˜ · · · Q ˜ (S). Then b1

τ ) + c2 (" S) = χ(P1 ) · χ(C

c  j =1

b2

bn

τj ) − χ(C τ )), (χ(C

36

1 Fundamentals of Algebraic Surfaces

where χ(V ) denotes the Euler number of V . Since o˜ j is the unique singular τj which is a node, we have χ(C τj ) − χ(C τ ) = 1. Hence c2 (" point of C S) = 2(2 − 2g) + c. On the other hand, since the second Betti number increases by one after a quadratic transformation, we have c2 (S) = c2 (" S) − n. Hence, we obtain (b) c2 (S) = 4 − 4g + c − n.

(II) Polar curves. ∂f Recall M : f (w, x, y, z) = 0. We put Mx : ∂x f = 0 (where ∂x f = ∂x , polynomial of degree n − 1). Since we have ∂x f (P ) = 0 when P ∈ , there is a curve Dx satisfying Mx ∩ M =  ∪ Dx .

Definition 4.13.3 Dx is called a polar curve of M. Now let (ζ0 , ζ1 , ζ2 , ζ3 ) be a system of homogeneous coordinates on P3 and write ζν = aν0 w + aν1 x + aν2 y + aν3 z. We regard aνk as parameters. Since ∂x f = the aν1 ’s, {Dx } forms a linear system.

3 3   ∂ζν ∂f ∂f = aν1 is linear in ∂x ∂ζν ∂ζν ν=0

ν=0

4.13.4 The set of base points of {Dx } coincides with the set of all cuspidal points qj .

1.4 Riemann–Roch Theorem

37

Proof (i) Obviously, p ∈  is not a base point of {Dx }. So we consider points of .

(ii) We consider a cuspidal point qj . Recall that λ : (u, v) → (X, Y, Z) = (u2 , v, uv). We claim that qj is a base point of {Dx }. This can be seen as follows. Assume that w = 0 at qj . Putting w = 1 and considering (x, y, z) as a system of inhomogeneous coordinates on P3 , we have f (1, x, y, z) = XY 2 − Z 2 . Hence ∂x f = (∂x X) · Y 2 + (2∂x Y ) · XY − (2∂x Z) · Z and we have λ∗ (∂x f )(u, v) = ∂x X · v 2 + 2∂x Y · u2 v − 2∂x Z · uv = v(∂x X · x =  = λ−1 [] : v = 0, D v + ∂x Y · u2 + ∂x Z · u). It follows that  −1 2 x λ [Dx ] : ∂x X · v + ∂x Y · u + ∂x Z · u = 0. This implies that every D passes through q˜ j = (0, 0). Hence qj ∈ Dx for all x. We also remark that qj is a non-singular point of a general Dx . (For a general x, since we have ∂x X = 0, x is as in the above figure.) ∂x Y = 0, ∂x Z = 0 at qj , the picture of D (iii) If p ∈  but p = qj , then p ∈ Dx for a general x. This can be seen as follows. If p is a double point, then taking coordinates as in (ii), we have f (1, x, y, z) = Y · Z. Hence ∂x f = (∂x Y ) · Z + (∂x Z) · Y . For a general x, we can assume that ∂x Y = 0 and ∂x Z = 0 in a neighborhood of p. Therefore, if we consider ∂x f = 0, then Y = 0 ⇔ Z = 0. This implies that Mx ∩ M =  in a neighborhood of p, and Dx does not pass through p. If p is a triple point, then we can write f (1, x, y, z) = XY Z and argue similarly. By (i), (ii), and (iii), the set of base points of {Dx } coincides with the set of all cuspidal points {qj }.   {Dx } does not have a base point outside {qj }. Moreover, as we have already seen, qj is a simple point of a general Dx . It follows from Bertini’s theorem that a general Dx is non-singular. Now, we let L be a plane in P3 and put E =#[L]. Then#[Mx ] = (n − 1)E by definition. On the other hand, since we have Mx M =  Dx , we obtain (c)

x ] = (n − 1) +D E (where  E = λ∗ E). [

Next, we fix a general x and study the intersection points of Dx and . From the above discussion, we know that Dx ∩  contains all the cuspidal points q1 , . . . , qγ of M. At a general point of , f (1, x, y, z) = Y Z, fx = Yx Z + Zx Y = 0 and, hence (∂x f )Z=0 = Zx Y , (∂x f )Y =0 = Yx Z. Then for p ∈ Dx ∩  ⊂ Mx ∩ M, we

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1 Fundamentals of Algebraic Surfaces

have either Zx (p) = 0, Yx (p) = 0 or Zx (p) = 0, Yx (p) = 0. In the former case, for example, a neighborhood of p is  on Y = 0, and  + Dx on Z = 0. Therefore, if we let r1 , . . . , rσ be the points of intersection of Zx Yx = 0 and x ∩   = , then we can write Dx ∩  = q1 + · · · + qγ + r1 + · · · + rσ and D q˜ 1 + · · · + q˜ γ + r˜1 + · · · + r˜δ (the latter follows from the former). By Bertini’s theorem, r˜k is a simple point of D˜ x . We obtain

(d)

x ·   = γ + σ. D

(III) Second polar curves. We put Mxx : ∂x2 f = 0. Mxx is of degree n − 2. (i) At a cuspidal point qj ∈ , we have f = f (1, x, y, z) = XY 2 − Z 2 and, since (w, x, y, z) can be chosen generally, we can assume that fxx (qj ) = −2Zx2 = 0, i.e., Mxx  qj . (ii) At a triple point ti ∈ , we have f = XY Z and ∂x2 f = 2∂x X∂x Y ·Z+∂x Y ∂x Z· X + ∂x Z∂x X · Y + ∂x2 X · Y Z + · · · , and we can assume that ∂x X∂x Y ∂x Z = 0 at ti . Then Mxx and  meet normally at ti . (iii) At a general point p ∈ , we have f = Y Z and fxx = Yx Zx + Yxx Z + Zxx Y . Hence the necessary and sufficient condition to have p ∈ Mxx ∩  is fxx (p) = Yx (p)Zx (p) = 0, that is, p = ri . On the other hand, since Dx and  meet normally at ri , we have Zxx (ri ) = 0 and, hence, Mxx and  meet normally at ri . In sum, Mxx ∩  = r1 + · · · + rσ + t1 + · · · + tt and, at each ri or ti , Mxx has a simple point and meets every component of  normally. (e)

Mxx ·  = σ + 3t.

1.4 Riemann–Roch Theorem

39

(IV) Finally, we consider another polar curve Dy : ∂y f = 0.

 We have Dx ∩ Dy ∩  = qi . We consider a point p ∈ Dx ∩ Dy with p ∈ . That is, we assume that fx (p) = fy (p) = 0 and p ∈ . Since the tangent plane Tp (M) at a smooth point p is given by Tp (M) : fw (p)w + fx (p)x + fy (p)y + fz (p)z = 0, we have Tp (M) = Lτ at p ∈ Dx ∩ Dy , where τ = − ffwz (p) (p) . (Recall that Lτ : w − τ z = 0 and that oj is the point at which Lτj has a contact to M in the notation of (I).) Hence p = oj for some j . Conversely, when p = oj , then oj ∈ Dx ∩ Dy , since fx (oj ) = fy (oj ) = 0. Hence we obtain x ∩ D y = q˜ 1 + · · · + q˜ γ + o˜ 1 + · · · + o˜ c D and (f)

x2 = D x · D y = γ + c. D

τ ],  τ2 = C τ · C τ  = n, Recall that we have E = [Lτ ],  E = λ∗ E = [C E2 = C 2 2      E + K · E = Cτ + K · Cτ = 2π(Cτ ) − 2 = 2g − 2. Then it follows from (a) that (g)  E2 = n, (h) K ·  E = n(n − 4) − 2m. Now, we infer from 4.9.1 and (c) that  = (n − 4) [] E − K, x ] = (n − 1) ] = 3 [D E − [ E + K. These together with (d) and (h) give us x ·   = (3 γ +σ =D E + K)((n − 4) E − K) = 3n(n − 4) + (n − 7)K  E − K2 = n(n2 − 8n + 16) − (2n − 14)m − c12 .

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1 Fundamentals of Algebraic Surfaces

We also have σ = (n − 2)m − 3t from (e), since the degrees of Mxx and  are n − 2 and m, respectively. Hence (i) c12 = n(n − 4)2 − (3n − 16)m + 3t − γ . Next, it follows from (f), (g), and (h) that x2 = (3 E + K) = n(6n − 15) − 12m + c12 . γ +c =D Hence, from this and (a), (b), and (i), (j) c2 = n(n2 − 4n + 6) − (3n − 8)m + 3t − 2γ . We have only to compute γ . "  and there ˜ be the non-singular model of .  Each ˜tiν is a double point of  Let   " ˜ = π()  − 3t.  = i ˜tiν . Hence π() are 3t such points in total. We have Sing() ν " ˜ →  " is a double covering with branch points q˜ j (= " q˜ j )’s. On the other hand,  " ˜ as " =  " , we can compute the Euler number of  Since  j

" ˜ = 2(2 − 2π( ")) − γ 2 − 2π()  "j )) − γ . (2 − 2π( =2 j

" ˜ = π( ) − 3t, we get From this and π() −γ = 2

 "j ) − 2) − 2π( ) + 2 + 6t. (2π( j

 ) − 2 = (  + K) ·   = (n − 4)  = (n − 4) · 2m, π() − 1 = Since 2π( E·  " j (π(j ) − 1) + 2t, we get (k) −γ = 4π() − 4 − (2n − 8)m − 2t.

1.4 Riemann–Roch Theorem

41

Consequently, we infer from (i) and (j) that (l) c12 = n(n − 4)2 − (5n − 24)m + t + 4π() − 4, (m) c2 = n(n2 − 4n + 6) − (7n − 24)m − t + 8π() − 8. Example 4.14 Enriques surface. Let (z0 , z1 , z2 , z3 ) be a system of homogeneous coordinates on P3 and put M : f (z0 , z1 , z2 , z3 ) = a0 (z0 z1 z2 )2 + a1 (z1 z2 z3 )2 + a2 (z2 z3 z0 )2 + a3 (z3 z0 z1 )2 + z0 z1 z2 z3 ψ(z0 , z1 , z2 , z3 ) = 0, where ψ is a quadratic form. Definition 4.14.1 The non-singular model S of a general M is called an Enriques surface. 4.14.2 An Enriques surface is the first example of a non-rational surface with pa = pg = 0. First of all, we investigate  = Sing(M) for a general M. Varying ai and ψ, we get a linear system  = {M}. Its general member M has no singular points outside the base locus by Bertini’s theorem.  Putting λν : zλ = zν = 0, the set of all base points of {M} clearlycoincides with 0λ 0, there exists at least one "i satisfying

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K · "i > 0. Putting C1 = "i , we have K · C1  1. We write D = C1 + Y . Then we have Y  0.If Y = 0, then we are done. If Y > 0, then C1 · Y  1 by 7.8. Hence for Y = j =i "j , we can find "j satisfying C1 "j > 0. We put C2 = "j . Then C1 · C2 = D1 · C2  1. We argue by induction. Suppose that we have chosen C1 = "ν1 , C2 = "ν2 , . . . , Ci−1 = "i−1 so that Dj −1 Cj  1 (j = 2, 3, . . . , i − 1), where Dj −1 = C1 + C2 + · · · + Cj −1 . Writing D = Di−1 + Y  , if Y  = 0 then we are done; if Y  > 0, then Di−1 · Y  1 by 7.8 and we can find "λ < Y  satisfying Di−1 · "λ  1. Hence, letting Ci = "λ , we obtain Di−1 · Ci  1.   Remark 7.9.1 From the proof, if we are given " < D such that K · "  1, then we can put C1 = " in finding a composition series D = ni=1 Ci satisfying (α). Lemma 7.10 Assume that K · Ep = K · Eq = Ep · Eq = 0 (7.6). For D ∈ |eK|, if D = X + Y + Ep + Eq , X > 0, Y > 0, K · X > 0 and K · Y > 0, then X · Y  0. Proof We put r=

K ·X , K2

r1 =

Ep · X Ep · X , =− Ep2 2

r2 =

Eq · X Eq · X =− Eq2 2

and ξ = X − rK − r1 Ep − r2 Eq (recall that we have Ep2 = Eq2 = −2 by 7.3). Then K · ξ = Ep · ξ = Eq · ξ = 0 by hypothesis. Similarly, if we put s=

K ·Y , K2

s1 = −

Ep · Y , 2

s2 = −

Eq · Y 2

and η = Y − sK − s1 Ep − s2 Eq , then K · η = Ep · η = Eq · η = 0. On the other hand, we have r + s = e and r1 + s1 + 1 = r2 + s2 + 1 = 0 from K·(X+Y ) = K·D−K·Ep −K·Eq = eK 2 , Ep ·(X+Y ) = Ep ·D−Ep2 −Ep ·Eq = 2 and Eq · (X + Y ) = 2. Hence D = X + Y + Ep + Eq = (r + s)K + (r1 + s1 + 1)Ep + (r2 + s2 + 1)Eq + ξ + η = eK + ξ + η

and it follows that ξ + η ∼ 0. We also have X · Y = rsK 2 + r1 s1 Ep2 + r2 s2 Eq2 + ξ · η. Since rK 2 = sK 2 = K ·X  1, we obtain rsK 2  12 (r +s) = 2e from rsK 2  r, s. Moreover, r1 s1 Ep2 = −2r1 s1 = 2s1 (s1 +1) = 2(s1 + 12 )2 − 12  − 12 and, similarly, r2 s2 Eq2  − 12 . Finally, we have ξ · η = −ξ 2  0 (because ξ + η ∼ 0 and ξ 2  0 from K · ξ = 0 by 7.2). Therefore, X · Y  2e − 12 − 12 − ξ 2  2e − 1 > −1, that is, X · Y  0.   Remark 7.11 In the expression E = E1 + · · · + Eλ + · · · + E ρ as in Notation 7.6, if D ∈ |eK| meets Eλ , then Eλ < D. In fact, write Eλ = Ei (K · Ei = 0, π(Ei ) = 0, Ei2 = −2). Then it follows from D ∼ eK that D · Ei = eK · Ei = 0. Since D > 0, this implies that Ei < D when D ∩ Ei = ∅. Since Eλ is connected, we have Eλ < D.

2.7 Composition Series

59

Lemma 7.12 If D meets Eλ and Eν (λ = ν) (this is equivalent to saying that nEλ < D and Eν < D by 7.11), then there exists a composition series D = i=1 Ci satisfying (β) with Cn < Eν and Cn−1 < Eλ . Proof We take Ep < Eλ and Eq < Eν . Since K · D = eK 2 > 0, we can find an irreducible component " < D satisfying K · " > 0 and put C1 = ". Therefore, K · C1  1. Next, assume that we have chosen components C1 , C2 , . . . , Ci (i  2) of D so that  (Pi ) : D = C1 + C2 + · · · + Cj −1 + Cj + · · · + Ci−1 + Xi + Ep + Eq , Xi  0, (βi ) : Dj −1 · Cj  0, K · Cj + Dj −1 · Cj  1 (1  j  i − 1), where Dj −1 = C1 + · · · + Cj −1 . When i = 2, these merely say K · C1  1. In general, we have K · Xi  0 by 7.3. We shall show that there exists i0  i satisfying (Pi0 ), (βi0 ) and K · Xi0 = 0. In fact, if K · Xi > 0, then writing D = Di−1 + Xi + Ep + Eq , K · Di−1  K · C1  1 (7.3). Then we infer from 7.10 that Di−1 · Xi  0. There are the following two possibilities: 

(a) There exists a component C < X with Di−1 · C > 0. (b) Di−1 · C = 0 for all C < Xi .

If (a) is the case, then, putting Ci = C, Xi+1 = Xi − C, Di−1 · Ci  1. If (b) is the case, then, by K · Xi > 0, there exists a component C < Xi satisfying K · C > 0. We put Ci = C. Then, by (b), we have Di−1 · Ci = 0 and K · Ci + Di−1 · Ci  1. Therefore, in either case, we can choose Ci so that we have (Pi+1 ), (βi+1 ), K · Xi+1  0. Moreover, we obviously have K · Xi  K · Xi+1 ( 0). Since we can continue the procedure until we arrive at i0 with K · Xi0 = 0, we conclude that there exists i0  i( 2) satisfying ⎧ ⎨ (Pi0 ) : D = C1 + C2 + · · · + Ci0 −1 + Xi0 + Ep + Eq , Xi0  0, (β ) : Dj −1 · Cj  0, K · Cj + Dj −1 · Cj  1 (1  j  i0 − 1), ⎩ i0 and K · Xi0 = 0. As in the proof of 7.9, using 7.8, we can also choose components Ci0 +1 , . . . , Cn of Xi0 + Ep + Eq so that 

D = C1 + · · · + Ci0 + Ci0 +1 + · · · + C + · · · + Ch + · · · + Cn (C = Ep , Ch = Eq ), (α) : Dj −1 · Cj = (C1 + · · · + Cj −1 ) · Cj  1 (2  j  n).

We shall show that we can change the numbering of the Ci ’s so that Dj −1 · Cj  1

(2  j  n), Cn−1 < Eλ , Cn < Eν

if necessary. We remark that, if Ci · Ci+1 = 0, we can exchange them to get (α), because putting (αi ): (C1 +· · ·+Ci−1 )·Ci  1, we infer from Ci ·Ci+1 = 0 and (αi )

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2 Pluri-Canonical Systems on Algebraic Surfaces of General Type

that (C1 + · · · + Ci−1 + Ci+1 ) · Ci  1 and, from Ci · Ci+1 = 0, (αi+1 ) and (αi+2 ) that (C1 + · · · + Ci−1 ) · Ci+1  1 and (C1 + · · · + Ci−1 + Ci+1 + Ci ) · Ci+2  1. Note that every Ci (i > i0 ) is a component of E , because K · Xi0 = 0. Since two curves in different connected components do not meet, we can apply the argument just given and gather curves in the same connected component of E , assigning a consecutive number to them. Therefore, after Ci0 , we can choose the numbering so that components of Eλ , Eν are separated as D = C1 + · · · + Ci0 + Ci0 +1 + · · · + Ep + · · · + Cs + · · · + Eq + · · · + Cn . * +, - * +, < Eλ

< Eν

(In D, there are no components of Eλ , Eν other than indicated.) Furthermore, because we can move Cs to the position just before Cn similarly, it is possible to set Cs = Cn−1 .   Lemma 7.13 Given Eλ meeting D ∈ |eK|, there exists a composition series D = n C i=1 i of D which satisfies Cn < Eλ and the condition (β).  

Proof Similar to the proof of 7.12.

2.8 Conclusions Notation 8.0 We have chosen and fixed a positive integer e satisfying  dim |eK|  0 (7.7). Taking a member D ∈ |eK| and a composition series D = ni=1 Ci of D, we choose and fix an integer m with m > e and put Fi = mK − [Zi ]

(where Zi = Ci + · · · + Cn ).

(We shall employ this notation to the end.) Recall that Di = C1 + · · · + Ci (7.7). 8.1 Fi+1 · Ci − Ci2 − K · Ci = (m − e − 1)K · Ci + Di−1 · Ci . Proof Clear from Fi+1 − [Ci ] − K = (m − 1)K − [Ci + Zi+1 ] ∼ (m − 1 − e)K + [D − Zi ] = (m − 1 − e)K + [Di−1 ].   8.2 Take integers h, k  0. For x, y ∈ D (where D ∈ |eK|), if the multiplicities of D at x, y are  h,  k, respectively, then we put i = O(mK − Zi − hx − ky). (In what follows, for i , such conditions are assumed to be satisfied for preassigned h, k, x and y.)

2.8 Conclusions

61

In particular, 1 = O(mK − D − hx − ky) = O(mK − D) ∼ = O((m − e)K). 8.2.1 1 ⊂ 2 ⊂ · · · ⊂ i ⊂ · · · ⊂ n+1 , where n+1 = O(mK − hx − ky). Assumption 8.3 For h, k as above, we let hi , ki be integers determined by i+1 /i ∼ = O(Fi+1 − hi x − ki y)Ci . 8.3.1 If h = k = 1, x, y ∈ Ci and x, y ∈ Zi+1 , then hi = ki = 1. This can be seen as follows. In this case, we have i+1 = O(mK − Zi+1 − x − y) ∼ = O(Fi+1 − x − y), i = O(mK − Zi+1 − Ci ) ∼ = O(Fi+1 − Ci − x − y) and, hence, i+1 /i = O(Fi+1 − x − y)/O(Fi+1 − x − y − Ci ) ∼ = O(Fi+1 − x − y)Ci .

Lemma 8.4 If (m − e − 1)K · Ci + Di−1 · Ci > 14 (hi + 1)2 + 14 (ki + 1)2 , then dim H 1 (S, O((m − e)K))  dim H 1 (S, i ) (2  i  n + 1). Proof For α  0, we have 14 (hi + 1)2  (hi − α + 1)+ α. (Because: it is clear when hi −α+1 < 0. When hi −α+1  0, (hi −α+1)+ α = −(α− hi 2+1 )2 + 14 (hi +1)2  1 2 4 (hi + 1) .) We also have a similar inequality for ki . Then by hypothesis and 8.1 we infer from 6.1 that H 1 (Ci , O(Fi+1 − hi x − ki y)Ci ) = 0. From the exact sequence 0 → i → i+1 → O(Fi+1 − hi x − ki y)Ci → 0, we have the exact sequence · · · → H 1 (S, i ) → H 1 (S, i+1 ) → 0 and, hence, dim H 1 (S, i )  dim H 1 (S, i+1 ). On the other hand, we know 1 = O((m − e)K) (8.2).   Lemma 8.5 (Zariski [9]) There is an integer m0 such that, for any m  m0 , dim H 1 (S, O((m − e)K)) = dim H 1 (S, O(mK)).  Proof For D ∈ |eK|, we take a composition series D = ni=1 Ci of D satisfying condition (α) (7.7 and 7.9). If m  e+2, then, since m−e−1  1, from 7.3 and (α),

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2 Pluri-Canonical Systems on Algebraic Surfaces of General Type

we get (m−e−1)K ·Ci +Di−1 ·Ci  1. Putting h = k = 0, if we put i = O(mK − Zi ), then hi = ki = 0 (here, one can take any x, y ∈ D). Therefore, the assumption of 8.4 is satisfied and we obtain dim H 1 (S, O((m − e)K))  dim H 1 (S, O(mK)), If we put f (m) = dim H 1 (S, O(mK)), then f (m)  f (m + e)  f (m + 2e)  · · ·  f (m + ke)  · · · . Since, however, f (m) is a non-negative integer, there exists an integer 0 such that f (m + e) is constant for any   0 . Hence, we can find m0 as desired.   Theorem 8.6 Assume that Pe  2, eK 2  2, m  e + 2 and m  m0 . Then the following hold : (a) For any x ∈ S, the following sequence is exact: 0 → H 0 (S, O(mK − x)) → H 0 (S, O(mK)) → C → 0. (b) The m-canonical system |mK| has no base points and mK is a holomorphic map. Proof Firstly, recall that we have the following exact sequence for any x ∈ S (5.0.1): 0 → O(mK − x) → O(mK) → Cx → 0. Hence, we have the exact sequence τ

0 → H 0 (S, O(mK − x)) → H 0 (S, O(mK)) −→ C λ

→ H 1 (S, O(mK − x)) −→ H 1 (S, O(mK)) → 0,

(*)

where τ : ϕ → ϕ(x) for ϕ ∈ H 0 (S, O(mK)). We shall show that τ is surjective (hence (a)). Then (b) follows immediately. Since dim |eK|  1, we can find D ∈ |eK| satisfying x ∈ D. We consider two cases (I) x ∈ E , and (II) x ∈ E , separately. n (I) x ∈ E : By 7.9, we can take a composition series D = i=1 Ci of D satisfying (α)

KC1  1, Di−1 · Ci  1

(i  2).

If there exists a component " of D satisfying x ∈ " and K · "  1, then we can assume that C1 = " (7.9.1). We assume that x ∈ C , x ∈ C+1 + · · · + Cn . Then we have K · C  1 + δ1 . (This can be seen as follows. Since C < E , we have K · C  1. If  = 1, then x ∈ C1 and x ∈ C2 + · · · + Cn . Hence, from the remark on how to choose C1 in the composition series, we get K · Ci = 0 (2  i  n) and K · C1 = K · D = eK 2  2.) In 8.2, if we put h = 1, k = 0

2.8 Conclusions

63

and i = O(mK − Zi − x), then i+1 /i ∼ = O(Fi+1 − δi x)Ci

(1  i  n)

(**)

holds. In fact, (i) for i   − 1, we have i+1 ∼ = O(Fi+1 ), i ∼ = O(Fi+1 − Ci ) ∼ and i+1 /i = O(Fi+1 )Ci ; (ii) for i = , we have +1 ∼ = O(F+1 − x),  ∼ = O(F+1 − x − C ) and +1 / ∼ = O(F+1 − x)C ; (iii) for i   + 1, we have i+1 ∼ = O(Fi+1 − x), i ∼ = O(Fi+1 − x − Ci ) and i+1 /i ∼ = O(Fi+1 − x)Ci = O(Fi+1 )Ci . Hence (**) holds and we get hi = δi and ki = 0. Since m − e − 1  1 by hypothesis and KC  1 + δ1 , we get (m − e − 1)K · Ci + Di−1 · Ci  1 + δi >

1 1 (1 + δi )2 + 4 4

by (α). We infer from 8.4 that dim H 1 (S, O((m − e)K))  dim H 1 (S, n+1 ). Since m  m0 , however, we get dim H 1 (S, O((m − e)K)) = dim H 1 (S, O(mK)). Moreover, n+1 = O(mK − x) (8.2.1). In sum, we obtain dim H 1 (S, O(mK))  dim H 1 (S, O(mK − x)). We see from (∗) that λ is an isomorphism and, hence, τ is surjective. This completes the proof for the case (I). (II) x ∈ E : We assume that x ∈ Eλ . By 7.13, we can take a composition series D = ni=1 Ci of D satisfying Cn < Eλ

and

(β) K · Ci + Di−1 · Ci  1.

Since Cn < Eλ < E , we get K · Cn = 0. Therefore, K is trivial on Cn and O(mK)Cn ∼ = OCn . We have the exact sequence 0 → O(mK − Cn ) → O(mK) → OCn → 0. Since Cn ∼ = P1 and H 1 (Cn , OCn ) = 0, we get τ

0 → H 0 (S, O(mK − Cn )) → H 0 (S, O(mK)) −→ H 0 (Cn , OCn ) σ → H 1 (S, O(mK − Cn )) −→ H 1 (S, O(mK)) → 0.

(***)

In 8.2, we put h = k = 0 and i = O(mK − Zi ). Therefore, we have i+1 /i ∼ = O(Fi+1 )Ci and hi = ki = 0 (8.3). By hypothesis, m − e − 1  1. Since (m − e − 1)K · Ci + Di−1 · Ci  K · Ci + Di−1 · Ci  1 > 14 + 14 , we infer from 8.4 that dim H 1 (S, O(m − e)K)  dim H 1 (S, n ). On the other hand, since m  m0 , we have dim H 1 (S, O(m − e)K) = dim H 1 (S, O(mK)) and n = O(mK − Cn ). Therefore, dim H 1 (S, O(mK))  dim H 1 (S, O(mK − Cn )). We see that σ is an isomorphism in (***) and, hence, τ  is surjective. Since K is trivial on Cn , if we take ϕ ∈ H 0 (S, O(mK)), then ϕ takes the constant value ϕ(Cn ) on Cn . In fact, we have τ  : ϕ → ϕ(Cn ). Moreover,

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2 Pluri-Canonical Systems on Algebraic Surfaces of General Type

K is trivial on Eλ and ϕ as above is constant on Eλ . Since x ∈ Eλ , we have ϕ(Cn ) = ϕ(x). Therefore, returning to (∗∗), we see that τ = τ  and τ is surjective.   Theorem 8.7 H 1 (S, O(mK)) = 0 for m  2. Proof Taking e sufficiently large, we have Pe  2 by the Riemann–Roch theorem, and eK 2  2. If we put F = (m − 1)K and take an integer n satisfying n(m − 1)  e + 2 + m, then |nF | has no base points by 8.6. On the other hand, F 2 = (m − 1)2 K 2 > 0. Hence we infer from 6.3 that H 1 (S, O(K + F )) = H 1 (S, O(mK)) = 0.   Corollary 8.8 Pm (m  1) is a topological invariant of S. In fact, 

Pm = 12 m(m − 1)c12 + pg − q + 1 P1 = pg .

(m  2),  

Proof 4.2 and 8.7. Summing up, we obtain:

Theorem 8.9 If Pe  2, eK 2  2 and m  e + 2, then the m-canonical system |mK| has no base points. Therefore, mK is a holomorphic map. Proof By 8.7, we can take m0 = e + 2 in 8.5. Therefore, it is sufficient to assume m  e + 2 in order to assure the conditions on m in 8.6.   8.10 We next consider whether mK is one-to-one (onto the image). For E = E1 + · · ·+Eλ +· · ·+Er (7.6), since K is trivial on Eλ , any ϕ ∈ H 0 (S, O(mK)) is constant on Eλ . Hence mK (Eλ ) consists of one point, and mK fails to be one-to-one. Our answer to the problem will be given in 8.11. Definition 8.10.1 (a) If x = y and, if x and y are not contained in the same Eλ , then we say that x = y mod E (x and y are different modulo E ). (b) The holomorphic map  : S → Pn is said to be 1 : 1 mod E (one-to-one modulo E ) if  satisfies the condition −1 ((z)) =



z Eλ

(z ∈ E ), (z ∈ Eλ ).

(⇐⇒ If x = y mod E , then (x) = (y).) Remark 8.10.2 For x = y, if H 1 (S, O(mK − x − y)) = 0, then mK (x) = mK (y).

2.8 Conclusions

65

Proof We have the following exact sequence (5.0 and 5.0.1): 0 → O(mK − x − y) → O(mK) → Cx



Cy → 0.

Hence by hypothesis, we get the exact sequence λ

· · · → H 0 (S, O(mK)) −→ C2 → 0, where λ : ϕ → (ϕ(x), ϕ(y)) for ϕ ∈ H 0 (S, O(mK)). Since λ is surjective, we can find ϕ0 , ϕ1 ∈ H 0 (S, O(mK)) satisfying (1, 0) = (ϕ0 (x), ϕ0 (y)) and (0, 1) = (ϕ1 (x), ϕ1 (y)). We complete a basis {ϕ0 , ϕ1 , . . . } for H 0 (S, O(mK)) by extending ϕ0 , ϕ1 . Then mK : z → (ϕ0 (z), ϕ1 (z), ϕ2 (z), . . . ) ∈ Pn (modulo biholomorphic maps), and we have mK (x) = (1, 0, . . . ) and mK (y) = (0, 1, . . . ). Hence mK (x) = mK (y).   Theorem 8.11 Assume that Pe  3, eK 2  2 and m  e + 3. Then the following hold: (a) If x = y mod E , then H 1 (S, O(mK − x − y)) = 0. (b) mK is holomorphic and one-to-one modulo E . Proof If (a) is shown, then (b) follows from it, 8.9 and 8.10.2. We shall show (a). We separately consider three cases: (I) x, y ∈ E , (II) x, y ∈ E , and (III) x ∈ E but y ∈ E. (I) x, y ∈ E : Since dim H 0 (S, O(eK))  3, we can find D ∈ |eK| satisfying x, y ∈ D (which can be seen as follows: Letting {ϕ0 , ϕ1 , ϕ2 , . . . , ϕPe −1 } Pe −1 be a basis for H 0 (S, O(eK)), we can find a solution ϕ = ν=0 aν ϕν , (a0 , a1 , a2 , . . . ) = (0, 0, . . . , 0), of the simultaneous linear equation ϕ(x) = ϕ(y) = 0. We take such ϕ ∈ H 0 (S, O(eK)) and put D = (ϕ).) We infer n from 7.9 that there exists a composition series D = i=1 Ci of D which satisfies (α)

K · C1  1,

Di−1 · Ci  1

(i  2).

In particular, if we have a component " of D satisfying x ∈ " and k · "  1, then we put C1 = " and complete the composition series (7.9.1). In 8.2, we put h = k = 1 and i = O(mK − Zi − x − y). We assume that x ∈ C , x ∈ Z+1 , y ∈ Cj and y ∈ Zj +1 in the series D = ni=1 Ci . We may assume that   j . Then, i+1 /i ∼ = O(Fi+1 − δi x − δij y)Ci

(1  i  n).

(*)

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2 Pluri-Canonical Systems on Algebraic Surfaces of General Type

This can be seen as follows. If: ∼ O(Fi+1 ), i (Fi+1−C ) and i+1 /i ∼ (i) i   − 1, then i+1 = = i O(Fi+1 )Ci . (ii) i = , then +1 ∼ = O(F+1 − x),  ∼ = O(F+1 − C ) and +1 / ∼ = O(F+1 − x)C (since x ∈ C ). (iii)  + 1  i  j − 1, then i+1 ∼ = O(Fi+1 − x), i ∼ = O(Fi+1 − x − Ci ) and i+1 /i ∼ = O(Fi+1 − x)Ci = O(Fi+1 )Ci (since x ∈ Ci ). (iv) i = j , then j +1 ∼ = O(Fj +1 − y), j ∼ = O(Fj +1 − y − Cj ) and j +1 /j ∼ = O(Fj +1 − y)Cj . (v) i  j + 1, then i+1 ∼ = O(Fi+1 − x − y), i ∼ = O(Fi+1 − x − y − Ci ) and i+1 /i ∼ = O(Fi+1 − x − y)Ci = O(Fi+1 )Ci (since x, y ∈ Ci ). Hence we get (∗). Now, we put (m − e − 1)K · Ci + Di−1 · Ci  2K · Ci + Di−1 · Ci =: ri . Since C , Cj < E , we get KC  1 and KCj  1. In particular, paying special attention to that we have KC  1 + δ1 as in 8.6 by the remark on the first curve of the composition series, we have: if i =  = 1, then ri  4; if i =   2 or i = j > , then ri  3 by (α); if i = , j , then ri  1 in view of (α). Therefore, we get (m − e − 1)K · Ci + Di−1 · Ci  1 + δi + δij . Since we have hi = δi and ki = δij by (∗) (see 8.3), we get 1 1 1 3 3 (hi + 1)2 + (ki + 1)2 = + δi + δij < 1 + δi + δij . 4 4 2 4 4 Hence, we infer from 8.4 that dim H 1 (S, O((m − e)K))  dim H 1 (S, n+1 ). On the other hand, since m − e  3 by hypothesis, it follows from 8.7 that H 1 (S, O(m − e)K) = 0. Therefore, since we have n+1 = O(mK − x − y) (8.2.1), we obtain H 1 (mK − x − y) = 0. (II) x, y ∈ E : We assume x ∈ Eλ , y ∈ Eν . Since x = y mod E , we have λ = ν. We take D ∈ |eK| such that x, y ∈ D (this is possible by dim |eK|  2 (see (I)  for example)). We infer from 7.12 that there is a composition series D = ni=1 Ci of D satisfying Cn−1 < Eλ ,

Cn < Eν

and

(β) K · Ci + Di−1 · Ci  1

(1  i  n).

In 8.2, we put h = k = 0 and i = O(mK−Zi ). Then i+1 /i = O(Fi+1 )Ci and hi = ki = 0 for 1  i  n (8.3). On the other hand, since (m − e − 1)K · Ci + Di−1 · Ci  K · Ci + Di−1 · Ci  1 >

1 1 + , 4 4

it follows from 8.4 that dim H 1 (S, O((m − e)K)  dim H 1 (S, n−1 ). Since m − e  3, we have H 1 (S, O((m − e)K)) = 0 by 8.7 and, therefore, we get H 1 (S, n−1 ) = 0. On the other hand, since K · Cn−1 = 0 and K · Cn = 0, we see that K is trivial on Cn−1 , Cn and we have O(mK)Cn−1 ∼ = OCn−1 ,

2.8 Conclusions

67

O(mK)Cn ∼ = OCn . Hence, we obtain the following exact sequence (note that Cn−1 ∩ Cn = ∅): 0 → O(mK − Cn−1 − Cn ) → O(mK) → OCn−1



OCn → 0.

Since n−1 = O(mK − Cn−1 − Cn ), we get the exact sequence λ

0 → H 0 (S, O(mK − Cn−1 − Cn )) → H 0 (S, O(mK)) −→ C



C → 0,

where λ : ϕ → (ϕ(Cn−1 ), ϕ(Cn )), noting that K is trivial both on Eλ (> Cn−1 ) and Eν (> Cn ), and ϕ ∈ H 0 (S, O((mK))) takes the constant values ϕ(Cn−1 ), ϕ(Cn ) on Eλ , Eν , respectively. Now, since we have x ∈ Eλ and y ∈ Eν , we see that ϕ(Cn−1 ) = ϕ(x) and ϕ(Cn ) = ϕ(y). Hence λ : ϕ → (ϕ(x), ϕ(y)). On the other hand, we have the following exact sequences: 0 → O(mK − x − y) → O(mK) → Cx



Cy → 0.

 λ 0→ − x − y)) → −→ C C τ → H 1 (S, O(mK − x − y)) −→ H 1 (S, O(mK)) → 0, H 0 (S, O(mK

H 0 (S, O(mK))

where λ : ϕ → (ϕ(x), ϕ(y)) for ϕ ∈ H 0 (S, O(mK)). Therefore, λ = λ and it is surjective. Hence we know that τ is an isomorphism. Since H 1 (S, O(mK)) = 0 by 8.7, we obtain H 1 (S, O(mK − x − y)) = 0. (III) x ∈ E , y ∈ E : This case can be done  similarly to the above. Let x ∈ Eλ and choose a composition series D = ni=1 Ci of D satisfying Cn < Eλ and (β) (7.13). Put i = O(mK −Zi −y). If y ∈ C but y ∈ Z+1 , then K ·C  1 and i /i+1 ∼ = O(Fi+1 −δi y)Ci . We have (m−e−1)K ·Ci +Di−1 ·Ci  1+δi and, hence, H 1 (S, mK −Cn −y) = 0 by 8.4 and 8.7. Then 0 → H 0 (S, mK − 0 0 Cn − y) → H (S, mK) → H (Cn , OCn ) C → 0 is exact. From this, we infer readily that the restriction map H 0 (S, mK) → C2 (ϕ → (ϕ(x), ϕ(y))) is surjective, since x ∈ Eλ and En < Eλ .   8.12 Next, we study whether mK is biholomorphic on S \ E . Since we know that mK is 1:1 on S \ E under the assumption of 8.11, it is sufficient to see that the rank of the Jacobian matrix for mK at each point x ∈ S \E is always 2.

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2 Pluri-Canonical Systems on Algebraic Surfaces of General Type

Remark 8.12.1 If H 1 (S, O(mK − 2x)) = 0, then mK is biholomorphic in a neighborhood of x. Proof We have the following exact sequences (5.0 and 5.0.1 and the assumption): 0 → O(mK − 2x) → O(mK) → C3x → 0, λ

· · · → H 0 (S, O(mK)) −→ C3 → 0, where, letting (z1 , z2 ) be a system of local coordinates on S with the center x and writing ψ ∈ H 0 (S, O(mK)) as ψ = ψ0 + ψ1 z1 + ψ2 z2 + ψ3 z12 + ψ4 z1 z2 + ψ5 z22 + · · · in a neighborhood of x, we have λ : ψ → (ψ0 , ψ1 , ψ2 ). Since λ is surjective, there are ϕ0 , ϕ1 , ϕ2 ∈ H 0 (S, O(mK)) satisfying (ϕ00 , ϕ01 , ϕ02 ) = (1, 0, 0), (ϕ10 , ϕ11 , ϕ12 ) = (0, 1, 0) and (ϕ20 , ϕ21 , ϕ22 ) = (0, 0, 1). Completing the basis for H 0 (S, O(mK)) as {ϕ0 , ϕ1 , ϕ2 , . . . }, we may assume that mK (z) = (ϕ0 (z), ϕ1 (z), ϕ2 (z), . . . ) = (1 + · · · , z1 + · · · , z2 + · · · , ϕ3 (z), . . . ), where · · · in the first three components mean terms of order  2 in z1 , z2 . Therefore, the Jacobian matrix for mK is of rank 2 at x.   Theorem 8.13 If Pe  4, eK 2  2 and m  e + 3, then the following hold: (a) H 1 (S, O(mK − 2x)) = 0 for x ∈ E . (b) mK is holomorphic and biholomorphic modulo E (that is, biholomorphic on S \ E ). In particular, it is a birational holomorphic map. Proof If (a) is shown, then (b) follows from it, 8.9 and 8.12.1. We shall show (a). Since dim H 0 (S, O(eK)) = Pe  4, we can find non-zero ϕ ∈ H 0 (S, O(eK)) ∂ϕ ∂ϕ satisfying ϕ(x) = 0, ( ∂z )(x) = 0 and ( ∂z )(x) = 0 (e.g., the argument in 8.11), 1 2 where (z1 , z2 ) denotes a system of local coordinates on S with the center x. We put D = (ϕ) ∈ |eK|. Note that x is a multiple point of D. We infer from 7.9 that there is a composition series D = ni=1 Ci satisfying (α)

K · C1  1,

Di−1 · Ci  1

(2  i  n).

In particular, if there is a component " of D satisfying x ∈ " and K · "  1, then we can take C1 = " (7.9.1). In 8.2, put h = 2, k = 0 and i = O(mK − Zi − 2x). We assume that x ∈ C , x ∈ Z+1 in D = ni=1 Ci . Then K · C  1 + δ1 (by the same argument as in 8.6). We separately consider two cases: (I) x is a multiple point of C , and (II) x is a simple point of C .

2.8 Conclusions

69

If (I) is the case, then i+1 /i ∼ = O(Fi+1 − 2δi x)Ci .

(*)

This can be seen as follows. If (i) i   − 1, then x is a multiple point of Zi , Zi+1 and, hence, i+1 ∼ = O(Fi+1 ), i ∼ = O(Fi+1 − Ci ) and i+1 /i ∼ = O(Fi+1 )Ci . If ∼ (ii) i = , then +1 = O(F+1 − 2x),  ∼ = O(F+1 − 2x − C ) and +1 / ∼ = O(F+1 − 2x)C . If (iii) i   + 1, then similarly i+1 /i ∼ = O(Fi+1 − 2x)Ci = O(Fi+1 )Ci since x ∈ Ci . Hence we get (*). If we put (m − e − 1)K · Ci + Di−1 · Ci  2K · Ci + Di−1 · Ci =: ri , then: if i =  = 1, then ri  4 from K · C  1 + δ1 ; if i =  > 1, then ri  3 from K · C  1 and (α); if i = , then ri  1 from (α). Therefore, we get (m − e − 1)K · Ci + Di−1 · Ci  1 + 2δi . Since hi = δi and ki = 0 by (∗) (8.3), we get 14 (hi + 1)2 + 14 (ki + 1)2 = 12 + 2δi < 1 + 2δi . Then we infer from 8.4 that dim H 1 (S, O((m − e)K))  dim H 1 (S, n+1 ). On the other hand, we have H 1 (S, O((m − e)K)) = 0 by 8.7 and n+1 = O(mK − 2x). It follows that H 1 (S, O(mK − 2x)) = 0 and we finish (I). If (II) is the case, we can find another index j <  with x ∈ Cj and x ∈ Cj +1 + · · · + C−1 . In fact, since x is a simple point of C , we can find such a Cj by the fact that x is a multiple point of D. We have i+1 /i ∼ = O(Fi+1 − (δij + 2δi )x)Ci

(1  i  n).

(**)

This can be seen as follows. If (i) i  j − 1, then x is a multiple point of Zi+1 , Zi and, hence, we have i+1 = O(mK − Zi+1 ), i = O(mK − Zi+1 − Ci ) and i+1 /i ∼ = O(Fi+1 )Ci . If (ii) i = j , then x is a simple point of Zj +1 and j +1 ∼ = O(Fj +1 −x). To see this, it suffices to show that (j +1 )x ∼ = O(Fj +1 −x)x . For this purpose, we take a system of local coordinates (z1 , z2 ) on S with the center x so that C is defined by z2 = 0 in a neighborhood of x.

Then, the necessary and sufficient condition for ϕ ∈ O(mK − Zj +1 )x to be in O(mK − Zj +1 − 2x)x is that ϕ can be expressed as ϕ(z1 , z2 ) = z2 ψ(z1 , z2 ) with ψ ∈ O(Fj +1 )x satisfying ψ(0, 0) = 0, that is, ψ ∈ O(Fj +1 − x)x . Hence j +1 = O(mK − Zj +1 − 2x) ∼ = O(Fj +1 − x). On the other hand, since x is a multiple point of Zj , we have j ∼ = O(Fj +1 − Cj ) = O(Fj +1 − x − Cj ). Therefore, j +1 /j ∼ = O(Fj +1 − x). If (iii) j < i < , then it is not hard to see

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2 Pluri-Canonical Systems on Algebraic Surfaces of General Type

i+1 /i ∼ = O(Fi+1 )Ci . If (iv) i = , then +1 / ∼ = O(F+1 − 2x)C . If (v) i > , then i+1 /i ∼ = O(Fi+1 )Ci . Hence we get (∗∗). If we put (m − e − 1)K · Ci + Di−1 · Ci  2K · Ci + Di−1 · Ci =: ri , then, using K · Cj  1, K · C  1 (by Cj , C < E ) and (α), we can show that ri  1 when i = j, ; ri  2 when i = j ; ri  3 when i = . Therefore, (m − e − 1)K · Ci + Di−1 · Ci  1 + δij + 2δi . On the other hand, we infer from (∗∗) that hi = δij + 2δi , ki = 0 (8.3). It follows that 14 (hi + 1)2 + 14 (ki + 1)2 = 1 3 2 + 4 δij + 2δi < 1 + δij + 2δi . Hence, as before, we use 8.4 and 8.7 to see 1   H (S, O(mK − 2x)) = 0. 8.14 Finally, we consider what is the smallest possible value of e satisfying 8.9, 8.11, and 8.13 etc. Our answer will be found in 8.16. For this purpose, we need the following lemma. Lemma 8.15 If K 2 = 1, then pg  2 and q  1. Proof (a) We shall show pg  2. Assume that pg  2. Since dim |K| = pg − 1  1, we take a generic member D  ∈ |K| and write D = C1 + · + Cn (n  1). n Since 1 = K 2 = K · D = i=1 KCi , we can assume that K · C1 = 1, K · Ci = 0 (i  2). Therefore, for i  2, we have Ci < E and π(Ci ) = 0, Ci2 = −2 (7.3).  We claim that n = 1 and D = C1 . This can be seen as follows. If we put X = ni=2 Ci ( 0), then D = C1 + X and X is the fixed part of |K| and C12  0. We obtain C12  1 from 2π(C1 ) − 2 = C12 + K · C1 . Then 1 = K · C1 = D · C1 = C12 + C1 X, C1 X  0 and it follows that C12 = 1, C1 · X = 0, which shows X = 0 in view of 7.8. Hence D = C1 . In this way, we know that a generic member of |K| is an irreducible curve C with C 2 = 1. Then C has no singular points. (In fact, by Bertini’s theorem, C does not have a singular point other than base points of |K|. On the other hand, choose any base point p of |K|. If C  (= C) is another generic member of |K|, then p ∈ C  . However, C · C  = C 2 = 1 implies that p is a simple point of C.) We have π(C) = 12 (K · C + C 2 ) + 1 = 2. Since K = [C], we have the following exact sequences: 0 → O → O(K) → O(K)C → 0, 0 → H 0 (S, O) → H 0 (S, O(K)) → H 0 (C, O(KC )) → · · · . ∼ C and dim H 0 (C, O(KC ))  1, we get pg = Since H 0 (S, O) = 0 dim H (S, O(K))  2. Hence pg = 2. (b) We shall show q  1: We have q = K 2 + pg − P2 + 1 (8.8). Since K 2 = 1 and P2  pg , we get q  2. We assume that q = 2 and derive a contradiction. Let ϕ1 , ϕ2 be a basis for the space of holomorphic 1-forms on S. Then dϕν = 0 (ν = 1, 2). Let{γ1 , . . . , γ4 } be a Betti basis for 1-cycles on S (b1 (S) = 2q = 4). Put ωj ν = γj ϕν (1  j  4, 1  ν  2), ωj = (ωj 1 , ωj 2 ) ∈ C2 ,

2.8 Conclusions

# = {

71



mj ωj | mj ∈ Z} and A =  C2 /#, and consider the Albanese map z z   : S → A. We have  : z → (z) = p0 ϕ1 , p0 ϕ2 (mod #). We consider two cases: (1) ϕ1 ∧ ϕ2 = 0, (2) ϕ1 ∧ ϕ2 = 0.

(1) If ϕ1 ∧ ϕ2 = 0, then (S) = A and D = (ϕ1 ∧ ϕ 2 ) ∈ |K|. As we did in (a), using K · D = K 2 = 1, we can write D = C + ni=2 Ei , where K · C = 1, K · Ei = 0, π(Ei ) = 0 and Ei2 = −2. Then (Ei ) is a point. We claim that the restrictions ϕ1C , ϕ2C of ϕ1 , ϕ2 to C are linearly independent over C. [This  z can be seen  z as follows. If (a1 ϕ1 + a2 ϕ2 )D = 0 for (a1 , a2 ) = (0, 0), then a1 ϕ1 + a2 ϕ2 = constant for z ∈ C and we see that (C) ⊂ A is an elliptic curve. If we take a curve satisfying  ∩ (C) = ∅,  ∩ (Ei ) = ∅, then  −1 () does not meet C+ Ei and, hence, K· −1 () = 0. (Since (C) is the image of a line in C2 , if we take a line parallel to it, then we can get such  on A.) This implies, however, that every component "j of ψ −1 () satisfies K · " j = 0 and π("j ) = 0. Hence ("j ) is a point, which contradicts that  = j ("j ). We have shown that ϕ1C and ϕ2C are linearly independent.] On the other hand, C is non-singular and π(C) = 2, D = C. [Because: Let  be the non-singular model of C. By what we have just seen, there are two C  Thus π(C)  linearly independent holomorphic 1-forms ϕ˜1 and ϕ˜2 on C.   2. On  − 2  2π(C) − 2 = C 2 + K · C = (D − Ei ) · C + the other hand, 2  2π(C)  = 2 and, therefore, C = C.  K · C  2K · C= 2. This shows π(C) = π(C)  Moreover, C · ( Ei ) = 0. Then, by 7.8, we get Ei = 0 and conclude D = C.] Therefore, we get (ϕ1 ∧ ϕ2 ) = C. We have ϕ1 (z) = 0 when z ∈ C. We remark that ϕ1C has two zeros on C, since C is a non-singular curve with π(C) = 2. We can say the same thing for ϕ2C . We may assume that they have two simple zeros and put (ϕ1C ) = x + y, (ϕ2C ) = u + v.

We regard the holomorphic 1-form ϕ1 as a covariant vector field. Recall that c2 is the Euler number of S. We use the formula c2 = Ix (ϕ1 ) + Iy (ϕ1 ), where Ix (ϕ1 ), Iy (ϕ1 ) denotes the algebraic index of the covariant vector field ϕ1 at its singular point x, y. We can choose a system of local coordinates (w, z) with the center x on S so that dz = ϕ2 , ϕ1 ∧ ϕ2 = wdw ∧ dz (see figure above). Since then we have (ϕ1 − wdw) ∧ dz = 0, we can write ϕ1 − wdw = f dz. Since ∂f dϕ1 = 0, however, we get df ∧ dz = 0, that is, ∂w = 0. This implies that f is a holomorphic function only in z and we can write f = f (z). Consequently, we have ϕ1C = f (z)dz = zg(z)dz (g(z) = 0) and, hence, ϕ1 = wdw + gzdz (g = 0). Then we have Ix (ϕ1 ) = 1. Similarly, Iy (ϕ1 ) = 1. Therefore, from the above formula, we get c2 = 2. Recall that Noether’s formula 4.4 implies that c12 + c2 = 12(pg − q + 1) ≡ 0 (12), which is impossible when c12 = K 2 = 1, c2 = 2.

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2 Pluri-Canonical Systems on Algebraic Surfaces of General Type

(2) ϕ1 ∧ ϕ2 = 0: The Jacobian matrix for  is of rank 1 everywhere, (S) =  ⊂ A is a non-singular curve of genus 2, and "u =  −1 (u) is an irreducible non-singular curve for a general u ∈ . We have pg  1. (This is because we have 12pg = 7 + b2 from Noether’s formula 4.4, c12 = K 2 = 1, c2 = 2 − 4q + b2 and q = 2.) Then, from dim |K| = pg − 1  0, we can take n 2 = D 2 = 1. If we write D = |K|  D  0. We have D > 0 by K i=1 Ci , n 2 then 1 = K = K · D = i=1 K · Ci , K · Ci  0. Hence we may assume  that K · C1 = 1, K · Ci = 0 (i  2). We rewrite it as D = C + ni=2 Ei as before (where K · C = 1, K · Ei = 0, π(Ei ) = 0, Ei2 = −2 (7.3)). Note that (Ei ) is one point (∈ ). Therefore, "u ∩ Ei = ∅ (because, if "u ∩ Ei = ∅, then "u = Ei . This contradicts that "2u = 0, Ei2 = −2.) We have K · "u > 0 by "2u = 0 and 7.3. Since 2π("u ) − 2 = K · "u , we get K · "u  2. Hence C · "u = D · "u = K · "u  2.

 denote the non-singular model of C. Then C  is a (branched) covering of Let C . If we denote the covering degree by m and the degree of the ramification divisor  = mχ() − b (where χ(∗) denotes the Euler characteristic), that by b, then χ(C)  = m(2 − 2π()) − b. Since m = C · "u  2, we have 2π(C)  −2 = is, 2 − 2π(C)   3. On the other hand, m(2π() − 2) + b  2m  4 by π() = 2. Hence π(C) we have  Ei ) 2π(C) − 2 = K · C + C 2 = K · C + C · (D − = 2K · C −



i

C · Ei

 2K · C = 2

(by D ≈ K) (by CEi  0),

  π(C). which contradicts that π(C) In both cases of (1) and (2), we are led to a contradiction. Therefore, q  1.

 

Theorem 8.16 P2  2 and P3  4. Proof We infer from 8.8 that P2 = K 2 + pg − q + 1 (K 2 = c12 ) and P3 = 3K 2 + pg − q + 1 = 2K 2 + P2 . Since we have K 2  1 by Assumption 7.0, if we can show P2  2, then P3  4 follows. We shall show that P2  2. We consider the three cases pg  2, pg = 1, pg = 0.

2.8 Conclusions

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(a) pg  2: Since P2  P1 = pg , we get P2  2. (b) pg = 1: By Noether’s formula 4.4, we have q  2 (by pa = pg − q, K 2 = c12 , c2 = b2 ). We have K 2 − q  0 by 8.15, and P2 = pg + 1 + K 2 − q  2. (c) pg = 0: We have q = 0 (Enriques). Hence P2 = K 2 + 1  2.   Theorem 8.17 (a) If m  4, then mK is a holomorphic map. (b) If m  6, then mK is biholomorphic modulo E (i.e., biholomorphic on S \ E ) and, therefore, it is a birational holomorphic map. Proof In view of 8.16, (a) follows from 8.9 and, (b) follows from 8.13.

 

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