Counting Curves with Tangencies
Interpreting tangency as a limit of two transverse intersections, we obtain a concrete formula to enumerate smooth degree plane curves tangent to a given line at multiple points with arbitrary order of tangency. Extending that idea, we then enumerate curves with one node with multiple tangencies to...
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| Опубліковано в: : | Symmetry, Integrability and Geometry: Methods and Applications |
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| Цитувати: | Counting Curves with Tangencies. Indranil Biswas, Apratim Choudhury, Ritwik Mukherjee and Anantadulal Paul. SIGMA 21 (2025), 012, 50 pages |
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Digital Library of Periodicals of National Academy of Sciences of Ukraine| _version_ | 1860271457944207360 |
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| author | Biswas, Indranil Choudhury, Apratim Mukherjee, Ritwik Paul, Anantadulal |
| author_facet | Biswas, Indranil Choudhury, Apratim Mukherjee, Ritwik Paul, Anantadulal |
| citation_txt | Counting Curves with Tangencies. Indranil Biswas, Apratim Choudhury, Ritwik Mukherjee and Anantadulal Paul. SIGMA 21 (2025), 012, 50 pages |
| collection | DSpace DC |
| container_title | Symmetry, Integrability and Geometry: Methods and Applications |
| description | Interpreting tangency as a limit of two transverse intersections, we obtain a concrete formula to enumerate smooth degree plane curves tangent to a given line at multiple points with arbitrary order of tangency. Extending that idea, we then enumerate curves with one node with multiple tangencies to a given line of any order. Subsequently, we enumerate curves with one cusp that are tangent to first order to a given line at multiple points. We also present a new way to enumerate curves with one node; it is interpreted as a degeneration of a curve tangent to a given line. That method is extended to enumerate curves with two nodes, and also curves with one tacnode are enumerated. In the final part of the paper, it is shown how this idea can be applied in the setting of stable maps and perform a concrete computation to enumerate rational curves with first-order tangency. A large number of low-degree cases have been worked out explicitly.
|
| first_indexed | 2026-03-21T11:38:30Z |
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Symmetry, Integrability and Geometry: Methods and Applications SIGMA 21 (2025), 012, 50 pages
Counting Curves with Tangencies
Indranil BISWAS a, Apratim CHOUDHURY b, Ritwik MUKHERJEE c
and Anantadulal PAUL d
a) Department of Mathematics, Shiv Nadar University,
NH91, Tehsil Dadri, Greater Noida, Uttar Pradesh 201314, India
E-mail: indranil.biswas@snu.edu.in, indranil29@gmail.com
b) Institut für Mathematik, Humboldt-Universität zu Berlin,
Unter den Linden 6, Berlin 10099, Germany
E-mail: apratim.choudhury@hu-berlin.de
c) School of Mathematical Sciences, National Institute of Science Education and Research,
Bhubaneswar, An OCC of Homi Bhabha National Institute, Khurda 752050, Odisha, India
E-mail: ritwikm@niser.ac.in
d) International Center for Theoretical Sciences, Survey No. 151, Hesaraghatta,
Uttarahalli Hobli, Sivakote, Bangalore 560089, India
E-mail: anantadulal.paul@icts.res.in, paulanantadulal@gmail.com
Received May 04, 2024, in final form February 07, 2025; Published online February 23, 2025
https://doi.org/10.3842/SIGMA.2025.012
Abstract. Interpreting tangency as a limit of two transverse intersections, we obtain a con-
crete formula to enumerate smooth degree d plane curves tangent to a given line at multiple
points with arbitrary order of tangency. Extending that idea, we then enumerate curves
with one node with multiple tangencies to a given line of any order. Subsequently, we
enumerate curves with one cusp, that are tangent to first order to a given line at multiple
points. We also present a new way to enumerate curves with one node; it is interpreted as
a degeneration of a curve tangent to a given line. That method is extended to enumerate
curves with two nodes, and also curves with one tacnode are enumerated. In the final part
of the paper, it is shown how this idea can be applied in the setting of stable maps and per-
form a concrete computation to enumerate rational curves with first-order tangency. A large
number of low degree cases have been worked out explicitly.
Key words: enumeration of curves; tangency; nodal curve; cusp
2020 Mathematics Subject Classification: 14N35; 14J45; 53D45
1 Introduction
A prototypical question in enumerative geometry is as follows: what is the characteristic number
of curves in a linear system that have certain prescribed singularities and are tangent to a given
divisor of various orders at multiple points? The curves are, of course, required to meet further
insertion conditions so that the ultimate answer is a finite number. For example, in P2 there
are exactly 2 conics passing through 4 generic points that are tangent to a given line, and there
are 36 nodal cubics in P2 through 7 generic points tangent to a given line. These are some special
cases of the famous Caporaso–Harris formula [4], which addresses the following question: How
many degree d curves are there in P2 that pass through j generic points, having δ nodes that
are tangent to a given line at r distinct points, with the orders of tangency being k1, k2, . . . , kr,
where
j :=
d(d+ 3)
2
− (δ + (k1 + k2 + · · ·+ kr)).
indranil.biswas@snu.edu.in
indranil29@gmail.com
mailto:apratim.choudhury@hu-berlin.de
mailto:ritwikm@niser.ac.in
mailto:anantadulal.paul@icts.res.in
mailto:paulanantadulal@gmail.com
https://doi.org/10.3842/SIGMA.2025.012
2 I. Biswas, A. Choudhury, R. Mukherjee and A. Paul
We have defined here the order of tangency to be k if the order of contact is k + 1 (i.e., the
intersection multiplicity of the curve and the line is k + 1, hence the transversal intersection is
of contact order 1).
The purpose of this paper is to give a new way to think about the question of tangency.
The main idea can be summarized in one picture: This interpretation of tangency allows us to
effortlessly enumerate smooth curves with multiple tangencies of any order. With a little more
effort, this method can also be applied to enumerate 1-nodal curves with tangencies (henceforth,
curves with δ nodes are referred to as δ-nodal curves). We then go on to show that the method
can also be applied to enumerate 1-cuspidal curves (i.e., curves with one cusp) with first-order
tangencies at multiple points. We note that the Caporaso–Harris formula counts curves with
only nodal singularities.
Our idea can also be applied to study the question of enumerating stable maps tangent to
a given divisor. The difficulty of applying this idea in the context of stable maps has been
discussed by Gathmann in his paper [8, p. 41]. Subsequently, the idea has also been discussed
in the more recent paper by Dusa McDuff and Kyler Siegel [14, pp. 1179–1180]. The discussion
in [14] illustrates that for applying this idea in the context of stable maps to enumerate curves
with tangencies, we need to compute the characteristic number of curves with an m-fold point;
[2] precisely does the latter. Hence, using the results of [2] and by interpreting tangency as a limit
of points lying on a line (as illustrated by Figure 1), it is possible to enumerate stable maps with
tangencies. In Section 10, a very concrete computation based on this idea is worked out.
x2
x1 = x2x1
Figure 1. Tangency as a limit of transverse intersection.
Finally, in Section 10.3 this idea is pursued again, but with a difference. We make the points
in the domain come together, not in the image. This directly gives us rational curves with
tangencies. The idea is implemented using the equality of divisors in M0,4. In this way, we are
able to count rational curves with first-order tangencies without making use of the psi classes
in any way; only a knowledge of primary Gromov–Witten invariants is required.
2 Main results
We now state the main results of this paper. Let k1, k2, . . . , kn be positive integers and d be an
integer greater than or equal to k+n, where k := k1+ · · ·+kn. Define Nd(Tk1 . . .Tkn) to be the
number of degree d curves in P2, passing through the requisite number of generic points, and
that are tangent to a given line at n distinct points with orders k1, k2, . . . , kn. Note that aside
from these n points of contact, the curve also intersects the line transversally at d− (k + n)
further points. Define
Nd(A1Tk1 . . .Tkn) and Nd(A2Tk1 . . .Tkn)
to be the number of 1-nodal (respectively 1-cuspidal) degree d curves in P2, passing through the
requisite number of generic points and that are tangent to a given line at n distinct points with
orders k1, k2, . . . , kn. The first main result of this paper is as follows.
Counting Curves with Tangencies 3
Main Result 2.1. By interpreting tangency as a limit of transverse intersections, we are able
to compute
Nd(Tk1 . . .Tkn), Nd(A1Tk1 . . .Tkn) and Nd(A2 T1 . . .T1︸ ︷︷ ︸
n-times
),
provided d > n+ k − 1, d > n+ k and d > 2n+ 2, respectively.
Denote by Nd(A1) (respectively, Nd(A3)) the number of 1-nodal (respectively, 1-tacnodal)
plane degree d curves, passing through the right number of generic points. Also, denote Nd(A1A1)
by the number of 2-nodal plane degree d curves, passing through the right number of generic
points. Our next result is as follows.
Main Result 2.2. By interpreting a nodal point as a degeneration of a first-order tangency, we
obtain a new method to compute Nd(A1) and Nd(A1A1), provided d > 1 and d > 3, respectively.
Furthermore, by interpreting a tacnode as a limit of two nodal points, we obtain a new method
to compute Nd(A3), provided d > 3.
To see what we mean by a node arising as a degeneration of a tangency point, consider the
polynomial ft(x, y) := ty + y2 − x2. We note that the zero set of this curve is tangent to the
x-axis at the origin. The origin is also a smooth point of the curve when t ̸= 0. However, in the
limit when t goes to zero, we see that this becomes a nodal curve.
Our final result is on counting stable maps (modulo automorphisms of the domain). Let NT1
d
denote the number of rational degree d stable maps (modulo automorphisms) passing through
3d− 2 generic points and that are tangent to a given line (to first order). Our third result is as
follows.
Main Result 2.3. By interpreting tangency as a limit of transverse intersections, we obtain
two new ways to compute the number NT1
d . The first method consists of moving the two points
on the images of the stable maps, while the second method consists of moving the two points in
the domain of the stable map.
The bound we impose on d for Main Results 2.1 and 2.2 is not optimal. It is a sufficient
condition for our formula to be valid, it is not necessary. The bound is imposed to prove
transversality, which involves constructing curves satisfying certain conditions. If d is sufficiently
large, then one can easily construct such curves. In Section 11, we do several low degree checks.
We observe in that section that even when d is smaller than what is required by our result, the
values agree with the expected values.
We have written a Mathematica program to implement the formulas in Sections 6, 7, 8, 9
and 10. Also a Python program is written to implement the Caporaso–Harris formula.1
3 Comparison with other methods
Let us compare our method with the Caporaso–Harris formula. The central idea of the Capora-
so–Harris formula is a degeneration argument, which can briefly be described as follows. They
argue that the number of degree d plane curves with prescribed tangencies and passing through j
generic points is equal to the number of plane degree d curves with the same prescribed tangencies
passing through j−1 generic points and also intersecting the line transversally at one more fixed
point, plus an excess contribution. The excess contribution comes from
1The programs are available at https://www.sites.google.com/site/ritwik371/home. The reader is invited
to use the program and verify the assertions.
https://www.sites.google.com/site/ritwik371/home
4 I. Biswas, A. Choudhury, R. Mukherjee and A. Paul
� Degree d curves with the same prescribed tangencies, but where one of the free tangency
points gets replaced by a fixed tangency point.
� Curves of lower degree with possibly higher order tangencies.
The total excess contribution is a sum over all these terms.
This will be explained more precisely by working out the following question: How many
degree d curves are there in P2, that pass through δd − 1 generic points and are tangent to
a given line, where δd := d(d+3)
2 . After we explain how Caporaso and Harris solve this problem,
we then explain how we solve this question in this paper. This one example clearly illustrates
the difference between the two methods.
We make a few definitions. First of all, define Nd(T1,m) to be the number of degree d curves,
passing through δd − 1 −m generic points that are tangent to first order to a given line at an
unspecified point and intersecting the line transversally at m fixed points. Notice that requiring
the curve to intersect the line at a fixed point is what imposes a genuine constraint. Requiring the
curve to intersect the line transversally at some unspecified point would not impose any condition
(because any curve does intersect a line somewhere). Our goal is to compute Nd(T1, 0) (which
we often abbreviate as Nd(T1)).
Next, we define Nd
(
Tpt
1 ,m
)
to be the number of degree d curves, passing through δd − 2−m
generic points that are tangent to first order to a given line at a fixed point and that also
intersects the line at m fixed points. As before, we often abbreviate Nd
(
Tpt
1 , 0
)
as Nd
(
Tpt
1
)
.
Also, define Nd(S) to be the number of degree d curves passing through δd generic points. Of
course, this number is equal to one, but never the less denote it by the symbol Nd(S), which
make the geometric ideas behind the subsequent formulas more transparent. A special case of
the Caporaso–Harris formula [4, p. 348, Theorem 1.1] is as follows:
Nd(T1,m) = 2Nd
(
Tpt
1 ,m
)
+ Nd(T1,m+ 1) ∀d ≥ m+ 2 and (3.1)
Nd
(
Tpt
1 ,m
)
= Nd
(
Tpt
1 ,m+ 1
)
∀d ≥ m+ 3. (3.2)
Furthermore,
Nd
(
Tpt
1 , d− 2
)
= Nd−1(S) ∀d ≥ 2. (3.3)
Using the fact that Nd(S) = 1, equations (3.1), (3.2) and (3.3) recursively give us the value
of Nd(T1) for all d ≥ 2.
This recursive formula will be analysed. The structure of the formula is as follows: one gets
a recursive formula for Nd(T1) in terms of Nd−1(T1). Furthermore, the underlying geometric
principle is as follows: the point constraints that are imposed on the curves (to get a finite
number) are one by one moved to the line and the corresponding numerical invariants are
compared. As we keep moving the points on the line, the curve is forced to break into a reducible
curve (one of whose components is the line) and a curve of lower degree. Unwinding this
geometric phenomenon in terms of numbers ultimately gives us equations (3.1), (3.2) and (3.3).
Next, we explain how Caporaso–Harris enumerate curves with one node. We make a few
definitions first. Define Nd(A1,m) to be the number of degree d curves with one node, passing
through δd − 1 − m generic points and m generic points on the line. Our goal is to com-
pute Nd(A1, 0) (which as before, is abbreviated as Nd(A1)). With these notations, a special case
of the Caporaso–Harris formula [4, p. 348, Theorem 1.1] is as follows:
Nd(A1) = Nd(A1,m) ∀d ≥ m+ 1, (3.4)
Nd(A1, d− 1) = Nd(A1, d) + (d− 1)Nd−1(S) and (3.5)
Nd(A1, d) = Nd−1(A1) + dNd−1(S) + 2Nd−1(T1). (3.6)
Counting Curves with Tangencies 5
Equations (3.4), (3.5) and (3.6) give us that
Nd(A1) = (2d− 1) + Nd−1(A1) + 2Nd−1(T1). (3.7)
Since Nd−1(T1) can be computed, equation (3.7) enables us to compute Nd(A1).
Let us again try to understand the underlying geometric reason behind the formulas. Equa-
tion (3.4) says that the number of 1-nodal degree d curves passing through δd− 1 generic points
is the same as the number of 1-nodal degree d curves through δd − 1 − (d − 1) generic points
and d− 1 points on a line (apply the equation form = d−1, for which it is valid). Equation (3.5)
on the other hand says that once we move another point on the line, the two numbers are no
longer the same. There is an excess contribution from a curve of degree d−1 (because the curve
can now break). The two numbers Nd(A1, d − 1) and Nd(A1, d) are related via equation (3.5).
Finally, equation (3.6) gives us a formula for Nd(A1, d) in terms of counts of curves of lower
degree.
More generally, consider the computation of the number Nd
(
Aδ
1Tk1 . . .Tkn
)
, the number
of δ-nodal degree d curves tangent to a given line at n points, with order of tangency be-
ing k1, k2, . . . , kn (and also passing through the right number of generic points). The Caporaso–
Harris formula give a recursive formula for this number in terms of Nd′
(
Aδ′
1 Tl1 . . .Tlm
)
, where
d′ ≤ d and δ′ ≤ δ. The geometric principle based on which this is obtained is the same: the
point constraints are moved to the line and as a result the curve breaks, giving contributions to
the intersection from curves of lower degree.
We now briefly explain our method (details are of course worked out in the subsequent
sections). Consider the question of computing Nd(T1). We view this as an intersection number
in the ambient space D1×Dd×P2×P2, where D1 and Dd refer to the space of lines and degree d
curves in P2 respectively. On this ambient space, we define the subspace of degree d curves
and a line and two marked points, such that the two marked points lie on the line and the
curve. We find an expression for the homology class represented by the closure of this space.
Now we impose the condition that the two marked points come together (this is where we use
the collision lemma). That gives us the space of curves tangent to a given line. Finally, we
intersect it with the point constraints that are necessary to get a finite number. That gives us
the number Nd(T1).
Next, consider the question of computing Nd(A1). The underlying idea is as follows. We view
this as an intersection number on D1 ×Dd × P2. On this ambient space, we define the subspace
of degree d curves and a line and one marked point, such that the curve is tangent to the line
at the marked point. On the closure of this space, impose the condition that the directional
derivative of the polynomial defining the curve in the normal direction to the line vanishes. That
gives us the subspace of line, a curve and a marked point, such that the marked point is a nodal
point of the curve and it lies on the line. In order to compute Nd(A1), we simply make the curve
pass through δd − 1 generic points, and make the line pass through one point (the second point
through which the line will pass is the nodal point of the curve).
We now make a few remarks. First and foremost, the underlying geometric mechanism
that governs our formula is completely different from what Caporaso–Harris are doing. Our
formula for Nd(T1) is governed by the principle that when two transverse points coincide, they
become a point of tangency. A generalization of that principle is used in the computation of
Nd(Tk1 . . .Tkn). The Caporaso–Harris formula is based on the principle that when the point
constraints of an enumerative problem are repeatedly moved to a divisor (a line in this case),
the curve eventually has to break.
The second point is as follows: our formula is not a recursion on d. We directly get a formula
in terms of d. Our formula does involve recursion on other quantities, such as the number of
points of tangency and the orders of tangency, but not the degree of the curve. It is therefore
completely straightforward to see that our formula produces a polynomial in d (this is explained
6 I. Biswas, A. Choudhury, R. Mukherjee and A. Paul
clearly at the end of Section 6). It is however far from obvious that a recursion on d produce
a polynomial in d. To see why that is so, write down the full Caporaso–Harris formula [4, p. 348,
Theorem 1.1]
Nd,δ(α, β) =
∑
k:βk>0
kNd,δ(α+ ek, β − ek)
+
∑
Iβ−β′
(
α
α′
)(
β′
β
)
Nd−1,δ′(α′, β′), (3.8)
where the second sum is taken over all α′, β′ and δ′, such that
α′ ≤ α, β ≤ β′, δ′ ≤ δ and δ − δ′ + |β − β′| = d− 1.
The reader can refer to [4] for the relevant notation; for the purpose of this discussion only
the structure of the formula is important. From the structure of the formula, we can see that
it is far from clear that the characteristic numbers Nd,δ(α, β) are polynomials in d is because
they are expressed recursively in terms of characteristic numbers of lower degree. Seeing the
formula (3.8), we can not rule out the possibility that the characteristic numbers behave as
factorial or exponential functions of d. In light of Göttsche like conjectures (which broadly
speaking says that the solutions to enumerative problems involving degree d curves is given by
universal polynomials), this is perhaps an interesting point.
The final difference we would like to mention is that we enumerate curves with cuspidal
singularity that are tangent to a line at multiple points. As of now, Caporaso–Harris has only
been worked out for nodal curves with tangencies. However, in our paper [3], we have shown
how to extend the idea of Caporaso–Harris to enumerate curves with one cusp.
Finally, we mention another approach to enumerate curves with tangencies, which is taken
by the fourth author in [15, 16]. Again consider the computation of Nd(T1). Instead of viewing
tangency as a limit of two transverse intersections, only one point is considered. The geometric
condition imposed here is that the directional derivative of the function, defining the curve,
along the direction of the line is zero. This can be suitably interpreted as the Euler class of
a bundle. Using this approach, the author has been able to compute the characteristic number
of singular curves, tangent to first order to a given line at one point.
4 A few remarks on intersection theory
In this section, we summarize a few facts about intersection theory. The ambient space M inside
which intersection theory is done will be a product of projective spaces. Whenever we talk about
an open set, we mean open set with the analytic topology of the projective space; we are not
talking about the Zariski topology.
Let α and β be two homology classes in M, i.e., α, β ∈ H∗(M,R). We define α · β the
topological intersection of α and β to be the unique homology class whose Poincaré dual is the
cup product of the Poincaré duals of α and β.
We now follow a standard abuse of notation. Given a homology class α, we denote its Poincaré
dual (a cohomology class) by the same letter α. Hence, from the point of view of cohomology,
topological intersection is simply the cup product.
The homology classes that we encounter are going to be obtained by taking the closure of
certain algebraic varieties. More precisely, our situation is as follows: we have a set S, which is
a smooth complex submanifold of the ambient space M, but it is not closed. The closure of S
(inside M) need not be smooth. Nevertheless, S does define a homology class. The reason for
this is as follows. The singularities of the closure are of complex codimension one and hence,
of real codimension two. Hence, by Stokes theorem for analytic varieties [11, p. 33] integration
Counting Curves with Tangencies 7
over S makes sense. Hence, any analytic subvariety of a compact complex manifold M, always
defines a homology class in H∗(M,R) as explained in [11, pp. 33 and 61]. We often refer to S as
the open part of the cycle S.
We often be dealing with zero sets of sections of certain complex vector bundles. It is
a standard fact that when the section is transverse to zero, the cycle represented by the zero
set is Poincaré dual to the top Chern class of the vector bundle. The top Chern class is also
referred to as the Euler class of the bundle; we usually use the latter terminology.
5 The collision lemma
This section contains the central lemma that we will be using throughout the paper, which is
be referred to as the collision lemma. Before stating the lemma, we need to introduce a few
notations.
Let D1 be the space of lines in P2; it is the dual projective space P̌2. DefineM := D1×X1×X2,
where Xi is a copy of P2 for i = 1, 2. The pullback to M of the hyperplane classes in X1, X2
and D1 are denoted by a1, a2 and y1, respectively. Define
X := {(H, q1, q2) ∈M | q1, q2 ∈ H} and Y := {(H, q1, q2) ∈ X | q1 = q2}.
Lemma 5.1 (collision lemma). The cohomology class (a1 + a2 − y1) restricted to X is equal to
the Poincaré dual of Y in X.
Remark 5.2. Let ∆L
1,2 be the following divisor class on M
∆L
1,2 := a1 + a2 − y1. (5.1)
The geometric content of the collision lemma is that the intersection of X with the class ∆L
1,2 is
equivalent to the class obtained by making the two points come together in X.
Proof. First of all, we note that the cohomology of X is generated by a1, a2 and y1; this follows
from the Leray–Hirsch theorem. Since Y is a codimension one submanifold of X, we conclude
that the Poincaré dual of Y in X is given by
PDX [Y ] = (Aa1 +Ba2 + Cy1)|X , (5.2)
for some numbers A, B and C that are to be determined. To prove the lemma, it suffices to
show that A = 1, B = 1 and C = −1. This, we show by computing three different intersection
numbers.
First of all, we note that the Poincaré dual of X in M is given by
PDM [X] = (y1 + a1) · (y1 + a2). (5.3)
To see why this is so, first define Z1 as Z1 := {(H, q1) ∈ D1 × X1 | q1 ∈ H}. To prove (5.3),
it suffices to show that [Z1] = (y1 + a1). To justify this, note that [Z1] = ny1 + ma1 for
some numbers n and m. Next, we note that n = [Z1] · y1a21 and m = [Z1] · y21a1. Geometrically,
[Z1] ·y21a1 is the number of lines passing through two points
(
which corresponds to the factor y21
)
and a marked point on the line that intersects another generic line (which corresponds to the
factor a1). This is clearly 1. Similarly, [Z1] · y1a21 is geometrically the number of lines through
two points (the first point corresponds to the factor a21 and the second point corresponds to the
factor y1). This number is also equal to 1. Hence, [Z1] = y1 + a1, thereby proving (5.3).
Next, we note that Y can also be described as a codimension three submanifold of M in the
following way
Y := {(H, q1, q2) ∈M | q1 ∈ H, q1 = q2}.
8 I. Biswas, A. Choudhury, R. Mukherjee and A. Paul
Hence, the Poincaré dual of Y in M is given by PDM [Y ] = (y1 + a1) · [∆12], where ∆12 is the
subspace of points (H1, q1, q2) in M , such that q1 = q2. Using the standard fact that the class
of the diagonal [∆12] is equal to
(
a21 + a1a2 + a22
)
, we conclude that
PDM [Y ] = (y1 + a1) ·
(
a21 + a1a2 + a22
)
. (5.4)
Let µ be a class in M of degree three. By equations (5.2), (5.3) and (5.4), we conclude that
(y1 + a1) · (a21 + a1a2 + a22) · µ = (Aa1 +Ba2 + Cy1) · (y1 + a1) · (y1 + a2) · µ.
By suitably choosing µ, we can determine A, B and C. Choosing µ := a1a
2
2, a2a
2
1 and a2y
2
1, we
have A+C = 0, B+C = 0 and A = 1, respectively. This precisely implies that A = B = 1 and
C = −1. ■
6 Counting smooth curves with multiple tangencies
In this section, we use the collision lemma to derive our Main Result 2.1. Before we get into the
details, a brief outline of our method is described.
Let Dd denote the space of degree d curves in P2. This is a complex projective space of
dimension
δd :=
d(d+ 3)
2
. (6.1)
We now try to answer the following question: How many degree d curves are there in P2 passing
through δd − 1 generic points and that are tangent to a given line?
We approach this problem in the following way: First, consider the space of curves with
two marked points x1 and x2, such that both the points lie on the curve and the line. Now
impose the condition that x1 becomes equal to x2; this is precisely where we use the collision
lemma. Once we impose the condition that the points x1 and x2 have become equal, the curve
becomes tangent to this line. We now implement this idea precisely. Let X1 denote a copy of
the projective plane. Define the incidence variety
Id := {(Hd, x1) ∈ Dd ×X1 | x1 ∈ Hd}. (6.2)
Elements of the incidence variety consists of degree d-curve Hd and a marked point x1 that lies
on this curve. Let a1 and yd denote the divisor classes on Dd ×X1 obtained by pulling back the
hyperplane classes on X1 and Dd, respectively. Then we have
[Id] = yd + da1. (6.3)
Indeed, this follows immediately from the fact that the restrictions of (6.3) to both {point}×CP2
and Dd × {point} are valid.
Next, we study subspaces of D1×Dd×X1. First, define IL (respectively, IC) to be the subspace
of D1 ×Dd ×X1 consisting of all (H1, Hd, x1) ∈ D1 ×Dd ×X1 such that x1 ∈ H1 (respectively,
x1 ∈ Hd). Next, define
T0 (6.4)
to be the subspace of D1 ×Dd ×X1 consisting of all (H1, Hd, x1) ∈ D1 ×Dd ×X1 such that H1
and Hd intersect transversally at x1. Note that the closure T0 consists of (H1, Hd, x1) ∈ D1 ×
Dd ×X1 such that x1 ∈ H1
⋂
Hd.
For making the notation easier to read, we denote the homology class represented by the
closure by the notation [T0]
(
as opposed to more cumbersome
[
T0
])
.
Counting Curves with Tangencies 9
Lemma 6.1. The class [T0] ∈ H2(δd+2)(D1 × Dd ×X1,R) represented by the cycle T0 ⊂ D1 ×
Dd ×X1 (see (6.1), (6.4) and (6.1)) is the following:
[T0] = [IL] · [IC ], (6.5)
where [IL] and [IC ] are the homology classes of IL and IC, respectively.
Proof. Consider the divisor I1 ⊂ D1 × X1 in (6.2). Let f : I1 × Dd ↪→ D1 × Dd × X1 be the
natural map defined by ((H1, x1), Hd) 7−→ (H1, Hd, x1), where (H1, x1) ∈ I1 and Hd ∈ Dd. Let
p : D1 ×Dd ×X1 −→ Dd ×X1 (6.6)
be the natural projection. Note that T0 ⊂ I1 ×Dd (see (6.4)) and T0 is a divisor.
To prove the lemma, it suffices to show the following:
f∗p∗ODd×X1(Id) = OI1×Dd
(
T0
)
. (6.7)
We now describe three subvarieties S1, S2 and S3 of I1 ×Dd. Fix a point z0 ∈ I1 and define
S1 := {z0} × Dd ⊂ I1 ×Dd. (6.8)
Fix curves (H1, Hd) ∈ D1 ×Dd in X1, and we have
S2 := {(H1, Hd, x1) ∈ D1 ×Dd ×X1 | x1 ∈ H1} ⊂ I1 ×Dd. (6.9)
So S2 is identified with the line H1. Fix a point x0 ∈ X1 and also an element Hd ∈ Dd. Now
define
S3 := {(H,Hd, x0) ∈ D1 ×Dd ×X1 | x0 ∈ H} ⊂ I1 ×Dd. (6.10)
Consequently, S3 is identified with the pencil of lines in X1 containing x0.
Given two cohomology classes L1,L2 ∈ H2(I1×Dd,R), to show that L1 = L2, it is enough to
prove that L1
∣∣
Sj
= L2
∣∣
Sj
for j = 1, 2, 3 (see (6.8), (6.9), (6.10)).
In view of the above criterion, it is now straightforward to prove (6.7) using (6.3). ■
Note that using (6.3), we can rewrite equation (6.5) as
[T0] = (y1 + a1) · (yd + da1), (6.11)
where · denotes topological intersection. Generalizing the notion of T0, given any nonnegative
integer k, we define Tk to be the subspace of D1 ×Dd ×X1 consisting of all points (H1, Hd, x1)
such that
� The points x1 is a smooth point of the curve Hd.
� The line H1 intersects the curve Hd at the points x1 with the order of tangency precisely
equal to k.
Notice that as per its definition, T1 is not a subset of T0, but a subset of the closure T0.
Next, generalizing D1×Dd×X1, define Mn := D1×Dd×(X1×· · ·×Xn), where Xj , 1 ≤ j ≤ n,
is a copy of P2. Let S1,S2, . . . ,Sn be subvarieties of D1 × Dd × P2. Denote S1S2 . . . Sn by S.
Then, S ⊂ Mn consists of (H1, Hd, x1, . . . , xn) such that
� (H1, Hd, xi) ∈ Si for all i = 1 to n.
� The points x1, . . . , xn are all distinct.
10 I. Biswas, A. Choudhury, R. Mukherjee and A. Paul
As an example, consider the set T1T2. This comprises of the set of curves along with two
distinct marked points on a line, where the curve is tangent to first order to the line at the first
marked point and is tangent to second order to the line at the second marked point. Similarly,
T1T2 denotes a slightly bigger space, where the curve is at least as degenerate as being tangent
to the line to first order at the first marked point and is at least as degenerate as being tangent
to the line to second order at the second marked point. The curve could be tangent to second
order at the first marked point. The curve could even have a nodal point at the first marked
point, since these both lie in the closure T1. However, the two marked points have to be distinct.
In particular, T1T2 is not the set-theoretic intersection of T1 and T2, since the latter includes
the locus where the two marked points are equal. Finally, in the space T1T2, the two marked
points need not be distinct; this denotes the closure of the space T1T2 and it includes the locus
where the two marked points coincide.
We denote the homology class defined by the closure of S by the notation
[
S
]
as opposed to
the more cumbersome [S]; this makes some of the computations and statements easier to read.
Finally, let
π : Mn+1 −→ Mn (6.12)
be the projection that sends any (H1, Hd, x1, . . . , xn, xn+1) to (H1, Hd, x1, . . . , xn). Let
πn+1 : Mn+1 −→ M1 (6.13)
be the projection that sends any (H1, Hd, x1, . . . , xn, xn+1) to (H1, Hd, xn+1). For any 1 ≤ i ≤ n,
let ∆i,n+1 ⊂ Mn+1 be the locus of all (H1, Hd, x1, . . . , xn, xn+1) such that xi = xn+1. We are
now ready to state the main results to enumerate smooth curves with tangencies.
Theorem 6.2. Let n be a positive integer and k1, k2, . . . , kn nonnegative integers with k :=
k1 + k2 + · · ·+ kn. Then the following equality of elements of H2(δd+2−k)(Mn+1,R) holds:
π∗[Tk1 . . .Tkn ] · π∗n+1[T0] = [Tk1 . . .TknT0] +
n∑
i=1
(ki + 1)π∗[Tk1 . . .Tkn ] · [∆i,n+1] (6.14)
(see (6.12), (6.13)) provided d > k + n.
Note that implicit in Theorem 6.2 is the assertion that the closure of Tk1 . . .Tkn in Mn and
the closure of Tk1 . . .TknT0 in Mn+1 actually define homology classes. We will justify that
assertion as well. In order to prove Theorem 6.2, we first prove an intermediate statement which
is interesting in its own right.
Proposition 6.3. Let n be a positive integer and k1, k2, . . . , kn nonnegative integers with k :=
k1 + k2 + · · ·+ kn. Then Tk1Tk2 . . .Tkn is a smooth submanifold of Mn, provided d ≥ k + n.
Remark 6.4. We are not claiming that Tk1Tk2 . . .Tkn
is a smooth manifold. Notice the differ-
ence between Tk1Tk2 . . .Tkn and Tk1Tk2 . . .Tkn
; in the former space all the marked points are
distinct, while in the latter space, that is not necessarily true.
Proof. We start by proving the proposition for n = 1 and k1 = 0, i.e., we show that T0 is
a smooth submanifold of M1.
Let Fd be the space of polynomials in two variables of degree at most d. This is a vector
space of dimension d(d+3)
2 + 1, because an element of Fd can be viewed as
f(x, y) := f00 + f10x+ f01y +
f20
2
x2 + f11xy +
f02
2
y2 + · · ·+ f0d
d!
yd,
Counting Curves with Tangencies 11
and hence f can be identified with the vector (f00, f10, f01, . . . , f0d). Let F+
d denote the space
of nonzero polynomials of degree d. We note that the projectivization of F+
d is Dd.
Consider the projection map
π+ : F+
1 ×F+
d ×X1 −→ D1 ×Dd ×X1.
We note that the projection map is a submersion. Hence, it suffices to show that T̃0 := π−1
+
(
T0
)
is a smooth submanifold of F+
1 ×F+
d ×X1. We prove that now.
Let (g, f, p1) ∈ T̃0. By choosing a chart around the point p1, we can identify an open neigh-
bourhood of p1 (in X1) by C2. Hence, in order to show that T̃0 is a smooth submanifold
of F+
1 ×F+
d ×X1, it suffices to show that (0, 0) is a regular value of the map
φ : F+
1 ×F+
d × C2 −→ C2, given by φ(g, f, (x, y)) := (g(x, y), f(x, y)).
Let us prove that now. Assume that φ(g, f, (a, b)) = 0. We need to show that the differential
of φ, evaluated at (g, f, (a, b)) is surjective. In order to prove that, consider the two curves
γ1, γ2 : (−ε, ε) −→ F+
1 ×F+
d × C2, given by
γ1(t) := (g + tη1, f, (a, b)) and γ2(t) := (g, f + tη2, (a, b)),
where η1 and η2 are as yet, unspecified elements of F+
1 and F+
d . We now note that
{dφ|(g,f(a,b))}(γ′1(0)) = (η1(a, b), 0) and {dφ|(g,f,(a,b))}(γ′2(0)) = (0, η2(a, b)).
Hence, to prove that the differential is surjective, we simply need to produce η1 ∈ F+
1 and
η2 ∈ F+
d such that η1(a, b) ̸= 0 and η2(a, b) ̸= 0. That is easily achieved: we simply define both
of them to be the constant functions taking the value 1.
Before proceeding further, we make a couple of simplifications that make the subsequent
proofs easier. We showed that if φ(g, f, (a, b)) = 0, then the differential of φ is surjective.
We claim that by making a suitable change of coordinates, we can always assume that (a, b)
is the origin and the line is the x-axis. To see why this is so, assume that the line is given
by g00 + g10x+ g01y = 0. Assuming that g01 ̸= 0, define the new coordinates X and Y by
X := x−a and Y := g00+g10x+g01y. If g01 = 0, then defineX := y−b and Y := g00+g10x+g01y.
Define
F (X,Y ) := f(x(X,Y ), y(X,Y )).
This is the polynomial f written in the new coordinates X and Y . Define Fij :=
∂i+jF (X,Y )
∂Xi∂Y j
∣∣
(0,0)
.
In these new coordinates, the point under consideration is the origin and the line is the X-axis.
Furthermore, the coefficients of the polynomial are given by
Fij
i!j! . Hence, what we have shown
is that T̃0 is a fibre bundle over the incidence variety J (where J is defined to be the subset
of F+
1 ×X1 where the point lies on the line) and the fibre over (g, p) ∈ J can be identified
with
(
T0
)
Aff
, where
(
T0
)
Aff
:= {f ∈ F+
d : f00 = 0}. The map f going to F is a trivialization of
this fibre bundle. Henceforth, we set the line to be the x-axis and the point to be the origin;
this makes the calculations simpler.
We now show that Tk1 is a smooth submanifold of M1 for all k1. We use induction on k1.
Assume that we have proved the assertion till k1 − 1. Hence Tk1−1 is a smooth submanifold
of M1. Hence, T̃k1−1 := π−1
+
(
Tk1−1
)
is a smooth submanifold of F+
1 ×F+
d ×X1.
We now show that T̃k1 is a smooth submanifold of T̃k1−1. Define
(
Tk1−1
)
Aff
as follows:(
Tk1−1
)
Aff
:=
{
f ∈ F+
d | f00, f10 . . . , fk1−1,0 = 0
}
.
We note that T̃k1−1 is a fibre bundle over F+
1 ×X1 whose fibres can be identified with
(
Tk1−1
)
Aff
.
In order to show that T̃k1 is a smooth submanifold of F+
1 ×F+
d ×X1, it suffices to show that zero is
12 I. Biswas, A. Choudhury, R. Mukherjee and A. Paul
a regular value of the map φ :
(
Tk1−1
)
Aff
−→ C, given by φ(f) := fk1,0. Suppose f ∈
(
Tk1−1
)
Aff
.
Let γ : (−ε, ε) −→
(
Tk1−1
)
Aff
be a curve, given by γ(t) := f+tη, where η is as yet an unspecified
polynomial. We now note that {dφ}(γ′(0)) = ηk1,0. Now choose η as follows η(x, y) := xk1 .
Since f ∈
(
Tk1−1
)
Aff
, f + tη also belongs to
(
Tk1−1
)
Aff
for all t nonzero but small. Furthermore,
ηk1,0 ̸= 0. This proves the claim.
Next, for multiple points, we now use induction on n. If kn ≥ 1, define(
Tk1Tk2 . . .Tkn−1Tkn−1
)
Aff
to be the following subset of F+
d × Cn−1: it is the collection of all (f,a1, . . . ,an−1), such that
� The numbers a1,a2, . . .an−1 are all distinct from each other and different from zero.
� All the derivatives of f with respect to x at ai up to order ki are zero, for i = 1 to n− 1.
� All the derivatives of f with respect to x at (0, 0) up to order kn − 1 are zero.
It is also convenient to define
(
Tk1Tk2 . . .Tkn−1T−1
)
Aff
as the following subset of F+
d ×Cn−1: it
is the collection of all (f,a1, . . . ,an−1), such that
� The numbers a1,a2, . . . ,an−1 are all distinct from each other and different from zero.
� All the derivatives of f with respect to x at ai up to order ki are zero, for i = 1 to n− 1.
Arguing as before, it suffices to show that for all kn ≥ 0, zero is a regular value of the map
φ :
(
Tk1Tk2 . . .Tkn−1Tkn−1
)
Aff
−→ C given by φ(f, x1, . . . , xn−1) := fkn,0.
Let us prove that now. Suppose
(f,a1, . . . ,an−1) ∈
(
Tk1Tk2 . . .Tkn−1Tkn−1
)
Aff
.
Let γ : (−ε, ε) −→
(
Tk1Tk2 . . .Tkn−1Tkn−1
)
Aff
be a curve, given by γ(t) := (f + tη,a1, . . . ,an−1)
where η is as yet an unspecified polynomial. We now note that {dφ}(γ′(0)) = ηkn,0. Now
choose η as follows
η(x, y) := (x− a1)
k1+1 · · · (x− an−1)
kn−1+1(x− 0)kn .
Since (f,a1, . . . ,an−1) belongs to
(
Tk1Tk2 . . .Tkn−1Tkn−1
)
Aff
, γ(t) also lies there for all t. Fur-
thermore, ηkn,0 ̸= 0, because the ai are all different from zero. This proves the proposition. ■
We are now ready to prove Theorem 6.2.
Proof of Theorem 6.2. We first show that (6.14) is valid on the set-theoretic level. Consider
the first term on the left-hand side, namely π∗[Tk1 . . .Tkn ]. It is represented by the closure
of the following space: a line, a curve and n + 1 distinct points (x1, . . . , xn+1), such that the
curve is tangent to the line at the points (x1, . . . , xn) to orders k1, . . . , kn respectively; the last
point xn+1 is free (it does not have to lie on either the line or the curve).
Now consider second factor on the left-hand side of (6.14), namely π∗n+1[T0]. This is simply
represented by the following space: a line, a curve and n + 1 points (x1, . . . , xn+1), such that
the points (x1, . . . , xn) are free, while the last point xn+1 lies on the line and the curve.
Consider the set-theoretic intersection of the above two spaces. There are two possibilities.
The first possibility is that the point xn+1 is distinct from all the other points (x1, . . . , xn). The
closure of that space represents the first term on the right-hand side of (6.14). But there is
another possibility. The point xn+1 could be equal to one of the xi (for i ∈ {1, . . . , n}). That
precisely gives us the second term on the right-hand side of (6.14).
Counting Curves with Tangencies 13
To see that equation (6.14) is valid on the level of homology, we need to do the following. To
justify the first term on the right-hand side of (6.14), we need to show that the intersections are
transverse; this follows from the proof of Proposition 6.3. To justify the second term, we need
to justify the multiplicity of the intersection.
Consider the situation of x1 coinciding with xn+1. We take a chart that sends the point xn+1
to be the origin and sets the line to be the x-axis. The situation now is that we have a curve f
that is tangent to the x-axis to order k1 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 f00, f10, . . . , fk10 all
vanish. It is given by
f(x, y) =
fk1+1,0
(k1 + 1)!
xk1+1 +
fk1+2,0
(k1 + 2)!
xk1+2 + · · ·+
fd,0
d!
xd + yR(x, y).
Now consider the evaluation map
φ(f, x) := f(x, 0) =
fk1+1,0
(k1 + 1)!
xk1+1 +
fk1+2,0
(k1 + 2)!
xk1+2 + · · ·+
fd,0
d!
xd.
The order of vanishing of φ is clearly k1 + 1, provided fk1+1,0 ̸= 0 (the values of fk1+2,0, fk1+3,0,
. . . , fd,0 are not relevant for the order of vanishing in a neighbourhood of the origin if fk1+1,0 is
non-zero).
The assumption fk1+1,0 ̸= 0 is valid, because to compute the order of vanishing, we will
be intersecting with cycles that correspond to constraints being generic. Hence, the order of
vanishing is k1 + 1. This proves (6.14) on the level of homology. ■
We are now ready to prove our next result.
Theorem 6.5. Let n be a positive integer, k1, k2, . . . , kn−1 nonnegative integers and kn a pos-
itive integer. Define k := k1 + k2 + · · · + kn. Then the following equality of elements of
H2(δd−k)(Mn+1,R) holds:
[Tk1 . . .Tkn−1Tkn−1T0] ·
[
∆L
n,n+1
]
= π∗[Tk1 . . .Tkn ] · [∆n,n+1], (6.15)
provided d > k + n− 1.
Before we prove Theorem 6.5, a few things are explained. Consider the special case of this
theorem, when n = 1 and k1 = 1. In this case, (6.15) simplifies to
[T0T0] ·
[
∆L
12
]
= π∗[T1] · [∆12]. (6.16)
If we draw an analogy with equation (7.6), it might seem that the right-hand side of equa-
tion (6.16) has a missing term, namely a term that corresponds to nodal curves lying on a line.
However, that is not the case; there are no missing terms. The term that seems to be missing is
actually present: it is present inside the closure T1. However, since this locus is one codimension
higher, when we intersect equation (6.16) with a class of complementary dimension, we do not
get any contribution. We explain this more precisely. Define µ := y21y
δd−1
d . Now intersect both
sides of equation (6.16) with µ. That gives us [T0T0] ·
[
∆L
12
]
· µ = [T1] · µ. The term that one
might be worried that one has missed out, namely nodal curves with the node lying on the line,
giving empty intersection with µ; this is because we are making the curve pass through δd − 1
generic points
(
since we are intersecting with yδd−1
)
.
In contrast, look at equation (7.6), namely[
AF
1T0T0
]
·
[
∆L
12
]
= π∗
[
AF
1T1
]
· [∆12] + 2π∗
[
AL
1
]
· [∆12] ·
[
∆1
2
]
.
14 I. Biswas, A. Choudhury, R. Mukherjee and A. Paul
The first geometric fact we note that AL
1 is not a subset of the closure AF
1T1. What is true
is that AF
1A
L
1 is a subset of the closure (which for dimensional reasons, does not contribute
when we intersect with a class of complimentary dimension). In order to extract numbers,
define µ := y21y
δd−2
d . Intersecting both sides of equation (7.6) with µ gives us[
AF
1T0T0
]
· µ =
[
AF
1T1
]
· µ+ 2
[
AL
1
]
· µ.
In contrast to the earlier case, the intersection of
[
AL
1
]
with µ is nonzero (or at least not neces-
sarily zero) because we are making the curve pass through δd − 2 points; that is precisely the
right number of points to enumerate 1-nodal curves, with the node lying on a line.
Proof of Theorem 6.5. We first prove the theorem for n = 1 and k1 = 1, namely we prove
equation (6.16). The main set-theoretic statement that we need to prove is as follows: consider
the component of the closure T0T0. where the two marked points are equal. Then the first
derivative of the polynomial (defining the curve) along the direction of the line (evaluated at
the marked point) is zero. In other words, if (H1, Hd, p, p) ∈ T0T0, then (H1, Hd, p) ∈ T1. Let
us prove this assertion.
We continue with the set-up of the proof of Proposition 6.3 and Theorem 6.2. Choose
coordinate where the designated line stays the x-axis. Let ft be a curve that passes through
the origin and also through the point (t, 0). The expression for ft is of the form ft(x, y) =
φt(x) + yRt(x, y), where φt(x) =
(
ft10x+
ft20
2 x2 + · · ·
)
. Since the curve passes through (t, 0),
it follows that φt(x) = (Kt(x))x(x − t) for some function Kt(x). Denote f0 by f . Hence,
f(x, y) = (K0(x))x
2 + yR0(x, y). It is a simple check to see that f00 and f10 are both zero.
Hence, if two T0 points collide, then we get a point which is at least as degenerate as a T1 point
(it could be even more degenerate). This proves the assertion we made.
We now prove a more general statement. We claim the following: consider the component of
the closure TkT0 where the two marked points are equal. Then the (k + 1)-th derivative of the
polynomial (defining the curve) along the direction of the line (evaluated at the marked point)
is zero. In other words, if (H1, Hd, p, p) ∈ TkT0, then (H1, Hd, p) ∈ Tk+1.
In order to prove the above assertion, put the Tk point at (0, 0) and the T0 point at (t, 0).
The expression for ft is going to be of the form
ft(x, y) = φt(x) + yRt(x, y), where φt(x) = ft10x+
ft20
2
x2 +
ft30
6
x3 + · · · .
Since the curve is tangent of the x-axis to order k and it also passes through (t, 0), it follows
that φt(x) = (Kt(x))x
k+1(x− t) for some function Kt(x). To see what happens in the limit as t
goes to zero, denote f0 by f . Hence,
f(x, y) = (K0(x))x
k+2 + yR0(x, y).
It is a simple check that f00, f10, . . . , fk+1,0 are all zero. Hence, if a Tk and a T0 point collide,
then we get a point which is at least as degenerate as a Tk+1 point (it could be even more
degenerate).
We now need to prove the converse of the above assertion. We claim the following: if
(H1, Hd, p) ∈ Tk+1, then (H1, Hd, p, p) ∈ TkT0.
First we note that to prove the above claim, it is sufficient to prove the following: if
(H1, Hd, p) ∈ Tk+1, then (H1, Hd, p, p) ∈ TkT0. This is because if A is a subset of B, then
closure of A is a subset of closure of B. Hence, if B is a closed set, then to show that closure
of A is a subset of B, it is sufficient to show that A is a subset of B.
We start by proving the claim for k = 0, namely, we show that every T1 point can be obtained
as a limit to two T0 points.
Counting Curves with Tangencies 15
Let f ∈ (T1)Aff. This means that f00 and f10 are both equal to zero. Hence, f is given by
f(x, y) =
(
f20
2
x2 +
f30
6
x3 + · · ·
)
+ yR(x, y).
It will be shown that there exists a point (ft, (t, 0)) close to (f, (0, 0)), such that ft passes through
the origin and (t, 0). Note that since ft passes through the origin, it is of the form
ft(x, y) = ft10x+
(
ft20
2
x2 +
ft30
6
x3 + · · ·
)
+ yRt(x, y).
Furthermore, ft10 has to be small (since ft is close to f and f10 is zero). Impose the condition
that ft(t, 0) = 0. Plugging in (t, 0) inside ft and using the fact that t ̸= 0, it follows that
ft10 = −ft20
2
t+O
(
t2
)
. (6.17)
Hence, we have constructed this nearby curve ft and a marked point (t, 0) different from the
origin that lies on the curve and the line.
To find the multiplicity of the intersection, we note that using equation (6.17), using the fact
that ft20 ̸= 0 and the implicit function theorem, we can rewrite it as
t = − 2
ft20
ft10 +O
(
f2t10
)
. (6.18)
Setting the two points to be equal is the same as setting t to be equal to zero. By equation (6.18),
the order of vanishing of t is one. This justifies the multiplicity.
We now prove the assertion for a general k, i.e., we show that every Tk+1 curve is in the limit
of a Tk and T0 point. Let f ∈ (Tk+1)Aff. This means that f00, f10, . . . , fk+1,0 are all equal to
zero. Hence, f is given by
f(x, y) =
(
fk+2,0
(k + 2)!
xk+2 +
fk+3,0
(k + 3)!
xk+3 + · · ·
)
+ yR(x, y).
It will be shown that there exists a point (ft, (t, 0)) close to (f, (0, 0)), such that ft ∈ (Tk)Aff
and (t, 0). Since ft ∈ (Tk)Aff, it is of the form
ft(x, y) =
ftk+1,0
(k + 1)!
xk+1 +
(
ftk+2,0
(k + 2)!
xk+2 +
ftk+3,0
(k + 3)!
xk+3 + · · ·
)
+ yRt(x, y).
Furthermore, ftk+1,0
has to be small (since ft is close to f and fk+1,0 is zero). Impose the
condition that ft(t, 0) = 0. Plugging in (t, 0) inside ft and using the fact that t ̸= 0, it follows
that
ftk+1,0
= −
ftk+2,0
(k + 2)
t+O
(
t2
)
. (6.19)
Hence, we have constructed this nearby curve ft and a marked point (t, 0) different from the
origin that lies on the curve and the line.
To find the multiplicity of the intersection, we note that using equation (6.19), using the fact
that ftk+2,0
̸= 0 (which is true because after we make the curves pass through generic points)
and the implicit function theorem, we can rewrite it as
t = − 2
ftk+2,0
ftk+1,0
+O
(
f2tk+1,0
)
. (6.20)
16 I. Biswas, A. Choudhury, R. Mukherjee and A. Paul
Setting the two points to be equal is the same as setting t to be equal to zero. By equation (6.20),
the order of vanishing of t is one. This justifies the multiplicity.
We summarize the set-theoretic statement we have just proved. We have shown that (H1, Hd,
p, p) belongs to TkT0 if and only if (H1, Hd, p) belongs to Tk+1.
We now examine what happens when there are more than two points involved. We explain
with the help of a single example; the general case follows in a similar way. Consider the following
assertion:
[T2T0T0] ·
[
∆L
23
]
= π∗[T2T1] · [∆23].
The closure claim that we need to prove is as follows: (H1, Hd, q, p, p) belongs to T2T0T0
if and only if (H1, Hd, p) belongs to T1. One direction is the same as before namely that
if (H1, Hd, q, p, p) belongs to T2T0T0, then (H1, Hd, p) belongs to T1. The fact that (H1, Hd, q)
belongs to T2 makes no difference in the proof. It is the converse that requires a little bit more
argument. We need to show that every T2T1 point can be obtained as a limit of T2T0T0. First
see what we did when we had to show every T1 point can be obtained as a limit of T0T0. We
constructed the curve as given by equation (6.17). The problem with equation (6.17) is that
if we set ft10 as given by (6.17) and the remaining ftij to be complex numbers close to fij ,
then this curve does not satisfy the T2 condition at q. The problem is the ftij are not all free.
However, we have shown through the proof of Proposition 6.3 that restricted to the submani-
fold T2T0, the section induced by f10 is transverse to zero. Hence, f10 is an actual coordinate
on the submanifold T2T0. Hence, (6.17) defines a curve lying in T2T0 that converges to an
element of T2T1 as t goes to zero. The multiplicity computation is the same. The general case
follows from equation (6.19) and using the fact that fk+1,0 is a coordinate on T2Tk. When
there are more than three points involved, the same argument holds via the proof of Proposi-
tion 6.3, namely that fkn,0 is a coordinate on Tk1Tk2 . . .Tkn−1Tkn−1. This completes the proof
of Theorem 6.5. ■
Remark 6.6. The pullback to Mn of the hyperplane classes in Xi for i = 1, . . . , n are denoted
by a1, . . . , an. Note that (6.14) can be rewritten as
[Tk1Tk2 . . .TknT0] = π∗[Tk1Tk2 . . .Tkn ] · π∗n+1[T0]
−
n∑
i=1
(1 + ki)π
∗[Tk1Tk2 . . .Tkn ] · [∆i,n+1]. (6.21)
Also, [∆i,n+1] = a2i + aian+1 + a2n+1. Hence, using (6.11) and (6.21) we can recursively compute
all the intersection numbers involving the class [Tk1Tk2 . . .TknT0]. Next, let α be a class in Mn.
Then (6.15) implies that
[Tk1 . . .Tkn ] · α = [Tk1 . . .Tkn−1Tkn−1T0] ·
[
∆L
n,n+1
]
· α. (6.22)
Hence, using equations (6.22) and (5.1) and using the fact that all intersection numbers involving
the class [Tk1Tk2 . . .TknT0] are computable, we conclude that all intersection numbers involving
the class [Tk1Tk2 . . .Tkn ] are computable. From the procedure to compute, it is clear that these
intersection numbers are all polynomials in d.
7 Counting 1-nodal curves with multiple tangencies
We now consider plane curves with singularities. Consider the following question: How many 1-
nodal degree d curves are there in P2 that pass through δd−2 (see (6.1)) generic points and that
are tangent to a given line? For this, consider the space of curves with three distinct points p1, x1,
Counting Curves with Tangencies 17
and x2 such that the curve has a node at p1 and intersects the given line transversally at x1
and x2. The closure of this space represents a cycle. Next, impose the condition that x1 becomes
equal to x2 (which again, represents a cycle). We might naively expect that the intersection of
these two cycles give us the space of one nodal curves tangent to a given line, as it would be
suggested by the following picture:
x1 = x2
p1
x1
x2
p1
But there is an extra object that occurs. There is also the space of curves with one node
lying on the line, as shown by the following picture:
x1
x2
p1
p1 = x1 = x2
This results in an excess contribution to the intersection. The same thing happens if the
curve has a more degenerate singularity. We have been able to compute the excess contribution
to the intersection in the following cases:
• When the degree d plane curve has a node and is tangent to a given line at multiple points
of any order.
• When the degree d plane curve has a cusp and is tangent to a given line at multiple points
(only tangency of order one).
When the singularity is a node, our answers agree with those predicted by the Caporaso–Harris
formula. When the singularity is a cusp, our results are new (to the best of our knowledge).
We expect that this idea can be pursued further to enumerate curves with multiple nodes and
also enumerate singular curves tangent to a given line, when the singularities are even more
degenerate than a cusp. We hope to pursue these questions in future.
We now implement the idea which has just been described. For that, recall a standard
definition about singularities.
Definition 7.1. Let U be an open neighbourhood of the origin in C2 (open with respect to
the usual topology of C2 ≈ R4 given by the Euclidean metric) and let f : (U,0) −→ (C, 0) be
a holomorphic function. A point q ∈ f−1(0) has an Ak-singularity if there exists a coordinate
system (x, y) : (V,0) −→
(
C2,0
)
such that f−1(0) ∩ V is given by y2 + xk+1 = 0. An A1-
singularity is also called a node, an A2-singularity is called a cusp while an A3-singularity is also
called a tacnode.
Define
Mm
n := D1 ×Dd ×
(
X1 × · · · ×Xm
)
× (X1 × · · · ×Xn),
18 I. Biswas, A. Choudhury, R. Mukherjee and A. Paul
where each copy ofXi andX
j is P2; the hyperplane classes are denoted by ai and bj , respectively.
Let W and S1,S2, . . . ,Sn be subsets of D1×Dd×P2 (here m = 1). We define WS1S2 . . . Sn ⊂ M1
n
as follows: it consists of all (H1, Hd, p, x1, . . . , xn) such that
� (H1, Hd, p) ∈ W,
� (H1, Hd, xi) ∈ Si for all i ∈ {1, . . . , n},
� The points p, x1, . . . , xn are all distinct.
As before, we denote the homology class represented by the closure by putting a square bracket,
but without putting the cumbersome bar.
Next, define a few subsets of M1
0 (we are setting here m to be equal to 1). First of all, we
define AF
1 ⊂ M1
0 to be the subset of all (H1, Hd, p) ∈ M1
0 such that Hd has a node at p. The
letter F is there to remind us that the nodal point is free, i.e., it does not have to lie on the line.
Similarly, we define AL
1 ⊂ M1
0 to be the subset of all (H1, Hd, p) ∈ M1
0 such that
� The curve Hd has a node at p.
� The point p lies on the line H1.
Finally, given any nonnegative integer r, we define P(r)A1⊂M1
0 to be the subset of all (H1, Hd, p)∈
M1
0 such that
� The curve Hd has a node at p.
� The point p lies on the line H1.
� One of the branches of the node is tangent to the line H1 to order r.
We note that the closure of P(0)A1 is same as the closure of AL
1
(
in M1
0
)
. Note that[
AL
1
]
=
[
AF
1
]
· (y1 + b1); (7.1)
this is because intersecting with (y1 + b1) corresponds to the point p lying on the line.
We now prove a few transversality results that we use to enumerate nodal curves with
tangencies. In the following propositions, n, k1, k2, . . . , kn and r are nonnegative integers
and k := k1 + k2 + · · ·+ kn.
Proposition 7.2. The space P(r)A1Tk1 . . .Tkn is a smooth submanifold of M1
n, provided d ≥
k + r + n+ 2.
Proposition 7.3. The space AF
1Tk1 . . .Tkn is a smooth submanifold of M1
n, provided d ≥ k+n+1.
Proof of Proposition 7.2. We show that the space P(r)A1Tk1 . . .Tkn is a codimension one
submanifold of Tr+1Tk1 . . .Tkn . The proof is very similar to that of Proposition 6.3, where we
show that Tr+2Tk1 . . .Tkn is a smooth codimension one submanifold of Tr+1Tk1 . . .Tkn .
We switch to affine space at set the Tr+1 point to be the origin (and the line to be the x-axis).
To show that Tr+2Tk1 . . .Tkn is a smooth submanifold, we show that the section induced by
taking the (r+ 2)-th derivative (i.e., fr+2,0) is transverse to zero. The procedure for doing that
was as follows: we considered the polynomial η, given by
η(x, y) := (x− a1)
k1+1 · · · (x− an−1)
kn−1+1(x− an)
kn+1(x− 0)r+2.
Here (ai, 0) is the point at which the degree d curve is tangent to the x-axis to order ki. Using
this polynomial η, we are able to construct a tangent vector, such that the differential of the
section (that is induced by taking the (r + 2)-th derivative) evaluated on this tangent vector is
nonzero. This proves transversality.
Counting Curves with Tangencies 19
In a similar way, we can show that the section induced by taking the first derivative in the y
direction (i.e., f01) is transverse to zero. To do that, we define the polynomial η, given by
η(x, y) := (x− a1)
k1+1 · · · (x− an−1)
kn−1+1(x− an)
kn+1(x− 0)r+1(y − 0).
Using this η, we can use a similar argument to compute the differential and prove transversality.
Hence, P(r)A1Tk1 . . .Tkn is a smooth codimension one submanifold of Tr+1Tk1 . . .Tkn . Note
that the bound on d is required because to apply our argument, we need Tr+1Tk1 . . .Tkn to be
a smooth manifold; this where the bound on d is required (which is bigger than what is required
to simply construct the given η). ■
Proof of Proposition 7.3. Start with n = 0, i.e., we show that AF
1 is a smooth submanifold
of M1
0 of codimension three. The assertion is proved in [13, pp. 216–217], but for the convenience
of the reader, we include the proof here. Switching to affine space, we consider the map
φ : F+
d × C2 −→ C3, (f, (x, y)) 7−→ (f(x, y), fx(x, y), fy(x, y)).
It suffices to show that (0, 0, 0) is a regular value of φ. Suppose φ(f, (a, b)) = 0. We need to show
that the differential of φ, evaluated at (f, (a, b)) is surjective. Consider the polynomials ηij(x, y)
given by η00(x, y) := 1, η10(x, y) := (x − a), and η01(x, y) := (y − b). Let γij(t) be the curve
given by γij(t) := (f + tηij , (x, y)). We now note that
{dφ|(f,(a,b))}(γ′00(0)) = (1, 0, 0), {dφ|(f,(a,b))}(γ′10(0)) = (0, 1, 0) and
{dφ|(f,(a,b))}(γ′01(0)) = (0, 0, 1).
This shows that the differential is surjective.
Next, assume that n = 1 and k1 = 0. We show that AF
1T0 is a smooth submanifold of M1
1.
We switch to affine space and set the designated line to be the x-axis. Define
(
AF
1
)
Aff
to be(
AF
1
)
Aff
:=
{
(f, (x, y)) ∈ F+
d × C2 | f(x, y) = 0, fx(x, y) = 0, fy(x, y) = 0
}
. (7.2)
Consider the map ψ :
(
AF
1
)
Aff
−→ C, given by defined by (f, (x, y) 7−→ f(0, 0). We need to
show if ψ(f, (a, b)) = 0 and (a, b) ̸= (0, 0), then the differential of ψ is surjective. Note that
we are setting here the T0 point to be the origin. Let γ(t) be the curve in
(
AF
1
)
Aff
given
by γ(t) := (f + tη, (a, b)), where is η is as yet unspecified. We note that {dψ|(f,(a,b))}(γ′(0)) =
η(0, 0). We now see what is our requirement on η. First, we need that η(0, 0) ̸= 0. Second of
all, we need the curve γ(t) to lie in
(
AF
1
)
Aff
. For this, it is sufficient if the value of η and both
its first partial derivatives evaluated at (a, b) are equal to zero. Define θ(x, y) as follows:
θ(x, y) :=
{
(y − b)2 if b ̸= 0,
(x− a)2 if b = 0, a ̸= 0.
(7.3)
Define η(x, y) := θ(x, y). It is a simple check to see that η satisfies all the required condi-
tions. Note that we separately define θ, because it is used later on for further purposes (in the
subsequent proofs, our definition of η keep changing by multiplying θ with appropriate factors).
Hence, we have shown that AF
1T0 is a smooth submanifold of M1
1. We now show that AF
1Tk1
is a smooth submanifold of M1
1. We show that AF
1Tk1 is a smooth submanifold of AF
1Tk1−1 of
codimension one. The proof is as before; the required curve is given by η(x, y) := θ(x, y)xk1+1.
This proves the claim when n = 1.
Finally, suppose there are more than one point of tangency (in addition to the nodal point).
The nodal point is at (a, b) and suppose that the tangency points are at (a1, 0), (a2, 0), . . . ,
(an−1, 0) and (0, 0). The assertion is proved by considering the polynomial
η(x, y) := θ(x, y)(x− a1)
k1+1 · · · (x− an−1)
kn−1+1(x− 0)kn .
This completes the proof. ■
20 I. Biswas, A. Choudhury, R. Mukherjee and A. Paul
We are now ready to state and prove the results about enumerating 1-nodal curves with tan-
gencies. In the following theorems, it is always be understood that whenever there is a collection
of numbers k1, k2, . . . , kn, then k is defined to be k :=
∑n
i=1 ki.
We also recall the projection maps and the diagonal subspaces that we will be encountering.
We denote π : M1
n+1 −→ M1
n to be the projection that forgets the last marked point. We also
denote πn+1 : M
1
n+1 −→ M1 to be the map that forgets all the marked points, except the last
one. Next, for any 1 ≤ i ≤ n, we denote ∆i,n+1 to be the following subset of M1
n+1: It is the
locus of all (H1, Hd, p, x1, . . . , xn, xn+1) such that xi = xn+1. Finally, we denote ∆1
n+1 to be the
following subset of M1
n+1: It is the locus of all (H1, Hd, p, x1, . . . , xn, xn+1) such that p = xn+1.
Theorem 7.4. Let n, k1, k2, . . . , kn be nonnegative integers. Then the following equality of
homology classes in H∗
(
M1
n+1;R
)
holds:
π∗
[
AF
1Tk1 . . .Tkn
]
· π∗n+1[T0] =
[
AF
1Tk1 . . .TknT0
]
+
n∑
i=1
(1 + ki)π
∗[AF
1Tk1 . . .Tkn
]
· [∆i,n+1], (7.4)
provided d > k + n+ 1.
Proof. This is simply a straightforward generalization of Theorem 6.2; the proof is identical. ■
Next, we generalize Theorem 6.5.
Theorem 7.5. Let n be a positive integer, k1, k2, . . . , kn−1 nonnegative integers and kn a positive
integer. Define
mkn :=
{
2 if kn = 1,
1 if kn > 1.
Then the following equality of homology classes in H∗
(
M1
n+1;R
)
holds:[
AF
1Tk1 . . .Tkn−1Tkn−1T0
]
·
[
∆L
n,n+1
]
= π∗
[
AF
1Tk1 . . .Tkn
]
· [∆n,n+1]
+mknπ
∗[P(kn−1)A1Tk1 . . .Tkn−1
]
· [∆n,n+1] ·
[
∆1
n+1
]
, (7.5)
provided d > k + n.
Remark 7.6. Let us see how to extract numbers from this. On the (n+ 1)-pointed space M1
n,
let α and β be the following classes:
α := yr1y
s
db
ν1
1 a
ε1
1 . . . aεnn and β := yr1y
s
db
ν1+εn
1 aε11 . . . a
εn−1
n−1 .
Intersecting both sides of equation (7.5) with α, gives us[
AF
1Tk1 . . .Tkn−1Tkn−1T0
]
·
[
∆L
n,n+1
]
· α
=
[
AF
1Tk1 . . .Tkn
]
· α+mkn
[
P(kn−1)A1Tk1 . . .Tkn−1
]
· β.
Proof of Theorem 7.5. We first prove the special case where n = 1 and kn = 1. In that case,
equation (7.5) simplifies to[
AF
1T0T0
]
·
[
∆L
12
]
= π∗
[
AF
1T1
]
· [∆12] + 2π∗
[
AL
1
]
· [∆12] ·
[
∆1
2
]
. (7.6)
The proof of (7.6) builds on what was already shown in Theorem 6.5, namely when two T0
points collide the first derivative along the line vanishes. Now there are two possibilities. The
Counting Curves with Tangencies 21
first one is that the limiting point is a smooth point of the curve. This corresponds to the
locus AF
1T1. There is another possibility that the limiting point is a singular point of the curve.
This corresponds to the locus AL
1. On the set-theoretic level, this argument shows that the left-
hand side of (7.6) is a subset of its right-hand side. To show that the right-hand side is a subset
of the left-hand side, we need to show that every element of AF
1T1 and AL
1 can be obtained as
a limit of elements in AF
1T0T0. It was shown in the proof of Theorem 6.5 that every element
of T1 arises as a limit of elements in T0T0. To complete the proof here, it is enough to show
that every element of AF
1T1 can be obtained as a limit of elements in AF
1T0T0. The argument
for it is the same as how we showed (at the end of the proof of Theorem 6.5) that every element
of T2T1 arises as a limit of elements of T2T0T0. The crucial fact that was used there is that f10
is indeed a local coordinate on the space T2T0. The proof of Proposition 7.3 shows that f10
is a local coordinate on AF
1T2T0. This completes the proof about why every element AF
1T1 can
be obtained as a limit of elements in AF
1T0T0. In particular, this justifies the first term on the
right-hand side of equation (7.6).
The new thing we need to do for completing the proof of equation (7.6) (on the set-theoretic
level) is to show that every element of AL
1 can be obtained as a limit of elements in AF
1T0T0.
To prove this assertion, switch to affine space. Let f belong to (AL
1)Aff. As before, the line is
the x-axis. For convenience, set the nodal point to be the origin. Hence, the expression for f is
given by
f(x, y) =
f20
2
x2 + f11xy +
f02
2
y2 +R(x, y),
where the remainder term R(x, y) is of degree three or higher.
We now try to construct a curve ft close to f , such that ft passes through the origin, passes
through (t, 0) and has a nodal point close to (0, 0). The expression for ft is given by
ft(x, y) = ft10x+ ft01y +
ft20
2
x2 + ft11xy +
ft02
2
y2 + · · · .
First of all, ft10 and ft01 are small. It is required that ft has a nodal point close to the origin.
Hence, we need to find (u, v) ̸= (0, 0) but small, such that
ft(u, v) = 0, (ft)x(u, v) = 0 and (ft)y(u, v) = 0. (7.7)
To solve (7.7), using the facts that (ft)x(u, v) = 0 and (ft)y(u, v) = 0 it is deduced that
ft10 = −vft11 − uft20 −Rx(u, v) and ft01 = −uft11 − vft02 −Ry(u, v). (7.8)
Plugging in these values for ft10 and ft01 from (7.8), and using the fact that ft(u, v) = 0, it
follows that
ft20
2
u2 + ft11uv +
ft02
2
v2 +R2(u, v) = 0, (7.9)
where R2(u, v) := −2R(u, v) + 2uRx(u, v) + 2vRy(u, v).
We now try to solve u in terms of v using (7.9). Since the curve has a genuine node at the
origin, it may be assumed that the hessian is non-degenerate; in other words, ft20ft02 − f2t11 is
nonzero. Hence, after making a change of coordinates and using the fact that the remainder
terms R2(u, v) is of order three, it is deduced that there are two solutions to equation (7.9)
given by
u =
−ft11 +
√
f2t11 − ft20ft02
ft20
v +O
(
v2
)
or u =
−ft11 −
√
f2t11 − ft20ft02
ft20
v +O
(
v2
)
.
22 I. Biswas, A. Choudhury, R. Mukherjee and A. Paul
Here
√
f2t11 − ft20ft02 is a specific branch of the square root, which exists because f2t11 − ft20ft02
is nonzero. Hence, the above solution can be re-written as
u = Av +O
(
v2
)
, where A :=
−ft11 ±
√
f2t11 − ft20ft02
ft20
. (7.10)
Now impose the condition that the curve passes through the point (t, 0). Using (7.10) and the
condition that ft(t, 0) = 0, it follows that
t = 2
(
ft11 +Aft20
ft20
)
v +O
(
v2
)
. (7.11)
Hence, the condition of making the two points equal (namely setting t = 0) has a multiplicity,
which is is given by (7.11). For each value of A, the multiplicity is one. Since there are
two possible values of A (corresponding to the branch of the square-root chosen), the total
multiplicity is two. Hence, when we intersect with the class [q1 = q2], each branch of AL
1
contributes with a multiplicity of 1 resulting in a total multiplicity is 2. By a branch, we refer to
each distinct solution to equation (7.7) of a neighbourhood of AF
1T0T0 inside A
L
1. This property
of multiplicity finally shows that (7.6) is true on the level of homology.
We now prove the next case where n = 1 and kn = k − 1 with k ≥ 3. In that case,
equation (7.5) simplifies to[
AF
1Tk−2T0
]
·
[
∆L
12
]
=
[
AF
1Tk−1
]
· [∆12] +
[
P(k−2)A1
]
· [∆12] ·
[
∆1
2
]
. (7.12)
In order to prove (7.12), we build on what has already been proved in Theorem 6.5. The
justification for the first term on the right-hand side of (7.12) is the same as in the proof of
Theorem 6.5. In order to show that every AF
1Tk−1 point can be obtained as a limit of AF
1Tk−1,
we use the fact that fk−1,0 is indeed a local coordinate on AF
1Tk−2 as seen in the proof of
Proposition 7.3.
The new thing needed is to justify the second term on the right-hand side of (7.12). In
particular, it is needed to show that every element of P(k−2)A1 can be obtained as a limit of
elements in AF
1T0Tk−2. To prove this assertion, restrict, as before, to affine space.
Let f be a curve that belongs to P(k−2)A1; the line is set to be the x-axis and the nodal point
is set to be the origin. We try to construct a curve ft that is tangent to the x-axis at the origin
to order k − 2 (i.e., it is an element of Tk−2). It is given by
ft(x, y) = uy + Pxk−1 + ft11xy +
ft02
2
y2 + Qxk +R(x, y), where
R(x, y) := xk+1A(x) + yx2B(x) + y2C(x, y) and C(0, 0) = 0.
Notice that ft01 is written as u. Assume that Q ̸= 0. Next, impose the condition that the curve
also passes through (t, 0). In other words, ft(t, 0) = 0. Using this equation and dividing out
by t, it follows that P = −Qt + O
(
t2
)
. Now impose the condition that the curve has a node
at (x, y). Hence, ft(x, y) = 0, (ft)x(x, y) = 0 and (ft)y(x, y) = 0. We are looking for solutions
where u, t, x and y are small and (x, y) ̸= (0, 0). Using the equation (ft)y(x, y) = 0, it follows
that
x = − u
ft11
− yft02
ft11
+ E2(u, y), (7.13)
where the error term E2(u, y) is of second order in (u, y). Next, plug this in the equation
(ft)x(x, y) = 0 and solve for y in terms of u and t. This produces
(−1)ky =
kQ
fkt11
uk−1 +
(k − 1)Q
fk−1
t11
uk−2t+ Ek(u, t), (7.14)
Counting Curves with Tangencies 23
where the error term Ek(u, t) is of order k in (u, t). Plugging all this in ft(x, y) = 0 gives an
implicit relationship between u and t. Note that u cannot be zero. This is because if u were
zero, then y would be zero (equation (7.14)) and as a result x could be zero (equation (7.13)).
This is a contradiction, since we are looking for solutions where x and y are not both equal to
zero.
Next, notice that the expression for ft(x, y) contains a factor of u3. Since u ̸= 0, we can
cancel off the factor of u3 and get a simplified implicit expression for u and t. Now we can
directly solve for u in terms of t and conclude that u = −ft11t+O
(
t2
)
. Plugging this back into
the expression for x, gives
x = t+O
(
t2
)
. (7.15)
These solutions are the only solutions. Hence, from the expression for x (namely (7.15)), it
follows that the multiplicity of the intersection is one. This proves equation (7.12).
Notice that the nearby curve was constructed by specifying the value of u (which is f01) and P
(which, up to a constant factor of (k− 1)! is fk−1,0). Now we note that on top of Tk−2, both f01
and fk−1,0 are indeed local coordinates (as seen in the proofs of Propositions 6.3 and 7.2). The
general case of Theorem 7.5 (when n > 1) now follows in an identical way. ■
These next few results enable us to enumerate curves tangent to a line at multiple points and
one node lying on the line, such that one of the branches of the node is tangent to the line to
some given order. We state the first one of these results.
Theorem 7.7. Let n, r and k1, k2, . . . , kn be nonnegative integers. Then the following equality
of homology classes of M1
n+1 holds:
π∗
[
P(r)A1Tk1Tk2 . . .Tkn
]
· π∗n+1[T0]
=
[
P(r)A1Tk1Tk2 . . .TknT0
]
+
n∑
i=1
(ki + 1)π∗
[
P(r)A1Tk1Tk2 . . .Tkn
]
· [∆i,n+1]
+ (r + 2)π∗
[
P(r)A1Tk1Tk2 . . .Tkn
]
·
[
∆1
n+1
]
, (7.16)
provided d > n+ k + r + 2.
Proof. This is a generalization of Theorem 6.2. The new thing needed is to justify the third
term on the right-hand side of (7.16). The special case of the theorem where n = 0 will be
proved first. In this special case (7.16) simplifies to
π∗
[
P(r)A1
]
· π∗1[T0] =
[
P(r)A1T0
]
+ (r + 2)π∗
[
P(r)A1
]
·
[
∆1
1
]
. (7.17)
On the set-theoretic level, the justification is the same as before (the first term corresponds to
when the two marked points are distinct, while the second term corresponds to the case where
the two marked points coincide). The reason that the first term on the right-hand side of (7.17)
appears with a multiplicity of one is because the intersections are transverse (this is the content
of Proposition 7.2).
The new thing that we have to justify is the multiplicity of (r + 2) for the second term on
the right-hand side; let us justify that. As before, we switch to affine space. Set the P(r)A1
point to be the origin. The situation now is that we have a nodal curve such that one of the
branches of the node is tangent to the x-axis to order r (at he origin). Hence, the curve f is
such that f00, f10, f01, f20, . . . , fr+1,0 all vanish. The function f is given by
f(x, y) =
fr+2,0
(r + 2)!
xr+2 + yR(x, y),
24 I. Biswas, A. Choudhury, R. Mukherjee and A. Paul
where yR(x, y) is a remainder term. Now consider the evaluation map
φ(f, x) := f(x, 0) =
fr+2,0
(r + 2)!
xr+2.
The order of vanishing of φ is clearly r + 2, provided fr+2,0 ̸= 0. But that assumption is valid,
since to compute the order of vanishing, we will be intersecting with cycles that correspond to
constraints being generic. Hence, the order of vanishing is r+2. This proves (7.17) on the level
of homology.
The general statement of Theorem 7.7 is now similar to how Theorem 6.2 is proved. ■
We prove the next theorem.
Theorem 7.8. Let n be a positive integer, k1, k2, . . . , kn−1 nonnegative integers and kn a positive
integer. Then the following equality of homology classes holds in H∗
(
M1
n+1;R
)
:[
P(r)A1Tk1 . . .Tkn−1Tkn−1T0
]
·
[
∆L
n,n+1
]
= π∗
[
P(r)A1Tk1 . . .Tkn ] · [∆n,n+1
]
, (7.18)
provided d > n+ k + r + 1.
Proof. This is a generalization of Theorem 6.5; the proof is the same. ■
The final result is as follows.
Theorem 7.9. Let r be a positive integer. Then the following equality of homology classes hold
in H∗
(
M1
1;R
)
:[
P(r−1)A1T0
]
·
[(
∆1
1
)L]
=
[
P(r)A1
]
·
[
∆1
1
]
, (7.19)
provided d > r + 1 and where the class
[(
∆1
1
)L]
is defined as[(
∆1
1
)L]
:= b1 + a1 − y1.
Proof. First (7.19) will be proved on the set-theoretic level. We switch to affine space. As
before, the line is the x-axis. Set the P(r−1)A1 point to be the origin and the T0 point to
be (t, 0). Let ft be a curve that has a P(r−1)A1 point at the origin and a T0 point at (t, 0). The
former condition says that the first r derivatives with respect to x vanish at the origin. The fact
that the curve also passes through (t, 0) tells that the (r + 1)-th derivative is given by
fr+1,0 = −
(
fr+2,0
r + 2
)
t+O
(
t2
)
.
Hence, as t goes to zero, fr+1,0 vanishes. Furthermore, the curve has a node at the origin. Hence,
in the limit, the curve belongs to P(r)A1.
To complete the proof on the set-theoretic level, it suffices to show that every element of P(r)A1
arises as a limit of elements of P(r−1)A1T0. The proof is exactly the same as how we show
every element of Tr+1 is a limit of elements of TrT0. The multiplicity of the intersection also
follows in the same way. This proves (7.19) on the level of homology and completes the proof
Theorem 7.9. ■
Finally, we note that the expression for the homology class
[
AF
1
]
can be computed using the
results of [1]. We give a new way to derive that expression in Section 9. For now, assume that
the expression for
[
AF
1
]
is known (which in Section 9.1, given by (9.1)).
Counting Curves with Tangencies 25
We now explain how to compute all the characteristic numbers involving the class
[
AF
1Tk1 . . .
Tkn
]
. Using equations (7.4) and (7.18), we can reduce it to a question of computing char-
acteristic numbers involving the classes
[
AF
1Tk1 . . .Tkn−1
]
and
[
P(kn−1)A1Tk1 . . .Tkn−1
]
. Using
equations (7.16) and (7.18), this ultimately reduces to the computation of all intersection num-
bers involving the class
[
P(r)A1
]
. Using equations (7.19) and (7.16), this finally reduces to
the computation of
[
AL
1
]
; this can be computed from equations (9.1) and (7.1). Hence, all the
intersection numbers can be computed.
8 Counting 1-cuspidal curves
with multiple tangencies of first order
In this section, we show how to enumerate one cuspidal curves with first-order tangencies. We
continue with the set up and notation of Section 7. In addition, we need to define two new
spaces. Define AF
2 ⊂ M1
0 to be the subset of all (H1, Hd, p) ∈ M1
0 such that Hd has a cusp at p.
Similarly, define AL
2 ⊂ M1
0 to be the subset of all (H1, Hd, p) ∈ M1
0 such that
� The curve Hd has a cusp at p.
� The point p lies on the line H1.
Note that[
AL
2
]
=
[
AF
2
]
· (y1 + b1); (8.1)
this is because intersecting with (y1 + b1) corresponds to the point p lying on the line. Let
us now prove a few transversality results. In the following propositions, n, k1, k2, . . . , kn are
nonnegative integers and k := k1 + k2 + · · ·+ kn.
Proposition 8.1. The space AF
2Tk1 . . .Tkn is a smooth submanifold of M1
n, provided d ≥ k+n+2.
Proposition 8.2. The space AL
2Tk1 . . .Tkn is a smooth submanifold of M1
n, provided d ≥ k+n+2.
Proof of Proposition 8.1. We show that AF
2Tk1 . . .Tkn is a codimension one submanifold
of AF
1Tk1 . . .Tkn . We first prove it for n = 0, i.e., we show that AF
2 is smooth codimension
one submanifold of AF
1 . As before, we switch to affine space. Let
(
AF
1
)
Aff
be as defined in
equation (7.2). Define the map
φ :
(
AF
1
)
Aff
−→ C, (f, (x, y)) 7−→
(
fxxfyy − f2xy
)
(x, y).
It suffices to show that whenever (a, b) is a genuine cuspidal point of f , the differential of φ is
surjective.
In order to prove that claim, assume that (f, (a, b)) ∈ φ−1(0) and that (a, b) is a genuine
cuspidal point of f . In that case, fxx(a, b) and fyy(a, b) both can’t be zero, because otherwise,
even fxy(a, b) would be zero, making (a, b) a triple point of f (i.e., it is not a genuine cusp).
Assume that fyy(a, b) ̸= 0. Define the polynomial η given by η(x, y) := (x − a)2. Define the
curve γ given by γ(t) := (f + tη, x, y). We note that γ(t) lies in
(
AF
1
)
Aff
since t is nonzero but
small. We now note that
{dφ|(f,(a,b))}(γ′(0)) = fyy(a, b)ηxx(a, b).
Since fyy(a, b) ̸= 0 by assumption and ηxx(a, b) ̸= 0 by our construction of η, we conclude that
the differential is surjective. Note that if fyy(a, b) = 0 (which means in turn that fxx(a, b) ̸= 0),
then we would have defined η(x, y) := (y − b)2. In that case, we would get
{dφ|(f,(a,b))}(γ′(0)) = fxx(a, b)ηyy(a, b).
This is again nonzero. This completes the proof of the proposition for n = 0.
26 I. Biswas, A. Choudhury, R. Mukherjee and A. Paul
The rest of the proof (for n ≥ 1) is now identical to the proof of Proposition 7.3. The only
change we need to make is in equation (7.3) where we define θ. That definition is replaced by
θ(x, y) :=
{
(y − b)3 if b ̸= 0,
(x− a)3 if b = 0, a ̸= 0.
(8.2)
Modulo that redefinition of θ, the proof is identical. ■
Proof of Proposition 8.2. We show that AL
2Tk1 . . .Tkn is a codimension one submanifold
of AL
1Tk1 . . .Tkn . We first prove it for n = 0, i.e., we show that AL
2 is smooth codimension
one submanifold of AL
1. This is identical to the first part of the proof of Proposition 8.1, where
we show that AF
2 is smooth codimension one submanifold of AF
1 .
The proof for n ≥ 1 is also identical to the second part of the proof of Proposition 8.1. The
only point to note here is that now since all the points lie on the line, θ is unambiguously defined
via equation (8.2), i.e., θ(x, y) := (x− a)3. ■
We are now ready to present our main results. The following result enumerates all one
cuspidal curves with tangencies at multiple points but all are of first order.
Theorem 8.3. Let n be a nonnegative integer. Then the following equality of classes in
H∗
(
M1
n+1,R
)
holds:
π∗
[
AF
2 T1T1 . . .T1︸ ︷︷ ︸
n
]
· π∗n+1[T0] =
[
AF
2 T1T1 . . .T1︸ ︷︷ ︸
n
T0
]
+
n∑
i=1
2π∗
[
AF
2 T1T1 . . .T1︸ ︷︷ ︸
n
]
· [∆i,n+1],
provided d > 2n+ 2.
Proof. This is simply a generalization of Theorems 6.2 and 7.4; the proof is identical. ■
Theorem 8.4. Let n be a nonnegative integer. Then the following equality of classes in
H∗
(
M1
n+2,R
)
holds:
π∗
[
AF
2 T1T1 . . .T1︸ ︷︷ ︸
n
T0
]
· π∗n+2[T0]
=
[
AF
2 T1T1 . . .T1︸ ︷︷ ︸
n
T0T0
]
+
n∑
i=1
2π∗
[
AF
2 T1T1 . . .T1︸ ︷︷ ︸
n
T0
]
· [∆i,n+2]
+ π∗
[
AF
2 T1T1 . . .T1︸ ︷︷ ︸
n
T0
]
· [∆n+1,n+2],
provided d > 2n+ 3.
Proof. This is simply a generalization of Theorems 6.2 and 7.4; the proof is identical. ■
Theorem 8.5. Let n be a positive integer. Then the following equality of homology classes hold
in H∗
(
M1
n+1;R
)
:[
AF
2 T1 . . .T1︸ ︷︷ ︸
n−1
T0T0
]
·
[
∆L
n,n+1
]
=
[
AF
2 T1 . . .T1︸ ︷︷ ︸
n
]
· α+ 3π∗
[
AL
2 T1 . . .T1︸ ︷︷ ︸
n−1
]
· [∆n,n+1] ·
[
∆1
n+1
]
, (8.3)
provided d > 2n+ 2.
Counting Curves with Tangencies 27
Remark 8.6. Let us explain how to extract numbers. On M1
n, consider the following classes,
given by
α := yr1y
s
db
ν1
1 a
ε1
1 . . . aεnn and β := yr1y
s
db
ν1+εn
1 aε11 . . . a
εn−1
n−1 .
Intersecting both sides of equation (8.3) gives us[
AF
2 T1 . . .T1︸ ︷︷ ︸
n−1
T0T0
]
·
[
∆L
n,n+1
]
· α =
[
AF
2 T1 . . .T1︸ ︷︷ ︸
n
]
· α+ 3
[
AL
2 T1 . . .T1︸ ︷︷ ︸
n−1
]
· β.
Proof. This is a generalization of Theorems 6.5 and 7.5. We first prove the special case
where n = 1. Then (8.3) simplifies to[
AF
2T0T0
]
·
[
∆L
12
]
= π∗
[
AF
2T1
]
· [∆12] + 3π∗
[
AL
2
]
· [∆12] ·
[
∆1
1
]
. (8.4)
To prove (8.4), we proceed in a similar way to how we proved equation (7.6). The first term on
the right-hand side of (8.4) is justified in the same way as the first term on the right-hand side
of equation (7.6). Next, we justify the second term.
On the set-theoretic level, we need to show that every point of AL
2 arises as a limit of AF
2T0T0.
As before, we will be working in affine space. Take f ∈
(
AL
2
)
Aff
. As usual, the line is the x-
axis and the cuspidal point is the origin. Assume that f20 ̸= 0; here fij denotes the (i, j)-th
derivative at the origin. Let ft be a curve close to f that passes through the origin and also
passes through (t, 0). The Taylor expansion of ft is
ft(x, y) = uy + Px+
ft20
2
x2 + ft11xy +
1
2
(
s+
f2t11
ft20
)
y2 +
ft30
6
x3
+
ft21
2
x2y +
ft12
2
xy2 +
ft03
6
y3 + E4(x, y),
where the error term E4(x, y) is fourth order in (x, y). Here ft01 is denoted by u and ft10 is
denoted by P.
Now impose the condition that the curve passes through (t, 0), so ft(t, 0) = 0. This implies
that
P = −ft20
2
t+O
(
t2
)
.
Now impose the condition that the curve has a cuspidal point at (x, y). This implies that
ft(x, y) = 0, (ft)x(x, y) = 0, (ft)y(x, y) = 0 and
Hft(x, y) :=
(
(ft)xx(ft)yy − (ft)
2
xy
)
(x, y) = 0.
Using the conditions that (ft)x(x, y) = 0 and (ft)y(x, y) = 0, it follows that
ft20
2
t = yft11 + xft20 +
ft12
2
y2 + ft21xy +
ft30
2
x2 + E3(x, y) and
−u =
(
s+
f2t11
ft20
)
y + xft11 +
ft03
2
y2 + ft12xy +
ft21
2
x2 + Ẽ3(x, y). (8.5)
Next, use the condition Hft(x, y) = 0 to conclude that
−s =
(
ft12 +
f2t11ft30
f2t20
)
x+
(
ft30 −
2ft11ft12
ft20
+
f2t11ft21
f2t20
)
y.
28 I. Biswas, A. Choudhury, R. Mukherjee and A. Paul
Finally, plugging all these in ft(x, y) = 0, it follows that
−(ft20x+ ft11y)
2
2ft20
− ft30
3
x3 − ft21x
2y + xy2
(
f2t11ft30
2f2t20
− ft12
2
)
+
(
ft03
6
− ft11ft12
ft20
+
f2t11ft21
2f2t20
)
y3 = 0. (8.6)
Next, make the following change of coordinates:
x̂ := ft20x+ ft11y and ŷ := y. (8.7)
Note that this is a valid change of coordinate because ft20 ̸= 0. Under a further genericity
assumption on the third derivatives, we can make a change of coordinates (centred around the
origin) so that (8.6) can be rewritten as
x̂2 − ŷ3 = 0. (8.8)
Equation (8.8) has exactly one solution close to the origin, namely
x̂ = v3 and ŷ = v2. (8.9)
Using (8.5), (8.7) and (8.9), it follows that
t =
2
ft20
v3 +O
(
v4
)
. (8.10)
Since we are setting t to be equal to zero to obtain the AL
2 point, (8.10) implies that the
multiplicity of the intersection in the second term on the right-hand side of (8.4) is 3. This
proves (8.4). Note that the assumption ft20 ̸= 0 and the genericity assumption of the third
derivative is valid, since to compute the multiplicity of intersections, we will intersect with
a generic cycle. The general case now follows as before. ■
Theorem 8.7. Let n be a nonnegative integer. Then, the following equality of homology classes
hold in H∗
(
M1
n+2,R
)
:[
AL
2 T1 . . .T1︸ ︷︷ ︸
n
T0T0
]
·
[
∆L
n+1,n+2
]
= π∗
[
AL
2 T1 . . .T1︸ ︷︷ ︸
n+1
]
· [∆n+1,n+2], (8.11)
provided d > 2n+ 4.
Proof. This is a generalization of Theorems 6.5 and 7.8; the proof is the same. ■
Remark 8.8. Let us explain how to extract numbers. On M1
n+2 consider the following classes,
given by
α := yr1y
s
db
ν1
1 a
ε1
1 . . . a
εn+2
n+2 and β := yr1y
s
db
ν1
1 a
ε1
1 . . . a
εn+1+εn+2
n+1 .
Intersecting both sides of (8.11) with α gives us[
AL
2 T1 . . .T1︸ ︷︷ ︸
n
T0T0
]
·
[
∆L
n+1,n+2
]
· α =
[
AL
2 T1 . . .T1︸ ︷︷ ︸
n+1
]
· β.
The next theorem is as follows.
Counting Curves with Tangencies 29
Theorem 8.9. Let n be a nonnegative integer. Then the following equality of classes in
H∗(M
1
n+1,R) holds:
π∗
[
AL
2 T1T1 . . .T1︸ ︷︷ ︸
n
]
· π∗n+1[T0]
=
[
AL
2 T1T1 . . .T1︸ ︷︷ ︸
n
T0
]
+
n∑
i=1
2π∗
[
AL
2 T1T1 . . .T1︸ ︷︷ ︸
n
]
· [∆i,n+1]
+ 2π∗
[
AL
2 T1T1 . . .T1︸ ︷︷ ︸
n
]
·
[
∆1
n+1
]
, (8.12)
provided d > 2n+ 3.
Proof. This is a generalization of Theorem 7.7. The new thing we need to show is to justify
the third term on the right-hand side of (8.12).
We first prove the special case of Theorem 8.9 where n = 0. In this case, (8.12) simplifies to
π∗
[
AL
2
]
· π∗1[T0] =
[
AL
2T0
]
+ 2π∗
[
AL
2
]
·
[
∆1
1
]
. (8.13)
We prove equation (8.13) in a similar way to how we proved equation (7.17). The justification
for the first term on the right-hand side of (8.13) is same as the justification for the first term on
the right-hand side of equation (7.17). The new thing we need to do is justify the second term.
On the set-theoretic level, the second term is clear. What we need to do now is justify the
multiplicity of 2. For convenience, set the AL
2 point to be the origin. We are now going to study
the multiplicity with which the evaluation map vanishes at the origin. Note that the curve f is
such that f00, f10, f01, f20f02 − f211 all vanish. Further assume that f20 ̸= 0. Consequently, the
curve is given by
f(x, y) =
f20
2
x2 + f11xy +
f211
f02
y2 +R(x, y),
where R(x, y) is a remainder term of order three. The order of vanishing of the evaluation map
should be 2. To see why that is so, consider the evaluation map
φ(f, x) := f(x, 0) =
f20
2
x2 +O
(
x3
)
.
Since by assumption f20 ̸= 0, we conclude that the order of vanishing is 2. The assumption
that f20 ̸= 0 is valid, since to compute the multiplicity of intersections, we intersect with generic
cycles. This proves (8.13) on the level of homology. The general statement now follows simi-
larly. ■
Theorem 8.10. Let n be a nonnegative integer. Then the following equality of classes in
H∗
(
M1
n+2,R
)
holds:
π∗
[
AL
2 T1T1 . . .T1︸ ︷︷ ︸
n
T0
]
· π∗n+2[T0]
=
[
AL
2 T1T1 . . .T1︸ ︷︷ ︸
n
T0T0
]
+ 2π∗
[
AL
2 T1T1 . . .T1︸ ︷︷ ︸
n
T0
]
·
[
∆1
n+2
]
+ π∗
[
AL
2 T1T1 . . .T1︸ ︷︷ ︸
n
T0
]
·
[
∆n+1,n+2
]
+
n∑
i=1
2π∗
[
AL
2 T1T1 . . .T1︸ ︷︷ ︸
n
T0
]
· [∆i,n+2],
provided d > 2n+ 4.
30 I. Biswas, A. Choudhury, R. Mukherjee and A. Paul
Proof. The proof is the same as that of Theorem 8.9. ■
Finally, we note that the expression for the homology class
[
AF
2
] (
which is an element
of H∗
(
M1
0,R
))
can be computed using the results of [1], given by[
AF
2
]
=
(
12d2 − 36d+ 24
)
y2db
2
1 +
(
8d− 12
)
y3db1 + 2y4d.
Using equation (8.1), we can compute
[
AL
2
]
as well. Hence, using the same reasoning given at the
end of Section 7 and using the theorems proved in this section, we conclude that all intersection
numbers involving
[
AF
2 T1 . . .T1︸ ︷︷ ︸
n
]
can be computed.
9 Counting singular curves
In this section, we derive our Main Result 2.2. We begin by asking the following question:
How many 1-nodal degree d curves are there in P2 that pass through δd − 1 (see (6.1)) generic
points?
We solve the above question by treating it as a special case of the following more general
question: how many pairs are there, a line and a degree d curve, such that the line passes
through m generic points and the curve passes through δd−m points and the curve has a nodal
point lying on the line? Note that the answer to this question is automatically zero if m ≥ 3,
since a line does not pass through three generic points. Fixing the line corresponds to setting m
to be equal to 2 (since a unique line passes through two points). The answer to our original
question is obtained when m = 1.
In order to solve the above question, we first consider the space of curves with a marked
point p at which the curve is tangent to the line. On this space, we impose the condition that
the derivative of the curve in the normal direction vanishes. That precisely means that the curve
has a node at p.
p
Taking the derivative in the normal direction induces a section of an appropriate bundle.
This bundle is described in detail in Section 9.1 (see equation (9.8)). Hence, computation of the
Euler class of this relevant bundle yields the desired number.
We now explain how to enumerate curves with one tacnode. Before that there is small
digression. We first try to solve the question of enumerating curves with two nodes. However,
both the nodes are required to lie on a given line. Consider the space of curves with three
marked points x1, x2 and x3, such that all the three points lie on the curve and the line while
the curve has a node at the point x1. Pictorially, such a curve looks as follows:
x2x1
x3
Counting Curves with Tangencies 31
Now impose the condition that x2 and x3 become equal. This results in the following objects:
x1
x2 = x3
Next impose the condition that the derivative of the curve at x2 in the normal direction
is zero. This corresponds to the curve having a node at x2. However, there is a degenerate
contribution to the Euler class, which occurs when x1 and x2 come together, as is seen by the
following picture:
x1 = x2 = x3
In the above picture, one of the branches of the node is tangent to the given line of second
order. Subtracting off this degeneracy allows us to solve the problem. In particular, we can
count the number of pairs consisting of a line and a degree d curve, such that the line passes
through m points, the curve passes through n points with m+ n = δd − 2 (see (6.1)) while the
curve has two nodes lying on the line. When m = 2 and n = δd − 4, it corresponds to case
where the curve has two nodes on the same fixed line. On the other hand, setting m = 0 and
n = δd − 2 corresponds to the case where the two nodes are free.
For tacnodes, we use the fact that they occur precisely when two nodes collide. Consider the
space of curves with two nodes x1 and x2, both of them lying on a given line. Now impose the
condition that x1 = x2, and the following object is obtained:
x1 x2
x1 = x2
Figure 2. Two nodes on the line limiting to a tacnode.
Next assume that the line passes through m points and the curve passes through n points,
where m + n = δd − 3. The case where m = 0 and n = 3 gives the number of degree d curves
through δd − 3 generic points that have a tacnode.
To summarize, we have been able to solve the following questions about enumerating singular
curves:
• Plane curve of degree d having a node.
• Plane curve of degree d having two distinct nodes.
• Plane curve of degree d having a tacnode.
32 I. Biswas, A. Choudhury, R. Mukherjee and A. Paul
Let us now implement this idea precisely. Similar to AF
1 and AF
2 , define AF
3 ⊂ M1
0 to be the
locus (H1, Hd, p) such that Hd has a tacnode at p. Furthermore, define AF
1A
F
1 ⊂ M2
0, to be the
locus (H1, Hd, p1, p2) such that Hd has a node at p1 and p2 and p1 and p2 are distinct. The
following formulas will be proven in this section:[
AF
1
]
=
(
3d2 − 6d+ 3
)
ydb
2
1 + (3d− 3)y2db1 + y3d, (9.1)[
AF
3
]
=
(
50d2 − 192d+ 168
)
y3db
2
1 + (25d− 48)y4db1 + 5y5d and (9.2)[
AF
1A
F
1
]
=
(
9d4 − 36d3 + 12d2 + 81d− 66
)
y2db
2
1b
2
2
+
(
9d3 − 27d2 − d− 30
)
y3db1b
2
2 +
(
9d3 − 27d2 − d− 30
)
y3db
2
1b2
+
(
3d2 − 6d− 4
)
y4db
2
1 +
(
9d2 − 18d+ 2
)
y4db1b2 +
(
3d2 − 6d− 4
)
y4db
2
2
+ (3d− 3)y5db1 + (3d− 3)y5db2 + y6d. (9.3)
9.1 Counting 1-nodal curves
We prove (9.1). Recall that (7.1) says that the class
[
AL
1
]
can be computed by multiplying the
class
[
AF
1
]
with (yd + b1). Note that knowing
[
AL
1
]
does not – a priori – give
[
AF
1
]
. Nevertheless,
it will be shown that knowing
[
AL
1
]
in fact does give the class
[
AF
1
]
.
First of all, we note that
[
AF
1
]
is a codimension 3 class in M1
0. Hence, it is of the form[
AF
1
]
=
∑
i+j=3,
i,j≥0,
j≤2
Cijy
i
db
j
1.
There is no term involving y1, because it is the pullback of a class in Dd × X1. We also note
that j can not be greater than 2, since b31 is zero. Hence, it suffices to compute the following
three numbers C12, C21 and C30. Next, we note that
C12 =
[
AF
1
]
· y21y
δd−1
d , C21 =
[
AF
1
]
· y21y
δd−2
d b1 and C30 =
[
AF
1
]
· y21y
δd−3
d b21.
It is rather straightforward that C21 is the number of degree d curves passing through δd − 2
generic points with a node lying on a line; intersecting with yδdd − 2 makes the curve pass
through δd − 2 points and intersecting with b1 makes the nodal point lie on a line (the line
represents the class b1). But that can also be obtained as an intersection number involving the
class
[
AL
1
]
, namely
C21 =
[
AL
1
]
· y21y
δd−2
d . (9.4)
Hence, knowing
[
AL
1
]
gives the coefficient C21.
Next, observe that C12 is the number of degree d curves passing through δd−1 generic points
with a node. Hence, one concludes that
C12 =
[
AL
1
]
· y1yδd−1
d . (9.5)
Indeed, intersecting with yδd−1
d makes the nodal curve pass through δd−1 points. For each such
curve, the nodal point is now fixed. Intersecting with y1 now fixes a unique line. Hence, the
right-hand side of (9.5) gives us C2.
Finally, note that C30 is the number of degree d curves passing through δd − 3 generic points
with a node located at a fixed point. A similar argument gives that
C2 =
[
AL
1
]
· y21b1y
δd−3
d . (9.6)
Hence, in order to compute
[
AF
1
]
it suffices to compute
[
AL
1
]
.
Counting Curves with Tangencies 33
We do the intersection theory on M1
0. Think of T1 as a subspace of M1
0 := D1 × Dd × X1.
Let (H1, Hd, p) ∈ T1 and suppose that the curve Hd is given by the zero set of the polynomial fd.
Furthermore, suppose that the line is given by the zero set of f1. This gives us the following
short exact sequence
0 −→Ker∇f1|p −→ TX1|p −→ γ∗D1
⊗ γ∗X1 −→ 0. (9.7)
Here γD1 and γX1 denote the tautological line bundles over D1 and X1 respectively and γ∗D1
and γ∗X1 denote their duals. The first nontrivial map in equation (9.7) denotes the inclusion
map into the tangent space TX1|p. The second map denotes the vertical derivative ∇f1|p. This
map is surjective since the point p is not a singular point of the line f−1
1 (0) (all points of a line
are smooth). Define the line bundle L −→ T1, whose fiber over each point is Ker∇f1|p.
We now impose the condition that the derivative of fd in the normal direction to the line
is zero. This means that ∇fd|p(u) = 0, ∀u ∈
(
TX1/L
)
. Taking the derivative along u induces
a section of the vector bundle
V :=
(
TX1/L
)∗ ⊗ γ∗Dd
⊗ (γ∗X1)
d. (9.8)
Hence,[
AL
1
]
= [T1] · (yd − y1 + (d− 1)b1). (9.9)
The second term on the right-hand side of (9.9) is the Euler class of V, which can be computed
using (9.7). Using the results of Section 6, all intersection numbers involving the class [T1]
can be computed. Hence, (9.9) enables us to compute all intersection numbers involving the
class
[
AL
1
]
. As a result, using (9.4), (9.5) and (9.6) one obtains (9.1).
It remains to prove that the intersection in (9.9) is transverse. But this is precisely the
content of the proof of Proposition 7.2, with n = 0 and r = 0.
9.2 Counting 2-nodal curves
We now prove (9.3). First of all, note that
[
AF
1A
F
1
]
is a codimension 6 class in M2
0. Hence, it is
of the form[
AF
1A
F
1
]
=
∑
i+j+k=6,
i,j,k≥0,
j,k≤2
Cijky
i
db
j
1b
k
2. (9.10)
There is no term involving y1, because it is the pullback of a class in Dd×X1×X2. We also note
that j or k can not be greater than 2, since b31 and b
3
2 are both zero. We also note that Cijk = Cikj ;
this is because the map from AF
1A
F
1 to itself, that permutes the two marked points is a bijection.
Hence, it suffices to compute the following five numbers C222, C312, C420, C411, C510 and C600.
We perform intersection theory on M2
0. Define AL
1A
L
1 ⊂ M2
0 to be the locus (H1, Hd, p1, p2) such
that curve Hd has a node at the two distinct points p1 and p2. Furthermore, the two points p1
and p2 also lie on the lineH1. Let us for the moment assume that we can compute all intersection
numbers involving the class
[
AL
1A
L
1
]
. It will now be shown how to compute the numbers Cijk
from this information. In particular, it will be shown that
C222 =
[
AL
1A
L
1
]
· yδd−2
d , C312 =
[
AL
1A
L
1
]
· yδd−3
d b1, C420 =
[
AL
1A
L
1
]
· yδd−4
d b21,
C411 =
[
AL
1A
L
1
]
· yδd−4
d b1b2, C510 =
[
AL
1A
L
1
]
· yδd−5
d b1b
2
2 and
C600 =
[
AL
1A
L
1
]
· yδd−6
d b21b
2
2. (9.11)
34 I. Biswas, A. Choudhury, R. Mukherjee and A. Paul
We start by justifying the expression for C222. Note that by definition C222 =
[
AF
1A
F
1
]
· y21y
δd−2
d .
Hence, C222 is equal to the number of degree d curves passing through δd − 2 points and having
two (ordered) nodal points. This is the same as the right-hand side of the first equation of (9.11).
Let us see why this is true. Intersecting
[
AL
1A
L
1
]
with yδd−2
d corresponds to making the curve
pass through δd − 2 points. By definition of
[
AL
1A
L
1
]
, both the nodal points lie on the line. Since
there are two nodal points, the line is now fixed. This proves the first assertion of (9.11). The
remaining five assertions of (9.11) can be seen similarly.
Hence, we have shown that to prove (9.3), it suffices to compute all intersection numbers
involving the class
[
AL
1A
L
1
]
. To explain how to compute those numbers, think of AL
1T1 as a sub-
space of M2
0. Let (H1, Hd, p1, p2) ∈ AL
1T1. We now impose the condition that the derivative
of the polynomial defining the curve Hd in the normal direction to the line H1 vanishes at p2.
Analogous to (9.9), it is tempting to conclude that
[
AL
1T1
]
· (yd − y1 + (d− 1)b2) =
[
AL
1A
L
1
]
. Un-
fortunately, the above equation is incorrect. This is because when p1 and p2 collide in the closure
we get a P(1)A1, as can be seen intuitively by the following picture:
p1
p2
p1 p2
p1 = p2
yd − y1 + (d− 1)b2
In other words, we are claiming that on H∗
(
M2
0;R
)
, the following equality of homology classes
hold [
AL
1T1
]
· (yd − y1 + (d− 1)b2) =
[
AL
1A
L
1
]
+
[
P(1)A1
]
·
[
∆11
]
. (9.12)
Here ∆11 denotes the locus of points (H1, Hd, p1, p2) ∈ M2
0 such that p1 = p2. We first explain
how to extract numbers from this equation. Let α and β be classes in H∗
(
M2
0;R
)
given by
α := yr1y
s
db
ε1
1 b
ε2
2 and β := yr1y
s
db
ε1+ε2
1 .
Using equation (9.12), we conclude that[
AL
1T1
]
· (yd − y1 + (d− 1)b2) · α =
[
AL
1A
L
1
]
· α+
[
P(1)A1
]
· β (9.13)
=⇒
[
AL
1A
L
1
]
· α =
[
AL
1T1
]
· (yd − y1 + (d− 1)b2) · α−
[
P(1)A1
]
· β. (9.14)
By the results of Section 7, we can compute all intersection numbers involving the classes
[
AL
1T1
]
and
[
P(1)A1
]
. Hence, using (9.14) we can compute all intersection numbers involving the
class
[
AL
1A
L
1
]
. Hence, using (9.11) and (9.10), we get (9.3).
We now justify (9.12). First, it will be proved on the set-theoretic level. We switch to affine
space. As before, the line is the x-axis. We have already shown that
(
AL
1T1
)
Aff
is a smooth
submanifold of F+
d × C2. The argument that the intersection of the cycles on the open part is
transverse is similar to how we have shown the earlier transversality statements. This justifies
the first term on the right-hand side of (9.13) (on the level of homology).
The second term will now be justified. One needs to figure out what happens when the nodal
point p1 and the T1 point p2 coincide. The following fact has already been shown: if at p1 and p2
the first derivatives of the polynomial defining the curve vanish (along the direction of the line),
then when p1 and p2 coincide, the first and second derivatives of the polynomial vanish. This
was shown while proving that T1 and T1 collide to form a T3. Notice that one does not need p1
and p2 to be smooth points of the curve for the argument to work.
Now note that p1 is a singular point of the curve. Hence, when the two points collide, it will
continue to remain a singular point of the curve. Hence, we conclude the following: when p1
Counting Curves with Tangencies 35
and p2 coincide, the first two derivatives of the polynomial defining the curve vanish (along the
direction of the line). Furthermore, it is a singular point of the curve. Note that any element
of P(1)A1 satisfies these conditions. We now show that every curve in P(1)A1 actually lies in the
closure.
Let (f, (0, 0)) belong to
(
P(1)A1
)
Aff
. Note that we are setting the P(1)A1 point to be (0, 0).
Hence, the Taylor expansion of f is given by
f(x, y) = f11xy +
f02
2
y2 +
f21
2
x2y +
f12
2
xy2 +
f03
6
y3 +
f40
24
x4 + · · · .
We now try to find a nearby curve ft that has a nodal point at (0, 0) and has a T1 point at (t, 0).
The Taylor expansion of ft is given by
ft(x, y) =
ft20
2
x2 + ft11xy +
ft02
2
y2 +
ft30
6
x3 +
ft21
2
x2y +
ft12
2
xy2 +
ft03
6
y3 +
ft40
24
x4 + · · · .
We now impose the condition ft(t, 0) = 0 and (ft)x(t, 0) = 0. We can solve for this and get an
expression for ft20 and ft30 . Moreover, every solution is constructed by this procedure. Hence,
there is only one branch.
It remains to justify the multiplicity of the intersection. Consider the condition of taking
the derivative in the normal direction at the point (t, 0). This is given by (ft)y(t, 0). Written
explicitly, it is given by the map t −→ ft11t. Assuming that ft11 ̸= 0, the order of vanishing of
the above function at t = 0 is one. This is a valid assumption, since to compute the intersection
multiplicity, we will be intersecting with generic cycles. This proves (9.13) on the level of
homology and hence, completes the proof of (9.3).
9.3 Counting 1-tacnodal curves
We now prove (9.2). First of all, we note that
[
AF
3
]
is a codimension 5 class in M1
0. Hence, it is
of the form[
AF
3
]
=
∑
i+j=5,
i,j≥0,
j≤2
Cijy
i
db
j
1.
There is no term involving y1, because it is the pullback of a class in Dd × X1. We also note
that j can not be greater than 2, since b31 is zero. Hence, it suffices to compute the following three
numbers C32, C41 and C50. We make a small digression and discuss the condition of a singularity
being a tacnode. Recall Definition 7.1: the zero set of a holomorphic function f : U −→ C has
a tacnode at the origin if after a local (analytic) change of coordinates, the function can be
written as f(x, y) := y2 + x4. Here U is an open subset of C2 (with the usual topology of C).
A tacnode satisfies the condition that the kernel of the hessian is precisely one dimensional (i.e.,
the hessian is degenerate, but not identically zero). We call the kernel of the hessian of f to be
the distinguished direction of the tacnode. For the tacnode y2 + x4 = 0, the tangent vector ∂
∂x
is the distinguished direction.
We now define PA3 ⊂ M1
0 to be the locus (H1, Hd, p1) such that curve Hd has a tacnode at p1
and the distinguished direction of the tacnode is given by the line H1. Pictorially, it is denoted
by the following picture:
p1 H1
36 I. Biswas, A. Choudhury, R. Mukherjee and A. Paul
Notice the difference between the spaces AL
3 and PA3; the latter lies in the closure of the
former. The space AL
3 is pictorially represented as follows:
p1
It will now be shown that computing intersection numbers involving the class [PA3] enables
us to compute the numbers Cij . In particular, it will be shown that
C32 = [PA3] · yδd−3
d , C41 = [PA3] · yδd−4
d b1 and C50 = [PA3] · yδd−5
d b21. (9.15)
Let us justify the first term of (9.15), namely the computation of C32. Note that by definition,
C32 =
[
AF
3
]
· y21y
δd−3
d .
Hence, C32 denotes the number of degree d curves passing through δd − 3 generic points and
having a tacnode. We now note that intersecting [PA3] with yδd−3
d makes the curve pass
through δd − 3 points. By definition of PA3, the line is now fixed because there is a unique
line that passes through the tacnodal point and is the branch of the tacnode. This proves the
first equation of (9.15). The remaining two equations follow similarly.
Hence, it has been shown that to prove (9.2), it suffices to compute all intersection numbers
involving the class [PA3]. It will now be explained how to compute those numbers.
Although we are enumerating curves with one singularity, the intersection theory will be done
on the two pointed space M2
0. Let (H1, Hd, p1, p2) ∈ AL
1A
L
1. Now impose the condition that the
two points p1 and p2 come together. It will be shown shortly that when that happens, we get
a curve in PA3 (see Figure 2). In H∗
(
M2
0;R
)
, the following equality of homology classes holds:[
AL
1A
L
1
]
· (b1 + b2 − y1) = [PA3] ·
[
∆11
]
. (9.16)
We first explain how to extract numbers. Let α be a class in H∗
(
M1
0;R
)
. Intersecting both sides
of equation (9.16) with α, we get that
[PA3] · α =
[
AL
1A
L
1
]
· (b1 + b2 − y1) · α. (9.17)
Using the results of Section 9.2, we can compute all intersection numbers involving the class[
AL
1A
L
1
]
. Hence, using (9.17), we can compute all intersection numbers involving the class [PA3].
We now justify (9.16). First of all recall the proof of the fact that when a T1 point and
another T1 point collide, we get a T3 point. The proof in fact shows the following: suppose the
first derivative of f vanishes at p1 and p2, then when the two points coincide, the first, second
and third derivatives coincide. The proof does not in any way require the points to be smooth
points of the curve. Hence, when two nodal points lying on a line coincide, the first, second and
third derivatives along the line vanish. Furthermore, the point is a singular point of the curve.
Note that any curve in PA3 satisfies these conditions.
We now show that every element of PA3 lies in the closure. We switch to affine space. Let f
be a curve that has a PA3 point at the origin. Hence, the Taylor expansion of f is given by
f(x, y) = f11xy +
f02
2
y2 +
f21
2
x2y +
f12
2
xy2 +
f03
6
y3
+
f40
24
x4 +
f31
6
x3y +
f22
4
x2y2 +
f13
6
xy3 +
f04
24
y4 + · · · .
Counting Curves with Tangencies 37
We now try to construct a nearby curve ft that has a nodal point at (0, 0) and at (t, 0). We also
show that every nearby curve is of the type we have constructed.
The Taylor expansion of ft is given by
ft(x, y) =
ft20
2
x2 + ft11xy +
ft02
2
y2 +
ft30
6
x3 +
ft21
2
x2y +
ft12
2
xy2 +
ft03
6
y3
+
ft40
24
x4 +
ft31
6
x3y +
ft22
4
x2y2 +
f13
6
xy3 +
ft04
24
y4 + · · · .
We now impose the condition ft(t, 0) = 0, (ft)x(t, 0) = 0 and (ft)y(t, 0) = 0. Using the equa-
tion (ft)y(t, 0) = 0, we can uniquely solve for ft11 . Next, using the equation (ft)x(t, 0) = 0, we
can uniquely solve for ft20 . Plugging these two solutions in the equation ft(t, 0) = 0, we can
uniquely solve for ft30 .
This gives us a procedure to construct a curve ft, close to f , that has a nodal point at (0, 0)
and at (t, 0). Since our solution was unique, this implies that every nearby curve is of this type,
i.e., there is only one branch.
It remains to compute the multiplicity of the intersection. We are basically setting the x-
coordinate to be equal to zero. But the x-coordinate is t. Hence, the order of vanishing is one
(since there is exactly one branch). This completes the proof of (9.17).
10 Counting rational curves
Finally, in this section, we give an alternative approach to enumerate stable maps with first-order
tangency. Before getting into the details, the idea will be outlined.
First of all, it may be recalled that we are countingmaps, not zero sets of polynomials. We will
be considering the Kontsevich moduli space of maps of rational curves with two marked points,
i.e., we will be doing intersection theory on M0,2
(
P2, d
)
. Denote an element of M0,2
(
P2, d
)
as [u, y1, y2]. The letter u denotes a map from a possibly singular genus zero Riemann surface
to P2 while y1 and y2 are two distinct points on the domain. The square bracket is there since
we are looking at equivalence classes of such maps. Denote x1 := u(y1) and x2 := u(y2). Note
that x1 and x2 are points on the target space
(
i.e., P2
)
. Now consider the space of maps of
rational curves with two marked points and a line, such that the image of the curve evaluated
at those two points intersects the line. A pictorial representation of an element of this space is
as follows:
u(y1) = x1 u(y2) = x2
Figure 3. Stable maps intersecting a line at two points.
On this space, now impose the condition that the points x1 and x2 become equal. There are
two possibilities now that can be pictorially seen as follows: In the first case the corresponding
points on the domain also become equal, i.e., y1 = y2. This corresponds to the curve having
a tangency. In the second case, the corresponding points on the domains are not the same, i.e.,
y1 ̸= y2. This corresponds to the image of the curve having a self intersection, so the curve
has a node. Hence, imposing the condition x1 = x2, what we get can be summarized by the
following picture:
38 I. Biswas, A. Choudhury, R. Mukherjee and A. Paul
u(y1) = x1 u(y2) = x2
y1 6= y2
y1 = y2
Using the results of [2], all intersection numbers involving the second term on the right-hand
side of Figure 3 can be computed. Hence, we can compute the characteristic number of rational
curves tangent to a given line. Sections 10 and 11.4 contain a detailed computation along the
above line.
Finally, in Section 10.3, we pursue this idea again by making the points in the domain come
together. This is implemented by extending the idea behind the derivation of Kontsevich’s
recursion formula. We choose a suitable subspace of the four pointed moduli space M0,4
(
P2, d
)
and intersect it with the pullback of two divisors from M0,4. Equating those two intersection
numbers, gives a recursive formula for the characteristic number of rational curves tangent to
a given line.
We now implement these ideas precisely. But first, we make a digression and review how
Gathmann enumerates rational curves tangent to a divisor.
10.1 A review of Gathmann’s approach to count curves with tangencies
In his papers [7, 9] and [10], Andreas Gathmann gives a systematic approach to solve the
following question: Let Y be a hypersurface inside Pn. What are the characteristic number of
rational degree d curves in Pn that are tangent to Y at a given point to order k? Gathmann
successfully solves the above question for any ample hypersurface and any k. He goes on to use
this study to compute Gromov–Witten invariants of the quintic threefold.
Consider a special case of Gathmann’s result when Y is a line inside P2 and ask the following
question: How many rational degree d curves are there in P2, that pass through 3d− 2 generic
points and are tangent to a given line? In this subsection we recapitulate Gathmann’s approach
to solve this question. The next subsection describes an alternative approach to the question
based on applying Figure 1 in the setting of stable maps (which is also discussed in [8, pp. 41]
and [14, pp. 1179–1180]).
Let us now describe Gathmann’s idea. The setup is modified in order to solve a slightly more
general question. The question we solve is as follows: How many pairs – consisting of a line and
a rational degree d curve, passing through m points and n points respectively – are there such
that the line is tangent to the curve and m+n = 3d+1? The special case of m = 2 corresponds
to the line being fixed.
We start by describing the ambient space. Recall that M0,0
(
P2, d
)
is the compactification of
the Kontsevich moduli space of maps of rational curves (with no marked points). Let H denote
the divisor that corresponds to the subspace of curves that pass through a generic point. We
note that the intersection number
[
M0,0
(
P2, d
)]
· Hm is computable via Kontsevich’s recursion
formula. For dimensional reasons, the above number is nonzero only when m = 3d− 1. On the
zero pointed moduli space, this is the only intersection number that is relevant for our purposes;
it is also called a primary Gromov–Witten invariant.
Now consider M0,1
(
P2, d
)
, the one marked moduli space. As before, we have the divisor H
which corresponds to the subspace of curves whose image passes through a generic point. But
now, there are two other things as well. Denote the pullback (via the evaluation map) of the
hyperplane class in P2 by ev∗(b1). Finally, consider L −→ M0,1
(
P2, d
)
, the universal tangent
bundle, whose fibre over each point is the tangent space at that marked point. Denote the first
Chern class of the dual of this bundle by ψ, i.e., ψ := c1(L∗). It is a standard fact that all the
Counting Curves with Tangencies 39
intersection numbers[
M0,1
(
P2, d
)]
· Hm · ev∗(b1)n · ψθ (10.1)
are computable for any choice of m, n and θ. This can be seen from the paper [12, p. 311,
Proposition 2.2]. When θ is greater than zero, the above number is also called a descendant
Gromov–Witten invariant.
We now explain the geometric idea behind Gathmann’s method to enumerate rational curves
that are tangent to a fixed line, and how to modify his method when the line is not fixed but
is free to move in a family. Denote by M0,k
(
P2, d
)
the k marked moduli space. The pullback
of the hyperplane classes (via the evaluation map) are denoted by ev∗(bn1
1 ), . . . , ev∗(bnk
k ). Now
define M1 as
M1 := D1 ×M0,1
(
P2, d
)
× P2
1,
where P2
i denotes a copy of P2 and D1 denotes the space of lines in P2. The corresponding
hyperplane classes are denoted by ai and y1.
Note that an element of M1 consists of a line, a one pointed rational curve (namely an element
of M0,1
(
P2, d
)
), and a point of P2
1. The relevant classes that live in M1 are y1, H, ev∗(b1), ψ,
and a1. Since we can compute all the primary and descendant Gromov–Witten invariants (i.e.,
the numbers in equation (10.1)), we can compute all the following intersection numbers:
[M1] · Hm · ev∗(bn1
1 ) · ψθ · yr1 · as1. (10.2)
Now define (T0)st to be the following subspace of M1:
(T0)st := {([f1], [u, y1], x1) ∈ M1 | u(y1) = x1, f1(x1) = 0}.
Remark 10.1. Note the following fact: we are typically going to denote the marked point of
the domain by the letter yi. It is not going to cause any confusion with the other place where
the letter y1 is used, namely for the hyperplane class of D1.
Returning to the discussion, an element of (T0)st can be pictorially described as follows:
u(y1) = x1
Let us now see how we can go about describing the class [(T0)st]. First of all, note that
the condition f1(q1) = 0 is same as intersecting with the class (y1 + a1) (see equation (6.3)). It
remains to figure out how to express the condition u(y1) = q1. Consider the map
ev × idP2
1
: M1 −→ P2 × P2
1, ([f1], [u, y1], x1) 7−→ (u(y1), x1).
The condition u(y1) = x1 is same as intersecting with the pullback of the diagonal, namely
(ev × idP2
1
)∗(∆b1a1). Hence, we conclude that
[(T0)st] =
(
ev∗
(
b21
)
+ ev∗(b1)a1 + a21
)
· (y1 + a1). (10.3)
Since all the intersection numbers in equation (10.2) are computable, we conclude from equa-
tion (10.3) that all the intersection numbers
[(T0)st] · Hm · ev∗(bn1
1 ) · ψθ · yr1 · as1 (10.4)
are computable.
We now define (T1)st. It is the subspace of (T0)st, where the curve is tangent to the line at
the point x1. It is pictorially described as follows:
40 I. Biswas, A. Choudhury, R. Mukherjee and A. Paul
x1
Next, we explain Gathmann’s approach to compute intersection numbers involving the class
[(T1)st]. In the next subsection, we give an alternative approach to compute these intersection
numbers. Gathmann’s approach is best summarized by the following equation:
[(T1)st] = [(T0)st] · (ψ + y1 + a1). (10.5)
We explain why equation (10.5) is true. We note that for a rational curve [u, y1] to be tangent
to the line, the differential du|y1 should take values in the tangent space of the line. In other
words, the differential du|y1 has to vanish in the normal direction of the line. This condition is
interpreted as the vanishing of a section of an appropriate line bundle.
First of all, consider the line bundle L −→ (T0)st whose fibre over each point ([f1], [u, y1], x1)
is the tangent space of the line f−1
1 (0) at the point x1. This is basically the same line bundle
we defined in Section 9.1. For the convenience of the reader, we review the definition, namely
the short exact sequence into which the line bundle fits
0 −→L −→ TP2
1|x1 −→ γ∗D1
⊗ γ∗P2
1
−→ 0. (10.6)
The condition that [u, y1] is tangent to the line at x1 (namely that the differential du|y1 vanishes
in the normal direction to the line) can be interpreted as a section of the following line bundle L∗⊗
ev∗
(
TP2
1/L
)
. Using equation (10.6), we conclude that the Euler class of the above line bundle
is equal to (ψ + y1 + a1) which gives us equation (10.5).
Note that using equations (10.5), (10.3), and the fact that all the intersection numbers of
equation (10.2) are computable, we conclude that all the following intersection numbers [(T1)st] ·
Hm · ev∗(bn1
1 ) · ψθ · yr1 · as1 are computable.
This idea can be pushed further to enumerate rational curves with higher order tangency.
However, starting from second-order tangency, there is a non-trivial geometric phenomenon that
occurs. In the closure of curves tangent to a given line, there are bubble maps. By a bubble
map, we refer to the following stable maps u ∈M0,n
(
P2, d
)
of degree d (see [13, Section 5.1]):
• The domain has finite number of components Ci each isomorphic to P1 and they are
meeting each other at nodes.
• Let ui = u|Ci of degree di. Then the image of ui and uj intersect at a common point
in P2, which is the image of the nodal point and the total degree of u is d =
∑
i di. If ui
is constant for some i, then it is referred as ghost bubble or ghost component.
These bubble maps are in the zero locus of the section that computes the second derivative.
Hence, one has to analyse a degenerate locus and subtract off from the Euler class. Gathmann
does that successfully in his papers [7, 9, 10] and is able to enumerate rational curves tangent
to any order.
10.2 A new method to count rational curves with first-order tangency
In this subsection, we give an alternate approach to enumerate rational curves with tangencies
based on Figure 1. This idea has been discussed in [8] (see p. 41) and [14, pp. 1179–1180]. The
idea presented has an obvious difficulty (namely the formation of self intersection). We overcome
that difficulty using the result of our paper [2].
Counting Curves with Tangencies 41
We continue with the setup of Section 10.1. We perform intersection theory on M2, which is
defined as
M2 := D1 ×M0,2
(
P2, d
)
× P2
1 × P2
2.
The relevant classes that live in M2 are y1, H, ev∗(b1), ev
∗(b2), a1 and a1. Define the following
projection map π21 : M2 −→ M1,given by π21([f1], [u, y1, y2], x1, x2) := ([f1], [u, y1], x1). Basi-
cally, the map forgets the second marked point on M0,2
(
P2, d
)
. Furthermore, it forgets the
second factor P2
2. The cycles that have been defined in the one pointed moduli space M1 can be
pulled back to the two pointed moduli space M2 via this projection map.
Now define the following divisor in the two pointed moduli space M0,2
(
P2, d
)
. We denote
this divisor by the symbol [y1 = y2]. The above symbol allows us to guess the divisor we will be
talking about: this divisor is represented by the space of stable maps with two marked points,
where the two marked points have coincided. It can be pictorially represented as follows:
y1 = y2
d
d = 0
Usually, in the literature, it is denoted by D({∅}, 2; d, 0) (see [6]). Note that this space can
be identified with the one marked moduli space M0,1
(
P2, d
)
. We now explain how to intersect
with this divisor. It is a standard fact that all the primary intersection numbers[
M0,2
(
P2, d
)]
· Hm · ev∗(bn1
1 ) · ev∗(bn2
2 ) (10.7)
are computable.
Now define
(
T0T0
)
st
to be the following subspace of M2: it consists of a line and a rational
curve and two marked points, where the two marked points lie on the line (see Figure 3).
It will be shown that the following equality of classes holds in M2:(
ev∗
(
b21
)
+ ev∗(b1)a1 + a21
)
· (y1 + a1) ·
(
ev∗
(
b22
)
+ ev∗(b2)a2 + a22
)
· (y1 + a2)
= [(T0T0)st] + [(T0)st]. (10.8)
A slight abuse of notation is being made here. Equation (10.8) is an equality of classes in M2.
The second term on the right-hand side, namely [(T0)st] is a class in M1. However, we can
identify that as a class in M2 via the following process: we attach a ghost bubble at the marked
point and add one more marked point on the ghost component. This is the intended meaning
of the second term of the right-hand side.
The reason for why equation (10.8) is true is similar to why Theorem 6.2 is true (for the
special case n = 1 and k1 = 0). It has been shown in Section 10.1 that intersecting with(
ev∗
(
b21
)
+ ev∗(b1)a1 + a21
)
· (y1 + a1)
is imposing the condition that the first marked point lies on the line. Similarly, intersecting with(
ev∗
(
b22
)
+ ev∗(b2)a2 + a22
)
· (y1 + a2)
imposes the condition that the second marked point lies on a line. Intersecting with both of
them gives us the condition that both the marked points lie on the line. However, the two
42 I. Biswas, A. Choudhury, R. Mukherjee and A. Paul
marked points can coincide. That gives us the degenerate locus corresponding to the second
term on the right-hand side of equation (10.8). Rewriting equation (10.8), one concludes that
[(T0T0)st] =
(
ev∗
(
b21
)
+ ev∗(b1)a1 + a21
)
· (y1 + a1)
×
(
ev∗
(
b22
)
+ ev∗(b2)a2 + a22
)
· (y1 + a2)− [(T0)st]. (10.9)
Using equation (10.9), the fact that all the intersection numbers of equation (10.7) are com-
putable and the fact that all the intersection numbers in equation (10.4) are computable, one
concludes that all the intersection numbers
[(T0T0)st] · Hm · ev∗(bn1
1 ) · ev∗(bn2
2 ) · yr1 · a
n1
1 · an2
2 (10.10)
are computable.
Next, define
(
A1
)
st
to be the following subspace of the two pointed moduli space M0,2
(
P2, d
)
(A1)st :=
{
[u, y1, y2] ∈M0,2
(
P2, d
)
| u(y1) = u(y2)
}
.
Also, define
(
AL
1
)
st
to be the following subspace of M2(
AL
1
)
st
:= {([f1], [u, y1, y2], x1, x2) ∈ M2 | [u, y1, y2] ∈ (A1)st,
u(y1) = x1, u(y2) = x2, f1(x1) = 0, f1(x2) = 0}.
Pictorially, the space
(
AL
1
)
st
can be represented as follows:
x1
We now explain how to compute the characteristic number of rational curves with first-order
tangency. On the space
(
T0T0
)
st
, impose the additional condition that x1 = x2. By the collision
lemma, this is same as intersecting with (a1 + a2 − y1). Hence,
[(T0T0)st] · (a1 + a2 − y1) = [(T1)st] + 2
[(
AL
1
)
st
]
. (10.11)
Again, we are making an abuse of notation here. Equation (10.11) is an equality of classes in M2.
The first term on the right-hand side, namely [(T1)st] is a class in M1. However, we can identify
that as a class in M2 via the following process: we attach a ghost bubble at the marked point
and add one more marked point on the ghost component. This is the intended meaning of the
first term of the right-hand side.
We now see why (10.11) is true. We note that when we impose the condition x1 = x2, there
are two possibilities. The first possibility is that the corresponding points in the domain are also
the same and hence it corresponds to a point of tangency as given by the following picture:
x1 = x2
u(y1) = x1 u(y2) = x2
Counting Curves with Tangencies 43
This corresponds to the first term of the right-hand side of equation (10.11). But there is a
second possibility. The points in the domain can be distinct. This is a point of self intersection,
which is a nodal point. This corresponds to the second term in the right-hand side of (10.11).
The intersection occurs with a multiplicity of 2, because if a line intersects a nodal point of
a curve, then it contributes 2 to the intersection. Denote µ as follows µ := Hm ·ev∗(bn1
1 ) ·yr1 ·a
n1
1 .
Rewriting equation (10.11) and intersecting with µ, we conclude that
[(T1)st] · µ = [(T0T0)st] · (a1 + a2 − y1) · µ− 2
[(
AL
1
)
st
]
· µ. (10.12)
Since all the intersection numbers in (10.10) are computable, we conclude that the first term
on the right-hand side of (10.12) is computable. It will be shown in a moment that using the
result of [2], one can also compute the second term. Hence, the right-hand side of (10.12) is
computable for any µ. This gives us an alternative way to compute the characteristic number
of rational curves tangent to a given line to first order. The reader can refer to Section 11.4,
where we have tabulated explicit numbers.
This idea can actually be pursued further. Suppose we wish to enumerate curves with second-
order tangency. Then as expected, we define the space (T1T0)st. We then impose the condition
of the two points coming together. This results in two things. First of all we have curves with
second-order tangency. We also encounter curves with a node lying on the line, such that the
line is one of the branches of the node. Hence, we need to compute that number. In order to do
that, we define the space
(
AL
1T0
)
st
and require the points to come together. The two things we
get are, a nodal curve, with the line being one of the branches of the node. The second thing we
get is a curve with a triple point. To compute the latter, we can use the result of [2]. We have
actually carried out this entire computation. The details of the computation are available on
request. However, the analysis of the degenerate locus and its contribution to the intersection is
more non-trivial. We hope to pursue this approach in a more thorough way in future and figure
out how to enumerate rational curves with k-th order tangency.
It remains to be shown how to compute the second term on the right-hand side of (10.12).
First of all, we note that intersecting
[(
AL
1
)
st
]
with a1 or ev∗(b1) is the same thing. Hence,
it suffices to give a procedure to find intersection numbers with Hm · yr1 · an1 . Let α := (d, 2)
and let Nα(m,n) denote the intersection number defined in [2, pp. 5]. Here m and n are non-
negative integers. From the definition of Nα(m,n) and the proof of the correspondence result,
[2, pp. 13–16], we conclude that
[(
AL
1
)
st
]
· Hm · yr1 · an1 =
0 if r = 0, (10.13a)
Nα(m,n) if r = 1, (10.13b)
Nα(m, 1) if r = 2 and n = 0, (10.13c)
Nα(m, 2) if r = 2 and n = 1, (10.13d)
0 if r = 2 and n ≥ 2, (10.13e)
0 if r ≥ 3. (10.13f)
Let us see why this is true. The first case, (10.13a) follows from dimensional reasons.
We justify the next case, (10.13b). From the proof of the correspondence result, [2, pp. 13–
16], we conclude that Nα(m,n) denotes the number of rational curves passing through m generic
points and with a choice of a node lying at the intersection of n generic lines. Once such a nodal
curve is fixed, a unique line passes through one point and the nodal point. That precisely
corresponds to the left-hand side of the equation.
Next, we justify (10.13c). We start by unwinding the left-hand side. Intersecting with y21
corresponds to fixing a line. Now we intersect with Hm. This is equal to the number of rational
curves passing through m generic points and with a choice of a node lying on a line. That is
precisely equal to Nα(m, 1); again this follows from the proof of the correspondence result.
44 I. Biswas, A. Choudhury, R. Mukherjee and A. Paul
Next, let us justify (10.13d). Again, we begin by unwinding the left-hand side. Intersecting
with y21 corresponds to fixing a line. We recall that the nodal point lies on this line. Intersecting
with a1 is restricting the nodal point to lie on another line; in other words, we are restricting
the nodal point to lie on a fixed point. Now we intersect with Hm. This is equal to the
number of rational curves passing through m generic points and with a choice of a node lying
on a fixed point. That is precisely equal to Nα(m, 2); again this follows from the proof of the
correspondence result. The next case, (10.13e) follows immediately, since intersecting with y21
restricts the nodal point to lie on a line and intersecting with a21 is restricting the nodal point
to lie on a point; since a line and a point do not intersect, this number is clearly zero.
Finally, (10.13f) follows immediately since y31 is zero.
10.3 Another method to count rational curves with first-order tangency
We conclude this paper by illustrating that there is yet another way to exploit Figure 1 to
enumerate stable maps with tangencies. Recall that in the previous subsection, we made the
two points in the images come together. This results in two things; curves that are tangent to
the line and curves that have a point of self intersection lying on the line. The former occurs
when the points in the domain also coincide, while the latter occurs when the points in the
domain remain distinct.
What if we could simply make the points in the domain come together? In this final part of
the paper, we do precisely that. The idea will be implemented by extending Kontsevich’s idea
to enumerate rational curves, i.e., the WDVV equation. The details are as follows.
Recall that Kontsevich’s recursion formula gives an answer to the following question: How
many degree d rational curves are there in P2 that pass through 3d− 1 generic points? Denote
this number by nd. Also denote by NT1
d the number of rational degree d curves in P2 that pass
through 3d−2 and that are tangent to a given line. A recursive formula for NT1
d will be obtained
by using the WDVV equation. Note that for simplicity, the line will be kept fixed; the more
general case of keeping the line variable can be worked out with very little extra effort (the only
issue would be notational).
Consider M0,4
(
P2, d
)
, the moduli space of genus zero stable maps, with 4 marked points.
Let L be a fixed line inside P2. Define Xd to be the following subspace of M0,4
(
P2, d
)
: it is the
subspace of rational degree d curves where the image of the first two marked points lie on two
distinct points of the line L. It is pictorially represented as follows:
x1
x2
L
We denote Xd to be the closure of Xd inside M0,4
(
CP2, d
)
.
Following the idea behind Kontsevich’s recursion formula, we consider the forgetful map
π : M0,4
(
P2, d
)
−→M0,4. Let [(ij|kl)] denote the divisor in M0,4 corresponding to the wedge of
two spheres and where the marked points (yi, yj) lie on the one sphere and (yk, yl) lie on the
other sphere. In M0,4
(
P2, d
)
, define the class Z as Z := ev∗3(pt) · ev∗4(pt) · H3d−4. Since M0,4 is
isomorphic to P1, any two points determine the same divisor. Hence, [(12|34)] is equal to [(13|24)]
as divisors. Hence,[
Xd
]
· [π∗(12|34)] · Z =
[
Xd
]
· [π∗(13|24)] · Z. (10.14)
Counting Curves with Tangencies 45
Note that the left-hand side and right-hand side of (10.14) denote intersection numbers in
M0,4
(
P2, d
)
. We now unravel both the sides of the equation and get a recursive formula. Before
that, let us recapitulate a terminology about bubble maps. We define a bubble map to be of
type (d1, d2) if it is the following object: a holomorphic map from a wedge of two spheres (P1),
such that the map is of degree d1 on the first component and is of degree d2 on the second
component. Holomorphic here means that restricted to each of the components, the map is
holomorphic.
We unwind the left-hand side of (10.14) by looking at it geometrically. First, consider the
possibility when the two points y3 and y4 come together. That results in a bubble map of
type (d1, d2) such that
� The marked points y1 and y2 lie on the d1 component.
� The marked points y3 and y3 lie on the d2 component.
� The image of the marked points y1 and y2 intersect the line L.
� The image of y3 and y4 coincide with two generic points (this corresponds to intersection
with ev∗3(pt) and ev∗4(pt)).
� The entire configuration passes through 3d − 4 points
(
this corresponds to intersecting
with H3d−4
)
.
The total number of such bubble maps is given by
∑
d1+d2=d
(
3d− 4
3d1 − 1
)
nd1nd2(d1d2)d1(d1 − 1).
The factor d1d2 is to encode the number of choices for the bubble point. The factor d1(d1 − 1)
is there to encode the number of choices where the image of the points y1 and then y2 can lie.
We now see what happens when y1 and y2 come together. What happens is that we get
a rational curve, tangent to the given line. The point of tangency is given by the image of y1
(which is also the same as the image of y2). Furthermore, the image of y3 and y4 coincide with
two generic points (which corresponds to intersection with ev∗3(pt) and ev∗4(pt)) and the entire
configuration passes through 3d − 4 points
(
which corresponds to intersecting with H3d−4
)
.
The total number of such objects is precisely equal to the number of rational curves passing
through 3d− 2 generic points, tangent to a given line. That is precisely equal to NT1
d . Hence,
the left-hand side of (10.14) is
NT1
d +
∑
d1+d2=d
(
3d− 4
3d1 − 1
)
nd1nd2(d1d2)d1(d1 − 1). (10.15)
Let us now unwind what is the right-hand side of (10.14) by looking at it geometrically. When the
two points y1 and y3 come together (or y2 and y4 come together), a bubble map of type (d1, d2)
is obtained such that:
� The marked points y1 and y3 lie on the d1 component.
� The marked points y2 and y4 lie on the d2 component.
� The image of the marked point y1 intersects the line L.
� The image of the marked point y2 intersects the line L.
� The image of y3 coincides with a generic points (this corresponds to intersection with
ev∗3(pt)).
46 I. Biswas, A. Choudhury, R. Mukherjee and A. Paul
� The image of y4 coincides with a generic points (this corresponds to intersection with
ev∗4(pt)).
� The entire configuration passes through 3d − 4 points
(
this corresponds to intersecting
with H3d−4
)
.
The total number of such configurations is given by
∑
d1+d2=d
(
3d− 4
3d1 − 2
)
nd1nd2(d1d2)(d1)(d2). (10.16)
This is precisely the right-hand side of (10.14). Equating (10.15) and (10.16), we conclude that
NT1
d =
∑
d1+d2=d
((
3d− 4
3d1 − 2
)
d1d2 −
(
3d− 4
3d1 − 1
)
d1(d1 − 1)
)
nd1nd2d1d2. (10.17)
11 Low degree checks
In this section, we perform various non-trivial low degree checks which will be compared with
the existing known results. We remind the reader that the bound we impose on d to obtain our
results are sufficient conditions for our formulas to be valid; they are not necessary. Many of
the numbers we have computed are obtained by applying the formula where the value of d is
lower than what is required to apply our theorem. Nevertheless, we display these values and
point out to the reader that the numbers we obtain at the end agree with the expected values
(obtained by other means).
11.1 Counting smooth curves with tangencies:
confirmation with Caporaso–Harris
In Section 6, it was shown that the following numbers [Tk1 . . .Tkn ] · y21y
δd−k
d , can be computed.
We remind the reader that k := k1 + · · ·+ kn. A few values are displayed in the following table:
d 4 5 6 7 8 9
µ y21y
10
4 y21y
16
5 y21y
23
6 y21y
31
7 y21y
40
8 y21y
50
9
1
2
[
T1T1T2
]
· µ 0 0 0 36 144 360
Table 1.
Note that we have divided by a factor of 2, because in the computation of
[
T1T1T2
]
· µ all the
tangency points are ordered. It is more natural to consider the two T1 points as unordered.
The corresponding nonzero numbers obtained from the Caporaso–Harris by setting
δ := 0, α := (0) and β := (i, 2, 1), ∀i = 0, 1, 2.
are as follows:
d 7 8 9
Nd,δ(α, β) 36 144 360
Table 2.
Counting Curves with Tangencies 47
We are assuming that the reader is familiar with the notation developed by Caporaso–Harris in
their paper [4]; we have followed that notation in Table 2. Notice that the values in the last row
of Tables 1 and 2 are in agreement.
Next, the value of the following numbers [Tk1 . . .Tkn ] · y21y
δd−k−1
d a1, will be displayed for
certain values. This corresponds to fixing the line (intersecting with y21) and fixing the location
of the first tangency point (intersecting with a1). The remaining tangency points are free. The
values obtained by using the theorems of Section 6 are
d 7 8 9
µ y21y
30
7 a1 y21y
39
8 a1 y21y
49
9 a1[
T1T1T2
]
· µ 12 36 72
Table 3.
Note that in this case, there is no significance of dividing by 2, because the two T1 points are
different; the first T1 point lies on a fixed point, while the second T1 is free.
The corresponding numbers from Caporaso–Harris by setting
δ := 0, α := (0, 1) and β := (i, 1, 1), ∀i = 0, 1, 2,
are as follows:
d 7 8 9
Nd,δ(α, β) 12 36 72
Table 4.
The values in the last row of Tables 3 and 4 are in agreement.
11.2 Counting one nodal curves with tangencies:
confirmation with Caporaso–Harris
Using the theorems of Section 7, the following numbers
[
AF
1Tk1 . . .Tkn
]
· y21y
δd−k−1
d . can be
computed. A few values are tabulated below:
d 7 8
µ y21y
30
7 y21y
39
8
1
2
[
AF
1T1T1T2
]
· µ 3420 19404
Table 5.
The corresponding numbers from Caporaso–Harris by setting
δ := 1, α := (0), β := (0, 2, 1) and
δ := 1, α := (0), β := (1, 2, 1)
are as follows:
d 7 8
Nd,δ(α, β) 3420 19404
Table 6.
The values in the last row of Tables 5 and 6 are in agreement.
48 I. Biswas, A. Choudhury, R. Mukherjee and A. Paul
Next, the value of the following numbers
[
AF
1Tk1 . . .Tkn
]
· y21y
δd−k−2
d a1, will be displayed for
certain values. This corresponds to fixing the line (intersecting with y21) and fixing the location
of the first tangency point (intersecting with a1). The remaining tangency points are free. The
values obtained by using the theorems of Section 6 are
d 8
µ y21y
38
8 a1[
AF
1T1T1T2
]
· µ 4912
Table 7.
The corresponding numbers from Caporaso–Harris by setting
δ := 1, α := (0, 1) and β := (1, 1, 1)
are as follows:
d 8
Nd,δ(α, β) 4912
Table 8.
The values in the last row of Tables 7 and 8 are in agreement.
11.3 Counting one cuspidal cubics with tangencies:
confirmation with Ernström–Kennedy
Using the theorems of Section 8, one obtains that for d = 3,
[
AF
2T1
]
· y21y63 = 60. This number is
equal to the number of rational cuspidal cubics passing through 6 generic points that is tangent
to a given line. This is in agreement with the answer obtained by Ernström and Kennedy in [5].
11.4 Counting stable maps with tangencies: confirmation with Gathmann
Using the results of Section 10.2, the following numbers [(T1)st]·y21 ·H3d−2−nan1 can be computed.
A few numbers are tabulated in the following two tables:
d 3 4 5 6 7 8
µ y21H7 y21H10 y21H13 y21H16 y21H19 y21H22[
(T1)st
]
· µ 36 2184 335792 106976160 61739450304 58749399019136
Table 9.
d 3 4 5 6 7 8
µ y21H6a1 y21H9a1 y21H12a1 y21H15a1 y21H18a1 y21H21a1[
(T1)st
]
· µ 10 428 51040 13300176 6498076192 5362556317120
Table 10.
These are all in agreement with the numbers computed by Gathmann’s program GROWI (that
implements the formulas in [7, 9, 10]).
Finally, Gathmann’s approach is extend in Section 10.1 where the line can be varied. In
particular, the following intersection numbers can be computed [(T1)st] · yr1 · H3d−n−r · an1 . A few
numbers are tabulated in the following two tables:
Counting Curves with Tangencies 49
d 3 4 5 6 7 8
µ y1H8 y1H11 y1H14 y1H17 y1H20 y1H23[
(T1)st
]
· µ 48 3720 698432 263129760 175401698304 189360514383488
Table 11.
These values are all in agreement with the alternative approach given in Section 10.2 to compute
these intersection numbers.
Finally, for d = 3 one can compute (using the theorems of Section 7) that
[
AF
1T1
]
· y1y83 = 48.
This number is in agreement with the first number tabulated in Table 11.
Finally, we tabulate the values of NT1
d using (10.17):
d 3 4 5 6 7 8
NT1
d 36 2184 335792 106976160 61739450304 58749399019136
Table 12.
These are all in agreement with the numbers computed by Gathmann’s program GROWI.
Acknowledgements
We are very grateful to the referees for giving us constructive and detailed comments on the
earlier version of the manuscript. We are grateful to Chitrabhanu Chaudhuri for several use-
ful discussions related to this paper. We also thank Soumya Pal for writing a Python pro-
gram to implement Caporaso–Harris formula for verification. The first author is partially sup-
ported by a J.C. Bose Fellowship (JBR/2023/000003). The second author is funded by the
Deutsche Forschungsgemeinschaft (DFG, German Research Foundation) under Germany’s Ex-
cellence Strategy – The Berlin Mathematics Research Center MATH+ (EXC-2046/1, project
ID: 390685689). The fourth author would like to acknowledge the support of the Department
of Atomic Energy, Government of India, under project no. RTI4001.
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1 Introduction
2 Main results
3 Comparison with other methods
4 A few remarks on intersection theory
5 The collision lemma
6 Counting smooth curves with multiple tangencies
7 Counting 1-nodal curves with multiple tangencies
8 Counting 1-cuspidal curves with multiple tangencies of first order
9 Counting singular curves
9.1 Counting 1-nodal curves
9.2 Counting 2-nodal curves
9.3 Counting 1-tacnodal curve
10 Counting rational curves
10.1 A review of Gathmann's approach to count curves with tangencies
10.2 A new method to count rational curves with first-order tangency
10.3 Another method to count rational curves with first-order tangency
11 Low degree checks
11.1 Counting smooth curves with tangencies: confirmation with Caporaso-Harris
11.2 Counting one nodal curves with tangencies: confirmation with Caporaso-Harris
11.3 Counting one cuspidal cubics with tangencies: confirmation with Ernström-Kennedy
11.4 Counting stable maps with tangencies: confirmation with Gathmann
References
|
| id | nasplib_isofts_kiev_ua-123456789-212879 |
| institution | Digital Library of Periodicals of National Academy of Sciences of Ukraine |
| issn | 1815-0659 |
| language | English |
| last_indexed | 2026-03-21T11:38:30Z |
| publishDate | 2025 |
| publisher | Інститут математики НАН України |
| record_format | dspace |
| spelling | Biswas, Indranil Choudhury, Apratim Mukherjee, Ritwik Paul, Anantadulal 2026-02-13T13:49:49Z 2025 Counting Curves with Tangencies. Indranil Biswas, Apratim Choudhury, Ritwik Mukherjee and Anantadulal Paul. SIGMA 21 (2025), 012, 50 pages 1815-0659 2020 Mathematics Subject Classification: 14N35; 14J45; 53D45 arXiv:2312.10759 https://nasplib.isofts.kiev.ua/handle/123456789/212879 https://doi.org/10.3842/SIGMA.2025.012 Interpreting tangency as a limit of two transverse intersections, we obtain a concrete formula to enumerate smooth degree plane curves tangent to a given line at multiple points with arbitrary order of tangency. Extending that idea, we then enumerate curves with one node with multiple tangencies to a given line of any order. Subsequently, we enumerate curves with one cusp that are tangent to first order to a given line at multiple points. We also present a new way to enumerate curves with one node; it is interpreted as a degeneration of a curve tangent to a given line. That method is extended to enumerate curves with two nodes, and also curves with one tacnode are enumerated. In the final part of the paper, it is shown how this idea can be applied in the setting of stable maps and perform a concrete computation to enumerate rational curves with first-order tangency. A large number of low-degree cases have been worked out explicitly. We are very grateful to the referees for giving us constructive and detailed comments on the earlier version of the manuscript. We are grateful to Chitrabhanu Chaudhuri for several useful discussions related to this paper. We also thank Soumya Pal for writing a Python program to implement the Caporaso–Harris formula for verification. The first author is partially supported by a J.C. Bose Fellowship (JBR/2023/000003). The second author is funded by the Deutsche Forschungsgemeinschaft (DFG, German Research Foundation) under Germany’s Excellence Strategy– The Berlin Mathematics Research Center MATH+ (EXC-2046/1, project ID: 390685689). The fourth author would like to acknowledge the support of the Department of Atomic Energy, Government of India, under project no. RTI4001. en Інститут математики НАН України Symmetry, Integrability and Geometry: Methods and Applications Counting Curves with Tangencies Article published earlier |
| spellingShingle | Counting Curves with Tangencies Biswas, Indranil Choudhury, Apratim Mukherjee, Ritwik Paul, Anantadulal |
| title | Counting Curves with Tangencies |
| title_full | Counting Curves with Tangencies |
| title_fullStr | Counting Curves with Tangencies |
| title_full_unstemmed | Counting Curves with Tangencies |
| title_short | Counting Curves with Tangencies |
| title_sort | counting curves with tangencies |
| url | https://nasplib.isofts.kiev.ua/handle/123456789/212879 |
| work_keys_str_mv | AT biswasindranil countingcurveswithtangencies AT choudhuryapratim countingcurveswithtangencies AT mukherjeeritwik countingcurveswithtangencies AT paulanantadulal countingcurveswithtangencies |