2-Spanned surfaces of sectional genus six

Annali di Matematica pura ed applicata (IV), Vol. CLXV (1993), pp. 197-216

2-Spanned Surfaces of Sectional Genus Six(*). ANTONIO LANTERI

- Un fibrato lineare olomorfo su una superficie algebrica proiettiva complessa non singolare S ~ detto k-spanned se le sue sezioni separano opportuni O-cicli di S di lunghezza k + 1 - - da (k + 1)-uple di punti distinti sino a getti di ordine k - - e forniscono pertanto una immersione proiettiva di S di ordine superiore. Si fornisce un contributo alla classificazione proiettiva delle superfici determinando quelle di genere sezionale 6 immerse da un fibrato 2spanned e quelle di genere sezionale 7 immerse da un fibrato 3-spanned.

Sunto.

Introduction.

The notion of k-spanned line bundle L on a complex projective smooth surface S has been introduced in [BFS] in connection with the concept of higher order embedding. Essentially L is k-spanned if F(L) separates 0-cicles of length k § 1 on S, going from (k + 1)-tuples of distinct points up to jets of order k (for the precise definition see Sect. 0). We say that S is k-spanned to mean that it is embedded via a k-spanned line bundle. Afterwards this notion has been extensively studied in [BS]. In particular, in [BS] k-spanned surfaces (k I> 2) of sectional genus g ~ 2), then L . C 1> k for every smooth rational curve CoX. (0.3) LEMMA. - If L is 2-spanned then L . C ~>4 for every effective divisor C of X

with p~(C) = 1. PROOF. - If C is reducible the assertion follows from (0.2) for k = 2. So we can assume that C is an irreducible plane cubic of (X, L). But then taking a 0-cycle Z cut out on C by a line we see that (0.0.1) is not a surjection, contradicting the 2-spannedness of L. 9 Note that for C a smooth curve, (0.3) is a consequence of[BS, (0.5.2) and (1.4.1)]. The following result proven by BALLmO[Bal], [Ba2] is very important in the sequel. (0.4) If L is k-spanned (k I> 2), then h ~(L) I> k + 5 unless k = 2 and X embedded by ILl is either the Veronese surface or a general complete intersection of three quadrics of p s . k-spannedness results for some special Pezzo surfaces can be found in [BS, Sect. will refer to them throughout the paper, which we need for cases 7, 9 and 8 in the (0.7)].

classes of surfaces like pl-bundles and Del 3 and (0.8)] (see also [BFS, (2.6)]) and we Here we prove two 2-spannedness results Theorem. The first one is inspired by [BS,

(0.5) PROPOSITION. - Let X be one of the following surfaces polarized by the corresponding very ample line bundle H:

i) Bpl ..... p~(Fe), e ~< 1, H = r~*[2z + (e + 3 ) f ] | [ E 1 + ... + E6] - 1 , ii) Bp (Y), where Y is the Del Pezzo surface of degree Ky. Ky = 2 (p a general point), H = rc*Ky 2 | [E] -1 , where, in both cases 7: denotes the blowing up and E = r~-1 (p), Ei = =-1 (Pi). In both cases the line bundle L = H | K~ 1 is 2-spanned. PROOF. - To prove the 2-spannedness of L we use the main result of[BFS]. Assume that L can be written as (0.5.1)

L = Kx | M ,

with M numerically effective and such that M. M/> 13. Then, if L is not 2-spanned X must contain an effective divisor D satisfying the following numerical inequalities: (0.5.2)

M . D - 3 I - 3 K x ' D

in view of the spannedness of r:*f. Hence - K x ' D < 2 and so, the ampleness of K~/1 implies - K x ' D = 1. F r o m the first equality in (0.5.2) we have (0.5.3)

D . D >I M . D - 3 = - 3Kx" D + r~*f. D - 3 = rc*f" D >t 0 ;

furthermore, by the genus formula we see that (0.5.4)

D - D = 2pa (D) - 2 - Kx" D = 2pa (D) - 1

is odd. Putting together (0.5.3) and (0.5.4) we thus get D - D I> 1. On the other hand, since M . D ~< 5, by (0.5.2), the Hodge index theorem implies that 25 i> (M. D) ~ I> (M. M)(D-D) = 30D-D, giving D .D < 1, a contradiction. CASE ii). As K x = ~ v * K y | with

we have L = K x 2 |

1. Thus (0.5.1) holds

M = K x 8 | r:* K y 1 . Note that r:*Ky 1 is numerically effective and that K x 1 is ample, since p is general; so that M is ample; in addition M . M = 9Kx" K x + 6Kx" r:* K y + K y . K y = 23.

Assume that L is not 2-spanned; then the last inequality in (0.5.2) gives 6 > M.D

= - 3 K x ' D - r : * K y . D >I - 3 K x ' D ,

in view of the numerical effectivity of r:*K:71 . Therefore - K x ' D < 2 and then -Kx'D = 1, as before, since K x 1 is ample. The same argument as in case i) thus shows that D . D I> 1. Then, by the Hodge index theorem, 1 = (Kx" D) 2 >I (Kx" K x ) ( D " D) >I Kx" K x = 1,

so we get [D] - K x 1 (in fact [D] = K~/1, since Pic (X) is torsion free). Then, recalling

ANTONIO LANTERI: 2 - s p a n n e d s u r f a c e s o f s e c t i o n a l g e n u s s i x

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the first inequality in (0,5.2) we have 1 = K x ' K x = D . D >t M . D

- 3 = 3Kx'D - 3 + r~*K~'.D = =*Ky1.D.

On the other hand r:* K ~ I ' D

a contradiction.

= r~* K y ' K x =

r~* Ky.7~* K y = 2,

[]

Now consider the conic bundle (X, H), where (0.6.0) X = Bpl ..... pg(F~),

e~ 0, 2~ being numerically effective. Assume that 2~. C-- 0; then F would be an irreducible curve of degree m in (S, 2) having a point of multiplicity m. Then 0 ~I degr176 we conclude that degr = 1.

') - 2)

9

I conclude this section with the following Lemma, which will play a relevant role in Sects. 2, 3. (0.9) LEMMA.- Let L be a 2-spanned line bundle of genus g on a rational surface X with K x ' K x >I O. Let H = Kx | L and set d' = H . H . Then d ' ~< 2g - 6 + 16/(2g + 2), equality implying both that H and L are linearly dependent in Num (X) and L . Kx = = - 4 . In addition Kx. Kx >~ d' + 6 - 2g.

PROOF.- Since X is rational we have h 2 ( K ~ 1) = h~ theorem gives h~

= 0. So the Riemann-Roch

~) >1 z(g:~ t) = t + K x ' K x .

Therefore there exists an effective divisor D e IKxI]. Note that the arithmetic genus

ANTONIO LANTERI: 2 - s p a n n e d surfaces o f sectional g e n u s six

205

of D is 1. So by L e m m a (0.3), we have that (0.9.1)

4 I d d ' >1 d ' ( 2 g + 2) and this proves the former inequality. Moreover, i f equality holds, then both b, for some a , b ~ Z and equality holds in (0.9.1), i . e . L . K x = - 4 . Now,

L~-H as

d ' = Kx" K x + 2(2g - 2) - d,

we get from (0.9.1) 4 ! 20 - 3 d ' .

This, combined with (1.6.1), provides the inequality d' I> 9, which contradicts (1.6.3). It thus follows that (X, H) is a scroll. On the other hand there is a unique elliptic scroll in p4 (e.g. see [La]): X is the pl-bundle of invariant e = - 1 over an elliptic curve and H N [~ + 2f]. It turns out that L N [3~ + f ] , which is in fact 2-spanned, as one can

ANTONIO LANTERI: 2-spanned surfaces of sectional genus six

209

prove by using the same argument of[BS, (3.4)]. However, since d e g L : = 4, L cannot be 3-spanned, according to [BS, (1.4.1)]. So we have proved (1.6) For pg = 0, q = 1 and H very ample we get case 2 in Theorem 1.

2. - P r o o f of Theorem 1: second part. In this section we deal with the last and most intricated CASE D. - H is very ample and X embedded by IHI is a rational surface of pS. In view of (1.2) we know that d' ~< 9. Now let g' = g(H). F r o m (2.0.0)

g' = g + Kx" (Kx | L) = Kx" K x + 3g - 2 - d,

(2.0.1)

d'= (Kx| L).(Kx|

L) = K x ' K x + 4g - 4 - d,

we immediately get (2.0.2)

d' -- g' + 4.

Hence (X,/4) can only have the following characters: (d', g ' ) = (4, 0), (5, 1), (6. 2), (7, 3), (8, 4), (9, 5). Now we look at the various possibilities by using the classification of rational surfaces with small sectional genus [Lil], [Li2]. As a first thing, let (d',g') = (4, 0). In this case (X,H) is either (p2, Op2(2)) or a scroll of degree 4 in P~, i.e. (X, H ) = (F0, [z + 2f]) or (F2, [~ + 3f]). Since L = H | K / 1 , this gives

(p2, Op2 (5)), !

(Z, L) = i(F0, [3z + 4f]), [(F2, [3~ + 7f]). In the first case L is 5 spanned by (0.1). In the second case L = [a + f ] 2 | [~ + 2f], hence it is 3 spanned in view of (0.1). This gives cases 1 and 3 of Theorem 1. In the third case, note that d e g L : = 1, so L cannot be 2-spanned. Let (d', g') = (5, 1). In this case X is the Del Pezzo surface of degree 5, i.e. X = = Bpl ..... p4(p2) is the blow-up of p2 at 4 points in general position and H = K;/1 . Then L = K x e is 2-spanned by [BS, (0.8)]: this gives cases 6 of Theorem 1. Let (d',g') = (6,2). In this case X = B p l ..... p6(Fe), is the blow-up of Fe (e ~ 3. From the Hodge index theorem, recalling (2.0.2), we have (2.0.3)

(g' - 6) 2 = (2g' - 2 - d ,)2 = (H. Kx) 2 >t d' gz" Kx = (g' + 4) gx" g x .

On the other hand, by (2.0.1) we have s d ' = K x ' K x + 2 0 - d; recalling (1.0.2) and (2.0.2) this gives (2.0.4)

Kx" Kx >1g' - 6.

By combining (2.0.3), (2.0.4) we get (2.1.1)

-3i 0, equality holding iff (X, H) is a conic bundle [So, (2.1)]. Recalling (2.0.2) this yields (2.1.4)

K x ' K x >I 8 - 3g'

with = iff (X, H ) is a conic bundle.

So (2.1.1) can be replaced with the sharper inequality (2.1.1')

-I dd', and in the same way as in Sect. 1 we get (3.2) d' ~ O. Then' d ' < 9, by L e m m a (0.9), equality implying that L a - ( K x | b for some a,b e Z and L ' K x = - 4 . L e t t = (a - b)/b; then -

(3.6.1)

td = - 4.

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On the other hand, from (3.6.2)

9 = d ' = H . H = K x ' K x + 2(2g - 2) - d = K x ' K x + 24 - g,

we have (t 2 - 1)d = - 15.

(3.6.3)

Now (3.6.1), (3.6.3) give t = - 1/4 or 4. Of course, it cannot be t = 4, otherwise K x would be ample, contradicting the fact that X is a rational surface. Hence L - K~ 4 . In particular X is a Del Pezzo surface, L = Ks/4, since Pic (X) is torsion free, and (3.6.2) gives Kx" K x = 1. Recalling the general properties of Del Pezzo surfaces we thus conclude that IK~/~l contains a smooth curve F of genus 1. But F. L = 4Kx" K x = 4,

which contradicts the 3-spannedness of L in view of[BS, (1.4.1)]. So case d' = 9 cannot occur. 9 Now let (d',g') = (9,4). In this case Lemma (3.6) and (3.5.1) imply that K x . K x = - 1 and (X, H) is ruled by cubic. F r o m [Lie, p. 169] we see that X is not a Pl-bundle and so this case cannot occur in view of the following (3.7) REMARK. If L is 3-spanned and (X, H) is ruled by cubics, then X must be a PLbundle. Actually, were F = F1 + F2 a reducible fibre of the ruling of X, then for the irreducible component F~ satisfying H-F~ = 1 we would get L . F i = H ' F i - K x " 9Fi ~< 1 + 1 = 2, since Fi or another irreducible component of F is an exceptional curve. But this contradicts the 3-spannedness of L, by (0.2). -

Now, taking into account (3.0.1) and (3.3.1), we get from (2.0.1) Kx. K x = d ' + d - 24 ~ d' - 1 2 = g ' - 7.

L e m m a (3.6) and the above inequality show that (3.8)

If ( d ' , g ' ) = (10, 5) then K x ' K x = - 1

or - 2 , while if ( d ' , g ' ) = (11,6) then

gx.gx = -1.

Now let (d', g') = (10, 5). By (3.8) we have to look in [Li2, pp. 161, 169] only at surfaces with •o = 11 and 12 and we thus see that (X,H) alwasy contains a ( - 1)-line E. Then L . E = H . E - K x ' E = 2, which contradicts the 3-spannedness of L, by (0.2). Finally let (d', g') = (11, 6). By (3.8) we have to look in [Li2, pp. 161, 169] only at surfaces with ~'0 = 12. Then X = Bp, ..... pl0 (p2), with H = 7~* Op2 (9) | [3E1 + ... + 3E6 + 2Es +

216

ANTONIO LANTERI: 2-spanned surfaces of sectional genus six

+ ... + 2E10] -1 . However, for the corresponding line bundle L we have h~ = 7 [Li2, Case 19 in the table at p. 161], and this contradicts the 3-spannedness of L, by (O.4). This concludes the discussion of case D' and the proof of Theorem 2.

A d d e n d u m ( F e b r u a r y 1993). After the submission of this paper and its circulation as a preprint further progress on the subject was made. I r e f e r to the paper by M. ANDREATTA, Surface8 of sectional genus