\documentclass{beamer}%
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{array}
\usepackage{setspace}
%\usepackage{graphicx}
%\usepackage{etex}
%\usepackage{amsthm}
\usepackage{color}
%\usepackage{wasysym}
%\usepackage[all]{xy}
\usepackage{textpos}
\usepackage{color}
%\usepackage{epsfig,amsfonts,bbm,mathrsfs}
\usepackage{verbatim}
\usepackage{amsfonts}
\usepackage{tikz}
\usepackage{ytableau}
\usepackage{hyperref}%
\setcounter{MaxMatrixCols}{30}
\definecolor{grau}{rgb}{.5 , .5 , .5}
\definecolor{dunkelgrau}{rgb}{.35 , .35 , .35}
\definecolor{schwarz}{rgb}{0 , 0 , 0}
\definecolor{violet}{RGB}{143,0,255}
\definecolor{forestgreen}{RGB}{34, 100, 34}
\newcommand{\red}{\color{red}}
\newcommand{\grey}{\color{grau}}
\newcommand{\green}{\color{forestgreen}}
\newcommand{\violet}{\color{violet}}
\newcommand{\blue}{\color{blue}}
\let\sumnonlimits\sum
\let\prodnonlimits\prod
\let\cupnonlimits\bigcup
\let\capnonlimits\bigcap
\renewcommand{\sum}{\sumnonlimits\limits}
\renewcommand{\prod}{\prodnonlimits\limits}
\renewcommand{\bigcup}{\cupnonlimits\limits}
\renewcommand{\bigcap}{\capnonlimits\limits}
\definecolor{darkred}{rgb}{0.7,0,0} % darkred color
\newcommand{\defn}[1]{{\color{darkred}\emph{#1}}} % emphasis of a definition
\newcommand{\icast}{\circled{$\ast$}}
\newcommand{\id}{\operatorname{id}}
\newcommand{\exc}{\operatorname{exc}}
\newcommand{\rev}{\operatorname{rev}}
\newcommand{\ssum}{\operatorname{sum}}
\newcommand{\sub}{\operatorname{sub}}
\newcommand{\SYT}{\operatorname{SYT}}
\newcommand{\SSYT}{\operatorname{SSYT}}
\newcommand{\FSSYT}{\operatorname{FSSYT}}
\newcommand{\px}{\phantom{x}}
\newcommand{\ZZ}{\mathbb{Z}}
\newcommand{\NN}{\mathbb{N}}
\newcommand{\QQ}{\mathbb{Q}}
\newcommand{\calE}{\mathcal{E}}
\newcommand{\calF}{\mathcal{F}}
\newcommand{\bb}{\mathbf{b}}
\newcommand{\powset}[2][]{\ifthenelse{\equal{#2}{}}{\mathcal{P}\left(#1\right)}{\mathcal{P}_{#1}\left(#2\right)}}
\newcommand{\set}[1]{\left\{ #1 \right\}}
\newcommand{\abs}[1]{\left| #1 \right|}
\newcommand{\tup}[1]{\left( #1 \right)}
\newcommand{\ive}[1]{\left[ #1 \right]}
\newcommand{\floor}[1]{\left\lfloor #1 \right\rfloor}
\newcommand{\lf}[2]{#1^{\underline{#2}}}
\newcommand{\underbrack}[2]{\underbrace{#1}_{\substack{#2}}}
\newcommand{\horrule}[1]{\rule{\linewidth}{#1}}
\newcommand{\nnn}{\nonumber\\}
\newcommand{\sslash}{\mathbin{/\mkern-6mu/}}
\newcommand\arxiv[1]{\href{http://www.arxiv.org/abs/#1}{\texttt{arXiv:#1}}}
\usetheme{Frankfurt}
\usefonttheme[onlylarge]{structurebold}
\setbeamerfont*{frametitle}{size=\normalsize,series=\bfseries}
\setbeamertemplate{navigation symbols}{}
\setbeamertemplate{footline}[frame number]
\setbeamertemplate{itemize/enumerate body begin}{}
\setbeamertemplate{itemize/enumerate subbody begin}{\normalsize}
\beamersetuncovermixins{\opaqueness<1>{0}}{\opaqueness<2->{15}}
\newcommand{\STRUT}{\vrule width 0pt depth 8pt height 0pt}
\newcommand{\ASTRUT}{\vrule width 0pt depth 0pt height 11pt}
\theoremstyle{plain}
\newtheorem{conj}[theorem]{Conjecture}
\setbeamertemplate{headline}{}
\begin{document}

\title{The Pak--Postnikov and Naruse skew hook length formulas: a new proof}
\date{}
\begin{frame}
\titlepage

\begin{center}
\vspace{-5pc}
Nazar Korniichuk (NSL no. 145, Kyiv) \\
Kostiantyn Molokanov (NSL no. 145, Kyiv)\\
Severyn Khomych (Brucknergymnasium, Wels) \\
%(145 School, Kyiv)}
mentor: Darij Grinberg (Drexel University) \\
\vspace{1pc}14 June 2023, MIT
\end{center}

\textbf{slides:} {\color{red}
\url{http://www.cip.ifi.lmu.de/~grinberg/algebra/yd2023.pdf}}
\medskip\\
\textbf{paper:} {\color{red}
\url{https://arxiv.org/abs/2310.18275}} \medskip\\
\end{frame}

\begin{frame}
\frametitle{\ \ \ \ Young diagrams}

\begin{itemize}

\item A \defn{partition} is a weakly decreasing sequence $\lambda = \tup{\lambda_1 \geq \lambda_2 \geq \lambda_3 \geq \ldots}$ of nonnegative integers with only finitely many positive entries.

\pause

\item The \defn{Young diagram} of this partition $\lambda$ is a left-aligned table with $\lambda_i$ cells in row $i$ (indexed from the top). We call it \defn{$Y\tup{\lambda}$}. Formally:
\[
Y\tup{\lambda} = \set{ \tup{i,j} \mid i > 0 \text{ and } 0 < j \leq \lambda_i } .
\]

\pause

\only<3>{
\item \textbf{Example:} If $\lambda = \tup{4,2,2,0,0,0,\ldots} = \tup{4,2,2}$ (we omit zeroes), then
\[
Y\tup{\lambda} = \ydiagram{4,2,2}.
\]
}

\pause

\item Two partitions $\mu$ and $\lambda$ satisfy \defn{$\mu \subseteq \lambda$} if $Y\tup{\mu} \subseteq Y\tup{\lambda}$. In this case, the \defn{skew diagram $Y\tup{\lambda/\mu}$} is defined to be $Y\tup{\lambda} \setminus Y\tup{\mu}$.

\pause

\only<5>{
\item \textbf{Example:} If $\lambda = \tup{5,2,2}$ and $\mu = \tup{1,1}$, then
\[
Y\tup{\lambda/\mu} = \ydiagram{1+4,1+1,2}.
\]
}

\pause

\item More generally, any set of (square) cells is called a \defn{diagram}.

\item \textbf{Example:}
\[
\ytableaushort{\none& &\none& \\ & & \\ \none& }
\]

\end{itemize}

\vspace{10pc}

\end{frame}

\begin{frame}
\frametitle{\ \ \ \ Standard tableaux}

\begin{itemize}

\item Given a diagram $D$, we can fill it with the numbers $1,2,\ldots,n$. Such a filling is called a \defn{standard tableau} (of shape $D$) if
\begin{itemize}
    \item each of the numbers $1,2,\ldots,n$ appears exactly once;
    \item the numbers increase along each row;
    \item the numbers increase down each column.
\end{itemize}

\pause

\item If $D = Y\tup{\lambda}$, we let \defn{$\SYT\tup{\lambda}$} be the set of all standard tableaux of shape $Y\tup{\lambda}$.

\pause

\item Likewise \defn{$\SYT\tup{\lambda/\mu}$}.

\pause

\item \textbf{Example:} If $\lambda = \tup{5,4,3,3}$ and $\mu = \tup{2,1,1}$, then
\[
\begin{ytableau} \none&\none&1&3&9\\ \none &2&4&10\\ \none&5&6\\ 7&8&11 \end{ytableau}\ \ \in \SYT\tup{\lambda / \mu} .
\]

\pause

\item \textbf{Question:} Given a diagram $D$, how many standard tableaux of shape $D$ exist?
\end{itemize}

\end{frame}

\begin{frame}
\frametitle{\ \ \ \ Hooks}

\begin{itemize}

\only<1>{\item For $D = Y\tup{\lambda}$, the classical \defn{hook length formula} of Frame, Robinson and Thrall (1953) gives a beautiful answer in terms of the \textbf{hooks} of $\lambda$.}

\pause

\item If $c=\left(  i,j\right)  $ is a cell of a Young diagram $Y\left(  \lambda\right)  $, we let the \defn{hook $H_{\lambda}\left(  c\right)  $} be 
\begin{align*}
& \set{\text{all cells of $Y\tup{\lambda}$ lying due east of $c$}} \\
& \qquad \cup \set{\text{all cells of $Y\tup{\lambda}$ lying due south of $c$}} \cup \set{c} \\
&=\left\{  \left(  i,k\right)  \in Y\left(
\lambda\right)  \ \mid\ k\geq j\right\}  \cup\left\{  \left(  k,j\right)
\in Y\left(  \lambda\right)  \ \mid\ k\geq i\right\}  .
\end{align*}

\pause

\item The \defn{hook length} $h_{\lambda}(c)$ is defined to be $\left\vert H_{\lambda
}\left(  c\right)  \right\vert $, that is, the number of all cells in the hook
of $c$.

\pause

\item \textbf{Example:} If $\lambda = \tup{4,3,3,2}$, then
\[
H_{\lambda}(2, 2) = 
\begin{ytableau} 
\phantom{x} & & &\\ 
\phantom{x} & \ast& \ast\\ 
\phantom{x} &\ast &\\
\phantom{x} &\ast
\end{ytableau}
\quad \text{ and $h_{\lambda}(2, 2) = 4$.}
\]
\end{itemize}


\end{frame}

\begin{frame}
\frametitle{\ \ \ \ The hook length formula}

\begin{itemize}

\item The original \defn{hook length formula} says that
\[
\abs{\operatorname*{SYT}\left(  \lambda\right) }
=\dfrac
{n!}{\prod_{c\in Y\left(  \lambda\right)  }h_{\lambda}\left(  c\right)  },
\]
where $n$ is the number of cells in $Y\left(  \lambda\right)  $.

\pause

\item \textbf{Example:} If $\lambda = (3,2)$, then 
\[
|\SYT\tup{\lambda}| = \frac{5!}{1\cdot 3\cdot 4\cdot 1\cdot 2} = 5.
\]
\only<2>{
Here are the hooks of all five cells:
\[
\begin{ytableau} 
\phantom{x} & & \ast\\ 
\phantom{x} &  
\end{ytableau}
 \quad
\begin{ytableau} 
\phantom{x} & \ast& \ast\\ 
\phantom{x} &\ast  
\end{ytableau}
 \quad
\begin{ytableau} 
\ast & \ast& \ast\\ 
\ast &  
\end{ytableau}
\]
\[
\begin{ytableau} 
\phantom{x} & & \\ 
\phantom{x} &\ast  
\end{ytableau}
 \quad
\begin{ytableau} 
\phantom{x} & & \\\ 
\ast & \ast
\end{ytableau}
\]
}
\pause
\only<3>{
Here is $\SYT(\lambda)$:
\begin{align*}
\begin{ytableau} 
1 & 2 & 3\\ 
4 & 5 
\end{ytableau} ,
\quad
\begin{ytableau} 
1 & 2 & 4\\ 
3 & 5 
\end{ytableau} ,
\quad
\begin{ytableau} 
1 & 2 & 5\\ 
3 & 4 
\end{ytableau} ,
\\
\begin{ytableau} 
1 & 3 & 4\\ 
2 & 5 
\end{ytableau} ,
\quad
\begin{ytableau} 
1 & 3 & 5\\ 
2 & 4 
\end{ytableau} .
\end{align*}
}

\vspace{10pc}

\end{itemize}


\end{frame}

\begin{frame}
\frametitle{\ \ \ \ The hook length formula: example}

\begin{itemize}

\item \textbf{Example:} The number of Dyck paths from $\tup{0,0}$ to $\tup{2n,0}$ is the $n$-th \defn{Catalan number} $C_n = \dfrac{\tup{2n}!}{n!\tup{n+1}!}$.

This follows from the hook length formula, applied to $\lambda = \tup{n,n}$, and a simple bijection $\set{\text{Dyck paths}} \to \SYT\tup{\lambda}$:
\[
\begin{tikzpicture}[scale=0.8]
\draw(0, 0) -- (1, 1) -- (2, 2) -- (3, 1) -- (4, 0) -- (5, 1) -- (6, 0);
\fill
[green!60!black] (0, 0) -- (1, 1) -- (2, 2) -- (3, 1) -- (4, 0) -- (5, 1) -- (6, 0);
\filldraw(0, 0) circle [fill=red, radius=0.1];
\filldraw(1, 1) circle [fill=red, radius=0.1];
\filldraw(2, 2) circle [fill=red, radius=0.1];
\filldraw(3, 1) circle [fill=red, radius=0.1];
\filldraw(4, 0) circle [fill=red, radius=0.1];
\filldraw(5, 1) circle [fill=red, radius=0.1];
\filldraw(6, 0) circle [fill=red, radius=0.1];
\end{tikzpicture}
\quad\longleftrightarrow\quad
\ytableaushort{125,346} .
\]

\[
\begin{tikzpicture}[scale=0.8]
\draw(0, 0) -- (1, 1) -- (2, 2) -- (3, 1) -- (4, 2) -- (5, 1) -- (6, 0);
\fill
[green!60!black] (0, 0) -- (1, 1) -- (2, 2) -- (3, 1) -- (4, 2) -- (5, 1) -- (6, 0);
\filldraw(0, 0) circle [fill=red, radius=0.1];
\filldraw(1, 1) circle [fill=red, radius=0.1];
\filldraw(2, 2) circle [fill=red, radius=0.1];
\filldraw(3, 1) circle [fill=red, radius=0.1];
\filldraw(4, 2) circle [fill=red, radius=0.1];
\filldraw(5, 1) circle [fill=red, radius=0.1];
\filldraw(6, 0) circle [fill=red, radius=0.1];
\end{tikzpicture}
\quad\longleftrightarrow\quad
\ytableaushort{124,356} .
\]


\end{itemize}

\end{frame}

\begin{frame}
\frametitle{\ \ \ \ Excited moves}

\begin{itemize}

\item \defn{Naruse's skew hook length formula} (Naruse, 2014) expresses $\abs{\SYT\tup{\lambda / \mu}}$ in terms of excitations.
\vspace{10pc}
\end{itemize}

\end{frame}

\begin{frame}
\frametitle{\ \ \ \ Excited moves}

\begin{itemize}
\item An \defn{excited move} for a cell $c=(i,j)\in D$ means moving this cell from $(i,j)$ to $(i+1,j+1)$. \\
This is allowed only if the three cells marked $\times$ (that is, $\tup{i+1,j},\ \tup{i,j+1},\ \tup{i+1,j+1}$) are not in $D$.
% \item If $D$ is a diagram that contains some cell $c = \tup{i,j} \in D$, and contains none of $\tup{i+1,j}$, $\tup{i,j+1}$, $\tup{i+1,j+1}$, then we define the diagram \defn{$\operatorname*{exc}\nolimits_{c}D$} by

\vspace{1pc}
\[
\begin{tabular}[c]{cc}\cline{1-1}%
\multicolumn{1}{|c|}{$c$} & \multicolumn{1}{|c}{$\times$}\\\cline{1-1}%
$\times$ & $\times$
\end{tabular}
\rightarrow%
\begin{tabular}
[c]{cc}%
$\times$ & $\times$\\\cline{2-2}%
$\times$ & \multicolumn{1}{|c|}{$c$}\\\cline{2-2}%
\end{tabular}
\]

\pause

\only<2>{
\item \textbf{Example:}

\vspace{1pc}

\[
\begin{ytableau}
    \px & \px &\px \\ \px& c \\ \px
\end{ytableau}
\longmapsto
\begin{ytableau}
     \px & \px &\px \\ \px\\ \px&\none& c
\end{ytableau}
\]
}
\pause


\only<3>{
\textbf{However,}

\vspace{1pc}
\[
\begin{ytableau}
    \px & \px &\px \\ \px& c &\px\\ \px
\end{ytableau}
\not\longmapsto
\begin{ytableau}
     \px & \px &\px \\ \px&\none&\px\\ \px&\none& c
\end{ytableau}
\]
}
\pause


\only<4>{
\textbf{However,}

\vspace{1pc}
\[
\begin{ytableau}
    \px & \px &\px \\ \px& c \\ \px&\px
\end{ytableau}
\not\longmapsto
\begin{ytableau}
     \px & \px &\px \\ \px&\none\\ \px&\px& c
\end{ytableau}
\]
}

\pause
\vspace{1pc}

\only<5>{
\textbf{However,}

\vspace{1pc}
\[
\begin{ytableau}
    \px & \px &\px \\ \px& c\\ \px&\none&\px
\end{ytableau}
\not\longmapsto
\begin{ytableau}
     \px & \px &\px \\ \px&\none\\ \px&\none& c
\end{ytableau}
\]
}
\end{itemize}

\vspace{10pc}

\end{frame}

\begin{frame}
\frametitle{\ \ \ \ Excitations}
\begin{itemize}

% \item Let $c=(i,j)\in D$, where diagram $D$ does not contain cells $\tup{i+1,j}$, $\tup{i,j+1}$, $\tup{i+1,j+1}$, then \defn{$\operatorname*{exc}\nolimits_{c}D$} is defined as
% \[\operatorname*{exc}\nolimits_{c}D = \left(  D\setminus\left\{  (i, j)\right\}  \right)  \cup\left\{ 
% (i+1, j+1)\right\} .
% \]



% \pause

% \vspace{0.5pc}
% \only<2-3>{
% We say that \defn{$\operatorname*{exc}\nolimits_{c}D$} is obtained from $D$ by one \defn{excited move}.
% \pause
% \vspace{1pc}
% \item {\textbf{Example:} If $\mu = (3, 2)$, then
% \[
% \begin{array}{|c|c|c|}
% Y\tup{\mu}
% & \operatorname*{exc}\nolimits_{(1,3)}\tup{Y\tup{\mu}}
% & \operatorname*{exc}\nolimits_{(2, 2)}\tup{\operatorname*{exc}\nolimits_{(1,3)}\tup{Y\tup{\mu}}}
% \\
% \phantom{u}
% \begin{ytableau} 
% \phantom{x} & & \\ 
% \phantom{x} &  
% \end{ytableau}
% \phantom{u}
% &
% \phantom{u}
% \begin{ytableau} 
% \phantom{x} & & \none\\ 
% \phantom{x} & & \none &
% \end{ytableau}
% \phantom{u}
% &
% \phantom{u}
% \begin{ytableau} 
% \phantom{x} & & \none\\ 
% \phantom{x} &\none & \none &\\
% \none & \none & 
% \end{ytableau}
% \phantom{u}
% \end{array}
% \]
% }}

% \pause
% \only<4-9>{
% \vspace{0.3pc}

\item An \defn{excitation} of a diagram $D$ is a diagram obtained from $D$ by a sequence of excited moves.
% \\
% (In particular, $D$ itself qualifies.)



\pause
\vspace{2pc}
\only<2>{
\item

{\textbf{Example:} Original diagram $D$:
\[
\begin{ytableau}
    \px & \px \\ \px & \px \\ \px & \none & \none & \none
\end{ytableau}
\]
}}
\pause
\only<3>{\textbf{Example:} After a single excited move:
\[
\begin{ytableau}
    \px & \px \\ \px & \px \\ \none \\ \none & \px & \none & \none
\end{ytableau}
\]
}

\pause
\only<4>{
{\textbf{Example:} After two excited moves:
\[
\begin{ytableau}
    \px & \px \\ \px \\ \none & \none & \px \\ \none & \px & \none & \none
\end{ytableau}
\]
}}
\pause
\only<5>{\textbf{Example:} After three excited moves:
\[
\begin{ytableau}
    \px \\ \px & \none & \px \\ \none & \none & \px \\ \none & \px & \none & \none
\end{ytableau}
\]
}




\pause
\only<6>{
\textbf{Example:} After four excited moves:
\[
\begin{ytableau}
    \px \\ \px & \none & \px \\ \none \\ \none & \px & \none & \px
\end{ytableau}
\]
}




\pause

\item Now, for two partitions $\lambda$ and $\mu$, we define \defn{$\mathcal{E} \left(\lambda/\mu\right)$} to be the set of all excitations $E$ of $Y\left(  \mu\right)  $ that satisfy $E\subseteq Y\left(  \lambda\right)  $.



\vspace{10pc}

\end{itemize}

\end{frame}

\begin{frame}
\frametitle{\ \ \ \ Naruse's skew hook length formula}

\begin{itemize}

\item \defn{Naruse's skew hook length formula} says that 
\[
\left\vert \operatorname*{SYT}\left(  \lambda/\mu\right)  \right\vert
=n!\sum_{E\in\mathcal{E}\left(  \lambda/\mu\right)  }\ \ \prod_{c\in Y\left(
\lambda\right)  \setminus E}\dfrac{1}{h_{\lambda}\left(  c\right)  }
\]
if $\lambda$ and $\mu$ are two partitions with $\mu \subseteq \lambda$ with $\abs{Y\tup{\lambda/\mu}} = n$.

\pause

\item \textbf{Example:} If $\lambda = (2, 2, 2)$ and $\mu = (1, 1)$, then
\only<2>{
\[
\mathcal{E}
\left(\lambda/\mu\right)
= \set{
\begin{ytableau} 
\ast &\\ 
\ast &\\
\phantom{x} & \\
\end{ytableau} ,
\quad
\begin{ytableau} 
\ast &\\ 
\phantom{x} &\\
\phantom{x} & \ast\\
\end{ytableau} ,
\quad
\begin{ytableau} 
\phantom{x} &\\ 
\phantom{x} & \ast\\
\phantom{x} & \ast\\
\end{ytableau} }.
\]
Thus,
\begin{align*}
\left\vert \operatorname*{SYT}\left(  \lambda/\mu\right)  \right\vert = 4! \cdot \tup{\frac{1}{3\cdot 2\cdot1\cdot2} + \frac{1}{3\cdot 2\cdot3\cdot2} + \frac{1}{3\cdot 2\cdot3\cdot4}} = 3\\
\end{align*}}
\pause
% And indeed,
\[
\operatorname*{SYT}\left(  \lambda/\mu\right)
= \set{
\begin{ytableau} 
\none & 1\\ 
\none & 3\\
2 & 4\\
\end{ytableau} ,
\quad
\begin{ytableau} 
\none & 1\\ 
\none & 2\\
3 & 4\\
\end{ytableau} ,
\quad
\begin{ytableau} 
\none & 2\\ 
\none & 3\\
1 & 4\\
\end{ytableau}
\quad
}.
\]

\pause

\item Known proofs use algebraic geometry (Naruse) or complicated combinatorics (Morales/Pak/Panova and Konvalinka).


\end{itemize}

\end{frame}

\begin{frame}
\frametitle{\ \ \ \ The Pak--Postnikov generalization}

\begin{itemize}

\only<1>{
\item In 2001, Pak and Postnikov generalized the classical hook length formula in a different direction.
}

\pause

\item If $T$ is a standard tableau (of any shape), and if $k$ is a positive integer,
then \defn{$c_{T}\left(  k\right)$} shall denote the difference $j-i$, where
$\left(  i,j\right)  $ is the cell of $T$ that contains the entry $k$.

% \only<2-3>{
% \item The following filling shows the difference $j-i$ for each cell $\tup{i,j}$:
% \[
% \begin{ytableau}0 & 1 & 2 & 3 \\ -1 & 0 & 1
% \\ -2 & -1\end{ytableau}
% \]}
% \pause

\only<2>{
\item \textbf{Example:} If $
T = 
\begin{ytableau} 
1 & 3 & 4\\ 
2 & 5 
\end{ytableau} $,
then
\begin{align*}
    c_T(1) &= 1-1 = 0, \\
    c_T(2) &= 1-2 = -1, \\
    c_T(3) &= 2-1 = 1, \\
    c_T(4) &= 3-1 = 2, \\
    c_T(5) &= 2-2 = 0.
\end{align*}
}

\only<3>{
\item \textbf{Example:} If $
T = 
\begin{ytableau} 
1 & 3 & 4\\ 
{\blue 2} & 5 
\end{ytableau} $,
then
\begin{align*}
    c_T(1) &= 1-1 = 0, \\
    {\blue c_T(2)\ } & {= \blue 1-2 = -1}, \\
    c_T(3) &= 2-1 = 1, \\
    c_T(4) &= 3-1 = 2, \\
    c_T(5) &= 2-2 = 0.
\end{align*}
}

\only<4>{
\item \textbf{Example:} If $
T = 
\begin{ytableau} 
{\blue 1} & 3 & 4\\ 
2 & {\blue 5} 
\end{ytableau} $,
then
\begin{align*}
    {\blue c_T(1)\ } & {\blue = 1-1 = 0}, \\
    c_T(2) &= 1-2 = -1, \\
    c_T(3) &= 2-1 = 1, \\
    c_T(4) &= 3-1 = 2, \\
    {\blue c_T(5)\ } & {\blue = 2-2 = 0}.
\end{align*}
}

\only<5>{
\item \textbf{Example:} If $
T = 
\begin{ytableau} 
1 & {\blue 3} & 4\\ 
2 & 5
\end{ytableau} $,
then
\begin{align*}
    c_T(1) & = 1-1 = 0, \\
    c_T(2) &= 1-2 = -1, \\
    {\blue c_T(3)\ } & {\blue = 2-1 = 1}, \\
    c_T(4) &= 3-1 = 2, \\
    c_T(5) & = 2-2 = 0.
\end{align*}
}

\only<6>{
\item \textbf{Example:} If $
T = 
\begin{ytableau} 
1 & 3 & {\blue 4}\\ 
2 & 5 
\end{ytableau} $,
then
\begin{align*}
    c_T(1) &= 1-1 = 0, \\
    c_T(2) &= 1-2 = -1, \\
    c_T(3) &= 2-1 = 1, \\
    {\blue c_T(4)\ } & {\blue = 3-1 = 2}, \\
    c_T(5) &= 2-2 = 0.
\end{align*}
}

\pause \pause \pause \pause \pause
\item Let
$\ldots,z_{-2},z_{-1},z_{0},z_{1},z_{2},\ldots$ be
commuting indeterminates.

\pause
\item
For any cell $c=\left(  i,j\right)  $ of $Y\left(
\lambda\right)  $, we define the \defn{algebraic
hook length}\ $h_{\lambda}\left(  c;z\right)  $ by%
\[
h_{\lambda}\left(  c;z\right)  :=\sum_{\left(  i,j\right)
\in H_{\lambda}\left(  c\right)  }z_{j-i}.
\]

\pause

\item For any standard tableau $T$ with $n$ cells, we define the fraction%
\[
z_{T}:=\dfrac{1}{\prod_{k=1}^{n}\left(  z_{c_{T}\left(  k\right)  }%
+z_{c_{T}\left(  k+1\right)  }+\cdots+z_{c_{T}\left(  n\right)  }\right)  }%
\]

\pause

\item The \defn{Pak-Postnikov generalization of the hook length formula} states that
\[
\sum_{T\in\operatorname*{SYT}\left(  \lambda\right)  }z_{T}=\prod_{c\in
Y\left(  \lambda\right)  }\dfrac{1}{h_{\lambda}\left(  c;z\right)  }.
\]

\end{itemize}

\end{frame}

\begin{frame}
\frametitle{\ \ \ \ The Pak--Postnikov generalization: example}

\begin{itemize}

\item \textbf{Example:} For $\lambda = \tup{2,1}$, we have $\operatorname{SYT}\tup{\lambda} = \set{ \ytableaushort{12,3}, \quad \ytableaushort{13,2}}$, so the formula becomes
\begin{align*}
& \dfrac{1}{z_{-1}\tup{z_{-1}+z_1}\tup{z_{-1}+z_1+z_0}}
+
\dfrac{1}{z_1\tup{z_1+z_{-1}}\tup{z_1+z_{-1}+z_0}} \\
& =
\dfrac{1}{\tup{z_1+z_{-1}+z_0} z_1 z_{-1}}.
\end{align*}

\pause

\item Known proofs involve polytopes (Pak/Postnikov) or P-partitions and tropical RSK (Hopkins).

\end{itemize}

\end{frame}


\begin{frame}
\frametitle{\ \ \ \ Merging the streams: skew Pak--Postnikov}

\begin{itemize}

\only<1>{
\item We propose a generalization of the Pak--Postnikov formula to skew diagrams, thus extending Naruse's hook length formula as well.
}

\pause

\item \textbf{Main theorem.}
Let $\lambda$ and $\mu$ be two partitions with $\mu \subseteq \lambda$ such that the skew diagram $Y\tup{\lambda / \mu}$ has $n$ cells.
% Let $\ldots,z_{-2},z_{-1},z_{0},z_{1},z_{2},\ldots$ be commuting indeterminates.
\only<2>{

Define $z_T$ for $T \in \SYT\tup{\lambda/\mu}$ as before.

Define $h_{\lambda}\left(  c;z\right)$ for $c \in Y\tup{\lambda}$ as before (this does not depend on $\mu$!).}
\pause
Then,
\[
\sum_{T\in\operatorname*{SYT}\left(  \lambda/\mu\right)  }z_{T}=\sum
_{E\in\mathcal{E}\left(  \lambda/\mu\right)  }\ \ \prod_{c\in Y\left(
\lambda\right)  \setminus E}\dfrac{1}{h_{\lambda}\left(  c;z\right)
}.
\]

\pause

\begingroup
\scriptsize
\item \textbf{Example:} For $\lambda = (2,2)$ and $\mu = (1)$, we have 
\[
\operatorname{SYT}\tup{\lambda/\mu} = \set{ \ytableaushort{\none1,23}, \quad \ytableaushort{\none2,13}}
\quad \text{and} \quad
\mathcal{E}
\left(\lambda/\mu\right) = \set{ \begin{ytableau} 
\ast &\\ 
\phantom{x} &
\end{ytableau}, \quad \begin{ytableau} 
\phantom{x} &\\ 
\phantom{x} & \ast
\end{ytableau}},
\]
so the formula becomes
\begin{align*}
\frac{1}{z_0\cdot(z_0 + z_{-1})\cdot(z_0 + z_{-1} + z_{1})} + 
\frac{1}{z_0\cdot(z_{0} + z_1)\cdot(z_0 + z_{1} + z_{-1})} = \\
\frac{1}{(z_1 + z_0) \cdot z_0 \cdot (z_{-1} + z_0)} + \frac{1}{(z_1 + z_0) \cdot (z_0 + z_{-1} + z_1) \cdot (z_{-1} + z_0)} .
\end{align*}
\endgroup

\pause

\item This was first observed by Grinberg.
An intricate combinatorial proof was sketched by Konvalinka in 2019.

\end{itemize}

\end{frame}


\begin{frame}
\frametitle{\ \ \ \ Proof idea: the Konvalinka recursion, 1}

\begin{itemize}

\only<1-2>{\item We propose a new, elementary proof of this generalized formula.}

\pause

\only<2>{\item Induct on $\abs{Y\tup{\lambda/\mu}}$, increasing $\mu$ by one cell in the induction step.}

\pause

\item Let \defn{$f(\lambda/\mu)$} $= \sum_{T\in\operatorname*{SYT}\left(  \lambda\right)  }z_{T}$.

\pause

\item We easily obtain the recurrence
\[
z_{T} = \frac{1}{\sum\limits_{(i,j) \in Y\left( \lambda/\mu\right)} z_{j-i}}\cdot z_{T'},
\]
where $T'$ is the same tableau as $T$, with the entry $1$ removed and all other entries decreased by $1$.

\pause

\item \textbf{Example:} Let $\lambda = \tup{3,3,2}$ and $\mu = \tup{2,1}$.
\[
\begin{tabular}{c|c}
\text{If $T$ is ...} & \text{then $T'$ is ...} \\
$\ytableausetup{centertableaux}
\begin{ytableau}
\none & \none & 2 \\
\none & 1 & 3 \\
4 & 5 \\
\end{ytableau}
\in \SYT\tup{\lambda / \mu} $
&
$\ytableausetup{centertableaux}
\begin{ytableau}
\none & \none & 1 \\
\none & \none & 2 \\
3 & 4 \\
\end{ytableau}
\in \SYT\tup{\lambda / \nu}$
\end{tabular}
\]
for $\nu = \tup{2,2}$.
Thus,
$z_{T} = \frac{1}{z_{-1} + z_{-2} + z_{1}+z_{2}+z_0}\cdot z_{T'}$.

\end{itemize}

\end{frame}


\begin{frame}
\frametitle{\ \ \ \ Proof idea: the Konvalinka recursion, 2}

\begin{itemize}

\item Thus we get a recurrence for $f(\lambda/\mu)$:
\[
f\left(  \lambda/\mu\right)  =\frac{1}{\sum\limits_{(i,j)\in Y\left(
\lambda/\mu\right)  }z_{j-i}}\cdot\sum\limits_{\mu\lessdot\nu\subseteq\lambda
}f\left(  \lambda/\nu\right)  .
\]
\item Here, \defn{$\mu\lessdot\nu$} means that the partition $\nu$ is obtained by adding $1$ to some entry of $\mu$.

\pause

\item The induction step thus reduces to the following claim:

\item \textbf{Konvalinka recursion.} Let $\lambda/\mu$ be any skew partition, and let $x_{1},x_{2},x_3,\ldots$ and $y_{1},y_{2},y_3,\ldots$ be two infinite families of commuting indeterminates.
Then,
\begin{align*}
& \left(\sum_{\nexists i: \ \lambda_k-k = \mu_i-i} x_k+ \sum_{\nexists j: \ \lambda^t_p-p = \mu^t_j-j} y_p\right) \sum_{D \in \mathcal{E}\left(\lambda/\mu\right)}\ \ \prod_{(i,j)\in D}(x_{i}+y_{j}) \nonumber\\
& = \ \sum_{\mu \lessdot \nu \subseteq \lambda}\ \ \sum_{D \in \mathcal{E}\left(\lambda/\nu\right)}\ \ \prod_{(i,j)\in D}(x_{i}+y_{j}) .
\end{align*}

\end{itemize}

\end{frame}


\begin{frame}
\frametitle{\ \ \ \ Proof ingredient 1: Flagged SSYTs, 1}

\begin{itemize}

\item Let $D$ be a diagram. A \defn{semistandard tableau} (of shape $D$) means a filling of the cells of $D$ with positive integers such that
\begin{itemize}
    \item the numbers \textbf{weakly} increase along each row,
    \item the numbers \textbf{strictly} increase down each column.
\end{itemize} 
\only<1>{
\item 
\textbf{Example:} Here is a semistandard tableau for $\mu=(4,3,1)$:
\[\begin{ytableau} 1&1&1&2\\  2&3&3\\ 4 \end{ytableau}\]
}
%Can we make it a set with no flagging

\pause

\item A \defn{flagging} means a sequence $\bb:=\tup{b_1, b_2, b_3, \ldots}$ of positive integers.
% Sequence $\tup{b_1, b_2, b_3, \ldots}$ is \emph{weakly increasing}, so $b_1 \leq b_2 \leq b_3 \leq \cdots$. 

\pause
\item A \defn{flagged semistandard tableau} of shape $(\mu, \bb)$ is a semistandard tableau of shape $Y\tup{\mu}$ in which all entries in row $i$ are $\leq b_i$. \vspace{-1pc}
\[
\begin{tabular}
[c]{ccccc}\cline{1-2}%
\multicolumn{1}{|c}{\phantom{w}} & \multicolumn{1}{|c}{\phantom{w}} & \multicolumn{1}{|c}{\phantom{w}} & $\leq
b_{1}$ & \\\cline{1-2}%
\multicolumn{1}{|c}{} & \multicolumn{1}{|c}{} & \multicolumn{1}{|c}{} & $\leq
b_{2}$ & \\\cline{1-1}\cline{1-2}%
\multicolumn{1}{|c}{} & \multicolumn{1}{|c}{} &  & $\leq b_{3}$ &
\\\cline{1-1}
&  &  &  &
\end{tabular}
\]

\pause \vspace{-1pc}
\item For two partitions $\lambda$ and $\mu$, we define \defn{$\mathcal{F}(\lambda/\mu)$} to be the set of flagged semistandard tableaux of shape $(\mu, \bb)$, where $\bb:=\tup{b_1, b_2, b_3, \ldots}$ with
\[
b_i := \max\set{k \geq i \mid \lambda_k - k \geq \mu_i - i } \qquad \text{ for all $i\geq 1$.}
\]

\end{itemize}

\end{frame}


\begin{frame}
\frametitle{\ \ \ \ Proof ingredient 1: Flagged semistandard tableaux, 2}

\begin{itemize}

\item Now, there is a bijection from $\mathcal{E}(\lambda/\mu)$ to $\mathcal{F}(\lambda/\mu)$, defined as follows: Each excitation $D \in \mathcal{E}(\lambda/\mu)$ is sent to the flagged semistandard tableau $T$ of shape $(\mu, \bb)$, where
\[
T(i,j) = i + \tup{\# \text{ of excited moves that cell $(i, j)$ makes in $D$}}.
\]
Here $T(i, j)$ means the entry of $T$ in cell $(i, j)$.

\begingroup
\scriptsize
\vspace{0.5pc}
\item \textbf{Example:} For $\lambda = (3, 3, 3, 1)$ and $\mu = (2, 1)$:
\[
\begin{ytableau}
\ast & \ast &\\ 
\ast & &\\
\phantom{x} & &\\
\phantom{x}
\end{ytableau}
\longleftrightarrow
\begin{ytableau} 
1 & 1\\ 
2
\end{ytableau}
\quad 
\begin{ytableau} 
\ast & &\\ 
\ast & & \ast\\
\phantom{x} & &\\
\phantom{x}
\end{ytableau}
\longleftrightarrow
\begin{ytableau} 
1 & 2\\ 
2 
\end{ytableau}
\quad 
\begin{ytableau} 
\ast & \ast &\\ 
\phantom{x} & & \\
\phantom{x} & \ast &\\
\phantom{x}
\end{ytableau}
\longleftrightarrow
\begin{ytableau} 
1 & 1\\ 
3 
\end{ytableau}
\]
\[
\begin{ytableau} 
\ast & &\\ 
\phantom{x} & &\ast\\
\phantom{x} & \ast &\\
\phantom{x}
\end{ytableau}
\longleftrightarrow
\begin{ytableau} 
1 & 2\\ 
3 
\end{ytableau}
\quad 
\begin{ytableau} 
\phantom{x} & &\\ 
\phantom{x} & \ast & \ast\\
\phantom{x} & \ast &\\
\phantom{x}
\end{ytableau}
\longleftrightarrow
\begin{ytableau} 
2 & 2\\ 
3 
\end{ytableau}
\]
\endgroup

\pause


\item Thus, we can work with flagged semistandard tableaux instead of excited diagrams.

\end{itemize}

\end{frame}


\begin{frame}
\frametitle{\ \ \ \ Proof ingredient 2: a general Jacobi--Trudi formula, 1}

\begin{itemize}

\item \textbf{Theorem (generalized Jacobi--Trudi formula).} Let $\lambda = \tup{\lambda_1 \geq \lambda_2 \geq \cdots \geq \lambda_k}$ and
$\mu = \tup{\mu_1 \geq \mu_2 \geq \cdots \geq \mu_k}$ be two
partitions. Let $a_{1}\leq a_{2}\leq\cdots\leq a_{k}$ and $b_{1}\leq b_{2}%
\leq\cdots\leq b_{k}$ be positive integers. Let $u_{i,j}$ be a variable for each pair $\left(  i,j\right)  \in\mathbb{Z}^2$.

Then,
\begin{align*}
&  \sum_{\substack{T\text{ is a semistandard tableau}\\\text{of shape
}Y\left(  \lambda/\mu\right)  ;\\a_{i}\leq T\left(  i,j\right)  \leq
b_{i}\text{ for all }\left(  i,j\right)  }}\ \ \prod_{(i,j)\in Y(\lambda/\mu
)}u_{j-i,\ T\left(  i,j\right)  }\\
&  =\det\left(  \sum_{a_{i}\leq t_{\mu_{i}+1}  \leq t_{\mu_{i}+2}  \leq\cdots\leq t_{\lambda_{j}}  \leq b_{j}%
}\ \ \prod_{c=\mu_{i}+1}^{\lambda_{j}}u_{c-i,\ t_c} \right) _{i,j\in\left[
k\right]  }.
\end{align*}

\item This is implicit in a preprint of Gessel and Viennot 1989.

\end{itemize}

\end{frame}


\begin{frame}
\frametitle{\ \ \ \ Proof ingredient 2: a general Jacobi--Trudi formula, 2}

\begin{itemize}

\item If $\mu=\left(  0,0,\ldots,0\right)  $ and all $a_{i}$ are $0$ as well,
and if $u_{i,j}=x_{j}+y_{i+j}$, and if we rename $\lambda$ as $\mu$, then the
left hand side here becomes%
\[
\sum_{\substack{T \text{ is a flagged semistandard} \\ \text{ tableau of shape }(\mu,\mathbf{b})}}
\ \ \prod_{(i,j)\in Y(\mu)}(x_{T(i,j)}+y_{T(i,j)+j-i}),
\]
\pause
which equals the%
\[
\sum_{D\in\mathcal{E}\left(  \lambda/\mu\right)  }\ \ \prod_{\left(
i,j\right)  \in D}\left(  x_{i}+y_{j}\right)
\]
in the Konvalinka recursion.

\end{itemize}

\end{frame}


\begin{frame}
\frametitle{\ \ \ \ Proof ingredient 3: a determinantal identity}

\begin{itemize}

\only<1>{
\item Jacobi--Trudi transforms both sides of the Konvalinka recursion into sums of determinants.

\item After some nontrivial work, it becomes an easy determinantal identity:
}

\pause

\item \textbf{Theorem.} Let $M$ and ${\red N}$ be two $n\times n$-matrices. Then,
\begin{center}
$\sum_{k=1}^n \det( M \text{ with its $k$-th row replaced }$ \\
$\qquad \qquad \qquad \text{by the $k$-th row of ${\red N}$} ) $ \\
$\phantom{.}
= \sum_{k=1}^n \det( M \text{ with its $k$-th column replaced }$ \\
$ \qquad \qquad \qquad \text{by the $k$-th column of ${\red N}$} )$.
\end{center}

% \[
% t^1\text{-coefficient of }\det\left(A+tC\right)
% \]

\begingroup
\small
\item \textbf{Example:}
\begin{align*}
\vspace{-1pc}
& \det
\begin{pmatrix}
\red{A} & \red{B} & \red{C}\\
a^{\prime} & b^{\prime} & c^{\prime}\\
a^{\prime\prime} & b^{\prime\prime} & c^{\prime\prime}%
\end{pmatrix}
  +\det
\begin{pmatrix}
a & b & c\\
\red{A^{\prime}} & \red{B^{\prime}} & \red{C^{\prime}}\\
a^{\prime\prime} & b^{\prime\prime} & c^{\prime\prime}%
\end{pmatrix}
  +\det
\begin{pmatrix}
a & b & c\\
a^{\prime} & b^{\prime} & c^{\prime}\\
\red{A^{\prime\prime}} & \red{B^{\prime\prime}} & \red{C^{\prime\prime}}%
\end{pmatrix}
  \\
&=\det
\begin{pmatrix}
\red{A} & b & c\\
\red{A^{\prime}} & b^{\prime} & c^{\prime}\\
\red{A^{\prime\prime}} & b^{\prime\prime} & c^{\prime\prime}%
\end{pmatrix}
  +\det
\begin{pmatrix}
a & \red{B} & c\\
a^{\prime} & \red{B^{\prime}} & c^{\prime}\\
a^{\prime\prime} & \red{B^{\prime\prime}} & c^{\prime\prime}%
\end{pmatrix}
  +\det
\begin{pmatrix}
a & b & \red{C}\\
a^{\prime} & b^{\prime} & \red{C^{\prime}}\\
a^{\prime\prime} & b^{\prime\prime} & \red{C^{\prime\prime}}%
\end{pmatrix} .
\end{align*}
\endgroup

\end{itemize}

\end{frame}


\begin{frame}
\frametitle{\ \ \ \ Proof ingredient 4: Combinatorial lemmas}

\begin{itemize}

\item Two of the lemmas used along the way:

\item \textbf{Lemma 1.}
        Let $\lambda$ be a partition.
        Let $\lambda^t$ be its conjugate (i.e., Young diagram flipped across the main diagonal).
        
        Then, the sets
        \[ \set{\lambda_i - i\ \vert\ i\in \mathbb{N}}\quad\text{and}\quad \set{j - \lambda^t_j - 1\ \vert\ j\in \mathbb{N}}\] are disjoint and their union is $\mathbb{Z}$.

\item \textbf{Lemma 2.}
    Let $\bb = \tup{b_1, b_2, b_3, \ldots}$ be the flagging of $\lambda/\mu$.

    Let $\mu^{+i}$ be the partition obtained from $\mu$ by increasing the $i$-th entry by $1$.
    
     Let $\bb^{*i} = \tup{b^{*i}_1, b^{*i}_2, b^{*i}_3, \ldots}$ be the flagging induced by $\lambda/\mu^{+i}$.
    
     Then:
         \[
           -1 \leq b^{*i}_i-b_i \leq 0,
           \qquad \text{and} \qquad
           b^{*i}_k = b_k \text{ for all $k \neq i$}.
         \]

\end{itemize}

\end{frame}

\begin{frame}
\frametitle{\ \ \ \ Bibliography}

\begin{itemize}

\item \href{http://www.samuelfhopkins.com/docs/rsk.pdf}{Sam Hopkins, \textit{RSK via local transformations}}. {\red \url{http://www.samuelfhopkins.com/docs/rsk.pdf}}

\item \href{https://doi.org/10.1016/j.ejc.2020.103104}{Matja\v{z} Konvalinka,
\textit{A bijective proof of the hook length formula for skew shapes}, 2020.} Preprint:
{\red \url{https://www.fmf.uni-lj.si/~konvalinka/excited.pdf}}

\item \href{https://doi.org/10.1016/j.ejc.2019.103079}{Matja\v{z} Konvalinka,
\textit{Hook, line and sinker: a bijective proof of the skew shifted hook-length formula}, 2019.} Preprint: {\red \url{https://www.fmf.uni-lj.si/~konvalinka/excited_shifted.pdf}}

\item \href{https://doi.org/10.1016/j.jcta.2017.09.002}{Alejandro H. Morales, Igor Pak, Greta Panova, \textit{Hook formulas for skew shapes I. $q$-analogues and bijections}, 2018.} Preprint:
{\red \url{https://arxiv.org/abs/1512.08348v5}}.

\item Bruce E. Sagan, \textit{Combinatorics: The Art of Counting}, AMS 2020. {\red \url{https://users.math.msu.edu/users/bsagan/Books/Aoc/final.pdf}}

\item Richard P. Stanley, \textit{Enumerative Combinatorics, volume 2}, 2001. See {\red \url{http://math.mit.edu/~rstan/ec/}} for errata.

\end{itemize}

\end{frame}

\begin{frame}
\frametitle{\ \ \ \ Thanks to ...}

\begin{itemize}

\item the \textbf{Yulia's Dream} program and \textbf{Pavel Etingof}, \textbf{Slava Gerovich}, \textbf{Vasily Dolgushev}, \textbf{Dmytro Matvieievskyi} in particular.

\item {\red\textbf{Darij Grinberg}} for mentoring this project, and {\blue{{\textbf{Andrei Ionov}}}} for mentorship during summer.

\item \textbf{Alexander Postnikov}, \textbf{Matja\v{z} Konvalinka} for discussions on hook-length formulas.

\item our families for their support.

\end{itemize}

\end{frame}





\end{document}

\begin{frame}
\frametitle{\ \ \ \ Frame}

\begin{itemize}

\item 

\end{itemize}

\end{frame}


