% Title: Shuffles in the symmetric group algebra

% Abstract:
% Ever since the famous 1992 work of Bayer and Diaconis, it has been
% known that random shuffles of a deck of cards (with the back side up)
% can be modelled as elements of the group algebra $\mathbb{R}[S_n]$ of
% the symmetric group $S_n$. This viewpoint has spawned progress in both
% card shuffling and the representation theory of the symmetric group.
% In this talk, I will focus on two projects in the latter: one focusing
% on the "somewhere-to-below shuffles"
% $t_i := (i) + (i,i+1) + (i,i+1,i+2) + \cdots + (i,i+1,\ldots,n) \in
% \mathbb{R}[S_n]$
% for $1 \leq i \leq n$ (where the parenthesized expressions mean
% cycles; the $1$-cycle $(i)$ is the identity), and one focusing on the
% "$k$-random-to-random shuffles"
% $R_k := \sum_{1 \leq i_1 < i_2 < \cdots < i_k \leq n} \sum_{w \in S_n
% \text{ such that } w(i_1) < w(i_2) < \cdots < w(i_k)} w \in
% \mathbb{R}[S_n]$
% for $0 \leq k \leq n$.
% Both families have revealed a variety of unexpected properties. For
% instance, the $R_0, R_1, \ldots, R_n$ commute, whereas the $t_1, t_2,
% \ldots, t_n$ are simultaneously triangularizable (i.e., there is an --
% explicitly describable -- basis of $\mathbb{R}[S_n]$ on which right
% multiplication by each $t_i$ acts as a triangular matrix). In both
% cases, all eigenvalues are integers and can be explicitly described
% and assigned to Specht modules (irreducible representations of $S_n$).
% Many of these properties furthermore generalize to the (type-A)
% Iwahori-Hecke algebra.

% Due to the amount of results, this talk will be an overview with no proofs.

% Some of the above is joint work with Nadia Lafrenière, Sarah Brauner,
% Patricia Commins and Franco Saliola.

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\begin{document}

\title{Shuffles in the symmetric group algebra}
\author{Darij Grinberg (Drexel University)
\and \and \and \and \and joint work with Nadia Lafreni\`ere,
\and Sarah Brauner, \and Patricia Commins, \and Franco Saliola}
\date{Dartmouth College, 2025-10-14, \\
MIT, 2025-10-15}

\frame{\titlepage\textbf{slides:}
\redurl{http://www.cip.ifi.lmu.de/~grinberg/algebra/ne2025.pdf} \medskip
\newline\textbf{papers:}
\arxiv{2503.17580} \qquad \arxiv{2212.06274} \\
\quad \ \ \qquad \arxiv{2309.05340} \qquad \arxiv{2508.00752} }

\begin{frame}
\fti{Chapter 1}
\begin{center}
{\Huge \bf \sc Chapter 1} \\
\noindent\rule[0.5ex]{\linewidth}{1pt}
{\Large \bf Introduction}
\end{center}
\vspace{1cm}
References:
\begin{itemize}
\item \redhref{https://fardila.com/Clase/RepTh/sagan.pdf}{Bruce Sagan, \textit{The symmetric group}, 2nd edition 2001}.
\item \redhref{https://math.mit.edu/~etingof/reprbook.pdf}{Pavel Etingof et al., \textit{Introduction to representation theory}, AMS 2011}, \S\S 5.12--5.17.
\item \redhref{https://eudml.org/doc/287582}{Murray Bremner, Sara Madariaga, Luiz
A. Peresi, \textit{Structure theory for the group algebra of the symmetric
group, ...}, Commentationes Mathematicae Universitatis Carolinae, 2016}.
\item \redhref{https://archive.org/details/substitutionalan0000ruth/page/n5/mode/2up}{%
Daniel Edwin Rutherford, \textit{Substitutional Analysis}, Edinburgh
1948}.
\item \redhref{https://arxiv.org/abs/2507.20706}{Darij Grinberg, \textit{An introduction to the symmetric group algebra}, arXiv:2507.20706}.
\end{itemize}
\end{frame}

\begin{frame} \frametitle{\ \ \ \ Finite group algebras: Basics}

\begin{itemize}
\item[$\circled{$\ast$}$] Let $\mathbf{k}$ be any commutative ring. (Usually
$\mathbb{Z}$, $\mathbb{Q}$ or a polynomial ring.)

\item[$\circled{$\ast$}$] Let $G$ be a finite group. (We will only use
symmetric groups.)

\item[$\circled{$\ast$}$] Let $\defnm{\kk\ive{G}}$ be the group
algebra of $G$ over $\mathbf{k}$. Its elements are formal $\mathbf{k}$-linear
combinations of elements of $G$. The multiplication is inherited from $G$ and
extended bilinearly.

\pause

\item \textbf{Example:} Let $G$ be the symmetric group $S_{3}$ on the set
$\left\{  1,2,3\right\}  $. For $i\in\left\{  1,2\right\}  $, let $s_{i}\in
S_{3}$ be the simple transposition that swaps $i$ with $i+1$. Then, in
$\mathbf{k}\left[  G\right]  =\mathbf{k}\left[  S_{3}\right]  $, we have%
\begin{align*}
\left(  1+s_{1}\right)  \left(  1-s_{1}\right)   &  =
\textcolor{blue}{1}+\textcolor{red}{s_{1}}-\textcolor{red}{s_{1}}
-\textcolor{blue}{s_{1}^2}=0
\end{align*}
(since $\textcolor{blue}{s_1^2} = \textcolor{blue}{1}$);
\pause
\begin{align*}
\left(  1+s_{2}\right)  \left(  1+s_{1}+s_{1}s_{2}\right)   &  =1+s_{2}%
+s_{1}+s_{2}s_{1}+s_{1}s_{2}+s_{2}s_{1}s_{2} \\
&=\sum_{w\in S_{3}}w.
\end{align*}


\end{itemize}

\end{frame} \begin{frame} \frametitle{\ \ \ \ Finite group algebras: $L\left(a\right)$ and $R\left(a\right)$}

\begin{itemize}
\item[$\circled{$\ast$}$] For each $a\in\mathbf{k}\left[  G\right]  $, we
define two $\mathbf{k}$-linear maps%
\begin{align*}
\defnm{L\left(  a\right)}  :\mathbf{k}\left[  G\right]   &  \rightarrow\mathbf{k}%
\left[  G\right]  ,\\
x  &  \mapsto ax\ \ \ \ \ \ \ \ \ \ \left(  \text{\textquotedblleft left
multiplication by }a\text{\textquotedblright}\right)
\end{align*}
and%
\begin{align*}
\defnm{R\left(  a\right)}  :\mathbf{k}\left[  G\right]   &  \rightarrow\mathbf{k}%
\left[  G\right]  ,\\
x  &  \mapsto xa\ \ \ \ \ \ \ \ \ \ \left(  \text{\textquotedblleft right
multiplication by }a\text{\textquotedblright}\right)  .
\end{align*}
(So $L\left(  a\right)  \left(  x\right)  =ax$ and $R\left(  a\right)  \left(
x\right)  =xa$.)

\end{itemize}

\vspace{0.5pc}

\only<1>{\textcolor{green!70!black}{\textbf{Note:} The symbol \textcolor{blue}{$\circled{$\ast$}$} denotes important points.}}
\pause

\begin{itemize}

\item Both $L\left(  a\right)  $ and $R\left(  a\right)  $ are
endomorphisms of the free $\mathbf{k}$-module $\mathbf{k}\left[  G\right]  $.
Thus, they can be viewed as
$\left\vert G\right\vert \times\left\vert G\right\vert $-matrices.

\pause

\item Hence, $L\left(  a\right)  $ and $R\left(  a\right)  $
are ``matrix proxies''\ for $a$,
allowing to apply linear algebra to studying $a$.

{\small (The reason this works is that the maps%
$a \mapsto L\tup{a}$
and
$a \mapsto \tup{R\tup{a}}^T$
are two injective $\mathbf{k}$-algebra morphisms
from $\kk\ive{G}$ to the matrix ring
$\operatorname{End}_{\kk}\tup{\kk\ive{G}}
\cong \kk^{\abs{G} \times \abs{G}}$.)}
% (known as the left and right
% regular representations of the group $G$).)}
\end{itemize}

\vspace{9pc}

\end{frame} \begin{frame} \frametitle{\ \ \ \ Finite group algebras: Minimal polynomials}

\begin{itemize}
\item[$\circled{$\ast$}$] Each $a\in\mathbf{k}\left[  G\right]  $ has a
\defn{minimal polynomial}, i.e., a minimum-degree monic polynomial
$P\in\mathbf{k}\left[  X\right]  $ such that $P\left(  a\right)  =0$. It is
unique when $\mathbf{k}$ is a field.

The minimal polynomial of $a$ is also the minimal polynomial of the
endomorphisms $L\left(  a\right)  $ and $R\left(  a\right)  $.

\item When $\mathbf{k}$ is a field, we can also study the eigenvectors and
eigenvalues of $L\left(  a\right)  $ and $R\left(  a\right)  $.

\pause

\item \textbf{Theorem 1.1.} Assume that $\mathbf{k}$ is a field. Let
$a\in\mathbf{k}\left[  G\right]  $. Then, the two linear endomorphisms
$L\left(  a\right)  $ and $R\left(  a\right)  $ are conjugate in
$\operatorname*{End}\nolimits_{\mathbf{k}}\left(  \mathbf{k}\left[  G\right]
\right)  $ (that is, similar as matrices).

(Thus, they have the same eigenstructure.)

\item This is surprisingly nontrivial!
\end{itemize}

\end{frame} \begin{frame} \frametitle{\ \ \ \ Finite group algebras: The antipode}

\begin{itemize}
\item[$\circled{$\ast$}$] The \defn{antipode} of the group algebra
$\mathbf{k}\left[  G\right]  $ is defined to be the $\mathbf{k}$-linear map
\begin{align*}
S:\mathbf{k}\left[  G\right]   &  \rightarrow\mathbf{k}\left[  G\right]  ,\\
g  &  \mapsto g^{-1}\ \ \ \ \ \ \ \ \ \ \text{for each }g\in G.
\end{align*}
We shall write \defn{$a^{\ast}$} for $S\left(  a\right)  $.

\pause

\item[$\circled{$\ast$}$] \textbf{Proposition 1.2.} The antipode $S$ is an
involution:%
\[
a^{\ast\ast}=a\ \ \ \ \ \ \ \ \ \ \text{for all }a\in\mathbf{k}\left[
G\right]  ,
\]
and a $\mathbf{k}$-algebra anti-automorphism:%
\[
\left(  ab\right)  ^{\ast}=b^{\ast}a^{\ast}\ \ \ \ \ \ \ \ \ \ \text{for all
}a,b\in\mathbf{k}\left[  G\right]  .
\]

\end{itemize}

\end{frame} \begin{frame} \frametitle{\ \ \ \ Finite group algebras: Proof of Theorem 1.1}

\begin{itemize}
\item \textbf{Lemma 1.3.} Assume that $\mathbf{k}$ is a field. Let
$a\in\mathbf{k}\left[  G\right]  $. Then, $L\left(  a\right)  \sim L\left(
a^{\ast}\right)  $ in $\operatorname*{End}\nolimits_{\mathbf{k}}\left(
\mathbf{k}\left[  G\right]  \right)  $.

\item \textit{Proof:} Consider the standard basis $\left(  g\right)  _{g\in
G}$ of $\mathbf{k}\left[  G\right]  $. The matrices representing the
endomorphisms $L\left(  a\right)  $ and $L\left(  a^{\ast}\right)  $ in this
basis are mutual transposes. But
\redhref{https://math.stackexchange.com/a/596842/}{the Taussky--Zassenhaus
theorem} says that over a field, each matrix $A$ is similar to its transpose
$A^{T}$.

\pause

\item \textbf{Lemma 1.4.} Let $a\in\mathbf{k}\left[  G\right]  $. Then,
$L\left(  a^{\ast}\right)  \sim R\left(  a\right)  $ in $\operatorname*{End}%
\nolimits_{\mathbf{k}}\left(  \mathbf{k}\left[  G\right]  \right)  $.

\item \textit{Proof:} We have $R\left(  a\right)  =S\circ L\left(  a^{\ast
}\right)  \circ S$ and $S=S^{-1}$.

\pause

\item \textit{Proof of Theorem 1.1:} Combine Lemma 1.3 with Lemma 1.4.

\item \textbf{Remark (Martin Lorenz).} Theorem 1.1 generalizes to arbitrary
finite-dimensional Frobenius algebras.
\end{itemize}


\section{The symmetric group algebra}

\end{frame} \begin{frame} \frametitle{\ \ \ \ Symmetric groups: Notations}

\begin{itemize}
\item[$\circled{$\ast$}$] Let $\defnm{\mathbb{N}} :=\left\{  0,1,2,\ldots\right\}  $.

\item[$\circled{$\ast$}$] Let $\defnm{\left[  k\right]}  :=\left\{  1,2,\ldots
,k\right\}  $ for each $k\in\mathbb{N}$.

\pause

\item[$\circled{$\ast$}$] Now, fix a positive integer $n$, and let $S_{n}$ be
the \defn{$n$-th symmetric group}, i.e., the group of permutations of the
set $\left[  n\right]  $.

Multiplication in $S_{n}$ is composition:%
\begin{align*}
\left(  \alpha\beta\right)  \left(  i\right) 
&=\left(  \alpha\circ
\beta\right)  \left(  i\right)  =\alpha\left(  \beta\left(  i\right)  \right)
\\
&\ \ \ \ \ \ \ \ \ \ \text{for all }\alpha,\beta\in S_{n}\text{ and }%
i\in\left[  n\right]  .
\end{align*}


(\textbf{Warning:} SageMath has a different opinion!)
\end{itemize}

\end{frame} \begin{frame} \frametitle{\ \ \ \ Symmetric group algebras}

\begin{itemize}
\item What can we say about the group algebra $\mathbf{k}\left[  S_{n}\right]
$ that doesn't hold for arbitrary $\mathbf{k}\left[  G\right]  $?

\only<1>{
\item There is a classical theory (\textquotedblleft Young's seminormal
form\textquotedblright) of the structure of $\mathbf{k}\left[  S_{n}\right]  $
when $\mathbf{k}$ has characteristic $0$. See:

\begin{itemize}
\item \redhref{https://eudml.org/doc/287582}{Murray Bremner, Sara Madariaga, Luiz
A. Peresi, \textit{Structure theory for the group algebra of the symmetric
group, ...}, Commentationes Mathematicae Universitatis Carolinae, 2016}.
(Quick and to the point.)

\item \redhref{https://archive.org/details/substitutionalan0000ruth/page/n5/mode/2up}{%
Daniel Edwin Rutherford, \textit{Substitutional Analysis}, Edinburgh
1948}. (Dated but careful and quite readable; perhaps the best treatment.)

\item \redhref{https://doi.org/10.1007/978-3-030-58373-6}{Adriano M. Garsia,
\"{O}mer Egecioglu, \textit{Lectures in Algebraic Combinatorics}, Springer
2020}. (Messy but full of interesting things.)
\end{itemize}
}
\pause

\item \textbf{Theorem 2.1 (Artin--Wedderburn--Young).} If $\mathbf{k}$ is a
field of characteristic $0$, then%
\[
\mathbf{k}\left[  S_{n}\right]  \cong\prod_{\lambda\text{ is a partition of
}n}\ \ \ \underbrace{\operatorname*{M}\nolimits_{f^{\lambda}}\left(  \mathbf{k}%
\right)  }_{\text{matrix ring}}\ \ \ \ \ \ \ \ \ \ \left(  \text{as
}\mathbf{k}\text{-algebras}\right)  ,
\]
where $f^{\lambda}$ is the number of standard Young tableaux of shape
$\lambda$.

\only<2>{
\item \textit{Proof:} This follows from Young's seminormal form. For the
shortest readable proof, see Theorem 1.45 in Bremner/Madariaga/Peresi.

Or, for a different proof, see
\redhref{https://www.cip.ifi.lmu.de/~grinberg/t/24s/sga.pdf}{my
\textit{introduction to the symmetric group algebra}} (\S 5.14).
}

\pause

\item The structure of $\mathbf{k}\left[  S_{n}\right]  $ for
$0<\operatorname*{char}\mathbf{k}\leq n$ is far less straightforward. See, e.g.,

\begin{itemize}
\item
\redhref{https://pnp.mathematik.uni-stuttgart.de/lexmath/kuenzer/thesis.pdf}{Matthias
K\"{u}nzer, \textit{Ties for the integral group ring of the symmetric group},
thesis 1998.}
\end{itemize}
\pause

\item \textbf{Remark.} If $\mathbf{k}$ is a field of characteristic $0$, then
each $a\in\mathbf{k}\left[  S_{n}\right]  $ satisfies $a\sim a^{\ast}$ in
$\mathbf{k}\left[  S_{n}\right]  $.

But not for general $\mathbf{k}$.
\end{itemize}

\section{The Young--Jucys--Murphy elements}

\begin{itemize}
\item From now on, we shall focus on concrete elements in $\mathbf{k}\left[
S_{n}\right]  $.
\end{itemize}

\end{frame} \begin{frame} \frametitle{\ \ \ \ The YJM elements: Definition and commutativity}

\begin{itemize}
\item[$\circled{$\ast$}$] For any distinct elements $i_{1},i_{2},\ldots,i_{k}$
of $\left[  n\right]  $, let $\defnm{\operatorname*{cyc}\nolimits_{i_{1},i_{2}%
,\ldots,i_{k}}}$ be the permutation in $S_{n}$ that cyclically permutes
$i_{1}\mapsto i_{2}\mapsto i_{3}\mapsto\cdots\mapsto i_{k}\mapsto i_{1}$ and
leaves all other elements of $\left[  n\right]  $ unchanged.

\item \textbf{Note.} We have $\operatorname*{cyc}\nolimits_{i}%
=\operatorname*{id}$, whereas $\operatorname*{cyc}\nolimits_{i,j}$ is the
transposition $t_{i,j}$.

\pause

\item[$\circled{$\ast$}$] For each $k\in\left[  n\right]  $, we define the
\defn{$k$-th Young--Jucys--Murphy (YJM) element}%
\[
\defnm{J_{k}}
:=\operatorname*{cyc}\nolimits_{1,k}+\operatorname*{cyc}\nolimits_{2,k}%
+\cdots+\operatorname*{cyc}\nolimits_{k-1,k}\in\mathbf{k}\left[  S_{n}\right]
.
\]


\item \textbf{Note.} We have $J_{1}=0$. Also, $J_{k}^{\ast}=J_{k}$ for each
$k\in\left[  n\right]  $.

\pause

\item[$\circled{$\ast$}$] \textbf{Theorem 3.1.} The YJM elements $J_{1}%
,J_{2},\ldots,J_{n}$ commute: We have $J_{i}J_{j}=J_{j}J_{i}$ for all $i,j$.

\item \textit{Proof:} Easy computational exercise.
\end{itemize}

\end{frame} \begin{frame} \frametitle{\ \ \ \ The YJM elements: Eigenvalues}

\begin{itemize}
\item[$\circled{$\ast$}$] \textbf{Theorem 3.2.} The minimal polynomial of
$J_{k}$ over $\mathbb{Q}$ divides%
\[
\prod_{i=-k+1}^{k-1}\left(  X-i\right)  =\left(  X-k+1\right)  \left(
X-k+2\right)  \cdots\left(  X+k-1\right)  .
\]
(For $k\leq3$, some factors here are redundant.)

\pause

\only<2>{
\item \textit{First proof:} Study the action of $J_{k}$ on each Specht module
(simple $S_{n}$-module). See, e.g.,
\redhref{https://doi.org/10.1016/0021-8693(81)90205-2}{G. E. Murphy, \textit{A
New Construction of Young's Seminormal Representation ...}, 1981} for details.

\item \textit{Second proof (Igor Makhlin):} Some linear algebra does the
trick. Induct on $k$ using the facts that $J_{k}$ and $J_{k+1}$ are
simultaneously diagonalizable over $\mathbb{C}$ (since they are symmetric as
real matrices and commute) and satisfy $s_{k}J_{k+1}=J_{k}s_{k}+1$, where
$s_{k}:=\operatorname*{cyc}\nolimits_{k,k+1}$. See
\redurl{https://mathoverflow.net/a/83493/} for details.
}

\pause

\item Thus, the eigenvalues of $J_{k}$ are $-k+1,-k+2,\ldots,k-1$ (except for
$0$ when $k\leq3$). Their multiplicities can be computed in terms of standard
Young tableaux. Even better:

\pause

\only<4-5>{
\item \textbf{Theorem 3.3.} Assume that $\mathbf{k}$ is a field of
characteristic $0$. Then, there exists a basis $\left(  e_{S,T}\right)  $ of
$\mathbf{k}\left[  S_{n}\right]  $ indexed by pairs of standard Young tableaux
of the same (partition) shape called the \defn{seminormal basis}. This basis
has the property that%
\[
J_{k}e_{S,T}=\mathsf{c}_{S}\left(  k\right)  \cdot e_{S,T},
\]
where $\mathsf{c}_{S}\left(  k\right)  = j-i$ if the number $k$
lies in cell $\left(  i,j\right)  $ of $S$.

\pause

\item Moreover, each Specht module $\mathcal{S}^{\lambda}$ (= irreducible
representation of $S_{n}$) is spanned by part of the seminormal basis, and
thus we find the eigenvalues of $J_{k}$ on that $\mathcal{S}^{\lambda}$.
}

\pause

\item The seminormal basis exists only for $\operatorname*{char}\mathbf{k}=0$
(or, more generally, when $n!$ is invertible in $\mathbf{k}$).

But Theorem 3.2 and the algebraic multiplicities transfer automatically to all
rings $\mathbf{k}$.

\pause

\item \textbf{Question.} Is there a self-contained algebraic/combinatorial
proof of Theorem 3.2 without linear algebra or representation theory? (Asked
on MathOverflow: \redurl{https://mathoverflow.net/questions/420318/} .)
\end{itemize}

\vspace{9pc}

\end{frame} \begin{frame} \frametitle{\ \ \ \ Symmetric polynomials in the YJM elements, 1}

\begin{itemize}
\item \textbf{Theorem 3.4.} For each $k\in\mathbb{N}$, we can evaluate the
$k$-th elementary symmetric polynomial $e_{k}$ at the YJM elements
$J_{1},J_{2},\ldots,J_{n}$ to obtain%
\[
e_{k}\left(  J_{1},J_{2},\ldots,J_{n}\right)  =\sum_{\substack{\sigma\in
S_{n};\\\sigma\text{ has exactly }n-k\text{ cycles}}}\sigma.
\]

\item \textit{Proof:} Nice homework exercise (once stripped of the algebra).

\pause

\item There are formulas for other symmetric polynomials applied to
$J_{1},J_{2},\ldots,J_{n}$ (see Garsia/Egecioglu).

There is also a general fact:

\end{itemize}

\end{frame} \begin{frame} \frametitle{\ \ \ \ Symmetric polynomials in the YJM elements, 2}

\begin{itemize}

\item \textbf{Theorem 3.5 (Murphy).}
\begin{align*}
&  \left\{  f\left(  J_{1},J_{2},\ldots,J_{n}\right)  \ \mid\ f\in
\mathbf{k}\left[  X_{1},X_{2},\ldots,X_{n}\right]  \text{ symmetric}\right\}
\\
&  =\left(  \text{center of the group algebra }\mathbf{k}\left[  S_{n}\right]
\right)  .
\end{align*}

\pause

\item \textit{Proof:} See any of:

\begin{itemize}
\item
\redhref{https://www.ams.org/journals/tran/1992-332-01/S0002-9947-1992-1062873-1/}{Gadi
Moran, \textit{The center of }$\mathbb{Z}\left[  S_{n+1}\right]  $
\textit{...}, 1992}.

\item \redhref{https://doi.org/10.1016/0021-8693(83)90219-3}{G. E. Murphy,
\textit{The Idempotents of the Symmetric Group ...}, 1983}, Theorem 1.9 (for
the case $\mathbf{k}=\mathbb{Z}$, but the general case easily follows).

\item
\redhref{https://doi.org/10.1017/CBO9781139192361}{Ceccherini-Silberstein/Scarabotti/Tolli,
\textit{Representation Theory of the Symmetric Groups}, 2010}, Theorem 4.4.5
(for the case $\mathbf{k}=\mathbb{Q}$, but the proof is easily adapted to all
$\mathbf{k}$).

This book also has more on the $J_{1},J_{2},\ldots,J_{n}$ (but mind
\redhref{https://mathoverflow.net/questions/485958}{the errata}).
\end{itemize}
\end{itemize}

\vspace{9pc}

\end{frame} \begin{frame} \frametitle{\ \ \ \ The card shuffling point of view}

\begin{itemize}

\item Permutations are often visualized as shuffled decks of cards:

Imagine a deck of cards labeled $1,2,\ldots,n$.

A permutation $\sigma\in S_{n}$ corresponds to the \defn{state} in which the
cards are arranged $\sigma\left(  1\right)  ,\sigma\left(  2\right)
,\ldots,\sigma\left(  n\right)  $ from top to bottom.

\pause

\item A \defn{random state} is an element $\sum\limits_{\sigma\in S_{n}%
}a_{\sigma}\sigma$ of $\mathbb{R}\left[  S_{n}\right]  $ whose coefficients
$a_{\sigma}\in\mathbb{R}$ are nonnegative and add up to $1$. This is
interpreted as a distribution on the $n!$ possible states, where $a_{\sigma}$
is the probability for the deck to be in state $\sigma$.

\pause

\only<3-4>{
\item We drop the \textquotedblleft add up to $1$\textquotedblright%
\ condition, and only require that $\sum_{\sigma\in S_{n}}a_{\sigma}>0$. The
probabilities must then be divided by $\sum_{\sigma\in S_{n}}a_{\sigma}$.
}

\only<4>{
\item For instance, $1+\operatorname*{cyc}\nolimits_{1,2,3}$ corresponds to
the random state in which the deck is sorted as $1,2,3$ with probability
$\dfrac{1}{2}$ and sorted as $2,3,1$ with probability $\dfrac{1}{2}$.
}

\pause \pause

\item An $\mathbb{R}$-vector space endomorphism of $\mathbb{R}\left[
S_{n}\right]  $, such as $L\left(  a\right)  $ or $R\left(  a\right)  $ for
some $a\in\mathbb{R}\left[  S_{n}\right]  $, acts as a \defn{(random)
shuffle}, i.e., a transformation of random states. This is just the standard
way how Markov chains are constructed from transition matrices.

\pause

\item For example, if $k>1$, then the right multiplication $R\left(
J_{k}\right)  $ by the YJM element $J_{k}$ corresponds to swapping the $k$-th
card with some card above it (chosen uniformly at random).

\pause

\item Transposing such a matrix means time-reversing the random shuffle.
\end{itemize}

\end{frame} \begin{frame} \frametitle{\ \ \ \ Bottom-to-random and random-to-bottom shuffles: definitions}

\begin{itemize}
\item[$\circled{$\ast$}$] Another family of elements of $\mathbf{k}\left[
S_{n}\right]  $ are the \defn{$k$-bottom-to-random shuffles}%
\[
\mathcal{B}_{n,k}:=\sum_{\substack{\sigma\in S_{n};\\\sigma^{-1}\left(
1\right)  <\sigma^{-1}\left(  2\right)  <\cdots<\sigma^{-1}\left(  n-k\right)
}}\sigma
\]
defined for all $k\in\left\{  0,1,\ldots,n\right\}  $. Thus,%
\begin{align*}
\mathcal{B}_{n,n}  &  =\mathcal{B}_{n,n-1}=\sum_{\sigma\in S_{n}}\sigma;\\
\mathcal{B}_{n,1}  &  =\sum_{i=1}^{n}\operatorname*{cyc}%
\nolimits_{n,n-1,\ldots,i};\\
\mathcal{B}_{n,0}  &  =\operatorname*{id}.
\end{align*}
We set $\mathcal{B}_{n}:=\mathcal{B}_{n,1}$.

\pause

\only<2>{
\item As a random shuffle, $\mathcal{B}_{n,k}$ (to be precise, $R\left(
\mathcal{B}_{n,k}\right)  $) takes the bottom $k$ cards and moves them to
random positions.

Its antipode $\mathcal{B}_{n,k}^{\ast}$ takes $k$ random cards and moves them
to the bottom positions.
}

\pause

\item $\mathcal{B}_{n}:=\mathcal{B}_{n,1}$ is known as the
\defn{bottom-to-random shuffle} or the \defn{Tsetlin library}.

\vspace{9pc}

\end{itemize}

\end{frame} \begin{frame} \frametitle{\ \ \ \ Bottom-to-random and random-to-bottom shuffles: facts}

\begin{itemize}

\item \textbf{Theorem 5.1 (Diaconis, Fill, Pitman).} We have
\[
\mathcal{B}_{n,k+1}=\left(  \mathcal{B}_{n}-k\right)  \mathcal{B}%
_{n,k}\ \ \ \ \ \ \ \ \ \ \text{for each }k\in\left\{  0,1,\ldots,n-1\right\}
.
\]

\pause

\item \textbf{Corollary 5.2.} The $n+1$ elements $\mathcal{B}_{n,0}%
,\mathcal{B}_{n,1},\ldots,\mathcal{B}_{n,n}$ commute and are polynomials in
$\mathcal{B}_{n}$, namely%
\[
\mathcal{B}_{n,k}=\prod_{i=0}^{k-1}\left(  \mathcal{B}_{n}-i\right)
\ \ \ \ \ \ \ \ \ \ \text{for each }k\in\left\{  0,1,\ldots,n\right\}  .
\]

\pause


\item \textbf{Theorem 5.3 (Wallach).} The minimal polynomial of $\mathcal{B}%
_{n}$ over $\mathbb{Q}$ is%
\[
\prod_{i\in\left\{  0,1,\ldots,n-2,n\right\}  }\left(  X-i\right)  =\left(
X-n\right)  \prod_{i=0}^{n-2}\left(  X-i\right)  .
\]

\pause


\item These are not hard to prove in this order. See
\redurl{https://mathoverflow.net/questions/308536} for the details.

\end{itemize}

\end{frame} \begin{frame} \frametitle{\ \ \ \ Bottom-to-random and random-to-bottom shuffles: more}

\begin{itemize}

\item More can be said: in particular, the multiplicities of the eigenvalues
$0,1,\ldots,n-2,n$ of $R\left(  \mathcal{B}_{n}\right)  $ over $\mathbb{Q}$
are known.

\pause

\item The antipodes
\[
\mathcal{B}_{n,k}^{\ast}:=\sum_{\substack{\sigma\in S_{n};\\\sigma\left(
1\right)  <\sigma\left(  2\right)  <\cdots<\sigma\left(  n-k\right)  }}\sigma
\]
of $\mathcal{B}_{n,k}$ are known as the \defn{$k$-random-to-bottom shuffles}
and have the same properties (since $S$ is an algebra anti-automorphism).

\pause

\item Moreover, there are \defn{top-to-random} and \defn{random-to-top}
shuffles defined in the same way but with renaming $1,2,\ldots,n$ as
$n,n-1,\ldots,1$. They are just images of the $\mathcal{B}_{n,k}$ and
$\mathcal{B}_{n,k}^{\ast}$ under the automorphism $a\mapsto w_{0}aw_{0}^{-1}$
of $\mathbf{k}\left[  S_{n}\right]  $, where $w_{0}$ is the permutation with
one-line notation $\left(  n,n-1,\ldots,1\right)  $.

Thus, top vs. bottom is mainly a matter of notation.

\end{itemize}

\end{frame} \begin{frame} \frametitle{\ \ \ \ Bottom-to-random and random-to-bottom shuffles: references}

\begin{itemize}

\item Main references:

\begin{itemize}
\item \redhref{https://doi.org/10.2969/aspm/01410123}{Nolan R. Wallach,
\textit{Lie Algebra Cohomology and Holomorphic Continuation of Generalized
Jacquet Integrals}, 1988, Appendix}.

\item
\redhref{https://statweb.stanford.edu/~cgates/PERSI/papers/randomshuff92.pdf}{Persi
Diaconis, James Allen Fill and Jim Pitman, \textit{Analysis of Top to Random
Shuffles}, 1992}.
\end{itemize}
\end{itemize}


\section{Random-to-random shuffles}

\end{frame}

\begin{frame}
\fti{Chapter 2}
\begin{center}
{\Huge \bf \sc Chapter 2} \\
\noindent\rule[0.5ex]{\linewidth}{1pt}
{\Large \bf Random-to-random shuffles}
\end{center}
\vspace{1cm}
References:
\begin{itemize}
\item \redhref{https://arxiv.org/abs/1102.2460}{Victor Reiner, Franco Saliola,
Volkmar Welker, \textit{Spectra of Symmetrized Shuffling Operators},
arXiv:1102.2460}.
% \item \redhref{https://doi.org/10.1016/j.aim.2017.10.034}{A.B. Dieker, F.V.
% Saliola, \textit{Spectral analysis of random-to-random Markov chains}, 2018}.
% \item \redhref{https://arxiv.org/abs/1912.07718}{Nadia Lafreni\`{e}re,
% \textit{Valeurs propres des op\'{e}rateurs de m\'{e}langes sym\'{e}tris\'{e}s}%
% , thesis, 2019}.
\item \redhref{https://arxiv.org/abs/2407.08644}{Ilani Axelrod-Freed, Sarah
Brauner, Judy Hsin-Hui Chiang, Patricia Commins, Veronica Lang,
\textit{Spectrum of random-to-random shuffling in the Hecke algebra},
arXiv:2407.08644.}
\item \redhref{https://arxiv.org/abs/2503.17580}{Sarah
Brauner, Patricia Commins, Darij Grinberg, Franco Saliola, \textit{The
q-deformed random-to-random family in the Hecke algebra}, arXiv:2503.17580.}
\end{itemize}
\end{frame}

\begin{frame} \frametitle{\ \ \ \ Random-to-random shuffles: Definition}

\begin{itemize}
\item[$\circled{$\ast$}$] Here is a further family. For each $k\in\left\{
0,1,\ldots,n\right\}  $, we let%
\[
\defnm{\mathcal{R}_{n,k}}
:=\sum_{\sigma\in S_{n}}\operatorname*{noninv}\nolimits_{n-k}%
\left(  \sigma\right)  \cdot\sigma,
\]
where $\operatorname*{noninv}\nolimits_{n-k}\left(  \sigma\right)  $ denotes
the number of $\left(  n-k\right)  $-element subsets of $\left[  n\right]  $
on which $\sigma$ is increasing. This is called the
\defn{$k$-random-to-random shuffle}.

\pause

\only<2>{
\item \textbf{Example:} Writing permutations in one-line notation,%
\begin{align*}
\mathcal{R}_{4,2}  &
=6[1,2,3,4]+5[1,2,4,3]+5[1,3,2,4]+4[1,3,4,2]\\
&  \ \ \ \ \ \ \ \ \ \ +4[1,4,2,3]+3[1,4,3,2]+5[2,1,3,4]+4[2,1,4,3]\\
&  \ \ \ \ \ \ \ \ \ \ +4[2,3,1,4]+3[2,3,4,1]+3[2,4,1,3]+2[2,4,3,1]\\
&  \ \ \ \ \ \ \ \ \ \ +4[3,1,2,4]+3[3,1,4,2]+3[3,2,1,4]+2[3,2,4,1]\\
&  \ \ \ \ \ \ \ \ \ \ +2[3,4,1,2]+[3,4,2,1]+3[4,1,2,3]+2[4,1,3,2]\\
&  \ \ \ \ \ \ \ \ \ \ +2[4,2,1,3]+[4,2,3,1]+[4,3,1,2].
\end{align*}
}

\pause

\item \textbf{Note:} $\mathcal{R}_{n,0}=\operatorname*{id}$ and $\mathcal{R}%
_{n,n-1}=n\sum_{\sigma\in S_{n}}\sigma$ and $\mathcal{R}_{n,n}=\sum_{\sigma\in
S_{n}}\sigma$.

\pause

\item The card-shuffling interpretation of $\mathcal{R}_{n,k}$ is
\textquotedblleft pick any $k$ cards from the deck and move them to $k$
randomly chosen positions\textquotedblright.
\end{itemize}

\vspace{9pc}

\end{frame} \begin{frame} \frametitle{\ \ \ \ Random-to-random shuffles: Two surprises}

\begin{itemize}
\item[$\circled{$\ast$}$] \textbf{Theorem 6.1 (Reiner, Saliola, Welker).} The
$n+1$ elements $\mathcal{R}_{n,0},\mathcal{R}_{n,1},\ldots,\mathcal{R}_{n,n}$
commute (but are not polynomials in $\mathcal{R}_{n,1}$ in general).

\pause

\item[$\circled{$\ast$}$] \textbf{Theorem 6.2 (Dieker, Saliola,
Lafreni\`{e}re).} The minimal polynomial of each $\mathcal{R}_{n,k}$ over
$\mathbb{Q}$ is a product of $X-i$'s for distinct integers $i$. For example,
the one of $\mathcal{R}_{n,1}$ divides
\[
\prod_{i=0}^{n^{2}}\left(  X-i\right)  .
\]
The exact factors can be given in terms of certain statistics on Young diagrams.

\end{itemize}


\end{frame} \begin{frame} \frametitle{\ \ \ \ Random-to-random shuffles: References}

\begin{itemize}
\item Main references: the \textquotedblleft classics\textquotedblright

\begin{itemize}
\item \redhref{https://arxiv.org/abs/1102.2460}{Victor Reiner, Franco Saliola,
Volkmar Welker, \textit{Spectra of Symmetrized Shuffling Operators},
arXiv:1102.2460}.

\item \redhref{https://doi.org/10.1016/j.aim.2017.10.034}{A.B. Dieker, F.V.
Saliola, \textit{Spectral analysis of random-to-random Markov chains}, 2018}.

\item \redhref{https://arxiv.org/abs/1912.07718}{Nadia Lafreni\`{e}re,
\textit{Valeurs propres des op\'{e}rateurs de m\'{e}langes sym\'{e}tris\'{e}s}%
, thesis, 2019}.
\end{itemize}

and the two recent preprints

\begin{itemize}
\item \redhref{https://arxiv.org/abs/2407.08644}{Ilani Axelrod-Freed, Sarah
Brauner, Judy Hsin-Hui Chiang, Patricia Commins, Veronica Lang,
\textit{Spectrum of random-to-random shuffling in the Hecke algebra},
arXiv:2407.08644.}

\item \redhref{https://arxiv.org/abs/2503.17580}{Sarah
Brauner, Patricia Commins, Darij Grinberg, Franco Saliola, \textit{The
q-deformed random-to-random family in the Hecke algebra}, arXiv:2503.17580.}
\end{itemize}
\end{itemize}


\end{frame} \begin{frame} \frametitle{\ \ \ \ Random-to-random shuffles: What we do}

\begin{itemize}

\item The \textquotedblleft classical\textquotedblright\ proofs are
complicated, technical and long.

In this talk, I will outline some parts of the two recent preprints, including
a simpler proof of Theorem 6.1 and most of Theorem 6.2. (The full proof of
Theorem 6.2 is still long and hard.)
\end{itemize}

\end{frame} \begin{frame} \frametitle{\ \ \ \ $\mathcal{R}$ vs. $\mathcal{B}$}

\begin{itemize}
\item The first step is a formula that is easy to prove combinatorially:

\item[$\circled{$\ast$}$] \textbf{Proposition 6.3.} For each $k\in\left\{
0,1,\ldots,n\right\}  $, we have%
\[
\mathcal{R}_{n,k}=\dfrac{1}{k!}\cdot\mathcal{B}_{n,k}^{\ast}\,\mathcal{B}%
_{n,k}.
\]

\pause

\item However, the $\mathcal{B}_{n,k}$ do not commute with the $\mathcal{B}%
_{n,k}^{\ast}$, so this is not by itself an answer.
\end{itemize}


\section{Digression: The Hecke algebra $\mathcal{H}_{n}$}

\end{frame} \begin{frame} \frametitle{\ \ \ \ The Hecke algebra, 1}

\begin{itemize}
\item Let $q\in\mathbf{k}$ be a parameter.

The $n$-th \defn{Hecke algebra} (or \defn{Iwahori--Hecke algebra})
% to be more historically correct)
is a $q$-deformation of the group algebra
$\mathbf{k}\left[  S_{n}\right]  $.

It has generators $T_{1},T_{2}%
,\ldots,T_{n-1}$ and relations%
\begin{align*}
T_{i}^{2}  &  =\left(  q-1\right)  T_{i}+q\ \ \ \ \ \ \ \ \ \ \text{for all
}i\in\left[  n-1\right]  ;\\
T_{i}T_{j}  &  =T_{j}T_{i}\ \ \ \ \ \ \ \ \ \ \text{whenever }\left\vert
i-j\right\vert >1;\\
T_{i}T_{i+1}T_{i}  &  =T_{i+1}T_{i}T_{i+1}\ \ \ \ \ \ \ \ \ \ \text{for all
}i\in\left[  n-2\right]  .
\end{align*}
We call this algebra \defn{$\mathcal{H}_{n}$}.

\pause

\item For $q=1$, this is the group algebra $\mathbf{k}%
\left[  S_{n}\right]  $ (and the generator $T_{i}$ is the simple transposition
$s_{i}=\operatorname*{cyc}\nolimits_{i,i+1}$).

\pause

\item For general $q$, it still is a free $\mathbf{k}%
$-module of rank $n!$, with a basis $\left(  T_{w}\right)  _{w\in S_{n}}$
indexed by permutations $w\in S_{n}$. The basis vectors are defined by
$T_{w}:=T_{i_{1}}T_{i_{2}}\cdots T_{i_{k}}$, where
$s_{i_{1}}s_{i_{2}}\cdots s_{i_{k}}$ is a reduced expression for $w$.
For $q=1$, this $T_{w}$ is just $w$.
\end{itemize}


\section{The Hecke algebra $\mathcal{H}_{n}$}

\end{frame} \begin{frame} \frametitle{\ \ \ \ The Hecke algebra, 2}

\begin{itemize}
\item Much of the theory of $\mathbf{k}\left[
S_{n}\right]  $ exists in a subtler form for $\mathcal{H}_{n}$. Sometimes, the
added difficulty brings the best proofs to light.

\pause

\item[$\circled{$\ast$}$] \textbf{Almost all results of this talk} hold for the Hecke algebra $\mathcal{H}_n$
(occasionally requiring assumptions such as ``$q$ is not a root unity''\ for structural results).
The YJM elements must be $q$-deformed; integers become $q$-integers; etc.

\pause

\textbf{Main change:} The random-to-random shuffle must now be \textbf{defined} as
\[
\mathcal{R}_{n,k} := \dfrac{1}{\left[k\right]!_q}\cdot\mathcal{B}_{n,k}^{\ast}\,\mathcal{B}_{n,k}.
\]
Noninversions no longer work!

\pause

But we will stick to the $q = 1$ case in this talk.
\end{itemize}


\section{The recursion and commutativity}

\end{frame} \begin{frame} \frametitle{\ \ \ \ The recursion}

\begin{itemize}
\item[$\circled{$\ast$}$] \textbf{Theorem 8.1 (Brauner--Commins--G.--Saliola
2025, based on Axelrod-Freed--Brauner--Chiang--Commins--Lang 2024).} For any
$1\leq k\leq n$, we have
\[
\mathcal{B}_{n}\,\mathcal{R}_{n,k}
=
\underbrace{\left( \mathcal{R}_{n-1,k}
+ \left(  \left( n+1-k\right) + J_{n}\right)
\,\mathcal{R}_{n-1,k-1}\right)  }_{=:\mathcal{W}_{n,k}}\mathcal{B}_{n}.
\]


\pause

\item The proof takes about 5 pages, relying on some more elementary
computations from prior work (ca. 10--15 pages in total).

\pause

\item This recursion does not actually compute $\mathcal{R}_{n,k}$. But it
says enough about $\mathcal{R}_{n,k}$ to carry our proofs.

\item Note also that $\mathcal{R}_{n,k}\in\mathcal{B}_{n}^{\ast}%
\,\mathbf{k}\left[S_n\right]$ by its definition (when $k\geq1$). This makes the recursion
so useful.
\end{itemize}

\end{frame} \begin{frame} \frametitle{\ \ \ \ Commutativity of random-to-random}

\begin{itemize}
\item Theorem 8.1 leads fairly easily to a proof of commutativity (Theorem 6.1).

Indeed, inducting on $n$, we observe that the $\mathcal{W}_{n,k}$s all commute
by the induction hypothesis (and the easy fact that $J_{n}$ commutes with
everything in $\mathbf{k}\left[S_{n-1}\right]$). Thus, using
$\mathcal{B}_n\, {\blue \mathcal{R}_{n,k}}
= {\blue \mathcal{W}_{n,k}}\, \mathcal{B}_n$, we find
\begin{align*}
\mathcal{B}_{n}\,{\blue \mathcal{R}_{n,i}\,\mathcal{R}_{n,j}}
&  = {\blue \mathcal{W}_{n,i}}\,\mathcal{B}_{n}\,{\blue \mathcal{R}_{n,j}}
= {\blue \mathcal{W}_{n,i}\,\mathcal{W}_{n,j}}\,\mathcal{B}_{n}\\
&  = {\blue \mathcal{W}_{n,j}\,\mathcal{W}_{n,i}}\,\mathcal{B}_{n}
= {\blue\mathcal{W}_{n,j}}\,\mathcal{B}_{n}\,{\blue \mathcal{R}_{n,i}}
= \mathcal{B}_{n}\, {\blue \mathcal{R}_{n,j}\,\mathcal{R}_{n,i}}.
\end{align*}


Remains to get rid of the $\mathcal{B}_{n}$ factor at the front. Recall that
all $\mathcal{R}_{n,i}$ (except for the trivial $\mathcal{R}_{n,0}$) lie in
$\mathcal{B}_{n}^{\ast}\,\mathbf{k}\left[S_n\right]$. But we can WLOG
assume that $\mathbf{k} = \mathbb{Q}$, and then the equality
$\mathcal{B}_{n}\,\mathcal{B}_{n}^{\ast}\,a=0$ entails
$\mathcal{B}_{n}^{\ast}\,a=0$ (positivity trick! cf. linear algebra:
$\operatorname*{Ker}\left(  A^{T}A\right)  =\operatorname*{Ker}A$ for real
matrix $A$).

\item Alternatively, the trick can be avoided (see \arxiv{2503.17580}).
\end{itemize}


\section{\textquotedblleft Integrality\textquotedblright\ of eigenvalues}

\end{frame} \begin{frame} \frametitle{\ \ \ \ The approach to eigenvalues, 1}

\begin{itemize}
\item Now to Theorem 6.2:

The eigenvalues of $\mathcal{R}_{n,k}$ are \textbf{nonnegative reals}, since $\mathcal{R}_{n,k}$ is represented by a positive semidefinite symmetric matrix (Proposition 6.3).

But why are they \textbf{integers}?

\pause

\item We have a theory of \textquotedblleft split elements\textquotedblright%
\ that can help answer such questions in general. Here is an outline:

\pause

\item[$\circled{$\ast$}$] An element $a$ of a $\mathbf{k}$-algebra $A$ is said
to be \defn{split} (over $\mathbf{k}$) if there exist some scalars
$u_{1},u_{2},\ldots,u_{n}\in\mathbf{k}$ (not necessarily distinct) such that
$\prod\limits_{i=1}^{n}\left(  a-u_{i}\right)  =0$.

\pause

\item[$\circled{$\ast$}$] When $\mathbf{k}$ is an integral domain and $A$ is a
free $\mathbf{k}$-module of finite rank, this is the same as saying that
$R\left(  a\right)  $ has all eigenvalues in $\mathbf{k}$.

\pause

\item In particular, for $\mathbf{k}=\mathbb{Z}$ and
$A=\mathbf{k}\left[S_n\right]$, this means that all eigenvalues of $R\tup{a}$ are $\in
\mathbb{Z}$. This is what we want to show for $a=\mathcal{R}%
_{n,k}$.

\pause

\item So we must show that $\mathcal{R}_{n,k}$ is split over $\mathbb{Z}$.
\end{itemize}


\end{frame} \begin{frame} \frametitle{\ \ \ \ General theory of split elements, 1}

\begin{itemize}
\item We prove several general properties of split elements (nice exercises on
commutative algebra!):

\pause

\item[$\circled{$\ast$}$] \textbf{Theorem 9.1.} If two commuting elements
$a,b\in A$ are split, then both $a+b$ and $ab$ are split.

\item[$\circled{$\ast$}$] \textbf{Corollary 9.2.} A commutative subalgebra of
$A$ generated by split elements consists entirely of split elements.

\pause

\item[$\circled{$\ast$}$] \textbf{Theorem 9.3.} If $b,c,f$ are elements of $A$
such that $f$ is split and such that $bc=fb$ and $c\in Ab$, then $c$ is split.

\pause

\item Theorem 9.3 is tailored to our use:
\[%
\begin{tabular}
[c]{|c|c|}\hline
$bc=fb$ & $c\in Ab$\\\hline
$\mathcal{B}_{n}\,\mathcal{R}_{n,k}=\mathcal{W}_{n,k}\,\mathcal{B}_{n}$ &
$\mathcal{R}_{n,k}\in\mathbf{k}\left[S_n\right]$ $\mathcal{B}_{n}$\\\hline
\end{tabular}
\ \ \ \ .
\]
The splitness of $\mathcal{W}_{n,k}$ follows from the splitness of the
commuting elements $J_{n}$, $\mathcal{R}_{n-1,k-1}$ and $\mathcal{R}_{n-1,k}$
(induction!) by Corollary 9.2. We need the splitness of the YJM elements,
which was proved (e.g.) by Murphy.
\end{itemize}

\end{frame} \begin{frame} \frametitle{\ \ \ \ General theory of split elements, 2}

\begin{itemize}

\item Theorem 9.3 looks baroque, but in fact it easily decomposes into two
particular cases:

\textbf{Corollary 9.4.} If $ba$ is split, then $ab$ is also split.

\textbf{Corollary 9.5.} If $a$ is split and $b^{2}=ab$, then $b$ is split.

(Both times, $a,b\in A$ are arbitrary.)
\end{itemize}


\section{Formulas for eigenvalues}

\end{frame} \begin{frame} \frametitle{\ \ \ \ Formulas for eigenvalues, 1}

\begin{itemize}
\item The splitness theory proves easily that all eigenvalues of
$\mathcal{R}_{n,k}$ are integers, but it does not compute them explicitly.
Indeed, it produces
\textquotedblleft phantom eigenvalues\textquotedblright\ which do not actually
appear.

\pause

\item With a lot more work (Specht modules, seminormal basis,
Pieri rule, etc.), we have been able to compute the
eigenvalues with their multiplicities fully.

\item I only have time to state the main result.
\end{itemize}


\end{frame} \begin{frame} \frametitle{\ \ \ \ Formulas for eigenvalues, 2}

\begin{itemize}

{\small
\item \textbf{Theorem 10.1.} Let $n,k\geq0$. The eigenvalues of $R\left(
\mathcal{R}_{n,k}\right)  $ on $\ksn$ are the elements%
\[
\mathcal{E}_{\lambda\setminus\mu}(k)
:= \sum_{\textcolor{green!40!black}{j}<(\ell
_{1}<\ell_{2}<\cdots<\ell_{k})\leq n}\ \prod_{m=1}^{k} \left(\ell
_{m}+1-m
+\textcolor{blue}{\mathsf{c}}_{\textcolor{red}{\mathfrak{t}^{\lambda\setminus\mu}}}\left(  \ell
_{m}\right)  \right)
\]
for all horizontal strips $\lambda\setminus\mu$ that satisfy
$\lambda \vdash n$ and $\textcolor{violet}{d^{\mu}}\neq0$. Here,

\begin{itemize}
\item \small $\textcolor{violet}{d^{\mu}}$ denotes the number of
\defn{desarrangement tableaux} of
shape $\mu$ (that is, standard tableaux of shape $\mu$ whose smallest
non-descent is even);

\item \small $\textcolor{green!40!black}{j}$ is the size of $\mu$;

\item \small $\textcolor{red}{\mathfrak{t}^{\lambda\setminus\mu}}$ is the skew tableau of shape
$\lambda\setminus\mu$ obtained by filling in the boxes of $\lambda\setminus
\mu$ with $j+1,j+2,\ldots,n$ from top to bottom;

\item \small $\textcolor{blue}{\mathsf{c}_{\mathfrak{t}^{\lambda\setminus\mu}}\left(  p\right)}  =y-x$
if the cell of $\mathfrak{t}^{\lambda\setminus\mu}$ containing the entry $p$
is $\left(  x,y\right)  $.
\end{itemize}

Moreover, the multiplicity of each such eigenvalue $\mathcal{E}_{\lambda
\setminus\mu}(k)$ is $d^{\mu}f^{\lambda}$, where $f^{\lambda}$ is the number
of standard tableaux of shape $\lambda$ (unless there are collisions).
}
\end{itemize}

\end{frame} \begin{frame} \frametitle{\ \ \ \ Formulas for eigenvalues, 3}

\begin{itemize}

\item \small We have explicit formulas for specific shapes and strips:%
\begin{align*}
\mathcal{E}_{(n)\setminus\varnothing}(k)
&= k! \dbinom{n}{k}^2; \\
\mathcal{E}_{(n-1,1)\setminus(j,1)}(k)
&= k! \dbinom{n-j-1}{k}\dbinom{n+j}{k}
\qquad\text{ for all }j\in\left[  n-1\right]  .
\end{align*}
But there is no such nice formula for $\mathcal{E}_{(4,1,1) \setminus (1,1)}(1)$.
\end{itemize}

\end{frame} \begin{frame} \frametitle{\ \ \ \ Open questions}

\begin{itemize}

\item \textbf{Question:} Any nicer formulas for the eigenvalues $\mathcal{E}%
_{\lambda\setminus\mu}\left(  k\right)  $ ?

\item \textbf{Question (Reiner):} What is the dimension of the subalgebra of $\mathbb{Q}%
\left[  S_{n}\right]  $ generated by $\mathcal{R}_{n,0},\mathcal{R}%
_{n,1},\ldots,\mathcal{R}_{n,n}$ ?
{\footnotesize
\[
\hspace{-2.5pc}
\begin{tabular}
[c]{|c||c|c|c|c|c|c|c|c|c|c|c|c|}\hline
$n$ & $1$ & $2$ & $3$ & $4$ & $5$ & $6$ & $7$ & $8$ & $9$ & $10$ & $11$ &
$12$\\\hline
$\dim\left(  \text{subalgebra}\right)  $ & $1$ & $2$ & $4$ & $7$ & $15$ & $30$
& $54$ & $95$ & $159$ & $257$ & $400$ & $613$\\\hline
\end{tabular}
\ \ \ \
\]
(sequence not in the OEIS as of 2025-10-06).}

The same numbers hold for the $q$-deformation!
\end{itemize}

\section{Somewhere-to-below shuffles}

\end{frame}

\begin{frame}
\fti{Chapter 3}
\begin{center}
{\Huge \bf \sc Chapter 3} \\
\noindent\rule[0.5ex]{\linewidth}{1pt}
{\Large \bf Somewhere-to-below shuffles}
\end{center}
\vspace{1cm}
References:
\begin{itemize}
\item \redhref{https://arxiv.org/abs/2212.06274}{Darij Grinberg, Nadia Lafreni\`ere, \textit{The one-sided cycle shuffles in the symmetric group algebra}, Algebraic Combinatorics (2024), arXiv:2212.06274}.
\item \redhref{https://arxiv.org/abs/2309.05340}{Darij Grinberg, \textit{Commutator nilpotency for somewhere-to-below shuffles}, arXiv:2309.05340}.
\item \redhref{https://arxiv.org/abs/2508.00752}{Darij Grinberg, \textit{The representation theory of somewhere-to-below shuffles}, arXiv:2508.00752}.
\end{itemize}
\end{frame}

\begin{frame} \frametitle{\ \ \ \ Somewhere-to-below shuffles: introduction}

\begin{itemize}
\item Now to something completely different...

\pause

\item[$\circled{$\ast$}$] In 2021, Nadia Lafreni\`{e}re defined the
\defn{somewhere-to-below shuffles} $\mathbf{t}_{1},\mathbf{t}_{2}%
,\ldots,\mathbf{t}_{n}$ by setting%
\[
\defnm{\mathbf{t}_{\ell}}
:=\operatorname*{cyc}\nolimits_{\ell}+\operatorname*{cyc}%
\nolimits_{\ell,\ell+1}+\operatorname*{cyc}\nolimits_{\ell,\ell+1,\ell
+2}+\cdots+\operatorname*{cyc}\nolimits_{\ell,\ell+1,\ldots,n}\in
\mathbf{k}\left[  S_{n}\right]
\]
for each $\ell\in\left[  n\right]  $.
% (These $\mathbf{t}_{\ell}$ are called $t_{\ell}$ in my papers.)

\item[$\circled{$\ast$}$] Note: $\mathbf{t}_{n}=\operatorname*{id}$.

\pause

\item As a card shuffle, $\mathbf{t}_{\ell}$ takes the $\ell$-th card from the
top and moves it further down the deck.

\pause

\item $\mathbf{t}_{1}$ is called the \defn{top-to-random shuffle}.
Upon renaming $1,2,\ldots,n$ as $n,n-1,\ldots,1$, it becomes
$\mathcal{B}_{n,1}$.
(So it is conjugate to $\mathcal{B}_{n,1}$ by $w_0$.)

\end{itemize}

\end{frame} \begin{frame} \frametitle{\ \ \ \ Somewhere-to-below shuffles: non-commutativity}

\begin{itemize}

\item $\mathbf{t}_{1},\mathbf{t}_{2},\ldots,\mathbf{t}_{n}$ do
not commute for $n\geq3$. For $n=3$, we have%
\[
\left[  \mathbf{t}_{1},\mathbf{t}_{2}\right]  =\operatorname*{cyc}%
\nolimits_{1,2}+\operatorname*{cyc}\nolimits_{1,2,3}-\operatorname*{cyc}%
\nolimits_{1,3,2}-\operatorname*{cyc}\nolimits_{1,3}.
\]

\pause

\item However, they come pretty close to commuting!

\item[$\circled{$\ast$}$] \textbf{Theorem 20.1 (Lafreniere, G., 2022).} There
exists a basis of the $\mathbf{k}$-module $\mathbf{k}\left[  S_{n}\right]  $
in which all of the endomorphisms \newline$R\left(  \mathbf{t}_{1}\right)
,R\left(  \mathbf{t}_{2}\right)  ,\ldots,R\left(  \mathbf{t}_{n}\right)  $ are
represented by upper-triangular matrices.
\end{itemize}

\section{The descent-destroying basis}

\end{frame} \begin{frame} \frametitle{\ \ \ \ The descent-destroying basis, 1}

\begin{itemize}
\item This basis is not hard to define, but I haven't seen it before.

\item[$\circled{$\ast$}$] For each $w\in S_{n}$, we let%
\[
\defnm{\operatorname*{Des}w}
:=\left\{  i\in\left[  n-1\right]  \ \mid\ w\left(
i\right)  >w\left(  i+1\right)  \right\}.
\]
This is called the \defn{descent set} of $w$.
\pause


\item[$\circled{$\ast$}$] For each $i\in\left[  n-1\right]  $, we let
$\defnm{s_{i}}:=\operatorname*{cyc}\nolimits_{i,i+1}$.
\pause

\item[$\circled{$\ast$}$] For each $I\subseteq\left[  n-1\right]  $, we let
\[
\defnm{G\left(  I\right)}  :=\left(  \text{the subgroup of }S_{n}\text{ generated by
the }s_{i}\text{ for }i\in I\right)  .
\]
This is called a \defn{Young parabolic subgroup} of $S_n$.

\end{itemize}

\end{frame} \begin{frame} \frametitle{\ \ \ \ The descent-destroying basis, 2}

\begin{itemize}

\item[$\circled{$\ast$}$] For each $w\in S_{n}$, we let%
\[
\mathbf{a}_{w}:=\sum_{\sigma\in G\left(  \operatorname*{Des}w\right)  }%
w\sigma\in\mathbf{k}\left[  S_{n}\right]  .
\]
In other words, $\mathbf{a}_{w}$ is obtained by breaking up the word $w$ into
maximal decreasing factors and re-sorting each factor arbitrarily (without
mixing different factors).
% (The $\mathbf{a}_{w}$ are called $a_{w}$ in my papers.)
\pause

\item[$\circled{$\ast$}$] The family $\left(  \mathbf{a}_{w}\right)  _{w\in
S_{n}}$ is a basis of $\mathbf{k}\left[  S_{n}\right]  $ (by triangularity).
\pause

\item For instance, for $n=3$, we have
\begin{align*}
\mathbf{a}_{\left[  {\textcolor{blue}{1}}{\textcolor{red}{2}}%
{\textcolor{green!50!black}{3}}\right]  }  &  =\left[  {\textcolor{blue}{1}}%
{\textcolor{red}{2}}{\textcolor{green!50!black}{3}}\right]  ;\\
\mathbf{a}_{\left[  {\textcolor{blue}{1}}\textcolor{red}{32}\right]  }  &
=\left[  {\textcolor{blue}{1}}\textcolor{red}{32}\right]  +\left[
{\textcolor{blue}{1}}\textcolor{red}{23}\right]  ;\\
\mathbf{a}_{\left[  \textcolor{blue}{21}\textcolor{red}{3}\right]  }  &
=\left[  \textcolor{blue}{21}\textcolor{red}{3}\right]  +\left[
\textcolor{blue}{12}\textcolor{red}{3}\right]  ;\\
\mathbf{a}_{\left[  \textcolor{blue}{2}\textcolor{red}{31}\right]  }  &
=\left[  \textcolor{blue}{2}\textcolor{red}{31}\right]  +\left[
\textcolor{blue}{2}\textcolor{red}{13}\right]  ;\\
\mathbf{a}_{\left[  \textcolor{blue}{31}\textcolor{red}{2}\right]  }  &
=\left[  \textcolor{blue}{31}\textcolor{red}{2}\right]  +\left[
\textcolor{blue}{13}\textcolor{red}{2}\right]  ;\\
\mathbf{a}_{\left[  \textcolor{blue}{321}\right]  }  &  =\left[
\textcolor{blue}{321}\right]  +\left[  \textcolor{blue}{312}\right]  +\left[
\textcolor{blue}{231}\right]  +\left[  \textcolor{blue}{213}\right]  +\left[
\textcolor{blue}{132}\right]  +\left[  \textcolor{blue}{123}\right]  .
\end{align*}

\end{itemize}

\end{frame} \begin{frame} \frametitle{\ \ \ \ The descent-destroying basis, 3}

\begin{itemize}

\item[$\circled{$\ast$}$] \textbf{Theorem 14.1 (Lafreni\`{e}re, G.).} For any
$w\in S_{n}$ and $\ell\in\left[  n\right]  $, we have%
\[
\mathbf{a}_{w}\mathbf{t}_{\ell}=\mu_{w,\ell}\mathbf{a}_{w}+\sum
_{\substack{v\in S_{n};\\v\prec w}}\lambda_{w,\ell,v}\mathbf{a}_{v}%
\]
for some nonnegative integer $\mu_{w,\ell}$, some integers $\lambda_{w,\ell
,v}$ and a certain partial order $\prec$ on $S_{n}$.
\pause

Thus, the endomorphisms $R\left(  \mathbf{t}_{1}\right)  ,R\left(
\mathbf{t}_{2}\right)  ,\ldots,R\left(  \mathbf{t}_{n}\right)  $ are
upper-triangular with respect to the basis $\left(  \mathbf{a}_{w}\right)
_{w\in S_{n}}$.

\end{itemize}

\end{frame} \begin{frame} \frametitle{\ \ \ \ The descent-destroying basis, 4}

\begin{itemize}

\item \textbf{Example:} For $n=4$, we have%
\[
\mathbf{a}_{\left[  4312\right]  }\mathbf{t}_{2}=\mathbf{a}_{\left[
4312\right]  }+\underbrace{\mathbf{a}_{\left[  4321\right]  }-\mathbf{a}%
_{\left[  4231\right]  }-\mathbf{a}_{\left[  3241\right]  }-\mathbf{a}%
_{\left[  2143\right]  }}_{\text{subscripts are }\prec\left[  4312\right]  }.
\]
\pause

\item \textbf{Example:}
For $n=3$, the endomorphism $R\left(  \mathbf{t}_{1}\right)  $ is
represented by the matrix%
\[%
\begin{tabular}
[c]{c|cccccc}
& $\mathbf{a}_{\left[  321\right]  }$ & $\mathbf{a}_{\left[  231\right]  }$ &
$\mathbf{a}_{\left[  132\right]  }$ & $\mathbf{a}_{\left[  213\right]  }$ &
$\mathbf{a}_{\left[  312\right]  }$ & $\mathbf{a}_{\left[  123\right]  }%
$\\\hline
$\mathbf{a}_{\left[  321\right]  }$ & $3$ & $1$ & $1$ &  & $1$ & \\
$\mathbf{a}_{\left[  231\right]  }$ &  &  &  & $1$ & $-1$ & $1$\\
$\mathbf{a}_{\left[  132\right]  }$ &  &  &  & $1$ &  & \\
$\mathbf{a}_{\left[  213\right]  }$ &  &  &  & $1$ &  & \\
$\mathbf{a}_{\left[  312\right]  }$ &  &  &  &  & $1$ & \\
$\mathbf{a}_{\left[  123\right]  }$ &  &  &  &  &  & $1$%
\end{tabular}
\]
(empty cells = zero entries). For instance, the last column means
$\mathbf{a}_{\left[  123\right]  }\mathbf{t}_{1}=\mathbf{a}_{\left[
123\right]  }+\mathbf{a}_{\left[  231\right]  }$.
\end{itemize}

\end{frame} \begin{frame} \frametitle{\ \ \ \ Eigenvalues of somewhere-to-below shuffles, 1}

\begin{itemize}

\item \textbf{Corollary 14.2.} The eigenvalues of the endomorphisms
$R\left(  \mathbf{t}_{1}\right)  ,R\left(  \mathbf{t}_{2}\right)
,\ldots,R\left(  \mathbf{t}_{n}\right)  $ and of all their linear combinations%
\[
R\left(  \lambda_{1}\mathbf{t}_{1}+\lambda_{2}\mathbf{t}_{2}+\cdots
+\lambda_{n}\mathbf{t}_{n}\right)
\]
are integers as long as $\lambda_{1},\lambda_{2},\ldots,\lambda_{n}$ are.
\pause

\item How many different eigenvalues do they have?
\pause

\item $R\left(  \mathbf{t}_{1}\right) \cong R\left(  \mathcal{B}_{n,1}\right)  $ has
only $n$ eigenvalues: $0,1,\ldots,n-2,n$, as we have seen before. The other
$R\left(  \mathbf{t}_{\ell}\right)  $'s have even fewer.
\pause

\item But their linear combinations $R\left(  \lambda_{1}\mathbf{t}%
_{1}+\lambda_{2}\mathbf{t}_{2}+\cdots+\lambda_{n}\mathbf{t}_{n}\right)  $ can
have many more. How many?
\end{itemize}

\section{Lacunar sets and Fibonacci numbers}

\end{frame} \begin{frame} \frametitle{\ \ \ \ Lacunar sets and Fibonacci numbers}

\begin{itemize}
\item[$\circled{$\ast$}$] A set $S$ of integers is called \defn{lacunar} if
it contains no two consecutive integers (i.e., we have $s+1\notin S$ for all
$s\in S$).
\pause

\item[$\circled{$\ast$}$] \textbf{Theorem 15.1 (combinatorial interpretation of
Fibonacci numbers, folklore).} The number of lacunar subsets of $\left[
n-1\right]  $ is the \defn{Fibonacci number} $f_{n+1}$.

(Recall: $\qquad f_{0}=0,\qquad f_{1}%
=1,\qquad $ $f_{n}=f_{n-1}+f_{n-2}$.)
\pause

\item[$\circled{$\ast$}$] \textbf{Theorem 15.2.} When $\lambda_{1},\lambda
_{2},\ldots,\lambda_{n}\in\mathbb{C}$ are generic, the endomorphism
$R\left(  \lambda_{1}\mathbf{t}_{1}+\lambda_{2}\mathbf{t}%
_{2}+\cdots+\lambda_{n}\mathbf{t}_{n}\right)  $
is diagonalizable and has $f_{n+1}$ distinct eigenvalues.
\pause

\item Note that $f_{n+1}\ll n!$.

\end{itemize}

\end{frame} \begin{frame} \frametitle{\ \ \ \ The Fibonacci filtration, 1}

\begin{itemize}

\item[$\circled{$\ast$}$] We prove this by finding a filtration
\[
0=F_{0}\subseteq F_{1}\subseteq F_{2}\subseteq\cdots\subseteq F_{f_{n+1}%
}=\mathbf{k}\left[  S_{n}\right]
\]
of the $\mathbf{k}$-module $\mathbf{k}\left[  S_{n}\right]  $ such that each
$R\left(  \mathbf{t}_{\ell}\right)  $ acts as a \textbf{scalar} on each of its
quotients $F_{i}/F_{i-1}$. In matrix terms, this means bringing $R\left(
\mathbf{t}_{\ell}\right)  $ to a block-triangular form, with the diagonal
blocks being \textquotedblleft scalar times $I$\textquotedblright\ matrices.

\item It is only natural that the quotients should correspond to the lacunar
subsets of $\left[  n-1\right]  $.

\item Let us approach the construction of this filtration.

\end{itemize}

\end{frame} \begin{frame} \frametitle{\ \ \ \ The Fibonacci filtration, 2}

\begin{itemize}

\item[$\circled{$\ast$}$] For each $I\subseteq\left[  n\right]  $, we set%
\[
\defnm{\operatorname*{sum}I}
:=\sum_{i\in I}i
\]
\pause
and%
\[
\defnm{\widehat{I}}
:=\left\{  0\right\}  \cup I\cup\left\{  n+1\right\}
\ \ \ \ \ \ \ \ \ \ \left(  \text{\textquotedblleft
enclosure\textquotedblright\ of }I\right)
\]
and%
\[
\defnm{I^{\prime}}
:=\left[  n-1\right]  \setminus\left(  I\cup\left(  I-1\right)
\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{\textquotedblleft
non-shadow\textquotedblright\ of }I\right)
\]
and%
\[
\defnm{F\left(  I\right)}
:=\left\{  \mathbf{q}\in\mathbf{k}\left[  S_{n}\right]
\ \mid\ \mathbf{q}s_{i}=\mathbf{q}\text{ for all }i\in I^{\prime}\right\}
\subseteq\mathbf{k}\left[  S_{n}\right]  .
\]
In probabilistic terms, $F\left(  I\right)  $ consists of those random states
of the deck that do not change if we swap the $i$-th and $\left(  i+1\right)
$-st cards from the top as long as neither $i$ nor $i+1$ is in $I$. To put it
informally: $F\left(  I\right)  $ consists of those random states that are
\textquotedblleft fully shuffled\textquotedblright\ between any two
consecutive $\widehat{I}$-positions.

\end{itemize}

\end{frame} \begin{frame} \frametitle{\ \ \ \ The Fibonacci filtration, 3}

\begin{itemize}

\item \textbf{Example:} If $n=11$ and $I=\left\{  3,6,7\right\}  $, then
\[
\widehat{I}:=\left\{  0\right\}  \cup I\cup\left\{  n+1\right\}
=\left\{  0,3,6,7,12\right\}
\]
and
\[
I^{\prime}:=\left[  n-1\right]  \setminus\left(  I\cup\left(  I-1\right)
\right)
=\left\{ 1,4,8,9,10\right\}
\]
and
\[
F\left(  I\right)  =\left\{  \mathbf{q}\in\mathbf{k}\left[  S_{11}\right]
\ \mid\ \mathbf{q}s_{1}=\mathbf{q}s_{4}=\mathbf{q}s_{8}=\mathbf{q}%
s_{9}=\mathbf{q}s_{10}=\mathbf{q}\right\}  .
\]
Illustrating this:%
\[
\begin{tikzpicture}[scale=0.8]
\begin{scope}[every edge/.style={draw=black,thick}]
\filldraw[very thick, fill=blue, draw=green!50!black] (0,0) rectangle (1, 1);
\draw[very thick, draw=green!50!black] (1,0) rectangle (2, 1);
\filldraw
[very thick, fill=black!50!white, draw=green!50!black] (2,0) rectangle (3, 1);
\filldraw[very thick, fill=black, draw=green!50!black] (3,0) rectangle (4, 1);
\draw[very thick, draw=green!50!black] (4,0) rectangle (5, 1);
\filldraw
[very thick, fill=black!50!white, draw=green!50!black] (5,0) rectangle (6, 1);
\filldraw[very thick, fill=black, draw=green!50!black] (6,0) rectangle (7, 1);
\filldraw[very thick, fill=black, draw=green!50!black] (7,0) rectangle (8, 1);
\draw[very thick, draw=green!50!black] (8,0) rectangle (9, 1);
\draw[very thick, draw=green!50!black] (9,0) rectangle (10, 1);
\draw[very thick, draw=green!50!black] (10,0) rectangle (11, 1);
\filldraw
[very thick, fill=blue!50!white, draw=green!50!black] (11,0) rectangle (12, 1);
\filldraw
[very thick, fill=blue, draw=green!50!black] (12,0) rectangle (13, 1);
\node at (0.5, -0.5) {$0$};
\node at (1.5, -0.5) {$1$};
\node at (2.5, -0.5) {$2$};
\node at (3.5, -0.5) {$3$};
\node at (4.5, -0.5) {$4$};
\node at (5.5, -0.5) {$5$};
\node at (6.5, -0.5) {$6$};
\node at (7.5, -0.5) {$7$};
\node at (8.5, -0.5) {$8$};
\node at (9.5, -0.5) {$9$};
\node at (10.5, -0.5) {$10$};
\node at (11.5, -0.5) {$11$};
\node at (12.5, -0.5) {$12$};
\end{scope}
\end{tikzpicture}
\]
(black = $I$;\qquad grey = $I-1$;\qquad blue = $\widehat{I}\setminus I$;\\
lightblue = $n$;\qquad white = $I^{\prime}$).

\end{itemize}

\end{frame} \begin{frame} \frametitle{\ \ \ \ The Fibonacci filtration, 4}

\begin{itemize}


\item[$\circled{$\ast$}$] For any $\ell\in\left[  n\right]  $, we let
$m_{I,\ell}$ be the distance from $\ell$ to the next-higher element of
$\widehat{I}$. In other words,%
\[
m_{I,\ell}:=\left(  \text{smallest element of }\widehat{I}\text{ that is }%
\geq\ell\right)  -\ell\in\left\{  0,1,\ldots,n\right\}  .
\]


In our above example,%
\[
\left(  m_{I,1},\ m_{I,2},\ \ldots,\ m_{I,11}\right)  =\left(
2,\ 1,\ 0,\ 2,\ 1,\ 0,\ 0,\ 4,\ 3,\ 2,\ 1\right)  .
\]

\[
\begin{tikzpicture}[scale=0.8]
\begin{scope}[every edge/.style={draw=black,thick}]
\filldraw[very thick, fill=blue, draw=green!50!black] (0,0) rectangle (1, 1);
\draw[very thick, draw=green!50!black] (1,0) rectangle (2, 1);
\filldraw
[very thick, fill=black!50!white, draw=green!50!black] (2,0) rectangle (3, 1);
\filldraw[very thick, fill=black, draw=green!50!black] (3,0) rectangle (4, 1);
\draw[very thick, draw=green!50!black] (4,0) rectangle (5, 1);
\filldraw
[very thick, fill=black!50!white, draw=green!50!black] (5,0) rectangle (6, 1);
\filldraw[very thick, fill=black, draw=green!50!black] (6,0) rectangle (7, 1);
\filldraw[very thick, fill=black, draw=green!50!black] (7,0) rectangle (8, 1);
\draw[very thick, draw=green!50!black] (8,0) rectangle (9, 1);
\draw[very thick, draw=green!50!black] (9,0) rectangle (10, 1);
\draw[very thick, draw=green!50!black] (10,0) rectangle (11, 1);
\filldraw
[very thick, fill=blue!50!white, draw=green!50!black] (11,0) rectangle (12, 1);
\filldraw
[very thick, fill=blue, draw=green!50!black] (12,0) rectangle (13, 1);
\node at (0.5, -0.5) {$0$};
\node at (1.5, -0.5) {$1$};
\node at (2.5, -0.5) {$2$};
\node at (3.5, -0.5) {$3$};
\node at (4.5, -0.5) {$4$};
\node at (5.5, -0.5) {$5$};
\node at (6.5, -0.5) {$6$};
\node at (7.5, -0.5) {$7$};
\node at (8.5, -0.5) {$8$};
\node at (9.5, -0.5) {$9$};
\node at (10.5, -0.5) {$10$};
\node at (11.5, -0.5) {$11$};
\node at (12.5, -0.5) {$12$};
\node at (1.5, 0.5) {$2$};
\node at (2.5, 0.5) {$1$};
\node at (3.5, 0.5) {$\textcolor{white}0$};
\node at (4.5, 0.5) {$2$};
\node at (5.5, 0.5) {$1$};
\node at (6.5, 0.5) {$\textcolor{white}0$};
\node at (7.5, 0.5) {$\textcolor{white}0$};
\node at (8.5, 0.5) {$4$};
\node at (9.5, 0.5) {$3$};
\node at (10.5, 0.5) {$2$};
\node at (11.5, 0.5) {$1$};
\node at (12.5, 0.5) {$\textcolor{white}0$};
\end{scope}
\end{tikzpicture}
\]

% For another example, if $n=5$ and $I=\left\{  2,3\right\}  $, then
% $\widehat{I}=\left\{  0,2,3,6\right\}  $ and%
% \[
% \left(  m_{I,1},\ m_{I,2},\ m_{I,3},\ m_{I,4},\ m_{I,5}\right)  =\left(
% 1,\ 0,\ 0,\ 2,\ 1\right)  .
% \]


\item We note that, for any $\ell\in\left[  n\right]  $, we have the
equivalence%
\[
m_{I,\ell}=0\ \ \ \ \Longleftrightarrow\ \ \ \ \ell\in\widehat{I}%
\ \ \ \ \Longleftrightarrow\ \ \ \ \ell\in I.
\]

\end{itemize}

\end{frame} \begin{frame} \frametitle{\ \ \ \ The Fibonacci filtration, 5}

\begin{itemize}


\item[$\circled{$\ast$}$] \textbf{Crucial Lemma 16.1.} Let $I\subseteq\left[
n\right]  $ and $\ell\in\left[  n\right]  $. Then,%
\[
\mathbf{qt}_{\ell}\in m_{I,\ell}\mathbf{q}+\underbrace{\sum
_{\substack{J\subseteq\left[  n\right]  ;\\\operatorname*{sum}%
J<\operatorname*{sum}I}}F\left(  J\right)  }_{\substack{\text{Think of these
as}\\\text{\textquotedblleft lower-order terms\textquotedblright}%
}}\ \ \ \ \ \ \ \ \ \ \text{for each }\mathbf{q}\in F\left(  I\right)  .
\]


\item \textit{Proof:} Expand $\mathbf{qt}_{\ell}$ by the definition of
$\mathbf{t}_{\ell}$, and break up the resulting sum into smaller bunches using
the interval decomposition%
\[
\left[  \ell,n\right]  =\left[  \ell,i_{k}-1\right]  \sqcup\left[
i_{k},i_{k+1}-1\right]  \sqcup\left[  i_{k+1},i_{k+2}-1\right]  \sqcup
\cdots\sqcup\left[  i_{p},n\right]
\]
(where $i_{k}<i_{k+1}<\cdots<i_{p}$ are the elements of $I$ larger or equal to
$\ell$). The $\left[  \ell,i_{k}-1\right]  $ bunch gives the $m_{I,\ell
}\mathbf{q}$ term; the others live in appropriate $F\left(  J\right)  $'s.

See \arxiv{2212.06274} for the details.

\end{itemize}

\end{frame} \begin{frame} \frametitle{\ \ \ \ The Fibonacci filtration, 6}

\begin{itemize}


\item[$\circled{$\ast$}$] Thus, we obtain a filtration of $\mathbf{k}\left[
S_{n}\right]  $ if we label the subsets $I$ of $\left[  n\right]  $ in the
order of increasing $\operatorname*{sum}I$ and add up the respective $F\left(
I\right)  $s.

On each subquotient of this filtration, $\mathbf{t}_{\ell}$ acts as a scalar
$m_{I,\ell}$.

\item Unfortunately, this filtration has $2^{n}$, not $f_{n+1}$ terms.
\pause

\item[$\circled{$\ast$}$] Fortunately, that's because many of its terms are
redundant. The ones that aren't correspond precisely to the $I$'s that are
lacunar subsets of $\left[  n-1\right]  $:

\item \textbf{Lemma 16.2.} Let $k\in\mathbb{N}$. Then,%
\[
\sum_{\substack{J\subseteq\left[  n\right]  ;\\\operatorname*{sum}%
J<k}}F\left(  J\right)  =\sum_{\substack{J\subseteq\left[  n-1\right]  \text{
is lacunar;}\\\operatorname*{sum}J<k}}F\left(  J\right)  .
\]

\item \textit{Proof:}
If $J\subseteq\left[  n\right]  $ contains $n$ or fails
to be lacunar, then $F\left(  J\right)  $ is a submodule of some $F\left(
K\right)  $ with $\operatorname*{sum}K<\operatorname*{sum}J$. (Exercise!)

\end{itemize}

\end{frame} \begin{frame} \frametitle{\ \ \ \ The Fibonacci filtration, 7}

\begin{itemize}

\item Now, we let $\defnm{Q_{1},Q_{2},\ldots,Q_{f_{n+1}}}$
be the $f_{n+1}$ lacunar
subsets of $\left[  n-1\right]  $, listed in such an order that%
\[
\operatorname*{sum}\left(  Q_{1}\right)  \leq\operatorname*{sum}\left(
Q_{2}\right)  \leq\cdots\leq\operatorname*{sum}\left(  Q_{f_{n+1}}\right)  .
\]
Then, for each $i\in\left[  0,f_{n+1}\right]  $,
define a $\mathbf{k}$-submodule%
\[
\defnm{F_{i}}
:=F\left(  Q_{1}\right)  +F\left(  Q_{2}\right)  +\cdots+F\left(
Q_{i}\right)  \qquad \text{of }\mathbf{k}\left[  S_{n}\right]
\]
(so that $F_{0}=0$). The resulting
filtration%
\[
0=F_{0}\subseteq F_{1}\subseteq F_{2}\subseteq\cdots\subseteq F_{f_{n+1}%
}=\mathbf{k}\left[  S_{n}\right]
\]
(which we call the \defn{Fibonacci filtration} of $\mathbf{k}\left[
S_{n}\right]  $) satisfies the properties we need:

\end{itemize}

\end{frame} \begin{frame} \frametitle{\ \ \ \ The Fibonacci filtration, 8}

\begin{itemize}

\item \textbf{Theorem 16.3.} For each $i\in\left[  f_{n+1}\right]  $ and
$\ell\in\left[  n\right]  $, we have%
\[
F_{i}\cdot\left(  \mathbf{t}_{\ell}-m_{Q_{i},\ell}\right)  \subseteq F_{i-1}%
\]
(so that $R\left(  \mathbf{t}_{\ell}\right)  $ acts on $F_{i}/F_{i-1}$ as
multiplication by $m_{Q_{i},\ell}$).

\item \textit{Proof:} Lemma 16.1 + Lemma 16.2.
\pause

\item \textbf{Lemma 16.4.} The quotients $F_{i}/F_{i-1}$ are nontrivial for all
$i\in\left[  f_{n+1}\right]  $.

\item \textit{Proof:} See below.
\pause

\item[$\circled{$\ast$}$] \textbf{Corollary 16.5.} Let $\mathbf{k}$ be a field,
and let $\lambda_{1},\lambda_{2},\ldots,\lambda_{n}\in\mathbf{k}$. Then, the
eigenvalues of $R\left(  \lambda_{1}\mathbf{t}_{1}+\lambda_{2}\mathbf{t}%
_{2}+\cdots+\lambda_{n}\mathbf{t}_{n}\right)  $ are the linear combinations%
\[
\lambda_{1}m_{I,1}+\lambda_{2}m_{I,2}+\cdots+\lambda_{n}m_{I,n}%
\ \ \ \ \ \ \ \ \ \ \text{for }I\subseteq\left[  n-1\right]  \text{ lacunar.}%
\]
\pause

\item Theorem 15.2 easily follows by some linear algebra.

\item More generally, this holds not just for linear combinations $\lambda
_{1}\mathbf{t}_{1}+\lambda_{2}\mathbf{t}_{2}+\cdots+\lambda_{n}\mathbf{t}_{n}$
but for any noncommutative polynomials in $\mathbf{t}_{1},\mathbf{t}%
_{2},\ldots,\mathbf{t}_{n}$.
\end{itemize}

\section{Back to the basis}

\end{frame} \begin{frame} \frametitle{\ \ \ \ Back to the basis, 1}

\begin{itemize}

\item The descent-destroying basis $\left(  \mathbf{a}_{w}\right)  _{w\in
S_{n}}$ is compatible with our filtration:

\item[$\circled{$\ast$}$] \textbf{Theorem 17.1.} For each $I\subseteq\left[
n\right]  $, the family $\left(  \mathbf{a}_{w}\right)  _{w\in S_{n}%
;\ I^{\prime}\subseteq\operatorname*{Des}w}$ is a basis of the $\mathbf{k}%
$-module $F\left(  I\right)  $.
\pause

\item[$\circled{$\ast$}$] If $w\in S_{n}$ is any permutation, then the
\defn{$Q$-index} of $w$ is defined to be the \textbf{smallest} $i\in\left[
f_{n+1}\right]  $ such that $Q_{i}^{\prime}\subseteq\operatorname*{Des}w$. We
call this $Q$-index $\defnm{\operatorname*{Qind}w}$.

\item \textbf{Proposition 17.2.} Let $w\in S_{n}$ and $i\in\left[
f_{n+1}\right]  $. Then, $\operatorname*{Qind}w=i$ if and only if
$Q_{i}^{\prime}\subseteq\operatorname*{Des}w\subseteq\left[  n-1\right]
\setminus Q_{i}$.

\end{itemize}

\section{Back to the basis}

\end{frame} \begin{frame} \frametitle{\ \ \ \ Back to the basis, 2}

\begin{itemize}

\item \textbf{Note:} The numbering $Q_{1},Q_{2},\ldots,Q_{f_{n+1}}$ of the
lacunar subsets of $\left[  n-1\right]  $ is not unique; we just picked one.
The $Q$-index $i=\operatorname*{Qind}w$ of a $w\in S_{n}$ depends on this
numbering. However, the corresponding lacunar set $Q_{i}$ does not, since
Proposition 17.2 determines it canonically (it is the unique lacunar
$L\subseteq\left[  n-1\right]  $ satisfying $L^{\prime}\subseteq
\operatorname*{Des}w\subseteq\left[  n-1\right]  \setminus L$).

Thus, think of this set $Q_{i}$ as the \textquotedblleft
real\textquotedblright\ index of $w$.
We just found $i$ easier to work with.

(``Morally'', the Fibonacci filtration should be indexed
by a poset; then you need not choose any numbering.)

\end{itemize}

\section{Back to the basis}

\end{frame} \begin{frame} \frametitle{\ \ \ \ Back to the basis, 3}

\begin{itemize}

\item[$\circled{$\ast$}$] \textbf{Theorem 17.3.} For each $i\in\left[
0,f_{n+1}\right]  $, the $\mathbf{k}$-module $F_{i}$ is free with basis
$\left(  \mathbf{a}_{w}\right)  _{w\in S_{n};\ \operatorname*{Qind}w\leq i}$.
\pause

\item[$\circled{$\ast$}$] \textbf{Corollary 17.4.} For each $i\in\left[
f_{n+1}\right]  $, the $\mathbf{k}$-module $F_{i}/F_{i-1}$ is free with basis
$\left(  \overline{\mathbf{a}_{w}}\right)  _{w\in S_{n};\ \operatorname*{Qind}%
w=i}$.
\pause

\item This yields Lemma 9.4 and also leads to Theorem 7.1, made precise as follows:

\item[$\circled{$\ast$}$] \textbf{Theorem 17.5 (Lafreni\`{e}re, G.).} For any
$w\in S_{n}$ and $\ell\in\left[  n\right]  $, we have%
\[
\mathbf{a}_{w}\mathbf{t}_{\ell}=\mu_{w,\ell}\mathbf{a}_{w}+\sum
_{\substack{v\in S_{n};\\\operatorname*{Qind}v<\operatorname*{Qind}w}%
}\lambda_{w,\ell,v}\mathbf{a}_{v}%
\]
for some nonnegative integer $\mu_{w,\ell}$ and some integers $\lambda
_{w,\ell,v}$.

Thus, the endomorphisms $R\left(  \mathbf{t}_{1}\right)  ,R\left(
\mathbf{t}_{2}\right)  ,\ldots,R\left(  \mathbf{t}_{n}\right)  $ are
upper-triangular with respect to the basis $\left(  \mathbf{a}_{w}\right)
_{w\in S_{n}}$ as long as the permutations $w\in S_{n}$ are ordered by
increasing $Q$-index.
\end{itemize}

\section{The multiplicities}

\end{frame} \begin{frame} \frametitle{\ \ \ \ The multiplicities, 1}

\begin{itemize}

\item In Corollary 9.5, we found the eigenvalues of the endomorphism $R\left(
\lambda_{1}\mathbf{t}_{1}+\lambda_{2}\mathbf{t}_{2}+\cdots+\lambda
_{n}\mathbf{t}_{n}\right)  $. With Corollary 17.4, we can also find their
algebraic multiplicities. To state a formula for them, we need a definition:
\pause

\item[$\circled{$\ast$}$] For each $i\in\left[  f_{n+1}\right]  $, we set%
\[
\defnm{\delta_{i}}
:=\left(  \text{the number of all }w\in S_{n}\text{ satisfying
}\operatorname*{Qind}w=i\right)  .
\]


\item[$\circled{$\ast$}$] \textbf{Corollary 18.1 (informal version).}
Assume that $\mathbf{k}$ is a field. Let $\lambda_{1},\lambda_{2}%
,\ldots,\lambda_{n}\in\mathbf{k}$. Then, the endomorphism $R\left(
\lambda_{1}\mathbf{t}_{1}+\lambda_{2}\mathbf{t}_{2}+\cdots+\lambda
_{n}\mathbf{t}_{n}\right)  $ has eigenvalues%
\begin{align*}
& \lambda_{I} :=\lambda_{1}m_{I,1}+\lambda_{2}m_{I,2}+\cdots+\lambda_{n}%
m_{I,n}\\
&\qquad \ \ \ \ \ \ \ \ \ \ \text{for all lacunar }I\subseteq\left[
n-1\right] \\
& \text{ with respective multiplicities } \delta_i, \\
& \qquad
\text{ where }
i\in\left[  f_{n+1}\right] \text{ is such that } I=Q_{i}.
\end{align*}
(If some $\lambda_I$ happen to coincide, then their algebraic multiplicities must be
added together.)
\end{itemize}

\section{The multiplicities}

\end{frame} \begin{frame} \frametitle{\ \ \ \ The multiplicities, 2}

\begin{itemize}

\item Can we compute the $\delta_{i}$ explicitly? Yes!

\item[$\circled{$\ast$}$] \textbf{Theorem 18.2.} Let $i\in\left[
f_{n+1}\right]  $. Then:

\begin{enumerate}
\item[\textbf{(a)}] Write the set $Q_{i}$ in the form $Q_{i}=\left\{
i_{1}<i_{2}<\cdots<i_{p}\right\}  $, and set $i_{0}=1$ and $i_{p+1}=n+1$. Let
$j_{k}=i_{k}-i_{k-1}$ for each $k\in\left[  p+1\right]  $. Then,%
\[
\delta_{i}=\underbrace{\dbinom{n}{j_{1},j_{2},\ldots,j_{p+1}}}%
_{\substack{\text{multinomial}\\\text{coefficient}}}\cdot\prod_{k=2}%
^{p+1}\left(  j_{k}-1\right)  .
\]
\pause

\item[\textbf{(b)}] We have $\delta_{i}\mid n!$.
\end{enumerate}

\item \textbf{Note.} This reminds of the hook-length formula for standard
tableaux, but is much simpler.
\end{itemize}

\section{Variants}

\end{frame} \begin{frame} \frametitle{\ \ \ \ Variants}

\begin{itemize}
\item Most of what we said about the somewhere-to-below shuffles
$\mathbf{t}_{\ell}$ can be extended to their antipodes
$\mathbf{t}_{\ell}^\ast$ (the ``\defn{below-to-somewhere
shuffles}''). For instance:

\item \textbf{Theorem 19.1.} There exists a basis of the $\mathbf{k}$-module
$\mathbf{k}\left[  S_{n}\right]  $ in which all of the endomorphisms $R\left(
\mathbf{t}_{1}^\ast \right)  ,R\left(  S\mathbf{t}_{2}^\ast \right)  ,
\ldots,R\left(  \mathbf{t}_{n}^\ast \right)  $
are represented by upper-triangular matrices.

\item We can also use left instead of right multiplication:

\item \textbf{Theorem 19.2.} There exists a basis of the $\mathbf{k}$-module
$\mathbf{k}\left[  S_{n}\right]  $ in which all of the endomorphisms $L\left(
\mathbf{t}_{1}\right)  ,L\left(  \mathbf{t}_{2}\right)  ,\ldots,L\left(
\mathbf{t}_{n}\right)  $ are represented by upper-triangular matrices.

\item These follow from Theorem 14.1 using dual bases and
transpose matrices. No new combinatorics required!

% \item \textbf{Question.} Do we have $L\left(  \mathbf{t}_{\ell}\right)  \sim
% R\left(  \mathbf{t}_{\ell}\right)  $ in $\operatorname*{End}%
% \nolimits_{\mathbf{k}}\left(  \mathbf{k}\left[  S_{n}\right]  \right)  $ when
% $\mathbf{k}$ is not a field?

% \item \textbf{Remark.} The similarity $\mathbf{t}_{\ell}\sim S\left(
% \mathbf{t}_{\ell}\right)  $ in $\mathbf{k}\left[  S_{n}\right]  $ holds when
% $\operatorname*{char}\mathbf{k}=0$, but not for general fields $\mathbf{k}$.
% (E.g., it fails for $\mathbf{k}=\mathbb{F}_{2}$ and $n=4$ and $\ell=1$.)
% \end{itemize}

\end{itemize}

\section{Commutators}

\end{frame} \begin{frame} \frametitle{\ \ \ \ Commutators, 1}

\begin{itemize}
\item The simultaneous trigonalizability of the endomorphisms \newline%
$R\left(  \mathbf{t}_{1}\right)  ,R\left(  \mathbf{t}_{2}\right)
,\ldots,R\left(  \mathbf{t}_{n}\right)  $ yields that their pairwise
commutators are nilpotent. Hence, the pairwise commutators $\left[
\mathbf{t}_{i},\mathbf{t}_{j}\right]  $ are also nilpotent.

\item \textbf{Question.} How small an exponent works in $\left[
\mathbf{t}_{i},\mathbf{t}_{j}\right]  ^{\ast}=0$ ?

\item[$\circled{$\ast$}$] \textbf{Theorem 20.1.} We have $\left[
\mathbf{t}_{i},\mathbf{t}_{j}\right]  ^{j-i+1}=0$ for any $1\leq i\leq j\leq
n$.

\item[$\circled{$\ast$}$] \textbf{Theorem 20.2.} We have $\left[
\mathbf{t}_{i},\mathbf{t}_{j}\right]  ^{\left\lceil \left(  n-j\right)
/2\right\rceil +1}=0$ for any $i,j\in\left[  n\right]  $.

\item Depending on $i$ and $j$, one of the exponents is better than the other.

\textbf{Conjecture.} The better one is optimal! (Checked for all $n\leq12$.)

\end{itemize}

\end{frame} \begin{frame} \frametitle{\ \ \ \ Commutators, 2}

\begin{itemize}

\item[$\circled{$\ast$}$] Stronger results hold, replacing powers by products.

\item[$\circled{$\ast$}$] Several other curious facts hold: For example,%
\[
\hspace{-1pc}
\mathbf{t}_{i+1}\mathbf{t}_{i}=\left(  \mathbf{t}_{i}-1\right)  \mathbf{t}%
_{i}\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ \mathbf{t}_{i+2}\left(
\mathbf{t}_{i}-1\right)  =\left(  \mathbf{t}_{i}-1\right)  \left(
\mathbf{t}_{i+1}-1\right)
\]
and%
\[
\hspace{-0.5pc}
\mathbf{t}_{n-1}\left[  \mathbf{t}_{i},\mathbf{t}_{n-1}\right]
=0\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ \left[  \mathbf{t}%
_{i},\mathbf{t}_{n-1}\right]  \left[  \mathbf{t}_{j},\mathbf{t}_{n-1}\right]
=0
\]
for all $i$ and $j$.

\item All this is completely elementary but surprisingly hard to prove (dozens
of pages of manipulations with sums and cycles). The proofs can be found in
\arxiv{2309.05340}.


\item What is \textquotedblleft really\textquotedblright\ going on? No idea...
\end{itemize}

\section{Representation theory}

\end{frame} \begin{frame} \frametitle{\ \ \ \ Representations, 1}

\begin{itemize}
\item Two natural questions:

\begin{enumerate}
\item The $F\left(  I\right)  $ and the $F_{i}$ are left ideals of
$\mathbf{k}\left[  S_{n}\right]  $; how do they decompose into Specht modules?

\item How do $\mathbf{t}_{1},\mathbf{t}_{2},\ldots,\mathbf{t}_{n}$ act on a
given Specht module?
\end{enumerate}

\pause
\item We can answer these.

\item The answer uses symmetric functions, specifically:

\item Let $\defnm{s_{\lambda}}$ be the Schur function for a partition $\lambda$.

\item Let $\defnm{h_{m}} =s_{\left(  m\right)  }$ be the $m$-th complete homogeneous
symmetric function for each $m\geq0$.

\item Let $\defnm{z_{m}} =s_{\left(  m-1,1\right)  }=h_{m-1}h_{1}-h_{m}$ for each $m>1$.

\end{itemize}

\end{frame} \begin{frame} \frametitle{\ \ \ \ Representations, 2}

\begin{itemize}
\item For each lacunar subset $I$ of $\left[  n-1\right]  $, we define a
symmetric function
\[
\defnm{z_{I}}
:=h_{i_{1}-1}\prod_{j=2}^{k}z_{i_{j}-i_{j-1}}\ \ \ \ \ \ \ \ \ \ \left(
\text{over }\mathbb{Z}\right)  ,
\]
where $i_{1},i_{2},\ldots,i_{k}$ are the elements of $I\cup\left\{
n+1\right\}  $ in increasing order (so that $i_{k}=n+1$ and $I=\left\{
i_{1}<i_{2}<\cdots<i_{k-1}\right\}  $).

This is a skew Schur function corresponding to a disjoint union of hooks:
e.g., if $n=11$ and $I=\left\{  3,6,8\right\}  $, then it is
\[
\begin{tikzpicture}[scale=0.5]
\begin{scope}[every edge/.style={draw=black,thick}]
\filldraw[thick, fill=green!50!white] (6,0) rectangle (8,-1);
\filldraw
[thick, fill=yellow!50!white] (4,-1) -- (6, -1) -- (6, -2) -- (5, -2) -- (5, -3) -- (4, -3) -- cycle;
\filldraw[thick, fill=blue!50!white] (3, -3) rectangle (4, -5);
\filldraw
[thick, fill=red!50!white] (0, -5) -- (3, -5) -- (3, -6) -- (1, -6) -- (1, -7) -- (0, -7) -- cycle;
\end{scope}
\node at (-3, -0.5) {$i_1-1$};
\node at (-3, -1.5) {$i_2-i_1-1$};
\node at (-3, -2.5) {$1$};
\node at (-3, -3.5) {$i_3-i_2-1$};
\node at (-3, -4.5) {$1$};
\node at (-3, -5.5) {$i_4-i_3-1$};
\node at (-3, -6.5) {$1$};
\end{tikzpicture}
\]

\end{itemize}

\end{frame} \begin{frame} \frametitle{\ \ \ \ Representations, 3}

\begin{itemize}

\item For each lacunar $I\subseteq\left[  n-1\right]  $ and each partition
$\lambda$ of $n$, we let \defn{$c_{\lambda}^{I}$} be the coefficient of $s_{\lambda}$
in the Schur expansion of $z_{I}$.

This is a Littlewood--Richardson coefficient
(since $z_{I}$ is a skew Schur function), thus $\in \NN$.
\pause

\item \textbf{Theorem 21.1.} Let $\nu$ be a partition. Let $\lambda
_{1},\lambda_{2},\ldots,\lambda_{n}\in\mathbf{k}$. Then, the
shuffle $\lambda_{1}\mathbf{t}_{1}+\lambda_{2}\mathbf{t}_{2}+\cdots
+\lambda_{n}\mathbf{t}_{n}$ acts on the
Specht module $\mathcal{S}^{\nu}$ as a linear
map with eigenvalues%
\begin{align*}
&  \lambda_{1}m_{I,1}+\lambda_{2}m_{I,2}+\cdots+\lambda_{n}m_{I,n}\\
&  \ \ \ \ \ \ \ \ \ \ \text{for all lacunar }I\subseteq\left[  n-1\right]
\text{ satisfying }c_{\nu}^{I}\neq0,
\end{align*}
and the multiplicity of each such eigenvalue is $c_{\nu}^{I}$ in the generic
case.
% (i.e., if no two $I$'s produce the same linear combination; otherwise the
%multiplicities of colliding eigenvalues should be added together).

If all these linear combinations are distinct, then this linear map is diagonalizable.

\end{itemize}

\end{frame} \begin{frame} \frametitle{\ \ \ \ Representations, 4}

\begin{itemize}


\item \textbf{Theorem 21.2 (lazy version).} Let $\mathbf{k}$ be a field of
characteristic $0$. Let $i\in\left[  f_{n+1}\right]  $. As a representation of
$S_{n}$, the quotient module $F_{i}/F_{i-1}$ has Frobenius characteristic
$z_{Q_{i}}$.

\item \textbf{Theorem 21.2 (careful version, true in every characteristic).}
Let $i\in\left[  f_{n+1}\right]  $. Consider the lacunar subset $Q_{i}$ of
$\left[  n-1\right]  $. Let $i_{1},i_{2},\ldots,i_{k}$ be the elements of
$Q_{i}\cup\left\{  n+1\right\}  $ in increasing order. Then, as
representations of $S_{n}$, we have%
\[
F_{i}/F_{i-1}\cong
%\underbrace{
\mathcal{H}_{i_{1}-1}\ast\mathcal{Z}%
_{i_{2}-i_{1}}\ast\mathcal{Z}_{i_{3}-i_{2}}\ast\cdots\ast\mathcal{Z}%
_{i_{k}-i_{k-1}}
%}_{\substack{\text{the first factor is an }\mathcal{H}%
%\text{,}\\\text{while all others are }\mathcal{Z}\text{'s}}}
,
\]
where $\ast$ means induction product (that is, $U\ast V=\operatorname*{Ind}%
\nolimits_{S_{i}\times S_{j}}^{S_{i+j}}\left(  U\otimes V\right)  $), and
where $\mathcal{H}_{m}$ is the trivial $1$-dimensional representation of
$S_{m}$, whereas $\mathcal{Z}_{m}$ is the reflection representation of $S_{m}$
(that is, $\mathbf{k}^{m}$ modulo the span of $\left(  1,1,\ldots,1\right)  $).

\item Proofs appear in \arxiv{2508.00752}.
\end{itemize}

\section{Conjectures and questions}

\end{frame} \begin{frame} \frametitle{\ \ \ \ Conjectures and questions}

\begin{itemize}
\item \textbf{Question.} What can be said about the $\mathbf{k}$-subalgebra
$\mathbf{k}\left[  \mathbf{t}_{1},\mathbf{t}_{2},\ldots,\mathbf{t}_{n}\right]
$ of $\mathbf{k}\left[  S_{n}\right]  $ ? Note:%
\[%
\begin{tabular}
[c]{|c||c|c|c|c|c|c|c|c|}\hline
$n$ & $1$ & $2$ & $3$ & $4$ & $5$ & $6$ & $7$ & $8$\\\hline
$\dim\left(  \mathbb{Q}\left[  \mathbf{t}_{1},\mathbf{t}_{2},\ldots
,\mathbf{t}_{n}\right]  \right)  $ & $1$ & $2$ & $4$ & $9$ & $23$ & $66$ &
$212$ & $761$\\\hline
\end{tabular}
\ \
\]
(this sequence is not in the OEIS as of 2025-10-08).

\item \textbf{Question.} Do the results about commutators and representations generalize to the Hecke algebra?

(The Fibonacci filtration and descent-destroying basis definitely do.
Proofs forthcoming...)
\end{itemize}

\end{frame}

\begin{frame} \frametitle{\ \ \ \ I thank}

\begin{itemize}
\item \textbf{Sarah Brauner, Patricia Commins, Nadia Lafreni\`ere}
and \textbf{Franco Saliola} for obvious reasons.

\item the \textbf{Mathematisches Forschungsinstitut Oberwolfach} for the
Research in Pairs program.

\item \textbf{Martin Lorenz, Marcelo
Aguiar, Vic Reiner, Travis Scrimshaw, Theo Douvropoulos,
Volkmar Welker} for various ideas shared over the years.

\item \textbf{Alex Postnikov} for the invitation (MIT).

\item \textbf{GaYee Park} for the invitation (Dartmouth).

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

\end{frame}

\end{document}

