University of Southern California Dissertations and Theses

Applications on symmetric and quasisymmetric functions

(USC Thesis Other)

Applications on symmetric and quasisymmetric functions

PDF

Download

Open document

Flip pages

Download a page range

Download transcript

Copy asset link

Request this asset

Transcript (if available)

Content Applications on Symmetric and Quasisymmetric Functions by Shiyun Wang A Dissertation Presented to the FACULTY OF THE GRADUATE SCHOOL UNIVERSITY OF SOUTHERN CALIFORNIA In Partial Fulfillment of the Requirements for the Degree DOCTOR OF PHILOSOPHY (MATHEMATICS) August 2023 Copyright 2023 Shiyun Wang To my family ii Acknowledgements First and foremost, I am deeply indebted to my advisor Greta Panova for her generous support and continuous guidance. She has introduced me to the field of Algebraic Com- binatorics and helps me to develop my research ideas. I am grateful to her for providing insightful feedback, invaluable academic career guidance and opportunities, and consistent encouragement and patience. I would like to thank Jason Fulman and Jiapeng Zhang for serving on my dissertation committee and providing comments and advice on my papers. I would also like to thank them, together with Sami Assaf and Susan Montgomery for being the committee members of my oral qualifying exam. IhavegreatlybenefitedfrommycollaboratorsElizabethNiese,SheilaSundaram,Stephanie vanWilligenburgandJulianneVega. Thisdissertationwouldnothavebeenpossiblewithout their support and wisdom. I sincerely thank them for generously sharing their knowledge, being highly encouraging of my research and providing me with professional opportunities to pursue my academic career. I would like to thank all faculties and administrative staff of the Department of Mathe- matics for supporting my graduate studies at USC. I also would like to thank my best friends in these years: Trizzie Huynh, Jiaqi Liu, Xiaoying Pan, Yi Yu, Xiaoshu Zeng, Chenchen Zhao, and many others, for their support and companionship. Finally, I would like to thank my parents, who make me what I am today and always support me unconditionally. I have been fortunate enough to have my husband over the years, who always believe in me and never left my side. iii Table of Contents Dedication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ii Acknowledgements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iii Abstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . viii Chapter 1: Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.1 Symmetric functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.2 Quasisymmetric functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 1.3 The pertinent bases of quasisymmetric functions . . . . . . . . . . . . . . . . 4 Chapter2: Thee-positivityoftheChromaticSymmetricFunctionsandthe Inverse Kostka Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 2.1.1 Chromatic (quasi)symmetric functions and the (3+1)-free posets conjecture . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 2.1.2 Dyck path . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 2.1.3 Special rim hook tabloid . . . . . . . . . . . . . . . . . . . . . . . . . 14 2.1.4 Inverse Kostka matrix . . . . . . . . . . . . . . . . . . . . . . . . . . 16 2.2 The expansion of the chromatic symmetric functions in terms of the inverse Kostka number . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 2.3 The proof of the e-positivity of the chromatic symmetric functions . . . . . . 21 2.4 e-positivity for a class of chromatic symmetric functions for Dyck paths of bounce number three . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 2.5 Discussions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38 Chapter 3: Row-strict dual immaculate functions . . . . . . . . . . . . . . . 40 3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40 iv 3.2 Row-strict dual immaculate functions . . . . . . . . . . . . . . . . . . . . . . 41 3.2.1 A combinatorial definition . . . . . . . . . . . . . . . . . . . . . . . . 41 3.2.2 Creation operators and row-strict immaculate functions . . . . . . . . 47 3.2.3 Results obtained by using ψ . . . . . . . . . . . . . . . . . . . . . . . 52 3.3 Skew row-strict dual immaculate functions . . . . . . . . . . . . . . . . . . . 55 3.3.1 A combinatorial approach . . . . . . . . . . . . . . . . . . . . . . . . 55 3.3.2 Hopf algebra approach . . . . . . . . . . . . . . . . . . . . . . . . . . 61 3.3.3 Expansions of skew Schur functions . . . . . . . . . . . . . . . . . . . 64 3.4 Hook dual immaculate functions . . . . . . . . . . . . . . . . . . . . . . . . . 66 Chapter 4: 0-Hecke modules for row-strict dual immaculate functions . . . 71 4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71 4.2 A 0-Hecke algebra action forRS ∗ . . . . . . . . . . . . . . . . . . . . . . . . 75 4.3 A partial order from the 0-Hecke action . . . . . . . . . . . . . . . . . . . . . 83 4.4 The 0-Hecke module structure of ch − 1 (RS ∗ α ) . . . . . . . . . . . . . . . . . . 87 4.5 A row-strict analogue for extended Schur functions . . . . . . . . . . . . . . 104 4.6 New 0-Hecke modules from the dual immaculate action . . . . . . . . . . . . 120 4.7 A new descent set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154 List of Tables 4.1 Standardtableauxofcompositionshapeα ofn,withdistinctentries{1,2,...,n}, where all rows increase (strictly) left to right: . . . . . . . . . . . . . . 75 4.2 S ∗ α versus the new basisRS ∗ α . . . . . . . . . . . . . . . . . . . . . . . . . . 152 4.3 VariousfamiliesinQSymandassociatedH n (0)-modules; theyform a basis of QSym unless otherwise indicated . . . . . . . . . . . . . . . . . . . 153 v List of Figures 2.1 P ≃ P(d) for d = (4,6,6,6,6,7,8,8) with the corresponding partition τ = (6511). b(d) = (0,0),(0,4),(4,4),(4,6),(6,6),(6,8),(8,8). m = (4,6,8). S 1 ={1,2,3,4}; S 2 ={5,6}; S 3 ={7,8}. G=inc(P). . . . . . . . . . . . . . 15 2.2 Special rim hook tabloids T 1 (left) and T 2 (right). . . . . . . . . . . . . . . . 15 2.3 All special rim hook tabloids of shape µ ′ =(n− 2l− j,l+j,l) and type λ . . 17 2.4 The injective map α . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 2.5 The injective map f. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 2.6 The injective map g for case 1 (left) and case 2(a) (right). . . . . . . . . . . 25 2.7 The injective map g for case 2(b) (left) and case 2(c) (right). . . . . . . . . . 26 2.8 Examples of P-tableaux in set A (left) and set B (right). . . . . . . . . . . . 27 2.9 The injective map ϕ (left) and the adjusted map ϕ ′ (right). . . . . . . . . . . 28 2.10 Examples of the class of natural unit interval orders in Theorem 6.. . . . . . 29 2.11 The injective map ψ : Case 1. . . . . . . . . . . . . . . . . . . . . . . . . . . 31 2.12 The injective map ψ : Case 2a. . . . . . . . . . . . . . . . . . . . . . . . . . . 32 2.13 The injective map ψ : Case 2b (left) and Case 2c (right). . . . . . . . . . . . 32 2.14 The injective map ψ : Case 3. . . . . . . . . . . . . . . . . . . . . . . . . . . 33 2.15 The injective map ψ : Case 4a (left) and Case 4b (right). . . . . . . . . . . . 33 2.16 The injective map ψ : Case 4c (left) and Case 4d (right). . . . . . . . . . . . 34 2.17 The injective map ψ : Case 4d’. . . . . . . . . . . . . . . . . . . . . . . . . . 34 2.18 The injective map σ 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35 2.19 The injective map σ 2 : Case 1a (left) and Case 1b (right). . . . . . . . . . . . 36 2.20 The injective map σ 2 : Case 1b’. . . . . . . . . . . . . . . . . . . . . . . . . . 36 2.21 The injective map σ 2 : Case 2a (left) and Case 2a ′ (right). . . . . . . . . . . 37 2.22 The injective map σ 2 : Case 2b (left) and Case 2b ′ (right). . . . . . . . . . . . 38 vi 2.23 An example of the class of natural unit interval orders in Corollary 1. . . . . 39 2.24 A special rim hook tabloid of type (n− i,i). . . . . . . . . . . . . . . . . . . 39 3.1 The start of the posetP with edge labels. A horizontal 3-strip is shown in red and a vertical 3-strip is shown in blue. . . . . . . . . . . . . . . . . . . . . . . . . . 55 3.2 The path P has D(P) = {1,3,6} and A(P) = {2,4,5}, while Des S ∗ (T) = {1,3,6} and Des RS ∗ (T)={2,4,5}. . . . . . . . . . . . . . . . . . . . . . . . 57 3.3 The standard skew immaculate tableau T and its corresponding path can be decomposed into maximal horizontal strips (3,2,3) 3 → (3,2,2) 3 → (3,2,1), (3,2,1) 2 → (3,1,1) 3 → (3,1), and (3,1) 1 → (2,1). Alternatively, decompose P intomaximalverticalstrips(3,2,3) 3 →(3,2,2),(3,2,2) 3 →(3,2,1) 2 →(3,1,1), and (3,1,1) 3 →(3,1) 1 →(2,1). . . . . . . . . . . . . . . . . . . . . . . . . . . 57 3.4 An example of Ω n− k (T) and℧ k (T). . . . . . . . . . . . . . . . . . . . . . . . 63 4.1 The eight flavours of tableaux, their characteristics and H n (0)-modules, re- lated in pairs by the involution ψ , from the four descent sets. The double arrow-head indicates a quotient module, and the hooked arrow indicates a submodule. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74 4.2 The row-strict dual immaculate poset PRS ∗ 223 ; the red tableaux and S col 223 are in- creasing along rows and up columns, so are in SET(223)=[S col 223 ,S row 223 ]; the blue tableaux and S col 223 = S row∗ 223 are the elements in SIT ∗ (223)=[S 0 223 ,S row∗ 223 ], with the smallest elements in the first column. . . . . . . . . . . . . . . . . . . . . . . . 105 vii Abstract This thesis consists of three projects I worked on symmetric and quasisymmetric functions. The chromatic symmetric functions are analogies of the classical symmetric functions. After its discovery, chromatic quasisymmetric functions were defined, followed by various proper- ties of these functions and their relationships with different bases of symmetric functions, LLT polynomials and Macdonald polynomials. With associated graphs, they also provide us with an algebraic tool to study the properties of the graph. For example, a long-standing conjecture in algebraic combinatorics is that the chromatic symmetric functions distinguish non-isomorphic trees. Further, the chromatic symmetric functions are closely connected with the cohomology of regular semisimple Hessenberg varieties. Regarding quasisymmetric functions,numerousbasesofquasisymmetricfunctionswerediscoveredasanaturalnonsym- metric generalization of symmetric functions, and various analogies of classical symmetric functions were established. These functions and their properties have grown to be a flour- ishing area of research throughout algebraic combinatorics. Chapter 1 gives the background and basic definitions about symmetric and quasisym- metric functions. The first project (Chapter 2) [45] extends the current investigations on the (3+1)-free posets conjectures by Stanley-Stembridge and Shareshian-Wachs with the q-parameterized version. We expand the chromatic symmetric functions for Dyck paths of bounce number three in the elementary symmetric basis using a combinatorial interpreta- tion of the inverse of the Kostka matrix studied in E˜ gecio˜ glu-Remmel (1990). We construct sign-reversinginvolutionstoprovethatcertaincoefficientsinthisexpansionarepositive. We use a similar method to establish the e-positivity of chromatic symmetric functions for Dyck pathsofbouncenumberthreebeyondthe“hook-shape”caseofCho-Huh(2019). Ourresults viii provide more supportive evidence for the Stanley-Stembridge Conjecture by extending the e-positive class of the incomparability graph of natural unit interval orders. The second and third projects are joint work with Elizabeth Niese, Sheila Sundaram, Stephanie van Willigenburg, and Julianne Vega. In Chapter 3 [35], we define a new basis of quasisymmetric functions, the row-strict dual immaculate functions, as the generating function of a particular set of tableaux. We show that this definition gives a function that can also be obtained by applying the involution ψ to the dual immaculate functions of Berg, Bergeron, Saliola, Serrano, and Zabrocki (2014) and establish numerous combinatorial properties for our functions. We give an equivalent formulation of our functions via Bernstein-like operators, in a similar fashion to Berg et al. (2014). Weconcludethischapterbydefiningskewdualimmaculatefunctionsandhookdual immaculate functions, and establishing combinatorial properties for them. In Chapter 4 [34], we focus on the representation-theoretic aspects of the row-strict dual immaculate functions. We construct a cyclic, indecomposable 0-Hecke algebra module for these functions. More precisely, by identifying the descent set corresponding to the row- strict dual immaculate functions, one can determine a 0-Hecke algebra action on the set of standard immaculate tableaux, from where we can construct a 0-Hecke module whose quasisymmetriccharacteristicisexactlytherow-strictdualimmaculatefunctions. Wegivean explicitdescriptionoftheeffectof ψ ontheassociated0-Heckemodules,viatheposetinduced bythe0-Heckeactiononstandardimmaculatetableaux. Thisremarkableposetrevealsother 0-Hecke submodules and quotient modules, often cyclic and indecomposable, notably for a row-strict analogue of the extended Schur functions studied in Assaf-Searles (2019). We give a complete combinatorial and representation-theoretic picture by constructing 0-Hecke modules for the remaining variations on descent sets, and showing that all the possible variationsforgeneratingfunctionsoftableauxoccurascharacteristicsofthe0-Heckemodules determined by these descent sets. ix Chapter 1 Background 1.1 Symmetric functions A composition of a positive integer n is a sequence α =(α 1 ,...,α k ) such that P α i =n. We write α ⊨ n. We sometimes denote n by|α | and k by ℓ(α ), and α j 1 =··· =α j+m =i as i m . The diagram of α =(α 1 ,...,α k ) is a collection of left-justified boxes with α i boxes in row i, where row 1 is the bottom row. Example 1. For α =(3,1,4,2,5,1), the diagram is as follows. A partition λ of n is a positive sequence of integers λ = (λ 1 ,··· ,λ l ) summing to n such that λ 1 ≥···≥ λ l . Definition 1. [41] A homogeneous symmetric function of degree n over a commutative ring R (with identity) is a formal power series f(x)= X α c α x α 1 where 1. α is over all compositions of n; 2. c α ∈R; 3. x α =x α 1 1 x α 2 2 ··· ; 4. f(x ω(1) ,x ω(2) ,··· )=f(x 1 ,x 2 ,··· ) for all permutation ω ofP. Thesymmetricfunctionshavevariousbases. Forλ ⊢n,wedenotethemonomialsymmet- ric functions, elementary symmetric functions, complete homogeneous symmetric functions, power sum symmetric functions, Schur functions by m λ ,e λ ,h λ ,p λ ,s λ respectively. We give the definitions of pertinent bases used later in this chapter. Definition 2. [41] Define the elementary symmetric functions e λ for λ ⊢n as e n = X i 1 <··· <in x i 1 ··· x in ,n≥ 1 with e 0 =1, e λ =e λ 1 e λ 2 ··· . A semistandard Young tableaux (SSYT) of shape λ is an array T = (T ij ) of positive integers of shape λ (1 ≤ i ≤ l(λ ),1 ≤ j ≤ λ i ) that is weakly increasing in each row and strictly increasing in each column. Definition 3. [41] Let λ be a partition of n. The Schur functions s λ (x) of shape λ in the variables x=(x 1 ,x 2 ,··· ) is the formal power series s λ (x)= X T x T where T goes over all SSYTs of shape λ . 2 1.2 Quasisymmetric functions Compositions of n are in bijection with subsets of {1,2,...,n− 1}. Given a composition α =(α 1 ,α 2 ,...,α k )ofn, thecorrespondingsetisset(α )={α 1 ,α 1 +α 2 ,...,α 1 +··· +α k− 1 }. For α =(3,1,4,2,5,1) that is a composition of 16, set(α )={3,4,8,10,15}⊆{ 1,2,...,15}. Given a subset S ={s 1 ,s 2 ,...,s j } of{1,2,...,n− 1}, the corresponding composition of n is comp(S) = (s 1 ,s 2 − s 1 ,...,s j − s j− 1 ,n− s j ). For S ={2,3,5,9,10,14}⊆{ 1,2,...,15}, comp(S) = (2,1,2,4,1,4,2). The composition obtained by reversing the order of the parts of α , the reverse of α , is rev(α ) = (α k ,α k− 1 ,...,α 1 ). The complement of a composition α , denoted α c is the composition obtained from α by taking the complement of the set correspondingtoα . Thatis,α c =comp(set(α ) c ). Thetransposeofacompositionα ,denoted α t is the composition obtained from α by taking the complement of the set corresponding to the reverse of α . That is, α t =comp(set(rev(α )) c ). Forexample,ifα =(3,1,2,4),rev(α )=(4,2,1,3),set(rev(α ))={4,6,7},set(rev(α )) c = {1,2,3,5,8,9}, so α t =(1,1,1,2,3,1,1). We will use several different orders on compositions. For compositions α and β , we say α precedes β in lexicographic order, denoted by α ≤ ℓ β , if either α 1 <β 1 or there is a j > 1 such that α j < β j but α i = β i ,1≤ i≤ j− 1. We say that a composition β = (β 1 ,...,β m ) is a refinement of a composition α = (α 1 ,...,α k ), denoted β ≼ α , if each part of α can be obtained by adding consecutive parts of β . Equivalently, we say that α is a coarsening of β . For example, β = (1,2,1,1,3,2) is a refinement of α = (3,2,5). Finally, we use an order, defined in [9], where α ⊂ s β if 1. |β |=|α |+s, 2. α j ≤ β j , ∀ 1≤ j≤ ℓ(α ), and 3. ℓ(β )≤ ℓ(α )+1. 3 Note that the last two parts guarantee that ℓ(α ) ≤ ℓ(β ) ≤ ℓ(α )+1. If we have only the second condition then this is denoted α ⊆ β . A function f ∈Q[[x 1 ,x 2 ,...]] is quasisymmetric if the coefficient of x α 1 1 x α 2 2 ··· x α k k is the same as the coefficient of x α 1 i 1 x α 2 i 2 ··· x α k i k for every (α 1 ,α 2 ,...,α k ) and i 1 < i 2 < ··· < i k . The set of all quasisymmetric functions forms a Hopf algebra graded by degree, QSym = L n QSym n , whereeachQSym n isavectorspaceoverQ withbasesindexedbycompositions of n. 1.3 The pertinent bases of quasisymmetric functions The pertinent bases for our purposes include the monomial, fundamental, dual immaculate, and quasisymmetric Schur bases. Given a composition α =(α 1 ,α 2 ,...,α k ) of n, the monomial quasisymmetric function is M α = X (i 1 ,i 2 ,...,i k ) i 1 i n− j if j∈set(α ) x in ··· x i 1 =F α (x n ,...,x 1 ). Finallythetruthof (1.7)isevidentuponpassingtothefundamentalexpansionoff ∈QSym. There are corresponding involutions in NSym, denoted by the same letters, and defined on the noncommutative ribbon basis by ψ (r α )=r α c ψ (r α r β )=ψ (r α )ψ (r β ) (1.8) ρ (r α )=r rev(α ) ρ (r α r β )=ρ (r β )ρ (r α ) (1.9) ω(r α )=r α t ω(r α r β )=ω(r β )ω(r α ). (1.10) In NSym, ρ and ω are anti-automorphisms while ψ is an automorphism. We also have that ψ (h α )=e α , ρ (h α )=h rev(α ) and ω(h α )=e rev(α ) . Proposition 1. The pairing between NSym and QSym is invariant under the map ψ . That is, for F ∈QSym and g∈NSym, we have ⟨g,F⟩=⟨ψ (g),ψ (F)⟩. 7 Proof. It suffices to check that the equality holds for the noncommutative ribbon basis elementsg =r α and the basis of fundamental quasisymmetric functions F =F β , where α,β are compositions of n. But this is clear from the preceding definitions. The immaculate functionsS α are a basis of NSym formed by iterated creation opera- tors [9]. Their duals in QSym form the basis consisting of dual immaculate functions,S ∗ α . These functions can be defined combinatorially as the generating function for immaculate tableaux. Definition 4. Given a composition α , an immaculate tableau of shape α is a filling, D, of the cells of the diagram of α with positive integers such that 1. The leftmost column entries strictly increase from bottom to top. 2. The row entries weakly increase from left to right. An immaculate tableau of shape α ⊨ n is standard if it is filled with distinct entries taken from {1,2,...,n}. Given an immaculate tableau D, we form a content monomial, x D , by setting the exponent of x i to be d i , the number of i’s in the tableau D, namely, x D =x d 1 1 x d 2 2 ··· x d k k . Wecallthevector(d 1 ,d 2 ,...)thecontent ofthetableauD. Inparticular a standard tableau of shape α ⊨ n has content equal to the composition (1 n ). Definition 5. The dual immaculate function indexed by the composition α is S ∗ α = X D x D where the sum is over all immaculate tableaux of shape α . We can rewrite the dual immaculate functions in terms of the fundamental basis as a sum over standard immaculate tableaux. To do this, we first standardize each immaculate tableau and define a descent set on the standard immaculate tableaux. The reading word of an immaculate tableau D is obtained by reading the entries of D from left to right starting 8 with the top row. We can standardize a semi-standard tableau (repeated entries allowed) by replacing all the 1’s in the reading word by 1,2,..., in reading order, then the 2’s, etc. Example 2. Here is an immaculate tableau of shape α =(3,2,4,1,2) that has reading word 675344522112, and its standardization. T = 6 7 5 3 4 4 5 2 2 1 1 2 S = 11 12 9 6 7 8 10 3 4 1 2 5 Foracompositionα , letSIT(α )denotethesetofstandardimmaculatetableauxofshape α . Definition 6. [9, Definition 3.20] Given a standard immaculate tableau S, the descent set of S, denoted Des S *(S), is Des S *(S)={i:i+1 appears strictly above i in S}. WerefertoDes S *(S)astheS ∗ -descentsetofS,andtoitsassociatedcompositioncomp(Des S *(S)) as the S ∗ -descent composition of S. For the standard immaculate tableau in Example 2, Des S *(S)={2,5,8,10}. Then [9, Proposition 3.37] S ∗ α = X S∈SIT(α ) F comp(Des S * (S)) , (1.11) where the sum is over all standard immaculate tableaux of shape α . 9 Chapter 2 The e-positivity of the Chromatic Symmetric Functions and the Inverse Kostka Matrix 2.1 Introduction 2.1.1 Chromatic (quasi)symmetric functions and the (3+1)-free posets conjecture Chromatic symmetric functions were first introduced by Stanley [40] and have grown to be a fruitful area of research in relation to enumerative combinatorics, algebraic geometry, and representationtheory. Theycanformnumerousbasesforthealgebraofsymmetricfunctions, Λ = ⊕ n Λ n , and have a close relationship with Hessenberg varieties. Chromatic symmetric functions are an algebraic tool to study graphs. Let G be a finite graph with V(G) = {v 1 ,v 2 ,··· ,v n } andP = {1,2,3,···} . The chro- matic symmetric function X G (x) associated with G is defined as X G (x)=X G (x 1 ,x 2 ,··· )= X κ x κ (v 1 ) x κ (v 2 ) ··· x κ (vn) , where the sum ranges over all proper colorings κ : V → P. Stanley [40] considered the expansion of X G (x) into various symmetric function bases. For example, the expansion of X G (x) in terms of the monomial symmetric functions m λ can be considered as a generating 10 function for stable partitions into independent sets of the vertices of G. The coefficients of X G (x) in the expansion of the power sum symmetric functions p λ are related to the Mobius function of the lattice of contractions of G. If the chromatic symmetric function is expanded into the elementary symmetric function e λ as X G (x) = P λ c λ e λ , then the summation of the coefficients P λ ⊢V,l(λ )=j c λ is the number of acyclic orientations of G with j sinks. Further, if X G (x) can be written as a nonnegative linear combination of a basis b, then we say X G (x) is b-positive. Since X G (x) is m-positive and the image of X G (x) is p-positive under the automorphism ω in symmetric functions, it is natural to explore the combinatorial interpretation of the coefficients of e λ . Thus the famous Stanley-Stembridge conjecture is the following. LetP beafiniteposetandinc( P)beitsincomparabilitygraph,thatis,thegraphGwith vertices the elements of P where two vertices are connected by an edge if and only if the elements are incomparable in P. We say P is (a+b)-free if P does not contain an induced subposet isomorphic to a disjoint union of an a-element chain and a b-element chain. Conjecture 1. [40][43] If P is (3+1)-free, then X inc(P) is e-positive. Somecriticalresultsrelatedtothisconjecturehavebeendiscovered. Gasharov[19]proved theincomparabilitygraphsof(3+1)-freeposetsareSchur-positive. In[19],Gasharovdefined the P-tableau to expand X G (x) in terms of the Schur basis s λ . Let a i,j be the entry of a Young diagram at i th row and j th column. A P-tableau of shape λ is a filling of a Young diagram of shape λ with elements of the poset P =(P,≺ ) satisfying the following. 1. Each element in P appears once in the diagram. 2. a i,j ≺ a i,j+1 for all i,j. 3. a i+1,j ⊀ a i,j for all i,j. Theorem 1. [19] Let P be a (3+1)-free posets, and G = inc(P). If X G (x) = P λ c λ s λ , then c λ is the number of P-tableaux of shape λ . 11 Since each e λ expands positively into the Schur basis, the s-positivity of X G (x) leads to more confidence in Conjecture 1. Guay-Paquet [22] reduced the conditions in Conjecture 1 so that it is enough to prove the e-positivity of the incomparability graphs of natural unit interval orders, that is, posets which are both (3+1)-free and (2+2)-free. Shareshian-Wachs [39] introduced a quasisymmetric refinement of Stanley’s chromatic symmetric function with a variable q counting ascents in the colorings. For a finite graph G=(V,E), the chromatic quasisymmetric function is defined as X G (x,q)= X κ q asc(κ ) x κ , where the sum ranges over all proper colorings κ : V →P and asc(κ ) :=|{(v i ,v j )∈ E : i < j and κ (i)<κ (j)}|. For G the incomparability graph of a natural unit interval order, this quasisymmetric function X G (x,q) turns out to be symmetric. Shareshian-Wachs generalized the expansion of X G (x,q) in terms of various symmetric function bases. For example, The coefficients c λ (q)ofX G (x,q)intheexpansionintermsofthe e λ weregeneralizedas P λ ⊢|V|,l(λ )=j c λ (q)= P o∈O(G,j) q asc(o) , where O(G,j) is the set of acyclic orientations of G with j sinks and asc(o) isthenumberofdirectededges(a,b)ofosuchthataj and i appears above j in T. Let inv(T) denote the number of inversions of T and sh(T) be the shape of T. Theorem 2. [39] Let P be a natural unit interval order and G=inc(P). Then X G (x,q)= X T∈PT(G) q inv(T) s sh(T) . 12 NotethatthisexpansionisequivalenttoX G (x,q)= P λ B λ (q)s λ ,whereB λ = X T∈PT(G) sh(T)=λ q inv(T) . Furthermore, if X G (x,q) can be expanded into a basis b such that all coefficients are in N[q], then we say X G (x,q) is b-positive. Shareshian-Wachs obtained the following version of the Stanley-Stembridge conjecture. Conjecture 2. [39] Let G be the incomparability graph of a natural unit interval order, then X G (x,q) is e-positive. This long-time open (3 +1)-free posets conjecture and its quasisymmetric refinement were widely studied and proved for some special cases, such as cycle graphs [3] and complete graphs[18],lollipopgraphs[16],meltinglollipopgraphs[26],graphsgeneratedbyabelianDyck paths [14, 25, 1] and some certain classes of non-abelian Dyck paths [14, 13]. In particular, since the unit interval orders on [n] are in bijection with Dyck paths on an n× n board, Cho-Huh [14] proved Conjecture 2 for Dyck paths with bounce number two (abelian) and a ”hook-shape” case with bounce number three (non-abelian). The results of Harada-Precup [25] on the cohomology of abelian Hessenberg varieties also imply that Conjecture 2 holds in the abelian case. Cho-Hong [13] showed Conjecture 2 is true when the corresponding Dyck paths have bounce number≤ 3 and q = 1. Recently, Abreu-Nigro [2] proved that for q = 1 the coefficients of e λ for λ of length two are positive. Hamaker-Sagan-Vatter [24] claimed that the coefficient of e n in X G (x,q) is inN[q]. 2.1.2 Dyck path Recall that the natural unit interval orders are posets both (3+1)-free and (2+2)-free. Shareshian-Wachs [39] provided a helpful characterization of this family of posets. Let d := (d 1 ,d 2 ,··· ,d n− 1 ) be a sequence of positive integers satisfying d 1 ≤ d 2 ≤···≤ d n− 1 ≤ nandd i ≥ iforalli. disalsocalledaDyck pathonann× nboard,takingastaircase 13 walkfromthebottomleftcorner(0,0)tothetoprightcorner(n,n)thatliesabovethediago- nal. It is connecting (0,0),(0,d 1 ),(1,d 1 ),(1,d 2 ),(2,d 2 ),(2,d 3 ),(3,d 3 ),··· ,(n− 2,d n− 1 ),(n− 1,d n− 1 ),(n− 1,n),(n,n) with either horizontal or vertical steps. Let P(d) be the poset on [n] with the order relations given by i ≺ j if i < n and j∈{d i +1,d i +2,··· ,n}. i.e., the poset relations i≺ j in P(d) are given by all cells (i,j) that lie above the Dyck path d. Denote the Young diagram of all cells above d by τ . Then a bijection exists between the natural unit interval orders and the Dyck paths. Proposition 1. [39] Let P be a poset on [n], then P is a natural unit interval order iff P ≃P(d) for some d=(d 1 ,d 2 ,··· ,d n− 1 ). Thus, given a natural unit interval order P on [n], we can obtain its corresponding Dyck path d = (d 1 ,d 2 ,··· ,d n− 1 ) on an n× n board and the partition τ determined by d. Let G = inc(P), and m := (m 0 ,m 1 ,··· ,m l ) be a sequence of positive integers sat- isfying d 1 = m 0 < m 1 < ··· < m l− 1 < m l = n. For all i ∈ [1,l− 1], m i is deter- mined recursively by m i− 1 such that m i = d m i− 1 +1 . Let b(d) be the bounce path connecting (0,0),(0,m 0 ),(m 0 ,m 0 ),(m 0 ,m 1 ),(m 1 ,m 1 ),(m 1 ,m 2 ),(m 2 ,m 2 ),··· ,(n,n). Denotethebounce number,thatis,thenumberoftimesthebouncepathhittingthediagonalincludingtheend- point (n,n) by|m|. Figure2.1illustratesanexampleofanaturalunitintervalorder P =P(d)forDyckpath d = (4,6,6,6,6,7,8,8). Let {1,2,··· ,m 0 } be the set S 1 , {m 0 +1,··· ,m 1 } be the set S 2 , and{m 1 +1,··· ,m 2 } be the set S 3 . Denote the sizes of S 1 ,S 2 ,S 3 by c,b,a respectively. 2.1.3 Special rim hook tabloid Following [46], A rim hook with length l of a partition λ ⊢ n is a sequence of l connected cells in the Young diagram satisfying the following: 1. Any two adjacent cells have a common edge. 14 a b c 1 G: 2 3 4 5 6 7 8 Figure 2.1: P ≃ P(d) for d = (4,6,6,6,6,7,8,8) with the corresponding partition τ = (6511). b(d) = (0,0),(0,4),(4,4),(4,6),(6,6),(6,8),(8,8). m = (4,6,8). S 1 = {1,2,3,4}; S 2 ={5,6}; S 3 ={7,8}. G=inc(P). 2. The sequence starts from a cell in the diagram’s southwest boundary and travels along its northeast rim. 3. We can obtain a valid Young diagram with size n− l by removing the rim hook. A special rim hook H is a rim hook with at least one cell in the first column. Define the sign of H as sign(H)=(− 1) ht(H)− 1 , where ht(H) is the height of the rim hook. A special rim hook tabloid T of shape µ and type λ =(λ 1 ,··· ,λ l ) is a tiling of the Young diagram of shape µ with special rim hooks of sizes {λ 1 ,··· ,λ l }. Define the sign of T as sign(T)= Q H sign(H). Example 1. Let T be a special rim hook tabloid of shape µ = (4333) and type λ = (5431). Then we can obtain two such possible T as in Figure 2.2. Figure 2.2: Special rim hook tabloids T 1 (left) and T 2 (right). Hence sign(T 1 )=1· (− 1)· (− 1)· 1=1, and sign(T 2 )=1· 1· (− 1)· 1=− 1. 15 2.1.4 Inverse Kostka matrix Let K λ,µ be the Kostka number enumerating semistandard Young Tableau (SSYT) of shape λ and type µ . We have s λ = P ν K λ,ν m ν . Since⟨h µ ,m ν ⟩ = δ µν for the symmetric functions scalar product⟨,⟩, we can obtain that h µ = P λ K λ,µ s λ . E˜ gecio˜ glu-Remmel [46] introduced a combinatorial interpretation of the inverse of the Kostka matrix. Since both s λ and h λ formbasesforthesymmetricfunctions, wecaninverttheKostkamatrixas s µ = P λ K − 1 λ,µ h λ , where K − 1 are the inverse Kostka number. Recall that in symmetric functions, there is an involution ω :Λ →Λ such that ω(e λ )=h λ . Applying ω to both sides we obtain s µ = X λ K − 1 λ,µ ′ e λ , (2.1) where µ ′ denotes the conjugate partition. Then the inverse Kostka number can be computed as the following. Theorem 3. [46] K − 1 λ,µ = X T sign(T), where the summation is over all special rim hook tabloids of type λ and shape µ . Example 2. In Example 1, K − 1 (5431),(4333) =sign(T 1 )+sign(T 2 )=0. 2.2 Theexpansionofthechromaticsymmetricfunctions in terms of the inverse Kostka number In an n× n board, let P be an unit interval order and G = inc(P). Then P ≃ P(d) for d a Dyck path of bounce number |m|. By Theorem 2 and Equation (2.1), we can write the associated chromatic symmetric function as X G (x,q)= X µ X λ B µ (q)K − 1 λ,µ ′ e λ . (2.2) 16 Remark 1. In Equation (2.2), the coefficient B µ (q) is nonzero only if µ 1 ≤| m|. Since for B µ (q) = P T∈PT(G),sh(T)=µ q inv(T) , a P-tableau T of shape µ exists only if µ 1 ≤ the longest chain of relations in P. In a longest chain of P, we can have at most one entry chosen from {1,··· ,m 0 }, and at most one entry chosen from {m 0 +1,··· ,m 1 } and so forth. Thus the length of the chain can be at most|m|. From now and throughout the paper, we consider X G (x,q) when|m|=3. Following the notations in Figure 2.1, we let|S 1 |=c,|S 2 |=b,|S 3 |=a and let k =a+b. Lemma 1. Let P be an unit interval order on [n] such that P ≃ P(d) for Dyck path d and |m| = 3. For any l and j, if a P-tableau T with sh(T) = 3 l 2 j 1 n− 3l− 2j exists, i.e., B 3 l 2 j 1 n− 3l− 2j ̸=0, then 2l+j≤ k. Proof. This is immediate from Remark 1. There are in total 2l+j entries in the second and third column of T, which can at most exhaust all elements of P in S 2 and S 3 . Remark 2. In the context of Lemma 1, using Remark 1, we can reduce Equation (2.2) by finding all possible nonzero terms, that is, when µ =3 l 2 j 1 n− 3l− 2j for l∈[0,min{a,b,c}] and j∈[0,k− 2l]. The six tabloids in Figure 2.3 include all possible special rim hook tabloids when µ has at most three columns. l j l j l j λ =(n− 2l− j,l+j,l); λ = (n− 2l− j+2,l+j− 1,l− 1); λ =(n− 2l− j+1,l+j+1,l− 2); l j l j l j λ =(n− 2l− j,l+j+1,l− 1); λ = (n− 2l− j+1,l+j− 1,l); λ =(n− 2l− j+2,l+j,l− 2). Figure 2.3: All special rim hook tabloids of shape µ ′ =(n− 2l− j,l+j,l) and type λ . FollowingRemark2,Theorem3andFigure2.3,forl∈[0,min{a,b,c}]andj∈[0,k− 2l], wecancomputetheinverseKostkanumberK − 1 λ,µ ′ inEquation(2.2)foreachµ andλ asfollows: 17 • When µ =1 n :K − 1 λ, (n) =        1 if λ =(n) 0 otherwise • When µ =2 j 1 n− 2j :K − 1 λ, (n− j,j) =                1 for λ =(n− j,j) − 1 for λ =(n− j+1,j− 1) 0 otherwise • Whenµ =3 l 2 j 1 n− 3l− 2j :K − 1 λ, (n− 2l− j,l+j,l) =                                                1 for λ =(n− 2l− j,l+j,l) 1 for λ =(n− 2l− j+2,l+j− 1,l− 1) 1 for λ =(n− 2l− j+1,l+j+1,l− 2) − 1 for λ =(n− 2l− j,l+j+1,l− 1) − 1 for λ =(n− 2l− j+1,l+j− 1,l) − 1 for λ =(n− 2l− j+2,l+j,l− 2) 0 otherwise Note that the first two cases can be merged into the last one by restricting λ in the last case to only valid shape. Namely, if l = 0 or j = 0, then K − 1 λ,µ ′ ̸= 0 provided λ i ≥ 0 for each i. Now we can rewrite Equation (2.2) as X G (x,q) = min{a,b,c} X l=0 k− 2l X j=0 B 3 l 2 j 1 n− 3l− 2js 3 l 2 j 1 n− 3l− 2j = min{a,b,c} X l=0 k− 2l X j=0 B 3 l 2 j 1 n− 3l− 2j X λ K − 1 λ, (n− 2l− j,l+j,l) e λ = min{a,b,c} X l=0 k− 2l X j=0 B 3 l 2 j 1 n− 3l− 2j e (n− 2l− j,l+j,l) +e (n− 2l− j+2,l+j− 1,l− 1) +e (n− 2l− j+1,l+j+1,l− 2) 18 − e (n− 2l− j+1,l+j− 1,l) − e (n− 2l− j,l+j+1,l− 1) − e (n− 2l− j+2,l+j,l− 2) = min{a,b,c} X l=0 " k− 2l X j=0 B 3 l 2 j 1 n− 3l− 2je (n− 2l− j,l+j,l) − k− 2l X j=0 B 3 l 2 j 1 n− 3l− 2je (n− 2l− j+1,l+j− 1,l) # + min{a,b,c}− 1 X l=− 1 " k− 2(l+1) X j=0 B 3 l+1 2 j 1 n− 3l− 2j− 3e (n− 2l− j,l+j,l) − k− 2(l+1) X j=0 B 3 l+1 2 j 1 n− 3l− 2j− 3e (n− 2l− j− 2,l+j+2,l) # + min{a,b,c}− 2 X l=− 2 " k− 2(l+2) X j=0 B 3 l+2 2 j 1 n− 3l− 2j− 6e (n− 2l− j− 3,l+j+3,l) − k− 2(l+2) X j=0 B 3 l+2 2 j 1 n− 3l− 2j− 6e (n− 2l− j− 2,l+j+2,l) # = min{a,b,c} X l=0 " k− 2l− 1 X j=0 B 3 l 2 j 1 n− 3l− 2j− B 3 l 2 j+1 1 n− 3l− 2j− 2 e (n− 2l− j,l+j,l) +B 3 l 2 k− 2l 1 n− 2k+le (n− k,k− l,l) − B 3 l 1 n− 3le (n− 2l+1,l,l− 1) # + min{a,b,c}− 1 X l=− 1 " k− 2l− 2 X j=2 B 3 l+1 2 j 1 n− 3l− 2j− 3− B 3 l+1 2 j− 2 1 n− 3l− 2j+1 e (n− 2l− j,l+j,l) +B 3 l+1 1 n− 3l− 3e (n− 2l,l,l) +B 3 l+1 2 1 1 n− 3l− 5e (n− 2l− 1,l+1,l) − B 3 l+1 2 k− 2l− 3 1 n+l− 2k+3e (n− k+1,k− l− 1,l) − B 3 l+1 2 k− 2l− 2 1 n+l− 2k+1e (n− k,k− l,l) # + min{a,b,c}− 2 X l=− 2 " k− 2l− 2 X j=3 B 3 l+2 2 j− 3 1 n− 3l− 2j− B 3 l+2 2 j− 2 1 n− 3l− 2j− 2 e (n− 2l− j,l+j,l) +B 3 l+2 2 k− 2l− 4 1 n− 2k+l+2e (n− k+1,k− l− 1,l) − B 3 l+2 1 n− 3l− 6e (n− 2l− 2,l+2,l) # = min{a,b,c} X l=0 " k− 2l− 1 X j=0 B 3 l 2 j 1 n− 3l− 2j− B 3 l 2 j+1 1 n− 3l− 2j− 2 e (n− 2l− j,l+j,l) (2.3) +B 3 l 2 k− 2l 1 n− 2k+le (n− k,k− l,l) − B 3 l 1 n− 3le (n− 2l+1,l,l− 1) # + min{a,b,c} X l=0 " k− 2l− 2 X j=2 B 3 l+1 2 j 1 n− 3l− 2j− 3− B 3 l+1 2 j− 2 1 n− 3l− 2j+1 e (n− 2l− j,l+j,l) +B 3 l+1 1 n− 3l− 3e (n− 2l,l,l) +B 3 l+1 2 1 1 n− 3l− 5e (n− 2l− 1,l+1,l) − B 3 l+1 2 k− 2l− 3 1 n+l− 2k+3e (n− k+1,k− l− 1,l) − B 3 l+1 2 k− 2l− 2 1 n+l− 2k+1e (n− k,k− l,l) # 19 + min{a,b,c} X l=0 " k− 2l− 2 X j=2 B 3 l+2 2 j− 3 1 n− 3l− 2j− B 3 l+2 2 j− 2 1 n− 3l− 2j− 2 e (n− 2l− j,l+j,l) +B 3 l+2 2 k− 2l− 4 1 n− 2k+l+2e (n− k+1,k− l− 1,l) # = min{a,b,c} X l=0 " k− 2l− 2 X j=2 B 3 l 2 j 1 n− 3l− 2j− B 3 l 2 j+1 1 n− 3l− 2j− 2 +B 3 l+1 2 j 1 n− 3l− 2j− 3− B 3 l+1 2 j− 2 1 n− 3l− 2j+1 +B 3 l+2 2 j− 3 1 n− 3l− 2j− B 3 l+2 2 j− 2 1 n− 3l− 2j− 2 e (n− 2l− j,l+j,l) + B 3 l 1 n− 3l− B 3 l 2 1 1 n− 3l− 2 +B 3 l+1 1 n− 3l− 3 e (n− 2l,l,l) + B 3 l 2 1 1 n− 3l− 2− B 3 l 2 2 1 n− 3l− 4 +B 3 l+1 2 1 1 n− 3l− 5 e (n− 2l− 1,l+1,l) + B 3 l 2 k− 2l− 1 1 n+l− 2k+2− B 3 l 2 k− 2l 1 n+l− 2k− B 3 l+1 2 k− 2l− 3 1 n+l− 2k+3 +B 3 l+2 2 k− 2l− 4 1 n+l− 2k+2 e (n− k+1,k− l− 1,l) + B 3 l 2 k− 2l 1 n+l− 2k− B 3 l+1 2 k− 2l− 2 1 n+l− 2k+1 e (n− k,k− l,l) − B 3 l 1 n− 3le (n− 2l+1,l,l− 1) # = min{a,b,c} X l=0 " k− 2l X j=2 B 3 l 2 j 1 n− 3l− 2j− B 3 l 2 j+1 1 n− 3l− 2j− 2 +B 3 l+1 2 j 1 n− 3l− 2j− 3− B 3 l+1 2 j− 2 1 n− 3l− 2j+1 (2.4) +B 3 l+2 2 j− 3 1 n− 3l− 2j− B 3 l+2 2 j− 2 1 n− 3l− 2j− 2 e (n− 2l− j,l+j,l) + B 3 l 1 n− 3l− B 3 l 2 1 1 n− 3l− 2 +B 3 l+1 1 n− 3l− 3 e (n− 2l,l,l) + B 3 l 2 1 1 n− 3l− 2− B 3 l 2 2 1 n− 3l− 4 +B 3 l+1 2 1 1 n− 3l− 5− B 3 l+1 1 n− 3l− 3 e (n− 2l− 1,l+1,l) # Note that in the third line of Equation (2.3), we adjust the range of l since when l = − 1,e λ =0,andwhenl =min{a,b,c},thecorrespondingP-tableaudoesnotexistbyLemma 1, thus B 3 l+1 2 j 1 n− 3l− 2j− 3 =B 3 l+1 2 j− 2 1 n− 3l− 2j+1 =0. Similar reasoning can be applied to the last sum in this equation as well as in Equation (2.4), B 3 l 2 j+1 1 n− 3l− 2j− 2 =0 when j =k− 2l, and B 3 l+1 2 j 1 n− 3l− 2j− 3 =0 when j =k− 2l and j =k− 2l− 1. 20 2.3 Theproofofthee-positivityofthechromaticsymmetric functions We prove Conjecture 2 for certain coefficients when the bounce number is three. In partic- ular, for each coefficient of the elementary function in Equation (2.4), we construct a sign reversing involution such that each P-tableau with a negative sign is injectively mapped to one with a positive sign. Consequently, the remaining terms is a polynomial in q with positive coefficients. We first have the following lemma, which is critical in our insertion algorithms for con- structing such sign reversing involutions. Lemma 2. Given a natural unit interval order P(d) on [n] with |m| = 3. Let T be a P- tableau obtained from P, and let L be a vertical sequence of consecutive entries in column j of T. Denote the entry at the ith row in L by a i,j . For every i, there are at most two indices s such that s>i and a i,j ,a i+1,j ··· ,a s− 1,j ≺ a s,j . Proof. Following the notations defined in Figure 2.1, we have the fact that S 1 ≺ S 3 . Now suppose that for some i, there exist two indices s satisfying the chain relations described in the lemma, denote them by s 1 and s 2 . Without loss of generality, we can assume s 1 < s 2 . Then we have a i,j ,a i+1,j ··· ,a s 1 − 1,j ≺ a s 1 ,j and a i,j ,··· ,a s 1 ,j ,··· ,a s 2 − 1,j ≺ a s 2 ,j . Since |m| = 3, so the length of a poset chain is at most three by Remark 1. It is necessary that a i,j ,··· ,a s 1 − 1,j ∈ S 1 ,a s 1 ,j ∈ S 2 , and a s 2 ,j ∈ S 3 such that a i,j ≺ a s 1 ,j ≺ a s 2 ,j , thus no element is greater than a s 2 ,j in P. The result follows. Theorem4. Let P be a natural unit interval order on [n] such that P ≃P(d) for Dyck path d with |m| = 3, and G = inc(P). In the expansion of X G (x,q) in terms of the elementary symmetric functions, [e (n− 2l,l,l) ]X G (x,q) for l∈[0,min{a,b,c}] are inN[q]. Proof. We obtain the expansion of X G (x,q) in Equation (2.4). We prove the e-positivity for the term e (n− 2l,l,l) by showing that (B 3 l 1 n− 3l− B 3 l 2 1 1 n− 3l− 2 +B 3 l+1 1 n− 3l− 3) is a polynomial in 21 q with positive coefficients. In particular, we define an inversion preserving injective map α :{T − ∈ PT(G)|sh(T − ) = 3 l 2 1 1 n− 3l− 2 }7→{T + ∈ PT(G)|sh(T + ) = 3 l 1 n− 3l or 3 l+1 1 n− 3l− 3 } as follows. Let T be in the domain of α , and let a i,j be the entry at the ith row and jth column of T. For s ∈ [l+2,n− 2l− 1], if there exists an entry a s,1 in the first column such that a l+1,1 ,a l+2,1 ,··· ,a s− 1,1 ≺ a s,1 ≺ a l+1,2 , then we move a s,1 to the second position of row l+1 and move a l+1,2 right after a s,1 in the same row. See Figure 2.4 (left). Note that in this case, such a s,1 is unique (if exists) by Lemma 2, otherwise a l+1,1 ≺ a s 1 ,1 ≺ a s 2 ,1 ≺ a l+1,2 gives a 4-chain in P. Moreover, {a s− 1,1 } ∈ S 1 , {a l,2 ,a s,1 } ∈ S 2 , {a l,3 ,a l+1,2 } ∈ S 3 , we are guaranteed to obtain a valid P-tableau α (T). If such a s,1 does not exist, then we find the smallestm∈[l+2,n− 2l− 1]suchthata m,1 ⊀ a l+1,2 andinserta l+1,2 abovea m,1 . SeeFigure 2.4 (right). If no such a m,1 exists, that is, {a l+1,1 ,a l+2,1 ,··· ,a n− 2l− 1,1 } ≺ a l+1,2 , then we simply move a l+1,2 to the bottom of the first column. In all cases, for any two cells that are incomparable in the poset P, the relative position of above and below remains unchanged, preservingallinversionsthroughoutthemap. Moreover,fromthewayweconstructthismap, it is not hard to see that all resulting P-tableaux are distinct, thus the map is injective. a 1,1 a 1,2 a 1,3 . . . a l,1 a l,2 a l,3 a l+1,1 a l+1,2 . . . a s,1 . . . −→ a 1,1 a 1,2 a 1,3 . . . a l,1 a l,2 a l,3 a l+1,1 a s,1 a l+1,2 . . . a s− 1,1 a s+1,1 . . . a 1,1 a 1,2 a 1,3 . . . a l,1 a l,2 a l,3 a l+1,1 a l+1,2 . . . a m,1 . . . −→ a 1,1 a 1,2 a 1,3 . . . a l,1 a l,2 a l,3 a l+1,1 . . . a l+1,2 a m,1 . . . Figure 2.4: The injective map α . 22 Theorem5. Let P be a natural unit interval order on [n] such that P ≃P(d) for Dyck path d with |m| = 3, and G = inc(P). In the expansion of X G (x,q) in terms of the elementary symmetric functions, [e (n− 2l− 1,l+1,l) ]X G (x,q) for l∈[0,min{a,b,c}] are inN[q]. Proof. UsingEquation 2.4, wewanttoshowthat (B 3 l 2 1 1 n− 3l− 2− B 3 l 2 2 1 n− 3l− 4+B 3 l+1 2 1 1 n− 3l− 5− B 3 l+1 1 n− 3l− 3) is a polynomial in q with positive coefficients. Define two maps f : n T − ∈PT(G)|sh(T − )=3 l+1 1 n− 3l− 3 o 7→ n T + ∈PT(G)|sh(T + )=3 l 2 1 1 n− 3l− 2 o and g : n T + ∈PT(G)|sh(T + )=3 l+1 2 1 1 n− 3l− 5 o 7→ n T − ∈PT(G)|sh(T − )=3 l 2 2 1 n− 3l− 4 o Formapf,letT 1 beinthedomainoff,wefindthesmallest sfors∈[l+2,n− 2l− 2]such that a s,1 ⊀ a l+1,2 , then insert a l+1,2 above a s,1 , and move a l+1,3 left by one cell. See Figure 2.5. If there does not exist such a s,1 , then we simply move a l+1,2 to the bottom of the first column. Further, the resulting tableau f(T 1 ) is a valid P-tableau since a l+1,1 ≺ a l+1,3 ⊀ a l,2 . In addition, since a l+1,2 ∈ S 2 , inserting different a l+1,2 into the first column yields distinct P-tableaux. Thus map f is injective, and it is straightforward to check that all inversions are preserved. For map g, let T 2 be in the domain of g, we discuss two cases for entries of T 2 . • Case 1: a l+2,2 ≺ a l+1,3 . We have a l+1,1 ,a l+2,1 ∈ S 1 and a l+1,2 ,a l+2,2 ∈ S 2 and a l+1,3 ∈ S 3 , thus a l+2,1 ≺ a l+1,3 . Inthiscase, webumpa l+2,2 witha l+1,3 , andfindthesmallest s fors∈[l+3,n− 2l− 3] such that a s,1 ⊀ a l+2,2 , then insert a l+2,2 above a s,1 . See Figure 2.6 (left). If there does not exist such a s,1 , then we simply move a l+2,2 to the bottom of the first column. 23 a 1,1 a 1,2 a 1,3 . . . a l,1 a l,2 a l,3 a l+1,1 a l+1,2 a l+1,3 a l+2,1 . . . −→ a 1,1 a 1,2 a 1,3 . . . a l,1 a l,2 a l,3 a l+1,1 a l+1,3 . . . a s− 1,1 a l+1,2 a s,1 . . . Figure 2.5: The injective map f. Again all inversions are preserved. • Case 2: a l+2,2 ⊀ a l+1,3 . We consider three sub-cases. – Case 2a: If a l+1,2 ⊀ a l+2,2 and a l+2,1 ⊀ a l+1,2 , find the smallest s for s ∈ [l + 3,n− 2l− 3] such that a s,1 ⊀ a l+1,2 and insert a l+1,2 above a s,1 . Move a l+1,3 left by one cell. See Figure 2.6 (right). We can carefully check inversions in this case. Since a l+1,2 ≺ a l+1,3 while a l+2,2 ⊀ a l+1,3 , then a l+1,2 0 and f ∈NSym, F ⊥ (1 i ) (fh j )=F ⊥ (1 i ) (f)h j +F ⊥ (1 i− 1 ) (f)h j− 1 ; F ⊥ (i) (fh j )= min(i,j) X k=0 F ⊥ (i− k) (f)h j− k . In particular we have F ⊥ (i) (h j )=                0, i>j h j− i , 1≤ i≤ j h j , i=0; F ⊥ (1 i ) (h j )=                0, i>1 h j− 1 , i=1 h j , i=0. The next two definitions are made in [9]. Definition 4. [9, Definition 3.1] The noncommutative Bernstein operator B m is defined by B m = X i≥ 0 (− 1) i h m+i F ⊥ (1 i ) , and for α ∈Z m , B α =B α 1 ··· B α m . Note that when i = 0, (1 0 ) is the empty composition and thus F ⊥ (1 0 ) (f) = f = F ⊥ ∅ (f) for all f ∈NSym, since F ∅ =1 in QSym. Also F ⊥ (1 i ) (1)=F ⊥ (i) (1)=        0 i>0, 1 i=0. While we chose duality to define immaculate functions, the following is the original definition, which was proven to be equivalent in [9]. 49 Definition 5. [9, Definition 3.2] For any α ∈Z m , the immaculate functionS α ∈ NSym is given by S α =B α (1)=B α 1 ··· B α m (1). This definition was inspired by Bernstein’s original definition in the Hopf algebra of symmetric functions for a Schur function s α indexed by any m-tuple α ∈Z m . As observed in [9, Example 3.3], we have S (m) =B m (1)=h m , S (a,b) =B a (h b )=h a h b − h a+1 h b− 1 . Applying ψ to Lemma 1, and using Proposition 1 and the fact that ψ (F α )=F α c, so that ψ (F (1 i ) )=F (i) in NSym i , we obtain Lemma 2. For i,j >0 and f ∈NSym, F ⊥ (i) (fe j )=F ⊥ (i) (f)e j +F ⊥ (i− 1) (f)e j− 1 ; F ⊥ (1 i ) (fe j )= min(i,j) X k=0 F ⊥ (1 i− k ) (f)e j− k . In particular we have F ⊥ (1 i ) (e j )=                0, i>j e j− i , 1≤ i≤ j e j , i=0; F ⊥ (i) (e j )=                0, i>1 e j− 1 , i=1 e j , i=0. Now we define new operators as follows. Definition 6. Define the noncommutative Bernstein operator B rs m by B rs m = X i≥ 0 (− 1) i e m+i F ⊥ (i) , and for α ∈Z m , B rs α =B rs α 1 ··· B rs α m . 50 Note that when i = 0, this is the empty composition and F ∅ = 1 in QSym, and thus F ⊥ (0) (f)=f =F ⊥ ∅ (f) for all f ∈NSym. Furthermore we have the following. Lemma 3. For α ∈Z m , ψ (S α )=B rs α (1). Proof. From the above properties, it is clear that B rs m (1)=e m , ψ (S (a,b) )=B rs a (e b )=e a e b − e a+1 e b− 1 . Hence the result is true for m≤ 2. Let f ∈NSym. We claim that ψ (B m (f))=B rs m (ψ (f)). (3.6) We have ψ (B m (f))=ψ " X i≥ 0 h m+i F ⊥ (1 i ) (f) # = X i≥ 0 e m+i ψ [F ⊥ (1 i ) (f)] = X i≥ 0 e m+i F ⊥ (i) (ψ (f))=B rs m (ψ (f)), where the penultimate equality is thanks to Proposition 1. Since for α ∈Z m , B α (1)=B α 1 (f), f =B α 2 ··· B α m (1), the result now follows by induction. Theorem 4. The row-strict immaculate function RS α can be defined as the result of ap- plying a creation operator as follows: RS α =B rs α (1). Proof. Immediatefromtheprecedinglemma,sincewealreadyknowthatRS α =ψ (S α ). 51 Finally, just as left multiplication by h m can be expressed in terms of creation operators [9, Remark 3.6], we have the following. Lemma 4. Left multiplication by h m in NSym can be expressed as applying the operator h m = X i≥ 0 B m+1 F ⊥ (i) , and left multiplication by e m in NSym can be expressed as applying the operator e m = X i≥ 0 B rs m+1 F ⊥ (1 i ) . Proof. Immediate from Equation (3.6). 3.2.3 Results obtained by using ψ We can immediately obtain the row-strict analogue of many results in [9] by using the ψ involution. We list here the most pertinent for the remainder of the paper. We leave results for skew row-strict dual immaculate functions to the next section, as the combinatorial definition is not obviously equivalent. Theorem 5. 1. [9, Lemma 3.4] For s≥ 0,m∈Z and f ∈NSym, B m (f)h s =B m+1 (f)h s− 1 +B m (fh s ) ψ ⇐⇒ B rs m (f)e s =B rs m+1 (f)e s− 1 +B rs m (fe s ). 2. [9, Theorem 3.5] (Multiplicity-free right Pieri rule) S α h s = X α ⊂ s β S β ψ ⇐⇒ RS α e s = X α ⊂ s β RS β . 52 3. [9, Proposition 3.32] (Multiplicity-free right Pieri rule) For a composition α and s≥ 0, S α S (1 s ) =S α e s = X β S β ψ ⇐⇒ RS α RS (1 s ) =RS α h s = X β RS β , where the summation ranges over compositions ofβ of|α |+s such thatα i ≤ β i ≤ α i +1 and α i =0 for i>ℓ(α ). 4. [9, Corollary 3.31] S (1 n ) = X α ⊨ n (− 1) n− ℓ(α ) h α =e n ψ ⇐⇒ RS (1 n ) = X α ⊨ n (− 1) n− ℓ(α ) e α =h n . 5. [9, Theorem 3.27] (Jacobi-Trudi) For ℓ(α )=m, S α = X σ ∈Sm (− 1) sgn(σ ) h (α 1 +σ 1 − 1,α 2 +σ 2 − 2,...,α m+σ m− m) ψ ⇐⇒ RS α = X σ ∈Sm (− 1) sgn(σ ) e (α 1 +σ 1 − 1,α 2 +σ 2 − 2,...,α m+σ m− m) where S m is the symmetric group on m elements and (− 1) sgn(σ ) is the sign of σ . 6. [9, Corollary 3.31] S (1 n ) = X α ⊨ n (− 1) n− ℓ(α ) h α =e n ψ ⇐⇒ RS (1 n ) = X α ⊨ n (− 1) n− ℓ(α ) e α =h n . Also from [9, Lemma 2.5] and Equation (3.6), F ⊥ (1 r ) (S (1 n ) )=S (1 n− r ) , and for s>1,F ⊥ (s) (S (1 n ) )=0; ψ ⇐⇒ F ⊥ (r) (RS (1 n ) )=RS (1 n− r ) , and for s>1,F ⊥ (1 s ) (RS (1 n ) )=0. 53 7. [9, Proposition 3.16 and Corollary 3.18] h β = X α ≥ ℓ β K α,β S α ψ ⇐⇒ e β = X α ≥ ℓ β K α,β RS α and by Theorem 3 h β = X α ≥ ℓ β K ∗ α,β RS α ψ ⇐⇒ e β = X α ≥ ℓ β K ∗ α,β S α . 8. [9, Theorem 3.25] The ribbon function r β expands positively in both immaculate bases: r β = X α ≥ ℓ β L α,β S α ψ ⇐⇒ r β c = X α ≥ ℓ β L α,β RS α . 9. [9, Theorem 3.38] The Schur function s λ with ℓ(λ ) = k expands into the dual immac- ulate and row-strict dual immaculate bases as follows: s λ = X σ ∈S k (− 1) sgn(σ ) S ∗ σ (λ ) ψ ⇐⇒ s λ ′ = X σ ∈S k (− 1) sgn(σ ) RS ∗ σ (λ ) takingS ∗ σ (λ ) = 0 = RS ∗ σ (λ ) = 0 if σ and λ do not satisfy the condition below: for λ a partition and σ ∈ S ℓ(λ ) , we define σ (λ ) = (λ σ 1 +1− σ 1 ,...,λ σ k +k− σ k ) provided λ σ i +i− σ i >0 for each i. 10. [5, Theorem 1.1] For α a composition and c αβ ≥ 0, S ∗ α = X β c αβ ˆ S β ψ ⇐⇒ RS ∗ α = X β c αβ R ˆ S β , where ˆ S and R ˆ S are the Young quasisymmetric Schur and row-strict quasisymmetric Schur functions. 54 3.3 Skew row-strict dual immaculate functions 3.3.1 A combinatorial approach Following the work of Berg et. al. [9], we define the poset P of immaculate tableaux. The labelled posetP is on the set of all compositions. Place an arrow from α to β if α and β differ by a single box, denoted β ⊂ 1 α . The label of m on each cover α m −→ β denotes the row containing the single additional box. Denote the path from α to β inP by P =[α,β ]. ∅ 1 2 1 3 2 1 2 1 4 3 2 1 3 2 1 2 3 1 2 1 Figure 3.1: The start of the posetP with edge labels. A horizontal 3-strip is shown in red and a vertical 3-strip is shown in blue. To obtain a standard skew immaculate tableau from a path P = [α,β ], for each m i , 1≤ i≤ k, labeltherightmostunlabeledcellinrow m i ofα withk− i+1, seeExample4. In order to understand the combinatorial models for skew dual immaculate and skew row-strict dual immaculate functions we define two special types of paths. Definition 7. A path P ={α =β (0) m 1 −→ β (1) m 2 −→··· m k −→ β (k) =β } in the posetP is a • horizontal k-strip if m 1 ≤ m 2 ≤···≤ m k , and a • vertical k-strip if m 1 >m 2 >··· >m k . The horizontal 3-strip (red path) and vertical 3-strip (blue path) in Figure 3.1 give rise to the following tableaux. 55 1 2 3 3 2 1 horizontal strip vertical strip We can now make the following definition. Definition 8. A standard skew immaculate tableau of shape α/β is a filling of the shape α/β with the distinct positive integers{1,2,...,|α/β |}, such that rows strictly increase from left to right and the labels in α/β in cells that are in the first column of α must increase from bottom to top. For a path P = [α,β ] of length k, define the descent set of P to be D(P) = {k− i : m i > m i+1 } and the weak ascent set of P to A(P) = {k− i : m i ≤ m i+1 }. Each such path P = [α,β ] corresponds to a unique standard skew immaculate tableau T of shape α/β , and conversely. Furthermore, the descent set D(P) coincides with the descent set Des S ∗ (T) = {i : i + 1 appears strictly above i in T}, and similarly the ascent set A(P) coincides with the descent set Des RS ∗ (T)={i:i+1 appears weakly below i in T}. Example 4. For α/β =(3,2,3)/(1,1,2), T = 1 2 3 4 is a valid standard skew immaculate tableau. It corresponds to the path P = (3,2,3) 1 → (2,2,3) 1 → (1,2,3) 2 → (1,1,3) 3 → (1,1,2). Further, Des RS ∗ (T) = {1,2,3}, D(P) is empty, and A(P)={1,2,3}. GivenapathP =[α, ∅]correspondingtoastandardimmaculatetableauT, wehavethat Des S ∗ (T) = D(P) and Des RS ∗ (T) = A(P), by comparing the definitions, and is illustrated in Figure 3.2. 56 T = 4 7 2 3 5 1 6 P =(2,3,2) 3 →(2,3,1) 1 →(1,3,1) 2 →(1,2,1) 3 →(2,1) 2 →(1,1) 2 →(1) 1 →∅ Figure 3.2: The path P has D(P) = {1,3,6} and A(P) = {2,4,5}, while Des S ∗ (T) = {1,3,6} and Des RS ∗ (T)={2,4,5}. Note that given a skew immaculate tableau, it can be decomposed into horizontal or vertical strips in several ways. An example of decomposing a tableau into either horizontal or vertical strips is given in Figure 3.3. T = 2 4 5 3 1 P =(3,2,3) 3 →(3,2,2) 3 →(3,2,1) 2 →(3,1,1) 3 →(3,1) 1 →(2,1) Figure 3.3: The standard skew immaculate tableau T and its corresponding path can be decomposed into maximal horizontal strips (3,2,3) 3 → (3,2,2) 3 → (3,2,1), (3,2,1) 2 → (3,1,1) 3 → (3,1), and (3,1) 1 → (2,1). Alternatively, decompose P into maximal vertical strips (3,2,3) 3 →(3,2,2), (3,2,2) 3 →(3,2,1) 2 →(3,1,1), and (3,1,1) 3 →(3,1) 1 →(2,1). In [9] the poset P and horizontal strips are used to define the skew dual immaculate functions as follows. Definition 9. For {γ : β ⊆ γ ⊆ α } an interval in P, define the skew dual immaculate function to be S ∗ α/β = X γ ⟨S β h γ ,S ∗ α ⟩M γ . This can be rewritten in terms of both the fundamental basis and the dual immaculate basis. 57 Proposition 2. [9, Propositions 3.47 and 3.48] For{γ :β ⊆ γ ⊆ α } an interval inP, S ∗ α/β = X γ ⟨S β r γ ,S ∗ α ⟩F γ (3.7) = X γ ⟨S β S γ ,S ∗ α ⟩S ∗ γ (3.8) = X P=[β,α ]∈P F comp(D(P)) = X T a standard skew immaculate tableau of shape α/β F comp(Des S ∗ (T)) ; (3.9) in the last line, each path P from β to α corresponds to a unique standard skew immaculate tableau T of shape α/β . Note that the number of standard skew immaculate tableaux T of shape α/β with comp(Des S ∗ (T))=γ is⟨S β r γ ,S ∗ α ⟩. Definition 10. For {γ : β ⊆ γ ⊆ α } an interval in P, define the skew row-strict dual immaculate function to be RS ∗ α/β = X γ ⟨RS β h γ ,RS ∗ α ⟩M γ . We now quickly obtain the following. Theorem 6. For{γ :β ⊆ γ ⊆ α } an interval inP, RS ∗ α/β = X γ ⟨RS β r γ ,RS ∗ α ⟩F γ =ψ (S ∗ α/β ) = X γ ⟨RS β RS γ ,RS ∗ α ⟩RS ∗ γ = X P=[β,α ]∈P F comp(A(P)) = X T a standard skew immaculate tableau of shape α/β F comp(Des RS ∗ (T)) . 58 Proof. The first equality is immediate from Definition 10 by using (1.3) to expand h γ in terms of the ribbon basis, interchanging the order of summation, and finally using (1.1): RS ∗ α/β = X γ ⟨RS β X τ ≽γ r τ ,RS ∗ α ⟩M γ = X τ ⟨RS β r τ ,RS ∗ α ⟩ X γ ≼τ M γ ! = X τ ⟨RS β r τ ,RS ∗ α ⟩F τ . The second line now follows by applying ψ to the first equality in Proposition 2, and using the invariance of the pairing under ψ , which gives ψ (S ∗ α/β )= X γ ⟨ψ (S β ) ψ (r γ ),ψ (S ∗ α )⟩ ψ (F γ )= X γ ⟨RS ∗ β r γ c,RS ∗ α ⟩ F γ c, where we have used (3.1), (1.8) and (1.4). The last two lines are now immediate by applying ψ to the last two equations in Proposition 2, since A(P) and D(P) are complementary by definition, and each path P from β to α corresponds to a unique standard skew immaculate tableau T of shape α/β . Definition 11. Let α and β be compositions with β ⊆ α . Then a filling T of the diagram of α/β with positive integers is a skew immaculate tableau provided 1. the entries in the first column of α (if any remain in α/β ) are strictly increasing from bottom to top, and 2. rows weakly increase from left to right. Similarly, T is a skew row-strict immaculate tableau if 1. the entries in the first column of α (if any remain in α/β ) are weakly increasing from bottom to top, and 2. rows strictly increase from left to right. We now have the needed interpretation of the coefficients in Definitions 9 and 10 to rewriteS ∗ α/β andRS ∗ α/β as generating functions of skew immaculate tableaux. 59 Theorem 7. Let α and β be compositions with β ⊆ α . Then S ∗ α/β = X T x T where the sum is over all skew immaculate tableaux of shape α/β , and RS ∗ α/β = X T x T where the sum is over all skew row-strict immaculate tableaux of shape α/β . Proof. By Point (3) in Theorem 5, we know that for γ =γ 1 γ 2 ··· γ k , α can be obtained from β by a series of vertical strips of lengths γ 1 ,γ 2 ,...,γ k . Thus the coefficient ⟨RS β h γ ,RS ∗ α ⟩ represents the number of ways to add a sequence of vertical strips of lengths γ 1 ,γ 2 ,...,γ k from β to α , which counts the number of skew immaculate tableaux T of shape α/β such thatthedescentcompositionofT iscoarserthanγ ,sinceaddingaverticalstripafteranother one may or may not create a descent. See Example 5 below. Thus ⟨S β h γ ,S ∗ α ⟩ is the number of skew immaculate tableaux of shape α/β of content γ and ⟨RS β h γ ,RS ∗ α ⟩ is the number of skew row-strict immaculate tableaux of shape α/β of content γ . The result now follows immediately from the definitions. Example 5. Consider T = 1 4 3 2 60 and corresponding path P =(2,2,2) 3 →(2,2,1) 2 →(2,1,1) 1 →(1,1,1) 3 →(1,1). Note that T can be considered to be formed from vertical strips corresponding to γ =(1,3) or (1,1,2), or (1,2,1) or (1,1,1,1) since comp(Des RS ∗ (T))=(1,3) and is coarser than the listed options for γ. In analogy with the well-known Schur function identity [30, Eqn. (5.9)], s λ (X,Y)= X µ ⊂ λ s µ (X)s λ/µ (Y), (3.10) Theorem 7 immediately gives us the following: Theorem 8. Suppose we have two sets of variables, X and Y, ordered so that the alphabet X precedes the alphabet Y. Then S ∗ α (X,Y)= X β ⊂ α S ∗ β (X)S ∗ α/β (Y), (3.11) and RS ∗ α (X,Y)= X β ⊂ α RS ∗ β (X)RS ∗ α/β (Y). (3.12) 3.3.2 Hopf algebra approach We consider the Hopf algebra approach to defining skew dual immaculate functions and establish that it is equivalent to the previous definition. To start, we provide a brief intro- duction to the necessary Hopf algebra background. WehavethatNSymandQSymformdualHopfalgebrasusingthepairing⟨·,·⟩ :NSym⊗ QSym→ Q defined by ⟨h α ,M β ⟩=δ αβ where δ αβ =1 if α =β and 0 otherwise. 61 Given dual bases{B i } i∈I and{D i } i∈I , B i · B j = X k b k i,j B k ⇔ ∆ D k = X i,j b k i,j D i ⊗ D j D i · D j = X k d k i,j D k ⇔ ∆ B k = X i,j d k i,j B i ⊗ B j where· is the product and ∆ is the coproduct. For the fundamental quasisymmetric functions, we have that ∆ F α = X (β,γ ) with β ·γ =α or β ⊙ γ =α F β ⊗ F γ (3.13) whereforβ =(β 1 ,...,β k )andγ =(γ 1 ,...,γ n ), β · γ =(β 1 ,...,β k ,γ 1 ,...,γ n )isthe concate- nation of β and γ , and β ⊙ γ = (β 1 ,...,β k− 1 ,β k +γ 1 ,γ 2 ,...,γ n ) is the near-concatenation of β and γ . Following [11], we can define the coproduct ∆ S ∗ α in terms of skew elements ] S ∗ α/γ . Definition 12. Let α ⊨ n and define ∆ S ∗ α = X γ S ∗ γ ⊗ ] S ∗ α/γ . We show that ] S ∗ α/γ =S ∗ α/γ as described in Proposition 2. Lemma 5. ] S ∗ α/γ =S ∗ α/γ = X T F comp(Des S ∗ (T)) where the sum is over all standard skew immaculate tableaux T of shape α/γ . Proof. We use the technique of [11, Proposition 3.1]. Let T be a standard skew immaculate tableaux such that|T| = n. For any k with 0≤ k≤ n, let℧ k (T) be the standardization of the skew tableaux consisting of cells of T with entries{n− k+1,...,n}. Also let Ω k (T) be 62 T = 4 5 8 ∗ ∗ 6 7 ∗ ∗ 2 3 ∗ 1 9 Ω 4 (T)= 4 ∗ ∗ ∗ ∗ 2 3 ∗ 1 ℧ 5 (T)= ∗ 1 4 ∗ ∗ 2 3 ∗ ∗ ∗ ∗ ∗ ∗ 5 Figure 3.4: An example of Ω n− k (T) and℧ k (T). the skew tableaux consisting of the cells of T after removing the entries{k+1,...,n} as in Figure 3.4. NotethatifT isastandardimmaculatetableauofshapeα ,thenT =Ω n− k (T)∪(℧ k (T)+ (n− k))where℧ k (T)+(n− k)is℧ k (T)withn− kaddedtoeachentry. SupposeDes S ∗ (T)=α with|α |=n. Then we can rewrite (3.13) as ∆ F α = n X i=0 F β i ⊗ F γ i where |β i | = n− i, |γ i | = i, and either β i · γ i = α or β i ⊙ γ i = α . Observe that β i = comp(Des S ∗ (Ω n− i (T))) and γ i =comp(Des S ∗ (℧ i (T))). Then ∆ S ∗ α =∆ X T F comp(Des S ∗ (T)) ! = X T ∆ F comp(Des S ∗ (T)) = X T n X i=0 F β i ⊗ F γ i where T is a standard immaculate tableau of shape α . Further, by Definition 12 we have ∆ S ∗ α = X δ S ∗ δ ⊗ ] S ∗ α/δ = X δ X S F comp(Des S ∗ (S)) ⊗ ] S ∗ α/δ 63 where S is a standard immaculate tableau of shape δ . For a fixed S of shape δ with|δ |=n− k for some k, there exists a standard immaculate tableau T of shape α such that S = Ω n− k (T). Then ℧ k (T) has shape α/δ . Similarly, given astandardimmaculatetableau T ofshapeα , T =Ω n− k (T)∪(℧ k (T)+(n− k))where Ω n− k (T) has shape δ with|δ |=n− k and℧ k (T) has shape α/δ . Thus ] S ∗ α/δ = X T F comp(Des S ∗ (T)) =S ∗ α/δ where T is a standard skew immaculate tableau of shape α/δ . It follows by Theorem 6 that we have ∆ RS ∗ α = X β RS ∗ β ⊗R S ∗ α/β . 3.3.3 Expansions of skew Schur functions WecanalsouseaHopfalgebraapproachtoestablishskewversionsofPoint(9)inTheorem5, fromwherewerecallthatforλ apartitionandσ ∈S ℓ(λ ) ,define σ (λ )=(λ σ 1 +1− σ 1 ,...,λ σ k + k− σ k ) provided λ σ i +i− σ i >0 for each i. Also recall that s λ/µ =det(h λ i − µ j − i+j ). If we consider compositions α ⊆ λ , we can define s λ/α = det(h λ i − α j − i+j ). Note that if there exists some α j − j = α k − k for some j ̸= k, s λ/α =0 since two columns of the matrix will be equal. If no such pair j,k exists, then there exists a unique permutation τ such that τ (α )=(α τ 1 +1− τ 1 ,...,α τ k +k− τ k )=µ where µ is a partition. In this case, s λ/µ =(− 1) sgn(τ ) s λ/α . (3.14) Theorem 9. Let λ and µ be partitions with µ ⊆ λ . Then s λ/µ = X σ ∈S ℓ(λ ) (− 1) sgn(σ )+sgn(τ ) S ∗ σ (λ )/τ (µ ) 64 for any choice of τ such that τ (µ ) is a composition. Proof. Recall that ∆( s λ ) = P µ s λ/µ ⊗ s µ = P µ s µ ⊗ s λ/µ because the Hopf algebra of symmetric functions is cocommutative. We can rewrite ∆( s λ ) using Theorem 5, Point (9). Then ∆( s λ )=∆   X σ ∈S ℓ(λ ) (− 1) sgn(σ ) S ∗ σ (λ )   = X σ ∈S ℓ(λ ) (− 1) sgn(σ ) ∆ S ∗ σ (λ ) = X σ ∈S ℓ(λ ) (− 1) sgn(σ ) X β S ∗ β ⊗ S ∗ σ (λ )/β ! = X β S ∗ β ⊗   X σ ∈S ℓ(λ ) (− 1) sgn(σ ) S ∗ σ (λ )/β   . On the other hand, X µ s µ ⊗ s λ/µ = X µ   X τ ∈S ℓ(µ ) (− 1) sgn(τ ) S ∗ τ (µ )   ⊗ s λ/µ = X µ X τ ∈S ℓ(µ ) (− 1) sgn(τ ) S ∗ τ (µ ) ⊗ s λ/µ = X β S ∗ β ⊗   X τ ∈S ℓ(β ) (− 1) sgn(τ ) s λ/τ − 1 (β )   where β is a composition and τ − 1 (β ) is a partition. Thus for a fixed choice of β , X σ ∈S ℓ(λ ) (− 1) sgn(σ ) S ∗ σ (λ )/β = X τ ∈S ℓ(β ) (− 1) sgn(τ ) s λ/τ − 1 (β ) . 65 Note that for each β , there is at most one τ ∈ S ℓ(β ) such that s λ/τ − 1 (β ) = s λ/µ ̸= 0 for a partition µ . Thus s λ/µ = X σ ∈S ℓ(λ ) (− 1) sgn(σ )+sgn(τ ) S ∗ σ (λ )/τ (µ ) for any valid choice of τ . Choosing τ as the identity gives the following corollary. Corollary 2. For partitions λ and µ with µ ⊆ λ , s λ/µ = X σ ∈S ℓ(λ ) (− 1) sgn(σ ) S ∗ σ (λ )/µ . Applying ψ to both sides of Theorem 9 gives us an expansion in terms of the row-strict dual immaculate functions. Corollary 3. For partitions λ and µ with µ ⊆ λ and τ ∈S ℓ(µ ) such that τ (µ ) is a composi- tion, s λ ′ /µ ′ = X σ ∈S ℓ(λ ) (− 1) sgn(σ )+sgn(τ ) RS ∗ σ (λ )/τ (µ ) . 3.4 Hook dual immaculate functions Now that we have skew row-strict dual immaculate functions, we can define hook dual im- maculate functions in a combinatorial manner analogous to the hook Schur functions [37] and hook quasisymmetric Schur functions [32]. Definition 13. LetA={1,2,...,ℓ} andA ′ ={1 ′ ,2 ′ ,...,k ′ } be two alphabets with 1<2< ··· <ℓ<1 ′ <2 ′ <··· <k ′ . Then a semistandard hook immaculate tableau of shape α is a filling of the diagram of α such that 1. the first column increases from bottom to top with the increase strict in A and weak in A ′ , and 66 2. each row increases from left to right, weakly inA and strictly inA ′ . Denote the set of all semistandard hook immaculate tableaux of shape α by HI α . The content monomial of a hook tableau T is a monomial in two alphabets, x 1 ,...,x ℓ and y 1 ,...,y k , where z T = Y i∈A∪A ′ z # of i’s in T i where z i =x i if i∈A and z i =y i if i∈A ′ . Example 6. Let α = (3,1,2,4,3). Then T, as shown below, is a hook immaculate tableau with content monomial z T =x 2 1 x 2 x 2 3 y 3 1 y 2 y 3 y 2 4 y 5 . T = 1 ′ 2 ′ 4 ′ 1 ′ 3 ′ 4 ′ 5 ′ 3 1 ′ 2 1 1 3 Definition 14. The hook dual immaculate function indexed by α is HS ∗ α (X,Y)=HS ∗ α (x 1 ,...,x l ,y 1 ,...,y k )= X T∈HIα z T . It follows immediately from the definition that HS ∗ α (X,Y)= X γ ⊆ α S ∗ γ (X)RS ∗ α/γ (Y). (3.15) We can also expandHS ∗ α (X,Y) in terms of the super fundamental quasisymmetric func- tions. We use the definition in [32]. Definition 15. For α ⊨ n, ˜ Q α (X,Y)= X a 1 ≤ a 2 ≤···≤ an a i =a i+1 ∈A⇒i/ ∈set(α ) a i =a i+1 ∈A ′ ⇒i∈set(α ) z a 1 z a 2 ··· z an , 67 where z a =x a if a∈A and z a ′ =y a for a ′ ∈A ′ . Theorem 10. [32, Theorem 4.1] For α ⊨ n, ˜ Q α (X,Y)= n X i=0 F β (X)F γ (Y) where β · γ =α if i∈set(α ) and β ⊙ γ =α if i / ∈set(α ). As usual, we must have a standardization procedure for hook dual immaculate tableaux and an appropriate descent set to index the super fundamental quasisymmetric functions. To standardize a hook dual immaculate tableaux H, first replace the entries of H fromA by scanning unprimed entries from left to right, starting with the top row, replacing 1s as they are encountered in this reading order, followed by 2s, etc. Next continue with the entries of A ′ by scanning from right to left starting with the bottom row. Example7. The reading word of T, as shown below, is 3,2,1,1,3,1 ′ ,5 ′ ,4 ′ ,3 ′ ,1 ′ ,4 ′ ,2 ′ , giving rise to stdz(T) below. T = 1 ′ 2 ′ 4 ′ 1 ′ 3 ′ 4 ′ 5 ′ 3 1 ′ 2 1 1 3 stdz(T)= 8 9 12 7 10 11 13 4 6 3 1 2 5 Note that the standardization of a hook dual immaculate tableau is a standard dual immaculate tableau. Recall that the descent set of a standard dual immaculate tableau S is Des S ∗ (S) ={i : i+1 is strictly above i in S}. The descent set for stdz(T) in Example 7 is Des S ∗ (stdz(T))={2,3,5,6,7,11}. From the definition of standardization, we note that if T is a hook immaculate tableau of shape α with T =S∪U where S is an immaculate tableau of shape β and U is a skew row-strict immaculate tableau of shape α/β , then Des S ∗ (stdz(T))=Des S ∗ (stdz(S))∪(Des RS ∗ (stdz(U)) c +|β |) 68 if|β |+1 is weakly lower than|β | in stdz(T) and Des S ∗ (stdz(T))=Des S ∗ (stdz(S))∪(Des RS ∗ (stdz(U)) c +|β |)∪{|β |} if|β |+1 appears strictly above|β | in stdz(T). Theorem 11. Let α ⊨ n. Then HS ∗ α (X,Y)= X S ˜ Q comp(Des S ∗ (S)) (X,Y) where the sum is over all standard dual immaculate tableaux of shape α . Proof. We show that each polynomial consists of the same monomials. Suppose x a 1 ··· x a k y b 1 ··· y bm is the content monomial associated with a hook immaculate tableau T of shape α with a 1 ≤ a 2 ≤ ··· ≤ a k and b 1 ≤ b 2 ≤ ··· ≤ b m . Note that if a i = a i+1 , then i / ∈ Des S ∗ (stdz(T)) by the standardization procedure. Similarly, if b i = b i+1 , i+k ∈ Des S ∗ (stdz(T)), since b ′ i must occur in a lower row of T than b ′ i+1 . Thus x a 1 ··· x a k y b 1 ··· y bm is a monomial in ˜ Q comp(Des S ∗ (stdz(T))) (X,Y). Now suppose x a 1 ··· x a k y b 1 ··· y bm is a monomial in ˜ Q comp(Des S ∗ (S)) (X,Y) for some stan- dard immaculate tableau S of shape α . We must show that there exists a hook immaculate tableau with content a 1 ,...,a k , b ′ 1 ,...,b ′ m . Do this by replacing n in S with b ′ m , n− 1 in S with b ′ m− 1 and so on. Since b i = b i+1 implies that i+k ∈ Des S ∗ (S), we have that each primed entry in a row is distinct and increasing from left to right. Similarly, if a i = a i+1 , then i / ∈ Des S ∗ (S), guaranteeing that the first column is increasing bottom to top and has distinct unprimed entries. Thus the result is a hook immaculate tableau of content x a 1 ··· x a k y b 1 ··· y bm . Berele and Regev [8] defined hook Schur functions indexed by a partition λ as Hs λ (X,Y)= X µ ⊆ λ s µ (X)s λ ′ /µ ′(Y). 69 We have the following analogue of Theorem 5, Point (9). Theorem 12. Let λ be a partition. Then Hs λ (X,Y)= X τ ∈S ℓ(λ ) (− 1) sgn(τ ) HS ∗ τ (λ ) (X,Y). Proof. Let λ be a partition. Then Hs λ (X,Y)= X µ ⊆ λ s µ (X)s λ ′ /µ ′(Y) = X µ ⊆ λ   X σ ∈S ℓ(µ ) (− 1) sgn(σ ) S ∗ σ (µ ) (X)s λ ′ /µ ′(Y)   = X µ ⊆ λ   X σ ∈S ℓ(µ ) (− 1) sgn(σ ) S ∗ σ (µ ) (X)(− 1) sgn(σ ) s λ ′ /σ (µ ) ′(Y)   by (3.14) = X µ ⊆ λ   X σ ∈S ℓ(µ ) S ∗ σ (µ ) (X) X τ ∈S ℓ(λ ) (− 1) sgn(τ ) RS ∗ τ (λ )/σ (µ ) (Y)   = X τ ∈S ℓ(λ ) (− 1) sgn(τ )   X µ ⊆ λ X σ ∈S ℓ(µ ) S ∗ σ (µ ) (X)RS ∗ τ (λ )/σ (µ ) (Y)   . (3.16) Note that the only terms σ (µ ) that appear in (3.16) are those such that σ (µ ) = β for a composition β . We rewrite (3.16) as Hs λ (X,Y)= X τ ∈S ℓ(λ ) (− 1) sgn(τ )   X µ ⊆ λ X σ ∈S ℓ(µ ) S ∗ σ (µ ) (X)RS ∗ τ (λ )/σ (µ ) (Y)   = X τ ∈S ℓ(λ ) (− 1) sgn(τ ) X β S ∗ β (X)RS ∗ τ (λ )/β (Y) = X τ ∈S ℓ(λ ) (− 1) sgn(τ ) HS ∗ τ (λ ) (X,Y). 70 Chapter 4 0-Hecke modules for row-strict dual immaculate functions 4.1 Introduction A recent flourishing area of research is that of Schur-like functions, whose properties are analogous to the ubiquitous Schur functions that arise in many areas, from enumerative combinatorics where they are generating functions for Young tableaux, to representation theory where they are the irreducible representations for the general linear groups, as well as being intimately connected to representations of the symmetric group. TheareaofSchur-likefunctionsbeganwithquasisymmetricSchurfunctions[23],followed bydiscoveriesofrow-strictquasisymmetricSchurfunctions[33],YoungquasisymmetricSchur functions [29, 31], noncommutative Schur functions [11] and immaculate functions [9], qua- sisymmetric Schur Q-functions [28], quasisymmetric Macdonald polynomials [15], and Schur functions in noncommuting variables [4]. Inthisproject,ourfocusisthestudyoftheassociated0-Heckealgebramodules,whichwe defineandanalyseinordertodeveloptheanalogywithSchurfunctionsintherepresentation theory context. This programme was first carried out for the dual immaculate functions [10], the quasisymmetric Schur functions [44], and subsequently for the row-strict Young quasisymmetric functions [7] and the extended Schur functions [38]. 71 The descent set determines a 0-Hecke algebra action on the set SIT(α ) of standard immaculatetableauxofcompositionshapeα ,yieldingacyclicindecomposablemodulewhose quasisymmetric characteristic isRS ∗ α . The action defines a partial order on the set SIT( α ) which turns out to be dual to the partial order in [10], see Lemma 4. The resulting graded poset PRS ∗ (α ), which we call the immaculate Hecke poset, has remarkable properties, leading to the discovery of several other 0-Hecke modules with interesting quasisymmetric characteristics. AmongtheseisananalogueoftheextendedSchurfunctionsdefinedbyAssaf and Searles [6]. An examination of the poset (see Figure 4.2) reveals various subposets that are closed under either our action or the dual immaculate action of [10]. We investigate the associated submodules. One of the key contributions of this paper, then, is the careful scrutiny of the partial order resulting originally from the 0-Hecke algebra action defined in [10], and the discovery that hidden within it are special standard immaculate tableaux, which generate interesting 0-Heckealgebrasubmodulesandquotientmodules, forboththedualimmaculateactionand ournewrow-strictdualimmaculateaction. Thedualityintheposetreflectstheactionofthe involutionψ onQSym: weprovefirstthattheposetisgraded,andhasuniquetopandbottom elements, see Definition 3 and Definition 4. The cyclic generator for the dual immaculate module was shown to be the top element in [10]; for the row-strict dual immaculate module, we show that the cyclic generator is the bottom element of the poset. Similarly, the cyclic generators for the extended Schur Hecke-module of [38] and our row-strict extended Schur Hecke-module (see Theorem 10) are the top and bottom elements of the interval [S col α ,S row α ], the column superstandard and row superstandard tableaux of Definition 7 and Definition 4, respectively. Finally we show how another special standard tableau S row∗ α , see Definition 9, also generates a cyclic submodule of the dual immaculate 0-Hecke algebra module. The posetdualityphenomenoncanalsobeusedtoexplain, forexample, thepassagebetweenthe modules in [44] and [7], see Section 4.6. 72 Our proofs are technical, relying heavily on two straightening algorithms which produce saturated chains in the poset PRS ∗ (α ). We prove indecomposability by following the pio- neering work in [44] and [10], with considerable technical modifications. This chapter is organised as follows. In Section 4.2, we briefly review the necessary facts about 0-Hecke algebras, and then define a new 0-Hecke algebra action on standard immac- ulate tableaux. Section 4.3 describes a partial order on these tableaux, which, by standard arguments, leads to a filtration showing that our 0-Hecke module has quasisymmetric char- acteristic equal to the row-strict dual immaculate function. Section 4.4 is devoted to showing that our new 0-Hecke module is cyclic (Theorem 4) and indecomposable, culminating in Theorem 6. The technicalities here hinge on two key straightening algorithms, described in Propositions 2 and 3, which we use to establish that the poset of Section 4.3 is bounded, with a unique minimal and maximal element. The minimal element, defined in Definition 3, is shown to be the cyclic generator of our module. The maximal element is the cyclic generator of the module of [10]. 1 The straightening algorithms play an important role in the technical lemmas leading to the indecomposability proof. The work of this section also reveals a remarkable connection, see Lemma 4, between our partial order and that of [10], showing how the map ψ between the quasisymmetric characteristics of the dual immaculate and row-strict dual immaculate modules, is reflected in the duality of the posets. The motivation for Section 4.5 and Section 4.6 comes from a closer examination of the poset of standard immaculate tableaux defined by the 0-Hecke action, and the remarkable propertiestowhichwehavealludedabove. InSection4.5weshowhowtheposetrevealsa0- Hecke module whose characteristic is the row-strict analogue of the extended Schur function definedin[6],forwhicha0-Heckemodulewasconstructedin[38]. Infactweshowthatthere are not one but two related modules, a submodule (Theorem 10) and a quotient module (Theorem 11) of the row-strict dual immaculate action, both cyclic and indecomposable. 1 It is not explicitly established in [10] that the poset of standard immaculate tableaux defined by the 0-Hecke action has a maximal element. 73 To complete this analogy, in Section 4.6 we show once again how the poset of standard immaculate tableaux leads to the discovery of more cyclic modules, a submodule and two quotient modules of the (original) dual immaculate action, as well as a quotient module of the extended row-strict dual immaculate module (Theorem 13, Theorem 14). Figure 4.2 indicates the essential representation-theoretic properties of the immaculate Hecke poset. We conclude in Section 9 by considering the two remaining choices for the descent set of a standard immaculate tableau, apart from the two which determine the dual immacu- late and row-strict dual immaculate functions. Although the corresponding quasisymmetric functions are no longer independent, interestingly, both of the new variants come with as- sociated 0-Hecke actions. Furthermore, these actions determine the same immaculate Hecke poset, and consequently our straightening algorithms apply, giving cyclic 0-Hecke modules as in Section 4.4, generated respectively by S 0 α and S row α , as well as submodules generated respectivelybyS col α andS row∗ α asinSection4.5andSection4.6. Ourfinalresult,Theorem20, shows that these modules complete the combinatorial picture of tableaux considered in this paper, by accounting for all the possible variations on increasing rows and columns. This information is captured in Figure 4.1. T α (1st col<,rows<) : ¯ A α (Section 4.7) T α (1st col<,rows≤ ) :S ∗ α [10] T α (cols<,rows≤ ) :E α [38], [12] T α (1st col≤ ,rows<) :RS ∗ α (Section 3.2) T α (1st col≤ ,rows≤ ) :A α (Section 4.7) T α (cols≤ ,rows<) :RE α (Section 4.5) ψ T α (cols<,rows<) : ¯ A SET(α ) T α (cols≤ ,rows≤ ) :A SET(α ) Figure 4.1: The eight flavours of tableaux, their characteristics and H n (0)-modules, related in pairs by the involution ψ , from the four descent sets. The double arrow-head indicates a quotient module, and the hooked arrow indicates a submodule. Table 4.1 below provides a list of tableaux acronyms used in this paper. 74 Table4.1: Standardtableauxofcompositionshapeα ofn,withdistinctentries{1,2,...,n}, where all rows increase (strictly) left to right: NSET(α ) at least one column does NOT increase bottom to top SIT(α ) first column increases bottom to top SIT ∗ (α ) first column filled bottom to top consecutively with 1 ,2,... SET(α ) all columns increase bottom to top Table 4.2 provides a summary of our results, as well as comparison with prior work. Table 4.3 compiles the known quasisymmetric bases to date, together with their 0-Hecke modules, as well as the new quasisymmetric functions, all of which expand positively in the fundamental basis, arising from the 0-Hecke submodules and quotient modules discovered in this paper. 4.2 A 0-Hecke algebra action for RS ∗ Recall that the symmetric group S n can be defined via generators s i ,1 ≤ i ≤ n− 1, the adjacent transpositions, subject to the relations s i 2 =1; s i s i+1 s i =s i+1 s i s i+1 ; s i s j =s j s i , |i− j|≥ 2. Definition 1. [36] Let K be any field. The 0-Hecke algebra H n (0) is the K-algebra with generators π i ,1≤ i≤ n− 1 and relations π i 2 =π i ; π i π i+1 π i =π i+1 π i π i+1 ; π i π j =π j π i , |i− j|≥ 2. 75 The algebra H n (0) has dimension n! over K, with basis elements {π σ : σ ∈ S n }, where π σ =π i 1 ...π im if σ =s i 1 ...s im . This is well-defined by standard Coxeter group theory. From [36], the 0-Hecke algebra admits 2 n− 1 irreducible one-dimensional modules [L α ] = Span(v α ), one for each composition α ⊨ n, carrying an action defined by π i (v α )=        0, i∈set(α ), v α , otherwise. (4.1) Here set(α ) = {α 1 ,α 1 +α 2 ,...,α 1 +...+α k− 1 } is the subset of [n− 1] associated to the composition α =(α 1 ,...,α k ) of n of length k. Definition2. [17]ThequasisymmetriccharacteristicchisanisomorphismfromtheGrothen- dieck ring of H n (0) with respect to the induction product, to the ring of quasisymmetric functions QSym, which sends the 0-Hecke algebra irreducible module [L α ] indexed by the composition α ⊨ n to the fundamental quasisymmetric function indexed by α : ch([L α ])=F α . The following is our restatement of the main result of [10]. Theorem 1. [10, Theorem 3.12] There is an indecomposable cyclic 0-Hecke algebra module W α whose quasisymmetric characteristic is the dual immaculate functionS ∗ α , ch(W α )=S ∗ α . The module W α has dimension equal to the number of standard immaculate tableaux of shape α. 76 The S ∗ -action of the 0-Hecke algebra generator π i on the set of standard immaculate tableaux of shape α , for α ⊨ n, may be described as follows: π S ∗ i (T)=                        T, if i+1 is in a row weakly below i, 0, if i,i+1 are in column 1 of T, s i (T), if i+1 is strictly above i in T, and i,i+1 are NOT in column 1, (4.2) where s i (T) is the standard immaculate tableau obtained from T by swapping i and i+1. Furthermore, thedualimmaculatefunctionS ∗ α expandspositivelyinthefundamentalqua- sisymmetric functions as follows. Define the descent set Des S ∗ (T) of a standard immaculate tableau T to be the set Des S ∗ (T):={i:i+1 is strictly above i in T}. Then ch(W α )=S ∗ α = X T F comp(Des S ∗ (T)) , where the sum runs over all standard immaculate tableaux T of shape α. Our goal in this section is to define, for each composition α ⊨ n of n, a moduleV α whose image under the characteristic map is the quasisymmetric functionRS ∗ α . Following [10], we consider the vector space V α whose basis vectors are the standard immaculate tableaux of 77 shape α. Define, for each 1 ≤ i≤ n− 1 and each standard immaculate tableau T of shape α, theRS ∗ -action of the generator π i on T to be π RS ∗ i (T)=                        T, if i / ∈Des RS ∗ (T), 0, i∈Des RS ∗ (T) and swapping i and i+1 in T does NOT result in a standard immaculate tableau, s i (T), otherwise, (4.3) where s i (T) is the standard immaculate tableau obtained from T by swapping i and i+1. As in [7], we say an entry j in a tableau is right-adjacent to an entry i if i,j are in the same row and in adjacent columns, and j is to the right of i. Note that from the definition of the descent set, if i ∈ Des RS ∗ (T) and i,i+1 are in the same row, then i+1 must be right-adjacent to i. To avoid cumbersome notation, we write simply π i (T) for the row-strict immaculate action, using π RS ∗ i and π S ∗ i only when there is explicit need to distinguish between the actionsof (4.3)and(4.2). WerefertotheseastheRS ∗ -actionandtheS ∗ -actionrespectively. Likewise we may refer to the resulting H n (0)-modules as the RS ∗ -Hecke module and the S ∗ -Hecke module respectively. Lemma 1. Let T be a standard immaculate tableau and let i∈Des RS ∗ (T). Then 1. i,i+1 cannot both be in the leftmost column of T; 2. i and i+1 are in the same row ⇐⇒ i+1 is right-adjacent to i ⇐⇒ π i (T)=0; 3. if s i (T) is a standard immaculate tableau, then i / ∈Des RS ∗ (s i (T)). Proof. The first claim is immediate from the definitions. Forthesecondclaim,onedirectionisclear: ifi+1isright-adjacenttoi,clearlyswapping i and i+1 will make a non-increasing row, so π i (T)=0. 78 Now suppose i∈Des RS ∗ (T) and i,i+1 are not in the same row, so that i+1 is in a row strictly below i. Here it is clear that swapping i,i+1 does not violate row-increase (since a < i < b ⇐⇒ a < i+1 < b for a,b / ∈ {i,i+1}). Since by the first claim, i,i+1 are not both in the leftmost column, the latter is still increasing by the same argument. Hence s i (T) is also an immaculate tableau, and π i (T)=s i (T)̸=0. Note from above that s i (T) is an immaculate tableau ⇐⇒ i+1 is strictly below i in T, and hence i+1 is strictly above i in s i (T). This verifies the third and final claim. In particular, if 1∈Des RS ∗ (T), then π 1 (T)=0. Example 1. Consider the standard row-strict immaculate tableau S = 6 4 5 8 10 3 7 1 2 9 . Then Des RS ∗ (S) = {1,4,6,8} and Des S ∗ (T) = {2,3,5,7,9} (the complement in the set {1,2,...,9}). Hence we have π RS ∗ i (T)=T for i∈{2,3,5,7,9}, π RS ∗ 1 (T)=0=π RS ∗ 4 (T), and π RS ∗ 6 (T)=s 6 (T)= 7 4 5 8 10 3 6 1 2 9 , π RS ∗ 8 (T)=s 8 (T)= 6 4 5 9 10 3 7 1 2 8 . We can therefore reformulate the description of the action of π RS ∗ i on T more succinctly as follows: 79 π i (T)=π RS ∗ i (T)=                T, if i+1 is strictly above i , 0, if i+1 is right-adjacent to i, s i (T), if i+1 is strictly below i in T. (4.4) Theorem2. The operators π RS ∗ i define an action of the 0-Hecke algebra on the vector space V α . Proof. Clearly from the preceding analysis, π RS ∗ i (T) = π i (T) ∈ V α for every standard im- maculatetableauT ofshapeα. Wemustverifythattheoperatorssatisfythe0-Heckealgebra relations. To show π 2 i (T) =π i (T), we need only check the case when i+1 is strictly below i in T. In this case π i (T) = s i (T), and i is now strictly below i+1. Hence π i (s i (T)) = s i (T) and we are done. Let 1≤ i,j≤ n− 1 with|i− j|≥ 2. Then{i,i+1}∩{j,j+1}=∅, so the actions of π i and π j are independent of each other, and hence commute. It remains to show that π i π i+1 π i (T)=π i+1 π i π i+1 (T). (4.5) We examine four separate cases. Case 1: Assume i / ∈ Des RS ∗ (T),i+1 / ∈ Des RS ∗ (T): Then π i (T) = T,π i+1 (T) = T, and the claim is clear. Case 2: Assume i∈Des RS ∗ (T), but i+1 / ∈Des RS ∗ (T): Then π i+1 (T)=T. If π i (T)=0, then the left-hand side of (4.5) is 0, and so is the right-hand side. 80 If π i (T)̸= 0, then π i (T) = s i (T), and the left-hand side of (4.5) equals π i π i+1 (s i (T)), while the right-hand side is π i+1 (s i (T)). Hence (4.5) becomes π i π i+1 (s i (T))=π i+1 (s i (T)), (4.6) which we need to verify. Assume i+1 / ∈ Des RS ∗ (s i (T)). The left-hand side then equals π i (s i (T)) = s i (T) by Lemma 1, and this is also the right-hand side. Finally assume i+1∈Des RS ∗ (s i (T)). We now have i+1 strictly below i in T, so that i+1 is strictly above i in s i (T), and i+2 weakly below i+1 in s i (T). Also recall that i+2 was strictly above i+1 in T. It follows that In s i (T),i+2 is now strictly above i and weakly below i+1. (4.7) Ifi+2isright-adjacentto i+1,thenπ i+1 (s i (T))=0andEquation(4.6)isimmediate. If not, then In s i (T),i+2 is now strictly above i and strictly below i+1. (4.8) This implies π i+1 (s i (T)) = s i+1 (s i (T)), and in the latter we now have i below i+1, whichinturnisbelowi+2.Inparticulariisnotadescentofπ i+1 (s i (T))=s i+1 (s i (T)), and hence the latter tableau is fixed by π i . Equation (4.6) is thus verified. Case 3: Assume i / ∈Des RS ∗ (T), but i+1∈Des RS ∗ (T): Then π i (T)=T and the left-hand side of Equation (4.5) is π i (π i+1 (T)) =        0, i+2 is right-adjacent to i+1 in T, π i (s i+1 (T)), otherwise. 81 The right-hand side of Equation (4.5) is π i+1 π i π i+1 (T) =        0, i+2 is right-adjacent to i+1 in T, π i+1 (π i (s i+1 (T))), otherwise. Thus we need only consider the case when i+2 is not right-adjacent to i+1, so that necessarily i+2 is strictly below i+1 in T. Also i is strictly BELOW i+1 in T. That is, i+1 is strictly above both i and i+2 in T, and thus i+2 is strictly above both i,i+1 in s i+1 (T). Hence we have two possibilities for s i+1 (T): Either i is below i+1, which is below i+2, and hence π i and π i+1 both fix s i+1 (T), or i+1 is below i, and i is below i+2. In the latter case, applying π i to s i+1 (T) switches i and i+1, so that in π i (s i+1 (T)) we now have i below i+1, and i+1 (still) below i+2. But then π i+1 fixes π i (s i+1 )(T). Equation (4.5) has been established. Case 4: Assume i,i+1∈Des RS ∗ (T): First suppose i+1 is right-adjacent to i, so that π i (T) equals zero, and so does the left-hand side of Equation (4.5). If π i+1 (T)=0, we are done; if not π i+1 (T)=s i+1 (T), and the right-hand side of Equation (4.5) is π i+1 π i (s i+1 (T)). Since in T we have i+1 right-adjacent to i and i+2 strictly below both, this means in s i+1 (T) we have i+2 right-adjacent to i and i+1 strictly below them, forcing i to be a descent of s i+1 (T). Hence in π i (s i+1 (T)), we have i+2 right-adjacent to i+1 and i strictly below. But then π i+1 (π i (s i+1 (T)))=0, as desired. Finally, suppose π i (T) ̸= 0, so that i+1 is strictly below i in T and i+2 is weakly below i+1. For clarity we consider two sub-cases: 82 Case 4a: i + 2 is right-adjacent to i + 1 in T. This immediately makes the right-hand side of Equation (4.5) equal to 0, since π i+1 (T) = 0. Then the left-hand side of Equation (4.5) equals π i π i+1 s i (T)=π i s i+1 (s i (T)) since i+1 is now above i+2 in s i (T). But now i+1 is right-adjacent to i in s i+1 (s i (T)), so π i s i+1 (s i (T)) reduces to 0, as desired. Case 4b: i + 2 is strictly below i + 1, which is strictly below i in T, so that π i+1 (T) ̸= 0,π i (T)̸=0. Then Equation (4.5) becomes s i s i+1 s i (T)=s i+1 s i s i+1 (T), (4.9) and it is easy to see that this is indeed true. We have verified Equation (4.5) in all cases, thereby completing the proof that the action of the generators π i extends to an action of H n (0) onV α . 4.3 A partial order from the 0-Hecke action Letα ⊨ n, andSIT(α )denotethesetofallstandardimmaculatetableauxofshape α . Given T ∈ SIT(α ), let σ (T) ∈ S n be the permutation in one-line notation obtained from T by reading the entries of T from right to left in each row, and the rows from top to bottom. Example 2. If T = 4 5 2 1 3 then σ (T)=5 4 2 3 1. Given σ ∈S n recall that its inversion set is Inv(σ )={(p,q)|1≤ pσ (q)}. The number of inversions is denoted by inv(σ )=|Inv(σ )|. 83 Example 3. If σ (T)=5 4 2 3 1, then inv(σ )=9 from Inv(σ )={(1,2),(1,3),(1,4),(1,5),(2,4),(2,5),(3,4),(3,5),(4,5)}. Observe by our definition of π i that if T 1 ,T 2 ∈SIT(α ) and π i (T 1 )=T 2 with T 1 ̸=T 2 this means that 1. in T 1 we have that i appears strictly above i+1, so in σ (T 1 ) we have that i appears left of i+1. 2. in T 2 = π i (T 1 ) we have that i appears strictly below i+1, so in σ (T 2 ) we have that i appears right of i+1. Consequently, since all other entries other than i and i+1 are fixed by π i , we have that if T 2 =π i (T 1 ) with T 1 ̸=T 2 , then inv(σ (T 2 ))=inv(σ (T 1 ))+1. (4.10) Proposition 1. Let α ⊨ n and T 1 ,T 2 ∈SIT(α ). Define ≼ RS ∗ α on SIT(α ) by T 1 ≼ RS ∗ α T 2 if and only if there exists a permutation s i 1 ··· s i ℓ ∈S n such that π i 1 ··· π i ℓ (T 1 )=T 2 . Then≼ RS ∗ α is a partial order on SIT(α ). 84 Proof. That ≼ RS ∗ α is reflexive and transitive is immediate from the definition. To prove antisymmetry, if T 1 ≼ RS ∗ α T 2 then by Equation (4.10) inv(σ (T 1 ))≤ inv(σ (T 2 )) and if T 2 ≼ RS ∗ α T 1 then by Equation (4.10) inv(σ (T 2 ))≤ inv(σ (T 1 )). Thusinv(σ (T 1 ))=inv(σ (T 2 )).However, byEquation(4.10)weknoweachnon-zeroactionof aπ i thatisnottheidentityincreasesthenumberofinversions, hence T 1 =T 2 asdesired. We will now use our partial order to define an H n (0)-module indexed by a composition α ⊨ n, whosequasisymmetriccharacteristicistherow-strictdualimmaculatefunctionRS ∗ α . More precisely, given a composition α ⊨ n extend the partial order≼ RS ∗ α on SIT(α ) to an arbitrary total order on SIT(α ), denoted by≼ t RS ∗ α . Let the elements of SIT(α ) under≼ t RS ∗ α be {τ 1 ≼ t RS ∗ α ··· ≼ t RS ∗ α τ m }. Now letV τ i be theC-linear span V τ i =span{τ j |τ i ≼ t RS ∗ α τ j } for 1≤ i≤ m andobservethatthedefinitionof ≼ t RS ∗ α impliesthatπ i 1 ··· π i ℓ V τ i ⊆V τ i foranys i 1 ··· s i ℓ ∈S n . Thisobservationcombinedwiththefactthattheoperators{π i } n− 1 i=1 satisfythesamerelations as H n (0) by Theorem 2 gives the following result. Lemma 2. V τ i is an H n (0)-module. 85 Given the above construction, define V τ m+1 to be the trivial 0 module, and consider the following filtration of H n (0)-modules. V τ m+1 ⊂V τ m ⊂···⊂V τ 2 ⊂V τ 1 Then the quotient modules V τ i− 1 /V τ i for 2 ≤ i ≤ m+1 are 1-dimensional H n (0)-modules spanned by τ i− 1 . Furthermore, they are irreducible modules. We can identify which H n (0)- module they are by looking at the action of π j onV τ i− 1 /V τ i for 1≤ j≤ n− 1. We have π j (τ i− 1 ) =      0 j∈Des RS ∗ (τ i− 1 ) τ i− 1 otherwise. Thus, as an H n (0)-representation, V τ i− 1 /V τ i is isomorphic to the irreducible representation [L β ] where β is the composition corresponding to the descent set Des RS ∗ (τ i− 1 ). Given an H n (0)-module M, every time we have a short exact sequence of H n (0)-modules, say 0 → M ′ →M →M ′′ →0, then ch([M]) = ch([M ′ ])+ch([M ′′ ]). Thus, ch([V τ i− 1 /V τ i ])=F β 86 and ch([V τ 1 ]) = m+1 X i=2 ch([V τ i− 1 /V τ i ]) = m+1 X i=2 F comp(Des RS ∗ (τ i− 1 )) = X τ ∈SIT(α ) F comp(Des RS ∗ (τ )) = RS ∗ α . Consequently, we have established the following. Theorem 3. Let α ⊨ n and τ 1 ∈ SIT(α ) be the minimal element under the total order ≼ t RS ∗ α . Then V τ 1 is an H n (0)-module whose quasisymmetric characteristic is the row-strict dual immaculate functionRS ∗ α . 4.4 The 0-Hecke module structure of ch − 1 (RS ∗ α ) Our next goal is to analyse the structure of the moduleV τ 1 in Theorem 3. We will show that our partial order, and hence any linear extension of it, has a unique bottom element S 0 α . We will also show thatV τ 1 is cyclic, and generated by the standard tableau S 0 α . Lemma 3. Let α ⊨ n and T ∈ SIT(α ). If π i (T) = s i (T),π j (T) = s j (T) ∈ SIT(α ) and π i (T)=π j (T), then necessarily i=j. Proof. Ifi̸=j,wemayassumei<j.Supposej+1occupiescellxinT.Inπ i (T),j+1>i+1 is unchanged from its position in T. However, in π j (T) = s j (T), j +1 is now strictly above j, and cell x is now occupied by j. It follows that π i (T),π j (T), differ at least in cell x, a contradiction. The cover relation for our poset PRS ∗ α for row-strict dual immaculate tableaux of shape α ⊨ n is S≺ RS ∗ α T ⇐⇒ ∃i such that T =π RS ∗ i (S), (4.11) 87 with respect to the row-strict 0-Hecke action defined by Theorem 2 . On the other hand, the cover relation for the poset PS ∗ α for dual immaculate tableaux of shape α ⊨ n, as described in [10], is S≺ S ∗ α T ⇐⇒ ∃i such that S =π S ∗ i (T). (4.12) with respect to the dual immaculate 0-Hecke action defined by Theorem 1 . NotethatbyLemma3, ineachcase, acoverrelationisdeterminedbyauniquegenerator of the 0-Hecke algebra. Lemma 4. Let α ⊨ n and S,T ∈SIT(α ). Then S≺ S ∗ α T ⇐⇒ S≺ RS ∗ α T. Hence the two posets PS ∗ α and PRS ∗ α are isomorphic. Proof. We have S≺ S ∗ α T ⇐⇒ π S ∗ i (T)=S, where S̸=T,S̸=0 ⇐⇒ i+1 is strictly above i in T, i,i+1 not both in column 1, by (4.2) ⇐⇒ i+1 is strictly BELOW i in S ⇐⇒ π RS ∗ i (S)=s i (S)=T, by (4.4) ⇐⇒ S≺ RS ∗ α T. The claim follows. From Equation (4.10) and remarks preceding Proposition 1, it follows that the posets PS ∗ α ≃ PRS ∗ α are graded by the number of inversions. We call the poset PS ∗ α ≃ PRS ∗ α the immaculate Hecke poset associated to the composition α . 88 Next we show that the poset has a unique bottom element S 0 α . From this we will show that the unique bottom element S 0 α is the cyclic generator for the moduleV τ 1 in Theorem 3, whose quasisymmetric characteristic is RS ∗ α . Later we will also show that it has a unique top element S row α , see Definition 4. Definition 3. Let α ⊨ n be of length ℓ=ℓ(α ). Define S 0 α to be the standard tableau of shape α with entries 1,...,ℓ in column 1, increasing bottom to top, and then fill the remaining rows, top to bottom, left to right with consecutive integers starting at ℓ+1 and ending at n. Thus 1. the top row, row ℓ, contains the entry ℓ followed by the interval [ℓ+1,ℓ+α ℓ − 1]; note that n= P ℓ i=1 α i ≥ (ℓ− 1)+α ℓ ; 2. the next row, row ℓ− 1, contains the entry ℓ− 1 followed by the interval [ℓ+α ℓ ,ℓ+ α ℓ +α ℓ− 1 − 2]; 3. row i (from the bottom) contains the entry i followed by the interval [ℓ+α ℓ +α ℓ− 1 + ··· +α i+1 − (ℓ− i− 1),ℓ+α ℓ +··· +α i − (ℓ− i+1)] (note that there are α i − (ℓ− i+ 1)+(ℓ− i− 1)+1+1=α i entries; also note that n= P ℓ i=1 α i ≥ (i− 1)+ P ℓ j=i α i ); 4. the bottom row, row 1, contains 1 followed by the interval [ℓ+α ℓ +α ℓ− 1 +··· +α 2 − (ℓ− 2),ℓ+α ℓ +··· +α 1 − ℓ]. Example 4. We have S 0 43423 = 5 6 7 4 8 3 9 10 11 2 12 13 1 14 15 16 , S 0 43411 = 5 4 3 6 7 8 2 9 10 1 11 12 13 . The complement of the descent set of S 0 α is precisely [ℓ− 1]; all entries not in column 1 are descents, and ℓ is also a descent if ℓ̸=n. Furthermore, the entries i at the end of a row 89 of size 2 or more, are the only ones for which π RS ∗ i (S 0 α ) = s i (S 0 α ) ̸= 0. The following facts are a consequence of (4.4), and are illustrated by the above examples. Lemma 5. Let α ⊨ n and ℓ=ℓ(α ). Then 1. π RS ∗ i (S 0 α )=S 0 α ⇐⇒ i∈[ℓ− 1]. (4.13) 2. Assume α ℓ ≥ 2. Then π RS ∗ ℓ (S 0 α )=0, and (4.14) π RS ∗ j (S 0 α ) / ∈{S 0 α ,0} ⇐⇒ j∈{ℓ+α ℓ +··· +α i − (ℓ− i+1),2≤ i≤ ℓ}. (4.15) (Note that i=1 corresponds to j =n.) 3. Assume α ℓ =1 and k≤ ℓ− 1 is maximal such that α k ≥ 2. Then π RS ∗ j (S 0 α ) / ∈{S 0 α ,0} ⇐⇒ j =ℓ or j∈{ℓ+α ℓ +··· +α i − (ℓ− i+1), 2≤ i≤ k}. (4.16) 4. If there is no j such that π j (S 0 α )=0, then α ℓ =1 and α k ≤ 2 for all k≤ ℓ− 1. 5. Finally if α is a hook of the form (1 ℓ− 1 ,n− ℓ+1),1≤ ℓ≤ n, then S 0 α is the unique standard tableau of shape α. To avoid trivialities, unless otherwise stated, in everything that follows we will assume that α /∈{(1 ℓ− 1 ,n− ℓ+1):1≤ ℓ≤ n}. Proposition 2. Let α ⊨ n and consider the standard immaculate tableau S 0 α . Then for any T ∈ SIT(α ),T ̸= S 0 α , there is a sequence of generators π j i , and tableaux T i ∈ SIT(α ), 90 i = 1,...,r, such that π j i (T i ) = T i− 1 ,i = 1,2,...,r, where we set T 0 = T and T r = S 0 α . Hence we conclude S 0 α ≼ RS ∗ α T and T =π j 1 π j 2 ··· π jr (S 0 α ). In particular S 0 α is the unique minimal element of the poset PRS ∗ α . The proof consists of a straightening algorithm which we first illustrate with an example. LetT ∈SIT(α )̸=S 0 α .WedescribehowtoworkbackwardsfromT toS 0 α intheposetalonga saturated chain. This can be viewed as a“straightening” ofthe tableau T, whichtransforms it into S 0 α . Example5. Let α =223, so that S 0 α = 3 4 5 2 6 1 7 , and let T = 5 6 7 2 4 1 3 . First we straighten column 1 of T to match column 1 of S 0 α : Start with the lowest entry a j in column 1 of T such that a j ̸= j, and exchange it with a j − 1. Continue in this manner until column 1 has the entries 3,2,1 from top to bottom. T =T 0 π 4 ← − T 1 = 4 6 7 2 5 1 3 π 3 ← − T 2 = 3 6 7 2 5 1 4 . Next we work on the top row of T 2 , starting with the smallest entry which differs from the corresponding entry in the same cell of S 0 α , and continuing until the top rows match: T 2 π 5 ←− T 3 = 3 5 7 2 6 1 4 π 4 ←− T 4 = 3 4 7 2 6 1 5 π 6 ←− T 5 = 3 4 6 2 7 1 5 π 5 ←− T 6 = 3 4 5 2 7 1 6 . Now move down to the next row from the top, and proceed in the same manner, finding the smallest entry that differs from the corresponding entry in S 0 α : T 6 π 6 ←− T 7 = 3 4 5 2 6 1 7 =S 0 α . Hence we have T =π 4 π 3 π 5 π 4 π 6 π 5 π 6 (S 0 α ). 91 Proof of Proposition 2. As illustrated by Example 5, the following algorithm identifies a unique saturated chain from S 0 α to T. In what follows, when an integer a occupies row i and column j of T, by the counterpart of a in S 0 α we will mean the integer occupying the same cell, row i and column j, in S 0 α . Step 1: We begin by making the first column of T match the first column of S 0 α . Find the least j, 2 ≤ j ≤ ℓ, such that the entry x in cell (j,1) is not equal to j. Then x− 1 is in a lower row, not in column 1 by minimality of j. Hence T = π x− 1 (T 1 ), such that in T 1 ∈SIT(α ), x− 1 is now a descent strictly higher than x. Now repeat this procedure until x is replaced by j. Then continue with the next entry in column 1 of T which doesnotmatchinS 0 α .ClearlythisprocessendswithatableauT r =π xr π x r− 1 ...π x 1 (T), whose first column matches column 1 of S 0 α . Note that T =T r if T and S 0 α already agree in the first column. Step 2: First observe that T r and S 0 α now agree for all entries less than or equal to ℓ = ℓ(α ). Nowconsiderthetop-mostrowoflengthgreaterthan1,sayrowk. Findtheleastentry, say y, in this row of T r which differs from the corresponding entry in S 0 α . Note that y is then necessarily larger than its counterpart in S 0 α , by definition of the latter. Then y≥ ℓ+1 and y− 1 is strictly below y in T r . Hence T r =π y− 1 (T r+1 ) for T r+1 ∈SIT(α ), such that y− 1 is a descent in T r+1 strictly higher than y. We repeat this step until y has been replaced by its counterpart in S 0 α . Step 3: Continue in this manner to the end of the row. We now have a sequence of operators π i j andtableauxT j ∈SIT(α )suchthatT j− 1 =π i j (T j ),andthefinaltableaux T s agrees with S 0 α for all entries≤ ℓ+(α k − 1). Step 4: Proceed downwards to the next row where an entry in T s differs from its counterpart in S 0 α , and repeat Steps 2 and 3, until all rows are exhausted. 92 Since at the end of each iteration of Step 3, the number of entries that are in agreement with S 0 α increases, we see that the algorithm produces a saturated chain from S 0 α to T in the poset PRS ∗ α as claimed. We now immediately have: Theorem 4. The moduleV(T 1 )=ch − 1 (RS ∗ α ) of Theorem 3 is cyclic, and generated by the unique minimal element S 0 α of the poset PRS ∗ α . Proof. This is clear since Proposition 2 shows that T ∈SIT(α ),T ̸=S 0 α ⇒S 0 α ≼ RS ∗ α T, and also that there is a permutation σ ∈S n such that T =π σ (S 0 α ). Henceforth we shall writeV α for the cyclic module generated by S 0 α . BeforeproceedingwithouranalysisofthestructureofV α ,someremarksontheanalogous resultsin[10]areinorder. TheretheexistenceoftheuniquetopelementofPS ∗ α isimplicitly deducedviaamapbetweenananalogueofYamanouchiwordswhichtheycallY-words,which are in bijection with standard tableaux. The authors show indirectly that the top element of the poset PS ∗ α ≃PRS ∗ α is a cyclic generator for the module ch − 1 (S ∗ α ), by appealing to a parent module generated by a special Yamanouchi word; they invoke the fact that ch − 1 (S ∗ α ) is a quotient of this module. In order to give a complete and self-contained analysis of the immaculate Hecke poset PS ∗ α ≃PRS ∗ α ,wewillexplicitlyestablishtheexistenceoftheuniquetopelementbymeans of a straightening algorithm similar to Proposition 2. We will also be able to deduce that the H n (0)-module ch − 1 (S ∗ α ) is cyclically generated by this top element. Definition 4. Define S row α to be the row superstandard tableau in SIT(α ), whose entries bottom to top, left to right, are 1,2,...,n− 1,n in consecutive order. Note that if α is a hook of the form (1 ℓ− 1 ,n− ℓ+1),1≤ ℓ≤ n, then SIT(α ) has cardinality one and S 0 α =S row α . 93 Example 6. We have S row 43423 = 1415 16 1213 8 9 10 11 5 6 7 1 2 3 4 , S row 43411 = 13 12 8 9 10 11 5 6 7 1 2 3 4 . Proposition 3. Let α ⊨ n and consider the standard immaculate tableau S row α . Then for any T ∈ SIT(α ),T ̸= S row α , there is a sequence of generators π j i , and tableaux T i ∈ SIT(α ), i = 1,...,r, such that π j i (T i− 1 ) = T i ,i = 1,2,...,r, where we set T 0 = T and T r = S row α . Hence we conclude T ≼ RS ∗ α S row α and π jr π j r− 1 ··· π j 1 (T)=S row α . In particular S row α is the unique maximal element of the poset PRS ∗ α . We illustrate the straightening algorithm with an example. Example 7. We have S row 223 = 5 6 7 3 4 1 2 . Let T = T 0 = 3 4 7 2 6 1 5 . Start with the largest entry in the top-most row of T that differs from its counterpart in S row 223 , and repeat: T 0 π 4 −→ 3 5 7 2 6 1 4 =T 1 π 5 −→ 3 6 7 2 5 1 4 =T 2 π 3 −→ 4 6 7 2 5 1 3 =T 3 π 4 −→ 5 6 7 2 4 1 3 =T 4 . Once the top row coincides in both, move down to the next row where there is disagreement, again starting with the largest such element in that row: T 4 π 2 −→ 5 6 7 3 4 1 2 =S row α . Hence π 2 π 4 π 3 π 5 π 4 (T)=S row α . Proof of Proposition 3. In this case the algorithm is simpler to describe. We start with the top-most row of T which differs from S row α , and find the largest entry, say x, in that row 94 whichdiffersfromitscounterpartin S row α .Thennecessarilyx+1isinalowerrowof T =T 0 , and hence T 1 = π x (T 0 ) for the tableau T 1 ∈ SIT(α ) obtained from T by switching x,x+1. We continue in this manner along each row, top to bottom, right to left. Note that at the start of a new row j, the tableau T k agrees with S row α for all entries in the complement of [α 1 +··· +α j ]. ItisclearthatthisalgorithmterminatesinthetableauS row α ,producingasaturatedchain in the poset from T to S row α . In view of Lemma 4, and combined with the filtration in [10], analogous to our filtration of Lemma 2, we recover Theorem 1, specifically the results of [10, Theorem 3.5, Lemma 3.10]: Theorem 5. [10, Lemma 3.10] The moduleW α =ch − 1 (S ∗ α ) whose quasisymmetric charac- teristic is equal toS ∗ α , is cyclically generated by the standard immaculate tableau S row α . Comparing the two filtrations which give the H n (0)-modules, we may summarise the situation as follows: We may take the total order on the poset PS ∗ α ≃ PRS ∗ α to be the same in both cases, thanks to Lemma 4: S 0 α =T 1 ≼ t T 2 ≼ t ··· ≼ t T m =S row α , where m=|SIT(α )|. ForW α =ch − 1 (S ∗ α ), in [10] the authors use the filtration (0)⊂ span([S 0 α ,T 1 ])⊂···⊂ span([S 0 α ,T i ])⊂···⊂ span([S 0 α ,S row α ]), while forV α =ch − 1 (RS ∗ α ), our filtration (see Lemma 2) is (0)⊂ span([T m ,S row α ])⊂···⊂ span([T i ,S row α ])⊂···⊂ span([S 0 α ,S row α ]). 95 Returning to the row-strict dual immaculate functions, our final task in this section is to show that the cyclic moduleV α generated by S 0 α is in fact indecomposable. Since in any ring R with unity, if e is an idempotent we have the decomposition R = eR⊕ (1− e)R, to show indecomposability, it is enough to show that if f is an idempotent endomorphism of the module V, then f =0 or the identity [27, Proposition 3.1]. Theorem 6. LetV α =ch − 1 (RS ∗ α ) be the cyclic H n (0)-module generated by S 0 α . ThenV α is indecomposable. The proof of this theorem will require a series of technical lemmas. Lemma 6. [10, Proof of Theorem 3.12] Let P ∈ SIT(α ) and let j ∈ [n− 1] such that π j (P)̸=P. Then P cannot equal π j (T) for any tableau T ∈SIT(α ). Proof. ImmediatesinceotherwisewewouldhaveP =π j (T)⇒π j (P)=π 2 j (T)=π j (T)=P, a contradiction. Definition 5. For a composition α ⊨ n of length ℓ, denote by SIT ∗ (α ) the set of standard immaculate tableaux whose first (left-most) column consists of the integers 1,...,ℓ. Equiva- lently, SIT ∗ (α ) is the set of standard immaculate tableaux such that ℓ=ℓ(α ) is in row ℓ, the top row of α. Lemma 7. Suppose f is an idempotent endomorphism of the H n (0)-moduleV α , with f(S 0 α )= X T∈SIT(α ) a T T. Let P ∈ SIT(α ),P ̸= S 0 α , and suppose there is a j such that j ∈ Des RS ∗ (P)\Des RS ∗ (S 0 α ). Then a P =0. As a consequence, we have f(S 0 α )= X T∈SIT ∗ (α ) a T T. 96 Proof. From Equation (4.4), we have the following implications: j∈Des RS ∗ (P)⇒π j (P)̸=P, and j / ∈Des RS ∗ (S 0 α )⇒π j (S 0 α )=S 0 α . Hence f(S 0 α )=f(π j (S 0 α ))=π j (f(S 0 α ))= X T a T π j (T), and by Lemma 6, this does not contain P in its expansion. That is, f(S 0 α ) = P T a T T does not contain P in its expansion, and hence a P = 0. It follows that a P = 0 unless Des RS ∗ (P) is contained in Des RS ∗ (S 0 α ), which equals the complement of [ℓ− 1] from Equation (4.13). In particular, a P =0 unless 1,...,ℓ− 1 are NOT descents of P. But then it is easy to see (since rows must increase left to right) that 1,...,ℓ must occupy the first column of P. Hence we have a P ̸=0⇒1,...,ℓ occupy the first column of P, which is the claim. Lemma 8. Let α ⊨ n, ℓ(α )=ℓ, and let P ∈SIT ∗ (α ) be such that ℓ+1 is right-adjacent to i for some 1≤ i≤ ℓ. Then either π ℓ (P)=0 or there is an i∈[ℓ− 1] such that P ≺ RS ∗ α P 1 =π ℓ (P)≺ RS ∗ α P 2 =π ℓ− 1 (P 1 )≺ RS ∗ α ···≺ RS ∗ α P ℓ− i =π i+1 (P ℓ− i− 1 ) is a saturated chain in the poset PRS ∗ α , with i+1 right-adjacent to i in P ℓ− i . In either case, there is a permutation σ ∈S n such that π σ (P)=0. Proof. Since π ℓ (P) = 0 ⇐⇒ ℓ+1 is right-adjacent to ℓ, we may assume ℓ+1 is in row i≤ ℓ− 1. Thus ℓ+1 is strictly below ℓ, and hence π ℓ (P) = s ℓ (P) swaps ℓ and ℓ+1. That is, in P 1 =π ℓ (P), ℓ is now right-adjacent to i≤ ℓ− 1 If i=ℓ− 1, the argument stops here. 97 If not, now ℓ is strictly below ℓ− 1, and hence applying π ℓ− 1 produces a tableau s ℓ− 1 (P 1 ) in whichℓ− 1isnowright-adjacentto i≤ ℓ− 2. Clearlywecancontinuethisprocess until i+1 becomesright-adjacentto i,withtheappropriatereplacementsinthefirstcolumnabovethe entry i, producing the saturated chain as claimed. Nowapplyingπ i resultsin0,andhencetakingπ σ =π i π i+1 π i+2 ...π ℓ givesπ σ (P)=0. Example 8. Let ℓ = 5, α = 23223. We have the following saturated chain beginning at P ∈SIT ∗ (α ): P = 5 7 8 4 10 3 9 2 6 12 1 11 π 5 −→ P 1 = 6 7 8 4 10 3 9 2 5 12 1 11 π 4 −→ P 2 = 6 7 8 5 10 3 9 2 4 12 1 11 π 3 −→ P 3 = 6 7 8 5 10 4 9 2 3 12 1 11 , and for π σ =π 2 π 3 π 4 π 5 , π σ (P)=π 2 (P 3 )=0. Lemma 9. Let α ⊨ n, ℓ(α )=ℓ, and let P,Q∈SIT ∗ (α ) be such that ℓ+1 is • right-adjacent to i in P for some 2≤ i≤ ℓ, • right-adjacent to j in Q for some 1≤ j≤ i− 1. Let σ = π i π i+1 π i+2 ··· π ℓ . Then π σ (P) = 0,π σ (Q) ̸= 0, and rank(π σ (Q)) = (ℓ− i + 1) + rank(Q). Proof. This is clear from the preceding proof, since it takes (ℓ− i) consecutive applications of the π j to make i+1 right-adjacent to i in P; applying π ℓ ,π ℓ− 1 ,...,π i successively to Q changes the first column, and, in column 2, it changes only the entry in row j right-adjacent to j, diminishing the latter by 1 at each step. Hence after applying the (ℓ− i+1) factors in π σ ,, the entry right-adjacent to j in Q is i, and this is greater than j by hypothesis. This means that one final application of π i gives π σ (P)=0, but π σ (Q)̸=0. The statement about the ranks is clear since at each step we are applying a s k to the standard tableau to produce another standard tableau. 98 Example 9. Let ℓ=5, α =23223, take P as in Example 8 and consider the saturated chain beginning at Q∈SIT ∗ (α ): Q= 5 7 8 4 10 3 9 2 11 12 1 6 π 5 −→ Q 1 = 6 7 8 4 10 3 9 2 11 12 1 5 π 4 −→ Q 2 = 6 7 8 5 10 3 9 2 11 12 1 4 π 3 −→ Q 3 = 6 7 8 5 10 4 9 2 11 12 1 3 π 2 −→ Q 4 = 6 7 8 5 10 4 9 2 11 12 1 2 ; hence for π σ =π 2 π 3 π 4 π 5 , π σ (Q)=Q 4 ̸=0. Before proceeding with the proof of the next lemma, it is instructive to work through two more examples. Example 10. Let α =33223. Then S 0 α = 5 6 7 4 8 3 9 2 1011 1 12 13 ; we take P = 5 6 7 4 8 3 9 2 12 13 1 10 11 . Here p=10 is right-adjacent to 2 in S 0 α and to 1 in P. Applying π p− 1 ,π p− 2 ,... in succession we have P π 9 −→ 5 6 7 4 8 3 10 2 12 13 1 9 11 π 8 −→ 5 6 7 4 9 3 10 2 12 13 1 8 11 π 7 −→ 5 6 8 4 9 3 10 2 12 13 1 7 11 π 6 −→ 5 7 8 4 9 3 10 2 12 13 1 6 11 , so that ℓ+1=6 ends up right-adjacent to 1 in π 6 π 7 π 8 π 9 (P). 99 However, it is clear that the same sequence applied to S 0 α results in S 0 α π 9 −→ 5 6 7 4 8 3 10 2 9 11 1 12 13 π 8 −→ 5 6 7 4 9 3 10 2 8 11 1 12 13 π 7 −→ 5 6 8 4 9 3 10 2 7 11 1 12 13 π 6 −→ 5 7 8 4 9 3 10 2 6 11 1 12 13 Hence 6 ends up right-adjacent to 1 in π 6 π 7 π 8 π 9 (P) but to 2 in π 6 π 7 π 8 π 9 (S 0 α ). Example 11. This is the case p− 1=ℓ+1 in the lemma below. Let S 0 α = 9 8 7 6 10 5 11 4 12 3 2 13 1 , P = 9 8 7 6 10 5 13 4 12 3 2 11 1 . Then we have P π 10 −→ 9 8 7 6 11 5 13 4 12 3 2 10 1 . Again, 10 ends up right-adjacent to 2 in π 10 (P), but is right-adjacent to a larger entry, 6, in S 0 α . We are now ready to precisely formulate and prove the facts illustrated by the above examples. Lemma 10. Let α ⊨ n, ℓ(α ) = ℓ, and let P ∈ SIT ∗ (α ),P ̸= S 0 α be such that for all j, whenever j,j +1 are right-adjacent in S 0 α , they are also right-adjacent in P. Let p be the smallest entry in S 0 α , say in cell x, which differs from the entry in cell x of P. Then 1. p≥ ℓ+1; 2. S 0 α and P coincide in the top-most row, row ℓ; 3. the cell x is located in column 2 and row i for some i such that α i ≥ 2; in Cartesian coordinates, cell x is in position (i,2). 4. S 0 α and P coincide in all rows above the row i containing cell x, and hence all entries less than p occupy the same cells in both tableaux; 100 5. the entry p must occur in P in column 2 and in a lower row j where j j; this is because ℓ+1 is already in a lower row in P than in S 0 α . Proof. Note first that the hypothesis includes the case when there is no j such that j,j+1 are right-adjacent in S 0 α . This forces all parts of α to be at most 2. Item(1)isclear. Item(2)followsbyhypothesis,sincealltheentriesofS 0 α areconsecutive beginning in column 1 for the top row. Rows of length≥ 3 have consecutive entries from column 2 onwards in both tableaux, by hypothesis. The entry in column 2 of such a row is uniquely determined by any other entry in that row, and hence the least entry that differs must occur in column 2. This is Item (3). Item (4) is clear by minimality of the entry p in cell x, since all entries in S 0 α occurring to the left of cell x, or higher than it, must be less than p. Also rows increase left to right, so in P, if the entry p were not in column 2, there would be an entry q, ℓ+1 ≤ q < p, immediately to its left in its row. But this contradicts the fact that all entries strictly less than p occupy exactly the same cells in both tableaux. This establishes Item (5). Now note that when p=ℓ+1, Item (6) is immediate for τ equal to the identity permu- tation. See Example 11 with P replaced with π 10 (P). Thus we may assume p≥ ℓ+2. ItisnowclearfromExample10whathappens; applying π p− 1 putsp− 1intotheposition formerly occupied by p, in cell (j,2), of P, j < i, and moves p UP to the cell formerly occupied by p− 1. But in S 0 α , it puts p− 1 in cell (i,2), and moves p to the same cell as in P. This is by virtue of Item (4). Similarly, applying π p− 2 next will move p− 2 into the position (j,2), and p− 1 will now move UP. In S 0 α , it puts p− 2 into cell (i,2) and moves p− 1 into the same cell as in P. 101 In fact a comparison of the two tableaux after each application of π k shows that they coincide in all rows above the ith row and they differ in rows i and j, in column 2. The first step of our induction is depicted in Example 12 below. Subsequent steps are completely analogous, with p diminished by 1 at each step. Assume by induction that after applying π p− k , (p− k) was in cell (j,2) of P and in cell (i,2) of S 0 α , (recall j < i) and the two tableaux coincide in rows above row i. By induction hypothesis, p− k− 1 occupies the same cell in each tableau, since it is in a row higher than row i. Now apply π p− k− 1 . This swaps p− k− 1 and p− k in each tableau, putting p− k− 1 into cell (j,2) of P and in cell (i,2) of S 0 α , and moving p− k UP to the same cell in both tableaux. In P, cell (j,2) changes p to ℓ+1. In S 0 α , cell (i,2) changes from p to ℓ+1. We have thus established Item (6). Example 12. Let S 0 α = ℓ ℓ+1 ... ... ... p− 1 ... i p ...... j y>p ... ...... 2 1 , P = ℓ ℓ+1 ... ... ... p− 1 ... i z>p ...... j p ... ...... 2 1 . Then π p− 1 (P)= ℓ ℓ+1 ... ... ... p ... i z>p ...... j p− 1 ... ...... 2 1 . Lemma 11. Let P ̸=S 0 α , such that P ∈SIT ∗ (α ). Then there is a permutation σ =s i 1 ··· s ir such that 1. π σ (S 0 α )=0; 2. π σ (P) is nonzero and has rank r+rank(P) in the poset PRS ∗ α . 102 Proof. By hypothesis, P and S 0 α have identical first columns. Suppose there is a j ≤ n− 1 such that j,j + 1 are right adjacent in S 0 α , but NOT right-adjacent in P. Then from Equation (4.4), π j (S 0 α )=0 but π j (P)̸=0, so we are done. Otherwise, whenever j,j +1 are right-adjacent in S 0 α , they are also right-adjacent in P. Note that the case when j is never adjacent to j +1 is included here. But now Lemma 10, followed by Lemma 9, establish the claim. Lemma 12. Let T 1 ,T 2 ∈ SIT(α ) and let σ = s i 1 ··· s ir be a permutation in S n such that π RS ∗ σ (T 1 )=π RS ∗ σ (T 2 )=U, and rank(U)=r+rank(T 1 )=r+rank(T 2 ). Then T 1 =T 2 . Proof. By hypothesis, each application of π s i j produces a tableau in SIT(α ). The proof of Lemma 4 shows that when S,T ∈SIT(α ), π RS ∗ k (S)=T ⇐⇒ S =π S ∗ k (T). Hence we have T 1 =π S ∗ s ir π S ∗ s i r− 1 ··· π S ∗ s i 1 (U)=π S ∗ σ − 1(U)=T 2 , as claimed. Now the argument is nearly identical to [7, Proof of Theorem 4.1]. Proof of Theorem 6. From Lemma 7, we have the equation f(S 0 α )= X T∈SIT ∗ (α ) a T T. In the above sum, let ˆ T be a tableau of maximal rank in the poset PRS ∗ α such that a ˆ T ̸=0. Assume ˆ T ̸=S 0 α , and let σ =s i 1 ··· s ir be the permutation guaranteed by Lemma 11 for the tableau ˆ T. Let π σ ( ˆ T)=T ′ ̸=0. In particular, rank(T ′ )=r+rank( ˆ T). We have 0=f(π σ (S 0 α ))=π σ (f(S 0 α ))= X T∈SIT ∗ (α ) a T π σ (T). (4.17) 103 Suppose a T ̸=0 and π σ (T)=T ′ =π σ ( ˆ T). Comparing ranks, we obtain r+rank( ˆ T)=rank(T ′ )=rank(π σ (T))≤ r+rank(T), the last inequality being due to the fact that some π s j can fix a tableau. This implies rank( ˆ T)≤ rank(T). If the ranks are unequal, however, and a T ̸=0, we have a contradiction to the maximality of rank( ˆ T). Hence rank( ˆ T)=rank(T). By Lemma 12, this forces T = ˆ T. It follows that the coefficient of T ′ = π σ ( ˆ T) in the right-hand side of Equation (4.17) is a ˆ T , which must therefore be zero, a contradiction. Hence f(S 0 α )=a S 0 α S 0 α ; since f is idempotent, this implies a S 0 α =0 or 1, finishing the proof. 4.5 Arow-strictanalogueforextendedSchurfunctions This section is motivated by [38] and a closer examination of the immaculate Hecke poset PRS ∗ α . We consider again the H n (0)-moduleV α =ch − 1 (RS ∗ α ). Definition 6. Let SET(α ) be the set of all standard immaculate tableaux of shape α in which ALL columns increase from bottom to top. Equivalently, SET(α ) is simply the set of all standard tableaux of shape α with strictly increasing rows, left to right, and strictly increasing columns, bottom to top. Thus SET(α ) ⊂ SIT(α ). When the composition α is in fact a partition of n, the set SET(α ) coincides with the set of standard Young tableaux of shape α. In [38], the tableaux in SET(α ) are called standard extended tableaux. 104 S 0 223 = 3 4 5 2 6 1 7 3 4 6 2 5 1 7 3 4 5 2 7 1 6 3 5 6 2 4 1 7 3 4 7 2 5 1 6 3 4 6 2 7 1 5 4 5 6 2 3 1 7 3 5 7 2 4 1 6 3 4 7 2 6 1 5 3 5 6 2 7 1 4 4 5 7 2 3 1 6 3 6 7 2 4 1 5 3 5 7 2 6 1 4 4 5 6 2 7 1 3 4 6 7 2 3 1 5 S col 223 = 3 6 7 2 5 1 4 4 5 7 2 6 1 3 4 5 6 3 7 1 2 5 6 7 2 3 1 4 4 6 7 2 5 1 3 4 5 7 3 6 1 2 5 6 7 2 4 1 3 4 6 7 3 5 1 2 S row 223 = 5 6 7 3 4 1 2 π 5 π 6 π 4 π 6 π 5 π 3 π 6 π 4 π 5 π 6 π 4 π 6 π 3 π 5 π 4 π 6 π 3 π 5 π 3 π 4 π 5 π 3 π 6 π 2 π 4 π 3 π 5 π 2 π 6 π 3 π 4 π 2 π 5 π 2 π 4 Figure4.2: Therow-strictdualimmaculateposetPRS ∗ 223 ;theredtableauxandS col 223 areincreasing along rows and up columns, so are in SET(223)=[S col 223 ,S row 223 ]; the blue tableaux and S col 223 =S row∗ 223 are the elements in SIT ∗ (223)=[S 0 223 ,S row∗ 223 ], with the smallest elements in the first column. LetZ α :=span⟨SET(α )⟩. Figure 4.2 shows the elements of SET(223) in red; notice that they form a closed interval [S col α ,S row α ] of PRS ∗ 223 with respect to our RS ∗ -Hecke action defined in (4.3). This is no accident. Lemma 13. Z α is an H n (0)-submodule ofV α for theRS ∗ -action. Proof. It is enough to verify that T ∈SET(α )⇒π RS ∗ i (T)∈SET(α ). From Equation (4.4), we need only consider the case when i+1 is strictly below i in T. Note that i+1 is not in the same column as i, by defnition of SET. The two possible configurations are: T 1 = ...... x ... i ......... ...... i+1 ... (y) ... and T 2 = ...... i ... (x) ......... ...... y ... i+1 ... . 105 The parentheses indicate the possibility that there may be no cell under i in T 1 , and there may be no cell above (i+1) in T 2 . But the definition of SET( α ) eliminates T 1 , since it forces the contradiction i+1 < x < i. Hence T 2 is the only possibility, implying y < i(< x) and y j+1, and x− 1 is strictly to the right, and hence also below, x inT. HenceT =π x− 1 (T ′ ),T ′ ∈SET(α ) by Lemma 14, and now the first disagreement between T ′ and S col α has been diminished by 1. This inductive step is repeated until the original entry x is replaced with j +1, so that now our tableau coincides with S col α for all entries i≤ j+1. Since the number of entries in agreement with S col α increases at the end of each such step, continuing in this manner produces the saturated chain in [S col α ,T] as claimed. The statement of the lemma now follows. 109 The subposet SET(223) in Figure 4.2 is very small. We illustrate the preceding straight- ening algorithm with two larger examples, using α =232. Example 13. Here S col 232 = 3 6 2 5 7 1 4 ; let T = S row 232 = 6 7 3 4 5 1 2 . We have the sequence of operators T = 6 7 3 4 5 1 2 π 2 ← T 1 = 6 7 2 4 5 1 3 π 5 ← T 2 = 5 7 2 4 6 1 3 π 4 ← T 3 = 4 7 2 5 6 1 3 π 3 ← T 4 = 3 7 2 5 6 1 4 π 6 ← T 5 = 3 6 2 5 7 1 4 =S col 232 ,and thus T =π 2 π 5 π 4 π 3 π 6 (S col 232 ). Example 14. Here S col 232 = 3 6 2 5 7 1 4 ; let T = 5 7 3 4 6 1 2 . We have the sequence of operators T = 5 7 3 4 6 1 2 π 2 ← T 1 = 5 7 2 4 6 1 3 π 4 ← T 2 = 4 7 2 5 6 1 3 π 3 ← T 3 = 3 7 2 5 6 1 4 π 6 ← T 4 = 3 6 2 5 7 1 4 =S col 232 , and thus T =π 2 π 4 π 3 π 6 (S col 232 ). Lemma 16. Let P ∈SET(α ), P ̸=S col α . Then there is a j∈[n− 1] such that π j (S col α )=S col α but π j (P)̸=P. Proof. By definition of the RS ∗ -action, for any T ∈ SET(α ), we have π j (T)̸= T ⇐⇒ j∈ Des RS ∗ (T). In particular, the definition of S col α implies that π j (S col α )̸=S col α ⇐⇒ j is at the top of a column in S col α . Let P ∈ SET(α ),P ̸= S col α . Let j be the largest entry such that P and S col α coincide for all entries i ≤ j. Then j ≤ n− 2, and P and S col α coincide for all entries weakly below or 110 to the left of j in S col α . Hence, in P, j +1 must be strictly to the right of j; it cannot be in the same column as j by maximality of j and the fact that P is increasing in rows and in columns. This also forces j +1 to be either strictly below j or right-adjacent to j in P. In particular j∈Des RS ∗ (P), and hence π j (P)=s j (P)̸=P or π j (P)=0̸=P. If j ∈ Des RS ∗ (S col α ) then j must be at the top of some column of S col α , but in that case, since rows and columns are increasing in SET(α ), j +1 must be at the bottom of the next column in both S col α and P. This contradicts the maximality of j. Hence j / ∈ Des RS ∗ (S col α ) and π j (S col α )=S col α . Theorem 10. The cyclic H n (0)-submoduleZ α ofV α is indecomposable. Proof. Let f be an idempotent H n (0)-module endomorphism ofZ α . Our starting point here is the equation f(S col α )= X T∈SET(α ) a T T. Lemma 16 is the analogue of [10, Lemma 3.11]. To make our work self-contained, we reproduce the brief argument in [10, Theorem 3.12]. Let P ∈ SET(α ),P ̸= S col α ; we claim that a P = 0. Let j be the integer guaranteed by Lemma 16 for this P, and apply π j to the preceding equation. We then have, since π j fixes S col α , f(S col α )=f(π j (S col α ))= X T∈SET(α ) a T π j (T), but P ̸= π j (P), so P does not appear in the right-hand side by Lemma 6. Hence P does not appear in the expansion of f(S col α ), i.e. a P =0. We conclude that f(S col α ) equals a scalar multiple of S col α , which can only be 0 or 1 since f is idempotent. Next we examine the quotient moduleV α /Z α arising from the submoduleZ α ofV α . Definition 8. Let NSET(α ) be the set of all standard tableaux of shape α in which all rows increase left to right, but at least one column does NOT increase from bottom to top. 111 Lemma17. The quotient module ¯V α :=V α /Z α is also anH n (0)-module for theRS ∗ -action, with basis NSET(α )∩SIT(α ). The module ¯V α is nonzero if and only if α has at least two parts of size greater than or equal to 2. When it is nonzero, it is cyclically generated by S 0 α . Proof. The first statement is clear, since a basis for the quotient module is the complement of SET(α ) in SIT(α ), which is empty if α has at most one part of size greater than or equal 2; recall that the first column is always increasing by definition of SIT( α ). The second statement follows since S 0 α generates all ofV α . It is helpful to record the extreme cases of ¯V α . Lemma 18. Let α ⊨ n. 1. If α has at most one part greater than 1, then ¯V α =(0). 2. If α has exactly two parts greater than 1, and both of these have size 2, then ¯V α is one-dimensional, and hence irreducible and indecomposable. Proof. Item (1) is clear since in this case any T ∈ SIT(α ) has only one column of length greaterthan1, thefirst, andsoallitscolumnsareincreasing. HenceNSET( α )∩SIT(α )=∅. For Item (2), column 2 of any T ∈ SIT(α ) has only two cells, and hence can decrease bottom to top in only one way. For the remainder of this section we will assume NSET(α )∩SIT(α )̸=∅. Equivalently, by Lemma 18, we assume α has at least two parts greater than or equal to 2. We define a relation on NSET( α )∩SIT(α ) by setting, for S,T ∈ NSET(α )∩SIT(α ), S≼ NSET RS ∗ T if there is a permutation σ such that T = π RS ∗ σ (S). This is simply the relation induced on the subposet of PRS ∗ α consisting of the elements in NSET(α )∩SIT(α ). For any S,T ∈ NSET(α )∩SIT(α ), since SET(α ) is invariant under the H n (0)-action, the intervals [S,T]inSIT(α )andNSET(α )∩SIT(α )coincide,andhencetheinducedsubposetNSET(α )∩ SIT(α ) is also ranked. However, it has no top element since S row α / ∈NSET(α ). See the Hasse diagram of PRS ∗ α for α =223, Figure 4.2. 112 We therefore immediately have Lemma 19. The relation ≼ NSET RS ∗ is a partial order on NSET(α )∩ SIT(α ), with minimal element S 0 α . Given a composition α ⊨ n, extend the partial order ≼ NSET RS ∗ on NSET(α )∩SIT(α ) to an arbitrary total order on NSET(α )∩ SIT(α ), denoted by ≼ NSET t RS ∗ . Let the elements of NSET(α )∩SIT(α ) under≼ NSET t RS ∗ be {S 0 α =τ 1 ≼ NSET t RS ∗ ··· ≼ NSET t RS ∗ τ m }. Now let ¯V τ i be theC-linear span ¯V τ i =span{τ j |τ i ≼ NSET t RS ∗ α τ j } for 1≤ i≤ m and observe that the definition of ≼ NSET t RS ∗ implies that π RS ∗ i 1 ··· π RS ∗ i ℓ ¯V τ i ⊆ ¯V τ i for any s i 1 ··· s i ℓ ∈ S n . This observation combined with the fact that the operators {π RS ∗ i } n− 1 i=1 satisfy the same relations as H n (0) by Theorem 2 gives the following result. Lemma 20. ¯V τ i is an H n (0)-module, and gives us a filtration of H n (0)-modules 0⊂ ¯V τ m ⊂···⊂ ¯V τ 2 ⊂ ¯V τ 1 . Set ¯V α := ¯V τ 1 . Essentially the same arguments leading to Theorem 3 now give us: Theorem 11. Let α ⊨ n be such that α has at least two parts of size greater than or equal to 2, and let τ 1 =S 0 α ∈ SIT(α ) be the unique minimal element of NSET(α )∩SIT(α ). Then 113 ¯V α is an H n (0)-module, cyclically generated by S 0 α , whose quasisymmetric characteristic is the quasisymmetric function RE α = X T∈NSET(α )∩SIT(α ) F comp(Des RS ∗ (T)) . (4.19) Equivalently, RS ∗ α − RE α = X τ ∈SET(α ) F comp(Des RS ∗ (τ )) . (4.20) Proof. In Lemma 20, letting ¯V τ m+1 = 0, the quotient modules ¯V τ i− 1 / ¯V τ i for 2 ≤ i≤ m+1 are 1-dimensional H n (0)-modules spanned by τ i− 1 . Since π j (τ i− 1 ) =      0, j∈Des RS ∗ (τ i− 1 ), τ i− 1 , otherwise, as an H n (0)-module, ¯V τ i− 1 / ¯V τ i is isomorphic to the irreducible module [L β ] where β is the composition corresponding to the descent set Des RS ∗ (τ i− 1 ). Hence ch( ¯V τ i− 1 / ¯V τ i ) = F comp(Des RS ∗ (τ i− 1 )) and ch( ¯V α ) = m+1 X i=2 ch( ¯V τ i− 1 / ¯V τ i )= m+1 X i=2 F comp(Des RS ∗ (τ i− 1 )) = X τ ∈NSET(α )∩SIT(α ) F comp(Des RS ∗ (τ )) =RE α , as claimed. Equation (4.20) follows from the fact that RS ∗ α = X τ ∈SIT(α ) F comp(Des RS ∗ (τ )) . The statement about the cyclic generator was established in Lemma 17. 114 We now show that the module ¯V α is indecomposable. This will require a careful analysis of the technical lemmas in Section 4.4. In particular, we need to show that the analogue of Lemma 11 holds. Lemma 21. Let P ̸= S 0 α , such that P ∈ NSET(α )∩SIT ∗ (α ). Assume α has at least three parts of size greater than or equal to 2. Then there is a permutation σ =s i 1 ··· s ir such that 1. π σ (S 0 α )=0; 2. π σ (P)isnonzeroandhasrankr+rank(P)inthesubposetPRS ∗ α ∩NSET(α )∩SIT ∗ (α ). Proof. We follow the argument of Lemma 11: again, we may assume that whenever j,j+1 are adjacent in S 0 α , they are also adjacent in P. Now we need the analogues of Lemma 8, Lemma 9 and Lemma 10. It is enough to show that each of these lemmas holds when P ∈NSET(α )∩SIT ∗ (α ). In that case, note first that if column 2 of P increases bottom to top, then there must be a non-increasing column, column j, for j≥ 3. The important observation is that in each of the three lemmas, Lemma 8, Lemma 9 and Lemma 10, the algorithms described in the proofs affect only the first two columns of the tableau P. Hence we need only address the casewhencolumn2ofP (istheonlycolumnwhich)doesNOTincreasefrombottomtotop. Consider Lemma 8 and the algorithm which produces the saturated chain in the state- ment. It is clear that if ℓ+1 is right-adjacent to i≤ ℓ− 1 in P, then the algorithm affects only columns 1 and 2, and furthermore in column 2 it changes only the entry in the cell (i,2), diminishing it each time by 1. Clearly if i ≥ 2, all entries below row i in column 2 are greater than ℓ+1, and hence the algorithm preserves the fact that column 2 is NOT increasing bottom to top. If i = 1, then in column 2 above the first row there must be entries x < y such that x is in a higher row than y. But these entries are untouched by the algorithm, so the intermediate tableaux in the saturated chain are all in NSET(α ). The analogous analysis for Lemma 9 brings us to the same conclusion. 115 Finally we examine Lemma 10. This is somewhat more intricate. Note that we have assumedα hasatleastthreepartsgreaterthan1. ItisalsoclearfromtheproofofLemma10 that once again, the algorithm only affects the first two columns. As before, we assume column 2 of P is NOT increasing. Let p be the smallest entry in S 0 α where there is disagreement with P, and say it is in cell (i,2). Then P has an entry z, z > p in this cell, (i,2), and p appears in P in cell (j,2) for j < i. Again as in the proof of Lemma 10, we may assume p ≥ ℓ+2. The algorithm of Lemma 10 successively replaces P with π j (P) for some j. Our claim is that the second column of π j (P) continues to be non-increasing, i.e. at each step, π j (P)∈NSET(α ). Consulting Example 12, we note that all rows above row i coincide in both S 0 α and in P, and that if the second column of P is non-decreasing, clearly the same is true of π p− 1 (P). At each step the entry in cell (j,2) of P is diminished by 1, and hence continues to be smaller than the entry z in cell (i,2), which is unchanged. Also, at each step, entries in P above cell (i,2) continue to be smaller than p. It follows that column 2 is always non-increasing. Our argument is now complete. To finish the proof that ¯V α is indecomposable, we note first that Lemma 12 still holds, as does the following analogue of Lemma 7: Lemma 22. Suppose f is an idempotent endomorphism of the H n (0)-module ¯V α , with f(S 0 α )= X T∈NSET(α )∩SIT(α ) a T T. Let P ∈ NSET(α )∩SIT(α ),P ̸= S 0 α , and suppose there is a j such that j ∈ Des RS ∗ (P)\ Des RS ∗ (S 0 α ). Then a P =0. In particular, we have f(S 0 α )= X T∈NSET(α )∩SIT ∗ (α ) a T T. Proof. The proof of Lemma 7 goes through verbatim. 116 ItcannowbeverifiedthattheargumentfollowingLemma12carriesthroughunchanged, thanks to Lemma 21. Combined with Lemma 17, this gives us the following: Theorem 12. The H n (0)-module ¯V α is nonzero and indecomposable if α has at least two parts greater than or equal to 2, and is zero otherwise. Remark 1. The functions RE α cannot form a basis for QSym, since they equal zero when α ⊨ n has at most one part greater than 1. (Such compositions are called diving boards in [12].) In fact the set {RE α : α has at least two parts of size≥ 2} need not be linearly independent. For instance,RE 122 =F 311 =RE 212 , because NSET(122)∩SIT(122) has only one tableau 3 4 2 5 1 , as does NSET(212)∩SIT(212): 3 4 2 1 5 . Next we observe that each of the quasisymmetric functions of this section is in fact the generating function for an appropriate class of tableaux. Call a row-strict immaculate tableau T a row-strict semistandard immaculate tableau if ALLthecolumnsofT areweaklyincreasing,bottomtotop(andnecessarilyalltherowsofT are strictly increasing, left to right). Similarly, call an immaculate tableau S a column-strict semistandard immaculate tableau if ALL the columns of S are strictly increasing, bottom to top, (and necessarily all the rows of S are weakly increasing, left to right). Recall (Definition 3.2.1) that RS ∗ α is the generating function for all tableaux of shape α with weakly increasing first column, bottom to top, and strict increase along rows, left to right, whileS ∗ α (Definition 5) is the generating function for all tableaux of shape α with strictly increasing first column, bottom to top, and weak increase along rows, left to right. Proposition 5. Let α ⊨ n. For any tableau D of shape α , as in Definition 5, let d i be the number of entries equal to i in D. 1. RE α = P x d 1 1 x d 2 2 ··· where the sum is over all row-strict semistandard immaculate tableaux D of shape α . In particular when α is a partition λ ,RE λ =s λ t. 2. RE α = P x d 1 1 x d 2 2 ··· where the sum is over all row-strict immaculate tableaux of shape α with at least one column that is NOT increasing. 117 3. [12, Definition 8, Theorem 11, and Section 7] E α = P x d 1 1 x d 2 2 ··· where the sum is over all column-strict semistandard immaculate tableaux D of shape α . In particular when α is a partition λ ,E λ =s λ . Proof. Part(2)isimmediatefromPart(1)andthefactthatRS ∗ α −RE α =RE α . Part(3)is due to [12], where the authors define the shin basis ש α of NSym; their notation for ourE α is ש ∗ α . Their proof arrives at our definition of column-strict semistandard immaculate tableaux using Pieri rules. However, one can also adapt the following proof of Part (1). Our proof follows [42, Ex. 7.90(a)]. From Proposition 4 we have RE α = X T∈SET(α ) F comp(Des RS ∗ (T)) . It suffices to show that for any β ⊨ n, the coefficient of the monomial x β 1 1 x β 2 2 ··· in the right- hand side is the number of row-strict semistandard immaculate tableaux of shape α and content β , i.e. with β i i’s. To see this, we recall [29] that for a composition β , the monomial x β 1 1 x β 2 2 ··· appears in F γ if and only if β is a finer composition than γ , or equivalently, if and only if set(γ ) ⊆ set(β ). Hence the monomial x β 1 1 x β 2 2 ··· appears in RE α if and only if Des RS ∗ (T)⊆ set(β ). We claim that |{T ∈ SET(α ) : Des RS ∗ (T) ⊆ set(β )}| equals the number of row-strict semistandard immaculate tableaux of shape α with β i entries equal to i, for all i≥ 1. By definition, in a row-strict semistandard immaculate tableau τ , the entries 1,2,...,i appear in the first i columns of τ from the left. Fixacompositionβ ⊨ n. Givensuchatableauτ withβ i i’s,weassociatetotauatableau T ∈ SET(α ) obtained by replacing the β i entries equal to i with the β i consecutive entries in the interval [β 1 +··· +β i− 1 ,β 1 +··· +β i− 1 +β i ], where we read all the entries equal to i from bottom to top. Since the columns increase weakly and the rows increase strictly, no two i’s are in the same row, so this implies that the descents in T can occur only where i’s 118 change to (i+1)’s as we read up the columns of τ , starting with the leftmost column. Hence Des RS ∗ (T)⊆ set(β ). Conversely, suppose we have T ∈ SET(α ) such that Des RS ∗ (T)⊆ set(β ). To recover τ , replace1,2,...,β 1 with1’s,β 1 +1,...,β 1 +β 2 with2’sandsoon. TheconditionDes RS ∗ (T)⊆ set(β ) guarantees that for all i, the first β i entries appear in the first i columns of τ , and hence we have a tableau with weakly increasing columns, and strictly increasing rows. This finishes the proof. Inhispaper,SearlesshowsthatthesetNSET(α )constitutesanH n (0)-invariantsubmod- ule of a larger parent module for theS ∗ -action; he then shows that the resulting quotient module has the following properties: it has basis SET(α ); it has quasiymmetric charac- teristic E α , specialising to the Schur function s λ when α is the partition λ ; it is cyclically generated by the top element S row α (the row superstandard tableau) of the poset PS ∗ α ; it is indecomposable; the following difference expands positively in the fundamental basis for QSym: S ∗ α −E α = X T∈NSET(α )∩SIT(α ) F comp(Des S ∗ (T)) . By contrast, in this section we have shown that SET(α ) constitutes an H n (0)-invariant subposet of PRS ∗ α for theRS ∗ -action, and therefore spans an H n (0)-submoduleZ α ofV α with the following properties: it has basis SET(α ); it has quasisymmetric characteristic RE α =ψ (E α ), specialising to the Schur function s λ t when α is the partition λ ; it is cyclically generated by the column superstandard element S col α of the poset PRS ∗ α , which is also the bottomelementofthesubposetSET(α ); itisindecomposable; theresultingquotientmodule V α /Z α , with basis NSET(α )∩SIT(α ) and characteristicRE α , is cyclically generated by S 0 α ; it is indecomposable if α has at least two parts greater than or equal to 2, and is zero otherwise; the following difference expands positively in the fundamental basis for QSym: RS ∗ α −RE α =RE α = X T∈NSET(α )∩SIT(α ) F comp(Des RS ∗ (T)) . 119 We now observe that, similarly, the poset PRS ∗ (α ) gives a quotient module of W α with characteristic equal to the extended Schur functionE α . In Figure 4.2, by reversing the arrows, one sees that the tableaux that are NOT in SET(α ) form a closed subset under the S ∗ -action. Proposition 6. The set NSET(α )∩SIT(α ) is a basis for a submodule Y α of W α for the S ∗ -action. The resulting quotient module W α /Y α has basis SET(α ) with characteristic E α , and is cyclically generated by S row α . It is indecomposable. Proof. Lemma 13 shows that NSET(α )∩SIT(α ) spans a submoduleY α ofW α for theS ∗ - action, since, by Lemma 4, for T ∈ SIT(α ), π S ∗ i (T) = S ∈ SIT(α ) ⇐⇒ T = π RS ∗ i (S). Clearly the resulting quotient module has basis SET(α ), characteristicE α , and is cyclically generatedbyS row α . Itremainstocheckthatthisquotientmoduleisindecomposable. Butthis follows by invoking [9, Lemma 3.11], which states that for P ̸= S row α there is an i such that π S ∗ i fixes the generator S row α but not P. The argument is now identical to [9, Theorem 3.12] and the proof of Theorem 10. 4.6 New 0-Hecke modules from the dual immaculate action In this section we re-examine the dual immaculate actionS ∗ described in Theorem 1, orig- inally defined in [10], on the vector space with basis SIT( α ). Recall from Section 4.4 that SIT ∗ (α )isthesetofstandardimmaculatetableauxwhosefirstcolumnconsistsoftheintegers in [ℓ(α )]. In Figure 4.2, these are the blue tableaux lying in the interval [S 0 α ,S col α ]. In [38], Searles constructs the H n (0)-module (for the dual immaculate action) whose characteristicistheextendedSchurfunctionbytakingaquotientofalargermodule,different fromW α .WeshowedinProposition6howthisH n (0)-modulecanalsobeobtainedbytaking aquotientofW α .Inthissectionwewillconsideradifferentsubmodule,andresultingquotient 120 module of W α itself, for the dual immaculate action. This continues the analogy with the previous section, where we constructed both a submodule and quotient module ofV α for the row-strict dual immaculate action. Note that thanks to Lemma 4, the S ∗ -Hecke action on the poset PS ∗ (α ) is obtained from the RS ∗ -Hecke action on the poset PRS ∗ (α ) simply by reversing the arrows in the Hasse diagram. In particular the labels remain unchanged. Before stating the theorem, we extract the two main techniques used in this paper for establishing indecomposability of a cyclic H n (0)-moduleH. The technique of Part (1) below was originally used to prove indecomposability in [44, Lemma 7.7, Theorem 7.8], for the immaculate H n (0)-moduleW α in [10, Lemma 3.11, Theorem 3.12], as well as the extended immaculate module of [38, Theorem 3.13]; in the present paper it is applied to the row- strict extended moduleZ α in Lemma 16 and Theorem 10 (in the special case where σ v is a consecutive transposition). The technique of Part (2) is used in [7, Proof of Theorem 4.1] and in this paper for the row-strict Hecke moduleV α , in the proofs of Theorem 6, as well as in Lemma 22 and Theorem 12. Proposition 7. LetH be an H n (0)-module cyclically generated by u 0 , with K-basis{v} v∈B for some finite set B such that u 0 ∈ B. Let f be an endomorphism of H. Either of the following two conditions implies that f is a scalar multiple of the identity. In particular, if f is idempotent, then it must be either the identity or the zero map. 1. For every basis element v∈B, v̸=u 0 , there is a permutation σ v in S n such that π σ v (u 0 )=u 0 , but π σ v (v)̸=v. Here we also need to know that v̸= π σ v (v)⇒ v̸= π σ v (w) for any other basis element w. The latter holds automatically when σ v is a consecutive transposition, see Lemma 6. 121 2. For every basis element v∈B, v̸=u 0 , there is a permutation σ v in S n such that π σ v (u 0 )=0, but π σ v (v)̸=0. Here we also need to know that π σ (v)=π σ (v ′ )⇒v =v ′ . Definition 9. Define S row∗ α to be the standard immaculate tableau in SIT(α ) whose first column consists of 1,2,...,ℓ(α ), and whose remaining cells are filled with the entries ℓ+ 1,...,n in consecutive order along rows, bottom to top and left to right. Clearly S row∗ α has strictly increasing columns bottom to top, and hence S row∗ α ∈ SIT ∗ (α )∩SET(α ). Compare this with the definition of S row α in Definition 4. For example, we have S row∗ 221 =S col 221 = 3 6 7 2 5 1 4 , S row∗ 332 = 3 8 2 6 7 1 4 5 ̸=S col 332 , S row∗ 1323 = 4 8 9 3 7 2 5 6 1 ̸=S col 1323 . Lemma 23. The set SIT ∗ (α ) is a bounded interval of the immaculate Hecke posetPS ∗ (α ), the interval [S 0 α ,S row∗ α ]. Furthermore, for any T ∈ SIT ∗ (α ), there is a permutation σ ∈ S n such that, for theS ∗ -action, T =π S ∗ σ (S row∗ α ). Proof. Let ℓ=ℓ(α ), the length of α . The invariance of SIT ∗ (α ) under theS ∗ -action is clear from the fact (see Theorem 1) that π S ∗ i (T)=s i (T) / ∈{T,0} ⇐⇒ i,i+1 are NOT in column 1 of T and i+1 is strictly above i in T, and hence the first column, consisting of {1,2,...,ℓ(α )}, is preserved under the action. 122 Also Des RS ∗ (S row∗ α ) = {ℓ}. Hence either S row∗ α = S 0 α , which occurs if α ℓ is the only part greater than 1, or there is exactly one j such that π RS ∗ j (S row∗ α ) = T ∈ SIT(α ), namely j = ℓ. Moreover in that case T / ∈ SIT ∗ (α ). This shows that for theS ∗ -action, if for any j, π S ∗ j (T)=S row∗ α ,T ∈SIT(α ), then necessarily T / ∈SIT ∗ (α ). For clarity and consistency with the previous sections, we revert to the partial order defined in Section 4.3 and the poset PRS ∗ (α ). The lemma will then follow upon invoking Lemma 4. It suffices to apply the straightening algorithm of Proposition 3 for the RS ∗ -action. Recall that the algorithm produces a saturated chain from any tableau T ∈ SIT(α ) to the top element, the row superstandard tableau S row α . We straighten a tableau T ∈ SIT ∗ (α ) to the tableau S row∗ α in exactly the same manner, namely, by working on the rows from top to bottom, starting with the largest entry of T that differs from its counterpart in S row α , and moving to the next largest entry. This ensures that the entries{1,2,...,ℓ(α )} are preserved in the first column. Exactly as in the proof of Proposition 3, it follows that if T ∈ SIT ∗ (α ), then there is a sequence of operators π j i , and tableaux T i ∈ SIT ∗ (α ), i = 1,...,r, such that π j i (T i− 1 )=T i ,i=1,2,...,r, where we set T 0 =T and T r =S row∗ α . Hence we conclude T ≼ RS ∗ α S row∗ α and π jr π j r− 1 ··· π j 1 (T)=S row∗ α . In particular S row∗ α is the unique maximal element of the subposet SIT ∗ (α ) of the poset PRS ∗ α . Thanks to Lemma 4, by reversing the arrows, this establishes the result for the dual immaculateS ∗ -action as well. We illustrate the straightening with two examples. 123 Example 15. Let α = 431 so that S row∗ α = 3 2 7 8 1 4 5 6 . Indicating in bold the largest top-most entry that differs from that of S row∗ α , we have, for T =S 0 α = 3 2 4 5 1 6 7 8 , T 0 π 5 −→ 3 2 4 6 1 5 7 8 =T 1 π 6 −→ 3 2 4 7 1 5 6 8 =T 2 π 7 −→ 3 2 4 8 1 5 6 7 =T 3 π 4 −→ 3 2 5 8 1 4 6 7 =T 4 , and T 4 π 5 −→ 3 2 6 8 1 4 5 7 =T 5 π 6 −→ 3 2 7 8 1 4 5 6 =T 6 =S row∗ α . Similarly for U =S col α we have S col α = 3 2 5 7 1 4 6 8 π 7 −→ 3 2 5 8 1 4 6 7 =U 1 π 5 −→ 3 2 6 8 1 4 5 7 =U 2 π 6 −→ 3 2 7 8 1 4 5 6 =U 3 =S row∗ α . Theorem13. Letα ⊨ n,andletW α betheH n (0)-modulewithbasisSIT(α )asinTheorem1. Then 1. SIT ∗ (α ) = [S 0 α ,S row∗ α ] is a basis for an H n (0)-submodule X α of W α for theS ∗ -action, of dimension n− ℓ(α ) α 1 − 1,α 2 − 1,...,α ℓ − 1 . 2. X α is cyclically generated by S row∗ α ; it has characteristic ch(X α )= X T∈SIT ∗ (α ) F comp(Des S ∗ (T)) . Ifα has at most one part greater than 1, thenSIT ∗ (α ) has cardinality 1, and the module X α is the irreducible H n (0)-module indexed by the composition α. 124 3. Assume α has at least two parts of size greater than 1. Then the quotient module W α /X α is nonzero and cyclically generated by S row α , and furthermore it is indecompos- able. It has characteristic ch(W α /X α )= X T∈SIT(α )\SIT ∗ (α ) F comp(Des S ∗ (T)) . If α has at most one part greater than 1, SIT(α ) = SIT ∗ (α ) and the quotient module is zero. Proof of Part (1): ThefirststatementfollowsfromLemma23. Thedimensioncountisclear, since, ignoring the constant first column, the tableau is uniquely determined by a sequence of subsets S 1 ,...,S ℓ , whose union is [n]\[ℓ]; S i consists of the entries in row i excluding the first column, and thus |S i |=α i − 1. Proof of Part (2): Lemma23givesthecyclicgeneratoraswell. Thecharacteristicfollows fromthenowfamiliarargument: imposealinearextensionontheinterval[S 0 α ,S row∗ α ],induc- ing an H n (0)-equivariant filtration whose quotients are one-dimensional S ∗ -Hecke modules generated by T ∈[S 0 α ,S row∗ α ], and then sum up the individual quotient characteristics. The last statement is clear, since SIT ∗ (α ) has cardinality 1 if and only if α has at most one part of size greater than 1. Proof of Part (3): As above, it is clear that the quotient module is nonzero if and only if α has at least two parts of size greater than 1. The statement about the cyclic generator follows from the fact that the parent module W α is itself cyclically generated by the row superstandard tableau S row α . A basis for the quotient is given by the complement of the set SIT ∗ (α ) in SIT(α ), and the characteristic is computed via the usual filtration arising from a linear extension. 125 Inordertoproveindecomposability,weappealtothemethodinPart(1)ofProposition7. FollowingtheproofofTheorem10,forthequotientmoduleourstartingpointistheequation f(S row α )= X T∈SIT(α )\SIT ∗ (α ) a T T. We need the analogue of Lemma 16, but this is precisely the content of [10, Lemma 3.11], namely that for every tableau P ̸= S row α , there is a j ∈ [n− 1] such that π j fixes S row α but does NOT fix P. The rest of the proof is identical to the proof of Theorem 10. Thefunctionsch(X α )arenotindependent,asthefollowingsimplecounterexampleshows: Example 16. Let n = 3 and consider the compositions α = 12,β = 21. Then SIT ∗ (α ) = 2 3 1 ,SIT ∗ (β )= 2 1 3 . Clearly ch(X α )=ch(X β ), since both tableaux have the same S ∗ -descent set, namely the set{1}. We have the following pleasing expression for the characteristic of the moduleX α . Proposition8. Letα ⊨ n have lengthℓ, and let ¯α be the composition(α 1 − 1,α 2 − 1,...,α ℓ − 1) of n− ℓ, where we omit any part that is zero. Then ch(X α )= X k≥ ℓ e ℓ− 1 (x 1 ,...,x k− 1 )x k h ¯α (x k ,x k+1 ,...), where e r is the rth elementary symmetric function, and h β =h β 1 h β 2 ··· is the product of the homogeneous symmetric functions indexed by the parts of the composition β. Proof. Recall from Theorem 9 that for a partition λ we have the Schur function expansion s λ = X T∈SET(λ ) F comp(Des S * (T)) 126 andthisnaturallygeneralizes(see[42, Theorem7.19.7])forpartitions λ,µ totheskewSchur function expansion s λ/µ = X T∈SET(λ/µ ) F comp(Des S * (T)) where λ/µ is the diagram of λ with that of µ removed from the bottom left corner. See [21] for further details. Now note that the first column of every tableau T ∈SIT ∗ (α ) is in natural bijection with theuniqueSETofshape(1 ℓ ). AlsonotethatT withtheleftmostcolumnremovedandevery remainingtableauentry(thatisatleastℓ+1)reducedbyℓisinnaturalbijectionwithaSET denoted by D T of shape consisting of disconnected rows of length α 1 − 1,α 2 − 1,...,α ℓ − 1 from bottom to top. Moreover, {i− ℓ|i∈Des S *(T),i>ℓ}={i|i∈Des S *(D T )}. Furthermore,wehavethatℓ− 1∈Des S *(T)butℓ̸∈Des S *(T),sointhemonomialsappearing in ch(X α ) we have that the (ℓ− 1)th variable does not equal the ℓth variable, but the ℓth variable may equal the (ℓ+1)th variable. Hence if the ℓth variable is x k , then comparing the expressions for ch(X α ) and s λ/µ , the first column of T contributes s (1 ℓ− 1 ) (x 1 ,...,x k− 1 )x k =e ℓ− 1 (x 1 ,...,x k− 1 )x k and the remainder of the shape contributes s D T (x k ,x k+1 ,...)=h ¯α (x k ,x k+1 ,...). Here we exploit the fact that the skew shape λ/µ consists of disjoint pieces. Summing over all k≥ ℓ completes the proof. 127 It is not clear how to investigate the indecomposability of the module X α . The method in Part (2) of Proposition 7 fails, since π S ∗ j (P)=0 ⇐⇒ j,j+1 are in column 1 of P ⇐⇒ j∈[ℓ− 1], but this is trivially true for every P ∈ SIT ∗ (α ). Similarly, there are many examples which show that the technique of Part (1) of Proposition 7 also fails. Recall from Theorem 1 that for the dual immaculate action, we have π S ∗ j (P)=P ⇐⇒ j+1 is weakly below j in P. Let α =223 as in Figure 4.2, so that S row∗ α = 3 6 7 2 5 1 4 . Then π S ∗ j (S row∗ α )=S row∗ α ⇐⇒ j∈{3,6}. However, the tableau P = 3 5 6 2 7 1 4 is also fixed by both π S ∗ 3 and π S ∗ 6 , as well as by π S ∗ 5 . 128 RecallfromtheprevioussectionthatNSET(α )isthesetofallstandardfillingsof α with increasing rows, but with at least one column that is not increasing bottom to top. We now consider two more modules, one for theS ∗ -action, and one for theRS ∗ -action, whose basis will be the set SET(α )∩SIT ∗ (α ). Theorem 14. Let α ⊨ n be of length ℓ. 1. SET(α )∩SIT ∗ (α ) is the closed interval [S col α ,S row∗ α ] of the poset PRS ∗ (α )≃PS ∗ (α ). 2. The set D(α )=NSET(α )∩SIT ∗ (α ) is a basis for aS ∗ -invariant submoduleY α ofX α . 3. The quotient moduleX α /Y α , under theS ∗ -action, has basis SET(α )∩SIT ∗ (α ), and is cyclically generated by S row∗ α . 4. The set RD(α ) = SET(α )\SIT ∗ (α ) is a basis for an RS ∗ -invariant submodule RY α of the row-strict extended moduleZ α whose basis is SET(α ). 5. The quotient module Z α /RY α , under the RS ∗ -action, has basis SET(α )∩ SIT ∗ (α ), and is cyclically generated by S col α . Furthermore, it is indecomposable. 6. The quasisymmetric characteristic of the quotientS ∗ -moduleX α /Y α is ch(X α /Y α )= X k≥ ℓ e ℓ− 1 (x 1 ,...,x k− 1 )x k E α (x k ,...), (4.21) where α =(α 1 − 1,α 1 − 1,...) is the composition obtained from α by diminishing each part by 1, and discarding parts equal to zero, andE β is the extended Schur function of Theorem 8, Equation (4.18). 7. The quasisymmetric characteristic of the quotientRS ∗ -moduleZ α /RY α is ch(Z α /RY α )= X k≥ ℓ h ℓ− 1 (x 1 ,...,x k )x k RE α (x k+1 ,...) (4.22) 129 where α = (α 1 − 1,α 1 − 1,...) is as above, and RE β is the row-strict extended Schur function of Proposition 4. Proof. For Part (1), let P ∈ SET(α )∩SIT ∗ (α ). The straightening argument of Lemma 15 shows that P can be obtained by applying a sequence of operators π RS ∗ i to S col α , which preserves the first column. This is because the first columns already match, so we can proceedtotheinductivestep, whichonlyaffectstheremainingcolumns. Similarly, theproof of Lemma 23 shows that P can be obtained by applying a sequence of operators π S ∗ i to S row∗ α , again without affecting the first column. It remains to check that the sequence of operators produces tableaux lying entirely within SET(α ). But if theS ∗ -straightening went from S row∗ α to T and then T to P, where T / ∈ SET(α ), then by reversing the arrows this wouldmeanthat,withrespecttotheRS ∗ -action,T canbeobtainedbyapplyingasequence ofRS ∗ -operators to P. SinceP is in SET(α ), this contradicts Lemma 15, which established that SET(α ) is closed under theRS ∗ -action. For Part (2), if S ∈ NSET(α ) and π S ∗ i (S) = T ∈ SIT ∗ (α ), then S = π RS ∗ i (T). Hence T / ∈SET(α ), otherwisebyLemma15, S wouldalsobeinSET(α ). Part(4)followssimilarly, and Part (3), as well as the first statement of Part (5), are then immediate, since S row∗ α generatesX α and S col α generatesZ α . To see that the quotient moduleZ α /RY α in Part (5) is indecomposable under theRS ∗ - action, we observe that the proof of Theorem 10 goes through, although we now start with the equation f(S col α )= X T∈[S col α ,S row∗ α ] a T T; this is because, as is easily verified, Lemma 16 also applies here. For Part (6), the filtration argument of Section 4.3 gives ch(X α /Y α )= X T∈SET(α )∩SIT ∗ (α ) F comp(Des S ∗ (T)) . 130 The first column with 1 ,...,ℓ gives us e ℓ− 1 as in Proposition 8. Also clearly we do not have a descent at ℓ. Since columns and rows all increase in SET(α ), the remainder of the tableau must have ℓ+1 in the bottom left corner and numbers ℓ+2,...,n such that the columns increaseandsodotherows. Thisgivestableauxforthegeneratingfunctionfortheextended Schur functionE α . Similarly for Part (7), the filtration gives ch(Z α /RY α )= X T∈SET(α )∩SIT ∗ (α ) F comp(Des RS ∗ (T)) . This is clearly just the image of ch(X α /Y α ) under the involution ψ . Since only ℓ at the top of the first column is in Des RS ∗ (T) for every T ∈ SET(α )∩SIT ∗ (α ), the first column contributesh ℓ− 1 (x 1 ,...,x k )x k . Sinceℓ∈Des RS ∗ (T), theremustbeastrictincreaseinindex ℓ,sotheremainingshape,beingcolumnstrictineverycolumn,generatesRE α (x k+1 ,...). We have observed that Lemma 4 allowed us to interpret the passage between the dual immaculateandrow-strictdualimmaculatefunctions, viathemap ψ , intermsofthepartial orders induced by the respective Hecke actions on standard immaculate tableaux. We con- clude by pointing out that precisely the same relationship holds between the H n (0)-actions defined in [44] and [7]. Remark 2. The H n (0)-action on standard reverse composition tableaux (SRCT), as defined in [44], bears the same relationship to the action defined by Bardwell-Searles [7] for row- strict Young quasisymmetric functions, in the sense of Lemma 4. First we recall the actions that are involved in these two situations. We also need to recall the passage between standard reverse composition tableaux and standard Young composition tableaux of size n, as described 131 in[29, Chapter4, Chapter5], namelyreplaceeachiwithn+1− iandreversethecomposition. The H n (0)-action on standard reverse composition tableaux defined in [44] is: π TvW i (T)=                T, if i is strictly right of i+1, 0, if i is strictly NW of, or in the same column as, i+1, s i (T), if i is strictly SW of i+1. (4.23) The second action is the one defined by Bardwell-Searles for row-strict Young tableaux [7], but with left and right swapped to reflect using reverse tableaux when considering row-strict quasisymmetric Schur functions: π revBS i (T)=                T, if i is weakly left of i+1, 0, if i is right-adjacent to i+1, s i (T), if i is strictly right of i+1 but not in the same row. (4.24) Then we have the following analogue of Lemma 4: Lemma 24. Let S and T be standard reverse composition tableaux. If S̸=T, S̸=0 then π TvW i (T)=S ⇐⇒ π revBS i (S)=T. Proof. π TvW i (T)=S ⇐⇒ i is strictly SW of i+1 in T ⇐⇒ i is strictly NE of i+1 in S ⇐⇒ π revBS i (S)=s i (S)=T. The argument is now complete. 132 A summary of our results for row-strict dual immaculate functions appears in Table 4.2 below. Table 4.3 summarises the families of quasisymmetric functions, with corresponding 0-Hecke modules, that have appeared in the literature to date. 4.7 A new descent set The work of this section is motivated by a desire to complete the analysis of 0-Hecke actions on the set SIT(α ), by considering all the possible variations on the descent set definitions in Definition 6. There are four possible relative positions of i and i+1: (strictly/weakly) and (above/below), two of which give the dual immaculate and row-strict dual immaculate H n (0)-modules. AsinSection4.2, foreachcomposition α ofn, letV α denotethevectorspacewhosebasis is the set SIT(α ) of standard immaculate tableaux of shape α . In this section we define two new descent sets which are complementary, and construct 0-Hecke modules for each one. This in turn gives two new families of quasisymmetric functions which are related by the involution ψ . The general framework is tantalisingly similar to Section 4.2, but the details aresufficientlydifferentthatonceagainacarefulanalysisisrequiredtoensurethatonedoes indeed have a 0-Hecke action in each case. Definition 10. For T ∈SIT(α ), define Des A ∗ (T):={i:i+1 is strictly below i in T}. Now we turn to the complement of this descent set. Definition 11. For T ∈SIT(α ), define Des ¯ A ∗ (T):={i:i+1 is weakly above i in T}. We will prove the following. Theorem 15. There is a cyclic H n (0)-module A α , generated by the bottom element S 0 α of the poset PRS ∗ (α ), whose quasisymmetric characteristic is ch(A α )= X T∈SIT(α ) F comp(Des A ∗ (T)) . 133 Here Des A ∗ (T)={i:i+1 is strictly below i in T}. Similarly: Theorem 16. There is a cyclic H n (0)-module ¯ A α , generated by the top element S row α of the poset PRS ∗ (α ), whose quasisymmetric characteristic is ch( ¯ A α )= X T∈SIT(α ) F comp(Des ¯ A ∗ (T)) . Here Des ¯ A ∗ (T)={i:i+1 is weakly above i in T}. Note that, as is the case withS ∗ α and RS ∗ α , the two characteristics are related by the involution ψ :ψ (ch( ¯ A α ))=ch(A α ). The H n (0)-actions we define in Theorem 18 and Theorem 19, will also establish the following. The first two parts are analogues of results from Section 4.5; see Theorem 7 and Proposition6. ThethirdistheanalogueofProposition8inSection4.6. Onceagainwehave a pleasing expression for the quasisymmetric characteristic of the action of ¯ A α on SIT ∗ (α ). Proposition 9. The vector space with basis 1. SET(α ) is a cyclic H n (0)-submoduleA SET(α ) ofA α , generated by S col α , with character- istic X T∈SET(α ) F comp(Des A ∗ (T)) ; 2. SET(α ) is also a quotient module ¯ A SET(α ) of ¯ A α , cyclically generated by S row α , with characteristic X T∈SET(α ) F comp(Des ¯ A ∗ (T)) . 3. SIT ∗ (α ) is a cyclic H n (0)-submodule ¯ A SIT ∗ (α ) of ¯ A α generated by S row∗ α , with charac- teristic X T∈SIT ∗ (α ) F comp(Des ¯ A ∗ (T)) = X k≥ ℓ e ℓ− 1 (x 1 ,··· ,x k− 1 )x k e β (x k ,...)e α ℓ − 1 (x k+1 ,...). 134 where α has length ℓ and β =(α 1 − 1,...,α ℓ− 1 − 1) ignoring any parts of size 0. Proof of the quasisymmetric characteristic in Part (3): Note the similarity with the proof of Proposition 8. Recall from Theorem 9 that for a partition λ we have the Schur function expansion s λ = X T∈SET(λ ) F comp(Des S * (T)) andthisnaturallygeneralizes(see[42, Theorem7.19.7])forpartitions λ,µ totheskewSchur function expansion s λ/µ = X T∈SET(λ/µ ) F comp(Des S * (T)) where λ/µ is the diagram of λ with that of µ removed from the bottom left corner. See [21] for further details. Now note that the first column of every tableau T ∈SIT ∗ (α ) is in natural bijection with the unique SET of shape (1 ℓ ). Also note that T with the leftmost column removed and every remaining tableau entry (that is at least ℓ+1) reduced by ℓ is in natural bijection with a SET denoted by ˜ D T ∪C of shape ˜ D T consisting of disconnected columns of length α 1 − 1,α 2 − 1,...,α ℓ− 1 − 1 from bottom to top, and a further shape C at the top consisting of a column of length α ℓ − 1. Moreover, {i− ℓ|i∈Des ¯ A ∗ (T),i>ℓ}={i|i∈Des S *( ˜ D T ∪C)}. It is worth noting that if i+1 is strictly above i (so in a different row), or in the same row as i in T, then this descent gets maintained in ˜ D T ∪C when we switch rows to columns (with entries increasing bottom to top) so these descents all get transferred from T to ˜ D T ∪C. Furthermore, we have that ℓ− 1∈ Des ¯ A ∗ (T), so in the monomials appearing in ch(X α ) we have that the (ℓ− 1)th variable does not equal the ℓth variable, but the ℓth variable may or may not equal the (ℓ+1)th variable. In particular, it may, unless in T the entry ℓ+1 appearsinrowℓ, soℓ∈Des ¯ A ∗ (T), inwhichcasetheℓthvariablemaynotequalthe(ℓ+1)th 135 variable. Henceiftheℓthvariableisx k , thencomparingtheexpressionsforch(X α )ands λ/µ , the first column of T contributes s (1 ℓ− 1 ) (x 1 ,...,x k− 1 )x k =e ℓ− 1 (x 1 ,...,x k− 1 )x k and the remainder of the shape contributes s ˜ D T (x k ,...)s (1 α ℓ − 1 ) (x k+1 ,...)=e β (x k ,...)e α ℓ − 1 (x k+1 ,...). Again we exploit the fact that the skew shape λ/µ consists of disjoint pieces. Summing over all k≥ ℓ completes the proof. two more modules, one for theA-action, and one for the ¯ A-action, whose basis is the set SET(α )∩SIT ∗ (α ). Theorem 17. Let α ⊨ n be of length ℓ. 1. SET(α )∩SIT ∗ (α ) is the closed interval [S col α ,S row∗ α ] of the poset PRS ∗ (α )≃PS ∗ (α ). 2. The set D(α ) = NSET(α )∩SIT ∗ (α ) is a basis for a ¯ A-invariant submodule ¯ AY α of ¯ A SIT ∗ (α ) . 3. The quotient module ¯ A SIT ∗ (α ) / ¯ AY α , under the ¯ A-action, has basis SET(α )∩SIT ∗ (α ), and is cyclically generated by S row∗ α . 4. The set RD(α )=SET(α )\SIT ∗ (α ) is a basis for anA-invariant submodule RAY α of the moduleA SET(α ) whose basis is SET(α ). 5. The quotient moduleA SET(α ) /RAY α , under theA-action, has basis SET(α )∩SIT ∗ (α ), and is cyclically generated by S col α . 136 6. Thequasisymmetriccharacteristicofthequotient ¯ A-module ¯ A SIT ∗ (α ) / ¯ AY α isch( ¯ A SIT ∗ (α ) / ¯ AY α ) =        P k≥ ℓ e ℓ− 1 (x 1 ,...,x k− 1 )x k ch( ¯ A SET(α ) )(x k ,x k+1 ,...), α ̸=(1 m ,n− m), e n , otherwise, (4.25) where α =(α 1 − 1,α 1 − 1,...) is the composition obtained from α by diminishing each part by 1, and discarding parts equal to zero. 7. ThequasisymmetriccharacteristicofthequotientA-moduleA SET(α ) /RAY α isch(A SET(α ) /RAY α ) =        P k≥ ℓ h ℓ− 1 (x 1 ,...,x k )x k ch(A SET(α ) )(x k+1 ,...), α ̸=(1 m ,n− m), h n , otherwise. (4.26) where α =(α 1 − 1,α 1 − 1,...) is as above. Proof. We comment only on the last two parts, since the proofs are otherwise similar to Theorem 14. For Part (6), when α = (1 m ,n− m), Des ¯ A (T) ={1,2,...,n− 1} and hence ch( ¯ A SIT ∗ (α ) / ¯ AY α )=e n . For Part (7), note that when α = (1 m ,n− m), theA ∗ -descent set is always empty, and hence ch(A SET(α ) /RAY α )=h n . The example below shows that the quasisymmetric functions ch(A α ) do not form a basis for QSym, since they fail to be independent: ch(A α ) = F (n) for α ∈{(1 ℓ− 1 ,n− ℓ+1),1≤ ℓ≤ n}. Example 17. Let α = (1 ℓ− 1 ,n− ℓ+1),1≤ ℓ≤ n. There is only one tableau T in SIT(α ). The descent sets are Des S ∗ (T)=[ℓ− 1] or∅ if ℓ=1,Des RS ∗ (T)={ℓ,ℓ+1,...,n− 1} or∅ if ℓ=n,Des A (T)= ∅,Des ¯ A (T)=[n− 1]. 137 Hence the quasisymmetric characteristics are S ∗ α =F (1 ℓ− 1 ,n− ℓ+1) =F α , RS ∗ α =F (ℓ,1 n− ℓ ) , ch(A α )=F (n) , ch( ¯ A α )=F (1 n ) . Remark 3. Since Des A ∗ (T)⊆ Des RS ∗ (T), Des ¯ A ∗ (T)⊇ Des S ∗ (T), the fundamental quasisymmetric functions in the expansion of RS ∗ α (resp. S ∗ α ) correspond to compositions that are refinements of (resp. are coarser than) those appearing in the fundamentalexpansionofch(A α )(resp. ch( ¯ A α )). Forexample, thethreeelementsinSIT(13) are T 1 = 2 1 3 4 ,T 2 = 3 1 2 4 ,T 3 = 4 1 2 3 , with respective descent sets {2,3}, {1,3} and {1,2} for RS ∗ 31 , and {2}, {3} and ∅ for ch(A 31 ). Hence RS ∗ 31 = F 211 +F 121 +F 112 , ch(A 31 )=F 22 +F 31 +F 4 . We turn now to constructing the appropriate H n (0)-modules for each new descent set. The standard model in the literature for doing this is the following. We are given some basis that is a subset ST(α ), say, of the set of all standard tableaux of shape α , and some definition of descents. For T ∈ ST(α ), let s i (T) be the operator switching i and i+1 in T. This may or may not produce a basis element in ST(α ). Define the action of the generator π i on T ∈ST(α ) by π i (T)=                T if i is NOT a descent of T, s i (T) if i is a descent of T and s i (T)∈ST(α ), 0 otherwise. (4.27) Undersuitableconditions,thegeneratorswillsatisfytheHeckerelations,therebydefining a 0-Hecke module. We will show that these propitious circumstances occur forA and ¯ A, in 138 addition toS ∗ andRS ∗ , and that in each case, the partial order resulting from the 0-Hecke action gives the immaculate Hecke poset PRS ∗ α ≃PS ∗ α . For Definition 10, (4.27) gives us the following. Define, for each 1 ≤ i≤ n− 1 and each standard immaculate tableau T of shape α, the action of the generator π i on T to be π A ∗ i (T)=                T, if i / ∈Des A ∗ (T) ⇐⇒ i+1 strictly above or right-adjacent to i in T, s i (T), i∈Des A ∗ (T) ⇐⇒ i+1 strictly below i in T. where s i (T) is the standard immaculate tableau obtained from T by swapping i and i+1. The following analogue of Lemma 1 then follows by definition. Lemma 25. Let T be a standard immaculate tableau and let i∈Des A ∗ (T). Then 1. i,i+1 cannot both be in the leftmost column of T; 2. if s i (T) is a standard immaculate tableau, then i / ∈Des A ∗ (s i (T)). Theorem 18. The operators π A ∗ i define an action of the 0-Hecke algebra on the vector space V α . Proof. As usual for simplicity we will simply write π i for π A ∗ i . Clearly from the preceding analysis, π i (T)∈V α for every standard immaculate tableau T of shape α. We must verify that the operators satisfy the 0-Hecke algebra relations. To show π 2 i (T) = π i (T), we need only check the case when i+1 is strictly below i in T. In this case π i (T) = s i (T), and i is now strictly below i+1. Hence π i (s i (T)) = s i (T) and we are done. Let 1≤ i,j≤ n− 1 with|i− j|≥ 2. Then{i,i+1}∪{j,j+1}=∅, so the actions of π i and 139 π j are independent of each other, and hence commute. It remains to show that π i π i+1 π i (T)=π i+1 π i π i+1 (T). (4.28) We examine four separate cases. Case 1: Assume i / ∈ Des A ∗ (T),i+1 / ∈ Des A ∗ (T): Then π i (T) = T,π i+1 (T) = T, and the claim is clear. Case 2: Assume i∈Des A ∗ (T), but i+1 / ∈Des A ∗ (T): Then π i+1 (T)=T, π i (T)=s i (T). Hence (4.28) becomes π i π i+1 (s i (T))=π i+1 (s i (T)), (4.29) which we need to verify. Assume i + 1 / ∈ Des A ∗ (s i (T)). The left-hand side then equals π i (s i (T)) = s i (T) by Lemma 25, and this is also the right-hand side. Finally assume i+1∈ Des A ∗ (s i (T)). We now have i+1 strictly below i in T, so that i+1 is strictly above i in s i (T), and i+2 strictly below i+1 in s i (T). Also recall that i+2 was weakly above i+1 in T. It follows that In s i (T),i+2 is now weakly above i and strictly below i+1. (4.30) This implies π i+1 (s i (T))=s i+1 (s i (T)), and in the latter we now have i (weakly) below i+1, which is strictly below i+2. In particular i is not a descent of π i+1 (s i (T)) = s i+1 (s i (T)),andhencethelattertableauisfixedby π i .Equation(4.29)isthusverified. Case 3: Assume i / ∈Des A ∗ (T), but i+1∈Des A ∗ (T): Now (4.28) becomes π i (s i+1 (T))=π i+1 π i (s i+1 (T)), (4.31) 140 which we need to verify. Thus i+2 is strictly below i+1 in T. Also i is weakly BELOW i+1 in T. Suppose i and i+1 are NOT right-adjacent in T. Then i+1 is strictly above both i and i+2 in T, and thus i+2 is strictly above both i,i+1 in s i+1 (T). Here we have two possibilities for s i+1 (T): • Eitheriisbelowi+1,whichisbelowi+2,andhenceπ i andπ i+1 bothfix s i+1 (T); • or i+1 is below i, and i is below i+2. In the latter case, applying π i to s i+1 (T) switches i and i+1, so that in π i (s i+1 (T)) we now have i below i+1, and i+1 (still) below i+2. But then π i+1 fixes π i (s i+1 (T)). Now suppose i and i+1 ARE right-adjacent in T. Since i+2 is strictly below i+1 in T, the only possibility here for s i+1 (T) is that i,i+2 are right-adjacent, lying strictly above i+1. But then in π i (s i+1 (T)), i+1,i+2 are right-adjacent, lying strictly above i. Hence π i+1 fixes π i (s i+1 (T)), and Equation (4.31) has been established. Case 4: Assume i,i+1∈Des A ∗ (T): This means i is strictly above i+1 which is strictly above i+2 in T. Then Equation (4.28) becomes s i s i+1 s i (T)=s i+1 s i s i+1 (T), (4.32) and it is easy to see that this is indeed true. We have verified Equation (4.28) in all cases, thereby completing the proof that the action of the generators π i extends to an action of H n (0) onV α . ThisH n (0)-actiononV α inducesapartialorder≼ A ∗ α onSIT(α ),exactlyasinSection4.3. In fact it produces the same poset PRS ∗ α , since clearly π A ∗ i (T) / ∈{T,0} ⇐⇒ π RS ∗ i (T) / ∈{T,0}. 141 Moreover, we have the following: Proposition 10. The moduleA α is cyclically generated by the standard immaculate tableau S 0 α . Proof. Examining Proposition 2, we see that, since we have π A ∗ i (T)=s i (T) ⇐⇒ π RS ∗ i (T)=s i (T) ⇐⇒ i+1 is strictly below i in T, the straightening algorithm goes through without change, giving the same conclusion. Proof of Theorem 15. As in Section 4.3, extend the partial order ≼ A ∗ α on SIT(α ) to an arbitrary total order on SIT(α ), denoted by≼ t A ∗ α . Let the elements of SIT(α ) under≼ t A ∗ α be {τ 1 ≼ t A ∗ α ··· ≼ t A ∗ α τ m }. The induced filtration, as in Section 4.4, is 0⊂ span([T m ,S row α ])⊂···⊂ span([T i ,S row α ])⊂···⊂ span([S 0 α ,S row α ]). The key observation here is that π A ∗ i (T)=T ⇐⇒ i / ∈Des A ∗ (T), which guarantees that the successive quotients in the filtration are one-dimensional irre- ducible modules. Theorem 15 now follows. 142 ConsidernextDefinition11. Define,foreach1 ≤ i≤ n− 1andeachstandardimmaculate tableau T of shape α, the action of the generator π i on T to be ¯π i (T)=π ¯ A ∗ i (T)=                                T, i / ∈Des ¯ A ∗ (T) ⇐⇒ i+1 strictly below i in T, 0 i+1 right-adjacent to i in T or i,i+1 both in 1st column, s i (T), i+1 strictly above i in T, and i,i+1 not both in 1st column. where s i (T) is the standard immaculate tableau obtained from T by swapping i and i+1. Lemma 26. Suppose i,i+1 are not both in column 1 of T ∈ SIT(α ), and i∈ Des ¯ A ∗ (T). If ¯π i (T)=s i (T), then i cannot be in column 1 of T. Proof. This follows since rows must increase left to right, and i + 1 is strictly above i in T. Theorem 19. The operators ¯π i define an action of the 0-Hecke algebra on the vector space V α . Proof. Clearly from the preceding analysis, ¯π i (T) ∈ V α for every standard immaculate tableau T of shape α. We must verify that the operators satisfy the 0-Hecke algebra re- lations. That ¯π 2 i (T) = ¯π i (T) holds is clear if ¯π i (T) ∈ {T,0}, so we need only check the case when i+1 is strictly above i in T, and i,i+1 not in column 1. In this case ¯π i (T) = s i (T), and i is now strictly below i+1 in s i (T). Hence ¯π i (s i (T))=s i (T) and we are done. Let 1≤ i,j≤ n− 1 with|i− j|≥ 2. Then{i,i+1}∪{j,j+1}=∅, so the actions of ¯π i and ¯π j are independent of each other, and hence commute. It remains to show that ¯π i ¯π i+1 ¯π i (T)= ¯π i+1 ¯π i ¯π i+1 (T). (4.33) 143 Again there are four separate cases. Case 1: i,i+1 / ∈Des ¯ A ∗ (T): This case is clear as before, since ¯π i ,¯π i+1 both fix T. Case 2: i∈Des ¯ A ∗ (T),i+1 / ∈Des ¯ A ∗ (T): thus ¯π i+1 fixes T and i+2 is strictly below i+1. If ¯π i (T) = 0, we are done, so assume ¯π i (T) = s i (T). By Lemma 26, i cannot be in column 1, but i+1 could be. Note that Equation (4.33) becomes ¯π i ¯π i+1 s i (T)= ¯π i+1 s i (T). (4.34) Now i is strictly below i+1 in T, so either i+2 is below both of them, or i+2 is above i and below i+1. In s i (T) = ¯π i (T), we have, in the first case, i+2 below i+1 below i and so ¯π i+1 (as well as ¯π i ) fixes s i (T); and in the second case i+1 below i+2 below i, so ¯π i+1 (s i (T)) has i+2 below i+1 below i. Thus in either case, i is NOT a descent of ¯π i+1 (s i (T)), which is thus fixed by ¯ π i . This verifies (4.34). Case 3: Assume i / ∈Des ¯ A ∗ (T), but i+1∈Des ¯ A ∗ (T): This is Case 2 with the roles of i and i+1 interchanged, and the argument follows mutatis mutandis. Case 4: Assume i,i+1∈ Des ¯ A ∗ (T): This means i+2 is weakly above i+1 which is weakly above i in T. If both relations are strict, and neither of the pairs i+2,i+1 nor i+1,i is in column 1, Equation (4.33) becomes s i s i+1 s i (T)=s i+1 s i s i+1 (T), (4.35) and it is easy to see that this is indeed true. 144 If ¯π i (T)=0= ¯π i+1 (T), each side of Equation (4.33) is 0. Otherwise, we have two sub-cases. Case 4a: ¯π i+1 (T)̸=0 and ¯π i (T)=0: Thismakestheleftsideof (4.33)equalto0. Wemustshowthattherightsideof (4.33) is also zero. We have ¯π i+1 (T) = s i+1 (T), and i+2,i+1 are not both in column 1. The right side of (4.33) is then ¯π i+1 ¯π i (s i+1 (T)). Since ¯π i (T) = 0, either i,i+1 are right-adjacent in T with i+2 strictly above them or i,i+1 are in the first column with i+2 above them, not in the first column. In the first case, applying s i+1 makes i and i+2 right-adjacent with i+1 above them, and this followed by ¯π i makes i+1 and i+2 right-adjacent. This tableau is thus sent to 0 by ¯π i+1 . In the second case, applying s i+1 makes i and i+2 adjacent in column 1 with i+1 above them and strictly to the right, and following this with ¯π i puts both i+1 and i+2 in column 1. Again, this tableau is sent to 0 by ¯π i+1 . Case 4b: ¯π i+1 (T)=0 but ¯π i (T)̸=0: This is Case 4a with the roles of i and i+1 interchanged, and the argument follows mutatis mutandis. We have verified Equation (4.33) in all cases, thereby completing the proof that the action of the generators ¯π i extends to an action of H n (0) onV α . The analogue of Proposition 10 is Proposition 11. The module ¯ A α is cyclically generated by the standard immaculate tableau S row α . 145 Proof. Examining Proposition 3, we see that, since we have π ¯ A ∗ i (T)=s i (T) ⇐⇒ π S ∗ i (T)=s i (T) ⇐⇒ i+1 is strictly above i in T, and i,i+1 are not both in column 1, the straightening algorithm goes through without change, with the same conclusion. Proof of Theorem 16. This is proved using the filtration of H n (0)-modules 0⊂ span([S 0 α ,T 1 ])⊂···⊂ span([S 0 α ,T i ])⊂···⊂ span([S 0 α ,S row α ]), induced by a linear extension {τ 1 ≼ t ¯ A ∗ α ··· ≼ t ¯ A ∗ α τ m } of the poset, exactly as before. Again the key observation is that π ¯ A ∗ i (T)=T ⇐⇒ i / ∈Des ¯ A ∗ (T), guaranteeing that the successive quotients in the filtration are one-dimensional irreducible modules. Finally, we have: Proof of Proposition 9. ThisisprovedbyexaminingtheargumentsinLemma15andLemma23, which in turn rely on the straightening algorithms of Proposition 2 and Proposition 3, re- spectively, and observing that they go through unchanged. Remark 4. It is not clear how to approach the question of indecomposability for the module A α . The method of Part (2) of Proposition 7 fails since the action of the generators is never 0 by definition. As for Part (1), consider the basis element v = S row α , and note that Des A ∗ (S row α )istheemptyset; thusπ A ∗ i fixesthetopelement S row α foralli. Sincethegenerator 146 of the module is the bottom element S 0 α , Part (1) of Proposition 7 fails for the basis element S row α . This discussion also applies to the moduleA SET(α ) , since S row α ∈SET(α ). For the module ¯ A α , our only recourse is Part (2) of Proposition 7, because the cyclic generator is now S row α , and its ¯ A-descent set is [n− 1]. An example shows that this fails as well. Let α =3121. It is easy to check that the Hecke generators sending S row α to 0 coincide with those sending P =π ¯ A 6 (S row α ) to 0. While Example 17 showed that the quasisymmetric characteristics arising from these modules are not independent, we conclude the paper by showing that they are nonetheless combinatorially interesting. In analogy withS ∗ α andRS ∗ α , we will establish that these new characteristics are the generating functions for appropriately defined sets of tableaux with possibly repeated entries. See Figure 4.1 for a schematic summary. Let α ⊨ n. Given a filling T of the diagram of α with entries {1,2,...}, let cm(T), the content monomial of T, denote the monomial x d 1 1 x d 2 2 ··· , where d i is the number of entries equal to i in T. We then refer to the composition (d 1 ,d 2 ,...) of n as the content of the tableau T. Define T α (1st col<,rows≤ ) to be the set of tableaux of shape α whose first column entries increase strictly bottom to top, and whose rows all increase weakly left to right. Similarly,define T α (1st col≤ ,rows<)tobethesetoftableauxofshapeα whosefirstcolumn entries increase weakly bottom to top, and whose rows all increase strictly left to right. From Definition 5 and Definition 3.2.1, we have S ∗ α = X T∈Tα (1st col<,rows≤ ) cm(T), andRS ∗ α = X T∈Tα (1st col≤ ,rows<) cm(T). Define T α (cols<,rows≤ ) to be the set of column-strict tableaux of shape α , whose columns 147 ALL increase strictly bottom to top, and whose rows all increase weakly left to right. From Proposition 5, the tableau generating function for the extended Schur function is E α = X T∈Tα (cols<,rows≤ ) cm(T). Define T α (cols≤ ,rows<) to be the set of row-strict tableaux of shape α whose columns ALL increase weakly bottom to top, and whose rows all increase strictly left to right. From Proposition 5, the tableau generating function for the row-strict extended Schur function is RE α = X T∈Tα (cols≤ ,rows<) cm(T). Our final result, captured in Figure 4.1, is: Theorem 20. Let α ⊨ n. 1. Let T α (1st col≤ ,rows≤ ) denote the set of tableaux of shape α whose first column entries increase weakly bottom to top, and whose rows all increase weakly left to right. Then ch(A α )= X T∈SIT(α ) F comp(Des A ∗ (T)) = X T∈Tα (1st col≤ ,rows≤ ) cm(T). 2. Let T α (1st col<,rows<) denote the set of tableaux of shape α whose first column en- tries increase strictly bottom to top, and whose rows all increase strictly left to right. Then ch( ¯ A α )= X T∈SIT(α ) F comp(Des ¯ A ∗ (T)) = X T∈Tα (1st col<,rows<) cm(T). 148 3. LetT α (cols≤ ,rows≤ ) be the set of tableaux of shape α having weakly increasing entries in ALL columns, bottom to top, and all rows, left to right. Its generating function is the characteristic of the submoduleA SET(α ) ofA α . Equivalently, ch(A SET(α ) )= X T∈SET(α ) F comp(Des A ∗ (T)) = X T∈Tα (cols≤ ,rows≤ ) cm(T). 4. LetT α (cols<,rows<)bethesetoftableauxofshapeα havingstrictlyincreasingentries in ALL columns, bottom to top, and all rows, left to right. Its generating function is the characteristic of the quotient module ¯ A SET(α ) of ¯ A α . Equivalently, ch( ¯ A SET(α ) )= X T∈SET(α ) F comp(Des ¯ A ∗ (T)) = X T∈Tα (cols<,rows<) cm(T). Proof. Recall that a composition β is finer than α if and only if set(β )⊇ set(α ). We prove (1) by showing that, for a fixed β ⊨ n, the cardinality of the set S 1 ={T ∈SIT(α ):Des A ∗ (T)⊆ set(β )} equals the cardinality of the set S 2 ={T ∈T α (1st col≤ ,rows≤ ):T has shape α and content β }. Given U ∈ S 2 , replace the 1’s in U left to right, bottom to top with the consecutive entries 1,...,β 1 , then the 2’s with the next β 2 consecutive entries, again left to right, bottom to top, and so on. It is clear that this produces increasing entries going up the first column of the resulting standard tableau T, as well as along the rows. Also the descents can occur only where the last i of U in this reading order (left to right, bottom to top) changes to an i+1, so set(β )⊇ Des A ∗ (T). Conversely let T ∈ S 1 . The entries of the standard tableau T constitute a labelling of the diagram of α . We can then fill this diagram with β 1 1’s, β 2 ’s, etc. consecutively following 149 the order of the labels in T. If the cell labelled m is filled with entry i, then the cell labelled m+1 can be filled with i or i+1, but it must be filled with the label i+1 if m is a descent ofT. Suchafillingisnecessarilyweaklyrow-increasingandalsoincreasinginthefirstcolumn. In order to prove (2) we must show that, for a fixed β ⊨ n, the sets S 3 = {T ∈ SIT(α ) : Des ¯ A ∗ (T) ⊆ set(β )} and S 4 = {T ∈ T α (1st col<,rows<) : T has shape α and content β } are of the same cardinality. The argument is similar, except for the change in reading order. GivenU ∈S 4 , notefirstthatthereisatmostone iineachrow, inviewofthestrictincrease. Inparticular, becauseofthestrictlyincreasingfirstcolumn, allentriesequalto igofromleft to right and top to bottom in U. Now we replace the 1’s in U left to right,top to bottom, withtheconsecutiveentries1,...,β 1 ,thenthe2’swiththenextβ 2 consecutiveentries,again left to right, top to bottom, and so on. To go backwards from a standard tableau in S 3 to onewithrepeatedentries,notethatwheni+1isweaklyabovei,thenimustbeinset(β ),soif iwasreplacedbyanentryj,theni+1mustbereplacedbyj+1,guaranteeingrow-strictness. The reading order for (3) is the same as for (1): left to right, bottom to top, and like- wise, the reading order for (4) is the same as for (2): left to right, top to bottom. We omit the details for the otherwise identical arguments. See also the proof of Proposition 5. Example 18. Let α = 122. Then the immaculate Hecke poset PRS ∗ (α ) is a 3-element chain. Here SET(α ) consists of only two tableaux, S col α = 3 5 2 4 1 and S row α = 4 5 2 3 1 , while SIT(α ) has one additional tableau, S 0 α = 3 4 2 5 1 . We have Des A ∗ (S row α )=∅,Des A ∗ (S col α )={3},Des A ∗ (S 0 α )={4}, and hence ch(A SET(α ) )= F 5 +F 32 , while ch(A α )=F 5 +F 41 +F 32 . Consider ch(A SET(α ) ). Now F 5 will contribute all monomials with exponents β ⊨ 5, while F 32 will contribute only monomials with exponentsβ finer than the composition 32. One such 150 monomial that would appear in both expansions is x 1 x 2 2 x 2 3 , corresponding to the composition 122. There are exactly two tableaux of content 122 with weakly increasing rows and weakly increasing columns. Following the reading order in the above proof, left to right, bottom to top, we see that S col α = 3 5 2 4 1 maps to T 1 = 2 3 2 3 1 , while S row α = 4 5 2 3 1 maps to T 2 = 3 3 2 2 1 . Note that these are weakly increasing in all rows and all columns. Hence T 1 comes from F 32 , and T 2 comes from F 5 . Now consider the module ¯ A α . Here we have Des ¯ A ∗ (S row α ) = {1,2,3,4},Des ¯ A ∗ (S col α ) = {1,2,4}, and Des ¯ A ∗ (S 0 α ) = {1,2,3}, and hence ch( ¯ A SET(α ) ) = F 1 5 +F 1121 , while ch( ¯ A α ) = F 1 5 +F 1121 +F 1112 . F 1 5 can only contribute the unique square-free monomial of degree 5, corresponding to the tableau S row α . Now F 1121 contributes to ch( ¯ A α ) the monomial x 1 x 2 x 2 3 x 4 ; following the reading order for ¯ A ∗ , namely left to right and top to bottom, we see that S col α = 3 5 2 4 1 maps to T 3 = 3 4 2 3 1 . Note that T 3 is a non-standard tableau that is strictly increasing in all rows and all columns; also, it is the unique such tableau of content 1121. Similarly F 1112 contributes the monomial x 1 x 2 x 3 x 2 4 ; again following the reading order for ¯ A ∗ , namely left to right and top to bottom, we see that S 0 α = 3 4 2 5 1 maps to T 4 = 3 4 2 4 1 . Note that T 4 is strictly increasing in all rows, but only strictly increasing in the first column. Again, it is (necessarily) the unique such tableau of content 1112. 151 Table 4.2: S ∗ α versus the new basisRS ∗ α Immaculate Tableaux→ S ∗ α [10] RS ∗ α Dual immaculate Row-strict dual imm. 1st Col bottom to top strict↗ weak↗ Rows left to right weak↗ strict↗ Descents (for fund. expansion) {i:i+1 strictly above i} {i:i+1 weakly below i} Action of ψ in QSym S ∗ α (x 1 ,...,x n ) RS ∗ α =ψ (S ∗ α ) 0-Hecke action on SIT(α ) Note that standard tableaux are the same π i (T)=T i+1 weakly below i i+1 strictly above i π i (T)=0 i,i+1 in 1st column i,i+1 in same row T,s i (T) standard, i+1 strictly above i, i+1 strictly below i and π i (T)=s i (T) i,i+1 NOT both in 1st column Partial order on SIT(α ) Poset PS ∗ (α )=[S 0 α ,S row α ] Poset PRS ∗ (α )≃PS ∗ (α ) Cover relation S≺ S ∗ α T ⇐⇒ S =π S ∗ i (T) S≺ RS ∗ α T ⇐⇒ T =π RS ∗ i (S) Imm. module generated by top element: W α =⟨S row α ⟩ bottom element: V α =⟨S 0 α ⟩ Indecomposable? Yes [10] Yes Extended Schur fn basis E α ,E λ =s λ [6] ψ (E α )=RE α ,RE λ =s λ t Module, Extended Schur fn cyclic⟨S row α ⟩, indecomp. [38] cyclic⟨S col α ⟩, indecomp. Basis SET(α ) (quotient of larger module) submodule ofV α Quotient module, Ext Schur fn None Yes, cyclic⟨S row α ⟩, indecomp. Basis NSET(α )∩SIT(α ) quotient ofV α 152 Table 4.3: Various families in QSym and associated H n (0)-modules; they form a basis of QSym unless otherwise indicated Quasisymmetric function indexed by α ⊨ n H n (0)-module Fundamental F α Irreducible, one-dimensional [21] [17] Dual immaculate Cyclic (S row α )=W α , indecomp. S ∗ α = P T∈SIT(α ) F comp(Des S ∗ (T)) acts on standard immaculate tableaux SIT(α ) [9] [10] Row-strict dual immaculate Cyclic (S 0 α )=V α , indecomp. RS ∗ α =ψ (S ∗ α )= P T∈SIT(α ) F comp(Des RS ∗ (T)) acts on SIT(α ) Quasisymmetric Schur Indecomp. iff α is simple; ˇ S α = P T∈SRCT(α ) F comp(DesˇS (T)) acts on standard reverse DesˇS (T):={i:i+1 weakly right of i} composition tableaux SRCT(α ) [23] [44] Row-strict quasisym Schur ψ ( ˇ S α )=R ˇ S α = P T∈SRCT(α ) F comp(Des R ˇS (T)) Des R ˇS (T):={i:i+1 strictly left of i} [29, Chapter 4], [7] [33] (Column-strict) Young quasisym function ˆ S α =ρ ( ˇ S α r) via ρ (F α )=F α r [29, Chapter 4], [44] [29] Row-strict Young quasisym function Indecomp. iff α is simple; R ˆ S α =ψ ( ˆ S α ) acts on standard Young row-strict tableaux [31] [7] Extended Schur function Cyclic⟨S row α ⟩, quotient ofW α , indecomp.; E α = P T∈SET(α ) F comp(Des S ∗ (T)) acts on SET(α )=[S col α ,S row α ] [6] [38] Row-strict Extended Schur Cyclic⟨S col α ⟩, submodule ofV α , indecomp.; RE α =ψ (E α )= P T∈SET(α ) F comp(Des RS ∗ (T)) acts on SET(α )=[S col α ,S row α ] Row-strict Extended Quotient Schur Cyclic⟨S 0 α ⟩, quotient ofV α , indecomp.; RE α = P T∈NSET(α )∩SIT(α ) F comp(Des RS ∗ (T)) acts on NSET(α )∩SIT(α ) *NOT a basis* ch of Dual immaculate submodule Cyclic⟨S row∗ α ⟩, submodule ofW α ; ch(X α )= P T∈SIT ∗ (α ) F comp(Des S ∗ (T)) X α acts on [S 0 α ,S row∗ α ]=SIT ∗ (α ) *NOT a basis* ch of Dual immaculate quotient module Cyclic⟨S row α ⟩, quotient ofW α ; ch(W α /X α )= P T∈SIT(α )\SIT ∗ (α ) F comp(Des S ∗ (T)) W α /X α acts on SIT(α )\SIT ∗ (α ) *NOT a basis* The five functions in Section 9: Cyclic modules⟨S 0 α ⟩,⟨S row α ⟩ acting on SIT(α ); Theorem 15, Theorem 16, Proposition 9, submodule⟨S col α ⟩ ofA α acting on SET(α ); Theorem 20 and Figure 4.1 quotient module⟨S row α ⟩ of ¯ A α acting on SET(α ); *NOT a basis* submodule⟨S row∗ α ⟩ of⟨S row α ⟩ acting on SIT ∗ (α ) 153 Bibliography [1] Alex Abreu and Antonio Nigro. Chromatic symmetric functions from the modular law. Journal of Combinatorial Theory, Series A, 180:105407, 05 2021. [2] Alex Abreu and Antonio Nigro. Parabolic lusztig varieties and chromatic symmetric functions. ArXiv:2212.13497, 2022. [3] PerAlexanderssonandGretaPanova. Lltpolynomials,chromaticquasisymmetricfunc- tions and graphs with cycles. Discrete Mathematics, 341(12):3453–3482, 2018. [4] Farid Aliniaeifard, Shu Xiao Li, and Stephanie van Willigenburg. Schur functions in noncommuting variables. Adv. Math., 406:Paper No. 108536, 37, 2022. [5] EdwardE.Allen, JoshuaHallam, andSarahK.Mason. Dualimmaculatequasisymmet- ricfunctions expandpositively intoYoung quasisymmetricSchurfunctions. J. Combin. Theory Ser. A, 157:70–108, 2018. [6] Sami Assaf and Dominic Searles. Kohnert polynomials. Experimental Mathematics, 31(1):93–119, 2022. [7] Joshua Bardwell and Dominic Searles. 0-Hecke modules for Young row-strict quasisym- metric Schur functions. European Journal of Combinatorics, 102:103494, 2022. [8] A. Berele and A. Regev. Hook Young diagrams with applications to combinatorics and to representations of Lie superalgebras. Adv. in Math., 64(2):118–175, 1987. [9] Chris Berg, Nantel Bergeron, Franco Saliola, Luis Serrano, and Mike Zabrocki. A lift of the Schur and Hall-Littlewood bases to non-commutative symmetric functions. Canad. J. Math., 66(3):525–565, 2014. [10] Chris Berg, Nantel Bergeron, Franco Saliola, Luis Serrano, and Mike Zabrocki. Inde- composable modules for the dual immaculate basis of quasi-symmetric functions. Proc. Amer. Math. Soc., 143(3):991–1000, 2015. [11] C.Bessenrodt,K.Luoto,andS.vanWilligenburg.SkewquasisymmetricSchurfunctions and noncommutative Schur functions. Adv. Math., 226(5):4492–4532, 2011. [12] JohnCampbell,KarenFeldman,JenniferLight,PavelShuldiner,andYanXu. ASchur- like basis of NSym defined by a Pieri rule. Electron. J. Combin., 21(3):Paper 3.41, 19, 2014. 154 [13] Soojin Cho and Jaehyun Hong. Positivity of chromatic symmetric functions associated with hessenberg functions of bounce number 3. Electron. J. Comb., 29, 2022. [14] Soojin Cho and JiSun Huh. On e-positivity and e-unimodality of chromatic quasi- symmetric functions. SIAM J. Discret. Math., 33:2286–2315, 2019. [15] Sylvie Corteel, Jim Haglund, Olya Mandelshtam, Sarah Mason, and Lauren Williams. Compact formulas for Macdonald polynomials and quasisymmetric Macdonald polyno- mials. Selecta Math. (N.S.), 28(32), 2022. [16] Samantha Dahlberg and Stephanie van Willigenburg. Lollipop and lariat symmetric functions. SIAM Journal on Discrete Mathematics, 32(2):1029–1039, 2018. [17] G´ erard Duchamp, Daniel Krob, Bernard Leclerc, and Jean-Yves Thibon. Fonctions quasi-sym´ etriques, fonctions sym´ etriques non commutatives et alg` ebres de Hecke ` a q = 0. C. R. Acad. Sci. Paris S´ er. I Math. , 322(2):107–112, 1996. [18] Brittney Ellzey. On Chromatic Quasisymmetric Functions of Directed Graphs. PhD thesis, University of Miami, 2018. [19] VesselinGasharov. Incomparabilitygraphsof(3+1)-freeposetsares-positive. Discrete Mathematics, 157(1):193–197, 1996. [20] Israel M. Gelfand, Daniel Krob, Alain Lascoux, Bernard Leclerc, Vladimir S. Retakh, andJean-YvesThibon. Noncommutativesymmetricfunctions. Adv. Math.,112(2):218– 348, 1995. [21] Ira M. Gessel. Multipartite P-partitions and inner products of skew Schur functions. In Combinatorics and algebra (Boulder, Colo., 1983), volume 34 of Contemp. Math., pages 289–317. Amer. Math. Soc., Providence, RI, 1984. [22] Mathieu Guay-Paquet. A modular relation for the chromatic symmetric functions of (3+1)-free posets. ArXiv: 1306.2400, 2013. [23] J. Haglund, K. Luoto, S. Mason, and S. van Willigenburg. Quasisymmetric Schur functions. J. Combin. Theory Ser. A, 118(2):463–490, 2011. [24] Zachary Hamaker, Bruce Sagan, and Vincent Vatter. Chromatic symmetric functions and sign-reversing involutions, In preparation. [25] MegumiHaradaandMarthaE.Precup. ThecohomologyofabelianHessenbergvarieties andtheStanley–Stembridgeconjecture. AlgebraicCombinatorics,2(6):1059–1108,2019. [26] JiSunHuh,Sun-YoungNam,andMeesueYoo. Meltinglollipopchromaticquasisymmet- ric functions and schur expansion of unicellular llt polynomials. Discrete Mathematics, 343(3):111728, 2020. [27] Nathan Jacobson. Basic algebra. II. W. H. Freeman and Company, New York; Dover reprint, second edition, 1989. 155 [28] Naihuan Jing and Yunnan Li. A lift of Schur’s Q-functions to the peak algebra. J. Combin. Theory Ser. A, 135:268–290, 2015. [29] Kurt Luoto, Stefan Mykytiuk, and Stephanie van Willigenburg. An introduction to quasisymmetric Schur functions. SpringerBriefs in Mathematics. Springer, New York, 2013. [30] I.G. Macdonald. Symmetric Functions and Hall Polynomials. Oxford University Press, 1995. [31] Sarah K. Mason and Elizabeth Niese. Skew row-strict quasisymmetric Schur functions. J. Algebraic Combin., 42(3):763–791, 2015. [32] SarahK.MasonandElizabethNiese. Quasisymmetric(k,l)-hookSchurfunctions. Ann. Comb., 22(1):167–199, 2018. [33] SarahK.MasonandJeffreyRemmel. Row-strictquasisymmetricSchurfunctions. Ann. Comb., 18(1):127–148, 2014. [34] Elizabeth Niese, Sheila Sundaram, Stephanie van Willigenburg, Julianne Vega, and Shiyun Wang. 0-Hecke modules for row-strict dual immaculate functions. ArXiv: 2202.00708, 2022. [35] Elizabeth Niese, Sheila Sundaram, Stephanie van Willigenburg, Julianne Vega, and Shiyun Wang. Row-strict dual immaculate functions. ArXiv:2202.00706, 2022. [36] P. N. Norton. 0-Hecke algebras. J. Austral. Math. Soc. Ser. A, 27(3):337–357, 1979. [37] Jeffrey B. Remmel. The combinatorics of ( k,l)-hook Schur functions. In Combinatorics andalgebra(Boulder, Colo., 1983),volume34ofContemp.Math.,pages253–287.Amer. Math. Soc., Providence, RI, 1984. [38] Dominic Searles. Indecomposable 0-Hecke modules for extended Schur functions. Proc. Amer. Math. Soc., 148(5):1933–1943, 2020. [39] JohnShareshianandMichelleL.Wachs.Chromaticquasisymmetricfunctions. Advances in Mathematics, 295:497–551, 2016. [40] Richard P. Stanley. A symmetric function generalization of the chromatic polynomial of a graph. Advances in Mathematics, 111(1):166–194, 1995. [41] Richard P. Stanley. Enumerative Combinatorics, volume 2 of Cambridge Studies in Advanced Mathematics. Cambridge University Press, 1999. [42] RichardP.Stanley. Enumerative combinatorics. Vol. 2,volume62ofCambridge Studies in Advanced Mathematics. Cambridge University Press, Cambridge, 1999. With a foreword by Gian-Carlo Rota and appendix 1 by Sergey Fomin. 156 [43] Richard P. Stanley and John R. Stembridge. On immanants of jacobi-trudi matrices and permutations with restricted position. Journal of Combinatorial Theory, Series A, 62(2):261–279, 1993. [44] Vasu V. Tewari and Stephanie van Willigenburg. Modules of the 0-Hecke algebra and quasisymmetric Schur functions. Adv. Math., 285:1025–1065, 2015. [45] Shiyun Wang. The e-positivity of the chromatic symmetric functions and the inverse kostka matrix. ArXiv:2210.07567, 2022. [46] ¨ Omer E˜ gecio˜ glu and Jeffrey B. Remmel. A combinatorial interpretation of the inverse kostka matrix. Linear and Multilinear Algebra, 26(1-2):59–84, 1990. 157

Abstract (if available)

Abstract This thesis consists of three projects I worked on symmetric and quasisymmetric functions. The chromatic symmetric functions are analogies of the classical symmetric functions. After its discovery, chromatic quasisymmetric functions were defined, followed by various properties of these functions and their relationships with different bases of symmetric functions, LLT polynomials and Macdonald polynomials. With associated graphs, they also provide us with an algebraic tool to study the properties of the graph. For example, a long-standing conjecture in algebraic combinatorics is that the chromatic symmetric functions distinguish non-isomorphic trees. Further, the chromatic symmetric functions are closely connected with the cohomology of regular semisimple Hessenberg varieties. Regarding quasisymmetric functions, numerous bases of quasisymmetric functions were discovered as a natural nonsymmetric generalization of symmetric functions, and various analogies of classical symmetric functions were established. These functions and their properties have grown to be a flourishing area of research throughout algebraic combinatorics.

Chapter 1 gives the background and basic definitions of symmetric and quasisymmetric functions. The first project extends the current investigations on the (3+1)-free posets conjectures by Stanley-Stembridge and Shareshian-Wachs with the q-parameterized version. We expand the chromatic symmetric functions for Dyck paths of bounce number three in the elementary symmetric basis using a combinatorial interpretation of the inverse of the Kostka matrix studied in Egecioglu-Remmel (1990). We construct sign-reversing involutions to prove that certain coefficients in this expansion are positive. We use a similar method to establish the e-positivity of chromatic symmetric functions for Dyck paths of bounce number three beyond the “hook-shape” case of Cho-Huh (2019). Our results provide more supportive evidence for the Stanley-Stembridge Conjecture by extending the e-positive class of the incomparability graph of natural unit interval orders.

The second and third projects are joint work with Elizabeth Niese, Sheila Sundaram, Stephanie van Willigenburg, and Julianne Vega.

In Chapter 2, we define a new basis of quasisymmetric functions, the row-strict dual immaculate functions, as the generating function of a particular set of tableaux. We show that this definition gives a function that can also be obtained by applying the involution $\psi$ to the dual immaculate functions of Berg, Bergeron, Saliola, Serrano, and Zabrocki (2014) and establish numerous combinatorial properties for our functions. We give an equivalent formulation of our functions via Bernstein-like operators, in a similar fashion to Berg et al. (2014). We conclude this chapter by defining skew dual immaculate functions and hook dual immaculate functions, and establishing combinatorial properties for them.

In Chapter 3, we focus on the representation-theoretic aspects of the row-strict dual immaculate functions. We construct a cyclic, indecomposable 0-Hecke algebra module for these functions. More precisely, by identifying the descent set corresponding to the row-strict dual immaculate functions, one can determine a 0-Hecke algebra action on the set of standard immaculate tableaux, from where we can construct a 0-Hecke module whose quasisymmetric characteristic is exactly the row-strict dual immaculate functions. We give an explicit description of the effect of $\psi$ on the associated 0-Hecke modules, via the poset induced by the 0-Hecke action on standard immaculate tableaux. This remarkable poset reveals other 0-Hecke submodules and quotient modules, often cyclic and indecomposable, notably for a row-strict analogue of the extended Schur functions studied in Assaf-Searles (2019). We give a complete combinatorial and representation-theoretic picture by constructing 0-Hecke modules for the remaining variations on descent sets, and showing that all the possible variations for generating functions of tableaux occur as characteristics of the 0-Hecke modules determined by these descent sets.

Linked assets

University of Southern California Dissertations and Theses

Conceptually similar

PDF

Advances in the combinatorics of shifted tableaux

PDF

Colors, Kohnert's rule, and flagged Kostka coefficients

PDF

On the Kronecker product of Schur functions

PDF

Categorical operators and crystal structures on the ring of symmetric functions

PDF

Characters of flagged Schur modules and compatible slides

Asset Metadata

Creator Wang, Shiyun (author)

Core Title Applications on symmetric and quasisymmetric functions

School College of Letters, Arts and Sciences

Degree Doctor of Philosophy

Degree Program Mathematics

Degree Conferral Date 2023-08

Publication Date 07/26/2023

Defense Date 04/28/2023

Publisher University of Southern California. Libraries (digital)

Tag 0-Hecke algebra,dual immaculate functions,Hopf algebras,inde-composable module,inverse Kostka matrix,OAI-PMH Harvest,quasisymmetric functions,skew Schur functions,symmetric functions,tableaux combinatorics.

Language English

Contributor Electronically uploaded by the author (provenance)

Advisor Panova, Greta (committee chair), Fulman, Jason (committee member), Zhang, Jiapeng (committee member)

Creator Email shirleywangusc@gmail.com,shiyunwa@usc.edu

Permanent Link (DOI) https://doi.org/10.25549/usctheses-oUC113290664

Unique identifier UC113290664

Identifier etd-WangShiyun-12157.pdf (filename)

Legacy Identifier etd-WangShiyun-12157

Document Type Dissertation

Rights Wang, Shiyun

Internet Media Type application/pdf

Type texts

Source 20230728-usctheses-batch-1075 (batch), University of Southern California (contributing entity), University of Southern California Dissertations and Theses (collection)

Access Conditions The author retains rights to his/her dissertation, thesis or other graduate work according to U.S. copyright law. Electronic access is being provided by the USC Libraries in agreement with the author, as the original true and official version of the work, but does not grant the reader permission to use the work if the desired use is covered by copyright. It is the author, as rights holder, who must provide use permission if such use is covered by copyright.

Repository Name University of Southern California Digital Library

Repository Location USC Digital Library, University of Southern California, University Park Campus, Los Angeles, California 90089, USA

Repository Email cisadmin@lib.usc.edu