A non-vanishing theorem for Q-divisors on surfaces

22  Download (0)

Full text



A non-vanishing theorem for Q-divisors on surfaces

Jungkai Alfred Chen


, Meng Chen


, De-Qi Zhang


aDepartment of Mathematics, National Taiwan University, Taipei, 106, Taiwan bMathematics Division, National Center for Theoretical Science at Taipei, Taiwan

cInstitute of Mathematics, Fudan University, Shanghai, 200433, PR China dDepartment 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


We prove a non-vanishing theorem of the cohomology H0of the adjoint divisor KX+ 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.



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) KX+ L is nef, and

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

Then H0(X, KX+ 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 KXis 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 h0(X, KX+ L). To elaborate, the Kawamata–Viehweg vanishing [KV,Vi1] implies that h0(X, KX+ L) = χ(KX+ 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 henceL.(KX+ 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 H0(X, KX+ (−2KX)).

Despite of the observations above, in [Xi] it is proved that H0(X, KX+ (D − KX))= 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 H0(X, KX+ L) = 0 or H0(X, KX+ 4Lred)= 0.

The second conclusion may occur when KX 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 H0(X, KX+ L) is equivalent to that of H0(X, KX+ Lred), 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 H0(X, KX+ 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 KX+ 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 KX+ L are nef and big) but with H0(X, KX+ 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 H0(X, KX+ M) for big divisor M, the “bigness” assumption on the fundamental group π1(X) is necessary, because in (1) the M:= L  L is big and π1(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− (KX+ B) nef and big) in Kawamata’s conjecture [Ka]

for the non-vanishing of H0(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 − (KX+ B) = L − B = L is nef and big, D = KX+ L, and D is not nef for D.Di= −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 KV + L is nef. The well-known positivity says that fV /C ⊗ OV(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

ωV /B⊗ OV


= h0

F, KF+ L|F


F, KF+ L|F

= 0.

The positivity of fV /B⊗ OV(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=1Dj be a reduced connected divisor on a non-singular pro- jective surface X. Then D.(KX+ D)  −2 and the equality holds if and only if D is a rational tree.

Proof. Note that

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

We calculate:

D.(KX+ D) =


KX.Dj+ 2




2pa(Dj)− 2

+ 2(n − 1)  −2.

The lemma follows. 2

Lemma 2.3. Suppose that X is a non-singular projective surface with χ (OX)= 1 and D (= 0) a connected reduced divisor such that H0(X, KX+ 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 π1(X)= (1).

Proof. The Serre duality and Riemann–Roch theorem imply 0= h0(X, KX+ D) = h1(X, KX + D) + 12(KX + D).D + χ(OX) 0 + (−1) + 1 by Lemma 2.2. Thus D.(KX+ D) = −2 and hence D is a connected rational tree by the same lemma. So π1(D)= (1). Suppose that D supports on a nef and big effective divisor. Then the sur- jective map π1(D)→ π1(X) in Nori [No, Corollary 2.3] infers π1(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 H0(X, KX+ Lred)= 0. Then χ(OX)= 1, Lredis a connected rational tree and X is simply connected.

Proof. Note that pg(X) h0(X, KX+ Lred)= 0. So χ(OX)= 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, pg(X)= 0, and π1alg(X)= (1) (this is true if π1(X)= (1)). Let π : X→ P1be the unique elliptic fibration with F a general fibre. The following statements are true.


(1) π has exactly two multiple fibres F1, F2, and their multiplicities m1, m2are coprime.

In particular, if E is horizontal then E.F = m1m2m3( 6) for some positive integer m3.

(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, (m1, m2)= (2, 3) and E.F = 6 for some E in D (see [BPV, Chapter V, §7], for notation of singular fibres).

Proof. (1) Since π1(X)alg= (1), we have H1(X, Z)= (0) and hence q(X) = 0. So χ (OX)= 1. Since κ(X) = 1, there is an elliptic fibration π : X → π(X) = P1, where the image is P1 because q(X)= 0. Let Fi (1 i  t) be all multiple fibres of π, with multiplicity mi. If m= gcd(m1, m2) 2, then the relation m(F1/m− F2/m)∼ 0 induces an unramified Galois Z/(m)-cover of X, contradicting the assumption π1(X)alg= (1). If t 3, then by Fox’s solution to Fenchel’s conjecture (see [Fo,Ch]), there is a base change B→ P1ramified exactly over π(Fi) (1 i  t) and with ramification index mi. Then the normalization Y of the fibre product X×P1 B is an unramified cover of X (so that the induced fibration Y → B has no multiple fibres), again contradicting the assumption that π1(X)alg= (1).

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

KX= π(KP1)+ χ(OX)F1+

t i=1

(mi− 1)(Fi)redQ

−1 +

t i=1

 1− 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= KX = Γ2= 0. Hence Γ is a multiple of a fibre of π . Since the support of Γ (< D) is a tree, it is of type In, II, IIIor 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 IIand 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 H0(X, KX+

L) = 0. Indeed, we will see that both L and KX+ 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 P2) is dual to a smooth plane cubic (always having 9 inflections). Let X→ P2be 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 A2. Let X be the minimal resolution. According to Barth [Ba], these 9A2 are 3-divisible. That is, for some integral divisor G, we have 3G∼9

i=1(Ci + 2Di)



(Ci + Di) is a disjoint union of the 9 intersecting P1 (i.e., the 9A2). Let H be the pull back of a general line away from the 9 cusps on C. Then H2= 2 and H is disjoint from the 9A2, so H.G= 0. We can also calculate that G2= −6. Now let L= H + G − 139

i=1(Ci + 2Di). Then L = H + G and L2= −4. Clearly, KX+ L = L ≡ H is nef and big. However, by the Kawamata–Viehweg vanishing, and Riemann–Roch theorem, we have h0(X, KX+ L) = 12L2+ 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 : XAlb(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 KX+ L is nef. Then H0(X, K+ L) = 0.

To see this, we need the following lemma:

Lemma 3.2. LetF = 0 be a IT0 sheaf on an abelian variety A, i.e., for every i > 0 we have Hi(A,F ⊗ P ) = 0 for all P ∈ Pic0(A). Then H0(A,F) = 0.

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

Proof. Suppose on the contrary that H0(A,F) = 0. Since F is IT0, the Fourier–Mukai transform ofF is a locally free sheaf of rank = h0(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 fL is nef and big with simple normal crossing support. Let ∆:= fL − fL, 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 Ak:= fL1kN is ample for all k 0. We fix k such that (X , ∆+1kN ) is klt. Now we can write Ak= (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 ∆ := ∆ +1kN+n1E+n1B. Then we have, where P = (alb ◦f )P with P ∈ Pic0(A):

KX + fL

+ P ≡ KX +(alb◦f )M n + ∆ .

Let F := albfOX (KX + fL). By Kollar’s relative vanishing theorem (cf. [Ko, 10.19.2]), one sees thatF is IT0.


We claim thatF = 0.

Grant this claim for the time being. By the above lemma, it follows that h0

X , KX + fL 

= h0(A,F) = 0.

Since KX + fL = f(KX+ L) + Γ , where Γ is an exceptional divisor (possibly non-effective). It’s easy to see that fOX (Γ )⊂ OX. By the projection formula, one has:

0= H0

X , KX + fL 

= H0 X,OX


⊗ fOX (Γ )

⊂ H0

X, KX+ 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 thatF = 0. If dim(alb(X)) = 1. Let F be a general fiber of alb ◦f . Then we have:

rank(F) = h0 F,

KX + fL 


= h0

F, KF+ fL|F


Since fL is big, fL.F > 0. It follows that deg(fL|F) > 0.

If g(F ) > 0, then we have h0(F, KF + fL|F) > 0 already. If g(F ) = 0, note that KX+ L is nef. Note also that (KX + fL).F = (KX+ L).f (F ) since F is general. This implies that

deg KF+



KX + fL .F+


− fL .F

= (KX+ L).f (F ) + fL

− fL .F 0.

Hence h0(F, KF + LF) > 0. We conclude that F = 0 and hence the required non- vanishing that h0(X, KX+ 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= h0(X, KX+ Lred) ( pg(X)). So X is the blow up of an Enriques surface by the classification theory. On the other hand, π1(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= P2or P1× P1, 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 surfaceFd

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

We first fix some notations. Let π :Fd→ P1be the ruling. Let F be a general fibre and C the only negative curve (a cross-section, indeed) onFd. So C2= −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= Fd for some d 1. Let L be a nef and big effective Q-divisor such that KX+ 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 KX+ L = KX+ L + E is nef; then the Serre duality and Riemann–Roch theorem for Cartier divisor imply that h0(X, KX+ L)  12L(KX+ L) + χ(OX) 0 + 1.

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

Write L=


fjFj where C1= C, the Ci’s are distinct horizontal compo- nents and Fj’s are distinct fibres, where ci> 0, fj> 0.

Suppose on the contrary that H0(X, KX+ L) = 0. Then by Lemma 2.3, Lred is a connected rational tree. Hence one of the following cases occurs:

(i) L= c1C1+k

j=1fjFj(k 0);

(ii) L=k

i=1ciCi+ f1F1(k 2), and Lredis comb-shaped, i.e., Ci’s are disjoint cross- sections;

(iii) L=k

i=1ciCi (k 2).

Recall that KX∼ −2C1− (d + 2)F . The nefness of KX+ L implies:

0 (KX+ L).F = −2 +

ci(Ci.F ),

0 (KX+ L).C1= d − 2 − dc1+


ci(Ci.C1)+ fj,

ci(Ci.F ) 2,

fj 2 + (c1− 1)d −




In case (i), the above inequalities imply c1 2 and 

fj  2 + (c1− 1)d  d + 2, whence

L = c1C1+

fjFj∼ c1C1+ 

fj F

 2C1+  fj

F  2C1+ (d + 2)F1∼ −KX.

Hence H0(X, KX+ L) = 0.

Consider case (ii). Then one sees easily that k= 2 and C2∼ C1+ dF1(see [Ha, Chap- ter V, §2]). By the displayed inequalities, we have c1 2 − c2and f1 2 + (c1− 1)d. If c2> 1 thenL  C1+ 2C2+ F1>−KX, whence H0(X, KX+ L) = 0. So we may assume that c2 1. Then c1 1 and f1 2. Thus L  C1+ C2+ 2F1∼ −KX, whence H0(X, KX+ L) = 0.

Consider case (iii). Since L is a connected tree, we may assume that C1.C2= 1. So C2∼ n(C1+ dF ) + F for some integer n  1. Since Ci (i 2) is irreducible, we have Ci  C1+ dF by [Ha]. If k  3 or n  2, then we see that L k

i=1Ci>−KX. So assume that k= 2 and n = 1. By the inequalities displayed above, we have c1 2 − c2and c2 2 + (c1− 1)d. If c2> 1 thenL  C1+ 2C2>−KX. So assume that c2 1. Then c1 2 − 1 and c2 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 = P1. By the canonical divisor formula, we see that π has at most one fibre F0with multiplicity m 2; moreover, such F0(if exists) is of Kodaira type In(n 0), and −KX= (F0)red.

We show that Conjecture 1.1 is true if KX+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 KX+ L is nef and big. Then H0(X, KX+ L) = 0.

Proof. By Theorem 3.1, we may assume that q(X)= 0, so B = P1 and X is a rational surface.

Suppose that the Q-divisor L is nef and big and KX+ L is nef. Let F0= m(F0)redbe the multiple fiber. We set m= 1 and let F0 be a general (smooth) fibre, if π is multiple fibre free. Then KX∼ −(F0)red. Let a > 0. Consider the exact sequence:

0→ OX

KX+ aL − (F0)red

→ OX

KX+ aL

→ O(F0)red

KX+ aL|(F0)red

→ 0.


Let us find the condition for aL− (F0)redto be nef and big. Note that aL− (F0)redaL+ KX= (a − 1)L + (KX+ L). So aL − (F0)redis nef and big if either

(i) a > 1, or

(ii) a= 1 and KX+ L is nef and big.

Assume that either (i) or (ii) is satisfied. Then Hi(X, KX+ aL − (F0)red)= 0 = Hi(X, KX+ aL) for all i > 0, by Sakai’s vanishing for surfaces. For the integral divisor M:= KX+ aL and the reduced divisor C := (F0)redon X, the above exact sequence implies that



= χ


− χ

OX(M− C)

= C.M − C.(KX+ C)/2, where we applied the Riemann–Roch theorem for bothOX(M) andOX(M− C). Now C.(KX+ C) = 0 and C.M  (F0)red.(KX+ aL) > 0 (for 0 = C being nef and KX+ aL nef and big), so χ (OC(M|C)) > 0. By the vanishing above,


X, KX+ aL

= χ


= χ

OX(M− C) + χ


= h0

X, KX+ aL − (F0)red + χ


> 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 KX+ L is nef. Then H0(X, KX+ aL) = 0 provided that either a > 1, or a = 1 and KX+ 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 KX

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 En ) if its weighted dual graph is of Dynkin type An(respectively Dn, or En) but its weights may not all be (−2).

(2) Γ is of type In(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 In (respectively II, or III, or IV) if Γ is equal to an elliptic fibre of type In (respectively II, or III, or IV), including coefficients, but the self intersections of components of Γ may not all be (−2). E.g., Γ = 2n


j=n+1Cj is of type In , where Ci+ C0+ C1+

· · ·+Cn+Cjis 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 C2= −2 − C.KX −2. The others are clear.

Lemma 7.2.

(1) If C is a smooth rational curve on X, then C2 −2.

(2) If Γ is of type A n, D nor E nthen it is negative definite, i.e., the intersection matrix of components in Γ is negative definite.

(3) If Γ is one of types In, II, III and IV (respectively In , 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 I0. 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 I0. 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 C0be the irreducible component that meets k other components with the largest k. Then Γ − C0has exactly k connected components{∆i}.

We may assume that k 3. Let Ci be the irreducible component of ∆i that meets C0. 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 − C4, then now Γ has at least two connected components: C0+ C1+ C2+ C3and 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 C3meets at least 3 components, we take Γ = Γ − C3. Then Γ has at least 3 connected components and each one has length 5.

If C3meets 2 components, say C0, C4, then we take Γ = Γ − C4. Again, each connected component of Γ has at most 5 irreducible components. This proves the lemma. 2

Lemma 7.4. Suppose that q(X)= pg(X)= 0 and π1alg(X)= (1) (these are satisfied in the situation of Proposition 7.10; see its proof ).

(1) We have ρ(X) 10 − KX2  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, Lredsupports an effective divisor C of type I0 whose central and three of the tip components are all (−2)-curves.

Proof. (1) follows from: ρ(X) b2(X)= c2(X)− 2 + 4q(X) = 12χ(OX)− KX2 − 2 = 10− KX2  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 Lred> Supp(Γ ) (for L being nef and big), Γ is of type IIand Supp(E+ Γ ) ( Lred) supports a I0 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=0Di be a reduced divisor on X. Suppose that D− D0has a negative definite n× n intersection matrix (Di.Dj)1i,jnand D supports a divisor with positive self intersection.

(1) We have det(Dk.D)0k,n> 0 (respectively < 0) if n is even (respectively odd).

(2) Assign formally Gi:= Di and define Gi.Gj := Di.Dj (i= j) and G2i := −xi. Sup- pose that

(∗) the n × n matrix (Gi.Gj)1i,jnis negative definite.

If G2i  Di2for all 0 i  n, then

(∗∗) det(Gk.G)0k,n> 0 (respectively < 0) if n is even (respectively odd).

(3) Suppose that D2i  −2 for all 0  i  n. In (2) above for 0  k  n, choose the largest positive integer mk (if exists) such that (∗) and (∗∗) in (2) are satisfied for Gi with G2k= −mkand G2i = −2 (i = k). Then Dk2 −mk.

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 Gi has positive self intersec- tion. By the assumption some divisor ∆=

biDi has positive self intersection, then so is Γ =

biGi because Γ2=


bibjDi.Dj= ∆2> 0. The (3) follows from (2). 2

Let D=n

i=0Dibe 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 P2> 0.


In Lemmas 7.7 and 7.8 below, we do not need the bigness of P ; in Lemma 7.7, KXis irrelevant.

Lemma 7.7.

(1) Write P =n

i=0piDi. Then 0 pi  1, and pi < 1 holds if and only if Di  Supp(N ).

(2) Write Supp(N )=s

i=0Di after relabelling. Then (p0, . . . , ps) is the unique solution of the linear systemn

i=0xi(Di.Dj)= 0 (j = 0, . . . , s), where we set xj= 1 (j > s).

(3) Assign formally Gi := Di and Gi.Gj= Di.Dj (i= j). Suppose that for α  i  β, we assign G2i such that−2  G2i  D2i and (Gi.Gj)αi,jβ is negative definite. Let (xi = bi| α  i  β) be the unique solution of the linear systemn

i=0xiGi.Gj= 0 (α j  β), where we set xj = bj = pj if j < α or j > β. Then bi  pi for all α i  β.

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

We prove (3). It suffices to show that (∗∗∗) the sumβ

i(bi− pi)Gi.Gj 0 for all α  j  β.

Indeed, write

(bi− pi)Gi= A − B with A  0, B  0 and with no common compo- nents in A and B; then the condition (∗∗∗) implies that A.B − B2=

(bi− pi)Gi.B 0;

this and A.B  0 and B2 0 imply that B2= 0 and hence B = 0 by the negative- definiteness of (Gi.Gj).

Coming to the sum in (∗∗∗) above, it is equal ton


i=0piGi.Gj 0−n

i=0piDi.Dj 0. This proves the lemma. 2

Lemma 7.8. Suppose that Γ = D1+ · · · + Dmis an ordered linear chain contained in D such that Γ.(D− Γ ) = 1. Let Dt  Γ and Dm+1 D − Γ such that Dt.Dm+1= 1. If either t= m or D2t  −3, then Γ  Supp(N).

Proof. Write P =

jpjDj. If t= m, we set G2i = −2 (1  i  m) in Lemma 7.7 and obtain

pi bi=

 i m+ 1

pm+1< 1

and hence Γ  Supp(N). If Dt2 −3, we have only to show that pt < 1 because we already have pj < 1 for every 1 j  m with j = t, by the previous case. Now 0 P.Dt= ptD2t + pt−1+ pt+1+ pm+1<−3pt+ 3, whence pt < 1. This proves the lemma. 2


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

P =

n i=0


Then 0 < pi  1. Note that pj = 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 p0= 1.

Now we state the main result of the section.

Proposition 7.10. Let X be a minimal non-singular projective surface (i.e., KXis nef ) with pg(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=0Ci  Lredwith r 0 (after relabelling) such that Lred.r

i=0Ci 2.

(B) Supp(L) supports an effective divisor C of type in{In, III, IV} and the weights of the multiplicity 2 components of C are all (−2), so C.(KX+ C) = 0. Also the type IIIoccurs only when Lredis 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 KX and those of the 9 curves in Lred=8

i=0Ci; C0 meets exactly C1, C2, C3; C2+ C4+ C6 and C3+ C5+ C7+ C8 are linear chains;

C62= −3 and Ci2= −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, π1(X)= (1). In particular, q(X) = 0 and χ(OX)= 1.

Since p0= 1 by the additional assumption, C0 is not in Supp(N ). Since 0 P.C0= C02+

pj and C20 −2, where j runs in the set so that Cj meets C0, this C0meets at least two components of Supp(P )− C0. 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 C0 meets exactly two components of Supp(P )− C0. Then Proposition 7.10 is true.


Proof. Suppose that C0meets only C1and C2in Supp(P )− C0. Then 0 P.C0= C02+ p1+p2and C02 −2 imply that p1= p2= 1 and C20= −2. Inductively, we can prove that there is an ordered linear chain (after relabelling)b

i=aCiin Supp(P ) such that pi= 1 and Ci2= −2 for all a  i  b and Ca(respectively Cb) meets Ca−1and Ca−2(respectively Cb+1and Cb+2) such that

Ca−2+ Ca−1+ 2

b i=a

Ci+ Cb+1+ Cb+2

is of type Ib−aand Proposition 7.10(B) is true. 2

Lemma 7.12. Suppose that C0 meets at least four components of Supp(P )− C0. Then Proposition 7.10 is true.

Proof. Suppose that C0 meets Ci (1 i  k) with k  4. Let ∆i (1 i  k) be the connected component of Pred− C0 containing Ci. Set ni := #∆i. Assume that for only 1 j  s the divisor C0+ ∆j is a linear chain. By the proof of Lemma 7.8, we have pj nj/(nj+ 1) (j  s).

If Pred.C0= C02+ k  2, then Proposition 7.10(A) is true. So assume that C02 1 − k 

−3. Note that 0 P.C0= C02+

k i=1

pi C02+ (k − s) +

s i=1

pi 1 − s +

s i=1

pi 1 −

s i=1

1 ni+ 1. Suppose that #∆i= 1 for 1  i  s1and #∆i 2 for i  s1+ 1. Then


s i=s1+1


ni+ 1 1 −s1 2.

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

3k− s − s1= s1+ 2(s − s1)+ 3(k − s)  #P − 1  8.

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

Note that C0meets the mid-component Cjof ∆j(j= 3, 4). By the proof of Lemma 7.8, for every j with j= 0, 3, 4, we have pj  1/2. Thus 0  P.C0= C02+4

i=1pi −3 + 1/2+ 1/2 + 1 + 1 = 0, so C02= −3 and p3= p4= 1. Now 0  P.C3 C32+ p0+ 1/2 + 1/2 implies C32= −2. So Pred.(C0+ C3)= 2 and Proposition 7.10(A) is true. 2

Now we assume that C0meets exactly three components Ci (i= 1, 2, 3) of Supp(P ) − C0. Let ∆i be the connected component of Supp(P )− C0containing Ci. Set ni := #∆i. Then3

i=1ni= #P − 1  8. We may assume that n1 n2 n3. Then n3 6 and n1 2, so C0+ ∆1is a linear chain. By the proof of Lemma 7.8, we have p1 n1/(n1+ 1) < 1.




Related subjects :