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DOI 10.1007/s00209-006-0028-9

Mathematische Zeitschrift

Pluricanonical systems on irregular 3-folds of general type

Jungkai A. Chen · Christopher D. Hacon

Received: 2 February 2006 / Accepted: 19 May 2006 / Published online: 23 August 2006

© Springer-Verlag 2006

Abstract In this paper we prove that if X is an irregular 3-fold withχ(ωX) > 0, then

|mKX| is birational for all m ≥ 5.

1 Introduction

Given a non-singular variety of general type, by definition, the pluricanonical system

|mKX| defines a birational map ϕmto a projective space, for all sufficiently big inte- gers m. It is a natural question to try and find a reasonable effective bound for such integers m. When dimX= 1, it is classically known that ϕmis an embedding for m≥ 3.

For surfaces, by a result of [2],ϕmis birational for m ≥ 5. Recently, following ideas of Tsuji (cf. [9, 14, 17]), it has been shown that for any positive integer n> 0, there exists a positive integer rn> 0 such that ϕmis birational for m≥ rnfor all varieties of general type and of dimension n. Tsuji [18] has also shown, that r3 ≤ 18(29· 37)!.

Recently Todorov [15], has shown that for all but finitely many families of 3-folds of general type, the 5-th canonical map is birational.

The purpose of this note is to study pluricanonical systems on irregular varieties, i.e.

varieties with q(X) := h0(X, X) ≥ 0. In view of recent results on irregular varieties,

The first author was partially supported by NSC and NCTS of Taiwan. The second author was partially supported by NSF research grant no: 0456363.

J. A. Chen

Department of Mathematics, National Taiwan University, Taipei 10617, Taiwan

J. A. Chen (

B

)

National Center for Theoretical Sciences, Taipei Office, Taipei, Taiwan e-mail: jkchen@math.ntu.edu.tw

C. D. Hacon

Department of Mathematics, University of Utah, 155 South 1400 East, JWB 233, Salt Lake City, UT 84112-0090, USA

e-mail: hacon@math.utah.edu

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one might hope that their geometry behaves similarly to that of surfaces. In this article we are in fact able to prove the following:

Theorem 1.1 Let X be an irregular smooth projective threefold of general type with χ(ωX) > 0. Then |mKX+ P| is birational for m ≥ 5 and for all P ∈ Pic0(X).

It is easy to see that this result is optimal. Consider in fact a surface S of gen- eral type such that |4KS| is not birational and a curve C of general type. For all P∈ Pic0(C × S) ∼= Pic0(C) one sees that |4KC×S+ P| is not birational.

The main technical ingredient is the technique developed in [3], where the Fourier–Mukai transform and vanishing theorems are used to study adjoint linear series on irregular varieties. For the reader’s convenience, in Sect. 2 we recall the main result from [3]. Its application to pluricanonical system is summarized in The- orem 2.8. One finds a very sharp bound in Proposition 2.9 when the Albanese fiber dimension is small. Starting in Sect. 3, we will work on irregular varieties over curves.

The case when X maps to a curve with genus g≥ 2 is handled in Sect. 3 by a relatively straightforward application of the weak positivity of the relative dualizing sheaves.

The case when X maps to an elliptic curve is more subtle. We are unable to recover a general result in arbitrary dimension. However, by a careful study of the positivity of vector bundles over elliptic curves in Sect. 4, we are able to prove our main theorem for threefolds in Sect. 5.

1.1 Notation

We work over the field of complex numbersC. AQ-Cartier divisor D on a normal variety X is nef if D· C ≥ 0 for any curve C ⊂ X. We say that twoQ-divisors D1, D2 areQ-linearly equivalent (D1QD2) if there exists an integer m> 0 such that mDi

are linearly equivalent. Numerical equivalence is denoted≡. And we write D1≥ D2if D1− D2is effective. For any projective variety X, Pic0(X) denotes the abelian variety parametrizing the topologically trivial line bundles on X. If A is an abelian variety, then we can identify ˆA with Pic0(A). We will denote byP, the normalized Poincaré line bundle on A× ˆA.

2 Linear series on irregular varieties

We begin by recalling several important consequences of the theory of Fourier–Mukai transforms that will be needed throughout the paper.

Let A be an abelian variety. Recall that there is a functor ˆS from the category of OA-modules to the category ofOˆA-modules defined by

Sˆ(M) = (pˆA)(P⊗ pAM).

Here pA and pˆA denote the projections of A× ˆA to A and ˆA. Similarly we define S(N) = (pA)(P⊗ pˆAN). Let RS (resp. R ˆS) be the derived functor ofS (resp. ˆS) between the corresponding derived categories. By a result of Mukai (cf. [13] Theorem 2.2), we have:

Theorem 2.1 There exist isomorphisms of functors

R ˆS◦ RS= (−1ˆA)[−g] and RS◦ R ˆS= (−1A)[−g].

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A coherent sheafFon an abelian variety A is said to be IT0if Hi(A,F⊗ P) = 0 for all i> 0 and all P ∈ Pic0(A).

Lemma 2.2 LetF be a coherent IT0 sheaf on an abelian variety A. IfF = 0, then H0(A,F⊗ P) = 0 for all P ∈ Pic0(A).

Proof Please see [3] Lemma 2.1.

Lemma 2.3 LetF= 0 be an IT0coherent sheaf on an abelian variety A. Suppose that there is a non-zero mapF →C(z). Then the induced map H0(F⊗P) → H0(C(z)) is non-zero for general P∈ Pic0(A).

Proof Please see [3] Proposition 2.3.

Here are some useful consequences:

Corollary 2.4 Let X be a smooth projective variety and a : X → Z = a(X) ⊂ A a morphism to an abelian variety A. Let F be a general fiber and x a point on F. Let L be a divisor on X such that L ≡ aH+  for a nef and big Q-divisor H and a Q-divisor ≥ 0 so that  has simple normal crossings support and  = 0. Suppose furthermore that aOX(KX+ L) = 0. Then we have:

(1) If x∈ Bs|(KX+ L)|F|, then x ∈ Bs|KX+ L + aP| for a general P ∈ Pic0(A).

(2) Let x1, x2 be two points on different general fibers F1, F2 respectively. If xiBs|(KX + L)|Fi| for i = 1, 2, and one of the points xi, say x1, does not belong to Bs|KX+ L + aP| for all P ∈ Pic0(A), then |KX+ L + aP| separates x1, x2for a general P∈ Pic0(A).

(3) Let x1, x2 be two distinct points on a general fiber F. If xi ∈ Bs|(KX + L)|F| for i= 1, 2, |(KX+ L)|F| separates x1, x2, and one of xi, say x1,∈ Bs|KX+ L + aP| for all P∈ Pic0(A), then |KX+L+aP| separates x1, x2for a general P∈ Pic0(A).

Proof Let z∈ Z be a general point and F the fiber over z. By z being general, we mean that

aOX(KX+ L) ⊗C(z) ∼= H0(F,OF(KX+ L)).

We first look at the short exact sequence, obtained by evaluated at x:

(∗) 0 →OX(KX+ L) ⊗IxOX(KX+ L) →C(x) → 0.

Pushing forward to Z, we get

0→ a(OX(KX+ L) ⊗Ix) → aOX(KX+ L) →C(z) → · · · Since x∈ Bs|(KX+ L)|F| and

aOX(KX+ L) ⊗C(z) ∼= H0(F,OF(KX+ L)), one sees that the induced map

aOX(KX+ L) → aOX(KX+ L) ⊗C(z) →C(z) is non-zero, whence surjective.

By Corollary 10.15 of [12], aOX(KX+ L) is IT0. Since aOX(KX + L) = 0 by assumption, applying Lemma 2.3 to the sheaf aOX(KX+ L), we get (1).

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Before we move forward to the proof of the next two statements, we remark that the assumption that x1∈ Bs|KX+L+aP| for all P implies that a(OX(KX+L)⊗Ix1) is IT0(by tensoring(∗) with aP and replacing x by x1).

To see(3), we begin by looking at the exact sequence

0→ a(OX(KX+ L)⊗Ix1,x2) → a(OX(KX+ L)⊗Ix1) →C(z) → · · · The last map factors as

a(OX(KX+ L)⊗Ix1) → a(OX(KX+ L)⊗Ix1)⊗C(z)

→ aOX(KX+ L)⊗C(z) →C(z).

It is obtained by evaluating at x2. The assumption that|(KX+ L)|F| separates x1, x2 shows that this is surjective. Again, by applying Lemma 2.3 to the sheaf a(OX(KX+ L)⊗Ix1), we are done.

Finally we consider(2). Again we have (assuming now that x2lies over z):

0→ a(OX(KX+ L)⊗Ix1,x2) → a(OX(KX+ L)⊗Ix1) →C(z) → · · · But now we look at the map

a(OX(KX+ L)⊗Ix1) → a(OX(KX+ L)⊗Ix1)⊗C(z)

→ aOX(KX+ L)⊗C(z) →C(z)

which is obtained by evaluating at x2. Since a(x1) = a(x2), it follows that a(OX(KX+ L)⊗Ix1)⊗C(z) ∼= aOX(KX + L)⊗C(z) and hence the above map is surjective. By

Lemma 2.3 we are done.

We now fix some notation and conventions. For any P∈ Pic0(X), we defineBm,Pto be the fixed component of|mKX+aP| andBm(resp.Bm,F) to be the fixed component of|mKX| (resp. |mKF|). After replacing X by an appropriate birational model, we may and we usually do assume thatBm,P,Bm,Fhave simple normal crossing support.

We need the following:

Lemma 2.5 Let g : X → W be a morphism with a general fiber F. Suppose that κ(W) ≥ 0 and KXis W-big, i.e. sKX ≥ gL for some ample divisor L and some integer s 0 . Suppose further that Pm(F) ≥ 0 for some m ≥ 2. Then after replacing X by an appropriate birational model if necessary, there exist positive integers a, b and there is a normal crossing divisor B∈ |ab(m − 1)KX− gbH| such that

B ab



|F≤Bm,F,

B ab



≤Bm,P,

for all P∈ Pic0(W). Here H is a given nef and big divisor on W and a, b are sufficiently large integers depending on H and KX.

Proof Please see [3], Lemma 5.1.

Definition 2.6 Let a : X → A be a morphism to an abelian variety and D a Cartier divisor on X. Let

V0(D) :=

P∈ Pic0(A)|h0(X,OX(D) ⊗ aP) = 0 . We say that D is full if V0(D) = Pic0(A).

Note that if aOX(D) is IT0and non-zero, then by Lemma 2.2, D is full.

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2.7 General setting

We can now work on a more general setting. Let a : X → Z = a(X) ⊂ A be a morphism to an Abelian variety A. Usually, we take the Albanese map of X. Let ν : W → Z be a desingularization of the Stein factorization over Z. Replacing X by an appropriate birational model, we may assume that there is a morphism g : X→ W with connected fibres. We assume that X is of general type. It follows that KX is W-big. Let F be the general fiber of g. In the following discussion, we will assume that Pm(F)0 for some m ≥ 2. We then take Bm = abB with B∈ |ab(m − 1)KX − gbH| as in Lemma 2.5 and

Lm:= (m − 1)KX− Bm ≡ 1

agH+ {Bm}

where Bm = Bm + {Bm}. By [12] Theorem 10.15, since H is the pull back of a nef and big divisor on Z, one sees that

Hi(A, aOX(KX + Lm) ⊗ P) ∼= Hi(W, gOX(KX+ Lm)⊗νP) = 0

for all P ∈ Pic0(A) and all i > 0. In particular, aOX(KX + Lm) is IT0. Recall that by Lemma 2.5, one has Bm|F ≤ Bm,F. Thus, H0(F,OF((KX + Lm)|F)) ∼= H0(F,OF(mKF)). In particular, aOX(KX+ Lm) = 0. By Lemma 2.2, KX+ Lmis full and hence so is mKX.

Assume that for some integers m, n≥ 1, mKX and nKXare full and consider the morphism

|mKX + aP1| + |nKX+ aP2| → |(m + n)KX+ a(P1+ P2)|

= |KX + Lm+n+ a(P1+ P2)|,

for all P1, P2. If x∈ X is a general point, then for general P ∈ Pic0(A), x is neither a base point of mKX + aP nor a base point of nKX+ aP. Therefore x is not a base point of|(m + n)KX + aP| for all P ∈ Pic0(A). And it follows that x is not a base point of|KX+ Lm+n+ aP| since x is general.

By Corollary 2.4, we can now deduce the main result of this section.

Theorem 2.8 Let X be a smooth projective variety of general type, a : X → Z = a(X) ⊂ A a non-trivial morphism to an abelian variety, F a general fiber of a and m, n, t positive integers. If mKX, nKX, tKX are full, then:

(1) |(m + n)KX+ aP| separates two general points on two distinct general fibers for general P∈ Pic0(A).

(2) |(m + n + t)KX+ aP| separates two general points on two distinct general fibers for all P∈ Pic0(A).

(3) If(m + n)KFis birational, then|(m + n)KX+ aP| separates two general points on a general fiber for a general P∈ Pic0(A).

(4) If(m+n)KFis birational, then|(m+n+t)KX+aP| separates two general points on a general fiber for all P∈ Pic0(A).

Proof (1) follows from Corollary 2.4 (2) since for two general points x1and x2, we have seen that xi∈ Bs|KX+Lm+n+aP| for any P ∈ Pic0(A) and so xi∈ Bs|(KX+Lm+n)|F|.

(3) follows from Corollary 2.4 (3) since we assumed that|(m + n)KF| is birational and so, since by Lemma 2.5 we have(Bm+n)|F ≤ Bm+n,F, then|(KX + Lm+n)|F| is also birational.

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(2) and (4) now follow by considering the morphism

|(m + n)KX+ aP1| + |tKX+ aP2| → |(m + n + t)KX+ a(P1+ P2)|.

Since|(m + n)KX+ aP1| separates x1, x2for general P1∈ Pic0(A) and |tKX+ aP2| does not vanish along x1, x2 for general P2 ∈ Pic0(A), then |(m + n + t)KX + aP|

separates x1, x2for all P∈ Pic0(A).

Let F be an irreducible component of a general fibre of the Albanese map a= alb : X→ a(X). Turning to a more detailed discussion, we now distinguish varieties according to their Albanese fiber dimension, i.e. f := dimX − dima(X). We would like to remark that an analogous result holds for arbitrary non-trivial morphism to an abelian variety.

Proposition 2.9 Let X be a smooth projective variety of general type with q(X) > 0.

(1) If f = 0, then

(a) |mKX+ P| is birational for a general (resp. all) P ∈ Pic0(X) for m ≥ 4 (resp.

m≥ 6).

(b) If moreover KX is full, then|mKX+ P| is birational for a general (resp. all) P∈ Pic0(X) for m ≥ 2 (resp. m ≥ 3).

(2) If f = 1, then

(a) |mKX+ P| is birational for a general (resp. all) P ∈ Pic0(X) and for m ≥ 4 (resp. m≥ 6).

(b) If moreover KX is full, then|mKX+ P| is birational for a general (resp. all) P∈ Pic0(X) and for m ≥ 3 (resp. m ≥ 4).

(3) If f = 2, then

(a) |mKX+ P| is birational for a general (resp. all) P ∈ Pic0(X) and for m ≥ 5 (resp. m≥ 7).

(b) If moreover KX is full, then|mKX+ P| is birational for a general (resp. all) P∈ Pic0(X) and for m ≥ 5 (resp. m ≥ 6).

Proof If f = 0, then X is generically finite over its Albanese variety. By Lemma 2.2 KX + L2 is full and hence so is 2KX. Clearly|mKF| separates points on fibers for m≥ 1, so (a) follow from Theorem 2.8, letting m = n = t = 2. If we assume that KX

is full, then we let m= n = t = 1.

If f ≥ 1, one sees that for m ≥ 2, mKXis full if and only if Pm(F) = 0. Since F is of general type, for f ≤ 2 it is well known that Pm(F) > 0 for all m ≥ 2. Therefore mKX

is full for m≥ 2.

If f = 1, mKF is birational for m≥ 3. By Theorem 2.8, we have (a), by taking m= n = t = 2. If we assume that KX is full, then let m= 2, n = t = 1.

Finally, if f= 2, then mKFis birational for m≥ 5. By Theorem 2.8, we have (a) by letting m= 3, n = t = 2. If we assume that KX is full, then we let m= 4, n = t = 1.

3 Varieties over curves of genus at least 2

In this section, we study the case in which the Albanese image is a curve of g≥ 2.

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Proposition 3.1 Let a : X → C be a morphism from a smooth projective variety of general type to a curve of genus g(C) ≥ 2 with a (connected) general fiber F. Sup- pose that for some m≥ 4, |mKF| is birational. Then |mKX+ aP| is birational for all P∈ Pic0(C).

Proof Fix an integer k 0 such that P2k(F) > 0. We write

2kKX − 3kF = 2kKX/C+ 2kaKC− 3kF = 2kKX/C+ aH,

for some ample line bundle H of degree (4g(C) − 7)k. By the weak positivity of a2kX/C) (cf. [19,20]), one has that the restriction map

H0(X,OX(2nKX− 3nF)) → H0(F,OF(2nKF)),

is surjective for all n  k sufficiently large and divisible by k and by m. There- fore, a2nX(−3nF)) is a vector bundle of rank P2n(F) and degree at least P2n(F) (4g(C) − 7)n. Since F is of general type, by Riemann-Roch, one sees that 2KX− 3F is big. We write |2nKX − 3nF| = |Nn| + Bn where Bn is the fixed part. We have 2nKX ∼ Mn+ Bn= (Nn+ 3nF) + Bn. After replacing X by an appropriate birational model, we may assume that|Nn|, |Mn| are free and that Bn is a divisor with normal crossing support. Note that Mnis nef and big. We write|2nKF| = |(Mn)|F| +B2n,F. By the above surjection, we have thatB2n,F= (Bn)|F.

We have(m − 3)KX(m−3)M2n n+(m−3)B2n n and 2KX − 3F ≡Nnn+Bnn. Let Lm:= (m − 1)KX

(m − 1)Bn

2n

 . Since m divides n, we have that BnmnBm. Therefore,

KX + Lm≥ mKX

m− 1 2n

n mBm



≥ mKX

1 2Bm

 . Similarly,

KF+ Lm|F≥ mKF

m− 1 2n B2n,F



≥ mKF

1 2B2m,F

 , and|KF+ Lm|F| ∼= |mKF| = ∅. Note that

Lm(m − 3)Mn

2n +Nn

n +

(m − 1)Bn

2n

 + 3F.

By Kawamata–Viehweg vanishing, one has that

H1(X,OX(KX + Lm+ aP− sF)) = 0 for s≤ 3, P ∈ Pic0(C).

Consider the short exact sequence:

0→OX(KX+ Lm+ aP− F1− F2) →OX(KX+ Lm+ aP)

OF1(KF1+ Lm|F1) ⊕OF2(KF2+ Lm|F2) → 0.

One sees that|KX+ Lm+ aP| separates general points on distinct general fibers F1

and F2.

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Similarly, one has the surjection

H0(X,OX(KX+ Lm+ aP)) → H0(F,OF(KF+ Lm)).

If|mKF| is birational then |KX+ Lm+ aP| separates general points on F and so does

|mKX+ aP|. This completes the proof.

4 Varieties over elliptic curve

In this section, we consider varieties X of general type admitting a morphism to an elliptic curve a : X→ C. We will use the results and the notation of [1].

Lemma 4.1 For any integer m≥ 2, letI:=OX(−Bm) be as in (2.7). Then amX) = amXI) is an IT0vector bundle of rank Pm(F).

Proof Clearly, we have an inclusion amXI) → aXm). As in (2.7), one sees that they have the same rank. LetQ be the cokernel. Then Q has generic rank 0 and hence is supported on finitely many points. IfQ= 0, then h0(C,Q) = 0. However, amXI) is IT0and h0(aXmI)) = h0(amX)). This is a contradiction. Therefore,

Q= 0.

For simplicity, we will write Em:= amX).

Definition 4.2 Let E be a vector bundle on an elliptic curve C. We define d(E) := max{deg(L)|L ⊂ E is a rank one subbundle}.

Recall that a maximal splitting of a vector bundle E of rank r (on an elliptic curve C) is a sequence of vector bundles(L1, L2,. . . , Lr) such that L1 is a line bundle of maximum degree contained in E and (L2,. . . , Lr) is a maximal splitting of E/L1. Therefore d(E) = deg(L1). Recall also that by a result of Atiyah [1], given an inde- composable vector bundle E of rank r on an elliptic curve, there is a maximal splitting (L1, L2,. . . , Lr) such that L1 ≤ L2 ≤ · · · ≤ Lr are line bundles, where Li ≤ Li+1

denotes that there is an injection Li → Li+1. Moreover, if E is a vector bundle of rank r and degree 0≤ s ≤ r, then E has a maximal splitting with L1= · · · = Lr−s=O, and Lr−s+1= · · · = Lr= L for some line bundle L of degree 1. Notice that

µ(E) = d(E) = deg(L1).

We also remark, that by a result of Tu [16], any indecomposable vector bundle is semistable.

Lemma 4.3 Let E be an indecomposable vector bundle on an elliptic curve, then the following are equivalent

(1) deg(E) > 0, (2) E is IT0, (3) E is ample.

Proof (1) ⇒ (2). If deg(E) > 0 and h1(E ⊗ P) = 0 for some P ∈ Pic0(E), then h0(E⊗ P) = 0 and hence there is a non-zero homomorphism P → E. This is impossible as by Tu [16], Eis semi-stable of negative slope.

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(2) ⇒ (3). Let E be an IT0vector bundle. By a result of Hartshorne [8] (Theorem 1.3), it suffices to show that every quotient line bundle has degree≥ 1. Let L be a quotient line bundle of E. It is easy to see that E being IT0implies that L is IT0and hence has degree≥ 1.

(3) ⇒ (1). If E is ample, then det(E) is an ample line bundle. So deg(E) =

deg(det(E)) > 0.

Lemma 4.4 Let E be an indecomposable vector bundle on an elliptic curve with d(E) ≥ 1. Then E is generically generated by its global sections.

Proof This can be easily proven by induction on rank. If the rank of E is 1, then the statement follows from the fact that h0(E) > 0. Suppose that the statement is true when the rank is< r. Given E of rank r, the quotient bundle F = E/L1has a decom- position F= ⊕Fiwhere each Fiis indecomposable with d(Fi) ≥ 1. It follows from the inductive hypothesis that each Fiand hence also F is generically generated by its global sections. Since H1(L1) = 0, it follows that 0 → H0(L1) → H0(E) → H0(F) → 0 is exact. Therefore we have a commutative diagram

0 −−−−→ H0(L1) ⊗OC −−−−→ H0(E) ⊗OC −−−−→ H0(F) ⊗OC −−−−→ 0

α β γ

0 −−−−→ L1 −−−−→ E −−−−→ F −−−−→ 0

Sinceα and γ are surjective on an open set then so is β.

Definition 4.5 For any vector bundle E, we write E= ⊕Eifor its decomposition in to indecomposable vector bundles. We defineν(E) := min µ(Ei).

Theorem 4.6 Let a : X→ C be a morphism from a smooth projective variety of gen- eral type to an elliptic curve, F a general fiber and P∈ Pic0(C). If |mKF| is birational for some m≥ 2 and ν(aωmX) ≥ 1, then |mKX+ aP| separates points on F.

Proof This follows easily from the above Lemma 4.4.

Lemma 4.7 Let E, F be IT0vector bundles on an elliptic curve, then E⊗F is IT0. Proof By Lemma 4.3, one sees that for a vector bundle, being IT0is equivalent to being ample. Moreover, if E, F are ample, then E⊗F is ample. The Lemma now follows

easily.

Lemma 4.8 Let E1, E2 be indecomposable vector bundles on an elliptic curve. Then Hom(E1, E2) = 0 implies that µ(E2) ≥ µ(E1).

Proof See [16].

Lemma 4.9 Let E, F be vector bundles on an elliptic curve. Thenν(E⊗F) ≥ ν(E) + ν(F).

This is presumably well known, but as we could not find a reference, we include a sketch of the proof.

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Proof It suffices to show that if E and F are indecomposable, thenν(E⊗F) ≥ µ(E) + µ(F). Recall that by Lemma 24 of [1] we may write E ∼= EFh and F ∼= FFh

where(rank(E), deg(E)) = 1, (rank(F), deg(F)) = 1, h, h≥ 1 andFkdenotes the indecomposable vector bundle of degree 0 and rank k. By Lemma 18 of [1],FhFh

decomposes as a direct sum of indecomposable vector bundle of degree 0 sayFhi. It follows that

ν(E⊗F) = ν(E⊗ FFhFh) = min{ν(E⊗ FFhi)} = ν(E⊗ F).

Here we have used the fact that if G is an indecomposable vector bundle then ν(G⊗Fhi) = ν(G), which follows easily from Lemmas 24 and 18 of [1].

Sinceµ(E) = µ(E) and µ(F) = µ(F), it suffices to show that ν(E⊗F) ≥ µ(E) + µ(F).

If we write rank(E) = pk11· · · pknnand rank(F) = ql11· · · qlnmwhere pi(resp. qi) are distinct primes and ki(resp. li) are positive integers, then By Lemma 29 of [1], we have E= E1⊗ · · · ⊗Enand F= F1⊗ · · · ⊗Fmwhere the Ei(resp. Fi) are indecomposable vector bundles of rank pkii(resp. qlii) and(deg(Ei), pi) = 1 (resp. (deg(Fi), qi) = 1).

By Theorems 13 and 14 of [1], one sees that if pi= qjthen Ei⊗Fjsplits as a direct sum of irreducible vector bundles whose rank is a power of pi= qjand whose slope is also given byν(Ei⊗Fj) = µ(Ei) + µ(Fj). Repeatedly applying Lemma 28 of [1], one sees that E⊗ Fdecomposes as a product of indecomposable vector bundles of slope

µ(E) + µ(F).

Lemma 4.10 Let E be an IT0vector bundle on an elliptic curve which admits a short exact sequence

0→ F → E → Q → 0 such that Q has generic rank≤ 1. Then ν(E) ≥ min{1, ν(F)}.

Proof We distinguish the follows cases:

1. Q is torsion.

For every indecomposable component of E, say Ei, there exists a non-zero map from an indecomposable component Fjof F. By Lemma 4.8,µ(Ei) ≥ µ(Fj). So ν(E) ≥ ν(F). Notice that this step did not require E to be IT0.

2. Q is torsion-free.

Then Q is a line bundle. Since E is IT0, it follows that Q is IT0, henceµ(Q) ≥ 1.

For any Ei, there exists a non-zero map from a component Fjof F unless Ei= Q.

In either case, we haveµ(Ei) ≥ min{1, ν(F)}.

3. In general, let L := Q∗∗, and K := ker(E → L). Then it is clear that K fits into a short exact sequence

0→ F → K → Qtor→ 0.

Thusν(K) ≥ ν(F) by (1). By (2), ν(E) ≥ min{1, ν(F)}.

5 Proof of the main theorem

Proof of Theorem 1.1 We distinguish the follows cases according to their Albanese fiber dimension:

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1. f := dim X − dim alb(X) = 0.

It follows from generic vanishing theorem (cf. [6, 7]) that ifχ(ωX) > 0, then KX

is full. Therefore, by Proposition 2.9, we are done.

2. f= 1.

By [11] Corollary 3.2,χ(ωX) = χ(aωX) − χ(R1aωX) and by [10] Proposition 7.6, R1aωX = ωW. Since W is of maximal Albanese dimension, by Theorem 1 of [6],χ(ωW) ≥ 0. Hence χ(aωX) ≥ χ(ωX) > 0. By the generalized generic vanishing theorem of [4], Proposition 2.2, one sees that aωX is full and hence so is KX. Therefore, by Proposition 2.9, we are done with this case.

3. f= 2.

Let a : X → C ⊂ Alb(X) be the Albanese map. Note that a has connected fibers. Consider in fact its Stein factorization X→ W → C. Since W is normal, it is a smooth curve. There are induced maps of abelian varieties Alb(X) → Alb(W) → Alb(C) which are easily seen to be isomorphisms, so that W = C.

3.a. If g(C) ≥ 2.

This case follows from Proposition 3.1.

3.b. g(C) = 1.

We may assume that q(X) = 1 and thus pg(X) ≥ 1. We have that χ(X, ωX) = χ(C, aωX) − χ(C, R1aωX) + χ(C, R2aωX).

Moreover, by [10] Proposition 7.6, R2aωX= ωC=OCand soχ(R2aωX) = 0. Now, h1(C, R1aωX) = q(X) − h0(C, R2aωX) = 0 implies that χ(C, R1aωX) ≥ 0. Hence, one hasχ(C, aωX) ≥ 0. So by the generalized generic vanishing of [4], Proposition 2.2, we see that KXis full as in Case 2.

Let F be the general fiber of a and F0be its minimal model. If|4KF| is birational, then we are done by Theorem 2.8. So by Bombieri’s [2] classification, it remains to consider the case that(K2F0, pg(F0)) = (1, 2). We recall that for this surface, q = 0,

|KF0| has a unique base point and |2KF0| is base point free.

Claim Let Em= aXm). Then ν(Em) ≥ 1 for m ≥ 5.

Grant this for the time being. Then, by Lemma 4.4, for any P∈ Pic0(C) ∼= Pic0(X), every indecomposable component of Em⊗P for m ≥ 5 is generically generated by global sections. That is, there exists an open set U⊂ C and a surjection

H0(Em⊗ P) ⊗OU → (Em⊗ P)|OU. In particular, there is a surjection

H0(X, ωmX⊗ aP) → H0(Em⊗ P) ⊗OU → Em⊗ P ⊗ k(y) ∼= H0mFy), for general y∈ U. If follows that |mKX+ P| separates points on a general fiber Fyfor m≥ 5. Since |mKX+ P| also separates points on different general fibers for m ≥ 3 (cf. Theorem 2.8), we are done.

Therefore it remains to verify the Claim.

Let Rm := H0mF), we first show that R1Rm⊂ Rm+1has codimension≤ 1 for all m≥ 0. To see this, we may assume that F = F0. Let Y ∈ |KF| be a general curve. We have that Y is non-singular, for otherwise Y is singular at the base point of|KF| and hence 1= Y· Y ≥ 2 for Y= Y ∈ |KF|.

We claim that the cokernel of the homomorphism H0Fm) ⊗ H0F) → H0Fm+1) has dimension 1 for m= 1 and m ≥ 3 and dimension 0 for m = 2. Consider in fact

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the exact sequence:

0→ ωm−1F → ωmF ⊗ H0F) → ωm+1F →C(z) → 0

where z is the unique base point of|KF|. If m ≥ 3, all higher cohomologies vanish and so we get an exact sequence of global sections. An easy spectral sequence argument shows that the sequence

0→ H0m−1F ) → H0mF) ⊗ H0F) → H0m+1F )

is also exact for m= 1, 2. By an easy dimension count, the above statement follows.

We now consider the multiplication map

ϕm: aωX⊗ aωXm→ aωXm+1.

By the above claim, one sees thatϕmhas cokernel of generic rank≤ 1 for all m ≥ 1 and it is generically surjective for m= 2. Notice that Em= aωmXis IT0for m≥ 2 by Lemma 4.1. By Lemma 4.7, deg Em> 0 for m ≥ 2

We now distinguish the following cases:

1. E1 := aωXis indecomposable.

Since KX is full, so is E1. By the semipositivity of aωX (cf. [5]), we have that deg(E1) ≥ 0. If deg(E1) = 0, then E1= U ⊗ P for some vector bundle U with maximal splitting(OC,OC) and some P ∈ Pic0(C). However, U ⊗ P is not full.

Thus we have deg(E1) > 0 and hence µ(E1) ≥ rk(E)1 = 12. By Lemma 4.9, one sees thatν(E1⊗ E1) ≥ ν(E1)+ν(E1) ≥ 1 and hence ν(imϕ1) ≥ 1 by Lemma 4.8.

Moreover, the cokernel ofϕ1is IT0and has rank≤ 1, so ν(E2) ≥ 1 by Lemma 4.10. Repeatedly applying this argument, one sees thatν(Em) ≥ 1 for m ≥ 5.

2. E1is decomposable.

Then it is of the form L⊕ P with deg(L) ≥ 1, deg(P) ≥ 0. Note that ϕ1 factors through Sym2(E1) = P2⊕ L ⊗ P ⊕ L2. Moreover, im1) has rank ≥ 3 because ϕ1has cokernel of rank≤ 1. It follows that Sym2(E1) ∼= im(ϕ1).

2.a. E2is indecomposable.

By Lemma 4.8, µ(E2) ≥ 2 for there is a non-zero map from L2 to E2. By consideringϕ2, ...,ϕm−1and Lemma 4.9, one has thatν(Em) ≥ 1 for m ≥ 5.

2.b. E2 =E1E2with ranks 3, 1 respectively.

Thenµ(E1) ≥ 1 for there is a non-zero map from L or L2 in to im1) toE1. Since E2is IT0, so isE2. Thusµ(E2) ≥ 1 for it is a line bundle of IT0. Hence we haveν(E2) ≥ 1. Proceeding as above, it follows that ν(Em) ≥ 1 for m ≥ 5.

2.c. E2 = ⊕Eifactors into 4 bundles of rank 1.

Since E2 is IT0, so are the bundlesEi. Thusµ(Ei) ≥ 1 and hence ν(E2) ≥ 1.

Proceeding as above, it again follows thatν(Em) ≥ 1 for m ≥ 5.

2.d. E2decomposes into bundles of rank(2, 1, 1) or (2, 2).

Since each component is IT0, we haveν(E2) ≥ 12. We then consider the multi- plication map

ψ : aω2X⊗ aω2X → aω4X.

One can check that R2R2 ⊂ R4has codimension≤ 1 (its image in fact contains the image of R1R1R2and hence it contains the image of R1R3which has codi- mension at most 1). By Lemma 4.9,ν(E2⊗ E2) ≥ ν(E2) + ν(E2) ≥ 1 and hence ν(E4) ≥ 1 by Lemma 4.10. Then, by considering ϕ4,. . . , ϕm−1, it follows that ν(Em) ≥ 1 for m ≥ 5. This completes the proof.

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Acknowledgments We are indebted to Meng Chen, De-Qi Zhang, Yujiro Kawamata, I-Hsun Tsai, Hélène Esnault, Eckart Viehweg and Chin-Lung Wang for useful conversations and comments on this subject and to the referee for many helpful suggestions.

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