% 60-min talk

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\begin{document}

\title{Littlewood--Richardson coefficients and birational combinatorics}
\author{\href{http://www.cip.ifi.lmu.de/~grinberg/}{Darij Grinberg}}
\date{14 October 2020 \\% [corrected version] \\
Drexel University, Philadelphia, PA}

\frame{\titlepage\textbf{slides: \color{red}
\url{http://www.cip.ifi.lmu.de/~grinberg/algebra/drexel2020.pdf}}
\newline\textbf{paper: {\red \arxiv{2008.06128}} aka \color{red}
\url{http://www.cip.ifi.lmu.de/~grinberg/algebra/lrhspr.pdf}}
\newline}

\begin{frame}
\frametitle{\ \ \ \ \ Manifest}

\begin{itemize}
\item I shall review the Littlewood--Richardson coefficients
and some of their classical properties. \pause

\item I will then state a ``hidden symmetry'' conjectured
by Pelletier and Ressayre (\arxiv{2005.09877}) and outline
how I proved it. \pause

\item The proof is a nice example of
{\blue birational combinatorics}: the use of birational
transformations in elementary combinatorics (specifically,
here, in finding and proving a bijection).
\end{itemize}
\end{frame}

\begin{frame}
\fti{Chapter 1}
\begin{center}
{\Huge \bf \sc Chapter 1} \\
\noindent\rule[0.5ex]{\linewidth}{1pt}
{\Large \bf Littlewood--Richardson coefficients}
\end{center}

\vspace{0.2cm}

References (among many):
\begin{itemize}
\item \href{http://www-math.mit.edu/~rstan/ec/}{\red Richard Stanley, \textit{Enumerative Combinatorics, vol. 2}, Chapter 7}.

\item \href{http://arxiv.org/abs/1409.8356}{\red Darij Grinberg, Victor Reiner, \textit{Hopf Algebras in Combinatorics}, arXiv:1409.8356}.

\item \href{https://arxiv.org/abs/2004.04995}{\red Emmanuel Briand, Mercedes Rosas, \textit{The 144 symmetries of the Littlewood-Richardson coefficients of $SL_3$}, arXiv:2004.04995}.

\item \href{https://arxiv.org/abs/math/0407170}{\red Igor Pak, Ernesto Vallejo, \textit{Combinatorics and geometry of Littlewood-Richardson cones}, arXiv:math/0407170.}

\item \href{https://arxiv.org/abs/1410.8017}{\red Emmanuel Briand, Rosa Orellana, Mercedes Rosas, \textit{Rectangular symmetries for coefficients of symmetric functions}, arXiv:1410.8017}.
\end{itemize}
\end{frame}



\begin{frame}
\fti{Reminder on symmetric functions}

\begin{itemize}

\item Fix a commutative ring $\defnm{\textbf{k}}$ with unity. We shall do everything over $\kk$.

\item Consider the ring $\defnm{\kk\left[\left[x_1,x_2,x_3,\ldots\right]\right]}$ of formal power series in countably many indeterminates.

\pause

\item A formal power series $f$ is said to be \defn{bounded-degree} if the monomials it contains are bounded (from above) in degree.

\pause

\item A formal power series $f$ is said to be \defn{symmetric} if it is invariant under permutations of the indeterminates.
%\\ Equivalently, if its coefficients in front of $x_{i_1}^{a_1} x_{i_2}^{a_2} \cdots x_{i_k}^{a_k}$ and $x_{j_1}^{a_1} x_{j_2}^{a_2} \cdots x_{j_k}^{a_k}$ are equal whenever $i_1, i_2, \ldots, i_k$ are distinct and $j_1, j_2, \ldots, j_k$ are distinct.

\only<1-3>{\item For example:
\begin{itemize}
\item $1 + x_1 + x_2^3$ is bounded-degree but not symmetric.
\item $\left(1+x_1\right)\left(1+x_2\right)\left(1+x_3\right) \cdots $ is symmetric  but not bounded-degree.
\end{itemize}}
\pause

\item Let \defn{$\Lambda$} be the set of all symmetric bounded-degree power series in $\kk\left[\left[x_1,x_2,x_3,\ldots\right]\right]$. This is a $\kk$-subalgebra, called the \defn{ring of symmetric functions} over $\kk$. \\ It is also known as $\operatorname{Sym}$.

\vspace{3cm}
\end{itemize}

\end{frame}

\begin{frame}
\fti{Schur functions, part 1: Young diagrams}

\begin{itemize}

\item Let $\lambda = \left(\lambda_1, \lambda_2, \lambda_3, \ldots\right)$
be a \defn{partition}
(i.e., a weakly decreasing sequence of nonnegative integers such that
$\lambda_i = 0$ for all $i \gg 0$).

We commonly omit trailing zeroes: e.g., the partition
$\left(4,2,2,1,0,0,0,0,\ldots\right)$ is identified
with the tuple $\left(4,2,2,1\right)$.

\pause

The \defn{Young diagram} of
$\lambda$ is like a matrix, but the rows have different lengths, and
are left-aligned; the $i$-th row has $\lambda_i$ cells.
\\
\textbf{Examples:}
\begin{itemize}
\item The Young diagram of $\left(3,2\right)$ has the form
\[
\ydiagram{3, 2} .
\]
\item The Young diagram of $\left(4,2,1\right)$ has the form
\[
\ydiagram{4, 2, 1}.
\]
\end{itemize}

\end{itemize}

\vspace{9cm}

\end{frame}

\begin{frame}
\fti{Schur functions, part 2: semistandard tableaux}

\begin{itemize}

\item A \defn{semistandard tableau} of shape $\lambda$ is the
Young diagram of $\lambda$, filled with positive integers, such that
\begin{itemize}
\item the entries in each \textbf{row} are \textbf{weakly} increasing;
\item the entries in each \textbf{column} are \textbf{strictly}
increasing.
\end{itemize}
\textbf{Examples:}
\only<1>{
\begin{itemize}
\item A semistandard tableau of shape $\left(3,2\right)$ is
\[
\begin{ytableau}
2 & 3 & 3 \\
3 & 5
\end{ytableau} .
\]
\item A semistandard tableau of shape $\left(4,2,1\right)$ is
\[
\begin{ytableau}
2 & 2 & 3 & 4 \\
3 & 4 \\
5
\end{ytableau} .
\]
\end{itemize}
}
\only<2>{
\begin{itemize}
\item The semistandard tableaux of shape $\left(3,2\right)$ are the arrays of
the form
\[
\begin{ytableau}
a & b & c \\
d & e
\end{ytableau}
\]
with $a \leq b \leq c$ and $d \leq e$ and $a < d$ and $b < e$.
\end{itemize}
}

\end{itemize}

\vspace{9cm}

\end{frame}

\begin{frame}
\fti{Schur functions, part 3: definition of Schur functions}

\begin{itemize}

\item Given a partition $\lambda$, we define the
\defn{Schur function} $\defnm{s_\lambda}$ as the power series
\[
s_\lambda = \sum\limits_{\substack{T \text{ is a semistandard} \\
\text{tableau of shape } \lambda}} \xx_T, \qquad
\text{where } \xx_T = \prod\limits_{p \text{ is a cell of }T} x_{T\left(p\right)}
\]
(where $T\left(p\right)$ denotes the entry of $T$ in $p$).

\item \textbf{Examples:}

\begin{itemize}

\only<1>{
\item
\[
s_{\left(3,2\right)} = \sum\limits_{\substack{a \leq b \leq c,\ d \leq e,\\
a < d,\  b < e}} x_a x_b x_c x_d x_e ,
\]
because the semistandard tableau
\[
T = \begin{ytableau}
a & b & c \\
d & e
\end{ytableau}
\]
contributes the addend $\xx_T = x_a x_b x_c x_d x_e$.
}
\only<2-3>{
\item For any $n \geq 0$, we have
\[
s_{\left(n\right)} = \sum_{i_1 \leq i_2 \leq \cdots \leq i_n} x_{i_1} x_{i_2} \cdots x_{i_n} ,
\]
since the semistandard tableaux of shape $\left(n\right)$
are the fillings
\[
T = \begin{ytableau}
i_1 & i_2 & \none[\cdots] & \none[\cdots] & i_n
\end{ytableau}
\]
with $i_1 \leq i_2 \leq \cdots \leq i_n$.
}
\only<3>
{
\\
This symmetric function $s_{\left(n\right)}$ is commonly called $\defnm{h_n}$.
}
\only<4-5>{
\item For any $n \geq 0$, consider the partition
$\left(1^n\right) := \left(1,1,\ldots,1\right)$
(with $n$ entries). Then,
\[
s_{\left(1^n\right)} = \sum_{i_1 < i_2 < \cdots < i_n} x_{i_1} x_{i_2} \cdots x_{i_n} ,
\]
since the semistandard tableaux of shape $\left(1^n\right)$
are \vspace{-0.2pc}
\[
\text{the fillings}
\qquad
T =
\scalebox{0.6}{
\begin{ytableau}
i_1 \\
i_2 \\
\none[\vdots] \\
\none[\vdots] \\
i_n
\end{ytableau}
}
\qquad \text{with $i_1 < i_2 < \cdots < i_n$.}
\]
}
\only<5>
{
\\\vspace{-0.2pc}
This symmetric function $s_{\left(1^n\right)}$ is commonly called $\defnm{e_n}$.
}

\end{itemize}

\end{itemize}

\vspace{9cm}

\end{frame}


\begin{frame}
\fti{Schur functions, part 4: classical properties}

\begin{itemize}

\item \textbf{Theorem:} The Schur function $s_\lambda$ is
a symmetric function (= an element of $\Lambda$)
for any partition $\lambda$.

\item \textbf{Theorem:} The family
$\left(s_\lambda\right)_{\lambda \text{ is a partition}}$ is a basis of the
$\kk$-module $\Lambda$.

\pause

\item \textbf{Theorem:} Fix $n \geq 0$.
Let $\lambda = \left(\lambda_1, \lambda_2, \ldots, \lambda_n\right)$
be a partition with at most $n$ nonzero entries.
Then,
\begin{align*}
&s_\lambda\left(x_1, x_2, \ldots, x_n\right) \\
&= \underbrace{\det\left(\left(x_i^{\lambda_j + n - j}\right)_{1 \leq i, j \leq n} \right)}_{\text{this is called an \defn{alternant}}}
   \diagup
  \underbrace{\det\left(\left(x_i^{n - j}\right)_{1 \leq i, j \leq n} \right)}_{\substack{= \prod_{1 \leq i < j \leq n} \left(x_i - x_j\right) \\ \text{ (= the Vandermonde determinant)}}} .
\end{align*}
Here, for any $f \in \Lambda$, we let
$\defnm{f\left(x_1, x_2, \ldots, x_n\right)}$ denote
the result of substituting $0$ for $x_{n+1}, x_{n+2}, x_{n+3}, \ldots$
in $f$; this is a symmetric \textbf{polynomial}
in $x_1, x_2, \ldots, x_n$.

\pause

\item For proofs, see any text
on symmetric functions (e.g., Stanley's EC2, or
Grinberg-Reiner, or
{\red \href{http://www.ma.rhul.ac.uk/~uvah099/Maths/Sym/SymFuncs2020.pdf}{Mark Wildon's notes}}).

\end{itemize}

\vspace{9cm}

\end{frame}

\begin{frame}
\fti{Littlewood--Richardson coefficients: definition}

\begin{itemize}

\item If $\mu$ and $\nu$ are two partitions, then
$s_\mu s_\nu$ belongs to $\Lambda$
(since $\Lambda$ is a ring) \pause
and thus can be written in the form
\[
s_\mu s_\nu
= \sum_{\lambda \text{ is a partition}}
\defnm{c^\lambda_{\mu, \nu}} s_\lambda
\]
for some $c^\lambda_{\mu, \nu} \in \kk$
(since the $s_\lambda$ form a basis of $\Lambda$). \pause

\item The coefficients $c^\lambda_{\mu, \nu}$ are integers,
and are called the
\defn{Littlewood--Richardson coefficients}. \pause

\item \textbf{Example:}
\only<4>{
\begin{align*}
s_{\left(2,1\right)} s_{\left(3,1\right)}
&= s_{\left(3,2,1,1\right)} + s_{\left(3,2,2\right)} + s_{\left(3,3,1\right)} \\
& \qquad + s_{\left(4,1,1,1\right)} + 2s_{\left(4,2,1\right)} + s_{\left(4,3\right)} \\
& \qquad + s_{\left(5,1,1\right)} + s_{\left(5,2\right)}
\end{align*}
}
\only<5->{
\begin{align*}
s_{\left(2,1\right)} s_{\left(3,1\right)}
&= s_{\left(3,2,1,1\right)} + s_{\left(3,2,2\right)} + {\blue s_{\left(3,3,1\right)}} \\
& \qquad + s_{\left(4,1,1,1\right)} + {\violet 2s_{\left(4,2,1\right)}} + s_{\left(4,3\right)} \\
& \qquad + s_{\left(5,1,1\right)} + s_{\left(5,2\right)} ,
\end{align*}
so {\violet $c^{\left(4,2,1\right)}_{\left(2,1\right),\left(3,1\right)} = 2$} and
{\blue $c^{\left(3,3,1\right)}_{\left(2,1\right),\left(3,1\right)} = 1$}.
}

\pause \pause

\item \textbf{Theorem:} The coefficients $c^{\lambda}_{\mu, \nu}$ are \textbf{nonnegative integers}.
Various combinatorial interpretations (``\defn{Littlewood--Richardson rules}'') for them are known.

\end{itemize}

\vspace{9cm}

\end{frame}

\begin{frame}
\fti{Why Littlewood--Richardson coefficients? 1}

\begin{itemize}

\only<1>{
\item Before we say more about Littlewood--Richardson coefficients, let us see where else they appear.
}
\pause

\item For $\kk = \ZZ$, the cohomology ring
\[
\operatorname{H}^{\ast}\tup{\Gr\tup{k, n}}
\]
of the complex Grassmannian $\Gr\tup{k, n}$ (of $k$-subspaces
in $\CC^n$) is isomorphic to
\begin{align*}
 \Lambda \diagup\left(  h_{n-k+1},h_{n-k+2},\ldots,h_{n},
e_{k+1}, e_{k+2}, e_{k+3}, \ldots\right)
_{\operatorname*{ideal}} .
\end{align*}
The cohomology classes corresponding to the Schur
functions $s_\lambda$ are the \emph{Schubert classes} --
the classes of the \emph{Schubert varieties}.
Roughly speaking, these subdivide $\Gr\tup{k, n}$
according to the positions of the pivots in the row-reduced
echelon form.
\pause

\item Thus, the Littlewood--Richardson coefficients
$c^\lambda_{\mu, \nu}$
are intersection multiplicities of these Schubert varieties.
\pause

\item For details, see:
\begin{itemize}
\item {\red \href{https://bookstore.ams.org/smfams-6/}{Laurent Manivel, \textit{Symmetric Functions,
Schubert Polynomials and Degeneracy Loci},
AMS/SMF 1998.}}
\end{itemize}

\end{itemize}

\vspace{9cm}

\end{frame}

\begin{frame}
\fti{Why Littlewood--Richardson coefficients? 2}

\begin{itemize}

\only<1>{
\item Here is another interpretation of Littlewood--Richardson
coefficients, also related to subspaces of a vector space.}
\pause

\item Let $V$ be a finite-dimensional vector space.

\item The \defn{Jordan type $J\tup{A}$} of a nilpotent
endomorphism $A \in \End V$ is the partition
$\tup{\lambda_1, \lambda_2, \lambda_3, \ldots}$ with
$\lambda_i$ being the size of the $i$-th largest Jordan
block of $A$.
\pause

\item Pick a nilpotent endomorphism $A \in \End V$,
and let $\lambda = J\tup{\lambda}$ be its Jordan type.
Let $\mu$ and $\nu$ be two further partitions.

When is there an $A$-invariant vector subspace $W \subseteq V$
with
\[
J\tup{A} = \lambda, \qquad
J\tup{A\mid_W} = \mu, \qquad
J\tup{A/_W} = \nu ?
\]
($A/_W$ is the endomorphism of $V/W$ induced by $A$.)
\pause
\\
Precisely when $c^\lambda_{\mu, \nu} \neq 0$.
\pause

\item Moreover, the set of all such $W$ is a
subvariety of $\Gr\tup{k, n}$, and has
$c^\lambda_{\mu, \nu}$ irreducible components.

\item For details, see:
\begin{itemize}
\item {\red \href{http://wwwmathlabo.univ-poitiers.fr/~maavl/pdf/geometry.pdf}{Marc van Leeuwen, \textit{Flag Varieties and
Interpretations of Young Tableau Algorithms}.}}
\end{itemize}

\end{itemize}

\vspace{9cm}

\end{frame}

% \begin{frame}
% \fti{Why Littlewood--Richardson coefficients? 3}

% \begin{itemize}

% \only<1>{
% \item More linear algebra!
% }
% \pause

% \item Given three partitions $\lambda, \mu, \nu$.
% When are there two Hermitian matrices $A$ and $B$
% (of the same size, possibly large)
% with
% \[
% \Spec\tup{A+B} = \lambda, \qquad
% \Spec A = \mu, \qquad
% \Spec B = \nu \qquad ?
% \]
% \pause
% Precisely when $c^\lambda_{\mu, \nu} \neq 0$.

% \item

% \item For details, see:
% \begin{itemize}
% \item {\red \href{http://wwwmathlabo.univ-poitiers.fr/~maavl/pdf/geometry.pdf}{Marc van Leeuwen, \textit{Flag Varieties and
% Interpretations of Young Tableau Algorithms}.}}
% \end{itemize}

% \end{itemize}

% \vspace{9cm}

% \end{frame}


\begin{frame}
\fti{Why Littlewood--Richardson coefficients? 3}

\begin{itemize}

\item Fix an $N \geq 0$.
The irreducible polynomial representations
$V_\lambda$ of the group
$\GL\tup{N} := \GL\tup{N, \CC}$ are indexed
by partitions having $\leq N$ entries.
\pause

\item Their \emph{characters}
are the Schur functions $s_\lambda$.
\pause

\item The Littlewood--Richardson coefficients tell
how to decompose the tensor product of two such
representations:
\[
V_\mu \otimes V_\nu
= \bigoplus_\lambda V_\lambda^{\oplus c^\lambda_{\mu, \nu}} .
\]

\item For details, see:
\begin{itemize}
\item {\red \href{https://doi.org/10.1017/CBO9780511626241}{William Fulton, \textit{Young Tableaux}, CUP 1997.}}
\end{itemize}

\end{itemize}

\vspace{9cm}

\end{frame}


\begin{frame}
\fti{Skew semistandard tableaux}

\begin{itemize}

\item In order to formulate the classic (or, at least, best known) Littlewood--Richardson rule, we need a

\item \textbf{Definition:}
\begin{itemize}

\item
Two partitions $\lambda = \left(\lambda_1, \lambda_2, \lambda_3, \ldots\right)$
and $\mu = \left(\mu_1, \mu_2, \mu_3, \ldots\right)$ are said to satisfy
$\defnm{\mu \subseteq \lambda}$ if each $i \geq 1$ satisfies
$\mu_i \leq \lambda_i$. \\
\only<1>{(Equivalently: if the Young diagram of $\mu$ is contained in that of $\lambda$.)}
\pause

\item
A \defn{skew partition} is a pair $\left(\lambda, \mu\right)$ of two partitions
satisfying $\mu \subseteq \lambda$.
Such a pair is denoted by $\defnm{\lambda / \mu}$.
\pause

\item
If $\lambda / \mu$ is a skew partition, then the \defn{Young diagram}
of $\lambda / \mu$ is obtained from the Young diagram $\lambda$ when
all cells of the Young diagram of $\mu$ are removed.
\\
\only<3>{\textbf{Example:} The Young diagram of $\left(4,2,1\right) / \left(1,1\right)$ is
\[
\ydiagram{1+3,1+1,0+1}
\]
} \pause

\item
\defn{Semistandard tableaux} of shape $\lm$ are defined just as
ones of shape $\lambda$, except that we are now only filling the cells
of $\lm$.

\end{itemize}

\end{itemize}

\vspace{9cm}

\end{frame}


\begin{frame}
\fti{Littlewood--Richardson rule: the classical version}

\begin{itemize}

\item \textbf{Littlewood--Richardson rule:}
Let $\lambda$, $\mu$ and $\nu$ be three partitions.
Then, $c^\lambda_{\mu, \nu}$ is the number of
semistandard tableaux $T$ of shape $\lm$ such that
$\cont T = \nu$ and such that
$\cont\left(T\mid_{\operatorname{cols} \geq \, j}\right)$
is a partition for each $j$.
Here,
\begin{itemize}
\item $\cont T$ denotes the sequence $\left(c_1, c_2, c_3, \ldots\right)$,
where $c_i$ is the number of entries equal to $i$ in $T$;
\item $T\mid_{\operatorname{cols} \geq \, j}$ is what obtained from $T$
when the first $j-1$ columns are deleted.
\end{itemize}

\item \textbf{Example:}
$c^{\left(4,2,1\right)}_{\left(2,1\right),\left(3,1\right)} = 2$
due to the two tableaux
\[
\begin{ytableau}
\none & \none & 1 & 1 \\
\none & 1 \\
2
\end{ytableau}
\qquad \text{ and } \qquad
\begin{ytableau}
\none & \none & 1 & 1 \\
\none & 2 \\
1
\end{ytableau} .
\]
\pause

\item The shortest proof is due to Stembridge (using ideas by Gasharov);
see {\red \href{https://doi.org/10.37236/1666}{John R. Stembridge,
\textit{A Concise Proof of the Littlewood-Richardson Rule},
2002}}, or Section 2.6 in Grinberg-Reiner.

\end{itemize}

\vspace{9cm}

\end{frame}


\begin{frame}
\fti{Basic properties of Littlewood--Richardson coefficients}

\begin{itemize}

\item \textbf{Gradedness:}
$c^{\lambda}_{\mu, \nu} = 0$ unless $\abs{\lambda} = \abs{\mu} + \abs{\nu}$,
where $\abs{\kappa}$ denotes the \defn{size} (i.e., the sum of the entries)
of a partition $\kappa$. \\
(This is because $\Lambda$ is a graded ring and the $s_\lambda$
are homogeneous.)
\pause

\item \textbf{Transposition symmetry:}
$c^{\lambda}_{\mu, \nu} = c^{\lambda^t}_{\mu^t, \nu^t}$,
where $\kappa^t$ denotes the \defn{transpose} of a partition $\kappa$
(i.e., the partition whose Young diagram is obtained from that
of $\kappa$ by flipping across the main diagonal). \\
\only<2>{\textbf{Example:}
\[
\ydiagram{5,2,1}^t = \ydiagram{3,2,1,1,1}
\]
}
\pause

\item \textbf{Commutativity:}
$c^{\lambda}_{\mu, \nu} = c^{\lambda}_{\nu, \mu}$. \\
(Obvious from the definition, but hard to prove combinatorially
using the Littlewood--Richardson rule.)

\end{itemize}

\vspace{9cm}

\end{frame}


\begin{frame}
\fti{Littlewood--Richardson coefficients: more symmetries}

\begin{itemize}

\item Fix $n \in \NN$.
Let $\defnm{\Par[n]}$ be the set of all partitions having at most
$n$ nonzero entries. \pause

\item If $\lambda = \left(\lambda_1, \lambda_2, \ldots, \lambda_n\right)
\in \Par[n]$, and if $k \geq 0$ is such that all entries of $\lambda$
are $\leq k$, then $\defnm{\lambda^{\vee k}}$ shall denote the
partition
\[
\tup{k - \lambda_n, k - \lambda_{n-1}, \ldots, k - \lambda_1 } \in \Par[n] .
\]
This is called the \defn{$k$-complement} of $\lambda$.

\only<2>{
\textbf{Example:} If $n = 5$, then
\begin{align*}
\tup{3, 1, 1}^{\vee 7}
&= \tup{3, 1, 1, 0, 0}^{\vee 7}
= \tup{7-0, 7-0, 7-1, 7-1, 7-3} \\
&= \tup{7, 7, 6, 6, 4}.
\end{align*}
}
\only<3>{\textbf{Illustration:} If $n = 3$, then
\[
\tup{3,2}^{\vee 4}
= \ydiagram{3,2}^{\vee 4} \;\;
\]
}
\only<4>{\textbf{Illustration:} If $n = 3$, then
\[
\tup{3,2}^{\vee 4}
= \ydiagram[*(blue)]{3,2}^{\vee 4}  \;\;
\]
}
\only<5>{\textbf{Illustration:} If $n = 3$, then
\[
\tup{3,2}^{\vee 4}
= \begin{ytableau}
*(blue) & *(blue) & *(blue) & *(yellow) \\
*(blue) & *(blue) & *(yellow) & *(yellow) \\
*(yellow) & *(yellow) & *(yellow) & *(yellow) \\
\end{ytableau}
\]
}
\only<6>{\textbf{Illustration:} If $n = 3$, then
\[
\tup{3,2}^{\vee 4}
= \begin{ytableau}
\none & \none & \none & *(yellow) \\
\none & \none & *(yellow) & *(yellow) \\
*(yellow) & *(yellow) & *(yellow) & *(yellow) \\
\end{ytableau}
\]
}
\only<7>{\textbf{Illustration:} If $n = 3$, then
\[
\tup{3,2}^{\vee 4}
= \ydiagram[*(yellow)]{4,2,1}
\]
}
\only<8>{\textbf{Illustration:} If $n = 3$, then
\[
\tup{3,2}^{\vee 4}
= \tup{4,2,1} .
\]
}
\pause \pause \pause \pause \pause \pause \pause

\item \textbf{Complementation symmetry I:}
Let $\lambda, \mu, \nu \in \Par[n]$ and $k \geq 0$
be such that all entries of $\lambda, \mu, \nu$
are $\leq k$.
Then,
\[
c_{\mu,\nu}^{\lambda} = c_{\nu,\mu}^{\lambda}
= c_{\lambda^{\vee k},\nu}^{\mu^{\vee k}} = c_{\nu,\lambda^{\vee k}}^{\mu^{\vee k}}
= c_{\mu,\lambda^{\vee k}}^{\nu^{\vee k}} = c_{\lambda^{\vee k},\mu}^{\nu^{\vee k}} .
\]
\only<10>{(This can be proved by applying skew
Schur functions to $x_1^{-1}, x_2^{-1}, \ldots,
x_n^{-1}$, or by interpreting Schur functions
as fundamental classes in the cohomology of the
Grassmannian. See Exercise 2.9.15 in Grinberg-Reiner
for the former proof.)}
\pause
\pause

\item \textbf{Complementation symmetry II:}
Let $\lambda, \mu, \nu \in \Par[n]$ and $q, r \geq 0$
be such that all entries of $\mu$ are $\leq q$,
and all entries of $\nu$ are $\leq r$. Then:
\begin{itemize}
\item If all entries of $\lambda$ are $\leq q+r$,
then
$c_{\mu,\nu}^{\lambda}
= c_{\mu^{\vee q},\nu^{\vee r }}^{\lambda^{\vee \left(q+r\right)}}$.
\item If not, then $c_{\mu,\nu}^{\lambda} = 0$.
\end{itemize}
(See, e.g., Exercise 2.9.16 in Grinberg-Reiner.)
\end{itemize}

\vspace{9cm}

\end{frame}


\begin{frame}
\fti{The Briand--Rosas symmetry}

\begin{itemize}

\item In {\red \arxiv{2004.04995}}, Emmanuel Briand and
Mercedas Rosas have used a computer (and prior work
of Rassart, Knutson and Tao, which made the problem
computable) to classify all such
``symmetries'' of Littlewood--Richardson coefficients
$c^\lambda_{\mu, \nu}$
with $\lambda, \mu, \nu \in \Par[n]$
for fixed $n \in \set{3, 4, \ldots, 7}$.

\pause

\item For $n \in \set{4, 5, \ldots, 7}$, they only
found the complementation symmetries above, as well
as the trivial translation symmetries
(adding $1$ to each entry of $\lambda$ and $\nu$
does not change $c^{\lambda}_{\mu, \nu}$; nor does
adding $1$ to each entry of $\lambda$ and $\mu$).

\pause

\item For $n = 3$, they found an extra symmetry:
\[
c^{\tup{\lambda_1, \lambda_2, \lambda_3}}_{\tup{\mu_1, \mu_2}, \tup{\nu_1, \nu_2}}
= c^{\tup{\lambda_1, \nu_1, \lambda_3}}_{\tup{\mu_1 + \nu_1 - \lambda_2, \mu_2 + \nu_1 - \lambda_2}, \tup{\lambda_2, \nu_2}} \ .
\]
(Read the right hand side as $0$ if the tuples are not partitions.) \pause \\
\textbf{Question:} Is there a non-computer proof? What is the meaning of this identity?

\end{itemize}

\vspace{9cm}

\end{frame}

\begin{frame}
\fti{Chapter 2}
\begin{center}
{\Huge \bf \sc Chapter 2} \\
\noindent\rule[0.5ex]{\linewidth}{1pt}
{\Large \bf The Pelletier--Ressayre symmetry}
\end{center}

\vspace{0.2cm}

References (among many):
\begin{itemize}
\item \href{http://www.cip.ifi.lmu.de/~grinberg/algebra/lrhspr.pdf}{\red Darij Grinberg, \textit{The Pelletier--Ressayre hidden
symmetry for Littlewood--Richardson coefficients}}, {\red \arxiv{2008.06128}}.

\item \href{https://arxiv.org/abs/2005.09877}{\red Maxime Pelletier, Nicolas Ressayre, \textit{Some unexpected properties of
Littlewood-Richardson coefficients}, arXiv:2005.09877}.

\item \href{http://dx.doi.org/10.1088/1751-8113/46/34/349601}{\red Robert Coquereaux, Jean-Bernard Zuber, \textit{On sums of tensor and fusion multiplicities}, 2011}.

\end{itemize}
\end{frame}


\begin{frame}
\fti{Inspiration: The Coquereaux--Zuber sum identity, 1}

\begin{itemize}

\item \textbf{Theorem (\href{http://dx.doi.org/10.1088/1751-8113/46/34/349601}{\red Coquereaux and Zuber, 2011}):}
Let $n \geq 0$ and $\mu, \nu \in \Par[n]$.
Let $k \geq 0$
be such that all entries of $\mu$ are $\leq k$.
Then,
\[
\sum_{\lambda \in \Par[n]} c^\lambda_{\mu, \nu}
=
\sum_{\lambda \in \Par[n]} c^\lambda_{\mu^{\vee k}, \nu} .
\]
\only<1>{(See {\red \url{https://mathoverflow.net/a/236220/}} for a hint at a combinatorial proof.)}
\pause

\item This can be interpreted in terms of Schur
\textbf{polynomials}.
For any $\lambda \in \Par[n]$, the
\defn{Schur polynomial} $s_\lambda\left(x_1, x_2, \ldots, x_n\right)$
is the symmetric polynomial
\only<2>{
\begin{align*}
&s_\lambda\left(x_1, x_2, \ldots, x_n\right) \\
&= \underbrace{\det\left(\left(x_i^{\lambda_j + n - j}\right)_{1 \leq i, j \leq n} \right)}_{\text{this is called an \defn{alternant}}}
   \diagup
  \underbrace{\det\left(\left(x_i^{n - j}\right)_{1 \leq i, j \leq n} \right)}_{\substack{= \prod_{1 \leq i < j \leq n} \left(x_i - x_j\right) \\ \text{ (= the Vandermonde determinant)}}}
\end{align*}
in $x_1, x_2, \ldots, x_n$
obtained by setting $x_{n+1} = x_{n+2} = x_{n+3} = \cdots = 0$
in $s_\lambda$.
}
\only<3->{
\[
s_\lambda\left(x_1, x_2, \ldots, x_n\right) .
\]
}
\pause

\item The family $\tup{s_\lambda\left(x_1, x_2, \ldots, x_n\right)}_{\lambda \in \Par[n]}$
is a basis of the $\kk$-module of symmetric polynomials
in $x_1, x_2, \ldots, x_n$. We call it the \defn{Schur basis}.

\end{itemize}

\vspace{9cm}

\end{frame}

\begin{frame}
\fti{Inspiration: The Coquereaux--Zuber sum identity, 2}

\begin{itemize}

\item The theorem of Coquereaux and Zuber says that
\begin{align*}
&\operatorname{coeffsum} \tup{s_\mu\tup{x_1, x_2, \ldots, x_n} s_\nu\tup{x_1, x_2, \ldots, x_n}} \\
&= \operatorname{coeffsum} \tup{s_{\mu^{\vee k}}\tup{x_1, x_2, \ldots, x_n} s_\nu\tup{x_1, x_2, \ldots, x_n}} ,
\end{align*}
where $\operatorname{coeffsum} f$ denotes the sum of all coefficients in the expansion of a symmetric polynomial $f$ in the Schur basis.

\pause

\item So the products
\begin{align*}
&s_\mu\tup{x_1, x_2, \ldots, x_n} s_\nu\tup{x_1, x_2, \ldots, x_n} \\
\text{and } & s_{\mu^{\vee k}}\tup{x_1, x_2, \ldots, x_n} s_\nu\tup{x_1, x_2, \ldots, x_n}
\end{align*}
have the same sum of coefficients when expanded in the Schur basis. Do they also have the same multiset of coefficients?

\pause
\textbf{No}. \\
(Counterexample: $n = 5$ and $\mu = \tup{5, 2, 1}$ and $\nu = \tup{4, 2, 2}$.)
\pause
\\
\textbf{Question:} Does this hold for $n \leq 4$ ? (Proved for $n = 3$.)

\end{itemize}

\vspace{9cm}

\end{frame}


\begin{frame}
\fti{The Pelletier--Ressayre conjecture}

\begin{itemize}

\item \textbf{Conjecture (Pelletier and Ressayre, 2020):} It does hold when
$\mu$ is \defn{near-rectangular} -- i.e., when $\mu=\left(  a+b,a^{n-2}%
\right)  $ for some $a,b\geq0$.
Here, $\defnm{a^{n-2}}$ means $\underbrace{a,a,\ldots
,a}_{n-2\text{ times}}$.

In this case, for $k=a+b$, we have $\mu^{\vee k}=\left(  a+b,b^{n-2}\right)
$. (Taking $k$ higher makes no real difference.) \pause

\item In other words:

\textbf{Conjecture (Pelletier and Ressayre, 2020):} Let $n\geq0$ and
$\nu\in\operatorname*{Par}\left[  n\right]  $. Let $a,b\geq0$. Let
$\alpha=\left(  a+b,a^{n-2}\right)  $ and $\beta=\left(  a+b,b^{n-2}\right)
$. Then,%
\[
\left\{  c_{\alpha,\nu}^{\lambda} \mid \lambda \in \Par\ive{n}\right\}  _{\operatorname*{multiset}%
}=\left\{  c_{\beta,\nu}^{\lambda} \mid \lambda \in \Par\ive{n}\right\}  _{\operatorname*{multiset}}.
\]
\pause

\item This means that there should be a bijection $\varphi:\operatorname*{Par}%
\left[  n\right]  \rightarrow\operatorname*{Par}\left[  n\right]  $ such that
\[
c_{\alpha,\nu}^{\lambda}=c_{\beta,\nu}^{\varphi\left(  \lambda\right)
}\ \ \ \ \ \ \ \ \ \ \text{for each }\lambda\in\operatorname*{Par}\left[
n\right]  .
\]
\end{itemize}
\vspace{9cm}
\end{frame}

\begin{frame}
\frametitle{\ \ \ \ \ The Pelletier--Ressayre conjecture, restated}

\begin{itemize}
\item \textbf{Conjecture (Pelletier and Ressayre, 2020):} Let $n\geq0$ and
$\nu\in\operatorname*{Par}\left[  n\right]  $. Let $a,b\geq0$. Let
$\alpha=\left(  a+b,a^{n-2}\right)  $ and $\beta=\left(  a+b,b^{n-2}\right)
$. Then, there is a bijection $\varphi:\operatorname*{Par}\left[  n\right]
\rightarrow\operatorname*{Par}\left[  n\right]  $ such that
\[
c_{\alpha,\nu}^{\lambda}=c_{\beta,\nu}^{\varphi\left(  \lambda\right)
}\ \ \ \ \ \ \ \ \ \ \text{for each }\lambda\in\operatorname*{Par}\left[
n\right]  .
\]
\pause

\item \textbf{Theorem (G., 2020):} This is true. Moreover, this bijection
$\varphi$ can more or less be defined explicitly in terms of maxima of sums of
entries of $\lambda$ and $\nu$.

(``More or less'' means that we find a
bijection $\varphi:\mathbb{Z}^{n}\rightarrow\mathbb{Z}^{n}$, not
$\varphi:\operatorname*{Par}\left[  n\right]  \rightarrow\operatorname*{Par}%
\left[  n\right]  $, where we set $c_{\alpha,\nu}^{\lambda}=c_{\beta,\nu
}^{\lambda}=0$ for all $\lambda\in\mathbb{Z}^{n}\setminus\operatorname*{Par}%
\left[  n\right]  $.) \pause

\item The rest of this talk will sketch how this bijection $\varphi$ was found.
\end{itemize}

\vspace{9cm}
\end{frame}

\begin{frame}
\frametitle{\ \ \ \ \ Snakes}

\begin{itemize}
\item First, we notice that
\begin{align*}
\alpha & =\left(  a+b,a^{n-2}\right)  =\left(  a+b,a^{n-2},0\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{as }n\text{-tuple}\right)  \\
& =\left(  b,0^{n-2},-a\right)  +a
\end{align*}
(where ``$+a$'' means ``add $a$ to each entry''). \pause
\\
Likewise, $\beta=\left(  a,0^{n-2},-b\right)  + b$. \pause

\item This suggest allowing \textquotedblleft partitions with negative
entries\textquotedblright. We call them \textbf{snakes}. \pause

\item Formally: A \defn{snake} will mean an $n$-tuple $\left(  \lambda
_{1},\lambda_{2},\ldots,\lambda_{n}\right)  \in\mathbb{Z}^{n}$ with
$\lambda_{1}\geq\lambda_{2}\geq\cdots\geq\lambda_{n}$. Thus,%
\[
\operatorname*{Par}\left[  n\right]  \subseteq\left\{  \text{snakes}\right\}
\subseteq\mathbb{Z}^{n}.
\]
\pause\vspace{-1.1pc}

\only<5>{
\item Snakes index rational representations of $\operatorname*{GL}%
\left(  n\right)  $: See
{\red \href{https://doi.org/10.1016/0097-3165(87)90077-X}{John R. Stembridge, \textit{Rational tableaux and the tensor algebra of $\mathfrak{gl}_{n}$}}, 1987.}}\pause

\item If $\lambda\in\mathbb{Z}^{n}$ is any $n$-tuple, then

\begin{itemize}
\item we let $\defnm{\lambda_i}$ denote the $i$-th entry of $\lambda$ (for any $i$);

\item we let $\defnm{\lambda+a}$ denote the $n$-tuple $\left(  \lambda_{1}%
+a,\lambda_{2}+a,\ldots,\lambda_{n}+a\right)  $;

\item we let $\defnm{\lambda-a}$ denote the $n$-tuple $\left(  \lambda_{1}%
-a,\lambda_{2}-a,\ldots,\lambda_{n}-a\right)  $.
\end{itemize}
\end{itemize}

\vspace{9cm}
\end{frame}

\begin{frame}
\frametitle{\ \ \ \ \ Schur Laurent polynomials}

\begin{itemize}
\item We have defined a Schur polynomial $s_{\lambda}\left(  x_{1}%
,x_{2},\ldots,x_{n}\right)  \in \kk\left[  x_{1},x_{2},\ldots
,x_{n}\right]  $ for any $\lambda\in\operatorname*{Par}\left[  n\right]  $. We
now denote it by $\defnm{\overline{s}_{\lambda}}$. \pause

\item It is easy to see that%
\[
\overline{s}_{\lambda+a}=\left(  x_{1}x_{2}\cdots x_{n}\right)  ^{a}%
\overline{s}_{\lambda}\ \ \ \ \ \ \ \ \ \ \text{for any }\lambda\in\operatorname*{Par}%
\left[  n\right]  \text{ and }a\geq0.
\]
\pause

\item This allows us to extend the definition of $\overline{s}_{\lambda}$ from
the case $\lambda\in\operatorname*{Par}\left[  n\right]  $ to the more general
case $\lambda\in\left\{  \text{snakes}\right\}  $:

If $\lambda$ is a snake, then we choose some $a\geq0$ such that $\lambda
+a\in\operatorname*{Par}\left[  n\right]  $, and define%
\[
\overline{s}_{\lambda}=\left(  x_{1}x_{2}\cdots x_{n}\right)  ^{-a}%
\overline{s}_{\lambda+a}.
\]
This is a Laurent polynomial in
$\kk\left[  x_{1}^{\pm1},x_{2}^{\pm1},\ldots,x_{n}^{\pm1}\right]  $.
\pause

\item Alternatively, we can define $\overline{s}_{\lambda}$ explicitly by{%
\[
\overline{s}_{\lambda}=\det\left(  \left(  x_{i}^{\lambda_{j}+n-j}\right)
_{1\leq i,j\leq n}\right)  \diagup\det\left(  \left(  x_{i}^{n-j}\right)
_{1\leq i,j\leq n}\right)
\]
(same formula as before).}
\end{itemize}

\vspace{9cm}
\end{frame}

\begin{frame}
\frametitle{\ \ \ \ \ $\overline{s}_\alpha$ and $\overline{s}_\beta$ revealed}

\begin{itemize}
\item \only<1>{For any $k\geq0$, define the two Laurent polynomials%
\begin{align*}
h_{k}^{+} &  =h_{k}\left(  x_{1},x_{2},\ldots,x_{n}\right)  ,\\
h_{k}^{-} &  =h_{k}\left(  x_{1}^{-1},x_{2}^{-1},\ldots,x_{n}^{-1}\right)  .
\end{align*}
(Recall: $h_{k}=s_{\left(  k\right)  }=\sum_{i_{1}\leq i_{2}\leq\cdots\leq
i_{k}}x_{i_{1}}x_{i_{2}}\cdots x_{i_{k}}$.)}

\only<2->{For any $k\geq0$, define the two Laurent polynomials%
\begin{align*}
h_{k}^{+} &  =h_{k}\left(  x_{1},x_{2},\ldots,x_{n}\right)  =\sum_{1\leq
i_{1}\leq i_{2}\leq\cdots\leq i_{k}\leq n}x_{i_{1}}x_{i_{2}}\cdots x_{i_{k}%
},\\
h_{k}^{-} &  =h_{k}\left(  x_{1}^{-1},x_{2}^{-1},\ldots,x_{n}^{-1}\right)
=\sum_{1\leq i_{1}\leq i_{2}\leq\cdots\leq i_{k}\leq n}x_{i_{1}}^{-1}x_{i_{2}%
}^{-1}\cdots x_{i_{k}}^{-1}.
\end{align*}
}
\pause \pause \vspace{-0.5pc}


\item \textbf{Proposition:} Let $a,b\geq0$. Then,%
\[
\overline{s}_{\left(  b,0^{n-2},-a\right)  }=h_{a}^{-}h_{b}^{+}-h_{a-1}%
^{-}h_{b-1}^{+}.
\]
\pause \vspace{-0.8pc}

\item \textbf{Corollary:} Let $a,b\geq0$. Let $\alpha=\left(  a+b,a^{n-2}%
\right)  $ and $\beta=\left(  a+b,b^{n-2}\right)  $. Then,%
\begin{align*}
\overline{s}_{\alpha}  & =\left(  x_{1}x_{2}\cdots x_{n}\right)  ^{a}%
\cdot\left(  h_{a}^{-}h_{b}^{+}-h_{a-1}^{-}h_{b-1}^{+}\right)  ;\\
\overline{s}_{\beta}  & =\left(  x_{1}x_{2}\cdots x_{n}\right)  ^{b}%
\cdot\left(  h_{b}^{-}h_{a}^{+}-h_{b-1}^{-}h_{a-1}^{+}\right)  .
\end{align*}
\pause

\item Thus, if we \textquotedblleft know how to multiply by\textquotedblright%
\ $h_{k}^{-}$ and $h_{k}^{+}$, then we \textquotedblleft know how to multiply
by\textquotedblright\ $\overline{s}_{\alpha}$ and $\overline{s}_{\beta}$.
\end{itemize}

\vspace{9cm}
\end{frame}

\begin{frame}
\frametitle{\ \ \ \ \ Multiplying by $h_k^+$: the $h$-Pieri rule, 1}

\begin{itemize}
\item \textbf{Theorem (}$h$\textbf{-Pieri rule):} Let $\lambda$ be a
partition. Let $k\in\mathbb{Z}$. Then,%
\[
h_{k}\cdot s_{\lambda}=\sum_{\substack{\mu\text{ is a partition;}\\\left\vert
\mu\right\vert -\left\vert \lambda\right\vert =k;\\\mu_{1}\geq\lambda_{1}%
\geq\mu_{2}\geq\lambda_{2}\geq\cdots}}s_{\mu}.
\]
Here:

\begin{itemize}
\item We let $h_{k}=0$ if $k<0$. (And we recall that $h_{0}=1$.)

\item We let $\left\vert \kappa\right\vert $ denote the \defn{size} (i.e.,
the sum of the entries) of any partition $\kappa$. 

\item The $i$-th entry of a partition $\kappa$ is denoted by $\kappa_{i}$. \pause
\end{itemize}

\only<2>{\item Note that the chain of inequalities $\mu_{1}\geq\lambda_{1}\geq
\mu_{2}\geq\lambda_{2}\geq\cdots$ is saying that the diagram $\mu/\lambda$ is
a \defn{horizontal strip} (i.e., has no two cells in the same column). For
example,%
\[
\ydiagram{3+2,2+1,0+1}
\]
}
\pause

\only<3>{\item The Pieri rule is actually a particular case of
the Littlewood--Richardson rule (exercise!).}
\pause

\item By evaluating both sides at $x_{1},x_{2},\ldots,x_{n}$ (and recalling
that $s_{\mu}\left(  x_{1},x_{2},\ldots,x_{n}\right)  =0$ whenever $\mu$ is a
partition with more than $n$ nonzero entries), we obtain:
\end{itemize}

\vspace{9cm}
\end{frame}

\begin{frame}
\frametitle{\ \ \ \ \ Multiplying by $h_k^+$: the $h$-Pieri rule, 2}

\begin{itemize}
\item \textbf{Theorem (}$h^+$\textbf{-Pieri rule for symmetric polynomials):}
Let $\lambda\in\operatorname*{Par}\left[  n\right]  $. Let $k\in\mathbb{Z}$.
Then,%
\[
h_{k}^{+}\cdot\overline{s}_{\lambda}=\sum_{\substack{\mu\in\operatorname*{Par}%
\left[  n\right]  \text{;}\\\left\vert \mu\right\vert -\left\vert
\lambda\right\vert =k;\\\mu_{1}\geq\lambda_{1}\geq\mu_{2}\geq\lambda_{2}%
\geq\cdots\geq\mu_{n}\geq\lambda_{n}}}\overline{s}_{\mu}.
\]
Here:

\begin{itemize}
\item We let $\left\vert \kappa\right\vert $ denote the \defn{size} (i.e.,
the sum of the entries) of any $n$-tuple $\kappa$.

\item The $i$-th entry of an $n$-tuple $\kappa$ is denoted by $\kappa_{i}$. \pause
\end{itemize}

\item We can easily extend this from $\operatorname*{Par}\left[  n\right]  $
to $\left\{  \text{snakes}\right\}  $, and obtain the following:
\end{itemize}

\vspace{9cm}
\end{frame}

\begin{frame}
\frametitle{\ \ \ \ \ Multiplying by $h_k^+$: the $h$-Pieri rule, 3}

\begin{itemize}
\only<1>{
\item \textbf{Theorem (}$h^+$\textbf{-Pieri rule for Laurent polynomials):} Let
$\lambda\in\left\{  \text{snakes}\right\}  $. Let $k\in\mathbb{Z}$. Then,%
\[
h_{k}^{+}\cdot\overline{s}_{\lambda}=\sum_{\substack{\mu\in\left\{
\text{snakes}\right\}  \text{;}\\\left\vert \mu\right\vert -\left\vert
\lambda\right\vert =k;\\\mu_{1}\geq\lambda_{1}\geq\mu_{2}\geq\lambda_{2}%
\geq\cdots\geq\mu_{n}\geq\lambda_{n}}}\overline{s}_{\mu}.
\]
}
\only<2->{
\item \textbf{Theorem (}$h^+$\textbf{-Pieri rule for Laurent polynomials):} Let
$\lambda\in\left\{  \text{snakes}\right\}  $. Let $k\in\mathbb{Z}$. Then,%
\[
h_{k}^{+}\cdot\overline{s}_{\lambda}=\sum_{\substack{\mu\in\left\{
\text{snakes}\right\}  \text{;}\\\left\vert \mu\right\vert -\left\vert
\lambda\right\vert =k;\\\mu\rightharpoonup\lambda}}\overline{s}_{\mu}.
\]
}
Here:

\begin{itemize}
\item We let $\left\vert \kappa\right\vert $ denote the \defn{size} (i.e.,
the sum of the entries) of any $n$-tuple $\kappa$.

\item The $i$-th entry of an $n$-tuple $\kappa$ is denoted by $\kappa_{i}$. \pause

\only<2->{
\item The notation $\defnm{\mu\rightharpoonup\lambda}$ stands for
$\mu_{1}\geq\lambda_{1}\geq\mu_{2}\geq\lambda_{2}%
\geq\cdots\geq\mu_{n}\geq\lambda_{n}$. \\
(Note that if $\lambda, \mu \in \ZZ^n$ satisfy
$\mu \rightharpoonup \lambda$, then $\lambda$ and $\mu$
are snakes automatically.)
}
\end{itemize}
\pause

\item So we know how to multiply $\overline{s}_{\lambda}$ by $h_{k}^{+}$. What
about $h_{k}^{-}$ ?

\end{itemize}

\vspace{9cm}
\end{frame}


\begin{frame}
\frametitle{\ \ \ \ \ Multiplying by $h_k^-$: the reversed $h$-Pieri rule}

\begin{itemize}
\item \textbf{Theorem (}$h^-$\textbf{-Pieri rule for Laurent polynomials):} Let
$\lambda\in\left\{  \text{snakes}\right\}  $. Let $k\in\mathbb{Z}$. Then,%
\[
h_{k}^{-}\cdot\overline{s}_{\lambda}=\sum_{\substack{\mu\in\left\{
\text{snakes}\right\}  \text{;}\\\left\vert \lambda\right\vert -\left\vert
\mu\right\vert =k;\\\lambda\rightharpoonup\mu}}\overline{s}_{\mu}.
\]
\pause

\item This follows from the $h^+$-Pieri rule by substituting
$x_1^{-1}, x_2^{-1}, \ldots, x_n^{-1}$ for $x_1, x_2, \ldots, x_n$,
using the following fact:

\textbf{Proposition:} For any snake $\lambda$, we have
\[
\overline{s}_{\lambda^{\vee}}
= \overline{s}_\lambda \tup{x_1^{-1}, x_2^{-1}, \ldots, x_n^{-1}} .
\]
Here, $\defnm{\lambda^{\vee}}$ denotes the snake
$\tup{-\lambda_n, -\lambda_{n-1}, \ldots, -\lambda_1}$
(formerly denoted by $\lambda^{\vee 0}$, but now defined for any
snake $\lambda$). \pause

\item So we now know how to multiply $\overline{s}_{\lambda}$ by $h_{k}^{-}$.
\end{itemize}

\vspace{9cm}
\end{frame}


\begin{frame}
\fti{Back to the conjecture}

\begin{itemize}

\item A consequence of the above: \\
\textbf{Corollary:}
Let $\mu$ be a snake. Let $a,b\in\mathbb{Z}$. Then,%
\[
h_{a}^{-}h_{b}^{+}\overline{s}_{\mu}=\sum_{\gamma\text{ is a snake}}\left\vert
R_{\mu,a,b}\left(  \gamma\right)  \right\vert \overline{s}_{\gamma},
\]
where
$\defnm{R_{\mu,a,b}\left(  \gamma\right) }$ is the set of all snakes
$\nu$ satisfying
\[
\mu\rightharpoonup\nu \quad \text{and}%
 \quad \left\vert \mu\right\vert -\left\vert \nu\right\vert
=a \quad \text{and} \quad \gamma\rightharpoonup
\nu \quad \text{and} \quad \left\vert \gamma
\right\vert -\left\vert \nu\right\vert =b.
\]
\pause \vspace{-0.5pc}

\item \textbf{Corollary:}
Let $\nu\in\operatorname*{Par}\left[  n\right]  $. Let
$a,b\geq 0$. Define the partition $\alpha=\left(  a+b,a^{n-2}\right)
$. Then, every $\lambda\in\mathbb{Z}^{n}$ satisfies%
\[
c_{\alpha,\nu}^{\lambda}=\left\vert R_{\nu,a,b}\left(  \lambda-a\right)
\right\vert -\left\vert R_{\nu,a-1,b-1}\left(  \lambda-a\right)  \right\vert .
\]
Here, we understand $c_{\alpha,\nu}^{\lambda}$ to mean $0$ if $\lambda$ is not
a partition (i.e., not a snake with all entries nonnegative).
\pause

\item Recall that we want a bijection
$\varphi : \ZZ^n \to \ZZ^n$ such that
\[
c_{\alpha,\mu}^{\lambda}=c_{\beta,\mu}^{\varphi\left(  \lambda\right)
}\ \ \ \ \ \ \ \ \ \ \text{for each }\lambda\in\operatorname*{Par}\left[
n\right]  .
\]


\end{itemize}

\vspace{9cm}

\end{frame}


\begin{frame}
\fti{Closing in on the bijection, 1}

\begin{itemize}

\item
\only<1>{
So we want a bijection
$\varphi : \ZZ^n \to \ZZ^n$ such that
\begin{align*}
&\left\vert R_{\mu,a,b}\left(  \lambda-a\right)
\right\vert -\left\vert R_{\mu,a-1,b-1}\left(  \lambda-a\right)  \right\vert  \\
&= \left\vert R_{\mu,b,a}\left(  \varphi\tup{\lambda}-b\right)
\right\vert -\left\vert R_{\mu,b-1,a-1}\left(  \varphi\tup{\lambda}-b\right)  \right\vert
\end{align*}
for all $\lambda \in \ZZ^n$.
}
\only<2->{
So we want a bijection
$\ff : \ZZ^n \to \ZZ^n$ such that
\begin{align*}
&\left\vert R_{\mu,a,b}\left(  \gamma \right)
\right\vert -\left\vert R_{\mu,a-1,b-1}\left(  \gamma\right)  \right\vert  \\
&= \left\vert R_{\mu,b,a}\left(  \ff\tup{\gamma} \right)
\right\vert -\left\vert R_{\mu,b-1,a-1}\left( \ff\tup{\gamma} \right)  \right\vert
\end{align*}
for all $\gamma \in \ZZ^n$.
}
\pause\pause

\item
It clearly suffices to find a bijection $\ff : \ZZ^n \to \ZZ^n$ such that
\[
\abs{R_{\mu,a,b}\left(  \gamma\right)}
= \abs{R_{\mu,b,a}\left(  \ff\tup{\gamma}\right) }
\qquad \text{for all $\gamma \in \ZZ^n$},
\]
as long as this $\ff$ is independent on $a$ and $b$.

\pause

\item In other words, if $\ff\tup{\gamma} = \eta$, then
we want
\[
\abs{R_{\mu,a,b}\left(  \gamma\right)}
= \abs{R_{\mu,b,a}\left(  \eta\right) }.
\]
\end{itemize}

\vspace{9cm}

\end{frame}


\begin{frame}
\fti{Closing in on the bijection, 2}

\begin{itemize}

\item In other words, if $\ff\tup{\gamma} = \eta$, then
we want there to be a bijection from the snakes
$\nu$ satisfying
\[
\mu\rightharpoonup\nu \quad \text{and}%
 \quad \left\vert \mu\right\vert -\left\vert \nu\right\vert
=a \quad \text{and} \quad \gamma\rightharpoonup
\nu \quad \text{and} \quad \left\vert \gamma
\right\vert -\left\vert \nu\right\vert =b
\]
to the snakes $\zeta$ satisfying
\[
\mu\rightharpoonup\zeta \quad \text{and}%
 \quad \left\vert \mu\right\vert -\left\vert \zeta\right\vert
=b \quad \text{and} \quad \eta\rightharpoonup
\zeta \quad \text{and} \quad \left\vert \eta
\right\vert -\left\vert \zeta\right\vert =a.
\]
\vspace{-0.5pc}\pause

\item Forget at first about the size conditions
($\left\vert \mu\right\vert -\left\vert \nu\right\vert
=a$, etc.). \\
Then the former snakes satisfy
\begin{align*}
& \quad \mu\rightharpoonup\nu \quad \text{and}%
 \quad  \gamma\rightharpoonup \nu \\
 \Longleftrightarrow & \quad
\left( \mu_i \geq \nu_i \text{ for all } i \leq n \right)
\wedge
\left( \nu_i \geq \mu_{i+1} \text{ for all } i < n \right) \\
  & \quad \wedge
\left( \gamma_i \geq \nu_i \text{ for all } i \leq n \right)
\wedge
\left( \gamma_i \geq \gamma_{i+1} \text{ for all } i < n \right) \\
 \Longleftrightarrow & \quad
 \left( \min\set{\mu_i, \gamma_i} \geq \nu_i \text{ for all } i \leq n \right) \\
& \quad \wedge \left( \nu_i \geq \max\set{\mu_{i+1}, \gamma_{i+1}} \text{ for all } i < n \right)
 \\
 \Longleftrightarrow & \quad
 \left( \nu_i \in \left[\max\set{\mu_{i+1}, \gamma_{i+1}}, \min\set{\mu_i, \gamma_i}\right] \text{ for all } i < n \right) \\
& \quad \wedge \left( \min\set{\mu_n, \gamma_n} \geq \nu_n \right) .
\end{align*}

\end{itemize}

\vspace{9cm}

\end{frame}

\begin{frame}
\fti{Closing in on the bijection, 3}

\begin{itemize}

\item Compare the condition
 \[
 \nu_i \in \left[\max\set{\mu_{i+1}, \gamma_{i+1}}, \min\set{\mu_i, \gamma_i}\right] \text{ for all } i < n
 \]
with the analogous condition
 \[
 \zeta_i \in \left[\max\set{\mu_{i+1}, \eta_{i+1}}, \min\set{\mu_i, \eta_i}\right] \text{ for all } i < n
 \]
on $\zeta$.
\pause

\item It is thus reasonable to hope for
\[
\min\set{\mu_i, \gamma_i} - \max\set{\mu_{i+1}, \gamma_{i+1}}
= \min\set{\mu_i, \eta_i} - \max\set{\mu_{i+1}, \eta_{i+1}}
\]
for all $i < n$.
\pause

\item Size conditions also suggest that we should have
\[
\abs{\eta} - \abs{\mu} = \abs{\mu} - \abs{\gamma}.
\]
\pause

\item These conditions do not suffice to determine
$\ff\tup{\gamma} = \eta$ (nor probably to guarantee
$\abs{R_{\mu,a,b}\left(  \gamma\right)}
= \abs{R_{\mu,b,a}\left(  \eta\right) }$),
but let's see what they tell us.

\end{itemize}

\vspace{9cm}

\end{frame}


\begin{frame}
\fti{Closing in on the bijection: the case $n = 3$}

\begin{itemize}

\item Let $n = 3$.
We want $\ff\tup{\gamma} = \eta$ to satisfy
\begin{align*}
\only<1-3>{
\min\set{\mu_1, \gamma_1} - \max\set{\mu_2, \gamma_2}
&= \min\set{\mu_1, \eta_1} - \max\set{\mu_2, \eta_2};
}
\only<4->{
\min\set{\mu_1, \gamma_1} + \min\set{-\mu_2, -\gamma_2}
&= \min\set{\mu_1, \eta_1} + \min\set{-\mu_2, -\eta_2};
}
 \\
\only<1-3>{
\min\set{\mu_2, \gamma_2} - \max\set{\mu_3, \gamma_3}
&= \min\set{\mu_2, \eta_2} - \max\set{\mu_3, \eta_3};
}
\only<4->{
\min\set{\mu_2, \gamma_2} + \min\set{-\mu_3, -\gamma_3}
&= \min\set{\mu_2, \eta_2} + \min\set{-\mu_3, -\eta_3};
}
 \\
\only<1>{
\abs{\eta} - \abs{\mu} &= \abs{\mu} - \abs{\gamma}.
}
\only<2>{
\abs{\gamma} + \abs{\eta} &= 2 \abs{\mu} .
}
\only<3->{
\tup{\gamma_1 + \gamma_2 + \gamma_3}
+ \tup{\eta_1 + \eta_2 + \eta_3}
&= 2 \tup{\mu_1 + \mu_2 + \mu_3}
}
\end{align*}
\pause\pause\pause
(here we used $\max\tup{u, v} = - \min\tup{-u, -v}$).
\pause

\item This is a system of equations that only involves
the operations $+$, $-$ and $\min$. (Recall: $2a = a+a$.)
\pause

\item There is a trick for studying such systems:
\textbf{detropicalization}.

\end{itemize}

\vspace{9cm}

\end{frame}


\begin{frame}
\fti{Detropicalization in a nutshell}

\begin{itemize}

\item A \defn{semifield} is defined in the same way as
a field, but
\begin{itemize}
\item additive inverses and a zero element
are \textbf{not} required, and
\item \textbf{every} element (not just every nonzero
element) must have a multiplicative inverse.
\end{itemize}
\pause

\item \textbf{Example:} The set $\QQ_+$ of all positive
rationals is a semifield. \pause

\item \textbf{Example:}
The set $\ZZ$, equipped with the binary operation $\min$
as \textbf{addition} and the binary operation $+$ as
\textbf{multiplication} is a semifield (with the number
$0$ as \textbf{unity}).
\pause
This is called the \defn{min tropical semifield} of $\ZZ$.
We denote it $\defnm{\ZZ_{\operatorname{trop}}}$.
\pause
\only<5>{\\
The same construction works for any totally ordered abelian
group instead of $\ZZ$. }
\pause

\item If you see a system of equations using only $+$
and $\min$, you can thus
\begin{itemize}
\item view it as a system of
\textbf{polynomial} equations over $\ZZ_{\operatorname{trop}}$; \pause
%the min tropical semifield;
\item then solve it over the semifield $\QQ_+$ instead
\only<7>{(or any other ``normal'' semifield);} \pause
\only<8->{;}
\item then check if your solution still works over $\ZZ_{\operatorname{trop}}$.
%the min tropical semifield.
\end{itemize}
This strategy is known as \defn{detropicalization}.
\pause

\item It is particularly useful if you just want
\textbf{one} solution (rather than all of them).
Often, solutions over $\QQ_+$ are unique, while
those over the min tropical semifield are not.
%Detropicalization picks out the ``right'' one.

\end{itemize}

\vspace{9cm}

\end{frame}

\begin{frame}
\frametitle{\ \ \ \ \ Detropicalizing our system ($n=3$), 1}

\begin{itemize}
\item Recall our system%
\begin{align*}
{\min\left\{  \mu_{1},\gamma_{1}\right\}  +\min\left\{  -\mu_{2},-\gamma
_{2}\right\}  } &  {=\min\left\{  \mu_{1},\eta_{1}\right\}  +\min\left\{
-\mu_{2},-\eta_{2}\right\}  ;}\\
{\min\left\{  \mu_{2},\gamma_{2}\right\}  +\min\left\{  -\mu_{3},-\gamma
_{3}\right\}  } &  {=\min\left\{  \mu_{2},\eta_{2}\right\}  +\min\left\{
-\mu_{3},-\eta_{3}\right\}  ;}\\
{\left(  \gamma_{1}+\gamma_{2}+\gamma_{3}\right)  +\left(  \eta_{1}+\eta
_{2}+\eta_{3}\right)  } &  {=2\left(  \mu_{1}+\mu_{2}+\mu_{3}\right)  }%
\end{align*}
(where $\eta_{1},\eta_{2},\eta_{3}$ are unknown).
\pause

\item
Detropicalization transforms this into%
\begin{align*}
\left(  \mu_{1}+\gamma_{1}\right)  \left(  \dfrac{1}{\mu_{2}}+\dfrac{1}%
{\gamma_{2}}\right)    & =\left(  \mu_{1}+\eta_{1}\right)  \left(  \dfrac
{1}{\mu_{2}}+\dfrac{1}{\eta_{2}}\right)  ;\\
\left(  \mu_{2}+\gamma_{2}\right)  \left(  \dfrac{1}{\mu_{3}}+\dfrac{1}%
{\gamma_{3}}\right)    & =\left(  \mu_{2}+\eta_{2}\right)  \left(  \dfrac
{1}{\mu_{3}}+\dfrac{1}{\eta_{3}}\right)  ;\\
\left(  \gamma_{1}\gamma_{2}\gamma_{3}\right)  \left(  \eta_{1}\eta_{2}%
\eta_{3}\right)    & =\left(  \mu_{1}\mu_{2}\mu_{3}\right)  ^{2}.
\end{align*}

\end{itemize}

\vspace{9cm}

\end{frame}

\begin{frame}
\frametitle{\ \ \ \ \ Detropicalizing our system ($n=3$), 2}

\begin{itemize}
\item 
\only<1>{So we now need to solve the system
\begin{align*}
\left(  \mu_{1}+\gamma_{1}\right)  \left(  \dfrac{1}{\mu_{2}}+\dfrac{1}%
{\gamma_{2}}\right)    & =\left(  \mu_{1}+\eta_{1}\right)  \left(  \dfrac
{1}{\mu_{2}}+\dfrac{1}{\eta_{2}}\right)  ;\\
\left(  \mu_{2}+\gamma_{2}\right)  \left(  \dfrac{1}{\mu_{3}}+\dfrac{1}%
{\gamma_{3}}\right)    & =\left(  \mu_{2}+\eta_{2}\right)  \left(  \dfrac
{1}{\mu_{3}}+\dfrac{1}{\eta_{3}}\right)  ;\\
\left(  \gamma_{1}\gamma_{2}\gamma_{3}\right)  \left(  \eta_{1}\eta_{2}%
\eta_{3}\right)    & =\left(  \mu_{1}\mu_{2}\mu_{3}\right)  ^{2}.
\end{align*}
}

\only<2->{
Let us rename $\mu,\gamma,\eta$ as $u,x,y$. Then, this becomes%
\begin{align*}
\left(  u_{1}+x_{1}\right)  \left(  \dfrac{1}{u_{2}}+\dfrac{1}{x_{2}}\right)
& =\left(  u_{1}+y_{1}\right)  \left(  \dfrac{1}{u_{2}}+\dfrac{1}{y_{2}%
}\right)  ;\\
\left(  u_{2}+x_{2}\right)  \left(  \dfrac{1}{u_{3}}+\dfrac{1}{x_{3}}\right)
& =\left(  u_{2}+y_{2}\right)  \left(  \dfrac{1}{u_{3}}+\dfrac{1}{y_{3}%
}\right)  ;\\
\left(  x_{1}x_{2}x_{3}\right)  \left(  y_{1}y_{2}y_{3}\right)    & =\left(
u_{1}u_{2}u_{3}\right)  ^{2}.
\end{align*}
}
\pause\pause
\item This is a system of polynomial equations, so we can give it to a
computer. The answer is:

\end{itemize}
\vspace{9cm}
\end{frame}

\begin{frame}
\frametitle{\ \ \ \ \ Solving the detropicalized system ($n=3$)}

\begin{itemize}
\item
\only<1-2>{
\textit{Solution 1:}
\begin{align*}
y_{1} &  =\dfrac{u_{1}\left(  u_{1}u_{2}u_{3}+x_{1}u_{2}u_{3}+x_{1}x_{2}%
u_{3}+x_{1}x_{2}x_{3}\right)  }{u_{1}x_{2}u_{3}-x_{1}x_{2}x_{3}},\\
y_{2} &  =\dfrac{-u_{1}u_{2}u_{3}}{x_{1}x_{3}},\\
y_{3} &  =\dfrac{u_{2}u_{3}\left(  x_{1}x_{3}-u_{1}u_{3}\right)  }{u_{1}%
u_{2}u_{3}+x_{1}u_{2}u_{3}+x_{1}x_{2}u_{3}+x_{1}x_{2}x_{3}}.
\end{align*}
}
\only<3->{
{\grey
\textit{Solution 1:}
\begin{align*}
y_{1} &  =\dfrac{u_{1}\left(  u_{1}u_{2}u_{3}+x_{1}u_{2}u_{3}+x_{1}x_{2}%
u_{3}+x_{1}x_{2}x_{3}\right)  }{u_{1}x_{2}u_{3}-x_{1}x_{2}x_{3}},\\
y_{2} &  =\dfrac{-u_{1}u_{2}u_{3}}{x_{1}x_{3}},\\
y_{3} &  =\dfrac{u_{2}u_{3}\left(  x_{1}x_{3}-u_{1}u_{3}\right)  }{u_{1}%
u_{2}u_{3}+x_{1}u_{2}u_{3}+x_{1}x_{2}u_{3}+x_{1}x_{2}x_{3}}.
\end{align*}
}
}

\item \textit{Solution 2:}
\begin{align*}
y_{1} &  =\dfrac{u_{1}u_{3}\left(  u_{1}u_{2}+x_{1}u_{2}+x_{1}x_{2}\right)
}{x_{2}\left(  u_{1}u_{3}+u_{1}x_{3}+x_{1}x_{3}\right)  },\\
y_{2} &  =\dfrac{u_{1}u_{2}\left(  u_{2}u_{3}+x_{2}u_{3}+x_{2}x_{3}\right)
}{x_{3}\left(  u_{1}u_{2}+x_{1}u_{2}+x_{1}x_{2}\right)  },\\
y_{3} &  =\dfrac{u_{2}u_{3}\left(  u_{1}u_{3}+u_{1}x_{3}+x_{1}x_{3}\right)
}{x_{1}\left(  u_{2}u_{3}+x_{2}u_{3}+x_{2}x_{3}\right)  }.
\end{align*}

\pause

\item
\only<2>{Solution 1 is useless, since we want $y_1, y_2, y_3 \in \QQ_+$.}
 \pause
But Solution 2 looks promising. \pause
Note in particular the (unexpected) \textbf{cyclic symmetry}!

\end{itemize}

\vspace{9cm}
\end{frame}



\begin{frame}
\fti{The map $\ff$: definition}

\begin{itemize}

\item Reverse-engineering Solution 2, we come up with the following \\
\textbf{Definition:} Let $\KK$ be a semifield, let
$n \geq 1$, and let $u \in \KK^n$.
We define a map $\ff : \KK^n \to \KK^n$ as follows:

Let $x\in\mathbb{K}^{n}$ be an $n$-tuple. For each $j\in\mathbb{Z}$ and
$r\geq 0$, define an element $t_{r,j}\in\mathbb{K}$ by
\[
t_{r,j}=\sum_{k=0}^{r}\underbrace{x_{j+1}x_{j+2}\cdots x_{j+k}}_{=\prod
_{i=1}^{k}x_{j+i}}\cdot\underbrace{u_{j+k+1}u_{j+k+2}\cdots u_{j+r}}%
_{=\prod_{i=k+1}^{r}u_{j+i}}.
\]
(Here and in the following, all indices are cyclic modulo $n$.)

Define $y\in\mathbb{K}^{n}$ by setting%
\[
y_{i}=u_{i}\cdot\dfrac{u_{i-1}t_{n-1,i-1}}{x_{i+1}t_{n-1,i+1}}%
\ \ \ \ \ \ \ \ \ \ \text{for each }i\in\left\{  1,2,\ldots,n\right\}  .
\]
Set $\ff\left(  x\right)  =y$.
\pause

\item Note that $\ff$ depends on $u$ (whence I call it $\ff_u$ in
the paper).

\end{itemize}

\vspace{9cm}

\end{frame}


\begin{frame}
\fti{The map $\ff$: main properties}

\begin{itemize}

\item \textbf{Theorem.} Let $\KK$ be a semifield,
$n \geq 1$ and $u \in \KK^n$. Then:

\begin{enumerate}
\item[\textbf{(a)}] The map $\ff$ is an involution (i.e., we have
$\ff \circ \ff =\operatorname*{id}$). \pause

\item[\textbf{(b)}] Let $x\in\mathbb{K}^{n}$ and $y\in\mathbb{K}^{n}$ be such
that $y=\ff\left(  x\right)  $. Then,%
\[
\left( y_{1}y_{2}\cdots y_{n} \right)
\cdot \left( x_{1}x_{2}\cdots x_{n} \right)
=\left(  u_{1}u_{2}\cdots u_{n}\right)  ^{2}.
\]
 \vspace{-0.8pc}\pause

\item[\textbf{(c)}] Let $x\in\mathbb{K}^{n}$ and $y\in\mathbb{K}^{n}$ be such
that $y=\ff\left(  x\right)  $. Then,%
\[
\left(  u_{i}+x_{i}\right)  \left(  \dfrac{1}{u_{i+1}}+\dfrac{1}{x_{i+1}%
}\right)  =\left(  u_{i}+y_{i}\right)  \left(  \dfrac{1}{u_{i+1}}+\dfrac
{1}{y_{i+1}}\right)
\]
for each $i\in\mathbb{Z}$.
(Recall that indices are cyclic modulo $n$.)
\pause

\item[\textbf{(d)}] Let $x\in\mathbb{K}^{n}$ and $y\in\mathbb{K}^{n}$ be such
that $y=\ff\left(  x\right)  $. Then,%
\[
\prod_{i=1}^{n}\dfrac{u_{i}+x_{i}}{x_{i}}=\prod_{i=1}^{n}\dfrac{u_{i}+y_{i}%
}{u_{i}}.
\]
\end{enumerate}
\pause

\item \only<5>{In short: $\ff\tup{x}$ solves our system and more.
(Note that the $i=n$ case of part \textbf{(c)} is not part
of our original system!)}

\only<6>{The proof is heavily computational but not too
hard
(various auxiliary identities had to be discovered).}

\end{itemize}

\vspace{9cm}

\end{frame}


\begin{frame}
\fti{Back to snakes}

\begin{itemize}

\item Recall that we were looking for
a bijection $\ff : \ZZ^n \to \ZZ^n$ (independent on
$a$ and $b$) such that
\[
\abs{R_{\mu,a,b}\left(  \gamma\right)}
= \abs{R_{\mu,b,a}\left(  \ff\tup{\gamma}\right) }
\qquad \text{for all $\gamma \in \ZZ^n$}.
\]
\pause

\item The map $\ff$ constructed above, applied to
$\KK = \ZZ_{\operatorname{trop}}$ and
$u = \tup{\mu_1, \mu_2, \ldots, \mu_n}$, does the trick.
(This is not hard to prove using the above Theorem.)
\pause

\item Shifting by $a$ and $b$ thus produces the
bijection $\varphi$ needed for the Pelletier--Ressayre
conjecture. Explicitly:

\end{itemize}

\vspace{9cm}

\end{frame}


\begin{frame}
\fti{The Pelletier--Ressayre hidden symmetry, 1}

\begin{itemize}

\item \textbf{Theorem (G., 2020):} Assume that $n\geq2$. Let $a,b\geq 0$,
and set $\alpha=\left(  a+b,a^{n-2}\right)  $ and
$\beta=\left(  a+b,b^{n-2}\right)  $.

Fix any partition $\mu\in\operatorname*{Par}\left[  n\right]  $.

Define a map $\varphi:\mathbb{Z}^{n}\rightarrow\mathbb{Z}^{n}$ as follows:

Let $\omega\in\mathbb{Z}^{n}$. Set $\nu = \omega - a \in \ZZ^n$.
For each $j\in\mathbb{Z}$, set
\begin{align*}
\tau_{j}
&  =\min\left\{  \left(  \nu_{j+1}+\nu_{
j+2}+\cdots+\nu_{j+k}\right)  \right. \\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left.  +\left(  \mu_{
j+k+1}+\mu_{j+k+2}+\cdots+\mu_{
j+n-1}\right)  \right. \\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left.  \ \mid\ k\in\left\{
0,1,\ldots,n-1\right\} \right\},
\end{align*}
where (unusually for partitions!) all indices are cyclic modulo $n$.

Define an $n$-tuple $\eta=\left(  \eta_{1},\eta_{2},\ldots,\eta_{n}\right)
\in\mathbb{Z}^{n}$ by setting%
\[
\eta_{i}=\mu_{i}+\left(  \mu_{i-1}+\tau_{
i-1}\right)  -\left(  \nu_{i+1}+\tau_{
i+1}\right)  \ \ \ \ \ \ \ \ \ \ \text{for each }i.
\]


Let $\varphi\left(  \omega\right)  $ be the $n$-tuple $\eta + b
\in\mathbb{Z}^{n}$. Thus, we have
defined a map $\varphi:\mathbb{Z}^{n}\rightarrow\mathbb{Z}^{n}$.

\end{itemize}

\vspace{9cm}

\end{frame}



\begin{frame}
\fti{The Pelletier--Ressayre hidden symmetry, 2}

\begin{itemize}

\item \textbf{Theorem (cont'd):} Then:

\begin{itemize}
\item[\textbf{(a)}] The map $\varphi$ is a bijection.

\item[\textbf{(b)}] We have%
\[
c_{\alpha,\mu}^{\omega}=c_{\beta,\mu}^{\varphi\left(  \omega\right)
}\ \ \ \ \ \ \ \ \ \ \text{for each }\omega\in\mathbb{Z}^{n}.
\]
Here, we are using the convention that every $n$-tuple $\omega\in
\mathbb{Z}^{n}$ that is not a partition satisfies $c_{\alpha,\mu}^{\omega}=0$
and $c_{\beta,\mu}^{\omega}=0$.
\end{itemize}
\pause

\item This proves the conjecture.
\pause

\item \textbf{Question:} Does $\varphi$ have a more mainstream
combinatorial interpretation?
\pause

\item \textbf{Question:} Can $\varphi$ be written as a
composition of ``toggles'' (i.e., ``local'' transformations,
each affecting only one entry of the tuple)?


\end{itemize}

\vspace{9cm}

\end{frame}



\begin{frame}
\fti{Uniqueness questions, 1}

\begin{itemize}

\item
\textbf{Question:}
Given a semifield $\KK$ and $n \geq 2$ and $u \in \KK^n$.
Assume that $x \in \KK^n$ and $y \in \KK^n$ satisfy
\[
\left(  u_{i}+x_{i}\right)  \left(  \dfrac{1}{u_{i+1}}+\dfrac{1}{x_{i+1}%
}\right)  =\left(  u_{i}+y_{i}\right)  \left(  \dfrac{1}{u_{i+1}}+\dfrac
{1}{y_{i+1}}\right)
\]
for each $i\in\mathbb{Z}$.
Is it true that $y = \ff\tup{x}$
\pause or $y = x$ ? \pause

\item Yes if $\KK = \QQ_+$ (or, more generally, $\KK$
is a subsemifield of an integral domain).
\pause

\item No if $\KK = \ZZ_{\operatorname{trop}}$.

\end{itemize}

\vspace{9cm}

\end{frame}



\begin{frame}
\fti{Uniqueness questions, 2}

\begin{itemize}

\item
\textbf{Question:}
Given a semifield $\KK$ and $n \geq 2$ and $u \in \KK^n$.
Assume that $x \in \KK^n$ and $y \in \KK^n$ satisfy
\[
\left( y_{1}y_{2}\cdots y_{n} \right)
\cdot \left( x_{1}x_{2}\cdots x_{n} \right)
=\left(  u_{1}u_{2}\cdots u_{n}\right)  ^{2}
\]
and
\[
\left(  u_{i}+x_{i}\right)  \left(  \dfrac{1}{u_{i+1}}+\dfrac{1}{x_{i+1}%
}\right)  =\left(  u_{i}+y_{i}\right)  \left(  \dfrac{1}{u_{i+1}}+\dfrac
{1}{y_{i+1}}\right)
\]
for each $1 \leq i < n$.
(This is our detropicalized system.)

Is it true that $y = \ff\tup{x}$ ? \pause

\item Yes if $\KK = \QQ_+$. (Nice exercise!)
\pause

\item No if $\KK = \ZZ_{\operatorname{trop}}$. \pause

\item Thus, detropicalization has made the solution
unique by removing the ``extraneous'' solutions.
\end{itemize}

\vspace{9cm}

\end{frame}



\begin{frame}
\frametitle{\ \ \ \ Thank you}

\begin{itemize}
\item \textbf{Maxime Pelletier} and \textbf{Nicolas Ressayre}
for the conjecture.

\item \textbf{Georgi Medvedev} for the invitation.

\item \textbf{Tom Roby and Grigori Olshanski} for enlightening discussions.

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


\end{document}