%% The class cedram-ALCO is just a wrapper around amsart.cls (version 2)
%% implementing the layout of the journal, and some additionnal
%% administrative commands.
%% You can place one option:
%% * "Unicode" if the file is UTF-8 encoded.
\documentclass[ALCO,Unicode]{cedram}
%% Here you might want to add some standard packages if those
%% functionnalities are required.
%\usepackage[matrix,arrow,tips,curve]{xy}
% ...
%% The production will anyway use amsmath (all ams utilities except
%% amscd for commutative diagrams which you need to load explicilty if
%% required), hyperref, graphicx, mathtools, enumitem...
\usepackage{tikz}
\usepackage{colortbl}
\usepackage{cancel}
\equalenv{defn}{defi}
\equalenv{remark}{rema}
\equalenv{example}{exam}
\equalenv{proposition}{prop}
\equalenv{cor}{coro}
\equalenv{corollary}{cor}
%% User definitions if necessary... Such definitions are forbidden
\DeclareMathOperator{\PQT}{Tab}
\DeclareMathOperator{\PDS}{PDS}
\DeclareMathOperator{\Sym}{Sym}
\DeclareMathOperator{\Skip}{skip}
\DeclareMathOperator{\Flat}{Flat}
\DeclareMathOperator{\perm}{perm}
\DeclareMathOperator{\Coskip}{coskip}
\DeclareMathOperator{\ST}{ST}
\DeclareMathOperator{\MQT}{MQT}
\DeclareMathOperator{\PQ}{PQ}
\DeclareMathOperator{\Rev}{RevPQ}
\DeclareMathOperator{\MMLQ}{PQ}
\DeclareMathOperator{\dg}{dg}
\DeclareMathOperator{\T}{\mathcal{T}}
\DeclareMathOperator{\Z}{\mathbb{Z}}
\DeclareMathOperator{\skydg}{\diagram'}
\DeclareMathOperator{\augdg}{\widehat{\diagram}}
\DeclareMathOperator{\arm}{arm}
\DeclareMathOperator{\leg}{leg}
\DeclareMathOperator{\maj}{maj}
\DeclareMathOperator{\comaj}{comaj}
\DeclareMathOperator{\inv}{inv}
\DeclareMathOperator{\inc}{inc}
\DeclareMathOperator{\Des}{Des}
\DeclareMathOperator{\coinv}{coinv}
\DeclareMathOperator{\ribbon}{ribbon}
\DeclareMathOperator{\Inv}{Inv}
\DeclareMathOperator{\wt}{wt}
\DeclareMathOperator{\quinvL}{\quinv^{(\ell)}}
\DeclareMathOperator{\quinv}{quinv}
\DeclareMathOperator{\Bot}{Bot}
\DeclareMathOperator{\dinv}{dinv}
\DeclareMathOperator{\MP}{MP}
\DeclareMathOperator{\shape}{\bm \lambda}
\DeclareMathOperator{\TAZRP}{TAZRP}
\DeclareMathOperator{\Tab}{Tab}
\DeclareMathOperator{\South}{South}
\DeclareMathOperator{\SSYT}{SSYT}
\DeclareMathOperator{\LLT}{LLT}
\DeclareMathOperator{\rw}{rw}
\DeclareMathOperator{\boldnu}{\widehat{\boldsymbol{\nu}}}
\DeclareMathOperator{\invtwo}{\widehat{\inv}}
\DeclareMathOperator{\armtwo}{\widehat{\arm}}
\newcommand{\qbinom}[2]{\bgroup\renewcommand*{\arraystretch}{1}\begin{bmatrix} #1 \\ #2\end{bmatrix} \egroup}
\newcommand{\cQ}{\mathcal{Q}}
\newcommand{\tcr}[1]{#1}%{\textcolor{red}{#1}}
\newcommand{\tcb}[1]{\textcolor{blue}{#1}}
\newcommand\marginal[1]{\marginpar[\raggedleft\tiny #1]{\raggedright\tiny #1}}
\renewcommand{\AA}[1]{\marginal{\textcolor{red}{AA: #1}}}
\newcommand{\OM}[1]{\marginal{\textcolor{magenta}{OM: #1}}}
\newcommand{\JBM}[1]{\marginal{\textcolor{blue}{JBM: #1}}}
%%% tableau macros %%%
\newlength\cellsize \setlength\cellsize{12\unitlength}
\savebox2{%
\begin{picture}(12,12)
\put(0,0){\line(1,0){12}}
\put(0,0){\line(0,1){12}}
\put(12,0){\line(0,1){12}}
\put(0,12){\line(1,0){12}}
\end{picture}}
\newcommand\cellify[1]{\def\thearg{#1}\def\nothing{}%
\ifx\thearg\nothing
\vrule width0pt height\cellsize depth0pt\else
\hbox to 0pt{\usebox2\hss}\fi%
\vbox to 12\unitlength{
\vss
\hbox to 12\unitlength{\hss$#1$\hss}
\vss}}
\newcommand\tableau[1]{\vtop{\let\\=\cr
\setlength\baselineskip{-16000pt}
\setlength\lineskiplimit{16000pt}
\setlength\lineskip{0pt}
\halign{&\cellify{##}\cr#1\crcr}}}
\savebox3{%
\begin{picture}(12,12)
\put(0,0){\line(1,0){12}}
\put(0,0){\line(0,1){12}}
\put(12,0){\line(0,1){12}}
\put(0,12){\line(1,0){12}}
\end{picture}}
\newcommand\expath[1]{%
\hbox to 0pt{\usebox3\hss}%
\vbox to 12\unitlength{
\vss
\hbox to 12\unitlength{\hss$#1$\hss}
\vss}}
\newcommand\cell[3]{
\def\i{#1} \def\j{#2} \def\entry{#3}
\draw (\j-1,-\i)--(\j,-\i)--(\j,-\i+1)--(\j-1,-\i+1)--(\j-1,-\i);
\node at (\j-.5,-\i+.5) {\entry};
}
\newcommand{\qtrip}[3]{
\begin{tikzpicture}[scale=0.5]
\cell{1}{0}{#1} \cell{2}{0}{#2} \cell{2}{2.7}{#3}
\node at (1,-1.5) {$\cdots$};
\end{tikzpicture}
}
\newcommand{\trip}[3]{
\resizebox{!}{.5cm}{
\begin{tikzpicture}[scale=0.5]
\cell{1}{0}{#1} \cell{2}{0}{#2} \cell{2}{1.2}{#3}
\end{tikzpicture}
}
}
\newcommand\cellL[4]{
\def\i{#1} \def\j{#2} \def\entry{#3} \def\labl{#4}
\draw (\j-1,-\i)--(\j,-\i)--(\j,-\i+1)--(\j-1,-\i+1)--(\j-1,-\i);
\node at (\j-.5,-\i+.5) {\entry};
\node at (\j-1.3,-\i+.8) {\tiny\labl};
}
\newcommand\graycell[3]{
\def\i{#1} \def\j{#2} \def\entry{#3}
\draw[fill=gray!60](\j-1,-\i)--(\j,-\i)--(\j,-\i+1)--(\j-1,-\i+1)--(\j-1,-\i);
\node at (\j-.5,-\i+.5) {\entry};
}
\newcommand\circleT[5]{
\def \n {5}
\def \radius {.5cm}
\def \margin {20}
\edef\s{0}
\pgfmathparse{\s+1};
\foreach \en in {#1,#2,#3,#4,#5} {%
\node at ({360/\n * (\s - 1)}:\radius) {$\en$};
\draw[>=latex] ({360/\n * (\s - 1)+\margin}:\radius)
arc ({360/\n * (\s - 1)+\margin}:{360/\n * (\s)-\margin}:\radius);
\pgfmathparse{\s+1}
\xdef\s{\pgfmathresult}
}
}
\newcommand\circleS[4]{
\def \n {4}
\def \radius {.5cm}
\def \margin {20}
\edef\s{0}
\pgfmathparse{\s+1};
\foreach \en in {#1,#2,#3,#4} {%
\node at ({360/\n * (\s - 1)}:\radius) {$\en$};
\draw[>=latex] ({360/\n * (\s - 1)+\margin}:\radius)
arc ({360/\n * (\s - 1)+\margin}:{360/\n * (\s)-\margin}:\radius);
\pgfmathparse{\s+1}
\xdef\s{\pgfmathresult}
}
}
\DeclarePairedDelimiter\abs{\lvert}{\rvert}
\datereceived{2021-08-09}
\daterevised{2022-03-15}
\dateaccepted{2022-06-14}
\begin{document}
\title
[Modified Macdonald polynomials and the multispecies TAZRP: I]
{Modified Macdonald polynomials and the multispecies zero-range process: I}
\author[\initial{A} \lastname{Ayyer}]{\firstname{Arvind} \lastname{Ayyer}}
\address{Department of Mathematics\\
Indian Institute of Science\\
Bangalore 560 012, India}
\email{arvind@iisc.ac.in}
\thanks{AA was partially supported by the UGC Centre for Advanced Studies and by Department of Science and Technology grant EMR/2016/006624.}
\author[\initial{O} \lastname{Mandelshtam}]{\firstname{Olya} \lastname{Mandelshtam}}
\address{Department of Combinatorics and Optimization\\
University of Waterloo\\Waterloo, ON, Canada}
\email{omandels@uwaterloo.ca}
\thanks{OM was supported by NSF grants DMS-1704874 and DMS-1953891, and NSERC}
\author[\initial{J} \middlename{B} \lastname{Martin}]{\firstname{James} \middlename{B} \lastname{Martin}}
\address{Department of Statistics\\ University of Oxford\\ UK}
\email{martin@stats.ox.ac.uk}
\keywords{modified Macdonald polynomials, TAZRP, tableaux, zero range process}
\subjclass[2010]{05E05, 05A10, 05A19, 05A05, 33D52}
\begin{abstract}
In this paper we prove a new combinatorial formula for the modified Macdonald polynomials $\widetilde{H}_\lambda(X;q,t)$, motivated by connections to the theory of interacting particle systems from statistical mechanics. The formula involves a new statistic called \emph{queue inversions} on fillings of tableaux. This statistic is closely related to the \emph{multiline queues} which were recently used to give a formula for the Macdonald polynomials $P_\lambda(X;q,t)$. In the case $q=1$ and $X=(1,1,\dots,1)$, that formula had also been shown to compute stationary probabilities for a particle system known as the \emph{multispecies ASEP} on a ring, and it is natural to ask whether a similar connection exists between the modified Macdonald polynomials and a suitable statistical mechanics model. In a sequel to this work, we demonstrate such a connection, showing that the stationary probabilities of the \emph{multispecies totally asymmetric zero-range process} (mTAZRP) on a ring can be computed using tableaux formulas with the queue inversion statistic. This connection extends to arbitrary $X=(x_1,\dots, x_n)$; the $x_i$ play the role of site-dependent jump rates for the mTAZRP.
\end{abstract}
\maketitle
%\setcounter{tocdepth}{1}
%\tableofcontents
\section{Introduction}
The theory of symmetric functions has a long history, having its origins in invariant theory, Galois theory, group theory and, of course, combinatorics. As Stanley~\cite[Notes in Chapter~7]{stanley-ec2} remarks, the first published work on symmetric functions was a derivation of the well-known Newton--Girard identity by A. Girard~\cite{girard-1884} in 1629. The reason for the name is that it was independently rediscovered by Newton around 1666.
On the other hand, the theory of interacting particle systems is relatively modern. It was an influential paper of Spitzer~\cite{spitzer-1970} in 1970 that initiated the subject and set out the important questions in the field. In particular, the simple exclusion process and the zero-range process were first defined there.
The focus of this paper is a particular family of symmetric functions,
the \emph{modified Macdonald polynomials}, motivated by a new link
to an interactive particle system known as the \emph{multispecies totally
asymmetric zero-range process}.
Over the last couple of decades, the theory of special functions and symmetric functions have found unexpected connections to diverse interacting particle systems. The asymmetric simple exclusion process (ASEP) has played a central role in this connection. Building on work of Uchiyama, Sasamoto, and Wadati~\cite{USW,sasamoto-1999}, Corteel and Williams~\cite{CW11} found that the partition function of the single species ASEP with open boundaries is related to moments of Askey--Wilson polynomials, and discovered combinatorial formulas for those quantities. Generalizing this work, Corteel and Williams~\cite{CW18} and also Cantini~\cite{cantini-2017} showed that the partition function of the two-species ASEP with open boundaries is related to certain moments of Koornwinder polynomials, and soon after, the second author along with Corteel and Williams~\cite{CMW-2017} found the associated combinatorial formulas. More generally, Cantini, Garbali, de Gier and Wheeler~\cite{cantini-etal-2016} found that the partition function of a multispecies variant of the ASEP with open boundaries is a specialization of a Koornwinder polynomial, but finding formulas for these quantities is still an outstanding open question.
The multispecies ASEP on a ring is another version of the ASEP that has been found to have deep connections to orthogonal polynomials. Cantini, de Gier and Wheeler~\cite{CGW-2015} showed that the stationary probabilities of the ASEP on a ring are related to the nonsymmetric Macdonald polynomials, and that the partition function of the ASEP on a circle is equal to the symmetric Macdonald polynomial $P_{\lambda}(X;q,t)$ evaluated at $x_1=\cdots=x_n=q=1$. In the totally asymmetric case (at $t=0$), Ferrari and the third author~\cite{FM07} introduced a recursive formulation which expresses
the stationary probabilities as sums over combinatorial objects known as \emph{multiline queues}. Recently, the third author generalized these multiline queues to compute probabilities for the full ASEP~\cite{martin-2020}. Building upon that result, the second author together with Corteel and Williams further generalized these multiline queues to incorporate the parameters $x_1,\ldots,x_n,q$ to obtain formulas for $P_{\lambda}(X;q,t)$ and the related nonsymmetric Macdonald polynomials \cite{CMW18}. These formulas interpolate between probabilities of the ASEP on a ring and the Macdonald polynomials.
There are several natural bases for symmetric polynomials. The \emph{monomial symmetric polynomials}, \emph{elementary symmetric polynomials}, \emph{complete homogeneous symmetric polynomials} and \emph{power sum symmetric polynomials} are classical. Probably the single most important family is the family of \emph{Schur polynomials}, for which there are several generalizations.
The symmetric Macdonald polynomials are a remarkable two-parameter generalization of Schur polynomials that were introduced by Macdonald in 1987 \cite{Mac88, Macdonald}. They are indexed by partitions $\lambda$ and are denoted $P_{\lambda}(X; q,t)$, where $X=(x_1,x_2,\ldots)$ is the alphabet and $q$ and $t$ are parameters; coefficients of $P_{\lambda}(X;q,t)$ are in $\mathbb{Q}(q,t)$. They can be characterized as the unique basis for the ring of symmetric functions in $q$ and $t$ satisfying certain orthogonality and triangularity conditions.
The modified Macdonald polynomials $\widetilde{H}_{\lambda}(X;q,t)$ defined by Garsia and Haiman \cite{garsia-haiman-1993} are a special form of the $P_{\lambda}$'s with coefficients in $\mathbb{Z}[q,t]$ obtained through a formal operation called \emph{plethysm} from a scalar multiple of $P_{\lambda}(X;q,t)$. Understanding the combinatorics of these polynomials has been a fundamental area of interest in algebraic combinatorics. The most commonly used combinatorial description of $\widetilde{H}_{\lambda}(X;q,t)$ is a tableau formula due to Haglund, Haiman, and Loehr \cite{HHL05}:
\begin{equation}
\label{eq:HHLformula}
\widetilde{H}_{\lambda}(X;q,t) = \sum_{\sigma:\dg(\lambda) \rightarrow \mathbb{Z}} t^{\inv(\sigma)}q^{\maj(\sigma)}x^{\sigma}
\end{equation}
where the sum is over all fillings of a diagram $\dg(\lambda)$ whose shape is the partition $\lambda$. The \emph{inversion} statistic $\inv(\sigma)$ and the \emph{major index} statistic $\maj(\sigma)$ are explained in Section \ref{sec:definitions}. Note that the form we give in \eqref{eq:HHLformula} uses a different convention than that of \cite{HHL05} (in which the $\inv$ and $\maj$ statistics are exchanged and the shape of the diagram is transposed). This is justified because of the well-known identity $\widetilde{H}_{\lambda}(X;q,t) = \widetilde{H}_{\lambda'}(X;t,q)$~\cite[Prop. 2.6]{Hai99}.
Recently, a different combinatorial model for these polynomials has been given by Garbali and Wheeler~\cite{garbali-wheeler-2020}. Independently, the second author together with Corteel, Haglund, Mason, and Williams found formulas for $\widetilde{H}_{\lambda}(X;q,t)$ that were based on the combinatorial interpretation of plethysm applied to multiline queues \cite{compactformula}. One such formula was a compact version of the original tableaux formula of Haglund, Haiman, and Loehr. The authors of \cite{compactformula} also conjectured a formula related
to \eqref{eq:HHLformula} in which the inversion statistic is replaced by a variant
form which we call the \emph{queue inversion} statistic.
Our main result is a proof of this conjectural formula:
\begin{thm*}
The modified Macdonald polynomial may be written as
\begin{equation}\label{eq:MainResult}
\widetilde{H}_{\lambda}(X;q,t) = \sum_{\sigma:\dg(\lambda) \rightarrow \mathbb{Z}} t^{\quinv(\sigma)}q^{\maj(\sigma)}x^{\sigma},
\end{equation}
where the definition of the queue inversion statistic $\quinv(\sigma)$ is
again given in Section \ref{sec:definitions}.
\end{thm*}
Both the statistic $\inv(\sigma)$ appearing in \eqref{eq:HHLformula}
and the statistic $\quinv(\sigma)$ appearing in \eqref{eq:MainResult}
are defined in terms of the content of particular triples of cells in the filling $\sigma$.
Even though these two results look tantalizingly similar, we do
not know of a bijection translating between the two statistics even in very simple
cases (for instance tableaux with just two rows).
A particular motivation for our new representation is a link between the modified Macdonald polynomials at $q=1$ and the multispecies totally asymmetric zero-range process (TAZRP) on the ring, which we establish in a companion paper \cite{AMM22}.
The stationary probabilities of a TAZRP on a ring with $n$ sites, with site-specific
parameters $x_1, x_2, \dots, x_n$ and a global parameter $t$, and with
particle-types determined by a partition $\lambda$,
can be written (suitably normalized) as polynomials in $x_1,\dots, x_n$ and $t$ whose
sum is $\widetilde{H}_{\lambda}(X;1,t)$. Moreover, under
the probability distribution on fillings $\sigma$ of $\dg(\lambda)$
proportional to the weights $t^{\quinv(\sigma)} x^\sigma$ appearing on the right-hand side
of \eqref{eq:MainResult}, the distribution of the bottom row of the tableau
projects to the stationary distribution of the TAZRP.
See Section~\ref{sec:mTAZRP} for further details of this connection.
The plan of the article is as follows. In Section~\ref{sec:main}, we introduce necessary definitions and notation and formally state the main result mentioned above. We also give details of the link to the multispecies totally asymmetric zero-range process. Section~\ref{sec:background} gives some further background which will be required for the proofs. In Section~\ref{sec:sym}, we prove the symmetry of the function defined by
the right-hand side of \eqref{eq:MainResult}. The modified Macdonald polynomials are characterized by the three properties stated in Section~\ref{sec:axioms}. The third property will turn out to be immediate from our definition. In Section~\ref{sec:A1m}, we prove the first of these properties. In Section~\ref{sec:A2m}, we prove the second property in a certain nondegenerate case. Settling the second property in the degenerate case proves to be the main hurdle. This is taken care of in Section~\ref{sec:Phi-degenerate} using a bijection on certain tableaux that is described in Section~\ref{sec:bijPQT}. The latter is in turn obtained via a generating function identity on words respecting a coinversion-type statistic. This identity might be of independent interest and is proved in Section~\ref{sec:bijwords}. In Section~\ref{sec:conclusion}, we conclude the proof of our main result and mention various related questions.
\section{Main result and related work}
\label{sec:main}
\subsection{Definitions and main result}
\label{sec:definitions}
Let $\lambda=(\lambda_1,\ldots,\lambda_k)$ be a partition. The \emph{diagram of type $\lambda$}, which we denote $\dg(\lambda)$, consists of
the cells $\{(r,i), 1\leq i\leq k, 1\leq r\leq \lambda_i\}$, which we depict using
$k$ bottom-justified columns, where the $i$'th column from left to right has $\lambda_i$ boxes. See Figure~\ref{fig:readingorder} for an illustration of $\dg((3,2,1,1))$.
The cell $(r,i)$ corresponds to the cell in the $i$'th column in the $r$'th row of $\dg(\lambda)$, where rows are labeled from bottom to top.
We warn the reader that this is the opposite convention of labelling points in the plane using cartesian coordinates, but is somewhat standard in the literature. We will use this convention throughout the article.
A \emph{filling} of type $(\lambda,n)$ is a function $\sigma:\dg(\lambda) \rightarrow [n]$ defined on the cells of $\dg(\lambda)$ (where $[n]=\{1,2,\dots, n\}$).
We also refer to a diagram together with a filling of it as a \emph{tableau}.
Let $\PQT(\lambda,n)$ be the set of fillings of type $(\lambda, n)$,
and $\PQT(\lambda)$ the set of functions $\sigma:\dg(\lambda) \rightarrow \mathbb{Z}^+$.
\begin{center}
\begin{figure}[h]
\[
\tableau{1\\3&2\\7&6&5&4}
\]
\caption{The diagram $\dg((3,2,1,1))$ with the cells labeled according to the reading order in Definition~\ref{def:readingorder}. Our convention is that the cell labelled $5$ has coordinates $(1,3)$.}
\label{fig:readingorder}
\end{figure}
\end{center}
For $\sigma$ in $\PQT(\lambda, n)$ or in $\PQT(\lambda)$, and a cell $x=(r,i)$ of $\dg(\lambda)$, we call $\sigma(x)$ the \emph{content}
of the cell $x$ in the filling $\sigma$.
For convenience, we also define $\sigma((\lambda_i+1,i))=0$, where $(\lambda_i+1,i)$ is the nonexistent cell above column $i$.
When $r>1$, define $\South(x)$ to be the cell $(r-1,i)$ directly below cell $x$ in the same column.
\begin{defn}
\label{def:readingorder}
Define the \emph{reading order} on the cells of a tableau to be along the rows from right to left,
taking the rows from top to bottom.
\end{defn}
See Figure~\ref{fig:readingorder} for an illustration of the reading order.
We now define various statistics on fillings of tableaux.
\begin{defn}\label{def:maj}
Let $\sigma\in\PQT(\lambda)$. Define the \emph{leg} of a cell $(r,i)$ to be the number of cells in column $i$ above row $r$, i.e. $\leg((r,i))= \lambda_i - r$. Define the \emph{major index} of the filling $\sigma$ to be:
\[
\maj(\sigma) = \sum_{x\,:\,\sigma(x)>\sigma(\South(x))} (\leg(x)+1)
\]
\end{defn}
\begin{defn}\label{def:quinv}
Given a diagram $\dg(\lambda)$, a \emph{triple} consists of either
\begin{itemize}
\item three cells $(r+1,i)$, $(r,i)$ and $(r,j)$ with $i\rho^{(j)}(v)$ and either
\begin{enumerate}[label=\roman*.]
\item $ij$ and $d(u)=d(v)+1$.
\end{enumerate}
\end{defn}
We denote the number of inversions of $\boldsymbol{\rho}$ by $\inv(\boldsymbol{\rho})$, and
the monomial in $x$ corresponding to the content of $\boldsymbol{\rho}$ by
$x^{\boldsymbol{\rho}} = \prod_i \prod_{u\in \nu^{(i)}} x_{\rho^{(i)}(u)}$.
\begin{example}
Figure~\ref{fig:LLT1} shows a semistandard filling $\boldsymbol{\rho}=(\rho^{(1)},\rho^{(2)},\rho^{(3)})$ of the tuple of skew diagrams $\boldsymbol{\nu}$ where $\nu^{(1)}=(1,1)/\emptyset$, $\nu^{(2)}=(1,1)/\emptyset$, and $\nu^{(3)}=(2,2,2)/(2,1)$. There are five inversions in $\boldsymbol{\rho}$: the 3 in diagonal 3 forms an inversion with the 2 in diagonal 2, the 3 and the 2 in diagonal 2 form an inversion, the 2 in diagonal 2 forms an inversion with the bottommost 1 in diagonal 1, and the 4 in diagonal 2 forms an inversion with both 1's in diagonal 1. Thus $\inv(\boldsymbol{\rho})=5$ and $x^{\boldsymbol{\rho}}=x_1^2x_2^2x_3^2x_4$.
\end{example}
\begin{figure}[h]
\centering
\begin{tikzpicture}[node distance=2cm]
\def \d {0.4};
\def \h{0};
\draw[dashed,gray] (\h+3.1+\d,-1.5+\d)--(\h+6.1,1.5);
\node at (\h+3+\d,-1.7+\d) {$3$};
\draw[dashed,gray] (\h+3.1+2*\d,-1.5+\d)--(\h+6.1+\d,1.5);
\node at (\h+3+2*\d,-1.7+\d) {$2$};
\draw[dashed,gray] (\h+3.1+3*\d,-1.5+\d)--(\h+6.1+2*\d,1.5);
\node at (\h+3+3*\d,-1.7+\d) {$1$};
\node at (\h+2.5,-1.7+\d) {$d(u)$};
\node at (\h+4.2+\d,-1+\d) {$\tableau{3\\1}$};
\node at (\h+4+3*\d,-1.+2.5*\d) {$\tableau{2\\1}$};
\node at (\h+4+5*\d,-.8+4.5*\d) {$\tableau{3&4\\&2}$};
\end{tikzpicture}
\caption{For $\boldsymbol{\nu}=((1,1),(1,1),(2,2,2)/(2,1))$, we show a filling $\boldsymbol{\rho}\in\SSYT(\boldsymbol{\nu})$, with $\inv(\boldsymbol{\rho})=5$ and $x^{\boldsymbol{\rho}}=x_1^2x_2^2x_3^2x_4$.}
\label{fig:LLT1}
\end{figure}
\begin{defn}\label{def:LLT}
Let $\boldsymbol{\nu}$ be a tuple of skew diagrams. The LLT polynomial indexed by $\boldsymbol{\nu}$ is
\[
G_{\boldsymbol{\nu}}(X;t) = \sum_{\boldsymbol{\rho}\in\SSYT(\boldsymbol{\nu})} t^{\inv(\boldsymbol{\rho})}x^{\boldsymbol{\rho}}.
\]
\end{defn}
\begin{theorem}[\cite{LLT,LLT2}]\label{thm:LLT}
The polynomial $G_{\boldsymbol{\nu}}(X;t)$ is symmetric in the variables $x$.
\end{theorem}
\begin{remark} Our Theorem~\ref{thm:symmetry} relies on Theorem~\ref{thm:LLT}. The original proof for Theorem~\ref{thm:LLT} in \cite{LLT} uses a construction in the representation theory of affine Hecke algebras, and a purely combinatorial proof of the symmetry is given in \cite[Section 10: Appendix]{HHL05}.
\end{remark}
\begin{defn}\label{def:ribbon}
Define a \emph{ribbon} to be a connected skew diagram with no $2\times2$ squares. For a ribbon $\nu$ with $k=|\nu|$, we label its boxes from northwest to southeast by $1,\ldots,k$, and define its
{\em descent set}, denoted $\Des(\nu)$, to be the set of labels in $\{1,\ldots,k-1\}$ corresponding to boxes that have a box below in the same column.
Let $w$ be a word with entries in $\mathbb{Z}_+$. Denote the set of locations of descents in $w$ by $\Des(w) = \{i\,:\,w_i>w_{i+1}\}$. Let $\nu$ be the ribbon with the same descent set as $w$, i.e. $\Des(\nu)=\Des(w)$. We define a ribbon corresponding to $w$ to be a filling of the cells of $\nu$ with the entry $w_i$ in cell $i$ for $i=1,\ldots,k$. We call $\ribbon(w)$ the ribbon corresponding to $w$.
\end{defn}
It is immediate from the definition that $\ribbon(w)$ is a SSYT.
\begin{example}
Consider the word $w=(4,3,3,4,5,3,2,1,2,2)$, which has descent set $\Des(w)= \{1,5,6,7\}$. The ribbon $\nu=(6,4,4,4,1)/(3,3,3)$ is the unique ribbon with the same descent set: $\Des(\nu)=\Des(w)$. Below we show $\nu$ with its cells labeled from northwest to southeast, and the corresponding $\SSYT$ $\ribbon(w)$ of shape $\nu$ with its boxes filled by the entries of $w$.
\begin{center}
\begin{tikzpicture}[scale=0.4]
\node at (-4,-3) {$\nu=$};
\cell10{$1$}\cell20{$2$} \cell21{$3$} \cell22{$4$} \cell23{$5$} \cell33{$6$} \cell43{$7$} \cell53{$8$} \cell54{$9$} \cell55{$10$}
\node at (16,-3) {$\ribbon(w)=$};
\cell1{20}{$4$}\cell2{20}{$3$} \cell2{21}{$3$} \cell2{22}{$4$} \cell2{23}{$5$} \cell3{23}{$3$} \cell4{23}{$2$} \cell5{23}{$1$} \cell5{24}{$2$} \cell5{25}{$2$}
\end{tikzpicture}
\end{center}
\end{example}
\subsection{Super-alphabets and quasisymmetric function expansion}\label{sec:superization}
With $\maj$, $\quinv$ and $\PQT(\lambda)$ as defined in Section~\ref{sec:main}, we define:
\begin{equation}\label{eq:C}
C_{\lambda}(X;q,t) = \sum_{\sigma\in \PQT(\lambda)} q^{\maj(\sigma)} t^{\quinv(\sigma)}x^{\sigma}.
\end{equation}
After showing $C_{\lambda}(X;q,t)$ is symmetric in the variables $x_i$ in Section~\ref{sec:sym}, we will show it satisfies \eqref{A1m} and \eqref{A2m}. We do this by modifying the proof of the HHL formula in \cite[Section 4]{HHL05} for the setting of $\PQT(\lambda)$, where we will consider the \emph{superization} of $C_{\lambda}$. In this subsection, we adapt the well-known properties of superization to this new setting.
Let $n\geq 0$ and fix a subset $D\subseteq \{1,\ldots,n-1\}$. Define \emph{Gessel's quasisymmetric function} $Q_{n,D}(X)$ in the variables $X=x_1,x_2,\ldots$ by
\[
Q_{n,D}(X) = \sum_{\substack{a_1\leq a_2\leq \cdots \leq a_n\\a_i=a_{i+1} \implies i\not\in D}} x_{a_1}x_{a_2}\cdots x_{a_n},
\]
where the indices are $a_i\in\mathbb{Z}_+$. Define the ``super-alphabet''
\[
\mathcal{A}=\mathbb{Z}_+ \cup \mathbb{Z}_- = \{\bar{1},1,\bar{2},2,\ldots\}
\]
consisting of positive and ``negative'' letters $i,\bar{i}$ of our original alphabet. One can consider any total ordering on $\mathcal{A} \cup \{0\}$; in our proofs we will use the two total orderings
\begin{align*}
(\mathcal{A}\cup\{0\},<_1)& = \{0<1<\bar{1}<2<\bar{2}<\cdots\},\\
(\mathcal{A}\cup\{0\},<_2)& = \{0<1<2<3<\cdots<\bar{3}<\bar{2}<\bar{1}\}.
\end{align*}
For any fixed total ordering $<$ on $\mathcal{A}$, we define the more general ``super'' quasiymmetric function $\widetilde{Q}_{n,D}(X,Y)$ in the variables $X=x_1,x_2,\ldots$ and $Y=y_1,y_2,\ldots$ by
\begin{equation}
\widetilde{Q}_{n,D}(X,Y) = \sum_{\substack{a_1\leq a_2\leq \cdots \leq a_n\\a_i=a_{i+1}\in\mathbb{Z}_+\implies i\not\in D\\a_i=a_{i+1}\in\mathbb{Z}_-\implies i\in D}} z_{a_1}z_{a_2}\cdots z_{a_n},
\end{equation}
where the indices are $a_i \in \mathcal{A}$, and $z_{i}=x_{i}$ when $i\in\mathbb{Z}_+$ and $z_{i}=y_{i}$ when $i\in\mathbb{Z}_-$.
For $a,b \in \mathcal{A}\cup\{0\}$ and any given total ordering $<$, we will use the notation $I(a,b)$:
\[
I(a,b) = \begin{cases} 1, & a>b \mbox{ or } a=b \in \mathbb{Z}_-,\\
0, & a** 0$ and either $m < 0$ or $m > n$, $\qbinom {0}m = \delta_{m,0}$ and $\qbinom {-1}0 = 1$.
The {\em $q$-multinomial coefficient} is defined similarly. Suppose
$\alpha = (\alpha_1,\dots, \allowbreak \alpha_n)$ is a tuple of nonnegative integers.
Recall that $|\alpha| = \alpha_1 + \cdots + \alpha_n$. Then
\[
\qbinom{|\alpha|}{\alpha_1,\dots,\alpha_n} \coloneqq \frac{[|\alpha|]!}{[\alpha_1]! \cdots [\alpha_n]!}.
\]
For $\alpha = (\alpha_1,\dots, \alpha_n)$ a tuple of positive integers, let $W_\alpha$
be the set of words in the alphabet $[n]$ with $\alpha_i$ occurrences of the letter $i$,
$1 \leq i \leq n$. For a word $w\in W_{\alpha}$, define $\coinv(w) = |\{(i,j)\ :\ i \max(m,n)$. In particular, if $k > m+n$, then both sides are zero.
\end{theorem}
The $q$-Chu--Vandermonde identity is also valid when $m,n$ are negative
integers. In that case, a formulation useful for us will be the following.
\begin{cor}
\label{cor:q-dualchu}
For $m,n,k$ nonnegative integers such that $m \geq n \geq k$,
\[
\sum_{i=k}^{m-n+k} \qbinom {i}{k} \qbinom {m-i}{n-k} q^{i(n-k+1)} = \qbinom {m+1}{n+1} q^{k(n-k+1)}.
\]
\end{cor}
The last is a telescoping sum, which can be derived from the fundamental
recurrence \eqref{qbinom-recur} for the $q$-binomial coefficients.
\begin{proposition}
\label{prop:q-telescope}
For $m,n,k$ nonnegative integers such that $m+1 \geq n \geq k$,
\[
\sum_{i=k}^n \qbinom {m-i}{n-i} q^{i(m-n+1)} = \qbinom {m-k+1}{n-k} q^{k(m-n+1)}.
\]
Note that both sides are equal to $1$ when $m = n-1$, since by convention, $\qbinom {-1}0 = 1$.
\end{proposition}
\section{Proof that $C_{\lambda}(X,q,t)$ is symmetric}
\label{sec:sym}
This section is devoted to proving the following theorem.
\begin{theorem}\label{thm:symmetry}
The polynomial $C_{\lambda}(X; q,t)$ is symmetric in the variables $x_i$.
\end{theorem}
We will prove Theorem~\ref{thm:symmetry} by expanding $C_{\lambda}$ in terms of the LLT polynomials $G_{\boldsymbol{\nu}}(X;t)$
defined in Section~\ref{sec:LLT}.
Let $\lambda$ be a partition with $k$ parts, let $\widehat{\dg}(\lambda)=\{(r,j) \in \dg(\lambda): r>1\}$
be the cells in $\dg(\lambda)$ not contained in the bottom row, and let $D\subseteq \widehat{\dg}(\lambda)$ be any subset. We define $\boldnu(\lambda,D)=(\nu^{(k)},\ldots,\nu^{(2)},\nu^{(1)})$ to be a tuple of $k$ ribbons (see Definition~\ref{def:ribbon}) such that
\[
\Des(\nu^{(j)})=\{\lambda_j-r+1\ :\ (r,j)\in D\},
\]
where the ribbons are arranged from southwest to northeast, and such that the southeast-most cell of each ribbon is aligned on diagonal 1. In other words, the $j$'th ribbon has length $\lambda_j$ and its descent set $\Des(\nu^{(j)})$ corresponds to the restriction of $D$ to column $j$ of $\dg(\lambda)$ when read from top to bottom. See Example~\ref{ex:LLT}.
\begin{example}\label{ex:LLT}
Let $\lambda=(3,2,2)$ with the subset of descents chosen to be\linebreak[4] $D=\{(2,1),(2,3)\}$, indicated by the shaded boxes in the figure below. Then $\Des(\nu^{(1)}) = \{2\}$, $\Des(\nu^{(2)})=\emptyset$, $\Des(\nu^{(3)})=\{1\}$, and $\boldnu(\lambda,D)=\{\nu^{(3)},\nu^{(2)},\nu^{(1)}\}$ is the tuple of three ribbons shown below from left to right, with the cells of each ribbon labeled in reading order and descents marked by shaded boxes.
\begin{center}
\raisebox{1cm}{
\begin{tikzpicture}[scale=.45]
\node at (-3,-1) {$\dg(\lambda)=$};
\cell00{\ }
\graycell10{\ }\cell11{\ }\graycell12{\ }
\cell20{\ }\cell21{\ }\cell22{\ }
\end{tikzpicture}
}
\qquad\qquad
\begin{tikzpicture}
\def \d {0.4};
\def \h{0};
\def \v{-2};
\node at (\h+3,\v) {$\boldnu(\lambda,D)=$};
\draw[dashed,gray] (\h+3.1+\d,\v-1.5+\d)--(\h+6.1,\v+1.5);
\node at (\h+3+\d,\v-1.7+\d) {$3$};
\draw[dashed,gray] (\h+3.1+2*\d,\v-1.5+\d)--(\h+6.1+\d,\v+1.5);
\node at (\h+3+2*\d,\v-1.7+\d) {$2$};
\draw[dashed,gray] (\h+3.1+3*\d,\v-1.5+\d)--(\h+6.1+2*\d,\v+1.5);
\node at (\h+3+3*\d,\v-1.7+\d) {$1$};
\node at (\h+2.5,\v-1.7+\d) {row};
\fill[gray!60] (\h+4.41+4*\d,\v-.99+\d+4*\d)--(\h+4.41+\d+4*\d,\v-.99+\d+4*\d)--(\h+4.41+\d+4*\d,\v-.99+2*\d+4*\d)--(\h+4.41+4*\d,\v-.99+2*\d+4*\d)--(\h+4.41+4*\d,\v-.99+\d+4*\d);
\fill[gray!60] (\h+4.4,\v-.99+\d)--(\h+4.4+\d,\v-.99+\d)--(\h+4.4+\d,\v-.99+2*\d)--(\h+4.4,\v-.99+2*\d)--(\h+4.4,\v-.99+\d);
\node at (\h+4.2+\d,\v-1+\d) {$\tableau{1\\2}$};
\node at (\h+4.2+3*\d,\v-1.+.2+2.5*\d) {$\tableau{1&2}$};
\node at (\h+4+5*\d,\v-.8+4.5*\d) {$\tableau{1&2\\&3}$};
\end{tikzpicture}
\end{center}
\end{example}
We now refine \eqref{eq:C} by splitting it into fillings with a given descent set.
\begin{defn}\label{def:invtwo}
Let $\lambda$ be a partition and $\sigma\in\PQT(\lambda)$. A pair of cells $u=(r,i)$ and $v=(r',j)$ is said to be \emph{attacking} if
\begin{enumerate}[label=\roman*.]
\item $r=r'$ and $i>j$, or
\item $r=r'+1$ and $i\sigma(v)$ for a pair of attacking cells $u,v$, they form an \emph{attacking inversion}. If $\sigma(u)\neq\sigma(v)$ for every pair of attacking cells $u,v\in\dg(\lambda)$, we call $\sigma$ a \emph{non-attacking} filling.
Denote the number of attacking inversions in $\sigma\in\PQT(\lambda)$ by $\invtwo(\sigma)$. For a cell $u=(r,i)\in \dg(\lambda)$, we define $\armtwo(u)$ to be the number of cells $(r-1,j)\in\dg(\lambda)$ such that $j>i$, i.e. the shaded cells belong to $\armtwo$ of the cell labeled $a$ below.
\begin{center}
\begin{tikzpicture}[scale=.45]
\cell00{\ }\cell01{$a$}\cell02{\ }
\cell10{\ }\cell11{\ }\graycell12{\ }\graycell13{\ }\graycell14{\ }
\cell20{\ }\cell21{\ }\cell22{\ }\cell23{\ }\cell24{\ }\cell25{\ }
\end{tikzpicture}
\end{center}
\end{defn}
\begin{remark}
The $\,\widehat{\cdot}\,$ symbol will help differentiate $\widehat{\inv}$ and $\widehat{\arm}$ from the traditional definitions of the notions of inversions and arms that appear in the proofs of the corresponding HHL formulas in \cite[Section 3]{HHL05}. Note that we have called $\invtwo$ ``attacking inversions'' because they correspond to inversions occurring in ``attacking cells'' with respect to the reading order, to parallel the terminology used in \cite{HHL05} when referring to their version of attacking cells and inversions.
\end{remark}
We will be grouping tableaux by their descent sets, indexed by subsets $D\subseteq \widehat{\dg}(\lambda)$. For each subset $D\subseteq \widehat{\dg}(\lambda)$, define
\begin{equation}\label{eq:F_D}
F_{\lambda,D}(X;t) = \sum_{\substack{\sigma\in\PQT(\lambda)\\ \Des(\sigma)=D}} t^{\invtwo(\sigma)} x^{\sigma}.
\end{equation}
Now, we rewrite $C_{\lambda}$ in terms of the $F_{\lambda,D}$'s.
\begin{defn}
Let $U\subseteq \dg(\lambda)$ be a subset of cells. Define
\[
\maj(U) = \sum_{u\in U} (\leg(u)+1)
\]
and
\[
\armtwo(U)=\sum_{u\in U} \armtwo(u).
\]
\end{defn}
\begin{lemma}
\[
C_{\lambda}(X; q,t) = \sum_{D\subseteq \widehat{\dg}(\lambda)} q^{\maj(D)} t^{-\armtwo(D)} F_{\lambda,D}(X;t).
\]
\end{lemma}
\begin{proof}
We write \eqref{eq:C} as a sum over descent sets:
\[
C_{\lambda}(X;q,t)=\sum_{D\subseteq \widehat{\dg}(\lambda)} q^{\maj(D)}\sum_{\substack{\sigma\in\PQT(\lambda)\\\Des(\sigma)=D}} t^{\quinv(\sigma)}x^{\sigma}.
\]
Now suppose $\sigma\in\PQT(\lambda)$ with $D\coloneqq\Des(\sigma)$. We will show that
\begin{equation}\label{invtwo}
\quinv(\sigma)=\invtwo(\sigma)-\sum_{u\in D} \armtwo(u).
\end{equation}
Consider a triple $(x,y,z)\in\dg(\lambda)$ with respective contents $a\coloneqq\sigma(x)$, $b\coloneqq\sigma(y)$, $c\coloneqq\sigma(z)$ in the configuration
\[
\begin{tikzpicture}[scale=.5]
\cellL0{.5}{$a$}{$x$}\cellL1{.5}{$b$}{$y$}\cellL1{3.5}{$c$}{$z$}
\node at (1.3,-.5) {$\cdots$};
\end{tikzpicture}
\ .
\]
(The triple may be degenerate, in which case $a=0$.)
By comparing to \eqref{cQdef}, $(a,b,c)\in\cQ$ if and only if exactly one of the following is true:
\begin{equation}\label{quinv_conditions}
\left\{\begin{matrix}x \in D,\\(y,z) \text{ do not form an attacking inversion},\\ (z,x) \text{ do not form an attacking inversion}\end{matrix}\right\}.
\end{equation}
First suppose $x \not\in D$. If $(a,b,c)\in\cQ$, then exactly one of $(y,z)$ or $(z,x)$ contributes to $\invtwo(\sigma)$ according to \eqref{quinv_conditions}. If $(a,b,c)\not\in\cQ$, then neither $(y,z)$ or $(z,x)$ contributes to $\invtwo(\sigma)$. The contribution of the triple $(x,y,z)$ to the left hand side and the right hand side of \eqref{invtwo} match in each case.
Now suppose $x\in D$. If $(a,b,c)\in\cQ$, then both $(y,z)$ and $(z,x)$ contribute to $\invtwo(\sigma)$, but since $z\in\armtwo(x)$, the total contribution to the right hand side in \eqref{invtwo} is 1, which matches the contribution to the left hand side. If $(a,b,c)\not\in\cQ$, then exactly one of $(y,z)$ or $(z,x)$ contributes to $\invtwo(\sigma)$, but since $z\in\armtwo(x)$, the total contribution to the right hand side in \eqref{invtwo} is zero, again matching the contribution to the left hand side.
Every pair of cells $u,v$ where $u\in D$ and $v\in\armtwo(u)$, and every attacking pair that contributes to $\invtwo(\sigma)$, corresponds to some triple in $\sigma$. Thus every term in the left hand side is accounted for, and \eqref{invtwo} follows.
\end{proof}
We will now describe a weight-preserving map $\widehat{\LLT}$ from $\PQT(\lambda)$ with a fixed descent set $D$ to fillings $\SSYT(\boldnu(\lambda,D))$.
\begin{defn}\label{def:LLTtwo}
For each descent set $D \subseteq \{(r,j) \in \widehat{\dg}(\lambda)\}$, define the map
\[
\widehat{\LLT}: \{\sigma\in \PQT(\lambda):\Des(\sigma)=D\} \rightarrow \SSYT(\boldnu(\lambda,D))
\]
as follows. Let $\sigma\in\PQT(\lambda)$, and let $c_1,\ldots,c_k$ be its columns from left to right, where each column contains the entries in $\sigma$ from top to bottom: $c_i = (\sigma(\lambda_i,i),\linebreak[0],\ldots,\sigma(2,i),\sigma(1,i))$.
Set $\rho^{(i)} = \ribbon(c_i)$.
Define $\boldsymbol{\rho}=(\rho^{(k)},\rho^{(k-1)},\ldots,\rho^{(1)})$ to be the tuple of ribbons corresponding to the columns in reverse order, such that the last boxes of each ribbon are on the same diagonal, and set $\widehat{\LLT}(\sigma) = \boldsymbol{\rho}$. See Figure~\ref{fig:LLT_T} for an example.
\end{defn}
\begin{figure}[h]
\centering
\begin{tikzpicture}[node distance=2cm]
\node at (-1.5,0) {$\sigma=$};
\node at (0,0) {$\tableau{3\\1&2&3\\2&1&1}$};
\def \d {0.4};
\def \h{3};
\node at (\h+3,0) {$\widehat{\LLT}(\sigma)=$};
\draw[dashed,gray] (\h+3.1+\d,-1.5+\d)--(\h+6.1,1.5);
\node at (\h+3+\d,-1.7+\d) {$3$};
\draw[dashed,gray] (\h+3.1+2*\d,-1.5+\d)--(\h+6.1+\d,1.5);
\node at (\h+3+2*\d,-1.7+\d) {$2$};
\draw[dashed,gray] (\h+3.1+3*\d,-1.5+\d)--(\h+6.1+2*\d,1.5);
\node at (\h+3+3*\d,-1.7+\d) {$1$};
\node at (\h+2.5,-1.7+\d) {row};
\node at (\h+4.2+\d,-1+\d) {$\tableau{3\\1}$};
\node at (\h+4+3*\d,-1.+2.5*\d) {$\tableau{2\\1}$};
\node at (\h+4+5*\d,-.8+4.5*\d) {$\tableau{3\\1&2}$};
\end{tikzpicture}
\caption{The filling $\widehat{\LLT}(\sigma)$ corresponding to $\sigma\in\PQT(\lambda)$. The numbering of the diagonals is shown on the bottom. One can check that $\inv(\widehat{\LLT}(\sigma))=\invtwo(\sigma) = 5$. Moreover, the pairs of inversions in $\widehat{\LLT}(\sigma)$
correspond precisely to the attacking inversions in $\sigma$.
}
\label{fig:LLT_T}
\end{figure}
\begin{remark}
The careful reader may observe that the map $\widehat{\LLT}$ is identical to the corresponding map described in the proof of \cite[Proposition 3.4]{HHL05} if the columns of the filling of $\dg(\lambda)$ are taken in reverse order to be mapped to ribbons.
\end{remark}
From the definitions of LLT inversions in Definition~\ref{def:LLTinversions} and attacking inversions in fillings of diagrams in Definition~\ref{def:invtwo}, the following result is immediate. See Figure~\ref{fig:LLT_T} for an example.
\begin{lemma}\label{lem:LLTtwo}
Let $\sigma\in\PQT(\lambda)$ and $\boldsymbol{\rho}=\widehat{\LLT}(\sigma)$. Then
\[
\invtwo(\sigma) = \inv(\boldsymbol{\rho}).
\]
\end{lemma}
Thus we have defined a weight preserving bijection from $\PQT(\lambda)$ with descent set $D \subseteq \{(r,j) \in \widehat{\dg}(\lambda)\}$ to $\SSYT(\boldnu(\lambda,D))$. (The reverse map follows easily from reversing Definition~\ref{def:LLTtwo}.) From this we obtain our final key lemma.
\begin{lemma}\label{lem:FG}
Let $\lambda$ be a partition, and let $D \subseteq \{(r,j) \in \widehat{\dg}(\lambda)\}$. Then
\[
F_{\lambda,D}(X;t) = G_{\boldnu(\lambda,D)}(X;t).
\]
\end{lemma}
Theorem~\ref{thm:symmetry} follows directly from Theorem~\ref{thm:LLT} and Lemma~\ref{lem:FG}.
\section{Proof that $C_{\lambda}(X;q,t)$ satisfies \eqref{A1m}}
\label{sec:A1m}
In this section we will use the ordering $<_1$ on $\mathcal{A}$ and construct a sign-reversing, weight-preserving involution $\Psi$ on super fillings $\widetilde{\PQT}(\lambda)$, which will cancel out all terms involving $x^{\mu}$ if $\mu \not\leq \lambda$.
We begin by defining the following map.
\begin{defn}\label{def:Phi_u}
Let $u\in\dg(\lambda)$. For $\sigma\in\widetilde{\PQT}(\lambda)$, define the map $\Phi_u$ by
\[
\Phi_u(\sigma(w))=\begin{cases} \sigma(w),& w \neq u,\\[0.2cm]
\overline{\sigma(w)},& w=u.
\end{cases}
\]
\end{defn}
In other words, $\Phi_u$ is the map that negates the content of the cell $u$.
We first recall Definition~\ref{def:invtwo} for attacking cells in our setting. A pair of cells $u=(r,i)$ and $v=(r',j)$ is attacking if $r=r'$, or $r=r'+1$ and $ia$, if such an $a$ exists. We call such an $a$ the \emph{distinguished label} of $\sigma$.
If $a$ exists, we call the first cell in $\PQT$ reading order whose absolute value is $a$ the \emph{distinguished cell} of $\sigma$, denoted by $u(\sigma)$, and we define
\[\Upsilon\coloneqq\Phi_{u(\sigma)}.
\]
Moreover, $\sigma$ falls into one of the following three categories:
\begin{enumerate}[label=(\alph*),leftmargin=20pt]
\item if no such $a$ exists, we say $\sigma$ is \emph{$\Phi$-trivial}.
\item if the distinguished cell $u(\sigma)$ does not belong to any degenerate triples \textup{(}regardless of whether they are $\quinv$ triples\textup{)}, we say $\sigma$ is \emph{$\Phi$-nondegenerate}.
\item if the cell $u(\sigma)$ is part of any degenerate triples \textup{(}regardless of whether they are $\quinv$ triples\textup{)}, we say $\sigma$ is \emph{$\Phi$-degenerate}. If $r$ is the row containing $u(\sigma)$, we call the set of cells in row $r$ that form degenerate triples the \emph{degenerate segment}.
\end{enumerate}
See Figure~\ref{fig:degenerate} for examples of all three.
\end{defn}
\begin{figure}[h]
\centering
\begin{tikzpicture}[scale=0.5]
\cell003
\cell10{$\bar3$}\cell11{$\bar2$}\cell123
\cell201\cell212\cell22{$\bar1$}
\node at (-2,-0.5) {(a)};
\end{tikzpicture}
\quad
\begin{tikzpicture}[scale=0.5]
\graycell001
\cell10{$\bar3$}\cell11{$\bar1$}\cell123
\cell201\cell212\cell22{$\bar1$}
\node at (-2,-0.5) {(b)};
\end{tikzpicture}
\quad
\begin{tikzpicture}[scale=0.5]
\cell002
\cell10{$\bar1$}\cell11{$\bar2$}\graycell121
\cell201\cell212\cell22{$\bar1$}
\node at (-2,-0.5) {(c)};
\end{tikzpicture}
\caption{(a) $\Phi$-trivial, (b) $\Phi$-nondegenerate, and (c) $\Phi$-degenerate. The grey box marks the distinguished cell if such exists. The degenerate segment in (c) is composed of the two cells at the tops of columns 2 and 3 in row 2.}\label{fig:degenerate}
\end{figure}
Importantly, $\Phi$-trivial fillings satisfy the following restriction on their content.
\begin{lemma}\label{lem:phitrivialm}
If $\sigma$ is a $\Phi$-trivial filling of $\dg(\lambda)$ with partition content $\mu$, then $\mu \leq \lambda'$.
\end{lemma}
\begin{proof}
Let $(1^{\mu_1},2^ {\mu_2},\cdots)$ be the content of $|\sigma|$ where $\mu_1\geq \mu_2\geq \cdots$. If $\sigma$ is $\Phi$-nondegenerate, then for each $j \geq 1$, all $j$'s and $\bar{j}$'s must be contained in rows 1 through $j$, and so $\mu_1+\cdots+\mu_j \leq \lambda_1'+\cdots+\lambda_j'$ (the length of row $j$ of $\dg(\lambda)$ is $\lambda_j'$). This is precisely the condition that $\mu\leq \lambda'$.
\end{proof}
The main result in this section will be the following theorem.
\begin{theorem}\label{thm:A2FP}
\begin{align}
C_{\lambda}[X(q-1);q,t]& = \sum_{\substack{\sigma\in\widetilde{\PQT}(\lambda)\\\sigma\ \text{is}\ \Phi\text{-trivial}}} (-1)^{m(\sigma)} q^{p(\sigma)+\maj(\sigma)}t^{\quinv(\sigma)}x^{|\sigma|}\label{eq:AF1}\\
&=\sum_{\mu \leq \lambda'} c_{\mu\lambda}(q,t)m_{\mu}\label{eq:AF2}
\end{align}
for some coefficients $c_{\mu\lambda}(q,t)\in\mathbb{Z}(q,t)$.
\end{theorem}
The outline of our proof is as follows. First we show that $\Upsilon$ is an involution on $\Phi$-nondegenerate fillings which will cancel the terms coming from those fillings in \eqref{C2}. This mirrors the corresponding involution from \cite[Section 5.2]{HHL05}. Next, in Section~\ref{sec:Phi-degenerate} we will prove the existence of a bijection on $\Phi$-degenerate fillings, which will cancel the terms coming from those fillings in \eqref{C2}. By Theorem~\ref{prop:Phi-degenerate}, only terms arising from $\Phi$-trivial fillings remain, giving \eqref{eq:AF1}, which is equivalent to \eqref{eq:AF2} by Lemma~\ref{lem:phitrivialm}.
Recall the map $\Phi_u$ from Definition~\ref{def:Phi_u}. As we will see in the following lemma, under total ordering $<_2$, for certain $u$'s, the action of $\Phi_u$ on $\sigma$ has the property that it increases $\maj$ by exactly one, and all triples that are not contained in the degenerate segment contribute to $\quinv(\sigma)$ if and only if they contribute to $\quinv(\Phi_u(\sigma))$.
When $\sigma$ is not $\Phi$-trivial, the following two lemmas present two important properties of $\Phi_u(\sigma)$ that hold for certain choices of the cell $u$.
\begin{lemma}\label{lem:Phi_u:maj}
Suppose $\sigma$ is not $\Phi$-trivial with distinguished label $a$, let $r$ be the row containing the distinguished cell, and let $u$ be any cell in row $r$ with $|\sigma(u)|=a$.
Then,
\begin{equation}
\maj(\Phi_u(\sigma))=\maj(\sigma)+\begin{cases}1,&\mbox{if}\ \sigma(u)=a,\\-1,&\mbox{if}\ \sigma(u)=\bar{a}.\end{cases}\end{equation}
\end{lemma}
\begin{lemma}\label{lem:Phi_u:quinv}
Suppose $\sigma$ is not $\Phi$-trivial with distinguished label $a$, and let $r$ be the row containing the distinguished cell. Then
\begin{enumerate}
\item if $\sigma$ is $\Phi$-nondegenerate and $u$ is the distinguished cell, or
\item if $\sigma$ is $\Phi$-degenerate and $u$ is any cell in the degenerate segment of row $r$ with $|\sigma(u)|=a$,
\end{enumerate}
the following holds: every nondegenerate triple containing $u$ in $\Phi_u(\sigma)$ is a $\quinv$ triple if and only if it is also a $\quinv$ triple in $\sigma$.
\end{lemma}
Lemmas~\ref{lem:Phi_u:maj} and~\ref{lem:Phi_u:quinv} will be proved later in this section.
\begin{example}
Let $\sigma$ be the filling in Figure~\ref{fig:degenerate}(b). The distinguished label is $a=1$ and $u=(3,1)$ is the distinguished cell (i.e. the $1$ in the topmost row), then $\Phi_u(\sigma)$ is the filling
\begin{center}
\begin{tikzpicture}[scale=0.5]
\cell00{$\bar1$}
\cell10{$\bar3$}\cell11{$\bar1$}\cell123
\cell201\cell212\cell22{$\bar1$}
\end{tikzpicture}
\end{center}
Moreover, $\maj(\sigma)=3$, $\maj(\Phi_u(\sigma))=4$, and $\quinv(\sigma)=\quinv(\Phi_u(\sigma))=3$.
\end{example}
\begin{remark}
The knowledgeable reader may observe that a similar involution $\Phi_{u'}$ was
defined in \cite[Section 5.2]{HHL05}, where $u'$ was chosen to be the topmost, \emph{leftmost} cell of $\dg(\lambda)$ whose absolute value matches the distinguished label of $\sigma$.
\end{remark}
We make a key observation.
\begin{lemma}\label{lem:obs}
Let $a$ be the distinguished label and $r$ the row containing the distinguished cell in a filling $\sigma$. Let $x$ be any cell in either row $r-1$ or $r$, and let $y$ be any cell in row $r+1$ or higher. Then $|\sigma(x)|\geq a$ and $|\sigma(y)|>a$.
\end{lemma}
\begin{proof}
Since $a$ is by definition the smallest positive integer such that there exists a cell in row $j>a$ with absolute value $a$, a row greater than or equal to $r-1$ cannot contain any cells with absolute value strictly less than $a$. Moreover, since $u$ is chosen such that $r$ is maximal, rows $r+1$ and higher cannot contain any cells with absolute value less than or equal to $a$.
\end{proof}
\begin{proof}[Proof of Lemma~\ref{lem:Phi_u:maj}]
The proof is identical to that of \cite[Lemma 5.2]{HHL05}. Let $u$ be any cell in row $r$ with $|\sigma(u)|=a$. Without loss of generality, assume $\sigma(u)=a$. The only possible change in the descent set of $\sigma$ by the action of $\Phi_u$ can occur at the cell $u$ or the cell directly above $u$ if such exists.
Consider the cells $x$ and $y$ with contents $c\coloneqq\sigma(x)$ and $b\coloneqq\sigma(y)$ directly above and below $u$, respectively, if such exist.
Then we obtain a transition exactly analogous to \eqref{psi-action-maj}.
By Lemma~\ref{lem:obs}, $|b| \geq a$ and $|c| > a$. Thus $I_2(c,a)=I_2(\bar{a},b)=1$ and $I_2(c,\bar{a})=I_2(a,b)=0$, so necessarily $x\in\Des(\sigma)$ and $u\not\in\Des(\sigma)$, while $x\not\in\Des(\Phi_u(\sigma))$ and $u\in\Des(\Phi_u(\sigma))$, implying that $\maj(\Phi_u(\sigma))=\maj(\sigma)+1$. A similar argument proves the statement in the case where the cell $x$ may not exist.
\end{proof}
\begin{proof}[Proof of Lemma~\ref{lem:Phi_u:quinv}]
We adapt the proof of \cite[Lemma 5.2]{HHL05} to the setting of $\widetilde{\PQT}$. Let $u$ be a cell in row $r$ with $|\sigma(u)|=a$, such that either
\begin{enumerate}
\item $\sigma$ is $\Phi$-nondegenerate and $u$ is the distinguished cell, or
\item $\sigma$ is $\Phi$-degenerate and $u$ is in the degenerate segment of row $r$.
\end{enumerate}
Without loss of generality, assume $\sigma(u)=a$.
Clearly the only triples that may be affected by $\Phi_u$ are those containing the cell $u$. Let us inspect the three possible \emph{nondegenerate} configurations of such triples. Let $x,y$ be the other two cells in the triple, with contents $b\coloneqq\sigma(y)$ and $c\coloneqq\sigma(x)$; since the triples we examine are nondegenerate, both $x$ and $y$ must exist. We invoke Lemma~\ref{lem:obs}.
\begin{flushleft}
\begin{tikzpicture}[scale=0.5]
\node at (-4.5,0) {\emph{Case 1}:};
\cellL0{.5}{$a$}{$u$}\cellL1{.5}{$c$}{$x$}\cellL1{3.5}{$b$}{$y$}
\node at (1.3,-.5) {$\cdots$};
\node at (5.5,0) {$\longrightarrow$};
\node at (5.5,.7) {$\Phi_u$};
\cellL0{8.5}{$\bar{a}$}{$u$}\cellL1{8.5}{$c$}{$x$}\cellL1{11.5}{$b$}{$y$}
\node at (9.3,-.5) {$\cdots$};
\end{tikzpicture}
\end{flushleft}
Both $|b|\geq a$ and $|c|\geq a$. Then $I_2(a,c)=I_2(a,b)=0$ and $I_2(\bar{a},c)=I_2(\bar{a},b)=1$. Thus the triple $(u,x,y)$ is a $\quinv$ triple both in $\sigma$ and in $\Phi(\sigma)$ if and only if $I_2(b,c)=1$.
\begin{flushleft}
\begin{tikzpicture}[scale=0.5]
\node at (-4.5,0) {\emph{Case 2}:};
\cellL0{.5}{$c$}{$x$}\cellL1{.5}{$b$}{$y$}\cellL1{3.5}{$a$}{$u$}
\node at (1.3,-.5) {$\cdots$};
\node at (5.5,0) {$\longrightarrow$};
\node at (5.5,.7) {$\Phi_u$};
\cellL0{8.5}{$c$}{$x$}\cellL1{8.5}{$b$}{$y$}\cellL1{11.5}{$\bar{a}$}{$u$}
\node at (9.2,-.5) {$\cdots$};
\end{tikzpicture}
\end{flushleft}
Since $|c|>a$, we have $I_2(c,a)=1$ and $I_2(c,\bar{a})=0$. Since $|b|\geq a$, we have $I_2(a,b)=0$ and $I_2(\bar{a},b)=1$. Thus the triple $(x,y,u)$ is a $\quinv$ triple both in $\sigma$ and in $\Phi(\sigma)$ if and only if $I_2(c,b)=0$.
\begin{flushleft}
\begin{tikzpicture}[scale=0.5]
\node at (-4.5,0) {\emph{Case 3}:};
\cellL0{.5}{$b$}{$y$}\cellL1{.5}{$a$}{$u$}\cellL1{3.5}{$c$}{$x$}
\node at (1.3,-.5) {$\cdots$};
\node at (5.5,0) {$\longrightarrow$};
\node at (5.5,.7) {$\Phi_u$};
\cellL0{8.5}{$b$}{$y$}\cellL1{8.5}{$\bar{a}$}{$u$}\cellL1{11.5}{$c$}{$x$}
\node at (9.2,-.5) {$\cdots$};
\end{tikzpicture}
\end{flushleft}
This case can only occur if $\sigma$ is $\Phi$-nondegenerate, since otherwise $u$ is necessarily in the degenerate segment, contradicting the existence of the cell $y$. Thus we assume $u$ is the distinguished cell, which means it is the first cell in reading order with content equal to $a$ or $\bar{a}$. Consequently $|c|>a$ because $x$ precedes $u$ in the reading order, and so $I_2(c,a)=1$ and $I_2(c,\bar{a})=0$. Since $|b|>a$, we have $I_2(b,a)=1$ and $I_2(b,\bar{a})=0$. Thus, again, the triple $(y,u,x)$ is a $\quinv$ triple both in $\sigma$ and in $\Phi(\sigma)$ if and only if $I_2(b,c)=1$.
In each of these cases, all nondegenerate triples containing $u$ contribute to $\quinv(\sigma)$ if and only if they also contribute to $\quinv(\Phi_u(\sigma))$, as desired.
\end{proof}
\begin{corollary}\label{cor:nondegenerate}
Let $\sigma\in\widetilde{\PQT}(\lambda)$ be $\Phi$-nondegenerate.
\begin{enumerate}[label=(\roman*)]
\item $\Upsilon$ is an involution on $\widetilde{\PQT}(\lambda)$, and
\item $x^{|\Upsilon(\sigma)|} q^{p(\Upsilon(\sigma))+\maj(\Upsilon(\sigma))}t^{\quinv(\Upsilon(\sigma))} = x^{|\sigma|} q^{p(\sigma)+\maj(\sigma)}t^{\quinv(\sigma)}$.
\end{enumerate}
\end{corollary}
\begin{proof}
The fact that $\Upsilon$ is an involution is immediate, since $|\Upsilon(\sigma)|=|\sigma|$, and so both fillings have the same distinguished label. Hence $u(\Upsilon(\sigma))=u(\sigma)$, meaning that the same distinguished cell is chosen for $\Upsilon(\sigma)$ as for $\sigma$.
For part (ii), $|\Upsilon(\sigma)|=|\sigma|$ by definition. Without loss of generality, assume $\sigma(u)\in\mathbb{Z}_+$. Then by Lemma~\ref{lem:Phi_u:maj} and Lemma~\ref{lem:Phi_u:quinv}, $p(\Upsilon(\sigma))=p(\sigma)-1$, $\maj(\Upsilon(\sigma)) = \maj(\sigma)+1$, and $\quinv(\sigma)=\quinv(\Upsilon(\sigma))$ since $\sigma$ is $\Phi$-nondegenerate, so the equality follows.
\end{proof}
Unfortunately, when $\sigma\in\widetilde{\PQT}(\lambda)$ is $\Phi$-degenerate, it may be the case that no choice of the cell $u$ for which $|\sigma(u)|=a$ will be such that $\Phi_u$ preserves $\quinv$ and changes $\maj$ by 1. For example, going back to the filling $\sigma$ in Figure~\ref{fig:degenerate}(c), which has $\maj=3$ and $\quinv=3$, we list below the outputs of $\Phi_u(\sigma)$ where we set $u$ to be each of the four cells whose content is $1$ or $\bar{1}$, with the highlighted cell indicating the choice of $u$.
\begin{center}
\begin{tikzpicture}[scale=0.5]
\cell002
\graycell10{$1$}\cell11{$\bar2$}\cell12{1}
\cell201\cell212\cell22{$\bar1$}
\node at (0.5,-3) {$\substack{\maj=2\\\quinv=2}$};
\end{tikzpicture}
\quad
\begin{tikzpicture}[scale=0.5]
\cell002
\cell10{$\bar1$}\cell11{$\bar2$}\graycell12{$\bar1$}
\cell201\cell212\cell22{$\bar1$}
\node at (0.5,-3) {$\substack{\maj=4\\\quinv=4}$};
\end{tikzpicture}
\quad
\begin{tikzpicture}[scale=0.5]
\cell002
\cell10{$\bar1$}\cell11{$\bar2$}\cell121
\graycell20{$\bar1$}\cell212\cell22{$\bar1$}
\node at (0.5,-3) {$\substack{\maj=3\\\quinv=2}$};
\end{tikzpicture}
\quad
\begin{tikzpicture}[scale=0.5]
\cell002
\cell10{$\bar1$}\cell11{$\bar2$}\cell121
\cell201\cell212\graycell22{$1$}
\node at (0.5,-3) {$\substack{\maj=3\\\quinv=2}$};
\end{tikzpicture}
\end{center}
Thus we need to do something more subtle. In the following section, we will describe with some technical manipulations how to define a bijection on $\Phi$-degenerate fillings that preserves $\quinv$ and changes $\maj$ by 1.
\section{A bijection on $\Phi$-degenerate tableaux}
\label{sec:Phi-degenerate}
In this section, we \emph{demonstrate the existence} of a bijection $\Theta$ on $\Phi$-degenerate super fillings with the following properties. Let $\sigma\in\widetilde{\PQT}$.
\begin{enumerate}[label=\roman*.]
\item $x^{|\Theta(\sigma)|}=x^{|\sigma|}$ and $p(\Theta(\sigma))=p(\sigma) \pm 1$,
\item $\maj(\Theta(\sigma))=\maj(\sigma)\mp 1$, and
\item $\quinv(\Theta(\sigma))=\quinv(\sigma)$.
\end{enumerate}
Note that we do not succeed in constructing such an bijection, and it remains an open question to find an explicit bijection.
\begin{defn}For a $\Phi$-degenerate filling $\sigma$, define the \emph{degenerate word} to be the content of the degenerate segment containing the distinguished cell, read from left to right. For example, in Figure~\ref{fig:degenerate}(c), the degenerate word is $(\bar{2},1)$.
\end{defn}
The main result of this section is the following.
\begin{theorem}
\label{prop:Phi-degenerate}
For a partition $\lambda$, the sum over the $\Phi$-degenerate subset of fillings in $\widetilde{\PQT}(\lambda)$ in the right hand side of \eqref{C2}, is zero.
\[
\sum_{\substack{\sigma\in\widetilde{\PQT}(\lambda)\\\sigma\ \text{is}\ \Phi\text{-degenerate}}} (-1)^{m(\sigma)} q^{p(\sigma)+\maj(\sigma)}t^{\quinv(\sigma)}x^{|\sigma|} =0.
\]
\end{theorem}
The outline of our proof is as follows. Let $a$ be the distinguished label of a $\Phi$-degenerate filling $\sigma\in\widetilde{\PQT}$. We will show in Proposition~\ref{thm:Wbijection} that there exists a $\quinv$-preserving map $\Theta:\widetilde{\PQT}\rightarrow \widetilde{\PQT}$ that preserves the content of the filling in absolute value (in particular $\Theta$ preserves the content in all the letters that are not $a$ or $\bar{a}$), such that $\Theta(\sigma)$ has exactly one more or one less $\bar{a}$ as $\sigma$. Next in Definition~\ref{def:tau} we define entry-swapping operators $\{\tau_j\}$ on $\widetilde{\PQT}$ that swap the topmost entries in columns $j,j+1$ (i.e. they swap entries in the degenerate word) while preserving the $\maj$ and $\quinv$ in the rest of the filling (Lemmas~\ref{lem:tau:maj} and~\ref{lem:tau:quinv}). Suppose the distinguished cell $u(\sigma)$ lies in the degenerate segment with degenerate word $w$. There is then a unique product of operators $\tau_w=\tau_{i_1}\tau_{i_2}\ldots\tau_{i_s}$ that can be applied to $\Upsilon(\sigma)=\Phi_{u(\sigma)}(\sigma)$ such that
\begin{itemize}
\item the degenerate word of $\tau_w(\Upsilon(\sigma))$ matches $\phi(w)$, and
\item $\widehat{\sigma} = \tau_w(\Upsilon(\sigma))$ satisfies
\[q^{p(\widehat{\sigma})+\maj(\widehat{\sigma})}t^{\quinv(\widehat{\sigma})} = q^{p(\sigma)+\maj(\sigma)}t^{\quinv(\sigma)}.\]
\end{itemize}
\subsection{The operators $\tau_j$}
We first define the entry-swapping operator $\tau_j$ that swaps the topmost entries (if they are distinct) in columns $j,j+1$ of equal height while preserving certain statistics. This operator is modified for use in our setting from its original definition in \cite{LoehrNiese}.
\begin{defn}[Definition of the operator $\tau_j$]\label{def:tau}
Suppose two columns $j$ and $j+1$ have equal height, i.e.
$\lambda_j=\lambda_{j+1}=k$. We define an operator $\tau_j$ which
exchanges contents of certain cells between columns $j$ and $j+1$.
Write $\sigma(r,j)=a_r$ and $\sigma(r,j+1)=b_r$ for $r=1,\dots, k$.
Define $r_{\max}$ to be the largest $r\in\{2,\dots,k\}$ with the following property:
either $(a_r, a_{r-1}, b_{r-1})$ and $(b_r, a_{r-1}, b_{r-1})$ are both in $\cQ$,
or both are not in $\cQ$. \textup{(}That is, we look for the largest value of $r$
such that exchanging the entries $a_r$ and $b_r$
makes no difference to whether $((r,j), (r-1,j), (r-1, j+1))$ is a quinv triple.\textup{)}
If there is no such $r$, let $r_{\max}=1$.
Now let the operator $\tau_j$ swap the entries between columns $j$ and $j+1$
in rows $i$ with $r_{\max}\leq i\leq k$; i.e. for those values of $i$,
\[
\tau_j(\sigma)(i,j)=b_i,
\qquad
\tau_j(\sigma)(i,j+1)=a_i,
\]
while all other entries are the same in $\sigma$ and in $\tau_j(\sigma)$,
as in the picture below, where we denote $\ell\coloneqq r_{\max}$.
\begin{equation*}
\renewcommand{\arraystretch}{1} % only needed because for some reason we reset this value globally to 2, default is 1. Can adjust for wider spacing as desired.
\setlength{\arraycolsep}{1pt} %default is 6pt. Can adjust for wider spacing as desired
\begin{array}{l|c|c|}
\multicolumn{1}{c}{}&
\multicolumn{2}{c}{\sigma}
\\
\cline{2-3} % an alternative is \hhline{~--} which works better with shaded cells
% \hhline requires the hhline package
\text{row } k & a_k & b_k\\
\cline{2-3} % \hhline{~--}
\,\,\,\vdots & \vdots & \vdots \\
\cline{2-3}
\text{row } \ell & a_{\ell} & b_{\ell}\\
\cline{2-3} % \hhline{~--}
\text{row } \ell-1\,\,\,\,\,\, & a_{\ell-1} & b_{\ell-1}\\
\cline{2-3}
\,\,\,\vdots & \vdots & \vdots \\
\cline{2-3}
\multicolumn{1}{c}{}&
\multicolumn{1}{c}{\scriptstyle j}&
\multicolumn{1}{c}{\scriptstyle j+1}
\end{array}
\,\,\xrightarrow{\makebox[1.5cm]{$\tau_j$}}\,\,
\begin{array}{|c|c|}
\multicolumn{2}{c}{\tau_j(\sigma)}
\\
\hline
b_k & a_k\\
\hline
\vdots & \vdots \\
\hline
b_{\ell} & a_{\ell}\\
\hline
a_{\ell-1} & b_{\ell-1}\\
\hline
\vdots & \vdots \\
\hline
\multicolumn{1}{c}{\scriptstyle j}&
\multicolumn{1}{c}{\scriptstyle j+1}
\end{array}
\end{equation*}
Put another way: we swap the pair in row $k$, and iteratively, if we have swapped the pair in row $i$ and this made a difference to whether the triple $(i,j), (i-1, j), (i-1, j+1)$ is a quinv
triple, then we swap in row $i-1$ also.
\end{defn}
\begin{example}\label{ex:tau}
Suppose $\sigma$ has columns $j,j+1$ as shown below. Then $k=5$ and $r_{\max}=3$ since both $(2,3,4)\in\cQ$ and $(3,3,4)\in\cQ$. Thus applying the operator $\tau_j$ gives the following. The cells whose content was swapped are shown in grey.
\begin{equation*}
\renewcommand{\arraystretch}{1}
\setlength{\arraycolsep}{1pt}
\begin{array}{|c|c|}
\multicolumn{2}{c}{\sigma}
\\
\hline
3 & 4\\
\hline
2 & 3 \\
\hline
2 & 3 \\
\hline
3 & 4\\
\hline
1 & 3 \\
\hline
\multicolumn{1}{c}{\scriptstyle \phantom{+} j\phantom{1}}&
\multicolumn{1}{c}{\scriptstyle j+1}
\end{array}
\,\,\xrightarrow{\makebox[1.5cm]{$\tau_j$}}\,\,
\begin{array}{|c|c|}
\multicolumn{2}{c}{\tau_j(\sigma)}
\\
\hline
\cellcolor{gray!50}4 & \cellcolor{gray!50} 3\\
\hline
\cellcolor{gray!50}3 & \cellcolor{gray!50}2 \\
\hline
\cellcolor{gray!50}3 & \cellcolor{gray!50}2 \\
\hline
3 & 4\\
\hline
1 & 3 \\
\hline
\multicolumn{1}{c}{\scriptstyle \phantom{+} j\phantom{1}}&
\multicolumn{1}{c}{\scriptstyle j+1}
\end{array}
\end{equation*}
\end{example}
It is straightforward to check that $\tau_j$ is an involution: let $\sigma$ be a filling with $\sigma'=\tau_j(\sigma)$ such that the entries in rows $\ell,\cdots,k$ are swapped between columns $j$ and $j+1$, meaning that the $r_{\max}$ for $\sigma$ is $\ell$. One verifies that for any $c\in\mathcal{A}$, if $a_k\neq b_k$, then $(c,a_{k},b_{k})\in\cQ$ if and only if $(c,b_k,a_k)\not\in\cQ$. On the other hand, if $a_k= b_k\in\mathbb{Z}_+$, then $(c,a_{k},b_{k}) \not\in\cQ$, and if $a_k= b_k\in\mathbb{Z}_-$, then $(c,a_{k},b_{k})\in\cQ$, and so, necessarily, $r_{\max}\geq k+1$. Therefore, if a swap at row $k+1$ changes the contribution to $\quinv$ of the triple of cells $(x_{k+1},x_k,y_k)$, then $a_k\neq b_k$, and hence this change will be reversed by a swap at row $k$. Thus it is immediate that $\sigma'$ has the same value for $r_{\max}$, and so $\tau_j$ acting on $\sigma'$ reverses the swaps of the cells in rows $\ell,\dots,k$, recovering $\sigma$.
Proving the following Lemmas~\ref{lem:tau:maj} and ~\ref{lem:tau:quinv} regarding the properties of the action of $\tau_j$ on super fillings $\widetilde{\PQT}$ is also straightforward, but technical.
\begin{lemma}\label{lem:tau:maj}
Let $\sigma\in\widetilde{\PQT}$ and let $j$ be such that the columns $j,j+1$ are of equal height $k$. Then
\begin{equation}
\maj(\tau_j(\sigma))=\maj(\sigma).
\end{equation}
\end{lemma}
\begin{lemma}\label{lem:tau:quinv}
Let $\sigma\in\widetilde{\PQT}$ and let $j$ be such that the columns $j,j+1$ are of equal height $k$. Let $x=(k,j)$ and $y=(k,j+1)$ be the cells at the tops of columns $j$ and $j+1$. Then
\begin{equation}
\quinv(\tau_j(\sigma))=\quinv(\sigma) + \begin{cases} 1,& \sigma(x)>\sigma(y),\\
-1,&\sigma(x)<\sigma(y),\\
0,&\sigma(x)=\sigma(y). \end{cases}
\end{equation}
\end{lemma}
\begin{remark}
We point out that our definition of $\tau_j$ is very similar to that of a similar operator in \cite[Definition 3.7]{compactformula}, with two important differences. First, $\tau_j$ necessarily swaps the topmost entries in the columns if they are different and acts trivially if they are identical, instead of swapping the first pair of non-identical entries (from the bottom) as in \cite{compactformula}. Second, in our case swaps propagate downwards from row $k$, whereas in the corresponding operator defined in \cite{compactformula}, the swaps propagate upwards. This has to do with the fact that a $\quinv$ triple has configuration
\[
\qtrip{\ }{\ }{\ }
\]
whereas the corresponding triple studied in \cite{compactformula} has configuration
\[
\begin{tikzpicture}[scale=0.5]\cell10{\ }\cell20{\ } \cell1{2.7}{\ }
\node at (1,-0.5) {$\cdots$};
\end{tikzpicture}.
\]
\end{remark}
\begin{proof}[Proof of Lemma~\ref{lem:tau:maj}] The proof is almost identical to that of \cite[Lemma 3.10]{compactformula}, but we present it here as a warm-up for Lemma~\ref{lem:tau:quinv}, in the more nuanced setting of super~fillings.
If $k=1$, there is no contribution to $\maj$ from columns $j,j+1$, so assume $k>1$. Consider the entries in columns $j,j+1$ in the two rows $\ell,\ell-1$, such that $1 < \ell \leq k$, as shown below.
\begin{center}
\begin{tikzpicture}[scale=.5]
\cell01{$a$} \cell02{$b$}
\cell11{$c$} \cell12{$d$}
\node at (-2.65,.5) {\rm row $\ell$};
\node at (-2,-.5) {\rm row $\ell-1$};
\node at (.5, 1.5) {{\tiny$j$}};
\node at (2, 1.5) {{\tiny$j+1$}};
\end{tikzpicture}
\end{center}
If $a,b$ are not swapped by $\tau_j$, there is nothing to check. If both $a,b$ and $c,d$ are swapped by $\tau_j$, the contribution to $\maj$ from the cells $(\ell,j),(\ell,j+1)$ remains the same. Thus the only case we need to check is when $\ell=r_{\max}$ so that the swapping procedure terminates at row $\ell$ and $a,b$ are swapped, but $c,d$ are not. Thus assume $\tau_j$ sends \raisebox{.1in}{\tableau{a&b\\c&d}} to \raisebox{.1in}{\tableau{b&a\\c&d}}.
For the remainder of this section, set $I=I_2$. Since $\ell=r_{\max}$, we have that the triples $(a,c,d)$ and $(b,c,d)$ are are either both in $\cQ$, or neither is in $\cQ$. Following the condition in Definition~\ref{def:super} of a triple to be in $\cQ$, let us consider the sets
\[
ACD = \left\{\begin{matrix}I(a,c)=1\\I(a,d)=0\\I(d,c)=0 \end{matrix} \right\}
\quad \mbox{and} \quad
BCD = \left\{\begin{matrix}I(b,c)=1\\I(b,d)=0\\I(d,c)=0 \end{matrix} \right\},
\]
where we know that either both $ACD$ and $BCD$ contain exactly one condition that is true, or they both contain exactly two that are true. Recall that a pair of cells $u,v$ in $\sigma$ with $v=\South(u)$ forms a descent if $I(\sigma(u),\sigma(v))=1$. We consider two cases based on whether or not $I(d,c)=0$.
\emph{Case 1: Suppose $I(d,c)=0$.} If the two triples are in $\cQ$, both of the remaining conditions in $ACD$ and $BCD$ must be false, and so we must have $I(a,c)=I(b,c)=0$ and $I(a,d)=I(b,d)=1$. This implies the cells $(\ell,j),(\ell,j+1)$ make the same contribution to $\maj$ in $\sigma$ and $\tau_j(\sigma)$. On the other hand, if the two triples are not in $\cQ$, exactly one of the two remaining conditions in both $ACD$ and $BCD$ must be true, and so we must have $I(a,c)=I(a,d)$ and $I(b,c)=I(b,d)$. This also results in the same contribution to $\maj$ from the cells $(\ell,j),(\ell,j+1)$ in $\sigma$ and $\tau_j(\sigma)$.
\emph{Case 2: Suppose $I(d,c)=1$.} This case is similar. If the two triples are in $\cQ$, we must have $I(a,c)=I(a,d)$ and $I(b,c)=I(b,d)$ since exactly one of the two remaining conditions must be true. If the two triples are not in $\cQ$, both of the two remaining conditions must be true, and so we must have $I(a,c)=I(b,c)=1$ and $I(a,d)=I(b,d)=0$. Both cases result in the same contribution to $\maj$ from the cells $(\ell,j),(\ell,j+1)$.
We see finally that even though the locations of the descents may change after applying $\tau_j$, the total contribution to $\maj$ in columns $j,j+1$ remains constant.
\end{proof}
\begin{proof}[Proof of Lemma~\ref{lem:tau:quinv}]
The proof follows by a similar argument as that of \cite[Lemma 3.11]{compactformula}, though the details are a bit nuanced when we work with super fillings.
Consider the columns $j,j+1$ of $\sigma$ shown below with $x=(k,j)$, $y=(k,j+1)$, and $a_k=\sigma(x)$, $b_k=\sigma(y)$. Assume $a_k\neq b_k$, since otherwise $\tau_j$ is trivial. Let $\ell\coloneqq r_{\max}$, so $\tau_j$ swaps all pairs of entries in rows $j,j+1$ from row $k$ down to row $\ell$. If $\ell=1$, then the columns $j,j+1$ simply switched places, which trivially proves the claim. Thus assume $\ell>1$, and let $c,d$ be the cells in row $\ell-1$ in columns $j,j+1$ respectively.
\begin{center}
\begin{tikzpicture}[scale=0.55]
\node at (2,1.7){$\sigma$};
\node at (9,1.7){$\tau_j(\sigma)$};
\node at (-3.05,0.5){row $k$};
\node at (-3.1,-1.5){row $\ell$};
\node at (-2.5,-2.5){row $\ell-1$};
\draw[->](5,-1)--(6.5,-1) node[midway,above] {$\tau_j$};
\cell01{$a_k$}\cell02{$b_k$}\cell0{4.5}{$f_1$}
\cell11{$\vdots$}\cell12{$\vdots$}\cell1{4.5}{$\vdots$}
\cell21{$a_{\ell}$}\cell22{$b_{\ell}$}\cell2{4.5}{$f_{\ell}$}
\cell31{$c$}\cell32{$d$}\cell3{4.5}{$f$}
%%
\cell08{$b_k$}\cell0{9}{$a_k$}\cell0{11.5}{$f_1$}
\cell18{$\vdots$}\cell1{9}{$\vdots$}\cell1{11.5}{$\vdots$}
\cell28{$b_{\ell}$}\cell2{9}{$a_{\ell}$}\cell2{11.5}{$f_{\ell}$}
\cell38{$c$}\cell3{9}{$d$}\cell3{11.5}{$f$}
\node at (0.5,-3.5){{\tiny $j$}};
\node at (2,-3.5){{\tiny $j+1$}};
\node at (4,-3.5){{\tiny $s$}};
\node at (7.5,-3.5){{\tiny $j$}};
\node at (9,-3.5){{\tiny $j+1$}};
\node at (11,-3.5){{\tiny $s$}};
\end{tikzpicture}
\end{center}
Since only columns $j,j+1$ are affected by $\tau_j$, we only need to examine the contribution in $\quinv(\sigma)$ and $\quinv(\tau_j(\sigma))$ of triples with at least one entry in one of those columns. By construction, with the exception of the degenerate triple formed by the entries $(a_k,b_k)$, all triples in $\sigma$ in columns $j,j+1$ make the same contribution in $\quinv(\sigma)$ as the corresponding triples in the same locations of $\tau_j(\sigma)$ do in $\quinv(\tau_j(\sigma))$. Moreover, triples containing cells in columns to the left of $j$ will not be affected either, nor will any triples with all cells below row $\ell$. It remains for us to examine the triples containing cells in rows greater than or equal to $\ell-1$ in one of the columns $j,j+1$, and a cell in a column $s$ for some $s>j+1$. Both $\sigma$ and $\tau_j(\sigma)$ have the triples $((i+1,j),(i,j),(i,s))$ and $((i+1,j+1),(i,j+1),(i,s))$ with the same contents $(a_{i+1},a_i,f_i)$ and $(b_{i+1},b_i,f_i)$ for $\ell\leq i \leq k$ (per our convention we set $\sigma(a_{k+1})=\sigma(b_{k+1})=0$). Therefore the only triples whose content was altered are the triples $((\ell,j),(\ell-1,j),(\ell-1,s))$ and $((\ell,j+1),(\ell-1,j+1),(\ell-1,s))$ with respective contents $(a_{\ell},c,f)$ and $(b_{\ell},d,f)$ in $\sigma$, which ended up with respective content $(b_{\ell},c,f)$ and $(a_{\ell},d,f)$ in $\tau_j(\sigma)$. We will now check that the total number of $\quinv$ triples is preserved in each of these pairs. To simplify notation, we will drop the subscripts and write $a=a_{\ell}$ and $b=b_{\ell}$.
We begin with a key observation. Since $\ell=r_{\max}$, we have that either both $(a,c,d)$ and $(b,c,d)$ are in $\cQ$, or neither is in $\cQ$. Equivalently, this means that in the sets
\[
ACD = \left\{\begin{matrix}I(a,c)=1\\I(a,d)=0\\I(d,c)=0 \end{matrix} \right\}
\quad \mbox{and} \quad
BCD = \left\{\begin{matrix}I(b,c)=1\\I(b,d)=0\\I(d,c)=0 \end{matrix} \right\},
\]
either exactly one condition is true in both, or exactly two are true in both. By studying the cases in the proof of Lemma~\ref{lem:tau:maj}, we get that either both $I(a,c)=I(a,d)$ and $I(b,c)=I(b,d)$, or both $I(a,c)\neq I(a,d)$ and $I(b,c)\neq I(b,d)$.
Let us now consider the triples $(a,c,f)$, $(b,d,f)$, and $(b,c,f)$, $(a,d,f)$, respectively corresponding to the sets
{\scriptsize
\[
ACF = \left\{\begin{matrix}I(a,c)=1\\I(a,f)=0\\I(f,c)=0 \end{matrix} \right\},\
BDF = \left\{\begin{matrix}I(b,d)=1\\I(b,f)=0\\I(f,d)=0 \end{matrix} \right\}
\; \mbox{v/s} \;
BCF = \left\{\begin{matrix}I(b,c)=1\\I(b,f)=0\\I(f,c)=0 \end{matrix} \right\},\
ADF = \left\{\begin{matrix}I(a,d)=1\\I(a,f)=0\\I(f,d)=0 \end{matrix} \right\}.
\]
}
We split our argument into two cases. To eliminate extra notation, let us identify the triples with their corresponding sets.
In the first case, let $I(a,c)=I(a,d)$ and $I(b,c)=I(b,d)$. If $I(f,c)=I(f,d)$, then $ACF$ (\emph{resp.} $BDF$) has the same number of true conditions as $ADF$ (\emph{resp.} $BCF$), which means the pairs make the same contribution to $\quinv(\sigma)$ as to $\quinv(\tau_j(\sigma))$. Thus let us assume $I(f,c)\neq I(f,d)$. In the table below, we examine the cases recalling (from Definition~\ref{def:super}) that there is no contradiction in either $ACF$ or $ADF$ arising from the latter having either all conditions true or all conditions false.
Our notation is as follows: when we write, say, $\text{Set} \Rightarrow X$, this means that $X$ must be true in order to avoid a contradiction in $\text{Set}$.
\begin{center}
{\renewcommand{\arraystretch}{1.2}
\flushleft
\footnotesize
\begin{tabular}{l|c|c|}
& $I(f,c)=0$ and $I(f,d)=1$ & $I(f,c)=1$ and $I(f,d)=0$\\ \hline
$I(a,c)=I(a,d)=1$ & $ACF \Rightarrow I(a,f)=1$ & $ADF \Rightarrow I(a,f)=1$ \\ \hline
$I(a,c)=I(a,d)=0$ & $ADF \Rightarrow I(a,f)=0$ & $ACF \Rightarrow I(a,f)=0$\\ \hline
\end{tabular}
}
\end{center}
This table implies that necessarily, $I(a,c)=I(a,d)=I(a,f)$.
An analogous argument yields that necessarily, $I(b,d)=I(b,f)=I(b,c)$. From here it is straightforward to check that for any choice of $I(f,c)$ and $I(f,d)$, $ACF$ and $BDF$ make the same contribution to $\quinv(\sigma)$, as BCF and ADF to $\quinv(\tau_j(\sigma))$, and so the total contribution of the pair $ACF, BDF$ to $\quinv(\sigma)$ matches the contribution of the pair $BCF, ADF$ to $\quinv(\tau_j(\sigma))$.
In the second case, let $I(a,c)\neq I(a,d)$ and $I(b,c)\neq I(b,d)$. Because $\maj$ is preserved, we must also have that $I(a,c)\neq I(b,d)$, implying that $I(a,c)=I(b,c)$ and $I(b,d)=I(a,d)$. We must also consider all choices for $I(f,c)$ and $I(f,d)$. We complete the argument by carefully considering all possible cases in the table below. We use some additional shorthand notation: when we write, say, $\{\cancel{ACF}, ADF\}$, this means $ACF$ is not a $\quinv$ triple, but $ADF$ is one. When we write, say, $ACF=ADF$ \emph{dep.~on} $I(a,f)$, this means that the value of $I(a,f)$ determines both $ACF$ and $ADF$ (and they are equal in either case).
{\renewcommand{\arraystretch}{1.2}
\flushleft
\footnotesize
\begin{tabular}{l|c|c|c|c|}
& $I(f,c)=0,$ & $I(f,c)=0,$ & $I(f,c)=1,$ & $I(f,c)=1,$\\
& $I(f,d)=0$ & $I(f,d)=1$ & $I(f,d)=0$ & $I(f,d)=1$\\ \hline
%%
$I(a,c)=1$, & $ACF \Rightarrow I(a,f)=1$ & $ACF \Rightarrow$ & $ACF=ADF$ & $ADF \Rightarrow I(a,f)=0$ \\
$I(b,c)=1$,& $\Rightarrow\{\cancel{ACF},\ ADF\}$ & $I(a,f)=1$, &dep.~on~$I(a,f)$ & $\Rightarrow\{\cancel{ACF}, ADF\}$ \\
$I(b,d)=0$, & $BCF \Rightarrow I(b,f)=1$ & contradicts & $BCF=BDF$ & $BDF \Rightarrow I(b,f)=0$ \\
$I(a,d)=0$& $\Rightarrow \{BDF,\ \cancel{BCF}\}$ & $ADF$ &dep.~on~$I(b,f)$ & $\Rightarrow \{BDF,\ \cancel{BCF}\}$ \\\hline
%%
$I(a,c)=0$, & $ADF \Rightarrow I(a,f)=1$ & $ACF=ADF$ & $ACF \Rightarrow$ & $ACF \Rightarrow I(a,f)=0$ \\
$I(b,c)=0$,& $\Rightarrow \{ACF,\ \cancel{ADF}\}$ &dep.~on~$I(a,f)$ & $I(a,f)=0$, & $\Rightarrow \{ACF, \cancel{ADF}\}$ \\
$I(b,d)=1$, & $BDF \Rightarrow I(b,f)=1$ & $BCF=BDF$ & contradicts & $BCF \Rightarrow I(b,f)=0$ \\
$I(a,d)=1$& $\Rightarrow\{\cancel{BDF},\ BCF\}$ &dep.~on~$I(b,f)$ &
$ADF$ & $\Rightarrow\{\cancel{BDF},\ BCF\}$ \\\hline
\end{tabular}
}
\medskip
Again, in every possible case, the contribution of the pair $ACF, BDF$ to $\quinv(\sigma)$ matches the contribution of the pair $BCF, ADF$ to $\quinv(\tau_j(\sigma))$. Thus we conclude that the only triple whose contribution to $\quinv(\sigma)$ is different from its contribution to $\quinv(\tau_j(\sigma))$ is the degenerate triple $(x,y)$ at the top of the columns $j,j+1$. If $\sigma(x)=\sigma(y)$, $\tau_j(\sigma)=\sigma$. If $\sigma(x)\neq \sigma(y)$ and if $(x,y)$ was a degenerate $\quinv$ triple before the swap, swapping them will cause $\tau_i(\sigma)$ to lose one $\quinv$ triple, and gain one otherwise.
\end{proof}
\begin{remark}\label{rem:PDS}
We will be working with permutations, and corresponding reduced expressions of those permutations to products of the operators $\tau_j$. Unfortunately, the operators do not satisfy braid relations, i.e. in general $\tau_j\tau_{j+1}\tau_j(\sigma) \neq \tau_{j+1}\tau_j\allowbreak \tau_{j+1}(\sigma)$. Thus we will need to choose a canonical reduced expression for a given permutation to make our construction well-defined. One way to do this uses the \emph{positive distinguished subexpression}, or PDS, of a reduced expression, defined in \cite{MarshRietsch}. We denote by $\PDS(\pi)$ the PDS corresponding to a permutation $\pi$. We will refer the reader to \cite[Section 3]{compactformula} for a detailed treatment of this notion; in our setting it suffices to assert that $\PDS(\pi)$ is a unique reduced expression for any permutation $\pi$.
\end{remark}
\section{From a bijection on words to a bijection on
super fillings}
\label{sec:bijPQT}
In this section, we prove Proposition~\ref{thm:Wbijection}, which states the existence of a bijection on the set of $\Phi$-degenerate fillings $\widetilde{\PQT}$, using the super-alphabet $\mathcal{A}$ and total ordering $<_2$ on that alphabet. Proposition~\ref{thm:Wbijection} is needed in the proof of Theorem~\ref{prop:Phi-degenerate}.
We begin with some definitions. Let $n \geq 3$ be a positive integer and let $\beta = (\beta_2,\dots, \beta_{n-1})$ be a tuple of positive integers that will represent the fixed content in the letters $\{2,\ldots,n-1\}$ of the words we consider.
Denote $|\beta| = \sum_i \beta_i$ and let $N > |\beta|$ be a positive integer. For convenience, let $L = N - |\beta|$.
We will consider words of length $N$ in the alphabet $[n] = \{1,\dots, n\}$ as follows. For $0 \leq k \leq L$, define
\begin{equation}
\label{wk-def}
W_k \equiv W^{(\beta,L)}_k = \left\{ w \in [n]^N
\left|
\begin{matrix}
c_n(w) = k, c_1(w) = L-k, \\
c_i(w) = \beta_i \text{ for } 2 \leq i \leq n-1
\end{matrix}
\right. \right\},
\end{equation}
where $c_i(w)$ counts the number of times the letter $i$ appears in $w$. We say such words are of type $\beta$.
\begin{example}
\label{eg:Wk}
For example, with $n=3, N = 4, L = 3$ and $\beta = (1)$, we have
\begin{equation}
\label{Wk-eg}
\begin{split}
W_0 &= \{1112, 1121, 1211, 2111\}, \\
W_1 &= \{1132, 1123, 1213, 2113,1312, 1321, 1231, 2131,3112, 3121, 3211, 2311\}, \\
W_2 &= \{1332, 1323, 1233, 2133,3132, 3123, 3213, 2313,3312, 3321, 3231, 2331\},\\
W_3 &= \{3332, 3323, 3233, 2333\}.
\end{split}
\end{equation}
\end{example}
Recall that a {\em coinversion} of a word $w$ is a pair $(i,j)$, $i < j$, such that $w_i < w_j$, and $\coinv(w)$ is the number of such pairs. In our applications, we define the map $\rho_{\alpha}$ that maps a word $v$ with content $\alpha$ in the super-alphabet $\mathcal{A}=\{1,2,\ldots,\bar{2},\bar{1}\}$ under total ordering $<_2$ to a word $w$ in the letters $\{1,2,\ldots,n\}$, such that the order of the letters is preserved under the mapping, and such that the smallest letter $a=\min |\alpha|$ is mapped to 1 and $\bar{a}$ is mapped to $n$, as follows. Set $\ell=\max |\alpha|-\min |\alpha|+2$, and let $n=2(\ell-1)$. The map $\rho_{\alpha}$ will send the unbarred letters in $v$ to the letters $1,\ldots,\ell-1$ and the barred letters to $\ell, \dots, n$ by mapping $j \in v$ as follows:
\[
\rho_{\alpha}(j)= \begin{cases}
j-a + 1 & \mbox{if}\ j\in\mathbb{Z}_+,\\
n-|j|+a & \mbox{if}\ j\in\mathbb{Z}_-.
\end{cases}
\]
Now, for a fixed $1 \leq \ell \leq n$, let $w\in W_k$ (i.e.~$c_n(w)=k$). Define
\begin{equation}
\label{word-quinv-def}
\quinvL(w) = \coinv(w) + \binom{c_\ell(w)}{2} + \cdots + \binom{c_{n-1}(w)}{2} +\binom{k}{2}.
\end{equation}
Thus if the preimage $v=\rho_{\alpha}^{-1}(w)$ in $\mathcal{A}$ corresponds to the degenerate word of some $\sigma \in \widetilde{\PQT}$, $\quinvL(w)$ is equal to the contribution from the degenerate word $v$ to $\quinv(\sigma)$ (under the ordering $<_2$). We may slightly abuse notation by writing $\quinvL(v)\coloneqq
\quinvL(\rho_{\alpha}(v))$ to describe this contribution to $\quinv(\sigma)$.
\begin{example}
Let $v=(\bar{5},\bar{2},3,3,\bar{3},\bar{2},3,5,\bar{3})$ be a word in the alphabet $\mathcal{S}=\{3,5,\bar5,\bar3,\bar2\}$ with content $\alpha=\{\bar{2},\bar{2},3,3,3,\bar{3},\bar{3},5,\bar{5}\}$. Then $a=2$, $\ell=5-2+2=5$, $n=8$, and $w=\rho_{\alpha}(v)=(5,8,2,2,7,8,2,4,7)$ with $k=2, c_5(w)=1, c_6(w)=2$, and $c_7(w)=0$. Indeed, $\quinv(v)=18$ according to $<_2$, $\coinv(w)=16$, and $\quinvL(w)=16+{1 \choose 2}+{2 \choose 2}+{0 \choose 2}+{2 \choose 2}=18$.
\end{example}
The proof of Proposition~\ref{thm:Wbijection} relies on the existence of a $\quinvL$-preserving bijection $\phi$ on $W \equiv \cup_{k=0}^L W_k$ that satisfies the desired properties. This result is stated below, but first we need some additional notation. For $0 \leq k \leq L$, partition $W_k$ from \eqref{wk-def} into two subsets as follows. For $w \in W_k$, let $p_n(w)$ be the position of the leftmost $n$ in $w$ from the left and $p_1(w)$ be the position of the rightmost $1$ from the right, ignoring all $n$'s. If $c_j(w)=0$, set $p_j(w)=\infty$ for $j=1,n$. Then define
\begin{align*}
W_k^{\leq}& = \{w\in W_k\ |\ p_n(w) \leq p_1(w)\} \\
W_k^>&= \{w\in W_k\ |\ p_n(w) > p_1(w)\}
\end{align*}
Note that by our convention, $W_0^>=W_0$ and $W_L^\leq=W_L$, with $W_0^\leq=W_L^>=\emptyset$.
\begin{example}\label{Wkleq-eg}
Let $n=3, N=4, L=3$, and $\beta=(1)$. Then
\begin{align*}
W_0^>&=\{11{\bf1}2,112{\bf1},121{\bf1},211{\bf1}\},\\
W_1^\leq&=\{1{\bf31}2,{\bf3}1{\bf1}2,{\bf3}12{\bf1}, {\bf3}21{\bf1}\},\\
W_1^>&=\{1{\bf13}2, 1{\bf1}2{\bf3},12{\bf13},21{\bf13}, 1{\bf3}2{\bf1},12{\bf31},21{\bf31},2{\bf3}1{\bf1} \}, \\
W_2^\leq &= \{{\bf1 3} 2 3 , {\bf1 3} 3 2 , {\bf3 1} 2 3, {\bf3 1} 3 2 , {\bf3} 2 {\bf1} 3 , {\bf3} 2 3 {\bf1} , {\bf3} 3 {\bf1} 2 , {\bf3} 3 2 {\bf1} \}, \\
W_2^> &= \{{\bf1}2{\bf3}3, 2{\bf13}3, 2{\bf31}3, 2{\bf3}3{\bf1}\},\\
W_3^\leq &= \{{\bf3}332,{\bf3}323,{\bf3}233,2{\bf3}33\}.
\end{align*}
The leftmost $n$ and the rightmost $1$ are marked in bold. For instance, the pairs $(p_n(w),p_1(w))$ for the words in $W_2^>$ are $\{(3,2), (3,2), (2,1), (2,1)\}$, and for the words in $W_2^\leq$, they are $\{(2,2), (2,2),(1,2),(1,2),(1,1),(1,1),(1,2),(1,1)\}$.
\end{example}
Let $W^\leq = \bigcup_k W_k^\leq$ and $W^> = \bigcup_k W_k^>$.
\begin{theorem}\label{thm:invol}
There exists a bijection $\phi : W^> \to W^\leq$ satisfying the
following~conditions:
\begin{enumerate}[label=(\arabic*)]
\item \label{it1vvt1} $\phi$ maps $W_k^>$ to $W_{k+1}^\leq$ bijectively for $0 \leq k \leq L-1$.
\item \label{it1vvt2} $\quinvL(w) = \quinvL(\phi(w))$ for all $w \in W^>$, and
\item \label{it1vvt3} The subword of $w$ in the letters $2, \dots, n-1$ is preserved by
$\phi$.
\end{enumerate}
\end{theorem}
Section~\ref{sec:bijwords} is devoted to the proof of Theorem~\ref{thm:invol}, as it is quite technical.
Now, fix the content $\alpha$, which determines $n,\ell,L$. For $0\leq k\leq L$, define the sets
\[
V_k = \{\rho_\alpha^{-1}(w):w\in W_k\},\quad V_k^{\leq} = \{\rho_\alpha^{-1}(w):w\in W_k^{\leq}\},\quad V_k^{>} = \{\rho_\alpha^{-1}(w):w\in W_k^{>}\},
\]
\[
V^\leq = \bigcup_k V_k^\leq,\quad V^> = \bigcup_k V_k^>.
\]
Define also the bijection $\phi_\alpha:V^>\rightarrow V^{\leq}$ by $\phi_\alpha(v) \coloneqq \rho_\alpha^{-1}\circ\phi\circ\rho_\alpha(v)$. It is immediate that $\phi_\alpha$ satisfies the conditions~\ref{it1vvt1}--\ref{it1vvt2} from Theorem~\ref{thm:invol} when all $W$'s replaced by $V$'s. It also satisfies the analog of condition~\ref{it1vvt3}: the subword of $v$ in the letters between $a+1$ and $\overline{a+1}$ is preserved (where $a=\min|\alpha|$).
Let $\sigma$ be a $\Phi$-degenerate filling with degenerate word $v$ that contains the distinguished label. We split the set of $\Phi$-degenerate fillings into two disjoint sets that are denoted by $\mathcal{V}^>$ and $\mathcal{V}^{\leq}$:
\begin{itemize}
\item $\sigma\in \mathcal{V}^{>}$ if $v\in V^{>}$, and
\item $\sigma\in \mathcal{V}^{\leq}$ if $v\in V^{\leq}$.
\end{itemize}
\begin{example}\label{ex:u_choice}
Consider the set of words
\[
\{\bar1\bar12,\bar12\bar1, 2\bar1\bar1,\bar112,\bar121, 2\bar11,1\bar12,12\bar1,21\bar1,112,121,211\}.
\]
Here $V^{>}=\{211,121,2\bar11,112,21\bar1, \allowbreak 12\bar1\}$ with $\quinvL$-generating functions $1+q+q^2$ and $q+q^2+q^3$ for $V^{>}_0, V^{\leq}_1$ and for $V^{>}_1,V^{\leq}_2$, respectively. The bijection $\phi_\alpha$ from Theorem~\ref{thm:invol} is unique on the corresponding sets $W^{>}$ and $W^{\leq}$, thus the elements of $V^{>}$ are mapped to the elements of $V^{\leq}$ as follows:
\begin{align*}
\phi_\alpha(211) &= \bar121,\ \quinvL=0, \quad \phi_\alpha(121) = \bar112,\ \quinvL=1,\\
\phi_\alpha(2\bar11) &= \bar1\bar12,\ \quinvL=1, \quad \phi_\alpha(112)=1\bar12,\ \quinvL=2, \\
\quad \phi_\alpha(21\bar1) &=\bar12\bar1,\ \quinvL=2, \quad \phi_\alpha(12\bar1)=2\bar1\bar1,\ \quinvL=3.
\end{align*}
Consider the fillings $\sigma_1, \sigma_2$ shown below. In both, the degenerate word contains the distinguished label:
\[
\sigma_1=\raisebox{0.2in}{\tableau{2\\1&2&1&\bar1\\3&1&\bar2&\bar1\\2&1&2&1&\bar2}},
\quad
\sigma_2=\raisebox{0.2in}{\tableau{2\\1&1&\bar1&2\\3&1&\bar2&\bar1\\2&1&2&1&\bar2}}
\]
The filling $\sigma_1$ belongs to $\mathcal{V}^{>}$ since $21\bar1\in V^{>}$, and the filling $\sigma_2$ belongs to $\mathcal{V}^{\leq}$ since $1\bar12\in V^{\leq}$.
\end{example}
We are now ready to state the main result of this section.
\begin{proposition}\label{thm:Wbijection}
There exists a map $\Theta$ on the subset $\mathcal{V}^{>}$ of $\Phi$-degenerate fillings $\sigma$ such that
\begin{enumerate}[label=(\arabic*)]
\item \label{it2vvt1}$\Theta\ :\ \mathcal{V}^{>} \longrightarrow \mathcal{V}^{\leq}$ is a bijection,
\item \label{it2vvt2} $\maj(\Theta(\sigma)) = \maj(\sigma) +1$,
\item \label{it2vvt3} $\quinv(\Theta(\sigma))=\quinv(\sigma)$,
\item \label{it2vvt4} $x^{|\Theta(\sigma)|}=x^{|\sigma|}$ and $p(\Theta(\sigma))=p(\sigma)-1$.
\end{enumerate}
\end{proposition}
\begin{proof}
Let $\sigma$ be a $\Phi$-degenerate filling of $\dg(\lambda)$ with distinguished label $a$ contained in degenerate word $v$. We will define the map $\Theta$ acting on $\sigma\in \mathcal{V}^{>}$ with the following steps.
\begin{enumerate}[label=(\roman*)]
\item\label{item:u} Set $u(\sigma)$ to be the first cell in reading order with content $a$ in $\sigma$ (the distinguished cell).
\item Let $\widehat{v}$ be the degenerate word of $\Upsilon(\sigma)=\Phi_{u(\sigma)}(\sigma)$. ($\widehat{v}$ is equal to $v$, except that the rightmost $a$ is flipped to become $\bar{a}$.)
\item Let $\alpha$ be the content of $v$.
Let $\pi$ be a canonical permutation acting on positions to get from $\widehat{v}$ to $\phi_{\alpha}(v)$, i.e. $\widehat{v}\cdot \pi=\phi_\alpha(v)$. (For instance, $\pi$ could be canonically defined as follows: begin by fixing the leftmost entry of $\phi_\alpha(v)$ that doesn't match $\widehat{v}$ by a series of adjacent transpositions. Continue in this way until no mismatched entries remain.) Let $\PDS(\pi)=s_{i_1}s_{i_2}\ldots s_{i_m}$ be the unique reduced word which is the PDS corresponding to $\pi$ (see Remark~\ref{rem:PDS}).
\item Let $d$ be the number of columns in $\dg(\lambda)$ strictly to the left of the degenerate segment. Set
$\Theta = \Upsilon\circ \tau_{i_1+d}\circ \tau_{i_2+d} \circ\cdots \circ \tau_{i_m+d}$ acting on the right, so that
\[
\Theta(\sigma) = \tau_{i_m+d}\circ \cdots\circ\tau_{i_2+d}\circ \tau_{i_1+d}\circ\Upsilon(\sigma).
\]
\end{enumerate}
We prove that $\Theta$ satisfies the properties \ref{it2vvt1}--\ref{it2vvt4} in order.
\emph{$\Theta$ satisfies \ref{it2vvt1}:} The degenerate word of $\Theta(\sigma)$ is by construction equal to $\phi_\alpha(v)$. Thus $v\in V^{>}$ implies that $\Theta(\sigma)\in \mathcal{V}^{\leq}$. Next we note that the cell $u(\sigma)$ that is flipped in $\sigma\in \mathcal{V}^{>}$ by $\Upsilon$ is determined uniquely, as is the product $ \tau_{i_m+d}\circ\cdots\circ \tau_{i_2+d} \circ \tau_{i_1+d}$ of operators applied to $\Upsilon(\sigma)$. The operators $\tau_j$ are involutions, so this construction is well-defined. Thus we obtain the reverse map $\Theta^{-1}$ from $\mathcal{V}^{\leq}$ to $\mathcal{V}^{>}$ by simply reversing the operators in the map $\Theta$.
Let $\sigma\in \mathcal{V}^{\leq}$ be a filling of $\dg(\lambda)$, and again let $a$ be the distinguished label and $v\in V^{\leq}$ the degenerate word of $\sigma$. Call $\widehat{u}(\sigma)$ the cell in the diagram $\dg(\lambda)$ that contains the rightmost entry $a$ in $\phi^{-1}_\alpha(v)\in V^>$. Call $v^{\ast}$ the word obtained by flipping the rightmost $a$ in $\phi(v)$ to become $\bar{a}$. Then define the permutation $\pi$ such that $v^{\ast}\pi=v$, and let $\PDS(\pi)=s_{i_1}s_{i_2}\ldots s_{i_m}$. The reader may now check that the map from $\sigma\in W^{\leq}$ to $\Theta^{-1}(\sigma)$ given by
\[
\Theta^{-1}(\sigma) = \Phi_{\widehat{u}(\sigma)}\circ \tau_{i_1+d}\circ \tau_{i_2+d} \circ\cdots\circ \tau_{i_m+d}(\sigma),
\]
is the inverse of the map from $\sigma\in \mathcal{V}^{>}$ to $\mathcal{V}^{\leq}$, since $\widehat{u}(\sigma)$ is by design precisely equal to $u(\Theta^{-1}(\sigma))$ as defined in ~\ref{item:u}. Hence the construction of $\Theta$ is an injection.
Finally, by Theorem~\ref{thm:invol} we have that for each $k\geq 0$, the number of elements in $\mathcal{V}^{>}$ with exactly $k$ $\bar{a}$'s, the desired content, and a fixed number of $\quinv$ triples equals the number of elements in $\mathcal{V}^{\leq}$ with exactly $k+1$ $\bar{a}$'s, from which we conclude that $\Theta$ is also a surjection.
\emph{$\Theta$ satisfies \ref{it2vvt2}:}
By Lemma~\ref{lem:Phi_u:maj}, when $u$ is any cell in the degenerate segment of $\sigma$ with $\sigma(u)=a$, the operator $\Phi_u$ acts on $\sigma$ by increasing the $\maj$ by exactly one. By Lemma~\ref{lem:tau:maj} the operators $\tau_j$ leave the $\maj$ unchanged.
\emph{$\Theta$ satisfies \ref{it2vvt3}:} By Lemma~\ref{lem:Phi_u:quinv}, when $u$ is any cell in the degenerate segment of $\sigma$ with $|\sigma(u)|=a$, all triples with the exception of degenerate triples containing $u$ contribute to $\quinv(\sigma)$ if and only if they also contribute to $\quinv(\Phi_u(\sigma))$. Observe that the degenerate triples containing $u$ are precisely the elements in $v$, the degenerate word of $\sigma$. For $\ell$ defined as in ~\ref{sec:bijPQT} for the degenerate word $v$, the contribution to $\quinv(\sigma)$ from the degenerate triples in $v$ is precisely equal to $\quinvL(v)$ of \eqref{word-quinv-def}. Therefore let us write $\quinv(\sigma)=\quinv'(\sigma)+\quinvL(v)$, where $\quinv'(\sigma)$ is the contribution to $\quinv(\sigma)$ coming from all triples but the degenerate triples in $v$. We have $\quinv'(\Phi_u(\sigma))=\quinv'(\sigma)$ by Lemma~\ref{lem:Phi_u:quinv}, and $\quinv'(\tau_j(\sigma))=\quinv'(\sigma)$ by Lemma~\ref{lem:tau:quinv} for all $\tau_j$ acting on columns that contain the degenerate word $v$. By construction we have that $\Theta(\sigma)$ has degenerate word $\phi_\alpha(v)$, and since $\quinvL(v)=\quinvL(\phi_\alpha(v))$ by Theorem~\ref{thm:invol}, we have thus $\quinv(\Theta(\sigma))=\quinv(\sigma)$, which completes our argument.
\emph{$\Theta$ satisfies \ref{it2vvt4}:} The content of $\sigma$ is changed only by the operator $\Upsilon$, which flips exactly one entry from positive to negative, thus reducing $p(\sigma)$ by 1, while leaving the absolute value of the content of $\sigma$ unchanged.
\end{proof}
Theorem~\ref{prop:Phi-degenerate} follows as an immediate corollary.
\begin{example}
Consider
$\sigma_1$
from Example~\ref{ex:u_choice}. The degenerate word is $v=21\bar1\in V^{>}$, $\widehat{v}=2\bar1\bar1$, and $\phi_\alpha(v)=\bar12\bar1$ (here $\alpha=(1)$). Thus the permutation to get from $\widehat{v}$ to $\phi_\alpha(v)$ is $\pi=213$, which has corresponding PDS $\pi=s_1$, and $u(\sigma_1)$ is the first cell with content 1 in reading order. Thus we apply $\tau_1$ to $\sigma_1$ after flipping the 1 to obtain
\[
\Theta(\sigma_1)=\tau_1\circ\Phi_{u(\sigma_1)}(\sigma_1)=\raisebox{0.2in}{\tableau{2\\1&\bar1&2&\bar1\\3&\bar2&1&\bar1\\2&1&2&1&\bar2}}\in \mathcal{V}^{\leq}.
\]
For another example, consider
\[
\sigma_3=
\raisebox{0.2in}{\tableau{2\\1&1&2&1\\3&\bar2&1&\bar1\\2&1&2&1&\bar2}}
\in \mathcal{V}^{>}.
\]
Then $v= 121, \widehat{v}=12\bar{1}$, and $\phi_\alpha(v) = 12\bar{1}$, so that the right-acting permutation $\pi$ such that $\widehat{v}\pi=\phi(v)$ is $\pi=s_2 s_1$. The number of columns left of the degenerate segment is $\ell=1$, and the cell $u(\sigma_3)=(3,4)$. Thus $\Theta(\sigma_2)=\tau_2\circ\tau_3\circ\Phi_{(3,4)}(\sigma_3)$:
\[
\sigma_3 \xrightarrow{\Phi_{(3,4)}}
\raisebox{0.2in}{\tableau{2\\1&1&2&\bar1\\3&\bar2&1&\bar1\\2&1&2&1&\bar2}} \xrightarrow{\tau_3}
\raisebox{0.2in}{\tableau{2\\1&1&\bar1&2\\3&\bar2&\bar1&1\\2&1&2&1&\bar2}} \xrightarrow{\tau_2}
\raisebox{0.2in}{\tableau{2\\1&\bar1&1&2\\3&\bar2&\bar1&1\\2&1&2&1&\bar2}} = \Theta(\sigma_3)\in \mathcal{V}^{\leq}.
\]
\end{example}
%
%
%
%
%
\section{A bijection on words}
\label{sec:bijwords}
The aim of this section is to prove Theorem~\ref{thm:invol}: the existence of the bijection $\phi: W^>\rightarrow W^\leq$ for words in the alphabet $[n]$ with a fixed type $\beta=(\beta_2,\ldots,\beta_{n-1})$ in the letters $\{2,\ldots,n-1\}$. We denote the set of such words by $W=W(\beta)$.
For the proofs, we ask the reader to refer to the terminology and basic results on $q$-series from ~\ref{sec:qseries}.
First off, for such a map $\phi$ to exist, the $\quinvL$ generating function of $W$ must be a multiple
of 2. To see this, recall the definition of $\quinvL$ from \eqref{word-quinv-def} and write
\begin{align*}
\sum_{w \in W} q^{\quinvL(w)} =& \sum_{k=0}^L \sum_{w \in W_k} q^{\coinv(w) + \binom{\beta_\ell}{2} + \cdots + \binom{\beta_{n-1}}{2} + \binom{k}{2} } \\
=&\ q^{\binom{\beta_\ell}{2} + \cdots + \binom{\beta_{n-1}}{2}} \sum_{k=0}^L q^{\binom{k}{2}} \qbinom N{L-k,\beta_2,\dots, \beta_{n-1},k} \\
=&\ q^{\binom{\beta_\ell}{2} + \cdots + \binom{\beta_{n-1}}{2}} \qbinom{N}{L,\beta_2,\dots,\beta_{n-1}} \sum_{k=0}^L q^{\binom{k}{2}} \qbinom Lk,
\end{align*}
where we have used Proposition~\ref{prop:coinv-gf} in the second line. Recall that $W$ is the set of words of length $N$ in the alphabet $\{1,\ldots,n\}$ with content given by $(L-k,\beta_2,\ldots,\beta_{n-1},k)$ for $L=N-(\beta_2+\cdots+\beta_{n-1})$ and $k$ ranging from $0$ to $L$. For $j=2,\ldots,n-1$, the content in the letters $j$ is fixed to be $c_j(w)=\beta_j$ for all words $w\in W$, while $k$ corresponds to the number of letters $n$ in words of $W$. Now, the sum is a direct application
of the $q$-binomial theorem, Proposition~\ref{prop:qbinom}, with $x=1$ and we find that
\[
\sum_{w \in W} q^{\quinvL(w)} = 2 q^{\binom{\beta_\ell}{2} + \cdots + \binom{\beta_{n-1}}{2}} \qbinom{N}{L,\beta_2,\dots,\beta_{n-1}} \prod_{j=1}^{L-1} (1 + q^j),
\]
as claimed.
The following theorem shall lead us to a proof of Theorem~\ref{thm:invol}.
\begin{theorem}
\label{thm:quinv-gf}
For $n, N, L, \beta$, $k$ and $\ell$ as above, we have the identities
\begin{align}
\label{quinv-gf>}
\sum_{w \in W_k^>} q^{\quinvL(w)} =& q^{\binom{\beta_\ell}{2} + \cdots + \binom{\beta_{n-1}}{2} + \binom{k+1}{2}} \qbinom{N}{L,\beta_2,\dots,\beta_{n-1}} \qbinom{L-1}{k}, \\
\label{quinv-gf<}
\sum_{w \in W_k^\leq} q^{\quinvL(w)} =& q^{\binom{\beta_\ell}{2} + \cdots + \binom{\beta_{n-1}}{2} + \binom{k}{2}} \qbinom{N}{L,\beta_2,\dots,\beta_{n-1}} \qbinom{L-1}{k-1}.
\end{align}
\end{theorem}
\begin{proof}[Proof of Theorem~\ref{thm:invol}]
As a consequence of Theorem~\ref{thm:quinv-gf}, the $\quinvL$ generating functions of $W_{k}^>$ and $W_{k+1}^\leq$ are equal, thus proving~\ref{it1vvt2}.
To show~\ref{it1vvt3}, let $w = (w_1, \dots, w_N) \in W_k$ and let $w'$ be obtained from $w$ by replacing all
occurrences of the letters from $2$ through $n-1$ by $2$, and all the occurrences of $n$ by $3$.
Let $w''$ be the subword of $w$ consisting of the letters from $2$ through $n-1$ in $w$.
All we need to do is to show that the $\quinvL$ generating functions for every fixed $w''$ are the same.
First of all, it is clear that $\coinv(w) = \coinv(w') + \coinv(w'')$. Moreover, the question of whether $w$
belongs to $W_k^\leq$ or $W_k^>$ is the same as that of $w'$.
Thus we may refine the sums appearing in \eqref{quinv-gf>} and \eqref{quinv-gf<} by fixing the subword $w''$ and summing \eqref{word-quinv-def} over all words $w'$ of length $N$ with $N-L$ $2$'s.
This is implicitly done in the proof of Theorem~\ref{thm:quinv-gf}. Comparing \eqref{quinv-gf>} for $k$ and \eqref{quinv-gf<} for $k+1$, we see that the common $q$-multinomial coefficient corresponds to the sum of $\coinv(w'')$ over all possible subwords $w''$. The other factors match for $W_k^>$ and $W_{k+1}^\leq$, completing the proof of~\ref{it1vvt3}.
\end{proof}
\begin{example}
\label{eg:W<>}
As an illustration of Theorem~\ref{thm:quinv-gf}, consider Example~\ref{Wkleq-eg} and let $L= 3$. We compute the $\quinvL$ generating functions with $\ell = 2$ for the sets $W_k^>$ and $W_k^\leq$ for $0\leq k\leq 3$, denoted by $g_k^>(q)$ and $g_k^\leq(q)$, respectively:
\begin{equation*}
\begin{split}
g_0^>(q)=g_1^\leq(q)&=1+q+q^2+q^3,\\
g_1^>(q)=g_2^\leq(q)&=q+2q^2+2q^3+2q^4+q^5,\\
g_2^>(q)=g_3^\leq(q)&=q^3+q^4+q^5+q^6.
\end{split}
\end{equation*}
The bijection $\phi$ claimed in Theorem~\ref{thm:invol} is unique for the sets $W_0^> \xrightarrow{\phi} W_1^\leq$ (namely, the words $1112,1121,1211,2111$ are mapped to $1312,3112,3121,3211$, respectively) and $W_2^> \xrightarrow{\phi} W_3^\leq$ (the words $1233,2133,2313,2331$ are mapped to\linebreak $2333,3233,3323,3332$, respectively). However the bijection $\phi$ is not unique for the sets $W_1^> \xrightarrow{\phi} W_2^\leq$.
\end{example}
Although Theorem~\ref{thm:quinv-gf} looks as if it should be known, we have not seen this in the literature before. The result follows from a more general result, which we now state.
For $w \in W_k$, recall the definition of $p_1(w)$ and $p_n(w)$ given before Example~\ref{Wkleq-eg}.
\begin{theorem}
\label{thm:quinv-refined-gf}
Let $N, L, \beta$, $k$, and $\ell$ be as above.
Then
\begin{align}
\label{quinv-ref-gf>}
\sum_{\substack{ w \in W_k^> \\ p_1(w) = i}} q^{\quinvL(w)} &= q^{\binom{\beta_\ell}{2} + \cdots + \binom{\beta_{n-1}}{2} + \binom{k+1}{2} + (i-1)L}
\qbinom{N-L}{\beta_2,\dots,\beta_{n-1}} \cr
& \times \qbinom{N-i}{k, N-L-i+1, L-k-1}, \\
\label{quinv-ref-gf<}
\sum_{\substack{ w \in W_k^\leq \\ p_n(w) = j}} q^{\quinvL(w)} &= q^{\binom{\beta_\ell}{2} + \cdots + \binom{\beta_{n-1}}{2}+\binom{k}{2} + (j-1) L}
\qbinom{N-L}{\beta_2,\dots,\beta_{n-1}} \cr
& \times \qbinom{N-j}{k-1, N-L-j+1, L-k}.
\end{align} \end{theorem}
These two identities imply the corresponding ones in Theorem~\ref{thm:quinv-gf}.
\begin{proof}[Proof of Theorem~\ref{thm:quinv-gf}]
We use the telescoping sum identity in Proposition~\ref{prop:q-telescope} to sum \eqref{quinv-ref-gf>} and \eqref{quinv-ref-gf<} over $i$ and $j$, to obtain the identities \eqref{quinv-gf>} and \eqref{quinv-gf<}, respectively.
\end{proof}
We prove Theorem~\ref{thm:quinv-refined-gf} using the lemmas stated after, which are in turn
proved in the following two subsections.
\begin{proof}[Proof of Theorem~\ref{thm:quinv-refined-gf}]
It will suffice to prove Theorem~\ref{thm:quinv-refined-gf} for the case of $n=3$.
To see this, let $w = (w_1, \dots, w_N) \in W_k$ and $w'$ be obtained from $w$ by replacing all
occurrences of the letters from $2$ through $n-1$ by $2$, and all the occurrences of $n$ by $3$.
Let $w''$ be the subword of $w$ consisting of the letters from $2$ through $n-1$ in $w$.
Then, it is clear that $\coinv(w) = \coinv(w') + \coinv(w'')$. Moreover, the question of whether $w$
belongs to $W_k^\leq$ or $W_k^>$ is the same as that of $w'$.
We can then rewrite \eqref{quinv-ref-gf>} as
\[
\sum_{\substack{ w \in W_k^> \\ p_1(w) = i}} q^{\quinvL(w)} =
q^{\binom{\beta_\ell}{2} + \cdots + \binom{\beta_{n-1}}{2} + \binom{k}{2}}
\qbinom{N-L}{\beta_2,\dots,\beta_{n-1}}
\sum_{w'\in V_k^{>}} q^{\coinv(w')},
\]
where $V_k^> = \big(W^{(N-L),L}_k \big)^>$.
We now restrict to the case $n=3$ and $\beta = (\beta_2) = (N-L)$.
We first consider \eqref{quinv-ref-gf>}.
By summing the two cases in Lemma~\ref{lem:quinv>}, we see that, for $1 \leq i \leq N +1 - L$,
\begin{align*}
\sum_{\substack{ w \in W_k^> \\ p_1(w) = i}} q^{\coinv(w)}
=&\ q^{ iL + k - L} \qbinom{N-k-i}{L-k-1} \qbinom {N-i}k \
=&\ q^{ iL + k - L} \qbinom{L-1}k \qbinom{N-i}{L-1}.
\end{align*}
Now, the $i$-sum can be performed using Proposition~\ref{prop:q-telescope}. We then simplify to obtain the result.
For \eqref{quinv-ref-gf<}, the argument is similar. Summing the two cases in Lemma~\ref{lem:quinv<}
shows that, for $1 \leq j \leq N +1 - L$,
\begin{align*}
\sum_{\substack{ w \in W_k^\leq \\ p_3(w) = j}} q^{\coinv(w)}
=&\ q^{ j L - L} \qbinom{N-j-k+1}{L-k} \qbinom {N-j}{k-1} \\
=&\ q^{ j L - L} \qbinom{L-1}{k-1} \qbinom{N-j}{L-1}.
\end{align*}
Again, the $j$-sum can be performed using Proposition~\ref{prop:q-telescope}. This completes the proof.
\end{proof}
Let $p'_1(w)$ be the actual position of the rightmost $1$ from the right in $w$.
For both lemmas, there are two subcases to consider, either $p_3(w) < N+1 - p'_1(w)$ or not.
That is to say, the leftmost 3 is to the left of the rightmost 1 or not. The first lemma is for the subset $W_k^>$.
\begin{lemma}
\label{lem:quinv>}
Let $n=3$, and $N, L$ and $k$ be as above. Then,
\begin{enumerate}[label=(\arabic*)]
\item \label{it3vvt1}
The generating function
\begin{multline}
\label{quinv>l}
\sum_{\substack{ w \in W_k^> \\ p_1(w) = i \\ p_3(w) < N+1 - p'_1(w)}} q^{\coinv(w)} = q^{ (i-1) L}
\qbinom{N-k-i}{L-k-1} \Bigg( \qbinom {N-i}k q^k \\
- \qbinom{k+i-1}k q^{k(N-k-2i+2)} \Bigg),
\end{multline}
if $1 \leq i \leq \lfloor (N-k)/2 \rfloor$, and zero otherwise.
\item \label{it3vvt2}
For $1 \leq i \leq \lfloor (N-k)/2 \rfloor$,
\begin{equation}
\label{quinv>r1}
\sum_{\substack{ w \in W_k^> \\ p_1(w) = i \\ p_3(w) > N+1 - p'_1(w)}} q^{\coinv(w)} = q^{ (i-1) L + k(N-k-2i+2)}
\qbinom{N-k-i}{L-k-1} \qbinom{k+i-1}k,
\end{equation}
and for $\lfloor (N-k)/2 \rfloor + 1 \leq i \leq N +1 - L$,
\begin{equation}
\label{quinv>r2}
\sum_{\substack{ w \in W_k^> \\ p_1(w) = i \\ p_3(w) > N+1 - p'_1(w)}} q^{\coinv(w)} = q^{ (i-1) L + k}
\qbinom{N-k-i}{L-k-1} \qbinom {N-i}k.
\end{equation}
\end{enumerate}
\end{lemma}
The second lemma is for the subset $W_k^\leq$.
\begin{lemma}
\label{lem:quinv<}
Let $n=3$, and $N, L$ and $k$ be as above. Then,
\begin{enumerate}[label=(\arabic*)]
\item \label{it4vvt1}
The generating function
\begin{multline}
\label{quinv N+1 - p'_1(w)}} q^{\coinv(w)} = q^{(j-1)k + (L-k)(N-j-k+1)}
\qbinom{j-1}{L-k} \qbinom{N-j}{k-1},
\end{equation}
and for $\lfloor (N-k+1)/2 \rfloor + 1 \leq j \leq N +1 - L$,
\begin{equation}
\label{quinv N+1 - p'_1(w)}} q^{\coinv(w)} = q^{ (j-1) L}
\qbinom{N-j-k+1}{L-k} \qbinom {N-j}{k-1}.
\end{equation}
\end{enumerate}
\end{lemma}
\subsection{Proof of Lemma~\ref{lem:quinv>}}
We will now prove the cases of Lemma~\ref{lem:quinv>} separately.
We first begin with the second case, since it is conceptually simpler.
\begin{proof}[Proof of Lemma~\ref{lem:quinv>}\ref{it3vvt2}]
Suppose $w \in W_k^>$ with $p_3(w) = j$, $p'_1(w) = i+k$ and $j > N+1 - i-k$. Then
$w$ can be written as $w = w_1 1 2^{i+j+k-N-2} 3 w_2$, where $w_1 \in \{1,2\}^{N-i-k}$ with $L-k-1$ $1$'s
and $w_2 \in \{2,3\}^{N-j}$ with $k-1$ $3$'s. As a result, $p_1(w) = i$ and
\[
\coinv(w) = \coinv(w_1) + \coinv(w_2) + (L-k)(i+k-1) + k(j+k-L-1).
\]
Notice that the smallest (resp. largest) possible value of $i$ is $1$ (resp. $N-L+1$), namely when
$j = N+2-i-k$ and $w_2$ contains no $2$'s
(resp. $w_1$ contains no $2$'s).
Let the left hand sides of \eqref{quinv>r1} and \eqref{quinv>r2} be denoted $f(i)$.
By definition of $W_k^>$, we have $j > i$. Therefore,
\[
f(i) = \sum_{j=(i+1) \vee (N-i-k+2)}^{N-k+1} \sum_{w_1,w_2} q^{\coinv(w_1) + \coinv(w_2) + (L-k)(i+k-1) + k(j+k-L-1)},
\]
where $a \vee b$ denotes the maximum of $a$ and $b$,
and the sums over words $w_1$ and $w_2$ are performed using Proposition~\ref{prop:coinv-gf} to give
\[
f(i) = \sum_{j=(i+1) \vee (N-i-k+2)}^{N-k+1} \qbinom {N-i-k}{L-k-1} \qbinom{N-j}{k-1} q^{ (L-k)(i+k-1) + k(j+k-L-1)}.
\]
Now, the $j$-sum can be performed using Proposition~\ref{prop:q-telescope}. However, the result depends on
the lower limit. If $1 \leq i \leq \lfloor (N-k)/2 \rfloor$, then the lower limit is $N-i-k+2$, and otherwise,
it is $i-k+1$. These two different cases prove \eqref{quinv>r1} and \eqref{quinv>r2}.
\end{proof}
\begin{proof}[Proof of Lemma~\ref{lem:quinv>}\ref{it3vvt1}]
Suppose $w \in W_k^>$ with $p_3(w) = j$, $p'_1(w) = i+b_3$ and $j < N+1 - i-b_3$. Then
$w$ can be written as $w = w_1 3 w_2 1 w_3$, where $w_1 \in \{1,2\}^{j-1}$, $w_2 \in \{1,2,3\}^{N-i-b_3-j}$
and $w_3 \in \{2,3\}^{i+b_3-1}$ with a total of $L-k-1$ $1$'s in $w_1$ and $w_2$ and a total of
$k-1$ $3$'s in $w_2$ and $w_3$.
Suppose $a_1$ is the number of $1$'s in $w_1$ and $b_3$ is the number of $3$'s in $w_3$, so $p_1(w) = i$. To get an upper bound on the number of $2$'s in $w_3$ (which is equal to $i-1$), notice that $a_1 \leq L-k-1$, and there are $N-L+a_1-i-j+2$ $2$'s in $w_2$, so in particular, $(N-L+a_1+1) -(i+j-1) \geq 0$. By definition of $W_k^>$, we have $i < j$ and therefore
\[
2i \leq i+j-1 \leq N - L + a_1 + 1 \leq N - k,
\]
from which the condition follows.
After this split, the number of coinversions of $w$ is
\begin{multline*}
\coinv(w) = \coinv(w_1) + \coinv(w_2) +\coinv(w_3) + (L-k-a_1)(i+b_3-1) \\
+ k(j-1) + a_1(N-j-L+1+a_1) + b_3(N-L-i-j+a_1+2).
\end{multline*}
Let the left hand side of \eqref{quinv>l} be denoted $f(i)$.
We now follow the strategy of the proof of Lemma~\ref{lem:quinv>}\ref{it3vvt2}. The sums over all
words $w_1, w_2, w_3$ will give us appropriate $q$-multinomial coefficients by
Proposition~\ref{prop:coinv-gf}. Then we will obtain
\begin{multline*}
f(i) = \sum_{b_3 = 0}^{(k-1) \wedge (N-i)} \sum_{j=i+1}^{N-i-b_3} \sum_{a_1 = 0}^{(j-1) \wedge (L-k-1)}
\qbinom {j-1}{a_1} \qbinom{i+b_3-1}{b_3}\\
\times \qbinom {N-i-j-b_3}{L-k-a_1-1, N-i-j-L+a_1+2,k-1-b_3} \\
\times q^{(L-k-a_1)(i+b_3-1) + k(j-1) + a_1(N-j-L+1+a_1) + b_3(N-L-i-j+a_1+2)},
\end{multline*}
where $a \wedge b$ means the minimum of $a$ and $b$.
Let us clarify the limits of the sums first.
The $b_3$-sum is bounded above by the minimum
of the available number of $3$'s and the fact that $w_2$ is of nonnegative length.
Now, $j$ is at least $i+1$ by definition of the set
$W_k^>$ and is at most $N-i-b_3$ again because of $w_2$.
Then, $a_1$ is at most the minimum of the available number of $1$'s and the length
of $w_1$.
It is common parlance to say that the (upper and lower) limits of a hypergeometric sum are \emph{natural} if the terms vanish outside the range. We can see that the lower limits of the $b_3$ and $a_1$ sums are natural. Since both upper limits of the $b_3$-sum as well as the $a_1$-sum are natural, the answer is independent of which of them is larger. Therefore, we will only write one of them.
The $a_1$-sum can now be performed using Theorem~\ref{thm:q-chuvan} and we end up with
\begin{multline*}
f(i) = \sum_{b_3 = 0}^{k-1} \sum_{j=i+1}^{N-i-b_3} q^{(L-k)(i+b_3-1) + k(j-1) + b_3(N-L-i-j+2)} \qbinom{i+b_3-1}{b_3} \\
\times \qbinom {N-i-b_3-j}{k-1-b_3} \qbinom {N-i-k}{L-k-1} .
\end{multline*}
We now find that the $j$-sum can be performed using Proposition~\ref{prop:q-telescope}. The result is then
\[
f(i) = \sum_{b_3 = 0}^{k-1} q^{(i-1)L + b_3(N-k-2i+1) + k}
\qbinom{i+b_3-1}{i-1} \qbinom {N-2i-b_3}{N-k-2i}
\qbinom {N-i-k}{L-k-1} .
\]
Now, notice that if the upper limit of the $b_3$-sum were $k$, we could have applied
Corollary~\ref{cor:q-dualchu} after replacing $b_3$ by $b_3+i-1$. We thus add and subtract
the term for $b_3 = k$, perform the $b_3$-sum and simplify to obtain the result.
\end{proof}
\subsection{Proof of Lemma~\ref{lem:quinv<}}
We will now prove the cases of Lemma~\ref{lem:quinv<} separately.
We again begin with the second case first, since it is conceptually simpler.
\begin{proof}[Proof of Lemma~\ref{lem:quinv<}\ref{it4vvt2}]
Suppose $w \in W_k^\leq$ with $p_3(w) = j$, $p'_1(w) = i+k$ and $j > N+1 - i-k$. Then
$w$ can be written as $w = w_1 1 2^{i+j+k-N-2} 3 w_2$, where $w_1 \in \{1,2\}^{N-i-k}$ with $L-k-1$ $1$'s
and $w_2 \in \{2,3\}^{N-j}$ with $k-1$ $3$'s. As a result, $p_1(w) = i$ and
\[
\coinv(w) = \coinv(w_1) + \coinv(w_2) + (L-k)(i+k-1) + k(j+k-L-1).
\]
Notice that the smallest possible value of $j$ is $L-k+1$, namely when $i = N+2-j-k$ and $w_1$ contains no $2$'s. To see the largest possible value of $j$, note that $j \leq i$ by definition of $W_k^\leq$ and the maximum value of $i$ is $N-L+1$ when $w_1$ contains no $2$'s. Let the left hand sides of \eqref{quinvl} be denoted $f(j)$.
We now follow the strategy of the proof of Lemma~\ref{lem:quinv<}\ref{it4vvt2}.
The sums over all
words $w_1, w_2, w_3$ will give us appropriate $q$-multinomial coefficients by
Proposition~\ref{prop:coinv-gf}. Then we will obtain
\begin{multline*}
f(j) = \sum_{i = j}^{N-k-j+1}\ \sum_{b_3 = 0}^{(k-1) \wedge (N-i-j)}\ \sum_{a_1 = 0}^{(j-1) \wedge (L-k-1)}
\qbinom {j-1}{a_1} \qbinom{i+b_3-1}{b_3} \\
\times\qbinom {N-i-j-b_3}{L-k-a_1-1, N-i-j-L+a_1+2,k-1-b_3} \\
\times q^{(L-k-a_1)(i+b_3-1) + k(j-1) + a_1(N-j-L+1+a_1) + b_3(N-L-i-j+a_1+2)}.
\end{multline*}
Again, we need to clarify the limits. The lower limit of $i$ is $j$ by definition of $W_k^\leq$
and the upper limit of $i$ comes from the fact that the number of $3$'s in $w_2$ is upper bounded
by $N-i-j-b_3$. The upper limits of $a_1$ and $b_3$ are explained by the same reasoning
as the similar sum in the proof of Lemma~\ref{lem:quinv>}\ref{it3vvt1}.
The $a_1$-sum can now be performed using Theorem~\ref{thm:q-chuvan} and we end up with
\begin{multline*}
f(j) = \sum_{i = j}^{N-k-j+1} \sum_{b_3 = 0}^{k-1} q^{(L-k)(i+b_3-1) + k(j-1) + b_3(N-L-i-j+2)} \qbinom{i+b_3-1}{b_3} \\
\times \qbinom {N-i-b_3-j}{k-1-b_3} \qbinom {N-i-k}{L-k-1} .
\end{multline*}
We now find that the $b_3$-sum can be performed using Corollary~\ref{cor:q-dualchu} after replacing $b_3$
by $b_3+i-1$.
The result is then
\[
f(j) = \sum_{i = j}^{N-k-j+1} q^{(i-1)(L-k) + k(j-1)} \qbinom {N-j}{k-1}\qbinom {N-i-k}{L-k-1} ,
\]
We now rewrite the limits of the $i$-sum as $\sum_{i = j}^{N-L+1} - \sum_{i = N-k-j+2}^{N-L+1}$,
use Proposition~\ref{prop:q-telescope} and simplify to complete the proof.
\end{proof}
\section{Conclusion and further questions}\label{sec:conclusion}
We are finally equipped to prove the main result of this paper, Theorem~\ref{thm:mainconj}, stating that $\widetilde{H}_{\lambda}(X;q,t) = C_{\lambda}(X;q,t)$ where
\[
C_{\lambda}(X;q,t) = \sum_{\sigma\in\PQT(\lambda)}x^{\sigma}t^{\quinv(\sigma)}q^{\maj(\sigma)}.
\]
\begin{proof}[Proof of Theorem~\ref{thm:mainconj}]
By Theorem~\ref{thm:symmetry}, $C_{\lambda}(X;q,t)$ is symmetric in the variables $x_i$. Axioms \eqref{eq:A1}, \eqref{eq:A2}, and \eqref{eq:A3} uniquely characterize and define the modified Macdonald polynomials $\widetilde{H}_{\lambda}(X;q,t)$. Axiom \eqref{eq:A3} is equivalent to the condition that the coefficient of $m_{(n)}$ in $\widetilde{H}_{\lambda}(X;q,t)$ is $1$, since $s_{(n)}=h_{(n)}$ and $\langle h_{\lambda},m_{\mu}\rangle=\delta_{\lambda\mu}$. This is immediate for $C_{\lambda}(X;q,t)$, as a filling $\sigma$ such that $x^{\sigma}=x_j^n$ consists of all equal entries $j$ and has $\maj(\sigma)=\quinv(\sigma)=0$. Axioms \eqref{eq:A1} and \eqref{eq:A2} are written equivalently as \eqref{A1m} and \eqref{A2m}, respectively.
Those can in turn be rewritten in terms of $\widetilde{C}_{\lambda}(X;q,t)$, the superization of $C_{\lambda}(X;q,t)$, as \eqref{C1} and \eqref{C2}, respectively. \eqref{C1} is true by Theorem~\ref{lem:A1FP}. \eqref{C2} is true by Theorem~\ref{thm:A2FP}. This completes the proof.
\end{proof}
Our work leads to several natural questions, which we mention below.
\begin{question}
From \eqref{eq:HHLformula} and \eqref{eq:MainResult}, it follows that the modified Macdonald polynomial $\widetilde{H}_\lambda(X; q, t)$ can be expressed as a sum over tableaux either of queue inversion weights or of HHL weights.
In these weights, the terms for the content and the major index are the same. The only difference is in the notion of a ``triple'', as discussed at the end of Section \ref{sec:definitions}. Therefore it is natural to ask for a bijective proof of the equality of these sums.
\end{question}
We conjecture something stronger: there is a bijection from $\PQT(\lambda, n)$
to itself that preserves the row content of the fillings and the major index, and sends the $\quinv$ statistic to the HHL $\inv$ statistic. To formally state this, we define row-equivalency classes for the fillings.
\begin{defn}
Let $\sigma,\tau$ be fillings of $\dg(\lambda)$. We say $\sigma$ and $\tau$ are \emph{row-equivalent} if for every row of $\sigma$, its entries are a permutation of the entries of the corresponding row in $\tau$. We write $\sigma \sim \tau$ when $\sigma$ and $\tau$ are row-equivalent, and we denote by $[\sigma]$ the class of row-equivalent fillings that $\sigma$ belongs to.
\end{defn}
The following conjecture is sufficient to prove Theorem~\ref{thm:mainconj} via \cite[Theorem 2.2]{HHL05}, since it would give a weight-preserving bijection from queue inversion weights to HHL weights.
\begin{conjecture}\label{conj:bij}
Given $\lambda$ and $n$, there exists a bijection $\delta:\PQT(\lambda,n) \rightarrow \PQT(\lambda,n)$ such that for all $\sigma$,
\begin{enumerate}[label=\roman*.]
\item $\delta(\sigma) \sim \sigma$,
\item $\maj(\delta(\sigma)) = \maj(\sigma)$,
\item $\inv(\delta(\sigma)) = \quinv(\sigma)$.
\end{enumerate}
\end{conjecture}
Conjecture~\ref{conj:bij} implies, for a row-equivalency class $[T]$, that
\[
\sum_{\sigma\in[T]} t^{\quinv(\sigma)}q^{\maj(\sigma)} = \sum_{\sigma\in[T]} t^{\inv(\sigma)}q^{\maj(\sigma)}.
\]
\begin{example}
Consider the row-equivalency class $[T]=(\{1\},\{2,2,3\},\{1,1,2\})$ for $\lambda=(3,2,2)$. In Figure~\ref{fig:rowequiv}, we show the tableaux $F\in[T]$ with their queue inversion weights $t^{\quinv(F)}q^{\maj(F)}$ above and HHL weights $t^{\inv(F)}q^{\maj(F)}$ below. Observe that the sums of the weights are equal. Moreover, there is a bijection $\delta$ such that the queue inversion weight of $F$ matches the HHL weight of $\delta(F)$.
\end{example}
\begin{figure}[h!]
\begin{center}
\resizebox{\textwidth}{!}{
\begin{tikzpicture}[node distance=2cm]
\def \sh {1.6};
\node at (-1.2,-1.25) {quinv};
\node at (-1.2,-2) {HHL};
\node at (0,0) {$\tableau{1\\2&2&3\\1&1&2}$};
\node at (0,-1.25) {$t^2q^4$};
\node at (0,-2) {$q^4$};
\node at (\sh,0) {$\tableau{1\\2&2&3\\1&2&1}$};
\node at (\sh,-1.25) {$t^3q^3$};
\node at (\sh,-2) {$t^2q^3$};
\node at (2*\sh,0) {$\tableau{1\\2&2&3\\2&1&1}$};
\node at (2*\sh,-1.25) {$t^4q^2$};
\node at (2*\sh,-2) {$t^3q^2$};
\node at (3*\sh,0) {$\tableau{1\\2&3&2\\1&1&2}$};
\node at (3*\sh,-1.25) {$t^2q^3$};
\node at (3*\sh,-2) {$tq^3$};
\node at (4*\sh,0) {$\tableau{1\\2&3&2\\1&2&1}$};
\node at (4*\sh,-1.25) {$tq^4$};
\node at (4*\sh,-2) {$tq^4$};
\node at (5*\sh,0) {$\tableau{1\\2&3&2\\2&1&1}$};
\node at (5*\sh,-1.25) {$t^3q^2$};
\node at (5*\sh,-2) {$t^4q^2$};
\node at (6*\sh,0) {$\tableau{1\\3&2&2\\1&1&2}$};
\node at (6*\sh,-1.25) {$tq^3$};
\node at (6*\sh,-2) {$t^2q^3$};
\node at (7*\sh,0) {$\tableau{1\\3&2&2\\1&2&1}$};
\node at (7*\sh,-1.25) {$t^2q^3$};
\node at (7*\sh,-2) {$t^3q^3$};
\node at (8*\sh,0) {$\tableau{1\\3&2&2\\2&1&1}$};
\node at (8*\sh,-1.25) {$q^4$};
\node at (8*\sh,-2) {$t^2q^4$};
\end{tikzpicture}
}
\end{center}
\caption{For $\lambda=(3,2,2)$ and $n=3$, all fillings of type $(\lambda,n)$ in the row-equivalency class $[T]=(\{1\},\{2,2,3\},\{1,1,2\})$. Below are the corresponding queue
inversion weights and HHL weights.}\label{fig:rowequiv}
\end{figure}
\begin{question}
In Section~\ref{sec:bijwords}, we demonstrated in Theorem~\ref{thm:invol} the existence of a quinv-preserving bijection between sets of words respecting certain conditions.
We have not been able to find such a bijection and we think it would be interesting to find one.
\end{question}
\longthanks{We thank the anonymous referees for many useful comments. This work was inspired by OM's discussion with Lauren Williams and Sylvie Corteel, in particular Lauren Williams' suggestion of connecting modified Macdonald polynomials to a particle process via tableaux and the queue inversion statistic. OM also gratefully acknowledges conversations with Jim Haglund and Sarah Mason. This material is based upon work supported by the Swedish Research Council under grant no. 2016-06596 while the authors were in residence at Institut Mittag-Leffler in Djursholm, Sweden during the spring semester of 2020.}
%\let\thefootnote\relax\footnotetext{}
\smallskip
\noindent\emph{Note added in proof:} Recently, Loehr has announced a bijective proof of Theorems~\ref{thm:quinv-gf} and~\ref{thm:quinv-refined-gf} in \cite{Loehr}.
%: https://arxiv.org/abs/2210.11286
\bibliographystyle{mersenne-plain}
\bibliography{ALCO_Ayyer_626}
\end{document}
**