# On a fiber bundle version of the Caporaso-Harris formula

Indranil Biswas  Department of Mathematics, Shiv Nadar University, NH91,
Tehsil Dadri, Greater Noida, Uttar Pradesh 201314, India
<EMAIL_ADDRESS><EMAIL_ADDRESS>,  Apratim Choudhury
Institut für Mathematik, Humboldt-Universität zu Berlin, Unter den Linden 6,
Berlin- 10099, Germany<EMAIL_ADDRESS>,  Nilkantha Das  Stat-
Math Unit, Indian Statistical Institute, B.T.Road, Kolkata-700108, India.
<EMAIL_ADDRESS>and  Ritwik Mukherjee  School of Mathematical
Sciences, National Institute of Science Education and Research, Bhubaneswar,
An OCC of Homi Bhabha National Institute, Khurda 752050, Odisha, India.
<EMAIL_ADDRESS>
###### Abstract.

The Caporaso-Harris formula gives a recursive algorithm to enumerate
$\delta$-nodal degree $d$ curves in $\mathbb{P}^{2}$. The recursion is
obtained in terms of curves of lower degree that are tangent to a given
divisor. This paper presents two generalizations of this method. The first
result is on enumeration of one cuspidal curves on $\mathbb{P}^{2}$, and the
second result is an extension to the fiber bundle setting. We solve the
question of counting the characteristic number planar nodal cubics in
$\mathbb{P}^{3}$ by extending the idea in [2].

###### 2010 Mathematics Subject Classification:

14N35, 14J45

###### Contents

  1. 1 Introduction
  2. 2 Main Results
  3. 3 Caporaso-Harris formula for one nodal curves
    1. 3.1 Cycle version
    2. 3.2 Numerical invariants: enumerating curves with a free node
    3. 3.3 Numerical invariants: enumerating curves with the node lying on a line
  4. 4 Extension of Caporaso-Harris to one cuspidal curves
  5. 5 Extension of Caporaso-Harris for a family of nodal cubics
  6. 6 Computation of the characteristic number of conics in $\mathbb{P}^{3}$
    1. 6.1 Smooth conics
    2. 6.2 Nodal conics
    3. 6.3 Smooth conics tangent to a line
  7. 7 Acknowledgements

##  1\. Introduction

The Caporaso-Harris formula was a landmark achievement in the field of
enumerative geometry. Very simply stated, it solves the following problem: how
many $\delta$-nodal degree $d$ curves are there on $\mathbb{P}^{2}$ that pass
through $\frac{d(d+3)}{2}-\delta$ generic points? Their formula [2, p. 348,
Theorem 1.1] is a recursion in terms of curves of lower degree which can be
tangent (of higher order) to a given line. More generally, they obtain a
formula for the characteristic number of curves in $\mathbb{P}^{2}$, having
$\delta$-nodes, that intersect a given line tangentially at multiple points.

The underlying geometric idea behind the Caporaso-Harris formula is a
degeneration argument. The idea is to move the constraints (points) to the
line and compare the resulting numbers. Moving the point to the line usually
results in a degenerate locus involving a curve of a lower degree. Iterating
this process, one obtains a recursive formula.

A natural question is to count curves with higher singularities. Addressing
this, there are all completely different approaches and have nothing to do
with the degeneration argument of [2]. It is therefore natural to ask: can the
degeneration argument of [2] be used to enumerate curves with higher
singularities?

The goal of this article is twofold. This paper answers the above question
when the singularity is a cusp. We expect that with further effort, the
degeneration argument can be carried forward to enumerate curves with even
more degenerate singularities (such as tacnodes, triple-points etc).

The second result of this paper is explained with some background to put it in
perspective. A typical question in enumerative geometry is about the number of
objects (curves, for example) in a fixed space satisfying certain constraints.
In [5], [6] and [7], Kleiman and Piene introduce a very natural generalization
to this question. Suppose we have a family of varieties (i.e., a fiber
bundle). How many curves are there in this fiber bundle such that each curve
lies inside a single fiber and satisfies certain constraints? A prototypical
example to keep in mind is as follows: enumeration of planar degree $d$-curves
in $\mathbb{P}^{3}$ (i.e., curves, whose image lies inside some
$\mathbb{P}^{2}$), that satisfy certain constraints. In [9], Ties Laarakker
obtains a formula for the number of $\delta$-nodal planar degree $d$ curves in
$\mathbb{P}^{3}$, that intersect the correct number of generic lines. He
solves that question by extending the idea of Kool, Shende and Thomas, [8],
where they solve the question of enumerating $\delta$-nodal curves in a fixed
$\mathbb{P}^{2}$.

With this in the background, it is natural to ask whether the degeneration
method of Caporaso-Harris can be used to recover the numbers obtained by Ties
Laarakker.

The main problem in extending this idea to count nodal planar cubics in
$\mathbb{P}^{3}$ is as follows: what exactly would be the analogue of the
process of “moving points on a line” in the case of a moving family of
$\mathbb{P}^{2}$? The solution lies in interpreting the Caporaso-Harris
formula on the level of cycles in a suitable space (something more general
than [2, p. 351, Theorem 1.2]). Once we interpret the formula suitably on the
level of cycles, extending it to a family version is conceptually
straightforward.

We show that by extending the degeneration argument of Caporaso-Harris, it is
possible to compute the characteristic number of nodal planar cubics in
$\mathbb{P}^{3}$. This is the second result of this paper. The numbers
obtained agree with those in [9] and also agree with the numbers obtained by
the third and fourth author by alternative means [3], [10]. We hope that the
simple example of nodal planar cubics worked out in this paper will enable one
in future to solve the question of obtaining all the numbers computed by Ties
Laarakker in [9].

##  2\. Main Results

Let us now state the two main results of this paper. Let
$\mathsf{N}_{d}(\mathsf{A}_{1})$ denote the number of nodal degree $d$ curves
in $\mathbb{P}^{2}$ passing through $\delta_{d}-1$ generic points, where

$\delta_{d}\,\,:=\,\,\frac{d(d+3)}{2}.$ (2.1)

Denote by $\mathsf{N}_{d}(\mathsf{T}_{1})$ the number of degree $d$ curves in
$\mathbb{P}^{2}$, tangent to a given line (of first order) and passing through
$\delta_{d}-1$ generic points. A special case of [2, p. 348, Theorem 1.1] is

$\displaystyle\mathsf{N}_{d}(\mathsf{A}_{1})$
$\displaystyle\,=\,\,(2d-1)+\mathsf{N}_{d-1}(\mathsf{A}_{1})+2\mathsf{N}_{d-1}(\mathsf{T}_{1}).$
(2.2)

Denote by $\mathsf{N}_{d}(\mathsf{A}_{1}^{\mathsf{L}})$ to be the number of
degree $d$ curves with a node lying on a generic line and passing through
$\delta_{d}-2$ generic points. Also, let
$\mathsf{N}_{d}(\mathsf{T}_{1}^{\mathsf{Pt}})$ denote the number of degree $d$
curves in $\mathbb{P}^{2}$ tangent to a given line at a given point and
passing through $\delta_{d}-2$ generic points. Denote the number of degree $d$
curves in $\mathbb{P}^{2}$ having a cusp and passing through $\delta_{d}-2$
generic points by $\mathsf{N}_{d}(\mathsf{A}_{2})$. Finally, let
$\mathsf{N}_{d}(\mathsf{T}_{2})$ denote the number of degree $d$ curves in
$\mathbb{P}^{2}$ tangent to a given line (to second order) and passing through
$\delta_{d}-2$ generic points.

The first main result (it is proved in Section 3):

######  Theorem 2.1.

With the notations as above, the following two identities hold for $d\geq 3$:

$\displaystyle\mathsf{N}_{d}(\mathsf{A}_{1}^{\mathsf{L}})$
$\displaystyle\,\,=\,\,1+\mathsf{N}_{d-1}(\mathsf{A}_{1}^{\mathsf{L}})+2\mathsf{N}_{d-1}(\mathsf{T}_{1}^{\mathsf{Pt}}),$
(2.3) $\displaystyle\mathsf{N}_{d}(\mathsf{A}_{2})$
$\displaystyle\,=\,3\mathsf{N}_{d-1}(\mathsf{T}_{1})+3d\mathsf{N}_{d-1}(\mathsf{T}_{1}^{\mathsf{Pt}})+\mathsf{N}_{d-1}(\mathsf{A}_{2})+3\mathsf{N}_{d-1}(\mathsf{A}_{1}^{\mathsf{L}})+2\mathsf{N}_{d-1}(\mathsf{T}_{2}).$
(2.4)

First of all, note that $\mathsf{N}_{d}(\mathsf{T}_{k})$ and
$\mathsf{N}_{d}(\mathsf{T}_{k}^{\mathsf{Pt}})$ can be computed from the
Caporaso-Harris formula for all $d\,\geq\,1$ combined with eqs. 2.3 and 2.4.

Next note that $\mathsf{N}_{2}(\mathsf{A}_{1}^{\mathsf{L}})\,=\,3$. To see why
this is so, observe that $\mathsf{N}_{2}(\mathsf{A}_{1}^{\mathsf{L}})$ is the
number of nodal conics passing through three generic points, with the node
lying on a generic line. This is simply counting the number of pairs of lines
passing through three generic points, with the point of intersection lying on
a generic line. We can place one of the lines through two of the three points.
There are three ways to do that. Once we place the first line through two of
the three given points, the second line gets fixed (since it is the unique
line passing through the third remaining point and the point of intersection
of the first line with the line on which the nodal point has to lie). Hence,
$\mathsf{N}_{2}(\mathsf{A}_{1}^{\mathsf{L}})$ is equal to $3$.

Finally, it can shown that $\mathsf{N}_{2}(\mathsf{A}_{2})\,=\,0$. To see
this, we note that $\mathsf{N}_{2}(\mathsf{A}_{2})$ is the number of conics
passing through three generic points having a cusp. Since a double line does
not pass through three generic points, it follows that the number is zero.

These initial conditions enable us to use eq. 2.4 to recursively compute
$\mathsf{N}_{d}(\mathsf{A}_{2})$ for all $d$. For the convenience of the
reader we display the vales of $\mathsf{N}_{d}(\mathsf{A}_{2})$ for a the
first few values of $d$

$d$ | $3$ | $4$ | $5$ | $6$ | $7$ | $8$
---|---|---|---|---|---|---
$\mathsf{N}_{d}(\mathsf{A}_{2})$ | $24$ | $72$ | $144$ | $240$ | $360$ | $504$

Table 1.

It may be mentioned that although the computation of
$\mathsf{N}_{d}(\mathsf{A}_{2})$ has been done earlier by other means (see for
example the approach in Eisenbud’s book [4, Section 11.4]), the degeneration
formula eq. 2.4 itself is new. The numbers displayed in Table 1 agree with
those obtained by the formula given in [4, Proposition 11.4, Page 416].

The second result (it is proved in Section 5):

######  Theorem 2.2.

Let $\mathsf{N}_{3}^{\mathsf{Pl}}(\mathsf{A}_{1})(r,s)$ denote the number of
nodal planar cubics in $\mathbb{P}^{3}$ that intersect $r$ generic lines and
$s$ generic points, where $r+2s\,=\,11$. By extending the degeneration
argument of Caporaso-Harris, this number can be computed and is found to be

$(r,s)$ | $(11,0)$ | $(9,1)$ | $(7,2)$ | $(5,3)$
---|---|---|---|---
$\mathsf{N}_{3}^{\mathsf{Pl}}(\mathsf{A}_{1})(r,s)$ | $12960$ | $1392$ | $144$ | $12$

Table 2.

Note that $\mathsf{N}_{3}^{\mathsf{Pl}}(\mathsf{A}_{1})(r,s)$ is automatically
zero if $s$ is greater than or equal to four because four generic points do
not lie in a plane. The values given in Table 2 are in agreement with what is
obtained in the papers [3], [9] and [10]. We have written a mathematica
program to implement the computations of Section 5. The program is available
on the fourth author’s homepage

https://www.sites.google.com/site/ritwik371/home

##  3\. Caporaso-Harris formula for one nodal curves

In this section the following question is addressed:

How many degree $d$ curves are there in $\mathbb{P}^{2}$ that have a node and
pass through $\delta_{d}-1$ generic points? (see eq. 2.1)

Let $\mathsf{N}_{d}(\mathsf{A}_{1},k)$ be the number of nodal degree $d$
curves, passing through $\delta_{d}-1-k$ generic points and $k$ points on a
line; also, $\mathsf{N}_{d}(\mathsf{A}_{1},0)$ is written as
$\mathsf{N}_{d}(\mathsf{A}_{1})$ for short.

It is implied by [2, p. 348, Theorem 1.1] that

$\displaystyle\mathsf{N}_{d}(\mathsf{A}_{1})$
$\displaystyle\,=\,\mathsf{N}_{d}(\mathsf{A}_{1},1),$
$\displaystyle\mathsf{N}_{d}(\mathsf{A}_{1},1)$
$\displaystyle\,=\,\mathsf{N}_{d}(\mathsf{A}_{1},2),$ $\displaystyle\ldots$
$\displaystyle\mathsf{N}_{d}(\mathsf{A}_{1},d-2)$
$\displaystyle\,=\,\mathsf{N}_{d}(\mathsf{A}_{1},d-1).$ (3.1)

In other words,

$\displaystyle\mathsf{N}_{d}(\mathsf{A}_{1})$
$\displaystyle\,\,=\,\,\mathsf{N}_{d}(\mathsf{A}_{1},d-1).$ (3.2)

Hence, the number of nodal degree $d$ curves passing through $\delta_{d}-1$
generic points coincides with the number of nodal degree $d$ curves passing
through $\delta_{d}-d$ generic points and $(d-1)$ collinear points.
Intuitively, this is saying that if we move some of the point constraints to a
line, the resulting count of curves is still the same.

Unwinding the formula [2, p. 348, Theorem 1.1] further, it follows that

$\displaystyle\mathsf{N}_{d}(\mathsf{A}_{1},d-1)$
$\displaystyle\,\neq\,\mathsf{N}_{d}(\mathsf{A}_{1},d).$ (3.3)

Let us try to understand this via the example of cubics. Equation (3.2) says
that the number of nodal cubics through $8$ generic points is the same as the
number of nodal cubics through $6$ generic points and $2$ points on a line.
Equation (3.3) on the other hand says that this number is not the same as the
number of nodal cubics through $5$ generic points and $3$ points on a line.
The underlying geometric reason for eq. 3.3 is as follows: when the
$d^{\textnormal{th}}$ point is moved to the line, we get a degenerate
contribution from a curve of degree $d-1$. How then are the two numbers
$\mathsf{N}_{d}(\mathsf{A}_{1},d-1)$ and $\mathsf{N}_{d}(\mathsf{A}_{1},d)$
related?

Denote the number of smooth degree $d-1$ curves passing through $\delta_{d-1}$
generic points by $\mathsf{N}_{d-1}(\mathsf{S})$. This number is obviously
$1$, but let us answer the earlier question in terms of
$\mathsf{N}_{d-1}(\mathsf{S})$ (which will make the equation geometrically
more transparent). Unwinding [2, p. 348, Theorem 1.1], it follows that

$\displaystyle\mathsf{N}_{d}(\mathsf{A}_{1},d-1)$
$\displaystyle\,=\,\mathsf{N}_{d}(\mathsf{A}_{1},d)+(d-1)\mathsf{N}_{d-1}(\mathsf{S}).$
(3.4)

Using eqs. 3.2 and 3.4, the question of computing
$\mathsf{N}_{d}(\mathsf{A}_{1})$ has got reduced to a question of computing
$\mathsf{N}_{d-1}(\mathsf{S})$ and $\mathsf{N}_{d}(\mathsf{A}_{1},d)$. Since
we know that $\mathsf{N}_{d-1}(\mathsf{S})\,=\,1$, the question has now been
reduced to the computation of $\mathsf{N}_{d}(\mathsf{A}_{1},d)$.

Recall that $\mathsf{N}_{d-1}(\mathsf{T}_{1})$ is defined to be the number of
smooth degree $d-1$ curves passing through $\delta_{d-1}-1$ generic points and
tangent to a line. Unwinding [2, Theorem 1.1] further, we get that

$\displaystyle\mathsf{N}_{d}(\mathsf{A}_{1},d)$
$\displaystyle\,=\,\mathsf{N}_{d-1}(\mathsf{A}_{1})+d\mathsf{N}_{d-1}(\mathsf{S})+2\mathsf{N}_{d-1}(\mathsf{T}_{1}).$
(3.5)

Equations 3.2, 3.4 and 3.5 combined with the fact that
$\mathsf{N}_{d-1}(\mathsf{S})\,=\,1$ give eq. 2.2.

Let $\mathsf{N}_{d}(\mathsf{A}_{1}^{\mathsf{L}},k)$ denote the number of nodal
degree $d$ curves, with the nodal point lying on a line and passing through
$\delta_{d}-2-k$ generic points and $k$ points on a line. Note that the line
on which the nodal point lies and the line on which the $k$ points lie are
different. Again, when $k\,=\,0$, we omit writing the $k$, i.e.,
$\mathsf{N}_{d}(\mathsf{A}_{1}^{\mathsf{L}})\,:=\,\mathsf{N}_{d}(\mathsf{A}_{1}^{\mathsf{L}},0)$.
Analogous to how eq. 3.1 is proved, it will similarly be shown that

$\displaystyle\mathsf{N}_{d}(\mathsf{A}_{1}^{\mathsf{L}})$
$\displaystyle\,=\,\mathsf{N}_{d}(\mathsf{A}_{1}^{\mathsf{L}},1),$
$\displaystyle\mathsf{N}_{d}(\mathsf{A}_{1}^{\mathsf{L}},1)$
$\displaystyle\,=\,\mathsf{N}_{d}(\mathsf{A}_{1}^{\mathsf{L}},2),$
$\displaystyle\ldots$
$\displaystyle\mathsf{N}_{d}(\mathsf{A}_{1}^{\mathsf{L}},d-2)$
$\displaystyle\,=\,\mathsf{N}_{d}(\mathsf{A}_{1}^{\mathsf{L}},d-1).$ (3.6)

In other words,

$\displaystyle\mathsf{N}_{d}(\mathsf{A}_{1}^{\mathsf{L}})$
$\displaystyle\,\,=\,\,\mathsf{N}_{d}(\mathsf{A}_{1}^{\mathsf{L}},d-1).$ (3.7)

Next, similar to how eq. 3.4 is proved, it will be shown that

$\displaystyle\mathsf{N}_{d}(\mathsf{A}_{1}^{\mathsf{L}},d-1)$
$\displaystyle\,\,=\,\,\mathsf{N}_{d}(\mathsf{A}_{1}^{\mathsf{L}},d)+\mathsf{N}_{d-1}(\mathsf{S}).$
(3.8)

Finally, similar to how eq. 3.5 is proved, it will be shown that

$\displaystyle\mathsf{N}_{d}(\mathsf{A}_{1}^{\mathsf{L}},d)$
$\displaystyle\,\,=\,\,\mathsf{N}_{d-1}(\mathsf{A}_{1}^{\mathsf{L}})+2\mathsf{N}_{d-1}(\mathsf{T}_{1}^{\textsf{Pt}}).$
(3.9)

Equations 3.7, 3.8 and 3.9 combined together with the fact that
$\mathsf{N}_{d-1}(\mathsf{S})\,=\,1$, produce eq. 2.3.

Let us now explain this entire degeneration process in greater detail. Denote
by $\mathcal{D}_{d}$ the projective space of dimension $\delta_{d}$ given by
the curves of degree $d$ on the projective plane. Denote the hyperplane class
in $\mathcal{D}_{d}$ by $y_{d}$. Let $X^{1}$ be a copy of $\mathbb{P}^{2}$;
its hyperplane class will be denoted by $b_{1}$. Define

$\displaystyle\mathsf{A}_{1}$
$\displaystyle\,\,\subset\,\,\mathcal{D}_{d}\times X^{1}$

to be the locus of all $(H_{d},\,p)$ such that the curve $H_{d}$ has a node at
$p$. Let $\big{[}\mathsf{A}_{1}\big{]}$ denote the homology class represented
by the closure of $\mathsf{A}_{1}$ in $\mathcal{D}_{d}\times X^{1}$. Note that
$\big{[}\mathsf{A}_{1}\big{]}$ is a polynomial in $y_{d}$ and $b_{1}$; knowing
this polynomial will enable us to compute all the characteristic number of
nodal degree $d$-curves.

For example, note that

$\big{[}\mathsf{A}_{1}\big{]}\cdot y_{d}^{\delta_{d}-1}$

is precisely equal to $\mathsf{N}_{d}(\mathsf{A}_{1})$, the number of nodal
degree $d$ curves in $\mathbb{P}^{2}$ that pass through $\delta_{d}-1$ generic
points. However, knowing $\big{[}\mathsf{A}_{1}\big{]}$ also enables us to
compute some other numbers. For example,

$\big{[}\mathsf{A}_{1}\big{]}\cdot y_{d}^{\delta_{d}-2}\cdot b_{1}$

is equal to $\mathsf{N}_{d}(\mathsf{A}_{1}^{\mathsf{L}})$, the number of nodal
degree $d$ curves in $\mathbb{P}^{2}$ that pass through $\delta_{d}-2$ generic
points with the nodal point lying on a line.

Instead of focussing on computing the number $\mathsf{N}_{d}(\mathsf{A}_{1})$,
we will focus on computing the homology class represented by the closure of
the cycle $\mathsf{A}_{1}$. It may be mentioned that this class is actually
not computed by Caporaso-Harris. Before proceeding further, let us define a
few spaces that will be useful later.

Let $d$ be a positive integer and $n,m$ be non-negative integers. Define

$\mathsf{M}_{n}^{m}(d)\,\,:=\,\,\mathcal{D}_{1}\times\mathcal{D}_{d}\times(X^{1}\times\cdots\times
X^{m})\times(X_{1}\times\cdots\times X_{n}),$

where each $X^{j}$ or $X_{i}$ is $\mathbb{P}^{2}$. The hyperplane classes of
$\mathcal{D}_{1}$, $\mathcal{D}_{d}$, $X^{j}$ and $X_{i}$ will be denoted by
$y_{1},y_{d},b_{j}$ and $a_{i}$ respectively. For notational ease,
$\mathsf{M}_{n}^{m}(d)$ will be abbreviated by $\mathsf{M}_{n}^{m}$. Given a
subspace $\mathsf{Z}$ of $\mathsf{M}_{n}^{m}$, the homology class represented
by the closure of $\mathsf{Z}$ in $\mathsf{M}_{n}^{m}$ will be denoted by
$\big{[}\mathsf{Z}\big{]}$.

Define $\mathsf{A}_{1}^{\mathsf{F}}\,\,\subset\,\,\mathsf{M}_{0}^{1}$ to be
the locus of all $(H_{1},\,H_{d},\,p)$ such that the curve $H_{d}$ has a node
at $p$ and $p$ does not lie on $H_{1}$. Let

$\displaystyle\mathsf{A}_{1}^{\mathsf{F}}\underbrace{\mathsf{T}_{0}\cdots\mathsf{T}_{0}}_{n-\textnormal{times}}\,\,\subset\,\,\mathsf{M}_{n}^{1}$
(3.10)

be the locus of all $(H_{1},\,H_{d},\,p,\,x_{1},\,\ldots,\,x_{n})$ such that

  * •

the curve $H_{d}$ has a node at $p$,

  * •

the curve $H_{d}$ intersects $H_{1}$ at $x_{1},\,\cdots,\,x_{n}$
transversally,

  * •

the points $x_{1},\,\cdots,\,x_{n}$ are all distinct, and

  * •

the nodal point $p$ does not lie on $H_{1}$.

The space can be pictorially described as follows:

Figure 1. $\mathsf{A}_{1}^{\mathsf{F}}\mathsf{T}_{0}\cdots\mathsf{T}_{0}$

Similarly, define

$\displaystyle\mathsf{A}_{1}^{\mathsf{F}}\underbrace{\mathsf{T}_{0}\cdots\mathsf{T}_{0}}_{n-\textnormal{times}}\mathsf{F}\,\,\subset\,\,\mathsf{M}_{n+1}^{1}$
(3.11)

to be the locus of all $(H_{1},\,H_{d},\,p,\,x_{1},\ldots\,,x_{n},\,x_{n+1})$
such that

  * •

the curve $H_{d}$ has a node at $p$,

  * •

the curve $H_{d}$ intersects $H_{1}$ at $x_{1},\,\ldots,\,x_{n}$
transversally,

  * •

the points $x_{1},\,x_{2},\,\ldots,\,x_{n},\,x_{n+1}$ are all distinct, and

  * •

the nodal point $p$ does not lie on $H_{1}$.

The space can be pictorially described as follows:

Figure 2.
$\mathsf{A}_{1}^{\mathsf{F}}\mathsf{T}_{0}\cdots\mathsf{T}_{0}\mathsf{F}$

Next, define

$\big{(}x_{i}\in\mathsf{L}\big{)}\,\,\subset\,\,\mathsf{M}^{1}_{n}$

to be the locus of all $(H_{1},\,H_{d},\,p,\,x_{1},\,\cdots,\,x_{n})$ such
that $x_{i}$ lies on the line $H_{1}$. Similarly, define

$\big{(}x_{i}\in\mathcal{C}_{d}\big{)}\,\,\subset\,\,\mathsf{M}^{1}_{n}$

to be the locus of all $(H_{1},\,H_{d},\,p,\,x_{1},\,\cdots,\,x_{n})$ such
that $x_{i}$ lies on the curve $H_{d}$.

Finally, for $i\,\neq\,j$, define

$\big{(}x_{i}=x_{j}\big{)}\,\,\subset\,\,\mathsf{M}^{1}_{n}$

to be the locus of all $(H_{1},\,H_{d},\,p,\,x_{1},\,\cdots,\,x_{n})$ such
that $x_{i}$ is equal to $x_{j}$. Similarly, define

$\big{(}p=x_{j}\big{)}\,\,\,\subset\,\,\,\mathsf{M}^{1}_{n}$

to be the locus of all $(H_{1},\,H_{d},\,p,\,x_{1},\,\cdots,\,x_{n})$ such
that $p$ is equal to $x_{j}$.

Let us now recall a few standard facts; the following equalities of homology
classes in $\mathsf{M}^{1}_{n}$ hold:

$\displaystyle\big{[}\big{(}x_{i}=x_{j}\big{)}\big{]}$
$\displaystyle\,=\,a_{i}^{2}+a_{i}a_{j}+a_{j}^{2},\qquad\big{[}\big{(}p=x_{j}\big{)}\big{]}\,=\,b_{1}^{2}+b_{1}a_{j}+a_{j}^{2},$
$\displaystyle\big{[}\big{(}x_{i}\in\mathsf{L}\big{)}\big{]}$
$\displaystyle\,=\,y_{1}+a_{i},\qquad\textnormal{and}\qquad\big{[}\big{(}x_{i}\in\mathcal{C}_{d}\big{)}\big{]}\,=\,y_{d}+da_{i}.$
(3.12)

###  3.1. Cycle version

We are now ready to state the main results on the level of cycles. We will
first state and prove all the statements on cycle level. Enumerative
information are extracted from them in Section 3.2.

######  Theorem 3.1.

The following equality of homology classes hold in $\mathsf{M}^{1}_{1}(d)$:

$\displaystyle\big{[}\mathsf{A}_{1}^{\mathsf{F}}\mathsf{F}\big{]}\cdot\big{[}\big{(}x_{1}\in\mathsf{L}\big{)}\big{]}\cdot\big{[}\big{(}x_{1}\in\mathcal{C}_{d}\big{)}\big{]}$
$\displaystyle\,\,=\,\,\big{[}\mathsf{A}_{1}^{\mathsf{F}}\mathsf{T}_{0}\big{]},$

where $d\,\in\,\mathbb{Z}^{\geq 2}$.

3.1 is a special case of the following theorem when $n\,=\,0$. We will prove
this directly.

######  Theorem 3.2.

Let $d\,\in\,\mathbb{Z}^{\geq 2}$, and let $n$ be an integer such that

$0\ \leq\ n\ \leq\ d-2.$

Then, on $\mathsf{M}^{1}_{n+1}(d)$ the following equality of homology classes
hold:

$\displaystyle\big{[}\mathsf{A}_{1}^{\mathsf{F}}\underbrace{\mathsf{T}_{0}\cdots\mathsf{T}_{0}}_{n-\textnormal{times}}\mathsf{F}\big{]}\cdot\big{[}\big{(}x_{n+1}\in\mathsf{L}\big{)}\big{]}\cdot\big{[}\big{(}x_{n+1}\in\mathcal{C}_{d}\big{)}\big{]}\,\,$
$\displaystyle=\,\,\big{[}\mathsf{A}_{1}^{\mathsf{F}}\underbrace{\mathsf{T}_{0}\cdots\mathsf{T}_{0}}_{(n+1)-\textnormal{times}}\big{]}$
$\displaystyle+\sum_{i=1}^{n}\big{[}\mathsf{A}_{1}^{\mathsf{F}}\underbrace{\mathsf{T}_{0}\cdots\mathsf{T}_{0}}_{n-\textnormal{times}}\mathsf{F}\big{]}\cdot\big{[}\big{(}x_{i}=x_{n+1}\big{)}\big{]}.$
(3.13)

######  Remark 3.3.

Let us first examine the validity of 3.2 on set-theoretic level. Consider the
first term on the left-hand side, namely

$\big{[}\mathsf{A}_{1}^{\mathsf{F}}\underbrace{\mathsf{T}_{0}\cdots\mathsf{T}_{0}}_{n-\textnormal{times}}\mathsf{F}\big{]}.$

It is represented by the closure of the following space: a line, a nodal curve
with the node not lying on the line and $n+1$ distinct points
$(x_{1},\,\cdots,\,x_{n+1})$, such that the curve intersects the line
transversally at the points $(x_{1},\,\cdots,\,x_{n})$. The last point
$x_{n+1}$ is free (it does not have to lie on either the line or the curve).

Now consider the second and third factor on the left-hand side of 3.2, namely

$\big{[}\big{(}x_{n+1}\in\mathsf{L}\big{)}\big{]}\cdot\big{[}\big{(}x_{n+1}\in\mathcal{C}_{d}\big{)}\big{]}.$

It is represented by the following space: a line, a curve and $n+1$ points
$(x_{1},\,\cdots,\,x_{n+1})$, such that the points $(x_{1},\,\cdots,\,x_{n})$
are free, while the last point $x_{n+1}$ lies on the line and the curve.

The set-theoretic intersection of the above two spaces has two possibilities.
The first case to consider is when the nodal point does not lie on the line.
Corresponding to this, there are two sub-cases. The first sub-case is that the
point $x_{n+1}$ is distinct from all the other points
$(x_{1},\,\cdots,\,x_{n})$. The closure of that space represents the first
term on the right-hand side of 3.2.

The second sub-case is that the point $x_{n+1}$ could be equal to one of the
$x_{i}$ (for $i\,\in\,\\{1,\,\ldots,\,n\\}$). The closure of that space gives
us the second term on the right-hand side of 3.2.

The second possibility to consider is that the nodal point lies on the line.
It will be shown that this component has a high codimension and hence will not
intersect a cycle of complementary dimension.

First note that $\mathsf{M}^{1}_{n+1}(d)$ is a manifold of dimension
$\delta_{d}+2n+6$. 3.2 is an equality of classes of codimension $2n+5$. This
is because $\mathsf{A}_{1}^{\mathsf{F}}$ is a codimension three class.
Furthermore, imposing the condition that the point lies on both the line and
the curve is a codimension $2$ condition. Hence, the total codimension is

$3+2(n+1)=2n+5.$

Therefore, the dimension of the homology cycles representing each side of 3.2
is given by

$\displaystyle a$ $\displaystyle\,:=\,\delta_{d}+2n+6-(2n+5)$
$\displaystyle\,=\,\frac{d^{2}}{2}+\frac{3d}{2}+1.$

Now the dimension of the component where the nodal point lies on the line will
be computed. More precisely, we are looking at the space of degree $d$ curves,
with a nodal point on the designated line and $n+1$ points lying on the line.
Assume further that the curve is either irreducible or even if it is
reducible, the designated line is not one of the components of the curve
(i.e., there exists points on the line that do not lie on the curve). The
nodal point lying on the line is a codimension $4$ condition. Each of the
$(n+1)$ marked points lying on the line and curve is a codimension $2$
condition. Hence, the total codimension of this space is $2n+6$. This is
strictly greater than $2n+5$ and hence it does not contribute to the
intersection.

Now look at the space of degree $d$ curves, with a nodal point on the
designated line and $n+1$ points lying on the line. Furthermore, assume that
the curve is reducible, with the line being one of the components of the
curve. The dimension of this space is given by

$\displaystyle b$ $\displaystyle\ :=\ 2+\delta_{d-1}+(n+1).$ $\displaystyle\
=\ \frac{d^{2}}{2}+\frac{d}{2}+n+2.$

To see why this is so, note that the space of lines is two-dimensional and the
space of degree $d-1$ curves is of dimension $\delta_{d-1}$. We are now
allowed to place $n+1$ points on the line; each point has one degree of
freedom. Hence, the total dimension of the space is $2+\delta_{d-1}+(n+1)$.

Note that

$\displaystyle b-a$ $\displaystyle\ =\ n-(d-1).$ (3.14)

Hence, if $n\,<\,d-1$, then $b\,<\,a$; this implies that the space of
reducible curves with the line being one of the components does not contribute
to the intersection if $n\,<\,d-1$. Notice the importance of the inequality
$n\,<\,d-1$ for this argument (as we will see in the next theorem, this
component does contribute to the intersection).

###### Proof of 3.2.

Following the discussion in 3.3, it suffices to show that the relevant
intersections are transverse and that the multiplicities of the intersections
for the second term on the right-hand side of 3.2 are all one.

We claim that the open part of each of the cycles is actually a smooth
manifold of the expected dimension. This can be proved by showing that
$\mathsf{A}_{1}^{\mathsf{F}}$ is a smooth submanifold of $\mathsf{M}^{1}_{0}$
of codimension three (i.e., it has the expected dimension). This is a local
statement, we switch to affine spaces. Let $\mathcal{F}_{d}$ be the space of
polynomials in two variables of degree at most $d$. This is a vector space of
dimension $\frac{d(d+3)}{2}+1$, because an element of $\mathcal{F}_{d}$ can be
viewed as

$f(x,\,y)\,:=\,f_{00}+f_{10}x+f_{01}y+\frac{f_{20}}{2}x^{2}+f_{11}xy+\frac{f_{02}}{2}y^{2}+\cdots+\frac{f_{0d}}{d!}y^{d},$

and hence $f$ can be identified with
$(f_{00},\,f_{10},\,f_{01},\,\ldots,\,f_{0d})$. Let $\mathcal{F}_{d}^{*}$
denote the space of all non-zero polynomials of degree $d$.

Define $(\mathsf{A}_{1}^{\mathsf{F}})_{\textsf{Aff}}$ to be the following
subspace of $\mathcal{F}_{d}^{*}\times\mathbb{C}^{2}$: it consists of a curve
$f$ and a pair of complex numbers $(x,\,y)\,\in\,\mathbb{C}^{2}$ such that the
curve has a node at $(x,\,y)$. It will be shown that
$(\mathsf{A}_{1}^{\mathsf{F}})_{\textsf{Aff}}$ is a smooth submanifold of
$\mathcal{F}_{d}^{*}\times\mathbb{C}^{2}$.

Consider the map

$\varphi\,\,:\,\,\mathcal{F}_{d}^{*}\times\mathbb{C}^{2}\,\,\longrightarrow\,\,\mathbb{C}^{3},\
\ \,(f,\,(x,\,y))\,\longmapsto\,(f(x,\,y),\,f_{x}(x,\,y),\,f_{y}(x,\,y)).$

It suffices to show that $0$ is a regular value of $\varphi$. Consider the
polynomials $\eta_{ij}(x,\,y)$ given by

$\displaystyle\eta_{00}(x,\,y)$
$\displaystyle\,:=\,1,\qquad\eta_{10}(x,\,y)\,:=\,x,\qquad\textnormal{and}\qquad\eta_{01}(x,\,y)\,:=\,y.$

Let $\gamma_{ij}(t)$ be the curve given by

$\gamma_{ij}(t)\,\,:=\,\,(f+t\eta_{ij},\,x).$

Computing the differential of $d\varphi$ on these curves proves
transversality.

Next, we will show that $\mathsf{A}_{1}^{\mathsf{F}}\mathsf{T}_{0}$ is a
smooth submanifold of $\mathsf{M}^{1}_{1}$ of codimension five. Let us now fix
the $\mathtt{x}-$axis as the designated line. Define
$(\mathsf{A}_{1}^{\mathsf{F}}\mathsf{T}_{0})_{\textsf{Aff}}$ to be the
following subspace of
$\mathcal{F}_{d}^{*}\times\mathbb{C}^{2}\times\mathbb{C}$: it consists of a
curve $f$ and a triple of complex numbers
$(x,\,y,\,x_{1})\,\in\,\mathbb{C}^{3}$ such that $f$ has a node at $(x,\,y)$
and it intersects the $\mathtt{x}-$axis at $(x_{1},\,0)$ transversally.
Furthermore, we have $y\,\neq\,0$ (this corresponds to the condition that the
nodal point does not lie on the line). It will be shown that
$(\mathsf{A}_{1}^{\mathsf{F}}\mathsf{T}_{0})_{\textsf{Aff}}$ is a smooth
submanifold of $\mathcal{F}_{d}^{*}\times\mathbb{C}^{3}$. For that, define the
map

$\varphi\,:\,\mathsf{A}_{1}^{\mathsf{F}}\times\mathbb{C}\,\longrightarrow\,\mathbb{C},\
\ \ \,(f,\,(x,y),\,x_{1})\,\longmapsto\,f(x_{1},\,0).$

In order to show that
$(\mathsf{A}_{1}^{\mathsf{F}}\mathsf{T}_{0})_{\textsf{Aff}}$ is a smooth
submanifold of $\mathsf{A}_{1}^{\mathsf{F}}\times\mathbb{C}$, it suffices to
prove that $0$ is a regular value of $\varphi$. To show that $0$ is a regular
value of $\varphi$, assume that $\varphi(f,\,(x_{0},\,y_{0}),\,x_{1})\,=\,0$.
For computing the differential of $\varphi$ at
$(f,\,(x_{0},\,y_{0}),\,x_{1})$, consider the following curve

$\gamma(t)\,\,:=\,\,(f+t\eta,\,(x_{0},\,y_{0}),\,x_{1}),$

where $\eta$ is as yet an unspecified polynomial. We need to choose $\eta$ in
such a way that

$\\{d\varphi|_{(f,\,(x_{0},y_{0}),\,x_{1})}\\}(\gamma^{\prime}(0))$

is non-zero. Furthermore, $f+t\eta$ should have a node at $(x_{0},\,y_{0})$
for all $t$. Since $f$ has a node at $(x_{0},\,y_{0})$, this is equivalent to
the condition that

$\displaystyle\eta(x_{0},\,y_{0})$
$\displaystyle\,=\,0,\qquad\eta_{x}(x_{0},\,y_{0})\,=\,0\qquad\textnormal{and}\qquad\eta_{y}(x_{0},\,y_{0})\,=\,0.$
(3.15)

Now note that the differential of $\varphi$ is given by

$\left\\{d\varphi\big{|}_{(f,(x,y),x_{1})}\right\\}(\gamma^{\prime}(0))\,\,=\,\,\,\lim_{t\rightarrow
0}\frac{(f+t\eta)(x_{1},\,0)-f(x_{1},\,0)}{t}\,\,=\,\,\eta(x_{1},\,0).$

Hence, in addition to the conditions of eq. 3.15, it is required that
$\eta(x_{1},\,0)\,\neq\,0$. We now show that such an $\eta$ exists. Let

$\eta(x,\,y)\ :=\ (y-y_{0})^{2}.$

Since $y_{0}\,\neq\,0$, it follows that all the four conditions are satisfied.
This proves that $(\mathsf{A}_{1}^{\mathsf{F}}\mathsf{T}_{0})_{\textsf{Aff}}$
is a smooth submanifold of $\mathsf{A}_{1}^{\mathsf{F}}\times\mathbb{C}$.

Next, we will show that
$\mathsf{A}_{1}^{\mathsf{F}}\mathsf{T}_{0}\mathsf{T}_{0}$ is a smooth
submanifold of $\mathsf{M}^{1}_{2}$ of codimension seven. As before, we switch
to affine space and define
$(\mathsf{A}_{1}^{\mathsf{F}}\mathsf{T}_{0}\mathsf{T}_{0})_{\textsf{Aff}}$ as
follows:

$(\mathsf{A}_{1}^{\mathsf{F}}\mathsf{T}_{0}\mathsf{T}_{0})_{\textsf{Aff}}\,\,:=\,\,\\{(f,\,(x_{0},y_{0}),\,x_{1},\,x_{2})\,\in\,(\mathsf{A}_{1}^{\mathsf{F}}\mathsf{T}_{0})_{\textsf{Aff}}\times\mathbb{C}\,\,\big{|}\,\,f(x_{2},0)\,=\,0,\,\,x_{1}\,\neq\,x_{2}\\}.$

To show that
$(\mathsf{A}_{1}^{\mathsf{F}}\mathsf{T}_{0}\mathsf{T}_{0})_{\textsf{Aff}}$ is
a smooth submanifold of
$(\mathsf{A}_{1}^{\mathsf{F}}\mathsf{T}_{0})_{\textsf{Aff}}\times\mathbb{C}$,
define the map

$\varphi\,:\,(\mathsf{A}_{1}^{\mathsf{F}}\mathsf{T}_{0})_{\textsf{Aff}}\times\mathbb{C}\,\longrightarrow\,\mathbb{C},\
\ \,(f,\,(x_{0},\,y_{0}),\,x_{1},\,x_{2})\,\longmapsto\,f(x_{2},\,0).$

It suffices to show that the differential of $\varphi$ is nonzero when
$x_{1}\,\neq\,x_{2}$ and when $y_{0}\,\neq\,0$. To compute the differential of
$\varphi$ at $(f,\,(x_{0},\,y_{0}),\,x_{1},\,x_{2})$, consider the curve

$\gamma(t)\,\,:=\,\,(f+t\eta,\,(x_{0},\,y_{0}),\,x_{1},\,x_{2}),$

where $\eta$ is as yet an unspecified polynomial. It should be so that

$\\{d\varphi|_{(f,x_{1},x_{2})}\\}(\gamma^{\prime}(0))$

is non-zero. Furthermore, the curve $\gamma(t)$ is required to lie in
$(\mathsf{A}_{1}^{\mathsf{F}}\mathsf{T}_{0})_{\textsf{Aff}}\times\mathbb{C}$
for all $t$. This means that eq. 3.15 must be satisfied and furthermore

$\displaystyle\eta(x_{1},0)$ $\displaystyle\ =\ 0.$ (3.16)

The differential is given by

$\\{d\varphi|_{(f,(x_{0},y_{0}),x_{1},x_{2})}\\}(\gamma^{\prime}(0))\,\,=\,\,\lim_{t\rightarrow
0}\frac{(f+t\eta)(x_{2},\,0)-f(x_{2},\,0)}{t}\,\,=\,\,\eta(x_{2},\,0).$

Hence, $\eta$ should satisfy eqs. 3.15 and 3.16, and
$\eta(x_{2},\,0)\,\neq\,0$. Such an $\eta$ exists; for example consider

$\eta(x,\,y)\,\,:=\,\,(y-y_{0})^{2}(x-x_{1}).$

Notice that the conditions $y_{0}\,\neq\,0$ and $x_{1}\,\neq\,x_{2}$ are
crucial. We also need $d\,\geq\,3$ for the argument to work (since the
polynomial $\eta$ is of degree $3$).

Using a similar argument, it is seen that

$\big{[}\mathsf{A}_{1}^{\mathsf{F}}\underbrace{\mathsf{T}_{0}\cdots\mathsf{T}_{0}}_{n-\textnormal{times}}\big{]}$

is a smooth submanifold of $\mathsf{M}^{1}_{n}$ of codimension $3+2n$,
provided $n\,\leq\,d-2$. The bound on $d$ enables one to construct the desired
curve $\eta$ that is needed to prove the transversality.

Now the multiplicities of the intersections will be computed. The above
transversality shows that the first term in the right-hand side of 3.2 occurs
with multiplicity one. To justify the intersection multiplicities of the
remaining terms, consider the situation where $x_{1}$ coincides with
$x_{n+1}$. For convenience, set $x_{n+1}$ to be equal to zero (i.e., the last
point is the origin); in that case, $f$ intersects the $\mathtt{x}$–axis
transversally at the origin. We are now going to study the multiplicity with
which the evaluation map vanishes at the origin. Hence, $f$ is such that
$f_{00}$ vanishes. It is given by

$f(x,\,y)\,\,=\,\,f_{10}x+f_{01}y+\frac{f_{20}}{2}x^{2}+f_{11}xy+\frac{f_{02}}{2}y^{2}+\cdots.$

Now consider the evaluation map

$\varphi(f,\,x)\,\,:=\,\,f(x,\,0)\,\,=\,\,f_{10}x+\frac{f_{20}}{2}x^{2}+\cdots.$

The order of vanishing of $\varphi$ is clearly $1$, provided
$f_{10}\,\neq\,0$. But that assumption is valid, since $f$ intersects the
$\mathtt{x}$–axis transversally at the origin. Hence, the order of vanishing
is $1$. This proves the equality in 3.2. ∎

3.2 is false, when $n\,=\,d-1$. Before stating the correct statement for
$n\,=\,d-1$, we need to develop some more notation. Define

$\displaystyle\mathsf{R}_{n}(d)\ \subset\ \mathsf{M}^{1}_{n}(d)$ (3.17)

to be the locus of all $(H_{1},\,H_{d},\,p,\,x_{1},\,\cdots,\,x_{n})$ such
that

  * •

the curve $H_{d}$ is the union of the line $H_{1}$ and a degree $d-1$ curve,

  * •

the degree $d-1$ curve is smooth,

  * •

the degree $d-1$ curve intersects $H_{1}$ transversally at all the points.

  * •

the point $p$ is one of the points of intersection of the $d-1$ curve and
$H_{1}$, and

  * •

the points $x_{1},\,\cdots,\,x_{n}$ are all distinct and lie on $H_{1}$; they
are also distinct from $p$.

The space can be pictorially described as follows:

Figure 3. $\mathsf{R}_{n}(d)$

######  Remark 3.4.

Note that the black and orange lines in figure 3 are actually supposed to
represent the same line. We distinguish them to keep in mind that the line is
a part of a reducible curve, as well as a line in its own moduli space. This
remark also applies to the rest of the pictures in this section.

Insertion of one more point on both the line and the curve would yield the
following result.

######  Theorem 3.5.

Let $d\in\mathbb{Z}^{\geq 3}$. Then, the following equality of homology
classes holds in $\mathsf{M}_{{d}}^{1}(d)$

$\displaystyle\big{[}\mathsf{A}_{1}^{\mathsf{F}}\underbrace{\mathsf{T}_{0}\cdots\mathsf{T}_{0}}_{(d-1)-\textnormal{times}}\mathsf{F}\big{]}\cdot\big{[}\big{(}x_{d}\in\mathsf{L}\big{)}\big{]}\cdot\big{[}\big{(}x_{d}\in\mathcal{C}_{d}\big{)}\big{]}\,\,$
$\displaystyle=\,\,\big{[}\mathsf{A}_{1}^{\mathsf{F}}\underbrace{\mathsf{T}_{0}\cdots\mathsf{T}_{0}}_{d-\textnormal{times}}\big{]}$
$\displaystyle+\sum_{i=1}^{d-1}\big{[}\mathsf{A}_{1}^{\mathsf{F}}\underbrace{\mathsf{T}_{0}\cdots\mathsf{T}_{0}}_{(d-1)-\textnormal{times}}\mathsf{F}\big{]}\cdot\big{[}\big{(}x_{i}=x_{d}\big{)}\big{]}+\big{[}\mathsf{R}_{d}(d)\big{]}.$
(3.18)

######  Remark 3.6.

Let us first analyze the following set theoretic question: consider the
component of the closure of

$\mathsf{A}_{1}^{\mathsf{F}}\underbrace{\mathsf{T}_{0}\cdots\mathsf{T}_{0}}_{(d-1)-\textnormal{times}}$

when the nodal point $p$ lies on the line. If the nodal point is equal to one
of the other points $x_{1},\,x_{2},\,\cdots,\,x_{d-1}$, then a dimension
counting argument shows that this component will not contribute to the
intersection. Now assume that this point is different from all the other
marked points. This is possible only if the curve is reducible and the line is
one of the components of the curve. To see why this is so, note that if the
line is not one of the components, then the nodal point contributes at least
$2$ to the intersection and there are $d-1$ other distinct points. The curve
intersects the line at $d+1$ points; this is impossible, since the curve has
degree $d$.

Hence the only configuration that can potentially arise (that contributes to
the intersection) is a reducible curve, with one of the components the line
and the remaining points $x_{1},\,x_{2},\,\cdots,\,x_{d-1}$ lying on the line,
with the nodal point being distinct from all of them. Recall from eq. 3.14
that since $n$ is now equal to $d-1$, dimensional considerations do not rule
out this configuration. 3.5 states that this configuration and $x_{d}$
coincides with any of the $x_{i}$’s are the only possibility that can occur in
the degeneration.

###### Proof of 3.5.

We continue with the setup of 3.2. The first and second terms on the right-
hand side are justified as before. Let us now justify the third term, which is
the new thing that occurs. It will be shown that this configuration actually
arises. In other words, we need to show that

$\displaystyle\overline{\mathsf{R}}_{d}(d)$
$\displaystyle\subset\overline{\mathsf{A}_{1}^{\mathsf{F}}\underbrace{\mathsf{T}_{0}\cdots\mathsf{T}_{0}}_{(d-1)-\textnormal{times}}\mathsf{F}}.$
(3.19)

It suffices to show that

$\displaystyle\mathsf{R}_{d}(d)$
$\displaystyle\subset\overline{\mathsf{A}_{1}^{\mathsf{F}}\underbrace{\mathsf{T}_{0}\cdots\mathsf{T}_{0}}_{(d-1)-\textnormal{times}}\mathsf{F}}.$
(3.20)

Let $(H_{1},\,H_{d},\,p,\,x_{1},\,\cdots,\,x_{d-1},\,x_{d})$ be an arbitrary
point of $\mathsf{R}_{d}(d)$. We will switch to affine space as before. Assume
the line $H_{1}$ is the $\mathtt{x}$–axis and the nodal point $p$ is the
origin $(0,\,0)$. Also assume that $x_{1},\,x_{2},\,\cdots,\,x_{d-1}$ are
given by $(a_{1},\,0)$, $(a_{2},\,0),\,\cdots,\,(a_{d-1},\,0)$. Since these
points are all distinct and also not equal to the nodal point, it follows that
$a_{1},\,a_{2},\,\cdots,\,a_{d-1}$ and $0$ are all distinct.

Since $H_{d}$ is a reducible curve, with the line $H_{1}$ being one of the
reducible components, it is given in the affine coordinate by a polynomial $f$
of the form

$\displaystyle f(x,y)$ $\displaystyle\ =\ y\varphi(x,y),$

where $\varphi(x,\,y)$ is a polynomial of degree $d-1$. Let us write down
$\varphi(x,\,y)$ explicitly as

$\displaystyle\varphi(x,y)$ $\displaystyle\ :=\
b_{00}+b_{10}x+b_{01}y+\cdots+\frac{b_{0,d-1}}{(d-1)!}y^{d-1}.$

Recall that the origin is a nodal point of $f$. By definition, $\phi$ passes
through the origin and intersect the $\mathtt{x}$–axis transversally at
origin. Hence, $b_{00}\,=\,0$ and $b_{10}\,\neq\,0$.

Therefore, we have

$\displaystyle\Big{(}f,\,(0,\,0),\,a_{1},\,\cdots,\,a_{d-1},\,a_{d}\Big{)}$
$\displaystyle\,\in\,\big{(}\mathsf{R}_{d}(d)\big{)}_{\mathsf{Aff}}\,\subset\,\mathcal{F}_{d}^{*}\times\mathbb{C}^{2}\times\mathbb{C}^{d}.$

To prove eq. 3.20, it suffices to show the following:
given any sufficiently small open set $U$ of
$\mathcal{F}_{d}^{*}\times\mathbb{C}^{2}\times\mathbb{C}^{d}$ containing
$\left(f,\,(0,\,0),\,a_{1},\,\cdots,\,a_{d-1},\,a_{d}\right)$, there exists an
element

$\Big{(}F,\,(s,\,t),\,A_{1},\,\cdots,\,A_{d-1},\,A_{d}\Big{)}\ \in\ U$

such that

$\displaystyle\Big{(}F,\,(s,\,t),\,A_{1},\,\cdots,\,A_{d-1},\,A_{d}\Big{)}\
\in\
\Big{(}\mathsf{A}_{1}^{\mathsf{F}}\underbrace{\mathsf{T}_{0}\cdots\mathsf{T}_{0}}_{(d-1)-\textnormal{times}}\mathsf{F}\Big{)}_{\mathsf{Aff}}.$
(3.21)

This element in (3.21) will be constructed explicitly. First of all, note that
the curve given by $F$, if exists, passes through the points
$(A_{1},\,0),\,\cdots,\,(A_{d-1},\,0)$. Hence, $F$ has to be of the form

$\displaystyle F(x,y)$ $\displaystyle\ =\
(x-A_{1})\cdots(x-A_{d-1})(C_{0}+C_{1}x)+y\Phi(x,y),$

where $\Phi$ is a polynomial of degree $d-1$. Write down $\Phi$ explicitly as

$\displaystyle\Phi(x,y)$ $\displaystyle\ :=\
B_{00}+B_{10}x+B_{01}y+\cdots+\frac{B_{0,d-1}}{(d-1)!}y^{d-1}.$

Since $U$ is a sufficiently small open neighbourhood of
$\left(f,\,(0,\,0),\,a_{1},\,\cdots,\,a_{d-1},\,a_{d}\right)$, it follows that
$C_{0},\,C_{1}$ and $B_{00}$ are small and $B_{10}$ is non-zero. Furthermore,
$A_{i}$ are all close to $a_{i}$. Hence, $A_{1},\,\cdots,\,A_{d-1}$ are all
distinct and different from $0$. We now wish to find a point $(s,\,t)$ close
to the origin, such that $t\,\neq\,0$ (i.e., the point $(s,\,t)$ does not lie
on the $\mathtt{x}$–axis) and is a nodal point of $F$. Hence, it is needed to
solve the following set of equations

$\displaystyle F(s,t)$ $\displaystyle\,=\,0,\qquad
F_{x}(s,t)\,=\,0\qquad\textnormal{and}\qquad F_{y}(s,t)\,=\,0,$
$\displaystyle\qquad\textnormal{where}\qquad t$
$\displaystyle\,\neq\,0\qquad\textnormal{and}\qquad(s,\,t)\quad\textnormal{is
small}.$ (3.22)

To make the computations more convenient, rewrite $F(x,\,y)$ as follows:

$\displaystyle F(x,y)$
$\displaystyle\,=\,Q(x)(C_{0}+C_{1}x)+y(B_{00}+P(x,y)),\qquad\textnormal{where}$
$\displaystyle Q(x)$
$\displaystyle\,:=\,(x-A_{1})\cdots(x-A_{d-1}),\qquad\textnormal{and}$
$\displaystyle P(x,y)$
$\displaystyle\,:=\,B_{10}x+B_{01}y+\cdots+\frac{B_{0,d-1}}{(d-1)!}y^{d-1}.$
(3.23)

Using eq. 3.22, we can solve for $C_{0},\,C_{1}$ and $B_{00}$, which produce

$\displaystyle B_{00}$ $\displaystyle\,=\,-P(s,t)-tP_{y}(s,t),$ $\displaystyle
C_{0}$
$\displaystyle\,=\,-\frac{-t^{2}Q(s)P_{y}(s,t)-st^{2}Q^{\prime}(s)P_{y}(s,t)-stQ(s)P_{x}(s,t)}{Q(s)^{2}},\qquad\textnormal{and}$
$\displaystyle C_{1}$
$\displaystyle\,=\,-\frac{t^{2}Q^{\prime}(s)P_{y}(s,t)+tQ(s)P_{x}(s,t)}{Q(s)^{2}}.$
(3.24)

Since the $A_{i}$ are all distinct from zero, it follows that $Q(0)\,\neq\,0$.
Since $s$ is close to $0$, $Q(s)\,\neq\,0$. Hence, we are able to construct
the desired solution given by eq. 3.24. This proves eq. 3.20, and hence proves
eq. 3.19.

Next, to compute the multiplicity of the intersection, first note that every
curve close to $f$ is given by eq. 3.24. To compute the multiplicity of the
intersection, it is necessary to compute $F(a_{d},\,0)$, where $a_{d}$ is
distinct from $0,\,a_{1},\,\cdots,\,a_{d-1}$. Using eqs. 3.23 and 3.24, it
follows that

$\displaystyle F(a_{d},0)$ $\displaystyle\ =\ Q(a_{d})(C_{0}+C_{1}a_{d})$
$\displaystyle\ =\ \\-\frac{Q(a_{d})a_{d}B_{10}}{Q(0)}t+O(t^{2}).$ (3.25)

Since $a_{d}$ is distinct from the other $a_{i}$, it follows (using eq. 3.23
and the fact that $A_{i}$’s are close to $a_{i}$) that $Q(a_{d})$ is non-zero.
Furthermore, $a_{d}\,\neq\,0$, since $(a_{d},\,0)$ is not the nodal point of
the curve (which is the origin). Since $Q(a_{d})\,\neq\,0$, $a_{d}\,\neq\,0$,
$B_{10}\,\neq\,0$ and $Q(0)\,\neq\,0$, we conclude from eq. 3.25 that the
multiplicity of the intersection is $1$. The multiplicity computation
completes the proof of 3.5. ∎

The statement of 3.5 is also false when $n\,=\,d$. Before stating the correct
statement for $n\,=\,d$, we need to develop yet more notation. Define

$\displaystyle\big{(}\mathsf{R}\mathsf{T}_{1}\big{)}_{n}(d)\ \subset\
\mathsf{M}^{1}_{n}(d)$ (3.26)

to be the locus of all $(H_{1},\,H_{d},\,p,\,x_{1},\,\cdots,\,x_{n})$ such
that

  * •

the curve $H_{d}$ is the union of the line $H_{1}$ and a degree $d-1$ curve,

  * •

the degree $d-1$ curve is smooth,

  * •

the degree $d-1$ curve intersects $H_{1}$ tangentially to first order at the
point $p$; the remaining points of intersection are transverse, and

  * •

the points $x_{1},\,\cdots,\,x_{n}$ are all distinct and lie on $H_{1}$; they
are also distinct from $p$.

The space can be pictorially described as follows:

Figure 4. $\big{(}\mathsf{R}\mathsf{T}_{1}\big{)}_{n}(d)$

Also define

$\displaystyle\big{(}\mathsf{R}\mathsf{A}_{1}^{\mathsf{F}}\big{)}_{n}(d)\
\subset\ \mathsf{M}^{1}_{n}(d)$ (3.27)

to be the locus of all $(H_{1},\,H_{d},\,p,\,x_{1},\,\cdots,\,x_{n})$ such
that

  * •

the curve $H_{d}$ is the union of the line $H_{1}$ and a degree $d-1$ curve,

  * •

the degree $d-1$ curve is has a nodal point at $p$,

  * •

the nodal point $p$ does not lie on $H_{1}$,

  * •

the degree $d-1$ curve intersects $H_{1}$ transversally, and

  * •

the points $x_{1},\,\cdots,\,x_{n}$ are all distinct and lie on $H_{1}$.

The space can be pictorially described as follows:

Figure 5. $\big{(}\mathsf{R}\mathsf{A}_{1}^{\mathsf{F}}\big{)}_{n}(d)$

Finally, define

$\displaystyle\mathsf{R}_{n}^{i}(d)\ \subset\ \mathsf{M}^{1}_{n}(d)$ (3.28)

to be the locus of all $(H_{1},\,H_{d},\,p,\,x_{1},\,\cdots,\,x_{n})$ such
that

  * •

the curve $H_{d}$ is the union of $H_{1}$ and a degree $d-1$ curve,

  * •

the degree $d-1$ curve is smooth,

  * •

the degree $d-1$ curve intersects $H_{1}$ transversally at all the points,

  * •

the point $p$ is one of the points of intersection of the $d-1$ curve and
$H_{1}$,

  * •

the points $x_{1},\,\cdots,\,x_{n}$ are all distinct and lie on $H_{1}$, and

  * •

The point $p$ is equal to $x_{i}$.

The space can be pictorially described as follows:

Figure 6. $\mathsf{R}_{n}^{1}(d)$

######  Theorem 3.7.

Let $d\,\in\,\mathbb{Z}^{\geq 3}$. Then, the following equality of homology
classes in $\mathsf{M}_{{d+1}}^{1}(d)$ holds:

$\displaystyle\big{[}\mathsf{A}_{1}^{\mathsf{F}}\underbrace{\mathsf{T}_{0}\cdots\mathsf{T}_{0}}_{d-\textnormal{times}}\mathsf{F}\big{]}\cdot\big{[}\big{(}x_{d+1}\in\mathsf{L}\big{)}\big{]}\cdot\big{[}\big{(}x_{d+1}\in\mathcal{C}_{d}\big{)}\big{]}\,\,$
$\displaystyle=\,\,\sum_{i=1}^{d}\big{[}\mathsf{A}_{1}^{\mathsf{F}}\underbrace{\mathsf{T}_{0}\cdots\mathsf{T}_{0}}_{d-\textnormal{times}}\mathsf{F}\big{]}\cdot\big{[}\big{(}x_{i}=x_{d+1}\big{)}\big{]}$
$\displaystyle+\sum_{i=1}^{d}\big{[}\mathsf{R}^{i}_{d+1}(d)\big{]}$
$\displaystyle+\big{[}\big{(}\mathsf{R}\mathsf{A}_{1}^{\mathsf{F}}\big{)}_{d+1}(d)\big{]}+2\big{[}\big{(}\mathsf{R}\mathsf{T}_{1}\big{)}_{d+1}(d)\big{]}.$
(3.29)

###### Proof of Theorem 3.7.

We continue with the set-up of 3.2 and 3.5. The first term on the right-hand
side of 3.7 is justified as before. Let us now justify the second, third and
fourth terms. Let us first analyze the following situation: consider the
component of the closure of

$\mathsf{A}_{1}^{\mathsf{F}}\underbrace{\mathsf{T}_{0}\cdots\mathsf{T}_{0}}_{d-\textnormal{times}}$

when the nodal point $p$ lies on the line. Unlike before, it is the component
of the closure where the nodal point is equal to one of the other points
$x_{1},x_{2},\ldots,x_{d}$ will actually contribute to the intersection. This
precisely corresponds to the second term on the right-hand side of 3.7.

Next, let us analyze the component of the closure

$\mathsf{A}_{1}^{\mathsf{F}}\underbrace{\mathsf{T}_{0}\cdots\mathsf{T}_{0}}_{d-\textnormal{times}}$

when the nodal point $p$ lies on the line, but is not equal to any of the
other points $x_{1},\,x_{2},\,\cdots,\,x_{d}$. In such a case, it will be
shown that, in the closure, the line is actually going to be tangent to one of
the branches of the node. This corresponds to the third term on the right hand
side of 3.7; as it will be shown, it contributes with a multiplicity of $2$.

The component of the closure

$\mathsf{A}_{1}^{\mathsf{F}}\underbrace{\mathsf{T}_{0}\cdots\mathsf{T}_{0}}_{d-\textnormal{times}}$

when the nodal point $p$ does not lie on the line corresponds to the second
term the right-hand side of 3.7.

For the closure and multiplicity statements, we switch to affine spaces as
usual. First make a general convention that works for all the spaces involved
here.

######  Convention 1.

Given any point

$(H_{1},\,H_{d},\,p,\,x_{1},\,\cdots,\,x_{d},\,x_{d+1})\,\in\,\overline{\mathsf{A}_{1}^{\mathsf{F}}\underbrace{\mathsf{T}_{0}\cdots\mathsf{T}_{0}}_{d-\textnormal{times}}\mathsf{F}}\cap(x_{d+1}\in
L)$

in the affine coordinate, the line $H_{1}$ is taken to be the
$\mathtt{x}$–axis, the nodal point $p$ is assumed to be the point $(a,\,b)$,
and the induced element in the affine space is denoted by

$\Big{(}f,\,(a,\,b),\,a_{1},\,\cdots,\,a_{d},\,a_{d+1}\Big{)}\,\in\,\mathcal{F}_{d}^{*}\times\mathbb{C}^{2}\times\mathbb{C}^{d+1}.$

Here, $x_{1},\,x_{2},\,\cdots,\,x_{d}$ are given by $(a_{1},\,0)$,
$(a_{2},\,0),\,\cdots,\,(a_{d},\,0)$. Since these points are all distinct, it
follows that $a_{1},\,a_{2},\,\cdots,\,a_{d}$ are all distinct. In case, $p$
lines on $H_{1}$, tactically it is assumed to be the origin.

Consider the second term on the right hand side of 3.7. For the closure
statement, we need to show that for all $1\,\leq\,i\,\leq\,d$,

$\displaystyle\overline{\mathsf{R}_{d+1}^{i}(d)}$ $\displaystyle\ \subset\
\overline{\mathsf{A}_{1}^{\mathsf{F}}\underbrace{\mathsf{T}_{0}\cdots\mathsf{T}_{0}}_{d-\textnormal{times}}\mathsf{F}}.$
(3.30)

As before, we observe that it suffices to show that

$\displaystyle\mathsf{R}_{d+1}^{i}(d)$ $\displaystyle\ \subset\
\overline{\mathsf{A}_{1}^{\mathsf{F}}\underbrace{\mathsf{T}_{0}\cdots\mathsf{T}_{0}}_{d-\textnormal{times}}\mathsf{F}}.$
(3.31)

It is enough to consider the case $i\,=\,d$; other assertions are symmetric.
Let $(H_{1},\,H_{d},\,p,\,x_{1},\,\cdots,\,x_{d},\,x_{d+1})$ be an arbitrary
point of $\mathsf{R}_{d+1}^{i}(d)$. Switching to the affine spaces, the above
convention will be followed. Since $x_{d}$ is equal to the nodal point, it
follows that $a_{d}\,=\,0$.

Since $H_{d}$ is a reducible curve, with the line $H_{1}$ being one of the
reducible components, it is given by the zero set of a polynomial $f$, which
is of the type

$\displaystyle f(x,y)$ $\displaystyle\ =\ y\varphi(x,\,y),$

where $\varphi(x,\,y)$ is a polynomial of degree $d-1$. Write down
$\varphi(x,y)$ explicitly as

$\displaystyle\varphi(x,y)$ $\displaystyle\ :=\
b_{00}+b_{10}x+b_{01}y+\cdots+\frac{b_{0,d-1}}{(d-1)!}y^{d-1}.$

Recall that the origin is a nodal point of $f$. The degree $d-1$ component
passes through the origin and intersect the $\mathtt{x}$–axis transversally at
origin. Hence, $b_{00}\,=\,0$ and $b_{10}\,\neq\,0$.

Now note that

$\displaystyle\Big{(}f,\,(0,\,0),\,a_{1},\,\cdots,\,a_{d},\,a_{d+1}\Big{)}$
$\displaystyle\,\in\,\Big{(}\mathsf{R}_{d+1}^{i}(d)\Big{)}_{\mathsf{Aff}}\,\subset\,\mathcal{F}_{d}^{*}\times\mathbb{C}^{2}\times\mathbb{C}^{d+1}.$

To prove eq. 3.31 (equivalently, eq. 3.30), it suffices to show the following:
given any sufficiently small open set $U$ of
$\mathcal{F}_{d}^{*}\times\mathbb{C}^{2}\times\mathbb{C}^{d+1}$ containing
$\left(f,\,(0,\,0),\,a_{1},\,\cdots,\,a_{d},\,a_{d+1}\right)$, there exists an
element

$\displaystyle\Big{(}F,\,(s,\,t),\,A_{1},\,\cdots,\,A_{d},\,A_{d+1}\Big{)}\,\in\,\Big{(}\mathsf{A}_{1}^{\mathsf{F}}\underbrace{\mathsf{T}_{0}\cdots\mathsf{T}_{0}}_{d-\textnormal{times}}\mathsf{F}\Big{)}_{\mathsf{Aff}}\cap
U.$

This is shown using the idea of the proof of 3.5. Since the curve $F$ passes
through the points $(A_{1},\,0),\,\cdots,\,(A_{d-1},\,0)$ and $(A_{d},\,0)$,
it is of the form

$\displaystyle F(x,\,y)$
$\displaystyle\,=\,(x-A_{1})\cdots(x-A_{d-1})(x-A_{d})C_{0}+y\Phi(x,\,y),$

where $\Phi$ is a polynomial of degree $d-1$. Write $\Phi$ explicitly as

$\displaystyle\Phi(x,y)$
$\displaystyle\,:=\,B_{00}+B_{10}x+B_{01}y+\cdots+\frac{B_{0,d-1}}{(d-1)!}y^{d-1}.$

Since $U$ is a sufficiently small open neighbourhood of
$\left(f,\,\left(0,\,0\right),\,a_{1},\,\cdots,\,a_{d-1},\,a_{d},\,a_{d+1}\right)$,
it is deduced that $C_{0}$, $A_{d}$ and $B_{00}$ are small and $B_{10}$ is
non-zero (recall that $a_{d}\,=\,0$). Furthermore, $A_{1},\,\cdots,\,A_{d-1}$
are all distinct and different from $0$. We now wish to find a point $(s,\,t)$
close to the origin, such that $t\,\neq\,0$ (that is, the point $(s,\,t)$ does
not lie on the $\mathtt{x}$–axis) and is a nodal point of $F$. Hence, it is
necessary to solve the following set of equations

$\displaystyle F(s,\,t)$ $\displaystyle\,=\,0,\qquad
F_{x}(s,\,t)\,=\,0\qquad\textnormal{and}\qquad F_{y}(s,\,t)\,=\,0,$
$\displaystyle\qquad\textnormal{where}\qquad t$
$\displaystyle\,\neq\,0\qquad\textnormal{and}\qquad(s,\,t)\quad\textnormal{is
small}.$ (3.32)

To make the computations more convenient, rewrite $F(x,\,y)$ as follows:

$\displaystyle F(x,\,y)$
$\displaystyle\,=\,Q(x)(x-A_{d})C_{0}+y(B_{00}+P(x,\,y)),$ where
$\displaystyle Q(x)\,:=\,(x-A_{1})\cdots(x-A_{d-1}),$ and $\displaystyle
P(x,\,y)\,:=\,B_{10}x+B_{01}y+\cdots+\frac{B_{0,d-1}}{(d-1)!}y^{d-1}.$ (3.33)

Using eq. 3.32, solve for $C_{0}$, $A_{d}$ and $B_{00}$, and conclude that

$\displaystyle B_{00}$ $\displaystyle\,=\,-P(s,\,t)-tP_{y}(s,\,t),$
$\displaystyle C_{0}$
$\displaystyle\,=\,-\frac{t^{2}Q^{\prime}(s)P_{y}(s,\,t)+tQ(s)P_{x}(s,\,t)}{Q(s)^{2}},$
$\displaystyle A_{d}$ $\displaystyle\,=\,s+\frac{B_{01}}{B_{10}}t+O(t^{2}).$
(3.34)

Note that $B_{00},\,C_{0}$ and $A_{d}$ are indeed small (i.e., they go to zero
as $(s,\,t)$ goes to $(0,\,0)$). Hence, we have constructed solutions to eq.
3.33, which proves eq. 3.30.

To compute the multiplicity of the intersection, first note that every curve
close to $f$ is given by eq. 3.34. In order to compute the multiplicity of the
intersection, we simply need to compute $F(a_{d+1},\,0)$, where $a_{d+1}$ is
distinct from $0,\,a_{1},\,\cdots,\,a_{d-1}$. Hence, using eqs. 3.33 and 3.34
it follows that

$\displaystyle F(a_{d+1},0)$
$\displaystyle\,=\,Q(a_{d+1})(a_{d+1}-A_{d})C_{0}$
$\displaystyle=\,Q(a_{d+1})(a_{d+1}-A_{d})\frac{B_{10}}{Q(0)}t+O(t^{2}).$
(3.35)

Since $a_{d+1}$ is distinct from the other $a_{i}$, it follows using eq. 3.33
(and using the fact that $A_{i}$’s are close to $a_{i}$) that $Q(a_{d+1})$ is
non-zero. Furthermore, $a_{d+1}\,\neq\,0$, because $(a_{d+1},\,0)$ is not the
nodal point of the curve (which is the origin). Since

$Q(a_{d+1})\,\neq\,0,\qquad\big{(}a_{d+1}-A_{d}\big{)}\,\neq\,0,\qquad
B_{10}\,\neq\,0\qquad\textnormal{and}\qquad Q(0)\,\neq\,0,$

it follows from eq. 3.35 that the multiplicity of the intersection is $1$. The
multiplicity computation completes the justification for the second term on
the right-hand side of 3.7 (on the level of cycles).

Next, the fourth term on the right-hand side of 3.7 will be justified. As
before, it suffices to show that

$\displaystyle\big{(}\mathsf{R}\mathsf{T}_{1}\big{)}_{d+1}(d)$ $\displaystyle\
\subset\
\overline{\mathsf{A}_{1}^{\mathsf{F}}\underbrace{\mathsf{T}_{0}\cdots\mathsf{T}_{0}}_{d-\textnormal{times}}\mathsf{F}}.$
(3.36)

Let $(H_{1},\,H_{d},\,p,\,x_{1},\,\cdots,\,x_{d},\,x_{d+1})$ be an arbitrary
point of $\big{(}\mathsf{R}\mathsf{T}_{1}\big{)}_{d+1}(d)$. As before, switch
to affine spaces. Following 1, assume that $p$ — the point of tangency of $f$
— is the origin. Since $H_{d}$ is a reducible curve, with the line $H_{1}$
being one of the reducible components, it follows that $H_{d}$ is given by the
zero set of a polynomial $f$, which is of the type

$\displaystyle f(x,y)$ $\displaystyle\ =\ y\varphi(x,y),$

where $\varphi(x,\,y)$ is a polynomial of degree $d-1$. Write $\varphi(x,\,y)$
explicitly as

$\displaystyle\varphi(x,\,y)$
$\displaystyle\,:=\,b_{00}+b_{10}x+b_{01}y+\cdots+\frac{b_{0,d-1}}{(d-1)!}y^{d-1}.$

Since the degree $d-1$ component is tangent of first order to the line at the
origin, it follows that $b_{00}\,=\,0$, $b_{10}\,=\,0$ and $b_{20}\,\neq\,0$.
Furthermore, that the point is a smooth point of the degree $d-1$ curve
implies that $b_{01}\,\neq\,0$.

Now note that

$\displaystyle\Big{(}f,(0,0),a_{1},\ldots,a_{d},a_{d+1}\Big{)}$
$\displaystyle\,\in\,\Big{(}\big{(}\mathsf{R}\mathsf{T}_{1}\big{)}_{d+1}(d)\Big{)}_{\mathsf{Aff}}\,\subset\,\mathcal{F}_{d}^{*}\times\mathbb{C}^{2}\times\mathbb{C}^{d+1}.$

To prove eq. 3.36, it suffices to show the following: given small a
neighbourhood $U$ of
$\left(f,\,(0,\,0),\,a_{1},\,\cdots,\,a_{d},\,a_{d+1}\right)$, there exists an
element

$\displaystyle\Big{(}F,\,(s,\,t),\,A_{1},\,\cdots,\,A_{d},\,A_{d+1}\Big{)}\,\in\,\Big{(}\mathsf{A}_{1}^{\mathsf{F}}\underbrace{\mathsf{T}_{0}\cdots\mathsf{T}_{0}}_{d-\textnormal{times}}\mathsf{F}\Big{)}_{\mathsf{Aff}}\cap
U.$

We will follow the previous idea. Note that $F$ passes through the points
$(A_{1},\,0),\,\cdots,\,(A_{d},\,0)$. Hence, $F$ has to be of the form

$\displaystyle F(x,y)$ $\displaystyle\ =\
(x-A_{1})\cdots(x-A_{d-1})(x-A_{d})C_{0}+y\Phi(x,\,y),$

where $\Phi$ is a polynomial of degree $d-1$. Write $\Phi$ explicitly as

$\displaystyle\Phi(x,\,y)$
$\displaystyle\,:=\,B_{00}+B_{10}x+B_{01}y+\cdots+\frac{B_{0,d-1}}{(d-1)!}y^{d-1}.$

Since $U$ is a sufficiently small open neighbourhood of
$\left(f,\,\left(0,\,0\right),\,a_{1},\,\cdots,\,a_{d-1},\,a_{d},\,a_{d+1}\right)$,
it follows that $C_{0}$, $B_{00}$ and $B_{10}$ are small while $B_{20}$ and
$B_{01}$ are non-zero. Furthermore, $A_{1},\,\cdots,\,A_{d}$ are all distinct
and different from $0$. We now wish to find a point $(s,\,t)$ close to the
origin, such that $t\,\neq\,0$ (i.e., the point $(s,\,t)$ does not lie on the
$\mathtt{x}$–axis) and is a nodal point of the curve defined by the zero set
of $F$. Hence, we need to solve the following set of equations

$\displaystyle F(s,\,t)$ $\displaystyle\,=\,0,\qquad
F_{x}(s,\,t)\,=\,0\qquad\textnormal{and}\qquad F_{y}(s,\,t)\,=\,0,$
$\displaystyle\qquad\textnormal{where}\qquad t$
$\displaystyle\,\neq\,0\qquad\textnormal{and}\qquad(s,\,t)\quad\textnormal{is
small}.$ (3.37)

To make the computations more convenient, rewrite $F(x,\,y)$ as follows:

$\displaystyle F(x,\,y)$
$\displaystyle\,=\,Q(x)C_{0}+y(B_{00}+B_{10}x+P(x,\,y)),$ where $\displaystyle
Q(x)\,:=\,(x-A_{1})\cdots(x-A_{d}),$ and $\displaystyle
P(x,\,y)\,:=\,B_{01}y+\frac{B_{20}}{2}x^{2}+B_{11}xy+\frac{B_{02}}{2}y^{2}+\cdots+\frac{B_{0,d-1}}{(d-1)!}y^{d-1}.$
(3.38)

Using eqs. 3.37 and 3.38, solve for $C_{0}$, $B_{00}$ and $B_{10}$ to conclude
that

$\displaystyle C_{0}$ $\displaystyle\,=\,\frac{t^{2}P_{y}(s,t)}{Q(s)},$
$\displaystyle B_{00}$
$\displaystyle\,=\,-\Big{(}\frac{P(s,t)Q(s)+tQ(s)P_{y}(s,t)-stQ^{\prime}(s)P_{y}(s,t)-sQ(s)P_{x}(s,t)}{Q(s)}\Big{)},\quad\textnormal{and}$
$\displaystyle B_{10}$
$\displaystyle\,=\,-\Big{(}\frac{tQ^{\prime}(s)P_{y}(s,t)+Q(s)P_{x}(s,t)}{Q(s)}\Big{)}.$
(3.39)

This proves eq. 3.36.

Next, to compute the multiplicity of the intersection, first note that every
curve close to $f$ is given by eq. 3.39. To compute the multiplicity of the
intersection, we have to compute $F(a_{d+1},\,0)$, where $a_{d+1}$ is distinct
from $0,\,a_{1},\,\cdots,\,a_{d}$. Using eqs. 3.38 and 3.39 it follows that

$\displaystyle F(a_{d+1},0)$
$\displaystyle\,=\,Q(a_{d+1})\frac{B_{01}}{Q(0)}t^{2}+O(t^{3}).$ (3.40)

Since

$Q(a_{d+1})\,\neq\,0,\qquad Q(0)\,\neq\,0\qquad\textnormal{and}\qquad
B_{01}\,\neq\,0,$

it follows from eq. 3.40 that the multiplicity of the intersection is $2$. The
multiplicity computation completes the justification for the fourth term on
the right-hand side of (3.7) (on the level of cycles).

It remains to justify the third term on the right-hand side of 3.7. We need to
show that

$\displaystyle\big{(}\mathsf{R}\mathsf{A}_{1}^{\mathsf{F}}\big{)}_{d+1}(d)$
$\displaystyle\ \subset\
\overline{\mathsf{A}_{1}^{\mathsf{F}}\underbrace{\mathsf{T}_{0}\cdots\mathsf{T}_{0}}_{d-\textnormal{times}}\mathsf{F}}.$
(3.41)

Let $(H_{1},\,H_{d},\,p,\,x_{1},\,\cdots,\,x_{d},\,x_{d+1})$ be an arbitrary
point of $\big{(}\mathsf{R}\mathsf{A}_{1}^{\mathsf{F}}\big{)}_{d+1}(d)$.
Similarly, given a small enough neighbourhood of the point, we need to find an
element of it such that it belongs to the open part of the rightmost space of
eq. 3.41 as well. We now descend to the affine space. Following 1, assume that
$(a,b)$ is a nodal point. Since the nodal point does not lie on the line, it
follows that $b\,\neq\,0$. Since $H_{d}$ is a reducible curve, with the line
$H_{1}$ being one of the reducible components, the curve $f$ is given by

$\displaystyle f(x,\,y)$ $\displaystyle\ =\ y\varphi(x,\,y),$

where $\varphi(x,\,y)$ is a polynomial of degree $d-1$. Write down
$\varphi(x,y)$ explicitly as

$\displaystyle\varphi(x,\,y)$ $\displaystyle\ :=\
b_{00}+b_{10}x+b_{01}y+\cdots+\frac{b_{0,d-1}}{(d-1)!}y^{d-1}.$

Note that $(a,\,b)$ is a nodal point of $\varphi$. Hence,

$\displaystyle\varphi(a,\,b)$
$\displaystyle\,=\,0,\qquad\varphi_{x}(a,\,b)\,=\,0\qquad\textnormal{and}\qquad\varphi_{y}(a,\,b)\,=\,0.$
(3.42)

As before, given an open neighbourhood $U$ of
$\left(f,\,(a,\,b),\,a_{1},\,\cdots,\,a_{d},\,a_{d+1}\right)$, we will find an
element

$\displaystyle\Big{(}F,\,(A,\,B),\,A_{1},\,\cdots,\,A_{d},\,A_{d+1}\Big{)}\,\in\,\Big{(}\mathsf{A}_{1}^{\mathsf{F}}\underbrace{\mathsf{T}_{0}\cdots\mathsf{T}_{0}}_{d-\textnormal{times}}\mathsf{F}\Big{)}_{\mathsf{Aff}}\cap
U.$

The curve $F$ passes through the points $(A_{1},\,0),\,\cdots,\,(A_{d},\,0)$.
Hence, $F$ has to be of the form

$\displaystyle F(x,y)$ $\displaystyle\ =\
(x-A_{1})\cdots(x-A_{d-1})(x-A_{d})C_{0}+y\Phi(x,\,y),$

where $\Phi$ is a polynomial of degree $d-1$. Write down $\Phi$ explicitly as

$\displaystyle\Phi(x,\,y)$ $\displaystyle\ :=\
B_{00}+B_{10}x+B_{01}y+\cdots+\frac{B_{0,d-1}}{(d-1)!}x^{d-1}.$

Since $U$ is a sufficiently small open neighbourhood of
$\left(f,\,(a,\,b),\,a_{1},\,\cdots,\,a_{d},\,a_{d+1}\right)$, it follows that
$C_{0}$ $B_{00}$, $B_{10}$ and $B_{01}$ are small. Furthermore,
$A_{1},\,\cdots,\,A_{d}$ are all distinct. We now wish to find a point
$(A,\,B)$ close to $(a,\,b)$, such that it is a nodal point of the curve
defined by the zero set of $F$. Hence, we need to solve the following set of
equations

$\displaystyle F(A,\,B)\,=\,0,\qquad F_{x}(A,\,B)$
$\displaystyle\,=\,0,\qquad\textnormal{and}\qquad F_{y}(A,\,B)\,=\,0,$
$\displaystyle\textnormal{where $(A,\,B)$ is close to $(a,\,b)$}.$ (3.43)

Rewrite $F(x,\,y)$ as follows:

$\displaystyle F(x,\,y)$
$\displaystyle\,=\,Q(x)C_{0}+y(B_{00}+B_{10}x+B_{01}y+\alpha(x)x^{2}+\beta(y)y^{2}+\gamma(x,\,y)xy),$
where $\displaystyle Q(x)\,:=\,(x-A_{1})\cdots(x-A_{d}),$
$\displaystyle\alpha(x)\,:=\,\frac{B_{20}}{2}+\frac{B_{30}x}{6}+\cdots,$
$\displaystyle\beta(y)\,:=\,\frac{B_{02}}{2}+\frac{B_{03}y}{6}+\cdots,$
$\displaystyle\gamma(x,\,y)\,:=\,B_{11}+\frac{B_{21}}{2}x+\frac{B_{12}}{2}y+\cdots.$
(3.44)

Using eqs. 3.43 and 3.44, solve for $B_{00}$, $B_{10}$ and $B_{01}$ to
conclude that

$\displaystyle B_{00}\,=\,$ $\displaystyle
A^{2}\alpha(A)+B^{2}\beta(B)+AB\gamma(A,\,B)-\frac{2C_{0}Q(A)}{B}$
$\displaystyle+A^{3}\alpha^{\prime}(A)+B^{3}\beta^{\prime}(B)+\frac{AC_{0}Q^{\prime}(A)}{B}$
$\displaystyle+AB^{2}\gamma_{y}(A,B)+A^{2}B\gamma_{x}(A,\,B),$ $\displaystyle
B_{10}\,=\,$
$\displaystyle-2A\alpha(A)-B\gamma(A,\,B)-A^{2}\alpha^{\prime}(A)-\frac{C_{0}Q^{\prime}(A)}{B}-AB\gamma_{x}(A,\,B),\quad\textnormal{and}$
$\displaystyle B_{01}\,=\,$
$\displaystyle-2B\beta(B)-A\gamma(A,B)+\frac{C_{0}Q(A)}{B^{2}}-B^{2}\beta^{\prime}(B)-AB\gamma_{y}(A,\,B).$
(3.45)

Note that $B\,\neq\,0$ was needed to solve the equations. Substituting
$C_{0}\,=\,0$ in the expression for $B_{00},B_{10}$ and $B_{01}$, we precisely
get the expression for $b_{00},b_{10}$ and $b_{01}$, with $b_{ij}$ replaced by
$B_{ij}$. This follows from the fact that $(a,\,b)$ is a nodal point of
$\varphi$ (i.e., by using eq. 3.42). Hence, the $B_{00},B_{10}$ and $B_{01}$
we have constructed are indeed close to $b_{00},b_{10}$ and $b_{01}$. This
proves eq. 3.41.

Next, let us compute the multiplicity of the intersection. Note that every
curve close to $f$ is given by eq. 3.45. For intersection multiplicity here,
we need to compute $F(a_{d+1},\,0)$, where $a_{d+1}$ is distinct from
$a_{1},\,\cdots,\,a_{d}$. Using eq. 3.44, it follows that

$\displaystyle F(a_{d+1},0)$ $\displaystyle\,=\,Q(a_{d+1})C_{0}$
$\displaystyle=\,Q(a_{d+1})C_{0}+O(C_{0}^{2}).$ (3.46)

Since $Q(a_{d+1})\,\neq\,0$, the required intersection multiplicity is $1$ as
follows from eq. 3.46.

The third term on the right-hand side of 3.7 is now justified. This completes
the proof of 3.7. ∎

###  3.2. Numerical invariants: enumerating curves with a free node

Using the theorems proved above eqs. 3.1, 3.4 and 3.5 will be derived.

Using eq. 3.12, 3.1 can be rewritten as

$\displaystyle\big{[}\mathsf{A}_{1}^{\mathsf{F}}\mathsf{F}\big{]}\cdot(y_{1}+a_{1})\cdot(y_{d}+da_{d})$
$\displaystyle\,=\,\,\big{[}\mathsf{A}_{1}^{\mathsf{F}}\mathsf{T}_{0}\big{]}.$
(3.47)

Given a positive integer $t$, define the following cycle

$\displaystyle\mu_{t}\,\,:=\,\,y_{1}^{2}y_{d}^{\delta_{d}-(t+1)}a_{1}.$ (3.48)

Now intersect both sides of eq. 3.47 with $\mu_{t}$ with $t=1$. Using the fact
that $y_{1}^{3}\,=\,0$ and $a_{1}^{3}\,=\,0$, the equality simplifies to

$\displaystyle\big{[}\mathsf{A}_{1}^{\mathsf{F}}\mathsf{F}\big{]}\cdot
y_{1}^{2}y_{d}^{\delta_{d}-1}a_{1}^{2}$
$\displaystyle\,=\,\,\big{[}\mathsf{A}_{1}^{\mathsf{F}}\mathsf{T}_{0}\big{]}\cdot
y_{1}^{2}y_{d}^{\delta_{d}-2}a_{1}.$ (3.49)

The left-hand side of eq. 3.49 is an intersection number inside
$\mathsf{M}^{1}_{1}$. We are counting a line, a degree $d$ curve and two
points $p$ and $x_{1}$, such that the point $p$ is a nodal point of the curve
and the point $x_{1}$ is free (it need not lie on the line or the cubic).
Intersecting with $y_{1}^{2}$ fixes the line. Intersecting with $a_{1}^{2}$
fixes the free point. Intersecting with $y_{d}^{\delta_{d}-1}$ gives a nodal
curve of degree $d$. Hence, the left-hand side of eq. 3.49 is precisely equal
to $\mathsf{N}_{d}(\mathsf{A}_{1})$.

Consider the right-hand side of eq. 3.49. Intersecting with $y_{1}^{2}$ fixes
the line. However, now we are intersecting with $a_{1}$. This fixes the
location of the point $x_{1}$ to be on the line. Finally, intersecting with
$y_{d}^{\delta_{d}-2}$ gives the number of nodal degree $d$ curves passing
through $\delta_{d}-2$ generic points. Hence, the number on the right-hand
side is precisely equal to $\mathsf{N}_{d}(\mathsf{A}_{1},1)$. Consequently,
eq. 3.49 is exactly the same as the first equality of eq. 3.1.

As before, 3.2 can be rewritten as follows

$\displaystyle\big{[}\mathsf{A}_{1}^{\mathsf{F}}\underbrace{\mathsf{T}_{0}\ldots\mathsf{T}_{0}}_{n-\textnormal{times}}\mathsf{F}\big{]}\cdot\big{(}y_{1}+a_{n+1}\big{)}\cdot\big{(}y_{d}+da_{n+1}\big{)}\,\,$
$\displaystyle=\,\,\big{[}\mathsf{A}_{1}^{\mathsf{F}}\underbrace{\mathsf{T}_{0}\ldots\mathsf{T}_{0}}_{(n+1)-\textnormal{times}}\big{]}$
$\displaystyle+\sum_{i=1}^{n}\big{[}\mathsf{A}_{1}^{\mathsf{F}}\underbrace{\mathsf{T}_{0}\ldots\mathsf{T}_{0}}_{n-\textnormal{times}}\mathsf{F}\big{]}\cdot\big{(}a_{i}^{2}+a_{i}a_{n+1}+a_{n+1}^{2}\big{)}.$
(3.50)

Now intersect both sides of section 3.2 with $\mu_{n+1}$ (cf. eq. 3.48) and
simplify. This yields

$\displaystyle\big{[}\mathsf{A}_{1}^{\mathsf{F}}\underbrace{\mathsf{T}_{0}\ldots\mathsf{T}_{0}}_{n-\textnormal{times}}\mathsf{F}\big{]}\cdot
y_{1}^{2}y_{d}^{\delta_{d}-(n+1)}\cdot(a_{1}\ldots a_{n})\cdot a_{n+1}^{2}$
$\displaystyle=\big{[}\mathsf{A}_{1}^{\mathsf{F}}\underbrace{\mathsf{T}_{0}\ldots\mathsf{T}_{0}}_{(n+1)-\textnormal{times}}\big{]}\cdot
y_{1}^{2}y_{d}^{\delta_{d}-(n+2)}\cdot(a_{1}\ldots a_{n+1}).$ (3.51)

Consider the left-hand side of eq. 3.51. Recall that this is an intersection
number inside $\mathsf{M}^{1}_{n+1}$. We are counting a line, a degree $d$
curve and $(n+2)$ points, namely $p$, $x_{1},\,x_{2},\,\cdots,\,x_{n+1}$, such
that the point $p$ is a nodal point of the curve, the points $x_{1}$ to
$x_{n}$ are transverse points of intersection of the line and the curve while
the point $x_{n+1}$ is free.

Note that intersecting with $y_{1}^{2}$ fixes the line, and intersecting with
$a_{n+1}^{2}$ fixes the free point. Intersecting with
$y_{d}^{\delta_{d}-(n+1)}$ makes the nodal curve pass through
$\delta_{d}-(n+1)$ generic point. Finally, intersecting with $(a_{1}\ldots
a_{n})$ fixes the location of the points $x_{1},\cdots x,\,_{n}$ on the line.
Hence, the left-hand side of eq. 3.51 is precisely equal to
$\mathsf{N}_{d}(\mathsf{A}_{1},n)$, the number of nodal degree $d$ curves
passing through $\delta_{d}-(n+1)$ generic points and $n$ generic points on
the line. Similarly, the right-hand side of eq. 3.51 is equal to
$\mathsf{N}_{d}(\mathsf{A}_{1},n)$. Hence, eq. 3.51 implies — as we vary $n$
from $0$ to $d-2$ — all the equalities of eq. 3.1.

Equation 3.4 can be obtained using 3.5. Intersect both sides of 3.5 with
$\mu_{d}$ (cf. eq. 3.48). After simplifying, this gives us

$\displaystyle\big{[}\mathsf{A}_{1}^{\mathsf{F}}\underbrace{\mathsf{T}_{0}\ldots\mathsf{T}_{0}}_{(d-1)-\textnormal{times}}\mathsf{F}\big{]}\cdot
y_{1}^{2}y_{d}^{\delta_{d}-d}\cdot(a_{1}\ldots a_{d-1})\cdot a_{d}^{2}$
$\displaystyle\,=\,\big{[}\mathsf{A}_{1}^{\mathsf{F}}\underbrace{\mathsf{T}_{0}\ldots\mathsf{T}_{0}}_{d-\textnormal{times}}\big{]}\cdot
y_{1}^{2}y_{d}^{\delta_{d}-(d+1)}\cdot(a_{1}\ldots a_{d})$
$\displaystyle+\big{[}\mathsf{R}_{d}(d)\big{]}\cdot
y_{1}^{2}y_{d}^{\delta_{d}-(d+1)}a_{1}a_{2}\ldots a_{d}.$ (3.52)

As before, the left-hand side of eq. 3.52 is
$\mathsf{N}_{d}(\mathsf{A}_{1},d-1)$. Similarly, the first term on the right-
hand side of eq. 3.52 is $\mathsf{N}_{d}(\mathsf{A}_{1},d)$. Now analyze the
second term on the right-hand side, namely

$\big{[}\mathsf{R}_{d}(d)\big{]}\cdot
y_{1}^{2}y_{d}^{\delta_{d}-(d+1)}a_{1}a_{2}\ldots a_{d}.$

As before, intersecting with $y_{1}^{2}$ fixes the line. Note that

$\delta_{d}-(d+1)=\delta_{d-1}.$

Hence, intersecting with $y_{d}^{\delta_{d-1}}$ fixes the degree $(d-1)$
component of the curve (the line component has already been fixed, since we
have intersected with $y_{1}^{2}$). However, corresponding to each such degree
$(d-1)$ curve, there are $(d-1)$ choices for the nodal point $p$ (it is one of
the points of intersection of the curve with the line). Hence, this number is
precisely equal to

$(d-1)\mathsf{N}_{d-1}(\mathsf{S}).$

This gives us the degeneration formula of eq. 3.4.

Equation 3.5 is obtained by intersecting both sides of 3.7 with $\mu_{d+1}$
(cf. eq. 3.48). After simplifying, this yields the following:

$\displaystyle\big{[}\mathsf{A}_{1}^{\mathsf{F}}\underbrace{\mathsf{T}_{0}\ldots\mathsf{T}_{0}}_{d-\textnormal{times}}\mathsf{F}\big{]}\cdot
y_{1}^{2}y_{d}^{\delta_{d}-(d+1)}\cdot(a_{1}\ldots a_{d})\cdot a_{d+1}^{2}$
$\displaystyle=\sum_{i=1}^{d}\big{[}\mathsf{R}^{i}_{d+1}(d)\big{]}\cdot\mu_{4}$
$\displaystyle+\big{[}\big{(}\mathsf{R}\mathsf{A}_{1}^{\mathsf{F}}\big{)}_{d+1}(d)\big{]}\cdot\mu_{4}+2\big{[}\big{(}\mathsf{R}\mathsf{T}_{1}\big{)}_{d+1}(d)\big{]}\cdot\mu_{4}.$
(3.53)

As before, the left-hand side of eq. 3.53 is simply
$\mathsf{N}_{d}(\mathsf{A}_{1},d)$. Now analyze the right-hand side of eq.
3.53. After simplification it is obtained that the first term on the right-
hand side is

$\displaystyle\sum_{i=1}^{d}\big{[}\mathsf{R}^{i}_{d+1}(d)\big{]}\cdot\mu_{4}$
$\displaystyle\,=\,d\mathsf{N}_{d-1}(\mathsf{S}).$

Next, after simplification, it is obtained that the second term on the right-
hand side is equal to

$\displaystyle\big{[}\big{(}\mathsf{R}\mathsf{A}_{1}^{\mathsf{F}}\big{)}_{d+1}(d)\big{]}\cdot\mu_{4}$
$\displaystyle\,=\,\mathsf{N}_{d-1}(\mathsf{A}_{1}).$

Finally, note that

$\displaystyle\big{[}\big{(}\mathsf{R}\mathsf{T}_{1}\big{)}_{d+1}(d)\big{]}\cdot\mu_{4}$
$\displaystyle=\mathsf{N}_{d-1}(\mathsf{T}_{1}).$

Combining these facts, we obtain the degeneration formula given by eq. 3.5.

###  3.3. Numerical invariants: enumerating curves with the node lying on a
line

Now we will prove eq. 2.3 (first part of 2.1). It is enough to establish the
equalities of eqs. 3.6, 3.8 and 3.9.

Given an integer $t\geq 0$, define the following cycle:

$\displaystyle\mu_{t}^{\mathsf{L}}\ :=\
y_{1}^{2}y_{d}^{\delta_{d}-(t+2)}b_{1}a_{1}a_{2}\cdots a_{n}a_{n+1}.$

Intersecting both sides of section 3.2 with $\mu_{n+1}^{\mathsf{L}}$, and a
similar analysis as above, produce eq. 3.6.
In order to get eq. 3.8, intersect both sides of 3.5 with
$\mu_{d}^{\mathsf{L}}$; that will give eq. 3.8.

Similarly, to get eq. 3.9, intersect both sides of 3.7 with
$\mu_{d+1}^{\mathsf{L}}$. This produces the required equality.

##  4\. Extension of Caporaso-Harris to one cuspidal curves

We will now extend the degeneration arguments presented in Section 3 and solve
the following problem:

How many degree $d$ curves are there in $\mathbb{P}^{2}$ that have a cusp and
pass through $\delta_{d}-2$ generic points?

Let $\mathsf{N}_{d}(\mathsf{A}_{2},k)$ denote the number of cuspidal degree
$d$ curves, passing through $\delta_{d}-2-k$ generic points and $k$ points on
a line. When $k\,=\,0$, we abbreviate and omit writing the $k$, i.e.,
$\mathsf{N}_{d}(\mathsf{A}_{2})\,:=\,\mathsf{N}_{d}(\mathsf{A}_{2},0)$. We
will show that

$\displaystyle\mathsf{N}_{d}(\mathsf{A}_{2})$
$\displaystyle\,=\,\mathsf{N}_{d}(\mathsf{A}_{2},\,1),$
$\displaystyle\mathsf{N}_{d}(\mathsf{A}_{2},1)$
$\displaystyle\,=\,\mathsf{N}_{d}(\mathsf{A}_{2},\,2),$ $\displaystyle\ldots$
$\displaystyle\mathsf{N}_{d}(\mathsf{A}_{2},d-2)$
$\displaystyle\,=\,\mathsf{N}_{d}(\mathsf{A}_{2},\,d-1).$ (4.1)

In other words,

$\displaystyle\mathsf{N}_{d}(\mathsf{A}_{2})$
$\displaystyle\,=\,\mathsf{N}_{d}(\mathsf{A}_{2},\,d-1).$ (4.2)

Hence, the number of cuspidal degree $d$ curves passing through $\delta_{d}-2$
generic points is equal to the number of cuspidal degree $d$ curves passing
through $\delta_{d}-d-1$ generic points and $(d-1)$ collinear points. We
further show that

$\displaystyle\mathsf{N}_{d}(\mathsf{A}_{2},d-1)$
$\displaystyle\,=\,\mathsf{N}_{d}(\mathsf{A}_{2},d)+3\mathsf{N}_{d-1}(\mathsf{T}_{1});$
(4.3) $\displaystyle\mathsf{N}_{d}(\mathsf{A}_{2},d)$
$\displaystyle\,=\,\mathsf{N}_{d-1}(\mathsf{A}_{2})+3\mathsf{N}_{d-1}(\mathsf{A}_{1}^{\mathsf{L}})+3d\mathsf{N}_{d-1}(\mathsf{T}_{1}^{\mathsf{Pt}})+2\mathsf{N}_{d-1}(\mathsf{T}_{2}).$
(4.4)

Equations (4.2), (4.3) and (4.4) together imply eq. 2.4. Let us now prove
these assertions. As before, certain cycle level statements are proved first;
then we intersect them with cycles of complementary dimensions to get equality
of numbers. The cycle level statements are analogs of 3.1, 3.2, 3.5 and 3.7.
Here we consider cuspidal curves instead of the nodal ones.

In Section 3, a few subspaces of $\mathsf{M}_{n}^{1}$ (cf. eqs. 3.10, 3.11,
3.17, 3.26, 3.27 and 3.28) are considered. In all the definitions there
$\mathsf{A}_{1}$ stands for curves having only one node. Here, we consider
curves having one cusp only (that is, $\mathsf{A}_{2}$ singularity). We can
consider similar subspaces for cuspidal curves, and replace $A_{1}$ by $A_{2}$
in the notation. For example,

$\mathsf{A}_{2}^{\mathsf{F}}\underbrace{\mathsf{T}_{0}\cdots\mathsf{T}_{0}}_{n-\textnormal{times}}\,\,\subset\,\,\mathsf{M}_{n}^{1}.$

The definitions remain the same, but only the nodal condition will be replaced
by the cuspidal one.

The following result is analogous to 3.1.

######  Theorem 4.1.

The following equality of homology classes hold in $\mathsf{M}^{1}_{1}(d)$:

$\displaystyle\big{[}\mathsf{A}_{2}^{\mathsf{F}}\mathsf{F}\big{]}\cdot\big{[}\big{(}x_{1}\in\mathsf{L}\big{)}\big{]}\cdot\big{[}\big{(}x_{1}\in\mathcal{C}_{d}\big{)}\big{]}$
$\displaystyle\,=\,\big{[}\mathsf{A}_{2}^{\mathsf{F}}\mathsf{T}_{0}\big{]},$

where $d\,\in\,\mathbb{Z}^{\geq 3}$.

4.1 is a special case of the following analog of 3.2.

######  Theorem 4.2.

Take $d\,\in\,\mathbb{Z}^{\geq 3}$, and let $n$ be an integer such that

$0\,\leq\,n\,\leq\,d-2.$

Then, on $\mathsf{M}^{1}_{n+1}(d)$ the following equality of homology classes
hold:

$\displaystyle\big{[}\mathsf{A}_{2}^{\mathsf{F}}\underbrace{\mathsf{T}_{0}\cdots\mathsf{T}_{0}}_{n-\textnormal{times}}\mathsf{F}\big{]}\cdot\big{[}\big{(}x_{n+1}\in\mathsf{L}\big{)}\big{]}\cdot\big{[}\big{(}x_{n+1}\in\mathcal{C}_{d}\big{)}\big{]}$
$\displaystyle\,\,=\,\,\big{[}\mathsf{A}_{2}^{\mathsf{F}}\underbrace{\mathsf{T}_{0}\cdots\mathsf{T}_{0}}_{(n+1)-\textnormal{times}}\big{]}$
$\displaystyle+\sum_{i=1}^{n}\big{[}\mathsf{A}_{2}^{\mathsf{F}}\underbrace{\mathsf{T}_{0}\cdots\mathsf{T}_{0}}_{n-\textnormal{times}}\mathsf{F}\big{]}\cdot\big{[}\big{(}x_{i}=x_{n+1}\big{)}\big{]}.$
(4.5)

######  Remark 4.3.

Equation 4.1 can be obtained from 4.2 by intersecting both sides with

$\mu_{n+1}:=y_{1}^{2}y_{d}^{\delta_{d}-(n+3)}a_{1}a_{2}\cdots a_{n}a_{n+1}.$

The reasoning is identical to how 3.2 implies eq. 3.1 described in Section
3.2.

###### Proof of Theorem 4.2.

The proof is almost the same as the proof of 3.2. First, we need to show that
if $d\,\geq\,2$, then $\mathsf{A}_{2}^{\mathsf{F}}$ is a smooth manifold of
codimension $4$. This has been proved in [1, p. 22, Proposition 6.1].

The rest of the proof is the same if $n\,\leq\,d-3$. When $n\,=\,d-2$, there
is a small point we need to observe. By a dimension counting argument, we
can’t rule out a reducible curve, one of whose component is the line. However,
when we intersect with generic constraints, we will require the degree $d-1$
curve to intersect the line transversally. This is not possible for a simple
geometric reason; a cuspidal point of the degree $d$ curve can’t become a
nodal point (it can become a tacnode, but not a node). Hence, the reducible
configuration is not possible, even when $n\,=\,d-2$. ∎

######  Theorem 4.4.

Let $d\,\in\,\mathbb{Z}^{\geq 3}$. The following equality of homology classes
holds in $\mathsf{M}_{{d}}^{1}(d)$:

$\displaystyle\big{[}\mathsf{A}_{2}^{\mathsf{F}}\underbrace{\mathsf{T}_{0}\cdots\mathsf{T}_{0}}_{(d-1)-\textnormal{times}}\mathsf{F}\big{]}\cdot\big{[}\big{(}x_{d}\in\mathsf{L}\big{)}\big{]}\cdot\big{[}\big{(}x_{d}\in\mathcal{C}_{d}\big{)}\big{]}$
$\displaystyle\,\,=\,\,\big{[}\mathsf{A}_{2}^{\mathsf{F}}\underbrace{\mathsf{T}_{0}\cdots\mathsf{T}_{0}}_{d-\textnormal{times}}\big{]}$
$\displaystyle+\sum_{i=1}^{d-1}\big{[}\mathsf{A}_{2}^{\mathsf{F}}\underbrace{\mathsf{T}_{0}\cdots\mathsf{T}_{0}}_{(d-1)-\textnormal{times}}\mathsf{F}\big{]}\cdot\big{[}\big{(}x_{i}=x_{d}\big{)}\big{]}$
$\displaystyle+3\big{[}\big{(}\mathsf{R}\mathsf{T}_{1}\big{)}_{d}(d)\big{]}.$
(4.6)

######  Remark 4.5.

As before, intersecting both sides of 4.4 with

$\mu_{d}\ :=\ y_{1}^{2}y_{d}^{\delta_{d}-(d+2)}a_{1}a_{2}\cdots a_{d},$

eq. 4.3 can be obtained.

###### Proof of 4.4.

The first and second terms on the right-hand side of 4.4 are justified as
before. The third term needs to be justified. We continue with the set-up of
the proof of Theorem 3.5. Following the same thread of reasoning, it suffices
to show the following:

$\displaystyle\big{(}\mathsf{R}\mathsf{T}_{1}\big{)}_{d}(d)$ $\displaystyle\
\subset\
\overline{\mathsf{A}_{2}^{\mathsf{F}}\underbrace{\mathsf{T}_{0}\ldots\mathsf{T}_{0}}_{(d-1)-\textnormal{times}}\mathsf{F}}.$
(4.7)

Let $\left(H_{1},\,H_{d},\,p,\,x_{1},\,\cdots,\,x_{d-1},\,x_{d}\right)$ be an
arbitrary point of $\big{(}\mathsf{R}\mathsf{T}_{1}\big{)}_{d}(d)$. It will be
shown that every open neighbourhood of
$(H_{1},H_{d},x_{1},\ldots,x_{d-1},x_{d})$ intersects

$\mathsf{A}_{2}^{\mathsf{F}}\underbrace{\mathsf{T}_{0}\cdots\mathsf{T}_{0}}_{(d-1)-\textnormal{times}}\mathsf{F}.$

Switching to an affine space, assume $H_{1}$ to be the $\mathtt{x}$–axis, and
the point of tangency $p$ is the origin. Furthermore, assume that
$x_{1},\,x_{2},\,\cdots,\,x_{d-1}$ are given by $(a_{1},\,0)$,
$(a_{2},\,0),\,\cdots,\,(a_{d-1},\,0)$. Since these points are all distinct
and also not equal to $p$, it follows that $a_{1},\,a_{2},\,\cdots,\,a_{d-1}$
and $0$ are all distinct.

Note that $H_{1}$ being a component, the reducible curve $H_{d}$ is given by a
polynomial of the form

$\displaystyle f(x,\,y)$ $\displaystyle\ =\ y\varphi(x,\,y),$

where $\varphi(x,\,y)$ is a polynomial of degree $d-1$. Let us write down
$\varphi(x,\,y)$ explicitly as

$\displaystyle\varphi(x,y)\,\,$
$\displaystyle:=\,\,b_{00}+b_{10}x+b_{01}y+\cdots+\frac{b_{0,d-1}}{(d-1)!}y^{d-1}.$

Recall that the degree $d-1$ component is tangent of first order to the line
at the origin, and the origin is a smooth point of the degree $d-1$ component.
Hence, $b_{00}\,=\,0$, $b_{10}\,=\,0$; but $b_{20}\,\neq\,0$ and
$b_{01}\,\neq\,0$. Note that

$\displaystyle\Big{(}f,\,(0,\,0),\,a_{1},\,\cdots,\,a_{d-1},\,a_{d}\Big{)}$
$\displaystyle\,\in\,\Big{(}\big{(}\mathsf{R}\mathsf{T}_{1}\big{)}_{d}(d)\Big{)}_{\mathsf{Aff}}\,\subset\,\mathcal{F}_{d}^{*}\times\mathbb{C}^{2}\times\mathbb{C}^{d}.$

Given any sufficiently open neighbourhood $U$ of
$\left(f,\,(0,\,0),\,a_{1},\,\cdots,\,a_{d-1},\,a_{d}\right)$, we have to show
the existence of an element

$\displaystyle\Big{(}F,\,(s,\,t),\,A_{2},\,\cdots,\,A_{d-1},\,A_{d}\Big{)}\,\in\,\Big{(}\mathsf{A}_{2}^{\mathsf{F}}\underbrace{\mathsf{T}_{0}\cdots\mathsf{T}_{0}}_{(d-1)-\textnormal{times}}\mathsf{F}\Big{)}_{\mathsf{Aff}}\cap
U.$

The curve $F$, if it exists, passes through the points
$(A_{1},\,0),\,\cdots,\,(A_{d-1},\,0)$. Hence, it has to be of the form

$\displaystyle F(x,\,y)\,$
$\displaystyle=\,Q(x)(C_{0}+C_{1}x)+y(B_{00}+B_{10}x+x^{2}\alpha(x)+y\beta(x,\,y)),\qquad\textnormal{where}$
$\displaystyle Q(x)\,$ $\displaystyle:=\,(x-A_{1})(x-A_{2})\cdots(x-A_{d-1}),$
$\displaystyle\alpha(x)\,$
$\displaystyle:=\,\frac{B_{20}}{2}x^{2}+\frac{B_{30}}{6}x^{3}+\cdots\qquad\textnormal{and}\qquad\beta(x,\,y)\,:=\,B_{01}+B_{11}x+\cdots.$

We note that $C_{0},\,C_{1},\,B_{00}$ and $B_{10}$ are small. Furthermore,
$\alpha(0)\,=\,\frac{B_{20}}{2}$ and $\beta(0,\,0)\,=\,B_{11}$ are non-zero.
Also, $F$ satisfies the following system of equations:

$\displaystyle F(s,\,t)$ $\displaystyle\,=\,0,\qquad F_{x}(s,\,t)\,=\,0,\qquad
F_{y}(s,\,t)\,=\,0,\qquad\textnormal{and}\qquad(F_{xx}F_{yy}-F_{xy}^{2})(s,\,t)\,=\,0.$

Using the first three equations, we can solve for $C_{0},\,C_{1}$ and
$B_{00}$. Plugging that into the last equation, we can solve for $t$ in terms
of $B_{10}$ using the implicit function theorem. Therefore, such an $F$
exists. This proves eq. 4.7.

Now the multiplicity of the intersection will be computed. A straightforward
calculation yields

$\displaystyle F(a_{d},\,0)\,$
$\displaystyle=\,-\frac{Q(a_{d})a_{d}}{4\alpha(0)\beta(0,\,0)Q(0)}B_{10}^{3}+O(B_{10}^{4}).$

Since the coefficient of $B_{10}$ is non-zero, it follows that the
multiplicity of the intersection is $3$. This completes the proof of 4.4. ∎

######  Theorem 4.6.

Let $d\,\in\,\mathbb{Z}^{\geq 3}$. The following equality of homology classes
in $\mathsf{M}_{{d+1}}^{1}(d)$ holds:

$\displaystyle\big{[}\mathsf{A}_{2}^{\mathsf{F}}\underbrace{\mathsf{T}_{0}\cdots\mathsf{T}_{0}}_{d-\textnormal{times}}\mathsf{F}\big{]}\cdot\big{[}\big{(}x_{d+1}\in\mathsf{L}\big{)}\big{]}\cdot\big{[}\big{(}x_{d+1}\in\mathcal{C}_{d}\big{)}\big{]}$
$\displaystyle\,\,=\,\,\sum_{i=1}^{d}\big{[}\mathsf{A}_{2}^{\mathsf{F}}\underbrace{\mathsf{T}_{0}\cdots\mathsf{T}_{0}}_{d-\textnormal{times}}\mathsf{F}\big{]}\cdot\big{[}\big{(}x_{i}=x_{d+1}\big{)}\big{]}$
$\displaystyle+\sum_{i=1}^{d}3\big{[}\big{(}\mathsf{R}\mathsf{T}_{1}\big{)}_{d+1}^{i}(d)\big{]}+\big{[}\big{(}\mathsf{R}\mathsf{A}_{2}^{\mathsf{F}}\big{)}_{d+1}(d)\big{]}$
$\displaystyle+3\big{[}\big{(}\mathsf{R}\mathsf{A}_{1}^{\mathsf{L}}\big{)}_{d+1}(d)\big{]}+2\big{[}\big{(}\mathsf{R}\mathsf{T}_{2}\big{)}_{d+1}(d)\big{]}.$
(4.8)

######  Remark 4.7.

As before, eq. 4.4 is obtained by intersecting both sides of 4.6 with

$\mu_{d+1}\ :=\ y_{1}^{2}y_{d}^{\delta_{d}-(d+3)}a_{1}a_{2}\cdots
a_{d}a_{d+1}.$

Finally, eqs. 4.2, 4.3 and 4.4 is proved. All of them together produce eq. 2.4
(that is, the second part of 2.1).

###### Proof of 4.6.

The first term on the right-hand side of 4.6 is justified as before. The
remaining terms need to be justified.

Start with the second term. The justification of the closure and multiplicity
is similar to that of the third term of 4.4 (see the proof of 4.4 for
details).

Next, consider the third term. Its justification is similar to that of 3.7
(see proof of 3.7).

To justify the fifth term on the right hand side of 4.6, pass to affine
coordinates. Let

$\displaystyle F(x,\,y)\,$
$\displaystyle=\,Q(x)(C_{0})+y(B_{00}+B_{10}x+\frac{B_{20}}{2}x^{2}+\frac{\alpha(x)}{6}x^{3}+\beta(y)y+\gamma(x,\,y)xy),$
where $\displaystyle Q(x)\,:=\,(x-A_{1})\cdots(x-A_{d}),$
$\displaystyle\alpha(x)\,:=\,B_{30}+\frac{B_{40}}{4}x+\cdots,$
$\displaystyle\beta(y)\,:=\,B_{01}+\frac{B_{02}}{2}y+\cdots,$
$\displaystyle\gamma(x,y)\,:=\,B_{11}+\frac{B_{21}}{2}x+\frac{B_{12}}{2}y+\cdots;$

here $C_{0},\,B_{00},\,B_{10}$ and $B_{20}$ are small and
$\alpha(0)\,=\,B_{30}$ and $\beta(0)\,=\,B_{01}$ are non-zero. Again, we try
to solve for

$\displaystyle F(s,\,t)$ $\displaystyle\,=\,0,\quad F_{x}(s,\,t)\,=\,0,\quad
F_{y}(s,\,t)\,=\,0,\quad\textnormal{and}\quad(F_{xx}F_{yy}-F_{xy}^{2})(s,\,t)\,=\,0.$

Using the first three equations, we can solve for $C_{0},\,B_{00}$ and
$B_{10}$. Plugging these into the last equation, $B_{20}$ can be solved in
terms of $t$ using the implicit function theorem. Now evaluate $F$ at
$(a_{d+1},0)$ and get that

$\displaystyle F(a_{d+1},0)$
$\displaystyle\,=\,-\frac{Q(a_{d+1})\beta(0)}{Q(0)}t^{2}+O(t^{3}).$

The multiplicity of the intersection is clearly $2$.

To justify the fourth term on the right-hand side of 4.6, switch to affine
coordinates. Let

$\displaystyle F(x,\,y)\,$
$\displaystyle=\,Q(x)(C_{0})+y(B_{00}+B_{10}x+B_{01}y+\frac{\alpha(x)}{2}x^{2}+\frac{\beta(y)}{2}y^{2}+\gamma(x,\,y)xy),$

where $C_{0},\,B_{00},\,B_{10}$ and $B_{01}$ are small and
$\alpha(0)\beta(0)-\gamma(0)^{2}$ is non-zero. Furthermore, assume that
$\alpha(0)\,=\,B_{20}\,\neq\,0$. The latter can be assumed because the
$\mathtt{x}$–axis is not one of the branches of the node. Again, we try to
solve for

$F(s,\,t)\,=\,0,\quad F_{x}(s,\,t)\,=\,0,\quad
F_{y}(s,\,t)\,=\,0\quad\textnormal{and}\quad(F_{xx}F_{yy}-F_{xy}^{2})(s,t)\,=\,0.$

Using the first three equations, we can solve for $C_{0},\,B_{00}$ and
$B_{10}$. Plugging these into the last equation, $B_{01}$ can be solved in
terms of $t$ using the implicit function theorem. Now, evaluate $F$ at
$(a_{d+1},\,0)$ and get that

$\displaystyle F(a_{d+1},\,0)\ =\
Q(a_{d+1})\Big{(}\frac{\alpha(0)\beta(0)-\gamma(0)^{2}}{2\alpha(0)Q(0)}\Big{)}t^{3}+O(t^{4}).$

The multiplicity of the intersection is clearly $3$. This completes the proof
of 4.6. ∎

##  5\. Extension of Caporaso-Harris for a family of nodal cubics

In this section the following question will be addressed:

What is $N(r,s)$, the number of nodal planar cubics in $\mathbb{P}^{3}$ that
intersect $r$ generic lines and pass through $s$ generic points, where
$r+2s=11$?

Notation: $N(r,\,s)\,=\,0$ unless $r+2s\,=\,11$.

Let $\widehat{\mathbb{P}}^{3}$ denote the dual of $\mathbb{P}^{3}$; it is the
space of linear $\mathbb{P}^{2}$ inside $\mathbb{P}^{3}$. An element of
$\widehat{\mathbb{P}}^{3}$ can be thought of as a non-zero linear functional
$\eta\,:\,\mathbb{C}^{4}\,\longrightarrow\,\mathbb{C}$ up to scaling. Given
such an $\eta$, define the projectivization of its kernel as
$\mathbb{P}^{2}_{\eta}$. In other words,

$\mathbb{P}^{2}_{\eta}\ :=\ \mathbb{P}(\eta^{-1}(0)).$

Hence, an element of $\widehat{\mathbb{P}}^{3}$ determines a plane in
$\mathbb{P}^{3}$ and vice versa. Define
$\big{(}\mathbb{P}^{2}\big{)}_{\textsf{Fb}}$ as follows:

$\big{(}\mathbb{P}^{2}\big{)}_{\textsf{Fb}}\ :=\
\\{([\eta],\,q)\,\in\,\widehat{\mathbb{P}}^{3}\times\mathbb{P}^{3}\,\big{|}\,\,\eta(q)\,=\,0\\}.$

So $\big{(}\mathbb{P}^{2}\big{)}_{\textsf{Fb}}$ is a fiber bundle over
$\widehat{\mathbb{P}}^{3}$ whose fiber over any point $[\eta]$ is
$\mathbb{P}^{2}_{\eta}$.

Recall from the previous section that $\mathcal{D}_{1}$ is the space of lines
in $\mathbb{P}^{2}$. Let us now make a more pedantic notation and define
$\mathcal{D}_{1}(\eta)$ to be the space of lines in $\mathbb{P}^{2}_{\eta}$.
Similarly, define $\mathcal{D}_{3}(\eta)$ to be the space of cubics in
$\mathbb{P}^{2}_{\eta}$. Now define the following fiber bundles over
$\widehat{\mathbb{P}}^{3}$:

$\big{(}\mathcal{D}_{1}\big{)}_{\textsf{Fb}}\,\longrightarrow\,\widehat{\mathbb{P}}^{3}\
\,\text{ and }\,\
\big{(}\mathcal{D}_{3}\big{)}_{\textsf{Fb}}\,\longrightarrow\,\widehat{\mathbb{P}}^{3}.$
(5.1)

The fibers over any point $[\eta]$ are $\mathcal{D}_{1}(\eta)$ and
$\mathcal{D}_{3}(\eta)$, respectively. Finally, define

$\Big{(}\mathcal{D}_{1}\times\mathcal{D}_{3}\Big{)}_{\mathsf{Fb}}\
\longrightarrow\ \widehat{\mathbb{P}}^{3}$

to be the fiber product of the two fiber bundles in (5.1). Its fiber over any
$[\eta]$ is $\mathcal{D}_{1}(\eta)\times\mathcal{D}_{3}(\eta)$.

Next, define

$\mathfrak{M}^{m}_{n}(3)\ :=\
\Big{(}\mathcal{D}_{1}\times\mathcal{D}_{3}\Big{)}_{\mathsf{Fb}}\times\big{(}Y^{1}\times\ldots\times
Y^{m}\big{)}\times\big{(}Y_{1}\times\ldots\times Y_{n}\big{)},$

where each $Y^{i}$ and $Y_{j}$ is a copy of $\mathbb{P}^{3}$. The hyperplane
class of $Y^{i}$ and $Y_{j}$ will be denoted by $B_{i}$ and $H_{j}$
respectively. The first Chern classes of the hyperplane bundles of
$\big{(}\mathcal{D}_{1}\big{)}_{\mathsf{Fb}}$ and
$\big{(}\mathcal{D}_{3}\big{)}_{\mathsf{Fb}}$ will be denoted by $\lambda_{1}$
and $\lambda_{3}$ respectively. The hyperplane class of
$\widehat{\mathbb{P}}^{3}$ will be denoted by $a$. Since the entire discussion
is for cubics, we are denoting the space by $\mathfrak{M}^{m}_{n}$ and not
$\mathfrak{M}^{m}_{n}(3)$; this will make reading easier.

Next, define

$\displaystyle\big{(}\mathsf{M}^{m}_{n}\big{)}_{\textsf{Fb}}\ $
$\displaystyle\longrightarrow\ \widehat{\mathbb{P}}^{3}$ (5.2)

to be the fiber bundle over $\widehat{\mathbb{P}}^{3}$ whose fiber over each
point $[\eta]$ is $\mathsf{M}^{m}_{n}(\eta)$.

Note that on $\mathfrak{M}^{m}_{n}$, the following equality of cycles hold:

$\displaystyle\Big{[}\big{(}\mathsf{M}^{m}_{n}\big{)}_{\textsf{Fb}}\Big{]}\ $
$\displaystyle=\
(a+B_{1})\cdot\ldots\cdot(a+B_{m})\cdot(a+H_{1})\cdot\ldots\cdot(a+H_{n}).$
(5.3)

To see why this is true, denote an element of $\mathfrak{M}^{m}_{n}$ by

$\Big{(}\Pi,\,\mathsf{L},\,\mathcal{C}_{3},\,p_{1},\,\ldots,\,p_{m},\,x_{1},\,\cdots,\,x_{n}\Big{)};$

here $\Pi$ denotes a plane in $\mathbb{P}^{3}$. Furthermore, $\mathsf{L}$ and
$\mathcal{C}_{3}$ denote a line and a cubic respectively in the plane $\Pi$.
Whereas $\big{(}\mathsf{M}^{m}_{n}\big{)}_{\textsf{Fb}}$ is the locus of all

$\Big{(}\Pi,\,\mathsf{L},\,\mathcal{C}_{3},\,p_{1},\,\cdots,\,p_{m},\,x_{1},\,\cdots,\,x_{n}\Big{)}$

in $\mathfrak{M}^{m}_{n}$ such that

  * •

The points $p_{1},\,\cdots,\,p_{m}$ lie on the plane $\Pi$;

  * •

the points $x_{1},\,\cdots,\,x_{n}$ lie on the plane $\Pi$.

Note that the condition that $p_{i}$ belongs to the plane $\Pi$ corresponds to
intersecting with the class $a+B_{i}$. Similarly, the condition that $x_{j}$
belongs to the plane $\Pi$ corresponds to intersecting with the class
$a+H_{j}$. This justifies eq. 5.3.

Another notation: Suppose for each variety $Z$ ismorphic to $\mathbb{P}^{2}$
we have a space $S(Z)$. Then $\big{(}S\big{)}_{\mathsf{Fb}}$ is the fiber
bundle over $\widehat{\mathbb{P}}^{3}$ whose fiber over each $[\eta]$ is
$S(\mathbb{P}^{2}_{\eta})$. To give an example, recall the space
$\mathsf{A}_{1}^{\mathsf{F}}\mathsf{T}_{0}$ defined in section 3. It is the
locus of all $(\mathsf{L},\,\mathcal{C}_{3},\,p,\,x_{1})$ in
$\mathsf{M}^{1}_{1}$ such that

  * •

the cubic $\mathcal{C}_{3}$ has a node at $p$, and

  * •

The nodal point $p$ does not lie on the line $\mathsf{L}$.

Now define
$\big{(}\mathsf{A}_{1}^{\mathsf{F}}\mathsf{T}_{0})_{\mathsf{Fb}}\,\subseteq\,\big{(}\mathsf{M}^{1}_{1}\big{)}_{\mathsf{Fb}}$
to be the locus of all $(\Pi,\,\mathsf{L},\,\mathcal{C}_{3},\,p,\,x_{1})$ such
that

  * •

the cubic $\mathcal{C}_{3}$ has a node at $p$, and

  * •

the nodal point $p$ does not lie on the line $\mathsf{L}$.

Pictorially this looks like:

Figure 7. $\big{(}\mathsf{A}_{1}^{\mathsf{F}}\mathsf{T}_{0})_{\mathsf{Fb}}$

All the other spaces defined in section 3 can be analogously defined.

Next consider the cohomology ring of $\mathfrak{M}^{m}_{n}$. Its generators
are

$a,\,\lambda_{1},\,\lambda_{3},\,B_{1},\,\cdots,\,B_{m},\,H_{1},\,\cdots,\,H_{n}.$

The ring structure is given by the following relations:

$\displaystyle\lambda_{1}^{3}\,$
$\displaystyle=\,-\lambda_{1}^{2}a-\lambda_{1}a^{2}-a^{3},$
$\displaystyle\lambda_{3}^{10}\,$
$\displaystyle=\,-10\lambda_{3}^{9}a-55\lambda_{3}^{8}a^{2}-220\lambda_{3}^{7}a^{3},$
$\displaystyle B_{i}^{4}\,$
$\displaystyle=\,0\,\,\,\forall\,\,i\,=\,1,\,\cdots,\,m,\quad
H_{j}^{4}\,=\,0\,\,\,\forall\,\,j\,=\,1,\,\cdots,\,n,\quad\textnormal{and}\quad
a^{4}\,=\,0.$ (5.4)

This fact is justified in [3, pp. 7–8]. We are now ready to state the main
results that will be used in enumerating the nodal planar cubics in
$\mathbb{P}^{3}$. The statements can be considered as family version of
Theorems 3.1, 3.2, 3.5 and 3.7 with $d\,=\,3$. The spaces involved here are
fiber bundle analogue of the spaces involved in the respective statements in
Section 3. We will use the fiber bundle version of the spaces without defining
them.

######  Theorem 5.1.

The following equality of homology classes in
$\big{(}\mathsf{M}^{1}_{1}\big{)}_{\mathsf{Fb}}$ holds:

$\displaystyle\big{[}\big{(}\mathsf{A}_{1}^{\mathsf{F}}\mathsf{F}\big{)}_{\mathsf{Fb}}\big{]}\cdot\big{[}\big{(}x_{1}\in\mathsf{L}\big{)}_{\mathsf{Fb}}\big{]}\cdot\big{[}\big{(}x_{1}\in\mathcal{C}_{3}\big{)}_{\mathsf{Fb}}\big{]}$
$\displaystyle\ =\
\big{[}\big{(}\mathsf{A}_{1}^{\mathsf{F}}\mathsf{T}_{0}\big{)}_{\mathsf{Fb}}\big{]}.$
(5.5)

This is the family version of 3.1; the proof is identical.

Now the goal is to use eq. 5.5 and extract numbers. Here it is more convenient
to do intersection theory on the ambient space $\mathfrak{M}^{1}_{1}$, which
we recall is given by

$\mathfrak{M}^{1}_{1}:=\Big{(}\mathcal{D}_{1}\times\mathcal{D}_{3}\Big{)}_{\mathsf{Fb}}\times
Y^{1}\times Y_{1}.$

Let $\alpha$ be a class in $\big{(}\mathsf{M}^{1}_{1}\big{)}_{\mathsf{Fb}}$,
and let $\mu$ be a class in $\mathfrak{M}^{1}_{1}$. Denote the restriction of
$\mu$ to $\big{(}\mathsf{M}^{1}_{1}\big{)}_{\mathsf{Fb}}$ by the same symbol
$\mu$. Then, eq. 5.3 immediately yields the following:

$\alpha\cdot\mu\ =\ \alpha\cdot(a+B_{1})\cdot(a+H_{1})\cdot\mu.$ (5.6)

The left-hand side of equation (5.6) is an intersection number in
$\big{(}\mathsf{M}^{1}_{1}\big{)}_{\mathsf{Fb}}$, while the right-hand side is
an intersection number in $\mathfrak{M}^{1}_{1}$. (If two classes are not of
complementary dimension, then the corresponding intersection number is defined
to be zero.)

As before, the goal is to intersect both sides of eq. 5.5 with a suitable
$\mu$, and extract numbers using eq. 5.6. The objective is to compute
$N(r,\,s)$, the number of nodal planar cubics in $\mathbb{P}^{3}$ that
intersect $r$ generic lines and $s$ generic points. Let us start by explaining
how equation (5.5) will help us compute $N(r,\,s)$.

Similar to the space $\mathsf{A}_{1}^{\mathsf{F}}$ defined in Section 3,
define the analogous fiber bundle
$\big{(}\mathsf{A}_{1}^{\mathsf{F}}\big{)}_{\mathsf{Fb}}$ over
$\widehat{\mathbb{P}^{3}}$. An element of this space is a triple consisting of
a line, a cubic and a point such that all of them lie on a single plane in
$\mathbb{P}^{3}$.

We will compute

$\displaystyle\big{[}\big{(}\mathsf{A}_{1}^{\mathsf{F}}\big{)}_{\mathsf{Fb}}\big{]}\cdot\lambda_{1}^{m}\lambda_{3}^{j}a^{l},.$
(5.7)

It will be shown that it is enough to compute (5.7).

Using eq. 5.5, we will show that

$\displaystyle\big{[}\big{(}\mathsf{A}_{1}^{\mathsf{F}}\big{)}_{\mathsf{Fb}}\big{]}\cdot\lambda_{1}^{m}\lambda_{3}^{j}a^{l}\
$ $\displaystyle=\
\frac{1}{3}\big{[}\big{(}\mathsf{A}_{1}^{\mathsf{F}}\mathsf{T}_{0}\big{)}_{\mathsf{Fb}}\big{]}\cdot\lambda_{1}^{m}\lambda_{3}^{j}a^{l}.$
(5.8)

Hence, computation of

$\displaystyle\big{[}\big{(}\mathsf{A}_{1}^{\mathsf{F}}\mathsf{T}_{0}\big{)}_{\mathsf{Fb}}\big{]}\cdot\lambda_{1}^{m}\lambda_{3}^{j}a^{l}$
(5.9)

would give eq. 5.7.

To justify these assertions, we begin by showing that $N(r,s)$ can be obtained
as a suitable intersection number involving the class
$\big{[}\big{(}\mathsf{A}_{1}^{\mathsf{F}}\big{)}_{\mathsf{Fb}}\big{]}$. The
reader is reminded that this will be an intersection number inside the space
$\big{(}\mathsf{M}^{1}_{0}\big{)}_{\mathsf{Fb}}$.

The homology classes represented by the subspace of cubics that intersect a
generic line and intersect a generic point will be denoted by
$\mathcal{H}^{\mathcal{C}_{3}}_{l}$ and $\mathcal{H}^{\mathcal{C}_{3}}_{p}$
respectively. Similarly, the homology classes represented by the subspace of
lines that intersect a generic line and intersect a generic point will be
denoted by $\mathcal{H}^{\mathsf{L}}_{l}$ and $\mathcal{H}^{\mathsf{L}}_{p}$
respectively. It is shown in [3, pp. 10–11] that

$\displaystyle\mathcal{H}^{\mathsf{L}}_{l}$
$\displaystyle\,=\,\lambda_{1}+a,\quad\mathcal{H}^{\mathsf{L}}_{p}\,=\,\lambda_{1}a,\quad\mathcal{H}^{\mathcal{C}_{3}}_{l}\,=\,\lambda_{3}+3a\quad\textnormal{and}\quad\mathcal{H}^{\mathcal{C}_{3}}_{p}\,=\,\lambda_{3}a.$
(5.10)

Now observe that

$\displaystyle N(r,\,s)\ $ $\displaystyle=\
\big{[}\big{(}\mathsf{A}_{1}^{\mathsf{F}}\big{)}_{\mathsf{Fb}}\big{]}\cdot\big{(}\mathcal{H}^{\mathcal{C}_{3}}_{l}\big{)}^{r}\cdot\big{(}\mathcal{H}^{\mathcal{C}_{3}}_{p}\big{)}^{s}\cdot\big{(}\mathcal{H}^{\mathsf{L}}_{l}\big{)}^{2}.$
(5.11)

To see why this is so, first assume that $r+2s\,=\,11$; otherwise both the
sides of eq. 5.11 are equal to zero because of the convention (namely that if
two classes are not of complementary dimensions, then their intersection
number is zero).

Since $r+2s\,=\,11$, intersecting
$\big{[}\big{(}\mathsf{A}_{1}^{\mathsf{F}}\big{)}_{\mathsf{Fb}}\big{]}$ with
$\big{(}\mathcal{H}^{\mathcal{C}_{3}}_{l}\big{)}^{r}\cdot\big{(}\mathcal{H}^{\mathcal{C}_{3}}_{p}\big{)}^{s}$
fixes the cubic (and also the plane inside which the cubic lies). Now
intersecting with $\big{(}\mathcal{H}^{\mathsf{L}}_{l}\big{)}^{2}$ does the
following: the two lines will intersect the plane at two points and through
these two points on the plane a unique line passes. That fixes the line. This
proves equation (5.11). Using equations (5.11), (5.10) and (5.4), we conclude
that $N(r,s)$ is a sum of intersection numbers of the type given by eq. 5.7.

It remains to prove eq. 5.8. We will rewrite eq. 5.5 as an equality of
homology classes in $\mathfrak{M}^{1}_{1}$. A couple of definitions are needed
for this. Define
$\big{(}\mathsf{A}_{1}^{\mathsf{F}}\big{)}_{\mathsf{Fb}}\mathsf{F}$ to be the
locus of points $(\Pi,\,\mathsf{L},\,\mathcal{C}_{3},\,p,\,x_{1})$ in
$\mathfrak{M}^{1}_{1}$ such that
$(\Pi,\,\mathsf{L},\,\mathcal{C}_{3},\,p)\,\in\,\big{(}\mathsf{A}_{1}^{\mathsf{F}}\big{)}_{\mathsf{Fb}}$.
There is no restriction on the point $x_{1}$; it need not have to lie on the
plane $\Pi$. In other words,

$\displaystyle\big{(}\mathsf{A}_{1}^{\mathsf{F}}\big{)}_{\mathsf{Fb}}\mathsf{F}\
$ $\displaystyle=\ \big{(}\mathsf{A}^{\mathsf{F}}\big{)}_{\mathsf{Fb}}\times
Y_{1}.$ (5.12)

Note that $\big{(}\mathsf{A}_{1}^{\mathsf{F}}\big{)}_{\mathsf{Fb}}\mathsf{F}$
is not the same as
$\big{(}\mathsf{A}_{1}^{\mathsf{F}}\mathsf{F}\big{)}_{\mathsf{Fb}}$; in the
latter case the point $x_{1}$ is required to lie on the plane $\Pi$. Denote by
$(x_{1}\in\Pi)$ the locus of all points
$(\Pi,\,\mathsf{L},\,\mathcal{C}_{3},\,p,\,x_{1})$ in $\mathfrak{M}^{1}_{1}$
such that the point $x_{1}$ lies on the plane $\Pi$.

Note that eq. 5.5 is equivalent to the fact that on $\mathfrak{M}^{1}_{1}$ the
following equality of homology classes holds:

$\displaystyle\big{[}\big{(}\mathsf{A}_{1}^{\mathsf{F}}\big{)}_{\mathsf{Fb}}\mathsf{F}\big{]}\cdot\big{[}(x_{1}\in\Pi)\big{]}\cdot\big{[}\big{(}x_{1}\in\mathsf{L}\big{)}\big{]}\cdot\big{[}\big{(}x_{1}\in\mathcal{C}_{3}\big{)}\big{]}$
$\displaystyle\ =\
\big{[}\big{(}\mathsf{A}_{1}^{\mathsf{F}}\mathsf{T}_{0}\big{)}_{\mathsf{Fb}}\big{]}.$
(5.13)

A subtle point needs explanation. Consider the left-hand side of eq. 5.13.
Strictly speaking, we have not actually defined the third and fourth factors
on the left-hand side. We define $\big{(}x_{1}\in\mathsf{L}\big{)}$ to be a
subspace of $\mathfrak{M}^{1}_{1}$ (not
$\big{(}\mathsf{M}^{1}_{1}\big{)}_{\mathsf{Fb}}$), where the point $x_{1}$ is
required to lie on the line. Notice the difference between
$\big{(}x_{1}\in\mathsf{L}\big{)}$ and
$\big{(}x_{1}\in\mathsf{L}\big{)}_{\mathsf{Fb}}$; the latter is a subspace of
$\big{(}\mathsf{M}^{1}_{1}\big{)}_{\mathsf{Fb}}$. However, restricted to
$\big{(}x_{1}\in\Pi\big{)}$, these two subspaces are the same. The same remark
applies for the space $\big{(}x_{1}\in\mathcal{C}_{3}\big{)}$.

Note that in $\mathfrak{M}^{1}_{1}$, the following equalities of homology
classes hold:

$\displaystyle\big{[}x_{1}\in\Pi\big{]}$
$\displaystyle\,=\,a+H_{1},\quad\big{[}\big{(}x_{1}\in\mathsf{L}\big{)}_{\mathsf{Fb}}\big{]}\,=\,\lambda_{1}+H_{1}\quad\textnormal{
and
}\quad\big{[}\big{(}x_{1}\in\mathcal{C}_{3}\big{)}_{\mathsf{Fb}}\big{]}\,=\,\lambda_{3}+3H_{1}.$
(5.14)

Next, using eq. 5.12 it follows that

$\displaystyle\big{[}\big{(}\mathsf{A}_{1}^{\mathsf{F}}\big{)}_{\mathsf{Fb}}\big{]}\cdot\lambda_{1}^{m}\lambda_{3}^{j}a^{l}$
$\displaystyle\ =\
\big{[}\big{(}\mathsf{A}_{1}^{\mathsf{F}}\big{)}_{\mathsf{Fb}}\mathsf{F}\big{]}\cdot\lambda_{1}^{m}\lambda_{3}^{j}a^{l}H_{1}^{3}.$
(5.15)

Finally, the following identity is obtained:

$\displaystyle\big{[}\big{(}\mathsf{A}_{1}^{\mathsf{F}}\big{)}_{\mathsf{Fb}}\mathsf{F}\big{]}\cdot\lambda_{1}^{m}\lambda_{3}^{j}H_{1}^{n}a^{l}$
$\displaystyle\ =\
\begin{cases}\frac{1}{3}\big{[}\big{(}\mathsf{A}_{1}^{\mathsf{F}}\mathsf{T}_{0}\big{)}_{\mathsf{Fb}}\big{]}\cdot\lambda_{1}^{m}\lambda_{3}^{j}a^{l}\qquad\textnormal{if}\qquad
n\,=\,3\\\ 0\qquad\qquad\qquad\qquad\qquad\quad\textnormal{if}\qquad
n\,\neq\,3.\end{cases}$ (5.16)

The second case of eq. 5.16 follows from eq. 5.12. The first case of eq. 5.16
follows from equations (5.14), (5.13) and (5.12). Finally, note that eq. 5.16,
combined with eq. 5.15 gives eq. 5.8.

To summarize what has been established so far, the goal is to compute the
number $N(r,s)$. It is enough to compute all the intersection numbers given by
eq. 5.7. It is then established that it is enough to compute all intersection
numbers given by eq. 5.9; this is done through eq. 5.8.

The following theorem is used in computing the intersection numbers given by
eq. 5.9.

######  Theorem 5.2.

The following equality of homology classes in
$\big{(}\mathsf{M}^{1}_{2}\big{)}_{\mathsf{Fb}}$ holds:

$\displaystyle\big{[}\big{(}\mathsf{A}_{1}^{\mathsf{F}}\mathsf{T}_{0}\mathsf{F}\big{)}_{\mathsf{Fb}}\big{]}\cdot\big{[}\big{(}x_{2}\in\mathsf{L}\big{)}_{\mathsf{Fb}}\big{]}\cdot\big{[}\big{(}x_{2}\in\mathcal{C}_{3}\big{)}_{\mathsf{Fb}}\big{]}\,\,$
$\displaystyle=\,\,\big{[}\big{(}\mathsf{A}_{1}^{\mathsf{F}}\mathsf{T}_{0}\mathsf{T}_{0}\big{)}_{\mathsf{Fb}}\big{]}$
$\displaystyle+\big{[}\big{(}\mathsf{A}_{1}^{\mathsf{F}}\mathsf{T}_{0}\mathsf{T}_{0}\mathsf{F}\big{)}_{\mathsf{Fb}}\big{]}\cdot\big{[}\big{(}x_{1}=x_{2}\big{)}_{\mathsf{Fb}}\big{]}.$
(5.17)

This is the family version of 5.2 when $n=1$; the proof is identical.

Let us now explain how this Theorem will help us compute the intersection
numbers given by eq. 5.9. Note that

$\displaystyle\big{[}\big{(}\mathsf{A}_{1}^{\mathsf{F}}\mathsf{T}_{0}\big{)}_{\mathsf{Fb}}\big{]}\cdot\lambda_{1}^{m}\lambda_{3}^{j}a^{l}\
$ $\displaystyle=\
\big{[}\big{(}\mathsf{A}_{1}^{\mathsf{F}}\mathsf{T}_{0}\big{)}_{\mathsf{Fb}}\mathsf{F}\big{]}\cdot\lambda_{1}^{m}\lambda_{3}^{j}a^{l}H_{2}^{3},\qquad$
$\displaystyle\textnormal{where}\qquad\big{(}\mathsf{A}_{1}^{\mathsf{F}}\mathsf{T}_{0}\big{)}_{\mathsf{Fb}}\mathsf{F}\
$ $\displaystyle=\
\big{(}\mathsf{A}_{1}^{\mathsf{F}}\mathsf{T}_{0}\big{)}_{\mathsf{Fb}}\times
Y_{2}.$ (5.18)

The following will be shown:

$\displaystyle\big{[}\big{(}\mathsf{A}_{1}^{\mathsf{F}}\mathsf{T}_{0}\big{)}_{\mathsf{Fb}}\mathsf{F}\big{]}\cdot\lambda_{1}^{m}\lambda_{3}^{j}H_{2}^{n}a^{l}\
$ $\displaystyle=\
\begin{cases}\frac{1}{2}\big{[}\big{(}\mathsf{A}_{1}^{\mathsf{F}}\mathsf{T}_{0}\mathsf{T}_{0}\big{)}_{\mathsf{Fb}}\big{]}\cdot\lambda_{1}^{m}\lambda_{3}^{j}a^{l}\qquad\textnormal{if}\qquad
n\,=\,3\\\ 0\qquad\qquad\qquad\qquad\qquad\quad\textnormal{if}\qquad
n\,\neq\,3.\end{cases}$ (5.19)

The second case of eq. 5.19 follows immediately from the definition of
$\big{(}\mathsf{A}_{1}^{\mathsf{F}}\mathsf{T}_{0}\big{)}_{\mathsf{Fb}}\mathsf{F}$.
Let us now justify the first case, namely when $n\,=\,3$. Note that 5.2 is
equivalent to the fact that on $\mathfrak{M}^{1}_{2}$ the following equality
of homology classes hold:

$\displaystyle\big{[}\big{(}\mathsf{A}_{1}^{\mathsf{F}}\mathsf{T}_{0}\big{)}_{\mathsf{Fb}}\mathsf{F}\big{]}\cdot\big{[}\big{(}x_{2}\in\Pi\big{)}\big{]}\cdot\big{[}\big{(}x_{2}\in\mathsf{L}\big{)}\big{]}\cdot\big{[}\big{(}x_{2}\in\mathcal{C}_{3}\big{)}\big{]}\,\,$
$\displaystyle=\,\,\big{[}\big{(}\mathsf{A}_{1}^{\mathsf{F}}\mathsf{T}_{0}\mathsf{T}_{0}\big{)}_{\mathsf{Fb}}\big{]}$
$\displaystyle+\big{[}\big{(}\mathsf{A}_{1}^{\mathsf{F}}\mathsf{T}_{0}\mathsf{T}_{0}\big{)}_{\mathsf{Fb}}\mathsf{F}\big{]}\cdot\big{[}\big{(}x_{1}=x_{2}\big{)}\big{]}.$
(5.20)

Now note that in $\mathfrak{M}^{1}_{2}$, the following equality of homology
classes holds:

$\displaystyle\big{[}\big{(}x_{2}\in\Pi\big{)}\big{]}\,=\,a+H_{2},\quad\big{[}\big{(}x_{2}\in\mathsf{L}\big{)}\big{]}\,$
$\displaystyle=\,\lambda_{1}+H_{2},\quad\big{[}\big{(}x_{2}\in\mathcal{C}_{3}\big{)}\big{]}\,=\,\lambda_{3}+3H_{2}$
$\displaystyle\textnormal{and }\quad\big{[}\big{(}x_{1}=x_{2}\big{)}\big{]}\,$
$\displaystyle=\,H_{1}^{3}+H_{1}^{2}H_{2}+H_{1}H_{2}^{2}+H_{2}^{3}.$ (5.21)

The first case of eq. 5.19 now follows immediately from equations (5) and
(5.21).

Using equations (5.19) and (5.18) it follows that in order to compute the
intersection numbers given by eq. 5.9, it is enough to compute the following
intersection number:

$\displaystyle\big{[}\big{(}\mathsf{A}_{1}^{\mathsf{F}}\mathsf{T}_{0}\mathsf{T}_{0}\big{)}_{\mathsf{Fb}}\big{]}\cdot\lambda_{1}^{m}\lambda_{3}^{j}a^{l}.$
(5.22)

it was also shown that to compute $N(r,s)$, it is sufficient to compute the
numbers given by eq. 5.9. Hence, to compute $N(r,s)$, it suffices to compute
the numbers given by eq. 5.22. The following theorem will be used for it.

######  Theorem 5.3.

The following equality of homology classes in
$\big{(}\mathsf{M}_{{3}}^{1}\big{)}_{\mathsf{Fb}}$ holds:

$\displaystyle\big{[}\big{(}\mathsf{A}_{1}^{\mathsf{F}}\mathsf{T}_{0}\mathsf{T}_{0}\mathsf{F}\big{)}_{\mathsf{Fb}}\big{]}\cdot\big{[}\big{(}x_{3}\in\mathsf{L}\big{)}_{\mathsf{Fb}}\big{]}\cdot\big{[}\big{(}x_{3}\in\mathcal{C}_{3}\big{)}_{\mathsf{Fb}}\big{]}\,\,$
$\displaystyle=\,\,\big{[}\big{(}\mathsf{A}_{1}^{\mathsf{F}}\mathsf{T}_{0}\mathsf{T}_{0}\mathsf{T}_{0}\big{)}_{\mathsf{Fb}}\big{]}$
$\displaystyle+\big{[}\big{(}\mathsf{A}_{1}^{\mathsf{F}}\mathsf{T}_{0}\mathsf{T}_{0}\mathsf{F}\big{)}_{\mathsf{Fb}}\big{]}\cdot\big{[}\big{(}x_{1}=x_{3}\big{)}_{\mathsf{Fb}}\big{]}$
$\displaystyle+\big{[}\big{(}\mathsf{A}_{1}^{\mathsf{F}}\mathsf{T}_{0}\mathsf{T}_{0}\mathsf{F}\big{)}_{\mathsf{Fb}}\big{]}\cdot\big{[}\big{(}x_{2}=x_{3}\big{)}_{\mathsf{Fb}}\big{]}$
$\displaystyle+\big{[}\big{(}\mathsf{R}_{3}\big{)}_{\mathsf{Fb}}\big{]}.$
(5.23)

This the family version of 3.5; the proof is identical.

To explain how this helps in computing the intersection numbers in eq. 5.22,
note that

$\displaystyle\big{[}\big{(}\mathsf{A}_{1}^{\mathsf{F}}\mathsf{T}_{0}\mathsf{T}_{0}\big{)}_{\mathsf{Fb}}\big{]}\cdot\lambda_{1}^{m}\lambda_{3}^{j}a^{l}$
$\displaystyle=\big{[}\big{(}\mathsf{A}_{1}^{\mathsf{F}}\mathsf{T}_{0}\mathsf{T}_{0}\big{)}_{\mathsf{Fb}}\mathsf{F}\big{]}\cdot\lambda_{1}^{m}\lambda_{3}^{j}a^{l}H_{3}^{3},$
$\displaystyle\textnormal{where}\qquad\big{(}\mathsf{A}_{1}^{\mathsf{F}}\mathsf{T}_{0}\mathsf{T}_{0}\big{)}_{\mathsf{Fb}}\mathsf{F}$
$\displaystyle=\big{(}\mathsf{A}_{1}^{\mathsf{F}}\mathsf{T}_{0}\mathsf{T}_{0}\big{)}_{\mathsf{Fb}}\times
Y_{3}.$

It will be shown that

$\displaystyle\big{[}\big{(}\mathsf{A}_{1}^{\mathsf{F}}\mathsf{T}_{0}\mathsf{T}_{0}\big{)}_{\mathsf{Fb}}\mathsf{F}\big{]}\cdot\lambda_{1}^{m}\lambda_{3}^{j}H_{3}^{n}a^{l}$
$\displaystyle=\begin{cases}\big{[}\big{(}\mathsf{A}_{1}^{\mathsf{F}}\mathsf{T}_{0}\mathsf{T}_{0}\mathsf{T}_{0}\big{)}_{\mathsf{Fb}}\big{]}\cdot\lambda_{1}^{m}\lambda_{3}^{j}a^{l}\\\
\qquad\quad+\big{[}\big{(}\mathsf{R}_{3}\big{)}_{\mathsf{Fb}}\big{]}\cdot\lambda_{1}^{m}\lambda_{3}^{j}a^{l}\qquad\textnormal{if}\qquad
n\,=\,3\\\ 0\qquad\qquad\qquad\qquad\qquad\quad\textnormal{if}\qquad
n\,\neq\,3.\end{cases}$ (5.24)

The second case of eq. 5.24 is immediate from the definition of
$\big{(}\mathsf{A}_{1}^{\mathsf{F}}\mathsf{T}_{0}\mathsf{T}_{0}\big{)}_{\mathsf{Fb}}\mathsf{F}$.
To justify the first case, namely when $n\,=\,3$, note that 5.3 is equivalent
to the fact that on $\mathfrak{M}^{1}_{3}$ the following equality of homology
classes holds:

$\displaystyle\big{[}\big{(}\mathsf{A}_{1}^{\mathsf{F}}\mathsf{T}_{0}\mathsf{T}_{0}\big{)}_{\mathsf{Fb}}\mathsf{F}\big{]}\cdot\big{[}\big{(}x_{3}\in\Pi\big{)}\big{]}\cdot\big{[}\big{(}x_{3}\in\mathsf{L}\big{)}\big{]}\cdot\big{[}\big{(}x_{3}\in\mathcal{C}_{3}\big{)}\big{]}\,\,$
$\displaystyle=\,\,\big{[}\big{(}\mathsf{A}_{1}^{\mathsf{F}}\mathsf{T}_{0}\mathsf{T}_{0}\mathsf{T}_{0}\big{)}_{\mathsf{Fb}}\big{]}$
$\displaystyle+\big{[}\big{(}\mathsf{A}_{1}^{\mathsf{F}}\mathsf{T}_{0}\mathsf{T}_{0}\big{)}_{\mathsf{Fb}}\mathsf{F}\big{]}\cdot\big{[}\big{(}x_{1}=x_{3}\big{)}\big{]}$
$\displaystyle+\big{[}\big{(}\mathsf{A}_{1}^{\mathsf{F}}\mathsf{T}_{0}\mathsf{T}_{0}\big{)}_{\mathsf{Fb}}\mathsf{F}\big{]}\cdot\big{[}\big{(}x_{2}=x_{3}\big{)}\big{]}$
$\displaystyle+\big{[}\big{(}\mathsf{R}_{3}\big{)}_{\mathsf{Fb}}\big{]}.$
(5.25)

Note that in $\mathfrak{M}^{1}_{3}$, the following equalities of homology
classes hold:

$\displaystyle\big{[}\big{(}x_{3}\in\Pi\big{)}\big{]}\,=\,a+H_{3},\qquad\big{[}\big{(}x_{3}\in\mathsf{L}\big{)}\big{]}$
$\displaystyle\,=\,\lambda_{1}+H_{3},\qquad\big{[}\big{(}x_{3}\in\mathcal{C}_{3}\big{)}\big{]}\,=\,\lambda_{3}+3H_{3},$
$\displaystyle\big{[}\big{(}x_{1}=x_{3}\big{)}\big{]}$
$\displaystyle\,=\,H_{1}^{3}+H_{1}^{2}H_{2}+H_{1}H_{2}^{2}+H_{2}^{3}\qquad\textnormal{and}$
$\displaystyle\qquad\big{[}\big{(}x_{2}=x_{3}\big{)}\big{]}$
$\displaystyle\,=\,H_{2}^{3}+H_{2}^{2}H_{3}+H_{2}H_{3}^{2}+H_{3}^{3}.$ (5.26)

The first case of eq. 5.24 now follows immediately from equations (5) and
(5.26). Hence, we conclude, using eq. 5.24, that in order to compute the
intersection numbers given by eq. 5.22, it is sufficient to compute the
following intersection numbers

$\displaystyle\big{[}\big{(}\mathsf{A}_{1}^{\mathsf{F}}\mathsf{T}_{0}\mathsf{T}_{0}\mathsf{T}_{0}\big{)}_{\mathsf{Fb}}\big{]}\cdot\lambda_{1}^{m}\lambda_{3}^{j}a^{l}\qquad\textnormal{and}\qquad\big{[}\big{(}\mathsf{R}_{3}\big{)}_{\mathsf{Fb}}\big{]}\cdot\lambda_{1}^{m}\lambda_{3}^{j}a^{l}.$

We explain in Section 6 how to compute
$\big{[}\big{(}\mathsf{R}_{3}\big{)}_{\mathsf{Fb}}\big{]}\cdot\lambda_{1}^{m}\lambda_{3}^{j}a^{l}$.
Assuming that those numbers can be computed, we conclude that in order to
compute $N(r,s)$, it is enough to compute

$\displaystyle\big{[}\big{(}\mathsf{A}_{1}^{\mathsf{F}}\mathsf{T}_{0}\mathsf{T}_{0}\mathsf{T}_{0}\big{)}_{\mathsf{Fb}}\big{]}\cdot\lambda_{1}^{m}\lambda_{3}^{j}a^{l}.$
(5.27)

The next theorem is used for it.

######  Theorem 5.4.

The following equality of homology classes in
$\big{(}\mathsf{M}_{{4}}^{1}\big{)}_{\mathsf{Fb}}$ holds:

$\displaystyle\big{[}\big{(}\mathsf{A}_{1}^{\mathsf{F}}\mathsf{T}_{0}\mathsf{T}_{0}\mathsf{T}_{0}\mathsf{F}\big{)}_{\mathsf{Fb}}\big{]}\cdot\big{[}\big{(}x_{4}\in\mathsf{L}\big{)}_{\mathsf{Fb}}\big{]}\cdot\big{[}\big{(}x_{4}\in\mathcal{C}_{3}\big{)}_{\mathsf{Fb}}\big{]}\,\,$
$\displaystyle=\,\,\big{[}\big{(}\mathsf{A}_{1}^{\mathsf{F}}\mathsf{T}_{0}\mathsf{T}_{0}\mathsf{T}_{0}\mathsf{F}\big{)}_{\mathsf{Fb}}\big{]}\cdot\big{[}\big{(}x_{1}=x_{4}\big{)}_{\mathsf{Fb}}\big{]}$
$\displaystyle+\,\,\big{[}\big{(}\mathsf{A}_{1}^{\mathsf{F}}\mathsf{T}_{0}\mathsf{T}_{0}\mathsf{T}_{0}\mathsf{F}\big{)}_{\mathsf{Fb}}\big{]}\cdot\big{[}\big{(}x_{2}=x_{4}\big{)}_{\mathsf{Fb}}\big{]}$
$\displaystyle+\,\,\big{[}\big{(}\mathsf{A}_{1}^{\mathsf{F}}\mathsf{T}_{0}\mathsf{T}_{0}\mathsf{T}_{0}\mathsf{F}\big{)}_{\mathsf{Fb}}\big{]}\cdot\big{[}\big{(}x_{3}=x_{4}\big{)}_{\mathsf{Fb}}\big{]}$
$\displaystyle+\,\,\big{[}\big{(}\mathsf{R}^{1}_{4}\big{)}_{\mathsf{Fb}}\big{]}+\big{[}\big{(}\mathsf{R}^{2}_{4}\big{)}_{\mathsf{Fb}}\big{]}+\big{[}\big{(}\mathsf{R}^{3}_{4}\big{)}_{\mathsf{Fb}}\big{]}$
$\displaystyle+\,\,\big{[}\big{(}\big{(}\mathsf{R}\mathsf{A}_{1}^{\mathsf{F}}\big{)}_{4}\big{)}_{\mathsf{Fb}}\big{]}+2\big{[}\big{(}\big{(}\mathsf{R}\mathsf{T}_{1}\big{)}_{4}\big{)}_{\mathsf{Fb}}\big{]}.$
(5.28)

This is the family version of 3.7; the proof is identical.

In contrast to the earlier cases, it is now a more non-trivial task to see how
this theorem helps us compute the intersection numbers given by eq. 5.27. We
start as usual by observing that 5.4 is equivalent to the fact that on
$\mathfrak{M}^{1}_{4}$, the following equality of homology classes holds:

$\displaystyle\big{[}\big{(}\mathsf{A}_{1}^{\mathsf{F}}\mathsf{T}_{0}\mathsf{T}_{0}\mathsf{T}_{0}\big{)}_{\mathsf{Fb}}\mathsf{F}\big{]}\cdot\big{[}x_{4}\in\Pi\big{]}\cdot\big{[}\big{(}x_{4}\in\mathsf{L}\big{)}\big{]}\cdot\big{[}\big{(}x_{4}\in\mathcal{C}_{3}\big{)}\big{]}\,\,$
$\displaystyle=\,\,\big{[}\big{(}\mathsf{A}_{1}^{\mathsf{F}}\mathsf{T}_{0}\mathsf{T}_{0}\mathsf{T}_{0}\big{)}_{\mathsf{Fb}}\mathsf{F}\big{]}\cdot\big{[}\big{(}x_{1}=x_{4}\big{)}\big{]}$
$\displaystyle+\big{[}\big{(}\mathsf{A}_{1}^{\mathsf{F}}\mathsf{T}_{0}\mathsf{T}_{0}\mathsf{T}_{0}\big{)}_{\mathsf{Fb}}\mathsf{F}\big{]}\cdot\big{[}\big{(}x_{2}=x_{4}\big{)}\big{]}$
$\displaystyle+\big{[}\big{(}\mathsf{A}_{1}^{\mathsf{F}}\mathsf{T}_{0}\mathsf{T}_{0}\mathsf{T}_{0}\big{)}_{\mathsf{Fb}}\mathsf{F}\big{]}\cdot\big{[}\big{(}x_{3}=x_{4}\big{)}\big{]}$
$\displaystyle+\,\,\big{[}\big{(}\mathsf{R}^{1}_{4}\big{)}_{\mathsf{Fb}}\big{]}+\big{[}\big{(}\mathsf{R}^{2}_{4}\big{)}_{\mathsf{Fb}}\big{]}+\big{[}\big{(}\mathsf{R}^{3}_{4}\big{)}_{\mathsf{Fb}}\big{]}$
$\displaystyle+\,\,\big{[}\big{(}\big{(}\mathsf{R}\mathsf{A}_{1}^{\mathsf{F}}\big{)}_{4}\big{)}_{\mathsf{Fb}}\big{]}+2\big{[}\big{(}\big{(}\mathsf{R}\mathsf{T}_{1}\big{)}_{4}\big{)}_{\mathsf{Fb}}\big{]}.$
(5.29)

Next note that in $\mathfrak{M}^{1}_{4}$, the following equalities of homology
classes hold:

$\displaystyle\big{[}x_{4}\in\Pi\big{]}\,=\,a+H_{4},\quad\big{[}\big{(}x_{4}\in\mathsf{L}\big{)}\big{]}\,$
$\displaystyle=\,\lambda_{1}+H_{4},\quad\big{[}\big{(}x_{4}\in\mathcal{C}_{3}\big{)}\big{]}\,=\,\lambda_{3}+3H_{4},$
$\displaystyle\big{[}\big{(}x_{i}=x_{4}\big{)}\big{]}\,$
$\displaystyle=\,H_{i}^{3}+H_{i}^{2}H_{4}+H_{i}H_{4}^{2}+H_{4}^{3},\quad
i\,=\,1,\,2,\,3.$ (5.30)

To compute eq. 5.27, we first need to compute the following intersection
numbers:

$\displaystyle\big{[}\big{(}\mathsf{R}^{1}_{4}\big{)}_{\mathsf{Fb}}\big{]}\cdot\lambda_{1}^{m}\lambda_{3}^{j}a^{l},\qquad\big{[}\big{(}\mathsf{R}^{2}_{4}\big{)}_{\mathsf{Fb}}\big{]}\cdot\lambda_{1}^{m}\lambda_{3}^{j}a^{l},$
$\displaystyle\qquad\big{[}\big{(}\mathsf{R}^{3}_{4}\big{)}_{\mathsf{Fb}}\big{]}\cdot\lambda_{1}^{m}\lambda_{3}^{j}a^{l},$
$\displaystyle\big{[}\big{(}\big{(}\mathsf{R}\mathsf{A}_{1}^{\mathsf{F}}\big{)}_{4}\big{)}_{\mathsf{Fb}}\big{]}\cdot\lambda_{1}^{m}\lambda_{3}^{j}a^{l}\qquad\textnormal{and}$
$\displaystyle\qquad\big{[}\big{(}\big{(}\mathsf{R}\mathsf{T}_{1}\big{)}_{4}\big{)}_{\mathsf{Fb}}\big{]}\cdot\lambda_{1}^{m}\lambda_{3}^{j}a^{l}.$
(5.31)

These numbers are computed in Section 6. For now, let us assume that the
intersection numbers given in eq. 5.31 can be computed.

Next, define

$\displaystyle\mu$
$\displaystyle:=\lambda_{1}^{m}\lambda_{3}^{13-(m+n_{1}+n_{2}+n_{3}+n_{4}+l)}\cdot
H_{1}^{n_{1}}\cdot H_{2}^{n_{2}}\cdot H_{3}^{n_{3}}\cdot H_{4}^{n_{4}}a^{l},$
(5.32) $\displaystyle T(m,n_{1},n_{2},n_{3},n_{4},l)$
$\displaystyle:=\,\,\Big{(}\big{[}\big{(}\mathsf{R}^{1}_{4}\big{)}_{\mathsf{Fb}}\big{]}+\big{[}\big{(}\mathsf{R}^{2}_{4}\big{)}_{\mathsf{Fb}}\big{]}+\big{[}\big{(}\mathsf{R}^{3}_{4}\big{)}_{\mathsf{Fb}}\big{]}$
$\displaystyle\,\,\,\,\,+\big{[}\big{(}\big{(}\mathsf{R}\mathsf{A}_{1}^{\mathsf{F}}\big{)}_{4}\big{)}_{\mathsf{Fb}}\big{]}+2\big{[}\big{(}\big{(}\mathsf{R}\mathsf{T}_{1}\big{)}_{4}\big{)}_{\mathsf{Fb}}\big{]}\Big{)}\cdot\mu,\qquad\textnormal{and}$
(5.33) $\displaystyle\nu$
$\displaystyle:=\lambda_{1}^{m}\lambda_{3}^{16-(m+n_{1}+n_{2}+n_{3}+n_{4}+l)}\cdot
H_{1}^{n_{1}}\cdot H_{2}^{n_{2}}\cdot H_{3}^{n_{3}}\cdot H_{4}^{n_{4}}a^{l},$
(5.34) $\displaystyle\varphi(m,n_{1},n_{2},n_{3},n_{4},l)$
$\displaystyle:=\big{[}\big{(}\mathsf{A}_{1}^{\mathsf{F}}\mathsf{T}_{0}\mathsf{T}_{0}\mathsf{T}_{0}\big{)}_{\mathsf{Fb}}\mathsf{F}\big{]}\cdot\nu.$
(5.35)

Intersect both sides of section 5 with the class $\mu$ (as defined by eq.
5.32). It follows using equations (5.30), (5.33) and (5.35) that

$\displaystyle-\varphi(m,n_{1}+3,n_{2},n_{3},n_{4},l)$
$\displaystyle-\varphi(m,n_{1},n_{2}+3,n_{3},n_{4},l)$
$\displaystyle-\varphi(m,n_{1},n_{2},n_{3}+3,n_{4},l)$
$\displaystyle-\varphi(m,n_{1}+2,n_{2},n_{3},n_{4}+1,l)$
$\displaystyle-\varphi(m,n_{1},n_{2}+2,n_{3},n_{4}+1,l)$
$\displaystyle-\varphi(m,n_{1},n_{2},n_{3}+2,n_{4}+1,l)$
$\displaystyle+3\varphi(m,n_{1},n_{2},n_{3},n_{4}+2,l+1)$
$\displaystyle-\varphi(m,n_{1}+1,n_{2},n_{3},n_{4}+2,l)$
$\displaystyle-\varphi(m,n_{1},n_{2}+1,n_{3},n_{4}+2,l)$
$\displaystyle-\varphi(m,n_{1},n_{2},n_{3}+1,n_{4}+2,l)$
$\displaystyle+3\varphi(m+1,n_{1},n_{2},n_{3},n_{4}+1,l+1)$
$\displaystyle+3\varphi(m+1,n_{1},n_{2},n_{3},n_{4}+2,l)$
$\displaystyle+\varphi(m,n_{1},n_{2},n_{3},n_{4}+1,l+1)$
$\displaystyle+\varphi(m,n_{1},n_{2},n_{3},n_{4}+2,l)$
$\displaystyle+\varphi(m+1,n_{1},n_{2},n_{3},n_{4},l+1)$
$\displaystyle+\varphi(m+1,n_{1},n_{2},n_{3},n_{4}+1,l)=T(m,n_{1},n_{2},n_{3},n_{4},l).$
(5.36)

We will now show that the quantity
$\varphi(m,\,n_{1},\,n_{2},\,n_{3},\,n_{4},\,l)$ is computable in terms of the
function $T$. Write down eq. 5.36 three times; once with $n_{4}\,=\,1$, then
with $n_{4}\,=\,2$ and finally with $n_{4}\,=\,3$. We will refer to these as
equations one, two and three respectively. Now do the following: equation one
minus $3$ times equation two plus $9$ times equation three. After
simplification, we get

$\displaystyle-\varphi(m,n_{1}+1,n_{2},n_{3},3,l)$
$\displaystyle-\varphi(m,n_{1},n_{2}+1,n_{3},3,l)$
$\displaystyle-\varphi(m,n_{1},n_{2},n_{3}+1,3,l)$
$\displaystyle+\varphi(m,n_{1},n_{2},n_{3},3,l)$
$\displaystyle+3\varphi(m,n_{1}+2,n_{2},n_{3},3,l)$
$\displaystyle+3\varphi(m,n_{1},n_{2}+2,n_{3},n_{4},l)$
$\displaystyle+3\varphi(m,n_{1}+3,n_{2},n_{3},3,l)$
$\displaystyle-9\varphi(m,n_{1},n_{2}+3,n_{3},3,l)$
$\displaystyle-9\varphi(m,n_{1},n_{2},n_{3}+3,3,l)$
$\displaystyle=T(m,n_{1},n_{2},n_{3},1,l)$
$\displaystyle-3T(m,n_{1},n_{2},n_{3},2,l)+9T(m,n_{1},n_{2},n_{3},3,l).$
(5.37)

Next, note that

$\displaystyle\varphi(m,n_{1},n_{2},n_{3},n_{4},l)\ $ $\displaystyle=\
\begin{cases}0\qquad\textnormal{if}\quad n_{4}\neq 3,\\\
0\qquad\textnormal{if}\quad n_{1},n_{2}~{}~{}\textnormal{or}~{}~{}n_{3}\geq
4.\end{cases}$ (5.38)

Finally, we note that

$\displaystyle\varphi(m,n_{1},n_{2},n_{3},n_{4},l)\ $ $\displaystyle=\
\varphi(m,\sigma(n_{1}),\sigma(n_{2}),\sigma(n_{3}),n_{4},l),$ (5.39)

where $\sigma$ is any element of the permutation group $S_{3}$.

Using all these information, we can now solve the recursion in a brute-force
manner. Start with the following equality, which is obtained by putting
$n_{1}\,=\,n_{2}\,=\,n_{3}\,=\,3$ in eq. 5.37, and using the fact that
$\varphi$ vanishes if $n_{i}\,\geq\,4$ for $i\,\in\,\\{1,\,2,\,3\\}$:

$\displaystyle\varphi(m,3,3,3,3,l)\,=\,T(m,3,3,3,1,l)-3T(m,3,3,3,2,l)+9T(m,3,3,3,3,l)$

Since the right-hand side is only about conics, those numbers can be computed
and hence $\varphi(m,3,3,3,3,l)$ can be computed. Using this,
$\varphi(m,2,3,3,3,l)$ can be computed by putting
$n_{1}\,=\,2,\,n_{2}\,=\,n_{3}\,=\,3$ in eq. 5.37:

$\displaystyle\varphi(m,2,3,3,3,l)\,=\,\varphi(m,3,3,3,3,l)+T(m,2,3,3,1,l)-3T(m,2,3,3,2,l)+9T(m,2,3,3,3,l).$

We can continue in this manner and compute $\varphi(m,n_{1},3,3,3,l)$ for all
values of $n_{1}$ (again, keeping in mind that $\varphi$ is possibly non-
trivial only for values $n_{1}\,\in\,\\{0,\,1,\,2,\,3\\}$).

We would then want to compute $\varphi(m,3,2,3,3,l)$, but by the symmetry
property eq. 5.39, that is the same as $\varphi(m,2,3,3,3,l)$ which has been
computed. Similarly, the numbers $\varphi(m,3,n_{2},3,3,l)$ are obtained for
any $n_{2}$ by using symmetry.

Next, compute $\varphi(m,2,2,3,3,l)$ by putting
$n_{1}\,=\,2,\,n_{2}\,=\,2,\,n_{3}\,=\,3$ in eq. 5.37:

$\varphi(m,2,2,3,3,l)\ =\
\varphi(m,3,2,3,3,l)+\varphi(m,2,3,3,3,l)+T(m,2,2,3,1,l)$
$-3T(m,2,2,3,2,l)+9T(m,2,2,3,3,l).$

Proceeding similarly, $\varphi(m,n_{1},n_{2},3,3,l)$ can be computed for all
values of $n_{1},\,n_{2}$.

We can continue this process, i.e., put in the required values of $n_{i}$ in
eq. 5.37, and use eq. 5.38, eq. 5.39, as and when required. This process will
end since there are only a finite number of steps going from $n_{i}\,=\,3$ to
$n_{i}\,=\,0$. This solves the recursion problem.

##  6\. Computation of the characteristic number of conics in $\mathbb{P}^{3}$

The aim is to compute the characteristic number of conics in $\mathbb{P}^{3}$
that were needed to complete the computation in the last section. We will
start by computing the characteristic number of smooth conics. The
intersection space being considered is

$\displaystyle\mathfrak{M}^{1}_{n}(2)\ =\
\big{(}\mathcal{D}_{1}\times\mathcal{D}_{2}\big{)}_{\mathsf{Fb}}\times
Y^{1}\times Y_{1}\times\cdots\times Y_{n},$

where each $Y^{i}$ and $Y_{j}$ is a copy of $\mathbb{P}^{3}$. As before, the
hyperplane class of $Y^{i}$ and $Y_{j}$ are denoted by $B_{i}$ and $H_{j}$
respectively. The first Chern classes of the hyperplane bundles of
$\big{(}\mathcal{D}_{1}\big{)}_{\mathsf{Fb}}$ and
$\big{(}\mathcal{D}_{2}\big{)}_{\mathsf{Fb}}$ are denoted by $\lambda_{1}$ and
$\lambda_{2}$, respectively, and the hyperplane class of
$\widehat{\mathbb{P}}^{3}$ is denoted by $a$. Similar to eq. 5.2, we can have
the fiber bundle

$\displaystyle\big{(}\mathsf{M}^{1}_{n}(2)\big{)}_{\textsf{Fb}}$
$\displaystyle\longrightarrow\widehat{\mathbb{P}}^{3},$

where an element of the bundle is a tuple

$\left(\Pi,\,\mathsf{L},\,\mathcal{C}_{2},\,p,x_{1},\,\cdots,\,x_{n}\right)$

such that $\mathsf{L},\,\mathcal{C}_{2}$ and all the points lie on $\Pi$. A
similar argument to that of eq. 5.3 shows that the cycle of
$\left[\left(\mathsf{M}^{1}_{n}(2)\right)_{\textsf{Fb}}\right]$ inside
$\mathfrak{M}^{1}_{n}(2)$ is given by

$\displaystyle\left[\left(\mathsf{M}^{1}_{n}(2)\right)_{\textsf{Fb}}\right]$
$\displaystyle\ =\ (a+B_{1})\cdot(a+H_{1})\cdot\cdots\cdot(a+H_{n}).$

Note that the spaces involving conics in the previous section are actually
cubics. Indeed, they are given by a conic and the prescribed line. Thus, we
can think $\left(\mathcal{D}_{1}\times\mathcal{D}_{2}\right)_{\mathsf{Fb}}$ as
a subspace of $\left(\mathcal{D}_{3}\right)_{\mathsf{Fb}}$. Thus restricted to
$\left(\mathcal{D}_{1}\times\mathcal{D}_{2}\right)_{\mathsf{Fb}}$, we have

$\displaystyle\lambda_{3}$ $\displaystyle\ =\lambda_{1}+\lambda_{2}.$ (6.1)

Following the discussion in [3, pp. 7–8], we conclude that the cohomology ring
of $\mathfrak{M}^{1}_{n}(2)$ is generated by
$a,\,\lambda_{1},\,\lambda_{2},\,B_{1},\,\cdots,\,B_{m},\,H_{1},\,\cdots,\,H_{n}$,
together with the relations

$\displaystyle\lambda_{1}^{3}\,$
$\displaystyle=\,-\lambda_{1}^{2}a-\lambda_{1}a^{2}-a^{3},$
$\displaystyle\lambda_{2}^{6}\,$
$\displaystyle=\,-4\lambda_{2}^{5}a-10\lambda_{2}^{4}a^{2}-20\lambda_{2}^{3}a^{3},$
$\displaystyle B_{i}^{4}\,$
$\displaystyle=\,0\,\,\,\forall\,\,i\,=\,1,\,\cdots,\,m,\quad
H_{j}^{4}\,=\,0\,\,\,\forall\,\,j\,=\,1,\,\cdots,\,n,\quad\textnormal{and}\quad
a^{4}\,=\,0.$ (6.2)

###  6.1. Smooth conics

Let us compute the following intersection numbers:

$\displaystyle\big{[}\big{(}\mathsf{R}_{3}\big{)}_{\mathsf{Fb}}\big{]}\cdot\lambda_{1}^{m}\lambda_{3}^{j}a^{l}.$
(6.3)

Recall that $\left(\mathsf{R}_{3}\right)_{\mathsf{Fb}}$ is the fiber bundle
analogue of $\mathsf{R}_{3}(3)$ (cf. eq. 3.17). It can be thought of as a
subbundle of $\left(\mathsf{M}^{1}_{n}(2)\right)_{\textsf{Fb}}$. If

$\big{(}\Pi,\,\mathsf{L},\,\mathcal{C}_{2},\,p,\,x_{1},\,x_{2},\,x_{3}\big{)}\,\in\,\big{(}\mathsf{R}_{3}\big{)}_{\mathsf{Fb}},$

then $p$ lies on $\Pi$, $\mathsf{L}$ and $\mathcal{C}_{2}$ while $x_{i}$’s lie
on $\mathsf{L}$. The incidences correspond to the cycles $a+B_{1}$,
$\lambda_{1}+B_{1}$, $\lambda_{2}+2B_{1}$ and $\lambda_{1}+H_{i}$
respectively. The class of $\big{(}\mathsf{R}_{3}\big{)}_{\mathsf{Fb}}$ inside
$\mathfrak{M}^{1}_{n}(2)$ is given by

$\big{[}\big{(}\mathsf{R}_{3}\big{)}_{\mathsf{Fb}}\big{]}\ =\
(a+B_{1})\cdot(\lambda_{1}+B_{1})\cdot(\lambda_{2}+2B_{1})\cdot\prod_{i=1}^{3}\left[(a+H_{i})(\lambda_{1}+H_{i})\right].$

Intersection numbers of eq. 6.3 can now be computed using eq. 6.1.

Next, let us consider
$\left[\left(\mathsf{R}^{1}_{4}\right)_{\mathsf{Fb}}\right]$. It is a class
inside $\mathfrak{M}^{1}_{4}(2)$. The cycle $[p=x_{1}]$ is given by

$H_{1}^{3}+H_{1}^{2}B_{1}+H_{1}B_{1}^{2}+B_{1}^{3}.$

A similar argument would yield that the cycle
$\left[\left(\mathsf{R}^{1}_{4}\right)_{\mathsf{Fb}}\right]$ inside
$\mathfrak{M}^{1}_{4}(2)$ is given by

$\displaystyle\big{[}\big{(}\mathsf{R}^{1}_{4}\big{)}_{\mathsf{Fb}}\big{]}$
$\displaystyle=(a+B_{1})\cdot(\lambda_{1}+B_{1})\cdot(\lambda_{2}+2B_{1})\cdot(H_{1}^{3}+H_{1}^{2}B_{1}+H_{1}B_{1}^{2}+B_{1}^{3})\cdot\prod_{\begin{subarray}{c}i=1\\\
i\neq 1\end{subarray}}^{4}\left[(a+H_{i})(\lambda_{1}+H_{i})\right].$

Similarly,

$\displaystyle\big{[}\big{(}\mathsf{R}^{2}_{4}\big{)}_{\mathsf{Fb}}\big{]}$