Classifying the Near-Equality ofRibbon Schur FunctionsbyFoster TomB.Sc., The University of British Columbia, 2016A THESIS SUBMITTED IN PARTIAL FULFILLMENT OFTHE REQUIREMENTS FOR THE DEGREE OFMASTER OF SCIENCEinThe Faculty of Graduate and Postdoctoral Studies(Mathematics)THE UNIVERSITY OF BRITISH COLUMBIA(Vancouver)August 2018c© Foster Tom 2018iiAbstractWe consider the problem of when the difference of two ribbon Schur functions is a single Schurfunction. We prove that this near-equality phenomenon occurs in fourteen infinite families andwe conjecture that these are the only possible cases. Towards this converse, we prove that undercertain additional assumptions the only instances of near-equality are among our fourteen families.In particular, we prove that our first ten families are a complete classification of all cases where thedifference of two ribbon Schur functions is a single Schur function whose corresponding partitionhas at most two parts at least 2. We also provide a framework for interpreting the remaining fourfamilies and we explore some ideas towards resolving our conjecture in general. We also determinesome necessary conditions for the difference of two ribbon Schur functions to be Schur-positive.iiiLay SummaryWe study Schur functions and a variant of them called ribbon Schur functions. Schur functionswere introduced in 1901 and have since found numerous connections from representation theory andalgebraic geometry to quantum physics. Because of these applications, there has been keen interestin determining when the difference of two ribbon Schur functions is a sum of Schur functions.In this thesis, we consider the problem of when the difference of two ribbon Schur functions is asingle Schur function. We call this relationship a near-equality of ribbon Schur functions. We findfourteen cases in which this near-equality occurs and we work towards showing that these are theonly possible cases.ivPrefaceThis dissertation is original, unpublished, independent work by the author, F. Tom.vTable of ContentsAbstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iiiLay Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ivPreface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . vTable of Contents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . viList of Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . viiAcknowledgements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . viii1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32.1 Compositions and partitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32.2 Diagrams and ribbons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52.3 Schur functions and ribbon Schur functions . . . . . . . . . . . . . . . . . . . . . . . 82.4 The Littlewood–Richardson rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102.5 The Jacobi–Trudi determinant . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122.6 Near-equality of ribbon Schur functions . . . . . . . . . . . . . . . . . . . . . . . . . 163 Fourteen Families of Near-Equality . . . . . . . . . . . . . . . . . . . . . . . . . . . 194 Near-Equality with Different Parts . . . . . . . . . . . . . . . . . . . . . . . . . . . . 335 Near-Equality with Different Ends . . . . . . . . . . . . . . . . . . . . . . . . . . . . 365.1 The ends of a composition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 365.2 Littlewood–Richardson tableaux with different ends . . . . . . . . . . . . . . . . . . 365.3 Adjacent pairs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 435.4 Proof of Theorem 5.1.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 466 Near-Equality with Two Large Parts . . . . . . . . . . . . . . . . . . . . . . . . . . 497 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 517.1 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 517.2 Adjacent pairs revisited . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 527.3 Partial sums . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59viList of Notationcα,ν Littlewood–Richardson coefficient Section 2.4e(α) ends of α Definition 5.1.1h complete homogeneous symmetric function Section 2.5k number of 1’s of α Definition 2.2.2`(α) length of α Section 2.1mM (x) multiplicity of element x in multiset M Definition 2.5.4p, p′ profile and modified profile of α Definition 2.2.3q, q′ quasi-profile and modified quasi-profile of α Definition 2.2.3rα ribbon Schur function Section 2.3sν Schur function Section 2.3z non-1 parts of α Definition 2.2.3Eα,M set of lattice points Lemma 5.2.7N size of α, β, and ν Section 2.1M(α) multiset of coarsenings of α Definition 2.5.4P (α) partial sums of α Definition 7.3.1R length of α and β Section 2.1T,U tableaux Section 2.3α, β composition of size N and length R Section 2.1γ composition Section 2.1δα number of end 1’s of α Definition 2.2.2(α) tuple associated to α Definition 5.2.4λ(α) partition determined by α Section 2.1Λ algebra of symmetric functions [21, Chapter 7]µ, τ partitions Section 2.1ν partition of size N Section 2.1χ(P ) indicator of proposition P Definition 2.2.2ω involution on the algebra of symmetric functions Section 2.6ap(α) adjacent pairs of α Definition 5.3.1AP (α) generalized adjacent pairs of α Definition 7.2.5cont(T ) content of T Section 2.3LRα set of LR tableaux of shape α Section 2.4α∗, αt reverse of α, transpose of α Section 2.1ν ′ conjugate of ν Section 2.1α · β, α β, α ◦ β concatenation, near-concatenation, composition Section 2.1≥lex, ≥dom, ≥coar lexicographic, dominance, coarsening orders Section 2.1viiAcknowledgementsI would like to thank my friends and family for their encouragement and support.I would like to thank Omer Angel, Richard Anstee, Dragos Ghioca, Malabika Pramanik, LiorSilberman, and Jozsef Solymosi for the mathematical knowledge and wisdom they have given me.I would like to thank Andrew Rechnitzer for reading my thesis and for providing helpful com-ments.Most of all, I would like to thank my superb supervisor, Stephanie van Willigenburg, for herenthusiasm during many fun hours of research, her thorough scrutiny as I developed my writing,and her advice and guidance through difficult times these last two years. She was the one whofirst introduced me to algebraic combinatorics and so without her none of this would have beenpossible.viiiChapter 1IntroductionWe investigate Schur functions, which form the most esteemed basis for the algebra of symmetricfunctions. Schur functions appear in representation theory both as the characters of the irreduciblerepresentations of the symmetric group under the Frobenius map, and as the characters of the irre-ducible representations of the general linear group [5]. In algebraic geometry, Schur functions and inparticular, their structure constants, the Littlewood–Richardson coefficients, arise in the cup prod-uct of cohomology classes of the Grassmannian. Consequently, problems such as determining thenumber of linear subspaces of Cn that nontrivially intersect a given collection of linear subspaces ingeneral position can be formulated and calculated using Schur functions [5]. Littlewood–Richardsoncoefficients are also connected to eigenvalues of sums of Hermitian matrices [6]. There has beenkeen interest in determining when a symmetric function is Schur-positive, meaning a nonnegativelinear combination of Schur functions; in representation theory, such a function arises from thecorresponding direct sum of irreducible characters. This question has been studied for chromaticsymmetric functions of certain graphs [8] and for generating functions of sets of permutations [4].An especially notorious problem is to classify when the difference of two skew Schur functions isSchur-positive. While partial results exist [1, 11, 12, 15], even the question of determining whentwo skew Schur functions are equal remains unsolved.Fortunately, more progress has been made in the case of ribbon Schur functions, that is, skewSchur functions whose skew shape is connected with no 2× 2 subdiagram. Ribbon Schur functionswere first studied by MacMahon [14] in connection with descents of permutations with repeatedelements, and recently generating functions for permutations with prescribed descent sets havebeen calculated by applying ring homomorphisms to ribbon Schur function identities [20]. Rela-tionships between ribbon Schur functions and Schur functions have also been established based ondescent sets of tableaux [9] and on decompositions of skew diagrams [13]. Billera, Thomas, and vanWilligenburg classified when two ribbon Schur functions are equal in terms of operations on theirunderlying compositions [3], inspiring work on equality of skew Schur functions in terms of similardiagrammatic operations [2, 17, 19]. Necessary and sufficient conditions exist for the difference oftwo ribbon Schur functions to be Schur-positive [16, 22] and the sets of nonzero coefficients in theSchur function expansion are fairly well understood [7, 18].This thesis studies the special case of when the difference of two ribbon Schur functions is a sin-gle Schur function. After the classification of equality of ribbon Schur functions, this near-equality1Chapter 1. Introductionphenomenon is the next most elementary algebraic relationship to investigate and exhibits a min-imal nonzero Schur-positive difference. In representation theory, a near-equality of ribbon Schurfunctions manifests as two representations which differ by a single irreducible, and combinatoriallywe find a relationship between numbers of tableaux.This thesis is structured as follows. In Chapter 2 we introduce the necessary definitions andmachinery, along with worked exercises to gain familiarity. In Chapter 3 we prove fourteen infinitefamilies of near-equalities of ribbon Schur functions, which we conjecture to be the only ones.Working towards this converse, we show that these families are the only near-equalities undercertain additional assumptions. In Chapter 4 we classify near-equalities between ribbon Schurfunctions whose compositions have different parts, allowing us to interpret our first four families.In Chapter 5 we classify near-equalities between ribbon Schur functions whose compositions havedifferent end parts and thereby also those with a different distribution of parts of size 1, explainingthe following six families. In Chapter 6 we prove that these near-equalities account for all of thosewhere the partition in the difference has at most two parts at least 2. These results are summarizedin Chapter 7, after which we define a new statistic on compositions towards explaining the finalfour families. We conclude with further ideas towards a complete classification of near-equality ofribbon Schur functions and some example formulations of the problem of near-equality for moregeneral classes of symmetric functions.2Chapter 2Background2.1 Compositions and partitionsA composition is a finite sequence of positive integers α = α1 · · ·αR. The integers αi are called theparts of α. For convenience we concatenate the parts rather than write (α1, . . . , αR). The length ofa composition α, denoted by `(α), is the number of parts R of α and the size of α is the sum of itsparts N =∑Ri=1 αi. When α has m consecutive equal parts αi+1 = · · · = αi+m = j we will oftenabbreviate this subsequence as jm. A composition α is called a partition if its parts are weaklydecreasing, that is, α1 ≥ · · · ≥ αR. A composition α determines a unique partition λ(α) given byreordering its parts in weakly decreasing order.The reverse of a composition α = α1 · · ·αR is the composition α∗ = αR · · ·α1 given by readingthe parts of α in reverse order. We say that α is symmetric if α = α∗ and nonsymmetric otherwise.The set of compositions of size N is in bijection with the set of subsets of {1, . . . , N −1} as follows.Given a composition α = α1 · · ·αR of size N , define the subsetset(α) = {α1, α1 + α2, . . . , α1 + · · ·+ αR−1} ⊂ {1, . . . , N − 1}, and set(N) = ∅.Conversely, given a set A = {a1, . . . , as} ⊂ {1, . . . , N − 1}, where a1 < · · · < as, define thecomposition of size Ncomp(A) = a1(a2 − a1) · · · (as − as−1)(N − as), and comp(∅) = N.Now we define the transpose of a composition α, denoted αt, as the compositionαt = (comp((set(α))C))∗given by reversing the composition associated to the complement in {1, . . . , N − 1} of the setassociated to α. We will give an example shortly and in Section 2.2, we will present a more intuitiveway of interpreting and calculating the transpose. We also define the conjugate of a partition ν tobe the partition ν ′ whose parts areν ′j = |{i : νi ≥ j}|,that is, ν ′j is the number of parts of ν that are at least j. Note that if a composition α happens32.1. Compositions and partitionsto be a partition, we will see below that we do not have αt = α′ in general, so we distinguish thetranspose and the conjugate.Example 2.1.1. The compositions α = 313 and β = 412 have length 3 and size 7. The compositionα determines the partition ν = λ(α) = 331. Because α∗ = 313 = α, α is symmetric, while becauseβ∗ = 214 6= β, β is nonsymmetric. Let us now compute the transposes of α, β, and ν. We havethatset(α) = {3, 4} ⊂ {1, . . . , 6}, (set(α))C = {1, 2, 5, 6}, comp((set(α))C) = 11311, αt = 11311.Similarly,βt = (comp((set(412))C))∗ = (comp(({4, 5})C))∗ = (comp({1, 2, 3, 6}))∗ = (11131)∗ = 13111andνt = (comp((set(331))C))∗ = (comp(({3, 6})C))∗ = (comp({1, 2, 4, 5}))∗ = (11212)∗ = 21211.The conjugate of ν is ν ′ = 322 because ν has three parts at least 1, two parts at least 2, and twoparts at least 3. Note that νt 6= ν ′.Let α = α1 · · ·αR and β = β1 · · ·βR′ be compositions. We define three compositions arisingfrom α and β. The concatenation of α and β is the compositionα · β = α1 · · ·αRβ1 · · ·βR′ .The near-concatenation of α and β is the compositionα β = α1 · · ·αR−1(αR + β1)β2 · · ·βR′ .Finally, the composition of α and β [3, Section 3.1] is the compositionα ◦ β = βα1 · · · · · βαR ,where βαi denotes the near-concatenation of αi copies of β. We define three partial orders oncompositions. We define the lexicographic order by α ≥lex β if either α = β or αi > βi at thesmallest index i at which αi 6= βi. We define the dominance order by α ≥dom β ifα1 + · · ·+ αi ≥ β1 + · · ·+ βifor every i, where by convention αi = 0 for i > R and βi = 0 for i > R′. If α ≥dom β, then α ≥lex β[21, Page 289]. Finally, we say that α is a coarsening of β, denoted α ≥coar β, if α can be obtainedfrom β by summing adjacent parts of β.42.2. Diagrams and ribbonsExample 2.1.2. Let α = 313, β = 412, and ν = 331 as in Example 2.1.1. The concatenation,near-concatenation, and composition of α and β are given by α · β = 313412, α β = 31712, andα ◦ β = (412)3 · (412) · (412)3 = 4161612 · 412 · 4161612.We have that β >lex α because β1 = 4 > 3 = α1 and β >dom α because in additionβ1 + β2 = 5 > 4 = α1 + α2 and β1 + β2 + β3 = 7 = α1 + α2 + α3.We also have that β >coar βt because β can be obtained from βt as β = 412 = (1 + 3)1(1 + 1).Note that β and ν are incomparable in dominance order because we have β1 = 4 > 3 = ν1 andβ1 + β2 = 5 < 6 = ν1 + ν2.Throughout this thesis, the letters α and β will always denote compositions of size N and lengthR and the letter ν will always denote a partition of size N .2.2 Diagrams and ribbonsWe define the diagram of ν to be the left-justified array of cells with νi cells in the i-th row. Weuse the English convention, where rows are counted from the top. We define the ribbon diagram ofα to be the array of cells with αi cells in the i-th row and where the rightmost cell of the (i+ 1)-throw is directly below the leftmost cell of the i-th row.Example 2.2.1. The diagrams of ν = 331 and ν ′ = 322 and the ribbon diagrams of α = 313,αt = 11311, β = 412, and βt = 13111 are shown below.Taking the conjugate of a partition corresponds to reflecting its diagram across the diagonal andsimilarly taking the transpose of a composition corresponds to reflecting its ribbon diagram acrossthe diagonal from the top left corner towards the bottom right.We now define some useful parameters that describe the distribution of the parts of α equal to1. One application will be to provide a formula for αt in Proposition 2.2.5.Definition 2.2.2. Let k = |{i : αi = 1}| be the number of parts of α that are equal to 1. Let δαdenote the number of end parts of α that are equal to 1; to be precise,δα = χ(α1 = 1) + χ(αR = 1),52.2. Diagrams and ribbonswhere for a proposition P , we define the indicatorχ(P ) =1 P is true0 P is false.Throughout this thesis, the letter k will always denote the number of parts of α that are equal to1. We will always assume that k ≤ R − 1, meaning that α 6= 1R, because as we will see in Section2.6, this case does not produce any near-equalities and so is not needed for our purposes.Definition 2.2.3. Write α asα = 1p1z11p2z2 · · · 1pR−kzR−k1pR−k+1where the pi ≥ 0 and the zi ≥ 2. Now define the following sequences of nonnegative integers. Thenon-1 parts of α isz(α) = z1 · · · zR−k.The profile of α isp(α) = p1 · · · pR−k+1.The quasi-profile of α isq(α) = q0q1 · · · , where qj = |{i : pi = j}|is the number of occurrences of the integer j in the profile of α. We also define the modified profileof α by subtracting the first and last integers of the profile of α by 1, that isp′(α) = p′1p′2 · · · p′R−kp′R−k+1 = (p1 − 1)p2 · · · pR−k(pR−k+1 − 1)and we define the modified quasi-profile of α to beq′(α) = q′0q′1 · · · , where q′j = |{i : p′i = j}|.The quasi-profile and modified quasi-profile of α have an infinite tail of zeroes, which we omit forbrevity. Sometimes we will write zi(α), pi(α), qi(α), p′i(α), or q′i(α) to stress the dependence on α.Throughout this thesis, the letters z, p, q, p′, and q′ will always refer to the non-1 parts, profile,quasi-profile, modified profile, and modified quasi-profile of α respectively.Example 2.2.4. Let α = 313 = 10 3 11 3 10. Then we have thatk = 1, δα = 0, z(α) = 33, p(α) = 010, p′(α) = (−1)1(−1), q(α) = 21, and q′(α) = 01.We can now provide an easy formula for the transpose of a composition.62.2. Diagrams and ribbonsProposition 2.2.5. The transpose of α is given byαt = (p′R−k+1 + 2)1zR−k−2(p′R−k + 2) · · · (p′2 + 2)1z1−2(p′1 + 2).Proof. Recalling thatα = 1p1z11p2 · · · 1pR−kzR−k1pR−k+1 ,we have thatset(α) = {1, 2, . . . , p1, p1 + z1, p1 + z1 + 1, . . . , p1 + z1 + p2, p1 + z1 + p2 + z2, . . .}=⇒ (set(α))C = {p1 + 1, p1 + 2, . . . , p1 + z1 − 1, p1 + z1 + p2 + 1, p1 + z1 + p2 + 2, . . .}=⇒ comp(set(α))C = (p1 + 1)1z1−2(p2 + 2) · · · (pR−k + 2)1zR−k−2(pR−k+1 + 1)=⇒ αt = (pR−k+1 + 1)1zR−k−2(pR−k + 2) · · · (p2 + 2)1z1−2(p1 + 1)=⇒ αt = (p′R−k+1 + 2)1zR−k−2(p′R−k + 2) · · · (p′2 + 2)1z1−2(p′1 + 2), as desired.Example 2.2.6. Let β = 412 = 10 4 11 2 10. Then the transpose of β is given byβt = (−1 + 2) 12−2 (1 + 2) 14−2 (−1 + 2) = 13111.We will now use our formula in Proposition 2.2.5 to deduce several results relating the transpose,modified quasi-profile, and properties of the ribbon diagram of α.Proposition 2.2.7. We have the following.1. δαt = 2− δα2. λ(αt) = λ(1N−2R+k+2−δα 2q′0 3q′1 · · · )3. q′j is the number of columns of the ribbon diagram of α of length (j + 2)4.∑j≥0 qj = R− k + 1 and∑j≥0 q′j = R− k − 1 + δαProof.1. Note that p′i + 2 ≥ 2 except possibly for i = 1 and i = R− k+ 1, for which p′i + 2 = 1 exactlywhen αR = zR−k ≥ 2 and α1 = z1 ≥ 2 respectively. So by Proposition 2.2.5,δαt = χ(p′R−k+1 + 2 = 1) + χ(p′1 + 2 = 1) = χ(αR 6= 1) + χ(α1 6= 1) = 2− δα.72.3. Schur functions and ribbon Schur functions2. By Proposition 2.2.5, the number of 1’s in αt is(zR−k − 2) + · · ·+ (z1 − 2) + χ(p′1 = −1) + χ(p′R−k+1 = −1)= (z1 + · · ·+ zR−k)− 2(R− k) + (2− δα)= (N − k)− 2(R− k) + (2− δα) = N − 2R+ k + 2− δα.Meanwhile, for j ≥ 2 the number of j’s in αt is precisely the number of i such that p′i+ 2 = j,namely q′j−2.3. The number of columns of the ribbon diagram of α of length (j + 2) is equal to the numberof rows of the ribbon diagram of αt of length (j + 2), which is q′j by (2).4. We have∑j≥0 qj = |{i : pi ≥ 0}| = R−k+1 because we are including all of the pi. However,for∑j≥0 q′j , we are excluding when p′1 = −1 and when p′R−k+1 = −1, so∑j≥0q′j = (R−k+1)−χ(p′1 = −1)−χ(p′R−k+1 = −1) = (R−k+1)− (2−δα) = R−k−1+δα.2.3 Schur functions and ribbon Schur functionsWe now introduce tableaux, from which we define our two main objects of study, namely Schurfunctions and ribbon Schur functions.A semistandard Young tableau (SSYT) of shape ν is a filling of the cells of the diagram of νwith positive integers so that the integers in every row are weakly increasing from left to right andthe integers in every column are strictly increasing from top to bottom. Similarly, an SSYT ofribbon shape α is a filling of the cells of the ribbon diagram of α with positive integers so that rowsare weakly increasing and columns are strictly increasing as before.Given an SSYT T of shape corresponding to a partition or of ribbon shape corresponding toa composition, we use Ti,j to refer to the integer in the i-th row and j-th column of T . We alsodefine the content of T to be the sequencecont(T ) = cont1(T )cont2(T ) · · ·where conti(T ) is the number of i’s in T . Again we omit the tail of zeroes.Example 2.3.1. Below are four SSYTs of shape ν = 331 and two SSYTs T and U of ribbon shape82.3. Schur functions and ribbon Schur functionsα = 313. We have T2,3 = 2, cont(T ) = 421, U1,5 = 2, and cont(U) = 241.1 1 22 2 461 1 12 2 461 2 23 3 461 2 32 3 46T = 1 1 121 2 3U = 1 2 221 2 3Now for a partition ν we define the Schur function sν to be the formal power series in infinitelymany commuting variables (x1, x2, . . .) bysν =∑T an SSYT of shape νxcont1(T )1 xcont2(T )2 · · · .Fact 2.3.2. [21, Corollary 7.10.6] The Schur functions {sν : ν a partition} form a basis for thealgebra of symmetric functions Λ.Definition 2.3.3. A symmetric function F ∈ Λ is Schur-positive if when expanded in the Schurfunction basis asF =∑νcνsν ,all of the coefficients cν are nonnegative. For symmetric functions F,G ∈ Λ, we writeF ≥s Gif the difference F −G is Schur-positive.Similarly, for a composition α we define the ribbon Schur function rα byrα =∑T an SSYT of ribbon shape αxcont1(T )1 xcont2(T )2 · · · .Fact 2.3.4. [21, Theorem 7.10.2] The ribbon Schur function rα belongs to Λ.Example 2.3.5. Below are the terms of s331 corresponding to the four SSYTs of shape 331 fromExample 2.3.1.s331 = · · ·+ x21x32x4x6 + · · ·+ x31x22x4x6 + · · ·+ 2x1x22x23x4x6 + · · · .Below are the two terms of r313 corresponding to the two SSYTs of ribbon shape 313 from Example2.3.1.r313 = · · ·+ x41x22x3 + · · ·+ x21x42x3 + · · · .In Sections 2.4 and 2.5 we will present the two main tools that we use to calculate with Schurfunctions and ribbon Schur functions, namely the Littlewood–Richardson rule and the Jacobi–Trudideterminantal identity.92.4. The Littlewood–Richardson ruleBecause the Schur functions form a basis for the algebra of symmetric functions, we of coursehave sν = sµ only if ν = µ. However, this is not the case for ribbon Schur functions. Before weproceed, we address the question of when two ribbon Schur functions rα and rβ corresponding totwo different compositions are in fact equal. Billera, Thomas, and van Willigenburg provide thefollowing classification in terms of factorizations as compositions of compositions.Theorem 2.3.6.1. [3, Proposition 3.3] The set of compositions is a monoid under the operation ◦. Therefore, at-fold composition of compositions α(1) ◦ · · · ◦ α(t) is well-defined.2. [3, Proposition 3.9] For compositions α and β we have(α ◦ β)∗ = α∗ ◦ β∗.3. [3, Theorem 4.1] Two compositions α and β satisfy rα = rβ if and only if for some t we canfactorizeα = α(1) ◦ · · · ◦ α(t) and β = β(1) ◦ · · · ◦ β(t)where for each 1 ≤ i ≤ t, either β(i) = α(i) or β(i) = (α(i))∗. In particular, rα = rα∗ .Definition 2.3.7. A composition α is simple if the equality rα = rβ holds only for β = α andβ = α∗.We will see in Proposition 3.1.6 that all of the compositions we will study are simple. Therefore,we will frequently view compositions up to equivalence under reversal when considering their ribbonSchur functions.2.4 The Littlewood–Richardson ruleWe now introduce the first main combinatorial tool that we will use to calculate with ribbon Schurfunctions.Definition 2.4.1. Let T be an SSYT of ribbon shape α. The reverse reading word of T is thesequence of entries of T taken from right to left and top to bottom. A sequence of integers a1 · · · anis a lattice permutation if for every initial subsequence a1 · · · aj the number of i’s is at least thenumber of (i+ 1)’s for every i ≥ 1.Now T is a Littlewood–Richardson (LR) tableau of shape α if the reverse reading word of T isa lattice permutation. This condition is called the lattice word condition. We denote by LRα theset of all LR tableaux of shape α.102.4. The Littlewood–Richardson ruleExample 2.4.2. The tableau T from Example 2.3.1 is an LR tableau because its reverse readingword 1112321 is a lattice permutation. The tableau U from Example 2.3.1 is not an LR tableaubecause its reverse reading word 2212321 is not a lattice permutation as the initial substring 2contains more 2’s than 1’s. Note that the lattice word condition ensures us that the content of anyLR tableau is a partition.To gain familiarity with LR tableaux, let us examine some of their basic properties.Proposition 2.4.3. Let T be an LR tableau. Then the following hold.1. The number of i’s in the first j rows of T is at least the number of (i+ 1)’s in the first (j+ 1)rows of T .2. If there is an i in the j-th row of T , then i ≤ j.3. The rightmost column of length at least 2 is filled with the integers 1 through j in increasingorder from top to bottom, where j is the length of this column.4. The top row of length at least 2 is filled with (αj − 1) 1’s followed by a j from left to right,where this is the j-th row from the top.Proof.1. Because each row of T is weakly increasing, the (i + 1)’s in row (j + 1) occur earlier in thereverse reading word of T than the i’s in row (j + 1). Therefore, if a1 · · · an is the reversereading word of T and as is the leftmost (i+ 1) entry in the (j + 1)-th row of T , then in theinitial subsequence a1 · · · as, the number of i’s is the number of i’s in the first j rows of T ,which must be at least the number of (i + 1)’s, which is the number of (i + 1)’s in the first(j + 1) rows of T .2. This holds when i = 1. For i ≥ 2, if there is an i in the j-th row of T then by (1) there is an(i− 1) in the first (j − 1) rows of T , so i− 1 ≤ j − 1 by induction on i.3. The top entry of the rightmost column of T is in the first row of T and so must be a 1, thenbecause the entries of the column must be strictly increasing, the i-th entry of this column isat least i by induction on i and at most i by (2).4. For 2 ≤ i ≤ j there has been only one (i − 1) in the first (j − 1) rows of T by (3) so therecan be at most one i in the first j rows of T by (1), which is already present in the rightmostcolumn of T again by (3). Therefore, the first (αj − 1) entries of the j-th row must all be 1’sand the αj-th entry of the j-th row is an αj .112.5. The Jacobi–Trudi determinantExample 2.4.4. The three LR tableaux of shape α = 313 and the two LR tableaux of shapeβ = 412 are given below.1 1 121 1 31 1 121 2 31 1 122 2 31 1 1 121 31 1 1 122 3We now state the Littlewood–Richardson rule.Theorem 2.4.5. [21, Theorem A1.3.3] We have the following identity.rα =∑T∈LRαscont(T )Example 2.4.6. From the LR tableaux enumerated in Example 2.4.4, we see thatr313 = s511 + s421 + s331 and r412 = s511 + s421.Note thatr313 − r412 = s331.By collecting LR tableaux by content, Theorem 2.4.5 can equivalently be stated asrα =∑νcα,νsν ,where the Littlewood–Richardson (LR) coefficient cα,ν is the number of LR tableaux of shape αand content ν. Because cα,ν ≥ 0, we see that rα is Schur-positive.2.5 The Jacobi–Trudi determinantOur second main combinatorial tool explores the relationship between Schur functions, ribbon Schurfunctions, and the basis of complete homogeneous symmetric functions, which we now introduce.For an integer n define the n-th complete homogeneous symmetric function hn ashn =∑i1≤···≤inxi1 · · ·xin .Note that h0 = 1 and hn = 0 when n < 0. Now for a partition ν we define the complete homogeneoussymmetric function hν ashν = hν1 · · ·hν`(ν) .122.5. The Jacobi–Trudi determinantFact 2.5.1. [21, Corollary 7.6.2] The complete homogeneous symmetric functions {hν : ν a partition}form a basis for the algebra of symmetric functions Λ.We now have the following determinantal identity, which describes how to expand a Schurfunction in the complete homogeneous symmetric function basis, hereafter abbreviated as the h-basis.Theorem 2.5.2. [21, Theorem 7.16.1] We have the following identity.sν = det(hνi−i+j)i,jExample 2.5.3. For ν = 331 we haves331 = deth3 h4 h5h2 h3 h40 1 h1 = h331 − h421 − h43 + h52.There is also a similar determinantal identity for ribbon Schur functions, but it is more conve-niently expressed in terms of coarsenings.Definition 2.5.4. The multiset of coarsenings of α, denoted byM(α), is the multiset of partitionsdetermined by all coarsenings of α, that is,M(α) = {λ(β) : β ≥coar α}.For any multiset M we denote by mM (x) the multiplicity of x in M .Example 2.5.5. For α = 313 and β = 412 we haveM(α) = {331, 43, 43, 7}, M(β) = {421, 43, 52, 7}, and mM(α)(43) = 2.To clarify our understanding of this concept, let us prove the following proposition.Proposition 2.5.6.1. The multiset of coarsenings M(α) has (R−1R−`) = (R−1`−1) elements (with multiplicity) of length`. In particular, the total number of elements is |M(α)| = 2R−1.2. The partition µ = λ(z1 · · · zR−k21k−2) has multiplicity mM(α)(µ) = 2k −R− 1 + q0.Proof.1. We can visualize a coarsening of α by inserting a sequence of (R− 1) symbols, each either a‘+’ or a ‘·’, between the parts of α. A coarsening of length ` is (R − `) parts shorter than αso must have (R − `) symbols that are chosen to be a ‘·’ and (`− 1) that are chosen to be a‘+’ .132.5. The Jacobi–Trudi determinant2. The partition µ arises from α precisely by joining a pair of adjacent 1’s. A sequence of pi 1’sgives rise to exactly max{pi− 1, 0} such adjacent pairs, which is (pi− 1) except when pi = 0,in which case we add one to compensate. ThereforemM(α)(µ) =R−k+1∑i=1(pi − 1 + χ(pi = 0)) =R−k+1∑i=1(pi − 1) + |{i : pi = 0}|=k − (R− k + 1) + q0 = 2k −R− 1 + q0.Example 2.5.7. Let α = 1116311. The coarsening γ = 275 can be visualized byγ = 1 + 1 · 1 + 6 · 3 + 1 + 1.The partition µ = 632111 arises from joining a pair of adjacent 1’s in α and has multiplicity2 + 0 + 1 = 3. Indeed, 2k −R− 1 + q0 = 10− 7− 1 + 1 = 3.We can now expand a ribbon Schur function in the h-basis as follows.Theorem 2.5.8. [3, Equation (2.6)] We have the following identity.rα =∑ν∈M(α)(−1)R−`(ν)hνExample 2.5.9. From the multisets of coarsenings found in Example 2.5.5, we see thatr313 = h331 − 2h43 + h7 and r412 = h421 − h43 − h52 + h7.Note that from Example 2.5.3, we see again thatr313 − r412 = h331 − h421 − h43 + h52 = s331.Let us now make some observations regarding the terms in the h-basis expansions of Schurfunctions and ribbon Schur functions.Proposition 2.5.10.1. We can writesν = hν +∑µ>domνaµ(ν)hµfor some coefficients aµ(ν). In other words, hν is the unique dominance-minimal, and thereforelexicographically least term of the h-basis expansion of sν .142.5. The Jacobi–Trudi determinant2. We can writehν = sν +∑µ>domνa′µ(ν)sµfor some coefficients a′µ(ν).3. We can writerα = hλ(α) +∑µ>domλ(α)bµ(α)hµfor some coefficients bµ(α).4. We can writerα = sλ(α) +∑µ>domλ(α)b′µ(α)sµfor some coefficients b′µ(α).Proof.1. By Theorem 2.5.2, we have thatsν = det(hνi−i+j)i,j =∑w∈S`(ν)(−1)sgn(w)`(ν)∏i=1hνi−i+w(i) =∑w∈S`(ν)(−1)sgn(w)hλ((ν1−1+w(1))··· ).Now for each permutation w and i the partition µw = λ((ν1 − 1 + w(1)) · · · )) satisfiesµw1 + · · ·+ µwi ≥ (ν1 − 1 + w(1)) + · · ·+ (νi − i+ w(i)) ≥ ν1 + · · ·+ νiwith equality if and only if w({1, . . . , i}) = {1, . . . , i}; if we have this for all i then w is theidentity so (−1)sgn(w) = 1 and we have the term hν .2. Write hν in the Schur basis and let a′ν0sν0 be any dominance-minimal term, so thathν = a′ν0sν0 +∑µdomν0a′µsµ.Now use (1) to write the right hand side ashν = a′ν0sν0+∑µdomν0a′µsµ = a′ν0hν0+a′ν0∑µ0>domν0aµ0(ν0)hµ0+∑µdomν0a′µ(hµ+∑τ>domµaτ (µ)hτ ).Now the dominance conditions ensure that there is no hν0 term in any of the summations tocancel the a′ν0hν0 term, so we must have a′ν0 = 1 and ν0 = ν, that is, any dominance-minimalterm in the Schur function expansion of hν is specifically sν .152.6. Near-equality of ribbon Schur functions3. We have mM(α)(λ(α)) = 1. On the other hand, note thatλ(α1 · · ·αj−1(αj + αj+1)αj+2 · · ·αR) >dom λ(α1 · · ·αR)because for any i parts on the right hand side, the left hand side has (i−1) or i parts with anequal or greater sum. Therefore, if β >coar α, then by induction on the number of adjacentparts of α that were summed, we have that λ(β) >dom λ(α) so the result follows by Theorem2.5.2.4. This now follows directly from (2) and (3).2.6 Near-equality of ribbon Schur functionsThe equationr313 − r412 = s331from Examples 2.4.6 and 2.5.9 exhibits the curious situation of two ribbon Schur functions thatdiffer by a single Schur function. Our goal is to classify all those α, β, and ν for which we haverα − rβ = sν . (2.1)Note that if (2.1) holds we would have in particular that the difference rα−rβ is Schur-positive.By definition of ribbon Schur functions and by [18, Lemma 3.8], α and β must have the same sizeN and the same length R, which is why we make these assumptions, as well as that α 6= 1R.By the Littlewood–Richardson rule, (2.1) implies thatcα,µ =cβ,µ µ 6= νcβ,µ + 1 µ = ν. (2.2)So one way of studying near-equality is by enumerating LR tableaux.Additionally, if we expand (2.1) in the h-basis using Theorems 2.5.2 and 2.5.8, we see that sνprovides a prescription of exactly those coarsenings at which M(α) and M(β) differ, and withwhat multiplicities. To be precise, if cµ is the coefficient of hµ in the h-basis expansion of sν , thenwe must havemM(α)(µ)−mM(β)(µ) = (−1)R−`(µ)cµ. (2.3)Moreover, the dominance relations of Proposition 2.5.10 tell us that ν ≥dom λ(α) and that ν is theunique dominance-minimal partition at which the multisets of coarsenings differ.162.6. Near-equality of ribbon Schur functionsFinally, we state our third tool from symmetric function theory.Theorem 2.6.1. [21, Theorem 7.15.6] There is an involutive isomorphism ω on the algebra Λ ofsymmetric functions, which satisfiesω(rα) = rαt and ω(sν) = sν′ .In particular, from a near-equalityrα − rβ = sν ,we can apply the ω involution to derive a new near-equalityrαt − rβt = sν′ .Example 2.6.2. By applying the ω involution to the equationr313 − r412 = s331from Examples 2.4.6 and 2.5.9, we find that using Example 2.1.1 we haver11311 − r13111 = s322.We can use this involution to show the following.Corollary 2.6.3. If α is simple, then αt is simple.Proof. Let α be simple and suppose that rαt = rβ. Applying the ω involution, we have thatrα = rβt , so because α is simple, we have either that βt = α, in which case β = αt, or βt = α∗, inwhich case β = (α∗)t = (αt)∗ by the formula of Proposition 2.2.5. Therefore αt is simple.It turns out that conjugation reverses the dominance order on partitions [21, Page 288]. There-fore, applying the ω involution to Proposition 2.5.10, Part 4 for αt, we find thatrα = s(λ(αt))′ +∑µ′<dom(λ(αt))′b′µ(αt)sµ′ . (2.4)We conclude this section by making some observations about this dominance-maximal term s(λ(αt))′ .Proposition 2.6.4. We have (λ(αt))′1 = N −R+ 1 and for j ≥ 2(λ(αt))′j = R− k − 1 + δα −j−3∑i=0q′i.172.6. Near-equality of ribbon Schur functionsProof. This follows directly from Proposition 2.2.7, Parts 2 and 4.Corollary 2.6.5. Suppose that λ(α) = λ(β) and rα ≥s rβ. Then δα ≥ δβ. Moreover, if δα = δβ,then q′(α) ≤lex q′(β).Proof. We must have that (λ(αt)′) ≥lex (λ(βt)′), otherwise by (2.4) rα − rβ will contain the term−s(λ(βt)′), with a negative coefficient. Note that because λ(α) = λ(β), α and β have the samenumber of parts that are equal to 1. Now by Proposition 2.6.4, we have (λ(αt)′)1 = N − R + 1 =(λ(βt)′)1, so we must have that(λ(αt)′)2 = R− k − 1 + δα ≥ R− k − 1 + δβ = (λ(βt)′)2,and so δα ≥ δβ. Moreover, if δα = δβ, then letting j be the smallest index such that q′j(α) 6= q′j(β),we have by Proposition 2.6.4 that (λ(αt)′)j′ = (λ(βt)′)j′ for j′ ≤ j + 2, so we must have that(λ(αt)′)j+3 = R− k − 1 + δα −j∑i=0q′i(α) ≥ R− k − 1 + δβ −j∑i=0q′i(β) = (λ(βt)′)j+3,and so q′j(α) < q′j(β).18Chapter 3Fourteen Families of Near-EqualityWe begin by presenting our known fourteen cases of near-equality.Theorem 3.1.1. We have rα − rβ = sν in the following cases, where a ≥ b ≥ 2, c ≥ 1, and d ≥ 0.In Cases (3.8) through (3.10), c ≥ 2, and in Cases (3.5) through (3.14), b ≥ 3.α = b1da β = (b− 1)1d(a+ 1) ν = ab1d (3.1)α = ab1d β = (b− 1)(a+ 1)1d ν = ab1d (3.2)α = 1c+da1c β = 1c+d+1a1c−1 ν = a2c1d (3.3)α = (a− 1)1c−121c+d β = (a− 1)1c+d21c−1 ν = a2c1d (3.4)α = 1d+1a(b− 1) β = 1d+1(b− 1)a ν = ab1d (3.5)α = 1a1d(b− 1) β = 1(b− 1)1da ν = ab1d (3.6)α = (b− 1)1db(a− b+ 1) β = b1d(b− 1)(a− b+ 1) ν = ab1d (3.7)α = 1c−121c+d−1a β = 1c+d21c−2a ν = a2c1d (3.8)α = 1c−1a1c+d−12 β = 1c+da1c−22 ν = a2c1d (3.9)α = 1d21c−1a1c−1 β = 1d21c−2a1c ν = a2c1d(3.10)α = 1c+d+1a(b− 1)1c β = 1c+d+1(b− 1)a1c ν = ab2c1d(3.11)α = 1ca(b− 1)1c−121d β = 1c(b− 1)a1c−121d ν = ab2c1d(3.12)α = (b− 1)1c−121c+da β = (b− 1)1c+d21c−1a ν = ab2c1d(3.13)α = (a− b+ 1)(b− 1)1c−121c+d(b− 1) β = (a− b+ 1)(b− 1)1c+d21c−1(b− 1) ν = ab2c1d(3.14)In order to prove Theorem 3.1.1, we will first prove a convenient ribbon difference identity,generalizing [22, Theorem 22]. We make some preliminary definitions.Definition 3.1.2. Let i be minimal with αi ≥ 2 and let i < j ≤ R and 1 ≤ t ≤ αi− 1. Now definethe composition Mj,t(α) byMj,t(α) = α1 · · ·αi−1(αi − t)αi+1 · · ·αj−1(αj + t)αj+1 · · ·αR;19Chapter 3. Fourteen Families of Near-Equalitythat is, Mj,t(α) is formed from α by decrementing the i-th part by t and incrementing the j-th partby t.In addition, define Aj,t(α) to be the set of LR tableaux T of shape α such that for somei ≤ j′ ≤ j − 1 the number of 1’s in the first j′ rows of T does not exceed the number of 2’s in thefirst (j′ + 1) rows of T by ≥ t.Finally, define Bj,t(α) to be the set of LR tableaux U of shape Mj,t(α) such thatif Uj,j1 = · · · = Uj,j1+t−1 = 1, then j ≤ R− 1 and Uj,j1+t ≥ Uj+1,j1 ,where the leftmost cell of row j in U is in column j1. In other words, if the j-th row of U beginswith t 1’s, then the (t+1)-th entry of this row must be greater than or equal to the rightmost entryof the row below.We are now ready to state our ribbon difference identity. We will work through an examplebefore supplying the proof.Theorem 3.1.3. Let i be minimal with αi ≥ 2 and let i < j ≤ R and 1 ≤ t ≤ αi − 1. Then wehave the following identity.rα − rMj,t(α) =∑T∈Aj,t(α)scont(T ) −∑U∈Bj,t(α)scont(U)Example 3.1.4. Let α = 1116311, so that i = 4, and let j = 5 and t = 3, so that β = M5,3(α) =1113611.The first four rows of any T ∈ LRα must be filled as follows.1231 1 1 1 1 4Now Aj,t(α) is the set of such T for which the number of 1’s in the first four rows does notexceed the number of 2’s in the first five rows by at least three. As there are presently six 1’s inthe first four rows of T and one 2 in the first five rows, the fifth row must be filled with all 2’s. TheLR tableaux of Aj,t(α) are enumerated below.20Chapter 3. Fourteen Families of Near-Equality1231 1 1 1 1 42 2 2341231 1 1 1 1 42 2 2351231 1 1 1 1 42 2 256The first four rows of any U ∈ LRβ must be filled as follows.1231 1 4Now Bj,t(α) is the set of such U for which if U5,1 = U5,2 = U5,3 = 1, then U5,4 ≥ U6,1. Becausethe fifth row of U can have at most three numbers at least 2; namely two 2’s and one 5, then weindeed have U5,1 = U5,2 = U5,3 = 1 and U5,4 = 2 and so U6,1 = 2. The LR tableaux of Bj,t(α) areenumerated below.1231 1 41 1 1 2 2 5231231 1 41 1 1 2 2 526Finally, by Theorem 3.1.3, the difference rα − rβ isr1116311 − r1113611 =∑T∈Aj,t(α)scont(T ) −∑U∈Bj,t(α)scont(U)= (s6422 + s64211 + s641111)− (s64211 + s641111) = s6422.We will now prove Theorem 3.1.3.Proof of Theorem 3.1.3. For ease of notation, set β = Mj,t(α), A = Aj,t(α) and B = Bj,t(α).We will construct a content-preserving bijectionf : (LRα \A)→ (LRβ \B),21Chapter 3. Fourteen Families of Near-Equalityfrom which it immediately follows thatrα − rβ =∑T∈LRαscont(T ) −∑U∈LRβscont(U) =∑T∈Ascont(T ) −∑U∈Bscont(U).Given an LR tableau T ∈ LRα \A, we construct f(T ) as the tableau of shape β where the i-throw is filled with (βi − 1) 1’s followed by an i, as it must be by Proposition 2.4.3, Part 4; the j-throw is filled with t 1’s followed by the entries in the j-th row of T , and all other rows are filled asin T . Informally, we remove t 1’s from the i-th row of T and append them to the front of the j-throw of T to create f(T ).As an illustration, if α = 1116311, j = 5, and t = 3 as in Example 3.1.4, and T is the tableauto the left, then f(T ) is the tableau to the right.1231 1 1 1 1 41 2 5361231 1 41 1 1 1 2 536We first check that f(T ) ∈ LRβ \ B. By construction, f(T ) is of shape β and has the samecontent as T . The j-th row of f(T ) is still weakly increasing because the 1’s were added to thefront. In the case that i ≥ 2 and t = αi− 1, we need to check that the rightmost column of f(T ) isstill strictly increasing. However, because T /∈ A, the number of 1’s in the first i rows of T , namelyαi, must exceed the number of 2’s in the first (i+ 1) rows of T by at least t = αi − 1. So there isat most one 2 in the first (i+ 1) rows of T , which is in the second row, and so the rightmost entryof the (i + 1)-th row is not a 2 and must be an (i + 1). To check the lattice word condition, notethat since the reading word of f(T ) differs from that of T , which is a lattice word, only by movingt 1’s from the i-th to the j-th row, it suffices to check that there are not too many 2’s in this range.Now again since T /∈ A, the number of 1’s in T exceeds the number of 2’s by at least t from thei-th row to the j-th row, and so indeed f(T ) ∈ LRβ. Finally, f(T ) /∈ B because the first t entriesof the j-th row of f(T ) are all 1’s, and if j < R, then the (t+ 1)-th entry of the j-th row, which inT was directly above the rightmost entry of the (j+1)-th row, can not be greater than or equal to it.Conversely, given an LR tableau U ∈ LRβ \ B, then we construct f−1(U) as the tableau ofshape α where the i-th row is filled with (αi− 1) 1’s followed by an i, the j-th row is filled with therightmost (βj−t) entries of U , and all other rows are filled as in U . Because the first t entries of thej-th row of U are all 1’s and the (t+ 1)-th entry of this row is strictly smaller than the rightmostentry of the row below, and because t 1’s were moved to the i-th row we have f−1(U) ∈ LRα \ A.22Chapter 3. Fourteen Families of Near-EqualityBy construction, f−1(f(T )) = T and f(f−1(U)) = U , so f is a bijection.Now that we have Theorem 3.1.3 at our disposal, it is routine to prove Theorem 3.1.1.Proof of Theorem 3.1.1. We use Theorem 3.1.3 to first prove Cases (3.1), (3.2), (3.5), (3.6),(3.7), (3.11), and (3.12). Then the remaining cases follow by applying the ω involution, using theformula of Proposition 2.2.5, and relabelling. Throughout the proof we will include diagrams toillustrate the concrete case when a = 6, b = 4, c = 2, and d = 2.Case (3.1): α = b1da, β = (b− 1)1d(a+ 1), ν = ab1d.We have that β = Md+2,1(α). Any tableau T ∈ Ad+2,1(α) has its column of length (d + 2)filled with the integers 1 through (d + 2) and has all 1’s in its top row by Proposition 2.4.3,Parts 3 and 4, and must have 1’s and 2’s in the remaining cells. The number of 2’s must beat most b, the number of 1’s before the bottom row of T , by Proposition 2.4.3, Part 1, andmust be at least b by definition of Ad+2,1(α), so there is a unique T ∈ Ad+2,1(α), and it hascontent cont(T ) = ab1d.T = 1 1 1 1231 1 2 2 2 4U = 1 1 1234Any tableau U ∈ Bd+2,1(α) has its column of length (d+ 2) filled with the integers 1 through(d+ 2) and has all 1’s in its top row by Proposition 2.4.3, Parts 3 and 4, and must have 1’sand 2’s in the remaining cells. The number of 2’s must be at most (b − 1), the number of1’s before the bottom row of U , by Proposition 2.4.3, Part 1, and so there must be a 1 inthe leftmost cell because (b− 1) + 1 = b < a+ 1. However, by definition of Bd+2,1(α), therecan not be a 1 in the leftmost cell because d + 2 = R, so Bd+2,1(α) is empty. Therefore, byTheorem 3.1.3, we have thatrb1da − r(b−1)1d(a+1) =∑T∈Ad+2,1(α)scont(T ) −∑U∈Bd+2,1(α)scont(U) = sab1d .Case (3.2): α = ab1d, β = (b− 1)(a+ 1)1d, ν = ab1d.We have that β = M2,a−b+1(α). Any tableau T ∈ A2,a−b+1(α) has all 1’s in its first row byProposition 2.4.3, Part 4, and must have all 2’s in its second row in order for the number of1’s in the first row, a, not to exceed the number of 2’s in the second row by at least (a−b+1).Then for j ≥ 3 the entry in the j-th row must be at least j for the leftmost column to beincreasing and at most j by Proposition 2.4.3, Part 2, so there is a unique T ∈ A2,a−b+1(α),and it has content cont(T ) = ab1d.23Chapter 3. Fourteen Families of Near-EqualityT = 1 1 1 1 1 12 2 2 234U = 1 1 12Any tableau U ∈ B2,a−b+1(α) must have all 1’s in the first row by Proposition 2.4.3, Part 4,and can have at most (b−1) 2’s in the second row by Proposition 2.4.3, Part 1, so the secondrow must have at least (a− b+2) 1’s, as this row may only contain 1’s and 2’s by Proposition2.4.3, Part 2. However, by definition of B2,a−b+1(α), because the leftmost (a− b+ 1) entriesof the second row are 1’s, the following entry, another 1, must be at least the leftmost entryof the third row, which is impossible because the leftmost column must be strictly increasing,so B2,a−b+1(α) is empty. Therefore, by Theorem 3.1.3, we have thatrab1d − r(b−1)(a+1)1d =∑T∈A2,a−b+1(α)scont(T ) −∑U∈B2,a−b+1(α)scont(U) = sab1d .Case (3.5): α = 1d+1a(b− 1), β = 1d+1(b− 1)a, ν = ab1d.We have that β = Md+3,a−b+1(α). Any tableau T ∈ Ad+3,a−b+1(α) has its rightmost columnfilled with the integers 1 through (d + 2) and the remaining (a − 1) entries of its (d + 2)-throw are 1’s by Proposition 2.4.3, Parts 3 and 4. Then the (d+ 3)-th row of T must be all 2’sin order for the number of 1’s in the first (d + 2) rows of T not to exceed the total numberof 2’s by at least (a − b + 1), so there is a unique T ∈ Ad+3,a−b+1(α), and it has contentcont(T ) = ab1d.T = 1231 1 1 1 1 42 2 2U = 1231 1 4Any tableau U ∈ Bd+3,a−b+1(α) has its rightmost column filled with the integers 1 through(d + 2) and the remaining (b − 2) entries of its (d + 2)-th row are 1’s by Proposition 2.4.3,Parts 3 and 4. Then the entries of the (d + 3)-th row of U other than 1’s consist of atmost (b − 2) 2’s and at most one (d + 3) in order to satisfy the lattice word condition, sobecause (b − 2) + 1 = b − 1 < a there is a 1 in the leftmost cell. However, by definition ofBd+3,a−b+1(α), there can not be a 1 in the leftmost cell because d+ 3 = R, so Bd+3,a−b+1(α)is empty. Therefore, by Theorem 3.1.3, we have thatr1d+1a(b−1) − r1d+1(b−1)a =∑T∈Ad+3,a−b+1(α)scont(T ) −∑U∈Bd+3,a−b+1(α)scont(U) = sab1d .24Chapter 3. Fourteen Families of Near-EqualityCase (3.6): α = 1a1d(b− 1), β = 1(b− 1)1da, ν = ab1d.We have that β = Md+3,a−b+1(α). Any tableau T ∈ Ad+3,a−b+1(α) must have a 1 and a 2 inits rightmost column and the remaining (a−1) entries of its second row are 1’s by Proposition2.4.3, Parts 3 and 4. Then in order for the number of 1’s in the first (d+ 2) rows of T not toexceed the number of 2’s in the first (d+ 3) rows by at least (a− b+ 1), there must be a 2 ineach of the leftmost (b− 1) columns of T , so that the column of length (d+ 2) is filled withthe integers 1 through (d+ 2) and the remaining (b− 2) entries of the bottom row are all 2’s.Therefore, there is a unique T ∈ Ad+3,a−b+1(α), and it has content cont(T ) = ab1d.T = 11 1 1 1 1 2232 2 4U = 11 1 2Any tableau U ∈ Bd+3,a−b+1(α) must have a 1 and a 2 in its rightmost column and theremaining (b − 2) entries of its second row are 1’s by Proposition 2.4.3, Parts 3 and 4. Theremaining (d+ 1) entries of the column of length (d+ 2) are either filled with the integers 2through (d+ 2), in which case the entries of the (d+ 3)-th row of U other than 1’s consist ofat most (b − 3) 2’s, at most one 3, and the (d + 2); or are filled with the integers 3 through(d + 3), in which case the entries of the (d + 3)-th row of U other than 1’s consist of atmost (b− 2) 2’s and the (d+ 3) in order to satisfy the lattice word condition. In either case,because (b − 2) + 1 = b − 1 < a, there is a 1 in the leftmost cell. However, by definition ofBd+3,a−b+1(α), there can not be a 1 in the leftmost cell because d+ 3 = R, so Bd+3,a−b+1(α)is empty. Therefore, by Theorem 3.1.3, we have thatr1a1d(b−1) − r1(b−1)1da =∑T∈Ad+3,a−b+1(α)scont(T ) −∑U∈Bd+3,a−b+1(α)scont(U) = sab1d .Case (3.7): α = (b− 1)1db(a− b+ 1), β = b1d(b− 1)(a− b+ 1), ν = ab1d.We have that α = Md+2,1(β). Note that in this case the roles of α and β are reversed. Anytableau T ∈ Ad+2,1(β) must have its rightmost column of length (d+2) filled with the integers1 through (d+ 2) and has all 1’s in its top row by Proposition 2.4.3, Parts 3 and 4. Now bydefinition of Ad+2,1(β), the number of 1’s in the first (d+ 1) rows, namely b, must not exceedthe number of 2’s in the first (d + 2) rows by at least 1, but there is only space for (b − 2)additional 2’s, so Ad+2,1(β) is empty.25Chapter 3. Fourteen Families of Near-EqualityT = 1 1 1 1234U = 1 1 1231 2 2 41 1 2Any tableau U ∈ Bd+2,1(β) must have its rightmost column of length (d + 2) filled with theintegers 1 through (d+ 2) and has all 1’s in its top row by Proposition 2.4.3, Parts 3 and 4.Now the remaining (a − b + 1) entries of the (d + 2)-th row of U are 1’s and 2’s and thereare at most (b− 2) 2’s in order to satisfy the lattice word condition, so the leftmost entry ofthis row is a 1. Then by definition of Bd+2,1(β), the entry immediately to the right must bea 2 in order to be at least the entry below, which must also be a 2. Finally, again to satisfythe lattice word condition, the remaining (b− 1) entries of the bottom row of U must be all1’s. Therefore, there is a unique U ∈ Bd+2,1(β), and it has content cont(U) = ab1d, and soby Theorem 3.1.3, we have thatr(b−1)1db(a−b+1) − rb1d(b−1)(a−b+1) = −∑T∈Ad+2,1(β)scont(T ) +∑U∈Bd+2,1(β)scont(U) = sab1d .Case (3.11): α = 1c+d+1a(b− 1)1c, β = 1c+d+1(b− 1)a1c, ν = ab2c1d.We have that β = Mc+d+3,a−b+1(α). Any tableau T ∈ Ac+d+3,a−b+1(α) has its rightmostcolumn filled with the integers 1 through (c + d + 2) and the remaining (a − 1) entries ofits (c + d + 2)-th row are all 1’s by Proposition 2.4.3, Parts 3 and 4. Then in order for thenumber of 1’s in the first (c + d + 2) rows of T , namely a, not to exceed the number of 2’sin the first (c+ d+ 3) rows of T by at least (a− b+ 1), the (c+ d+ 3)-th row of T must beall 2’s. To satisfy the lattice word condition, the leftmost column of T is filled with integers2 through x, followed by (c + d + 3) through (2c + d − x + 4) for some 2 ≤ x ≤ c + 2, afterwhich cont(T ) = ab2x−212c+d−2x+4, and so∑T∈Ac+d+3,a−b+1(α)scont(T ) =c+2∑x=2sab2x−212c+d−2x+4 . (3.15)26Chapter 3. Fourteen Families of Near-EqualityT = 123451 1 1 1 1 62 2 2U = 123451 1 61 1 1 2 2 72Any tableau U ∈ Bc+d+3,a−b+1(α) has its rightmost column filled with the integers 1 through(c+d+ 2) and the remaining (b− 2) entries of its (c+d+ 2)-th row are all 1’s by Proposition2.4.3, Parts 3 and 4. Then the entries of the (c+ d+ 3)-th row of U other than 1’s consist ofat most (b− 2) 2’s and at most one (c+ d+ 3) in order to satisfy the lattice word condition,so the leftmost (a − b + 1) entries of the (c + d + 3)-th row are all 1’s, and by definition ofBc+d+3,a−b+1(α) we must indeed have (b − 2) 2’s and one (c + d + 3) in the (c + d + 3)-throw of U , and a 2 in the (c+ d+ 4)-th row. Again to satisfy the lattice word condition, theleftmost column of T is filled with the integers 1 through x, followed by (c+ d+ 4) through(2c+ d− x+ 4) for some 2 ≤ x ≤ c+ 1, after which cont(U) = ab2x−212c+d−2x+4, and so∑U∈Bc+d+3,a−b+1(α)scont(T ) =c+1∑x=2sab2x−212c+d−2x+4 . (3.16)Now subtracting (3.16) from (3.15), the terms with x 6= c + 2 cancel and by Theorem 3.1.3we have thatr1c+d+1a(b−1)1c − r1c+d+1(b−1)a1c = sab2c+2−212c+d−2(c+2)+4 = sab2c1d .Case (3.12): α = 1ca(b− 1)1c−121d, β = 1c(b− 1)a1c−121d, ν = ab2c1d.We have that β = Mc+2,a−b+1(α). Any tableau T ∈ Ac+2,a−b+1(α) must have its rightmostcolumn filled with the integers 1 through (c + 1), and the remaining entries of its (c + 1)-throw are all 1’s by Proposition 2.4.3, Parts 3 and 4. Then in order for the number of 1’s in thefirst (c+ 1) rows of T , namely a, not to exceed the number of 2’s in the first (c+ 2) rows ofT by at least (a− b+ 1), the (c+ 2)-th row of T must be all 2’s. To satisfy the lattice wordcondition, the second-leftmost column of T is filled with the integers 2 through x, followedby (c + 2) through (2c − x + 3) for some 2 ≤ x ≤ c + 1, after which the content so far isab2x−212c−2x+3. Then the leftmost column of T is filled with strictly increasing integers y1through yd+1 with y1 ≤ 2c− x+ 3 and such that ab2x−212c−2x+3 + η(y) is a partition, whereη(y)j = 1 if some yi = j and 0 otherwise. Therefore, with the convention that sσ = 0 if σ is27Chapter 3. Fourteen Families of Near-Equalitynot a partition, we have∑T∈Ac+2,a−b+1(α)scont(T ) =c+1∑x=2∑y1<···<yd+1y1≤2c−x+3sab2x−212c−2x+3+η(y). (3.17)T = 121 1 1 1 1 32 2 2U = 121 1 31 1 1 2 2 42Any tableau U ∈ Bc+2,a−b+1(α) must have its rightmost column filled with the integers 1through (c+1), and the remaining (b−2) entries of its (c+1)-th row are all 1’s by Proposition2.4.3, Parts 3 and 4. Then the entries of the (c+ 2)-th row of U other than 1’s consist of atmost (b− 2) 2’s and at most one (c+ 2) in order to satisfy the lattice word condition, so theleftmost (a− b+ 1) entries of the (c+ 2)-th row are all 1’s, and by definition of Bc+2,a−b+1(α)we must indeed have (b − 2) 2’s and one (c + 2) in the (c + 2)-th row of U , and a 2 in the(c + 3)-th row. Again to satisfy the lattice word condition, the second-leftmost column ofU is filled with the integers 1 through x, followed by (c + 3) through (2c − x + 3) for some2 ≤ x ≤ c + 1, after which the content so far is ab2x−212c−2x+3. Then the leftmost columnof U is filled with strictly increasing integers y1 through yd+1 with y1 ≤ 2c− x+ 3 and suchthat ab2x−212c−2x+3 + η(y) is a partition, where η(y)j = 1 if some yi = j and 0 otherwise.Additionally, if x = c + 1, then the bottom entry of the second-leftmost column of U is a(c+1), meaning that in fact y1 6= c+2 because the row must be weakly increasing. Therefore,with the convention that sσ = 0 if σ is not a partition, we have∑U∈Bc+2,a−b+1(α)scont(U) =c+1∑x=2∑y1<···<yd+1y1≤2c−x+3if x=c+1, then y1 6=c+2sab2x−212c−2x+3+η(y). (3.18)Now subtracting (3.18) from (3.17), the terms with x 6= c + 1 cancel because the innersummations are identical, and when x = c + 1, the terms with y1 6= c + 2 also cancel.What remains is the single term from (3.17) where x = c + 1 with a (c + 2) below it at thebottom of the column. Then y1 = c + 2, for which we must have yj = c + j + 1 in order forab2x−212c−2x+3 + η(y) to be a partition. So by Theorem 3.1.3 we have thatr1ca(b−1)1c−121d − r1c(b−1)a1c−121d = sab2c+1−212c−2(c+1)+3+η(y) = sab2c1d .28Chapter 3. Fourteen Families of Near-EqualityFinally, we apply the ω involution, using Proposition 2.2.5, to deduce the remaining cases.Case (3.3): α = 1c+da1c, β = 1c+d+1a1c−1, ν = a2c1d.Applying the ω involution to the equationrb′1d′a′ − r(b′−1)1d′ (a′+1) = sa′b′1d′of Case (3.1) producesr1a′−1(d′+2)1b′−1 − r1a′ (d′+2)1b′−2 = s(d′+2)2b′−11a′−b′ .Setting a′ = c+ d+ 1, b′ = c+ 1, and d′ = a− 2 produces Case (3.3).Case (3.4): α = (a− 1)1c−121c+d, β = (a− 1)1c+d21c−1, ν = a2c1d.Applying the ω involution to the equationra′b′1d′ − r(b′−1)(a′+1)1d′ = sa′b′1d′of Case (3.2) producesr(d′+1)1b′−221a′−1 − r(d′+1)1a′−121b′−2 = s(d′+2)2b′−11a′−b′ .Setting a′ = c+ d+ 1, b′ = c+ 1, and d′ = a− 2 produces Case (3.4).Case (3.8): α = 1c−121c+d−1a, β = 1c+d21c−2a, ν = a2c1d.Applying the ω involution to the equationr1d′+1a′(b′−1) − r1d′+1(b′−1)a′ = sa′b′1d′of Case (3.5) producesr1b′−221a′−2(d′+2) − r1a′−121b′−3(d′+2) = s(d′+2)2b′−11a′−b′ .Setting a′ = c+ d+ 1, b′ = c+ 1, and d′ = a− 2 produces Case (3.8).Case (3.9): α = 1c−1a1c+d−12, β = 1c+da1c−22, ν = a2c1d.Applying the ω involution to the equationr1a′1d′ (b′−1) − r1(b′−1)1d′a′ = sa′b′1d′29Chapter 3. Fourteen Families of Near-Equalityof Case (3.6) producesr1b′−2(d′+2)1a′−22 − r1a′−1(d′+2)1(b′−3)2 = s(d′+2)2b′−11a′−b′ .Setting a′ = c+ d+ 1, b′ = c+ 1, and d′ = a− 2 produces Case (3.9).Case (3.10): α = 1d21c−1a1c−1, β = 1d21c−2a1c, ν = a2c1d.Applying the ω involution to the equationr(b′−1)1d′b′(a′−b′+1) − rb′1d′ (b′−1)(a′−b′+1) = sa′b′1d′of Case (3.7) producesr1a′−b′21b′−2(d′+2)1b′−2 − r1a′−b′21b′−3(d′+2)1b′−1 = s(d′+2)2b′−11a′−b′ .Setting a′ = c+ d+ 1, b′ = c+ 1, and d′ = a− 2 produces Case (3.10).Case (3.13): α = (b− 1)1c−121c+da, β = (b− 1)1c+d21c−1a, ν = ab2c1d.Applying the ω involution to the equationr1c′+d′+1a′(b′−1)1c′ − r1c′+d′+1(b′−1)a′1c′ = sa′b′2c′1d′of Case (3.11) producesr(c′+1)1b′−321a′−2(c′+d′+2) − r(c′+1)1a′−221b′−3(c′+d′+2) = s(c′+d′+2)(c′+2)2b′−21a′−b′ .Setting a′ = c+ d+ 2, b′ = c+ 2, c′ = b− 2, and d′ = a− b produces Case (3.13).Case (3.14): α = (a − b + 1)(b − 1)1c−121c+d(b − 1), β = (a − b + 1)(b − 1)1c+d21c−1(b − 1),ν = ab2c1d.Applying the ω involution to the equationr1c′a′(b′−1)1c′−121d′ − r1c′ (b′−1)a′1c′−121d′ = sa′b′2c′1d′of Case (3.12) producesr(d′+1)(c′+1)1b′−321a′−2(c′+1) − r(d′+1)(c′+1)1a′−221b′−3(c′+1) = s(c′+d′+2)(c′+2)2b′−21a′−b′ .Setting a′ = c+ d+ 2, b′ = c+ 2, c′ = b− 2, and d′ = a− b produces Case (3.14).30Chapter 3. Fourteen Families of Near-EqualityWe now show that the compositions in Theorem 3.1.1 are simple. We first synthesize Theorem2.3.6 to derive conditions for simplicity that are easy to check.Lemma 3.1.5. Suppose that α is not simple. Then1. we can factorize α as a t-fold composition of compositionsα = α(1) ◦ · · · ◦ α(t)where at least two of the factors are nonsymmetric2. we can factorize α = β ◦ γ where `(β), `(γ) ≥ 2 and β and γ have a part at least 23. α1 + αR is a part of α4. α has at least three parts that are at least 2.Proof.1. Because α is not simple, there is a composition β such that rα = rβ but β 6= α and β 6= α∗and so by Theorem 2.3.6, Part 3, we can factorizeα = α(1) ◦ · · · ◦ α(t) and β = β(1) ◦ · · · ◦ β(t)where for each 1 ≤ i ≤ t, either β(i) = α(i) or β(i) = (α(i))∗.Now if every factor α(i) is symmetric, then we would have each β(i) = α(i) and so β = α. Ifexactly one factor α(j) is nonsymmetric, then either β(j) = α(j), in which case again β = α,or β(j) = (α(j))∗, in which case by Theorem 2.3.6, Part 2, we haveα∗ = (α(1) ◦ · · · ◦α(j) ◦ · · · ◦α(t))∗ = α(1) ◦ · · · ◦ (α(j))∗ ◦ · · · ◦α(t) = β(1) ◦ · · · ◦β(j) ◦ · · · ◦β(t) = β.Therefore at least two of the factors are nonsymmetric.2. By (1), we can factorizeα = α(1) ◦ · · · ◦ α(t)where at least two of the factors α(j1) and α(j2), 1 ≤ j1 < j2 ≤ t, are nonsymmetric andtherefore have length at least 2 and are not all 1’s so have a part at least 2. Now note that ifα(j) has length at least 2 and a part α(j)s at least 2 and α(i) is any composition, then α(i) ◦α(j)is a concatenation of near-concatenations of α(j) so also has length at least 2 and a part atleast 2, and α(j) ◦ α(i) is a concatenation of at least two near-concatenations of α(i) so haslength at least 2 and contains (α(i))α(j)s , which itself has the part α(i)1 + α(i)`(α(i))≥ 2. Nowthe result follows by setting β = α(1) ◦ · · · ◦ α(j2−1) and γ = α(j2) ◦ · · · ◦ α(t).31Chapter 3. Fourteen Families of Near-Equality3. By (2), we can factorize α = β ◦ γ, where β has a part βs at least 2 and therefore α containsγβs , which itself has the part γ1 + γ`(γ). Because `(γ) ≥ 2, we have α1 = γ1 and αR = γ`(γ),so this part is α1 + αR.4. By (2), we can factorize α = β ◦ γ, where β has a part βs at least 2 and γ has a part at least2, so α contains γ(βs), which has at least two parts at least 2. In addition, because `(β) ≥ 2,there is another (possibly trivial) near-concatenation of γ in α, which has one more part atleast 2.Proposition 3.1.6. In each case of Theorem 3.1.1, the compositions α and β are simple.Proof. In Cases (3.1), (3.2), (3.3), (3.4), (3.5), (3.6), (3.8), (3.9), (3.10), and (3.11), α and β donot have at least three parts that are at least 2 so are simple by Lemma 3.1.5, Part 4. In Case(3.14), α1 + αR = a > b− 1 ≥ 2 is not a part of α and β1 + βR = a > b− 1 ≥ 2 is not a part of βso these compositions are simple by Lemma 3.1.5, Part 3. Finally, in Case (3.7), Case (3.12), andCase (3.13), by Corollary 2.6.3, the ribbons α and β are simple because they are the transposes ofthe simple ribbons of Case (3.10), Case (3.14), and Case (3.11) respectively.We conclude this section by stating our conjecture that these are all possible near-equalities.Conjecture 3.1.7. Suppose that rα − rβ = sν . Then α, β, and ν are (up to reversal of α and β)as in one of the fourteen infinite families of Theorem 3.1.1.Our goal for the remainder of this thesis is to work towards this converse. One compellingpattern that Theorem 3.1.1 suggests is that ν must have at most two parts that are at least 3.Conjecture 3.1.8. Suppose that rα − rβ = sν . Then ν3 ≤ 2.32Chapter 4Near-Equality with Different PartsWe now begin to explore conditions under which near-equality can be possible. Our first resulttells us that there are no near-equalities rα − rβ = sν for which ν does not have at least two partsthat are at least 2.Theorem 4.1.1. Suppose that rα− rβ = sν . Then the partition ν can not be of the form ν = a1dfor a ≥ 1 and d ≥ 0. In other words, we must have ν2 ≥ 2.In order to illustrate our two combinatorial tools, we present two proofs of Theorem 4.1.1, oneusing the Littlewood–Richardson rule and one using the Jacobi–Trudi determinantal identity.Proof using the Littlewood–Richardson rule. Let T be an LR tableau of shape α or β andcontent ν = a1d. We must have R ≥ d+1 in order to place the (d+1) by Proposition 2.4.3, Part 2.On the other hand, for i ≥ 2 the rightmost cell of the i-th row is directly below a cell, specificallythe leftmost cell of the (i− 1)-th row, and so can not be filled with a 1, so R − 1 ≤ d. Therefore,we must have R = d + 1, and the d integers at least 2 must fill the rightmost cells of the d rowsbelow the top row in increasing order, thus determining T . Socα,ν = χ(R = d+ 1) = cβ,ν ,violating (2.2), and so we can not have rα − rβ = sν .Proof using the Jacobi–Trudi determinantal identity. Assume that ν = a1d. By Theorem 2.5.2,the h-basis expansion of sν = sa1d has exactly one term, namely hν , with exactly (d + 1) parts.Therefore, if µ is a partition with `(µ) = d+ 1, then by (2.3)mM(α)(µ) =mM(β)(µ) + (−1)R−d−1 µ = νmM(β)(µ) µ 6= ν. (4.1)However, by Proposition 2.5.6, Part 1, α and β both have the same total number of coarsenings,namely(R−1d), with exactly (d+ 1) parts, violating (4.1), and so we can not have rα− rβ = sν .Our second result will classify all instances of near-equality where the compositions α and βdetermine different partitions. This will allow us to understand Cases (3.1), (3.2), (3.3), and (3.4).33Chapter 4. Near-Equality with Different PartsTheorem 4.1.2. Suppose that rα−rβ = sν and λ(α) 6= λ(β). Then α, β, and ν are (up to reversalof α and β) as in Case (3.1) or Case (3.2), namelyCase (3.1): α = b1da, β = (b− 1)1d(a+ 1), ν = ab1dCase (3.2): α = ab1d, β = (b− 1)(a+ 1)1d, ν = ab1dfor some a ≥ b ≥ 2 and d ≥ 0.Proof. By Proposition 2.5.10, Part 4, we must have ν = λ(α) <dom λ(β), and by Theorem 4.1.1,we must have ν2 ≥ 2. Now if ν3 ≥ 2, then the h-basis expansion of sν has at least four terms withR parts, namelyhν , −hλ(ν1(ν2+1)(ν3−1)···νR), −hλ((ν1+1)(ν2−1)ν3···νR), and hλ((ν1+2)(ν2−1)(ν3−1)···νR),which is impossible because rα − rβ has at most two terms by Theorem 2.5.8. So ν = λ(α) = ab1dfor some a ≥ b ≥ 2 and d ≥ 0, and because the h-basis expansion of sν now contains the term−h(a+1)(b−1)1d with R parts, we have λ(β) = (a+ 1)(b− 1)1d. If d = 0 this determines α and β upto reversal, so suppose that d ≥ 1.The partition µ = ab21d−2 <lex (a+ 1)(b−1)1d = λ(β) so does not appear inM(β), so becauseby Theorem 2.5.2 the h-basis expansion of sν contains the term −(d− 1)hµ, we must have by (2.3)that mM(α)(µ) = d− 1. By Proposition 2.5.6, Part 2, we haved− 1 = mM(α)(µ) = 2k −R− 1 + q0 = 2d− (d+ 2)− 1 + q0 = d− 3 + q0,and so q0(α) = 2, or in other words in α all of the 1’s are together and so α is up to rever-sal one of b1da, ab1d, or ba1d. If a = b, then α = b1da or ab1d. If a 6= b, then the partitionτ = a(b+ 1)1d−1 <lex λ(β) so does not appear in M(β), so because by Theorem 2.5.2 the h-basisexpansion of sν contains the term −hτ , we must have by (2.3) that mM(α)(τ) = 1, so the b is nextto a 1, meaning that again α = b1da or ab1d.Finally, if α = b1da then rβ = rα − sν = r(a+1)1d(b−1) by Theorem 3.1.1, (3.1), and so β =(b− 1)1d(a+ 1) up to reversal by Proposition 3.1.6. If α = ab1d then rβ = rα − sν = r(b−1)(a+1)1dby Theorem 3.1.1, (3.2), and so β = (b− 1)(a+ 1)1d up to reversal by Proposition 3.1.6.By applying the ω involution, we find the following.Corollary 4.1.3. Suppose that rα − rβ = sν and λ(αt) 6= λ(βt). Then α, β, and ν are (up toreversal of α and β) as in Case (3.3) or Case (3.4), namelyCase (3.3): α = 1c+da1c, β = 1c+d+1a1c−1, ν = a2c1dCase (3.4): α = (a− 1)1c−121c+d, β = (a− 1)1c+d21c−1, ν = a2c1d34Chapter 4. Near-Equality with Different Partsfor some a ≥ 2, c ≥ 1, and d ≥ 0.Remark 4.1.4. Because we have now classified all cases of near-equality for which λ(α) 6= λ(β)or λ(αt) 6= λ(βt), we will frequently assume in what follows that λ(α) = λ(β) and λ(αt) = λ(βt).In such a situation, we also have the following.Proposition 4.1.5. Suppose that λ(α) = λ(β). Then λ(αt) = λ(βt) if and only if q′(α) = q′(β).Moreover, in such a case, we also have δα = δβ.Proof. This follows immediately from Proposition 2.2.7, Parts 2 and 4.35Chapter 5Near-Equality with Different Ends5.1 The ends of a compositionIn this chapter, we investigate Cases (3.5), (3.6), (3.7), (3.8), (3.9), and (3.10) in more detail. Thefollowing concept will be critical to this study.Definition 5.1.1. The ends of α is the multisete(α) = {α1, αR}.Our goal is to prove the following theorem, which classifies all cases of near-equality for whichλ(α) = λ(β) and λ(αt) = λ(βt), but e(α) 6= e(β). Along the way, we will prove a necessarycondition for Schur-positivity of ribbon differences rα − rβ, and we will investigate another keystatistic.Theorem 5.1.2. Suppose that rα − rβ = sν , λ(α) = λ(β), λ(αt) = λ(βt), and e(α) 6= e(β). Thenα, β, and ν are (up to reversal of α and β) as in Case (3.5), Case (3.6), or Case (3.7), namelyCase (3.5): α = 1d+1a(b− 1), β = 1d+1(b− 1)a, ν = ab1dCase (3.6): α = 1a1d(b− 1), β = 1(b− 1)1da, ν = ab1dCase (3.7): α = (b− 1)1db(a− b+ 1), β = b1d(b− 1)(a− b+ 1), ν = ab1dfor some a ≥ b ≥ 3 and d ≥ 0.5.2 Littlewood–Richardson tableaux with different endsOur first task is to count LR coefficients in Lemma 5.2.7, which we will show in Lemma 5.2.9 aresensitive to the ends of a composition. The following concept will help us construct LR tableaux.Definition 5.2.1. A cell x of the ribbon diagram of α is deep if there are at least two cells of theribbon diagram of α in the same column as x and above x.365.2. Littlewood–Richardson tableaux with different endsExample 5.2.2. Let α = 31311515. The four marked cells of the ribbon diagram of α are deep.Any LR tableau of shape α must have integers at least 3 in the deep cells.××××Proposition 5.2.3. There are∑j≥1 q′j columns of the ribbon diagram of α that contain deep cellsand there are (k − δα) deep cells in total.Proof. Deep cells arise precisely in the columns of the ribbon digram of α of length at least 3,which by Proposition 2.2.5 arise when p′i ≥ 1, and so there are∑i: p′i≥1 1 =∑j≥1 q′j columns thatcontain deep cells. A column of length (p′i + 2) gives rise to exactly max{p′i, 0} deep cells, which isp′i except when p′i = −1, in which case we add one to compensate. Therefore, by Proposition 2.2.5,the number of deep cells isR−k+1∑i=1p′i + χ(p′1 = −1) + χ(p′R−k+1 = −1) = (k − 2) + (2− δα) = k − δα.We now identify the relevant partitions to our study.Definition 5.2.4. Recall that z(α) = z1 · · · zR−k is the list of non-1 parts of α. We define thesequence of nonnegative integers(α) = 0(α) · · · R−k−1(α)= (z1 − 2 + χ(p1 = 0))(z2 − 2) · · · (zR−k−1 − 2)(zR−k − 2 + χ(pR−k+1 = 0)).Let S′ =∑j≥1 q′j(α). Then for 0 ≤M ≤ 12(N − 2R+ k+ 2− δα) and 1 ≤ u ≤ S′, unless S′ = 0, inwhich case u = 0, we define the partition of size Nµ(M,u) = (N −R+ 1−M)(R− k − 1 + δα +M)u1k−δα−u.Example 5.2.5. Let α = 31311515. Then N = 20, R = 8, k = 4, δα = 0, e(α) = {3, 5},(α) = 2134, p′(α) = (−1)121(−1), q′(α) = 021 and S′ = 2 + 1 = 3. Now for 0 ≤ M ≤12(20− 16 + 4 + 2− 0) = 5 and 1 ≤ u ≤ S′ = 3 we have the partitionµ(M,u) = (13−M)(M + 3)u14−u.375.2. Littlewood–Richardson tableaux with different endsRemark 5.2.6. The bound on M simply assures us that µ(M,u)1 ≥ µ(M,u)2. Note that ifS′ = 0, then by Proposition 5.2.3 there are no columns of the ribbon diagram of α with deep cells,so k − δα = 0 and indeed µ(M, 0) is a partition, where by convention we omit the trailing zero.Except for one step in the proof of Theorem 5.1.2, it suffices to consider when u = 1 if S′ ≥ 1, thatis, u = χ(S′ ≥ 1).We now count the number of LR tableaux of content µ(M,u).Lemma 5.2.7. If 0 ≤M ≤ 0(α), then the number of LR tableaux of shape α and content µ(M,u)iscα,µ(M,u) =(S′ − 1u− 1)|Eα,M |,where Eα,M ⊂ ZR−k−1 is the set of lattice pointsEα,M = {x1 · · ·xR−k−1 :R−k−1∑i=1xi = M, 0 ≤ xi ≤ i(α) for 1 ≤ i ≤ R− k − 1}.We present how the proof works in an example before diving into the details.Example 5.2.8. Let α = 31311515. By Example 5.2.5, we have k − δα = 4, S′ = 3, (α) = 2134,and we are considering LR tableaux of shape α and contentµ(M,u) = (13−M)(M + 3)u14−ufor 0 ≤M ≤ 0(α) = 2 and 1 ≤ u ≤ 3.In Example 5.2.2, we saw that there are exactly k − δα = 4 deep cells, which occupy S′ = 3columns, and which must be filled by the u + (4 − u) = 4 available entries at least 3. Of the u3’s, one must occupy the first deep cell and no two may appear in the same column, so there are(S′−1u−1)=(2u−1)choices of in which columns to place the remaining (u − 1) 3’s, after which theremaining (4− u) deep cells must be filled with the (4− u) entries at least 4 in increasing order.We must then place the 1’s and 2’s. The top two cells of the R − k − 1 + δα = 3 columns oflength at least 2 must be filled with a 1 above a 2 and the first row must be filled with 1’s byProposition 2.4.3, Part 4. An example filling with u = 2 is shown below.385.2. Littlewood–Richardson tableaux with different ends1 1 121 3241 523We now have M 2’s left to place. There are presently 0(α) = 2 more 1’s than 2’s placed and sobecause M ≤ 2 we do not need to worry about the lattice word condition. Because the rows mustbe weakly increasing, we need only specify how many 2’s will occupy each of our remaining threerows of length at least 2. The i-th such row from the top has i(α) vacant cells and so can be filledwith xi additional 2’s, where 0 ≤ xi ≤ i(α), and so the number of ways to place the remaining M2’s is exactly|{x1x2x3 :3∑i=1xi = M, 0 ≤ x1 ≤ 1, 0 ≤ x2 ≤ 3, 0 ≤ x3 ≤ 4}| = |Eα,M |. (5.1)For an example where M = 2, we have that |Eα,2| = |{110, 101, 020, 011, 002}| = 5.Now the proof of Lemma 5.2.7 works exactly as in this example.Proof of Lemma 5.2.7. Consider an LR tableaux T of shape α and content µ(M,u). ByProposition 5.2.3, there are exactly (k − δα) deep cells, which occupy S′ columns, and which mustbe filled by the u + (k − δα − u) available entries at least 3. Of the u 3’s, one must occupy thefirst deep cell and no two may appear in the same column, so there are(S′−1u−1)choices of in whichcolumns to place the remaining (u− 1) 3’s, after which the remaining (k− δα − u) deep cells mustbe filled with the (k − δα − u) entries at least 4 in increasing order.We must then place the 1’s and 2’s. By Proposition 2.2.7, Parts 3 and 4 there are∑j≥0q′j(α) = R− k − 1 + δαcolumns of length at least 2. The top two cells of these columns must be filled with a 1 above a 2and the rest of the top row of length at least two must be filled with 1’s by Proposition 2.4.3, Part4. We now have M 2’s left to place. There are presently (z1 − 2) more 1’s than 2’s placed, unlessp1 = 0, in which case there are (z1 − 1), and so becauseM ≤ 0(α) = z1 − 2 + χ(p1 = 0),395.2. Littlewood–Richardson tableaux with different endswe do not need to worry about the lattice word condition. Because the rows must be weaklyincreasing, we need only specify how many 2’s will occupy each of the remaining (R−k−1) rows oflength at least 2. The i-th such row from the top has its first and last cells occupied so has (zi− 2)vacant cells, unless i = R − k and pR−k+1 = 0, in which case only its last cell is occupied and sohas (zi − 1) vacant cells. Therefore, the number of additional 2’s xi that can be placed in this rowsatisfies0 ≤ xi ≤ i(α),and so the number of ways to place the remaining M 2’s is exactly |Eα,M |. Putting this all together,we have that the number of LR tableaux of shape α and content µ(M,u) is cα,µ(M,u) =(S′−1u−1)|Eα,M |,as desired.Now we will start looking at compositions in pairs in order to show that indeed the LR coefficientabove informs us about the ends of a composition.Lemma 5.2.9. Suppose that λ(α) = λ(β) and δα = δβ. Assume that α1 ≤ αR and β1 ≤ βR.1. If α1 < β1, then |Eα,0(α)| = |Eβ,0(α)|+ 1 + χ(α1 = αR).2. If α1 = β1 and αR < βR, then |Eα∗,0(α∗)| = |Eβ∗,0(α∗)|+ 1.3. If α1 = β1 and αR = βR, then |Eα,M | = |Eβ,M | for every M ≤ 12(N − 2R+ k + 2− δα).Remark 5.2.10. Because rα = rα∗ by Theorem 2.3.6, the assumptions α1 ≤ αR and β1 ≤ βR areof little concern.We illustrate with an example before examining the technicalities.Example 5.2.11. Let α = 31311515 as in Example 5.2.8 and let β = 51113315. Then (β) = 4114and so since M = 0(α) = 2, we haveEβ,0(α) = {x1x2x3 :3∑i=1xi = 2, 0 ≤ x1 ≤ 1, 0 ≤ x2 ≤ 1, 0 ≤ x3 ≤ 4} = {110, 101, 011, 002}.Comparing this with Equation (5.1), we see that the only difference is the constraint on x2. Therewe had the constraint x2 ≤ 3, which was superfluous because x1 + x2 + x3 = M ≤ 2, while herewe have the constraint x2 ≤ 1, which specifically excludes the single tuple 020 and as a result,|Eα,2| = |Eβ,2|+ 1.Proof of Lemma 5.2.9. The idea is to compare the sets of lattice points, finding that becauseλ(α) = λ(β), the non-1 parts z(α) and z(β) of α and β are permuted so the constraints are verysimilar. Most are permuted, which does not change the size of the set, and a difference in the endscauses one or two for β to be slightly stronger.405.2. Littlewood–Richardson tableaux with different ends1. Note that because δα = δβ, we must have α1 6= 1 and so δα = δβ = 0, 0(α) = α1 − 1 <β1 − 1 = 0(β), R−k−1(α) = αR − 1, and R−k−1(β) = βR − 1.Because λ(α) = λ(β), the non-1 parts z(α) and z(β) of α and β are permuted and so thereare i, j, i′, j′ such that z1(α) = zi(β), zR−k(α) = zj(β), z1(β) = zi′(α), and zR−k(β) = zj′(α).Now excluding these parts, we have that{1(α), . . . , R−k−2(α)} \ {i′(α), j′(α)} = {1(β), . . . , R−k−2(β)} \ {i(β), j(β)}as multisets; re-enumerate these as {1, . . . , R−k−4}. By permuting the constraints, whichdoes not change the sizes of the sets, we now have|Eα,0(α)| = |{x1 · · ·xR−k−1 :R−k−1∑i=1xi = α1 − 1, 0 ≤ xi ≤ i for 1 ≤ i ≤ R− k − 4,0 ≤ xR−k−3 ≤ β1 − 2, 0 ≤ xR−k−2 ≤ βR − 2, 0 ≤ xR−k−1 ≤ αR − 1}||Eβ,0(α)| = |{x1 · · ·xR−k−1 :R−k−1∑i=1xi = α1 − 1, 0 ≤ xi ≤ i for 1 ≤ i ≤ R− k − 4,0 ≤ xR−k−3 ≤ α1 − 2, 0 ≤ xR−k−2 ≤ αR − 2, 0 ≤ xR−k−1 ≤ βR − 1}|.We see that the constraints 0 ≤ xi ≤ i for 1 ≤ i ≤ R − k − 4 are identical for the two sets.Because α1−1 ≤ β1−2 ≤ βR−2 and α1−1 ≤ αR−1 by hypothesis, the last three constraintsfor Eα,0(α) are superfluous; similarly, because α1−1 ≤ βR−1, the last constraint for Eβ,0(α)is too. However, the constraint xR−k−3 ≤ α1 − 2 for Eβ,0(α) specifically excludes the singletuple 0 · · · 0(α1 − 1)00 and if α1 = αR, the constraint xR−k−2 ≤ αR − 2 specifically excludesthe single tuple 0 · · · 0(α1−1)0. Each of these appear in the first set. Therefore, we have that|Eα,0(α)| = |Eβ,0(α)|+ 1 + χ(α1 = αR).2. In this case, we apply the same analysis to α∗ and β∗, and in fact it is simpler because α∗R = β∗R.Note that because δα = δβ, we must have α∗1 6= 1 and so 0(α∗) = α∗1 − 1 < β∗1 − 1 = 0(β∗).Because λ(α∗) = λ(β∗), the non-1 parts z(α∗) and z(β∗) are permuted and so there are i, jsuch that z1(α∗) = zi(β∗) and z1(β∗) = zj(α∗). Also note that either α∗R = β∗R = 1, in whichcase R−k−1(α∗) = zR−k(α∗) − 2 and R−k−1(β∗) = zR−k(β∗) − 2, meaning that we subtract2 from the last non-1 parts just as with all the other non-1 parts, or α∗R = β∗R ≥ 2, in whichcase R−k−1(α∗) = α∗R − 1 = β∗R − 1 = R−k−1(β∗). In either case, we have that{1(α∗), . . . , R−k−1(α∗)} \ {j(α∗)} = {1(β∗), . . . , R−k−1(β∗)} \ {i(β∗)}415.2. Littlewood–Richardson tableaux with different endsas multisets; re-enumerate these as {1, . . . , R−k−2}. By permuting the constraints, whichdoes not change the sizes of the sets, we now have|Eα∗,0(α∗)| = |{x1 · · ·xR−k−1 :R−k−1∑i=1xi = α∗1 − 1, 0 ≤ xi ≤ i for 1 ≤ i ≤ R− k − 2,0 ≤ xR−k−1 ≤ β∗1 − 2}||Eβ∗,0(α∗)| = |{x1 · · ·xR−k−1 :R−k−1∑i=1xi = α∗1 − 1, 0 ≤ xi ≤ i for 1 ≤ i ≤ R− k − 2,0 ≤ xR−k−1 ≤ α∗1 − 2}|.We see that the constraints 0 ≤ xi ≤ i for 1 ≤ i ≤ R−k−2 are identical for the two sets. Be-cause α∗1−1 ≤ β∗1−2, the constraint xR−k−1 ≤ β∗1−2 for Eα∗,0(α∗) is superfluous. However, theconstraint xR−k−1 ≤ α∗1− 2 for Eβ∗,0(α∗) specifically excludes the single tuple 0 · · · 0(α∗1− 1),which appears in the first set. Therefore, we have that |Eα∗,0(α∗)| = |Eβ∗,0(α∗)|+ 1.3. Again we apply the same analysis, and in fact it is even simpler. Because λ(α) = λ(β), thenon-1 parts z(α) and z(β) are permuted. Also note that either αR = βR = 1, in which caseR−k−1(α) = zR−k(α) − 2 and R−k−1(β) = zR−k(β) − 2, meaning that we subtract 2 fromthe last non-1 parts just as with all the other non-1 parts, or αR = βR ≥ 2, in which caseR−k−1(α) = αR − 1 = βR − 1 = R−k−1(β). In either case, we have that{1(α), . . . , R−k−1(α)} = {1(β), . . . , R−k−1(β)}as multisets. By permuting the constraints, which does not change the sizes of the sets, wenow have|Eα,M | = |{x1 · · ·xR−k−1 :R−k−1∑i=1xi = M, 0 ≤ xi ≤ i for 1 ≤ i ≤ R− k − 1}| = |Eβ,M |.Now Lemma 5.2.7 and Lemma 5.2.9 allow us to identify specific partitions at which the LRcoefficients for α and β differ. One consequence is the following necessary condition for Schur-positivity of a difference rα − rβ, which generalizes [22, Theorem 40].Theorem 5.2.12. Suppose that λ(α) = λ(β). Let us also assume that α1 ≤ αR and β1 ≤ βR.Then the compositions α1αR and β1βR satisfyif rα ≥s rβ, then α1αR ≤lex β1βR.425.3. Adjacent pairsProof. Before we can apply Lemma 5.2.9, we must first address the case where δα 6= δβ. By Corol-lary 2.6.5, we must have δα > δβ, from which it follows that α1αR ≤lex β1βR, as desired. Now wemay assume that δα = δβ.If β1 < α1, then by Lemma 5.2.7 and Lemma 5.2.9, Part 1, we would have thatcβ,µ(0(α),χ(S′≥1)) = |Eβ,0(α)| > |Eα,0(α)| = cα,µ(0(α),χ(S′≥1)),contradicting rα ≥s rβ by Theorem 2.4.5, and so α1 ≤ β1. Similarly, if α1 = β1 and βR > αR, thenby Lemma 5.2.7 and Lemma 5.2.9, Part 2, we would have thatcβ∗,µ(0(α∗),χ(S′≥1)) = |Eβ∗,0(α∗)| > |Eα∗,0(α∗)| = cα∗,µ(0(α∗),χ(S′≥1)),contradicting rα ≥s rβ by Theorem 2.4.5 because rα = rα∗ by Theorem 2.3.6, and so αR ≤ βR, asdesired.Before we prove Theorem 5.1.2, we will investigate one more extremely valuable statistic, whichwill greatly simplify the argument.5.3 Adjacent pairsDefinition 5.3.1. The adjacent pairs of α is the multiset of multisetsap(α) = {{α1, α2}, {α2, α3}, . . . , {αR−1, αR}}.Because coarsenings of α with length (R−1) are precisely of the form α1 · · ·αi−1(αi+αi+1) · · ·αRfor some i, we see that ap(α) encodes these coarsenings, the longest and hence dominance-minimalones after λ(α). We now show why this statistic is related to the ends.Proposition 5.3.2. Suppose that λ(α) = λ(β). If e(α) 6= e(β), then ap(α) 6= ap(β).Proof. Every part αi belongs to two adjacent pairs, namely {αi−1, αi} and {αi, αi+1}, with theexception of α1 and αR, which belong to only one. Therefore, we can read off e(α) from how manytimes each integer of λ(α) is present in ap(α). To be concrete, we havee(α) = {α1, . . . , αR} ∪ {α1 . . . , αR} \ (∪x∈ap(α)x)where ∪ denotes the multiset union.Example 5.3.3. Let α = 1116311. Thenap(α) = {{1, 1}, {1, 1}, {1, 1}, {1, 3}, {1, 6}, {3, 6}}.435.3. Adjacent pairsThose partitions in M(α) of length 6 are{632111, 632111, 632111, 641111, 731111, 911111}.We can calculate the ends by ∪x∈ap(α)x = {1, 1, 1, 1, 1, 1, 1, 1, 3, 3, 6, 6} ande(α) = {1, 1, 1, 6, 3, 1, 1} ∪ {1, 1, 1, 6, 3, 1, 1} \ {1, 1, 1, 1, 1, 1, 1, 1, 3, 3, 6, 6} = {1, 1}.We now determine more specific information on the partitions determined by near-equal com-positions for which the adjacent pairs differ. By cross-referencing with Lemma 5.2.7 and Lemma5.2.9 of the previous section, we deduce a great deal on near-equal compositions with different ends,allowing us to prove Theorem 5.1.2 in the following section.Lemma 5.3.4. Suppose that rα − rβ = sν , λ(α) = λ(β), and ap(α) 6= ap(β). Then either• ν = ab1d, and λ(α) = a(b− 1)1d+1 or λ(α) = λ((a− b+ 1)b(b− 1)1d)• ν = ab21d and λ(α) = a(b− 1)21d+1for some a ≥ b ≥ 2 and d ≥ 0.Proof. Because ap(α) 6= ap(β), the multisets M(α) and M(β) differ at a partition with (R − 1)parts determined by a coarsening that arises from joining an adjacent pair of parts in α or β.Therefore, the dominance-minimal such partition ν has `(ν) = R − 1. Moreover, each partitionµ with `(µ) = R − 1 that occurs in the h-basis expansion of sν must be determined by such acoarsening, and so for some i, j we haveµ = λ((αi + αj)α1 · · ·αi−1αi+1 · · ·αj−1αj+1 · · ·αR). (5.2)Note that ν2 ≥ 2 by Theorem 4.1.1. By Theorem 2.5.2, the h-basis expansion of sν contains theterm −hν(1) , whereν(1) = λ((ν1 + 1)(ν2 − 1)ν3 · · · νR−1).By Proposition 2.5.10, Part 4, we must have λ(α) ≤dom ν, so (ν1 + 1) can not be a part of λ(α),and so by (5.2) taking µ = ν(1) we haveλ(α) = λ(xy(ν2 − 1)ν3ν4 · · · νR−1) (5.3)for some nonzero x, y with x+ y = ν1 + 1.Let us first suppose that ν3 ≤ 1, so that ν = ab1d for some a ≥ b ≥ 2 and d ≥ 0 andλ(α) = λ(xy(b − 1)1d). By (5.2) taking µ = ν, the (b − 1) must be summed with another part ofλ(α) to make either the b, in which case by (5.3) we have{x, y} = {a, 1} and λ(α) = a(b− 1)1d+1;445.3. Adjacent pairsor the a, in which case by (5.3) we have{x, y} = {a− b+ 1, b} and λ(α) = λ((a− b+ 1)b(b− 1)1d).Now suppose that ν3 ≥ 2. We show that in fact ν3 = 2. If ν3 ≥ 3, then by Theorem 2.5.2, theh-basis expansion of sν also contains the term hν(2) , whereν(2) = λ((ν1 + 2)(ν2 − 1)(ν3 − 1)ν4 · · · νR−1).As before, by Proposition 2.5.10, Part 4, we must have λ(α) ≤dom ν, so (ν1 + 2) can not be a partof λ(α), and so by (5.2) taking µ = ν(2) we haveλ(α) = λ(x′y′(ν2 − 1)(ν3 − 1)ν4 · · · νR−1) (5.4)for some nonzero x′, y′ with x′ + y′ = ν1 + 2.Comparing (5.3) and (5.4), we see thatλ(α) = λ((ν1 − ν3 + 2)(ν2 − 1)ν3(ν3 − 1)ν4 · · · νR−1).Again by (5.2) taking µ = ν, we see that two of (ν1− ν3 + 2), (ν2− 1), and (ν3− 1) are summedto make either ν1 or ν2, and the third is the other of ν1 and ν2. Because ν1 is larger than all threeof these, it must be the sum, and because ν2 > ν2−1 ≥ ν3−1, we specifically have ν2 = ν1−ν3+2.Also note that ν1 ≥ ν2 + 1 since ν1 = ν2 + ν3 − 2 and ν3 ≥ 3.By Theorem 2.5.2 the h-basis expansion of sν also contains the term −hν(3) , whereν(3) = λ(ν1(ν2 + 1)(ν3 − 1)ν4 · · · νR−1).However, taking µ = ν(3), we can not satisfy (5.2) because ν(3) has two parts at least (ν2 + 1) andλ(α) has none. Therefore, we can not have ν3 ≥ 3, and so ν3 = 2.Finally, if ν4 = 2, then by Theorem 2.5.2 the h-basis expansion of sν also contains the term−hν(4) , whereν(4) = λ((ν1 + 3)(ν2 − 1)(ν3 − 1)(ν4 − 1)ν5 · · · νR−1).However, taking µ = ν(4), we can not satisfy (5.2) because by (5.3), two of x, y, ν3 = 2, and ν4 = 2must be summed to make (ν1 + 3) but x + y = ν1 + 1, x + 2, y + 2 ≤ ν1 + 2 since x and y arenonzero, and 2 + 2 < ν1 + 3, so this is impossible. Therefore, ν4 = 1 so that ν = ab21d for somea ≥ b ≥ 2 and d ≥ 0, and by (5.3) we have that λ(α) = a(b− 1)21d+1.455.4. Proof of Theorem 5.1.25.4 Proof of Theorem 5.1.2We are now abundantly prepared to prove Theorem 5.1.2.Proof of Theorem 5.1.2. By reversing α and β if necessary, we may assume without loss ofgenerality that α1 ≤ αR and β1 ≤ βR. Also recall that because λ(α) = λ(β) and λ(αt) = λ(βt), wehave by Proposition 4.1.5 that q′(α) = q′(β) and δα = δβ. By Theorem 5.2.12, we must have thatα1αR ≤lex β1βR, and more specifically because δα = δβ and e(α) 6= e(β), we have that either2 ≤ α1 < β1, or α1 = β1 and 2 ≤ αR < βR.Therefore, takingM = 0(α) = α1 − 1 < β1 − 1 or M = 0(α∗) = αR − 1 < βR − 1respectively, we have by Lemma 5.2.7 and Lemma 5.2.9 that the LR coefficientscα,µ(M,u) 6= cβ,µ(M,u),for 1 ≤ u ≤ S′, unless S′ = 0, in which case u = 0. Note that if S′ ≥ 2, then cα,µ(M,1) 6= cβ,µ(M,1)and cα,µ(M,2) 6= cβ,µ(M,2), violating (2.2), and so we must have that S′ =∑j≥1 q′j ≤ 1, meaningthat all of the 1’s are together, with possibly the exception of lone 1’s on the end. Now we havethat cα,µ(M,χ(S′≥1)) 6= cβ,µ(M,χ(S′≥1)), and so by (2.2)ν = µ(M,χ(S′ ≥ 1)) = (N −R+ 1−M)(R− k − 1 + δα +M)1k−δα ,which is a partition of the form ν = ab1d with a ≥ b ≥ 2 and d ≥ 0, where a = N − R + 1 −M ,b = R− k − 1 + δα +M , and d = k − δα.Again because e(α) 6= e(β), we have by Proposition 5.3.2 that ap(α) 6= ap(β), so by Lemma5.3.4 and because ν = ab1d, we have that λ(α) = a(b− 1)1d+1 or λ(α) = λ((a− b+ 1)b(b− 1)1d).Note that in either case α has length R = d+ 3, so we have thatb = R− k − 1 + δα +M = (d+ 3)− d− 1 +M , so b− 1 = M + 1,and because either M = α1 − 1 < β1 − 1 or M = αR − 1 < βR − 1, we see that in either caseb− 1 ∈ e(α) \ e(β). Note that because δα = δβ, we must have that b ≥ 3.If λ(α) = a(b−1)1d+1, the only possibilities of α and β (up to reversal) satisfying our conditionsS′ ≤ 1, q′(α) = q′(β), δα = δβ, and b− 1 ∈ e(α) \ e(β) are• α = 1d+1a(b− 1), β = 1d+1(b− 1)a465.4. Proof of Theorem 5.1.2• α = 1a1d(b− 1), β = 1(b− 1)1da.These are Cases (3.5) and (3.6), as desired.On the other hand, if λ(α) = λ((a − b + 1)b(b − 1)1d), then note that because the number of1’s of α is k = d and d = k − δα, we must have δα = δβ = 0, meaning there are no end 1’s andindeed all the 1’s are together. Additionally, because those terms in the h-basis expansion of sνwith (R − 1) parts are precisely hab1d and h(a+1)(b−1)1d , and as we know these coarsenings ariseprecisely from joining adjacent pairs in α or β, we have by (2.3) thatmap(α)({b, a− b+ 1}) = map(β)({b, a− b+ 1}) + 1 (5.5)map(β)({b− 1, a− b+ 1}) = map(α)({b− 1, a− b+ 1}) + 1map(α)({x, y}) = map(β)({x, y}) otherwise.Now the only possibility of α and β (up to reversal) satisfying our conditions S′ ≤ 1, q′(α) =q′(β), δα = δβ = 0, b− 1 ∈ e(α) \ e(β), and (5.5) is• α = (b− 1)1db(a− b+ 1), β = b1d(b− 1)(a− b+ 1).This is Case (3.7), as desired.By applying the ω involution, we find the following.Corollary 5.4.1. Suppose that rα−rβ = sν , λ(α) = λ(β), λ(αt) = λ(βt), and e(αt) 6= e(βt). Thenα, β, and ν are (up to reversal of α and β) as in Case (3.8), Case (3.9), or Case (3.10), namelyCase (3.8): α = 1c−121c+d−1a, β = 1c+d21c−2a, ν = a2c1dCase (3.9): α = 1c−1a1c+d−12, β = 1c+da1c−22, ν = a2c1dCase (3.10): α = 1d21c−1a1c−1, β = 1d21c−2a1c, ν = a2c1dfor some a ≥ 2, c ≥ 2, and d ≥ 0.Remark 5.4.2. Because we have now classified all cases of near-equality for which λ(α) 6= λ(β),λ(αt) 6= λ(βt), e(α) 6= e(β), or e(αt) 6= e(βt), we will frequently assume that all of these are equalfor α and β:Hypothesis 5.4.3.λ(α) = λ(β), λ(αt) = λ(βt), e(α) = e(β), e(αt) = e(βt).In such a situation, we also have the following.Proposition 5.4.4. Suppose that q′(α) = q′(β). Then e(αt) = e(βt) if and only if q(α) = q(β).475.4. Proof of Theorem 5.1.2Proof. By Proposition 2.2.5, we have that e(αt) = {p′1 + 2, p′R−k+1 + 2}. Let ξi = 0 · · · 0(−1)10 bethe tuple with a (−1) in the (i − 1)-th place, a 1 in the i-th place, and 0 elsewhere. Now addingtuples pointwise we can express q(α) in terms of e(αt) asq(α) = q′(α) + ξp1(α) + ξpR−k+1(α)and so the result follows because the ξi are linearly independent.Therefore, by Proposition 4.1.5 and Proposition 5.4.4, Hypothesis 5.4.3 also implies thatq′(α) = q′(β), δα = δβ, q(α) = q(β), {p1(α), pR−k+1(α)} = {p1(β), pR−k+1(β)}. (5.6)48Chapter 6Near-Equality with Two Large PartsIn this chapter, we show that we have found all near-equalities for which ν = ab1d for some a ≥ b ≥ 2and d ≥ 0.Theorem 6.1.1. Suppose that rα − rβ = sν and e(α) = e(β). Then the partition ν can not be ofthe form ν = ab1d for some a ≥ b ≥ 2 and d ≥ 0.Proof. Note that by Theorem 2.5.2, the h-basis expansion of sab1d contains exactly one term,namely (−1)d+1h(a+d+1)(b−1), that has an (a+ d+ 1) part. Therefore if µ is a partition of N withan (a+ d+ 1) part, meaning in particular that µ1 = a+ d+ 1 becausea+ d+ 1 > a+ d =2a+ 2d2≥ a+ b+ d2=N2,then we havemM(α)(µ) =mM(β)(µ) + (−1)R−d−1 µ = (a+ d+ 1)(b− 1)mM(β)(µ) µ 6= (a+ d+ 1)(b− 1). (6.1)Our strategy will be to count the total multiplicities of all partitions µ determined by a coarseningof α or β that have an (a+ d+ 1) part and show that the difference∑µ: µ1=a+d+1mM(α)(µ)−∑µ: µ1=a+d+1mM(β)(µ)is even, contradicting (6.1).A coarsening γ ≥coar α with an (a + d + 1) part arises from a string of consecutive partsαi+1 · · ·αR−j in α with sum (a+ d+ 1), and then from further summing some of the i parts to theleft and some of the j parts to the right. As we have seen in Proposition 2.5.6, Part 1, there are2max{i−1,0} ways to sum some of the i parts to the left and 2max{j−1,0} ways to sum some of the jparts to the right. Also note that because a + d + 1 > N2 , i and j can be determined from γ sothere will be no double-counting. Therefore, settingSα = {(i, j) : αi+1 + · · ·+ αR−j = a+ d+ 1},49Chapter 6. Near-Equality with Two Large Partswe have that the total number of coarsenings of α with an (a+ d+ 1) part is∑µ: µ1=a+d+1mM(α)(µ) =∑(i,j)∈Sα2max{i−1,0}+max{j−1,0}.Separating those (i, j) with i, j ≤ 1 if they appear, we have that∑µ: µ1=a+d+1mM(α)(µ) =∑(i,j)∈Sα:i≥2 or j≥22max{i−1,0}+max{j−1,0}+ χ((0, 0) ∈ Sα) + χ((1, 0) ∈ Sα) + χ((0, 1) ∈ Sα) + χ((1, 1) ∈ Sα)and similarly for β. Now we can not have (0, 0) ∈ Sα because N = a + b + d > a + d + 1 sinceb ≥ 2. Meanwhile, (1, 0) ∈ Sα if and only if α1 = b − 1, (0, 1) ∈ Sα if and only if αR = b − 1,and (1, 1) ∈ Sα if and only if α1 + αR = b− 1. So because e(α) = e(β), the sum of these terms isidentical for α and β, and so∑µ: µ1=a+d+1mM(α)(µ)−∑µ: µ1=a+d+1mM(β)(µ)=∑(i,j)∈Sα:i≥2 or j≥22max{i−1,0}+max{j−1,0} −∑(i,j)∈Sβ :i≥2 or j≥22max{i−1,0}+max{j−1,0},which is even because all the terms in both summations are even. This produces the desiredcontradiction.By applying the ω involution, we have the following.Corollary 6.1.2. Suppose that rα − rβ = sν and e(αt) = e(βt). Then the partition ν can not beof the form ν = a2c1d for a ≥ 2, c ≥ 1, and d ≥ 0.Therefore, we see that the ten near-equalities (3.1), (3.2), (3.3), (3.4), (3.5), (3.6), (3.7), (3.8),(3.9), and (3.10) that we have studied so far are a complete classification of all cases where ν is ofthe form ν = ab1d or ν = a2c1d.50Chapter 7Conclusion7.1 SummaryLet us summarize the characterizations of the ten near-equalities we have seen so far.Theorem 7.1.1. Suppose that rα − rβ = sν . Then:1. ν 6= a1d2. The following are equivalent.(a) λ(α) 6= λ(β)(b) (3.1) or (3.2)3. Now suppose that λ(α) = λ(β). The following are equivalent.(a) λ(αt) 6= λ(βt)(b) (3.3) or (3.4)(c) q′(α) 6= q′(β)4. Now suppose also that λ(αt) = λ(βt). The following are equivalent.(a) e(α) 6= e(β)(b) (3.5) or (3.6) or (3.7)(c) ν = ab1d5. Finally, now suppose also that e(α) = e(β). The following are equivalent.(a) e(αt) 6= e(βt)(b) (3.8) or (3.9) or (3.10)(c) q(α) 6= q(β)(d) ν = a2c1dProof.1. This is Theorem 4.1.1.2. This is Theorem 4.1.2.517.2. Adjacent pairs revisited3. This is Corollary 4.1.3 and Proposition 4.1.5.4. This is Theorem 5.1.2 and Theorem 6.1.1.5. This is Corollary 5.4.1, Proposition 5.4.4, and Corollary 6.1.2.To be concise, we can alternatively summarize these results as follows.Theorem 7.1.2. Suppose that rα − rβ = sν . The following are equivalent.(a) λ(α) 6= λ(β), q′(α) 6= q′(β), e(α) 6= e(β), or q(α) 6= q(β)(b) (3.1), (3.2), (3.3), (3.4), (3.5), (3.6), (3.7), (3.8), (3.9), or (3.10)(c) ν = ab1d or ν = a2c1dIn the remainder of this chapter, we explore some ideas towards describing our four remainingcases, namely (3.11), (3.12), (3.13), and (3.14); and towards resolving Conjecture 3.1.7 in general.We end by proposing some new formulations of the problem of near-equality.7.2 Adjacent pairs revisitedWe begin by revisiting the remaining possibility of Lemma 5.3.4, in which ν = ab21d. By Theorem6.1.1, we exclude the case where ν = ab1d by assuming Hypothesis 5.4.3.Theorem 7.2.1. Suppose that rα − rβ = sν , α and β satisfy Hypothesis 5.4.3, and that ap(α) 6=ap(β). Then α, β, and ν are (up to reversal of α and β) as in Case (3.12) or Case (3.13) specializedto c = 1, namely• α = 1a(b− 1)21d, β = 1(b− 1)a21d, ν = ab21d• α = (b− 1)21d+1a, β = (b− 1)1d+12a, ν = ab21dfor some a ≥ b ≥ 3 and d ≥ 0.Because the proof of Theorem 7.2.1 includes an analysis of several cases, we first address thembelow so as not to interrupt our proof.Lemma 7.2.2. Suppose that rα − rβ = sab21d for some a ≥ b ≥ 2, d ≥ 0. ThenmM(α)(λ((N − x)x))−mM(β)(λ((N − x)x)) =(−1)R−d−1 x = b or a+ d+ 2(−1)R−d x = b+ d+ 1 or a+ 10 otherwise.(7.1)527.2. Adjacent pairs revisitedProof. This follows directly from (2.3) because by Theorem 2.5.2 the terms hµ appearing in theh-basis expansion of sab21d with `(µ) = 2 are precisely(−1)d−1h(a+d+2)b and (−1)dhλ((a+1)(b+d+1)).Lemma 7.2.3. We do not have rα−rβ = sab21d for the following cases, where a ≥ b ≥ 3 and d ≥ 1.1. α = (b− 1)21a1d, β = (b− 1)12a1d, a = b+ 12. α = a12(b− 1)1d, β = a21(b− 1)1d3. α = a1(b− 1)21d, β = a21(b− 1)1d4. α = 1d+1a(b− 1)2, β = 1d+1(b− 1)a2Proof. In each case we use Lemma 7.2.2 with R = d + 4 and see that (7.1) is violated: in (1) wetake x = b + 1, in (2) and (4) we take x = b, and in (3) we take x = b if d ≥ 2 and x = 3 ifd = 1.With this out of the way we now prove Theorem 7.2.1.Proof of Theorem 7.2.1. By Lemma 5.3.4 and Theorem 6.1.1, we have that λ(α) = a(b−1)21d+1and ν = ab21d for some a ≥ b ≥ 2 and d ≥ 0. By Corollary 6.1.2, we must have b ≥ 3 in order tosatisfy Hypothesis 5.4.3.By considering the (R− 1)-part partitionsν, (a+ 1)(b− 1)21d, (a+ 2)(b− 1)1d+1, and λ(a(b+ 1)1d+1),which arise by joining an adjacent pair in α or β, we have by Theorem 2.5.2 and (2.3) thatmap(β)({b− 1, 1}) = map(α)({b− 1, 1}) + 1 (7.2)map(α)({a, 1}) = map(β)({a, 1}) + 1map(β)({a, 2}) = map(α)({a, 2}) + 1map(α)({b− 1, 2}) = map(β)({b− 1, 2}) + 1.In other words, in α the a is next to a 1 and the (b − 1) is next to the 2, while in β the a isnext to the 2 and the (b− 1) is next to a 1 but not the 2. In α, if the a is next to the 2, then wemust have b− 1 = 2. Note that if d = 0, then the only possibilities of α and β (up to reversal) thatsatisfy (7.2) and e(α) = e(β) are• α = 1a(b− 1)2, β = 1(b− 1)a2537.2. Adjacent pairs revisited• α = (b− 1)21a, β = (b− 1)12aas desired. So assume that d ≥ 1.We now show that map(α)({a, 1}) = 1. Suppose that map(α)({a, 1}) = 2. By Theorem 2.5.2, theh-basis expansion of sν contains the term (1 + χ(a = b+ 1))hµ, whereµ = (a+ 1)(b+ 1)1d.If a 6= b+ 1, then µ can only be determined by a coarsening of α or β where the a is joined to a 1and the (b− 1) is joined to the 2. By (7.2), we havemM(β)(µ) = 0 and mM(α)(µ) = map(α)({a, 1}) = 2,a contradiction. If a = b+ 1, then µ could also arise from a coarsening of α or β where the (b− 1),the 2, and a 1 are joined. Because in α the (b − 1) is next to the 2 and neither is next to the asince map(α)({a, 1}) = 2, this happens once if 2 or (b− 1) is in e(α) and twice otherwise. Becausein β the (b− 1) is not next to the 2, this happens at most once. Therefore, in order to havemM(α)(µ)−mM(β)(µ) = 2as required by (2.3), we must indeed have that 2 or (b− 1) is in e(α) and that the (b− 1), a 1, andthe 2 in β can be joined to make µ. The only possibility (up to reversal) that satisfies (5.6) and(7.2) is now• α = (b− 1)21a1d, β = (b− 1)12a1d.However, by Lemma 7.2.3, Part 1, we do not have rα − rβ = sν in this case. So we must havemap(α)({a, 1}) = 1. Consequently, by (7.2), map(β)({a, 1}) = 0.By Theorem 2.5.2, the h-basis expansion of sν contains the term (d− 1)hτ , whereτ = (a+ 1)(b− 1)221d−2,which arises by joining the a to a 1 and then joining a pair of adjacent 1’s. Because map(β)({a, 1}) =0, we have mM(β)(τ) = 0. Now letting α˜ >coar α be the coarsening given by replacing the adjacenta and 1 in α by an (a+ 1), we have by Proposition 2.5.6, Part 2 thatmM(α)(τ) = mM(α˜)(τ) = 2(k − 1)− (R− 1)− 1 + q0(α˜) = d− 4 + q0(α˜) = d− 1by (2.3). Therefore q0(α˜) = 3, meaning that all the 1’s in α˜ must be together. So α is a concatena-tion of a1 (or 1a), (b−1)2 (or 2(b−1)), and 1d in some order. There are ostensibly twelve possiblities(up to reversal) for the choice of order and for the two choices, but because map(α)({a, 1}) = 1 and547.2. Adjacent pairs revisitedbecause if the a is next to the 2 in α we must have b− 1 = 2, the only possibilities (up to reversal)that satisfy (5.6) are now• α = 1a(b− 1)21d, β = 1(b− 1)a21d• α = (b− 1)21d+1a, β = (b− 1)1d+12a• α = a12(b− 1)1d, β = a21(b− 1)1d• α = a1(b− 1)21d, β = a21(b− 1)1d• α = 1d+1a(b− 1)2, β = 1d+1(b− 1)a2The first two cases are Case (3.12) and Case (3.13) specialized to c = 1, as desired. By Lemma7.2.3, Parts 2, 3, and 4, we do not have rα − rβ = sν in the other cases.By applying the ω involution, we have the following.Corollary 7.2.4. Suppose that rα− rβ = sν , α and β satisfy Hypothesis 5.4.3, and that ap(αt) 6=ap(βt). Then α, β, and ν are (up to reversal of α and β) as in Case (3.11) or Case (3.14) specializedto b = 3, namely• α = 1c+d+1a21c, β = 1c+d+12a1c, ν = a32c1d• α = (a− 2)21c−121c+d2, β = (a− 2)21c+d21c−12, ν = a32c1dfor some a ≥ 3, c ≥ 1, and d ≥ 0.In order to extend Theorem 7.2.1 to describe Cases (3.12) and (3.13) when c ≥ 2, we wish togeneralize our adjacent pairs statistic in a way that captures how respectively in α the (b− 1) partis adjacent to a string of (c − 1) 1’s and the a part is adjacent to a string of (c + d) 1’s, while inβ the a part is adjacent to the string of (c − 1) 1’s and the (b − 1) part is adjacent to a string of(c+ d) 1’s. We make the following definition.Definition 7.2.5. Recall that we write our composition α asα = 1p1z11p2z2 · · · 1pR−kzR−k1pR−k+1 .Now define the generalized adjacent pairs of α to be the multiset of pairsAP (α) = {(zi, pi), (zi, pi+1) : 1 ≤ i ≤ R− k}.Example 7.2.6. Let α = 1116311 and β = 1113611. ThenAP (α) = {(6, 3), (6, 0), (3, 0), (3, 2)} and AP (β) = {(3, 3), (3, 0), (6, 0), (6, 2)}.557.3. Partial sumsWe believe that the generalized adjacent pairs are precisely what we need to understand ourfour remaining cases.Conjecture 7.2.7. Suppose that rα−rβ = sν , α and β satisfy Hypothesis 5.4.3, and that AP (α) 6=AP (β) or AP (αt) 6= AP (βt). Then α, β, and ν are (up to reversal of α and β) as in Case (3.11),Case (3.12), Case (3.13), or Case (3.14), namelyCase (3.11): α = 1c+d+1a(b− 1)1c, β = 1c+d+1(b− 1)a1c, ν = ab2c1dCase (3.12): α = 1ca(b− 1)1c−121d, β = 1c(b− 1)a1c−121d, ν = ab2c1dCase (3.13): α = (b− 1)1c−121c+da, β = (b− 1)1c+d21c−1a, ν = ab2c1dCase (3.14): α = (a− b+ 1)(b− 1)1c−121c+d(b− 1), β = (a− b+ 1)(b− 1)1c+d21c−1(b− 1), ν = ab2c1dfor some a ≥ b ≥ 3, c ≥ 1, and d ≥ 0.We then suspect that these four remaining cases account for all those where ν = ab2c1d.Conjecture 7.2.8. Suppose that rα−rβ = sν , α and β satisfy Hypothesis 5.4.3, and that AP (α) =AP (β) and AP (αt) = AP (βt). Then the partition ν can not be of the form ν = ab2c1d for somea ≥ b ≥ 3, c ≥ 1, and d ≥ 0.7.3 Partial sumsThe final hurdle in proving Conjecture 3.1.7 would then be to show that there are no near-equalitiesrα− rβ = sν for which ν has at least three parts that are at least 3, and so ν3 ≤ 2. We provide thefollowing interpretation of the inequality ν3 ≤ 2.Definition 7.3.1. Define the partial sums of α to be the multiset unionP (α) = {α1 + · · ·+ αi : α1 + · · ·+ αi ≤ N2} ∪ {αR−j + · · ·+ αR : αR−j + · · ·+ αR < N2}.Example 7.3.2. Let α = 1116311. ThenP (α) = {1, 2, 3} ∪ {1, 2, 5} = {1, 1, 2, 2, 3, 5}.Note that those elements of M(α) with exactly two parts are{(13)1, (13)1, (12)2, (12)2, (11)3, 95} = {(N − x)x : x ∈ P (α)}.Proposition 7.3.3. Suppose that rα − rβ = sν . Then P (α) 6= P (β) if and only if ν3 ≤ 2.Proof. Note that coarsenings of α with exactly two parts arise precisely from summing the firstseveral parts α1 + · · ·+ αi and the last several parts αi+1 + · · ·+ αR. Because either the first sum567.3. Partial sumsis at most N2 or the second sum is less thanN2 , those elements ofM(α) with exactly two parts areprecisely{(N − x)x : x ∈ P (α)}.Therefore, P (α) 6= P (β) if and only if mM(α)(µ) 6= mM(β)(µ) for some partition µ with exactly twoparts. By (2.3) this occurs if and only if the h-basis expansion of sν contains such a term, whichby Theorem 2.5.2 occurs if and only if ν3 ≤ 2.As one application, we have the following result for the more general question of for which α,β, and ν do we have a relationrα − rβ = Csνfor some C ≥ 2.Proposition 7.3.4. Suppose that rα − rβ = Csν for some C ≥ 2. If ν3 ≤ 2, then C = 2.Proof. If ν3 ≤ 2, then by Theorem 2.5.2 the h-basis expansion of sν contains a term hµ with`(µ) = 2, and therefore we must have that by (2.3)|mM(α)(µ)−mM(β)(µ)| = C.However, because `(µ) = 2, α and β each have at most two coarsenings that determine µ, and so0 ≤ mM(α)(µ),mM(β)(µ) ≤ 2.Therefore C ≤ 2.We believe that the techniques we presented to analyze the Littlewood–Richardson rule and theJacobi–Trudi determinantal identity can be applied to show that there are no instances of such anear-near-equality, particularly when ν3 ≤ 2 and Proposition 7.3.4 applies.Conjecture 7.3.5. We do not have rα − rβ = Csν for any C ≥ 2.Another natural generalization of our near-equality question is to ask for which F and G be-longing to various classes of symmetric functions do we have a near-equality relationF −G = sν .The following are some example formulations of this question.Question 1. For which skew shapes λ/µ and σ/τ do we have a near-equality of skew Schurfunctionssλ/µ − sσ/τ = sν?577.3. Partial sumsQuestion 2. For which Schur functions do we have a near-equality of products of Schur functionssλsµ − sσsτ = sν?Question 3. For which graphs G and H do we have a near-equality of chromatic symmetricfunctionsXG −XH = sν?Question 4. For which sets of permutations A and B do we have a near-equality of the corre-sponding quasisymmetric functions defined in [4] and [9]Q(A)−Q(B) = sν?We hope that our results may provide some insight in this study of near-equality of symmetricfunctions.58Bibliography[1] C. Ballantine and R. Orellana, Schur-positivity in a square. Electron. J. of Combin. 21(2014) P3.46.[2] F. Barekat and S. van Willigenburg, Composition of transpositions and equality ofribbon Schur Q-functions. Electron. J. of Combin. 16 (2009) R110.[3] L. Billera, H. Thomas, and S. van Willigenburg, Decomposable compositions, symmet-ric quasisymmetric functions and equality of ribbon Schur functions. Adv. Math. 204 (2006)204–240.[4] S. Elizalde and Y. Roichman, Schur-positive sets of permutations via products of gridclasses. J. Algebraic Combin. 45 (2017) 363–405.[5] W. Fulton, Young Tableaux: With Applications to Representation Theory and Geometry.Cambridge University Press (1997).[6] W. Fulton, Eigenvalues, invariant factors, highest weights, and Schubert calculus. Bull. Am.Math. Soc. 37 (1999) 209–240.[7] M. Gaetz, W. Hardt, and S. Sridhar, Support equalities among ribbon Schur functions.(2017) 19pp. https://arxiv.org/pdf/1709.03011.pdf.[8] V. Gasharov, Incomparability graphs of (3+1)-free posets are s-positive. Discrete Math. 157(1996) 193–197.[9] I. Gessel, Multipartite P-partitions and inner products of skew Schur functions. Contemp.Math. 34 (1984), 289–301.[10] V. Guo, On Jensen’s and related combinatorial identities. Appl. Anal. Discrete Math. 5 (2011)201–211.[11] R. King, T. Welsh, and S. van Willigenburg, Schur positivity of skew Schur functiondifferences and applications to ribbons and Schubert classes. J. Algebraic Combin. 28 (2008)139–167.[12] T. Lam, A. Postnikov, and P. Pylyavskyy, Schur positivity and Schur log-concavity.Amer. J. Math. 129 (2007) 1611–1622.59Bibliography[13] A. Lascoux and P. Pragacz, Ribbon Schur functions. European J. of Combin. 9 (1988)561–574.[14] P. MacMahon, Combinatory Analysis. Cambridge University Press, (1917).[15] P. McNamara, Necessary conditions for Schur-positivity. J. Algebraic Combin. 28 (2008)495–507.[16] P. McNamara and S. van Willigenburg, Positivity results on ribbon Schur functiondifferences. European J. of Combin. 30 (2009) 1352–1369.[17] P. McNamara and S. van Willigenburg, Towards a combinatorial classification of skewSchur functions. Trans. Amer. Math. Soc. 361 (2009) 4437–4470.[18] P. McNamara and S. van Willigenburg, Maximal supports and Schur-positivity amongconnected skew shapes. European J. of Combin. 33 (2012) 1190–1206.[19] V. Reiner, K. Shaw, and S. van Willigenburg, Coincidences among skew Schur func-tions. Adv. Math. 216 (2007) 118–152.[20] J. Remmel and M. Riehl, Generating functions for permutations which contain a givendescent set. Electron. J. of Combin. 17 (2010) R27.[21] R. Stanley, Enumerative Combinatorics. Vol. 2, Cambridge University Press (1999).[22] F. Tom and S. van Willigenburg, Necessary conditions for Schur-maximality Electron. J.of Combin. 25 (2018) P2.30.60
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Classifying the near-equality of ribbon Schur functions Tom, Foster 2018
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Title | Classifying the near-equality of ribbon Schur functions |
Creator |
Tom, Foster |
Publisher | University of British Columbia |
Date Issued | 2018 |
Description | We consider the problem of when the difference of two ribbon Schur functions is a single Schur function. We prove that this near-equality phenomenon occurs in fourteen infinite families and we conjecture that these are the only possible cases. Towards this converse, we prove that under certain additional assumptions the only instances of near-equality are among our fourteen families. In particular, we prove that our first ten families are a complete classification of all cases where the difference of two ribbon Schur functions is a single Schur function whose corresponding partition has at most two parts at least 2. We also provide a framework for interpreting the remaining four families and we explore some ideas towards resolving our conjecture in general. We also determine some necessary conditions for the difference of two ribbon Schur functions to be Schur-positive. |
Genre |
Thesis/Dissertation |
Type |
Text |
Language | eng |
Date Available | 2018-08-10 |
Provider | Vancouver : University of British Columbia Library |
Rights | Attribution-NoDerivatives 4.0 International |
DOI | 10.14288/1.0370959 |
URI | http://hdl.handle.net/2429/66726 |
Degree |
Master of Science - MSc |
Program |
Mathematics |
Affiliation |
Science, Faculty of Mathematics, Department of |
Degree Grantor | University of British Columbia |
GraduationDate | 2018-09 |
Campus |
UBCV |
Scholarly Level | Graduate |
Rights URI | http://creativecommons.org/licenses/by-nd/4.0/ |
AggregatedSourceRepository | DSpace |
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