Sectional Invariants of Scrolls Over a Smooth Projective Variety
YOSHIAKIFUKUMA(*)
ABSTRACT- L etXbe a smooth complex projective variety of dimensionnand letEbe an ample vector bundle of rankronX. Then we calculate theith sectional Euler numberei(PX(E);H(E)) and theith sectional Betti numberbi(PX(E);H(E)) for i2n 3 or i1, and the ith sectional Hodge number of type (j;i j) hij;i j(PX(E);H(E)) fori2n 1and0ji, wherePX(E) is the projective space bundle associated withEandH(E) is its tautological line bundle. Moreover we define a new invariantv(X;E) of (X;E) forrn 1. This invariant is thought to be a generalization of curve genus. We will investigate some properties of this invariant.
1. Introduction.
LetX be a projective variety of dimensionndefined over the field of complex numbers, and letLbe an ample (resp. a nef and big) line bundle on X. Then (X;L) is called apolarized(resp. quasi-polarized)variety. IfXis smooth, then we say that (X;L) is a polarized (resp. quasi-polarized) manifold. In order to study polarized varieties, it is important to use an invariant of (X;L). There are the following three invariants of (X;L) which are well-known.
The degreeLn.
The sectional genusg(L).
TheD-genusD(L).
By using these invariants, many authors studied polarized varieties. In particular, P. Ionescu classified polarized manifolds by the degree under the assumption thatLis very ample withLn8 ([13], [14], and [15]), and
(*) Indirizzo dell'A.: Department of Natural Science, Faculty of Science, Kochi University, Akebono-cho, Kochi 780-8520, Japan.
E-mail: fukuma@kochi-u.ac.jp
T. Fujita classified polarized varieties by the D-genus and the sectional genus ([3]).
In [5], in order to study polarized varieties more deeply, the author introduced the notion of theith sectional geometric genusgi(X;L) of (X;L) for every integeriwith 0in. This is a generalization of the degree and the sectional genus of (X;L). Namely g0(X;L)Ln and g1(X;L)g(L).
Here we recall the reason why this invariant is called theith sectional geometric genus. Let (X;L) be a polarized manifold of dimension n2 with BsjLj ;, where BsjLjis the base locus of the complete linear system jLj. L etibe an integer with 1in. L etXn ibe the transversal inter- section of general n i members of jLj. In this case Xn i is a smooth projective variety of dimension i. Then we can prove that gi(X;L)hi(OXn i), that is,gi(X;L) is the geometric genus ofXn i.
Hence we can expect that gi(X;L) has analogous properties of the geometric genus ofi-dimensional varieties.
In [6] and [7], we introduced the notion of theith sectional H-arith- metic genusxHi (X;L)of(X;L). By definition we can prove that if BsjLj ;, then xHi (X;L)x(OXn i), where Xn i is the transversal intersection of general n i members of jLj. Namely xHi (X;L) is the Euler-Poincare characteristic of the structure sheaf ofXn i. (x(OXn i) is called the arith- metic genus ofXn iin the sense of Hirzebruch. (See [12, 15.5, Section 15, Chapter IV]. We also callx(OXn i) the H-arithmetic genus ofXn i.)
In [8], we also introduced someith sectional invariants of (X;L), that is, the ith sectional Euler number ei(X;L), the ith sectional Betti number bi(X;L) and theith sectional Hodge numberhj;i ji (X;L) of type (j;i j) for every integer j with 0ji, and we investigated some properties of these. In particular we proved that polarized manifolds' version of the Hodge duality and the Hodge decomposition hold (see [8, Theorem 3.1]).
In this paper we consider theith sectional Euler number and theith sectional Betti number of scrolls over a smooth projective variety. In this paper we say that a polarized manifold (P;H) is ascroll over a smooth projective variety X if there exists an ample vector bundleEonXsuch that (P;H)(PX(E);H(E)), whereH(E) is the tautological line bundle on PX(E).
In section 3, we calculate ei(PX(E);H(E)) and bi(PX(E);H(E)) for i2n 3 and i1 (see Theorems 3.1 and 3.2). We also calculate hj;i ji (PX(E);H(E)) for i2n 1 and 0ji(see Theorem 3.3). In par- ticular, by using these results, we can calculate these invariants of (PX(E);H(E)) completely forn1 or 2 (see Corollaries 3.1, 3.3 and 3.4).
In section 4, we will define a new invariant of generalized polarized manifolds. Here a generalized polarized manifoldmeans the pair (X;E) whereXis a smooth projective variety andEis an ample vector bundle on X. L etr:rank(E). Here we state the history of invariants of (X;E). First in [2], Fujita introduced the c1-sectional genus and the O(1)-sectional genus of (X;E). Next, in [1], for the case wherern 1, Ballico defined an invariant of (X;E) which is called thecurve genus cg(X;E)of(X;E) (see Definition 2.3), and several authors (in particular Lanteri, Maeda, Som- mese, and so on) studied this invariant.
As a generalization of the curve genus, for any ample vector bundleE withrn 1, Ishihara ([16, Definition 1.1]) defined an invariantg(X;E), which is called thecr-sectional genus of(X;E), and in [9] we investigated some properties aboutg(X;E). We note that ifn r1, theng(X;E) is the curve genus. This invariant means the following: If a general element of H0(E) has a zero locusZwhich is smooth of expected dimensionn r, then g(X;E)g(Z;detEjZ), that is,g(X;E) is the sectional genus of (Z;detEjZ).
Recently Fusi and Lanteri generalized this invariant. See [11] for detail.
In this paper, we will introduce a new invariantv(X;E) of generalized polarized manifolds (X;E) withrn 1, which is defined by using a result in section 3 (see Definition 4.1). Here we note thatv(X;E) is equal to the curve genus ifrn 1. We will investigatev(X;E) and give some results about this invariant (see Theorems 4.1 and 4.2, and Proposition 4.1).
The author would like to thank the referee for giving useful comments and suggestions.
Notation and Conventions.
We say thatXis avarietyifXis an integral separated scheme of finite type. In particularXis irreducible and reduced ifXis a variety. Varieties are always assumed to be defined over the field of complex numbers. In this article, we shall study mainly a smooth projective variety. The words
``line bundles'' and ``Cartier divisors'' are used interchangeably. The tensor products of line bundles are denoted additively.
O(D): invertible sheaf associated with a Cartier divisorDonX.
OX: the structure sheaf ofX.
x(F): the Euler-Poincare characteristic of a coherent sheafF. hi(F):dimHi(X;F) for a coherent sheafF onX.
hi(D):hi(O(D)) for a Cartier divisorD.
q(X)(h1(OX)): the irregularity ofX.
bi(X):dimHi(X;C).
KX: the canonical divisor ofX.
Pn: the projective space of dimensionn.
Qn: a smooth quadric hypersurface inPn1. (or): linear equivalence.
det (E): ^rE, whereEis a vector bundle of rankronX.
PX(E): the projective space bundle associated with a vector bundleEonX.
H(E): the tautological line bundle onPX(E).
E_: the dual of a vector bundleE.
ci(E): thei-th Chern class of a vector bundleE.
ci(X):ci(TX), where TX is the tangent bundle of a smooth projective varietyX.
For a real numbermand a non-negative integern, let [m]n: m(m1) (mn 1) if n1;
1 if n0:
(
[m]n: m(m 1) (m n1) if n1;
1 if n0:
(
Then fornfixed, [m]nand [m]nare polynomials inmwhose degree aren.
For any non-negative integern,
n! : [n]n if n1;
1 if n0:
Assume thatmandnare integers withn0. Then we put m
n :[m]n n!
We note that m
n 0 if 0m<n, and m 0 1.
2. Preliminaries.
NOTATION2.1. (1) LetYbe a smooth projective variety of dimensioni, letTY be the tangent bundle ofY, and letVY(V1Y) be the dual bundle of TY andVYj : ^jVY. For every integerjwith 0ji, we put
hi;j(c1(Y); ;ci(Y)):x(VjY)
Z
Y
ch(VjY)Td(TY):
(Here ch(VjY) (resp. Td(TY)) denotes the Chern character ofVjY(resp. the Todd class ofTY). See [10, Example 3.2.3 and Example 3.2.4].)
(2) Let (M;L) be a polarized manifold of dimensionm. For every in- tegersiandjwith 0jim, we put
Cji(M;L):Xj
l0
( 1)l m il 1 l
cj l(M)Ll;
wji(M;L):hi;j(C1i(M;L); ;Cii(M;L))Ln i:
(3) Let M be a smooth projective variety of dimensionm. For every integersiandjwith 0jim, we put
H1(i;j):
X
i j 1 s0
( 1)shs(VjM) ifj6i;
0 if ji;
8>
><
>>
:
H2(i;j):
Xj 1
t0
( 1)i tht(Vi jM ) if j60;
0 if j0:
8>
><
>>
:
DEFINITION 2.1. (See [8, Definition 3.1].) Let (M;L) be a polarized manifold of dimensionm, and letiandjbe integers with 0imand 0ji.
(1) Theith sectional Euler number ei(M;L)of(M;L) is defined by the following:
ei(M;L):Xi
l0
( 1)l m il 1 l
ci l(M)Lm il:
(2) Theith sectional Betti number bi(M;L)of(M;L) is defined by the following:
bi(M;L):
e0(M;L) if i0;
( 1)i ei(M;L) Xi 1
j0
2( 1)jbj(M)
!
if 1im:
8>
>>
<
>>
>:
(3) The ith sectional Hodge number hij;i j(M;L) of type (j;i j) of
(M;L) is defined by the following:
hij;i j(M;L):( 1)i jnwji(M;L) H1(i;j) H2(i;j)o :
REMARK2.1. (1) Ifi0, then
e0(M;L)b0(M;L)h0;00 (M;L)Lm: (2) Ifi1, then
e1(M;L)2 2g(L);
b1(M;L)2g(L);
h1;01 (M;L)h0;11 (M;L)g(L):
(3) Ifim, then
em(M;L)e(M);
bm(M;L)bm(M);
hmj;m j(M;L)hj;m j(M);
hm j;jm (M;L)hm j;j(M):
PROPOSITION2.1. Let(M; L)be a polarized manifold of dimensionm.
(1)For every integer i with0im, the following hold:
(1.1) bi(M;L)Pi
k0hk;i ki (M;L).
(1.2) hj;i ji (M;L)hi j;i j(M;L).
(1.3) hi;0i (M;L)h0;ii (M;L)gi(M;L).
(2)Assume that L is base point free. Then for every integers i and j with 1im and0ji the following hold.
(2.1) bi(M;L)bi(M).
(2.2) hj;i ji (M;L)hj;i j(M).
PROOF. See [8, Theorem 3.1 (3.1.1), (3.1.3), (3.1.4) and Proposition 3.3 (2) and (3)]. p
PROPOSITION2.2. For every integersa,k,landrwith0l, Xl
j0
( 1)j rj a j
r k l j
( 1)l k al l
:
PROOF. See [8, Proposition 2.5]. p
NOTATION2.2. LetXbe a smooth projective variety of dimensionn1 and letE be an ample vector bundle of rankronX. We put P:PX(E), H:H(E) andm:dimP. Thenmnr 1. In this paper we assume thatr2.
REMARK2.2. LetX,E,P,H,m,nandrbe as in Notation 2.2.
(1) By [18, (2.1) Proposition], we have
bj(P)bj(X)bj 2(X) bj 2r2(X):
(2) Leti be an integer withim. Thennr 1iand we obtain ri n1.
(2.1) Ifi2n 2 andi 1j, then
j 2r2(i 1) 2(i n1)2
2n 1 i 1:
So by (1) above, ifi2n 2 andi 1j, then by (1) we have
bj(P) Xl
k0
bj 2k(X) if j2l;
Xl
k0
bj 2k(X) if j2l1:
8>
>>
>>
><
>>
>>
>>
:
By the same argument as this, ifi2n 1, then we see that
bi(P) Xl
k0
bi 2k(X) if i2l;
Xl
k0
bi 2k(X) if i2l1:
8>
>>
>>
><
>>
>>
>>
:
(2.2) Assume that im 1. If i2n 3 and i 1j, then j 2r22. If this equality holds, then inr 1m. But this contradicts the assumption. Hencej 2r21, and by (1) above we have
bj(P) Xl
k0
bj 2k(X) if j2l;
Xl
k0
bj 2k(X) if j2l1:
8>
>>
>>
<
>>
>>
>:
By the same argument as this, if im 1, n1 (resp. n2) and i2n 2 (resp.i2n 3), then we see thatb2n 2(P)Pn 1
k0b2n 2 2k(X) (resp.b2n 3(P)Pn 2
k0b2n 3 2k(X)).
DEFINITION2.2. LetX be a smooth projective variety of dimensionn and letEbe a vector bundle of rankronX.
(1) TheChern polynomial ct(E) is defined by the following:
ct(E)1c1(E)tc2(E)t2 :
(2) For every integerjwithj0, thejth Segre class sj(E)ofEis defined by the following equation:ct(E_)st(E)1, wherect(E_) is the Chern poly- nomial ofE_andst(E)P
j0sj(E)tj.
REMARK2.3. (1) LetXbe a smooth projective variety and letF be a vector bundle onX. L et~sj(F) be the Segre class which is defined in [10, Chapter 3]. Thensj(F)~sj(F_).
(2) For every integer i with 1i,si(F) can be written by using the Chern classes cj(F) with 1ji. (For example, s1(F)c1(F), s2(F)c1(F)2 c2(F), and so on.)
DEFINITION2.3. LetX be a smooth projective variety of dimensionn and letE be an ample vector bundle of rankn 1 on X. Then thecurve genus cg(X;E)of(X;E) is defined as follows:
cg(X;E)11
2(KXc1(E))cn 1(E):
3. Sectional invariants of scrolls over a smooth projective manifold.
THEOREM3.1. Let X,E, P, H, m, n and r be as in Notation2.2. Then the following hold.
(3.1.1)If i2n 1, then ei(P;H)(i n1)cn(X).
(3.1.2)If n2, then e2n 2(P;H)(n 1)cn(X)cn(E).
(3.1.3)If n3, then e2n 3(P;H)(n 2)(cn(X) cn(E)) cn 1(E)(KXc1(E)).
(3.1.4)e1(P;H) (n 2)sn(E) (c1 E KX)sn 1(E).
(3.1.5)e0(P;H)sn(E).
PROOF. By [10, Example 3.2.11] and Remark 2.3 (1), for every integerl with 0lm
cl(P)Xl
j0
Xj
k0
r k j k
ck pE_
H(E)j kcl j pTX:
(Herep:P!Xdenotes the projection.) Hence ei(P;H)
Xi
j0
( 1)j m ij 1 j
ci j(P)Hm ij
Xi
j0
( 1)j m ij 1 j
Xi j
s0
Xs
u0
r u s u
cu pE_
H(E)m ij usci j s pTX
( )
X
0kti0k;t
X
i t k j0
( 1)j rjn i 2 j
r k i t k j
!
ck pE_
ct pTXH(E)m k t: By Proposition 2.2 and [3, (3.7) in Chapter 0], we get
ei(P;H) 1
X
0kti0k;t
( 1)i t k k(n i 2)i t k i t k
ck pE_
ct pTXH(E)m k t
X
0kti0k;t
( 1)i t k n t 2 i t k
ck pE_
ct pTXH(E)m k t
X
0kti0k;t
( 1)i t k n t 2 i t k
ck E_
ct TXsn k t(E):
Here we put
E(i;k;t):( 1)i t k n t 2 i t k
ck E_
ct TXsn k t(E):
In order to calculate ei(P;H), we have only to consider the case where E(i;k;t)60. We note that ifkt>n, thenck E_
ct TXsn k t(E)0. So we may assume thatktn.
(a) The case wherei2n 1.
First we note that if (n;i)(1;1), then
ei(P;H)e1(P;H)2 b1(P;H)2 2g1(P;H)2 2q(X)c1(X):
So we assume that (n;i)6(1;1).
Here we note that (n t 2) (i t k)n i 2k nk 1 1 and kti. Hence n t 2
i t k
60 if and only if one of the following holds.
(a.1)i t k0.
(a.2)i t k>0 andn t 2<0.
(a.1) Ifi t k0, then 2n 1itkn. Hencen1 andi1.
But this contradicts the assumption here.
(a.2) If i t k>0 and n t 2<0, then tn 1 or n because ktnandk0. Hence (t;k)(n 1;0), (n 1;1) or (n;0).
If (t;k)(n 1;0), then n t 2
i t k
1
i n1
( 1)i n1: If (t;k)(n 1;1), then
n t 2 i t k
1 i n
( 1)i n: If (t;k)(n;0), then
n t 2 i t k
2 i n
( 1)i n(i n1):
Hence if (n;i)6(1;1), then ei(P;H)
( 1)2(i n1)cn 1(TX)s1(E)( 1)2(i n)c1(E_)cn 1(TX)
( 1)2(i n)(i n1)cn(TX)
(i n1)cn(X):
Therefore in any case
ei(P;H)(i n1)cn(X) ifi2n 1.
(b) The case wherei2n 2. Here we note thatn2 in this case.
Then n t 2
i t k
n t 2
2n t 2 k
. Here we note that (n t 2)2n t 2 k. Hence n t 2
2n t 2 k
60 if and only if one of the following holds.
(b.1)kn.
(b.2)k<nandn t 2<0.
(b.1) If kn, then t0 because tkn and t0, and n t 2
2n t 2 k
1. Here we note thatn t 20 in this case because n2 andt0. Hence
E(2n 2;n;0)( 1)2n 2 ncn(E_)cn(E):
(b.2) Ifk<nandn t 2<0, thentn 1 orn. Hence (t;k)(n 1;0), (n 1;1) or (n;0). By the same argument as (a.2), we obtain
E(2n 2;0;n 1)E(2n 2;1;n 1)E(2n 2;0;n)
( 1)2(2n 2 n1)cn 1(X)s1(E)( 1)2(2n 2 n)c1(E_)cn 1(X)
( 1)2(2n 2 n)(n 1)cn(X)
(n 1)cn(X):
Hence we get
e2n 2(P;H)(n 1)cn(X)cn(E):
(c) Assume that i2n 3. Here we note that n3 in this case. Then n t 2
2n t k 3
60 if and only if one of the following holds.
(c.1)kn.
(c.2)kn 1.
(c.3)kn 2 andt>n 2.
First we consider the case (c.1). Thent0 becausektnandt0.
So we have
E(2n 3;n;0)( 1)2n 3 0 n n 0 2
2n 3 0 n
cn(E_)
n 2 n 3
cn(E)
(n 2)cn(E):
Next we consider the case (c.2). Then t0 or 1, and n t 2
2n t k 3
1. Hence we haveE(2n 3;n 1;0) cn 1(E)s1(E)
cn 1(E)c1(E) andE(2n 3;n 1;1)cn 1(E)c1(X).
Finally we consider the case (c.3). Then (k;t)(0;n 1), (1;n 1) or (0;n). Hence by the same argument as above
E(2n 3;0;n 1)E(2n 3;1;n 1)E(2n 3;0;n)(n 2)cn(X):
Therefore e2n 3(P;H)
(n 2)cn(E) cn 1(E)c1(E)cn 1(E)c1(X)(n 2)cn(X)
(n 2)(cn(X) cn(E))cn 1(E)(c1(X) c1(E))
(n 2)(cn(X) cn(E)) cn 1(E)(KXc1(E)):
(d) The case wherei1. Then by (1) we get e1(P;H)
X
0kt10k;t
( 1)1 t k n t 2
1 t k
ck E_
ct TXsn k t(E)
(n 2)sn(E)(c1 E_
c1 TX)sn 1(E)
(n 2)sn(E) (c1 E KX)sn 1(E):
(e) The case wherei0. Then by (1) we gete0(P;H)sn(E).
We get the assertion of Theorem 3.1. p
By Theorem 3.1, we get the following forn1 and 2.
COROLLARY3.1. Let X,E, P, H and n be as in Notation2.2.
(3.1.1)Assume that n1. Then we get the following:
ei(P;H) i(2 2g(X)) if i1;
degE if i0:
(
(3.1.2)Assume that n2. Then we get the following:
ei(P;H)
(i 1)c2(X) if i3;
c2(X)c2(E) if i2;
(c1 E KX)c1(E) if i1;
s2(E) if i0:
8>
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><
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:
THEOREM3.2. Let X,E, P, H, m, n and r be as in Notation2.2. Then the following hold.
(3.2.1)If mi2n 1, then bi(P;H)bi(P).
(3.2.2)If n2and m>2n 2, then b2n 2(P;H)b2n 2(P)cn(E) 1.
(3.2.3)If n3and m>2n 3, then
b2n 3(P;H)b2n 3(P)(n 2)cn(E)cn 1(E) KXc1(E) 2 2q(X):
(3.2.4)b1(P;H)2(n 2)sn(E) c1(E)KXsn 1(E).
(3.2.5)b0(P;H)sn(E).
PROOF. Sinceei(P;H) has been calculated in Theorem 3.1, we need to studyXi 1
k0
( 1)kbk(P). By using Remark 2.2, we calculate this.
(a) The case where i2n 1. Since bm(P;H)bm(P), we assume that i<m.
(a.1) Assume thati is even. Theni2n. Here we note thatbj(X)0 if j>2n. Then
Xi 1
k0
( 1)kbk(P)
Xn
k0
i
2 k
b2k(X) Xn
k1
i
2 k1
b2k 1(X)
Xn
k0
n k1
b2k(X) Xn
k1
(n k2)b2k 1(X)i 2n 2 2 e(X):
Here we note that by Remark 2.2 (2.1), we havebi(P)Xn
k0
b2k(X) because i2nin this case. Sinceiis even, we have
bi(P;H) bi(P)
( 1)i ei(P;H) 2Xi 1
j0
( 1)jbj(P)
! Xn
k0
b2k(X)
(i n1)cn(X) 2Xn
k0
(n k1)b2k(X)
2Xn
k1
(n k2)b2k 1(X) (i 2n 2)e(X) Xn
k0
b2k(X)
(n3)e(X)Xn
k0
(2k 2n 3)b2k(X)Xn
k1
(2n 2k4)b2k 1(X)
Xn
k0
(2k n)b2k(X) Xn
k1
(2k n 1)b2k 1(X):
Here we prove the following.
CLAIM3.1. (1)Xn
k0
(2k n)b2k(X)0.
(2)Xn
k1
(2k n 1)b2k 1(X)0.
PROOF. (1) Assume thatn2l. Then we note that (2k n)b2k(X)0 if kl. So by Poincare duality
Xn
k0
(2k n)b2k(X)
Xl 1
k0
(2k n)b2k(X) Xn
kl1
(2k n)b2k(X)
Xl 1
k0
(2k n)b2k(X)Xl 1
k0
(n 2k)b2n 2k(X)
Xl 1
k0
(2k n)b2k(X) Xl 1
k0
(2k n)b2k(X)
0:
Ifn2l1, then by Poincare duality Xn
k0
(2k n)b2k(X)
Xl
k0
(2k n)b2k(X) Xn
kl1
(2k n)b2k(X)
Xl
k0
(2k n)b2k(X)Xl
k0
(n 2k)b2n 2k(X)
Xl
k0
(2k n)b2k(X) Xl
k0
(2k n)b2k(X)
0:
Hence we obtain the assertion of (1).
(2) This can be proved by the same argument as (1). p By Claim 3.1,bi(P;H)bi(P) ifi2n 1 andiis even.
(a.2) Assume thatiis odd. Then Xi 1
j0
( 1)jbj(P)
Xn
k0
i1
2 k
b2k(X) Xn
k1
i 1
2 k1
b2k 1(X)
Xn
k0
(n k1)b2k(X) Xn
k1
(n k1)b2k 1(X)
i1
2 n 1
e(X):
By Remark 2.2 (2.1), we havebi(P)Xn
k1
b2k 1(X) becausei2n 1 in this case. Hence by Theorem 3.1 and Claim 3.1, we have
bi(P;H) bi(P)
(i n1)e(X) 2Xi 1
j0
( 1)jbj(P)
! bi(P)
(n2)e(X)Xn
k0
(2n 2k2)b2k(X) Xn
k1
(2n 2k3)b2k 1(X)
Xn
k0
(n 2k)b2k(X) Xn
k1
(n 2k1)b2k 1(X)
0:
Thereforebi(P;H)bi(P) ifi2n 1 andiis odd.
In any case, ifmi2n 1, thenbi(P;H)bi(P).
(b) The case wherei2n 2 and 2n 2<m. Then by Remark 2.2 (2.1) we obtain
bj(P) Xl
k0
bj 2k(X) if j2l, Xl
k0
bj 2k(X) if j2l1 8>
>>
>>
<
>>
>>
>:
for every integerjwithj<2n 2:Hence X
2n 3 j0
( 1)jbj(P)Xn 2
k0
(n k 1)b2k(X) Xn 1
k1
(n k)b2k 1(X):
By assumption, we havenr 1mi12n 1. So by Remark 2.2 (2.2) we obtain
b2n 2(P)Xn 1
k0
b2k(X):
Therefore
b2n 2(P;H) b2n 2(P)
e2n 2(P;H) 22nX3
j0
( 1)jbj(P)
!
b2n 2(P)
(n 1)e(X)cn(E) Xn 2
k0
(2n 2k 2)b2k(X)
Xn 1
k1
(2n 2k)b2k 1(X) Xn 1
k0
b2k(X)
(n 1)e(X)cn(E) Xn 2
k0
(2n 2k 1)b2k(X)
Xn 1
k1
(2n 2k)b2k 1(X) b2n 2(X)
(n 1)e(X)cn(E)e(X) 1 Xn 1
k0
(2n 2k)b2k(X)
Xn 1
k1
(2n 2k1)b2k 1(X)b2n 1(X)
ne(X)cn(E) 1 Xn 1
k0
(2n 2k)b2k(X)
Xn
k1
(2n 2k1)b2k 1(X):
Since
Xn 1
k0
(2n 2k)b2k(X)Xn
k0
(2n 2k)b2k(X);
we obtain
b2n 2(P;H) b2n 2(P)
ne(X)cn(E) 1 Xn
k0
(2n 2k)b2k(X)
Xn
k1
(2n 2k1)b2k 1(X)
cn(E) 1Xn
k0
(2k n)b2k(X) Xn
k1
(2k n 1)b2k 1(X):
By Claim 3.1, we get
b2n 2(P;H)b2n 2(P)cn(E) 1:
(c) The case where i2n 3 and i<m. Then by Remark 2.2 (2.2) we obtain
bj(P) Xl
k0
bj 2k(X) if j2l, Xl
k0
bj 2k(X) if j2l1 8>
>>
>>
<
>>
>>
>:
for every integerjwithj<2n 3. Hence X
2n 4 j0
bj(P)Xn 2
k0
(n k 1)b2k(X) Xn 2
k1
(n k 1)b2k 1(X):
By assumption we havenr 1mi12n 2. Hence by Remark 2.2 (2.2) we obtain
b2n 3(P)Xn 1
k1
b2k 1(X):
Therefore
b2n 3(P;H) b2n 3(P)
e2n 3(P;H)22nX4
j0
( 1)jbj(P) b2n 3(P)
(n 2) e(X) cn(E) cn 1(E) KXc1(E)
Xn 2
k0
(2n 2k 2)b2k(X) Xn 2
k1
(2n 2k 2)b2k 1(X) Xn 1
k1
b2k 1(X)
(n 2) e(X) cn(E) cn 1(E) KXc1(E)
Xn 2
k0
(2n 2k 2)b2k(X) Xn 2
k1
(2n 2k 1)b2k 1(X) b2n 3(X)
(n 2) e(X) cn(E) cn 1(E) KXc1(E)
2e(X)2b2n 2(X)2b2n(X) 3b2n 3(X) 2b2n 1(X)
Xn 2
k0
(2n 2k)b2k(X) Xn 2
k1
(2n 2k1)b2k 1(X)
(n 2) e(X) cn(E) cn 1(E) KXc1(E)
Xn
k0
(2n 2k)b2k(X) Xn
k1
(2n 2k1)b2k 1(X) 2e(X)2b2n(X) b2n 1(X)
(n 2)e(X)(n 2)cn(E)cn 1(E) KXc1(E)
Xn
k0
(n 2k)b2k(X) Xn
k1
(n 2k1)b2k 1(X)
(n 2)e(X)2b2n(X) b2n 1(X):
Sinceb2n(X)1 andb2n 1(X)b1(X)2q(X), by Claim 3.1 we obtain b2n 3(P;H) b2n 3(P)
(n 2)cn(E)cn 1(E) KXc1(E) 2 2q(X):
(d) The case wherei1.
Then by Theorem 3.1 (3.1.4) and the definition ofb1(P;H) b1(P;H) e1(P;H)2b0(P)
2(n 2)sn(E) c1(E)KXsn 1(E):
(e) The case wherei0.
Then by Theorem 3.1 (3.1.5) and the definition ofb0(P;H), b0(P;H)e0(P;H)sn(E):
Therefore we get the assertion. p
COROLLARY3.2. Let X,E, P, H, m and n be as in Notation2.2. If n2 and m>i2n 2, then bi(P;H)bi(P).
PROOF. Ifi2n 1, then by Theorem 3.2 (3.2.1) we get the assertion.
Next we consider the case where i2n 2. SinceE is ample, we have cn(E)1. Therefore by Theorem 3.2 (3.2.2), we seeb2n 2(P;H)b2n 2(P).
This completes the proof. p
COROLLARY3.3. Let X,E, P, H, m and n be as in Notation2.2.
(3.3.1)Assume that n1. Then we get the following:
bi(P;H) bi(P) if mi1;
degE if i0:
(
(3.3.2)Assume that n2. Then we get the following:
bi(P;H)
bi(P) if mi3;
b2(P)c2(E) 1 if i2;
c1(E)(c1 E KX)2 if i1;
s2(E) if i0:
8>
>>
<
>>
>:
PROOF. By Theorem 3.2, we get the assertion. (Here we note that if n2, then 2n 22<nr 1mbecause we assumer2 in this
paper.) p
REMARK 3.1. Since E is ample, we see that if n1 or 2, then bi(P;H)0 for anyi.
Here we calculatehj;i ji (P;H) for the case wheremi2n 1.
THEOREM 3.3. Let X, E, P, H, m and n be as in Notation 2.2. If mi2n 1and0ji, then hj;i ji (P;H)hj;i j(P).
PROOF. First we note that we can take an ample line bundleAonX such thatE Atis ample and spanned for every positive integert. Hence H(E At) is ample and spanned. We also note that there exists an iso- morphismf:PX(E At)!PwithH(E At)f(Hp(At)), where p:P!X is the projection. Therefore Hp(At) is also ample and spanned. By thisf, we identifyPX(E At) andP. By Theorem 3.2 we have
bi(P;H(E At))bi(P):
Hence
bi(P;Hp(At))
bi(P;H(E At))
bi(P):
On the other hand by Proposition 2.1 (1.1) and (2) we have hj;i ji (P;Hp(At))hj;i j(P)
because Hp(At) is ample and spanned. But since F(t):
:hj;i ji (P;Hp(At)) hi;i j(P) is a polynomial in t and F(t)0 for every positive integert, we see thatF(0)0, that is,
hj;i ji (P;H)hi;i j(P):
This completes the proof. p
COROLLARY3.4. Let X,E, P, H, m and n be as in Notation2.2.
(3.4.1)Assume that n1. Then we get the following:
hij;i j(P;H) hj;i j(P) if mi1and0ji;
degE if i0and j0:
(
(3.4.2)Assume that n2. Then we get the following:
hij;i j(P;H)
hj;i j(P) if mi3 and0ji;
h0;2(P) if i2 and j0;
h2;0(P) if i2 and j2;
h1;1(P)c2(E) 1 if i2 and j1;
12(c1(E)(c1 E KX))1 if i1 and j0;1;
s2(E) if i0 and j0:
8>
>>
>>
>>
>>
<
>>
>>
>>
>>
>:
PROOF. If n1, then by Corollary 3.3 and Theorem 3.3 we get the assertion.
Assume thatn2. Then by Corollary 3.3 and Theorem 3.3 we get the assertion for the case wheremi3 and 0ji. If (i;j)(2;0) (resp.
(2;2)), then by Proposition 2.1 (1.3) and [5, Example 2.10 (8)] we have h0;22 (P;H)g2(P;H)h2(OP)h0;2(P) (resp. h2;02 (P;H)g2(P;H)
h2(OP)h0;2(P)h2;0(P)). Moreover by Corollary 3.3 (3.3.2) and Pro- position 2.1 (1.1) we geth1;12 (P;H)h1;1(P)c2(E) 1.
Assume that i1. Then by Proposition 2.1 (1.1) we have b1(P;H)h1;01 (P;H)h0;11 (P;H). Moreover by Proposition 2.1 (1.2) we haveh1;01 (P;H)h0;11 (P;H). Therefore by Corollary 3.3 we get the asser- tion for the casei1.
By Corollary 3.3 we get the assertion for the case wherei0. p REMARK 3.2. Since E is ample, we see that if n1 or 2, then hj;i ji (P;H)0 for anyiandjwith 0jim.
4. Anew invariant of generalized polarized manifolds.
Here we use Notation 2.2. Assume that i2n 3, n3 andi<m.
Then by Theorem 3.2 (3.2.3) we see that (n 2)cn(E)cn 1(E)(KXc1(E)) is even because b2n 3(P;H) and b2n 3(P) are even (see [8, Theorem 3.1 (3.1.2)]). We put
v(X;E):11
2 (n 2)cn(E)cn 1(E)(KXc1(E)):
Here we note thatrn 1 sincenr 1mi12n 2.
Ifrn 1, then
v(X;E)11
2 KXc1(E)cn 1(E):
Sov(X;E) is thought to be a generalization of the curve genuscg(X;E) of (X;E) (see Definition 2.3). Here we definev(X;E) again.
DEFINITION4.1. Let (X;E) be a generalized polarized manifold of di- mensionn3. Assume thatrn 1. Then the invariantv(X;E) of (X;E) is defined as follows.
v(X;E):11
2 (n 2)cn(E)cn 1(E)(KXc1(E)):
Sinceb2n 3(P;H) b2n 3(P)2v(X;E) 2q(X) by Theorem 3.2 (3.2.3), we can propose the following conjecture.
CONJECTURE 4.1. Let (X;E) be a generalized polarized manifold of dimension n3. Assume that rn 1. Then v(X;E)q(X).
Here we study the non-negativity ofv(X;E).
THEOREM 4.1. Let (X;E)be a generalized polarized manifold of di- mension n3. Assume that rn 1. Then v(X;E)0.
PROOF. If rn 1, then v(X;E) is the curve genus and then v(X;E)0 by [19, Theorem 1]. So we may assume thatrn.
IfrnandKXc1(E) is nef, then (KXc1(E))cn 1(E)0 becauseEis ample. Furthermorecn(E)1. So we obtainv(X;E)2.
IfrnandKXc1(E) is not nef, then (X;E)(Pn;OPn(1)n) by [22, Theorem 1 and Theorem 2]. In this case (KXc1(E))cn 1(E) n and cn(E)1. Hencev(X;E)0. Therefore we get the assertion. p
THEOREM 4.2. Let (X;E)be a generalized polarized manifold of di- mension n3. Assume that rn 1.
(1)If v(X;E)0, then(X;E)is one of the following.
(1.1) (Pn;OPn(1)n).
(1.2) (Pn;OPn(1)n 1).
(1.3) (Pn;OPn(1)n 2 OPn(2)).
(1.4) (Qn;OQn(1)n 1).
(1.5) XPP1(F)for some vector bundleF of rank n onP1, and E nj11(H(F)pOP1(bj)), where p:X!P1 is the bundle projection.
(2) If v(X;E)1, then(X;E)is one of the following.
(2.1) XPC(F)for some vector bundleF of rank n on a smooth elliptic curve C, andEF OPn1(1)n 1for any fiber F of the bundle projection X!C.
(2.2) KXdet(E) OX andrn 1.
PROOF. If v(X;E)1, then by the proof of Theorem 4.1, one of the following cases occurs:
(4.2.1)rn 1.
(4.2.2)rnandKXc1(E) is not nef.
If rn 1 and v(X;E)0 (resp. v(X;E)1), then 0v(X;E)
cg(X;E) (resp. 1v(X;E)cg(X;E)) and by [19, Theorem 1 and Theo- rem 2] we get the above types.
IfrnandKXc1(E) is not nef, then (X;E)(Pn;OPn(1)n) by [22, Theorem 1 and Theorem 2] and then v(X;E)0. Hence we get the as- sertion. p
REMARK 4.1. (1) If (X;E) is the type (2.2) in Theorem 4.2, then a classification of (X;E) has been obtained by [20].
(2) Ifv(X;E)1, then we see thatv(X;E)q(X).
PROPOSITION 4.1. Let (X;E) be a generalized polarized manifold of dimension n3. Assume that rn 1 and E is spanned. Then
v(X;E)q(X).
PROOF. Set (P;H):(PX(E);H(E)). ThenHis ample and spanned. So by Proposition 2.1 (2.1) we obtain
b2n 3(P;H)b2n 3(P):
On the other hand
b2n 3(P;H) b2n 3(P)
2v(X;E) 2q(X):
Hence we get the assertion. p
REMARK4.2. (1) By Theorem 4.2 we can determine the type of (X;E) such thatv(X;E)q(X) andq(X)1.
(2) For every integer q2, there exists an example of (X;E) with v(X;E)qq(X) (This was given by the referee.): Let C be a smooth projective curve withg(C)q2. We note that there exist vector bundlesF andGonC with rankF nand rankG n 1 such thatE:H(F)p(G) is an ample vector bundle of rankn 1 onX, whereX:PC(F),H(F) is the tautological line bundle ofF and p:X!C is the bundle projection. Then we can easily check thatv(X;E)cg(X;E)q2.
(3) We see that if E is ample and spanned, rankE n 1 and v(X;E)q(X)2, then (X;E) is isomorphic to the type in (2) above.
This has been proved by [17, Theorem].
Here we give the following conjecture which was pointed out by the referee:
CONJECTURE4.2. Let X be a smooth projective variety of dimension n3and letEbe an ample vector bundle on X withrankE r. Assume that rn 1 and q(X)2. If E is spanned and v(X;E)q(X), then rn 1.
REMARK4.3. We consider the case wherern3. (The following (a) and (b) were also pointed out by the referee.)
(a) IfKXc1(E) is not ample, then by [4, Main Theorem] (X;E) is one of the following 6 types and we can calculatev(X;E) andq(X):
(a.1) (Pn;OPn(1)n1). In this case (v(X;E);q(X))(1(1=2) (n 2)(n1);0).
(a.2) (Pn;OPn(1)n). In this case (v(X;E);q(X))(0;0).
(a.3) XPC(F) for a vector bundleF of ranknon a smooth curve C, and E H(F)p(G) for a vector bundle G on C with rankG n, wherep:X!Cis the bundle projection. In this case (v(X;E);q(X))(g(C)(n 1)(g(C) 1c1(F)c1(G));
g(C)).
(a.4) (Pn;OPn(1)n 1 OPn(2)). In this case (v(X;E);q(X))
(n 1;0).
(a.5) (Pn;TPn). In this case (v(X;E);q(X))(1(1=2)(n 2) (n1);0).
(a.6) (Qn;OQn(1)n). In this case (v(X;E);q(X))(n 1;0).
Here we consider the case where (X;E) is the type (a.3) above. Then 1cn(E)c1(F)c1(G) becauseEis ample. Henceg(C) 1c1(F)
c1(G)0 and we get v(X;E)g(C)q(X). Moreover if g(C)1, then we see thatv(X;E)q(X)(n 1)>q(X).
We also note that ifEis spanned by global sections, thencn(E)2 by [21, (3.4) Theorem] and we obtainv(X;E)q(X)(n 1)>q(X).
(b) Next we assume that KXc1(E) is ample. Then we can prove the following:
PROPOSITION 4.2. (1) If KXc1(E) is ample, then v(X;E)1
1
2(n 1).
(2)If KXc1(E)is ample andEis spanned, then v(X;E)n.
PROOF. (1) By assumption, we have cn(E)1 and (KXc1(E))cn 1(E)1. Hence we get the assertion of (1).
(2) Since Eis spanned and KXc1(E) is ample, we see thatcn(E)2 by [21, (3.4) Theorem]. Hence (n 2)cn(E)(KXc1(E))cn 1(E) 2(n 2)2 because the term on the left is even. Therefore we get the assertion of (2). p
The above results suggest that for the case where rn and v(X;E)6q(X) there are some gaps for the value of v(X;E) q(X), de- pending onn. We will investigate this in a future paper.
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Manoscritto pervenuto in redazione il 10 luglio 2007.