Combinatorial Aspects of Generalizations of Schur Functions

Combinatorial aspects of generalizations of Schur 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

Abstract Combinatorial aspects of generalizations of Schur functions Derek Heilman Jennifer 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, sometimes 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 polynomials. 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 functions 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 reverse 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 discovered 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 functions, and tools for Schur functions from algebraic combinatorics. The theory of symmetric 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 functions. 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 GENERAL BACKGROUND 4

2.1 Symmetric functions

For an m-tuple of non-negative integers, γ = (γ1, γ2, ··· , γm), and n independent variables, (x1, x2, . . . , xn), deﬁne

γ (γ1,γ2,...,γm) γ1 γ2 γm x = x = x1 x2 . . . xm .

For an infinite sequence of non-negative integers ,γ = (γ1, γ2,...), and infinite independent variables, (x1, x2,...), define

γ (γ1,γ2,...) γ1 γ2 x = x = x1 x2 ··· .

(2,1,3,0,4) 2 1 3 0 4 2 3 4 Example. x = x1x2x3x4x5 = x1x2x3x5.

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

2 3 variables, whereas x1, x1 + 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, m X λ = (λ1, λ2, . . . , λm): λ1 ≥ λ2 ≥ · · · ≥ λm > 0, where λi = n. λ is a partition i=1 2 GENERAL BACKGROUND 5

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

X ω(λ) mλ = x , (2.1.1) ω

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)).

2 2 2 2 Example. For n = 3, m(1) = x1 + x2 + x3 and m(2,1) = x1x2 + x1x2 + x1x3 + x1x3 +

2 2 x2x3 + x2x3.

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. X Precisely, h = x x ··· x . k l1 l2 lk 1≤l1≤l2≤···≤lk≤n

Given a partition λ, hλ is deﬁned by: hλ = hλ1 hλ2 ··· 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 GENERAL BACKGROUND 6

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)}. 2 GENERAL BACKGROUND 7

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 2 GENERAL BACKGROUND 8

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 is S = 2 and S(1, 2) = 1. An example of a semi-standard Young tableau with N 1 1 x2 x2 alphabet {x1, y1, x2, y2,...} is T = . For the alphabet 3 = {1, 2, 3} with the x1 y1 N natural ordering, SSYT(2, 1) = 2 , 3 , 2 , 3 , 2 , 3 , 3 , 3 . 1 1 1 1 1 2 1 2 1 3 1 3 2 2 2 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 , wt 2 = (2, 1, 0). N3 1 1 Deﬁnition. The Schur polynomial of degree n indexed by partition λ is deﬁned by

X wt(T ) sλ = x , (2.2.1) T ∈SSYT(λ)

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,

wt 2 wt 3 wt 2 wt 3 wt 2 1 1 1 1 1 2 1 2 1 3 s2,1 = x + x + x + x + x wt 3 wt 3 wt 3 + x 1 3 + x 2 2 + x 2 3 ;

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

2 2 2 2 2 2 s2,1 = x1x2 + x1x3 + x1x2 + x1x2x3 + x1x2x3 + x1x3 + x2x3 + x2x3; 2 GENERAL BACKGROUND 9

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 A 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, hmλ, hµi = δλ,µ   1 if λ = µ (Kronecker delta), where δλ,µ = .  0 otherwise

A remarkable and useful result is that the Schur functions, {sλ}, are dual to themselves. That is to say hsλ, sµi = 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

X ν X ν if ∆fν = dλ,µfλ ⊗ fµ then gλgµ = dλ,µgν. λ,µ ν This powerful result above will be used later in Section 3. 2 GENERAL BACKGROUND 10

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:

X ν sλsµ = cλ,µsν. (2.4.1) ν

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

ν theory, and algebraic geometry. The coefficients cλ,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]:

X ν sλs(t) = cλ,(t)sν. (2.4.2) ν

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 2 GENERAL BACKGROUND 11

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].

4 Example. For α = 1 and R = 3 3 4 1 1 2 2 4 → 4 4 4 3 → 4 3 3 3 3 4 2 → 3 3 4 2 3 4 . 2 3 4 2 3 4 1 → 1 1 2 2 1 1 1 2 1 1 1 2 1 1 1 2 1 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 insertion 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].

4 Example. For R = 3 and removable cell (4, 1) 2 3 4 1 1 1 2 4 → 4 4 4 4 3 → 3 2 3 4 3 3 4 3 3 4 2 3 4 → 2 1 1 1 2 1 1 1 2 1 1 2 2 1 1 1 2 → 1 2 GENERAL BACKGROUND 12

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 resulting 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).

3 Example. For (α , α , α ) = (1, 1, 2) = T = 1 1 2 and R = , 1 2 3 2 2 3 3 ! 3 Ins 1, = (3, 1), 2 , 2 2 3 1 2 3 3 ! 3 ! Ins 1, 2 = (2, 2), 2 2 , 1 2 3 1 1 3 3 ! 3 ! Ins 2, 2 2 = (2, 3), 2 2 3 , 1 1 3 1 1 2 1 3 ! 3 RSK (1, 1, 2), = RSK (T,R) = 2 3 , 2 2 3 . 2 2 3 1 1 2 Since the RSK algorithm is constructed by iterating row insertion, which is an invertible 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. 2 GENERAL BACKGROUND 13

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, results 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). 2 GENERAL BACKGROUND 14

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 ). 2 GENERAL BACKGROUND 15

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. 3 3 RSK[ 1 1 2 , = 2 2 3 , 2 2 3 1 1 2 3 ! 3 and InvRSK\ 2 2 3 , = 1 1 2 , . 2 2 3 1 1 2 One of the most useful applications of the RSK insertion algorithm is the combinatorial proof of the Pieri rule for Schur functions.

Theorem 1. X sλs(t) = sµ (2.4.3) µ∈A(λ,t)

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

X X X X xwt(T1) xwt(T2) = xwt(T ) . (2.4.4)

T1∈SSYT((t)) T2∈SSYT(λ) µ∈A(λ,t) T ∈SSYT(µ)

Or equivalently,

X X X xwt(T1)+wt(T2) = xwt(T ) . (2.4.5)

(T1,T2)∈SSYT((t))×SSYT(λ) µ∈A(λ,t) T ∈SSYT(µ) 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 partitions 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

X ν ∆Gν = dλ,µGλ ⊗ Gµ. (3.1.1) λ,µ 3 THE PIERI RULE FOR THE DUAL GROTHENDIECK POLYNOMIALS 17

ν Buch gives a combinatorial definition for these coefficients, dλ,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

r(λ/ν) dν = (−1)t−|ν/λ| , (3.1.2) λ,(t) t − |ν/λ|

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

ν dλ,(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

X wt(T ) gλ = x , (3.2.1) T ∈RPP(λ)

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. 3 THE PIERI RULE FOR THE DUAL GROTHENDIECK POLYNOMIALS 18

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) = 1 , 2 , 1 , 2 , 2 , 1 1 1 1 1 2 1 2 2 2 wt 1 wt 2 wt 1 wt 2 wt 2 1 1 1 1 1 2 1 2 2 2 g2,1 = x + x + x + x + x ,

(2,0) (2,1) (1,1) (1,2) (0,2) 2 2 2 2 g2,1 = x + x + x + x + x = x1 + x1x2 + x1x2 + x1x2 + x2.

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].   2 3 3   Example. EF ((3, 3, 2, 1)/(3, 1)) = 1 2 , 1 2 , 2 2 . 1 1 1 1 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 3 THE PIERI RULE FOR THE DUAL GROTHENDIECK POLYNOMIALS 19

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. 1 1 2 2 3 Given λ = (4, 4, 3) and µ = (4, 2), if S = and E = 1 1 , then (S,E) ∈ 1 2 2 3 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. 1 1 2 2 3 If S = and E = 1 1 , then (S,E) ∈ SPE(λ). 1 2 2 3

1 1 2 So (S,E) can be written as T , where T = 2 3 1 1 . 1 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. 3 THE PIERI RULE FOR THE DUAL GROTHENDIECK POLYNOMIALS 20

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. 1 1 2 2 3 For S = and E = 1 1 from the example above, 1 2 2 3

1 1 2 −1 Ψ(S,E) = T , where T = 2 3 1 1 , and Ψ (T ) = (S,E). 1 2 2 3 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,ˆ 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].

2  3  1 2 2 3  2 2 2 3  Example. Given T = 1 2 2 , then Φ(ˆ T ) = , 1 1 1 .  1 1 1 2 2 2  1 1 1 3   1 1 1 2 2 2 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.  3  3 2 3 2 3  2 2 2 3  Ψ , 1 1 1 = 1 1 1 .  1 1 1 2 2 2    2 2 2 3 1 1 1 2 2 2 3 THE PIERI RULE FOR THE DUAL GROTHENDIECK POLYNOMIALS 21

The function Φ extends Φˆ 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(ˆ 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(ˆ 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 polynomials (3.1.1) gives the Littlewood-Richardson rule for the dual Grothendieck polynomials. 3 THE PIERI RULE FOR THE DUAL GROTHENDIECK POLYNOMIALS 22

Corollary 2. [3]

X ν gλgµ = dλ,µgν. (3.4.1)

ν Corollary 1 gave a formula for the coeﬃcients of the form dλ,(t). This formula gives the following:

Corollary 3.

X r(λ/ν) g g = (−1)t−|ν/λ| g . (3.4.2) λ (t) t − |ν/λ| ν (ν/λ) is a horizontal strip

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 functions (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.

Lemma 6. For any horizontal strip ν/λ,

r(λ/ν) = the number of Y cells for (ν, λ). (3.4.3) 3 THE PIERI RULE FOR THE DUAL GROTHENDIECK POLYNOMIALS 23

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 =ν ¯i < 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

number of Y cells for (µ, λ) k |XO(µ, λ, j)|= = , (3.4.4) j j

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

! ( O ) XO , , 1 = , 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,

t X X X t−j gλ gt = (−1) gµ. (3.4.5) j=0 µ∈A(λ,j) F ∈XO(µ,λ,t−j)

Proof. Lemma 6 and (3.4.2) give that

X number of Y cells for (ν, λ) g g = (−1)t−|ν/λ| g . (3.4.6) λ t t − |ν/λ| ν ν/λ is a horizontal strip

number of Y cells for (ν, λ) Notice, that if |ν/λ|> t, then = 0. Therefore (3.4.6) t − |ν/λ| can be written as a sum over horizontal j-strips, from j = 0 to j = t,

t X X number of Y cells for (ν, λ) g g = (−1)t−j g . (3.4.7) λ t t − j ν j=0 ν/λ is a j-horizontal strip

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 t X X X t−j gλ g(t) = (−1) gµ (4.0.1) j=0 µ∈A(λ,j) F ∈XO(µ,λ,t−j) 4 INSERTION PROOF OF THE DUAL GROTHENDIECK PIERI RULE 25

begins by substituting the dual Grothendieck polynomials with their weight generat- ing functions over reverse plane partitions. The left hand side becomes

   

X wt(R) X wt(R) X wt(R1)+wt(R2) gλgt =  x   x  = x ,

R∈RPP(λ) R∈RPP((t)) (R1,R2)∈RPP(t)×RPP(λ) and the right hand side is

t t X X X t−j X X X t−j X wt(R) (−1) gµ = (−1) x . j=0 µ∈A(λ,j) F ∈XO(µ,λ,t−j) j=0 µ∈A(λ,j) F ∈XO(µ,λ,t−j) R∈RPP(µ) (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- 4 INSERTION PROOF OF THE DUAL GROTHENDIECK PIERI RULE 26

mials.

2 X X X 2−j g(2,1)g(2) = (−1) gµ; j=0 µ∈A ,j F ∈XO µ, ,t−j

X X 2 X X 1 = (−1) gµ + (−1) gµ µ∈A ,0 F ∈XO µ, ,2 µ∈A ,1 F ∈XO µ, ,1

X X 0 + (−1) gµ; µ∈A ,2 F ∈XO µ, ,0

X X 2 X X 1 = (−1) gµ + (−1) gµ n o ( ) µ∈ F ∈XO µ, ,2 F ∈XO µ, ,1 µ∈ , ,

X X 0 + (−1) gµ; ( ) F ∈XO µ, ,0 µ∈ , , , X X X X = (−1)1g + (−1)1g n o µ= F =X µ= F ∈ ,X X X O O X X X X + (−1)1g + (−1)1g

X O O µ= F = µ= F = X X X X X + (−1)0g + (−1)0g

µ= F = µ= F = O O O O X X X X + (−1)0g + (−1)0g ;

O O µ= F = µ= F = O O

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. 4 INSERTION PROOF OF THE DUAL GROTHENDIECK PIERI RULE 27

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

1 X X X 1−j X wt(R) g1g1 = (−1) x ; j=0 µ∈A( ,j) F ∈XO(µ, ,1−j) R∈RPP(µ) X X X wt(R) X X X wt(R) g1g1 = (−1) x + x

µ= F =X R∈RPP( ) µ= F = O R∈RPP( ) X X X + xwt(R); µ= F =O R∈RPP

g1g1 = −g1 + g1,1 + g2;

RPP(1) = { 1 , 2 } ;

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

wt( 1 ) wt( 2 ) wt( 1 1 ) wt( 1 2 ) wt( 2 2 ) g1g1 = −x − x + x + x + x wt 1 wt 2 wt 2 + x 1 + x 1 + x 2 ;

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

2 2 g1g1 = −x1 − x2 + x1 + x1x2 + x2 + x1 + x1x2 + x2;

2 2 g1g1 = x1 + 2x1x2 + x2.

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 4 INSERTION PROOF OF THE DUAL GROTHENDIECK PIERI RULE 28

RPP((d)) × RPP(λ) by

−1 IDG(R1,R2) = Φ ◦ RSK(Φ([ R1), Φ(R2)).

3  2  2 1 2   2 2 Example. IDG 1 1 2 3 , 1 1 2 = ,   2 2 2  1 1 1 3  1 1 1 2 2 1 1 1 2 2 2 1 1 1 1 1 2 2 3 since

Φ( 1 1 2 3 ) = 1 1 2 3 and  2  3  1 2  2 3 Φ  1 1 2  = 1 1 1  1 1 1 3  2 2 2 3 1 1 1 2 2 2 1 1 1 2 2 2 and 3  3  2 2 3   1 3 RSK[  1 1 2 3 , 1 1 1  = 3 1 1  2 2 2 3  2 2 2 2 2 1 1 1 2 2 2 1 1 1 1 1 2 2 3 and  3  3  2  2 −1   1 2 Φ  1 3  = .  3 1 1  1 2 2  2 2 2 2 2  1 1 1 2 2 1 1 1 1 1 2 2 3 1 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

−1 partition R2 ∈ RPP(λ), IDG(R1,R2) is deﬁned as follows: Φ (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 4 INSERTION PROOF OF THE DUAL GROTHENDIECK PIERI RULE 29

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 RPP(ν). ν∈A(λ,d)

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

−1 Φ (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

−1 sh(Φ (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 . 4 INSERTION PROOF OF THE DUAL GROTHENDIECK PIERI RULE 30

Example. For R = 1 Φ(R) = 1 , the reverse plane partition 1 1 1 1 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 SRPP(ν, λ). ν∈A(λ,d)

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

−1 still needs to be shown that R3 ∈ SRPP(µ, λ). Deﬁne T = Φ (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)) ⊆ SRPP(ν, λ). ν∈A(λ,d) It remains to show that [ IDG(RPP((d), RPP(λ)) ⊇ SRPP(ν, λ). Let µ ∈ A(λ, d) and ν∈A(λ,d) T ∈ SRPP(µ, λ). Define Tˆ = Φ−1(T ), then define (D,ˆ Sˆ) = InvRSK(\ Tˆ , λ). This can ˆ ˆ be done, since T ∈ SSYT(µ) with alphabet Nn. Notice, since D ∈ SSYT((d)), then Remark 4 implies that D = Φ−1(Dˆ) = Dˆ. Sˆ(y, x) = Tˆ(y, x) or it was modified via the unbumping procedure, this can be seen from Lemma 1. Notice, that only cells contained in µ/λ are the origin for unbumping. If cell (y, x) ∈ µ/λ, then Tˆ(y, x) ≤ y − 2, since T ∈ SRPP(µ, λ). Let cell (i, αi) be the modified cell in row i resulting from the ˆ 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 4 INSERTION PROOF OF THE DUAL GROTHENDIECK PIERI RULE 31

ˆ ˆ ˆ (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) ∈ Sˆ Sˆ(k, j) ≤ k − 1. This means that if S = Φ−1(Sˆ), 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)) ⊇ SRPP(ν, λ) ν∈A(λ,d) 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.

    X wt(R) X wt(R) X X wt(R) gλgt =  x   x  = x . (4.2.1) R∈RPP(λ) R∈RPP((t)) ν∈A(λ,t) R∈SRPP(ν,λ)

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

Proof.

X X X X xwt(R1) xwt(R2) = xwt(T ), (4.2.2)

R1∈RPP((t)) R2∈RPP(λ) ν∈A(λ,t) T ∈SRPP(ν)

the equation above can be written equivalently as the following:

X X X xwt(R1)+wt(R2) = xwt(T ) . (4.2.3)

(R1,R2)∈RPP((t))×RPP(λ) ν∈A(λ,t) T ∈SRPP(ν) 4 INSERTION PROOF OF THE DUAL GROTHENDIECK PIERI RULE 32

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 polynomials 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) = ∅. 4 INSERTION PROOF OF THE DUAL GROTHENDIECK PIERI RULE 33

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) = RPP(ν) × XO(ν, λ, d − i). i=0,...,d ν∈A(λ,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.

2 O 1 Example. For R = 1 1 1 and γ = O : Φ(R) = 2 1 1 1 1 1 2 X 1 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) = 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∈A(λ,d) 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 changeable 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 4 INSERTION PROOF OF THE DUAL GROTHENDIECK PIERI RULE 34 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 Rˆ = 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). ˆ ˆ ˆ Define S as follows: S(y, x) = R(y, x) ∀(y, x):(y, x) 6= (i + 1, νi+1), and ˆ ˆ ¯ ˆ S(i + 1, νi+1) = i. S ∈ SSYT(ν) with alphabet Nn since sh(S) = sh(ν) ˆ ˆ ˆ 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 4 INSERTION PROOF OF THE DUAL GROTHENDIECK PIERI RULE 35

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 Rˆ.

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

(µ1, . . . , µi − 1, . . . , µ`(µ)). ν is a valid partition, because (i, j) is a removable 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 Rˆ(i, j) = i − 1 and the east-most such cell was chosen. Deﬁne Sˆ as follows: Sˆ(y, x) = Rˆ(y, x) ∀(y, x):(y, x) 6= (i, j), where Sˆ(i, j) ˆ ¯ ˆ is removed. S ∈ SSYT(ν) with alphabet Nn, since sh(S) = sh(ν) and Rˆ ∈ 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 Rˆ. Then place an X in γ in the allowable cell

(i − 1, λi−1) in the row below.

3.Π( R, γ) = (Φ−1(Sˆ), ω), where Sˆ 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.   2 2 X 3 3 Let (R, γ) =  2 2 3 , O  Φ(R) = Rˆ = 3 1 2 .  1 2 2  2 2 1 1 1 1 3 O 1 1 2 3 Step 1: (3, 3) is the southeast most changeable cell. 4 INSERTION PROOF OF THE DUAL GROTHENDIECK PIERI RULE 36

  3 3 X Step 2: (3, 3) is an O cell so (S,ˆ ω) =  3 1 , .  2 2 1 X  1 1 2 3 O   2 3 X Step 3: Π(R, γ) = (S, ω) = (Φ−1(Sˆ), ω) =  2 3 , .  1 2 2 X  1 1 2 3 O Example. Below is an example that shows that Π is its own inverse. Again, let λ = (3, 3, 2, 2) and t = 3.   2 3 X 3 3 Let (R, γ) =  2 3 , O  Φ(R) = Rˆ = 3 1 .  1 2 2  2 2 1 1 1 2 3 O 1 1 2 3 Step 1: (2, 3) is the southeast most changeable cell.   3 3 X Step 2: (2, 3) is an X cell so (S,ˆ ω) =  3 1 2 , .  2 2 1 X  1 1 2 3 O   2 2 X Step 3: Π(R, γ) = (S, ω) = (Φ−1(Sˆ), ω) =  2 2 3 , O .  1 2 2  1 1 1 3 O So from the two examples above the following was shown:     2 2 X 2 3 X Π  2 2 3 , O  =  2 3 ,   1 2 2   1 2 2 X  1 1 1 3 O 1 1 2 3 O and     2 3 X 2 2 X Π  2 3 ,  =  2 2 3 , O .  1 2 2 X   1 2 2  1 1 2 3 O 1 1 1 3 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.  2  3  1 2  2 3 Let (R, γ) =  1 2 2 , O , Rˆ = Φ(R) = 1 1 1 .  1 1 1 3 O  2 2 2 3 1 1 1 2 2 2 O O 1 1 1 2 2 2 There are no changeable cells, therefore there are no changes made.  2   2   1 2   1 2  Π  1 2 2 , O  =  1 2 2 , O .  1 1 1 3 O   1 1 1 3 O  1 1 1 2 2 2 O O 1 1 1 2 2 2 O O 4 INSERTION PROOF OF THE DUAL GROTHENDIECK PIERI RULE 37

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 INSERTION PROOF OF THE DUAL GROTHENDIECK PIERI RULE 38

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 Sˆ(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). 4 INSERTION PROOF OF THE DUAL GROTHENDIECK PIERI RULE 39

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,ˆ γˆ) 6= (R, γ), then

ˆ |sh(R)|−|sh(R)| = 1. 4 INSERTION PROOF OF THE DUAL GROTHENDIECK PIERI RULE 40

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,ˆ γˆ) 6= (R, γ), then (−1)d−|sh(R)/λ|xwt(R) + (−1)d−|sh(Rˆ)/λ|xwt(Rˆ) = 0.

Lemma 18. X (−1)|sh(T )/λ|xwt(T ) = 0. (T,D)∈M(λ,t)/SM(λ,d)

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-

P t−|sh(T )/λ| wt(T ) cause Π is a bijection (Lemma 14) imply that (T,D)∈P (λ,t) (−1) x + P t−|sh(T )/λ| wt(T ) (T,D)∈N(λ,t) (−1) x = 0. Obviously P (λ, t)∪N(λ, t) = M(λ, t)/SM(λ, t). P t−|sh(T )/λ| wt(T ) Therefore (T,D)∈M(λ,t)/SM(λ,d) (−1) x = 0. 4 INSERTION PROOF OF THE DUAL GROTHENDIECK PIERI RULE 41

The sign changing involution Π and the lemmas above concerning Π allow the following elegant combinatorial proof.

4.4 Combinatorial proof of the dual Grothendieck Pieri rule

Before proving the rule one simple remark is needed.

Remark 5.

X X xwt(R) + (−1)t−|sh(R)/λ|xwt(R) (R,γ)∈SM(λ,t) (R,D)∈M(λ,t)/SM(λ,d) X = (−1)t−|sh(R)/λ|xwt(R). (R,γ)∈M(λ,t)

What follows is a combinatorial proof of (3.4.5).

t X X X t−j gλ gt = (−1) gµ. j=0 µ∈A(λ,j) F ∈XO(λ,µ,t−j)

Proof. In Theorem 4 it was proved that the product of two dual Grothendieck polynomials, one is indexed by an arbitrary partition λ, and the other is indexed by a nonnegative integer can be written as

X X wt(R) gλgt = x . ν∈A(λ,t) R∈SRPP(ν,λ)

[ Using the fact from Lemma 12 that SM(λ, d) = SRPP(ν, λ) × {O(ν/λ)}, the ν∈A(λ,d) product above can be rewritten as

X wt(R) gλgt = x . (R,γ)∈SM(λ,t) 5 FACTORIAL SCHUR FUNCTIONS AND THEIR EXPANSION 42

X Lemma 18 states that (−1)t−|sh(R)/λ|xwt(R) = 0. This combined with (R,D)∈M(λ,t)/SM(λ,d) Remark 5 allows the following to be concluded,

X t−|sh(R)/λ| wt(R) gλgt = (−1) x . (R,γ)∈M(λ,t)

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

t X X X t−j wt(R) gλ gt = (−1) x j=0 µ∈A(λ,j) (R,F )∈RPP(µ)×XO(µ,λ,t−j) t X X X X = (−1)t−jxwt(R). j=0 µ∈A(λ,j) F ∈XO(µ,λ,t−j) R∈RPP(ν)

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

t X X X t−j gλ gt = (−1) gµ. j=0 µ∈A(λ,j) F ∈XO(µ,λ,t−j)

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 alphabet, 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- 5 FACTORIAL SCHUR FUNCTIONS AND THEIR EXPANSION 43 posing tensor products of representation using a particular bases: X Y bl(λ) = xS(i,j) + i − j − 1 [2]. These polynomials would later be gen- S∈SSYT(λ) (i,j)∈S eralized by replacing the fixed term to an new arbitrary set of variables aj−i: X Y fs(λ) = xS(i,j) − ai−j. These new factorial Schur functions are special S∈SSYT(λ) (i,j)∈S cases of double Schubert polynomials for Grassmannian permutations [14; 13]. Later factorial Schur functions were shown by Knutson and Tao to be the equivariant cohomology 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 mathematical 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 coefficients 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.

4 Example. For a tableau T = 3 4 the corresponding Ferrers diagram ﬁlled 2 3 3 1 2 2 3 5 FACTORIAL SCHUR FUNCTIONS AND THEIR EXPANSION 44

−3 with the content of each cell is −2−1 . −1 0 1 0 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:

Y fsp(T ) = (xT (z) − ac(z)+T (z)). z∈T

3 Example. If T = 2 3 , then 1 2

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.

X X Y fsλ = fsp(T ) = (xT (Z) − ac(Z)+T (Z)). (5.1.1) T ∈SSYT(λ) T ∈SSYT(λ) Z∈T

Example. For n = 3 and λ = (2, 1) = , the semi-standard Young tableaux of the shape λ can be written as follows

SSYT(λ) = 2 , 2 , 3 , 3 , 3 , 3 , 3 , 2 , 1 1 1 2 1 1 1 3 2 2 2 3 1 2 1 3 5 FACTORIAL SCHUR FUNCTIONS AND THEIR EXPANSION 45 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

2 2 2 2 fsλ = x1x2 − a2x1x2 − a1x1x2 − a1x1 + a1a2x2 + a1a2x1 + a1x1 − a1a2

2 2 2 2 + x1x2 − a3x1x2 − a1x2 − a1x1x2 + a1a3x2 + a1a3x1 + a1x2 − a1a3

2 2 2 2 + x1x3 − a2x1x3 − a1x1x3 − a2x1 + a1a2x3 + a2x1 + a1a2x1 − a1a2

2 2 + x1x3 − a4x1x3 − a1x3 − a2x1x3 + a1a4x3 + a2a4x1 + a1a2x3 − a1a2a4

2 2 2 2 + x2x3 − a3x2x3 − a2x2x3 − a2x2 + a2a3x3 + a2a3x2 + a2x2 − a2a3

2 2 2 2 + x2x3 − a4x2x3 − a2x3 − a2x2x3 + a2a4x3 + a2a4x2 + a2x3 − a2a4

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

2 2 + x1x2x3 − a4x1x2 − a1x2x3 − a1x1x3 + a1a4x2 + a1a4x1 + a1x3 − a1a4.

Some terms in the expansion need to be rearranged.

2 2 2 2 2 2 fsλ = x1x2 + x1x2 + x1x3 + x1x3 + x2x3 + x2x3 + x1x2x3 + x1x2x3

2 2 − a1x1 − a2x1 − a1x1x2 − a2x1x2 − a1x1x3 − a2x1x3

2 2 2 2 − a1x2 − a2x2 − a1x2x3 − a2x2x3 − a2x3 − a1x3

− a1x1x3 − a2x1x3 − a3x1x3 − a4x1x3 5 FACTORIAL SCHUR FUNCTIONS AND THEIR EXPANSION 46

− a1x2x3 − a2x2x3 − a3x2x3 − a4x2x3

2 2 + a1x1 + a1a2x1 + a1a2x1 + a1a3x1 + a1a4x1 + a2x1 + a2a3x1 + a2a4x1

2 2 + a1x2 + a1a2x2 + a1a2x2 + a1a3x2 + a1a4x2 + a2x2 + a2a3x2 + a2a4x2

2 2 + a1x3 + a1a2x3 + a1a2x3 + a1a3x3 + a1a4x3 + a2x3 + a2a3x3 + a2a4x3

2 2 2 2 2 2 − a1a2 − a1a3 − a1a2 − a1a2a4 − a2a3 − a2a4 − a1a2a3 − a1a4.

In the expression above certain terms clearly can be factored.

2 2 2 2 2 2 fsλ = (1)(x1x2 + x1x2 + x1x3 + x1x3 + x2x3 + x2x3 + x1x2x3 + x1x2x3)

2 2 3 − (a1 + a2)(x1 + x1x2 + x1x3 + x2 + x2x3 + x3)

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

2 2 + (a1 + a1a2 + a1a2 + a1a3 + a1a4 + a2 + a2a3 + a2a4)(x1 + x2 + x3)

2 2 2 2 2 2 − (a1a2 + a1a2 + a1a3 + a1a2a3 + a2a3 + a1a4 + a1a2a4 + a2a4)(1).

Recall that for n = 3

2 2 2 2 2 2 sλ = s = x1x2 + x1x2 + x1x3 + x1x3 + x2x3 + x2x3 + x1x2x3 + x1x2x3

2 2 3 s(2) = s = x1 + x1x2 + x1x3 + x2 + x2x3 + x3

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

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

s(0) = s∅ = (1). 5 FACTORIAL SCHUR FUNCTIONS AND THEIR EXPANSION 47

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

2 2 + (a1 + a1a2 + a1a2 + a1a3 + a1a4 + a2 + a2a3 + a2a4)s

2 2 2 2 2 2 − (a1a2 + a1a2 + a1a3 + a1a2a3 + a2a3 + a1a4 + a1a2a4 + a2a4)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 5 FACTORIAL SCHUR FUNCTIONS AND THEIR EXPANSION 48

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) = 1 , 2 , 1 , 2 , 2 , 1 , 2 , 2 . 1 2 1 2 1 3 1 3 2 3 1 4 1 4 2 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 ass ˆλ/µ(u)), which is the sum of the enumerations over the set FYT(λ/µ)

X D(λ, µ) = (−1)|λ/µ|awt(T ), T ∈FYT(λ/µ) 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

wt 1 wt 2 wt 1 wt 2 D(λ, ∅) = (−1)3a 1 2 + (−1)3a 1 2 + (−1)3a 1 3 + (−1)3a 1 3 wt 2 wt 1 wt 2 wt 2 + (−1)3a 2 3 + (−1)3a 1 4 + (−1)3a 1 4 + (−1)3a 2 4 ,

or equivalently

2 2 2 2 2 2 D(λ, ∅) = −(a1a2) − (a1a2) − (a1a3) − (a1a2a3) − (a2a3) − (a1a4) − (a1a2a4) − (a2a4).

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]. X fsλ = D(λ, µ)sµ. (5.2.1) µ⊆λ 5 FACTORIAL SCHUR FUNCTIONS AND THEIR EXPANSION 49

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

D(λ, λ) = 1;

1 2 D(λ, (2)) = (−1)1awt + (−1)1awt , or

D(λ, (2)) = (−1)(a1 + a2); wt wt wt wt D(λ, (1, 1)) = (−1)1a 1 + (−1)1a 2 + (−1)1a 3 + (−1)1a 4 , or

D(λ, (1, 1)) = (−1)(a1 + a2 + a3 + a4); wt 1 wt 1 wt 1 wt 1 D(λ, (1)) = (−1)2a 1 + (−1)2a 2 + (−1)2a 3 + (−1)2a 4 wt 2 wt 2 wt 2 wt 2 + (−1)2a 1 + (−1)2a 2 + (−1)2a 3 + (−1)2a 4 , or

2 2 D(λ, (1)) = (a1 + a1a2 + a1a3 + a1a4 + a2a1 + a2 + a2a3 + a2a4); wt 1 wt 2 wt 1 wt 2 D(λ, ∅) = (−1)3a 1 2 + (−1)3a 1 2 + (−1)3a 1 3 + (−1)3a 1 3 wt 2 wt 1 wt 2 wt 2 + (−1)3a 2 3 + (−1)3a 1 4 + (−1)3a 1 4 + (−1)3a 2 4 , or

2 2 2 2 2 2 D(λ, ∅) = (−1)(a1a2 + a1a2 + a1a3 + a1a2a3 + a2a3 + a1a4 + a1a2a4 + a2a4).

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)

2 2 + (a1 + a1a2 + a1a3 + a1a4 + a2a1 + a2 + a2a3 + a2a4)s(1)

2 2 2 2 2 2 − (a1a2 + a1a2 + a1a3 + a1a2a3 + a2a3 + a1a4 + a1a2a4 + a2a4)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 5 FACTORIAL SCHUR FUNCTIONS AND THEIR EXPANSION 50

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).

−1 For a partition λ, let pλ be the number i, such that pi = λ. Following the previous −1 −1 −1 −1 example: p∅ = 1, p(1) = 2, p(1,1,1) = 5, p(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.

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

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

 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))  A =   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 5 FACTORIAL SCHUR FUNCTIONS AND THEIR EXPANSION 51

 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 A =   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

 1 0 0 0 0 0 0

−(a1 + a2 + a3) 1 0 0 0 0 0  2 2   a1 + a1a2 + a2 −(a1 + a2) 1 0 0 0 0   A =  a1a2 + a1a3 + a1a4 + a2a3 + a2a4 −(a1 + a2 + a3 + a4) 0 1 0 0 0 3  a1 a1a2 −a1 0 1 0 0 2 2 2 2  −1(a1a2+a1a2+a1a3+a1a2a3 a1+a1a2+a1a3a1a4 −(a1+a2  2 2 2 2 +a +a ) −(a1 + a2) 0 1 0  +a2a3+a1a4+a1a2a4+a2a4) a2a1+a2+a2a3+a2a4 3 4  −1(a1a2a4+a1a2a4+a2a3a4+a1a2a5 a1a2+a1a3+a2a3+a1a4+a2a4 −(a1+a2+a3 a a +a a +a a +a a +a a 0 0 0 1 a1a3a5+a2a3a5+a1a4a5+a2a4a5+a3a4a5 3 4 1 5 2 5 3 5 4 4 +a4+a5)

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.   sp1      sp   2  −1 For a ﬁxed n let S = , the matrix S is a p by 1 matrix, since pp−1 = (n).  .  (n) (n)  .      s(n)   fsp1      fsp   2  Also let FS =  . .  .      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 1 0 ··· 0 sp1              fsp  D((1), ∅) 1 ··· 0  sp   2  =    2 .  .   . . .   .   .   . . .. 0  .              fs(n) D((n), ∅) D((n), (1)) ··· 1 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 coefficients of these expansions are positive polynomials of a. Kreiman discusses these coefficients and refers to rather complicated methods of finding them [10]. This section 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 combinatorial 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). 6 A REVERSE CHANGE OF BASIS 53

Example. Let n = 3 and λ = (2, 1) and µ = ∅. Then the set HYT(λ/µ) is

HYT(λ) = 1 , 1 , 1 , 1 , 1 , 2 , 2 , 2 . 2 1 2 2 3 1 3 2 3 3 3 1 3 2 3 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

X H(λ, µ) = awt(T ), T ∈HYT(λ/µ)

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

wt 1 wt 1 wt 1 wt 1 H(λ, µ) = H(λ, ∅) = a 2 1 + a 2 2 + a 3 1 + a 3 2 wt 1 wt 2 wt 2 wt 2 + a 3 3 + a 3 1 + a 3 2 + a 3 3 , or equivalently

2 2 2 2 2 2 H(λ, ∅) = a1a2 + a1a2 + a1a3 + a1a2a3 + a1a3 + a1a2a3 + a2a3 + a2a3.

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.

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

by,x = H(py, px).

Example. For n = 3 6 A REVERSE CHANGE OF BASIS 54

 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))  B =  . 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

 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 B =  . 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

 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 a2+a a +a a +a a +a2  1 1 2 1 3 1 4 2 a + a + a + a 0 1 0 0 0  +a a +a a +a2+a a 1 2 3 4  B =  2 3 2 4 3 3 4 . a1a2a3 a1a2 a1 0 1 0 0  2 2 2 2   a1a2+a1a2+a1a3+a1a2a3 a1+a1a2+a1a3+a1a4 a1+a2  2 2 2 2 +a +a a1 + a2 0 1 0 +a1a3+a1a2a3+a2a3+a2a3 +a2a1+a2+a2a3+a2a4 3 4  3 2 2 3 2 2 2  a1+a1a2+a1a2+a2+a1a3 a1+a1a2+a2+a1a3+a2a3 a1+a2+a3 2 2 2 3 2 2 0 +a +a 0 0 1 +a1a2a3+a2a3+a1a3+a2a3+a3 +a3+a1a4+a2a4+a3a4+a4 4 5

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) = FYT(λ, ν) × HYT(ν, µ) ν:µ⊆ν⊆λ for partitions λ and µ, such that µ ⊆ λ and λ 6= µ. 6 A REVERSE CHANGE OF BASIS 55

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. 1 2 3 2 1 Let S = 3 , and T = . Then t = 1, since it is the smallest letter in S and 2 1 T . And z = (1, 3) in S, since it is southeast most cell containing the letter 1.

1 2 3 ! 1 2 3 ! 2 1 2 1 Then ∆ 3 , = 3 , . 2 2 1 1

Next, consider the Case II. 1 2 3 2 1 Let S = 3 , and T = . Then t = 1 since it is the smallest letter in S and 2 1 6 A REVERSE CHANGE OF BASIS 56

T . And z is cell located at spot (1, 3) in T , since it is southeast most cell containing the letter 1.

1 2 3 ! 1 2 3 ! 2 1 2 1 Then ∆ 3 , = 3 , . 2 1 2 1 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,ˆ Tˆ). If (i, j) ∈ sh(λ) and (i, j) 6= (y, x) then either S(i, j) = Sˆ(i, j) or T (i, j) = Tˆ(i, j). Thus, the smallest letter in either Sˆ or Tˆ is t. This means, that cell z = (y, x) is still the southeast-most cell containing the letter t in Sˆ or Tˆ by the deﬁnition of sets FYT(λ, ν) and HYT(ν, µ). Therefore the map ∆ switches z ∈ (S,ˆ Tˆ) to it’s original position in (S,T ). This proves that ∆(∆(S,T )) = ∆(S,ˆ Tˆ) = (S,T ).

Lemma 22. For (S,T ) ∈ C(λ, µ), if ∆(S,T ) = (S,ˆ Tˆ), then

(−1)|sh(S)|awt(S)awt(T ) = (−1)(−1)|sh(Sˆ)|awt(Sˆ)awt(Tˆ).

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(Sˆ)awt(Tˆ). To prove the statement of the Lemma it remains to show that ||sh(S)|−|sh(Sˆ))||= 1. Consider two cases. Case I: ∆ moves a cell from S to T . Then sh(S) = λ − µ − sh(T ) and sh(Sˆ) = λ − µ − sh(Tˆ) = λ − µ − (sh(T ) + 1), so |sh(S) − sh(Sˆ)|= 1. Case II: ∆ moves a cell from T to S. Then sh(S) = λ − µ − sh(T ) and sh(Sˆ) = λ − µ − sh(Tˆ) = λ − µ − (sh(T ) − 1), so |sh(S) − sh(Sˆ)|= 1. 6 A REVERSE CHANGE OF BASIS 57

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.

−1 p(n) X ∀j, i cj,i = D(pj, pk)H(pk, pi). (6.3.1) k=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:

X cj,i = D(pj, ν)H(ν, pi). (6.3.2)

ν:pi⊆ν⊆pj

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:

   

X |pj /ν| X wt(S) X wt(T ) cj,i = (−1) a   a  . (6.3.3)

ν:pi⊆ν⊆pj S∈FYT(pj ,ν) T ∈HYT(ν,pi)

Finally, using the deﬁnition of C(pj, pi), (6.3.3) can be rewritten as

X |pj /sh(T )| wt(S) wt(T ) cj,i = (−1) a a . (6.3.4)

(S,T )∈C(pj ,pi)

It was proven in Lemma 21 that ∆ is an involution, therefore for every (S,T ) ∈ ˆ ˆ ˆ ˆ C(pj, pi) there is exactly one (S, T ) ∈ C(pj, pi) : ∆(S,T ) = (S, T ). Applying Lemma 6 A REVERSE CHANGE OF BASIS 58

22 yields (−1)|sh(S)|awt(S)awt(T ) + (−1)|sh(Sˆ)|awt(Sˆ)awt(Tˆ) = 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.

j X Corollary 4. spj = H(pj, pi)fspi . i=1 Proof. Recall FS = A ∗ S. Theorem 5 states AB = I, therefore B ∗ FS = B ∗ A ∗ S = I ∗ S = S or S = B ∗ FS

      sp1 1 0 ··· 0 fsp1              sp  H((1), ∅) 1 ··· 0  fsp   2  =    2   .   . . .   .   .   . . .. 0  .              s(n) H((n), ∅) H((n), (1)) ··· 1 fs(n)

j X For row j we have spj on the left and the right side gives H(pj, pi)fspi or spj = i=1 i X H(pj, pi)fspi . i=1

This is the expansion of Schur functions in terms of factorial Schur functions. The

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