C OMPOSITIO M ATHEMATICA
Y OSHIYUKI K URAMOTO
On the logarithmic plurigenera of algebraic surfaces
Compositio Mathematica, tome 43, n
o3 (1981), p. 343-364
<http://www.numdam.org/item?id=CM_1981__43_3_343_0>
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343
ON THE LOGARITHMIC PLURIGENERA OF ALGEBRAIC SURFACES
Yoshiyuki
Kuramoto@ 1981 Sijthoff & Noordhofl International Publishers - Alphen aan den Rijn Printed in the Netherlands
Introduction
Let S be a
nonsingular algebraic
surfaces over thecomplex
numberfield C. We
study
thelogarithmic plurigenera Prn(S)
under theassumption
that there is asurjective morphism f:S~0394
to a non-singular
curve A whosegeneral
fiber Cw is an irreducible curve such that thelogarithmic
Kodaira dimension03BA(0394)
isnon-negative.
Such asituation can arise from the
quasi-Albanese mapping
as of S. If03BA(CW) = -~, i.e.
Cw ~ Pl orAl,
thenPrn(S)
= 0 for all m > 0. Thus we assume that03BA(Cw) ~
0. Thenby
the addition formula forlogarithmic
Kodaira dimension
([6]),
we have some m > 0 such thatPm(S) ~
1. Inthis paper we shall look for
integers m
such thatPm(S) ~
1.We use the
following
notations:S:
anonsingular complete algebraic
surface which contains S as a Zariski open set,D =
S - S :
thecomplement
of S inS.
We assume that D hasonly
normal
crossings
and every irreducible component of D isnonsingular.
,A: a
nonsingular complete algebraic
curve which contains àas a Zariski open set,
f:S~0394:
a rationalmapping
definedby f.
We can assume thatf
isa
morphism by taking
a suitableS.
Cw:
ageneral
fiber of1,
Ks: the canonical divisor of
S, pg(S):
thegeometric
genus ofS,
0010-437X/81/060343-22$00.20/0
q(S):
theirregularity
ofS, Pm(S):
the m-genus ofS,
pg(S):
thelogarithmic geometric
genus ofS, q(S):
thelogarithmic irregularity
ofS,
Pm(S):
thelogarithmic
m-genus ofS,
03BA(V):
the Kodaira dimension of anonsingular complete
al-gebraic variety V,
03BA(V):
thelogarithmic
Kodaira dimension of anonsingular
al-gebraic variety V,
g(C):
the genus of anonsingular complete algebraic
curveC, 7T(C) = (1/2)(Ks
+C, C)
+1,
if C is a divisor onS,
~: the linear
equivalence
ofdivisors,
Di a D2 : means
Dl -
D2 is effective or 0 if Di and D2 are divisors.The
following
are our results.THEOREM 1: Under the above notations
if 03BA(0394) ~
0 andR(C,,) 0,
thenP4(S) ~
1 orP6(S) ~
1.COROLLARY: Let
S=A2-V(~)
where cp EC[x, y]
and cp is irre- ducible.If P4(S)
=P6
=0,
then S ~A’
XGm.
THEOREM 2: Let S be a
nonsingular algebraic surface
over C.If 03BA(S) ~
0 andq(S) ~ 1,
thenP4(S) ~
1 orP6(S) ~
1.If
S
is neither ruled norrational,
thenP4(S) ~
1 orP6(S) ~
1 fromthe
theory
ofcomplete algebraic
surfaces. Thus to prove Theorem1,
it suffices to treat the case thatS
is ruled or rational.In § 1,
wegive preliminary
results. In§2-§7,
we prove Theorem 1. Thefollowing
theorem due to Tsunoda
plays
an essential role in §7.THEOREM 3:
(Tsunoda)
LetS, S and
D be as above.If i«S)
= 2 and the intersection matrixof
D is notnegative definite,
thenP4(S) a
1 orP6(S)~1.
In §8 we prove Theorem 3. In §9 we prove
Corollary
and Theorem 2.The author would like to express his
hearty
thanks to Professor S.Iitaka for his kind advices and encouragement, and to Mr. S. Tsunoda who
proved
Theorem 3.§1. Preliminaries
LEMMA 1: Let
S, S and
D be as in the introduction.Suppose
thatthere is an
exceptional
curveof
thefirst
kind E onS
such that(E, KS+D)~0.
Let03BC:S~Sb
be the contractionof
E. Then03BC(D)
is adivisor with
only
normalcrossings
andPm(S)
=Pm (Sb -IL(D».
PROOF:
Easy
and omitted.Especially
ifD0
E and(D, E)
=1,
thenSb-03BC(D)
is called a halfpoint
detachment fromS,
and if D D E and(D - E, E)
=2, then li
is called a canonicalblowing
down.(c.f. [5])
LEMMA 2: Let
S, S and
D be as in the introduction. Let D = 03A3Dj
bethe irreducible
decomposition of
D. Leth(r(D»
be thecyclotomic
number
of
the dualgraph r(D) of
D, i.e.h(0393(D)) = (number of
connected componentsof 0393(D))
-
(number of
verticesof r(D»
+
(number of 1-simplices of 0393(D)).
Then we have
Where t is the dimension
of
the kernelof
the canonical homomor-phism H1(S, ~S) ~ H’(D, CD). (c.f. [9;
Theorem2.2]).
LEMMA 3: Let 7T: V~ W be an r-sheeted Galois
covering,
where Vand W are
complete
normalalgebraic
varieties over C. Let D be aCartier divisor on W. Then we have
a)
dimH°(V, ~V(03C0*D))~
1 ~ dimH°(W, ~W(rD)) ~
1,b)
dimH°(V, ~V(03C0*D))~
2 ~ dimH°(W, ~w(rD))~
2.PROOF:
Using
the same argument as[2; §6]
or[4; §10.11],
we canprove the above lemma. Details are omitted.
In that follows we
tacitly
use the notation in the introduction.§2. The case
03BA(Cw) = -~
and03BA(0394) = -~
PROPOSITION 1:
If 03BA(Cw) ~ 0, 03BA(0394) ~ 0, 03BA(Cw) = -~
and03BA(0394) = -~,
then
P2(S) ?
1.PROOF:
By
theassumption, S
is rationaland 0394~P1.
Hence thereexists a
composition
of a finite sequence ofquadratic
transformations03BC:S~
where:~0394
is aP’-bundle over â
andf=°03BC.
Since03BA(0394)~ 0, à
C C* and so D contains two reduced fibers off
which wedenote
by
FI and F2. If D contains two horizontal components with respect tof,
thenby
Lemma 2pg(S)~
1. Thus we assume that Dcontains
only
one horizontal component which we denoteby
H. If(H, F1)~2,
thenby
Lemma 2pg(S)~
1. Hence we assume(H, FI)
= 1.Let
CW
be ageneral
fiber off
andFI
be the fiber such thatFi =
supp(Fl).
SinceK(CW) ? 0,
it follows that(Cw, H) = (FI, H)~
2.Hence F1 ~ Fi and Fi contains an
exceptional
curveEi.
IfEt
fl H =Ø,
we can contract El. Thus we may assume(El, H)
= 1. If El is anedge
component of Fi(i.e. (E1,
F1 -El)
=1),
then(E1,
D -E1) =
2 andso
by
Lemma 1 we can contract Ei. Thus we may assume that FI contains two componentsCi
and C2 such that(CI, El) = (C2, E,)
= 1.Let
i£’: S~S’
be the contraction of E,. Put03BC’*(C1)
=CI, 03BC*’(C2) = C2
and
03BC’*(H)
= H’. Since 2Fi àCi
+ C2 +2E,,
we haveOn the other hand
by
the Riemann-Roch theorem we haveThus we have
Similarly,
dimH’(9, eJs(Ks
+ H +2F2)) ~
1.Hence, dim
H°(S, ~S(2KS
+ 2H +2F,
+2F2))~
1.Therefore, P2(S)
= dimH°(S, ~S(2(KS
+D)))~
1.Q.E.D.
§3. The case
03BAa(Cw) = -~
and03BA(0394)
= 0PROPOSITION 2:
If 03BA(Cw) ~ 0, 03BA(0394) ~ 0, K(Cw) = -00
and03BA(0394)
=0,
thenP2(S) -
1.PROOF:
By
theassumption
we have thefollowing
commutativediagram:
Where 0394
is anonsingular elliptic
curve,:~0394
is aP1-bundle,
IL :
S ~ S
is acomposition
of a finite number ofquadratic
trans-formations and
S -
S = D is a divisor with normalcrossings.
We may assume that all irreducible components of D are horizontal with respect tof.
Put D == 1Hj,
where theHj
are irreducible components.Since the
Hj
are horizontalg(Hj)~g(0394)=1.
Henceif 03A3g(Hj)~2,
then
by
Lemma 2 we havepg(S) ?
1. Thus we assume that D is anirreducible horizontal curve such that
g(D) = l.
We put 03BC = 03BC1° 03BC2°···° IL" where iii :Si
-Si-,
is aquadratic
transformation with center pi,So = S
andSr
=S.
Let E; =03BC-1i(pi).
For the sake of sim-plicity
we use the same letter Ei for(03BCi+1°···° 03BCr)*Ei
also. Put03BC(D)
=D.
Let vi be themultiplicity
of the proper transform ofD
toSi-,
at p;. Then we haveLet F be a fiber of
f : ~0394.
We put(F, D)
= d. Since every excep- tional curve onS
is contained in some fiber off,
it follows that(F, b) = (Cw, D). By
theassumption
that03BA(Cw)~ 0,
we have(ëw,
D)~
2. Hence d à 2.Since S
is a ruled surface of genus1,
there is anon-negative integer
b suih thatwhere ~ means numerical
equivalence.
Sinceg(D)
=1,
we haveThus we have
First we treat the case that Vi = 1 for all i. In this case b = 0 and
D
+(d/2)Ks =
0. We may assumeS
=S.
Then we haveIf
d > 2,
thenH0(, Ose - (m - 1) - (m - 1)K))
= 0 for m ~2,
becauseF2
= 0. Thus from the exact sequencewe get the exact sequence
Hence,
dimH1(, Os(m(D
+K)))
= dimH1(, Ús(-mD -Cm -1)Ks»
~ 1.
SinceH2(S, ~(m (
+K)))
~H0(, Os(-mD - (m - 1)K))
=0, apply- ing
the Riemann-Rochtheorem,
we obtainEspecially,
we knowP2(S) ?
1 in this case.Now we assume that d = 2. Put D =
JL Corresponding to
themorphism f|D:D~0394,
we get thehomomorphism 03C8:~0394
of 1-dimensional Abelian varieties. We denote the kernel
of tp by
G. LetS
=S
xJJ
be the fiberproduct.
Then we have thefollowing
com-mutative
diagram:
Where
03A8:~S and :~
areprojections. Since 03C8:~0394
is a2-sheeted unramified Galois
covering,
so is ’fi:É - S.
And we know that:~
is also a ruled surface of genus 1. LetD
be a cross-section of
f
definedby D
={(a, a)la
eD}.
Then we haveand
a(D)
~O"’(D) = ep
if 03C3~ 0"’.Hence 03A8-1(D)
consists of 2 connected components and each component is anonsingular elliptic
curve. Thusapplying
Lemma2,
we obtainSince 03A8*D +
Ks
=111 *(D
+Ks),
we canapply
Lemma3a)
and obtainThus we
complete
the casethat Vi
= 1 for all i.Now we assume that 03A3ri=1
(vi - 1)
> 0. It is obvious that vi ~ d for all i.Suppose
that thereis Vi
such that vi = d. ThenD
has ad-ple point
po. Let
Fo
be the fiber of1
which contains po. Since(D, Êo)
=d,
weknow
D
nFo
={po},
andD
andFo
have no common tangent line at po.The
quadratic
transformation with center po appears in IL = 03BC1°···°03BCr. We denote itby
03BC0. Let F’ 0 and D’ be proper transforms ofPo
andD by
1L0,respectively.
Then we haveFl
nD,
and(Fü)2
= -1. And the proper transform ofF’
toS
is also anexceptional
curve which does not intersect with D.
By
Lemma1,
we may assume that there is not such a curve onS.
Thus we assume that vi d for all i.If d =
2,
then wehave Vi
= 1 for all i. But this contradicts theassumption.
Therefore we assume that d ~ 3.Suppose
that thereis Vi
suchthat Vi
= d - 1. Then has a(d - 1)-
ple point
po. LetFo
be the fiberof f
which contains po. Let 03BC0 be thequadratic
transformation with center po, Fi and D’ be the proper transforms ofFo
andD by
bto,respectively.
IfD
flFo - {p0} = ep,
thenD
andFo
have no common tangent line at po and intersectsimply
atsome other
point.
IfD
nPo
=ipoi,
then D’ andF’ 0
intersectsimply
atone
point
of03BC-10(p0).
In each case, we have(F’, D’)
= 1 and(F’0)2 =
-1.
Thus,
there is anexceptional
curve E onS
such that(E, D)
= 1.By
Lemma1,
we may assume that there is no such curves onS.
Hence we reduce the
problem
to the case that d ~ 3and vi
~ d - 2 for ail i.Since D + KS ~
03BC*(
+K) - 03A3 (Vi - I)Ei,
we havei=1
On the other
hand, by
the Riemann-Roch theorem we haveCombining (2)
and(3),
we calculateEspecially
we get dimHO(S, Ùg(2(D
+KS))) ~ (d - 1)-1
> 0. Thereforewe have
P2(S) ~
1.Q.E.D.
§4. The case
03BA(Cw) = -~
and03BA(0394)
= 1PROPOSITION 3:
I f 03BA(Cw)~0, 03BA(0394)~0, 03BA(Cw)-~
an d03BA(0394)=1,
then
pg(S) ~
1 andP2~
2.PROOF: Let
D = 03A3ri=1 Ci
be the irreducibledecomposition.
Fromthe
assumption
we know(Cw, D) a
2. We may assume that allCi
are horizontal with respect tof.
Sinceg(Ci) ~ g(A) = q(S) ~ 2, by
virtueof Lemma 2 we have
pg(S) ~ Li=l g(Ci) - q(S) ~ (r - I)q(S).
Thus ifr ~
2,
we havepg(S) ~
2. Hence we assume that r = 1. Putflct = il.
Applying
Hurm,itz’ formulato 03C8:
Ci -0394, we
getwhere
e(p)
is the ramification indexof gi
at p. Sincedeg 03C8=(C1,
Cw) = (D, Cw)~ 2,
we seeHence
pg(S)~ g(Cl) - q(S)~ 1,
and ifpg(S)
=1,
then we haveq(S) =
2, deg t/1
=2, lpEC, (e(p) - 1)
= 0 andg(CI)
= 3. Thus ifpg(S)
=1,
then03C8:C1~0394
is a 2-sheeted unramifiedcovering. Let S
=S
0394C1 be the fiberproduct. Let. 03A8:~S and :~C1
denote the first and thesecond
projections respectively.
Then 11, is a 2-sheeted unramifiedcovering
andq(S)
=g(C1)
= 3. Put03A8-1(C1) = 03A3si=1 Ci
where theCi
are irreducible components.Applying
Hurwitz’ formula to03A8|C’i:C’i~C1,
we get
Thus
by
Lemma 2 we haveSince
P*(D
+KS) ~ 03A8-1(C1)
+03A8*KS
=03A8-1C1
+ Ks,applying
Lemma3b),
we infer thatP2(S)~
2.Q.E.D.
REMARK:
pg(S)~
1 wasproved by Miyanishi-Sugie
in[12; §2].
§5. The case
03BA(Cw)~0, 03BA(0394)=-~
andS
is an irrational ruled surfacePROPOSITION 4:
I f 03BA(Cw)~0, 03BA(0394)~0, K (Cw) à 0, 03BA(0394)=-~,
S iSruled and
q(S)~ 1,
thenpg(S) ?
1.PROOF: Let a :
S~B:
=a (S)
4Alb(S)
be the Albanesemapping
ofS,
then B is anonsingular
curve of genusq(S),
and theruling
ofS
isgiven by
a.By
theassumption,
we know that D =S -
S contains two reduced fibers off
which we denoteby
Fi and F2. For ageneral point
p of
S,
the fiber off
which contains p is anonsingular
curve of genus greater than 0 which we denoteby Ci,
and the fiber of a which contains p is anonsingular
rational curve which we denoteby C2.
Then
C1 ~ C2
and Ci nC2 ~ 0,
hence(CI, C2) ~
1. Therefore for anyu E B and w
~ 0394,
we have(f*(w), a*(u)) à
1.Especially,
Fi and F2 contain horizontal components with respect to a which we denoteby
Hl and H2respectively.
Then from Lemma 2 we infer thatpg(S) ~
g(H1)
+g(H2) - q (9) q(S) ~
1.Q.E.D.
§6. The case
03BA(Cw) = 0, 03BA(0394) = -~,
andS
is rationalPROPOSITION 5 :
I f 03BA(Cw) ~ 0, 03BA(0394) ~
0,03BA(Cw)
=0, 03BA(0394)=-~
andS
is
rational,
thenP3(S) -
1 orP4(S) ~
1.PROOF:
By assumption,
D contains two reduced fibers off
whichwe denote
by
Fi and F2. We may assume D = Fi + F2. We contract allexceptional
curves contained in fibers off
and then we have thefollowing
commutativediagram:
where
:~0394
is a rationalelliptic
surface which has noexceptional
curves contained in fibers
and S~ S
iscomposed
of a finite number ofquadratic
transformations. LetFI
andF2
be fibers off
such that Fi = supp
03BC*1
andF2
= supp03BC*2.
Thenby
the canoni- cal bundle formula[10],
we knowwhere the mvPv’s are all
multiple
fibers ofi
andÊ
is ageneral
fiber ofÎ. Since S
has anexceptional
curve E which is horizontal with respect tof ,
we haveHence
:~0394
has at most onemultiple
fiber which we denoteby moPo. Putting
mo = 1if f
is free frommultiple fibers,
we haveNow we use the classification of
singular
fibers ofelliptic
surfacesby
Kodaira
[11].
Sincepg()
=q(S)
=0,
we havewhere
v(T)
is the number of thesingular
fibers ofS
of type T.LEMMA 4: If
Fi
is asingular
fiber of type Ib orI*b
or II or III orIV,
thenIKs
+2Fi| ~ 0.
PROOF: If
Fi
is of type Ib,by
Lemma 2 wehave 1 Ks
+Fi| ~ 0.
IfÊi
is of type
It,
then 2 suppi ~ Êi
and we may assume03BC*i
containsno
exceptional
curves. Hence we haveIf Fi is of type II i.e. a rational curve with one cusp, then Fi contains
an
exceptional
curve E andnonsingular
rational curvesCI, C2
andC3
such that(CI, E) = (C2, E) = (C3, E) =
1. Let03BC’:S~S’
be the con-traction of E and denote
03BC’*(Cj) by C; for j
=1, 2,
3. Then we infer thatAnd
by
the Riemann-Roch theorem we haveThus we have dim
H’(9, Cg(Ks
+2Fi)) ~
1. IfFi
is of type III i.e. twononsingular
rational curvesintersecting
at onepoint
withmultiplicity
two, or type IV i.e. three
nonsingular
rational curvesintersecting
onepoint,
then Fi contains also anexceptional
curve E andnonsingular
rational curves
CI, C2
andC3
such that(CI, E) = (C2, E) = (C3, E)
= 1. Hence
similarly
we haveBy
Lemma 4 we infer that if bothFI
andF2
are of type Ib orIt
orII or IV, then
12(Ks
+ FI +F2)1 D |KS
+2Fi + IKs
+2F2| ~ ep,
and henceP2(S) ~
1.Now we assume that one of
Êi
is of type II* or III* or IV*.By
theassumption,
the functional invariantof S (if S
has amultiple fiber,
the functional invariant of thecorresponding elliptic
surface free frommultiple fibers)
is not constant. Therefore we know thatEb (v(Ib)
+v(I*b)) ~
1. Hence ifv(II*) ~
1, then we infer from(4)
thatXb bv(Ib) =
2 and
remaining v(T)
are 0. Thus one ofÊi
is of type Ih and thereforeby
Lemma 2 we havepg(S) ~
1. Thus we assume that one ofFi
is oftype III* or IV*.
Case 1. First we consider the case that one of
Êi
is of type III*.We assume that
FI
is of type III*. We may assume thatF2
is not oftype
Ib by
Lemma 2. Then from(4)
we know that F2 is of type II. We putwhere the
Bi
arenonsingular
rational curves with self-intersection number - 2 and C is a rational curve with one cusp.Since 03BC : S~
gives
rise to anisomorphism
on aneighborhood
of FI, we havewhere we denote
03BC*0398i by
the same letterOi.
On the other hand wehave
where il
=ILl 0 03BC2 °
03BC3 and C"’ is the proper transform of Cby gc
and Ei are theexceptional
curves of thequadratic
transformations gi and E"1 and E2 are the proper transforms of El and E2 toS, respectively.
We denote the total transform of Ei
by
the same letter E;. Then we haveSince m0P0 ~
C,
we infer thatOn the other
hand,
Hence we have
Therefore we get
P4(S) -
1.Case 2. Now we consider the case that one of Fi is of type IV*.
We assume FI is of type IV*. We may assume F2 is not of type
Ib.
Then from
(4)
we know that F2 is of type II or III. We putwhere the
Oi
arenonsingular
rational curves with self-intersection number -2.Since IL gives
rise to anisomorphism
on aneighborhood
of
F,,
we denote03BC*0398i by
the same letterOi.
Case 2.1 : First we consider the case that
F2
is of type II. ThenF2
and F2 are the same as in the case 1.
Quite similarly
we haveKs + F2 + E3 à 0. On the other
hand,
Hence we have
Therefore we get
È3(S) -
1.Case 2.2: Now we consider the case where
Ê2
is of type III. Thenwe have
Ê2
= Ci +C2,
where Ci andC2
arenonsingular
rationalcurves, Ci n
C2
is onepoint
and(CI, C2)
= 2. We put CIn C2
={p}.
Let
03BC1:S1~
be thequadratic
transformation with center p. We denote theexceptional
curve of 1£1by
El and the proper transforms ofCi
and C2by
C i andC2 respectively.
Let 1L2:S2 ~ Si
be thequadratic
transformation with center Ci ~
C2.
We denote theexceptional
curve of1L2
by E2
and the proper transforms ofCI, C2
andEl by C"1, C2
andEÍ respectively.
Then we may assume thatS
=S2 and ji
= ILl 0 1L2. And wehave
Hence
Thus we infer that
Hence we have
Therefore we obtain
P3(S) ~
1.Q.E.D.
Figure 1.
REMARK: We can construct rational
elliptic
surfaces free frommultiple
fiber of case1,
case 2.1 and case 2.2 as follows.Construction
of
case 1Let
Co
be aquartic
curve inP2
definedby
where
(X :
Y :Z)
is ahomogenuous
coodinate inP2.
ThenCo
has a tacnode at(0 : 0 : 1)
and a cusp at(0 : 1 : 0).
Let To be the tangent line ofCo
at(0 : 0 : 1).
We can resolve thesingularity
ofCo
at(0 : 0 : 1) by
the
quadratic
transformations asFigure
1.Then we know
T"0
is anexceptional
curve andT"0
rlC"0 = ep.
Wecontract
Ty
and denote theresulting
surfaceby
So. LetÊ2
be thedirect
image
of E2. Then we have(Ê2)2
= 0,(El)2
= -2 andC"
4Ê2
+ 2El. Hence0 ~ 03A32
and C"0 e12M + 4l|,
where M is the minimalsection and 1 is a fiber. We put
Ê2
=lo
andCl
nlo
=fp, ql.
Weperform quadratic
transformations at p and q asFigure
2.Let IL :
~ 80
be thecomposition
of thesequadratic
transformations.Let C and
03
be the proper transform ofCo’
and10 by 1£, respectively.
Put E13 -
E14
=Oo,
E23 - E24 =07,
E12 - E13 =01,
E22 -E23
=06, En - E12 = 03982, E21 - E22 = 03984
andM = 03985,
then theOi
arenonsingular
rational curves with self-intersection number -2. Put
Then
Éi - Ê2
and03A6|C|:~
P1 is a rationalelliptic
surface withsingular
fibers of type III* and of type II. Since E14 and E24 are cross-sections of
03A6|C|,
is free frommultiple
fibers. We put S= S -
suppÊ, - F2.
Thenwe infer that
P2(S)
=P3(S)
=0, P4(S)
= 1 andP12(S)
= 2.Figure 2.
Construction
of
case 2.2Let C be a conic and 11 a line in
P2
such that C and Il intersect atone
point
q withmultiplicity
2. Let lo be a line which does not contain q and intersects with Csimply
at twopoint
p2 and P3. Putl0~l1 =
pi.We
perform quadratic
transformations at PI, p2 and p3 as follows:Figure 3.
Let 03BC : ~ P2
be thecomposition
of thesequadratic
transformations.We denote the proper transforms of
lo,
l1 and Cby l’,
11" and C"’. We putl’0
=02,
E12 - E13 =Oo,
En - E12 =01,
E22 - E23 =05, E21 -
E22 -@3,
E32 - E33 -06
and E31 - E32 -04.
And we put1=03980+203981+303982+203983+203984+03985+03986,2=l"’1+C"’.
Then
Ê, - Ê2
and03A6|F1|:~P1
is a rationalelliptic
surface free frommultiple
fibers and withsingular
fibers of type IV* and of type III.Put S
= S -
suppÊl - F2.
Then we infer thatÊ2(S)
=0, P3(S)
= 1 andP12(s)
= 2.Construction
of
case 2.1In the above construction of case
2.2,
wereplace
C +l1 by
a cubiccurve with one cusp and the
remaining
part goesquite similarly.
ThenP2(S)
=0, P3(S)
= 1 andP6(S)
= 2.§7. The case
03BA(Cw) = 1, 03BA(0394) = -~ and 9
is rationalPROPOSITION 6: If
03BA(Cw)=1, 03BA(0394)~0, 03BA(0394)=-~
andS
isrational,
thenP4(S) ~
1 orP6(S) ~
1.PROOF: From the addition formula
([6]),
we know03BA(S) ~
1.By assumption
D contains two reduced fibers off
which we denoteby
Fi and F2. Hence the intersection matrix of D is notnegative
definite.Therefore if
K (S)
=2,
thenby
Theorem 3P4(S) ?
1 orP6(S) ~
1. Thuswe assume that
03BA(S) =
1. Thenby [7; (2.3)-(2.8)]
we have afibering
g :
S ~ P1
1 such that ageneral
fiber ofg 1 s
is a C* or anelliptic
curve.Since the genus of
Cw
is greater than1,
Fi and F2 contain horizontal components with respect to g. Therefore ageneral
fiber ofgis
is a C*.Hence
using [7; (2.6)
and(2.8)],
we reduce theproblem
to thefollowing:
Under the condition
that -2+03A3i (1-m-1i)>0,
look for theinteger
m such that-2m+li [m(1-m-1i)]~0,
where mi areintegers
or 00 and[ ]
is theintegral
part.We infer that for any such mi, m = 4 or 6 is sufficient.
Q.E.D.
§8. Proof of Theorem 3
First we note that when we contract an
exceptional
curve E onS
such that
(E,
Ks +D)~
0, the condition on the intersection matrix of D is alsopreserved.
Henceby
Lemma 1 we may assume that there isno such
exceptional
curve on S.By Proposition 2, 3,
4 we mayassume that
S
is rational andby
Lemma 2 we may assume that each connected component of D is a tree ofnonsingular
rational curves.Let Ks + D =
(Ks
+D)+
+(Ks
+D)-
be the Zariskidecomposition
of 0-divisor
(c.f. [7]).
We put(Ks
+D)- =03A3ri=1 aici
where the ai arerational numbers such that 0 ai ~ 1 and the
Ci
are irreducible curves.Suppose
thatCi~
D for 1 ~ i ~ t andCi
C D for t + 1 ~ i ~ r. If(Ks, Ci) ~
0 for 1 ~i ~ t,
thenThis contradicts to the
negative
definiteness of(Ks
+D)-.
Hence(Ks, Cio)
0 for some io such that 1 ~ i0 ~ t. Since(Ci,,, Cio) 0, Ci0
is anexceptional
curve such that(Ks + D, Ci0) ~ 0,
which is a contradic-tion. Therefore
Ci ~ D
for 1~i~r. Thus we have(Ks
+D)+ =
Ks +
Dm,
where Dm = D -(Ks
+D)-
is an effective Q-divisor. Since the intersection matrix of D is notnegative definite,
there are someirreducible components of D which don’t occur in
(Ks
+D)-.
Hence the part of Dm with coefficient 1 is an effectiveintegral
divisor whichwe denote
by
Do. Then we havewhere the
Ci
are irreducible curves and thedi
are rational number such that0di
1. It is easy to see that ifCi
intersects withDo, di=1-m-1i
where mi is apositive integer.
Sincen(KS+D)~
nKS-[-(n-1)Dm]+D0~nkS+[nDm],
we havePn(S) = dim H’(9,
Os(nKs - [-(n - 1)Dm]
+D0))
for n ~2,
where[ ]
is theintegral part
ofa Q-divisor. Since dim
H2(S, Os(nKs - [- (n - 1)Dm]
+D0)) = 0, by
the Riemann-Roch theorem we haveBy
Kawamata’svanishing
theorem[8]
we havefor n ~ 2. Hence we have
Suppose
thatP2(S)
= 0. Then we have(Do, 3Ks - 2[-Dm]
+D0) ~
0 Hence3(Do, Ks
+Do)
+2(Do, Li=l Ci) ~
0. LetDo, j
=1,...,
u be con-nected components of Do. Then we have
(Do, Kg+ Do)= - 2u.
Thuswe have
On the other
hand,
from(Ks
+ Dm,Do,)
0 we knowTherefore we have
Since 0
di 1,
we obtainFrom
(6)
and(8)
we haveSince every connected components of D is a tree of
nonsingular
rational curves, we know from
(9)
that 3 irreducible components of03A3si=1 Ci
intersect withDo,
for 1~j~
u. We denote the coefficients of these 3 components in Drnby
1 -a-1j,
1 -b il
and 1 -cil, where ai, bj
and Cj are
positive integers
suchthat aj ~ bj ~
Cj.Suppose
moreover thatP3(S)
= 0. Then from(5)
we infer thatHence for some
jo
we haveSince we have
a-1j+b-1j+c-1j~1
from(7),
we getaj0=2, bj0~3
andcjo ~
3,
and therefore we haveHence we have
Thus we deduce that
Suppose
moreover thatP4(S)
= 0. Then we can deduceby
similarway that
bj
= 3 andCj ~
6 for1~j~
u. Then we have§9. Conclusion
By Proposition
1- 6,
wecomplete
theproof
of Theorem 1.PROOF oF THEOREM 2: Let as : S - B
~AS
be thequasi-Albanese mapping
where B is the closure ofas(S)
in thequasi-Albanese variety
As. Thenby
theassumption
we may assume that dim B = 1.Then
by
the property of thequasi-Albanese mapping ([3]), 03BA(B)~0
and a
general
fiber of as is irreducible. Since03BA(S) ~
0, we have K(a general
fiber of03B1S) ~
0. Thus we canapply
Theorem 1 to as : S ~ B.Q.E.D.
PROOF oF COROLLARY To THEOREM 1: We have a
fibering
~ : S~C*. LetCu
be the affine curve definedby
cp = u.Then
by
Theorem 1 we getK-(Cu) ==
-00 for ageneral
u e C*. SinceCu
isaffine,
we haveCu * A’.
Thusby [4;
Theorem9.7]
wecomplete
theproof. Q.E.D.
REMARK: We can prove more
sharpened
result apart from Theorem 1 as follows. But to do this thedeeper
result[1]
is neces-sary.
PROPOSITION 7: Let
S = A2 - V(~)
where~~C[x, y]
and cp is irreducible. IfP2(S)
=0,
then S ~A1
x C*.PROOF: Let C be the closure of
V(~)
inP2
and Hm =P2 - A 2.
ThenS =
p2 _ (C
UH~).
We put C U Hm = D. If there are at least twopoints
at which D is not normal
crossing,
it is easy to see thatP2(S) ?
1.Hence C has at most one cusp. We may assume
deg
C~ 2. Sincepg(S)
=0,
C n Hoc is onepoint,
C is rational and C hasonly
cuspsingularity.
Put C n H~ ={p}.
If p is not the cusp ofC,
then p and the cusp of C are twopoints
at which D is not normalcrossing.
Thus C hasone cusp at p and so
V(~)
= C -{p} ~ A1.
Therefore we canapply [1]
and
complete
theproof. Q.E.D.
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(Oblatum 22-IX-1980) Dept. of Math.
Faculty of Science University of Tokyo Tokyo, Japan