ALGEBRAIC COMBINATORICS ON PARTIALLY ORDERED SETS AND GRAPHS

By

Jamie Kimble

A DISSERTATION

Submitted to
Michigan State University
in partial fulfillment of the requirements
for the degree of

Mathematics—Doctor of Philosophy

2025

ABSTRACT

This thesis considers three algebraically motivated combinatorics questions on partially ordered

sets (posets) and graphs. In the process, we consider rooted tree posets, inflated rooted tree posets,

shoelace posets, (3 + 1)-free posets, as well as incomparability graphs of a given poset.

Rooted trees are posets whose Hasse diagram is a graph-theoretic tree having a unique minimal

element. We study rowmotion on antichains and lower order ideals of rooted trees. Recently

Elizalde, Roby, Plante, and Sagan considered rowmotion on fences which are posets whose Hasse

diagram is a path (but permitting any number of minimal elements). They showed that in this case,

the orbits could be described in terms of tilings of a cylinder. They also defined a new notion

called homometry where a statistic takes a constant value on all orbits of the same size. This is a

weaker condition than the well-studied concept of homomesy which requires a constant value for

the average of the statistic over all orbits. Rowmotion on fences is often homometric for certain

statistics, but not homomesic. We introduce a tiling model for rowmotion on rooted trees. We

use it to study various specific types of trees and show that they exhibit homometry, although not

homomesy, for certain statistics.

We also study Defant and Kravitz’s generalization of Schützenberger’s promotion operator to

arbitrary labelings of finite posets. Defant and Kravitz showed that applying the promotion operator

𝑛 − 1 times to a labeling of a poset on 𝑛 elements always gives a natural labeling of the poset and

called a labeling tangled if it requires the full 𝑛 − 1 promotions to reach a natural labeling. They

also conjectured that there are at most (𝑛 − 1)! tangled labelings for any poset on 𝑛 elements.

We propose a strengthening of their conjecture by partitioning tangled labelings according to the

element labeled 𝑛 − 1 and prove that this stronger conjecture holds for inflated rooted forest posets

and a new class of posets called shoelace posets. We also introduce sorting generating functions

and cumulative generating functions for the number of labelings that require 𝑘 applications of the

promotion operator to give a natural labeling. We prove that the coefficients of the cumulative

generating function of the ordinal sum of antichains are log-concave and obtain a refinement of the

weak order on the symmetric group.

We also consider (3 + 1)-free posets, motivated by a reduction of the Stanley-Stembridge

conjecture posited by Foley, Hoàng, and Merkel (2019), stating that the twinning operation on

graphs preserves 𝑒-positivity of the chromatic symmetric function. A counterexample to this

general conjecture was given by Li, Li, Wang, and Yang (2021). We prove that 𝑒-positivity is

preserved by the twinning operation on cycles, by giving an 𝑒-positive generating function for the

chromatic symmetric function, as well as an 𝑒-positive recurrence. We derive similar 𝑒-positive

generating functions and recurrences for twins of paths. Our methods make use of the important

triple deletion formulas of Orellana and Scott (2014), as well as new symmetric function identities.

Copyright by
JAMIE KIMBLE
2025

ACKNOWLEDGEMENTS

I want to start by thanking Dr. Bruce E. Sagan for taking me on as his ultimate student. You

rekindled my love of math and have provided steadfast support for me the last four years. Your

compassion, kindness, and absolute dedication to everything you do inspire me to be a better person

and mathematician. You are the best advisor a student could ask for, and I am eternally grateful for

your presence in my life.

To my committee members, Drs. Teena Gerhardt, Peter Magyar, Michael Shapiro, and Avery

St. Dizier, thank you for your kind words of encouragement and advice. Thank you especially to

Dr. St. Dizier for the many coffee shop collaborations.

An enormous thank you to my collaborators: Pranjal Dangwal, Zach Stewart, Herman Chau,

Mark Denker, Owen Goff, Yi-Lin Lee, Megan Chang-Lee, John Lentfer, and Drs. Jianzhi Lou,

Margaret Bayer, Kyle Celano, Laura Colmenarejo, Esther Banaian, and Sheila Sundaram. Working

with you all proved to me that mathematics is simply more fun when explored with others. This

dissertation would not exist without each and every one of you, and I am so happy to have had the

privilege of working with you.

To Dr. Jinting Liang, my academic sister and collaborator, thank you for your constant en-

couragement and unwavering enthusiasm. I am a better mathematician because of your faith in

me.

Thank you to all of my friends. To my cohort, thank you for getting me through first year. It

was arguably one of the most difficult years of my life, and I would quite literally not be here if

not for your help. Aldo, Owen, Val, Gokul, Nick, David, Aaron, and Ivan, even though we were

scattered across the world, we managed to find ways to support each other. I am proud to have

you all as friends. To Lys, Brianna, Wyn, Colton, and Kenzie, thank you for the virtual parallel

work sessions and constant support, and specifically to Lys for a second set of critical eyes. To all

of the countless others who have lifted me up these last five years, thank you for the coffeeshop

study dates, impromptu vacations, and the inspiration I drew from you all. Your successes and

achievements consistently pushed me to be a better version of myself.

v

I started knitting and crocheting in 2020, when I started my Ph.D. In the past five years, I have

used more than 70,000 yards of yarn, making sweaters, hats, blankets, socks, and mittens as a very

effective creative outlet. Thank you to the folks at Woven Art Yarn Shop for welcoming me to East

Lansing with open arms. The Thursday night social stitch-ins were a staple in my life that helped

to ground me every week.

To my parents, John and Michelle, and my sister, Hopper, thank you for always encouraging

my curiosity and dreams. I went through many possible career paths, and you’ve supported each

one with equal passion. Thank you for giving me the freedom to figure myself out, and for your

unconditional love.

To my fiancé Cole, thank you for seeing me at every step of this journey, no matter how difficult,

and loving me through it. Thank you for reminding me that life outside of math is just as important

as the life within it.

Portions of Chapter 2 appear in “Dangwal, P., Kimble, J., Liang, J., Lou, J., Sagan, B.E., &

Stewart, Z. (2023). Rowmotion on Rooted Trees. Séminaire Lothringien de Combinatoire.” [18]

and “Dangwal, P., Kimble, J., Liang, J., Lou, J., Sagan, B.E., & Stewart, Z. (2022). Rowmotion on

Rooted Trees.” [17] and are reprinted here under a CC BY 4.0 License.

Portions of Chapter 3 appear in “Bayer, M., Chau, H., Denker, M., Goff, O., Kimble, J., Lee,

Y., & Liang, J. (2024). Promotion, Tangled Labelings, and Sorting Generating Functions.” [6] and

are reprinted here under a perpetual, non-exclusive ArXiV 1.0 License.

Portions of Chapter 4 appear in “Banaian, E., Celano, K., Chang-Lee, M., Colmenarejo, L.,

Goff, O., Kimble, J., Kimpel, L., Lentfer, J., Liang, J., & Sundaram, S. (2024). The 𝑒-Positivity

of the Chromatic Symmetric Function for Twinned Paths and Cycles.” [5] and are reprinted here

under a perpetual, non-exclusive ArXiv 1.0 License.

vi

CHAPTER 1

INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . . . .

.

1

TABLE OF CONTENTS

CHAPTER 2

.
.
. .

6
ROWMOTION ON ROOTED TREES . . . . . . . . . . . . . . . . . . .
.
7
.
.
2.1 Tilings .
. 14
2.2 Stars .
.
. .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
2.3 Trees with Three Leaves
2.4 Combs and Zippers .
. 24
.
2.5 Comments and Open Questions . . . . . . . . . . . . . . . . . . . . . . . . . . 29

. . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . .

.
.
. .

.
.

.
.

CHAPTER 3

EXTENDED PROMOTION . . . . . . . . . . . . . . . . . . . . . . . . 34
3.1 Definitions and Properties of Extended Promotion . . . . . . . . . . . . . . . . 36
Inflated Rooted Forest Posets . . . . . . . . . . . . . . . . . . . . . . . . . . . 44
3.2
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52
3.3 Shoelace Posets .
3.4 Generating Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58
3.5 Ordinal Sum of Antichains . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68
. 76
3.6 Future Work .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. .

.

.

.

.

.

.

CHAPTER 4

TWINNING AND THE CHROMATIC SYMMETRIC FUNCTION . . . 77
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80
𝑒-positivity via Generating Functions . . . . . . . . . . . . . . . . . . . . . . . 87
𝑒-positivity via Recurrences
. . . . . . . . . . . . . . . . . . . . . . . . . . . 108
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118

4.1 Preliminaries
4.2
4.3
4.4 Future Directions . .

. .

.

.

BIBLIOGRAPHY .

.

.

.

.

. .

. .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119

vii

CHAPTER 1

INTRODUCTION

This dissertation explores several attributes of enumerative and algebraic combinatorics, on both

partially ordered sets (posets) and graphs. A brief outline of the chapters is as follows. We start by

exploring posets through an algebraic lens, focusing on a group action called rowmotion and how

the orbits of this action can demonstrate nice combinatorial properties when examining a particular

family of posets. Then, we shift to studying an operation called extended promotion on labelings

of particular families of posets, strengthening a conjecture posited last year. Finally, we transition

to more questions related to graphs, motivated by incomparability graphs of posets. We evaluate

how the twinning operation on graphs affects the 𝑒-positivity of a graph, determining several new

results in a rich field of combinatorics. We now give a more detailed outline of each chapter.

In Chapter 2, we will start by investigating the group action rowmotion on rooted tree posets,

analyzing the orbits for particular combinatorial properties known as homomesy and homometry.

The action of rowmotion has been rediscovered and renamed many times, appearing in the literature

as the Fon-der-Flaass action [42], the Panyushev action and complement [4, 7], among other

references [1, 9, 21, 26, 52, 38]. We will follow the conventions of Striker and Williams in [56] and

refer to the action as rowmotion due to its nature of working across “rows” of posets. This action

relates to many other mathematical objects, such as flag simplicial complexes, representation finite

algebras, trim lattices, Auslander–Reiten translation on certain quivers, Zamolodchikov periodicity,

and totally symmetric self-complementary plane partitions, among many others [36] [57][56]. We

will consider the action on both antichains and lower order ideals of rooted trees, specifically

searching for examples of homomesy and homometry under different statistics.

Homomesy has been studied rather extensively in recent years; the term was coined in 2013

by Jim Propp and Tom Roby [39], but an example was conjectured in 2009 by Panyushev and

proven in 2013 by Armstrong, Stump, and Thomas [38][4]. A statistic on a combinatorial object

is homomesic if, for a given group action on these objects, the average value of the statistic is the

same over all orbits. In this chapter, we use rowmotion as our group action and examine cardinality

1

statistics on rooted tree posets. Furthermore, we dive into the newer, more broad phenomenon

known as homometry. Given a group action on a set of combinatorial objects, a statistic on these

objects is considered homometric if its value is the same over all orbits of the same cardinality.

Elizalde, Roby, Plante, and Sagan introduced this new concept in 2021 and found many examples

of homometry demonstrated by the cardinality statistic on orbits of rowmotion on fence posets [23].

We present several new results concerning homomesy and homometry by restricting ourselves

to a well-known family of posets. A tree can be defined as a graph in which any two vertices are

connected by exactly one path. A rooted tree is a tree where one vertex is designated as the “root.”

Rooted trees have been well-studied in graph theory, dating as far back as 1857 [11]. A rooted tree

can be specialized further by assigning the edges of the rooted tree with a natural orientation, either

towards or away from this root. More recently, these types of trees have heavily influenced data

science methods through decision trees, tree data structures, and mathematical modeling, among

other applications. Orienting the edges of a rooted tree naturally turns our graph into a poset, where

the edges represent the covering relations of the partial order and the vertices represent the objects

of that partial order. In Chapter 2, we will orient our rooted trees away from the root, where the

root becomes the minimum element in our poset. This is called an arborescence, or out-tree. In

our case, we will just call them rooted trees.

We conclude that for rooted tree posets, we can visualize the orbits of rowmotion by using a

tiling model, and we can therefore count cardinality statistics using that tiling model. This tool

allows us to present several natural homometry results concerning specific families of rooted tree

posets, such as stars, trees with three leaves, combs, and zippers. We also provide an example of a

rooted tree where the cardinality statistics do not exhibit any homometry or homomesy.

In Chapter 3, we transition to an investigation of extended promotion on labelings of posets. A

labeling of a poset 𝑃 with 𝑛 elements is a bijection from 𝑃 to {1, 2, . . . , 𝑛}. A labeling is considered

natural if it respects the partial order of 𝑃.

In 1972, Schützenberger introduced the promotion

operator on natural labelings of posets [44]. The motivation for the promotion operator comes from

an earlier paper of Schützenberger [43], in which he defines a related operator, evacuation, to study

2

the celebrated RSK algorithm. Promotion and evacuation were subsequently studied by Stanley

in relation to Hecke algebra products [52], by Rhoades in relation to cyclic sieving phenomena

[40], and by Striker and Williams in relation to rowmotion and alternating sign matrices [56].

Traditionally, promotion was only considered on the set of natural labelings of posets. In their

2023 paper, Defant and Kravitz introduced the notion of extended promotion, which acts on the

set of all labelings of a poset [19]. They determined that extended promotion will eventually turn

any labeling of a poset into a natural labeling, sorting the labeling with respect to the partial order.

They showed that any labeling of an 𝑛 element poset will become a natural after 𝑛 − 1 applications

of extended promotion, and we call labelings that take exactly that long tangled. They conjectured

that for a poset 𝑃 on 𝑛 elements, there are no more than (𝑛 − 1)! tangled labelings. They proved

this conjecture for inflated rooted forests, which is a large class of posets related to rooted trees. In

this case, we will orient the edges of a rooted tree towards the root. This orientation transforms the

root into the maximum element in our poset, an example of an anti-arborescence, or in-tree [20].

We will also refer to these as rooted trees, with orientation made clear by context.

We refine this conjecture and prove our refinement for both inflated rooted forests as well as

a new family of posets, called shoelaces. Additionally, we follow the lead of both [32] and [19]

in investigating properties of the sorting time of various labelings. We count labelings by the

number of extended promotion steps needed to yield a natural labeling, and we define two related

generating functions on 𝑃 in order to examine how these generating functions change if we attach

some minimal elements to 𝑃. Our result provides a simple and unified proof of enumerating tangled

labelings and quasi-tangled labelings in [19] and [32].

In Chapter 4, we will shift our attention to a different type of generating function related to a

graph called its chromatic symmetric function, and examine the twinning operation in relation to

this formal power series. To fully introduce the chromatic symmetric function, we must start by

defining a proper coloring of a graph. A graph 𝐺 = (𝑉, 𝐸) with vertex set 𝑉 and edge set 𝐸 is

colored when one assigns labels (called colors) to each vertex in 𝑉. A coloring on 𝐺 is proper

if no two vertices connected by an edge share the same color. The chromatic number of 𝐺 is the

3

smallest number of colors that can be used in a proper coloring. The chromatic number of a graph

is one of the most well-studied invariants in graph theory. The Four Color Theorem states that if

a graph 𝐺 can be drawn in the plane without any edge crossings, then its chromatic number is at

most four [3]. Famously, this theorem was a conjecture for over 100 years, and was one of the first

theorems proven by using extensive computer assistance.

The chromatic polynomial of a graph is a closely related function that enumerates proper

colorings of a graph. Birkhoff defined the chromatic polynomial, 𝑃(𝐺; 𝑡), to be the number of

proper colorings of 𝐺 with 𝑡 colors [8]. This polynomial has various properties that seem a bit

miraculous at first glance. For example, Stanley [50] proved theat if 𝐺 has 𝑛 vertices, then

𝑃(𝐺, −1) = (−1)𝑛 (the number of acyclic orientations of 𝐺)

It is not intuitively clear what it means to color a graph with −1 colors, but this result (among

others) implies a deep mathematical significance to the polynomial.

Generalizing the chromatic polynomial further, Stanley defined the chromatic symmetric func-

tion of a graph 𝐺 = (𝑉, 𝐸) to be

𝑋𝐺 (x) =

∑︁

(cid:214)

𝑥𝜅(𝑣),

𝜅
where x = {𝑥1, 𝑥2, . . .} is a countably infinite set of variables, and the sum is over all proper

𝑣∈𝑉

colorings 𝜅 : 𝑉 → Z>0 of 𝐺 by positive integers [47]. This made it possible to make new and

unexpected connections betweeen graph coloring, the theory of symmetric functions, and even

algebraic geometry (Hessenberg varieties). Stanley proved that the chromatic symmetric functions

of paths and cycles are 𝑒-positive, that is, their expansion in the basis of elementary symmetric

functions has nonnegative coefficients. The result for paths is originally due to Carlitz, Scoville, and

Vaughan in a different context [10, p.242]. More generally, much of the research on the chromatic

symmetric function has centered around the incomparability graph Inc(𝑃) of a (3 + 1)-free poset 𝑃,

defined as a poset containing no induced subposet isomorphic to the disjoint union of a 3-chain and

a 1-chain. This direction is motivated by the famous Stanley-Stembridge Conjecture, stating that

if 𝑃 is a (3 + 1)-free poset, then 𝑋Inc(𝑃) (x) is 𝑒-positive. This conjecture had been standing since

4

1993, though Hikita recently proved it in his preprint [31]. The work in Chapter 4 was completed

prior to the appearance of this proof.

Given a graph 𝐺 and a vertex 𝑣, the twin of 𝐺 at 𝑣 is the graph, denoted by 𝐺𝑣, obtained by

adding a new vertex 𝑣′ and connecting 𝑣′ to 𝑣 and to all of its neighbors. We refer to this operation

as the twinning of a graph and to the resulting graph 𝐺𝑣 as the twinned graph. Twinning is a natural

operation considered often in graph theory, usually aiding in evaluating graph isomorphisms and

subgraph inclusion. It is then reasonable to ask how twinning a graph might affect its chromatic

symmmetric function.

Specifically, we investigate the change in 𝑋𝐺 (x) when one twins a vertex 𝑣 of a graph 𝐺.

We determine explicit 𝑒-positive formulas for the generating function of the chromatic symmetric

function of four types of twinned graphs, as well as 𝑒-positive recurrence relations for five different

graph families.

5

CHAPTER 2

ROWMOTION ON ROOTED TREES

Let 𝑆 be a set with #𝑆 finite where the hash symbol denotes cardinality. A statistic on 𝑆 is a function

st : 𝑆 → Z where Z is the integers. We extend st to subsets 𝑅 ⊆ 𝑆 by letting

𝑟∈𝑅
Now suppose that 𝐺 is a finite group acting on 𝑆. Statistic st is said to be homomesic if, for any

st 𝑅 =

∑︁

st 𝑟.

orbit O of 𝐺, we have

st O
#O

= 𝑐

for some constant 𝑐. To be more specific, we say in this case that this statistic is 𝑐-mesic. Homomesy

is a well-studied property; see the survey articles of Roby [41] or Striker [55]. Recently Elizalde,

Roby, Plante, and Sagan [23] introduced a weaker notion which is exhibited by certain actions

and statistics. We say that a statistic st is homometric if for any two orbits O1 and O2 of the

same cardinality we have st O1 = st O2. We will see numerous examples of statistics which are

homometric but not homomesic in the present work.

Now consider a finite partially ordered set, often abbreviated to poset, (𝑃, ≤). An antichain of

𝑃 is a 𝐴 ⊆ 𝑃 such that no two elements of 𝐴 are comparable. We denote the set of all antichains as

A (𝑃) = {𝐴 ⊆ 𝑃 | 𝐴 is an antichain}.

A lower order ideal of 𝑃 is 𝐿 ⊆ 𝑃 such that if 𝑦 ∈ 𝐿 and 𝑥 ≤ 𝑦 then 𝑥 ∈ 𝐿. We will use the notation

L (𝑃) = {𝐿 ⊆ 𝑃 | 𝐿 is a lower order ideal}.

The lower order ideal generated by any 𝑄 ⊆ 𝑃 is

𝑄 ↓ = {𝑥 ∈ 𝑃 | 𝑥 ≤ 𝑦 for some 𝑦 ∈ 𝑄}.

We also let min 𝑄 and max 𝑄 be the sets of minimal and maximal elements of 𝑄, respectively. We

now define rowmotion on antichains to be the action generated by 𝜌 : A (𝑃) → A (𝑃) where

𝜌( 𝐴) = min{𝑥 ∉ ( 𝐴 ↓)}.

6

Similarly, rowmotion on ideals has generator ˆ𝜌 : L (𝑃) → L (𝑃) with

ˆ𝜌(𝐿) = 𝜌(max 𝐿) ↓ .

We will usually use a hat to distinguish a notation on ideals from the corresponding one on

antichains. More information about rowmotion can be found in the aforementioned survey articles.

The paper of Elizalde et al. was devoted to the study of rowmotion on fences. A fence is a

poset whose Hasse diagram is a path. They showed that the antichain orbits can be modeled using

certain tilings of a cylinder. This tool permitted them to prove a number of homometries which

were not homomesies. In the present work we will consider rowmotion on rooted trees. In this

chapter, we will orient our poset away from a minimum element. We consider a poset 𝑇 as a rooted

tree if its Hasse diagram is a tree in the graph theory sense of the term and it has a unique minimal

element called the root and denoted ˆ0. Note that these posets are more general than fences in that

the tree need not be a path, but also more restricted in that fences can have any number of minimal

elements. We will assume all our trees are rooted.

The rest of this chapter is structured as follows. In the next section we will show that rowmotion

on antichains of a rooted tree can also be viewed in terms of certain cylindrical tilings. The

following three sections will apply this tiling model to three different families of trees: stars, trees

with three leaves, and finally combs and zippers. We end with a section with comments and open

questions.

2.1 Tilings

We will show that rowmotion orbits on antichains can be more easily viewed as certain tilings

of a cylinder. Given a rooted tree 𝑇 we will fix an embedding of the Hasse diagram of 𝑇 in the

plane and label its leaves (maximal elements) as 1, 2, . . . , 𝑛 from left to right. See the tree on the

left of Figure 2.1 for an example where 𝑛 = 5.

For nonnegative integers 𝑚, 𝑛 we use interval notation

[𝑚, 𝑛] = {𝑚, 𝑚 + 1, . . . , 𝑛}

7

2

3

5

( [2, 2], 3)

( [3, 3], 2)

( [5, 5], 2)

1

4

𝑦

𝑥

𝑇

( [1, 1], 2)

( [4, 4], 1)

( [1, 2], 1)

𝑦 = 𝑥 [3,5],1

( [3, 5], 2)
𝑥 = 𝑥 [3,5],2

( [1, 5], 2)

I (𝑇)

Figure 2.1 The intervals, branches, and 𝛽-values of a tree 𝑇

and abbreviate [𝑛] = [1, 𝑛]. Associate with each vertex 𝑥 in 𝑇 the set of all labels of leaves 𝑧 such

that 𝑧 ≥ 𝑥. Note that by our choice of labeling, this set will be an interval 𝐼. And the set of all

𝑥 with interval 𝐼 form a path called the branch corresponding to 𝐼 and denoted 𝐵𝐼. On the right

in Figure 2.1, 𝑇 has been decomposed into branches with each labeled by a pair where the first

component is the interval 𝐼 of 𝐵𝐼. For example, nodes 𝑥 and 𝑦 are exactly the ones below all three

leaves 3, 4, 5. So their associated interval is 𝐼 = [3, 5] and 𝐵[3,5] = {𝑥, 𝑦}. We will also label the

vertices on the branch for 𝐼 as 𝑥𝐼,1, 𝑥𝐼,2, . . . starting with the maximal element and working down.

Returning to our example, 𝑦 = 𝑥 [3,5],1 and 𝑥 = 𝑥 [3,5],2. Note the following two simple but important

properties of this family of intervals.

(I1) The singleton intervals [𝑖, 𝑖] are in this family for all 𝑖 ∈ [𝑛].

(I2) The family is nested in the sense that if 𝐼, 𝐽 are in the family with #𝐼 ≤ #𝐽 then either 𝐼 ⊆ 𝐽

or 𝐼 ∩ 𝐽 = ∅.

Given an interval 𝐼, let

𝛽𝐼 = 𝛽𝐼 (𝑇) = #𝐵𝐼 .

8

2

3

5

1

𝑢

𝑣

4

𝑦

𝑥

1

2

3

4

5

𝜌
↦→

=

𝑇

{𝑢, 𝑥}

{𝑣, 𝑦}

Figure 2.2 Rowmotion on antichains in terms of tilings

Returning to our usual example, for 𝐼 = [3, 5] we saw that 𝐵[3,5] = {𝑥, 𝑦} which implies 𝛽[3,5] = 2.

A crucial tool in defining the tilings will be the set

I (𝑇) = {(𝐼, 𝛽𝐼) | 𝐼 is the interval of some branch of nodes in 𝑇 }.

On the right in Figure 2.1, the elements of I (𝑇) are displayed next to their corresponding branches.

We will abuse notation and write 𝐼 ∈ I (𝑇) to mean that (𝐼, 𝛽𝐼) ∈ I (𝑇).

We will need to consider partitions of intervals. A partition of an interval 𝐼 is a collection of

nonempty subintervals 𝐼1, . . . , 𝐼𝑘 whose disjoint union is 𝐼. We say that another partition 𝐽1, . . . , 𝐽𝑙

of 𝐼 is a refinement of the first if for every 𝐽 𝑗 there is an 𝐼𝑖 with 𝐽 𝑗 ⊆ 𝐼𝑖. The refinement is proper if

the two collections of subintervals are not the same. Refinement is a partial order on partitions. If

all the intervals of the partition come from I (𝑇) then it is called an I (𝑇)-partition.

We now describe the procedure to produce a tiling 𝜏 from an orbit O of rowmotion on antichains

of a rooted tree 𝑇. Consider a column of 𝑛 boxes where the 𝑖th box corresponds to the leaf labeled

𝑖 in the embedding of 𝑇. The first column in Figure 2.2 is so labeled. Given an antichain 𝐴, we

take each 𝑥 ∈ 𝐴 and consider the interval 𝐼 of its branch. The boxes labeled by the elements of 𝐼

are then covered by a black tile. All other boxes are covered by a single yellow tile. Note that these

boxes are exactly the ones in rows 𝑖 such that there is no element of 𝐴 below the leaf labeled 𝑖.

Returning to Figure 2.2, consider the antichain 𝐴 = {𝑢, 𝑥} and the leftmost column of tiles. Since

9

𝑢 has interval [1, 1], the box in row 1 gets a black tile. Similarly, 𝑥’s interval is [3, 5] so the rows

for this interval also receive a black tile. The remaining square in row 2 then receives a yellow

tile. The reader should now not find it hard to verify that 𝜌( 𝐴) = {𝑣, 𝑦} and that this antichain

corresponds to the second column in the figure. We now paste the columns for all antichains in the

orbit O together in the same order that they appear in the orbit to get a tiling 𝜏 = 𝜏(O) of a cylinder.

Note that when pasting, if there are two consecutive columns with black tiles coming from the same

interval 𝐼 then these tiles are combined into one. Returning to our perennial example, the two tiles

for 𝐼 = [3, 5] become one tile as seen in the final diagram. And if there were more elements on the

branch corresponding to 𝐼, they would fatten the tile further. Three tilings corresponding to full

orbits are shown in Figure 2.3. The vertical sides of these rectangles are identified to make them

into cylinders. Also note that, in the middle tiling, a black tile in the second row stretches over this

boundary as indicated by having it protrude beyond the sides of the rectangle.

We wish to characterize the possible 𝜏(O). In the definition below, an 𝐼 × 𝑏 tile is a tile which

covers the rows indexed by 𝐼 and 𝑏 columns. Also, the maximal partitions used are maximal with

respect to the refinement order. They exist because property (I1) implies that any interval 𝐼 has a

partition using intervals in I (𝑇) since all singletons are intervals. And property (I2) guarantees

that among all such partitions of 𝐼 there is a maximal one.

Definition 2.1.1. Given a rooted tree 𝑇, an I (𝑇)-tiling is a tiling of a cylinder using 𝐼 × 𝛽𝐼 black

tiles and 𝐼 × 1 yellow tiles if #𝐼 = 1, satisfying the following two properties.

(t1) An 𝐼 × 𝛽𝐼 black tile is followed by a yellow tile if #𝐼 = 1, or by black tiles corresponding to

the intervals in a maximal proper I (𝑇)-partition of 𝐼 if #𝐼 ≥ 2.

(t2) If 𝐽 is a maximal interval of yellow tiles in a column, then they are followed by black tiles

corresponding to the intervals in a maximal I (𝑇)-partition of 𝐽.

Theorem 2.1.2. Given a rooted tree, 𝑇, the map O ↦→ 𝜏(O) is a bijection between the antichain

rowmotion orbits of 𝑇 and the possible I (𝑇)-tilings.

10

Proof. We must first show that this map is well defined in that 𝜏 = 𝜏(O) has tiles satisfying (t1) and

(t2) and of the correct shape. We will do this by studying how rowmotion affects various elements

of 𝑇.

Consider 𝐴 ∈ O and any 𝑥 ∈ 𝐴 which is not maximal in its branch and let 𝐼 be the associated

interval. Then there is a unique element 𝑦 which covers 𝑥 and it is in the same branch. Furthermore

𝑦 ∈ 𝜌( 𝐴). Since 𝑥 and 𝑦 correspond to the same interval 𝐼, it follows that the tile covering those

rows in the column for 𝐴 extends into the column for 𝜌( 𝐴). By induction, this tile extends into a

column for an antichain containing the maximal element on the branch.

Now suppose that 𝑥 ∈ 𝐴 is maximal in its branch. If #𝐼 = 1 then 𝑥 is maximal in 𝑇. So in 𝜌( 𝐴)

the branch will be empty and the algorithm will place a yellow tile in the corresponding row and

column. This proves the first case in (t1). On the other hand, if #𝐼 ≥ 2 then 𝑥 is covered by at least

two elements 𝑦1, . . . , 𝑦𝑘 . So the column for 𝜌( 𝐴) will contain tiles in the corresponding intervals

𝐼1, . . . , 𝐼𝑘 which is a proper I (𝑇)-partition of 𝐼 since 𝑘 ≥ 2. And it is maximal since if there is

some 𝐽 ∈ I (𝑇) containing two or more of the 𝐼𝑖 then there would have to be at least one element

between 𝑥 and the corresponding 𝑦𝑖’s. This completes the proof of (t1).

For (t2), we will assume for simplicity that 1, 𝑛 ∉ 𝐽 where 𝑛 is the number of leaves of 𝑇. The

cases when 𝐽 contains one or both of these special values is similar. Say 𝐽 = [𝑚, 𝑛]. Then by our

assumption, there are black tiles covering rows 𝑚 − 1 and 𝑛 + 1 in the column for 𝐽. Let 𝑥 and 𝑦

be the corresponding elements of 𝐴. Removing the ˆ0–𝑥 and ˆ0–𝑦 paths from 𝑇 breaks the lower

order ideal generated by the leaves in 𝐽 into rooted subtrees. Let 𝑧1, . . . , 𝑧𝑘 be their roots with

corresponding intervals 𝐼1, . . . , 𝐼𝑘 . Then 𝜌( 𝐴) contains these 𝑧𝑖 and so its column contains tiles for

the intervals 𝐼𝑖 which form a partition of 𝐽. Maximality is obtained by the same argument as in the

previous paragraph.

To complete showing that 𝜏 is well defined, we must check the shape of the tiles. Yellow tiles

are of the correct shape by definition of the algorithm. As far as the black tiles, they cover rows

indexed by intervals in I (𝑇) by definition. So it suffices to show that a tile in the rows indexed by

𝐼 has the correct length. From the previous two paragraphs we see that the tiles in the partitions

11

following the maximal element of a black tile or following an interval of yellow tiles all begin with

the minimal elements of their respective branches. And by the second paragraph, such a tile will

extend to the maximal element on its branch. So the tile will have length 𝛽𝐼, the length of the

branch.

To show that this map is a bijection, we construct its inverse. So given an I (𝑇)-tiling 𝜏, we

must construct a corresponding orbit O. For each column of 𝜏 we form an antichain 𝐴 as follows.

For each interval 𝐼 covered by a black tile, suppose the given column is the 𝑖th in that tile. Then

add the 𝑖th smallest element on the branch for 𝐼 to 𝐴. Now arrange the antichains in the same

order as the columns of the tiling to get an orbit. The demonstration that this map is well defined

is similar to the one just given. And the two functions are inverses since the algorithms described

are step-by-step reversals. This completes the proof.

□

We will often call the tiles of shape 𝐼 × 𝛽𝐼 simply 𝐼-tiles. As a first application of the tiling

model, we will use it to compute various statistics on rowmotion orbits. It will also give us a simple

proof of our first homomesy. Given 𝑥 ∈ 𝑇 we have the statistic on antichains 𝐴 ∈ A (𝑇) given by

𝜒𝑥 ( 𝐴) =

1 if 𝑥 ∈ 𝐴,

0 if 𝑥 ∉ 𝐴.





If we want to count the size of antichains we use the statistic

𝜒( 𝐴) =

∑︁

𝑥∈𝑇

𝜒𝑥 ( 𝐴) = #𝐴.

The corresponding statistics for ideals are denoted ˆ𝜒𝑥 and ˆ𝜒. Given a I (𝑇)-tiling 𝜏 we will use the

notation

𝑚𝐼 = 𝑚𝐼 (𝜏) = number of 𝐼-tiles in 𝜏.

Corollary 2.1.3. Let 𝑇 be a rooted tree and 𝜏 be a I (𝑇)-tiling corresponding to a rowmotion orbit

O on 𝑇. The following hold.

(a) If 𝑥 ∈ 𝑇 has interval 𝐼 then

𝜒𝑥 (O) = 𝑚𝐼 .

12

(b) We have

(c) If 𝑥 = 𝑥𝐼, 𝑗 then

𝜒(O) =

∑︁

𝛽𝐼 𝑚𝐼 .

𝐼∈I (𝑇)

ˆ𝜒𝑥 (O) = 𝑗 · 𝑚𝐼 + 𝑐𝐼

where 𝑐𝐼 is the number of columns of 𝜏 intersecting a 𝐽-tile for 𝐽 ⊂ 𝐼.

(d) We have

ˆ𝜒(O) =

∑︁

𝐼∈I (𝑇)

(cid:19)

(cid:20) (cid:18)𝛽𝐼 + 1
2

(cid:21)

𝑚𝐼 + 𝛽𝐼 𝑐𝐼

(e) If 𝑥, 𝑦 are in the same branch then 𝜒𝑥 − 𝜒𝑦 is 0-mesic.

Proof. (a) This follows from the fact that 𝑣 is represented by a single column in each 𝐼-tile of 𝜏.

(b) Since 𝐼-tiles have length 𝛽𝐼 = #𝐵𝐼 we get by summing (a)

𝜒(O) =

∑︁

𝜒𝑥 (O)

𝑥∈𝑇
∑︁

𝐼∈I (𝑇)
∑︁

𝐼∈I (𝑇)

=

=

𝑚𝐼

∑︁

𝑥∈𝐵𝐼

𝛽𝐼 𝑚𝐼 .

(c) For a lower order ideal 𝐿 we have that 𝑥 ∈ 𝐿 if and only if 𝑥 ≤ 𝑦 for some 𝑦 ∈ 𝐴 where

𝐴 = max 𝐿. Note also that if 𝑦 has interval 𝐽 then 𝑦 ≥ 𝑥 implies 𝐽 ⊆ 𝐼. If 𝐽 = 𝐼 then there are 𝑗

choices for 𝑦 and so 𝑗 · 𝑚𝐼 counts the total number of columns containing such an element. And 𝑐𝐼

accounts for the columns intersection some 𝐽-tile where 𝐽 ⊂ 𝐼.

(d) This result follows from (c) in much the same way that (b) followed from (a). So the proof

is left to the reader.

(e) Let the common branch be 𝐵𝐼. Using (a) one last time we get

𝜒𝑥 (O) − 𝜒𝑦 (O) = 𝑚𝐼 − 𝑚𝐼 = 0

which implies the homomesy.

□

13

1

2

3

𝑆(3, 3, 2)

Figure 2.3 The star 𝑆(3, 3, 2) and its tilings

We end this section with a recursive formula for the number of antichains in a rooted tree 𝑇

which will be useful in the sequel. We use 𝑇 \ { ˆ0} for the forest of rooted trees obtained by removing

ˆ0 from 𝑇.

Lemma 2.1.4. Let 𝑇 be a rooted tree. If #𝑇 = 1 then #A (𝑇) = 2. If #𝑇 ≥ 2 then let 𝑇1, . . . , 𝑇𝑘 be

the rooted tree components of 𝑇 \ { ˆ0}. In this case

#A (𝑇) = 1 +

𝑘
(cid:214)

𝑖=1

#A (𝑇𝑖).

Proof. If #𝑇 = 1 then 𝑇 has antichains ∅ and { ˆ0}. When #𝑇 ≥ 2, let 𝐴 be an antichain of 𝑇. Either

𝐴 = { ˆ0}, corresponding to the 1 is the sum, or 𝐴 ⊆ ⊎𝑖𝑇𝑖. In the latter case the restriction 𝐴𝑖 of 𝐴

to 𝑇𝑖 is an antichain and the product counts the possible 𝐴𝑖.

□

2.2 Stars

A star, 𝑆, is a rooted tree with 𝑛 leaves and

I (𝑆) = {( [1, 1], 𝛽1), . . . , ( [𝑛, 𝑛], 𝛽𝑛), ( [𝑛], 1)}

where we are using the abbreviation 𝛽𝑖 = 𝛽[𝑖,𝑖]. We will use the same abbreviation for other notation

involving a subscript [𝑖, 𝑖], for example 𝑥𝑖, 𝑗 = 𝑥 [𝑖,𝑖], 𝑗 . So 𝑆 is the result of taking 𝑛 chains of length

𝛽1, . . . , 𝛽𝑛 and identifying their minimal elements. Note that all tiles in a corresponding tiling will

only cover one row, except for the tile corresponding to ˆ0. We denote this star by 𝑆(𝛼1, . . . , 𝛼𝑛)

where 𝛼𝑖 = 𝛽𝑖 + 1 for 𝑖 ∈ [𝑛]. The reason for this change of variables is because it will make our

results easier to state since 𝛼𝑖 is the length of a black tile followed by a yellow tile in row 𝑖. The

14

star 𝑆(3, 3, 2) and its tilings are found in Figure 2.3. Given an orbit O we will use the notation

𝛿 =





1 if ˆ0 ∈ O,

0 if ˆ0 ∉ O.

Theorem 2.2.1. Consider the star 𝑆 = 𝑆(𝛼1, . . . , 𝛼𝑛) and an orbit O of rowmotion on 𝑆. Let

𝑙 = lcm(𝛼1, . . . , 𝛼𝑛).

(a) We have

#O = 𝑙 + 𝛿

and the number of orbits is 𝛼1 · · · 𝛼𝑛/𝑙.

(b) For any 𝑥 ∈ 𝑆,

(c) We have

𝜒𝑥 (O) =

𝑙/𝛼𝑖

if 𝑥 ∈ 𝐵𝑖,

𝛿

if 𝑥 = ˆ0.





𝜒(O) = 𝛿 +

𝑛
∑︁

𝑖=1

𝑙
𝛼𝑖

(𝛼𝑖 − 1).

Thus 𝜒 is homometric but not homomesic.

(d) For any 𝑥 ∈ 𝑆

(e) We have

ˆ𝜒𝑥 (O) =

𝑗𝑙/𝛼𝑖

if 𝑥 = 𝑥𝑖, 𝑗

𝑙

if 𝑥 = ˆ0.





ˆ𝜒(O) = 𝑙 +

𝑛
∑︁

𝑖=1

(cid:19)

.

𝑙
𝛼𝑖

(cid:18)𝛼𝑖
2

Thus ˆ𝜒 is homometric but not homomesic.

Proof. (a) Consider the tiling 𝜏 = 𝜏(O). For all 𝑖 ∈ [𝑛] the corresponding interval 𝐼 = [𝑖, 𝑖] has

#𝐼 = 1. So, by condition (t1) in Definition 2.1.1, each black tile in that row is followed by a yellow

tile. And this pair of tiles has length 𝛽𝑖 + 1 = 𝛼𝑖.

15

Now consider the case when ˆ0 ∉ O. So no tile spans more than one row. Now the previous

paragraph and (t2) imply that the black and yellow tiles alternate in row 𝑖. So the length of that row

is divisible by 𝛼𝑖. Since this is true for all 𝑖 we must have that 𝑙 divides #O. But since 𝑙 is the least

common multiple, a given column will recur after 𝑙 steps. So we must have #O = 𝑙. When ˆ0 ∈ O

then the same reasoning as above applies to the tiling once the column for ˆ0 is removed. So in this

case #O = 𝑙 + 1.

Now let 𝑘 be the number of orbits. From what we have just proved, #A (𝑆) = 1 + 𝑘𝑙. Also, it

follows easily from Lemma 2.1.4 that #A (𝑆) = 1 + 𝛼1 · · · 𝛼𝑛. Equating the two expressions results

in the desired count.

(b) We will consider the case 𝑥 ∈ 𝐵𝑖 as the other is trivial. Consider the tiling 𝜏 = 𝜏(O). From

the proof of (a), we see that row 𝑖 has 𝑙 columns which are tiled by a pair of consecutive black and

yellow tiles of combined length 𝛼𝑖. So the number of black tiles in that row is

𝑚𝑖 = 𝑙/𝛼𝑖.

(2.1)

We are now done by Corollary 2.1.3 (a).

(c) Using part (b) and Corollary 2.1.3 (b) we obtain

𝜒(O) = 𝛽[𝑛]𝑚 [𝑛] +

𝑛
∑︁

𝑖=1

𝛽𝑖𝑚𝑖 = 𝛿 +

𝑛
∑︁

𝑖=1

𝑙
𝛼𝑖

(𝛼𝑖 − 1).

(d) Again, this is easy to see if 𝑥 = ˆ0. If 𝑥 = 𝑥𝑖, 𝑗 then there is no 𝐽 ⊂ [𝑖, 𝑖] in I (𝑆). So by

Corollary 2.1.3 (c) and equation (2.1)

ˆ𝜒𝑥 (O) = 𝑗 · 𝑚𝑖 = 𝑗𝑙/𝛼𝑖.

(e) It suffices to calculate the terms in the sum of Corollary 2.1.3 (d). We will do the case when

ˆ0 ∉ O as the unique orbit when ˆ0 ∈ O is done similarly. We first look at the term for 𝐼 = [𝑛].

In this case 𝛽[𝑛] = 1 and 𝑚 [𝑛] = 0 by the choice of O. Since [𝑖, 𝑖] ⊂ [𝑛] for all 𝑖 and there is no

column for the empty antichain we have 𝑐 [𝑛] = 𝑙, the number of columns of the tiling. So the term

16

1

2

3

𝑆2(3, 3, 2)

Figure 2.4 The extended star 𝑆2(3, 3, 2) and its tilings

for 𝐼 = [𝑛] reduces to 𝑙. Now consider the summand for [𝑖, 𝑖]. We have 𝛽𝑖 + 1 = 𝛼𝑖 and 𝑚𝑖 = 𝑙/𝛼𝑖

by equation (2.1). Furthermore, there is no 𝐽 ⊂ [𝑖, 𝑖] so 𝑐𝑖 = 0. Thus the term for 𝐼 = [𝑖, 𝑖] is the

𝑖th one in the sum given in (e), as desired.

□

Stars exhibit a number of homomesies. The following results are all gotten by simple manipu-

lation of the formulas for 𝜒 and ˆ𝜒 in the previous theorem, so we suppress the demonstration.

Corollary 2.2.2. Consider the star 𝑆 = 𝑆(𝛼1, . . . , 𝛼𝑛).

(a) If 𝑥 ∈ 𝐵𝑖, then 𝛼𝑖 𝜒𝑥 + 𝜒ˆ0 is 1-mesic.

(b) If 𝑥 ∈ 𝐵𝑖 and 𝑦 ∈ 𝐵 𝑗 then 𝛼𝑖 𝜒𝑥 − 𝛼 𝑗 𝜒𝑦 is 0-mesic.

(c) If 𝑥 = 𝑥𝑖,𝑘 then 𝛼𝑖 ˆ𝜒𝑥 − 𝑘 ˆ𝜒ˆ0 is 0-mesic.

(d) If 𝑥 = 𝑥𝑖,𝑘 and 𝑦 = 𝑥 𝑗,𝑘 , then 𝛼𝑖 ˆ𝜒𝑥 − 𝛼 𝑗 ˆ𝜒𝑦 is 0-mesic.

□

It is easy to generalize Theorem 2.2.1 to the case where 𝑏 [𝑛] > 1 so that one has a fatter [𝑛]-tile.

More generally, we will describe what happens to any tree where ˆ0 is covered by a single element.

An example can be obtained by comparing Figures 2.3 and 2.4.

Proposition 2.2.3. Suppose 𝑇 \ { ˆ0} = 𝑇 ′ is a rooted tree with 𝑛 leaves. Let the I (𝑇)-tilings be

𝜏1, 𝜏2, . . . , 𝜏𝑘 where 𝜏1 is the tiling for the orbit of ˆ0. Then the I (𝑇 ′)-tilings are 𝜏′
1
1 is obtained from 𝜏1 by widening the [𝑛]-tile by one column.
𝜏′

, 𝜏2, . . . , 𝜏𝑘 where

17

Proof. Since 𝑇 \ { ˆ0} = 𝑇 ′, the intervals of 𝑇 and 𝑇 ′ are the same. Also

𝛽𝐼 (𝑇 ′) =





𝛽𝐼 (𝑇)

if 𝐼 ≠ [𝑛],

𝛽[𝑛] (𝑇) + 1 if 𝐼 = [𝑛].

Definition 2.1.1 now shows that the tilings transform as desired.

□

For a positive integer 𝑏 the 𝑏-extended star, 𝑆𝑏 (𝛼1, . . . , 𝛼𝑛), is the rooted tree with

I (𝑆𝑏) = {([1, 1], 𝛽1), . . . , ( [𝑛, 𝑛], 𝛽𝑛), ( [𝑛], 𝑏)}

and 𝛼𝑖 = 𝛽𝑖 + 1 for 𝑖 ∈ [𝑛]. So we recover ordinary stars when 𝑏 = 1. We see 𝑆2(3, 3, 2) in

Figure 2.4. The next result follows easily from Theorem 2.2.1 and Proposition 2.2.3 and so the

proof is omitted.

Corollary 2.2.4. Consider the extended star 𝑆𝑏 = 𝑆𝑏 (𝛼1, . . . , 𝛼𝑛) and an orbit O of rowmotion on

𝑆𝑏. Let 𝑙 = lcm(𝛼1, . . . , 𝛼𝑛).

(a) We have

#O = 𝑙 + 𝛿𝑏

and the number of orbits is 𝛼1 · · · 𝛼𝑛/𝑙.

(b) For any 𝑥 ∈ 𝑆,

(c) We have

𝜒𝑥 (O) =

𝑙/𝛼𝑖

if 𝑥 ∈ 𝐵𝑖,

𝛿

if 𝑥 ∈ 𝐵[𝑛] .





𝜒(O) = 𝛿𝑏 +

𝑛
∑︁

𝑖=1

𝑙
𝛼𝑖

(𝛼𝑖 − 1).

Thus 𝜒 is homometric but not homomesic.

(d) For any 𝑥 ∈ 𝑆

ˆ𝜒𝑥 (O) =





𝑗𝑙/𝛼𝑖

if 𝑥 = 𝑥𝑖, 𝑗

𝑙 + 𝛿( 𝑗 − 1)

if 𝑥 = 𝑥 [𝑛], 𝑗 .

18

𝑇 ′ = 𝑇 ′′

1 = 𝜏′′
𝜏′

1

2 = 𝜏′′
𝜏′

2

𝜏1,1
1

𝜏1,1
2

𝜏1,1
3

𝜏2,2
1

𝜏2,2
2

𝑇

(e) We have

𝜏1,2
1

𝜏2,1
1

Figure 2.5 The trees 𝑇 ′, 𝑇 ′′, 𝑇 and their tilings

ˆ𝜒(O) = 𝑙𝑏 + 𝛿

(cid:18)𝑏

(cid:19)

2

+

𝑛
∑︁

𝑖=1

(cid:19)

.

𝑙
𝛼𝑖

(cid:18)𝛼𝑖
2

Thus ˆ𝜒 is homometric but not homomesic.

□

2.3 Trees with Three Leaves

The special case 𝑛 = 3 of Corollary 2.2.4 gives information about the rowmotion orbits on trees

that have three leaves whose branches have minimal elements covering a single vertex of the tree.

Up to isomorphism, there is only one other arrangement of branches in a tree with three leaves and

this section is devoted to studying this case. First, we will prove a result about removing the branch

containing ˆ0 from a certain type of tree.

Proposition 2.2.3 describes the tilings of a tree 𝑇 whose ˆ0 is covered by a single element. We

will determine what happens when it is covered by two elements or, more generally, when removing

the branch of ˆ0 leaves exactly two rooted trees remaining. It is possible to derive a similar result

for any number of rooted subtrees, but the notation becomes cumbersome and we will only need

the case of two subtrees in the sequel.

19

In order to state our result we will need some notation. Let 𝑇 be a rooted tree such that

𝑇 \ 𝐵 = 𝑇 ′ ⊎ 𝑇 ′′ where 𝐵 is the branch of ˆ0 and 𝑇 ′, 𝑇 ′′ are rooted trees. Suppose that 𝑇 ′ has 𝑛′ leaves

and tilings 𝜏′
1
Further, let 𝑐′

, . . . , 𝜏′

𝑠 where 𝜏′

𝑖 be the number of columns of 𝜏′

1 is corresponds to the orbit containing ˆ0′, the minimal element of 𝑇 ′.
𝑖 for 𝑖 ∈ [𝑠]. Notation used previously for 𝑇 will be

given a single prime when applied to 𝑇 ′. Similarly, let 𝑇 ′′ have 𝑛′′ leaves and tilings 𝜏′′
1

, . . . , 𝜏′′
𝑡

with the same conventions about the tilings and other notation except with a double prime. An

example of this construction can be found in Figure 2.5.

Theorem 2.3.1. Let 𝑇 be a rooted tree with 𝑇 \ 𝐵 = 𝑇 ′ ⊎ 𝑇 ′′ as above.

(a) The tilings of 𝑇 can be described as follows. For all (𝑖, 𝑗) ∈ [𝑛′] × [𝑛′′] there are tilings

𝜏𝑖, 𝑗
𝑚 for 1 ≤ 𝑚 ≤ 𝑔𝑖, 𝑗 := gcd(𝑐′

𝑖, 𝑐′′

𝑗 ). Unless 𝑖 = 𝑗 = 𝑚 = 1, we have that 𝜏𝑖, 𝑗

𝑚 consists of

consecutive copies of 𝜏′

and has 𝑙𝑖, 𝑗 := 𝑙𝑐𝑚(𝑐′

𝑖 in the first 𝑛′ rows, consecutive copies of 𝜏′′
𝑗
𝑗 ) columns. Tiling 𝜏1,1

𝑖, 𝑐′′

1

is as in the previous sentence except that

in the last 𝑛′′ rows,

one copy of 𝜏′

1 and one of 𝜏′′

1 align so that their columns of all yellow tiles coincide, and an
[𝑛′ + 𝑛′′] × 𝑏 black tile is inserted directly after that column to make the total length of the

orbit 𝑙1,1 + 𝑏 where 𝑏 = #𝐵.

(b) Let O′

𝑖 , O′′

𝑗 , and O

𝑖, 𝑗
𝑚 be the orbits corresponding to tilings 𝜏′

𝑖 , 𝜏′′

𝑗 , and 𝜏𝑖, 𝑗

𝑚 , respectively. For

any 𝑥 ∈ 𝑇

(c) We have

𝜒𝑥 (O

𝑖, 𝑗
𝑚 ) =

𝑙𝑖, 𝑗 𝜒𝑥 (O′

𝑖 )/𝑐′
𝑖

if 𝑥 ∈ 𝑇 ′,

𝑙𝑖, 𝑗 𝜒𝑥 (O′′

𝑗 )/𝑐′′
𝑗

if 𝑥 ∈ 𝑇 ′′,

𝛿

if 𝑥 ∈ 𝐵.






𝜒(O

𝑖, 𝑗
𝑚 ) = 𝛿𝑏 + 𝑙𝑖, 𝑗 𝜒(O′

𝑖 )/𝑐′

𝑖 + 𝑙𝑖, 𝑗 𝜒(O′′

𝑗 )/𝑐′′
𝑗 .

20

(d) For any 𝑥 ∈ 𝑇

ˆ𝜒𝑥 (O

𝑖, 𝑗
𝑚 ) =

𝑙𝑖, 𝑗 ˆ𝜒𝑥 (O′

𝑖 )/𝑐′
𝑖

if 𝑥 ∈ 𝑇 ′,

𝑙𝑖, 𝑗 ˆ𝜒𝑥 (O′′

𝑗 )/𝑐′′
𝑗

if 𝑥 ∈ 𝑇 ′′,

𝑙𝑖, 𝑗 + 𝛿( 𝑗 − 1)

if 𝑥 = 𝑥 [𝑛′+𝑛′′], 𝑗 .






(e) We have

ˆ𝜒(O

𝑖, 𝑗
𝑚 ) = 𝑙𝑖, 𝑗 𝑏 + 𝛿

(cid:18)𝑏

(cid:19)

2

+ 𝑙𝑖, 𝑗 ˆ𝜒(O′

𝑖 )/𝑐′

𝑖 + 𝑙𝑖, 𝑗 ˆ𝜒(O′′

𝑗 )/𝑐′′
𝑗 .

Proof. We will only prove (a), as once this is established then the other parts of the theorem follow

from straight-forward computations similar to those already seen in Theorem 2.2.1. Let O be an

antichain orbit of 𝑇. Pick an antichain 𝐴 in O which does not contain an element of 𝐵, so that it

can be written as 𝐴 = 𝐴′ ⊎ 𝐴′′ where 𝐴′ = 𝐴 ∩ 𝑇 ′ and 𝐴′′ = 𝐴 ∩ 𝑇 ′′. Let O′ and O′′ be the orbits

of 𝐴′ and 𝐴′′ in 𝑇 ′ and 𝑇 ′′, respectively.

First consider the case when (at least) one of O′ and O′′ does not contain the empty antichain.

It follows that as 𝜌 is applied to 𝐴, the antichains 𝐴′ and 𝐴′′ will describe their respective orbits

O′ and O′′ in 𝑇 ′ and 𝑇 ′′. If 𝑐′ = #O′ and 𝑐′′ = #O′′ then, in order for both orbits to return to

𝐴′ and 𝐴′′ at the same time, we must have #O = lcm(𝑐′, 𝑐′′). And since there are 𝑐′𝑐′′ ways to

pair an antichain in O′ with one in O′′, the total number of orbits obtained from such pairs is
𝑐′𝑐′′/lcm(𝑐′, 𝑐′′) = gcd(𝑐′, 𝑐′′). This description matches the one given for the tilings 𝜏𝑖, 𝑗
𝑘

for as

long as we do not have 𝑖 = 𝑗 = 1.

In the case when both O′ and O′′ contain the empty antichain, the argument of the previous

paragraph goes through with one exception. Suppose the elements of O′ and O′′ are repeated in

O in such a way that at some point the empty antichain of 𝑇 is reached. Then ∅ will be followed

by the elements of 𝐵 in increasing order. This, in turn, will be followed by the antichain { ˆ0′, ˆ0′′}
which will cause the orbits O′ and O′′ to continue. This orbit corresponds to the tiling 𝜏1,1
1

and

completes our description of the orbits and their tilings.

□

Now consider a tree 𝑇 with three leaves which is not an extended star. It follows that, using a

21

1

2

3

𝑇3
Figure 2.6 The tree 𝑇3

suitable embedding, we will have

I (𝑇) = {([3], 𝑎), ([2], 𝑏), ( [1, 1], 𝑐), ( [2, 2], 𝑑), ( [3, 3], 𝑒)}

for 𝑎, 𝑏, 𝑐, 𝑑, 𝑒 ≥ 1. A particular tree of this form is shown in Figure 2.6. Although we can use

the previous theorem to calculate the orbits and their statistic values for arbitrary 𝑎, 𝑏, 𝑐, 𝑑, 𝑒 the

resulting formulas are not very enlightening. So we will concentrate on a specific tree of this type.

Define the three-leaf tree 𝑇𝑘 to be the one with

I (𝑇) = {([3], 𝑘), ([2], 𝑘), ( [1, 1], 𝑘 − 1), ( [2, 2], 𝑘 − 1), ( [3, 3], 𝑘 − 1)}.

The tree in Figure 2.6 is 𝑇3.

Theorem 2.3.2. The orbits of rowmotion on 𝑇𝑘 can be partitioned by length into three sets S (for

small), M (for medium), and L (for large) with the following properties.

(a) We have

#S = 𝑘 (𝑘 − 1),

#M = 𝑘 − 1,

#L = 1,

22

and

(b) We have

#O =

𝑘

2𝑘

3𝑘






if O ∈ S,

if O ∈ M,

if O ∈ L.

𝜒(O) =




3𝑘 − 3

if O ∈ S,

5𝑘 − 4

if O ∈ M,

6𝑘 − 4

if O ∈ L.


Thus 𝜒 is homometric but not homomesic.

(c) We have

ˆ𝜒(O) =

7
2

𝑘 2 − 3
2

𝑘

if O ∈ S,

11
2

𝑘 2 − 5
2

𝑘

if O ∈ M,

6𝑘 2 − 3𝑘

if O ∈ L.






Thus ˆ𝜒 is homometric but not homomesic.

Proof. (a) Let 𝐵 = 𝐵[3] and 𝑏 = #𝐵 = 𝑘. Then 𝑇𝑘 \ 𝐵 = 𝑆𝑘 (𝑘, 𝑘) ⊎ 𝑆(𝑘) is a disjoint union of

two (extended) stars. Clearly 𝑇 ′′ = 𝑆(𝑘) has only one orbit which contains ˆ0. By Corollary 2.2.4,

𝑇 ′ = 𝑆𝑘 (𝑘, 𝑘) has orbits of size lcm(𝑘, 𝑘) = 𝑘 and the total number of orbits is (𝑘 · 𝑘)/𝑘 = 𝑘. So one

of these orbits contains ˆ0 and the other 𝑘 − 1 do not, and they will have lengths given by 𝑘 + 𝛿𝑘. It

follows that the latter will be of length 𝑘 while the former is of length 2𝑘. Applying Theorem 2.3.1,
𝑇𝑘 will have 𝑘 (𝑘 − 1) orbits O𝑖,1
orbits in S. There will also be the orbits O1,1
𝑚. Here the length is lcm(2𝑘, 𝑘) = 2𝑘. These are the orbits in M. Finally, the unique orbit O1,1
1

𝑚 with 𝑖 ≠ 1 and these will have length lcm(𝑘, 𝑘) = 𝑘. These are the

𝑚 for 𝑚 ∈ [2, 𝑘] which gives 𝑘 − 1 possible values for

is

of length 2𝑘 + 𝑏 = 2𝑘 + 𝑘 = 3𝑘 and this describes L.

23

1

2

1

2

3

4

3

𝐶3

4

𝐶3,2

Figure 2.7 The comb 𝐶3 and extended comb 𝐶3,2

(b) We will do the case of O1,1

1 , the unique element of L, as the others are similar. Applying

Corollary 2.2.4 (c) to orbit O′

1 of 𝑇 ′ = 𝑆𝑘 (𝑘, 𝑘) gives

𝜒(O′

1) = 𝑘 +

2
∑︁

𝑖=1

𝑘
𝑘

(𝑘 − 1) = 3𝑘 − 2

Similarly, for O′′

1 in 𝑇 ′′ = 𝑆(𝑘) we have

𝜒(O′′

1 ) = 𝑘 − 1.

Now applying Theorem 2.3.1 (c) with 𝑙1,1 = lcm(2𝑘, 𝑘) = 2𝑘 yields

𝜒(O1,1
1

) = 𝑘 + 2𝑘 (3𝑘 − 2)/(2𝑘) + 2𝑘 (𝑘 − 1)/𝑘 = 6𝑘 − 4.

(c) The computations are like those in (b) except using Theorem 2.3.1 (e), so the details are

omitted.

2.4 Combs and Zippers

□

Combs are a particularly simple type of binary tree. They are useful in understanding the

structure of the free Lie algebra as shown, for example, in the work of Wachs [60]. In this section

we will compute the orbit structure of combs, combs with an extended backbone, and zippers which

are constructed by pasting together combs.

24

It will be convenient to consider combs which have 𝑛 + 1 leaves. Specifically, the comb, 𝐶𝑛, is

the rooted tree with

I (𝐶𝑛) = {([𝑛 + 1], 1), ([𝑛], 1), . . . , ( [2], 1), ( [1, 1], 1), ( [2, 2], 1), . . . , ( [𝑛 + 1, 𝑛 + 1], 1)}.

The comb 𝐶3 is shown on the left in Figure 2.7.

Theorem 2.4.1. The orbits of rowmotion on 𝐶𝑛 can be partitioned into two sets S and L having

the following properties.

(a) We have

and

(b) We have

#S = 2𝑛−1,

#L = 1,

#O =





2

if O ∈ S,

2𝑛+1 − 1

if O ∈ L.

𝜒(O) =

𝑛 + 1

if O ∈ S,

(2𝑛 + 1)2𝑛−1

if O ∈ L.

Thus 𝜒 is homometric but not homomesic.

(c) We have

ˆ𝜒(O) =

3𝑛 + 1

if O ∈ S,

2𝑛−1(6𝑛 − 5) + 3

if O ∈ L.

Thus ˆ𝜒 is homometric but not homomesic.









Proof. (a) We induct on 𝑛 where the result is easy to check if 𝑛 = 1. Assume the orbits are as stated

for 𝐶𝑛 and that the unique orbit in L is the one containing ˆ0. We see that 𝐶𝑛+1 \ { ˆ0} = 𝐶𝑛 ⊎ {𝑣}

where 𝑣 is the leaf labeled 𝑛 + 2. We will subscript notation with 𝑛 or 𝑛 + 1 to make it clear which

comb is meant.

25

Now 𝑇 ′′ = {𝑣} has only one orbit of length 2. By Theorem 2.3.1, this combines with each of

the orbits in S𝑛 to give orbits of length lcm(2, 2) = 2. Also, there will be gcd(2, 2) = 2 orbits in
S𝑛+1 for every one in S𝑛 for a total of 2 · 2𝑛−1 = 2𝑛 orbits. Thus the information about S𝑛+1 is as

desired.

The one orbit in L𝑛 will combine with the one for {𝑣} to give gcd(2, 2𝑛+1 − 1) = 1 orbit which

must be the one containing ˆ0. So its length will be lcm(2, 2𝑛+1 − 1) + 1 = 2𝑛+2 − 1, which finishes

the induction.

(b) Again we induct, only providing details for the orbit of ˆ0 in L. Using the notation for

Theorem 2.3.1 we have 𝑐′

1 = 2𝑛+1 − 1 and 𝑐′′

1 = 2. So 𝑙1,1 = 𝑐′

1

𝑐′′
1 and the formula in part (c) of that

theorem becomes

as it should be.

𝜒(O) = 1 + 2(2𝑛 + 1)2𝑛−1 + (2𝑛+1 − 1) · 1 = (2𝑛 + 3)2𝑛

(c) This demonstration is similar to that of (b) above using Theorem 2.3.1 (d) and so is

omitted.

□

We can generalize these comb results as follows. The backbone of a comb is the set of elements

which are not leaves. So 𝐶𝑛 has an 𝑛-element backbone and each element is an interval in I (𝐶𝑛).

We will extend each of these intervals, except for the one corresponding to ˆ0, so that they have 𝑘

elements. Formally, the extended comb, 𝐶𝑛,𝑘 , is defined as the tree with

I (𝐶𝑛) = {([𝑛 + 1], 1), ([𝑛], 𝑘), . . . , ( [2], 𝑘), ( [1, 1], 1), ( [2, 2], 1), . . . , ( [𝑛 + 1, 𝑛 + 1], 1)}.

On the right in Figure 2.7 is the extended comb 𝐶3,2. Note that 𝐶𝑛,1 = 𝐶𝑛.

Theorem 2.4.2. The orbits of the extended comb 𝐶𝑛,𝑘 can be partitioned into two sets S and L

when 𝑘 is odd, and into 𝑛 + 1 sets S1, S2, . . . , S𝑛 and L when 𝑘 is even. The orbits have properties

given by the following tables for 𝑘 odd:

26

𝑘 odd

#O

number of O

𝜒(O)

ˆ𝜒(O)

and for 𝑘 even:

𝑘 even

#O

number of O

𝜒(O)

ˆ𝜒(O)

S

2

2𝑛−1

𝑛 + 1

L

(𝑘 + 1)2𝑛 − 2𝑘 + 1

1

((𝑘 + 1)𝑛 + 1)2𝑛−1 − 𝑘 + 1

(2𝑘 + 1)𝑛 − 2𝑘 + 3

(2𝑘 + 1) (𝑘 + 1)𝑛2𝑛−1 − (5𝑘 2 + 3𝑘 − 3)2𝑛−1 + 3𝑘 2

S𝑖 for 𝑖 ∈ [𝑛]

𝑘 (𝑖 − 1) + 2

2𝑛−𝑖

L

𝑘 (𝑛 − 1) + 3

1

𝑘 (𝑖−1)+2
2

𝑛 − 𝑘

4 (𝑖2 − 5𝑖 + 4) + 1

𝑘
4

𝑛2 + 3𝑘+4
4

𝑛 − 𝑘 + 2

(2𝑘+1)(𝑘 (𝑖−1)+2)
2

𝑛 − 𝑘 (2𝑘+1)

4

𝑖2 + 3𝑘
4

𝑖 + (cid:0)𝑘−2
2

(cid:1)

𝑘 (2𝑘+1)
4

𝑛2 − 4𝑘 2−9𝑘−4

4

𝑛 + (cid:0)𝑘−2
2

(cid:1)

Thus 𝜒 and ˆ𝜒 are homometric on 𝐶𝑛,𝑘 .

Proof. We will just verify the orbit structure as, once that is done, the calculation of 𝜒 and ˆ𝜒 are

routine using Proposition 2.2.3 and Theorem 2.3.1. We will induct on 𝑛 where the base case is

easy. Note that 𝐶𝑛+1,𝑘 \ { ˆ0} = 𝐶′

𝑛,𝑘 is 𝐶𝑛,𝑘 with
its ˆ0-interval replaced by one with 𝑘 elements. It follows from Proposition 2.2.3 that the orbits

𝑛,𝑘 ⊎ {𝑣} where 𝑣 is the leaf labeled 𝑛 + 2 and 𝐶′

of these two posets are identical except for the orbit of ˆ0 whose [𝑛 + 1]-tile has been widened by

adding 𝑘 − 1 columns.

We now consider what happens when 𝑘 is odd. The orbits of length 2 for 𝐶′

𝑛,𝑘 combine with

the orbit of length 2 for {𝑣} in exactly the same way as in the proof of Theorem 2.4.1. As far as the

orbit containing ˆ0 in 𝐶′

𝑛,𝑘 , by induction and the last sentence of the previous paragraph it has length

[(𝑘 + 1)2𝑛 − 2𝑘 + 1] + 𝑘 − 1 = (𝑘 + 1)2𝑛 − 𝑘

which is odd by the parity of 𝑘. So, by Theorem 2.3.1, the orbit containing ˆ0 in 𝐶𝑛+1,𝑘 has length

lcm((𝑘 + 1)2𝑛 − 𝑘, 2) + 1 = 2[(𝑘 + 1)2𝑛 − 𝑘] + 1 = (𝑘 + 1)2𝑛+1 − 2𝑘 + 1

27

Figure 2.8 The zipper 𝑍3

which is the desired quantity.

When 𝑘 is even we have, by induction, that all the orbits of 𝐶𝑛,𝑘 have even length except for the

orbit of ˆ0 whose length is odd. It follows that all the orbits of 𝐶′

𝑛,𝑘 are of even length. So, when

each non-ˆ0 is combined with 𝑣’s orbit of length 2, this will result in two orbits of the same length.

This accounts for the orbits in S𝑖 of 𝐶𝑛+1,𝑘 for 𝑖 < 𝑛. The ˆ0-orbit of 𝐶′

𝑛,𝑘 will have length

[𝑘 (𝑛 − 1) + 3] + 𝑘 − 1 = 𝑘𝑛 + 2.

Since this is even, when it combines with 𝑣’s orbit it will produce gcd(𝑘𝑛 + 2, 2) = 2 orbits for

𝐶𝑛+1,𝑘 . One of these will be of size lcm(𝑘𝑛 + 2, 2) = 𝑘𝑛 + 2 and that one will take care of S𝑛. The

other will have length one more and will be the orbit in L.

□

Another way to modify combs is by combining them together.

If 𝑇 is a rooted tree and

𝑇 \ { ˆ0} = 𝑇 ′ ⊎ 𝑇 ′′ then we will also write 𝑇 = 𝑇 ′ ⊕ 𝑇 ′′. Define the zipper, 𝑍𝑛, to be

𝑍𝑛 = 𝐶𝑛 ⊕ 𝐶𝑛

A picture of 𝑍3 will be found in Figure 2.8.

Theorem 2.4.3. The orbits of 𝑍𝑛 can be partitioned into four sets S, M, L, and G (for gigantic).

The properties of the orbits is summarized in the following table:

28

S

2

#O

number of O 22𝑛−1

M

2𝑛+1 − 1

2𝑛+1 − 2

L

2𝑛+1

1

G

2𝑛+2 − 2

2𝑛

𝜒(O)

ˆ𝜒(O)

2𝑛 + 2

2𝑛 (2𝑛 + 1)

2𝑛 (2𝑛 + 1) + 1

2𝑛 (4𝑛 + 3) − 𝑛 − 1

6𝑛 + 4

3 · 2𝑛 (2𝑛 − 1) + 5

3 · 2𝑛 (2𝑛 − 1) + 5

2𝑛−1(51𝑛 − 25) + 3

Thus 𝜒 and ˆ𝜒 are homometric on 𝑍𝑛.

Proof. As usual, we will just give details about the orbit structure. Since 𝑍𝑛 \ { ˆ0} is a disjoint union

of two copies of 𝐶𝑛, we use Theorems 2.4.1 and 2.3.1. Let S′ and L′ refer to the orbit partition of

𝐶𝑛 and use unprimed notation for 𝑍𝑛.

Combining two orbits from S′ gives gcd(2, 2) = 2 orbits of 𝑍𝑛 of length lcm(2, 2) = 2. Since

#S′ = 2𝑛−1, the total number of orbits formed in this way is

2 · 2𝑛−1 · 2𝑛−1 = 22𝑛−1.

These are the orbits of S.

Putting together an orbit from S′ with the unique orbit in L′ results in gcd(2, 2𝑛+1 − 1) = 1

orbit of size lcm(2, 2𝑛+1 − 1) = 2𝑛+2 − 2. Now the total number of orbits is

2 · 2𝑛−1 · 1 = 2𝑛

and they are the orbits in G.

Finally, the combination of the orbit in L′ with itself gives gcd(2𝑛+1 − 1, 2𝑛+1 − 1) = 2𝑛+1 − 1

orbits. All of these orbits will have length lcm(2𝑛+1 − 1, 2𝑛+1 − 1) = 2𝑛+1 − 1 except for the one

containing ˆ0 which will have one more element. These orbits are precisely the ones in M ⊎ L, and

so we are done.

2.5 Comments and Open Questions

2.5.1 Other Trees

□

The trees considered in the previous section had such nice homometry properties that one

might ask if the same is true for other binary trees. In particular, one could consider the complete

29

𝑥

𝑦

𝑧

Figure 2.9 A complete binary tree

binary trees which are those all of whose leaves are at the same rank. Such a tree is displayed

in Figure 2.9. Unfortunately, homometry fails for this example tree. Consider the orbit O which

contains the antichain {𝑥, 𝑦} as well as the one O′ which contains {𝑧}. Then it is easy to verify that

#O = #O′ = 4. But

𝜒(O) = 15 ≠ 14 = 𝜒(O′) and ˆ𝜒(O) = 35 ≠ 26 = ˆ𝜒(O′).

As mentioned in the introduction, Elizalde et al. [23] considered fences whose Hasse diagrams

are paths with any number of minimal elements. Here we have concentrated on arbitrary trees,

but insisted that there be a unique minimal element. In Chapter 3, we define a family of posets

called shoelaces, which are a generalization of fences. It would be interesting to study homomesy

in shoelaces, or other general poset structures.

2.5.2 Piecewise-linear and Birational Rowmotion

There are two generalizations of rowmotion which have also been studied and which have

consequences for trees. We first need to describe rowmotion in terms of toggles. In the discussion

which follows we will just write ideal for lower order ideal.

If (𝑃, ≤𝑃) is a finite poset and 𝑥 ∈ 𝑃 then the corresponding toggle map is 𝑡𝑥 : L (𝑃) → L (𝑃)

defined by

𝑡𝑥 (𝐿) =

𝐿△{𝑥}

if 𝐿△{𝑥} ∈ L (𝑃),

𝐿

else





where △ denotes symmetric difference of sets. A linear extension of 𝑃 is a listing of 𝑃’s elements

𝑥1, 𝑥2, . . . , 𝑥 𝑝 such that 𝑥𝑖 ≤𝑃 𝑥 𝑗 implies 𝑥𝑖 is weakly left of 𝑥 𝑗 in the sequence, that is, 𝑖 ≤ 𝑗.

30

Cameron and Fon-Der-Flaass showed that rowmotion on ideals can be broken into a sequence of

toggles. In what follows we compose functions right to left.

Theorem 2.5.1 ([9]). For any finite poset 𝑃 and any linear extension 𝑥1, 𝑥2, . . . , 𝑥 𝑝 of 𝑃 we have

ˆ𝜌 = 𝑡𝑥1

𝑡𝑥2 · · · 𝑡𝑥 𝑝 .

□

Stanley [51] introduced the order polytope as a way to use geometry to study posets. Poset

𝑃 = {𝑥1, . . . , 𝑥 𝑝} has order polytope

Π(𝑃) = {( 𝑓 (𝑥1), . . . , 𝑓 (𝑥 𝑝)) ∈ [0, 1] 𝑝 | 𝑥𝑖 ≤𝑃 𝑥 𝑗 implies 𝑓 (𝑥) ≤ 𝑓 (𝑦)}.

So Π(𝑃) is a subpolytope of the 𝑝-dimensional unit cube. Also note that every ideal 𝐿 of 𝑃 has a

corresponding point of Π(𝑃) defined by the function

𝑓 (𝑥) =

0 if 𝑥 ∉ 𝑃,

1 if 𝑥 ∈ 𝑃.





Einstein and Propp [22] extended rowmotion to Π(𝑃). Write 𝑥 ⋖ 𝑦 if 𝑥 is covered by 𝑦 in 𝑃, that is

𝑥 <𝑃 𝑦 and there is no 𝑧 with 𝑥 <𝑃 𝑧 <𝑃 𝑦. If 𝑓 ∈ Π(𝑃) and 𝑥 ∈ 𝑃 then define the piecewise-linear

toggle 𝜎𝑥 of 𝑓 at 𝑥 to be 𝑔 = 𝜎𝑥 𝑓 ∈ Π(𝑃) where

using the notation

𝑔(𝑣) =





𝑀 + 𝑚 − 𝑓 (𝑥)

if 𝑣 = 𝑥,

𝑓 (𝑣)

if 𝑣 ≠ 𝑥

𝑀 = max
𝑦⋖𝑥

𝑓 (𝑦) and 𝑚 = min
𝑧⋗𝑥

𝑓 (𝑧).

(2.2)

(2.3)

It is not hard to verify from the definitions that 𝑔 ∈ Π(𝑃). One can also show that 𝜎𝑥 is an

involution just like 𝑡𝑥, and 𝜎𝑥 is also piecewise-linear as a function. Finally, one defines piecewise-

linear rowmotion, 𝜌PL : Π(𝑃) → Π(𝑃), by

𝜌PL = 𝜎𝑥1

𝜎𝑥2 · · · 𝜎𝑥 𝑝

where 𝑥1, 𝑥2, . . . , 𝑥 𝑝 is a linear extension of 𝑃. It is true, but not obvious from the equation just

given, that 𝜌PL is well defined in that it does not depend on the chosen linear extension. Since Π(𝑃)

31

has an infinite number of points, it is very possible for orbits of 𝜌PL to be infinite. However, in

certain cases the orbits are nice. Take, for example, the poset [ 𝑝] × [𝑞] which is the poset product

of a 𝑝-element chain and a 𝑞-element chain.

Theorem 2.5.2 ([22]). The order of 𝜌PL on [ 𝑝] × [𝑞] is 𝑝 + 𝑞.

□

One can extend piecewise-linear rowmotion even further to the birational realm by detropi-

calizing as done by Grinberg and Roby [29, 28]. This means that in equations (2.2) and (2.3)

sum becomes product, difference becomes quotient, and maximum become sum. To take care

of the minimum, we use the previous dictionary and the fact that for any set 𝑆 of real numbers

min 𝑆 = − max(−𝑆) where −𝑆 = {−𝑠 | 𝑠 ∈ 𝑆}. Now let 𝑃 be a finite poset and let ˆ𝑃 be 𝑃 with

a minimum element ˆ0 and a maximum element ˆ1 added. Let F be a field and consider a function

𝑓 : ˆ𝑃 → F. The birational toggle of 𝑓 at 𝑥 ∈ 𝑃 is 𝑔 = 𝑇𝑥 𝑓 where

(cid:205)𝑦⋖𝑥 𝑓 (𝑦)
𝑓 (𝑥) (cid:205)𝑧⋗𝑥 𝑓 (𝑧)−1

𝑓 (𝑣)

if 𝑣 = 𝑥,

if 𝑣 ≠ 𝑥.

𝑔(𝑣) =





One can verify that 𝑇𝑥 is an involution, is a birational function, and that the following is well defined.

Define birational rowmotion on functions 𝑓 : ˆ𝑃 → F as

𝜌B = 𝑇𝑥1

𝑇𝑥2 · · · 𝑇𝑥 𝑝

where, as usual, 𝑥1, 𝑥2, . . . , 𝑥 𝑝 is a linear extension of 𝑃. It is even more surprising when birational

orbits are finite. Indeed, 𝜌B being of finite order implies this is true for 𝜌PL. Again, everything

works well for rectangular posets.

Theorem 2.5.3 ([28]). The order of 𝜌B on [ 𝑝] × [𝑞] is 𝑝 + 𝑞.

□

Call a poset 𝑃 graded if all chains from a minimal element of 𝑃 to a maximal element have

the same length. Grinberg and Roby consider a class of inductively defined posets which they

call skeletal and includes graded rooted forests, that is, disjoint unions of rooted trees such that all

leaves have the same rank. In this context, they prove the following result.

32

Theorem 2.5.4 ([29]). If 𝑃 is a skeletal poset then 𝜌B has finite order.

They also give a formula for order of 𝜌𝐵 in the case that 𝑃 is a graded rooted forest which

agrees with the results in Corollary 2.2.4 for graded extended stars. A natural question is whether

𝜌B has finite order for any rooted trees which are not graded. Computer experiments suggest that

this is not the case, although we have not been able to provide a proof. Specifically, 200 trials were

run on 16 posets, and in all but one case the orbit had not repeated after 1, 000, 000 iterations of

rowmotion.

33

CHAPTER 3

EXTENDED PROMOTION

A labeling of a poset 𝑃 with 𝑛 elements is a bijection from 𝑃 to [𝑛]. 𝑃 is naturally labeled if the

labeling respects the ordering on elements of 𝑃. In 1972, Schützenberger introduced the promotion

operator on natural labelings of posets [44].

As originally defined, promotion applies only to natural labelings of posets. Defant and Kravitz

generalized the notion of promotion to operate on arbitrary poset labelings and referred to their

generalization as extended promotion [19]. Given a labeling 𝐿 of a poset, the extended promotion

of 𝐿 is denoted 𝜕𝐿. A key property of extended promotion is that applying it to a labeling yields

a new labeling that is closer to a natural labeling. This property is quantified precisely in the

following theorem.

Theorem 3.0.1 ([19, Theorem 2.8]). For any labeling 𝐿 of an 𝑛-element poset, the labeling 𝜕𝑛−1𝐿

is a natural labeling.

When applied to an arbitrary poset labeling, extended promotion will always result in a natural

labeling after a maximum of 𝑛−1 applications. Applied to a natural labeling of a poset, the extended

promotion will always produce another natural labeling. Defant and Kravitz [19] define a tangled

labeling of an 𝑛-element poset as a labeling that requires 𝑛 − 1 promotions to give a natural labeling.

Intuitively, the tangled labelings of a poset are those that are furthest from being sorted by extended

promotion; they require the full 𝑛 − 1 applications of extended promotion in theorem 3.0.1. Defant

and Kravitz studied the number of tangled labelings of a poset and conjectured the following upper

bound on the number of tangled labelings.

Conjecture 3.0.2 ([19, Conjecture 5.1]). An 𝑛-element poset has at most (𝑛 − 1)! tangled labelings.

Defant and Kravitz proved an enumerative formula for a large class of posets known as inflated

rooted forest posets (see section 3.2 for details). This formula was used by Hodges to show

34

conjecture 3.0.2 holds for all inflated rooted forest posets. Furthermore, Hodges conjectured a

stronger version of conjecture 3.0.2.

Conjecture 3.0.3 ([32, Conjecture 31]). An 𝑛-element poset with 𝑚 minimal elements has at most

(𝑛 − 𝑚)(𝑛 − 2)! tangled labelings.

Both [19] and [32] also considered counting labelings by the number of extended promotion

steps needed to yield a natural labeling.

In the preprint [19], Defant and Kravitz proposed the

following, listed as Conjecture 5.2. Hodges further examined this conjecture.

Conjecture 3.0.4 ([32, Conjecture 29]). Let 𝑃 be an 𝑛-element poset, and let 𝑎𝑘 (𝑃) denote the

number of labelings of 𝑃 requiring exactly 𝑘 applications of the extended promotion to be a natural

labeling. Then the sequence 𝑎0(𝑃), . . . , 𝑎𝑛−1(𝑃) is unimodal.

In this chapter, we study the number of tangled labelings of posets by partitioning tangled

labelings according to which poset element has label 𝑛 − 1. We propose the following new

conjecture.

Conjecture 3.0.5 (The (𝑛 − 2)! Conjecture). Let 𝑃 be an 𝑛-element poset with 𝑛 ≥ 2. For all

𝑥 ∈ 𝑃, let |T𝑥 (𝑃)| denote the number of tangled labelings of 𝑃 such that 𝑥 is labeled 𝑛 − 1. Then

|T𝑥 (𝑃)| ≤ (𝑛 − 2)! with equality if and only if there is a unique minimal element 𝑦 ∈ 𝑃 such that

𝑦 <𝑃 𝑥.

By results in section 3.1, both conjecture 3.0.2 and conjecture 3.0.3 follow from the (𝑛 − 2)!

conjecture. In theorem 3.2.14 and theorem 3.3.4, we prove that the (𝑛 − 2)! conjecture holds for

inflated rooted forest posets and for a new class of posets that we call shoelace posets. Furthermore,

the conjecture has been computationally verified on all posets with nine or fewer elements.

Following [32], we also consider the sorting time for labelings that are not tangled and introduce

associated generating functions.

In remark 3.5.4, we give a poset on six elements that is a

counterexample to conjecture 3.0.4. Our results completely determine the generating functions

for ordinal sums of antichains. We introduce a related generating function called the cumulative

35

generating function and prove log-concavity of the cumulative generating function for ordinal sums

of antichains.

In section 3.1 we review the basic properties of extended promotion. In section 3.2 we prove

that inflated rooted forest posets satisfy the (𝑛 − 2)! conjecture. In section 3.3 we prove that inflated

shoelace posets satisfy the (𝑛 − 2)! conjecture and give an exact enumeration for the number of

tangled labelings of a particular type of shoelace poset called a 𝑊-poset. In section 3.4 we study

the generating function of the sorting time of labelings of the ordinal sum of a poset 𝑃 with the

antichain 𝑇𝑘 on 𝑘 elements. In section 3.5 we show that the cumulative generating function for

ordinal sums of antichains are log-concave and use the cumulative generating functions to introduce

a new partial order on the symmetric group 𝔖𝑛. In section 3.6 we propose future directions to

explore.

3.1 Definitions and Properties of Extended Promotion

In this section, we review and prove some properties of the extended promotion operator that

will be used in later sections. Many of the definitions and results in this section come from [19]

and are cited appropriately.

3.1.1 Notation and Terminology

Let [𝑛] = {1, 2, . . . , 𝑛}. For a partially ordered set (or poset) 𝑃, the partial order on 𝑃 will be

denoted ≤𝑃. An element 𝑦 ∈ 𝑃 is said to cover 𝑥 ∈ 𝑃, denoted 𝑥 ⋖𝑃 𝑦, if 𝑥 <𝑃 𝑦 and there does

not exist an element 𝑧 ∈ 𝑃 such that 𝑥 <𝑃 𝑧 <𝑃 𝑦. A lower (resp. upper) order ideal of 𝑃 is a set

𝑋 ⊆ 𝑃 with the property that if 𝑦 ∈ 𝑋 and 𝑥 <𝑃 𝑦 (resp. 𝑥 >𝑃 𝑦) then 𝑥 ∈ 𝑋 also. For an element

𝑦 ∈ 𝑃, the principal lower order ideal of 𝑦 is denoted ↓ 𝑦 = {𝑥 ∈ 𝑃 : 𝑥 ≤𝑃 𝑦}. A poset 𝑃 is said

to be connected if its Hasse diagram is a connected graph. In this chapter, we only consider finite

posets and assume the reader is familiar with standard results on posets as can be found in [53,

Chapter 3].

A labeling of a poset 𝑃 with 𝑛 elements is a bijection from 𝑃 to [𝑛]. A labeling 𝐿 of 𝑃 is a natural

labeling if the sequence 𝐿−1(1), 𝐿−1(2), . . . , 𝐿−1(𝑛) is a linear extension of 𝑃. Equivalently, for

any elements 𝑥, 𝑦 ∈ 𝑃, if 𝑥 <𝑃 𝑦 then 𝐿(𝑥) < 𝐿(𝑦). Given a poset 𝑃, the set of all labelings of 𝑃

36

will be denoted Λ(𝑃). The set of all natural labelings (equivalently, linear extensions) of 𝑃 will be

denoted L (𝑃).

Definition 3.1.1 ([19, Definition 2.1]). Let 𝑃 be an 𝑛-element poset and 𝐿 ∈ Λ(𝑃). The extended

promotion of 𝐿, denoted 𝜕𝐿, is obtained from 𝐿 by the following algorithm:

1. Repeat until the element labeled 1 is maximal: Let 𝑥 be the element labeled 1 and let 𝑦 be

the element with the smallest label such that 𝑦 >𝑃 𝑥. Swap the labels of 𝑥 and 𝑦.

2. Simultaneously replace the label 1 with 𝑛 and replace the label 𝑖 with 𝑖 − 1 for all 𝑖 > 1.

In what follows, we will refer to extended promotion simply as promotion. For 𝑖 ≥ 0, the

notations 𝐿𝑖 and 𝜕𝑖 𝐿 are used interchangeably to denote the 𝑖th promotion of 𝐿. By convention,

𝐿0 and 𝜕0𝐿 denote the original labeling 𝐿. Promotion can be loosely thought of as “sorting” a

labeling 𝐿 so that 𝜕𝐿 is closer to being a natural labeling.

Definition 3.1.2 ([19, Section 2]). Let 𝐿 ∈ Λ(𝑃). The promotion chain of 𝐿 is the ordered set

of elements of 𝑃 whose labels are swapped in the first step of definition 3.1.1. The order of the

promotion chain is the order in which the labels were swapped in the first step of definition 3.1.1.

Example 3.1.3. fig. 3.1 shows the promotion algorithm applied to a labeling 𝐿 of a 6-element poset

𝑃. The promotion chain of 𝐿 is the ordered set {𝐿−1(1), 𝐿−1(2), 𝐿−1(5)}. A sequence of five

promotions of 𝐿 is shown in fig. 3.2. Observe that 𝐿𝑖 is not a natural labeling for 0 ≤ 𝑖 < 5 but

𝐿5 is a natural labeling. Since the poset 𝑃 has six elements and it takes five promotions to reach a

natural labeling, the labeling 𝐿 is tangled.

3

1

5
2

4

3

→

6

𝐿

5
1

4

3

→

2

6

2

6

1
5

4

2

→

6
4

3

1

5

𝜕𝐿

Figure 3.1 One promotion of the labeling 𝐿 on poset 𝑃. Swapped labels are shown in red

37

Definition 3.1.4 ([19, Section 1.1]). Let 𝑃 be an 𝑛-element poset and 𝐿 ∈ Λ(𝑃). The order

or sorting time of 𝐿, denoted or(𝐿), is the smallest integer 𝑘 ≥ 0 such that 𝐿 𝑘 ∈ L (𝑃).

If

or(𝐿) = 𝑛 − 1, then 𝐿 is a tangled labeling. The set of all tangled labelings of 𝑃 is denoted T (𝑃).

6
4

3

6

→

1

4

2

1

5

5
3

2

→

5

2

3

6
4

1

→

4

1

2

6
5

→

3

6

3

𝜕𝐿

𝜕2𝐿

𝜕3𝐿

𝜕4𝐿

2

5
4

1

𝜕5𝐿

Figure 3.2 Promotions of the labeling 𝐿 in fig. 3.1. Elements enclosed in a box are frozen

Definition 3.1.5. Let 𝑃 be an 𝑛-element poset and 𝑥 ∈ 𝑃. A labeling 𝐿 of 𝑃 is said to be an

𝑥-labeling if 𝐿 (𝑥) = 𝑛 − 1. The set of all tangled 𝑥-labelings of 𝑃 is denoted T𝑥 (𝑃).

For a poset 𝑃, the set of tangled labelings T (𝑃) is the disjoint union of T𝑥 (𝑃) as 𝑥 ranges over

elements in 𝑃. Thus, the number of tangled labelings of 𝑃 is equal to the sum

𝑥∈𝑃
We shall see that no tangled labeling has label 𝑛 − 1 on a minimal element of 𝑃. Thus, it follows

|T (𝑃)| =

∑︁

|T𝑥 (𝑃)|.

(3.1)

that the (𝑛 − 2)! conjecture implies conjecture 3.0.2. While investigating this conjecture, we will

occasionally want to consider a labeling restricted to a subposet.

Definition 3.1.6 ([19, Section 1.3]). Let 𝑃 be an 𝑛-element poset, 𝑄 be an 𝑚-element subposet of

𝑃, and 𝐿 ∈ Λ(𝑃). The standardization of 𝐿 on 𝑄 is the unique labeling st(𝐿) : 𝑄 → [𝑚] such

that st(𝐿)(𝑥) < st(𝐿)(𝑦) if and only if 𝐿 (𝑥) < 𝐿(𝑦) for all 𝑥, 𝑦 ∈ 𝑄.

Definition 3.1.7 ([19, Section 2]). Let 𝑃 be an 𝑛-element poset and 𝑥 ∈ 𝑃. The element 𝑥 is said

to be frozen with respect to a labeling 𝐿 ∈ Λ(𝑃) if 𝐿−1({𝑎, 𝑎 + 1, . . . , 𝑛}) is an upper order ideal

for every 𝑎 such that 𝐿 (𝑥) ≤ 𝑎 ≤ 𝑛. The set of frozen elements of 𝐿 will be denoted F (𝐿).

Equivalently, if 𝑥 is frozen, then the standardization of 𝐿 on the subposet 𝐿−1({𝐿 (𝑥), 𝐿(𝑥) +

1, . . . , 𝑛}) is a natural labeling. Thus, 𝐿 is a natural labeling of 𝑃 if and only if F (𝐿) = 𝑃. Observe

38

is a maximal element of 𝑃. More generally, by [19, Lemma 2.7], 𝐿−1

that by definition 3.1.1, for any labeling 𝐿 of an 𝑛-element poset 𝑃, the element labeled 𝑛 in 𝐿1 = 𝜕𝐿
𝑗+1(𝑛 − 𝑗) is frozen, so the
elements of 𝑃 with labels {𝑛 − 𝑗, 𝑛 − 𝑗 + 1, . . . , 𝑛} are “sorted.” The standardization of 𝐿 𝑗+1 on the

subposet of 𝑃 whose elements have 𝐿 𝑗+1-labels in {𝑛 − 𝑗, 𝑛 − 𝑗 + 1, . . . , 𝑛} is a natural labeling.

Example 3.1.8. In fig. 3.2, the frozen elements of each labeling are enclosed in boxes. Observe

that once an element is frozen, it remains frozen in subsequent promotions. Figure 3.3 shows a

subposet 𝑄 and the standardization of the labeling 𝐿 in fig. 3.1 on 𝑄.

2

3

1

4

Figure 3.3 The standardization of the labeling 𝐿 in fig. 3.1 on the subposet in the dotted box

We conclude this subsection by introducing funnels and basins. The basin elements of a poset

are a subset of its minimal elements. In proposition 3.1.17, we will see that for tangled labelings,

basins are the appropriate subset of minimal elements to pay attention to.

Definition 3.1.9. Let 𝑥 ∈ 𝑃 be a minimal element. The funnel of 𝑥 is

fun(𝑥) = {𝑦 ∈ 𝑃 : 𝑥 <𝑃 𝑦 and 𝑥 is the unique minimal element in ↓ 𝑦}.

Definition 3.1.10. A minimal element 𝑥 ∈ 𝑃 is a basin if fun(𝑥) ≠ ∅.

Example 3.1.11. Let 𝑃 be the poset with Hasse diagram in fig. 3.4. The basin elements in 𝑃 are 𝑔

and 𝑖. Their funnels are fun(𝑔) = {𝑑} and fun(𝑖) = { 𝑓 , 𝑐}, respectively. There are two basins 𝑔, 𝑖

in the lower order ideal ↓ 𝑎 and a single basin 𝑖 in the lower order ideal ↓ 𝑐.

In the terminology of this section, Defant’s and Kravitz’s characterization of tangled labelings

is as follows.

Theorem 3.1.12 ([19, Theorem 2.10]). A poset 𝑃 has a tangled labeling if and only if 𝑃 has a

basin.

39

𝑎

ℎ

𝑒

𝑐

𝑖

𝑓

𝑑

𝑏

𝑔

Figure 3.4 A poset with two basin elements 𝑔 and 𝑖

3.1.2 Properties of Extended Promotion

In this subsection, we provide some general lemmas on extended promotion and tangled label-

ings. We begin with a lemma implicit in [19] that gives a useful criterion for checking whether or

not a labeling is tangled.

Lemma 3.1.13. Let 𝑃 be a poset on 𝑛 elements and 𝐿 ∈ Λ(𝑃). The labeling 𝐿 is tangled if and

only if both of the following conditions are met:

1. 𝐿−1(𝑛) is minimal in 𝑃,

2. 𝐿−1(𝑛) <𝑃 𝐿−1

𝑛−2(1).

Proof. First, we will prove that conditions (1) and (2) together are sufficient for 𝐿 to be tangled. Let

𝑥 denote 𝐿−1(𝑛). By condition (1), 𝑥 is minimal so 𝐿𝑖+𝑟 (𝑥) = 𝐿𝑖 (𝑥) − 𝑟 whenever 𝐿𝑖 (𝑥) > 𝑟. Since
𝑛−2(2) = 𝐿−1(𝑛). Substituting into condition (2)

𝐿 (𝑥) = 𝑛, it follows that 𝐿𝑛−2(𝑥) = 2 and hence 𝐿−1

yields 𝐿−1

𝑛−2(2) <𝑃 𝐿−1

𝑛−2(1). Thus, 𝐿𝑛−2 is not yet sorted, and so 𝐿 is tangled.

By [19, Lemma 3.8], condition (1) is necessary for 𝐿 to be tangled. Thus, it remains to show

that condition (2) follows from assuming that 𝐿 is tangled and that condition (1) holds. By [19,

Lemma 2.7], 𝐿−1

𝑛−2(3), . . . , 𝐿−1

not sorted, which may occur only if 𝐿−1

𝑛−2(𝑛) are frozen with respect to 𝐿𝑛−2. Since 𝐿 is tangled, 𝐿𝑛−2 is
𝑛−2(1). Because 𝐿−1(𝑛) is minimal, we may
□

𝑛−2(2) <𝑃 𝐿−1

substitute 𝐿−1(𝑛) = 𝐿−1

𝑛−2(2) to yield condition (2).

As a consequence of condition (2), the element labeled 𝑛 − 1 cannot be minimal in a tangled

labeling of 𝑃. If an 𝑛-element poset 𝑃 has 𝑚 minimal elements, then conjecture 3.0.5 would imply

40

that the number of tangled labelings of 𝑃 is at most (𝑛 − 𝑚) (𝑛 − 2)!. Therefore, conjecture 3.0.5

also implies conjecture 3.0.3.

Lemma 3.1.14. Let 𝑃 be a poset on 𝑛 elements and 𝐿 ∈ Λ(𝑃). Then for all 2 ≤ 𝑖 ≤ 𝑛 and

0 ≤ 𝑗 ≤ 𝑛 − 1,

𝑗+1(𝑖 − 1) ≤𝑃 𝐿−1
𝐿−1

𝑗 (𝑖).

Proof. If 𝑖 is not the label of an element in the promotion chain of 𝐿 𝑗 , then the element 𝐿−1

𝑗 (𝑖) will

be labeled 𝑖 − 1 in 𝐿 𝑗+1, so 𝐿−1

𝑗+1(𝑖 − 1) = 𝐿−1
denote the element immediately preceding 𝐿−1

𝑗 (𝑖). If 𝐿−1

𝑗 (𝑖) is in the promotion chain of 𝐿 𝑗 , let 𝑥

𝑗 (𝑖) in the promotion chain of 𝐿 𝑗 . Such an element

exists since 𝑖 ≥ 2 so 𝐿−1

𝑗 (𝑖) cannot be the first element in the promotion chain. It follows that

𝐿−1
𝑗+1(𝑖 − 1) = 𝑥 ≤𝑃 𝐿−1

𝑗 (𝑖).

A consequence of lemma 3.1.14 is that for all 2 ≤ 𝑖 ≤ 𝑛,

𝑖−1(1) ≤𝑃 . . . ≤𝑃 𝐿−1
𝐿−1

1 (𝑖 − 1) ≤𝑃 𝐿−1(𝑖).

Setting 𝑖 = 𝑛 − 1 gives, in particular,

𝑛−2(1) ≤𝑃 𝐿−1
𝐿−1

𝑛−3(2) ≤𝑃 . . . ≤𝑃 𝐿−1

1 (𝑛 − 2) ≤𝑃 𝐿−1(𝑛 − 1).

□

(3.2)

(3.3)

Corollary 3.1.15. Let 𝑃 be a poset on 𝑛 elements and let 𝐿 ∈ L (𝑃) be a tangled labeling. For

𝑟 = 0, 1, . . . , 𝑛 − 2,

𝑟 (𝑛 − 𝑟) <𝑃 𝐿−1
𝐿−1

𝑟 (𝑛 − 1 − 𝑟).

In particular, 𝐿−1(𝑛) <𝑃 𝐿−1(𝑛 − 1).

Proof. By lemma 3.1.14, 𝐿−1

𝑟 (𝑛 − 𝑟) ≤𝑃 𝐿−1(𝑛), and by (2) in Lemma 3.1.13, 𝐿−1(𝑛) <𝑃 𝐿−1

𝑛−2(1).
𝑟 (𝑛 − 1 − 𝑟). Combining these

Additionally, by eq. (3.3), 𝐿−1

𝑛−2(1) ≤𝑃 𝐿−1

𝑛−3(2) ≤𝑃 · · · ≤𝑃 𝐿−1

inequalities yields the desired result 𝐿−1

𝑟 (𝑛 − 𝑟) <𝑃 𝐿−1

𝑟 (𝑛 − 1 − 𝑟). If we set 𝑟 = 0, then we see

that 𝐿−1(𝑛) <𝑃 𝐿−1(𝑛 − 1).

□

41

In [19, Corollary 3.7], Defant and Kravitz showed that any poset with a unique minimal element

satisfies conjecture 3.0.2. We strengthen this result to show that posets with any number of minimal

elements—but only one basin—also satisfy conjecture 3.0.2. We will need the following lemma

that is the key tool in Defant and Kravitz’s proof of theorem 3.0.1.

Lemma 3.1.16 ([19, Lemma 2.6]). Let 𝑃 be an 𝑛-element poset and let 𝐿 ∈ Λ(𝑃) \ L (𝑃). Then

F (𝐿) ⊊ F (𝜕𝐿).

Proposition 3.1.17. If 𝐿 is a tangled labeling of 𝑃, then 𝐿−1(𝑛) is a basin. In particular, if 𝑃 has

exactly one basin, then |T (𝑃)| ≤ (𝑛 − 1)!.

Proof. We first show that for any minimal element 𝑥 ∈ 𝑃 that is not a basin, there is no tangled

labeling 𝐿 with 𝐿 (𝑥) = 𝑛. Suppose to the contrary that there exists such a tangled labeling 𝐿. Let

𝑤 = 𝐿−1

𝑛−2(1). By lemma 3.1.13, 𝑥 <𝑃 𝑤. Since 𝑥 is not a basin, fun(𝑥) = ∅. Hence, there exists a

minimal element 𝑧 ≠ 𝑥 such that 𝑧 <𝑃 𝑤.

Since 𝑤 = 𝐿−1

𝑛−2(𝑚) for some 𝑚 ≥ 3.
𝑛−2(𝑛) are frozen as a consequence of lemma 3.1.16. Recall that the
set of frozen elements is an upper order ideal. Since 𝑧 is a frozen element and 𝑧 <𝑃 𝑤, 𝑤 must also

𝑛−2(1) and 𝑥 = 𝐿−1(𝑛) = 𝐿−1
𝑛−2(3), . . . , 𝐿−1

𝑛−2(2), it follows that 𝑧 = 𝐿−1

The elements 𝐿−1

be a frozen element, which is a contradiction since 𝐿𝑛−2 is not a natural labeling. Therefore if 𝐿 is

a tangled labeling and 𝐿−1(𝑛) is a minimal element of 𝑃, then 𝐿−1(𝑛) must be a basin.

Finally, suppose 𝑃 has a unique basin 𝑥. Then any tangled labeling 𝐿 of 𝑃 must satisfy 𝐿(𝑥) = 𝑛.

There are (𝑛 − 1)! labelings 𝐿 that satisfy 𝐿(𝑥) = 𝑛, so |T (𝑃)| ≤ (𝑛 − 1)!.

□

The following two lemmas relate tangled labelings and funnels of posets. They will be used in

section 3.3 to prove that shoelace posets satisfy the (𝑛 − 2)! conjecture.

Lemma 3.1.18. Let 𝑥 be a basin of 𝑃 and let 𝐿 be a labeling such that 𝐿−1(𝑛) = 𝑥 and 𝐿−1(𝑛 − 1) ∈

fun(𝑥). Then 𝐿 is tangled.

Proof. It is clear from the definition of basins that condition (1) of lemma 3.1.13 is satisfied. So it

suffices to show that 𝐿−1(𝑛) <𝑃 (𝐿𝑛−2)−1(1). From eq. (3.3) and the condition that 𝐿−1(𝑛 − 1) ∈

42

fun(𝑥),

Furthermore,

𝑥 ≤𝑃 (𝐿𝑛−2)−1(1) ≤𝑃 𝐿−1(𝑛 − 1).

𝑥 = 𝐿−1(𝑛) = (𝐿𝑛−2)−1(2) ≠ (𝐿𝑛−2)−1(1).

Thus, we have the strict inequality 𝑥 = 𝐿−1(𝑛) <𝑃 (𝐿𝑛−2)−1(1), which is precisely condition (2) of

lemma 3.1.13.

□

Lemma 3.1.19. Let 𝑃 be a poset on 𝑛 elements and 𝐿 a tangled labeling of 𝑃. Let 𝑥, 𝑦 ∈ 𝑃 such

that 𝑥 is a minimal element and 𝑥 <𝑃 𝑦. If 𝐿(𝑥) = 𝑛 and 𝐿 (𝑦) = 𝑛 − 1, then there exists 𝑧 ∈ fun(𝑥)

such that 𝑧 ≤𝑃 𝑦.

Proof. Let 𝑧 = 𝐿−1

𝑛−2(1). By eq. (3.3), 𝑧 = 𝐿−1

tangled labeling, lemma 3.1.13 implies that 𝑥 = 𝐿−1

𝑛−2(1) ≤𝑃 𝐿−1(𝑛 − 1) = 𝑦. Thus, 𝑧 ≤𝑃 𝑦. Since 𝐿 is a
𝑛−2(1) = 𝑧. There are at least 𝑛 − 2
frozen elements with respect to 𝐿𝑛−2, but 𝑥 and 𝑧 are not frozen with respect to 𝐿𝑛−2. Since the set

𝑛−2(2) <𝑃 𝐿−1

of frozen elements with respect to a labeling form an upper order ideal, it follows that 𝑧 covers 𝑥

and no other elements. Hence, 𝑧 ∈ fun(𝑥).

□

Lemma 3.1.20. Let 𝑃1 be a poset with 𝑛1 elements and 𝑃2 a poset with 𝑛2 elements. If conjec-

ture 3.0.5 holds for 𝑃1 and 𝑃2, then conjecture 3.0.5 also holds for the disjoint union 𝑃1 ⊔ 𝑃2.

Proof. Let 𝑥 ∈ 𝑃1 ⊔ 𝑃2 and 𝐿 be an 𝑥-labeling of 𝑃 (i.e., 𝐿(𝑥) = 𝑛 − 1). If 𝑥 ∈ 𝑃1 and 𝑛1 ≥ 2, then

by [19, Theorem 3.4], 𝐿 is tangled if and only if 𝐿−1(𝑛) ∈ 𝑃1 and st(𝐿|𝑃1) ∈ T (𝑃1). Thus, the
tangled 𝑥-labelings of 𝑃1 ⊔ 𝑃2 are enumerated by a choice of one of the |T𝑥 (𝑃1)| tangled 𝑥-labelings
of 𝑃1, one of the (cid:0)𝑛1+𝑛2−2
(cid:1) assignments of the labels 𝐿−1(𝑃1) \ {𝑛, 𝑛−1}, and one of the 𝑛2! labelings
𝑛1−2
on 𝑃2. Since 𝑃1 satisfies conjecture 3.0.5, |T𝑥 (𝑃1)| ≤ (𝑛1 − 2)!. Therefore,

|T𝑥 (𝑃1 ⊔ 𝑃2)| = |T𝑥 (𝑃1)| · 𝑛2! ·

(cid:19)

(cid:18)𝑛1 + 𝑛2 − 2
𝑛1 − 2
(cid:18)𝑛1 + 𝑛2 − 2
𝑛1 − 2

(cid:19)

(3.4)

≤ (𝑛1 − 2)! · 𝑛2! ·

= (𝑛1 + 𝑛2 − 2)!.

43

If 𝑥 ∈ 𝑃1 and 𝑛1 < 2, then by the contrapositive of corollary 3.1.15, 𝐿 is not tangled.

In

this case, tangled 𝑥-labelings of 𝑃1 ⊔ 𝑃2 do not exist, so |T𝑥 (𝑃1 ⊔ 𝑃2)| ≤ (𝑛1 + 𝑛2 − 2)! clearly.

Equality in eq. (3.4) holds if and only if |T𝑥 (𝑃1)| = (𝑛1 − 2)!. Since 𝑃1 satisfies conjecture 3.0.5,

|T𝑥 (𝑃1)| = (𝑛1 − 2)! if and only if there is a unique minimal element 𝑧 ∈ 𝑃1 such that 𝑧 <𝑃1
follows that equality in eq. (3.4) holds if and only if there is a unique minimal element 𝑧 ∈ 𝑃1 ⊔ 𝑃2

𝑥. It

𝑥. If 𝑥 ∈ 𝑃2, then by an identical argument, |T𝑥 (𝑃1 ⊔ 𝑃2)| ≤ (𝑛1 + 𝑛2 − 2)!, with

such that 𝑧 <𝑃1⊔𝑃2
equality if and only if there is a unique minimal element 𝑧 ∈ 𝑃2 such that 𝑧 <𝑃1⊔𝑃2
𝑃1 ⊔ 𝑃2 satisfies conjecture 3.0.5.

𝑥. Therefore,

□

By lemma 3.1.20, it suffices to show the (𝑛 − 2)! conjecture for connected posets. Thus, for the

remainder of the paper, we will assume our posets are connected.

3.2

Inflated Rooted Forest Posets

In [19], a large class of posets known as inflated rooted forest posets was introduced and it

was shown in [32] that conjecture 3.0.2 holds for inflated rooted forest posets. In this section, we

strengthen this result by showing that conjecture 3.0.5 holds for inflated rooted forest posets.

Definition 3.2.1 ([19, Definition 3.2]). Let 𝑃, 𝑄 be finite posets. The poset 𝑃 is an inflation of 𝑄

if there exists a surjective map 𝜑 : 𝑃 → 𝑄 that satisfies the following two properties:

1. For any 𝑥 ∈ 𝑄, the preimage 𝜑−1(𝑥) has a unique minimal element in 𝑃.

2. For any 𝑥, 𝑦 ∈ 𝑃 such that 𝜑(𝑥) ≠ 𝜑(𝑦), 𝑥 <𝑃 𝑦 if and only if 𝜑(𝑥) <𝑄 𝜑(𝑦).

Such a map 𝜑 is called an inflation map.

Example 3.2.2. In fig. 3.5 the poset 𝑃 is an inflation of the poset 𝑄. The inflation map 𝜑 is constant

on each colored box in 𝑃 and maps to the corresponding element in 𝑄 pointed to by the arrow.

For example, the element labeled 𝑢1,1 in 𝑄 corresponds to the subposet 𝜑−1(𝑢1,1) in 𝑃 outlined

in green. In general, the preimage of an element in 𝑄 must have a unique minimal element, by

definition, but may have multiple maximal elements.

44

Figure 3.5 A rooted tree 𝑄 and its inflation 𝑃. The inflation map 𝜑 is represented by the arrows
from 𝑃 to 𝑄

Definition 3.2.3 ([19, Definition 3.1]). A rooted tree poset 𝑄 is a finite poset satisfying the following

two properties:

1. There is a unique maximal element of 𝑄 called the root of 𝑄.

2. Every non-root element in 𝑄 is covered by exactly one element.

Notice here that we are taking the convention of an anti-arborescence, where the root is a

maximal element, the opposite orientation of a rooted tree in Chapter 2. A rooted forest poset is

defined to be a finite poset that can be written as the disjoint union of rooted tree posets. The posets

𝑃 and 𝑄 in fig. 3.5 are examples of an inflated rooted tree poset and a rooted tree poset, respectively.

Throughout the rest of this section, unless otherwise specified, 𝑄 will denote a rooted tree poset

and 𝑃 will denote an inflation of 𝑄 with inflation map 𝜑.

The following definitions on inflated rooted tree posets can be found in [19]. We reproduce

them here for the reader’s convenience and to state lemma 3.2.5 precisely. Let 𝑟 be the root of 𝑄

and let 𝑥 be a non-root element of 𝑄. The unique element 𝑦 that covers 𝑥 in 𝑄 is called the parent

of 𝑥. The minimal elements of 𝑄 are called leaves. A rooted tree poset is said to be reduced if

every non-leaf element covers at least 2 elements. By [19, Remark 3.3], every inflated rooted tree

45

ℓ1=u1,0ℓ2=u2,0ℓ3ℓ4u1,1=u2,1r=u1,2=u2,2=u3,1=u4,1ϕ−1(ℓ1)ϕ−1(ℓ2)ϕ−1(ℓ3)ϕ−1(ℓ4)ϕ−1(u1,1)ϕ−1(r)PQposet can be obtained as an inflation of a reduced rooted tree poset, so in the following we will

generally restrict ourselves to reduced rooted tree posets.

Let ℓ1, . . . , ℓ𝑚 denote the leaves of 𝑄, where 𝑚 is the number of leaves. For each 𝑖 ∈ [𝑚], we

have a unique maximal chain from ℓ𝑖 to 𝑟

ℓ𝑖 = 𝑢𝑖,0 ⋖𝑄 𝑢𝑖,1 ⋖𝑄 · · · ⋖𝑄 𝑢𝑖,𝜔𝑖 = 𝑟,

(3.5)

where 𝜔𝑖 denotes the length of the chain. Recall that 𝑢𝑖,0 ⋖𝑄 𝑢𝑖,1 means that 𝑢𝑖,1 covers 𝑢𝑖,0 in 𝑄.

For 𝑖 ∈ [𝑚] and 𝑗 ∈ [𝜔𝑖], define the two quantities

𝑏𝑖, 𝑗 =

∑︁

|𝜑−1(𝑣)|,

𝑐𝑖, 𝑗 =

𝑣≤𝑄𝑢𝑖, 𝑗 −1
∑︁

|𝜑−1(𝑣)|.

𝑣<𝑄𝑢𝑖, 𝑗

(3.6)

The fraction

𝑏𝑖, 𝑗
𝑐𝑖, 𝑗 therefore represents the fraction of elements in 𝑃 below the minimal element of
𝜑−1(𝑢𝑖, 𝑗 ) that lie on the preimage of the maximal chain from ℓ𝑖 to 𝑟. When it is necessary to specify

the rooted tree poset 𝑄, we shall do so by indicating 𝑄 in parentheses. For example, we will write

𝑢𝑖, 𝑗 (𝑄) instead of 𝑢𝑖, 𝑗 or 𝜔𝑖 (𝑄) instead of 𝜔𝑖.

Example 3.2.4. The vertices of 𝑄 in fig. 3.5 are labeled in accordance with our definitions above.

For example, the maximal chain from ℓ1 to 𝑟 is ℓ1 = 𝑢1,0⋖𝑢1,1⋖𝑢1,2 = 𝑟. The length of this maximal

chain is 𝜔1 = 2. As another example, the maximal chain from ℓ2 to 𝑟 is ℓ2 = 𝑢2,0 ⋖ 𝑢2,1 ⋖ 𝑢2,2 = 𝑟.

Notice that 𝑢𝑖, 𝑗 may refer to the same element in 𝑄 for distinct 𝑖 and 𝑗. For example, 𝑢2,1 = 𝑢1,1 in

fig. 3.5, and the root 𝑟 is equal to 𝑢1,2, 𝑢2,2, 𝑢3,1, and 𝑢4,1.

The quantity 𝑏1,1 can be computed by

𝑏1,1 =

∑︁

𝑣≤𝑄𝑢1,0

|𝜑−1(𝑣)| = |𝜑−1(𝑢1,0)| = 4.

Similarly, the quantity 𝑐1,1 can be computed by

𝑐1,1 =

∑︁

𝑣<𝑄𝑢1,1

|𝜑−1(𝑣)| = |𝜑−1(𝑢1,0)| + |𝜑−1(𝑢2,0)| = 6.

46

Therefore

𝑏1,1
𝑐1,1

of 𝜑−1(ℓ1).

= 4

6 of the elements in 𝑃 below the minimal element of 𝜑−1(𝑢1,1) lie in the direction

The following technical lemma provides a useful bound for the formula in theorem 3.2.13. The

left side of eq. (3.7) appears in [19, Theorem 3.5], and a similar term also appears in [32, Theorem

9].

Lemma 3.2.5. Let 𝑄 be a reduced rooted tree poset with 𝑚 leaves and let 𝑃 be an inflation of 𝑄

with 𝑛 elements. Then

𝑚
∑︁

𝜔𝑖 (𝑄)
(cid:214)

𝑖=1

𝑗=1

𝑏𝑖, 𝑗 (𝑄) − 1
𝑐𝑖, 𝑗 (𝑄) − 1

≤

1

if 𝑛 = 1,

𝑛−𝑚
𝑛−1

otherwise.





(3.7)

Proof. We will prove the bound by inducting on ℎ(𝑄) = max{𝜔1(𝑄), . . . , 𝜔𝑚 (𝑄)}. The base case

is when ℎ(𝑄) = 0. In this case, there is a single leaf so 𝑚 = 1 and 𝜔1(𝑄) = 0. Thus, the left

side of the inequality is the sum of a single empty product which is equal to 1. The right side is 1

regardless of whether 𝑛 = 1 or 𝑛 > 1, so the inequality holds when ℎ(𝑄) = 0.

Now, suppose ℎ(𝑄) > 0 and that the lemma holds for all rooted tree posets 𝑄′ with ℎ(𝑄′) <

ℎ(𝑄). Since ℎ(𝑄) > 0, 𝑛 > 1. Now, let 𝑟 denote the root of 𝑄 and let 𝑞1, . . . , 𝑞𝑡 be the elements

covered by 𝑟. Recall that for an element 𝑥 in a poset, ↓ 𝑥 denotes the set of elements less than

or equal to 𝑥. The subposets 𝑄 𝑘 = ↓ 𝑞𝑘 are all rooted tree posets with ℎ(𝑄 𝑘 ) ≤ ℎ(𝑄) − 1, and

𝑃𝑘 = 𝜑−1(𝑄 𝑘 ) is an inflation of 𝑄 𝑘 . Let 𝑛𝑘 = |𝜑−1(𝑄 𝑘 )| so that 𝑛 − |𝜑−1(𝑟)| = 𝑛1 + · · · + 𝑛𝑡, and let

𝑚𝑘 denote the number of leaves of 𝑄 𝑘 so that 𝑚 = 𝑚1 + · · · + 𝑚𝑡. For convenience, let 𝑀𝑘 denote

the 𝑘th partial sum 𝑚1 + · · · + 𝑚𝑘 and let 𝑀0 = 0. Without loss of generality, order the leaves

ℓ1, . . . , ℓ𝑚 of 𝑄 such that the leaves of 𝑄 𝑘 are ℓ𝑀𝑘−1+1, . . . , ℓ𝑀𝑘 .

Observe that for 𝑀𝑘−1 + 1 ≤ 𝑖 ≤ 𝑀𝑘 , 𝜔𝑖 (𝑄 𝑘 ) = 𝜔𝑖 (𝑄) − 1, and for 1 ≤ 𝑗 ≤ 𝜔𝑖 (𝑄 𝑘 ), 𝑏𝑖, 𝑗 (𝑄 𝑘 ) =

𝑏𝑖, 𝑗 (𝑄) and 𝑐𝑖, 𝑗 (𝑄 𝑘 ) = 𝑐𝑖, 𝑗 (𝑄). Additionally, 𝑏𝑖,𝜔𝑖 (𝑄) (𝑄) = 𝑛𝑘 and 𝑐𝑖,𝜔𝑖 (𝑄) (𝑄) = 𝑛 − |𝜑−1(𝑟)|.

47

Thus,

𝑚
∑︁

𝜔𝑖 (𝑄)
(cid:214)

𝑖=1

𝑗=1

𝑏𝑖, 𝑗 (𝑄) − 1
𝑐𝑖, 𝑗 (𝑄) − 1

=

=

=

𝑡
∑︁

𝑘=1

𝑡
∑︁

𝑘=1

𝑡
∑︁

𝑘=1

𝑀𝑘∑︁

(cid:169)
(cid:173)
𝑖=𝑀𝑘−1+1
(cid:171)
𝑀𝑘∑︁

(cid:169)
(cid:173)
𝑖=𝑀𝑘−1+1
(cid:171)

𝜔𝑖 (𝑄)
(cid:214)

𝑗=1

𝑏𝑖, 𝑗 (𝑄) − 1
𝑐𝑖, 𝑗 (𝑄) − 1

𝑛𝑘 − 1
𝑛 − |𝜑−1(𝑟)| − 1

(cid:170)
(cid:174)
(cid:172)

·

𝑛𝑘 − 1
𝑛 − |𝜑−1(𝑟)| − 1

𝑀𝑘∑︁

· (cid:169)
(cid:173)
𝑖=𝑀𝑘−1+1
(cid:171)

For each 1 ≤ 𝑘 ≤ 𝑡, if 𝑛𝑘 = 𝑚𝑘 = 1, then we clearly have

𝜔𝑖 (𝑄)−1
(cid:214)

𝑗=1

𝜔𝑖 (𝑄 𝑘)
(cid:214)

𝑗=1

𝑏𝑖, 𝑗 (𝑄) − 1
𝑐𝑖, 𝑗 (𝑄) − 1

𝑏𝑖, 𝑗 (𝑄 𝑘 ) − 1
𝑐𝑖, 𝑗 (𝑄 𝑘 ) − 1

(cid:170)
(cid:174)
(cid:172)

(cid:170)
(cid:174)
(cid:172)

.

𝑛𝑘 − 1
𝑛 − |𝜑−1(𝑟)| − 1

𝑀𝑘∑︁

· (cid:169)
(cid:173)
𝑖=𝑀𝑘−1+1
(cid:171)

𝜔𝑖 (𝑄 𝑘)
(cid:214)

𝑗=1

𝑏𝑖, 𝑗 (𝑄 𝑘 ) − 1
𝑐𝑖, 𝑗 (𝑄 𝑘 ) − 1

(cid:170)
(cid:174)
(cid:172)

≤

𝑛𝑘 − 𝑚𝑘
𝑛 − |𝜑−1(𝑟)| − 1

,

as both sides of the inequality are 0. If 𝑛𝑘 > 1, then by the inductive hypothesis we also have

𝑛𝑘 − 1
𝑛 − |𝜑−1(𝑟)| − 1

𝑀𝑘∑︁

· (cid:169)
(cid:173)
𝑖=𝑀𝑘−1+1
(cid:171)

𝜔𝑖 (𝑄 𝑘)
(cid:214)

𝑗=1

𝑏𝑖, 𝑗 (𝑄 𝑘 ) − 1
𝑐𝑖, 𝑗 (𝑄 𝑘 ) − 1

(cid:170)
(cid:174)
(cid:172)

≤

=

𝑛𝑘 − 1
𝑛 − |𝜑−1(𝑟)| − 1

·

𝑛𝑘 − 𝑚𝑘
𝑛𝑘 − 1

𝑛𝑘 − 𝑚𝑘
𝑛 − |𝜑−1(𝑟)| − 1

.

Thus, we conclude that

𝑚
∑︁

𝜔𝑖 (𝑄)
(cid:214)

𝑖=1

𝑗=1

𝑏𝑖, 𝑗 (𝑄) − 1
𝑐𝑖, 𝑗 (𝑄) − 1

≤

=

≤

𝑡
∑︁

𝑛𝑘 − 𝑚𝑘
𝑛 − |𝜑−1(𝑟)| − 1

𝑘=1
𝑛 − |𝜑−1(𝑟)| − 𝑚
𝑛 − |𝜑−1(𝑟)| − 1
𝑛 − 𝑚
𝑛 − 1

.

(3.8)

□

Remark 3.2.6. Since |𝜑−1(𝑟)| > 0, the final inequality in eq. (3.8) is strict for 𝑚 > 1. If 𝑚 = 1,

then there is only one leaf in 𝑄, so 𝑏1, 𝑗 (𝑄) = 𝑐1, 𝑗 (𝑄) for 1 ≤ 𝑗 ≤ 𝜔1(𝑄) and equality holds. In

particular, the upper bound in lemma 3.2.5 is never sharp for 𝑚 > 1.

Definition 3.2.7. Let 𝑃 be an 𝑛 element poset and 𝑋 ⊆ 𝑃. A partial labeling of 𝑃 is an injective

map 𝑀 : 𝑋 → [𝑛]. A labeling 𝐿 : 𝑃 → [𝑛] is an extension of 𝑀 if 𝐿| 𝑋 = 𝑀. The set of extensions

of 𝑀 is denoted Λ(𝑃, 𝑀).

48

Definition 3.2.8. Let 𝑃 be a poset and 𝑥 ∈ 𝑃. The element 𝑥 is lower order ideal complete

(LOI-complete) if any element that is comparable to some element in ↓ 𝑥 is also comparable to 𝑥

itself.

𝑎

𝑒

𝑑

𝑔

𝑏

ℎ

𝑐

𝑃

𝑓

𝑗

𝑖

𝑄

Figure 3.6 A rooted tree poset 𝑄 and an inflation 𝑃 of 𝑄. The LOI-complete elements in 𝑃 are
colored black

Example 3.2.9. Consider the rooted tree poset 𝑄 and its inflation 𝑃 in fig. 3.6. In 𝑃, the elements

𝑐, 𝑓 , 𝑔, and 𝑗 are all LOI-complete, since for each of those elements, all elements comparable to

↓ 𝑐, ↓ 𝑓 , ↓ 𝑔, and ↓ 𝑗 are also comparable to 𝑐, 𝑓 , 𝑔, and 𝑗, respectively. The elements 𝑎, 𝑏, 𝑑, 𝑒,

ℎ, and 𝑖, colored in red, are not LOI-complete. For example, 𝑏 is not LOI-complete because the

element 𝑎 is comparable to 𝑐 ∈↓ 𝑏 but 𝑎 is not comparable to 𝑏.

Lemma 3.2.10. Let 𝑄 be a rooted tree poset and let 𝑃 be an inflation of 𝑄 with inflation map

𝜑 : 𝑃 → 𝑄. For any 𝑞 ∈ 𝑄, the unique minimal element of 𝜑−1(𝑞) is LOI-complete in 𝑃.

Proof. Denote the unique minimal element of 𝜑−1(𝑞) by 𝑥. Let 𝑦 ∈↓ 𝜑−1(𝑞) and suppose 𝑧 ∈ 𝑃

is comparable to 𝑦. If 𝑦 = 𝑥 then 𝑧 and 𝑥 are comparable by definition. Otherwise, if 𝑦 ≠ 𝑥, then

𝜑(𝑦) <𝑄 𝜑(𝑥) = 𝑞 since 𝑥 is the unique minimal element of 𝜑−1(𝑞). Since 𝑧 ∈ 𝑃 is comparable to

𝑦 and 𝑃 is a rooted tree poset, 𝜑(𝑧) and 𝜑(𝑦) are comparable and hence 𝜑(𝑧) and 𝑞 is comparable.

If 𝜑(𝑧) <𝑄 𝑞, then 𝑧 <𝑃 𝑥 by definition 3.2.1. If 𝜑(𝑧) = 𝑞, then 𝑥 ≤𝑃 𝑧 since 𝑥 is the unique

minimal element of 𝜑−1(𝑞).

If 𝑞 <𝑄 𝜑(𝑧), then 𝑥 <𝑃 𝑧.

In each case, 𝑧 is comparable to 𝑥.

Therefore 𝑥 is LOI-complete in 𝑃.

□

49

We will need the following probability lemmas from [19], so we have reproduced them for

convenience.

Lemma 3.2.11 ([19, Lemma 3.10]). Let 𝑃 be an 𝑛 element poset and 𝑥 ∈ 𝑃 be LOI-complete. Let

𝑋 = ↓ 𝑥 \ {𝑥}. For 𝐿 ∈ Λ(𝑃) and 𝑘 ≥ 0, the set 𝐿 𝑘 (𝑋) depends only on the set 𝐿 (𝑋) and the

restriction 𝐿|𝑃\𝑋. It does not depend on the way in which labels in 𝐿(𝑋) are distributed among the

elements of 𝑋.

Lemma 3.2.12 ([19, Lemma 3.11]). Let 𝑃 be an 𝑛 element poset and 𝑥 ∈ 𝑃 be LOI-complete. Let

𝑋 = ↓ 𝑥 \ {𝑥} and suppose that 𝑋 ≠ ∅. Let 𝐴 ⊆ 𝑋 have the property that no element of 𝐴 is

comparable with any element in 𝑋 \ 𝐴 and let 𝑀 : 𝑃 \ 𝑋 → [𝑛] be a partial labeling such that

𝑛−1(1) ∈ 𝑋 for every extension 𝐿 of 𝑀. If a labeling 𝐿 is chosen uniformly at random from the
𝐿−1
extensions in Λ(𝑃, 𝑀), then the probability that 𝐿−1

𝑛−1(1) ∈ 𝐴 is | 𝐴|
|𝑋 | .

By suitably modifying the proof of [19, Theorem 3.5], one can strengthen it to obtain theo-

rem 3.2.13. The following proof is self-contained, but the interested reader may wish to refer to

[19, Section 3] for further details.

Theorem 3.2.13. Let 𝑄 be a reduced rooted tree poset with 𝑚 leaves and let 𝑃 be an inflation

of 𝑄 with 𝑛 elements, with inflation map 𝜑 : 𝑃 → 𝑄. For a nonminimal element 𝑥 ∈ 𝑃, let

ℓ(𝑥) = {𝑖 ∈ [𝑚] : ℓ𝑖 ≤𝑄 𝜑(𝑥)} and 𝜔𝑖,𝑥 = max{ 𝑗 : 𝑢𝑖, 𝑗 ≤𝑄 𝜑(𝑥)}. Then the number of tangled

𝑥-labelings of 𝑃 is given by

|T𝑥 (𝑃)| = (𝑛 − 2)!

𝜔𝑖,𝑥
(cid:214)

∑︁

𝑖∈ℓ(𝑥)

𝑗=1

𝑏𝑖, 𝑗 (𝑄) − 1
𝑐𝑖, 𝑗 (𝑄) − 1

.

Proof. Fix a leaf ℓ𝑖 of 𝑄 and let 𝑥0 be the unique minimal element of 𝜑−1(ℓ𝑖). We will count the

number of tangled labelings 𝐿 such that 𝐿−1(𝑛) = 𝑥0 and 𝐿−1(𝑛 − 1) = 𝑥. By corollary 3.1.15, if

𝐿 is tangled, then 𝑥0 = 𝐿−1(𝑛) <𝑃 𝐿−1(𝑛 − 1) = 𝑥. Thus, we need only consider leaves ℓ𝑖 such that

ℓ𝑖 ≤𝑄 𝜑(𝑥). Furthermore, since 𝑄 is reduced, 𝐿 is tangled if and only if 𝐿−1

𝑛−2(1) ∈ 𝜑−1(ℓ𝑖).

If 𝜔𝑖,𝑥 = 0, then the product (cid:206)𝜔𝑖,𝑥
𝑗=1

𝑏𝑖, 𝑗 (𝑄)−1
𝑐𝑖, 𝑗 (𝑄)−1 is the empty product 1. In this case, 𝑥 ∈ 𝜑−1(ℓ𝑖) so

all 𝑥-labelings 𝐿 such that 𝐿−1(𝑛) = 𝑥0 are tangled.

50

Now, assume 𝜔𝑖,𝑥 ≥ 1 and choose a labeling 𝐿 ∈ Λ(𝑃) uniformly at random among the (𝑛 − 2)!

labelings that satisfy 𝐿−1(𝑛) = 𝑥0 and 𝐿−1(𝑛 − 1) = 𝑥. We will proceed to compute the probability
(cid:101)𝑃. For 1 ≤ 𝑗 ≤ 𝜔𝑖,𝑥, let 𝑥 𝑗 be the unique minimal
that 𝐿 is tangled. Let (cid:101)𝑃 = 𝑃 \ {𝑥0} and (cid:101)
element of (cid:101)

𝜑−1(𝑢𝑖, 𝑗 ), and define the sets

𝜑 = 𝜑|

𝑋 𝑗 =↓ 𝑥 𝑗 \ {𝑥 𝑗 } and 𝐴 𝑗 =

(cid:216)

𝑣≤𝑄𝑢𝑖, 𝑗 −1

𝜑−1(𝑣).
(cid:101)

The sizes of the sets are |𝑋 𝑗 | = 𝑏𝑖, 𝑗 (𝑄) − 1 and | 𝐴 𝑗 | = 𝑐𝑖, 𝑗 (𝑄) − 1.

For any partial labeling 𝑀 : (cid:101)𝑃 \ 𝑋𝜔𝑖,𝑥 → [𝑛 − 1] such that 𝑀 (𝑥) = 𝑛 − 1 and any extension 𝐿 of
𝑛−2(1) ∈ 𝑋𝜔𝑖,𝑥 holds since 𝑥 ∈ 𝜑−1(𝑢𝑖,𝜔𝑖,𝑥 ). Furthermore, since 𝑃 is an inflated
𝜑−1(𝑢𝑖, 𝑗 ), 𝑥 𝑗 is LOI-complete, and no

𝑀, the condition 𝐿−1
rooted forest poset and 𝑥 𝑗 is the unique minimal element of (cid:101)
element of 𝐴 𝑗 is comparable with any element of 𝑋 𝑗 \ 𝐴 𝑗 . Thus, the poset (cid:101)𝑃, the subsets 𝑋𝜔𝑖,𝑥 and
𝐴𝜔𝑖,𝑥 , and the partial labeling 𝑀 satisfy the conditions in lemma 3.2.12. Applying the lemma tells

us that the probability that 𝐿−1

𝑛−2(1) ∈ 𝐴𝜔𝑖,𝑥 is
| 𝐴𝜔𝑖,𝑥 |
|𝑋𝜔𝑖,𝑥 |

=

𝑏𝑖,𝜔𝑖,𝑥 (𝑄) − 1
𝑐𝑖,𝜔𝑖,𝑥 (𝑄) − 1

.

Furthermore, lemma 3.2.11 tells us that the occurrence of this event only depends on 𝐿|

(cid:101)𝑃\𝐴𝜔𝑖, 𝑥

.

This process can be continued for 𝑗 = 𝜔𝑖,𝑥 − 1, . . . , 1 to deduce that the probability that

𝐿−1
𝑛−2(1) ∈ 𝜑−1(𝑢𝑖,0) is the product

𝜔𝑖,𝑥
(cid:214)

𝑗=1

𝑏𝑖, 𝑗 (𝑄) − 1
𝑐𝑖, 𝑗 (𝑄) − 1

.

Summing over all the leaves such that ℓ𝑖 ≤𝑄 𝜑(𝑥) yields the result.

□

Theorem 3.2.14. If 𝑃 is an inflated rooted forest poset on 𝑛 elements and 𝑥 ∈ 𝑃, then |T𝑥 (𝑃)| ≤

(𝑛 − 2)!. Equality holds if and only if there is a unique minimal element 𝑧 ∈ 𝑃 such that 𝑧 <𝑃 𝑥.

Proof. We first consider the case of an inflated rooted tree poset. Let 𝑄 be a reduced rooted tree

poset and 𝑃 an inflation of 𝑄 with |𝑃| = 𝑛. For an element 𝑥 of 𝑃, theorem 3.2.13 implies that

|T𝑥 (𝑃)| = (𝑛 − 2)!

𝜔𝑖,𝑥
(cid:214)

∑︁

𝑖∈ℓ(𝑥)

𝑗=1

𝑏𝑖, 𝑗 (𝑄) − 1
𝑐𝑖, 𝑗 (𝑄) − 1

.

51

The subposet (cid:101)𝑄 :=↓ 𝜑(𝑥) is also a rooted tree poset. Let (cid:101)𝑃 := 𝜑−1( (cid:101)𝑄) and (cid:101)
Then (cid:101)𝑃 is an inflated rooted tree poset, so lemma 3.2.5 gives the upper bound

𝜑 be the restriction 𝜑|

(cid:101)𝑃.

𝜔𝑖,𝑥
(cid:214)

∑︁

𝑖∈ℓ(𝑥)

𝑗=1

𝑏𝑖, 𝑗 (𝑄) − 1
𝑐𝑖, 𝑗 (𝑄) − 1

≤ 1.

(3.9)

Therefore, |T𝑥 (𝑃)| ≤ (𝑛 − 2)! in the case of an inflated rooted tree poset.

Let 𝑚 denote the number of leaves in the subposet (cid:101)𝑄. By remark 3.2.6, the inequality in eq. (3.9)
is strict if and only if 𝑚 > 1. The number of leaves in the subposet (cid:101)𝑄 is precisely the number of
minimal elements in (cid:101)𝑃. By definition of (cid:101)𝑃, the minimal elements in (cid:101)𝑃 are precisely the minimal
elements 𝑧 ∈ 𝑃 that satisfy 𝑧 <𝑃 𝑥. Thus, equality in eq. (3.9) holds if and only if there is a unique

minimal element 𝑧 ∈ 𝑃 that satisfies 𝑧 <𝑃 𝑥.

The general case of an inflated rooted forest poset follows from lemma 3.1.20, since an inflated

rooted forest poset is a disjoint union of inflated rooted tree posets.

□

3.3 Shoelace Posets

In this section, we will study tangled labelings on a new family of posets called shoelace posets

and show that the (𝑛 − 2)! conjecture holds for them. The proof involves a careful analysis of

the number of tangled labelings where a fixed element in the poset is labeled 𝑛 − 1. We note that

in general, shoelace posets are not the inflation of any rooted forest poset. We will also examine

a specific subset of shoelace posets called 𝑊-posets, and enumerate the exact number of tangled

labelings of these posets.

Definition 3.3.1. A shoelace poset 𝑃 is a connected poset defined by a set of minimal elements

{𝑥1, . . . , 𝑥ℓ}, a set of maximal elements {𝑦1, . . . , 𝑦𝑚}, and a set S(𝑃) ⊆ {𝑥1, . . . , 𝑥ℓ} × {𝑦1, . . . , 𝑦𝑚}

such that the following three conditions hold:

1. For every (𝑖, 𝑗) ∈ [ℓ] × [𝑚], the elements 𝑥𝑖 and 𝑦 𝑗 are comparable in 𝑃 if and only if

(𝑥𝑖, 𝑦 𝑗 ) ∈ S(𝑃).

2. For every (𝑥𝑖, 𝑦 𝑗 ) ∈ S(𝑃), the open interval (𝑥𝑖, 𝑦 𝑗 )𝑃 is a (possibly empty) chain, denoted

𝐶 𝑗
𝑖 .

52

3. For distinct pairs (𝑥𝑖, 𝑦 𝑗 ), (𝑥𝑖′, 𝑦 𝑗 ′) ∈ S(𝑃), the chains 𝐶 𝑗

𝑖 and 𝐶 𝑗 ′

𝑖′ are disjoint.

We will use the following notation

S 𝑗 (𝑃) = {𝑥𝑖 : (𝑥𝑖, 𝑦 𝑗 ) ∈ S(𝑃)},

S𝑖 (𝑃) = {𝑦 𝑗 : (𝑥𝑖, 𝑦 𝑗 ) ∈ S(𝑃)}.

The funnels of a shoelace poset can be described fairly simply. The funnel of a minimal element

𝑥𝑖 consists of the elements in 𝐶 𝑗
that satisfy S 𝑗 (𝑃) = {𝑥𝑖}.

𝑖 for 𝑦 𝑗 ∈ S𝑖 (𝑃), along with the maximal elements 𝑦 𝑗 for 𝑦 𝑗 ∈ S𝑖 (𝑃)

Example 3.3.2. fig. 3.7 depicts a shoelace poset 𝑃 with 3 minimal elements and 4 maximal

elements. In this example,

S(𝑃) = {(𝑥1, 𝑦2), (𝑥1, 𝑦3), (𝑥1, 𝑦4), (𝑥2, 𝑦1), (𝑥2, 𝑦3), (𝑥3, 𝑦3), (𝑥3, 𝑦4)}.

The elements of the chain 𝐶4

3 are highlighted in red and the chain 𝐶3

1 is empty. Notice also that

S1(𝑃) = {𝑦2, 𝑦3, 𝑦4} and S2(𝑃) = {𝑥1}.

𝑦1

𝑦2

𝑦3

𝑦4

𝐶4
3

𝑥1

𝑥2

𝑥3

Figure 3.7 An example of a shoelace poset

In order to prove that shoelace posets satisfy the (𝑛 − 2)! conjecture, we will partition labelings

according to the location of the label 𝑛 − 1, and bound |T𝑥 (𝑃)| for the various elements 𝑥.

For the following lemma, we use the following notation: for 𝑆 a set and 𝑓 a function whose

codomain is well-ordered, argmin𝑆 𝑓 is the element 𝑥 ∈ 𝑆 such that 𝑓 (𝑥) is minimal.

Lemma 3.3.3. Let 𝑃 be a shoelace poset with minimal elements 𝑥1, . . . , 𝑥ℓ and maximal elements

𝑦1, . . . , 𝑦𝑚. Let 𝐿 ∈ T (𝑃), 𝑖 ∈ [ℓ], and 𝑗 ∈ [𝑚] such that 𝐿 (𝑦 𝑗 ) = 𝑛 − 1 and 𝐿(𝑥𝑖) = 𝑛.
|S 𝑗 (𝑃)| ≥ 2, then 𝑥𝑖 ∈ S 𝑗 (𝑃), 𝐶 𝑗

𝑖 ≠ ∅, and

If

𝐿 ∈ 𝐶 𝑗
𝑖 .

argmin
↓𝑦 𝑗 \S 𝑗 (𝑃)

53

Proof. Since 𝐿 is tangled, 𝐿−1(𝑛) <𝑃 𝐿−1(𝑛 − 1) by corollary 3.1.15. Therefore, 𝑥𝑖 <𝑃 𝑦 𝑗 , which

implies 𝑖 ∈ S 𝑗 (𝑃). By lemma 3.1.19, there exists 𝑧 ∈ fun(𝑥𝑖) such that 𝑧 ≤𝑃 𝑦 𝑗 . By the assumption
that |S 𝑗 (𝑃)| ≥ 2, we observe that 𝑦 𝑗 ∉ fun(𝑥𝑖). Therefore, 𝑥𝑖 <𝑃 𝑧 <𝑃 𝑦 𝑗 , so 𝐶 𝑗

𝑖 ≠ ∅.

Next, let 𝑟 be the smallest positive integer such that the 𝑟th promotion chain ends in 𝑦 𝑗 . Denote

the 𝑟th promotion chain by (𝑧1, . . . , 𝑧𝑛, 𝑦 𝑗 ). Since 𝑟 is the smallest such positive integer, 𝑦 𝑗 does
𝑟−1(𝑛 − 1 − (𝑟 − 1)) = 𝑦 𝑗 . Then, after the

not lie on the 𝑞th promotion chain for 𝑞 < 𝑟, and hence 𝐿−1

𝑟th promotion, 𝐿−1

𝑟 (𝑛 − 1 − 𝑟) = 𝑧𝑛. Since 𝐿 is a tangled labeling, corollary 3.1.15 implies that

𝑥𝑖 = 𝐿−1

𝑟 (𝑛 − 𝑟) <𝑃 𝐿−1

𝑟 (𝑛 − 1 − 𝑟) = 𝑧𝑛.

Therefore, 𝑧𝑛 ∈ 𝐶 𝑗
promotion chain are also on 𝐶 𝑗
𝑖 .

𝑖 . Since 𝑧1 <𝑃 . . . <𝑃 𝑧𝑛−1, the remaining elements 𝑧1, . . . , 𝑧𝑛−1 in the 𝑟th

Now, let 𝑧 ∈ ↓ 𝑦 𝑗 \ S 𝑗 (𝑃) and let 𝑡 = 𝐿 (𝑧). Then either 𝐿−1

ends in 𝑦 𝑗 , or 𝐿−1
it follows that 𝑟 ≤ 𝑡. Since the starting element of the 𝑟th promotion chain lies in 𝐶 𝑗

𝑡−1(1) = 𝑧 and the 𝑡th promotion chain
𝑡−1(1) <𝑃 𝑧 and the 𝑡′-th promotion chain ends in 𝑦 𝑗 for some 𝑡′ < 𝑡. In either case,
𝑖 , we conclude

that argmin
↓𝑦 𝑗 \S 𝑗 (𝑃)

𝐿 ∈ 𝐶 𝑗
𝑖 .

□

Essentially, if a labeling on a shoelace poset is tangled, and 𝐿(𝑦 𝑗 ) = 𝑛 − 1, then the element with

smallest label in ↓ 𝑦 𝑗 \ 𝑆 𝑗 (𝑃) must be above the element labeled 𝑛. This is therefore a necessary

condition for a labeling on a shoelace poset to be tangled. This will be instrumental in proving the

following theorem.

Theorem 3.3.4. If 𝑃 is a shoelace poset on 𝑛 elements and 𝑧 ∈ 𝑃, then |T𝑧 (𝑃)| ≤ (𝑛 − 2)!. Equality

holds if and only if there is a unique minimal element 𝑥 <𝑃 𝑧.

Proof. Let 𝑥1, . . . , 𝑥ℓ be minimal elements of 𝑃, and 𝑦1, . . . , 𝑦𝑚 be maximal elements of 𝑃. The
element 𝑧 can either be a minimal element, an element on a chain 𝐶 𝑗
𝑖 for some 𝑖 and 𝑗, or a maximal
element. There is a unique minimal element 𝑥 <𝑃 𝑧 only if 𝑧 ∈ 𝐶 𝑗
𝑖 or if 𝑧 is one of the maximal
elements 𝑦 𝑗 and |S 𝑗 (𝑃)| = 1. For convenience, we set 𝑠 := |S 𝑗 (𝑃)|. Below, we separate the cases

mentioned above and claim that equality holds only in Case 2 and Case 3.

54

Case 1: Suppose 𝑧 is a minimal element. In this case, it is impossible to find an element

labeled 𝑛 such that 𝐿−1(𝑛) <𝑃 𝐿−1(𝑛 − 1) = 𝑧. So by corollary 3.1.15, |T𝑧 (𝑃)| = 0.

Case 2: Suppose 𝑧 lies on some chain 𝐶 𝑗

𝑖 . In this case there is a unique basin 𝑥𝑖 that in ↓ 𝑧.
Any tangled labeling 𝐿 ∈ T𝑧 (𝑃) must satisfy 𝐿(𝑧) = 𝑛 − 1 and 𝐿 (𝑥𝑖) = 𝑛. There are at most (𝑛 − 2)!

such labelings, and by lemma 3.1.18 all such labelings are tangled so |T𝑧 (𝑃)| = (𝑛 − 2)!.

Case 3: Suppose 𝑧 is a maximal element 𝑦 𝑗 and 𝑠 = 1. Since 𝑠 = 1, for any tangled labeling

𝐿, 𝐿−1(𝑛) must be the unique 𝑥𝑖 satisfying 𝑥𝑖 <𝑃 𝑦 𝑗 = 𝑧. There are (𝑛 − 2)! such labelings, and by

lemma 3.1.18 all such labelings are tangled. Thus, |T𝑧 (𝑃)| = (𝑛 − 2)!.

Case 4: Suppose 𝑧 is a maximal element 𝑦 𝑗 and 𝑠 ≥ 2. Partition Λ(𝑃) into equivalence

classes, where two labelings 𝐿 and 𝐿′ belong to the same equivalence class if and only if they

restrict to the same labeling on 𝑃 \ S 𝑗 (𝑃). Labelings in T𝑧 (𝑃) require 𝑦 𝑗 to be labeled 𝑛 − 1 and

some element in S 𝑗 (𝑃) to be labeled 𝑛. The number of equivalence classes where this is possible

is (𝑛 − 2)(𝑛 − 3) · · · 𝑠. In each such equivalence class, the tangled labelings 𝐿 have only one choice

of 𝐿−1(𝑛) according to lemma 3.3.3. Therefore, at most (𝑠 − 1)! labelings in each equivalence class

are tangled. Consequently, |T𝑧 (𝑃)| ≤ (𝑛 − 2) (𝑛 − 3) · · · 𝑠(𝑠 − 1)! = (𝑛 − 2)!.

With a little more careful analysis, one can conclude that at least one of the equivalence classes

has strictly fewer than (𝑠 − 1)! labelings. Consider an equivalence class where the label 1 is in

S 𝑗 (𝑃) and the label 2 is in ↓ 𝑦 𝑗 \ S 𝑗 (𝑃). Then in this equivalence class, there is the additional

restriction 𝐿−1(1) ≮𝑃 𝐿−1(2). Thus, there are strictly fewer than (𝑠 − 1)! tangled labelings, so

|T𝑧 (𝑃)| < (𝑛 − 2)!.

□

Notice that theorem 3.3.4 shows that shoelaces satisfy conjecture 3.0.5, and therefore also

satisfy conjecture 3.0.3 and conjecture 3.0.2.

We have proven an upper bound on the number of tangled labelings of shoelaces, but we are

also able to enumerate the exact number of tangled labelings for a specific subfamily of shoelace

posets called 𝑊-posets. In general, few explicit formulas for tangled labelings are known. The

proof of this formula will also involve counting the number of tangled labelings by fixing the label

55

𝑛 − 1.

Definition 3.3.5. Given 𝑎, 𝑏, 𝑐, 𝑑 ∈ Z≥0, the 𝑊-poset 𝑊𝑎,𝑏,𝑐,𝑑 is a poset on 𝑎 + 𝑏 + 𝑐 + 𝑑 + 3

elements: 𝛼1, . . . , 𝛼𝑎, 𝛽1, . . . , 𝛽𝑏, 𝛾1, . . . , 𝛾𝑐, 𝛿1, . . . , 𝛿𝑑, 𝑥, 𝑦, 𝑧. The partial order has covering

relations 𝛼𝑖 ⋖𝑊 𝛼𝑖+1, 𝛽𝑖 ⋖𝑊 𝛽𝑖+1, 𝛾𝑖 ⋖𝑊 𝛾𝑖+1, 𝛿𝑖 ⋖𝑊 𝛿𝑖+1, 𝑥 ⋖𝑊 𝛼1, 𝑥 ⋖𝑊 𝛽1, 𝛽𝑏 ⋖𝑊 𝑦, 𝛾𝑐 ⋖𝑊 𝑦,

𝑧 ⋖𝑊 𝛾1, and 𝑧 ⋖𝑊 𝛿1.

The poset 𝑊𝑎,𝑏,𝑐,𝑑 can be viewed as the shoelace poset with the set of minimal elements {𝑥, 𝑧},

the set of maximal elements {𝛼𝑎, 𝑦, 𝛿𝑑 } and the relations S(𝑃) = {(𝑥, 𝛼𝑎),(𝑥, 𝑦),(𝑧, 𝑦),(𝑧, 𝛿𝑑)}.

Example 3.3.6. The Hasse diagram for 𝑊2,2,1,1 is shown in fig. 3.8. There are 34,412 tangled

labelings of this poset.

𝑦

𝛼2

𝛽2

𝛼1

𝛽1

𝛾1

𝛿1

𝑥

𝑧

Figure 3.8 The poset 𝑊2,2,1,1

Theorem 3.3.7. Let 𝑎, 𝑏, 𝑐, 𝑑 be four positive integers and 𝑛 = 𝑎 + 𝑏 + 𝑐 + 𝑑 + 3. Let

𝑋 =

𝑍 =

(cid:18)𝑛 − 2
𝑎

(cid:18)𝑛 − 2
𝑑

(cid:19) 𝑏−1
∑︁

𝑑
∑︁

𝑖=0
(cid:19) 𝑐−1
∑︁

𝑗=0
𝑎
∑︁

𝑖=0

𝑗=0

(𝑑 − 𝑗 + 1)

(cid:18)𝑖 + 𝑗 + 𝑐 − 1
𝑖, 𝑗, 𝑐 − 1

(cid:19)

, and

(𝑎 − 𝑗 + 1)

(cid:18)𝑖 + 𝑗 + 𝑏 − 1
𝑖, 𝑗, 𝑏 − 1

(cid:19)

.

Then the number of tangled labelings of 𝑊𝑎,𝑏,𝑐,𝑑 is given by (𝑛 − 2) (𝑛 − 2)! − 𝑎!𝑏!𝑐!𝑑!(𝑋 + 𝑍).

Proof. Fix 𝑎, 𝑏, 𝑐, 𝑑 and write 𝑊 = 𝑊𝑎,𝑏,𝑐,𝑑. By eq. (3.1), it suffices to compute |T𝑝 (𝑊)| as 𝑝

ranges over elements of 𝑊. If 𝑝 = 𝑥 or 𝑝 = 𝑧, then |T𝑝 (𝑊)| = 0 due to Case 1 in the proof of

theorem 3.3.4. If 𝑝 = 𝛼𝑖 or 𝑝 = 𝛽𝑖, then this belongs to Cases 2 and 3 in the proof of theorem 3.3.4,

56

and so |T𝑝 (𝑊)| = (𝑛 − 2)!. Similarly, if 𝑝 = 𝛾𝑖 or 𝑝 = 𝛿𝑖, then |T𝑝 (𝑊)| = (𝑛 − 2)!. With the

exception of 𝑝 = 𝑦, we have counted (𝑎 + 𝑏 + 𝑐 + 𝑑) (𝑛 − 2)! = (𝑛 − 3) (𝑛 − 2)! tangled labelings.

Let us now count the number of tangled labelings 𝐿 that satisfy 𝐿(𝑦) = 𝑛 − 1. Observe that

permuting the labels 𝐿 (𝛼1), . . . , 𝐿 (𝛼𝑎) does not change whether or not 𝐿 is tangled. Similarly, per-

muting the labels 𝐿 (𝛽1), . . . , 𝐿 (𝛽𝑏), the labels 𝐿(𝛾1), . . . , 𝐿 (𝛾𝑐), and the labels 𝐿 (𝛿1), . . . , 𝐿 (𝛿𝑑)

among themselves does not change whether or not 𝐿 is tangled. Thus, we will additionally im-

pose the conditions 𝐿 (𝛼1) < · · · < 𝐿(𝛼𝑎), 𝐿(𝛽1) < · · · < 𝐿(𝛽𝑏), 𝐿 (𝛾1) < · · · < 𝐿(𝛾𝑐), and

𝐿 (𝛿1) < · · · < 𝐿(𝛿𝑑). To obtain the total number of tangled labelings, we will count the number

of such tangled labelings 𝐿 satisfying these conditions and then multiply by 𝑎!𝑏!𝑐!𝑑!.

We split into two cases. The first case is where 𝐿 (𝛽1) < 𝐿 (𝛾1). Let 𝑚 𝛽 = 𝐿 (𝛽1). In this

case, a necessary condition for 𝐿 to be tangled is that 𝐿(𝑥) = 𝑛. To see this, suppose otherwise

that 𝐿(𝑧) = 𝑛. Then note that 𝐿−1

𝑚𝛽 (𝑛 − 1 − 𝑚 𝛽) ∈ [𝑥, 𝑦). This is because for the first 𝑚 𝛽
promotions, the only promotion chains ending in 𝑦 are those that begin with some element in [𝑥, 𝑦)

and furthermore, there exists at least one promotion chain ending in 𝑦, namely the 𝑚 𝛽-th one. It

follows that 𝐿−1

𝑛−2(1) ≯𝑊 𝑧 so 𝐿 cannot be tangled if 𝐿 (𝑧) = 𝑛 (lemma 3.1.13).

Now, the total number of labelings that satisfy all these conditions is given by 1
2

(cid:0)

𝑛−2
𝑎,𝑏,𝑐,𝑑,1

(cid:1), since

it amounts to choosing 𝑎 of the labels in [𝑛 − 2] for 𝛼1, . . . , 𝛼𝑎, 𝑏 of the labels for the 𝛽s and so on.

To account for the condition 𝐿 (𝛽1) < 𝐿(𝛾1), we divide by 2 because there is an involution swapping

𝐿(𝛽1) and 𝐿 (𝛾1). We will now subtract the number of labelings satisfying these conditions that

are not tangled.

Given that 𝐿 satisfies all the conditions above, 𝐿 is not tangled if and only if 𝐿(𝑧) < 𝐿(𝛽1)

and there do not exist 𝛿𝑖 such that 𝐿 (𝛽1) < 𝐿(𝛿𝑖) < 𝐿(𝛾1). To see this, observe that 𝐿 is not

tangled if and only if there is some 𝑗 < 𝑚 𝛽 where the 𝑗th promotion chain begins with an element

in [𝑧, 𝑦) and ends in 𝑦. Since 𝐿 (𝛽1) < 𝐿(𝛾1), this can occur only if 𝐿 (𝑧) < 𝐿(𝛽1). Now,

let 𝛿𝑖 <𝑊 𝛿𝑖+1 <𝑊 · · · <𝑊 𝛿 𝑗 be all the 𝛿’s with labels in between 𝐿 (𝑧) and 𝐿(𝛾1). Then the

𝐿 (𝑧), 𝐿(𝛿𝑖), . . . , 𝐿 (𝛿 𝑗−1)th promotion chains would all begin with 𝑧 and end with some 𝛿𝑘 , and the

𝐿 (𝛿 𝑗 )th promotion chain would begin with 𝑧 and end with 𝑦. Thus, in order for 𝐿 to not be tangled

57

we must have 𝐿 (𝛿 𝑗 ) < 𝐿(𝛽1). And conversely, if we do have 𝐿(𝛿 𝑗 ) < 𝐿(𝛽1) then 𝐿 is not tangled

since the 𝐿(𝛿 𝑗 )th promotion chain would start with 𝑧 and end with 𝑦.

Now, we wish to count the number of such labelings 𝐿. To do so, observe that the labels of the

𝛼’s are subject to no constraints. We will suppose that 𝐿(𝛿1) < · · · < 𝐿 (𝛿𝑑− 𝑗 ) < 𝐿 (𝛽1) < · · · <

𝐿 (𝛽𝑏−𝑖) < 𝐿(𝛾1) and sum over 0 ≤ 𝑖 ≤ 𝑏 − 1 and 0 ≤ 𝑗 ≤ 𝑑.

For each 𝑖, 𝑗 there are (𝑑 − 𝑗 + 1) choices of what 𝐿(𝑧) could be and (cid:0)𝑖+ 𝑗+𝑐−1
𝑖, 𝑗,𝑐−1

(cid:1) choices for the

labels greater than 𝐿(𝛾1). This yields

𝑋 =

(cid:18)𝑛 − 2
𝑎

(cid:19) 𝑏−1
∑︁

𝑑
∑︁

𝑖=0

𝑗=0

(𝑑 − 𝑗 + 1)

(cid:18)𝑖 + 𝑗 + 𝑐 − 1
𝑖, 𝑗, 𝑐 − 1

(cid:19)

.

By a similar argument, if 𝐿 (𝛾1) < 𝐿(𝛽1) then a necessary condition for 𝐿 to be tangled is
(cid:1) and the number of

𝐿 (𝑧) = 𝑛. The number of labelings satisfying these conditions is 1
2

𝑛−2
𝑎,𝑏,𝑐,𝑑,1

(cid:0)

these labelings that are not tangled is

𝑍 =

(cid:18)𝑛 − 2
𝑑

(cid:19) 𝑐−1
∑︁

𝑎
∑︁

𝑖=0

𝑗=0

(𝑎 − 𝑗 + 1)

(cid:18)𝑖 + 𝑗 + 𝑏 − 1
𝑖, 𝑗, 𝑏 − 1

(cid:19)

.

Let 𝐸 = 𝑎!𝑏!𝑐!𝑑!. Then, the number of tangled labelings 𝐿 that satisfy 𝐿 (𝑦) = 𝑛 − 1 is

𝐸

(cid:18)

(cid:18) 1
2

𝑛 − 2
𝑎, 𝑏, 𝑐, 𝑑, 1

(cid:19)

− 𝑋 +

(cid:18)

(cid:19)

𝑛 − 2
𝑎, 𝑏, 𝑐, 𝑑, 1

1
2

(cid:19)

− 𝑍

= 𝐸

(cid:18)

(cid:19)

𝑛 − 2
𝑎, 𝑏, 𝑐, 𝑑, 1

− 𝐸 (𝑋 + 𝑍)

= (𝑛 − 2)! − 𝐸 (𝑋 + 𝑍).

Adding this to the (𝑛 − 3)(𝑛 − 2)! tangled labelings where 𝐿−1(𝑛 − 1) ≠ 𝑦 yields the desired

formula.

□

In principle, one could compute the exact number of tangled labelings for various subsets of

shoelace posets in this way. Even for the class of 𝑊-posets, however, the computations appear

rather unwieldy.

3.4 Generating Functions

In the previous sections, we focused on counting the number of tangled labelings of various

posets and analyzed their upper bounds. In this section, we are interested in exploring the number

58

of labelings of a poset 𝑃 on 𝑛 elements that have a fixed order 𝑘. Recall that the order of a labeling

𝐿 is the minimal integer 𝑘 ≥ 0 such that 𝐿 𝑘 is sorted. Such labelings we will call 𝑘-sorted; see

definition 3.4.1. Dual to 𝑘-sorted labelings are 𝑘-tangled labelings that have order 𝑛 − 𝑘 − 1.

We define two kinds of generating functions (definition 3.4.2) on 𝑃 and investigate how these

generating functions change if we attach some minimal elements to 𝑃. Our result provides a simple

and unified proof of enumerating tangled labelings and quasi-tangled labelings in [19] and [32]

(see remark 3.4.11).

Definition 3.4.1. Let 𝑃 be an 𝑛-element poset. A labeling 𝐿 ∈ Λ(𝑃) is said to be 𝑘-sorted if

or(𝐿) = 𝑘 and is said to be 𝑘-tangled if or(𝐿) = 𝑛 − 𝑘 − 1.

Observe that natural labelings are synonymous with 0-sorted labelings and tangled labelings

are synonymous with 0-tangled labelings. Quasi-tangled labelings introduced in [32] correspond

exactly to 1-tangled labelings.

Definition 3.4.2. Let 𝑃 be an 𝑛-element poset. The sorting generating function of 𝑃 is defined to

be

𝑓𝑃 (𝑞) :=

∑︁

𝑞or(𝐿) =

𝐿∈Λ(𝑃)

𝑛−1
∑︁

𝑖=0

𝑎𝑖𝑞𝑖,

where 𝑎𝑖 counts the number of 𝑖-sorted labelings of 𝑃. The cumulative generating function of 𝑃 is

defined to be

𝑔𝑃 (𝑞) :=

𝑛−1
∑︁

𝑖=0

𝑏𝑖𝑞𝑖,

where 𝑏𝑖 := 𝑎0 + 𝑎1 + · · · + 𝑎𝑖 is the partial sum of 𝑎𝑖’s. In particular, 𝑏𝑛−1 = 𝑛!.

Example 3.4.3. We list all the six labelings and their orders of the Λ-shaped poset 𝑃 in table 3.1.

The sorting generating function and cumulative generating function of 𝑃 are given by 𝑓𝑃 (𝑞) = 2+4𝑞

and 𝑔𝑃 (𝑞) = 2 + 6𝑞 + 6𝑞2.

We now define precisely what it means to attach 𝑘 minimal elements to a poset. The operation

we need is the ordinal sum of two posets 𝑃 and 𝑄.

59

1

1

3

3

1

1

2

1

2

1

3

3

2

1

1

1

3

0

2

2

3

0

1

2

Labeling
Order

Table 3.1 The six labelings and their corresponding orders for the Λ-shaped poset

Definition 3.4.4. Let 𝑃 and 𝑄 be two posets. The ordinal sum of 𝑃 and 𝑄 is the poset 𝑃 ⊕ 𝑄 on

the elements of the disjoint union 𝑃 ⊔ 𝑄 such that 𝑠 ≤ 𝑡 in 𝑃 ⊕ 𝑄 if and only if at least one of the

following conditions hold:

1. 𝑠, 𝑡 ∈ 𝑃 and 𝑠 ≤𝑃 𝑡, or

2. 𝑠, 𝑡 ∈ 𝑄 and 𝑠 ≤𝑄 𝑡, or

3. 𝑠 ∈ 𝑃 and 𝑡 ∈ 𝑄.

The 𝑛-element chain will be denoted 𝐶𝑛 and the 𝑘-element antichain will be denoted 𝑇𝑘 . In

the language of ordinal sums, we can view 𝐶𝑛 as the ordinal sum of 𝑛 copies of 𝐶1’s and we can

view attaching 𝑘 minimal elements to a poset 𝑃 as the ordinal sum 𝑇𝑘 ⊕ 𝑃. Our main result in this

section provides a way to compute the sorting generating function 𝑓𝑇𝑘 ⊕𝑃 (𝑞) from 𝑓𝑃 (𝑞).

Define a lower-triangular 𝑛 × 𝑛 matrix 𝑋𝑛 (𝑘) whose (𝑖, 𝑗) entry 𝑥𝑖 𝑗 is given by

𝑥𝑖 𝑗 :=

𝑘!(cid:0)𝑘+𝑖−2
𝑘−1

(cid:1)

𝑘!(cid:0)𝑘+𝑖−1
𝑘

(cid:1)

if 𝑖 > 𝑗,

if 𝑖 = 𝑗,

0

otherwise.






Recall that given a labeling on a poset, the standardization of the restricted labeling on a

subposet 𝑄 shifts the labels to those from 1 to |𝑄|; see definition 3.1.6.

Theorem 3.4.5. Let 𝑃 be an 𝑛-element poset and 𝑓𝑃 (𝑞) = (cid:205)𝑛−1
𝑖=0
function of 𝑃. Write the sorting generating function of 𝑇𝑘 ⊕ 𝑃 as 𝑓𝑇𝑘 ⊕𝑃 (𝑞) = (cid:205)𝑛+𝑘−1
𝑣 = (𝑎0, 𝑎1, . . . , 𝑎𝑛−1)⊺ be the column vector of the coefficients of 𝑓𝑃 (𝑞) and 𝑣′ = (𝑎′
0
the column vector of the first 𝑛 coefficients of 𝑓𝑇𝑘 ⊕𝑃 (𝑞). Then

𝑎𝑖𝑞𝑖 be the sorting generating
𝑖𝑞𝑖. Let
𝑎′
𝑛−1)⊺

, . . . , 𝑎′

, 𝑎′
1

𝑖=0

60

1. 𝑋𝑛 (𝑘)𝑣 = 𝑣′,

2. 𝑎′

𝑛 = 𝑛!𝑘!(cid:0)𝑛+𝑘−1
𝑘−1

(cid:1), and

3. 𝑎′

𝑖 = 0 for 𝑖 = 𝑛 + 1, 𝑛 + 2, . . . , 𝑛 + 𝑘 − 1.

Proof. Let 𝑥1, 𝑥2, . . . , 𝑥𝑘 be the elements of 𝑇𝑘 . Since the roles of the 𝑥𝑖’s are symmetrical, it

follows that permuting the labels of the 𝑥𝑖’s on any labeling 𝐿 ∈ Λ(𝑇𝑘 ⊕ 𝑃) doesn’t change or(𝐿).

Therefore, we will compute the number of labelings that satisfy 𝐿 (𝑥1) < 𝐿(𝑥2) < · · · < 𝐿(𝑥𝑘 ) and

then multiply by 𝑘!.

Now, we will define a procedure that, given a labeling 𝐿 ∈ Λ(𝑃) and a 𝑘-tuple of distinct
numbers 𝐼 = (𝑖1, . . . , 𝑖𝑘 ) ∈ [𝑛 + 𝑘] 𝑘 , produces a labeling 𝐿 𝐼 ∈ Λ(𝑇𝑘 ⊕ 𝑃) such that 𝐿 𝐼 (𝑥𝑠) = 𝑖𝑠 for
1 ≤ 𝑠 ≤ 𝑘. Since we are counting labelings where the labels of the 𝑥𝑖’s are increasing, we will

assume that 𝑖1 < 𝑖2 < · · · < 𝑖𝑘 for the rest of the proof.

To obtain 𝐿 𝐼, first define labelings of 𝐿0, 𝐿1, . . . , 𝐿 𝑘 of 𝑃, where 𝐿0 := 𝐿 and for 𝑠 = 1, . . . , 𝑘,

recursively define 𝐿𝑠 by

𝐿𝑠 (𝑥) :=





𝐿𝑠−1(𝑥) + 1

if 𝐿𝑠−1(𝑥) ≥ 𝑖𝑠,

𝐿𝑠−1(𝑥)

otherwise.

Then define 𝐿 𝐼 on 𝑇𝑘 ⊕ 𝑃 by

𝐿 𝐼 (𝑥) :=

𝑖𝑠

if 𝑥 = 𝑥𝑠,

𝐿 𝑘 (𝑥)

if 𝑥 ∈ 𝑃.





(3.10)

In fig. 3.9, we give an example of defining 𝐿 𝐼 of 𝑇3 ⊕ 𝑃 on a 7-element poset 𝑃 and with
𝐼 = (2, 4, 7). The labeling of 𝑃 is given in the left figure, and the middle three figures illustrate the
process mentioned above. The right figure is the resulting labeling 𝐿 𝐼 of 𝑇3 ⊕ 𝑃.

One can check that at each step 𝑠 = 1, . . . , 𝑘 the standardization st(𝐿𝑠) is precisely 𝐿. Therefore,

the standardization of 𝐿 𝐼 |𝑃 is st(𝐿 𝐼 |𝑃) = st(𝐿 𝑘 ) = 𝐿. In other words, 𝐿 𝐼 is the unique labeling in

61

4

3

5
1

6
2

𝐿0 ∈ Λ(𝑃)

4

5

6
1

𝐿1

5

7
3

6

7
1

𝐿2

5

8
3

6

8
1

𝐿3

9
3

6

5

8
1
4
𝐿 𝐼 ∈ Λ(𝑇3 ⊕ 𝑃)

9
3
7

2

Figure 3.9 Defining 𝐿 𝐼 of 𝑇3 ⊕ 𝑃 with 𝐼 = (2, 4, 7)

Λ(𝑇𝑘 ⊕ 𝑃) that assigns the label 𝑖𝑠 to 𝑥𝑠 for 𝑠 = 1, . . . , 𝑘 and whose standardization when restricted

to 𝑃 is 𝐿. As a consequence, the set of labelings Λ(𝑇𝑘 ⊕ 𝑃) can be partitioned as

Λ(𝑇𝑘 ⊕ 𝑃) =

(cid:26)

(cid:196)

𝐿 [𝐼] : 𝐼 ∈

𝐿∈Λ(𝑃)

(cid:19)(cid:27)

,

(cid:18)[𝑛]
𝑘

(3.11)

where 𝐿 [𝐼] contains the labeling 𝐿 𝐼 and the labelings obtained from 𝐿 𝐼 by permuting all the labels

of the 𝑥𝑖’s.

Next, we proceed with the following two claims.

Claim 1. Given 𝐿 ∈ Λ(𝑃) and 𝐼 = (𝑖1, . . . , 𝑖𝑘 ), the standardization of 𝐿 𝐼 |𝑃 is preserved under a

sequence of promotions:

st((𝐿 𝐼

𝑗 )|𝑃) = 𝐿 𝑗 , for all 𝑗 ∈ Z≥0.

(3.12)

Proof of Claim 1. We will show Claim 1 by induction. When 𝑗 = 0, the identity holds by the

definition of 𝐿 𝐼. Suppose it holds for some 𝑗 and consider 𝐿 𝐼

𝑗+1. If 𝐿 𝐼

𝑗 (𝑥𝑠) > 1 for all 𝑠 = 1, 2, . . . , 𝑘,

then these minimal elements 𝑥𝑠’s are not in the ( 𝑗 + 1)-th promotion chain and the claim holds.

On the other hand, if there exists an 𝑠 such that 𝐿 𝐼

𝑗 (𝑥𝑠) = 1, then the ( 𝑗 + 1)-th promotion begins

at 𝑥𝑠. Since 𝑥𝑠 ≤ 𝑥 for all 𝑥 ∈ 𝑃, the next element in the promotion chain is (𝐿 𝐼

𝑗 )−1(𝑦), where

𝑦 = min{𝐿 𝐼

𝑗 (𝑧) : 𝑧 ∈ 𝑃}. This element is exactly (𝐿 𝐼

𝑗 )−1(𝑦) = (𝐿 𝑗 )−1(1). From this point on,

the rest of the promotion chain is the same in 𝐿 𝐼

𝑗 and 𝐿 𝑗 . Therefore, st((𝐿 𝐼

𝑗 )|𝑃) = 𝐿 𝑗 for all

𝑗 ∈ Z≥0.

Claim 2. Given 𝐿 ∈ Λ(𝑃) and 𝐼 = (𝑖1, . . . , 𝑖𝑘 ), the order of 𝐿 𝐼 is given by

or(𝐿 𝐼) = max(𝑖𝑘 − 𝑘, or(𝐿)).

62

□

(3.13)

Proof of Claim 2. We observe that for some nonnegative integer 𝑗, 𝐿 𝐼

𝑗 is a natural labeling if and

only if two conditions are satisfied:

1. the set of labels {𝐿 𝐼

𝑗 (𝑥1), . . . , 𝐿 𝐼

𝑗 (𝑥𝑘 )} is [𝑘], and

2. (𝐿 𝐼

𝑗 )|𝑃 is a natural labeling.

By eq. (3.12), the second condition is satisfied if and only if 𝑗 ≥ or(𝐿). On the other hand, we

show below that the first condition is satisfied if and only if 𝑗 ≥ 𝑖𝑘 − 𝑘.

To see this, we notice that the first 𝑖1 − 1 promotions only decrement the labels of 𝑥1, . . . , 𝑥𝑘 .

Let S𝑗 := {𝐿 𝐼

𝑗 (𝑥1), . . . , 𝐿 𝐼

𝑗 (𝑥𝑘 )} and let 𝑠 𝑗 be the maximum value (possibly 0) such that [𝑠 𝑗 ] ⊆ S𝑗 .

Then the minimum label in 𝐿 𝐼

𝑗 |𝑃 is 𝑠 𝑗 + 1 and in the ( 𝑗 + 1)-th promotion, (𝐿 𝐼

𝑗 )−1(𝑠 𝑗 + 1) is part

of the promotion chain, so S𝑗+1 = [𝑠 𝑗 ] ∪ {𝑦 − 1 : 𝑦 ∈ S𝑗 \ [𝑠 𝑗 ]}. Note that 𝑠 𝑗+1 > 𝑠 𝑗 if and only

if 𝑠 𝑗 + 1 ∈ {𝑦 − 1 : 𝑦 ∈ S𝑗 \ [𝑠 𝑗 ]}. Thus, it follows by an inductive argument that 𝑠 𝑗 ≥ 𝑡 if and

only if 𝑗 ≥ 𝑖𝑡 − 𝑡 which yields the desired result. Combining these two conditions implies that

or(𝐿 𝐼) = max(𝑖𝑘 − 𝑘, or(𝐿)).

□

We are now ready to prove the first statement, in which we show that for 1 ≤ 𝑠 ≤ 𝑛, 𝑘! times

the number of labelings in Λ(𝑇𝑘 ⊕ 𝑃) with order 𝑚 − 1 is equal to the 𝑚th row of 𝑋𝑛 (𝑘)𝑣. By
eq. (3.11), we can sum over all labelings 𝐿 ∈ Λ(𝑃) and count the number of 𝐼 ∈ (cid:0)[𝑛]
(cid:1) such that
𝑘
or(𝐿 𝐼) = 𝑚 − 1. We proceed by cases analysis of or(𝐿).

• Suppose or(𝐿) < 𝑚 − 1. Then in order for or(𝐿 𝐼) = max(𝑖𝑘 − 𝑘, or(𝐿)) = 𝑚 − 1 to hold,
(cid:1) ways to choose

it must be that 𝑖𝑘 − 𝑘 = 𝑚 − 1. Fixing 𝑖𝑘 = 𝑘 + 𝑚 − 1, there are (cid:0)𝑘+𝑚−2
𝑘−1
𝑖1, . . . , 𝑖𝑘−1 such that or(𝐿 𝐼) = 𝑚 − 1.

• Suppose or(𝐿) = 𝑚 − 1. Then in order for or(𝐿 𝐼) = max(𝑖𝑘 − 𝑘, or(𝐿)) = 𝑚 − 1 to hold,
(cid:1) ways to choose

it must be that 𝑖𝑘 − 𝑘 ≤ 𝑚 − 1. Thus, 𝑖𝑘 ≤ 𝑘 + 𝑚 − 1 so there are (cid:0)𝑘+𝑚−1
𝑖1, . . . , 𝑖𝑘 such that or(𝐿 𝐼) = 𝑚 − 1.

𝑘

• Suppose or(𝐿) > 𝑚 − 1. Then or(𝐿 𝐼) = max(𝑖𝑘 − 𝑘, or(𝐿)) > 𝑚 − 1 so there are no choices

of 𝐼 that yield or(𝐿 𝐼) = 𝑚 − 1.

63

After multiplying by 𝑘! to account for the fact that permuting the labels of 𝑥1, . . . , 𝑥𝑘 do not change

the order of a labeling of 𝑇𝑘 ⊕ 𝑃, the first case yields the (𝑚, 𝑗) entry of 𝑋𝑛 (𝑘) when 𝑗 < 𝑚, the

middle case yields the (𝑚, 𝑚) entry of 𝑋𝑛 (𝑘), and the last case yields the (𝑚, 𝑗) entry of 𝑋𝑛 (𝑘)

when 𝑗 > 𝑚. This completes the proof of the first statement.

To prove the second statement, observe that since or(𝐿) ≤ 𝑛 − 1 for any 𝐿 ∈ Λ(𝑃), then
or(𝐿 𝐼) = max(𝑖𝑘 − 𝑘, or(𝐿)) = 𝑛 if and only if 𝑖𝑘 − 𝑘 = 𝑛. Fixing 𝑖𝑘 = 𝑘 + 𝑛, there are (cid:0)𝑛+𝑘−1
𝑘−1

(cid:1)

choices for 𝑖1, . . . , 𝑖𝑘−1, regardless of or(𝐿). Multiplying by 𝑘! to account for permuting the labels

of 𝑥1, . . . , 𝑥𝑘 yields

𝑎′
𝑛 = 𝑘!

(cid:19)

(cid:18)𝑛 + 𝑘 − 1
𝑘 − 1

(𝑎0 + 𝑎1 + · · · + 𝑎𝑛−1) = 𝑛!𝑘!

(cid:18)𝑛 + 𝑘 − 1
𝑘 − 1

(cid:19)

.

This completes the proof of the second statement.

Finally to prove the last statement, first observe that 𝑖𝑘 ≤ 𝑘 +𝑛 since there are only 𝑘 +𝑛 elements

in 𝑇𝑘 ⊕ 𝑃. Thus, 𝑖𝑘 − 𝑘 ≤ 𝑛. In addition, any labeling 𝐿 ∈ Λ(𝑃) has or(𝐿) ≤ 𝑛 − 1. It follows that
or(𝐿 𝐼) ≤ 𝑛 for any choice of 𝐿 ∈ Λ(𝑃) and 𝐼 ∈ (cid:0)[𝑛+𝑘]
(cid:1). Thus, there do not exist labelings 𝑇𝑘 ⊕ 𝑃

𝑘

with order greater than 𝑛 and hence 𝑎′

𝑖 = 0 for 𝑖 = 𝑛 + 1, 𝑛 + 2, . . . , 𝑛 + 𝑘 − 1. This completes the

proof of the last statement.

□

We would like to point out that if 𝑘 ≥ 2, then 𝑇𝑘 ⊕ 𝑃 has no tangled labelings.

Example 3.4.6. Let 𝑃 be as in example 3.4.3. The sorting generating function of 𝑃 is given by

𝑓𝑃 (𝑞) = 2 + 4𝑞. Let 𝑣 be the column vector (2, 4, 0)⊺. We show below how to obtain the sorting

generating function of posets shown in fig. 3.10 from theorem 3.4.5.

For 𝑇1 ⊕ 𝑃,

𝑋3(1)𝑣 =

1 0 0
(cid:170)
(cid:174)
(cid:174)
1 2 0
(cid:174)
(cid:174)
(cid:174)
1 1 3
(cid:172)

(cid:169)
(cid:173)
(cid:173)
(cid:173)
(cid:173)
(cid:173)
(cid:171)

2
(cid:170)
(cid:174)
(cid:174)
4
(cid:174)
(cid:174)
(cid:174)
0
(cid:172)

(cid:169)
(cid:173)
(cid:173)
(cid:173)
(cid:173)
(cid:173)
(cid:171)

=

(cid:169)
(cid:173)
(cid:173)
(cid:173)
(cid:173)
(cid:173)
(cid:171)

2

(cid:170)
(cid:174)
(cid:174)
10
(cid:174)
(cid:174)
(cid:174)
(cid:172)

6

and 𝑎′

3 = 1(2 + 4 + 0) = 6.

64

Then 𝑓𝑇1⊕𝑃 (𝑞) = 2 + 10𝑞 + 6𝑞2 + 6𝑞3. For 𝑇2 ⊕ 𝑃,

2 0

0

4

𝑋3(2)𝑣 =

(cid:169)
(cid:173)
(cid:173)
(cid:173)
(cid:173)
(cid:173)
(cid:171)
Then 𝑓𝑇2⊕𝑃 (𝑞) = 4 + 32𝑞 + 36𝑞2 + 48𝑞3. Finally, for 𝑇3 ⊕ 𝑃,

(cid:170)
(cid:174)
(cid:174)
(cid:174)
(cid:174)
(cid:174)
6 6 12
(cid:172)

(cid:170)
(cid:174)
(cid:174)
32
(cid:174)
(cid:174)
(cid:174)
36
(cid:172)

and 𝑎′

(cid:169)
(cid:173)
(cid:173)
(cid:173)
(cid:173)
(cid:173)
(cid:171)

(cid:169)
(cid:173)
(cid:173)
(cid:173)
(cid:173)
(cid:173)
(cid:171)

4 6

=

0

2
(cid:170)
(cid:174)
(cid:174)
4
(cid:174)
(cid:174)
(cid:174)
0
(cid:172)

3 = 8(2 + 4 + 0) = 48.

6

0

0

𝑋3(3)𝑣 =

(cid:169)
(cid:173)
(cid:173)
(cid:173)
(cid:173)
(cid:173)
(cid:171)

0

18 24

(cid:170)
(cid:174)
(cid:174)
(cid:174)
(cid:174)
(cid:174)
36 36 60
(cid:172)

2
(cid:170)
(cid:174)
(cid:174)
4
(cid:174)
(cid:174)
(cid:174)
0
(cid:172)

(cid:169)
(cid:173)
(cid:173)
(cid:173)
(cid:173)
(cid:173)
(cid:171)

=

12

132

216

(cid:169)
(cid:173)
(cid:173)
(cid:173)
(cid:173)
(cid:173)
(cid:171)

(cid:170)
(cid:174)
(cid:174)
(cid:174)
(cid:174)
(cid:174)
(cid:172)

and 𝑎′

3 = 60(2 + 4 + 0) = 360.

Then 𝑓𝑇3⊕𝑃 (𝑞) = 12 + 132𝑞 + 216𝑞2 + 360𝑞3.

𝑃

𝑇1 ⊕ 𝑃

𝑇2 ⊕ 𝑃

𝑇3 ⊕ 𝑃

Figure 3.10 The posets obtained from 𝑃 by attaching 1, 2 and 3 minimal elements

An analogous result for the cumulative generating function 𝑔𝑇𝑘 ⊕𝑃 (𝑞) is stated below.

Theorem 3.4.7. Let 𝑃 be an 𝑛-element poset and 𝑔𝑃 (𝑞) = (cid:205)𝑛−1
𝑖=0
function of 𝑃. Assume 𝑔𝑇𝑘 ⊕𝑃 (𝑞) = (cid:205)𝑛+𝑘−1

𝑏𝑖𝑞𝑖 the cumulative generating
𝑖𝑞𝑖. Let 𝑤 = (𝑏0, 𝑏1, . . . , 𝑏𝑛−1)⊺ be the column vector of
𝑏′
𝑛−1)⊺ be the column vector of the first 𝑛 coefficients

, . . . , 𝑏′

𝑖=0

the coefficients of 𝑔𝑃 (𝑞) and 𝑤′ = (𝑏′
0

, 𝑏′
1

of 𝑔𝑇𝑘 ⊕𝑃 (𝑞). Then

1. 𝑌𝑛 (𝑘)𝑤 = 𝑤′, where 𝑌𝑛 (𝑘) is the 𝑛 × 𝑛 diagonal matrix, the 𝑖th diagonal entry given by

(𝑘+𝑖−1)!
(𝑖−1)!

.

2. 𝑏′

𝑖 = (𝑛 + 𝑘)! for 𝑖 = 𝑛, 𝑛 + 1, . . . , 𝑛 + 𝑘 − 1.

65

Proof. Let 𝑅𝑛 be the lower triangular matrix of size 𝑛 whose lower triangular entries (including the
diagonal entries) are 1. If 𝑣 = (𝑎0, 𝑎1, . . . , 𝑎𝑛−1)⊺ is the column vector of the coefficients of 𝑓𝑃 (𝑞),
then it is easy to see that 𝑅𝑛𝑣 = 𝑤. One can also check that 𝑌𝑛 (𝑘)𝑅𝑛 = 𝑅𝑛 𝑋𝑛 (𝑘).

By theorem 3.4.5, the first part of the statement follows from the identities below.

𝑌𝑛 (𝑘)𝑤 = 𝑌𝑛 (𝑘)𝑅𝑛𝑣 = 𝑅𝑛 𝑋𝑛 (𝑘)𝑣 = 𝑅𝑛𝑣′ = 𝑤′.

Since 𝑎′

𝑖 = 0 for 𝑖 = 𝑛 + 1, 𝑛 + 2, . . . , 𝑛 + 𝑘 − 1, this implies that 𝑏′

𝑖 = (𝑛 + 𝑘)! for 𝑖 = 𝑛, 𝑛 + 1, . . . , 𝑛 +

𝑘 − 1.

□

We close this section with a special family of posets which are obtained from a given 𝑛-element

poset 𝑃 by attaching the chain with ℓ elements below 𝑃, that is,

(cid:16)(cid:201)ℓ

𝑖=1

(cid:17)

𝑇1

⊕ 𝑃. For convenience,

we denote it by 𝑃(ℓ). Note that 𝑃(ℓ) has 𝑛 + ℓ elements.

We assume that the sorting and cumulative generating functions of 𝑃(ℓ) are written as 𝑓𝑃 (ℓ ) (𝑞) =

(cid:205)𝑛+ℓ−1
𝑖=0

𝑎 (ℓ)
𝑖 𝑞𝑖 and 𝑔𝑃 (ℓ ) (𝑞) = (cid:205)𝑛+ℓ−1

𝑖=0

𝑏 (ℓ)
𝑖 𝑞𝑖, respectively. Two propositions are stated below.

Proposition 3.4.8. Let 𝑃 be an 𝑛-element poset and 𝑃(ℓ) the poset obtained from 𝑃 by attaching

the chain with ℓ elements below 𝑃. The last ℓ + 1 coefficients of the cumulative generating function

𝑔𝑃 (ℓ ) (𝑞) are given by

for 0 ≤ 𝑟 ≤ ℓ.

𝑏 (ℓ)
𝑛+ℓ−(𝑟+1) = (𝑛 + ℓ − 𝑟)𝑟 (𝑛 + ℓ − 𝑟)!,

Moreover, 𝑃(ℓ) satisfies conjecture 3.0.2 if and only if

𝑏 (ℓ)
𝑛−2

≥ (𝑛 − 1)ℓ+1(𝑛 − 1)!.

Proof. Applying theorem 3.4.7 with 𝑘 = 1 repeatedly, we obtain

𝑛+ℓ−(𝑟+1) = (𝑛 + ℓ − 𝑟)𝑏 (ℓ−1)
𝑏 (ℓ)

𝑛+ℓ−(𝑟+1) = · · · = (𝑛 + ℓ − 𝑟)𝑡 𝑏 (ℓ−𝑡)

𝑛+ℓ−(𝑟+1)

(3.14)

(3.15)

,

for 0 ≤ 𝑟 ≤ ℓ and for some non-negative integer 𝑡. When (𝑛 + ℓ − (𝑟 + 1)) − (ℓ − 𝑡) = 𝑛 − 1, that is,
when 𝑡 = 𝑟, 𝑏 (ℓ−𝑡)
Therefore, 𝑏 (ℓ)

𝑛+ℓ−(𝑟+1) is the leading coefficient of the 𝑔𝑃 (ℓ −𝑡 ) (𝑞). Hence, 𝑏 (ℓ−𝑡)
𝑛+ℓ−(𝑟+1) = (𝑛 + ℓ − 𝑟)𝑟 (𝑛 + ℓ − 𝑟)!.

𝑛+ℓ−(𝑟+1) = (𝑛 + ℓ − 𝑟)!.

66

conjecture 3.0.2 implies that 𝑎𝑛−1 ≤ (𝑛 − 1)!. Since 𝑏𝑛−1 = 𝑏𝑛−2 + 𝑎𝑛−1,

𝑏𝑛−2 = 𝑏𝑛−1 − 𝑎𝑛−1 ≥ 𝑛! − (𝑛 − 1)! = (𝑛 − 1) (𝑛 − 1)!.

We again apply theorem 3.4.7 with 𝑘 = 1 repeatedly, then

𝑛−2 = (𝑛 − 1)𝑏 (ℓ−1)
𝑏 (ℓ)

𝑛−2 = · · · = (𝑛 − 1)ℓ𝑏𝑛−2 ≥ (𝑛 − 1)ℓ+1(𝑛 − 1)!.

The converse statement is argued in a similar way and will be omitted here.

□

We then state below the counterpart result of Proposition 3.4.8.

Proposition 3.4.9. Let 𝑃 be an 𝑛-element poset. For 0 ≤ 𝑟 ≤ ℓ − 1, the number of 𝑟-tangled

labelings of 𝑃(ℓ) is given by

𝑎 (ℓ)
𝑛+ℓ−(𝑟+1) =

(cid:16)

(𝑛 + ℓ − 𝑟)𝑟+1 − (𝑛 + ℓ − (𝑟 + 1))𝑟+1(cid:17)

(𝑛 + ℓ − (𝑟 + 1))!.

(3.16)

Moreover, 𝑃(ℓ) satisfies conjecture 3.0.2 if and only if

𝑎 (ℓ)
𝑛−1

≤

(cid:16)

𝑛ℓ+1 − (𝑛 − 1)ℓ+1(cid:17)

(𝑛 − 1)!.

(3.17)

Proof. Notice that 𝑏 (ℓ)

𝑛+ℓ−(𝑟+1) − 𝑏 (ℓ)

𝑛+ℓ−(𝑟+2) = 𝑎 (ℓ)

𝑛+ℓ−(𝑟+1) for 0 ≤ 𝑟 ≤ ℓ − 1. Then eq. (3.16) follows

immediately from eq. (3.14).

By eq. (3.14) with 𝑟 = ℓ, 𝑏 (ℓ)

𝑛−1 = 𝑛ℓ𝑛!. Then eq. (3.17) is obtained from 𝑎 (ℓ)
eq. (3.15). The converse statement can be argued similarly and is omitted here.

𝑛−1 = 𝑏 (ℓ)
𝑛−1

− 𝑏 (ℓ)

𝑛−2 and
□

We next show that our poset 𝑃(ℓ) satisfies [32, Conjecture 23]. This conjecture states that for an

𝑛-element poset 𝑃, the number of labelings 𝐿 ∈ Λ(𝑃) such that 𝐿𝑛−3 ∉ L (𝑃) has an upper bound

3(𝑛 − 1)!.

Corollary 3.4.10. Let 𝑃 be an 𝑛-element poset and ℓ ≥ 1. The number of labelings 𝐿 ∈ Λ(𝑃(ℓ))
such that 𝐿𝑛+ℓ−3 ∉ L (𝑃(ℓ)), that is, the total number of tangled and quasi-tangled labelings of 𝑃(ℓ),

equals

3(𝑛 + ℓ − 1)! − (𝑛 + ℓ − 2)! ≤ 3(𝑛 + ℓ − 1)!,

67

Proof. By proposition 3.4.9 with 𝑟 = 0, the number of tangled labelings of 𝑃(ℓ) is

𝑎 (ℓ)
𝑛+ℓ−1 = (𝑛 + ℓ − 1)!.

Take 𝑟 = 1 in proposition 3.4.9, we obtain the number of quasi-tangled labelings of 𝑃(ℓ), which is

given by

𝑎 (ℓ)
𝑛+ℓ−2 =

(cid:16)

(𝑛 + ℓ − 1)2 − (𝑛 + ℓ − 2)2(cid:17)

(𝑛 + ℓ − 2)!

= (2(𝑛 + ℓ − 1) − 1) (𝑛 + ℓ − 2)!

= 2(𝑛 + ℓ − 1)! − (𝑛 + ℓ − 2)!.

Summing these two numbers gives the desired result.

□

Remark 3.4.11. Let 𝑃 be an 𝑛-element poset. We are able to give a simple and unified proof of

some results given by Defant and Kravitz in [19] and by Hodges in [32].

• Take ℓ = 1, the poset 𝑃(1) has one minimal element. By proposition 3.4.9 with 𝑟 = 0, the

number of tangled labelings of 𝑃(1) is given by

𝑎 (1)
𝑛 = (𝑛 + 1 − 𝑛) 𝑛! = 𝑛!.

This gives an alternative proof of [19, Corollary 3.7] (for a connected poset).

• Take ℓ = 2, the poset 𝑃(2) has one minimal element and this minimal element has exactly

one parent. By proposition 3.4.9 with 𝑟 = 1, the number of quasi-tangled labelings of 𝑃(2) is

given by

𝑎 (2)
𝑛 =

(cid:16)

(𝑛 + 1)2 − 𝑛2(cid:17)

𝑛! = (2𝑛 + 1)𝑛! = 2(𝑛 + 1)! − 𝑛!.

This gives a simpler proof of [32, Corollary 10].

3.5 Ordinal Sum of Antichains

In this section, we consider a family of posets consisting of the ordinal sum of antichains. Let

𝐶 = (𝑐1, 𝑐2, . . . , 𝑐𝑟) be an ordered sequence of 𝑟 positive integers. Throughout this section, we

68

write 𝑃𝐶 = 𝑇𝑐𝑟 ⊕ 𝑇𝑐𝑟 −1 ⊕ · · · ⊕ 𝑇𝑐1 for the ordinal sum of antichains of 𝐶. We completely determine

the cumulative generating function of this family of posets. We also show various properties and a

poset structure of its cumulative generating function.

The cumulative generating function of the 𝑘-element antichain 𝑇𝑘 is 𝑔𝑇𝑘 (𝑞) = 𝑘!(1 + 𝑞 + 𝑞2 +
· · · + 𝑞𝑘−1). To find 𝑔𝑃𝐶 (𝑞), we start from the antichain 𝑇𝑐1 and let 𝑤 = (𝑐1!, . . . , 𝑐1!)⊺ be the
column vector consisting of the coefficients of 𝑔𝑇𝑐
(𝑞). We next attach 𝑐2 minimal elements to 𝑇𝑐1;
(𝑞) is obtained by theorem 3.4.7. Recall that 𝑌𝑐1 (𝑐2)

the cumulative generating function 𝑔𝑇𝑐
⊕𝑇𝑐
1
denotes the 𝑐1 × 𝑐1 diagonal matrix whose 𝑖th diagonal entry is given by (𝑐2+𝑖−1)!
(𝑖−1)!
multiplication 𝑌𝑐1 (𝑐2)𝑤 gives the first 𝑐1 coefficients of 𝑔𝑇𝑐
(𝑞) and the rest of coefficients are
given by (𝑐1 + 𝑐2)!. As a consequence, we can obtain 𝑔𝑃𝐶 by applying theorem 3.4.7 repeatedly in

. The matrix

⊕𝑇𝑐
2

1

1

2

this way. The explicit formula of 𝑔𝑃𝐶 (𝑞) is summarized in the following proposition.

Proposition 3.5.1. Let 𝑃𝐶 be the ordinal sum of antichains of 𝐶, where 𝐶 = (𝑐1, 𝑐2, . . . , 𝑐𝑟) is
an ordered sequence of 𝑟 positive integers. Write 𝑔𝑃𝐶 (𝑞) = (cid:205)𝑐1+···+𝑐𝑟 −1
𝑏𝑠𝑞𝑠 for the cumulative

𝑠=0

generating function of 𝑃𝐶. For each 0 ≤ 𝑠 < 𝑐1 + · · · + 𝑐𝑟, let 𝑗 ∈ [𝑟] be the unique integer such

that

Then

𝑗−1
∑︁

𝑘=1

𝑐𝑘 ≤ 𝑠 <

𝑗
∑︁

𝑘=1

𝑐𝑘 .

𝑏𝑠 = (𝑐1 + 𝑐2 + · · · + 𝑐 𝑗 )!

𝑟
(cid:214)

𝑚= 𝑗+1

(𝑐𝑚 + 𝑠)!
𝑠!

.

(3.18)

We now present the following symmetry property for the poset 𝐵𝑛,𝑘 = 𝑇𝑛 ⊕ 𝐶𝑘+1, where 𝐶𝑘+1

is the chain of 𝑘 + 1 elements and 𝑛, 𝑘 ∈ Z≥0. This poset is sometimes called a broom.

Proposition 3.5.2. Let 𝑛, 𝑘 ∈ Z≥0. Write 𝑓𝐵𝑛,𝑘 (𝑞) = (cid:205)𝑛+𝑘
𝑠=0
function of 𝐵𝑛,𝑘 . Then

𝑎𝑠 (𝑛, 𝑘)𝑞𝑠 for the sorting generating

𝑎𝑠 (𝑛, 𝑘) =






(𝑛 + 𝑠)!(𝑠 + 1) 𝑘+1−𝑠 − (𝑛 + 𝑠 − 1)!𝑠𝑘+2−𝑠,

for 𝑠 = 0, 1, . . . , 𝑘 + 1,

(3.19)

0,

for 𝑠 = 𝑘 + 2, 𝑘 + 3, . . . , 𝑛 + 𝑘.

69

In particular, we have the symmetry property

𝑎𝑘 (𝑛, 𝑘) = 𝑎𝑛 (𝑘, 𝑛), for 0 ≤ 𝑛 ≤ 𝑘

(3.20)

Proof. By proposition 3.5.1 with 𝑐1 = 𝑐2 = · · · = 𝑐𝑘+1 = 1 and 𝑐𝑘+2 = 𝑛, the cumulative generating
function of 𝑇𝑛 ⊕ 𝐶𝑘+1 is given by 𝑔𝑇𝑛⊕𝐶𝑘+1 (𝑞) = (cid:205)𝑛+𝑘
𝑠=0

𝑏𝑠 (𝑛, 𝑘)𝑞𝑠, where

𝑏𝑠 (𝑛, 𝑘) = (𝑠 + 1)!(𝑠 + 1) 𝑘−𝑠 (𝑛 + 𝑠)!

𝑠!

= (𝑛 + 𝑠)!(𝑠 + 1) 𝑘+1−𝑠,

for 𝑠 = 0, 1, . . . , 𝑘. We also have 𝑏𝑠 (𝑛, 𝑘) = (𝑛 + 𝑘 + 1)! for 𝑠 = 𝑘 + 1, 𝑘 + 2, . . . , 𝑛 + 𝑘.

Then eq. (3.19) follows immediately from the fact that 𝑎𝑠 (𝑛, 𝑘) = 𝑏𝑠 (𝑛, 𝑘) − 𝑏𝑠−1(𝑛, 𝑘). The

symmetry property (eq. (3.20)) can be verified directly using eq. (3.19). This completes the proof

of proposition 3.5.2.

□

We next study problems originally proposed by Defant and Kravitz1. Given an 𝑛-element poset

𝑃, are the coefficients of the sorting generating function 𝑓𝑃 (𝑞) and the cumulative generating func-

tion 𝑔𝑃 (𝑞) unimodal or log-concave? We prove that the coefficients of the cumulative generating

function are log-concave for the ordinal sum of antichains and provide a counterexample to the

conjecture that the coefficients of the sorting generating function of a general poset are unimodal.

Recall that a sequence of real numbers (𝑎𝑖)𝑛

𝑎0 ≤ 𝑎1 ≤ 𝑎2 ≤ · · · ≤ 𝑎 𝑗 ≥ 𝑎 𝑗+1 ≥ · · · ≥ 𝑎𝑛. We say this sequence is log-concave if 𝑎2

𝑖=0 is called unimodal if there is an index 𝑗 such that
𝑖 ≥ 𝑎𝑖−1𝑎𝑖+1

for 𝑖 = 1, 2, . . . , 𝑛 − 1. Note that a positive sequence is log-concave implies that this sequence is

unimodal.

We show below that the coefficients of the cumulative generating function of 𝑃𝐶 are log-concave.

Proposition 3.5.3. Let 𝑃𝐶 be the ordinal sum of antichains of 𝐶, where 𝐶 = (𝑐1, 𝑐2, . . . , 𝑐𝑟) is
a sequence of 𝑟 positive integers. Let 𝑔𝑃𝐶 (𝑞) = (cid:205)𝑐1+···+𝑐𝑟 −1
𝑏𝑠𝑞𝑠 be the cumulative generating
function of 𝑃𝐶. Then the sequence (𝑏𝑠)𝑐1+...+𝑐𝑟 −1

is log-concave.

𝑠=0

𝑠=0

1The problems are stated as Conjecture 5.2 and Problem 5.3 in their 2020 preprint, but not in the published version

[19].

70

𝑏2
𝑠
𝑏𝑠−1𝑏𝑠+1

Proof. We will show that
eq. (3.18). For 𝑗 = 1, 2, . . . , 𝑟, let I𝑗 = {𝑠 : (cid:205) 𝑗−1
𝑘=1
following four cases of the index 𝑠. We present the calculation for the first two cases below; the

≥ 1 for 𝑠 = 1, 2, . . . , 𝑐1 + · · · + 𝑐𝑟 − 2 by direct computation using

𝑐𝑘 }. The proof is based on the

𝑐𝑘 ≤ 𝑠 < (cid:205) 𝑗

𝑘=1

other two cases can be proved similarly and we leave them to the reader.

Case 1: 𝑠 − 1, 𝑠, 𝑠 + 1 ∈ I𝑗 for some 𝑗. In this case

𝑏2
𝑠
𝑏𝑠−1𝑏𝑠+1

=

=

=

(cid:16)

(𝑐1 + . . . + 𝑐 𝑗 )! (cid:206)𝑟

𝑚= 𝑗+1

(cid:17) 2

(𝑐𝑚+𝑠)!
𝑠!

𝑚= 𝑗+1

(𝑐𝑚+𝑠−1)!(𝑐𝑚+𝑠+1)!
(𝑠−1)!(𝑠+1)!

(cid:0)(𝑐1 + . . . + 𝑐 𝑗 )!(cid:1) 2 (cid:206)𝑟
𝑟
(𝑠 + 1) (𝑐𝑚 + 𝑠)
(cid:214)
𝑠(𝑐𝑚 + 𝑠 + 1)

𝑚= 𝑗+1
𝑟
(cid:214)

𝑚= 𝑗+1

𝑠𝑐𝑚 + 𝑠2 + 𝑐𝑚 + 𝑠
𝑠𝑐𝑚 + 𝑠2 + 𝑠

≥ 1,

since 𝑐𝑚 and 𝑠 are positive integers and thus the denominator is always smaller than the numerator.

Case 2: 𝑠 − 1, 𝑠 ∈ I𝑗 and 𝑠 + 1 ∈ I𝑗+1 for some 𝑗. In this case, 𝑠 = (cid:205) 𝑗

𝑘=1

𝑐𝑘 − 1, and

𝑏2
𝑠
𝑏𝑠−1𝑏𝑠+1

=

(cid:16)

(𝑐1 + . . . + 𝑐 𝑗 )! (cid:206)𝑟

(𝑐1 + . . . + 𝑐 𝑗 )!
(𝑐1 + . . . + 𝑐 𝑗+1)!

·

(cid:16)

(𝑐1 + . . . + 𝑐 𝑗 )! (cid:206)𝑟
(𝑐𝑚+𝑠−1)!
(𝑠−1)!

(𝑐𝑚+𝑠)!
𝑚= 𝑗+1
𝑠!
(𝑐1 + . . . + 𝑐 𝑗+1)! (cid:206)𝑟

(cid:17) (cid:16)

(cid:17) 2

𝑚= 𝑗+1
(𝑐 𝑗+1 + 𝑠)!(𝑐 𝑗+1 + 𝑠)!(𝑠 − 1)!
(𝑐 𝑗+1 + 𝑠 − 1)!𝑠!𝑠!

(cid:17)

(𝑐𝑚+𝑠+1)!
(𝑠+1)!

𝑚= 𝑗+2

𝑟
(cid:214)

·

𝑚= 𝑗+2

(𝑠 + 1)(𝑐𝑚 + 𝑠)
𝑠(𝑐𝑚 + 𝑠 + 1)

=

=

=

=

(𝑐1 + . . . + 𝑐 𝑗 )!
(𝑐1 + . . . + 𝑐 𝑗+1)!

(𝑐 𝑗+1 + 𝑠)! · (𝑐 𝑗+1 + 𝑠)
𝑠! · 𝑠

(𝑠 + 1)(𝑐𝑚 + 𝑠)
𝑠(𝑐𝑚 + 𝑠 + 1)

(𝑠 + 1)!
(𝑠 + 1 + 𝑐 𝑗+1)!

(𝑐 𝑗+1 + 𝑠)! · (𝑐 𝑗+1 + 𝑠)
𝑠! · 𝑠

·

(𝑠 + 1)(𝑐𝑚 + 𝑠)
𝑠(𝑐𝑚 + 𝑠 + 1)

𝑟
(cid:214)

·

𝑚= 𝑗+2
𝑟
(cid:214)

𝑚= 𝑗+2

(𝑠 + 1)(𝑐 𝑗+1 + 𝑠)
𝑠(𝑐 𝑗+1 + 𝑠 + 1)

·

𝑟
(cid:214)

𝑚= 𝑗+2

(𝑠 + 1)(𝑐𝑚 + 𝑠)
𝑠(𝑐𝑚 + 𝑠 + 1)

≥ 1

by similar reasoning as in Case 1.

We omit the calculation of showing

𝑏2
𝑠
𝑏𝑠−1𝑏𝑠+1

≥ 1 for the last two cases, since they can be proved

similarly.

Case 3: 𝑠 − 1 ∈ I𝑗 and 𝑠, 𝑠 + 1 ∈ I𝑗+1 for some 𝑗. In this case, 𝑠 = (cid:205) 𝑗
Case 4: 𝑠 − 1 ∈ I𝑗 , 𝑠 ∈ I𝑗+1 and 𝑠 + 1 ∈ I𝑗+2 for some 𝑗.

𝑐𝑘 .

𝑘=1

In this case, 𝑐 𝑗+1 = 1 and

𝑠 = (cid:205) 𝑗+1
𝑘=1

𝑐𝑘 .

□

71

Remark 3.5.4. For the poset 𝑃 = 𝑇2 ⊕ 𝑇2 ⊕ 𝑇2 the sorting generating function is 𝑓𝑃 (𝑞) = 8 +

64𝑞 + 216𝑞2 + 192𝑞3 + 240𝑞4 and the cumulative generating function is 𝑔𝑃 (𝑞) = 8 + 72𝑞 + 288𝑞2 +

480𝑞3 + 720𝑞4 + 720𝑞5. One can see that the coefficients of 𝑓𝑃 (𝑞) are not unimodal, which gives a

counterexample to [32, Conjecture 29] (see also Conjecture 5.2 in the 2020 preprint of [19]). One

can also check that the coefficients of 𝑔𝑃 (𝑞) are log-concave.

We close this section with a new direction for studying the cumulative generating function of

the ordinal sum of antichains 𝑃𝐶. One can ask: how do the cumulative generating functions 𝑔𝑃𝐶 (𝑞)

and 𝑔𝑃𝐶′ (𝑞) compare when 𝐶′ is a permutation of elements of 𝐶? Given an ordered sequence of

𝑟 distinct positive integers 𝐶 = (𝑐1, 𝑐2, . . . , 𝑐𝑟) and a permutation 𝜋 in the symmetric group on

𝑟 elements 𝔖𝑟, define 𝜋(𝐶) = (𝑐𝜋(1), 𝑐𝜋(2), . . . , 𝑐𝜋(𝑟)). The collection of the coefficients of the

cumulative generating function of 𝑃𝜋(𝐶) for all 𝜋 ∈ 𝔖𝑟 is defined to be

(cid:40)

B(𝐶) =

b 𝜋 = (𝑏0, 𝑏1, . . . , 𝑏 |𝐶 | −1) : 𝑔𝑃𝜋 (𝐶) (𝑞) =

𝑏𝑖𝑞𝑖 for 𝜋 ∈ 𝔖𝑟

(cid:41)

,

|𝐶 | −1
∑︁

𝑖=0

where |𝐶 | = (cid:205)𝑟

𝑖=1

𝑐𝑖. A natural partial order on B(𝐶) is given by the following dominance relation.

Definition 3.5.5. For a pair of integer sequences b = (𝑏0, 𝑏1, . . . , 𝑏𝑛) and b′ = (𝑏′
0

, 𝑏′
1

, . . . , 𝑏′

𝑛),

we say b′ dominates b, denoted by b ⪯ b′, if 𝑏𝑖 ≤ 𝑏′

𝑖 for 𝑖 = 0, 1, . . . , 𝑛.

If b and b′ denote the coefficients of the cumulative generating function of 𝑃 and 𝑃′ respectively,

then the relation b ⪯ b′ can be interpreted as saying that the labelings of 𝑃′ require fewer promotions

to be sorted compared to those of 𝑃. It is easy to check that ⪯ is a partial order on the set B(𝐶).

Example 3.5.6. For 𝐶 = (1, 2, 3), the cumulative generating functions 𝑃𝜋(𝐶) for 𝜋 ∈ 𝔖3 are

computed and their coefficients listed below:

b123 = (12, 144, 360, 720, 720, 720),

b132 = (12, 144, 288, 480, 720, 720),

b213 = (12, 96, 360, 720, 720, 720),

b231 = (12, 96, 360, 480, 600, 720),

b312 = (12, 72, 216, 480, 720, 720),

b321 = (12, 72, 216, 480, 600, 720).

72

The Hasse diagram of (B(𝐶), ⪯) is shown in the left of fig. 3.11. Observe that the subgraph

consisting of all the black edges forms the Hasse diagram of the dual to the weak order on 𝔖3 (see

for instance [53, Exercises 3.183 and 3.185] for the definition of weak and strong order on 𝔖𝑛).

The red edge (b312 ⪯ b213) shows a new cover relation which does not occur in the weak order on

𝔖3.

Moreover, we draw the Hasse diagram of (B(𝐶), ⪯) where 𝐶 = (1, 2, 3, 4) in the right picture

of fig. 3.11. Similarly, the subgraph consisting of black edges forms the Hasse diagram of the

dual to the weak order on 𝔖4 while the red edges show new cover relations in our poset structure

compared to the weak order of 𝔖4. We formulate this observation more generally in the following

theorem.

Figure 3.11 The Hasse diagram of (B(𝐶), ⪯) where 𝐶 = (1, 2, 3) (left) and 𝐶 = (1, 2, 3, 4)
(right), which contains a subposet (shown as a subgraph consisting of all the black edges) that is
isomorphic to the weak order of 𝔖3 (left) and 𝔖4 (right). The new cover relations in our poset
structure compared to the weak order of 𝔖3 (left) and 𝔖4 (right) are drawn in red

Theorem 3.5.7. Given an ordered sequence of 𝑟 distinct positive integers 𝐶 = (𝑐1, 𝑐2, . . . , 𝑐𝑟). Let

(cid:40)

B(𝐶) =

b 𝜋 = (𝑏0, 𝑏1, . . . , 𝑏 |𝐶 | −1) : 𝑔𝑃𝜋 (𝐶) (𝑞) =

𝑏𝑖𝑞𝑖 for 𝜋 ∈ 𝔖𝑟

(cid:41)

,

|𝐶 | −1
∑︁

𝑖=0

73

b123b132b213b312b231b321b1234b2134b1324b1243b2314b3124b2143b1342b1423b2341b3214b3142b2413b4123b1432b3241b2431b3412b4213b4132b3421b4231b4312b4321where |𝐶 | = (cid:205)𝑟

𝑖=1

𝑐𝑖. Then the poset (B(𝐶), ⪯) is isomorphic to a refinement of the poset (𝔖𝑟, ≤),

where ≤ is the weak order on 𝔖𝑟.

We first prove the following lemma which will be used to prove theorem 3.5.7.

Lemma 3.5.8. Given an ordered sequence of 𝑟 distinct positive integers 𝐶 = (𝑐1, 𝑐2, . . . , 𝑐𝑟). Let

𝜋 = (𝑖, 𝑖 + 1) ∈ 𝔖𝑟 be a transposition. Let b = (𝑏0, . . . , 𝑏𝑐1+...+𝑐𝑟 −1) and b𝜋 = (𝑏′
𝑐1+...+𝑐𝑟 −1)
0
be the coefficients of the cumulative generating functions 𝑔𝑃𝐶 (𝑞) and 𝑔𝑃 𝜋 (𝐶 ) (𝑞), respectively. If
𝑐𝑖 < 𝑐𝑖+1, then b𝜋 ⪯ b for 𝑖 = 1, 2, . . . , 𝑟 − 1.

, . . . , 𝑏′

Proof. For convenience, we write 𝜋(𝐶) = (𝑑1, 𝑑2, . . . , 𝑑𝑟), where 𝑑𝑖 = 𝑐𝑖+1, 𝑑𝑖+1 = 𝑐𝑖 and 𝑑𝑘 = 𝑐𝑘
for 𝑘 ≠ 𝑖, 𝑖 + 1. For 𝑗 = 1, 2, . . . , 𝑟, let I𝑗 = {𝑠 : (cid:205) 𝑗−1
𝑘=1
𝑠 < (cid:205) 𝑗

𝑑𝑘 }. Since 𝑐𝑘 = 𝑑𝑘 only differ for 𝑘 = 𝑖 and 𝑘 = 𝑖 + 1, proposition 3.5.1 implies that

𝑗 = {𝑠 : (cid:205) 𝑗−1
𝑘=1

𝑐𝑘 ≤ 𝑠 < (cid:205) 𝑗

𝑐𝑘 } and I′

𝑑𝑘 ≤

𝑘=1

𝑘=1
𝑠 for 𝑠 ∈ I𝑗 where 𝑗 ≠ 𝑖, 𝑖 + 1.

𝑏𝑠 = 𝑏′

Notice that I𝑖 ∪ I𝑖+1 = I′

𝑖 ∪ I′

𝑖+1 and I𝑖 ⊆ I′

𝑖 , so it remains to check 𝑏′

𝑠/𝑏𝑠 ≤ 1 holds for the

following three cases: (1) 𝑠 ∈ I𝑖, (2) 𝑠 ∈ I′

𝑖 \ I𝑖, and (3) 𝑠 ∈ I′

𝑖+1. This will imply that b𝜋 ⪯ b.

For the last case, we obtain the equality 𝑏𝑠 = 𝑏′

𝑠 by proposition 3.5.1 immediately. The calculation

for the first two cases is presented below.

Let 𝑥𝑛 = (cid:206)𝑛

𝑘=1(𝑥 + 𝑘 − 1) denote the rising factorial of 𝑥.

Case 1: 𝑠 ∈ I𝑖. We may write 𝑠 = 𝑐1 + . . . + 𝑐𝑖−1 + 𝑡, where 0 ≤ 𝑡 ≤ 𝑐𝑖 − 1. Then for each such

𝑡,

𝑏′
𝑠
𝑏𝑠

=

=

=

=

𝑚=𝑖+1

𝑚=𝑖+1

(𝑑𝑚+𝑠)!
𝑠!
(𝑐𝑚+𝑠)!
𝑠!

(𝑑1 + . . . + 𝑑𝑖)! (cid:206)𝑟
(𝑐1 + . . . + 𝑐𝑖)! (cid:206)𝑟
(𝑐1 + . . . + 𝑐𝑖−1 + 𝑐𝑖+1)!(𝑐𝑖 + 𝑠)!
(𝑐1 + . . . + 𝑐𝑖−1 + 𝑐𝑖)!(𝑐𝑖+1 + 𝑠)!
(𝑐1 + . . . + 𝑐𝑖−1 + 𝑐𝑖+1)!(𝑐1 + . . . + 𝑐𝑖−1 + 𝑐𝑖 + 𝑡)!
(𝑐1 + . . . + 𝑐𝑖−1 + 𝑐𝑖)!(𝑐1 + . . . + 𝑐𝑖−1 + 𝑐𝑖+1 + 𝑡)!
(𝑐1 + . . . + 𝑐𝑖−1 + 𝑐𝑖 + 1)𝑡
(𝑐1 + . . . + 𝑐𝑖−1 + 𝑐𝑖+1 + 1)𝑡

≤ 1,

because 𝑐𝑖 < 𝑐𝑖+1.

74

Case 2: 𝑠 ∈ I′

𝑖 \ I𝑖. We may write 𝑠 = 𝑐1 + . . . + 𝑐𝑖 + 𝑡, where 0 ≤ 𝑡 ≤ 𝑐𝑖+1 − 𝑐𝑖 − 1. Then for

each such 𝑡,

𝑏′
𝑠
𝑏𝑠

=

=

=

=

𝑚=𝑖+1

(𝑑1 + . . . + 𝑑𝑖)! (cid:206)𝑟
(𝑐1 + . . . + 𝑐𝑖+1)! (cid:206)𝑟
(𝑐1 + . . . + 𝑐𝑖−1 + 𝑐𝑖+1)!(𝑐𝑖 + 𝑠)!
(𝑐1 + . . . + 𝑐𝑖+1)!𝑠!

(𝑑𝑚+𝑠)!
𝑠!
(𝑐𝑚+𝑠)!
𝑠!

𝑚=𝑖+2

(𝑐1 + . . . + 𝑐𝑖−1 + 𝑐𝑖+1)!(𝑐1 + . . . + 𝑐𝑖 + 𝑐𝑖 + 𝑡)!
(𝑐1 + . . . + 𝑐𝑖 + 𝑐𝑖+1)!(𝑐1 + . . . + 𝑐𝑖 + 𝑡)!

(𝑐1 + . . . + 𝑐𝑖−1 + 𝑐𝑖 + 𝑡 + 1)𝑐𝑖+1−𝑐𝑖−𝑡
(𝑐1 + . . . + 𝑐𝑖 + 𝑐𝑖 + 𝑡 + 1)𝑐𝑖+1−𝑐𝑖−𝑡

≤ 1,

by the same reasoning in Case 1. This completes the proof of lemma 3.5.8.

□

Proof of theorem 3.5.7. Without loss of generality, we assume the elements of 𝐶 are written in the

increasing order, 𝑐1 < 𝑐2 < · · · < 𝑐𝑟. The permutations 𝜋 ∈ 𝔖𝑟 in this proof will be written in the

one-line notation 𝜋 = 𝑝1 𝑝2 · · · 𝑝𝑟.

Define the map 𝜑 : (𝔖𝑟, ≤) → (B(𝐶), ⪯) by sending a permutation 𝜋 = 𝑝1 𝑝2 . . . 𝑝𝑟 to brev(𝜋),

where rev(𝜋) = 𝑝𝑟 𝑝𝑟−1 . . . 𝑝1 is the reverse of 𝜋, and brev(𝜋) is the sequence of the coefficients of
𝑔𝑃rev( 𝜋 ) (𝐶 ) (𝑞). Let 𝜎 be the adjacent transposition that swapped the elements at positions 𝑖 and 𝑖 + 1.
Let 𝜋 ∈ 𝔖𝑟 be a permutation such that 𝜋 ≤ 𝜎𝜋 in the weak order. One may write 𝜋 = 𝑝1 𝑝2 . . . 𝑝𝑟

with 𝑝𝑖 < 𝑝𝑖+1, and 𝜎𝜋 = 𝑝1 . . . 𝑝𝑖−1 𝑝𝑖+1 𝑝𝑖 𝑝𝑖+2 . . . 𝑝𝑟.

We show that if 𝜋 ≤ 𝜎𝜋 in (𝔖𝑟, ≤), then 𝜑(𝜋) ⪯ 𝜑(𝜎𝜋) in (B(𝐶), ⪯). Intuitively, rev(𝜋) (𝐶) =

{𝑐 𝑝𝑟 , . . . , 𝑐 𝑝1 } and rev(𝜎𝜋)(𝐶) = {𝑐 𝑝𝑟 , . . . , 𝑐 𝑝𝑖+2
, . . . , 𝑐 𝑝1 }. Since 𝑝𝑖 < 𝑝𝑖+1 and
𝑐 𝑝𝑖 < 𝑐 𝑝𝑖+1 (by the assumption that 𝑐𝑖’s are increasing as 𝑖 increases), by lemma 3.5.8, we obtain

, 𝑐 𝑝𝑖 , 𝑐 𝑝𝑖+1

, 𝑐 𝑝𝑖−1

brev(𝜋) ⪯ brev(𝜎𝜋).

Therefore, 𝜑(𝜋) = brev(𝜋) ⪯ brev(𝜎𝜋) = 𝜑(𝜎𝜋). The poset (B(𝐶), ⪯) is thus isomorphic to a

refinement of (𝔖𝑟, ≤).

□

We would like to point out that (B(𝐶), ⪯) is not a subposet of the strong order of 𝔖𝑛 in general.

Take 𝐶 = (1, 2, 3, 4) as an example (see the right picture of fig. 3.11 again); the cover relation

75

b4123 ⪯ b3214, under the inverse of the map 𝜑 defined in the proof of theorem 3.5.7, does not relate

in the strong order of 𝔖4. One can also check that (B(𝐶), ⪯) is not graded in general.

3.6 Future Work

We present some future directions from this work. In this chapter, we propose the (𝑛 − 2)!

conjecture (conjecture 3.0.5), stating that the number of tangled 𝑥-labelings (the label of 𝑥 fixed as

𝑛 − 1) of an 𝑛-element poset 𝑃 is bounded by (𝑛 − 2)!. In section 3.2 and section 3.3, we prove

that inflated rooted forest posets and shoelace posets satisfy the (𝑛 − 2)! conjecture. We also obtain

the exact enumeration of tangled labelings of the 𝑊-poset (a special case of the shoelace poset)

in theorem 3.3.7. One can define inflated shoelace posets in analogy with inflated rooted forest

posets. An interesting question would be to investigate whether inflated shoelace posets satisfy the

(𝑛 − 2)! conjecture. Other general classes of posets that would be of interest to study include posets

related to Young tableaux.

In section 3.4, we explicitly determine the number of 𝑘-sorted labelings of the poset 𝑇𝑠 ⊕ 𝑃 from

𝑃 (attach 𝑠 minimal elements to 𝑃) via the matrix multiplication stated in theorem 3.4.5. However,

obtaining the number of 𝑘-sorted labelings of the poset 𝑃 ⊕ 𝑇𝑠 from 𝑃 (attach 𝑠 maximal elements

to 𝑃) does not seem to have such a nice pattern. There may exist some other ways to express them.

We leave this direction to be pursued by the interested reader.

In section 3.5, we introduce the new poset structure (B(𝐶), ⪯) and show that it contains a

subposet which is isomorphic to the weak order of the symmetric group (theorem 3.5.7). It would

be an interesting follow-up to fully characterize our poset (B(𝐶), ⪯) as a poset on permutations.

76

CHAPTER 4

TWINNING AND THE CHROMATIC SYMMETRIC FUNCTION

The chromatic symmetric function of a graph 𝐺 = (𝑉, 𝐸) is defined by Stanley [47] to be

𝑋𝐺 (x) =

∑︁

(cid:214)

𝜅

𝑣∈𝑉

𝑥𝜅(𝑣),

where the sum is over all proper colorings 𝜅 : 𝑉 → Z>0 of 𝐺 by positive integers. The goal of this

chapter is to study the effect that a small change to the graph 𝐺 has on 𝑋𝐺 (x). Specifically, we

look at the change in 𝑋𝐺 (x) when one twins (or clones) a vertex 𝑣 of a graph 𝐺, that is, when one

adds a vertex 𝑣′ incident to 𝑣 and all its neighbors, to produce a new graph 𝐺𝑣. Precise definitions

of this operation and related terms appear in Section 4.1.

Question 4.0.1. For a given vertex 𝑣 of a graph 𝐺, how are the polynomials 𝑋𝐺 𝑣 (x) and 𝑋𝐺 (x)

related?

In the seminal paper [47], Stanley proved that the chromatic symmetric functions of paths and

cycles are 𝑒-positive, that is, their expansion in the basis of elementary symmetric functions has

nonnegative coefficients. As observed in [47], the result for paths is originally due to Carlitz,

Scoville, and Vaughan in a different context [10, p.242]. More generally, spurred by the following

conjecture of Stanley and Stembridge, much of the research on the chromatic symmetric function

has centered around the incomparability graph Inc(𝑃) of a (3 + 1)-free poset 𝑃, defined as a poset

containing no induced subposet isomorphic to the disjoint union of a 3-chain and a 1-chain. We

note that Hikita did very recently prove the Conjecture in [31].

Conjecture 4.0.2 ([47, 49]). If 𝑃 is a (3 + 1)-free poset, then 𝑋Inc(𝑃) (x) is 𝑒-positive.

To twin a poset 𝑃 at a vertex 𝑣, producing 𝑃𝑣, is to add an element 𝑣′ such that 𝑣′ is incomparable

to 𝑤 if and only if either 𝑤 = 𝑣 or 𝑤 is incomparable to 𝑣. Note that if 𝐺 = Inc(𝑃), then

Inc(𝑃𝑣) = 𝐺𝑣. This next simple lemma is the main motivation for considering the twinning

operation. Its proof is immediate from the fact that if 𝑢 < 𝑣, then 𝑢 < 𝑣′, and if 𝑣 < 𝑤 then 𝑣′ < 𝑤.

77

Lemma 4.0.3. The twin of a (3 + 1)-free poset is (3 + 1)-free.

One can therefore make a weakened version of the Stanley–Stembridge conjecture, first appear-

ing in the work of Foley, Hoàng, and Merkel [24].

Conjecture 4.0.4 ([24]). If 𝑃 is (3+1)-free and 𝑋Inc(𝑃) (x) is 𝑒-positive, then 𝑋Inc(𝑃𝑣) (x) is 𝑒-positive

for any 𝑣 ∈ 𝑃.

Remark 4.0.5. Li, Li, Wang, and Yang [33, Theorem 3.6] prove that the twinning operation on

graphs does not always preserve 𝑒-positivity. They give an example of a graph 𝐺 [33, Theorem

4.1] that is not an incomparability graph of a (3 + 1)-free poset, but whose chromatic symmetric

function 𝑋𝐺 (x) is 𝑒-positive, and show that for a certain vertex 𝑣 of 𝐺, the chromatic symmetric

function for the twinned graph 𝐺𝑣 does not expand positively even in the Schur basis, and so it

cannot be 𝑒-positive. This suggests that there is something special about the twinning operation on

(3 + 1)-free posets.

In 2001, Gebhard and Sagan [27] introduced the stronger notion of (𝑒)-positivity of chromatic

symmetric functions in noncommuting variables. This implies 𝑒-positivity for chains of complete

graphs [27, Corollary 7.7], and includes twins of paths as a special case. Later, Dahlberg and van

Willigenburg [14] gave a direct proof of 𝑒-positivity for lollipop graphs, which are a special case

of [27, Corollary 7.7], and which again include paths twinned at a leaf.

Throughout the chapter, we refer to a property of the chromatic symmetric function of the graph

𝐺 as being a property of the graph 𝐺. For instance, we use interchangeably the phrases “the

generating function of the chromatic symmetric function of a graph” and “the generating function

of a graph”. Similarly, we use “the chromatic symmetric function of the graph 𝐺 is 𝑒-positive” and

“the graph 𝐺 is 𝑒-positive” interchangeably.

This chapter studies the effect of twinning on the 𝑒-expansions of the chromatic symmetric

function of certain graphs. We specifically look at twins of path and cycle graphs, a few of which

are shown in Figure 4.2. A summary of our progress on Question 4.0.1 follows.

78

Our first main contribution is a series of explicit 𝑒-positive formulas for the generating function

of the following families of twinned graphs:

1. The path twinned at one leaf (Proposition 4.2.9)

2. The path twinned at both leaves (Theorem 4.2.14)

3. The path twinned at an interior vertex (Theorem 4.2.24)

4. The cycle twinned at a vertex (Theorem 4.2.29)

The fourth family, examined in detail in Section 4.2.3, and culminating in Theorem 4.2.29, was not

known to be 𝑒-positive until now. The first three families appear in [27] and were shown to have

the stronger property of (𝑒)-positivity of their chromatic symmetric functions with noncommuting

variables [27, Theorems 7.6 and 7.8]. The 𝑒-positivity of the first graph was later re-established

directly in [14]. The explicitly 𝑒-positive expressions for the generating functions that we give

in Proposition 4.2.9, Theorem 4.2.14 and Theorem 4.2.24 are special cases of K-chains and slightly

melting K-chains considered by Foster Tom in [58]. In Corollary 4.2.4 we provide a new 𝑒-positive

expression for the generating function of the path that isolates the terms containing 𝑒1. Our

derivations make crucial use of the triple deletion formula of Orellana and Scott [37].

For all but the third family, the expression we obtain for the generating function has the form

𝑋𝐺 𝑛 𝑧𝑛 =

∑︁

𝑛≥0

𝑓𝐺

1 − (cid:205)𝑖≥2(𝑖 − 1)𝑒𝑖𝑧𝑖 + ℎ𝐺

where 𝑓𝐺 and ℎ𝐺 are some 𝑒-positive functions depending on the family and ℎ𝐺 has finite degree.

The third family has ℎ𝐺 with infinite degree. Note that the denominator in the rational expression

above coincides with the denominator in the generating function for both the path and the cycle

(see Theorem 4.2.1). This allows us to easily obtain explicit formulas for the coefficients of the

elementary symmetric functions (see Corollary 4.2.10, Corollary 4.2.15, Corollary 4.2.30). We

state our formulas for the coefficients using a new statistic 𝜀(𝜆) associated with a partition 𝜆

(see Section 4.1.3). This statistic appears naturally when computing the 𝑒-coefficients of the path

and cycle, and appears to be of independent interest.

79

The identities presented in Section 4.2.1, particularly in Lemma 4.2.3 and its proof, are the

starting point for our 𝑒-positivity results. They also seem interesting in their own right.

Our second main contribution is an 𝑒-positive recurrence relation for each of the families listed

above, as well as a graph appearing in the computation for the twinned cycle that we call the moose

graph, which has been shown to be 𝑒-positive as a special case of hat graphs [63, Theorem 3.9].

1. The path twinned at one leaf (Proposition 4.3.2)

2. The path twinned at both leaves (Proposition 4.3.3)

3. The path twinned at an interior vertex (Theorem 4.3.4)

4. The cycle twinned at a vertex (Theorem 4.3.6)

5. The moose graph (Proposition 4.3.7)

4.1 Preliminaries

In this section, we define the basic notions used throughout the chapter, as well as discuss

previous results. We assume a familiarity with symmetric functions as in [48, Chapter 7] or [35].

A graph 𝐺 is a pair of sets (𝑉, 𝐸) where 𝑉 is the set of vertices and 𝐸 is a set of 2-element

subsets of vertices, called edges. We denote edges by {𝑢, 𝑣} or simply by 𝑢𝑣. We assume that 𝑉

and 𝐸 are both finite, and that the graph is simple (i.e., there are no loops and no multiple edges).

A leaf of a graph is a vertex contained within exactly one edge. An internal vertex is a vertex

contained within at least two edges. Two graphs that are important for this chapter are the path 𝑃𝑛,

which has vertex set 𝑉 = [𝑛] = {1, . . . , 𝑛} and edge set 𝐸 = {{𝑖, 𝑖 + 1} | 𝑖 ∈ [𝑛 − 1]}, and the cycle

𝐶𝑛, which also has vertex set 𝑉 = [𝑛] and edge set 𝐸 = {{𝑖, 𝑖 + 1} | 𝑖 ∈ [𝑛 − 1]} ∪ {1, 𝑛}. We

illustrate them in Figure 4.1.

A proper coloring of a graph 𝐺 = (𝑉, 𝐸) is a function 𝜅 : 𝑉 → Z>0 such that if 𝑢𝑣 ∈ 𝐸,

then 𝜅(𝑢) ≠ 𝜅(𝑣). Let x = (𝑥1, 𝑥2, . . . ) be an infinite set of commuting variables. The chromatic

80

•

•

•

•

•

•

•

· · ·

(b) Cycle 𝐶𝑛

•

•

•

•

· · ·

•

(a) Path 𝑃𝑛

Figure 4.1 The path and cycle graphs

symmetric function of a graph 𝐺 = (𝑉, 𝐸) is defined to be

𝑋𝐺 := 𝑋𝐺 (x) =

∑︁

(cid:214)

𝜅

𝑣∈𝑉

𝑥𝜅(𝑣),

where the sum is over all proper colorings 𝜅 of 𝐺. This symmetric function was first introduced

by Stanley in [47] and has been studied by numerous authors since then. One central goal has

been to characterize graphs 𝐺 for which 𝑋𝐺 is 𝑒-positive, i.e., 𝑋𝐺 has nonnegative coefficients in

the elementary symmetric function basis. We give an overview of the previous 𝑒-positivity results

in Section 4.1.2.

4.1.1 Graph Operations

We begin by defining the two operations on graphs that appear in this chapter.

Given an edge 𝜖 of a graph 𝐺, the deletion of 𝜖 in 𝐺 is the graph, denoted by 𝐺 − 𝜖, obtained by

removing the edge 𝜖 from 𝐺. The following formulas of Orellana and Scott are used extensively in

our arguments and we refer to them as the triple deletion arguments.

Proposition 4.1.1 (Triple Deletion Formula [37, Theorem 3.1]). Let 𝐺 be any graph. Suppose

edges 𝜖1, 𝜖2, 𝜖3 form a triangle in 𝐺. Then,

𝑋𝐺 = 𝑋𝐺−𝜖1 + 𝑋𝐺−𝜖2 − 𝑋𝐺−{𝜖1,𝜖2}.

Notice that Proposition 4.1.1 requires the graph to contain a triangle. However, one can use this

formula to derive other relations for graphs that do not necessarily contain a triangle. An example

of such a relation is the following.

81

Corollary 4.1.2 ([37, Corollary 3.2]). Let 𝜖1 = 𝑣𝑣1 ∈ 𝐸, 𝜖2 = 𝑣𝑣2 ∈ 𝐸 and suppose 𝜖3 = 𝑣1𝑣2 ∉ 𝐸.

Then

𝑋𝐺 = 𝑋(𝐺−𝜀1)∪𝜖3 + 𝑋𝐺−𝜖2 − 𝑋(𝐺−{𝜖1,𝜖2})∪𝜖3

.

We now introduce the main operation studied in this chapter.

Definition 4.1.3. Given a graph 𝐺 and a vertex 𝑣, the twin of 𝐺 at 𝑣 is the graph, denoted by 𝐺𝑣,

obtained by adding a new vertex 𝑣′ and connecting 𝑣′ to 𝑣 and to all of its neighbors. We refer to

this operation as the twinning of a graph and to the resulting graph 𝐺𝑣 as the twinned graph. By

extension, 𝐺𝑣,𝑤 denotes the graph 𝐺 twinned at the vertices 𝑣 and 𝑤 in succession.

A simple example is the complete graph 𝐺 = 𝐾𝑛 on 𝑛 vertices. For any vertex 𝑣, (𝐾𝑛)𝑣 is the

complete graph 𝐾𝑛+1. We illustrate in Figure 4.2 the twinned path at a leaf and at an interior vertex,

and the twinned cycle.

𝑣′
•

•
𝑣

•
𝑤

•

· · ·

•

•

𝑣′
•

•
𝑣

•
𝑤

· · ·

•

•

· · ·

•
𝑢

(a) Twin paths 𝑃𝑛,𝑣

𝑢

•

•

•

𝑣
•

•
𝑣′

· · ·

(b) 𝐶𝑛,𝑣

𝑤

•

•

•

Figure 4.2 Twinning of the path and cycle graphs

Twinning a non-isolated vertex always produces a triangle, so the triple deletion argument is a

natural method to reduce the twinned graph back to the original one, as the following result shows.

Corollary 4.1.4. Let 𝐻 be a graph on 𝑛 vertices and let 𝑢 be a vertex of 𝐻. Let 𝐻′ be the graph

obtained by adding a new vertex 𝑣 and the edge 𝑢𝑣 to 𝐻 and let 𝐻′′ be the graph obtained by adding

a new vertex 𝑤 and the edge 𝑣𝑤 to 𝐻′. Finally let 𝐻′

𝑣 be the graph 𝐻′ twinned at vertex 𝑣, with 𝑣′

denoting the new vertex. Then

𝑋𝐻′

𝑣 = 2(𝑋𝐻′′ − 𝑒2𝑋𝐻).

82

Proof. This is clear by the triple deletion argument using the edges 𝑢𝑣, 𝑢𝑣′ of the triangle {𝑢, 𝑣, 𝑣′}

as shown in Figure 4.3. Note also that 𝑋𝑃2 = 2𝑒2.

□

𝑣′
•

•
𝑣

𝑢
•

𝐻′
𝑣

𝑢
•

𝑤
•

•
𝑣

𝑢
•

𝑤
•

•
𝑣

𝐻′′

𝐻 ⊔ 𝑃2

Figure 4.3 The triple deletion argument used in Corollary 4.1.4

4.1.2 Known 𝑒-positivity Results

Stanley defined the chromatic symmetric function 𝑋𝐺 of a graph 𝐺 in 1995. Since then, many

families of graphs have been examined. We provide an extensive, but by no means exhaustive, list

of known 𝑒-positivity results as of May 2024 in Table 4.1. We do not define these classes of graphs,

but instead provide references containing their definitions as well as proofs of their 𝑒-positivity

classification. By convention, a family of graphs listed as “not 𝑒-positive” means that there is at

least one graph in that class that is not 𝑒-positive. The table is roughly sorted chronologically by

reference, and it is condensed so that some subclasses of other results are omitted.

83

Graph

Paths
Cycles
Complete graphs
Co-triangle-free graphs
𝐾𝛼-chains
Diamond and path chains
(claw, 𝑃4)-free graphs
(claw, diamond)-free graphs
(claw, co-claw)-free graphs
(claw, 𝐾4)-free graphs
(claw, 4𝐾1)-free graphs
(claw, 2𝐾2)-free graphs
(claw, 𝐶4)-free graphs
(claw, paw)-free graphs
(claw, co-paw)-free graphs
Generalized bull graphs
Lollipops and lariats
𝑃3-free graphs
(claw, 𝐾3)-free graphs
(claw, co-𝑃3)-free graphs
(co-claw)-free unit interval graphs
Generalized pyramid graphs
2𝐾2-free unit interval graphs
Triangular ladders
Star graphs
Saltire and augmented saltire graphs
Triangular tower graphs
Tadpole graphs
Line graphs of tadpole graphs
Cycle-chord graphs
Kayak paddle graphs
Generalized nets
Melting 𝐾𝛼-chains

Positivity
𝑒-positive
𝑒-positive
𝑒-positive
𝑒-positive
𝑒-positive
𝑒-positive
𝑒-positive
not 𝑒-positive
not 𝑒-positive
not 𝑒-positive
not 𝑒-positive
not 𝑒-positive
not 𝑒-positive
𝑒-positive
𝑒-positive
𝑒-positive
𝑒-positive
𝑒-positive
𝑒-positive
𝑒-positive
𝑒-positive
𝑒-positive
𝑒-positive
𝑒-positive
not 𝑒-positive
not 𝑒-positive
not 𝑒-positive
𝑒-positive
𝑒-positive
𝑒-positive
𝑒-positive
not 𝑒-positive
𝑒-positive

Reference

[47, Proposition 5.3]
[47, Proposition 5.4]
[49, Equation 3.1]
[49, Theorem 4.3]
[27, Corollary 7.7]
[27, Theorem 7.8]
[59, Theorem 1.4]
[30, Lemma 7]
[30, Lemma 7]
[30, Lemma 7]
[30, Lemma 7]
[30, Lemma 7]
[30, Lemma 7]
[30, Theorem 3]
[30, Theorem 4]
[12, Theorem 3.7]
[14, Theorem 9]
[24, Theorem 5]
[24, Theorem 5]
[24, Theorem 5]
[24, Theorem 18]
[34, Theorem 7]
[34, Theorem 13]
[13, Theorem 22]
[16, Example 11]
[15, Lemmas 4.4, 4.9]
[15, Lemma 5.4]
[33, Theorem 3.1]
[62, Corollary 3.3]
[61], [62, Corollary 4.6]
[2, Proposition 6.7]
[25, Theorem 1]
[58, Corollary 4.18]

Table 4.1 Known 𝑒-positivity results and their references

84

4.1.3 A New Statistic on Partitions

In this subsection, we introduce a new statistic on the set of partitions that will allow us to

describe 𝑒-coefficients more compactly.

Recall that a partition 𝜆 of 𝑛, written 𝜆 ⊢ 𝑛, is a weakly decreasing sequence of positive
integers 𝜆 = (𝜆1, . . . , 𝜆ℓ) that sum to 𝑛, that is, (cid:205)𝑖 𝜆𝑖 = 𝑛. We write ℓ = ℓ(𝜆) for the length

of the partition, that is, the number of entries in the sequence. A partition 𝜆 of 𝑛 can also be

written as 𝜆 = ⟨1𝑚1, 2𝑚2, . . . , 𝑛𝑚𝑛⟩, where 𝑚𝑖 = 𝑚𝑖 (𝜆) ≥ 0 denotes the multiplicity of the part

𝑖 in 𝜆. The support of 𝜆, denoted supp(𝜆), is the set of distinct parts appearing in 𝜆, that is,

supp(𝜆) := {𝑖 ∈ Z>0 : 𝑚𝑖 (𝜆) ≥ 1}.

Now we are ready to introduce the new statistic.

Definition 4.1.5. For a partition 𝜆, define 𝜀(𝜆) to be the quantity

𝜀(𝜆) := ℓ(𝜆)!

(cid:214)

𝑗 ∈supp(𝜆)

( 𝑗 − 1)𝑚 𝑗 (𝜆)
𝑚 𝑗 (𝜆)!

with

𝜀(∅) = 1.

(4.1)

Moreover, for a partition 𝜆 of 𝑛 and a part 𝑎 such 𝑚𝑎 (𝜆) ≥ 1, let 𝜆 − 𝑎 denote the partition of

𝑛 − 𝑎 obtained by deleting one part equal to 𝑎 from 𝜆. By convention, if 𝑎 is not a part of 𝜆, we set

𝜀(𝜆 − 𝑎) = 0.

For example, 𝜀((𝑛)) = 𝑛 − 1 and 𝜀((2𝑛)) = 1 for any positive integer 𝑛. Additionally, 𝜖 (𝜆) = 0

if 𝜆 contains a 1. In Table 4.2, we include several examples of partitions 𝜆 together with their

statistic 𝜀(𝜆).

𝜆
𝜀(𝜆)

𝜆
𝜀(𝜆)

(2)
1

(7)
6

(3)
2

(4)
3

(2,2)
1

(5)
4

(3,2)
4

(6)
5

(4,2)
6

(3,3)
4

(2,2,2)
1

(5,2)
8

(4,3)
12

(3,2,2)
6

(8)
7

(6,2)
10

(5,3)
16

(4,4)
9

(4,2,2)
9

(3,3,2)
12

(2,2,2,2)
1

Table 4.2 Examples of 𝜀(𝜆) for some partitions 𝜆

Remark 4.1.6. Note that in (4.1),

ℓ(𝜆)!
𝑚 𝑗 (𝜆)!

(cid:18)

=

ℓ(𝜆)
𝑚1(𝜆), . . . , 𝑚𝑛 (𝜆)

(cid:19)

,

85

which is always a nonnegative integer. Thus, 𝜀(𝜆) is also a nonnegative integer. In fact, 𝜀(𝜆) = 0

if and only if 1 is a part in 𝜆, i.e., 𝑚1(𝜆) ≥ 1.

Next, we present other properties of 𝜀(𝜆).

Lemma 4.1.7. Let 𝜆 and 𝜇 be partitions of 𝑛 and 𝑚, respectively. Then, we have the following:

1. For 𝑗 ∈ supp(𝜆), ( 𝑗 − 1)𝜀(𝜆 − 𝑗) = 𝑚 𝑗 (𝜆)

𝜀(𝜆)
ℓ(𝜆)

;

2. 𝜀(𝜆) =

∑︁

𝑗 ∈supp(𝜆)

( 𝑗 − 1)𝜀(𝜆 − 𝑗); and

3. 𝜀(𝜆)𝜀(𝜇) = 𝜀(𝜆 ∪ 𝜇)

(cid:18)ℓ(𝜆 ∪ 𝜇)
ℓ(𝜆)

(cid:19) −1 (cid:214)

(cid:19)

(cid:18)𝑚 𝑗 (𝜆 ∪ 𝜇)
𝑚 𝑗 (𝜆)

where 𝜆 ∪ 𝜇 is the partition of

𝑗 ∈supp(𝜆∪𝜇)

𝑛 + 𝑚 formed by listing the parts of 𝜆 and 𝜇 together in decreasing order.

Proof.

1. Note first that both sides are identically zero if 1 ∈ supp(𝜆). For 𝑗 ∈ supp(𝜆) with

𝑗 ≠ 1, this identity follows from the definition by noticing that

𝜀(𝜆) = ( 𝑗 − 1)

ℓ(𝜆)
𝑚 𝑗 (𝜆)

(cid:169)
(cid:173)
(cid:171)

( 𝑗 − 1)𝑚 𝑗 (𝜆)−1 (cid:214)

(𝑙 − 1)𝑚𝑙 (𝜆)

𝑙∈supp(𝜆),𝑙≠ 𝑗

(ℓ(𝜆) − 1)!
(𝑚 𝑗 (𝜆 − 1) (cid:206)𝑙≠ 𝑗 𝑚𝑙 (𝜆)!

.

(cid:170)
(cid:174)
(cid:172)

2. This identity follows from the definition of 𝜀(𝜆), using

∑︁

𝑗 ∈supp(𝜆)

𝑚 𝑗 (𝜆) = ℓ(𝜆).

3. This identity follows by expanding 𝜀(𝜆 ∪ 𝜇), using 𝑚 𝑗 (𝜆 ∪ 𝜇) = 𝑚 𝑗 (𝜆) +𝑚 𝑗 (𝜇) and ℓ(𝜆 ∪ 𝜇) =

ℓ(𝜆) + ℓ(𝜇).

□

Remark 4.1.8. Intuitively, the formula for 𝜀(𝜆) can be interpreted as the number of pairs (𝑤, 𝑓 )

of words 𝑤 on the set {1, . . . , ℓ(𝜆)} of type 𝜆, i.e. with 𝜆𝑖 occurrences of the letter 𝑖, together with

a function 𝑓 : {1, . . . , ℓ(𝜆)} → Z satisfying 1 ≤ 𝑓 ( 𝑗) ≤ 𝜆 𝑗 − 1 for each 𝑗 ∈ {1, . . . , ℓ(𝜆)}. These

are exactly the codes of Stembridge [54] with no fixed points and can be used to prove Lemma 4.1.7

combinatorially. For example, the right-hand side of part (𝑏) can be interpreted as the number of

ways of making a code of type 𝜆 from a code whose type has length ℓ(𝜆) − 1.

86

4.2

𝑒-positivity via Generating Functions

For given family of graphs 𝐺 = {𝐺𝑛}𝑛≥0, one can show 𝑒-positivity of 𝑋𝐺 𝑛 by showing that its

generating function

can be written in the form

X𝐺 (𝑧) =

𝑋𝐺 𝑛 𝑧𝑛

∑︁

𝑛≥0

X𝐺 (𝑧) =

𝑃(𝑧)
1 − 𝑄(𝑧)

,

(4.2)

where 𝑃(𝑧) and 𝑄(𝑧) are 𝑒-positive formal power series in 𝑧. For the path 𝑃𝑛 and the cycle 𝐶𝑛, it

is known from Stanley’s original paper [47] that this can be done. (See also [10, p.242] for paths.)

Theorem 4.2.1 ([47, Propositions 5.3 and 5.4]).

X𝑃 (𝑧) :=

X𝐶 (𝑧) :=

𝑋𝑃𝑛 𝑧𝑛 =

𝑋𝐶𝑛 𝑧𝑛 =

∑︁

𝑛≥0
∑︁

𝑛≥2

(cid:205)𝑖≥0

𝑒𝑖𝑧𝑖

1 − (cid:205)𝑖≥1(𝑖 − 1)𝑒𝑖𝑧𝑖
𝑖(𝑖 − 1)𝑒𝑖𝑧𝑖
(cid:205)𝑖≥2
1 − (cid:205)𝑖≥1(𝑖 − 1)𝑒𝑖𝑧𝑖

,

.

Note in particular that 𝑋𝑃0 = 1.

In this section, we establish identities of the form (4.2) for several families of twinned graphs by

applying generating function techniques to the relations obtained from the triple deletion argument.

It is useful to convert the preceding result to a recurrence relation for the chromatic symmetric

function as follows. We will use this formulation several times in this chapter, notably in the proofs

of Lemma 4.2.23 and Proposition 4.2.25, as well as in Section 4.3 .

Proposition 4.2.2. We have the following recurrence relations:

1. For 𝑛 ≥ 3,

𝑋𝑃𝑛 = 𝑛𝑒𝑛 +

= 𝑛𝑒𝑛 +

𝑛−1
∑︁

𝑗=2
𝑛−2
∑︁

𝑖=1

( 𝑗 − 1)𝑒 𝑗 𝑋𝑃𝑛− 𝑗

(𝑛 − 𝑖 − 1)𝑒𝑛−𝑖 𝑋𝑃𝑖 ,

with initial conditions 𝑋𝑃0 = 1, 𝑋𝑃1 = 𝑒1 and 𝑋𝑃2 = 2𝑒2.

87

2. For 𝑛 ≥ 4,

𝑋𝐶𝑛 = 𝑛(𝑛 − 1)𝑒𝑛 +

𝑛−2
∑︁

𝑗=2

( 𝑗 − 1)𝑒 𝑗 𝑋𝐶𝑛− 𝑗 ,

with initial conditions 𝑋𝐶1 = 0, 𝑋𝐶2 = 2𝑒2, and 𝑋𝐶3 = 6𝑒3.

4.2.1 Symmetric Function Identities and Technical Lemmas

In this section, we examine more closely the relationship between the generating function 𝐸 (𝑧)

for the elementary symmetric functions, and the generating functions X𝑃 (𝑧) and X𝐶 (𝑧) for the

chromatic symmetric functions of the path and the cycle. We also present some formulas for

several families of coefficients appearing in the 𝑒-expansion of X𝑃 (𝑧) and X𝐶 (𝑧). We start by

introducing some definitions and notation to facilitate our study.

Let 𝐸 (𝑧) := (cid:205)𝑖≥0

𝑒𝑖𝑧𝑖 be the generating function for the elementary symmetric functions and

define

𝐷 (𝑧) := 𝐸 (𝑧) − 𝑧𝐸′(𝑧) = 1 −

(𝑖 − 1)𝑒𝑖𝑧𝑖.

∑︁

𝑖≥2

Theorem 4.2.1 can then be rewritten as:

X𝑃 (𝑧) =

𝐸 (𝑧)
𝐷 (𝑧)

and

X𝐶 (𝑧) =

𝑧2𝐸′′(𝑧)
𝐷 (𝑧)

.

(4.3)

It will be useful for our study to collect here the definitions of several 𝑒-positive series and their

truncations and tails. Considering 𝑘 ≥ 2 whenever it appears, we define

𝐸≥𝑘 (𝑧) =

𝐾≥𝑘 (𝑧) =

𝑒𝑖𝑧𝑖,

𝑖𝑒𝑖𝑧𝑖,

∑︁

𝑖≥𝑘

∑︁

𝑖≥𝑘

(𝑖 − 1)𝑒𝑖𝑧𝑖,

𝐺 ≥𝑘 (𝑧) =

𝐺 ≤𝑘 (𝑧) =

∑︁

𝑖≥𝑘

∑︁

(𝑖 − 1)𝑒𝑖𝑧𝑖,

(𝑖 − 1)𝑒𝑖𝑧𝑖 = 𝐺 (𝑧) −𝐺 ≥𝑘+1(𝑧).

2≤𝑖≤𝑘

𝐾 (𝑧) =

𝑖𝑒𝑖𝑧𝑖,

∑︁

𝑖≥2

𝐺 (𝑧) = 1−𝐷 (𝑧) =

∑︁

𝑖≥2

1
𝐷 (𝑧)

=

∑︁

𝑖≥0

𝐺 (𝑧)𝑖,

(4.4)

The next lemma collects some 𝑒-positivity results concerning the generating functions intro-

duced above.

88

Lemma 4.2.3.

1. The following expressions are 𝑒-positive:

a) 𝑧2𝐸′′(𝑧) − 𝑧𝐸′(𝑧) + 𝑒1𝑧;

b) 2𝑧2𝐸′′(𝑧) − 3𝑧𝐸′(𝑧) + 3𝑒1𝑧 + 2𝑒2𝑧2; and

c) 𝑧2𝐸′′(𝑧) − 3𝑧𝐸′(𝑧) + 3𝐸 (𝑧) + 𝑒2𝑧2.

2. The following expressions can be written as rational functions with 𝑒-positive numerators:

a) X𝑃 (𝑧) − (1 + 𝑒1𝑧); and

b) (1 + 𝑒1𝑧)X𝐶 (𝑧) − X𝑃 (𝑧) + 1 + 𝑒1𝑧.

Proof.

1. The 𝑒-positivity results follow, respectively, from the identities:

a) 𝑧2𝐸′′(𝑧) − 𝑧𝐸′(𝑧) = −𝑒1𝑧 + (cid:205)𝑖≥3

𝑖(𝑖 − 2)𝑒𝑖𝑧𝑖;

b) 2𝑧2𝐸′′(𝑧) − 3𝑧𝐸′(𝑧) = −3𝑒1𝑧 − 2𝑒2𝑧2 + (cid:205)𝑖≥3(2𝑖2 − 5𝑖)𝑒𝑖𝑧𝑖; and

c) 𝑧2𝐸′′(𝑧) − 3𝑧𝐸′(𝑧) + 3𝐸 (𝑧) = 3 − 𝑒2𝑧2 + (cid:205)𝑖≥4(𝑖 − 1) (𝑖 − 3)𝑒𝑖𝑧𝑖.

2. For the results concerning 𝑒-positive numerators, we have that:

a) By (4.3),

X𝑃 (𝑧) − (1 + 𝑒1𝑧) =

=

𝑧𝐸′(𝑧) + 𝑒1𝑧[𝑧𝐸′(𝑧) − 𝐸 (𝑧)]
𝐷 (𝑧)

(cid:205)𝑖≥2

𝑖𝑒𝑖𝑧𝑖 + 𝑒1𝑧 (cid:205)𝑖≥2(𝑖 − 1)𝑒𝑖𝑧𝑖

𝐷 (𝑧)

.

b) By the previous item,

(1 + 𝑒1𝑧)X𝐶 (𝑧) − X𝑃 (𝑧) + 1 + 𝑒1𝑧 =

=

89

(1 + 𝑒1𝑧) [𝑧2𝐸′′(𝑧) − 𝑧𝐸′(𝑧)] + 𝑒1𝑧𝐸 (𝑧)
𝐷 (𝑧)
(1 + 𝑒1𝑧)𝐹1(𝑧) + 𝑒1𝑧(𝐸 (𝑧) − 1 − 𝑒1𝑧)
𝐷 (𝑧)

,

where 𝐹1(𝑧) = (cid:205)𝑖≥3

𝑖(𝑖 − 2)𝑒𝑖𝑧𝑖 and 𝐸 (𝑧) − 1 − 𝑒1𝑧 = (cid:205)𝑖≥2

𝑒𝑖𝑧𝑖 are 𝑒-positive.

□

Using Lemma 4.2.3, we get an 𝑒-positive expression for the generating function for paths that

isolates those terms containing 𝑒1 and that is different from the one in Theorem 4.2.1.

Corollary 4.2.4. We have

X𝑃 (𝑧) =

𝐾 (𝑧)
𝐷 (𝑧)

+ 𝑒1𝑧

𝐺 (𝑧)
𝐷 (𝑧)

+ (1 + 𝑒1𝑧).

We now analyze these generating functions to extract closed formulas for the coefficients in the

𝑒-expansions. Recall the statistic on partitions 𝜀(𝜆) introduced in Section 4.1.3.

We start with a result that shows the relation between the coefficients of 𝐺 (𝑧) 𝑘 and

1
𝐷 (𝑧)

in

their 𝑒-expansion and 𝜀(𝜆).

Lemma 4.2.5. The coefficient of 𝑒𝜆𝑧|𝜆| in 𝐺 (𝑧) 𝑘 is 𝜀(𝜆) and hence

1
𝐷 (𝑧)

∑︁

=

𝜀(𝜆)𝑒𝜆𝑧|𝜆|,

𝜆

where the sum is over all partitions 𝜆.

Proof. This follows by manipulating the formal series directly:

𝐺 (𝑧) 𝑘 =

(cid:33) 𝑘

(𝑖 − 1)𝑒𝑖𝑧𝑖

(cid:32)

∑︁

𝑖≥2

∑︁

=

𝜆
ℓ(𝜆)=𝑘

𝑒𝜆𝑧|𝜆| (cid:214)
𝑖≥2

(𝑖 − 1)𝑚𝑖 (𝜆) =

∑︁

𝜆
ℓ(𝜆)=𝑘

𝜀(𝜆)𝑒𝜆𝑧|𝜆|.

□

We end this subsection by showing that several families of coefficients in the 𝑒-expansion of

X𝑃 (𝑧) and X𝐶 (𝑧) can be expressed compactly in terms of 𝜀(𝜆). (See also [64].)

Proposition 4.2.6. Given a graph 𝐺, let 𝑐𝜆 be the coefficient of 𝑧|𝜆|𝑒𝜆 in X𝐺, that is, X𝐺 = (cid:205) 𝑐𝜆𝑧|𝜆|𝑒𝜆.

Then, we have the following:

1. For 𝐺 = 𝑃𝑛, 𝑐𝜆 = 𝜀(𝜆) +

∑︁

𝜀(𝜆 − 𝑎) =

∑︁

𝑎 𝜀(𝜆 − 𝑎).

𝑎∈supp(𝜆)
In particular, if 𝜆 = 1 ∪ 𝜇 for some partition 𝜇, then 𝑐𝜆 =

𝑎∈supp(𝜆)

(𝑎 − 1)𝜀(𝜇 − 𝑎).

∑︁

𝑎∈supp(𝜇)
𝑎≥2

90

Moreover, we can also extract particular coefficients like

𝑐(𝑛) = 𝑛,

𝑐(𝑛−1,1) = 𝑛 − 2,

𝑐(2𝑘) = 2,

and

𝑐(2𝑘,1) = 1.

2. For 𝐺 = 𝐶𝑛, we have that

∑︁

𝑐𝜆 =

𝑎∈supp(𝜆)

𝑎(𝑎 − 1) 𝜀(𝜆 − 𝑎).

Proof. We use the generating functions in (4.3).

1. The first expression comes directly from the path generating function X𝑃 (𝑧) and the second

expression also follows from Lemma 4.2.3. The equality of the two expressions and the case

when 𝜆 = 1 ∪ 𝜇 follow using Lemma 4.1.7.

2. The formula for this coefficient comes directly from the cycle generating function X𝐶 (𝑧). □

4.2.2 Generating Functions for Twinned Paths

In this section, we focus on studying the various ways to twin a path. The following is a key

result.

Lemma 4.2.7. For 𝑘 ≥ 2, the rational function

1 − 𝐺 ≤𝑘 (𝑧)
𝐷 (𝑧)

and the function X𝑃 (𝑧) (1 − 𝐺 ≤𝑘 (𝑧))

are 𝑒-positive.

Proof. For the rational function, we have that

1 − 𝐺 ≤𝑘 (𝑧)
𝐷 (𝑧)

=

1 − 𝐺 (𝑧) + 𝐺 ≥𝑘+1(𝑧)
1 − 𝐺 (𝑧)

= 1 +

𝐺 ≥𝑘+1(𝑧)
𝐷 (𝑧)

.

(4.5)

This is 𝑒-positive since 𝐺 ≥𝑘+1(𝑧) =

∑︁

(𝑖 − 1)𝑒𝑖𝑧𝑖 and

expands 𝑒-positively in powers

1
𝐷 (𝑧)

𝑖≥𝑘+1
of 𝐺 (𝑧). The 𝑒-positivity of the second function follows from (4.3) and the identity

X𝑃 (𝑧) (1 − 𝐺 ≤𝑘 (𝑧)) = 𝐸 (𝑧) + X𝑃 (𝑧) 𝐺 ≥𝑘+1(𝑧).

□

91

4.2.2.1 Paths Twinned at a Leaf

The recurrence for the chromatic symmetric function of twinned paths at a leaf (i.e., a vertex of

degree 1) appears in Dahlberg and van Willigenburg [14, Equation 5], where the graph 𝑃𝑛,𝑣 with 𝑣

a leaf is called the lariat graph 𝐿𝑛+3. Its chromatic symmetric function had been considered earlier

by Wolfe in [64], and 𝑒-positivity was first established by Gebhard and Sagan in [27, Corollary

7.7].

Proposition 4.2.8. Let 𝑣 be a leaf of the path 𝑃𝑛. The generating function for the chromatic

symmetric function of the twin 𝑃𝑛,𝑣 of a path on 𝑛 vertices satisfies the following identity:

2 + 2𝑒1𝑧 +

∑︁

𝑛≥1

𝑋𝑃𝑛,𝑣 𝑧𝑛+1 = 2(1 − 𝑒2𝑧2)X𝑃 (𝑧).

Proof. By [14, Equation 5], the chromatic symmetric function of 𝑃𝑛,𝑣, with 𝑛 ≥ 1, is given by

𝑋𝑃𝑛,𝑣 = 2𝑋𝑃𝑛+1 − 𝑋𝑃2

𝑋𝑃𝑛−1

.

The proof now follows by using the generating function X𝑃 (𝑧).

(4.6)

□

Now we are ready to derive a generating function for paths twinned at a leaf. Although the

𝑒-positivity was established in [27, Corollary 7.7] and again in [14], as mentioned earlier, our

contribution here is to give the manifestly 𝑒-positive generating function below for 𝑋𝑃𝑛,𝑣 , using

only symmetric functions, which enables a more efficient coefficient extraction.

Proposition 4.2.9. Let X𝑃𝑣 (𝑧) be the generating function for the twinned path at a leaf, that is,
X𝑃𝑣 := (cid:205)𝑛≥1

𝑋𝑃𝑛,𝑣 𝑧𝑛+1. Then

1
2

X𝑃𝑣 (𝑧) = 𝐾 (𝑧)

𝐺 ≥3(𝑧)
𝐷 (𝑧)

+ 𝑒1𝑧𝐺 (𝑧)

𝐺 ≥3(𝑧)
𝐷 (𝑧)

+ 𝑒2𝑧2 +

∑︁

𝑖≥3

𝑖𝑒𝑖𝑧𝑖 + 𝑒1𝑧𝐺 ≥3(𝑧).

In particular 𝑋𝑃𝑛,𝑣 is 𝑒-positive, and the initial values are

𝑋𝑃1,𝑣 = 2𝑒2,

𝑋𝑃2,𝑣 = 2(3𝑒3),

𝑋𝑃3,𝑣 = 2(4𝑒4 + 2𝑒1𝑒3),

𝑋𝑃4,𝑣 = 2(4𝑒2𝑒3 + 3𝑒1𝑒4 + 5𝑒5).

An 𝑒-positive expression without denominators in terms of the path generating function X𝑃 is

1
2

X𝑃𝑣 (𝑧) = X𝑃 (𝑧)𝐺 ≥3(𝑧) +

𝑒𝑖𝑧𝑖.

∑︁

𝑖≥2

92

Proof. By Lemma 4.2.7 and Corollary 4.2.4, we have that

1
2

X𝑃𝑣 (𝑧) = (1 − 𝑒2𝑧2)X𝑃 − (1 + 𝑒1𝑧) = 𝐾 (𝑧)

1 − 𝑒2𝑧2
𝐷 (𝑧)

+ 𝑒1𝑧𝐺 (𝑧)

1 − 𝑒2𝑧2
𝐷 (𝑧)

− 𝑒2𝑧2(1 + 𝑒1𝑧)

= (𝐾 (𝑧) + 𝑒1𝑧𝐺 (𝑧))

= (𝐾 (𝑧) + 𝑒1𝑧𝐺 (𝑧))

𝐺 ≥3(𝑧)
𝐷 (𝑧)
𝐺 ≥3(𝑧)
𝐷 (𝑧)

+ (𝐾 (𝑧) + 𝑒1𝑧𝐺 (𝑧)) − 𝑒2𝑧2(1 + 𝑒1𝑧)

+ (𝐾 (𝑧) − 𝑒2𝑧2) + 𝑒1𝑧(𝐺 (𝑧) − 𝑒2𝑧2).

Since 𝐾 (𝑧) − 𝑒2𝑧2 = 𝑒2𝑧2 + (cid:205)𝑖≥3

𝑖𝑒𝑖𝑧𝑖 and 𝐺 (𝑧) − 𝑒2𝑧2 = (cid:205)𝑖≥3(𝑖 − 1)𝑒𝑖𝑧𝑖, the result follows.

The second expression is obtained from the first by rewriting the formula in Corollary 4.2.4 as

follows:

X𝑃 (𝑧) =

𝐾 (𝑧) + 𝑒1𝑧𝐺 (𝑧)
𝐷 (𝑧)

+ (1 + 𝑒1𝑧).

□

Corollary 4.2.10. Let 𝑐𝜆 be the coefficient of 𝑒𝜆𝑧|𝜆| in X𝑃𝑣 , that is X𝑃𝑣 = (cid:205) 𝑐𝜆𝑒𝜆𝑧|𝜆|, where 𝑣 is a

leaf of the path 𝑃𝑛. The following is a list of closed formulas for all the coefficients 𝑐𝜆 involved in

the general expression of X𝑃𝑣 (𝑧):

1. 𝑐(𝑘) = 2𝑘, 𝑘 ≥ 3, and 𝑐(2) = 2;

2. 𝑐(𝑘−1,1) = 2(𝑘 − 2), 𝑘 ≥ 4;

3. 𝑐(𝑘−2,2) = 4(𝑘 − 3), 𝑘 ≥ 5;

4. 𝑐(𝑖, 𝑗) = 2𝑖( 𝑗 − 1) + 2 𝑗 (𝑖 − 1) = 2(2𝑖 𝑗 − 𝑖 − 𝑗), 𝑖 > 𝑗 ≥ 3;

5. 𝑐(𝑖,𝑖) = 2𝑖(𝑖 − 1), 𝑖 ≥ 3.

6. 𝑐(3,2𝑘) = 8 and 𝑐(3,2𝑘,1) = 4, 𝑘 ≥ 2.

7. If 𝑐1∪𝜇 ≠ 0 and ℓ(𝜇) ≥ 2, then 1 ∉ supp(𝜇) and there exists 𝑎 ≥ 3 such that 𝑎 ∈ supp(𝜇).

In particular 𝑐(2𝑘,1) = 0. The coefficient of 𝑒1∪𝜇 is equal to twice the coefficient of 𝑒𝜇 in
𝐺 (𝑧)𝐺 ≥3(𝑧) 𝐺 (𝑧)ℓ(𝜇)−2, and it equals

(𝑎 − 1) (𝑏 − 1)𝜀((𝜇 − 𝑎) − 𝑏).

∑︁

2

(𝑎,𝑏)
𝑎,𝑏∈supp(𝜇)
𝑎≥2,𝑏≥3

93

8. Assume 1 ∉ supp(𝜆) and ℓ(𝜆) ≥ 2. If 𝑐𝜆 ≠ 0, then 𝜆 contains at least one part of size at

least 3. In particular 𝑐(2𝑘) = 0. The coefficient 𝑐𝜆 is equal to twice the coefficient of 𝑒𝜆 in
𝐾 (𝑧)𝐺 ≥3(𝑧) 𝐺 (𝑧)ℓ(𝜆)−2, and it equals

𝑎(𝑏 − 1)𝜀((𝜆 − 𝑎) − 𝑏).

∑︁

2

(𝑎,𝑏)
𝑎,𝑏∈supp(𝜆)
𝑎≥2,𝑏≥3

Note that cases (c)-(f) are particular cases of (g) and (h).

4.2.2.2 Paths Twinned at Both Leaves

In this section, we consider the twinned path 𝑃𝑛,𝑤,𝑣 at both leaves, which we label with 𝑤 and

𝑣. The 𝑒-positivity of its chromatic symmetric function is a consequence of [27, Corollary 7.7],

whose proof relies on the theory of symmetric functions in noncommutating variables. Here we

derive an 𝑒-positive generating function using only symmetric function identities.

Unlike the other families of graphs, here one needs to pay special attention to the smaller

values of 𝑛. We consider the special case of the path on two vertices first. Twinning both vertices

produces the twin of the cycle graph 𝐶3 at one vertex, which is also the complete graph 𝐾4, as

shown in Figure 4.4, and therefore we have the following.

Lemma 4.2.11. For the path 𝑃2 twinned at both vertices, 𝑋𝑃2,𝑣,𝑤 = 𝑋𝐶3,𝑣 = 24𝑒4.

𝑣′
•

𝑣

•

𝑤

•

𝑣

•

𝑣′
•

•
𝑤′

𝑤

•

𝑣

•

𝑃2,𝑣

𝑃2,𝑣,𝑤

𝑤

•

𝑣′
•

•
𝑢

𝐶3,𝑣

Figure 4.4 Demonstration that 𝑃2,𝑣,𝑤 = 𝐶3,𝑣

For the general case, we start with a consequence of the triple deletion argument.

94

Corollary 4.2.12. Let 𝑣, 𝑤 be the two leaves of the path 𝑃𝑛, and let 𝑃𝑛,𝑣,𝑤 be the path twinned at

both leaves. Then, for 𝑛 ≥ 3,

𝑋𝑃𝑛,𝑣,𝑤 = 2𝑋𝑃𝑛+1,𝑣 − 2𝑒2𝑋𝑃𝑛−1,𝑣 = 4(𝑋𝑃𝑛+2 − 2𝑒2𝑋𝑃𝑛 + 𝑒2
2

𝑋𝑃𝑛−2).

(4.7)

This relation allows us to give the following generating function identity.

Proposition 4.2.13. For the graph 𝑃𝑛,𝑣,𝑤 , we have

1
4

∑︁

𝑛≥3

𝑋𝑃𝑛,𝑣,𝑤 𝑧𝑛+2 = (1 − 𝑒2𝑧2)

1
2

X𝑃𝑣 +

𝛼,

1
2

(4.8)

where 𝛼 = 2𝑒2
2

𝑧4 − (8𝑒4𝑧4 + 4𝑒3𝑒1𝑧4 + 6𝑒3𝑧3 + 2𝑒2𝑧2).

Proof. Multiply both sides of the first equality in Corollary 4.2.12 by 𝑧𝑛+2 and sum for 𝑛 ≥ 3:

𝑋𝑃𝑛,𝑣,𝑤 𝑧𝑛+2 = 2

∑︁

𝑛≥3

= 2

∑︁

𝑛≥3
∑︁

𝑛≥4

𝑋𝑃𝑛+1,𝑣 𝑧𝑛+2 − 2𝑒2

∑︁

𝑋𝑃𝑛−1,𝑣 𝑧𝑛+2

𝑛≥3
𝑋𝑃𝑛,𝑣 𝑧𝑛+1 − 2𝑒2𝑧2 ∑︁
𝑛≥2

𝑋𝑃𝑛,𝑣 𝑧𝑛+1

= 2(1 − 𝑒2𝑧2)X𝑃𝑣 − 2(𝑋𝑃3,𝑣 𝑧4 + 𝑋𝑃2,𝑣 𝑧3 + 𝑋𝑃1,𝑣 𝑧2 − 2𝑒2𝑧2𝑋𝑃1,𝑣 𝑧2)

= 2(1 − 𝑒2𝑧2)X𝑃𝑣 − 2[(8𝑒4 + 4𝑒3𝑒1)𝑧4 + 6𝑒3𝑧3 + 2𝑒2𝑧2 − 2𝑒2
2

𝑧4]

where the computations for 𝑋𝑃𝑛,𝑣 follow from Proposition 4.2.9.

□

The next theorem follows from Corollary 4.2.12 and manipulation of the formal series.

Theorem 4.2.14. The generating function

1
4

∑︁

𝑛≥3

𝑋𝑃𝑛,𝑣,𝑤 𝑧𝑛+2 has the following 𝑒-positive expansion:

1
4

∑︁

𝑛≥3

𝑋𝑃𝑛,𝑣,𝑤 𝑧𝑛+2 = (𝐾 (𝑧) + 𝑒1𝑧𝐺 (𝑧))

𝐺 ≥3(𝑧)2
𝐷 (𝑧)

+ 𝑒1𝑧𝐺 ≥3(𝑧)2

(cid:32)

(cid:33)

𝑖𝑒𝑖𝑧𝑖 + 𝑒1𝑧 ∑︁
𝑖≥4
An 𝑒-positive expression without denominators in terms of the path generating function X𝑃 is

(𝑖 − 1)𝑒𝑖𝑧𝑖 + 𝑒2𝑧2 ∑︁
𝑖≥3

(𝑖 − 2)𝑒𝑖𝑧𝑖

𝐺 ≥3(𝑧)

𝑖𝑒𝑖𝑧𝑖.

∑︁

∑︁

𝑖≥5

𝑖≥3

+

+

X𝑃 (𝑧)𝐺2

≥3(𝑧) + 𝐾≥5(𝑧) + 𝐺 ≥3(𝑧)

∑︁

𝑖≥3

𝑒𝑖𝑧𝑖 + 𝑒1𝑧𝐺 ≥4(𝑧) + 𝑒2𝑧2 ∑︁
𝑖≥3

(𝑖 − 2)𝑒𝑖𝑧𝑖.

95

Proof. We use the generating function in Proposition 4.2.9 to expand 1

2 (1 − 𝑒2𝑧2)X𝑃𝑛,𝑣 as

1
2

(1 − 𝑒2𝑧2)X𝑃𝑣 =(1 − 𝑒2𝑧2) (𝐾 (𝑧) + 𝑧𝑒1𝐺 (𝑧))

𝐺 ≥3(𝑧)
𝐷 (𝑧)

(cid:32)

+ (1 − 𝑒2𝑧2)

𝑒2𝑧2 +

𝑖𝑒𝑖𝑧𝑖 + 𝑧𝑒1𝐺 ≥3(𝑧)

(cid:33)

.

∑︁

𝑖≥3

By (4.5),

1 − 𝑒2𝑧2
𝐷 (𝑧)

= 1 +

𝐺 ≥3(𝑧)
𝐷 (𝑧)

, and we can rewrite the above expression as

1
2

(1 − 𝑒2𝑧2)X𝑃𝑣 = (𝐾 (𝑧) + 𝑧𝑒1𝐺 (𝑧)) 𝐺 ≥3(𝑧)

(cid:18)

1 +

(cid:19)

𝐺 ≥3(𝑧)
𝐷 (𝑧)

(cid:32)

+ (1 − 𝑒2𝑧2)

𝑒2𝑧2 +

𝑖𝑒𝑖𝑧𝑖 + 𝑧𝑒1𝐺 ≥3(𝑧)

(cid:33)

.

∑︁

𝑖≥3

Next, we arrange the expression above so that the term − 1
2

𝛼 appears:

1
2

(1 − 𝑒2𝑧2)X𝑃𝑣 = (𝐾 (𝑧) + 𝑧𝑒1𝐺 (𝑧))

𝐺 ≥3(𝑧)2
𝐷 (𝑧)

−

𝛼 +

∑︁

1
2

𝑖≥5
+ (𝐾 (𝑧) + 𝑧𝑒1𝐺 (𝑧)) 𝐺 ≥3(𝑧) − 𝑒2𝑧2 ∑︁
𝑖≥3

(𝑖 − 1)𝑒𝑖𝑧𝑖

𝑖𝑒𝑖𝑧𝑖 + 𝑒1𝑧 ∑︁
𝑖≥4
𝑖𝑒𝑖𝑧𝑖 − 𝑒2𝑧2(𝑧𝑒1)𝐺 ≥3(𝑧).

Thus, we have that

1
4

∑︁

𝑛≥3

𝑋𝑃𝑛,𝑣,𝑤 𝑧𝑛+2 =

1
2

(1 − 𝑒2𝑧2)X𝑃𝑣 +

= (𝐾 (𝑧) + 𝑧𝑒1𝐺 (𝑧))

𝛼

1
2
𝐺 ≥3(𝑧)2
𝐷 (𝑧)

+

∑︁

𝑖≥5

𝑖𝑒𝑖𝑧𝑖 + 𝑒1𝑧 ∑︁
𝑖≥4

(𝑖 − 1)𝑒𝑖𝑧𝑖

(4.9)

+ (𝐾 (𝑧) + 𝑧𝑒1𝐺 (𝑧)) 𝐺 ≥3(𝑧) − 𝑒2𝑧2 ∑︁
𝑖≥3

𝑖𝑒𝑖𝑧𝑖 − 𝑒2𝑧2(𝑧𝑒1)𝐺 ≥3(𝑧),

(4.10)

where the terms in line (4.9) are 𝑒-positive. Thus, we only need to show that the terms in line (4.10)

are also 𝑒-positive. Note that

(cid:32)

𝐾 (𝑧)𝐺 ≥3(𝑧) =

2𝑒2𝑧2 +

(cid:33)

𝑖𝑒𝑖𝑧𝑖

∑︁

𝑖≥3

𝐺 ≥3(𝑧) = 2𝑒2𝑧2𝐺 ≥3(𝑧) + 𝐺 ≥3(𝑧)

𝑖𝑒𝑖𝑧𝑖.

∑︁

𝑖≥3

96

Together with the fact that 𝐺 (𝑧) − 𝑒2𝑧2 = 𝐺 ≥3(𝑧), line (4.10) can be written as

(𝐾 (𝑧) + 𝑧𝑒1𝐺 (𝑧)) 𝐺 ≥3(𝑧) − 𝑒2𝑧2 ∑︁
𝑖≥3

𝑖𝑒𝑖𝑧𝑖 − 𝑒2𝑧2(𝑧𝑒1)𝐺 ≥3(𝑧)

(cid:32)

= 𝐺 ≥3(𝑧)

= 𝐺 ≥3(𝑧)

∑︁

𝑖≥3
∑︁

𝑖≥3

𝑖𝑒𝑖𝑧𝑖 + 𝑒2𝑧2

∑︁

2

(𝑖 − 1)𝑒𝑖𝑧𝑖 −

∑︁

𝑖𝑒𝑖𝑧𝑖

𝑖≥3
(𝑖 − 2)𝑒𝑖𝑧𝑖 + 𝑒1𝑧𝐺 ≥3(𝑧)2.

𝑖≥3

𝑖𝑒𝑖𝑧𝑖 + 𝑒2𝑧2 ∑︁
𝑖≥3

(cid:33)

+ 𝑒1𝑧(𝐺 (𝑧) − 𝑒2𝑧2)𝐺 ≥3(𝑧)

Since the expression in (4.11) is also 𝑒-positive, the result follows.

The second expression involving X𝑃 follows as in the proof of Proposition 4.2.9.

In particular, we can extract the following formulas for the coefficients.

(4.11)

□

Corollary 4.2.15. Let 𝑐𝜆 be the coefficient of 𝑒𝜆𝑧|𝜆| in X𝑃𝑣,𝑤 , that is, X𝑃𝑣,𝑤 = (cid:205) 𝑐𝜆𝑒𝜆𝑧|𝜆|. We have

the following list of closed formulas:

1. For 𝑘 ≥ 3, 𝑐(𝑘+2) = 4(𝑘 + 2), 𝑐(𝑘,2) = 4(𝑘 − 2), and 𝑐(𝑘+1,1) = 4𝑘.

2. For 𝑖 ≥ 3, 𝑐(𝑖,𝑖) = 4(𝑖 − 1)𝑖, and for 𝑖, 𝑗 ≥ 3, 𝑖 ≠ 𝑗, 𝑐(𝑖, 𝑗) = 4( 𝑗 − 1)𝑖 + 4(𝑖 − 1) 𝑗.

3. For 𝑖, 𝑗 ≥ 3, 𝑖 ≠ 𝑗, 𝑐(𝑖,𝑖,1) = 4(𝑖 − 1)2, 𝑐(𝑖, 𝑗,1) = 8(𝑖 − 1) ( 𝑗 − 1), 𝑖, 𝑗 ≥ 3, 𝑖 ≠ 𝑗, and zero

otherwise.

4. If 𝑐1∪𝜇 ≠ 0, then 1 ∉ supp(𝜇).

5. For all 𝑘 ≥ 0, 𝑐(32,2𝑘+1) = 32 and 𝑐(32,2𝑘+1,1) = 16.

4.2.2.3 Paths Twinned at an Interior Vertex

In this section, we establish an 𝑒-positive generating function for the path 𝑃𝑛,ℓ twinned at an

interior vertex ℓ, where we label the vertices of 𝑃𝑛 by 1, 2, . . . , 𝑛 from left to right. As stated in

the introduction, the 𝑒-positivity can also be deduced from [27, Theorem 7.8].

As in the preceding section, we first derive a triple deletion formula for the chromatic symmetric

function of 𝑃𝑛,ℓ, (Proposition 4.2.19), and then deduce an 𝑒-positive generating function for its

chromatic symmetric function (Theorem 4.2.24). We begin with some definitions.

97

Definition 4.2.16. For 𝑛 ≥ 2 and 1 ≤ ℓ ≤ 𝑛 − 1, let ˜𝑇𝑛,ℓ (T for triangle) denote the graph obtained

from the path graph 𝑃𝑛 by adding a vertex adjacent to both ℓ and ℓ + 1. For 𝑛 ≥ 1 and 1 ≤ ℓ ≤ 𝑛,

let 𝐹𝑛,ℓ (F for flagpole) denote the graph obtained from 𝑃𝑛 by adding a vertex adjacent to ℓ.

By the triple deletion argument illustrated in Figure 4.5, we have the following result.

Lemma 4.2.17. For 𝑛 ≥ 3 and 2 ≤ ℓ ≤ 𝑛 − 1, we have

𝑋𝑃𝑛,ℓ = 2𝑋 ˜𝑇𝑛,ℓ −1

− 𝑋 ˜𝑇ℓ,ℓ −1

𝑋𝑃𝑛−ℓ .

𝑃𝑛,ℓ

· · ·

(cid:101)𝑇𝑛,ℓ−1
· · ·

•
1

•
1

(cid:101)𝑇ℓ,ℓ−1 ⊔ 𝑃𝑛−ℓ
· · ·

•
1

•
ℓ − 1

•
ℓ − 1

•
ℓ − 1

ℓ′
•
𝜖3
•
ℓ

ℓ′
•

•
ℓ

ℓ′
•

•
ℓ

𝜖1
𝜖2

•
ℓ + 1

•
ℓ + 1

•
ℓ + 1

· · ·

· · ·

· · ·

•
𝑛

•
𝑛

•
𝑛

Figure 4.5 The triple deletion argument applied as in Lemma 4.2.17

By carefully applying the triple deletion argument to various 𝑃𝑛,ℓ, we can deal with the triangles

(cid:101)𝑇𝑛,ℓ by “shifting” them around. Note that (cid:101)𝑇𝑛,ℓ has a triangle, and so the triple deletion argument

applies to two different sets of edges, to which we refer as left and right shifts. We illustrate them

on the left-hand side and right-hand side of Figure 4.6, respectively.

Lemma 4.2.18 (Left and Right Shift Lemma). For 𝑛 ≥ 3 and 2 ≤ ℓ ≤ 𝑛 − 1, we have

(cid:101)𝑇𝑛,ℓ = 𝑋𝐹𝑛,ℓ + 𝑋𝑃𝑛+1 − 𝑋𝑃ℓ+1
𝑋

𝑋𝑃𝑛−ℓ

(cid:101)𝑇𝑛,ℓ = 𝑋𝐹𝑛,ℓ+1 + 𝑋𝑃𝑛+1 − 𝑋𝑃ℓ 𝑋𝑃𝑛−ℓ+1
𝑋

Left shift

Right shift

Our next step is to use Lemma 4.2.18 to obtain a formula equivalent to that in Lemma 4.2.17

which does not involve twinning paths.

98

(cid:101)𝑇𝑛,ℓ
· · ·

𝐹𝑛,ℓ

· · ·

𝑃𝑛+1
· · ·

•
1

•
1

•
1

𝑃ℓ+1 ⊔ 𝑃𝑛−ℓ
· · ·
•
1

𝜖3
𝜖2

•

(ℓ + 1)′
𝜖1

•
ℓ + 1

· · ·

•

(ℓ + 1)′

•
ℓ + 1

· · ·

•

(ℓ + 1)′

•
ℓ + 1

· · ·

•

(ℓ + 1)′

•
ℓ

•
ℓ

•
ℓ

•
𝑛

•
𝑛

•
𝑛

•
ℓ

•
ℓ + 1

· · ·

•
𝑛

(cid:101)𝑇𝑛,ℓ
· · ·

𝐹𝑛,ℓ+1
· · ·

𝑃𝑛+1
· · ·

•
1

•
1

•
1

𝑃ℓ ⊔ 𝑃𝑛−ℓ+1
· · ·
•
1

𝜖1
𝜖2

•

(ℓ + 1)′
𝜖3

•
ℓ + 1

· · ·

•

(ℓ + 1)′

•
ℓ + 1

· · ·

•

(ℓ + 1)′

•
ℓ + 1

· · ·

•

(ℓ + 1)′

•
ℓ

•
ℓ

•
ℓ

•
𝑛

•
𝑛

•
𝑛

•
ℓ

•
ℓ + 1

· · ·

•
𝑛

Figure 4.6 Illustration of Lemma 4.2.18

Proposition 4.2.19. For 𝑛 ≥ 3 and 2 ≤ ℓ ≤ 𝑛 − 1, we have

𝑋𝑃𝑛,ℓ = −2𝑋𝑃ℓ −1

𝑋𝑃𝑛−ℓ+2 + 2𝑒1𝑋𝑃𝑛 + 4𝑋𝑃𝑛+1 − 2𝑋𝑃ℓ 𝑋𝑃𝑛−ℓ+1 + 2𝑒2𝑋𝑃ℓ −1

𝑋𝑃𝑛−ℓ − 2𝑋𝑃ℓ+1

𝑋𝑃𝑛−ℓ .

Proof. Applying the Left and Right Shift Lemmas at ℓ − 𝑘 − 1 implies that

𝑋𝐹𝑛,ℓ −𝑘 = 𝑋𝐹𝑛,ℓ −𝑘−1 + 𝑋𝑃ℓ −𝑘−1

𝑋𝑃𝑛−ℓ+𝑘+2 − 𝑋𝑃ℓ −𝑘 𝑋𝑃𝑛−ℓ+𝑘+1

.

(4.12)

By applying (4.12) repeatedly, we get that

𝑋𝐹𝑛,ℓ = 𝑋𝐹𝑛,1 +

ℓ−2
∑︁

𝑘=0

(cid:0)𝑋𝑃ℓ −𝑘−1

𝑋𝑃𝑛−ℓ+𝑘+2 − 𝑋𝑃ℓ −𝑘 𝑋𝑃𝑛−ℓ+𝑘+1

(cid:1) .

(4.13)

Since 𝐹𝑛,1 is precisely 𝑃𝑛+1, we can telescope the sum in (4.13) to obtain

𝑋𝐹𝑛,ℓ = 𝑋𝑃𝑛+1 + 𝑋𝑃1

𝑋𝑃𝑛 − 𝑋𝑃ℓ 𝑋𝑃𝑛−ℓ+1

.

(4.14)

Recall the formula in Lemma 4.2.17:

𝑋𝑃𝑛,ℓ = 2𝑋

(cid:101)𝑇𝑛,ℓ −1

− 𝑋

(cid:101)𝑇ℓ,ℓ −1

𝑋𝑃𝑛−ℓ .

We apply the Left Shift Lemma to 𝑋

(cid:101)𝑇𝑛,ℓ −1

and the Right Shift Lemma to 𝑋

(cid:101)𝑇ℓ,ℓ −1

, and obtain

𝑋𝑃𝑛,ℓ = 2 (cid:0)𝑋𝐹𝑛,ℓ −1 + 𝑋𝑃𝑛+1 − 𝑋𝑃ℓ 𝑋𝑃𝑛−ℓ+1

(cid:1) − (cid:0)𝑋𝐹ℓ,ℓ + 𝑋𝑃ℓ+1 − 𝑋𝑃ℓ −1

𝑋𝑃2

(cid:1) 𝑋𝑃𝑛−ℓ .

99

Since 𝐹ℓ,ℓ is 𝑃ℓ+1, we can rewrite the last equation as:

𝑋𝑃𝑛,ℓ = 2 (cid:0)𝑋𝐹𝑛,ℓ −1 + 𝑋𝑃𝑛+1 − 𝑋𝑃ℓ 𝑋𝑃𝑛−ℓ+1

(cid:1) − (cid:0)2𝑋𝑃ℓ+1 − 𝑋𝑃ℓ −1

𝑋𝑃2

(cid:1) 𝑋𝑃𝑛−ℓ

= 2𝑋𝐹𝑛,ℓ −1 + (cid:2)2𝑋𝑃𝑛+1 − 2𝑋𝑃ℓ 𝑋𝑃𝑛−ℓ+1 + 𝑋𝑃ℓ −1

𝑋𝑃2

𝑋𝑃𝑛−ℓ − 2𝑋𝑃ℓ+1

𝑋𝑃𝑛−ℓ

(cid:3) .

Substituting in (4.14) for 𝑋𝐹𝑛,ℓ −1, we have:

𝑋𝑃𝑛,ℓ =2 (cid:0)𝑋𝑃𝑛+1 + 𝑋𝑃1

𝑋𝑃𝑛 − 𝑋𝑃ℓ −1
+ (cid:2)2𝑋𝑃𝑛+1 − 2𝑋𝑃ℓ 𝑋𝑃𝑛−ℓ+1 + 𝑋𝑃ℓ −1

𝑋𝑃𝑛−ℓ

(cid:1)

𝑋𝑃2

𝑋𝑃𝑛−ℓ − 2𝑋𝑃ℓ+1

𝑋𝑃𝑛−ℓ

(cid:3) .

Finally, the formula in the statement follows by collecting all the terms and evaluating 𝑋𝑃1 = 𝑒1
□

and 𝑋𝑃2 = 2𝑒2.

Now we investigate the generating function of 𝑋𝑃𝑛,ℓ . For this, we introduce two families of

polynomials in the variable 𝑧 with coefficients in the ring of symmetric functions.

Definition 4.2.20.

1. For ℓ ≥ 2, we define the following polynomial of degree ℓ + 1 in 𝑧:

𝑓ℓ (𝑧) := 2 + 𝑒1𝑧 − 𝑋𝑃ℓ −1

𝑧ℓ−1(1 − 𝑒2𝑧2) − 𝑋𝑃ℓ 𝑧ℓ − 𝑋𝑃ℓ+1

𝑧ℓ+1.

2. For ℓ ≥ 2, we define the following polynomial of degree ℓ + 1 :

𝑔ℓ (𝑧) := −

ℓ
∑︁

𝑗=0

𝑋𝑃 𝑗 𝑧 𝑗 − (1 + 𝑒1𝑧)

ℓ−2
∑︁

𝑗=0

𝑋𝑃 𝑗 𝑧 𝑗 −(𝑋𝑃ℓ+1 − 𝑒2𝑋𝑃ℓ −1)𝑧ℓ+1.

The following result gives an identity for the generating function for the chromatic symmetric

function of the twinned path in terms of the generating function for the chromatic symmetric

function of the path and the new families of polynomials introduced.

Proposition 4.2.21. Let 2 ≤ ℓ ≤ 𝑛 − 1. The generating function for the chromatic symmetric

function of the twinned path 𝑃𝑛,ℓ, twinned at vertex ℓ, can be written in terms of the path generating

function X𝑃 as follows:

𝑋𝑃𝑛,ℓ 𝑧𝑛+1 = 2X𝑃 (𝑧) 𝑓ℓ (𝑧) + 2𝑔ℓ (𝑧).

(4.15)

∑︁

𝑛≥ℓ+1

100

Proof. We reorder the terms appearing in the recurrence in Proposition 4.2.19 to make the source

of the factor 𝑓ℓ (𝑧) accompanying X𝑃 more transparent:

𝑋𝑃𝑛,ℓ = 4𝑋𝑃𝑛+1 + 2𝑒1𝑋𝑃𝑛 − 2𝑋𝑃ℓ −1 (𝑋𝑃𝑛−ℓ+2 − 𝑒2𝑋𝑃𝑛−ℓ ) − 2𝑋𝑃ℓ 𝑋𝑃𝑛−ℓ+1 − 2𝑋𝑃ℓ+1

𝑋𝑃𝑛−ℓ .

(4.16)

Multiplying by 𝑧𝑛+1 and summing over 𝑛 ≥ ℓ + 1 gives

∑︁

𝑋𝑃𝑛,ℓ 𝑧𝑛+1

𝑛≥ℓ+1
= 2(cid:2)2 + 𝑒1𝑧 − 𝑋𝑃ℓ −1

− 4

ℓ
∑︁

𝑗=0

𝑗=0

𝑧ℓ−1(1 − 𝑒2𝑧2) − 𝑋𝑃ℓ 𝑧ℓ − 𝑧ℓ+1𝑋𝑃ℓ +1
ℓ−1
∑︁

(cid:3)X𝑃

𝑋𝑃 𝑗 𝑧 𝑗 − 2𝑧𝑒1

𝑋𝑃 𝑗 𝑧 𝑗 + 2𝑧ℓ−1(1 + 𝑧𝑒1) 𝑋𝑃ℓ −1 + 2𝑋𝑃ℓ 𝑧ℓ−2(𝑋𝑃ℓ+1 − 𝑒2𝑋𝑃ℓ −1)𝑧ℓ+1.

Here we have made the substitutions (cid:205)2

𝑗=0

𝑋𝑃 𝑗 𝑧 𝑗 = 1 + 𝑒1𝑧 + 2𝑒2𝑧2, (cid:205)1

𝑗=0

𝑋𝑃 𝑗 𝑧 𝑗 = 1 + 𝑒1𝑧 and

𝑋𝑃0 = 1.

The expression for 𝑓ℓ (𝑧) follows immediately from the first line above.

Now rewrite the second line as

− 4𝑋𝑃ℓ 𝑧ℓ − 4𝑋𝑃ℓ −1

𝑧ℓ−1 − 4

ℓ−2
∑︁

𝑗=0

𝑋𝑃 𝑗 𝑧 𝑗 − 2𝑧ℓ𝑒1𝑋𝑃ℓ −1 − 2𝑧𝑒1

ℓ−2
∑︁

𝑗=0

𝑋𝑃 𝑗 𝑧 𝑗

+ 2𝑋𝑃ℓ −1

𝑧ℓ−1 + 2𝑒1𝑋𝑃ℓ −1

𝑧ℓ + 2𝑋𝑃ℓ 𝑧ℓ−2(𝑋𝑃ℓ+1 − 𝑒2𝑋𝑃ℓ −1)𝑧ℓ+1,

which in turn yields the expression for 𝑔ℓ (𝑧) in Definition 4.2.20.

□

Although 𝑓ℓ (𝑧) is not 𝑒-positive, we can conclude the following.

Corollary 4.2.22. The 𝑒-positivity of 𝑋𝑃𝑛,ℓ is equivalent to the 𝑒-positivity of X𝑃 (𝑧) 𝑓ℓ (𝑧).

Proof. The degree of 𝑔ℓ (𝑧) as a polynomial in 𝑧 is ℓ +1, while the left-hand side of (4.15) has lowest

degree ℓ + 2 in 𝑧. We conclude that all terms in 𝑔ℓ (𝑧) are necessarily canceled out by identical

terms in X𝑃 (𝑧) 𝑓ℓ (𝑧).

□

Our next result rewrites 𝑓ℓ (𝑧) as a positive expansion of other functions that were introduced

in (4.4).

101

Lemma 4.2.23. For ℓ ≥ 2, we have

𝑓ℓ (𝑧) =

ℓ+1
∑︁

𝑖=3

(𝑖 − 2)𝑒𝑖𝑧𝑖 + 2(𝐷 + 𝐺 ≥ℓ+2) +

ℓ−2
∑︁

𝑖=1

(𝐷 + 𝐺 ≥ℓ+2−𝑖) 𝑋𝑃𝑖 𝑧𝑖.

(4.17)

Proof. By Definition 4.2.20,

𝑓ℓ (𝑧) = 2 + 𝑒1𝑧 + 𝑋𝑃ℓ −1

𝑒2𝑧ℓ+1 − 𝑋𝑃ℓ −1

𝑧ℓ−1 − 𝑋𝑃ℓ 𝑧ℓ − 𝑋𝑃ℓ+1

𝑧ℓ+1.

For ℓ = 2, recall that 𝑋𝑃3 = 𝑒2𝑒1 + 3𝑒3, and 𝑋𝑃1 = 𝑒1 and 𝑋𝑃2 = 2𝑒2. Then we have

𝑓2(𝑧) = 2 − 2𝑒2𝑧2 − 3𝑒3𝑧3 = 2(1 − 𝑒2𝑧2 − 2𝑒3𝑧3) + 𝑒3𝑧3

= 2(1 − 𝐺 (𝑧) + 𝐺 ≥4(𝑧)) + 𝑒3𝑧3 = 2(𝐷 (𝑧) + 𝐺 ≥4(𝑧)) + 𝑒3𝑧3.

Let 𝜑ℓ denote the right-hand side of (4.17). We show that 𝜑ℓ and 𝑓ℓ satisfy the same recurrence

relation. It is straightforward to see that for ℓ ≥ 2,

𝑓ℓ+1 − 𝑓ℓ = 𝑋𝑃ℓ 𝑒2𝑧ℓ+2 − 𝑋𝑃ℓ+2

𝑧ℓ+2 − 𝑋𝑃ℓ −1

𝑒2𝑧ℓ+1 + 𝑋𝑃ℓ −1

𝑧ℓ−1.

(4.18)

Next we look at 𝜑ℓ+1 − 𝜑ℓ. Observe from (4.4) that 𝐺 ≥𝑚+1 − 𝐺 ≥𝑚 = −(𝑚 − 1)𝑒𝑚𝑧𝑚. We therefore

have

𝜑ℓ+1 =

𝜑ℓ =

ℓ+2
∑︁

𝑖=3
ℓ+1
∑︁

𝑖=3

(𝑖 − 2)𝑒𝑖𝑧𝑖 + 2(𝐷 + 𝐺 ≥ℓ+3) +

(𝑖 − 2)𝑒𝑖𝑧𝑖 + 2(𝐷 + 𝐺 ≥ℓ+2) +

ℓ−1
∑︁

𝑖=1
ℓ−2
∑︁

𝑖=1

(𝐷 + 𝐺ℓ+3−𝑖) 𝑋𝑃𝑖 𝑧𝑖

(𝐷 + 𝐺ℓ+2−𝑖) 𝑋𝑃𝑖 𝑧𝑖

and hence we obtain, for ℓ ≥ 2,

𝜑ℓ+1 − 𝜑ℓ = ℓ𝑒ℓ+2𝑧ℓ+2 − 2(ℓ + 1)𝑒ℓ+2𝑧ℓ+2 −

ℓ−2
∑︁

𝑖=1

(ℓ + 1 − 𝑖)𝑒ℓ+2−𝑖𝑧ℓ+2−𝑖 𝑋𝑃𝑖 𝑧𝑖 + (𝐷 + 𝐺 ≥4) 𝑋𝑃ℓ −1

𝑧ℓ−1.

The path recurrence relation in Proposition 4.2.21 tells us that

𝑋𝑃ℓ+2 = (ℓ + 2)𝑒ℓ+2 +

ℓ−2
∑︁

𝑗=1

(ℓ + 1 − 𝑗)𝑒ℓ+2− 𝑗 𝑋𝑃 𝑗 + 2𝑒3𝑋𝑃ℓ −1 + 𝑒2𝑋𝑃ℓ .

Together with 𝐷 + 𝐺 ≥4 = 1 − 𝐺 ≤3 = 1 − 𝑒2𝑧2 − 2𝑒3𝑧3, this gives

𝜑ℓ+1 − 𝜑ℓ = (−𝑋𝑃ℓ+2 + 2𝑒3𝑋𝑃ℓ −1 + 𝑒2𝑋𝑃ℓ )𝑧ℓ+2 + (1 − 𝑒2𝑧2 − 2𝑒3𝑧3) 𝑋𝑃ℓ −1

𝑧ℓ−1.

102

The terms containing 𝑒3𝑋𝑃ℓ −1 cancel, and the remaining expression coincides with the one for
𝑓ℓ+1 − 𝑓ℓ in (4.18). Hence 𝜑ℓ and 𝑓ℓ satisfy the same recurrence relation. Since their initial values

also coincide, the claim follows.

□

The preceding efforts culminate in the following 𝑒-positivity result, as announced at the start of

this section.

Theorem 4.2.24. For ℓ ≥ 2, we have the 𝑒-positive expansion
ℓ−2
∑︁

ℓ+1
∑︁

(𝑖 − 2)𝑒𝑖𝑧𝑖 + 2(𝐸 + X𝑃𝐺 ≥ℓ+2) +

X𝑃 𝑓ℓ = X𝑃

(𝐸 + X𝑃𝐺 ≥ℓ+2−𝑖) 𝑋𝑃𝑖 𝑧𝑖.

(4.19)

Hence the generating function (cid:205)𝑛≥ℓ+1

𝑖=3

𝑖=1
𝑋𝑃𝑛,ℓ 𝑧𝑛+1 is 𝑒-positive.

Proof. The expression for 𝑓ℓ in Lemma 4.2.23 immediately allows us to conclude (4.19), using the
fact that X𝑃 (𝑧)𝐷 (𝑧) = 𝐸 (𝑧). It then suffices to observe that the generating function (cid:205)𝑛≥ℓ+1
is comprised precisely of all the terms of degree ≥ ℓ + 2 in the 𝑒-positive rational expression

𝑋𝑃𝑛,ℓ 𝑧𝑛+1

2X𝑃 𝑓ℓ =

=

2𝐸 𝑓ℓ
𝐷

2𝐸

(cid:32) ℓ+1
∑︁

𝑖=3

(𝑖 − 2)𝑒𝑖𝑧𝑖 + 𝐺 ≥ℓ+2 +

ℓ−2
∑︁

𝐺 ≥ℓ+2−𝑖 𝑋𝑃𝑖 𝑧𝑖

(cid:33)

𝑖=0
1 − (cid:205)𝑖≥2(𝑖 − 1)𝑒𝑖𝑧𝑖

+ 2(1 + 𝐸)

ℓ−2
∑︁

𝑖=0

𝑋𝑃𝑖 𝑧𝑖.

□

From Theorem 4.2.24 and a tedious computation of X𝑃 𝑓ℓ + 𝑔ℓ, we obtain the following
(cid:205)𝑛≥ℓ+1

cancellation-free 𝑒-positive expression for 1
2

(Note that the sum is zero if the

𝑋𝑃𝑛,ℓ 𝑧𝑛+1.

range of summation is empty.)

Proposition 4.2.25. For integers 𝑛 ≥ 3 and 2 ≤ ℓ ≤ 𝑛 − 1, the twin 𝑃𝑛,ℓ of the path 𝑃𝑛 at the degree

2 vertex ℓ is 𝑒-positive. In particular, we have

1
2

∑︁

𝑛≥ℓ+1

𝑋𝑃𝑛,ℓ 𝑧𝑛+1 = ℓ𝑒ℓ+1𝑧ℓ+1

(cid:32) ℓ−2
∑︁

𝑖=1

𝑋𝑃𝑖 𝑧𝑖

+ 𝐸≥ℓ+2

ℓ−2
∑︁

𝑖=0

𝑋𝑃𝑖 𝑧𝑖 +

(cid:33)

(cid:32)

+

ℓ
∑︁

𝑖=3

∑︁

𝑖≥ℓ−1

𝑖−4
∑︁

(𝑖 − 1)𝑒𝑖𝑧𝑖 (cid:169)
(cid:173)
(cid:171)
(cid:33) (cid:32) ℓ+1
∑︁
(𝑖 − 2)𝑒𝑖𝑧𝑖

𝑋𝑃𝑖 𝑧𝑖

𝑗=0

𝑋𝑃ℓ −2− 𝑗 𝑧ℓ−2− 𝑗 (cid:170)
(cid:174)
(cid:172)

(cid:33)

𝑖=2

+ 𝐸≥ℓ+2

+ 2X𝑃𝐺 ≥ℓ+2 + X𝑃

ℓ−2
∑︁

𝑖=1

𝐺 ≥ℓ+2−𝑖 𝑋𝑃𝑖 𝑧𝑖.

103

4.2.3 Generating Function for Twinned Cycles

In this section, we establish a new result, the 𝑒-positivity of the chromatic symmetric function

of the twinned cycle. Again, our goal of obtaining an 𝑒-positive generating function for the twinned

cycle will begin with a formula for the chromatic symmetric function of the twinned cycle, which

is derived using the triple deletion formula.

We start by introducing two more families of graphs. Consider the twinned cycle graph 𝐶𝑛,𝑣

where 𝑣′ is the twinned vertex of 𝑣 and 𝑢 and 𝑤 are the adjacent vertices to 𝑣 and 𝑣′. Let 𝐷𝑛+1 be

the graph obtained from 𝐶𝑛,𝑣 by removing the edge 𝑢𝑣 and let Tad𝑛+1 be the graph obtained from

𝐶𝑛,𝑣 by removing the edges 𝑢𝑣 and 𝑣𝑣′. We illustrate these two definitions in Figure 4.7.

𝑢

•

•

•

𝑣
•

•
𝑣′

· · ·

(a) 𝐶𝑛,𝑣

𝑤

•

•

•

𝑣
•

•
𝑣′

𝑢

•

•

𝑤

•

•

•

•

· · ·
(b) 𝐷 𝑛+1
Figure 4.7 𝐶𝑛,𝑣, 𝐷𝑛+1, and Tad𝑛+1

𝑢

•

•

•

𝑣
•

•
𝑣′

· · ·

(c) Tad𝑛+1

𝑤

•

•

•

Lemma 4.2.26. For 𝑛 ≥ 3:

𝑋𝐶𝑛,𝑣 = 4𝑋𝐶𝑛+1 + 2𝑒1𝑋𝐶𝑛 − 6𝑋𝑃𝑛+1 + 2𝑒2𝑋𝑃𝑛−1

.

Proof. Consider 𝑛 ≥ 3. By the triple deletion argument applied to 𝜖1 = 𝑢𝑣 and 𝜖2 = 𝑢𝑣′, we get

that

𝑋𝐶𝑛,𝑣 = 2𝑋𝐷𝑛+1 − 𝑋𝑃𝑛,𝑣 = 2𝑋𝐷𝑛+1 − 2𝑋𝑃𝑛+1 + 𝑋𝑃2

𝑋𝑃𝑛−1

.

In 𝐷𝑛+1, applying the triple deletion argument to 𝜖1 = 𝑣𝑤 and 𝜖2 = 𝑣𝑣′ gives

𝑋𝐷 𝑛+1 = 2𝑋Tad𝑛+1 − 𝑒1𝑋𝐶𝑛,

while applying the triple deletion argument to 𝜖1 = 𝑣𝑤 and 𝜖2 = 𝑣′𝑤 gives

(4.20)

(4.21)

𝑋𝐷𝑛+1 = 𝑋Tad𝑛+1 + 𝑋𝐶𝑛+1 − 𝑋𝑃𝑛+1

.

104

Subtracting both expressions for 𝑋𝐷 𝑛+1 we obtain that

𝑋Tad𝑛+1 = 𝑋𝐶𝑛+1 + 𝑒1𝑋𝐶𝑛 − 𝑋𝑃𝑛+1

,

and therefore

𝑋𝐷𝑛+1 = 2𝑋𝐶𝑛+1 + 𝑒1𝑋𝐶𝑛 − 2𝑋𝑃𝑛+1

.

(4.22)

Finally, putting together (4.20) and (4.22), we have

𝑋𝐶𝑛,𝑣 = 4𝑋𝐶𝑛+1 + 2𝑒1𝑋𝐶𝑛 − 6𝑋𝑃𝑛+1 + 2𝑒2𝑋𝑃𝑛−1

,

as claimed.

□

Let X𝐶𝑣 be the generating function for the twinned cycle, that is, X𝐶𝑣 (𝑧) := (cid:205)𝑛≥3

𝑋𝐶𝑛,𝑣 𝑧𝑛+1.

By Lemma 4.2.26 we have the following expression for X𝐶𝑣 .

Corollary 4.2.27. The generating function of the twinned cycle can be written as

X𝐶𝑣 (𝑧) = 2(2 + 𝑒1𝑧)X𝐶 − 2(3 − 𝑒2𝑧2)X𝑃 + 6(1 + 𝑒1𝑧) + 2𝑒2𝑧2 − 6𝑒3𝑧3.

Proof. The generating function follows by multiplying the formula in Lemma 4.2.26 by 𝑧𝑛+1 and

summing over all 𝑛 ≥ 3. In particular, taking into account the initial terms that do not appear and

using the initial values 𝑋𝐶1 = 0, 𝑋𝑃1 = 𝑒1, 𝑋𝐶2 = 2𝑒2 = 𝑋𝑃2

, 𝑋𝐶3 = 6𝑒3, and 𝑋𝑃3 = 𝑒2𝑒1 + 3𝑒3, we

get the following expressions in terms of the generating functions for the cycle and the path:

1. 4 (cid:205)𝑛≥3

𝑋𝐶𝑛+1

𝑧𝑛+1 = 4(X𝐶 − 𝑧2𝑋𝐶2 − 𝑧3𝑋𝐶3),

2. 2(𝑒1𝑧) (cid:205)𝑛≥3

𝑋𝐶𝑛 𝑧𝑛 = 2𝑒1𝑧(X𝐶 − 𝑧2𝑋𝐶2),

3. 6 (cid:205)𝑛≥3

𝑋𝑃𝑛+1

𝑧𝑛+1 = 6(X𝑃 − 1 − 𝑧𝑋𝑃1 − 𝑧2𝑋𝑃2 − 𝑧3𝑋𝑃3), and

4. 2𝑒2𝑧2 (cid:205)𝑛≥3

𝑋𝑃𝑛−1

𝑧𝑛−1 = 2𝑒2𝑧2(X𝑃 − 1 − 𝑧𝑋𝑃1).

Putting all this together gives the generating function as stated.

□

105

Consider the following 𝑒-positive generating functions that appear in the proof of Lemma 4.2.3:

𝐹2 =

∑︁

𝑖≥3

(2𝑖2 − 5𝑖)𝑒𝑖𝑧𝑖

and

𝐹3 =

[(𝑖 − 1) (𝑖 − 3)]𝑒𝑖𝑧𝑖.

∑︁

𝑖≥4

Our goal is to show that the expression given in Corollary 4.2.27 is indeed 𝑒-positive.

Lemma 4.2.28. The twinned cycle generating function, scaled by 1

2, can be written as

1
2

X𝐶𝑣 =

1
𝐷 (𝑧)

[𝐹2 + 𝑒1𝑧𝐹3 + 𝑒2𝑧2(𝐸 − 1 − 𝑒1𝑧)] −

𝑒2𝑧2
𝐷 (𝑧)

+ 𝑒2𝑧2 − 3𝑒3𝑧3.

Proof. By Corollary 4.2.27,

1
2

X𝐶𝑣 = (2 + 𝑒1𝑧)X𝐶 − 3(X𝑃 − 1 − 𝑒1𝑧) + 𝑒2𝑧2(X𝑃 + 1) − 3𝑒3𝑧3.

(4.23)

From the proof of Lemma 4.2.3, we have the following

X𝑃 − (1 + 𝑒1𝑧) =

𝑧(1 + 𝑒1𝑧)𝐸′(𝑧) − 𝑒1𝑧𝐸 (𝑧)
𝐷 (𝑧)

and X𝐶 =

𝑧2𝐸′′(𝑧)
𝐷 (𝑧)

,

and by definition X𝑃 (𝑧) =

𝐸 (𝑧)
𝐷 (𝑧) and X𝐶 (𝑧) =

𝑧2𝐸 ′′ (𝑧)
𝐷 (𝑧)

. Substituting these into (4.23), we get

1
2

X𝐶𝑣 =

=

=

1
𝐷 (𝑧)
1
𝐷 (𝑧)
1
𝐷 (𝑧)

[(2 + 𝑒1𝑧)𝑧2𝐸′′(𝑧) − 3(𝑧𝐸′(𝑧) + 𝑧2𝑒1𝐸′(𝑧) − 𝑒1𝑧𝐸 (𝑧)) + 𝑒2𝑧2𝐸 (𝑧)] + 𝑒2𝑧2 − 3𝑒3𝑧3

[(2𝑧2𝐸′′ − 3𝑧𝐸′) + 𝑒1𝑧(𝑧2𝐸′′ − 3𝑒1𝑧𝐸′ + 3𝐸) + 𝑒2𝑧2𝐸 (𝑧)] + 𝑒2𝑧2 − 3𝑒3𝑧3

[𝐹2 − 3𝑒1𝑧 − 2𝑒2𝑧2 + 𝑒1𝑧(𝐹3 − 𝑒2𝑧2 + 3) + 𝑒2𝑧2𝐸 (𝑧)] + 𝑒2𝑧2 − 3𝑒3𝑧3,

where the final equality comes from Lemma 4.2.3. The statement then follows after further algebraic

manipulations.

□

Theorem 4.2.29. The generating function for 1
2

𝑋𝐶𝑛,𝑣 has the following 𝑒-positive rational expres-

sion:

1
2

∑︁

𝑛≥3

𝑋𝐶𝑛,𝑣 𝑧𝑛+1 =

∑︁

𝑖≥4

(2𝑖2 − 5𝑖)𝑒𝑖𝑧𝑖 +

Proof. Let 𝐸≥2 := 𝐸 − 1 − 𝑒1𝑧 = (cid:205)𝑖≥2

𝑒1𝑧𝐹3 + 𝐹2

(cid:205)𝑖≥3(𝑖 − 1)𝑒𝑖𝑧𝑖 + 𝑒2𝑧2 (cid:205)𝑖≥3(2𝑖2 − 6𝑖 + 2)𝑒𝑖𝑧𝑖

1 − (cid:205)𝑖≥2(𝑖 − 1)𝑒𝑖𝑧𝑖
𝑒𝑖𝑧𝑖. Then by Lemma 4.2.28, we have

.

1
2

X𝐶𝑣 =

(cid:16)

1
𝐷 (𝑧)

𝐹2 + 𝑒1𝑧𝐹3 + 𝑒2𝑧2𝐸≥2

(cid:17)

−

= 𝐹2 − 3𝑒3𝑧3 +

𝑒1𝑧𝐹3
𝐷 (𝑧)

(cid:34)

+

𝐹2

(cid:18) 1
𝐷 (𝑧)

𝑒2𝑧2
𝐷 (𝑧)
(cid:19)

− 1

+ 𝑒2𝑧2 − 3𝑒3𝑧3

+

𝑒2𝑧2𝐸≥2
𝐷 (𝑧)

−

𝑒2𝑧2
𝐷 (𝑧)

(cid:35)

+ 𝑒2𝑧2

.

(4.24)

106

The first two terms in (4.24) can be written as:

𝐹2 − 3𝑒3𝑧3 =

(2𝑖2 − 5𝑖)𝑒𝑖𝑧𝑖.

∑︁

𝑖≥4

(4.25)

For the generating function in the brackets of (4.24), we have

𝐹2

(cid:18) 1
𝐷 (𝑧)
∑︁

(cid:19)

+

− 1

𝑒2𝑧2𝐸≥2
𝐷 (𝑧)
𝐺 𝑘 + 𝑒2𝑧2𝐸≥2

= 𝐹2

𝑘 ≥1
∑︁

= 𝐹2

𝐺 𝑘 + 𝑒2𝑧2𝐸≥2

−

+ 𝑒2𝑧2

𝑒2𝑧2
𝐷 (𝑧)
𝐺 𝑘 − 𝑒2𝑧2 ∑︁
𝑘 ≥1
𝐺 𝑘−1 − 𝑒2𝑧2 ∑︁
𝑘 ≥1

∑︁

𝑘 ≥0
∑︁

𝑘 ≥1

𝐺 𝑘

𝐺 𝑘

𝐺𝐹2 − 𝑒2𝑧2(𝐺 − 𝐸≥2)

(cid:17)

𝑘 ≥1
𝐺 𝑘−1 (cid:16)

∑︁

𝑘 ≥1

∑︁

𝑘 ≥1

∑︁

(cid:32)

(cid:32)

𝐺 𝑘−1

𝐺 𝑘−1

=

=

=

=

𝐺𝐹2 − 𝑒2𝑧2𝐹2 + 𝑒2𝑧2𝐹2 − 𝑒2𝑧2 ∑︁
𝑖≥3

(cid:33)

(𝑖 − 2)𝑒𝑖𝑧𝑖

(cid:33)

𝐹2𝐺 ≥3 + 𝑒2𝑧2 ∑︁
𝑖≥3

𝑘 ≥1
𝐹2𝐺 ≥3 + 𝑒2𝑧2 (cid:205)𝑖≥3(2𝑖2 − 6𝑖 + 2)𝑒𝑖𝑧𝑖
𝐷 (𝑧)

(2𝑖2 − 6𝑖 + 2)𝑒𝑖𝑧𝑖

,

(4.26)

where the penultimate equality follows from the definitions of 𝐺 ≥3 and 𝐹2. Combining (4.24), (4.25),

and (4.26), we obtain the desired expression.

□

From the generating function in Theorem 4.2.29, we can readily extract the 𝑒-coefficients of

𝑋𝐶𝑛,𝑣 .

Corollary 4.2.30. Let 𝜆 be a partition of 𝑘 ≥ 3, 𝜆 = ⟨1𝑚1, 2𝑚2, . . . , 𝑘 𝑚𝑘 ⟩, and let 𝑐𝜆 be the

coefficient of 𝑒𝜆𝑧|𝜆| in 1

2X𝐶𝑣 . We have the following list of expressions for the coefficients:

1. 𝑐(𝑘) = 𝑘 (2𝑘 − 5).

2. If 𝑚1 > 1, then 𝑐𝜆 = 0.

3. If 𝑚1 = 1 and 𝜆 = 𝜇 ∪ 1 (so that 𝑚1(𝜇) = 0), then

∑︁

𝑐𝜆 =

(𝑖 − 1) (𝑖 − 3)𝜀(𝜇 − 𝑖).

𝑖≥4
𝑖∈supp(𝜇)

107

Note that this is 0 unless 𝜆 has a part of size at least 4.

4. If 𝑚1 = 𝑚2 = 0, then

∑︁

𝑐𝜆 =

𝑎∈supp(𝜆)

(2𝑎2 − 5𝑎)𝜀(𝜆 − 𝑎).

5. If 𝑚1 = 0 and 𝑚2 = ℓ(𝜆), then 𝑐𝜆 = 0.

6. If 𝑚1 = 0 and 1 ≤ 𝑚2 < ℓ(𝜆), then

𝑐𝜆 =

∑︁

𝑎,𝑏≥3
(𝑎,𝑏)∈supp(𝜆)

𝜀(𝜆 − 𝑎 − 𝑏) (2𝑎2 − 5𝑎) (𝑏 − 1) +

𝜀(𝜆 − 𝑐 − 2) (2𝑐2 − 6𝑐 + 2)

∑︁

𝑐≥3
𝑐∈supp(𝜆)

where (𝑎, 𝑏) ∈ supp(𝜆) means both 𝑎 and 𝑏 are in supp(𝜆) if 𝑎 ≠ 𝑏, and 𝑚𝑎 ≥ 2 if 𝑎 = 𝑏.

4.3

𝑒-positivity via Recurrences

In this section, we reprove several 𝑒-positivity results for certain classes of graphs by exhibiting

an 𝑒-positive recurrence relation. The recurrence relations for paths and cycles from Proposi-

tion 4.2.2 serve as the model for those of this section, and in fact will play a key role in our

derivations. We will also need explicit expressions for some coefficients, which are readily ex-

tracted from Proposition 4.2.2. These are recorded in the next result.

Corollary 4.3.1. Given a graph 𝐺, let [𝑒𝜆] 𝑋𝐺 denote the coefficient of 𝑒𝜆 in the chromatic symmetric

function of 𝐺, 𝑋𝐺.

• For 𝑛 ≥ 2, [𝑒𝑛] 𝑋𝑃𝑛 = 𝑛, [𝑒𝑛−1𝑒1] 𝑋𝑃𝑛 = 𝑛 − 2, and [𝑒𝑛] 𝑋𝐶𝑛 = 𝑛(𝑛 − 1).

• For 𝑛 ≥ 5, [𝑒𝑛−2𝑒2] 𝑋𝑃𝑛 = 3𝑛 − 8 and [𝑒𝑛−2𝑒2] 𝑋𝐶𝑛 = 𝑛(𝑛 − 3).

• For 𝑘 ≥ 2 and 𝑟 ≥ 1, [(𝑒𝑘 )𝑟] 𝑋𝑃𝑘𝑟 = 𝑘 (𝑘 − 1)𝑟−1 and [𝑒(𝑘𝑟 )] 𝑋𝐶𝑘𝑟 = 𝑘 (𝑘 − 1)𝑟.

• [𝑒2

2] 𝑋𝐶4 = 2.

The rest of this section follows the structure of Section 4.2.

108

4.3.1 Recurrences for Twinned Paths

In this section we derive recurrence formulas for the chromatic symmetric function for a path

twinned at one or both leaves or at an internal vertex.

4.3.1.1 Paths Twinned at a Leaf

In the next proposition, we give formulas for the chromatic symmetric function 𝑋𝑃𝑛,𝑣 of the path

twinned at a leaf, one in terms of the path chromatic symmetric function, and the other a recurrence

in the spirit of Proposition 4.2.2. The recurrence given below makes the 𝑒-positivity transparent.

Proposition 4.3.2. Let 𝑣 be a leaf of the path 𝑃𝑛. Then, for 𝑛 ≥ 4,

𝑋𝑃𝑛,𝑣 = 2(𝑛 + 1)𝑒𝑛+1 + 2

𝑛
∑︁

𝑗=3

( 𝑗 − 1)𝑒 𝑗 𝑋𝑃𝑛+1− 𝑗 .

Thus 𝑋𝑃𝑛,𝑣 is 𝑒-positive. Moreover, for 𝑛 ≥ 4, 𝑋𝑃𝑛,𝑣 satisfies the 𝑒-positive recurrence

𝑋𝑃𝑛,𝑣 =

𝑛−2
∑︁

𝑗=2

( 𝑗 − 1)𝑒 𝑗 𝑋𝑃𝑛− 𝑗,𝑣 + 2(𝑛 + 1)𝑒𝑛+1 + 2(𝑛 − 1)𝑒𝑛𝑒1 + 2(𝑛 − 3)𝑒𝑛−1𝑒2,

with initial values 𝑋𝑃1,𝑣 = 2𝑒2, 𝑋𝑃2,𝑣 = 6𝑒3, and 𝑋𝑃3,𝑣 = 8𝑒4 + 4𝑒3𝑒1.

Proof. The first expression follows from Proposition 4.2.21 and Proposition 4.2.8, noting that

𝑋𝑃𝑛,𝑣 = 2𝑋𝑃𝑛+1 − 𝑋𝑃2

𝑋𝑃𝑛−1 = 2(𝑛 + 1)𝑒𝑛+1 + 2

𝑛
∑︁

𝑗=2

( 𝑗 − 1)𝑒 𝑗 𝑋𝑃𝑛+1− 𝑗 − 2𝑒2𝑋𝑃𝑛−1

= 2(𝑛 + 1)𝑒𝑛+1 + 2

𝑛
∑︁

𝑗=3

( 𝑗 − 1)𝑒 𝑗 𝑋𝑃𝑛+1− 𝑗 + 2𝑒2𝑋𝑃𝑛−1 − 2𝑒2𝑋𝑃𝑛−1

.

For the second recurrence, we apply the triple deletion argument to 𝑃𝑛,𝑣 followed by Proposi-

109

tion 4.2.21 to both terms. Thus,

𝑋𝑃𝑛,𝑣 = 2𝑋𝑃𝑛+1 − 𝑋𝑃2

𝑋𝑃𝑛−1 = 2𝑋𝑃𝑛+1 − 2𝑒2𝑋𝑃𝑛−1

= 2(𝑛 + 1)𝑒𝑛+1 + 2

𝑛
∑︁

𝑗=2

( 𝑗 − 1)𝑒 𝑗 𝑋𝑃𝑛+1− 𝑗 − 2𝑒2

= 2(𝑛 + 1)𝑒𝑛+1 − 2(𝑛 − 1)𝑒𝑛−1𝑒2 + 2

𝑛−2
∑︁

𝑗=2

(𝑛 − 1)𝑒𝑛−1 +

𝑛−2
∑︁

𝑗=2








( 𝑗 − 1)𝑒 𝑗 𝑋𝑃𝑛−1− 𝑗








( 𝑗 − 1)𝑒 𝑗 (cid:2)2𝑋𝑃𝑛+1− 𝑗 − 2𝑒2𝑋𝑃𝑛−1− 𝑗

(cid:3)

+ 2(𝑛 − 2)𝑒𝑛−1𝑋𝑃2 + 2(𝑛 − 1)𝑒𝑛 𝑋𝑃1

= 2(𝑛 + 1)𝑒𝑛+1 + 2

𝑛−2
∑︁

𝑗=2

( 𝑗 − 1)𝑒 𝑗 𝑋𝑃𝑛− 𝑗,𝑣 + 2(𝑛 − 1)𝑒𝑛𝑒1 + 2(𝑛 − 3)𝑒𝑛−1𝑒2,

where the final step follows by the triple deletion argument applied to 𝑃𝑛− 𝑗,𝑣. As this is an 𝑒-positive

recursion with 𝑒-positive initial conditions, by induction it follows that 𝑋𝑃𝑛,𝑣 is 𝑒-positive for all

𝑛.

□

4.3.1.2 Paths Twinned at Both Leaves

Our new contribution is the 𝑒-positive recurrence below.

Proposition 4.3.3. For 𝑛 ≥ 6, the chromatic symmetric function 𝑋𝑃𝑛,𝑣,𝑤 for the path 𝑃𝑛 twinned at

both leaves 𝑣, 𝑤 satisfies the recurrence

1
4

𝑋𝑃𝑛,𝑣,𝑤 =

1
4

𝑛−3
∑︁

𝑗=3

( 𝑗 − 1)𝑒 𝑗 𝑋𝑃𝑛− 𝑗,𝑣,𝑤

+ (𝑛 + 2)𝑒𝑛+2 + 𝑛 𝑒𝑛+1𝑒1 + 3(𝑛 − 2)𝑒𝑛−1𝑒3 + 2(𝑛 − 3)𝑒𝑛−2𝑒3𝑒1 + 4(𝑛 − 3)𝑒𝑛−2𝑒4

+ 𝑒2

(cid:20) 1
4

𝑋𝑃𝑛−2,𝑣,𝑤 − 2𝑒𝑛 − (𝑛 − 4)𝑒𝑛−2𝑒2 − (𝑛 − 2)𝑒𝑛−1𝑒1

(cid:21)

,

with the initial conditions

𝑋𝑃2,𝑣,𝑤 = 24𝑒4,
𝑋𝑃3,𝑣,𝑤 = 4𝑒3𝑒2 + 12𝑒4𝑒1 + 20𝑒5,

𝑋𝑃4,𝑣,𝑤 = 24𝑒2
3 + 8𝑒4𝑒2 + 16𝑒5𝑒1 + 24𝑒6,
𝑋𝑃5,𝑣,𝑤 = 16𝑒3𝑒3𝑒1 + 68𝑒4𝑒3 + 12𝑒5𝑒2 + 20𝑒6𝑒1 + 28𝑒7.

Moreover, despite the negative terms, the expression is 𝑒-positive.

110

Proof. By the triple deletion argument, we have that

𝑋𝑃𝑛,𝑣,𝑤 = 2𝑋𝑃𝑛+1,𝑣 − 2𝑒2𝑋𝑃𝑛−1,𝑣 .

Applying the triple deletion argument again to both twinned terms, we have that 𝑋𝑃𝑛+1,𝑣 = 2𝑋𝑃𝑛+2 −
2𝑒2𝑋𝑃𝑛 and 𝑋𝑃𝑛−1,𝑣 = 2𝑋𝑃𝑛 − 2𝑒2𝑋𝑃𝑛−2. Thus, for 𝑛 ≥ 3,

1
4

𝑋𝑃𝑛,𝑣,𝑤 = 𝑋𝑃𝑛+2 + 𝑒2
2

𝑋𝑃𝑛−2 − 2𝑒2𝑋𝑃𝑛 .

(4.27)

Now we prove the recurrence relation by strong induction on 𝑛. The initial conditions are

checked directly. For 𝑛 ≥ 6, using repeatedly (4.27) and Proposition 4.2.21, we write

1
4

𝑋𝑃𝑛,𝑣,𝑤 = 𝑋𝑃𝑛+2 − 2𝑒2𝑋𝑃𝑛 + 𝑒2
2

𝑋𝑃𝑛−2

= (𝑛 + 2)𝑒𝑛+2 +

𝑛+1
∑︁

𝑗=2

( 𝑗 − 1)𝑒 𝑗 𝑋𝑃𝑛+2− 𝑗 + (𝑛 − 2)𝑒𝑛−2𝑒2

2 + 𝑒2
2

𝑛−3
∑︁

𝑗=2

( 𝑗 − 1)𝑒 𝑗 𝑋𝑃𝑛−2− 𝑗

− 2𝑛𝑒𝑛𝑒2 − 2𝑒2

𝑛−1
∑︁

𝑗=2

( 𝑗 − 1)𝑒 𝑗 𝑋𝑃𝑛− 𝑗

=

1
4

𝑛−3
∑︁

𝑗=2

( 𝑗 − 1)𝑒 𝑗 𝑋𝑃𝑛− 𝑗,𝑣,𝑤

+ (𝑛 + 2)𝑒𝑛+2 + 𝑛𝑒𝑛+1𝑒1 + 3(𝑛 − 2)𝑒𝑛−1𝑒3 + 2(𝑛 − 3)𝑒𝑛−2𝑒3𝑒1 + 4(𝑛 − 3)𝑒𝑛−2𝑒4

− 2𝑒𝑛𝑒2 − (𝑛 − 4)𝑒𝑛−2𝑒2

2 − (𝑛 − 2)𝑒𝑛−1𝑒2𝑒1.

To rearrange this into the announced form, we peel off the 𝑗 = 2 term from the sum and group it

with the negative terms:

1
4

𝑋𝑃𝑛,𝑣,𝑤 =

1
4

𝑛−3
∑︁

𝑗=3

( 𝑗 − 1)𝑒 𝑗 𝑋𝑃𝑛− 𝑗,𝑣,𝑤

+ (𝑛 + 2)𝑒𝑛+2 + 𝑛𝑒𝑛+1𝑒1 + 3(𝑛 − 2)𝑒𝑛−1𝑒3 + 2(𝑛 − 3)𝑒𝑛−2𝑒3𝑒1 + 4(𝑛 − 3)𝑒𝑛−2𝑒4

+ 𝑒2

(cid:20) 1
4

𝑋𝑃𝑛−2,𝑣,𝑤 − 2𝑒𝑛 − (𝑛 − 4)𝑒𝑛−2𝑒2 − (𝑛 − 2)𝑒𝑛−1𝑒1

(cid:21)

.

Next we prove that the term within brackets

𝑋𝑃𝑛−2,𝑣,𝑤 − 2𝑒𝑛 − (𝑛 − 4)𝑒𝑛−2𝑒2 − (𝑛 − 2)𝑒𝑛−1𝑒1

(cid:21)

(cid:20) 1
4

(4.28)

111

is 𝑒-positive. This follows by comparing the coefficients of the 𝑒-functions involved. Con-

sider (4.27) applied to 1
4

𝑋𝑃𝑛−2,𝑣,𝑤 , and use the coefficients described in Corollary 4.3.1. We obtain

the following formulas for the coefficients:

[𝑒𝑛] 1
4

𝑋𝑃𝑛−2,𝑣,𝑤 = 𝑛,

[𝑒𝑛−1𝑒1] 1
4

𝑋𝑃𝑛−2,𝑣,𝑤 = 𝑛 − 2,

and

[𝑒𝑛−2𝑒2] 1
4

𝑋𝑃𝑛−2,𝑣,𝑤 = 𝑛 − 4.

In particular, notice that the coefficients of 𝑒𝑛 and 𝑒𝑛−1𝑒1 are zero and that the coefficient of 𝑒𝑛𝑒2

is 𝑛 − 2, which is positive for 𝑛 ≥ 6. Thus, the negative terms appearing in (4.28) are absorbed by

terms in 1
4

𝑋𝑃𝑛−2,𝑣,𝑤 , and (4.28) is 𝑒-positive.

Finally, as this is an 𝑒-positive recurrence with 𝑒-positive initial conditions, by induction it

follows that 𝑋𝑃𝑛,𝑣,𝑤 is 𝑒-positive for all 𝑛.

□

4.3.1.3 Paths Twinned at an Interior Vertex

Next we provide an 𝑒-positive recurrence relation for path graphs twinned at an interior vertex.

Theorem 4.3.4. The chromatic symmetric function 𝑋𝑃𝑛,ℓ for the path 𝑃𝑛 twinned at the interior

vertex ℓ satisfies the 𝑒-positive recurrence for ℓ ≥ 2, 𝑛 ≥ ℓ + 1, and 𝑛 ≥ 4,

𝑋𝑃𝑛,ℓ =

𝑛−ℓ−1
∑︁

𝑗=2

+ 4

( 𝑗 − 1)𝑒 𝑗 𝑋𝑃𝑛− 𝑗,ℓ + 4(𝑛 + 1)𝑒𝑛+1 + 2𝑛𝑒1𝑒𝑛 + 2𝑒1

𝑛−1
∑︁

𝑗=𝑛−ℓ+2

( 𝑗 − 1)𝑒 𝑗 𝑋𝑃𝑛− 𝑗

𝑛
∑︁

( 𝑗 − 1)𝑒 𝑗 𝑋𝑃𝑛+1− 𝑗 + 2

𝑗=𝑛−ℓ+3

𝑛−ℓ+2
∑︁

𝑗=𝑛−ℓ+1

( 𝑗 − 2)𝑒 𝑗 𝑋𝑃𝑛+1− 𝑗 + (𝑛 − ℓ − 2)𝑒𝑛−ℓ 𝑋𝑃ℓ,ℓ .

Thus, for 𝑛 ≥ 3 and 2 ≤ ℓ ≤ 𝑛 − 1, 𝑋𝑃𝑛,ℓ is 𝑒-positive.

Proof. Fix ℓ ≥ 2, and consider 𝑛 ≥ ℓ + 1 with 𝑛 ≥ 4. We start with the recurrence relation in the

statement of Proposition 4.2.19. This can be rewritten as

𝑋𝑃𝑛,ℓ = 4𝑋𝑃𝑛+1 + 2𝑒1𝑋𝑃𝑛 + 2𝑒2𝑋𝑃ℓ −1

𝑋𝑃𝑛−ℓ

− 2𝑋𝑃ℓ+1

𝑋𝑃𝑛−ℓ − 2𝑋𝑃ℓ 𝑋𝑃𝑛−ℓ+1 − 2𝑋𝑃ℓ −1

𝑋𝑃𝑛−ℓ+2

.

(4.29)

112

From (4.29) and using Proposition 4.2.21, we have that

𝑋𝑃𝑛,ℓ = 4(𝑛 + 1)𝑒𝑛+1 + 4

𝑛
∑︁

𝑗=2

( 𝑗 − 1)𝑒 𝑗 𝑋𝑃𝑛+1− 𝑗 + 2𝑒1

𝑛 𝑒𝑛 +

𝑛−1
∑︁

𝑗=2

( 𝑗 − 1)𝑒 𝑗 𝑋𝑃𝑛− 𝑗















+ 2𝑒2𝑋𝑃ℓ −1








− 2𝑋𝑃ℓ+1

− 2𝑋𝑃ℓ








− 2𝑋𝑃ℓ −1

(𝑛 − ℓ)𝑒𝑛−ℓ +

𝑛−ℓ−1
∑︁

𝑗=2








( 𝑗 − 1)𝑒 𝑗 𝑋𝑃𝑛−ℓ − 𝑗

(𝑛 − ℓ)𝑒𝑛−ℓ +

𝑛−ℓ−1
∑︁

𝑗=2

( 𝑗 − 1)𝑒 𝑗 𝑋𝑃𝑛−ℓ − 𝑗






















(𝑛 + 1 − ℓ)𝑒𝑛+1−ℓ +

( 𝑗 − 1)𝑒 𝑗 𝑋𝑃𝑛+1−ℓ − 𝑗

𝑛−ℓ
∑︁

𝑗=2

(𝑛 + 2 − ℓ)𝑒𝑛+2−ℓ +

𝑛−ℓ+1
∑︁

𝑗=2








( 𝑗 − 1)𝑒 𝑗 𝑋𝑃𝑛+2−ℓ − 𝑗

(4.30)

(4.31)

.








Notice that, in the six summands above, for each fixed 𝑗, the terms attached to the factor ( 𝑗 − 1)𝑒 𝑗 ,

when collected together, match the six terms in the right-hand side of the recurrence (4.29) applied

to 𝑋𝑃𝑛− 𝑗,ℓ . Grouping the remaining terms into an expression 𝑌 if they have a positive sign, or 𝑍 if

they have a negative sign, we obtain

𝑋𝑃𝑛,ℓ =

𝑛−ℓ−1
∑︁

𝑗=2

( 𝑗 − 1)𝑒 𝑗 𝑋𝑃𝑛− 𝑗,ℓ + 𝑌 −𝑍.

The positive terms 𝑌 are given by

𝑌 = 4(𝑛 + 1)𝑒𝑛+1 + 2𝑛𝑒𝑛𝑒1 + 2𝑒2𝑋𝑃ℓ −1 (𝑛 − ℓ)𝑒𝑛−ℓ

+ 4

𝑛
∑︁

𝑗=𝑛−ℓ

( 𝑗 − 1)𝑒 𝑗 𝑋𝑃𝑛+1− 𝑗 + 2𝑒1

𝑛−1
∑︁

𝑗=𝑛−ℓ

( 𝑗 − 1)𝑒 𝑗 𝑋𝑃𝑛− 𝑗

= 4(𝑛 + 1)𝑒𝑛+1 + 2𝑛𝑒𝑛𝑒1 + 4

𝑛
∑︁

( 𝑗 − 1)𝑒 𝑗 𝑋𝑃𝑛+1− 𝑗 + 2𝑒1

𝑗=𝑛−ℓ+3

𝑛−1
∑︁

𝑗=𝑛−ℓ+2

( 𝑗 − 1)𝑒 𝑗 𝑋𝑃𝑛− 𝑗

𝑛−ℓ+2
∑︁

𝑛−ℓ+1
∑︁

+ 4

( 𝑗 − 1)𝑒 𝑗 𝑋𝑃𝑛+1− 𝑗 + 2𝑒1
𝑗=𝑛−ℓ
𝑗=𝑛−ℓ
(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)
(cid:124)

( 𝑗 − 1)𝑒 𝑗 𝑋𝑃𝑛− 𝑗 + 2𝑒2𝑋𝑃ℓ −1 (𝑛 − ℓ)𝑒𝑛−ℓ
(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)

,

(cid:125)

(cid:123)(cid:122)
𝑌1

113

where the last line is obtained by splitting the summations. The negative terms 𝑍 are given by

𝑍 = 2𝑋𝑃ℓ+1 (𝑛 − ℓ)𝑒𝑛−ℓ + 2𝑋𝑃ℓ (𝑛 + 1 − ℓ)𝑒𝑛+1−ℓ + 2𝑋𝑃ℓ −1 (𝑛 + 2 − ℓ)𝑒𝑛+2−ℓ

+ 2𝑋𝑃ℓ (𝑛 − ℓ − 1)𝑒𝑛−ℓ 𝑋𝑃1 + 2𝑋𝑃ℓ −1

𝑛−ℓ+1
∑︁

𝑗=𝑛−ℓ

( 𝑗 − 1)𝑒 𝑗 𝑋𝑃𝑛+2−ℓ − 𝑗 ,

where the last two terms come from the sums in (4.30) and (4.31). We rewrite 𝑍 so that 𝑌 − 𝑍 is

easier to analyze.

𝑍 = 2

𝑛−ℓ+2
∑︁

𝑗=𝑛−ℓ

𝑗 𝑒 𝑗 𝑋𝑃𝑛+1− 𝑗

+ 2𝑋𝑃ℓ (𝑛−ℓ−1)𝑒𝑛−ℓ 𝑋𝑃1 + 2𝑋𝑃ℓ −1 (𝑛 − ℓ)𝑒𝑛−ℓ+1𝑋𝑃1 + 2𝑋𝑃ℓ −1 (𝑛−ℓ−1)𝑒𝑛−ℓ 𝑋𝑃2
𝑛−ℓ+2
∑︁

𝑛−ℓ+1
∑︁

( 𝑗 − 1)𝑒 𝑗 𝑋𝑃𝑛− 𝑗 + 2𝑋𝑃2

𝑋𝑃ℓ −1 (𝑛−ℓ−1)𝑒𝑛−ℓ

𝑗 𝑒 𝑗 𝑋𝑃𝑛+1− 𝑗 + 2𝑒1

= 2

𝑗=𝑛−ℓ

𝑗=𝑛−ℓ

Using 𝑋𝑃2 = 2𝑒2, we have

𝑌1 − 𝑍 = 2

𝑛−ℓ+2
∑︁

𝑗=𝑛−ℓ

( 𝑗 − 2)𝑒 𝑗 𝑋𝑃𝑛+1− 𝑗 − 2𝑋𝑃ℓ −1 (𝑛 − ℓ − 2)𝑒𝑛−ℓ𝑒2,

because the terms with the factor 𝑒1 = 𝑋𝑃1 can be seen to vanish identically. By splitting the sum,
𝑌1 − 𝑍 can be rewritten as

𝑌1 − 𝑍 = 2

𝑛−ℓ+2
∑︁

𝑗=𝑛−ℓ+1

( 𝑗 − 2)𝑒 𝑗 𝑋𝑃𝑛+1− 𝑗 + 2(𝑛 − ℓ − 2)𝑒𝑛−ℓ (𝑋𝑃ℓ+1 − 𝑒2𝑋𝑃ℓ −1).

𝑋𝑃ℓ −1 = 𝑋𝑃ℓ,ℓ
In the last term on the right-hand side, the factor of 2(𝑋𝑃ℓ+1 − 𝑒2𝑋𝑃ℓ −1) = 2𝑋𝑃ℓ+1 − 𝑋𝑃2
is precisely the chromatic symmetric function of the path 𝑃ℓ twinned at a leaf. Thus, putting all of

this together, we obtain the recurrence relation from the statement.

Finally, we can deduce the 𝑒-positivity. For the initial values 𝑛 = ℓ + 1, ℓ + 2, ℓ + 3, 𝑋𝑃𝑛,ℓ

is 𝑒-positive by Proposition 4.2.25. We proceed by strong induction on 𝑛 to show the claimed

𝑒-positivity for 𝑋𝑃𝑛,ℓ for 𝑛 ≥ ℓ + 4. Our induction hypothesis is that 𝑋𝑃𝑚,ℓ is 𝑒-positive for all

𝑚 < 𝑛, 𝑚 ≥ ℓ + 1. We only need to look at 𝑌1 − 𝑍 since that is the part containing negative

terms. By Proposition 4.3.2 we know that 𝑌1 − 𝑍 is 𝑒-positive. Hence, by induction the proof is

complete.

□

114

As an application of the preceding recurrence, we have the following corollary.

Corollary 4.3.5 (See also [27, Theorem 7.8]). Consider the path 𝑃𝑛 on 𝑛 vertices, with 𝑛 ≥ 4. Let

2 ≤ ℓ ≤ 𝑛 − 2 and let 𝑣 be the leaf 𝑛. Then the chromatic symmetric function of 𝑃𝑛,ℓ,𝑣 is 𝑒-positive.

Proof. The triple deletion argument implies that

𝑋𝑃𝑛,ℓ,𝑣 = 2𝑋𝑃𝑛+1,ℓ − 𝑋𝑃2

𝑋𝑃𝑛−1,ℓ = 2(𝑋𝑃𝑛+1,ℓ − 𝑒2𝑋𝑃𝑛−1,ℓ ).

Examining the recurrence for 𝑋𝑃𝑛,ℓ in Theorem 4.3.4 one sees that, when 𝑛 − ℓ − 1 ≥ 2, the initial

term in the first sum in the expression for 𝑋𝑃𝑛,ℓ is 𝑒2𝑋𝑃𝑛−2,ℓ , making 𝑋𝑃𝑛,ℓ − 𝑒2𝑋𝑃𝑛−2,ℓ 𝑒-positive.
□
Replacing 𝑛 by 𝑛 + 1 now gives 𝑒-positivity of 𝑋𝑃𝑛+1,ℓ − 𝑒2𝑋𝑃𝑛−1,ℓ for 𝑛 − ℓ ≥ 2.

4.3.2 Recurrence for Twinned Cycles

In this section we derive an 𝑒-positive recursive formula for the twinned cycle, analogous to

those for the twinned path from the last section. We give a similar formula for another family of

graphs that we call moose graphs.

4.3.2.1 Twinned Cycles

We start with the cycle graph.

Theorem 4.3.6. The chromatic symmetric function 𝑋𝐶𝑛,𝑣 for the cycle 𝐶𝑛 twinned at a vertex 𝑣 is

𝑒-positive. For 𝑛 ≥ 5, it satisfies the 𝑒-positive recurrence

𝑋𝐶𝑛,𝑣 =

𝑛−2
∑︁

𝑘=3

(𝑘 − 1)𝑒𝑘 𝑋𝐶𝑛−𝑘,𝑣 + 2(𝑛 + 1) (2𝑛 − 3)𝑒𝑛+1 + 2(𝑛 − 1) (𝑛 − 3)𝑒𝑛𝑒1

+ 𝑒2

(cid:2)𝑋𝐶𝑛−2,𝑣 − 2(𝑛 − 3)𝑒𝑛−1

(cid:3) ,

with initial conditions

𝑋𝐶1,𝑣 = 2𝑒2,

𝑋𝐶2,𝑣 = 6𝑒3,

𝑋𝐶3,𝑣 = 24𝑒4,

and 𝑋𝐶4,𝑣 = 50𝑒5 + 6𝑒4𝑒1 + 4𝑒3𝑒2.

Proof. We proceed by induction on 𝑛. The initial cases 𝑛 ≤ 4 are verified by direct computation

using Theorem 4.2.1 and Lemma 4.2.26. Note that these initial terms are all 𝑒-positive.

115

Assume we have shown the claim to be true for 𝑋𝐶𝑚,𝑣 for all 𝑚 < 𝑛. Now we rewrite the

expression in Lemma 4.2.26 using Proposition 4.2.2 to obtain:

𝑋𝐶𝑛,𝑣 = 4(𝑛 + 1)𝑛𝑒𝑛+1 + 4

𝑛−1
∑︁

𝑘=2

(𝑘 − 1)𝑒𝑘 𝑋𝐶𝑛+1−𝑘

(cid:34)

+ 2𝑒1

𝑛(𝑛 − 1)𝑒𝑛 +

𝑛−2
∑︁

(𝑘 − 1)𝑒𝑘 𝑋𝐶𝑛−𝑘

(cid:34)

− 6

(𝑛 + 1)𝑒𝑛+1 +

𝑘=2
𝑛
∑︁

(𝑘 − 1)𝑒𝑘 𝑋𝑃𝑛+1−𝑘

𝑘=2

(cid:35)

(cid:35)

(cid:34)

+ 2𝑒2

(𝑛 − 1)𝑒𝑛−1 +

(cid:35)

(𝑘 − 1)𝑒𝑘 𝑋𝑃𝑛−1−𝑘

.

𝑛−2
∑︁

𝑘=2

(4.32)

Applying Lemma 4.2.26 again, we can collect the four summations above into one sum and

three additional terms as follows:

𝑛−2
∑︁

𝑘=2

(𝑘 − 1)𝑒𝑘 (cid:2)4𝑋𝐶𝑛+1−𝑘 + 2𝑒1𝑋𝐶𝑛−𝑘 − 6𝑋𝑃𝑛+1−𝑘 + 2𝑒2𝑋𝑃𝑛−1−𝑘

(cid:3)

+ 4(𝑛 − 2)𝑒𝑛−1𝑋𝐶2 − 6(𝑛 − 2)𝑒𝑛−1𝑋𝑃2 − 6(𝑛 − 1)𝑒𝑛 𝑋𝑃1
𝑛−2
∑︁

(𝑘 − 1)𝑒𝑘 𝑋𝐶𝑛−𝑘,𝑣 + 4(𝑛 − 2)𝑒𝑛−1𝑋𝐶2 − 6(𝑛 − 2)𝑒𝑛−1𝑋𝑃2 − 6(𝑛 − 1)𝑒𝑛 𝑋𝑃1

.

=

𝑘=2

Combining this expression with the remaining terms from (4.32), we obtain

𝑋𝐶𝑛,𝑣 =

𝑛−2
∑︁

𝑘=2

(𝑘 − 1)𝑒𝑘 𝑋𝐶𝑛−𝑘,𝑣 + 2(𝑛 + 1) (2𝑛 − 3)𝑒𝑛+1 + 2(𝑛 − 1) (𝑛 − 3)𝑒𝑛𝑒1

− 2(𝑛 − 3)𝑒𝑛−1𝑒2.

(4.33)

We isolate the term 𝑘 = 2 from the summation and regroup it with the last term in (4.33), so

that it becomes

𝑋𝐶𝑛,𝑣 =

𝑛−2
∑︁

𝑘=3

(𝑘 − 1)𝑒𝑘 𝑋𝐶𝑛−𝑘,𝑣 + 2(𝑛 + 1) (2𝑛 − 3)𝑒𝑛+1 + 2(𝑛 − 1) (𝑛 − 3)𝑒𝑛𝑒1

+ 𝑒2

(cid:2)𝑋𝐶𝑛−2,𝑣 − 2(𝑛 − 3)𝑒𝑛−1

(cid:3) ,

as stated in the theorem.

116

Now we want to show 𝑒-positivity. By the induction hypothesis, only the last term requires

scrutiny. Using Corollary 4.3.1 and Lemma 4.2.26, the coefficient of 𝑒𝑛−1 in 𝑋𝐶𝑛−2,𝑣 is 2(𝑛−1)(2𝑛−
7). Therefore, the coefficient of 𝑒𝑛−1 in 𝑋𝐶𝑛−2,𝑣 − 2(𝑛 − 3)𝑒𝑛−1 is 2(𝑛 − 1) (2𝑛 − 7) − 2(𝑛 − 3) =
4(𝑛2 − 5𝑛 + 5), which is nonnegative for 𝑛 ≥ 4. Thus, by the induction hypothesis, the formula in

the statement for 𝑋𝐶𝑛,𝑣 is indeed an 𝑒-positive recurrence for the twinned cycles for 𝑛 ≥ 5.

□

4.3.2.2 The Moose Graph

We define the moose graph 𝐴𝑛+2 to be the graph on 𝑛 + 2 vertices and 𝑛 + 1 edges, obtained

from the cycle graph 𝐶𝑛 by attaching a leaf to each of the vertices 𝑣, 𝑤 of an edge 𝑣𝑤 in 𝐶𝑛,

•

•

𝑣

•

𝑤

•

•

•

•

•

· · ·
Figure 4.8 The moose graph 𝐴𝑛+2

We provide an 𝑒-positive recurrence relation for the chromatic symmetric function of the moose

graph. This graph was shown to be 𝑒-positive as a special case in [63, Theorem 3.9]. We omit the

proof.

Proposition 4.3.7. For 𝑛 ≥ 2, the chromatic symmetric function of the moose graph 𝐴𝑛+2 is

𝑒-positive. For 𝑛 ≥ 4, it satisfies the 𝑒-positive recurrence

𝑛−2
∑︁

𝑋𝐴𝑛+2 = (cid:169)
(cid:173)
(cid:171)
+ (𝑛 + 2)(𝑛 − 1)𝑒𝑛+2 + 2𝑒1𝑒𝑛+1(𝑛2 − 𝑛 − 1) + (𝑛 − 1) (𝑛 − 2)𝑒2
1

( 𝑗 − 1)𝑒 𝑗 𝑋𝐴𝑛+2− 𝑗 (cid:170)
(cid:174)
(cid:172)

𝑗=2

𝑒𝑛 + 2𝑒2𝑒𝑛,

117

with initial values

𝑋𝐴4 = 𝑋𝑃4 = 2𝑒2

2 + 2𝑒3𝑒1 + 4𝑒4,

𝑋𝐴5 = 2𝑒3𝑒2

1 + 2𝑒3𝑒2 + 10𝑒4𝑒1 + 10𝑒5,

𝑋𝐴6 = 2𝑒3

2 + 2𝑒3𝑒2𝑒1 + 6𝑒4𝑒2

1 + 6𝑒4𝑒2 + 22𝑒5𝑒1 + 18𝑒6.

4.4 Future Directions

This work and the work of Tom [58] provide explicit 𝑒-positive generating functions for the

chromatic symmetric function of twinned paths and cycles, and suggests that it may be worthwhile

to undertake a similar study for twins of other graph families.

More generally, an examination of Table 4.1 shows that much of the recent literature focuses on

establishing Gebhard and Sagan’s (𝑒)-positivity of the chromatic symmetric function in noncom-

muting variables. Although (𝑒)-positivity implies 𝑒-positivity as a symmetric function in ordinary

commuting variables, in such cases an explicit 𝑒-positive generating function or recurrence would

be desirable. We propose the following future investigations in this direction:

1. For the twinned cycle graph, is the chromatic symmetric function in noncommuting variables

(𝑒)-positive?

2. Are there pleasing 𝑒-positive symmetric function expansions for those families whose 𝑒-

positivity is known only via the stronger (𝑒)-positivity property? Specific examples that may

admit nice generating functions are the triangular ladder [13, 48], the kayak paddle graphs

[2] and the tadpole graph [27, 33].

It would also be interesting to examine twinning for the chromatic quasisymmetric function of

Shareshian and Wachs [45, 46] since the main class of posets of study for these, whose incom-

parability graphs are unit interval graphs, is also closed under the appropriately defined twinning

operation for labeled graphs.

118

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