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The “Young” and “reverse” dichotomy of polynomials

Sarah Mason Department of Mathematics, Wake Forest University, Winston-Salem, NC 27109, U.S.A. [email protected]  and  Dominic Searles Department of Mathematics and Statistics, University of Otago, 730 Cumberland St., Dunedin 9016, New Zealand [email protected]
Abstract.

A “flip-and-reversal” involution arising in the study of quasisymmetric Schur functions provides a passage between what we term “Young” and “reverse” variants of bases of polynomials or quasisymmetric functions. Building on this perspective, which has found recent application in the study of qq-analogues of combinatorial Hopf algebras and generalizations of dual immaculate functions, we develop and explore Young analogues of well-known bases for polynomials. We prove several combinatorial formulas for the Young analogue of the key polynomials, show that they form the generating functions for left keys, and provide a representation-theoretic interpretation of Young key polynomials as traces on certain modules. We also give combinatorial formulas for the Young analogues of Schubert polynomials, including their crystal graph structure. We moreover determine the intersections of (reverse) bases and their Young counterparts, further clarifying their relationships to one another.

Key words and phrases:
Key polynomials, quasisymmetric Schur polynomials, Young quasisymmetric Schur polynomials
2010 Mathematics Subject Classification:
Primary 05E05

1. Introduction

Tableau models provide an indispensable framework for giving explicit positive combinatorial formulas for important families of polynomials and their relationships to one another. The celebrated Schur polynomials, which form a basis for the ring Symn\mathrm{Sym}_{n} of symmetric polynomials in nn variables, are famously realized as the weight generating functions of semistandard Young tableaux: tableaux of partition shape whose entries weakly increase from left to right in each row and strictly increase from bottom to top in each column. In fact, this definition may be reversed and Schur polynomials may alternatively be realized as the weight generating functions of semistandard reverse tableaux, whose entries weakly decrease from left to right along rows rows and strictly decrease from bottom to top in each column.

Symn\mathrm{Sym}_{n} is a subring of the ring QSymn\mathrm{QSym}_{n} of quasisymmetric polynomials. Basis elements of QSymn\mathrm{QSym}_{n} are indexed by compositions (sequences of positive integers) with at most nn parts. The semistandard reverse tableau model used in Symn\mathrm{Sym}_{n} naturally extends to produce tableaux of composition shape. The diagram D(α)D(\alpha) of a composition α\alpha, written in French notation, is the diagram consisting of left-justified rows of boxes whose ithi^{th} row from the bottom contains αi\alpha_{i} boxes. A tableau (of shape α\alpha) is a filling of D(α)D(\alpha) with positive integers. A reverse composition tableau is a tableau with entries no larger than nn, so that entries weakly decrease from left to right along rows.

Imposing different choices of further restrictions on the entries produces collections of reverse composition tableaux whose weight generating functions are, for example, the quasisymmetric Schur polynomial [HLMvW11], the fundamental quasisymmetric polynomial [Ges84], or the monomial quasisymmetric polynomial [Ges84] corresponding to α\alpha. On the other hand, certain other bases of QSymn\mathrm{QSym}_{n} are naturally described instead by restrictions of Young composition tableaux, where entries weakly increase from left to right along rows. Examples include the dual immaculate polynomials [BBS+14], the Young quasisymmetric Schur polynomials [LMvW13], and the extended Schur polynomials [AS19].

Extending further, reverse fillings provide a combinatorial framework that naturally generalizes the model of reverse composition tableaux to the ring Polyn\mathrm{Poly}_{n} of polynomials in nn variables. Basis elements of Polyn\mathrm{Poly}_{n} are indexed by weak compositions: sequences of nonnegative integers. The diagram D(a)D(a) of a weak composition aa is the diagram in ×\mathbb{N}\times\mathbb{N} having aia_{i} boxes in row ii, left-justified. A filling (of shape aa) is an assignment of positive integers, no larger than nn, to the boxes of D(a)D(a). A reverse filling is a filling in which entries weakly decrease from left to right along each row.

By imposing further restrictions on the entries, one can obtain a set of reverse fillings of D(a)D(a) whose weight generating function is, for example, the key polynomial [RS95], the quasi-key polynomial [AS18b], the Demazure atom [Mas09], or the fundamental slide polynomial [Sea20] corresponding to aa. At present, the majority of well-studied bases for Polyn\mathrm{Poly}_{n} are described in terms of reverse fillings, i.e., with decreasing rows.

As noted earlier, Schur polynomials may be realized in terms of either semistandard Young tableaux or semistandard reverse tableaux. This coincidence can be understood in terms of an involution on tableaux whose entries are at most nn, namely, replacing each entry ii with n+1in+1-i. This bijectively maps semistandard Young tableaux to semistandard reverse tableaux and vice versa. While this map is weight-reversing rather than preserving, the fact that Schur polynomials are symmetric means that the multiset of weights of semistandard Young tableaux is equal to the multiset of weights of semistandard reverse tableaux.

This map inspires a closely-related flip-and-reverse map on composition tableaux, defined by reversing the order of the rows (reverse) and replacing every entry ii with n+1in+1-i (flip). This weight-reversing map changes decreasing rows to increasing rows and vice versa. As is the case for Schur polynomials, the flip-and-reverse map preserves both the monomial and fundamental bases of QSymn\mathrm{QSym}_{n}. However, bases of QSymn\mathrm{QSym}_{n} are not preserved in general. In particular, the reverse composition tableaux that generate the quasisymmetric Schur polynomial corresponding to α\alpha are mapped to precisely the Young composition tableaux that generate the Young quasisymmetric Schur polynomial corresponding to rev(α){\rm rev}(\alpha), the composition obtained by reading α\alpha in reverse. Typically a Young quasisymmetric Schur polynomial is not also a quasisymmetric Schur polynomial; we characterize their coincidences in Section 2.

The flip-and-reverse map extends naturally to fillings of weak composition diagrams, giving two parallel constructions of bases for Polyn\mathrm{Poly}_{n}, one (reverse) defined by reverse fillings and one (Young) defined by Young fillings, i.e., fillings in which entries increase from left to right along rows. The fillings obtained by applying the flip-and-reverse map to those reverse fillings that generate a particular basis of Polyn\mathrm{Poly}_{n} generate a Young analogue of that basis.

Young analogues of the quasi-key and fundamental slide bases and a reverse analogue of the dual immaculate functions were introduced in [MS20] and properties of these bases were developed including a number of useful applications. In particular, these analogues were used to extend a result of [AHM18] on positive expansions of dual immaculate functions to the full polynomial ring, to establish properties of stable limits of these polynomials and their expansions, and to uncover a previously-unknown connection between dual immaculate functions and Demazure atoms. These results necessitated repeated passage between reverse and Young analogues. In particular, reverse analogues were needed to study stable limits for a polynomial ring analogue of the dual immaculate functions, whereas Young analogues were needed to connect to established results in QSymn\mathrm{QSym}_{n} from [AHM18]. In a similar vein, Young analogues of pre-existing reverse bases of QSymn\mathrm{QSym}_{n} were applied in the study of qq-analogues of combinatorial Hopf algebras [Li15] and skew variants of quasisymmetric bases [MN15] to take advantage of classical combinatorics in Symn\mathrm{Sym}_{n} concerning Schur functions and Young tableaux. This type of relabelling is also used in [PR21] (there called “shifting”) to simplify arguments relating to the equivariant cohomology of Springer fibers for GLn()GL_{n}(\mathbb{C}).

We are motivated by the utility of the flip-and-reverse perspective to explore and develop further Young analogues of bases of Polyn\mathrm{Poly}_{n} and establish structural results. The Young analogue of the key polynomials is of particular interest and forms a primary focus. In fact, this Young basis has already found application: this variant of the key polynomials is used in [HRS18] to obtain the Hilbert series of a generalization of the coinvariant algebra. In Section 3 we establish a connection with left and right keys of semistandard Young tableaux, proving in Theorem 3.18 that the Young key polynomials are in fact a generating function for semistandard Young tableaux whose left key is greater than a fixed key. We establish an analogous result for the Young analogue of the Demazure atom basis. We also provide a representation-theoretic construction for the Young key polynomials as traces of the action of a diagonal matrix on certain modules. Moreover, in addition to the Young skyline filling model arising from the flip-and-reverse map, we detail several other constructions and interpretations of the Young key polynomials and Young atoms, including divided difference operators, crystal graphs, and compatible sequences.

In Section 4 we provide a new formula for the expansion of a key polynomial into fundamental slide polynomials as well as a new combinatorial construction of the fundamental particle basis for polynomials [Sea20] in terms of flag-compatible sequences. We describe Young analogues for additional families of polynomials, classify which of these Young bases expand positively in one another, and explain different behaviour exhibited by Young and reverse versions including stable limits and embedding into larger polynomial rings. We also completely determine the intersection of the Young and reverse versions of all bases we consider. As a result, we find that when the Young and reverse versions of such a basis of Polyn\mathrm{Poly}_{n} extend a given basis of Symn\mathrm{Sym}_{n} or QSymn\mathrm{QSym}_{n}, the intersection of the Young and reverse basis of Polyn\mathrm{Poly}_{n} is exactly the original basis. For example, we show that the intersection of the Young key polynomials and the key polynomials is exactly the Schur polynomials, and the intersection of the fundamental slide and Young fundamental slide polynomials is exactly the fundamental quasisymmetric polynomials.

Finally in Section 5, we introduce a Young analogue of the famous Schubert polynomials, extending this perspective further. We describe how to generate the Young Schubert polynomials using pipe dreams and divided difference operators and detail how Young Schubert polynomials expand into Young key polynomials. Interestingly, unlike the case for Young analogues of other polynomial bases, there is no basis of Polyn\mathrm{Poly}_{n} consisting of Young Schubert polynomials. We also describe the crystal graph structure for Young Schubert polynomials (analogous to the crystal graph structure for Young key polynomials), as Demazure subcrystals of the crystal on reduced factorizations introduced in [MS16], using methods that were developed on a flipped and reversed version of this crystal in [AS18a].

2. Background

Throughout the following, we denote permutations in one-line notation and allow the transposition sis_{i} to act on the right by swapping the entries in the iith and (i+1)(i+1)th positions. For a weak composition aa, let sort(a){\rm sort}(a) denote the partition obtained by recording the entries of aa in weakly decreasing order. We refer to assignments of integers to diagrams of compositions as tableaux and assignments of integers to diagrams of weak compositions as fillings. For any tableau or filling TT, the weight wt(T){\rm wt}(T) denotes the weak composition whose iith entry is the number of occurrences of ii in TT.

2.1. Quasisymmetric polynomials

Let α\alpha be a composition with at most nn parts. The fundamental quasisymmetric polynomial Fα(x1,,xn)F_{\alpha}(x_{1},\ldots,x_{n}) was originally introduced through the enumeration of PP-partitions [Ges84]. Although there are several different ways to generate the fundamental quasisymmetric polynomials, we describe them as generating functions for certain tableau-like objects which we call fundamental reverse composition tableaux to align with other definitions to follow. Fundamental reverse composition tableaux are those reverse composition tableaux (i.e., entries decrease from left to right in each row) satisfying the additional condition that if i<ji<j, then every entry in row ii is strictly smaller than every entry in row jj. It is straightforward to check that this definition is equivalent to the definition of the fundamental quasisymmetric polynomials as generating functions of ribbon tableaux (see for example [Hua16, Section 4.1]).

In this way, Fα(x1,,xn)F_{\alpha}(x_{1},\ldots,x_{n}) is the sum of all monomials xwt(T)x^{{\rm wt}(T)}, where TT ranges over fundamental reverse composition tableaux of shape α\alpha and largest entry at most nn.

Example 2.1.

We have F13(x1,x2,x3)=x013+x103+x112+x121+x130F_{13}(x_{1},x_{2},x_{3})=x^{013}+x^{103}+x^{112}+x^{121}+x^{130}, as witnessed by the following fundamental reverse composition tableaux.

3 3 3
2
 
3 3 3
1
 
3 3 2
1
 
3 2 2
1
 
2221
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The monomial quasisymmetric polynomial Mα(x1,,xn)M_{\alpha}(x_{1},\ldots,x_{n}) is the generating function of what we call monomial reverse composition tableaux, which are those fundamental reverse composition tableaux in which all entries in the same row are equal.

Example 2.2.

We have M13(x1,x2,x3)=x013+x103+x130M_{13}(x_{1},x_{2},x_{3})=x^{013}+x^{103}+x^{130}, as witnessed by the following monomial reverse composition tableaux.

3 3 3
2
 
3 3 3
1
 
2 2 2
1
 
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One may also define fundamental Young composition tableaux and monomial Young composition tableaux, by replacing the decreasing row condition with the corresponding increasing row condition in the definitions of fundamental (respectively, monomial) reverse composition tableaux. One could then define Young fundamental quasisymmetric polynomials and Young monomial quasisymmetric polynomials to be the generating functions of fundamental (respectively, monomial) Young composition tableaux. In this case, however, the polynomials remain the same.

Proposition 2.3.

The generating function of the fundamental Young composition tableaux of shape α\alpha is Fα(x1,,xn)F_{\alpha}(x_{1},\ldots,x_{n}) and the generating function of the monomial Young composition tableaux of shape α\alpha is Mα(x1,,xn)M_{\alpha}(x_{1},\ldots,x_{n}).

Proof.

By definition, the monomial reverse composition tableaux are exactly the monomial Young composition tableaux. Since every entry in any row of a fundamental reverse composition tableaux is strictly smaller than any entry in the row above, reversing the entries of every row is a weight-preserving bijection between fundamental reverse composition tableaux and fundamental Young composition tableaux of the same shape. ∎

We turn our attention to the quasisymmetric Schur polynomials 𝒮α\operatorname{\mathscr{S}}_{\alpha} and the Young quasisymmetric Schur polynomials 𝒮^α\operatorname{\hat{\mathscr{S}}}_{\alpha}, where we will see a distinction between the reverse and the Young models. To define quasisymmetric Schur polynomials, we first define triples in reverse composition tableaux. These are collections of three boxes in D(α)D(\alpha) with two adjacent in a row and either (Type A) the third box above the right box with the lower row weakly longer, or (Type B) the third box below the left box with the higher row strictly longer. A triple of either type is said to be an inversion triple if it is not the case that zyxz\geq y\geq x.

yzxType Alower row weakly longerzxyType Bhigher row strictly longer\begin{array}[]{l}\vbox{ \halign{&\tableaucell{#}\cr\vbox to9.68747pt{\vfil\hrule width=9.68747pt,height=0.0pt}&\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$y$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}\crcr}}\\ \\ \vbox{ \halign{&\tableaucell{#}\cr\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$z$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}&\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$x$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}\crcr}}\\ \mbox{Type A}\\ \mbox{lower row weakly longer}\end{array}\hskip 29.06241pt\begin{array}[]{l}\hskip 29.06241pt\vbox{ \halign{&\tableaucell{#}\cr\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$z$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}&\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$x$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}\crcr}}\\ \\ \hskip 29.06241pt\vbox{ \halign{&\tableaucell{#}\cr\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$y$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}&\vbox to9.68747pt{\vfil\hrule width=9.68747pt,height=0.0pt}\crcr}}\\ \hskip 29.06241pt\mbox{Type B}\\ \mbox{higher row strictly longer}\end{array}

Figure 1. Triples for reverse composition tableaux.

Define the semistandard reverse composition tableaux RCT(α)\mathrm{RCT}(\alpha) for α\alpha to be the fillings of D(α)D(\alpha) satisfying the following conditions.

  1. (1)

    Entries in each row weakly decrease from left to right.

  2. (2)

    Entries strictly increase from bottom to top in the first column.

  3. (3)

    All type A and type B triples are inversion triples.

Then 𝒮α(x1,,xn)\operatorname{\mathscr{S}}_{\alpha}(x_{1},\ldots,x_{n}) is the generating function of RCT(α)\mathrm{RCT}(\alpha) [HLMvW11].

Example 2.4.

We have 𝒮13(x1,x2,x3)=x013+x022+2x112+x103+x202+x121+x211+x130+x220\operatorname{\mathscr{S}}_{13}(x_{1},x_{2},x_{3})=x^{013}+x^{022}+2x^{112}+x^{103}+x^{202}+x^{121}+x^{211}+x^{130}+x^{220}, as witnessed by the semistandard reverse composition tableaux in Figure 2.

3 3 3
2
 
3 3 2
2
 
3 3 1
2
 
3 3 3
1
 
3 3 2
1
 
3 3 1
1
 
3 2 2
1
 
3 2 1
1
 
2 2 2
1
 
2 2 1
1
 
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Figure 2. The ten elements of RCT(13)\mathrm{RCT}(13) with entries at most 33.

A Young triple is a collection of three boxes with two adjacent in a row such that either (Type I) the third box is below the right box and the higher row is weakly longer, or (Type II) the third box is above the left box and the lower row is strictly longer (Figure 3). A Young triple of either type is said to be a Young inversion triple if it is not the case that xyzx\geq y\geq z.

zxyType Ihigher row weakly longeryzxType IIlower row strictly longer\begin{array}[]{l}\vbox{ \halign{&\tableaucell{#}\cr\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$z$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}&\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$x$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}\\\vbox to9.68747pt{\vfil\hrule width=9.68747pt,height=0.0pt}&\vbox to9.68747pt{\vfil\hrule width=9.68747pt,height=0.0pt}&\vbox to9.68747pt{\vfil\hrule width=9.68747pt,height=0.0pt}\\\vbox to9.68747pt{\vfil\hrule width=9.68747pt,height=0.0pt}&\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$y$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}\crcr}}\\ \mbox{Type I}\\ \mbox{higher row weakly longer}\end{array}\hskip 29.06241pt\begin{array}[]{l}\hskip 29.06241pt\vbox{ \halign{&\tableaucell{#}\cr\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$y$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}\\\vbox to9.68747pt{\vfil\hrule width=9.68747pt,height=0.0pt}&\vbox to9.68747pt{\vfil\hrule width=9.68747pt,height=0.0pt}&\vbox to9.68747pt{\vfil\hrule width=9.68747pt,height=0.0pt}\\\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$z$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}&\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$x$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}\crcr}}\\ \hskip 29.06241pt\mbox{Type II}\\ \mbox{lower row strictly longer}\end{array}
Figure 3. Young triples for Young composition tableaux.

Define the Young semistandard reverse composition tableaux YCT(α)\mathrm{YCT}(\alpha) for α\alpha to be the fillings of D(α)D(\alpha) satisfying the following conditions.

  1. (1)

    Entries in each row weakly increase from left to right.

  2. (2)

    Entries strictly increase from bottom to top in the first column.

  3. (3)

    All type I and type II Young triples are Young inversion triples.

Then the Young quasisymmetric Schur polynomial 𝒮^α(x1,,xn)\operatorname{\hat{\mathscr{S}}}_{\alpha}(x_{1},\ldots,x_{n}) is the generating function of YCT(α)\mathrm{YCT}(\alpha) [LMvW13].

Remark 2.5.

Young quasisymmetric Schur polynomials are most often defined in terms of a single triple condition; e.g [LMvW13], [AHM18]. While this is more compact, it does not extend appropriately to define a Young analogue of key polynomials. The proof that these definitions are equivalent is analogous to the corresponding proof for reverse composition tableaux given in [HLMvW11].

Example 2.6.

We have 𝒮^13(x1,x2,x3)=x130+x121+x112+x103+x013\operatorname{\hat{\mathscr{S}}}_{13}(x_{1},x_{2},x_{3})=x^{130}+x^{121}+x^{112}+x^{103}+x^{013}, as witnessed by the semistandard Young composition tableaux in Figure 4.

2 2 2
1
 
2 2 3
1
 
2 3 3
1
 
3 3 3
1
 
3 3 3
2
 
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Figure 4. The five elements of YCT(13)\mathrm{YCT}(13) with entries at most 33.

Notice that 𝒮^13(x1,x2,x3)𝒮13(x1,x2,x3)\operatorname{\hat{\mathscr{S}}}_{13}(x_{1},x_{2},x_{3})\neq\operatorname{\mathscr{S}}_{13}(x_{1},x_{2},x_{3}); indeed, they have a different number of terms. However, quasisymmetric Schur and Young quasisymmetric Schur polynomials are related by the following formula.

Proposition 2.7.

[LMvW13] Let α\alpha be a composition with at most nn parts. Then

𝒮^α(x1,,xn)=𝒮rev(α)(xn,,x1).\operatorname{\hat{\mathscr{S}}}_{\alpha}(x_{1},\ldots,x_{n})=\operatorname{\mathscr{S}}_{{\rm rev}(\alpha)}(x_{n},\ldots,x_{1}).
Remark 2.8.

As mentioned in the introduction, the flip-and-reverse map on composition tableaux which reverses the order of the rows and exchanges entries i(n+1i)i\leftrightarrow(n+1-i) is a weight-reversing bijection between YCT(α)\mathrm{YCT}(\alpha) and RCT(rev(α))\mathrm{RCT}({\rm rev}(\alpha)), implying Proposition 2.7. In particular, reversing the order of the rows ensures the increasing first column condition is preserved.

To illustrate this, we compute the Young quasisymmetric Schur polynomial 𝒮^31(x1,x2,x3)\operatorname{\hat{\mathscr{S}}}_{31}(x_{1},x_{2},x_{3}); compare this to the computation of 𝒮13(x1,x2,x3)\operatorname{\mathscr{S}}_{13}(x_{1},x_{2},x_{3}) in Example 2.4.

Example 2.9.

We have 𝒮^23(x1,x2,x3)=x310+x220+2x211+x301+x202+x121+x112+x031+x022\operatorname{\hat{\mathscr{S}}}_{23}(x_{1},x_{2},x_{3})=x^{310}+x^{220}+2x^{211}+x^{301}+x^{202}+x^{121}+x^{112}+x^{031}+x^{022}, as witnessed by the semistandard Young composition tableaux in Figure 5.

2
1 1 1
 
2
1 1 2
 
2
1 1 3
 
3
1 1 1
 
3
1 1 2
 
3
1 1 3
 
3
1 2 2
 
3
1 2 3
 
3
2 2 2
 
3
2 2 3
 
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Figure 5. The ten elements of YCT(31)\mathrm{YCT}(31) with entries at most 33.

Notice this involution preserves monomial and fundamental quasisymmetric polynomials: Mα(x1,x2,,xn)=Mrev(α)(xn,,x2,x1)M_{\alpha}(x_{1},x_{2},\ldots,x_{n})=M_{{\rm rev}(\alpha)}(x_{n},\ldots,x_{2},x_{1}) and Fα(x1,x2,,xn)=Frev(α)(xn,,x2,x1)F_{\alpha}(x_{1},x_{2},\ldots,x_{n})=F_{{\rm rev}(\alpha)}(x_{n},\ldots,x_{2},x_{1}).

Proposition 2.10.

[HLMvW11, LMvW13] Quasisymmetric Schur and Young quasisymmetric Schur polynomials expand positively in the fundamental quasisymmetric basis, and

𝒮^α(x1,,xn)=βcβαFβ(x1,,xn)\operatorname{\hat{\mathscr{S}}}_{\alpha}(x_{1},\ldots,x_{n})=\sum_{\beta}c_{\beta}^{\alpha}F_{\beta}(x_{1},\ldots,x_{n})

if and only if

𝒮rev(α)(x1,,xn)=βcβαFrev(β)(x1,,xn).\operatorname{\mathscr{S}}_{{\rm rev}(\alpha)}(x_{1},\ldots,x_{n})=\sum_{\beta}c_{\beta}^{\alpha}F_{{\rm rev}(\beta)}(x_{1},\ldots,x_{n}).

For example, 𝒮^31(x1,x2,x3)=F31(x1,x2,x3)+F22(x1,x2,x3)\operatorname{\hat{\mathscr{S}}}_{31}(x_{1},x_{2},x_{3})=F_{31}(x_{1},x_{2},x_{3})+F_{22}(x_{1},x_{2},x_{3}), whereas 𝒮13(x1,x2,x3)=F13(x1,x2,x3)+F22(x1,x2,x3).\operatorname{\mathscr{S}}_{13}(x_{1},x_{2},x_{3})=F_{13}(x_{1},x_{2},x_{3})+F_{22}(x_{1},x_{2},x_{3}).

A remarkable property of the quasisymmetric Schur and Young quasisymmetric Schur polynomials is that they both positively refine Schur polynomials:

Proposition 2.11.

[LMvW13]

sλ(x1,,xn)=sort(α)=λ𝒮α(x1,,xn)=sort(α)=λ𝒮^α(x1,,xn)s_{\lambda}(x_{1},\ldots,x_{n})=\sum_{{\rm sort}(\alpha)=\lambda}\operatorname{\mathscr{S}}_{\alpha}(x_{1},\ldots,x_{n})=\sum_{{\rm sort}(\alpha)=\lambda}\operatorname{\hat{\mathscr{S}}}_{\alpha}(x_{1},\ldots,x_{n})
Remark 2.12.

As noted in the introduction, Schur polynomials may be described in terms of either decreasing or increasing semistandard tableaux. Therefore Schur polynomials and “Young Schur polynomials” are the same (provided we consider a partition and its reversal to be the same), so from this perspective it makes sense that Schur polynomials expand positively into both the quasisymmetric Schur and Young quasisymmetric Schur bases. Similarly, the fact that both quasisymmetric Schur and Young quasisymmetric Schur polynomials expand positively in fundamental quasisymmetric polynomials (Proposition 2.10) makes sense due to the fact that fundamental quasisymmetric polynomials may also be described in terms of either increasing or decreasing tableaux (Proposition 2.3), and thus are the same as “Young fundamental quasisymmetric polynomials”.

Typically a Young quasisymmetric Schur polynomial is not equal to any quasisymmetric Schur polynomial. However, we can classify their coincidences. We delay the proof to the appendix.

Theorem 2.13.

𝒮^α(x1,,xn)=𝒮β(x1,,xn)\operatorname{\hat{\mathscr{S}}}_{\alpha}(x_{1},\ldots,x_{n})=\operatorname{\mathscr{S}}_{\beta}(x_{1},\ldots,x_{n}) if and only if α=β\alpha=\beta and either α\alpha has all parts the same, or all parts of α\alpha are 11 or 22, or n=(α)n=\ell(\alpha) and consecutive parts of α\alpha differ by at most 11.

2.2. Key polynomials and Demazure atoms

We now shift our attention to the ring Polyn=[x1,,xn]\mathrm{Poly}_{n}=\mathbb{Z}[x_{1},\ldots,x_{n}] of all polynomials in nn variables. This ring possesses a variety of bases important in geometry and representation theory. A principal example is the basis of key polynomials, which are characters of (type A) Demazure modules [Dem74, LS90, RS95] and which also arise as specializations of nonsymmetric Macdonald polynomials. Closely related is the basis of Demazure atoms, originally introduced as standard bases in [LS90]. Demazure atoms were shown in [Mas09] to also be a specialization of nonsymmetric Macdonald polynomials. They are equal to the smallest non-intersecting pieces of type AA Demazure characters and can be obtained through a truncated application of divided difference operators. Intuitively, one can build the Demazure atoms by starting with a monomial and partially symmetrizing, keeping only the monomials not appearing in the previous iteration of this process.

2.2.1. Semi-skyline fillings

Both key polynomials and Demazure atoms are defined in terms of reverse fillings that are often referred to as semi-skyline fillings. To define the key polynomial corresponding to a weak composition aa of length nn, first note that the definition of type A and B triples extends verbatim from composition diagrams to weak composition diagrams. We need to include a basement column, an extra 0th column in the diagram: for our purposes the basement entry of row ii is n+1in+1-i. Basement entries do not contribute to the weight of a filling. Define the key fillings KSSF(a)\mathrm{KSSF}(a) for aa to be the fillings of D(rev(a))D({\rm rev}(a)) (note the reversal) satisfying the following conditions.

  1. (1)

    Entries in each row, including basement entries, weakly decrease from left to right.

  2. (2)

    Entries do not repeat in any column.

  3. (3)

    All type A and type B triples, including triples containing basement entries, are inversion triples.

We use the following as definitional for key polynomials.

Theorem 2.14.

[HHL08, Mas09], Let aa be a weak composition of length nn. Then

κa=TKSSF(a)xwt(T),\operatorname{\kappa}_{a}=\sum_{T\in\mathrm{KSSF}(a)}x^{{\rm wt}(T)},

where only the non-basement entries contribute to the weight.

For example, we have κ032=x032+x122+x212+x302+x311+x320+x131+x221+x230\operatorname{\kappa}_{032}=x^{032}+x^{122}+x^{212}+x^{302}+x^{311}+x^{320}+x^{131}+x^{221}+x^{230}, which is computed using the elements of KSSF(032)\mathrm{KSSF}(032) shown in Figure 6 below.

1
2 2 2 2
3 3 3
 
1
2 2 2 1
3 3 3
 
1
2 2 1 1
3 3 3
 
1
2 1 1 1
3 3 3
 
1
2 1 1 1
3 3 2
 
1
2 1 1 1
3 2 2
 
1
2 2 2 2
3 3 1
 
1
2 2 1 1
3 3 2
 
1
2 2 2 2
3 1 1
 
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Figure 6. The 9 key fillings of shape 032032. (Basement entries in bold.)

The definition of the Demazure atoms in terms of semi-skyline fillings comes from specializing the diagram fillings used to generate the nonsymmetric Macdonald polynomials [HHL08]. Define the atom fillings ASSF(a)\mathrm{ASSF}(a) for aa to be the fillings of D(a)D(a) (no basement) satisfying the following conditions.

  1. (1)

    Entries weakly decrease from left to right in each row.

  2. (2)

    Entries do not repeat in any column.

  3. (3)

    The first entry of each row is equal to its row index.

  4. (4)

    All type A and type B triples are inversion triples.

We use the following as definitional for Demazure atoms.

Theorem 2.15.

[Mas09] Let aa be a weak composition of length nn. Then

𝒜a=TASSF(a)xwt(T).\operatorname{\mathcal{A}}_{a}=\sum_{T\in\mathrm{ASSF}(a)}x^{{\rm wt}(T)}.

2.2.2. Left and right keys

The eponymous formula for the key polynomial κa\operatorname{\kappa}_{a} is given in terms of right keys. A semistandard Young tableau (or SSYT\mathrm{SSYT}) TT is a tableau of partition shape such that entries weakly increase along rows and strictly increase up columns. For a partition λ\lambda, let SSYT(λ)\mathrm{SSYT}(\lambda) denote the set of all SSYT\mathrm{SSYT} of shape λ\lambda, and SSYTn(λ)\mathrm{SSYT}_{n}(\lambda) the subset of SSYT(λ)\mathrm{SSYT}(\lambda) whose entries are at most nn. A semistandard Young tableau TT is a key if the entries appearing in the (i+1)(i+1)th column of TT are a subset of the entries appearing in the ithi^{th} column of TT, for all ii. For a weak composition aa, define key(a)\mathrm{key}(a) to be the unique key of weight aa. For any semistandard Young tableau TT, there are two keys of the same shape as TT associated to TT, called the right key of TT, denoted K+(T)K_{+}(T), and the left key of TT, denoted K(T)K_{-}(T)

We now describe procedures for computing right and left keys, which will be illustrated in Example 2.17 below. There are several different methods for computing keys (see, for example [Mas09][Wil13]) but we use the classical method presented in [RS95] as it involves several tools we will need later. Two words 𝐛{\bf b} and 𝐜{\bf c} in {1,2,n}\{1,2,\ldots n\} are said to be Knuth-equivalent, written 𝐛𝐜{\bf b}\sim{\bf c}, if one can be obtained from the other by a series of the following local moves:

𝐝xzy𝐞𝐝zxy𝐞\displaystyle{\bf d}xzy{\bf e}\sim{\bf d}zxy{\bf e}  for xy<z\displaystyle\quad\mbox{ for }\quad x\leq y<z
𝐝yxz𝐞𝐝yzx𝐞\displaystyle{\bf d}yxz{\bf e}\sim{\bf d}yzx{\bf e}  for x<yz\displaystyle\quad\mbox{ for }\quad x<y\leq z

for words 𝐝{\bf d} and 𝐞{\bf e} and letters x,y,zx,y,z.

Define the column word factorization of a word vv to be the decomposition of vv into subwords v=v(1)v(2)v=v^{(1)}v^{(2)}\cdots by starting a new subword between every weak ascent. Then the column form of vv (denoted colform(v)\operatorname{colform}(v)) is the composition whose parts are the lengths of the subwords appearing in the column word factorization. Let λ\lambda be the shape of the SSYT\mathrm{SSYT} obtained when Schensted insertion (see, e.g., [Ful97, Sag13, Sta99]) is applied to vv. The word vv is said to be column-frank if colform(v)\operatorname{colform}(v) is a rearrangement of the nonzero parts of λ\lambda^{\prime}, where λ\lambda^{\prime} denotes the conjugate shape of λ\lambda obtained by reflecting the diagram of λ\lambda across the line y=xy=x. Let TSSYT(λ)T\in\mathrm{SSYT}(\lambda). Then the right key (resp. left key) of TT is the key of shape λ\lambda whose jthj^{th} column is equal to the last (resp. first) subword in any column-frank word which is Knuth equivalent to the column word col(T)\operatorname{col}(T) of TT (obtained by reading the entries of TT down columns from left to right) and whose last (resp. first) subword has length λj\lambda_{j}^{\prime}.

Notice the difference in the construction of left and right keys. The weight of the left key is usually not a reversal of the weight of the right key; the subtle connection between left and right keys is explored in Section 3.3, wherein we also define polynomials naturally associated to left keys.

Theorem 2.16.

[LS90, RS95] Let aa be a weak composition of length nn. Then

κa=TSSYTn(sort(a))K+(T)key(a)xwt(T),\operatorname{\kappa}_{a}=\sum_{\begin{subarray}{c}T\in\mathrm{SSYT}_{n}({\rm sort}(a))\\ K_{+}(T)\leq\mathrm{key}(a)\end{subarray}}x^{{\rm wt}(T)},

where K+(T)key(a)K_{+}(T)\leq\mathrm{key}(a) if each entry of K+(T)K_{+}(T) is weakly smaller than the corresponding entry of key(a)\mathrm{key}(a).

Example 2.17.

Let a=032a=032. Then key(a)=33222\mathrm{key}(a)=\vbox{ \halign{&\tableaucell{#}\cr\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$3$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}&\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$3$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}\\\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$2$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}&\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$2$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}&\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$2$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}\crcr}}, which is a tableau of shape λ=32\lambda=32. The nine tableaux whose right keys are smaller than or equal to key(a)\mathrm{key}(a) are

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To illustrate the process of finding right (and left) keys, let TT be the last tableau in the list above. Then col(T)=21322\operatorname{col}(T)=21322. The words that are Knuth-equivalent to 2132221322 (listed with vertical bars indicating the column word factorizations) are {21|32|2, 21|2|32, 2|21|32, 2|2|31|2}\{21|32|2,\;21|2|32,\;2|21|32,\;2|2|31|2\}. The column form of the first three words is a rearrangement of 221221, the shape of λ\lambda^{\prime}, so these three words are column-frank. The fourth is not column-frank so we ignore it. Looking at the rightmost subword in each column-frank word, the first of these words tells us that the column of K+(T)K_{+}(T) of length 11 consists of a single 22, and the second (or third) word tells us that the columns of K+(T)K_{+}(T) of length 22 each contain a 22 and a 33.

Thus K+(T)=33222K_{+}(T)=\vbox{ \halign{&\tableaucell{#}\cr\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$3$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}&\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$3$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}\\\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$2$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}&\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$2$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}&\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$2$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}\crcr}}. Similarly, via leftmost subwords, we obtain K(T)=22112K_{-}(T)=\vbox{ \halign{&\tableaucell{#}\cr\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$2$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}&\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$2$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}\\\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$1$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}&\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$1$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}&\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$2$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}\crcr}}.

One may also use right keys to define the Demazure atoms. Given a weak composition aa of length nn, the Demazure atom 𝒜a\operatorname{\mathcal{A}}_{a} can also be given by

(2.1) 𝒜a=TSSYTn(λ(a))K+(T)=key(a)xwt(T).\displaystyle\operatorname{\mathcal{A}}_{a}=\sum_{\begin{subarray}{c}T\in\mathrm{SSYT}_{n}(\lambda(a))\\ K_{+}(T)=\mathrm{key}(a)\end{subarray}}x^{{\rm wt}(T)}.

From this construction and Theorem 2.16, it is apparent that key polynomials expand positively in Demazure atoms. In particular,

(2.2) κa=ba𝒜b,\displaystyle\operatorname{\kappa}_{a}=\sum_{b\leq a}\operatorname{\mathcal{A}}_{b},

where bab\leq a if and only if sort(b)=sort(a){\rm sort}(b)={\rm sort}(a) and the permutation ww such that w(sort(b))=bw({\rm sort}(b))=b is less than or equal to the permutation vv such that v(sort(a))=av({\rm sort}(a))=a in the Bruhat order.

2.2.3. Divided differences and crystal graphs

Key polynomials can be defined in terms of divided difference operators. Given a positive integer ii, where 1i<n1\leq i<n, define an operator i\partial_{i} on [x1,,xn]\mathbb{Z}[x_{1},\ldots,x_{n}] by

i(f)=fsi(f)xixi+1\partial_{i}(f)=\frac{f-s_{i}(f)}{x_{i}-x_{i+1}}

where sis_{i} exchanges xix_{i} and xi+1x_{i+1}. Now define another operator πi\pi_{i} on [x1,,xn]\mathbb{Z}[x_{1},\ldots,x_{n}] by

πi(f)=i(xif).\pi_{i}(f)=\partial_{i}(x_{i}f).

For a permutation ww, define πw=πi1πir\pi_{w}=\pi_{i_{1}}\cdots\pi_{i_{r}}, where si1sirs_{i_{1}}\cdots s_{i_{r}} is any reduced word for ww. (This definition is independent of the choice of reduced word because the πi\pi_{i} satisfy the commutation and braid relations for the symmetric group.) Recall that sort(a){\rm sort}(a) is the rearrangement of the entries of aa into decreasing order. For a weak composition aa let waw_{a} be the minimal length permutation that sends aa to sort(a){\rm sort}(a) acting on the right. Then the key polynomial is given by

κa=πwaxsort(a).\operatorname{\kappa}_{a}=\pi_{w_{a}}x^{{\rm sort}(a)}.
Example 2.18.

Let a=032a=032. Then the minimal length permutation taking aa to sort(a)=320{\rm sort}(a)=320 is s1s2s_{1}s_{2}. We compute

π1π2(x13x22)\displaystyle\pi_{1}\pi_{2}(x_{1}^{3}x_{2}^{2}) =π1x13x23x13x33x2x3\displaystyle=\pi_{1}\frac{x_{1}^{3}x_{2}^{3}-x_{1}^{3}x_{3}^{3}}{x_{2}-x_{3}}
=π1(x13x22+x13x2x3+x13x32)\displaystyle=\pi_{1}(x_{1}^{3}x_{2}^{2}+x_{1}^{3}x_{2}x_{3}+x_{1}^{3}x_{3}^{2})
=(x14x22x12x24)+(x14x2x3x1x24x3)+(x14x32x24x32)x1x2\displaystyle=\frac{(x_{1}^{4}x_{2}^{2}-x_{1}^{2}x_{2}^{4})+(x_{1}^{4}x_{2}x_{3}-x_{1}x_{2}^{4}x_{3})+(x_{1}^{4}x_{3}^{2}-x_{2}^{4}x_{3}^{2})}{x_{1}-x_{2}}
=x13x22+x12x23+x13x2x3+x12x22x3+x1x23x3+x13x32+x12x2x32+x1x22x32+x23x32\displaystyle=x_{1}^{3}x_{2}^{2}+x_{1}^{2}x_{2}^{3}+x_{1}^{3}x_{2}x_{3}+x_{1}^{2}x_{2}^{2}x_{3}+x_{1}x_{2}^{3}x_{3}+x_{1}^{3}x_{3}^{2}+x_{1}^{2}x_{2}x_{3}^{2}+x_{1}x_{2}^{2}x_{3}^{2}+x_{2}^{3}x_{3}^{2}
=κ032.\displaystyle=\operatorname{\kappa}_{032}.

Demazure atoms can also be described in terms of divided difference operators. In particular, let πi¯=πi1\overline{\pi_{i}}=\pi_{i}-1. Then (see  [Mas09])

𝒜a=π¯waxsort(a).\operatorname{\mathcal{A}}_{a}=\overline{\pi}_{w_{a}}x^{{\rm sort}(a)}.

The action of the divided difference operators can be realised in terms of Demazure crystals. A crystal graph is a directed and colored graph whose edges are defined by Kashiwara operators [Kas91, Kas93, Kas95] eie_{i} and fif_{i}. See [HK02] for a detailed introduction to the theory of quantum groups and crystal bases and [BS17] for a more combinatorial exploration of crystals.

For a partition λ\lambda, the type AnA_{n} highest weight crystal B(λ)B(\lambda) of highest weight λ\lambda has vertices indexed by SSYTn(λ)\mathrm{SSYT}_{n}(\lambda). The character of B(λ)B(\lambda) is

ch(B(λ))=TB(λ)xwt(T),\operatorname{ch}(B(\lambda))=\sum_{T\in B(\lambda)}x^{{\rm wt}(T)},

which is equal to the Schur polynomial sλ(x1,,xn)s_{\lambda}(x_{1},\ldots,x_{n}), reflecting the fact that Schur polynomials are characters for irreducible highest weight modules for GLnGL_{n}. See Figure 7 below for B(21)B(21) when n=3n=3, in which the arrows index the Kashiwara operators f1f_{1} and f2f_{2}. Precise definitions of the fif_{i} can be found in e.g. [BS17]; in particular we note that fi(b)=0f_{i}(b)=0 if there is no ii-arrow emanating from vertex bb, and the eie_{i} are defined by ei(b)=be_{i}(b)=b^{\prime} if fi(b)=bf_{i}(b^{\prime})=b, and ei(b)=0e_{i}(b)=0 otherwise.

221111
331111
221122
221133
331122
331133
332222
332233
1111111122222222
Figure 7. Crystal graph B(21)B(21) for n=3n=3.

A Demazure crystal is a subset of B(λ)B(\lambda) whose character is a key polynomial [Lit95, Kas93], obtained by a truncated action of the Kashiwara operators. Specifically, given a subset XX of B(λ)B(\lambda), define operators 𝔇i\mathfrak{D}_{i} for 1i<n1\leq i<n by

𝔇iX={bB(λ)|eir(b)X for some r0}.\mathfrak{D}_{i}X=\{b\in B(\lambda)|e_{i}^{r}(b)\in X\textrm{ for some }r\geq 0\}.

Given a permutation ww with reduced word w=si1si2sikw=s_{i_{1}}s_{i_{2}}\cdots s_{i_{k}}, define

Bw(λ)=𝔇i1𝔇i2𝔇ik{uλ},B_{w}(\lambda)=\mathfrak{D}_{i_{1}}\mathfrak{D}_{i_{2}}\cdots\mathfrak{D}_{i_{k}}\{u_{\lambda}\},

where uλu_{\lambda} is the highest weight element in B(λ)B(\lambda), i.e., ei(uλ)=0e_{i}(u_{\lambda})=0 for all 1i<n1\leq i<n. If b,bBw(λ)B(λ)b,b^{\prime}\in B_{w}(\lambda)\subseteq B(\lambda) and fi(b)=bf_{i}(b)=b^{\prime} in B(λ)B(\lambda), then the crystal operator fif_{i} is also defined in Bw(λ)B_{w}(\lambda). The character of a Demazure crystal Bw(λ)B_{w}(\lambda) is defined as

chBw(λ)=bBw(λ)x1wt(b)1xnwt(b)n,\operatorname{ch}B_{w}(\lambda)=\sum_{b\in B_{w}(\lambda)}x_{1}^{{\rm wt}(b)_{1}}\cdots x_{n}^{{\rm wt}(b)_{n}},

which is equal to κa\operatorname{\kappa}_{a} when ww is of shortest length such that w(a)=λw(a)=\lambda [Lit95, Kas93]. The repeated actions of the 𝔇i\mathfrak{D}_{i} starting with uλu^{\lambda} precisely mirrors the repeated action of the divided difference operators πi\pi_{i} starting with the monomial xλx^{\lambda}.

Example 2.19.

Let a=102a=102. Then the shortest length ww such that w(a)=sort(a)=210w(a)={\rm sort}(a)=210 is w=s2s1w=s_{2}s_{1}. Therefore, the crystal graph for κ102\operatorname{\kappa}_{102} is the subgraph of B(21)B(21) consisting of all vertices that can be obtained from the highest weight 22 11 11 by first applying a sequence of f1f_{1}’s and then a sequence of f2f_{2}’s. In Figure 7, these are the tableaux of weight 210210, 201201, 120120, 111111 (the leftmost such) and 102102. Hence κ102=x12x2+x1x22+x12x3+x1x2x3+x1x32\operatorname{\kappa}_{102}=x_{1}^{2}x_{2}+x_{1}x_{2}^{2}+x_{1}^{2}x_{3}+x_{1}x_{2}x_{3}+x_{1}x_{3}^{2}.

2.2.4. Compatible Sequences

Key polynomials can also be constructed using compatible sequences as follows. Let 𝐛=b1b2bp{\bf b}=b_{1}b_{2}\cdots b_{p} be a word in the alphabet {1,2,n}\{1,2,\ldots n\}. A word 𝐰=w1w2wp{\bf w}=w_{1}w_{2}\cdots w_{p} is 𝐛{\bf b}-compatible if

  1. (1)

    1w1w2wpn1\leq w_{1}\leq w_{2}\leq\cdots\leq w_{p}\leq n,

  2. (2)

    wk<wk+1w_{k}<w_{k+1} whenever bk<bk+1,b_{k}<b_{k+1}, for all 1k<p1\leq k<p, and

  3. (3)

    wkbkw_{k}\leq b_{k} for all 1kp1\leq k\leq p (flag condition).

Theorem 2.20.

[RS95] Let aa be a weak composition of length nn. Then

κa=rev(b)col(key(a)),w is b-compatiblexcomp(w),\operatorname{\kappa}_{a}=\sum_{{\rm rev}(b)\sim\operatorname{col}(\mathrm{key}(a)),\;w\textrm{ is $b$-compatible}}x^{\mathrm{comp}(w)},

where comp(w)\mathrm{comp}(w) is the weak composition whose ithi^{th} entry counts the incidences of ii in ww.

Example 2.21.

Let a=032a=032. We have key(032)=33222{\rm key}(032)=\vbox{ \halign{&\tableaucell{#}\cr\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$3$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}&\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$3$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}\\\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$2$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}&\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$2$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}&\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$2$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}\crcr}}, and col(key(032))=32322\operatorname{col}({\rm key}(032))=32322. The set of words Knuth-equivalent to 3232232322 is {32322,33222,32232,23232,23322}\{32322,33222,32232,23232,23322\}. Reversing these gives the set {22323,22233,23223,23232,22332}.\{22323,22233,23223,23232,22332\}. We compute the set of compatible sequences for each of these:

Word Compatible sequences
22323 11223
22233 22233 12233 11233 11133 11123 11122
23223 12223
23232
22332 11222
Figure 8. Compatible sequences.

Observe there are 99 compatible sequences, each having the weight comp(w)\mathrm{comp}(w) of a monomial of κ032\operatorname{\kappa}_{032}. In Proposition 4.1, we interpret the fundamental slide expansion of a key polynomial in terms of Knuth equivalence classes.

3. Young key polynomials

We now introduce the Young key polynomial basis for polynomials. This basis has proved useful in computing the Hilbert series of a generalization of the coinvariant algebra, specifically, in constructing a Gröbner basis for the ideal In,k=x1k,x2k,,xnk,en,en1,,enk+1I_{n,k}=\langle x_{1}^{k},x_{2}^{k},\ldots,x_{n}^{k},e_{n},e_{n-1},\ldots,e_{n-k+1}\rangle [HRS18]. However, the combinatorial and representation-theoretic properties of the Young key polynomials have not, to our knowledge, been explored previously, nor has the connection to the overall flip-and-reverse perspective. We begin by providing a combinatorial description of the Young key polynomial basis analogous to that of the Young version of the quasisymmetric Schur polynomials.

Note that the definition of Young triples extends verbatim to weak composition diagrams. As in the definition of key polynomials, we append a basement column to diagrams. Given a weak composition aa of length nn, define the Young key fillings YKSSF(a)\mathrm{YKSSF}(a) for aa to be the fillings of D(rev(a))D({\rm rev}(a)) (note the reversal) with entries from {1,,n}\{1,\ldots,n\} satisfying the following conditions.

  1. (1)

    Entries in each row, including basement entries, weakly increase from left to right.

  2. (2)

    Entries do not repeat in any column.

  3. (3)

    All type I and type II Young triples, including triples using basement entries, are Young inversion triples.

Define the Young key polynomial κ^a\operatorname{\hat{\kappa}}_{a} by

κ^a=TYKSSF(a)xwt(T),\operatorname{\hat{\kappa}}_{a}=\sum_{T\in\mathrm{YKSSF}(a)}x^{{\rm wt}(T)},

where only the non-basement entries contribute to the weight.

For example, we have κ^230=x230+x221+x212+x203+x113+x023+x131+x122+x032\operatorname{\hat{\kappa}}_{230}=x^{230}+x^{221}+x^{212}+x^{203}+x^{113}+x^{023}+x^{131}+x^{122}+x^{032}, which is computed by the elements of YKSSF(230)\mathrm{YKSSF}(230) shown in Figure 9.

1 1 1
2 2 2 2
3
 
1 1 1
2 2 2 3
3
 
1 1 1
2 2 3 3
3
 
1 1 1
2 3 3 3
3
 
1 1 2
2 3 3 3
3
 
1 2 2
2 3 3 3
3
 
1 1 3
2 2 2 2
3
 
1 1 2
2 2 3 3
3
 
1 3 3
2 2 2 2
3
 
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Figure 9. The 9 Young key fillings of shape 230230. (Basement entries are in bold.)

Note that the definition immediately implies that

(3.1) κ^a(x1,x2,,xn)=κrev(a)(xn,xn1,,x1).\operatorname{\hat{\kappa}}_{a}(x_{1},x_{2},\ldots,x_{n})=\operatorname{\kappa}_{{\rm rev}(a)}(x_{n},x_{n-1},\ldots,x_{1}).
Proposition 3.1.

The Young key polynomials are a basis for Polyn\mathrm{Poly}_{n}, containing the Schur polynomials. In particular, if aa is decreasing then

κ^a=sλ(x1,,xn),\operatorname{\hat{\kappa}}_{a}=s_{\lambda}(x_{1},\ldots,x_{n}),

where λ\lambda is aa with trailing zeros removed.

Proof.

The Young key polynomials are equinumerous with the key polynomials. Any polynomial can be expressed as a linear combination of key polynomials (since key polynomials are a basis of Polyn\mathrm{Poly}_{n}), and thus as a linear combination of Young key polynomials by (3.1). Hence the Young key polynomials are a basis of Polyn\mathrm{Poly}_{n}.

We have κrev(a)=sa\operatorname{\kappa}_{{\rm rev}(a)}=s_{a} [Mac91], hence κ^a=sa=sλ\operatorname{\hat{\kappa}}_{a}=s_{a}=s_{\lambda} by (3.1) and because Schur polynomials are symmetric, hence invariant under exchanging variables. ∎

In this way, both the key and Young key polynomials extend the Schur polynomials to Polyn\mathrm{Poly}_{n}. This is in fact their only coincidence.

Theorem 3.2.

The polynomials that are both key polynomials and Young key polynomials are exactly the Schur polynomials.

Proof.

Suppose sλ(x1,,xn)s_{\lambda}(x_{1},\ldots,x_{n}) is a Schur polynomial in nn variables. Then

sλ(x1,,xn)=κ0n(λ)×rev(λ)=κ^λ×0n(λ),s_{\lambda}(x_{1},\ldots,x_{n})=\operatorname{\kappa}_{0^{n-\ell(\lambda)}\times{\rm rev}(\lambda)}=\operatorname{\hat{\kappa}}_{\lambda\times 0^{n-\ell(\lambda)}},

where 0m×b0^{m}\times b (respectively, b×0mb\times 0^{m}) denotes bb with mm zeros prepended (respectively, appended).

For the converse, note that for any weak composition aa, the key polynomial κa\operatorname{\kappa}_{a} has the monomial xsort(a)x^{{\rm sort}(a)} as a term; this follows from the divided difference definition. But the only Young key polynomial containing xsort(a)x^{{\rm sort}(a)} as a term is κ^sort(a)\operatorname{\hat{\kappa}}_{{\rm sort}(a)} itself, which is a Schur polynomial. So if κa\operatorname{\kappa}_{a} is not a Schur polynomial it cannot be equal to any Young key polynomial. ∎

We also define a Young analogue of the Demazure atoms. Let aa be a weak composition of length nn. Define the Young atom fillings YASSF(a)\mathrm{YASSF}(a) for aa to be the fillings of D(a)D(a) (no basement) with entries from {1,,n}\{1,\ldots,n\} satisfying the following conditions.

  1. (1)

    Entries weakly increase from left to right in each row.

  2. (2)

    Entries do not repeat in any column.

  3. (3)

    All type I and type II Young triples are Young inversion triples.

  4. (4)

    The first entry of each row is equal to its row index.

Define the Young atom 𝒜^a\operatorname{\hat{\mathcal{A}}}_{a} by

𝒜^a=TYASSF(a)xwt(T).\operatorname{\hat{\mathcal{A}}}_{a}=\sum_{T\in\mathrm{YASSF}(a)}x^{{\rm wt}(T)}.

The definition immediately implies that 𝒜^a(x1,x2,,xn)=𝒜rev(a)(xn,xn1,,x1).\operatorname{\hat{\mathcal{A}}}_{a}(x_{1},x_{2},\ldots,x_{n})=\operatorname{\mathcal{A}}_{{\rm rev}(a)}(x_{n},x_{n-1},\ldots,x_{1}). Similar to Proposition 3.1, the Young atoms form a basis of Polyn\mathrm{Poly}_{n}. We can establish the coincidences between Demazure atoms and Young atoms, as we did in Theorem 3.2 for keys and Young keys. Note the condition for coincidence is less restrictive than that for coincidence of quasisymmetric Schur and Young quasisymmetric Schur polynomials (Theorem 2.13), due to elements of YASSF(a)\mathrm{YASSF}(a) and ASSF(a)\mathrm{ASSF}(a) necessarily having identical first column.

Theorem 3.3.

The polynomials that are both Demazure atoms and Young atoms are precisely the 𝒜^a\operatorname{\hat{\mathcal{A}}}_{a} such that |aiai+1|1|a_{i}-a_{i+1}|\leq 1 for all 1i<n1\leq i<n.

Proof.

First we show that if 𝒜^a=𝒜b\operatorname{\hat{\mathcal{A}}}_{a}=\operatorname{\mathcal{A}}_{b} then a=ba=b. Suppose max(a)>max(b)\max(a)>\max(b), where max(a)\max(a) is the largest part of aa. Then since entries cannot repeat in any column for either YASSF\mathrm{YASSF} or ASSF\mathrm{ASSF}, 𝒜^a\operatorname{\hat{\mathcal{A}}}_{a} has terms where some xix_{i} has degree max(a)\max(a), but 𝒜b\operatorname{\mathcal{A}}_{b} cannot have any such term. Hence if 𝒜^a=𝒜b\operatorname{\hat{\mathcal{A}}}_{a}=\operatorname{\mathcal{A}}_{b}, the longest row(s) in D(a)D(a) and D(b)D(b) must have the same length. By a similar argument, the next-longest rows must then have the same length, etc. Thus if 𝒜^a=𝒜b\operatorname{\hat{\mathcal{A}}}_{a}=\operatorname{\mathcal{A}}_{b}, then bb must be a rearrangement of aa.

Now suppose bb rearranges aa. Let TYASSF(a)T\in\mathrm{YASSF}(a) be such that all entries in the jjth row (for each jj) are equal to jj, and suppose there exists SASSF(b)S\in\mathrm{ASSF}(b) with the same weight as TT. By definition, the first entry in each row jj of SS is jj. Because the rows of bb rearrange those of aa, the number of boxes in each column of D(b)D(b) is the same as that for each column of D(a)D(a). It follows that the set of entries in each column of SS must be the same as that in the corresponding column of TT, since TT has aja_{j} instances of each entry jj, and entries cannot repeat in any column of TT or SS.

Now consider the entries in the second column of SS, which are a subset of the entries in the first column for both SS and TT. None of these entries can go in a row above the row that contains that entry in the first column, else the two copies of that entry must violate one of the triple conditions. Nor can they go in a row below, since entries must decrease along each row. So each entry must go immediately adjacent to the same entry in the first column of SS. Continuing thus, we obtain S=TS=T, so in particular a=ba=b.

Now suppose aiai+12a_{i}-a_{i+1}\geq 2 for some ii. Let TYASSF(a)T\in\mathrm{YASSF}(a) be such that all entries in each row jj are jj, and let TT^{\prime} be obtained by changing the rightmost ii in TT to i+1i+1. Since aiai+12a_{i}-a_{i+1}\geq 2, this new i+1i+1 is not in the first column, and is at least two columns to the right of any other i+1i+1, so no YASSF\mathrm{YASSF} properties are affected by this change and TYASSF(a)T^{\prime}\in\mathrm{YASSF}(a). But there is no SASSF(a)S\in\mathrm{ASSF}(a) with weight equal to TT^{\prime}: in rows i+1i+1 and above, entries in SS must agree with entries in TT^{\prime}, and then there is nowhere the new i+1i+1 could be placed in SS. Hence 𝒜^a𝒜a\operatorname{\hat{\mathcal{A}}}_{a}\neq\operatorname{\mathcal{A}}_{a}. A similar argument shows that if ai+1ai2a_{i+1}-a_{i}\geq 2, then 𝒜a𝒜^a\operatorname{\mathcal{A}}_{a}\neq\operatorname{\hat{\mathcal{A}}}_{a}.

Conversely, it is straightforward to observe that if |aiai+1|1|a_{i}-a_{i+1}|\leq 1 for all 1i<n1\leq i<n, then both 𝒜^a\operatorname{\hat{\mathcal{A}}}_{a} and 𝒜a\operatorname{\mathcal{A}}_{a} are equal to the single monomial xax^{a}. ∎

3.1. Compatible sequences

The Young key polynomials may also be described in terms of compatible sequences. For a word ww in {1,2,,n}\{1,2,\ldots,n\} define the flip of ww to be the word f(w)f(w) in {1,2,,n}\{1,2,\ldots,n\} obtained by replacing each entry wiw_{i} with n+1win+1-w_{i}. Also define the flip-reverse of ww, denoted frev(w)\mathrm{frev}(w), to be the word f(rev(w))f({\rm rev}(w)), or equivalently rev(f(w)){\rm rev}(f(w)).

Example 3.4.

If n=6n=6 and w=2446154w=2446154, then f(w)=5331623f(w)=5331623 and frev(w)=3261335\mathrm{frev}(w)=3261335.

Let TT be an SSYT\mathrm{SSYT}. Define the right-to-left column reading word colR(T)\operatorname{col}_{R}(T) to be the word obtained by reading the entries in each column of TT from top to bottom starting with the rightmost column and moving from right to left.

Lemma 3.5.

Let aa be a weak composition. Then frev(col(key(a)))=colR(key(rev(a)))\mathrm{frev}(\operatorname{col}(\mathrm{key}(a)))=\operatorname{col}_{R}(\mathrm{key}({\rm rev}(a))).

Proof.

First of all, key(a)\mathrm{key}(a) and key(rev(a))\mathrm{key}({\rm rev}(a)) have the same shape. To see this, note that the height of the ithi^{th} column of key(a)\mathrm{key}(a) is equal to the number of entries aja_{j} in aa such that ajia_{j}\geq i. This number is the same for aa and rev(a){\rm rev}(a).

This also shows that for any given column, the entries of that column in key(a)\mathrm{key}(a) are the flips of the entries of that column in key(rev(a))\mathrm{key}({\rm rev}(a)). Hence when the word for key(rev(a))\mathrm{key}({\rm rev}(a)) is reversed, the column breaks line up and the word in each column is the flip-reverse of the word in that column of key(a)\mathrm{key}(a). The statement follows. ∎

Example 3.6.

Let a=(2,4,0,3)a=(2,4,0,3). We have

key(a)=442241122 and key(rev(a))=443331113.\mathrm{key}(a)=\vbox{ \halign{&\tableaucell{#}\cr\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$4$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}&\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$4$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}\\\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$2$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}&\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$2$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}&\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$4$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}\\\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$1$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}&\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$1$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}&\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$2$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}&\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$2$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}\crcr}}\qquad\mbox{ and }\qquad\mathrm{key}({\rm rev}(a))=\vbox{ \halign{&\tableaucell{#}\cr\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$4$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}&\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$4$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}\\\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$3$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}&\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$3$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}&\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$3$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}\\\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$1$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}&\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$1$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}&\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$1$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}&\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$3$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}\crcr}}.

Here col(key(a))\operatorname{col}(\mathrm{key}(a)) is 421|421|42|2421|421|42|2 and colR(key(rev(a)))\operatorname{col}_{R}(\mathrm{key}({\rm rev}(a))) is 3|31|431|4313|31|431|431, which is indeed equal to frev(col(key(a)))\mathrm{frev}(\operatorname{col}(\mathrm{key}(a))) (column-breaks included for emphasis).

The following lemma is fairly well-known [Ful97, Appendix A.1]; we include a proof here for completeness and to illustrate the flip-and-reverse procedure.

Lemma 3.7.

Let w,ww,w^{\prime} be words in {1,,n}\{1,\ldots,n\}. Then www\sim w^{\prime} if and only if frev(w)frev(w)\mathrm{frev}(w)\sim\mathrm{frev}(w^{\prime}).

Proof.

It is enough to show this for the case that ww and ww^{\prime} are related by a single Knuth move. For xx a letter in ww, let x¯\overline{x} denote n+1xn+1-x. Suppose ww contains the sequence xzy\ldots xzy\ldots where xy<zx\leq y<z. Then one may perform a Knuth move to obtain w=zxyw^{\prime}=\ldots zxy\ldots. In frev(w)\mathrm{frev}(w) we have y¯z¯x¯\ldots\overline{y}\overline{z}\overline{x}\ldots where z¯<y¯x¯\overline{z}<\overline{y}\leq\overline{x}. Then one may perform a Knuth move to obtain the word y¯x¯z¯\ldots\overline{y}\overline{x}\overline{z}\ldots, which is indeed frev(w)\mathrm{frev}(w^{\prime}). Now suppose ww contains the sequence yxz\ldots yxz\ldots where x<yzx<y\leq z. Then one may perform a Knuth move to obtain w=yzxw^{\prime}=\ldots yzx\ldots. In frev(w)\mathrm{frev}(w) we have x¯z¯y¯\ldots\overline{x}\overline{z}\overline{y}\ldots where z¯y¯<x¯\overline{z}\leq\overline{y}<\overline{x}. Then one may perform a Knuth move to obtain the word z¯x¯y¯\ldots\overline{z}\overline{x}\overline{y}\ldots, which is indeed frev(w)\mathrm{frev}(w^{\prime}).

Therefore, frev(w)frev(w)\mathrm{frev}(w)\sim\mathrm{frev}(w^{\prime}) whenever www\sim w^{\prime}. The converse direction is immediate from the fact that frev\mathrm{frev} is an involution. ∎

Proposition 3.8.

Let aa be a weak composition. Then col(key(a))\operatorname{col}(\mathrm{key}(a)) is Knuth-equivalent to colR(key(a))\operatorname{col}_{R}(\mathrm{key}(a)).

Proof.

It suffices to show that the word colR(key(a))\operatorname{col}_{R}(\mathrm{key}(a)) inserts to key(a)\mathrm{key}(a).

Suppose TT is a key. Let ww be a word that contains the entries in the leftmost column of TT and let TT^{\prime} be the key obtained by removing the leftmost column of TT. We will show that inserting the entries of ww into TT^{\prime} in order from largest to smallest yields another key, namely TT^{\prime} with the column whose entries are the entries of ww adjoined on the left. This is the key TT and the conclusion then follows by induction, the base case where TT is empty being trivial.

We will establish that insertion of the iith entry of ww causes (a copy of) the (i1)(i-1)th entry of ww to be bumped from the first into the second row, the (i2)(i-2)nd entry of ww to be bumped from the 2nd to the 3rd row, etc, culminating in the first entry of ww arriving at the end of the iith row. This is clearly true for i=1i=1, as the largest entry of ww is weakly larger than any entry of TT (due to the key condition) so it is inserted at the end of the first row. Suppose this is true for all entries up to the (i1)(i-1)th entry of ww. Now, when the iith entry of ww is inserted, it bumps (a copy of) the (i1)(i-1)th entry of ww from row 11, since there is no entry xx in the tableau such that wi<x<wi1w_{i}<x<w_{i-1} by the key condition. Then the (i1)(i-1)th entry of ww must bump (a copy of) the (i2)(i-2)nd entry of ww (which is in row 22 by the inductive hypothesis), since again there is no entry yy in the tableau such that wi1<y<wi2w_{i-1}<y<w_{i-2} by the key condition. Continuing thus, w1w_{1} is eventually bumped into row ii, and comes to rest at the end of row ii since it is weakly larger than any other entry in the tableau.

Hence the insertion process results in a new entry wiw_{i} in each row |w|+1i|w|+1-i. There is a unique such semistandard Young tableau, and by the key condition each entry wiw_{i} (or a copy of this entry) must appear as the first entry of row |w|+1i|w|+1-i for every ii. Therefore the result is TT^{\prime} with the column determined by ww appended, as required. ∎

We now give a formula for Young key polynomials in terms of compatible sequences.

Theorem 3.9.

Let aa be a weak composition of length nn. Then

κ^a=f(c)col(key(a)),w is c-compatiblexcomp(f(w)).\operatorname{\hat{\kappa}}_{a}=\sum_{f(c)\sim\operatorname{col}(\mathrm{key}(a)),\,w\textrm{ is $c$-compatible}}x^{\mathrm{comp}(f(w))}.
Proof.

The set XX of words Knuth-equivalent to col(key(rev(a)))\operatorname{col}(\mathrm{key}({\rm rev}(a))) is equal to the set of words Knuth-equivalent to colR(key(rev(a)))\operatorname{col}_{R}(\mathrm{key}({\rm rev}(a))) by Proposition 3.8, which is equal to the set of words Knuth-equivalent to frev(col(key(a)))\mathrm{frev}(\operatorname{col}(\mathrm{key}(a))) by Lemma 3.5. Then by Lemma 3.7, the flip-reverses of the words in XX form the set YY of words Knuth-equivalent to col(key(a))\operatorname{col}(\mathrm{key}(a)). Since Y={frev(x):xX}Y=\{\mathrm{frev}(x):x\in X\}, we have {f(y):yY}={rev(x):xX}\{f(y):y\in Y\}=\{{\rm rev}(x):x\in X\}. By Theorem 2.20, κrev(a)(x1,,xn)\operatorname{\kappa}_{{\rm rev}(a)}(x_{1},\ldots,x_{n}) is generated by the compatible sequences for {rev(x):xX}\{{\rm rev}(x):x\in X\}, and thus also generated by the compatible sequences for {f(y):yY}\{f(y):y\in Y\}. Since κ^a(xn,,x1)=κrev(a)(x1,,xn)\operatorname{\hat{\kappa}}_{a}(x_{n},\ldots,x_{1})=\operatorname{\kappa}_{{\rm rev}(a)}(x_{1},\ldots,x_{n}), the compatible sequences for {f(y):yY}\{f(y):y\in Y\} generate κ^a(xn,,x1)\operatorname{\hat{\kappa}}_{a}(x_{n},\ldots,x_{1}), i.e.,

κ^a(xn,,x1)=f(c)col(key(a)),w is c-compatiblexcomp(w).\operatorname{\hat{\kappa}}_{a}(x_{n},\ldots,x_{1})=\sum_{f(c)\sim\operatorname{col}(\mathrm{key}(a)),\,w\textrm{ is $c$-compatible}}x^{\mathrm{comp}(w)}.

Finally, flipping each compatible sequence in the formula above yields κ^a(x1,,xn)\operatorname{\hat{\kappa}}_{a}(x_{1},\ldots,x_{n}). ∎

Example 3.10.

Let a=230a=230. Then key(a)=22112\mathrm{key}(a)=\vbox{ \halign{&\tableaucell{#}\cr\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$2$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}&\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$2$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}\\\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$1$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}&\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$1$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}&\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$2$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}\crcr}}; its column word is 2121221212. The set of words Knuth-equivalent to 2121221212 is {22121,22211,21221,21212,22112}\{22121,22211,21221,21212,22112\}. We compute the set of compatible sequences for the flips of each of these words.

Word flip Compatible sequences Flips of Compatible sequences
22121 22323 11223 33221
22211 22233 11122   11123   11133 33322   33321   33311
11233   12233   22233 33211   32211   22211
21221 23223 12223 32221
21212 23232
22112 22332 11222 33222
Figure 10. Compatible sequences and their flips

The corresponding monomials indeed sum up to κ^230\operatorname{\hat{\kappa}}_{230}; compare this example to Example 2.21 computing κ032\operatorname{\kappa}_{032} in terms of compatible sequences.

3.2. Divided differences and Demazure crystals

Young key polynomials may also be described in terms of divided difference operators. Given a weak composition aa, let revsort(a){\rm revsort}(a) be the rearrangement of aa into increasing order. Let w^a\hat{w}_{a} be the permutation of shortest length rearranging aa to revsort(a){\rm revsort}(a). For 1i<n1\leq i<n define an operator

π^i=ixi+1,\hat{\pi}_{i}=-\partial_{i}x_{i+1},

and for a permutation ww, define π^w=π^1π^r\hat{\pi}_{w}=\hat{\pi}_{1}\cdots\hat{\pi}_{r}, where s1srs_{1}\cdots s_{r} is any reduced word for ww.

Lemma 3.11.

Let ff be a polynomial in [x1,,xn]\mathbb{Z}[x_{1},\ldots,x_{n}]. We have

I(πif)=π^niI(f)I({\pi}_{i}f)=\hat{\pi}_{n-i}I(f)

where I(f)I(f) is the polynomial obtained by exchanging variables xjxn+1jx_{j}\leftrightarrow x_{n+1-j}.

Proof.

By linearity, it suffices to show this is true for a monomial f=xbf=x^{b}, where bb is a weak composition of length nn. We compute

I(πixb)\displaystyle I(\pi_{i}x^{b}) =I(x1b1xibi+1xi+1bi+1xnbnx1b1xibi+1xi+1bi+1xnbnxixi+1)\displaystyle=I\left(\frac{x_{1}^{b_{1}}\cdots x_{i}^{b_{i}+1}x_{i+1}^{b_{i+1}}\cdots x_{n}^{b_{n}}-x_{1}^{b_{1}}\cdots x_{i}^{b_{i+1}}x_{i+1}^{b_{i}+1}\cdots x_{n}^{b_{n}}}{x_{i}-x_{i+1}}\right)
=xnb1xn+1ibi+1xnibi+1x1bnxnb1xn+1ibi+1xnibi+1x1bnxn+1ixni\displaystyle=\frac{x_{n}^{b_{1}}\cdots x_{n+1-i}^{b_{i}+1}x_{n-i}^{b_{i+1}}\cdots x_{1}^{b_{n}}-x_{n}^{b_{1}}\cdots x_{n+1-i}^{b_{i+1}}x_{n-i}^{b_{i}+1}\cdots x_{1}^{b_{n}}}{x_{n+1-i}-x_{n-i}}

and

π^niI(xb)\displaystyle\hat{\pi}_{n-i}I(x^{b}) =π^ni(x1bnxnibi+1xn+1ibixnb1)\displaystyle=\hat{\pi}_{n-i}(x_{1}^{b_{n}}\cdots x_{n-i}^{b_{i+1}}x_{n+1-i}^{b_{i}}\cdots x_{n}^{b_{1}})
=x1bnxnibi+1xn+1ibi+1xnb1x1bnxnibi+1xn+1ibi+1xnb1xn+1ixni\displaystyle=\frac{x_{1}^{b_{n}}\cdots x_{n-i}^{b_{i+1}}x_{n+1-i}^{b_{i}+1}\cdots x_{n}^{b_{1}}-x_{1}^{b_{n}}\cdots x_{n-i}^{b_{i}+1}x_{n+1-i}^{b_{i+1}}\cdots x_{n}^{b_{1}}}{x_{n+1-i}-x_{n-i}}

as required. ∎

Lemma 3.12.

π^w\hat{\pi}_{w} is well-defined.

Proof.

Since the πi\pi_{i}’s satisfy the commutativity and braid relations of SnS_{n}, it follows from Lemma 3.11 that the π^i\hat{\pi}_{i}’s also do. ∎

Theorem 3.13.

Let aa be a weak composition of length nn. Then κ^a=π^w^axrevsort(a).\operatorname{\hat{\kappa}}_{a}=\hat{\pi}_{\hat{w}_{a}}x^{{\rm revsort}(a)}.

Proof.

First observe that if wa=si1sikw_{a}=s_{i_{1}}\cdots s_{i_{k}} is the minimal length permutation sending aa to sort(a){\rm sort}(a), then sni1sniks_{n-i_{1}}\cdots s_{n-i_{k}} is the minimal length permutation sending rev(a){\rm rev}(a) to revsort(rev(a)){\rm revsort}({\rm rev}(a)), i.e., is w^rev(a)\hat{w}_{{\rm rev}(a)}.

Therefore, by Lemma 3.11 and the fact that I(xsort(a))=xrevsort(a)=xrevsort(rev(a))I(x^{{\rm sort}(a)})=x^{{\rm revsort}(a)}=x^{{\rm revsort}({\rm rev}(a))}, we have

π^w^rev(a)xrevsort(rev(a))=π^w^rev(a)I(xsort(a))=I(πwa(xsort(a)))=I(κa)=κ^rev(a).\hat{\pi}_{\hat{w}_{{\rm rev}(a)}}x^{{\rm revsort}({\rm rev}(a))}=\hat{\pi}_{{\hat{w}_{{\rm rev}(a)}}}I(x^{{\rm sort}(a)})=I(\pi_{w_{a}}(x^{{\rm sort}(a)}))=I(\operatorname{\kappa}_{a})=\operatorname{\hat{\kappa}}_{{\rm rev}(a)}.\qed
Example 3.14.

Let a=230a=230. Then the minimal length permutation taking aa to revsort(a)=023{\rm revsort}(a)=023 is s2s1s_{2}s_{1}. We compute

π^2π^1(x22x33)\displaystyle\hat{\pi}_{2}\hat{\pi}_{1}(x_{2}^{2}x_{3}^{3}) =π^2x23x33x13x33x2x1\displaystyle=\hat{\pi}_{2}\frac{x_{2}^{3}x_{3}^{3}-x_{1}^{3}x_{3}^{3}}{x_{2}-x_{1}}
=π^2(x2x33+x1x22x33+x13x33)\displaystyle=\hat{\pi}_{2}(x_{2}x_{3}^{3}+x_{1}x_{2}^{2}x_{3}^{3}+x_{1}^{3}x_{3}^{3})
=(x22x34x24x32)+(x1x2x34x1x24x3)+(x12x34x12x24)x3x2\displaystyle=\frac{(x_{2}^{2}x_{3}^{4}-x_{2}^{4}x_{3}^{2})+(x_{1}x_{2}x_{3}^{4}-x_{1}x_{2}^{4}x_{3})+(x_{1}^{2}x_{3}^{4}-x_{1}^{2}x_{2}^{4})}{x_{3}-x_{2}}
=x22x33+x23x32+x1x2x33+x1x22x32+x1x23x3+x12x33+x12x2x32+x12x22x3+x12x23\displaystyle=x_{2}^{2}x_{3}^{3}+x_{2}^{3}x_{3}^{2}+x_{1}x_{2}x_{3}^{3}+x_{1}x_{2}^{2}x_{3}^{2}+x_{1}x_{2}^{3}x_{3}+x_{1}^{2}x_{3}^{3}+x_{1}^{2}x_{2}x_{3}^{2}+x_{1}^{2}x_{2}^{2}x_{3}+x_{1}^{2}x_{2}^{3}
=κ^230.\displaystyle=\operatorname{\hat{\kappa}}_{230}.

Recall the Demazure crystal structure for key polynomials described in Section 2.2.3. The Young key polynomials may be realized as characters of crystals that are obtained via Demazure truncations beginning from the lowest weight of the crystal B(λ)B(\lambda) rather than the highest. For a subset XX of B(λ)B(\lambda), define 𝔇^iX={bB(λ)|fir(b)X for some r0}.\hat{\mathfrak{D}}_{i}X=\{b\in B(\lambda)|f_{i}^{r}(b)\in X\textrm{ for some }r\geq 0\}.

Theorem 3.15.

Let aa be a weak composition of length nn and let ww be of shortest length such that w(a)=revsort(a)w(a)={\rm revsort}(a). Then the Young key polynomial κ^a\operatorname{\hat{\kappa}}_{a} is the character of the subcrystal of B(sort(a))B({\rm sort}(a)) obtained by

𝔇^i1𝔇^ik{u^λ},\hat{\mathfrak{D}}_{i_{1}}\cdots\hat{\mathfrak{D}}_{i_{k}}\{\hat{u}_{\lambda}\},

where si1siks_{i_{1}}\cdots s_{i_{k}} is a reduced word for ww and u^λ\hat{u}_{\lambda} is the lowest weight element of B(λ)B(\lambda).

Proof.

Recall that the shortest permutation sending rev(a){\rm rev}(a) to sort(rev(a)){\rm sort}({\rm rev}(a)) is sni1sniks_{n-i_{1}}\cdots s_{n-i_{k}}. Performing the Lusztig involution [Lus10] \star on B(λ)B(\lambda) exchanges each fif_{i} with enie_{n-i} and eie_{i} with fnif_{n-i}, and reverses the weight of each vertex. Hence, applying a Demazure truncation with sni1sniks_{n-i_{1}}\cdots s_{n-i_{k}} from the highest weight of B(λ)B(\lambda)^{\star} yields κrev(a)\operatorname{\kappa}_{{\rm rev}(a)} with variables reversed, which is equal to κ^a\operatorname{\hat{\kappa}}_{a} by (3.1). The statement follows. ∎

Observe that the repeated actions of the 𝔇^i\hat{\mathfrak{D}}_{i} starting with u^λ\hat{u}_{\lambda} precisely mirrors the repeated action of the divided difference operators π^i\hat{\pi}_{i} starting with the monomial x0n(λ)×rev(λ)x^{0^{n-\ell(\lambda)}\times{\rm rev}(\lambda)}.

Example 3.16.

Let a=201a=201, and recall B(21)B(21) from Figure 7. The shortest-length ww such that w(a)=revsort(a)w(a)={\rm revsort}(a) is w=s1s2w=s_{1}s_{2}. Therefore, the crystal graph for κ^201\operatorname{\hat{\kappa}}_{201} is the subgraph of B(21)B(21) consisting of all vertices that can be obtained from the lowest weight 33 22 33 by first applying a sequence of e2e_{2}’s and then a sequence of e1e_{1}’s. Hence κ^201=x2x32+x22x3+x1x32+x1x2x3+x12x3\operatorname{\hat{\kappa}}_{201}=x_{2}x_{3}^{2}+x_{2}^{2}x_{3}+x_{1}x_{3}^{2}+x_{1}x_{2}x_{3}+x_{1}^{2}x_{3}.

3.3. Young key polynomials as generators for left keys

Recall Theorem 2.16 states that a key polynomial can be described as the generating function for the set of all SSYT\mathrm{SSYT} with bounded right key. In this section we provide an analogous description of Young key polynomials as well as the corresponding description of Young Demazure atoms.

Given a semistandard Young tableau TT, let frev(T)\mathrm{frev}(T) denote the filling obtained by flipping all entries in TT and reversing the order of the resulting column entries. Compare this to the definition of frev\mathrm{frev} applied to a word at the beginning of Section 3.1. It is a straightforward observation that when TT is a key, frev(T)\mathrm{frev}(T) is the key whose entries in each column are the flip-reverses of the entries in the corresponding column of TT. (However, if TT is not a key then frev(T)\mathrm{frev}(T) is not necessarily even a semistandard Young tableau.)

We need to establish a weight-reversing bijection between semistandard Young tableaux with a given right key UU and semistandard Young tableaux with left key frev(U)\mathrm{frev}(U). This is done in the following lemma, which can also be understood in terms of the evacuation operation on semistandard Young tableaux. Recall that a word ww is Knuth equivalent to a semistandard Young tableau TT if and only if Schensted insertion of the word ww produces the tableau TT.

Lemma 3.17.

Let TT be a semistandard Young tableau. Then the left key of the tableau obtained via Schensted insertion of frev(col(T))\mathrm{frev}(\operatorname{col}(T)) is frev(K+(T))\mathrm{frev}(K_{+}(T)).

Proof.

Let TT have shape λ\lambda and let UU be the semistandard Young tableau obtained by Schensted insertion of frev(col(T))\mathrm{frev}(\operatorname{col}(T)). Consider any column index jj. Consider any word ww^{\prime} that is Knuth equivalent to col(T)\operatorname{col}(T), has column form a rearrangement of λ\lambda^{\prime}, and whose rightmost maximal decreasing subsequence has length λj\lambda_{j}^{\prime}. Then the entries in column jj of K+(T)K_{+}(T) are the entries of the rightmost maximal decreasing subsequence of ww^{\prime}. Now, the column form of frev(w)\mathrm{frev}(w^{\prime}) is the reversal of the column form of ww^{\prime} (thus also a rearrangement of λ\lambda^{\prime}), and Lemma 3.7 implies that frev(col(T))\mathrm{frev}(\operatorname{col}(T)) is Knuth equivalent to frev(w)\mathrm{frev}(w^{\prime}). Therefore the leftmost maximal decreasing subsequence of frev(w)\mathrm{frev}(w^{\prime}) is the flip-reverse of the rightmost maximal decreasing subsequence of ww^{\prime}, and hence the entries in the the jthj^{th} column of the left key of UU are precisely the flip-reverses of the entries in the jthj^{th} column of the right key of TT. ∎

Theorem 3.18.

The Young Demazure atoms and Young key polynomials are generated by the left keys of semistandard Young tableaux as follows:

𝒜^a=TSSYTn(λ(a))K(T)=key(a)xwt(T) and κ^a=TSSYTn(λ(a))K(T)key(a)xwt(T),\operatorname{\hat{\mathcal{A}}}_{a}=\sum_{\begin{subarray}{c}T\in\mathrm{SSYT}_{n}(\lambda(a))\\ K_{-}(T)=\mathrm{key}(a)\end{subarray}}x^{{\rm wt}(T)}\qquad\mbox{ and }\qquad\operatorname{\hat{\kappa}}_{a}=\sum_{\begin{subarray}{c}T\in\mathrm{SSYT}_{n}(\lambda(a))\\ K_{-}(T)\geq\mathrm{key}(a)\end{subarray}}x^{{\rm wt}(T)},

where \geq means entrywise comparison and n=(a)n=\ell(a).

Proof.

Consider the first expansion. Recall that 𝒜^a(x1,,xn)=𝒜rev(a)(xn,,x1)\operatorname{\hat{\mathcal{A}}}_{a}(x_{1},\ldots,x_{n})=\operatorname{\mathcal{A}}_{{\rm rev}(a)}(x_{n},\ldots,x_{1}) and that (by Equation 2.1) 𝒜rev(a)\operatorname{\mathcal{A}}_{{\rm rev}(a)} is generated by the set of all semistandard Young tableaux whose right key equals key(rev(a))\mathrm{key}({\rm rev}(a)). It is therefore enough to exhibit a weight-reversing bijection between the set of all semistandard Young tableaux whose right key equals key(rev(a))\mathrm{key}({\rm rev}(a)) and the set of all semistandard Young tableaux whose left key is key(a)\mathrm{key}(a).

We know from Lemma 3.17 that if TT is a semistandard Young tableau such that K+(T)=key(rev(a))K_{+}(T)=\mathrm{key}({\rm rev}(a)), then the semistandard Young tableau SS obtained via Schensted insertion of frev(col(T))\mathrm{frev}(\operatorname{col}(T)) has K(S)=frev(K+(T))=frev(key(rev(a)))=key(a)K_{-}(S)=\mathrm{frev}(K_{+}(T))=\mathrm{frev}(\mathrm{key}({\rm rev}(a)))=\mathrm{key}(a). This process is clearly invertible, hence bijective, and the application of frev\mathrm{frev} to col(T)\operatorname{col}(T) ensures it is weight-reversing.

For the second expansion, we recall that κ^a(x1,,xn)=κrev(a)(xn,,x1)\operatorname{\hat{\kappa}}_{a}(x_{1},\ldots,x_{n})=\operatorname{\kappa}_{{\rm rev}(a)}(x_{n},\ldots,x_{1}) and that by Theorem 2.16 κrev(a)\operatorname{\kappa}_{{\rm rev}(a)} is generated by the set of all semistandard Young tableaux whose right key is less than or equal to key(rev(a))\mathrm{key}({\rm rev}(a)). It is straightforward to check that if K+(T)key(rev(a))K_{+}(T)\leq\mathrm{key}({\rm rev}(a)), then the semistandard Young tableau SS obtained via Schensted insertion of frev(col(T))\mathrm{frev}(\operatorname{col}(T)) has K(S)frev(K+(T))=key(a)K_{-}(S)\geq\mathrm{frev}(K_{+}(T))=\mathrm{key}(a). The second expansion then follows by applying the same argument used to prove the first expansion. ∎

Example 3.19.

Let T=34112T=\vbox{ \halign{&\tableaucell{#}\cr\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$3$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}&\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$4$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}\\\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$1$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}&\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$1$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}&\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$2$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}\crcr}}, which has right key K+(T)=44222K_{+}(T)=\vbox{ \halign{&\tableaucell{#}\cr\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$4$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}&\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$4$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}\\\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$2$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}&\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$2$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}&\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$2$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}\crcr}}. We have col(T)=31412\operatorname{col}(T)=31412. Schensted insertion of frev(col(T))=34142\mathrm{frev}(\operatorname{col}(T))=34142 produces the semistandard Young tableau 33 44 11 22 44 which indeed has left key 33113=frev(K+(T))\vbox{ \halign{&\tableaucell{#}\cr\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$3$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}&\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$3$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}\\\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$1$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}&\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$1$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}&\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$3$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}\crcr}}=\mathrm{frev}(K_{+}(T)).

3.4. Row-frank words

Our next aim is to realize Young key polynomials as traces on modules. For this, we first adapt a formula of [LS90] expressing key polynomials in terms of row-frank words. The first condition below is equivalent to the condition of being row-frank; see [RS95] for details. The standardization of a semistandard Young tableau TT, denoted std(T)\operatorname{std}(T), is the standard Young tableau obtained by replacing the 11’s in TT from left to right by 1,2,,γ11,2,\ldots,\gamma_{1}, the 22’s by γ1+1,γ1+2,,γ1+γ2\gamma_{1}+1,\gamma_{1}+2,\ldots,\gamma_{1}+\gamma_{2}, and so on, where γi\gamma_{i} equals the number of times the entry ii appears in TT. Given a word uu in positive integers, its row-word factorization is u(2)u(1)\cdots u^{(2)}u^{(1)}, where each row-word u(i)u^{(i)} is a weakly increasing subsequence of maximal length.

For a weak composition aa, let 𝒲(a)\operatorname{\mathcal{W}}(a) be the set of all words u=u(2)u(1)u=\cdots u^{(2)}u^{(1)} with each u(i)u^{(i)} having aia_{i} letters, satisfying the following conditions.

  1. (1)

    The word uu maps to a pair (P,std(key(a)))(P,\operatorname{std}(\mathrm{key}(a))) under the column insertion described in [RS95].

  2. (2)

    No letter of u(i)u^{(i)} exceeds ii.

Theorem 3.20.

[LS90] The key polynomials are generated using words in 𝒲(a)\operatorname{\mathcal{W}}(a) as follows:

κa=u𝒲(a)xu.\operatorname{\kappa}_{a}=\sum_{u\in\operatorname{\mathcal{W}}(a)}x_{u}.

We now provide the analogue of this generating function for Young key polynomials. For a weak composition aa, let 𝒲^(a)\operatorname{\hat{\mathcal{W}}}(a) be the set of all words u=u(2)u(1)u=\cdots u^{(2)}u^{(1)} with each u(i)u^{(i)} having aia_{i} letters, satisfying the following conditions.

  1. (1)

    The word frev(u)\mathrm{frev}(u) maps to a pair (P,std(key(rev(a))))(P,\operatorname{std}(\mathrm{key}({\rm rev}(a)))) under column insertion.

  2. (2)

    For each letter jj of u(i)u^{(i)}, we have ij(a)i\leq j\leq\ell(a).

Example 3.21.

We have

𝒲(032)={33|222|,33|122|,33|112|,33|111|,23|111|,23|112|,23|122|,22|111|,22|112|}\operatorname{\mathcal{W}}(032)=\{33|222|,33|122|,33|112|,33|111|,23|111|,23|112|,23|122|,22|111|,22|112|\}

and

𝒲^(230)={|222|11,|223|11,|233|11,|333|11,|333|12,|233|12,|223|12,|333|22,|233|22},\operatorname{\hat{\mathcal{W}}}(230)=\{|222|11,|223|11,|233|11,|333|11,|333|12,|233|12,|223|12,|333|22,|233|22\},

where the vertical bars denote the row word factorization (including empty row-words).

Theorem 3.22.

The Young key polynomials are generated using the words in 𝒲^(a)\operatorname{\hat{\mathcal{W}}}(a) as follows:

κ^a=w𝒲^(a)xw.\operatorname{\hat{\kappa}}_{a}=\sum_{w\in\operatorname{\hat{\mathcal{W}}}(a)}x_{w}.
Proof.

Consider a word uu in 𝒲(a)\operatorname{\mathcal{W}}(a) and let w=frev(u)w=\mathrm{frev}(u). Then ww satisfies condition (1) for 𝒲^(a)\operatorname{\hat{\mathcal{W}}}(a) by construction. Consider a letter bb in u(i)u^{(i)}. By definition, bib\leq i. The flip nb+1n-b+1 of bb appears in the (ni+1)th(n-i+1)^{th} row-word of ww, and bib\leq i implies nb+1ni+1n-b+1\geq n-i+1. So ww satisfies both the conditions to be in the set 𝒲^(a)\operatorname{\hat{\mathcal{W}}}(a). Since flipping and reversing is an invertible process, we have that the words in 𝒲^(a)\operatorname{\hat{\mathcal{W}}}(a) are exactly the flip-reverses of the words in 𝒲(rev(a))\operatorname{\mathcal{W}}({\rm rev}(a)). Then since the monomials appearing in κ^a(x1,xn)\operatorname{\hat{\kappa}}_{a}(x_{1},\ldots x_{n}) are the flips of those appearing in κrev(a)(xn,,x1)\operatorname{\kappa}_{{\rm rev}(a)}(x_{n},\ldots,x_{1}), it follows from Theorem 3.20 that 𝒲^(a)\operatorname{\hat{\mathcal{W}}}(a) generates κ^a\operatorname{\hat{\kappa}}_{a}. ∎

3.5. Young key polynomials as traces on modules

In [RS95], generalized flagged Schur modules and key modules are defined. The key polynomials are realized as traces on key modules, which are a special case of generalized flagged Schur modules. In this section we modify the Reiner-Shimozono approach to construct modules so that the Young key polynomials are realised as traces on these modules.

As in [RS95], a diagram DD is a finite subset of the Cartesian product ×\mathbb{P}\times\mathbb{P} of the positive integers with itself, where every element of ×\mathbb{P}\times\mathbb{P} in DD is thought of as being a box. A filling of shape DD is a map T:DT:D\rightarrow\mathbb{P} assigning a positive integer to each box in DD (note this is called a tableau of shape DD in [RS95]).

Let 𝔽\mathbb{F} be a field of characteristic 0, and let 𝒯Dn\mathcal{T}^{n}_{D} be the vector space over 𝔽\mathbb{F} with basis the set of all fillings TT of shape DD whose largest entry does not exceed nn. Fix an order 𝔟1,𝔟2,\mathfrak{b}_{1},\mathfrak{b}_{2},\ldots on the boxes of DD, and identify the filling TT with the tensor product ϵT(𝔟1)ϵT(𝔟2)\epsilon_{T(\mathfrak{b}_{1})}\otimes\epsilon_{T(\mathfrak{b}_{2})}\otimes\cdots, where ϵi\epsilon_{i} is the iith standard basis vector. Then an action of GLn(𝔽)GL_{n}(\mathbb{F}) on 𝒯Dn\mathcal{T}^{n}_{D} is defined by letting GLn(𝔽)GL_{n}(\mathbb{F}) act on each ϵi\epsilon_{i} as usual and extending this action linearly.

The row group R(D)R(D) (respectively column group C(D)C(D)) is the set of all permutations of the boxes of DD which fixes the rows (resp. columns) in which the boxes appear. These groups act on 𝒯Dn\mathcal{T}^{n}_{D} by permuting the positions of the entries within a filling. As in [RS95], define

eT=αR(D),βC(D)sgn(β)Tαβ,e_{T}=\sum_{\alpha\in R(D),\,\,\beta\in C(D)}{\rm sgn}(\beta)T\alpha\beta,

where TαβT\alpha\beta is the filling obtained by acting first by α\alpha and then by β\beta.

Define the Young generalized flagged Schur module 𝔉𝔖^Dn\operatorname{\widehat{\mathfrak{FS}}}^{n}_{D} for an arbitrary diagram DD (with nn at least the maximum row index of DD) to be the subspace of 𝒯Dn\mathcal{T}^{n}_{D} spanned by the set {eT}\{e_{T}\} as TT runs over all fillings of shape DD whose entries in row ii are not smaller than ii. It is straightforward that 𝔉𝔖^Dn\operatorname{\widehat{\mathfrak{FS}}}^{n}_{D} is a BB-module, where BB is the Borel subgroup of GLn(𝔽)GL_{n}(\mathbb{F}) consisting of lower-triangular matrices.

Remark 3.23.

The construction of the generalized flagged Schur module 𝔉𝔖D\operatorname{\mathfrak{FS}}_{D} described in [RS95] is similar, but serves to illustrate an important difference in the behaviors of Young and reverse families of polynomials. In [RS95] 𝒯D\mathcal{T}_{D} is defined to be the vector space with basis consisting of all fillings of shape DD, with no restriction on the size of the entries. In this way, 𝒯D\mathcal{T}_{D} is a GL(𝔽)GL_{\infty}(\mathbb{F})-module. Then 𝔉𝔖D\operatorname{\mathfrak{FS}}_{D} is spanned by the set {eT}\{e_{T}\} as TT runs over all fillings of shape DD whose entries in row ii are not larger than ii, which is finite even though 𝒯D\mathcal{T}_{D} is infinite-dimensional. In this way, 𝔉𝔖D\operatorname{\mathfrak{FS}}_{D} is a module for the opposite Borel subgroup BB_{-} consisting of upper-triangular elements of GL(𝔽)GL_{\infty}(\mathbb{F}). The dependence on nn in the Young case is reflected in the fact that appending zeros to a weak composition does not change the corresponding key polynomial, but does change the Young key polynomial.

Example 3.24.

Let a=032a=032. Then if T = 
2 3
1 2 2
T\mbox{ $=$ }\vline\vbox{ \halign{&\tableaucell{#}\cr\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$2$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}&\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$3$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}\\\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$1$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}&\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$2$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}&\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$2$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}\\\vbox to9.68747pt{\vfil\hrule width=9.68747pt,height=0.0pt}\\\hline\cr}}
, applying elements of the row group to TT yields the following:

223122223212223221232122232212232221\begin{array}[]{c@{\hskip 1.5\cellsize}c@{\hskip 1\cellsize}c@{\hskip 1.5\cellsize}c@{\hskip 1.5\cellsize}c@{\hskip 1.5\cellsize}c@{\hskip 1.5\cellsize}c}2\,\,\vline\vbox{ \halign{&\tableaucell{#}\cr\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$2$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}&\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$3$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}\\\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$1$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}&\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$2$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} 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where the coefficients are 22 because there are two distinct permutations yielding each ordering of 1,2,21,2,2. It is easy to see that for any filling SS with repeated entries in any column, we have βC(D)sgn(β)Sβ=0\sum_{\beta\in C(D)}{\rm sgn}(\beta)S\beta=0, hence only the first and fifth fillings above contribute to eTe_{T}. Applying the column group to each of these and summing the resulting fillings yields

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Define the key module 𝒦a\operatorname{\mathcal{K}}_{a} for the weak composition aa to be the BB_{-}-module 𝔉𝔖D(a)\operatorname{\mathfrak{FS}}_{D(a)}.

Theorem 3.25.

[RS95] For u=u(2)u(1)u=\cdots u^{(2)}u^{(1)} in 𝒲(a),\operatorname{\mathcal{W}}(a), let T(u)T(u) be the filling of shape D(a)D(a) obtained by placing u(j)u^{(j)} in row jj. Then {eT(u):u𝒲(a)}\{e_{T(u)}:u\in\operatorname{\mathcal{W}}(a)\} is a basis for the key module 𝒦a\operatorname{\mathcal{K}}_{a}.

We now describe the variation on the Reiner-Shimozono construction that needed to describe the Young key polynomials as characters. Let aa be a weak composition of length nn, and define the Young key module 𝒦^a\operatorname{\hat{\mathcal{K}}}_{a} for the weak composition aa to be the BB-module 𝔉𝔖^D(a)\operatorname{\widehat{\mathfrak{FS}}}_{D(a)}. Here we may drop nn from the notation, since nn is determined by the weak composition aa.

Corollary 3.26.

For u=u(1)u(2)u=u^{(1)}u^{(2)}\cdots in 𝒲^(a)\operatorname{\hat{\mathcal{W}}}(a), let T(u)T(u) be the filling of shape D(a)D(a) obtained by placing u(j)u^{(j)} in row jj. Then {eT(u):u𝒲^(a)}\{e_{T(u)}:u\in\operatorname{\hat{\mathcal{W}}}(a)\} is a basis for the Young key module 𝒦^a\operatorname{\hat{\mathcal{K}}}_{a}.

Proof.

The flip-and-reverse map on fillings extends linearly to an involution ψ\psi, hence an isomorphism, on 𝒯D(a)n\mathcal{T}^{n}_{D(a)}. Moreover, ψ\psi sends a filling whose entries are at least their row index to a filling whose entries are at most their row index, and vice versa. In particular, by the proof of Theorem 3.22, ψ\psi carries {eT(u):u𝒲^(a)}\{e_{T(u)}:u\in\operatorname{\hat{\mathcal{W}}}(a)\} to the basis {eT(frev(u)):frev(u)𝒲(rev(a))}\{e_{T(\mathrm{frev}(u))}:\mathrm{frev}(u)\in\operatorname{\mathcal{W}}({\rm rev}(a))\} of 𝒦rev(a)\operatorname{\mathcal{K}}_{{\rm rev}(a)} given by Theorem 3.25.

Therefore, {eT(u):u𝒲^(a)}\{e_{T(u)}:u\in\operatorname{\hat{\mathcal{W}}}(a)\} is a linearly independent set, since any linear dependence in this set would imply, via ψ\psi, a linear dependence in the linearly independent set {eT(frev(u)):frev(u)𝒲(rev(a))}\{e_{T(\mathrm{frev}(u))}:\mathrm{frev}(u)\in\operatorname{\mathcal{W}}({\rm rev}(a))\}. Similarly {eT(u):u𝒲^(a)}\{e_{T(u)}:u\in\operatorname{\hat{\mathcal{W}}}(a)\} is spanning: suppose eT𝒦^ae_{T}\in\operatorname{\hat{\mathcal{K}}}_{a}. Then ψ(eT)𝒦rev(a)\psi(e_{T})\in\operatorname{\mathcal{K}}_{{\rm rev}(a)}, hence is in the span of the spanning set {eT(frev(u)):frev(u)𝒲(rev(a))}\{e_{T(\mathrm{frev}(u))}:\mathrm{frev}(u)\in\operatorname{\mathcal{W}}({\rm rev}(a))\} of 𝒦rev(a)\operatorname{\mathcal{K}}_{{\rm rev}(a)}, and applying ψ\psi again yields eTe_{T} as a linear combination of {eT(u):u𝒲^(a)}\{e_{T(u)}:u\in\operatorname{\hat{\mathcal{W}}}(a)\}. ∎

Remark 3.27.

The order in which entries of u(j)u^{(j)} are placed in row jj does not matter, since fillings with any given ordering of u(j)u^{(j)} in each row jj appear in eTe_{T} due to the action of the row group. In Example 3.29, we represent eTe_{T} by the filling TT with entries increasing from left to right in each row, which agrees with the choices of representatives for key modules in [RS95].

Let xx be the diagonal matrix whose diagonal entries are x1,x2,,xnx_{1},x_{2},\ldots,x_{n}. We immediately obtain the following (compare to Corollary 14 in [RS95]).

Corollary 3.28.

The Young key polynomial κ^a\operatorname{\hat{\kappa}}_{a} is the trace of xx acting on the Young key module 𝒦^a\operatorname{\hat{\mathcal{K}}}_{a}.

Example 3.29.

The Young key module 𝒦^230\operatorname{\hat{\mathcal{K}}}_{230} has basis {eT}\{e_{T}\} for the following fillings TT.

2 2 2
1 1
 
2 2 3
1 1
 
2 3 3
1 1
 
3 3 3
1 1
 
2 3 3
1 2
 
2 2 3
1 2
 
3 3 3
2 2
 
23322.
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4. Other polynomial families and intersections

In this section, we provide a new formula in terms of Knuth equivalence for the fundamental slide expansion of a key polynomial, and interpret compatible sequences in terms of the fundamental particle basis, introduced in [Sea20] as a common refinement of the fundamental slide and Demazure atom bases. As we did for Young key polynomials and Young atoms, we also determine the intersections of further reverse bases and their Young analogues.

4.1. The fundamental and monomial slide bases

For a weak composition aa, define the fundamental fillings FF(a)\mathrm{FF}(a) for aa [Sea20] to be the (reverse) fillings of D(a)D(a) satisfying the following conditions.

  1. (1)

    Entries weakly decrease from left to right in each row.

  2. (2)

    No entry in row ii is greater than ii.

  3. (3)

    If a box with label bb is in a lower row than a box with label cc, then b<cb<c.

The fundamental slide polynomial 𝔉a\operatorname{\mathfrak{F}}_{a} [AS17] is the generating function of FF(a)\mathrm{FF}(a):

𝔉a=TFF(a)xwt(T).\operatorname{\mathfrak{F}}_{a}=\sum_{T\in\mathrm{FF}(a)}x^{{\rm wt}(T)}.

For example, 𝔉103=x103+x112+x121+x130\operatorname{\mathfrak{F}}_{103}=x^{103}+x^{112}+x^{121}+x^{130}, computed by FF(103)\mathrm{FF}(103) below.

3 3 3
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The monomial slide basis can also be described using reverse fillings. Given a weak composition aa, the monomial fillings MF(a)\mathrm{MF}(a) [Sea20] are the subset of FF(a)\mathrm{FF}(a) for which all entries in the same row are equal. The monomial slide polynomial 𝔐a\operatorname{\mathfrak{M}}_{a} [AS17] is

𝔐a=TMF(a)xwt(T).\operatorname{\mathfrak{M}}_{a}=\sum_{T\in\mathrm{MF}(a)}x^{{\rm wt}(T)}.

For example, 𝔐103=x103+x130\operatorname{\mathfrak{M}}_{103}=x^{103}+x^{130}.

Various formulas have been given [AS18b], [Ass17], [MPS19] for the fundamental slide expansion of a key polynomial. Here we provide another, more in keeping with the theme of the previous section.

Proposition 4.1.
κa=rev(b)col(key(a))𝔉maxcomp(b).\operatorname{\kappa}_{a}=\sum_{{\rm rev}(b)\sim\operatorname{col}(\mathrm{key}(a))}\operatorname{\mathfrak{F}}_{\mathrm{maxcomp}(b)}.

where maxcomp(b)\mathrm{maxcomp}(b) is the weak composition associated to the compatible sequence for bb whose entries are maximum possible. (If bb has no compatible sequences, we declare 𝔉maxcomp(b)=0\operatorname{\mathfrak{F}}_{\mathrm{maxcomp}(b)}=0.)

Proof.

We need to establish 𝔉maxcomp(b)=w is b-compatiblexw\displaystyle{\operatorname{\mathfrak{F}}_{\mathrm{maxcomp}(b)}=\sum_{\textrm{$w$ is $b$-compatible}}x^{w}}; the statement then follows from Theorem 2.20. The compatible sequence for a word bb whose entries are maximum possible is found as follows. First, partition bb into (weakly) decreasing runs b=(r1|r2||rk)b=(r_{1}|r_{2}|\ldots|r_{k}). Let b(i)b^{(i)} denote the rightmost (i.e. smallest) entry of bb in the ithi^{th} run rir_{i}. We proceed right-to-left, at each step replacing every entry in a run rir_{i} with a certain number cic_{i}. To begin, replace every element in rkr_{k} with b(k)b^{(k)}, i.e., we set ck=b(k)c_{k}=b^{(k)}. Proceeding leftwards, replace every entry in rir_{i} with cimin{b(i),ci+11}c_{i}\coloneqq\min\{b^{(i)},c_{i+1}-1\}. This process is a variant of the construction of the weak descent composition of a word in [Ass17], [MS20].

Every compatible sequence ww for bb can be obtained from the maximal one by decrementing parts as long as we still have wi<wi+1w_{i}<w_{i+1} whenever bi<bi+1b_{i}<b_{i+1}. In exactly the same way, every fundamental filling for maxcomp(b)\mathrm{maxcomp}(b) can be obtained from the filling that has every entry equal to its row index by decrementing entries as long as entries in a given row remain strictly larger than entries in any lower row. This gives a weight-preserving bijection between the compatible sequences for bb and the fundamental fillings for maxcomp(b)\mathrm{maxcomp}(b). ∎

For example, suppose b=435254b=435254. Then the partition into weakly decreasing runs gives 43|52|5443|52|54. We replace each entry in the last run with 44, obtaining 43|52|𝟒𝟒43|52|{\bf 44}. Next, we replace each entry in the next run with min{2,41}=2\min\{2,4-1\}=2, obtaining 43|𝟐𝟐|𝟒𝟒43|{\bf 22}|{\bf 44}. Finally, we replace each entry in the first run with min{3,21}=1\min\{3,2-1\}=1, obtaining 𝟏𝟏|𝟐𝟐|𝟒𝟒{\bf 11}|{\bf 22}|{\bf 44}. The largest compatible sequence for bb is thus 112244112244.

Example 4.2.

From the table in Figure 8, we compute κ032=𝔉221+𝔉032+𝔉131+ 0+𝔉230\operatorname{\kappa}_{032}=\operatorname{\mathfrak{F}}_{221}+\operatorname{\mathfrak{F}}_{032}+\operatorname{\mathfrak{F}}_{131}+\,0+\operatorname{\mathfrak{F}}_{230}. The only compatible sequence for b=23223b=23223 is 1222312223, so maxcomp(23223)=131\mathrm{maxcomp}(23223)=131.

This yields a formula for the Young fundamental slide expansion of Young key polynomials, proved similarly to Theorem 3.9.

Proposition 4.3.
κ^a=f(b)col(key(a))𝔉^rev(maxcomp(b)).\operatorname{\hat{\kappa}}_{a}=\sum_{f(b)\sim\operatorname{col}(\mathrm{key}(a))}\operatorname{\hat{\mathfrak{F}}}_{{\rm rev}(\mathrm{maxcomp}(b))}.

4.2. Quasi-key polynomials and fundamental particles

For a weak composition aa, define the quasi-key fillings QF(a)\mathrm{QF}(a) to be the (reverse) fillings of D(a)D(a) satisfying the following conditions.

  1. (1)

    Entries weakly decrease from left to right in each row.

  2. (2)

    No entry in row ii is greater than ii.

  3. (3)

    Entries strictly increase up the first column, and entries in any column are distinct.

  4. (4)

    All type A and type B triples are inversion triples.

The quasi-key polynomial is

𝔔a=TQF(a)xwt(T).\operatorname{\mathfrak{Q}}_{a}=\sum_{T\in\mathrm{QF}(a)}x^{{\rm wt}(T)}.

Quasi-key polynomials were first defined in [AS18b] as a lift of the quasisymmetric Schur functions to a basis of Polyn\mathrm{Poly}_{n}. The above formula is due to [MPS19]. For example, we have 𝔔103=x103+x112+x202+x121+x211+x130+x220\operatorname{\mathfrak{Q}}_{103}=x^{103}+x^{112}+x^{202}+x^{121}+x^{211}+x^{130}+x^{220} which is computed by the quasi-key fillings shown in Figure 11 below.

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Figure 11. The 7 quasi-key fillings of shape 103103.

The set of fillings ASSF(a)\mathrm{ASSF}(a) generating Demazure atoms is exactly the subset of QF(a)\mathrm{QF}(a) consisting of those fillings whose entries in the leftmost column are equal to their row index. For example, 𝒜103=x103+x112+x202+x121+x211\operatorname{\mathcal{A}}_{103}=x^{103}+x^{112}+x^{202}+x^{121}+x^{211}, which is computed by those fillings in Figure 11 whose leftmost column entries are 11 and 33.

Finally, define the particle fillings LF(a)\mathrm{LF}(a) to be the subset of ASSF(a)\mathrm{ASSF}(a) consisting of those fillings such that whenever i<ji<j, all entries in row ii are strictly smaller than all entries in row jj. Then the fundamental particle 𝔏a\operatorname{\mathfrak{L}}_{a} [Sea20] is defined to be

𝔏a=TLF(a)xwt(T).\operatorname{\mathfrak{L}}_{a}=\sum_{T\in\mathrm{LF}(a)}x^{{\rm wt}(T)}.

For example, 𝔏103=x103+x112+x121\operatorname{\mathfrak{L}}_{103}=x^{103}+x^{112}+x^{121}, by the 1st, 2nd, and 4th fillings in Figure 11.

We give a new formula for 𝔏a\operatorname{\mathfrak{L}}_{a} in terms of compatible sequences. Let S={p1,p2,,pk}S=\{p_{1},p_{2},\ldots,p_{k}\} be the set of the partial sums of the entries in aa with duplicate entries (obtained when an entry of aa is 0) removed. Then we say that a compatible sequence ww for the word formed by writing aia_{i} instances of ii consecutively is aa-flag compatible if for all piSp_{i}\in S, the letter in position pip_{i} of ww is equal to the row index of the ithi^{th} nonzero entry in aa.

Theorem 4.4.

Let aa be a weak composition of length nn, Then

𝔏a=w is a-flag compatiblexcomp(w).\operatorname{\mathfrak{L}}_{a}=\sum_{w\textrm{ is a-flag compatible}}x^{\mathrm{comp}(w)}.
Proof.

The statement follows from the fact that the aa-flag compatible sequences correspond to LF(a)\mathrm{LF}(a) via the following bijection. Let ww be an aa-flag compatible sequence and let w~(i)\tilde{w}^{(i)} be the subword of ww corresponding to the subword a(i)a^{(i)}. Construct the ithi^{th} row of a LF\mathrm{LF} by writing w~(i)\tilde{w}^{(i)} in weakly decreasing order. Conditions (1),(2), and (3) in the definition of a QF\mathrm{QF} are satisfied by construction. Condition (4) is satisfied since the entries in a given row are all smaller than all of the entries in any higher row. The flag condition guarantees that these fillings are in ASSF(a)\mathrm{ASSF}(a), and further, the fact that the entries in a given row are all smaller than all of the entries in any higher row implies these fillings are in LF(a)\mathrm{LF}(a). To obtain an aa-flag compatible sequence from an element of LF(a)\mathrm{LF}(a), record the entries in each row from right to left (to force them to be weakly increasing), reading rows from bottom to top. ∎

Figure 12 below shows how the bases discussed here expand into one another. An arrow indicates that the basis at the tail expands positively in the basis at the head. This figure is taken from that in [Sea20].

κa{\operatorname{\kappa}_{a}}𝔔a{\operatorname{\mathfrak{Q}}_{a}}𝔉a{\operatorname{\mathfrak{F}}_{a}}𝔐a{\operatorname{\mathfrak{M}}_{a}}𝒜a{\operatorname{\mathcal{A}}_{a}}𝔏a{\operatorname{\mathfrak{L}}_{a}}xa{x^{a}}[AS18]\scriptstyle{\rm[AS18]}[AS18]\scriptstyle{\rm[AS18]}[Sea20]\scriptstyle{\rm[Sea20]}[AS17]\scriptstyle{\rm[AS17]}[Sea20]\scriptstyle{\rm[Sea20]}[Sea20]\scriptstyle{\rm[Sea20]}
Figure 12. Positive expansions between bases defined by reverse fillings.

4.3. Young bases and intersections

Young analogues may be defined for all the families described above. Indeed, Young analogues of the fundamental slide polynomials and the quasi-key polynomials were introduced and utilized in [MS20]. In addition to the Young key polynomials and Young Demazure atoms studied in Section 3, Young analogues of the monomial slide polynomials and fundamental particles may be defined similarly, and these families can be shown (by utilizing Lemma 4.6 below) to exhibit positive expansions in Figure 13 analogous to those shown in Figure 12.

κ^a{\operatorname{\hat{\kappa}}_{a}}𝔔^a{\operatorname{\hat{\mathfrak{Q}}}_{a}}𝔉^a{\operatorname{\hat{\mathfrak{F}}}_{a}}𝔐^a{\operatorname{\hat{\mathfrak{M}}}_{a}}𝒜^a{\operatorname{\hat{\mathcal{A}}}_{a}}𝔏^a{\operatorname{\hat{\mathfrak{L}}}_{a}}xa{x^{a}}
Figure 13. Positive expansions between bases defined by Young fillings.
Remark 4.5.

All of the families of Young polynomials listed in Figure 13 are bases for Polyn\mathrm{Poly}_{n}, since their reverse analogues are bases for Polyn\mathrm{Poly}_{n} and the flip-and-reverse process is an involution on Polyn\mathrm{Poly}_{n} that preserves both cardinality and linear independence, cf. Proposition 3.1.

Lemma 4.6.

Let aa be a weak composition of length nn, and let Filla\mathrm{Fill}_{a} denote the set of all possible fillings of D(a)D(a) with entries from 1,,n1,\ldots,n, one entry per box. Define θ:FillaFillrev(a)\theta:\mathrm{Fill}_{a}\rightarrow\mathrm{Fill}_{{\rm rev}(a)} by letting θ(T)\theta(T) be the filling obtained by moving all boxes in row ii to row n+1in+1-i and replacing every entry jj with n+1jn+1-j, for all 1i,jn1\leq i,j\leq n. Then the following statements are true.

  1. (1)

    The map θ\theta is an involution.

  2. (2)

    If TT has weight bb then θ(T)\theta(T) has weight rev(b){\rm rev}(b).

  3. (3)

    The relative order of entries in row ii of TT is the reverse of the relative order of entries in row ii of θ(T)\theta(T).

  4. (4)

    The relative order of entries in any column of TT is the same as the relative order of entries in the same column of θ(T)\theta(T).

  5. (5)

    A triple of boxes in TT is an inversion triple if and only if the image of those boxes is a Young inversion triple in θ(T)\theta(T).

Proof.

The first four properties are immediate from the definition of θ\theta. Since the relative order of entries in the boxes of a triple in TT is the reverse of the relative order of entries in the images of those boxes in θ(T)\theta(T), it follows from the definition of inversion triples and Young inversion triples that the image of an inversion triple (of type A, respectively B) in TT must be a Young inversion triple (of type I, respectively II) in θ(T)\theta(T). Likewise, the images of non-inversion triples in TT are Young non-inversion triples in θ(T)\theta(T). ∎

Given a weak composition aa of length nn, define the Young fundamental fillings YFF(a)\mathrm{YFF}(a) of aa to be the fillings of D(a)D(a) with entries from 1,,n1,\ldots,n satisfying the following conditions.

  1. (1)

    Entries weakly increase from left to right in each row

  2. (2)

    No entry in row ii is less than ii

  3. (3)

    If a box with label bb is in a lower row than a box with label cc, then b<cb<c.

In particular, YFF(a)\mathrm{YFF}(a) is the image of FF(rev(a))\mathrm{FF}({\rm rev}(a)) under θ\theta. The Young fundamental slide polynomial 𝔉^a\operatorname{\hat{\mathfrak{F}}}_{a} [MS20] is the generating function of YFF(a)\mathrm{YFF}(a):

𝔉^a=TYFF(a)xwt(T).\operatorname{\hat{\mathfrak{F}}}_{a}=\sum_{T\in\mathrm{YFF}(a)}x^{{\rm wt}(T)}.

For example, we have 𝔉^301=x301+x211+x121+x031\operatorname{\hat{\mathfrak{F}}}_{301}=x^{301}+x^{211}+x^{121}+x^{031}, which is computed by the elements of YFF(301)\mathrm{YFF}(301) shown below.

3
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3
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3
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For a weak composition aa of length nn, define the Young monomial fillings YMF(a)\mathrm{YMF}(a) to be the subset of YFF(a)\mathrm{YFF}(a) for which all entries in any row are equal. Define the Young monomial slide polynomial 𝔐^a\operatorname{\hat{\mathfrak{M}}}_{a} to be the generating function of YMF(a)\mathrm{YMF}(a):

𝔐^a=TYMF(a)xwt(T).\operatorname{\hat{\mathfrak{M}}}_{a}=\sum_{T\in\mathrm{YMF}(a)}x^{{\rm wt}(T)}.

For example, we have 𝔐^301=x301+x031\operatorname{\hat{\mathfrak{M}}}_{301}=x^{301}+x^{031}.

Proposition 4.7.

The Young fundamental slide and the Young monomial slide bases of [x1,,xn]\mathbb{Z}[x_{1},\ldots,x_{n}] contain (respectively) the fundamental quasisymmetric and monomial quasisymmetric bases of quasisymmetric polynomials in nn variables. Specifically, if aa is a weak composition of length nn such that all zero entries are to the right of all nonzero entries, then

𝔉^a=Fflat(a)(x1,,xn) and 𝔐^a=Mflat(a)(x1,,xn),\operatorname{\hat{\mathfrak{F}}}_{a}=F_{{\rm flat}(a)}(x_{1},\ldots,x_{n})\qquad\mbox{ and }\qquad\operatorname{\hat{\mathfrak{M}}}_{a}=M_{{\rm flat}(a)}(x_{1},\ldots,x_{n}),

where flat(a){\rm flat}(a) is the composition obtained by deleting all 0 parts of aa.

Proof.

This is shown in [MS20] for Young fundamental slides. For monomial slides, since all nonzero entries of aa occur before all zero entries, the flag condition on YMF\mathrm{YMF} is always satisfied whenever the other conditions are satisfied. Hence the YMF\mathrm{YMF} are exactly the monomial Young composition tableaux (Proposition 2.3). ∎

Theorem 4.8.

The polynomials in [x1,,xn]\mathbb{Z}[x_{1},\ldots,x_{n}] that are both a fundamental (respectively, monomial) slide polynomial and a Young fundamental (respectively, monomial) slide polynomial are exactly the fundamental (respectively, monomial) quasisymmetric polynomials in nn variables.

In other words, {𝔉a}{𝔉^b}={Fα(x1,,xn)}\{\operatorname{\mathfrak{F}}_{a}\}\cap\{\operatorname{\hat{\mathfrak{F}}}_{b}\}=\{F_{\alpha}(x_{1},\ldots,x_{n})\} and {𝔐a}{𝔐^b}={Mα(x1,,xn)}\{\operatorname{\mathfrak{M}}_{a}\}\cap\{\operatorname{\hat{\mathfrak{M}}}_{b}\}=\{M_{\alpha}(x_{1},\ldots,x_{n})\}.

Proof.

We prove this in the fundamental case; the monomial case is completely analogous. First, let α\alpha be a composition of length (α)n\ell(\alpha)\leq n. Then

Fα(x1,xn)=𝔉0n(α)×α=𝔉^α×0n(α).F_{\alpha}(x_{1},\ldots x_{n})=\operatorname{\mathfrak{F}}_{0^{n-\ell(\alpha)}\times\alpha}=\operatorname{\hat{\mathfrak{F}}}_{\alpha\times 0^{n-\ell(\alpha)}}.

For the other direction, let 𝔉a\operatorname{\mathfrak{F}}_{a} be a fundamental slide polynomial that is not equal to Fα(x1,,xn)F_{\alpha}(x_{1},\ldots,x_{n}) for any composition α\alpha. This implies aa has a zero entry to the right of a nonzero entry ([AS17]). Let aja_{j} be the earliest such zero entry, so aj1a_{j-1} is nonzero. Let a¯\overline{a} denote the weak composition obtained by exchanging the entries aj1a_{j-1} and aja_{j}. Then xa𝔉ax^{a}\in\operatorname{\mathfrak{F}}_{a} and xa¯𝔉ax^{\overline{a}}\notin\operatorname{\mathfrak{F}}_{a}. However, if a Young fundamental slide polynomial contains xax^{a}, it must also contain xa¯x^{\overline{a}}. Hence 𝔉a\operatorname{\mathfrak{F}}_{a} is not equal to any Young fundamental slide polynomial. ∎

For a weak composition aa of length nn, define the Young quasi-key fillings YQF(a)\mathrm{YQF}(a) to be the (Young) fillings of D(a)D(a) obtained by applying θ\theta to QF(rev(a))\mathrm{QF}({\rm rev}(a)). Specifically, these are the fillings such that entries increase along rows, entries are at least their row index, entries strictly increase up the first column and entries in any column are distinct, and all type I and II Young triples are Young inversion triples. These generate the Young quasi-key polynomial 𝔔^a\operatorname{\hat{\mathfrak{Q}}}_{a} [MS20]. Unsurprisingly, the conditions governing the intersections of quasi-key and Young quasi-key polynomials are similar to those governing the intersections of the quasisymmetric bases that they extend (Theorem 2.13).

Theorem 4.9.

The polynomials that are both quasi-key and Young quasi-key polynomials are precisely the 𝔔^a\operatorname{\hat{\mathfrak{Q}}}_{a} such that aa is a number of equal parts followed by zeros, or a sequence of 11’s and 22’s followed by zeros, or aa has no zero parts and consecutive parts differ by at most 11.

Proof.

For any aa, the polynomial 𝔔^a\operatorname{\hat{\mathfrak{Q}}}_{a} contains the monomial xax^{a}, realised by TYQF(a)T\in\mathrm{YQF}(a) whose entries in row jj are all jj. Suppose a quasi-key polynomial 𝔔b\operatorname{\mathfrak{Q}}_{b} contains xax^{a}, realised by some SQF(b)S\in\mathrm{QF}(b). Suppose aa has a zero entry preceding a nonzero entry, e.g., ai=0a_{i}=0 but ai+1a_{i+1} is nonzero. Create SS^{\prime} by changing the rightmost i+1i+1 in SS to an ii. Since we change the rightmost i+1i+1, entries of SS^{\prime} still decrease along rows, and since no other ii’s exist in SS, entries still strictly increase up the first column of SS^{\prime} and do not repeat in any column of SS^{\prime}, and the relative order of the entries in any triple in SS remains unchanged. Hence SQF(b)S^{\prime}\in\mathrm{QF}(b), but there is no element of YQF(a)\mathrm{YQF}(a) that has this weight since all entries of TT are already minimal possible. Therefore 𝔔^a𝔔b\operatorname{\hat{\mathfrak{Q}}}_{a}\neq\operatorname{\mathfrak{Q}}_{b} for any bb.

It follows that for 𝔔^a\operatorname{\hat{\mathfrak{Q}}}_{a} to be equal to 𝔔b\operatorname{\mathfrak{Q}}_{b}, aa must consist of an interval of nonzero entries, followed by zero entries. But then 𝔔^a=𝒮^α(x1,,xn)\operatorname{\hat{\mathfrak{Q}}}_{a}=\operatorname{\hat{\mathscr{S}}}_{\alpha}(x_{1},\ldots,x_{n}) by [MS20]. The quasi-key polynomials that are quasisymmetric are exactly the quasisymmetric Schur polynomials: 𝔔b=𝒮β(x1,,xn)\operatorname{\mathfrak{Q}}_{b}=\operatorname{\mathscr{S}}_{\beta}(x_{1},\ldots,x_{n}) where bb is an interval of zero entries followed by an interval β\beta of nonzero entries [AS18]. Then, by Theorem 2.13, 𝒮^α(x1,,xn)\operatorname{\hat{\mathscr{S}}}_{\alpha}(x_{1},\ldots,x_{n}) is equal to 𝒮β(x1,,xn)\operatorname{\mathscr{S}}_{\beta}(x_{1},\ldots,x_{n}) exactly when α\alpha has all parts the same, or all parts of α\alpha are 11 or 22, or (α)=n\ell(\alpha)=n (so a=αa=\alpha has no zero parts) and consecutive parts differ by at most 11. ∎

Similarly, define the Young particle fillings YLF(a)\mathrm{YLF}(a) to be the image of LF(rev(a))\mathrm{LF}({\rm rev}(a)) under θ\theta. These Young fillings, which are the YASSF(a)\mathrm{YASSF}(a) such that any entry in a lower row is strictly smaller than any entry in a higher row, generate the Young fundamental particle 𝔏^a\operatorname{\hat{\mathfrak{L}}}_{a}.

Theorem 4.10.

The polynomials that are both fundamental particles and Young fundamental particles are precisely the 𝔏^a\operatorname{\hat{\mathfrak{L}}}_{a} such that aa has no zero part adjacent to a part of size at least 22.

Proof.

The LF\mathrm{LF} (respectively, YLF\mathrm{YLF}) obey all the conditions on ASSF\mathrm{ASSF} (respectively YASSF\mathrm{YASSF}), hence the same argument used in the proof of Theorem 3.3 shows that if 𝔏^a=𝔏b\operatorname{\hat{\mathfrak{L}}}_{a}=\operatorname{\mathfrak{L}}_{b} then a=ba=b.

If ai+1=0a_{i+1}=0 and ai2a_{i}\geq 2 for some ii, then let TYLF(a)T\in\mathrm{YLF}(a) such that all entries in each row jj are jj. Let also TYLF(a)T^{\prime}\in\mathrm{YLF}(a) be obtained by changing the rightmost ii to i+1i+1. Then there is no SLF(a)S\in\mathrm{LF}(a) with the same weight as TT^{\prime}, since the entries in SS above row ii must agree with those in TT^{\prime} above row ii, and then there is nowhere the new i+1i+1 could be placed in SS. Hence 𝔏^a𝔏a\operatorname{\hat{\mathfrak{L}}}_{a}\neq\operatorname{\mathfrak{L}}_{a}. A similar argument shows that if ai+12a_{i+1}\geq 2 and ai=0a_{i}=0 then 𝔏a𝔏^a\operatorname{\mathfrak{L}}_{a}\neq\operatorname{\hat{\mathfrak{L}}}_{a}.

Straightforwardly, 𝔏^a=𝔏a=xa\operatorname{\hat{\mathfrak{L}}}_{a}=\operatorname{\mathfrak{L}}_{a}=x^{a} if aa has no zero part next to a part of size at least 22. ∎

Remark 4.11.

While the Young and reverse analogues of a given basis have similar definitions, they have important structural differences. Unlike the reverse families, for each family of Young polynomials, the basis of Young polynomials of Polyn\mathrm{Poly}_{n} does not embed into Polyn+1\mathrm{Poly}_{n+1}. For example, 𝔉^0101=x2x4+x3x4Poly4\operatorname{\hat{\mathfrak{F}}}_{0101}=x_{2}x_{4}+x_{3}x_{4}\in\mathrm{Poly}_{4} is not a Young fundamental slide polynomial in Poly5\mathrm{Poly}_{5}. Because of this, we cannot use the typical definition of a weak composition as an infinite sequence of nonnegative integers (almost all zero); the number of entries in the sequence matters and the value of nn must be specified.

Remark 4.12.

Stable limits are obtained for all families in the top row of Figure 12 by prepending zeros to the weak composition aa; they are equal to the appropriate quasisymmetric function for flat(a){\rm flat}(a) ([Sea20, Theorem D]). While stable limits for the Young analogues of these families can be defined (by appending zeros to the weak composition aa), they are not equal to the quasisymmetric functions for flat(a){\rm flat}(a), except in the case that all nonzero entries of aa are to the left of all zero entries of aa. A polynomial satisfying this condition is in fact already quasisymmetric, and indeed limits to the expected quasisymmetric function.

For example, the stable limit of the Young fundamental slide polynomial 𝔉^230\operatorname{\hat{\mathfrak{F}}}_{230} (which is equal to F23(x1,x2,x3)F_{23}(x_{1},x_{2},x_{3})) is the (Young) fundamental quasisymmetric function F23F_{23}. However, the stable limit of 𝔉^203\operatorname{\hat{\mathfrak{F}}}_{203} is not F23F_{23}.

5. Young Schubert polynomials

Schubert polynomials were first introduced in [LS82] to represent Schubert classes in the cohomology of the flag manifold. Schubert polynomials are typically indexed by permutations. However, every permutation corresponds to a weak composition called a Lehmer code, which may also be used to index the Schubert polynomial. For each nn there is a \mathbb{Z}-basis for Polyn\mathrm{Poly}_{n} consisting of Schubert polynomials, however unlike the previously-discussed bases of Polyn\mathrm{Poly}_{n}, the indexing compositions of the Schubert basis elements are not compositions of length nn but of arbitrary length. It is a long-standing open problem to find a positive combinatorial formula for the structure constants of the Schubert basis. See [Mac91, Man98] for more details about the geometry, algebra, and combinatorics of Schubert polynomials.

We will take the combinatorial “pipe dreams” model introduced in [BB93] as our definition of Schubert polynomials. Consider a permutation wSnw\in S_{n}. The Lehmer code of ww is the weak composition L(w)L(w) of length nn whose ithi^{th} term equals the number of pairs (i,j)(i,j) with i<ji<j such that wi>wjw_{i}>w_{j}. For example, if w=31254w=31254 then L(w)=(2,0,0,1,0)L(w)=(2,0,0,1,0). A (reduced) pipe dream is a tiling of the first quadrant of ×\mathbb{Z}\times\mathbb{Z} with elbows and crosses so that any two of the resulting strands (called pipes) cross at most once. The associated permutation can be read from the diagram by following the pipes from the yy-axis to the xx-axis. Let PD(w)\mathrm{PD}(w) denote the set of pipe dreams for ww. The five pipe dreams in PD(31524)\mathrm{PD}(31524) are shown in Figure 14.

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Figure 14. The 55 pipe dreams associated to the permutation 3152431524.

Let wSnw\in S_{n}. The Schubert polynomial 𝔖w=𝔖w(x1,,xn)\operatorname{\mathfrak{S}}_{w}=\operatorname{\mathfrak{S}}_{w}(x_{1},\ldots,x_{n}) is given by

𝔖w=PPD(w)xwt(P),\operatorname{\mathfrak{S}}_{w}=\sum_{P\in\mathrm{PD}(w)}x^{{\rm wt}(P)},

where wt(P){\rm wt}(P) is the weak composition whose ithi^{th} term counts the crosses in the ithi^{th} row of PP.

For example, by Figure 14 the Schubert polynomial indexed by the permutation 3152431524 is

𝔖31524=x13x2+x12x22+x13x3+x12x2x3+x12x32.\operatorname{\mathfrak{S}}_{31524}=x_{1}^{3}x_{2}+x_{1}^{2}x_{2}^{2}+x_{1}^{3}x_{3}+x_{1}^{2}x_{2}x_{3}+x_{1}^{2}x_{3}^{2}.

Let Red(w)\mathrm{Red}(w) denote the set of reduced words for a permutation ww. Every Schubert polynomial can be written as a positive sum of key polynomials according to the following theorem.

Theorem 5.1 ([RS95, LS89]).
𝔖w=col(T)Red(w1)κwt(K0(T)),\operatorname{\mathfrak{S}}_{w}=\sum_{\operatorname{col}(T)\in\mathrm{Red}(w^{-1})}\operatorname{\kappa}_{{\rm wt}(K^{0}_{-}(T))},

where the sum is over semistandard Young tableaux TT, and K0(T)K^{0}_{-}(T) is the left nil key of TT, obtained via a modification of Knuth equivalence called nilplactic equivalence.

5.1. Young pipe dreams

Towards giving a combinatorial construction of the Young analogue of Schubert polynomials, we define a Young analogue of pipe dreams. Relabel the row indices (on the yy-axis) with nn as the bottom row, n1n-1 as the second row, and so on. Then read the “reversal” of the permutation by following the pipes from the yy-axis to the xx-axis. This reversal is the permutation ww read from right to left (in one-line notation), which we denote rev(w){\rm rev}(w). This new diagram is called the Young pipe dream corresponding to the permutation obtained by reading the pipes in this manner, and the set of all Young pipe dreams for a permutation ww is denoted YPD(w)\mathrm{YPD}(w). Let the Young Lehmer code of a permutation wSnw\in S_{n}, denoted (w)\mathcal{L}(w), be the weak composition of length nn whose ithi^{th} term is the number of pairs (i,j)(i,j) with i>ji>j such that wi>wjw_{i}>w_{j}. The Young weight ywt(P){\rm ywt}(P) of a Young pipe dream PP is the weak composition whose ithi^{th} part is the number of crosses in the ithi^{th} row from the top.

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Figure 15. The 55 elements of YPD(42513)\mathrm{YPD}(42513).

Let wSnw\in S_{n}. Then the Young Schubert polynomial 𝔖^w=𝔖^w(x1,,xn)\operatorname{\widehat{\mathfrak{S}}}_{w}=\operatorname{\widehat{\mathfrak{S}}}_{w}(x_{1},\ldots,x_{n}) is given by

𝔖^w=PYPD(w)xywt(P).\operatorname{\widehat{\mathfrak{S}}}_{w}=\sum_{P\in\mathrm{YPD}(w)}x^{{\rm ywt}(P)}.

For example, the Young Schubert polynomial associated to the permutation 4251342513 can be calculated by reading the Young weights of the Young pipe dreams in Figure 15 as follows:

𝔖^42513=x4x53+x42x52+x3x53+x3x4x52+x32x52.\operatorname{\widehat{\mathfrak{S}}}_{42513}=x_{4}x_{5}^{3}+x_{4}^{2}x_{5}^{2}+x_{3}x_{5}^{3}+x_{3}x_{4}x_{5}^{2}+x_{3}^{2}x_{5}^{2}.

It is straightforward to check that (rev(w))=rev(L(w))\mathcal{L}({\rm rev}(w))={\rm rev}(L(w)). It follows that

(5.1) 𝔖^w(x1,,xn)=𝔖rev(w)(xn,,x1).\operatorname{\widehat{\mathfrak{S}}}_{w}(x_{1},\ldots,x_{n})=\operatorname{\mathfrak{S}}_{{\rm rev}(w)}(x_{n},\ldots,x_{1}).
Remark 5.2.

Schubert polynomials have a well-known stability property, namely, for wSnw\in S_{n}, 𝔖w=𝔖in(w)\operatorname{\mathfrak{S}}_{w}=\operatorname{\mathfrak{S}}_{i_{n}(w)}, where in:SnSn+1i_{n}:S_{n}\rightarrow S_{n+1} is the embedding in which SnS_{n} acts on the first nn letters. The same is not true for Young Schubert polynomials, for example 𝔖^132=x2x3\operatorname{\widehat{\mathfrak{S}}}_{132}=x_{2}x_{3} but 𝔖^1324=x2x3x43\operatorname{\widehat{\mathfrak{S}}}_{1324}=x_{2}x_{3}x_{4}^{3}. Analogous to Remark 4.11, a Young Schubert polynomial in Polyn\mathrm{Poly}_{n} is not a Young Schubert polynomial in Polyn+1\mathrm{Poly}_{n+1}. Similarly, the stable limit of a Schubert polynomial (on prepending zeros to the Lehmer code) exists, and was shown in [Mac91] to be a Stanley symmetric function. Analogous to Remark 4.12, there is no corresponding stable limit for Young Schubert polynomials.

Remark 5.3.

The analogue of Remark 4.5 fails in this case: despite the fact that Schubert polynomials form a basis for Polyn\mathrm{Poly}_{n}, no collection of Young Schubert polynomials forms a basis for Polyn\mathrm{Poly}_{n}. This is due to the fact that the exponent of xix_{i} in a monomial in a Young Schubert polynomial is bounded by i1i-1. For Schubert polynomials this “staircase” condition goes the opposite way: the exponent of xix_{i} is bounded by mim-i when the indexing permutation is in SmS_{m}. Hence, by increasing mm as needed, one can find a Schubert polynomial in Polyn\mathrm{Poly}_{n} containing any given monomial in Polyn\mathrm{Poly}_{n}.

Remark 5.4.

For completeness, we note that no polynomial is both a Schubert and a Young Schubert polynomial. All Schubert polynomials have at least one monomial divisible by x1x_{1}, but no Young Schubert polynomials do.

A permutation ww is said to be vexillary if for every sequence a<b<c<da<b<c<d of indices, one never has wb<wa<wd<wcw_{b}<w_{a}<w_{d}<w_{c}. That is, ww is vexillary if and only if ww avoids the pattern 21432143. For ww vexillary, we have [LS90]

𝔖w=κL(w).\operatorname{\mathfrak{S}}_{w}=\operatorname{\kappa}_{L(w)}.

Thus the Young Schubert polynomials indexed by permutations whose reversal is vexillary are the Young key polynomials indexed by Young Lehmer codes of 34123412-avoiding permutations.

Theorem 5.1 and (3.1) yield the following formula for writing any Young Schubert polynomial as a positive sum of Young key polynomials.

𝔖^w=col(T)Red((rev(w))1)κ^rev(wt(K0(T))).\operatorname{\widehat{\mathfrak{S}}}_{w}=\sum_{\operatorname{col}(T)\in{\rm\mathrm{Red}}(({\rm rev}(w))^{-1})}\operatorname{\hat{\kappa}}_{{\rm rev}({\rm wt}(K^{0}_{-}(T)))}.

Other combinatorial descriptions of Schubert polynomials can similarly be translated into descriptions of Young Schubert polynomials.

Schubert polynomials were initially defined in terms of divided difference operators so that

𝔖w(x1,x2,,xn)=w1w0(x1n1x2n2xn1),\operatorname{\mathfrak{S}}_{w}(x_{1},x_{2},\ldots,x_{n})=\partial_{w^{-1}w_{0}}(x_{1}^{n-1}x_{2}^{n-2}\cdots x_{n-1}),

where w0=nn12 1w_{0}=n\;n-1\;\cdots 2\;1 is the longest permutation of an nn-element set and i(f)=fsi(f)xixi+1\partial_{i}(f)=\frac{f-s_{i}(f)}{x_{i}-x_{i+1}}. There is a natural way to describe Young Schubert polynomials in terms of divided difference operators, which we establish below. For wSnw\in S_{n}, let frev(w)\mathrm{frev}(w) be the permutation w0ww0w_{0}ww_{0}. It is straightforward to see that in one-line notation, frev(w)\mathrm{frev}(w) is obtained from ww by reversing the entries of ww and replacing each entry ii with n+1in+1-i, e.g. frev(31542)=42153\mathrm{frev}(31542)=42153.

Lemma 5.5.

Let si1sirs_{i_{1}}\cdots s_{i_{r}} be a reduced word for wSnw\in S_{n}. Then frev(w)=sni1snir\mathrm{frev}(w)=s_{n-i_{1}}\cdots s_{n-i_{r}}.

Proof.

We induct on the length of ww. If ww has length 0, then w=frev(w)=idw=\mathrm{frev}(w)=id and the statement holds. Now suppose the statement holds for all ww of length rr, for some r0r\geq 0. Suppose ww has an ascent in position jj, i.e. w(j)<w(j+1)w(j)<w(j+1). Then wsjws_{j} has length r+1r+1, and is obtained by exchanging the jjth and (j+1)(j+1)th entries of ww. We have wsj=si1sirsjws_{j}=s_{i_{1}}\cdots s_{i_{r}}s_{j}; we need to show frev(wsj)=sni1snirsnj\mathrm{frev}(ws_{j})=s_{n-i_{1}}\cdots s_{n-i_{r}}s_{n-j}. But sni1snirs_{n-i_{1}}\cdots s_{n-i_{r}} is equal to frev(w)\mathrm{frev}(w) by the inductive hypothesis, and therefore sni1snirsnjs_{n-i_{1}}\cdots s_{n-i_{r}}s_{n-j} is obtained from frev(w)\mathrm{frev}(w) by exchanging the entries in the (nj)(n-j)th and (nj+1)(n-j+1)th positions. This permutation is exactly frev(wsj)\mathrm{frev}(ws_{j}). ∎

Lemma 5.6.

Let ff be a polynomial in x1,,xnx_{1},\ldots,x_{n}, and let I(f)I(f) be defined as in Lemma 3.11. Then I(i1ir(f))=(1)rni1nir(I(f))I(\partial_{i_{1}}\cdots\partial_{i_{r}}(f))=(-1)^{r}\partial_{n-i_{1}}\cdots\partial_{n-i_{r}}(I(f)).

Proof.

We show that I(i(f))=ni(I(f))I(\partial_{i}(f))=-\partial_{n-i}(I(f)); after which repeated iteration establishes the result. To see this, consider the monomial xiaxi+1bx_{i}^{a}x_{i+1}^{b} where a>ba>b. (The case where a<ba<b is similar and if a=ba=b then i(xiaxi+1b)=0\partial_{i}(x_{i}^{a}x_{i+1}^{b})=0.)

I(i(xiaxi+1b))=I(xiaxi+1bxibxi+1axixi+1)\displaystyle I(\partial_{i}(x_{i}^{a}x_{i+1}^{b}))=I\left(\frac{x_{i}^{a}x_{i+1}^{b}-x_{i}^{b}x_{i+1}^{a}}{x_{i}-x_{i+1}}\right) =xn+1iaxnibxn+1ibxniaxn+1ixni\displaystyle=\frac{x_{n+1-i}^{a}x_{n-i}^{b}-x_{n+1-i}^{b}x_{n-i}^{a}}{x_{n+1-i}-x_{n-i}}
=ni(xnibxn+1ia)=ni(I(xiaxi+1b)).\displaystyle=-\partial_{n-i}(x_{n-i}^{b}x_{n+1-i}^{a})=-\partial_{n-i}(I(x_{i}^{a}x_{i+1}^{b})).\qed

We are now ready to establish a divided difference formula for 𝔖^w\operatorname{\widehat{\mathfrak{S}}}_{w}. The power of 1-1 appearing in the formula below is due solely to the fact that since we begin with x2x32xnn1x_{2}x_{3}^{2}\cdots x_{n}^{n-1}, we apply i\partial_{i} to a polynomial whose power of xix_{i} is smaller than its power of xi+1x_{i+1} in each monomial. The power of 1-1 could be defined away by replacing the denominator with xi+1xix_{i+1}-x_{i} in the definition of i\partial_{i}.

Theorem 5.7.

Let wSnw\in S_{n}. Then 𝔖^w(x1,x2,,xn)=(1)(w)w1(x2x32xnn1).\operatorname{\widehat{\mathfrak{S}}}_{w}(x_{1},x_{2},\ldots,x_{n})=(-1)^{\ell(w)}\partial_{w^{-1}}(x_{2}x_{3}^{2}\cdots x_{n}^{n-1}).

Proof.

Let si1sirs_{i_{1}}\cdots s_{i_{r}} be a reduced word for w1w^{-1}. Combining Lemmas 5.5 and 5.6, we have

I(frev(w1)(𝔖w0))=I(ni1nir(𝔖w0))=(1)ri1ir(I(𝔖w0))=(1)rw1(I(𝔖w0)).I(\partial_{\mathrm{frev}(w^{-1})}(\operatorname{\mathfrak{S}}_{w_{0}}))=I(\partial_{n-i_{1}}\cdots\partial_{n-i_{r}}(\operatorname{\mathfrak{S}}_{w_{0}}))=(-1)^{r}\partial_{i_{1}}\cdots\partial_{i_{r}}(I(\operatorname{\mathfrak{S}}_{w_{0}}))=(-1)^{r}\partial_{w^{-1}}(I(\operatorname{\mathfrak{S}}_{w_{0}})).

Recall that 𝔖^w=I(𝔖rev(w))\operatorname{\widehat{\mathfrak{S}}}_{w}=I(\operatorname{\mathfrak{S}}_{{\rm rev}(w)}), and in particular 𝔖^id=I(𝔖w0)=I(x1n1x2n2xn1)\operatorname{\widehat{\mathfrak{S}}}_{id}=I(\operatorname{\mathfrak{S}}_{w_{0}})=I(x_{1}^{n-1}x_{2}^{n-2}\cdots x_{n-1}). Note also that w01=w0w_{0}^{-1}=w_{0}, that (w)=(w1)\ell(w)=\ell(w^{-1}), and that ww0=rev(w)ww_{0}={\rm rev}(w). We therefore have

𝔖^w=I(𝔖rev(w))\displaystyle\operatorname{\widehat{\mathfrak{S}}}_{w}=I(\operatorname{\mathfrak{S}}_{{\rm rev}(w)}) =I((rev(w))1w0(𝔖w0))\displaystyle=I(\partial_{({\rm rev}(w))^{-1}w_{0}}(\operatorname{\mathfrak{S}}_{w_{0}}))
=I((ww0)1w0(𝔖w0))\displaystyle=I(\partial_{(ww_{0})^{-1}w_{0}}(\operatorname{\mathfrak{S}}_{w_{0}}))
=I(w0w1w0(𝔖w0))\displaystyle=I(\partial_{w_{0}w^{-1}w_{0}}(\operatorname{\mathfrak{S}}_{w_{0}}))
=I(frev(w1)(𝔖w0))\displaystyle=I(\partial_{\mathrm{frev}(w^{-1})}(\operatorname{\mathfrak{S}}_{w_{0}}))
=(1)(w)w1(I(𝔖w0))\displaystyle=(-1)^{\ell(w)}\partial_{w^{-1}}(I(\operatorname{\mathfrak{S}}_{w_{0}}))
=(1)(w)w1(𝔖^id)=(1)(w)w1(x2x32xnn1).\displaystyle=(-1)^{\ell(w)}\partial_{w^{-1}}(\operatorname{\widehat{\mathfrak{S}}}_{id})=(-1)^{\ell(w)}\partial_{w^{-1}}(x_{2}x_{3}^{2}\cdots x_{n}^{n-1}).\qed
Example 5.8.

Let w=2314=s1s2w=2314=s_{1}s_{2}. Then w1=3124=s2s1w^{-1}=3124=s_{2}s_{1} and we have

𝔖^2314=(1)(2314)(2314)1(x2x32x43)\displaystyle\operatorname{\widehat{\mathfrak{S}}}_{2314}=(-1)^{\ell(2314)}\partial_{(2314)^{-1}}(x_{2}x_{3}^{2}x_{4}^{3}) =(1)2(3124)(x2x32x43)\displaystyle=(-1)^{2}\partial_{(3124)}(x_{2}x_{3}^{2}x_{4}^{3})
=21(x2x32x43)\displaystyle=\partial_{2}\partial_{1}(x_{2}x_{3}^{2}x_{4}^{3})
=2(x2x32x43x1x32x43x1x2)\displaystyle=\partial_{2}\left(\frac{x_{2}x_{3}^{2}x_{4}^{3}-x_{1}x_{3}^{2}x_{4}^{3}}{x_{1}-x_{2}}\right)
=2(x32x43)\displaystyle=\partial_{2}(-x_{3}^{2}x_{4}^{3})
=x32x43x22x43x2x3=((x3+x2)x43)=x3x43+x2x43.\displaystyle=-\frac{x_{3}^{2}x_{4}^{3}-x_{2}^{2}x_{4}^{3}}{x_{2}-x_{3}}=-(-(x_{3}+x_{2})x_{4}^{3})=x_{3}x_{4}^{3}+x_{2}x_{4}^{3}.

Compare this to 𝔖rev(2314)=𝔖4132\operatorname{\mathfrak{S}}_{{\rm rev}(2314)}=\operatorname{\mathfrak{S}}_{4132}, which is equal to x13x2+x13x3x_{1}^{3}x_{2}+x_{1}^{3}x_{3}.

5.2. Demazure crystal structure

We use the recently developed crystal structure for Stanley symmetric functions [MS16] and the Demazure crystal structure for Schubert polynomials [AS18a] to generate the Demazure crystal structure for Young Schubert polynomials.

Let wSnw\in S_{n}. Following [MS16], a reduced factorization for ww is a partition of a reduced word for ww into blocks (possibly empty) of consecutive entries such that entries decrease from left to right within each block; let RF(w)\mathrm{RF}^{\ell}(w) denote the set of all reduced factorisations of ww with \ell blocks. In [MS16], a crystal structure is defined on RF(w)\mathrm{RF}^{\ell}(w). Precise definitions of the eie_{i} and fif_{i} operators may be found in [MS16, Section 3.2]. See Figure 16 for the crystal structure on RF3(21534)\mathrm{RF}^{3}(21534), with arrows fif_{i} labelled. For our purposes, we need to define the weight wt(r){\rm wt}(r) of rRF(w)r\in\mathrm{RF}^{\ell}(w) to be the weak composition of length nn given by (0,,0,|r|,|r1||r1|)(0,\ldots,0,|r^{\ell}|,|r^{\ell-1}|\ldots|r^{1}|) (as opposed to (|r|,|r1||r1|)(|r^{\ell}|,|r^{\ell-1}|\ldots|r^{1}|) used in [MS16]). In particular we define wt(r){\rm wt}(r) to begin with nn-\ell zeros, e.g., for (41)()(3)RF3(21534)(41)()(3)\in\mathrm{RF}^{3}(21534), we have n=5n=5, =3\ell=3 and wt((41)()(3))=00102{\rm wt}((41)()(3))=00102.

Let \ell be the position of the rightmost descent in ww. Define the reduced factorisations with Young cutoff for ww, denoted RFYC(w)\mathrm{RFYC}(w), to be those elements of RF(w)\mathrm{RF}^{\ell}(w) such that the smallest entry in the ithi^{th} block from the left is at least ii. See Figure 16, in which the elements of RFYC(21534)\mathrm{RFYC}(21534) are bolded. Compare this to the reduced factorisations with cutoff defined in [AS18a].

Theorem 5.9.

The Young Schubert polynomial 𝔖^w\operatorname{\widehat{\mathfrak{S}}}_{w} is equal to rRFYC(rev(w))xwt(r)\sum_{r\in\mathrm{RFYC}({\rm rev}(w))}x^{{\rm wt}(r)}. Moreover, RFYC(w)\mathrm{RFYC}(w) is a union of Demazure crystals, under the convention that we begin with the lowest weight rather than the highest and use the fif_{i} operators.

Proof.

In [AS18a], a crystal structure isomorphic to that of [MS16] is obtained by reversing each reduced factorisation for ww (thus obtaining reduced factorisations for w1w^{-1} partitioned into increasing blocks), and exchanging the roles of fif_{i} with enie_{n-i} and eie_{i} with fnif_{n-i}. Restricting this isomorphism to RFYC(w)\mathrm{RFYC}(w) gives the set of reduced factorisations with cutoff for w1w^{-1}, of which the weight generating function is 𝔖w\operatorname{\mathfrak{S}}_{w} [AS18a]. Since this isomorphism is weight-reversing, it follows from (5.1) that the weight generating function of RFYC(w)\mathrm{RFYC}(w) is 𝔖^rev(w)\operatorname{\widehat{\mathfrak{S}}}_{{\rm rev}(w)}. By [AS18a, Theorem 5.11], reduced factorisations with cutoff have a Demazure crystal structure, and the isomorphism implies RFYC(w)\mathrm{RFYC}(w) is a union of Demazure truncations of the components of RF(w)\mathrm{RF}(w), starting with the lowest weight. ∎

()()(431)()()(431)()(4)(31)()(4)(31)(4)()(31)(4)()(31)()(43)(1)()(43)(1)(4)(3)(1)(4)(3)(1)()(431)()()(431)()(43)()(1)(43)()(1)(4)(31)()(4)(31)()(43)(1)()(43)(1)()(𝟒𝟑𝟏)()(){\bf(431)()()}()(1)(43)()(1)(43)(𝟏)()(𝟒𝟑){\bf(1)()(43)}()()(41)(3)()()(41)(3)(𝟏)(𝟒)(𝟑){\bf(1)(4)(3)}(4)(1)(3)(4)(1)(3)(𝟒𝟏)()(𝟑){\bf(41)()(3)}(𝟏)(𝟒𝟑)(){\bf(1)(43)()}(𝟒𝟏)(𝟑)(){\bf(41)(3)()}1111111111111111111122222222222222222222
Figure 16. The crystal on RF3(21534)\mathrm{RF}^{3}(21534) and the subcrystal RFYC(21534)\mathrm{RFYC}(21534) (bold).

The Demazure crystal structure provides another method for expanding Young Schubert polynomials in Young key polynomials, cf. [AS18a, Corollary 5.12].

Example 5.10.

Figure 16 demonstrates that 𝔖^43512=κ^00003+κ^00201\operatorname{\widehat{\mathfrak{S}}}_{43512}=\operatorname{\hat{\kappa}}_{00003}+\operatorname{\hat{\kappa}}_{00201}, where κ^00003=x53\operatorname{\hat{\kappa}}_{00003}=x_{5}^{3} is the bolded Demazure truncation of the left component and κ^00201=x4x52+x42x5+x3x52+x3x4x5+x32x5\operatorname{\hat{\kappa}}_{00201}=x_{4}x_{5}^{2}+x_{4}^{2}x_{5}+x_{3}x_{5}^{2}+x_{3}x_{4}x_{5}+x_{3}^{2}x_{5} is the bolded Demazure truncation of the right component.

Acknowledgements

We thank Sami Assaf and Anne Schilling for suggesting a connection with the Demazure crystal structure for Schubert polynomials, and Martha Precup and Brendon Rhoades for pointing out further recent appearances of the Young/reverse dichotomy for polynomials and tableaux. We also thank Vic Reiner for pointing out a connection to evacuation in Section 3.

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Appendix A Intersections of polynomial families

In this appendix we determine the polynomials that are both quasisymmetric Schur and Young quasisymmetric Schur polynomials. Throughout, let \ell be the length of α\alpha and nn\geq\ell the number of variables.

Lemma A.1.

If 𝒮^α(x1,,xn)=𝒮β(x1,,xn)\operatorname{\hat{\mathscr{S}}}_{\alpha}(x_{1},\ldots,x_{n})=\operatorname{\mathscr{S}}_{\beta}(x_{1},\ldots,x_{n}), then α=β\alpha=\beta.

Proof.

By the same argument in the proof of Theorem 3.3, if 𝒮^α=𝒮β\operatorname{\hat{\mathscr{S}}}_{\alpha}=\operatorname{\mathscr{S}}_{\beta}, then β\beta must be a rearrangement of α\alpha. Therefore suppose β\beta rearranges α\alpha and the length of α\alpha (and thus of β\beta) is \ell. Let TYCT(α)T\in\mathrm{YCT}(\alpha) be such that the entries in each row jj are all jj. Suppose SRCT(β)S\in\mathrm{RCT}(\beta) has the same weight as TT. Since the first column of SS must increase strictly from top to bottom, and we must use all entries 11 through \ell in SS, the first entry in each row jj of SS is forced to be jj. By the same argument in the proof of Theorem 3.3, the set of entries in each column of SS must be the same as that in the corresponding column of TT.

Suppose βα\beta\neq\alpha, and let ii be the largest index such that βiαi\beta_{i}\neq\alpha_{i}. Consider rows \ell down to i+1i+1, where the row lengths are identical in α\alpha and β\beta. Since entries of SS must decrease along rows, the \ell’s can only go in the \ellth row of SS, and thus completely fill the \ellth row of SS. By the same reasoning, all 1\ell-1’s must go in row 1\ell-1 of SS, and so forth down to (and including) row i+1i+1. Now, if βi<αi\beta_{i}<\alpha_{i}, it is impossible to place αi\alpha_{i} many ii’s in row ii of SS, but ii’s cannot go in any lower row of SS since entries must decrease along rows, and cannot go in any higher row of SS since all boxes above row ii are occupied, so we cannot construct SS of the same weight as TT. So assume βi>αi\beta_{i}>\alpha_{i}. Then we must place αi\alpha_{i} many ii’s in the first αi\alpha_{i} boxes of the iith row. The next entry placed in this row (in column αi+1\alpha_{i}+1) is some x<ix<i. Since the column sets of TT and SS must agree and each column set of TT is a subset of the previous one, there must be an xx in column αi\alpha_{i} of SS. Since all boxes weakly above row ii in this column are occupied by entries at least ii, xx must be strictly below row ii in this column. But then these two copies of xx must violate one of the triple conditions in SS. It follows that if αβ\alpha\neq\beta, then there is no SRCT(β)S\in\mathrm{RCT}(\beta) with the same weight as TYCT(α)T\in\mathrm{YCT}(\alpha), and thus 𝒮^α𝒮β\operatorname{\hat{\mathscr{S}}}_{\alpha}\neq\operatorname{\mathscr{S}}_{\beta}. ∎

Therefore, the question reduces to determining when 𝒮^α(x1,,xn)=𝒮α(x1,,xn)\operatorname{\hat{\mathscr{S}}}_{\alpha}(x_{1},\ldots,x_{n})=\operatorname{\mathscr{S}}_{\alpha}(x_{1},\ldots,x_{n}).

Lemma A.2.

Let α\alpha be a composition of length \ell. If there are i<ki<k such that

  1. (1)

    αiαk2\alpha_{i}\leq\alpha_{k}-2 and there is no i<j<ki<j<k such that αj=αk1\alpha_{j}=\alpha_{k}-1, or

  2. (2)

    αiαk+2\alpha_{i}\geq\alpha_{k}+2 and there is no i<j<ki<j<k such that αj=αi1\alpha_{j}=\alpha_{i}-1

then 𝒮^α(x1,,xn)𝒮α(x1,,xn)\operatorname{\hat{\mathscr{S}}}_{\alpha}(x_{1},\ldots,x_{n})\neq\operatorname{\mathscr{S}}_{\alpha}(x_{1},\ldots,x_{n}).

Proof.

For (1), create SRCT(α)S\in\mathrm{RCT}(\alpha) by letting all entries be equal to their row index, except the last entry of row kk is ii. The condition that there is no i<j<ki<j<k such that αj=αk1\alpha_{j}=\alpha_{k}-1 ensures SS does not violate the triple condition (B). Then wt(S)=(α1,αi+1,,αk1,α,0,,0){\rm wt}(S)=(\alpha_{1},\ldots\alpha_{i}+1,\ldots,\alpha_{k}-1,\ldots\alpha_{\ell},0,\ldots,0). One cannot create TYCT(α)T\in\mathrm{YCT}(\alpha) with weight equal to that of SS. The first column of TT must contain the entries 11 through \ell from bottom to top, i.e., the first entry of each row is the row index. Then since entries must increase along rows of TT, all α1\alpha_{1} 11’s must be in row 11, α2\alpha_{2} 22’s in row 22, etc, but then one cannot place αi+1\alpha_{i}+1 ii’s in row ii, since its length is αi\alpha_{i}. Hence 𝒮^α𝒮α\operatorname{\hat{\mathscr{S}}}_{\alpha}\neq\operatorname{\mathscr{S}}_{\alpha}. The proof of (2) is similar, starting by creating TYCT(α)T\in\mathrm{YCT}(\alpha) whose entries in each row are equal to their row index, except the last entry of row ii is kk. ∎

It follows from Lemma A.2 that the only α\alpha where 𝒮^α(x1,,xn)\operatorname{\hat{\mathscr{S}}}_{\alpha}(x_{1},\ldots,x_{n}) could possibly be equal to 𝒮α(x1,,xn)\operatorname{\mathscr{S}}_{\alpha}(x_{1},\ldots,x_{n}) are those α\alpha such that for each ii, |αiαi+1|1|\alpha_{i}-\alpha_{i+1}|\leq 1.

Lemma A.3.

Let α\alpha be a composition of length \ell and n>n>\ell. If 2αi<αi+12\leq\alpha_{i}<\alpha_{i+1} or 2αi+1<αi2\leq\alpha_{i+1}<\alpha_{i} for some ii, then 𝒮^α(x1,,xn)𝒮α(x1,,xn)\operatorname{\hat{\mathscr{S}}}_{\alpha}(x_{1},\ldots,x_{n})\neq\operatorname{\mathscr{S}}_{\alpha}(x_{1},\ldots,x_{n}).

Proof.

Suppose 2αi<αi+12\leq\alpha_{i}<\alpha_{i+1}. Construct TYCT(α)T\in\mathrm{YCT}(\alpha) by letting all entries be equal to their row index in the first i+1i+1 rows, except the last entry of row ii is i+2i+2, and then all entries of each row rr for r>i+1r>i+1 are r+1r+1. Since αi<αi+1\alpha_{i}<\alpha_{i+1}, the triple condition (II) is not violated. Now attempt to construct SRCT(α)S\in\mathrm{RCT}(\alpha) with weight equal to that of TT. All +1\ell+1’s must go in row \ell, then all \ell’s in row 1\ell-1, down to and including row i+2i+2. The sole i+2i+2 must be the first entry in row i+1i+1, since all boxes above row i+1i+1 are occupied. The i+1i+1’s can’t all fit in row i+1i+1, so necessarily the first entry in row ii must be i+1i+1 if all i+1i+1’s are to be placed. This means all i+1i+1’s must be placed in row i+1i+1 or row ii. Since αi<αi+1\alpha_{i}<\alpha_{i+1}, they cannot all be placed in row ii; at least one must be in row i+1i+1, immediately following the entry i+2i+2. But then the i+1i+1 in row ii, column 11, the i+2i+2 in row i+1i+1, column 11, and the i+1i+1 in row i+1i+1, column 22 violate the triple condition (B).

For α\alpha satisfying 2αi+1<αi2\leq\alpha_{i+1}<\alpha_{i}, a similar argument works by letting SRCT(α)S\in\mathrm{RCT}(\alpha) be such that entries are equal to to their row index in the first i1i-1 rows, then all entries of each row rr for rir\geq i are r+1r+1, except the last entry of row i+1i+1 is ii. ∎

Proof of Theorem 2.13: It follows from Lemmas A.2 and A.3 that the only α\alpha where 𝒮^α\operatorname{\hat{\mathscr{S}}}_{\alpha} could possibly be equal to 𝒮α\operatorname{\mathscr{S}}_{\alpha} are those α\alpha whose parts are all the same, those α\alpha whose parts are all 11 or 22, or (only when n=(α)n=\ell(\alpha)) those α\alpha whose consecutive parts differ by at most one.

If all parts of α\alpha are the same, then 𝒮α\operatorname{\mathscr{S}}_{\alpha} and 𝒮^α\operatorname{\hat{\mathscr{S}}}_{\alpha} are both equal to the Schur function sαs_{\alpha} by Proposition 2.11).

If all parts of α\alpha are 11 or 22, define a map ψ\psi on tableaux of shape α\alpha by swapping the entries in each row of length 22, and then reordering the rows so the first column is increasing from top to bottom. We will show that ψ\psi restricts to a bijection between YCT(α)\mathrm{YCT}(\alpha) and RCT(α)\mathrm{RCT}(\alpha). First we observe ψ\psi maps each TYCT(α)T\in\mathrm{YCT}(\alpha) to a tableau of shape α\alpha: if a row of length 11 is above a row of length 22 in TT, then the entry in the row of length 11 must be larger than both entries of the row of length 22, the first due to the increasing first column, and the second due to the triple condition (II). If a row of length 11 is below a row of length 22, then the entry in the row of length 11 must be smaller that both entries of the row of length 22, due to the increasing first column and the fact that entries increase along rows. Hence re-ordering occurs only amongst rows of length 22 that do not have a row of length 11 between them. In particular, re-ordering never exchanges a row of length 11 and a row of length 22.

Next we show that if TYCT(α)T\in\mathrm{YCT}(\alpha), then ψ(T)\psi(T) has no repeated entries in any column. Suppose there are two instances of the same entry ii in TT. The ii in column 22 cannot be strictly above the ii in column 11 because entries increase along rows and strictly increase up the first column. Also, the ii in column 2 cannot be strictly below the ii in column 1, or these two instances of ii would violate one of the triple conditions. Therefore, the ii’s must be in the same row of TT, and so cannot be in the same column of ψ(T)\psi(T).

Now we show ψ(T)RCT(α)\psi(T)\in\mathrm{RCT}(\alpha). By definition, entries decrease along rows of ψ(T)\psi(T) and increase up the first column. First consider type B triples in ψ(T)\psi(T), in which case the lower row in the triple has length 11. All entries above a given row of length 11 in TT are strictly larger than that entry, since entries increase along rows and up the first column. So the same is true in ψ(T)\psi(T), and the type B triple rule is satisfied. Now consider type A triples in ψ(T)\psi(T). Then both rows in the triple have length 22. If these rows are not swapped under ψ\psi, then in TT the second entry in the higher row is larger than the second entry in the lower row. Combining this with the triple condition in TT and the increasing first column, both entries of the higher row must be strictly larger than both entries of the lower row in TT. This implies the same is true in ψ(T)\psi(T), hence the type A triple rule is satisfied. If they are swapped, we have

Txyzwwzyxψ(T)T\ni\vbox{ \halign{&\tableaucell{#}\cr\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$x$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}&\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$y$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}\\\vbox to9.68747pt{\vfil\hrule width=9.68747pt,height=0.0pt}&\vbox to9.68747pt{\vfil\hrule width=9.68747pt,height=0.0pt}&\vbox to9.68747pt{\vfil\hrule width=9.68747pt,height=0.0pt}\\\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$z$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}&\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$w$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}\crcr}}\mapsto\vbox{ \halign{&\tableaucell{#}\cr\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$w$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}&\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$z$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}\\\vbox to9.68747pt{\vfil\hrule width=9.68747pt,height=0.0pt}&\vbox to9.68747pt{\vfil\hrule width=9.68747pt,height=0.0pt}&\vbox to9.68747pt{\vfil\hrule width=9.68747pt,height=0.0pt}\\\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$y$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}&\begin{picture}(1.0,1.0)\put(0.0,0.22){\makebox(1.0,1.0)[b]{$x$}} \put(0.0,0.0){\line(1,0){1.0}} \put(0.0,1.0){\line(1,0){1.0}} \put(0.0,0.0){\line(0,1){1.0}} \put(1.0,0.0){\line(0,1){1.0}} \end{picture}\crcr}}\in\psi(T)

where z<xz<x and y<wy<w. Note that z<yz<y, since x<yx<y, but also z<xz<x, so the triple involving x,y,zx,y,z in ψ(T)\psi(T) satisfies the type A triple rule. Hence the map ψ\psi sends YCT(α)\mathrm{YCT}(\alpha) to RCT(α)\mathrm{RCT}(\alpha). A similar argument shows ψ\psi sends RCT(α)\mathrm{RCT}(\alpha) to YCT(α)\mathrm{YCT}(\alpha) and that ψψ\psi\circ\psi is the identity when restricted to either RCT(α)\mathrm{RCT}(\alpha) or YCT(α)\mathrm{YCT}(\alpha), so ψ:YCT(α)RCT(α)\psi:\mathrm{YCT}(\alpha)\rightarrow\mathrm{RCT}(\alpha) is a bijection. Since ψ\psi is also weight-preserving, this implies 𝒮^α=𝒮α\operatorname{\hat{\mathscr{S}}}_{\alpha}=\operatorname{\mathscr{S}}_{\alpha}.

Finally if consecutive parts of α\alpha differ by at most one and n=(α)n=\ell(\alpha), the only element of YCT(α)\mathrm{YCT}(\alpha) and RCT(α)\mathrm{RCT}(\alpha) is the tableau whose entries in each row ii are all ii, and thus 𝒮^α=xα=𝒮α\operatorname{\hat{\mathscr{S}}}_{\alpha}=x^{\alpha}=\operatorname{\mathscr{S}}_{\alpha}. We proceed by induction on the number of columns of D(α)D(\alpha). Suppose TYCT(α)T\in\mathrm{YCT}(\alpha); certainly in the first column the entry in each row ii must be ii. Suppose this is true for the first cc columns. Consider the boxes in column c+1c+1 from highest to lowest. If the highest box 𝔟\mathfrak{b} is in the top row (i.e., row nn) then it must have entry nn by the increasing row condition. If it is in row i<ni<n, then row i+1i+1 must be one box shorter than row ii (since 𝔟\mathfrak{b} is highest in its column and consecutive parts of α\alpha differ by at most one), and the box in row i+1i+1, column cc must have entry i+1i+1 (by assumption). Then 𝔟\mathfrak{b} cannot have entry greater than ii or the triple condition (II) is violated, so 𝔟\mathfrak{b} must have entry ii by the increasing row condition and the fact that (by assumption) the box immediately left of 𝔟\mathfrak{b} has entry ii.

Now suppose the highest kk boxes in column cc have entry equal to their row index, and suppose the (k+1)(k+1)th highest box 𝔟\mathfrak{b} is in row ii. If every row above row ii has a box in column c+1c+1, then by assumption these boxes all have entry equal to their row index, and then 𝔟\mathfrak{b} must have entry ii since its entry is at least ii, and entries cannot repeat in a column. Otherwise, consider the lowest row ii^{\prime} above row ii that does not have a box in column c+1c+1. Since consecutive parts of α\alpha differ by at most one, the rightmost box in row ii^{\prime} must be in column cc, and thus by assumption it has entry ii^{\prime}. Then 𝔟\mathfrak{b} must have entry strictly smaller than ii^{\prime}, otherwise triple condition (II) is violated by 𝔟\mathfrak{b}, the box immediately left of 𝔟\mathfrak{b} (which has entry ii), and the rightmost box in row ii^{\prime}. But since ii^{\prime} was the lowest row above row ii without a box in column c+1c+1, there are boxes in rows i+1,,i1i+1,\ldots,i^{\prime}-1 and column c+1c+1 with entry equal to their row index. Therefore, since entries cannot repeat in a column, the entry in 𝔟\mathfrak{b} must be ii. It follows that all boxes in column c+1c+1 have entry equal to their row index, and then that all boxes in TT have entry equal to their row index. A similar argument shows that TT is also the only element of RCT(α)\mathrm{RCT}(\alpha). ∎