Enumerating Multiplicity-Free Labeled FloorDiagrams for Rational Curves

Kevin HuNaperville Central High School

Naperville, Illinois

under the direction of

Mr. John Lesieutre

Massachusetts Institute of Technology

November 2010

Abstract

Labeled floor diagrams are connected, weighted, directed graphs with edges respecting vertex

order such that the sum of the weights of all edges exiting each node is at most one more than the

sum of the weights of all edges entering the node. This final condition is known as the Divergence

Condition. Manipulation of the enumeration of such graphs with degree d counts the number of

tropical curves with certain properties, which correspond to and are therefore equinumerous to al-

gebraic curves that pass through d points in general configuration in the complex projective plane.

In 2010, G. Mikhalkin and S. Fomin asked the question: how many multiplicity-free labeled floor

diagrams of genus zero are there? In other words, how many labeled floor diagrams are there with

d nodes and d− 1 edges, each of weight 1? They suggested the answer may be A029768 (as cata-

logued in the Online Encyclopedia of Integer Sequences). In this paper, Enumerating Multiplicity-

Free Labeled Floor Diagrams for Rational Curves, we show that A029768 is in fact not the correct

answer. We then categorize labeled floor diagrams by examining each nodes divergence: the sum

of outgoing edge weights, minus the sum of incoming edge weights. Using recursion, we analyze

and deduce formulas for several specific types of multiplicity-free labeled floor diagrams, based

on the number of nodes with divergence 1. The categorization method we use suggests two novel

classification methods for labeled floor diagrams using the distributions and the locations of nodes

divergences, so we explore these and relate the number of diagram types in each classification

method to two well-known functions: the Partition Function and the Catalan Numbers.

1

Summary

Imagine that there are several points arranged on an imaginary line from left to right. There

are also curved line segments with these points as endpoints. The number of line segments is one

less than the number of points. By traveling only on the curved line segments, it is possible to start

from any point and end up at any other point. This is the definition of connectedness. Take each

curved segment and turn it into an arrow pointing from left to right, so that each arrow exits one

point and enters another point on its right. Then we are left with a directed graph.

There is one other condition that makes this arrangement special. At each point, the number of

outgoing arrows is at most one more than the number of incoming arrows. If this condition is also

satisfied, the arrangement we have is a multiplicity-free labeled floor diagram for a rational curve.

Recently, mathematicians discovered that the counts of labeled floor diagrams are intricately

related to the counts of certain three-dimensional surfaces that pass through fixed configurations of

points. Furthermore, these surfaces are important to modern physics and string theory; thus, they

are both mathematically and practically interesting and important to examine. In 2010, S. Fomin

and G. Mikhalkin posed the problem: enumerate the multiplicity-free labeled floor diagrams for

rational curves that have a certain number of points.

This paper, Enumerating Multiplicity-Free Labeled Floor Diagrams for Rational Curves, will

explore these objects. First, we identify that the answer suggested by Fomin and Mikhalkin is in

fact incorrect. We then enumerate the objects with specific conditions. We also introduce two novel

methods of classifying the objects. We determine conditions that produce all possible classes, and

then prove that there are no empty classes, therefore showing the exact number of classes for the

objects. These classifications and their conditions open up two new ways of looking at labeled

floor diagrams, and two new ways of attacking the original open problem, by breaking down the

enumeration of multiplicity-free labeled floor diagrams into smaller and more specific cases.

2

Contents

1 Introduction 5

2 Enumerating ωd,d−2 7

2.1 Case 1: The other node of divergence 1 has two outgoing edges and one incoming

edge . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

2.2 Case 2: The other node of divergence 1 has one outgoing edge and no incoming

edges . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

2.3 Closed form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

3 Enumerating ωd,2 12

3.1 Case 1: Remaining set contains an ωx,1-graph . . . . . . . . . . . . . . . . . . . . 13

3.2 Case 2: Remaining set contains an ωy,2-graph . . . . . . . . . . . . . . . . . . . . 14

3.3 Closed form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

4 Holistic divergence vectors and the Partition function 15

5 Node divergence vectors and the Catalan numbers 16

6 Construction of labeled floor diagrams from node divergence vectors 19

7 Conclusion 21

7.1 Applications in detail . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

8 Acknowledgments 23

3

List of Figures

1 An example of a multiplicity-free LFD . . . . . . . . . . . . . . . . . . . . . . . . 6

2 Previously known values of ωd . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

3 An example of an LFD from Case 2.1 . . . . . . . . . . . . . . . . . . . . . . . . 8

4 An M -enumerated LFD with nodes appended to the left. . . . . . . . . . . . . . . 9

5 The corresponding M -enumerated LFD with degree d− 1. . . . . . . . . . . . . . 9

6 An example of an LFD from Case 2.2 . . . . . . . . . . . . . . . . . . . . . . . . 10

7 An LFD described by Lemma 1. . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

8 The remaining set from Case 3.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

9 The generated LFD. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

10 Monotonic paths, counted by the Catalan numbers . . . . . . . . . . . . . . . . . . 18

4

1 Introduction

How many lines are determined by two points? How many conic sections are determined by five

points? How many cubics are determined by eight points? These questions and their answers are,

by now, well-known in the mathematical community. A more general question is of particular

interest to algebraic geometers: How many curves of genus g defined by polynomials of degree d

pass through a fixed generic configuration of 3d+ g − 1 points on CP2?

The above examples ask the question for genus-zero curves, and degrees of 1, 2, and 3, re-

spectively. These cases are relatively simple. At higher degrees and higher genera, the problem

becomes much more difficult, and we enter the study of Gromov-Witten invariants.

These counts of curves are closely related to modern physics. String theory models particles as

vibrating loops; thus, the paths taken by particles, and more interestingly particle interactions, can

be modeled with curves in CP2.

The search for Gromov-Witten invariants was revitalized in 2008 by Brugalle and Mikhal-

kin [1], who introduced the labeled floor diagram (LFD). They determined that, when complex

algebraic curves in CP2 are transformed into tropical curves via tropicalization, the resulting curves

are combinatorially related to LFDs.

A labeled floor diagram of degree d and genus g is a weighted directed graph with d nodes and

d+ g − 1 edges satisfying the following conditions:

1. The nodes are collinear from left to right.

2. Each edge is directed from left to right. In this paper, any edges between nodes in figures are

assumed to be directed from left to right.

3. At each vertex, the sum of outgoing edge weights is at most one more than the sum of

incoming edge weights.

The third condition involves a quantity called the divergence, which is defined as the sum of

5

outgoing edge weights, minus the sum of incoming edge weights. We will call this condition the

divergence condition. In this paper, any edges without labels are assumed to have a weight of 1,

and is thus “multiplicity-free.”

Figure 1: An LFD with d = 6, g = 0, and divergences (from left to right) of 1, 1,−1, 1,−1,−1.

In 2009, Fomin and Mikhalkin [2] further explored LFDs and their enumerations. They showed

that LFDs of genus zero are equinumerous to labeled trees, and can thus be counted by Cayley’s

Formula; in particular, the number of LFDs of genus zero and degree d is dd−2. Floor diagram

methods also made possible a new form for a celebrated formula of Kontsevich: that, for d ≥ 2,

the enumeration of the genus-zero degree-d Gromov-Witten invariants of CP2, or Nd, is defined

by the recursion

Nd =∑k+l=d

NkNlk2l

[l

(3d− 4

3k − 2

)− k(

3d− 4

3k − 1

)].

In the final sections of their paper, Fomin and Mikhalkin propose several open problems. One

of these is to enumerate multiplicity-free labeled floor diagrams of degree d and genus 0; i.e., those

diagrams in which the weight of every edge is 1.

We explore this problem. Let ωd be the number of multiplicity-free labeled floor diagrams of

degree d and genus 0.

As Fomin and Mikhalkin note, the first few values are equal to those in the sequence A029768

as indexed by the On-Line Encyclopedia of Integer Sequences [3], which enumerates the number

of increasing rooted trees with cyclically-ordered branches, i.e. increasing mobiles (see Figure 2).

However, using a computational algorithm to analyze the adjacency matrices of floor diagrams, we

determine that ω7 = 2081, while the next term in A029768 is 2076.

Instead of examining the relation between multiplicity-free LFDs and increasing mobiles, we

decompose ωd by categorizing labeled floor diagrams by their holistic divergence vectors, or char-

6

d ωd A0297681 1 12 1 13 2 24 7 75 36 366 245 245

Figure 2: Values of ωd for d ≤ 6, from Fomin and Mikhalkin [2].

acteristic vectors with exactly ~v · ∆̂i nodes of divergence 2 − i, where ∆̂i is the unit vector in the

di direction (e.g. ∆̂2 = [0, 1, 0, 0, . . .]). Our decomposition breaks ωd into the sum of ω~v, where ω~v

is the number of LFDs with holistic divergence vector ~v.

We define the following notation: ωa,b is equal to ω~v where ~v satisfifesa+1∑i=2

~v · ∆̂i = b.

We first prove two formulas for specific cases of LFDs.

ωd,d−2 = 2d−3d− d+ 1

ωd,2 = 2d−4(d2 − 7d+ 22)− d− 1

We then show that the number of possible holistic divergence vectors for an LFD of degree d is

equal to P (d)−1, with P being the partition function. Next, we introduce nodal divergence vectors

and show that the number of possible nodal divergence vectors for an LFD of degree d is equal to

C(d−1), with C being the Catalan function. Finally, we show that, given any vector satisfying the

conditions for a nodal divergence vector, we can construct an LFD with the corresponding nodal

divergence vector. This gives formulas for both the number of holistic divergence vectors and the

number of nodal divergence vectors for LFDs.

2 Enumerating ωd,d−2

Theorem 1. ωd,d−2 = 2d−3d− d+ 1.

7

Proof. Let V1, V2, . . . be the nodes, numbered from left to right. There are d − 2 nodes with non-

positive divergence, so there are exactly 2 nodes with divergence equal to 1, one of which is the

first node. Thus, there is one other node with divergence 1. We proceed to analyze this in several

cases: either the other node of divergence 1 has two outgoing edges and one incoming edge, or

one outgoing edge and no incoming edges. If there were more outgoing and incoming edges, then

there would necessarily be more than two nodes of divergence 1.

2.1 Case 1: The other node of divergence 1 has two outgoing edges and one

incoming edge

All graphs with this condition must have edges VjVj+1 for 1 ≤ j ≤ k and div(Vi) = 1 for some

i ≤ k; e.g. Figure 3. We show that there are 2d−2 − d+ 1 such graphs.

First, we prove that if k = 2, then there are 2d−3−1 graphs. If d = 3, we must have edges V1V2

and V2V3, and div(V2) = 1, which is impossible.

If d = 4, there is one such graph: its edge set is {V1V2, V2V3, V2V4}, and there are two nodes

with negative divergence: namely, V3 and V4.

Let M(d) be the number of multiplicity-free labeled floor diagrams of degree d and genus

0 with edges V1V2 and V2V3 and with div(V2) = 1. We define an M -enumerated LFD to be a

multiplicity-free LFD with edges V1V2 and V2V3 with div(V2) = 1.

Now, suppose for sake of induction that for some natural number k ≥ 2, M(k) = 2k−3 − 1,

and that each of the enumerated LFDs has two nodes with negative divergence. We show that

M(k + 1) = 2k−2 − 1.

Note that, for every M -enumerated LFD of degree d, we can remove Vd and a corresponding

Figure 3: An example of the case with edges VjVj+1 for 1 ≤ j ≤ k and div(Vi) = 1 for some i ≤ k

8

edge, and we form either an M -enumerated LFD of degree d − 1 or a multiplicity-free LFD with

edges ViVi+1 for 1 ≤ i ≤ d− 2. Thus, these two types of graphs generate all M -enumerated LFD

of degree d. Conversely, given any M -enumerated LFD of degree k, we can add a node Vk+1 and

add an edge from an initial node of the original LFD to Vk+1. We produce an M -enumerated LFD

only if this initial node has negative divergence. There are two nodes with negative divergence in

the original LFD, so for eachM -enumerated LFD of degree k, twoM -enumerated LFDs of degree

k + 1 can be generated. Additionally, we generate one M -enumerated LFD of degree k + 1 from

a multiplicity-free LFD with edges ViVi+1 for 1 ≤ i ≤ k − 1 by adding a node on the right of the

initial LFD. The new LFD has the edge V2Vk+1 and the edges ViVi+1 for all 1 ≤ i ≤ k − 1.

Thus, for d ≥ 4, M(d+ 1) = 2M(d) + 1. Since M(4) = 1, we find that M(d) = 2d−3 − 1.

We must also count LFDs that can be formed when new nodes are appended to the left of

existent M -enumerated LFDs of lesser degrees, so that there are edges between every pair of

adjacent new nodes. Then the new leftmost node has divergence 1 while the original leftmost node

has divergence 0. These LFDs are thus equinumerous to M -enumerated LFDs of lesser degrees.

For example, the LFD with edge set {V1V2, V2V3, V3V4, V3V5} has degree 5 and corresponds to the

M -enumerated, degree-4 LFD with edge set {V1V2, V2V3, V2V4} (see Figures 4 and 5).

Figure 4: An M -enumerated LFD with nodes appended to the left.

Figure 5: The corresponding M -enumerated LFD with degree d− 1.

There are LFDs of degree d that correspond to M -enumerated, degree d − i LFDs for each

i ≤ d − 3, so in fact the number of LFDs with edges VjVj+1 for 1 ≤ j ≤ k and div(Vk) = 1 is

9

equal to M(3) +M(4) + · · ·+M(d− 1) +M(d). We find the closed formula

M(d) +M(d− 1) + · · ·+M(3) +M(2) = (2d−3 − 1) + (2d−4 − 1) + · · ·+ (20 − 1)

= (2d−3 + 2d−4 + · · ·+ 20)− (d− 2)

= (2d−2 − 1)− (d− 2)

= 2d−2 − d+ 1.

2.2 Case 2: The other node of divergence 1 has one outgoing edge and no

incoming edges

Now we consider LFDs in which the two nodes with divergence of 1 have one edge exiting and no

edges entering; e.g. Figure 6.

Suppose these two nodes are V1 and V2. We remove V1 and its corresponding edge. Then the

remaining diagram is either a multiplicity-free LFD with edges ViVi+1 for 1 ≤ i ≤ d− 2 or it is a

diagram enumerated by ωd−1,d−3. Thus, we can conversely generate all degree-d LFDs of Case 2

satisfying div(V1) = div(V2) = 1 starting from these two situations.

We generate such a Case 2 LFD from the multiplicity-free LFD with edges ViVi+1 for 1 ≤ i ≤

d− 2 as follows. Add a node to the left of the original LFD, and renumber the nodes. Then, add an

edge from V1 to Vx where x is any integer satisfying 3 ≤ x ≤ d. There are d− 2 choices.

We also generate Case 2 LFDs with div(V1) = div(V2) = 1 from the LFDs enumerated by

ωd−1,d−3 as follows. Add a node to the left of the original LFD, and renumber the nodes. There

is no edge V1V2; or else we do not generate a diagram with V1, V2 each having one edge exiting

Figure 6: An example of the case in which the other node of divergence 1 has one outgoing edge and noincoming edges

10

and no edges entering. We now must add an edge from V1 to another node. If there is no edge

V1V3, then there will be three nodes of divergence 1 — namely, V1, V2, V3. Thus, we must have the

edge V1V3. Then for each graph enumerated by ωd−1,d−3, we can generate one graph enumerated

by ωd,d−2.

The two nodes with divergence 1 need not be V1 and V2. One must be V1. If Vβ is the other, it

is necessarily true that there are edges between each pair of adjacent nodes from V1, V2, . . . , Vβ−1.

Then we can consider these nodes to be appended to the left of a Case 2 LFD of lesser degree.

Thus, Case 2 consists of [(d−2) +ωd−1,d−3] + [(d−3) +ωd−2,d−4] + · · ·+ [2 +ω3,1] + [1 +ω2,0] =

(d− 2)(d− 1)

2+

d−3∑i=0

ωi+2,i LFDs.

2.3 Closed form

These two major cases determine a recursion between ωd,d−2 and ωd−1,d−3. Namely, we have

ωd,d−2 = (2d−2 − d+ 1)︸ ︷︷ ︸Case 1

+(d− 2)(d− 1)

2+ ωd−1,d−3 + ωd−2,d−4 + · · ·+ ω3,1︸ ︷︷ ︸

Case 2

.

Now, we prove that ωd,d−2 = 2d−3d−d+1 through induction. The base case of d = 3 is trivial:

there is one graph with edge set {V1V3, V2V3}.

Suppose that ωd,d−2 = 2d−3d− d+ 1 when d = ` for some natural number ` ≥ 3. Then, by our

recursion, we know

ω`+1,`−1 = 2`−1 − (`+ 1) + 1 +(`− 1)(`)

2+ ω`,`−2 + ω`−1,`−3 + · · ·+ ω3,1.

11

We also know that

ω`,`−2 = 2`−2 − `+ 1 +(`− 2)(`− 1)

2+ ω`−1,`−3 + ω`−2,`−4 + · · ·+ ω3,1

ω`,`−2 − 2`−2 + `− 1− (`− 2)(`− 1)

2= ω`−1,`−3 + ω`−2,`−4 + · · ·+ ω3,1.

Then we can use substitution to obtain

ω`+1,`−1 = 2`−1 − (`+ 1) + 1 +(`− 1)(`)

2+ ω`,`−2 + ω`,`−2 − 2`−2 + `− 1− (`− 2)(`− 1)

2

= 2ω`,`−2 + 2`−2 + `− 2.

By our inductive hypothesis, this is equal to

2(2`−3`− `+ 1) + 2`−2 + `− 2

= 2(`+1)−3(`+ 1)− (`+ 1) + 1.

We make the replacement d = `+1 to see ωd,d−2 = 2d−3d−d+1, so our induction is complete,

and the statement is true.

3 Enumerating ωd,2

Theorem 2. ωd,2 = 2d−4(d2 − 7d+ 22)− d− 1.

Proof.

Lemma 1. There are(k−12

)multiplicity-free LFDs of degree d and genus 0 that have 2 nodes of

nonpositive divergence and exactly one node of divergence equal to 0.

Proof. If Vi is the node such that div(Vi) = 0, then all other nodes have an edge to Vd except for a

node Vj , where j < i, and Vj has an edge to Vi. Then j ranges from 1 to i− 1, giving i− 1 graphs

12

Figure 7: An LFD described by Lemma 1.

for each value of i. Since i ranges from 2 to d−1, the number of multiplicity-free LFDs of degree d

and genus 0 that have 2 nodes of nonpositive divergence and exactly one node of divergence equal

to 0 is equal tod−1∑i=2

i− 1 =

(d− 1

2

).

Definition 1. A disjoint node is a node that has no outgoing or incoming edges.

Suppose we have a multiplicity-free LFD of degree d+ 1 and genus 0 with 2 nodes of nonpos-

itive divergence. Then it is counted by ωd+1,2. We can remove Vd+1 and all its edges, and we are

left with two cases. We prove that either the remaining set of nodes and edges consists of an LFD

counted by ωx,1 and several disjoint nodes, or it consists of an LFD counted by ωy,2 and several

disjoint nodes, for some integers x, y ≤ d− 1.

If the remaining set of nodes and edges contains an LFD counted by ωj,z where z > 2, then

replacing the node Vd+1 would result in a diagram with more than d − 1 nodes of divergence 1.

Since there must be d− 1 nodes of divergence 1, this is a contradiction.

The remaining set of nodes and edges contains either exactly one diagram counted by ωx,1

where x ≥ 2 or exactly one diagram counted by ωy,2 where y ≥ 3. If the remaining set contains

more than one of these diagrams, then replacing the node Vd+1 would result in a diagram with at

least d nodes of divergence 1; this is a contradiction.

3.1 Case 1: Remaining set contains an ωx,1-graph

In this case, for each possible value of s, we can choose s nodes from the d available nodes to place

an ωs,1-graph. Then there is one way to make an edge from the ωs,1-graph to Vd+1, and one way

13

Figure 8: The remaining set contains an ωx,1-graph.

Figure 9: The generated LFD.

to connect each of the disjoint nodes to Vd+1. Since s ranges from 2 to d, the number of diagrams

generated from this case is equal to

d∑s=2

(d

s

)= 2d − d− 1.

3.2 Case 2: Remaining set contains an ωy,2-graph

If the ωy,2-graph does not have a node with divergence equal to 0, then it must necessarily have two

nodes of negative divergence. Attaching an edge to a node with negative divergence would generate

an ωz,3-graph. Then the only ωy,2-graphs that generate multiplicity-free labeled floor diagrams of

degree d + 1 and genus 0 with 2 nodes of nonpositive divergence are those with one node of

divergence equal to 0. Since the sum of the divergences of all nodes is equal to 0 (because each

edge contributes 1 to the divergence of its left endpoint and −1 to the divergence of its right

endpoint), an ωy,2-graph can only have at most one node of divergence zero; if there are two

nodes of divergence zero, there are no other nodes with nonpositive divergence, so the sum of the

divergences of all nodes would be negative. Then each graph with a node of divergence zero can

generate one diagram.

Lemma 1 states that there are(b−12

)multiplicity-free labeled floor diagrams of degree b and

genus 0 with exactly one node with divergence zero. For each b we can choose b points to place

the ωy,2-graph on, so for each b we can generate(b−12

)(db

)graphs. Since b ranges from 3 to d, the

14

number of graphs that can be generated by this case is equal to

d∑b=3

(b− 1

2

)(d

b

)=2d−2

((d− 2

2

)+ 1

)− 1.

3.3 Closed form

We total the number of graphs that can be generated by the cases to find the total number of

diagrams.

The number of diagrams of degree d+ 1 is equal to

(2d − d− 1) +

(2d−2

[(d− 2

2

)+ 1

]− 1

)= 2d − d− 1 + 2d−2

(d− 2

2

)+ 2d−2 − 1

= 2d−2

[(d− 2)(d− 3)

2+ 5

]− d− 2

= 2d−2

[d2 − 5d+ 16

2

]− d− 2

= 2d−3(d2 − 5d+ 16)− d− 2.

Then we have the closed form

ωd,2 = 2d−4(d2 − 7d+ 22)− d− 1.

4 Holistic divergence vectors and the Partition function

In an LFD, every edge serves as an outgoing edge from one node and an incoming edge from

another node. Thus, its contribution to the sum of the divergences of each node in the LFD is zero.

Therefore,

15

Theorem 3. The sum of the divergences of the nodes in any LFD is equal to zero.

How many vectors are there so that, if we interpret them as holistic divergence vectors, they

satisfy Theorem 3 and they have d nodes? We later refer to these two conditions as the holistic

divergence vector conditions.

Theorem 4. The number of vectors ~v that, when interpreted as holistic divergence vectors with d

nodes, satisfy Theorem 3, is equal to P (d)− 1, where P is the partition function.

Proof. Suppose that the vector ~v is [v1, v0, v−1, v−2, . . .]. Then we know

v1 + v0 + v−1 + . . . = d

(1)v1 + (0)v0 + (−1)v( − 1) + . . . = 0.

If we subtract the second equation from the first, we get

(1)v0 + (2)v−1 + (3)v−2 + . . . = d

The vi are nonnegative integer variables. Then this becomes a familiar representation of the

partition function [4]; we partition d into v0 1’s, v−1 2’s, and so on. Then the number of ~v is

equal to the partition function of d. However, in a labeled floor diagram, the first node must have

divergence 1, so we must discount ~v = [0, d, 0, 0, . . .]. Then the number of possible vectors that

can serve as holistic divergence vectors for an LFD is equal to P (d)− 1.

Note that, conversely, all holistic divergence vectors satisfy Theorem 3.

5 Node divergence vectors and the Catalan numbers

We introduce the node divergence vector. A node divergence vector ~u of an LFD has as its j th entry

the divergence of the j th node from the left. Let us define three conditions, which we later refer to

16

as the node divergence vector conditions.

1. The sum of the entries of ~u equals zero.

2. The sum of the first k entries of ~u is a positive integer for all k < d.

3. Each entry of ~u is an integer, at most one.

Any LFD’s node divergence vector must satisfy these three conditions. The first is from The-

orem 3. The third condition is the divergence condition. The second comes from the following

observations:

1. Any edges between the first k nodes of an LFD contribute zero to their total divergence sum.

2. Any edges from one of these first nodes to a node outside this group of nodes must be

outgoing from a node in the group, and thus contributes one to the group’s divergence sum.

3. There are no nodes before the first k nodes, so there are no incoming edge contributions.

Theorem 5. The number of vectors ~u that satisfy Conditions 1, 2, and 3 is equal to C(d − 1),

where C is the Catalan function.

Proof. Let the entries of ~u be [u1, u2, . . . , ud−1, ud]. Then, by the divergence condition, we know

1 ≥ u1 ≥ 1

2 ≥ u1 + u2 ≥ 1

3 ≥ u1 + u2 + u3 ≥ 1

... ≥ ... ≥ ...d− 1 ≥ u1 + u2 + · · ·+ ud−1 ≥ 1.

We also know

0 = u1 + u2 + · · ·+ ud.

17

Let µi = 1− ui. The above inequalities and equation then become

0 ≥ µ1 ≥ 0

1 ≥ µ1 + µ2 ≥ 0

2 ≥ µ1 + µ2 + µ3 ≥ 0

... ≥ ... ≥ ...d− 2 ≥ µ1 + µ2 + · · ·+ µd−1 ≥ 0

d = µ1 + µ2 + · · ·+ µd.

Notice that for any j, by the divergence condition, uj ≤ 1, so we know µj ≥ 0. Thus we can

strengthen the inequalities to

0 ≥ µ1 ≥ 0

1 ≥ µ1 + µ2 ≥ µ1

2 ≥ µ1 + µ2 + µ3 ≥ µ1 + µ2

... ≥ ... ≥ ...d− 2 ≥ µ1 + µ2 + · · ·+ µd−1 ≥ µ1 + µ2 + · · ·+ µd−2.

Imagine a d × d grid. For every configuration of the µk, we draw the unique points (0, µ1),

(1, µ1 +µ2), (2, µ1 +µ2 +µ3), . . ., (d− 2, µ1 +µ2 + · · ·+µd−1), and (d− 1, d− 1). Then we can

draw a monotonic path through these points [5], since the y-coordinates are non-decreasing as the

x-coordinates increase, and each point satisfies y ≤ x.

Figure 10: Monotonic paths, counted by the Catalan numbers. In this case, C(4) counts the number ofmonotonic paths on a 4× 4 grid.

Monotonic paths are counted by the Catalan numbers, so we have determined that the number

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of configurations of the µk, which is clearly equal to the number of configurations of the uk, is

equal to C(d− 1).

6 Construction of labeled floor diagrams from node divergence

vectors

We culminate our discussion of LFD classification by showing that there is an LFD for every node

divergence vector, and furthermore for every holistic divergence vector. We use strong induction.

Proof. For a base case, it is trivial to show that the vector [1,−1] can be the node divergence vector

for an LFD. Now, suppose that for some k ∈ N, each vector with at most k−1 entries that satisfies

the three node divergence vector conditions corresponds to at least one LFD. Then we examine

vectors that satisfy the three node divergence vector conditions with k entries.

Examine the penultimate entry of any such vector. The entry must be an integer at most 1.

If the penultimate entry is equal to 1, create a new vector by removing the penultimate entry

from the original and adding 1 to the final entry. This creates a vector with k−1 entries that satisfies

the three node divergence vector conditions, so this new vector corresponds to at least one LFD.

Take any one of the corresponding LFDs and construct a node between the k− 2 and k− 1 nodes.

Construct the edge from the new node to the k − 1 node. Then we are left with an LFD with k

nodes that corresponds to our original node divergence vector.

If the penultimate entry is equal to 0, create a new vector by removing the penultimate entry

from the original. This creates a vector with k − 1 entries that satisfies the three node divergence

vector conditions, so this new vector corresponds to at least one LFD. Examine any one of the

corresponding LFDs. There is at least one edge going into the k − 1 node; pick one such edge,

and suppose it is outgoing from the j node. Construct a node between the k − 2 and k − 1 nodes.

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Remove the edge from the j node to the k − 1 node, and draw an edge from the j node to the

newly-constructed node. Draw an edge from the newly-constructed node to the k − 1 node. Then

we are left with an LFD with k nodes that corresponds to our original node divergence vector.

If the penultimate entry is equal to −n where n is a positive integer, then the corresponding

LFD, if it exists, has either n + 1 edges incoming to the penultimate node and one edge from the

penultimate node to the final node, or there are n edges incoming to the penultimate node and no

edge from the penultimate node to the final node.

If we can show a series of steps that respect the conditions on LFDs while transforming a

hypothetical LFD with k nodes into an LFD with k−1 nodes, then we can simply reverse the steps

to construct from an LFD with k nodes from a (not-necessarily unique) corresponding LFD with

k − 1 nodes.

If the corresponding LFD has n+1 edges incoming to the penultimate node and one edge from

the penultimate node to the final node, we take the hypothetical LFD and remove the edge from

the k − 1 node to the k node. For each other edge that was incoming to the k node, we redraw

them to be incoming to the k− 1 node. This step, when reversed, produces multiple corresponding

LFDs; hence, when we reverse the steps, the LFD we produce will not be unique, but we are not

concerned since we only wish to show that at least one original LFD exists. Now we remove the k

node. We are left with an LFD with k − 1 nodes, so we can reverse the steps to construct at least

one LFD with the original vector as its node divergence vector.

If the corresponding LFD has n edges incoming to the penultimate node and no edge from the

penultimate node to the final node, we take the hypothetical LFD, remove an edge incoming to

the k node, and redraw the remaining edges that were incoming to the k node to be incoming to

the k − 1 node instead. We remove the k node. Then we have an LFD with k − 1 nodes, so we

can reverse the steps to construct at least one LFD with the original vector as its node divergence

vector.

We have addressed every possible penultimate entry for a node divergence vector, so we have

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shown that any vector that satisfies the node divergence vector conditions corresponds to at least

one LFD. Then the number of node divergence vectors for LFDs of degree d is equal to C(d− 1),

since there are C(d− 1) vectors satisfying the node divergence vector conditions.

Furthermore, notice that from any node divergence vector, we can calculate the corresponding

LFD’s holistic divergence vector. Then, since any vector satisfying the holistic divergence vector

conditions corresponds to at least one node divergence vector, we can conclude that any vector

satisfying the holistic divergence vector conditions corresponds to at least one LFD. Then the

number of holistic divergence vectors for LFDs of degree d is equal to P (d), since there are P (d)

vectors satisfying the holistic divergence vector conditions.

7 Conclusion

Through decomposition, we have explored multiplicity-free labeled floor diagrams and their clas-

sifications. First, we used recursion to enumerate LFDs with certain conditions. Next, we examined

two novel classification methods for LFDs: holistic divergence vectors and node divergence vec-

tors. We were able to identify sets of conditions that would produce holistic divergence vectors

and node divergence vectors; namely, the holistic divergence vector conditions and the node diver-

gence vector conditions. Next, we enumerated the vectors that satisfied each set of conditions. We

then showed that any vector that satisfies the holistic divergence vector conditions corresponds to a

labeled floor diagram, and that any vector that satisfies the node divergence vector conditions cor-

responds to a labeled floor diagram. This implies that the two sets of conditions we described are

in fact definitions of holistic divergence vectors and node divergence vectors, and that our enumer-

ations of the two classes of vectors are actually enumerations of the number of classes for labeled

floor diagrams.

Future work may be done with these two methods of classification, now that we have deter-

mined the number of classes and a mathematically-rigorous definition for these classes. There are

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many areas to explore regarding holistic divergence vectors and node divergence vectors.

Problem 1. How many LFDs are in each class of holistic/node divergence vectors?

These two problems would effectively solve Fomin and Mikhalkin’s original problem [2], and

are thus particularly important.

Problem 2. Do holistic divergence vectors and/or node divergence vectors classify the correspond-

ing algebraic curves as well?

This problem approaches the practical side of the field, and may be interesting because it in-

volves both the combinatorics of labeled floor diagrams and the algebraic geometry of rational

curves in CP2. If a relation between classes of rational curves and holistic divergence vectors

and/or node divergence vectors is established, perhaps other strategies for enumerating the alge-

braic curves can be explored.

7.1 Applications in detail

Closed A model topological string theory requires six space-time dimensions. These dimensions

produce a symplectic manifold, and the world-sheets must be parameterizable by pseudoholomor-

phic curves with finite-dimensional moduli spaces. Since Gromov-Witten invariants are integrals

over these moduli spaces, they are also path integrals of the theory. Specifically, at genus g, the

free energy of the A model is the generating function of the genus-g Gromov-Witten invariants.

Therefore, the problem we explore may be able to determine the free energy of the A model at

genus 0, characterizing the trace paths and interactions of string particles. [6]

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8 Acknowledgments

First and foremost, I would like to thank my mentor, Mr. John Lesieutre from the Massachusetts

Institute of Technology, for his experience and expertise, for his encouragement, and for his dedi-

cation to helping me and teaching me.

I am also grateful to the Center for Excellence in Education, the Research Science Institute

(RSI), and the Massachusetts Institute of Technology and its Mathematics Department for provid-

ing for my research.

Thanks to Dr. John Rickert from the Rose-Hulman Institute of Technology. As my tutor at the

Research Science Institute, he helped me with proofreading, ideas, and presentation skills.

I would like to thank Professor Sergey Fomin from the University of Michigan for his corre-

spondence regarding his paper.

Thanks to Kartik Venkatram from the Massachusetts Institute of Technology for patiently edit-

ing this paper.

Thank you to Jacob Hurwitz from the Massachusetts Institute of Technology, who helped me

with editing, technology, and LATEX.

I would also like to thank Dr. Tanya Khovanova from the Massachusetts Institute of Technology

for her guidance as the main coordinator for RSI mathematics, as well as her assistance with this

paper.

Thanks also to Professor David Jerison from the Massachusetts Institute of Technology for

organizing the RSI mathematics program as general supervisor.

I am grateful to Mr. Steven Long and Mrs. Luella Chavez D’Angelo, Program Manager and

President of the Western Union Foundation, and to Mr. and Mrs. Fred Kunik for sponsoring my

work at RSI.

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arXiv:0812.3354v2 [math.AG], (2008/12/17)

[2] S. Fomin, G. Mikhalkin: Labeled Floor Diagrams for Plane Curves, arXiv:0906.3828v4[math.AG], (2010/01/18)

[3] A029768, OEIS: The Online Encyclopedia of Integer Sequences,http://www.research.att.com/˜njas/sequences/A029768, (2010/06/25)

[4] Wolfram Mathworld, Partition Function P,http://mathworld.wolfram.com/PartitionFunctionP.html

[5] C. Vasudev: Theory and Problems of Combinatorics, New Age International, 2005

[6] L. Surhone, M. Timpledon, S. Marseken: Topological String Theory: String Theory, Topolog-ical Quantum Field Theory, Supersymmetry, Chern-Simons Theory, Gromov-Witten Invari-ants, Mirror Symmetry, Introduction to M-theory, Quantum Topology, Betascript Publishers,2009

[7] G. Mikhalkin: Enumerative Tropical Algebraic Geometry in R2, arXiv:math/0312530v4[math.AG], (2010/07/20)

[8] F. Bergeron, G. Labelle, P. Leroux: Combinatorial Species and Tree-like Structures, Cam-bridge University Press, 1998

[9] E. Brugalle, G. Mikhalkin: Enumeration of Curves via Floor Diagrams, arXiv:0706.0083[math.AG], (2007/06/01)

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