Vector bundles on ruled Fano 3-folds

manuscripta math. 100, 335–349 (1999) © Springer-Verlag 1999

Laura Costa

Vector bundles on ruled Fano 3-folds

Received: 16 February 1999 / Revised version: 2 July 1999

Abstract. Let X = P(E) → P1 × P

1 be a ruled Fano 3-fold. The goal of this paper is tocompute the dimension, prove the irreducibility and smoothness and describe the structure ofthe moduli spaceML(2; c1, c2) of L-stable, rank 2 vector bundlesE onX with certain Chernclasses and for a suitable polarizationLclosely related toc2. More precisely, we will cover thestudy of some moduli spacesML(2; c1, c2) such that the generic point[E] ∈ ML(2; c1, c2)

is given as a non-trivial extension of line bundles. This work nicely reflects the generalphilosophy that moduli spaces inherits a lot of geometrical properties of the underlyingvariety.

1. Introduction

LetX be ann-dimensional, smooth, irreducible, algebraic variety over the complexfield and letLbe an ample divisor onX.We will denote byMX,L(r; c1,· · · ,cmin(r,n))

the moduli space of rankr, L-stable (in the sense of Mumford-Takemoto) vectorbundlesE onX with fixed Chern classesci(E) = ci ∈ H 2i (X, Z).

Moduli spaces for stable vector bundles on smooth, irreducible, algebraic pro-jective varieties were constructed in the 70′s. Many detailed and interesting resultshave been proved regarding these moduli spaces when the underlying variety isa surface and very little is known if the variety has dimension greater or equalthan three. Up to now there are no general results about these moduli spaces con-cerning the number of connected components, dimension, smoothness, rationality,topological invariants, etc...

It was a major result in the theory of vector bundles on an algebraic surfaceS theproof that for largec2 the moduli spaceMS,L(r; c1, c2) of rankr, L-stable, vectorbundles onS with fixed c1 and fixed polarization,L, is irreducible, genericallysmooth and of the expected dimension 2rc2−(r−1)c2

1−(r2−1)χ(OS). For modulispaces of vector bundles on a higher dimensional variety,X, the situation changesdrastically. The smoothness and irreducibility turn to be false when dimX ≥ 3.See for instance, [BMR97]; Theorem 0.1, and [MROL97].

L. Costa: Dept. Algebra i Geometria, Facultat de Matemàtiques, Universitat de Barcelona,E-08007 Barcelona, Spain.e-mail: [email protected]

Mathematics Subject Classification (1991):14F05, 14D20, 14J45

336 L. Costa

Let X = P(E) → P1 × P

1 be a ruled Fano 3-fold. The goal of this paper is tocompute the dimension, prove the irreducibility and smoothness and describe thestructure of the moduli spaceML(2; c1, c2) of L-stable, rank 2 vector bundlesE onX with certain Chern classes and for a suitable polarizationL closely related toc2.More precisely, we will cover the study of some moduli spacesML(2; c1, c2) suchthat the generic point[E] ∈ ML(2; c1, c2) is given as a non-trivial extension of linebundles (see Theorem 3.5). This work nicely reflects the general philosophy thatmoduli spaces inherits a lot of geometrical properties of the underlying variety. Foranalogous results on moduli spaces of vector bundles on rational normal scrolls oron P

d -bundles,P(E) → C, beingC a smooth projective curve of arbitrary genus,the reader can see [Cos98] and [CMR98].

Next, we outline the ideas used for proving our results and the structure of thepaper. In section 2, we recall some generalities on ruled Fano 3-folds and we provesome lemmas that will be very useful for us later on. We start section 3 provingthe Proposition 3.1 that will be the key point for our result. Namely, the existenceof sections of a suitable twist of a rank 2 vector bundleE on a ruled Fano 3-fold,whose scheme of zeros(s)0 has codimension bigger or equal than two. Then weprove our main result on moduli spaces, namely the irreducibility, smoothness andstructure of the moduli spaceML(2; c1, c2) of L-stable, rank 2 vector bundles on aruled Fano 3-fold,X = P(E) → P

1 ×P1, with certain Chern classes and a suitable

polarizationL. Our approach is to writeL-stable, rank 2 vector bundlesE on X

as an extension of two line bundles. A well known result for vector bundles overcurves is that any vector bundle of rankr ≥ 2, can be written as an extension oflower rank vector bundles. For higher dimensional varieties we may not to be ableto get such a nice result. (For instance it is not true forX = P

n, n ≥ 2). However, itturns out to be true for certainL-stable, rank 2 vector bundlesE over a ruled Fano3-fold X = P(E) → P

1 × P1. We finish this section computing the Picard group

and the Kodaira dimension of these moduli spaces.

2. Generalities on ruled Fano 3-folds

Let E be a rankr ≥ 2 vector bundle on a smooth complex projective varietyY andconsider

X = P(E) = Proj(SymE)π−→Y

the projectivized vector bundle associated toE with the natural projectionπ . Thetautological line bundleξ = ξE on X = P(E) is uniquely determined by theconditions

ξ|F ∼= OF (1), π∗ξ ∼= E

beingF ∼= Pr−1 a fiber ofπ . The Picard group ofX can be expressed as

Pic (X) ∼= ZξE ⊕ π∗Pic(Y ). (1)

Vector bundles on ruled Fano 3-folds 337

For any line bundleL onY ,

P(E ⊗ L) ∼= P(E) with ξE⊗L ∼= ξE ⊗ π∗L.

In addition,E is generated by its global sections (resp. ample) if and only ifOX(ξE )

is generated by its global sections (resp. ample).The Theorem of Leray and Hirsch yields that in the cohomology ring ofX the

following holds

0 =r∑

i=0

(−1)iπ∗ci(E)ξ r−i . (2)

We say that a vector bundleE on Y is Fano ifP(E) is a Fano manifold. RuledFano 3-foldsX = P(E) → Y defined by a rank two vector bundleE on a smooth,complex, projective surfaceY , were classified in [SW90]. Let us recall now theclassification theorem

Theorem 2.1 ((M. Szurek; A. Wisniewski)).There are 21 types of ruled Fano 3-foldsV = P(E) with a rank two vector bundleE on a surfaceS. Up to a twist, theyare:

(1–7) S = P2 andE one of the following

O(1) ⊕ O(−1), O ⊕ O(1), O ⊕ O,TP2(−2), 0 → O → E → Ix → 0, x ∈ P

2

E stable withc1 = 0, c2 = 2, i.e.0 → O(−1)2 → O4 → E(1) → 0E stable withc1 = 0, c2 = 3, such that there is an exact sequence

0 → O(−2) → O3 → E(1) → 0

(8–12) S = P1 × P

1 andE one of the followingO(−1, −1) ⊕ O, O ⊕ O, O(0, 1) ⊕ O, O(0, −1) ⊕ O(−1, 0)

E stable withc1 = (−1, −1), c2 = 2, i.e.0 → O(−1, −1) → O3 → E(1, 1) → 0

(13–15)S is the Hirzebruch surfaceF1, with a blow-down mapβ : F1 → P2 and

E one of the followingβ∗(O ⊕ O(−1)), O ⊕ O β∗(TP2(−2))

(16–21)S is a Del Pezzo surface obtained by blow-upk ≤ 8 points ofP2 andE = O ⊕ O.

In this work, we turn our attention to study moduli spaces of rank 2, stablevector bundles on ruled Fano 3-folds overP

1 × P1. So, from now on,Y will be the

quadric surfaceP1 × P1 and we will denote its Picard group by

Pic(Y ) =< l1, l2 > .

Hence, the canonical divisor isKY = −2l1 − 2l2.LetX = P(E) → Y be a ruled Fano 3-fold defined by a rank two vector bundle

E on Y with c1(E) = d1l1 + d2l2 andc2(E) = c2. According to Theorem 2.1 and

338 L. Costa

the fact thatP(E) ∼= P(E ⊗ OY (D)) for any divisorD ∈ Pic(Y ), we can assumewithout loss of generality thatE is one of the following:

OY (l1 + l2) ⊕ OY , (d1, d2) = (1, 1) c2 = 0OY ⊕ OY , (d1, d2) = (0, 0) c2 = 0OY (li) ⊕ OY , i = 1, 2, (d1, d2) = (1, 0), (0, 1) c2 = 0OY (l1) ⊕ OY (l2) (d1, d2) = (1, 1) c2 = 1

E given by

0 → OY (−l1 − l2) → O3Y → E → 0

with (d1, d2) = (1, 1), c2 = 2.Notice that in each of the above casesE is generated by its global sections. We

have

Pic(X) =< ξ, H1, H2 >

−KX = 2ξ + π∗(−KY − c1E) = 2ξ + (2 − d1)H1 + (2 − d2)H2

whereξ = ξE andHi = π∗li , i = 1, 2. Moreover we have the following relation

0 = ξ2 − π∗c1(E)ξ + π∗c2(E) = ξ2 − (d1H1 + d2H2)ξ + c2H1H2. (3)

Sincel2i = 0, we haveH 2i = 0. MoreoverH1H2ξ = 1 and from (3) we get

0 = ξ2H1 − (d1H1 + d2H2)H1ξ + c2H21 H2 = ξ2H1 − d2,

0 = ξ2H2 − (d1H1 + d2H2)H2ξ + c2H1H22 = ξ2H2 − d1,

0 = ξ3 − (d1H1 + d2H2)ξ2 + c2H1H2ξ = ξ3 − 2d1d2 + c2.

Therefore, the intersection product onX is given by

H1H2ξ = 1, H 2i = 0, ξ2H1 = d2, ξ2H2 = d1, ξ3 = 2d1d2 − c2.

Lemma 2.2.LetOX(aξ + π∗D) be a line bundle onX = P(E) with D ∈ Pic(Y ).Then,

Hi(X, OX(aξ + π∗D)) ={

0 if −2 < a < 0Hi(Y, Sa(E) ⊗ OY (D)) if a ≥ 0

beingSa(E) the a-th symmetric power ofE . In addition,H 0(X, OX(aξ +π∗D)) =0 for a < 0.

Proof. For a line bundleOX(aξ + π∗D) we have

OX(aξ + π∗D) = OP(E )(a) ⊗ π∗OY (D).

By the projection formula

Riπ∗OX(aξ + π∗D) = Riπ∗OP(E )(a) ⊗ OY (D).


By [Har77]; III, Ex. 8.3, we have

π∗OP(E )(a) = SaE for a ≥ 0R1π∗OP(E )(a) = 0 for a > −2π∗OP(E )(a) = 0 for a < 0.

On the other hand, since the fiber ofπ is F ∼= P1, for anyy ∈ Y , we have

Hi(F, (OP(E )(a))y) = 0 for i ≥ 2.

Therefore, from the cohomology base change

Riπ∗OP(E )(a) = 0 for i ≥ 2.

Thus putting altogether,

Riπ∗OP(E )(a) = 0 for i > 0 anda > −2.

By the degeneration of the Leray Spectral sequence, fora > −2 we get

Hi(X, OX(aξ + π∗D)) = Hi(Y, π∗OP(E )(a) ⊗ OY (D))

which gives us what we want. The last statement follows from the fact that ifa < 0we have

H 0OX(aξ + π∗D) = H 0π∗(OX(aξ + π∗D)) = 0

since in this caseπ∗(OX(aξ + π∗D)) = 0. utIn the next two Lemmas we characterize ample (resp. effective) divisors onX.

Lemma 2.3.The divisorL = aξ + b1H1 + b2H2 onX is ample fora ≥ 1, b1 ≥ 2andb2 ≥ 2.

Proof. SinceE is generated by global sectionsE ⊗OY (2l1+2l2) is an ample vectorbundle. So,

ξ + 2H1 + 2H2

is an ample divisor. Therefore, for any ample divisor onY , αl1 + βl2 with α > 0andβ > 0,

ξ + (2 + α)H1 + (2 + β)H2

is ample. Finally, sinceξ is generated by global sections

aξ + b1H1 + b2H2

is ample fora ≥ 1 andbi ≥ 2. ut

340 L. Costa

Lemma 2.4.Let C = α0ξ + α1H1 + α2H2 be a non-zero effective divisor onX.Then

α0 ≥ 0 α0(2 + d1) + α1 ≥ 0 α0(2 + d2) + α2 ≥ 0

and ifα0 = 0 thenα1 > 0 or α2 > 0.

Proof. The inequalityα0 ≥ 0 follows from Lemma 2.2. Let us prove the otherinequalities. Fort � 0, Ht := ξ + tH2 + 2H2 is an ample divisor on X (seeLemma 2.3). Thus sinceC is an effective divisor we haveCH 2

t > 0, which isequivalent to

0 < α0ξ3 + 2tα0d2 + 4α0d1 + 4tα0 + α1d2 + 4α1 + α2d1 + 2tα2.

Hence, fort � 0, we get

α0(2 + d2) + α2 ≥ 0.

In the same way using the ample divisorLt := ξ + 2H1 + tH2 (see Lemma 2.3)we get the other inequality.ut

3. Moduli spaces of vector bundles on ruled Fano 3-folds

Throughout this sectionX will be a ruled Fano 3-fold

X = P(E) → P1 × P

1

defined over a quadric surface and we will keep the notations introduced in thesecond section of this work. In addition, we fix integersα, e1, e2 ∈ {1, 0}, 2c >

e2 − d2, b � 0 and an ample divisor onX

L = (2 − α)ξ − (2b − β)H1 − bH2

beingβ = e1 + 2e2 − 4c − (2 − α)(d1 + d2) − 1. Notice thatβ is independentof b.

Let us start with a result that will be very useful for us in the sequel.

Proposition 3.1.Let X = P(E) → P1 × P

1 be a ruled Fano 3-fold andE a ranktwo vector bundle onX with Chern classes

c̃1 := c1E = αξ + e1H1 + e2H2

c̃2 := c2E = (α − 1)ξ2 + (−2b + bα + e1)ξH1 + (−2c + cα + e2)ξH2+(−2bc + be2 + ce1)H1H2

beingα, e1, e2 ∈ {0, 1}. Then

E(−ξ − bH1 − cH2)

has a non-zero section whose scheme of zeros has codimension≥ 2.


Remark 3.2.In spite of we have the relationξ2 = (d1H1 + d2H2)ξ − c2H1H2,during all the work, we keepξ2 in order to simplify the notation.

Proof of Proposition 3.1.Let F ∼= P1 be the generic fiber of the rulingπ . Then

E|P1 = OP1(a) ⊕ OP1(α − a)

and sincec1(E|P1) = α ≥ 0 we may assume thata > 0 if α = 1 anda ≥ 0 ifα = 0. The injection

OP1(a) ↪→ E|P1

induces an injection

OX(aξ + n1H1 + n2H2) ↪→ E.

Take 0 6= s ∈ H 0E(−aξ − n1H1 − n2H2) and letZ be its scheme of zeros. LetA

be the maximal effective divisor contained inZ. s can be regarded as a section ofE(−aξ − n1H1 − n2H2 − A) and its scheme of zeros has codimension≥ 2. Thus,

E(−lξ − m1H1 − m2H2)

with l > 0 if α = 1 andl ≥ 0 if α = 0 has a non-zero section whose scheme ofzeros has codimension≥ 2. Let us see that

l = 1, m1 = b, m2 = c.

First of all we will see that ifα = 0 thenl > 0. Let us assume thatα = l = 0and we will get a contradiction. SinceE(−m1H1 − m2H2) has a non-zero sectionwhose scheme of zeros has codimension≥ 2

c2E(−m1H1 − m2H2)Lt ≥ 0

beingLt := ξ + tH1 + 2H2 with t � 0 an ample divisor onX. Thus, fort � 0we have

0 ≤ c2E(−m1H1 − m2H2)tH1 = t (−d2 − 2c + e2).

Therefore, 2c ≤ e2 − d2 which contradicts the assumption 2c > e2 − d2. Thus,from now on we can assumel > 0.

Claim.

mi ≥ (1 − α)di − l(l − α)di + (2 − α)γi + (l − 1)ei

2l − α

with γ1 = b andγ2 = c. In particular, forl = 1, m1 ≥ b andm2 ≥ c.

342 L. Costa

Proof of the Claim.We consider the ample divisorLt := ξ + tH1 +2H2 onX witht � 0. SinceE(−lξ − m1H1 − m2H2) has a non-zero section whose scheme ofzeros has codimension≥ 2, we have

0 ≤ c2E(−lξ − m1H1 − m2H2)Lt .

Hence, sincet � 0 we obtain

0 ≤ c2E(−lξ−m1H1−m2H2)tH1 = t ((α − 1)d2 − 2c + cα + e2)

+t (−lαd2 − le2 − m2α + l2d2 + 2lm2)

which, sincet > 0, is equivalent to

(2l − α)m2 ≥ (1 − α)d2 − l(l − α)d2 + (2 − α)c + (l − 1)e2

that gives us one of the desired inequalities.In the same way, using the ample divisorMt := ξ + 2H1 + tH2 onX we get

(2l − α)m1 ≥ (1 − α)d1 − l(l − α)d1 + (2 − α)b + (l − 1)e1

and the Claim follows.SinceE(−lξ − m1H1 − m2H2) has a non-zero section whose scheme of zeros

has codimension≥ 2, if we consider the ample divisorRt := tξ + 2H1 + 2H2 onX with t � 0 we get

0 ≤ c2E(−lξ − m1H1 − m2H2)ξ

= (α − 1)ξ3 + (−2b + bα + e1)d2 + (−2c + cα + e2)d1

− 2bc + be2 + ce1 − lαξ3 − le1d2 − le2d1− m1αd2 − m1e2 − m2αd1 − m2e1

+ l2ξ3 + 2lm1d2 + 2lm2d1 + 2m1m2.

(4)

Finally, sinceOX(lξ + m1H1 + m2H2) ↪→ E andE is L-stable we have

(lξ + m1H1 + m2H2)L2 <

c1(E)L2

2= (αξ + e1H1 + e2H2)L

2

2

which is equivalent to

0 > (l − α2 )(2 − α)2ξ3 + (2l − α)(2 − α)(β − 2b)d2 − (2l − α)(2 − α)bd1

−(2l − α)(β − 2b)b + (2 − α)2d2m1 − 2b(2 − α)m1 + (2 − α)2d1m2

+2(2 − α)(−2b + β)m2 − e12 (2 − α)2d2 + e1(2 − α)b − e2

2 (2 − α)2d1−(2 − α)(β − 2b)e2.

Dividing by (2 − α) > 0 this is equivalent to

M := (l − α2 )(2 − α)ξ3 + (2l − α)(β − 2b)d2 − (2l − α)bd1

− 2l−α2−α

(β − 2b)b + (2 − α)d2m1

− 2bm1 + (2 − α)d1m2 + 2(−2b + β)m2− e1

2 (2 − α)d2 + e1b − e22 (2 − α)d1 − (β − 2b)e2 < 0.


Using the Claim for the bounds ofm1 andm2 we get

0 > M ≥ (l − α2 )(2 − α)ξ3 + (2l − α)(β − 2b)d2 − (2l − α)bd1

− 2l−α2−α

(β − 2b)b

+ ((2 − α)d2 − 2b)((1−α)d1−l(l−α)d1+(2−α)b+(l−1)e1

2l−α)

+ ((2 − α)d1 + 2(−2b + β))((1−α)d2−l(l−α)d2+(2−α)c+(l−1)e2

2l−α)

− e12 (2 − α)d2 + e1b − e2

2 (2 − α)d1 − (β − 2b)e2 = N.

Notice that the coefficient ofb2 in N is

2(2l − α)

2 − α− 2(2 − α)

2l − α.

If l > 1, 2l − α ≥ 4 − α and the coefficient ofb2 is strictly positive, which givesus a contradiction. Hence,l = 1.

Moreover it follows from the claim thatm1 ≥ b andm2 ≥ c. We will see nowthat we have equalities.

If m1 > b, using the fact thatl = 1 andm2 ≥ c we get

0 > M ≥ (l − α2 )(2 − α)ξ3 + (2 − α)(β − 2b)d2 − (2 − α)bd1

− (β − 2b)b + (b + 1)((2 − α)d2 − 2b)

+ c(2 − α)d1 + 2c(−2b + β)

− e12 (2 − α)d2 + e1b − e2

2 (2 − α)d1 − (β − 2b)e2 = P.

Notice that inP the coefficient ofb2 is zero and that the coefficient ofb is

−2(2 − α)d2 − (2 − α)d1 − β + (2 − α)d2 − 2 − 4c + e1 + 2e2 = −1

Thus, sinceb � 0, P > 0 and we get a contradiction. Therefore,m1 = b. Finally,using (4) and the fact thatm2 ≥ c we obtain

0 ≤ (m2 − c)((2 − α)d1 − e1 + 2b) ≤ 0

which implies thatm2 = c and the proposition follows.utOne way to study rank two vector bundles over an algebraic varietyX is to use

extensions of line bundles. Using this idea we construct the following families.

Construction 3.3.Let

P = P(Ext1(OX((α −1)ξ−(b − e1)H1−(c − e2)H2), OX(ξ + bH1 + cH2))∗)

and pX : P × X → X, pP : P × X → P the projections. There is a universalextension onP × X

0 → p∗X(OX(ξ + bH1 + cH2)) → F

→ p∗X(OX((α − 1)ξ − (b − e1)H1 − (c − e2)H2)) → 0 (5)

whereb, c were fixed before andα, e1, e2 ∈ {0, 1}.

344 L. Costa

Proposition 3.4.LetX = P(E) → P1 ×P

1 be a ruled Fano 3-fold. For any vectorbundleE = Fx with x ∈ P we have

(a) H 0E((1 − α)ξ + (b − e1)H1 + (c − e2)H2) = 0.(b) E is a rank two,L-stable vector bundle with Chern classes

c̃1 = c1E = αξ + e1H1 + e2H2

c̃2 = c2E = (α − 1)ξ2 + (−2b + bα + e1)ξH1 + (−2c + cα + e2)ξH2+(−2bc + be2 + ce1)H1H2.

(c) If x and x′ are distinct points ofP, then Fx and Fx′ are not isomorphic.Moreover,

dimP = h1OX((2 − α)ξ + (2b − e1)H1 + (2c − e2)H2) − 1.

Proof. (a) We consider the cohomology exact sequence

0 → H 0OX((2 − α)ξ + (2b − e1)H1 + (2c − e2)H2)

→ H 0E((1 − α)ξ + (b − e1)H1 + (c − e2)H2) →H 0OX

δ−→H 1OX((2 − α)ξ + (2b − e1)H1 + (2c − e2)H2) → · · · (6)

associated to the exact sequence (5). Sinceb � 0, by Lemma 2.4

H 0OX((2 − α)ξ + (2b − e1)H1 + (2c − e2)H2) = 0.

On the other hand, sinceE is given by a non-trivial extension and

H 1OX((2 − α)ξ + (2b − e1)H1 + (2c − e2)H2) ∼=Ext1(OX((α − 1)ξ − (b − e1)H1 − (c − e2)H2), OX(ξ + bH1 + cH2))

δ(1) is the element ofExt1(OX((α − 1)ξ − (b − e1)H1 − (c − e2)H2), OX(ξ +bH1 + cH2)) corresponding to the extension; andδ is an injection. Therefore by(6) we get

H 0E((1 − α)ξ + (b − e1)H1 + (c − e2)H2) = 0

which proves (a).(b) SinceE ∈ F we have

c1E = (ξ + bH1 + cH2) + ((α − 1)ξ − (b − e1)H1 − (c − e2)H2)

= αξ + e1H1 + e2H2c2E = (ξ + bH1 + cH2)((α − 1)ξ − (b − e1)H1 − (c − e2)H2)

= (α − 1)ξ2 + (−2b + bα + e1)ξH1+ (−2c + cα + e2)ξH2 + (−2bc + be2 + ce1)H1H2.

Now we will see thatE is L-stable, i.e. that for any rank one vector bundle

OX(D) ↪→ E


we haveDL2 <c1(E)L2

2 . SinceE ∈ F , for any OX(D) ↪→ E we have twopossibilities:

(1) OX(D) ↪→ OX(ξ + bH1 + cH2)

(2) OX(D) ↪→ OX((α − 1)ξ − (b − e1)H1 − (c − e2)H2).

In the first case,

DL2 ≤ (ξ + bH1 + cH2)L2 <

c1(E)L2

2if, and only if,

0 > (1 − α2 )(2 − α)2ξ3 + (2 − α)2(β − 2b)d2 − (2 − α)2bd1

− (2 − α)(β − 2b)b + (2 − α)2d2b − 2b(2 − α)b

+ (2 − α)2d1c + 2c(2 − α)(−2b + β)

− e12 (2 − α)2d2 + e1(2 − α)b − e2

2 (2 − α)2d1 − (2 − α)(β − 2b)e2.


0 > (1 − α2 )(2 − α)ξ3 + b(−(2 − α)d2 − (2 − α)d1 − β − 4c + e1 + 2e2)

+ β((2 − α)d2 + 2c − e2) + c(2 − α)d1 − e12 (2 − α)d2 − e2

2 (2 − α)d1.

Notice that sinceβ = e1 + 2e2 − 4c − (2− α)(d1 + d2) − 1, the coefficient ofb is equal to 1. Therefore, sinceb � 0 the above inequality is satisfied.

Let us assume now that

OX(D) ↪→ OX((α − 1)ξ − (b − e1)H1 − (c − e2)H2).

By (a),D = (α − 1)ξ − (b − e1)H1 − (c − e2)H2 − C where

C = α0ξ + α1H1 + α2H2

is a non-zero effective divisor. Hence

DL2 <c1(E)L2

2if, and only if,

0 > (α2 − 1)(2 − α)2ξ3 − (2 − α)2(β − 2b)d2 + (2 − α)2bd1

+ (2 − α)(β − 2b)b − (2 − α)2(b − e1)d2 + 2(2 − α)b(b − e1)

− (2 − α)2d1(c − e2) − 2(2 − α)(−2b + β)(c − e2) − e12 (2 − α)2d2

+ e1(2 − α)b − e22 (2 − α)2d1 − (2 − α)(β − 2b)e2 − α0(2 − α)2ξ3

− 2α0(2 − α)(β − 2b)d2 + 2α0(2 − α)bd1

+ 2α0(β − 2b)b − α1(2 − α)2d2 + 2α1(2 − α)b

− α2(2 − α)2d1 − 2α2(2 − α)(−2b + β).


0 > (α2 − 1)(2 − α)ξ3 + b((2 − α)d2 + (2 − α)d1 + β − e1 + 4c − 2e2)

− β((2 − α)d2 + 2c − e2) − c(2 − α)d1 + e12 (2 − α)d2 + e2

2 (2 − α)d1

− α0(2 − α)ξ3 − 2α0(β − 2b)d2 + 2α0bd1

+ 2α02−α

(β − 2b)b − α1(2 − α)d2

+ 2α1b − α2(2 − α)d1 − 2α2(−2b + β) = J.

346 L. Costa

If we write

γ = (α2 − 1)(2 − α)ξ3 − β((2 − α)d2 + 2c − e2)

− c(2 − α)d1 + e12 (2 − α)d2 + e2

2 (2 − α)d1

and we substitute inJ the value ofβ, the above inequality is equivalent to

J = γ − b − α0(2 − α)ξ3 − 2α0(β − 2b)d2 + 2α0bd1 + 2α02−α

(β − 2b)b

− α1(2 − α)d2 + 2α1b − α2(2 − α)d1 − 2α2(−2b + β).

If α0 > 0, by Lemma 2.4 we have

−α1 ≤ α0(2 + d1) − α2 ≤ α0(2 + d2).

Using this two inequalities, we get

J ≤ γ − b − α0(2 − α)ξ3 − 2α0(β − 2b)d2 + 2α0bd1

+ 2α02−α

(β − 2b)b + α0(2 + d1)(2 − α)d2 − 2α0b(2 + d1)

+ α0(2 + d2)(2 − α)d1 + 2α0(2 + d2)(−2b + β) < 0

where the last inequality follows from the fact that the coefficient ofb2 is 4α0α−2 < 0.

Assume nowα0 = 0. In this case we have to see

R = γ − b − α1(2 − α)d2 + 2α1b − α2(2 − α)d1 − 2α2(−2b + β) < 0.

If α1 > 0, sinceα2 ≥ 0 we get

R ≤ γ − b − α1(2 − α)d2 + 2α1b < 0

since the coefficient ofb is 2α1 − 1 > 0 andb � 0.Finally, if α0 = α1 = 0, sinceα2 > 0 we get

R = γ − b − α2(2 − α)d1 − 2α2(−2b + β) < 0

since the coefficient ofb is 4α2 − 1 > 0 andb � 0.Therefore,E is L-stable and we have finished the proof of (b).

(c) It follows from (a). In addition,

dimP = dim Ext1(OX((α − 1)ξ − (b − e1)H1− (c − e2)H2), OX(ξ + bH1 + bH2)) − 1

= h1OX((2 − α)ξ + (2b − e1)H1 + (2c − e2)H2) − 1. utNow we can state and prove the main result of this work.

Theorem 3.5.Let X = P(E) → P1 × P

1 be a ruled Fano 3-fold. Fix integersα,e1, e2 ∈ {0, 1}, 2c > e2 − d2, b � 0, andL the ample divisor onX

L = (2 − α)ξ − (2b − β)H1 − bH2

beingβ = e1 + 2e2 − 4c − (2 − α)(d1 + d2) − 1. Consider

c̃1 = αξ + e1H1 + e2H2

c̃2 = (α − 1)ξ2 + (−2b + bα + e1)ξH1 + (−2c + cα + e2)ξH2+ (−2bc + be2 + ce1)H1H2.

Then the moduli spaceML(2; c̃1, c̃2) is a smooth, irreducible, rational, projectivevariety of dimensionh1OX((2 − α)ξ + (2b − e1)H1 + (2c − e2)H2) − 1.


Proof. Using Proposition 3.4 we have defined a morphism

φ : P → ML(2; c̃1, c̃2)

which is an open immersion. Indeed, it follows from the fact that sinceα − 2 < 0,we have

Hom(OX(ξ + bH1 + cH2), OX((α − 1)ξ − (b − e1)H1 − (c − e2)H2))

∼= h0OX((α − 2)ξ − (2b − e1)H1 − (2c − e2)H2) = 0

(see Lemma 2.2). Moreover, since for all[E] ∈ ML(2; c̃1, c̃2), E(−ξ − bH1 −cH2) has a non-zero section whose scheme of zeros has codimension≥ 2 (seeProposition 3.1) andc2E(−ξ − bH1 − cH2) = 0, φ is a surjection.

Now we will see that

dimT[E]ML(2; c̃1, c̃2) = dimP.

By deformation theory,

dimT[E]ML(2; c̃1, c̃2) = dim Ext1(E, E).

Sinceφ is surjective, any[E] ∈ ML(2; c̃1, c̃2) is given by a non-trivial extension

0 → OX(ξ + bH1 + cH2) → E

→ OX((α − 1)ξ − (b − e1)H1 − (c − e2)H2) → 0. (7)

Applying the functor Hom(·, E) to this exact sequence we get

0 → H 0E((1 − α)ξ + (b − e1)H1 + (c − e2)H2) → Hom(E, E)

→ H 0E(−ξ − bH1 − cH2) → H 1E((1 − α)ξ + (b − e1)H1 + (c − e2)H2)

→ Ext1(E, E) → H 1E(−ξ − bH1 − cH2) → · · · (8)

Now we consider the exact cohomology sequence

0 → H 0OX → H 0E(−ξ − bH1 − cH2)

→ H 0OX((α − 2)ξ − (2b − e1)H1 − (2c − e2)H2)

→ H 1OX → H 1E(−ξ − bH1 − cH2)

→ H 1OX((α − 2)ξ − (2b − e1)H1 − (2c − e2)H2) → · · ·associated to the exact sequence (7).

Sinceα − 2 < 0, we get

h0E(−ξ − bH1 − cH2) = h0OX = h0OY = 1.

Claim.

H 1OX((α − 2)ξ − (2b − e1)H1 − (2c − e2)H2) = 0.

348 L. Costa

Proof of the Claim..By duality we have

h = h1OX((α − 2)ξ − (2b − e1)H1 − (2c − e2)H2)

= h2OX(−αξ + (2b − e1 + d1 − 2)H1 + (2c − e2 + d2 − 2)H2).

If α = 1, by Lemma 2.2 we geth = 0. If α = 0, also applying Lemma 2.2 we have

h = h2OY ((2b − e1 + d1 − 2)l1 + (2c − e2 + d2 − 2)l2)

= h0OY ((−2b + e1 − d1)l1 + (−2c + e2 − d2)l2) = 0

where the last equality follows from the fact−2c + e2 − d2 < 0. Thus

h1OX((α − 2)ξ − (2b − e1)H1 − (2c − e2)H2) = 0

which finish the proof of the Claim.

From the Claim and the fact thath1OX = h1OY = 0 we get

H 1E(−ξ − bH1 − cH2) = 0.

Finally, we consider the cohomology exact sequence associated to (7)

0 → H 0OX((2 − α)ξ + (2b − e1)H1 + (2c − e2)H2)

→ H 0E((1 − α)ξ + (b − e1)H1 + (c − e2)H2)

→ H 0OXδ−→H 1OX((2 − α)ξ + (2b − e1)H1 + (2c − e2)H2)

→ H 1E((1 − α)ξ + (b − e1)H1 + (c − e2)H2) → H 1OX → · · · .

By Proposition 3.4; (a) we have

H 0E((1 − α)ξ + (b − e1)H1 + (c − e2)H2) = 0.

Hence,

h1E((1 − α)ξ + (b − e1)H1 + (c − e2)H2)

= h1OX((2 − α)ξ + (2b − e1)H1 + (2c − e2)H2) − 1.

Putting all this results together and using the exact sequence (8) we obtain

dim Ext1(E, E) = dim Hom(E, E) − h0E(−ξ − bH1 − cH2)

+ h1E((1 − α)ξ + (b − e1)H1 + (c − e2)H2)

= h1OX((2 − α)ξ + (2b − e1)H1 + (2c − e2)H2) − 1= dimP

where the last equality follows from Proposition 3.4.Since dimT[E]ML(2; c̃1, c̃2) = dimP, the moduli spaceML(2; c̃1, c̃2) is

smooth. Moreover, we have

ML(2; c̃1, c̃2) ∼= P(Ext1(OX((α − 1)ξ − (b − e1)H1−(c − e2)H2), OX(ξ + bH1 + bH2))).

Therefore, the moduli spaceML(2; c̃1, c̃2) is a non-empty, smooth, irreducible,rational, projective variety of dimensionh1OX(2 − α)ξ + (2b − e1)H1 + (2c −e2)H2) − 1, which proves the Theorem.ut


Remark 3.6.Using Bogomolov’s inequality

(c1(E)2 − 4c2(E))L2 < 0

the reader can see that the hypothesisb � 0 is not too restrictive.

We will end this work computing the Picard group and the Kodaira dimensionof these moduli spaces.

Corollary 3.7. Let X = P(E) → P1 × P

1 be a ruled Fano 3-fold. Fix integersα,e1, e2 ∈ {0, 1}, 2c > e2 − d2, b � 0, andL the ample divisor onX

L = (2 − α)ξ − (2b − β)H1 − bH2

beingβ = e1 + 2e2 − 4c − (2 − α)(d1 + d2) − 1. Consider

c̃1 = αξ + e1H1 + e2H2

c̃2 = (α − 1)ξ2 + (−2b + bα + e1)ξH1 + (−2c + cα + e2)ξH2+ (−2bc + be2 + ce1)H1H2.

Then

Pic(ML(2; c̃1, c̃2)) ∼= Z,

Kod(ML(2; c̃1, c̃2)) = ∞Proof. It follows from the fact that

ML(2; c̃1, c̃2) ∼= P(Ext1(OX((α − 1)ξ − (b − e1)H1− (c − e2)H2), OX(ξ + bH1 + bH2))). ut

Acknowledgements.I am very grateful to Rosa Maria Miró-Roig for giving me many goodideas and for her interesting comments that have been very useful for me. I was partiallysupported by DGICYT PB97-0893.

References

[BMR97] E. Ballico and R.M. Miró-Roig: A lower bound for the number of componentof the moduli schemes of stable rank 2 vector bundles on projective 3-folds.Preprint, 1997

[CMR98] L. Costa and R.M. Miró-Roig: Moduli spaces of vector bundles on higherdimensional varieties. Preprint, 1998

[Cos98] L. Costa: Phd. thesis. University of Barcelona, 1998[Har77] R. Hartshorne: Algebraic Geometry. GTM 52, 1977[MROL97] R.M. Miró-Roig and J.A. Orus-Lacort: On the smoothness of the moduli space

of mathematical instanton bundles. Compositio Math.105, 109–119, 1997[SW90] M. Szurek and A. Wisniewski. Fano bundles of rank 2 on surfaces. Compositio

Math.76, 295–305, 1990

Vector bundles on ruled Fano 3-folds

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