Maximal Mordell-Weil lattices of fibred surfaces with p
g q 0.
SHINYAKITAGAWA(*)
Dedicated to Professor Yoshihara Hisao on his sixtieth birthday
ABSTRACT -Slope inequalities are given for fibred regular surfaces with geometric genus zero according as the Clifford index of a general fibre.
In the case where the slopes attain the minimums, such fibred surfaces are rational, whose constructions are described. Furthermore, the max- imal Mordell-Weil lattices of the fibred rational surfaces are completely determined.
1. Introduction.
We shall work over the complex number field C. LetX be a smooth projective surface withpgq0, wherepgandqrespectively denote as usual the geometric genus and the irregularity ofX. Letf :X!P1 be a relatively minimal fibration whose general fibreFis a smooth projective curve of genusgand of Clifford indexc. After Shioda ([13]) introduced and developed the theory of the Mordell-Weil lattice, several attempts have been made to clarify the Mordell-Weil lattices for higher genus fibrations.
For example, the cases where c0;1 and 2 are respectively studied in [12], [11] and [8], where the maximal Mordell-Weil lattices are completely determined (see also [7]). The present article is an extension of them.
THEOREM1.1. Keep the same notation as above. Assume that c3.
Let r be the Mordell-Weil rank of f :X!P1. Then the following hold:
(*) Indirizzo dell'A.: Department of Mathematics, Graduate School of Science, Osaka University, Machikaneyama 1-16, Toyonaka, Osaka 560-0043, Japan.
E-mail: [email protected]
Supported by The 21st Century COE Program named ``Towards a new basic science: depth and synthesis.
(1) When g>(c2)(c3)=2,
r2(c2)(gc1)=(c1):
If the right-hand side is an integer, then there exists f which attains the equality except when c is odd, g is a multiple of(c1)and g<(c1)2.
(2) When(c1)(c2)=2g(c2)(c3)=2, r3g (c3)(c 4)=2:
For all g, there exists f which attains the equality.
Furthermore, in both cases, X is a rational surface if r attains the above maximum.
We consider fibred rational surfaces whose Mordell-Weil ranks attain the maximums as in Theorem 1.1. Then we have an explicit description of the NeÂron-Severi group NS(X). Thus, for such fibrations, the Mordell-Weil lattices are completely determined and the corresponding Dynkin dia- grams are expressed in terms ofcandg. In particular, for the case (1), they are certain extensions ofE8(see Theorem 3.7).
In order to show Theorem 1.1, we apply the same method as in [8], where we take a reduction (Y;G) of (X;F) and give a lower bound for (KYG)2. In Theorem 2.3, we have two slope inequalities, which give upper bounds of the Mordell-Weil rank as in Theorem 1.1. Ifrattains the bound, then any fibres off must be irreducible. In the first of §3, we show such fibrations are indeed constructed from (Y;G) with the minimal (KYG)2under the suitable conditions.
Wheng<(c1)(c2)=2, which Theorem 1.1 does not cover, we have a loose bound r((2c8)g6c8)=(c2) in the same way as in Theorem 1.1. The proof of Theorem 2.3 shows that the case where g<(c1)(c2)=2 is complicated. Ifc4 andg(c1)(c2)=2, then there in fact exist two kinds of fibred rational surfaces with the minimal slopes whose structures are distinct for the plane curve models ofF. One of them is a plane curve of degree (c4) having (c2) double points as its singularity. AnotherGis near to a multi-anti-canonical divisor onYwhich is obtained by blowingP2up at five points.
Acknowledgments. The author would like to express his heartfelt gratitude to Professor Kazuhiro Konno for his valuable advice, guidance and encouragement. Thanks are also due to Professors Hisashi Usui and Viet Nguyen Khac for interesting discussions. The author thanks Pro- fessor Atsushi Noma for sending me a preprint [10].
2. Slope inequalities
LetXbe a smooth projective surface withpgq0 andf :X!P1a relatively minimal fibration of curves of genus g2. We briefly review basic notation and results of such fibred surfaces while considering the adjoint bundle KXF according to [8, §1]. It is known that KX=P1KX2F is nef and we have (KX2F):C0 for an irreducible curve C onX if and only ifC is a ( 2)-curve contained in a fibre off. Since pgq0, we have that KXFKX=P1 F is also nef and h0(X;KXF)g. Remark that X is automatically a rational surface when (KXF)2<2g 2 (see [8, Lemma 1.1]).
LEMMA 2.1 [cf. [8, Lemma 1.2]]. Let C be an irreducible curve on X such that(KXF):C0. If (KXF)2>0, then C is a smooth rational curve satisfying one of the following:
(1) C is a( 2)-curve contained in a fibre.
(2) C is a( 1)-section, i.e., a( 1)-curve with F:C1.
Suppose that there exists a ( 1)-curveEwith (KXF):E0 and let m1:X!X1 be its contraction. SinceF:E1,F1:(m1)F is smooth on X1. Furthermore, we havem1(KX1F1)KXF. If there exists a ( 1)- curveE1 with (KX1F1):E10, then, by contracting it, we get the pair (X2;F2) withF2 smooth andKX2F2 pulls back toKXF. We can con- tinue the procedure until we arrive at a pair (Xn;Fn) such that we cannot find a ( 1)-curve En with (KXnFn):En0. We put Y:Xn and G:Fn. Ifm:X!Ydenotes the natural map, thenm(KYG)KXF and GmF is a smooth curve isomorphic to F. The original fibration f :X!P1 corresponds to a pencilLf jGjwith at most simple (but not necessarily transversal) base points. When (KX F)2>0,KXF is nef and big. This implies that,Y is the minimal resolution of singularities of the surface Proj(R(X;KX F)), which has at most rational double points by Lemma 2.1, where R(X;KXF)L
n0H0(X;n(KXF)). Therefore, such a model is uniquely determined. We call the pair (Y;G) thereduction of (X;F).
We consider the rational map FjKXFj:X!Pg 1 defined by jKXFj. If jKXFj is composed of a pencil, thenf :X!P1 is a hy- perelliptic fibration (cf. [8, Lemma 1.3]). Furthermore, we have the following:
PROPOSITION 2.2 [cf. [8, Lemma 1.3 and Proposition 1.1]]. Assume
that0<(KXF)22g 5. ThenFjKXFjdefines a birational morphism
onto the image, which factors through the reduction of (X; F).
Before proceeding further, we recall here two important invariants of a smooth projective curveCof genusg(C)4. The gonality gon(C) ofCis given as the minimum of the degrees of surjective morphisms ofCtoP1. The existence theorem on special divisors (e.g. [1]) implies that gon(C)(g(C)3)=2. The Clifford index ofCis defined as
Cliff(C)minfdegL 2h0(L)2 jL2Pic(C); h0(L)>1; h1(L)>1g:
These two invariants are closely related to each other. It is known that Cliff(C)gon(C) 2 or gon(C) 3 and in the latter case we have infinitely manyg1gon(C)'s onC. For this and further properties, see [2].
We now return to the situation we are interested in. The slope off is defined bylf K2X=P1=degfKX=P1. We have the following lower bounds:
THEOREM2.3. Let X be a smooth projective surface with pgq0 and f :X!P1a relatively minimal fibration of genus g and of Clifford index c3. Let(Y;G)denote the reduction of(X;F). Then the following hold:
When g>(c2)(c3)=2;
lf 5(c 1)=(c1) (2c4)=g:
(2:1)
Iflf attains the equality in(2:1), then Y is the Hirzebruch surfaceSdof degree d.
When(c1)(c2)=2g(c2)(c3)=2, lf 5(c3)(c 4)=(2g):
(2:2)
If lf attains the equality in (2:2), then there exists a blow-down y0:(Y;G)!(P2;G0)such that G0is one of the following:
(i) a curve of degree(3c6)=2with four(c2)=2-ple points and a c=2-ple point, where c is even and g(c1)(c2)=2,
(ii) a curve of degree(3c5)=2with five(c1)=2-ple points, where c is odd, c5and g(c1)(c2)=2,
(iii) a curve of degree nine with four triple points, where c4and g16,
(iv) a curve of degree (c4) with (n1) double points, where g(c1)(c4)=2 n and 1nc1.
In particular, X is a rational surface iflf attains the equality in(2:1) or(2:2).
In order to show Theorem 2.3, from glf (KXF)24(g 1) and Proposition 2.2, we only have to prove the following:
PROPOSITION2.4. LetY be a smooth rational surface andGa smooth irreducible curve of genusgand of Clifford indexc3onY. Assume that (KY G):E1for any( 1)-curvesEonY. Then the following hold:
When g>(c2)(c3)=2;
(KYG)22c(g c 1)=(c1):
If the equality sign holds, then Y Sd. When(c1)(c2)=2g(c2)(c3)=2;
(KYG)2g(c2 c 4)=2:
If(KYG)2attains the equality, then there exists a blow-downy0:(Y;G)!
!(P2;G0)such that G0is a curve with the same degree and singularity as in (i)-(iv)ofTheorem 2:3.
First of all, in the caseY P2, it is well-known thatGis a smooth plane curve of degree (c4). Therefore, we have g(c2)(c3)=2 and (KYG)2(c1)2g(c2 c 4)=2. Next, in the case YSd, we describe explicitly as follows:
LEMMA 2.5. Let G be a smooth irreducible curve of genus g and of Clifford index c3onSd. LetD0be a section withD20 d andGa fibre of Sd. Assume thatD0:G2when d1andD0:GG:G when d0. Then
G(c2)D0((c2)d=21g=(c1))G;
where the symbolmeans the linear equivalence of divisors. Furthermore, (KSdG)22c(g c 1)=(c1) holds and (c;g;d) satisfies one of the following:
(i) c is even, g(c1)2, g is divided by (c1) and 0d 2(gc1)=((c1)(c2)).
(ii) c is odd, g(c1)2, g is divided by (c1)=2 and 0d 2(gc1)=((c1)(c2))with d2g=(c1) mod 2.
(iii) c is even,(c1)(c4)=2gc(c1), g is divided by(c1) and d1.
(iv) c is odd, (c1)(c4)=2g(c1)(2c1)=2, g is an odd multiple of(c1)=2and d1.
PROOF. Under the assumptions,G:Ggon(G)c2 holds (see [9]).
Remark thatD0:G0 sinceGis irreducible. The rest of statements follow
from a standard calculation. p
LetVbe a smooth rational surface which is neitherP2norSdandDa smooth irreducible curve of genus g7 onV. Assume that Cliff(D)3 and (KVD):E1 for any ( 1)-curvesEonV. Then we can find at least one base-point-free pencil of rational curves onV. We choose among them a pencil jRj of rational curves with R20 in such a way that a:(KVD):Ris minimal. We callatheminimal ruling degree of(V;D).
Note that we haveD:Ra25 sinceKV:R 2. Letc:V!P1be the morphism defined byjRj. We take a relatively minimal model ofVwith respect tocand consider the image ofD. Then we perform a succession of elementary transformations ([5]) at singular points of the image curve to arrive at a particular relatively minimal model (V#;D#), called a #-mini- mal model in [6], enjoying several nice properties which we collect below.
The natural mapy:(V;D)!(V#;D#) is a minimal succession of blowing- ups which resolves the singular points ofD#. We assume thatV#'Sdand D#(a2)D0bG. Let pi, 1in, be the singular points ofD# in- cluding infinitely near ones, and let mi be the multiplicity of D# atpi. Assume for simplicity thatm1m2 mn 2. SincejRjis chosen so that (KVD):Ris minimal, we can assume that the following are satisfied (see [5] and [6]):
(#1)b(a2)dwhend>0, andba2 whend0, (#2)ba2m1whend1,
(#3)m1(a2)=2 andm1minf(a2)=2; b (a2)gwhend1.
We say that (V#;D#) is of general type if 2b (a2)d2(a2).
Otherwise, i.e., whend1 and 2b<3(a2), the pair is called ofspecial type. If this is the case, by contracting the minimal section, we get a model ofDwhich is a plane curve of degreebwith a (b a 2)-ple point andn other singular points of respective multiplicitiesmi(b a 2).
At first, consider the case where (V#;D#) is of general type. Set b0b (d2)(a2)=2, where b0maxf0;(d 2)(a2)=2g, and where b02Zorb02Z[1=2] according asais even or odd. Then we have
D#(a2)D0(b0(d2)(a2)=2)G:
(2:3)
LEMMA 2.6. Keep the notation and assumptions as above. Then the following hold:
When g(a1)2,
(KVD)2>a(g a 1)=(a1):
If a is even and g<(a1)2, then
(KVD)2 2a(g 1)=(a2):
When the equality sign holds here, b00and mn(a2)=2.
If a is odd and g<(a1)2, then
(KV D)22g(a 1)=(a1)2;
When the equality sign holds, b00and mn (a1)=2.
PROOF. From a standard calculation, we have (KVD)2
2a(ab0) Pn
i1(mi 1)2and
2g2(a1)(a1b0) Xn
i1
mi(mi 1):
(2:4)
Therefore, we have (KV D)2 2(m1 1)
m1 (g (a1)(a1b0)) 2a(ab0)
Xn
i1
(mi 1) 1 mi m1
;
which is non-negative. Hence (KVD)22(m1 1)
m1 (g (a1)(a1b0))2a(ab0):
(2:5)
If the equality sign holds here, thenm1 mn. On the other hand, 2(m1 1)
m1 (g (a1)(a1b0))2a(ab0) 2a
a1g 2a
2(m1 a 1)(g (a1)(a1b0)) m1(a1)
is positive from (#3) and (2:4). Thus (KV D)2>2a(g a 1)=(a1) holds.
In the rest of the proof, we assume thatg<(a1)2. We restrict our- selves to the case wherea is even, since the other case is quite similar.
Then (#3) and (2:4) imply that 2(m1 1)
m1 (g (a1)(a1b0))2a(ab0) 2a
a2(gb0 1)
2(2m1 a 2)(g (a1)(a1b0)) m1(a2)
is non-negative. Therefore we have
(KV D)22a(gb0 1)=(a2) (2:6)
from (2:5). Thus (KV D)22a(g 1)=(a2). If the equality sign holds,
thenb00 andm1 mn(a2)=2. p
Next, we consider the case where (V#;D#) is of special type. Let m0b (a2). Then we have
D#(a2)D0(a2m0)G; 2m0<(a2)=2:
(2:7)
LEMMA 2.7. Keep the notation and assumptions as above. Then the following hold:
When(a1)(a4)=2ga(a1)or(a1)(2a1)=2according as a is even or odd,
(KVD)2>2a(g a 1)=(a1):
If a(a3)=2g<(a1)(a4)=2and a is even, then (KVD)2g(a2 a 4)=2:
When the equality sign holds, (m0;n)(2;(a1)(a4)=2 g) or (a;n;m5)(6;5;3).
If g<a(a3)=2and a is even, then
(KVD)22g(a 2)=a4:
When the equality sign holds, mna=2and n>44=(a 2).
If(a1)(a2)=2g<(a1)(a4)=2and a is odd, then (KVD)2g(a2 a 4)=2:
When the equality sign holds, (m0;n)(2;(a1)(a4)=2 g) or (n;m4)(4;(a1)=2).
If g<(a1)(a2)=2and a is odd, then
(KV D)22g(a 1)=(a1)1:
When the equality sign holds, mn(a1)=2and n>4.
PROOF. From a standard calculation, we have (KVD)2 a22(m0 1)a Pn
i1(mi 1)2; 2g (a1)(a2m0) Pn
i1mi(mi 1):
(2:8)
In the same way of (2:5), we have (KVD)22(m0 1)
m0 ga2(1 m0)a 2m202m0
m0 :
(2:9)
If the equality sign holds here, thenm0m1 mn. By an argument similar to Lemma 2.6, we have (KV D)2>2a(g a 1)=(a1).
Wheng(a2a4m0)=2, 2(m0 1)
m0 ga2(1 m0)a 2m202m0
m0 ga2 a 4
2
(2:10)
(2g a2 a 4m0)(m0 2) 2m0
is non-negative. Remark that (a2a4m0)=2 increases as m0 grows.
Consider the case where ga(a3)=2 or (a1)(a2)=2 according as a is even or odd. From (2:10), (#3) and (2:9), we have (KV D)2g(a2 a 4)=2. Furthermore, ifm02, then (KVD)2 g(a2 a 4)=2 and n(a1)(a4)=2 g. When m062 and (KV D)2g(a2 a 4)=2 hold, m0m1 mn and 2g
a2a4m0. Hence, then, we havem0 mna=2 or (a1)=2 sincega(a3)=2 or (a1)(a2)=2 according to the parity ofa. These and (2:8) implyn44=(a 2) or 4 according asais even or odd.
In the rest of the proof, we may consider the case whereg<a(a3)=2 andais even, since the other case is similar. Recall thatm0a=2. When ga(a1m0)=2,
2(m0 1)
m0 ga2(1 m0)a 2m202m0 m0
2a 4 a g4
(2g a(a1m0))(2m0 a) am0
is non-negative. Sincea(a3)=2a(a1m0)=2 for anym0, it follows from (2:9) that (KVD)22g(a 2)=a4. If (KVD)2 attains the equality, thenm0 mnand 2m0 a0. p
As a corollary of Lemmas 2.6 and 2.7, we have the following.
LEMMA2.8. Let V be a smooth rational surface which is neitherP2nor Sd and D a smooth irreducible curve of genus g7on V. Assume that (KVD):E1for any( 1)-curves E on V and the minimal ruling de- gree of(V;D)is a3. Then the following hold:
When g(a1)(a4)=2,
(KVD)2>2a(g a 1)=(a1):
When(a1)(a2)=2g<(a1)(a4)=2, (KVD)2g(a2 a 4)=2:
If (KVD)2 attains the equality, then there exists a blow-down y0:V !P2contracting( 1)-curves Eisuch that D is one of the following:
(i) D((3a6)=2)y0OP2(1) ((a2)=2)P3
i0Ei (a=2)E4, where a is even and g(a1)(a2)=2,
(ii) D((3a5)=2)y0OP2(1) ((a1)=2)P4
i0Ei, where a is odd, a5and g(a1)(a2)=2,
(iii) D9y0OP2(1) 3P3
i0Ei, where a4and g16, (iv) D(a4)y0OP2(1) 2Pn
i0Ei, where n(a1)(a4)=2 g) and(a1)(a2)=2g<(a1)(a4)=2:
Furthermore,Cliff(D)a if(V;D)is one of them.
When g<(a1)(a2)=2,
(KVD)2 2a(g 1)=(a2) if a is even;
2g(a 1)=(a1)1 otherwise:
( (2:11)
PROOF. Wheng(a1)(a4)=2, from Lemmas 2.6 and 2.7, we have (KVD)2 >2a(g a 1)=(a1). When g<(a1)(a2)=2, the in- equalities (2:11) also follow immediately. Suppose that (V;D) satisfies one of (i)-(iv). Then we have Cliff(D)afrom [8, Proposition 2.2], sincea3 andD2>(a2)2.
For the case where (a1)(a2)=2g<(a1)(a4)=2, we may restrict ourselves to the case whereais even, since the other case is similar and simpler. Wheng(a23a4)=2, we have
2a(g 1)=(a2)g(a2 a 4)=2;
where the equality sign holds if and only if g(a23a4)=2. Thus Lemmas 2.6 and 2.7 imply the statement for g>(a23a4)=2. Fur- thermore, if
(g;(KVD)2)((a23a4)=2;a2a); ((a1)(a2)=2;a2a 1) and (V#;D#) is of special type, then (m0;n)(2;(a1)(a4)=2 g) from Lemma 2:7.
In what follows, we consider the case where (V#;D#) is of general type.
At first, we concentrate on the case where g(a23a4)=2. Then we only have to determine the pair (V;D) satisfying (KVD)2a2a. It follows from Lemma 2.6 that such a pair satisfies b00 and mn (a2)=2. Furthermore we haven4 4=afrom (2:4). Thusa4 and n3 follow. Therefore D6yD0(63d)yG 3P3
i1Ei and we haved2 becauseyD0:D6 3d0. Remark thatD#has no singular points on the minimal sectionD0ofV#in the case whered2. Hence the case whered1 covers the cases whered0;2.
We next consider the case where g(a1)(a2)=2. It follows from (2:6) that
(KVD)2a2a2a(b0 1)=(a2)>a2a 1
when b01. Hence we can assume b00. If D# has at least four (a2)=2-ple points, then (2:4) shows g<(a1)(a2)=2. In the same way as in Lemma 2.6, we have (KVD)2a2a2 n0 for (V;D) whose D# has just n0 (a2)=2-ple points with n00;1;2;3. If (KV D)2a2a2 n0 holds, then mn01 mna=2. Fur- thermore, (KVD)2 attains the minimum a2a 1 at n03. In fact, (2:4) and (g;(KVD)2)((a1)(a2)=2;a2a 1) imply thatD#has exactly three (a2)=2-ple points and a a=2-ple point as its singularity.
Thus
D(a2)yD0(d2)(a2)
2 yG a2 2
X3
i1
Ei a 2E4
and we haved2 sinceyD0:D(a2)(1 d=2)0. We can show that
the case whered1 covers the cases where d0;2 similarly as in the
previous case. p
Keep the situation as in Lemma 2.8. Furthermore, we consider the case where 4c1a2c 2. When g(c1)(c2)=2, we have (KVD)2 >g(c2 c 4)=2 from Lemma 2.8. We also have (KVD)2 >2c(g c 1)=(c1) when
g<(a1)(a4)=2 (a c)(2c 2 a)=(c 1) org>(c1)(a1). For (V;D) satisfying
(KVD)22c(g c 1)=(c1) (2:12)
and
(a1)(a4)=2 (a c)(2c 2 a)=(c 1)g(c1)(a1);
(2:13)
we have the following:
LEMMA 2.9. Keep the notation and assumptions as above. Then Cliff(D)a.
PROOF. Assume that (V;D) satisfies (2:12). Then it follows from Lemma 2.6 that (V;D) has a #-minimal model of special type. We recall (2:7). First of all, we show that (V;D) attains maximum ofnatm02 as follows:Take a smooth irreducible curveB2 jDPn
i1(mi 2)Eijfor a fixed (V;D). We consider (V;B) and its#-minimal model (V#;B#). Remark that B# has just n double points as its singularity. Then (KVB)2
(KVD)2Pn
i1mi(mi 2) and 2g(B)2gPn
i1(mi1)(mi 2). These and (2:12) imply that
2c
c1g(B) 2c (KVB)2 1 c1
Xn
i1
(c mi)(mi 2):
Sincec(a2)=2>m0m1 mn 2, the right-hand side is non- negative. Hence (V;B) also satisfies (KVB)22c(g(B) c 1)=(c1).
From (2:12) and (2:8) with m12, we have n(a c)(2c 2 a2(2 m0))=(c 1). In the end, atm02,
n(a c)(2c 2 a)=(c 1):
(2:14)
We now return to the proof of Lemma 2.9. We have D242c2g 2g=(c1) n from (2:12) and (KVD)2
2(KVD):DKV2 D24g 48 n D2. Therefore (2:13) and (2:14) imply that
D2 (a2)24(a2 (a c)(2c 2 a)=(c 1))>0:
Hence Lemma 2.9 follows from [8, Proposition 2.2]. p PROOF OFPROPOSITION2.4. By Lemma 2.5 and the argument just before Lemma 2.5, we can assume thatYis a smooth rational surface which is nei- therP2norSd. Now the assumption implies that the minimal ruling degreea of (Y;G) is greater than or equal toc. From Lemma 2.9, in the case where a>c, we have (KYG)2>maxfg(c2 c 4)=2;2c(g c 1)=(c1)g when g(c1)(c2)=2. For the case where ac, Lemma 2.8 implies
Proposition 2.4. p
3. Mordell-Weil lattices
We briefly review the theory of the Mordell-Weil lattice due to Shioda.
Let X be a smooth projective surface withpgq0 andf :X!P1 a relatively minimal fibration of genusg. We assume that it has a section. Via f, we can regardXas a smooth projective curve of genusgdefined over the rational function field KC(P1). Let JF be the Jacobian variety of a generic fibreF off. The Mordell-Weil group offis the group ofK-rational pointsJF(K). It is a finitely generated abelian group, sinceXis a regular surface. The rank of the group, which we denote byr, is called theMordell- Weil rank. It follows from [13, Theorem 3] thatris given by
rr(X) 2 X
t2P1
(vt 1);
wherer(X) denotes the Picard number, that is, the rank of NS(X), andvt
denotes the number of irreducible components of the fibref 1(t). In par- ticular, we haverr(X) 2 iff has irreducible fibres only.
There exists a natural one-to-one correspondence between the set ofK- rational pointsF (K) and the set of sections off. ForP2F (K) we denote by (P) the section corresponding toPwhich is regarded as a curve onX.
We specify a section (O) corresponding to the origin ofJF(K) and call it thezero section. Shioda's main idea in [13] is to regard JF(K) as a Eu- clidean lattice endowed with a natural pairing induced by the intersection
form onH2(X). LetTbe the subgroup of NS(X) generated by (O) and all the irreducible components of fibres off. With respect to the intersection pairing, the sublatticeTis called thetrivial sublattice and its orthogonal complement LT?NS(X) is called the essential sublattice. Via the natural isomorphism of groupsJF(K)'NS(X)=Tin [13, Theorem 3], we obtain a symmetric bilinear formh;ionJF(K) which induces the struc- ture of a positive-definite lattice onJF(K)=JF(K)tor (see [13, Theorem 7]). The lattice (JF(K)=JF(K)tor;h;i) is called theMordell-Weil latticeof the fibrationf :X!P1. Shioda shows that if all the fibres off are irre- ducible, then the Mordell-Weil lattice off is isomorphic toL , where the opposite lattice L is defined from L by putting the minus sign on the intersection pairing onL(see [13, Theorems 3 and 8]).
As to fibred surfaces withpgq0 of Clifford indexc, we can bound the Mordell-Weil rank by Theorem 2.3 in the same way as in [8, Lem- ma 3.1] (see also [12, Theorem 2.8], [11, Proposition 2.2] and [7, Theo- rem 3.4]):
LEMMA 3.1. Let X be a smooth projective surface with pgq0, f :X!P1a relatively minimal fibration of genus g and of Clifford index c3. Let r be the Mordell-Weil rank of f . Then the following hold:
(1) r2(c2)(gc1)=(c1)when g>(c2)(c3)=2.
(2) r3g (c3)(c 4)=2when(c1)(c2)=2g(c2)(c3)=2.
If r attains the maximum, then all fibres of f are irreducible and the reduction of(X;F)is obtained by FjKXFj. In particular, X is a rational surface when r attains the maximum.
In what follows, we restrict ourselves to the case when r attains the maximum. Namely, we letf :X!P1be a fibred rational surface of genus g and of Clifford index c3 whose Mordell-Weil rank attains the max- imum value given in Lemma 3.1. Suppose that we are given such a fibra- tionf :X!P1and let (Y;G) be the reduction of (X;F). Thenf gives us a pencilLf jGjall of whose members must be irreducible. Note also that all the base points of Lf must be transversal ones since KX=P1 is ample.
Conversely, if we are given a pencil L jGj which has only transversal base points and all fo whose members are irreducible, then we get a fi- bration with the desired properties by blowing up BsL.
Let the situation be as in (1) of Lemma 3.1. Then Y Sd from Theorem 2.3. We show that the case (c1)(c2)d2(gc1) is impossible. This can be seen as follows:If (c1)(c2)d2(gc1),
then D0:G0. Hence the pull-back to X of D0 is a rational curve con- tained a fibre of f, which is inadequate in our situation. Therefore, we have (c1)(c2)d<2(gc1).
Let (Sd;G) be as in Lemma 2.5 with g>(c2)(c3)=2 and (c1)(c2)d<2(gc1). ThenGis very ample. Hence we can find a pencilL jGjwhose members are all irreducible and which has exactlyG2 transversal base points. In fact, we can take it as a Lefschetz pencil (see [4]). Then the fibration f :X!P1 obtained by Lhas Mordell-Weil rank which attains the equality as in (1) of Lemma 3.1.
DEFINITION 3.2. Let (c;g;d) be as in (i)-(iv) of Lemma 2:5 with g>(c2)(c3)=2 and d<2(gc1)=((c1)(c2)). A fibration f :X!P1of genusgand of Clifford indexc3 obtained by blowing upSd as above is called a fibration of type (c;g;d;0).
Let the situation be as in (2) of Lemma 3.1. From Theorem 2.3, there exists the blow-down y0:(Y;G)!(P2;G0) constracting (n1) ( 1)- curves E0;E1;. . .;En such thatG0 is as in (i)-(iv) of Theorem 2.3, where n4 for (i) or (ii), and where n3 for (iii). Putpjy0(Ej). Ifp1 is an infinitely near point ofp0, then the pull-back toXofE0 E1is a rational curve contained in a fibre offfrom (E0 E1):G0, which contradicts the irreducibility off. Furthermore, in the quite same argument, we have the following:
PROPOSITION3.3. Keep the notation and assumptions as above. If
n1, then a configuration of (n1) points p0;. . .; pn satisfies two
conditions as follows: The first condition is that p0;. . .; pn expect the
c=2-ple point p4 as in (i) of Theorem 2.3 are not infinitely near
points. For each case of (i)-(iv), the second condition is the following:
(i) Any three points of p0;. . .;p3are not colinear, though p4may be an infinitely near point.
(ii) Any four points of p0;. . .;p4are not colinear.
(iii) p0;. . .;p3 are in general position, that is, any three points of them are not colinear.
(iv) Any (c4)=2or(c5)=2points of p0;. . .;pn are not colinear according as c is even or odd.
We use the following theorem in order to show the next lemma.
THEOREM 3.4 ([10]). Let SPdeg (S) g(S)1 be a nondegenerate, line- arly normal, smooth regular surface of degree deg (S) and of sectional genus g(S). Let x0;. . .;xk be distinct (k1) points of S for kdeg (S) 2g(S) 2. Let s be a blowing up at x0;. . .;xk of S and Ejs 1(xj). Set HsOS(1) Pk
i0Ei. Then H is very ample if and only if for all j with1jk1, any distinct j points offx0;. . .;xkgdo not lie on any irreducible reduced curve of degree j on S.
LEMMA3.5. Let D0 be a plane curve with the same degree and singu- larity as in(i)-(iv) of Theorem2:3. Assume that a configuration of sin- gular points p0;. . .;pn of D0 satisfies the two conditions as in Proposi- tion3.3. Let y0:(V;D)!(P2;D0) denote the composite of blow-ups at n1points p0;. . .;pn. Then D is very ample.
PROOF. PutEjy01(pj). At first, we consider the case (i). We remark thatj KVjis free from base points, becausep0;. . .;p4are in almost gen- eral position (see [3, Theorem 1 of III in p. 39]). Consider the surface ob- tained by blowingP2up atp0;. . .;p3in general position. Then the anti-bi- canonical map is an embedding toP15. LetSbe the embedded surface toP15 ands:V!Sthe blow-down contractingE4. Since no line is onS,
Hs( 2KS) E46y0OP2(1) 2X3
i0
Ei E4
is very ample from Theorem 3.4. Therefore DH(c=2 1)( KV) is also. We can treat the case (ii) in the same way by setting H4y0OP2(1) P4
i0Ei. For the case (iii), it is obvious.
Let the situation be as in (iv). Since the case wherecis even is similar and simpler, we restrict ourselves to the case wherecis odd. LetSbeP2 embedded by jOP2((c5)=2)j. Then irreducible reduced curves whose degree is at most (n1) come from only lines on P2. Therefore Theo- rem 3.4 implies thatHy0OP2((c5)=2) Pn
i0Eiis very ample. Hence we only have to show that
jD Hj c3
2 y0OP2(1) Xn
i0
Ei
is free from base points. If any (c3)=2 points of p0;. . .;pn are not co- linear, then we can see similarly thatD His also very ample by applying
Theorem 3.4. Thus we suppose thatp0;. . .;p(c1)=2lie on a linel0;...;(c1)=2. Furthermore, we concentrate on the case wherenc1, since the other cases are similar and simpler. Remark that any pi, (c3)=2ic1 are not lie on l0;...;(c1)=2 from the second condition (iv) in Proposition 3.3.
Letlibe a line throughpi. In particular, we denote the line throughpiand pi(c1)=2 byli;i(c1)=2, 1i(c1)=2. Consider the strict transform of l0(c1)=2P
i1 li;i(c1)=2
by y0. It is disjoint from the strict transform of l0;...;(c1)=2ifl0is distinct froml0;...;(c1)=2. Furthermore,jD Hjis free from base points on E0 since we can choosel0 for any directions at p0. In the same way, jD Hj is also on the whole of y0l0;...;(c1)=2. Next, consider l0;...;(c1)=2 c1P
i(c3)=2li. Then jD Hj is free from base points on Vny0l0;...;(c1)=2 since we can chooselifor any directions atpi. p From Lemma 3.5, for any cases as in Proposition 3.3, we can show the existence of the fibrations similarly as in the previous case. Hence we can take a pencilL jDjenjoying the desired properties, as a Lefschetz pencil for example. In particular, L has (3g (c3)(c 4)=2 n) transversal base points.
DEFINITION 3.6. Let (V;D) be as in Lemma 3:5. If V is obtained by blowing up n1 points ofP2, the corresponding fibrationf :X!P1 of genus g and of Clifford index c as above is called a fibration of type (c;g;1;n).
3.1 ±Type(c;g;d;0)
Here we determine the Mordell-Weil lattices for fibrationsf :X!P1 of type (c;g;d;0). For this purpose, we use the following notation. We denote the pull-backs toXofD0andGby the same symbols. Furthermore, we denote bye1;e2;. . .;erthe disjoint ( 1)-sections off coming from the base points of Lf, where r2(c2)(gc1)=(c1). Then we have NS(X)'ZD0ZGLr
i1Zeiand F(c2)D0 (c2)d
2 1 g c1
G Xr
i1
ei: (3:15)
We takeeras the zero section (O). The sublatticeT(c;g;d;0) NS(X) gener- ated by er andF is the trivial sublattice. LetL(c;g;d;0) be the orthogonal complement ofT(c;g;d;0). Then the Mordell-Weil lattice (JF(K);h;i) is iso- morphic toL(c;g;d;0) .
Recall then that the degreedof the Hirzebruch surface is an invariant of the fibration.
THEOREM3.7. For a fibration of type(c;g;d;0), the lattice L(c;g;d;0) is isomorphic to a positive-definite unimodular lattice of rank r2(c2)(gc1)=(c1) whose Dynkin diagram is given by the following:
Figure 1 in the case when(c2)d=2 g=(c1)0(modc2) Figure 2 in the case when(c2)d=2 g=(c1)1(modc2) Figure 3 in the case when(c2)d=2 g=(c1)1`(modc2)with 1`c 1,
Figure 4 in the case when(c2)d=2 g=(c1)c1(modc2)
Fig. 1
Fig. 2
Fig. 3
Fig. 4
In particular, L(c;g;d;0) depends on only 2g=(c1) mod (c2) in the case when c is odd, and on the combination of g=(c1) mod (c2)and the parity of d in the case when c is even. Here the numbers in the circles denote the self-parings of elements, and a line between two circles shows that the paring of the corresponding two elements is equal to( 1). Fur- thermore, L(c;g;d;0) is an odd lattice in the case when c is odd(see Table1) and the parity of the lattice is the same as that of(dg1)or of(g1) respectively in the case when c2 mod 4 (see Tables 2 and 3) or c0 mod 4 (see Tables4and 5). In particular, even and odd lattices both occur for a fixed g(c1)2in the case when c2 mod 4.
Table 1. ±L(c;g;d;0) is an odd lattice in the case whenc1 mod 2.
2g=(c1) modc2 Dynkin diagram
0 Figure 1
1 Figure 3 with`(c 1)=2
2 Figure 4
2i1 (i1;2;. . .;(c 3)=2) Figure 3 with`(c 1)=2 i
2j (j2;3;. . .;(c 1)=2) Figure 3 with`c1 j
c Figure 2
c1 Figure with`(c1)=2
Table 2. ±L(c;g;d;0) is an even lattice in the case whenc2 mod 4:
g=(c1) modc2 d Dynkin diagram
0 odd Figure 3 with`c=2
1 even Figure 4
h (h2;4;. . .;c=2 1) odd Figure 3 with`c=2 h
i (i3;5;. . .;c=2) even Figure 3 with`c1 i
c=21 odd Figure 1
c=21j (j1;3;. . .;c=2 2) even Figure 3 with`c=2 j
c=21k (k2;4;. . .;c=2 1) odd Figure 3 with`c1 k
c1 even Figure 2
Table 3. ±L(c;g;d;0) is an odd lattice in the case whenc2 mod 4:
g=(c1) modc2 d Dynkin diagram
0 even Figure 1
h (h1;3;. . .;c=2 2) odd Figure 3 with`c=2 h
i (i2;4;. . .;c=2 1) even Figure 3 with`c1 i
c=2 odd Figure 2
c=21 even Figure 3 with`c=2
c=22 odd Figure 4
c=22j (j1;3;. . .;c=2 2) even Figure 3 with`c=2 1 j
c=22k (k2;4;. . .;c=2 1) odd Figure 3 with`c k
Table 4. ±L(c;g;d;0) is an even lattice in the case whenc0 mod 4:
g=(c1) modc2 d Dynkin diagram
1 even Figure 4
h (h1;3;. . .;c=2 1) odd Figure 3 with`c=2 h
i (i3;5;. . .;c=21) even Figure 3 with`c1 i
c=21 odd Figure 1
c=21j (j2;4;. . .;c=2 2) even Figure 3 with`c=2 j
c=21k (k2;4;. . .;c=2) odd Figure 3 with`c1 k
c1 even Figure 2
Table 5. ±L(c;g;d;0) is an odd lattice in the case whenc0 mod 4:
g=(c1) modc2 d Dynkin diagram
0 even Figure 1
h (h0;2;. . .;c=2 2) odd Figure 3 with`c=2 h
i (i2;4;. . .;c=2) even Figure 3 with`c1 i
c=2 odd Figure 2
c=22 even Figure 3 with`c=2 1
c=22 odd Figure 4
c=22j (j2;4;. . .;c=2 2) even Figure 3 with`c=2 1 j
c=22k (k2;4;. . .;c=2 2) odd Figure 3 with`c k
PROOF. Let us keep the notation as above. In particular,Fis given by (3:15) and (O)er. Take the following elements fromL(c;g;d;0):
jr 1G Xc2
i1
ei; jiei ei1 (1ir 2; i6r c 1):
We takejrandjr c 1fromL(c;g;d;0) according to the following rule:
(1) Ifc1d=2 g=((c1)(c2))2Z, then put jrD0c1G e1; jr c 1er c 1 er c: (2) Ifc2d=2 g=((c1)(c2)) 1=(c2)2Z, then put
jrD0c2G; jr c 1er c 1 er c:
(3) Ifc3d=2 g=((c1)(c2)) (1`)=(c2)2Z, then put jrD0c3G Xr 1
ir `
ei; jr c 1er c 1 er c:
(4) Ifc4d=2 g=((c1)(c2))1=(c2)2Z, then put jrD0c4G e1 F(O); jr c 1F(O) er c:
Here the numbering of theji's corresponds to that of the vertices in Figures 1, 2, 3, 4 according to 2g=(c1) mod (c2) withd2g=(c1) mod 2 in the case whencis odd and to the combination ofg=(c1) mod (c2) and the parity ofdin the case whencis even.
Then these together withF, (O) clearly form a basis for NS(X) overQ in either case. While we see that fj1;j2;. . .;jrg forms a Z-basis for L(c;g;d;0) , we divide our argument between the case (4) and the other cases.
At first, we restrict ourselves to the case (3), since the cases (1) and (2) are quite similar. Consider the matrix representing the base change from (D0;G;e1;e2;. . .;er) to (jr;jr 1;j1;j2;. . .;jr 2;F;(O)). Then it is easy to see that, off the (r1)-th row, it is an integral triangular matrix all of whose diagonal entries are equal to one, and we have
Fer 1(c2)jr 2g
c12`
jr 1 2g
c11` Xc2
k1
kjk
rX` 1
kc3
(r`(c2) k)jk Xr 2
kr `
((c3)(r k) c 2)jk (O):
For (4), we only have to note that the matrix representing the base
change from (D0;G;e1;e2;. . .;er) to (jrF(O);jr 1;j1;j2;. . .;jr c 2; er c 1;er c;. . .;er 1;(O)) is an integral triangular matrix, all of whose diagonal entries are equal to one, and we have
F er c 1(c2)(jrF(O))
2g
c1jr 1Xc2
k1
2gk
c1c2 k
jkr cX3
kc3
(r c 2 k)jk Xr 1
kr c
ek (O);
jr c 1 F (O) er c; ji ei1ei (ir c;. . .;r 2):
Hence in either casefj1;j2;. . .;jrgforms aZ-basis forL(c;g;d;0) and we obtain the corresponding Dynkin diagrams.
As to the last statement, we consider the case whenc2 mod 4 only, since the other cases are similar. We note that the self-pairing numbers of ji's are always even except forjr, while the parity of that ofjrvaries even if we fixg(c1)2. Consider for example the case wheng is divisible by (c1)(c2). Ifdis even thenjr22g=((c1)(c2))1 is odd, and ifd is odd thenjr22g=((c1)(c2))(c2)=2 is even with`c=2. p
3.2 ±Type(c;g;1;n)
We shall determine the Mordell-Weil lattices for fibrationsf :X!P1 of type (c;g;1;n) with (c1)(c2)=2g(c2)(c3)=2. Whenn0, the reductionYis obtained by blowingP2up atp0;. . .;pncorresponding to singular points ofG0as in Theorem 2.3. Furthermore, the configuration is as in Proposition 3:3. Leteibe the inverse image onXofpi(i0;. . .;n).
We denote bylthe pull-back toXof a line onP2and letei,n1ir, be the disjoint ( 1)-sections of f coming from base points of Lf. Then NS(X)'ZlLr
i0Zei. We take the last ( 1)-sectioneras the zero section (O). Then (O) andF generate the trivial sublatticeT(c;g;1;n)NS(X). Let L(c;g;1;n)be the essential sublattice obtained as the orthogonal complement ofT(c;g;1;n).
We first consider the case whereG0is as in (iv) of Theorem 2.3. Then F(c4)l 2Xn
i0
ei Xr
in1
ei:
THEOREM3.8. Let(c1)(c2)=2g(c2)(c3)=2. For a fibra- tion of type(c;g;1;n)with n(c1)(c4)=2 g, the lattice L(c;g;1;n) is
isomorphic to a positive-definite odd unimodular lattice of rank r3g (c3)(c 4)=2whose Dynkin diagram is as in Figure5.
Fig. 5
PROOF. Take the following elements fromL(c;g;1;n), whose numbering corresponds to that of the vertices in Figure 5:
jiei 1 ei (1ir 1; i6n1);
jn1en en1 en2; jr l Xr 1
ir c 4
ei:
Then we can show that the above elements form aZ-basis forL(c;g;1;n)in the same way as in the case (3) in the proof of Theorem 3.7. p Secondly, we shall determine the Mordell-Weil lattices for fibrations of type (c;(c1)(c2)=2;1;4). Ifcis even, then
F3c6
2 l c2 2
X3
i0
ei c 2e4
X
c25c9
i5
ei; (O)ec25c9: THEOREM3.9. For a fibration of type(c;(c1)(c2)=2;1;4)with an even number c, the lattice L(c;(c1)(c2)=2;1;4) is isomorphic to a positive- definite odd unimodular lattice of rank c25c9whose Dynkin diagram is as in Figure6.
Fig. 6
PROOF. Take the following elements from L(c;(c1)(c2)=2;1;4), whose numbering corresponds to that of the vertices in Figure 6:
jiei 1 ei (i1;2;3;6;7;. . .;c25c8);
j4e3 e4 e5; j5e4 c=24X
i5
ei; jc25c9l e0 e1 e2:
Then we can show that the above elements form a Z-basis for L(c;(c1)(c2)=2;1;4) in the same way as in the case (3) in the proof of Theo-
rem 3.7. p
Next, we consider the case wherec is odd,g(c1)(c2)=2,d1 andn4. Then
F3c5
2 l c1 2
X4
i0
ei c2X5c9
i5
ei; (O)ec25c9:
THEOREM3.10. For a fibration of type(c;(c1)(c2)=2;1;4)with an odd number c, the lattice L(c;(c1)(c2)=2;1;4) is isomorphic to a positive-de- finite odd unimodular lattice of rank c25c9whose Dynkin diagram is as in Figure7.
Fig. 7
PROOF. Take the following elements from L(c;(c1)(c2)=2;1;4), whose numbering corresponds to that of the vertices in Figure 7:
jiei 1 ei (1ic25c8; i65; (2c27c15)=2);
j5e4
X
(c9)=2 i5
ei; j(2c27c15)=2e(2c27c13)=2 (F(O));
jc25c9l e0 e1 e2 e(2c27c15)=2:
Then we can show that the above elements form a Z-basis for L(c;(c1)(c2)=2;1;4) in the same way as in the case (4) in the proof of Theo-
rem 3.7 p
In the last, we shall determine the Mordell-Weil lattices for fibrations of type (4;16;1;3). Then
F9l 3X3
i0
ei X48
i4
ei; (O)e48:
THEOREM3.11. For a fibration of type(4;16;1;3), the lattice L(4;16;1;3)
is isomorphic to a positive-definite even unimodular lattice of rank 48 whose Dynkin diagram is as in Figure 8.
Fig. 8
PROOF. Take the following elements fromL(4;16;1;3), whose numbering corresponds to that of the vertices in Figure 8:
jiei 1 ei (i1;2;3;5;6;. . .;47); j4e3 e4 e5 e6; j48l e0 e1 e2:
Then we can show that the above elements form aZ-basis forL(4;16;1;3)in the same way as in the case (3) in the proof of Theorem 3.7. p
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