\documentclass[finalversion]{FPSAC2020}
\articlenumber{8}
\addbibresource{fpsac.bib}

\usepackage[utf8]{inputenc}


\usepackage[labelfont={normalsize}]{caption,subfig}
\usepackage{units}
\usepackage{commath}

\usepackage{subfiles}

\usepackage[backgroundcolor=yellow]{todonotes}

\usepackage{tikz}
\usetikzlibrary{decorations.markings}

\definecolor{Set1-3-1}{RGB}{228,26,28}
\definecolor{Set1-3-A}{RGB}{228,26,28}
\definecolor{Set1-3-2}{RGB}{55,126,184}
\definecolor{Set1-3-B}{RGB}{55,126,184}
\definecolor{Set1-3-3}{RGB}{77,175,74}
\definecolor{Set1-3-C}{RGB}{77,175,74}

\theoremstyle{plain}
\newtheorem{thm}{Theorem}[section]
\newtheorem{lem}[thm]{Lemma}
\newtheorem{proposition}[thm]{Proposition}
\newtheorem{corollary}[thm]{Corollary}
\newtheorem{conjecture}[thm]{Conjecture}


\theoremstyle{definition}
\newtheorem{definition}[thm]{Definition}
\newtheorem{remark}[thm]{Remark}
\newtheorem{example}[thm]{Example}


\newcommand{\on}{\operatorname}
\newcommand{\pos}{\on{pos}}
\newcommand{\sh}{\on{sh}}
\newcommand{\Pp}{\mathbb{P}_N}
\newcommand{\Pt}{\widetilde{\mathbb{P}}_N}
\newcommand{\sq}{\square}
\newcommand{\ORD}{\widetildetableaux_{\sq_N}}

\newcommand{\E}{\mathbb{E}}
\newcommand{\N}{\mathbb{N}}
\newcommand{\R}{\mathbb{R}}
\newcommand{\C}{\mathbb{C}}
\newcommand{\Young}{\mathbb{Y}}

\newcommand{\tableaux}{\mathcal{T}}
\newcommand{\widetildetableaux}{\widetilde{\mathcal{T}}}

\newcommand{\tab}{T}
\newcommand{\multi}{M}

\newcommand{\ttt}{\mathbf{T}'_N}
\newcommand{\ttilde}{\mathbf{M}'_N}
\newcommand{\www}{\mathbf{W}'_N}

\newcommand{\water}{w}

\newcommand{\tim}{t}

\newcommand{\point}{P}
\newcommand{\Longitude}{\Psi}
\newcommand{\LongitudeTilde}{\widetilde{\Psi}}
\newcommand{\longitude}{\psi}
\newcommand{\pat}{\mathbf{q}}

\newcommand{\Sym}{\mathfrak{S}}
\newcommand{\mynumber}{n}
\newcommand{\mysquare}{[0,1]^2}
\newcommand{\mypoint}{X}
\newcommand{\JM}{Z}

\newcommand{\jdtcomplete}{J}
\newcommand{\pr}{\mu}

\title[Limit shapes of evacuation and jeu de taquin paths in random square tableaux]{Limit shapes  of evacuation and jeu de taquin paths in
	random square tableaux}

\author[\L ukasz Ma\'slanka and Piotr \'Sniady]{%
	\L ukasz Ma\'slanka%
	\thanks{\href{mailto:lmaslanka@impan.pl}{lmaslanka@impan.pl}.}%
	\addressmark{1} \and Piotr \'Sniady%
	\thanks{\href{mailto:psniady@impan.pl}{psniady@impan.pl}% \newline 
	%Supported by Grant 2017/26/A/ST1/00189 of Narodowe Centrum Nauki.}%
	.}
	\addressmark{1}}

\address{\addressmark{1} Institute of Mathematics, Polish Academy of Sciences, Warszawa, Poland}

\received{\today}

\abstract{We consider large random square Young tableaux and look for typical
	(in the sense of probability) \emph{jeu de taquin paths} and \emph{evacuation
		paths} in the asymptotic setting. 
	We show that the probability distribution of such paths converges to a random
	\emph{meridian} connecting the opposite corners of the square.}

\keywords{random Young tableaux, limit shape, jeu de taquin, Sch\"utzenberger's
	evacuation, square Young diagrams}

\begin{document}

\maketitle

\section{Introduction}

A full version of this extended abstract 
will be published in \cite{Maslanka2019}. 

\subsection{Notation}
For any $n \in \N$ let $\Young_n$ denote the \emph{set of Young diagrams} with $n$ boxes.
By $\sq_N\in\Young_{N^2}$ we~denote \emph{the square Young diagram} of side $N$.
For a Young diagram $\lambda \in \Young_n$ let $|\lambda| := n$ be the \emph{size} of diagram $\lambda$ 
and by $\tableaux_\lambda$ denote the \emph{set of standard
	Young tableaux of~shape~$\lambda$}. 
On~the~set~$\tableaux_{\sq_N}$ of square
tableaux with $N^2$ boxes we consider \emph{the uniform probability
	measure}~$\Pp$. 

We draw Young tableaux in the \emph{French notation}. 
For a tableau $\tab$, by $\on{pos}_k(\tab) = (x_k, y_k)$ we denote the
\emph{Cartesian coordinates of entry~$k$}, i.e.~$x_k$ is its column and $y_k$
is its row. The~difference $u_k^\tab:=x_k- y_k$ will be called \emph{the
    $u$-coordinate} of the box $k$ (in the literature the name \emph{content} is
also used).

The irreducible representation of the symmetric group $\Sym_n$ which corresponds
to \mbox{$\lambda\in\Young_n$} will be denoted by $V^\lambda$
and its character by $\chi^\lambda$.

\subsection{Jeu de taquin, jeu de taquin path, evacuation path}

\bigskip

\textbf{\emph{Jeu de taquin}} is a transformation of tableaux that was
introduced by Sch\"utzenberger~\cite{Schuetzenberger1977}. It is a fundamental
tool for studying the combinatorics of Young tableaux,
Robinson--Schensted--Knuth (RSK) algorithm, and Littlewood--Richardson
coefficients 
\cite{Fulton1997}.

Jeu de taquin acts as follows (see~\cref{fig:jdt}\subref{fig:jdtA} and
\subref{fig:jdtB}): we start with the bottom-left corner box of the given
tableau $\tab$, we erase it and obtain a~\emph{hole} in its place. Then we look
at the neigboring two boxes: the one to the right and the one  above the hole,
and choose the smaller one. We slide the selected box into the hole; as a
result the hole moves one box to the right or up. We continue sliding  as long as there
is some box to right or up to the hole. The resulting tableau will be denoted
by $j(\tab)$. The zig-zag path traversed by~the~hole will be called
\textbf{\emph{jeu de taquin path}}.

For a given tableau $\tab\in \tableaux_\lambda$ with $n=|\lambda|$ boxes we may
iterate jeu de taquin $n$ times until we end with the empty tableau. During
each iteration the box with the biggest number $n$ either moves one node left
or down, or stays put. Its trajectory 
\begin{equation}
\label{eq:evacuation}
 \pos_n(\tab),\ 
\pos_n\big( j(\tab) \big),\ 
\pos_n\big( j^2(\tab) \big),\ \dots,\
\pos_n\big( j^{n-1}(\tab) \big) 
\end{equation}
will be called \textbf{\emph{evacuation path}}.

\begin{figure}[b]

    \centering
    \subfloat[]{
			\begin{tikzpicture}[scale=0.65]
					\fill[blue!20] (0,0) rectangle +(1,1);
					\fill[blue!20] (0,1) rectangle +(1,1);
					\fill[blue!20] (1,1) rectangle +(1,1);
					\fill[blue!20] (2,1) rectangle +(1,1);
					\fill[blue!20] (2,2) rectangle +(1,1);
					
					
					\begin{scope}
					\clip[](5,0) -- (5,1) -- (4,1) -- (4,2) -- (3,2) -- (3,3) -- (1,3) -- (1,4) -- (0,4) -- (0,0);
					\draw (0,0) grid (10,10);
					\end{scope}
					\draw[thick] (5,0) -- (5,1) -- (4,1) -- (4,2) -- (3,2) -- (3,3) -- (1,3) -- (1,4) -- (0,4) -- (0,0) -- cycle;
					\draw (0.5,0.5) node {$1$};
					\draw (0.5,1.5) node {$2$};
					\draw (1.5,0.5) node {$3$};
					\draw (2.5,0.5) node {$7$};
					\draw (3.5,0.5) node {$10$};
					\draw (4.5,0.5) node {$13$};
					\draw (1.5,1.5) node {$4$};
					\draw (0.5,2.5) node {$5$};
					\draw (2.5,1.5) node {$6$};
					\draw (1.5,2.5) node {$8$};
					\draw (0.5,3.5) node {$11$};
					\draw (2.5,2.5) node {$9$};
					\draw (3.5, 1.5) node {$12$};
									

					\end{tikzpicture}        
\label{fig:jdtA}
} 
\hfill
\subfloat[]{
			\begin{tikzpicture}[scale=0.65]
					\fill[blue!20] (0,0) rectangle +(1,1);
					\fill[blue!20] (0,1) rectangle +(1,1);
					\fill[blue!20] (1,1) rectangle +(1,1);
					\fill[blue!20] (2,1) rectangle +(1,1);
					
					\begin{scope}
					\clip[](5,0) -- (5,1) -- (4,1) -- (4,2) -- (2,2) -- (2,3) -- (1,3) -- (1,4) -- (0,4) -- (0,0);
					\draw (0,0) grid (10,10);
					\end{scope}
					\draw[thick] (5,0) -- (5,1) -- (4,1) -- (4,2) -- (2,2) -- (2,3) -- (1,3) -- (1,4) -- (0,4) -- (0,0) -- cycle;
					\draw (0.5,0.5) node {$2$};
					\draw (0.5,1.5) node {$4$};
					\draw (1.5,0.5) node {$3$};
					\draw (2.5,0.5) node {$7$};
					\draw (3.5,0.5) node {$10$};
					\draw (4.5,0.5) node {$13$};
					\draw (1.5,1.5) node {$6$};
					\draw (0.5,2.5) node {$5$};
					\draw (2.5,1.5) node {$9$};
					\draw (1.5,2.5) node {$8$};
					\draw (0.5,3.5) node {$11$};
					\draw (3.5, 1.5) node {$12$};
			\end{tikzpicture}
			\label{fig:jdtB}
		}        
\hfill
\subfloat[]{ 
	\begin{tikzpicture}[scale=0.65]
	\fill[blue!20] (0,0) rectangle +(1,1);
	\fill[blue!20] (0,1) rectangle +(1,1);
	\fill[blue!20] (1,1) rectangle +(1,1);
	\fill[blue!20] (2,1) rectangle +(1,1);
	\fill[blue!40, green!60] (2,2) rectangle +(1,1);
	
	
	\begin{scope}
	\clip[](5,0) -- (5,1) -- (4,1) -- (4,2) -- (3,2) -- (3,3) -- (1,3) -- (1,4) -- (0,4) -- (0,0);
	\draw (0,0) grid (10,10);
	\end{scope}
	\draw[thick] (5,0) -- (5,1) -- (4,1) -- (4,2) -- (3,2) -- (3,3) -- (1,3) -- (1,4) -- (0,4) -- (0,0) -- cycle;
	\draw (0.5,0.5) node {$1$};
	\draw (0.5,1.5) node {$3$};
	\draw (1.5,0.5) node {$2$};
	\draw (2.5,0.5) node {$6$};
	\draw (3.5,0.5) node {$9$};
	\draw (4.5,0.5) node {$12$};
	\draw (1.5,1.5) node {$5$};
	\draw (0.5,2.5) node {$4$};
	\draw (2.5,1.5) node {$8$};
	\draw (1.5,2.5) node {$7$};
	\draw (0.5,3.5) node {$10$};
	\draw (3.5, 1.5) node {$11$};
	\draw (2.5, 2.5) node {$13$};
	\end{tikzpicture}        
	\label{fig:jdtC} } \caption{\protect\subref{fig:jdtA} Standard Young tableau
$\tab$ of shape $\lambda$. The highlighted boxes form the~\emph{jeu de taquin
    path}. \protect\subref{fig:jdtB} The outcome $j(\tab)$ of jeu de taquin
transformation. \protect\subref{fig:jdtC} A bijection on $\tableaux_\lambda$,
\emph{modified jeu de taquin $\jdtcomplete(\tab)$}, is defined by creating an
additional box with~entry~\mbox{$|\lambda|+1$} at~the~final position of the
hole and then decreasing all entries by 1.} \label{fig:jdt}
\end{figure}

\subsection{The problem}
\label{sec:problems}

If we draw the boxes of a given square tableau $\tab_N\in \tableaux_{\sq_N}$ as little squares of
size $\frac{1}{N}$ then both the corresponding jeu de taquin path as well as the
evacuation path is a zig-zag line connecting the opposite corners of the unit square $[0,1]^2$.
We consider the case when $\tab_N\in \tableaux_{\sq_N}$ is a \emph{random} square tableau
sampled with the uniform distribution~$\Pp$. 
\textbf{The~goal of the current
	paper is to find asymptotics of the corresponding scaled random zig-zag
	lines in the limit as $N\to\infty$, see \cref{fig:simulation-jdt}.}

\begin{figure}
\centering
\hfill
\subfloat[]{
\subfile{evacuation-figures/kwadrat-jdt-duzo-100.tex}

\label{fig:simulation-jdt}
}
\hfill
\subfloat[]{
	\begin{tikzpicture}[scale=0.45]
	\tiny
	\clip[](10,0) -- (10,10) -- (0,10) -- (0,0);
	\draw[black!20] (0,0) grid (10,10);
	\draw[Set1-3-3,ultra thick] plot[smooth,scale=10] file {circle025.txt};
	\draw[Set1-3-2,ultra thick] (0,10) -- (10,0);
	\draw[Set1-3-1,ultra thick] plot[smooth,scale=10] file {circle075.txt};
	
	\draw[line width=1mm,Set1-3-1,dashed,rounded corners](10,4) -- (8,4) -- (8,5) -- (7,5) -- (7,7) -- (6,7) -- (6,8) -- (5,8) -- (5,9) -- (2,9) -- (2,10); 
	\draw[line width=1mm,Set1-3-2,dashed,rounded corners](10,2) -- (8,2) -- (8,3) -- (7,3) -- (7,4) -- (5,4) -- (5,5) -- (4,5) -- (4,6) -- (2,6) -- (2,8) -- (1,8) -- (1,10); 
	\draw[line width=1mm,Set1-3-3,dashed,rounded corners](9,0) -- (9,1) -- (6,1) -- (6,2) -- (3,2) -- (3,4) -- (2,4) -- (2,5) -- (1,5) -- (1,7) -- (0,7); 
	\begin{scope}
	\end{scope}
	\draw (0.5,0.5) node {$1$};
	\draw (1.5,0.5) node {$2$};
	\draw (2.5,0.5) node {$3$};
	\draw (0.5,1.5) node {$4$};
	\draw (1.5,1.5) node {$5$};
	\draw (3.5,0.5) node {$6$};
	\draw (0.5,2.5) node {$7$};
	\draw (4.5,0.5) node {$8$};
	\draw (5.5,0.5) node {$9$};
	\draw (0.5,3.5) node {$10$};
	\draw (1.5,2.5) node {$11$};
	\draw (2.5,1.5) node {$12$};
	\draw (2.5,2.5) node {$13$};
	\draw (1.5,3.5) node {$14$};
	\draw (2.5,3.5) node {$15$};
	\draw (3.5,1.5) node {$16$};
	\draw (0.5,4.5) node {$17$};
	\draw (4.5,1.5) node {$18$};
	\draw (5.5,1.5) node {$19$};
	\draw (1.5,4.5) node {$20$};
	\draw (6.5,0.5) node {$21$};
	\draw (7.5,0.5) node {$22$};
	\draw (0.5,5.5) node {$23$};
	\draw (8.5,0.5) node {$24$};
	\draw (0.5,6.5) node {$25$};
	\draw (0.5,7.5) node {$26$};
	\draw (3.5,2.5) node {$27$};
	\draw (4.5,2.5) node {$28$};
	\draw (0.5,8.5) node {$29$};
	\draw (3.5,3.5) node {$30$};
	\draw (6.5,1.5) node {$31$};
	\draw (2.5,4.5) node {$32$};
	\draw (4.5,3.5) node {$33$};
	\draw (5.5,2.5) node {$34$};
	\draw (6.5,2.5) node {$35$};
	\draw (1.5,5.5) node {$36$};
	\draw (1.5,6.5) node {$37$};
	\draw (3.5,4.5) node {$38$};
	\draw (2.5,5.5) node {$39$};
	\draw (3.5,5.5) node {$40$};
	\draw (5.5,3.5) node {$41$};
	\draw (1.5,7.5) node {$42$};
	\draw (7.5,1.5) node {$43$};
	\draw (0.5,9.5) node {$44$};
	\draw (8.5,1.5) node {$45$};
	\draw (9.5,0.5) node {$46$};
	\draw (6.5,3.5) node {$47$};
	\draw (4.5,4.5) node {$48$};
	\draw (7.5,2.5) node {$49$};
	\draw (9.5,1.5) node {$50$};
	\draw (4.5,5.5) node {$51$};
	\draw (8.5,2.5) node {$52$};
	\draw (2.5,6.5) node {$53$};
	\draw (9.5,2.5) node {$54$};
	\draw (5.5,4.5) node {$55$};
	\draw (2.5,7.5) node {$56$};
	\draw (3.5,6.5) node {$57$};
	\draw (7.5,3.5) node {$58$};
	\draw (5.5,5.5) node {$59$};
	\draw (4.5,6.5) node {$60$};
	\draw (6.5,4.5) node {$61$};
	\draw (8.5,3.5) node {$62$};
	\draw (1.5,8.5) node {$63$};
	\draw (5.5,6.5) node {$64$};
	\draw (3.5,7.5) node {$65$};
	\draw (4.5,7.5) node {$66$};
	\draw (2.5,8.5) node {$67$};
	\draw (3.5,8.5) node {$68$};
	\draw (1.5,9.5) node {$69$};
	\draw (4.5,8.5) node {$70$};
	\draw (5.5,7.5) node {$71$};
	\draw (6.5,5.5) node {$72$};
	\draw (9.5,3.5) node {$73$};
	\draw (6.5,6.5) node {$74$};
	\draw (7.5,4.5) node {$75$};
	\draw (7.5,5.5) node {$76$};
	\draw (8.5,4.5) node {$77$};
	\draw (7.5,6.5) node {$78$};
	\draw (8.5,5.5) node {$79$};
	\draw (5.5,8.5) node {$80$};
	\draw (6.5,7.5) node {$81$};
	\draw (2.5,9.5) node {$82$};
	\draw (9.5,4.5) node {$83$};
	\draw (8.5,6.5) node {$84$};
	\draw (7.5,7.5) node {$85$};
	\draw (6.5,8.5) node {$86$};
	\draw (3.5,9.5) node {$87$};
	\draw (9.5,5.5) node {$88$};
	\draw (9.5,6.5) node {$89$};
	\draw (4.5,9.5) node {$90$};
	\draw (7.5,8.5) node {$91$};
	\draw (8.5,7.5) node {$92$};
	\draw (5.5,9.5) node {$93$};
	\draw (9.5,7.5) node {$94$};
	\draw (8.5,8.5) node {$95$};
	\draw (6.5,9.5) node {$96$};
	\draw (7.5,9.5) node {$97$};
	\draw (8.5,9.5) node {$98$};
	\draw (9.5,8.5) node {$99$};
	\draw (9.5,9.5) node {$100$};
	
	\draw[ultra thick](0,0) -- (0,10) -- (10,10) -- (10,0) -- cycle ;
	\end{tikzpicture}
\label{fig:level-curves}
}
\hfill \

	\caption{ \protect\subref{fig:simulation-jdt} Zigzag lines are sample jeu de
    taquin paths for random square tableaux of size $N=100$, 
    selected so they cross the anti-diagonal near the corresponding
    meridians (smooth thick curves) with the longitudes
    $\longitude\in\{\nicefrac{1}{10},\nicefrac{2}{10},\dots,\nicefrac{9}{10}\}$.
    The grey lines are the meridians with the longitudes
    $\longitude\in\{\nicefrac{2}{100},\nicefrac{4}{100},\dots,\nicefrac{98}{100}\}$. \protect\subref{fig:level-curves} Sample random square tableau of size $N=10$. Zigzag lines are the level curves for $\alpha\in\{ \nicefrac{1}{4} ,\nicefrac{1}{2}, \nicefrac{3}{4} \}$. The smooth lines are the corresponding circles of~latitude~$g_\alpha$.}
\end{figure}

\section{Asymptotics of evacuation paths}\label{sec:circles}

It is easy to change the definition of jeu de taquin so that
it becomes a permutation $\jdtcomplete\colon \tableaux_\lambda\to \tableaux_\lambda$ on the set
of standard tableaux of prescribed shape, see \cref{fig:jdtC}. 

We keep notations from \cref{sec:problems} and consider the evacuation path
\eqref{eq:evacuation} for $\tab=\tab_N$ and $n=N^2$. The position of each of
the boxes in the evacuation path coincides with the~position of a specific box
in the standard Young tableau obtained by iterating $\jdtcomplete$:
\begin{equation}
\label{eq:happy-box} 
\pos_{N^2}\big( j^{i}(\tab_N)\big)  = \pos_{N^2-i} \big( \jdtcomplete^{i}(\tab_N)\big).
\end{equation}
The latter standard tableaux $\jdtcomplete^i(\tab_N)$ is a
uniformly random square Young tableau.

Some light on the problem of finding the right-hand side of
\eqref{eq:happy-box} is provided by the~work of Pittel and~Romik; one of their results 
\cite[Theorem~2]{Pittel2007} gives an explicit family of~curves~$(g_\alpha)$
indexed by $\alpha\in[0,1]$ which fit inside the square $\mysquare$, see
\cref{fig:level-curves}. 
Each~curve~$g_\alpha$ turns out to be  one of the level
curves of (scaled down) random standard Young tableau of shape $\sq_N$ which
separates the boxes with entries $\leq \alpha N^2$ from the~boxes with entries
bigger than this threshold
(asymptotically, as $N\to\infty$, except for an~event of negligible probability).
We will refer to the curves $(g_\alpha)$ as \textbf{\emph{circles of latitude}}.

It follows that \textbf{asymptotically the (scaled down) position of the box
    \eqref{eq:happy-box} in the evacuation path is very close to some point on the
    circle of latitude $g_{1-\tim}$ where $\tim = \nicefrac{i}{N^2}$. \emph{The
        remaining difficulty is to pinpoint a specific location of this point on the
        curve}. For~this purpose we need a convenient parametrization on each circle
    of~latitude~$g_\alpha$.}

\subsection{Longitude. Geographic coordinates on the square}

The work of Pittel and Romik \cite[Theorem~2]{Pittel2007} additionally gives the
limit distribution of~the~(scaled down) position of the~box~$\lfloor \alpha
N^2\rfloor$ in a random square tableau with $N^2$ entries.
This probability distribution is supported on the curve $g_\alpha$ and will be denoted by $\pr_\alpha$.
We will use $\pr_\alpha$ to construct \emph{geographical coordinate system} on the unit square $\mysquare$.

For any point $p \in \mysquare$ there is exactly one curve $g_\alpha$ on which it lies. We say that 
\emph{the~latitude of $p$ is equal to $\alpha$}. The \textbf{\emph{longitude} of $p$}, denoted by $\longitude(p)\in
[0,1]$, is defined as the~measure~$\pr_\alpha$ of the set of points on the curve
$g_\alpha$ which have their $u$-coordinate smaller than the~\mbox{$u$-coordinate} of~$p$. 
For given $\alpha,\longitude\in [0,1]$ we denote by $\point_{\alpha, \longitude}$
the unique point of~the~square~$\mysquare$ with the appropriate latitude and
longitude.

\subsection{The first main result: asymptotics of evacuation paths}

For a given tableau $\tab_N\in \tableaux_{\sq_N}$ and $\tim\in[0,1]$ we denote
by
\[\mypoint_\tim= \mypoint_\tim(\tab_N) = \frac{1}{N} \pos_{N^2}\left( j^{\lfloor \tim N^2 \rfloor}(\tab_N) \right) \in \mysquare \]
the scaled down point from the evacuation path, cf.~\eqref{eq:evacuation}.

Our first main result states that, asymptotically, \emph{the scaled evacuation
	path in a random square tableau is a random meridian}, i.e.~a curve which
consists of points in the square $\mysquare$ with equal longitude $\longitude$.
The probability distribution of this longitude $\longitude$ is the uniform
distribution on the interval $[0,1]$.

\begin{thm}
\label{mainthm}

For each $N\in \N$ there exists a map~$\Longitude_N\colon \tableaux_{\sq_N} \to [0,1]$ such
that for each $\varepsilon > 0$
\begin{equation}
\label{eq:main-theorem}
\lim_{N \to \infty} \Pp \Big\{ \tab_N \in \tableaux_{\sq_N}: \sup_{\tim \in [0,1]} 
			\left| \mypoint_\tim(\tab_N) -
			\point_{1-\tim, \Longitude_N(\tab_N)}  \right| > \varepsilon \Big\} = 0.
\end{equation}
The probability distribution of the random variable $\Longitude_N$ converges,
as $N\to\infty$, to the uniform distribution on the unit interval $[0,1]$.
\end{thm}

\section{Asymptotics of jeu de taquin paths}

Let $\tab$ be a standard Young tableau with $n$ boxes. We define
$\pat(\tab)=(\pat_1,\dots,\pat_n) \subset \N^2$ to be the corresponding \emph{jeu de taquin
	path in the lazy parametrization}. More specifically, $\pat_i$~is~defined as~the~last box 
	along the jeu de taquin path corresponding to $\tab$
(cf.~\cref{fig:jdtA}) which contains a number $\leq i$.

Our second main result states that, asymptotically, 
\emph{the scaled jeu de taquin path in~a~random square tableau is a random meridian}.

\begin{thm}\label{jdtthm}
For each $N\in \N$ there exists a map~$\LongitudeTilde_N\colon
\tableaux_{\sq_N} \to [0,1]$ such that for each $\varepsilon > 0$
\[\lim_{N \to \infty} \Pp \Big\{ \tab_N \in \tableaux_{\sq_N}: \sup_{\tim \in [0,1]} 
\left| 
 \frac{1}{N} \pat_{\lceil \tim N^2\rceil}(\tab_N)  -
P_{\tim,\LongitudeTilde(\tab_N)} \right| > \varepsilon \Big\} = 0. \]
The probability distribution of the random variable $\LongitudeTilde_N$
converges, as $N\to\infty$, to the uniform distribution on the unit interval
$[0,1]$.
\end{thm}

An analogous problem was studied by Romik and \'Sniady \cite{Romik2015a} for
random tableaux obtained by applying Robinson--Schensted correspondence to a
random permutation (\emph{``Plancherel measure''}). The proof
presented there is not applicable in our context.
 

\begin{proof}
We show equivalence of \cref{mainthm,jdtthm} by
comparing positions of~boxes in lazy jeu de taquin path and the evacuation path
in pairs of corresponding tableaux.

In the proof we use: 
(1.) a~\emph{Robinson-Schensted-Knuth algorithm} which is a bijection which applied to a permutation $\sigma = (\sigma_1, \ldots, \sigma_n) \in \Sym_n$ 
returns a pair $(P(\sigma),Q(\sigma))$ of~tableaux of~the~same shape; 
(2.) a~\emph{Sch\"utzenberger involution} which is a~map 
\[\Sym_n \ni \sigma \mapsto \varepsilon^*(\sigma):=(n+1-\sigma_n, \ldots, n+1-\sigma_1);\]
and (3.) the identity~\cite[Section~2]{Romik2015a} in which \mbox{$s(\sigma) = s(\sigma_1, \sigma_2, \ldots, \sigma_n) := (\sigma_2, \ldots, \sigma_n)$}:
\begin{equation}\label{shift}
\left(j \circ Q\right)(\sigma) = \left(Q \circ \; s\right) (\sigma), \ \ \ \sigma = (\sigma_1, \ldots, \sigma_n) \in \Sym_n
\end{equation}
(\eqref{shift} holds true up to renumbering of the boxes on the left-hand side.)

Let $\sigma \in \Sym_{N^2}$. In the lazy jeu de taquin path $\pat(T) := \pat(Q(\sigma))$ we have by \eqref{shift}:
\[\pat_n(T) = Q\left(\sigma_1, \sigma_2 \ldots, \sigma_{n} \right) \setminus Q\left(\sigma_2, \ldots, \sigma_{n} \right).\]
On the other hand, using \eqref{shift} many times, in the evacuation path
\eqref{eq:evacuation} in the tableau $T^* := Q(\varepsilon^*(\sigma))$,
the~position of~the~box with entry $N^2$ in $j^{N^2-p}(T^*)$ is
\[Q\left(N^2+1-\sigma_{p}, \ldots, N^2+1-\sigma_2, N^2+1-\sigma_{1}\right) \setminus Q\left(N^2+1-\sigma_{p}, \ldots, N^2+1-\sigma_{2}\right).\]
By Greene theorem, this position is equal to $Q\left(\sigma_1, \sigma_2 \ldots, \sigma_{p} \right) \setminus Q\left(\sigma_2, \ldots, \sigma_{p} \right) = \pat_p(T)$. 
 \end{proof}

\section{Sketch of proof of \cref{mainthm}}

\begin{proof}[Proof of \cref{mainthm}]
In order to investigate the point $X_t$ 
we will pass to its geographic coordinates $\alpha(X_t)$ and $\longitude(X_t)$;
our goal is to show that for each $c>0$ and $\varepsilon > 0$
\begin{equation}
\label{eq:latitudelongitude}
\sup_{\tim \in [0,1]}  \big|
\alpha(X_t)- (1-t)   
\big|  < \varepsilon 
\qquad \text{and} \qquad
\sup_{\tim \in [c,1-c]}  \big|
\longitude(X_t) - \Longitude_N(\tab_N)   
\big|  < \varepsilon
\end{equation}
hold true except for $T_N$ in a set which has asymptotically negligible
probability $o(1)$.

The discussion from the beginning of \cref{sec:circles} shows
that  the first statement in 
\eqref{eq:latitudelongitude} indeed holds true
(except for an event of negligible 
probability, for $N\to\infty$).

If the second statement in \eqref{eq:latitudelongitude} instead of a supremum
would involve only a fixed value of $\tim_0\in (0,1)$, we could simply define
$\Longitude_N(\tab_N):=\longitude(X_{\tim_0})$ to be longitude of $X_{\tim_0}$.
Unfortunately, we need to justify that this choice of $\Longitude_N(\tab_N)$ is
also good for other values of $\tim\neq \tim_0$. This kind of result is
provided by \cref{prop:all-the-same}.
\end{proof}

\begin{proposition}
\label{prop:all-the-same}
Assume that $0<\tim_1 < \tim_2 <1$. Then  for each $\varepsilon>0$
\[ \lim_{N\to\infty} 
\Pp \left\{ \tab_N \in\tableaux_{\sq_N} : 
    \big| \longitude(X_{\tim_1}) - \longitude(X_{\tim_2}) \big| > \varepsilon \right\} = 0.\]

	
\end{proposition}	

\section{Sketch of proof of \cref{prop:all-the-same}. Surfers on the sink}

\cref{prop:all-the-same} is equivalent to the conjuction of the following two statements for $\varepsilon>0$:
\begin{align}
\lim_{N\to\infty} \Pp \left\{ \tab_N \in\tableaux_{\sq_N} : 
\longitude(X_{\tim_1}) - \longitude(X_{\tim_2})  > \varepsilon \right\} &= 0,
\label{eq:sink-A} 
\\ 
\lim_{N\to\infty} \Pp \left\{ \tab_N \in\tableaux_{\sq_N}
:  \longitude(X_{\tim_1}) - \longitude(X_{\tim_2})  < -\varepsilon \right\} &=
0. \label{eq:sink-B}
\end{align}
We present the proof of \eqref{eq:sink-A}; the other claim
\eqref{eq:sink-B} can be proved in an analogous way.

\subsection{Single surfer scenario}
\label{sec:single}

We present the story of the evacuation path in a different light. We will speak
about a~\emph{square pool of side $N$} (=the square Young diagram $\sq_N$)
filled with \emph{$N^2-1$ particles of~water} (=the Young tableau $\tab_N$ with
the largest entry removed), a passive \emph{surfer} (=the box with the biggest
entry) and its behaviour when the pool is drained (=jeu de taquin). Our~goal~in
\cref{mainthm} is to show that, when pool is big enough, the surfer has some
typical paths along which he/she moves. 

\textbf{In our proof of \cref{prop:all-the-same} we start our analysis at time $\tim_1$
when jeu de taquin was already applied $m=\lfloor \tim_1 N^2 \rfloor$ times.}
Our starting point is therefore the tableau 
\[ \tab'_N:=j^m(\tab_N)\]
with $N^2-m$ boxes. Let us renumber the entries of this tableau so it becomes
standard. We denote $\water=N^2-m-1$. In this way the boxes with numbers $1,\dots,\water$
correspond to~the~\emph{water} and the box with the number $\water+1$ to the
\emph{surfer}. By removing the box with the surfer
\[ W_N':= \tab_N' \setminus \{\water+1\}\]
we get a standard Young tableau which encodes \emph{the configuration of the water}.

\subsection{Pieri tableaux. Multisurfer scenario}
\label{sec:multi}

Let $k=\lfloor \sqrt[4]{N} \rfloor$. 
Let $\multi$ be a tableau in which the $k$ largest entries are numbered
by~consecutive integers $l+1,\dots,l+k$. We say that $\multi$ is a \emph{Pieri
    tableau} if these $k$ largest boxes are in \emph{an increasing order} from
north-west to south-east:
\[ u^{\multi}_{l+1} < \cdots < u^{\multi}_{l+k}.\]
By $\widetildetableaux_{\sq_N}$ we denote the set of standard Young tableaux of
shape $\sq_N$ which are Pieri, and~by~$\Pt$~the~uniform distribution
on~the~set~$\widetildetableaux_{\sq_N}$.

It is easy to check that if $\multi$ has at least $k+1$ boxes then $\multi$ is~a~Pieri tableau 
if and only if $j(\multi)$ is a Pieri tableau.

\medskip

To pinpoint the position of the surfer in random water, we will need a point of~reference. 
We introduce \textbf{the multisurfer scenario} in which
we consider the square pool filled with $N^2-k$ \emph{particles of water} on
which are surfing $k$ \emph{surfers} (=$k$~boxes with the biggest entries). We
assume that the multisurfers are in an increasing order; it~follows that this
scenario corresponds to a Pieri tableau $\multi_N\in
\widetildetableaux_{\sq_N}$ We assume that $\multi_N$ is a random
tableau sampled with the probability distribution $\Pt$.

\textbf{We start our analysis when jeu de taquin was already applied $m+1-k$ times.}
Our~starting point is therefore the tableau
\[ \multi'_N:=j^{m+1-k}(\multi_N)\] 
with $N^2+k-m-1=\water+k$ boxes. Let us renumber the entries of this tableau so
it~becomes  standard. In this way, just as in the single surfer scenario, the
boxes with numbers $1,\dots,\water$ correspond to the \emph{water}. On the
other hand, the boxes with the numbers $\water+1,\dots,\water+k$ correspond to
the \emph{multisurfers}. By removing the multisurfers
\begin{equation} 
\label{eq:water} \widetilde{W}_N':= \multi_N' \setminus
\{\water+1,\dots,\water+k\} = J^{m+1-k} ( \multi_n ) \setminus
\{\water+1,\dots, N^2\}
\end{equation}
we get a standard Young tableau which encodes the configuration of the water.

\subsection{Counting multisurfers gives longitude}
\label{multiplesketch-0}

For $0\leq i \leq N^2-k$ we consider the situation after jeu de taquin was
applied $i$ times in~the~multisurfer scenario, i.e.~$\multi'_N =
j^i(\multi_N)$, and for $u\in [-1,1]$ we define the random variable
$\overline{\Longitude}_i(u)$ to~be~the~fraction of the number of multisurfers 
which have scaled \mbox{$u$-coordinates} smaller than~$u$,~i.e.:~
\begin{equation} 
\label{eq:psi-via-multi}	
	\overline{\Longitude}_i(u) := \frac{1}{k} \max\left\{ p\in\{1,\dots,k\}  : 
	\frac{1}{N} u^{\multi_N'}_{\water+p} \leq u \right\}.  \end{equation}

The following result states that $\overline{\Longitude}_i$ gives a good
approximation of the longitude.

\begin{proposition}
\label{prop:longitude-experimental}
Let $\alpha\in(0,1)$ be fixed.	
For a given $\longitude\in[0,1]$ we set  $(x,y)=\point_{\alpha,\longitude}$ to
be the~Cartesian coordinates of the point with geographic coordinates
$\alpha,\longitude$ and define $U_{\alpha,\longitude}:=x-y$.

We set $i:=\lfloor (1-\alpha) N^2\rfloor$. The random variable
$\overline{\Longitude}_{i}\left( U_{\alpha,\longitude} \right)$ converges in
probability to $\psi$ and~this convergence is uniform over $\longitude$; in
other words for each $\varepsilon>0$
\[ 
\widetilde{\mathbb{P}}_N\Big\{ 
\multi_N\in\widetildetableaux_{\sq_N} :
\sup_{\longitude\in [0,1]}
\left| \longitude - \overline{\Longitude}_{i} \big( U_{\alpha,\longitude} \big) \right| > \varepsilon  
	   \Big\}  = O\left( \frac{1}{\sqrt{N}} \right). \]
\end{proposition}
\begin{proof}
	The scaled $u$-coordinates of the multisurfers may be encoded by a random
probability measure on $\R$
	\[ \mu := \frac{1}{k} \sum_{1\leq p\leq k} \delta\left(\frac{1}{N}\ u_{\water+p}^{\multi_N'}  \right),  \]
where $\delta(x)$ denotes the point measure concentrated in $x$. Our goal is to
prove that the cumulative probability function of $\mu$ converges uniformly to
some explicit limit; the~convergence should hold true in probability.

In order to achieve this goal we investigate the \emph{moments} of the measure $\mu$:
\[ \mathcal{M}_r = \int_{\R} z^r \dif \mu = 
\frac{1}{k} \sum_{1\leq p\leq k} \left(\frac{1}{N}\ u_{\water+p}^{\multi_N'}  \right)^r \qquad \text{for } r\in\{1,2,\dots\}.
\]
By applying \cref{expvalue} for some specific symmetric polynomials $W$ we are
able to find explicit values for the mean value and the variance of the random
variable $\mathcal{M}_r$. It turns out that, asymptotically for $N\to\infty$,
the mean value $\E \mathcal{M}_r$ converges to the `right' value and the
variance $\operatorname{Var} \mathcal{M}_r$ tends to zero which implies that
the convergence of measures \emph{in the sense of moments} indeed holds true in
probability. 

Since the limit measure is compatly supported, the convergence
\emph{in the sense of moments} implies also convergence of measures \emph{in
	the weak topology of probability measures}, as~required.
\end{proof}

\textbf{\cref{expvalue} provides a link between the statistical properties of
	the multisurfers and~the~representation theory of the symmetric groups.} The
problem of understanding (the~$u$-coordinates of) the positions of the
multisurfers after $i$ applications of jeu de taquin is equivalent to finding
(the  $u$-coordinates~of) the boxes with numbers from the~set~$M:=\{a,\dots,b\}:=\{N^2+1-k-i,\dots,N^2-i\}$ 
in the random tableau $S:=J^i(\multi_N)$.

We will view the symmetric group $\Sym_k\subset \Sym_{N^2}$ as the group of 
permutations of the set $M$ and define an element of the symmetric group
algebra \mbox{$p_{\Sym_k} = \frac{1}{k!} \sum_{\sigma \in \Sym_k} \sigma \in
\C \Sym_{N^2}$}. We~also consider Jucys--Murphy elements $\JM_s :=
\sum_{i=1}^{s-1} (i, s)\in\C \Sym_{N^2}$   (see~\cite{Jucys1974}).

\begin{lem}\label{expvalue}
	Let $W(x_1, \dots, x_k)$ be a symmetric polynomial in $k$ variables. Let $S$
be a random element (sampled with the uniform distribution) of the set of
tableaux $T\in\tableaux_{\sq_N}$ with the additional property that
\[ u_{a}^T < \cdots < u_{b}^T.\] 

Then
	\begin{equation}\label{exp}
	\E\ W\left(u^S_{a},\ \dots,\ u^S_{b}\right) = \frac{
\chi^{\sq_N}\!\big(p_{\Sym_k} \cdot W \left( \JM_{a},\ \dots,\ \JM_{b}\right) \big)}{ \chi^{\sq_N}\!\left(p_{\Sym_k} \right) } .
	\end{equation}
\end{lem}
\begin{proof}
The proof has the following ingredients.

Firstly, the observation that the vector space $V^{\sq_N}$
has a linear basis $(e_{\tab})$ indexed by~standard Young tableaux
$\tab\in\tableaux_{\sq_N}$ in which the action of Jucys--Murphy elements is~diagonal, 
with the eigenvalue equal to the $u$-coordinate of the appropriate
box:
\[ \JM_s\  e_\tab = u^{\tab}_{s}\ e_\tab.\]

Secondly, with the help of Littlewood--Richardson coefficients, we investigate
the~decomposition of~the~restricted representation $V^{\sq_N}
\big\downarrow^{\Sym_{N^2}}_{\Sym_{N^2-i-k} \times \Sym_{k} \times \Sym_i}$
into irreducible components
\end{proof}

\subsection{Presence of multisurfers does not influence the water}
\label{podobienstwowody}

We will show that the probability distribution of water in the single surfer
scenario is~very close to its multisurfer  counterpart.

\begin{lem}
\label{probdiv}
For any standard Young tableau $S$ with shape $\lambda:=\sh S$ and $\water$ boxes
\begin{equation}
\label{Pp=Pt}
\frac{\Pp\Big\{ \tab_N\in\tableaux_{\sq_N} : W_N'= S \Big\} }%
{\Pt\left\{ \multi_N\in\widetildetableaux_{\sq_N} : \widetilde{W}_N' = S \right\} }
=
\frac{\Pp\Big\{ \tab_N\in\tableaux_{\sq_N} : \sh W_N'= \lambda \Big\} }%
{\Pt\left\{ \multi_N\in\widetildetableaux_{\sq_N} : \sh \widetilde{W}_N' = \lambda \right\} }
= 1 + O\left(\frac{1}{\sqrt{N}}\right).
\end{equation}
\end{lem}
\begin{proof} 
The first equality follows from the observation that the conditional
probability distribution of $W_N'$ under the event that $\sh W_N'=\lambda$ is
the uniform measure on $\tableaux_\lambda$ and~the~analogous result for
$\widetilde{W}_N'$.

Since
$\Pt(\cdot) = \Pp(\cdot | \ORD)$ is a conditional probability, 
Bayes rule can be applied and~the~middle part of \eqref{Pp=Pt} is equal to
\begin{equation}
\label{eq:Bayes}
\frac{\Pp\left(\ORD\right)}{\Pp\left(\ORD \big| \sh W_N' = \lambda \right) }.
\end{equation}

We will show the following estimate for the denominator:
\begin{equation} 
\label{eq:cond}
\Pp\left(\ORD \big| \sh W_N' = \lambda \right) = \frac{1 + O \left( \frac{1}{\sqrt{N}} \right)}{k!}.
\end{equation}
Equation \eqref{eq:water} shows that under condition $\sh W_N' = \lambda$ the
conditional probability distribution of $U:=\jdtcomplete^{m+1-k}
(\multi_N ) \setminus \{1,\dots,\water\}$ coincides with the uniformly
random tableau with the skew shape $\sq_N \setminus \lambda$. The conditional
probability \eqref{eq:cond} is thus equal to the probability that the smallest
$k$ boxes of $U$ (or, equivalently, the smallest $k$ boxes of the rectified
tableau $V:=\operatorname{rect} U$) are ordered from north-west to south-east
\emph{\'a la Pieri}. We denote by~$\mu$ the~shape of $V$. It follows that 
\begin{equation}
\label{eq:average} 
 \Pp\left(\ORD \big| \sh W_N' = \lambda \right) = 
\sum_{{\mu\in\Young_{N^2-\water} \text{ such that } \mu \subset \sq_N }} 
\mathbb{P}_N\left\{ \tab_N \in \tableaux_{\sq_N} :V = \mu \right\} \cdot P_\mu, 
\end{equation}
where $P_\mu$ denotes the probability that $V$ (viewed as a uniformly random
tableau sampled from $\tableaux_\mu$) has entries $1,\dots,k$ ordered from
north-west to south-east or, equivalently, 
the shape of the restricted tableau $\operatorname{sh} \left( V|_{1,\dots,k} \right) = (k)$
is the one-row Young diagram with~$k$~boxes. 

The link between standard Young tableaux and combinatorics of  irreducible
representations of the symmetric groups implies that
\begin{multline} 
\label{eq:ortho}
P_\mu = \frac{\text{multiplicity of $V^{(k)}$ in $\left( V^{\mu}\big\downarrow^{\Sym_{N^2-\water}}_{\Sym_k}\right)$}}%
{\operatorname{dim} V^{\mu}} =
\frac{
	\left\langle \chi^{\mu} \big\downarrow^{\Sym_{N^2-\water}}_{\Sym_k}, \chi^{(k)} \right\rangle 
}{
 \operatorname{dim} V^{\mu}
}	
	= \\  \frac{1}{k!} \left( 1 + 
\sum_{\pi\in \Sym_k \text{ and } \pi \neq \on{id}} 
\frac{ \chi^{\mu}(\pi) }{ 
			\operatorname{dim} V^{\mu}
		} \right),
\end{multline}
where the second equality follows from the orthogonality of irreducible characters.

Let $C>0$ be fixed. We say that a Young diagram $\mu$ is $C$-balanced 
\label{balanced}
if both
its number of~rows and its number of columns are $\leq C\sqrt{|\lambda|}$.
Each tableau $\mu$ which contributes to~\eqref{eq:average} is $C$-balanced for 
$C=\frac{N}{\sqrt{t_1 N^2}}=\frac{1}{\sqrt{t_1}}$.
The
result of Fer\'ay and \'Sniady \cite[Theorem~1]{Feray2011} gives a good upper
bound for the character ratios on the right-hand side of \eqref{eq:ortho} for
$\mu$ in~the~class of~$C$-balanced diagrams which implies that
\[ P_\mu = \frac{1 + O \left( \frac{1}{\sqrt{N}} \right)}{k!} .\]
The latter estimate combined with \eqref{eq:average} implies that \eqref{eq:cond} indeed holds true.

\medskip

An analogous (but simpler) reasoning shows that the numerator  from \eqref{eq:Bayes} fulfills
\begin{equation} 
\label{eq:cond2}
\Pp\left(\ORD\right) = \frac{1+O\left( \frac{1}{\sqrt{N}} \right)}{k!}.
\end{equation}

The estimates \eqref{eq:cond} and \eqref{eq:cond2} for the terms in \eqref{eq:Bayes} complete the proof.
\end{proof}

\subsection{Single- and multisurfer scenario on the same water}
\label{eq:single-multi}

\textbf{In \cref{sec:single,sec:multi} we considered two random tableaux: $\tab_N'$ and
$\multi_N'$ defined on~two~different probability spaces. Our goal is
to define variants of these two random tableaux $\ttt$ and $\ttilde$ on
\emph{the same} probability space in such a way that the corresponding
configurations of water would coincide: $\www:=W_N'= \widetilde{W}_N'$.}

In order to achieve this goal we start by randomly sampling the tableau $\www$
(=\emph{water}) with the probability distribution of $W_N'$ from
\cref{sec:single}. %
Once $\www$ is selected we randomly choose the tableaux $\ttt$ (=\emph{single
	surfer}) and the Pieri tableau $\ttilde$ (=\emph{multisurfers}) according to
the conditional probability distributions (conditioned by choosing $\www$)
\begin{align*} 
\mathbb{P}(\ttt=S)&:= \mathbb{P}_N\left( \tab_N'=S | W_N'=\www \right), \\
\mathbb{P}(\ttilde=S )&:= \widetilde{\mathbb{P}}_N\left( \multi_N'=S | \widetilde{W}_N'=\www  \right) 
\end{align*}
for an arbitrary standard tableau $S$.

The probability distribution of $\ttt$ coincides with the distribution of
$\tab'_N$ from \cref{sec:single}. On the other hand \cref{probdiv} shows that
the total variation distance between the distribution of $\ttilde$ and the
distribution of $\multi'_N$ from \cref{sec:multi} is small, of~order~$O\left( \frac{1}{\sqrt{N}} \right)$.

\subsection{Ghosts of multisurfers do not overtake the surfer from right to left}

Let us fix a pair of tableaux $\ttt$ and $\ttilde$ given by the above
construction. For any $0\leq q \leq \water$
we will iteratively apply jeu de taquin $q$ times to both tableaux and compare the outcomes.
(Informally speaking, with the notations of \cref{multiplesketch-0}, this means that jeu de taquin was applied $i=\lfloor \tim_1 N^2\rfloor+q$ times to the tableau $\tab_N$.) 

A simple inductive argument shows that after $q$ steps 
the configurations of water in~$j^q(\ttt)$ and~$j^q(\ttilde)$ coincide:
\[ j^q(\ttt) \setminus \{\water+1\} = j^q(\ttilde) \setminus\{\water+1,\dots,\water+k\}.\]
\textbf{In other words, we are dealing with two alternative universes in which
	the history of~water is the same, and differ only by the number of surfers on
	top. Since the~dynamics of the surfer and the multisurfers depends only on the
	water, there is no interaction between the surfer and the 
	multisurfers (they are like ghosts).}

\medskip

We define $\widetilde{\psi}_q$ as the fraction of the multisurfers which are to
the left of the surfer:
\[ \widetilde{\psi}_q := \frac{1}{k} \max\left\{ p\in\{1,\dots,k\}  : 
 u^{j^q(\ttilde)}_{\water+p} \leq u^{j^q(\ttt)}_{\water+1} \right\}.  \] 
This quantity is a modification of \eqref{eq:psi-via-multi} in which the role
of~$j^i(\multi_N)$ is played by~$j^q(\ttilde)$ evaluated at $u=\frac{1}{N}
u^{j^q(\ttt)}_{\water+1}$. The discussion from \cref{eq:single-multi} shows
that for the conclusion of \cref{prop:longitude-experimental} remains valid
also for this modification. It follows that both at time $\tim_1$ as~well
as~at~time $\tim_2$ we have
\begin{align}
\label{eq:almostA} 
\left| \psi(X_{\tim_1}) - \widetilde{\psi}_0 \right| &< \nicefrac{\varepsilon}{2}, \\
\label{eq:almostB}
\left| \psi(X_{\tim_2}) - \widetilde{\psi}_{\lfloor \tim_2 N^2\rfloor - \lfloor \tim_1 N^2\rfloor} \right| &< \nicefrac{\varepsilon}{2}.
\end{align} 
hold true, except for an event of negligible probability. 
 
By comparing the action of jeu de taquin on $j^q(\ttilde)$ with its action on $j^q(\ttt)$ it~follows that
the sequence $(\widetilde{\psi}_q)$ is weakly decreasing:
\begin{equation} 
\label{eq:decreasing}
\widetilde{\psi}_0 \geq \cdots \geq \widetilde{\psi}_{\water};
\end{equation}
in other words, \textbf{if we compare the \emph{relative} positions of~the~surfer 
and the ghosts of~the~multisurfers in the alternative universe, the
	multisurfers can only move from the left of the surfer to the right, but not in
	the opposite direction}.

By combining \eqref{eq:almostA}, \eqref{eq:almostB} and \eqref{eq:decreasing}
it follows that \eqref{eq:sink-A} indeed holds true, as we claimed. This
concludes the proof of \cref{prop:all-the-same}.

 \acknowledgements{The authors were supported by Grant 2017/26/A/ST1/00189 of Narodowe Centrum Nauki.}

\printbibliography

\end{document}
