C OMPOSITIO M ATHEMATICA
S HELDON K ATZ
On the finiteness of rational curves on quintic threefolds
Compositio Mathematica, tome 60, n
o2 (1986), p. 151-162
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151
ON THE FINITENESS OF RATIONAL CURVES ON
QUINTIC
THREEFOLDS
Sheldon Katz
C Martinus Nijhoff Publishers, Dordrecht - Printed in the Netherlands
Introduction
This paper addresses Clemens’
conjecture [2]
that ageneric quintic
threefold contains a finite non-zero number of smooth rational curves of
’any given degree.
The paper is
organized
as follows.In
§1,
theconjecture
is reduced toproducing
aninfinitesimally rigid
curve, and
proving
theirreducibility
of the incidencecorrespondence
ofsmooth rational curves on
quintics. Irreducibility
is provenfor d
7.The construction of a rational curve on a
generic quintic
threefold(Theorem 2.1) proceeds by finding
one on aquartic
K3surface,
thenusing
Clemens’argument given
in[1]
to deform the curve to ageneric quintic.
As acorollary (2.4),
Clemens’conjecture
is true ford
7.The number
(609,250)
of conics on ageneric quintic
iscomputed
in§3.
Finally,
in threeappendices,
weexplicitly compute
the normal bundle of any rational curve C ofdegree d
3 on anyquintic threefold,
smoothalong
C. Thisexplicitly
exhibits a curvefor d
3 whose existence wasguaranteed by
Theorem 2.1.I’d like to express my thanks here to Ron
Donagi
and Bob Friedman forhelpful conversations,
to the referee forsuggesting
asimplification
ofTheorem
2.1,
andespecially
to Herb Clemens forsharing
his numerousinsights
andsuggestions
with me.1. Formulation of the
problem
and a method of attackIn this paper, the term
’quintic
threefold’ refers to ahypersurface
ofdegree
5 inP4C.
As stated in the
introduction,
this paper isprimarily
concerned with theCONJECTURE 1.1: Let d be a
positive integer.
Then the schemeof
smoothrational curves
of degree
d on ageneric quintic threefold
isfinite,
non-empty,
and reduced.REMARK:
Here, ’generic’
means that theaccompanying
statement is truefor all
quintics except
thoseparameterized by
a propersubvariety
of themoduli space of
quintics.
Let C c
p4
be an irreducible rational curve ofdegree
d. Then C canbe
parameterized by
5 forms03B10(t, u),..., 03B14(t, u), homogeneous
ofdegree d
onP1.
Let
Iffd
be the moduli space of smooth rational curves ofdegree
d.Taking
into account the abovedescription
and theambiguity arising
from the
GL(2)
action ofP1,
we see thatMd
is irreducible of dimension5(d+1)-4=5d+1.
Let P =
PH0(P4 OP4(5))
be theprojective
space ofquintics
inP4.
Let Y denote the incidence
variety
with
projection
maps i7l:F~Md, 03C02 :
Je P. If we use the same letter Fto describe the
equation
of thequintic,
then Y is defined insideJI! d
X Pby
theequation
The left hand side of
[1.2]
is apolynomial
ofdegree 5d,
hence itsvanishing imposes
at most 5d + 1 conditions onMd
X Pcomparing
withdim
Md
= 5d + 1gives
that dim F dim P.CLAIM 1.3:
Conjecture [1.2]
is trueprovided
that(a)
S is irreducible(b)
There exists a smooth rational curve C ofdegree d
on somesmooth
quintic
threefold F with normal bundle N =Ne/p = OC(-1)
~OC(-1).
REMARK:
OC(k)
denotes theunique
invertible sheaf ofdegree
k onP1.
In
particular,
this notation does not refer to theprojective embedding
C
p 4.
PROOF of claim: Define the open set
fo
c Yby
together
withKF ~ OF
anddeg T. = 2
shows thatdeg NCIF - -
2.Thus,
either
Nc / F * OC(-1) ~ OC(-1),
orH0(NC/F) ~
0.Let
L= {(C, F, p ) EJo
XP4 | p ~ C)
be the " universalcurve",
andF= {(C, F, p) E Jo
XP4 | p ~ F}
be the " universalquintic".
ThenNrc/.:F
is coherent and flat over
Jo’
so that dimH0(NC/F)
is an upper semicon- tinuous function onJo’
This says thatJo
has the stratificationJ0 ~ J1
D ..., where
Now
(b)
says that5i
is a propersubvariety
of50,
Suppose
that03C02(J1)
were dense in P. Then forgeneric F ~ P,
7’2 l(F) ~ J1
would be anon-empty
propersubvariety
of03C02-1(F) ~ J0
of
positive
codimension. But03C02-1(F) n 50 - 03C02-1(F) n 51
isfinite,
sinceany
(C, F)
in this set would haveH0(NC/F)
=0,
which says that C isrigid
in F. But then03C0-12(F) ~ J1
could not be anonempty
propersubvariety
ofpositive
codimension. This contradiction proves finiteness for any smooth F in the dense open subset P --u2
The non-emp- tiness followsby (b)
andby dim J0
dim P. Reducedness follows since thetangent
space isH0(N)
= 0.We conclude this section
by showing
thathypothesis (a)
of(1.3)
isvalid for
all d
7.LEMMA 1.4: 5 in irreducible
if d
7.PROOF:
By [5],
ifC c P 4
is rational ofdegree d,
then C is6-regular,
hence
H0(P4, OP4(5)) ~ H0(C, OC(5)) is surjective.
So the fibersof ff,
are
(irreducible) projective
spaces of constant dimension 124-5d. SinceMd
is alsoirreducible,
we conclude that 5 is irreducible.2. Smooth rational curves of
arbitrary degree
The
goal
of this section is to prove thathypothesis (b)
of(1.3)
is valid for all d.THEOREM 2.1: Let F be a
generic quintic threefold,
d apositive integer.
Then there exists a smooth rational curve C c F
of degree d,
withNCIF
=OC(-1) ~ OC(-1).
REMARK: Theorem
(2.1)
was proven forinfinitely
manyintegers d
in[1].
We first assert that there exists a smooth
quartic
surface S cP3
whichcontains a smooth rational curve D of
degree
d. This isjust
aspecial
case of a theorem of Mori
[11],
which says that you can find such aconfiguration
D c S ifg = genus(D) d2/8, (g, d) ~ (3, 5).
The
proof
concludesby applying
the deformationargument
of[1]
tothe curve D and
quartic
surface S toyield
a curve with therequired properties
on thegeneric quintic
threefold.REMARK: For d =
1, 2, 3, explicit examples
of C c Fhaving
the proper- ties of Theorem(2.1)
will begiven
in theappendices.
It ishoped
that thecomputational
methodsgiven
there will findapplication
elsewhere.COROLLARY 2.2: Clemens’
conjecture (1.1)
is truefor d
7.3. The number of conics on the
generic quintic
threefoldTHEOREM 3.1: Let F be a
generic quintic threefold.
Then F contains609,250
smooth conics.REMARK: The formula for conics on
quartic
threefoldsanalogous
tocll( B ) given
below can be found in[3].
A different line ofreasoning
isgiven here,
whichgives
an intrinsicinterpretation
of the vector bundle B.We review some of the notation and results of
[3].
Let vif be the moduli space of conics in
p4.
Then -4Y may be identified withP(S2U*),
where U denotes the universal bundle on the Grassmanian G =G(3, 5)
ofplanes
inP4,
andP( E )
denotes theprojec-
tive bundle of lines in the vector bundle E.
We describe the Chow
ring
of vif. Let z EA1(M)
denote the class ofC9p(E)(l).
Then the Chowring
of vif isgenerated
over the Chowring
ofG
by z.
The Chow
ring
of G is well-known. It isgenerated multiplicatively by
Schubert
cycles.
Let x = class inA1(G)
of ai = set of2-planes meeting
agiven line, and y =
class inA2(G)
of a2 = set of2-planes passing through
agiven point. By
the canonicalinjection
ofA*(G)
inA*(M),
we can view x
and y
as classes inA*(M).
Thus every element ofA*(M)
can berepresented
aspolynomial
in x, y, z.For enumerative purposes, it is necessary to work with
A11(M)
= Z ·[point].
The non-zero monomials ofweight
11 in x, y, and z are[3]:
Next,
we let W be the scheme of conicson F,
andproceed
tocompute
the class of W inA*(M).
Let 03C0: M ~ G denote the natural map. Then an
equation
for Finduces a section of
S5U*,
hence of03C0*S5U*.
At apoint
of vit repre-senting
a conicC,
this sectionrepresents
theplane quintinc
cut outby
the
2-plane supporting
C. C will lie on F if andonly
if thisquintic
factors into C and a cubic. The set of
quintics factoring
in this wayglobalizes
to the vector bundle T ~03C0*S3U*,
where T is thetautological
bundle on -4Y =
P(S2U*) (whose
sheaf of sections isO(-1)).
There is anatural inclusion of
T~03C0*S3U*
into03C0*S5U*;
consider thequotient
bundle B =
03C0*S5U*/(T ~ 03C0*S3U*).
Then W isprecisely
the zero locusof the section of B induced
by
anequation
for F. Since F isgeneric,
Wis finite.
Thus,
the number of conics on F(including multiplicity)
iscll( B). Using
standardtechniques
forcalculating
Chern classes(and
acomputer),
this class isThis number
accurately represents
the number of smooth conics onF,
since:(a)
There are nopairs
of lines on F that meet, hence there are noreducible conics on F: Let I be the
variety
ofpairs
of distinct intersect-ing
lines inp4.
I is 10 dimensional.Let Q
be theprojective
space of allquintics.
Let C ={(~1, e2, f) ~ I Q |~1 ~ ~2 ~ F}.
Since allpairs
ofdistinct
intersecting
lines areequivalent
underPGL(5),
we may choose coordinates(X0, ... , X4)
forp4
sothat t1
is definedby Xl = X3
=X4
=0, and ?2
is definedby X2
=X3 = X4
= 0. A necessary and sufficient condition for aquintic
F to lie in03C0-11(~1, e2)
is that F have anequation
of the form
where
f4, f4
arequartics,
andf3
is a cubic. This shows that the fibers of iri are all 114-dimensional. Thus C is 124 dimensional.Since Q
is125-dimensional,
thegeneric quintic
cannot lie in theimage
of 03C02.(b)
The scheme of lines on F isreduced,
hence here are no non-re-duced conics.
(c)
The scheme of conics on F isreduced,
hence each conic countsexactly
once in the above Chern classcomputation.
Appendix
A. Lines onquintic
threefoldsIn these
appendices,
wecompute
the normal bundle of rational curveswith d
3 on aquintic threefold,
andgive
anexample
of a curve withNC/F ~ OC(-1) ~ OC(-1).
In thisappendix,
we consider d = 1.Let L c F be a line on a smooth
quintic
threefold. We can assumethat L is
given parametrically by (t,
u,0, 0, 0),
withhomogeneous
coordinates
(X0,..., X4)
onP4,
and(t, u )
onL ~ P1.
Since L is thecomplete
intersection ofhyperplanes X2 = X3 = X4 = 0,
we can write theequation
of F aswhere
the f,
arehomogeneous
inthe Xt (
=0, ... , 4)
ofdegree
4.To
compute
N =NLIF,
we use the standard exact sequencesvalid for any C c
F,
where T denotes atangent
sheaf. We know that every vector bundle on Lsplits
into a sum of line bundles[4]. By
thesyzygy theorem
[10] applied
toP1,
we know thatgiven generators (h)
ofan
ideal, homogeneous
in twovariables,
we can find a set of relationsamong the
ft
whichgenerate
all of therelations,
withouthigher syzygies.
This
technique
will allow us tocompute
all of the bundles in[4.2].
We start with
[4.2a].
The mapOL ~ OP4(1)5 |L
has matrixt(X0 X1 X2 X3 X4) = t(t
u 0 00).
The cokernel must be describedby
the rela-tions,
which aregiven by
the matrixComparing degrees,
we see thatTP4 | L ~ (fl L (2)
~OL(1)
~OL(1)
~OL(1).
Turning
next to[4.2b],
we observe thatTP4 | L ~ O(5) | L
is inducedby
££8 /8 £ - 03A3~~F/~Xi.
NowaF jaXo 1
L=
~F/~X1 L
=0; ~F/~Xi L
= fi | L
for i =2, 3,
4. We abuse notationby omitting
the restriction toL, writing fi
inplace of fl 1 LI thinking
ofthe f,
asquartics in t,
u. So ifTP4 | L ~ O(5) | L
has matrix(a1
a 2 a3a4)
in the basis forTP4 | L just described,
then we haveSolving,
ai =0, and a = fi
f or i =2, 3,
4. So weget
the matrixAgain,
the syzygy theoremgives
the matrix of the kernelTF | L ~ T p4 L.
Because of the 0 in the matrix[4.5]
there is the relationgiven by
’(1
0 00),
but there must also be two others among thefi.
There are
always
at least twoindependent
sextic relations03A3qifl
=0,
with
the qi quadratic;
this is becausethe qi depend
on 3 - 3 = 9 parame- ters, while sexticsdepend
on 7. So if there are no relations of lowerweight
than 6(the generic case),
these two relations willcomplete
thedescription
of the kernel. In this case,TF | L ~ OL (2) ED OL(-1)
~OL(-1).
If there
happens
to be aquintic relation E tih
= 0 with linear?’i,
but noquartic relation,
then there will be a factor ofOL
inTF L. Considering degrees
in[4.2b]
shows that detTF | L
=deg TP4 | L - deg (QL(5)
= 5 - 5=
0,
hence in this case,TF | L ~ CL (2) ~ OL
~OL(-2). Finally,
if there isa
quartic relation Y- cifi
= 0 withsçalars
cl , weget TF | L ~ OL (2) ED L(1)
~
(9L ( - 3).
Having computed TF 1 LI
we now turn to[4.2c].
We couldcompute
explicitly
asbefore,
but wemerely
observe that sinceTL = (QL(2),
thenthe matrix of
TL ~ TF | L
must contain zeros at the locationscorrespond- ing
to factors of(QL(a)
inTF | L,
with a 2. With thisobservation,
weconclude that
Now choose F so that
f2 = t4, f3 = t2u2, f4 = u4.
Then there areclearly
no relations ofweight
5 or 4 among thefi,
henceN ~ L(-1)~
L(-1) by [4.6].
REMARK: The number of lines on a
generic
F has beencomputed
to be2875
[6].
Since N =(!)( -1)
~O(-1),
it follows thatHO(N)
=0,
so the Lhas no first order deformations in
F, implying
that L hasmultiplicity
one in F. This
gives
anotherproof
that all lines occur withmultiplicity
one.
Appendix
B. Conics onquintic
threefoldsLet C c F be a smooth conic on a smooth
quintic
threefold. We can assume that C isgiven parametrically by (t2,
tu,u2, 0, 0).
Since C is definedby
theequations X12 - X0X2 = X3 = X4 = 0,
we can write theequation
of thequintic
F in the formFor some cubic
f,
andquartics f3, f4.
Now the matrix of
OC ~ O(1)5|C
in[4.2a]
is1(t2 tu u2
00),
so thecokernel may be described
by
the matrix of relationsso that
TP4 | C ~ OC(3)
~OC(3)
~Oc (2)
~OC(2).
Turning
next to[4.2b]
the matrix(a b
cd )
ofTP4 | C - O(5) | C ~ OC(10)
is such thatwhere the
right
hand side is the vector ofpartial
derivatives of F.Here,
we think of
f
as asextic,
andf3, f4
andoctics,
in thevariables t,
u.Solving [5.3]
for( a b c d ),
weget
the matrixso
TF 1 e
iscomputed by finding
relations among them. For now, assume thatf ~
0. The relationt ( - uf )
+u ( tf )
= 0gives
anme(2)
factor. Look-ing
for additional relations isequivalent
tolooking
for relations betweenf, f3,
andf4.
There arealways
at least 2independent
relations ofweight
11
qf + c3f3
+c4f4
= 0with q
aquintic,
and cicubics,
since(q,
c3,c4) depends
on 6 + 4 + 4 = 14parameters,
whiledegree
11polynomials depend
on 12parameters;
if these are the relations of lowestweight,
thisgives TF | C ~ OC(2) ~ OC(-1) ~ OC(-1); a
relation ofweight
10 wouldgive TF | C ~ OC(2) ~ OC ~ OC(- 2);
a relation ofweight
9 wouldgive TF | C ~ OC(2) ~ OC(1)
~OC(- 3);
a relation ofweight
8 wouldgive TF |c
=
me(2)
~OC(2) ~ OC(-4).
Finally, considering [4.2c],
we conclude as before that iff ~
0Now,
let F be such thatf = t6, f3
=t3u5, f4
=U8. By [5.5],
N =OC(-1)
~
C9c( -1).
REMARK: If
f = 0,
then theplane
P definedby X3 = X4
= 0 istangent
toF
along
C. In that case, C deforms to first order to Fby moving
in any direction inP,
andNC/F ~ OC(4) ~ (9c(- 6).
Appendix
C. Twisted cubics onquintic
threefoldsThe case of the twisted cubic is
complicated
somewhatby
the fact that it is not acomplete
intersection.Let C c F be a twisted cubic curve on a
quintic
threefold. We can assume that C isgiven parametrically by (t3, t2u, tu2, u3, 0).
LetThen C is
given by
theequations QO = Ql = Q2 = X4 =
0. We can writethe
equation
of thequintic
F in the formwhere
the f
arecubics,
and g is aquartic.
Asbefore,
we consider[4.2a]
and compute
a matrix of relations fort(t3 t2u tu 2 U3 0) to get
giving T4 | C ~ Oc (4)
~OC(4)
~OC(4)
~OC(3),
which in turnyields
as amatrix for the map
TP4 | C ~ O(5) | C ~ OC(15)
of[4.2b]
Here
deg a
=deg b
=deg
c =11, deg
g = 12. SoTF 1 c
iscomputed by finding
relations among a,b,
c, g. The relationt2a + tub + u2c = 0,
which has
weight 13, gives
anOC(2)
factor. This relation may reduct to lowerweight,
inspecial
cases. Rather than include the exhaustiveanaly-
sis
here,
we content ourselves with a few observations.1. a, b, c
cannot be all zero. For then the threeweight
11 relations1 · a = 1 · b = 1 · c = 0 would
give TF | C ~ OC(4)
~(9c(4)
~OC(4),
whichcontradicts
deg TF 1 c
= 0.2. There cannot be two
independent weight
11 relations. For if there were, then there would be a non-zeropolynomial
a ofdegree
11 andscalars
a’, b’, c’,
not all zero, such that a =a’«, b
=b’a,
c = c’a. Butthen
t2a
+ tub+ u 2c
= 0 becomes(a’t2 +
b’tu+ c’u2)03B1
=0,
which isimpossible.
3. There are
always
at least twoindependent weight
16 relations ql a +q2b
+ q3c + qg =0,
wherethe qi
arequintics and q
is aquartic.
Thisis because the data
( ql,
q2, q3lq) depends
on 23parameters,
but needs to be reduced to 19 because of the set of trivial relationset 2a
+ etub +eu2c
= 0 for any cubic e, whiledegree
16polynomials depend
on 17parameters.
Summarizing,
weget
thefollowing
chart for the relations of lowestweight
REMARK: It would be very
interesting
to know what the number of twisted cubics on ageneric quintic
is. A more detailedunderstanding
ofthe moduli space of twisted cubics is necessary. In
[9],
we see that thisnumber is divisible
by 5, by studying
thedegeneration
of thequintic
to aunion of 5
hyperplanes, along
the lines of[8].
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(Oblatum 6-VI-1985 & 4-X-1985) Department of Mathematics
University of Oklahoma Norman, OK 73019 USA