Abacus-histories and the combinatorics of creation operators
aa r X i v : . [ m a t h . C O ] S e p Abacus-histories and the combinatorics of creation operators
Nicholas A. Loehr ∗ Dept. of MathematicsVirginia TechBlacksburg, VA 24061-0123 [email protected]
Gregory S. Warrington † Dept. of Mathematics and StatisticsUniversity of VermontBurlington, VT 05401 [email protected]
September 16, 2020
Abstract
Creation operators act on symmetric functions to build Schur functions, Hall–Littlewoodpolynomials, and related symmetric functions one row at a time. Haglund, Morse, Zabrocki,and others have studied more general symmetric functions H α , C α , and B α obtained byapplying any sequence of creation operators to 1. We develop new combinatorial models forthe Schur expansions of these and related symmetric functions using objects called abacus-histories. These formulas arise by chaining together smaller abacus-histories that encode theeffect of an individual creation operator on a given Schur function. We give a similar treat-ment for operators such as multiplication by h m , h ⊥ m , ω , etc., which serve as building blocksto construct the creation operators. We use involutions on abacus-histories to give bijec-tive proofs of properties of the Bernstein creation operator and Hall–Littlewood polynomialsindexed by three-row partitions. Keywords:
Hall–Littlewood polynomials; Bernstein operators; Jing operators; HMZ opera-tors; creation operators; Schur functions; semistandard tableaux; abaci; abacus histories; latticepaths.
Creation operators are an important technical tool in the study of the Schur polynomials s µ ,the Hall–Littlewood polynomials H µ , and related symmetric functions. Let Λ denote the ringof symmetric functions with coefficients in the field F = Q ( q ), where q is a formal variable. Foreach integer b , the Bernstein creation operator S b is an F -linear operator on Λ. These operators“create” the Schur symmetric functions, one row at a time, in the following sense. Given anyinteger partition µ = ( µ ≥ µ ≥ · · · ≥ µ L ), we have s µ = S µ ◦ S µ ◦ · · · ◦ S µ L (1) . (1)Similarly, the Jing creation operators [7] are F -linear operators H b on Λ that create the Hall–Littlewood symmetric functions H µ [16, Chpt. III]. Specifically, for any integer partition µ , H µ = H µ ◦ H µ ◦ · · · ◦ H µ L (1) . (2) ∗ This work was supported by a grant from the Simons Foundation/SFARI ( † This work was supported by a grant from the Simons Foundation/SFARI (
Abacus-histories and creation operators arsia, Haglund, Morse, Xin, and Zabrocki [4, 5] have studied variations of the Jing creationoperators, denoted C b and B b , that play a crucial role in the study of q, t -Catalan numbers,diagonal harmonics modules, and the Bergeron–Garsia nabla operator. Replacing each H µ i in (2) by C µ i or B µ i produces symmetric functions that are closely related to Hall–Littlewoodpolynomials. More generally, we can consider operators indexed by arbitrary compositions ratherthan restricting to partitions. For any finite sequence of integers α = ( α , α , . . . , α L ), we candefine symmetric functions S α = S α ◦ S α ◦ · · · ◦ S α L (1); (3) H α = H α ◦ H α ◦ · · · ◦ H α L (1); (4) C α = C α ◦ C α ◦ · · · ◦ C α L (1); (5) B α = B α L ◦ B α L − ◦ · · · ◦ B α (1) . (6)On one hand, as we explain in Section 2.2, each S α is either 0 or ± s µ for some partition µ , where µ can be found from α by repeated use of the commutation rule S m ◦ S n = − S n − ◦ S m +1 ( m, n ∈ Z ) . (7)On the other hand, the H α , C α , and B α are more complicated symmetric functions that maybe regarded as generalized Hall–Littlewood polynomials. For general α , the Schur coefficients ofthese symmetric functions are polynomials in q (possibly multiplied by a fixed negative powerof q ) containing a mixture of positive and negative coefficients.The primary goal of this paper is to develop explicit combinatorial formulas for the Schurexpansions of H α , C α , and B α based on signed, weighted collections of non-intersecting latticepaths called abacus-histories . Along the way, we develop concrete formulas giving the Schurexpansion of the image of any Schur function under a single operator S b , H b , C b , B b , or any finitecomposition of such operators. We also give a similar treatment for some simpler operators suchas ω , multiplication by h b , h ⊥ b , etc., which serve as building blocks for constructing the moreelaborate creation operators.Some related work appears in a paper by Jeff Remmel and Meesue Yoo [18]. Our approachfeatures two key innovations leading to new and explicit combinatorial formulas. First, we use abacus diagrams rather than Ferrers diagrams as a means of visualizing the indexing partition µ for a Schur function s µ . This lets us record a particular Schur coefficient using a one-dimensionalpicture instead of a two-dimensional picture. Second, we utilize the second dimension of ourpicture to show the evolution of the abacus over time as various operators are applied to ourinitial Schur function. We thereby generate collections of non-intersecting lattice paths (abacus-histories) that represent the individual terms in the Schur expansion of the desired symmetricfunction. In some instances, we can define involutions on abacus-histories that cancel out neg-ative objects, thereby proving Schur-positivity or related identities.The rest of this paper is organized as follows. Section 2 reviews the needed backgroundon symmetric functions and covers definitions and algebraic properties of various creation op-erators. Section 3 develops combinatorial versions of the creation operators, showing how toimplement each operator by acting on an abacus diagram for one or two time steps. We usethis combinatorics to reprove (from first principles) some creation operator identities such as (1)and (7). Section 4 iterates our constructions to develop abacus-history formulas for the Schurexpansions of H α , C α , B α , and related symmetric functions. As a sample application of this Abacus-histories and creation operators echnology, Section 4.2 gives an elementary combinatorial proof of the Schur-positivity of three-row Hall–Littlewood polynomials, leading to a simple formula for the Schur coefficients of theseobjects. We conclude by presenting some open problems and directions for further work. We assume readers are familiar with basic background material on symmetric functions, in-cluding definitions and facts concerning integer partitions, the elementary symmetric functions e k , the complete homogeneous symmetric functions h k , the Schur symmetric functions s µ , theinvolution ω , and the Hall scalar product h· , ·i on Λ. In particular, the Schur functions s µ (with µ ranging over all integer partitions) form an orthonormal basis of Λ relative to the Hall scalarproduct, and ω is an involution, ring isomorphism, and isometry on Λ sending each s µ to s µ ′ .(See standard texts on symmetric functions such as [14, 16, 19] for more information.)Our initial definitions of the creation operators (following [4, 5]) utilize plethystic notation ,but readers need not have any detailed prior knowledge of plethystic notation to understand thispaper. In fact, one of our goals here is to offer an alternative, highly concrete and combinatorialtreatment of creation operators to complement the plethystic computations that appear in muchof the existing literature on this topic. Thus, each plethystic definition is immediately followedby an equivalent algebraic definition not using plethystic notation. Familiarity with plethysticnotation is required in only one (optional) section that proves the equivalence of the two defini-tions. The paper [15] has a detailed exposition of plethystic notation containing all facts neededhere. Later in the paper, we develop completely combinatorial definitions of creation operators(and related operators) in terms of abacus-histories. Recall that F is the field Q ( q ), and Λ is the F -algebra of symmetric functions with coefficients in F . For any symmetric function f ∈ Λ, define the linear operator M f : Λ → Λ to be multiplicationby f : M f ( P ) = f P ( P ∈ Λ) . (8)We frequently take f to be h c (the complete homogeneous symmetric function) or e c (the ele-mentary symmetric function).The Pieri Rules [14, Sec. 9.11] show how M h c and M e c act on the Schur basis. First, let HS( c )be the set of all skew shapes λ/µ consisting of a horizontal strip of c cells. For all partitions µ , M h c ( s µ ) = h c s µ = X λ : λ/µ ∈ HS( c ) s λ . (9)Pictorially, we apply M h c to s µ by adding a horizontal strip of size c to the Ferrers diagram of µ in all possible ways and summing the Schur functions indexed by the new diagrams.Second, let VS( c ) be the set of all skew shapes λ/µ consisting of a vertical strip of c cells.For all partitions µ , M e c ( s µ ) = e c s µ = X λ : λ/µ ∈ VS( c ) s λ . (10)This time, we compute M e c ( s µ ) by adding a vertical strip of size c to the diagram of µ in allpossible ways and summing the resulting Schur functions. Abacus-histories and creation operators or any linear operator G on Λ, let G ⊥ denote the operator on Λ that is adjoint to G relativeto the Hall scalar product. So, G ⊥ is the unique linear map on Λ satisfying h G ⊥ ( P ) , Q i = h P, G ( Q ) i for all P, Q ∈ Λ. (11)When G is a multiplication operator M f , we define f ⊥ = ( M f ) ⊥ for brevity. Thus, h f ⊥ ( P ) , Q i = h P, f Q i for all f, P, Q ∈ Λ. (12)We mostly use h ⊥ c and e ⊥ c acting on the Schur basis. Since the Schur basis is orthonormal relativeto the Hall scalar product, it follows from (12) and (9) that h ⊥ c ( s µ ) = X ν : µ/ν ∈ HS( c ) s ν . (13)In other words, h ⊥ c acts on s µ by removing a horizontal c -strip from the Ferrers diagram of µ inall possible ways and summing the resulting Schur functions. Similarly, e ⊥ c ( s µ ) = X ν : µ/ν ∈ VS( c ) s ν . (14)So e ⊥ c acts on s µ by removing a vertical c -strip from the Ferrers diagram of µ in all possibleways and summing the resulting Schur functions.As a convention, when c is a negative integer, we define the operators M h c , M e c , h ⊥ c , and e ⊥ c to be the zero operator. S m As in [5, pg. 834], we give a plethystic formula defining the Bernstein creation operators S m .More information on these operators (which can be combined into a single operator denoted S or Γ( z )) appears in earlier works by Thibon et al. [20, 21, 22]. For any integer m and anysymmetric function P , set S m ( P ) = ( P (cid:20) X − z (cid:21) ∞ X k =0 h k z k )(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) z m . (15)To explain this formula briefly: we first compute P [ X − (1 /z )] by expressing P (uniquely) as apolynomial in the power-sum symmetric functions p n with coefficients in F , then replacing each p n by p n − (1 /z n ). Next we multiply by the formal power series P k ≥ h k z k to obtain a formalLaurent series in z with coefficients in Λ. Taking the coefficient of z m in this series gives us S m ( P ).Here is an equivalent algebraic definition of S m not using plethystic notation: S m = ∞ X c =0 ( − c M h m + c ◦ e ⊥ c . (16)(This definition appears in [16, Ex. 29, pp. 95–97], but Macdonald uses the notation B m forour S m . We use the notation S m from [5] to avoid confusion with the operator B m below.) Weprove the equivalence of definitions (15) and (16) in Section 5. Abacus-histories and creation operators m ( s µ ) = + s ( m, for all m ≥ S ( s µ ) = − s (777744221) , S ( s µ ) = − s (777644221) , S ( s µ ) = − s (777544221) , S ( s µ ) = − s (777444221) , S − ( s µ ) = − s (777333221) , S − ( s µ ) = − s (777332221) , S − ( s µ ) = − s (777331111) , S − ( s µ ) = + s (777331100) , S m ( s µ ) = 0 for all other m ∈ Z .Table 1: Values of S m ( s (88844221) ) for all m ∈ Z .By combining the Pieri formulas and dual Pieri formulas discussed above, we can give acombinatorial prescription for computing S m ( s µ ) based on Ferrers diagrams. Starting with thediagram of µ , do the following steps in all possible ways. First choose a nonnegative integer c .Then remove a vertical strip of c cells from µ to get some shape ν . Then add a horizontal stripof m + c cells to ν to get a new shape λ . Record ( − c s λ as one of the terms in S m ( s µ ).Now, there is a much simpler way of computing S m ( s µ ) based on formulas (1), (3), and (7).Given a partition µ = ( µ ≥ · · · ≥ µ L ) and integer m , start with the list ( m, µ , . . . , µ L ).If m ≥ µ , then output the Schur function indexed by this new list. Otherwise, repeatedlyperform the following steps on the list (with infinitely many zero parts appended). Initialize asign variable ǫ = +1. Look for the unique ascent a < b in the current list. If b = a + 1, thenreturn zero as the answer. Otherwise replace the sublist ( a, b ) by ( b − , a + 1), replace ǫ by − ǫ , and continue. We eventually return zero or obtain a weakly decreasing list of nonnegativeintegers. In the latter case, return ǫ times the Schur function indexed by this list.This algorithm is a version of Littlewood’s method for straightening Jacobi–Trudi determi-nants [12]. Given any list of integers α = ( α , . . . , α L ), let D ( α ) be the determinant of the L × L matrix with i, j -entry h α i + j − i . For an integer partition µ , D ( µ ) is the Schur function s µ by theJacobi–Trudi formula. For any α , D ( α ) is either 0 or ± s ν for some partition ν . We can find ν byrepeatedly interchanging rows i and i + 1 of the matrix where α i < α i +1 . Each such interchangecauses a sign change and replaces parts α i and α i +1 in α by α i +1 − α i + 1, respectively.Comparing to the previous paragraph, we see that S m ( s µ ) is none other than D ( m, µ , . . . , µ L ).The straightening process can also be performed visually on composition diagrams using the slinky rule , as illustrated in [1]. Example 1.
Given µ = (8 , , , , , , ,
1) and m = −
2, we compute S − ( s µ ) = S ( − , , , , , , , , = − S (7 , − , , , , , , , = + S (7 , , , , , , , , = − S (7 , , , , , , , , = + S (7 , , , , , , , , = − S (7 , , , , , , , , = − s (777333221) . By similar calculations, we find the values of S m ( s µ ) shown in Table 1.It is not obvious that the two methods we have described for computing S m ( s µ ) always givethe same result. We prove this fact later using abacus-histories, and we will also give directcombinatorial proofs of (1) and (7). Abacus-histories and creation operators .3 Jing’s Creation Operators H m Here is a plethystic definition of Jing’s creation operators H m . As in [5, (2.2)], for all m ∈ Z and P ∈ Λ, let H m ( P ) = ( P (cid:20) X + q − z (cid:21) ∞ X k =0 h k z k )(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) z m . (17)In this case, P [ X + ( q − /z ] denotes the image of P under the plethystic substitution sendingeach p n to p n + ( q n − /z n .Alternatively, we could define H m = X c ≥ q c S m + c ◦ h ⊥ c . (18)We prove the equivalence of definitions (17) and (18) in Section 5.We can compute H m ( s µ ) via Ferrers diagrams as follows. Starting with the diagram of µ ,do the following steps in all possible ways. First choose a nonnegative integer c . Then removea horizontal strip of c cells from µ to get some shape ν . Compute S m + c ( ν ) as described earlierto obtain zero or a signed Schur function ± s λ . Record q c times the answer as one of the termsin the Schur expansion of H m ( s µ ). By iterating this description, it is evident that for everylist of integers α , the Schur coefficients of H α are polynomials in q with integer coefficients.Jing [7] proved that when α is a partition µ , H µ = H µ ◦ · · · ◦ H µ L (1) is none other than theHall–Littlewood symmetric function indexed by µ . C m and B m We use [5, Remark 3.7, pg. 835] as our plethystic definition of the creation operator C m . For m ∈ Z and P ∈ Λ, let C m ( P ) = ( ( − q − ) m − P (cid:20) X + q − − z (cid:21) ∞ X k =0 h k z k )(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) z m . (19)This operator is a variation of H m obtained by replacing q by 1 /q in H m , and then multiplyingthe output by a global factor ( − /q ) m − . So (18) leads at once to the following alternativedefinition of C m : C m = ( − q − ) m − X c ≥ q − c S m + c ◦ h ⊥ c . (20)Proposition 3.6 of [5] proves an inverse version of this identity, namely S m = ( − q ) m − X i ≥ C m + i ◦ e ⊥ i . Creation operators satisfy some useful commutation relations. For example, Proposition 3.2of [5] shows that for m, n ∈ Z , q C m ◦ C n − C m +1 ◦ C n − = C n ◦ C m − q C n − ◦ C m +1 , and in particular q C m ◦ C m +1 = C m +1 ◦ C m . Analogous relations for H m were proved muchearlier by Jing (see (1.1) in [7] or (0.18) in [8]). Abacus-histories and creation operators inally, we define the creation operator B m by conjugating H m by ω (see [5, pg. 829]): B m = ω ◦ H m ◦ ω. (21)Recall that ω is the linear operator on Λ sending each Schur function s λ to s λ ′ , where λ ′ is thepartition conjugate to λ obtained by transposing the Ferrers diagram of λ . Proposition 3.5 of [5]shows that for m + n > B n ◦ C m = q C m ◦ B n . ω Let C ω denote conjugation by ω , which sends any operator G on Λ to C ω ( G ) = ω ◦ G ◦ ω . Wenow give some useful identities involving C ω . First, C ω ( M f ) = M ω ( f ) for all f ∈ Λ. (22)To check this, recall that ω is a ring homomorphism on Λ and an involution ( ω ◦ ω = id). So forany P ∈ Λ, C ω ( M f )( P ) = ω ( M f ( ω ( P ))) = ω ( f · ω ( P )) = ω ( f ) · ω ( ω ( P )) = ω ( f ) · P = M ω ( f ) ( P ) . Second, ω ⊥ = ω . This follows since ω is an involution and an isometry (relative to the Hallscalar product), which means that for all P, Q ∈ Λ, h ω ( P ) , Q i = h P, ω ( Q ) i = h ω ⊥ ( P ) , Q i .Third, C ω ( f ⊥ ) = ( ω ( f )) ⊥ for all f ∈ Λ. (23)To see this, we use the first two facts and the adjoint property ( F ◦ G ) ⊥ = G ⊥ ◦ F ⊥ to compute: C ω ( f ⊥ ) = ω ◦ ( M f ) ⊥ ◦ ω = ω ⊥ ◦ ( M f ) ⊥ ◦ ω ⊥ = ( ω ◦ M f ◦ ω ) ⊥ = ( M ω ( f ) ) ⊥ = ( ω ( f )) ⊥ . Fourth, using (16) and C ω ( F ◦ G ) = C ω ( F ) ◦ C ω ( G ), we find that C ω ( S m ) = X c ≥ ( − c C ω ( M h m + c ) ◦ C ω ( e ⊥ c ) = X c ≥ ( − c M e m + c ◦ h ⊥ c . Fifth, using this result and (18), we get B m = C ω ( H m ) = X d ≥ q d C ω ( S m + d ) ◦ C ω ( h ⊥ d ) = X c ≥ X d ≥ q d ( − c M e m + d + c ◦ h ⊥ c ◦ e ⊥ d . So, we can compute B m ( s µ ) via Ferrers diagrams as follows. Starting with the diagram of µ ,do the following steps in all possible ways. First, choose integers c, d ≥
0. Remove a verticalstrip of d cells from µ , then remove a horizontal strip of c cells, then add a vertical strip of m + d + c cells. Record q d ( − c times the Schur function indexed by the new shape as one of theterms in the Schur expansion of B m ( s µ ). Later, we use abacus-histories to find a more efficientcombinatorial rule for computing this Schur expansion. Abacus-histories and creation operators Combinatorial Development of Creation Operators
This section develops combinatorial formulas for the Schur expansions of G ( s µ ), where s µ isany Schur function and G is one of the operators M h m , h ⊥ m , M e m , e ⊥ m , ω , S m , H m , C m , or B m .These formulas are based on the abacus model for representing integer partitions. James andKerber [6] introduced abaci to prove facts about k -cores and k -quotients of integer partitions.Abaci with labeled beads can be used to prove many fundamental facts about Schur functions,including the Pieri Rules for expanding s µ h k and s µ e k and the Littlewood–Richardson Rule [13].In our work here, it suffices to consider unlabeled abaci. We introduce the new ingredient oftracking the evolution of the abacus over time to model compositions of operators applied toa given Schur function. This leads to new combinatorial objects called abacus-histories thatmodel the Schur expansions of H α , C α , B α , and other symmetric functions built by composingcreation operators. First we review the correspondence between partitions and abaci. Suppose
N > µ = ( µ , µ , . . . , µ N ) is an integer partition (weakly decreasing sequence) consisting of N nonnegative parts. Let δ ( N ) = ( N − , N − , . . . , , , µ to µ + δ ( N ) =( µ + N − , µ + N − , . . . , µ N ) is a bijection from the set of weakly decreasing sequences of N nonnegative integers onto the set of strictly decreasing sequences of N nonnegative integers.We visualize the sequence µ + δ ( N ) by drawing an abacus with positions numbered 0 , , , . . . ,and placing a bead in position µ i + N − i for 1 ≤ i ≤ N . We use µ + δ ( N ) rather than µ becauseeach position can hold at most one bead. To formalize this concept, we define an N -bead abacus to be a word w = w w w · · · with all w i ∈ { , } and w i = 1 for exactly N indices i . Here w i = 1 means the abacus has a bead in position i , while w i = 0 means the abacus has a gap inposition i . For example, if N = 10 and µ = (8 , , , , , , , , , w = 11010110011000011100000 · · · . (24)In the theory of symmetric functions, we usually identify two partitions that differ only byadding or deleting zero parts. In fact, it is often most convenient to regard an integer partitionas an infinite weakly decreasing sequence ending in infinitely many zeroes. To model such asequence µ as an abacus, we use a doubly-infinite word w = ( w i : i ∈ Z ) such that w i = 1 forall i < w = 0, and w i = 1 for only finitely many indices i ≥
0. The nonzero parts of µ can be recovered from the abacus w by counting the number of gaps to the left of each bead onthe positive side of the abacus. The convention of putting the leftmost gap at position 0 is notessential; any left-shift or right-shift of the word w leads to the same combinatorics.We can also construct the abacus for a partition µ by following the frontier of the Ferrersdiagram of µ . As illustrated in Figure 1, the frontier consists of infinitely many north steps(corresponding to the infinitely many zero parts at the end of the sequence µ ), followed by asequence of east and north steps that follow the edge of the diagram, followed by infinitely manyeast steps at the top edge. Replacing each north step with a bead, replacing each east stepwith a gap, and declaring the first east step to have index 0 produces the doubly-infinite abacusassociated with µ . The singly-infinite abacus w in (24) is the ten-bead version of the full abacusshown in Figure 1, obtained by discarding all beads to the left of the tenth bead from the rightand shifting the origin to this location. Abacus-histories and creation operators .. ...0 1 2 ... Figure 1: The Ferrers diagram of µ and the doubly-infinite abacus built from the frontier of µ .For convenience, we mostly use N -bead abaci in this paper, which leads to identities valid forsymmetric polynomials in N variables. However, for certain abacus moves to work, we must besure to pad µ with enough zero parts (corresponding to beads at the far left end of the abacus)since these beads might participate in the move. This reflects the algebraic fact that somesymmetric function identities are only true provided the number of variables is large enough. M h m and h ⊥ m Now we describe how to compute G ( s µ ) using abaci, for various operators G . Here and below, werepresent the input s µ as an N -bead abacus drawn in the top row of a diagram. Each operator G acts by moving beads on the abacus according to certain rules, producing several possiblenew abaci that may be multiplied by signs or weights (powers of q ). Each abacus stands for theSchur function indexed by the partition corresponding to the abacus. We make a diagram foreach possible new abacus produced by G , where the output abacus appears in the second row(see the figures below for examples). Moving downward through successive rows represents thepassage of time as various operators are applied to the initial abacus. This convention lets us usea two-dimensional picture to display the evolution of an abacus as a whole sequence of operatorsare performed. It is much more difficult to visualize such an operator sequence using Ferrersdiagrams, particularly when some operators act by adding cells and others act by removing cells.As our first example, consider the computation of M h m ( s µ ) = h m s µ using abaci. By thePieri Rule (9), we know h m s µ is the sum of all s λ where λ is obtained by adding a horizontalstrip of m cells to the diagram of µ in all possible ways. Using the correspondence betweenthe frontier of µ and the abacus for µ , it is routine to check that adding such a horizontal stripcorresponds to moving various beads right a total of m positions on the abacus. (See [13] or [14,Sec. 10.10] for more details.) A given bead may move more than once, but no bead may moveinto a position originally occupied by another bead.To record this move in an abacus-history diagram, we start in row 1 with an N -bead abacusfor µ , where µ must end in at least one part equal to 0. We draw the beads as dots located atlattice points in row 1. Next, in all possible ways, we draw a total of m east steps starting atvarious beads, then move every bead 1 step south to represent the passage of time. For example,Figure 2 shows the abacus-histories encoding the computation M h ( s (3110) ) = s (5110) + s (4210) + s (4111) + s (3310) + s (3211) . Abacus-histories and creation operators igure 2: Computing M h ( s (3110) ) using abacus-histories.Figure 3: Computing h ⊥ ( s (332) ) using abacus-histories.Note that the extra zero part is needed to accommodate horizontal strips that use cells in therow below the last nonzero part of µ .Next consider how to compute h ⊥ m ( s µ ). Recall that the answer is the sum of all s ν where ν can be obtained by removing a horizontal strip of m cells from the Ferrers diagram of µ . Toexecute this action on an abacus, we move various beads west a total of m positions, avoidingcollisions with the original locations of the beads. Then we move every bead south one step torepresent the passage of time. For example, Figure 3 shows the abacus-histories encoding thecomputation h ⊥ ( s (332) ) = s (330) + s (321) . Note that µ need not be padded with zero parts when using this rule. M e m and e ⊥ m Now we describe abacus implementations of M e m and e ⊥ m . Recall M e m ( s µ ) is the sum of all s λ where λ is obtained by adding a vertical strip of m cells to the Ferrers diagram of µ . Comparingthe frontiers of µ and λ , we see that adding such a vertical strip corresponds to simultaneouslymoving m distinct beads east one step each. Beads cannot collide on the new abacus, but abead is allowed to move into a position vacated by another bead.To record this move in an abacus-history diagram, start with an N -bead abacus for µ where µ is padded with at least m zero parts. In all possible ways, pick a set of m beads that eachmove southeast one step, while the remaining beads move south one step with no collisions. Forexample, Figure 4 shows the abacus-histories encoding the computation M e ( s (41100) ) = s (41111) + s (42110) + s (51110) + s (42200) + s (52100) . Note that the m zero parts are needed to accommodate a vertical strip that we might add belowthe last nonzero part of µ .Next, e ⊥ m ( s µ ) is the sum of all s ν where ν is obtained from the Ferrers diagram of µ byremoving a vertical strip of m cells. Starting with the abacus for µ , we pick a set of m beadsthat each move southwest one step, while the remaining beads move south one step with nocollisions. For example, Figure 5 shows the abacus-histories encoding the computation e ⊥ ( s (442) ) = s (431) + s (332) . Note that µ need not be padded with zero parts when using this rule. Abacus-histories and creation operators igure 4: Computing M e ( s (41100) ) using abacus-histories.Figure 5: Computing e ⊥ ( s (442) ) using abacus-histories. ω and C ω We know ω ( s µ ) = s µ ′ , where the Ferrers diagram of µ ′ is found by transposing the diagram of µ . This transposition interchanges the roles of north and east steps on the frontier of µ andreverses the order of these steps. So, ω acts on the doubly-infinite abacus for µ by interchangingbeads and gaps and reversing the abacus. Let us call this move an abacus-flip .Now suppose we know a description of an operator G in terms of moves on an abacus. Then C ω ( G ) = ω ◦ G ◦ ω acts on the abacus for s µ by doing an abacus-flip, then doing the moves for G , then doing another abacus-flip. Therefore, we obtain a description of the operator C ω ( G )from the given description of G by interchanging the roles of beads and gaps, and interchangingthe roles of east and west.For example, consider G = M h m and C ω ( G ) = M e m . We know G acts on the abacus for s µ bymoving some beads m steps east , avoiding the original positions of all beads . Therefore, we cansay that C ω ( G ) acts on the abacus for s µ by moving some gaps m steps west avoiding the originalpositions of all gaps . One may check that this description of M e m (involving gap motions) isequivalent to the description given earlier (involving bead motions). When composing severaloperators to build bigger abacus-histories, it is often easier to work with descriptions that alwaysmove beads rather than gaps. S m We develop an initial abacus-history implementation of the operator S m based on formula (16),which will subsequently be simplified using a sign-reversing involution on abacus-histories. Givenany partition µ , we know by (16) that S m ( s µ ) = X c ≥ ( − c h m + c e ⊥ c ( s µ ) . To compute this with an abacus-history, start with the abacus for µ (padding µ with a zero partif needed) and perform the following steps in all possible ways. Choose an integer c ≥
0. Inthe first time step, apply e ⊥ c by moving c distinct beads one step southwest while the remainingbeads move one step south with no collisions. In the second time step, apply h m + c by movingsome beads a total of m + c steps east, never moving into a position occupied by a bead atthe start of this time step; then move all beads one step south. Record a term ( − c times the Abacus-histories and creation operators chur function corresponding to the final abacus. For example, Figure 6 shows how to compute S ( s (3110) ) via abacus-histories. When c = 0, we obtain the three positive objects labeled A, B,C; when c = 1, we obtain the nine negative objects labeled D through L; and so on. Adding upthe 23 signed Schur functions encoded by these abaci, there is massive cancellation leading tothe answer − s (2211) . UQMIEA BFJNRV CGKOSW TPLDH
Figure 6: Initial computation of S ( s (3110) ) using abacus-histories.We now introduce a sign-reversing involution on abacus-histories that explains the cancel-lation in the last example. Suppose an abacus-history appearing in the computation of S m ( s µ )contains a bead that moves southwest, then moves east i > e denotes a gap onthe abacus that no bead visits during the second time interval. Conversely, if a bead initiallymoves south and then has a gap to its left that no bead visits, then we can replace this initialsouth step with a southwest step followed by an east step. These path modifications change c by 1 and hence change the sign of the abacus-history, while preserving the requirement of taking m + c total east steps in the second time interval. The involution acts on an abacus-history byscanning for the leftmost occurrence of one of the patterns in Figure 7 and replacing it with theother pattern. It is clear that doing the involution twice restores the original object. The fixedpoints of the involution (which could be negative) consist of all abacus-histories avoiding bothpatterns in Figure 7. For example, applying the involution to the objects in Figure 6 producesthe following matches: A ↔ F, B ↔ H, C ↔ K, D ↔ S, E ↔ R, G ↔ Q, I ↔ O, J ↔ P, M ↔ V, N ↔ U, T ↔ W. We are left with the single negative fixed point L , confirming that S ( s (3110) ) = − s (2211) .We can now prove a formula for S m ( s µ ) that interprets S m as a bead-creation operator for abacus-histories. As consequences of this formula, we can finally justify equations (1), (7),and the technique for computing S m ( s µ ) used in Example 1. To state the formula, we need somepreliminary definitions. For any partition µ , assign a label and a sign to each gap on the abacus Abacus-histories and creation operators ast steps e east steps0 or more1 or more Figure 7: Cancellation move for abacus-histories appearing in S m ( s µ ).for µ as follows. Given a gap with g gaps strictly to its left and b beads to its right, let this gaphave label g − b and sign ( − b . Here is another way to compute the gap labels. The gap to theright of the rightmost bead on the abacus for µ has label µ . Any gap i positions to the right(resp. left) of this gap on the abacus has label µ + i (resp. µ − i ), as is readily checked. Theorem 2.
For any partition µ and integer m , S m ( s µ ) is if no gap in the abacus for µ haslabel m . Otherwise, we compute S m ( s µ ) by filling the unique gap labeled m with a new bead,then multiplying the Schur function for the new abacus by the sign of this gap.Proof. First compute S m ( s µ ) by generating a collection of signed abacus-histories, as describedat the start of this section. Next apply the sign-reversing involution to cancel out pairs ofobjects. We must now analyze the structure of the fixed points that remain. All fixed pointsavoid occurrences of both patterns shown in Figure 7. From our abacus-history characterizationof the action of S m ( s µ ), there are only two other possible move patterns for a bead starting inthe abacus for s µ . The first is for a bead to move south and then east zero or more east steps,but there must be another bead southwest of this bead’s starting point (i.e., in the positionmarked e on the right side of Figure 7). The second is for a bead to move southwest and thensouth with no intervening east steps. In the second case, we refer to this pair of steps as a zig-down move and say that the bead zigs down .Given a fixed point, suppose there is a bead Q on the input abacus that makes a zig-downmove. On one hand, suppose there is a bead P immediately to the left of Q (i.e., with no gapsin between). Then P must also zig down, since there is no room to do anything else. On theother hand, suppose R is the next bead somewhere to the right of Q (if any). There must bea vacancy immediately southwest of R ’s initial position (since Q zigs down). It follows that R must also zig down to avoid the two forbidden patterns. Define a block of beads on an abacus tobe a maximal set of beads with no gaps between any pair of them. Iterating our two precedingobservations, we conclude that if some bead zigs down in a fixed point, then all beads to its rightand all beads to its left in its block must also zig down .Next consider a bead S on the input abacus that has g > T that does not zig down. On one hand, S does not zig down (or else T wouldtoo). On the other hand, for T to avoid the second pattern in Figure 7, bead S must movesouth, then g steps east, then south. We now see that all fixed points for S m ( s µ ) must have thefollowing structure. There are zero or more beads at the right end that all zig down, startingwith some bead Q at the beginning of a block of beads. The motions of all remaining beads areuniquely determined (they move east as far as possible), except for the next bead P to the leftof Q . If beads P and Q are separated by g > P has the optionof moving south, then i steps east, then south, for any i satisfying 0 ≤ i < g . As a special case,if no beads zig down, then P is the rightmost bead on the abacus, which can move i steps east Abacus-histories and creation operators =−3:m=10:m=3: m=−8:m=−6:m=1:m=−2: Some fixed points:Input abacus (aligned with output) with gap labels:
Figure 8: Fixed points in the computation of S m ( s (888442210) ) for various values of m . Newlycreated beads are circled.for any i ≥
0. For example, Figure 8 illustrates some of the fixed points for S m ( s (888442210) ) forvarious choices of m (compare to Example 1).It is visually evident from this example that, for general µ and any fixed point of S m ( s µ ),the output abacus arises from the input abacus by shifting every bead one position to the left(which corresponds to deleting a zero part from the end of µ ) and then filling one gap with anew bead. Suppose this gap has g gaps strictly to its left and b beads to its right on the abacusfor µ . The label of this gap is, by definition, g − b . To complete the proof, we need only confirmthat g − b = m . By our characterization of fixed points, there are b southwest steps in timeinterval 1 (since all beads to the right of this gap zig down). By (16) and our observations at thebeginning of this section, these southwest steps arise from the action of e ⊥ b . As such, there mustbe m + b east steps arising from the subsequent action of h m + b . But, as illustrated in Figure 8,these east steps are in bijection with the gaps to the left of the new bead. So the number ofgaps g is m + b . It follows that g − b = ( m + b ) − b = m , as needed.Using the second method of computing gap labels, we see that for any partition µ and anyinteger m ≥ µ , S m ( s µ ) = + s ( m,µ ) . Iterating this result starting with s = 1, we obtain (1).Similarly, formula (7) can be deduced quickly from Theorem 2 by the following abacus-basedproof. The left side of (7) acts on s µ by first creating a new bead in the gap labeled n , thencreating a new bead in the gap now labeled m (returning zero if either gap does not exist).One readily checks that creating a new bead in the gap labeled n has the effect of decrementingall remaining gap labels. Thus, S m ◦ S n acts by filling the gap labeled n , then filling the gap originally labeled m + 1, if these gaps exist. Similarly, S n − ◦ S m +1 acts by filling the gap labeled m + 1, then filling the gap originally labeled n , if these gaps exist. These actions are the same Abacus-histories and creation operators xcept for the order in which the two new beads are created, which causes the two answers todiffer by a sign change. As a special case, when n = m + 1, both sides of (7) output zero becausethe second operator on each side tries to fill a gap that no longer exists. This completes the proofof (7). Finally, the computations in Example 1 are now justified by combining formulas (1), (3),and (7).By applying the results in Section 3.4, we also obtain a dual theorem characterizing theaction of C ω ( S m ) as a “gap-creation operator” or a “bead-destruction operator.” Specifically,given a bead with b beads strictly to its right and g gaps to its left, let this bead have label b − g and sign ( − g . (Note that this labeling of beads is related to, but different from, our priorlabeling scheme for gaps .) If the abacus for µ has a bead labeled m , then C ω ( S m )( s µ ) is the signof this bead times s ν , where we get the abacus for ν by replacing the bead labeled m by a gap.If the abacus for µ has no bead labeled m , then C ω ( S m )( s µ ) is zero. H m and C m With Theorem 2 in hand, we can describe how to compute H m ( s µ ) and C m ( s µ ) using abacus-histories. Our implementation of H m is based on formula (18). Starting with the abacus for µ ,do the following steps in all possible ways. Choose an integer c ≥
0. In time step 1, apply h ⊥ c by moving some beads a total of c steps west (avoiding original bead positions), then movingall beads one step south. In time step 2, apply S m + c by creating a new bead in the gap nowlabeled m + c (if any). The Schur function corresponding to the new abacus is weighted by q c times the sign of the gap where the new bead was created. Figure 9: Computing H − ( s (31) ) using abacus-histories.For example, by drawing the objects in Figure 9, we find H − ( s (31) ) = + s (200) − q s (200) − q s (110) + q s (110) . (25)In these abacus-histories, we have included gap labels in the middle row and circled the newbeads created by S m + c . By making similar pictures, one can check that H ( s (31) ) = − s (221) + q s (221) + q s (320) + q s (311) + q s (410) . (26)Note that these answers are neither Schur-positive nor Schur-negative. We observe that everyoriginal bead takes two consecutive south steps in these abacus-histories. By combining thesesteps into a single south step, we can shorten the time needed for the H m operator from twotime steps to one time step. We do this from now on.We can compute C m ( s µ ) by generating exactly the same collection of abacus-histories usedfor H m ( s µ ). The only difference is that each weight q c is replaced by q − c , and the final answeris multiplied by the global factor ( − /q ) m − . For example, C − ( s (31) ) = − q s (200) + q s (200) + q s (110) − q s (110) ; C ( s (31) ) = − s (221) + q − s (221) + q − s (320) + q − s (311) + q − s (410) . Abacus-histories and creation operators .7 Abacus Version of B m Finally, we describe how to compute B m ( s µ ) using abacus-histories. Since B m = C ω ( H m ) = ω ◦ H m ◦ ω , one approach is to calculate H m ( s µ ′ ) as described earlier, then conjugate all partitionsin the resulting Schur expansion. For example, using (25) and (26), we compute: B − ( s (211) ) = + s (11) − q s (11) − q s (2) + q s (2) ; B ( s (211) ) = − s (32) + q s (32) + q s (221) + q s (311) + q s (2111) . Alternatively, we can use the formula B m = X d ≥ q d C ω ( S m + d ) ◦ e ⊥ d proved in Section 2.5. Starting with the doubly-infinite abacus for µ , do the following steps inall possible ways. Choose an integer d ≥
0. In time step 1, apply e ⊥ d by moving d distinct beadsone step southwest while the remaining beads move one step south with no collisions. In timestep 2, apply C ω ( S m + d ) by destroying the bead with label m + d (if any). The Schur functioncorresponding to the new abacus is weighted by q d times the sign of the destructed bead. H α , C α , B α , etc. Now that we have abacus-history interpretations for the operators S m , H m , C m , B m , h ⊥ m , etc.,we can build abacus-history models giving the Schur expansion when any finite sequence ofthese operators is applied to any Schur function. We simply concatenate the diagrams forthe individual operators in all possible ways and sum the signed, weighted Schur functionscorresponding to the final abaci. We illustrate this process here by describing combinatorialformulas for H α , C α , B α , and the analogous symmetric functions H α ( s µ ), C α ( s µ ), and B α ( s µ )obtained by replacing 1 by s µ in (4), (5), and (6).Fix a sequence of integers α = ( α , α , . . . , α L ). We compute H α = H α ◦ · · · ◦ H α L (1) usingabacus-histories that take L time steps. We start with an empty abacus (corresponding to theinput s (0) = 1) where the gap in each position i ≥ i . In time step 1, we cannotmove any beads west, so we create a new bead in the gap labeled α L . In time step 2, we choose c ≥
0, move the lone bead west c steps and south once, then create a new bead in the gapnow labeled α L − + c . In time step 3, we choose c ≥
0, move the two beads west a total of c steps and south, then create a new bead in the gap now labeled α L − + c . And so on. Ifat any stage there is no gap with the required label, then that particular diagram disappearsand contributes zero to the answer. If the diagram survives through all L time steps, then itsfinal abacus contributes a Schur function weighted by q c + c + ··· + c L times the signs arising fromall the bead creation steps. Thus, the final power of q is the total number of west steps takenby all the beads, while the final sign is − H α ( s µ ) is the same, except now westart with the abacus for µ instead of an empty abacus. Here we might move some beads c ≥ α L + c . When µ = 0 we must have c = 0. Abacus-histories and creation operators AI B C DFE G HLJ KQM N O P
Figure 10: Computing H (123) using abacus-histories.The following remarks can aid in generating the diagrams for H α . The default startingpositions for the new beads are α L , α L − + 1 , α L − + 2 , . . . , α + L −
1. These are the positions(not gap labels) on the initial abacus where new beads would appear if all c i were zero. (Thisfollows since gap labels coincide with position numbers on an empty abacus, and each beadcreation decrements all current gap labels.) The actual starting positions for the new beads are α L + c , α L − + 1 + c , α L − + 2 + c , . . . , α + L + c L ;these are obtained by moving each default starting position east by the number of west steps inthe preceding row. Example 3.
Figure 10 uses abacus-histories to compute H (123) = H ( H ( H (1))). For brevity,we omit the top rows, which have no beads in any positive position. All three new beads havedefault starting position 3. There are 18 objects in all, but we find two pairs of objects thatcancel (C with F, and I with N). We are left with H (123) = q s (600) + ( q + q ) s (510) + ( q + q + q − q ) s (420) + q s (411) + q s (330) + ( q + q − q ) s (321) + ( q − q ) s (222) . We compute C α (resp. C α ( s µ )) by making exactly the same abacus-histories used for H α (resp. H α ( s µ )). The only difference is that the q -weight of each abacus-history is now q − ( c + ··· + c L ) and the final answer must be multiplied by ( − /q ) | α |− ℓ ( α ) , where | α | = α + · · · + α L and ℓ ( α ) = L .Since B m = C ω ( H m ) for every integer m , we can compute B α ◦ · · · ◦ B α L ( s µ ) by applying H α to s µ ′ as described above, then conjugating all partitions in the resulting Schur expansion.Alternatively, we can chain together the moves for the B α i described at the end of Section 3.7.Beware that B α (as defined in [5]) is found by starting with 1 and applying B α , B α , . . . , B α L in this order. Remark 4.
Python and SageMath code for computing with creation operators and abacus-histories is available on the second author’s website [23]. For k = 2 , , , ,
6, the computation of H α for α = (3 k ) involves 4, 27, 338, 6262, and 168312 abacus-histories, respectively. The timerequired (in seconds) was 0 .
04, 0 .
05, 0 .
08, 0 .
68, and 24 .
61, respectively. As another example, C (5 , , , , , has 16682 terms and takes 5 seconds to compute. Abacus-histories and creation operators .2 Application to Three-Row Hall–Littlewood Polynomials As we have seen, for general α the Schur expansion of H α has a mixture of positive and negativeterms. But for partitions ν , a celebrated theorem of Lascoux and Sch¨utzenberger [11] showsthat all Schur coefficients of H ν are polynomials in q with nonnegative integer coefficients (the Kostka–Foulkes polynomials ). According to this theorem, the coefficient of s λ in H ν is the sumof q charge( T ) over all semistandard Young tableaux T of shape λ and content ν , where chargeis computed from T by an explicit combinatorial rule (see [2], [16, pg. 242], and [17, Sec. 1.7]for more details). Kirillov and Reshetikhin [9, 10] gave another combinatorial formula for theKostka–Foulkes polynomials as a sum over rigged configurations weighted by a suitable chargestatistic. A detailed survey of combinatorial formulas for Hall–Littlewood polynomials and theirapplications to representation theory may be found in [3].Our combinatorial formula for H α based on abacus-histories holds for general integer se-quences α , but it is not manifestly Schur-positive when α happens to be an integer partition.On the other hand, our q -statistic (the total number of west steps in the object) is much simplerto work with compared to the complicated charge statistic on tableaux. As an application of ourabacus-history model, we now give a simple bijective proof of the Schur-positivity of H ν when ν is a partition with at most three parts.If ν has only one part, then it is immediate that H ν = H ν (1) = s ν . Next suppose ν =( ν ≥ ν ) has two parts. When computing H ν via abacus-histories, the default starting positionsof the two beads are ν and ν + 1 > ν . When the second bead is created, the first bead hasmoved to a column ν − c for some c ≥
0, and the second bead actually starts in column ν + 1 + c > ν − c . This means that all abacus-histories for H ν have positive sign, and ourformula is already Schur-positive in this case.Now let ν = ( ν ≥ ν ≥ ν ) have three parts. As before, since ν ≥ ν , the second beadalways gets created to the right of the first bead’s current column. For similar reasons, thebead created third must start to the right of the first bead in the lowest row. But it is possiblethat this third bead appears to the left of the second bead’s position in that row, leading to anegative object that must be canceled.We cancel these objects using the involution suggested in Figure 11. Given a negative objectas just described, let k be the distance between the new bead in the lowest row and the beadto its right. Let a, b, c ≥ a SW b S and bead 2’s path is W c S. The involution acts by replacing a by a − k and b by b + k , which causes bead 2’s actual starting position to move left k columns and bead 3’s actualstarting position to move right k columns. This action matches the given negative object witha positive object having the same number of west steps (hence the same q -power) and the samebead positions on the final abacus (hence the same Schur function).Going the other way, consider a positive object (as shown on the right in Figure 11) wherethe new bead in the bottom row is k columns to the right of the second bead, bead 1’s pathis W a ′ SW b ′ S, and bead 2’s path is W c ′ S. If k ≤ b ′ , then the involution acts by replacing a ′ by a ′ + k and b ′ by b ′ − k , causing the other two paths to switch places as before. If k > b ′ , thenthis positive object is a fixed point of the involution.To see that this involution really works, we must check a few items. Fix an arbitrary negativeobject with notation as in Figure 11. First, we must show k ≤ a , since a ′ is not allowed to benegative. On one hand, the actual starting position of the new bead in the bottom row is ν + 2 + b + c . On the other hand, the bead created in the middle row starts in position ν + 1 + a Abacus-histories and creation operators need k<=b’) Negative object: kab c
Positive object: ka’b’ c’
Figure 11: An involution on abacus-histories with three beads.and ends in the bottom row in position ν + 1 + a − c . Therefore, k = ( ν + 1 + a − c ) − ( ν + 2 + b + c ) = a − ( b + 2 c + 1 + ν − ν ) . (27)Since b, c ≥ ν ≥ ν , the quantity subtracted from a is strictly positive, so we actuallyhave k < a .Second, we show that applying the involution to a negative object does not cause the firsttwo beads to collide in the middle row. After replacing a by a ′ = a − k , the first bead movesfrom the top row to the middle row in position ν − a ′ = ν − a + k . The bead created in themiddle row now starts in position ν + 1 + a ′ = ν + 1 + a − k and moves left c ′ = c steps toposition ν + 1 + a − k − c before moving down to the bottom row. Therefore, to avoid a beadcollision, we need ν − a + k < ν + 1 + a − k − c , or equivalently 2( a − k ) > c + ν − ν − a − k , we need 2( b + 2 c + 1 + ν − ν ) > c + ν − ν −
1, whichrearranges to 2 b + 3 c + 3 + 2 ν − ν − ν >
0. This is true, since b, c ≥ ν ≥ ν ≥ ν .Third, we show that applying the involution twice restores the original object. This followssince the value of k is the same for the object and its image, and b ′ = b + k automatically satisfies k ≤ b ′ . We have now proved that our involution is well-defined and cancels all negative objects.By analyzing the fixed points of this involution more closely, we can prove the followingexplicit formula for the Schur coefficients of H ν when ν is a three-part partition. Theorem 5.
For all partitions ν = ( ν ≥ ν ≥ ν > and λ = ( λ ≥ λ ≥ λ ≥ such that | λ | = | ν | , the coefficient of s λ in H ν is min( λ − λ ,λ − λ ,ν − λ ,λ − ν ) X b =0 q ν − λ + λ − ν − b . (28) Proof.
We fix λ, ν as in the theorem statement and enumerate the positive fixed points of theinvolution. To obtain an uncanceled term s λ in the computation of H ν , the beads in the abacus-history must move as follows. The first bead starts in position ν and ends in position λ aftermoving along some path W a SW b S. (We switch here to unprimed parameters for the positivefixed point.) The second bead starts in position ν + 1 + a and ends in position λ + 1 (since allnegative objects cancel) after moving along some path W c S. The third bead starts in position ν + 2 + b + c and ends (without moving) in position λ + 2. We deduce that λ = ν − a − b , λ + 1 = ν + 1 + a − c , and λ + 2 = ν + 2 + b + c . Since λ and ν are fixed, the entire object isuniquely determined once we select the value of b . Eliminating a and c , we see that the q -weightof the object is q a + b + c = q ν − λ + λ − ν − b . Abacus-histories and creation operators hich choices of b are allowed? We certainly need b ≥
0, as well as a ≥ c ≥
0. Using a = ν − λ − b and c = λ − ν − b , the conditions on a and c are equivalent to b ≤ ν − λ and b ≤ λ − ν . We also need the first two beads not to collide in the middle row, so we need ν − a < ν + 1 + a − c . This condition is equivalent to b + λ < λ + 1 and, hence, to b ≤ λ − λ .Finally, letting k = ( λ + 2) − ( λ + 1) be the distance between the two rightmost beads in thebottom row, we need k > b for this positive object to be a fixed point of the involution. Thiscondition rearranges to b ≤ λ − λ . Combining the five conditions on b leads to the summationin the theorem statement.It is also possible to derive (28) starting from the charge formula for the Schur expansion of H ν . But such a proof is quite tedious, requiring a messy case analysis due to the complicateddefinition of the charge statistic.One might ask if the involution for three-row shapes extends to partitions ν with morethan three parts. While the same involution certainly applies to the first three rows of largerabacus-histories, more moves are needed to eliminate all negative objects. Even in the case offour-row shapes, the new cancellation moves are much more intricate than the move describedhere. We leave it as an open question to reprove the Schur-positivity of H ν for all partitions ν via an explicit involution on abacus-histories. It would also be interesting to find a specificweight-preserving bijection between the set of fixed points for such an involution and the set ofsemistandard tableaux. This appendix proves the equivalence of the plethystic definitions and the algebraic definitionsof S m and C m . We require just three basic plethystic identities (see [15] for a detailed expositionof plethystic notation including proofs of these facts). First, for any alphabets A and B and anypartition µ , we have the plethystic addition rule s µ [ A + B ] = X ν : ν ⊆ µ s ν [ A ] s µ/ν [ B ] . Second, for formal variables q and z and partitions ν ⊆ µ , s µ/ν [ q/z ] = ( ( q/z ) | µ/ν | if µ/ν is a horizontal strip;0 otherwise.Third, for all ν ⊆ µ , s µ/ν [ − /z ] = ( − | µ/ν | s µ ′ /ν ′ [1 /z ] = ( ( − /z ) | µ/ν | if µ/ν is a vertical strip;0 otherwise.Recall that HS( c ) (resp. VS( c )) is the set of horizontal (resp. vertical) strips with c cells.We prove the equivalence of the definitions (15) and (16) for S m by showing that the twoformulas have the same action on the Schur basis. Taking P = s µ in (15) and using the rules Abacus-histories and creation operators bove, we compute: S m ( s µ ) = s µ [ X − (1 /z )] X k ≥ h k z k (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) z m = X ν ⊆ µ s ν [ X ] s µ/ν [ − /z ] X k ≥ h k z k (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) z m = X c ≥ X ν ⊆ µ : µ/ν ∈ VS( c ) ( − /z ) c X k ≥ s ν h k z k (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) z m = X c ≥ X ν ⊆ µ : µ/ν ∈ VS( c ) ( − c s ν h m + c = X c ≥ ( − c M h m + c ( e ⊥ c ( s µ )) . This agrees with (16).Now we prove the equivalence of (17) and (18). Applying (17) to P = s µ gives: H m ( s µ ) = s µ [( X − /z ) + q/z ] X k ≥ h k z k (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) z m = X ν ⊆ µ s ν [ X − /z ] s µ/ν [ q/z ] X k ≥ h k z k (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) z m = X c ≥ X ν ⊆ µ : µ/ν ∈ HS( c ) ( q/z ) c s ν [ X − /z ] X k ≥ h k z k (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) z m = X c ≥ q c X ν ⊆ µ : µ/ν ∈ HS( c ) s ν [ X − /z ] X k ≥ h k z k (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) z m + c = X c ≥ q c X ν ⊆ µ : µ/ν ∈ HS( c ) S m + c ( s ν )= X c ≥ q c S m + c X ν ⊆ µ : µ/ν ∈ HS( c ) s ν = X c ≥ q c S m + c ( h ⊥ c ( s µ )) . This agrees with (18).
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