ALFRED JUNGKAI CHEN, MENG CHEN, AND DE-QI ZHANG
Abstract. We prove a non-vanishing theorem of the cohomology H0 of the adjoint divisor KX+ dLe where dLe is the round up of a nef and big Q-divisor L.
1. Introduction
We work over the complex numbers 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:
Conjecture 1.1. Let X be a nonsingular 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+ dLe) 6= 0, where dLe is the round up of L.
This kind of non-vanishing problem has been considered by Ambro [Am], A.
Chen-Hacon [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 KX is nef).
Conjecture 1.1 is slightly different from that of Kawamata’s in [Ka]. It is some- what general in the sense that the divisor L in question is not assumed to have integral coefficients. It is precisely this non-Cartierness of L that causes a lot of trouble when estimating h0(X, KX+ dLe). To elaborate, the Kawamata-Viehweg vanishing ([KV], [Vi1]) implies that h0(X, KX + dLe) = χ(KX + dLe) when the fractional part of L is of normal crossings. However, the formula for χ may not be effective because dLe may not be nef and hence dLe.(KX+ dLe) may not be non-negative. 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, (D − KX) + KX) 6= 0 for Picard number one Gorenstein del Pezzo surface X and nef and big Q-Cartier Weil divisor 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]).
2000 Mathematics Subject Classification. Primary 14F17; Secondary 14J29.
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.
1
Theorem 1.2. Let X be a nonsingular 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+ dLe) 6= 0 or H0(X, KX+ 4Lred) 6= 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 whenever L is nef and big. So the coefficients of L can be as small as one likes. Thus the non-vanishing of H0(X, KX+ dLe) is equivalent to that of H0(X, KX+ Lred), which is stonger 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 Theorem 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.
Theorem 1.3. Let X and L be as in Conjecture 1.1 satisfying the first two condi- tions only. Then H0(X, KX+ dLe) 6= 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+ dLe) = 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 := dLe ≥ 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 have dLe = L + B with B a simple normal crossing effective divisor so that [B] = 0, whence (X, B) is klt. To be precise, let D := dLe. Then D −(KX+B) = dLe−B = L is nef and big, D = KX+ dLe, 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 nonsingular 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 f∗(ωV /C ⊗ OV(dLe)) 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(dLe))) = h0(F, KF+ dLe|F) ≥ h0(F, KF+ dL|Fe) 6= 0.
The positivity of f∗(ωV /B⊗ OV(dLe)) has direct applications in studying prop- erties 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 problem.
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 =Pn
j=1Dj be a reduced connected divisor on a nonsingular projective surface X. Then D.(KX+ D) ≥ −2 and the equality holds if and only if D is a rational tree.
Proof. Note that P
k<jDk.Dj ≥ n − 1 and the equality holds if and only if D is a tree. We calculate: D.(KX + D) = P
Dj2+P
KX.Dj + 2P
k<jDk.Dj ≥ P
j(2pa(Dj) − 2) + 2(n − 1) ≥ −2. The lemma follows. ¤ Lemma 2.3. Suppose that X is a nonsingular projective surface with χ(OX) = 1 and D (6= 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 connected). 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 surjective map π1(D) → π1(X) in Nori [No, Cor. 2.3] infers π1(X) = (1). ¤
The next result is a very important restriction on X and L in Theorem 1.2.
Proposition 2.4. Let X be a nonsingular 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, Lred is 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. ¤
The result below is used in the subsequent sections.
Lemma 2.5. Suppose that X is a minimal nonsingular 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 → P1 be 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, m2 are 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 contains 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.
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 → P1 ramified exactly over π(Fi) (1 ≤ i ≤ t) and with ramification index mi. Then the normalization Y of the fibre product X ×P1B 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+Pt
i=1(mi−1)(Fi)red∼Q (−1+Pt
i=1(1−m1
i))F1(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.Γ = 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∗, 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). ¤
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 nonsingular 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+ dLe) = 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 singularities. This C (regarded as a curve in the dual plane P2∗) is dual to a smooth plane cubic (always having 9 inflectins). Let X → P2 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 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 ∼P9
i=1(Ci+ 2Di) where`
(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 − 13P9
i=1(Ci+ 2Di).
Then dLe = H + G and dLe2 = −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+ dLe) = 12dLe2+ 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 nonsingular 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 nonsingular 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 + dLe) 6= 0.
To see this, we need the following lemma:
Lemma 3.2. Let F 6= 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) 6= 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 of F 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. ¤
Proof of Theorem 3.1. Let f : X0 → X be an embedded resolution for (X, L). It is clear that f∗L is nef and big with simple normal crossing support. Let ∆ :=
df∗Le − f∗L, then ∆ is klt. By a property of nef and big divisor (see e.g. [La], ex 2.2.17), there is an effective divisor N such that Ak := f∗L − 1kN is ample for all k À 0. We fix k such that ∆ +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 X0. Pick irreducible divisor B ∈ |(n − 1)A| for n À 0 such that ∆0:= ∆ +1kN +n1E +1nB is again klt. Then we have, where P0= (alb ◦ f )∗P with P ∈ Pic0(A):
KX0+ df∗Le ⊗ P0 ≡ KX0+(alb ◦ f )∗M n + ∆0.
Let F := alb∗f∗OX(KX0 + df∗Le). By Koll´ar’s relative vanishing theorem (cf.
[Ko], 10.19.2), one sees that F is IT0. We claim that F 6= 0.
Grant this claim for the time being. By the above lemma, it follows that h0(X0, KX0+ df∗Le) = h0(A, F) 6= 0.
Since KX0+df∗Le = f∗(KX+dLe)+∆, where ∆ is an exceptional divisor (possibly non-effective). It’s easy to see that f∗OX0(∆) ⊂ OX. By the projection formula, one has:
0 6= H0(X0, KX0+ df∗Le) = H0(X, KX+ dLe)⊗f∗OX0(∆)) ⊂ H0(X, KX+ dLe).
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 6= 0. If dim(alb(X)) = 1. Let F be a general fiber of alb ◦ f . Then we have:
rank(F) = h0(F, (KX0+ df∗Le)|F) = h0(F, KF+ df∗L|Fe).
Since f∗L is big, f∗L.F > 0. It follows that deg(df∗L|Fe) > 0.
If g(F ) > 0, then we have h0(F, KF + df∗LFe) > 0 already. If g(F ) = 0, note that KX+ L is nef. Note also that (KX0 + f∗L).F = (KX+ L).f (F ) since F is general. This implies that
deg(KF + df∗LFe) = (KX0+ f∗L).F + (df∗Le − f∗L).F
= (KX+ L).f (F ) + (dLe − L).F ≥ 0.
Hence h0(F, KF + dLFe) > 0. We conclude that F 6= 0 and hence the required
non-vanishing that h0(X, KX+ dLe) 6= 0. ¤
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 the Conjecture 1.1 in the Introduction is true for surfaces X (not necessarily minimal) with Kodaira dimenion κ(X) = 0.
Theorem 4.1. Suppose that X is a nonsingular projective surface (not necessarily minimal) 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. ¤
5. Surfaces with negative κ, Part I : ruled surfaces
In this section, we prove Conjecture 1.1 for relatively minimal surfaces X of Kodaira dimension κ(X) = −∞. By Theorem 3.1, we may assume that q(X) = 0, so X is a relatively minimal rational surface. If X = P2 or 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 surface Fd of degree d ≥ 1 (though, F1 is not relatively minimal).
We first fix some notations. Let π : Fd→ P1 be the ruling. Let F be a general fibre and C the only negative curve (a cross-section, indeed) on Fd. 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 := dLe − L is effective and nef; so dLe = L+E is nef and big and KX+dLe = KX+L+E is nef; then the Serre duality and Riemann-Roch theorem for Cartier divisor imply that h0(X, KX + dLe) ≥
1
2dLe(KX+ dLe) + χ(OX) ≥ 0 + 1. Theorefore, we may assume that Supp(L) contains C.
Write L = P
iciCi+P
fjFj where C1 = C, the Ci’s are distinct horizontal components and Fj’s are distinct fibres, where ci> 0, fj> 0.
Suppose on the contrary that H0(X, KX+ dLe) = 0. Then by Lemma 2.3, Lred
is a connected rational tree. Hence one of the following cases occurs:
Case (i). L = c1C1+Pk
j=1fjFj (k ≥ 0), Case (ii). L =Pk
i=1ciCi+ f1F1 (k ≥ 2), and Lred is comb-shaped, i.e., Ci’s are disjoint cross-sections.
Case (iii). L =Pk
i=1ciCi (k ≥ 2).
Recall that KX ∼ −2C1− (d + 2)F . The nefness of KX+ L implies:
0 ≤ (KX+ L).F = −2 +X
ci(Ci.F ), 0 ≤ (KX+ L).C1= d − 2 − dc1+X
i≥2
ci(Ci.C1) +X fj, Xci(Ci.F ) ≥ 2,
Xfj ≥ 2 + (c1− 1)d −X
i≥2
ci(Ci.C1).
In Case (i), the above inequalities imply c1≥ 2 andP
fj ≥ 2 + (c1− 1)d ≥ d + 2, whence dLe = dc1eC1 +P
dfjeFj ∼ dc1eC1+ (P
dfje)F ≥ 2C1 + (P
fj)F ≥ 2C1+ (d + 2)F1∼ −KX. Hence H0(X, KX+ dLe) 6= 0.
Consider Case (ii). Then one sees easily that k = 2 and C2 ∼ C1+ dF1 (see [Ha, Chapter V, §2]). By the displayed inequalities, we have c1 ≥ 2 − c2 and f1 ≥ 2 + (c1− 1)d. If c2 > 1 then dLe ≥ C1+ 2C2 + F1 > −KX, whence H0(X, KX+ dLe) 6= 0. So we may assume that c2≤ 1. Then c1 ≥ 1 and f1 ≥ 2.
Thus dLe ≥ C1+ C2+ 2F1∼ −KX, whence H0(X, KX+ dLe) 6= 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 dLe ≥Pk
i=1Ci>
−KX. So assume that k = 2 and n = 1. By the inequalities displayed above, we have c1 ≥ 2 − c2 and c2≥ 2 + (c1− 1)d. If c2 > 1 then dLe ≥ C1+ 2C2 > −KX. So assume that c2≤ 1. Then c1≥ 2 − 1 and c2≥ 2 + 0d, a contradiction. ¤
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 the 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 F0 with 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+ dLe) 6= 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)red
be 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+daLe−(F0)red) → OX(KX+daLe) → O(F0)red(KX+daLe|(F0)red) → 0.
Let us find the condition for aL − (F0)red to be nef and big. Note that aL − (F0)red∼ aL + 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 H1(X, KX+daLe−(F0)red) = 0 = H1(X, KX+ daLe) by Sakai’s vanishing for surfaces. For the integral divisor M :=
KX+ daLe and the reduced divisor C := (F0)red on X, the above exact sequence implies that χ(OC(M |C)) = χ(OX(M )) − χ(OX(M − C)) = C.M − C.(KX+ C)/2, where we applied the Riemann-Roch theorem for both OX(M ) and OX(M − C).
Now C.(KX + C) = 0 and C.M ≥ (F0)red.(KX + aL) > 0 (for 0 6= C being nef and KX + aL nef and big), so χ(OC(M |C)) > 0. By the vanishing above, h0(X, KX+ daLe) = χ(OX(M )) = χ(OX(M − C)) + χ(OC(M |C)) = h0(X, KX+ daLe − (F0)red)) + χ(OC(M |C)) > 0 + 0. This proves the theorem. ¤ Remark 6.2. The above argument actually proved the following: 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. Then H0(X, KX+ daLe) 6= 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 nonsingular projective surface with KXnef 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. (1) Let Γ be a connected rational tree on X. This Γ is of type A0n (resp. D0n, or En0) if its weighted dual graph is of Dynkin type An (resp. Dn, or En) but its weights may not all be (−2).
(2) Let Γ be a connected effective integral divisor on X so that Supp(Γ) is a connected rational tree. Γ is of type In∗ (resp. 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∗’ (resp. II∗’, or III∗’, or IV∗’) if Γ is equal to an elliptic fibre of type In∗ (resp. II∗, or III∗, or IV∗), including coefficients, but the self intersections of components of Γ may not all be (−2). E.g. Γ = 2Pn
i=0Ci+Pn+4
j=n+1Cj is of type In∗’, where Ci+ C0+ C1+ · · · + Cn + Cj is an ordered linear chain for all i ∈ {n + 1, n + 2} and j ∈ {n + 3, n + 4}.
The assertion(1) below follows from 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 An’, Dn’ or En’ then 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∗ (resp. In∗’, II∗’, III∗’ and IV∗’, but at least one component of Γ is not a (−2)-curve), then Γ is negative semi-definite (resp. negative definite).
(4) If Γ is a connected rational tree with (the number of irreducible components)
#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. Let Γ be a connected rational tree whose (−2)-components do not support a divisor of type I0∗. Let r = min{9, #Γ−1}. Then there is a subgraph Γ0 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 C0 be the irreducible component that meets k other components with the largest k. Then Γ − C0 has exactly k connected components {∆i}. We may assume that k ≥ 3. Let Cibe 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 Γ0 =P
∆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 Γ0 = Γ − C4, then now Γ0 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 Γ0 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 C3 meets at least 3 components, we take Γ0 = Γ − C3. Then Γ0 has at least 3 connected components and each one has length ≤ 5. If C3 meets 2 components, say C0, C4, then we take Γ0= Γ − C4. Again, each connected component of Γ0 has at most 5 irreducible components .
This proves the lemma. ¤
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 effective 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 II∗ and 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. ¤
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 =Pn
i=0Di be a reduced divisor on X. Suppose that D − D0
has a negative definite n × n intersection matrix (Di.Dj)1≤i,j≤n and D supports a divisor with positive self intersection.
(1) We have det(Dk.D`)0≤k,`≤n> 0 (resp. < 0) if n is even (resp. odd).
(2) Assign formally Gi := Di and define Gi.Gj := Di.Dj (i 6= j) and G2i := −xi. Suppose that (∗) the n × n matrix (Gi.Gj)1≤i,j≤n is negative definite. If G2i ≥ D2i for all 0 ≤ i ≤ n, then (∗∗) det(Gk.G`)0≤k,`≤n > 0 (resp. < 0) if n is even (resp.
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 = −mk and G2i = −2 (i 6= k). Then D2k ≥ −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 intersection. By the assumption some divisor ∆ = P
biDi has positive self intersection, then so is Γ =P
biGi because Γ2 =P
bibjGi.Gj ≥P
bibjDi.Dj =
∆2> 0. The (3) follows from (2). ¤
Let D 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 P2> 0.
In Lemmas 7.7 and 7.8 below, we do not need the bigness of P ; in Lemma 7.7, KX is irrelevant.
Lemma 7.7. (1) Write P =Pn
i=0piDi. Then 0 ≤ pi≤ 1, and pi< 1 holds if and only if Di≤ Supp(N ).
(2) Write Supp(N ) = Ps
i=0Di after relabelling. Then (p0, . . . , ps) is the unique solution of the linear system Pn
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 6= j). Suppose that for α ≤ i ≤ β, we assign G2i such that −2 ≥ G2i ≥ Ci2 and (Gi.Gj)α≤i,j≤β is negative definite. Let (xi = bi| α ≤ i ≤ β) be the unique solution of the linear system Pn
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 (1 ≤ j ≤ s) and that N has negative definite (and hence invertible) intersection matrix.
We prove (3). It suffices to show that (*) the sumPβ
i=α(bi− pi)Gi.Gj≤ 0 for all α ≤ j ≤ β.
Indeed, writeP
(bi− pi)Gi = A − B with A ≥ 0, B ≥ 0 and with no common components in A and B; then the condition (*) implies that A.B − B2=P
(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 toPn
i=0biGi.Gj−Pn
i=0piGi.Gj≤ 0 −Pn
i=0piDi.Dj = 0. This proves the lemma. ¤
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 =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 D2t ≤ −3, we have only to show that pt< 1 because we already have pj < 1 for every 1 ≤ j ≤ m with j 6= t, by the previous case. Now 0 ≤ P.Dt = ptDt2+ pt−1+ pt+1+ pm+1<
−3pt+ 3, whence pt< 1. This proves the lemma. ¤
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 ≥ L. So Pred= Lred. Write
P = Xn i=0
piCi,
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. Additional assumption : L = P and p0= 1 (and also #P ≤ 9).
Now we state the main result of the section.
Proposition 7.10. Let X be a minimal nonsingular projective surface (i.e., KX
is nef) with pg(X) = 0. Suppose that a Q-divisor L is nef and big and 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 =Pr
i=0Ci≤ Lred with r ≥ 0 (after relabelling) such that Lred.Pr
i=0Ci≥ 2.
(B) Supp(L) supports an effective divisor C of type in { In∗’, III∗’, IV∗’ } but the weights of the multiplicity ≥ 2 components of C are all (−2), so C.(KX+ C) = −2.
Also the type III∗’ occurs only when Lred is given as follows:
Case (B1). κ(X) = 1 and ρ(X) = 10; det(P ic(X)) = −1, and P ic(X) is generated by the divisor class of KX and those of the 9 curves in Lred=P8
i=0Ci; C0 meets exactly C1, C2, C3; C2+ C4+ C6 and C3+ C5+ C7+ C8are linear chains; C62= −3 and Ci2= −2 (i 6= 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, Cor. 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+P
pj and C02≤ −2, where j runs in the set so that Cj meets C0, this C0 meets at least two components of Supp(P ) − C0. Now the proposition follows
from the lemmas below. ¤
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 C02= −2. Inductively, we can prove that there is an ordered linear chain (after relabelling) Pb
i=aCi in Supp(P ) such that pi= 1 and Ci2= −2 for all a ≤ i ≤ b and Ca (resp. Cb) meets Ca−1 and Ca−2(resp. Cb+1 and Cb+2) such that Ca−2+ Ca−1+ 2Pb
i=aCi+ Cb+1+ Cb+2 is of type Ib−a∗ ’ and Proposition 7.10 (B) is true. ¤ 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− C0containing 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+Pk
i=1pi ≤ C02+ (k − s) +Ps
i=1pi≤ 1 − s +Ps
i=1pi ≤ 1 −Ps
i=11/(ni+ 1). Suppose that #∆i= 1 for 1 ≤ i ≤ s1 and
#∆i≥ 2 for i ≥ s1+ 1. Then 0 ≤
Xs i=s1+1
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 C0 meets the mid-component Cj of ∆j (j = 3, 4). By the proof of Lemma 7.8, for every j with j 6= 0, 3, 4, we have pj ≤ 1/2. Thus 0 ≤ P.C0 = C02+P4
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. ¤
Now we assume that C0 meets 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. ThenP3
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. This and 0 ≤ P.C0= C02+ p1+ p2+ p3, together with C02≤ −2, imply that C02= −2. We shall apply Lemma 7.6 frequently, where G0can be chosen as C0 or C3.
Lemma 7.13. Suppose that #∆i = 1 for i = 1 and 2 (this is true if #∆3 = 6).
Then Proposition 7.10 is true.
Proof. By the proof of Lemma 7.8, we have pi ≤ 1/2 for i = 1 and 2. Now 0 ≤ P.C0 = C02 + p1 + p2+ p3 (and C02 = −2) imply p3 = 1 and pi = 1/2 (i = 1, 2). By Lemma 7.11 and 7.12 (applied to C3), we may assume that C3
meets exactly three components C0, C4, C5 of Supp(P ) − C3. If C32 = −2, then Pred.(C0+ C3) = 2 and Proposition 7.10 (A) is true. Suppose that C32≤ −3. Then as above C32+ p0+ p4+ p5= P.C3≥ 0 implies that C32= −3 and p4= p5= 1. (Of course, p0= 1 is always assumed). Again by the same Lemmas we may assume that Ci(i = 4, 5) meets exactly three components (one of which is C3). Then #P ≥ 10, a contradiction to the additional assumption #P ≤ 9. ¤ Lemma 7.14. Suppose that C0+ ∆i is a linear chain for all i = 1, 2, 3. Then Proposition 7.10 is true.
Proof. Note that P3
i=1ni = #P − 1 ≤ 8. By Lemma 7.13, we may assume that n3≤ 5. Except the cases below, P is negative (semi-) definite by Lemma 7.2 (which is impossible):
(n1, n2, n3) = (1, 3, 4), (2, 2, 4), (2, 3, 3), (2, 2, 3).
In the first (resp. the last three) cases, Supp(P ) supports a divisor D of type III∗’ (resp. IV∗’). We need to show that the coefficient ≥ 2 components of D are (−2)-curves and that Pred = Lred is given as in Proposition 7.10 (B1) in the first case. These follow from Lemma 7.6 applied to all 0 ≤ k ≤ 8. For instance, in notation of Proposition 7.10 (B1), if we set −2 ≥ G2k = −xk (k = 6, 8) and
