Combinatorial aspects of generalizations of Schur …...homogeneous symmetric polynomials are two bases for the ring of symmetric func-tions. The Hall inner product is de ned using

Combinatorial aspects of generalizations ofSchur functions

A Thesis

Submitted to the Faculty

of

Drexel University

by

Derek Heilman

in partial fulfillment of the

requirements for the degree

of

Doctor of Philosophy

March 2013

CONTENTS ii

Contents

Abstract iv

1 Introduction 1

2 General background 3

2.1 Symmetric functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

2.2 Schur functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

2.3 The Hall inner product . . . . . . . . . . . . . . . . . . . . . . . . . . 9

2.4 The Pieri rule for Schur functions . . . . . . . . . . . . . . . . . . . . 10

3 The Pieri rule for the dual Grothendieck polynomials 16

3.1 Grothendieck polynomials . . . . . . . . . . . . . . . . . . . . . . . . 16

3.2 Dual Grothendieck polynomials . . . . . . . . . . . . . . . . . . . . . 17

3.3 Elegant fillings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

3.4 Pieri rule for the dual Grothendieck polynomials . . . . . . . . . . . . 21

4 Insertion proof of the dual Grothendieck Pieri rule 24

4.1 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

4.2 Insertion algorithm . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

4.3 Sign changing involution on reverse plane partitions and XO-diagrams 32

4.4 Combinatorial proof of the dual Grothendieck Pieri rule . . . . . . . . 41

5 Factorial Schur functions and their expansion 42

5.1 Definition of a factorial Schur polynomial . . . . . . . . . . . . . . . . 43

5.2 The expansion of factorial Schur functions in terms of Schur functions 47

6 A reverse change of basis 52

6.1 Change of basis coefficients . . . . . . . . . . . . . . . . . . . . . . . . 52

CONTENTS iii

6.2 A combinatorial involution . . . . . . . . . . . . . . . . . . . . . . . . 55

6.3 Reverse change of basis . . . . . . . . . . . . . . . . . . . . . . . . . . 57

References 59

ABSTRACT iv

AbstractCombinatorial aspects of generalizations of Schur functions

Derek HeilmanJennifer Morse, Ph.D

The understanding of the space of symmetric functions is gained through the study

of its bases. Certain bases can be defined by purely combinatorial methods, some-

times enabling important properties of the functions to fall from carefully constructed

combinatorial algorithms. A classic example is given by the Schur basis, made up of

functions that can be defined using semi-standard Young tableaux. The Pieri rule for

multiplying an important special case of Schur functions is proven using an insertion

algorithm on tableaux that was defined by Robinson, Schensted, and Knuth. Further-

more, the transition matrices between Schur functions and other symmetric function

bases are often linked to representation theoretic multiplicities. The description of

these matrices can sometimes be given combinatorially as the enumeration of a set of

objects such as tableaux.

A similar combinatorial approach is applied here to a basis for the symmetric

function space that is dual to the Grothendieck polynomial basis. These polynomials

are defined combinatorially using reverse plane partitions. Bijecting reverse plane

partitions with a subset of semi-standard Young tableaux over a doubly-sized alphabet

enables the extension of RSK-insertion to reverse plane partitions. This insertion,

paired with a sign changing involution, is used to give the desired combinatorial

proof of the Pieri rule for this basis. Another basis of symmetric functions is given by

the set of factorial Schur functions. While their expansion into Schur functions can

ABSTRACT v

be described combinatorially, the reverse change of basis had no such formulation. A

new set of combinatorial objects is introduced to describe the expansion coefficients,

and another sign changing involution is used to prove that these do in fact encode

the transition matrices.

1 INTRODUCTION 1

1 Introduction

Symmetric functions play a large role in many mathematical fields including group

theory, Lie algebras, and algebraic geometry. There are many different bases for the

ring of symmetric functions, one of the most fundamental is the Schur functions. Schur

functions are indexed by partitions and are directly connected to other mathematical

fields including geometry and representation theory. The most usual definition of a

Schur function is the combinatorial one (Section 2). This definition uses combinatorial

objects allowing many instrumental mathematical proofs. These combinatorial proofs

can be seen visually acting on these objects. One example is the RSK insertion proof,

which proves the Pieri rule for Schur functions. Another set of functions that form

a basis for the ring of symmetric functions are the Grothendieck polynomials, they

have many similar properties to those of Schur functions [6; 8; 15; 18].

Lascoux and Schutzenberger introduced Grothendieck polynomials [14]. These

are inhomogeneous polynomials representing classes of structure sheaves of Schubert

varieties in the Grothendieck ring of the flag varieties. Fomin and Kirillov continued

the study of these polynomials giving a combinatorial construction of Grothendieck

polynomials in terms of rc-graphs [5]. Buch continued work on these polynomials and

developed the Littlewood-Richardson rule for them [3]. Similar to the Littlewood-

Richardson rule for Schur functions, this rule defines the coefficients for the expansion

of the product of two Grothendieck polynomials in terms of Grothendieck polynomi-

als. The set of polynomials that are dual to the Grothendieck polynomials is known

as the dual Grothendieck polynomials. The Littlewood-Richardson coefficients for

these dual polynomials are also derived in Buch’s paper by means of the coproduct

[3].

The dual Grothendieck polynomials were first studied directly and called dual

stable Grothendieck polynomials [11]. They can be defined combinatorially using

1 INTRODUCTION 2

objects known as reverse plane partitions. Recall that the Pieri rule for Schur func-

tions has a very elegant proof using RSK insertion. However, there was no such

elegant proof for the dual Grothendieck polynomials due to the structure of the re-

verse plane partitions. There does exists a bijection between reverse plane partitions

and pairs of semi-standard Young tableaux and elegant fillings. Lam and Pylyavskyy

defined one bijection and Bandlow and Morse defined another slightly more intuitive

approach [11; 1]. These topics will be covered in Section 3. Using these pairs of

semi-standard Young tableaux and elegant fillings, a similar insertion method will be

used to construct an insertion based proof for the Pieri rule for the dual Grothendieck

polynomials (Section 4). The proof will also require a sign changing involution to be

defined to account for the remaining terms (section 4).

Schur functions, when limited to n variables, can be generalized to a different

type of functions. One set of these functions are the factorial Schur functions. They

are a generalization of Schur functions introducing another set of variables a. When

Biedenharn and Louck first discovered them, they fixed the values of the variables a to

ai = 1− i [2]. This was done to decompose tensor products of representations when

using particular bases. Factorial Schur functions are special cases of double Schu-

bert polynomials for Grassmannian permutations [14; 13]. Knutson and Tao also

showed that factorial Schur functions are the equivariant cohomology of Grassmanni-

ans [9]. Chen and Louck gave new foundations based on divided difference operators

[4]. Goulden and Hamel further developed the analogy between Schur functions and

factorial Schur functions [7]. Molev and Sagan found a Littlewood-Richardson rule

for factorial Schur functions as well as other useful results [17]. Kreiman later dis-

covered more interesting facts for factorial Schur functions [10]. One of these facts

is the the change of basis formula for expanding factorial Schur functions in terms

of Schur functions, as well as Schur functions in terms of factorial Schur functions.

Molev also gave an easy combinatorial method for computing the coefficients of the

2 GENERAL BACKGROUND 3

expansion of factorial Schur functions in terms of Schur functions, which is covered in

Section 5 [16]. Although this method can be used to compute the reverse expansion,

it requires more mechanics. Section 6 will give a simple combinatorial method of

describing these reverse change of basis coefficients and an elegant proof.

2 General background

This section will describe symmetric functions, bases for the ring of symmetric func-

tions, and tools for Schur functions from algebraic combinatorics. The theory of sym-

metric functions applies to enumerative combinatorics. These applications branch out

to many other fields of mathematics including group theory, Lie algebras, algebraic

geometry, and representation theory. Partitions and Ferrers diagrams will play an

important role in indexing and describing bases for the ring of symmetric functions.

Fillings of Ferrers diagrams and particularly semi-standard Young tableaux are the

primary combinatorial objects used to both define Schur functions and appear as a

tool in various combinatorial proofs. Monomial symmetric functions and the complete

homogeneous symmetric polynomials are two bases for the ring of symmetric func-

tions. The Hall inner product is defined using the the monomial symmetric functions

and the complete homogeneous symmetric polynomials. This Hall inner product is

used to define duality between two bases for the ring of symmetric functions. The

Schur functions are the only set of functions that is dual to itself. Expanding products

of Schur functions in terms of Schur functions was a classical problem in the field of

algebraic combinatorics. A simplified version of this problem is the Pieri rule for the

Schur functions. The classical proof of the Pieri rule for Schur functions was done

using the RSK algorithm [6; 8; 15; 18].


2.1 Symmetric functions

For an m-tuple of non-negative integers, γ = (γ1, γ2, · · · , γm), and n independent

variables, (x1, x2, . . . , xn), define

xγ = x(γ1,γ2,...,γm) = xγ11 xγ22 . . . xγmm .

For an infinite sequence of non-negative integers ,γ = (γ1, γ2, . . .), and infinite inde-

pendent variables, (x1, x2, . . .), define

xγ = x(γ1,γ2,...) = xγ11 xγ22 · · · .

Example. x(2,1,3,0,4) = x21x12x

33x

04x

45 = x21x2x

33x

45.

For a set of n independent variables x = (x1, x2, . . . , xn), given f(x) =∑cαx

α, where

α ranges over n-tuples of non-negative integers and cα ∈ R, f(x) is a symmetric

function of n variables if f(x) = f(ω(x)) ∀ω ∈ Sn, where Sn is the symmetric group

of degree n. Therefore f(x) is invariant by any permutation on the variables x. Λn

is the ring formed from all the symmetric functions of n variables. The limit of Λn

as n → ∞ is defined to be Λ. Elements of Λ are symmetric functions in an infinite

variable setting.

Example. x1 + x2 + x3, x1x2 + x1x3 + x2x3, and x1x2x3 are symmetric functions of

3 variables, whereas x1, x21 + x2 + x3, and x1x2 + x2x3 are not symmetric functions of

3 variables.

There are many different bases for the ring of symmetric functions both in the finite

as well as the infinite cases. The simplest is the monomial basis, which is indexed by

partitions. A partition of n is a set of non-increasing positive integers,

λ = (λ1, λ2, . . . , λm) : λ1 ≥ λ2 ≥ · · · ≥ λm > 0, wherem∑i=1

λi = n. λ is a partition


of n is denoted as λ ` n. The length of a partition, `(λ), is the number of parts the

partition consists of. If λ = (λ1, . . . , λm), then `(λ) = m. For `(λ) = m, λm+k is

defined to be 0 for all k > 0.

Example. (4, 3, 1, 1) is a partition of 9 with length 4.

Definition. For a finite n, the monomial symmetric function is defined by

mλ =∑ω

xω(λ), (2.1.1)

where ω ranges over all distinct permutations ω = (ω1, ω2, . . . , ωn) ∈ Sn of the entries

of λ = (λ1, . . . , λ`(λ)). For the infinite case ω ranges over all distinct permutations

ω = (ω1, ω2, . . .) of the entries of λ = (λ1, . . . , λ`(λ)).

Example. For n = 3, m(1) = x1 + x2 + x3 and m(2,1) = x21x2 + x1x22 + x21x3 + x1x

23 +

x22x3 + x2x23.

The monomial symmetric functions, mλ, are obviously symmetric and form a basis

for Λn in the finite case, and for Λ in the infinite case.

Definition. The complete homogeneous symmetric polynomial of degree k,

written as hk, of variables x1, . . . , xn is the sum of all monomials of total degree k.

Precisely, hk =∑

1≤l1≤l2≤···≤lk≤n

xl1xl2 · · ·xlk .

Given a partition λ, hλ is defined by: hλ = hλ1hλ2 · · ·hλ`(λ) .

The complete homogeneous symmetric polynomials also form a basis for the ring

of symmetric functions both in the finite and infinite cases. Another basis for Λ is

the Schur functions which will be described in the following section, however more

combinatorial objects are needed to understand them.


2.2 Schur functions

A Ferrers diagram is a method of displaying a partition. The diagram is created

by placing left (west) justified rows of boxes equal in number to the corresponding

part in order bottom (south) to top (north). The above notation is known as French

notation and will be used through out this paper. English notation places the rows in

the opposite order. A cell (y, x) denotes the box located at row y and column x. A cell

(y, x) is contained in a partition λ if there is a box at that location. A cell C = (y, x)

contained by λ will be denoted by C ∈ λ or (y, x) ∈ λ. (y, x) ∈ λ ⇒ `(λ) ≥ y and

λy ≥ x.

Example. Cell (2,3) has been marked with a thick frame in the Ferrers diagram of

the partition (4,3,1,1).

A partition and the partition’s Ferrers diagram will be used interchangeably.

A t-row shaped partition is a partition with only one part of length t. If λ is a

t-row shaped partition, then λ = (λ1) = (t). A t-row shaped partition will be written

as t or (t) interchangeably in this paper.

Example. Let λ = (4). λ is a 4-row shaped partition, whose Ferrers diagram is

.

A partition λ contains another partition µ if the Ferrers diagram for λ contains the

Ferrers diagram for µ. λ contains µ is denoted as λ ⊇ µ and µ is contained by λ is

denoted as µ ⊆ λ.

µ ⊆ λ⇒ `(λ) ≥ `(µ) and λi ≥ µi ∀i ∈ 1, . . . , `(µ).


Example. Let λ = (6, 4, 2, 1, 1) and µ = (3, 2, 1, 1). λ contains µ. This can be seen

below from the Ferrers diagram of λ, where µ has been marked with thick frames.

A skew partition is formed from two partitions λ and µ, where λ ⊇ µ. It is denoted

as λ/µ. The Ferrers diagram of a skew partition is the Ferrers diagram of λ with the

cells of µ removed.

Example. Let λ = (6, 4, 2, 1, 1) and µ = (3, 2, 1, 1). The skew Ferrers diagram for

λ/µ is

.

A horizontal t-strip is a skew partition with a Ferrers diagram of t cells, where at

most one cell occurs in each column.

Example. The previous example is not a horizontal strip whereas (4, 3, 1, 1)/(3, 1, 1)

is a horizontal 4-strip with skew Ferrers diagram .

An alphabet is a well ordered list of elements called letters (either finite or infi-

nite). The simplest alphabet is the set of natural numbers N with the natural or-

dering. Here 0 /∈ N. Given two infinite set of integers x1, x2, . . . and y1, y2, . . .,

x1, y1, x2, y2, . . . is another example of an alphabet.

A filling of shape λ/µ is a Ferrers diagram of shape λ/µ, where each cell is filled

with one letter from a fixed alphabet. For a filling, S, define S(y, x) to be the letter

in cell (y, x). Also if (y, x) /∈ λ/µ then define S(y, x) = ∅. For a fixed filling S denote

sh(S) to be the shape or the skew shape of the filling.

A semi-standard Young tableau of shape λ/µ is a filling of shape λ/µ such that


rows are nondecreasing increasing and columns are increasing. SSYT(λ/µ) is the set

of all semi-standard Young tableaux of shape λ/µ with a fixed alphabet.

Example. An example of a semi-standard Young tableau of shape (2, 1) and alphabet

N is S = 21 1

and S(1, 2) = 1. An example of a semi-standard Young tableau with

alphabet x1, y1, x2, y2, . . . is T =x2 x2

x1 y1. For the alphabet N3 = 1, 2, 3 with the

natural ordering,

SSYT(2, 1) =

21 1

, 31 1

, 21 2

, 31 2

, 21 3

, 31 3

, 32 2

, 32 3

.

wt(T ) is the weight of a semi-standard Young tableau, T , which is an n-tuple of

non-negative integers (in the infinite case it is an infinite sequence of non-negative

integers), where the ith entry is defined as the number of cells containing the ith

letter in the ordering. When using the alphabet N, the ith spot simply counts the

number of i’s in the tableau.

Example. For the alphabet N3, wt

(21 1

)= (2, 1, 0).

Definition. The Schur polynomial of degree n indexed by partition λ is defined

by

sλ =∑

T∈SSYT(λ)

xwt(T ), (2.2.1)

where the fixed alphabet is Nn = 1, 2, . . . , n with the natural ordering. A Schur

function is defined for the fixed alphabet N. It can be thought of as the limit of the

Schur polynomial, as n approaches infinity.

Example. For n = 3,

s2,1 = xwt(

21 1

)+ x

wt(

31 1

)+ x

wt(

21 2

)+ x

wt(

31 2

)+ x

wt(

21 3

)

+ xwt(

31 3

)+ x

wt(

32 2

)+ x

wt(

32 3

);

s2,1 = x(2,1,0) + x(2,0,1) + x(1,2,0) + x(1,1,1) + x(1,1,1) + x(1,0,2) + x(0,2,1) + x(0,1,2);

s2,1 = x21x2 + x21x3 + x1x22 + x1x2x3 + x1x2x3 + x1x

23 + x22x3 + x2x

23;


Cell (y, x) is an addable cell to a partition λ if cell (y, x) /∈ λ and cells (y −

1, x), (y, x− 1) ∈ λ. Cells (`(λ), 1) and (1, λ1 + 1) are also considered to be addable.

A cell (y, x) is a removable cell to a partition λ, if (y, x) ∈ λ and cells (y +

1, x), (y, x+ 1) /∈ λ.

Example. For λ = (4, 3, 1, 1) the addable cells are (5, 1), (3, 2), (2, 4), (1, 5) and the

removable cells are (4, 1), (2, 3), (1, 4). Below is a Ferrers diagram of λ with the

addable cells marked with A’s and the removable cells are marked with thick frames:

A

AA

A

2.3 The Hall inner product

The ring of symmetric functions, Λ, has a useful inner product known as the Hall inner

product. It is defined by making mλ and hµ dual bases. That is, 〈mλ, hµ〉 = δλ,µ

(Kronecker delta), where δλ,µ =

1 if λ = µ

0 otherwise

.

A remarkable and useful result is that the Schur functions, sλ, are dual to them-

selves. That is to say 〈sλ, sµ〉 = δλ,µ. This inner product also leads to many useful

properties regarding the relationship between two bases that are dual to each other.

One very useful result from linear algebra is as follows:

Proposition 1. If fλ and gµ are dual bases to each other, and

if ∆fν =∑λ,µ

dνλ,µfλ ⊗ fµ then gλgµ =∑ν

dνλ,µgν.

This powerful result above will be used later in Section 3.


2.4 The Pieri rule for Schur functions

The Schur functions form a basis for Λ therefore any element in Λ can be expressed

as a sum of Schur functions. For sλ, sµ ∈ Λ, sλsµ ∈ Λ. Hence the product of two

Schur functions can be expanded in terms of the Schur basis:

sλsµ =∑ν

cνλ,µsν . (2.4.1)

This expansion is one of great importance in algebraic combinatorics, representation

theory, and algebraic geometry. The coefficients cνλ,µ are know as the Littlewood-

Richardson coefficients and play a large role in the mathematical fields listed above.

The case of limiting µ to be a row shape presents a simpler yet still very important

problem of determining the coefficients of the expansion below [12; 15; 18]:

sλs(t) =∑ν

cνλ,(t)sν . (2.4.2)

This is known as the Pieri rule, named after the Italian geometer Mario Pieri. This

rule plays a large role in Schubert calculus. It can also be iteratively used to expand

a complete homogeneous function, hλ, in terms of the Schur basis. There is an

elegant combinatorial method of describing the Pieri rule for Schur functions using

the insertion algorithm [12; 15; 18]. A useful combinatorial procedure is the row

insertion algorithm. It is a method of inserting a letter α into a semi-standard Young

tableau S.

Procedure 1. Start on the first row, find the leftmost letter that is greater than α.

If there are none, then place the letter α at the end of this row. If there is such a

letter that is greater than α, then replace (or bump as it is called) the leftmost one.

Insert the letter that was bumped into the row above. Continue this process until the

letter is added to the end of the row or inserted into an empty row in which case it


just is added to the first spot in that row. The final result of row insertion is that

a single addable cell for the original semi-standard Young tableau’s shape has been

added [6; 18].

Example. For α = 1 and R =43 3 41 1 2 2

1→

43 3 41 1 2 2

2→43 3 41 1 1 2

3→ 42 3 41 1 1 2

4→32 3 41 1 1 2

432 3 41 1 1 2

.

Row insertion can be seen as a mapping from a letter paired with a semi-standard

Young tableau (α,R) to a cell paired with a new semi-standard Young tableau

((y, x), T ), where T is the result of inserting α into R and cell (y, x) is the cell added

to R. Denote this mapping as Ins(α,R). Ins(α,R) = ((y, x), T ), where sh(T )/sh(R)

is a skew partition containing only one cell (y, x), and (y, x) is an addable cell to

sh(R) [6].

Row insertion has an inverse procedure which removes a letter from the semi-standard

tableau by “unbumping” letters one row at a time, it is called the inverse row inser-

tion algorithm. Given any semi-standard Young tableau T , any removable cell (y, x)

for sh(T ) can be removed from T using the inverse row insertion algorithm [6].

This procedure can be seen as a mapping from a cell paired with a semi-standard

Young tableau ((y, x), T ), where (y, x) is a removable cell for sh(T ), to a letter

paired with a new semi-standard Young tableau (α,R). This mapping is denoted

as Rem((y, x), T ). Hence Rem((y, x), T ) = (α,R) where α is the final letter that is

unbumped out, (y, x) is the single cell of sh(T )/sh(R), and Ins(α,R) = ((y, x), T ) [6].

Example. For R =

432 3 41 1 1 2

and removable cell (4, 1)

432 3 41 1 1 2

→ 442 3 41 1 1 2

→ 343 3 41 1 1 2

→ 2

43 3 41 1 2 2 → 1


Row insertion is often done for “words”, an ordered set of individual letters. This

procedure is done by inserting the first letter, then inserting the second into the re-

sulting semi-standard Young tableau. Insertion is repeated until all the letters have

been inserted. This procedure is known as the Robinson-Schensted-Knuth algorithm

or RSK for short [8].

This procedure can be seen as a mapping from a ordered set of individual letters

(α1, α2, . . . , αm) paired with a semi-standard Young tableau, T , to a semi-standard

tableau Q, which records the added cells in the order they were added and the semi-

standard Young tableau P , which results from inserting the letters into R. This map-

ping is denoted as RSK((α1, α2, . . . , αm), T ). Therefore, if Ins(αi, Ti−1) = ((yi, xi), Ti),

then RSK((α1, α2, . . . , αm), T ) = (Q,P ), where Q(yi, xi) = i and P = Tm. Note, if

αi+1 ≥ αi ∀1 ≤ i ≤ m− 1, then (α1, α2, . . . , αm) can be seen as a semi-standard

Young tableau, so for sh(T1) = (d),

RSK(S1, S2) = RSK((R1(1, 1), (R1(1, 2), . . . , (R1(1, d))), R2).

Example. For (α1, α2, α3) = (1, 1, 2) = T = 1 1 2 and R = 32 2 3

,

Ins

(1, 3

2 2 3

)=

((3, 1),

321 2 3

),

Ins

(1,

321 2 3

)=

((2, 2),

32 21 1 3

),

Ins

(2,

32 21 1 3

)=

((2, 3),

32 2 31 1 2

),

RSK

((1, 1, 2), 3

2 2 3

)= RSK (T,R) =

(1

2 3 ,32 2 31 1 2

).

Since the RSK algorithm is constructed by iterating row insertion, which is an invert-

ible mapping, then the RSK algorithm is also invertible by iterating the Rem((y, x), T )

mapping denoted by InvRSK(Q,P ). If RSK((α1, α2, . . . , αm), R) = (Q,P ), then

InvRSK(Q,P ) = ((α1, α2, . . . , αm), R) [6; 18]. Row insertion and the RSK algorithm

have many useful properties that will be used later in this paper.


Lemma 1. (Fulton)[6]

Given a semi-standard tableau, S0, with alphabet A, letters α, β ∈ A, Ins(α, S0) =

((y1, x1), S1), and Ins(β, S1) = ((y2, x2), S2), α ≤ β if and only if the following

statements are true:

1. y1 ≤ y0 and x1 ≥ x0.

2. If S0(i, ai) 6= S1(i, ai) and S1(i, bi) 6= S2(i, bi), then ai ≤ bi.

This lemma is proved by Fulton [6] on p.10.

Definition. Let A(λ, t) be the set of the partitions formed by adding a horizontal

t-strip to λ. So if µ ∈ A(λ, t) then |µ/λ|= t and µ/λ is a horizontal strip.

Example. For λ = (2, 1) and t = 2

A(λ, t) =

, , ,

= (4, 1), (3, 2), (3, 1, 1), (2, 2, 1).

Lemma 2. For fixed semi-standard tableaux, R and D, with alphabet A, where

sh(D) = (t), if RSK(D,R) = (Q,P ) then sh(Q) ∈ A(sh(R), t).

Inserting a t-row shaped semi-standard tableau into a semi-standard tableau, R, re-

sults in a new semi-standard tableau, P . The shape of P is the original shape of R

plus a horizontal t-strip.

Proof. Recall that sh(Q) = sh(P )/sh(R). The equality |sh(Q)|= t is obvious since

t letters are being inserted into R. D(1, i + 1) ≥ D(1, i) ∀1 ≤ i ≤ t, since D ∈

SSYT((t)). From this fact Lemma 1 implies that the cell added by inserting letter

D(1, i + 1) must be to the right (east) of the added cell created by inserting letter

D(1, i), therefore Q can not have two cells in the same column.

Remark 1. For a fixed semi-standard tableau, P , with alphabet A, and a horizontal

t-strip λ/µ, where λ = sh(P ), ∃!Q such that InvRSK(Q,P ) = (D,R), where sh(D) =

(t).


Proof. Since λ/µ is a horizontal t-strip, there is at most one cell in each column.

Q is created such that sh(Q) = λ/µ and the letter i is placed in the ith leftmost

(westmost) cell. InvRSK(Q,P ) = ((α1, α2, . . . , αt), R) for some α1, α2, . . . , αt ∈ A

and R ∈ SSYT(µ). Letter i is to the left of the letter i + 1 in Q, thus Lemma 1

implies that αi ≤ αi+1; therefore RSK((α1, α2, . . . , αt), R) = RSK(D,R) = (Q,P ),

where D(1, i) = αi. The uniqueness of Q follows directly from the fact that the RSK

algorithm is a bijection.

Lemma 3. For fixed semi-standard tableaux, R and D, with alphabet A, where

sh(D) = (t), if RSK(D,R) = (Q,P ), then the following statements are true:

1. ∀(y, x) ∈ R, P (y, x) ≤ R(y, x);

2. ∀(y, x) ∈ P : (y, x) /∈ R, P (y, x) ≤ D(1, t) if y = 1, or P (y, x) ≤ R(y −

1, sh(R)y−1).

Notice: if y > 1, then R(y − 1, sh(R)y−1) is the last letter in row y − 1 of R.

Proof. Both statements are a direct result of row insertion. The first statement follows

from the fact that if a letter belonging to R is changed from row insertion then it

can only change by being bumped, and it can only be bumped by a letter before it

in the alphabet. The second statement has a condition on y. First, if a cell (1, x) is

added to R, then it must be one of the inserted letters (1, i) ∈ D, and D(1, t) is the

greatest of these letters in the alphabet. If y > 1, then this letter must have been

bumped from the row below and the rightmost cell in a row is the greatest letter in

the alphabet, since R is a semi-standard Young tableau. Notice that when the row

insertion is iterated these properties continue to hold.

Definition. RSK(D,R) = Proj2(RSK(D,R)), where D ∈ SSYT((t)), R ∈ SSYT(λ),

and Proj2(A1, A2) = A2, in our case Proj2(RSK(D,R)) = P since RSK(D,R) =

(Q,P ).


Lemma 4. For a fixed t ∈ N and fixed partition λ, RSK is a bijection from SSYT((t))×

SSYT(λ) to its image.

Proof. For a fixed D ∈ SSYT((t)) and R ∈ SSYT(λ) remark 1 states that ∃!Q :

RSK(D,R) = (Q,P ) where P = RSK(D,R). Therefore ∃!P such that InvRSK(Q,P ) =

(D,R). Hence RSK is a one to one mapping.

Remark 2. wt(D) + wt(R) = wt(RSK(D,R)).

Proof. This is obvious, because there are the same letters in D and R as there are in

RSK(D,R), since each letter in D was inserted into R.

Since RSK is a bijection then it has an inverse bijection. The inverse is denoted as

InvRSK. Notice, that InvRSK does require the original fixed partition (λ above).

Example.

RSK

(1 1 2 ,

32 2 3

)=

32 2 31 1 2

,

and InvRSK

(32 2 31 1 2

,

)=

(1 1 2 ,

32 2 3

).

One of the most useful applications of the RSK insertion algorithm is the combina-

torial proof of the Pieri rule for Schur functions.

Theorem 1.

sλs(t) =∑

µ∈A(λ,t)

sµ (2.4.3)

Proof. Using definition (2.2.1) (definition of Schur functions), it suffices to show that

∑T1∈SSYT((t))

xwt(T1)∑

T2∈SSYT(λ)

xwt(T2) =∑

µ∈A(λ,t)

∑T∈SSYT(µ)

xwt(T ) . (2.4.4)

Or equivalently,

∑(T1,T2)∈SSYT((t))×SSYT(λ)

xwt(T1)+wt(T2) =∑

µ∈A(λ,t)

∑T∈SSYT(µ)

xwt(T ) . (2.4.5)

3 THE PIERI RULE FOR THE DUAL GROTHENDIECK POLYNOMIALS 16

This identity follows from the fact that, for fixed partition λ and integer t > 0, there

is a weight preserving bijection between the set of tableaux with shape µ ∈ A(λ, t)

and the set of pairs of tableaux (T1, T2), for which sh(T1) = (t) and sh(T2) = λ. In

fact, this weight preserving bijection is none other than RSK.

3 The Pieri rule for the dual Grothendieck poly-

nomials

3.1 Grothendieck polynomials

Schur functions are a useful homogeneous basis for the ring of symmetric functions.

The Grothendieck polynomials are inhomogeneous polynomials related to Schubert

varieties. Whereas the Schur functions are dual to themselves, the Grothendieck

polynomials are not. They are dual to a set of polynomials known as the dual

Grothendieck polynomials. This section will focus on these dual polynomials and the

combinatorics involved to define and use them. As Schur functions can be defined

using semi-standard Young tableaux, dual Grothendieck polynomials can be defined

using combinatorial objects known as reverse plane partitions. Reverse plane parti-

tions have a one-to-one correspondence with pairs of semi-standard Young tableaux

and elegant fillings, another combinatorial object. Similar to Schur functions there

are many different rules and properties describing their behavior. Buch developed

the rule for the coproduct of the Grothendieck polynomials which in turn gives the

coefficients for the Pieri rule for the dual Grothendieck Polynomials [3; 11; 14].

Theorem 2. (Buch) [3] p.20

∆Gν =∑λ,µ

dνλ,µGλ ⊗Gµ. (3.1.1)


Buch gives a combinatorial definition for these coefficients, dνλ,µ, which is similar

to that of the Littlewood-Richardson coefficients for Grothendieck polynomials. If

µ is restricted to be a row shape partition, then the definition for the coefficients is

simplified drastically. Let

ν = (ν2, . . . , ν`(ν)),

notice, ν is equal to ν with v1 removed.

Corollary 1. (Buch) [3] p.20 If ν/λ is a horizontal strip and t ≥ 0 is an integer, then

dνλ,(t) = (−1)t−|ν/λ|(r(λ/ν)

t− |ν/λ|

), (3.1.2)

where r (λ/ν) is the number of rows in λ/ν. If ν/λ is not a horizontal strip then

dνλ,(t) = 0.

3.2 Dual Grothendieck polynomials

Unlike the Schur polynomials, the Grothendieck polynomials are not self dual. The

polynomials which are dual to them are known as the dual Grothendieck polynomials.

Dual Grothendieck polynomials have a combinatorial definition similar to that of the

Schur polynomials.

Definition. The dual Grothendieck polynomial gλ, is defined as

gλ =∑

T∈RPP(λ)

xwt(T ), (3.2.1)

where RPP(λ) are reverse plane partitions of shape λ, which are fillings of shape

λ with a fixed alphabet with nondecreasing rows and columns. wt(T ) is the weight

of T , a set of non-negative integers, where the ith spot is defined as the number of

columns that contain the ith letter.


Notice, that SSYT(λ) ⊆ RPP(λ). All semi-standard Young tableaux of shape λ are

also reverse plane partitions of shape λ. Also notice, that the definition of weight is

the same if restricted to semi-standard Young tableaux, since a letter can only occur

in a column once.

Example. For fixed alphabet N2 = 1, 2 with the standard ordering

RPP(2, 1) =

11 1

, 21 1

, 11 2

, 21 2

, 22 2

,

g2,1 = xwt(

11 1

)+ x

wt(

21 1

)+ x

wt(

11 2

)+ x

wt(

21 2

)+ x

wt(

22 2

),

g2,1 = x(2,0) + x(2,1) + x(1,1) + x(1,2) + x(0,2) = x21 + x21x2 + x1x2 + x1x22 + x22.

3.3 Elegant fillings

Dual Grothendieck polynomials can be seen as sums of reverse plane partitions. Re-

verse plane partitions do not have a simple row insertion due to their column structure.

However, reverse plane partitions of a fixed shape can be bijected to pairs of elegant

fillings and semi-standard Young tableaux. This allows reverse plane partitions to

be mapped to new combinatorial objects. These new combinatorial objects allow a

simple row insertion method, and thus a word insertion algorithm very similar to the

RSK algorithm.

Definition. Elegant fillings of shape λ/µ, EF(λ/µ), are fillings of skew Ferrer

diagram of shape λ/µ with the fixed alphabet N, where row i is restricted to letters

1, . . . , i− 1, rows are nondecreasing, and columns are increasing [11].

Example. EF ((3, 3, 2, 1)/(3, 1)) =

21 2

1 1,

31 2

1 1,

32 2

1 1

.

For a filling of a skew Ferrers diagram, the part that has been removed will often be

left simply as empty boxes. This is done to help the reader see the shape of the skew

Ferrers diagram. Let S ∈ SSYT(µ) and E ∈ EF(λ/µ). The pair (S,E) is called a


pair of semi-standard tableau and elegant filling of shapes λ and µ. SEP(λ) is the set

of all such pairs of a semi-standard Young tableau of shape µ and an elegant filling

of shape λ/µ. Notice that µ1 = λ1, since elegant fillings can not have any letters in

the first row.

Example.

Given λ = (4, 4, 3) and µ = (4, 2), if S = 2 31 2 2 3

and E =1 1 2

1 1 , then (S,E) ∈

SPE(λ).

Define Nn to be the alphabet 1, 1, 2, 2, . . . n, n ordered by

1 < 2 < · · · < n < 1 < 2 < · · · < n .

For any pair (S,E) ∈ SPE(λ) the elegant filling is of shape λ/µ and the semi-standard

Young tableau is of shape µ. They both can be placed inside the Ferrer diagram of

shape λ. The letters of the elegant filling are marked by barring them (a line placed

above the letter), and thus are selected from the second part of the alphabet Nn.

Notice, that n has to be significantly large such that n ≥ max(S(y, x)) ∀(y, x) ∈ S,

and n ≥ max(E(y, x)) ∀(y, x) ∈ E. It is assumed that n is always chosen to be large

enough throughout this paper.

Example.

If S = 2 31 2 2 3

and E =1 1 2

1 1 , then (S,E) ∈ SPE(λ).

So (S,E) can be written as T , where T =1 1 22 3 1 11 2 2 3

.

Observation. For any (S,E) ∈ SEP(λ) ∃!T ∈ SSYT(λ) with alphabet Nn,

such that cell S(y, x) = T (y, x) ∀(y, x) ∈ S and E(y, x) = T (y, x) ∀(y, x) ∈ E,

where E(y, x) is the letter E(y, x) barred.


This observation follows directly from the example above. Denote Ψ to be the map

from a pair (S,E) ∈ SEP(λ) to the semi-standard Young tableau T with alphabet Nn.

Ψ is a one-to-one mapping, thus there is an inverse mapping, Ψ−1 : Ψ(SEP(λ)) →

SEP(λ). Notice, if T ∈ Ψ(SEP(λ)), then T ∈ SSYT(λ) with alphabet Nn and the

barred letters in row i are restricted to letters 1, 2, . . . , i− 1.

Example.

For S = 2 31 2 2 3

and E =1 1 2

1 1 from the example above,

Ψ(S,E) = T , where T =1 1 22 3 1 11 2 2 3

, and Ψ−1(T ) = (S,E).

Throughout this paper, pairs (S,E) ∈ SEP(λ) will be used interchangeably with the

semi-standard young tableaux, T = Ψ(S,E), where T ∈ SSYT(λ) with alphabet

Nn. There exists a bijection from these semi-standard Young tableaux of shape λ

to reverse plane partitions of shape λ. Lam and Pylyavskyy found a bijection that

maps reverse plane partitions of shape λ to a pairs of semi-standard tableau of shapes

contained by λ and an elegant filling of shape λ/µ, where µ is the shape of the semi-

standard tableau [11]. Later Bandlow and Morse created another bijection, Φ, which

uses the jeu de taquin transformation to preserve weight between the reverse plane

partitions and the semi-standard tableaux [1]. For details on jeu de taquin refer to

[6] and for details on this bijection refer to [1].

Example. Given T =

21 21 2 21 1 1 31 1 1 2 2 2

, then Φ(T ) =

2 2 2 31 1 1 2 2 2

,

32 31 1 1

.

Recall that a pair (S,E) ∈ SPE(λ) can be mapped to a semi-standard Young tableau,

T ∈ SSYT(λ) with alphabet Nn, using the map Ψ.

Ψ

2 2 2 31 1 1 2 2 2

,

32 31 1 1

=

32 31 1 12 2 2 31 1 1 2 2 2

.


The function Φ extends Φ to be a bijection from reverse plane partitions to a subset of

semi-standard Young tableaux of the same shape with alphabet Nn, or more precisely

Φ(R) = Ψ(Φ(R)).

Lemma 5. Φ is a bijection.

Proof. Φ is a composition of two bijections. Therefore Φ is a bijection.

From Lemma 5 it can be deduced that there is a reverse bijection Φ−1, which

maps a subset of semi-standard Young tableaux with alphabet Nn to reverse plane

partitions. If (S,E) ∈ SPE(λ) and Ψ(S,E) = T , then Φ−1(T ) = R, where R ∈

RPP(λ) and Φ(R) = (T ). The weight of a semi-standard Young tableau T with

alphabet Nn is an n-tuple of non-negative integers, such that the ith entry counts the

number of i’s in the tableau. Notice, the barred letters do not affect the weight.

Remark 3. Given a reverse plane partition, R, wt(Φ(R)) = wt(R).

Proof. Given Φ(R) = (S,E). It was shown in [1] that wt(R) = wt(S) and clearly

wt(Ψ(S,E)) = wt(S), since the elegant filling portion is ignored when measuring the

weight.

Remark 4. ∀R ∈ RPP((d)) the following is true: R = Φ(R), that is R(y, x) =

Φ(R)(y, x) ∀(y, x) ∈ sh(R).

This is clear, due to the fact that (d) is a row shape, therefore EF (λ/µ) = ∅ ∀µ,

due to the restriction of the letters in each row.

3.4 Pieri rule for the dual Grothendieck polynomials

Proposition 1 implies that the expansion of the coproduct of the Grothendieck poly-

nomials (3.1.1) gives the Littlewood-Richardson rule for the dual Grothendieck poly-

nomials.


Corollary 2. [3]

gλgµ =∑

dνλ,µgν. (3.4.1)

Corollary 1 gave a formula for the coefficients of the form dνλ,(t). This formula

gives the following:

Corollary 3.

gλg(t) =∑

(ν/λ) is a horizontal strip

(−1)t−|ν/λ|(r(λ/ν)

t− |ν/λ|

)gν. (3.4.2)

The proof of the Pieri rule for Schur functions (2.4.3) gives the motivation for the

insertion proof of the Pieri rule for the dual Grothendieck polynomials in Section 4.

To start, the above formula (3.4.1) is reformulated as a sum over a certain family

of combinatorial objects so that it is in the spirit of the Pieri rule for Schur func-

tions (2.4.3). The next part of this section introduces new combinatorial notations

necessary for the proof in Section 4.

Definition. Given two partitions µ and λ (or the pair (µ, λ)), where λ ⊆ µ, a

Y cell for (µ, λ) is a removable cell in λ that has no cells above it in µ. Notice,

that a removable cell (y, x) of λ is a Y cell for (µ, λ) if it satisfies the following:

`(µ) = y or µy+1 < x .

Example. The only Y cell of (µ, λ) for µ = (6, 4, 2, 1, 1) and λ = (4, 3, 1, 1) is (2, 3).

See the diagram below, where removable cells of λ are marked with thick frames.

Y

Lemma 6. For any horizontal strip ν/λ,

r(λ/ν) = the number of Y cells for (ν, λ). (3.4.3)


Proof. First notice, that since a Y cell for shapes (ν, λ) is a removable corner of λ,

there can be at most one in each row of λ. Precisely, a row 1 ≤ i ≤ `(λ) of λ has a

Y cell, when λi > νi+1 (recall ν`(ν)+1 = 0), since a Y cell has no cells of ν above it.

On the other hand, if λ/ν is a valid skew shape, then row 1 ≤ i ≤ `(λ) of λ/ν is

non-empty when νi+1 = νi < λi. To this end, it will be shown that λ/ν is a horizontal

strip by proving that λi+1 ≤ νi ≤ λi. Clearly, λi+1 ≤ νi+1 = νi since λ ⊆ ν. By

contradiction, if νi+1 > λi then column i + 1 of ν/λ has at least 2 cells. But since

ν/λ is a horizontal strip, νi+1 ≤ λi.

Given partitions µ and λ, a diagram of µ is an XO-diagram of shape (µ, λ), when

every cell of µ/λ contains an O and a subset (possibly empty) of the Y cells of (µ, λ)

contains X’s. The reason for marking the O cells will become apparent later.

Denote the set of all XO-diagrams of shape (µ, λ) that contain exactly j X’s by

XO(µ, λ, j). Notice, that if k is the number of Y cells of (µ, λ), then

|XO(µ, λ, j)|=(

number of Y cells for (µ, λ)

j

)=

(k

j

), (3.4.4)

when j ≤ k, and is zero otherwise.

Example. For λ = (2, 1) and j = 1, for each µ ∈ A(λ, 1), the XO-diagrams of shape

(µ, λ) with only 1 X are

XO

(, , 1

)=

O

X

,

XO

(, , 1

)=

X O

,

XO

(, , 1

)=

X

O,

X O

.

4 INSERTION PROOF OF THE DUAL GROTHENDIECK PIERI RULE 24

Theorem 3. For any partition λ and integer t > 0,

gλ gt =t∑

j=0

∑µ∈A(λ,j)

∑F∈XO(µ,λ,t−j)

(−1)t−j gµ. (3.4.5)

Proof. Lemma 6 and (3.4.2) give that

gλgt =∑

ν/λ is a horizontal strip

(−1)t−|ν/λ|(

number of Y cells for (ν, λ)

t− |ν/λ|

)gν . (3.4.6)

Notice, that if |ν/λ|> t, then

(number of Y cells for (ν, λ)

t− |ν/λ|

)= 0. Therefore (3.4.6)

can be written as a sum over horizontal j-strips, from j = 0 to j = t,

gλgt =t∑

j=0

∑ν/λ is a j-horizontal strip

(−1)t−j(

number of Y cells for (ν, λ)

t− j

)gν . (3.4.7)

From the definition of A(λ, j) it follows that summing over horizontal j-strips ν/λ

amounts to summing over elements of A(λ, j), and (3.4.4) then implies (3.4.5).

The proof above shows that Theorem 3 is simply another way of writing a simplified

version of (3.4.2). In the next section a combinatorial proof of Theorem 3 will be

given.

4 Insertion proof of the dual Grothendieck Pieri

rule

Similar to the proof of the Pieri rule for Schur functions, the combinatorial approach

to proving that

gλ g(t) =t∑

j=0

∑µ∈A(λ,j)


(−1)t−jgµ (4.0.1)


begins by substituting the dual Grothendieck polynomials with their weight generat-

ing functions over reverse plane partitions. The left hand side becomes

gλgt =

∑R∈RPP(λ)

xwt(R)

∑R∈RPP((t))

xwt(R)

=∑

(R1,R2)∈RPP(t)×RPP(λ)

xwt(R1)+wt(R2) ,

and the right hand side is

t∑j=0

∑µ∈A(λ,j)


(−1)t−jgµ =t∑

j=0

∑µ∈A(λ,j)


(−1)t−j∑

R∈RPP(µ)

xwt(R).

(4.0.2)

An insertion algorithm that gives a bijection between the set of pairs in this

summand and a certain set of reverse plane partitions, SRPP, will be introduced in

this section. However, there is an intricacy here that does not arise in the proof of

the Schur Pieri rule, namely, there is an alternating sign in the right hand side of

(4.0.2). This requires a sign-reversing involution to be created on the objects arising

in the right hand summand of (4.0.2). Fixed points of this involution are precisely

the elements of SRPP.

4.1 Examples

The following two examples are shown to clarify the problem.

Example. Below is the expansion of the product of two dual Grothendieck polyno-


mials.

g(2,1)g(2) =2∑j=0

∑µ∈A

(,j)

∑F∈XO

(µ, ,t−j

) (−1)2−jgµ;

=∑

µ∈A(

,0)

∑F∈XO

(µ, ,2

) (−1)2gµ +∑

µ∈A(

,1)

∑F∈XO

(µ, ,1

) (−1)1gµ

+∑

µ∈A(

,2)

∑F∈XO

(µ, ,0

) (−1)0gµ;

=∑

µ∈

∑F∈XO

(µ, ,2

) (−1)2gµ +∑

µ∈

, ,

∑

F∈XO(µ, ,1

) (−1)1gµ

+∑

µ∈

, , ,

∑

F∈XO(µ, ,0

) (−1)0gµ;

=∑µ=

∑F=X

X

(−1)1g +∑

µ=

∑F∈

XO,X

O

(−1)1g

+∑µ=

∑F=XO

(−1)1g +∑µ=

∑F=

O

X

(−1)1g

+∑

µ=

∑F=

OO

(−1)0g +∑

µ=

∑F= O

O

(−1)0g

+∑

µ=

∑F=

O

O

(−1)0g +∑µ=

∑F=

OO

(−1)0g ;

g2g2,1 = g2,1 − g3,1 − g3,1 − g2,2 − g2,1,1 + g4,1 + g3,2 + g3,1,1 + g2,2,1.

The polynomial g3,1 has two −1 terms with different XO-diagrams. The importance of

separating these individual XO-diagrams will be shown later. When the polynomials

are expanded into the individual x terms, most of them will cancel out due to the

change of signs. This will be shown in the next example.


Example. Let N2 = 1, 2 be the fixed alphabet with the usual ordering.

g1g1 =1∑j=0

∑µ∈A( ,j)

∑F∈XO(µ, ,1−j)

(−1)1−j∑

R∈RPP(µ)

xwt(R);

g1g1 = (−1)∑µ=

∑F=X

∑R∈RPP( )

xwt(R) +∑µ=

∑F= O

∑R∈RPP( )

xwt(R)

+∑µ=

∑F=O

∑R∈RPP

( )xwt(R);

g1g1 = −g1 + g1,1 + g2;

RPP(1) = 1 , 2 ;

RPP(2) = 1 1 , 1 2 , 2 2 ;

RPP(1, 1) =

11, 21, 22

;

g1g1 = −xwt( 1 ) − xwt( 2 ) + xwt( 1 1 ) + xwt( 1 2 ) + xwt( 2 2 )

+ xwt(

11

)+ x

wt(

21

)+ x

wt(

22

);

g1g1 = −x(1,0) − x(0,1) + x(2,0) + x(1,1) + x(0,2) + x(1,0) + x(1,1) + x(0,1);

g1g1 = −x1 − x2 + x21 + x1x2 + x22 + x1 + x1x2 + x2;

g1g1 = x21 + 2x1x2 + x22.

From this example it is clear that cancellation does occur. Later it will be shown that

all negative terms are canceled out and only the shapes with the maximum number

of cells have any remaining terms.

4.2 Insertion algorithm

Recall, that for a given reverse plane partition, R, Φ(R) is a semi-standard Young

tableau with the new alphabet, Nn. Hence the RSK insertion algorithm can be

preformed.

Definition. Given positive integer d and partition λ, let IDG be the map defined on


RPP((d))× RPP(λ) by

IDG(R1, R2) = Φ−1 RSK(Φ(R1),Φ(R2)).

Example. IDG

1 1 2 3 ,

21 21 1 21 1 1 31 1 1 2 2 2

=

322 22 2 21 1 1 2 21 1 1 1 1 2 2 3

,

since

Φ ( 1 1 2 3 ) = 1 1 2 3

and

Φ

21 21 1 21 1 1 31 1 1 2 2 2

=

32 31 1 12 2 2 31 1 1 2 2 2

and

RSK

1 1 2 3 ,

32 31 1 12 2 2 31 1 1 2 2 2

=

321 33 1 12 2 2 2 21 1 1 1 1 2 2 3

and

Φ−1

321 33 1 12 2 2 2 21 1 1 1 1 2 2 3

=

321 21 2 21 1 1 2 21 1 1 1 1 2 2 3

.

Lemma 7. Given R1 ∈ RPP((d)), R2 ∈ RPP(λ), the following is true for some

partition, µ : IDG(R1, R2) ∈ RPP(µ). Or equivalently, IDG maps a pair of reverse

plane partitions, the first having a row shape and the second’s shape is arbitrary, to

another reverse plane partition of a different shape.

Proof. Given a reverse plane partition R1 ∈ RPP((d)) and another reverse plane

partition R2 ∈ RPP(λ), IDG(R1, R2) is defined as follows: Φ−1(RSK(Φ(R1),Φ(R2))).

For Φ−1(T ) to be a valid reverse plane partition it is required that T is a valid semi-

standard tableau, and that T (i, x) ≤ i− 1 ∀1 ≤ i ≤ `(sh(T )), or equivalently the


letters in row i are less than the letter i. Let T = RSK(Φ(R1),Φ(R2)). T is the result

of using the RSK algorithm to insert Φ(R1) into Φ(R2). Since Φ(R) ∈ SSYT(λ)

∀R ∈ RPP(λ), then T is a semi-standard Young tableau resulting from the RSK

algorithm. The only remaining requirement is that letters in row i are restricted

to 1, 2, . . . , n, 1, 2, . . . , i− 1. Recall that Φ(R) maps R ∈ RPP(λ) to T ∈ SSYT(λ)

with alphabet Nn, such that T (y, x) ≤ y − 1 ∀(y, x) ∈ T . From Lemma 3 we have

T (y, x) ≤ Φ(R1)(1, d) for y = 1, Remark 4 states that Φ(R1) = R1. R1(1, i) can

not be barred, since no barred letters can occur in the first row. Now for cell (y, x),

where y > 1, lemma 3 implies that T (y, x) ≤ Φ(R2)(y − 1, sh(T )y−1). Φ(R2)(y −

1, sh(T )y−1) ≤ y − 1, since Φ(R2) ∈ Ψ(SEP(sh(R2))). Therefore T is a valid semi-

standard tableau and Φ−1(T ) ∈ RPP(µ), where µ is a valid partition.

Lemma 8. The map IDG on RPP(d)× RPP(λ) is injective into⋃

ν∈A(λ,d)

RPP(ν).

Proof. Given R1 ∈ RPP((d)) and R2 ∈ RPP(λ), IDG(R1, R2) is defined as follows:

Φ−1(RSK(Φ(R1),Φ(R2))). Lemma ?? implies that IDG(R1, R2) ∈ RPP(µ). All that

remains to be shown, is that µ ∈ A(λ, d). Φ(R) maps R ∈ RPP(µ) to T ∈ SSYT(µ)

with alphabet Nn and Remark 4 states that Φ(R1) = R1. RSK(Φ(R1),Φ(R2)) =

RSK(R1, T ), where R1 ∈ SSYT((d)) and T ∈ SSYT(λ), RSK(R1, T ) = P , where

sh(P )/sh(T ) ∈ A(λ, d), shown in Lemma 2. This shows that

sh(Φ−1(RSK(Φ(R1),Φ(R2)))) ∈ A(λ, d).

To determine the image of the map IDG, it is necessary to introduce an important

subset of reverse plane partitions.

Definition. Given partitions λ and µ, a reverse plane partition R of shape λ is

defined to be special, when no cell of λ/µ in Φ(R) contains an i− 1 in row i, for all

i. The set of all such elements is denoted by SRPP(λ, µ). Or equivalently,

SRPP(λ, µ) =R ∈ RPP(λ) : If Φ(R) = T then T (y, x) < y − 1 ∀(y, x) ∈ λ/µ

.


Example. For R = 11 1

Φ(R) = 11 1

, the reverse plane partition

R ∈ SRPP

(,

)but R /∈ SRPP

(,

).

Lemma 9. The map IDG is a bijection.

Proof. From Lemma 5 it follows that Φ−1 is a bijection, and from Lemma 4 RSK is

a bijection. Thus IDG is a composition of two bijections, hence it is a bijection.

Lemma 10. The image of IDG(RPP((d),RPP(λ)) is⋃

ν∈A(λ,d)

SRPP(ν, λ).

Proof. Given partition λ and positive integer d, consider (R1, R2) ∈ RPP(d)×RPP(λ).

Define R3 = IDG(R1, R2). Lemma 8 implies that R3 ∈ RPP(µ), where µ ∈ A(λ, d). It

still needs to be shown that R3 ∈ SRPP(µ, λ). Define T = Φ−1(R3). Remark 4 states

Φ(R1) = R1 therefore T = RSK(R1,Φ(R2))). Let (y, x) ∈ µ/λ. If y = 1 then, Lemma

3 states that T (y, x) ≤ R1(1, d) and R1(1, d) < 1. If y > 1, then Lemma 3 states

that T (y, x) ≤ Φ(R2)(y − 1, sh(Φ(R2))y−1), and Φ(R2)(y − 1, sh(Φ(R2))y−1) ≤ y − 2

since R2 ∈ RPP(λ). Therefore, if (y, x) ∈ µ/λ, then T (y, x) < y − 1, hence R3 ∈

SRPP(µ, λ). The above implies that IDG(RPP((d),RPP(λ)) ⊆⋃

ν∈A(λ,d)

SRPP(ν, λ).

It remains to show that

IDG(RPP((d),RPP(λ)) ⊇⋃

ν∈A(λ,d)

SRPP(ν, λ). Let µ ∈ A(λ, d) and

T ∈ SRPP(µ, λ). Define T = Φ−1(T ), then define (D, S) = InvRSK(T , λ). This can

be done, since T ∈ SSYT(µ) with alphabet Nn. Notice, since D ∈ SSYT((d)), then

Remark 4 implies that D = Φ−1(D) = D. S(y, x) = T (y, x) or it was modified via the

unbumping procedure, this can be seen from Lemma 1. Notice, that only cells con-

tained in µ/λ are the origin for unbumping. If cell (y, x) ∈ µ/λ, then T (y, x) ≤ y − 2,

since T ∈ SRPP(µ, λ). Let cell (i, αi) be the modified cell in row i resulting from the

unbumping of (y, x). ∀1 ≤ i ≤ y − 1, (i, αi) = T (i + 1, αi+1), where αy = x. This is

a direct result from the definition of row insertion and Lemma 1. Recall, since cell


(i, αi) was unbumped, then T (i, αi) < T (i+1, αi+1) and the inequality T (y, x) ≤ y − 2

implies that ∀(k, j) ∈ S S(k, j) ≤ k − 1. This means that if S = Φ−1(S), then

S ∈ RPP(λ) and RSK(D,S) = T . Since µ and T are arbitrary the following can

be concluded: ∀µ ∈ A(λ, d) ∀T ∈ SRPP(µ/λ) ∃!D ∈ RPP((d)), S ∈ RPP(λ) :

IDG(D,S) = T . This implies that IDG(RPP((d),RPP(λ)) ⊇⋃

ν∈A(λ,d)

SRPP(ν, λ)

thus completing the proof.

Lemma 11. For S ∈ RPP(λ) and D ∈ RPP((d)), wt(S) + wt(D) = wt(IDG(D,S)).

Proof. This follows from the fact that both the bijection Φ and the RSK preserve

weight. Let S ∈ RPP(λ), D ∈ RPP((d)), then

wt(IDG(S,D)) = wt(Φ−1(RSK(Φ(D),Φ(S)))). Remark 3 implies that

wt(Φ−1(RSK(Φ(D),Φ(S)))) = wt(RSK(Φ(D),Φ(S))). Remark 2 implies that

wt(RSK(Φ(D),Φ(S))) = wt(Φ(D) + wt(Φ(S)). And again Remark 3 implies that

wt(Φ(D),Φ(S)) = wt(D) + wt(S).

Theorem 4.

gλgt =

∑R∈RPP(λ)

xwt(R)

∑R∈RPP((t))

xwt(R)

=∑

ν∈A(λ,t)

∑R∈SRPP(ν,λ)

xwt(R). (4.2.1)

The proof is very similar to the classical proof of the Pieri rule for Schur functions.

Proof.

∑R1∈RPP((t))

xwt(R1)∑

R2∈RPP(λ)

xwt(R2) =∑

ν∈A(λ,t)

∑T∈SRPP(ν)

xwt(T ), (4.2.2)

the equation above can be written equivalently as the following:

∑(R1,R2)∈RPP((t))×RPP(λ)

xwt(R1)+wt(R2) =∑

ν∈A(λ,t)

∑T∈SRPP(ν)

xwt(T ) . (4.2.3)


This identity follows by showing for fixed partition λ and integer t > 0, that there

is a weight preserving bijection between the set of special reverse plane partitions

with shape ν ∈ A(λ, t) and the set of pairs of reverse plane partitions (R1, R2), where

sh(R1) = (t) and sh(R2) = λ. In fact, this weight preserving bijection is none other

than IDG.

Recall the Pieri rule for dual Grothendieck polynomials (3.4.5) defined g(d)gλ as

a sum of dual Grothendieck polynomials. In the example before it was shown that

certain terms cancels out and Theorem 4 shows which terms do not cancel out. The

next step for the combinatorial proof of the Pieri rule for dual Grothendieck poly-

nomials is creating a sign changing involution. This involution will be used to show

how the remaining terms cancel out.

4.3 Sign changing involution on reverse plane partitions and

XO-diagrams

Denote by NRPP (λ, µ) the set of all reverse plane partitions of shape λ which do

not belong to SRPP(λ, µ). Or equivalently,

NRPP (λ, µ) = R ∈ RPP(λ) : R /∈ SRPP(λ, µ .

It is clear that

SRPP(λ, µ) ∪NRPP (λ, µ) = RPP(λ) and SRPP(λ, µ) ∩NRPP (λ, µ) = ∅.


Also let M(λ, d) be the set of pairs of all reverse plane partitions of ν and XO-diagrams

of (ν, λ) (where ν/λ is a horizontal strip of length at most d). Or equivalently,

M(λ, d) =⋃

i=0,...,dν∈A(λ,i)

RPP(ν)× XO(ν, λ, d− i).

A cell (y, x) is defined to be a changeable cell for a pair of a reverse plane partition

and an XO-diagram, (R, γ), if γ(y, x) = X or γ(y, x) = O and Φ(E)(y, x) = y − 1.

Example. For R =21 1 11 1 1 2

and γ =O

OX

: Φ(R) =12 1 11 1 1 2

(1, 4) is a changeable cell since it is an X cell.

(2, 3) is a changeable cell since it is an O cell and Φ(R)(2, 3) = 1.

(3, 1) is NOT a changeable cell since it is an O cell and Φ(R)(3, 1) = 1.

Define SM(λ, d) to be the set of pairs of all reverse plane partitions and XO-diagrams

of M(λ, d), such that they contain no changeable cells. Or equivalently,

SM(λ, d) = (R, γ) ∈M(λ, d) : (R, γ) contain no changeable cells .

Lemma 12. SM(λ, d) =⋃

ν∈A(λ,d)

SRPP(ν, λ)× XO(ν, λ, d) =⋃ν∈A(λ,d)

SRPP(ν, λ)× O(ν/λ), where O(ν/λ) ∈ XO(ν, λ, 0), making it the filling of

ν/λ, with each cell in the skew-shape being an O cell.

Proof. Let (R, γ) ∈M(λ, d), it is necessary to determine when (R, γ) has no change-

able cells. Or equivalently, when (R, γ) ∈ SM(λ, d). First, if sh(R) /∈ A(λ, d),

then γ must contain an X cell, since |sh(R)/λ|< d. An X cell is a changeable cell,

therefore (R, γ) /∈ SM(λ, d). If sh(R) ∈ A(λ, d) and R /∈ SRPP(sh(R), λ), then

(R, γ) must have at least one cell in ν/λ that contains the letter i− 1, where i is the

row of that cell. This cell is an O cell, hence it is also a changeable cell, implying


that (R, γ) /∈ SM(λ, d). Therefore (R, γ) ∈ M(λ, d), where sh(R) ∈ A(λ, d) and

R ∈ SRPP(ν, λ). It is important to see that γ is fixed to one XO-diagram by ν and

λ, since ν/λ is a horizontal d-strip, therefore γ is simply ν/λ filled with O cells, or

γ = O(ν/λ).

For a fixed partition λ and a fixed positive number d, Π is a mapping from the

set M(λ, d) to itself. For (R, γ) ∈ M(λ, d) denote µ = sh(R). This implies that all

the X cells in γ are Y cells for (µ, λ), and the O cells form the shape µ/λ, which is

horizontal strip of some length t (where 0 ≤ t ≤ d). Π is defined using the following

steps:

1. Find the southeast-most changeable cell and record (i, j) to be that cell.

2. Denote R = Φ(R)

(a) If (i, j) is an X cell, then define ν = (µ1, . . . , µi+1 + 1, . . . , µ`(µ)). ν is a

valid partition, because µi > µi+1, due to the fact that (i, j) is an X cell,

thus a Y cell for (µ, λ).

Define S as follows: S(y, x) = R(y, x) ∀(y, x) : (y, x) 6= (i + 1, νi+1), and

S(i + 1, νi+1) = i. S ∈ SSYT(ν) with alphabet Nn since sh(S) = sh(ν)

and R ∈ SSYT(µ) and S(i, νi+1) ≤ i− 1 and S(i + 1, µi+1) ≤ i from the

definition of Φ.

Define ω as follows: ω(y, x) = γ(y, x) ∀(y, x) : (y, x) 6= (i, j) or (y, x) 6=

(i+1, νi+1). ω(i, j) is empty and ω(i+1, νi+1+1) = O. ω ∈ XO(sh(ω), λ, d−

|µ/λ|−1), since sh(ω) ∈ A(λ, |µ/λ|+1), because (i + 1, νi+1 + 1) is an

addable cell to µ, due to the fact that (i, j) is a Y cell for (µ, λ).

Verbally this step can be understood as follows: remove the X cell (i, j)

from γ and insert an O cell in γ at the end of the row above it at cell

(i+ 1, sh(R)i+1 + 1). If there is no row above it, then Π just places the O


cell in the first location, or cell (i+1, 1). Now insert an i as a newly added

cell in the same location as the newly added O cell into R.

(b) If (i, j) is an O cell, then i > 1, since R(i, j) = i− 1. Define ν =

(µ1, . . . , µi − 1, . . . , µ`(µ)). ν is a valid partition, because (i, j) is a re-

movable cell for µ, due to the fact that it is an O cell, which implies that

(i, j) ∈ µ/λ, where µ/λ ∈ A(λ, |µ/λ|). Also, (i, j) is the east-most O cell

in row i, because R(i, j) = i− 1 and the east-most such cell was chosen.

Define S as follows: S(y, x) = R(y, x) ∀(y, x) : (y, x) 6= (i, j), where S(i, j)

is removed. S ∈ SSYT(ν) with alphabet Nn, since sh(S) = sh(ν) and

R ∈ SSYT(µ) and (i, j) is a removable cell for µ.

Define ω as follows: ω(y, x) = γ(y, x) ∀(y, x) : (y, x) 6= (i, j) or (y, x) 6=

(i−1, λi−1). ω(i, j) is removed and ω(i−1, λi−1) = X. ω ∈ XO(sh(ω), λ, d−

|µ/λ|+1), since sh(ω) ∈ A(λ, |µ/λ|−1), because (i, j) is a removable cell

of µ. Also, (i − 1, λi−1) is a Y cell for (ν, λ), since νi < νi−1, but not an

X cell because if γ(i − 1, λi−1) = X, then it would be a changeable cell,

which is more southeast than (i, j).

Verbally this step can be understood as follows: Remove both the O cell

from γ and the i− 1 from R. Then place an X in γ in the allowable cell

(i− 1, λi−1) in the row below.

3. Π(R, γ) = (Φ−1(S), ω), where S and ω are defined in the step above.

If there are no changeable cells for (R, γ), then (R, γ) is a fixed point for Π.

Example. Here is an example of using the mapping Π where λ = (3, 3, 2, 2) and

t = 3.

Let (R, γ) =

2 22 2 31 2 21 1 1 3

,

XO

O

Φ(R) = R =

3 33 1 22 2 11 1 2 3

.

Step 1: (3, 3) is the southeast most changeable cell.


Step 2: (3, 3) is an O cell so (S, ω) =

3 33 12 2 11 1 2 3

,

X

XO

.

Step 3: Π(R, γ) = (S, ω) = (Φ−1(S), ω) =

2 32 31 2 21 1 2 3

,

X

XO

.

Example. Below is an example that shows that Π is its own inverse. Again, let

λ = (3, 3, 2, 2) and t = 3.

Let (R, γ) =

2 32 31 2 21 1 2 3

,

XO

O

Φ(R) = R =

3 33 12 2 11 1 2 3

.

Step 1: (2, 3) is the southeast most changeable cell.

Step 2: (2, 3) is an X cell so (S, ω) =

3 33 1 22 2 11 1 2 3

,

X

XO

.

Step 3: Π(R, γ) = (S, ω) = (Φ−1(S), ω) =

2 22 2 31 2 21 1 1 3

,

XO

O

.

So from the two examples above the following was shown:

Π

2 22 2 31 2 21 1 1 3

,

XO

O

=

2 32 31 2 21 1 2 3

,

X

XO

and

Π

2 32 31 2 21 1 2 3

,

X

XO

=

2 22 2 31 2 21 1 1 3

,

XO

O

.

In this next example, the reverse plane partition and the XO-diagram are a fixed

point for Π.

Example. For λ = (4, 3, 2, 2, 1) and t = 4.

Let (R, γ) =

21 21 2 21 1 1 31 1 1 2 2 2

, OO

O O

, R = Φ(R) =

32 31 1 12 2 2 31 1 1 2 2 2

.

There are no changeable cells, therefore there are no changes made.

Π

21 21 2 21 1 1 31 1 1 2 2 2

, OO

O O

=

21 21 2 21 1 1 31 1 1 2 2 2

, OO

O O

.


Π mapped (R, γ) to itself, hence (R, γ) is a fixed point for Π.

Lemma 13. Given (R, γ) ∈ M(λ, d), if Π(R, γ) = (S, ω), then wt(R) = wt(S). Or

equivalently, Π is a weight preserving mapping.

Proof. Remark 3 implies that wt(R) = wt(Φ(R)) and wt(S) = wt(Φ(S)). It remains

to show that wt(Φ(R)) = wt(Φ(S)). Φ(S)(y, x) = Φ(R)(y, x) ∀(y, x) except one

cell (i, j), from the definition of Π. Φ(S)(i, j) = i− 1 and Φ(R)(i, j) is blank or

Φ(R)(i, j) = i− 1 and Φ(S)(i, j) is blank, either way this letter does not affect the

weight, hence wt(Φ(R)) = wt(Φ(S)).

Lemma 14. Given (R, γ) ∈ M(λ, d), Π(Π(R, γ)) = (R, γ). Or equivalently, Π is an

involution.

Proof. Given (R, γ) ∈ M(λ, d), let sh(R) = µ. There are three different cases for

(R, γ).

Case I: (i, j) is the southeast-most changeable cell and it is an X cell, then

Π(R, γ) = (S, ω), where sh(S) = ν = (µ1, . . . , µi+1 + 1, . . . , µ`(µ)).

Φ(S) = Φ(R) ∀(y, x) : (y, x) 6= (i+ 1, νi+1), where Φ(S)(i+ 1, νi+1) = i.

ω(y, x) = γ(y, x) ∀(y, x) : (y, x) 6= (i, j) or (y, x) 6= (i+1, νi+1), where ω(i, j) is empty

and ω(i+ 1, νi+1 + 1) = O.

The cell (i+1, νi+1) is the southeast-most changeable cell for (S, ω). This can be seen

from the following facts:

1. There is no X cell south of (i, j), otherwise it would have been chosen for (R, γ).

2. Row i of ω does not contain an X cell, since an x cell (i, j) was removed from

γ and there can only be one Y cell per row.

3. There are no changeable O cells southeast of (i+ 1, νi+1), otherwise they would

have been chosen for (R, γ).


4. Lastly, (i + 1, νi+1) is a changeable cell for (S, ω) since ω(i + 1, νi+1) = O and

Φ(S)(i+ 1, νi+1) = i.

Let Π(S, ω) = (T, θ). Because (i+ 1, νi+1) is an O cell, the following is true:

1. sh(T ) = κ = (ν1, . . . , νi+1 − 1, . . . , ν`(ν)) = (µ1, . . . , µi+1 + 1 − 1, . . . , µ`(µ)),

therefore κ = µ.

2. Φ(T )(y, x) = Φ(S)(y, x) ∀(y, x) : (y, x) 6= (i+1, νi+1) and Φ(S) = Φ(R) ∀(y, x) :

(y, x) 6= (i + 1, νi+1), therefore Φ(T )(y, x) = Φ(R)(y, x) ∀(y, x) : (y, x) 6= (i +

1, νi+1) and Φ(T )(i+ 1, νi+1) is removed, so Φ(T ) = Φ(R), thus T = R.

3. θ(y, x) = ω(y, x) ∀(y, x) : (y, x) 6= (i + 1, νi+1) or (y, x) 6= (i, λi) and ω(y, x) =

γ(y, x) ∀(y, x) : (y, x) 6= (i, j) or (y, x) 6= (i+ 1, νi+1), therefore θ(y, x) = γ(y, x)

∀(y, x) : (y, x) 6= (i, j) or (y, x) 6= (i+1, νi+1), because cell λi = j. θ(i+1, νi+1))

is removed and θ(i, λi−1) = X, thus θ = γ.

It was shown that T = R and θ = γ, this proves that Π(S, ω) = (R, γ), therefore

Π(Π(R, γ)) = (R, γ).

Case II: (i, j) is the southeast-most changeable cell and it is an O cell, then Π(R, γ) =

(S, ω), where sh(S) = ν = (µ1, . . . , µi − 1, . . . , µ`(µ)).

Φ(S) = Φ(R) ∀(y, x) : (y, x) 6= (i, j), where S(i, j) is removed.

ω(y, x) = γ(y, x) ∀(y, x) : (y, x) 6= (i, j) or (y, x) 6= (i − 1, λi−1), where ω(i, j) is

removed and ω(i− 1, λi−1) = X.

The cell (i − 1, λi−1) is the southeast-most changeable cell for (S, ω). This can be

seen from the following facts:

1. There are no X cells south of (i − 1, λi−1), otherwise they would have been

chosen for (R, γ).


2. There are no changeable O cells southeast of (i− 1, λi−1), otherwise they would

have been chosen for (R, γ).

3. Lastly, (i− 1, λi−1) is a changeable cell for (S, ω), since ω(i− 1, λi−1) = X.

Let Π(S, ω) = (T, θ). Because (i− 1, λi−1) is an X cell, the following is true:

1. sh(T ) = κ = (ν1, . . . , νi + 1, . . . , ν`(ν)) = (µ1, . . . , µi − 1 + 1, . . . , µ`(µ)), therefore

κ = µ.

2. Φ(T )(y, x) = Φ(S)(y, x) ∀(y, x) : (y, x) 6= (i, νi), Φ(S) = Φ(R) ∀(y, x) : (y, x) 6=

(i, j), and j = νi, therefore Φ(T )(y, x) = Φ(R)(y, x) ∀(y, x) : (y, x) 6= (i, j), and

Φ(T )(i, j) is i− 1, so Φ(T ) = Φ(R), thus T = R.

3. θ(y, x) = ω(y, x) ∀(y, x) : (y, x) 6= (i, νi) or (y, x) 6= (i − 1, λi−1), ω(y, x) =

γ(y, x) ∀(y, x) : (y, x) 6= (i, j) or (y, x) 6= (i + 1, νi+1), and j = νi, therefore

θ(y, x) = γ(y, x) ∀(y, x) : (y, x) 6= (i, j) or (y, x) 6= (i+ 1, νi+1). θ(i+ 1, νi+1)) is

removed and θ(i, λi−1) = X, thus θ = γ.

It was shown that T = R and θ = γ, this proves that Π(S, ω) = (R, γ), therefore

Π(Π(R, γ)) = (R, γ).

Case III: (R, γ) does not contain any changeable cells. Then (R, γ) is a fixed point

for Π, thus Π(R, γ) = (R, γ), implying that Π(Π(R, γ)) = (R, γ).

Lemma 15. Π(R, γ) = (R, γ) if and only if (R, γ) ∈ SM(λ, d).

Proof. If (R, γ) /∈ SM(λ, d), then (R, γ) must contain a changeable cell by definition

of SM(λ, d). If (R, γ) ∈ SM(λ, d), then (R, γ) does not contain a changeable cell and

therefore (R, γ) is a fixed point for Π.

Lemma 16. For (R, γ) ∈M(λ, d), If Π(R, γ) = (R, γ) 6= (R, γ), then∣∣∣|sh(R)|−|sh(R)|∣∣∣ = 1.


Proof. Lemma 15 states that if (R, γ) is not a fixed point for Π, then R /∈ SM(λ, d).

This means that there is at least one changeable cell for (R, γ). Π will change the

lowest changeable cell either by removing an O cell and adding an X cell or by

removing an X cell and adding an O cell. Therefore Π will either remove or add a

single cell to the shape of R.

Lemma 17. Let (R, γ) ∈ Ω(λ, d). If Π(R, γ) = (R, γ) 6= (R, γ), then

(−1)d−|sh(R)/λ|xwt(R) + (−1)d−|sh(R)/λ|xwt(R) = 0.

Proof. From Lemma 16 it is obvious that∣∣∣|sh(R)/λ|−|sh(R)/λ|

∣∣∣ = 1. Lemma 13

states that wt(R) = wt(R). Therefore (−1)d−|sh(R)/λ|xwt(R) = (−1)(−1)d−|sh(R)/λ|xwt(R).

Lemma 18. ∑(T,D)∈M(λ,t)/SM(λ,d)

(−1)|sh(T )/λ|xwt(T ) = 0.

Proof. Define P (λ, t) and N(λ, t) as follows:

P (λ, t) =⋃

(T,D) ∈M(λ, t)/SM(λ, t) : (−1)t−|sh(T )/λ| = 1,

N(λ, t) =⋃

(T,D) ∈M(λ, t)/SM(λ, t) : (−1)t−|sh(T )/λ| = −1.

Lemma 16 implies that if (R, γ) ∈ P (λ, t), then Π(R, γ) ∈ N(λ, t), or conversely if

(R, γ) ∈ N(λ, t), then Π(R, γ) ∈ P (λ, t). This follows from the fact that (R, γ) /∈

SM(λ, t). Lemma 17 implies that if (R, γ) ∈ M(λ, t)/SM(λ, t) and Π(R, γ) =

(S, ω), then (−1)t−|sh(R)/λ|xwt(R) + (−1)t−|sh(S)/λ|xwt(S) = 0. These facts and be-

cause Π is a bijection (Lemma 14) imply that∑

(T,D)∈P (λ,t) (−1)t−|sh(T )/λ|xwt(T ) +∑(T,D)∈N(λ,t) (−1)t−|sh(T )/λ|xwt(T ) = 0. Obviously P (λ, t)∪N(λ, t) = M(λ, t)/SM(λ, t).

Therefore∑

(T,D)∈M(λ,t)/SM(λ,d) (−1)t−|sh(T )/λ|xwt(T ) = 0.


The sign changing involution Π and the lemmas above concerning Π allow the follow-

ing elegant combinatorial proof.

4.4 Combinatorial proof of the dual Grothendieck Pieri rule

Before proving the rule one simple remark is needed.

Remark 5.

∑(R,γ)∈SM(λ,t)

xwt(R) +∑

(R,D)∈M(λ,t)/SM(λ,d)

(−1)t−|sh(R)/λ|xwt(R)

=∑

(R,γ)∈M(λ,t)

(−1)t−|sh(R)/λ|xwt(R).

What follows is a combinatorial proof of (3.4.5).

gλ gt =t∑

j=0

∑µ∈A(λ,j)

∑F∈XO(λ,µ,t−j)

(−1)t−j gµ.

Proof. In Theorem 4 it was proved that the product of two dual Grothendieck poly-

nomials, one is indexed by an arbitrary partition λ, and the other is indexed by a

nonnegative integer can be written as

gλgt =∑

ν∈A(λ,t)

∑R∈SRPP(ν,λ)

xwt(R).

Using the fact from Lemma 12 that SM(λ, d) =⋃

ν∈A(λ,d)

SRPP(ν, λ)× O(ν/λ), the

product above can be rewritten as

gλgt =∑

(R,γ)∈SM(λ,t)

xwt(R).

5 FACTORIAL SCHUR FUNCTIONS AND THEIR EXPANSION 42

Lemma 18 states that∑

(R,D)∈M(λ,t)/SM(λ,d)

(−1)t−|sh(R)/λ|xwt(R) = 0. This combined with

Remark 5 allows the following to be concluded,

gλgt =∑

(R,γ)∈M(λ,t)

(−1)t−|sh(R)/λ|xwt(R).

The definition of M(λ, t) develops this further.

gλ gt =t∑

j=0

∑µ∈A(λ,j)

∑(R,F )∈RPP(µ)×XO(µ,λ,t−j)

(−1)t−jxwt(R)

=t∑

j=0

∑µ∈A(λ,j)


∑R∈RPP(ν)

(−1)t−jxwt(R).

Finally, from the definition of dual Grothendieck polynomials the following is reached

gλ gt =t∑

j=0

∑µ∈A(λ,j)


(−1)t−jgµ.

The elegant proof of the dual Grothendieck Pieri rule used the bijection Φ to

convert reverse plane partitions to semi-standard Young tableaux with a new alpha-

bet, then the RSK algorithm paired with the sign changing involution Π was used to

complete the proof.

5 Factorial Schur functions and their expansion

When limited to n variables, Schur functions can be generalized to the factorial

Schur functions. Factorial Schur functions are defined with two sets of variables

x1, x2, . . . , xn and a1, a2, . . . and are defined using a product of differences. Fac-

torial Schur functions were first discovered by Biedenharn and Louck when decom-


posing tensor products of representation using a particular bases:

bl(λ) =∑

S∈SSYT(λ)

∏(i,j)∈S

xS(i,j) + i− j − 1 [2]. These polynomials would later be gen-

eralized by replacing the fixed term to an new arbitrary set of variables aj−i:

fs(λ) =∑

S∈SSYT(λ)

∏(i,j)∈S

xS(i,j) − ai−j. These new factorial Schur functions are special

cases of double Schubert polynomials for Grassmannian permutations [14; 13]. Later

factorial Schur functions were shown by Knutson and Tao to be the equivariant co-

homology of Grassmannians [9]. Chen and Louck, and later Goulden and Hamel

continued discovering further uses for factorial Schur functions [4; 7].

Factorial Schur functions, as shown above, play an important role in many math-

ematical fields, and because of this it is important to study these polynomials and

to truly understand their structure. Just as examining the results of multiplying

two Schur functions is important, understanding the product of two factorial Schur

function is also very useful, particularly when trying to understand the equivariant

cohomology of Grassmannians [9]. Molev and Sagan found a Littlewood-Richardson

rule for factorial Schur functions as well as other useful results [17]. The change of

basis formula for expanding factorial Schur functions in terms of Schur functions and

many more interesting properties of factorial Schur functions were later discovered

by Krieman [10]. There is an easy combinatorial method for computing the coeffi-

cients for expanding factorial Schur functions in terms of Schur functions which Molev

discovered [16].

5.1 Definition of a factorial Schur polynomial

For a cell z = (y, x) in a tableau, define the content of the cell to be c(z) = x− y.

Example. For a tableau T =

43 42 3 31 2 2 3

the corresponding Ferrers diagram filled


with the content of each cell is

−3−2−1−1 0 10 1 2 3

.

For a semi-standard Young tableau T ∈ SSYT(λ), a enumeration of T that will used

in the definition of factorial Schur functions is as follows:

fsp(T ) =∏z∈T

(xT (z) − ac(z)+T (z)).

Example. If T =32 31 2

, then

fsp(T ) = (x3 − a3−2)(x2 − a2−1)(x3 − a3+0)(x1 − a1+0)(x2 − a2+1)

= (x3 − a1)(x2 − a1)(x3 − a3)(x1 − a1)(x2 − a3).

Definition. Factorial Schur function is defined as a sum of the enumerations

described above over all semi-standard Young tableaux with a fixed alphabet Nn.

fsλ =∑

T∈SSYT(λ)

fsp(T ) =∑

T∈SSYT(λ)

∏Z∈T

(xT (Z) − ac(Z)+T (Z)). (5.1.1)

Example. For n = 3 and λ = (2, 1) = , the semi-standard Young tableaux of

the shape λ can be written as follows

SSYT(λ) =

21 1

, 21 2

, 31 1

, 31 3

, 32 2

, 32 3

, 31 2

, 21 3

,


and the factorial Schur function is

fsλ = (x2 − a1)(x1 − a1)(x1 − a2) + (x2 − a1)(x1 − a1)(x2 − a3)

+ (x3 − a2)(x1 − a1)(x1 − a2) + (x3 − a2)(x1 − a1)(x3 − a4)

+ (x3 − a2)(x2 − a2)(x2 − a3) + (x3 − a2)(x2 − a2)(x3 − a4)

+ (x3 − a2)(x1 − a1)(x2 − a3) + (x2 − a1)(x1 − a1)(x3 − a4).

After expanding the products fsλ takes the form

fsλ = x21x2 − a2x1x2 − a1x1x2 − a1x21 + a1a2x2 + a1a2x1 + a21x1 − a21a2

+ x1x22 − a3x1x2 − a1x22 − a1x1x2 + a1a3x2 + a1a3x1 + a21x2 − a21a3

+ x21x3 − a2x1x3 − a1x1x3 − a2x21 + a1a2x3 + a22x1 + a1a2x1 − a1a22

+ x1x23 − a4x1x3 − a1x23 − a2x1x3 + a1a4x3 + a2a4x1 + a1a2x3 − a1a2a4

+ x22x3 − a3x2x3 − a2x2x3 − a2x22 + a2a3x3 + a2a3x2 + a22x2 − a22a3

+ x2x23 − a4x2x3 − a2x23 − a2x2x3 + a2a4x3 + a2a4x2 + a22x3 − a22a4

+ x1x2x3 − a3x1x3 − a1x2x3 − a2x1x2 + a1a3x3 + a2a3x1 + a1a2x2 − a1a2a3

+ x1x2x3 − a4x1x2 − a1x2x3 − a1x1x3 + a1a4x2 + a1a4x1 + a21x3 − a21a4.

Some terms in the expansion need to be rearranged.

fsλ = x21x2 + x1x22 + x21x3 + x1x

23 + x22x3 + x2x

23 + x1x2x3 + x1x2x3

− a1x21 − a2x21 − a1x1x2 − a2x1x2 − a1x1x3 − a2x1x3

− a1x22 − a2x22 − a1x2x3 − a2x2x3 − a2x23 − a1x23

− a1x1x3 − a2x1x3 − a3x1x3 − a4x1x3


− a1x2x3 − a2x2x3 − a3x2x3 − a4x2x3

+ a21x1 + a1a2x1 + a1a2x1 + a1a3x1 + a1a4x1 + a22x1 + a2a3x1 + a2a4x1



− a21a2 − a21a3 − a1a22 − a1a2a4 − a22a3 − a22a4 − a1a2a3 − a21a4.

In the expression above certain terms clearly can be factored.

fsλ = (1)(x21x2 + x1x22 + x21x3 + x1x

23 + x22x3 + x2x

23 + x1x2x3 + x1x2x3)

− (a1 + a2)(x21 + x1x2 + x1x3 + x22 + x2x3 + x33)

− (a1 + a2 + a3 + a4)(x1x2 + x1x3 + x2x3)

+ (a21 + a1a2 + a1a2 + a1a3 + a1a4 + a22 + a2a3 + a2a4)(x1 + x2 + x3)

− (a21a2 + a1a22 + a21a3 + a1a2a3 + a22a3 + a21a4 + a1a2a4 + a22a4)(1).

Recall that for n = 3

sλ = s = x21x2 + x1x22 + x21x3 + x1x

23 + x22x3 + x2x

23 + x1x2x3 + x1x2x3

s(2) = s = x21 + x1x2 + x1x3 + x22 + x2x3 + x33

s(1,1) = s = x1x2 + x1x3 + x2x3

s(1) = s = x1 + x2 + x3

s(0) = s∅ = (1).


Therefore the factorial Schur function fsλ can be rewritten in terms of Schur functions.

fsλ = (1)s

− (a1 + a2)s

− (a1 + a2 + a3 + a4)s

+ (a21 + a1a2 + a1a2 + a1a3 + a1a4 + a22 + a2a3 + a2a4)s

− (a21a2 + a1a22 + a21a3 + a1a2a3 + a22a3 + a21a4 + a1a2a4 + a22a4)s∅.

Factorial Schur functions form a basis for Λn and hence there exists a set of coefficients

known as the change of basis coefficients. A single factorial Schur function can be

expressed as an expansion of Schur functions using these coefficients. In the example

above the expansion was an alternating sum of Schur functions whose coefficients

are polynomials of a. It is shown in [16] (4.5) that the coefficients of any expansion

of factorial Schur functions are sign alternating polynomials of a. The next section

introduces a new combinatorial object, which is necessary for computing the change

of basis coefficients.

5.2 The expansion of factorial Schur functions in terms of

Schur functions

Definition. [16] A the set of all fillings of skew-shape λ/µ with letters from the

alphabet N is denoted by FYT(λ/µ) if fillings in the set are created using the following

rules: columns are nondecreasing and rows are increasing. Column i can be filled

with the following letters: 1, . . . , n + c(z(i)), where z(i) is the highest cell contained

in column i and c(z) is the content of cell z.

Example. For n = 3, and λ = (2, 1) = the FYT(λ) is the set below. Notice,

that z(1) is the cell (2, 1) and c(z(1)) = −1, therefore column one can only contain


the letters 1 and 2, and z(2) is the cell (1, 2) and c(z(2)) = 1, therefore column two

can contain the letters 1 through 4.

FYT(λ) =

11 2

, 21 2

, 11 3

, 21 3

, 22 3

, 11 4

, 21 4

, 22 4

.

The last step in defining coefficients in factorial Schur function expansion requires

introduction of a polynomial D(λ, µ) (in Molev’s paper it is written as sλ/µ(u)),

which is the sum of the enumerations over the set FYT(λ/µ)

D(λ, µ) =∑

T∈FYT(λ/µ)

(−1)|λ/µ|awt(T ),

where wt(T ) is the weight of T as described for semi-standard young tableaux. D(λ, λ)

is defined to be 1. D(λ, µ) = 0 if µ * λ.

Example. For n = 3 and λ = (2, 1) = the D(λ, µ) is written as follows

D(λ, ∅) = (−1)3awt(

11 2

)+ (−1)3a

wt(

21 2

)+ (−1)3a

wt(

11 3

)+ (−1)3a

wt(

21 3

)

+ (−1)3awt(

22 3

)+ (−1)3a

wt(

11 4

)+ (−1)3a

wt(

21 4

)+ (−1)3a

wt(

22 4

),

or equivalently

D(λ, ∅) = −(a21a2)− (a1a22)− (a21a3)− (a1a2a3)− (a22a3)− (a21a4)− (a1a2a4)− (a22a4).

This polynomial D(λ, ∅) was the coefficient in front of s∅ when expanding fs(2,1) in

terms of Schur functions. In fact, all the polynomials D(λ, µ) are the coefficients in

the expansion of factorial Schur functions fsλ in terms of Schur functions sµ (Molev

4.5) [16].

fsλ =∑µ⊆λ

D(λ, µ)sµ. (5.2.1)


Example. For n = 3 and λ = (2, 1) = all the coefficients D(λ, µ) are listed

below:

D(λ, λ) = 1;

D(λ, (2)) = (−1)1awt(

1)

+ (−1)1awt(

2), or

D(λ, (2)) = (−1)(a1 + a2);

D(λ, (1, 1)) = (−1)1awt(

1

)+ (−1)1a

wt(

2

)+ (−1)1a

wt(

3

)+ (−1)1a

wt(

4

), or

D(λ, (1, 1)) = (−1)(a1 + a2 + a3 + a4);

D(λ, (1)) = (−1)2awt(

11

)+ (−1)2a

wt(

12

)+ (−1)2a

wt(

13

)+ (−1)2a

wt(

14

)

+ (−1)2awt(

21

)+ (−1)2a

wt(

22

)+ (−1)2a

wt(

23

)+ (−1)2a

wt(

24

), or

D(λ, (1)) = (a21 + a1a2 + a1a3 + a1a4 + a2a1 + a22 + a2a3 + a2a4);

D(λ, ∅) = (−1)3awt(

11 2

)+ (−1)3a

wt(

21 2

)+ (−1)3a

wt(

11 3

)+ (−1)3a

wt(

21 3

)

+ (−1)3awt(

22 3

)+ (−1)3a

wt(

11 4

)+ (−1)3a

wt(

21 4

)+ (−1)3a

wt(

22 4

), or

D(λ, ∅) = (−1)(a21a2 + a1a22 + a21a3 + a1a2a3 + a22a3 + a21a4 + a1a2a4 + a22a4).

Therefore, the expansion is

fsλ = D(λ, λ)sλ +D(λ, (2))s(2) +D(λ, (1, 1))s(1,1) +D(λ, (1))s(1) +D(λ, ∅)s∅

= (1)s(2,1) − (a1 + a2)s(2) − (a1 + a2 + a3 + a4)s(1,1)

+ (a21 + a1a2 + a1a3 + a1a4 + a2a1 + a22 + a2a3 + a2a4)s(1)

− (a21a2 + a1a22 + a21a3 + a1a2a3 + a22a3 + a21a4 + a1a2a4 + a22a4)s∅.

The introduction of FYT(λ/µ) gives a nice combinatorial method of expressing fsλ in

terms of Schur functions without the need of the algebraic approach of expanding the

products and factoring the Schur functions as was shown in previous section. There


is a matrix form of the expansion of factorial Schur functions. This form will be

used later to find the reverse change of basis coefficients. In order to write matrix

equivalent of factorial Schur expansion, it is necessary to define an ordering on the

set of partitions. Let λ and µ both be partitions and λ 6= µ. If |λ|< |µ| then λ < µ.

If |λ|= |µ| then for the smallest i, such that λ1 + λ2 + · · · + λi 6= µ1 + µ2 + · · · + µi,

if λ1 + λ2 + · · · + λi < µ1 + µ2 + · · · + µi then λ < µ else λ > µ. Notice, that this

ordering is well defined as long as λ 6= µ then either λ < µ or λ > µ.

Example. Below is the ordering for all partitions of 4 or less:

∅ < (1) < (1, 1) < (2) < (1, 1, 1) < (2, 1) < (3) < (1, 1, 1, 1) < (2, 1, 1) < (2, 2) <

(3, 1) < (4).

This ordering can be used as an index for all partitions. Let pi be the ith partition in

the order above. In the previous example p1 = ∅, p2 = (1), p5 = (1, 1, 1), p10 = (2, 2).

For a partition λ, let p−1λ be the number i, such that pi = λ. Following the previous

example: p−1∅ = 1, p−1(1) = 2, p−1(1,1,1) = 5, p−1(2,2) = 10. The ordering defined above

will now be used to construct a coefficient matrix of the expansion of factorial Schur

functions in terms of Schur functions.

Definition. The matrix A = (ay,x) is a p−1(n) by p−1(n) matrix defined pointwise by

ay,x = D(py, px).

Example. If n = 3, then p−1(n) = p−1(3) = 7. The matrix of coefficients of factorial Schur

expansion is

A =

D(∅, ∅) D(∅, (1)) D(∅, (1, 1)) D(∅, (2)) D(∅, (1, 1, 1)) D(∅, (2, 1)) D(∅, (3))

D((1), ∅) D((1), (1)) D((1), (1, 1)) D((1), (2)) D((1), (1, 1, 1)) D((1), (2, 1)) D((1), (3))

D((1, 1), ∅) D((1, 1), (1)) D((1, 1), (1, 1)) D((1, 1), (2)) D((1, 1), (1, 1, 1)) D((1, 1), (2, 1)) D((1, 1), (3))

D((2), ∅) D((2), (1)) D((2), (1, 1)) D((2), (2)) D((2), (1, 1, 1)) D((2), (2, 1)) D((2), (3))

D((1, 1, 1), ∅) D((1, 1, 1), (1)) D((1, 1, 1), (1, 1)) D((1, 1, 1), (2)) D((1, 1, 1), (1, 1, 1)) D((1, 1, 1), (2, 1)) D((1, 1, 1), (3))

D((2, 1), ∅) D((2, 1), (1)) D((2, 1), (1, 1)) D((2, 1), (2)) D((2, 1), (1, 1, 1)) D((2, 1), (2, 1)) D((2, 1), (3))

D((3), ∅) D((3), (1)) D((3), (1, 1)) D((3), (2)) D((3), (1, 1, 1)) D((3), (2, 1)) D((3), (3))

By using the definition of D(λ, λ) = 1 and D(λ, µ) = 0 if µ * λ A can be rewritten


as

A =

1 0 0 0 0 0 0

D((1), ∅) 1 0 0 0 0 0

D((1, 1), ∅) D((1, 1), (1)) 1 0 0 0 0

D((2), ∅) D((2), (1)) 0 1 0 0 0

D((1, 1, 1), ∅) D((1, 1, 1), (1)) D((1, 1, 1), (1, 1)) 0 1 0 0

D((2, 1), ∅) D((2, 1), (1)) D((2, 1), (1, 1)) D((2, 1), (2)) 0 1 0

D((3), ∅) D((3), (1)) 0 D((3), (2)) 0 0 1

After writing explicitly each polynomial D(λ, µ), the matrix A takes the form

A =

1 0 0 0 0 0 0

−(a1 + a2 + a3) 1 0 0 0 0 0

a21 + a1a2 + a22 −(a1 + a2) 1 0 0 0 0

a1a2 + a1a3 + a1a4 + a2a3 + a2a4 −(a1 + a2 + a3 + a4) 0 1 0 0 0

a31 a1a2 −a1 0 1 0 0

−1(a21a2+a1a2

2+a21a3+a1a2a3

+a22a3+a2

1a4+a1a2a4+a22a4)

a21+a1a2+a1a3a1a4

a2a1+a22+a2a3+a2a4

−(a1+a2+a3+a4)

−(a1 + a2) 0 1 0

−1(a1a2a4+a1a2a4+a2a3a4+a1a2a5a1a3a5+a2a3a5+a1a4a5+a2a4a5+a3a4a5

a1a2+a1a3+a2a3+a1a4+a2a4a3a4+a1a5+a2a5+a3a5+a4a4

0−(a1+a2+a3

+a4+a5)0 0 1

Lemma 19. Matrix A for any set n is lower triangular and has 1s on the diagonal.

Proof. ai,i = D(pi, pi) = 1 from the definition of D(λ, λ). Entries ai,j = D(pi, pj), if

j > i then pj > pi by the definition of the ordering. Consider two possible cases:

Case I: |pj|> |pi|. Then pj * pi and therefore D(pi, pj) = 0;

Case II: |pj|= |pi|. But then pj * pi since pi 6= pj and therefore D(pi, pj) = 0.

For a fixed n let S =

sp1

sp2...

s(n)

, the matrix S is a p−1(n) by 1 matrix, since pp−1

(n)= (n).

Also let FS =

fsp1

fsp2...

fs(n)

.

Then the matrix form of the factorial Schur expansion can be written as FS = A∗S,

6 A REVERSE CHANGE OF BASIS 52

or equivalently

fsp1

fsp2...

fs(n)

=

1 0 · · · 0

D((1), ∅) 1 · · · 0

......

. . . 0

D((n), ∅) D((n), (1)) · · · 1

sp1

sp2...

s(n)

.

6 A reverse change of basis

Since factorial Schur functions form a basis, it follows that Schur functions can be

expressed as a sum of factorial Schur functions. What is surprising is that the co-

efficients of these expansions are positive polynomials of a. Kreiman discusses these

coefficients and refers to rather complicated methods of finding them [10]. This sec-

tion presents a simple method of describing the reverse change of basis coefficients as

well as an elegant proof. This is accomplished by first introducing a new combinato-

rial object very similar to a counter semi-standard Young tableaux. These objects are

enumerated and then added together to yield the entries of a new matrix B. A sign

reserving involution is then created to show that Matrix B is the inverse of matrix

A. Matrix multiplication is used to finish the proof giving the explicit expansion of

Schur functions in terms of factorial Schur functions.

6.1 Change of basis coefficients

Definition. HYT(λ/µ) is the set of all fillings of shape λ/µ, such that T ∈ HYT(λ/µ)

is constructed following these rules: rows are non increasing and columns are decreas-

ing, and a cell z = (y, x) can only contain letters: 1, . . . , n+ c(z).


Example. Let n = 3 and λ = (2, 1) and µ = ∅. Then the set HYT(λ/µ) is

HYT(λ) =

12 1

, 12 2

, 13 1

, 13 2

, 13 3

, 23 1

, 23 2

, 23 3

.

Similarly to the polynomial D(λ, µ), defined in the previous section, the sum of the

enumerations over the set HYT(λ/µ) is defined as

H(λ, µ) =∑

T∈HYT(λ/µ)

awt(T ),

where wt(T ) is the weight of T as described for semi-standard young tableaux.

H(λ, λ) is defined to be 1. H(λ, µ) = 0 if µ * λ.

Example. Let n = 3 and λ = (2, 1) and µ = ∅, then

H(λ, µ) = H(λ, ∅) = awt(

12 1

)+ a

wt(

12 2

)+ a

wt(

13 1

)+ a

wt(

13 2

)

+ awt(

13 3

)+ a

wt(

23 1

)+ a

wt(

23 2

)+ a

wt(

23 3

),

or equivalently

H(λ, ∅) = a21a2 + a1a22 + a21a3 + a1a2a3 + a1a

23 + a1a2a3 + a22a3 + a2a

23.

The coefficients H(λ, µ) are used as entries of the matrix, which will be proven to be

the inverse of the matrix of coefficients of factorial Schur expansion, A.

Definition. The matrix B = (by,x) is a p−1(n) by p−1(n) matrix defined pointwise by

by,x = H(py, px).

Example. For n = 3


B =

H(∅, ∅) H(∅, (1)) H(∅, (1, 1)) H(∅, (2)) H(∅, (1, 1, 1)) H(∅, (2, 1)) H(∅, (3))

H((1), ∅) H((1), (1)) H((1), (1, 1)) H((1), (2)) H((1), (1, 1, 1)) H((1), (2, 1)) H((1), (3))

H((1, 1), ∅) H((1, 1), (1)) H((1, 1), (1, 1)) H((1, 1), (2)) H((1, 1), (1, 1, 1)) H((1, 1), (2, 1)) H((1, 1), (3))

H((2), ∅) H((2), (1)) H((2), (1, 1)) H((2), (2)) H((2), (1, 1, 1)) H((2), (2, 1)) H((2), (3))

H((1, 1, 1), ∅) H((1, 1, 1), (1)) H((1, 1, 1), (1, 1)) H((1, 1, 1), (2)) H((1, 1, 1), (1, 1, 1)) H((1, 1, 1), (2, 1)) H((1, 1, 1), (3))

H((2, 1), ∅) H((2, 1), (1)) H((2, 1), (1, 1)) H((2, 1), (2)) H((2, 1), (1, 1, 1)) H((2, 1), (2, 1)) H((2, 1), (3))

H((3), ∅) H((3), (1)) H((3), (1, 1)) H((3), (2)) H((3), (1, 1, 1)) H((3), (2, 1)) H((3), (3))

.

Using the definition of H(λ, λ) = 1 and H(λ, µ) = 0 if µ * λ, B can be written

as

B =

1 0 0 0 0 0 0

H((1), ∅) 1 0 0 0 0 0

H((1, 1), ∅) H((1, 1), (1)) 1 0 0 0 0

H((2), ∅) H((2), (1)) 0 1 0 0 0

H((1, 1, 1), ∅) H((1, 1, 1), (1)) H((1, 1, 1), (1, 1)) 0 1 0 0

H((2, 1), ∅) H((2, 1), (1)) H((2, 1), (1, 1)) H((2, 1), (2)) 0 1 0

H((3), ∅) H((3), (1)) 0 H((3), (2)) 0 0 1

.

Each polynomial H(λ, µ) can be written explicitly in terms of a, therefore B takes

the following form

B =

1 0 0 0 0 0 0

a1 + a2 + a3 1 0 0 0 0 0

a1a2 + a1a3 + a2a3 a1 + a2 1 0 0 0 0

a21+a1a2+a1a3+a1a4+a2

2+a2a3+a2a4+a2

3+a3a4

a1 + a2 + a3 + a4 0 1 0 0 0

a1a2a3 a1a2 a1 0 1 0 0

a21a2+a1a2

2+a21a3+a1a2a3

+a1a23+a1a2a3+a2

2a3+a2a23

a21+a1a2+a1a3+a1a4

+a2a1+a22+a2a3+a2a4

a1+a2+a3+a4

a1 + a2 0 1 0

a31+a2

1a2+a1a22+a3

2+a21a3

+a1a2a3+a22a3+a1a2

3+a2a23+a3

3

a21+a1a2+a2

2+a1a3+a2a3+a2

3+a1a4+a2a4+a3a4+a24

0a1+a2+a3+a4+a5

0 0 1

.

Lemma 20. Matrix B for any set n is lower triangular and has 1s on the diagonal.

Proof. This proof is identical to the proof of Lemma 19.

For the matrix B to be the reverse change of basis matrix, it needs to be the inverse

of matrix A. To prove that this is indeed the case it is necessary to construct an involu-

tion on a set of pairs (S, T ) ∈ C(λ, µ), where C(λ, µ) =⋃

ν:µ⊆ν⊆λ

FYT(λ, ν)× HYT(ν, µ)

for partitions λ and µ, such that µ ⊆ λ and λ 6= µ.


6.2 A combinatorial involution

For fixed partitions λ and µ, such that µ ⊆ λ, define a mapping from C(λ, µ) to itself,

∆, using the following procedure.

Let (S, T ) ∈ C(λ, µ), this means that S ∈ FYT(λ, ν) for some ν : µ ⊆ ν ⊆ λ and

T ∈ HYT(ν, µ); let t be the smallest letter in either S and T ; and let z = (y, x) be

the southeast-most cell in S or T that contains the letter t. Consider two cases.

Case I: z ∈ S. In this case (y − 1, x) /∈ S, because otherwise S(y − 1, x) ≤ S(y, x)

by the definition of S ∈ FYT(λ, ν), and z can not be the southeast-most cell in S.

Also from the definition of S ∈ FYT(λ, ν), (y, x− 1) /∈ S. Therefore z is an addable

cell for ν. Moreover, since z contains the letter t, then by definition of HYT(ν, µ), if

(y − 1, x) ∈ T , then T (y − 1, x) ≥ t, and if (y, x− 1) ∈ T then T (y − 1, x) > t, thus

z can be placed into T . Therefore ∆ switches the cell z from S to T .

Case II: z ∈ T . Then (y + 1, x) /∈ T , otherwise, by the definition of HYT(ν, µ),

T (y + 1, x) < T (y, x), and (y, x+ 1) /∈ T , otherwise T (y, x+ 1) ≤ T (y, x). Therefore

z is a removable cell for ν. Also, since z contains the letter t, it can be placed into S,

since if (y+ 1, x) ∈ S, then S(y+ 1, x) ≤ t, and if (y, x+ 1) ∈ S, then S(y, x+ 1) < t.

Therefore ∆ switches the cell z from T to S.

Example. Let n = 3, λ = (3, 3, 3), µ = (1), ν = (2, 2). First, consider the situation,

described in the Case I.

Let S =1 2 3

31

, and T = 2 12

. Then t = 1, since it is the smallest letter in S and

T . And z = (1, 3) in S, since it is southeast most cell containing the letter 1.

Then ∆

(1 2 3

31

, 2 12

)=

(1 2 3

3 ,2 1

2 1

).

Next, consider the Case II.

Let S =1 2 3

3 , and T = 2 12 1

. Then t = 1 since it is the smallest letter in S and


T . And z is cell located at spot (1, 3) in T , since it is southeast most cell containing

the letter 1.

Then ∆

(1 2 3

3 ,2 1

2 1

)=

(1 2 3

31

, 2 12

).

Lemma 21. ∆ is an involution. Or equivalently, ∆(∆(S, T )) = (S, T ) ∀(S, T ) ∈

C(λ, µ).

Proof. Let (S, T ) ∈ C(λ, µ). Let t be the smallest letter in either S or T , and cell

z = (y, x) be the southeast-most cell containing the letter t. Denote ∆(S, T ) = (S, T ).

If (i, j) ∈ sh(λ) and (i, j) 6= (y, x) then either S(i, j) = S(i, j) or T (i, j) = T (i, j).

Thus, the smallest letter in either S or T is t. This means, that cell z = (y, x) is

still the southeast-most cell containing the letter t in S or T by the definition of sets

FYT(λ, ν) and HYT(ν, µ). Therefore the map ∆ switches z ∈ (S, T ) to it’s original

position in (S, T ). This proves that ∆(∆(S, T )) = ∆(S, T ) = (S, T ).

Lemma 22. For (S, T ) ∈ C(λ, µ), if ∆(S, T ) = (S, T ), then

(−1)|sh(S)|awt(S)awt(T ) = (−1)(−1)|sh(S)|awt(S)awt(T ).

Proof. First, notice that ∆ only shifts one cell from S to T or T to S and does not

change the letter contained in that cell. Therefore awt(S)awt(T ) = awt(S)awt(T ). To

prove the statement of the Lemma it remains to show that ||sh(S)|−|sh(S))||= 1.

Consider two cases.

Case I: ∆ moves a cell from S to T . Then sh(S) = λ − µ − sh(T ) and sh(S) =

λ− µ− sh(T ) = λ− µ− (sh(T ) + 1), so |sh(S)− sh(S)|= 1.

Case II: ∆ moves a cell from T to S. Then sh(S) = λ − µ − sh(T ) and sh(S) =

λ− µ− sh(T ) = λ− µ− (sh(T )− 1), so |sh(S)− sh(S)|= 1.


6.3 Reverse change of basis

Theorem 5. Matrix B is the inverse of the matrix A. Or equivalently, AB = BA =

I.

Proof. Let C = AB. First, notice that from Lemma 19 and Lemma 20 it follows that

C is a lower triangular matrix and has 1s on the diagonal, hence ci,i = 1 and cj,i = 0

if i > j. All that is remaining to show is that cj,i = 0 if i < j. This follows from the

definition of matrix multiplication and the definitions of A and B.

∀j, i cj,i =

p−1(n)∑k=1

D(pj, pk)H(pk, pi). (6.3.1)

Here k is varying over all partitions of size n and below (including ∅). Since D(λ, µ) =

H(λ, µ) = 0 if µ * λ, then (6.3.1) can be rewritten as the following:

cj,i =∑

ν:pi⊆ν⊆pj

D(pj, ν)H(ν, pi). (6.3.2)

Notice if µ * λ, then there are no ν : µ ⊆ ν ⊆ λ, therefore cj,i = 0.

Using the definitions of D(λ, µ) and H(λ, µ) (6.3.2) takes the form:

cj,i =∑

ν:pi⊆ν⊆pj

(−1)|pj/ν|∑

S∈FYT(pj ,ν)

awt(S)

∑T∈HYT(ν,pi)

awt(T )

. (6.3.3)

Finally, using the definition of C(pj, pi), (6.3.3) can be rewritten as

cj,i =∑

(S,T )∈C(pj ,pi)

(−1)|pj/sh(T )|awt(S)awt(T ). (6.3.4)

It was proven in Lemma 21 that ∆ is an involution, therefore for every (S, T ) ∈

C(pj, pi) there is exactly one (S, T ) ∈ C(pj, pi) : ∆(S, T ) = (S, T ). Applying Lemma


22 yields (−1)|sh(S)|awt(S)awt(T ) + (−1)|sh(S)|awt(S)awt(T ) = 0. This proves that

cj,i = 0. (6.3.5)

An immediate result from Theorem 5 is an expansion of Schur functions in terms

of factorial Schur functions.

Corollary 4. spj =

j∑i=1

H(pj, pi)fspi.

Proof. Recall FS = A ∗ S.

Theorem 5 states AB = I, therefore B ∗ FS = B ∗A ∗ S = I ∗ S = S or S = B ∗ FS

sp1

sp2...

s(n)

=

1 0 · · · 0

H((1), ∅) 1 · · · 0

......

. . . 0

H((n), ∅) H((n), (1)) · · · 1

fsp1

fsp2...

fs(n)

For row j we have spj on the left and the right side gives

j∑i=1

H(pj, pi)fspi or spj =

i∑i=1

H(pj, pi)fspi .

This is the expansion of Schur functions in terms of factorial Schur functions. The

coefficients are always positive polynomials of the variables (a1, . . . , a2n−1) defined

using the combinatorial objects HYT(λ, µ). These combinatorial objects and the

involution ∆ allowed us to construct the reverse change of basis coefficients.

REFERENCES 59

References

[1] J. Bandlow and J. Morse. Combinatorial expansions in k-theoretic bases.preprint, 2011.

[2] L. C. Biedenharn and J. D. Louck. A new class of symmetric polynomials definedin terms of tableaux. Adv. in Appl. Math., 10:396–438, 1989.

[3] A. Buch. A littlewood-richardson rule for the k-theory of grassmannians. ActaMath, 189:36–78, 2002.

[4] W. Y. C. Chen and J. Louck. The factorial schur function. J. Math. Phys.,34:4144–4160, 1993.

[5] S. Fomin and A. N. Kirillov. Grothendieck polynomials and the yang-baxterequation. Proceedings of the 6th Conference on Formal Power Series and Alge-braic Combinatorics (DIMACS, 1994), page 183190, 1994.

[6] W. Fulton. Young Tableaux. Cambridge University Press, 1997.

[7] I. P. Goulden and A. M. Hamel. Shift operators and factorial symmetric func-tions. J. Combin. Theory Ser. A, 69:51–60, 1995.

[8] D. E. Knuth. Art of Computer Programming Volume 3: Sorting and Searching.Addison Wesley Professional, 1998.

[9] A. Knutson and T. Tao. Puzzles and (equivariant) cohomology of grassmannians.Duke Math. J., 119:221–260, 2003.

[10] V. Kreiman. Products of factorial schur functions. Electron. J. Combin., 15,2008.

[11] T. Lam and P. Pylyavskyy. Combinatorial hopf algebras and k-homology ofgrassmannians. Int. Math. Res. Not. IMRN, 2007.

[12] A. Lascoux. Symmetric Functions and Combinatorial Operators on Polynomials.American Mathatical Society, 2003.

[13] A. Lascoux and M.-P. Schutzenberger. Polynomes de schubert. C.R. Acad SciParis Ser I Math, 294:447–450, 1982.

[14] A. Lascoux and M.-P. Schutzenberger. Structure de hopf de lanneau de coho-mologie et de lanneau de grothendieck dune variete de drapeaux. C.R. Acad SciParis Ser I Math, 283:629633, 1982.

[15] I. G. Macdonald. Symmetric Functions and Hall Polynomials. Clarendon Press,1979.

REFERENCES 60

[16] A. I. Molev. Littlewood-richardson polynomials. Journal of Algebra, 321:3450–3468, 2009.

[17] A. I. Molev and B. E. Sagan. A littlewood-richardson rule for factorial schurfunctions. Trans. Amer. Math. Soc. 351, 351:4429–4443, 1999.

[18] R. P. Stanley. Enumerative Combinatorics Vol. 2. Cambrige University Press,1999.

Documents

Combinatorial aspects of generalizations of Schur …...homogeneous symmetric polynomials are two bases for the ring of symmetric func-tions. The Hall inner product is de ned using