www.elsevier.com/locate/jalgebra

**A non-vanishing theorem for Q-divisors** on surfaces

^{✩}

### Jungkai Alfred Chen

^{a,b}### , Meng Chen

^{c}

### , De-Qi Zhang

^{d,}^{∗}

a*Department of Mathematics, National Taiwan University, Taipei, 106, Taiwan*
b*Mathematics Division, National Center for Theoretical Science at Taipei, Taiwan*

c*Institute of Mathematics, Fudan University, Shanghai, 200433, PR China*
d*Department of Mathematics, National University of Singapore,*

*2 Science Drive 2, Singapore 117543, Singapore*
Received 29 September 2004
Available online 8 June 2005
Communicated by Michel Broué

**Abstract**

*We prove a non-vanishing theorem of the cohomology H*^{0}*of the adjoint divisor K*_{X}*+ L where*

**L is the round up of a nef and big Q-divisor L.**

2005 Published by Elsevier Inc.

**1. Introduction**

**We work over the complex number field C. The motivation of this note is to find an**
effective version of the famous non-vanishing theorem of Kawamata and Shokurov (see
[KMM,Sh]). We propose the following:

✩ The first author was partially supported by NSC, Taiwan. The second author was supported by the National Natural Science Foundation of China (No. 10131010). The third author was supported by an Academic Research Fund of NUS.

* Corresponding author.

*E-mail addresses: jkchen@math.ntu.edu.tw (J.A. Chen), mchen@fudan.edu.cn (M. Chen),*
matzdq@nus.edu.sg (D.-Q. Zhang).

0021-8693/$ – see front matter 2005 Published by Elsevier Inc.

doi:10.1016/j.jalgebra.2005.05.005

**Conjecture 1.1. Let X be a non-singular projective variety. Let L be a Q-divisor on X***satisfying the conditions below:*

*(1) L is nef and big,*
*(2) K*_{X}*+ L is nef, and*

*(3) either L is a Cartier integral divisor, or L is effective.*

*Then H*^{0}*(X, K*_{X}*+ L) = 0, where L is the round up of L.*

This kind of non-vanishing problem has been considered by Ambro [Am], Chen and Ha-
*con [CH], Kawamata [Ka], Kollar [Ko], Takayama [Ta], and others. When L is an integral*
*Cartier divisor, Kawamata [Ka] has proved the above Conjecture 1.1 if either dim X*= 2,
*or dim X= 3 and X is minimal (i.e., the canonical divisor K**X*is nef).

Conjecture 1.1 is slightly different from that of Kawamata’s in [Ka]. It is somewhat
*general in the sense that the divisor L in question is not assumed to have integral coeffi-*
*cients. It is precisely this non-Cartierness of L that causes a lot of trouble when estimating*
*h*^{0}*(X, K*_{X}*+ L). To elaborate, the Kawamata–Viehweg vanishing [KV,Vi1] implies that*
*h*^{0}*(X, K*_{X}*+ L) = χ(K**X**+ L) when the fractional part of L is of normal crossings.*

*However, the Riemann–Roch formula for χ may not be effective becauseL may not be*
nef and hence*L.(K**X**+ L) may not be non-negative when X is a non-singular surface.*

**The worse thing is that as remarked in a recent paper of [Xi], there are Q-Fano 2-folds and**
*3-folds (see [Fl]) with vanishing H*^{0}*(X, K*_{X}*+ (−2K**X**)).*

*Despite of the observations above, in [Xi] it is proved that H*^{0}*(X, K**X**+ (D − K**X**))*= 0
**for Picard number one Gorenstein del Pezzo surface X and nef and big Q-Cartier Weil di-***visor D. In this note we shall prove the following which is a consequence of Theorems 4.1,*
*5.1, 8.1 and 8.2 (for the case of integral Cartier L, see [Ka]).*

**Theorem 1.2. Let X be a non-singular projective surface. Suppose that X and L satisfy***the conditions (1)–(3) in Conjecture 1.1. Suppose further that X is relatively minimal. Then*
*either H*^{0}*(X, K*_{X}*+ L) = 0 or H*^{0}*(X, K*_{X}*+ 4L*red*)= 0.*

*The second conclusion may occur when K** _{X}* is nef (and the Kodaira dimension

*κ(X)*1). In this case, the conditions in Conjecture 1.1 are automatically satisfied when-

**ever L is nef and big. So if L is an effective Q-divisor with all coefficients less than 1, then***the non-vanishing of H*

^{0}

*(X, K*

_{X}*+ L) is equivalent to that of H*

^{0}

*(X, K*

_{X}*+ L*red

*), which*is stronger than our conclusion. Remark 8.4 shows that it is hard to replace the coefficient

“4” in the theorem above by “1.”

In Sections 3 and 6 (Theorems 3.1 and 6.1), we prove the following non-vanishing results without assuming the condition (3) in Conjecture 1.1, and the proof presented for the first assertion is applicable to higher dimensional varieties. The Fourier–Mukai transforms are applied in the proof.

**Theorem 1.3. Let X and L be as in Conjecture 1.1 satisfying the first two conditions only.**

*Then H*^{0}*(X, K*_{X}*+ L) = 0 if either*

*(i) X is a surface with irregularity q(X) > 0, or*

*(ii) X is a relatively minimal elliptic surface with κ(X)= −∞ and K**X**+ L nef and big.*

**Remark 1.4. (1) In Example 2.6, we construct an example of pair (X, L) satisfying the***conditions (1) and (2) in Conjecture 1.1 (indeed, both L and K*_{X}*+ L are nef and big) but*
*with H*^{0}*(X, K*_{X}*+ L) = 0. So an extra condition such as the (3) in Conjecture 1.1 is*
necessary.

(2) The same example shows that in Kollar’s result [Ko] on non-vanishing of
*H*^{0}*(X, K**X**+ M) for big divisor M, the “bigness” assumption on the fundamental group*
*π*_{1}*(X) is necessary, because in (1) the M:= L L is big and π*1*(X)= (1).*

(3) The example also shows the necessity to assume the nefness of the Cartier integral
*divisor D (with (X, B) klt and D− (K**X**+ B) nef and big) in Kawamata’s conjecture [Ka]*

*for the non-vanishing of H*^{0}*(X, D). Indeed, in the example, we haveL = L + B with*
*B a simple normal crossing effective divisor so that[B] = 0, whence (X, B) is klt. To be*
*precise, let D:= L. Then D − (K**X**+ B) = L − B = L is nef and big, D = K**X**+ L,*
*and D is not nef for D.D** _{i}*= −1 with the notation in the example.

We end the introduction with:

**Remark 1.5. Consider a fibred space f : V***→ C where V is a non-singular projective*
**variety and C a complete curve. Assume L is a nef and big normal crossing Q-divisor***such that K**V* *+ L is nef. The well-known positivity says that f*∗*(ω**V /C* *⊗ O**V**(L)) is*
*positive whenever it is not equal to 0. Pick up a general fibre F of f . The induction of the*
*non-vanishing problem on F may imply that*

rk
*f*_{∗}

*ω**V /B**⊗ O**V*

*L*

*= h*^{0}

*F, K**F**+ L|**F*

* h*^{0}

*F, K**F**+ L|**F*

*= 0.*

*The positivity of f*_{∗}*(ω*_{V /B}*⊗ O**V**(L)) has direct applications in studying properties of*
the moduli schemes for polarized manifolds. Please refer to [Vi2] for more details.

The above remark shows one aspect of the importance of the effective non-vanishing
**for Q-divisors.**

**2. Some preparations and an example**
We begin with:

**Definition 2.1. A reduced connected divisor Γ , with only simple normal crossings, is a***rational tree if every component of Γ is a rational curve and the dual graph of Γ is a tree*
(i.e., it contains no loops).

Before proving Proposition 2.4 below, we need two lemmas in advance.

* Lemma 2.2. Let D*=

_{n}*j*=1*D*_{j}*be a reduced connected divisor on a non-singular pro-*
*jective surface X. Then D.(K*_{X}*+ D) −2 and the equality holds if and only if D is a*
*rational tree.*

**Proof. Note that**

*k<j**D**k**.D**j* * n − 1 and the equality holds if and only if D is a tree.*

We calculate:

*D.(K**X**+ D) =*

*D*_{j}^{2}+

*K**X**.D**j*+ 2

*k<j*

*D**k**.D**j*

*j*

*2p*_{a}*(D*_{j}*)*− 2

*+ 2(n − 1) −2.*

The lemma follows. 2

**Lemma 2.3. Suppose that X is a non-singular projective surface with χ (**O*X**)= 1 and*
*D (= 0) a connected reduced divisor such that H*^{0}*(X, K*_{X}*+ D) = 0. Then the following*
*statements are true.*

*(1) D is a connected rational tree.*

*(2) Suppose further that D supports a nef and big divisor (so D is automatically con-*
*nected). Then π*_{1}*(X)= (1).*

**Proof. The Serre duality and Riemann–Roch theorem imply 0***= h*^{0}*(X, K*_{X}*+ D) =*
*h*^{1}*(X, K*_{X}*+ D) +* ^{1}_{2}*(K*_{X}*+ D).D + χ(O**X**) 0 + (−1) + 1 by Lemma 2.2. Thus*
*D.(K*_{X}*+ D) = −2 and hence D is a connected rational tree by the same lemma. So*
*π*_{1}*(D)= (1). Suppose that D supports on a nef and big effective divisor. Then the sur-*
*jective map π*_{1}*(D)→ π*1*(X) in Nori [No, Corollary 2.3] infers π*1*(X)= (1). 2*

*The next result is a very important restriction on X and L in Theorem 1.2.*

**Proposition 2.4. Let X be a non-singular projective surface with q(X)**= 0 and L a nef**and big effective Q-divisor such that H**^{0}*(X, K*_{X}*+ L*red*)= 0. Then χ(O**X**)= 1, L*red*is a*
*connected rational tree and X is simply connected.*

**Proof. Note that p**_{g}*(X) h*^{0}*(X, K*_{X}*+ L*red*)= 0. So χ(O**X**)*= 1. Now the proposition
follows from Lemma 2.3. 2

The result below is used in the subsequent sections.

**Lemma 2.5. Suppose that X is a minimal non-singular projective surface with Kodaira***dimension κ(X)= 1, p**g**(X)= 0, and π*_{1}^{alg}*(X)= (1) (this is true if π*1*(X)= (1)). Let*
*π : X***→ P**^{1}*be the unique elliptic fibration with F a general fibre. The following statements*
*are true.*

*(1) π has exactly two multiple fibres F*_{1}*, F*_{2}*, and their multiplicities m*_{1}*, m*_{2}*are coprime.*

*In particular, if E is horizontal then E.F* *= m*1*m*_{2}*m*_{3}(* 6) for some positive integer*
*m*_{3}*.*

*(2) Suppose further that a reduced connected divisor D on X is a rational tree and con-*
*tains strictly the support of an effective Γ of elliptic fibre type. Then Γ is a full fibre*
*of π and of type II*^{∗}*, (m*_{1}*, m*_{2}*)= (2, 3) and E.F = 6 for some E in D (see [BPV,*
*Chapter V, §7], for notation of singular fibres).*

**Proof. (1) Since π**_{1}*(X)*^{alg}*= (1), we have H*1**(X, Z)**= (0) and hence q(X) = 0. So*χ (O**X**) = 1. Since κ(X) = 1, there is an elliptic fibration π : X → π(X) = P*

^{1}, where

**the image is P**

^{1}

*because q(X)= 0. Let F*

*i*(1

*i t) be all multiple fibres of π, with*

*multiplicity m*

_{i}*. If m= gcd(m*1

*, m*

_{2}

*) 2, then the relation m(F*1

*/m− F*2

*/m)*∼ 0 induces

**an unramified Galois Z/(m)-cover of X, contradicting the assumption π**_{1}

*(X)*

^{alg}

*= (1). If*

*t*3, then by Fox’s solution to Fenchel’s conjecture (see [Fo,Ch]), there is a base change

*B*

**→ P**

^{1}

*ramified exactly over π(F*

_{i}*) (1 i t) and with ramification index m*

*i*. Then the

*normalization Y of the fibre product X*×

_{P}^{1}

*B is an unramified cover of X (so that the*

*induced fibration Y*

*→ B has no multiple fibres), again contradicting the assumption that*

*π*

_{1}

*(X)*

^{alg}

*= (1).*

On the other hand, by the canonical divisor formula, we have

*K*_{X}*= π*^{∗}*(K*** _{P}**1

*)+ χ(O*

*X*

*)F*

_{1}+

*t*
*i*=1

*(m*_{i}*− 1)(F**i**)*_{red}∼**Q**

−1 +

*t*
*i*=1

1− 1

*m*_{i}

*F*_{1}

*(so π is the only elliptic fibration on X). Since κ(X)= 1, we see that t 2. Now the*
lemma follows from the results above.

*(2) Since Γ is of elliptic fibre type, 0= K**X**.Γ* *= Γ*^{2}*= 0. Hence Γ is a multiple of a*
*fibre of π . Since the support of Γ (< D) is a tree, it is of type I*^{∗}* _{n}*, II

^{∗}, III

^{∗}or IV

^{∗}, whence

*Γ is a full fibre (and is not a multiple fibre). By the assumption, there is an E in D such*

*that Supp(E+ Γ ) is a connected rational tree. Thus E.Γ 6 and the equality holds if and*

*only if Γ is of type II*

^{∗}

*and E meets the coefficient-6 component of Γ . Now (2) follows*from (1). 2

The example below shows that an assumption like the condition (3) in Conjecture 1.1 might be necessary.

**Example 2.6. We shall construct a non-singular projective surface X and a Q-divisor L***such that the conditions (1) and (2) in Conjecture 1.1 are satisfied, but that H*^{0}*(X, K** _{X}*+

*L) = 0. Indeed, we will see that both L and K**X***+ L are nef and big Q-divisors.**

*Let C be a sextic plane curve with 9 ordinary cusps (of type (2, 3)) and no other sin-*
**gularities. This C (regarded as a curve in the dual plane P**^{2}^{∗}) is dual to a smooth plane
*cubic (always having 9 inflections). Let X***→ P**^{2}*be the double cover branched at C. Then*
*X is a normal K3 surface with exactly 9 Du Val singularities (lying over the 9 cusps) of*
*Dynkin type A*_{2}*. Let X be the minimal resolution. According to Barth [Ba], these 9A*_{2}
*are 3-divisible. That is, for some integral divisor G, we have 3G*∼_{9}

*i*=1*(C*_{i}*+ 2D**i**)*

where

*(C*_{i}*+ D**i***) is a disjoint union of the 9 intersecting P**^{1} *(i.e., the 9A*_{2}*). Let H*
*be the pull back of a general line away from the 9 cusps on C. Then H*^{2}= 2 and
*H is disjoint from the 9A*2*, so H.G= 0. We can also calculate that G*^{2}= −6. Now
*let L= H + G −* ^{1}_{3}9

*i*=1*(C**i* *+ 2D**i**). Then* *L = H + G and L*^{2}= −4. Clearly,
*K*_{X}*+ L = L ≡ H is nef and big. However, by the Kawamata–Viehweg vanishing, and*
*Riemann–Roch theorem, we have h*^{0}*(X, K*_{X}*+ L) =* ^{1}_{2}*L*^{2}+ 2 = 0.

A similar example can be constructed, if one can find a quartic surface with 16 nodes (i.e., a normal Kummer quartic surface).

**3. Irregular surfaces**

In this section, we shall show that Conjecture 1.1 holds true (with only the first two
*conditions there but not the last condition) for surfaces X with positive irregularity q(X).*

*To be precise, let X be a non-singular projective surface with q(X) > 0 and let alb : X*→
*Alb(X) be the Albanese map. Then we have:*

**Theorem 3.1. Let X be a non-singular projective surface with q(X) > 0. Let L be a nef****and big Q-divisor such that K***X**+ L is nef. Then H*^{0}*(X, K+ L) = 0.*

To see this, we need the following lemma:

**Lemma 3.2. Let**F = 0 be a IT^{0} *sheaf on an abelian variety A, i.e., for every i > 0 we*
*have H*^{i}*(A,F ⊗ P ) = 0 for all P ∈ Pic*^{0}*(A). Then H*^{0}*(A,F) = 0.*

The proof can be found in [CH], but we reprove it here.

**Proof. Suppose on the contrary that H**^{0}*(A,F) = 0. Since F is IT*^{0}, the Fourier–Mukai
transform of*F is a locally free sheaf of rank = h*^{0}*(A,F), hence the zero sheaf. The only*
sheaf that transforms to the zero sheaf is the zero sheaf, which is a contradiction. 2
**Proof of Theorem 3.1. Let f : X**^{}*→ X be an embedded resolution for (X, L). It is clear*
*that f*^{∗}*L is nef and big with simple normal crossing support. Let ∆:= f*^{∗}*L − f*^{∗}*L, then*
*(X*^{}*, ∆) is Kawamata log terminal (klt for short; for its definition and property, see [KMM,*
Definition 0-2-10]). By a property of nef and big divisor (see, e.g., [La, Example 2.2.17]),
*there is an effective divisor N such that A**k**:= f*^{∗}*L*−^{1}_{k}*N is ample for all k* 0. We
*fix k such that (X*^{}*, ∆*+^{1}_{k}*N ) is klt. Now we can write A**k**= (alb ◦f )*^{∗}*M+ E for some*
**ample Q-divisor M on A**:= Alb(X) and effective divisor E on X^{}. Pick irreducible divisor
*B∈ |(n − 1)A| for n 0 such that (X*^{}*, ∆*^{}*) is klt, where ∆*^{}*:= ∆ +*^{1}_{k}*N*+_{n}^{1}*E*+_{n}^{1}*B. Then*
*we have, where P*^{}*= (alb ◦f )*^{∗}*P with P* ∈ Pic^{0}*(A):*

*K** _{X}*+

*f*

^{∗}

*L*

*+ P*^{}*≡ K**X*^{}+*(alb◦f )*^{∗}*M*
*n* *+ ∆*^{}*.*

Let *F := alb*_{∗}*f*_{∗}*O**X*^{}*(K*_{X}*+ f*^{∗}*L). By Kollar’s relative vanishing theorem (cf. [Ko,*
10.19.2]), one sees that*F is IT*^{0}.

We claim that*F = 0.*

Grant this claim for the time being. By the above lemma, it follows that
*h*^{0}

*X*^{}*, K** _{X}*+

*f*

^{∗}

*L*

*= h*^{0}*(A,F) = 0.*

*Since K*_{X}*+ f*^{∗}*L = f*^{∗}*(K*_{X}*+ L) + Γ , where Γ is an exceptional divisor (possibly*
*non-effective). It’s easy to see that f*_{∗}*O**X*^{}*(Γ )⊂ O**X*. By the projection formula, one has:

0*= H*^{0}

*X*^{}*, K** _{X}*+

*f*

^{∗}

*L*

*= H*^{0}
*X,O**X*

*K*_{X}*+ L*

*⊗ f*∗*O**X**(Γ )*

*⊂ H*^{0}

*X, K*_{X}*+ L*
*.*

This is the required non-vanishing.

*To see the claim, if dim(alb(X))= 2, then alb ◦f is generically finite. Hence it is clear*
that*F = 0. If dim(alb(X)) = 1. Let F be a general fiber of alb ◦f . Then we have:*

*rank(F) = h*^{0}
*F,*

*K** _{X}*+

*f*

^{∗}

*L*

*F*

*= h*^{0}

*F, K**F*+
*f*^{∗}*L*|*F*

*.*

*Since f*^{∗}*L is big, f*^{∗}*L.F > 0. It follows that deg(f*^{∗}*L*|*F**) > 0.*

*If g(F ) > 0, then we have h*^{0}*(F, K*_{F}*+ f*^{∗}*L*|*F**) > 0 already. If g(F ) = 0, note that*
*K*_{X}*+ L is nef. Note also that (K**X*^{}*+ f*^{∗}*L).F* *= (K**X**+ L).f (F ) since F is general. This*
implies that

deg
*K** _{F}*+

*f*^{∗}*L*|*F*

=

*K*_{X}*+ f*^{∗}*L*
*.F*+

*f*^{∗}*L*

*− f*^{∗}*L*
*.F*

*= (K**X**+ L).f (F ) +*
*f*^{∗}*L*

*− f*^{∗}*L*
*.F 0.*

*Hence h*^{0}*(F, K**F* *+ L**F**) > 0. We conclude that F = 0 and hence the required non-*
*vanishing that h*^{0}*(X, K**X**+ L) = 0. 2*

**Remark 3.3. In the proof of Theorem 3.1, without taking log-resolution at the beginning,**
one can apply Sakai’s lemma [Sa] for surfaces to get the vanishing of higher cohomology.

However, our argument here works for higher dimensional situation. It shows that non- vanishing for general fiber gives the non-vanishing.

**4. Surfaces of Kodaira dimension 0**

*In this section, we show that Conjecture 1.1 in the introduction is true for surfaces X*
*(not necessarily minimal) with Kodaira dimension κ(X)*= 0.

**Theorem 4.1. Suppose that X is a non-singular projective surface (not necessarily mini-***mal) of Kodaira dimension κ(X) = 0. Then Conjecture 1.1 is true for effective Q-divisor L.*

* Proof. By Theorem 3.1, we may assume that q(X)*= 0. We may also assume that
0

*= h*

^{0}

*(X, K*

_{X}*+ L*red

*) ( p*

*g*

*(X)). So X is the blow up of an Enriques surface by the*

*classification theory. On the other hand, π*

_{1}

*(X)= (1) by Proposition 2.4, a contradiction.*

This proves the theorem. 2

**5. Surfaces with negative κ , part I: Ruled surfaces**

*In this section, we prove Conjecture 1.1 for relatively minimal surfaces X of Kodaira di-*
*mension κ(X)= −∞. By Theorem 3.1, we may assume that q(X) = 0, so X is a relatively*
*minimal rational surface. If X***= P**^{2}**or P**^{1}**× P**^{1}, it is easy to verify that Conjecture 1.1 is
*true since effective divisor is then nef. We thus assume that X is the Hirzebruch surface*F*d*

*of degree d* 1 (though, F1is not relatively minimal).

*We first fix some notations. Let π :*F*d***→ P**^{1}*be the ruling. Let F be a general fibre and*
*C the only negative curve (a cross-section, indeed) on*F*d**. So C*^{2}*= −d.*

**Theorem 5.1. Let X be a relative minimal surface of Kodaira dimension κ(X)**= −∞.

**Then Conjecture 1.1 holds for effective Q-divisor L.**

* Proof. As mentioned above, we assume that X*= F

*d*

*for some d 1. Let L be a nef*

**and big effective Q-divisor such that K**

_{X}*+ L is nef. If Supp(L) does not contain the*

*negative curve C, then E:= L − L is effective and nef; so L = L + E is nef and big*

*and K*

*X*

*+ L = K*

*X*

*+ L + E is nef; then the Serre duality and Riemann–Roch theorem*

*for Cartier divisor imply that h*

^{0}

*(X, K*

_{X}*+ L)*

^{1}

_{2}

*L(K*

*X*

*+ L) + χ(O*

*X*

*)*0 + 1.

*Therefore, we may assume that Supp(L) contains C.*

*Write L*=

*i**c*_{i}*C** _{i}*+

*f*_{j}*F*_{j}*where C*_{1}*= C, the C**i*’s are distinct horizontal compo-
*nents and F*_{j}*’s are distinct fibres, where c*_{i}*> 0, f**j**> 0.*

*Suppose on the contrary that H*^{0}*(X, K**X**+ L) = 0. Then by Lemma 2.3, L*red is a
connected rational tree. Hence one of the following cases occurs:

*(i) L= c*1*C*_{1}+_{k}

*j*=1*f*_{j}*F*_{j}*(k* 0);

*(ii) L*=_{k}

*i*=1*c*_{i}*C*_{i}*+ f*1*F*_{1}*(k 2), and L*red*is comb-shaped, i.e., C** _{i}*’s are disjoint cross-
sections;

*(iii) L*=*k*

*i*=1*c**i**C**i* *(k* 2).

*Recall that K*_{X}*∼ −2C*1*− (d + 2)F . The nefness of K**X**+ L implies:*

0* (K**X**+ L).F = −2 +*

*c**i**(C**i**.F ),*

0* (K**X**+ L).C*1*= d − 2 − dc*1+

*i*2

*c*_{i}*(C*_{i}*.C*_{1}*)*+
*f*_{j}*,*

*c**i**(C**i**.F ) 2,*

*f**j** 2 + (c*1*− 1)d −*

*i*2

*c**i**(C**i**.C*1*).*

*In case (i), the above inequalities imply c*_{1} 2 and

*f*_{j}* 2 + (c*1*− 1)d d + 2,*
whence

*L = c*1*C*1+

*f**j**F**j**∼ c*1*C*1+

*f**j*
*F*

* 2C*1+
*f*_{j}

*F* * 2C*1*+ (d + 2)F*1*∼ −K**X**.*

*Hence H*^{0}*(X, K*_{X}*+ L) = 0.*

*Consider case (ii). Then one sees easily that k= 2 and C*2*∼ C*1*+ dF*1(see [Ha, Chap-
*ter V, §2]). By the displayed inequalities, we have c*_{1}* 2 − c*2*and f*_{1}* 2 + (c*1*− 1)d. If*
*c*_{2}*> 1 thenL C*1*+ 2C*2*+ F*1*>−K**X**, whence H*^{0}*(X, K*_{X}*+ L) = 0. So we may*
*assume that c*_{2}* 1. Then c*1* 1 and f*1* 2. Thus L C*1*+ C*2*+ 2F*1*∼ −K**X*, whence
*H*^{0}*(X, K*_{X}*+ L) = 0.*

*Consider case (iii). Since L is a connected tree, we may assume that C*_{1}*.C*_{2}= 1. So
*C*_{2}*∼ n(C*1*+ dF ) + F for some integer n 1. Since C**i* *(i* 2) is irreducible, we have
*C*_{i}* C*1*+ dF by [Ha]. If k 3 or n 2, then we see that L *_{k}

*i*=1*C*_{i}*>−K**X*. So
*assume that k= 2 and n = 1. By the inequalities displayed above, we have c*1* 2 − c*2and
*c*_{2}* 2 + (c*1*− 1)d. If c*2*> 1 thenL C*1*+ 2C*2*>−K**X**. So assume that c*_{2} 1. Then
*c*_{1}* 2 − 1 and c*2* 2 + 0d, a contradiction. 2*

**6. Surfaces with negative κ , part II: Relatively minimal elliptic**

*In this section we consider relatively minimal elliptic surface π : X→ B with Kodaira*
*dimension κ(X)*= −∞. As far as Conjecture 1.1 is concerned, we may assume that the
*irregularity q(X) = 0 by virtue of Theorem 3.1. So X is a rational surface and B = P*

^{1}.

*By the canonical divisor formula, we see that π has at most one fibre F*

_{0}with multiplicity

*m 2; moreover, such F*0(if exists) is of Kodaira type I

*n*

*(n 0), and −K*

*X*

*= (F*0

*)*

_{red}.

*We show that Conjecture 1.1 is true if K**X**+L is nef and big (but without the assumption*
*of the effectiveness of L):*

**Theorem 6.1. Let π : X**→ B be a relatively minimal elliptic surface with κ(X) = −∞.

**Suppose that L is a nef and big Q-divisor such that K**_{X}*+ L is nef and big. Then*
*H*^{0}*(X, K*_{X}*+ L) = 0.*

**Proof. By Theorem 3.1, we may assume that q(X)****= 0, so B = P**^{1} *and X is a rational*
surface.

**Suppose that the Q-divisor L is nef and big and K**_{X}*+ L is nef. Let F*0*= m(F*0*)*_{red}be
*the multiple fiber. We set m= 1 and let F*0 *be a general (smooth) fibre, if π is multiple*
*fibre free. Then K*_{X}*∼ −(F*0*)*_{red}*. Let a > 0. Consider the exact sequence:*

0*→ O**X*

*K*_{X}*+ aL − (F*0*)*_{red}

*→ O**X*

*K*_{X}*+ aL*

*→ O**(F*_{0}*)*_{red}

*K*_{X}*+ aL|**(F*_{0}*)*_{red}

*→ 0.*

*Let us find the condition for aL− (F*0*)*_{red}*to be nef and big. Note that aL− (F*0*)*_{red}∼
*aL+ K**X**= (a − 1)L + (K**X**+ L). So aL − (F*0*)*_{red}is nef and big if either

*(i) a > 1, or*

*(ii) a= 1 and K**X**+ L is nef and big.*

*Assume that either (i) or (ii) is satisfied. Then H*^{i}*(X, K*_{X}*+ aL − (F*0*)*_{red}*)*= 0 =
*H*^{i}*(X, K*_{X}*+ aL) for all i > 0, by Sakai’s vanishing for surfaces. For the integral divisor*
*M:= K**X**+ aL and the reduced divisor C := (F*0*)*_{red}*on X, the above exact sequence*
implies that

*χ*

*O**C**(M|C)*

*= χ*

*O**X**(M)*

*− χ*

*O**X**(M− C)*

*= C.M − C.(K**X**+ C)/2,*
where we applied the Riemann–Roch theorem for both*O**X**(M) andO**X**(M− C). Now*
*C.(K*_{X}*+ C) = 0 and C.M (F*0*)*_{red}*.(K*_{X}*+ aL) > 0 (for 0 = C being nef and K**X**+ aL*
*nef and big), so χ (O**C**(M|C)) > 0. By the vanishing above,*

*h*^{0}

*X, K**X**+ aL*

*= χ*

*O**X**(M)*

*= χ*

*O**X**(M− C)*
*+ χ*

*O**C**(M|C)*

*= h*^{0}

*X, K*_{X}*+ aL − (F*0*)*_{red}
*+ χ*

*O**C**(M|C)*

*> 0+ 0.*

This proves the theorem. 2

**Remark 6.2. The above argument actually proved the following: Let π : X**→ B be a rela-*tively minimal elliptic surface with κ(X) = −∞. Suppose that L is a nef and big Q-divisor*

*such that K*

*X*

*+ L is nef. Then H*

^{0}

*(X, K*

_{X}*+ aL) = 0 provided that either a > 1, or a = 1*

*and K*

_{X}*+ L is nef and big.*

**7. Preparations for surfaces with κ****= 1 or 2**

*Throughout this section, we assume that X is a non-singular projective surface with K**X*

*nef and Kodaira dimension κ(X)*= 1 or 2. The main result is Proposition 7.10 to be used
in the next section.

* Definition 7.1. Up to Lemma 7.3, we let Γ be a connected effective integral divisor on X*
which consists of smooth rational curves and has a (rational) tree as its dual graph.

*(1) We say that Γ is of type A*^{}_{n}*(respectively D*^{}_{n}*, or E*_{n}^{}) if its weighted dual graph is of
*Dynkin type A*_{n}*(respectively D*_{n}*, or E*_{n}*) but its weights may not all be (−2).*

*(2) Γ is of type I*^{∗}* _{n}*(respectively II

^{∗}, or III

^{∗}, or IV

^{∗}

*) if Γ is of the respective elliptic fibre*

*type (hence Supp(Γ ) is a union of (−2)-curves). Γ is of type I*

^{∗ }

*n*(respectively II

^{∗ }, or III

^{∗ }, or IV

^{∗ }

*) if Γ is equal to an elliptic fibre of type I*

^{∗}

*(respectively II*

_{n}^{∗}, or III

^{∗}, or IV

^{∗}

*), including coefficients, but the self intersections of components of Γ may not*

*all be (−2). E.g., Γ = 2*

_{n}*i*=0*C** _{i}*+

_{n}_{+4}

*j**=n+1**C** _{j}* is of type I

^{∗ }

_{n}*, where C*

_{i}*+ C*0

*+ C*1+

*· · ·+C**n**+C**j**is an ordered linear chain for all i∈ {n+1, n+2} and j ∈ {n+3, n+4}.*

*(3) For a divisor D on X, we denote by #D the number of irreducible components of D.*

*The assertion (1) below follows from the fact that C*^{2}*= −2 − C.K**X* −2. The others
are clear.

**Lemma 7.2.**

*(1) If C is a smooth rational curve on X, then C*^{2}* −2.*

*(2) If Γ is of type A*^{}_{n}*, D*^{}_{n}*or E*^{}_{n}*then it is negative definite, i.e., the intersection matrix of*
*components in Γ is negative definite.*

*(3) If Γ is one of types I*^{∗}_{n}*, II*^{∗}*, III*^{∗} *and IV*^{∗} *(respectively I*^{∗ }_{n}*, II*^{∗ }*, III*^{∗ } *and IV*^{∗ }*, but*
*at least one component of Γ is not a (−2)-curve), then Γ is negative semi-definite*
*(respectively negative definite).*

*(4) If #Γ 5, then Γ is negative definite, unless Γ supports a divisor of type I*^{∗}_{0}*.*
The Picard number can be estimated in the following way:

**Lemma 7.3. Suppose that the (**−2)-components of Γ do not support a divisor of type I^{∗}_{0}*.*
*Let r= min{9, #Γ −1}. Then there is a subgraph Γ*^{}*of r components with negative definite*
*intersection matrix. In particular, ρ(X) r + 1. Also if ρ(X) 9 then #Γ 9.*

**Proof. We have only to prove the first assertion. By taking a subgraph, we may assume**
*that #Γ* 10.

*If Γ is a linear chain, then it has negative definite intersection matrix, and we are done.*

Thus we may assume that there exists an irreducible component which meets more than
*two other irreducible components. Let C*_{0}*be the irreducible component that meets k other*
*components with the largest k. Then Γ* *− C*0*has exactly k connected components{∆**i*}.

*We may assume that k 3. Let C**i* *be the irreducible component of ∆*_{i}*that meets C*_{0}.
*By Lemma 7.2, if #∆*_{i}* 5 for all i then each ∆**i* *is negative definite. By taking Γ*^{}=

*∆** _{i}*, we are done.

*The remaining cases of (#∆*1*, . . . , #∆**k**) are* *{(1, 1, 6), (1, 1, 7), (1, 2, 6), (1, 1, 1, 6)}.*

*For the case (1, 1, 1, 6), we take Γ*^{}*= Γ − C*4*, then now Γ*^{} has at least two connected
*components: C*0*+ C*1*+ C*2*+ C*3and others. It is clear that each connected component has
*at most 5 irreducible components. Hence Γ*^{}*is negative definite. For the cases (1, 1, 6) and*
*(1, 2, 6), similar argument works.*

*It remains to work with the case (1, 1, 7). If C*_{3}meets at least 3 components, we take
*Γ*^{}*= Γ − C*3*. Then Γ*^{}has at least 3 connected components and each one has length 5.

*If C*_{3}*meets 2 components, say C*_{0}*, C*_{4}*, then we take Γ*^{}*= Γ − C*4. Again, each connected
*component of Γ*^{}has at most 5 irreducible components. This proves the lemma. 2

**Lemma 7.4. Suppose that q(X)**= p*g**(X)= 0 and π*_{1}^{alg}*(X)= (1) (these are satisfied in the*
*situation of Proposition 7.10; see its proof ).*

*(1) We have ρ(X) 10 − K*_{X}^{2} * 10, and ρ(X) = 10 holds only when κ(X) = 1.*

*(2) For L in Proposition 7.10, suppose that some (−2)-components of L support an effec-*
*tive divisor Γ of elliptic fibre type. Then Γ is of type II*^{∗}*, κ(X)= 1 and ρ(X) = 10 *

*#L. Moreover, L*_{red}*supports an effective divisor C of type I*^{∗ }_{0} *whose central and three*
*of the tip components are all (−2)-curves.*

* Proof. (1) follows from: ρ(X) b*2

*(X)= c*2

*(X)− 2 + 4q(X) = 12χ(O*

*X*

*)− K*

_{X}^{2}− 2 = 10

*− K*

_{X}^{2}10 (Noether’s equality).

*(2) Since a surface of general type does not contain such Γ , we have κ(X)*= 1. By
*Lemma 2.5 and its notation and noting that L*_{red}*> Supp(Γ ) (for L being nef and big),*
*Γ is of type II*^{∗}*and Supp(E+ Γ ) ( L*red) supports a I^{∗ }_{0} *as described in (2). Also #L*

*#Γ+1 = 10 and ρ(X) 2+(#Γ −1) = 10. Thus ρ(X) = 10. This proves the lemma. 2*
By the lemma above and Lemma 7.3, to prove Proposition 7.10, we may assume:

* Remark 7.5. Assumption: #L 9, and the (−2)-components of L do not support a divisor*
of elliptic fibre type.

We need three more lemmas in proving Proposition 7.10.

* Lemma 7.6. Let D*=

_{n}*i*=0*D*_{i}*be a reduced divisor on X. Suppose that D− D*0*has a*
*negative definite n× n intersection matrix (D**i**.D*_{j}*)*_{1}_{i,jn}*and D supports a divisor with*
*positive self intersection.*

*(1) We have det(D*_{k}*.D*_{}*)*_{0}_{k,n}*> 0 (respectively < 0) if n is even (respectively odd).*

*(2) Assign formally G*_{i}*:= D**i* *and define G*_{i}*.G*_{j}*:= D**i**.D*_{j}*(i= j) and G*^{2}_{i}*:= −x**i**. Sup-*
*pose that*

*(∗) the n × n matrix (G**i**.G*_{j}*)*_{1}_{i,jn}*is negative definite.*

*If G*^{2}_{i}* D*_{i}^{2}*for all 0 i n, then*

*(∗∗) det(G**k**.G*_{}*)*_{0}_{k,n}*> 0 (respectively < 0) if n is even (respectively odd).*

*(3) Suppose that D*^{2}_{i}* −2 for all 0 i n. In (2) above for 0 k n, choose the largest*
*positive integer m**k* *(if exists) such that (∗) and (∗∗) in (2) are satisfied for G**i* *with*
*G*^{2}_{k}*= −m**k**and G*^{2}_{i}*= −2 (i = k). Then D*_{k}^{2}* −m**k**.*

**Proof. For (1), suppose that the matrix in (1) is similar (over Q) to a diagonal matrix J .***Then the condition implies that J has one positive and n negative diagonal entries. So (1)*
follows.

*For (2), we have only to show that a linear combination of G** _{i}* has positive self intersec-

*tion. By the assumption some divisor ∆*=

*b*_{i}*D** _{i}* has positive self intersection, then so

*is Γ*=

*b**i**G**i* *because Γ*^{2}=

*b**i**b**j**G**i**.G**j*

*b**i**b**j**D**i**.D**j**= ∆*^{2}*> 0. The (3) follows*
from (2). 2

*Let D*=_{n}

*i*=0*D*_{i}*be a reduced divisor and let D= P +N be the Zariski decomposition*
**with P the nef and N the negative part so that P and N are effective Q-divisor with***P .N= 0 (see [Fu1,Fu2,Mi]). D supports a nef and big divisor if and only if P*^{2}*> 0.*

*In Lemmas 7.7 and 7.8 below, we do not need the bigness of P ; in Lemma 7.7, K** _{X}*is
irrelevant.

**Lemma 7.7.**

*(1) Write P* =_{n}

*i*=0*p*_{i}*D*_{i}*. Then 0 p**i* * 1, and p**i* *< 1 holds if and only if D**i*
*Supp(N ).*

*(2) Write Supp(N )*=_{s}

*i*=0*D*_{i}*after relabelling. Then (p*_{0}*, . . . , p*_{s}*) is the unique solution*
*of the linear system*_{n}

*i*=0*x*_{i}*(D*_{i}*.D*_{j}*)= 0 (j = 0, . . . , s), where we set x**j**= 1 (j > s).*

*(3) Assign formally G**i* *:= D**i* *and G**i**.G*_{j}*= D**i**.D*_{j}*(i= j). Suppose that for α i β,*
*we assign G*^{2}_{i}*such that−2 G*^{2}_{i}* D*^{2}_{i}*and (G*_{i}*.G*_{j}*)*_{α}_{i,jβ}*is negative definite. Let*
*(x*_{i}*= b**i**| α i β) be the unique solution of the linear system*_{n}

*i*=0*x*_{i}*G*_{i}*.G** _{j}*= 0

*(α j β), where we set x*

*j*

*= b*

*j*

*= p*

*j*

*if j < α or j > β. Then b*

_{i}*p*

*i*

*for all*

*α i β.*

**Proof. For (1), see [Fu1] or [Mi]. (2) follows from the fact that P .D***j**= 0 (0 j s) and*
*that N has negative definite (and hence invertible) intersection matrix.*

We prove (3). It suffices to show that
*(∗∗∗) the sum**β*

*i**=α**(b*_{i}*− p**i**)G*_{i}*.G*_{j}* 0 for all α j β.*

Indeed, write

*(b*_{i}*− p**i**)G*_{i}*= A − B with A 0, B 0 and with no common compo-*
*nents in A and B; then the condition (∗∗∗) implies that A.B − B*^{2}=

*(b**i**− p**i**)G**i**.B* 0;

*this and A.B* * 0 and B*^{2}* 0 imply that B*^{2}*= 0 and hence B = 0 by the negative-*
*definiteness of (G*_{i}*.G*_{j}*).*

*Coming to the sum in (∗∗∗) above, it is equal to*_{n}

*i*=0*b**i**G**i**.G**j*−_{n}

*i*=0*p**i**G**i**.G**j*
0−*n*

*i*=0*p*_{i}*D*_{i}*.D** _{j}* 0. This proves the lemma. 2

**Lemma 7.8. Suppose that Γ***= D*1*+ · · · + D**m**is an ordered linear chain contained in D*
*such that Γ.(D− Γ ) = 1. Let D**t* * Γ and D**m*+1* D − Γ such that D**t**.D*_{m}_{+1}*= 1. If*
*either t= m or D*^{2}_{t}* −3, then Γ Supp(N).*

* Proof. Write P* =

*j**p*_{j}*D*_{j}*. If t= m, we set G*^{2}_{i}*= −2 (1 i m) in Lemma 7.7 and*
obtain

*p**i** b**i*=

*i*
*m*+ 1

*p**m*+1*< 1*

*and hence Γ* * Supp(N). If D*_{t}^{2}* −3, we have only to show that p**t* *< 1 because*
*we already have p**j* *< 1 for every 1 j m with j = t, by the previous case. Now*
0* P.D**t**= p**t**D*^{2}_{t}*+ p**t*−1*+ p**t*+1*+ p**m*+1*<−3p**t**+ 3, whence p**t* *< 1. This proves the*
lemma. 2

*For L in Proposition 7.10, let L*_{red}*= P + N be the Zariski decomposition, so P 0*
*and N 0. By the maximality of P , we have L*red* P εL for a suitable small ε > 0*
*(one can take ε such that 1/ε is the maximum of coefficients in L). So P*_{red}*= L*red. Write

*P* =

*n*
*i*=0

*p*_{i}*C*_{i}*.*

*Then 0 < p*_{i}* 1. Note that p**j* *= 1 for some j for otherwise Supp(L) = Supp(P ) ⊆*
*Supp(N ) would be negative definite. So we assume the following (after relabelling):*

**Remark 7.9. In order to prove the proposition below, we may and will assume that L**= P*and p*_{0}= 1.

Now we state the main result of the section.

**Proposition 7.10. Let X be a minimal non-singular projective surface (i.e., K**_{X}*is nef ) with*
*p*_{g}*(X) = 0. Suppose that L is a nef and big effective Q-divisor supported by a rational tree.*

*Then X is simply connected and Supp(L) is connected. Moreover, either (the number of*
*irreducible components) #L 10 = ρ(X) and κ(X) = 1, or #L 9 and (A) or (B) below*
*is true.*

*(A) There is a linear chain C*=_{r}

*i*=0*C*_{i}* L*red*with r 0 (after relabelling) such that*
*L*_{red}*.*_{r}

*i*=0*C*_{i}* 2.*

*(B) Supp(L) supports an effective divisor C of type in*{I^{∗ }_{n}*, III*^{∗ }*, IV*^{∗ }*} and the weights of*
*the multiplicity 2 components of C are all (−2), so C.(K**X**+ C) = 0. Also the type*
III^{∗ }*occurs only when L*_{red}*is given as follows:*

*(B1) κ(X)= 1 and ρ(X) = 10; det(Pic(X)) = −1, and Pic(X) is generated by the*
*divisor class of K*_{X}*and those of the 9 curves in L*_{red}=_{8}

*i*=0*C*_{i}*; C*_{0} *meets*
*exactly C*_{1}*, C*_{2}*, C*_{3}*; C*_{2}*+ C*4*+ C*6 *and C*_{3}*+ C*5*+ C*7*+ C*8 *are linear chains;*

*C*_{6}^{2}*= −3 and C*_{i}^{2}*= −2 (i = 6).*

**Proof. Since L is nef and big and a rational tree, κ(X)**= 1, 2. Since L is nef and big,*a positive multiple of L is Cartier and 1-connected. By [No, Corollary 2.3] or the proof of*
*Lemma 2.3, π*_{1}*(X)= (1). In particular, q(X) = 0 and χ(O**X**)*= 1.

*Since p*_{0}*= 1 by the additional assumption, C*0 *is not in Supp(N ). Since 0 P.C*0=
*C*_{0}^{2}+

*p*_{j}*and C*^{2}_{0}* −2, where j runs in the set so that C**j* *meets C*_{0}*, this C*_{0}meets at
*least two components of Supp(P )− C*0. Now the proposition follows from the lemmas
below. 2

By Lemma 7.4, to prove the above proposition, we only need to consider the case

*#P* 9.

**Lemma 7.11. Suppose that C**_{0} *meets exactly two components of Supp(P )− C*0*. Then*
*Proposition 7.10 is true.*

**Proof. Suppose that C**_{0}*meets only C*_{1}*and C*_{2}*in Supp(P )− C*0. Then 0* P.C*0*= C*_{0}^{2}+
*p*_{1}*+p*2*and C*_{0}^{2}* −2 imply that p*1*= p*2*= 1 and C*^{2}_{0}= −2. Inductively, we can prove that
there is an ordered linear chain (after relabelling)_{b}

*i**=a**C*_{i}*in Supp(P ) such that p** _{i}*= 1 and

*C*

_{i}^{2}

*= −2 for all a i b and C*

*a*

*(respectively C*

_{b}*) meets C*

_{a}_{−1}

*and C*

_{a}_{−2}(respectively

*C*

_{b}_{+1}

*and C*

_{b}_{+2}) such that

*C*_{a}_{−2}*+ C**a*−1+ 2

*b*
*i**=a*

*C*_{i}*+ C**b*+1*+ C**b*+2

is of type I^{∗ }_{b}* _{−a}*and Proposition 7.10(B) is true. 2

**Lemma 7.12. Suppose that C**_{0} *meets at least four components of Supp(P )− C*0*. Then*
*Proposition 7.10 is true.*

**Proof. Suppose that C**_{0} *meets C** _{i}* (1

*i k) with k 4. Let ∆*

*i*(1

*i k) be the*

*connected component of P*

_{red}

*− C*0

*containing C*

_{i}*. Set n*

_{i}*:= #∆*

*i*. Assume that for only 1

*j s the divisor C*0

*+ ∆*

*j*is a linear chain. By the proof of Lemma 7.8, we have

*p*

_{j}*n*

*j*

*/(n*

_{j}*+ 1) (j s).*

*If P*_{red}*.C*_{0}*= C*_{0}^{2}*+ k 2, then Proposition 7.10(A) is true. So assume that C*_{0}^{2}* 1 − k *

−3. Note that
0* P.C*0*= C*0^{2}+

*k*
*i*=1

*p*_{i}* C*0^{2}*+ (k − s) +*

*s*
*i*=1

*p*_{i}* 1 − s +*

*s*
*i*=1

*p** _{i}* 1 −

*s*
*i*=1

1
*n** _{i}*+ 1

*.*

*Suppose that #∆*

*i*

*= 1 for 1 i s*1

*and #∆*

*i*

*2 for i s*1+ 1. Then

0

*s*
*i**=s*1+1

1

*n** _{i}*+ 1 1 −

*s*

_{1}2

*.*

*Note also that #∆**j** 3 for all s + 1 j k. Thus,*

*3k− s − s*1*= s*1*+ 2(s − s*1*)+ 3(k − s) #P − 1 8.*

*These two highlighted inequalities imply that s= 2 and (#∆*1*, . . . , #∆**k**)= (1, 1, 3, 3).*

*Note that C*_{0}*meets the mid-component C*_{j}*of ∆*_{j}*(j= 3, 4). By the proof of Lemma 7.8,*
*for every j with j= 0, 3, 4, we have p**j* * 1/2. Thus 0 P.C*0*= C*_{0}^{2}+4

*i*=1*p**i* −3 +
*1/2+ 1/2 + 1 + 1 = 0, so C*_{0}^{2}*= −3 and p*3*= p*4*= 1. Now 0 P.C*3* C*_{3}^{2}*+ p*0*+ 1/2 +*
*1/2 implies C*_{3}^{2}*= −2. So P*red*.(C*_{0}*+ C*3*)*= 2 and Proposition 7.10(A) is true. 2

*Now we assume that C*_{0}*meets exactly three components C*_{i}*(i= 1, 2, 3) of Supp(P ) −*
*C*_{0}*. Let ∆*_{i}*be the connected component of Supp(P )− C*0*containing C*_{i}*. Set n*_{i}*:= #∆**i*.
Then_{3}

*i*=1*n*_{i}*= #P − 1 8. We may assume that n*1* n*2* n*3*. Then n*_{3}* 6 and n*1 2,
*so C*_{0}*+ ∆*1*is a linear chain. By the proof of Lemma 7.8, we have p*_{1}* n*1*/(n*_{1}*+ 1) < 1.*