Introductionfaculty.missouri.edu/~qinz/paper/rk2.pdfIntroduction Recently there is a surge of...

STABLE RANK-2 BUNDLES ON CALABI-YAU MANIFOLDS

WEI-PING LI1 AND ZHENBO QIN2

1. Introduction

Recently there is a surge of research interest in the construction of stable vectorbundles on Calabi-Yau manifolds motivated by questions from string theory. Aninteresting aspect of the moduli spaces of stable sheaves on Calabi-Yau manifoldsis their relation to the higher dimensional gauge theory studied by Donaldson,R. Thomas and Tian et al. [D-T, Tho, Tia]. A holomorphic Casson invariantfor Calabi-Yau three-fold has been defined in [D-T, Tho]. This invariant requiresthe full description of the moduli spaces. Comparing with the work of stablesheaves on surfaces, our knowledge about stable sheaves on Calabi-Yau manifoldsis very limited. Examples of full moduli spaces are scarce. However, some ofsucsessful methods for stable bundles on surfaces have been extended to Calabi-Yau manifolds, for instance, the work of Friedman, Morgan and Witten [FMW] onelliptic fibrations, the work of Bridgeland and Maciocia [B-M] on Fourier-Mukaitransform for elliptic and K3-fibrations, and the work of Thomas [Tho] wherefull examples were computed by using K3-fibrations and the Serre construction,among others. The technique of chamber structures has been used quite often inthe theory of stable vector bundles on surfaces. In this paper, we make use ofchamber structures to construct full moduli spaces of stable sheaves with certainChern classes on some Calabi-Yau manifolds. As an application, we will computesome holomorphic Casson invariants.

To introduce our results, we recall the definitions of stabilities. Let L be anample line bundle on a smooth projective variety X of dimension n, and V be arank-r torsion-free sheaf on X. We say that V is (slope) L-stable if

c1(F ) · c1(L)n−1

rank(F )<

c1(V ) · c1(L)n−1

r

for any proper subsheaf F of V , and V is Gieseker L-stable if

χ(F ⊗ L⊗k)

rank(F )<

χ(V ⊗ L⊗k)

r, k � 0

for any proper subsheaf F of V . Similarly, we define L-semistability and GiesekerL-semistability by replacing the above strict inequalities < by inequalities ≤. For aclass c in the Chow group A∗(X), let ML(c) be the moduli space of L-stable rank-2

2000 Mathematics Subject Classification. Primary 14D20, 14J60; Secondary: 14F05, 14J32.Key words and phrases. Walls, chambers, vector bundles, Calabi-Yau manifolds.1Partially supported by the grants HKUST6170/99P and HKUST6114/02P.2Partially supported by an NSF grant.

1

2 WEI-PING LI AND ZHENBO QIN

bundles with total Chern class c, and let ML(c) be the moduli space of GiesekerL-semistable rank-2 torsion-free sheaves with total Chern class c.

Now let X = Pm × Pn where m and n are positive integers with (m + n) ≥ 3.Our main idea is to construct stable rank-2 vector bundles on X and study theirmoduli spaces, by using walls and chambers. Roughly speaking, the process goesas follows. We take a rank-2 bundle close to being unstable (in the sense of theBogomolov inequality), and move the polarization (i.e., the Kahler form) to a pointwhere the Bogomolov inequality fails, so that the bundle must have a destablizingsubsheaf. Using this, we get a presentation of the bundle which can be proved tobe stable for polarizations on the other side of an appropriate wall in the amplecone (i.e., the Kahler cone).

To put it in another way, given classes c1 ∈ A1(X) and c2 ∈ A2(X), we candefine chambers C of type (c1, c2) on the ample cone of X. For two ample linebundles L1 and L2 such that c1(L1) and c1(L2) are in the same chamber C, themoduli spaces ML1(c) and ML2(c) are isomorphic. On the other hand, if c1(L1)and c1(L2) lie in different chambers, one may be able to give a description of theset of stable rank-2 bundles which are stable with respect to L1 but unstable withrespect to L2.

In this paper, we consider two kinds of Calabi-Yau manifolds. The first one isa generic hyperplane section Y ⊂ P1 × P1 × Pn of type (2, 2, n + 1). Let π be therestriction to Y of the projection of P1 × P1 × Pn to the product of the last twofactors. Then, π : Y → P1×Pn is a double covering. Let (a, b) stand for the divisorc1(π

∗1OP1(a)⊗ π∗2OPn(b)) where π1 and π2 are the projections of P1 × Pn to its two

factors. Let LYr be the Q-line bundle OY (π∗(1, r)). For convenience, we adopt the

convention that e/0 = +∞ for e > 0. We shall prove the following result.

Theorem 1.1. Let n ≥ 2, ε1, ε2 = 0, 1, and k = (1 + ε1)(

n+2−ε2n

)− 1. Let

c = [1 + π∗(−1, 1)] · [1 + π∗(ε1 + 1, ε2 − 1)].

Then the moduli space MLYr(c) is empty when 0 < r < n(2 − ε2)/(2 + ε1), and is

isomorphic to Pk when n(2− ε2)/(2 + ε1) < r < n(2− ε2)/ε1.

It follows that when n = 2 and 2(2 − ε2)/(2 + ε1) < r < 2(2 − ε2)/ε1, theholomorphic Casson invariant λ(Y, LY

r , c, 2) of the Calabi-Yau three-fold Y is equalto (−1)k(k +1), where k = (1+ ε1)(4− ε2)(3− ε2)/2− 1. On the other hand, when0 < r < 2(2− ε2)/(2 + ε1), the holomorphic Casson invariant λ(Y, LY

r , c, 2) is zero.In particular, the holomorphic Casson invariants depend on polarizations.

The second kind of Calabi-Yau manifolds considered in this paper is a generichyperplane section S ⊂ P1 × Pn of type (2, n + 1). We obtain results for S similarto the above results described for Y . In particular, we show that the moduli spaceof Gieseker semistable rank-2 torsion-free sheaves on S for certain class c ∈ H∗(S)is empty with respect to some ample line bundles, and is isomorphic to a projectivespace with respect to other ample line bundles.

For the double cover Y of X, we pull back stable bundles on X to get LYr -stable

vector bundles on Y , as long as n(2 − ε2)/(2 + ε1) < r < n(2 − ε2)/ε1. A priori,

STABLE RANK-2 BUNDLES ON CALABI-YAU MANIFOLDS 3

these pulled-back bundles on Y form a subset of the Gieseker moduli space MLYr(c).

Our detailed analysis shows that this set of stable bundles is in fact the full modulispace on Y . Here the idea of chamber structures plays a crucial role again. For thecase of the hyperplane section S ⊂ X, we restrict stable bundles on X to S. Withsome careful examinations, we can show that these are all the Gieseker semistablesheaves on S with the given Chern classes.

R. Thomas kindly pointed to us that the method can also be applied to Calabi-Yau manifolds which are hyperplane sections of a Pn-bundle over P1.

The organization of the paper is as follows. In section 2, we study wall andchamber structures of X = Pm × Pn. In section 3, using chamber structures andstandard techniques, we construct stable vector bundles on X. In section 4, westudy the pulled-back stable bundles on the double cover Y of X and prove The-orem 1.1. In section 5, we study the restriction of the stable bundles constructedin section 3 to S and prove a theorem similar to Theorem 1.1. In section 6, wecompute the holomorphic Casson invariants for the Calabi-Yau three-folds.Acknowledgment: We thank R. Thomas and G. Tian for helpful discussions.The first author also thanks G. Tian, and the Mathematics departments in MITand the University of Missouri for their hospitalities during the spring and summerof 2001 where the major part of this work was done. We are grateful to the refereefor many valuable comments and suggestions.

2. Chamber structures of Pm × Pn

Let m and n be two positive integers with (m + n) ≥ 3. Let

X = Pm × Pn, (2.1)

and π1 and π2 be the two projections of X. Then, we have

A1(X) ∼= Num(X) = Z[H1 × Pn]⊕ Z[Pm ×H2] ∼= Z⊕ Z

where H1 and H2 stands for hyperplanes in Pm and Pn respectively. Moreover, wesee that the nef cone of X is the first quadrant in

H2(X; R) ∼= Num(X)⊗ R = R[H1 × Pn]⊕ R[Pm ×H2] ∼= R⊕ R.

Let Z be a codimension-2 cycle in X. Then, Z is of the form

Z = a[(H1)2 × Pn] + b[H1 ×H2] + c[Pm × (H2)

2], a, b, c ∈ Z. (2.2)

Let (d1, d2) and OX(d1, d2) denote formally d1(H1 × Pn) + d2(Pm × H2) andπ∗1OPm(d1)⊗π∗2OPn(d2) respectively. When d1 and d2 are rational numbers, (d1, d2)(respectively, OX(d1, d2)) is a Q-divisor (respectively, a Q-line bundle). Let

Lr = OX(1, r).


Then, Lr is ample if and only if r > 0, and we have the intersection number

(d1, d2) · c1(Lr)m+n−1

=[d1(H1 × Pn) + d2(Pm ×H2)

]·[(H1 × Pn) + r(Pm ×H2)

]m+n−1

=(m + n− 1)! rn−1

m! n!(md1r + nd2). (2.3)

Let V be a rank-2 torsion-free sheaf over X. Put c1 = c1(V ) and c2 = c2(V ).Assume that V is Lr1-semistable but Lr2-unstable, where r1, r2 > 0. Then V hasan Lr2-destablizing rank-1 subsheaf OX(F )⊗ IZ1 sitting in an exact sequence

0 → OX(F )⊗ IZ1 → V → OX(c1 − F )⊗ IZ2 → 0 (2.4)

where Z1 and Z2 are subschemes of codimension at least two. So F ·c1(Lr1)m+n−1 >

c1 ·c1(Lr1)m+n−1/2, i.e., 0 < (d1, d2)·c1(Lr2)

m+n−1 where (d1, d2) denotes the divisor2F − c1. Since V is Lr1-semistable, we also have (d1, d2) · c1(Lr1)

m+n−1 ≤ 0.Moreover, since c2(IZ1) ≥ 0 and c2(IZ2) ≥ 0, we see from (2.4) that c2 = c2(V ) ≥F (c1 − F ). It follows that (4c2 − c2

1) ≥ −(2F − c1)2 = −(d1, d2)

2. In summary, wehave proved the following:

(d1, d2) ≡ c1 (mod 2), (2.5)

(d1, d2) · c1(Lr1)m+n−1 ≤ 0 < (d1, d2) · c1(Lr2)

m+n−1, (2.6)

−(d1, d2)2 ≤ (4c2 − c2

1). (2.7)

In particular, we see from (2.6) and (2.3) that

(md1r1 + nd2) ≤ 0 < (md1r2 + nd2).

So d1 and d2 are nonzero integers with opposite signs. Intersecting (2.7) with thenumerically effective cycle (H1)

m−1 × (H2)n−1, we obtain

−2d1d2 ≤ (4c2 − c21) · ((H1)

m−1 × (H2)n−1). (2.8)

Therefore, there are only finitely many choices of such divisors (d1, d2).

Definition 2.1. Let X = Pm × Pn. Fix c1 ∈ A1(X) and c2 ∈ A2(X).

(i) A wall of type (c1, c2) in H2(X; R) is of the form

W (d1,d2) = {(a, b)| a > 0, b > 0, md1(b/a) + nd2 = 0}where d1 and d2 are nonzero integers with opposite signs satisfying

(d1, d2) ≡ c1 (mod 2), −(d1, d2)2 ≤ (4c2 − c2

1);

(ii) A chamber of type (c1, c2) is a connected component in the ample cone ofX cut out by all the walls of type (c1, c2).

Let ξ = (d1, d2) define a (nonempty) wall. Then, the wall W ξ is the ray in thefirst quadrant with slope −nd2/(md1). For r1 > 0 and r2 > 0, we say that (r1, r2)lies above (respectively, below) the wall W ξ if r2/r1 > −nd2/(md1) (respectively,r2/r1 < −nd2/(md1)). A chamber C lies above (respectively, below) the wall W ξ ifthere exists (r1, r2) ∈ C lying above (respectively, below) the wall W ξ.

Lemma 2.2. Fix c1 ∈ A1(X) and c2 ∈ A2(X).


(i) There are only finitely many walls and chambers of type (c1, c2).(ii) If there exists a rank-2 torsion-free sheaf V with c1(V ) = c1 and c2(V ) = c2

such that V is semistable with respect to Lr1 but unstable with respect toLr2, then c1(Lr1) and c1(Lr2) are separated by a wall of type (c1, c2).

Proof. Follows from the paragraph preceding Definition 2.1. �

The following is a partial converse of Lemma 2.2 (ii).

Lemma 2.3. Fix c1 ∈ A1(X) and c2 ∈ A2(X). Let C− and C+ be two adjacentwalls of type (c1, c2) sharing a common wall W . Assume that W is represented bythe class ξ = (2F − c1), and that ξ ·Dm+n−1

− < 0 < ξ ·Dm+n−1+ for some D− ∈ C−

and D+ ∈ C+. Let V be a rank-2 bundle sitting in a nonsplitting extension:

0 → OX(F ) → V → OX(c1 − F )⊗ IZ → 0

where Z is a codimension-2 locally complete intersection. Assume c2(V ) = c2.Then V is stable with respect to OX(D−) but unstable with respect to OX(D+).

Proof. Follows from the argument in the proof of the Theorem 1.2.3 in [Qin]. �

Next, as an application of the concepts of walls and chambers for X, we study thesecond Chern classes of semistable rank-2 bundles on X. The results will be used inlater sections. Recall that if V is a semistable rank-2 torsion-free sheaf on a smoothprojective surface, then the Bogomolov inequality says that (4c2(V )− c1(V )2) ≥ 0.In general, we see from [MR1, MR2] that if V is an L-semistable rank-2 torsion-freesheaf on a smooth projective variety Y , then

(4c2(V )− c1(V )2) · c1(L)dim Y−2 ≥ 0. (2.9)

Lemma 2.4. Let n ≥ 2, and V be a rank-2 bundle on X = Pm × Pn with

(4c2(V )− c1(V )2) = a[(H1)2 × Pn] + b[H1 ×H2] + c[Pm × (H2)

2].

Assume that V is Lr1-semistable for some ample line bundle Lr1.

(i) If c < 0, then b ≥ 2 and c ≥ −b2/4.

(ii) If b < 0, then m ≥ 2, a > 0, c > 0, and b ≥ −2√

(m− 1)(n− 1)ac/(mn).

Proof. (i) Let r > 0. The intersection number (4c2(V )− c1(V )2) · c1(Lr)m+n−2 is

(m + n− 2)! rn−2

m! n![m(m− 1)r2a + mnrb + n(n− 1)c].

Since c < 0, letting r → 0+, we see from (2.9) that there exists r0 > 0 such thatthe bundle V is Lr0-unstable. So V is semistable with respect to some ample linebundle but unstable with respect to some other ample line bundle.

Now there is ξ = (d1, d2) such that ξ defines a wall of type (c1(V ), c2(V )) and

−(d1, d2)2 ≤ (4c2(V )− c1(V )2). (2.10)

Intersecting (2.10) with (H1)m × (H2)

n−2 and (H1)m−1 × (H2)

n−1 respectively, weobtain −d2

2 ≤ c and −2d1d2 ≤ b. Since ξ defines a wall, d1 and d2 are nonzero withopposite signs. So b ≥ 2, |d2| ≤ b/2, and c ≥ −d2

2 ≥ −b2/4.


(ii) Since (4c2(V ) − c1(V )2) · ((H1)m−1 × (H2)

n−1) = b < 0, we see from (2.8)and Definition 2.1 (i) that there exists no wall of type (c1(V ), c2(V )). So V issemistable with respect to every ample line bundle. By (2.9), for every r > 0,

(4c2(V )− c1(V )2) · (Lr)m+n−2 ≥ 0

which is equivalent to m(m− 1)r2a + mnrb + n(n− 1)c ≥ 0. Letting r → 0+, weobtain c ≥ 0. In fact, c 6= 0 since b < 0. So c > 0. Similarly, letting r → +∞, wesee that m ≥ 2 and a > 0. Consider the quadratic polynomial

f(r) = m(m− 1)a r2 + mnb r + n(n− 1)c

defined over R. We check that its minimal value is equal to

n(n− 1)c− mn2b2

4(m− 1)a

reached at n(−b)/(2(m− 1)a). Since n(−b)/(2(m− 1)a) > 0, we must have

n(n− 1)c− mn2b2

4(m− 1)a≥ 0.

Noting b < 0, we conclude that b ≥ −2√

(m− 1)(n− 1)ac/(mn). �

Remark 2.5. Let n ≥ 2 and c = 1+ c1 + c2 where c1 ∈ A1(X) and c2 ∈ A2(X). Theproof of Lemma 2.4 (i) shows that if the coefficient of [Pm × (H2)

n−2] of (4c2 − c21)

is negative, then MLr0(c) is empty for sufficiently small positive number r0.

3. Stable rank-2 bundles on Pm × Pn

3.1. The general setup and a theorem of Serre.Let m + n ≥ 3. As applications of Lemma 2.3, we are interested in constructing

stable rank-2 bundles V on X = Pm × Pn sitting in extensions of the form:

0 → OX(F ) → V → OX(c1 − F )⊗ IZ → 0 (3.1)

where Z is a codimension-2 locally complete intersection, and ξ = (2F − c1) =(d1, d2) defines a wall of type (c1, c2(V )).

Recall that we have the local-to-global exact sequence

0 → H1(X,Hom(OX(c1 − F )⊗ IZ ,OX(F )))

→ Ext1(OX(c1 − F )⊗ IZ ,OX(F ))

→ H0(X, Ext1(OX(c1 − F )⊗ IZ ,OX(F )))

→ H2(X,Hom(OX(c1 − F )⊗ IZ ,OX(F ))). (3.2)

Note that Hom(OX(c1 − F ) ⊗ IZ ,OX(F )) ∼= OX(2F − c1). Also, since Z is acodimension-2 locally complete intersection, we see from page 37 in [Fri] that

Ext1(OX(c1 − F )⊗ IZ ,OX(F )) ∼= det(NZ)⊗OX(2F − c1)


where NZ is the normal bundle of Z in X. So (3.2) can be simplified to

0 → H1(X,OX(ξ)) → Ext1(OX(c1 − F )⊗ IZ ,OX(F ))∂→ H0(X, det(NZ)⊗OX(ξ)) → H2(X,OX(ξ)). (3.3)

By a theorem of Serre (see the Theorem 8 in Chapter 2 of [Fri]), the sheaf Vin (3.1) corresponding to an extension class e ∈ Ext1(OX(c1 − F ) ⊗ IZ ,OX(F ))is locally free if and only if ∂(e) ∈ H0(X, det(NZ)⊗OX(ξ)) generates the torsionsheaf det(NZ) ⊗ OX(ξ) supported on Z. In the following, we shall present twoconstructions.

3.2. The case when Z is empty.First of all, we consider the easy case when Z is empty. We may assume that

m ≤ n. So n ≥ 2 since m + n ≥ 3. Now, by the Kunneth formula, we have

Ext1(OX(c1 − F ),OX(F )) ∼= H1(X,OX(ξ))∼= H1(Pm,OPm(d1))⊗H0(Pn,OPn(d2)).

So Ext1(OX(c1 − F ),OX(F )) 6= 0 if and only if m = 1, d1 ≤ −2 and d2 > 0.

Lemma 3.1. Fix a divisor c1 on X = P1×Pn. Let ξ = (d1, d2) ≡ c1 (mod 2) withd1 ≤ −2 and d2 > 0, F = (c1 + ξ)/2, and k = (−d1 − 1)

(n+d2

n

)− 1. Then,

(i) all the rank-2 bundles V sitting in nonsplitting extensions

0 → OX(F ) → V → OX(c1 − F ) → 0 (3.4)

form a Pk-family of nonisomorphic bundles on P1 × Pn with c1(V ) = c1,(4c2(V )− c1(V )2) = −ξ2, dim Ext1(V, V ) = k, and injections

0 → Ext2(V, V ) → H2(X,OX(−ξ)).

(ii) these bundles V are Lr-stable, where (1, r) ∈ C and C is the chamber of type(c1, c2(V )) such that (1, r)n · ξ < 0 and the closure C of C contains W ξ.

Proof. Note that Hom(OX(F ), V ) ∼= C and dim Ext1(OX(c1 − F ),OX(F )) =k + 1 ≥ 1. So the set of all these rank-2 bundles V (up to isomorphisms)are parametrized by Pk. Since ξ defines a wall of type (c1, c2(V )), we see fromLemma 2.3 that these bundles V are Lr-stable whenever (1, r) ∈ C.

Next, tensoring the exact sequence (3.4) by V ∗ ∼= V ⊗ OX(−c1) and takingcohomology, we obtain the long exact sequence

0 → H0(X, V ⊗OX(F − c1)) → Hom(V, V ) → H0(X, V ⊗OX(−F ))

→ H1(X, V ⊗OX(F − c1)) → Ext1(V, V ) → H1(X, V ⊗OX(−F ))

→ H2(X, V ⊗OX(F − c1)) → Ext2(V, V ) → H2(X, V ⊗OX(−F )). (3.5)

Since (3.4) is nonsplitting, H0(X, V ⊗ OX(F − c1)) = 0. Since V is stable andH0(X, V ⊗OX(−F )) ∼= C, the exact sequence (3.5) can be simplified to:

0 → H1(X, V ⊗OX(F − c1)) → Ext1(V, V ) → H1(X, V ⊗OX(−F ))

→ H2(X, V ⊗OX(F − c1)) → Ext2(V, V ) → H2(X, V ⊗OX(−F )). (3.6)


Tensoring (3.4) by OX(F − c1) and taking cohomology groups, we see that

H2(X, V ⊗OX(F − c1)) ∼= H2(X,OX(ξ)) = 0

and h1(X, V ⊗OX(F − c1)) = h1(X,OX(ξ))− 1 = k. Similarly, we have

H1(X, V ⊗OX(−F )) ∼= H1(X,OX(−ξ)) = 0

and an injection 0 → H2(X, V ⊗OX(−F )) → H2(X,OX(−ξ)). By (3.6),

dim Ext1(V, V ) = h1(X, V ⊗OX(F − c1)) = k,

and we have an injection 0 → Ext2(V, V ) → H2(X,OX(−ξ)). �

Remark 3.2. Let notations be as in Lemma 3.1. Assume further that either n ≥ 3,or n = 2 and d2 < 3. Then, H2(X,OX(−ξ)) = 0. Thus, Ext2(V, V ) = 0.

Theorem 3.3. Fix a divisor c1 on X = P1 × Pn with n ≥ 2. Let ξ = (d1, d2) ≡ c1

(mod 2) with d1 ≤ −2 and d2 > 0, c2 = (c21 − ξ2)/4, c = 1 + c1 + c2, and

k = (−d1 − 1)

(n + d2

n

)− 1.

(i) Let C be the chamber of type (c1, c2) such that (1, r)n · ξ < 0 for some(1, r) ∈ C and C ⊃ W ξ. When (1, r) ∈ C, MLr(c) is isomorphic to Pk andconsists of all the rank-2 bundles V sitting in nonsplitting extensions:

0 → OX((c1 + ξ)/2) → V → OX((c1 − ξ)/2) → 0. (3.7)

(ii) The moduli space MLr0(c) is empty when (1, r0) lies on or below W ξ.

Proof. First of all, we claim that every bundle V ∈ MLr(c) sits in a nonsplittingexact sequence of the form (3.7). Indeed, we have

(4c2 − c21) = −ξ2 = −2d1d2[p×H]− d2

2[P1 ×H2]

where p denotes a point in P1 and H denotes a hyperplane in Pn. So the coefficientof [P1 × H2] in (4c2 − c2

1) is negative. We see from the proof of Lemma 2.4 (i)that V is Lr0-unstable when the positive number r0 is sufficiently small. We mayfurther assume that r0 < r. Thus, V sits in a nonsplitting exact sequence:

0 → OX(G) → V → OX(c1 −G)⊗ IZ → 0 (3.8)

where Z is a codimension-2 locally complete intersection in X = P1 × Pn, andη = 2G− c1 defines a wall of type (c1, c2) such that

(1, r)n · η < 0 < (1, r0)n · η. (3.9)

Put η = (d1, d2). Then we see from (2.3) and (3.9) that

d1r + nd2 < 0 < d1r0 + nd2.

Since r0 < r, we get d1 < 0. So d2 > 0. Since −η2 ≤ (4c2 − c21), we obtain

−2d1d2 ≤ −2d1d2 and −(d2)2 ≤ −d2

2. So (−d1)d2 ≤ (−d1)d2 and d2 ≥ d2. Sincethe boundary of C contains W ξ and W η strictly separates (1, r) ∈ C and (1, r0)where r0 is sufficiently small, the slope of W ξ is at least equal to that of W η. Hencewe conclude from Definition 2.1 (i) that −(nd2)/d1 ≥ −(nd2)/d1, i.e., d2/(−d1) ≤


d2/(−d1). Combining with (−d1)d2 ≤ (−d1)d2 and d2 ≥ d2, we obtain d2 = d2 and

d1 = d1. Therefore, η = ξ, G = (c1 + ξ)/2, Z = ∅, and (3.8) becomes (3.4).Note that the preceding paragraph also shows that the moduli space MLr0

(c) is

empty whenever (1, r0) lies on or below the wall W ξ.In view of Lemma 3.1, there is a bijective morphism Φ : Pk → MLr(c). In

particular, the dimension of MLr(c) is k. By Lemma 3.1 (i), the Zariski tangentspace Ext1(V, V ) of MLr(c) at every point V has dimension k. Thus, MLr(c) issmooth. By Zariski’s Main Theorem, Φ must be an isomorphism. �

Next, we look at an example of ξ for c1 = (ε1, ε2) where ε1, ε2 = 0, 1. Forconvenience, we take the convention that e/0 = +∞ when e > 0.

Corollary 3.4. Let X = P1 × Pn with n ≥ 2, c1 = (ε1, ε2) where ε1, ε2 = 0, 1,c2 = (−1, d) · (ε1 + 1, ε2 − d) with d ≥ 1, and c = 1 + c1 + c2. Consider all thebundles V sitting in nonsplitting extensions of the form:

0 → OX(−1, d) → V → OX(ε1 + 1, ε2 − d) → 0. (3.10)

(i) We have c(V ) = c and dim Ext1(V, V ) = k := (1 + ε1)

(n + 2d− ε2

n

)− 1.

Moreover, Ext2(V, V ) = 0 when n ≥ 3, or n = 2 and d = 1.(ii) The moduli space MLr(c) is isomorphic to Pk and consists of all the rank-2

bundles V sitting in the nonsplitting extensions (3.10) when

n(2d− ε2)/(2 + ε1) < r < n(2d− ε2)/ε1. (3.11)

(iii) The moduli space MLr(c) is empty when 0 < r ≤ n(2d− ε2)/(2 + ε1).

Proof. Put ξ = (−2 − ε1, 2d − ε2). Then ξ satisfies the conditions in Lemma 3.1and Theorem 3.3. Now (i) follows from Lemma 3.1 (i) and Remark 3.2. Note thatW ξ is a wall of type (c1, c2), and has slope n(2d− ε2)/(2+ ε1). So (iii) follows fromTheorem 3.3 (ii). Moreover, letting C be the chamber in Theorem 3.3 (i), we seethat (ii) follows from Theorem 3.3 (i) if we can prove that (3.11) is equivalent to(1, r) ∈ C. In the following, we prove this only for c1 = (0, 0), (1, 0) since the case(0, 1) is similar to the case (0, 0) and the case (1, 1) is similar to the case (1, 0).

Let c1 = (0, 0) = 0. We claim that there is no wall lying between W ξ and thepositive side of the [P1×H]-axis in H2(X; R), i.e., the region between W ξ and thepositive side of the [P1 ×H]-axis is the chamber C, as illustrated below:

-

6

0

[P1 ×H]

[p× Pn]

W ξ

C

H2(X; R)


Indeed, suppose that ξ defines a wall of type (0, c2) lying between W ξ and the

positive side of the [P1 × H]-axis. Let ξ = (d1, d2). We may assume that d1 < 0

and d2 > 0. Since ξ ≡ c1 (mod 2), we have d1 ≤ −2. Comparing the slopes of W ξ

and W ξ, we see that d2/(−d1) > d, i.e., d2 > d(−d1). By comparing the coefficients

of [p×H] in 4c2 ≥ −ξ2, we obtain the contradiction 8d ≥ −2d1d2 > 2dd21 ≥ 8d.

Next, let c1 = (1, 0). Then, (−1, 2d) defines a wall of type (c1, c2) with slope

2nd. So it suffices to show that if W ξ is the wall of type (c1, c2) lying immediately

above W ξ, then ξ = (−1, 2d). This is illustrated below:

-

6

��

0

[P1 ×H]

[p× Pn]

W ξW ξ

C

H2(X; R)

Indeed, put ξ = (d1, d2) with d1 < 0 and d2 > 0. Since ξ ≡ c1 (mod 2), d1 is

odd and d2 is even. Comparing the slopes of W ξ and W ξ, we get d2 > 2d(−d1)/3.

Comparing the coefficients of [p×H] and [P1×H2] in (4c2− c21) ≥ −ξ2, we obtain

12d ≥ −2d1d2 > 4dd21/3 and −4d2 ≥ −d2

2. Thus, d1 > −3 and d2 ≥ 2d. So

d1 = −1. In view of 12d ≥ −2d1d2, we obtain 2d ≤ d2 ≤ 6d. Since W ξ is the walllying immediately above W ξ, we have d2 = 2d and ξ = (−1, 2d). �


3.3. A construction when Z is nonempty.This subsection is independent of the remaining sections. Here we consider a

situation when the codimension-2 cycle Z in (3.1) is nonempty. Let m = 1 (soX = P1 × Pn and n ≥ 2). Let s be a positive integer, p1, . . . , ps be distinct pointsin P1, and Y1, . . . , Ys be degree-d smooth hypersurfaces in Pn. For each i, pi×Yi isthe complete intersection of pi × Pn and P1 × Yi, and the normal bundle of pi × Yi

in X is isomorphic to the restriction of OX(1, 0) ⊕ OX(0, d) to pi × Yi. Let Z bethe disjoint union of these pi × Yi, 0 ≤ i ≤ s. Then,

NZ = (OX(1, 0)⊕OX(0, d))|Z .

Therefore, we have det(NZ) ∼= OX(1, d)|Z ∼= OX(0, d)|Z . For ξ = (d1, d2), we getOX(ξ)|Z = OX(d1, d2)|Z ∼= OX(0, d2)|Z and

H0(X, det(NZ)⊗OX(ξ)) ∼= H0(Z,OX(0, d2 + d)|Z). (3.12)

Proposition 3.5. Fix a divisor c1 on P1 × Pn and a positive integer s. Let ξ =(d1, d2) such that ξ ≡ c1 (mod 2), d1 > 0 and d2 < 0. Assume either n ≥ 3, orn = 2 and d2 > −3. Let k = s

(1 +

(n−d2

n

))− 1. Then,

(i) there exists a k-dimensional family of nonisomorphic rank-2 bundles V onP1 × Pn with c1(V ) = c1 and (4c2(V ) − c1(V )2) = −ξ2 + 4s(−d2)[p × H].Moreover, dim Ext1(V, V ) = (2s + d1 − 1)

(n−d2

n

)− 1, and Ext2(V, V ) = 0.

(ii) these bundles V are Lr-stable, where (1, r) ∈ C and C is the chamber of type(c1, c2(V )) such that (1, r)n · ξ < 0 and the closure C contains W ξ.

Proof. Let F = (c1 + ξ)/2, d = (−d2), and Z be chosen as at the beginningof this subsection. Then, 2F − c1 = ξ. Note that H1(X,OX(ξ)) = 0. Also,H2(X,OX(ξ)) = 0 since either n ≥ 3, or n = 2 and d2 > −3. By (3.12),

H0(X, det(NZ)⊗OX(ξ)) ∼= H0(Z,OZ) ∼=s⊕

i=1

C.

By (3.3) and Serre’s theorem, there exist rank-2 bundles V sitting in (3.1):

0 → OX(F ) → V → OX(c1 − F )⊗ IZ → 0. (3.13)

Since ξ defines a wall of type (c1, c2(V )), we conclude from Lemma 2.3 that thesebundles V are Lr-stable. Note that Hom(OX(F ), V ) ∼= C. Therefore, the numberof moduli of these stable rank-2 bundles is equal to:

(s− 1) + #(moduli of Z) = (s− 1) + s ·#(moduli of pi and Yi) = k.

It is clear that c1(V ) = c1 and (4c2(V )− c1(V )2) = −ξ2 + 4s(−d2)[p×H].Tensoring the exact sequence (3.13) by V ∗ ∼= V ⊗OX(−c1) and taking cohomol-

ogy, we obtain the long exact sequence:

0 → H0(X, V ⊗OX(F − c1)) → Hom(V, V ) → H0(X, V ⊗OX(−F )⊗ IZ)

→ H1(X, V ⊗OX(F − c1)) → Ext1(V, V ) → H1(X, V ⊗OX(−F )⊗ IZ)

→ H2(X, V ⊗OX(F − c1)) → Ext2(V, V ) → H2(X, V ⊗OX(−F )⊗ IZ).


Note that H0(X, V ⊗ OX(F − c1)) = 0, Hom(V, V ) ∼= C, and there exists aninjection 0 → H0(X, V ⊗ OX(−F ) ⊗ IZ) → H0(X, V ⊗ OX(−F )) ∼= C. So theabove long exact sequence can be simplified to:

0 → H1(X, V ⊗OX(F − c1)) → Ext1(V, V ) → H1(X, V ⊗OX(−F )⊗ IZ)

→ H2(X, V ⊗OX(F − c1)) → Ext2(V, V ) → H2(X, V ⊗OX(−F )⊗ IZ).

Next, we study H i(X, V ⊗ OX(F − c1)) for i = 1, 2. Since H1(X,OX(ξ)) =H2(X,OX(ξ)) = 0, we see from the exact sequence

0 → OX(ξ) → V ⊗OX(F − c1) → IZ → 0

that H1(X, V ⊗OX(F − c1)) ∼= H1(X, IZ) and

0 → H2(X, V ⊗OX(F − c1)) → H2(X, IZ).

¿From the exact sequence 0 → IZ → OX → OZ → 0, we get

h1(X, IZ) = h0(X,OZ)− h0(X,OX) = s− 1

and H2(X, IZ) ∼= H1(X,OZ) ∼= ⊕si=0H

1(Yi,OYi). ¿From the exact sequence

0 → OPn(d2) → OPn → OYi→ 0,

we obtain H1(Yi,OYi) ∼= H2(Pn,OPn(d2)) = 0. In summary, we have

h1(X, V ⊗OX(F − c1)) = s− 1

and H2(X, V ⊗OX(F − c1)) = 0. So we see from the previous paragraph that

dim Ext1(V, V ) = (s− 1) + h1(X, V ⊗OX(−F )⊗ IZ), (3.14)

0 → Ext2(V, V ) → H2(X, V ⊗OX(−F )⊗ IZ). (3.15)

Now we compute H i(X, V ⊗OX(−F )) for i = 1, 2. ¿From the exact sequence

0 → OX → V ⊗OX(−F ) → OX(−ξ)⊗ IZ → 0, (3.16)

we get H i(X, V ⊗OX(−F )) ∼= H i(X,OX(−ξ)⊗ IZ) for i = 1, 2. ¿From the exactsequence 0 → OX(−ξ)⊗ IZ → OX(−ξ) → OX(−ξ)|Z → 0, we obtain

0 → H0(X,OX(−ξ)|Z) → H1(X,OX(−ξ)⊗ IZ) → H1(X,OX(−ξ))

→ H1(X,OX(−ξ)|Z) → H2(X,OX(−ξ)⊗ IZ) → H2(X,OX(−ξ)). (3.17)

Note that H2(X,OX(−ξ)) = 0 and OX(−ξ)|Z ∼= ⊕si=0OYi

(−d2). ¿From the exactsequence 0 → OPn → OPn(−d2) → OYi

(−d2) → 0, we obtain H1(Yi,OYi(−d2)) = 0

and h0(Yi,OYi(−d2)) = h0(X,OPn(−d2))− 1. Combining with (3.17), we conclude

H2(X, V ⊗OX(−F )) ∼= H2(X,OX(−ξ)⊗ IZ) = 0, (3.18)

h1(X, V ⊗OX(−F )) = h1(X,OX(−ξ)⊗ IZ)

= s

((n− d2

n

)− 1

)+ h1(X,OX(−ξ)). (3.19)

To compute H i(X, V ⊗OX(−F )⊗ IZ) for i = 1 and 2, consider:

0 → V ⊗OX(−F )⊗ IZ → V ⊗OX(−F ) → V ⊗OX(−F )|Z → 0. (3.20)


Recall that IZ/I2Z = (NZ)∗ ∼= (OX ⊕ OX(0, d2))|Z . Restricting (3.16) to Z, we

obtain a surjection V ⊗OX(−F )|Z → IZ/I2Z ⊗OX(−ξ) → 0. So we must have

V ⊗OX(−F )|Z ∼= IZ/I2Z ⊗OX(−ξ) ∼=

s⊕i=0

(OYi(−d2)⊕OYi

).

Therefore, the exact sequence (3.20) can be rewritten as

0 → V ⊗OX(−F )⊗ IZ → V ⊗OX(−F ) →s⊕

i=0

(OYi(−d2)⊕OYi

) → 0. (3.21)

Recall H1(Yi,OYi) = H1(Yi,OYi

(−d2)) = 0 from above. Taking cohomology from(3.21) and using (3.18) and (3.19), we obtain H2(X, V ⊗OX(−F )⊗ IZ) = 0 and

h1(X, V ⊗OX(−F )⊗ IZ)

=s∑

i=1

h0(Yi,OYi(−d2)⊕OYi

) + h1(X, V ⊗OX(−F ))

= (2s + d1 − 1)

(n− d2

n

)− s.

Now our results for Ext1(V, V ) and Ext2(V, V ) follow from (3.14) and (3.15). �

Fix a divisor c1 on X = P1 × Pn with n ≥ 2. Then there exists ξ = (d1, d2)such that ξ ≡ c1 (mod 2), d1 > 0 and −3 < d2 < 0. By letting s → +∞ inProposition 3.5, we see that there exist stable rank-2 bundles V on P1 × Pn suchthat c1(V ) = c1, the coefficient of [p × H] in c2(V ) is arbitrarily large, while thecoefficient of [P1 ×H2] in c2(V ) is fixed.

Let notations be as in Proposition 3.5, c2 = (c21 − ξ2)/4 + s(−d2)[p × H], and

c = 1 + c1 + c2. Then the irreducible component M of the moduli space MLr(c)containing the bundles V is smooth at the points V with dimension

(2s + d1 − 1)

(n− d2

n

)− 1

which is larger than k = s(1 +

(n−d2

n

))− 1. It would be interesting to construct

bundles representing a nonempty Zariski open and dense subset of MLr(c).

4. Stable rank-2 bundles on Calabi-Yau manifolds via double covering

Let n ≥ 2. Take a generic divisor Y of type (2, 2, n + 1) in Z = P1 × P1 × Pn.Then, Y is a smooth Calabi-Yau (n+1)-fold. By the Lefschetz hyperplane theorem,Pic(Y ) ∼= Pic(P1 × P1 × Pn). Let πi be the projection from P1 × P1 × Pn to thei-th factor. Then the projection π = (π2 × π3)|Y is a double covering from Y toX = P1 × Pn. For simplicity, we use OY (a, b, c) to represent the restriction ofπ∗1OP1(a)⊗ π∗2OP1(b)⊗ π∗3OPn(c) to Y . Put LY

r = OY (0, 1, r) = π∗Lr.Note that π : Y → X = P1× Pn is a ramified double covering with the ramifica-

tion locus B ⊂ X being a smooth divisor of type (4, 2n + 2). In particular,

π∗OY∼= OX ⊕OX(−2,−n− 1). (4.1)


Lemma 4.1. H1(Y, π∗OX(a, b)) ∼= H1(X,OX(a, b)) when b < (n + 1).

Proof. By the projection formula and (4.1), we obtain

H1(Y, π∗OX(a, b)) ∼= H1(X,OX(a, b)⊗ π∗OY )∼= H1(X,OX(a, b)⊕OX(a− 2, b− n− 1)) ∼= H1(X,OX(a, b))

since H1(X,OX(a− 2, b− n− 1)) = 0 when n ≥ 2 and b < (n + 1). �

Now we study stable rank-2 torsion-free sheaves on Y with c1 = (0, ε1, ε2)|Y .

Lemma 4.2. Let n ≥ 2. Let ε1, ε2 = 0, 1, and r < n(2− ε2)/ε1. Assume that E isan LY

r -stable rank-2 torsion-free sheaf on Y with

c(E) = c := [1 + π∗(−1, 1)] · [1 + π∗(ε1 + 1, ε2 − 1)]. (4.2)

Then, r > n(2− ε2)/(2 + ε1) and E sits in a nonsplitting extension

0 → OY (0,−1, 1) → E → OY (0, ε1 + 1, ε2 − 1) → 0. (4.3)

Proof. Notice that c2(E) = π∗((2 + ε1 − ε2)[p × H] − (1 − ε2)[P1 × H2]

)and

c1(E) = π∗(ε1, ε2). So (4c2(E)− c1(E)2) · c1(LYr0

)n−1 is equal to

2(2− ε2)rn−20 [2(2 + ε1)r0 − (2− ε2)(n− 1)].

By (2.9), E is LYr0

-unstable if 0 < r0 < (2− ε2)(n− 1)/(2(2 + ε1)). Fix such an r0

with r0 < r. Since E is LYr -stable, there is a nonsplitting extension

0 → OY (a, b, c)⊗ IZ1 → E → OY (−a, ε1 − b, ε2 − c)⊗ IZ2 → 0 (4.4)

such that OY (a, b, c)⊗ IZ1 destablizes E with respect to LYr0

, where Z1 and Z2 arecodimension at least two subschemes of Y . Therefore, c1(OY (a, b, c)) · c1(L

Yr0

)n >c1(E) · c1(L

Yr0

)n/2, which can be simplified into

n[(2c− ε2) + (n + 1)a] + (2a + 2b− ε1)r0 > 0. (4.5)

On the other hand, since E is LYr -stable, we must have

n[(2c− ε2) + (n + 1)a] + (2a + 2b− ε1)r < 0. (4.6)

Calculating the second Chern class from the exact sequence (4.4), we get

OY (a, b, c) · OY (−a, ε1 − b, ε2 − c) ≤ c2(E) (4.7)

since c2(IZ1) and c2(IZ2) are effective cycles. Regarding (4.7) as an inequality ofcycles in Z and comparing the coefficients of [p× p×H] and [p× P1 ×H2] yield

[2a + (2b− ε1)](2c− ε2) + (n + 1)a(2b− ε1) ≥ −(ε1 + 2)(2− ε2), (4.8)

[(2c− ε2) + 2(n + 1)a](2c− ε2) ≥ (2− ε2)2. (4.9)

Since 0 < r0 < r, we see from (4.5) and (4.6) that (2c− ε2) + (n + 1)a > 0 and(2a+2b− ε1) < 0. By (4.9), (2c− ε2)+2(n+1)a and (2c− ε2) have the same sign,and so must be both positive. In particular, c ≥ 1. By (4.8),

(n + 1)a(2b− ε1) ≥ −[2a + (2b− ε1)](2c− ε2)− (ε1 + 2)(2− ε2). (4.10)

In the following, we consider the cases ε1 = 0 and ε1 = 1 separately.


Assume ε1 = 0. By (4.10), (n + 1)a(2b) ≥ 0. Together with (2a + 2b) < 0, thisimplies either a < 0 and b ≤ 0, or a = 0 and b < 0. If a < 0 and b ≤ 0, then we seefrom (4.8) that −(2−ε2) ≤ (a+b)(2c−ε2)+(n+1)ab = a(2c−ε2)+[(2c−ε2)+(n+1)a]b ≤ a(2c− ε2) ≤ −(2c− ε2) ≤ −(2− ε2). So a = −1 and c = 1, contradicting to(2c− ε2) + (n + 1)a ≥ 1 and n ≥ 2. If a = 0 and b < 0, then b(2c− ε2) ≥ −(2− ε2)by (4.8). Since b(2c−ε2) ≤ −(2c−ε2) ≤ −(2−ε2), we must have b = −1 and c = 1.By (4.2) and (4.4), we obtain c(OZ1(0,−1, 1))c(OZ2(0, 1, ε2−1)) = 1. Thus Z1 andZ2 are empty, and (4.4) becomes (4.3). Note from (4.6) that r > n(2− ε2)/2.

Next, assume ε1 = 1. By (4.10), (n + 1)a(2b − 1) ≥ −(2a + 2b − 1)(2c − ε2) −3(2 − ε2) ≥ 1 − 6 = −5. So a(2b − 1) ≥ −1 since n ≥ 2. If a(2b − 1) = −1,then we see from 2a + (2b − 1) < 0 that a = −1 and b = 1. By (4.10) again,(2c− ε2) ≤ 3(2− ε2)− (n + 1) ≤ 5− n contradicting to (2c− ε2) + 2(n + 1)a ≥ 1and n ≥ 2. Therefore, we must have a(2b − 1) ≥ 0. Since 2a + (2b − 1) < 0, weconclude that either a < 0 and (2b − 1) ≤ 0, or a = 0 and (2b − 1) < 0. As inthe previous paragraph, we see that a = 0, b = 0 or −1. If b = 0, then we obtainfrom (4.6) that r > n(2c − ε2) ≥ n(2 − ε2) contradicting to our assumption thatr < n(2 − ε2). Therefore, b = −1. As in the previous paragraph again, we verifythat c = 1, Z1 and Z2 are empty, (4.4) becomes (4.3), and r > n(2− ε2)/3. �

Remark 4.3. Let n ≥ 2, ε1, ε2 = 0, 1, and c be given by (4.2).

(i) Let n(2− ε2)/(2 + ε1) < r < n(2− ε2)/ε1, and E ∈ MLYr(c). Then E is LY

r -semistable. The same proof above also shows that E sits in a nonsplittingextension (4.3). In particular, by Lemma 4.5 below, E is LY

r -stable.(ii) Similarly, we see that MLY

r(c) is empty when 0 < r < n(2− ε2)/(2 + ε1).

Lemma 4.4. Let notations be the same as in Lemma 4.2. Then,

(i) E ∼= π∗V for a unique bundle V sitting in a nonsplitting extension

0 → OX(−1, 1) → V → OX(ε1 + 1, ε2 − 1) → 0. (4.11)

(ii) dim Ext1(E, E) = dim Ext1(V, V ) = (1 + ε1)

(n + 2− ε2

n

)− 1.

Proof. (i) Note that E sits in a nonsplitting extension (4.3). By Lemma 4.1,

Ext1(OY (0, ε1 + 1, ε2 − 1),OY (0,−1, 1)) ∼= Ext1(OX(ε1 + 1, ε2 − 1),OX(−1, 1)).

Thus E ∼= π∗V for a unique bundle V sitting in a nonsplitting extension (4.11).(ii) Since E∗ ∼= E ⊗ π∗OX(−ε1,−ε2) ∼= π∗(V ⊗OX(−ε1,−ε2)), we have

Ext1(E, E) ∼= H1(Y, π∗(V ⊗ V ⊗OX(−ε1,−ε2)))∼= H1

(X, (V ⊗ V ⊗OX(−ε1,−ε2))⊗ π∗OY

)∼= Ext1(V, V )⊕H1(X, V ⊗ V ⊗OX(−ε1 − 2,−ε2 − n− 1)). (4.12)

Tensoring (4.11) by OX(−ε1 − 3,−ε2 − n) and taking cohomology, we get

H i(X, V ⊗OX(−ε1 − 3,−ε2 − n)

)= 0, i = 1, 2. (4.13)


Similarly, H1(X, V ⊗OX(−1,−n− 2)

)= 0, and H2

(X, V ⊗OX(−1,−n− 2)

)= 0

when n > 2. We see from tensoring (4.11) by V ⊗OX(−ε1 − 2,−ε2 − n− 1) that

H1(X, V ⊗ V ⊗OX(−ε1 − 2,−ε2 − n− 1)

)= 0 (4.14)

H2(X, V ⊗ V ⊗OX(−ε1 − 2,−ε2 − n− 1)

)= 0, n > 2. (4.15)

So our formula follows from (4.12) and Lemma 3.1 (i) with ξ = (−2−ε1, 2−ε2). �

Lemma 4.5. Let n ≥ 2, ε1, ε2 = 0, 1, and n(2 − ε2)/(2 + ε1) < r < n(2 − ε2)/ε1.Then, every bundle E sitting in the nonsplitting extension (4.3) is LY

r -stable.

Proof. Let OY (a, b, c) be a sub-line bundle of E with torsion-free quotient. Then,we have (4.8) and (4.9). Also, note from (4.3) that OY (a, b, c) admits an injectionto either OY (0,−1, 1) or OY (0, ε1 + 1, ε2 − 1). In the former case,

c1(OY (a, b, c)) · c1(LYr )n ≤ c1(OY (0,−1, 1)) · c1(L

Yr )n = 2rn−1(n− r)

< rn−1(ε1r + ε2n) = c1(E) · c1(LYr )n/2.

Next, let OY (a, b, c) admit an injection to OY (0, ε1 + 1, ε2 − 1). Regardingc1(OY (−a, ε1 + 1 − b, ε2 − 1 − c)) as an effective cycle in Z and examining itscoefficients for [p×p×Pn], [p×P1×H], [P1×p×H] and [P1×P1×H2], we obtain

a + b ≤ ε1 + 1, (n + 1)a + (2c− ε2) ≤ ε2 − 2, (4.16)

(n + 1)b + (2c− ε2) ≤ (n + 1)(ε1 + 1) + ε2 − 2, c ≤ ε2 − 1. (4.17)

We claim that a+ b ≤ ε1. Assume a+ b = (ε1 +1). If c = ε2− 1, then a ≤ 0 andb ≤ (ε1 + 1) by (4.16) and (4.17). So a = 0 and b = (ε1 + 1). Thus (4.3) splits, acontradiction. Therefore c ≤ ε2 − 2. Since a + b = (ε1 + 1), we see from (4.8) that

(n + 1)a(2b− ε1) ≥ 2(ε1 + 2)(ε2 − 1− c) > 0.

So a and (2b−ε1) have the same sign. By a+b = (ε1+1) again, ε1 = a = b = 1. Since(2c− ε2) < 0, we get (2c− ε2)+2(n+1)a ≤ −1 from (4.9). So (2c− ε2) ≤ −2n−3.By (4.8), −3(2− ε2) ≤ 3(2c− ε2) + (n + 1) ≤ −5n− 8 which is absurd.

Therefore a+b ≤ ε1. Since (n+1)a+(2c−ε2) ≤ ε2−2 by (4.16) and r < n(2−ε2)when ε1 = 1, we conclude that c1(OY (a, b, c)) · c1(L

Yr )n = rn−1[2nc + n(n + 1)a +

2(a + b)r] < rn−1(ε1r + ε2n) = c1(E) · c1(LYr )n/2. �

Theorem 4.6. Let n ≥ 2, ε1, ε2 = 0, 1, and k = (1 + ε1)

(n + 2− ε2

n

)− 1. Let Y

be a generic smooth Calabi-Yau hyperplane section in P1 × P1 × Pn, and

c = [1 + π∗(−1, 1)] · [1 + π∗(ε1 + 1, ε2 − 1)].

(i) When n(2 − ε2)/(2 + ε1) < r < n(2 − ε2)/ε1, MLYr(c) is isomorphic to Pk

and consists of all the bundles V sitting in nonsplitting extensions (4.3).(ii) When 0 < r < n(2− ε2)/(2 + ε1), the moduli space MLY

r(c) is empty.


Proof. Note that (ii) follows from Remark 4.3 (ii). Next, we see from the proofof Lemma 4.4 (i) that dim Ext1(OY (0, ε1 + 1, ε2 − 1),OY (0,−1, 1)) is equal tok + 1. So Remark 4.3 (i) and Lemma 4.5 imply that there is a bijective morphismΦ : Pk → MLY

r(c). In particular, the dimension of MLY

r(c) is k. By Lemma 4.4 (ii),

the moduli space MLYr(c) is smooth. Hence Φ is an isomorphism. �

Remark 4.7. Our proofs and results in this section might work for more generalbranched double covers of P1 × Pn as well.

5. Stable rank-2 bundles on Calabi-Yau manifolds via restrictions

Let n ≥ 2, and X = P1 × Pn. Let S be a generic hyperplane section in X oftype (2, n + 1). Then, S is a smooth Calabi-Yau n-fold with Pic(S) ∼= Pic(X). Weuse OS(a, b) to denote OX(a, b)|S. Put LS

r = Lr|S = OX(1, r)|S. Let ε1, ε2 = 0, 1.

Lemma 5.1. Let V be a bundle sitting in the nonsplitting exact sequence

0 → OX(−1, 1) → V → OX(ε1 + 1, ε2 − 1) → 0. (5.1)

Then, the restriction V |S is stable with respect to LSr when r satisfies

(n2 − 1)(2− ε2)

[2(n− 1) + (n + 1)ε1 + 2ε2]< r <

(n2 − 1)(2− ε2)

[ε1 ·max(0, n + 2ε2 − 3)]. (5.2)

Proof. Restricting (5.1) to S yields the exact sequence

0 → OS(−1, 1) → V |S → OS(ε1 + 1, ε2 − 1) → 0. (5.3)

Let OS(a, b) be a sub-line bundle of V |S with torsion-free quotient. Then, we havec1(OS(a, b)) · c1(OS(ε1 − a, ε2 − b)) ≤ c2(V |S). Regarding this as an inequality ofcycles in X and examining the coefficients for [p×H2] and [P1 ×H3], we obtain

(n + 1)(ε2a + ε1b− 2ab) + 2b(ε2 − b) ≤ (n + 1)(2 + ε1 − ε2) + 2(ε2 − 1), (5.4)

b(ε2 − b) ≤ (ε2 − 1), n > 2. (5.5)

Note that if OS(a, b) admits a nonzero map to OS(−1, 1), then we have

c1(OS(a, b)) · c1(LSr )n−1 ≤ c1(OS(−1, 1)) · c1(L

Sr )n−1 < c1(V |S) · c1(L

Sr )n−1/2

by the first inequality in (5.2). In the following, we assume that OS(a, b) admits anonzero map to OS(ε1 + 1, ε2 − 1). Regarding c1(OS(ε1 + 1− a, ε2 − 1− b)) as aneffective cycle in X and examining its coefficients for [p×H] and [P1×H2], we get

(n + 1)(a− ε1 − 1) + 2(b− ε2 + 1) ≤ 0, b ≤ ε2 − 1. (5.6)

First of all, suppose b = ε2 − 1. Then, a ≤ ε1 + 1 by (5.6). Consider

0 → V ⊗OX(−ε1 − 3,−ε2 − n) → V ⊗OX(−ε1 − 1,−ε2 + 1)

→ V ⊗OX(−ε1 − 1,−ε2 + 1)|S → 0. (5.7)

Since (5.1) is nonsplitting, H0(X, V ⊗OX(−ε1−1,−ε2 +1)) = 0. Therefore we seefrom (5.7) and (4.13) that h0(S, V ⊗OS(−ε1 − 1,−ε2 + 1)) = 0. Thus, a 6= ε1 + 1,and so a ≤ ε1. If a ≤ 0, then we obtain from b = ε2 − 1 ≤ 0 that

c1(OS(a, b)) · c1(LSr )n−1 < c1(V |S) · c1(L

Sr )n−1/2. (5.8)


The remaining case 1 ≤ a ≤ ε1 forces ε1 = a = 1. By the second inequality in(5.2), we conclude that (5.8) holds as well.

Now suppose b ≤ ε2 − 2. If (n + 1)a + 2b ≤ ((n + 1)ε1 + 2ε2)/2, then we verifydirectly that (5.8) holds. So let (n + 1)a + 2b > ((n + 1)ε1 + 2ε2)/2. Then, a ≥ 1.If ε1 = 0, then we obtain the contradiction:

(n + 1)(2− ε2) ≤ (n + 1)a(−b) < (n + 1)a(−b) + ((n + 1)a + 2b)(ε2 − b)

≤ (n + 1)(2− ε2) + 2(ε2 − 1) ≤ (n + 1)(2− ε2)

where we have used (5.4) in the third step. Hence ε1 = 1, and

(n + 1)(3− ε2) ≥ (n + 1)(3− ε2) + 2(ε2 − 1)

≥ (n + 1)(a− 1)(−b) + ((n + 1)a + 2b)(ε2 − b)

= (n + 1)[(a− 1)(−b) + (ε2 − b)/2],

where we have used (n + 1)a + 2b > ((n + 1) + 2ε2)/2 ≥ (n + 1)/2 in the last step,and (5.4) in the second step. So (3 − ε2) > (a − 1)(−b) + (ε2 − b)/2. This forcesa < 2 since b ≤ ε2 − 2. So a = 1. Using b ≤ ε2 − 2 again, we have

2/rn−2 · [c1(OS(a, b)) · c1(LSr )n−1 − c1(V |S) · c1(L

Sr )n−1/2]

= (n + 1 + 4b− 2ε2)r + (n2 − 1)(2b− ε2)

≤ (n + 2ε2 − 7)r + (n2 − 1)(ε2 − 4). (5.9)

If (n + 2ε2 − 7) ≤ 0, then (5.8) holds. If (n + 2ε2 − 7) > 0, then we conclude from(5.9) and the second inequality in (5.2) that (5.8) holds. �

Lemma 5.2. Let n ≥ 2, and E sit in a nonsplitting extension

0 → OS(−1, 1) → E → OS(ε1 + 1, ε2 − 1) → 0 (5.10)

where ε1, ε2 = 0, 1, and k = (1 + ε1)

(n + 2− ε2

n

)− 1. Then,

(i) E ∼= V |S for a unique bundle V sitting in a nonsplitting extension (5.1),and dim Ext1(OS(ε1 + 1, ε2 − 1),OS(−1, 1)) = k + 1.

(ii) dim Ext1(E, E) = dim Ext1(V, V ) = k when n > 2.

Proof. (i) Since H i(X,OX(−ε1 − 4,−ε2 − n + 1)) = 0 for i = 1 and 2, we obtainH1(S,OS(−ε1−2,−ε2+2)) ∼= H1(X,OX(−ε1−2,−ε2+2)) from the exact sequence

0 → OX(−ε1 − 4,−ε2 − n + 1) → OX(−ε1 − 2,−ε2 + 2)

→ OS(−ε1 − 2,−ε2 + 2) → 0.

So Ext1(OS(ε1 + 1, ε2− 1),OS(−1, 1)) ∼= Ext1(OX(ε1 + 1, ε2− 1),OX(−1, 1)) withdimension (k + 1), and E ∼= V |S for a unique V sitting in a nonsplitting (5.1).

(ii) Since E ∼= V |S, we have Ext1(E, E) ∼= H1(X, (V ⊗ V ⊗ OX(−ε1,−ε2))|S

).

In view of (4.14) and (4.15), we conclude from the exact sequence

0 → V ⊗ V ⊗OX(−ε1 − 2,−ε2 − n− 1) → V ⊗ V ⊗OX(−ε1,−ε2)

→ (V ⊗ V ⊗OX(−ε1,−ε2))|S → 0


that H1(X, (V ⊗ V ⊗ OX(−ε1,−ε2))|S

) ∼= H1(X, V ⊗ V ⊗ OX(−ε1,−ε2)

). So

Ext1(E, E) ∼= Ext1(V, V ). Now our formula follows from Lemma 3.1 (i). �

Let n = 2. Then S is a smooth K3 surface in X = P1 × P2 of type (2, 3).Let r satisfy (5.2), and MS be the moduli space of LS

r -stable rank-2 torsion-freesheaves E on S with c2(E) = (−1, 1)|S · (ε1 + 1, ε2 − 1)|S = (3ε1 − ε2 + 4) andc1(E) = (ε1, ε2)|S. By Lemma 5.1, MS is nonempty. Also,

4c2(V |S)− c1(V |S)2 − 3χ(OS) = 10 + 12ε1 − 6ε2 − 6ε1ε2.

If ε1 = 1 or ε2 = 1, then (ε1, ε2)|S is indivisible by 2. By [Muk], MS is a smoothsymplectic manifold of complex dimension (10 + 12ε1 − 6ε2 − 6ε1ε2).

To state our next result, let’s recall a theorem due to Tyurin. Let X be a Fanothreefold, S ∈ |−KX | be a smooth K3 surface, and MX and MS be the componentsof the moduli spaces of stable vector bundles with fixed Chern classes c1 on X andc1|S on S respectively such that if V ∈ MX , then V |S ∈ MS. Let resS: MX → MS

be the restriction map.

Theorem 5.3. ([Tyu]) With the notations as above. Suppose that Ext2(V, V ) = 0for every V ∈ MX . Then resS(MX) ⊂ MS is a Lagrangian subvariety of MS.

Applying Tyurin’s result to our case, we obtain the following proposition.

Proposition 5.4. Let n = 2, ε1 = 1 or ε2 = 1, k = (5 + 6ε1 − 3ε2 − 3ε1ε2), and2(2 − ε2)/(2 + ε1) < r < 2(2 − ε2)/ε1. Then the (2k)-dimensional smooth modulispace MS contains a Lagrangian submanifold isomorphic to Pk.

Proof. Note that r satisfies (5.2). Let MX be the moduli space of Lr-stablerank-2 bundles on X with Chern classes (ε1, ε2) and (−1, 1) · (ε1 +1, ε2− 1). ApplyCorollary 3.4 to n = 2 and d = 1. We see that the moduli space MX is isomorphicto Pk, and that Ext2(V, V ) = 0 for every bundle V ∈ MX . Now our conclusionfollows from Theorem 5.3 and Lemma 5.1. �

In the following, we assume n ≥ 3. We shall prove results for Calabi-Yau hyper-plane sections S ⊂ P1×Pn parallel to Lemma 4.2 and Theorem 4.6 for Calabi-Yauhyperplane sections Y ⊂ P1×P1×Pn. Let notations be the same as in Lemma 5.1.Then we see from the proof of Lemma 5.2 (ii) that the Zariski tangent space atE = V |S is isomorphic to the Zariski tangent space at V . This indicates thatwe might be able to recover the entire moduli space containing E from the entiremoduli space containing V , and explains the reason behind our results below.

Lemma 5.5. Let n ≥ 3, ε1, ε2 = 0, 1, and r < (n2 − 1)(2 − ε2)/[ε1(n + 2ε2 − 3)].Let E be an LS

r -stable rank-2 torsion-free sheaf with

c(E) = [1 + (−1, 1)|S] · [1 + (ε1 + 1, ε2 − 1)|S].

Then, r > (n2− 1)(2− ε2)/[2(n− 1) + (n + 1)ε1 + 2ε2], and the rank-2 sheaf E sitsin a nonsplitting extension of the form (5.10).


Proof. Note that c2(E) = ((2 + ε1− ε2)[p×H]− (1− ε2)[P1×H2])|S and c1(E) =(ε1, ε2)|S. So (4c2(E)− c1(E)2) · c1(L

Sr0

)n−2 is equal to

(2− ε2)rn−30 [2((n + 1)(2 + ε1)− (2− ε2))r0 − (n− 2)(n + 1)(2− ε2)].

Thus, E is LSr0

-unstable if 0 < r0 < (n−2)(n+1)(2−ε2)/[2((n+1)(2+ε1)−(2−ε2))].Fix such an r0 with r0 < r. Since E is LS

r -stable, there is a nonsplitting extension

0 → OS(a, b)⊗ IZ1 → E → OS(ε1 − a, ε2 − b)⊗ IZ2 → 0 (5.11)

such that OS(a, b) ⊗ IZ1 destablizes E with respect to LSr0

, where Z1 and Z2 arecodimension at least two subschemes of Y . We have

(n2 − 1)(2b− ε2) + [(n + 1)(2a− ε1) + 2(2b− ε2)]r0 > 0, (5.12)

(n2 − 1)(2b− ε2) + [(n + 1)(2a− ε1) + 2(2b− ε2)]r < 0, (5.13)

(n + 1)(ε2a + ε1b− 2ab) + 2b(ε2 − b) ≤ (n + 1)(2 + ε1 − ε2) + 2(ε2 − 1), (5.14)

b(ε2 − b) ≤ (ε2 − 1) (5.15)

as in (4.5), (4.6), (5.4) and (5.5). Note that (5.14) can be written as

[(n + 1)(2a− ε1) + (2b− ε2)](2b− ε2)

≥ −[(n + 1)(2 + ε1)− (2− ε2)](2− ε2), (5.16)

while (5.15) is equivalent to (2b− ε2)2 ≥ (2− ε2)

2. So we get

|2b− ε2| ≥ (2− ε2). (5.17)

Since 0 < r0 < r, we see from (5.12) and (5.13) that (2b − ε2) > 0 and (n +1)(2a− ε1) + 2(2b− ε2) < 0. Hence (2a− ε1) < 0, and b ≥ 1 by (5.17). Thus,

−(2b− ε2)2 > [(n + 1)(2a− ε1) + 2(2b− ε2)](2b− ε2)− (2b− ε2)

2

= [(n + 1)(2a− ε1) + (2b− ε2)](2b− ε2) ≥ −[(n + 1)(2 + ε1)− (2− ε2)](2− ε2)

by (5.16). So (2b− ε2) <√

[(n + 1)(2 + ε1)− (2− ε2)](2− ε2). By (5.16) again,

[(n + 1)(2a− ε1) + (2b− ε2)] ≥ −[(n + 1)(2 + ε1)− (2− ε2)]. (5.18)

Therefore, we conclude from a straightforward computation that

(2a− ε1) =(n + 1)(2a− ε1) + (2b− ε2)

(n + 1)− (2b− ε2)

(n + 1)> −(2 + ε1)− 1.

Combining with (2a− ε1) < 0, we obtain either a = −1, or a = 0 and ε1 = 1.Assume a = −1. By (5.16), (2b−2)[(n+1)(−2−ε1)+(2b+2−2ε2)] ≥ 0. If b > 1,

then (2b−ε2) ≥ (n+1)(2+ε1)−(2−ε2). So√

[(n + 1)(2 + ε1)− (2− ε2)](2− ε2) >(n + 1)(2 + ε1) − (2 − ε2), which is impossible. Thus b = 1. Now (5.11) becomes(5.10), and r > (n2 − 1)(2− ε2)/[2(n− 1) + (n + 1)ε1 + 2ε2] by (5.13).

Finally, assume a = 0 and ε1 = 1. Then, we see from (5.13) and b ≥ 1 that

0 > (n2 − 1)(2b− ε2) + [2(2b− ε2)− (n + 1)]r

≥ (n2 − 1)(2− ε2)− (n + 2ε2 − 3)r

contradicting to r < (n2 − 1)(2− ε2)/(n + 2ε2 − 3) when ε1 = 1. �


Remark 5.6. Let n ≥ 3, ε1, ε2 = 0, 1 and c = [1 + (−1, 1)|S] · [1 + (ε1 + 1, ε2 − 1)|S].

(i) Let r satisfy (5.2), and E ∈ MLSr(c). Then the same proof above shows that

E sits in a nonsplitting extension (5.10). By Lemma 5.2, E is LSr -stable.

(ii) Similarly, we see that the moduli space MLSr(c) is empty when

0 < r <(n2 − 1)(2− ε2)

[2(n− 1) + (n + 1)ε1 + 2ε2].

Theorem 5.7. Let n ≥ 3, ε1, ε2 = 0, 1, and k = (1 + ε1)

(n + 2− ε2

n

)− 1. Let S

be a generic smooth Calabi-Yau hyperplane section in P1 × Pn. Let

c = [1 + (−1, 1)|S] · [1 + (ε1 + 1, ε2 − 1)|S].

(i) The moduli space MLSr(c) is isomorphic to Pk, and consists of all the rank-2

bundles E sitting in nonsplitting extensions (5.10) when

(n2 − 1)(2− ε2)

2(n− 1) + (n + 1)ε1 + 2ε2

< r <(n2 − 1)(2− ε2)

ε1(n + 2ε2 − 3). (5.19)

(ii) When 0 < r < (n2−1)(2−ε2)/[2(n−1)+(n+1)ε1 +2ε2], MLSr(c) is empty.

Proof. Follows from Lemma 5.1, Lemma 5.2, Lemma 5.5 and Remark 5.6. �

Remark 5.8. That some moduli spaces of (slope) stable bundles on Calabi-Yauthree-folds, which are hyperplane sections of a P3-bundle over P1, are isomorphicto projective spaces are also shown by Nakashima in [Nak]. The method is viaextensions of a line bundle by another line bundle as well.

6. Holomorphic Casson invariants

We recall from the Definition 3.54 and Corollary 3.39 in [Tho] that for a polarizedsmooth Calabi-Yau 3-fold (Y, L), a homolomorphic Casson invariant λ(Y, L, c, k) ∈Z can be defined via the moduli space of Gieseker L-semistable rank-k torsion-freesheaves with total Chern class c, provided that all the sheaves in this moduli spaceare actually stable. Moreover, when the moduli space is smooth, λ(Y, L, c, k) isequal to the top Chern class of the cotangent bundle of the moduli space.

Corollary 6.1. Let Y ⊂ P1 × P1 × P2 be a generic smooth Calabi-Yau hyperplanesection. Let c = [1 + π∗(−1, 1)] · [1 + π∗(ε1 + 1, ε2 − 1)] where ε1, ε2 = 0, 1, andπ : Y → P1 × P2 is the restriction to Y of the projection of P1 × P1 × P2 to theproduct of the last two factors. Let k = (1 + ε1)(4− ε2)(3− ε2)/2− 1. Then.

(i) λ(Y, LYr , c, 2) = (−1)k(k + 1) when 2(2− ε2)/(2 + ε1) < r < 2(2− ε2)/ε1;

(ii) λ(Y, LYr , c, 2) = 0 when 0 < r < 2(2− ε2)/(2 + ε1).

Proof. We apply the results in Section 4 to n = 2. By Remark 4.3, all the sheavesin MLY

r(c) are LY

r -stable. So our conclusions follow from Theorem 4.6. �

Corollary 6.2. Let S ⊂ P1×P3 be a generic smooth Calabi-Yau hyperplane section.Let c = [1 + (−1, 1)|S] · [1 + (ε1 + 1, ε2 − 1)|S] where ε1, ε2 = 0, 1.


(i) If 4(2− ε2)/(2 + 2ε1 + ε2) < r < 4(2− ε2)/(ε1ε2), then

λ(S, LSr , c, 2) = −(1 + ε1)

(5− ε2

3

).

(ii) λ(S, LSr , c, 2) = 0 when 0 < r < 4(2− ε2)/(2 + 2ε1 + ε2).

Proof. We apply the results in Section 5 to n = 3. By Remark 5.6, all the sheavesin MLS

r(c) are LS

r -stable. So our conclusions follow from Theorem 5.7. �

References

[B-M] T. Bridgeland and A. Maciocia, Fourier-Mukai transforms for K3 and elliptic fibrations,J. Alg. Geom. 11 (2002), 629–657.

[D-T] S.K. Donaldson and R.P. Thomas, Gauge theory in higher dimensions, In: The geomet-ric Universe (Oxford, 1996), Oxford University Press, Oxford, 1998, 31–47.

[Fri] R. Friedman, Algebraic surfaces and holomorphic vector bundles, Universitext. Springer-Verlag, New York, 1994.

[FMW] R. Friedman, J. Morgan and E. Witten, Vector bundles over elliptic fibrations, J. Alg.Geom. 8 (1999), 279–401.

[MR1] V.B. Mehta and A. Ramananthan, Semistable sheaves on projective varieties and theirrestriction to curves, Math. Ann. 258 (1982), 213–224.

[MR2] V.B. Mehta and A. Ramananthan, Restriction of stable sheaves and representations ofthe fundamental group, Invent. Math. 77 (1984), 163–172.

[Muk] S. Mukai, On the moduli space of bundles on K3 surfaces. I, Vector bundles on algebraicvarieties (Bombay, 1984), 341–413.

[Nak] T. Nakashima, Moduli spaces of stable bundles on K3 fibered Calabi-Yau threefold, Com-mun. Contemp. Math. 5 (2003), 119-126.

[Qin] Z. Qin, Equivalence classes of polarizations and moduli spaces of sheaves, J. Differ.Geom. 37 (1993), 397-415.

[Tho] R.P. Thomas, A holomorphic Casson invariant for Calabi-Yau 3-folds, and bundles onK3 fibrations, J. Differ. Geom. 54 (2000), 367–438.

[Tia] G. Tian, Gauge theory and calibrated geometry, I, Ann. Math. 151 (2000), 193-268.[Tyu] A.N. Tyurin, The moduli spaces of vector bundles on threefolds, surfaces and curves,

Math. Inst.Univ. Gottingen preprint.

Department of Mathematics, HKUST, Clear Water Bay, Kowloon, Hong KongE-mail address: [email protected]

Department of Mathematics, University of Missouri, Columbia, MO 65211, USAE-mail address: [email protected]

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