On higher syzygies of ruled surfaces III

Journal of Pure and Applied Algebra 219 (2015) 4653–4666

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Journal of Pure and Applied Algebra www.elsevier.com/locate/jpaa

On higher syzygies of ruled surfaces III Youngook Choi a , Euisung Park b,∗ a

Department of Mathematics Education, Yeungnam University, 280 Daehak-Ro Gyeongsan, Gyeongbuk 712-749, Republic of Korea b Department of Mathematics, Korea University, Seoul 136-701, Republic of Korea

a r t i c l e

i n f o

Article history: Received 28 May 2014 Available online 13 March 2015 Communicated by R. Vakil MSC: 13D02; 14J26; 14N05

a b s t r a c t In this paper, we study the minimal free resolution of homogeneous coordinate rings of a ruled surface S over a curve of genus g with the numerical invariant e < 0 and a minimal section C0 . Let L ∈ PicX be a line bundle in the numerical class of aC0 +bf such that a ≥ 1 and 2b − ae = 4g − 1 + k for some k ≥ max(2, −e). We prove that the Green–Lazarsfeld index index(S, L) of (S, L), i.e. the maximum p such that L satisfies condition N2,p , satisfies the inequalities    2g − 3 + ae − k k k ae + 3 − g ≤ index(S, L) ≤ − + max 0, . 2 2 2 4 Also if S has an effective divisor D ≡ 2C0 +  ef, then we obtain another upper bound of index(S, L), i.e., index(S, L) ≤ k + max 0,  2g−4−k  . This gives a better bound 2 in case b is small compared to a. Finally, for each e ∈ {−g, . . . , −1} we construct a ruled surface S with the numerical invariant e and a minimal section C0 which has an effective divisor D ≡ 2C0 + ef. © 2015 Elsevier B.V. All rights reserved.

1. Introduction In this article we study the minimal free resolution of homogeneous coordinate rings of ruled surfaces over a smooth projective irrational curve. Let L be a very ample line bundle on a projective variety X, which defines the linearly normal embedding X ⊂ PH 0 (X, L)∗ . Due to D. Eisenbud et al. [4], we will say that (X, L) satisfies condition N2,p for some p ≥ 1 if the homogeneous ideal of X in PH 0 (X, L)∗ is generated by quadrics and the syzygies among them are generated * Corresponding author. E-mail addresses: [email protected] (Y. Choi), [email protected] (E. Park). http://dx.doi.org/10.1016/j.jpaa.2015.02.037 0022-4049/© 2015 Elsevier B.V. All rights reserved.

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by linear syzygies until the (p − 1)-th step. The largest p such that (X, L) satisfies condition N2,p is called the Green–Lazarsfeld index of (X, L) and denoted by index(X, L) (cf. [1]). If X fails to satisfy property N2,1 , then we define index(X, L) = 0. Thus index(X, L) is an important invariant in order to describe the minimal free resolution of the homogeneous ideal of X embedded in PH 0 (X, L)∗ . A guiding principle concerning index(X, L) is that (X, L) satisfies condition N2,p for more and more large p as L is more and more “positive”. For example, if X is a smooth curve of genus g and if deg(L) ≥ 2g + 1 + k for some k ≥ 1, then (X, L) satisfies condition N2,k (cf. [7]). Our purpose in the present paper is to investigate index(S, L) in the case where S is a ruled surface over a smooth irrational curve and L is a very ample line bundle on S. To give precise statements, we use the following denotations and terminology, following R. Hartshorne’s book [8], V §2: • C: a smooth projective curve of genus g > 0. • E: vector bundle of rank 2 over C which is normalized, i.e., H 0 (C, E) = 0 while H 0 (C, E ⊗ OC (D)) = 0 for every divisor D of negative degree. 2 E and e = −deg(e). • e= • S = PC (E): the associated ruled surface with the projection morphism π : S → C. • C0 : a minimal section of S such that OS (C0 ) = OPC (E) (1). • For a line bundle L on S, we will denote by L|C0 the restriction of L to C0 . • For b ∈ Pic(C), we denote by bf the pullback of b by the projection morphism π. Thus any line bundle on S can be written as OS (aC0 + bf) with a ∈ Z and b ∈ Pic(C). Also any element of Num(S) can be written as aC0 + bf with a, b ∈ Z. Now, let L = OS (aC0 + bf) be a line bundle on S with a ≥ 1 and deg(b) = b. Then the positivity of L is measured by the so-called minimal slope μ− (π∗ L) of the vector bundle π∗ L on C, which is computed as −

μ (π∗ L) =



b − ae b − ae 2

if e ≥ 0, and if e < 0.

Namely, L is an ample line bundle if and only if μ− (π∗ L) is positive (cf. [10, Theorem 5.1]). Keeping these facts in mind, the following results illustrate how index(S, L) is related to the positivity of L in the case where the invariant e is non-negative and negative, respectively. Theorem 1.1. (See Theorems 1.2, 1.3, 1.5 and Corollary 1.4 in [12].) Let S be as above with e ≥ 0 and L = OS (aC0 + bf) a line bundle on S with a ≥ 1 and deg(b) = b. If b − ae = 2g + 1 + k for some k ≥ 1, then    g−k−3 . min(k, k + e + 2 − g) ≤ index(S, L) ≤ index(C0 , L|C0 ) ≤ k + max 0, 2 Moreover, the equality index(S, L) = k holds in the cases where (α) g = 1 or g = 2; (β) e ≥ g − 2 and C is a hyperelliptic; (γ) e ≥ g − 2 and k ≥ g − 3. A proof of Theorem 1.1 is provided at the beginning of Section 3. This result shows that as k increases the index of (S, L) approaches to k. Note that the integer b −ae is exactly the degree of the line bundle L|C0 . Also it is a well-known fact that the index of (C0 , L|C0 ) is always equal to k if k ≥ g − 3. Thus Theorem 1.1 enables us to conclude that index(S, L) is governed by the restriction of L to a minimal section of S.

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The main theorem of this paper is as follows. Theorem 1.2. Let S be as above with e < 0 and L = OS (aC0 + bf) a line bundle on S with a ≥ 1 and deg(b) = b. Suppose that 2b − ae = 4g − 1 + k for some integer k ≥ 2. Then (a) If k ≥ max(2, −e), then

   k ae + 3 2g − 3 + ae − k k − g ≤ index(S, L) ≤ − + max 0, . 2 2 2 4

(b) If there is an effective divisor D ≡ 2C0 +ef, then D is a smooth curve of genus 2g −1 and the restriction map f : D → C of π : S → C to D is an étale double covering. Also for a ≥ 2, it holds that    2g − 4 − k index(S, L) ≤ index(D, L|D ) ≤ k + max 0, 2 where L|D is the restriction of L to D. The proof of this result is provided in Section 3. The low bound of index(S, L) in Theorem 1.2 is already known (cf. [11, Theorem 1.7]). The upper bound of Theorem 1.2 is well understood when S is an elliptic ruled surface. More precisely, an elliptic ruled surface S with e = −1 always has a curve D numerically equivalent to 2C0 − f (cf. [9] and [6]) and it is shown in [12, Theorem 1.1] that the inequality index(S, L) ≤ min{index(C0 , L|C0 ), index(D, L|D )} holds. On the other hand, there are few results on the upper bound of index(S, L) when g ≥ 2. For a fixed positive integer a, if b is large (and hence k is large), then the value of index(S, L) approaches to k2 asymptotically by Theorem 1.2(a). On the other hand, if there is an effective divisor D ≡ 2C0 + ef and if b is small (and hence k is small), then D is more essential to bound the index of (S, L). That is, as in the case of elliptic ruled surface with e = −1, the index of (S, L) is governed by both the minimal section and the effective divisor D ≡ 2C0 − f (obviously when it does exist). This seems to be the main difference between the cases where e is non-negative and negative. The existence of an effective divisor D ≡ 2C0 + ef on a ruled surface S with e < 0 is unknown except the case of elliptic ruled surfaces. Along this line, for any smooth curve C of genus g ≥ 2 and any given integer −g ≤ e ≤ 0, we construct a ruled surface S over C with an invariant e such that S has an effective divisor D which is numerically equivalent to 2C0 + ef. For details, see Theorem 4.3 in Section 4. Finally, in Section 5, we investigate in detail the problem of which kind of invariants determines the index of (S, L) in the case where C is of genus 2 and S has an effective divisor D ≡ 2C0 + ef. Most of all, we obtain a necessary condition for index(S, L) ≥  for all . Namely, if (S, L) satisfies condition N2, then both (C0 , L|C0 ) and (D, L|D ) satisfy condition N2, , or equivalently, b − ae ≥ 5 + 

and 2b − ae ≥ 7 + .

See Fig. 1 in Section 3 and Theorem 5.2 in Section 5. 2. Preliminaries 2.1. Linear section Let X be a projective variety and L a very ample line bundle on X which defines a linearly normal embedding X ⊂ PH 0 (X, L)∗ = Pr

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of X. For an effective divisor D of X, consider its linear span D in Pr . Obviously we have D ⊆ X ∩ D. In this subsection we will give a proof of the probably well-known fact that if X ⊂ Pr is projectively normal then X ∩ D is the union of D and the base locus Z := Bs|L(−D)| of L(−D). Proposition 2.1. Keep the previous notation and assume that X ⊂ Pr is projectively normal. Then X ∩ D = D ∪ Z as sets. Moreover, if Z = ∅ then ID = IX + ID Proof. We may assume that H 0 (X, L(−D)) = 0. Consider the evaluation map σ : H 0 (X, L(−D)) ⊗ OX → L(−D). Let J be the image of σ. It is clear that the support of L(−D)/J is precisely the base locus Z := Bs|L(−D)| of L(−D). Moreover, the ideal sheaf I := J ⊗ L−1 ⊗ OX (D) ⊂ OX defines a scheme structure of Z. From now on, we regard Z as the closed subscheme of X defined by I. Let ID and ID denote respectively the ideal sheaf of D and the ideal sheaf of D in Pr . Then the intersection X ∩ D is scheme-theoretically defined by IX + ID . Therefore we get the desired assertion by showing that the quotient sheaf G := ID /(IX + ID ) is exactly supported on Z. To this aim, consider the following commutative diagram: 0

0

↓

↓

0 →IX →IX + ID →  0 →IX →

↓ ID

→0

↓ →OX (−D)→ 0

↓ G

F

↓ =

G

↓

↓

0

0

where F := (IX + ID )/IX . From the evaluation map σ : H 0 (X, L(−D)) ⊗ OX → L(−D), we consider the R-module E=



Ej

where Ej := H 0 (X, Lj (−D))

Ej

where Ej := H 0 (X, J ⊗ Lj−1 ).

j∈Z

and its submodule E  :=

j∈Z

Note that E1 = E1 and hence the submodule F := E1  of E is indeed contained in E  . From the first row, we can see that the graded R-module

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H∗0 (F) :=



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H 0 (X, F ⊗ Lj )

j∈Z

associated to F coincides with F . Now, in order to compare F with E  , we consider the exact sequence 0 → Ker → H 0 (X, L(−D)) ⊗ OX → J → 0 where Ker := Ker(σ). To this aim, we investigate the multiplication map  μj : H 0 (X, L(−D)) ⊗ H 0 (X, Lj ) → H 0 (X, J ⊗ Lj ) = Ej+1 .

Since X ⊂ Pr is a projectively normal variety, we have Im(μj ) = Fj+1 . Also for all j  0, we have  H 1 (X, Ker ⊗ Lj ) = 0 and hence Ej+1 = Fj+1 . That is, E  /F is a finite dimensional vector space and so E  and F have the same sheafification. Thus we have F = J ⊗ L−1 = I ⊗ OX (−D) where I is the ideal sheaf of Z. This implies that G = OX (−D)/F = OX (−D)/I ⊗ OX (−D) = OX (−D) ⊗ OZ and hence G is exactly supported on Z. 2 2.2. Elementary transformation Let S be a smooth surface and let  : Sx → S be the blowing up of S at x ∈ S. We will denote the exceptional divisor of Sx by E := −1 (x). Given a curve D in S its strict transform is the curve ˜ := −1 (D − x). Let π : S → C be a ruled surface over a smooth curve C of genus g and let x ∈ S with D π(x) = P . We denote the elementary transformation of S at x by S  = elm x (S). It is built blowing up S f by σ : Sx → S  . We denote the birational at x by  : Sx → S and contracting the exceptional generator P −1  ˜ If S  is an map σ ◦  by ν : S → S. Given a curve D on S we define the strict transform as D = σ∗ (D).  elementary transform of S at x, then S is the elementary transform of S at y, where y is the image by σ f and E on Sx . We call D by n-secant curve if D  f = n. of the intersection of the exceptional divisors P Proposition 2.2. Let π : S → C be a ruled surface and let S  = elm x (S) be its elementary transform at a point x ∈ S with π(x) = P . Then: (1) If b is a divisor on C, then ν ∗ (bf) = bf. (2) If D( P f) is a curve on S, then ν ∗ (D) = D + μx (D)P f, where μx (D) is a multiplicity of D at x. Proof. See Proposition 4.4 in [5].

2

Proposition 2.3. Let π : S → C be a ruled surface and let S  = elm x (S) be its elementary transform at a point x ∈ S with π(x) = P . If D1 is an n-secant curve on S and D2 is an m-secant curve, then D1  D2 = D1  D2 + nm − nμx (D2 ) − mμx (D1 ). Therefore, if D1 and D2 are uni-secant curves, then (1) If x ∈ D1 ∩ D2 , then D1  D2 = D1  D2 − 1. (2) If x ∈ / D1 ∪ D2 , then D1  D2 = D1  D2 + 1. (3) If x ∈ D1 but x ∈ / D2 , then D1  D2 = D1  D2 . Proof. See Proposition 4.4 in [5].

2

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Lemma 2.4. Let π : S → C be a ruled surface and let π  : S  → C be the elementary transformation of S at a point x ∈ P f. Let D be an irreducible uni-secant curve on S and let D be a strict transform of X. If OD (D)  OC (b) where b is a divisor on C, then OD (D )  OC (b + (1 − 2μx (D))P ). Proof. See Lemma 4.6 in [5]. 2 Theorem 2.5. Let π : S(:= P(E)) → C be a ruled surface. Let x ∈ S be a point with π(x) = P such that there is no minimal self-intersection curve passing through x. Let S  be the elementary transform of S at x. Then, S  is a ruled surface corresponding to a normalized sheaf E  with ∧2 E   OC (e ) satisfying e = e + P and hence e = e − 1. Furthermore, minimal self-intersection curves on S  are strict transforms of minimal self-intersection curves on S and uni-secant irreducible curves which pass through x and are numerically equivalent to C0 + f. Proof. For an irreducible uni-secant curve D on X, by Proposition 2.3, if x ∈ D, then D 2 = D2 − 1 and if x ∈ / D, then D 2 = D2 + 1 and for a minimal self-intersection curve E, we have E  2 = E 2 + 1 because x∈ / E. If D ∈ |C0 + bf| is an irreducible curve such that x ∈ / D, then D 2 = D2 + 1 = C02 + 2 deg b + 1 ≥ E  2

(2.1)

since deg b ≥ 0. If D ∈ |C0 + bf| contains a point x (and hence deg b ≥ 1), then D 2 = D2 − 1 = C02 + 2 deg b − 1 ≥ C02 + 1 = E  2 .

(2.2)

Therefore, E  is a minimal curve on S  and by Lemma 2.4, we have OC0 (C0 )  OC (e + P ) and e = e + P . In Eq. (2.1), the equality holds if and only if deg b = 0 and in Eq. (2.2), the equality holds if and only if deg b = 1. Therefore, minimal self-intersection curves on S  are strict transforms of minimal self-intersection curves on S and uni-secant irreducible curves which pass through x and are numerically equivalent to C0 + f. 2 3. Syzygies of a ruled surface The aim of this section is to provide a proof of Theorem 1.1 and Theorem 1.2. We begin with the following known result. Theorem 3.1. (See Theorem 8.8 and Theorem 8.9 in [3].) (a) If a variety X ⊂ Pr intersects a linear subspace Λ = P in a finite scheme of length at least  + 2, then index(X) ≤  − 1. (b) Let X be a smooth curve of genus ρ and L a line bundle of degree 2ρ + 1 + k, with k ≥ 0. Then   X ⊂ PH 0 (X, L)∗ has a (q + 3)-secant (q + 1)-plane for q = k + max 0,  ρ−k−3  . Thus 2  ρ−k−3   k ≤ index(X, L) ≤ k + max 0,  2 In particular, if k ≥ ρ − 3, then index(X, L) = k. We are ready to give a proof of Theorem 1.1.

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Proof of Theorem 1.1. The lower bound of index(S, L) is proved in [12, Theorem 1.2]. For the upper bound of index(S, L), assume that (S, L) satisfies condition N2, for some  ≤ k + g2 + 1. Then (C0 , L|C0 ) satisfies condition N2, by [12, Theorem 1.3]. On the other hand, Theorem 3.1(b) provides an upper bound of the value of index(C0 , L|C0 ). In consequence, we have    g−k−3 .  ≤ index(C0 , L|C0 ) ≤ k + max 0, 2    Now, observe that k + max 0, g−k−3 < k + g2 + 1. It follows that index(S, L) is at most equal to 2 index(C0 , L|C0 ). For the cases (α), (β) and (γ), the equality index(S, L) = k holds by [12, Corollary 1.4] and [12, Theorem 1.5]. 2 Now, we turn to the case where the invariant e of the ruled surface S is negative. Lemma 3.2. Keep the notations in Section 1. Let S = PC (E) be a ruled surface over a smooth curve C of genus g > 0 such that the invariant e is negative. Let L = OS (aC0 + bf) be a line bundle on S such that deg(b) = b. Then (a) (b) (c) (d) (e)

L is ample if and only if a ≥ 1 and 2b − ae ≥ 1. If a ≥ 0 and 2b − ae ≥ 4g − 3, then H 1 (S, L) = 0. If a ≥ 0 and 2b − ae ≥ 4g − 1, then L is globally generated. If a ≥ 1 and 2b − ae ≥ 4g + 1, then L is very ample. Assume that a ≥ 1 and 2b − ae ≥ 4g + 1. Let ML be the kernel of the evaluation homomorphism H 0 (S, L) ⊗ OS → L and let N ∈ Pic(S) be a line bundle numerically equivalent to sC0 + tf . Then m H 1 (S, ML ⊗ N ) = 0 if s ≥ min{m, a}

and

2t − se >

2m(2b − ae) + 4g − 4. 2b − ae − 2g

Proof. We refer the reader to [2, Lemma 5.4] for (a), to [2, Lemma 1.12] and [12, Lemma 2.4] for (b)–(d) and to [11, Lemma 3.1] for (e). 2 Theorem 3.3. (See Theorem 5.1A in [2] and Theorem 1.7 in [11].) Let S and L be as in Lemma 3.2. Then (a) (S, L) is projectively normal if 2b − ae ≥ 4g + 1. (b) (S, L) satisfies property N2, if 2b − ae ≥ 6g − 2 + 2. In particular, if a = 1 and b ≥ (3g − 1 + 2e ) +  then (S, L) satisfies property N2, . Now, we are ready to give a proof of Theorem 1.2. Proof of Theorem 1.2. (a): Theorem 3.3(b) shows that (S, L) satisfies condition N2, for  ≤ proves the inequality at the first place. For the inequality at the second place, observe that

k+1 2

− g, which

2b − (a − 1)e = 4g − 1 + k + e ≥ 4g − 1. Thus it follows by Lemma 3.2(b) and (c) that H 1 (S, L(−C0 )) = 0 and L(−C0 ) is base point free. Let L|C0 be the restriction of L to C0 . The natural map H 0 (S, L) → H 0 (C0 , L|C0 ) is surjective since H 1 (S, L(−C0 )) = 0. This implies that C0 in PH 0 (S, L)∗ is linearly normal. Now, let C0  be the span of C0 in PH 0 (S, L)∗ . Since

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deg(L|C0 ) = L  C0 = b − ae = 2g + 1 +

k ae + 3 − , 2 2

the curve C0 ⊂ C0  admits a (q + 3)-secant (q + 1)-plane, say Λ, where    k ae + 3 2g − 3 + ae − k + max 0, q= − 2 2 4 (cf. Theorem 3.1(b)). Also since L(−C0 ) is base point free, the intersection S ∩ C0  is exactly equal to C0 by Proposition 2.1. This shows that Λ is a (q + 3)-secant (q + 1)-plane to S, too. Therefore, index(S, L) is at most q by Theorem 3.1(a), which completes the proof. (b): Assume that an effective divisor D ≡ 2C0 + ef is reducible. Say D = Y1 ∪ Y2 , so Y1 ≡ C0 + e1 f and Y2 ≡ C0 + e2 f with e1 + e2 = e. Since e < 0, e1 or e2 is negative and hence Y1 or Y2 is not effective which gives a contradiction to the assumption that D is reducible. Therefore, D is an irreducible divisor. The map f : D → C is apparently a double covering of C. Hence, by Hurwitz formula, the genus of the normalization of D is not less than 2g −1. Let KC and KS be the canonical divisors of C and S, respectively. Also let pa (D) be the arithmetic genus of D. Since KS ≡ −2C0 + (KC + e)f, we get pa (D) = 2g − 1 from the adjunction formula 2pa (D) − 2 = D · (D + KS ). This means that the genus of the normalization of Y is not more than 2g − 1. Therefore the geometric genus of D is same with the arithmetic genus of D and the restriction map f : D → C is an unramified double covering of C. This completes the proof that D is a smooth curve of genus 2g − 1 and the map f : D → C is an etale covering. The line bundle L(−D) is numerically equivalent to (a − 2)C0 + (b + 2)f. Since a − 2 ≥ 0 and 2(b − e) − (a − 2)e = 2b − ae ≥ 4g − 1, it follows by Lemma 3.2(b) and (c) that H 1 (S, L(−D)) = 0 and L(−D) is base point free. Thus D in PH 0 (S, L)∗ is linearly normal. Also, letting D be the span of D in PH 0 (S, L)∗ , Proposition 2.1 implies that the intersection S ∩ D is exactly equal to D. The degree of L|D is given as deg(L|D ) = L.D = 2b − ae = 2(2g − 1) + 1 + k and hence Theorem 3.1(b) shows that index(D, L|D ) ≤ q  where    2g − 4 − k . q = k + max 0, 2 

This shows the inequality at the second place. Moreover, D ⊂ D admits a (q  + 3)-secant (q  + 1)-plane, say Λ . Since S ∩ D = D, it follows that Λ is a (q  + 3)-secant (q  + 1)-plane to S. This implies by Theorem 3.1(a) that index(S, L) is at most q  . Let IS and ID be respectively the ideal sheaves of S and D in the projective space P := PH 0 (S, L)∗ . Thus we have the exact sequence 0 → IS → ID → OS (−D) → 0.

(3.1)

Let M = ΩP (1) and let ML be the restriction of M to S. Consider the cohomology long exact sequence · · · → H 1 (P,

α 

M ⊗ IS (j)) → H 1 (P,

α 

M ⊗ ID (j)) → H 1 (S,

α 

ML ⊗ Lj ⊗ OS (−D)) → · · ·

(3.2)

which is induced from the above (3.1). We claim that if j ≥ 2 and α ≤ q  then the third term vanishes. Indeed, by Lemma 3.2(e), the cohomology group H 1 (S,

α 

ML ⊗ Lj ⊗ OS (−D)) = H 1 (S,

vanishes if the inequality

α 

ML ⊗ OS ((ja − 2)C0 + (2b + e)f))

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2(2b − e) − (ja − 2)e >

4661

α(2b − ae) + 4g − 4 2b − ae − 2g

is satisfied. This inequality is equivalent to α<

(4b − jae − 4g + 4)(2b − ae − 2g) 2(2b − ae)

and hence, in order to get the vanishing H 1 (S,

α 

ML ⊗ Lj ⊗ OS (−D)) = 0

for all j ≥ 2 and α ≤ q 

it is enough to show that q <

(2b − ae − 2g + 2)(2b − ae − 2g) (4b − 2ae − 4g + 4)(2b − ae − 2g) = . 2(2b − ae) (2b − ae)

(3.3)

One can easily check inequality (3.3) by hand. Now, we are ready to prove the inequality index(S, L) ≤ index(D, L|D ). Indeed, putting  := index(S, L), we know that  ≤ q  . Also since S ⊂ P is projectively normal (cf. Theorem 3.3(a)), the first term in (3.2) vanishes for all j ≥ 2 and α ≤  (cf. Theorem 5.8 in [3]). Since the third term in (3.2) also vanishes for all j ≥ 2 and α ≤ , it holds that H 1 (P,

α 

M ⊗ ID (j)) = 0 for all j ≥ 2 and α ≤ .

(3.4)

Then by using the two short exact sequences 0 → ID → OP → OD → 0

and 0 → M → H 0 (S, L) ⊗ OP → L → 0,

one can easily see that (3.4) implies the vanishing H 1 (P,

α 

M ⊗ (L|D )j ) = 0 for all j ≥ 1 and α ≤  + 1.

Therefore (D, L|D ) satisfies condition N2, (cf. Lemma 2.6 in [11]).

2

For the remaining part of this section, we focus on the case where C is a smooth curve of genus 2 and S is a ruled surface over C with e = −1 or e = −2. Let L = OS (aC0 + bf) be a line bundle on S and set b := deg(b). Corollary 3.4. Let (S, L) be as above and suppose that a ≥ 1 and 2b − ae ≥ 9. Then (a) The integer index(S, L) satisfies the inequalities b−5−

1 + ae ≤ index(S, L) ≤ b − ae − 5. 2

(3.5)

In particular, if a = 1 then b − e − 6 ≤ index(S, L) ≤ b − e − 5. (b) Assume that there exists an effective divisor D whose numerical class is 2C0 + ef. Then index(S, L) ≤ 2b − ae − 7.

(3.6)

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Fig. 1. Bound of index(S, L).

Proof. The assertions come immediately from Theorem 1.2.

2

Remark 3.5. Let S and L be as in Corollary 3.4. Let us fix a. Then Corollary 3.4(a) shows that the value of index(S, L) is bounded above by b − ae − 5. Also Corollary 3.4(b) shows that if there exists an effective divisor D ≡ 2C0 + ef then 2b − ae − 7 is another upper bound of index(S, L). Namely, we have b−5−

1 + ae ≤ index(S, L) ≤ min(b − ae − 5, 2b − ae − 7). 2

If b is large compared to a, then min(b − ae − 5, 2b − ae − 7) is equal to b − ae − 5. On the other hand, if b is small compared to a, then the upper bound of index(S, L) in Eq. (3.6) is sharper than that in (3.5) and hence D is more important to bound index(S, L). In Fig. 1, the possible numbers of index(S, L) are contained in the shadowed region. 4. Existence of a special divisor Let C be a smooth projective curve of genus g ≥ 2 and i ∈ {0, 1, . . . , g} an integer. In this section, we construct ruled surfaces πi : Si = PC (Ei ) → C with a minimal section Ci where Ei is a normalized rank 2 vector bundle on with deg(∧2 Ei ) = i such that there is a smooth irreducible curve Di which is numerically equivalent to 2Ci − if. These surfaces are constructed inductively from a ruled surface S0 and Lemma 4.2 is the beginning of this process. Assume that M is a non-trivial line bundle on C such that M 2  OC and set E0 := OC ⊕ M . Let π0 : S0 = PC (E0 ) → C be a ruled surface with a minimal section C0 . Lemma 4.1. Let C and M be as above, and let Zk = P1 + · · · + Pk ∈ Divk (C) be a general divisor of degree k with 1 ≤ k ≤ g − 1. Then H 0 (C, M (Zk )) = 0.

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Proof. Since M is a non-trivial line bundle of degree 0, we have h0 (C, M ) = 0, i.e., h0 (C, KC ⊗M −1 ) = g−1. By the Riemann–Roch theorem, we have h0 (C, M (Zk )) = 0

if and only if h1 (C, M (Zk )) = g − 1 − k.

Since h1 (C, M (Zk )) = h0 (C, KC ⊗ M −1 (−Zk )) and h0 (C, KC ⊗ M −1 ) = g − 1, we get the desired vanishing H 0 (C, M (Zk )) = 0 for a general divisor Zk with 1 ≤ k ≤ g − 1. 2 Lemma 4.2. Let π0 : S0 = PC (OC ⊕ M ) → C be as above. Then (1) There exists a smooth irreducible curve D0 of genus 2g − 1 which is linearly equivalent to 2C0 . (2) Let P ∈ C be a closed point. Then h0 (S0 , C0 + P f) ≤ 2. Furthermore, if P is general then h0 (S0 , C0 + P f) = 1. (3) Let P1 ∈ C be a general point and let x1 ∈ P1 f. Then there are at most finitely many irreducible curves which are numerically equivalent to C0 + f and which pass through x1 . Proof. (1) Since h0 (S0 , 2C0 ) = h0 (C, Sym 2 (OC ⊕ M )) = 2, there exists an effective divisor D0 ∼ 2C0 . Assume that D0 is reducible. Say D0 = Y1 ∪ Y2 , then Y1 ∼ C0 + bf and Y2 ∼ C0 − bf for some divisor b of degree 0. Since h0 (S0 , C0 + bf) = h0 (C, b ⊕ (b + e)) > 0, b must be linearly equivalent to OC or e. Hence we get that Y1 = Y2 = C0 or Y1 = Y2 = C0 + ef. So a general D0 of |2C0 | is irreducible. As in the proof of Theorem 1.2(b), D0 is a smooth curve of genus 2g − 1. (2) From the following exact sequences: 0 → OC (P ) → E0 (P ) → M (P ) → 0, we have h0 (C, E0 (P )) = h0 (S0 , C0 +P f) ≤ 2. If P is a general point, then h0 (C, E0 (P )) = h0 (S0 , C0 +P f) = 1 by Lemma 4.1. (3) By (2), we have h0 (S0 , C0 + P1 f) = 1 for a general point P1 ∈ C which implies that there is no irreducible curve in |C0 + P1 f|. Therefore, the set of point P ∈ C with h0 (S0 , C0 + P f) = 2 is finite. Let Φ := {R1 , . . . , Rs } be the set of all points in C such that h0 (S0 , C0 + P f) = 2 and define for 1 ≤ j ≤ s Aj := Bs|C0 + Rj f|. Since Aj is contained in the intersection of two members in |C0 +Rj f|, Aj is finite and hence ∪sj=1 Aj is finite. It gives that any point x ∈ / ∪sj=1 Aj is contained in finitely many irreducible curves which are numerically equivalent to C0 + f. 2 The following theorem tells that a ruled surface πi+1 : Si+1 → C of invariant i + 1 is constructed inductively from S0 by an elementary transform of a ruled surface πi : Si → C of invariant i at xi+1 ∈ Di \Ui with πi (xi+1 ) = Pi+1 for 1 ≤ i ≤ g − 1 where Di is a smooth curve which is numerically equivalent to 2Ci − if and Ui is the union of all minimum sections in Si such that Si+1 has a smooth curve Di+1 which is numerically equivalent to 2Ci+1 − (i + 1)f. Theorem 4.3. Assume that πi : Si (:= P(Ei )) → C is a ruled surface with a minimal section Ci which satisfies the following properties (Ai ) Si has finitely many minimal sections, (Bi ) det(Ei ) = M (P1 + · · · + Pi ) and there exists a smooth irreducible curve Di ≡ 2Ci − (P1 + · · · + Pi )f on Si ,

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(Ci ) for a general point xi+1 ∈ Di , there exists finitely many irreducible curves containing xi+1 which is numerically equivalent to Ci + f. Let Pi+1 be a general point on C. Assume that πi+1 : Si+1 → C is an elementary transform of Si at xi+1 ∈ Di \ Ui with πi (xi+1 ) = Pi+1 . Then Si+1 holds (Ai+1 ), (Bi+1 ) and (Ci+1 ) for i ≤ g − 3, Sg−1 holds (Ag−1 ) and (Bg−1 ) and Sg holds (Bg ). Proof. Note that, on S0 , there exist two minimal sections and a smooth irreducible curve D0 on S0 which is numerically equivalent to 2C0 by Lemma 4.2. Furthermore, by Lemma 4.2, there exists finitely many irreducible curves containing a point x1 ∈ D0 which is numerically equivalent to C0 + f. Therefore, the initial step of the induction is satisfied. Now, we assume that a ruled surface Si satisfies the conditions (Ai ), (Bi ) and (Ci ) for i ≤ g − 3. By Lemma 2.5, minimal sections on Si+1 are the strict transforms of minimal sections on Si and the strict transforms of uni-secant irreducible curves which are numerically equivalent to Ci + f containing xi+1 . Therefore Si+1 has finitely many minimal sections for i ≤ g −2. Also we have det(Ei+1 ) = M (P1 +· · ·+Pi+1 ) by Lemma 2.5. By Proposition 2.2, the strict transform Di+1 of Di is a smooth irreducible curve which is linearly equivalent to 2Ci+1 − (P1 + · · · + Pi+1 )f. It gives the condition (Bi+1 ) for a ruled surface Si+1 for i ≤ g − 1. Assume that i ≤ g − 3 and consider the exact sequence on C, 0 → OC (Pi+2 ) → Ei+1 (Pi+2 ) → Mi+1 (Pi+2 ) → 0, where Mi+1 = M (P1 + · · · + Pi+1 ) is a line bundle corresponding to a minimal section on Ci+1 on Si+1 . By Lemma 4.1, we have H 0 (C, Ei+1 (Pi+2 )) = H 0 (Si+1 , Ci+1 + Pi+2 f) = 1 for a general Pi+2 ∈ C. The same method in the proof of Lemma 4.2 gives that Si+1 has finitely many irreducible curves containing a general point xi+2 ∈ Di+1 \ Ui+1 which is numerically equivalent to Ci+1 + f and hence it gives the condition (Ci+1 ) for Si+1 for i ≤ g − 3. 2 5. Genus two case Throughout this section, let C be a smooth curve of genus 2 and π : S → C a ruled surface over C with an invariant e < 0. Also let L = OS (aC0 + bf) be a very ample line bundle on S (and hence a ≥ 1) and set b := deg(b). In this section, we investigate in detail the problems discussed in Section 3 on (S, L). Remark 5.1. (1) A line bundle on C is very ample if and only if its degree is bigger than or equal to 5. In our case, deg(L|C0 ) is equal to b − ae and hence we have b − ae ≥ 5. Moreover, Theorem 3.1 says that (C0 , L|C0 ) satisfies condition N2, for  ≥ 1 if and only if deg(L|C0 ) = b − ae ≥ 5 + . (2) If there exists an effective divisor D on S which is linearly equivalent to 2C0 + ef, then D is a smooth irreducible curve of genus 3 by Theorem 1.2(b). A line bundle L|D on D is very ample if and only if we have either (i) L|D = KD or (ii) deg(L|D ) = 2b − ae = 6 and H 0 (D, L|D − KD ) = 0 or (iii) deg(L|D ) = 2b − ae ≥ 7 (cf. Lemma 4.1 in [13]). Note that if L|D = KD , then KD imbeds D as a smooth plane quartic and if deg(L|D ) = 2b − ae = 6, then L|D imbeds D with a tri-secant line and hence L|D fails to satisfy N2,1 . Therefore, (D, L|D ) satisfies condition N2, for  ≥ 1 if and only if deg(L|D ) = 2b −ae ≥ 7 + by Theorem 3.1.

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Theorem 5.2. Let (S, L) be as above and suppose that there exists an effective divisor D whose numerical class is 2C0 + ef. If (S, L) satisfies condition N2, for some  ≥ 1, then both (C0 , L|C0 ) and (D, L|D ) satisfy condition N2, , or equivalently, b − ae ≥ 5 + 

and

2b − ae ≥ 7 + .

Proof. Since L is a very ample line bundle on S, its restriction L|D to D is also a very ample line bundle. Thus L|D is one of three cases in Remark 5.1(2). Assume that L|D = KD . Since KD imbeds D as a smooth plane quartic, the natural map H 0 (S, L) → H 0 (D, L|D ) is surjective and hence D in PH 0 (S, L)∗ is linearly normal. Since any line in the plane spanned by D ⊂ PH 0 (S, L)∗ is a 4-secant line and S does not contain a plane, S has a 4-secant line. Hence (S, L) fails to satisfy condition N2,1 which is a contradiction to our assumption. Therefore we can assume that deg(L|D ) = 2b − ae ≥ 6. Since π∗ L(−D) is a zero sheaf for a = 1 1 and μ− (L(−D)) = b − ae 2 ≥ 3 for a ≥ 2, we have h (S, L(−D)) = 0 by Lemma 3.2. It implies that D in 0 ∗ PH (S, L) is linearly normal. If deg(L|D ) = 2b − ae = 6, then D ⊂ D = P3 admits a tri-secant line. Otherwise, D has a smooth plane model. If deg(L|D ) = 2b − ae = 7, then D ⊂ D = P4 admits a tri-secant line by Theorem 3.1. This tri-secant line cannot be a fiber of π : S → C because D  f = 2. Since S ⊂ PH 0 (S, L)∗ does not contain a rational curve except fibers, S has a tri-secant line and hence (S, L) fails to hold condition N2,1 . Therefore, if (S, L) satisfies condition N2,1 , then 2b − ae ≥ 8. If deg(L|D ) = 2b − ae = 8, then by Theorem 3.1, D ⊂ P5 admits a 4-secant 2-plane, say Λ. Assume that S ∩ Λ = Y ∪ Y  where Y is a plane curve of degree d. If d ≥ 3, then S contains d-secant line and hence (S, L) fails to hold condition N2,1 . If d = 2, then a = 2 and hence Y is a fiber of π : S → C and Y  D = 2. Since Λ is a 4-secant 2-plane to D, Y  cannot be an empty set and hence S admits a tri-secant line. If d = 1, then a = 1 and Y is a ruling of π : S → C. In this case, deg Y  ≥ 2 and hence S has a tri-secant line. Therefore (S, L) fails to hold condition N2,1 . If dim(S ∩ Λ) = 0, then S contains a 4-secant 2-plane and hence (S, L) fails to hold condition N2,2 by [4, Theorem 1.1]. Hence, if (S, L) satisfies condition N2,2 , then 2b − ae ≥ 9. Assume that deg(L|D ) = 2b − ae = 9 + k

with k ≥ 0.

If a = 1, then Corollary 3.4(a) implies that b − e ≥ 5 +  and hence 2b − e ≥ 7 + . If a ≥ 2, then Theorem 1.2(b) shows that  ≤ index(S, L) ≤ k + 2. This completes the proof that 2b − ae ≥ 7 + .

(5.1)

Since L is a very ample line bundle on S, its restriction L|C0 to C0 is also very ample line bundle, i.e., b − ae ≥ 5. If we assume 2b − ae ≥ 9, then the assumption that (S, L) satisfies condition N2,l and Theorem 3.4(a) give the inequality l ≤ index(S, L) ≤ b − ae − 5. As a result, if (S, L) satisfies condition N2, and 2b − ae ≥ 9 which is always true for l ≥ 2 by (5.1), then (C0 , L|C0 ) satisfies condition N2, and b − ae ≥ 5 + . Now we assume that  = 1 and 2b − ae = 8. Except the cases e = −2, a = 1, b = 3 and e = −1, a = 2, b = 3, we have b − ae ≥ 6, i.e., C0 satisfies condition N2,1 . By the way, the exceptional cases, we have b − ae = 5 and hence C0 ⊂ P5 admits a tri-secant line. This tri-secant line cannot be contained in S since S does not contain a rational curve except fibers and D  f = 2. Hence (S, L) fails to satisfy condition N2,1 . It completes the proof of the theorem. 2

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Acknowledgements We would like to express our gratitude to Professor Bangere Purnaprajna for his encouragement and helpful advice. We are also glad to thank University of Kansas and KIAS for warm hospitality during our visiting there. This research was supported by Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education, Science and Technology (NRF-2011-0014465) and by the 2013 Yeungnam University Research Grant for the first named author. This research was supported by the Korea Research Foundation Grant funded by the Korean Government (NRF-2013R1A1A2008445) for the second named author. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13]

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