% Slides version

% Title: Tales from the descent algebra

% \begin{abstract}
% \textbf{Abstract.} The descent algebra (introduced by Solomon in 1976) is a
% remarkable $2^{n-1}$-dimensional subalgebra of the group algebra of the
% symmetric group $S_{n}$, spanned by sums of all permutations with a given
% descent set. In some aspects, it is similar to the symmetric group algebra,
% while in others, it is entirely opposite, such as its heavy
% non-semisimplicity. (It also makes for a natural example of an algebra not
% isomorphic to its opposite.)

% In this talk, I will present two new developments on the descent algebra:

% \begin{enumerate}
% \item A new basis of the symmetric group algebra represents the elements of
% the descent algebra as triangular matrices with combinatorially meaningful
% diagonal entries. This gives a new approach to Bidigare's eigenvalue formulas
% for descent-related shuffles.

% This is joint work with Ekaterina A. Vassilieva.

% \item Certain generators of the descent algebra generate right ideals of the
% group algebra that are Gelfand models of $S_{n}$: representations that contain
% each irreducible exactly once. This leads to a new view of the classical
% Gelfand model of Adin, Postnikov and Roichman as well as a new proof of one of
% the Reiner-Saliola-Welker commutativities; we also resolve recent questions of Lafreni\`{e}re.

% This is joint work with Sarah Brauner, Patricia Commins and Franco Saliola.

% \end{enumerate}
% \end{abstract}

\documentclass[hyperref={colorlinks,linkcolor=blue,filecolor=red,urlcolor=red}]{beamer}
\usepackage[all,cmtip]{xy}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsthm}
\usepackage{tikz}
\usepackage{needspace}
\usepackage{ytableau}
\usepackage{tabls}
\usepackage{comment}
%TCIDATA{OutputFilter=latex2.dll}
%TCIDATA{Version=5.50.0.2960}
%TCIDATA{LastRevised=Thursday, March 19, 2026 16:19:17}
%TCIDATA{SuppressPackageManagement}
%TCIDATA{<META NAME="GraphicsSave" CONTENT="32">}
%TCIDATA{<META NAME="SaveForMode" CONTENT="1">}
%TCIDATA{BibliographyScheme=Manual}
%TCIDATA{Language=American English}
%BeginMSIPreambleData
\providecommand{\U}[1]{\protect\rule{.1in}{.1in}}
%EndMSIPreambleData
\usetikzlibrary{arrows}
\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}
\definecolor{darkred}{rgb}{0.7,0,0} % darkred color
\newcommand{\red}{\color{red}}
\newcommand{\grey}{\color{grau}}
\newcommand{\green}{\color{forestgreen}}
\newcommand{\violet}{\color{violet}}
\newcommand{\blue}{\color{blue}}
\newcommand{\defn}[1]{{\color{darkred}\emph{#1}}}
\newcommand{\defnm}[1]{{\color{darkred}{#1}}}
\newenvironment{statement}{\begin{quote}}{\end{quote}}
\newenvironment{fineprint}{\begin{small}}{\end{small}}
\iffalse
\newenvironment{proof}[1][Proof]{\noindent\textbf{#1.} }{\ \rule{0.5em}{0.5em}}
\newenvironment{question}[1][Question]{\noindent\textbf{#1.} }{\ \rule{0.5em}{0.5em}}
\fi
\newcommand{\arinj}{\ar@{_{(}->}}
\newcommand{\arinjrev}{\ar@{^{(}->}}
\newcommand{\arsurj}{\ar@{->>}}
\newcommand{\arelem}{\ar@{|->}}
\newcommand{\arback}{\ar@{<-}}
\newcommand{\id}{\operatorname{id}}
\newcommand{\ed}{\end{document}}
\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}
\newcommand*\circled[1]{\tikz[baseline=(char.base)]{\node[shape=circle,fill,inner sep=2pt] (char) {\textcolor{white}{#1}};}}
\newcommand{\icast}{\circled{$\ast$}}
\newcommand\arxiv[1]{\href{http://www.arxiv.org/abs/#1}{\texttt{arXiv:#1}}}
\setlength\tablinesep{3pt}
\setlength\arraylinesep{3pt}
\setlength\extrarulesep{3pt}
\newenvironment{noncompile}{}{}
\excludecomment{noncompile}
\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}
\setbeamertemplate{headline}{}
\newcommand{\fti}[1]{\frametitle{\ \ \ \ \ #1}}
\newenvironment{iframe}[1][]{\begin{frame} \fti{[#1]} \begin{itemize}}{\end{frame}\end{itemize}}
\begin{document}

\title{Tales of the descent algebra}
\author{Darij Grinberg (Drexel University)
\and \and \and \and \and joint work with Ekaterina A. Vassilieva,
\and Sarah Brauner, \and Patricia Commins,
\and Franco Saliola}
\date{Richard P. Stanley Seminar in Combinatorics, \\ MIT, 2026-03-20}

\frame{\titlepage
\textbf{Slides of this talk:}
\begin{itemize}
\item \url{https://www.cip.ifi.lmu.de/~grinberg/algebra/da2026s.pdf}
\end{itemize}}

% \begin{frame}
% \fti{Where to find this work}

% \textbf{Preprints:}

% \begin{itemize}
% \item Sarah Brauner, Patricia Commins, Darij Grinberg, Franco Saliola,
% \textit{A left ideal Gelfand model for the symmetric group},\newline\url{https://www.cip.ifi.lmu.de/~grinberg/algebra/dyadic.pdf}
% \end{itemize}

% \noindent\textbf{Slides of this talk:}

% \begin{itemize}
% \item \url{https://www.cip.ifi.lmu.de/~grinberg/algebra/da2026s.pdf}

% \begin{noncompile}
% \url{https://darijgrinberg.gitlab.io/algebra/da2026s.pdf}
% \end{noncompile}
% \end{itemize}

\begin{frame}
\fti{Chapter 1}
\begin{center}
{\Huge \bf \sc Chapter 1} \\
\noindent\rule[0.5ex]{\linewidth}{1pt}
{\Large \bf The descent algebra: an introduction}
\end{center}
\vspace{1cm}
References:
\begin{itemize}
\item \href{https://doi.org/10.1016/0021-8693(76)90182-4}{Louis Solomon, \textit{A Mackey formula in the group ring of a Coxeter group}, 1976}.
\item \href{https://saliola.github.io/maths/publications/LectureNotes/DesAlgLectureNotes.pdf}{Franco Saliola, \textit{Hyperplane arrangements and descent algebras}, 2006} (\href{https://www.cip.ifi.lmu.de/~grinberg/algebra/saliola-errata.pdf}{errata}).
\item \href{https://doi.org/10.1016/0001-8708(89)90020-0}{A. M. Garsia, C. Reutenauer, \textit{A decomposition of Solomon's descent algebra}, 1989}.
\item \href{https://libgen.bz/edition.php?id=137349596}{Manfred Schocker, \textit{The descent algebra of the symmetric group}, 2003}.
\item \href{https://www.mat.univie.ac.at/~slc/opapers/s29laue.html}{D. Blessenohl, H. Laue, \textit{Algebraic Combinatorics Related to the Free Lie Algebra}, 1992} (\href{https://www.cip.ifi.lmu.de/~grinberg/algebra/blfl-errata.pdf}{errata}).
\end{itemize}
\end{frame}

\begin{frame} \frametitle{\ \ \ \ Notations}

\begin{itemize}
\item Let \defn{$\mathbf{k}$} be a field of characteristic $0$. (Some results hold
for $\mathbb{Z}$ as well.)

\pause
\item Let $\defnm{\mathbb{N}} = \left\{  0,1,2,\ldots\right\}  $.

\item Fix $\defnm{n} \in\mathbb{N}$.

\item Consider the symmetric group \defn{$S_{n}$} of the set
$\defnm{\left[  n\right]}
:=\left\{  1,2,\ldots,n\right\}  $.

\pause
\item Let $\defnm{\mathcal{A}}
:=\mathbf{k}\left[  S_{n}\right]  $ be its group algebra
over $\mathbf{k}$.

\pause
\item Any permutation $w\in S_{n}$ is written in
\defn{one-line notation} as a
bracketed word $\left[  w\left(  1\right)  \ w\left(  2\right)  \ \cdots
\ w\left(  n\right)  \right]  $.

\pause
\item The \defn{descents} of a permutation $w\in S_{n}$ are the $i\in\left[
n-1\right]  $ that satisfy $w\left(  i\right)  >w\left(  i+1\right)  $.

\item The \defn{descent set} of $w\in S_{n}$ is%
\[
\defnm{\operatorname*{Des}w}
:=\left\{  \text{descents of }w\right\}  \subseteq\left[
n-1\right]  .
\]

\end{itemize}

\end{frame}

\begin{frame} \frametitle{\ \ \ \ The descent algebra: definition}

\begin{itemize}
\item For each $I\subseteq\left[  n-1\right]  $, we define two elements
\[
\defnm{\mathbf{B}_{I}} =\sum_{\substack{w\in S_{n};\\\operatorname{Des}(w)\subseteq
I}}w\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ %
\defnm{\mathbf{D}_{I}}
=\sum_{\substack{w\in S_{n};\\\operatorname{Des}(w)=I}}w
\]
of $\mathcal{A}$.

\pause
\item \textbf{Proposition [2--]:}\only<2>{\footnote{Difficulty rating following
Stanley's EC standards.}} The families $\left(  \mathbf{B}_{I}\right)
_{I\subseteq\left[  n-1\right]  }$ and $\left(  \mathbf{D}_{I}\right)
_{I\subseteq\left[  n-1\right]  }$ are each linearly independent, and are
related by inclusion-exclusion:%
\[
\mathbf{B}_{I}=\sum_{J\subseteq I}\mathbf{D}_{J}%
;\ \ \ \ \ \ \ \ \ \ \mathbf{D}_{I}=\sum_{J\subseteq I}\left(  -1\right)
^{\left\vert I\setminus J\right\vert }\mathbf{B}_{J}.
\]

\pause
So they have the same span.
\pause

\item This span is called the \defn{descent algebra}%
\[
\defnm{\mathcal{D}}
:=\defnm{\mathcal{D}_{n}}
:=\operatorname*{span}\left\{  \mathbf{B}%
_{I}\ \mid\ I\subseteq\left[  n-1\right]  \right\}  =\operatorname*{span}%
\left\{  \mathbf{D}_{I}\ \mid\ I\subseteq\left[  n-1\right]  \right\}  .
\]
As a $\mathbf{k}$-vector space, it has dimension $2^{n-1}$ (or $1$ if $n=0$).
\end{itemize}

% \only<1,3,4>{\vspace{4pc}}

\end{frame}

\begin{frame} \frametitle{\ \ \ \ The Solomon Mackey formula}

\begin{itemize}
\item Why are we calling it \textquotedblleft descent \textbf{algebra}%
\textquotedblright?
\pause

\item For any $I\subseteq\left[  n-1\right]  $, define $W_{I}$ to be the
\defn{Young subgroup} of $S_{n}$ corresponding to $I$. It is the subgroup
generated by the simple transpositions $s_{i}:=t_{i,i+1}$ for $i\in I$.

\item Let $\defnm{\widetilde{I}}
:=\left[  n-1\right]  \setminus I$ for each
$I\subseteq\left[  n-1\right]  $.
\pause

\item \textbf{Solomon Mackey formula
(\href{https://doi.org/10.1016/0021-8693(76)90182-4}{\textbf{Louis Solomon,
1976}}) [3+]:} For any $I,J\subseteq\left[  n-1\right]  $, we have
\[
\mathbf{B}_{\widetilde{I}}\mathbf{B}_{\widetilde{J}}=\sum_{K\subseteq\left[
n-1\right]  }a_{K}^{I,J}\mathbf{B}_{\widetilde{K}},
\]
where $a_{K}^{I,J}\in\mathbb{N}$ is the \# of permutations $w\in S_{n}$ that
are minimal in their double coset $W_{I}wW_{J}$ and satisfy $w^{-1}W_{I}w\cap
W_{J}=W_{K}$.

\only<3>{
Here, \textquotedblleft minimal\textquotedblright\ means \textquotedblleft has
minimum length (= inversion number) in the double coset\textquotedblright, but
it is in fact also the unique minimum in the Bruhat order.}

\pause

\item \textbf{Corollary [1+]:} The descent algebra $\mathcal{D}$ is closed
under multiplication, thus really a subalgebra of $\mathcal{A}=\mathbf{k}%
\left[  S_{n}\right]  $.

\pause
\item Solomon proved the analogous fact for arbitrary finite Coxeter groups.
\end{itemize}

\end{frame}

\begin{frame} \frametitle{\ \ \ \ The Solomon Mackey formula: examples}

\begin{itemize}
\item For $n=3$, the descent algebra $\mathcal{D}$ has basis $\left(
\mathbf{B}_{\varnothing},\mathbf{B}_{\left\{  1\right\}  },\mathbf{B}%
_{\left\{  2\right\}  },\mathbf{B}_{\left\{  1,2\right\}  }\right)  $, where
(using one-line notation)%
\begin{align*}
\mathbf{B}_{\varnothing}  &  =\left[  123\right]  ,\\
\mathbf{B}_{\left\{  1\right\}  }  &  =\left[  123\right]  +\left[
213\right]  +\left[  312\right]  ,\\
\mathbf{B}_{\left\{  2\right\}  }  &  =\left[  123\right]  +\left[
132\right]  +\left[  231\right]  ,\\
\mathbf{B}_{\left\{  1,2\right\}  }  &  =\left[  123\right]  +\left[
132\right]  +\left[  213\right]  +\left[  231\right]  +\left[  312\right]
+\left[  321\right]  .
\end{align*}

\pause
\item Note that the permutations $\operatorname*{id}=\left[  123\cdots
n\right]  $ and $\defnm{w_{0}}=\left[  n\cdots321\right]  $ always lie in
$\mathcal{D}$, since%
\[
\operatorname*{id}=\mathbf{B}_{\varnothing}=\mathbf{D}_{\varnothing
}\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ w_{0}=\mathbf{D}_{\left[
n-1\right]  }.
\]

\pause
\item The \defn{Eulerian elements }$\mathbf{E}_{k}:=\sum_{\substack{w\in
S_{n};\\\left\vert \operatorname*{Des}w\right\vert =k}}w$ lie in
$\mathcal{D}$ as well, and form their own subalgebra, which is furthermore
commutative [3+].
\end{itemize}

\end{frame}

\begin{frame} \frametitle{\ \ \ \ Compositions, 1}

\begin{itemize}
\item A \defn{composition} of $n$ means a tuple of positive integers
summing to $n$. For instance, $\left(  2,1,5\right)  $ is a composition of $8$.

\item Let \defn{$\operatorname*{Comp}\nolimits_{n}$} be the set of all compositions
of $n$.
\pause

\item There is a bijection%
\begin{align*}
\defnm{D}
:\operatorname*{Comp}\nolimits_{n}  &  \rightarrow\left\{  \text{subsets of
}\left[  n-1\right]  \right\}  ,\\
\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{k}\right)   &  \mapsto\left\{
\alpha_{1}+\alpha_{2}+\cdots+\alpha_{i}\ \mid\ i\in\left[  k-1\right]
\right\}
\end{align*}
(the \textquotedblleft\defn{partial sum encoding}\textquotedblright\ of a
composition). For example,%
\[
D\left(  2,4,1,3\right)  =\left\{  2,6,7\right\}  .
\]
\end{itemize}

\end{frame}

\begin{frame} \frametitle{\ \ \ \ Compositions, 2}

\begin{itemize}
\item For any $\alpha=\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{k}\right)
\in\operatorname*{Comp}\nolimits_{n}$, we can break up the interval $\left[
n\right]  $ into $k$ intervals \defn{$I_{1}^{\alpha},I_{2}^{\alpha},\ldots
,I_{k}^{\alpha}$} of lengths $\alpha_{1},\alpha_{2},\ldots,\alpha_{k}$;
explicitly:%
\[
I_{j}^{\alpha}:=\left[  \alpha_{j-1}+1,\ \alpha_{j}\right]
\ \ \ \ \ \ \ \ \ \ \text{for all }j\in\left[  k\right]  .
\]
Visually:
\[
\begin{tikzpicture}[scale=0.5]
\node(a) at (0.7,5) {$1$};
\node(a) at (17.3,5) {$n$};
\draw(0,2) rectangle (5,4);
\draw(5,2) rectangle (8,4);
\draw(8,2) rectangle (12,4);
\draw(12,2) rectangle (18,4);
\node(A) at (2.5, 3) {$I_1^\alpha$};
\node(A) at (6.5, 3) {$I_2^\alpha$};
\node(A) at (10, 3) {$\cdots$};
\node(A) at (15, 3) {$I_k^\alpha$};
\node[blue] (a) at (2.5, 1.2) {$\alpha_1$};
\draw[->, blue] (0.1,1.2) -- (2,1.2);
\draw[->, blue] (4.9,1.2) -- (3,1.2);
\node[blue] (a) at (6.5, 1.2) {$\alpha_2$};
\draw[->, blue] (5.1,1.2) -- (6,1.2);
\draw[->, blue] (7.9,1.2) -- (7,1.2);
\node[blue] (a) at (15, 1.2) {$\alpha_k$};
\draw[->, blue] (12.1,1.2) -- (14.5,1.2);
\draw[->, blue] (17.9,1.2) -- (15.5,1.2);
\end{tikzpicture}
\]
For instance, if $\alpha=\left(  3,1,4\right)  \in\operatorname*{Comp}%
\nolimits_{8}$, then%
\[
I_{1}^{\alpha}=\left\{  1,2,3\right\}  ,\ \ \ \ \ \ \ \ \ \ I_{2}^{\alpha
}=\left\{  4\right\}  ,\ \ \ \ \ \ \ \ \ \ I_{3}^{\alpha}=\left\{
5,6,7,8\right\}  .
\]

\end{itemize}

\end{frame}

\begin{frame} \frametitle{\ \ \ \ The descent algebra: composition indexing}

\begin{itemize}

\item For any $\alpha\in\operatorname*{Comp}\nolimits_{n}$, set%
\[
\defnm{\mathbf{B}_{\alpha}}:=\mathbf{B}_{D\left(  \alpha\right)  }=\sum
_{\substack{w\in S_{n}\\\text{increases on each }I_{j}^{\alpha}}}w.
\]

\pause
\item So $\left(  \mathbf{B}_{\alpha}\right)  _{\alpha\in\operatorname*{Comp}%
\nolimits_{n}}$ is a basis of $\mathcal{D}$, just reindexing the basis
$\left(  \mathbf{B}_{I}\right)  _{I\subseteq\left[  n-1\right]  }$.

\end{itemize}

\end{frame}

\begin{frame} \frametitle{\ \ \ \ The Solomon Mackey formula: composition form}

\begin{itemize}

\item \textbf{Solomon Mackey formula in composition form
(\href{https://doi.org/10.1007/BF02582950}{\textbf{A. M. Garsia, J. Remmel,
1985}}) [3]:} For any compositions $\alpha=\left(  \alpha_{1},\alpha
_{2},\ldots,\alpha_{k}\right)  $ and $\beta=\left(  \beta_{1},\beta_{2}%
,\ldots,\beta_{\ell}\right)  $ of $n$, we have
\[
\mathbf{B}_{\alpha}\mathbf{B}_{\beta}=\sum_{\substack{M\in\mathbb{N}%
^{\ell\times k}\text{ has}\\\text{row sums }\beta_{1},\beta_{2},\ldots
,\beta_{\ell}\\\text{and column sums }\alpha_{1},\alpha_{2},\ldots,\alpha_{k}%
}}\mathbf{B}_{\operatorname*{read}M},
\]
where $\operatorname*{read}M$ is the reading word of $M$ (= the
concatenation of the rows of $M$ from top to bottom) with the zeroes removed.
Visually:\vspace{-0.9pc}
\[
\scalebox{0.9}{
\begin{tabular}
[c]{rcccc}
& $\alpha_{1}$ & $\alpha_{2}$ & $\cdots$ & $\alpha_{k}$\\
& $\downarrow$ & $\downarrow$ &  & $\downarrow$\\\cline{2-5}%
$\beta_{1}\rightarrow$ & \multicolumn{1}{|c}{r} & \multicolumn{1}{|c}{e} &
\multicolumn{1}{|c}{a} & \multicolumn{1}{|c|}{d}\\\cline{2-5}%
$\beta_{2}\rightarrow$ & \multicolumn{1}{|c}{i} & \multicolumn{1}{|c}{n} &
\multicolumn{1}{|c}{g} & \multicolumn{1}{|c|}{w}\\\cline{2-5}%
$\vdots$ & \multicolumn{1}{|c}{o} & \multicolumn{1}{|c}{r} &
\multicolumn{1}{|c}{d} & \multicolumn{1}{|c|}{$\cdots$}\\\cline{2-5}%
$\beta_{\ell}\rightarrow$ & \multicolumn{1}{|c}{} & \multicolumn{1}{|c}{} &
\multicolumn{1}{|c}{} & \multicolumn{1}{|c|}{}\\\cline{2-5}%
\end{tabular}
}
\]

\end{itemize}

\end{frame}

\begin{frame} \frametitle{\ \ \ \ The Solomon Mackey formula: example}

\begin{itemize}

\item \textbf{Example:} To compute $\mathbf{B}_{\left(  1,n-1\right)
}\mathbf{B}_{\left(  n-1,1\right)  }$, solve the nonogram%
\[%
\begin{tabular}
[c]{rcc}
& $1$ & $n-1$\\
& $\downarrow$ & $\downarrow$\\\cline{2-3}%
$n-1\rightarrow$ & \multicolumn{1}{|c}{$x$} & \multicolumn{1}{|c|}{$y$%
}\\\cline{2-3}%
$1\rightarrow$ & \multicolumn{1}{|c}{$z$} & \multicolumn{1}{|c|}{$w$%
}\\\cline{2-3}%
\end{tabular}
\ \ \ \ \ \ \ \ \ \ \left(  \text{for }x,y,z,w\in\mathbb{N}\right)  .
\]
\pause
Get%
\[%
\begin{tabular}
[c]{cccccc}
&  &  & $\left(
\begin{array}
[c]{cc}%
0 & n-1\\
1 & 0
\end{array}
\right)  $ &  & $\left(
\begin{array}
[c]{cc}%
1 & n-2\\
0 & 1
\end{array}
\right)  $\\
$\mathbf{B}_{\left(  1,n-1\right)  }$ & $\mathbf{B}_{\left(  n-1,1\right)  }$
& $=$ & $\mathbf{B}_{\left(  n-1,1\right)  }$ & $+$ & $\mathbf{B}_{\left(
1,n-2,1\right)  }.$%
\end{tabular}
\
\]

\end{itemize}

\end{frame}

\begin{frame} \frametitle{\ \ \ \ The descent algebra: properties, 1}

\begin{itemize}
\item Recall that $\mathcal{A}=\mathbf{k}\left[  S_{n}\right]  $ is semisimple
(Maschke). But $\mathcal{D}$ is anything but!

\pause
\item Let \defn{$\mathcal{Z}$} be the center of $\mathcal{A}$. As a $\mathbf{k}%
$-algebra, $\mathcal{Z}\cong\prod\limits_{\lambda\vdash n}\mathbf{k}$, with
each factor corresponding to a Specht module (irrep) of $S_{n}$.

\pause
\item \textbf{Theorem (still
\href{https://doi.org/10.1016/0021-8693(76)90182-4}{\textbf{Louis Solomon,
1976}}) [3--].} There is a surjective $\mathbf{k}$-algebra morphism%
\begin{align*}
\mathcal{D}  &  \rightarrow\mathcal{Z},\\
\mathbf{B}_{\alpha}  &  \mapsto\left(  \text{character of Young module for
}\alpha\right)  .
\end{align*}

\pause
\item \textbf{Theorem (still
\href{https://doi.org/10.1016/0021-8693(76)90182-4}{\textbf{Louis Solomon,
1976}}) [3]. }The kernel of this morphism is nilpotent! Thus it is the radical
of $\mathcal{D}$, while the semisimple quotient of $\mathcal{D}$ is
$\mathcal{Z}\cong\prod\limits_{\lambda\vdash n}\mathbf{k}$.

\pause
\item \textbf{Corollary [2--].} Each irreducible representation of
$\mathcal{D}$ is $1$-dimensional.

\pause
\item \textbf{Fun fact.} $\mathcal{D} \not\cong \mathcal{D}%
^{\operatorname*{op}}$ as algebras for $n=5$.

\end{itemize}

\end{frame}

\begin{frame} \frametitle{\ \ \ \ The descent algebra: properties, 2}

\begin{itemize}
\item Nowadays, the Solomon Mackey formula is usually proved using the face
algebra of the braid arrangement (combinatorially: a monoid of set
compositions): see
\href{https://saliola.github.io/maths/publications/LectureNotes/DesAlgLectureNotes.pdf}{Franco
Saliola's notes}
(\href{https://www.cip.ifi.lmu.de/~grinberg/algebra/saliola-errata.pdf}{errata}%
) or \href{https://radon.mat.univie.ac.at/~slc/s/s29laue.pdf}{Blessenohl/Laue}
(Proposition 4.3). The proof was found by
\href{https://www.proquest.com/docview/304356095}{Patrick Bidigare}.

Further developments by Ken Brown
(\textit{\href{https://arxiv.org/abs/math/0006145v1}{\textit{Semigroups, rings
and Markov chains}}} and
\textit{\href{https://web.archive.org/web/20150929050124/http://www.math.cornell.edu/~kbrown/papers/toronto.pdf}{\textit{Semigroup
methods}}}).

\pause
\item \href{https://doi.org/10.1016/0001-8708(89)90020-0}{Garsia and
Reutenauer 1989} construct idem-/nilpotent bases in $\mathcal{D}$ and relate
it to the free Lie algebra.

\end{itemize}

\end{frame}

\begin{frame} \frametitle{\ \ \ \ The descent algebra: properties, 3}

\begin{itemize}

\item $\mathcal{D}$ is wild in characteristic $p$, but some things are known
(e.g., \href{https://arxiv.org/abs/math/0408211}{Erdmann and Schocker 2004}).

\item The Ext-quiver of $\mathcal{D}$ is known. See
\href{https://libgen.bz/edition.php?id=137349596}{Schocker 2003} for survey on
all of the above.

\pause
\item The algebra $\mathcal{D}$ is isomorphic to $\operatorname*{NSym}%
\nolimits_{n}$ under the \emph{internal (Kronecker) product}. This is a lift
of the Kronecker product of symmetric functions.

\item For Hopf algebraists, $\operatorname*{NSym}$ consists of universal
endomorphisms of a graded connected cocommutative Hopf algebra. Internal
product = composition.

\item Generalization to non-cocommutative Hopf algebras:
\href{https://arxiv.org/abs/2401.14648}{arXiv:2401.14648}. But this does not
live in $\mathbf{k}\left[  S_{n}\right]  $ or anything like that.

\pause
\item And there is
\href{https://scholar.google.com/scholar?hl=en&q=solomon+"descent+algebra"}{much
more}.
\end{itemize}

\end{frame}

\begin{frame}
\fti{Chapter 2}
\begin{center}
{\Huge \bf \sc Chapter 2} \\
\noindent\rule[0.5ex]{\linewidth}{1pt}
{\Large \bf The LRM-basis of the symmetric group algebra}
\end{center}
\vspace{1cm}
References:
\begin{itemize}
\item \href{https://arxiv.org/abs/2601.02952}{Darij Grinberg, Ekaterina A.
Vassilieva, \textit{The left-to-right minima basis of the group algebra of the
symmetric group}, arXiv:2601.02952.}
\end{itemize}
\end{frame}

\begin{frame} \frametitle{\ \ \ \ The LRM basis: the question}

\begin{itemize}
\item Clearly, $\mathcal{A}$ is a left $\mathcal{D}$-module (since
$\mathcal{D}\subseteq\mathcal{A}$). But all irreps of $\mathcal{D}$ are
$1$-dimensional. So $\mathcal{A}$ has a Jordan-H\"{o}lder flag: a filtration
by left $\mathcal{D}$-submodules of dimensions $0,1,2,\ldots,n!$.

\pause
\textbf{Can we construct one explicitly?}

\pause
\item It is reasonable to build it from a basis of $\mathcal{A}$ that contains
(as subfamilies) bases of all the right ideals $\mathbf{B}_{I}\mathcal{A}%
=\mathbf{B}_{\alpha}\mathcal{A}$.

\item First step: understand these right ideals.
\end{itemize}

\end{frame}

\begin{frame} \frametitle{\ \ \ \ The left ideals $\mathbf{B}_{\alpha}\mathcal{A}$: definition and equality}

\begin{itemize}
\item \textbf{All of this chapter is joint work with Ekaterina A. Vassilieva
(\href{https://arxiv.org/abs/2601.02952}{arXiv:2601.02952}).}
\pause

\item For each composition $\alpha\in\operatorname*{Comp}\nolimits_{n}$, there
is a right ideal $\mathbf{B}_{\alpha}\mathcal{A}$ of $\mathcal{A}$. But some
of these are equal:%
\[
\mathbf{B}_{\left(  2,3\right)  }\mathcal{A}=\mathbf{B}_{\left(  3,2\right)
}\mathcal{A}\ \ \ \ \ \ \ \ \ \ \text{because }\mathbf{B}_{\left(  2,3\right)
}=\mathbf{B}_{\left(  3,2\right)  }\left[  45123\right]  .
\]
\pause
More generally:

\item \textbf{Proposition [2--].} If two compositions $\alpha,\beta
\in\operatorname*{Comp}\nolimits_{n}$ differ only in the order of their
entries, then $\mathbf{B}_{\alpha}\mathcal{A}=\mathbf{B}_{\beta}\mathcal{A}$.

\pause

\item For any composition $\alpha$, let \defn{$\widetilde{\alpha}$} be the partition
obtained by sorting the parts of $\alpha$ (in decreasing order). So we just showed:

\item \textbf{Proposition [2--].} If two compositions $\alpha,\beta
\in\operatorname*{Comp}\nolimits_{n}$ satisfy $\widetilde{\alpha
}=\widetilde{\beta}$, then $\mathbf{B}_{\alpha}\mathcal{A}=\mathbf{B}_{\beta
}\mathcal{A}$.

\pause

\item Thus the right ideals $\mathbf{B}_{\alpha}\mathcal{A}$ only depend on
the partitions $\widetilde{\alpha}$.

\end{itemize}

\end{frame}

\begin{frame} \frametitle{\ \ \ \ The left ideals $\mathbf{B}_{\alpha}\mathcal{A}$: inclusion from refinement}

\begin{itemize}
\item A partition $\lambda\vdash n$ is said to \defn{refine} a partition
$\mu\vdash n$ if we can obtain $\mu$ from $\lambda$ by merging parts (i.e., if
there exists a map $f$ such that each $j$ satisfies $\mu_{j}=\sum
\limits_{i;\ f\left(  i\right)  =j}\lambda_{i}$).

We write $\lambda\mathrel{\defnm{\preceq_{\pi}}}\mu$ for this (with
$\pi$ standing for \textquotedblleft
partition\textquotedblright). Not the same as refinement of compositions!
\pause

\item \textbf{Proposition [2+].} If two compositions $\alpha,\beta
\in\operatorname*{Comp}\nolimits_{n}$ satisfy $\widetilde{\alpha}\preceq_{\pi
}\widetilde{\beta}$, then $\mathbf{B}_{\alpha}\mathcal{A}\subseteq
\mathbf{B}_{\beta}\mathcal{A}$.
\pause

\item \textbf{Remark.} Over $\mathbb{Q}$, we also have $\mathbf{B}_{\alpha
}\mathcal{D}\subseteq\mathbf{B}_{\beta}\mathcal{D}$, but our above results
hold even over $\mathbb{Z}$.

\pause
\item So the right ideals $\mathbf{B}_{\alpha}\mathcal{A}$ form a
\textquotedblleft poset-indexed filtration\textquotedblright\ of $\mathcal{A}$
(indexed by the partitions of $n$, ordered by $\preceq_{\pi}$).

\pause
\item \textbf{Proposition [2].} Each $\mathbf{B}_{\alpha}\mathcal{A}$ is
closed under left $\mathcal{D}$-action, i.e., is a $\left(  \mathcal{D}%
,\mathcal{A}\right)  $-subbimodule of $\mathcal{A}$.

\end{itemize}

\end{frame}

\begin{frame} \frametitle{\ \ \ \ The left ideals $\mathbf{B}_{\alpha}\mathcal{A}$: subquotients}

\begin{itemize}

\item So our \textquotedblleft poset-indexed filtration\textquotedblright\ of
$\mathcal{A}$ is a filtration by left $\mathcal{D}$-modules, not just by right
ideals. How does $\mathcal{D}$ act on its subquotients?

\pause
\item \textbf{Proposition [2].} Fix $\alpha=\left(  \alpha_{1},\alpha
_{2},\ldots,\alpha_{k}\right)  $ and $\beta=\left(  \beta_{1},\beta_{2}%
,\ldots,\beta_{\ell}\right)  $ be two compositions of $n$. Let \defn{$\eta_{\beta
}(\alpha)$} be the number of ``ways how $\widetilde{\beta}$ refines
$\widetilde{\alpha}$'' (that is,
of functions $f:\left[  \ell\right]  \rightarrow
\left[  k\right]  $ such that each $j\in\left[  k\right]  $ satisfies
$\alpha_{j}=\sum_{i;\ f\left(  i\right)  =j}\beta_{i}$). Then left
multiplication by $\mathbf{B}_{\alpha}$ acts on the quotient $\mathbf{B}%
_{\beta}\mathcal{A}/\sum_{\widetilde{\gamma}\prec_{\pi}\widetilde{\beta}%
}\mathbf{B}_{\gamma}\mathcal{A}$ as multiplication by the scalar $\eta_{\beta
}(\alpha)$.

\pause
\item So we have a \textquotedblleft poset-indexed
filtration\textquotedblright\ of $\mathcal{A}$ by $\left(  \mathcal{D}%
,\mathcal{A}\right)  $-subbimodules, and $\mathcal{D}$ acts by scalars on the subquotients.

\pause
\item Thus there must be a basis of $\mathcal{A}$ that restricts to bases of
these subbimodules. In fact, many such bases. Let's find one!
\end{itemize}

\end{frame}

\begin{frame} \frametitle{\ \ \ \ Left-to-right minima}

\begin{itemize}
\item For any permutation $w\in S_{n}$, the \defn{left-to-right minima} of $w$
are the numbers that are smaller than everything before them in the one-line
notation of $w$. Formally, they are the $w\left(  i\right)  $ such that
$w\left(  1\right)  ,w\left(  2\right)  ,\ldots,w\left(  i-1\right)  $ are all
$>w\left(  i\right)  $.

\pause
Furthermore,

\begin{itemize}
\item we let $\defnm{\operatorname*{LRM}w}\subseteq\left[  n\right]  $ be the set of
all left-to-right minima of $w$ (including $1$ and $w\left(  1\right)  $);

\item we let $\defnm{\operatorname*{LRM}\nolimits^{\prime}w}\subseteq\left[
n-1\right]  $ be the set of $k-1$ for all $k\in\operatorname*{LRM}w$ distinct
from $1$;

\item we let $\defnm{\operatorname*{cLRM}\nolimits^{\prime}w}$ denote the composition
$\alpha\in\operatorname*{Comp}\nolimits_{n}$ such that $D\left(
\alpha\right)  =\operatorname*{LRM}\nolimits^{\prime}w$.
\end{itemize}

\pause
\only<3>{
\item \textbf{Example:} If $n=9$ and $w=\left[  946283517\right]  $, then%
\begin{align*}
\operatorname*{LRM}w  &  =\left\{  9,4,2,1\right\}  ;\\
\operatorname*{LRM}\nolimits^{\prime}w  &  =\left\{  8,3,1\right\}  ;\\
\operatorname*{cLRM}\nolimits^{\prime}w  &  =\left(  1,2,5,1\right)  .
\end{align*}
}

\pause
\only<4>{
\item \textbf{Exercise [2]:} For every $w\in S_{n}$, show
\[
\operatorname{LRM}%
\left(  w^{-1}\right)  =w^{-1}\left(  \operatorname{LRM}w\right) .
\]

I.e., the LRMs of $w^{-1}$ are the \textbf{positions} of the LRMs of $w$.
}

\pause
\item Know
\href{https://groupprops.subwiki.org/wiki/Foata_transformation}{Foata's
fundamental transformation} $\mathcal{F}$? The partition
$\widetilde{\operatorname*{cLRM}\nolimits^{\prime}w}$ (obtained by sorting
$\operatorname*{cLRM}\nolimits^{\prime}w$ in decreasing order) is the cycle
type of $\mathcal{F}\left(  w^{-1}\right)  $.
\end{itemize}

\vspace{9pc}

\end{frame}

\begin{frame} \frametitle{\ \ \ \ The LRM-basis of $\mathcal{A}$}

\begin{itemize}
\item \textbf{Theorem [3--]:} The family%
\[
\left(  \mathbf{B}_{\operatorname{LRM}^{\prime}(w)}\,w\right)  _{w\in S_{n}%
}=\left(  \mathbf{B}_{\operatorname*{cLRM}{}^{\prime}(w)}\,w\right)  _{w\in
S_{n}}%
\]
is a basis of $\mathcal{A}$. We call it the \defn{LRM-basis}.

\pause
\item \textbf{Proof idea:} Show that $\mathbf{B}_{\operatorname{LRM}^{\prime
}(w)}\,w=w+\left(  \text{lex-smaller permutations}\right)  $.

\pause
\item \textbf{Example:} For $n=3$, this basis consists of the six elements%
\begin{align*}
\mathbf{B}_{\operatorname{LRM}^{\prime}(\left[  123\right]  )}\,\left[
123\right]   &  =\mathbf{B}_{\left(  3\right)  }\ \left[  123\right]  =\left[
123\right]  ;\\
\mathbf{B}_{\operatorname{LRM}^{\prime}(\left[  132\right]  )}\,\left[
132\right]   &  =\mathbf{B}_{\left(  3\right)  }\ \left[  132\right]  =\left[
132\right]  ;\\
\mathbf{B}_{\operatorname{LRM}^{\prime}(\left[  213\right]  )}\,\left[
213\right]   &  =\mathbf{B}_{\left(  1,2\right)  }\ \left[  213\right]
=\left[  213\right]  +\left[  123\right]  +\left[  132\right]  ;\\
\mathbf{B}_{\operatorname{LRM}^{\prime}(\left[  231\right]  )}\,\left[
231\right]   &  =\mathbf{B}_{\left(  1,2\right)  }\ \left[  231\right]
=\left[  231\right]  +\left[  132\right]  +\left[  123\right]  ;\\
\mathbf{B}_{\operatorname{LRM}^{\prime}(\left[  312\right]  )}\,\left[
312\right]   &  =\mathbf{B}_{\left(  2,1\right)  }\ \left[  312\right]
=\left[  312\right]  +\left[  213\right]  +\left[  123\right]  ;\\
\mathbf{B}_{\operatorname{LRM}^{\prime}(\left[  321\right]  )}\,\left[
321\right]   &  =\mathbf{B}_{\left(  1,1,1\right)  }\ \left[  321\right] \\
&  =\left[  321\right]  +\left[  312\right]  +\left[  231\right]  +\left[
213\right]  +\left[  132\right]  +\left[  123\right]  .
\end{align*}
\end{itemize}

\vspace{9pc}

\end{frame}

\begin{frame} \frametitle{\ \ \ \ The LRM-basis of $\mathcal{A}$ is nice to the filtration}

\begin{itemize}
\item \textbf{Theorem [3+]:} This LRM-basis restricts to bases of all the
right ideals $\mathbf{B}_{\alpha}\mathcal{A}$. Namely,%
\[
\mathbf{B}_{\alpha}\mathcal{A}=\operatorname{span}\bigl\{\mathbf{B}%
_{\operatorname{LRM}^{\prime}(w)}\,w\;\big|\;\widetilde{\operatorname{cLRM}%
^{\prime}(w)}\preceq_{\pi}\widetilde{\alpha}\bigr\}
\]
for each $\alpha\in\operatorname*{Comp}\nolimits_{n}$.

\pause
\only<2>{
\item \textbf{Proof idea:} Let $\mathcal{A}$ act on the $n$-th degree
component of the free algebra in $n$ variables $x_{1},x_{2},\ldots,x_{n}$ from
the right by permuting positions (not entries!). For each $\beta=\left(
\beta_{1},\beta_{2},\ldots,\beta_{k}\right)  \in\operatorname*{Comp}%
\nolimits_{n}$, let $\mathbf{V}_{\beta}$ be a product of nested commutators of
lengths $\beta_{1},\beta_{2},\ldots,\beta_{k}$. Show that $\mathbf{V}_{\beta
}\,\mathbf{B}_{\gamma}=0$ whenever $\widetilde{\beta}\npreceq_{\pi
}\widetilde{\gamma}$, and compute $\mathbf{V}_{\beta}\,\mathbf{B}_{\gamma}$ in
the case $\ell\left(  \beta\right)  =\ell\left(  \gamma\right)  $. The crucial
tool is a formula of Garsia and Reutenauer (Theorem 2.1 in
\href{https://doi.org/10.1016/0001-8708(89)90020-0}{their 1989 paper}, or
Lemma 9.33 in
\href{https://www.amazon.com/Algebras-London-Mathematical-Society-Monographs/dp/0198536798}{Reutenauer's
\textit{Free Lie algebras}}). Finish with leading term arguments.
}

\pause
\item \textbf{Alternative proof idea:} Get the dimensions of the
$\mathbf{B}_{\alpha}\mathcal{A}$ from
\href{https://www.proquest.com/docview/304356095}{Bidigare's 1997 thesis},
then argue only $\supseteq$ part. Needs some tweaks to work over $\mathbb{Z}$.

\pause
\item Yes: all the results in this chapter (except for one remark) hold over
$\mathbb{Z}$ and over any commutative ring.

\pause
\item \textbf{Question:} Easier proof?
\end{itemize}

\vspace{9pc}
\end{frame}

\begin{frame} \frametitle{\ \ \ \ Eigenvalues recovered, 1}

\begin{itemize}
\item Using the LRM-basis, we can recover Bidigare's formulas for the
eigenvalues of an element of $\mathcal{D}$ acting on $\mathcal{A}$:
\pause

\item For any $a\in\mathcal{A}$, we define the $\mathbf{k}$-linear maps%
\begin{align*}
\defnm{L\left(  \mathbf{a}\right)}  :\mathcal{A}  &  \rightarrow\mathcal{A},\\
\mathbf{x}  &  \mapsto\mathbf{ax}%
\end{align*}
and%
\begin{align*}
\defnm{R\left(  \mathbf{a}\right)}  :\mathcal{A}  &  \rightarrow\mathcal{A},\\
\mathbf{x}  &  \mapsto\mathbf{xa}.
\end{align*}
These are the \defn{left} and \defn{right multiplication} by $\mathbf{a}$.
\pause

\item In general, their eigenvalues are a mess. However, if $\mathbf{a}%
\in\mathcal{D}$, then they all lie in $\mathbf{k}$, and have explicit expressions:
\end{itemize}

\vspace{9pc}
\end{frame}

\begin{frame} \frametitle{\ \ \ \ Eigenvalues recovered, 2}

\begin{itemize}

\item \textbf{Corollary [2].} Let $\mathbf{a}=\sum_{\alpha\in
\operatorname{Comp}_{n}}\lambda_{\alpha}\mathbf{B}_{\alpha}$ (with
$\lambda_{\alpha}\in\mathbf{k}$) be any element of $\mathcal{D}$. Then, the
eigenvalues of the map $L\left(  \mathbf{a}\right)  $ (with their algebraic
multiplicities) are the numbers $\sum_{\alpha\in\operatorname{Comp}_{n}%
}\lambda_{\alpha}\eta_{\operatorname{cLRM}^{\prime}(w)}\left(  \alpha\right)
$, where $w$ ranges over $S_{n}$, and where the integers $\eta_{\beta}\left(
\alpha\right)  $ were defined before.

Moreover, the same holds for $R\left(  \mathbf{a}\right)  $.
\pause

\only<2>{\item This corollary is responsible for many descriptions of spectra of
shuffles (uniform BHR-shuffles such as the Tsetlin library).}

\pause
\item \textbf{Proof.} Order the LRM-basis so that each of the $\left(
\mathcal{D},\mathcal{A}\right)  $-subbimodules $\mathbf{B}_{\alpha}%
\mathcal{A}$ is spanned by the first so-and-so many basis vectors.

Then, the map $L\left(  \mathbf{a}\right)  $ is represented by an
upper-triangular matrix, and its diagonal entries are the scalars by which it
acts on the subquotients of the $\mathbf{B}_{\alpha}\mathcal{A}$-filtration;
these are the $\sum_{\alpha\in\operatorname{Comp}_{n}}\lambda_{\alpha}%
\eta_{\beta}\left(  \alpha\right)  $.

Thus the claim about $L\left(  \mathbf{a}\right)  $ follows. As for $R\left(
\mathbf{a}\right)  $, it always has the same eigenvalues as $L\left(
\mathbf{a}\right)  $, since $\mathcal{A}$ is semisimple (or Frobenius).
\end{itemize}

\vspace{9pc}
\end{frame}

\begin{frame} \frametitle{\ \ \ \ Eigenvalues recovered, 3}

\begin{itemize}

\item \textbf{Remark.} If we acted on $\mathcal{D}$ instead of $\mathcal{A}$,
then $L\left(  \mathbf{a}\right)  $ and $R\left(  \mathbf{a}\right)  $ would
not have the same eigenvalues!

(At least not counting with multiplicities; the sets are always the same.)

\pause
\item Bidigare ((4.6) in \href{https://www.proquest.com/docview/304356095}{his
thesis}) states the above formulas in a different way. To move between the
two, recall Foata's fundamental transformation, which shows that any partition
$\lambda\vdash n$ satisfies%
\begin{align*}
&  \left(  \text{\# of }w\in S_{n}\text{ such that }%
\widetilde{\operatorname*{cLRM}\nolimits^{\prime}\left(  w\right)  }%
=\lambda\right) \\
&  =\left(  \text{\# of }w\in S_{n}\text{ with cycle type }\lambda\right)  .
\end{align*}

\pause
\item Cycles come out naturally from Bidigare's approach; LRMs from ours.
\end{itemize}

\vspace{9pc}
\end{frame}

\begin{frame} \frametitle{\ \ \ \ Eigenvalues recovered, 4: digression}

\begin{itemize}

\item Geometric multiplicities and semisimplicity ($\Longleftrightarrow$
diagonalizability) questions are subtler.

\textbf{Theorem.} Let $a\in\mathcal{D}$ be a linear
combination of the $\mathbf{B}_{\alpha}$ with \textbf{nonnegative}
coefficients. Then:

\begin{enumerate}
\item[\textbf{(a)}] (\href{https://doi.org/10.1023/A:1007822931408}{Brown,
2000}) The element $a$ (that is, any action of $a$ on $\mathcal{D}$ or on
$\mathcal{A}$ or on any left or right $\mathcal{D}$-module) is diagonalizable.

\item[\textbf{(b)}] (\href{https://arxiv.org/abs/2508.06740v1}{G. and Parlett,
2025}) The same holds for the elements $aw_{0}$ and $w_{0}a$, where $w_{0}$ is
the permutation $\left[  n\cdots321\right]  \in S_{n}$.
\end{enumerate}

This, of course, is all in characteristic $0$.
\end{itemize}

\end{frame}

\begin{frame} \frametitle{\ \ \ \ Questions}

\begin{itemize}
\item \textbf{Question 1.} $\mathcal{A}$ has a filtration with $1$-dimensional
subquotients not just as a left $\mathcal{D}$-module, but also as a $\left(
\mathcal{D},\mathcal{D}\right)  $-bimodule. Is there a good basis for the
latter structure, too? (So both $L\left(  \mathbf{a}\right)  $ and $R\left(
\mathbf{a}\right)  $ act triangularly when $\mathbf{a}\in\mathcal{D}$.)

General reasoning says \textquotedblleft yes\textquotedblright, but we don't
have a combinatorial construction or even a canonical choice.

\pause
\item \textbf{Question 2 (downstream from 1).} What are the eigenvalues of
$L\left(  \mathbf{a}\right)  +R\left(  \mathbf{b}\right)  $ when
$\mathbf{a},\mathbf{b}\in\mathcal{D}$ ?

\pause
\item \textbf{Question 3.} What about the other types? What are left-to-right minima?
\end{itemize}

\end{frame}

\begin{frame}
\fti{Chapter 3}
\begin{center}
{\Huge \bf \sc Chapter 3} \\
\noindent\rule[0.5ex]{\linewidth}{1pt}
{\Large \bf The dyadic shuffles and the dyadic Gelfand model}
\end{center}
\vspace{1cm}
References:
\begin{itemize}
\item Sarah Brauner, Patricia Commins, Darij Grinberg, Franco Saliola,
\textit{A left ideal Gelfand model for the symmetric group},
working draft.
\newline\url{https://www.cip.ifi.lmu.de/~grinberg/algebra/dyadic.pdf}
\end{itemize}
\end{frame}

\begin{frame} \frametitle{\ \ \ \ Antipodes and symmetrization}

\begin{itemize}
\item The \defn{antipode} of $\mathcal{A}=\mathbf{k}\left[  S_{n}\right]  $ is
the $\mathbf{k}$-linear map%
\begin{align*}
\mathcal{A}  &  \rightarrow\mathcal{A},\\
w  &  \mapsto w^{-1}\ \ \ \ \ \ \ \ \ \ \text{for all }w\in S_{n}.
\end{align*}
The image of some $a\in\mathcal{A}$ under the antipode will be denoted
\defn{$a^{\ast}$} here.

\pause
\item \textbf{Basic properties [2--]:} The antipode is an anti-involution of
$\mathcal{A}$: that is, all $a,b\in\mathcal{A}$ satisfy%
\[
a^{\ast\ast}=a\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ \left(
ab\right)  ^{\ast}=b^{\ast}a^{\ast}.
\]

\pause
\item The antipode of the descent algebra $\mathcal{D}$ is the
\textquotedblleft anti-descent algebra\textquotedblright\ $\mathcal{D}^{\ast}%
$. It does not commute with $\mathcal{D}$.

\end{itemize}

\end{frame}

\begin{frame} \frametitle{\ \ \ \ Defining the $\mathbf{S}_\alpha$}

\begin{itemize}

\item Still, for all $\alpha\in\operatorname*{Comp}\nolimits_{n}$, we can
define the \textquotedblleft symmetrized B-basis element\textquotedblright%
\[
\defnm{\mathbf{S}_{\alpha}}:=\mathbf{B}_{\alpha}\mathbf{B}_{\alpha}^{\ast}.
\]

\pause
\item These do not span a subalgebra.
\pause
But:

\item \textbf{Theorem (\href{https://arxiv.org/abs/1102.2460}{\textbf{Reiner,
Saliola, Welker 2011}};
\href{https://arxiv.org/abs/1509.08580}{\textbf{Dieker, Saliola 2015}};
\href{https://arxiv.org/abs/1912.07718}{\textbf{Lafreni\`{e}re 2019}};
\href{https://arxiv.org/abs/2407.08644}{\textbf{Axelrod-Freed, Brauner,
Chiang, Commins, Lang 2024}};
\href{https://arxiv.org/abs/2503.17580}{\textbf{Brauner, Commins, G, Saliola
2025}}) [4]:} The elements $\mathbf{S}_{\left(  k,1^{n-k}\right)  }$ for all
$k\in\left[  n\right]  $ (where $1^{n-k}$ denotes $\underbrace{1,1,\ldots
,1}_{n-k\text{ times}}$) commute. Moreover, each of them acts diagonalizably
(from left and right) on $\mathcal{A}$ with integer eigenvalues.

These are the \defn{$k$-random-to-random shuffles} and have been the subject
of \href{https://darijgrinberg.gitlab.io/algebra/ne2025.pdf}{a prior talk}.

\pause
\item In general, $\mathbf{S}_{\alpha}$ don't commute and don't have integer
eigenvalues (e.g., $\mathbf{S}_{\left(  3,2\right)  }$ has some irrational
eigenvalues). But Reiner, Saliola and Welker found another subfamily that do:

\end{itemize}

\end{frame}

\begin{frame} \frametitle{\ \ \ \ The second commuting subfamily}

\begin{itemize}

\item \textbf{Theorem (\href{https://arxiv.org/abs/1102.2460}{\textbf{Reiner,
Saliola, Welker, 2011}}) [4--]:} The elements $\mathbf{S}_{\left(
2^{k},1^{n-2k}\right)  }$ for all $k\leq n/2$ (where $2^{k}%
=\underbrace{2,2,\ldots,2}_{k\text{ times}}$ and $1^{n-2k}%
=\underbrace{1,1,\ldots,1}_{n-2k\text{ times}}$) commute. Moreover, each of
them acts diagonalizably (from left and right) on $\mathcal{A}$ with integer eigenvalues.

\pause
\item The two theorems look similar, but are anything but! The proofs have no
ideas in common.

\pause
\item We refer to the $\mathbf{S}_{\left(  2^{k},1^{n-2k}\right)  }$ (up to a
scalar factor) as the \defn{dyadic shuffles}.

\pause
\item \textbf{Question:} What are their eigenvalues?

\pause
\item We will give a very partial answer. (Formulas with $\pm$ signs are easy
to get; but the eigenvalues are nonnegative integers, and yet no combinatorial
expression is known.)

But we will learn a lot more about the dyadic shuffles.
\end{itemize}

\end{frame}

\begin{frame} \frametitle{\ \ \ \ The dyadic shuffles, 1}

\begin{itemize}
\item \textbf{All of this chapter is joint work with Sarah Brauner, Patricia
Commins and Franco Saliola (\href{https://www.cip.ifi.lmu.de/~grinberg/algebra/dyadic.pdf}{partial draft}).}

\pause
\item The easiest way to define the dyadic shuffles is different.

\pause
\item An \defn{edge} will mean a $2$-element subset of $\left[  n\right]  $.

\pause
\item A permutation $w\in S_{n}$ is said to \defn{increase} on an edge
$\left\{  i<j\right\}  $ if and only if $w\left(  i\right)  <w\left(
j\right)  $.

\pause
\item A set of $k$ disjoint edges will be called a \defn{$k$-matching}.

\pause
\item A permutation $w\in S_{n}$ is said to \defn{increase} on a $k$-matching
$M$ if $w$ increases on each edge $P\in M$.

\pause
\item For instance, the permutation $\left[  24513\right]  \in S_{5}$
increases on exactly four $2$-matchings, namely%
\begin{align*}
&  \text{on }\left\{  \left\{  1,2\right\}  ,\left\{  4,5\right\}  \right\}
\text{ since }{\color{red}{24}}5{\color{blue}{13}\color{black};}\\
&  \text{on }\left\{  \left\{  1,3\right\}  ,\left\{  4,5\right\}  \right\}
\text{ since }{\color{red}{2}}4{\color{red}{5}}{\color{blue}{13}\color{black}}%
;\\
&  \text{on }\left\{  \left\{  2,3\right\}  ,\left\{  4,5\right\}  \right\}
\text{ since }2{\color{red}{45}}{\color{blue}{13}\color{black}}\text{;}\\
&  \text{on }\left\{  \left\{  1,5\right\}  ,\left\{  2,3\right\}  \right\}
\text{ since }{\color{blue}{2}}{\color{red}{45}}%
1{\color{blue}{3}\color{black}.}%
\end{align*}
\end{itemize}

\end{frame}

\begin{frame} \frametitle{\ \ \ \ The dyadic shuffles, 2}

\begin{itemize}

\item Given a permutation $w\in S_{n}$ and an integer $k\geq0$, we let
\[
\defnm{\operatorname*{incmat}\nolimits_{k}w}
:=\left(  \text{\# of }k\text{-matchings
on which }w\text{ increases}\right)  .
\]

\pause
\item For any $k\in\mathbb{N}$, we define the \defn{dyadic shuffle}%
\[
\defnm{\mathcal{S}_{n,k}}
:=\sum_{w\in S_{n}}\operatorname*{incmat}\nolimits_{k}\left(
w\right)  w\in\mathcal{A}.
\]

\pause

\item \textbf{Examples:}

\begin{itemize}
\item We have $\mathcal{S}_{n,0}=\sum_{w\in S_{n}}w$.

\pause
\item We have $\mathcal{S}_{3,1}=3\left[  123\right]  +2\left[  132\right]
+2\left[  213\right]  +\left[  231\right]  +\left[  312\right]  $.

\pause
\item We have $\mathcal{S}_{n,k}=0$ for all $k>n/2$, for lack of $k$-matchings.
\end{itemize}

\end{itemize}

\end{frame}

\begin{frame} \frametitle{\ \ \ \ The dyadic shuffles, 3}

\begin{itemize}
\item \textbf{Proposition
(\href{https://arxiv.org/abs/1102.2460}{\textbf{Reiner, Saliola, Welker,
2011}}) [2+]:} Let $k\leq n/2$. If $\alpha\in\operatorname*{Comp}%
\nolimits_{n}$ is a composition consisting of $k$ many $2$'s and $n-2k$ many
$1$'s (for example, $\left(  2^{k},1^{n-2k}\right)  $), then
\[
\mathcal{S}_{n,k}=\frac{1}{k!\left(  n-2k\right)  !}\underbrace{\mathbf{S}%
_{\alpha}}_{=\mathbf{B}_{\alpha}\mathbf{B}_{\alpha}^{\ast}}\in
\underbrace{\mathbf{B}_{\alpha}\mathcal{A}}_{\substack{\text{our
favorite}\\\text{right ideal}}}\cap\underbrace{\mathcal{A}\mathbf{B}_{\alpha
}^{\ast}}_{\text{its antipode}}.
\]

\pause
\item We can now restate the result of Reiner, Saliola and Welker:

\item \textbf{Theorem (\href{https://arxiv.org/abs/1102.2460}{\textbf{Reiner,
Saliola, Welker, 2011}}) [4--]:} The elements $\mathcal{S}_{n,k}$ for all
$k\geq0$ commute. Moreover, each of them acts diagonalizably (from left and
right) on $\mathcal{A}$ with integer eigenvalues.

\pause
\item It turns out that something stronger holds:

\end{itemize}
\end{frame}

\begin{frame} \frametitle{\ \ \ \ The Gelfand model}

\begin{itemize}
\item \textbf{Theorem [3+]:} Consider the composition%
\[
\defnm{\alpha} :=%
\begin{cases}
\left(  2^{n/2}\right)  , & \text{if }n\text{ is even};\\
\left(  2^{\left(  n-1\right)  /2},1\right)  , & \text{if }n\text{ is odd}%
\end{cases}
\ \ \in\operatorname*{Comp}\nolimits_{n}.
\]
Then, the left ideal $\mathcal{A}\mathbf{B}_{\alpha}^{\ast}$ (or,
equivalently, the right ideal $\mathbf{B}_{\alpha}\mathcal{A}$) of
$\mathcal{A}$ is a \defn{Gelfand model} of $S_{n}$: that is, viewed as
representation of $S_{n}$, it is isomorphic to the direct sum of all irreps (=
irreducible representations)!

All the $\mathcal{S}_{n,k}$ lie in this ideal.

\pause
\only<2>{
\item \textbf{Proof idea:}

\begin{itemize}
\item Gelfand model: see below.

\item All $\mathcal{S}_{n,k}$ lie in the ideal: follows from the
$\mathbf{B}_{\alpha}\mathcal{A}\subseteq\mathbf{B}_{\beta}\mathcal{A}$ result
in the previous chapter.
\end{itemize}
}

\pause
\item \textbf{Corollary [2]:} The dimension of this ideal is%
\[
\sum_{\lambda\vdash n}\dim\left(  \mathcal{S}^{\lambda}\right)  =\left(
\text{\# of involutions in }S_{n}\right)  .
\]
This also follows from the previous chapter.

\pause
\item To recover the commutativity and nice eigenvalues of $\mathcal{S}_{n,k}$
from this theorem, we need to study general properties of Gelfand models --
and, more generally, of multiplicity-free representations.
\end{itemize}

\end{frame}

\begin{frame} \frametitle{\ \ \ \ Multiplicity-free representations, 1}

\begin{itemize}
\item What follows are general facts about multiplicity-free representations
of $S_{n}$. Some of them generalize to all finite groups $G$, or even to all
semisimple $\mathbf{k}$-algebras.

\item Except for fact 1, I have never seen them before, but I can't believe
they are new!

\pause
\item A representation $V$ of $S_{n}$ is called \defn{multiplicity-free} if it
contains no two isomorphic irreducible subrepresentations.

\end{itemize}

\end{frame}

\begin{frame} \frametitle{\ \ \ \ Multiplicity-free representations, 2}

\begin{itemize}

\item \textbf{Theorems [2+ to 3--]:} Let $J$ be a left ideal of $\mathcal{A}$
that is multiplicity-free as representation of $S_{n}$. Then:

\begin{enumerate}
\item The endomorphism ring $\operatorname*{End}\nolimits_{\mathcal{A}}J$ is commutative.

\item We have $J\left[  J,J\right]  =0$ (where \defn{$\left[  U,V\right]  $} means
the commutator space $\operatorname*{span}\left\{  uv-vu\ \mid\ u\in U\text{
and }v\in V\right\}  $).

\item We have $\left[  J^{\ast}J,\ J^{\ast}J\right]  =0$.

More generally, $\left[  K^{\ast}J,\ K^{\ast}J\right]  =0$ if $K$ is a further
multiplicity-free left ideal of $\mathcal{A}$.

\item We have $v^{\ast}w=w^{\ast}v$ for all $v,w\in J$.

\item We have $x^{\ast}=x$ for each $x\in J^{\ast}J$.

\item Each $a\in J$ acts (from left and right) on $\mathcal{A}$ with all
eigenvalues in $\mathbf{k}$.

\item Each irrep $\mathcal{S}^{\lambda}$ of $S_{n}$ satisfies $\dim\left(
J^{\ast}\mathcal{S}^{\lambda}\right)  \leq1$.

\item If $\mathbf{k}$ is an ordered field, then the subspace $J^{\ast}J$ of
$\mathcal{A}$ is a nonunital $\mathbf{k}$-subalgebra that is isomorphic to the
direct product $\mathbf{k}^{r}$ of $r$ copies of $\mathbf{k}$, where $r$ is
the \# of irreps in $J$. (So it is unital, but not with the unity of
$\mathcal{A}$.)
\end{enumerate}

\end{itemize}

\end{frame}

\begin{frame} \frametitle{\ \ \ \ Multiplicity-free representations, 3}

\begin{itemize}
\item \textbf{Proof ideas:} Representation basics (Maschke, Schur); geometric
irreducibility of irreps (true for $S_{n}$ over any characteristic-$0$ field);
existence of nondegenerate symmetric invariant bilinear form on each irrep of
$S_{n}$.
\end{itemize}

\end{frame}

\begin{frame} \frametitle{\ \ \ \ Why $\mathcal{A}\mathbf{B}_{\alpha}^{\ast}$ is a Gelfand model, 1}

\begin{itemize}
\item Let $\alpha$ be the composition $\left(  2^{n/2}\right)  $ or $\left(
2^{\left(  n-1\right)  /2},1\right)  $ as above.

Why is the left ideal $\mathcal{A}\mathbf{B}_{\alpha}^{\ast}$ a Gelfand model
of $S_{n}$ ?

\pause

\item The irreps of $S_{n}$ are the \defn{Specht modules}%
\[
\mathcal{S}^{\lambda}=\operatorname*{span}\left(  \text{all polytabloids
}\defnm{\mathbf{e}_{T}}\text{ of shape }\lambda\right)
\]
for all the partitions $\lambda\vdash n$. (A \defn{polytabloid} is a
column-antisymmetrized tabloid. A \defn{tabloid} is a row-equivalence class of
tableaux. A \defn{tableau} must have the entries $1,2,\ldots,n$, each exactly once.)

\pause

\item Thus a representation $V$ of $S_{n}$ is a Gelfand model if and only if
\[
\dim\operatorname*{Hom}\nolimits_{\mathcal{A}}\left(  \mathcal{S}^{\lambda
},V\right)  =1\ \ \ \ \ \ \ \ \ \ \text{for all }\lambda\vdash n
\]
(since $\dim\operatorname*{End}\nolimits_{\mathcal{A}}\left(  \mathcal{S}%
^{\lambda}\right)  =1$ for each $\lambda$).

\pause

\item \textbf{Proposition [2+].} For any left ideal $V$ of $\mathcal{A}$, we
have%
\[
\operatorname*{Hom}\nolimits_{\mathcal{A}}\left(  \mathcal{S}^{\lambda
},V\right)  \cong V^{\ast}\mathcal{S}^{\lambda}\ \ \ \ \ \ \ \ \ \ \text{as
vector spaces.}%
\]

\end{itemize}

\end{frame}

\begin{frame} \frametitle{\ \ \ \ Why $\mathcal{A}\mathbf{B}_{\alpha}^{\ast}$ is a Gelfand model, 2}

\begin{itemize}

\item Thus, in order to prove that $\mathcal{A}\mathbf{B}_{\alpha}^{\ast}$ is
multiplicity-free, it suffices to show that $\dim\left(  \left(
\mathcal{A}\mathbf{B}_{\alpha}^{\ast}\right)  ^{\ast}\mathcal{S}^{\lambda
}\right)  =1$ for all $\lambda\vdash n$. That is, we must show that
$\dim\left(  \mathbf{B}_{\alpha}\mathcal{S}^{\lambda}\right)  =1$ for all
$\lambda\vdash n$.

\pause

\item \textbf{Reasonable proof idea:} Find the one vector that spans
$\mathbf{B}_{\alpha}\mathcal{S}^{\lambda}$.

\pause

\item And indeed, there is a nice one:%
\[
\defnm{\mathbf{v}_{\lambda}}
:=\sum_{\substack{S\text{ is a column-standard}%
\\\text{tableau of shape }\lambda}}\mathbf{e}_{S}.
\]
(\textquotedblleft\defn{Column-standard}\textquotedblright\ means that the
entries increase down columns, but nothing is known about the rows.)

\pause

\item \textbf{Proposition [3+?]:} For any $\lambda\vdash n$, the vector
$\mathbf{v}_{\lambda}$ is nonzero and spans the vector space $\mathbf{B}%
_{\alpha}\mathcal{S}^{\lambda}$.

\pause

\item Once this is proved, Gelfandness follows.

\end{itemize}

\end{frame}

\begin{frame} \frametitle{\ \ \ \ Why $\mathcal{A}\mathbf{B}_{\alpha}^{\ast}$ is a Gelfand model, 3}

\begin{itemize}

\item The nonzeroness of $\mathbf{v}_{\lambda}$ is not too hard (either using
leading terms or using positivity on the determinantal avatar).

\pause

\item That $\mathbf{v}_{\lambda}\in\mathbf{B}_{\alpha}\mathcal{S}^{\lambda}$
follows from an explicit factorization, related to the Pfaffian of the
\textquotedblleft$1$ above the diagonal, $-1$ below it\textquotedblright%
\ alternating matrix.

\pause

\item The spanning is hard! We would expect a direct proof of $\mathbf{B}%
_{\alpha}\mathbf{e}_{T}=\left(  \operatorname*{scalar}\right)  \cdot
\mathbf{v}_{\lambda}$ for all $T$. But we don't know how!

\pause

\item Instead, we show that $\mathcal{A}\mathbf{B}_{\alpha}$ has the
right dimension to be a Gelfand model. By what we proved above, it contains
one as a subrepresentation, so it must be it.

\pause

\item This is very roundabout and uncombinatorial!

\textbf{Question.} How to prove $\mathbf{B}_{\alpha}\mathbf{e}_{T}=\left(
\operatorname*{scalar}\right)  \cdot\mathbf{v}_{\lambda}$ directly?

\end{itemize}

\end{frame}

\begin{frame} \frametitle{\ \ \ \ Why $\mathcal{A}\mathbf{B}_{\alpha}^{\ast}$ is a Gelfand model, 4: looking for a nice proof}

\begin{itemize}

\item
\textbf{Restatement of the question:} Choose $2k$ distinct cells $c_{1},c_{2},\ldots
,c_{k},d_{1},d_{2},\ldots,d_{k}$ in the Young diagram of a partition
$\lambda\vdash n$. Prove that%
\[
\sum_{\substack{T\text{ is a tableau of shape }\lambda;\\T\left(
c_{i}\right)  <T\left(  d_{i}\right)  \text{ for each }i}}\mathbf{e}%
_{T}=\left(  \operatorname*{scalar}\right)  \cdot\mathbf{v}_{\lambda
}\ \ \ \ \ \ \ \ \ \ \text{in }\mathcal{S}^{\lambda}.
\]

\item Some \textbf{examples} (each $\left\{  c_{m},d_{m}\right\}  $ is colored in some
color):%
\[
\hspace{-0.8pc}
\scalebox{0.9}{
%TCIMACRO{\TeXButton{TeX field}{\begin{ytableau}
%*(yellow) c_1 & *(blue!50!white) c_2 & *(green) c_3 &  \\
%& *(green) d_3 & *(blue!50!white) d_2 \\
%*(yellow) d_1 &  & \\
%\phantom{1}
%\end{ytableau}}}%
%BeginExpansion
\begin{ytableau}
*(yellow) c_1 & *(blue!50!white) c_2 & *(green) c_3 &  \\
& *(green) d_3 & *(blue!50!white) d_2 \\
*(yellow) d_1 &  & \\
\phantom{1}
\end{ytableau}%
%EndExpansion
\qquad%
%TCIMACRO{\TeXButton{TeX field}{\begin{ytableau}
%*(yellow) c_1 & *(blue!50!white) c_2 & *(green) c_3 &  *(red) d_4\\
%& *(green) d_3 & *(blue!50!white) d_2 \\
%*(yellow) d_1 & *(red) c_4 &
%\end{ytableau}}}%
%BeginExpansion
\begin{ytableau}
*(yellow) c_1 & *(blue!50!white) c_2 & *(green) c_3 &  *(red) d_4\\
& *(green) d_3 & *(blue!50!white) d_2 \\
*(yellow) d_1 & *(red) c_4 &
\end{ytableau}%
%EndExpansion
\qquad%
%TCIMACRO{\TeXButton{TeX field}{\begin{ytableau}
%*(yellow) c_1 & *(blue!50!white) c_2 & *(green) c_3 &  *(red) d_4\\
%& *(green) d_3 & *(blue!50!white) d_2 \\
%*(yellow) d_1 & *(red) c_4 & *(brown!70!white) d_5 \\
%*(brown!70!white) c_5
%\end{ytableau}}}%
%BeginExpansion
\begin{ytableau}
*(yellow) c_1 & *(blue!50!white) c_2 & *(green) c_3 &  *(red) d_4\\
& *(green) d_3 & *(blue!50!white) d_2 \\
*(yellow) d_1 & *(red) c_4 & *(brown!70!white) d_5 \\
*(brown!70!white) c_5
\end{ytableau}%
%EndExpansion
\qquad%
%TCIMACRO{\TeXButton{TeX field}{\begin{ytableau}
%*(yellow) c_1 & *(blue!50!white) c_2 & *(green) c_3 &  *(red) d_4\\
%*(brown!70!white) c_5 & *(green) d_3 & *(blue!50!white) d_2 \\
%*(yellow) d_1 & *(red) c_4 & *(brown!70!white) d_5
%\end{ytableau}}}%
%BeginExpansion
\begin{ytableau}
*(yellow) c_1 & *(blue!50!white) c_2 & *(green) c_3 &  *(red) d_4\\
*(brown!70!white) c_5 & *(green) d_3 & *(blue!50!white) d_2 \\
*(yellow) d_1 & *(red) c_4 & *(brown!70!white) d_5
\end{ytableau}%
%EndExpansion
\ \ .
}
\]
\textbf{Exercise.} Do you see why the first one gives $0$ ?

\pause
\item
\textbf{Bonus question:} If $k<\sum_{i\geq1}\left\lfloor \lambda_{i}%
^{t}/2\right\rfloor $, then the left hand side is $0$.

\end{itemize}

\end{frame}

\begin{frame} \frametitle{\ \ \ \ So what about the eigenvalues?}

\begin{itemize}
\item As a consequence of the above, each element of $\mathcal{A}%
\mathbf{B}_{\alpha}^{\ast}$ acts on each irrep $\mathcal{S}^{\lambda}$ of
$S_{n}$ as a matrix of rank $\leq1$.

\pause
\item So it has at most one nonzero eigenvalue, which is its character value
and thus $\in\mathbf{k}$.

\pause
\item For general reasons, the eigenvalues of $\mathcal{S}_{n,k}\in
\mathcal{A}\mathbf{B}_{\alpha}^{\ast}$ are nonnegative integers. \textbf{Can
we find them?}

\end{itemize}

\end{frame}

\begin{frame} \frametitle{\ \ \ \ Eigenvalues on hook irreps}

\begin{itemize}
\item We only have an answer for $\lambda$ of hook shape:

\item \textbf{Theorem [4].} Let $\lambda$ be the hook-shaped partition
$\left(  n-\ell,1^{\ell}\right)  $, where $0\leq\ell<n$. Let $k\in\mathbb{N}$.

Then (on the understanding that $p!:=\infty$ for any $p<0$, and that a
fraction with $\infty$ in its denominator is $0$):

\begin{enumerate}
\item If $\ell$ is even, then the only nonzero eigenvalue of $\mathcal{S}%
_{n,k}$ on $\mathcal{S}^{\lambda}$ is%
\[
\frac{n!\left(  \ell/2\right)  !\left(  n-\ell\right)  !}{\left(
\ell+1\right)  !\left(  n-2k\right)  !\left(  k-\ell/2\right)  !2^{2k-\ell}}.
\]


\item If $\ell$ is odd, then the only nonzero eigenvalue of $\mathcal{S}%
_{n,k}$ on $\mathcal{S}^{\lambda}$ is%
\[
\frac{\left(  n+1\right)  !\left(  \left(  \ell+1\right)  /2\right)  !\left(
n-\ell-1\right)  !}{\left(  \ell+2\right)  !\left(  n-2k\right)  !\left(
k-\left(  \ell+1\right)  /2\right)  !2^{2k-\ell-1}}.
\]

\end{enumerate}

\pause
\item The proof is surprisingly hard and uses various deep properties of the
seminormal form.
\end{itemize}

\end{frame}

\begin{frame} \frametitle{\ \ \ \ More on the dyadic shuffles}

\begin{itemize}

\item For other shapes $\lambda$, we have recursions connecting the
eigenvalues of $\mathcal{S}_{n,k}$ and $\mathcal{S}_{n,k-1}$. These are
obtained from the...

\textbf{Nice recursion [3]:} For any $k\geq0$, we have%
\begin{align*}
\binom{n-2\left(  k-1\right)  }{2}\mathcal{S}_{n,k-1}  &  =\mathcal{S}%
_{n,k}\left(  \mathcal{B}_{n}-\left(  n-2k\right)  \right) \\
&  =\left(  \mathcal{B}_{n}^{\ast}-\left(  n-2k\right)  \right)
\mathcal{S}_{n,k},
\end{align*}
where%
\[
\mathcal{B}_{n}=\mathbf{B}_{\left(  n-1,1\right)  }^{\ast}=\sum_{j=1}%
^{n}\operatorname*{cyc}\nolimits_{n,n-1,n-2,\ldots,j}%
\]
(and where $\mathcal{S}_{n,-1}:=0$).

\pause
\item \textbf{Remark [2+].} We have $w_{0}\mathcal{S}_{n,k}=\mathcal{S}%
_{n,k}w_{0}$ for each $k\geq0$.
\end{itemize}

\end{frame}

\begin{frame} \frametitle{\ \ \ \ Other Gelfand models}

\begin{itemize}
\item $\mathcal{A}\mathbf{B}_{\alpha}^{\ast}$ is not the first Gelfand model
of $S_{n}$ to be found.

\pause
\item Easy to construct others using Young symmetrizers.

\pause
\item \href{https://arxiv.org/abs/math/0402216v1}{Kodiyalam, Verma 2004} and
\href{https://arxiv.org/abs/0709.3962}{Adin, Postnikov, Roichman 2007}
construct one using involutions.

It is the associated graded module of $\mathcal{A}\mathbf{B}_{\alpha}^{\ast}$,
equipped with the filtration
\[
\mathcal{A}\mathbf{B}_{\left(  1^{n}\right)  }^{\ast}\subseteq\mathcal{A}%
\mathbf{B}_{\left(  2,1^{n-2}\right)  }^{\ast}\subseteq\mathcal{A}%
\mathbf{B}_{\left(  2,2,1^{n-4}\right)  }^{\ast}\subseteq\cdots.
\]


\pause
\item \href{https://doi.org/10.1081/AGB-100002136}{Aguado, Araujo 2006}
construct one using harmonic polynomials.

Some similarities with ours, but exact relation unclear.
\end{itemize}

\end{frame}

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

\begin{itemize}
\item \textbf{Sarah Brauner, Patricia Commins, Franco Saliola, Katya
Vassilieva} for obvious reasons.

\item \textbf{Ron Adin, Nadia Lafreni\`{e}re, Victor Reiner, Yuval Roichman}
for helpful conversations recent and not so recent.

\item \textbf{Sasha Postnikov} for the invitation.

\item \textbf{you} for your patience.
\end{itemize}

\end{frame}

\end{document}