Schur-Positivity of Differences of Augmented Staircase Diagrams
by
Matthew Morin
B.Sc., Simon Fraser University, 2003 M. Sc., University of British Columbia, 2005
A THESIS SUBMITTED IN PARTIAL FULFILMENT OF THE REQUIREMENTS FOR THE DEGREE OF
Doctor of Philosophy
in
The Faculty of Graduate Studies
(Mathematics)
The University of British Columbia (Vancouver) April 2010 © Matthew Morin 2010 Abstract
The Schur functions {s} and ubiquitous Littlewood-Richardson coefficients c are instrumental in describing representation theory, symmetric functions, and even certain areas of algebraic geometry. Determining when two skew diagrams D1, D2 have the same skew Schur function or determining when
the difference of two such skew Schur functions SD1 — 5D2 is Schur-positive reveals information about the structures corresponding to these functions. By defining a set of staircase diagrams that we can augment with other (skew) diagrams, we discover collections of skew diagrams for which the ques tion of Schur-positivity among each difference can be resolved. Furthermore, for certain Schur-positive differences we give explicit formulas for computing the coefficients of the Schur functions in the difference. We extend from simple staircases to fat staircases, and carry on to dia grams called sums of fat staircases. These sums of fat staircases can also be augmented with other (skew) diagrams to obtain many instances of Schur positivity. We note that several of our Schur-positive differences become equalities of skew Schur functions when the number of variables is reduced. Finally, we give a factoring identity which allows one to obtain many of the non-trivial finite-variable equalities of skew Schur functions.
11 Contents
Abstract ii
Table of Contents iii
Acknowledgements v
1 Introduction 1 1.1 Partitions, Compositions, Ferrers Diagrams, and Skew Diagrams 1 1.2 Combining and Decomposing Diagrams 4 1.3 Tableaux 5 1.4 Symmetric Functions 6 1.5 The Littlewood-Richardson Rule 9 1.6 Some Useful Identities for Schur Functions 10 1.7 A Brief History of Schur Functions 12 1.8 Overview of Schur-Positivity Results 15
2 Staircases with Bad Foundations 23 2.1 Regular Staircases 23 2.2 Fat Staircases 27 2.3 Sums of Fat Staircases 33
3 Hook Foundations 43
3.1 Schur Comparability / Incomparability . 43 3.2 Expressions for Schur-Positive Differences 60 3.3 Hasse Diagrams for k> 1 73
4 Differences of Transposed Foundations 91 4.1 Single Part Partitions 91 4.2 Two Part Partitions 95
111 5 Complements in a Rectangle . 111 5.1 Preliminaries 111 5.2 Staircases with Hook Complement Foundations 116 5.3 Sums of Fat Staircases 123 5.4 Miscellaneous Results 128
6 Reduction to Finite Variables 133 6.1 Differences of Transposed Foundations 133 6.2 Factoring in Finite Variables 136
7 Conclusion 144
Bibliography 146
iv Acknowledgements
I would like to thank my supervisor, Stephanie van Willigenburg, for her con tinual support, her remarkable ability to neatly turn a problem on its head, and her constant and vigilant attention to detail. Without her guidance, this thesis would be naught but a work-in-progress. I would also like to show my gratitude to all members of my thesis com mittee for accepting the responsibility of judging my dissertation. The will ingness to spend the time and effort for this endeavor is much appreciated. Financial credit must be given to NSERC for funding the past three years of my doctoral program. Life would have been much more difficult without the scholarship they generously awarded me. More thanks are due to UBC, for their support throughout my entire years of study and the many opportunities they have granted. The entire staff of the Mathematics department deserves praise for its ability to keep everything running smoothly and efficiently not just in the department as a whole, but down through each and every individual in the department. Finally, I must thank all my friends, family, and loved-ones for their non academic support. I could not be here now without your unwavering belief in myself and my abilities. Although you may not fully understand what it is I do, you have always shown pride in my desire to do it.
V Chapter 1
Introduction
1.1 Partitions, Compositions, Ferrers Diagrams, and Skew Diagrams
A partition A of a positive integer n, written A H n, is a sequence of weakly decreasing positive integers A = (A1, A2, , A,) with A = n. \1Ve call each A2 a part of A, and if A has exactly k parts we say A is of length k and write 1(A) = k. When A I— n we will also write IA n and say that the size of A is n. We shall use jT to denote the sequence j, j j consisting of r j’s. Under this notation, we shall write A = (k1, k — 17’k—1 . . , U’) for the partition which has r1 parts of size one, r2 parts of size two, ..., and rk parts of size k.
= . , we let A ,u Given two partitions A (A1, . . A,) and p = (h,. . . , ,Lim), U denote the partition that consists of the parts A1, .. . , A, /J1,. , ltm placed in weakly decreasing order.
We say a = (a1,a2,.. . , ak) is a composition of n, written a = n, if each aj is a positive integer and a = n. As with partitions, we call each a2 a part of a, write a = n for the size of a, and if a has exactly k parts we say a is of length k and write 1(a) = k. If we relax the conditions to consider sums of non-negative integers, that is, allowing some of the a, to be zero, then we obtain the concept of a weak composition of n. If a = n we obtain a partition of n by reordering the a into weakly decreasing order. We may sometimes find it useful to treat partitions and weak composi tions as vectors with non-negative integer entries. VThen we write the vector z = (z1, z2,z3,. . . , z) we shall mean the infinite vector z=(z1,z23,...,z,O,O,O,...).
We shall only consider vectors with finitely many non-zero entries. Hence we
1 _
CHAPTER 1. INTRODUCTION shall oniy display vectors with finite length. In this manner we may unam biguously add vectors of different lengths. Thus, we have defined addition among partitions and weak-compositions. Further, given a positive integer i, we shall let e denote the i-th standard basis vector. That is, the vector that has its i-th entry equal to 1 and all remaining entries equal to 0.
Given a partition A, we can represent it via the diagram of left justified rows of boxes whose i-th row contains A boxes. The diagrams of these type are called Ferrers diagrams, or just diagrams for short. We shall use the symbol A when refering to both the partition and its Ferrers diagram. Whenever we find a diagram D’ contained in a diagram D as a subset of boxes, we say that D’ is a subdiagram of D. Suppose partitions A
(A1,A2,...,A) F-n and fL = (t1,ii2,.. .jtk) F-rn, with m < n, k A for each i = 1, 2,.. . , k are given. Then t is a subdiagram of A, and a particular copy of is found at the top-left corner of A. We can form the skew diagram A/,u by removing that copy of p from A. Henceforth, when we say that D is a diagram, it is assumed that D is either the Ferrers diagram of some partition A, or D is the skew diagram A/j for some partitions A, ,u. Example Here we consider the partitions A = (4, 4, 2) and = (3, 1), and form the skew diagram A/si. [=iJr H For two boxes b1 and b2 in a diagram D, we define a path from b1 to b2 in D to be a sequence of steps either up, down, left, or right that begins at b, ends at b2, and at no time leaves the diagram D. We say that a diagram D is connected if for any two boxes b1 and b2 of D there is a path from b to b2 in D. If D is not connected we say it is disconnected. Given any diagram D, the 1800 rotation of a diagram D is denoted by D°. Further, the conjugate or transpose diagram Dt is obtained by simply transposing the array of boxes along the main diagonal. When D is the Ferrers diagram of a partition A then the conjugate diagram Dt is also the Ferrers diagram of a partition called the conjugate partition and denoted At. It must be noted that most authors use the notation D’ and A’ for the conjugate diagram of D and conjugate partition of A. We hope our notation causes no confusion. 2 CHAPTER 1. INTRODUCTION Example Here we consider the partitions A = (4, 4, 2) and i = (3, 1), and form the skew diagram A/1i. HH The conjugates, At = (3,3,2,2), = (2,1,1) and (A/ji)t = At/itt are shown below. Finally, we see the skew diagrams (A/i)° and ((A/j)t)° = ((A/)0)t. The number of boxes that appear in a given row or a given column of a diagram is called the length of that row or column. A hook is the Ferrers diagram corresponding to a partition A that satisfies A < 1 for all i > 1. Hence a hook has at most one row of length larger than 1. Example Here we see the hooks A = (4, 1, 1) and t = (5, 1). Li ‘A’ It 3 _ __ CHAPTER 1. INTRODUCTION 1.2 Combining and Decomposing Diagrams In this section we look at combining and decomposing skew diagrams in a few simple ways. Given two diagrams D1 and D2 the concatenation of D1 and D2, denoted • D2, is the skew diagram obtained by placing D1 and D2 so that the top- right box of D1 is immediately below the bottom-left box of D2. Similarly, the near-concatenation of D1 and D2, denoted D1 0 D2, is the skew diagram obtained by placing D1 and D2 so that the top-right box of D1 is immediately left of the bottom-left box of D2. Finally, if the last column of D1 and first column of D2 are both of length i, then the near-concatenation of depth i of D1 and D2, denoted D1 0 D2, is the is the skew diagram obtained by placing D1 and D2 so that the top-right box of D1 is one step left and i — 1 steps up from the bottom-left box of D2. We note that 0 = 0. Example Let D1 be the skew diagram (2, 2, 2)/(1, 1) and D2 be the Ferrers diagram (4,4,2). Here we show the diagrams D1 D2, D1 0 D2, D1 02 D2, and D1 0 D2. I_ D1 0 D2 D1 02 D2 D1 03 D2 Given a diagram D, the connected components of D are the maximally connected subdiagrams of D. Any skew diagram D decomposes into a finite number of connected components D1,D2,. . . , D,. Conversely, given con nected diagrams D1,D2,. . . , Dk, we define the diagram D = D, called the direct sum of the D, to be the skew diagram with connected components D1,D2,. . . , Dk such that the top-right box of D is one step left and one step down from the bottom-left box of D1 for i = 1, 2,. . . , k — 1. We note that =0o. 4 CHAPTER 1. INTRODUCTION Example With D1 = (2, 2, 2)/(1, 1) and D2 = (4, 4, 2), as in the previous example, we now show the diagram D1 D2. D12 1.3 Tableaux If D is a diagram, then a tableau—plural tableaux—T of shape D is the array obtained by filling the boxes of the D with the positive integers, where repetition is allowed. A tableau is said to be a semistandard Young tableau (or simply semistandard, for short) if each row gives a weakly increasing sequence of integers and each column gives a strictly increasing sequence of integers. We will often abbreviate “semistandard Young tableau” by SSYT and “semistandard Young tableaux” by SSYTx. When we wish to depict a certain tableau we will either show the under lying diagram with the entries residing in the boxes of the diagram or we may simply replace the boxes with the entries, so that the entries themselves depict the underlying shape of the tableau. The content of a tableau T is the weak composition given by v(T) = (v1,u2,...), where Vj is the number of i’s that appear in T. Example Here we consider a semistandard Young tableau 7Ej with shape D1 = (2, 2, 2)/(1, 1) and content (1,0, 1, 2), and a semistandard Young tableau 7 of shape D2 = (4,4,2) and content (2,2,3,2, 1). 1124 3 2335 144 34 7 72 We may sometimes find it useful to take the transpose of a tableau T so that we obtain a tableau of conjugate shape. In these cases we simply transpose the entries of T along with the underlying shape D. In this way, we obtain a tableau 7 of shape Dt with c(Tt) = c(T). 5 ______ CHAPTER 1. INTRODUCTION Similarly, given a tableau T of shape D, we may wish to focus on the entries of T that lie in some subdiagram D’ of D. In this way we obtain a subtableau of shape D’. Also, given tableau 7j and 7 of shapes D1 and D2 respectively, we may form the tableaux ‘Tj 7 of shape D1 D2, 7j 7 of shape D1 D2, 7 0 7 of shape D1 0 D2, and 7 O 7 of shape D1 ® D2 (when D1 0 D2 is defined) in the obvious way. Example For the SSYTx 7 and 7 in the previous example, we now create the tableaux 7 7, 7 0 7, 7 02 7, and 7 o 7. 1 2 1 1 2 4 3 2335 1124 134 12335 111124 3 334 323135 44 144 14434] 7i07 7027 7037; Of these tableaux, only ‘Jj 0 7 and 7 02 7 are semistandard. 1.4 Symmetric Functions In this section we consider an infinite set of variables x = (xi, x2,...) and the ring of formal power series C[x]. By S, we mean the group of permutations of the letters {1, 2,. . . , n}. We can let each ‘it E S act on elements f(x1,x2,...) C[xj by defining X2,...) f(x(i), X(2),. . . X), (1.1) n+1, Xn+2,. . We are interested in certain functions f(x1,x2,...) C[x] which are fixed by each ‘it e S, for every n E N. For instance, the n-th power symmetric function, pn(x) = j 1 is such a function. Two other functions with this property are the the n-th elementary symmetric function, e(x)= :: xilxi2 . . xfl and the n-th homogeneous symmetric function, h(x)= xilxi2 Xi. 6 CHAPTER 1. INTRODUCTION Given a partition A = )2,. ,A) by the power symmetric function, elementary symmetric function, and homogeneous symmetric function corre sponding to A, we shall mean p(x) Ho e(x) and = 11ex, = k h(x) Additionally, by the monomial symmetric function corresponding to A we mean m(x) = where the above sum is taken over all distinct monomial terms of the form • • • x. We note that m()(x) p(x) and m(ln)(x) = e(x). Each of p, e, h, and mA are invariant under the action described in Equation 1.1. Further, it turns out that each of the sets {p(x)j A I— n}, {eA(x) A I— n}, {hA(x)j A I— n}, and {mA(x) A I— n} are independent and span the same space of functions; that is, they are all bases of a common space. This space, denoted A7, is called the set of homogeneous symmetric functions of degree n and is usually defined as the span of the m, (x) for A F-n. For any tableau ‘T of content ii = (zi, v2,...) we have the weight xT defined by xT = JJx”. Given this weight function, the Schur function corresponding to A is defined to be s.x(x) = xT, (1.2) where the sum is taken over all SSYTx ‘T of shape A. If Al = n, then sA(x) is a homogeneous symmetric function of degree n. As with the other symmetric functions, the set {sA(x) A F- n} is a basis of A. Further details can be found in [63]. 7 CHAPTER 1. INTRODUCTION Finally, given a skew diagram A/ji, the skew Schur function corresponding to )/i is s1(x) = xT, (1.3) where the sum is taken over all SSYTx T of shape The skew Schur is homogeneous symmetric function of degree n function sA/(x) a = — Example For a single row = (n), the semistandard condition on a tableau of shape ) implies that the entries form a weakly increasing sequence. Thus we obtain s()(x) = xjx2 = h(x), the n-th homogeneous symmetric function. = (in), For a single column t the semistandard condition on a tableau of shape ji implies that the entries form a strictly increasing sequence. Thus we obtain S(in)(X) = = en, il Example We wish to compute the Schur function S(2,i) (x). Thus we need to inspect each SSYT of shape ) = (2, 1). Let ‘T be such a tableau and let i, j, and k be the entries of T as shown below. Since the columns of T must strictly increase, we require that i < k, and, since the rows of T weakly increase, we require i S(2,l)(X) = XXjXi4 i = (xixxk) — XiXjXk j i (xixk) — XXjXk = j i = e(l)(x)e(2)(x) — e(3)(x). 8 CHAPTER 1. INTRODUCTION We note that when x is taken to be a finite set of variables (for example, when setting all x = 0 for each i > n), then sA(x) is often called a Schur polynomial. S’Ve shall henceforth drop reference to the variables x and just write sA in place of sA (x), and similarly with the other symmetric functions. 1.5 The Littlewood-Richardson Rule Since the set {sjA F— n} is a basis of A, any homogeneous symmetric func tion of degree n can be written as a unique linear combination of the sA. In particular, for any partitions p and v, ss >csA, (1.4) Al—n where n = ji + v, for some coefficients The coefficients appear in many other expressions (see Section 1.7) and are called the Littlewood Richardson coefficients. The Littlewood-Richardson coefficients also appear as the coefficients of the skew Schur function 5A/p. That is, for any skew diagram A/u, sA/. = (1.5) In addition, the Littlewood-Richardson coefficients are where n = — i-. non-negative integers and there is a certain collection of SSYTx that they count. Given a tableau T, the reading word of T is the sequence of integers obtained by reading the entries of the rows of T from right to left, proceeding from the top row to the bottom. We say that a sequence r = r1,r2,. . . , is lattice if, for each j, when reading the sequence from left to right the number of j’s that we have read is never less than the number of j+ l’s that we have read. Theorem 1.5.1 (Littlewood-Richardson Rule) ([ 3]) For partitions A, , and ii, the Littlewood-Richardson coefficient is the number of SSYTx of shape A/i-i, content u, with lattice reading word. We note that the content of a tableau was defined as a weak composition, not a partition. Nevertheless, the lattice condition forces any tableau with lattice reading word to have a content that is in fact a partition. 9 ____ CHAPTER 1. INTRODUCTION Example Let us compute the skew Schur function S(4,3,2)/(2,2) by using Equa tion 1.5 and the Littlewood-Richardson rule. Thus we need to find all SSYTx (432) of shape (4, 3, 2)/(2, 2) with lattice reading word, then C(22)v is the number of these tableaux with content i’. We leave it to the reader to check that the following four tableaux are the only SSYTx of shape (4, 3, 2)/(2, 2) with lattice reading word. 1j2 v= (4,1) v= (3,2) v= (3,1,1) v= (2,2,1) Therefore S(4,3,2)/(2,2) = S(4,) + S(3,2) + 5(3,1,1) + 5(2,2,1). For any f = aAs.X E A, we say that f is Schur-positive, and write f 0, if each aA > 0. Further, we say that f is multiplicity-free if each aA E {0, 1}. Therefore, the Littlewood-Richardson rule shows that both s,s,1 and SA/ are Schur-positive. For f, g e A, we will be interested in whether or not the difference f — g is Schur positive. We shall write f > g whenever f — g is Schur-positive. If neither f — g nor g — f is Schur-positive we say that f and g are Schur-incomparable. Further, we will find it conveniellt to write D1 ‘- D2 if SD1 s 8D2• If we consider the relation )— on the set of all Schur-equivalent classes of diagrams (i.e. [D]3 = {D’sD = SD’}), then defines a partial ordering. This allows us to view the Hasse diagram for the relation on the set of these Schur-equivalent classes of skew diagrams. 1.6 Some Useful Identities for Schur Func tions In this section we list a few easy to state results that shall prove useful. The first two results are commonly know as Pieri rules. The origins of these result can be found in [54]. To state these two results we require the following two definitions. A row strip is a skew diagram with no two boxes in the same column and a column strip is a skew diagram with no two boxes in the same row. The Pieri rules can now be stated as follows. Theorem 1.6.1 We have SvS(n) = S) where the sum is over all partitions ) such that )/(v) is a row strip of size n. 10 CHAPTER 1. INTRODUCTION Theorem 1.6.2 We have SvS(ln) = where the sum is over all partitions ) such that A/(v) is a column strip of size n. The Pieri rules have been generalized in such work as [1], [5], and [52]. We now give two results that we shall use often. Theorem 1.6.3 [[69], Exercise 7.56(a)] Given a skew diagram D, = 5D0. (1.6) A result on the Littlewood-Richardson coefficients of conjugate diagrams can be found in [30]. For products, we have the following identity. Theorem 1.6.4 The Schur function of any disconnected skew diagram is reducible. If D = D1 D2, then we have = (1.7) Proof Any SSYT of shape D1 D gives rise to a SSYT of shape D1 and a SSYT of shape D2 by taking the obvious subtableaux. Conversely, any SSYTx 7 of shape D1 and 7 of shape D2 give rise to the tableau ‘Tj 7 of shape D1 D2, which is clearly semistandard. • Lastly, we mention another expression for the product 5D12 Theorem 1.6.5 [[5], Chapter 1.5, Example 2.1 (a)] For skew diagrams D1 and D2, we have SD12 = 5D12 + 5D1OD2 (1.8) Proof Given SSYTx 7j of shape D1 and 7 of shape D2 we let a be the entry of the top-right box of D1 and b be the entry of the bottom-left box of D2. If a < b then Ti 0 7 is a SSYT, and if a > b then Ti 7 is a SSYT, and clearly all SSYTx of shape D1 0 D2 and D1 . D2 arise in this fashion. I 11 CHAPTER 1. INTRODUCTION 1.7 A Brief History of Schur Functions Schur polynomials and Schur functions have appeared in various mathemat ical contexts during the last two centuries. Their first definition was given in terms of certain bialternants. That is, as a quotient of alternants. Namely, given a composition 1u = u2,. . . , tj), we may obtain the alternant Ihir(1) 1ir(l) a(xi,. . . ,x1) = sgn(r)x1 . rES1 where the S1 is the group of permutations of {1, 2, . .. , l}, and sgn(7r) is the sign of the permutation ir. Then, taking a partition A of length 1 and using 6 = (1 — 1, 1 — 2,. . . , 1, 0), the Schur polynomial can be defined as aA+o These functions first appeared in a paper of Cauchy [15], dating 1815, in which he was able to prove that this rational function was in fact a polyno mial. The next noteworthy appearance of Schur functions is thanks to Jacobi [32] in 1841, in which the Schur functions were again defined as above and a form of the now-famous Jacobi-Trudy identity was given. This identity gives another method of computing s, and can be stated as follows. Theorem 1.7.1 Let A = (A1, A2, . . . , A1), then s = and 5t = where At is the conjugate of A and both determinants are I x 1. Throughout this time the Schur functions had not yet been given the name that we presently know them by. Issai Schur, born in 1875, also used the bialternant definition of S), in his dissertation [66]. However, his interest in Schur functions arose from their connection to the study of representation theory, a topic whose origins can be traced to a collection of papers by Frobenius [20, 21, 22] dating from 1896 to 1897. Although the definition of representations did not appear until Frobenius’ paper of 1897, his series of papers in 1896 introduced the character 12 CHAPTER 1. INTRODUCTION theory of finite groups, which is a fundamental aspect of represenation theory. Schur was one of Frobenius’ doctoral students, and he, together with notable others, furthered the development of representation theory and the theory of characters in his doctoral dissertation and in his subsequent work. In the study of characters of Sn-modules, there is a bijection taking the irreducible character xA to the function It was by this type of connection between representations of the symmetric group and homogeneous symmetric functions that Schur came to study the functions sA, which were later called Schur functions in his honour. For a detailed discussion of the pioneering work on representation theory, it is recommended that the reader consult Curtis [17]. This source includes a rich bibliography of several mathematicians who helped shape this field as well as provides an excellent survey of the development of representation theory. Other sources for the representation theory behind Schur functions can be found in [3] and [63]. Details on the symmetric function side can be found in [45] and [69]. More on the history of group characters can be foulld in [31]. Our preferred description of Schur functions is the combinatorial defini tion given by Equation 1.2. This description, together with Equation 1.4 and Theorem 1.5.1, allows us to work with Schur functions through purely combinatorial arguments. The equivalence between the combinatorial and the original definition of 5A can be found in [12]. The first steps of this combinatorial description began with Kostka [36] decomposing the Schur functions into monomial symmetric functions via = Mitchell [50] was first to show that the Kostka numbers K were non negative. Finally Littlewood [42] posited that the Kostka numbers K counted the number of SSYTx of shape A and content i. The first version of the Littlewood-Richardson rule, Theorem 1.5.1, was given by Littlewood and Richardson [43]. The first complete proof of the statement was due to Schützenberger [67] in 1977. Short proofs can be found in [23], [60], and [77]. Some other objects that the Littlewood-Richardson coefficients enumerate can be found in [13], [55], and [80]. One of the early attempts to prove the Littlewood-Richardson rule led Robinson to an early form of the RSK (Robinson-Schensted-Knuth) algo rithm [61]. Schensted [65] independently developed the algorithm in order to enumerate permutations by the length of their longest increasing and de creasing sequences. Knuth [35] extended this algorithm to its commonly used 13 CHAPTER 1. INTRODUCTION form. Further extension to the RSK algorithm have been investigated and can be found in [46, 64, 79]. Further topics on the Littlewood-Richardson rule and the RSK algorithm are discussed in [72, 73, 75, 78]. Through the connection to representation theory, Equation 1.4 gives rise to the equation X X’1 = describing the multiplicities of the irreducible character xA in the product of the irreducible characters and ‘, and (S 0 SV) t= cS’, describing the multiplicities of the Specht module 5A in the induced tensor product of the Specht modules S” and 5i In the completely different setting of enumerative geometry, the same cal culations were being performed under the guise of Schubert Calculus, which was introduced by H. C. H. Schubert. In the cohomology ring of the Grass manian, the cup product of two Schubert classes cr1. and u,,, is given by g1 U U, = c,oA. This was first proved by Carrell in [14]. Some studies of Littlewood-Richardson coefficients from this area include [11], [16], and [56]. Results specializing on multiplicity-free Schubert calculus can be found in [27] and [74]. Thus we see how the structure of the Schur functions plays an impor tant role in each of these areas. Equally important is the ability to com pute the Littlewood-Richardson coefficients However, it is known that the problem of computing the Littlewood-Richardson coefficients is #P complete [51]. Therefore it is unlikely that there any efficient algorithms to compute these coefficients. An estimate on the size of the Littlewood Richardson coefficients is given in [58]. A discussion on the many connec tions of Littlewood-Richardson coefficients to tableaux and the operations and algorithms on tableaux is given in [19]. A recent look at the possibility of factoring Littlewood-Richardson coefficients in special cases is researched in [33]. Other products on the Schur functions have also been studied. The Kro necker product, for instance, is examined in [2, 10, 62] Despite the difficulty in computing Littlewood-Richardson coefficients, there is much research into Schur function and skew Schur functions. Many 14 CHAPTER 1. INTRODUCTION recent results on determining the equivalence classes [D]3 of Schur-equivalent skew Schur functions can be found in [8, 24, 26, 48, 59], and several results on the Schur-positivity of certain differences can be found in [6, 25, 34, 37, 47, 49, 57]. Each of these Schur-positive differences may be viewed as giving a set of inequalities that the corresponding Littlewood-Richardson coefficients must satisfy. In the next section we shall survey many of the known instances of Schur-positive differences and describe the Schur-positivity results that are proved in this thesis. Instances of Schur-positivity can lead to interesting results. For instance, a Schur-positivity result is used in [18] to characterize the eigenvalues of Hermitian matrix. However, the story does not end with Schur functions. There have been many analogues and generalisations of the Schur functions since their cre ation. There are the Hall-Littlewood polynomials, the Jack polynomials, and the Macdonald polynomials. Information on these polynomials can be found in [44] and [45]. There are generalizations into the realm of quasisymmetric functions. Some results in this area are discussed in [8], [29], and [39]. There are also Schur P-functions, Schur Q-functions, and k-Schur functions [71]. These new functions share many of the same properties as the original Schur functions. Also, we are interested in many of the same questions. For example, the possible equalities between a Schur and skew Schur function was given in [76] and the Schur Q-function analogue of this question is addressed in [4] and [28]. In [70], Stembridge determined which products of Schur func tions was multiplicity-free and Bessenrodt [7] classified the multiplicity-free products of Schur P-functions. Another multiplicity-free result on Schur P functions is given in [68], in which it is determined when a Schur P-function expressed in terms of regular Schur functions is multiplicity-free. Some re sults regarding the k-Schur functions can be found in [9] and [40]. Many new results of the classical Schur functions and of the more recent generalizations appear each year as attempts are made to better describe and understand their composition and their implications in the varied areas they appear. The breadth of open questions regarding these functions will serve to keep mathematical interest in this area open for a long time. 1.8 Overview of Schur-Positivity Results In this section we shall outline the results that are proved in the course of this thesis. However, we begin by surveying many of the Schur-positive results that have been proved in recent years. To state many of these results, 15 CHAPTER 1. INTRODUCTION we must make several definitions and discuss many ways of defining new partitions from given partitions. The definitions that are introduced in this section to describe these known Schur-positivity results will not be used in the remainder of this thesis. Before we begin, we shall first define a special type of diagram that will appear in several of these results. A ribbon is a connected skew diagram that does not contain the diagram (2, 2). Similarly, a weak ribbon is a skew diagram that does not contain the diagram (2, 2). We now begin by mentioning some of the known necessary conditions that are required for a difference of skew Schur functions to be Schur-positive. Given a skew diagram )/ with r rows. Then the k-th row overlap se quence is defined by rk(/) = (ar,. . . where a2 is the number of columns occupied in common by rows i, i+1,. . . , i+ k — 1. Then define rowsk)/) to be the partition obtained by placing the values of rk (A/it) in weakly decreasing order. Similarly COlSk (A/si) may be defined by inspecting the column overlaps of )/u. Also, one can define rectk,1)/L) to be the ilumber of k x 1 rectangular subdiagrams contained inside )/t. For partitions .A and u, we write ). ,u if for each i = 1, . . . , l(?), where is 0 for any i > l(t). The relation - defines a partial ordering on the set of partitions called the dominance order. In [47], the following necessary conditions for Schur-positive differences are proved. Theorem 1.8.1 (Corollary 3.10, [7]) >8 Let )/j and v/p be skew shapes. If sA/,1 — 5L’/p 0, then 1. rowsk(A/I1) rowsk(v/p) for all k, 2. colsj(?../i) colsj(v/p) for all 1, and 3. rectk,l)/,u) rectk,1(v/p) for all k, 1. Given a skew diagram )/p, the base partition of )/i is denoted B/t) and is defined as the intersection of all diagrams v such that 0. The cover partition of A/ri is denoted by C(X/) and is the union of all diagrams 16 CHAPTER 1. INTRODUCTION v such that L c 0. In [25], it is remarked that 5x/ Sv/p 0 requires that B(A/) ç B(v/p) and C(v/p) ç C(A/). These give two additional necessary conditions for the existence of a Schur positive difference. Alternative methods of computing the base and cover partitions are determined in [25], which gives a method of checking if these conditions are satisfied. Having looked at some necessary conditions, we now list a variety of instances of Schur-positivity. Many of these results are phrased as differences of products of Schur functions rather than differences of skew Schur functions. However, since any product of Schur functions can be realized as a skew Schur function via Theorem 1.6.4, these can all be restated as instances of Schur positive differences of skew Schur functions. We begin with two Schur-positivity results discussed in [6] involving the operation *. Conjecture 1.8.2 was first stated in [18] and the later general ized form, Conjecture 1.8.3, was put forth in [6]. Given an ordered pair of = (*, partitions ji), a new ordered pair (, u) ji*) is defined by taking — — Ak k + #{lI.u 1 “ — k} for each k, and — := — 1 + #{kPk k > 1u — l} for each 1. Conjecture 1.8.2 (Conjecture 5.1, /18]) For partitions A and we have S)*S,3* — Sj.S > 0. The definition of * can also be extended to pairs of skew shapes (/c, v/3) by taking v/ ( v)/( )* Conjecture 1.8.3 (Conjecture 2.9, /6]) For any skew partitions ji/c and ii//3, if (A,p) = (j/cv/J3)* then — S/aSvf5 >9 0. Some instances of Conjecture 1.8.2 and Conjecture 1.8.3 have been ad dressed in [6]. Namely, the following two results, Theorem 1.8.4 and The orem 1.8.5, have been proved. The first describes several pairs for which these Conjectures hold, while the second shows that the bounded height case reduces to checking Conjecture 1.8.2 for a finite number of pairs. 17 CHAPTER 1. INTRODUCTION Theorem 1.8.4 (Theorem 3.1, [6]) Conjecture 1.8.2 (or Conjecture 1.8.3) holds 1. For any pair A, i) of hook shapes. 2. For skew pairs of the form (/c, v//3), where ,a, v, c, are hooks with o=/3. 3. For skew pairs of the form (0, v//3), with v//3 a weak ribbon. Theorem 1.8.5 (Theorem 3.2, [6]) For any positive integer p, let v be a fixed partition with at most p parts. Then, the validity of Conjecture 1.8.2 for the infinite set of all pairs (j, v), with l(u) We shall now list several other operations on partitions that have given rise to Schur-positive differences. Given a partition A with each part even we define := (, ,. . Theorem 1.8.6 (Conjecture 1, [37]) For two skew shapes A/1i and v/p such that A + v and + p both have even parts, we have — S\//ASV/p >8 0. A generalization of this result may be stated as follows. Theorem 1.8.7 (Theorem 11, [37]) Let A/it and v/p be any two skew shapes. Then we have 5/ILSV/p S 0. — Given partitions A and ,u, we have the partition v = A U p obtained by listing all the parts of A and p together in weakly decreasing order. We then set sorti(A,p) = (vj,v3,v5,...) and sort2(A,p) = (v2,v46,. . Theorem 1.8.8 (Conjecture 2.7, [18]) Given partitions A and p, we have Ssortl(),)Ssort2()L) — 5Ap s 0. 18 CHAPTER 1. INTRODUCTION This result may be generalized to skew shapes as follows. Theorem 1.8.9 (Corollary 12, [37]) Let A/ti and v/p be skew shapes. Then we have p,v)/sorti 5sort2(,)/sort(,p) — 5A/LSv/p s 0. a partition A, let A”1 (A . In this way the parts Given Ai+n, Ai+2n, . .). A[i,?] of A are distributed among the partitions for i = 1,. . . Theorem 1.8.10 (Conjecture 6.4, [41]) For integers 1 m < n and a partition A, we have fl m fJ S[i,n] S[i,m] > 0. — A generalization of this result to skew shapes can also be given. Theorem 1.8.11 (Theorem 13, [37]) Let A(’)/i(1), ..., A()/1i(’), be n skew shapes, let A = U be the par tition obtained by the decreasing rearrangement of the parts in all and = U. Then we have > 0. Given two partitions A and ji, we can define A V ji and A A ji by A V i := (max{A1, max{A2, Ii2},...) and A A := (min{A1, ii}, min{A2, Ji2},. . Then, given skew diagrams A/1i and v/p, the operators V and A can be defined on pairs of skew diagrams via (A/ii) V (v/p) := (A V v)/( V p) and (A/1i) A (v/p) := (A A v)/(i A p). Theorem 1.8.12 (Conjecture 4.6, [38]) Let A/1u and v/p be any two skew shapes. Then we have SA/Sii/p 0. 19 CHAPTER 1. INTRODUCTION Theorem 1.8.6 was first conjectured in [53], Theorem 1.8.8 was first con jectured in [18], Theorem 1.8.10 was first conjectured in [41], and Theo rem 1.8.12 was first conjectured in [38]. In [37], Theorem 1.8.12 was proved and was then used to prove each of Theorem 1.8.7, Theorem 1.8.9, and The orem 1.8.11, which in turn imply each of Theorem 1.8.6, Theorem 1.8.8, and Theorem 1.8.10, respectively. We have seen many operations that construct examples of Schur-positive differences. A seperate problem to investigate involves finding families of skew diagrams for which all instances of Schur-positive differences within these families can be determined. One result of this type occurs in [49], in which the collection of multiplicity- free ribbons is inspected. Using results of [27], multiplicity-free ribbons are classified as those ribbons with at most two rows of length greater than one and at most two columns of length greater than one. Then, among the col lection of multiplicity-free ribbons of a given size, the Hasse diagram which describes all Schur-positive differences and Schur-incomparabilities among these diagrams is explicitly described. Moreover, the Hasse diagram is es sentially a product of two chains. We now begin to inspect another collection of ribbons for which the ques tion of Schur-positivity is completely answered. It is not difficult to see that the skew diagram )./i’ is uniquely determined by the row overaps rows1(/t) and rows2/t), thus we may identify the skew Schur function using the overlap notation = {rows1/i)rows2(A/t)}. In the case when A/si is a ribbon we have = {c11()_1}, where o is the composition given by the row lengths of )/1u. In this situation we use the notation := = {c11)_1} and call ra a ribbon Schur function. In [8], it was shown that the collection {rA},H forms a basis of A. In [34], the following theorem was proved, which identifies the Schur-positive differences among this collection. Theorem 1.8.13 (Theorem 3.3, [3]) 20 CHAPTER 1. INTRODUCTION Let A and be partitions of n, then r — r), 0 if and only if,u -< A and 1(A) = l(,u). More Schur-positive differences were discovered in [34], and are easiest to state in terms of this overlap notation. We begin by considering the following hypothesis on compositions a and r. Hypothesis 1.8.14 Let a and T be compositions such that 1(a) = s > 0 and 1(r) = t> 0, and let a- and ‘ be sequences of non-negative integers that satisfy the following conditions: 1. The lengths of a- and are s and t respectively; 2. as 1 when s > 0; 3. = 1 when t > 0; 4. ö Under this hypothesis, [34] proves that the following differences of skew Schur functions are Schur-positive. Theorem 1.8.15 (Theorem 3.1, [34]) Assume a, r, a-, and satisfy Hypothesis 1.8.14. If a b > 2 then we have {a,a,b,rja-,1,5} — {a,a+ 1,b— 1,rja-,1,’} >30 and 1n—1 1n—1 {a, a, ra-, — {a, a + 1, a2, a — 1, -ja-, } We now turn to describing the results and the flow of this thesis. We begin in Chapter 2 by defining staircases and fat staircases. We then describe how we augment these staircases with a skew diagram called the foundation by attaching the skew diagram to the staircase via a parame ter k. The augmented diagrams of this type are called staircases with bad foundations and will remain the focus for nearly all of this thesis. In Section 2.3 we define what it means for a diagram to be a sum of fat staircases and we prove Theorem 2.3.1 that, for each sum of fat staircases D, provides a collection of Schur-positivity results. Namely, we obtain an instance of Schur-positivity for each choice of the foundation. The remainder 21 CHAPTER 1. INTRODUCTION of Section 2.3 attempts to determine conditions for a diagram D to be a sum of fat staircases. In Chapter 3 we fix a fat staircase and a value k 0 and, for a given hook length, consider all possible hook foundations. We determine the structure of the entire Hasse diagram, which describes which pairs of the staircases with hook foundations have Schur-positive difference and which are Schur incomparable. In Section 3.1 we describe the Hasse diagram for values 0 k < 1 via Theorem 3.1.1, Theorem 3.1.2, Theorem 3.1.3, Theorem 3.1.4, and Theorem 3.1.5. In Section 3.3 we extend these results to describe the Hasse diagram for values k > 1 by proving Theorem 3.3.1, Theorem 3.3.2, Theorem 3.3.3, Theorem 3.3.4, Theorem 3.3.5, and Theorem 3.3.6. In Section 3.2 the expression for each Schur-positive difference for values 0 < k < 1 are computed by Theorem 3.2.2 and Theorem 3.2.3. In Chapter 4 we inspect the case of differences of fat staircases with bad foundations in which the foundations of the two diagrams are transposes of each other. In Theorem 4.1.1 and Theorem 4.2.1, we show that the differ ence is Schur-positive when one foundation is taken as either a single row or any two row partition. Also, when reducing to finitely many variables, we determine the largest number of variables for which this difference is zero. Most of the results of the earlier chapters are extended in Chapter 5 to yield results for fat staircases with complementary foundations. In particular, for a given fat staircase and hook length, Theorem 5.2.1 describes the Hasse diagram for the collection of fat staircases with hook complement foundations and the Schur-positive differences is calculated by Theorem 5.2.2 and The orem 5.2.3. Theorem 5.4.3 gives a collection of Schur-positive results from each sum of fat staircases D. Further, Theorem 5.3.1 and Theorem 5.3.2 allow us to construct many examples of sums of fat staircases. Finally, in Chapter 6 we once again consider reducing Schur functions to finitely many variables and prove several equalities of skew Schur functions that arise in this setting. The majority of our results make use of Lemma 2.2.2 and Lemma 2.2.3, which are proved in Section 2.2, and allow us to prove these results of Schur positivity by focusing solely on fillings of the foundation. These lemma pro vide much simplification while considering these skew Schur functions. 22 Chapter 2 Staircases with Bad Foundations 2.1 Regular Staircases A Ferrers diagram is a staircase if it is the Ferrers diagram of a partition of 1800 the form \ = (n,n — 1,n — 2,...,2,1) or if it is the rotation of such a diagram. Both these diagrams are referred to as staircases of length n and will be denoted by S, and respectively. For what follows we will also define _i = = 0. Example Here we see the two staircases of length 5: 65 If S, C ) as Ferrers diagrams then we say that \ contains a staircase of size n. In these cases we may create the skew diagram A/6. Example Here we see that the partition ) = (8, 5, 4. 2, 1) contains a staircase of size 3 and the corresponding skew diagram )/ö3. 23 CHAPTER 2. STAIRCASES WITH BAD FOUNDATIONS I I A In this example A actually contains a staircase of size 4 and a staircase of size 5, but the skew diagrams obtained by removing these staircases are not connected. Definition Given n 1, k 0, and a partition A containing a staircase 6k—l such that skew diagram A/t5k_1 is connected and k A1 ii + k, then we can create a skew diagram called a staircase with a bad foundation, denoted S(A, k, n), by placing A/6k1 immediately below L such that the rows of the two diagrams overlap in precisely A1 — k positions. We call the subdiagram A/ök_1 the foundation of S(A,k,n). We use the terminology “bad foundation” since if we imagine the diagram S(A, k, n) being pulled downward by a gravitational force then, for most choices of the foundation A, the foundation would fail to support S(A, k, n) properly, causing the diagram to tip over to the right. In the staircase with bad foundation S(A, k, n), k is the distance to the left of / that A extends before the copy of 6k1 is removed. The requirement k < A1, guarantees that the top row of the skew diagram A/Sk_l is non- empty. The diagram S(A, k, n) is connected if k = components, A/k_1 and , if k A1. We also note that the skew diagram S(A, k, n) can be written in the form P/6n+k—1, where a is a partition. Example For k = 0 we have 6k—1 = = 0, and and A overlap in A1 places. Thus S(A, k, n) consists of the foundation A left-justified with the stair case L. For example, with n = 5, k = 0, and A = (5, 4, 2) we have the following staircase with bad foundation: S(A,0,5) 24 CHAPTER 2. STAIRCASES WITH BAD FOUNDATIONS For k = 1 we have 6k1 = = 0, and and .\ overlap in — 1 places. Thus k, n) is the diagram SA, 0, n) with its foundation ,\ shifted one to the left. For example, with n = 5, k 1, and ) = (5, 4, 2) we have the following staircase with bad foundation: S,1,5) A Example Here we show the staircase with a bad foundation 8, 4,5), where = (8, 5, 4, 2, 1). S(A,4,5) In this instance the conjugate of A, At = (5,4, 3, 3, 2, 1, 1, 1), also gives rise to a staircase with a bad foundation S(At, 4, 5). S(At, 4, 5) 25 CHAPTER 2. STAIRCASES WITH BAD FOUNDATIONS One of the advantages in computing the skew Schur functions of staircases with bad foundations is that, when using the Littlewood-Richarson rule, the /S portion of the diagram can be filled in only one way. This can be seen algebraically from Theorem 1.6.3 since s = s = s, where s6 is a Schur function. The unique filling of Z?1 obeying the semistandard conditions and the lattice condition is easily found to be the filling shown below. 1 1 1 n—4n—3n—2 1 2 n—3n--2n--1 1 2 3 n-2n-1 fl Another advantage in computing the skew Schur function of a staircase with bad foundation is the following fact, which arises from the structure of the unique content, v = = (n, n — 1, . . . , 2, 1), just mentioned. Lemma 2.1.1 Let S(A, k, n) be a staircase with bad foundation and ‘T be a SSYT of shape S(\, k, n) whose reading word is lattice. Then the entries in the first row ofthe foundation A/ökl of T form a strictly increasing sequence. Proof Let ‘T be a SSYT of shape k, n) whose reading word is lattice. As discussed previously, the Littlewood-Richardson rule allows only one way of filling the / portion of the shape k, n). The content of this filling of is(n,n—1,...,2,1). Let R be the first row of the foundation, and suppose that R contains the value j twice. Then j 1 since the second value in R is immediately below a value that is greater than or equal to 1 and the columns of ‘T strictly increase. Also, if j > 1, then when reading ‘T the lattice condition will be violated once we have read both j’s. Therefore no value j can be repeated in R, and hence the first row of the foundation of ‘T contains no repeated values. 26 CHAPTER 2. STAIRCASES WITH BAD FOUNDATIONS 2.2 Fat Staircases In the previous section we saw that in creating any SSYT of shape S(A, k, n) with lattice reading word, the following two properties held. Namely, 1. the entries of / are uniquely determined, and 2. the entries in the first row of A/ok_i are distinct. In this section we look into what other shapes besides the staircase can we append a foundation, so that we still have the analogous properties. Instead of Z, we begin by looking at a general skew diagram D = p/ps. Also, instead of removing 0k—l from A, we consider removing an arbitrary partition ic A. Then we want to append the diagram A/ic to the bottom of D. Namely, for a given k, we place A/ic immediately below D such that the rows of the two diagrams overlap in precisely A — ic1 — k positions. Then k is the distance to the left that the first row of A/ic extends from the last row of D. We represent this new skew diagram by S(A, Ic, D; k). Now given S(A, ic, D; k), we are interested in what conditions will guarantee that, when creating any SSYT S(A, Ic, D; k) with lattice reading word, the filling requires that 1’. the entries D are uniquely determined, and 2’. the entries in the first row of A/ic are distinct. We now look at which skew diagrams S(A, ic, D; k) satisfy these properties. Suppose a given skew diagram S(A, ic, D; k) satisfies 1’ and 2’. Then, by 1’, any SSYT S(A, ic, D; k) with lattice reading word has the same content for the subtableau of shape D. Let ii be this unique content. Since any SSYT of shape D with lattice reading word can be extended to a SSYT of shape S(A, ic, D; k) with lattice reading word, 1’ implies that there is only one SSYT of shape D with lattice reading word, and this tableau has content v. Therefore 5D = s,, where v is a partition by the Littlewood-Richardson rule. We now make use of the following result. Theorem 2.2.1 [76, Theorem 2.1] For partitions A, u, ii sA/,1 = s if and only if A/p = 11 or 27 CHAPTER 2. STAIRCASES WITH BAD FOUNDATIONS Hence, if S(;k, D; k) satisfies 1’, then D = ii or D = 110 where v is a partition. We now inspect property 2’. The proof of property 2 (i.e. Lemma 2.1.1) relied on the the fact that the first row of )/ök.i extended at most 1 box to the left of the staircase and the fact that the content of was the partition 6, which is the sequence of consecutively decreasing integers (n, n — 1,. . . , 2, 1). In fact, if the first row of )/i extends at most 1 box to the left of the diagram D, where D = 11 or D = v° for v equal to any sequence of consecutively decreasing integers ending in 1 with repetitions allowed, then the same method proves that 2’ holds. That is, all partitions 1a1), 11 given by sequences of the form (n°, (n — 1)°—’,. . . ,2°’, where each cvj > 1. This partition arises as the content of the tableaux of shape D = (n°’,(n— 1)° 1,...,2a2,11) and D = (fln,(n_ 1)aTh_1,...,22,1a1)0. We shall denote these two diagrams by — 11) . . (n°, (n 1)°—’, . , 2°, and 1)n_1,...,2Q2,1)0, = (n°,(n— where c = (c, . . . , c) is a composition. This composition c has a = = j where j = l(O) = l(za) is the length of these diagrams. We have just showed that for diagrams D = Oa and D = /. and 0 < k < 1 we have that S(), i, D; k) satisfies 1’ and 2’. Conversely, if either k> 1 or if either D = ii or D v°, where 11j1 — v > 1 for some i, then ,, D; k) cannot satisfy 2’. This follows since the value 1 can be placed in the first row of .A/ic a total of k times and the value i can be placed in the first row of D = and D = are precisely a total of 11j1 — v times. Therefore O the diagrams in which we are interested. We call a skew diagram D a fat staircase if D = or D = L for some composition . The numbers c, count the number of rows of D with i boxes, for each i. Using this notation the regular staircases may be expressed as = 6(m) and = (1), respectively. We note that the unique semistandard filling of La (O, respectively) has c as its largest entry. We choose the terminology “fat staircase” for describing these staircases with repeated rows in analogy to the usage of “fat hook” in [70] in which a fat hook is defined as a diagram of the form A2’). 28 CHAPTER 2. STAIRCASES WITH BAD FOUNDATIONS Example Here we see the fat staircases 6(1,2,2) and 6(1,2,2) (3,1,2,3) Just as the diagram k, n) was of the form IL/ön+k_1, where 6n+k—1 is a staircase, it would be nice if S, ,c, c; k) was of the form i/67, where 67 is a fat staircase. This requires that 0 k 1 and ,ç = 6 for some com position 13. Thus the diagrams of this form can be written as S(\, 6, ; k) for 0 < k < 1. We shall be interested in the more general case of a founda tion )/t and k 0. Therefore we make the following definition and use the following simplified notation. > — Given a composition c, k 0, and partitions X, 4a with ) — k 1(a) we now define S\, , c; k) to be the diagram obtained by placing .A/i immediately below such that the rows of the two diagrams overlap in precisely )4 — — k positions. We call S\, t, c; k) a fat staircase with bad foundation. The subdiagram A/1i is called the foundation of S(\, ,u, a; k). If ,u = 6 for a composition /3, then we write /3, ; k). When we do not wish to cut out any diagram t, we shall write p = 0 and simply use ; k) in place of S(?, z, c; k). The reason we write /3, c; k) instead of writing S, /3, c; k) is to avoid confusion in the cases when the compositions o = c2, misinterpreted and /3 = (/3 /32,...) are weakly decreasing, and could be as representing diagrams of partitions. For example, consider the diagrams S((2, 1), (2, 2); 1) and S((2, 1), (2, 2)1; 1) shown below. The first uses the di agram D = (2, 2), whereas the second uses the diagram D = (2,2). S((2, 1), (2, 2); 1) S((2, 1), (2, 2)1; 1) 29 CHAPTER 2. STAIRCASES WITH BAD FOUNDATIONS Example Here we show that the partition A contains the fat staircase 5, for A (7, 7, 5, 3, 3, 2) and 3 (1,2,2). We also display the skew diagram A/6j. (5(1,2,2) ç A Now we show the fat staircase with bad foundation S(A, , c; 1) for A = (7, 7, 5, 3, 3, 2), = (1,2,2), and c = (3,1,2,3). S(A, c; 1) We have seen that the fat staircases with bad foundations S(A, t, ; k) satisfy 1’ and 2’ when 0 k 1. Moreover, we have the following result when considering any k 0. Lemma 2.2.2 Let S(A, ,u, ; k) be a fat staircase with bad foundation for some k > 0 and T be a SSYT of shape S(A, t, c; k) whose reading word is lattice. If = (cr1,. . . , cj, then the entries in the first row of the foundation of T consist of values taken from the set Rak={1+fl+1_i =12...n}U{ {} 30 CHAPTER 2. STAIRCASES WITH BAD FOUNDATIONS Furthermore, the value 1 can occur at most k times and the rest of the values can appear at most once. Proof Let R be the first row of the foundation of 7 and t e R. Since ‘T is a SSYT, the columns strictly increase. Thus t = 1 is allowed if and only if k 1 since it is precisely in that case that the first value in R is not below an entry of &. Furthermore, since there are only k boxes from the first row of the foundation of T that extend out from , there can be at most k l’s in R. If t > 1 then, when reading the row R from right to left, the lattice condition implies that there is at least one more t — 1 in than there , — 11), are t’s in zic. Since the content of & is (n°, (n 1)’,.. . the only instances when this occurs are when t = cn+ij for j = 1, 2,.. . , n. Therefore every entry of R is an element of RQ,k. Further, if a value t > 1 appeared twice in R, then the lattice condition would be violated. Hence each t t 1, can appear at most once in R. I The next result tells us when we may obtain a SSYT of shape S(A, i, c; k) with lattice reading word from a SSYT of shape )/i) /. with lattice read ing word, where D1 $ D denotes the disjoint union of the diagrams D1 and D2. Lemma 2.2.3 Let c be a composition, ). and be partitions, and k > 0 such that ) — — k < l(o). If T is a SSYT of shape )/t with lattice reading word such that there are at most k 1 ‘s in the first row of )/t, then the tableau of shape S\, ,u, c; k) obtained from T by shifting the foundation )/ to the right is also a SSYT with lattice reading word. Proof Let T be a SSYT of shape )./i & with lattice reading word and let T, be the tableau of shape S\, a, c<; k) obtained from T by shifting the foundation ) to the right. Since shifting A/ to the right does not affect the order in which the entries are read, T has a lattice reading word. Also, the rows of T weakly increase since they are the same as the rows of T. Further, to check that the columns of T, strictly increase, we need only check that they strictly increase at the positions where the two subdiagrams L and are joined. proof of Let R denote the first row of )/ and ci = (, . . . , aj. As in the Lemma 2.2.2, the lattice condition on T implies that the entries of R consist of values of R,k. Further, the value 1 can occur at most k times and the rest of the values of R are distinct. Let q be the number of times 1 appears in R, so that k > q. Further, let r1 r2 r be the entries of R. 31 CHAPTER 2. STAIRCASES WITH BAD FOUNDATIONS Consider the case k 1. Since r1 = = = rq = 1, we have rq = min(R,k) and for each 1 j n we have the (j + 1)-th smallest value of R,k = 1 + jrrl Since k > q, for each 1 j n we have rk rj+q 1 + n+1-i i=1 As illustrated in the diagram below, the entry r+k is beneath boxes. From the unique fillillg of , the entry of & directly above r+k is >lrl n+1—i j fl—i II r T2 rk ri+4 T2+) r3+, 32 CHAPTER 2. STAIRCASES WITH BAD FOUNDATIONS Thus the columns strictly increase. Therefore T is a SSYT with lattice reading word, as desired. Now consider the case when k = 0. Then for each 1 J r > j-th smallest value of R,k 1 + Also, the entry r3 is beneath precisely boxes, so the entry of L, directly above r3 is cn+i. Thus the columns strictly increase. Therefore This a SSYT with lattice reading word, as desired. • Example Let c = (2, 2, 1), = 0, ) = (3, 2), and k = 1. Consider the SSYT T of shape ) with lattice reading word and only one 1 in the first row of ) shown on the left. This gives rise to the SSYT of shape S, c; 1) with lattice reading word shown on the right. 1 2 1 3 2 4 1 3 5 2 6 3 2.3 Sums of Fat Staircases Suppose a diagram D = p/ic is such that SD = vfat::staircase That is, for every v that is not a fat staircase, we have c, = 0. When this happens we say that D is a sum of fat staircases. For each fat staircase v we 33 CHAPTER 2. STAIRCASES WITH BAD FOUNDATIONS let c(zi) be the composition such that = ii. That is, c(v) is the number of parts of i’ equal to i. Then using these compositions we can rewrite SD as 8D= v=fat staircase Example Let p=(4,3,3,3,3,3,3), ic= (2,2,2, 1, 1), and D=p/,. Then SD S(4,32,i,j,i) + S(3,2,i,i,i) + S(3,2,i,i) S(313) = (3,2,1,1) + + Thus, when computing the Schur function of D, we are interested in the following diagrams. D E (3,2,1,1) ‘(3,1,3) L(2,3) For a sum of fat staircases D and a value k 0, we consider the diagram , D; k). By assumption, we have CvS(). = zi=fat staircase When we compute the skew Schur function SS(A,,D;k) we notice that, since the subdiagram D is strictly above the foundation the contents that arise from the fillings of D are precisely the fat staircases 5, for each v with [sJ (SD) # 0. For each of these fillings of D, we then need to extend them by 34 CHAPTER 2. STAIRCASES WITH BAD FOUNDATIONS filling the foundation )/ such that the resulting tableaux are SSYTx with lattice reading words. Thus, when computing the Schur function of S(A, , D; k), we are inter ested in the possible fillings of the following diagrams. -I- The next result summarizes the relationship between these fillings. Theorem 2.3.1 Suppose a diagram D p/ii is such that SD = C,S11. v=fat staircase Then for each partition ) and t with u ), and for each k 0 we have SS(Aq,D;k) s Proof Since D is a sum of fat staircases, the content of each SSYT of shape D with lattice reading word is some fat staircase ii. To prove the iden tity, we consider the map that takes a SSYT T of shape S(A, i, D; k) with lattice reading word and c(D) = v to the tableau T’ of shape S, u, a(zi); k) obtained by filling the copy of with content ii and filling the copy of )/ identically to its filling in T. 35 CHAPTER 2. STAIRCASES WITH BAD FOUNDATIONS We claim that each tableau T’ is also a SSYT with lattice reading word. Given such a tableau T’, consider the corresponding tableau T’ of shape Then T’ is a SSYT since both of the subtableaux of shape A/ and are semistandard. Further T’ has a lattice reading word since checking the lattice condition within the subtableau of shape is trivial, and then, by the construction of T’, the lattice condition after this point simply becomes identical to the lattice condition for T. Therefore ‘T’ is a SSYT with lattice reading word. Further, the first row of the subtableau A/ of T’ coiltams at most k l’s since T was a SSYT. Hence, by Lemma 2.2.3, T’ is also a SSYT with lattice reading word, exactly as claimed. Thus, for each SSYT T of shape $,, D; k) with lattice reading word and c(D) = v we obtain a SSYT T’ of shape S(A, ,u, cv(v); k). Now fix an image T’ and suppose that T1 T are both SSYT of shape S(A, , D; k) with lattice reading word such that T = T’ = T. Let SX, p, c(v); k) be the shape of T = T’ = T. Then the filling of the foundation A/[L in T1 is the same as the filling of the foundation A/si in T2, and the content of the subdiagram D is v for both T1 and T2. Since there are only c SSYT of shape D with lattice reading word and content v, there are at most c, distinct SSYT of shape p, D; k) with lattice reading word that could map to T’. This implies that SS(Aqj,D;k) s > as we desired. I Example Let D=(2,2,2,2,l)/(1,l), )=(2,2),=ø, andk=l. Then = S(13) SD = 8(2,2,2,1) + S(2,2,l,l,l) + (32) is a sum of fat staircases. Here we display D, (1,3), and L(3,2). D L(i,3) (3,2) We are now interested in the diagrams S(\,D;l), S(A, (1,3)1; 1), and S(A, (3, 2)1; 1) for ) = (2, 2). 36 CHAPTER 2. STAIRCASES WITH BAD FOUNDATIONS S(A, D; 1) S(), (1, 3); 1) S(?, (3, 2); 1) We can compute that SS,D;1) = 5(3,3,2,2,1) + 5(3,3,2,1,1,1) + S(3,,i,i,i,i,i) + 5(3,2,2,2,2) + 5(3,2,2,2,1,1) + S(3,2,i,i,i,i) + 5(2,2,2,2,2,1) + 5(2,2,2,2,1,1,1) and SS(A,(1,3);1) + SS,(3,2)<;1) = 2s(3,2,l) + 25(3,3,2,1,1,1) + 5(3,3,1,1,1,1,1) + S(3,2) + 2S(3,2,l,l) + S(3,2,i,i,i,i) + 5(2,2,2,2,2,1) + S(2,,l,i,i), ®j so we see that ® SS(A,D;1) s SS,(1,3);l) + SS,(3,2)1;1), as predicted by Theorem 2.3.1. ®j We now proceed to examine which skew diagrams D are sums of fat staircases. The following results are concerned with adding or removing columns to a skew diagram on the left or right side of the diagram. We recall that D1 D2 denotes the near-concatenation of D1 and D2 of depth i. That is, if the last column of D1 and first column of D2 are both of length greater than or equal to i, then D1 D2 is the is the skew diagram obtained by placing D1 and D2 so that the top-right box of D1 is one step left and i — 1 steps up from the bottom-left box of D2. In other words, the last column of D1 and the first column of D2 overlap in exactly i boxes. Lemma 2.3.2 Let c be a column and D be a skew diagram that is not a sum of fat staircases and let D1 = D ® c and D2 = c D be obtained from D by the addition of a single column c. Then neither D1 nor D2 is a sum of fat staircases. 37 CHAPTER 2. STAIRCASES WITH BAD FOUNDATIONS Proof We shall begin by proving that D2 is not a sum of fat staircases. Since D is not a sum of fat staircases there is a SSYT T of shape D and lattice reading word whose content is v, where v is not a fat staircase. That is, v3 v+ + 2 for some j. Let n be the length of the column c. We create a tableau T’ of shape D2 and lattice reading word by filling c with the numbers 1,2,. . . , n and by filling the rest of D2 as D is filled in the tableau T. Then T’ is semistandard (in). and has a lattice reading word. Further, c(T’) = v + There are three cases to consider. If j < n, comparing the j-th and (j + 1)-th entries of this content gives c(T’) = Vj + 1 vj+i + 3 = c(T’)+i + 2. If j = n, comparing the j-th and (j + 1)-th entries of this content gives c(T’) = Vj + 1 vj+ + 3 c(T’)+i + 3. And if j > n, comparing the j-th and (j + 1)-th entries of this content gives = vj v1 + 2 = c(T’)+i + 2. In all cases we see that c(T’) is not a fat staircase. Hence D2 is not a sum of fat staircases. Then, since SDo = 5D is also not a sum of fat staircases, the above shows that c ® D° is not a sum of fat staircases. Now, since SD1 = 5DGc = S(DGc)o = ScoDo, we find that D1 is not a sum of fat staircases. • Corollary 2.3.3 If D is not a sum of fat staircases and D’ is obtained from D by the addition any number of columns, then D’ is not a sum of fat stair cases. Proof Let D’ be a diagram obtained from D by adding, in order, the columns C1, C2 . . . , Cn. Let D = D0 and, for each i = 1,.. . , n, let D be the subdiagram of D’ consisting of D and the columns c1,. . . , c. Then using Lemma 2.3.2 repeatedly we find that D is not a sum of fat staircases for each i = 1,. . . , n. Since D = D’, we are done. • Corollary 2.3.4 If D is a sum of fat staircases and D’ is a connected subdi agram of D obtained by removing columns, then D’ is a sum of fat staircases. 38 CHAPTER 2. STAIRCASES WITH BAD FOUNDATIONS Proof This is precisely the contrapositive of Corollary 2.3.3 with the roles of D and D’ reversed. I Therefore every diagram D that is a sum of fat staircases can be viewed as a column extension of a smaller diagram D’ that is also a sum of fat staircases. Knowing this, for a given diagram D that is a sum of fat staircases, we now consider what length of columns can be added to D and how much can a column overlap with D if we wish the new diagram to also be a sum of fat staircases. Towards this end, we have the following result. Lemma 2.3.5 Let c be a column, D = p/si be a sum of fat staircases, and D’ be given by either D’ = c® D or D’ = D ® c. If D’ is also a sum of fat staircases, then we have 1(c) + 1 R0(),l and i + 1 for each ii with c0. Proof We shall consider the case = c GD. As in the proof of Lemma 2.3.2, the second case = D Ø c can then be obtained by rotating diagrams by 180. Suppose 1(c) + 1 e R(),l for some ii with 0. Since c 0, there is a SSYT T of shape D with lattice reading word and content v. Now we create a tableau T’ of shape D’ by filling c with the values 1, 2,.. . 1(c) and filling D as in the tableau T. Then T’ is clearly a SSYT with lattice reading word. Further, c(T’) v + (11(c)). Since 1(c) + 1 e RQ(V),I, we have — 1o(v)0_, . ., 1(c) = c(v)+i_ for some k, where z’ (n(”)’,n Thus c(T’) = v + (i) — 1c(v)0_1, . . n+1k — = n , n + 1 — k n ka(v)0_k + (1(v)n(ii)_ 1(t’)0, — = (n + . . . , n + 2 — n kn_k,.. ., Thus c(T’) is not a fat staircase, and so D’ is not a sum of fat staircases. Therefore, if D’ is a sum of fat staircases, then we require that 1(c) + 1 0 Ra(v),l for each v with c,, $ 0. Similarly, suppose that i+1 E R(),l for some v with 0. Since c, 0, there is a SSYT T of shape D with lattice reading word and content v. We create a tableau T’ of shape D’ by filling c with the values 1, 2, . .. , i, 1(u) + 1,1(v) + 2,. . . , 1(v) + 1(c) — i and filling D as in the tableau T. Again, T’ is 11(c)_i). a SSYT with lattice reading word. Further, c(T’) = ii + (U) + (01(v), 39 CHAPTER 2. STAIRCASES WITH BAD FOUNDATIONS ii = Since i + 1 e we have i = for some k, where (nv)n, — 1(M),_1, 1a(v)i) Thus 11(c)_i) c(T’) = v + () + (O(1/)n 1n(v)1, — — ka(v)n+1_k, n — , n + 1 n — + (01(v), 11(c)_i) Q(l/)_i . . k(M)n+1_k, = (n + , n + 2 — — 2a(v)2, 1c(v)l+1(c)_i). Thus c(T’) is not a fat staircase, and so D’ is not a sum of fat staircases. Therefore, if D’ is a sum of fat staircases, then we require that i + 1 Ra(v),l for each v with c 0. • Example Consider the diagram D = (2, 2, 2, 1, 1, 1)/(1), seen below. We have S(2,,i,i,i,) S(42). SD = S(2,,2,l,1) + = (23) + Therefore, when adding a column, we must avoid lengths and overlaps that are one more that an element of R(2,3),i U R(4,2),i = {1, 4, 6} U {1, 3, 7}, if we wish our new diagram to also be a sum of fat staircases. For instance, if we add a column of length 6 to the diagram, then since 6 + 1 E R(2,3),i U R(4,2),i the resulting diagram is not a sum of fat staircases. We show a tableau that demonstrates this when the overlap i = 4. 40 CHAPTER 2. STAIRCASES WITH BAD FOUNDATIONS Since the content of this tableau is i’ = (3, 3, 2, 2, 2, 2), the new diagram = (16) ®. D is not a sum of fat staircases. It is important to note that although the conditions 1(c) + 1 R),i and ‘i + 1 ‘ Rc(v),1 are necessary in Lemma 2.3.5, they are by no means sufficient conditions. Theorem 2.3.6 If D is a sum of fat staircases then the columns of D have distinct lengths. Proof Let D be a sum of fat staircases. Suppose D has n columns c1, . . . , c and, for each j, let m3 be the number of columns of length j. We create a tableau T of shape D by filling each column c with the entries 1, 2,. . , l(cj). Then T is a SSYT with lattice reading word and content = 2(l1(c)) I, Thus, for each j, zi3 is the number of columns of D of length greater than or equal to j. Therefore we have = v3_1 + m, (2.1) for each j. Since D is a sum of fat staircases, ii is a fat staircase. Therefore Equation 2.1 implies that 0 < rn 1 for each j. In other words, the columns of D have distinct lengths. • Example Considerthe diagram D = (5,4,4,3,3,3,2, 1)/(3,2,2, 1,1), which has two columns of length 3. Filling each column c of D with the entries 1, 2, 3,. . . , 1(c) gives the following tableau of content i = (5, 4, 4, 2, 1). 41 CHAPTER 2. STAIRCASES WITH BAD FOUNDATIONS Since ii is not a fat staircase, D is not a sum of fat staircases. We finish this chapter by proving the converse of Theorem 2.3.6 in the case of a diagram with two columns. Lemma 2.3.7 If D is a connected diagram with two columns and these columns have distinct lengths, then D is a sum of fat staircases. Proof Let D be a connected diagram with two columns of distinct lengths. Let T be a SSYT of shape D with lattice reading word and let u be the content of T. Since there are only two columns in T and the entries of each column strictly increases we have v < 2 for each i. Hence, if D is not a sum of fat staircases then there must be a SSYT = T of shape D and lattice reading word with content i (2”), for some n. Since the columns of T strictly increase, this implies that both columns are of length n, contrary to assumption. Therefore D is a sum of fat staircases. I 42 Chapter 3 Hook Foundations In this chapter we consider fat staircases with bad foundations using two simple families for the foundations. In Section 3.1 we shall look at SA, ; k) for hook diagrams A. We shall be able to completely describe which pairs of these fat staircases with hook foundations D1, D2 satisfy 8D1 — SD2 0 and which pairs are Schur-incomparable. We also give a Schur-positivity result using a diagram D that is a sum of fat staircases. In Section 3.2 we will give an explicit formula for computing the Schur positive differences of fat staircases with hook foundations that were dis cussed in Section 3.1. 3.1 Schur Comparability / Incomparability Recall from the introduction, that we write D1 >— D2 whenever 5D1 — 8D2 0, and if we consider the relation ? on the set of all Schur-equivalent classes of diagrams (i.e. [D]8 = {D’jsD = SD’}), then defines a partial ordering. This allows us to view the Hasse diagram for the relation >- on the set of these Schur-equivalent classes. For the sake of convenience, we write D in place of [Dj8. 43 CHAPTER 3. HOOK FOUNDATIONS Example Here we show the Hasse diagram for on the collection of stair cases with bad foundations S(A, 0, 7) (or (i); 0) in the fat staircase notation), for ) varying over all hooks of size 7. A line drawn from a diagram D1 to a diagram D2 in an upwards direction indicates that SD1 — 5D2 0. We note that the diagrams along the top are all Schur-incomparable. That is, they form an anti-chain with regards to >-. Also, the diagrams along the right are all comparable. That is, they form a chain with regards to >—. 44 CHAPTER 3. HOOK FOUNDATIONS Now, if we consider hooks A of size 6, we will obtain the following Hasse diagram. Again, the diagrams along the top are all Schur-incomparable and the diagrams along the right are all comparable. When working with hooks, we shall find it convenient to describe each hook by its arm length and leg length. Hence, we let 1u be the hook (.ua, 1’’) and A be the hook (Aa, P”’). In this chapter we shall use a fixed fat stair , case where c = (cr1, €2, . . . , Thus n is the length of the bottom row of the staircase . 45 CHAPTER 3. HOOK FOUNDATIONS The following results summarise all the ? relationships between diagrams of the form S, ; k) when A is a hook of fixed size h < n+k and 0 < k < 1. The restriction h n + k is needed to guarantee that SA, c; k) is a skew diagram for every hook A of size h. For each pair of hooks A, 1u with Aa, ILa Ei, Theorem 3.1.1 and The orem 3.1.2 each prove one side of the Schur-incomparability of this pair, thus describing the antichain structure displayed along the top of the Hasse diagrams in the previous examples. For each pair of hooks A, u with Aa < Ita, Theorem 3.1.3 shows that S(A, c; k) Js S(, c; k). This describes the chain structure displayed along the right of the Hasse diagrams in the previous examples. Finally, for each pair of hooks A, [t with Aa, I-’i Theorem 3.1.5 show that Aa t’i if and only if S(A, c; k) ‘— S(i, ; k). This describes the relationships between the diagrams displayed on the right with the diagrams displayed along the top in the previous Hasse diagrams. Before we begin, we review some facts regarding any SSYT of shape S(A, c; k) with lattice reading word. Any such tableau will have & filled , with content 6a = (n°, (n — 1)-’,. . . 2, 1’). This content led us to define the set =1_k2_k...n} of size n + k and largest element ci + 1. This set gives the values of the foundation A which can be read immediately after reading . If we read some other value immediately after reading , then the lattice condition will be violated. Hence, when reading the entries of the foundation, in order to read some value x 0 Ra,k without violating the lattice condition, we must first read some value r e Rc,k and read each of r + 1, r + 2, . . . , x — 1 before reading x. Let us finally begin. We start by looking at the antichain structure. Theorem 3.1.1 Let A and it be distinct hooks with Aj = j,a = h n+k and Aa < ita and let 0 k 1. Then we have S(it c; k) S(A, a; k). Proof We shall show that there exists a SSYT T of shape S(A, c; k) with lattice reading word such that there is no SSYT of shape S(it c; k) with lattice reading word having the same content. This is sufficient to prove the theorem. 46 CHAPTER 3. HOOK FOUNDATIONS Since Aa < Ita and A IiI, we have A1 > . Let r1 < r < < rj be the values of R,k. We can create a SSYT of shape A by filling the boxes of A as follows. r1 r_1 + 1 c + A1 It is easy to check that the resulting tableau of shape A & has lattice reading word since each of the entries in the first row of A are from RQ,k. Thus Lemma 2.2.3 provides us with a SSYT ‘T of shape S(A, c; k) with lattice reading word, where A is filled as shown above. Since < A1, we have l(S(jt, a; k)) = oj + 1uj < k + A1. Therefore no SSYT of shape S(t, c<; k) with lattice reading word can contain the entry c + A1. Thus no SSYT of shape S(p,, a<; k) with lattice reading word can have the same content as T. Hence, it follows that SS(;k) — 8S(A,c;k) s 0. I Theorem 3.1.2 Let A and i be distinct hooks with A = = h < n+k and Aa Proof Let r1 < r2 < < rfl4 be the values of R,k. We can create a SSYT of shape i by filling the boxes of p as follows. r1 r2 ra + 1 Since we are using distinct values of R,k, it is easy to check that the resulting tableau of shape [LL has lattice reading word. Thus Lemma 2.2.3 provides us with a SSYT ‘T of shape S(, c; k) with lattice reading word, where is filled as shown above. We now wish to count all SSYT of shape S(i, c; k) (shape S(A, c; k), respectively) with content v = c(T). Since & has a unique way of being filled, we must find all semistandard fillings of i (A, resp.) with the values r1,r2, . . . , rk. Since r1 < r2 < < rh, the value r1 must appear in the top-left corner of j (A, resp.). Further, once we choose — 1 (Aa — 1, resp.) 47 ______ CHAPTER 3. HOOK FOUNDATIONS of the values r2 ,‘ (h—i = [t — 1 and the number of SSYTx of shape S(, c; k) = i’/p with lattice reading word and content v = c(T) is given by (h—i Cp,, i\ ?a1 Since )a < a < we have h + 1 > a + I-a Therefore we have > — h — ‘a Pa 1 and we obtain hAaj>/aij, (3.1) for each i. Therefore (h—i)! — (h — ia)!(iia 1)! aAa1 — (h_i). ri — (h)a)!(Aa1)! h\i = cpvx /ta1 > cv, where we have used Equation 3.10 in the final step. Since we have found a i’ content with c,, > we can conclude that ss(,<;k) — SS(,1;k) 0. I We now depart from looking at the hooks .)., t satisfying Aa < [La < ri. Instead, we turn to the hooks )., [L satisfying [ < /ta. The following theorem shows that we have the chain stucture that was evident in the examples. 48 CHAPTER 3. HOOK FOUNDATIONS Theorem 3.1.3 Let ). and ,u be hooks with j) = h < n + k and ri a cL, Having shown that this inequality holds for any content ii for which a SSYT of shape S(, c; k) with content v and lattice reading word exists, this will imply that ss(,a;k) SS(u,o;k) s 0. Let v be a content such that there is a SSYT 7j of shape S(t, c; k) with content v and lattice reading word. By Lemma 2.2.2 we know that the first row of ji contains a strictly increasing sequence a1 < a < < where each a e R,k. Also, since the columns of 7 strictly increase, the first column of ) contains a strictly increasing sequence a < a < < a, where a = a1. Since can only be filled in one way (in both shapes S(t, c; k) and a1; k)), in order to obtain a tableau of shape S\, a; k) and content i’ we only need to show how to place the values of {ai,a2,...,aa}U{a,a3,...,aj} in If 1(v) > Ic + 1 for this particular content ii then there are entries of 7 greater than a + 1. Since & has content öa, the lattice condition implies that c + 2 appears in [t. Since each a e Rc,k, we have a < c + 1 for a 2 for some > 1. The lattice condition fact each i and so = I°d + j and the that the column strictly increases gives that = a1 = a1 = aI+tj—j+2. Again, by the lattice condition, it is clear that any SSYT of shape a1; k) with lattice reading word and content v must also have these values a, a+1, 49 CHAPTER 3. HOOK FOUNDATIONS the last 1 entries of the first column of A. Since < /ta as — j + Aa 1 < 1. This gives, we have A1 — gives IM—j+lA1, (3.2) so these entries do fit in this column. Note that if 1(u) < a + 1, then this , sequence of values a, a+1,. . . a1 is empty and we do not have to worry about placing any entries larger than al + 1. Let M be the multiset {a1,a2,. . . , a} U {a, as,. . . , a_1}. Then M is the remaining entries that we still need to place in A to obtain a tableau of shape S(A, a; k) and content v. We have IM = ILa + j — 2 and max(M) = al + 1. Let R = {ai,a2,.. .,aa} 0 We note that R { a, a2,. . . , a,La}fl{a, a, . . . , a} since Lemma 2.2.2 shows aula E R,k, which implies aa < al + 2 = a. Thus R is the set of values that appear in both the first row and the first column of t. Since the values of R all appear in the first row of , Lemma 2.2.2 gives that R ç For any SSYT of shape S(A, c; k) with lattice reading word and content v, Lemma 2.2.2 shows that the values in the first row of A are distinct, so, when creating a filling of A, the values in R must also appear in both the first row of A and the first column of A. Consider A {a1,a2,.. . , aa}—R and A’ = {a, a,. . . , a_1}—R. Since we know that the values of R must appear in both the first row of A and first column of A, A U A’ contains the remaining values of M that need to be placed in A. In other words, A U A’ is the set of all values aj + 1 that can appear in exactly one of the first row of A or the first column of A. Since Aa > we have Aa > A1. Thus we have R by choosing Aa — R additional values from (AUA’)flR,k. There are enough values to choose from since there are Ita — > Aa — (3.4) values of (A U A’) 0 R,k present in the first row of i. The sequence of b’s is strictly increasing since (A U A’) n R = 0. — . , C — Now M {b1,.. b} M R contains w = jMj — Aa = [La +j —2— Aa distinct values, each no greater than l + 1. That is, they are an increasing sequence c1 A1 = [La+IMAa 50 CHAPTER 3. HOOK FOUNDATIONS — 1+(/a+j2Aa)+(jti_j+1) = so we may fill A as shown below. b1 b2 bAa Cl C2 cw a+1 By the definitions of R and the sequence of b, it is clear that b1 = a1 = aç. Thus, a1 < c1. Since the sequence of c is strictly increasing and since od+1 < we have That is, the first column of A is increasing. We also had so the first row of A is increasing. Hence this filling gives us a SSYT T of shape A L and content ii. We now check that T has a lattice reading word so that we may apply Lemma 2.2.3 to obtain the desired tableau 7 of shape S(A, a<; k). Suppose that T does not have a lattice reading word. Then, when reading the foun dation A of T, we must reach a point where the lattice condition failed. Let x be the value that, when read, caused the lattice condition to fail. The lat tice condition could not have failed when reading the first row of A since the values in the first row of A were distinct values chosen from R,k. Therefore this x appears somewhere in the first column of A. We consider the the two cases x E Ra,k and x RQ,k. Consider the first case, x e Rc,k. If a value of R,k appears only once in the foundation then reading this value cannot violate the lattice condition. Thus, since the lattice condition failed at this x, this x cannot be the first 51 CHAPTER 3. HOOK FOUNDATIONS time that x was read in A. Since the columns strictly increase, the previous x must have appeared in the first row of A and, since the values in the first row are distinct, this is the only other x in A. Since the content of A is the same as the content of p, these two x’s appear in p as well. Using the fact that ‘Jj has a lattice reading word, together with the content of , we find that the value x —1 appeared in p. Thus the value x —1 also appears in A. Now, since the rows and columns of A strictly increase, either the x — 1 appears in the first column of A above the entry x, or the x — 1 appears in the first row of A left of the entry x. In either case the x — 1 is read before the second x is read and the lattice condition will not fail when reading this second x, contrary to our assumption. We now look at the second case, where x R,k. Again, the x we are interested in appears in the first column of A. There cannot be a second x in A since the column strictly increases and x 0 R,k implies that no other x was placed in the first row of A. As before, x must have appeared in p and, in particular, it also appears somewhere in the first column. Since 7j has a lattice reading word we must read a sequence of values t, t + 1, t + 2, x — 2, x — 1 in p, where t E before we read the x, and we may assume that none of the values t + 1, t +2, ..., x — 2, x — 1 are from Ra,k. Hence each of the values t, t + 1, ..., x — 2, x — 1 also appear in A. None of the values t + 1, t + 2, ..., x — 1 can appear in the first row of A since the first row was . chosen from Ra,k. That is, each value t + 1, t + 2,. . , x — 1 appears in the first column of A. Also, since the rows and columns of A strictly increase, either the t appears either in the first column of A above the entry t + 1, or the t appears in the first row of A. In either case the entire sequence t, t + 1, ., x — 2, x — 1 is read before the x is read in A and the lattice condition does not fail at x, contradicting our assumption. Since T has a lattice reading word, we can now apply Lemma 2.2.3 to obtain the SSYT 7 of shape S(A, c; k) with lattice reading word and con tent xi. Therefore from any SSYT 7 of shape S(p, cv; k) with lattice reading word and content z-’ we can create a SSYT 7 of shape S(A, c; k) with lattice reading word and content xi. Let c be the number of SSYT of shape S(A, c; k) with lattice reading word and content xi, and be the number of SSYT of shape S(p, ; k) with lattice reading word and content xi. We shall show that the sets R and AuA’ that were described above are completely determined by the content xi. That is, without starting with a specific tableau, but only starting with the desired content of a SSYT of some fat staircase with hook foundation with lattice reading word, we show how to determine R, the set of values that 52 CHAPTER 3. HOOK FOUNDATIONS must appear in both the first row and first column of the foundation, and determine AU A’, the set of values less than or equal to c + 1 that can oniy appear in one of the first row or the first column of the foundation. Since L is uniquely filled, from ii we can determine the content of the foundation (A, respectively) needed to create a SSYT of shape S([I, a; k) (S(A, c<; k), resp.) with lattice reading word and content v. From the content of the foundation we can determine the values . a, a+1, . . greater than c+1. Since Lemma 2.2.2 shows that the entries in the first row of the foundation strictly increase, any value that appears twice in the foundation must appear in both the first row of u (A, resp.) and first column of (A, resp.). These values, together with the minimum value in the content of the foundation, give the set R. Then A U A’ is the values in the foundation that are less than or equal to c + 1 but are not in R. After determining the remaining entries of the first row of t (first row of A, resp.), the rest of the foundation is ulliquely determined. Now, to actually form a SSYT of shape S(i, c; k) (S(A, c<; k), resp.) with lattice reading word and content ii, we only need to choose the remaining — ILa RI (Aa — RI, resp.) values from the set (A U A’) fl R,k to place in the first row of ii (first row of A, resp.). Therefore the number of SSYTx of shape S(1, c; k) = ic’/p’ with lattice reading word and content v is given by (AUA’)flR,kI Cp_— I” [LaIRI and the number of SSYTx of shape S(A, c; k) = ic/p with lattice reading word and content v is given by (AUA’)flR,kI — ( AaIRI SinceAA1>jij>j—1,wehave 0 >j —1 Aa. (3.5) Thus, for each i we have [LaRj [1a_jR_+(j_1_Aa) (ILa_Rj)+(j_1_jR)_(AaR)_j IAI+IA’IAaIRI)j — U fl — (A A’) R,kI (Aa — RI) That is, — [La — RI (Au A’) fl R,kI (Aa RI) — i, (3.6) for each i. 53 CHAPTER 3. HOOK FOUNDATIONS Therefore — (AuA’)nR,kl! CPV — — — (l(AuA’) flR,kl — (Aa IRD)!(Aa IRD! — j(AuA’)nR,kl! - U A’) fl - - (j(A R,kl - (ILa l))!(Ita IRD! aa1 /ialRH X TT l(AUA’)flRck(AalRD i=O ILaAa1 ri ILaRli — CPII,X l(AUA’)flRc,kl(AatRDj > where we have used Equation 3.6 in the final step. Since this inequality holds for all contents ii for which there was a SSYT of shape S(p c<; k) with lattice reading word and content ii, we have SS(,;k) — SS(1;k) s 0. • Finally, we turn to the hooks A, satisfying Aa, ii ri. Theorem 3.1.4 Let A and p be hooks with Al = = h < n + k and 1. If then $(A, k) $(, k). Aa, ri and let 0 < k Aa > ; <; Proof We are given that Aa i’i and we wish to show that S(A, cv; k) >- S(jt, c; k). We claim that proof of Theorem 3.1.3, with a few equations verified under the current hypotheses, also proves this theorem. Given the SSYT of shape S(i, cv<; k) with lattice reading word and con tent v, in order to create the SSYT of shape S(A, c; k) with lattice reading word and content ii we needed to check that Equation 3.2, Equation 3.3 and 1Uj 1 RI > Aa — IRI. Since the first two equations were satisfied, we were able to fit the required values into the first row and first column of A. Further, — values fill first of since Ita — RI A RI, there were enough to the row A, and therefore we could construct the tableau with all the desired properties. In order to show Equation 3.2 holds for the assumptions of this theorem, we note that r i < A1. In order to show Equation 3.3 holds for the assumptions of this theorem, we note that RI < 1Uj < Aa. In order to show Eiation 3.4 holds for the assumptions of this theorem, we note that lta Aa. Therefore we can create a SSYT of shape S(A, ; k) with lattice reading word and content ii whenever there exists a SSYT of shape S(u, c; k) with lattice reading word and content ii. 54 CHAPTER 3. HOOK FOUNDATIONS Next, the proof of Theorem 3.1.3 checked that, for each of these con tents ii, the number of SSYT of shape S(A, c; k) with lattice reading word and content v is greater than or equal to the number of SSYT of shape SCu, c; k) with lattice reading word and content ii. To prove this, we first required that Equation 3.5 held. Namely, we required that Aa > j — 1. We used this equation to show that Equation 3.6 held for each i, which gave us the desired inequality for the Littlewood-Richardson numbers. In order to show Equation 3.5 holds for the assumptions of this theorem, we note that Aa > j > j — 1. Therefore the inequality for the Littlewood-Richardson numbers holds here as well, which proves that S5(A,o1;k) — 3S(,o;k) s 0. I Theorem 3.1.5 Let A and be hooks with A = = h < n + k and and let 0 k 1. IfS(A,;k) S(,;k), then Aa i. Proof Towards a contradiction, suppose Aa rh Since we are using distinct values of R,k, it is easy to check that the resulting tableau of shape jiL has lattice reading word. Thus Lemma 2.2.3 provides us with a SSYT T of shape o<; k) with lattice reading word, where p is filled as shown above. We now wish to count all SSYT of shape S(, c; k) (shape S(A, c; k), respectively) with content ii = c(’T). Since & has a unique way of being filled, we must find all semistandard fillings of (A, resp.) with the values r1,T2,. . . , rh. Since r1 < r < < Th, the value r1 must appear in the top-left corner of i (A, resp.). Further, once we choose — 1 (Aa — 1, resp.) of the values r2 < r < Therefore the number of SSYTx of shape S(j, c; k) = ti’/p’ with lattice reading word and content ji = c(’T) is given by K’ (h—i Cpu) = 111 — 1 55 CHAPTER 3. HOOK FOUNDATIONS and the number of SSYTx of shape S(A, o<; k) = ic/p with lattice reading word and content ii = c(T) is given by (h—i Since \a < < we have h + 1 > Aa + jig. Therefore we have — h — )‘a > p 1 and we obtain h)aj>iij, (3.7) for each i. Therefore — (h—i)! C,1? — (h — — 1)! — (hi). x — (ha)!(\ai)! 1Licz1 = u > c1?, where we have used Equation 3.7 in the final step. Since we have found a ii > content with c1? c1?, we can conclude that SS(A,;k) — SS(i,a;k) s 0, which is a contradiction. Therefore we have “a 1M I We finish this section with a result for the Ss(\t,D;1) difference — where D is a sum of fat staircases and ) is a single row. Theorem 3.1.6 Let D = p/si be such that SD = > C,S1?, uzfat staircase and ) be a single row. Then we have SS(t,D;1) — SS(,D;1) s Cv(SS(At,a(v);1) — SS((v);i)) s 0. 56 CHAPTER 3. HOOK FOUNDATIONS Proof Applying Theorem 2.3.1 shows that SS(A,D;) s CvSS(A,c(v)<;1), so we have S8(A,D;1) s — CySS(A,(v);l). (3.8) We now intend to show that SS(t,D;1) = CvSSt,c(v)1;1). (3.9) Then, adding Equation 3.8 and Equation 3.9 will give that SS(,D;1) — SS(A,D;1) s Cy(SS(At,(v);l) — SS((v);1)). To prove Equation 3.9, we first note that Theorem 2.3.1 shows that SSt,D;1) s > CvSS(At,(v)<;1). We now recall how the proof of Theorem 2.3.1 proceeded. The proof looked at the map that took a SSYT T of shape S(.At, D; 1) with lattice reading word \t and c(D) = v to the tableau T’ of shape S(\t, o(v); 1) whose foundation was filled identically as in T. Then for each such image T’ the proof showed that 1. T’ was a SSYT with lattice reading word and c(T’) = c(T), and 2. there are at most c, distinct SSYT of shape s(At, D; 1) that could map to T’. Further, the possible tableau of shape $t, D; 1) mentioned in 2 were the tableaux of shape S(At, D; 1) where the filling of ) was identical to the filling of )‘ in T’ and the filling of D was one of the c, semistandard fillings of D with lattice reading word and content ii. We now check that each of these preimages of T’ is a SSYT of shape S(At, D; 1) with lattice reading word. Let T be a preimage of T’. Then the subtableau of ‘T of shape D is semistandard and has lattice reading word since, as just mentioned, it is one of the c, semistandard fillings of D with lattice reading word and content 11. We also have that the foundation ) of T is semistandard since it is filled 57 CHAPTER 3. HOOK FOUNDATIONS indentically to the foundation of T’, which was semistandard. Further, since ) is a single column and k = 1, the foundation of S(At, D; 1) does not appear directly below any box of D. Therefore T is a SSYT. Also, ‘T has a lattice reading word since D has a lattice reading word and, since we have c(D) = c(/()), after reading D the lattice restrictions on T are identical to the lattice restrictions on T’, which has a lattice reading word. Hence, we have shown that each of these c preimages of T’ is a SSYT of shape $)t, D; 1) with lattice reading word. This proves Equation 3.9. That is, S5(At,D;1) = CvSS(At,(v)l;1). We may now combine Equation 3.8 and Equation 3.9 to obtain SS(,D;1) — SS(A,D;1) s Cj,(S8(t,a(j,)<;1) — SS(,a(v)<;1)). Now, using Theorem 3.1.4 on each pair Ss(At,(v)<;1), SS(A,(v);1) gives S8(At,a(v);1) — SS(\,(v)1;1) 0, for each fat staircase a(v). Therefore, we now have that S8,D;1) — SS(A,D;1) s — SS(),(v);l)) s 0. I Example Once again we consider the example with p = (4, 3,3, 3, 3,3,3), = (2, 2, 2, 1, 1), and D = p/n. We have previous seen that D is a sum of fat staircases. Namely, we have SD = S(4,32,i,i,i) + S(3,2,i,i,i) + S(3,3,2,2,2,1,1) = S(3211) + (313) + We now wish to apply Theorem 3.1.6 to this diagram D using ). = (3). We are interested in appending A to D and each of L\(3,21), L(3,i,), and (2,3,2) and also in appending At to D and each of L(3,i,), and L(2,3). Therefore, we are interested in the following diagrams. 58 CHAPTER 3. HOOK FOUNDATIONS Theorem 3.1.6 states that SS(t,D;1) S3(,D;1) s v(Ss(t,a(zi);1) — SS(A,(z,);1)) s 0. In fact if we compute both Cj, (Ss t,(v)<;1) — SS(,a(v) ;i)) 1 (Ss(At,(3,2,1,1);1) — S8(A,(3,2,1,1);1)) + 1 (sspt,(3,i,)i;i) — SS,(3,i ,3);1)) + 1 (ss(At,(2,3,2ya;1) — SS,(2,3,2y;1)) and SS(At,D;1) then the difference is given by (Ss(t,D;1) — SS(A,D;1)) C,(ss(At,(v)<;1) — — ( = S(5,432,i,i,i) + S(5,42.i,i) + S(5,42,i,1). 59 CHAPTER 3. HOOK FOUNDATIONS Further, one can also check that Cy(5S(,(ziy;l) — SS(A,c(,,))) s 0. 3.2 Expressions for Schur-Positive Differences We now wish extend the proof of Theorem 3.1.3 to give an exact formula for shall find the difference ssp,;k) — SS(u,c;k). In order to state this result we it helpful to think of weak compositions as vectors with non-negative integer entries. When we write the vector z = (z1, z2,z3,. . . , z,) we shall mean the infinite vector z = (z1, z2,z3,. . . ‘Zn, 0, 0,0, .. .). We shall only consider vectors with finitely many non-zero entries and hence we shall only display vectors with finite length, but in this way we unambiguously may add vectors of different lengths. Given a positive integer i, we let e denote the i-th standard basis vec tor. That is, the vector that has its i-th entry equal to 1 and all remaining entries equal to 0. Further, for A being a finite set of positive integers, we the indicator vector of A. Given C A, is called let eA = ZaEA ea be C C a maximal interval ofA ifC is of the form C = {c,c+ 1,c+ 2,...,c+j} where c — 1, c + j + 1 A. We say that two maximal intervals A1, A2 of A are disjoint if A1 fl A2 = 0. It is clear that any finite set A of non-negative integers can be decomposed into a finite collection of maximal disjoint inter vals {A1,A2,.. . , Am}, and that this collection is unique. Finally, when we write min(A) we mean the minimum of the set A in the usual sense. That is, min(A) is the smallest number in A. We begin with the following lemma. Lemma 3.2.1 Let a fat staircase, 0 < k < 1, and h 1. Then 1h-IBI-IC) zi(B, C) = ö + eB + ec + (01, is a partition for each pair of sets B, C satisfying • BC{2—k,3—k,...jnd+1}, CCBflRa,k—min(B), • +1EBifB+Cjh—1, • if B = U=1 B3 where the B3 are the maximal disjoint intervals of B, then min(B) E R,k for each j, and • if C = U1 C3 where the C3 are the maximal disjoint intervals of C, then min(Cj)—1 EB—Cforeachj. 60 CHAPTER 3. HOOK FOUNDATIONS Proof To begin with, 6 is a partition. We shall think of 6, as being the content of & and show that we can read the remaining h values of ii(B, C) while maintaining the lattice condition. This will prove the result. Adding the content eB corresponds to reading the values of B = U=1 B3. Since each J3 is a set of consecutively increasing numbers and each min(B3) e R,k, the lattice condition is maintained while reading each of these maximal disjoint intervals. Therefore the lattice condition is maintained while reading all of B. Next, adding the content ec corresponds to reading the values of C = U1 C3. Since C ç B, this is the second time we are reading each of these values. The fact that each value is in R,k allowed us to read all these values the first time. Now, since each C3 is a set of consecutively increasing numbers and we have previously read a min(C3) — 1 e B — C, the lattice condition is maintained while reading each of these maximal disjoint intervals. Therefore the lattice condition is maintained while reading all of C this time. We note that we have not yet added any values greater than al + 1. Finally, adding (O11+’, lh—IBH-IcI) corresponds to reading the sequence al + 2, al + 3,..., i + h — Bl — IC! + 1. This sequence is non-empty only if BI + CI Theorem 3.2.2 Let A and 1u be hooks with A = p = h < n + k and f1Aa<1ta and letO 8S,o;k) — S5(jy1;k) = aB,cs(B,C) where the coefficients aB,c are given by — C) 1 — C) 1 — ((B flRa,kl ((B flR,kl aB,c— AaC11 ) — aCH1 the partitions v(B, C) that arise are given by 11HBHI0I), zi(B, C) = 6 + eB + ec + (O’, and the sum is over all sets B, C such that 61 CHAPTER 3. HOOK FOUNDATIONS • BC{2—k,3—k,...,cI+1},CCBflR,k—min(B), • Cj + 1 • B fl Rc,ic, • BI+ICI then minC) — 1 B — C for each j. Proof From the proof of Theorem 3.1.3 we saw that, for a given content ii, a SSYT of shape <; k) with lattice reading word and content 1 exists whenever a SSYT of shape S(1i,; k) with lattice reading word and content v exists. Furthermore, when such tableaux exist, we saw that the number of SSYT of shape S(\, c; k) = tc/p with lattice reading word and content 11 is given by I j(AUA’)flR,kI AaIR and the number of SSYT of shape S(t, c; k) = ic’/p’ with lattice reading word and content ii is given by (AuA’)flRa,k — ( [tajR where, for the given content ii, R was the set of values that must appear in both the first row and first column of the foundation of each diagram, and A U A’ are the values less than or equal to °i + 1 that must also appear in the foundation of each diagram but are not in R. We recall that all but one element of R appear twice in the foundation. Namely, all but the smallest value of R, which is necessarily the top-left entry of each foundation. Let v be such that s,, appears in the difference with non-zero coefficient. We consider the following two cases. Case 1: There is a SSYT of shape S(t, c; k) and content ‘i with lattice reading word. Case 2: There is no SSYT of shape S(, c; k) and content v with lattice reading word. 62 CHAPTER 3. HOOK FOUNDATIONS Case 1 will give rise to the terms s where there is a SSYT with content xi and lattice reading word of both shapes S(p, c; k) and $(A, ; k). Case 2 will give rise to the terms s where there is a SSYT with content ii and lattice reading word of shape S, c; k) but not of shape S(p, ; k). We will see that in each case there exists sets B and C with the desired properties such that v(B, C) = xi and aB,c gives the correct coefficients in the difference. Further, we will see that in Case 1 we also have the properties that p C + 1 and Ita lB fl Ra,kI, and in Case 2 we have the property that either p < C + 1 or I.ta> lB fl R,kl. Therefore, in each case, we shall show that for each term s in the difference there are sets B and C with the properties listed in the statement of the theorem together with these extra properties. Then, conversely, in each case, we shall show that for each pair of sets B and C with the properties listed in the statement of the theorem together with these extra properties on p we will obtain a term Sv(B,c) in the difference. Case 1: Since there is a SSYT of shape S(p, a; k) with content xi and lattice read ing word, Theorem 3.1.3 implies that there is also SSYT of shape S, c; k) with lattice reading word and content xi. Since the content of /. is fixed, the contents of the foundations are equal. If we let C be the values less than or equal to I +1 that appear twice in the foundation we obtain RI = id + 1. Further, if we let B be the set of all values less than or equal to ic + 1 that appear in the foundation, we have B—R = AUA’. Since Lemma 2.2.2 shows that the first row of each foundation strictly increases, each value of C must appear in both the first row and the first column of each foundation. Thus for each content xi we obtain a pair B, C. We now show that the sets B and C have the properties listed in the statement of the theorem together with the two extra properties of this case. By the definition of B we have B ç {1, 2,..., ic + 1}. When k = 0 every box in the foundation occurs below some box of z. Hence, since the columns strictly increase, the value 1 cannot appear in the foundation for k = 0. Thus B C {2,3,...,ij+ 1} fork = 0. Therefore, forD < k < 1. we have BC{2—k,3—k,...,ial-+-1}. 63 CHAPTER 3. HOOK FOUNDATIONS Since the multiset of values B U C all appear in each foundation, we have B+IC Further, if jB + C h — 1, then there are entries in each foundation that do not belong to either B or C. By the definition of B and C, these entries must be greater than or equal to cj +2 and then the lattice condition implies that c + 1 must appear in the foundation. That is, ifB+IC jC +1 < A1 and jC + 1 III. Since Lemma 2.2.2 shows that each entry of the first row of A and the each entry of first row of p is in R,k we have BflRc,kl Aa and BflRa,k Pa. Decompose C = U1 C, for disjoint C,. Then, by the definition of C, for each C, min(C,) e C appears twice in the foundation but min(C,) — 1 C does not. The lattice condition implies that min(C,) — 1 appears in each foundation. Thus we have min(C,)—1 eB—Cforeachj. The smallest entry in the foundation is b = min(B), and, as previously mentioned, this value appears in the top-left corner of each foundation. Since both the first row and first column strictly increase, b occurs only once ill each foundation. Hence b C. Since every value of C appears in the first row of the foundation, Lemma 2.2.2 shows that C Ra,k. Also, the definition of B and C implies C C B. Thus we obtain Cc BflR,k—min(B). Now decompose B = U=1 B, for disjoint B. Since min(B,) — 1 B for each j, each value min(B,) — 1 does not appear in either foundation. Hence, the lattice condition requires that min(B,) e for each j. 64 CHAPTER 3. HOOK FOUNDATIONS Therefore for each content ii we obtain B and C with the desired proper > ties. Also, h — BI — Cl denotes the number of values c + 2 that appear in the foundation, and the lattice condition shows that these must be the . . ., — values kl + 2, c + 3, a + h — Bl Cl + 1. Altogether, the entries of /, B, C, and the values c + 2, ll + 3, ..., c + h — BI — CI + 1 make up the entire content v. Hence the content v can be expressed as ih-IBI-ICI) & + eB + cc + (0’, = ii(B, C). Conversely, for any sets B and C with the properties in the statement of the theorem together with the two extra properties, we show that there exists a SSYT of shape S(p, c; k) with lattice reading word and content 1h-IBI-ICI) v(B, C) = ö, + eB + ec + (01, To this end we begin by creating a filling of p with content eB + ec + (0101+1, iHBHIC). The steps of creating the tableau are the same as in the proof of Theorem 3.1.3, but are now phrased in terms of the sets B and C. First, we place the h— B— IC values kl+2, c+3,..., o+h— B — ICI+1 at the bottom of the first column of p. This adds (0kH’, lh—IBHCI) to our content. We then place min(B) in the top-left box of p and place the values of C in both the first row and first column of p, which we can do since p > p ICI+1. Then we choose PaICH1 values from (B—C—min(B))flR,k, which is possible since lB fl R,kj p and C U {min(B)} C to fill the — remaining Pa — CI 1 boxes in the first row of p. We place the remaining values of B — C — min(B) in increasing order in the remaining boxes of the first column of p. This gives a semistandard filling of p. Moreover, we have placed each entry of C twice and each entry of B — C once. This adds 2ec + eB_c = eB + ec to the content. Together with the unique filling of L, this gives a SSYT T of shape S(p, ; k) and content v(B, C). We now show that T has a lattice reading word. Suppose not. Then, when reading the foundation p of T, we must reach a point where the lattice condition failed. Let x be the entry that, when read, caused the lattice condition to fail. The lattice condition could not have failed when reading the first row of p since the values in the first row of p were distinct values chosen from Ra,k. Therefore this x appears somewhere in the first column of p. We consider the the two cases x e R,k and x RG,k. Consider the first case, x e Since the lattice condition failed at this x, this x cannot be the first x to appear in p. Thus there was a previous x in the first row of p. Hence x e C = U1 C3. So x e C, for some j. Let 65 CHAPTER 3. HOOK FOUNDATIONS t = min(C3). Then t — 1 e B — C appears exactly once in and each of the values t, t + 1, . . . , x C appear twice in Since the first row and first column of i strictly increase, we must read the t — 1 before reading the second copy of any of the values t, t + 1,. . . , x, but then the lattice condition does not fail at x, contrary to our assumption. Now consider the second case, x Rc,k. Since x R,k, Lemma 2.2.2 implies that there is no x in the first row. Since the tableau is semistandard this implies that there is only one x in the tableau. If x < + 1, then x E B = U=1B. Thus x B for some i. Let t = min(B) e Then each of the values value t, t+ 1, . . . , x —1 also appears in the tableau and, since the first row and first column of t strictly increase, they are read before x. Thus the lattice condition does not fail at x, contrary to assumption. If x > + 1, then B + C < h — 1. Therefore cj + 1 e B. Further, t — the values in greater than °d + 1 are precisely the h Bj — IC values + 2, c + 3,..., kI + h — B — C + 1, and these all appear at the bottom of the first column of p. Since the o + 1 is read prior to these, the lattice condition could not fail at x, contrary to assumption. Therefore, in Case 1, the contents v for which there exists a SSYT of shape S(, a; k) with content v and lattice reading word are enumerated pre cisely by the contents v(B, C), where B and C satisfy the properties listed in the statement of the theorem together with the two extra properties. Theo rem 3.1.3 showed that, for these v, there is also a SSYT of shape S, c<; k) with content v and lattice reading word. Furthermore, for S(ji, ; k) = ic’/p’ and ; k) = ic/p, we have that (AuA’)nR,k — (I’ - ILaIRI (B—R)nR,kI — ( 1LaH1 (B—C)nR,k{—1 — - ( JLaCH1 and I(AUA’)11?cx,k — ( Cp(B,c) — — I1 I(B1?)11?a,k — — ( ‘\ .aCH1 (BC)flRa,k11 — - ( )aICH1 66 CHAPTER 3. HOOK FOUNDATIONS This shows that aB,c are the correct coeffiecients for the terms Sv(B,C) that arise in Case 1. Case 2: In this case there is no SSYT of shape S(t, c; k) with lattice reading word and content v. However, we are assuming that content v is such that the Schur function s appears in the difference with ion-zero coefficient. Therefore there is a SSYT of shape S()i, c; k) with lattice reading word and content v. Following the same reasoning as in Case 1 we find that there exists sets B and C such that • BC{2—k,3—k,...,Io+i},CCBflRa,k—min(B), • + 1 • • IB+CH h, and Iod+1 EBif B+C< h—i, • if B = U1 B3 where the B3 are the maximal disjoint intervals of B, then min(B3) e R for each j, and • if C = IJL1 C, where the C, are the maximal disjoint intervals of C, then min(C,) — 1 e B — C for each j, Furthermore, given such sets, the only two ways there could not exist a SSYT of shape S(i, c; k) with lattice reading word and content v(B, C) is • there was not enough room in one of the first row or first column of t to fit all of C and min(B), or • after placing min(B) and the values of C in the first row of , there were not enough elements of (B — C — min(B)) fl R to fill the rest of the first row of j. Since I-ti < ILa, the first case implies that uj < C + 1. Since min(B) U C C R,k fl B, the second case implies that B fl Rc,k <1ta. Therefore all the desired properties are satisfied. Conversely, if we are given any sets B and C satisfying the properties listed in the statement of the theorem together with either < Cj + 1 or I BflRcx,k 67 CHAPTER 3. HOOK FOUNDATIONS there is a SSYT of shape $)j, c; k) with lattice reading word and content v(B, C). Moreover, since either p < Cl + 1 or lB fl Ra,kl < Pa, either there is not enough room in the first column of p to fit all of C and min(B) or Lemma 2.2.2 shows that there are not enough values of B to fill the first row of p while satisfying the conditions for a SSYT with lattice reading word. That is, there is no SSYT of shape S(p, ; k) with lattice reading word and content v(B, C). Therefore the contents v for which there exists a SSYT of shape c; k) with content v and lattice reading word but there is no SSYT of shape S(p, a; k) with content v and lattice reading word are enumerated precisely by the contents v(B, C), where B and C satisfy the desired properties of this case. Again, for S(p, c; k) = ic’/p’ and S(A, ; k) = ic/p, we have that — I I(AUA’)flRa,kl Cp(B,c) — — l1 l(BC)flRa,k11 — - (‘ palCH1 and — ( l(AUA’)flRa,kl Cp(B,c) — — RI I(B—C)flRk}—1 — - ( AaICH1 but now, since p > lB fl R,kl where C U {min(B)} c B fl R,k, we have iialCH1 > l(BC)flRa,k11, and so — I(B—C)flR,kI—1 CpI(B,c) ( —o — — — CI 1 — Pa This shows that aB,c are the correct coefficients for the terms Sv(B,C) that arise in Case 2. Thus the formula given for SS(,a;k) — 5Scu,a;k) is correct. • We now give the formula for the Schur-positive differences discussed in Theorem 3.1.4. Theorem 3.2.3 Let ,\ and p be hooks with Al = ll = h < n + k and Aa,pi and let 0< k <1. If Aa then SS(A,a4;k) — SS(,c;k) = aB,cs(B,c) 68 CHAPTER 3. HOOK FOUNDATIONS where the coefficients aB,c are given by — ( I(B—C)flR,kH1 (I(BC)flRa,k1 aB,c— AaCH1 aICH1 the partitions v(B, C) that arise are given by lh-IBI-C), v(B, C) öa + eB + cc + (O1’, and the sum is over all sets B, C such that • Bc{2—k,3—k,...,jc+1}, CcBnR,k—min(B), • C)+1 BflRa,i, • B+CIh, and a+1 e B ifB+C< h—i, • if B = U’=1 B3 where the B, are the maximal disjoint intervals of B, then min(B) E for each j, and • if C = U1 C, where the C, are the maximal disjoint intervals of C, then min(C,) — 1 e B — C for each j. Proof The proof follows as in the proof of Theorem 3.2.2. The only condition on B and C that is different in the statement of this theorem is that we now insist that C + 1 )a instead of jC + 1 This is because Aa < A and all the values of R = C U min(B) must appear in the first row of A. For each v that does appear in the difference, there is one of the two following cases. Case 1: There is a SSYT of shape S(ji, c; k) and content zi with lattice reading word. Case 2: There is no SSYT of shape S(p, c<; k) and content i’ with lattice reading word. As before, in each case there exists sets B and C with the desired prop erties such that v(B, C) = v. This time we can see that in Case 1 we must also have the properties that Pa C + 1 and I-ia B fl and in Case 2 we must also have that either [La < Cj + 1 or 1a> B fl R,kj. 69 CHAPTER 3. HOOK FOUNDATIONS Then the proof that, in each of Case 1 and Case 2, the sets B and C with these desired properties enumerate all partitions v that arise in the difference follows exactly as in the proof of Theorem 3.2.2 using Theorem 3.1.4 in place of Theorem 3.1.3 when appropriate. Also, it once again easy to see that, in each case, the numbers aB,c are the correct coefficients of ii(B, C) in the difference. • Example Let a = (1,1,1,1), ) = (2,1,1), = (3,1) and k = 0. We are interested in the following two diagrams. SQ, a; 0) S(t, a<; 0) For this a, k, we have R,k = {2, 3, 4, 5}. Thus Theorem 3.2.3 gives SS(A,1;O) — SS(,&;O) = aB,cs(B,c), where • B C {2, 3,4, 5}, C C {3, 4, 5}, • lCj+12Bl, • lBl+jCl<4,and5EBiflBl+lCI<3, • if B = U=1 B3 where the B3 are the maximal disjoint intervals of B, then min(B3) e R for each j, and • if C = U1 C3 where the C3 are the maximal disjoint intervals of C, then min(C3) — 1 E B — C for each j. The condition Cl + 1 < 2 gives Cl < 1, and this implies that C is empty, or C is a singleton. If C = {c} is a singleton then IBI + Cl < 4 gives that lBl<3.SinceCCBandmin(C)—1B,wehave{c—1,c}CB.Also, if we assume BI = 2 then BI + IC! < 3, so we require that 5 e B, which implies that C = {5}. Hence, for singleton sets C, we have either 1. C = {3} or C = {4} and BI = 3, or 70 CHAPTER 3. HOOK FOUNDATIONS 2. C = {5} and B <3. If C = {3}, then {2, 3} C B, so the possibilities for B are the sets {2, 3, 4} and {2, 3, 5}. If C = {4}, then {3, 4} B, so the possibilities for B are the sets {2, 3, 4} and {3, 4, 5}. In each of these four cases we can compute (B—C)nR,kJ—1 (j(B—C)nR,kj—l — ( aB,c — AaICH1 )_ taCH1 ( 2—1 ( 2—1 = 2—1—1)3—1—1 - (i (1 - =1-1 =0. If C = {5}, then {4, 5} C B, so the possibilities for B are the sets {4, 5}, {2,4,5} and {3,4,5}. When B is either {2,4,5} or {3,4,5} we once again obtain — ((B—C)nR,kI--1 I I(B—C)flR,k—1 aB,c — AaCH1 )_ /aCH1 2—1 2—1 — “ — ( ( 2—1—1) — 3—1—1 (1 - (1 - =1-1 =0. When B = {4, 5} we have (B—C)flR,k—1 — ( ( (B—C)nR,k—1 aB,c — aCH1 )_ aCH1 1—1 1—1 — 1 ‘\ ( — 2—1—1 )3—1—1 (0 - (o - =1-0 = 1. 71 CHAPTER 3. HOOK FOUNDATIONS This completes the cases when C is a singleton. The only other cases occurwhenC=O. Therefore the possible sets B are {2,5}, {3,5}, {4,5}, {2,3,5}, {2,4,5}, {3, 4, 5}, and {2, 3,4, 5}. When B is one of {2, 5}, {3, 5}, or {4, 5} we have — ((B—C)nR,k—1 N (I(B—C)flR,kH-1 aB,c — aCH1 iaICH1 2—1 2—1 — ( N 1 — 2—0—1}3—0—1 - (iN (1 - =1-0 when B is one of {2, 3, 5}, {2, 4, 5}, or {3, 4, 5} we have (B—C)nR0,k—1 — ( N (I(B—C)flR,k—1 aB,c — XaICH1 ialCH1 3—1 3—1 — ( N ( — 2—0—1)3_0_1 - (2N (2 - =2-1 =1, and when B is {2,3,4,5} we have — ( (B—C)nR,k—1 N I (B—C)flR,kI—1 aB,c — aICH1 i-’aCH1 4—’ 4—’ — I N 1 — 2—0—1)3—0—1 (3 - (3N - =3-3 =0. 72 CHAPTER 3. HOOK FOUNDATIONS Thus we have SS(,1;O) — SS(,c;O) = aB,cs(B,c) = 1S({4,5},{5}) + 1S({2,5},O) + 1S({3,5},O) + lSv({4,5},O) + 1S({2,35},O) + lSti({2,4,5},Ø) + 1S({3,4,5}.O) = S(4,3,2,2,1) + S(4,2,l,1,1,l) + S(4,3,3,l,,l,l) + S(4,3,2,2,l,l,l) +S(4,3,i,i,i) + S(4,,2,2,l,l) + S(4,3,3,2,,l), where we have omitted the details of computing v(B, C) for each B, C. For an example of this, we have zi({4, 5}, {5}) (1,1,1,1) + e + e + (Os, 14_2_1) bE{4,5} c{5} = (4,3,2,1)+ (o,O,o,1,1)+(o,o,o,o,1)+(o,o,o,o,o,1) = (4,3,2,2,2,1). 3.3 Hasse Diagrams for k> 1 In Sections 3.1 and 3.2, we required that 0 < k < 1. For these two values the Hasse diagram we obtained in Section 3.1 was identical and we could explic itly compute the Schur-positive differences in the diagram. As a byproduct, we could then express D as a skew shape of the form D = That is, the partition removed from 1u was a fat staircase. This fact will become necessary for our results in Chapter 5. However, we can generalize the hook foundation results of this chapter by relaxing the restriction on k. We return to the case of fat staircases with hook foundations. As before, we let ,tt be the hook ([ta, 1’), ) be the hook (a, 1’) and & be a fat staircase, where c = (ui, c2,. . . , Now we consider any k satisfying 0 < k h, where h is the size of the hooks under consideration. We shall summarise all the relationships between diagrams of the form S, c; k) when A is a hook of fixed size h < n + k and 0 < k < h. The restriction h < n + k is needed to guarantee that S(A, c; k) is a skew diagram for every hook A of size h. We impose the restriction k < h since, for k h, every diagram of the form S(A, ; k), when A is a hook of fixed size k, is disconnected. Thus, for each k > h, the skew Schur function of 73 CHAPTER 3. HOOK FOUNDATIONS these diagrams factor as SS(A,o1;k) = S)S. In particular SSf),cy1;k) = SS.,c;h) for all k so there is no change among the differences for k h. Further more, we shall see that for k h, none of the differences is Schur-positive. Let us begin with the following example. Example For each 0 k < 6 we show the Hasse diagrams for on the collection of staircases with bad foundations S(A, c; k) for some o = (,. . . , c) where n> 6, and A varying over all hooks of size 6. SS(,o;/c) 8S( ,o;k) S5(tJ,c;k) i i v;k) k=0,1 SS(H iii i,c;k) 11111 ,c;k) 74 CHAPTER 3. HOOK FOUNDATIONS S3(,&;/c) SS(JQi;k) 8S(fl,&;k) Ss( i i hc;k) k=2 I I I SS(j I I I I I k) 8S(,c;k) SS(J&a;k) SS(IJ,a1;k) I I k=3 SS(H III i,a;k) Ss(1 I I I I I 75 CHAPTER 3. HOOK FOUNDATIONS SS(,c;k) SS( ,cx;k) SS(J1,cx;k) I SS(U k=4 S.5(H ,c;k) SS(1 I I I ,c;k) SS(,cy1;k) SS(Jo1;k) SS(LJ,c1;k) I I a;k) k=5 SS(H iii i,;k) SS(1 liii ,c;k) 76 CHAPTER 3. HOOK FOUNDATIONS SS(,c1;k) SS(jc1;k) SS(D,c1;k) SS( I k>6 I I I i,&;k) SS(i 11111 For 0 k 1 we have the same structure of the Hasse diagrams that was discussed in Section 3.1. As k increases, fewer of the relations remain satisfied. Finally, when k 6, there are no Schur-positive differences among these diagrams. We note that the chain structure that was prevalent among the diagrams on the right for 0 k 1 also lost its structure as k increased. In particular, when k = 4, each of the three diagrams on the right still was comparable to at least one of the others, but the three no longer formed a chain. The following results summarise all the ? relationships between diagrams of the form S, c; k) when A is a hook of fixed size h < n+k and 0 < k < h. For each pair of hooks A, with Aa, ILa [ii, Theorem 3.3.1 and The orem 3.3.2 each prove one side of the Schur-incomparability of this pair, thus describing the antichain structure displayed along the top of the Hasse diagrams in the previous example. For each pair of hooks A, ,u with < Aa < ILa, Theorem 3.3.3 and Theorem 3.3.4 shows that S(A, ; k) >- S(i, c; k) if and only if Aa p + k — 1. This describes the relations among those diagrams displayed on the right of the Hasse diagrams in the previous example. Finally, for each pair of hooks A, i with Aa, ,u 77 CHAPTER 3. HOOK FOUNDATIONS only if S, c; k) S(1, <; k), and when k 0 we have Aa > if and 1; only if S(A, c; k) ?r S(j, k). This describes the relationships between the diagrams displayed on the right with the diagrams displayed along the top in the previous Hasse diagrams. Let us finally begin. We start by looking at the antichain structure. Theorem 3.3.1 Let A and t be distinct hooks with Aj = = h n+k and Since Aa < /La and Al = [II, we have A > Let r1 = = = rk = 1 and let rl+k < r2+k < .•• < Tfl be the values of greater than 1. We can create a SSYT of shape A by filling the boxes of A as follows. r1 rk rl+k r2+k rAa_1 + 1 Ic + A1 Using the unique filling of L, it is easy to check that the resulting tableau of shape A Li,- has lattice reading word since each of the entries in the first row of A are from Rc,k and the entry 1 appears k times. Thus Lemma 2.2.3 provides us with a SSYT T of shape S(A, c; k) with lattice reading word, where A is filled as shown above. Since p < A1, we have l(S(,u, c; k)) = c + j < c + A. Therefore no SSYT of shape S(u, c; k) with lattice reading word can contain the entry c + A1. Thus no SSYT of shape S(ji, c; k) with lattice reading word can have the same content as T. Hence, it follows that SS(L,;k) — S8(X,1;k) s 0. I Theorem 3.3.2 Let A and be distinct hooks with Al = = h n+k and Aa We let r1 = = rk = 1 and let r+k < r2+k < ... < rfl+k be the values of R,k greater than 1, where we note n + k h. We can create a SSYT of shape jt by filling the boxes of u as follows. 78 CHAPTER 3. HOOK FOUNDATIONS r1 rk rl+k r2+k Ta+l rh Since r = r2 = = 1 and rl+k < < < rh are distinct values of R1,k, it is easy to check that the resulting tableau of shape p & has lattice reading word. Thus Lemma 2.2.3 provides us with a SSYT T of shape S(p, c; k) with lattice reading word, where p is filled as shown above. We now wish to count all SSYTx of shape S(p, c; k) (shape $(A, c<; k), respectively) with content ii = c(’T). Since & has a unique way of being filled, we must find all semistandard fillings of p (A, resp.) with the values r1,r2,. . . , rh. Since r1 = = = rk 1 and T1+k < r2+k < < rh, the values r1,r2,. . . , rk must appear as the first k values of the first row of p (A, resp.). Further, once we choose p — k (Aa — k, resp.) of the values r1+k, r2+k, . , rh to appear in the first row of p (first row of A, resp.), then the remaining r’s must appear in the first column and the order of all these values is uniquely determined by the semistandard conditions. Therefore the number of SSYTx of shape S(p, c; k) = i’/p’ with lattice reading word and content v = c(T) is given by (h—k p’ii = ‘\ Pa — k and the number of SSYTx of shape S(A, ; k) = ic/p with lattice reading word and content ii = c(T) is given by (h—k CPv=k\Aa_k Since Aa hAai>pakj, (3.10) for each i. Therefore jçl (h—k)! = — (h — pa)!(pa k)! 79 CHAPTER 3. HOOK FOUNDATIONS IL LII — li). TT — — (h_Aa)!(Aa_k)!X [Lak = fjj/ ‘1 [Lak2 > cv, where we have used Equation 3.10 in the final step. Therefore SS(A,i;k) — SSu,1;k) s 0. I We now depart from looking at the hooks ;\qL satisfying a <[La 1. Instead, we turn to the hooks A, [L satisfying i 5 Aa <[La. The following theorems describe the relations among the diagramsr we displayed on the right in our examples. Theorem 3.3.3 Let A and t be hooks with = Ii < Al = [LI n + k and ri Aa <[La, and letO k h. 1fAa [Li+k—1 thenS(A,ci;k) >- S(i,c;k). Proof To prove the result, we shall consider any content v such that a SSYT of shape S([L, c; k) with content v and lattice reading word exists. First we shall show that there is also a SSYT of shape S(A, c; k) with content i’ and lattice reading word. Then, letting S(A, c; k) = ic/p and S([L, c; k) = i’c’/p’ for partitions ic, ic’, p, and p’, we shall show that the Littlewood-Richardson coefficients for these two diagrams and this content satisfy CL) : Having shown that this inequality holds for any content v for which a SSYT of shape S([L, c; k) with content v and lattice reading word exists, this will imply that SS(A,;k) — Ss(,a1;k) s 0. Let ii be a content such that there is a SSYT 7j of shape S([L, ; k) with content v and lattice reading word. By Lemma 2.2.2 we know that the first row of contains at most k l’s. Let q be the number of l’s in the first row of . In the case that q = 0, then we may follow the proof of Theorem 3.1.3 to show that the inequality holds on the Littlewood-Richardson coefficients on these contents. Thus we consider the case q > 1. We write a1 = a2 = = aq = 1. Then the rest of the first row of consists of a strictly increasing sequence aq+1 < aq+2 < < a where each aq+j Ra,k with aq+j > 1. Also, since the columns of 7 strictly increase, the first column of contains a strictly increasing sequence aç < a < < a1, where a = a1 = 80 CHAPTER 3. HOOK FOUNDATIONS 1. Since L can only be filled in one way (in both shapes S(i, c; k) and SEA, c; k)) and the values a1, ..., a must be placed in the first q positions in the first row of either foundation, in order to obtain a tableau of shape S, c; k) and content ii we only need to show how to place the values of a} {a, a,. . . , a1} { a+i, aq+2,. . . , U in If 1(v) > + 1 for this particular content 11 then there are entries of 7j greater than cij + 1. Since & has content öa, the lattice condition implies that + 2 appears in . Since each a E R,k, we have a2 < °i + 1 for each i and so a = c + 2 for some j. The lattice condition and the fact that the column strictly increases gives that = k+3 = Again, by the lattice condition, it is clear that any SSYT of shape 8,; k) with lattice reading word and content v must also have these values a, a+1, a1 the last 4u1 1 as — j + entries of the first column of A. Since Aa < ILa gives < we have A1, — Ii ,u j + 1 p—j+1Aj, (3.11) so these entries do fit in this column. Note that if 1(v) < aj + 1, then this , a1 sequence of values a, a+1,.. . is empty aild we do not have to worry about placing any entries larger than k + 1 into A. (In such a case we may consider j = ILi + 1.) Let M be the multiset {aq+i, aq+2,. . . , a,} U {a, as,. . . , a_}. Then M is the remaining entries that we still need to place in A to obtain a tableau — of shape S(A, c; k) and content v. We have Ml = ILa + j 2 — q and max(M) = cI + 1. Let R = {aq+1,aq+2,.. .,aa} fl ..,a_1}. We note that R = {aq+i, aq+2,. . . , a} fl {a, a, . . . , a1} since Lemma 2.2.2 shows a,a e R,k, which implies a, < c + 2 = a. Thus R is the set of values (except 1) that appear in both the first row and the first column of 1tt. Since the values of R all appear in the first row of , Lemma 2.2.2 gives that R c Rc,k. For any SSYT of shape $(A, <; k) with lattice reading word and content v, Lemma 2.2.2 shows that, besides 1, the values in the first row 81 CHAPTER 3. HOOK FOUNDATIONS of A are distinct, so, when creating a filling of A, the values in R must also appear in both the first row of A and the first column of A.