Parameter spaces for curves on surfaces and enumeration of … · 2011-10-03 · Compositio Mathematica 113: 155–208, 1998. 155 c 1998 Kluwer Academic Publishers. Printed in the

Compositio Mathematica 113: 155–208, 1998. 155c 1998 Kluwer Academic Publishers. Printed in the Netherlands.

Parameter spaces for curves on surfaces andenumeration of rational curves

LUCIA CAPORASO1 and JOE HARRIS21Mathematics department, Harvard University, Cambridge MA 02138, USA and Dipartimento diMatematica, Universita di Roma 2, Roma 00133, [email protected] department, Harvard University, Cambridge MA 02138, [email protected]

Received 23 April 1996; accepted in final form 13 May 1997

Abstract. Let be a smooth, minimal rational surface. The geometry of the Severi variety para-metrising irreducible, rational curves in a given linear system on is studied. The results obtained areapplied to enumerative geometry, in combination with ideas from Quantum Cohomology. Formulasenumerating rational curves are found, some of which generalise Kontsevich’s formula for planecurves.

Mathematics Subject Classifications (1991): 14M10, 14M45, 14J26, 14N10.

Keywords: rational surface, rational curve, enumerative geometry, Severi variety, cross-ratio method.

1. Introduction

In this paper we investigate the geometry of families of rational curves on anonsingular, rational surface . All varieties are assumed to be projective over .Let be an effective divisor class in and let be the set of all effective

divisors linearly equivalent to ; this is a projective space whose dimension wedenote by . Inside , we consider the locus of rational curves: we let

such that is an irreducible rational curve

This is a locally closed subset of ; we let be its closure. We callthe Severi variety of rational curves associated to the divisor class , and

we denote its dimension by 0 . We have in general 0with equality holding in all the cases that we shall study.The particular aspect of the geometry of of concern to us here is its

degree, which we denote by . This can be characterized directly: it is thenumber of irreducible rational curves that are linearly equivalent to and thatpass through 0 general points of . The principal results of this paper is thecomputation of in some cases. For simplicity, we define to be zero if

is empty.

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156 LUCIA CAPORASO AND JOE HARRIS

There are various approaches to the calculation of degrees of Severi varieties(see [CH] for a different technique). We call the one we take here the ‘cross-ratio’method; it is based on ideas of Kontsevich and Manin, expressed in the ‘FirstReconstruction Theorem’ of [KM]. In [KM] they describe a formula discoveredby Kontsevich, for the number of plane rational curves of given degree passingthrough the appropriate number of points (the first proofs of it appear in in [RT]and in [K] – [KM]). These ideas have also been used to give formulas for thedegrees of genus 0 Severi varieties on certain rational surfaces; see [CM], [DI] and[KP].For an illustration of how the cross-ratio method can be used to give a rather

simple proof of Kontsevich’s formula see [C] (see also [CH] for an even simplerproof).Let 1 1 be a Hirzebruch surface. The Picard group of

has rank 2, and we choose generators as follows

Pic

where 2 2 0 and 1. We denote by the unique curve ofnegative self intersection, so that 2 and .Let be any divisor class on the surface other than , and let :1 2 be any sequence of positive integers with . We

define the locally closed subvariety to be the locus of irreduciblerational curves such that, if : 1 is the normalization of , the pullbackdivisor

for some collection of distinct points 11, and we let be its

closure; for example, as we will see, if 1 1 1 , then .When contains a single integer greater than 1 (i.e. 1 1 1 ), wedenote these by and respectively. We set

0 dim

and

deg

We have for 1 , 1 and 010 .We define

to be zero if is empty.Similarly, let 1 be any collection of distinct

points. We let be the locus of irreducible rational curves suchthat, if : 1 is the normalization of , then for some collection of distinctpoints 1

1 we have

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PARAMETER SPACES AND ENUMERATION OF RATIONAL CURVES 157

and

and again let be its closure.Now we give a list of some formulas including all the ones that we prove in this

paper. They are very similar from a formal point of view. We state them in a waythat highlights the analogies.Fix two curves 3 and 4 on . For any pair of divisor classes 1 and 2 we

introduce the function

1 2 :

1 20 3

0 1 11 3 2 4

0 3

0 1 22 3 2 4

Using this notation, we state the following results

Recursion for 2 ([KM]) Let 3 and 4 be two fixed lines in the plane, then

1 2

1 2 1 2

Recursion for 1 1 ([KM], [DI], [KP]) Let 3 and 4 be two fixed elements ofthe two distinct rulings, then

1 2

1 2 1 2

The first new result of this paper is a recursion formula for the degrees of Severivarieties of rational curves on 2 . The recursion contains now a new term which isdue to the contribution of degenerate curves containing .

Recursion for 2 (Theorem 3.2) Let 3 and 4 be two fixed elements of the class, then

21 2

1 2 1 2

21 2

1 2 1 2

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Finally, on , the general reducible curves that are limits ofirreducible rational curves and contain have the property that each component

may have a point of tangency of order with – that is, will belong to, where is the divisor class of . Accordingly, we shall define later

(Section 3.4) a generalized version of the number 1 2 ; this will be a function1 2 1 2 depending recursively on the degrees . In these

terms, we give a formula expressing the degree of on in terms ofthe degrees of the tangential Severi varieties of smaller divisor classes.

A sample formula for (Theorem 3.4)

1 2

1 2 1 1 1 2

2 1 2 1

: 11 1

The difference here is that in case 3 this does not give a complete recursion:to be able to enumerate rational curves on such surfaces, we would need formulasfor the degrees of the ‘tangential’ Severi varieties as well, that is, we need formulasfor . The first case for which this occurs is that of 3 . Very possibly acomplete recursion could still be obtained using the cross-ratio method, althoughthe level of difficulty seems to us to get very high. Instead we found a differenttechnique that we successfully applied in a few cases; for example, we obtained acomplete set of recursions for the surface 3 . This different method is the subjectof another paper of ours (cf. [CH]); it also is heavily based on the deformationtheory results that are developed in the second chapter of this paper.Finally, we obtain a closed formula for the class 2 on any ruled surface .

Closed formula for 2C on (Theorem 3.3)

21

0

2 2 2

2. Degenerations of rational curves

2.1. THE BASIC SET-UP

We start with the complete linear system associated to a divisor class onthe ruled surface , and with the Severi variety . We thenchoose 0 1 general points 1 1 , and let be the intersectionof with the linear subspace of curves in passing through 1 1;

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we let be the corresponding family of curves over .Next, we let be the normalization of the base , and

the normalization of the pullback of the family to , so thatis a family whose general fiber is a smooth rational curve. If is a fiber of ,the notation will be used for a corresponding fiber of the family ,which may differ from the normalization of .Then we fix two curves 3 and 4 in , which will be linearly equivalent to .

We need to make a further base change , so that the points of intersectionof the curves in our family with 3 and 4 become rational over the base. We thuslet be any finite cover, unramified at the points with singular,and let be the pullback of the family to . (By Propositions2.1 and 2.5, the map introduced in Chapters 1 and 3 in order to definethe sections will indeed be unramified at the points of corresponding to thesingular fibers of .) Because the results of this chapter are all local in thebase of our family, however, we will not need to introduce this extra step in theconstruction. For the remainder of this chapter, accordingly, we will take ;and all of the results of the chapter describing the map will still holdafter the base change .Nextwe introduce the nodal reduction of the family . That is to say, after

making a base change and blowing up the pullback family ,we arrive at a family such that

(1) is a family all of whose fibers of are reduced curves havingonly nodes as singularities;

(2) the total space is smooth;(3) admits a regular birational map over .

In fact, most of our concerns with this definition will turn out in the end to beunnecessary: we will see below as a corollary of Propositions 2.6 and 2.7 that infact is already a family of nodal curves. Thus, in practice, we will nothave to make a base change at all at this stage, and will be simply the minimaldesingularization of . For this reason (and because is itself already an arbitraryfinite cover of the normalization of our original base ) we will abuse notationslightly and omit the tilde in , that is, we will speak of the family .One further remark: in the applications we will have four sections of the family

and will correspondingly want to consider this as a family of four-pointednodal curves. For this reason, we may want to make further blow-ups at pointswhere these sections cross. By Propositions 2.1 and 2.5, however, the sections inquestion will cross only at smooth fibers of and so this will not affect ourdescriptions of the singular fibers of the family.The final construction is one that we will use only in the following chapter,

but we mention it here just to have all the definitions in one place. After arrivingas above at a family of nodal curves with four disjoint sections , wemay then proceed to blow down ‘extraneous’ components of fibers of :

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that is, any component of that meets the other components of in only onepoint, and that meets at most one of the sections . Iterating this process untilthere are no extraneous components left, we arrive at what we will call theminimalsmooth semistable model of our family: that is, a family such that issmooth, the fibers are nodal, the sections are disjoint and is minimalwith respect to these properties. Note that the special fiber of must be a chainof rational curves 0 with two of the sections meeting each of the two endcomponents

(the case 0 is simply the case where is irreducible). Finally, we can blowdown the intermediate components 1 1 in this chain to arrive at a family

of 4-pointed stable curves, called the stable model of our family. Thespecial fiber of this family will have just two components (or one, if 0), with asingularity of type at the point of their intersection.In sum, we have the diagram of families and maps.

2.2. THE MAIN RESULTS FROM DEFORMATION THEORY

We give here a summary of the main results to be proved in this chapter.

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The first is Proposition 2.1 in which we consider the Severi varieties and, compute their dimension and describe the geometry of their general

point. In particular, we characterize the general fiber of the family .The results are unsurprising: for example, the general point ofcorresponds to a curve with only nodes as singularities; general points, of, respectively, and correspond to curves , that

intersect transversely.Then, in Proposition 2.5, we study the geometry of the general point of theboundary of . We do that by listing all types of reducible fibers thatoccur in the family . This result is not predictable on the basis ofa simple dimension count; in most linear systems on the subvarietycorresponding to reducible rational curves containing is larger-dimensionalthan ; so the question of which points of the former lie in the closure ofthe latter does not have an immediate answer.The third result is Proposition 2.6, which is specifically about the family

. We describe the geometry of the base in a neighborhood of eachpoint corresponding to a degenerate fiber . In particular, we say howmany branches has at and say how the nodes of the nearby irreduciblefibers approach the singularities of as we approach along each branchof .Finally we have Proposition 2.7, describing the singularities of the total spaceof the families and . This will be a crucial ingredient incalculating the multiplicities of zeroes of the cross-ratio function on the baseof our family.

One word of warning is in order. Many of both the statements and proofs ofthese propositions are just routine verifications of statements easily guessed on thebasis of naive dimension counts. At the same time, mixed in with these largelypredictable statements are some interesting phenomena. These are described in thesecond parts of Propositions 2.5, 2.6 and 2.7, in which we describe the geometryof the one-parameter families and in a neighborhood of thereducible fibers containing . Near such a curve, the local geometry of the universalfamily over the Severi variety is, to us, somewhat surprising.

2.3. GEOMETRY OF SEVERI VARIETIES

Here is the first result about the varieties defined in the introduction.

PROPOSITION 2.1. Let and be any linear series on the surface; let be any fixed curve not containing and let 1 2

be any given finite collection of points. Let 1 2 be any collection ofpositive integers with .

(1) If is nonempty, then it has pure dimension

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dim 1 1

(2) A general point of any component of corresponds to a curvehaving only nodes as singularities, smooth everywhere along ,

intersecting transversely and not containing for any .(3) If and are general points of irreducible components of and

respectively, then and intersect transversely, and none of theirpoints of intersection lie on or .

REMARK. Many of the techniques necessary to prove this statement are in [H].In fact, many of these assertions are proved there, but unfortunately with slightlydifferent hypotheses: they are proved first on a general rational surface , but onlyfor , that is, without the tangency condition (Proposition (2.1) of [H]); andthen with a single tangency condition, but only with respect to a line in the plane(Lemma (2.4) of [H]).

Proof.We start with the dimension statement. The assertion that the dimensionof is everywhere equal to 1 is standard deformation theory(and is well known; c.f. [K]). To see it, observe first that if is anypoint and : the normalization of the corresponding curve, thefirst-order deformations of the map are given by sections of the pullbackof the tangent bundle to . Now, the tangent bundle to the ruled surface isgenerated by its global sections everywhere except along ; since doesn’t contain, it will likewise be true that the pullback will be generically generated

by its global sections. Since 1, it follows in turn that 1 0.The deformations of the map are thus unobstructed, from which it follows thatthe space of such deformations is smooth of dimension

0 deg 2

2

If we mod out by automorphisms of the domain 1, we see that the space ofdeformations of the image curve as a rational curve has dimension

0 1 3 1

which is the same as the dimension of .We next establish the

CLAIM. The dimension of , and hence of , is everywhere at least0 1 .

To see this, set . Let be any point, an analyticneighborhood of in , the universal family of curvesover , and and the normalizations of and ; we may assume that themap : is smooth. Now let be the th symmetric fiber product of

. We then have a map

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:

Now, inside the symmetric product , the locus of divisors having pointsof multiplicities or more is irreducible of codimension 1 ; since

is an open subset of the inverse image 1 , it follows that it musthave dimension at least dim 1 everywhere.Note that an analytic neighborhood of any point of admits a map to, sending to the images ; the fibers of this map are analytic

open sets in the varieties . In particular, we have

dim dim

so that in order to prove the opposite inequality dim 01 , it is enough to show that the dimension of the variety is equal to0 for any subset 1 .To prove the remaining parts of the Proposition we first identify the projective

tangent space to the space of deformations of a given reduced curve preservingthe geometric genus of ; and then the subspaces corresponding to deformationsthat also preserve singularities other than nodes and/or tangencieswith fixed curves.This is the part that is in common with [H], and for the most part we simply recallhere the statements of the relevant results (Theorem 2.2 and Lemma 2.3). Then, toapply these, we need to estimate the dimension of these subspaces of ; this iscarried out in Lemma 2.4 and the following argument.We identify the tangent space to the linear series at with the character-

istic series

00

where 0 is the section vanishing along (this identification isnatural up to scalars; more precisely, the tangent space to 0 at

is

Hom0 0

Now suppose that we are given any subvariety of the linear series on. Let be a general point of . The following theorem of Zariski ([Z],Theorems 1 and 2) characterizes the tangent space to at .

THEOREM 2.2. (Zariski’s theorem) In terms of the identification of the tan-gent space to the linear series at with the characteristic series0 ,

(1) The tangent space is contained in the subspace 0 of0 , where is the adjoint ideal of ;

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(2) If has any singularities other than nodes, then is contained in asubspace 0 where is an ideal strictly contained in the adjointideal.

This characterizes the tangent space to at a general point . (If thefact that it does is not clear, it will be after Lemma 2.4 below.) Now, we have toconsider the additional information coming from the tangency with . To expressthis, note first that, if : is the normalization of and is anyideal contained in the adjoint ideal of , then the pullback map gives a naturalbijection between ideals contained in and ideals

. We will invoke this correspondence implicitly in our notation: ifis any point, and any ideal contained in the adjoint ideal of , wewill write to mean the ideal in whose pullback to is

. In these terms, we have the following.

LEMMA 2.3. Let be any fixed curve and a smooth point of . Letbe any subvariety of . If the general point of satisfies the condition:

there is a point such that and

mult

then the tangent space to at satisfies

0

Moreover, if has any singularities other than nodes, or is singular at the point, we have

0

where is an ideal strictly contained in the adjoint ideal.Proof. We will prove the Lemma by applying Zariski’s theorem to the proper

transform of on the surface obtained by blowing up a total of timesalong the curve . To carry this out, let 1 0 be the blow-up of 0 at thepoint , 1 1 the exceptional divisor of the blow-up and 1 1 the point ofintersection of 1 with the proper transform of in 1. Similarly, let 2 1 bethe blow-up of 1 at the point 1, 2 2 the exceptional divisor of the blow-upand 2 2 the point of intersection of 2 with the proper transform of in 1,and so on, until we arrive at the surface ; we will denote by : thecomposite of the blow-up maps, by the proper transform of in and bythe proper transform of in ; so that the pullback to of the divisor is givenby

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We denote by the branch of corresponding to the point , that is, theimage of an analytic neighborhood of in , by its proper transform in ,and by the point of lying over .Now, let be the proper transform of in , and let be the multiplicity of1 at the point 1; for each 1 we will set

1 2

Thus, for example, we have the equality of divisors

1

Similarly, we let be the proper transform on in , the multiplicity of1 at 1 and 1 . Note that for each ; and the requirement

that have intersection multiplicity with at is equivalent to the assertionthat

mult

so that we have in particular , with equality if and only if (locally) .We can also write the intersection number as

mult mult

so we see that one of three things occurs: eitheris smooth, 1 for all , and meets the last exceptional divisor

transversely; orpasses through the point 1 for some ; or

for some , meets the exceptional divisor at a point other than1 or 1, and has a point of intersection multiplicity 1

with .We now compare the adjoint ideal of with that of . The basic fact here

is that if is any curve on a smooth surface, a point of multiplicity, and the proper transform of in the blow-up : of at , the

adjoint ideals of and are related by the formula

1

where is the exceptional divisor. Applying this times to the curve , we have

Now, being general, any deformation of coming from the fami-ly preserves the multiplicities , and hence the decomposition

. It also preserves the geometric genus of , so that identifying the space

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0 of deformations of with a subspace of the deformations0 of via the pullback map, we have

0

0

0

0

0

0

Note that the inclusion in the last line of the above sequence is proper if .Now, suppose that is not smooth at . In this case, as we noted willeither be singular at or be tangent to there, or else will pass through thepoint 1 for some . In the first case, since has a unibranch singularity,its deformations correspond to sections of 0 for some ideal strictlycontained in the adjoint ideal ; while in the latter two cases the deformations cor-respond to sections of 0 vanishing at . In either case, the inclusion inthe first line of the equation above is strict. Thus 0 1unless is smooth at , and the remainder of the statement of the Lemma follows.

To conclude the proof of Proposition 2.1 we need one more fact. To state it,let be any irreducible rational curve, : the normalization and1 2 any points; suppose that the divisor has multiplicity at. Let be the adjoint ideal of , and set

Let be any ideal of index 2 or less in – that is, any ideal with0 2, or equivalently an ideal of the form

for some pair of points . We will need these ideals of index 2 inorder to see, for example, that a general curve does not have a node on. In these terms, our result is the

LEMMA 2.4. The ideal imposes independent conditions on the linear series, i.e.,

0 0 dim

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In particular, imposes independent conditions on , that is,

00

Proof. By the adjunction formula we have

Thus,

Now, 0, and on , we have

2

so that we have an inequality of divisor classes

2

Moreover, the divisor class 2 has intersection number at least 3 with anyirreducible curve not linearly equivalent to either or , so it follows that

deg 2 3 1

Thus

deg 1

so that 1 0, and the result follows.

We can now complete the proof of Proposition 2.1.We have already established,in the Claim above, that

dim 0 1 ;

but applying Lemmas 2.3 and 2.4 in turn we see that for any subset 1,

dim 0

0

and hence

dim dim

0 1

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so that equalitymust hold.Moreover, if a general point correspondedto a curve with singularities other than nodes, the second inequality abovewouldbe strict; so must be nodal, and smooth at its points of intersection with .We can eliminate all the other possible misbehaviors of our general curve

similarly. If the point is mapped to one of the points , we would have

dim 0

0 ;

and if the multiplicity of the pullback divisor at were 1 we wouldhave

dim 0 10

Suppose next that had a node on , with branches corresponding to a pairof points and the branch corresponding to transverse to . It wouldfollow that

0 0 1

since a section of vanishing at but not at would correspond to a defor-mation of in in which the two branches would meet in distinct points.Finally, to prove part 3 of Proposition 2.1, we simply let be a generalmember

of the family and apply the above to , including in andits points of intersection with and among the points .

The next Proposition is stated as a characterization of the reducible elementsof the one-parameter family , but in fact it is a characterization of thecodimension one components of the boundary of .

PROPOSITION 2.5. Let be any reducible fiber of the family .(1) If does not contain , then has exactly two irreducible components 1

and 2, with and 1 2 . Moreover is a generalpoint in .

(2) If does contain , then has irreducible components , 1 ,with and 1 . Moreover each isgeneral in for some collection 1 of positive integers suchthat 1 .

REMARK. Notice that by Proposition 2.1, the above result says that if does notcontain , then it has only nodes as singularities. And, if contains , away fromthe points of tangency of with the curves , has only nodes as singularities.

Proof. Assume first that does not contain . Write the divisor as1 where 0 and the are irreducible curves in . We claim first

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that since , all the curves must be rational. To see this, take anyone-parameter family of irreducible rational curves specializing to .Proceeding as in 2.1 we arrive at a family of nodal curves, with generalfiber 1, that admits a regular map . Now, since the fibers of arereduced curves of arithmetic genus 0, every component of every fiber of mustbe a rational curve. Thus every component of is dominated by a rational curveand so must be itself rational.Thus , where are divisor classes such that . On

the other hand, since is a general member of an 0 1 -dimensional family,we must have

10 0 1

11 2

12

Comparing the two sides, we see that

21

1 0

But 2 for any curve on other than ; sowemay conclude that all1 and that 2. Moreover, if 2 we have equality in the above inequality,

which says that the pair of curves 1 2 is general in 1 2 .We come now to the case where contains . The first thing we see here is

that the dimension-count argument we used above doesn’t work: since

2

the sums of rational curves may well move in a larger-dimensional family than itself.The key here is to look at the semistable reduction of a family of curves in

specializing to . This will allow us to limit the number of points of intersectionof the curves with , that is to say, to show that in fact the belong tofor suitable . This replaces the naive bound above on the dimension of the familyof such curves with a stronger one, which turns out to be sharp.Consider then the family obtained from as in Section 2.1. We

can thus assume that the total space of the family is smooth and every fiber ofis a union of smooth rational curves meeting transversely, and whose dual graph isa tree.Now, let be the special fiber of . We decompose into two parts:

we let be the union of the irreducible components of mapping to , and

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the union of the remaining components. Next, we decompose further intoparts, letting be the union of the components mapping to . Denote the

connected components of by , and for each let be the degree of themap : , so that . Similarly, let be the connectedcomponents of and the degree of the restriction : , so that

.Note that the inverse image of in is given by 1

1 .(where : is the natural map.)As we indicated, the essential new aspect of the argument in this case is keeping

track of the number of points of intersection of the with . To do this, we notethat, over any such point, there will be a point of intersection of a component ofwith the inverse image 1 ; which by the expression above for 1 will

be either a point of intersection of with or one of the points of intersectionof the with .It thus remains to bound the number of points of intersection of with the

remaining parts of . This we can do by using the fact that the dual graph of isa tree: this says that the number of pairwise points of intersection of the connectedcomponents of and the connected components of is equal to the totalnumber of all such connected components, minus one. Thus,

# # connected components of

# connected components of

Note that the degree 0 on each component of , so that


and similarly


Thus we can deduce in particular that

1

Now, say for each 1 . Let : be the nor-malization map. Choose any irreducible component 0 of dominating (andhence dominating the normalization ), and let : be the restrictionof to . Trivially, the total number of points of the pullback of tois

# # # 0

and hence

# # 0 #

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with strict inequality if any 1. But the sum of degrees of on the curvesis at least

deg

1

Comparing the number of points of the pullbacks of to the normalizationswith the degrees of these pullbacks, we conclude that there must be multiplicitiesin these divisors: specifically, the sum 1 of the multiplicities minus onemust be the difference of these numbers, so that

1 deg

1

1

1 1 1

This in turn allows us to bound the number of degrees of freedom of the curves: we have

dim 0 1

11 1

1

1 1

On the other hand, this must be at least equal to the dimension of minusone, that is,

0 1 2

2

2 2

In the end, then, we must have2 2

1

1 1

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We can (partially) cancel the 1 and 2 terms, and combine the termsinvolving to rewrite this as

1 1 1 1

or, in other words,

1 1 1 0

Now, we have already observed that 2 meets every curvestrictly positively, so that the sum in this last expression is nonnegative. We

conclude that 1, and (since any 1 would have led to strict inequality) thatall 1. Next, since there is a unique component of mapping to each , eachcurve will have at most one point of intersection multiplicity 1 with .Thus, finally, is a general member of the family for some collection ofintegers 1 with 1 , completing the proof of Proposition2.5.

Note that we have not said here that every reducible curve satisfying the condi-tions of the Proposition in fact lies in the closure of the locus of irreducible rationalcurves. This is true, and is not hard to see in the case of curves of types (1); but forcurves of type (2) it is a deeper fact, and we will require the proof of Proposition 2.7to establish it.Having characterized as a set the locus of curves in passing through

1 1, we now turn to a statement about the local geometry of aroundeach point.We introduce one bit of terminology here. Let be a fiber of ; and, in

case is locally reducible at the point , pick a branch of at (that is,a point of the normalization of lying over ). Let be a node of . Wethen make the following

DEFINITION. If is a limit of nodes of fibers of near in the chosenbranch–that is, if is in the closure of the singular locus of the map

–we will say that is an old node of . If is an isolatedsingular point of the map we will say that is a new nodeof .Equivalently, is an old node if the fiber of over is smooth at

the (two) points lying over ; if it is a new node, will have a single point lyingover , which will be a node of .Note that if is a singular point of other than a node, the situation is not so

black-and-white. For example, if is an -fold tacnode – that is, if the curvehas two smooth branches at with contact of order – then a priori, any number

of nodes of nearby fibers may approach along any branch of at ,with the result that the fiber of over the corresponding point will

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have an -fold tacnode over , or will be smooth over if . (Theproof of the relevant case 1 will emerge in the proof of Proposition 2.7.)In these terms, we can state

PROPOSITION 2.6. Let be a reducible fiber of the family . Keeping thenotations and hypotheses of Proposition 2 5,(1) If 1 2 does not contain , and 1 and 2 meet at 1 2

points 1 , then in a neighborhood of has smooth branches1 ; along the point is new, and all other nodes of are old.

(2a) If 1 , and meets transversely inpoints 1 , then in a neighborhood of consists of smoothbranches 1 . Along the points 1 1 are new, andall other nodes of are old.

(2b) If 1 , and meets transversely inpoints 1 for 2 , while 1 has a point of intersectionmultiplicity 2with , then in a neighborhood of consists of 2smooth branches 2 . Along the points 2 2 are new;all other nodes of are old; and exactly 1 nodes of nearby fibers willtend to .

REMARK 1. The proof of this Proposition will not be complete until the end ofthe following section. More precisely, we will postpone the proof of the existenceand smoothness of the branches of . Actually, cases 1 and 2 could very well beproved here, but it is more convenient do it later (that is, at the beginning of theproof of Proposition 2.7).

REMARK 2. We believe that an analogous description of the family maybe given without the assumption that the components of the curve other thanhave altogether at most one point of tangency with , and otherwise intersecttransversely in distinct points. The restricted statement above will suffice for ourpresent purposes. We hope to prove the general statement in the future.

REMARK 3. The statement of Proposition 2.6 can also be expressed in terms of thenormalized family , and indeed that is howwe will use it in the followingchapter. In these terms, the statements are:(1) If is a point of corresponding to a curve in our family not containing, then there will be 1 2 points of lying over , corresponding

naturally to the nodes of . The fibers of over these points will be thenormalizations of at all the nodes of 1 and 2 and at all but one of the pointsof intersection of 1 with 2.(2a) If 1 contains and the components intersecttransversely, then the fibers of over points lying over are

the curves obtained by normalizing at all nodes of the , at all the points ofpairwise intersection of the , and at all but one of the points of intersection of

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with each of the components . In other words, the fibers consist of the disjointunion of the normalizations of the curves , each attached to at one point.(2b) If 1 as before and one of the components 1 ofhas a smooth point of intersection multiplicity 2 with , then the fibersof corresponding to are the curves obtained by normalizingat all nodes of the , at all the points of pairwise intersection of the , at

all but one of the points of intersection of with each of the components for2 , at all the transverse points of intersection of 1 with , and finally

taking the partial normalization of at having an ordinary node over . (Thefact that each fiber of lying over has an ordinary node over followseither from the fact that the -invariant of the singularity is and that,along each branch, 1 nodes of nearby fibers tend to ; or – what is essentiallythe same thing – the fact that the arithmetic genus of the fibers of arezero. This will be verified independently in the course of the proof of Proposition2.7.) The picture is therefore similar to the preceding case: the fibers consist of thedisjoint union of the normalizations of the curves , each attached to at onepoint. The one difference is that, while for 2 the point of attachment ofthe normalizations with can lie over any of the points of intersection ofwith , the point of intersection of the normalization of 1 with can only be thepoint lying over .A typical picture of the original curve and its partial normalization is

this:

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Proof. Consider first of all a reducible curve in our family that does notcontain . By Proposition 2.5, this must be of the form 1 2 where isa general member of the family with 1 2 . In particular, is anirreducible rational curve with nodes, and 1 and 2 intersect transverselyin 1 2 points. Note that

21

so that the total number of nodes of will be

1 2 1 2 1

In other words, along any branch of , all but one of the nodes of will be limitsof nodes of nearby fibers (that is, will be old nodes), while one node of will bea new node. Note also that not any node of can be the new node: that must beone of the points of intersection of the two components 1 and 2; otherwise thefiber of the normalization would be disconnected.In case contains , the analogous computation yields that has

nodes (or nodes and one tacnode of order in case 2 ; hencehas new nodes (or, 1 in 2 ). Then the analysis in the proof of Proposition 2.5shows that in the normalization of the total space of the family, the correspondingfiber will consist of a curve mapping to , plus the normalizations of thecurves , each meeting in one point and disjoint from each other. In particular,all the nodes of arising from points of pairwise intersection of the componentsare old. As for the points of intersection of the components with , there

are two cases. First, if a component has a point of contact of order 1 with, that must be the image of the point ; and all the other points of

will be old nodes of on any branch. On the other hand, if a componentintersects transversely, any one of its points of intersection with can be a

new node.

2.4. SINGULARITIES OF THE TOTAL SPACE

We come to the fourth result, in which we describe the singularities of the totalspace of the normalized family along a given fiber . (Given a fiberover , we will fix a corresponding fiber throughout.)We keep a simplified form of the notation introduced in the statement of Propo-

sition 2.6: we denote by 1 the new nodes of along , coming fromtransverse points of intersection of other components of with ; and by (if itexists) one double point of other than a node, coming from a point of contact oforder 2 of with another component of . We recall that the nearby fibersof our family are smooth near , there will be one point of lying over each

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, which will be a node of , while the nearby fibers have 1 nodes tendingto the point , so that the partial normalization will again have one pointlying over , and that point will be a node of . With all this said, we have

PROPOSITION 2.7 (1) If does not contain , or if contains and the closureof intersects transversely, then is smooth along .(2) In case does contain and the closure of has a point

of intersection multiplicity 2 with , the point of lying over is asmooth point of ; the other nodes of will be singularities of type 1 of

.Proof.We start with the first statement, which is by far the easier. Recall that

by the two previous propositions , being a general point on a codimension-onelocus in , will have or 1 nodes, depending whetherdoes or doesn’t contain . Of these, will be old nodes and the remainingones are new nodes; if is contained in , then the new nodes all lie on . Let1 be the old nodes of and let be any fixed new node. The fiberof lying over will be the partial normalization of at 1 , so

that will certainly be smooth there, and we need only concern ourselves withthe point of lying over .Consider, in an analytic neighborhood of in , the locus of curves that

pass through the base points 1 0 1 and that preserve all of the old nodesof . The projective tangent space to at will be contained in the sub-linearseries of of curves passing through the old nodes of and through1 0 1. This gives a total of 0 1 1 pointswhich, by an argument analogous to the proof of Lemma 2.4, impose independentconditions on the linear series . We only exhibit the proof in case is acomponent of , the other case being similar and easier. Let be the ideal sheafof the subscheme of given by the old nodes 1 , and let : bethe normalization map. We have to show that 1 impose independentconditions on , which will follow (cf. Lemma 2.4) from

1 0

This, by the adjunction formula, is equivalent to

0 1 0

where is the adjoint ideal of . Now notice that the line bundle 1

has degree on the component of lying over , and degree 1 on every othercomponent. Since has degree 2 2 on and negative degree on ,the line bundle 1 cannot have any sections.

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We conclude that is smooth of dimension 1. Notice that this completes theproof of Proposition 2.6, parts (1) and (2a).To analyze the total space of we consider the map from to the versal

deformation space of the node . This has nonzero differential becauseis not a base point of the linear series of curves passing through 1 0 1and through the old nodes of (to see this, the argument above applied tothe ideal sheaf of the union of the old nodes of and will work). Thus thefamily has local equation 0 near ; in particular, it is smoothat .We turn now to the second part, which will occupy us for the remainder of

this chapter. We will start by carrying out a global analysis of the family in aneighborhood of the whole fiber , and then proceed to a local analysis around thepoint specifically. From the global picture wewill establish that, for some integer, the point will be a singularity of type and the points all singularities oftype . The local analysis will then show that in fact we have 1.To carry out the global analysis, we use the family and the map: (cf. Section 2.1), where is the minimal desingularization of thesurface . Since the singularities of the fiber of are all nodes, the totalspace will have singularities of type at each; let us say the point is an

singularity of , and the point an singularity. When we resolve thesingularity at we get a chain 1 1 of smooth rational curves; likewise,is replaced by a chain 1 1 of smooth rational curves. Denoting the

component of meeting at by and the component meeting at by(we are not assuming here that these are distinct irreducible components of ), wearrive at a picture of the relevant part of the fiber of .

We now look at the pull-back of from to . We can write it as

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where is a curve in that meets the fiber only along and , with1 and .

We can use what we know about the degree of this divisor on the variouscomponents of to impose conditions on the coefficients , and . First,since maps components and to points in ,

deg deg 0

Now, each of the curves and has self-intersection 2; so, setting0 and 0 0 , we get

1 2 1 0

for each 1 1; and similarly

1 2 1 0

for each 1 1–in other words, the sequences 0 and 0 ,are arithmetic progressions. On the other hand, the map restricted to the

component is transverse to at ; so the multiplicity at of therestriction to of the divisor is one. This says that 1 1; andsimilarly 1 . Following the arithmetic progression 0 up fromto , we arrive at and hence .The proof of the Proposition will be completed once we show that 1, that

is, that is a smooth point of .Note that this part of the analysis did not rely, except notationally, on the

hypothesis that all but one point of intersection of with the remaining componentsof are transverse. If the points were points of intersection multiplicityof with other components of , we could (always assuming that 1nodes of the general fiber of our family approach ) carry out the same analysisand deduce that for some integer , the point was a singularity of type –loosely speaking, the singularity of at is ‘inversely proportional’ to the orderof contact of with at . The remaining question then would be, is the numberas small as possible, that is, the least common multiple of the ? That is what

we will establish with the following local analysis, which does ultimately rely onthe hypothesis that all but one of the are one.

2.4.1. The versal deformation space of the tacnode.We now carry out the analysisaround the point . The versal deformation of has the vector space

as base,where is the Jacobian ideal of at . Choose local coordinatesfor centered at , so that the curve is given as 0 and the equation ofis

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2 0

The Jacobian ideal of this polynomial is 2 1 . The monomials2 2 and 1 2 1 form a basis for , so that

we can write down explicitly a versal deformation space: the base will bean analytic neighborhood of the origin in affine space 2 1 with coordinates0 1 2 and 0 1 1, and the deformation space will be thefamily , with 2 , given by the equation

20 1 2

2

0 1 22

11 0

Inside we look closely at the closures 1 and of the loci correspondingto curves with 1 and nodes, respectively. We have

LEMMA 2.8. (1) is given in by the equations 0 1 0; inparticular it is smooth of dimension 1.(2) 1 is irreducible of dimension , with sheets crossing transversely at

a general point of .Proof.We introduce the discriminant of the polynomial above, viewed as a

quadratic polynomial in :

22

1 02

4 11

1 0

Note that the map : to the space of monic polynomials of degree 2in with vanishing 2 1 term is an isomorphism of with a neighborhood ofthe origin in : given an equation

22

1 02 4 1

11 0

22 2

2 21 0

we can write

22 22 3

2 32

42 4

22

24 2 4

22 2

8

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and so on, recursively expressing the coefficients as polynomials in the coef-ficients 2 2 . We can then solve for the in terms of the remainingcoefficients 1 0, thus obtaining a polynomial inverse to the map .Now, since the equation above for is quadratic in , the fibers of

are expressed as double covers of the -line. The discriminant is a polynomial ofdegree 2 in , so that the general fiber of , viewed as a double cover ofthe -axis, will have 2 branch points near . To say that any fiber hasnodes is thus tantamount to saying that has double roots – thatis the square of a polynomial of degree . The locus of squares being smooth ofdimension 1 in , we see that is smooth of dimension 1; indeed, itis given simply by the vanishing 0 1 0.Similarly, to say that a fiber has 1 nodes amounts to saying that

has 1 double roots, i.e., that it can be written as a quadratic polynomial intimes the square of a polynomial of degree 1

12

21 0

2 21 0

The Lemma is then proved.

Now we consider the natural map from a suitable analytic neighborhoodof to . To set this up, let 1 be the old nodes of ; since all thesingularities of other than are nodes, this will consist of nodes onand nodes lying off where . Since 1 nodes of thegeneral curve of our family tend to , we have 1. Nowconsider, in an analytic neighborhood of the point , the locus of curvespassing through the 0 1 assigned points 1 0 1 and preserving thenodes 1 of – that is, such that the restriction of the family of curves

to is equisingular at each point of . Since this is a total of0 1 1 points and they impose independentconditions on the linear series , we see that is smooth of dimension at.We then get a natural map : such that 0. We will prove thatis an immersion and that the intersection of with 1 is the union of

with a smooth curve ; moreover and will have contact of order at theorigin. This will conclude the proof of Proposition 2.6; in fact the original family

will be the pullback to of the restriction to of the versal deformation.

To illustrate, here is a representation of the simplest case 2. This does notconvey the general picture, because 1 happens to be proper. Also, thepicture is inaccurate in at least one respect: the actual surface 1 in the deformationspace of a tacnode is also singular along the locus of curves with cusps.

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(Note that we see again locally the picture that we have already observed globallyin the linear series : the closure of the variety of irreducible rationalcurves has the expected dimension; but the locus of rational curves has anothercomponent of equal or larger dimension.)Now that we have a basic picture of the deformation space , the crux of our

argument will be to describe the pullback of the loci and 1 under the map: (or, equivalently, the intersection of with these loci). We start inthe following subsection by saying what we can about the geometry of the map .

2.4.2. The deformations coming from . Let : be as before, denoteby the subspace of given by 0 0. Then we have

LEMMA 2.9. The map is an immersion; the tangent space to the image at theorigin contains the plane 0 1 0 but is not contained in .

REMARK. It is important to note here, and throughout the following argument,that while the loci and 1 are well-defined subsets of the base of the

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deformation space of our tacnode, is not; it depends on the choice of coordinates.It is well-defined, however, as a hyperplane in the tangent spaceto at the origin: it corresponds to the quotientm of the maximalidealm .

Proof. The projective tangent space to at the point is the sublinear seriesof of curves passing through the points 1 and 1 0 1. Thekernel of the differential at of the map is thus the vector space of sectionsof the line bundle vanishing at 1 and 1 0 1 andlying in the subsheaf , where is as before the Jacobian idealof at . The zero locus of such a section will be a curve in the linear series

containing 1 1 0 1 and and so must contain , that is,must be of the form with . Moreover, from the descriptionabove of we see that must also have contact of order at least with atas well as pass through the nodes of lying off and the assigned points1 2 3 0 1. This represents a total of

0 1 1 1

conditions, so we need to show that they are independent to conclude that no suchcurve exists. But they are also a subset of the adjoint conditions of , hence imposeindependent conditions on the series 2 , and henceon the series .The remaining statements of the lemma, that the tangent space to the image

contains the plane 0 1 0 but is not contained in the hyperplane0 0, follow from the facts that the image contains the subvariety andthat not every curve in the linear series containing 1 1 0 1contains .

To complete the proof of Proposition 2.7, we thus have to establish the followinglemma about the geometry of the deformation space .LEMMA 2.10. Let be a smooth, -dimensional variety such that(1) contains ;(2) the tangent space to at the origin is not contained in .Then the intersection 1 consists of the union of and a smooth curvehaving contact of order with at the origin.Our proof of this Lemma is lengthy and roundabout; it occupies the remainder

of this chapter. We give here a summary of the four main steps:First, in 2.4.3 we treat a special case. In Lemma 2.11 we prove Lemma

2.10 by direct calculation when is the linear subspace of given by equations1 1 0. The results of 2.4.3 also appear in [R]; we include ourproof for the sake of completeness.

Secondly, in subsection 2.4.4, we introduce the the blow-up of along, and translate Lemma 2.11 into the statement that the proper transform 1

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of 1 in is smooth at the point Q, and has contact of order with theexceptional divisor of the blow up there.

Third, we use the automorphisms of the deformation space (and its blow-up ) to deduce that 1 is smooth everywhere along the open subset 0

of the fiber of the blow up over the origin, and has contact of orderwith the exceptional divisor (Lemma 2.15). Lemma 2.14 will say that 1

contains , and the global analysis will say that the intersection multiplicity of1 with at a general point of is at least . Then the special case, together

with Lemma 2.13, implies that the intersection multiplicity is at most , and henceexactly , at any point of 0; and that 1 is smooth along 0.

Finally, for any subvariety satisfying the hypotheses of Lemma 2.10,its proper transform in will intersect transversely at a point of 0, and thedesired result – that the intersection of 1 with consists of the union ofand a smooth curve having contact of order with at the origin – follows.

2.4.3. A special case.We will start by considering the intersection of 1 withthe simplest possible variety satisfying the hypotheses of Lemma 2.10, the plane0 given by 1 1 0. We obtain

LEMMA 2.11. The intersection of 1 with 0 consists of the union of withmultiplicity and a smooth curve having contact of order with at theorigin.

Proof. Restricting to 0, we can rewrite the equation of the family more simplyas

20 1 2

2 0

and the discriminant as

22

1 02 4

We need now to express the condition that has 1 double roots. One obviouslysufficient condition is that 0, so that is a square. If we assume 0,however, things get more interesting. To see the locus of 0 2 thatsatisfy this condition, set

22

1 0

and write

2 4 2 2

Now, if 0, the two factors in this last expression have no common factors; soif their product has 1 double roots, each must have a number of double rootsitself: 2 and 2 are polynomials of degree with a combinedtotal of 1 double roots. In fact, this uniquely characterizes and up to aone-parameter group of automorphisms of 1, as we will prove in the following.

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LEMMA 2.12. Let be a nonzero scalar, and let be a positive integer. There isa polynomial of degree , monic with no 1 term, such that

(1) if is odd, the polynomials and each have 1 2double roots.

(2) if is even, has 2 double roots and has 2 2double roots.

In both cases, is unique up to replacing by , where is an throot of unity.

Proof.Suppose that is a polynomial satisfying the conditions of the lemma.Take first the case of 2 1 odd, and consider the map : 1 1 givenby . This is a map of degree , sending the point to , and totallyramified there. In addition the hypotheses assert that over the points in thetarget we have ramification points. The point is, this accounts for a total of

1 2 1 2 2 ramification points, and these are all a map of degreefrom 1 to 1 will have. We have thus specified the covering up to a finite

number of coverings, and our principal claim is that in fact we have describeduniquely, up to automorphisms of the domain.This is combinatorial. The monodromy permutation around the point is

cyclic, while the the monodromy permutations and around and are eachproducts of disjoint transpositions. Our claim that there is a unique such coveringof 1 by 1 amounts then to the assertion that, up to the action of the symmetricgroup by conjugation, there is a unique pair of permutations and , each aproduct of disjoint transpositions, whose product is cyclic of order . Thisis an easy combinatorial exercise. To complete the proof of Lemma 2.12, considerthe effect on of automorphisms of the domain. The requirement that– that is, that is a polynomial! – restricts us to the group of automorphisms

; the requirement that have no 1 term limits us to automor-phisms of the form ; and the fact that is monic says that must be anth root of unity.

Note that, in case is odd, by uniqueness we must have ; that is,will be odd. Similarly, in case is even (where the two branch points have

different multiplicity) we must have , so that will be even. Thiswill not be logically relevant to the following calculation, but will be reflected inthe notation.Back to the proof of Lemma 2.11. Note that if we do not specify the value of

the polynomial will not be unique; we can replace it with for anynonzero scalar . Now, suppose first that 2 is even. Choose 1, and let

22

44

0

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be the polynomial satisfying the conditions of the lemma. Then any collection0 1 2 with 0 such that the discriminant

22

1 02 4

has 1 double roots, must be of the form

0 0 1 0

21

2 3 0

42

4

and so on, ending with 2 2; with finally 4. This is then aparametric representation of the closure of the intersection 0 1 .It is obviously a curve; the fact that it is smooth is visible from the coordinate

2 2; and we see that it has contact of order with from theexponent in the expression for .Finally, in case 2 1 is even we get a similar expression. Let

22

44

1

be the polynomial satisfying the conditions of the lemma for 1. Then anycollection 0 1 2 with 0 such that the discriminant

22

1 02 4

has 1 double roots must be of the form

0 0 1 1

2 0 31

3

ending with 2 2; again we have 4. So once more we see thatis a smooth curve having contact of order with at the origin.

Let us now prove Lemma 2.10 and Proposition 2.7 in this special case. First,in the case 2 even, the restriction of the family to hasequation

22

2 24

4 20 4 0

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We can think of the total space of this family as a double cover of the -plane, with branch divisor the zero locus of the discriminant

22 2

44 2

02

By hypothesis, for each value of the polynomial is the product of the square ofa polynomial of degree 1 and a quadratic polynomial . Since iseven, 2 and must each be; and given the homogeneity of with respect to andwe see that we can write

2 21

2 22

2 21

2 2

for suitable constants 1 1 and . For example, in case 2, the equationof is simply

2 2 2 4 0

and the discriminant is just 2 2 2 . In general, the branch divisor ofover the -plane will be simply a union of the -axis, with multiplicity 2; 1parabolas tangent to the -axis at the origin, each with multiplicity 2; and one moreparabola tangent to the -axis at the origin and appearing with multiplicity 1. Thedouble cover will thus be nodal over the double components of this branchdivisor, and smooth elsewhere.Finally, the normalization of the total space will be the double cover of

the -plane branched over the single component of multiplicity 1 in the branchdivisor; that is, it will have equation

2 2

and in particular, since the component 2 is smooth, will be smooth aswell, establishing Proposition 2.7 for this particular family.The picture in case 2 1 is odd is exactly the same: here has equation

22

2 24

4 21 4 0

with discriminant

22 2

44 2

12

21

2 22

2 2 2 2

for suitable constants 1 and . For example, in case 3, the equationof will be

2 3 3 3 0

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(we are scaling here to make the coefficients nicer), and the discriminant is just

3 3 2 4 3

6 6 4 9 2 2 4 3

2 2 2 4

In general, for odd the branch divisor of over the -plane will be simplya union of parabolas tangent to the -axis at the origin, each with multiplicity 2;and one more parabola tangent to the -axis at the origin and appearing with multi-plicity 1. As before, the normalization of the total space will be simply thedouble cover of the -plane branched over the single component 2 ofmultiplicity 1 in the branch divisor; and as before, since this component is smooth,

will be smooth as well, establishing Proposition 2.7 in this case.

2.4.4. The geometry of the locus 1. In order to focus on the essential aspectsof the geometry of 1, and in particular to remove the excess intersection of

1, we will work on the blow-up : Bl of along. To express our results, we have to introduce some notation. We will denote

by 1 the exceptional divisor of the blow up, and by 1 and theproper transforms of 1 and in .Our goalwill be to describe the intersection 1 : 1 . The fibers of

over are projective spaces 1 with homogeneous coordinates 0 1;we will denote the fiber 1 0 of over the origin by , by 0 the open setgiven by 0 0, and by the point of with coordinates 1 0 0 (this is thepoint of intersection of with in the example above).Note that there is a more intrinsic characterization of : the tangent space toat the origin is the subspace of of polynomials divisible by , so that

– the projectivization of the normal space – is just the space of polynomials inmodulo those vanishing to order at 0 0 and modulo scalars. In these

terms, 0 is simply the subspace of polynomials not vanishing at the origin andthe point corresponding to constants.To study 1 we will take advantage of an equivariant action of the mul-

tiplicative group on the map . Explicitly, for any , we defineautomorphisms of and of simultaneously by

2

Since and commutewith the projection , preserves the loci .

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In particular, since preserves , it lifts to an automorphism of ; and sincepreserves 1 the lifted action will preserve 1. We can read off from

the above expression the action of on the fiber of : over the origin:in terms of the homogeneous coordinates 0 1, we have

: 0 110

21 2 1

The key fact about this action, for our present purposes, follows immediately fromthis description:

LEMMA 2.13. Every orbit of the action of on that intersects 0 contains thepoint in its closure.We are now prepared to state and prove our main lemma on the geometry of1 and 1.

LEMMA 2.14. (1) The fibers of 1 over are unions of linear spaces.(2) For any arc in tending to the origin, the limiting position of the

fiber of 1 over is contained in the complement of 0.(3) itself is an irreducible component of 1.Proof. The proof is by induction on , using Lemma 2.11.First we introduce a natural stratification of the locus . Identifying

with the space of monic polynomials of degree in with no 1 term, welook at the loci of polynomials with roots of given multiplicity: for any partition

1 2 we define the locus 1 by

1 : 0 2 0 0 : 22

0

1 1 2 2 for some distinct 1

Note that the codimension of 1 in is 1 .Suppose is any point of other than the origin. Say lies in the stratum1 , and write the corresponding polynomial as

1 1 2 2

with 1 distinct. The fiber of over is a reducible curveconsisting of two branches, the -axis 0 and the curve 1 1

2 2 , which meet at the points 1 1 0 0with multiplicities 1 .Let be the versal deformation spaces of the singular points

. By the openness of versality the natural map from a neighborhoodof in to the product has surjective differential at (the fibers are theequisingular deformations of , in which only the locations of the points on the-axis vary). Let 1 and be the loci in analogous to 1and in , that is, the closures of the loci of deformations of the singular points

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with 1 and nodes near respectively. Then in the neighborhoodof , we have

11 2

and

11

11 1

In other words, the locus 1 will have branches in a neighborhood of , eachcontaining , along the th of which the fibers of will have nodestending to for each and 1 nodes tending to .We can use this description to give a more intrinsic characterization of the fiber

1 of over the point , analogous to the one given above for .Briefly, is the projectivization of the normal space to in at , which isthe product of the normal spaces to the in at the origin; this is just thespace of polynomials on the -axis modulo those vanishing to order at foreach .We may now apply the induction hypothesis to describe, in these terms, the

fiber of 1 over . By the statement of the Lemma for , the propertransform of the th branch of 1 will intersect in the linear subspace of

corresponding to polynomials vanishing to order at for each ; theintersection with with the proper transform of 1 itself will be the union ofthese linear subspaces.This establishes part (1) of the Lemma. Now say that is any arc in

tending to the origin; will lie in some stratum 1 for all small0. As goes to zero, the singular points of approach the point ,

so that the limiting position of the intersection with of the proper transformof the th branch of 1 will be simply the linear space of polynomials whoserestriction to the -axis vanishes to order at ; in particular, it is containedin the hyperplane 0 0 of polynomials vanishing at . We have thusproved parts (1) and (2) of the Lemma, given part (3) for all .Finally, we need to prove for each new value of that is an irreducible

component of 1. Now, by Lemma 2.11, the point 1 0 0 liesin 1. But we have completely described the closure in 1 of the inverseimage 1 0 of the complement of the origin, and is not on it. mustthus lie on an irreducible component of 1 not meeting 1 0 , thatis to say, an irreducible component of 1 contained in ; since 1 has puredimension 1, this irreducible component must be itself.

For example, here is a picture of 1 in the case 2. In this case 1 has onlytwo components, and a component finite of degree 2 over 2.

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Finally, we deduce

LEMMA 2.15. (1) 1 is smooth everywhere along 0.(2) The intersection multiplicity of 1 and along is .Proof. We use the analysis carried out in Lemma 2.11. Let 0 be the proper

transform of the linear space 0 in . Since no component of 1 other thanpasses through , the only component of the intersection 0 1 containingwill be the proper transform of the curve described in Lemma 2.11.

Since this is smooth, and the intersection 0 1 is proper in a neighborhoodof ( 0 and 1 each have dimension in the 2 1 -dimensional , andtheir intersection is locally a curve) it follows that 1 must be smooth at . ByLemma 2.13, then, it must be smooth at every point of 0.For the second statement, notice that Lemma 2.11 asserts that this is true when

restricted to the proper transform 0, and it follows that it is true on 1

End of the proof of Lemma 2 10 and Proposition 2 7. We shall now concludethat the intersection of 1 with any subvariety satisfying the hypothesesof Lemma 2.10 – and in particular, the image – is the union of and asmooth curve , such that has contact of order with at the origin. Noticethat this will conclude the proof of Proposition 2.6 as well. To begin with, theproper transform of in intersects in a section, crossing at some point ;we likewise have from Lemma 2.9 that 0. 1 is then smooth at . Sincethe tangent space to 1 at contains the tangent space to and the tangentspace to at is complementary to the tangent space to , 1 and intersecttransversely in a smooth curve in a neighborhood of ; since that curve is nottangent to at , its image 1 is again a smooth curve. Finally, theintersection number of with in will be the intersection number of 1,and at ; which by Lemma 2.15 will be . We have thus completed the proof

of Lemma 2.10 and hence that of Proposition 2.7.

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3. Formulas

Before we prove our formulas, we need a simple result on the order of zeroes andpoles of the cross-ratio function .

3.1. A REMARK ON THE CROSS-RATIO FUNCTION

Suppose we are given a family : over a smooth one-dimensional base, whose restriction : 1 to the complement 0 of a

point 0 is a family of smooth rational curves; and four sections : ,disjoint over . We get a map : 0 4, which then extends over ; and theproblem is to determine the coefficient of the point in the pullback via of theboundary components of 0 4. To put it another way, the cross-ratio of the foursections 1 3 2 4 defines a rational function on and hence on ; and we asksimply for the order of zero or pole of this function at 0.Wewill answer this in terms of any completion of our family to a family of nodal

rational curves. Recall first of all the set-up of Section 2.1: we have a resolutionof singularities of the total space of our family, such that is afamily of nodal curves and the extensions of the sections to are disjoint. Wethen proceed to blow down ‘extraneous’ components of to arrive at the minimalsmooth semistable model of our family: that is, a family such thatis smooth, the fibers are nodal, the sections are disjoint and isminimal with respect to these properties. Finally, we blow down the intermediatecomponents in this chain to arrive at a family of 4-pointed stable curves.The special fiber of this family will have just two components (or one, if 0),with a singularity of type at the point of their intersection.In these terms we prove

LEMMA 3.1. If the sections 1 and 2 (respectively, 1 and 3) meet the samecomponent of , then the point 0 is a zero (respectively, pole) of multiplicity ofthe function .

Proof.We will consider the case where 1 and 2 meet the same component of. Note first that if we blow down the component of meeting 1 and 2, we

arrive at a smooth family, that is (replacing if necessary by a neighborhood of 0in ), a product 1. (Equivalently, we could arrive at this family by blowingdown the component of meeting 1 and 2, then doing the same thing on theresulting surface, and so on times.) 3 and 4 will remain disjoint from each otherin this process, and disjoint from 1 and 2; but 1 and 2 will meet each other withcontact of order : in other words, we can choose an affine coordinate on 1 anda local coordinate on centered around 0 so that the sections are given by

1 ; 2 0; 3 1; and 4

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The cross-ratio function is then 1 , which takes on the value 0 withmultiplicity at 0.

3.2. THE RECURSION FOR 2

Let be any effective divisor class other than on the ruled surface 2 . Weare going to find a formula for the degree of the variety . Toset this up, we start by choosing as usual 0 1 general points on , whichwe label 1 2 3 0 1, and consider the one-parameter familyof curves passing through 1 2 3 0 1 . As before,we let be the normalization of and the normalization of the pullbackfamily. Next, we fix general curves 3 and 4 in the linear series , andadopt as usual the convention that we will choose points 3 and 4 on the curvesof our family lying on 3 and 4 respectively. Making the corresponding base

change, we arrive at a family ; as before, we will denote by the minimaldesingularization of and by the smooth semistable model.Then we calculate the degree of the cross-ratio map : 0 4

1 in twoways by equating the number of zeroes and poles of . We get one contributionto the degree of 0 immediately from the curves in our family that happento pass through either of the two points of intersection of 3 with 4; this gives atotal contribution of 2 to the degree of 0 .The remaining zeroes and poles of necessarily correspond to reducible curves

in the family . There are two types of these: those that contain and thosethat don’t. Consider first a reducible curve in our family that does not contain .By Proposition 2.5, this must be of the form 1 2 where is a generalmember of the family for some pair of divisor classes 1 and 2 addingup to . In particular, is an irreducible rational curve with nodes, and1 and 2 intersect transversely in 1 2 points. Moreover, by Proposition

2.6, the curve will consist of 1 2 smooth branches near the point ,corresponding to the points of intersection of 1 and 2; thus there are 1 2points in the normalization lying over each such point .How does such a fiber of the family contribute to the degrees of either0 or ? It depends on how the points are distributed. If three or four

lie on one component, it does not contribute to either, but if there are two on eachit may: for example, if 1 and 2 lie on the same component – say 1 – of , and3 and 4 on the other, we get a zero of . Now, as we observed in the proof ofProposition 2.5, each component of must contain exactly 0 of the points1 2 3 0 1. If 1 is to contain 1 and 2, it will contain 0 1 2 ofthe points , and 2 will contain the remaining 0 0 1 1 0 2 .Thus, to specify such a fiber, we have first to break the 0 3 points intodisjoint sets of 0 1 2 and 0 2 . The curve 1 can then be any of the 1irreducible rational curves in the linear series 1 passing through 1, 2 and the

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first set, while 2 can then be any of the 2 irreducible rational curves in thelinear series 2 passing through the second set. Altogether, then, we see that therewill be

1 20 30 1 2

points in of this type, and correspondingly

1 2 1 20 30 1 2

such points in the normalization . Finally, if a fiber of lying over sucha point of is to contribute to 0 , we have to choose 3 and 4 to lie on 2,that is, to be any of the 2 points of intersection of 2 with 3 and 4respectively. There are thus a total of 2

2 fibers of of this type lyingover each such point of .To complete the calculation of the contribution of fibers of this type to the degree

of 0 , we observe that the fiber of the normalization over such a point willhave two components, the normalizations of the curves , meeting at one point(the point of each lying over the new node). Moreover, by Proposition 2.7, the totalspace will be smooth at such a point; and it follows by Lemma 3.1 that thecorresponding point of will be a simple zero of . In sum, then, fibers ofof this type contribute a total of

1 2 1 20 30 1 2 2

2

to the degree of 0 .The contribution of such fibers to the degree of the divisor is found

analogously, the only difference being that, in order to get a pole of the cross-ratio,the points 1 and 3 must lie on one component – say 1 – of , while 2 and4 will lie on the other. Thus, instead of breaking the 0 3 points intosubsets of 0 1 2 and 0 2 , we divide them into subsets of 0 1 1and 0 2 1; and instead of 1 2

0 30 1 2 such points in of this

type we have 1 20 30 1 1 . Similarly, instead of choosing 3 among

the 2 points of 2 3, we choose it among the 1 points of 1 3;so that instead of 2

2 zeroes of the cross-ratio lying over each such point ofthere will be 1 2 . Again, each pole of the cross-ratio corresponding

to a fiber of this type will have multiplicity one; so the total contribution to thedegree of is

1 2 1 20 30 1 1 1 2

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It remains to add up the number of zeroes and poles of coming from membersof our family containing . Proposition 2.5 describes all such curves, and thedescription is particularly simple, given that we are on the surface 2 . There areonly two types: a degenerate member of our family must consist either of

(1) the union of and an irreducible rational nodal curve 1 , sim-ply tangent at one point (which will be a smooth point of 1) and meetingtransversely elsewhere; or

(2) the union of and two curves , which will correspond to generalpoints of the varieties for some pair of divisor classes 1 and 2 with1 2 . In particular, 1 and 2 will intersect each other and

transversely.

Now, we can forget about curves of the first type; in fact, since cannot containany of the points 1 4, these will be distinct points of 1. Hence the cross-ratio function will not be zero or infinite at such a point of . On the other hand,fibers of the second type may contribute. To see what our family looks like in aneighborhood of such a curve, recall first that by Proposition 2.6, as we approachalong any branch of , all the points of intersection of 1 and 2, as well as all

but one of the points of intersection of each curve with , will be old nodes;exactly one of the points of intersection of each with will be new. The fiberof the normalized family will thus consist of the normalizations of 1and 2, each meeting a copy of in one point and disjoint from each other

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Recall also that the total space of will be smooth along such a fiber.Again, can’t contain any of the points , and if three or four lie on either

curve the corresponding point of will be neither a zero nor a pole of ; butwe may get a contribution if two are on each . Specifically, if 1 and 2 lie onone component – say 1 – and 3 and 4 on the other, we get a zero of ; while if1 and 3 lie on a component – again, call this one 1 – and 2 and 4 on the other,we get a pole of . That said, we can count the number of such fibers exactly as inthe preceding case.We do the zeroes first. We begin by specifying a point in – that is, we

break the points into subsets of size 0 1 2 and 0 2 respectively, andchoose 1 among the 1 irreducible rational curves in 1 through 1, 2 andthe first set and 2 among the 2 irreducible rational curves in 2 throughthe second set. Next, a point in : we can take any of the 1 2 pointsof lying over . Lastly, we have to choose 3 and 4 among the 2points of intersection of 2 with 3 and 4 respectively. We have, in sum,

1 20 30 1 2 1 2 2

2

zeroes of of this type.The poles of the cross-ratio coming from such curves are counted in the same

way; the differences being exactly as in the preceding case: in specifying the pointwe have to choose a subset of 0 1 1 rather than 0 1 2 of the

points ; and 3 must be chosen among the 1 points of 1 3. There arethus a total of

1 20 30 1 1 1 2 1 2

poles of this type.There is one important difference between this case and the previous, however:

here, the fiber of the normalization has three components, with thecomponents 1 and 2 containing the points separated by the component .Since by Proposition 2.7 the total space is smooth, we see by Lemma 3.1 that

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such points will be double zeroes and poles of . The contribution to the degreesof these divisors coming from fibers of this type is thus twice the number of suchfibers.We can now calculate the degree of the divisors 0 and . We havedeg 0 2

1 21 2

1 20 30 1 2 1 2 2

2

21 21 2

1 20 30 1 2

1 2 22

Similarly,deg

1 21 2

1 20 30 1 1 1 2 1 2

21 21 2

1 20 30 1 1

1 2 1 2

To express the final result we introduce the notation

1 2 : 1 20 30 1 1 1 2

0 30 1 2 2

2

We now write deg 0 deg and solve the resulting equation forto arrive at the recursion formula for on 2 .

THEOREM 3.2 Let Pic 2 and let be the number of irreduciblerational curves in the linear series that pass through 0 general points of2 ; then we have

12

1 21 2

1 2 1 2

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1 21 2

1 2 1 2

3.3. THE CLASS 2 ON

We now analyze the linear series 2 on the ruled surface for any . We arriveat a closed-form expression for 2 rather than a recursion. This is clear: sinceevery linear series on with 2 that actually contains irreducible curveshas arithmetic genus 0, we can say immediately how many degenerate fibers ofeach type there are in our one-parameter family of curves in 2 .The dimension of 2 is 3 2. The arithmetic genus of the curves in the series

is 1, so that the expected dimension of the Severi variety is 0 2 2 3.This is in fact the actual dimension: any irreducible nodal curve 2 is bedisjoint from (if it met , it would contain it, having intersection number 0 withit); so that the nodes of impose independent conditions on 2 .So, we choose as usual 2 2 general points on , whichwe label 1 2 3 ,

2 2 and consider the one-parameter family of curves 2 passing through1 2 3 2 2 ; we denote this family . As before, we let be

the normalization of and the normalization of the pullback family.Next, we fix general curves 3 and 4 in the linear series , and adopt theconvention that we choose points 3 and 4 on the curves of our family lyingon 3 and 4 respectively. Making the corresponding base change, we arrive at afamily ; as before, we denote by the minimal desingularization of andby the smooth semistable model.Now we consider the cross-ratio map : 0 4

1 as before; we shallobtain a formula for from

deg 0 deg

Of course, we get one contribution to the degree of 0 from the curves in ourfamily that pass through any of the points of intersection of 3 with 4; thisgives a total contribution of 2 to the degree of 0 .The remaining zeroes and poles of correspond to reducible curves in the

family . As before we look first at curves that do not contain . They can onlybe of the form 1 2 where 1 and 2 are each linearly equivalent to .Such a fiber of the family can be either a pole or a zero of , depending

of course on how the points are distributed. Namely, if 1 and 2 lie on the samecomponent and 3 and 4 on the other, we get a zero. To specify such a fiber,we simply have to break the 2 points into disjoint sets of 1 and 1.The two components of the curve will be the (unique) curve in the series

containing 1, 2 and the first subset; and the unique curve in the seriescontaining the second subset.

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Next, we have to count the number of points of lying over each point ofcorresponding to such a curve. Since the general curve of our family has 1 nodes,and the nodes of 1 2 impose independent conditions on the series 2 ,the curve has (smooth) branches at the point ; thus the normalization ofhas points lying over (cf. Proposition 2.6). Moreover, for each of these pointsthere will be a point of for every possible choice of points 3 and 4; these can beany of the points of intersection of the component not containing1 or 2 with 3 and 4 respectively. There are thus a total of 2

12 fibers

of of this type.Now, for each such fiber of , the fiber of will be simply the

normalization of at the 1 old nodes.

In particular, it has just two components and is stable, and as we have seenalready is smooth at the node of such a fiber. Each such point is thus a simple zeroof ; so the total contribution to the degree of 0 of such curves is

21

3

Similarly, we could have 1 and 3 on the same component and 2 and 4on the other; in this case, we get a point of . The only difference in thiscase is that to specify such a fiber, we have to break the 2 points into twodisjoint sets of points apiece. The two components of the curve will be the(unique) curves in the series containing 1 and the first subset; and the uniquecurve in the series containing 2 and the second subset. The rest of the analysisis exactly the same – for each such curve, we get 3 points of , each of which isa simple pole of the cross-ratio function – so the total contribution to the degreeof of such curves is

2 3

The remainder of the calculation will be spent evaluating the contributions tothe degree of the pullbacks of the boundary components of 0 4 coming from thecurves in our original family containing . As we have indicated, these curves are

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of 1 types: for each 1 1 we have a finite number of curves inour family consisting of the sum of , fibers 1 of 1 and anirreducible curve linearly equivalent to , with having a singlepoint of -fold intersectionwith , as pictured below. For each of these types,there are a number of possibilities for the distribution of the points 1 4 onthe various components. For each such distribution corresponding to points inthe inverse image of a boundary component of 0 4, we consider the numberof fibers of that type and the coefficient with which the corresponding points

appear in the divisor ; in the end we will sum up the contributions toarrive at an expression for 2 on .

1 1 2 ; 3 4 . In such a curve, the fiber components must eachcontain one of the points . To specify such a curve, then, we must first choose asubset of of the 2 points and take the unique curve in the linearseries containing them. Next, we have to single out one of these points,and label the corresponding fiber . At this point 3 and 4 will be determined,as the unique points of intersection of with the curves 3 and 4 respectively.Finally, we choose a curve passing through the remaining2 of the points and having a point of -fold tangency with . (Notethat the ordering of the points chosen to lie on fibers does not matter; all thatcounts is which one is chosen to lie on .)Now, the linear series cuts on the curve 1 the complete

linear series 1 . This linear series is parametrized by the space ofpolynomials of degree on 1 modulo scalars; and in that projective space thedivisors consisting simply of times a single point – that is, th powersof linear forms – form a rational normal curve. In the linear series ,then, the locus of curves having a single point of -fold intersectionwith is a cone over a rational normal curve in (with vertex the subseries

2 of curves containing ); in particular, ithas degree . There are thus exactly curves linearly equivalent to

passing through 1, 2 and the remaining 2 of the pointsand having a point of -fold tangency with . In sum, the number of fibersof this type in our family is 2 .

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It remains to determine, for each such fiber of our family, the coefficient withwhich the corresponding point appears in the pullback divisor 0 . To dothis, we need to know the local geometry of the family near ; in particular,we need to have the picture of the corresponding fibers of the familiesand . For the first, the only thing we need to know is which of the singularpoints of the fiber are limits of nodes of nearby fibers and (in the case of the pointof intersection of with ) how many. The answer, as provided in Proposition2.7, is that the points of intersection of with the fibers are all limits of nodeson nearby curves; and the remaining 1 nodes of the general fiber ofthe family tend to the point of intersection of with . When we normalize thetotal space of the family, then, the curves and are pulled apart; and the pointof intersection of with becomes a node, so that the fiber of overconsists of a rational curve with fibers 1 and the curve attached.But as we also saw in Proposition 2.7, will not be smooth: at the point lying

over each point of intersection of with a fiber , will have a singularity of type1. Resolving each of these, we arrive at this picture of the fiber of

over

Finally, we can blow down the extraneous curves and for to arriveat the picture of the fiber of the family of semistable 4-pointed curveswith smooth total space

Inasmuch as there are rational curves in the chain connecting thecomponents and containing the pairs 1 2 and 3 4 , each such fiber

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represents a point of multiplicity 1 in the divisor 0 . In sum, then, thefibers of this type contribute a total of

1

1

2 1

to the degree of 0 .2 1 3 ; 2 4 or 2 4 ; 1 3 . In the first of these

cases we are simply exchanging the locations of 2 and 3 to obtain a fibercorresponding to a point in the inverse image ; this will affect the count ofthe number of such fibers, but not the final configuration on the semistable modelwith smooth total space, so the multiplicity of each such point in the divisor 0will be as in the preceding case 1.The difference here is that, because the fixed point 2 lies on one of the fiber

components, we can put the remaining fiber components through only 1 of thepoints ; at the same time, we can put the curve through 1 and the remaining2 1. To specify such a curve, then, we must first choose a subset of 1of the 2 points and take the unique curve in the linear seriescontaining them and 2; the component of containing 2 we call . As in thepreceding case, there will be exactly curves in the linear seriespassing through the remaining 2 1 points and the point 1 and having apoint of intersection multiplicity with ; the curve can be any of these.At this point 4 will be determined, as the unique point of intersection of withthe curve 4; while 3 can be taken to be any of the

3 2

points of intersection of with 3. The number of fibers of this type in ourfamily is thus 2

1 2 .As we said, each such fiber of our family is a pole of order 1 of the

cross-ratio function . Finally, since exchanging 1 with 4 (as in the second caseabove) yields an identical result, the total contribution of the fibers of these typesto the poles of is

21

1

21

2 1

3 1 2 ; 3 and 4 , . This case is very similar to the first;again, we have first to select a subset of of the 2 points and take theunique curve in the linear series containing them. We then have to single outtwo of these points, and label the corresponding fibers and , which in turndetermines the points 3 3 and 4 4. Finally, we take as beforeto be any of the curves in passing through 1 and 2 and the

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remaining 2 of the points and having a point of -fold tangencywith. Thus, the number of fibers of this type in our family is 2 1 .At this point, we see another difference with the preceding case: here, to arrive

at the fiber of the family of semistable 4-pointed curves with smooth total spacewe blow down the curves and for all including and , to arrive at thesimpler fiber

Since this is already semistable, each such fiber represents a simple zero of .In sum, then, the fibers of this type contribute a total of

1

1

21

to the degree of 0 .For the remaining cases we indicate only the distribution of the points and the

resulting contribution; the reader should find no difficulty in supplying the details.4 1 3 ; 2 and 4 , ; or 2 4 ; 1 and 3 ,. The fibers of this type contribute a total of

21

1

21

1 2

to the degree of .5 3 4 ; 1 and 2 , . Such fibers contribute a total of

1

1

22

2 2

to the degree of 0 .6 1 , 2 and 3 4 , ; or 1 , 2 and 3 4 ,. We get a contribution of

21

1

21

1 2

to the degree of 0 .

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7 1 , 3 and 2 4 , ; or 2 , 4 , and 1 3 ,. These fibers contribute a total of

21

1

21

1 2

to the degree of .8 3 , 1 and 2 4 , ; or 4 , 2 , and 1 3 ,. Contributing a total of

21

1

22

2 2

to the degree of .We come now to the last three cases, those in which none of the four points

lie on . The next one is the last to contribute to the degree of 0 .9 1 , 2 and 3 4 , . The total contribution of

such fibers is

1

1

22

2 2

10 1 3 and 2 4 ; or 2 4 and 1 3 ,. Counting both possible exchanges,we see that the total contribution

of such fibers to the degree of is

21

1

22

2 2

11 1 3 and 2 4 , The total contribution of such fibers tothe degree of is thus

1

1

22

2 2

We are now ready to add up all the contributions to the degrees of 0 and, equating the results and solving for 2 . We have

deg 0 2 3 21

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1

1

2 1 1

21

2 1

22

2 2 2

while on the other hand

deg

3 2 1

1

21

2 2 1

2 2 1 2 1

22

2 2

2 2 2

Combining these, we arrive at the expression

2 3 2 21

where1

1

2 21

2 2

22

(Note that we can now enlarge the formal limits of summation to include 0without affecting the sum; this will be convenient in the following calculations.)To reduce this further, we separate it into two terms: we write , where

1

04 2 2

1

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and1

0

22

21

22

The expression for telescopes nicely and we have simply1

0

2 2

As for , we can combine that with the remaining two terms in the expression for2 , and together they simplify. To start with, observe that

3 2 21

2 21

Now, combining this with the expression for above, we have

2 21

1

04 2 2

1

2 21

422

823

4 420

Now we use standard binomial identities to reduce this to

2 21

1

04 2 2

1

2 2 21

22 1

232 1

32 10

21

0

2 2

Finally, we can combine this and the expression above for : we have

2 3 2 21

21

0

2 2 1

0

2 2

1

0

2 2 2

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We have therefore proved the

THEOREM 3.3. Let 2 be the number of irreducible rational curves in thelinear series 2 on passing through 2 3 points, then

21

0

2 2 2

For example, on 2 we have 2 10; on 3 we have 2 69; and on4 we have 2 406 and so on.We now show how to arrive at an expression of 2 on as a coefficient

of a simple generating function. We simply write out the sum involved, and thentelescope it using the standard binomial relations as before: that is, we write

22 2

142 2

292 2

32 2 2

0

and use the relations

2 21

2 22

2 31

32 2

232 2

332 3

2

and so on to rewrite this as

22 3

132 3

262 3

3

12

2 30

1

0

12

2 3

We can also think of this as the coefficient of in the product of the power series

2 31 2 3

and

22

11 3

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so that we can write 2 on as the coefficient

21 2 3

1 3

3.4. A FORMULA FOR

We conclude our paper with a formula for the general ruled surface . Herewe define the function 1 1 giving the contribution to the cross-ratio corresponding to a given decomposition 1 2 or

1 2, with . Recall that the variety is the closure inof the locus of irreducible rational curves that have a point of contact of order

with the exceptional curve . We define

1 1 :

0 310 1 1 2

0 2 1 30 3

3

1

1

2

2 3

2

0 310 1 2 2

0 230 3

2

2 1 11

11 2

In these terms, we can state

THEOREM 3.4. Let be a divisor on the surface . Let be the num-ber of irreducible rational curves in that pass through 0 general pointsof . Then

1 2

1 2 1 1 1 2

2 1 2

1 : 11 1

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Acknowledgments

Our interest in these questions grew out of conversations with Enrico Arbarello,Ciro Ciliberto and Bill Fulton, to whom we are very grateful.

References[A] Artin, M.: Lectures on deformations of singularities. Tata Inst. Lecture Notes in math. and

phys. vol. 54. Springer-Verlag, Berlin and New York, 1976.[C] Caporaso, L.: Counting curves on surfaces: a guide to new techniques and results. To

appear on Proceedings of the Second European Congress of Mathematicians, Budapest1996.

[CH] Caporaso, L. and Harris, J.: Enumerating rational curves: the rational fibration method.1995, Preprint.

[CM] Crauder, B. and Miranda, R.:Quantum cohomology of rational surfaces. TheModuli Spaceof Curves, Progress in mathematics 129, Birkhauser, 1995, pp. 33–80.

[DI] Di Francesco, P. and Itzykson, C.:Quantum intersection rings. TheModuli space of curves,Progress in mathematics 129, Birkhauser, 1995, pp. 81–148.

[H] Harris, J.: On the Severi problem. Invent. Math. 84 (1986), pp. 445–461.[K] Kollar, J.: Rational Curves on Algebraic Varieties. Springer 1996.[KP] Kleiman, S. and Piene, R.: Private communication.[K] Kontsevich, M.: Enumeration of rational curves via torus action. The Moduli Space of

Curves, Progress in mathematics 129, Birkhauser, 1995 pp. 335–368.[KM] Kontsevich, M. and Manin, Y.: Gromov–Witten classes, Quantum Cohomology and Enu-

merative Geometry. Preprint MPI-94-21.[R] Ran, Z.: Enumerative geometry of singular plane curves. Invent. Math. 97 (1989), pp. 447–

465.[RT] Ruan, Y. and Tian, G.: A mathematical theory of quantum cohomology, J. Diff. Geom.

42(2) (1995), pp. 259–367.[Z] Zariski, O.: Dimension-theoretic characterization of maximal irreducible algebraic system

of plane nodal curves of a given order and with a given number of nodes. Am. J. Math.104 (1982), pp. 209–226.

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