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\title{Birational rowmotion over noncommutative rings}
\author{Darij
Grinberg\thanks{\href{mailto:darijgrinberg@gmail.com}{darijgrinberg@gmail.com}%
.}\addressmark{1}
\and Tom Roby\thanks{\href{mailto:tom.roby@uconn.edu}{tom.roby@uconn.edu}.}\addressmark{2}}

\address{\addressmark{1}Department of Mathematics, Drexel University, Philadelphia, USA\\ \addressmark{2}Department of Mathematics, University of Connecticut, USA}

\received{\today}

\abstract{We study the dynamics of birational rowmotion over an arbitrary  noncommutative
ring $\mathbb{K}$.  This generalizes the birational rowmotion map in the commutative
setting, which itself lifts
the well-studied combinatorial rowmotion map on a finite poset.  
When the underlying poset $P$ is a rectangle (i.e., a product of two chains), this operation has ``twisted
periodicity'' and  ``reciprocity'' properties, surprisingly similar to the commutative
setting.  We briefly outline proofs of these results (details
are on the arXiv) and discuss extensions and variants. In particular, we conjecture similar
results for the case when $P$ is a $\Delta$- or $\nabla$-shaped triangle or a trapezoid.
We also conjecture that the results remain valid
when $\mathbb{K}$ is a semiring. We further prove some elementary properties of birational
rowmotion for general $P$, and (for the sake of exposition) discuss connections to the
octahedron recurrence and Zamolodchikov periodicity (which are not new, but deserve better
circulation).}

\keywords{birational rowmotion, birational combinatorics, toggles, dynamical algebraic combinatorics, rowmotion, Zamolodchikov periodicity, posets}

\begin{document}

\maketitle

\section{\label{sec.intro}Introduction}

Let $P$ be a finite poset, and $\mathcal{J}\left(  P\right)  $ be the set of
order ideals of $P$. \emph{Combinatorial rowmotion} is an invertible map
$\rho:\mathcal{J}(P)\rightarrow\mathcal{J}(P)$ which takes each ideal
$I\in\mathcal{J}(P)$ to the order ideal generated by the minimal elements of
the complement of $I$ in $P$. It can also be viewed as a map on the antichains
$\mathcal{A}(P)$ of a poset, via the well-known bijection $\mathcal{J}\left(
P\right)  \rightarrow\mathcal{A}\left(  P\right)  $ that sends an order ideal
to its set of maximal elements. Rowmotion in this sense was studied in a
sporadic sequence of papers by Brouwer and Schrijver~\cite{brouwer1974period},
and Cameron and Fon-der-Flaass~\cite{CF95}, who realized it as a product of
\textquotedblleft toggling\textquotedblright\ involutions, one for each
element of the poset. Among the core results of these works were formulas for
the \emph{order} of $\rho$ for special types of posets, such as products of chains.

In 2012, Striker and Williams~\cite{strikerwilliams} defined a related notion
of \emph{order-ideal promotion} $\pi$, which generalizes some cases of
\emph{Sch\"{u}tzenberger promotion} on semistandard tableaux. Both $\rho$ and
$\pi$ live in the \emph{toggle group} of $P$ (the group of permutations of
$\mathcal{J}\left(  P\right)  $ generated by the \textquotedblleft
toggling\textquotedblright\ involutions), and are conjugate in it, which
allows properties of $\pi$ to be derived from properties of $\rho$ and vice
versa. Armstrong, Stump, and Thomas~\cite{ast} identified rowmotion for
\emph{root posets} with the \emph{Kreweras complementation map} studied by Lie
theorists. This rekindled interest in $\rho$; a flurry of activity followed,
opening up what is now known as \emph{dynamical algebraic combinatorics}. We
refer to \cite{strikerwilliams, robydac, Hopkins2020dyn} for some snapshots of
progress in this field.

\begin{noncompile}
DG: Removing this because we're not (really) discussing homomesies.

Propp and Roby~\cite{propp-roby} noticed that this conjecture was merely one
instance of a much broader phenomenon which they dubbed \emph{homomesy}. Given
a finite set $S$, a \textquotedblleft statistic\textquotedblright%
\ $f:S\rightarrow\mathbb{C}$, and an invertible map $\varphi$ on $S$, we call
$f$ \emph{homomesic} if the average value of $f$ is the same for every
$\varphi$-orbit $\mathcal{R}$, i.e., $\displaystyle\frac{1}{\#\mathcal{R}}%
\sum_{x\in\mathcal{R}}f(x)=c$, where $c$ is a constant not dependent on the
choice of orbit $\mathcal{R}$. The confluence of all this work was the
beginning of \emph{dynamical algebraic combinatorics} as a distinct area
within algebraic combinatorics (with antecedents going back to the
Robinson--Schensted--Knuth correspondence and related operations on Young
tableaux such as promotion, evacuation, and cyclage).

In the past decade, the area has grown in a number of directions, and the
study of rowmotion has been of continuing interest.
\end{noncompile}

A major avenue of study have been \textquotedblleft liftings\textquotedblright%
\ (i.e., generalizations) of $\rho$ and $\pi$. The map $\rho$ has been lifted
three times successively:

\begin{itemize}
\item first, to a \emph{piecewise-linear rowmotion} $\rho_{\mathcal{P}}$ on
the \emph{order polytope} $\mathcal{O}(P):=\{f:P\rightarrow\lbrack0,1]:f\text{
is order-preserving}\}$ (see \cite{einpropp}),

\item then by \textquotedblleft detropicalization\textquotedblright\ to a
\emph{birational rowmotion} map $R$ (also known as $\rho_{\mathcal{B}}$)
defined on vertex-labelings of $P$ by rational functions (in commuting
variables), as detailed by Einstein and Propp~\cite{einpropp},

\item and finally to \emph{noncommutative birational rowmotion}, where the
rational functions are replaced by elements of an arbitrary ring.
\end{itemize}

\emph{A priori}, there was no reason to expect periodicity at all for these
lifted maps; indeed, already $\rho_{\mathcal{P}}$ can have infinite order even
if the poset $P$ is rather \textquotedblleft nice\textquotedblright. However,
in \cite{GrRo16, GrRo15, Okada2021}, it was shown that for a few classes of
posets (e.g., root posets and minuscule posets associated with
finite-dimensional Lie algebras and \emph{skeletal posets}, built up
inductively as a special kind of series-parallel poset), the order of
birational rowmotion $R$ is finite and surprisingly small --- generally the
\textbf{same} as for combinatorial rowmotion $\rho$. For the case where
$P=[p]\times\lbrack q]$ is a product of two chains, proofs of periodicity
(i.e., finite order) of $R$ took significant effort
\cite{GrRo15,musiker-roby}. The result was extended to all minuscule
posets by Okada \cite{Okada2021} in a type-by-type way (with some computer
algebra), and the conceptual reasons for the periodicity remain a mystery,
although Garver, Patrias and Thomas \cite{Garver2018} have found an
explanation for the periodicity of the (less general) $\rho_{\mathcal{P}}$
using quiver representations. Periodicity of $R$ for all root posets of
coincidental types is still an open question \cite[Conj. 4.40]%
{hopkins2019minuscule}, as a proof for \textquotedblleft
trapezoid-shaped\textquotedblright\ posets \cite[Conj. 75]{GrRo15}
has not been found.\footnote{To keep the story simple, we are limiting our
discussion to $\rho$ (as opposed to $\pi$) and to periodicity (as opposed to
other properties, such as \emph{homomesies}). The promotion map $\pi$ has also
been lifted (along with the entire toggle group) to the birational and even to
the noncommutative realm \cite{Joseph2019,JosRob20,JosRob21}. Much of the work
has been focusing on the antichain variant of rowmotion, which however is
equivalent to the order-ideal variant $\rho$ via the \emph{transfer map}.}

\begin{noncompile}
DG: Removing this because I don't understand it and am not sure what to do
about it.

The question of why certain properties of $\rho$ on many families of posets
lift to the birational realm is still far from answered; however, Hopkins has
shown that this must happen for posets which satisfy the \emph{tCDE property}
and have a grid-like structure~\cite{hopkins2019minuscule}. For the root
minuscule posets associated with Lie algebras of various types, the theorems
regarding periodicity and refined homomesy generally require type-by-type
analysis to be proven, e.g., Rush and Wang's extension of \textquotedblleft
file homomesy\textquotedblright\ to all minuscule posets~\cite[Thm.~1.2]%
{rush-wang-homomesy-minuscule}, and its birational lifting by
S.~Okada~\cite{Okada2021}. Even the uniformly-stated equivariant bijection of
Armstrong, Stump, and Thomas~\cite{ast} that resolves a conjecture of
Panyushev resorts to a case-by-type proof.

Some properties of combinatorial rowmotion on the set of \emph{antichains},
$\mathcal{A}(P)$, of $P$, e.g., periodicity, are easily resolved via the
aforementioned (easy) equivariant bijection between order ideals of $I$ and
its set of maximal elements. Homomesies, however, are a different story since
the statistics involved are functions on distinct domains. There turns out to
be a parallel toggle group acting on $\mathcal{A}(P)$ (with different
relations from the order-ideal toggle group), in which antichain rowmotion can
be written as a \textquotedblleft Coxeter element\textquotedblright\ of
toggles (one for each element of $P$). The parallel lifting to
piecewise-linear and birational versions of rowmotion in this setting by
Joseph and Joseph--Roby is also of
interest~\cite{Joseph2019,JosRob20,JosRob21}. They extended the antichain
lifting all the way to the case of noncommutative variables, and showed that
resolving the periodicity question in the antichain setting was equivalent to
resolving it for order-ideal rowmotion.
\end{noncompile}

The noncommutative realm remained unexplored until recently. First results
were obtained by Joseph and Roby in \cite{JosRob20,JosRob21}, but (as with any
of the three levels of generality) some of the good behavior of $R$ is lost
when allowing noncommutativity. Our initial conjecture from 2014 was that the
periodicity property of $R$ on $\left[  p\right]  \times\left[  q\right]  $
remains true, up to a (predictable and simple) \textquotedblleft
twist\textquotedblright.  This resisted a number of different proof attempts until  
we finally resolved it~\cite{main}. 

By well-understood reductions, our proof perforce establishes
periodicity of $\rho_{\mathcal{P}}$ and $\rho$ as well (for $P=\left[
p\right]  \times\left[  q\right]  $). We feel that this proof is more
elementary and simpler than any that we have seen up until now at any level
above the combinatorial.
This abstract summarizes the results of \cite{main} and briefly outlines the
main steps of their proofs, sets them in context and adds some newer observations.

%An updated historical survey is available in Thomas and
%Williams~\cite[\S7]{rowmotion-in-slow-motion}.


\section{Definitions and examples}

Throughout this work, $P$ denotes a finite poset, and $\mathbb{K}$ denotes a
ring (associative and with $1$, but not necessarily commutative). We use the
notation $\overline{x}$ for the (multiplicative) inverse $x^{-1}$ of an
element $x\in\mathbb{K}$.

The simplest type of poset is a $k$\emph{-chain} (for $k\in\mathbb{N}$) --
i.e., the set $\left\{  1,2,\ldots,k\right\}  $ equipped with its usual total
order. We denote it by $\left[  k\right]  $.

Our favorite poset will be the $p\times q$\emph{-rectangle} (for
$p,q\in\mathbb{N}$) -- i.e., the Cartesian product $\left[  p\right]
\times\left[  q\right]  $ with entrywise partial order. Its Hasse diagram
looks like a rectangular grid.

We let $\widehat{P}$ denote the poset $P$ with two new elements $0$ and $1$
adjoined to it, arranged so that every $p\in\widehat{P}$ satisfies $0\leq
p\leq1$ in $\widehat{P}$.

A $\mathbb{K}$\emph{-labelling} (or, short, \emph{labelling}) of $P$ means a
map $f:\widehat{P}\rightarrow\mathbb{K}$. Its values $f\left(  p\right)  $ are
called its \emph{labels} at the \textquotedblleft points\textquotedblright%
\ $p\in\widehat{P}$, and we draw it by overlaying these labels at the
respective points on the Hasse diagram of $\widehat{P}$. For example, if $P$
is the $2\times2$-rectangle $\left[  2\right]  \times\left[  2\right]  $, then
the following pictures show $P$ itself (Fig. 1), the extended poset
$\widehat{P}$ (Fig. 2) and a labelling of $P$ (Fig. 3):%
\[%
\begin{tabular}
[c]{|c|c|c|}\hline
$\xymatrixrowsep{0.9pc}\xymatrixcolsep{-0.50pc}%
%TCIMACRO{\TeXButton{Fig 1}{\xymatrix{ \boxed{\text{Fig. 1}} & \\ & & \left
%(2,2\right) \ar@{-}[rd] \ar@{-}[ld] & \\ & \left(2,1\right) \ar@
%{-}[rd] & & \left(1,2\right) \ar@{-}[ld] \\ & & \left(1,1\right) & }}}%
%BeginExpansion
\xymatrix{ \boxed{\text{Fig. 1}} & \\ & & \left(2,2\right) \ar@{-}[rd] \ar@
{-}[ld] & \\ & \left(2,1\right) \ar@{-}[rd] & & \left(1,2\right) \ar@
{-}[ld] \\ & & \left(1,1\right) & }%
%EndExpansion
$ & $\xymatrixrowsep{0.9pc}\xymatrixcolsep{-0.50pc}%
%TCIMACRO{\TeXButton{Fig 2}{\xymatrix{ \boxed{\text{Fig. 2}} & & 1 \ar@
%{-}[d] & \\ & & \left(2,2\right) \ar@{-}[rd] \ar@{-}[ld] & \\ & \left
%(2,1\right) \ar@{-}[rd] & & \left(1,2\right) \ar@{-}[ld] \\ & & \left
%(1,1\right) \ar@{-}[d] & \\ & & 0 & }}}%
%BeginExpansion
\xymatrix{ \boxed{\text{Fig. 2}} & & 1 \ar@{-}[d] & \\ & & \left
(2,2\right) \ar@{-}[rd] \ar@{-}[ld] & \\ & \left(2,1\right) \ar@
{-}[rd] & & \left(1,2\right) \ar@{-}[ld] \\ & & \left(1,1\right) \ar@
{-}[d] & \\ & & 0 & }%
%EndExpansion
$ & $\xymatrixrowsep{0.9pc}\xymatrixcolsep{-0.50pc}%
%TCIMACRO{\TeXButton{Fig 3}{\xymatrix{ \boxed{\text{Fig. 3}} & & f(1) \ar@
%{-}[d] & \\ & & f\left(2,2\right) \ar@{-}[rd] \ar@{-}[ld] & \\ & f\left
%(2,1\right) \ar@{-}[rd] & & f\left(1,2\right) \ar@{-}[ld] \\ & & f\left
%(1,1\right) \ar@{-}[d] & \\ & & f(0) & }}}%
%BeginExpansion
\xymatrix{ \boxed{\text{Fig. 3}} & & f(1) \ar@{-}[d] & \\ & & f\left
(2,2\right) \ar@{-}[rd] \ar@{-}[ld] & \\ & f\left(2,1\right) \ar@
{-}[rd] & & f\left(1,2\right) \ar@{-}[ld] \\ & & f\left(1,1\right) \ar@
{-}[d] & \\ & & f(0) & }%
%EndExpansion
$\\\hline
\end{tabular}
\]


When $u$ and $v$ are two elements of $\widehat{P}$, we shall use the notation
\textquotedblleft$u\lessdot v$\textquotedblright\ (or, equivalently,
\textquotedblleft$v\gtrdot u$\textquotedblright) for the statement
\textquotedblleft$u<v$, and there is no $w\in\widehat{P}$ such that
$u<w<v$\textquotedblright.

We are now ready to define the main actor of our play: the \emph{birational
rowmotion} operation. While it is traditionally defined as a composition of
\emph{toggles} (see, e.g., \cite[Definition 3.16]{main}), we instead here use
a recursive equation (which is equivalent):

\begin{definition}
\label{def.R}\emph{Birational rowmotion} is the partial map $R:\mathbb{K}%
^{\widehat{P}}\dashrightarrow\mathbb{K}^{\widehat{P}}$ that transforms each
labelling $f$ of $P$ into a new labelling $Rf$ whose values are%
\begin{align*}
\left(  Rf\right)  \left(  0\right)   &  =f\left(  0\right)
,\ \ \ \ \ \ \ \ \ \ \left(  Rf\right)  \left(  1\right)  =f\left(  1\right)
,\ \ \ \ \ \ \ \ \ \ \text{and}\\
\left(  Rf\right)  \left(  v\right)   &  =\left(  \sum\limits_{\substack{u\in
\widehat{P};\\u\lessdot v}}f\left(  u\right)  \right)  \cdot\overline{f\left(
v\right)  }\cdot\overline{\sum\limits_{\substack{u\in\widehat{P};\\u\gtrdot
v}}\overline{\left(  Rf\right)  \left(  u\right)  }}%
\ \ \ \ \ \ \ \ \ \ \text{for all }v\in P.
\end{align*}

\end{definition}

Note that the formula for $\left(  Rf\right)  \left(  v\right)  $ involves
several inverses, which may or may not exist. If they don't, $Rf$ is
undefined, which is why $R$ is only a \textbf{partial} map. Also, this formula
is \textbf{recursive}: The value it assigns to $\left(  Rf\right)  \left(
v\right)  $ depends on the values $\left(  Rf\right)  \left(  u\right)  $ for
elements $u\in\widehat{P}$ satisfying $u\gtrdot v$. Since the poset
$\widehat{P}$ is finite, we can solve this recursion \textquotedblleft
starting from the top of $P$\textquotedblright\ (using $\left(  Rf\right)
\left(  1\right)  =f\left(  1\right)  $ to bootstrap the process). Making this
procedure explicit gives precisely the description of $R$ as a composition of
toggles that is used as a definition of $R$ in \cite{main}. (See
\cite[Proposition 3.18]{main} for details.)

The map $R$ is what we called \textquotedblleft noncommutative birational
rowmotion\textquotedblright\ in \cref{sec.intro} -- that is, the third
(and ultimate) generalization of combinatorial rowmotion $\rho$. The second
generalization (commutative birational rowmotion) is obtained when
$\mathbb{K}$ is a field. The first generalization (piecewise-linear rowmotion
$\rho_{\mathcal{P}}$) is recovered from the second by \emph{tropicalization}
(i.e., by setting $\mathbb{K}$ to be a tropical semifield\footnote{Of course,
a tropical semifield is not a field. However, properties of $R$ that hold for
all fields will automatically hold for all semifields (since they reduce to
subtraction-free polynomial identities), and thus we can allow $\mathbb{K}$ to
be a semifield.}). The original rowmotion $\rho$ on $\mathcal{J}\left(
P\right)  $ is the restriction of $\rho_{\mathcal{P}}$ to the vertices of the
order polytope $\mathcal{O}\left(  P\right)  $ (which are the indicator
functions of order ideals of $P$). Therefore, any (positive) result about
noncommutative birational rowmotion automatically yields results about all
lower levels (commutative, piecewise-linear and combinatorial).

Partial maps can be composed; in particular, $R$ can be composed with itself.
As usual, $R^{k}$ shall mean the composition $R\circ R\circ\cdots\circ R$ with
$k$ many $R$'s.

\begin{example}
\label{eg:2x2} Here are the first four iterations of $R$ for $P=[2]\times
\lbrack2]$, acting on an arbitrary labelling $f$:
\[%
\begin{tabular}
[c]{|c|c|c|}\hline
$\xymatrixrowsep{0.9pc}\xymatrixcolsep{0pc}%
%TCIMACRO{\TeXButton{f}{\xymatrix{
%\boxed{f}
%& & b \ar@{-}[d] & \\
%& & z \ar@{-}[rd] \ar@{-}[ld] & \\
%& x \ar@{-}[rd] & & y \ar@{-}[ld] \\
%& & w \ar@{-}[d] & \\
%& & a &
%}}}%
%BeginExpansion
\xymatrix{
\boxed{f}
& & b \ar@{-}[d] & \\
& & z \ar@{-}[rd] \ar@{-}[ld] & \\
& x \ar@{-}[rd] & & y \ar@{-}[ld] \\
& & w \ar@{-}[d] & \\
& & a &
}%
%EndExpansion
$ & $\xymatrixrowsep{0.9pc}\xymatrixcolsep{0pc}%
%TCIMACRO{\TeXButton{Rf}{\xymatrix{
%\boxed{Rf}
%& b \ar@{-}[d] & \\
%& (x+y)\overline{z}b \ar@{-}[rd] \ar@{-}[ld] & \\
%w\overline{x}(x+y)\overline{z}b \ar@{-}[rd] & & w\overline{y}(x+y)\overline
%{z}b \ar@{-}[ld] \\
%& a \overline{z} b \ar@{-}[d] & \\
%& a &
%}}}%
%BeginExpansion
\xymatrix{
\boxed{Rf}
& b \ar@{-}[d] & \\
& (x+y)\overline{z}b \ar@{-}[rd] \ar@{-}[ld] & \\
w\overline{x}(x+y)\overline{z}b \ar@{-}[rd] & & w\overline{y}(x+y)\overline
{z}b \ar@{-}[ld] \\
& a \overline{z} b \ar@{-}[d] & \\
& a &
}%
%EndExpansion
$ & $\xymatrixrowsep{0.9pc}\xymatrixcolsep{0pc}%
%TCIMACRO{\TeXButton{RRf}{\xymatrix{
%\boxed{R^2f}
%& b \ar@{-}[d] & \\
%& w\left(\overline{x}+\overline{y}\right)b \ar@{-}[rd] \ar@{-}[ld] & \\
%a\overline{y}b \ar@{-}[rd] & & a\overline{x}b \ar@{-}[ld] \\
%& a\overline{b}z\cdot\overline{x+y}\cdot b \ar@{-}[d] & \\
%& a &
%}}}%
%BeginExpansion
\xymatrix{
\boxed{R^2f}
& b \ar@{-}[d] & \\
& w\left(\overline{x}+\overline{y}\right)b \ar@{-}[rd] \ar@{-}[ld] & \\
a\overline{y}b \ar@{-}[rd] & & a\overline{x}b \ar@{-}[ld] \\
& a\overline{b}z\cdot\overline{x+y}\cdot b \ar@{-}[d] & \\
& a &
}%
%EndExpansion
$\\\hline
\end{tabular}
\]%
\[%
\begin{tabular}
[c]{|c|c|}\hline
$\xymatrixrowsep{0.9pc}\xymatrixcolsep{-2pc}%
%TCIMACRO{\TeXButton{RRRf}{\xymatrix{
%\boxed{R^3f}
%& b \ar@{-}[d] & \\
%& a\overline{w}b \ar@{-}[rd] \ar@{-}[ld] & \\
%a \overline{b}z\cdot\overline{x+y}\cdot y\overline{w}b \ar@{-}%
%[rd] & & a \overline{b}z\cdot\overline{x+y}\cdot x\overline{w}b \ar@{-}[ld] \\
%& a\overline{b}\cdot\overline{\overline{x}+\overline{y}}\cdot\overline
%{w}b \ar@{-}[d] & \\
%& a &
%}}}%
%BeginExpansion
\xymatrix{
\boxed{R^3f}
& b \ar@{-}[d] & \\
& a\overline{w}b \ar@{-}[rd] \ar@{-}[ld] & \\
a \overline{b}z\cdot\overline{x+y}\cdot y\overline{w}b \ar@{-}%
[rd] & & a \overline{b}z\cdot\overline{x+y}\cdot x\overline{w}b \ar@{-}[ld] \\
& a\overline{b}\cdot\overline{\overline{x}+\overline{y}}\cdot\overline
{w}b \ar@{-}[d] & \\
& a &
}%
%EndExpansion
$ & $\xymatrixrowsep{0.9pc}\xymatrixcolsep{-0.50pc}%
%TCIMACRO{\TeXButton{RRRRf}{\xymatrix{
%\boxed{R^4f}
%& b \ar@{-}[d] & \\
%& a\overline{b}z\overline{a}b \ar@{-}[rd] \ar@{-}[ld] & \\
%a\overline{b}x\overline{a}b \ar@{-}[rd] & & a\overline{b}y\overline{a}%
%b \ar@{-}[ld] \\
%& a\overline{b}w\overline{a}b \ar@{-}[d] & \\
%& a &
%}}}%
%BeginExpansion
\xymatrix{
\boxed{R^4f}
& b \ar@{-}[d] & \\
& a\overline{b}z\overline{a}b \ar@{-}[rd] \ar@{-}[ld] & \\
a\overline{b}x\overline{a}b \ar@{-}[rd] & & a\overline{b}y\overline{a}%
b \ar@{-}[ld] \\
& a\overline{b}w\overline{a}b \ar@{-}[d] & \\
& a &
}%
%EndExpansion
$\\\hline
\end{tabular}
\]


Here, some nontrivial algebra has been used to simplify some of the labels;
e.g., the $a\overline{b}y\overline{a}b$ in $R^{4}f$ was originally an
$a\overline{b}\cdot\overline{\overline{x}+\overline{y}}\cdot\overline
{x+y}\cdot y\left(  \overline{x}+\overline{y}\right)  \left(  x+y\right)
\overline{a}b$.
\end{example}

\section{The main theorems for rectangles}

\cref{eg:2x2} illustrates two surprising phenomena that apply to all
rectangular posets. One is a \textquotedblleft twisted
periodicity\textquotedblright, saying that the initial labels of $f$ reappear
in $R^{4}f$ framed by $a\overline{b}$ on the left and $\overline{a}b$ on the
right. (Note that this is not a conjugation, since $(a\overline{b}%
)^{-1}=b\overline{a}\neq\overline{a}b$ unless $a$ and $b$ commute.) For a
generic poset $P$, nothing like this holds (instead, the labels of $R^{k}f$
get increasingly complicated as $k$ grows, even if $\mathbb{K}$ is
commutative), but it turns out to be true whenever $P$ is a rectangle:

\begin{theorem}
[Periodicity for the $p\times q$-rectangle]\label{thm.rect.ord} Let $p$ and
$q$ be two positive integers. Let $P=\left[  p\right]  \times\left[  q\right]
$, and let $f\in\mathbb{K}^{\widehat{P}}$ be a $\mathbb{K}$-labelling such
that $R^{p+q}f$ is well-defined. Set $a=f\left(  0\right)  $ and $b=f\left(
1\right)  $. Then, $a$ and $b$ are invertible, and for any $x\in\widehat{P}$
we have
\begin{equation}
\left(  R^{p+q}f\right)  \left(  x\right)  =a\overline{b}\cdot f\left(
x\right)  \cdot\overline{a}b. \label{eq.thm.rect.ord.claim}%
\end{equation}

\end{theorem}

This periodicity actually follows from the following \textquotedblleft
antipodal reciprocity\textquotedblright\ property, which is the second
phenomenon seen in \cref{eg:2x2}. This property generalizes the
$a\overline{z}b$ label in $Rf$, the $a\overline{y}b$ and $a\overline{x}b$
labels in $R^{2}f$, and the $a\overline{w}b$ label in $R^{3}f$. In its general
form, it says that the inverse of any label in $R^{k}f$ appears as a label in
$R^{\ell}f$ (for some $\ell\geq k$), framed by $a$ on the left and by $b$ on
the right. The $\ell$ depends on the position of the label; the precise
statement is as follows:

\begin{theorem}
[Reciprocity for the $p\times q$-rectangle]\label{thm.rect.antip} Let $p$ and
$q$ be two positive integers. Let $P=\left[  p\right]  \times\left[  q\right]
$. Fix $\ell\in\mathbb{N}$, and let $f\in\mathbb{K}^{\widehat{P}}$ be a
$\mathbb{K}$-labelling such that $R^{\ell}f$ is well-defined. Set $a=f\left(
0\right)  $ and $b=f\left(  1\right)  $. Let $\left(  i,j\right)  \in P$
satisfy $\ell-i-j+1\geq0$. Then,%
\begin{equation}
\left(  R^{\ell}f\right)  \left(  i,j\right)  =a\cdot\overline{\left(
R^{\ell-i-j+1}f\right)  \left(  p+1-i,\ q+1-j\right)  }\cdot b.
\label{eq.thm.rect.antip.claim}%
\end{equation}

\end{theorem}

We conjectured these two theorems in 2014, after proving them in the
particular case when $\mathbb{K}$ is commutative (\cite[\S 4--\S 8]%
{GrRo15}).

\section{Connections which do not lift}

Strikingly, this noncommutative generalization has resisted all approaches
that have previously succeeded in the commutative case. The determinantal
computations involved in the proof in~\cite{GrRo15} can be extended to
the noncommutative setting using the \emph{quasideterminants} of Gelfand and
Retakh, but it seems impossible to make a rigorous proof out of them (lacking,
e.g., any useful notation of Zariski topology in this setting, it is not clear
what it means for a $\mathbb{K}$-labelling to be \textquotedblleft
generic\textquotedblright). The Musiker--Roby proof of commutative
periodicity~\cite{musiker-roby} (via a lattice-path formula for $\left(
R^{i}f\right)  \left(  v\right)  $) could not be generalized either. Thus the
noncommutative case remained an open problem.
%\footnote{This is not the first time that rational maps in algebraic
%combinatorics have been generalized to the noncommutative case; some other
%instances are \cite{IyuShk14, BerRet15, Rupel17, GonKon21}. Each time, the
%generalizations have been much harder to prove, not least because very little
%of the commutative groundwork is (currently?) available at the noncommutative
%level. For instance, it is insufficient to work over the ``free skew fields'',
%since an identity between rational expressions can be true in all skew fields
%yet fail in some noncommutative rings (such as the identity
%$x \left(yx\right)^{-1} y = 1$). For this reason, while natural from an
%algebraic point of view, the noncommutative setting is only recently and
%slowly getting explored.}


At some point, Glick and Grinberg noticed that the $Y$-variables in the
type-$AA$ Zamolodchikov periodicity theorem of Volkov~\cite{volkov} could be
written as ratios of labels under iterated birational rowmotion~\cite[\S ~4.4]%
{robydac}; this allows the periodicity in one setting to be derived from that
in the other (with some work). However, for noncommutative $\mathbb{K}$,
Zamolodchikov periodicity fails even in small examples such as $r=r^{\prime
}=2$ (no matter in which of the $120$ possible orders we multiply the
factors), while noncommutative birational rowmotion continues to exhibit
periodicity. This approach is therefore unavailable in the noncommutative case
as well.

Another recent proof of periodicity and reciprocity for commutative
$\mathbb{K}$ (Johnson and Liu, \cite{johnsonliu1}) uses a connection to the
octahedron recurrence; this connection, too, does not seem to generalize
readily to the noncommutative case (although a proper analysis of all options
has not been made yet), and anyway the proof in \cite{johnsonliu1} uses
determinants, which are not available for general $\mathbb{K}$.

\begin{verlong}
TODO:

\begin{itemize}
\item Check whether \cite[Lemma 2.14]{johnsonliu1} can be generalized. Note
that this may require a creative lift of the octahedron recurrence, maybe
related to the strange-looking noncommutative lift of the Pl\"{u}cker-Ptolemy
relations in Berenstein/Retakh, \textit{Noncommutative Marked Surfaces}.

\item Whitehouse \cite[$\sigma_{1}\sigma_{2}\cdots\sigma_{n-1}$]{Whiteh01} as
a linearization of $1\times n$-rectangle $R$. Doesn't seem to generalize to
$p\times q$-rectangles with $p,q>1$ however.
\end{itemize}
\end{verlong}

\section{\label{sec.proof}Proof outline}

We shall now outline the main steps of our proofs of \cref{thm.rect.ord,thm.rect.antip}. Full details are available in
\cite{main}, including justifications of all the inverses being taken (in the
following, we just assume they all exist) and proofs of all lemmas. \medskip

%%%% STEP 1


\textbf{Step 1:} \textit{Observe that \cref{thm.rect.ord} follows from
\cref{thm.rect.antip}.} Indeed, assume that
\cref{thm.rect.antip} is proved. Now, to prove
\cref{thm.rect.ord}, assume WLOG that $x\in P$ (as opposed to $x=0$ or
$x=1$, which cases are easy), and write $x=\left(  u,v\right)  $. Now, apply
\cref{thm.rect.antip} first to $\left(  \ell,i,j\right)  =\left(
p+q,\ u,\ v\right)  $. Then, apply \cref{thm.rect.antip} again to
$\left(  \ell,i,j\right)  =\left(  p+q-u-v+1,\ p+1-u,\ q+1-v\right)  $ in
order to rewrite the right hand side.

It thus remains to prove \cref{thm.rect.antip}.

\begin{verlong}
\textbf{Step 1:} \textit{Show that \cref{thm.rect.ord} (twisted
periodicity) follows from \cref{thm.rect.antip} (antipodal
reciprocity).} To do this set $i^{\prime}:=p+1-i$ and $j^{\prime}:=q+1-j$ so
that $(i,j)$ and $(i^{\prime},j^{\prime})$ are antipodal in $P=[p]\times
\left[  q\right]  $. One iteration of \cref{thm.rect.antip} gives:
\[
\left(  R^{i^{\prime}+j^{\prime}-1}f\right)  \left(  i^{\prime},j^{\prime
}\right)  =a\cdot\overline{f\left(  i,j\right)  }\cdot b
\]
while another gives
\[
\left(  R^{p+q}f\right)  \left(  i,j\right)  =a\cdot\overline{\left(
R^{i^{\prime}+j^{\prime}-1}f\right)  \left(  i^{\prime},j^{\prime}\right)
}\cdot b=a\cdot\overline{a\cdot\overline{f\left(  i,j\right)  }\cdot b}\cdot
b=a\overline{b}\cdot f(i,j)\cdot\overline{a}b
\]

\end{verlong}

\medskip

\textbf{Step 2:} \textit{Introduce notations:} Fix a $\mathbb{K}$-labelling
$f$ of $P=\left[  p\right]  \times\left[  q\right]  $. For any element
$x=\left(  i,j\right)  \in P$, we define its \emph{antipode} $x^{\sim}\in P$
by $\displaystyle x^{\sim}:=\left(  p+1-i,\ q+1-j\right)  $. For instance,
$\left(  1,1\right)  ^{\sim}=\left(  p,q\right)  $. For any $x\in\widehat{P}$
and $\ell\in\mathbb{N}$, we write%
\begin{equation}
x_{\ell}:=\left(  R^{\ell}f\right)  \left(  x\right)  . \label{eq.xl=}%
\end{equation}
Informally, we think of our subscripts as indicating the \textquotedblleft
time\textquotedblright\ of the label (i.e., number of times $R$ was applied to
$f$). For instance, in \cref{eg:2x2}, we have $\left(  1,2\right)
_{2}=\left(  R^{2}f\right)  \left(  1,2\right)  =a\overline{x}b$.

In this notation, \cref{thm.rect.antip} claims that
\begin{equation}
x_{\ell}=a\cdot\overline{x_{\ell-i-j+1}^{\sim}}\cdot b
\label{pf.thm.rect.antip.restate1}%
\end{equation}
for all $x=\left(  i,j\right)  \in P$ and all $\ell\geq i+j-1$. Clearly, it
suffices to prove this for $\ell=i+j-1$.

For each $v\in P$ and $\ell\in\mathbb{N}$, we define the two \emph{slack
values}\footnote{In both sums, $u$ ranges over elements of $\widehat{P}$.}
\[
\downslack_{\ell}^{v}:=v_{\ell}\cdot\overline{\sum\limits_{u\lessdot v}%
u_{\ell}}\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ \upslack_{\ell}%
^{v}:=\overline{\sum\limits_{u\gtrdot v}\overline{u_{\ell}}}\cdot
\overline{v_{\ell}}.
\]
Furthermore, when $v\in\left\{  0,1\right\}  $, we set $\downslack_{\ell}%
^{v}:=1$ and $\upslack_{\ell}^{v}:=1$ for all $\ell\in\mathbb{N}$. For
instance, in \cref{eg:2x2}, we have $\downslack_{0}^{\left(
2,2\right)  }=z\cdot\overline{y+x}$ and $\upslack_{0}^{\left(  1,1\right)
}=\overline{\overline{y}+\overline{x}}\cdot\overline{w}$ and $\upslack_{1}%
^{\left(  1,1\right)  }=\overline{\overline{w\overline{y}\left(  x+y\right)
\overline{z}b}+\overline{w\overline{x}\left(  x+y\right)  \overline{z}b}}%
\cdot\overline{a\overline{z}b}=w\cdot\overline{a}$ (after simplifications).

A \emph{path} will mean a sequence $\left(  v_{0},v_{1},\ldots,v_{k}\right)  $
of elements of $\widehat{P}$ satisfying $v_{0}\gtrdot v_{1}\gtrdot
\cdots\gtrdot v_{k}$. We will call it a \emph{path from }$v_{0}$\emph{ to
}$v_{k}$. For instance, there are two paths from $\left(  2,2\right)  $ to
$\left(  1,1\right)  $, namely $\left(  \left(  2,2\right)  ,\left(
1,2\right)  ,\left(  1,1\right)  \right)  $ and $\left(  \left(  2,2\right)
,\left(  2,1\right)  ,\left(  1,1\right)  \right)  $.

For any path $\mathbf{p}=\left(  v_{0},v_{1},\ldots,v_{k}\right)  $ and any
$\ell\in\mathbb{N}$, we define the \emph{slack products}
\[
\downslack_{\ell}^{\mathbf{p}}:=\downslack_{\ell}^{v_{0}}\downslack_{\ell
}^{v_{1}}\cdots\downslack_{\ell}^{v_{k}}\ \ \ \ \ \ \ \ \ \ \text{and}%
\ \ \ \ \ \ \ \ \ \ \upslack_{\ell}^{\mathbf{p}}:=\upslack_{\ell}^{v_{0}%
}\upslack_{\ell}^{v_{1}}\cdots\upslack_{\ell}^{v_{k}}%
\]


If $u$ and $v$ are elements of $\widehat{P}$, and if $\ell\in\mathbb{N}$, then
we define the \emph{slack sums}%
\[
\downslack_{\ell}^{u\rightarrow v}:=\sum_{\mathbf{p}\text{ is a path from
}u\text{ to }v}\downslack_{\ell}^{\mathbf{p}}\ \ \ \ \ \ \ \ \ \ \text{and}%
\ \ \ \ \ \ \ \ \ \ \upslack_{\ell}^{u\rightarrow v}:=\sum_{\mathbf{p}\text{
is a path from }u\text{ to }v}\upslack_{\ell}^{\mathbf{p}}.
\]
For instance,%
\[
\downslack_{\ell}^{\left(  2,2\right)  \rightarrow\left(  1,1\right)
}=\downslack_{\ell}^{\left(  \left(  2,2\right)  ,\left(  1,2\right)  ,\left(
1,1\right)  \right)  }+\downslack_{\ell}^{\left(  \left(  2,2\right)  ,\left(
2,1\right)  ,\left(  1,1\right)  \right)  }=\downslack_{\ell}^{\left(
2,2\right)  }\downslack_{\ell}^{\left(  1,2\right)  }\downslack_{\ell
}^{\left(  1,1\right)  }+\downslack_{\ell}^{\left(  2,2\right)  }%
\downslack_{\ell}^{\left(  2,1\right)  }\downslack_{\ell}^{\left(  1,1\right)
}.
\]


The following formula connects the slack values at \textquotedblleft time
$\ell$\textquotedblright\ with those at \textquotedblleft time $\ell
-1$\textquotedblright:

\begin{proposition}
[Transition equation in $\downslack$-$\upslack$-form]\label{prop.rect.trans}%
Let $v\in\widehat{P}$ and $\ell\geq1$. Then, $\upslack_{\ell}^{v}%
=\downslack_{\ell-1}^{v}$.
\end{proposition}

\begin{proof}
This is just the formula for $\left(  Rf\right)  \left(  v\right)  $ in
\cref{def.R}, rewritten.
\end{proof}

\medskip

\textbf{Step 3:} \textit{The path formulas.} The following formulas allow us
to recover the labels $u_{\ell}$ from the slack values $\downslack_{\ell}^{v}$
or $\upslack_{\ell}^{v}$:

\begin{theorem}
[path formulas for rectangle]\label{thm.rect.path}Let $\ell\in\mathbb{N}$.
Then, each $u\in P$ satisfies%
\begin{equation}
u_{\ell}=\overline{\upslack_{\ell}^{1\rightarrow u}}\cdot b=\overline
{\upslack_{\ell}^{\left(  p,q\right)  \rightarrow u}}\cdot b
\label{eq.thm.rect.path.up}%
\end{equation}
and%
\begin{equation}
u_{\ell}=\downslack_{\ell}^{u\rightarrow0}\cdot a=\downslack_{\ell
}^{u\rightarrow\left(  1,1\right)  }\cdot a. \label{eq.thm.rect.path.down}%
\end{equation}

\end{theorem}

\begin{proof}
Induction on $u$ (downwards for (\ref{eq.thm.rect.path.up}), upwards for
(\ref{eq.thm.rect.path.down})).
\end{proof}

Using (\ref{eq.thm.rect.path.down}) and (\ref{eq.thm.rect.path.up}) and
\cref{prop.rect.trans}, we can already prove that $\left(
1,1\right)  _{1}=a\overline{\left(  p,q\right)  _{0}}b$, which confirms
(\ref{pf.thm.rect.antip.restate1}) for $x=\left(  1,1\right)  $. The general
case, however, requires more work. \medskip

\begin{wrapfigure}{r}{0.45\textwidth}
\[
\begin{tikzpicture}[scale=0.7]
\draw (-3.3, -6.3) -- (-3.3, 3.3) -- (6.3, 3.3) -- (6.3, -6.3) -- cycle;
\coordinate (W) at (-3, 0);
\coordinate (N) at (0, 3);
\coordinate (E) at (6, -3);
\coordinate (S) at (3, -6);
\draw[thick] (W) -- (N) -- (E) -- (S) -- (W);
\coordinate [label=80:{$u$}] (u) at (1, 2);
\coordinate [label=80:{$u'$}] (u') at (1.5, 1.5);
\coordinate [label=190:{$d$}] (d) at (0.5, -3.5);
\coordinate [label=190:{$d'$}] (d') at (1, -4);
\fill (u) circle [radius=0.1];
\fill (u') circle [radius=0.1];
\fill (d) circle [radius=0.1];
\fill (d') circle [radius=0.1];
\draw[red, very thick] (u) -- (0.5, 1.5) -- (1, 1) -- (-0.5, -0.5) -- (0, -1) -- (-1, -2) -- (d);
\draw[blue, very thick] (u') -- (2.5, 0.5) -- (2, 0) -- (2.5, -0.5) -- (2, -1) -- (2.5, -1.5) -- (2, -2) -- (2.5, -2.5) -- (2, -3) -- (d');
\end{tikzpicture}
\]
\end{wrapfigure}

\textbf{Step 4:} \textit{The conversion lemma.} The following lemma is at the
core of our proof:\footnote{The figure on the right illustrates this lemma
(where the red path is one of the paths contributing to the sum
$\downslack_{\ell}^{u\rightarrow d}$, while the blue path contributes to
$\upslack_{\ell}^{u^{\prime}\rightarrow d^{\prime}}$).}

\begin{lemma}
[Conversion lemma]\label{lem.rect.conv}Let $u$ and $u^{\prime}$ be two
elements of the northeastern edge of $P$ satisfying $u\gtrdot u^{\prime}$
(that is, $u=\left(  k,q\right)  $ and $u^{\prime}=\left(  k-1,q\right)  $ for
some $k\in\left\{  2,3,\ldots,p\right\}  $). Let $d$ and $d^{\prime}$ be two
elements of the southwestern edge of $P$ satisfying $d\gtrdot d^{\prime}$
(that is, $d=\left(  i,1\right)  $ and $d^{\prime}=\left(  i-1,1\right)  $ for
some $i\in\left\{  2,3,\ldots,p\right\}  $). Then, for each $\ell\in
\mathbb{N}$, we have%
\begin{equation}
\downslack_{\ell}^{u\rightarrow d}=\upslack_{\ell}^{u^{\prime}\rightarrow
d^{\prime}}. \label{eq.lem.rect.conv.eq}%
\end{equation}

\end{lemma}

Note that both sides of this equality are slack sums at \textquotedblleft time
$\ell$\textquotedblright\ (i.e., defined purely in terms of the labeling
$R^{\ell}f$). Thus, this lemma does not involve birational rowmotion.
Nevertheless, its proof is far from trivial. It involves \textquotedblleft
interpolating\textquotedblright\ between $\downslack_{\ell}^{u\rightarrow d}$
and $\upslack_{\ell}^{u^{\prime}\rightarrow d^{\prime}}$ through a
\textquotedblleft mixed slack sum\textquotedblright. The latter is a sum over
\textquotedblleft\emph{path-jump-paths}\textquotedblright, which are like
paths but allow a \textquotedblleft jump\textquotedblright\ in the middle. We
refer to \cite[\S 9]{main} for the details. \medskip

\textbf{Step 5:} \textit{The southwestern edge.} If $\ell\geq1$, then
(\ref{eq.lem.rect.conv.eq}) can be rewritten as $\downslack_{\ell
}^{u\rightarrow d}=\downslack_{\ell-1}^{u^{\prime}\rightarrow d^{\prime}}$ (by
\cref{prop.rect.trans}). Applying this equality many times
(\textquotedblleft gliding down\textquotedblright\ the rectangle), we see that
each $i\in\left[  p\right]  $ satisfies $\downslack_{i-1}^{\left(
p,\ q\right)  \rightarrow\left(  i,\ 1\right)  }=\downslack_{0}^{\left(
p-i+1,\ q\right)  \rightarrow\left(  1,\ 1\right)  }$. However, \cref{prop.rect.trans} yields $\upslack_{i}^{\left(  p,\ q\right)
\rightarrow\left(  i,\ 1\right)  }=\downslack_{i-1}^{\left(  p,\ q\right)
\rightarrow\left(  i,\ 1\right)  }=\downslack_{0}^{\left(  p-i+1,\ q\right)
\rightarrow\left(  1,\ 1\right)  }$. In view of (\ref{eq.thm.rect.path.down})
and (\ref{eq.thm.rect.path.up}), we can readily use this to show that $\left(
i,1\right)  _{i}=a\cdot\overline{\left(  p+1-i,\ q\right)  _{0}}\cdot b$. This
proves (\ref{pf.thm.rect.antip.restate1}) for $x=\left(  i,1\right)  $. In
other words, (\ref{pf.thm.rect.antip.restate1}) is proved for all $x$ lying on
the southwestern edge of $P$ (speaking in terms of the Hasse diagram). A
symmetric argument applies when $x$ lies on the southeastern edge. \medskip

\textbf{Step 6:} \textit{Inducting up the poset.} It remains to prove
(\ref{pf.thm.rect.antip.restate1}) for all $x=\left(  i,j\right)  $ with
$i,j\geq2$. This is done by strong induction on $i+2j$, which allows us to
apply the induction hypothesis to the four pairs%
\[
m:=\left(  i,j-1\right)  ,\ \ \ \ \ \ \ \ \ \ u:=\left(  i+1,j-1\right)
,\ \ \ \ \ \ \ \ \ \ s:=\left(  i,j-2\right)  ,\ \ \ \ \ \ \ \ \ \ t:=\left(
i-1,j-1\right)
\]
in lieu of $x=\left(  i,j\right)  $ (to the extent that these pairs do belong
to $P$). As a result, we obtain%
\begin{align}
m_{\ell}  &  =a\cdot\overline{m_{\ell-k}^{\sim}}\cdot
b,\ \ \ \ \ \ \ \ \ \ s_{\ell-1}=a\cdot\overline{s_{\ell-k}^{\sim}}\cdot
b,\label{pf.thm.rect.antip.short.4}\\
t_{\ell-1}  &  =a\cdot\overline{t_{\ell-k}^{\sim}}\cdot
b,\ \ \ \ \ \ \ \ \ \ m_{\ell-1}=a\cdot\overline{m_{\ell-k-1}^{\sim}}\cdot
b,\label{pf.thm.rect.antip.short.5}\\
u_{\ell}  &  =a\cdot\overline{u_{\ell-k-1}^{\sim}}\cdot
b\ \ \ \ \ \ \ \ \ \ \left(  \text{where }k=i+j-2\right)  .
\label{pf.thm.rect.antip.short.5b}%
\end{align}
On the other hand, the formula for $\left(  Rf\right)  \left(  v\right)  $ in
\cref{def.R} (applied once to $v=m$ and once to $v=m^{\sim}$)
yields
\begin{align}
m_{\ell}  &  =\left(  s_{\ell-1}+t_{\ell-1}\right)  \cdot\overline{m_{\ell-1}%
}\cdot\overline{\overline{u_{\ell}}+\overline{x_{\ell}}}%
\ \ \ \ \ \ \ \ \ \ \text{and}\label{pf.thm.rect.antip.short.6}\\
m_{\ell-k}^{\sim}  &  =\left(  u_{\ell-k-1}^{\sim}+x_{\ell-k-1}^{\sim}\right)
\cdot\overline{m_{\ell-k-1}^{\sim}}\cdot\overline{\overline{s_{\ell-k}^{\sim}%
}+\overline{t_{\ell-k}^{\sim}}}. \label{pf.thm.rect.antip.short.7}%
\end{align}
Taking reciprocals on both sides of (\ref{pf.thm.rect.antip.short.7}), we
obtain%
\begin{equation}
\overline{m_{\ell-k}^{\sim}}=\left(  \overline{s_{\ell-k}^{\sim}}%
+\overline{t_{\ell-k}^{\sim}}\right)  \cdot m_{\ell-k-1}^{\sim}\cdot
\overline{u_{\ell-k-1}^{\sim}+x_{\ell-k-1}^{\sim}}.
\label{pf.thm.rect.antip.short.7rec}%
\end{equation}
In view of the five equations (\ref{pf.thm.rect.antip.short.4}%
)--(\ref{pf.thm.rect.antip.short.5b}), we see that (up to the $a$ and $b$
factors, which cancel out) each label appearing in
(\ref{pf.thm.rect.antip.short.6}) is the inverse of the corresponding label in
(\ref{pf.thm.rect.antip.short.7rec}), except for $x_{\ell}$ and $x_{\ell
-k-1}^{\sim}$. Since the equations (\ref{pf.thm.rect.antip.short.6}) and
(\ref{pf.thm.rect.antip.short.7rec}) can be uniquely solved for $x_{\ell}$ and
$x_{\ell-k-1}^{\sim}$, we conclude that $x_{\ell}$ must be $a\cdot
\overline{x_{\ell-k-1}^{\sim}}\cdot b$. This completes the induction step,
proving (\ref{pf.thm.rect.antip.restate1}) and with it \cref{thm.rect.antip}. Recalling Step 1, we have thus proved \cref{thm.rect.ord} too.

\section{Other posets}

The rectangles $P=\left[  p\right]  \times\left[  q\right]  $ may be the
simplest family of posets for which noncommutative birational rowmotion
exhibits a (twisted) periodicity.\footnote{Apart from the chains $\left[
m\right]  $, which are just $m\times1$-rectangles, and from the antichains,
which are trivial to analyze.} In the commutative case, the \textquotedblleft%
\emph{skeletal posets}\textquotedblright\ of \cite{GrRo16} are even
simpler, but their periodicity does not extend to the noncommutative case. For
example, if $P$ is the $4$-element \textquotedblleft claw\textquotedblright%
\ poset $%
%TCIMACRO{\TeXButton{tikz burn in hell}{\scalebox{0.2}{
%\begin{tikzpicture}
%\node(x) at (0, 0) {};
%\node(y1) at (-1, 1.5) {} edge[very thick] (x);
%\node(y2) at (0, 1.5) {} edge[very thick] (x);
%\node(y3) at (1, 1.5) {} edge[very thick] (x);
%\filldraw[fill=black] (x) circle (0.2);
%\filldraw[fill=black] (y1) circle (0.2);
%\filldraw[fill=black] (y2) circle (0.2);
%\filldraw[fill=black] (y3) circle (0.2);
%\end{tikzpicture}
%}}}%
%BeginExpansion
\scalebox{0.2}{
\begin{tikzpicture}
\node(x) at (0, 0) {};
\node(y1) at (-1, 1.5) {} edge[very thick] (x);
\node(y2) at (0, 1.5) {} edge[very thick] (x);
\node(y3) at (1, 1.5) {} edge[very thick] (x);
\filldraw[fill=black] (x) circle (0.2);
\filldraw[fill=black] (y1) circle (0.2);
\filldraw[fill=black] (y2) circle (0.2);
\filldraw[fill=black] (y3) circle (0.2);
\end{tikzpicture}
}%
%EndExpansion
$, then $R^{6}=\operatorname*{id}$ for commutative $\mathbb{K}$, but no
periodicity holds for noncommutative $\mathbb{K}$ (not even \textquotedblleft
twisted\textquotedblright; see \cite[\S 13.4]{main} for details).
Experimentally, things look better for \textquotedblleft
Lie-theoretical\textquotedblright\ families of posets, although nothing has
been proven so far.  The family of \emph{type-A positive root posets}, also known as
\emph{triangles }$\Delta\left(  p\right)  $, may be within reach. 

% One family of posets $P$ whose periodicity we appear close
% to proving is that of the \emph{type-A positive root posets}, also known as
% \emph{triangles }$\Delta\left(  p\right)  $.

We can easily define these triangles as pieces of a $p\times p$-rectangle.
Indeed, let $p$ be a positive integer. Define two subsets $\Delta\left(
p\right)  $ and $\nabla\left(  p\right)  $ of the $p\times p$-rectangle
$\left[  p\right]  \times\left[  p\right]  $ by%
\begin{align*}
\Delta\left(  p\right)   &  =\left\{  \left(  i,j\right)  \in\left[  p\right]
\times\left[  p\right]  \ \mid\ i+j>p+1\right\}  ;\\
\nabla\left(  p\right)   &  =\left\{  \left(  i,j\right)  \in\left[  p\right]
\times\left[  p\right]  \ \mid\ i+j<p+1\right\}  .
\end{align*}
These two subsets $\Delta\left(  p\right)  $ and $\nabla\left(  p\right)  $
inherit partial orders from $\left[  p\right]  \times\left[  p\right]  $. Now,
we claim:

\begin{conjecture}
[Periodicity for $\Delta\left(  p\right)  $ and $\nabla\left(  p\right)  $]Let
$p\geq2$ be an integer. Assume that $P$ is the poset $\Delta\left(  p\right)
$ or $\nabla\left(  p\right)  $. Let $f\in\mathbb{K}^{\widehat{P}}$ be a
$\mathbb{K}$-labelling such that $R^{p}f$ is well-defined. Let $a=f\left(
0\right)  $ and $b=f\left(  1\right)  $. Let $x\in\widehat{P}$. Let
$x^{\prime}\in\widehat{P}$ be defined as follows: If $x=\left(  i,j\right)  $,
then we set $x^{\prime}:=\left(  j,i\right)  $; otherwise (i.e., if $x=0$ or
$x=1$), we set $x^{\prime}:=x$. Then,
\[
\left(  R^{p}f\right)  \left(  x\right)  =a\overline{b}\cdot f\left(
x^{\prime}\right)  \cdot\overline{a}b.
\]

\end{conjecture}

\begin{verlong}
\begin{proof}
[Proof attempt.]The idea (as in the commutative case \cite[\S 18]%
{bir-row-arxiv}) is to extend $f$ to a labelling $\widehat{f}$ of the entire
rectangle $\left[  p\right]  \times\left[  p\right]  $. The extension should
be performed in an $R$-equivariant way, i.e., in such a way that
$\widehat{Rf}=R\widehat{f}$. If this can be done, then properties of $f$ will
follow from appropriate properties of $\widehat{f}$.

To define $\widehat{f}$, we set $\widehat{f}\left(  v\right)  =%
\begin{cases}
\overline{\left(  R^{p-i-j+1}f\right)  \left(  v^{\prime}\right)  }, &
\text{if }v\in\nabla\left(  p\right)  ;\\
x^{\left[  v\text{ is even}\right]  }, & \text{if }v\in\operatorname*{Eq}%
\left(  p\right)  ;\\
f\left(  v\right)  , & \text{if }v\in\Delta\left(  p\right)
\end{cases}
$ for all $v\in\left[  p\right]  \times\left[  p\right]  $, where
$\operatorname*{Eq}\left(  p\right)  $ is the middle layer of $\left[
p\right]  \times\left[  p\right]  $ (that is, the set of all $\left(
i,k\right)  \in\left[  p\right]  \times\left[  p\right]  $ satisfying
$i+k=p+1$), and where $v^{\prime}$ is the reflection of $v$ across this middle
layer. And we call a pair $v\in\operatorname*{Eq}\left(  p\right)  $
\emph{even} if its first entry is even. What is $x$ ? Well, we take the ring
$\mathbb{K}\left[  x^{-1},x\right]  ]$ of Laurent series in $x$ over
$\mathbb{K}$, and then all of $x$, $1+x$ and $1+x^{-1}$ are invertible.

OLD Sketch the proof: To obtain periodicity for the Delta or Nabla triangle,
fold as in the commutative case, but fill the middle layer not with
1,1,1,...,1 but rather with 1,x,1,x,1,x,... (alternatingly), where x is an
indeterminate. To ensure that everything we need to invert is invertible, we
have to work in the ring of Laurent series in x over our base ring. Then, both
x and 1+x and $1+x^{-1}$ are invertible.

TODO! Check that this really works!
\end{proof}
\end{verlong}

Further posets $P$ for which $R$ is conjecturally periodic (with a twist) are
\textquotedblleft right-half triangles\textquotedblright\ \cite[\S 13.2]{main}
and trapezoids \cite[\S 13.3]{main}.

\section{General posets}

As already mentioned, periodicity and reciprocity are features of certain
classes of posets, even though the complete identification of these classes is
still unsolved even at the piecewise-linear level\footnote{See
\cite[Conjecture 5.7]{Hopkins2020dyn} for a sequence of posets $V\left(
n\right)  $ for which $\rho_{\mathcal{P}}$ is conjectured to have finite
order, but $R$ seems to have infinite order even when $\mathbb{K}$ is
commutative.}. Nevertheless, the $\left(  i,j\right)  =\left(  1,1\right)  $
case of \cref{thm.rect.antip} can be generalized to arbitrary posets in
the following form (\cite[\S 14]{main}):

\begin{proposition}
\label{prop.R.bottom-top}Let $P$ be any finite poset. Let $f\in\mathbb{K}%
^{\widehat{P}}$ be a labeling of $P$ such that $Rf$ is well-defined and all
labels $\left(  Rf\right)  \left(  u\right)  $ are invertible. Let $a=f\left(
0\right)  $ and $b=f\left(  1\right)  $. Then,%
\[
b\cdot\sum_{\substack{u\in\widehat{P};\\u\gtrdot0}}\overline{\left(
Rf\right)  \left(  u\right)  }\cdot a=\sum_{\substack{u\in\widehat{P}%
;\\u\lessdot1}}f\left(  u\right)  .
\]

\end{proposition}

\begin{verlong}
\begin{proof}
Using the notations of \cref{sec.proof} (and omitting the
\textquotedblleft$u\in\widehat{P}$\textquotedblright\ under the summation
signs), our claim becomes $b\cdot\sum\limits_{u\gtrdot0}\overline{u_{1}}\cdot
a=\sum\limits_{u\lessdot1}u_{0}$. But this follows from%
\begin{align*}
b\cdot\sum_{u\gtrdot0}\overline{u_{1}}\cdot a &
\overset{\text{(\ref{eq.thm.rect.path.up})}}{=}b\cdot\sum_{u\gtrdot0}%
\overline{\overline{\upslack_{1}^{1\rightarrow u}}\cdot b}\cdot a=\sum
_{u\gtrdot0}\upslack_{1}^{1\rightarrow u}\cdot a=\upslack_{1}^{1\rightarrow
0}\cdot a\overset{\text{\cref{prop.rect.trans}}}{=}\downslack_{0}%
^{1\rightarrow0}\cdot a\\
&  =\sum_{u\lessdot1}\downslack_{0}^{u\rightarrow0}\cdot
a\overset{\text{(\ref{eq.thm.rect.path.down})}}{=}\sum_{u\lessdot1}%
u_{0}.\qquad\qedhere
\end{align*}

\end{proof}
\end{verlong}

\begin{noncompile}
If $P=\left[  p\right]  \times\left[  q\right]  $, then this is saying that
$b\cdot\overline{\left(  1,1\right)  _{1}}\cdot a=\left(  p,q\right)  _{0}$
(using the notation (\ref{eq.xl=})), which follows from
(\ref{eq.thm.rect.path.up}), (\ref{eq.thm.rect.path.down}) and \cref{prop.rect.trans}. It is not hard to adapt this argument to the general
case, even though there will usually be several $v\in\widehat{P}$ satisfying
$v\gtrdot0$ (resp. $v\lessdot1$).
\end{noncompile}

This adds to the (so far rather meager) list of properties of $R$ that have
been observed for general posets $P$. (The rest of the list are results by
Joseph and Roby \cite[\S 5]{JosRob20} that connect $R$ with \emph{birational
antichain rowmotion} and the \emph{birational toggle group}.)

Note that \cref{prop.R.bottom-top} can be used to generalize the
w-tuple periodicity \cite[Proposition 37]{GrRo16} to the case of
noncommutative $\mathbb{K}$ (again with a twist).

\begin{verlong}
A slightly neater result is obtained for \emph{graded posets} -- i.e., posets
$P$ such that all maximal chains $0=v_{0}\lessdot v_{1}\lessdot v_{2}%
\lessdot\cdots\lessdot v_{m}=1$ in $\widehat{P}$ have the same length. (There
are some other definitions of this concept as well.) {\color{violet}
\itshape
TWR: Can we do this more simply, e.g., just assume $\widehat{P}$ is ranked in
the very standard sense? I also find it hard to quickly recognize
\textquotedblleft deg\textquotedblright\ as an operator when it appears inline
surrounded by text in the same font.} When $\mathbb{K}$ is commutative, some
general properties of $R$ for graded posets $P$ have been shown in
\cite[\S 3--\S 7]{GrRo16}. Generalizing to noncommutative $\mathbb{K}$
moves us to a more barren landscape; however, at least one of the results of
\cite[\S 4]{bir-row-arxiv} remains valid.

To state this, we fix an $n$-graded poset $P$. This means a poset $P$ for
which there exists a surjective map $\deg:\widehat{P}\rightarrow\left\{
0,1,\ldots,n+1\right\}  $ such that $\deg0=0$ and $\deg1=n+1$ and such that
\[
\deg u=\deg v+1\ \ \ \ \ \ \ \ \ \ \text{for any }u,v\in\widehat{P}\text{
satisfying }u\gtrdot v.
\]
(This is equivalent to requiring that any maximal chain $0=v_{0}\lessdot
v_{1}\lessdot v_{2}\lessdot\cdots\lessdot v_{m}=1$ in $\widehat{P}$ has length
$m=n+1$; however, we need the map $\deg$ itself.) The map $\deg$ is unique in
this case, and is called the \emph{degree map} of $P$. For instance, the
rectangle $\left[  p\right]  \times\left[  q\right]  $ is $\left(
p+q-1\right)  $-graded, with $\deg\left(  i,j\right)  =i+j-1$.

Now, for each $\mathbb{K}$-labelling $f$ of $P$ and each $i\in\left\{
0,1,\ldots,n\right\}  $, we set%
\[
\mathbf{w}_{i}\left(  f\right)  :=\sum f\left(  x\right)  \overline{f\left(
y\right)  },
\]
where the sum ranges over all pairs $\left(  x,y\right)  $ of elements of
$\widehat{P}$ that satisfy $x\lessdot y$ and $\deg x=i$ (and therefore $\deg
y=i+1$). Then:

\begin{theorem}
\label{thm.wtup.cyc}Let $f$ be a labelling of $P$. Let $a=f\left(  0\right)  $
and $b=f\left(  1\right)  $. Then:

\textbf{(a)} We have $\mathbf{w}_{0}\left(  Rf\right)  =a\overline{b}%
\cdot\mathbf{w}_{n}\left(  f\right)  \cdot b\overline{a}$.

\textbf{(b)} We have $\mathbf{w}_{i}\left(  Rf\right)  =\mathbf{w}%
_{i-1}\left(  f\right)  $ for each $i>0$.
\end{theorem}

\begin{proof}
\textbf{(a)} follows from \cref{prop.R.bottom-top}.

\textbf{(b)} follows easily from the definition of $R$.
\end{proof}

Note that the right hand side in \cref{thm.wtup.cyc} ends with a
$b\overline{a}$ factor, not with an $\overline{a}b$ factor as in
(\ref{eq.thm.rect.ord.claim}).
\end{verlong}

\section{The semiring case}

The definition of $R$ is \emph{subtraction-free}: It involves only addition,
multiplication and division in $\mathbb{K}$. Thus, we can replace $\mathbb{K}$
by an arbitrary \emph{semiring} (i.e., a \textquotedblleft ring without
additive inverses\textquotedblright). It is thus natural to wonder:

\begin{question}
\label{quest.semiring}Do \cref{thm.rect.ord,thm.rect.antip}
still hold if $\mathbb{K}$ is only a semiring?
\end{question}

This question is subtler than it may seem. When $\mathbb{K}$ is commutative,
the answer is positive for general reasons: A polynomial identity true over
all commutative rings is automatically true over all commutative semirings.
However, such automatic transfer principles do not exist in the noncommutative
realm. Case in point: If $\mathbb{K}$ is a ring, then the identity
$a\cdot\overline{a+b}\cdot b=b\cdot\overline{a+b}\cdot a$ holds for all
$a,b\in\mathbb{K}$ for which $a+b$ is invertible; but this is no longer true
if $\mathbb{K}$ is merely a semiring (David Speyer, MathOverflow \#401273). A
negative answer to \cref{quest.semiring} would thus not be entirely unexpected.

The proofs of \cref{thm.rect.ord,thm.rect.antip} sketched
above rely on subtraction in Step 6, so they cannot be applied to semirings
unless $\min\left\{  p,q\right\}  \leq2$ (in which case there is nothing to
subtract). We suspect that the smallest \textquotedblleft
interesting\textquotedblright\ case is $\left[  3\right]  \times\left[
3\right]  $ for \cref{thm.rect.antip} and $\left[  3\right]
\times\left[  4\right]  $ for \cref{thm.rect.ord}. Noncommutative
semirings that cannot be embedded in rings yet have many multiplicative
inverses do exist\footnote{One such semiring consists of all weakly increasing
functions from $\mathbb{R}$ to $\mathbb{R}$. Its addition is pointwise
maximum, and its multiplication is composition.}
% Every continuous strictly
%increasing function from $\mathbb{R}$ to $\mathbb{R}$ is then an invertible
%element of this semiring.}
, but automated computation inside them is not
straightforward, which complicates experimentation. We nevertheless believe
that progress can be made in this direction.
\medskip

\acknowledgements{We are greatly indebted to the Mathematisches Forschungsinstitut Oberwolfach
for hosting us for three weeks of  Summer 2021, during which much of this paper was
conceived.
We are also grateful to Banff International Research Station for hosting a
hybrid workshop on dynamical algebraic combinatorics in November 2021, where
these results were first presented.
We further acknowledge our appreciation of Marcelo Aguiar,
Michael Joseph, Tim Campion, Max Glick, Maxim
Kontsevich, Gregg Musiker, Pace Nielsen, James Propp, Pasha Pylyavskyy, Bruce
Sagan, Roland Speicher, David Speyer, Hugh Thomas, and Jurij Volcic for
useful advice and conversations.}

\printbibliography

\end{document}
