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\usepackage{pgf,tikz}
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\newcommand{\SYT}{\mathop{\mathrm{SYT}}}
\newcommand{\A}{\mathop{\mathrm{area}}}
\newcommand{\h}{\mathop{\mathrm{ht}}}
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\title[Explicit formulas for Characters of the module of multivariate diagonal harmonics]{Explicit Combinatorial Formulas for  Some Irreducible Characters of the $GL_k\times \mathbb{S}_n$-module of multivariate diagonal harmonics}

%% define your authors in the usual way
%% use \addressmark{1}, \addressmark{2} etc for the institutions, and use \thanks{} for contact details
\author{Nancy Wallace\thanks{\href{mailto:wallace.nancy@courrier.uqam.ca}{wallace.nancy@courrier.uqam.ca}. Nancy Wallace is supported by a scholarship from the NSERC.}}

%% then use \addressmark to match authors to institutions here
\address{LaCIM, Universit\'e du Qu\'ebec \`a Montr\'eal, Montr\'eal (Qu\'ebec), Canada}

%% put the date of submission here
\received{15-11-2018}
%% leave this blank until submitting a revised version
\revised{27-03-2019}

%% put your English abstract here, or comment this out if you don't have one yet
%% please don't use custom commands in your abstract / resume, as these will be displayed online
%% likewise for citations -- please don't use \cite, and instead write out your citation as something like (author year)
\abstract{
We give an explicit combinatorial formula for some irreducible components of $GL_k\times \mathbb{S}_n$-modules of multivariate diagonal harmonics. To this end we introduce a new path combinatorial object $T_{n,s}$ allowing us to give the formula directly in terms of Schur functions. }

%% put your French abstract here, or comment this out if you don't have one
\resumetitle{R\'esum\'e}
\resume{Dans ce long r\'esum\'e, certaines composantes irr\'eductibles des caract\`eres de  $GL_k\times \mathbb{S}_n$-modules harmoniques diagonaux multivari\'es sont donn\'es sous forme d'une formule combinatoire explicite. \`A cette fin, nous introduisons un nouvel objet de la combinatoire des chemins, $T_{n,s}$, \`a partir duquel nous associons une fonction de Schur \`a chaque chemin.}

%% put your keywords here, or comment this out if you don't have them yet
\keywords{Multivariate diagonal harmonics, path combinatorics, Macdonald eigenoperators}

%% you can include your bibliography however you want, but using an external .bib file is STRONGLY RECOMMENDED and will make the editor's life much easier
%% regardless of how you do it, please use numerical citations, ie. [xx, yy] in the text

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\begin{document}

\maketitle


%%%%%%%%%%%%%%   Intro
\section{Introduction}
The aim of his paper is to describe some features of the characters of "rectangular" $GL_k\times\mathbb{S}_n$-modules, $\mathcal{E}_{m,n}^{\langle k\rangle}$, introduced by F. Bergeron in \cite{[B2018]}.

When $k=2$ and $m=n$, these modules are the modules of diagonal harmonics whose characters have been studied for many years. As shown in \cite{[GH1996a]} and \cite{[H2002]}, the Frobenius transformation of its graded characters may be expressed as $\nabla(e_n)$  where $\nabla$ is the Macdonald eigenoperator introduced in \cite{[BG1999]} and recalled in \cref{sec : car} and $e_n$ is the $n$-th elementary symmetric function. A combinatorial interpretation that became known as the Shuffle Conjecture was introduced in \cite{[HHLRU2005]} and proved recently by Carlson and Mellit \cite{[CM2015],[M2016]}. These characters also intervene in torus knot link homology and algebraic geometry; see, for instance \cite{[H2017],[K2007],[GNR2016],[OR2018]}. The case $k=3$ was studied in \cite{[BP2012]}. 

In recent work \cite{[B2018]} F. Bergeron made a breakthrough in the multivariate case ($k$ arbitrary). He found interesting relations between various irreducible characters of the modules. This allowed the study of the character of $\mathcal{E}_{n,n}^{\langle k\rangle}$ developed in the elementary symmetric functions for $n\leq 4$ in \cite{[BCP2018]} (in our notation this is the specialization $\sum c_{\lambda,\mu}s_\lambda(q,1,...,1,0,0,...)s_\mu(X)$, where there are $k-1$, $1$'s). These relations are also exploited here to obtain our main result. 

To state our result we briefly fix the required notation. We encode the characters of the irreducible $GL_k\times \mathbb{S}_n$ -modules  as products of Schur functions, $s_\lambda(\bm{Q})s_\mu(\bm{X})$ (see \cref{sec : car} for details). For easier reading we will also write  $s_\lambda\otimes s_\mu$ for $s_\lambda(\bm{Q})s_\mu(\bm{X})$. As shown in \cite{[B2013]} for the case $m=n$ there is a stability property that makes it possible to avoid mentioning $k$. Therefore the character of $\mathcal{E}_{m,n}^{\langle k \rangle}$ can be expressed in the form $\mathcal{E}_{m,n}=\sum_\mu \left(\sum_\lambda c_{\lambda,\mu}s_\lambda\right)\otimes s_\mu$. Our aim is to describe some features of $\mathcal{E}_{m,n}$, or equivalently, $\langle \mathcal{E}_{m,n},s_\mu\rangle =\sum_\lambda c_{\lambda,\mu}s_\lambda$. 

The main result of this paper is a combinatorial description of the multiplicity of $s_\lambda\otimes s_\mu$ in $\mathcal{E}_{n,
  n}$, when $\lambda$ is hook shaped. The result is constructive, in that the hooks are determined combinatorially by a standard Young tableau and certain paths in a staircase shaped grid. More precisely, we give the following combinatorial description (the basic combinatorial notations used in \cref{thm : main} are recalled in \cref{sec : tools}).


  \begin{thm}\label{thm : main}If  $r=1$ and $\mu\in\{(n), (n-1,1), (n-2,1,1),(1^n)\}$ then:
  \begin{equation}\label{eq : lift delta}\langle \mathcal{E}_{rn,n}, s_{\mu}\rangle|_{\hooks}=\sum_{\tau\in \SYT(\mu)} \sum_{\gamma \in T_{n, \dd(\tau')}} s_{\hook(\gamma)}
  \end{equation}
where the first sum is over all standard Young tableaux of shape $\mu$, the second sum is over paths $\gamma$ in $T_{n, \dd(\tau')}$ and $\hook(\gamma)=\left((r-1)\binom{n}{2} +\A(\gamma)+\h(\gamma)-\m(\tau')+1,1^{n-2-\h(\gamma) }\right)$.

Furthermore, when $\mu=1^n$, \eqref{eq : lift delta} holds for all positive integers $r$ that satisfy the equality $ \langle\mathcal{E}_{rn,n}, s_{k+1,1^{n-k-1}}\rangle=e^\perp_k \langle\mathcal{E}_{rn,n}, s_{1^{n}}\rangle$ for all $k$.
\end{thm}


The combinatorial object $T_{n,s}$ ($T_{n,\dd(\tau')}$ in \eqref{eq : lift delta}) represents a set of paths, these paths and their statistics ($\A$ and $\h$) will be defined in \cref{sec : tn}. This object was introduced by the author in order to eliminate alternating sums and obtain a Schur positive expression. When $s=0$, they afford a generating function correlated to the $q$-Pochhammer symbol $(-qz;z)_n$. Symmetric functions $s_\lambda$, $e_n$, the operators $\nabla$ and $\Delta'$ will be recollected in \cref{sec : car}. Finally, \cref{sec : main} will be dedicated to \eqref{eq : lift delta}, the definition of $|_{\hooks}$, of the operator  $e_k^\perp$  and a proposition restricting theorem 1 to the $GL_2\times \mathbb{S}_n$-characters mentioned above, which gives formulas in terms of the major index for $\langle \nabla^r(e_n), s_{\mu}\rangle|_{\hooks}$ and $\langle \Delta'_{e_k}(e_n),e_n\rangle|_{\hooks}$ .





%%%%%%%%%%%%%%   Tools
\section{Combinatorial Tools}\label{sec : tools}


A \begin{it}partition\end{it} of $n$ is a decreasing sequence of positive integers often represented as a Ferrer's diagram. For $\lambda$ a  \begin{it}Ferrer's diagram \end{it} of shape $\lambda=\lambda_1,\lambda_2,\ldots,\lambda_k$ is a left justified pile of boxes having $\lambda_i$ boxes in the $i$-th row. We will use the French notation so the second row lies on top of the first row (see \cref{fig : tableau}). We can see them as a subset of $\mathbb{N}\times\mathbb{N}$ if we put the bottom left corner of the diagram to the origin. In this setting, we can associate the bottom left corner of a box to the coordinate it lies on. We say a partition is  \begin{it}hook shaped\end{it} if it has the form $(a,1, \ldots,1)=(a,1^k)$, where $a, k \in \mathbb{N}$. If to each box of a diagram we associate an entry it is called a \begin{it}tableau\end{it}. A tableau is of shape $\lambda$ if it is a filling by integers of a diagram of shape $\lambda$. For $\lambda$ a partition of $n$, a tableau of shape $\lambda$ with distinct entries from $1$ to $n$ strictly increasing in rows and columns is a  \begin{it}standard Young tableau\end{it}. The set of all standard Young tableaux of shape $\mu$ is denoted $\SYT(\mu)$. The  \begin{it}descent\end{it} set of a tableau is the set of entries $i$ such that the entry $i+1$ lies in a row strictly above $i$. For a tableau $\tau$, the descent set is denoted $\DD(\tau)$ and the cardinality is denoted  $\dd(\tau)$. The sum of the elements of  $\DD(\tau)$ is called the  \begin{it}major index\end{it} and is denoted $\m(\tau)$ (see \cref{fig : descents}). The  \begin{it}conjugate \end{it} of a partition $\lambda$, (respectively a diagram, a tableau) is denoted $\lambda'$, and is its reflection through the line $x=y$ (see \cref{fig : conjugate}). 

\begin{figure}[ht!]
\begin{minipage}{5.13cm}
\centering
\begin{tikzpicture}[scale=.5]
\draw (0,0)--(4,0)--(4,1)--(2,1)--(2,3)--(1,3)--(1,5)--(0,5)--(0,0);
\draw (1,0)--(1,3);
\draw(2,0)--(2,1);
\draw (3,0)--(3,1);
\draw (4,0)--(4,1);
\draw (0,1)--(2,1);
\draw (0,2)--(2,2);
\draw (0,3)--(1,3);
\draw (0,4)--(1,4);
\end{tikzpicture}
\caption{$\lambda=42211$}
\label{fig : tableau}
\end{minipage}
\begin{minipage}{5.05cm}
\centering
\begin{tikzpicture}[scale=.5]
\draw (0,0)--(5,0)--(5,1)--(3,1)--(3,2)--(1,2)--(1,4)--(0,4)--(0,0);
\draw (1,0)--(1,2);
\draw(2,0)--(2,2);
\draw (3,0)--(3,1);
\draw (4,0)--(4,1);
\draw (0,1)--(3,1);
\draw (0,2)--(1,2);
\draw (0,3)--(1,3);
\draw (0,4)--(1,4);
\end{tikzpicture}
\caption{
$\lambda'=5311$}
\label{fig : conjugate}
\end{minipage}
\begin{minipage}{6.1cm}
\centering
\begin{tikzpicture}[scale=.5]
\draw (0,0)--(4,0)--(4,1)--(2,1)--(2,3)--(1,3)--(1,5)--(0,5)--(0,0);
\draw (1,0)--(1,3);
\draw(2,0)--(2,1);
\draw (3,0)--(3,1);
\draw (4,0)--(4,1);
\draw (0,1)--(2,1);
\draw (0,2)--(2,2);
\draw (0,3)--(1,3);
\draw (0,4)--(1,4);
\node(1) at (.5,.5){$\textcolor{black}{1}$};
\node(2) at (1.5,.5){$\textcolor{black}{2}$};
\node(3) at (.5,1.5){$3$};
\node(4) at (2.5,.5){$\textcolor{black}{4}$};
\node(5) at (.5,2.5){$\textcolor{black}{5}$};
\node(6) at (.5,3.5){$6$};
\node(7) at (1.5,1.5){$7$};
\node(8) at (3.5,.5){$8$};
\node(9) at (.5,4.5){$\textcolor{black}{9}$};
\node(10) at (1.5,2.5){$\textcolor{black}{10}$};
\end{tikzpicture}
\caption{In $\SYT(42211)$;\\$\DD(\tau)=\{2,4,5,8\}$,\\ $\dd(\tau)=4$, $\m(\tau)=19$}
\label{fig : descents}
\end{minipage}
\end{figure}


We end this section by recalling the definition of the Gaussian polynomials, $\left[ \genfrac{}{}{0pt}{}{n}{k}\right]_q$. Let:
\begin{equation*}[n]_q=1+q+q^2+\cdots+q^{n-1} \text{ , }~[n]!_q=\prod_{i=1}^n [i]_q
\text{ and }\begin{bmatrix} n\\k\end{bmatrix}_q=\frac{[n]!_q}{[n-k]!_q[k]!_q}.
\end{equation*}
So when $q=1$ this gives the usual binomial coefficient. It is well known that the Gaussian polynomials are related to the north-east paths of a $k \times (n-k)$ grid: if $\mathcal{C}_{k}^n$ denotes the set of such paths, and the \begin{it}area\end{it} of a path $\gamma$, denoted $\A(\gamma)$, is defined as the number of boxes beneath the path, then $\sum_{\gamma\in\mathcal{C}_k^n} q^{area(\gamma)}=\left[ \genfrac{}{}{0pt}{}{n}{k}\right]_q$. 


%%%%%%%%%%%%%%   The space

\section{The space and characters}\label{sec : car}

Before introducing our new combinatorial objects we provide more details about the modules $\mathcal{E}_{n,n}^{\langle k\rangle}$ and why they are interesting. (This section is not necessary for understanding our main result.) 

 Let $X=(x_{i,j})_{i,j}$ where $1\leq i\leq k$,  $1\leq j\leq n$ and let $R_n^{\langle k \rangle}=\mathbb{Q}[X]$ denote the polynomial ring in the variables $X$. For $(\tau,\sigma)$ in $GL_k\times \mathbb{S}_n$ the group $GL_k\times \mathbb{S}_n$ acts on $R_n^{\langle k \rangle}$ as follows:  
   \begin{equation*}
    (\tau,\sigma). F(X)=F(\tau \cdot X \cdot \sigma)
    \end{equation*}
With this action we can define $\mathcal{E}_{n,n}^{\langle k \rangle}$ as the smallest submodule of $R_n^{\langle k \rangle}$ that contains the Vandermonde determinant, is closed under all higher polarization operators $\sum_{j=0}^n x_{r,j}\partial_{x_{s,j}}$ and  is closed under all partial derivatives $ \partial_{x_{s,j}}$.


When we separate this modules into a direct sum of the smallest possible modules that preserve this action, we say that it is a decomposition into irreducible modules. A module can be encoded by its character. There is a one to one correspondence between irreducible modules and irreducible characters. It is known that the irreducible characters for $GL_k$ and the Frobenius transform of irreducible characters of $\mathbb{S}_n$ are the set of Schur functions, denoted $\{s_\lambda\}$, indexed over all partitions. 

Meanwhile, the \begin{it}ring of symmetric polynomials\end{it} is a set of polynomials which are invariant by permutation of the variables $Y=\{y_1,y_2, \ldots, y_n\}$. So $\Lambda$ is embedded in $\mathbb{Q}[Y]$. In other words for all $\sigma \in \mathbb{S}_n$ we have:
\begin{equation*}
f(y_1,y_2,\ldots, y_n)=f(y_{\sigma^{-1}(1)}, y_{\sigma^{-1}(2)}, \ldots, y_{\sigma^{-1}(n)}).
\end{equation*}  
 The \begin{it}ring of symmetric functions\end{it}, denoted $\Lambda$, can be thought of as the ring of symmetric polynomials in an infinite set of variables. It is a graded ring and has the Schur functions as a basis. Therefore the ring of symmetric functions is the right setting for our study. \begin{it}Elementary symmetric functions\end{it} are defined by $e_n(X)=\sum_{i_1<\cdots<i_n} x_{i_1}x_{i_2}\cdots x_{i_n}$ and $e_\lambda=e_{\lambda_1}e_{\lambda_2}\cdots e_{\lambda_k}$. They also form a basis for the ring of symmetric functions $\Lambda$.

If we start with the ring $\mathbb{Q}(q,t)[Y]$ we also get a ring with Schur functions as a basis. Additionally, the combinatorial Macdonald polynomials, denoted $H_\mu$, are a basis for $\mathbb{Q}(q,t)[Y]$. They appear as eigenvectors of special operators (see \cite{[B2009]} for more on this), $\nabla$ and the $\Delta'_{e_m}$, introduced in \cite{[BG1999]}, \cite{[BGHT1999]}. These Macdonald operators are defined as follows:
\begin{equation*}\nabla(H_\mu)=\prod_{(i,j)\in \mu}q^it^jH_\mu~ \text{ and }~\Delta'_{e_m}(H_\mu)=\left(\underset{|S|=m}{\sum_{S\subset \mu / (0,0)}} \prod_{(i,j)\in S}q^it^j\right)H_\mu.
\end{equation*}
By definition we have $\Delta'_{e_{n-1}}(e_n)=\nabla(e_n)$, which gives the character decomposition of  the $GL_2\times \mathbb{S}_n$ case stated in the introduction. It was proven that the coefficients are symmetric polynomials in the $q$ and $t$ variables, therefore, one could write the coefficients in the form $\sum_{\lambda,\mu} c_{\lambda,\mu}s_\lambda(q,t,0,\ldots)s_\mu(X)$. More generally, the $GL_k$ characters can be obtained by setting  $q_{k+1}=q_{k+2}=\cdots=0$. We can therefore use a more general notation $\mathcal{E}_{n,n}$.

We will also discuss characters of the form $\mathcal{E}_{rn,n}$ which are related to $\nabla^r(e_n)$ when we restrict to $GL_2$. They are constructed by adding a set of inert variables considered to be of degree zero. For more details see \cite{[B2018]}.



%%%%%%%%%%%%%%  T_n

\section{Defining {$T_{n,s}$}}\label{sec : tn}

Our formula will be formulated in terms of new combinatorial objects that we denote by $T_{n,s}$. It is the set of north-east paths in an $n-2$ staircase shaped grid lying in $\mathbb{N}^2$, starting at $(0,s)$ and ending at a point in the set $\{ (x,y)~|~x+y=n-2 , x\geq 0\text{ and } y\geq s\}$. For an example see \cref{fig : tns}. The relevant paths can be represented as words of length $n-s-2$ in the alphabet $\{N,E\}$. 

The \begin{it}area\end{it}, denoted $\A$, of a path is the number of boxes south-east of the path (see \cref{fig : aire}). The \begin{it}height\end{it} of a path, $\h$, is the $y$ coordinate of its end point (see \cref{fig : ht}). 


\begin{figure}[ht!]
\centering
\begin{minipage}{4.28cm}
\begin{tikzpicture}[scale=.38]
\draw [thin, gray](0,0)--(0,5)--(1,5)--(1,4)--(2,4)--(2,3)--(3,3)--(3,2)--(4,2)--(4,1)--(5,1)--(5,0)--(0,0);
\draw [thin, gray](1,0)--(1,4);
\draw [thin, gray](2,0)--(2,3);
\draw [thin, gray](3,0)--(3,2);
\draw [thin, gray](4,0)--(4,1);
\draw [thin, gray](0,1)--(4,1);
\draw [thin, gray](0,2)--(3,2);
\draw [thin, gray](0,3)--(2,3);
\draw [thin, gray](0,4)--(1,4);
\node(d) at (0,2){\textcolor{blue}{$\bullet$}};
\node(deb) at (-2.5,2){\textcolor{blue}{start}};
\draw[->, blue] (-1,2)--(-.4,2);
\node(f3) at (3,2){\textcolor{red}{$\bullet$}};
\node(f4) at (2,3){\textcolor{red}{$\bullet$}};
\node(f5) at (1,4){\textcolor{red}{$\bullet$}};
\node(f6) at (0,5){\textcolor{red}{$\bullet$}};
\node(fin) at (5.5,5.5){\textcolor{red}{finish}};
\draw[->, red] (fin)--(f3);
\draw[->, red] (fin)--(f4);
\draw[->, red] (fin)--(f5);
\draw[->, red] (fin)--(f6);
\draw[<->] (-0.5,0)--(-.5,1.8);
\node(p) at (-1,1){$s$};
\draw[<->] (0,-.5)--(5,-.5);
\node(n) at (2.5,-1.2){$n-2$};
\end{tikzpicture}
\caption{$T_{7,2}$}
\label{fig : tns}
\end{minipage}
\begin{minipage}{6.28cm}
\begin{tikzpicture}[scale=.5]
\filldraw[pink] (0,2)--(0,3)--(1,3)--(1,4)--(2,4)--(2,3)--(3,3)--(3,2)--(4,2)--(4,1)--(5,1)--(5,0)--(0,0);
\draw[thin, gray] (0,0)--(0,5)--(1,5)--(1,4)--(2,4)--(2,3)--(3,3)--(3,2)--(4,2)--(4,1)--(5,1)--(5,0)--(0,0);
\draw[thin, gray] (1,0)--(1,4);
\draw[thin, gray] (2,0)--(2,3);
\draw[thin, gray] (3,0)--(3,2);
\draw[thin, gray] (4,0)--(4,1);
\draw[thin, gray] (0,1)--(4,1);
\draw[thin, gray] (0,2)--(3,2);
\draw[thin, gray] (0,3)--(2,3);
\draw[thin, gray] (0,4)--(1,4);
\draw[blue, thick] (0,2)--(0,3)--(1,3)--(1,4);
\node(d) at (0,2){\textcolor{blue}{$\bullet$}};
\node(deb) at (-5,2){};
\node(f3) at (3,2){\textcolor{red}{$\bullet$}};
\node(f4) at (2,3){\textcolor{red}{$\bullet$}};
\node(f5) at (1,4){\textcolor{red}{$\bullet$}};
\node(f6) at (0,5){\textcolor{red}{$\bullet$}};
\node(arranger le bas) at (2.5,-.6){};
\end{tikzpicture}
\caption{ $\A(NEN)=13$}
\label{fig : aire}
\end{minipage}
\begin{minipage}{5.72cm}
\begin{tikzpicture}[scale=.5]
\node(fin) at (5.5,6){};
\draw[thin, gray] (0,0)--(0,5)--(1,5)--(1,4)--(2,4)--(2,3)--(3,3)--(3,2)--(4,2)--(4,1)--(5,1)--(5,0)--(0,0);
\draw[thin, gray] (1,0)--(1,4);
\draw[thin, gray] (2,0)--(2,3);
\draw[thin, gray] (3,0)--(3,2);
\draw[thin, gray] (4,0)--(4,1);
\draw[thin, gray] (0,1)--(4,1);
\draw[thin, gray] (0,2)--(3,2);
\draw[thin, gray] (0,3)--(2,3);
\draw[thin, gray] (0,4)--(1,4);
\draw[blue, thick] (0,2)--(0,3)--(1,3)--(1,4);
\node(d) at (0,2){\textcolor{blue}{$\bullet$}};
\node(deb) at (-1.5,2){};
\node(f3) at (3,2){\textcolor{red}{$\bullet$}};
\node(f4) at (2,3){\textcolor{red}{$\bullet$}};
\node(f5) at (1,4){\textcolor{red}{$\bullet$}};
\node(f6) at (0,5){\textcolor{red}{$\bullet$}};
\draw[dotted] (1,4)--(6,4);
\draw[<->] (6,0)--(6,4);
\node(h) at (7.5,2){$ht(\gamma)$};
\node(arranger le bas) at (2.5,-.6){};
\end{tikzpicture}
\caption{ $\h(NEN)=4$}
\label{fig : ht}
\end{minipage}
\end{figure}

When $s>n-1$ we set $T_{n,s}=\{N^{n-2}\}$ for reasons stated later on. Since
\begin{equation*}
\sum_{\gamma\in\mathcal{C}_k^n} q^{\A(\gamma)}=\begin{bmatrix} n\\k\end{bmatrix}_q,
\end{equation*}
 it can be observed that summing over the paths ending at height $j+s$ will contribute $q^{\binom{s+j+1}{2}+s(r-j)}\left[{\genfrac{}{}{0pt}{}{r}{j}}\right]_q$ to the generating function. We thus get the following result.

\begin{prop}Let $T_{n,s}(q,z)=\sum_{\gamma\in T_{n,s}} q^{\A(\gamma)}z^{\h(\gamma)}$. Then for $r=n-s-2$ we have:
\begin{equation*}
  T_{n,s}(q,z)=\sum_{j=0}^{r} q^{\binom{s+j+1}{2}+s(r-j)}\begin{bmatrix}r \\ j\end{bmatrix}_q z^{j+s}=T_{r,0}(q,z)z^{s}q^{rs+\binom{s+1}{2}}
 \end{equation*}
In particular, if $s=0$, we have:
\begin{equation*}
 T_{n,0}(q,z)=\sum_{j=0}^{n-2} q^{\binom{j+1}{2}}\begin{bmatrix}n-2 \\ j\end{bmatrix}_q z^{j}=(-qz;q)_{n-2}
 \end{equation*}
\end{prop}

The following result shows how this object is used to eliminate an alternating sum and make the relevant formula positive.

\begin{prop}Let $g_j:\mathbb{N}^*\rightarrow \mathbb{Z}$ be such that $g_j(k)-g_j(k-1)=k+j$ for all $k\geq 1$  and $g_j(0)-\binom{j+1}{2}=g_i(0)-\binom{i+1}{2}$ for all $i,j$. Then
\begin{equation*}\sum_{j=1}^{n-1}\sum_{k=0}^{n-j-1}(-1)^k\begin{bmatrix}n-1\\j+k\end{bmatrix}_qq^{-k+g_j(k)}z^{j-1}=T_{n,0}(q,z)q^{g_{j}(0)-\binom{j+1}{2}}.
\end{equation*}
In particular we have:
\begin{equation}\label{eq : alt->pos}\sum_{k=0}^{n-j-1}(-1)^k\begin{bmatrix}n-1\\j+k\end{bmatrix}_qq^{-k+g_j(k)}=\begin{bmatrix}n-2\\j-1\end{bmatrix}_qq^{g_j(0)}.
\end{equation}
\end{prop}

We will only give a sketch of  proof for $g_j(k)=\binom{j+k+1}{2}$. Notice that by the previous proposition we only need to prove the second part. For some fixed $k$ and $j$ we can represent the term $\left[{\genfrac{}{}{0pt}{}{n-1}{j+k}}\right]_qq^{-k+g_j(k)}$ by the set of paths in $\mathcal{C}^{n-1}_{j+k}$ to which we add $g_j(k)$ boxes and subtract $k$ boxes (see \cref{fig : g_j(k)}).
Then there is a bijection between  paths ending with a north step in $\mathcal{C}^{n-1}_{j+k+1}$ (blue in \cref{fig : boxes}) and paths ending with an east step in $\mathcal{C}^{n-1}_{j+k}$ (red in  \cref{fig : boxes}). We only need to change the last step. They both account for the same number of boxes (see \cref{fig : boxes} as an example). Since the terms have coefficient $(-1)^k$ they cancel out pairwise in the sum. So the only steps left are the ones when $k=0$ and the path in $\mathcal{C}^{n-1}_{j}$ ends with a north step. Eliminating the last north step doesn't affect the area if there are no east steps afterwards. Therefore we can consider the paths in $\mathcal{C}^{n-2}_{j-1}$ with the same statistic, which is what we needed.

\begin{figure}[ht!]
\begin{minipage}{8.cm}
\centering
\begin{tikzpicture}[scale=.5]
\filldraw[gray!50] (0,4)--(0,6)--(1,6)--(1,5)--(2,5)--(2,4)--(0,4);
\filldraw[pink] (2,0)--(2,4)--(3,4)--(3,3)--(4,3)--(4,2)--(5,2)--(5,1)--(6,1)--(6,0)--(2,0);
\draw[thin, gray] (0,0)--(0,6)--(1,6)--(1,5)--(2,5)--(2,4)--(3,4)--(3,3)--(4,3)--(4,2)--(5,2)--(5,1)--(6,1)--(6,0)--(0,0);
\draw[thin, gray] (1,0)--(1,5);
\draw[thin, gray] (2,0)--(2,4);
\draw[thin, gray] (3,0)--(3,3);
\draw[thin, gray] (4,0)--(4,2);
\draw[thin, gray] (5,0)--(5,1);
\draw[thin, gray] (0,1)--(5,1);
\draw[thin, gray] (0,2)--(4,2);
\draw[thin, gray] (0,3)--(3,3);
\draw[thin, gray] (0,4)--(2,4);
\draw[thin, gray] (0,5)--(1,5);
\draw[red] (0,0)--(2,0)--(2,4)--(0,4)--(0,0);
\node(d) at (0,0){\textcolor{black}{$\bullet$}};
\node(f4) at (2,4){\textcolor{red}{$\bullet$}};
\node(k) at (-1,3){$k$};
\draw[<->] (-.5,2)--(-.5,4);
\node(p) at (-1,1){$j$};
\draw[<->] (-.5,0)--(-.5,2);
\draw[<->] (0,-.5)--(6,-.5);
\draw (-1,2)--(6,2);
\node(n) at (3,-1){$n-1$};
\node(x1) at (3.5,2.5){X};
\node(x2) at (2.5,3.5){X};
\node(g) at (5,4){\textcolor{red!55}{$g_j(k)$}};
\draw[->] (g)--(3.75,1.5);
\end{tikzpicture}
\caption{Representation of the term $\left[{\genfrac{}{}{0pt}{}{n-1}{j+k}}\right]_qq^{-k+g_j(k)}$ in \eqref{eq : alt->pos}}
\label{fig : g_j(k)}
\end{minipage}
\begin{minipage}{8.4cm}
\centering
\begin{tikzpicture}[scale=.5]
\filldraw[blue!20] (1,0)--(2,0)--(2,5)--(1,5)--(1,0);
\filldraw[pink] (2,0)--(2,4)--(3,4)--(3,3)--(4,3)--(4,2)--(5,2)--(5,1)--(6,1)--(6,0)--(2,0);
\draw[thin, gray] (0,0)--(0,6)--(1,6)--(1,5)--(2,5)--(2,4)--(3,4)--(3,3)--(4,3)--(4,2)--(5,2)--(5,1)--(6,1)--(6,0)--(0,0);
\draw[thin, gray] (1,0)--(1,5);
\draw[thin, gray] (2,0)--(2,4);
\draw[thin, gray] (3,0)--(3,3);
\draw[thin, gray] (4,0)--(4,2);
\draw[thin, gray] (5,0)--(5,1);
\draw[thin, gray] (0,1)--(5,1);
\draw[thin, gray] (0,2)--(4,2);
\draw[thin, gray] (0,3)--(3,3);
\draw[thin, gray] (0,4)--(2,4);
\draw[thin, gray] (0,5)--(1,5);
\node(d) at (0,0){\textcolor{black}{$\bullet$}};
\node(f4) at (2,4){\textcolor{red}{$\bullet$}};
\draw[very thick,red] (1,4)--(2,4);
\draw[very thick,blue] (1,4)--(1,5);
\node(f5) at (1,5){\textcolor{blue}{$\bullet$}};
\node(k) at (-1,3){$k$};
\draw[<->] (-.5,2)--(-.5,4);
\node(p) at (-1,1){$j$};
\draw[<->] (-.5,0)--(-.5,2);
\draw (-1,2)--(6,2);
\node(x1) at (1.5,4.5){\textcolor{gray}{X}};
\node(x2) at (2.5,3.5){X};
\node(x3) at (3.5,2.5){X};
\end{tikzpicture}

\caption{Comparing a path ending with a North step in $\mathcal{C}^{n-1}_{j+k+1}$ and a path with an east step in $\mathcal{C}^{n-1}_{j+k}$}
\label{fig : boxes}
\end{minipage}
\end{figure}


%%%%%%%%%%%%%%   Formula

\section{The formula}\label{sec : main}

First let us recall that the dual Pieri rule describes the multiplication of a Schur function by $e_k$. The adjoint of the dual Pieri rule for the Hall scalar product, denoted $e_k^\perp$, is defined on the Schur basis and extended linearly. More precisely, $e_k^\perp s_\lambda$ is the sum  $\sum_\mu s_\mu$ over all partitions $\mu$ obtained by deleting $k$ boxes each lying in a different row (see \cref{fig : pieri}).


\begin{figure}[ht!]
\centering
\begin{tikzpicture}[scale=.5]
\draw (0,0)--(0,4)--(1,4)--(1,2)--(3,2)--(3,1)--(4,1)--(4,0)--(0,0);
\draw (0,1)--(4,1);
\draw (0,2)--(3,2);
\draw (0,3)--(1,3);
\draw (1,0)--(1,4);
\draw (2,0)--(2,2);
\draw (3,0)--(3,1);
\node(e) at (-1,2){$e_2^\perp$};
\end{tikzpicture}
\begin{tikzpicture}[scale=.5]
\draw (0,0)--(0,4)--(1,4)--(1,2)--(2,2)--(2,1)--(3,1)--(3,0)--(0,0);
\draw (0,1)--(3,1);
\draw (0,2)--(2,2);
\draw (0,3)--(1,3);
\draw (1,0)--(1,4);
\draw (2,0)--(2,2);
\draw (3,0)--(3,1);
\node(e) at (-1,2){$=$};
\end{tikzpicture}
\begin{tikzpicture}[scale=.5]
\draw (0,0)--(0,3)--(1,3)--(1,2)--(3,2)--(3,0)--(0,0);
\draw (0,1)--(3,1);
\draw (0,2)--(3,2);
\draw (1,0)--(1,3);
\draw (2,0)--(2,2);
\draw (3,0)--(3,1);
\node(e) at (-1,2){$+$};
\end{tikzpicture}
\begin{tikzpicture}[scale=.5]
\draw (0,0)--(0,3)--(1,3)--(1,2)--(2,2)--(2,1)--(4,1)--(4,0)--(0,0);
\draw (0,1)--(4,1);
\draw (0,2)--(2,2);
\draw (1,0)--(1,3);
\draw (2,0)--(2,2);
\draw (3,0)--(3,1);
\node(e) at (-1,2){$+$};
\end{tikzpicture}
\begin{tikzpicture}[scale=.5]
\draw (0,0)--(0,2)--(3,2)--(3,1)--(4,1)--(4,0)--(0,0);
\draw (0,1)--(4,1);
\draw (0,2)--(3,2);
\draw (1,0)--(1,2);
\draw (2,0)--(2,2);
\draw (3,0)--(3,1);
\node(e) at (-1,2){$+$};
\end{tikzpicture}
\caption{Example: $e_2^\perp s_{4311}=s_{3211}+s_{331}+s_{421}+s_{43}$}
\label{fig : pieri}

\end{figure}

For the following lemma we first consider the application $\psi:\Lambda\rightarrow \mathbb{Q}[q,t]$ which is defined on the Schur basis by $\psi(s_\lambda)=q^{\lambda_1}t^{\ell(\lambda)-1}$ and then extended linearly.  

The following lemma allows us to lift $\mathcal{E}^{\langle 2\rangle}_{m,n}$ to $\mathcal{E}_{m,n}$. Note that we can not obtain all the coefficients of $\mathcal{E}_{m,n}$ this way, but we know which are left out. If we consider the Schur decomposition of $\langle\mathcal{E}_{m,n},e_n\rangle$, the following lemma shows  how to obtain all the coefficients of the Schur functions indexed by partitions of shape  $(a,b+1,1^j)$, with $a$ and $j$ arbitrary and $b$ fixed. If $\sum_\lambda c_\lambda s_\lambda$ is a linear combination of Schur functions, and $\mathcal{V}$ is a set of shapes, then $\sum_\lambda c_\lambda s_\lambda|_\mathcal{V}=\sum_{\lambda \in \mathcal{V}} c_\lambda s_\lambda$. On a symmetric function in the Schur basis, we set the restriction $|_{1 \textrm{part}}$ (respectively  $|_{\hooks}$) to be the partial sum over the Schur functions indexed by partitions having only one part (respectively that are hook shaped). 
 
   \begin{lem}Let $b$ be a constant in $\mathbb{N}$ and $G \in \Lambda_Q$ be  a symmetric functions in the variables $Q=\{q_1,q_2,\ldots\}$. Let  $\mathcal{V}_b$ be the set of partitions of shape $(a,b,1^k)$ with $k$ and $a$ arbitrary. If $f_0(q)=G|_{\mathcal{V}_{b}}$ and
   \begin{equation*}
   f_i(q)=\psi((e_i^\perp G|_{\mathcal{V}_{b+1}})^{\langle 2\rangle}|_{\mathcal{V}_{b}})t^{-1},
   \end{equation*}  then:
   \begin{equation*} \psi(G|_{\mathcal{V}_{b+1}})(q,t)=\sum_{j\geq 1}\sum_{k=0}^j (-1)^k f_{j-k}(q)q^{-k}t^j
   \end{equation*}
 \end{lem}
 Note that when $b\geq2$, no formula is known for $\langle \nabla(e_n), s_{k,1^{n-k}}\rangle|_{\mathcal{V}_b}$ at this moment. Given this formula, the lemma gives a way to find the formulas for $\langle \mathcal{E}_n, s_{1^n}\rangle|_{\mathcal{V}_{b+1}}$.  
 
 We should also notice that $ \psi(G|_{\hooks})(q,t)$ is the sum of the restriction to $\mathcal{V}_1$ and the restriction to one parts ($|_{\mathcal{V}_0}$). This is the reason why, in \cref{sec : tn}, we used the convention that $T_{n,s}=\{N^{n-2}\}$ if $s>n-2$. Since the path $N^{n-2}$ relates to the restriction to one part.


We can now prove the main theorem piece by piece. Using the combinatorics of $m$-Schr\"oder paths, we found that for  $\psi( \langle\nabla^r(s_{1^n}), s_{j+1,1^{n-j}}\rangle|_{1 \pp})$ we have:
 \begin{equation*}f_j^{(r)}(q)=q^{r\binom{n}{2}-\binom{j}{2}}\begin{bmatrix}n-1\\ j \end{bmatrix}_{q^{-1}}
 \end{equation*}
 We can prove the next proposition with this, the previous lemma and a theorem in \cite{[B2018]} stating that $ \langle\mathcal{E}_{n,n}, s_{j+1,1^{n-j}}\rangle=e^\perp_j \langle\mathcal{E}_{n,n}, s_{1^{n}}\rangle$.
 \begin{prop}For $r=1$ we have:
\begin{equation}\label{eq : rn}\langle \mathcal{E}_{rn,n}, e_n\rangle|_{\hooks}= \sum_{\gamma \in T_{n}} s_{\hook1(\gamma)}
\end{equation}
where $\hook1(\gamma)$ is the partition $\left((r-1)\binom{n}{2}+\A(\gamma)+ht(\gamma)+1,1^{n-2-ht(\gamma) }\right)$.

Moreover, for all $r\in \mathbb{N}^*$ such that the equality $ \langle\mathcal{E}_{rn,n}, s_{j+1,1^{n-j}}\rangle=e^\perp_j \langle\mathcal{E}_{rn,n}, s_{1^{n}}\rangle$ is true, then \eqref{eq : rn} holds. 
\end{prop}

\cref{fig : E_4} gives an example for $r=1$ and $n=4$. One might want to take note that for $n=4$ there are only hooks so $\langle \mathcal{E}_{4,4}, e_4\rangle|_{hooks}=\langle \mathcal{E}_{4,4}, e_4\rangle$. This is why we can state in the introduction that \eqref{eq : lift delta} specializes to the equation of theorem 3.2.5 in \cite{[BCP2018]} using a different basis.

\begin{figure}[ht!]
\centering
\begin{tikzpicture}[scale=.5]
\node(w) at (-5.5,1){The elements of $T_{2,0}$: };
\node(a) at (-4.5,-1){$\A(\gamma):$ };
\node(h) at (-4.5,-2){$ht(\gamma): $};
\node(e) at (-4.5,-3){$hook(\gamma):$ };
\node(e) at (-4.5,-4){$s_{hook(\gamma)}:$ };
\filldraw[pink] (0,0)--(0,2)--(1,2)--(1,1)--(2,1)--(2,0)--(0,0);
\draw[thin,gray] (0,0)--(0,2)--(1,2)--(1,1)--(2,1)--(2,0)--(0,0);
\draw[thin,gray] (1,0)--(1,1)--(0,1);
\draw[blue, thick]  (0,0)--(0,2);
\node(f6) at (0,2){\textcolor{red}{$\bullet$}};
\node(d) at (0,0){\textcolor{blue}{$\bullet$}};
\node(a1) at (1,-1){$3$};
\node(h1) at (1,-2){$2$};
\node(H1) at (1,-3){$(6,1^{0})$};
\node(s) at (1,-4){$s_6$};
\end{tikzpicture}
\begin{tikzpicture}[scale=.5]
\node(s) at (-.5,.5){,};
\filldraw[pink] (0,0)--(0,1)--(2,1)--(2,0)--(0,0);
\draw[thin,gray] (0,0)--(0,2)--(1,2)--(1,1)--(2,1)--(2,0)--(0,0);
\draw[thin,gray] (1,0)--(1,1)--(0,1);
\draw[blue, thick]  (0,0)--(0,1)--(1,1);
\node(f6) at (1,1){\textcolor{red}{$\bullet$}};
\node(d) at (0,0){\textcolor{blue}{$\bullet$}};
\node(a1) at (1,-1){$2$};
\node(h1) at (1,-2){$1$};
\node(H1) at (1,-3){$(4,1^{1})$};
\node(s) at (1,-4){$s_{41}$};
\end{tikzpicture}
\begin{tikzpicture}[scale=.5]
\node(w) at (-.5,.5){,};
\filldraw[pink] (1,0)--(1,1)--(2,1)--(2,0)--(1,0);
\draw[thin,gray] (0,0)--(0,2)--(1,2)--(1,1)--(2,1)--(2,0)--(0,0);
\draw[thin,gray] (1,0)--(1,1)--(0,1);
\draw[blue, thick]  (0,0)--(1,0)--(1,1);
\node(f6) at (1,1){\textcolor{red}{$\bullet$}};
\node(d) at (0,0){\textcolor{blue}{$\bullet$}};
\node(a1) at (1,-1){$1$};
\node(h1) at (1,-2){$1$};
\node(H1) at (1,-3){$(3,1^{1})$};
\node(s) at (1,-4){$s_{31}$};
\end{tikzpicture}
\begin{tikzpicture}[scale=.5]
\node(w) at (-.5,.5){,};
\draw[thin,gray] (0,0)--(0,2)--(1,2)--(1,1)--(2,1)--(2,0)--(0,0);
\draw[thin,gray] (1,0)--(1,1)--(0,1);
\draw[blue, thick]  (0,0)--(2,0);
\node(f6) at (2,0){\textcolor{red}{$\bullet$}};
\node(d) at (0,0){\textcolor{blue}{$\bullet$}};
\node(a1) at (1,-1){$0$};
\node(h1) at (1,-2){$0$};
\node(H1) at (1,-3){$(1,1^{2})$};
\node(s) at (1,-4){$s_{111}$};
\end{tikzpicture}
  \begin{align*}\langle \mathcal{E}_{4,4}, e_4\rangle|_{hooks}&= s_{6}+s_{41}+s_{31}+s_{111}
 \end{align*}
\caption{Example of $\langle \mathcal{E}_{4,4}, e_4\rangle|_{hooks}$}
\label{fig : E_4}
\end{figure}


For those who are used to seeing $\langle \mathcal{E}^{\langle 2\rangle}_{n,n}, s_{\mu}\rangle=\langle \nabla(e_n), s_{\mu}\rangle$ in terms of Dyck paths and Schr\"oder paths note that each path in $T_{n,k}$ is associated to a subset of Schr\"oder paths.
The next proposition is obtained by restricting the formula to $\mathcal{E}^{\langle 2\rangle}_{n,n}$ and by using a bijection between a subset of Schr\"oder paths with $d$ diagonal steps and the set $\SYT(d+1,1^{n-d-1})\times\{ 1, 2, \ldots, n-d-1\}$. The subset is such that the paths end with a north step and the $\A$ statistic of that path is  equal to $1$.

 \begin{prop} If  $r=1$ and $\mu\in\{(k,1^{n-k})~|~1\leq k\leq n\}$ then we have:
  \begin{align}\label{eq : nabla}\langle \nabla^r(e_n), s_{\mu}\rangle|_{\hooks}=&\sum_{\tau\in \SYT(\mu)}  s_{r\binom{n}{2}-\m(\tau')}(q,t)+\sum_{i=2}^{\dd(\tau)} s_{r\binom{n}{2}-\m(\tau')-i,1}(q,t).
  \end{align}

Additionally, for all $k$ we have: 
\begin{equation}\label{eq : delta en}\langle \Delta'_{e_k}(e_n), e_n\rangle|_{\hooks}=\sum_{\tau\in \SYT((n-k,1^k))}  s_{\m(\tau)}(q,t)+\sum_{i=2}^{k} s_{\m(\tau)-i,1}(q,t).
\end{equation}

Furthermore for all $\mu\in\{(n), (n-1,1), (n-2,1,1),(1^n)\}$, $0\leq k < n-1$ we have:
\begin{equation}\label{eq : delta mu}e_{n-k-1}^\perp\left(\langle \mathcal{E}_{n,n},s_\mu\rangle|_{\hooks}\right)^{\langle 2\rangle}=\sum_{\tau\in \SYT(\mu)}\sum s_{(k-1+\A(\gamma)-\m(\tau'),1)}+\sum s_{(k+\A(\gamma)-\m(\tau'))}.
\end{equation}
Where the second sum is over paths in $T_{n,\dd(\tau')}$ of height $k-2$ and $k-1$  and the third sum is over paths in $T_{n,\dd(\tau')}$ of height $k-1$ and $k$.

Finally, if $t=0$ \eqref{eq : delta mu} holds for all $\mu$ and \eqref{eq : nabla} holds for all $r$ whenever $\mu$ is hook shaped. Equation \eqref{eq : delta en} already holds for all $q$ and $t$.
\end{prop}

This proposition proves that the main theorem holds in the case where $\mu$ is of shapes $(n-2,1,1)$, $(n-1,1)$ or $(n)$. This is because the height of the hooks is bounded by $2$ in these cases. Note that, when $\mu\not=1^n$ and $r>1$, computer experiments indicate that some hook shaped terms are not accounted for by \eqref{eq : lift delta} of \cref{thm : main}. 


It is shown in \cite{[B2018]} that $ \langle\mathcal{E}_{n,n}, s_{(k+1,1^{n-k-1})}\rangle=e^\perp_k \langle\mathcal{E}_{n,n}, s_{1^{n}}\rangle$. By \cref{prop : perp sur chemin} this can be done directly in terms of paths. It also supports the assumption that \eqref{eq : lift delta} is true for all hooked shape $\mu$ when $r=1$.

 Lets consider the three following sets of paths:
\begin{equation*}\bigcup_{\tau\in \SYT((k+1,1^{n-k-1}))}\{\gamma=N^jE\gamma'\in T_{n-2,\dd(\tau')}~|~j\geq n-k-\textrm{min}( \DD(\tau'))\}=T_{n,k}^+
\end{equation*}
 \begin{equation*}\bigcup_{\tau\in \SYT((k+1,1^{n-k-1}))}\{\gamma=N^jE\gamma'\in T_{n-2,\dd(\tau')}~|~j <n-k- \textrm{min}( \DD(\tau'))\}=T_{n,k}^-
 \end{equation*}
 and
\begin{equation*}T_n^k=\{\gamma\in T_{n,0} ~| ~ n-2-\h(\gamma)\geq k\}.
\end{equation*}
 
   For $k$ between $1$ and $n-1$, we will now define two families $\{\underline{e_{k-}^\perp}\}$ and $\{\underline{e_{k+}^\perp}\}$ of maps:
\begin{equation*} \underline{e_{k-}^\perp}:T_n^{k-1}\backslash\{E^{n-2}\} \rightarrow T_{n,k}^-~ \text{ and }~ \underline{e_{k+}^\perp}:T_n^{k} \rightarrow T_{n,k}^+
\end{equation*}

For $\gamma\in T_{n,0}$, lets consider the prefix of $\gamma$ ending with the $k$-th east step. The prefix exists by definition of $T_n^k$. Let  $p_1,\ldots,p_k$ denote the integers such that $p_i$ is the number of north steps before the $i$-th east step. To this we associate $\tau'$, the standard Young tableau such that $\DD(\tau')=\{ n-2-i+1-p_i-1~|~1\leq i\leq k\}$. Then $\underline{e_{k+}^\perp}(\gamma)$ is the path in $T_{n,\dd(\tau')}$ given by discarding all the $k$ first east steps of $\gamma$. (See \cref{fig : perp colonne})



\begin{figure}[ht!]
\begin{minipage}{7cm}
\begin{tikzpicture}[scale=.4]
\filldraw[pink] (-1,0)--(-1,3)--(0,3)--(0,0)--(-1,0);
\filldraw[pink]  (0,0)--(0,1)--(1,1)--(1,2)--(2,2)--(3,2)--(3,3)--(5,3)--(6,3)--(6,2)--(7,2)--(7,1)--(8,1)--(8,0)--(0,0);
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\draw[thin, gray] (5,0)--(5,3);
\draw[thin, gray] (6,0)--(6,2);
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\draw[thin, gray] (0,3)--(5,3);
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\end{minipage}
\begin{minipage}{8cm}
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\draw[thin, gray] (1,2)--(1,7);
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\draw[thin, gray] (3,2)--(3,5);
\draw[thin, gray] (4,2)--(4,4);
\draw[thin, gray] (5,2)--(5,3);
\draw[thin, gray] (6,1)--(6,2);
\draw[thin, gray] (7,0)--(7,1);
\draw[thin, gray] (5,1)--(7,1);
\draw[thin, gray] (0,2)--(6,2);
\draw[thin, gray] (0,3)--(5,3);
\draw[thin, gray] (0,4)--(4,4);
\draw[thin, gray] (0,5)--(3,5);
\draw[thin, gray] (0,6)--(2,6);
\draw[thin, gray] (0,7)--(1,7);
\draw[blue, thick]  (0,2)--(0,4)--(1,4)--(1,5)--(3,5);
\draw[pink] (-1,0)--(7,0)--(7,1)--(5,1)--(5,2)--(-1,2)--(-1,0);
\node(t) at (2,1){$\m(\tau')$};
\node(d) at (0,2){\textcolor{blue}{$\bullet$}};
\node(f1) at (8,0){\textcolor{red}{$\bullet$}};
\node(f2) at (7,1){\textcolor{red}{$\bullet$}};
\node(f3) at (6,2){\textcolor{red}{$\bullet$}};
\node(f4) at (5,3){\textcolor{red}{$\bullet$}};
\node(f5) at (4,4){\textcolor{red}{$\bullet$}};
\node(f6) at (3,5){\textcolor{red}{$\bullet$}};
\node(f4) at (2,6){\textcolor{red}{$\bullet$}};
\node(f5) at (1,7){\textcolor{red}{$\bullet$}};
\node(f6) at (0,8){\textcolor{red}{$\bullet$}};
\node(gamma) at (-4,4){$\underline{e_{2+}^\perp}(\gamma)=$};
\end{tikzpicture}
\end{minipage}
\caption{For $NENEENEE \in T_{9,0}$, ~$\underline{e_{2+}^\perp}(NENEENEE)=NNENEE\in T_{9,\dd(\tau')}$, where $\DD(\tau')=\{6,8\}$ }\label{fig : perp colonne}
\end{figure}

The map $\underline{e_{k-}^\perp}(\gamma)$ is defined in a similar way. We denote by $p_1,\ldots,p_{k-1}$ the integers such that $p_i$ is the number of north steps before the $i$-th east step. Let $h$ be the number of east steps before the first north step. We choose $\tau'$ to be the standard Young tableau such that $\DD(\tau')=\{ n-2-i+1-p_i+1~|~1\leq i\leq k-1\}\cup\{\textrm{max}(1,h-k+2)\}$. Then $\underline{e_{k-}^\perp}(\gamma)$ is the path in $T_{n,\dd(\tau')}$ given by discarding all the $k-1$ first east steps of and the first north step $\gamma$. (See \cref{fig : perp 1ere ligne})

\begin{figure}[ht!]
\begin{minipage}{7cm}
\begin{tikzpicture}[scale=.4]
\filldraw[pink] (-1,0)--(-1,3)--(0,3)--(0,0)--(-1,0);
\filldraw[pink]  (0,0)--(0,1)--(1,1)--(1,2)--(2,2)--(3,2)--(3,3)--(5,3)--(6,3)--(6,2)--(7,2)--(7,1)--(8,1)--(8,0)--(0,0);
\node(h) at (-.5,1.5){$\h$};
\draw[thin, gray] (0,0)--(0,8)--(1,8)--(1,7)--(2,7)--(2,6)--(3,6)--(3,5)--(4,5)--(4,4)--(5,4)--(5,3)--(6,3)--(6,2)--(7,2)--(7,1)--(8,1)--(8,0)--(0,0);
\draw[thin, gray] (1,0)--(1,7);
\draw[thin, gray] (2,0)--(2,6);
\draw[thin, gray] (3,0)--(3,5);
\draw[thin, gray] (4,0)--(4,4);
\draw[thin, gray] (5,0)--(5,3);
\draw[thin, gray] (6,0)--(6,2);
\draw[thin, gray] (7,0)--(7,1);
\draw[thin, gray] (0,1)--(7,1);
\draw[thin, gray] (0,2)--(6,2);
\draw[thin, gray] (0,3)--(5,3);
\draw[thin, gray] (0,4)--(4,4);
\draw[thin, gray] (0,5)--(3,5);
\draw[thin, gray] (0,6)--(2,6);
\draw[thin, gray] (0,7)--(1,7);
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\draw (0,-1)--(0,9);
\node(d) at (0,0){\textcolor{blue}{$\bullet$}};
\node(f1) at (8,0){\textcolor{red}{$\bullet$}};
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\node(f3) at (6,2){\textcolor{red}{$\bullet$}};
\node(f4) at (5,3){\textcolor{red}{$\bullet$}};
\node(f5) at (4,4){\textcolor{red}{$\bullet$}};
\node(f6) at (3,5){\textcolor{red}{$\bullet$}};
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\node(f5) at (1,7){\textcolor{red}{$\bullet$}};
\node(f6) at (0,8){\textcolor{red}{$\bullet$}};
\node(gamma) at (-3,4){$\gamma=$};
\end{tikzpicture}
\end{minipage}
\begin{minipage}{8cm}
\begin{tikzpicture}[scale=.4]
\filldraw[pink] (-1,2)--(-1,4)--(0,4)--(0,2)--(-1,2);
\node(h) at (-.5,3){$\h$};
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\draw[thin, gray] (0,2)--(0,8)--(1,8)--(1,7)--(2,7)--(2,6)--(3,6)--(3,5)--(4,5)--(4,4)--(5,4)--(5,3)--(6,3)--(6,2)--(7,2)--(7,1)--(8,1)--(8,0)--(7,0)--(7,1)--(0,1)--(0,2);
\draw[thin, gray] (1,1)--(1,7);
\draw[thin, gray] (2,1)--(2,6);
\draw[thin, gray] (3,1)--(3,5);
\draw[thin, gray] (4,1)--(4,4);
\draw[thin, gray] (5,1)--(5,3);
\draw[thin, gray] (6,1)--(6,2);
\draw[thin, gray] (7,0)--(7,1);
\draw[thin, gray] (5,1)--(7,1);
\draw[thin, gray] (0,2)--(6,2);
\draw[thin, gray] (0,3)--(5,3);
\draw[thin, gray] (0,4)--(4,4);
\draw[thin, gray] (0,5)--(3,5);
\draw[thin, gray] (0,6)--(2,6);
\draw[thin, gray] (0,7)--(1,7);
\draw[blue, thick]  (0,2)--(0,3)--(2,3)--(2,4)--(4,4);
\draw[pink] (-1,0)--(7,0)--(7,1)--(5,1)--(5,2)--(-1,2)--(-1,0);
\node(t) at (2,.5){$\m(\tau')$};
\node(d) at (0,2){\textcolor{blue}{$\bullet$}};
\node(f1) at (8,0){\textcolor{red}{$\bullet$}};
\node(f2) at (7,1){\textcolor{red}{$\bullet$}};
\node(f3) at (6,2){\textcolor{red}{$\bullet$}};
\node(f4) at (5,3){\textcolor{red}{$\bullet$}};
\node(f5) at (4,4){\textcolor{red}{$\bullet$}};
\node(f6) at (3,5){\textcolor{red}{$\bullet$}};
\node(f4) at (2,6){\textcolor{red}{$\bullet$}};
\node(f5) at (1,7){\textcolor{red}{$\bullet$}};
\node(f6) at (0,8){\textcolor{red}{$\bullet$}};
\node(gamma) at (-4,4){$\underline{e_{2-}^\perp}(\gamma)=$};
\end{tikzpicture}
\end{minipage}
\caption{For $NENEENEE \in T_{9,0}$, ~$\underline{e_{2-}^\perp}(NENEENEE)=NNEENEE \in T_{9,\dd(\tau')}$, where $\DD(\tau')=\{1,8\}$}
\label{fig : perp 1ere ligne}
\end{figure}
We will recall that for $r=1$ the hooks in \cref{thm : main} are given by: 
\begin{equation*}\hook(\gamma)=\left(\A(\gamma)+\h(\gamma)-\m(\tau')+1,1^{n-2-\h(\gamma) }\right)
\end{equation*}



\begin{lem} For all $k$ the maps $\underline{e_{k+}^\perp}$  is a well-defined bijection such that
\begin{equation*}\hook(\underline{e_{k+}^\perp}(\gamma))=\left(\A(\gamma)+\h(\gamma)+1,1^{n-2-\h(\gamma) -k}\right).
\end{equation*}
For all $k$ the map $\underline{e_{k-}^\perp}$ is a well defined  injective map such that
\begin{equation*}\hook(\underline{e_{k-}^\perp}(\gamma))=\left(\A(\gamma)+\h(\gamma),1^{n-1-\h(\gamma) -k}\right).
\end{equation*}
Furthermore,
\begin{equation*}e_k^\perp(s_{\hook(\gamma)})=s_{\hook(\underline{e_{k+}^\perp}(\gamma))}+s_{\hook(\underline{e_{k-}^\perp}(\gamma))}
\end{equation*}

\end{lem}


For the next proposition we extend our maps so that $\underline{e_{k+}^\perp}(\gamma)=\emptyset$, if $\gamma\in T_{n,0}\backslash T_n^k$, $\underline{e_{k-}^\perp}(\gamma)=\emptyset$ if $\gamma\in T_{n,0}\backslash T_n^{k-1}$ and $s_{\hook(\emptyset)}=0$.
\begin{prop}\label{prop : perp sur chemin} For all $k$, we have: 
\begin{equation*}e_k^\perp\left(\langle\mathcal{E}_{n,n}, e_n \rangle|_{\hooks}\right)=\sum_{\gamma \in T_{n,0}} s_{\hook(\underline{e_{k+}^\perp}(\gamma))}+s_{\hook(\underline{e_{k-}^\perp}(\gamma))},
\end{equation*}
where $\hook$ is defined as in \cref{thm : main}.

In addition, $\sum_{\tau\in \SYT(k+1,1^{n-k-1})} \sum_{\gamma \in T_{n, \dd(\tau')}} s_{\hook(\gamma)}-e_k^\perp\left(\langle\mathcal{E}_{n,n}, s_{1^{n}}\rangle|_{\hooks}\right)$ has a Schur positive expansion.
\end{prop}

This last proposition gives reason to believe that \eqref{eq : lift delta} holds for all $\mu$, a hook, since the missing terms should be obtained by the restriction to shapes having two columns. The following proposition starts the second column.

\begin{prop}
For $\mathcal{W}=\{(a,2,1^k)~|~ k\in\mathbb{N}, a\in \mathbb{N}_{\geq 2}\}$, we have:
\begin{equation*}\langle \mathcal{E}_n,e_n\rangle|_{\mathcal{W}}=\sum_{i=2}^{n-3}\sum s_{\shape U},
\end{equation*}
Where the second sum is over paths $\gamma\in T_{n,0}$ such that $\gamma\not=N\tilde{\gamma}N^{i-1}$ and $i\leq \h(\gamma)\leq n-3$. Additionally $\shape U$ is the partition given by $\left(\A(\gamma)+\h(\gamma)+1-i,2,1^{n-3-\h(\gamma)}\right)$.
\end{prop}
We haven't proven that the equation $e^\perp_1(\langle \mathcal{E}_n,s_{1^n}\rangle|_{\mathcal{W}})|_{\hooks}$ is equivalent to  the remainder of the remainder of $\sum_{\tau\in \SYT(2,1^{n-2})} \sum_{\gamma \in T_{n, \dd(\tau')}} s_{\hook(\gamma)}-e_1^\perp\left(\langle\mathcal{E}_{n,n}, s_{1^{n}}\rangle|_{\hooks}\right)$.  Such an equivalence would prove the case $\mu=(2,1^{n-2})$ of \eqref{eq : lift delta}.


%%%%%%%%%%%%%%   Conclusion

\section{Conclusion and further questions}

It would be interesting to show that the formula hold for all $\mu$ when $r=1$ and all $r$ when $\mu=1^n$. Has we mentioned previously, this could be obtained, for $\mu$ a hook, by having a formula for the restriction to Schur's having two columns. Writing the $q,t$ statistics of the $m$-Schr\"oder paths in term of Schur functions would give rise to a more general formula. Finally, the author is already working on showing how existing formulas for special cases are equivalent to the proposed formula. 




\acknowledgements{I would like to thank  Fran\c cois Bergeron for making available to me draft of \cite{[B2018]} and related useful discussions. I would also like to than Aram Dermenjian, Pauline Hubert and M\'elodie Lapointe for proofreading and an anonymous reviewer for beneficial comments.}

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