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\begin{document}

\title{\bf Iterative properties of birational rowmotion I: generalities and skeletal posets}
\author{Darij Grinberg\thanks{Supported by NSF grant 1001905.}\\{\small Department of Mathematics}\\[-0.8ex] {\small Massachusetts Institute of Technology}\\[-0.8ex] {\small Massachusetts, U.S.A.}\\{\small \texttt{darijgrinberg@gmail.com}}\\
\and Tom Roby\thanks{Supported by NSF grant 1001905.}\\{\small Department of Mathematics}\\[-0.8ex] {\small University of Connecticut}\\[-0.8ex] {\small Connecticut, U.S.A.}\\{\small \texttt{tom.roby@uconn.edu }}}


\date{\dateline{May 1, 2014}{Feb 7, 2016}{Feb 19, 2016}\\
{\small Mathematics Subject Classifications: 06A07, 05E99}}
\maketitle

\begin{abstract}
We study a birational map associated to any finite poset $P$. This map is a
far-reaching generalization (found by Einstein and Propp) of classical
rowmotion, which is a certain permutation of the set of order ideals of $P$.
Classical rowmotion has been studied by various authors (Fon-der-Flaass,
Cameron, Brouwer, Schrijver, Striker, Williams and many more) under different
guises (Striker-Williams promotion and Panyushev complementation are two
examples of maps equivalent to it). In contrast, birational rowmotion is new
and has yet to reveal several of its mysteries. In this paper, we set up the
tools for analyzing the properties of iterates of this map, and prove that
it has finite order for a certain class of posets which we call ``skeletal''.
Roughly speaking, these are graded posets constructed from one-element posets by
repeated disjoint union and ``grafting onto an antichain''; in particular,
any forest having its leaves all on the same rank is such a poset.
We also make a parallel analysis of classical rowmotion on this kind of posets,
and prove that the order in this case equals the order of birational rowmotion.

\bigskip\noindent\textbf{Keywords:} rowmotion; posets; order ideals;
trees; graded posets; tropicalization.

\end{abstract}

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\section{Introduction}

The present paper, and its continuation \cite{grinberg-roby-part2},
had originally been intended as companion papers to David
Einstein's and James Propp's work \cite{einstein-propp, einstein-propp-fpsac}, which
introduced piecewise-linear and birational rowmotion as extensions of the classical
concept of rowmotion on order ideals. While the present paper is
mathematically self-contained, it provides only a modicum of motivation and
applications for the results it discusses. For the latter, the reader
may consult \cite{einstein-propp}.

Let $P$ be a finite poset, and $J\left(  P\right)  $ the set of the order
ideals\footnote{An \textit{order ideal} of a poset $P$ is a subset $S$ of $P$
such that every $s\in S$ and $p\in P$ with $p\leq s$ satisfy $p\in S$.} of
$P$. \textit{Rowmotion} is a classical map $J\left(  P\right)  \rightarrow
J\left(  P\right)  $ which can be defined in various ways, one of which
is as follows: For every $v\in P$, let $\mathbf{t}_{v}:J\left(
P\right)  \rightarrow J\left(  P\right)  $ be the map sending every order
ideal $S\in J\left(  P\right)  $ to 
$$\left\{
\begin{array}
[c]{l}%
S\cup\left\{  v\right\}  \text{, if }v\notin S\text{ and }S\cup\left\{
v\right\}  \in J\left(  P\right)  ;\\
S\setminus\left\{  v\right\}  \text{, if }v\in S\text{ and }S\setminus\left\{
v\right\}  \in J\left(  P\right)  ;\\
S\text{, otherwise.}%
\end{array}
\right.  $$ These maps $\mathbf{t}_{v}$ are called \textit{classical
toggles}\footnote{or just \textit{toggles} in literature which doesn't occupy
itself with birational rowmotion}, since all they do is ``toggle'' an element
into or out of an order ideal. Let $\left(  v_{1},v_{2},\ldots,v_{m}\right)  $ be
a linear extension of $P$ (see Definition \ref{def.linext} for the meaning of
this). Then, (classical) rowmotion is defined as the composition
$\mathbf{t}_{v_{1}}\circ\mathbf{t}_{v_{2}}\circ\ldots\circ\mathbf{t}_{v_{m}}$
(which turns out to be independent of the choice of linear
extension $\left(  v_{1},v_{2},\ldots,v_{m}\right)  $). This rowmotion map has
been studied from various perspectives; in particular, it is
isomorphic\footnote{By this,
we mean that there exists a bijection $\phi$ from $J\left(P\right)$ to the
set of all antichains of $P$ such that rowmotion is $\phi^{-1} \circ f
\circ \phi$.} to the
map $f$ of Fon-der-Flaass \cite{fon-der-flaass}\footnote{Indeed, let
$\mathcal{A}\left(  P\right)  $ denote the set of all antichains of $P$. Then,
the map $J\left(  P\right)  \rightarrow\mathcal{A}\left(  P\right)  $ which
sends every order ideal $I\in J\left(  P\right)  $ to the antichain of the
maximal elements of $I$ is a bijection which intertwines rowmotion and
Fon-der-Flaass' map $f$.}, the map $F^{-1}$ of Brouwer and Schrijver
\cite{brouwer-schrijver}, and the map $f^{-1}$ of Cameron and Fon-der-Flaass
\cite{cameron-fon-der-flaass}\footnote{This time, the intertwining bijection
from rowmotion to the map $f^{-1}$ of \cite{cameron-fon-der-flaass} is given
by mapping every order ideal $I$ to its indicator function. This is a
bijection from $J\left(  P\right)  $ to the set of Boolean monotonic functions
$P\rightarrow\left\{  0,1\right\}  $.}. More recently, it has been studied
(and christened ``rowmotion'') in Striker and Williams \cite{striker-williams}%
, where further sources and context are also given. Since so much has already
been said about this rowmotion map, we will only briefly touch on its
properties in Section \ref{sect.classicalr}, while most of this paper will be
spent studying a much more general construction.

Among the questions that have been posed about rowmotion, the most prevalent
was probably that of its order: While it clearly has finite order (being a
bijective map from the finite set $J\left(  P\right)  $ to itself), it turns
out to have a much smaller order than one would naively expect when the
poset $P$ has certain ``special'' forms (e.g., a rectangle, a root poset, a
product of a rectangle with a $2$-chain, or -- apparently first considered in
this paper -- a forest). Most strikingly, when $P$ is the rectangle $\left[
p\right]  \times\left[  q\right]  $, then the
\mbox{$\left(  p+q\right)  $-th power} of the rowmotion operator is the
identity map. This is proven in \cite[Theorem 3.6]{brouwer-schrijver} and
\cite[Theorem 2]{fon-der-flaass}, and a proof can also be constructed from the
ideas given in \cite[\S 3.3.1]{propp-roby}. In Section
\ref{sect.classicalr} we give a simple algorithm to find the order of rowmotion
on graded forests and similar posets.

In \cite{einstein-propp}, David Einstein and James Propp
(inspired by
work of Arkady Berenstein and Anatol Kirillov)
have lifted the rowmotion map from the set $J\left(
P\right)  $ of order ideals to the progressively more general setups of:

\textbf{(a)} the order polytope $\mathcal{O}\left(  P\right)  $ of the poset
$P$ (as defined in \cite[Example 4.6.17]{stanley-ec1} or \cite[Definition
1.1]{stanley-polytopes}), and

\textbf{(b)} even more generally, the affine variety of $\mathbb{K}%
$-labellings of $P$ for $\mathbb{K}$ an arbitrary infinite field.

In case \textbf{(a)}, order ideals of $P$ are replaced by points in the order
polytope $\mathcal{O}\left(  P\right)  $, and the role of the map
$\mathbf{t}_{v}$ (for a given $v\in P$) is assumed by the map which reflects
the $v$-coordinate of a point in $\mathcal{O}\left(  P\right)  $ around the
midpoint of the interval of all values it could take without the point leaving
$\mathcal{O}\left(  P\right)  $ (while all other coordinates are considered
fixed). The operation of ``piecewise linear'' rowmotion is still defined as
the composition of these reflection maps in the same way as rowmotion is the
composition of the toggles $\mathbf{t}_{v}$. This ``piecewise linear''
rowmotion extends (interpolates, even) classical rowmotion, as order ideals
correspond to the vertices of the order polytope $\mathcal{O}\left(  P\right)
$ (see \cite[Corollary 1.3]{stanley-polytopes}). We will not study case
\textbf{(a)} here, since all of the results we could find in this case can be
obtained by tropicalization from similar results for case \textbf{(b)}.

In case \textbf{(b)}, instead of order ideals of $P$ one considers maps from
the poset $\widehat{P}:=\left\{  0\right\}  \oplus P\oplus\left\{  1\right\}
$ (where $\oplus$ stands for the ordinal sum; see Definition~\ref{def.Phat} %
% \footnote{More explicitly,
% $\widehat{P}$ is the poset obtained by adding a new element $0$ to $P$, which
% is set to be lower than every element of $P$, and adding a new element $1$ to
% $P$, which is set to be higher than every element of $P$ (and $0$).}
) to a
given infinite field $\mathbb{K}$ (or, to speak more graphically, of all
labellings of the elements of $P$ by elements of $\mathbb{K}$, along with two
additional labels \textquotedblleft at the very bottom\textquotedblright\ and
\textquotedblleft at the very top\textquotedblright). The maps $\mathbf{t}%
_{v}$ are then replaced by certain birational maps which we call
\textit{birational }$v$\textit{-toggles} (Definition \ref{def.Tv}); the
resulting composition is called \textit{birational rowmotion} and denoted by
$R$. By a careful limiting procedure (the tropical limit), we can
\textquotedblleft degenerate\textquotedblright\ $R$ to the \textquotedblleft
piecewise linear\textquotedblright\ rowmotion of case \textbf{(a)}, and thus
it can be seen as an even higher generalization of classical rowmotion. We
refer to the body of this paper for precise definitions of these maps. Note
that birational $v$-toggles (but not birational rowmotion) in the case of a
rectangle poset have also appeared in \cite[(3.5)]{osz-geoRSK}, but
(apparently) have not been composed there in a way that yields birational rowmotion.

As in the case of classical rowmotion on $J\left(  P\right)  $, the most
interesting question is the order of this map $R$, which in general no longer
has an obvious reason to be finite (since the affine variety of $\mathbb{K}%
$-labellings is not a finite set like $J\left(  P\right)  $). Indeed, for some
posets $P$ this order is infinite (examples of these can be found in
\cite[\S 12]{grinberg-roby-part2}). In this paper we will prove the following facts:

\begin{itemize}
\item Birational rowmotion (i.e., the map $R$) on any graded poset (in the
meaning of this word introduced in Definition \ref{def.graded.graded}) has a
very simple effect (namely, cyclic shifting) on the so-called ``w-tuple'' of a
labelling (a rather simple fingerprint of the labelling). This does not mean
$R$ itself has finite order (but turns out to be crucial in proving this in
several cases).

\item Birational rowmotion on graded forests and, slightly more generally,
skeletal posets (Definition \ref{def.skeletal}) has finite order (which can be
bounded from above by an iterative lcm, and also easily computed
algorithmically). Moreover, its order in these cases coincides with the order
of classical rowmotion (Section \ref{sect.classicalr}).
\end{itemize}

The following results will be proven in the second paper \cite{grinberg-roby-part2}:
\begin{itemize}
\item Birational rowmotion on a $p\times q$-rectangle has order $p+q$ and
satisfies a further symmetry property.
These results were originally conjectured by James Propp and the second
author, and can be used as an alternative route to certain properties of
(Sch\"{u}tzenberger's) promotion map on semistandard Young tableaux.

\item Birational rowmotion on certain triangle-shaped posets -- more
precisely, posets which can be obtained from the $p\times p$-square by
cutting it along either diagonal, or along both diagonals at once -- also
has finite order (usually $p$ or $2p$), except for the case when we cut
along both diagonals and $p$ is even (we conjecture that the order is $p$
in this case as well).
\end{itemize}

An extended (12-page) abstract \cite{grinberg-roby} of this paper and
\cite{grinberg-roby-part2} has been published in the proceedings of the FPSAC 2014
conference. A longer version for readers who would like fuller details and remarks is
available on the arXiv~\cite{grinberg-roby-arXiv}.  

\subsection{Leitfaden}

\begin{minipage}[t]{0.68\textwidth}
The Hasse diagram at right shows how the sections of this paper depend upon
each other.
A section $n$ depends substantially on a section $m$ if and only if $m>n$ in
the poset whose Hasse diagram is depicted at right. Only substantial dependencies
are shown; dependencies upon definitions do not count as substantial (e.g.,
Section \ref{sect.opposite} depends on Definition \ref{def.ord}, but this does not make it
substantially dependent on Section \ref{sect.ord}), and dependencies which are
only used in proving inessential claims do not count.

No section of this paper depends on the Introduction.
\end{minipage}
\begin{minipage}[t]{0.3\textwidth}
\vskip -40pt
\[
\xymatrixrowsep{0.68pc}
\xymatrixcolsep{0.68pc}
%\kern -50pt
\xymatrix{
& & & 1 \ar@{-}[d] & & \\
& & & 2 \ar@{-}[dddlll] \ar@{-}[d] & & \\
& & & 3 \ar@{-}[dl] \ar@{-}[d] & & \\
& & 4 \ar@{-}[ddl] & 5 \ar@{-}[dl] & & \\
8 \ar@{-}[ddr] & & 6 \ar@{-}[dl] & & & \\
& 7 \ar@{-}[d] & & & & \\
& 9 \ar@{-}[d] & & & & \\
& 10 & & & & \\
}
\]
\end{minipage}

\section{Linear extensions of posets}

This first section serves to introduce some general notions concerning posets
and their linear extensions. In particular, we highlight that the set of
linear extensions of any finite poset is non-empty and connected by a simple
equivalence relation (Proposition \ref{prop.linext.transitive}). This will be
used in subsequent sections for defining the basic maps that we consider
throughout the paper.

We start by defining general notation related to posets:

\begin{definition}
Let $P$ be a poset. Let $u\in P$ and $v\in P$. In this definition, we will use
$\leq$, $<$, $\geq$ and $>$ to denote the lesser-or-equal relation, the lesser
relation, the greater-or-equal relation and the greater relation,
respectively, of the poset $P$.

\textbf{(a)} The elements $u$ and $v$ of $P$ are said to be
\textit{incomparable} if we have neither $u\leq v$ nor $u\geq v$.

\textbf{(b)} We write $u\lessdot v$ if we have $u<v$ and there is no $w\in P$
such that $u<w<v$. (One often says that ``$u$ is covered by $v$'' to signify
that $u\lessdot v$.)

\textbf{(c)} We write $u\gtrdot v$ if we have $u>v$ and there is no $w\in P$
such that $u>w>v$. (Thus, $u\gtrdot v$ holds if and only if $v\lessdot u$.)
(One often says that ``$u$ covers $v$'' to signify that $u\gtrdot v$.)

\textbf{(d)} An element $u$ of $P$ is called \textit{maximal} if every $v\in
P$ satisfying $v\geq u$ satisfies $v=u$. It is easy to see that every nonempty
finite poset has at least one maximal element.

\textbf{(e)} An element $u$ of $P$ is called \textit{minimal} if every $v\in
P$ satisfying $v\leq u$ satisfies $v=u$. It is easy to see that every nonempty
finite poset has at least one minimal element.

When any of these notations becomes ambiguous because the elements involved
belong to several different posets simultaneously, we will disambiguate it by
adding the words \textquotedblleft in $P$\textquotedblright\ (where $P$ is the
poset which we want to use).     %%\footnotemark
\end{definition}


% \footnotetext{For instance, if $R$ denotes the poset $\mathbb{Z}$ endowed with
% the reverse of its usual order, then we say (for instance) that
% \textquotedblleft$1\lessdot0$ in $R$\textquotedblright\ rather than just
% \textquotedblleft$1\lessdot0$\textquotedblright.}


\begin{definition}
\label{def.linext}Let $P$ be a finite poset. A \textit{linear extension} of
$P$ will mean a list $\left(  v_{1},v_{2},\ldots,v_{m}\right)  $ of the elements
of $P$ such that every element of $P$ occurs exactly once in this list, and
such that any $i\in\left\{  1,2,\ldots,m\right\}  $ and $j\in\left\{
1,2,\ldots,m\right\}  $ satisfying $v_{i}<v_{j}$ in $P$ must satisfy $i<j$.
\end{definition}


% A brief remark on this definition is in order. Stanley, in \cite[one paragraph
% below the proof of Proposition 3.5.2]{stanley-ec1}, defines a linear extension
% of a poset $P$ as an order-preserving bijection from $P$ to the chain
% $\left\{  1,2,\ldots,\left|  P\right|  \right\}  $; this is equivalent to our
% definition (indeed, our linear extension $\left(  v_{1},v_{2},\ldots,v_{m}%
% \right)  $, whose length obviously is $m = \left|  P\right|  $, corresponds to
% the bijection $P\rightarrow\left\{  1,2,\ldots,\left|  P\right|  \right\}  $
% which sends each $v_{i}$ to $i$). Another widespread definition of a linear
% extension of $P$ is as a total order on $P$ compatible with the given order of
% the poset $P$; this is equivalent to our definition as well (the total order
% is the one defined by $v_{i}<v_{j}$ whenever $i<j$).



% Notice that if $\left(  v_{1},v_{2},\ldots,v_{m}\right)  $ is a linear extension
% of a nonempty finite poset $P$, then $v_{1}$ is a minimal element of $P$ and
% $v_{m}$ is a maximal element of $P$. The only linear extension of the empty
% poset $\varnothing$ is the empty list $\left(  {}\right)  $.

\begin{theorem}
\label{thm.linext.ex}Every finite poset $P$ has a linear extension.
\end{theorem}


Theorem \ref{thm.linext.ex} is a well-known fact, and can be proven, e.g., by
induction over $\left\vert P\right\vert $ (with the induction step consisting
of splitting off a maximal element $u$ of $P$ and appending it to a linear
extension of the residual poset $P\setminus\left\{  u\right\}  $).

The following proposition can be easily checked by the reader:

\begin{proposition}
\label{prop.linext.switch}Let $P$ be a finite poset. Let $\left(  v_{1}%
,v_{2},\ldots,v_{m}\right)  $ be a linear extension of $P$. Let $i\in\left\{
1,2,\ldots,m-1\right\}  $ be such that the elements $v_{i}$ and $v_{i+1}$ of $P$
are incomparable. Then $\left(  v_{1},v_{2},\ldots,v_{i-1},v_{i+1},v_{i}%
,v_{i+2},v_{i+3},\ldots,v_{m}\right)  $ (this is the tuple obtained from the
tuple $\left(  v_{1},v_{2},\ldots,v_{m}\right)  $ by interchanging the adjacent
entries $v_{i}$ and $v_{i+1}$) is a linear extension of $P$ as well.
\end{proposition}

\begin{definition}
\label{def.L(P)}Let $P$ be a finite poset. The set of all linear extensions of
$P$ will be called $\mathcal{L}\left(  P\right)  $. Thus, $\mathcal{L}\left(
P\right)  \neq\varnothing$ (by Theorem \ref{thm.linext.ex}).
\end{definition}

In our approach to birational rowmotion, we will use the following fact (which
is folklore and has applications in various contexts, including Young tableau theory):

\begin{proposition}
\label{prop.linext.transitive}Let $P$ be a finite poset. Let $\sim$ denote the
finest equivalence relation on $\mathcal{L}\left(  P\right)  $ with the
following property: For any linear extension $\left(  v_{1},v_{2}%
,\ldots,v_{m}\right)  $ of $P$ and any $i\in\left\{  1,2,\ldots,m-1\right\}  $ such
that the elements $v_{i}$ and $v_{i+1}$ of $P$ are incomparable, we have
$\left(  v_{1},v_{2},\ldots,v_{m}\right)  \sim\left(  v_{1},v_{2},\ldots,v_{i-1}%
,v_{i+1},v_{i},v_{i+2},v_{i+3},\ldots,v_{m}\right)  $ (noting that \newline$\left(
v_{1},v_{2},\ldots,v_{i-1},v_{i+1},v_{i},v_{i+2},v_{i+3},\ldots,v_{m}\right)  $ is
also a linear extension of $P$, because of Proposition
\ref{prop.linext.switch}).
%\footnote{More formally, $\sim$ is the reflexive and transitive
%closure of the symmetric relation described in Proposition~\ref{prop.linext.switch}.} 
% Here is a more formal way to restate this
% definition of $\sim$:
% \par
% We first introduce a binary relation $\equiv$ on the set $\mathcal{L}\left(
% P\right)  $ as follows: If $\mathbf{v}$ and $\mathbf{w}$ are two linear
% extensions of $P$, then we set $\mathbf{v}\equiv\mathbf{w}$ if and only if the
% list $\mathbf{w}$ can be obtained from the list $\mathbf{v}$ by interchanging
% two adjacent entries $v$ and $v^{\prime}$ which are incomparable in $P$. It is
% clear that this binary relation $\equiv$ is symmetric. Now, we
% define an equivalence relation $\sim$ on $\mathcal{L}\left(P\right)$
% as the reflexive and transitive closure of the binary relation
% $\equiv$.} Then, any two elements of $\mathcal{L}\left(  P\right)  $ are
% equivalent under the relation $\sim$.
\end{proposition}

This proposition is basic (it generalizes the fact that the symmetric
group $S_{n}$ is generated by the adjacent-element transpositions) and
classical, but a proof is hard to find in the literature. One proof is in
\cite[Proposition 4.1 (for the $\pi^{\prime}=\pi\tau_{j}$ case)]%
{ayyer-klee-schilling}; another is sketched in \cite[p. 79]{ruskey} and
presented in more detail in \cite[Lemma 1]{etienne}. We give a
fully-detailed proof in~\cite{grinberg-roby-arXiv}.
%, using the following lemma. 
% Once the lemma is proved by induction on $\ell$, Proposition~\ref{prop.linext.transitive} is
% easily proven by induction on $|P|$.
% 
% % For the
% % sake of completeness, we will prove it too. Our proof is based on the
% % following lemma (which is more or less a simple particular case of Proposition
% % \ref{prop.linext.transitive}):
% 
% \begin{lemma}
% \label{lem.linext.transitive}Let $P$ be a finite poset. Define the equivalence
% relation $\sim$ on $\mathcal{L}\left(  P\right)  $ as in Proposition
% \ref{prop.linext.transitive}. Let $a_{1}$, $a_{2}$, $\ldots$, $a_{k}$ be some
% elements of $P$. Let $b_{1}$, $b_{2}$, $\ldots$, $b_{\ell}$ be some further
% elements of $P$. Let $u$ be a maximal element of $P$. Assume that $\left(
% a_{1},a_{2},\ldots,a_{k},u,b_{1},b_{2},\ldots,b_{\ell}\right)  $ is a linear
% extension of $P$. Then, $\left(  a_{1},a_{2},\ldots,a_{k},b_{1},b_{2}%
%,\ldots,b_{\ell},u\right)  $ is a linear extension of $P$ satisfying $\left(
% a_{1},a_{2},\ldots,a_{k},u,b_{1},b_{2},\ldots,b_{\ell}\right)  \sim\left(
% a_{1},a_{2},\ldots,a_{k},b_{1},b_{2},\ldots,b_{\ell},u\right)  $.
% \end{lemma}
% 
% 
% \begin{proof}
% [Proof of Lemma \ref{lem.linext.transitive}.]We will show that
% every $i\in\left\{  0,1,\ldots,\ell\right\}  $ satisfies the following assertion:%
% \begin{equation}
% \left(
% \begin{array}
% [c]{c}%
% \text{The tuple }\left(  a_{1},a_{2},\ldots,a_{k},b_{1},b_{2},\ldots,b_{i}%
%,u,b_{i+1},b_{i+2},\ldots,b_{\ell}\right)  \text{ is a}\\
% \text{linear extension of }P\text{ satisfying}\\
% \left(  a_{1},a_{2},\ldots,a_{k},u,b_{1},b_{2},\ldots,b_{\ell}\right)  \sim\left(
% a_{1},a_{2},\ldots,a_{k},b_{1},b_{2},\ldots,b_{i},u,b_{i+1},b_{i+2},\ldots,b_{\ell
% }\right)
% \end{array}
% \right). \label{pf.linext.transitive.lem1}%
% \end{equation}


% \textit{Proof of (\ref{pf.linext.transitive.lem1}):} We will prove
% (\ref{pf.linext.transitive.lem1}) by induction over $i$:


% \textit{Induction base:} If $i=0$, then \newline$\left(  a_{1},a_{2}%
%,\ldots,a_{k},b_{1},b_{2},\ldots,b_{i},u,b_{i+1},b_{i+2},\ldots,b_{\ell}\right)
% =\left(  a_{1},a_{2},\ldots,a_{k},u,b_{1},b_{2},\ldots,b_{\ell}\right)  $. Hence,
% (\ref{pf.linext.transitive.lem1}) is a tautology for $i=0$, and the induction
% base is done.



% \textit{Induction step:} Let $I\in\left\{  1,2,\ldots,\ell\right\}  $. Assume
% that (\ref{pf.linext.transitive.lem1}) holds for $i=I-1$. We need to prove
% that (\ref{pf.linext.transitive.lem1}) holds for $i=I$.

% We have assumed that (\ref{pf.linext.transitive.lem1}) holds for $i=I-1$. In
% other words, the tuple \newline$\left(  a_{1},a_{2},\ldots,a_{k},b_{1}%
%,b_{2},\ldots,b_{I-1},u,b_{I-1+1},b_{I-1+2},\ldots,b_{\ell}\right)  $ is a linear
% extension of $P$ satisfying $\left(  a_{1},a_{2},\ldots,a_{k},u,b_{1}%
%,b_{2},\ldots,b_{\ell}\right)  \sim\left(  a_{1},a_{2},\ldots,a_{k},b_{1}%
%,b_{2},\ldots,b_{I-1},u,b_{I-1+1},b_{I-1+2},\ldots,b_{\ell}\right)  $.


% Denote the smaller relation of $P$ by $<$. Since the tuple $\left(
% a_{1},a_{2},\ldots,a_{k},u,b_{1},b_{2},\ldots,b_{\ell}\right)  $ is a linear
% extension of $P$, we cannot have $u\geq b_{I}$ (because $u$ appears strictly
% to the left of $b_{I}$ in this tuple). But we cannot have $u<b_{I}$ either
% (since $u$ is a maximal element of $P$). Thus, $u$ and $b_{I}$ are incomparable.




% Now,%
% \begin{align*}
% &  \left(  a_{1},a_{2},\ldots,a_{k},u,b_{1},b_{2},\ldots,b_{\ell}\right) \\
% &  \sim\left(  a_{1},a_{2},\ldots,a_{k},b_{1},b_{2},\ldots,b_{I-1},u,b_{I-1+1}%
%,b_{I-1+2},\ldots,b_{\ell}\right) \\
% &  =\left(  a_{1},a_{2},\ldots,a_{k},b_{1},b_{2},\ldots,b_{I-1},u,b_{I}%
%,b_{I+1},b_{I+2},\ldots,b_{\ell}\right) \\
% &  \sim\left(  a_{1},a_{2},\ldots,a_{k},b_{1},b_{2},\ldots,b_{I-1},b_{I}%
%,u,b_{I+1},b_{I+2},\ldots,b_{\ell}\right) \\
% &  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of the relation }%
% \sim\text{, since }u\text{ and }b_{I}\text{ are incomparable}\right) \\
% &  =\left(  a_{1},a_{2},\ldots,a_{k},b_{1},b_{2},\ldots,b_{I},u,b_{I+1}%
%,b_{I+2},\ldots,b_{\ell}\right).
% \end{align*}
% The proof of this equivalence also shows that its right hand side is a linear
% extension of $P$. Thus, (\ref{pf.linext.transitive.lem1}) holds for $i=I$.
% This completes the induction step, whence (\ref{pf.linext.transitive.lem1}) is proven.

% Lemma \ref{lem.linext.transitive} now follows by applying
% (\ref{pf.linext.transitive.lem1}) to $i=\ell$.
% \end{proof}


% \begin{proof}
% [Proof of Proposition \ref{prop.linext.transitive}.]We prove
% Proposition \ref{prop.linext.transitive} by induction over $\left\vert
% P\right\vert $. The induction base $\left\vert P\right\vert =0$ is trivial.
% For the induction step, let $N$ be a positive integer. Assume that Proposition
% \ref{prop.linext.transitive} is proven for all posets $P$ with $\left\vert
% P\right\vert =N-1$. Now, let $P$ be a poset with $\left\vert P\right\vert =N$.

% Let $\left(  v_{1},v_{2},\ldots,v_{N}\right)  $ and $\left(  w_{1},w_{2}%
%,\ldots,w_{N}\right)  $ be two elements of $\mathcal{L}\left(  P\right)  $. We
% are going to prove that $\left(  v_{1},v_{2},\ldots,v_{N}\right)  \sim\left(
% w_{1},w_{2},\ldots,w_{N}\right)  $.

% Let $u=v_{N}$. Then, $u$ is a maximal element of $P$ (since it comes last in
% the linear extension $\left(  v_{1},v_{2},\ldots,v_{N}\right)  $). Let $i$ be the
% index satisfying $w_{i}=u$.

% Consider the poset $P\setminus\left\{  u\right\}  $. This poset has size
% $\left\vert P\setminus\left\{  u\right\}  \right\vert =\underbrace{\left\vert
% P\right\vert }_{=N}-1=N-1$. Define a relation $\sim$ on $\mathcal{L}\left(
% P\setminus\left\{  u\right\}  \right)  $ in the same way as the relation
% $\sim$ on $\mathcal{L}\left(  P\right)  $ was defined. Recall that $u$ is a
% maximal element of $P$. Hence,%
% \begin{equation}
% \left(
% \begin{array}
% [c]{c}%
% \text{if }\left(  a_{1},a_{2},\ldots,a_{N-1}\right)  \text{ is a linear extension
% of }P\setminus\left\{  u\right\}  \text{, then}\\
% \left(  a_{1},a_{2},\ldots,a_{N-1},u\right)  \text{ is a linear extension of }P
% \end{array}
% \right). \label{pf.linext.transitive.0.short}%
% \end{equation}
% Moreover, just by recalling how the relations $\sim$ were defined, we can
% easily see that%
% \begin{equation}
% \left(
% \begin{array}
% [c]{c}%
% \text{if two linear extensions }\left(  a_{1},a_{2},\ldots,a_{N-1}\right)  \text{
% and }\left(  b_{1},b_{2},\ldots,b_{N-1}\right)  \text{ of }P\setminus\left\{
% u\right\} \\
% \text{satisfy }\left(  a_{1},a_{2},\ldots,a_{N-1}\right)  \sim\left(  b_{1}%
%,b_{2},\ldots,b_{N-1}\right)  \text{ in }\mathcal{L}\left(  P\setminus\left\{
% u\right\}  \right)  \text{, then}\\
% \left(  a_{1},a_{2},\ldots,a_{N-1},u\right)  \text{ and }\left(  b_{1}%
%,b_{2},\ldots,b_{N-1},u\right)  \text{ are two linear extensions}\\
% \text{of }P\text{ satisfying }\left(  a_{1},a_{2},\ldots,a_{N-1},u\right)
% \sim\left(  b_{1},b_{2},\ldots,b_{N-1},u\right)  \text{ in }\mathcal{L}\left(
% P\right)
% \end{array}
% \right)  \label{pf.linext.transitive.1.short}%
% \end{equation}
% (here, the fact that $\left(  a_{1},a_{2},\ldots,a_{N-1},u\right)  $ and $\left(
% b_{1},b_{2},\ldots,b_{N-1},u\right)  $ are linear extensions of $P$ follows from
% (\ref{pf.linext.transitive.0.short})).

% It is rather clear that $\left(  v_{1},v_{2},\ldots,v_{N-1}\right)  $ and
% $\left(  w_{1},w_{2},\ldots,w_{i-1},w_{i+1},w_{i+2},\ldots,w_{N}\right)  $ are two
% linear extensions of the poset $P\setminus\left\{  u\right\}  $ (since they
% are obtained from the linear extensions $\left(  v_{1},v_{2},\ldots,v_{N}\right)
% $ and $\left(  w_{1},w_{2},\ldots,w_{N}\right)  $ of $P$ by removing $u$). Since
% we can apply Proposition \ref{prop.linext.transitive} to this poset
% $P\setminus\left\{  u\right\}  $ in lieu of $P$ (by the induction hypothesis,
% since $\left\vert P\setminus\left\{  u\right\}  \right\vert =N-1$), we thus
% see that%
% \[
% \left(  v_{1},v_{2},\ldots,v_{N-1}\right)  \sim\left(  w_{1},w_{2},\ldots,w_{i-1}%
%,w_{i+1},w_{i+2},\ldots,w_{N}\right)
% \]
% in $\mathcal{L}\left(  P\setminus\left\{  u\right\}  \right)  $. By
% (\ref{pf.linext.transitive.1.short}), this yields that $\left(  v_{1}%
%,v_{2},\ldots,v_{N-1},u\right)  $ and \newline$\left(  w_{1},w_{2},\ldots,w_{i-1}%
%,w_{i+1},w_{i+2},\ldots,w_{N},u\right)  $ are two linear extensions of $P$
% satisfying
% \[
% \left(  v_{1},v_{2},\ldots,v_{N-1},u\right)  \sim\left(  w_{1},w_{2}%
%,\ldots,w_{i-1},w_{i+1},w_{i+2},\ldots,w_{N},u\right)
% \]
% in $\mathcal{L}\left(  P\right)  $.

% Now, we know that the tuple $\left(  w_{1},w_{2},\ldots,w_{N}\right)  $ is a
% linear extension of $P$. Since
% \begin{align*}
% &  \left(  w_{1},w_{2},\ldots,w_{N}\right) \\
% &  =\left(  w_{1},w_{2},\ldots,w_{i-1},\underbrace{w_{i}}_{=u},w_{i+1}%
%,w_{i+2},\ldots,w_{N}\right)  =\left(  w_{1},w_{2},\ldots,w_{i-1},u,w_{i+1}%
%,w_{i+2},\ldots,w_{N}\right),
% \end{align*}
% this rewrites as follows: The tuple $\left(  w_{1},w_{2},\ldots,w_{i-1}%
%,u,w_{i+1},w_{i+2},\ldots,w_{N}\right)  $ is a linear extension of $P$. Hence, we
% can apply Lemma \ref{lem.linext.transitive} to $k=i-1$, $\ell=N-i$,
% $a_{j}=w_{j}$ and $b_{j}=w_{i+j}$. As a result, we see that $\left(
% w_{1},w_{2},\ldots,w_{i-1},w_{i+1},w_{i+2},\ldots,w_{N},u\right)  $ is a linear
% extension of $P$ satisfying $\left(  w_{1},w_{2},\ldots,w_{i-1}%
%,u,w_{i+1},w_{i+2},\ldots,w_{N}\right)  \sim\left(  w_{1},w_{2},\ldots,w_{i-1}%
%,w_{i+1},w_{i+2},\ldots,w_{N},u\right)  $. Since the relation $\sim$ is symmetric
% (because $\sim$ is an equivalence relation), this yields%
% \[
% \left(  w_{1},w_{2},\ldots,w_{i-1},w_{i+1},w_{i+2},\ldots,w_{N},u\right)
% \sim\left(  w_{1},w_{2},\ldots,w_{i-1},u,w_{i+1},w_{i+2},\ldots,w_{N}\right).
% \]


% Altogether,%
% \begin{align*}
% \left(  v_{1},v_{2},\ldots,v_{N}\right)   &  =\left(  v_{1},v_{2},\ldots,v_{N-1}%
%,\underbrace{v_{N}}_{=u}\right)  =\left(  v_{1},v_{2},\ldots,v_{N-1},u\right) \\
% &  \sim\left(  w_{1},w_{2},\ldots,w_{i-1},w_{i+1},w_{i+2},\ldots,w_{N},u\right) \\
% &  \sim\left(  w_{1},w_{2},\ldots,w_{i-1},\underbrace{u}_{=w_{i}},w_{i+1}%
%,w_{i+2},\ldots,w_{N}\right) \\
% &  =\left(  w_{1},w_{2},\ldots,w_{i-1},w_{i},w_{i+1},w_{i+2},\ldots,w_{N}\right)
% =\left(  w_{1},w_{2},\ldots,w_{N}\right).
% \end{align*}
% We thus have shown that any two elements $\left(  v_{1},v_{2},\ldots,v_{N}%
% \right)  $ and $\left(  w_{1},w_{2},\ldots,w_{N}\right)  $ of $\mathcal{L}\left(
% P\right)  $ satisfy $\left(  v_{1},v_{2},\ldots,v_{N}\right)  \sim\left(
% w_{1},w_{2},\ldots,w_{N}\right)  $. In other words, Proposition
% \ref{prop.linext.transitive} is proven for $\left\vert P\right\vert =N$, so
% the induction step is complete, and Proposition \ref{prop.linext.transitive}
% is proven.
% \end{proof}


\section{Birational rowmotion}

In this section, we introduce the basic objects whose nature we will
investigate: labellings of a finite poset $P$ (by elements of a field) and a
birational map between them called ``birational rowmotion''. This map
generalizes (in a certain sense) the notion of ordinary rowmotion on the set
$J\left(  P\right)  $ of order ideals of $P$ to the vastly more general
setting of field-valued labellings. We will discuss the technical concerns
raised by the definitions, and provide an example and an alternative
description of birational rowmotion. A deeper study of birational rowmotion is
deferred to the following sections.

The concepts which we are going to define now go back to \cite{einstein-propp}
and earlier sources, and are often motivated there. The reader should be
warned that the notations used in \cite{einstein-propp} are not identical with
those used in the present paper (not to mention that \cite{einstein-propp} is
working over $\mathbb{R}_{+}$ rather than over fields as we do).

\medskip 
\noindent {\large \textbf{Notational conventions used throughout this paper:}}

Unless otherwise stated, we will consistently use the following notations. 

$\mathbb{N}$ denotes the set of all nonnegative integers.

$P$ will denote a finite poset (although a few definitions apply equally well to
infinite posets). In Sections~3--9, $P$ will denote an $n$-graded poset, where $n\in
\mathbb{N}$ is a nonnegative integer. (The meaning of this will be defined
in Section~3.)

$\mathbb{K}$ will denote a field, which we tacitly assume is either infinite or at least can
be enlarged when necessity arises. This assumption is needed in order to clarify the notions
of rational maps and generic elements of algebraic varieties over $\mathbb{K}$. We do not
require $\mathbb{K}$ to be algebraically closed.


\begin{definition}
\label{def.Phat}Let $P$ be a poset. Then, $\widehat{P}$ will denote the poset
% defined as follows: As a set, let $\widehat{P}$ be the disjoint union of the
% set $P$ with the two-element set $\left\{  0,1\right\}  $. The
% smaller-or-equal relation $\leq$ on $\widehat{P}$ will be given by%
whose ground set is the disjoint union $P\cupdot \{0,1 \}$, with partial order relation 
\[
\left(  a\leq b\right)  \Longleftrightarrow\left(  \text{either }\left(  a\in
P\text{ and }b\in P\text{ and }a\leq b\text{ in }P\right)  \text{ or
}a=0\text{ or }b=1\right)
\]
(where \textquotedblleft either/or\textquotedblright\ has a non-exclusive
meaning). Here and in the following, we regard the canonical injection of the
set $P$ into the disjoint union $\widehat{P}$ as an inclusion; thus, $P$
becomes a subposet of $\widehat{P}$. In the terminology of Stanley's
\cite[section 3.2]{stanley-ec1}, this poset $\widehat{P}$ is the ordinal sum
$\left\{  0\right\}  \oplus P\oplus\left\{  1\right\}  $.  Since the relation $<_{P}$ is a
restriction of $<_{\widehat{P}}$, we never need to distinguish these from one another.  
\end{definition}


% \begin{convention}
% Let $P$ is a finite poset, and let $u$ and $v$ be two elements of $P$. Then,
% $u$ and $v$ are also elements of $\widehat{P}$ (since we are regarding $P$ as
% a subposet of $\widehat{P}$). Thus, strictly speaking, statements like
% \textquotedblleft$u<v$\textquotedblright\ or \textquotedblleft$u\lessdot
% v$\textquotedblright\ are ambiguous because it is not clear whether they are
% referring to the poset $P$ or to the poset $\widehat{P}$. However, this
% ambiguity is irrelevant, because it is easily seen that the truth of each of
% the statements \textquotedblleft$u<v$\textquotedblright, \textquotedblleft%
% $u\leq v$\textquotedblright, \textquotedblleft$u>v$\textquotedblright,
% \textquotedblleft$u\geq v$\textquotedblright, \textquotedblleft$u\lessdot
% v$\textquotedblright, \textquotedblleft$u\gtrdot v$\textquotedblright\ and
% \textquotedblleft$u$ and $v$ are incomparable\textquotedblright\ is
% independent on whether it refers to the poset $P$ or to the poset
% $\widehat{P}$. We are going to therefore omit mentioning the poset in these
% statements, unless there are other reasons for us to do so.
% \end{convention}



\begin{definition}
\label{def.labelling} A
$\mathbb{K}$\textit{-labelling of }$P$ will mean a map $f:\widehat{P}%
\rightarrow\mathbb{K}$. Thus, $\mathbb{K}^{\widehat{P}}$ is the set of all
$\mathbb{K}$-labellings of $P$. If $f$ is a $\mathbb{K}$-labelling of $P$ and
$v$ is an element of $\widehat{P}$, then $f\left(  v\right)  $ will be called
the \textit{label of }$f$ \textit{at }$v$.
\end{definition}

% In the following, whenever we are working with a field $\mathbb{K}$, we will
% tacitly assume that $\mathbb{K}$ is either infinite or at least can
% be enlarged when necessity arises. This assumption is needed in order to
% clarify the notions of rational maps and generic elements of algebraic
% varieties over $\mathbb{K}$. (We will not require $\mathbb{K}$ to be
% algebraically closed.)

\begin{definition}
We will use the terminology of algebraic varieties and rational maps between
them, although the only algebraic varieties that we will be considering are
products of affine and projective spaces, as well as their open subsets. We
use the punctured arrow $\dashrightarrow$ to signify rational maps.
% (i.e., a rational map from a variety $U$ to a variety $V$ is written
% $U\dashrightarrow V$). 
A rational map $U\dashrightarrow V$ is said to be
\textit{dominant} if its image is dense in $V$ (with respect to the Zariski topology).

% The words \textquotedblleft generic\textquotedblright\ and \textquotedblleft
% almost\textquotedblright\ will always refer to the Zariski topology. For
The words \textit{generic} and 
\textit{almost} will always refer to the Zariski topology. For
example, if $U$ is a finite set, then an assertion saying that some statement
holds \textquotedblleft for almost every point $p\in\mathbb{K}^{U}%
$\textquotedblright\ is supposed to mean that there is a Zariski-dense open
subset $D$ of $\mathbb{K}^{U}$ such that this statement holds for every point
$p\in D$. A \textquotedblleft generic\textquotedblright\ point on an algebraic
variety $V$ (for example, this can be a \textquotedblleft generic
matrix\textquotedblright\ when $V$ is a space of matrices, or a
\textquotedblleft generic $\mathbb{K}$-labelling of a poset $P$%
\textquotedblright\ when $V$ is the space of all $\mathbb{K}$-labellings of
$P$) means a point lying in some fixed Zariski-dense open subset $S$ of $V$;
the concrete definition of $S$ can usually be inferred from the context
(often, it will be the subset of $V$ on which everything we want to do with
our point is well-defined), but of course should never depend on the actual
point. 
% (Note that one often has to read the whole proof in order to be able to
% tell what this $S$ is. This is similar to the use of the \textquotedblleft for
% $\epsilon$ small enough\textquotedblright\ wording in analysis, where it is
% often not clear until the end of the proof how small exactly the $\epsilon$
% needs to be.) 
We will sometimes abuse notation and say that an
equality holds \textquotedblleft for every point\textquotedblright\ instead of
\textquotedblleft for almost every point\textquotedblright\ when it is really
clear what the $S$ is. (For example, if we say that \textquotedblleft the
equality $\dfrac{x^{3}-y^{3}}{x-y}=x^{2}+xy+y^{2}$ holds for every
$x\in\mathbb{K}$ and $y\in\mathbb{K}$\textquotedblright, it is clear that $S$
has to be the set $\mathbb{K}^{2}\setminus\left\{  \left(  x,y\right)
\in\mathbb{K}^{2}\mid x=y\right\}  $, because the left hand side of the
equality makes no sense when $\left(  x,y\right)  $ is outside of this set.)
\end{definition}

\begin{remark}
Most statements that we make below work not only for fields, but also more
generally for semifields\footnotemark\ such as the semifield $\mathbb{Q}_{+}$
of positive rationals or the tropical semiring. Some (but not all!) statements
actually simplify when the underlying field is replaced by a semifield in
which no two nonzero elements add to zero (because in such cases, e.g., the
denominators in (\ref{def.Tv.def}) cannot become zero unless some labels of
$f$ are $0$). Thus, working with such semifields instead of fields would save
us the trouble of having things defined \textquotedblleft almost
everywhere\textquotedblright. Moreover, applying our results to the tropical
semifield would yield some of the statements about order polytopes made in
\cite{einstein-propp}. Nevertheless, we prefer to work with fields, for the
following reasons:

-- If we were to work
in semifields which \textbf{do} contain two nonzero elements summing up to
zero, then we would still have the issue of zero denominators, but we are not
aware of a theoretical framework in the spirit of Zariski topology for fields
to reassure us in this case that these issues are negligible.

-- If an identity between subtraction-free rational expressions (such as
$\dfrac{x^{3}+y^{3}}{x+y}+3xy=\left(  x+y\right)  ^{2}$) holds over every
field (as long as the denominators involved are nonzero), then it must hold
over every semifield as well (again as long as the denominators involved are
nonzero), even if a proof of this identity uses subtraction in its
intermediate steps (e.g., a proof of $\dfrac{x^{3}+y^{3}}{x+y}+3xy=\left(
x+y\right)  ^{2}$ over a field can begin by simplifying $\dfrac{x^{3}+y^{3}%
}{x+y}$ to $x^{2}-xy+y^{2}$, a technique not available over a semifield). This
is simply because every true identity between subtraction-free rational
expressions can be verified by multiplying by a common denominator 
%(an operation which does not introduce any subtractions) 
and comparing
coefficients. Since our main results (such as Proposition~\ref{prop.skeletal.ords})
can be construed as identities between subtraction-free
rational expressions, this yields that all these results hold over any
semifield (provided the denominators are nonzero) if they hold over every
field. So we are not losing any generality by restricting ourselves to
considering only fields.
\end{remark}

\footnotetext{The word \textquotedblleft\textit{semifield}\textquotedblright%
\ here means a commutative semiring in which each element other than $0$ has a
multiplicative inverse. (In contrast to other authors' conventions, our
semifields do have zeroes.) A \textit{semiring} is defined as a set with two
binary operations called \textquotedblleft addition\textquotedblright\ and
\textquotedblleft multiplication\textquotedblright\ and two elements $0$ and
$1$ which satisfies all axioms of a ring 
% (in particular, it must be associative and satisfy $0\cdot a=a\cdot0=0$ and $1\cdot
% a=a\cdot1=a$ for all $a$) 
except for having additive inverses.}

\begin{definition}
\label{def.Tv}Let
$v\in P$. We define a rational map $T_{v}:\mathbb{K}^{\widehat{P}%
}\dashrightarrow\mathbb{K}^{\widehat{P}}$ by%
\begin{equation}
\left(  T_{v}f\right)  \left(  w\right)  =\left\{
\begin{array}
[c]{l}%
f\left(  w\right),\ \ \ \ \ \ \ \ \ \ \text{if }w\neq v;\\
\dfrac{1}{f\left(  v\right)  }\cdot\dfrac{\sum\limits_{\substack{u\in
\widehat{P};\\u\lessdot v}}f\left(  u\right)  }{\sum\limits_{\substack{u\in
\widehat{P};\\u\gtrdot v}}\dfrac{1}{f\left(  u\right)  }}%
,\ \ \ \ \ \ \ \ \ \ \text{if }w=v
\end{array}
\right.  \ \ \ \ \ \ \ \ \ \ \text{for all }w\in\widehat{P}
\text{ and } f\in\mathbb{K}^{\widehat{P}}.
\label{def.Tv.def}
\end{equation}
%for all $f\in\mathbb{K}^{\widehat{P}}$. 
Note that this rational map $T_{v}$ is
well-defined, because the right-hand side of (\ref{def.Tv.def}) is
well-defined on a Zariski-dense open subset of $\mathbb{K}^{\widehat{P}}$.
(This follows from the fact that for every $v\in P$, there is at least one
$u\in\widehat{P}$ such that $u\gtrdot v$.) %\ \ \ \ \footnotemark.)
We call $T_{v}$ the $v$\textit{-toggle}.
\end{definition}


% \footnotetext{Indeed, either there is at least one $u\in P$ such that
% $u\gtrdot v$ in $P$ (and therefore also $u\gtrdot v$ in $\widehat{P}$), or
% else $v$ is maximal in $P$ and then we have $1\gtrdot v$ in $\widehat{P}$.}



The map $T_{v}$ that we have just introduced (but defined over the
semifield $\mathbb{R}_{+}$ instead of our field $\mathbb{K}$) is called a
\textquotedblleft birational toggle operation\textquotedblright\ in
\cite{einstein-propp} (where it is denoted by $\phi_{i}$ with $i$ being a
number indexing the elements $v$ of $P$; however, the same notation is used
for the \textquotedblleft tropicalized\textquotedblright\ version of $T_{v}$).
As is clear from its definition, it only changes the label at the element $v$.
Note also the following almost trivial fact:

\begin{proposition}
\label{prop.Tv.invo}For every $v\in P$, the rational map $T_{v}$ is an involution, i.e., the map
$T_{v}^{2}$ is well-defined on a Zariski-dense open subset of $\mathbb{K}%
^{\widehat{P}}$ and satisfies $T_{v}^{2}=\operatorname*{id}$ on this subset.
\end{proposition}

We are calling this \textquotedblleft almost trivial\textquotedblright%
\ because one subtlety is easily overlooked: We have to check that the map
$T_{v}^{2}$ is well-defined on a Zariski-dense open subset of $\mathbb{K}%
^{\widehat{P}}$; this requires observing that for every $v\in P$, there exists
at least one $u\in\widehat{P}$ such that $u\lessdot v$.


Proposition \ref{prop.Tv.invo} yields the following:

\begin{corollary}
\label{cor.Tv.dom}For every $v\in P$, the map $T_{v}$ is a dominant rational map.
\end{corollary}

% The reader should remember that 
Dominant rational maps (unlike general
rational maps) can be composed, and their compositions are still dominant
rational maps. 
%%% TR: Already explained above Definition~10. 
% Of course, we are brushing aside subtleties like the fact that
% dominant rational maps are defined only over infinite fields (unless we are
% considering them in a sufficiently formal sense); as far as this paper is
% concerned, it never hurts to extend the field $\mathbb{K}$ (say, by
% introducing a new indeterminate), so when in doubt the reader can assume that
% the field $\mathbb{K}$ is infinite.
%
%The following proposition is trivially obtained by rewriting (\ref{def.Tv.def}%
%); we are merely stating it for easier reference in proofs:
%
%\begin{proposition}
%\label{prop.Tv}Let
%$v\in P$. For every $f\in\mathbb{K}^{\widehat{P}}$ for which $T_{v}f$ is
%well-defined, we have:
%
%\textbf{(a)} Every $w\in\widehat{P}$ such that $w\neq v$ satisfies $\left(
%T_{v}f\right)  \left(  w\right)  =f\left(  w\right)  $.
%
%\textbf{(b)} We have%
%\[
%\left(  T_{v}f\right)  \left(  v\right)  =\dfrac{1}{f\left(  v\right)  }%
%\cdot\dfrac{\sum\limits_{\substack{u\in\widehat{P};\\u\lessdot v}}f\left(
%u\right)  }{\sum\limits_{\substack{u\in\widehat{P};\\u\gtrdot v}}\dfrac
%{1}{f\left(  u\right)  }}.
%\]
%
%\end{proposition}
It is very easy to check the following ``locality principle'':

\begin{proposition}
\label{prop.Tv.commute} For all $v,w\in P$, we have
$T_{v}\circ T_{w}=T_{w}\circ T_{v}$, unless
either $v \lessdot w$ or $w \lessdot v$.
\end{proposition}


\begin{proof}
[Proof of Proposition \ref{prop.Tv.commute}.]
%%% Changed this as per 2nd Referee Report
% Assume that neither $v
% \lessdot w$ nor $w \lessdot v$. Also, WLOG, assume that $v \neq w$, lest the
% claim of the proposition be obvious.
%
The action of $T_{v}$ on a labelling of $P$ merely changes the label at $v$
to a new value which depends on the label at $v$ and on the labels at the
elements $u\in\widehat{P}$ satisfying $u\lessdot v$ or $u \gtrdot v$.
% A similar thing can be said about the action of $T_{w}$. 
Thus, the actions of $T_{v}$ and $T_{w}$ don't
interfere with each other unless either
$v \lessdot w$ or $w \lessdot v$ or $v=w$. Hence,
$T_{v}\circ T_{w}=T_{w}\circ T_{v}$ unless 
$v \lessdot w$ or $w \lessdot v$ or $v=w$
(in which latter case, the proposition is obvious).  
% in the sense that the changes made by either of
% them are the same no matter whether the other has been applied before it or
% not. That is, $T_{v}\circ T_{w}=T_{w}\circ T_{v}$ (since
% we have neither $v\lessdot w$ nor $w\lessdot v$
% nor $v=w$).
\end{proof}

\begin{corollary}
\label{cor.Tv.commute}
Let $v$ and $w$ be two incomparable elements of $P$. Then,
$T_{v}\circ T_{w}=T_{w}\circ T_{v}$.
\end{corollary}

%%% TR: Omit as per 2nd referee. report
% \begin{proof}
% [\nopunct]This follows from Proposition \ref{prop.Tv.commute} since
% incomparable elements never cover each other.
% \end{proof}




Combining Corollary \ref{cor.Tv.commute} with Proposition
\ref{prop.linext.transitive}, we obtain:


\begin{corollary}
\label{cor.R.welldef}
Let $\left(  v_{1},v_{2},\ldots,v_{m}\right)  $ be a linear extension of $P$.
Then, the dominant rational map $T_{v_{1}}\circ T_{v_{2}}\circ\ldots\circ
T_{v_{m}}:\mathbb{K}^{\widehat{P}}\dashrightarrow\mathbb{K}^{\widehat{P}}$ is
well-defined and independent of the choice of the linear extension $\left(
v_{1},v_{2},\ldots,v_{m}\right)  $.
\end{corollary}


\begin{definition}
\label{def.rm}
\textit{Birational rowmotion} is defined as the dominant rational map
$T_{v_{1}}\circ T_{v_{2}}\circ\ldots\circ T_{v_{m}}:\mathbb{K}^{\widehat{P}%
}\dashrightarrow\mathbb{K}^{\widehat{P}}$, where $\left(  v_{1},v_{2}%
,\ldots,v_{m}\right)  $ is a linear extension of $P$. This rational map is
well-defined 
% (in particular, it does not depend on the linear extension
% $\left(  v_{1},v_{2},\ldots,v_{m}\right)  $ chosen) 
by Corollary~\ref{cor.R.welldef} and Theorem~\ref{thm.linext.ex};
% (and also because a linear extension of $P$ always exists;
% this is Theorem \ref{thm.linext.ex}). 
it will be denoted by $R$.
\end{definition}

The reason for the names ``birational toggle'' and ``birational rowmotion'' is
explained in \cite{einstein-propp}, in which birational rowmotion
(defined over $\mathbb{R}_{+}$ rather than over $\mathbb{K}$) is
denoted (serendipitously from the standpoint of the second author of this
paper) by $\rho_{\mathcal{B}}$.

\begin{example}
\label{ex.rowmotion.31}Let us demonstrate the effect of birational toggles and
birational rowmotion on a rather simple $4$-element poset. Namely, for this
example, we let $P$ be the poset $\left\{  p,q_{1},q_{2},q_{3}\right\}  $ with
order relation defined by setting $p<q_{i}$ for each $i\in\left\{
1,2,3\right\}  $. This poset has Hasse diagram
\[
\xymatrixrowsep{0.9pc}\xymatrixcolsep{2pc}\xymatrix{
q_1 \ar@{-}[rd] & q_2 \ar@{-}[d] & q_3 \ar@{-}[ld] \\
& p &
}
\]
The extended poset $\widehat{P}$ has Hasse diagram
\[
\xymatrixrowsep{0.9pc}\xymatrixcolsep{2pc}\xymatrix{
& 1 \ar@{-}[rd] \ar@{-}[d] \ar@{-}[ld] & \\
q_1 \ar@{-}[rd] & q_2 \ar@{-}[d] & q_3 \ar@{-}[ld] \\
& p \ar@{-}[d] & \\
& 0 &
}
\]
We can visualize a $\mathbb{K}$-labelling $f$ of $P$ by replacing, in the
Hasse diagram of $\widehat{P}$, each element $v\in\widehat{P}$ by the label $f\left(
v\right)  $. Let $f$ be a $\mathbb{K}$-labelling sending $0$, $p$, $q_{1}$,
$q_{2}$, $q_{3}$, and $1$ to $a$, $w$, $x_{1}$, $x_{2}$, $x_{3}$, and $b$,
respectively (all in $\mathbb{K}$); this $f$ is then visualized as follows:
\[
\xymatrixrowsep{0.9pc}\xymatrixcolsep{2pc}\xymatrix{
& b \ar@{-}[rd] \ar@{-}[d] \ar@{-}[ld]  & \\
x_1 \ar@{-}[rd] & x_2 \ar@{-}[d] & x_3 \ar@{-}[ld] \\
& w \ar@{-}[d] & \\
& a &
}
\]
%Now, recall the definition of birational rowmotion $R$ on our poset $P$. 
Since 
$\left(  p,q_{1},q_{2},q_{3}\right)  $ is a linear extension of $P$, we have
$R=T_{p}\circ T_{q_{1}}\circ T_{q_{2}}\circ T_{q_{3}}$. Let us track how this
transforms our labelling $f$.  
We first apply $T_{q_{3}}$, obtaining%
\[
\xymatrixrowsep{0.9pc}\xymatrixcolsep{2pc}\xymatrix{
& & b \ar@{-}[rd] \ar@{-}[d] \ar@{-}[ld]  & \\
T_{q_3} f = \  & x_1 \ar@{-}[rd] & x_2 \ar@{-}[d] & {\color{red} \frac{bw}{x_3}} \ar@{-}[ld] \\
& & w \ar@{-}[d] & \\
& & a &
}
\]
(where we colored the label at $q_{3}$ red to signify that it is the label at
the element which got toggled). Indeed, only the label at $q_3$
changes under $T_{q_{3}}$ becoming%
\[
\left(  T_{q_{3}}f\right)  \left(  q_{3}\right)  =\dfrac{1}{f\left(
q_{3}\right)  }\cdot\dfrac{\sum\limits_{\substack{u\in\widehat{P};\\u\lessdot
q_{3}}}f\left(  u\right)  }{\sum\limits_{\substack{u\in\widehat{P};\\u\gtrdot
q_{3}}}\dfrac{1}{f\left(  u\right)  }}=\dfrac{1}{f\left(  q_{3}\right)  }%
\cdot\dfrac{f\left(  p\right)  }{\left(  \dfrac{1}{f\left(  1\right)
}\right)  }=\dfrac{1}{x_{3}}\cdot\dfrac{w}{\left(  \dfrac{1}{b}\right)
}=\dfrac{bw}{x_{3}}.
\]
% Having applied $T_{q_{3}}$, we next apply $T_{q_{2}}$, obtaining%
% \[
% \xymatrixrowsep{0.9pc}\xymatrixcolsep{2pc}\xymatrix{
% & & b \ar@{-}[rd] \ar@{-}[d] \ar@{-}[ld]  & \\
% T_{q_2} T_{q_3} f = \  & x_1 \ar@{-}[rd] & {\color{red} \frac{bw}{x_2}} \ar@{-}[d] & \frac{bw}{x_3} \ar@{-}[ld] \\
% & & w \ar@{-}[d] & \\
% & & a &
% }.
% \]
% Next, we apply $T_{q_{1}}$, obtaining%
% \[
% \xymatrixrowsep{0.9pc}\xymatrixcolsep{2pc}\xymatrix{
% & & b \ar@{-}[rd] \ar@{-}[d] \ar@{-}[ld]  & \\
% T_{q_1} T_{q_2} T_{q_3} f = \  & {\color{red} \frac{bw}{x_1}} \ar@{-}[rd] & \frac{bw}{x_2} \ar@{-}[d] & \frac{bw}{x_3} \ar@{-}[ld] \\
% & & w \ar@{-}[d] & \\
% & & a &
% }.
% \]
% Finally, we apply $T_{p}$, resulting in
Now applying successively $T_{q_{2}}$, $T_{q_{1}}$, and $T_{p}$, we obtain
\[
\xymatrixrowsep{0.9pc}\xymatrixcolsep{2pc}\xymatrix{
& & b \ar@{-}[rd] \ar@{-}[d] \ar@{-}[ld]  & \\
R f = T_p T_{q_1} T_{q_2} T_{q_3} f = \  & \frac{bw}{x_1} \ar@{-}[rd] & \frac{bw}{x_2} \ar@{-}[d] & \frac{bw}{x_3} \ar@{-}[ld] \\
& & {\color{red} \frac{ab}{x_1+x_2+x_3}} \ar@{-}[d] & \\
& & a &
}
\]
since the birational $p$-toggle $T_{p}$ has changed the label at $p$ to%
\begin{align*}
%&
\left(  T_{p}T_{q_{1}}T_{q_{2}}T_{q_{3}}f\right)  \left(  p\right)
% \\
&  =\dfrac{1}{\left(  T_{q_{1}}T_{q_{2}}T_{q_{3}}f\right)  \left(  p\right)
}\cdot\dfrac{\sum\limits_{\substack{u\in\widehat{P};\\u\lessdot p}}\left(
T_{q_{1}}T_{q_{2}}T_{q_{3}}f\right)  \left(  u\right)  }{\sum
\limits_{\substack{u\in\widehat{P};\\u\gtrdot p}}\dfrac{1}{\left(  T_{q_{1}%
}T_{q_{2}}T_{q_{3}}f\right)  \left(  u\right)  }}\\
%&  =\dfrac{1}{\left(  T_{q_{1}}T_{q_{2}}T_{q_{3}}f\right)  \left(  p\right)
%}\cdot\dfrac{\left(  T_{q_{1}}T_{q_{2}}T_{q_{3}}f\right)  \left(  0\right)
%}{\dfrac{1}{\left(  T_{q_{1}}T_{q_{2}}T_{q_{3}}f\right)  \left(  q_{1}\right)
%}+\dfrac{1}{\left(  T_{q_{1}}T_{q_{2}}T_{q_{3}}f\right)  \left(  q_{2}\right)
%}+\dfrac{1}{\left(  T_{q_{1}}T_{q_{2}}T_{q_{3}}f\right)  \left(  q_{3}\right)
%}}\\
&  =\dfrac{1}{w}\cdot\dfrac{a}{\dfrac{1}{bw\diagup x_{1}}+\dfrac{1}{bw\diagup
x_{2}}+\dfrac{1}{bw\diagup x_{3}}}=\dfrac{ab}{x_{1}+x_{2}+x_{3}}.
\end{align*}
%We thus have computed $Rf$ (since $R = T_p T_{q_1} T_{q_2} T_{q_3} T_{q_4}$).
By repeating this procedure (or just substituting
the labels of $Rf$ obtained as variables), we can compute $R^{2}f$, $R^{3}f$
etc. Specifically, we obtain%
\begin{align*}
% &
% \xymatrixrowsep{0.9pc}\xymatrixcolsep{6pc}\xymatrix{ & & b \ar@{-}[rd] \ar@{-}[d] \ar@{-}[ld] & \\ R f = \  & \frac{bw}{x_1} \ar@{-}[rd] & \frac{bw}{x_2} \ar@{-}[d] & \frac{bw}{x_3} \ar@{-}[ld] \\ & & \frac{ab}{x_1+x_2+x_3} \ar@{-}[d] & \\ & & a & },\\
% & \\
&
\xymatrixrowsep{0.9pc}\xymatrixcolsep{3pc}\xymatrix{ & & b \ar@{-}[rd] \ar@{-}[d] \ar@{-}[ld] & \\ R^2 f = \  & \frac{abx_1}{w\left(x_1+x_2+x_3\right)} \ar@{-}[rd] & \frac{abx_2}{w\left(x_1+x_2+x_3\right)} \ar@{-}[d] & \frac{abx_3}{w\left(x_1+x_2+x_3\right)} \ar@{-}[ld] \\ & & \frac{ax_1x_2x_3}{w\left(x_2x_3+x_3x_1+x_1x_2\right)} \ar@{-}[d] & \\ & & a & }\\
& \\
&
\xymatrixrowsep{0.9pc}\xymatrixcolsep{3pc}\xymatrix{ & & b \ar@{-}[rd] \ar@{-}[d] \ar@{-}[ld] & \\ R^3 f = \  & \frac{x_2x_3\left(x_1+x_2+x_3\right)}{x_2x_3+x_3x_1+x_1x_2} \ar@{-}[rd] & \frac{x_3x_1\left(x_1+x_2+x_3\right)}{x_2x_3+x_3x_1+x_1x_2} \ar@{-}[d] & \frac{x_1x_2\left(x_1+x_2+x_3\right)}{x_2x_3+x_3x_1+x_1x_2} \ar@{-}[ld] \\ & & w \ar@{-}[d] & \\ & & a & }\\
\end{align*}

\begin{align*}
\ \xymatrixrowsep{0.9pc}\xymatrixcolsep{3pc}\xymatrix{ & & b \ar@{-}[rd] \ar@{-}[d] \ar@{-}[ld] & \\ R^4 f = \  & \frac{bw\left(x_2x_3+x_3x_1+x_1x_2\right)}{x_2x_3\left(x_1+x_2+x_3\right)} \ar@{-}[rd] & \frac{bw\left(x_2x_3+x_3x_1+x_1x_2\right)}{x_3x_1\left(x_1+x_2+x_3\right)} \ar@{-}[d] & \frac{bw\left(x_2x_3+x_3x_1+x_1x_2\right)}{x_1x_2\left(x_1+x_2+x_3\right)} \ar@{-}[ld] \\ & & \frac{ab}{x_1+x_2+x_3} \ar@{-}[d] & \\ & & a & }\\
& \\
\end{align*}%
\begin{align*}
&
\xymatrixrowsep{0.9pc}\xymatrixcolsep{3pc}\xymatrix{ & & b \ar@{-}[rd] \ar@{-}[d] \ar@{-}[ld] & \\ R^5 f = \  & \frac{abx_2x_3}{w\left(x_2x_3+x_3x_1+x_1x_2\right)} \ar@{-}[rd] & \frac{abx_3x_1}{w\left(x_2x_3+x_3x_1+x_1x_2\right)} \ar@{-}[d] & \frac{abx_1x_2}{w\left(x_2x_3+x_3x_1+x_1x_2\right)} \ar@{-}[ld] \\ & & \frac{ax_1x_2x_3}{w\left(x_2x_3+x_3x_1+x_1x_2\right)} \ar@{-}[d] & \\ & & a & }\\
& \\
&
R^6 f = f.
% \xymatrixrowsep{0.9pc}\xymatrixcolsep{6pc}\xymatrix{ & & b \ar@{-}[rd] \ar@{-}[d] \ar@{-}[ld] & \\ R^6 f = \  & x_1 \ar@{-}[rd] & x_2 \ar@{-}[d] & x_3 \ar@{-}[ld] \\ & & w \ar@{-}[d] & \\ & & a & }.
\end{align*}
There are several patterns here that catch the eye, some of which are related
to the very simple structure of $P$ and don't seem to generalize well.
However, the most striking observation here is that $R^{n}f=f$ for some
positive integer $n$ (namely, $n=6$). We will see in
Proposition \ref{prop.skeletal.ords} that this generalizes to a rather wide
class of posets, which we call \textquotedblleft skeletal
posets\textquotedblright\ (Definition~\ref{def.skeletal}), a class
of posets which contain (in particular) all graded forests such as our poset
$P$ here. 
% (See Definition \ref{def.skeletal} for the definitions of the
% concepts involved here.)
A different example is given in \cite{grinberg-roby-part2}.
\end{example}

Let us state another proposition, which describes birational rowmotion implicitly:

\begin{proposition}
\label{prop.R.implicit}
Let $v\in P$ and $f\in\mathbb{K}^{\widehat{P}}$. Then,%
\begin{equation}
\left(  Rf\right)  \left(  v\right)  =\dfrac{1}{f\left(  v\right)  }%
\cdot\dfrac{\sum\limits_{\substack{u\in\widehat{P};\\u\lessdot v}}f\left(
u\right)  }{\sum\limits_{\substack{u\in\widehat{P};\\u\gtrdot v}}\dfrac
{1}{\left(  Rf\right)  \left(  u\right)  }}. \label{prop.R.implicit.eq}%
\end{equation}

\end{proposition}

Here (and in statements further down this paper), we are taking the liberty to
leave assumptions such as \textquotedblleft Assume that $Rf$ is
well-defined\textquotedblright\ unsaid (for instance, such an assumption is
needed in Proposition \ref{prop.R.implicit}) because these assumptions are
satisfied when the parameters belong to some Zariski-dense open subset of their domains.


\begin{proof}
[Proof of Proposition \ref{prop.R.implicit}.]Fix a linear extension
$\left(  v_{1},v_{2},\ldots,v_{m}\right)  $ of $P$. Recall that $R$ has been
defined as the composition $T_{v_{1}}\circ T_{v_{2}}\circ\ldots\circ T_{v_{m}}$.
Hence, $Rf$ can be obtained from $f$ by traversing the linear extension
$\left(  v_{1},v_{2},\ldots,v_{m}\right)  $ in the order $v_{m}, v_{m-1},\dotsc $,
% from right to left (thus starting
% with the largest element $v_{m}$, then proceeding to $v_{m-1}$, etc.), and 
at every step toggling the element being traversed. When an element $v$ is being
toggled, the elements $u\in\widehat{P}$ satisfying $u\lessdot v$ have not yet
been toggled, whereas
those satisfying $u\gtrdot v$ have been toggled already. Denoting the state of
the $\mathbb{K}$-labelling \text{before} the $v$-toggle by $g$, we see that
the state \textit{after} the $v$-toggle will be $T_{v}g$ with%
\begin{equation}
\left(  T_{v}g\right)  \left(  w\right)  =\left\{
\begin{array}
[c]{l}%
g\left(  w\right),\ \ \ \ \ \ \ \ \ \ \text{if }w\neq v;\\
\dfrac{1}{g\left(  v\right)  }\cdot\dfrac{\sum\limits_{\substack{u\in
\widehat{P};\\u\lessdot v}}g\left(  u\right)  }{\sum\limits_{\substack{u\in
\widehat{P};\\u\gtrdot v}}\dfrac{1}{g\left(  u\right)  }}%
,\ \ \ \ \ \ \ \ \ \ \text{if }w=v
\end{array}
\right.  \ \ \ \ \ \ \ \ \ \ \text{for all }w\in\widehat{P}.
\label{pf.R.implicit.short.1}%
\end{equation}

But $g\left(  v\right)  =f\left(  v\right)  $ (since $v$ has not yet been
toggled at the time of $g$) and $\left(  T_{v}g\right)  \left(  v\right)
=\left(  Rf\right)  \left(  v\right)  $ (since $v$ has been toggled at the
time of $T_{v}g$, and is not going to be toggled again during the process
of computing $Rf$); moreover, all $u\in\widehat{P}$ satisfying $u\lessdot v$
satisfy $g\left(  u\right)  =f\left(  u\right)  $ (since these $u$ have not
yet been toggled), whereas all $u\in\widehat{P}$ satisfying $u\gtrdot v$
satisfy $g\left(  u\right)  =\left(  Rf\right)  \left(  u\right)  $ (since
these $u$ have already been toggled once and for all). Thus,
(\ref{pf.R.implicit.short.1}) (applied to $w=v$) transforms into
(\ref{prop.R.implicit.eq}), proving Proposition \ref{prop.R.implicit}.
\end{proof}

Here is a little triviality to complete the picture of Proposition
\ref{prop.R.implicit}:

\begin{proposition}
\label{prop.R.implicit.01} Let $f\in\mathbb{K}^{\widehat{P}}$. Then, $\left(  Rf\right)  \left(
0\right)  =f\left(  0\right)  $ and $\left(  Rf\right)  \left(  1\right)
=f\left(  1\right)  $. 
\end{proposition}


\begin{proof}
[\nopunct]This is clear since no toggle changes the labels at $0$ and~$1$.
\end{proof}

We will often use Proposition \ref{prop.R.implicit.01} tacitly. A trivial
corollary of Proposition \ref{prop.R.implicit.01} is:

\begin{corollary}
\label{cor.R.implicit.01} Let $f\in\mathbb{K}^{\widehat{P}}$. 
Then for all $\ \ell \in\mathbb{N}$, we have
$\left(  R^{\ell}f\right)  \left(  0\right)  =f\left(
0\right)  $ and $\left(  R^{\ell}f\right)  \left(  1\right)  =f\left(
1\right)  $.
\end{corollary}

% \footnotetext{Here and in the following, $\mathbb{N}$ denotes the set
% $\left\{  0,1,2,\ldots\right\}  $.}


We will also need a converse of Propositions \ref{prop.R.implicit} and
\ref{prop.R.implicit.01}:

\begin{proposition}
\label{prop.R.implicit.converse}Let $f, g\in\mathbb{K}^{\widehat{P}}$ 
satisfy $f\left(  0\right)  =g\left(  0\right)  $ and $f\left(  1\right)
=g\left(  1\right)  $. Assume that%
\begin{equation}
g\left(  v\right)  =\dfrac{1}{f\left(  v\right)  }\cdot\dfrac{\sum
\limits_{\substack{u\in\widehat{P};\\u\lessdot v}}f\left(  u\right)  }%
{\sum\limits_{\substack{u\in\widehat{P};\\u\gtrdot v}}\dfrac{1}{g\left(
u\right)  }}\ \ \ \ \ \ \ \ \ \ \text{for every }v\in P.
\label{prop.R.implicit.converse.eq}%
\end{equation}
(This means, in particular, that we assume that all denominators in
(\ref{prop.R.implicit.converse.eq}) are nonzero.) Then, $Rf$ is well-defined and $Rf = g$.
\end{proposition}


\begin{proof}
[Proof of Proposition \ref{prop.R.implicit.converse}.]It is clearly
enough to show that $g\left(  v\right)  =\left(  Rf\right)  \left(  v\right)
$ for every $v\in\widehat{P}$. Since this is clear for $v=0$,
%(since $g\left(0\right)  =f\left(  0\right)  =\left(  Rf\right)  \left(  0\right)  $)
we only need to consider the case when $v\in\left\{  1\right\}  \cup P$. In this
case, we can prove $g\left(  v\right)  =\left(  Rf\right)  \left(  v\right)  $
by descending induction over $v$ -- that is, we assume as an induction
hypothesis that $g\left(  u\right)  =\left(  Rf\right)  \left(  u\right)  $
holds for all elements $u\in\left\{  1\right\}  \cup P$ which are greater than
$v$ in $\widehat{P}$. The induction base ($v=1$) is clear, 
and the induction step follows by comparing (\ref{prop.R.implicit.eq}) with
(\ref{prop.R.implicit.converse.eq}). We leave the details (including a check
that $Rf$ is well-defined, which piggybacks on the induction) to the reader
(see also \cite{grinberg-roby-arXiv}).
\end{proof}


% As an aside, at this point we could give an alternative proof of Corollary
% \ref{cor.R.welldef}, foregoing the use of Proposition
% \ref{prop.linext.transitive}. In fact, the proofs of Propositions
% \ref{prop.R.implicit}, \ref{prop.R.implicit.01} and
% \ref{prop.R.implicit.converse} only used that $R$ is a composition $T_{v_{1}%
% }\circ T_{v_{2}}\circ\ldots\circ T_{v_{m}}$ for \textbf{some} linear extension
% $\left(  v_{1},v_{2},\ldots,v_{m}\right)  $ of $P$. Thus, starting with
% \textbf{any} linear extension $\left(  v_{1},v_{2},\ldots,v_{m}\right)  $ of $P$,
% we could have defined $R$ as the composition $T_{v_{1}}\circ T_{v_{2}}%
% \circ\ldots\circ T_{v_{m}}$, and then used Propositions \ref{prop.R.implicit},
% \ref{prop.R.implicit.01} and \ref{prop.R.implicit.converse} to characterize
% the image $Rf$ of a $\mathbb{K}$-labelling $f$ under this map $R$ in a unique
% way without reference to $\left(  v_{1},v_{2},\ldots,v_{m}\right)  $, and thus
% concluded that $R$ does not depend on $\left(  v_{1},v_{2},\ldots,v_{m}\right)
% $. The details of this derivation are left to the reader.

As an aside, at this point we could give an alternative proof of Corollary
\ref{cor.R.welldef}, foregoing the use of Proposition
\ref{prop.linext.transitive}. Indeed, Propositions
\ref{prop.R.implicit}, \ref{prop.R.implicit.01} and
\ref{prop.R.implicit.converse} characterize $R$ independent of the choice of linear
extension.  Details are left to the reader. 


On a related note, Proposition \ref{prop.R.implicit}, Proposition
\ref{prop.R.implicit.01} and Proposition \ref{prop.R.implicit.converse}
combined can be used as an alternative definition of birational rowmotion $R$,
which works even when the poset $P$ fails to be finite, as long as for every
$v\in P$, there exist only finitely many $u\in P$ satisfying $u>v$ and there
exist only finitely many $u\in P$ satisfying $u\lessdot v$ (provided that
some technicalities arising from Zariski topology on infinite-dimensional
spaces
are dealt with)\footnote{The asymmetry between the $>$ and $\lessdot$ signs in
this requirement is intentional. For instance, birational rowmotion can be
defined (but will not be invertible) for the poset $\left\{  0,-1,-2,-3,\ldots
\right\}  $ (with the usual order relation), but not for the poset $\left\{
0,1,2,3,\ldots\right\}  $ (again with the usual order relation).}. We will not
dwell on this.


Another general property of birational rowmotion concerns the question of what
happens if the birational toggles are composed not in the ``from top to
bottom'' order as in the definition of birational rowmotion, but the other way
round. It turns out that the result is the inverse of birational rowmotion, since we are
simply composing involutions in the opposite order.  

\begin{proposition}
\label{prop.R.inverse}Birational rowmotion $R$ on a finite poset $P$ is invertible (as a
rational map). Its inverse 
$R^{-1}$ is $T_{v_{m}}\circ T_{v_{m-1}}\circ\ldots\circ T_{v_{1}}:\mathbb{K}%
^{\widehat{P}}\dashrightarrow\mathbb{K}^{\widehat{P}}$, where $\left(
v_{1},v_{2},\ldots,v_{m}\right)  $ is a linear extension of $P$.
\end{proposition}

% \begin{proof}
% [Proof of Proposition \ref{prop.R.inverse}.]We know that $T_{w}$ is
% an involution for every $w\in P$. Thus, in particular, for every $w\in P$, the
% map $T_{w}$ is invertible and satisfies $T_{w}^{-1}=T_{w}$.

% Let $\left(  v_{1},v_{2},\ldots,v_{m}\right)  $ be a linear extension of $P$.
% Then, $R=T_{v_{1}}\circ T_{v_{2}}\circ\ldots\circ T_{v_{m}}$ (by the definition
% of $R$), so that $R^{-1}=T_{v_{m}}^{-1}\circ T_{v_{m-1}}^{-1}\circ\ldots\circ
% T_{v_{1}}^{-1}$ (this makes sense since the map $T_{w}$ is invertible for
% every $w\in P$). Since $T_{w}^{-1}=T_{w}$ for every $w\in P$, this simplifies
% to $R^{-1}=T_{v_{m}}\circ T_{v_{m-1}}\circ\ldots\circ T_{v_{1}}$. This proves
% Proposition \ref{prop.R.inverse}.
% \end{proof}

\section{Graded posets}

In this section, we restrict our attention to what we call \textit{graded
posets} (a notion that encompasses most of the posets we are interested in;
see Definition \ref{def.graded.n-graded}), and define (for this kind of
posets) a family of \textquotedblleft refined rowmotion\textquotedblright%
\ operators $R_{i}$ which toggle only the labels of the $i$-th degree of the
poset. These each turn out to be involutions, and their composition from top
to bottom degree is $R$ on the entire poset. We will later on use these
$R_{i}$ to get a better understanding of $R$ on graded posets.

Let us first introduce our notion of a graded poset:

\begin{definition}
\label{def.graded.n-graded}Let $P$ be a finite poset. Let $n$ be a nonnegative
integer. We say that the poset $P$ is $n$\textit{-graded} if there exists a
surjective map $\deg:P\rightarrow\left\{  1,2,\ldots,n\right\}  $ such that the
following three assertions hold:

\textit{Assertion 1:} Any two elements $u$ and $v$ of $P$ such that $u \gtrdot
v$ satisfy $\deg u=\deg v+1$.

\textit{Assertion 2:} We have $\deg u=1$ for every minimal element $u$ of $P$.

\textit{Assertion 3:} We have $\deg v=n$ for every maximal element $v$ of $P$.
\end{definition}

% Note that the word ``surjective'' in Definition \ref{def.graded.n-graded} is
% almost superfluous: Indeed, whenever $P\neq\varnothing$, then any map
% $\deg:P\rightarrow\left\{  1,2,\ldots,n\right\}  $ satisfying the Assertions 1, 2
% and 3 of Definition \ref{def.graded.n-graded} is automatically surjective
% (this is easy to prove). But if $P=\varnothing$, such a map exists (vacuously)
% for every $n$, whereas requiring surjectivity forced $n=0$.

\begin{example}
The poset $P$ studied in
Example \ref{ex.rowmotion.31} is $2$-graded. The empty poset is $0$-graded,
but not $n$-graded for any positive $n$. A chain with $k$ elements is $k$-graded.
\end{example}



\begin{definition}
\label{def.graded.graded}Let $P$ be a finite poset. We say that the poset $P$
is \textit{graded} if there exists an $n\in\mathbb{N}$ such that $P$ is
$n$-graded. % This $n$ is then called the \textit{height} of $P$.
\end{definition}

\begin{definition}
\label{def.graded.Phat}Let $n\in\mathbb{N}$. Let $P$ be an $n$-graded poset.
Then, there exists a surjective map $\deg:P\rightarrow\left\{
1,2,\ldots,n\right\}  $ that satisfies the Assertions 1, 2 and 3 of Definition
\ref{def.graded.n-graded}. A moment of thought reveals that such a map $\deg$
is also uniquely determined by $P$\ \ \ \ \footnotemark. Thus, we will call
$\deg$ the \textit{degree map} of $P$.

Moreover, we extend this map $\deg$ to a map $\widehat{P}\rightarrow\left\{
0,1,\ldots,n+1\right\}  $ by letting it map $0$ to $0$ and $1$ to $n+1$. This
extended map will also be denoted by $\deg$ and called the degree map. Notice
that this extended map $\deg$ still satisfies Assertion 1 of Definition
\ref{def.graded.n-graded} if $P$ is replaced by $\widehat{P}$ in that assertion.

For every $i\in\left\{  0,1,\ldots,n+1\right\}  $, we will denote by
$\widehat{P}_{i}$ the subset $\deg^{-1}\left(  \left\{  i\right\}  \right)  $
of $\widehat{P}$. For every $v\in\widehat{P}$, the number $\deg v$ is called
the \textit{degree} of $v$.
\end{definition}


\footnotetext{In fact, if $v\in P$, then it is easy to see that $\deg v$
equals the number of elements of any maximal chain in $P$ with highest element
$v$. This clearly determines $\deg v$ uniquely.}

The notion of an \textquotedblleft$n$-graded poset\textquotedblright\ we just
defined is identical with the notion of a \textquotedblleft graded finite
poset of rank $n-1$\textquotedblright\ as defined in \cite[\S 3.1]%
{stanley-ec1}; in particular, all maximal chains have the same length.  (There are several
other definitions of ``graded'' lurking in different corners of combinatorics.)  
The degree of an element $v$ of $P$ as defined in Definition
\ref{def.graded.Phat} is off by $1$ from the rank of $v$ in $P$ in the sense
of \cite[\S 3.1]{stanley-ec1}, but the degree $\deg v$ of an element $v$ of
$\widehat{P}$ equals its rank in $\widehat{P}$ in the sense of \cite[\S 3.1]%
{stanley-ec1}.

\textbf{\textit{From this point until \S 10, $P$ will always denote an $n$-graded poset for
some (fixed) $n\in \mathbb{N}$.}} 

The way we extended the map $\deg:P\rightarrow\left\{  1,2,\ldots,n\right\}  $ to
a map $\deg:\widehat{P}\rightarrow\left\{  0,1,\ldots,n+1\right\}  $ in
Definition \ref{def.graded.Phat}, of course, was not arbitrary. In fact, it
was tailored to make the following (easily proved) statements true:

\begin{proposition}
\label{prop.graded.Phat.why}
%Let $n\in\mathbb{N}$. Let $P$ be an $n$-graded poset. 
Let $u,v\in\widehat{P}$. Consider the map
$\deg:\widehat{P}\rightarrow\left\{  0,1,\ldots,n+1\right\}  $ defined in
Definition \ref{def.graded.Phat}.

\textbf{(a)} If $u\lessdot v$ in $\widehat{P}$, then $\deg u=\deg v-1$.

\textbf{(b)} If $u<v$ in $\widehat{P}$, then $\deg u<\deg v$.

\textbf{(c)} If $u<v$ in $\widehat{P}$ and $\deg u=\deg v-1$, then $u\lessdot
v$ in $\widehat{P}$.

\textbf{(d)} If $u\neq v$ and $\deg u=\deg v$, then $u$ and $v$ are
incomparable in $\widehat{P}$.
\end{proposition}

% \begin{proof}
% [\nopunct]The rather simple proofs of these facts are left to the reader.
% (Note that part \textbf{(a)} incorporates all three Assertions 1, 2 and 3 of
% Definition \ref{def.graded.n-graded}.)
% \end{proof}

% In words, Proposition \ref{prop.graded.Phat.why} \textbf{(d)} states that any
% two distinct elements of $\widehat{P}$ having the same degree are
% incomparable. We will use this several times below.

%One important observation is that any two distinct elements of a graded poset
%having the same degree are incomparable. Hence:

\begin{corollary}
\label{cor.Ri.welldef}
%Let $n\in\mathbb{N}$. Let $\mathbb{K}$ be a field. Let $P$ be an $n$-graded poset. 
Fix $i\in\left\{  1,2,\ldots,n\right\}  $, and let
$\left(  u_{1},u_{2},\ldots,u_{k}\right)  $ be any list of the elements of
$\widehat{P}_{i}$ with every element of $\widehat{P}_{i}$ appearing exactly
once in the list. Then, the dominant rational map $T_{u_{1}}\circ T_{u_{2}%
}\circ\ldots\circ T_{u_{k}}:\mathbb{K}^{\widehat{P}}\dashrightarrow
\mathbb{K}^{\widehat{P}}$ is well-defined and independent of the choice of the
list $\left(  u_{1},u_{2},\ldots,u_{k}\right)  $.
\end{corollary}

\begin{proof}
[Proof of Corollary \ref{cor.Ri.welldef}.]This is analogous to, but simpler than the
proof of Corollary \ref{cor.R.welldef}, because any two distinct elements of
$\widehat{P}_{i}$ are incomparable.
% (In place of the set $\mathcal{L}\left(
% P\right)  $ now serves the set of all lists of elements of $\widehat{P}_{i}$
% (with every element of $\widehat{P}_{i}$ appearing exactly once in the list).
% Any two elements of this latter set are equivalent under the relation $\sim$,
% because any two adjacent elements in such a list of elements of $\widehat{P}%
% _{i}$ are incomparable and can thus be switched.)
\end{proof}

\begin{definition}
\label{def.Ri}
Let $n$, $\mathbb{K}$ and $P$ be as in Corollary \ref{cor.Ri.welldef}.
Then, we define $R_i$ to be the dominant rational map
$T_{u_{1}}\circ T_{u_{2}}\circ\ldots\circ
T_{u_{k}}:\mathbb{K}^{\widehat{P}}\dashrightarrow\mathbb{K}^{\widehat{P}}$
described in Corollary \ref{cor.Ri.welldef}.
%Let $n\in\mathbb{N}$. Let $\mathbb{K}$ be a field. Let $P$ be an
%$n$-graded poset. Let $i\in\left\{  1,2,\ldots,n\right\}  $. Then, let $R_{i}$
%denote the dominant rational map $T_{u_{1}}\circ T_{u_{2}}\circ\ldots\circ
%T_{u_{k}}:\mathbb{K}^{\widehat{P}}\dashrightarrow\mathbb{K}^{\widehat{P}}$,
%where $\left(  u_{1},u_{2},\ldots,u_{k}\right)  $ is any list of the elements of
%$\widehat{P}_{i}$ with every element of $\widehat{P}_{i}$ appearing exactly
%once in the list. This map $T_{u_{1}}\circ T_{u_{2}}\circ\ldots\circ T_{u_{k}}$
%is well-defined (in particular, it does not depend on the list $\left(
%u_{1},u_{2},\ldots,u_{k}\right)  $) by Corollary \ref{cor.Ri.welldef}.
\end{definition}

The following two propositions show that for $n$-graded posets P, the $R_{i}$'s give
a way of writing $R$ as a product of fewer (but more complicated than toggles) involutions.  The proofs
involve simply writing everything as products of toggle maps and noting that each $R_{i}$ is
a product of \emph{commuting} involutions. 

\begin{proposition}
\label{prop.Ri.R}
% Let $n\in\mathbb{N}$. Let $\mathbb{K}$ be a field. Let $P$ be
% an $n$-graded poset. Then,%
The rowmotion operator on any $n$-graded poset $P$ can be factored as
\begin{equation}
R=R_{1}\circ R_{2}\circ\ldots\circ R_{n}. \label{def.Ri.R}%
\end{equation}

\end{proposition}

% \begin{proof}
% [Proof of Proposition \ref{prop.Ri.R}.]For every $i\in\left\{
% 1,2,\ldots,n\right\}  $, let $\left(  u_{1}^{\left[  i\right]  },u_{2}^{\left[
% i\right]  },\ldots,u_{k_{i}}^{\left[  i\right]  }\right)  $ be a list of the
% elements of $\widehat{P}_{i}$ with every element of $\widehat{P}_{i}$
% appearing exactly once in the list. Then, every $i\in\left\{
% 1,2,\ldots,n\right\}  $ satisfies $R_{i}=T_{u_{1}^{\left[  i\right]  }}\circ
% T_{u_{2}^{\left[  i\right]  }}\circ\ldots\circ T_{u_{k_{i}}^{\left[  i\right]  }%
% }$.

% But any listing of the elements of $P$ in order of increasing degree is a
% linear extension of $P$ (because any two distinct elements of a graded poset
% having the same degree are incomparable). Thus,
% \[
% \left(  u_{1}^{\left[  1\right]  },u_{2}^{\left[  1\right]  },\ldots,u_{k_{1}%
% }^{\left[  1\right]  },\ \ \ u_{1}^{\left[  2\right]  },u_{2}^{\left[
% 2\right]  },\ldots,u_{k_{2}}^{\left[  2\right]  },\ \ \ \ldots,\ \ \ u_{1}^{\left[
% n\right]  },u_{2}^{\left[  n\right]  },\ldots,u_{k_{n}}^{\left[  n\right]
% }\right)
% \]
% is a linear extension of $P$. Thus, by the definition of $R$, we have%
% \begin{align*}
% R  &  =\left(  T_{u_{1}^{\left[  1\right]  }}\circ T_{u_{2}^{\left[  1\right]
% }}\circ\ldots\circ T_{u_{k_{1}}^{\left[  1\right]  }}\right)  \circ\left(
% T_{u_{1}^{\left[  2\right]  }}\circ T_{u_{2}^{\left[  2\right]  }}%
% \circ\ldots\circ T_{u_{k_{2}}^{\left[  2\right]  }}\right)  \circ\ldots\circ\left(
% T_{u_{1}^{\left[  n\right]  }}\circ T_{u_{2}^{\left[  n\right]  }}%
% \circ\ldots\circ T_{u_{k_{n}}^{\left[  n\right]  }}\right) \\
% &  =R_{1}\circ R_{2}\circ\ldots\circ R_{n}%
% \end{align*}
% (since every $i\in\left\{  1,2,\ldots,n\right\}  $ satisfies $T_{u_{1}^{\left[
% i\right]  }}\circ T_{u_{2}^{\left[  i\right]  }}\circ\ldots\circ T_{u_{k_{i}%
% }^{\left[  i\right]  }}=R_{i}$). This proves Proposition \ref{prop.Ri.R}.
% \end{proof}

% We recall that birational rowmotion is a composition of toggle maps. As
% Proposition \ref{prop.Ri.R} shows, the operators $R_{i}$ are an
% \textquotedblleft intermediate\textquotedblright\ step between these toggle
% maps and birational rowmotion as a whole, though they are defined only when
% the poset $P$ is graded. They will be rather useful for us in our
% understanding of birational rowmotion (and the condition on $P$ to be graded
% doesn't prevent us from using them, since most of our results concern only
% graded posets anyway).

\begin{proposition}
\label{prop.Ri.invo}
% Let $n\in\mathbb{N}$. Let $\mathbb{K}$ be a field. Let $P$
% be an $n$-graded poset. 
For $i\in\left\{  1,2,\ldots,n\right\}$, the birational map $R_{i}$
is an involution (that is, $R_{i}^{2}=\operatorname*{id}$ on the set where
$R_{i}$ is defined).
\end{proposition}


% \begin{proof}
% [Proof of Proposition \ref{prop.Ri.invo}.]We defined $R_{i}$ as the
% composition $T_{u_{1}}\circ T_{u_{2}}\circ\ldots\circ T_{u_{k}}$ of the toggles
% $T_{u_{i}}$ where $\left(  u_{1},u_{2},\ldots,u_{k}\right)  $ is any list of the
% elements of $\widehat{P}_{i}$ with every element of $\widehat{P}_{i}$
% appearing exactly once in the list. These toggles are involutions and commute
% (the latter because any two distinct elements of $\widehat{P}_{i}$ are
% incomparable, having the same degree in $P$). Since a composition of commuting
% involutions is always an involution, this shows that $R_{i}$ is an involution.
% \end{proof}


Similarly to Proposition \ref{prop.R.implicit} one can then prove: 

\begin{proposition}
\label{prop.Ri.implicit}Let $n\in\mathbb{N}$. Let $P$ be an $n$-graded poset.
Let $i\in\left\{  1,2,\ldots,n\right\}  $. Let $\mathbb{K}$ be a field. Let
$v\in\widehat{P}$. Let $f\in\mathbb{K}^{\widehat{P}}$.

\textbf{(a)} If $\deg v\neq i$, then $\left(  R_{i}f\right)  \left(  v\right)
=f\left(  v\right)  $.

\textbf{(b)} If $\deg v=i$, then
\begin{equation}
\left(  R_{i}f\right)  \left(  v\right)  =\dfrac{1}{f\left(  v\right)  }%
\cdot\dfrac{\sum\limits_{\substack{u\in\widehat{P};\\u\lessdot v}}f\left(
u\right)  }{\sum\limits_{\substack{u\in\widehat{P};\\u\gtrdot v}}\dfrac
{1}{f\left(  u\right)  }}. \label{prop.Ri.implicit.eq}%
\end{equation}

\end{proposition}


% \begin{proof}
% [\nopunct]The proof of this proposition is very similar to that of Proposition
% \ref{prop.R.implicit} and therefore left to the reader.
% \end{proof}


Notice that using the proof of Proposition \ref{prop.Ri.implicit}, it is easy
to give an alternative proof of Corollary \ref{cor.Ri.welldef} (in the same
way as we saw that an alternative proof of Corollary \ref{cor.R.welldef} could
be given using the proofs of Propositions \ref{prop.R.implicit},
\ref{prop.R.implicit.01} and \ref{prop.R.implicit.converse}).


\section{w-tuples}

This section continues the study of birational rowmotion on graded posets by
introducing a ``fingerprint'' or ``checksum'' of a $\mathbb{K}$-labelling
called the w-tuple, defined by summing ratios of elements between successive
degrees (i.e., rows in the Hasse diagram). This w-tuple serves to extract some
information from a $\mathbb{K}$-labelling; we will later see how to make the
``rest'' of the labelling more manageable.

\begin{definition}
\label{def.wi}
% Let $n\in\mathbb{N}$. Let $\mathbb{K}$ be a field. Let $P$ be an
% $n$-graded poset. Let $f\in\mathbb{K}^{\widehat{P}}$. 
For $f\in\mathbb{K}^{\widehat{P}}$ and $i\in\left\{
0,1,\ldots,n\right\}  $, define $\mathbf{w}_{i}\left(  f\right) \in \mathbb{K} $ 
by
\[
\mathbf{w}_{i}\left(  f\right)  =\sum_{\substack{x\in\widehat{P}_{i}%
;\ y\in\widehat{P}_{i+1};\\y\gtrdot x}}\dfrac{f\left(  x\right)  }{f\left(
y\right)  }.
\]
(This element is not always defined, but is defined in the ``generic'' case
when $0\notin f\left(  \widehat{P}\right)  $.)
\end{definition}

Intuitively, one could think of $\mathbf{w}_{i}\left(  f\right)  $ as a kind
of \textquotedblleft checksum\textquotedblright\ for the labelling $f$, 
displaying how much its labels at degree $i+1$ differ from those at degree $i$.
Of course, in general, the knowledge of $\mathbf{w}_{i}\left(  f\right)  $ for
all $i\in\left\{  0,1,\ldots,n\right\}  $ is far from sufficient to reconstruct
the whole labelling $f$; however, in Definition \ref{def.hgeq}, we will
introduce the so-called homogenization of $f$, which will provide
\textquotedblleft complementary data\textquotedblright\ to these
$\mathbf{w}_{i}\left(  f\right)  $. 
For now we show that the 
$\mathbf{w}_{i}\left(  f\right)  $ behave in a rather simple way under the
maps $R$ and $R_{j}$.

\begin{definition}
% Let $n\in\mathbb{N}$. Let $\mathbb{K}$ be a field. Let $P$ be an $n$-graded
% poset. 
For $f\in\mathbb{K}^{\widehat{P}}$, we call the $\left(  n+1\right)  $-tuple
$\left(  \mathbf{w}_{0}\left(  f\right),\mathbf{w}_{1}\left(  f\right)
,\ldots,\mathbf{w}_{n}\left(  f\right)  \right)  $ the
\textit{w-tuple} of the $\mathbb{K}$-labelling $f$.
\end{definition}

\begin{proposition}
\label{prop.wi.Ri}
% Let $n\in\mathbb{N}$. Let $\mathbb{K}$ be a field. Let $P$
% be an $n$-graded poset. 
For every $i\in\left\{  1,2,\ldots,n\right\}  $ and every
$f\in\mathbb{K}^{\widehat{P}}$  we have
\begin{align*}
&  \left(  \mathbf{w}_{0}\left(  R_{i}f\right),\mathbf{w}_{1}\left(
R_{i}f\right),\ldots,\mathbf{w}_{n}\left(  R_{i}f\right)  \right) \\
&  =\left(  \mathbf{w}_{0}\left(  f\right),\mathbf{w}_{1}\left(  f\right)
,\ldots,\mathbf{w}_{i-2}\left(  f\right),\mathbf{w}_{i}\left(  f\right)
,\mathbf{w}_{i-1}\left(  f\right),\mathbf{w}_{i+1}\left(  f\right)
,\mathbf{w}_{i+2}\left(  f\right),\ldots,\mathbf{w}_{n}\left(  f\right)
\right).
\end{align*}
In other words, the map $R_{i}$ changes the w-tuple of a $\mathbb{K}%
$-labelling by interchanging its $\left(  i-1\right)  $-st entry with its
$i$-th entry (where the entries are labelled starting at $0$).
\end{proposition}

The proof of this is a straightforward computation, taking cases $j=i$, $j=i-1$, and $j\not
\in \{i-1,i \}$.  Details are left to the reader, or may be found in
\cite{grinberg-roby-arXiv}.  
% \begin{proof}
% [Proof of Proposition \ref{prop.wi.Ri}.]%

% Let $f\in\mathbb{K}^{\widehat{P}}$. We need to show that every $j\in\left\{
% 0,1,\ldots,n\right\}  $ satisfies%
% \begin{equation}
% \mathbf{w}_{j}\left(  R_{i}f\right)  =\mathbf{w}_{\tau_{i}\left(  j\right)
% }\left(  f\right), \label{pf.wi.Ri.1}%
% \end{equation}
% where $\tau_{i}$ is the permutation of the set $\left\{  0,1,\ldots,n\right\}  $
% which transposes $i-1$ with $i$ (while leaving all other elements of this set invariant).

% \textit{Proof of (\ref{pf.wi.Ri.1}):} Let $j\in\left\{  0,1,\ldots,n\right\}  $.
% We distinguish between three cases:

% \textit{Case 1:} We have $j=i$.

% \textit{Case 2:} We have $j=i-1$.

% \textit{Case 3:} We have $j\notin\left\{  i-1,i\right\}  $.


% Let us first consider Case 1. In this case, we have $j=i$. By the definition
% of $\mathbf{w}_{i}\left(  R_{i}f\right)  $, we have%
% \begin{align}
% \mathbf{w}_{i}\left(  R_{i}f\right)   &  =\sum_{\substack{x\in\widehat{P}%
% _{i};\ y\in\widehat{P}_{i+1};\\y\gtrdot x}}\dfrac{\left(  R_{i}f\right)
% \left(  x\right)  }{\left(  R_{i}f\right)  \left(  y\right)  }=\sum
% \limits_{x\in\widehat{P}_{i}}\left(  R_{i}f\right)  \left(  x\right)
% \sum\limits_{\substack{y\in\widehat{P}_{i+1};\\y\gtrdot x}}\left(
% \underbrace{\left(  R_{i}f\right)  \left(  y\right)  }_{\substack{=f\left(
% y\right)  \\\text{(by Proposition \ref{prop.Ri.implicit} \textbf{(a)})}%
% }}\right)  ^{-1}\nonumber\\
% &  =\sum\limits_{x\in\widehat{P}_{i}}\left(  R_{i}f\right)  \left(  x\right)
% \sum\limits_{\substack{y\in\widehat{P}_{i+1};\\y\gtrdot x}}\left(  f\left(
% y\right)  \right)  ^{-1}. \label{pf.wi.Ri.c1.short.1}%
% \end{align}
% But Proposition
% \ref{prop.Ri.implicit} \textbf{(b)} shows that
% every $x\in\widehat{P}_{i}$ satisfies
% \begin{align*}
% \left(  R_{i}f\right)  \left(  x\right)   &  =\dfrac{1}{f\left(  x\right)
% }\cdot\dfrac{\sum\limits_{\substack{u\in\widehat{P};\\u\lessdot x}}f\left(
% u\right)  }{\sum\limits_{\substack{u\in\widehat{P};\\u\gtrdot x}}\dfrac
% {1}{f\left(  u\right)  }}
% =\dfrac{1}{f\left(  x\right)  }\cdot\sum\limits_{\substack{u\in
% \widehat{P};\\u\lessdot x}}f\left(  u\right)  \cdot\left(  \sum
% \limits_{\substack{u\in\widehat{P};\\u\gtrdot x}}\left(  f\left(  u\right)
% \right)  ^{-1}\right)  ^{-1} \\
% &  =\dfrac{1}{f\left(  x\right)  }\cdot\sum\limits_{\substack{u\in
% \widehat{P}_{i-1};\\u\lessdot x}}f\left(  u\right)  \cdot\left(
% \sum\limits_{\substack{u\in\widehat{P}_{i+1};\\u\gtrdot x}}\left(  f\left(
% u\right)  \right)  ^{-1}\right)  ^{-1}%
% \end{align*}
% (here, we replaced $u\in\widehat{P}$ by $u\in\widehat{P}_{i-1}$ in the first
% sum (because every $u\in\widehat{P}$ satisfying $u\lessdot x$ must belong to
% $\widehat{P}_{i-1}$\ \ \ \ \footnote{since $x\in\widehat{P}_{i}$}) and we
% replaced $u\in\widehat{P}$ by $u\in\widehat{P}_{i+1}$ in the second sum (for
% similar reasons)) and thus
% \begin{equation}
% \left(  R_{i}f\right)  \left(  x\right)  \sum\limits_{\substack{u\in
% \widehat{P}_{i+1};\\u\gtrdot x}}\left(  f\left(  u\right)  \right)
% ^{-1}=\dfrac{1}{f\left(  x\right)  }\cdot\sum\limits_{\substack{u\in
% \widehat{P}_{i-1};\\u\lessdot x}}f\left(  u\right)  =\sum
% \limits_{\substack{u\in\widehat{P}_{i-1};\\u\lessdot x}}\dfrac{f\left(
% u\right)  }{f\left(  x\right)  }.
% \label{pf.wi.Ri.c1.short.1auv}
% \end{equation}
% Now, (\ref{pf.wi.Ri.c1.short.1}) (with the summation index $y$ renamed
% as $u$) becomes
% \begin{align}
% \mathbf{w}_{i}\left(  R_{i}f\right)   &  =\sum\limits_{x\in\widehat{P}_{i}%
% }\left(  R_{i}f\right)  \left(  x\right)  \sum
% \limits_{\substack{u\in\widehat{P}_{i+1};\\u\gtrdot x}}\left(  f\left(
% u\right)  \right)  ^{-1}
% =\sum\limits_{x\in\widehat{P}_{i}}\sum\limits_{\substack{u\in\widehat{P}%
% _{i-1};\\u\lessdot x}}\dfrac{f\left(  u\right)  }{f\left(  x\right)  }
% \ \ \ \ \ \ \ \ \ \ \ \left(\text{by (\ref{pf.wi.Ri.c1.short.1auv})}\right)
% \nonumber \\
% &=\sum\limits_{\substack{u\in\widehat{P}_{i-1};\ x\in\widehat{P}_{i};\\x\gtrdot
% u}}\dfrac{f\left(  u\right)  }{f\left(  x\right)  }
% %=\sum\limits_{\substack{x\in\widehat{P}_{i-1};\ y\in\widehat{P}%
% %_{i}\\y\gtrdot x}}\dfrac{f\left(  x\right)  }{f\left(  y\right)}
% %\nonumber\\
% %&\ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the indices }
% %u \text{ and } x \text{ as } x \text{ and } y \right) \nonumber\\
% %&
% =\mathbf{w}_{i-1}\left(  f\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the
% definition of }\mathbf{w}_{i-1}\left(  f\right)  \right)
% \label{pf.wi.Ri.short.main}\\
% &  =\mathbf{w}_{\tau_{i}\left(  i\right)  }\left(  f\right).\nonumber
% \end{align}
% In other words, (\ref{pf.wi.Ri.1}) holds for $j=i$. Thus, (\ref{pf.wi.Ri.1})
% is proven in Case 1.




% Let us now consider Case 2. In this case, $j=i-1$. Now, it can be shown that
% $\mathbf{w}_{i-1}\left(  R_{i}f\right)  =\mathbf{w}_{i}\left(  f\right)  $.
% This can be proven either in a similar way to how we proved $\mathbf{w}%
% _{i}\left(  R_{i}f\right)  =\mathbf{w}_{i-1}\left(  f\right)  $ (the details
% of this are left to the reader), or by noticing that%
% \begin{align*}
% \mathbf{w}_{i}\left(  f\right)   &  =\mathbf{w}_{i}\left(  R_{i}^{2}f\right)
% \ \ \ \ \ \ \ \ \ \ \left(
% \begin{array}
% [c]{c}%
% \text{since Proposition \ref{prop.Ri.invo} yields that }R_{i}^{2}%
% =\operatorname*{id}\text{,}\\
% \text{hence }\mathbf{w}_{i}\left(  R_{i}^{2}f\right)  =\mathbf{w}_{i}\left(
% \operatorname*{id}f\right)  =\mathbf{w}_{i}\left(  f\right)
% \end{array}
% \right) \\
% &  =\mathbf{w}_{i}\left(  R_{i}\left(  R_{i}f\right)  \right)  =\mathbf{w}%
% _{i-1}\left(  R_{i}f\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
% (\ref{pf.wi.Ri.short.main}), applied to }R_{i}f\text{ instead of }f\right).
% \end{align*}
% Either way, we end up knowing that $\mathbf{w}_{i-1}\left(  R_{i}f\right)
% =\mathbf{w}_{i}\left(  f\right)  $. Thus, $\mathbf{w}_{i-1}\left(
% R_{i}f\right)  =\mathbf{w}_{i}\left(  f\right)  =\mathbf{w}_{\tau_{i}\left(
% i-1\right)  }\left(  f\right)  $. In other words, (\ref{pf.wi.Ri.1}) holds for
% $j=i-1$. Thus, (\ref{pf.wi.Ri.1}) is proven in Case 2.



% Let us finally consider Case 3. In this case, $j\notin\left\{  i-1,i\right\}
% $. Hence, $\tau_{i}\left(  j\right)  =j$. On the other hand, by the definition
% of $\mathbf{w}_{j}\left(  R_{i}f\right)  $, we have%
% \begin{align*}
% \mathbf{w}_{j}\left(  R_{i}f\right)   &  =\sum_{\substack{x\in\widehat{P}%
% _{j};\ y\in\widehat{P}_{j+1};\\y \gtrdot x}}\dfrac{\left(  R_{i}f\right)
% \left(  x\right)  }{\left(  R_{i}f\right)  \left(  y\right)  }=\sum
% _{\substack{x\in\widehat{P}_{j};\ y\in\widehat{P}_{j+1};\\y \gtrdot x}}\left(
% \underbrace{\left(  R_{i}f\right)  \left(  y\right)  }_{\substack{=f\left(
% y\right)  \\\text{(by Proposition \ref{prop.Ri.implicit} \textbf{(a)})}%
% }}\right)  ^{-1}\cdot\underbrace{\left(  R_{i}f\right)  \left(  x\right)
% }_{\substack{=f\left(  x\right)  \\\text{(by Proposition
% \ref{prop.Ri.implicit} \textbf{(a)})}}}\\
% &  =\sum_{\substack{x\in\widehat{P}_{j};\ y\in\widehat{P}_{j+1};\\y \gtrdot
% x}}\left(  f\left(  y\right)  \right)  ^{-1}\cdot f\left(  x\right)
% =\sum_{\substack{x\in\widehat{P}_{j};\ y\in\widehat{P}_{j+1};\\y \gtrdot x
% }}\dfrac{f\left(  x\right)  }{f\left(  y\right)  }.
% \end{align*}
% Compared with $\mathbf{w}_{j}\left(  f\right)  =\sum\limits_{\substack{x\in
% \widehat{P}_{j};\ y\in\widehat{P}_{j+1};\\y \gtrdot x}}\dfrac{f\left(
% x\right)  }{f\left(  y\right)  }$ (by the definition of $\mathbf{w}_{j}\left(
% f\right)  $), this yields $\mathbf{w}_{j}\left(  R_{i}f\right)  =\mathbf{w}%
% _{j}\left(  f\right)  $. Since $j=\tau_{i}\left(  j\right)  $, this becomes
% $\mathbf{w}_{j}\left(  R_{i}f\right)  =\mathbf{w}_{\tau_{i}\left(  j\right)
% }\left(  f\right)  $. Hence, (\ref{pf.wi.Ri.1}) is proven in Case 3.

% We have thus proven (\ref{pf.wi.Ri.1}) in each of the three possible cases 1,
% 2 and 3. This completes the proof of (\ref{pf.wi.Ri.1}) and thus of
% Proposition \ref{prop.wi.Ri}.
% \end{proof}

From Proposition \ref{prop.wi.Ri}, and (\ref{def.Ri.R}), we conclude:

\begin{proposition}
\label{prop.wi.R} The map $R$ changes the w-tuple of any $f\in\mathbb{K}^{\widehat{P}}$ by 
shifting it cyclically: 
% Let $n\in\mathbb{N}$. Let $\mathbb{K}$ be a field. Let $P$ be
% an $n$-graded poset. Then, every $f\in\mathbb{K}^{\widehat{P}}$ satisfies%
For every $f\in\mathbb{K}^{\widehat{P}}$, we have
\[
\left(  \mathbf{w}_{0}\left(  Rf\right),\mathbf{w}_{1}\left(  Rf\right)
,\ldots,\mathbf{w}_{n}\left(  Rf\right)  \right)  =\left(  \mathbf{w}_{n}\left(
f\right),\mathbf{w}_{0}\left(  f\right),\mathbf{w}_{1}\left(  f\right)
,\ldots,\mathbf{w}_{n-1}\left(  f\right)  \right).
\]
\end{proposition}


% \begin{proof}
% [Proof of Proposition \ref{prop.wi.R}.]Proposition \ref{prop.Ri.R}
% yields $R=R_{1}\circ R_{2}\circ\ldots\circ R_{n}$. But for every $i\in\left\{
% 1,2,\ldots,n\right\}  $, recall from Proposition \ref{prop.wi.Ri} that the map
% $R_{i}$ changes the w-tuple of a $\mathbb{K}$-labelling by interchanging its
% $\left(  i-1\right)  $-st entry with its $i$-th entry (where the entries are
% labelled starting at $0$). Hence, the effect of the compound map $R=R_{1}\circ
% R_{2}\circ\ldots\circ R_{n}$ on the w-tuple is that of first interchanging the
% $\left(  n-1\right)  $-st entry with the $n$-th entry, then interchanging the
% $\left(  n-2\right)  $-st entry with the $\left(  n-1\right)  $-st entry, and
% so on, through to finally interchanging the $0$-th entry with the $1$-st
% entry. But this latter sequence of interchanges is equivalent to a cyclic
% shift of the w-tuple\footnote{Indeed, the composition $\left(  0,1\right)
% \circ\left(  1,2\right)  \circ\ldots\circ\left(  n-1,n\right)  $ of
% transpositions in the symmetric group on the set $\left\{  0,1,\ldots,n\right\}
% $ is the $\left(  n+1\right)  $-cycle $\left(  0,1,\ldots,n\right)  $.}. Hence,
% the map $R$ changes the w-tuple of a $\mathbb{K}$-labelling by shifting it
% cyclically.
% \end{proof}

As a consequence of Proposition \ref{prop.wi.R}, the map $R^{n+1}$ (for an
$n$-graded poset $P$) leaves the w-tuple of a $\mathbb{K}$-labelling fixed.

\section{Graded rescaling of labellings}

In general, birational rowmotion $R$ has something that one might call an
``avalanche effect'': If $f$ and $g$ are two $\mathbb{K}$-labellings of a
poset $P$ which differ from each other only in their labels at one single
element $v$, then the labellings $Rf$ and $Rg$ (in general) differ at all
elements covering $v$ and all elements beneath $v$, and further applications
of $R$ make the labellings even more different. Thus a change of just one
label in a labelling will often ``spread'' through a large part of the poset
when $R$ is repeatedly applied; the effect of such a change is hard to track
in general. Thus, knowing the behavior of one particular $\mathbb{K}%
$-labelling $f$ under $R$ does not help us at understanding the behaviors of
$\mathbb{K}$-labellings obtained from $f$ by changing labels at particular
elements. However, if $P$ is a \textbf{graded} poset and we simultaneously
multiply the labels at \textbf{all elements of a given degree} in a given
labelling of $P$ with a given scalar, then the changes this causes to the
behavior of the labelling under $R$ are rather predictable. We are going to
formalize this observation in this section, proving some explicit formulas for
how birational rowmotion $R$ and its iterates react to such rescalings. These
explicit formulas will be subsumed into slick conclusions in Section
\ref{sect.homogeneous}, where we will introducing a notion of
\emph{homogeneous equivalence} which formalizes the idea of a ``labelling
modulo scalar factors at each degree''.

The following definition formalizes the idea of multiplying the labels at all
elements of a certain degree with one and the same scalar factor:

\begin{definition}
\label{def.bemol}Let $P$ be
an $n$-graded poset. Let $f\in\mathbb{K}^{\widehat{P}}$, and
let\footnote{Here and in the following,
$\mathbb K^\times$ denotes the multiplicative group of nonzero elements
of $\mathbb K$.} $\left(  a_{0}%
,a_{1},\ldots,a_{n+1}\right)  \in\left(  \mathbb{K}^{\times}\right)  ^{n+2}$
be an $\left(  n+2\right)  $-tuple of nonzero elements of $\mathbb{K}$.
We
define a $\mathbb{K}$-labelling \newline
$\left(  a_{0},a_{1},\ldots,a_{n+1}\right)  \flat
f\in\mathbb{K}^{\widehat{P}}$ by%
\[
\left(  \left(  a_{0},a_{1},\ldots,a_{n+1}\right)  \flat f\right)  \left(
v\right)  =a_{\deg v}\cdot f\left(  v\right)  \ \ \ \ \ \ \ \ \ \ \text{for
every }v\in\widehat{P}.
\]
\end{definition}

We preserve the notations and the setting of this definition for the rest
of this section. Assuming that all applications of $R$ and
$R_{i}$ are well-defined, we have the following straightforward consequences
for $R_{i}$, $R$, and $R^{\ell}$.   


\begin{proposition}
\label{prop.Ri.scalmult}For every $i\in\left\{  1,2,\ldots,n\right\}  $,
we have
\[
 R_{i}\left(  \left(  a_{0},a_{1},\ldots,a_{n+1}\right)  \flat f\right)
  =\left(  a_{0},a_{1},\ldots,a_{i-1},\dfrac{a_{i+1}a_{i-1}}{a_{i}}%
,a_{i+1},a_{i+2},\ldots,a_{n+1}\right)  \flat\left(  R_{i}f\right)
\]

\end{proposition}

%A similar result can be obtained for $R$ instead of $R_{i}$:

\begin{proposition}
\label{prop.R.scalmult}Let $r=\dfrac{a_{n+1}}{a_{n}}$.  Then,
\[
R\left(  \left(  a_{0},a_{1},\ldots,a_{n+1}\right)  \flat f\right)  =\left(
a_{0},ra_{0},ra_{1},\ldots,ra_{n-1},a_{n+1}\right)  \flat\left(  Rf\right),
\]
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.R.scalmult}.]
We claim that every $j\in\left\{  1,2,\ldots,n+1\right\}  $ satisfies%
\begin{align}
&  \left(  R_{j}\circ R_{j+1}\circ\ldots\circ R_{n}\right)  \left(  \left(
a_{0},a_{1},\ldots,a_{n+1}\right)  \flat f\right) \nonumber\\
&  =\left(  a_{0},a_{1},a_{2},\ldots,a_{j-1},ra_{j-1},ra_{j},\ldots,ra_{n-1}%
,a_{n+1}\right)  \flat\left(  \left(  R_{j}\circ R_{j+1}\circ\ldots\circ
R_{n}\right)  f\right). \label{pf.R.scalmult.j}%
\end{align}
Indeed, (\ref{pf.R.scalmult.j}) is easily verified by reverse induction over
$j$ (that is, induction over $n+1-j$), using Proposition
\ref{prop.Ri.scalmult} in the step. Now, applying (\ref{pf.R.scalmult.j}) to
$j=1$ and recalling that $R = R_{1} \circ R_{2} \circ\ldots \circ R_{n}$, we
obtain Proposition \ref{prop.R.scalmult}.
\end{proof}

% We can go further and generalize Proposition \ref{prop.R.scalmult} to iterated
% birational rowmotion:

\begin{proposition}
\label{prop.Rl.scalmult}
For every $k, \ell \in\left\{  0,1,\ldots,n+1\right\}$, define an element 
$\widehat{a}_{k}^{\left(  \ell\right)  }\in\mathbb{K}^{\times}$ by%
\[
\widehat{a}_{k}^{\left(  \ell\right)  }=\left\{
\begin{array}
[c]{c}%
\dfrac{a_{n+1}a_{k-\ell}}{a_{n+1-\ell}},\ \ \ \ \ \ \ \ \ \ \  \ \text{if }%
k\geq\ell;\\
\dfrac{a_{n+1+k-\ell}a_{0}}{a_{n+1-\ell}},\ \ \ \ \ \ \ \ \ \ \text{if }k<\ell
\end{array}
\right..
\]
Then every $\ell \in \left\{0, 1, \ldots, n+1\right\}$ satisfies
% Let $f\in\mathbb{K}^{\widehat{P}}$ be a $\mathbb{K}$-labelling. Then, every
% $\ell\in\left\{  0,1,\ldots,n+1\right\}  $ satisfies%
\[
R^{\ell}\left(  \left(  a_{0},a_{1},\ldots,a_{n+1}\right)  \flat f\right)
=\left(  \widehat{a}_{0}^{\left(  \ell\right)  },\widehat{a}_{1}^{\left(
\ell\right)  },\ldots,\widehat{a}_{n+1}^{\left(  \ell\right)  }\right)
\flat\left(  R^{\ell}f\right)
\]
\end{proposition}


\begin{proof}
[Proof of Proposition \ref{prop.Rl.scalmult}.]This proof is a
completely straightforward induction over $\ell$, with the base case being
trivial and the induction step relying on Proposition \ref{prop.R.scalmult}.
To simplify the computations, notice that 
$\widehat{a}_{k}^{\left(  \ell\right)  }=\dfrac{a_{n+1+k-\ell}a_{0}%
}{a_{n+1-\ell}}$
if $k \leq \ell$.
\end{proof}



As a corollary, rescaling labellings commutes with 
$R^{n+1}$ for an $n$-graded poset $P$:

\begin{corollary}
\label{cor.Rl.scalmult} 
We have
$
R^{n+1}\left(  \left(  a_{0},a_{1},\ldots,a_{n+1}\right)  \flat f\right)
=\left(  a_{0},a_{1},\ldots,a_{n+1}\right)  \flat\left(  R^{n+1}f\right)
$.
%Let $n\in\mathbb{N}$. Let $\mathbb{K}$ be a field. Let
%$P$ be an $n$-graded poset.
%For every $\mathbb{K}$-labelling $f\in
%\mathbb{K}^{\widehat{P}}$ and any $\left(  n+2\right)  $-tuple $\left(
%a_{0},a_{1},\ldots,a_{n+1}\right)  \in\left(  \mathbb{K}^{\times}\right)  ^{n+2}%
%$, we define a $\mathbb{K}$-labelling $\left(  a_{0},a_{1},\ldots,a_{n+1}\right)
%\flat f\in\mathbb{K}^{\widehat{P}}$ as in Definition \ref{def.bemol}.
% Let $\left(  a_{0},a_{1},\ldots,a_{n+1}\right)  \in\left(  \mathbb{K}^{\times
% }\right)  ^{n+2}$. Let $f\in\mathbb{K}^{\widehat{P}}$ be a $\mathbb{K}%
% $-labelling. Then,%
% \[
% R^{n+1}\left(  \left(  a_{0},a_{1},\ldots,a_{n+1}\right)  \flat f\right)
% =\left(  a_{0},a_{1},\ldots,a_{n+1}\right)  \flat\left(  R^{n+1}f\right)
% \]
% %(provided that $R^{n+1}f$ is well-defined).
\end{corollary}


% \begin{proof}
% [\nopunct]We leave deriving Corollary \ref{cor.Rl.scalmult} from Proposition
% \ref{prop.Rl.scalmult} to the reader.
% \end{proof}

Finally, straightforward computations from the definitions show how rescaling a labeling by 
degree affects w-tuples: %(as defined in Definition \ref{def.wi}):

\begin{proposition}
\label{prop.w.scalmult}
% Let $n\in\mathbb{N}$. Let $\mathbb{K}$ be a field. Let
% $P$ be an $n$-graded poset.
%For every $\mathbb{K}$-labelling $f\in
%\mathbb{K}^{\widehat{P}}$ and any $\left(  n+2\right)  $-tuple $\left(
%a_{0},a_{1},\ldots,a_{n+1}\right)  \in\left(  \mathbb{K}^{\times}\right)  ^{n+2}%
%$, we define a $\mathbb{K}$-labelling $\left(  a_{0},a_{1},\ldots,a_{n+1}\right)
%\flat f\in\mathbb{K}^{\widehat{P}}$ as in Definition \ref{def.bemol}.
%
% Let $f\in\mathbb{K}^{\widehat{P}}$ be a $\mathbb{K}$-labelling of $P$. Let
% $\left(  a_{0},a_{1},\ldots,a_{n+1}\right)  \in\left(  \mathbb{K}^{\times
% }\right)  ^{n+2}$. Then, 
The w-tuple of the $\mathbb{K}$-labelling $\left(
a_{0},a_{1},\ldots,a_{n+1}\right)  \flat f$ is%
\[
\left(  \dfrac{a_{0}}{a_{1}}\mathbf{w}_{0}\left(  f\right),\dfrac{a_{1}%
}{a_{2}}\mathbf{w}_{1}\left(  f\right),\ldots,\dfrac{a_{n}}{a_{n+1}}%
\mathbf{w}_{n}\left(  f\right)  \right).
\]

\end{proposition}


% \begin{proof}
% [\nopunct]Proposition \ref{prop.w.scalmult} follows by computation using just
% the definitions of the notions involved.
% \end{proof}



\section{\label{sect.homogeneous}Homogeneous labellings}

In the previous section, we have quantified how the rescaling of all labels at
a given degree affects a labelling (of a graded poset) under birational
rowmotion. In this section, we will introduce a notion of \textquotedblleft
homogeneous labellings\textquotedblright\ which (roughly speaking) are
\textquotedblleft labellings up to rescaling at a given
degree\textquotedblright\ in the same way as a point in a projective space can
be regarded as \textquotedblleft a point in the affine
space up to rescaling the coordinates\textquotedblright. To be precise, we
will need to restrict ourselves to considering only \textquotedblleft
zero-free\textquotedblright\ labellings (a Zariski-dense open subset of all
labellings) for the same reason as we need to exclude $0$ when defining a
projective space. Once done with the definitions, we will see that birational
rowmotion (and the maps $R_{i}$) can be defined on homogeneous labellings 
(making use of the results of the previous section).

%Let us begin with the definitions:

\begin{definition}
%Let $\mathbb{K}$ be a field.

\textbf{(a)} For every $\mathbb{K}$-vector space $V$, let $\mathbb{P}\left(
V\right)  $ denote the projective space of $V$ (that is, the set
$V\setminus\left\{  0\right\}  $ modulo proportionality).

\textbf{(b)} For every $n\in\mathbb{N}$, we let $\mathbb{P}^{n}\left(
\mathbb{K}\right)  $ denote the projective space $\mathbb{P}\left(
\mathbb{K}^{n+1}\right)  $.
\end{definition}

\begin{definition}
\label{def.hgeq}
% Let $n\in\mathbb{N}$. Let $\mathbb{K}$ be a field. Let $P$ be
% an $n$-graded poset.
%
\textbf{(a)} Denote by $\overline{\mathbb{K}^{\widehat{P}}}$ the product
$\prod\limits_{i=1}^{n}\mathbb{P}\left(  \mathbb{K}^{\widehat{P}_{i}}\right)
$ of projective spaces. Notice that the product is just a Cartesian product of
algebraic varieties, and a reader unfamiliar with algebraic geometry can just
regard it as a Cartesian product of sets.\footnotemark\ 

We have $\overline{\mathbb{K}^{\widehat{P}}}=\prod\limits_{i=1}^{n}%
\mathbb{P}\left(  \mathbb{K}^{\widehat{P}_{i}}\right)  \cong\prod
\limits_{i=1}^{n}\mathbb{P}^{\left\vert \widehat{P}_{i}\right\vert -1}\left(
\mathbb{K}\right)  $ (since every $i \in \left\{1, 2, \ldots, n\right\}$
satisfies
$\mathbb{P}\left(  \mathbb{K}^{\widehat{P}_{i}}\right)
\cong \mathbb{P}^{\left\vert \widehat{P}_{i}\right\vert -1}\left(
\mathbb{K}\right)$). We denote the elements of $\overline{\mathbb{K}%
^{\widehat{P}}}$ as \textit{homogeneous labellings}.

Notice that $\overline{\mathbb{K}^{\widehat{P}}}=\prod\limits_{i=1}%
^{n}\mathbb{P}\left(  \mathbb{K}^{\widehat{P}_{i}}\right)  \cong%
\prod\limits_{i=0}^{n+1}\mathbb{P}\left(  \mathbb{K}^{\widehat{P}_{i}}\right)
$ (as algebraic varieties). This is because $\mathbb{K}^{\widehat{P}_{0}}$ and
$\mathbb{K}^{\widehat{P}_{n+1}}$ are $1$-dimensional vector spaces
(since $\left|\widehat{P}_{0}\right| = 1$ and
$\left|\widehat{P}_{n+1}\right| = 1$), and thus
the projective spaces $\mathbb{P}\left(  \mathbb{K}^{\widehat{P}_{0}}\right)
$ and $\mathbb{P}\left(  \mathbb{K}^{\widehat{P}_{n+1}}\right)  $ each consist
of a single point.

\textbf{(b)} A $\mathbb{K}$-labelling $f\in\mathbb{K}^{\widehat{P}}$ is said
to be \textit{zero-free} if for every $i\in\left\{  0,1,\ldots,n+1\right\}  $,
there exists some $v\in\widehat{P}_{i}$ satisfying $f\left(  v\right)  \neq0$
% (In other words, a $\mathbb{K}$-labelling $f\in\mathbb{K}^{\widehat{P}}$ is
% said to be zero-free if there exists no $i\in\left\{  0,1,\ldots,n+1\right\}  $
% such that $f$ is identically $0$ on all elements of $\widehat{P}$ having
% degree $i$.) 
(i.e., there exists no $i\in\left\{  0,1,\ldots,n+1\right\}  $
such that $f$ is identically $0$ on all elements of $\widehat{P}$ having
degree $i$.) 
Let $\mathbb{K}_{\neq0}^{\widehat{P}}$ be the set of all
zero-free $\mathbb{K}$-labellings. Clearly, this set $\mathbb{K}_{\neq
0}^{\widehat{P}}$ is a Zariski-dense open subset of $\mathbb{K}^{\widehat{P}}$.

\textbf{(c)} Identify the set $\mathbb{K}^{\widehat{P}}$ with $\prod
\limits_{i=0}^{n+1}\mathbb{K}^{\widehat{P}_{i}}$ in the obvious way 
% (since $\widehat{P}$, regarded as a set, is the disjoint union of the sets
% $\widehat{P}_{i}$ over all $i\in\left\{  0,1,\ldots,n+1\right\}  $).
(since, as a set, $\widehat{P} = \bigsqcup_{i=0}^{n+1} 
\widehat{P}_{i}$).


Using the identifications $\mathbb{K}^{\widehat{P}} \cong
\prod\limits_{i=0}^{n+1}\mathbb{K}^{\widehat{P}_{i}}$ and
$\overline{\mathbb{K}^{\widehat{P}}}
\cong \prod\limits_{i=0}^{n+1}
\mathbb{P}\left(  \mathbb{K}^{\widehat{P}_{i}}\right)$, we now
define a
rational map $\pi:\mathbb{K}^{\widehat{P}}\dashrightarrow\overline
{\mathbb{K}^{\widehat{P}}}$ as the product of the canonical projections
$\mathbb{K}^{\widehat{P}_{i}}\dashrightarrow\mathbb{P}\left(  \mathbb{K}%
^{\widehat{P}_{i}}\right)  $ (which are defined everywhere outside of the
$\left\{  0\right\}  $ subsets) over all $i\in\left\{  0,1,\ldots,n+1\right\}  $.
Notice that the domain of definition of this rational map $\pi$ is precisely
$\mathbb{K}_{\neq0}^{\widehat{P}}$. For every $f\in\mathbb{K}^{\widehat{P}}$,
we denote $\pi\left(  f\right)  $ as the \textit{homogenization} of the
$\mathbb{K}$-labelling $f$.

\textbf{(d)} Two zero-free $\mathbb{K}$-labellings $f\in\mathbb{K}%
^{\widehat{P}}$ and $g\in\mathbb{K}^{\widehat{P}}$ are said to be
\textit{homogeneously equivalent} if and only if they satisfy one of the
following equivalent conditions:

\textit{Condition 1:} For every $i\in\left\{  0,1,\ldots,n+1\right\}  $ and any
two elements $x$ and $y$ of $\widehat{P}_{i}$, we have $\dfrac{f\left(
x\right)  }{f\left(  y\right)  }=\dfrac{g\left(  x\right)  }{g\left(
y\right)  }$.

\textit{Condition 2:} There exists an $\left(  n+2\right)  $-tuple $\left(
a_{0},a_{1},\ldots,a_{n+1}\right)  \in\left(  \mathbb{K}^{\times}\right)  ^{n+2}$
such that
$g = \left(a_0, a_1, \ldots, a_{n+1}\right) \flat f$
(that is, such that
every $x\in\widehat{P}$ satisfies $g\left(  x\right)  =a_{\deg
x}\cdot f\left(  x\right)  $).

\textit{Condition 3:} We have $\pi\left(  f\right)  =\pi\left(  g\right)  $.

(The equivalence between these three conditions is very easy to check. We will
never actually use Condition 1.)
\end{definition}

\footnotetext{The structure of an algebraic variety will only be needed to define
the Zariski topology on $\overline{\mathbb{K}^{\widehat{P}}}$, which is more
or less obvious already (e.g., when we say that something holds ``for almost
every element $x$ of $\prod\limits_{i=1}^{n}\mathbb{P}\left(  \mathbb{K}%
^{\widehat{P}_{i}}\right)  $'', we could equivalently say that it holds ``for
$x=\operatorname*{proj}\left(  X\right)  $ for almost every element $X$ of
$\prod\limits_{i=1}^{n}\left(  \mathbb{K}^{\widehat{P}_{i}}\setminus\left\{
0\right\}  \right)  $'', where $\operatorname*{proj}$ is the canonical map
$\prod\limits_{i=1}^{n}\left(  \mathbb{K}^{\widehat{P}_{i}}\setminus\left\{
0\right\}  \right)  \rightarrow\prod\limits_{i=1}^{n}\mathbb{P}\left(
\mathbb{K}^{\widehat{P}_{i}}\right)  $ defined as the product of the
projections $\mathbb{K}^{\widehat{P}_{i}}\setminus\left\{  0\right\}
\rightarrow\mathbb{P}\left(  \mathbb{K}^{\widehat{P}_{i}}\right)  $).}

\begin{remark}
Clearly, homogeneous equivalence is an equivalence relation on the set
$\mathbb{K}_{\neq0}^{\widehat{P}}$ of all zero-free $\mathbb{K}$-labellings.
We can identify $\overline{\mathbb{K}^{\widehat{P}}}$ with the quotient of the
set $\mathbb{K}_{\neq0}^{\widehat{P}}$ modulo this relation. Then, $\pi$
becomes the canonical projection map $\mathbb{K}^{\widehat{P}}\dashrightarrow
\overline{\mathbb{K}^{\widehat{P}}}$.
\end{remark}

%One remark about the notion \textquotedblleft zero-free\textquotedblright:
Being zero-free is a very weak condition on a $\mathbb{K}$-labelling (indeed
the zero-free $\mathbb{K}$-labellings form a Zariski-dense open subset of the
space of all $\mathbb{K}$-labellings), and the $\mathbb{K}$-labellings which
don't satisfy this condition are rather useless for us (if $f$ is a
$\mathbb{K}$-labelling which is not zero-free, then $R^{2}f$ is not
well-defined, and usually not even $Rf$ is well-defined). We are almost never
giving up any generality if we require a labelling to be zero-free.

\begin{remark}
% Let $n\in\mathbb{N}$. Let $\mathbb{K}$ be a field. Let $P$ be an $n$-graded
% poset.
%For every $\mathbb{K}$-labelling $f\in\mathbb{K}^{\widehat{P}}$ and any
%$\left(  n+2\right)  $-tuple $\left(  a_{0},a_{1},\ldots,a_{n+1}\right)
%\in\left(  \mathbb{K}^{\times}\right)  ^{n+2}$, we define a $\mathbb{K}%
%$-labelling $\left(  a_{0},a_{1},\ldots,a_{n+1}\right)  \flat f\in\mathbb{K}%
%^{\widehat{P}}$ as in Definition \ref{def.bemol}.

Let $f\in\mathbb{K}_{\neq 0}^{\widehat{P}}$ 
and $\left(  a_{0},a_{1},\ldots,a_{n+1}\right)  \in\left(  \mathbb{K}%
^{\times}\right)  ^{n+2}$. Then $\left(
a_{0},a_{1},\ldots,a_{n+1}\right)  \flat f$ is also zero-free. (This follows
immediately from the definitions.)
\end{remark}

\begin{definition}
\label{def.hgeq.pii}
% Let $n\in\mathbb{N}$. Let $\mathbb{K}$ be a field. Let $P$
% be an $n$-graded poset. 
For every zero-free $f\in\mathbb{K}^{\widehat{P}}$
and every $i \in \left\{1,2,\ldots,n\right\}$,
the image of the restriction of $f : \widehat{P} \to \mathbb{K}$ to
$\widehat{P}_{i}$ under the canonical
projection $\mathbb{K}^{\widehat{P}_{i}}\dashrightarrow\mathbb{P}\left(
\mathbb{K}^{\widehat{P}_{i}}\right)  $ will be denoted by $\pi_{i}\left(
f\right)  $. This image $\pi_{i}\left(  f\right)  $ encodes the values of $f$
at the elements of $\widehat{P}$ of degree $i$ up to multiplying all these
values by a common nonzero scalar. Notice that
\begin{equation}
\pi\left(  f\right)  =\left(  \pi_{1}\left(  f\right),\pi_{2}\left(
f\right),\ldots,\pi_{n}\left(  f\right)  \right).
\label{def.hgeq.pii.eq}%
\end{equation}
% for every $f\in\mathbb{K}^{\widehat{P}}$. (Here, the right hand side of
% (\ref{def.hgeq.pii.eq}) is regarded as an element of $\overline{\mathbb{K}%
% ^{\widehat{P}}}$ because it belongs to $\prod\limits_{i=1}^{n}\mathbb{P}%
% \left(  \mathbb{K}^{\widehat{P}_{i}}\right)  =\overline{\mathbb{K}%
% ^{\widehat{P}}}$.)
\end{definition}

%We are next going to see:
We next note that birational rowmotion preserves homogeneous equivalence.  

\begin{corollary}
\label{cor.hgRi}
% Let $n\in\mathbb{N}$. Let $\mathbb{K}$ be a field. Let $P$ be
% an $n$-graded poset. Let $i\in\left\{  1,2,\ldots,n\right\}  $. 
If $f, g\in
\mathbb{K}^{\widehat{P}}$ are two
homogeneously equivalent zero-free $\mathbb{K}$-labellings, then $R_{i}f$ is
homogeneously equivalent to $R_{i}g$ (as long as $R_{i}f$ and $R_{i}g$ are zero-free).
\end{corollary}

\begin{corollary}
\label{cor.hgR}
% Let $n\in\mathbb{N}$. Let $\mathbb{K}$ be a field. Let $P$ be
% an $n$-graded poset. 
If $f, g\in\mathbb{K}^{\widehat{P}}$ are two homogeneously equivalent zero-free
$\mathbb{K}$-labellings, then $Rf$ is homogeneously equivalent to $Rg$ (as
long as $Rf$ and $Rg$ are zero-free).
\end{corollary}

Notice that Corollary \ref{cor.hgRi} would not be valid if we were to replace
$R_{i}$ by a single toggle $T_{v}$! So the operators $R_{i}$ in some sense
combine the nice properties of $T_{v}$ (like being an involution, cf.
Proposition \ref{prop.Ri.invo}) with the nice properties of $R$ (like having
an easily describable action on w-tuples, cf. Proposition \ref{prop.wi.Ri},
and respecting homogeneous equivalence, cf. Corollary \ref{cor.hgRi}).

\begin{proof}
[Proof of Corollary \ref{cor.hgRi}.]
% Let $f\in\mathbb{K}%
% ^{\widehat{P}}$ and $g\in\mathbb{K}^{\widehat{P}}$ be two homogeneously
% equivalent zero-free $\mathbb{K}$-labellings.
%We know that $f$ and $g$ are homogeneously equivalent.
By Condition 2 in
Definition \ref{def.hgeq} \textbf{(d)}, there exists an
$\left(  n+2\right)  $-tuple $\left(  a_{0},a_{1},\ldots,a_{n+1}\right)
\in\left(  \mathbb{K}^{\times}\right)  ^{n+2}$ such that
%\[
$g=\left(  a_{0},a_{1},\ldots,a_{n+1}\right)  \flat f$.
%\]
%Consider this $\left(  n+2\right)  $-tuple $\left(  a_{0},a_{1},\ldots,a_{n+1}%
%\right)  $. Since $g=\left(  a_{0},a_{1},\ldots,a_{n+1}\right)  \flat f$, we
%have
Thus, %This $\left(n+2\right)$-tuple thus satisfies
\begin{align*}
R_{i}g=R_{i}\left(  \left(  a_{0},a_{1},\ldots,a_{n+1}\right)  \flat f\right)
&  =\left(  a_{0},a_{1},\ldots,a_{i-1},\dfrac{a_{i+1}a_{i-1}}{a_{i}}%
,a_{i+1},a_{i+2},\ldots,a_{n+1}\right)  \flat\left(  R_{i}f\right)
\end{align*}
(by Proposition \ref{prop.Ri.scalmult}). 
Thus, $R_{i}f$ and $R_{i}g$ are homogenously equivalent.  
%
% Hence, there exists an $\left(
% n+2\right)  $-tuple $\left(  b_{0},b_{1},\ldots,b_{n+1}\right)  \in\left(
% \mathbb{K}^{\times}\right)  ^{n+2}$ such that
%\[
%$R_{i}g=\left(  b_{0},b_{1},\ldots,b_{n+1}\right)  \flat\left(  R_{i}f\right)$.
%\]
%(namely, $\left(  b_{0},b_{1},\ldots,b_{n+1}\right)  =\left(  a_{0}%
%,a_{1},\ldots,a_{i-1},\dfrac{a_{i+1}a_{i-1}}{a_{i}},a_{i+1},a_{i+2}%
%,\ldots,a_{n+1}\right)  $).
% In other words, there exists an $\left(  n+2\right)
% $-tuple $\left(  b_{0},b_{1},\ldots,b_{n+1}\right)  \in\left(  \mathbb{K}%
% ^{\times}\right)  ^{n+2}$ such that every $x\in\widehat{P}$ satisfies $\left(
% R_{i}g\right)  \left(  x\right)  =b_{\deg x}\cdot\left(  R_{i}f\right)
% \left(  x\right)  $. But this is precisely Condition 2 in Definition
% \ref{def.hgeq} \textbf{(d)}, stated for the labellings $R_{i}f$ and $R_{i}g$
% instead of $f$ and $g$. Hence, $R_{i}f$ and $R_{i}g$ are homogeneously
% equivalent. This proves Corollary \ref{cor.hgRi}.
\end{proof}

\begin{proof}
[Proof of Corollary \ref{cor.hgR}.]
%Let $f\in\mathbb{K}%
%^{\widehat{P}}$ and $g\in\mathbb{K}^{\widehat{P}}$ be two homogeneously
%equivalent zero-free $\mathbb{K}$-labellings. By iterative application of
%Corollary \ref{cor.hgRi}, we then conclude that the $\mathbb{K}$-labellings
%$\left(  R_{1}\circ R_{2}\circ\ldots\circ R_{n}\right)  f$ and $\left(
%R_{1}\circ R_{2}\circ\ldots\circ R_{n}\right)  g$ are homogeneously equivalent
%(if they are well-defined). Since $R_{1}\circ R_{2}\circ\ldots\circ R_{n}=R$ (by
%Proposition \ref{prop.Ri.R}), this rewrites as follows: The $\mathbb{K}%
%$-labellings $Rf$ and $Rg$ are homogeneously equivalent. This proves Corollary
%\ref{cor.hgR}.
Corollary \ref{cor.hgRi} shows that the map
$R_i$ (for every $1\leq i \leq n$) preserves
homogeneous equivalence of $\mathbb{K}$-labellings. Hence, so does
the composition $R_1 \circ R_2 \circ \ldots \circ R_n$ of these maps, which 
is $R$ (by Proposition \ref{prop.Ri.R}). 
\end{proof}


Let us introduce a general piece of notation:

\begin{definition}
% Let $S$ and $T$ be two sets. Let $\sim_{S}$ be an equivalence relation on the
% set $S$, and let $\sim_{T}$ be an equivalence relation on the set $T$. Let
% $\overline{S}$ be the quotient of the set $S$ modulo the equivalence relation
% $\sim_{S}$, and let $\overline{T}$ be the quotient of the set $T$ modulo the
% equivalence relation $\sim_{T}$. Let $\pi_{S}:S\rightarrow\overline{S}$ and
% $\pi_{T}:T\rightarrow\overline{T}$ be the canonical projections of a set on
% its quotient. Let $f:S\rightarrow T$ be a map. If $\overline{f}:\overline
% {S}\rightarrow\overline{T}$ is a map for which the diagram%
Let $S$ and $T$ be two sets, equipped with respective equivalence relations $\sim_{S}$ and
$\sim_{T}$. Let $\overline{S}=S/\sim_{S}$ and $\overline{T}=T/\sim_{T}$ be the
respective quotients, and $\pi_{S}:S\rightarrow\overline{S}$ and
$\pi_{T}:T\rightarrow\overline{T}$ the canonical projections.  
Let $f:S\rightarrow T$ be a map. If $\overline{f}:\overline
 {S}\rightarrow\overline{T}$ is a map for which the diagram%
\[
\xymatrix{
S \ar[r]^f \ar[d]_{\pi_S} & T \ar[d]^{\pi_T} \\
\overline{S} \ar[r]_{\overline{f}} & \overline{T}
}
\]
is commutative, then we say that \textquotedblleft the map $f$
\textit{descends} to the map $\overline{f}$\textquotedblright. It is easy to
see that there exists \textbf{at most one} map $\overline{f}:\overline
{S}\rightarrow\overline{T}$ such that the map $f$ descends to the map
$\overline{f}$ (for given $S$, $T$, $\sim_{S}$, $\sim_{T}$ and $f$). Moreover,
the existence of a map $\overline{f}:\overline{S}\rightarrow\overline{T}$ such
that the map $f$ descends to the map $\overline{f}$ is equivalent to the
statement that every two elements $x$ and $y$ of $S$ satisfying $x\sim_{S}y$
satisfy $f\left(  x\right)  \sim_{T}f\left(  y\right)  $.

The above statements are not literally true if we replace the map
$f:S\rightarrow T$ by a partial map $f:S\dashrightarrow T$. However, when $S$
and $T$ are two algebraic varieties and $\sim_{S}$ and $\sim_{T}$ are
algebraic equivalences (i.e., equivalence relations defined by polynomial
relations between coordinates of points) and $f:S\dashrightarrow T$ is a
\textbf{rational map}, then the above statements still are true (of course,
with $\overline{f}$ being a partial map).
\end{definition}

\begin{definition}
\label{def.hgRi}
% Let $n\in\mathbb{N}$. Let $\mathbb{K}$ be a field. Let $P$ be
% an $n$-graded poset.
Let $i\in\left\{  1,2,\ldots,n\right\}  $. 
Because of Corollary \ref{cor.hgRi}, the rational map $R_{i}:\mathbb{K}^{\widehat{P}%
}\dashrightarrow\mathbb{K}^{\widehat{P}}$ descends (through the projection
$\pi:\mathbb{K}^{\widehat{P}}\dashrightarrow\overline{\mathbb{K}^{\widehat{P}%
}}$) to a partial map $\overline{\mathbb{K}^{\widehat{P}}}\dashrightarrow
\overline{\mathbb{K}^{\widehat{P}}}$, which we denote by $\overline{R_{i}}$. Thus, the
diagram% 
\begin{equation}%
%TCIMACRO{\TeXButton{Ri commutes with pi}{\xymatrixcolsep{5pc}\xymatrix{
%\mathbb K^{\widehat P} \ar@{-->}[r]^{R_i} \ar@{-->}[d]_-{\pi} & \mathbb
%K^{\widehat P}
%\ar@{-->}[d]^-{\pi} \\
%\overline{\mathbb K^{\widehat P}} \ar@{-->}[r]_{\overline{R_i}} & \overline
%{\mathbb K^{\widehat P}}
%}} }%
%BeginExpansion
\xymatrixcolsep{5pc}\xymatrix{
\mathbb K^{\widehat P} \ar@{-->}[r]^{R_i} \ar@{-->}[d]_-{\pi} & \mathbb
K^{\widehat P}
\ar@{-->}[d]^-{\pi} \\
\overline{\mathbb K^{\widehat P}} \ar@{-->}[r]_{\overline{R_i}} & \overline
{\mathbb K^{\widehat P}}
}
%EndExpansion
\label{def.hgRi.commut}%
\end{equation}
is commutative.
\end{definition}

\begin{definition}
\label{def.hgR}
% Let $n\in\mathbb{N}$. Let $\mathbb{K}$ be a field. Let $P$ be
% an $n$-graded poset. 
Define the partial map $\ds \overline{R}:\overline
{\mathbb{K}^{\widehat{P}}}\dashrightarrow\overline{\mathbb{K}^{\widehat{P}}}$
by%
%\[
$\overline{R}=\overline{R_{1}}\circ\overline{R_{2}}\circ\ldots\circ\overline
{R_{n}}$.
%\]
Then, the diagram%
\begin{equation}%
%TCIMACRO{\TeXButton{R commutes with pi}{\xymatrixcolsep{5pc}\xymatrix{
%\mathbb K^{\widehat P} \ar@{-->}[r]^{R} \ar@{-->}[d]_-{\pi} & \mathbb
%K^{\widehat P}
%\ar@{-->}[d]^-{\pi} \\
%\overline{\mathbb K^{\widehat P}} \ar@{-->}[r]_{\overline{R}} & \overline
%{\mathbb K^{\widehat P}}
%}} }%
%BeginExpansion
\xymatrixcolsep{5pc}\xymatrix{
\mathbb K^{\widehat P} \ar@{-->}[r]^{R} \ar@{-->}[d]_-{\pi} & \mathbb
K^{\widehat P}
\ar@{-->}[d]^-{\pi} \\
\overline{\mathbb K^{\widehat P}} \ar@{-->}[r]_{\overline{R}} & \overline
{\mathbb K^{\widehat P}}
}
%EndExpansion
\label{def.hgR.commut}%
\end{equation}
is commutative\footnotemark. In other words, $\overline{R}$ is the partial map
$\overline{\mathbb{K}^{\widehat{P}}}\dashrightarrow\overline{\mathbb{K}%
^{\widehat{P}}}$ to which the partial map $R:\mathbb{K}^{\widehat{P}%
}\dashrightarrow\mathbb{K}^{\widehat{P}}$ descends (through the projection
$\pi:\mathbb{K}^{\widehat{P}}\dashrightarrow\overline{\mathbb{K}^{\widehat{P}%
}}$).
\end{definition}
%
\footnotetext{\textit{Proof.} We have $R=R_{1}\circ R_{2}\circ\ldots\circ R_{n}$
and $\overline{R}=\overline{R_{1}}\circ\overline{R_{2}}\circ\ldots\circ
\overline{R_{n}}$. Hence, the diagram (\ref{def.hgR.commut}) can be obtained
by stringing together the diagrams (\ref{def.hgRi.commut}) for all
$i\in\left\{  1,2,\ldots,n\right\}  $ and then removing the ``interior edges''.}
% Therefore, the diagram (\ref{def.hgR.commut}) is commutative (since the
% diagrams (\ref{def.hgRi.commut}) are commutative for all $i$), qed.}
%
The next result says roughly that a zero-free
$\mathbb{K}$-labelling $f\in\mathbb{K}^{\widehat{P}}$ is almost always
uniquely determined by three specifications: its w-tuple $\left(  \mathbf{w}_{0}\left(
f\right),\mathbf{w}_{1}\left(  f\right),\ldots,\mathbf{w}_{n}\left(
f\right)  \right)  $, its homogenization $\pi\left(  f\right)  $ and the value
$f\left(  0\right)  $. More precisely: 
%
\begin{proposition}
\label{prop.reconstruct}
Let $f,g\in \mathbb{K}_{\neq 0}^{\widehat{P}}$ satisfy
\begin{itemize}
\item $\pi\left(  f\right)  =\pi\left(
g\right)  $;
\item $f\left(  0\right)  =g\left(  0\right)  $; and
\item
$\left(  \mathbf{w}%
_{0}\left(  f\right),\mathbf{w}_{1}\left(  f\right),\ldots,\mathbf{w}%
_{n}\left(  f\right)  \right)  =\left(  \mathbf{w}_{0}\left(  g\right)
,\mathbf{w}_{1}\left(  g\right),\ldots,\mathbf{w}_{n}\left(  g\right)  \right)
\in \left(\mathbb{K}^{\times} \right)^{n+1}$.
\end{itemize}
Then, $f=g$.
\end{proposition}

Proposition \ref{prop.reconstruct} is easily proven by reconstructing $f$ and
$g$ \textquotedblleft bottom-up\textquotedblright\ along $\widehat{P}$.
Alternatively, we can 
use Proposition \ref{prop.w.scalmult}, as follows:

\begin{proof}
[Proof of Proposition \ref{prop.reconstruct}.]Since $\pi\left(
f\right)  =\pi\left(  g\right)  $, we know that $f$ and $g$ are homogeneously
equivalent. By Condition 2 in Definition \ref{def.hgeq} \textbf{(d)}, this
means that there exists an $\left(  n+2\right)  $-tuple $\left(  a_{0}%
,a_{1},\ldots,a_{n+1}\right)  \in\left(  \mathbb{K}^{\times}\right)  ^{n+2}$ such
that $g=\left(  a_{0},a_{1},\ldots,a_{n+1}\right)  \flat f $; hence,
\begin{align*}
\left(  \text{the w-tuple of }g\right)
%   &  =\left(  \text{the w-tuple of
%}\left(  a_{0},a_{1},\ldots,a_{n+1}\right)  \flat f\right) \\ &
=\left(  \dfrac{a_{0}}{a_{1}}\mathbf{w}_{0}\left(  f\right),\dfrac{a_{1}%
}{a_{2}}\mathbf{w}_{1}\left(  f\right),\ldots,\dfrac{a_{n}}{a_{n+1}}%
\mathbf{w}_{n}\left(  f\right)  \right)
\end{align*}
(by Proposition \ref{prop.w.scalmult}). Compared with%
\[
\left(  \text{the w-tuple of }g\right)  =\left(  \mathbf{w}_{0}\left(
g\right),\mathbf{w}_{1}\left(  g\right),\ldots,\mathbf{w}_{n}\left(
g\right)  \right)  =\left(  \mathbf{w}_{0}\left(  f\right),\mathbf{w}%
_{1}\left(  f\right),\ldots,\mathbf{w}_{n}\left(  f\right)  \right),
\]
this yields%
\[
\left(  \dfrac{a_{0}}{a_{1}}\mathbf{w}_{0}\left(  f\right),\dfrac{a_{1}%
}{a_{2}}\mathbf{w}_{1}\left(  f\right),\ldots,\dfrac{a_{n}}{a_{n+1}}%
\mathbf{w}_{n}\left(  f\right)  \right)  =\left(  \mathbf{w}_{0}\left(
f\right),\mathbf{w}_{1}\left(  f\right),\ldots,\mathbf{w}_{n}\left(
f\right)  \right).
\]
%In other words, $\dfrac{a_{i}}{a_{i+1}}\mathbf{w}_{i}\left(  f\right)
%=\mathbf{w}_{i}\left(  f\right)  $ for every $i\in\left\{  0,1,\ldots,n\right\}
%$.
Hence, $\dfrac{a_{i}}{a_{i+1}}=1$ for every $i\in\left\{
0,1,\ldots,n\right\}$. (Here we use the assumption that \newline
$\left(  \mathbf{w}%
_{0}\left(  f\right),\mathbf{w}_{1}\left(  f\right),\ldots,\mathbf{w}%
_{n}\left(  f\right)  \right)
\in \left(\mathbb{K}^{\times} \right)^{n+1}$.)
This forces $a_{0}=a_{1}=\ldots=a_{n+1}$, so the labels of 
$g$ are obtained from those of $f$ by multiplying by one single
constant. The assumption $f(0) = g(0)$ shows that this constant is
$1$, and thus $f=g$.
\end{proof}

\begin{definition}
In the following, if $S$ is a finite set, and $q$ is an element of a
projective space $\mathbb{P}\left(  \mathbb{K}^{S}\right)  $ of the free
vector space with basis $S$, and $k$ is an integer, then $q^{k}$ will denote
the element of $\mathbb{P}\left(  \mathbb{K}^{S}\right)  $ obtained by
replacing every homogeneous coordinate of $q$ by its $k$-th power. This is
well-defined (and will mostly be used for $k=-1$). In particular, this
definition applies to $S=\left\{  1,2,\ldots,n\right\}$
(in which case $\mathbb{K}^{S}=\mathbb{K}^{n}$).
\end{definition}

We can explicitly describe the action of the $\overline{R_{i}}$ when the
``structure of the poset $P$ between degrees $i-1$, $i$ and $i+1$'' is
particularly simple:

\begin{proposition}
\label{prop.hgRi.1}Fix $i\in\left\{  1,2,\ldots,n\right\}  $, and assume that
every $u\in\widehat{P}_{i}$ and every $v\in\widehat{P}_{i+1}$ satisfy
$u\lessdot v$. Assume further that every $u\in\widehat{P}_{i-1}$ and every
$v\in\widehat{P}_{i}$ satisfy $u\lessdot v$. Then,
\begin{align*}
&  \left(  \pi_{1}\left(  R_{i}f\right),\pi_{2}\left(  R_{i}f\right)
,\ldots,\pi_{n}\left(  R_{i}f\right)  \right) \\
&  =\left(  \pi_{1}\left(  f\right),\pi_{2}\left(  f\right),\ldots,\pi
_{i-1}\left(  f\right),\left(  \pi_{i}\left(  f\right)  \right)  ^{-1}%
,\pi_{i+1}\left(  f\right),\pi_{i+2}\left(  f\right),\ldots,\pi_{n}\left(
f\right)  \right).
\end{align*}

\end{proposition}

From this proposition, we obtain two corollaries:

\begin{corollary}
\label{cor.hgRi.1}Fix $i\in\left\{  1,2,\ldots,n\right\}  $. Assume that
every $u\in\widehat{P}_{i}$ and every $v\in\widehat{P}_{i+1}$ satisfy
$u\lessdot v$. Assume further that every $u\in\widehat{P}_{i-1}$ and every
$v\in\widehat{P}_{i}$ satisfy $u\lessdot v$. Let $\widetilde{f}=\left(
\widetilde{f}_{1},\widetilde{f}_{2},\ldots,\widetilde{f}_{n}\right)  \in
\overline{\mathbb{K}^{\widehat{P}}}$. Then,%
\[
\overline{R_{i}}\left(  \widetilde{f}\right)  =\left(  \widetilde{f}%
_{1},\widetilde{f}_{2},\ldots,\widetilde{f}_{i-1},\widetilde{f}_{i}%
^{-1},\widetilde{f}_{i+1},\widetilde{f}_{i+2},\ldots,\widetilde{f}_{n}\right).
\]
\end{corollary}

\begin{corollary}
\label{cor.hgRi.2}Assume that, for every $i\in\left\{
1,2,\ldots,n-1\right\}  $, every $u \in\widehat{P}_{i}$ and every $v
\in\widehat{P}_{i+1}$ satisfy $u \lessdot v$. Let $f\in\mathbb{K}%
^{\widehat{P}}$ be zero-free. Then,%
\[
\left(  \pi_{1}\left(  Rf\right),\pi_{2}\left(  Rf\right),\ldots,\pi
_{n}\left(  Rf\right)  \right)  =\left(  \left(  \pi_{1}\left(  f\right)
\right)  ^{-1},\left(  \pi_{2}\left(  f\right)  \right)  ^{-1},\ldots,\left(
\pi_{n}\left(  f\right)  \right)  ^{-1}\right).
\]
\end{corollary}

\section{\label{sect.ord}Order}

In this short section, we will relate the orders of the maps $R$ and
$\overline{R}$ for a graded poset $P$. The relation will later be used to better
understand both of these orders. 
We begin by defining the order of a partial map.  

\begin{definition}
\label{def.ord}Let $S$ be a set.

\textbf{(a)} If $\alpha$ and $\beta$ are two partial maps from the set $S$,
then we write \textquotedblleft$\alpha=\beta$\textquotedblright\ if and only
if every $s\in S$ for which both $\alpha\left(  s\right)  $ and $\beta\left(
s\right)  $ are well-defined satisfies $\alpha\left(  s\right)  =\beta\left(
s\right)  $. This is not a well-behaved notation \textit{per se}, e.g., it is
possible that three partial maps $\alpha$, $\beta$ and $\gamma$ satisfy
$\alpha=\beta$ and $\beta=\gamma$ but not $\alpha=\gamma$. However, we 
will only use this notation for rational maps and their quotients (and, of
course, total maps); in all of these cases, the notation \textbf{is}
well-behaved (e.g., if $\alpha$, $\beta$ and $\gamma$ are three rational maps
satisfying $\alpha=\beta$ and $\beta=\gamma$, then $\alpha=\gamma$, because
the intersection of two Zariski-dense open subsets is Zariski-dense and open).

\textbf{(b)} The \textit{order} of a partial map $\varphi:S\dashrightarrow S$
is defined to be the smallest positive integer $k$ satisfying $\varphi
^{k}=\operatorname*{id}\nolimits_{S}$, if such a positive integer $k$ exists,
and $\infty$ otherwise. Here, we are disregarding the fact that $\varphi$ is
only a partial map; we will be working only with dominant rational maps and
their quotients (and total maps), so nothing will go wrong.

We denote the order of a partial map $\varphi:S\dashrightarrow S$ as
$\operatorname*{ord}\varphi$.
\end{definition}

\begin{convention}
In the following, we are going to occasionally make arithmetical statements
involving the symbol $\infty$. We declare that $0$ and $\infty$ are divisible
by $\infty$, but no positive integer is divisible by $\infty$. We further
declare that every positive integer (but not $0$) divides $\infty$. We set
$\operatorname{lcm}\left(  a,\infty\right)  $ and $\operatorname{lcm}\left(
\infty,a\right)  $ to mean $\infty$ whenever $a$ is a positive integer.
\end{convention}

As a consequence of Proposition \ref{prop.reconstruct}, we have:

\begin{proposition}
\label{prop.ord-projord}Let $n\in\mathbb{N}$. Let $\mathbb{K}$ be a field. Let
$P$ be an $n$-graded poset. Then, $\operatorname*{ord}R=\operatorname{lcm}%
\left(  n+1,\operatorname*{ord}\overline{R}\right)  $. (Recall that
$\operatorname{lcm}\left(  n+1,\infty\right)  $ is to be understood as
$\infty$.)
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.ord-projord}.]%
The proof of this boils down to considering the effect of $R$ on the w-tuple
$\left(  \mathbf{w}_{0}\left(  f\right),\mathbf{w}_{1}\left(
f\right),\ldots,\mathbf{w}_{n}\left(  f\right)  \right)  $ and on the
homogenization $\pi\left(  f\right)  $ of a $\mathbb{K}$-labelling $f$. The
former is a cyclic shift (by Proposition \ref{prop.wi.R}),
with order $n+1$, and the latter is $\overline{R}$. It
is now easy to see (invoking Proposition \ref{prop.reconstruct}) that the
order of $R$ is the $\operatorname{lcm}$ of the orders of these two actions.
%% Details are left to the reader, or may be found in \cite{grinberg-roby-arXiv}.  
We outline the steps here; full details may be found in \cite{grinberg-roby-arXiv}.  

\textit{1st step:} Use commutativity of the diagram (\ref{def.hgR.commut})
to show $\overline{R}^{\ell}\circ \pi=\pi\circ R^{\ell}$ for all $\ell \in
\mathbb{N}$. Thus, $\overline{R}^{\ell} = \operatorname*{id}$ whenever
$R^{\ell} = \operatorname*{id}$, so $\operatorname*{ord}% 
\overline{R}\mid\operatorname*{ord}R$.  
% yields $\overline{R}\circ\pi=\pi\circ R$. Hence,
% \begin{equation}
% \text{every }\ell\in\mathbb{N}\text{ satisfies }\overline{R}^{\ell}\circ
% \pi=\pi\circ R^{\ell} \label{pf.ord-projord.Rl}%
% \end{equation}
% (this is clear by induction over $\ell$). Thus, if some $\ell\in\mathbb{N}$
% satisfies $R^{\ell}=\operatorname*{id}$, then it satisfies $\overline{R}%
% ^{\ell}=\operatorname*{id}$ as well\footnote{\textit{Proof.} Let $\ell
% \in\mathbb{N}$ be such that $R^{\ell}=\operatorname*{id}$. Then, $\overline
% {R}^{\ell}\circ\pi=\pi\circ\underbrace{R^{\ell}}_{=\operatorname*{id}}=\pi$.
% Since $\pi$ is right-cancellable (since $\pi$ is surjective), this yields
% $\overline{R}^{\ell}=\operatorname*{id}$, qed.}. Hence, 
% $\operatorname*{ord}% 
% \overline{R}\mid\operatorname*{ord}R$ (recall that every positive integer
% divides $\infty$, but only $0$ and $\infty$ are divisible by $\infty$). In
% particular, if $\operatorname*{ord}\overline{R}=\infty$, then
% $\operatorname*{ord}R=\infty$. Thus, Proposition \ref{prop.ord-projord} is
% obvious in the case when $\operatorname*{ord}\overline{R}=\infty$. Hence, for
% the rest of the proof of Proposition \ref{prop.ord-projord}, we WLOG
% assume that $\operatorname*{ord}\overline{R}\neq\infty$.

\textit{2nd step:} Assume $\ord \overline{R} = m \neq \infty$ (otherwise the statement
is easy), and let $\ell =\lcm
(n+1,m)$.  To show that $R^{\ell}=\id$, it suffices to show that
$R^\ell f = f$ for almost every
(Zariski) $f\in \KK_{\neq 0}^{\widehat{P}}$.  A simple argument shows that almost every $f \in
\KK^{\widehat{P}}$ is zero-free and has (well-defined) nonzero $\mathbf{w}_{i}(f)$ for each
$i$.  By Proposition~\ref{prop.wi.R}, the labellings $R^{\ell}f$ and $f$ have the same w-tuple
(since $n+1 \mid \ell$), and from $\ord \overline{R} = m \mid \ell$ we obtain
$\pi (R^{\ell}f) = \pi (f)$.  So using Corollary~\ref{cor.R.implicit.01} and applying
Proposition~\ref{prop.reconstruct} to $g = R^\ell f$, we get $R^{\ell}f=f$.
Thus, $R^{\ell}=\id$, so $\ord R \mid \ell$.  

% Since $\operatorname*{ord}\overline{R}\neq\infty$, we know
% that $\operatorname*{ord}\overline{R}$ is a positive integer. Let $m$ be this
% positive integer. Then, $m=\operatorname*{ord}\overline{R}$, so that
% $\overline{R}^{m}=\operatorname*{id}$.

% Let $\ell=\operatorname{lcm}\left(  n+1,m\right)  $. Then, $n+1\mid\ell$ and
% $m\mid\ell$. Since $\operatorname*{ord}\overline{R}=m\mid\ell$, we have
% $\overline{R}^{\ell}=\operatorname*{id}$. But from (\ref{pf.ord-projord.Rl}),
% we have $\pi\circ R^{\ell}=\underbrace{\overline{R}^{\ell}}%
% _{=\operatorname*{id}}\circ\pi=\pi$.

% We are now going to prove that $R^{\ell}=\operatorname*{id}$. In order to
% prove this, it is clearly enough to show that almost every (in the sense of
% Zariski topology) zero-free $\mathbb{K}$-labelling $f$ of $P$ satisfies
% $R^{\ell}f=\operatorname*{id}f$ (because $R^{\ell}f=\operatorname*{id}f$ is a
% polynomial identity in the labels of $f$). But it is easily shown that for
% almost every (in the sense of Zariski topology) zero-free $\mathbb{K}%
% $-labelling $f$ of $P$, the w-tuple $\left(  \mathbf{w}_{0}\left(  f\right)
%,\mathbf{w}_{1}\left(  f\right),\ldots,\mathbf{w}_{n}\left(  f\right)  \right)
% $ of $f$ consists of nonzero elements of
% $\mathbb{K}$\ \ \ \ \footnote{\textit{Proof.} We will prove a slightly
% better result: Almost every
% $f\in\mathbb{K}^{\widehat{P}}$ is a zero-free $\mathbb{K}$-labelling of $P$
% with the property that%
% \begin{equation}
% \left(  \mathbf{w}_{i}\left(  f\right)  \text{ is well-defined and nonzero for
% every }i\in\left\{  0,1,\ldots,n\right\}  \right)
%.\label{pf.ord-projord.short.step2.zari.1}%
% \end{equation}
% \par
% In fact, the condition (\ref{pf.ord-projord.short.step2.zari.1}) on an
% $f\in\mathbb{K}^{\widehat{P}}$ is a requirement saying that certain rational
% expressions in the values of $f$ do not vanish (namely, the denominators in
% $\mathbf{w}_{i}\left(  f\right)  $ and the sums $\mathbf{w}_{i}\left(
% f\right)  $ themselves). If we can prove that none of these expressions is
% identically zero, then it will follow that for almost every $f\in
% \mathbb{K}^{\widehat{P}}$, none of these expressions vanishes (because there
% are only finitely many expressions whose vanishing we are trying to avoid, and
% the infiniteness of $\mathbb{K}$ allows us to avoid them all if none of them
% is identically zero); thus (\ref{pf.ord-projord.short.step2.zari.1}) will
% follow and we will be done. Hence, it remains to show that none of these
% expressions is identically zero.
% \par
% Assume the contrary. Then, one of our rational expressions -- either a
% denominator in one of the $\mathbf{w}_{i}\left(  f\right)  $, or one of the
% sums $\mathbf{w}_{i}\left(  f\right)  $ -- identically vanishes. It must be
% one of the sums $\mathbf{w}_{i}\left(  f\right)  $, since the denominators in
% the $\mathbf{w}_{i}\left(  f\right)  $ cannot identically vanish (they are
% simply values $f\left(  y\right)  $). So there exists some $i\in\left\{
% 0,1,\ldots,n\right\}  $ such that every $\mathbb{K}$-labelling $f$ of $P$ (for
% which $\mathbf{w}_{i}\left(  f\right)  $ is well-defined) satisfies
% $\mathbf{w}_{i}\left(  f\right)  =0$. Consider this $i$. Notice that $i\leq n$
% and thus $1\notin\widehat{P}_{i}$.
% \par
% We have
% \[
% 0=\mathbf{w}_{i}\left(  f\right)  =\sum_{\substack{x\in\widehat{P}_{i}%
% ;\ y\in\widehat{P}_{i+1};\\y\gtrdot x}}\dfrac{f\left(  x\right)  }{f\left(
% y\right)  }=\sum_{x\in\widehat{P}_{i}}f\left(  x\right)  \sum_{\substack{y\in
% \widehat{P}_{i+1};\\y\gtrdot x}}\dfrac{1}{f\left(  y\right)  }.
% \]
% This forces the sum $\sum_{\substack{y\in\widehat{P}_{i+1};\\y\gtrdot
% x}}\dfrac{1}{f\left(  y\right)  }$ to be identically $0$ for every
% $x\in\widehat{P}_{i}$ (because these sums for different values of $x$ are
% prevented from canceling each other by the completely independent $f\left(
% x\right)  $ coefficients in front of them). Fix some $x\in\widehat{P}_{i}$
% (such an $x$ clearly exists since $\deg:\widehat{P}\rightarrow\left\{
% 0,1,\ldots,n+1\right\}  $ is surjective), and ponder what it means for the sum
% $\sum_{\substack{y\in\widehat{P}_{i+1};\\y\gtrdot x}}\dfrac{1}{f\left(
% y\right)  }$ to be identically $0$. It means that this sum is empty, i.e.,
% that there exists no $y\in\widehat{P}_{i+1}$ satisfying $y\gtrdot x$. But this
% can only happen when $x=1$, which is not the case in our situation (because
% $x\in\widehat{P}_{i}$ and $1\notin\widehat{P}_{i}$). So we have obtained a
% contradiction.}. Hence, in order to
% prove $R^{\ell}=\operatorname*{id}$, it is enough to show that every zero-free
% $\mathbb{K}$-labelling $f$ of $P$ for which the w-tuple $\left(
% \mathbf{w}_{0}\left(  f\right),\mathbf{w}_{1}\left(  f\right)
%,\ldots,\mathbf{w}_{n}\left(  f\right)  \right)  $ of $f$ consists of nonzero
% elements of $\mathbb{K}$ satisfies $R^{\ell}f=\operatorname*{id}f$.
% %  This is
% % what we are going to do now.

% So let $f$ be such a labelling.
% %We will prove that $R^{\ell}f=\operatorname*{id}f$.
% From Proposition \ref{prop.wi.R}, we know that the map $R$ changes the w-tuple
% of a $\mathbb{K}$-labelling by shifting it cyclically. Hence, for every
% $k\in\mathbb{N}$, the map $R^{k}$ changes the w-tuple of a $\mathbb{K}%
% $-labelling by shifting it cyclically $k$ times. If this $k$ is divisible by
% $n+1$, then this obviously means that the map $R^{k}$ preserves the w-tuple of
% a $\mathbb{K}$-labelling (because the w-tuple has $n+1$ entries, and thus
% shifting it cyclically for a multiple of $n+1$ times leaves it invariant).
% Hence, the w-tuple of $f$ equals the w-tuple of $R^{\ell}f$. Recalling the
% definition of a w-tuple, we can rewrite this as follows:%
% \[
% \left(  \mathbf{w}_{0}\left(  f\right),\mathbf{w}_{1}\left(  f\right)
%,\ldots,\mathbf{w}_{n}\left(  f\right)  \right)  =\left(  \mathbf{w}_{0}\left(
% R^{\ell}f\right),\mathbf{w}_{1}\left(  R^{\ell}f\right),\ldots,\mathbf{w}%
% _{n}\left(  R^{\ell}f\right)  \right).
% \]




% Moreover, by assumption, the w-tuple $\left(  \mathbf{w}_{0}\left(  f\right)
%,\mathbf{w}_{1}\left(  f\right),\ldots,\mathbf{w}_{n}\left(  f\right)  \right)
% $ of $f$ consists of nonzero elements of $\mathbb{K}$. In other words, no
% $i\in\left\{  0,1,\ldots,n\right\}  $ satisfies $\mathbf{w}_{i}\left(  f\right)
% =0$.
% %
% Furthermore $\pi\left(  R^{\ell}f\right)  =\underbrace{\left(  \pi\circ
% R^{\ell}\right)  }_{=\pi}f=\pi\left(  f\right)  $.
% %
% Also, Corollary \ref{cor.R.implicit.01} (applied to $k=\ell$) yields $\left(
% R^{\ell}f\right)  \left(  0\right)  =f\left(  0\right)  $.
% %
% We now can apply Proposition \ref{prop.reconstruct} to $g=R^{\ell}f$. As a
% result, we obtain $R^{\ell}f=f$. In other words, $R^{\ell}f=\operatorname*{id}%
% f$.

% %Now forget that we fixed $f$.
% We have thus shown that $R^{\ell}%
% f=\operatorname*{id}f$ for every zero-free $\mathbb{K}$-labelling $f$ of $P$
% for which the w-tuple $\left(  \mathbf{w}_{0}\left(  f\right),\mathbf{w}%
% _{1}\left(  f\right),\ldots,\mathbf{w}_{n}\left(  f\right)  \right)  $ of $f$
% consists of nonzero elements of $\mathbb{K}$. Therefore, we have shown that
% $R^{\ell}=\operatorname*{id}$ (by what we have said above). Thus,
% $\operatorname*{ord}R\mid\ell=\operatorname{lcm}\left(  n+1,\underbrace{m}%
% _{=\operatorname*{ord}\overline{R}}\right)  =\operatorname{lcm}\left(
% n+1,\operatorname*{ord}\overline{R}\right)  $.

\textit{3rd step:} To show the inverse divisibility
($\ell \mid \ord R$), first assume $\ord R = q \neq
\infty$ (else the claim is obvious).  
It is easy to see that for almost every 
zero-free $\mathbb{K}$-labelling $f$ of $P$, the entries of the w-tuple
$\left(  \mathbf{w}_{0}\left(  f\right),\mathbf{w}_{1}\left(  f\right)
,\ldots,\mathbf{w}_{n}\left(  f\right)  \right)  $ of $f$ are pairwise distinct. So there
exists $f\in \KK_{\neq 0}^{\widehat{P}}$ with this property such that $R^{k}f$ is well-defined for all
$k\in\left\{  0,1,\ldots,q\right\}  $.  Then $R^{q}f = f$, so they have the same w-tuple.  But
by pairwise-distinctness, this can only happen if $n+1 \mid q$.  Combined with the first
step, this gives $\ell \mid \ord R$.  
%
% **************************************************************************
% We now will show that $\operatorname{lcm}\left(
% n+1,\operatorname*{ord}\overline{R}\right)  \mid\operatorname*{ord}R$.
%
% In order to do that, we assume WLOG that $\operatorname*{ord}R\neq\infty$
% (else, the claim is obvious). Hence,
% $\operatorname*{ord}R$ is a positive integer. Denote this positive integer by
% $q$. So, $q=\operatorname*{ord}R$.
%
% It is easy to see that for almost every (in the sense of Zariski topology)
% zero-free $\mathbb{K}$-labelling $f$ of $P$, the entries of the w-tuple
% $\left(  \mathbf{w}_{0}\left(  f\right),\mathbf{w}_{1}\left(  f\right)
%,\ldots,\mathbf{w}_{n}\left(  f\right)  \right)  $ of $f$ are pairwise distinct.
% Hence, there exists a zero-free $\mathbb{K}$-labelling $f$ of $P$ such that
% the entries of the w-tuple $\left(  \mathbf{w}_{0}\left(  f\right)
%,\mathbf{w}_{1}\left(  f\right),\ldots,\mathbf{w}_{n}\left(  f\right)  \right)
% $ of $f$ are pairwise distinct and such that $R^{k}f$ is well-defined for all
% $k\in\left\{  0,1,\ldots,q\right\}  $. Consider such an $f$.
%
% Since $q=\operatorname*{ord}R$, we have $R^{q}=\operatorname*{id}$, so that
% $R^{q}f=f$.
%
%
% Recall once again (from the 2nd step) that for every $k\in\mathbb{N}$, the map
% $R^{k}$ changes the w-tuple of a $\mathbb{K}$-labelling by shifting it
% cyclically $k$ times.
% %In particular, the map $R^{q}$ changes the w-tuple of
% %the $\mathbb{K}$-labelling $f$ by shifting it cyclically $q$ times. In other
% %words,
% Therefore,
% the w-tuple of $R^{q}f$ is obtained from the w-tuple of $f$ by shifting
% it cyclically $q$ times. Since the w-tuple of $f$ is an
% $\left(  n+1\right)  $-tuple of pairwise distinct entries, we thus must
% have $R^q f \neq f$ unless $n+1 \mid q$. Since $R^q f = f$, we thus have
% $n+1\mid q$.
%
%
%
% Combining $n+1\mid q=\operatorname*{ord}R$ with $\operatorname*{ord}%
% \overline{R}\mid\operatorname*{ord}R$, we obtain $\operatorname{lcm}\left(
% n+1,\operatorname*{ord}\overline{R}\right)  \mid\operatorname*{ord}R$.
% Combining this with $\operatorname*{ord}R\mid\operatorname{lcm}\left(
% n+1,\operatorname*{ord}\overline{R}\right)  $, we obtain $\operatorname*{ord}%
% R=\operatorname{lcm}\left(  n+1,\operatorname*{ord}\overline{R}\right)  $.
% This proves Proposition \ref{prop.ord-projord}.
\end{proof}

\section{The opposite poset} \label{sect.opposite}

Before we move on to the first interesting class of posets for which we can
compute the order of birational rowmotion, let us prove an easy ``symmetry
property'' of birational rowmotion.  For this section, we only assume that $P$ is a finite
poset (not necessarily graded).  

\begin{definition}
\label{def.op}Let $P$ be a poset. Then, $P^{\operatorname*{op}}$ will denote
the poset defined on the same ground set as $P$ but with the order relation
defined by%
\[
\left(  \left(  a<_{P^{\operatorname*{op}}}b\text{ if and only if }%
b<_{P}a\right)  \text{ for all }a\in P\text{ and }b\in P\right)
\]
% (where $<_{P}$ denotes the smaller relation of the poset $P$, and where
% $<_{P^{\operatorname*{op}}}$ denotes the smaller relation of the poset
% $P^{\operatorname*{op}}$ which we are defining). 
The poset $P^{\operatorname*{op}}$ is called the \textit{opposite poset} (or \textit{dual}
poset~\cite{stanley-ec1}) of $P$.
\end{definition}

% Note that $P^{\operatorname*{op}}$ is called the \textit{dual} of the poset
% $P$ in \cite{stanley-ec1}.

% \begin{remark}
% It is clear that $\left(  P^{\operatorname*{op}}\right)  ^{\operatorname*{op}%
% }=P$ for any poset $P$. Also, if $n\in\mathbb{N}$, and if $P$ is an $n$-graded
% poset, then $P^{\operatorname*{op}}$ is an $n$-graded poset.
% \end{remark}

\begin{definition}
We denote the maps $R$
and $\overline{R}$ by $R_{P}$ and $\overline{R}_{P}$, respectively, so as to
make their dependence on $P$ explicit.
\end{definition}

We can now state a symmetry property of $\operatorname*{ord}R$ (as defined in
Definition \ref{def.ord}):

\begin{proposition}
\label{prop.op.ord}We have $\operatorname*{ord}%
\left(  R_{P^{\operatorname*{op}}}\right)  =\operatorname*{ord}\left(
R_{P}\right)  $ and $\operatorname*{ord}\left(  \overline{R}%
_{P^{\operatorname*{op}}}\right)  =\operatorname*{ord}\left(  \overline{R}%
_{P}\right)  $.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.op.ord}.]Define a rational map
$\kappa:\mathbb{K}^{\widehat{P}}\dashrightarrow\mathbb{K}%
^{\widehat{P^{\operatorname*{op}}}}$ by%
\[
\left(  \kappa f\right)  \left(  w\right)  =\left\{
\begin{array}
[c]{c}%
\dfrac{1}{f\left(  w\right)  },\ \ \ \ \ \ \ \ \ \ \text{if }w\in P;\\
\dfrac{1}{f\left(  1\right)  },\ \ \ \ \ \ \ \ \ \ \text{if }w=0;\\
\dfrac{1}{f\left(  0\right)  },\ \ \ \ \ \ \ \ \ \ \text{if }w=1
\end{array}
\right.  \ \ \ \ \ \ \ \ \ \ \text{for every }w\in
\widehat{P^{\operatorname*{op}}}\text{ for every }f\in\mathbb{K}
^{\widehat{P}}.
\]
This map $\kappa$ is a birational map. (Its inverse map is defined in the same way.)

We claim that $\kappa\circ R_{P}=R_{P^{\operatorname*{op}}}^{-1}\circ\kappa$.

Indeed, it is easy to see (by computation) that every element $v\in P$
satisfies
\begin{equation}
\kappa\circ T_{v}=T_{v}\circ\kappa, \label{pf.op.ord.1}%
\end{equation}
where the $T_{v}$ on the left hand side is defined with respect to the poset
$P$, and the $T_{v}$ on the right hand side is defined with respect to the
poset $P^{\operatorname*{op}}$. Now, let $\left(  v_{1},v_{2},\ldots,v_{m}%
\right)  $ be a linear extension of $P$. Then, $\left(  v_{m},v_{m-1}%
,\ldots,v_{1}\right)  $ is a linear extension of $P^{\operatorname*{op}}$, so
that Proposition \ref{prop.R.inverse} (applied to $P^{\operatorname*{op}}$ and
$\left(  v_{m},v_{m-1},\ldots,v_{1}\right)  $ instead of $P$ and $\left(
v_{1},v_{2},\ldots,v_{m}\right)  $) yields that $R_{P^{\operatorname*{op}}}%
^{-1}=T_{v_{1}}\circ T_{v_{2}}\circ\ldots\circ T_{v_{m}}:\mathbb{K}%
^{\widehat{P^{\operatorname*{op}}}}\dashrightarrow\mathbb{K}%
^{\widehat{P^{\operatorname*{op}}}}$. On the other hand, the definition of
$R_{P}$ yields $R_{P}=T_{v_{1}}\circ T_{v_{2}}\circ\ldots\circ T_{v_{m}%
}:\mathbb{K}^{\widehat{P}}\dashrightarrow\mathbb{K}^{\widehat{P}}$. Now, using
(\ref{pf.op.ord.1}), it is easy to see that%
\[
\kappa\circ\left(  T_{v_{1}}\circ T_{v_{2}}\circ\ldots\circ T_{v_{m}}\right)
=\left(  T_{v_{1}}\circ T_{v_{2}}\circ\ldots\circ T_{v_{m}}\right)  \circ\kappa.
\]
Since the $T_{v_{1}}\circ T_{v_{2}}\circ\ldots\circ T_{v_{m}}$ on the left hand
side equals $R_{P}$, and the $T_{v_{1}}\circ T_{v_{2}}\circ\ldots\circ T_{v_{m}}$
on the right hand side equals $R_{P^{\operatorname*{op}}}^{-1}$, this rewrites
as $\kappa\circ R_{P}=R_{P^{\operatorname*{op}}}^{-1}\circ\kappa$. Since
$\kappa$ is a birational map, this shows that $R_{P}$ and
$R_{P^{\operatorname*{op}}}^{-1}$ are birationally equivalent, so that
$\operatorname*{ord}\left(  R_{P}\right)  =\operatorname*{ord}\left(
R_{P^{\operatorname*{op}}}^{-1}\right)  =\operatorname*{ord}\left(
R_{P^{\operatorname*{op}}}\right)  $. Since $\kappa$ commutes with
homogenization, we also obtain the birational equivalence of the maps
$\overline{R}_{P}$ and $\overline{R}_{P^{\operatorname*{op}}}^{-1}$, whence
$\operatorname*{ord}\left(  \overline{R}_{P}\right)  =\operatorname*{ord}%
\left(  \overline{R}_{P^{\operatorname*{op}}}^{-1}\right)
=\operatorname*{ord}\left(  \overline{R}_{P^{\operatorname*{op}}}\right)  $.
%This proves Proposition \ref{prop.op.ord}.
\end{proof}

\section{Skeletal posets}

We will now introduce a class of posets which we call ``skeletal posets''.
Roughly speaking, these are graded posets built up inductively from the empty
poset by the operations of (1) disjoint union of two $n$-graded posets (with
the same $n$)  and (2) ``grafting'' on an antichain (generalizing the idea of grafting a tree on a
new root). In particular, this class includes all graded forests (oriented either away from the roots or towards
the roots) as well as various other posets. 

\begin{definition}
Let $P$ and $Q$ be two $n$-graded posets. We denote by
$PQ$ the disjoint union of the posets $P$ and $Q$ (denoted by $P+Q$ in \cite[\S
3.2]{stanley-ec1}). Its poset structure is defined 
in such a way that any element of $P$ and any element of $Q$ are incomparable,
while $P$ and $Q$ are subposets of $PQ$. Clearly, $PQ$ is also an $n$-graded poset.
\end{definition}

%Note that Stanley (e.g., in \cite[\S 3.2]{stanley-ec1}) denotes by $P+Q$ what
%we just called $PQ$.

\begin{definition}
Let $P$ be an $n$-graded poset. Let $k$ be a positive
integer. We denote by $B_{k}P$ the result of adding $k$ new elements to the
poset $P$, and declaring these $k$ elements to be smaller than each of the
elements of $P$ (but incomparable with each other). Clearly, $B_{k}P$ is an
$\left(  n+1\right)  $-graded poset.
\end{definition}

\begin{definition}
Let $n\in\mathbb{N}$. Let $P$ be an $n$-graded poset. Let $k$ be a positive
integer. We denote by $B_{k}^{\prime}P$ the result of adding $k$ new elements
to the poset $P$, and declaring these $k$ elements to be larger than each of
the elements of $P$ (but incomparable with each other). Clearly,
$B_{k}^{\prime}P$ is an $\left(  n+1\right)  $-graded poset.
\end{definition}

In the
notation of Stanley (\cite[\S 3.2]{stanley-ec1}),  $B_{k}P=A_{k}\oplus
P$ and $B_{k}^{\prime}P=P\oplus A_{k}$, where $A_{k}$ denotes the $k$-element antichain.
It is easy to see that $B_{k}P$ and $B_{k}^{\prime}P$ are ``symmetric''
notions with respect to taking the opposite poset (Def.~\ref{def.op}):

\begin{proposition}
\label{prop.skeletal.op}
Let $P$ be an $n$-graded poset.
Then, $B_{k}^{\prime}P=\left(  B_{k}\left(  P^{\operatorname*{op}}\right)
\right)  ^{\operatorname*{op}}$. 
% (Here, we are using the notation introduced
% in Definition \ref{def.op}.)
\end{proposition}

%We now define the notion of a skeletal poset:

\begin{definition}
\label{def.skeletal}We define the class of \textit{skeletal posets}
inductively by means of the following axioms:

-- The empty poset is skeletal.

-- If $P$ is an $n$-graded skeletal poset and $k$ is a positive integer, then
the posets $B_{k}P$ and $B_{k}^{\prime}P$ are skeletal.

-- If $n$ is a nonnegative integer and $P$ and $Q$ are two $n$-graded skeletal
posets, then the poset $PQ$ is skeletal.

Notice that every skeletal poset is graded, and that every graded
rooted forest (made into a poset by having every node smaller than its
children) is a skeletal poset. (Recall that our sense of ``graded'' means all maximal chains
have the same length, so not all trees are graded.)  Indeed, every graded rooted forest can be
constructed from $\varnothing$ using merely the operations $P\mapsto B_{1}P$
and $\left(  P,Q\right)  \mapsto PQ$. Also, every graded rooted arborescence
(i.e., the opposite poset of a graded rooted tree) is a skeletal poset (for a
similar reason). 
\end{definition}

\begin{example}
The rooted forest $%
%TCIMACRO{\TeXButton{x}{\xymatrix@C=1em@R=1em{
%\bullet\ar@{-}[dr] & & \bullet\ar@{-}[dl] & \bullet\ar@{-}[d] & \bullet\ar@
%{-}[d]\\
%& \bullet\ar@{-}[dr] & & \bullet\ar@{-}[dl] & \bullet\ar@{-}[d] \\
%& & \bullet& & \bullet}}}%
%BeginExpansion
\xymatrix@C=1em@R=1em{
\bullet\ar@{-}[dr] & & \bullet\ar@{-}[dl] & \bullet\ar@{-}[d] & \bullet\ar@
{-}[d]\\
& \bullet\ar@{-}[dr] & & \bullet\ar@{-}[dl] & \bullet\ar@{-}[d] \\
& & \bullet& & \bullet}%
%EndExpansion
$ is skeletal, and in fact can be written as $\left(  B_{1}\left(  \left(
B_{1}\left(  B_{2}\varnothing\right)  \right)  \left(  B_{1}\left(
B_{1}\varnothing\right)  \right)  \right)  \right)  \left(  B_{1}\left(
B_{1}\left(  B_{1}\varnothing\right)  \right)  \right)  $. (This form of
writing is not unique, since $B_{2}\varnothing=\left(  B_{1}\varnothing
\right)  \left(  B_{1}\varnothing\right)  $.)

The tree $%
%TCIMACRO{\TeXButton{x}{\xymatrix@C=1em@R=1em{
%& & \bullet\ar@{-}[d] \\
%\bullet\ar@{-}[dr] & & \bullet\ar@{-}[dl] \\
%& \bullet&
%}}}%
%BeginExpansion
\xymatrix@C=1em@R=1em{
& & \bullet\ar@{-}[d] \\
\bullet\ar@{-}[dr] & & \bullet\ar@{-}[dl] \\
& \bullet&
}%
%EndExpansion
$ can be written as $B_{1}\left(  \left(  B_{1}\varnothing\right)  \left(
B_{1}\left(  B_{1}\varnothing\right)  \right)  \right)  $, but is \textbf{not}
skeletal because $B_{1}\varnothing$ and $B_{1}\left(  B_{1}\varnothing\right)
$ are not $n$-graded with one and the same $n$.

The poset $%
%TCIMACRO{\TeXButton{x}{\xymatrix@C=1em@R=1em{
%\bullet\ar@{-}[dr] \ar@{-}[d] & \bullet\ar@{-}[d] \ar@{-}[dl] & & \bullet
%\ar@{-}[dl]\ar@{-}[dr] & \\
%\bullet\ar@{-}[drr] & \bullet\ar@{-}[dr] & \bullet\ar@{-}[d] & & \bullet
%\ar@{-}[dll] \\
%& & \bullet& &
%}}}%
%BeginExpansion
\xymatrix@C=1em@R=1em{
\bullet\ar@{-}[dr] \ar@{-}[d] & \bullet\ar@{-}[d] \ar@{-}[dl] & & \bullet
\ar@{-}[dl]\ar@{-}[dr] & \\
\bullet\ar@{-}[drr] & \bullet\ar@{-}[dr] & \bullet\ar@{-}[d] & & \bullet
\ar@{-}[dll] \\
& & \bullet& &
}%
%EndExpansion
$ is neither a tree nor an arborescence, but it has the form $B_{1}\left(
\left(  B_{2}\left(  B_{2}\varnothing\right)  \right)  \left(  B_{1}^{\prime
}\left(  B_{2}\varnothing\right)  \right)  \right)  $ and is skeletal.
\end{example}

Our main result on skeletal posets is the following:

\begin{proposition}
\label{prop.skeletal.ords}If $P$ is a skeletal poset, then $\operatorname*{ord}\left(
R_{P}\right)  $ and 
$\operatorname*{ord}\left(  \overline{R}_{P}\right)  $ are finite.
\end{proposition}

We first build up some
machinery for determining $\operatorname*{ord}\left(  R_{P}\right)  $ and
$\operatorname*{ord}\left(  \overline{R}_{P}\right)  $ given such orders in
smaller posets. Here is a very basic fact to get started:

\begin{proposition}
\label{prop.PQ.ord}Fix $n\in\mathbb{N}$, and let $P$ and $Q$ be two $n$-graded
posets. 
Then, we have $\operatorname*{ord}\left(
R_{PQ}\right)  =\operatorname{lcm}\left(  \operatorname*{ord}\left(
R_{P}\right),\operatorname*{ord}\left(  R_{Q}\right)  \right)  $.
\end{proposition}


\begin{proof}
[Proof of Proposition \ref{prop.PQ.ord}.]
The proof of this is as easy as it looks: a $\mathbb{K}$-labelling
of the disjoint union $PQ$ can be regarded as a pair of a $\mathbb{K}%
$-labelling of $P$ and a $\mathbb{K}$-labelling of $Q$ (with identical labels
at $0$ and $1$), and the map $R$ (as well as all $R_{i}$) acts on these
labellings independently.
\end{proof}



The analogue of Proposition \ref{prop.PQ.ord} with all $R$'s replaced by
$\overline{R}$'s is false. Instead, $\operatorname*{ord}\left(  \overline
{R}_{PQ}\right)  $ can be computed as follows:


\begin{proposition}
\label{prop.PQ.ord2}Fix $n\in\mathbb{N}$, and let $P$ and $Q$ be two $n$-graded
posets. Let $\mathbb{K}$ be a field. Then, $\operatorname*{ord}\left(
\overline{R}_{PQ}\right)  =\operatorname{lcm}\left(  \operatorname*{ord}%
\left(  R_{P}\right),\operatorname*{ord}\left(  R_{Q}\right)  \right)  $.
(As we know, this is equal to $\operatorname{ord}\left(R_{PQ}\right)$.)
\end{proposition}

%% TR: NEED TO SUMMARIZE THIS!
The most useful point here is the statement that 
$\operatorname*{ord}\left(
\overline{R}_{PQ}\right)  \mid  \operatorname{lcm}\left(  \operatorname*{ord}%
\left(  R_{P}\right),\operatorname*{ord}\left(  R_{Q}\right)  \right)  $.
This follows immediately from Propositions \ref{prop.ord-projord}
and \ref{prop.PQ.ord2}.
The reverse divisibility is mainly a technical result which can be
proved by splitting a labelling of $PQ$ into a labelling of $P$ and
one of $Q$ and comparing how $\overline{R}$ affects these labellings.
We refer to \cite[\S 9]{grinberg-roby-arXiv} for details.

% \begin{proof}
% [Proof of Proposition \ref{prop.PQ.ord2}.]Assume WLOG that $n\neq0$
% (else, everything is obvious). Hence, $P$ and $Q$ are nonempty (being $n$-graded).

% Proposition \ref{prop.ord-projord} yields $\operatorname*{ord}\left(
% R_{PQ}\right)  =\operatorname{lcm}\left(  n+1,\operatorname*{ord}\left(
% \overline{R}_{PQ}\right)  \right)  $.

% WLOG assume that $\operatorname*{ord}\left(  R_{P}\right)  $ and
% $\operatorname*{ord}\left(  R_{Q}\right)  $ are finite\footnote{Otherwise,
% $\operatorname{lcm}\left(  \operatorname*{ord}\left(  R_{P}\right)
%,\operatorname*{ord}\left(  R_{Q}\right)  \right)  $ is infinite, whence
% $\operatorname*{ord}\left(  R_{PQ}\right)  $ is infinite (by Proposition
% \ref{prop.PQ.ord}), whence $\operatorname*{ord}\left(  \overline{R}%
% _{PQ}\right)  $ is infinite (because $\operatorname*{ord}\left(
% R_{PQ}\right)  =\operatorname{lcm}\left(  n+1,\operatorname*{ord}\left(
% \overline{R}_{PQ}\right)  \right)  $), whence Proposition \ref{prop.PQ.ord2}
% is trivial.}. Then, Proposition \ref{prop.PQ.ord} shows that
% $\operatorname*{ord}\left(  R_{PQ}\right)  =\operatorname{lcm}\left(
% \operatorname*{ord}\left(  R_{P}\right),\operatorname*{ord}\left(
% R_{Q}\right)  \right)  $ is finite, so that $\operatorname*{ord}\left(
% \overline{R}_{PQ}\right)  $ is finite (because $\operatorname*{ord}\left(
% R_{PQ}\right)  =\operatorname{lcm}\left(  n+1,\operatorname*{ord}\left(
% \overline{R}_{PQ}\right)  \right)  $). Let $\ell$ be $\operatorname*{ord}%
% \left(  \overline{R}_{PQ}\right)  $. Then, $\ell$ is finite and satisfies
% $\overline{R}_{PQ}^{\ell}=\operatorname*{id}$. We will show that $n+1\mid\ell$.


% For every $\mathbb{K}$-labelling $f$ of $PQ$ and every $i\in\left\{
% 0,1,\ldots,n\right\}  $, define two elements $\mathbf{w}_{i}^{\left(  1\right)
% }\left(  f\right)  $ and $\mathbf{w}_{i}^{\left(  2\right)  }\left(  f\right)
% $ of $\mathbb{K}$ by%
% \[
% \mathbf{w}_{i}^{\left(  1\right)  }\left(  f\right)  =\sum_{\substack{x\in
% \widehat{P}_{i};\ y\in\widehat{P}_{i+1};\\y\gtrdot x}}\dfrac{f\left(
% x\right)  }{f\left(  y\right)  }\ \ \ \ \ \ \ \ \ \ \text{and}%
% \ \ \ \ \ \ \ \ \ \ \mathbf{w}_{i}^{\left(  2\right)  }\left(  f\right)
% =\sum_{\substack{x\in\widehat{Q}_{i};\ y\in\widehat{Q}_{i+1};\\y\gtrdot
% x}}\dfrac{f\left(  x\right)  }{f\left(  y\right)  }%
% \]
% (where, of course, $\widehat{P}_{j}$ and $\widehat{Q}_{j}$ are embedded into
% $\widehat{PQ}_{j}$ for every $j\in\left\{  0,1,\ldots,n+1\right\}  $ in the
% obvious way). These elements $\mathbf{w}_{i}^{\left(  1\right)  }\left(
% f\right)  $ and $\mathbf{w}_{i}^{\left(  2\right)  }\left(  f\right)  $ are
% defined not for every $f$, but for \textquotedblleft almost
% every\textquotedblright\ $f$ in the sense of Zariski topology. We denote the
% $\left(  n+1\right)  $-tuple
% \[
% \left(  \mathbf{w}_{0}^{\left(  1\right)  }\left(  f\right)  \diagup
% \mathbf{w}_{0}^{\left(  2\right)  }\left(  f\right),\ \mathbf{w}%
% _{1}^{\left(  1\right)  }\left(  f\right)  \diagup\mathbf{w}_{1}^{\left(
% 2\right)  }\left(  f\right),\ \ldots,\ \mathbf{w}_{n}^{\left(  1\right)
% }\left(  f\right)  \diagup\mathbf{w}_{n}^{\left(  2\right)  }\left(  f\right)
% \right)
% \]
% as the \textit{comparative w-tuple} of the labelling $f$. The advantage of
% comparative w-tuples over usual w-tuples is the following fact: If $f$ and $g$
% are two homogeneously equivalent $\mathbb{K}$-labellings of $PQ$, then%
% \begin{equation}
% \left(  \text{the comparative w-tuple of }f\right)  =\left(  \text{the
% comparative w-tuple of }g\right). \label{pf.PQ.ord2.hgeq.short}%
% \end{equation}
% (This is easy to check and has no analogue for regular w-tuples.)



% It is furthermore easy to see (in analogy to Proposition \ref{prop.wi.R}) that
% the map $R_{PQ}$ changes the comparative w-tuple of a $\mathbb{K}$-labelling
% by shifting it cyclically.

% But it is also easy to see (the nonemptiness of $P$ and $Q$ must be used here)
% that there exists some $f\in\mathbb{K}^{\widehat{PQ}}$ such that the ratios
% $\mathbf{w}_{i}^{\left(  1\right)  }\left(  f\right)  \diagup\mathbf{w}%
% _{i}^{\left(  2\right)  }\left(  f\right)  $ are well-defined and pairwise
% distinct for all $i\in\left\{  0,1,\ldots,n\right\}  $ and such that $R^{j}f$ is
% well-defined for every $j\in\left\{  0,1,\ldots,\ell\right\}  $. Consider such an
% $f$. The ratios $\mathbf{w}_{i}^{\left(  1\right)  }\left(  f\right)
% \diagup\mathbf{w}_{i}^{\left(  2\right)  }\left(  f\right)  $ are pairwise
% distinct for all $i\in\left\{  0,1,\ldots,n\right\}  $; that is, the comparative
% w-tuple of $f$ contains no two equal entries.


% Since $\overline{R}_{PQ}^{\ell}=\operatorname*{id}$, we have $\overline
% {R}_{PQ}^{\ell}\left(  \pi\left(  f\right)  \right)  =\pi\left(  f\right)  $.
% The commutativity of the diagram (\ref{def.hgR.commut}) yields $\overline
% {R}_{PQ}^{\ell}\circ\pi=\pi\circ R_{PQ}^{\ell}$. Now,%
% \[
% \pi\left(  f\right)  =\overline{R}_{PQ}^{\ell}\left(  \pi\left(  f\right)
% \right)  =\left(  \underbrace{\overline{R}_{PQ}^{\ell}\circ\pi}_{=\pi\circ
% R_{PQ}^{\ell}}\right)  \left(  f\right)  =\left(  \pi\circ R_{PQ}^{\ell
% }\right)  \left(  f\right)  =\pi\left(  R_{PQ}^{\ell}f\right).
% \]
% In other words, the labellings $f$ and $R_{PQ}^{\ell}f$ are homogeneously
% equivalent. Thus,%
% \begin{equation}
% \left(  \text{the comparative w-tuple of }f\right)  =\left(  \text{the
% comparative w-tuple of }R_{PQ}^{\ell}f\right)  \label{pf.PQ.ord2.2.short}%
% \end{equation}
% (by (\ref{pf.PQ.ord2.hgeq.short})).




% Now, recall that the map $R_{PQ}$ changes the comparative w-tuple of a
% $\mathbb{K}$-labelling by shifting it cyclically. Hence, for every
% $k\in\mathbb{N}$, the map $R_{PQ}^{k}$ changes the comparative w-tuple of a
% $\mathbb{K}$-labelling by shifting it cyclically $k$ times. Applying this to
% the $\mathbb{K}$-labelling $f$ and to $k=\ell$, we see that the comparative
% w-tuple of $R_{PQ}^{\ell}f$ is obtained from the comparative w-tuple of $f$ by
% an $\ell$-fold cyclic shift. Due to (\ref{pf.PQ.ord2.2.short}), this rewrites
% as follows: The comparative w-tuple of $f$ is obtained from the comparative
% w-tuple of $f$ by an $\ell$-fold cyclic shift. In other words, the comparative
% w-tuple of $f$ is invariant under an $\ell$-fold cyclic shift. But since the
% comparative w-tuple of $f$ consists of $n+1$ pairwise distinct entries, this
% is impossible unless $n+1\mid\ell$. Hence, we must have $n+1\mid\ell$.



% Now,%
% \[
% \operatorname*{ord}\left(  R_{PQ}\right)  =\operatorname{lcm}\left(
% n+1,\operatorname*{ord}\left(  \overline{R}_{PQ}\right)  \right)
% =\operatorname*{ord}\left(  \overline{R}_{PQ}\right)
% \]
% (since $n+1\mid\ell=\operatorname*{ord}\left(  \overline{R}_{PQ}\right)  $).
% Hence,%
% \[
% \operatorname*{ord}\left(  \overline{R}_{PQ}\right)  =\operatorname*{ord}%
% \left(  R_{PQ}\right)  =\operatorname{lcm}\left(  \operatorname*{ord}\left(
% R_{P}\right),\operatorname*{ord}\left(  R_{Q}\right)  \right).
% \]
% This proves Proposition \ref{prop.PQ.ord2}.
% \end{proof}

Now, let us track the effect of $B_{k}$ on the order of $\overline{R}$:

\begin{proposition}
\label{prop.Bk.ord}Let $P$ be an $n$-graded poset. 

\textbf{(a)} We have $\operatorname*{ord}\left(  \overline{R}_{B_{1}P}\right)
=\operatorname*{ord}\left(  \overline{R}_{P}\right)  $.

\textbf{(b)} For every integer $k>1$, we have $\operatorname*{ord}\left(
\overline{R}_{B_{k}P}\right)  =\operatorname{lcm}\left(  2,\operatorname*{ord}%
\left(  \overline{R}_{P}\right)  \right)  $.
\end{proposition}

%%% TR: NEED TO SUMMARIZE THIS! 


\begin{proof}
[Proof of Proposition \ref{prop.Bk.ord}.]
% We will be proving parts
% \textbf{(a)} and \textbf{(b)} together.
% Let $k$ be a positive integer.
% (this
% has to be $1$ for proving part \textbf{(a)}).
We only need to prove that
\begin{equation}
\operatorname*{ord}\left(  \overline{R}_{B_{k}P}\right)  =\left\{
\begin{array}
[c]{l}%
\operatorname{lcm}\left(  2,\operatorname*{ord}\left(  \overline{R}%
_{P}\right)  \right),\ \ \ \ \ \ \ \ \ \ \text{if }k>1;\\
\operatorname*{ord}\left(  \overline{R}_{P}\right)
,\ \ \ \ \ \ \ \ \ \ \text{if }k=1
\end{array}
\right.. \label{pf.Bk.ord.1}%
\end{equation}
% Proving this clearly will prove both parts \textbf{(a)} and \textbf{(b)} of
% Proposition \ref{prop.Bk.ord}.

Let us make some conventions:

\begin{itemize}
\item For any $n$-tuple $\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)
$ and any object $\beta$, let $\beta\rightthreetimes\alpha$ denote the
$\left(  n+1\right)  $-tuple $\left(  \beta,\alpha_{1},\alpha_{2}%
,\ldots,\alpha_{n}\right)  $.

\item We are going to identify $P$ with a subposet of $B_{k}P$ in the obvious
way. However, the degree maps of these posets differ and have to be
distinguished. We also embed $\widehat{P}$ into $\widehat{B_{k}P}$
by identifying the respective $0$s and the respective $1$s.
% But of course, the degree map of $B_{k}P$ restricted to $P$ is not
% identical with the degree map of $P$ (but rather differs from it by $1$), so
% we will have to distinguish between \textquotedblleft degree in $P$%
% \textquotedblright\ and \textquotedblleft degree in $B_{k}P$\textquotedblright%
%. We identify the elements $0$ and $1$ of $\widehat{P}$ with the elements $0$
% and $1$ of $\widehat{B_{k}P}$, respectively. Thus, $\widehat{P}$ becomes a
% subposet of $\widehat{B_{k}P}$. However, it is not generally true that every
% $u\lessdot v$ in $\widehat{P}$ must satisfy $u\lessdot v$ in $\widehat{B_{k}%
% P}$.

\item The rational maps $\pi:\mathbb{K}^{\widehat{P}}\dashrightarrow
\overline{\mathbb{K}^{\widehat{P}}}$ and $\pi:\mathbb{K}%
^{\widehat{B_{k}P}}\dashrightarrow\overline{\mathbb{K}^{\widehat{B_{k}P}}}$
are denoted by the same letter and
can be distinguished by their different domains. We will also use the letter
$\pi$ to denote the rational map $\mathbb{K}^{k}\dashrightarrow\mathbb{P}%
\left(  \mathbb{K}^{k}\right)  $ obtained from the canonical projection
$\mathbb{K}^{k}\setminus\left\{  0\right\}  \rightarrow\mathbb{P}\left(
\mathbb{K}^{k}\right)  $ of the nonzero vectors in $\mathbb{K}^{k}$ onto the
projective space.
\end{itemize}

Now the operation $B_{k}$ clearly raises the degree of every element of $P$
by $1$,
% \ \ \ \ \footnote{In terms of the Hasse diagram, this can be regarded as
% the $k$ new elements ``bumping up'' all existing elements of $P$ by $1$
% degree.}
while the $k$ newly added elements all obtain degree $1$ in
$B_{k}P$. Formally speaking, this means that $\widehat{B_{k}P}_{i}%
=\widehat{P}_{i-1}$ for every $i\in\left\{  2,3,\ldots,n+1\right\}  $, while
$\widehat{B_{k}P}_{1}$ is a $k$-element set. Moreover, for any $i\in\left\{
2,3,\ldots,n+1\right\}  $, any $u\in\widehat{B_{k}P}_{i}=\widehat{P}_{i-1}$ and
any $v\in\widehat{B_{k}P}_{i+1}=\widehat{P}_{i}$, we have
\[
u\lessdot v\text{ in }\widehat{B_{k}P}\text{ if and only if }u\lessdot v\text{
in }\widehat{P}\text{.}%
\]
% (This would not be true if we would allow $i=1$, $u\in\widehat{P}_{0}$ and
% $v\in\widehat{P}_{1}$.)


We have $\mathbb{K}^{\widehat{B_{k}P}_{i}}=\mathbb{K}^{\widehat{P}_{i-1}}$ for
every $i\in\left\{  2,3,\ldots,n+1\right\}  $,
% (since $\widehat{B_{k}P}%
%_{i}=\widehat{P}_{i-1}$ for every $i\in\left\{  2,3,\ldots,n+1\right\}  $),
whereas $\mathbb{K}^{\widehat{B_{k}P}_{1}}\cong\mathbb{K}^{k}$.
% (since
%$\widehat{B_{k}P}_{1}$ is a $k$-element set).
We will actually identify
$\mathbb{K}^{\widehat{B_{k}P}_{1}}$ with $\mathbb{K}^{k}$. Now,%
\begin{align}
\overline{\mathbb{K}^{\widehat{B_{k}P}}}  &  =\prod\limits_{i=1}%
^{n+1}\mathbb{P}\left(  \mathbb{K}^{\widehat{B_{k}P}_{i}}\right)
=\mathbb{P}\left(  \underbrace{\mathbb{K}^{\widehat{B_{k}P}_{1}}}%
_{=\mathbb{K}^{k}}\right)  \times\prod\limits_{i=2}^{n+1}\mathbb{P}\left(
\underbrace{\mathbb{K}^{\widehat{B_{k}P}_{i}}}_{=\mathbb{K}^{\widehat{P}%
_{i-1}}}\right) \nonumber\\
&  =\mathbb{P}\left(  \mathbb{K}^{k}\right)  \times\prod\limits_{i=2}%
^{n+1}\mathbb{P}\left(  \mathbb{K}^{\widehat{P}_{i-1}}\right)  =\mathbb{P}%
\left(  \mathbb{K}^{k}\right)  \times\underbrace{\prod\limits_{i=1}%
^{n}\mathbb{P}\left(  \mathbb{K}^{\widehat{P}_{i}}\right)  }_{=\overline
{\mathbb{K}^{\widehat{P}}}}=\mathbb{P}\left(  \mathbb{K}^{k}\right)
\times\overline{\mathbb{K}^{\widehat{P}}}. \label{pf.Bk.ord.decomp}%
\end{align}
Thus, the elements of $\overline{\mathbb{K}^{\widehat{B_{k}P}}}$ have the form
$\widetilde{p}\rightthreetimes\widetilde{g}$, where $\widetilde{p}%
\in\mathbb{P}\left(  \mathbb{K}^{k}\right)  $ and $\widetilde{g}\in
\overline{\mathbb{K}^{\widehat{P}}}$.

On the other hand, 
% recall that $\widehat{P}$ is a subposet of $\widehat{B_{k}%
% P}$. More precisely, 
$\widehat{P}$ is the subposet $\widehat{B_{k}P}%
\setminus\widehat{B_{k}P}_{1}$ of $\widehat{B_{k}P}$. Thus, we can define a
map $\Phi:\mathbb{K}^{k}\times\mathbb{K}^{\widehat{P}}\rightarrow
\mathbb{K}^{\widehat{B_{k}P}}$ by setting%
\[
\left(  \Phi\left(  p,g\right)  \right)  \left(  v\right)  =\left\{
\begin{array}
[c]{c}%
p\left(  v\right),\ \ \ \ \ \ \ \ \ \ \text{if }v\in\widehat{B_{k}P}_{1};\\
g\left(  v\right),\ \ \ \ \ \ \ \ \ \ \text{if }v\notin\widehat{B_{k}P}_{1}%
\end{array}
\right.  \ \ \ \ \ \ \ \ \ \ \text{for every }v\in\widehat{B_{k}P}%
\]
for every $\left(  p,g\right)  \in\mathbb{K}^{k}\times\mathbb{K}^{\widehat{P}%
}$. Here, the term $p\left(  v\right)  $ is to be understood by means of
regarding $p$ as an element of $\mathbb{K}^{\widehat{B_{k}P}_{1}}$ (since
$p\in\mathbb{K}^{k}=\mathbb{K}^{\widehat{B_{k}P}_{1}}$). Clearly, $\Phi$ is a
bijection. Moreover, it is easy to see that
\begin{equation}
\pi\left(  \Phi\left(  p,g\right)  \right)  =\pi\left(  p\right)
\rightthreetimes\pi\left(  g\right)  \ \ \ \ \ \ \ \ \ \ \text{for all }%
p\in\mathbb{K}^{k}\text{ and }g\in\mathbb{K}^{\widehat{P}}
\label{pf.Bk.ord.pitriv}%
\end{equation}
(where the $\pi$ on the left hand side is the map $\pi:\mathbb{K}%
^{\widehat{B_{k}P}}\dashrightarrow\overline{\mathbb{K}^{\widehat{B_{k}P}}}$,
whereas the $\pi$ in \textquotedblleft$\pi\left(  p\right)  $%
\textquotedblright\ is the map $\pi:\mathbb{K}^{k}\dashrightarrow
\mathbb{P}\left(  \mathbb{K}^{k}\right)  $, and the $\pi$ in \textquotedblleft%
$\pi\left(  g\right)  $\textquotedblright\ is the map $\pi:\mathbb{K}%
^{\widehat{P}}\dashrightarrow\overline{\mathbb{K}^{\widehat{P}}}$).


Now, we claim that every $\widetilde{p}\in\mathbb{P}\left(  \mathbb{K}%
^{k}\right)  $ and $\widetilde{g}\in\overline{\mathbb{K}^{\widehat{P}}}$
satisfy
\begin{equation}
\left(\overline{R_{i}}\right)_{B_{k}P}\left(
\widetilde{p}\rightthreetimes\widetilde{g}%
\right)  =\widetilde{p}\rightthreetimes\left(\overline{R_{i-1}}\right)_{P}\left(
\widetilde{g}\right)  \ \ \ \ \ \ \ \ \ \ \text{for all }i\in\left\{
2,3,\ldots,n+1\right\}  \label{pf.Bk.ord.Ri.big}%
\end{equation}
and%
\begin{equation}
\left(\overline{R_{1}}\right)_{B_{k}P}\left(
\widetilde{p}\rightthreetimes\widetilde{g}%
\right)  =\widetilde{p}^{-1}\rightthreetimes\widetilde{g}.
\label{pf.Bk.ord.Ri.small}%
\end{equation}



\textit{Proof of (\ref{pf.Bk.ord.Ri.big}) and (\ref{pf.Bk.ord.Ri.small}):} In
order to prove (\ref{pf.Bk.ord.Ri.big}), it is clearly enough to show that
every $p\in\mathbb{K}^{k}$ and $g\in\mathbb{K}^{\widehat{P}}$ satisfy%
\begin{equation}
\left(  R_{i}\right)  _{B_{k}P}\left(  p\rightthreetimes g\right)  \sim
p\rightthreetimes\left(  R_{i-1}\right)  _{P}\left(  g\right)
\ \ \ \ \ \ \ \ \ \ \text{for all }i\in\left\{  2,3,\ldots,n+1\right\},
\label{pf.Bk.ord.Ri.big.pf.short.1}%
\end{equation}
where the sign $\sim$ stands for homogeneous equivalence.

It is easy to prove the relation (\ref{pf.Bk.ord.Ri.big.pf.short.1}) for $i>2$
(because if $i>2$, then the elements of $\widehat{B_{k}P}$ having degrees
$i-1$, $i$ and $i+1$ are precisely the elements of $\widehat{P}$ having
degrees $i-2$, $i-1$ and $i$, and therefore toggling the elements of
$\widehat{B_{k}P}_{i}$ in $p\rightthreetimes g$ has precisely the same effect
as toggling the elements of $\widehat{P}_{i-1}$ in $g$ while leaving $p$
fixed, so that we even get the stronger assertion that $\left(  R_{i}\right)
_{B_{k}P}\left(  p\rightthreetimes g\right)  =p\rightthreetimes\left(
R_{i-1}\right)  _{P}\left(  g\right)  $). It is not much harder to check that
it also holds for $i=2$ (indeed, for $i=2$, the only difference between
toggling the elements of $\widehat{B_{k}P}_{i}$ in $p\rightthreetimes g$ and
toggling the elements of $\widehat{P}_{i-1}$ in $g$ while leaving $p$ fixed is
a scalar factor which is identical across all elements being toggled in either
poset
\footnote{because every $u\in\widehat{B_{k}P}_{1}$ and every
$v\in\widehat{B_{k}P}_{2}$ satisfy $u\lessdot v$}; therefore, the results are
the same up to homogeneous equivalence).

Finally, (\ref{pf.Bk.ord.Ri.small}) is trivial to check (e.g., using Corollary
\ref{cor.hgRi.1}).



But recall that $\overline{R}=\overline{R_{1}}\circ\overline{R_{2}}%
\circ\ldots\circ\overline{R_{n}}$ for any $n$-graded poset. Hence, $\overline
{R}_{B_{k}P}
=\left(\overline{R_{1}}\right)_{B_{k}P}
\circ\left(\overline{R_{2}}\right)_{B_{k}P}
\circ\left(\overline{R_{3}}\right)_{B_{k}P}\circ\ldots
\circ\left(\overline{R_{n+1}}\right)_{B_{k}P}$
(because $B_{k}P$ is an $\left(  n+1\right)  $-graded poset) and $\overline
{R}_{P}=\left(\overline{R_{1}}\right)_{P}\circ
\left(\overline{R_{2}}\right)_{P}\circ\ldots\circ
\left(\overline{R_{n}}\right)_{P}$ (because $P$ is an $n$-graded poset). Because of these
equalities, and by (\ref{pf.Bk.ord.Ri.big}) and (\ref{pf.Bk.ord.Ri.small}%
), it is now easy to see that every $\widetilde{p}\in\mathbb{P}\left(
\mathbb{K}^{k}\right)  $ and $\widetilde{g}\in\overline{\mathbb{K}%
^{\widehat{P}}}$ satisfy
\begin{equation}
\overline{R}_{B_{k}P}\left(  \widetilde{p}\rightthreetimes\widetilde{g}%
\right)  =\widetilde{p}^{-1}\rightthreetimes\overline{R}_{P}\left(
\widetilde{g}\right). \label{pf.Bk.ord.R}%
\end{equation}



Furthermore, every $\widetilde{p}\in\mathbb{P}\left(  \mathbb{K}^{k}\right)  $
and $\widetilde{g}\in\overline{\mathbb{K}^{\widehat{P}}}$ satisfy
\begin{equation}
\overline{R}_{B_{k}P}^{\ell}\left(  \widetilde{p}\rightthreetimes
\widetilde{g}\right)  =\widetilde{p}^{\left(  -1\right)  ^{\ell}%
}\rightthreetimes\overline{R}_{P}^{\ell}\left(  \widetilde{g}\right)
\ \ \ \ \ \ \ \ \ \ \text{for all }\ell\in\mathbb{N}. \label{pf.Bk.ord.Rl}%
\end{equation}
(This is proven by induction over $\ell$, using (\ref{pf.Bk.ord.R}).)

We know that the elements of $\overline{\mathbb{K}^{\widehat{B_{k}P}}}$ have
the form $\widetilde{p}\rightthreetimes\widetilde{g}$, where $\widetilde{p}%
\in\mathbb{P}\left(  \mathbb{K}^{k}\right)  $ and $\widetilde{g}\in
\overline{\mathbb{K}^{\widehat{P}}}$. Conversely, every element $\widetilde{p}%
\rightthreetimes\widetilde{g}$ with $\widetilde{p}\in\mathbb{P}\left(
\mathbb{K}^{k}\right)  $ and $\widetilde{g}\in\overline{\mathbb{K}%
^{\widehat{P}}}$ lies in $\overline{\mathbb{K}^{\widehat{B_{k}P}}}$. Hence,
for every $\ell\in\mathbb{N}$, we have the following equivalence of
assertions:%
\begin{align*}
&  \ \left(  \overline{R}_{B_{k}P}^{\ell}=\operatorname*{id}%
\right) 
  \Longleftrightarrow\ \left(  \text{every }\widetilde{p}\in\mathbb{P}\left(
\mathbb{K}^{k}\right)  \text{ and }\widetilde{g}\in\overline{\mathbb{K}%
^{\widehat{P}}}\text{ satisfy }\overline{R}_{B_{k}P}^{\ell}\left(
\widetilde{p}\rightthreetimes\widetilde{g}\right)  =\widetilde{p}%
\rightthreetimes\widetilde{g}\right) \\
&  \Longleftrightarrow\ \left(  \text{every }\widetilde{p}\in\mathbb{P}\left(
\mathbb{K}^{k}\right)  \text{ and }\widetilde{g}\in\overline{\mathbb{K}%
^{\widehat{P}}}\text{ satisfy }\widetilde{p}^{\left(  -1\right)  ^{\ell}%
}\rightthreetimes\overline{R}_{P}^{\ell}\left(  \widetilde{g}\right)
=\widetilde{p}\rightthreetimes\widetilde{g}\right) \text{ (by (\ref{pf.Bk.ord.Rl}))}\\
%&  \ \ \ \ \ \ \ \ \ \ \left(  \text{due to (\ref{pf.Bk.ord.Rl})}\right) \\
&  \Longleftrightarrow\ \left( \text{every }\widetilde{p}\in\mathbb{P}\left(
\mathbb{K}^{k}\right)  \text{ and }\widetilde{g}\in\overline{\mathbb{K}%
^{\widehat{P}}}\text{ satisfy }\widetilde{p}^{\left(  -1\right)  ^{\ell}%
}=\widetilde{p}\text{ and }\overline{R}_{P}^{\ell}\left(  \widetilde{g}%
\right)  =\widetilde{g}\right) \\
&  \Longleftrightarrow\ \left( \underbrace{ \text{every }\widetilde{p}%
\in\mathbb{P}\left(  \mathbb{K}^{k}\right)  \text{ satisfies }\widetilde{p}%
^{\left(  -1\right)  ^{\ell}}=\widetilde{p}}_{\text{this is equivalent to }
\left(2\mid \ell  \text{ if } k > 1 \right)}
\text{, and }
\underbrace{\text{every
}\widetilde{g}\in\overline{\mathbb{K}^{\widehat{P}}}\text{ satisfies
}\overline{R}_{P}^{\ell}\left(  \widetilde{g}\right)  =\widetilde{g}
}_{\text{this is equivalent to }\operatorname{ord} \left(\overline{R}_{P}\right)
\mid  \ell }
\right) \\
% &  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the sets }\mathbb{P}\left(
% \mathbb{K}^{k}\right)  \text{ and }\overline{\mathbb{K}^{\widehat{P}}}\text{
% are nonempty}\right) \\
% &  \Longleftrightarrow\ \left( \left(  2\mid\ell\text{ if
% }k>1\right)  \text{ and }\underbrace{\overline{R}_{P}^{\ell}%
% =\operatorname*{id}}_{\text{this is equivalent to }\operatorname*{ord}\left(
% \overline{R}_{P}\right)  \mid\ell}\right) \\
% &  \Longleftrightarrow\ \left( \left(  2\mid\ell\text{ if
% }k>1\right)  \text{ and }\operatorname*{ord}\left(  \overline{R}_{P}\right)
% \mid\ell\right) \\
% &  \Longleftrightarrow\ \left\{
% \begin{array}
% [c]{c}%
% \left( 2\mid\ell\text{ and }\operatorname*{ord}\left(
% \overline{R}_{P}\right)  \mid\ell\right),\ \ \ \ \ \ \ \ \ \ \text{if
% }k>1;\\
% \left( \operatorname*{ord}\left(  \overline{R}_{P}\right)
% \mid\ell\right),\ \ \ \ \ \ \ \ \ \ \text{if }k=1
% \end{array}
% \right. \\
% &  \Longleftrightarrow\ \left\{
% \begin{array}
% [c]{c}%
% \left( \operatorname{lcm}\left(  2,\operatorname*{ord}\left(
% \overline{R}_{P}\right)  \right)  \mid\ell\right)
%,\ \ \ \ \ \ \ \ \ \ \text{if }k>1;\\
% \left( \operatorname*{ord}\left(  \overline{R}_{P}\right)
% \mid\ell\right),\ \ \ \ \ \ \ \ \ \ \text{if }k=1
% \end{array}
% \right. \\
&  \Longleftrightarrow\ \left( \left\{
\begin{array}
[c]{l}%
\operatorname{lcm}\left(  2,\operatorname*{ord}\left(  \overline{R}%
_{P}\right)  \right),\ \ \ \ \ \ \ \ \ \ \text{if }k>1;\\
\operatorname*{ord}\left(  \overline{R}_{P}\right)
,\ \ \ \ \ \ \ \ \ \ \text{if }k=1
\end{array}
\right.  \mid\ell\right).
\end{align*}
Hence, for every $\ell\in\mathbb{N}$, we have the following equivalence of
assertions:%
\begin{align*}
\left( \operatorname*{ord}\left(  \overline{R}_{B_{k}%
P}\right)  \mid\ell\right)  \  &  \Longleftrightarrow\ \left(
\overline{R}_{B_{k}P}^{\ell}=\operatorname*{id}\right)
\Longleftrightarrow\ \left( \left\{
\begin{array}
[c]{l}%
\operatorname{lcm}\left(  2,\operatorname*{ord}\left(  \overline{R}%
_{P}\right)  \right),\ \ \ \ \ \ \ \ \ \ \text{if }k>1;\\
\operatorname*{ord}\left(  \overline{R}_{P}\right)
,\ \ \ \ \ \ \ \ \ \ \text{if }k=1
\end{array}
\right.  \mid\ell\right).
\end{align*}
Consequently, $\operatorname*{ord}\left(  \overline{R}_{B_{k}P}\right)
=\left\{
\begin{array}
[c]{l}%
\operatorname{lcm}\left(  2,\operatorname*{ord}\left(  \overline{R}%
_{P}\right)  \right),\ \ \ \ \ \ \ \ \ \ \text{if }k>1;\\
\operatorname*{ord}\left(  \overline{R}_{P}\right)
,\ \ \ \ \ \ \ \ \ \ \text{if }k=1
\end{array}
\right.  $. 
% This is exactly what (\ref{pf.Bk.ord.1}) claims. Thus,
% (\ref{pf.Bk.ord.1}) is proven, and with it Proposition \ref{prop.Bk.ord}.
\end{proof}

Here is an analogue of Proposition \ref{prop.Bk.ord}:

\begin{proposition}
\label{prop.B'k.ord}Let $P$ be an $n$-graded poset. 

\textbf{(a)} We have $\operatorname*{ord}\left(  \overline{R}_{B_{1}^{\prime
}P}\right)  =\operatorname*{ord}\left(  \overline{R}_{P}\right)  $.

\textbf{(b)} For every integer $k>1$, we have $\operatorname*{ord}\left(
\overline{R}_{B_{k}^{\prime}P}\right)  =\operatorname{lcm}\left(
2,\operatorname*{ord}\left(  \overline{R}_{P}\right)  \right)  $.
\end{proposition}

\begin{proof}
[\nopunct]The proof of this is very similar (though not exactly identical) to
that of Proposition \ref{prop.Bk.ord}. Alternatively, it is easy to deduce
Proposition \ref{prop.B'k.ord} from Proposition \ref{prop.Bk.ord} using
Proposition \ref{prop.op.ord} and Proposition \ref{prop.skeletal.op}.
\end{proof}

\begin{proof}
[Proof of Proposition \ref{prop.skeletal.ords}.]
For any
skeletal poset $T$, we can compute $\operatorname*{ord}\left(  R_{T}\right)  $
and $\operatorname*{ord}\left(  \overline{R}_{T}\right)  $ inductively using
Propositions \ref{prop.PQ.ord}--\ref{prop.B'k.ord} (and the fact that
$\operatorname*{ord}\left(  R_{\varnothing}\right)  =1$ and
$\operatorname*{ord}\left(  \overline{R}_{\varnothing}\right)  =1$). More precisely:

\begin{itemize}
\item If $T$ is the empty poset $\varnothing$, then $\operatorname*{ord}%
\left(  R_{T}\right)  =\operatorname*{ord}\left(  R_{\varnothing}\right)  =1 =\operatorname*{ord}\left(  \overline{R}_{T}\right)  =\operatorname*{ord}%
\left(  \overline{R}_{\varnothing}\right)$.

\item If $T$ has the form $B_{k}P$ for some $n$-graded skeletal poset $P$ and
some positive integer $k$, then Proposition \ref{prop.Bk.ord} yields
\[
\operatorname*{ord}\left(  \overline{R}_{T}\right)  =\operatorname*{ord}%
\left(  \overline{R}_{B_{k}P}\right)  =\left\{
\begin{array}
[c]{l}%
\operatorname{lcm}\left(  2,\operatorname*{ord}\left(  \overline{R}%
_{P}\right)  \right),\ \ \ \ \ \ \ \ \ \ \text{if }k>1;\\
\operatorname*{ord}\left(  \overline{R}_{P}\right)
,\ \ \ \ \ \ \ \ \ \ \text{if }k=1
\end{array}
\right.,
\]
and Proposition \ref{prop.ord-projord} yields $\operatorname*{ord}\left(
R_{T}\right)  =\operatorname{lcm}\left(  n+1,\operatorname*{ord}\left(
\overline{R}_{T}\right)  \right)  $.

\item Analogously one can compute $\operatorname*{ord}\left(  R_{T}\right)  $
and $\operatorname*{ord}\left(  \overline{R}_{T}\right)  $ if $T$ has the form
$B_{k}^{\prime}P$.

\item If $T$ has the form $PQ$ for two WLOG nonempty $n$-graded skeletal
posets $P$ and $Q$, then Proposition \ref{prop.PQ.ord} yields
$\operatorname*{ord}\left(  R_{PQ}\right)  =\operatorname{lcm}\left(
\operatorname*{ord}\left(  R_{P}\right),\operatorname*{ord}\left(
R_{Q}\right)  \right)  $, and Proposition \ref{prop.PQ.ord2} yields
$\operatorname*{ord}\left(  \overline{R}_{PQ}\right)  =\operatorname{lcm}%
\left(  \operatorname*{ord}\left(  R_{P}\right),\operatorname*{ord}\left(
R_{Q}\right)  \right)  $.
\end{itemize}

This gives an algorithm for inductively computing $\operatorname*{ord}\left(
R_{T}\right)  $ and $\operatorname*{ord}\left(  \overline{R}_{T}\right)  $ for
a skeletal poset $T$. It is clear that these orders as computed by this
algorithms will be finite.
\end{proof}

As the proof of Proposition~\ref{prop.skeletal.ords} shows,
we can recursively compute (rather than just bound
from the above) the orders of $R_{P}$ and $\overline{R}_{P}$ for any skeletal
poset $P$ without doing any computations in $\mathbb{K}$. (This also shows
that the orders of $R_{P}$ and $\overline{R}_{P}$ don't depend on the base
field $\mathbb{K}$ as long as $\mathbb{K}$ is infinite and $P$ is skeletal.)

In the case of forests and trees we %can also use this induction to establish a
obtain a concrete bound:

\begin{corollary}
\label{cor.for.ord}Let $P$ be an $n$-graded poset which is also  
a rooted forest (made into a poset by having every node smaller than its children).

\textbf{(a)} Then, $\operatorname*{ord}\left(  R_{P}\right)  \mid
\operatorname{lcm}\left(  1,2,\ldots,n+1\right)  $.

\textbf{(b)} Moreover, if $P$ is a tree, then $\operatorname*{ord}\left(
\overline{R}_{P}\right)  \mid\operatorname{lcm}\left(  1,2,\ldots,n\right)  $.
\end{corollary}

Corollary \ref{cor.for.ord} is also valid if we replace ``every node smaller
than its children'' by ``every node larger than its children'', and the proof
is exactly analogous.


\begin{proof}
[Proof of Corollary \ref{cor.for.ord}.] 
Use strong induction on $\left\vert P\right\vert $, applying
Propositions~\ref{prop.Bk.ord}~\textbf{(a)} and \ref{prop.ord-projord} when $P$ is a tree,
and Proposition~\ref{prop.PQ.ord} when $P$ is a forest.
(See \cite{grinberg-roby-arXiv} for details.)
\end{proof}
% \begin{proof}
% [Proof of Corollary \ref{cor.for.ord}.] Corollary
% \ref{cor.for.ord} can be proven by strong induction over
% $\left\vert P\right\vert $. Indeed, if $P$ is an $n$-graded poset and a rooted
% forest, then we must be in one of the following three cases:

% \textit{Case 1:} We have $P=\varnothing$.

% \textit{Case 2:} The rooted forest $P$ is a tree.

% \textit{Case 3:} The rooted forest $P$ is a disjoint union of more than one tree.

% The validity of Corollary \ref{cor.for.ord} \textbf{(a)} is trivial in
% Case 1, and in Case 3 it follows from the induction hypothesis using
% Proposition \ref{prop.PQ.ord}. (Corollary \ref{cor.for.ord} \textbf{(b)}
% makes no statement about these two Cases.)
% In Case 2, we have $P=B_{1}Q$ for some rooted forest $Q$, which is
% necessarily $\left(  n-1\right)  $-graded; thus, the induction hypothesis
% (applied to $Q$ instead of $P$) yields $\operatorname*{ord}\left(
% R_{Q}\right)  \mid\operatorname{lcm}\left(  1,2,\ldots,\left(  n-1\right)
% +1\right)  =\operatorname{lcm}\left(  1,2,\ldots,n\right)  $, and we obtain%
% \begin{align*}
% \operatorname*{ord}\left(  \overline{R}_{P}\right)   &  =\operatorname*{ord}%
% \left(  \overline{R}_{B_{1}Q}\right)  =\operatorname*{ord}\left(  \overline
% {R}_{Q}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by Proposition
% \ref{prop.Bk.ord} \textbf{(a)}}\right) \\
% &  \mid\operatorname{lcm}\left(  1,2,\ldots,n\right)
% \end{align*}
% and%
% \begin{align*}
% \operatorname*{ord}\left(  R_{P}\right)   &  =\operatorname{lcm}\left(
% n+1,\underbrace{\operatorname*{ord}\left(  \overline{R}_{P}\right)  }%
% _{\mid\operatorname{lcm}\left(  1,2,\ldots,n\right)  }\right)
% \ \ \ \ \ \ \ \ \ \ \left(  \text{by Proposition \ref{prop.ord-projord}%
% }\right) \\
% &  \mid\operatorname{lcm}\left(  n+1,\operatorname{lcm}\left(
% 1,2,\ldots,n\right)  \right)  =\operatorname{lcm}\left(  1,2,\ldots,n+1\right).
% \end{align*}
% Thus, the induction step is complete in each of the three Cases.
% \end{proof}



\section{\label{sect.classicalr}Postscript: Classical rowmotion on skeletal
posets}

The above results concerning birational rowmotion on skeletal posets suggest
the question of what can be said about \textbf{classical} rowmotion (on the
set of order ideals) on this class of posets. Indeed, while the classical
rowmotion map (as opposed to the birational one) has been the object of
several studies (e.g., \cite{striker-williams} and
\cite{cameron-fon-der-flaass}), it seems that this rather simple case has
never been explicitly covered. Let us bridge this
gap and derive the counterparts of Propositions \ref{prop.Bk.ord} and
\ref{prop.skeletal.ords} and Corollary \ref{cor.for.ord} for classical
rowmotion.  
% First, we define the maps involved.

% \begin{definition}
% Let $P$ be a poset.

% \textbf{(a)} An \textit{order ideal} of $P$ means a subset $S$ of $P$ such
% that every $s\in S$ and $p\in P$ with $p\leq s$ satisfy $p\in S$.

% \textbf{(b)} The set of all order ideals of $P$ will be denoted by $J\left(
% P\right)  $.
% \end{definition}

To start we define (classical) toggles on order ideals (an analogue of Definition
\ref{def.Tv}). Recall that an \textit{order ideal} of $P$ means a subset $S$ of $P$ such
that every $s\in S$ and $p\in P$ with $p\leq s$ satisfy $p\in S$.  We denote the set of all
order ideals of $P$ by $J(P)$.


\begin{definition}
\label{def.classical.tv}Let $P$ be a finite poset. Let $v\in P$. Define a map
$\mathbf{t}_{v}:J\left(  P\right)  \rightarrow J\left(  P\right)  $ by%
\[
\mathbf{t}_{v}\left(  S\right)  =\left\{
\begin{array}
[c]{l}%
S\cup\left\{  v\right\}  \text{, if }v\notin S\text{ and }S\cup\left\{
v\right\}  \in J\left(  P\right)  ;\\
S\setminus\left\{  v\right\}  \text{, if }v\in S\text{ and }S\setminus\left\{
v\right\}  \in J\left(  P\right)  ;\\
S\text{, otherwise}%
\end{array}
\right.  \ \ \ \ \ \ \ \ \ \ \text{for every }S\in J\left(  P\right).
\]
(This is clearly well-defined.) This map $\mathbf{t}_{v}$ will be called the
\textit{classical }$v$\textit{-toggle}.
\end{definition}

We can rewrite this definition in more \textquotedblleft
local\textquotedblright\ terms, by replacing the conditions \textquotedblleft%
$S\cup\left\{  v\right\}  \in J\left(  P\right)  $\textquotedblright\ and
\textquotedblleft$S\setminus\left\{  v\right\}  \in J\left(  P\right)
$\textquotedblright\ by the respectively equivalent conditions
\textquotedblleft every element $u\in P$ satisfying $u\lessdot v$ lies in
$S$\textquotedblright\ and \textquotedblleft no element $u\in P$ satisfying
$u\gtrdot v$ lies in $S$\textquotedblright\ (in fact, the equivalence of these
conditions is easily seen). Hence, we obtain the following analogue to
our definition of $T_v$:

\begin{proposition}
\label{prop.classical.Tv}Let $P$ be a finite poset. Let $v\in P$. For every
$S\in J\left(  P\right)  $, we have:

\textbf{(a)} If $w$ is an element of $P$ such that $w\neq v$, then we have
$w\in\mathbf{t}_{v}\left(  S\right)  $ if and only if $w\in S$.

\textbf{(b)} We have $v\in\mathbf{t}_{v}\left(  S\right)  $ if and only if
\begin{align*}
&  \left(  v\in S\text{ and not }\left(  \text{no element }u\in P\text{
satisfying }u\gtrdot v\text{ lies in }S\right)  \right) \\
&  \text{or }\left(  v\notin S\text{ and }\left(  \text{every element }u\in
P\text{ satisfying }u\lessdot v\text{ lies in }S\right)  \right).
\end{align*}

\end{proposition}

While the complicated logical statement in Proposition \ref{prop.classical.Tv}
\textbf{(b)} can be simplified, the form we have stated it in exhibits its
similarity to our definition of $T_v$ particularly well. This, in fact, is
more than a similarity: If we allow $\mathbb{K}$ to be a semifield rather than
a field, we can regard the classical $v$-toggle $\mathbf{t}_{v}$ as a
restriction of the birational toggle $T_{v}$ (when $\mathbb{K}$ is chosen
appropriately)\footnote{Here are the details: Let $\operatorname*{Trop}%
\mathbb{Z}$ be the tropical semiring over $\mathbb{Z}$, that is, the semiring
obtained by endowing the set $\mathbb{Z}\cup\left\{  -\infty\right\}  $ with
the binary operation $\left(  a,b\right)  \mapsto\max\left\{  a,b\right\}  $
as \textquotedblleft addition\textquotedblright\ and the binary operation
$\left(  a,b\right)  \mapsto a+b$ as \textquotedblleft
multiplication\textquotedblright\ (where the usual rules for sums involving
$-\infty$ apply). Then, $\operatorname*{Trop}\mathbb{Z}$ is a semifield, with
$\left(  a,b\right)  \mapsto a-b$ serving as \textquotedblleft
subtraction\textquotedblright, with $-\infty$ serving as \textquotedblleft
zero\textquotedblright\ and with the integer $0$ serving as \textquotedblleft
one\textquotedblright. Now, to every order ideal $S\in J\left(  P\right)  $,
we can assign a $\left(  \operatorname*{Trop}\mathbb{Z}\right)  $-labelling
$\operatorname*{tlab}S\in\left(  \operatorname*{Trop}\mathbb{Z}\right)
^{\widehat{P}}$, defined by%
\[
\left(  \operatorname*{tlab}S\right)  \left(  v\right)  =\left\{
\begin{array}
[c]{c}%
1,\text{ if }v\notin S\cup\left\{  0\right\}  ;\\
0,\ \text{if }v\in S\cup\left\{  0\right\}
\end{array}
\right..
\]
This yields a map $\operatorname*{tlab}:J\left(  P\right)  \rightarrow\left(
\operatorname*{Trop}\mathbb{Z}\right)  ^{\widehat{P}}$, obviously injective.
This map $\operatorname*{tlab}$ satisfies $T_{v}\circ\operatorname*{tlab}%
=\operatorname*{tlab}\circ\mathbf{t}_{v}$ for every $v\in P$. This allows us
to regard the classical toggles $\mathbf{t}_{v}$ as restrictions of the
birational toggles $T_{v}$, if we consider this map $\operatorname*{tlab}$ as
an inclusion. This reasoning goes back to Einstein and Propp
\cite{einstein-propp}.}. Hence, some theorems about birational toggles can be
used to derive analogous theorems about classical toggles\footnote{For
example, we could derive Proposition \ref{prop.classical.tv.commute} from
Proposition \ref{prop.Tv.commute} using this tactic. However, we could not
derive (say) Proposition \ref{prop.classical.skeletal.ords} from Proposition
\ref{prop.skeletal.ords} this way, because the order of a restriction of a
permutation could be a proper divisor of the order of the permutation.}. We
will not use this tactic in the following, because often it will be easier to
study the classical $v$-toggles on their own. However, many of the properties
of classical toggles (and classical rowmotion) that we are going to discuss
will have proofs that are parallel to the proofs of the analogous results
about birational toggles. We will omit these proofs when the analogy is
glaring enough.

We have the following easily-verified analogues of Proposition
\ref{prop.Tv.invo}, Proposition \ref{prop.Tv.commute} and Corollary
\ref{cor.R.welldef}:

\begin{proposition}
\label{prop.classical.tv.invo}Let $P$ be a finite poset. Let $v\in P$. Then,
the map $\mathbf{t}_{v}$ is an involution on $J\left(  P\right)  $ (that is,
we have $\mathbf{t}_{v}^{2}=\operatorname*{id}$).
\end{proposition}

\begin{proposition}
\label{prop.classical.tv.commute}Let $P$ be a finite poset. Let $v\in P$ and
$w\in P$. Then, $\mathbf{t}_{v}\circ\mathbf{t}_{w}=\mathbf{t}_{w}%
\circ\mathbf{t}_{v}$, unless we have either $v\lessdot w$ or $w\lessdot v$.
\end{proposition}

\begin{corollary}
\label{cor.classical.r.welldef}Let $P$ be a finite poset. Let $\left(
v_{1},v_{2},\ldots,v_{m}\right)  $ be a linear extension of $P$. Then, the map
$\mathbf{t}_{v_{1}}\circ\mathbf{t}_{v_{2}}\circ\ldots\circ\mathbf{t}_{v_{m}%
}:J\left(  P\right)  \rightarrow J\left(  P\right)  $ is well-defined and
independent of the choice of the linear extension $\left(  v_{1}%
,v_{2},\ldots,v_{m}\right)  $.
\end{corollary}

The three results above are observations made on \cite[page 546]%
{cameron-fon-der-flaass} (in somewhat different notation).

Two convenient advantages of the classical setup are that we don't have to
worry about denominators becoming zero, so our maps are actual maps rather
than partial maps, and that we don't have to pass to the poset $\widehat{P}$.
We can now define rowmotion in analogy to Definition \ref{def.rm}:

\begin{definition}
\label{def.classical.rm}Let $P$ be a finite poset. \textit{Classical
rowmotion} (simply called \textquotedblleft rowmotion\textquotedblright\ in
existing literature) is defined as the map $\mathbf{r}: = \mathbf{t}_{v_{1}}\circ
\mathbf{t}_{v_{2}}\circ\ldots\circ\mathbf{t}_{v_{m}}:J\left(  P\right)
\rightarrow J\left(  P\right)  $, where $\left(  v_{1},v_{2},\ldots,v_{m}\right)
$ is a linear extension of $P$. This map is well-defined
% (in particular, it does not depend on the linear extension $\left(  v_{1},v_{2},\ldots,v_{m}\right)
% $ chosen) 
by Corollary \ref{cor.classical.r.welldef}. 
% (and also by the
% fact that a linear extension of $P$ exists; this is Theorem
% \ref{thm.linext.ex}). This map will be denoted by $\mathbf{r}$.
\end{definition}

To highlight the similarities between the classical and birational cases, let
us state the analogue of Proposition \ref{prop.R.implicit}:

\begin{proposition}
\label{prop.classical.r.implicit}Let $P$ be a finite poset. Let $v\in P$. Let
$S\in J\left(  P\right)  $. Then, $v\in\mathbf{r}\left(  S\right)  $ holds if
and only if the following two conditions hold:

\textit{Condition 1:} Every $u\in P$ satisfying $u\lessdot v$ belongs to $S$.

\textit{Condition 2:} Either $v\notin S$, or there exists an $u\in
\mathbf{r}\left(  S\right)  $ satisfying $u\gtrdot v$. (The ``either/or'' is
meant non-exclusively.)
\end{proposition}

This proposition is easily seen to be equivalent to the following well-known
equivalent description of rowmotion (\cite[Lemma 1]{cameron-fon-der-flaass},
translated into our notation):

\begin{proposition}
\label{prop.classical.r.maxmin}Let $P$ be a finite poset. Let $S\in J\left(
P\right)  $. Then, the maximal elements of $\mathbf{r}\left(  S\right)  $ are
precisely the minimal elements of $P\setminus S$.
\end{proposition}

We record the analogue of Proposition \ref{prop.R.implicit.converse}:

\begin{proposition}
\label{prop.classical.r.implicit.converse}Let $P$ be a finite poset. Let $S$
and $T$ be two order ideals of $P$. Assume that for every $v\in P$, the
relation $v\in T$ holds if and only if Conditions 1 and 2 of Proposition
\ref{prop.classical.r.implicit} hold with $\mathbf{r}\left(  S\right)  $
replaced by $T$. Then, $T=\mathbf{r}\left(  S\right)  $.
\end{proposition}

In analogy to Proposition \ref{prop.R.inverse}, we have:

\begin{proposition}
\label{prop.classical.r.inverse}Let $P$ be a finite poset. Then, classical
rowmotion $\mathbf{r}$ is invertible. Its inverse $\mathbf{r}^{-1}$ is
$\mathbf{t}_{v_{m}}\circ\mathbf{t}_{v_{m-1}}\circ\ldots\circ\mathbf{t}_{v_{1}%
}:J\left(  P\right)  \rightarrow J\left(  P\right)  $, where $\left(
v_{1},v_{2},\ldots,v_{m}\right)  $ is a linear extension of $P$.
\end{proposition}

We can study graded posets again. In analogy to Corollary \ref{cor.Ri.welldef}%
, Definition \ref{def.Ri}, Proposition \ref{prop.Ri.R} and Proposition
\ref{prop.Ri.invo}, we have:

\begin{corollary}
\label{cor.classical.ri.welldef}
Let $P$ be an $n$-graded
poset. Fix $i\in\left\{  1,2,\ldots,n\right\}  $. Let $\left(  u_{1}%
,u_{2},\ldots,u_{k}\right)  $ be any list of the elements of $\widehat{P}_{i}$
with every element of $\widehat{P}_{i}$ appearing exactly once in the list.
\footnote{Note that $\widehat{P}_{i}$ is simply $\left\{  v\in P\ \mid\ \deg
v=i\right\}  $, because $i$ equals neither $0$ nor $n+1$. We are using the
notation $\widehat{P}_{i}$ despite not working with $\widehat{P}$ merely to
stress some analogies.} Then, the map $\mathbf{r}_{i}:=\mathbf{t}_{u_{1}}\circ\mathbf{t}%
_{u_{2}}\circ\ldots\circ\mathbf{t}_{u_{k}}:J\left(  P\right)  \rightarrow
J\left(  P\right)  $ is well-defined and independent of the choice of the list
$\left(  u_{1},u_{2},\ldots,u_{k}\right)  $.
\end{corollary}

% \begin{definition}
% \label{def.classical.ri}Let $P$ be an $n$-graded poset.
% Let $i\in\left\{  1,2,\ldots,n\right\}  $. Then, let $\mathbf{r}_{i}$ denote the
% map $\mathbf{t}_{u_{1}}\circ\mathbf{t}_{u_{2}}\circ\ldots\circ\mathbf{t}_{u_{k}%
% }:J\left(  P\right)  \rightarrow J\left(  P\right)  $, where $\left(
% u_{1},u_{2},\ldots,u_{k}\right)  $ is any list of the elements of $\widehat{P}%
% _{i}$ with every element of $\widehat{P}_{i}$ appearing exactly once in the
% list. This map $\mathbf{t}_{u_{1}}\circ\mathbf{t}_{u_{2}}\circ\ldots\circ
% \mathbf{t}_{u_{k}}$ is well-defined (in particular, it does not depend on the
% list $\left(  u_{1},u_{2},\ldots,u_{k}\right)  $) by Corollary
% \ref{cor.classical.ri.welldef}.
% \end{definition}

\begin{proposition}
\label{prop.classical.ri.r}
Let $P$ be an $n$-graded
poset. Then, $
\mathbf{r}=\mathbf{r}_{1}\circ\mathbf{r}_{2}\circ\ldots\circ\mathbf{r}_{n}.
$

\end{proposition}

\begin{proposition}
\label{prop.classical.ri.invo}Let $P$ be an $n$-graded
poset. Let $i \in \left\{1,2,\ldots,n\right\}$.
Then, $\mathbf{r}_{i}$ is an
involution on $J\left(  P\right)  $ (that is, $\mathbf{r}_{i}^{2}%
=\operatorname*{id}$).
\end{proposition}

A parody of w-tuples can also be defined. The following is analogous to
Definition \ref{def.wi}:

\begin{definition}
\label{def.classical.wi}
For $S\in J\left(  P\right)  $ and $i\in\left\{  1,2,\ldots,n\right\}  $, set
%$\mathbf{w}_{i}\left(  S\right)  $ will denote the integer%
\[
\mathbf{w}_{i}\left(  S\right) =
\left\{
\begin{array}
[c]{l}%
1,\text{ if }P_{i}\subseteq S\text{ and }P_{i+1}\cap S=\varnothing\\
0,\text{ otherwise}%
\end{array}
\right..
\]
Here, $P_{j}$ denotes the subset $\deg^{-1}\left(
\left\{  j\right\}  \right)  $ of $P$, with $P_{0}=P_{n+1}=\varnothing$.
% this subset is empty if $j=0$ and also
% empty if $j=n+1$.
\end{definition}

Analogues of Proposition \ref{prop.wi.Ri} and Proposition \ref{prop.wi.R} are
easily found:

\begin{proposition}
\label{prop.classical.wi.ri}
For any $S\in J\left(  P\right)  $ and $i\in\left\{  1,2,\ldots,n\right\}  $, we have
\begin{align*}
&  \left(  \mathbf{w}_{0}\left(  \mathbf{r}_{i}\left(  S\right)  \right)
,\mathbf{w}_{1}\left(  \mathbf{r}_{i}\left(  S\right)  \right)
,\ldots,\mathbf{w}_{n}\left(  \mathbf{r}_{i}\left(  S\right)  \right)  \right) \\
&  =\left(  \mathbf{w}_{0}\left(  S\right),\mathbf{w}_{1}\left(  S\right)
,\ldots,\mathbf{w}_{i-2}\left(  S\right),\mathbf{w}_{i}\left(  S\right)
,\mathbf{w}_{i-1}\left(  S\right),\mathbf{w}_{i+1}\left(  S\right)
,\mathbf{w}_{i+2}\left(  S\right),\ldots,\mathbf{w}_{n}\left(  S\right)
\right).
\end{align*}

\end{proposition}

\begin{proposition}
\label{prop.classical.wi.r}
For any $S\in J\left(  P\right)  $, we have
% Let $n\in\mathbb{N}$. Let $P$ be an $n$-graded
% poset. Then, every $S\in J\left(  P\right)  $ satisfies%
\[
\left(  \mathbf{w}_{0}\left(  \mathbf{r}\left(  S\right)  \right)
,\mathbf{w}_{1}\left(  \mathbf{r}\left(  S\right)  \right),\ldots,\mathbf{w}%
_{n}\left(  \mathbf{r}\left(  S\right)  \right)  \right)  =\left(
\mathbf{w}_{n}\left(  S\right),\mathbf{w}_{0}\left(  S\right)
,\mathbf{w}_{1}\left(  S\right),\ldots,\mathbf{w}_{n-1}\left(  S\right)
\right).
\]

\end{proposition}


However, the $\left(  n+1\right)  $-tuple $\left(  \mathbf{w}_{0}\left(
S\right),\mathbf{w}_{1}\left(  S\right),\ldots,\mathbf{w}_{n}\left(
S\right)  \right)  $ obtained from an order ideal $S$ is not particularly
informative. In fact, it is $\left(  0,0,\ldots,0\right)  $ for ``most'' order
ideals; here is what this means precisely:

\begin{definition}
\label{def.classical.level}
An order ideal of $P$ is said to be \textit{level} if and only if it
has the form $P_{1}\cup P_{2}\cup\ldots\cup P_{i}$ for some $i\in\left\{
0,1,\ldots,n\right\}  $.
\end{definition}

Easy properties of level order ideals are:

\begin{proposition}
\label{prop.classical.level}Let $P$ be an $n$-graded poset.

\textbf{(a)} There exist precisely $n+1$ level order ideals of $P$, and those
form an orbit under classical rowmotion $\mathbf{r}$. Namely, one has%
\[
\mathbf{r}\left(  P_{1}\cup P_{2}\cup\ldots\cup P_{i}\right)  =\left\{
\begin{array}
[c]{l}%
P_{1}\cup P_{2}\cup\ldots\cup P_{i+1},\ \ \ \ \ \ \ \ \ \ \text{if }i<n;\\
\varnothing,\ \ \ \ \ \ \ \ \ \ \text{if }i=n
\end{array}
\right..
\]


\textbf{(b)} If $S\in J\left(  P\right)  $, then $\left(  \mathbf{w}%
_{0}\left(  S\right),\mathbf{w}_{1}\left(  S\right),\ldots,\mathbf{w}%
_{n}\left(  S\right)  \right)  =\left(  0,0,\ldots,0\right)  $ unless $S$ is level.
\end{proposition}

Now, we can define a (toylike, but rather useful)
notion of homogeneous equivalence, in somewhat questionable analogy with
Definition \ref{def.hgeq}.

\begin{definition}
\label{def.classical.hgeq}
Two order ideals $S$ and $T$ of $P$ are said to be \textit{homogeneously
equivalent} if and only if either both $S$ and $T$ are level or we have $S=T$.
Clearly, being homogeneously equivalent is an equivalence relation. Let
$\overline{J\left(  P\right)  }$ denote the set of equivalence classes of
elements of $J\left(  P\right)  $ modulo this relation. Let $\pi$ denote the
canonical projection $J\left(  P\right)  \rightarrow\overline{J\left(
P\right)  }$. (We distinguish this map $\pi$ from the map $\pi$ defined in
Definition \ref{def.hgeq} by the fact that they act on different objects.)
\end{definition}

The following analogue of Corollary \ref{cor.hgR} is almost trivial:

\begin{corollary}
\label{cor.classical.hgr}
If $S$ and $T$ are two homogeneously equivalent order ideals of $P$, then
$\mathbf{r}\left(  S\right)  $ is homogeneously equivalent to $\mathbf{r}%
\left(  T\right)  $.
\end{corollary}

(An analogue of Corollary \ref{cor.hgRi} exists as well.) We also have the
following analogue of Proposition \ref{prop.reconstruct}:

\begin{proposition}
\label{prop.classical.reconstruct}
Let $S$ and $T$ be two order ideals of $P$ such that
$\pi\left(S\right) = \pi\left(T\right)$ and
$\left(  \mathbf{w}_{0}\left(  S\right),\mathbf{w}_{1}\left(  S\right)
,\ldots,\mathbf{w}_{n}\left(  S\right)  \right)  =\left(  \mathbf{w}_{0}\left(
T\right),\mathbf{w}_{1}\left(  T\right),\ldots,\mathbf{w}_{n}\left(
T\right)  \right)  $.
Then, $S=T$.
\end{proposition}

We can furthermore state analogues of Definitions \ref{def.hgRi} and
\ref{def.hgR}:

\begin{definition}
\label{def.classical.hgri}
Fix $i\in\left\{  1,2,\ldots,n\right\}  $. The map $\mathbf{r}%
_{i}:J\left(  P\right)  \rightarrow J\left(  P\right)  $ descends (through the
projection $\pi:J\left(  P\right)  \rightarrow\overline{J\left(  P\right)  }$)
to a map $\overline{J\left(  P\right)  }\rightarrow\overline{J\left(
P\right)  }$. We denote this map $\overline{J\left(  P\right)  }%
\rightarrow\overline{J\left(  P\right)  }$ by $\overline{\mathbf{r}_{i}}$.
Thus, the diagram%
\[%
%TCIMACRO{\TeXButton{ri commutes with pi}{\xymatrixcolsep{5pc}\xymatrix{
%J\left(P\right) \ar[r]^{\mathbf r_i} \ar[d]_-{\pi} & J\left(P\right)
%\ar[d]^-{\pi} \\
%\overline{J\left(P\right)} \ar[r]_{\overline{\mathbf r_i}}
%& \overline{J\left(P\right)}
%}}}%
%BeginExpansion
\xymatrixcolsep{5pc}\xymatrix{
J\left(P\right) \ar[r]^{\mathbf r_i} \ar[d]_-{\pi} & J\left(P\right)
\ar[d]^-{\pi} \\
\overline{J\left(P\right)} \ar[r]_{\overline{\mathbf r_i}}
& \overline{J\left(P\right)}
}%
%EndExpansion
\]
is commutative.
\end{definition}

\begin{definition}
\label{def.classical.hgr}
Define the map $\overline{\mathbf{r}}:\overline{J\left(  P\right)
}\rightarrow\overline{J\left(  P\right)  }$ by%
\[
\overline{\mathbf{r}}=\overline{\mathbf{r}_{1}}\circ\overline{\mathbf{r}_{2}%
}\circ\ldots\circ\overline{\mathbf{r}_{n}}.
\]
Then, the following diagram commutes:
\begin{equation}%
%TCIMACRO{\TeXButton{r commutes with pi}{\xymatrixcolsep{5pc}\xymatrix{
%J\left(P\right) \ar[r]^{\mathbf r} \ar[d]_-{\pi} & J\left(P\right)
%\ar[d]^-{\pi} \\
%\overline{J\left(P\right)} \ar[r]_{\overline{\mathbf r}} & \overline
%{J\left(P\right)}
%}} }%
%BeginExpansion
\xymatrixcolsep{5pc}\xymatrix{
J\left(P\right) \ar[r]^{\mathbf r} \ar[d]_-{\pi} & J\left(P\right)
\ar[d]^-{\pi} \\
\overline{J\left(P\right)} \ar[r]_{\overline{\mathbf r}} & \overline
{J\left(P\right)}.
}
%EndExpansion
\label{def.classical.hgr.commute}%
\end{equation}
% In other words, $\overline{\mathbf{r}}$ is the map
% $\overline{J\left(  P\right)  }\rightarrow\overline{J\left(  P\right)  }$ to
% which the map $\mathbf{r}:J\left(  P\right)  \rightarrow J\left(  P\right)  $
% descends (through the projection $\pi:J\left(  P\right)  \rightarrow
% \overline{J\left(  P\right)  }$).
\end{definition}

It might seem that the map $\overline{\mathbf{r}}$ is not worth considering,
since its cycle structure differs from the cycle structure of $\mathbf{r}$
only in the collapsing of an $\left(  n+1\right)  $-cycle (the one formed by
all level order ideals) to a point. However, triviality in combinatorics does
not preclude usefulness, and we will employ the ``projective'' version
$\overline{\mathbf{r}}$ of classical rowmotion as a stirrup in determining the
order of classical rowmotion $\mathbf{r}$ on skeletal posets.

We have the following simple relation between the orders of $\mathbf{r}$ and
$\overline{\mathbf{r}}$:

\begin{proposition}
\label{prop.classical.ord-projord}Let $P$ be an
$n$-graded poset. Then, $\operatorname*{ord}\mathbf{r}=\operatorname{lcm}%
\left(  n+1,\operatorname*{ord}\overline{\mathbf{r}}\right)  $. 
% (Here,
% $\operatorname{lcm}\left(  n+1,\infty\right)  $ is to be understood as
% $\infty$.)
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.classical.ord-projord}.]We know
that $\mathbf{r}$ is an invertible map $J\left(  P\right)  \rightarrow
J\left(  P\right)  $, thus a permutation of the finite set $J\left(  P\right)
$. Hence, $\operatorname*{ord}\mathbf{r}$ is the $\operatorname{lcm}$ of the
lengths of the cycles of this permutation $\mathbf{r}$. Similarly,
$\operatorname*{ord}\overline{\mathbf{r}}$ is the $\operatorname{lcm}$ of the
lengths of the cycles of the permutation $\overline{\mathbf{r}}$ of the finite
set $\overline{J\left(  P\right)  }$. Therefore, in order to prove
Proposition \ref{prop.classical.ord-projord}, it is enough to show that
there exists a 1-to-1 correspondence between the cycles of $\mathbf{r}$ and
the cycles of $\overline{\mathbf{r}}$ which preserves the length of a cycle
except for one particular length-$\left(n+1\right)$-cycle of $\mathbf{r}$
which it maps to a $1$-cycle of $\overline{\mathbf{r}}$.

Such a correspondence is easy to find:
The set $\overline{J\left(  P\right)  }$ is the quotient of
$J\left(  P\right)  $ modulo homogeneous equivalence. But homogeneous
equivalence merely identifies the $n+1$ level order ideals (which form a
cycle under $\mathbf{r}$), while all other
elements of $J\left(  P\right)  $ are pairwise non-equivalent.
Thus, the cycles of $\overline{\mathbf{r}}$ are in 1-to-1 correspondence
with the cycles of $\mathbf{r}$, and corresponding cycles have equal
lengths except for the length-$\left(n+1\right)$ cycle formed by the
$n+1$ level order ideals in $J\left(P\right)$ and its corresponding
length-$1$ cycle in $\overline{J\left(P\right)}$. This is the
correspondence we were looking for.
%
% Let $Z_{1}$, $Z_{2}$, $\ldots$, $Z_{k}$ be the cycles of the permutation
% $\mathbf{r}$ of $J\left(  P\right)  $. We assume WLOG that $Z_{1}$ is the
% cycle consisting of the $n+1$ level order ideals (because we know that they
% form a cycle). Thus, $\left\vert Z_{1}\right\vert =n+1$. Since
% $\operatorname*{ord}\mathbf{r}$ is the $\operatorname{lcm}$ of the lengths of
% the cycles of the permutation $\mathbf{r}$, we have $\operatorname*{ord}%
% \mathbf{r}=\operatorname{lcm}\left(  \left\vert Z_{1}\right\vert,\left\vert
% Z_{2}\right\vert,\ldots,\left\vert Z_{k}\right\vert \right)  $.
% 
% Now, let us recall that $\overline{J\left(  P\right)  }$ is the quotient of
% $J\left(  P\right)  $ modulo homogeneous equivalence. But homogeneous
% equivalence merely identifies the $n+1$ level order ideals, while all other
% elements of $J\left(  P\right)  $ are still pairwise non-equivalent. Hence,
% the cycles of the permutation $\overline{\mathbf{r}}$ of $\overline{J\left(
% P\right)  }$ are $\pi\left(  Z_{1}\right)  $, $\pi\left(  Z_{2}\right)  $,
% $\ldots$, $\pi\left(  Z_{k}\right)  $, and while $\pi\left(  Z_{2}\right)  $,
% $\pi\left(  Z_{3}\right)  $, $\ldots$, $\pi\left(  Z_{k}\right)  $ are isomorphic
% to $Z_{2}$, $Z_{3}$, $\ldots$, $Z_{k}$, respectively, the first cycle $\pi\left(
% Z_{1}\right)  $ (being the projection of the cycle of the level order ideals)
% now has length $1$. Now, $\operatorname*{ord}\overline{\mathbf{r}}$ is the
% $\operatorname{lcm}$ of the lengths of the cycles of the permutation
% $\overline{\mathbf{r}}$ of the finite set $\overline{J\left(  P\right)  }$.
% Since these cycles are $\pi\left(  Z_{1}\right)  $, $\pi\left(  Z_{2}\right)
% $, $\ldots$, $\pi\left(  Z_{k}\right)  $, this yields%
% \begin{align*}
% \operatorname*{ord}\overline{\mathbf{r}}  &  =\operatorname{lcm}\left(
% \left\vert \pi\left(  Z_{1}\right)  \right\vert,\left\vert \pi\left(
% Z_{2}\right)  \right\vert,\ldots,\left\vert \pi\left(  Z_{k}\right)  \right\vert
% \right)  =\operatorname{lcm}\left(  \underbrace{\left\vert \pi\left(
% Z_{1}\right)  \right\vert }_{=1},\left\vert \pi\left(  Z_{2}\right)
% \right\vert,\left\vert \pi\left(  Z_{3}\right)  \right\vert,\ldots,\left\vert
% \pi\left(  Z_{k}\right)  \right\vert \right) \\
% &  =\operatorname{lcm}\left(  1,\left\vert \pi\left(  Z_{2}\right)
% \right\vert,\left\vert \pi\left(  Z_{3}\right)  \right\vert,\ldots,\left\vert
% \pi\left(  Z_{k}\right)  \right\vert \right)  =\operatorname{lcm}\left(
% \left\vert \pi\left(  Z_{2}\right)  \right\vert,\left\vert \pi\left(
% Z_{3}\right)  \right\vert,\ldots,\left\vert \pi\left(  Z_{k}\right)  \right\vert
% \right) \\
% &  =\operatorname{lcm}\left(  \left\vert Z_{2}\right\vert,\left\vert
% Z_{3}\right\vert,\ldots,\left\vert Z_{k}\right\vert \right)
% \ \ \ \ \ \ \ \ \ \ \left(
% \begin{array}
% [c]{c}%
% \text{since }\pi\left(  Z_{2}\right)  \text{, }\pi\left(  Z_{3}\right)
% \text{, }\ldots\text{, }\pi\left(  Z_{k}\right)  \text{ are}\\
% \text{isomorphic to }Z_{2}\text{, }Z_{3}\text{, }\ldots\text{, }Z_{k}\text{,
% respectively}%
% \end{array}
% \right).
% \end{align*}
% Now,%
% \begin{align*}
% \operatorname*{ord}\mathbf{r}  &  =\operatorname{lcm}\left(  \left\vert
% Z_{1}\right\vert,\left\vert Z_{2}\right\vert,\ldots,\left\vert Z_{k}\right\vert
% \right)  =\operatorname{lcm}\left(  \underbrace{\left\vert Z_{1}\right\vert
% }_{=n+1},\underbrace{\operatorname{lcm}\left(  \left\vert Z_{2}\right\vert
%,\left\vert Z_{3}\right\vert,\ldots,\left\vert Z_{k}\right\vert \right)
% }_{=\operatorname*{ord}\overline{\mathbf{r}}}\right) \\
% &  =\operatorname{lcm}\left(  n+1,\operatorname*{ord}\overline{\mathbf{r}%
% }\right).
% \end{align*}
% This proves Proposition \ref{prop.classical.ord-projord}.
\end{proof}

Our goal is to make a statement about the order of classical rowmotion on
skeletal posets. Of course, the finiteness of these orders is obvious in this
case, because $J\left(  P\right)  $ is a finite set. However, we can make
stronger claims:

\begin{proposition}
\label{prop.classical.skeletal.ords}For any skeletal poset $P$,
we have
$\operatorname*{ord}\left(  R_{P}\right)
=\operatorname*{ord}\left(  \mathbf{r}_{P}\right)  $ and $\operatorname*{ord}%
\left(  \overline{R}_{P}\right)  =\operatorname*{ord}\left(  \overline
{\mathbf{r}}_{P}\right)  $.
\end{proposition}

% Here, we are using the following convention:

% \begin{definition}
% \label{def.classical.conv.rP}Let $P$ be a finite poset. We denote the maps
% $\mathbf{r}$ and $\overline{\mathbf{r}}$ by $\mathbf{r}_{P}$ and
% $\overline{\mathbf{r}}_{P}$, respectively, so as to make their dependence on
% $P$ explicit.
% \end{definition}

Proposition \ref{prop.classical.skeletal.ords} yields (in particular) that the
order of classical rowmotion coincides with the order of birational rowmotion
(whatever the base field) for skeletal posets. This was conjectured by James
Propp (private communication) for the case of $P$ a tree. We are going to
prove Proposition \ref{prop.classical.skeletal.ords} by exhibiting further
analogies between classical and birational rowmotion. First of all, the
following proposition is just as trivial as its birational counterpart
(Proposition \ref{prop.PQ.ord}):

\begin{proposition}
\label{prop.classical.PQ.ord}Fix $n\in\mathbb{N}$, and let $P$ and $Q$ be two
$n$-graded posets. Then, $\operatorname*{ord}\left(  \mathbf{r}_{PQ}\right)
=\operatorname{lcm}\left(  \operatorname*{ord}\left(  \mathbf{r}_{P}\right)
,\operatorname*{ord}\left(  \mathbf{r}_{Q}\right)  \right)  $.
\end{proposition}

We can show a simple counterpart of this proposition for $\operatorname*{ord}%
\left(  \overline{\mathbf{r}}_{PQ}\right)  $ (but still with
$\operatorname*{ord}\left(  \mathbf{r}_{P}\right)  $ and $\operatorname*{ord}%
\left(  \mathbf{r}_{Q}\right)  $ on the right hand side!):

\begin{proposition}
\label{prop.classical.PQ.ord2}Fix $n\in\mathbb{N}$, and let $P$ and $Q$ be
two\textbf{ }$n$-graded posets. Then, $\operatorname*{ord}\left(
\overline{\mathbf{r}}_{PQ}\right)  =\operatorname{lcm}\left(
\operatorname*{ord}\left(  \mathbf{r}_{P}\right),\operatorname*{ord}\left(
\mathbf{r}_{Q}\right)  \right)  $.
\end{proposition}

Again, the divisibility of the right hand side by the left one is
the most useful statement, and it follows trivially from the results
before. The opposite divisibility can be obtained by studying the
orbit of the order ideal $P$ of $PQ$ (which is not level and thus
does not collapse under $\pi$). See \cite[\S 10]{grinberg-roby-arXiv} for details.

% \begin{proof}
% [Proof of Proposition \ref{prop.classical.PQ.ord2}.]WLOG, assume
% that $n\neq0$ (else, the statement is trivial). Hence, $P$ and $Q$ are nonempty.

% Consider the order ideal $P$ of $PQ$. Then, one can easily see (by induction)
% that every $i\in\left\{  0,1,\ldots,n+1\right\}  $ satisfies%
% \begin{align*}
% &  \mathbf{r}_{PQ}^{i}\left(  P\right) \\
% &  =\left(  \left\{
% \begin{array}
% [c]{l}%
% P_{1}\cup P_{2}\cup\ldots\cup P_{i-1},\ \ \ \ \ \ \ \ \ \ \text{if }i>0;\\
% P,\ \ \ \ \ \ \ \ \ \ \text{if }i=0
% \end{array}
% \right.  \right)  \cup\left(  \left\{
% \begin{array}
% [c]{l}%
% Q_{1}\cup Q_{2}\cup\ldots\cup Q_{i},\ \ \ \ \ \ \ \ \ \ \text{if }i\leq n;\\
% \varnothing,\ \ \ \ \ \ \ \ \ \ \text{if }i=n+1
% \end{array}
% \right.  \right).
% \end{align*}
% From this, it follows that the smallest positive integer $k$ satisfying
% $\mathbf{r}_{PQ}^{k}\left(  P\right)  =P$ is $n+1$. Since $P$ is not level,
% this does not change under applying $\pi$; that is, the smallest positive
% integer $k$ satisfying $\overline{\mathbf{r}}_{PQ}^{k}\left(  P\right)  =P$ is
% still $n+1$. Hence, $n+1\mid\operatorname*{ord}\left(  \overline{\mathbf{r}%
% }_{PQ}\right)  $. But Proposition \ref{prop.classical.ord-projord} (applied to
% $PQ$ instead of $P$) yields%
% \[
% \operatorname*{ord}\left(  \mathbf{r}_{PQ}\right)  =\operatorname{lcm}\left(
% n+1,\operatorname*{ord}\left(  \overline{\mathbf{r}}_{PQ}\right)  \right)
% =\operatorname*{ord}\left(  \overline{\mathbf{r}}_{PQ}\right)
% \]
% (since $n+1\mid\operatorname*{ord}\left(  \overline{\mathbf{r}}_{PQ}\right)
% $), so that%
% \[
% \operatorname*{ord}\left(  \overline{\mathbf{r}}_{PQ}\right)
% =\operatorname*{ord}\left(  \mathbf{r}_{PQ}\right)  =\operatorname{lcm}\left(
% \operatorname*{ord}\left(  \mathbf{r}_{P}\right),\operatorname*{ord}\left(
% \mathbf{r}_{Q}\right)  \right)
% \]
% (by Proposition \ref{prop.classical.PQ.ord}). This proves Proposition
% \ref{prop.classical.PQ.ord2}.
% \end{proof}

More interesting is the analogue of Proposition \ref{prop.Bk.ord}, whose proof is also
analogous, though easier (full details in \cite{grinberg-roby-arXiv}). 

\begin{proposition}
\label{prop.classical.Bk.ord}Let $n\in\mathbb{N}$. Let $P$ be an $n$-graded poset.

\textbf{(a)} We have $\operatorname*{ord}\left(  \overline{\mathbf{r}}%
_{B_{1}P}\right)  =\operatorname*{ord}\left(  \overline{\mathbf{r}}%
_{P}\right)  $.

\textbf{(b)} For every integer $k>1$, we have $\operatorname*{ord}\left(
\overline{\mathbf{r}}_{B_{k}P}\right)  =\operatorname{lcm}\left(
2,\operatorname*{ord}\left(  \overline{\mathbf{r}}_{P}\right)  \right)  $.
\end{proposition}

% \begin{proof}
% [Proof of Proposition \ref{prop.classical.Bk.ord}.]
% %We will be
% %proving parts \textbf{(a)} and \textbf{(b)} together.
% Let $k$ be a positive
% integer.
% % (this has to be $1$ for proving part \textbf{(a)}).
% We need to prove
% that
% \begin{equation}
% \operatorname*{ord}\left(  \overline{\mathbf{r}}_{B_{k}P}\right)  =\left\{
% \begin{array}
% [c]{l}%
% \operatorname{lcm}\left(  2,\operatorname*{ord}\left(  \overline{\mathbf{r}%
% }_{P}\right)  \right),\ \ \ \ \ \ \ \ \ \ \text{if }k>1;\\
% \operatorname*{ord}\left(  \overline{\mathbf{r}}_{P}\right)
%,\ \ \ \ \ \ \ \ \ \ \text{if }k=1
% \end{array}
% \right.. \label{pf.classical.Bk.ord.1}%
% \end{equation}
% Proving this clearly will prove both parts \textbf{(a)} and \textbf{(b)} of
% Proposition \ref{prop.classical.Bk.ord}.

% Notice that $B_{k}P$ is an $\left(  n+1\right)  $-graded poset. For every
% $\ell\in\left\{  1,2,\ldots,n+1\right\}  $, let $\left(  B_{k}P\right)  _{\ell}$
% be the subset $\deg^{-1}\left(  \left\{  \ell\right\}  \right)  $ of $B_{k}P$.
% Thus, $\left(  B_{k}P\right)  _{\ell}=\left\{  v\in B_{k}P\ \mid\ \deg
% v=\ell\right\}  $. In particular, $\left(  B_{k}P\right)  _{1}$ is the set of
% all minimal elements of $B_{k}P$, so that $\left(  B_{k}P\right)  _{1}$ is an
% antichain of size $k$ (by the construction of $B_{k}P$). We also have $w<v$
% for every $w\in\left(  B_{k}P\right)  _{1}$ and $v\in P$.

% For every finite poset $Q$, the map $\overline{\mathbf{r}}_{Q}$ is an
% invertible map $\overline{J\left(  Q\right)  }\rightarrow\overline{J\left(
% Q\right)  }$, that is, a permutation of the finite set $\overline{J\left(
% Q\right)  }$. Hence, the order $\operatorname*{ord}\left(  \overline
% {\mathbf{r}}_{Q}\right)  $ is the $\operatorname{lcm}$ of the lengths of the
% orbits of this map $\overline{\mathbf{r}}_{Q}$. We are going to compare the
% orbits of the maps $\overline{\mathbf{r}}_{B_{k}P}$ and $\overline{\mathbf{r}%
% }_{P}$.

% Define a map $\phi:J\left(  P\right)  \rightarrow J\left(  B_{k}P\right)  $ by%
% \[
% \phi\left(  S\right)  =\left(  B_{k}P\right)  _{1}\cup
% S\ \ \ \ \ \ \ \ \ \ \text{for every }S\in J\left(  P\right).
% \]
% It is easy to see that this map $\phi$ is well-defined (that is, $\left(
% B_{k}P\right)  _{1}\cup S$ is an order ideal of $B_{k}P$ for every $S\in
% J\left(  P\right)  $), and that it sends level order ideals of $P$ to level
% order ideals of $B_{k}P$. Hence, it preserves homogeneous equivalence, so that
% it induces a map $\overline{J\left(  P\right)  }\rightarrow\overline{J\left(
% B_{k}P\right)  }$. Denote this map $\overline{J\left(  P\right)  }%
% \rightarrow\overline{J\left(  B_{k}P\right)  }$ by $\overline{\phi}$. Thus,
% $\overline{\phi}\circ\pi=\pi\circ\phi$.

% It is moreover easy to see that $\overline{\mathbf{r}}_{B_{k}P}\circ
% \overline{\phi}=\overline{\phi}\circ\overline{\mathbf{r}}_{P}$%
% \ \ \ \ \footnote{\textit{Proof.} In order to prove this, it is enough to show
% that for every $S\in J\left(  P\right)  $, the order ideals $\left(
% \mathbf{r}_{B_{k}P}\circ\phi\right)  \left(  S\right)  $ and $\left(
% \phi\circ\mathbf{r}_{P}\right)  \left(  S\right)  $ are homogeneously
% equivalent. This is clear in the case when $S$ is level (because both $\left(
% \mathbf{r}_{B_{k}P}\circ\phi\right)  \left(  S\right)  $ and $\left(
% \phi\circ\mathbf{r}_{P}\right)  \left(  S\right)  $ are level in this case),
% so let us WLOG assume that $S$ is not level. Then, we can actually show that
% $\left(  \mathbf{r}_{B_{k}P}\circ\phi\right)  \left(  S\right)  $ and $\left(
% \phi\circ\mathbf{r}_{P}\right)  \left(  S\right)  $ are identical. Indeed, it
% is easy to see that:
% \par
% \begin{itemize}
% \item for every $T\in J\left(  P\right)  $ and every $v\in P$, we have
% $\left(  \mathbf{t}_{v}\circ\phi\right)  \left(  T\right)  =\left(  \phi
% \circ\mathbf{t}_{v}\right)  \left(  T\right)  $;
% \par
% \item for every nonempty $T\in J\left(  P\right)  $ and every $w\in\left(
% B_{k}P\right)  _{1}$, we have $\left(  \mathbf{t}_{w}\circ\phi\right)  \left(
% T\right)  =\phi\left(  T\right)  $.
% \end{itemize}
% \par
% Using these facts, and the definition of classical rowmotion as a composition
% of classical toggle maps $\mathbf{t}_{v}$, we can then easily see that
% $\left(  \mathbf{r}_{B_{k}P}\circ\phi\right)  \left(  S\right)  =\left(
% \phi\circ\mathbf{r}_{P}\right)  \left(  S\right)  $. This completes the proof
% of $\overline{\mathbf{r}}_{B_{k}P}\circ\overline{\phi}=\overline{\phi}%
% \circ\overline{\mathbf{r}}_{P}$.}. Hence, the subset $\overline{\phi}\left(
% \overline{J\left(  P\right)  }\right)  $ is closed under application of the
% map $\overline{\mathbf{r}}_{B_{k}P}$.

% The map $\overline{\phi}$ also is injective (this is very easy to see again,
% since the only order ideals of $P$ which are mapped to level order ideals by
% $\phi$ are themselves level). Thus, $\operatorname*{ord}\left(  \overline
% {\mathbf{r}}_{B_{k}P}\mid_{\overline{\phi}\left(  \overline{J\left(  P\right)
% }\right)  }\right)  =\operatorname*{ord}\left(  \overline{\mathbf{r}}%
% _{P}\right)  $ (because the injectivity of $\overline{\phi}$ allows us to
% identify $\overline{J\left(  P\right)  }$ with $\overline{\phi}\left(
% \overline{J\left(  P\right)  }\right)  $ along the map $\overline{\phi}$, and
% then the equality $\overline{\mathbf{r}}_{B_{k}P}\circ\overline{\phi
% }=\overline{\phi}\circ\overline{\mathbf{r}}_{P}$ rewrites as $\overline
% {\mathbf{r}}_{B_{k}P}\mid_{\overline{\phi}\left(  \overline{J\left(  P\right)
% }\right)  }=\overline{\mathbf{r}}_{P}$, so that $\operatorname*{ord}\left(
% \overline{\mathbf{r}}_{B_{k}P}\mid_{\overline{\phi}\left(  \overline{J\left(
% P\right)  }\right)  }\right)  =\operatorname*{ord}\left(  \overline
% {\mathbf{r}}_{P}\right)  $).

% Let $H$ be the set of all nonempty proper subsets of $\left(  B_{k}P\right)
% _{1}$. It is clear that $H\subseteq J\left(  B_{k}P\right)  $. Notice that
% $H=\varnothing$ if $k=1$. Every $T\in H$ satisfies%
% \[
% \mathbf{r}_{B_{k}P}\left(  T\right)  =\left(  B_{k}P\right)  _{1}\setminus T
% \]
% (this is easy to see from any definition of classical rowmotion, or from
% Proposition \ref{prop.classical.r.implicit}). Hence, the set $H$ is closed
% under application of the map $\mathbf{r}_{B_{k}P}$, and this map
% $\mathbf{r}_{B_{k}P}$ maps every element of $H$ to its complement in $\left(
% B_{k}P\right)  _{1}$. In particular, this shows that $\operatorname*{ord}%
% \left(  \mathbf{r}_{B_{k}P}\mid_{H}\right)  =\left\{
% \begin{array}
% [c]{l}%
% 2,\ \ \ \ \ \ \ \ \ \ \text{if }k>1;\\
% 1,\ \ \ \ \ \ \ \ \ \ \text{if }k=1
% \end{array}
% \right.  $.

% We now use the map $\pi$ to identify the set $H$ with its projection
% $\pi\left(  H\right)  $ under $\pi$ (this is allowed because $\pi$ is
% injective on $H$). This identification entails $\left.  \overline{\mathbf{r}%
% }_{B_{k}P}\mid_{H}\right.  =\mathbf{r}_{B_{k}P}\mid_{H}$. In particular, the
% set $H$ is closed under application of the map $\overline{\mathbf{r}}_{B_{k}%
% P}$.

% But it is easy to see that $J\left(  B_{k}P\right)  $ is the union of the two
% subsets $H$ and $\phi\left(  J\left(  P\right)  \right)  $ (because every
% order ideal of $B_{k}P$ either contains the whole $\left(  B_{k}P\right)
% _{1}$, or it does not, in which case it cannot contain \textbf{any} element of
% degree $>1$). Hence, $\overline{J\left(  B_{k}P\right)  }$ is the union of the
% two subsets $\pi\left(  H\right)  =H$ and $\pi\left(  \phi\left(  J\left(
% P\right)  \right)  \right)  =\overline{\phi}\left(  \overline{J\left(
% P\right)  }\right)  $. Moreover, these two subsets are disjoint and each of
% them is closed under application of the map $\overline{\mathbf{r}}_{B_{k}P}$.
% Hence,%
% \begin{align*}
% \operatorname*{ord}\left(  \overline{\mathbf{r}}_{B_{k}P}\right)   &
% =\operatorname{lcm}\left(  \operatorname*{ord}\left(  \underbrace{\overline
% {\mathbf{r}}_{B_{k}P}\mid_{H}}_{=\mathbf{r}_{B_{k}P}\mid_{H}}\right)
%,\operatorname*{ord}\left(  \overline{\mathbf{r}}_{B_{k}P}\mid_{\overline
% {\phi}\left(  \overline{J\left(  P\right)  }\right)  }\right)  \right) \\
% &  =\operatorname{lcm}\left(  \underbrace{\operatorname*{ord}\left(
% \mathbf{r}_{B_{k}P}\mid_{H}\right)  }_{=\left\{
% \begin{array}
% [c]{l}%
% 2,\ \ \ \ \ \ \ \ \ \ \text{if }k>1;\\
% 1,\ \ \ \ \ \ \ \ \ \ \text{if }k=1
% \end{array}
% \right.  },\underbrace{\operatorname*{ord}\left(  \overline{\mathbf{r}}%
% _{B_{k}P}\mid_{\overline{\phi}\left(  \overline{J\left(  P\right)  }\right)
% }\right)  }_{=\operatorname*{ord}\left(  \overline{\mathbf{r}}_{P}\right)
% }\right) \\
% &  =\operatorname{lcm}\left(  \left\{
% \begin{array}
% [c]{l}%
% 2,\ \ \ \ \ \ \ \ \ \ \text{if }k>1;\\
% 1,\ \ \ \ \ \ \ \ \ \ \text{if }k=1
% \end{array}
% \right.,\operatorname*{ord}\left(  \overline{\mathbf{r}}_{P}\right)  \right)
% \\
% &  =\left\{
% \begin{array}
% [c]{l}%
% \operatorname{lcm}\left(  2,\operatorname*{ord}\left(  \overline{\mathbf{r}%
% }_{P}\right)  \right),\ \ \ \ \ \ \ \ \ \ \text{if }k>1;\\
% \operatorname*{ord}\left(  \overline{\mathbf{r}}_{P}\right)
%,\ \ \ \ \ \ \ \ \ \ \text{if }k=1
% \end{array}
% \right..
% \end{align*}
% This proves (\ref{pf.classical.Bk.ord.1}). Thus, the proof of Proposition
% \ref{prop.classical.Bk.ord} is complete.
% \end{proof}

We can also formulate an analogue of Proposition \ref{prop.B'k.ord}:

\begin{proposition}
\label{prop.classical.B'k.ord}Let $P$ be an $n$-graded poset.

\textbf{(a)} We have $\operatorname*{ord}\left(  \overline{\mathbf{r}}%
_{B_{1}^{\prime}P}\right)  =\operatorname*{ord}\left(  \overline{\mathbf{r}%
}_{P}\right)  $.

\textbf{(b)} For every integer $k>1$, we have $\operatorname*{ord}\left(
\overline{\mathbf{r}}_{B_{k}^{\prime}P}\right)  =\operatorname{lcm}\left(
2,\operatorname*{ord}\left(  \overline{\mathbf{r}}_{P}\right)  \right)  $.
\end{proposition}

\begin{proof}
[\nopunct]The proof of this is fairly similar to that of Proposition
\ref{prop.classical.Bk.ord}.
\end{proof}

We can now prove Proposition \ref{prop.classical.skeletal.ords}:

\begin{proof}
[Proof of Proposition \ref{prop.classical.skeletal.ords}.]In our proof
of Proposition~\ref{prop.skeletal.ords}, we gave an
algorithm for inductively computing $\operatorname*{ord}\left(
R_{T}\right)  $ and $\operatorname*{ord}\left(  \overline{R}_{T}\right)  $ for
a skeletal poset $T$. Using Propositions~\ref{prop.classical.PQ.ord}--\ref{prop.classical.B'k.ord}
% (and
% the fact that $\operatorname*{ord}\left(  \mathbf{r}_{\varnothing}\right)  =1$
% and $\operatorname*{ord}\left(  \overline{\mathbf{r}}_{\varnothing}\right)
% =1$)
instead of Propositions~\ref{prop.PQ.ord}--\ref{prop.B'k.ord}, we 
obtain the \textbf{same} algorithm for inductively computing
$\operatorname*{ord}\left(  \mathbf{r}_{T}\right)  $ and $\operatorname*{ord}%
\left(  \overline{\mathbf{r}}_{T}\right)  $ for a skeletal poset $T$. 
% And
% these two algorithms are \textbf{the same}, because of the direct analogy
% between the propositions that are used in the first algorithm and those used
% in the second one. 
Therefore, $\operatorname*{ord}\left(  R_{P}\right)
=\operatorname*{ord}\left(  \mathbf{r}_{P}\right)  $ and $\operatorname*{ord}%
\left(  \overline{R}_{P}\right)  =\operatorname*{ord}\left(  \overline
{\mathbf{r}}_{P}\right)  $. 
\end{proof}

Proposition \ref{prop.classical.skeletal.ords} does not generalize to
arbitrary graded posets. Counterexamples to such a generalization can be found
in \cite[\S 12]{grinberg-roby-part2}.

Finally, in analogy to Corollary \ref{cor.for.ord}, we can now show:

\begin{corollary}
\label{cor.classical.for.ord}Let $P$ be an $n$-graded
poset that is also a rooted forest (made into a poset by having every
node smaller than its children).

\textbf{(a)} Then, $\operatorname*{ord}\left(  \mathbf{r}_{P}\right)
\mid\operatorname{lcm}\left(  1,2,\ldots,n+1\right)  $.

\textbf{(b)} Moreover, if $P$ is a tree, then $\operatorname*{ord}\left(
\overline{\mathbf{r}}_{P}\right)  \mid\operatorname{lcm}\left(
1,2,\ldots,n\right)  $.
\end{corollary}

Corollary \ref{cor.classical.for.ord} is also valid if we replace
\textquotedblleft every node smaller than its children\textquotedblright\ by
\textquotedblleft every node larger than its children\textquotedblright, and
the proof is exactly analogous.

Let us notice that the algorithm described in the proof of Proposition
\ref{prop.classical.skeletal.ords} can be turned into an explicit formula (not
just an upper bound as in Corollary \ref{cor.classical.for.ord}), whose
inductive proof we leave to the reader:

\begin{proposition}
\label{prop.classical.for.ord.explicit}Let $P$ be an $n$-graded
poset that is also a rooted forest (made into a poset by having every
node smaller than its children).
 Notice that $\left\vert
\widehat{P}_{i}\right\vert \leq\left\vert \widehat{P}_{i+1}\right\vert $ for
every $i\in\left\{  0,1,\ldots,n-1\right\}  $. 
% (where $\widehat{P}_{i}$ and
% $\widehat{P}_{i+1}$ are defined as in Definition \ref{def.graded.Phat}). 
Then,%
\[
\operatorname*{ord}\left(  \overline{\mathbf{r}}_{P}\right)
=\operatorname{lcm}\left\{  n-i\ \mid\ i\in\left\{  0,1,\ldots,n-1\right\}
;\ \left\vert \widehat{P}_{i}\right\vert <\left\vert \widehat{P}%
_{i+1}\right\vert \right\}.
\]
Of course, $\operatorname*{ord}\left(  \mathbf{r}_{P}\right)  $ can now be
computed by $\operatorname*{ord}\left(  \mathbf{r}_{P}\right)
=\operatorname{lcm}\left(  n+1,\operatorname*{ord}\left(  \overline
{\mathbf{r}}_{P}\right)  \right)  $.
\end{proposition}

The same property therefore holds for birational rowmotion $R_{P}$ and its
homogeneous version $\overline{R}_{P}$.

\subsection*{Acknowledgments}

The notion of birational rowmotion is due to James Propp and Arkady
Berenstein. This paper owes James Propp also for a constant flow of
inspiration and useful suggestions.

Hugh Thomas corrected slips in our writing including an abuse of
Zariski topology and some accidental alternative history. Nathan
Williams noticed further typos.

The first author came to know birational rowmotion in Alexander Postnikov's
combinatorics pre-seminar at MIT. Postnikov also suggested veins of further study.

Jessica Striker helped the first author understand some of the past work on
this subject, in particular the labyrinthine connections between related
operators (rowmotion, Panyushev complementation, Striker-Williams promotion,
Sch\"{u}tzenberger promotion, etc.). The present paper explores merely one
corner of this labyrinth (the rowmotion corner).

We thank the two referees of our FPSAC abstract
\cite{grinberg-roby} for further helpful comments.
We also owe a number of improvements in this paper
to the suggestions of an anonymous EJC referee.

Both authors were partially supported by NSF grant \#1001905, and have
utilized the open-source CAS Sage (\cite{sage}, \cite{sage-combinat}) to
perform laborious computations. We thank Travis Scrimshaw, Fr\'{e}d\'{e}ric
Chapoton, Viviane Pons and Nathann Cohen for reviewing Sage patches relevant
to this project.





\begin{thebibliography}{99}                                                                                      %


\bibitem[AKSch12]{ayyer-klee-schilling}Arvind Ayyer, Steven Klee, Anne
Schilling, \textit{Combinatorial Markov chains on linear extensions}, 
J. Alg. Combin., September 2013, DOI 10.1007/s10801-013-0470-9.
Also at \arxiv{1205.7074v3}.

\bibitem[BrSchr74]{brouwer-schrijver}Andries E. Brouwer and A. Schrijver,
\textit{On the period of an operator, defined on antichains}, Math. Centr.
report ZW24, Amsterdam (Jun. 1974).\newline%
\url{http://www.win.tue.nl/~aeb/preprints/zw24.pdf}

\bibitem[CaFl95]{cameron-fon-der-flaass}Peter J. Cameron, Dmitry G.
Fon-der-Flaass, \textit{Orbits of Antichains Revisited}, Europ. J. 
Combin., vol. 16, Issue 6, November 1995, pp. 545--554.\newline%
\url{http://www.sciencedirect.com/science/article/pii/0195669895900365}. 

\bibitem[EiPr13]{einstein-propp}David Einstein, James Propp,
\textit{Combinatorial, piecewise-linear, and birational homomesy for products
of two chains}, \arxiv{1310.5294v1} (preliminary version), October 20,
2013.

\bibitem[EiPr14]{einstein-propp-fpsac}David Einstein, James Propp,
\textit{Piecewise-linear and birational toggling}, \arxiv{1404.3455v1}, 
DMTCS proc. FPSAC 2014.

\bibitem[Etienn84]{etienne}Gwihen Etienne,
\textit{Linear extensions of finite posets and a conjecture of G. Kreweras on permutations},
Discrete Mathematics, Volume 52, Issue 1, 1984, pp. 107--111.\newline
\url{http://www.sciencedirect.com/science/article/pii/0012365X84901080}

\bibitem[Flaa93]{fon-der-flaass}Dmitry G. Fon-der-Flaass, \textit{Orbits of
Antichains in Ranked Posets}, Europ. J. of Combin., vol. 14,
Issue 1, January 1993, pp. 17--22.
\url{http://www.sciencedirect.com/science/article/pii/S0195669883710036}.

\bibitem[GrRoArX]{grinberg-roby-arXiv}Darij Grinberg, Tom Roby, \textit{Iterative
properties of birational rowmotion}, \arxiv{1402.6178v3}, 2014.  (Full version with
detailed proofs.) 


\bibitem[GrRo13]{grinberg-roby}Darij Grinberg, Tom Roby, \textit{The order of
birational rowmotion}, DMTCS proc. FPSAC 2014. (Extended abstract of this paper.)
\newline%
\url{http://www.dmtcs.org/dmtcs-ojs/index.php/proceedings/article/viewArticle/dmAT0165}. 
% \texttt{\href{http://web.mit.edu/~darij/www/algebra/ipbrFPSAC6.pdf}{\texttt{http://web.mit.edu/\symbol{126}%
% darij/www/algebra/ipbrFPSAC6.pdf}}}

\bibitem[GrRo14]{grinberg-roby-part2}Darij Grinberg, Tom Roby, \textit{Iterative
properties of birational rowmotion II: rectangles and triangles}, 
{\it Electronic J. of Combin.}, {\bf 22(3)}, 2015, \#P3,40. 
\url{http://www.combinatorics.org/ojs/index.php/eljc/article/view/v22i3p40}. 

\bibitem[Haiman92]{haiman}Mark D. Haiman, \textit{Dual equivalence with
applications, including a conjecture of Proctor}, Discrete Mathematics, Volume
99, Issues 1--3, 2 April 1992, pp. 79--113.
\url{http://www.sciencedirect.com/science/article/pii/0012365X9290368P}


\bibitem[OSZ13]{osz-geoRSK}Neil O'Connell, Timo Sepp\"{a}l\"{a}inen, Nikos
Zygouras, \textit{Geometric RSK correspondence, Whittaker functions and
symmetrized random polymers}, Invent. Math., October 2013, DOI
10.1007/s00222-013-0485-9.
\url{http://dx.doi.org/10.1007/s00222-013-0485-9}.  An older preprint version at
\arxiv{1210.5126v2}.  

% \bibitem[PrRo13]{propp-roby}James Propp and Tom Roby, \textit{Homomesy in
% products of two chains}, DMTCS proc. FPSAC 2013,
% \texttt{\href{http://www.math.uconn.edu/~troby/ceFPSAC.pdf}{\texttt{http://www.math.uconn.edu/\symbol{126}%
% troby/ceFPSAC.pdf}}}

\bibitem[PrRo13]{propp-roby}James Propp and Tom Roby, \textit{Homomesy in
products of two chains}, 
{\it Electronic J. of Combin.}, {\bf 22(3)}, 2015, \#P3,4. 
\url{http://www.combinatorics.org/ojs/index.php/eljc/article/view/v22i3p4}. 
Also, \arxiv{arXiv:1310.5201v5}. 

\bibitem[Rusk92]{ruskey}Frank Ruskey, \textit{Generating Linear Extensions of
Posets by Transpositions}, J. of Combin. Thy, vol. 54, Issue 1,
January 1992, pp. 77--101.
\url{http://www.sciencedirect.com/science/article/pii/0095895692900678}

\bibitem[S{$^{+}$}09]{sage}W.\thinspace{}A. Stein et~al., \emph{{S}age
{M}athematics {S}oftware ({V}ersion 6.2.beta2)}, The Sage Development Team,
2014, \href{http://www.sagemath.org}{\texttt{http://www.sagemath.org}}.

\bibitem[Sage08]{sage-combinat}The Sage-Combinat community,
\textit{Sage-Combinat: enhancing Sage as a toolbox for computer exploration in
algebraic combinatorics}, 2008. \newline 
\url{http://combinat.sagemath.org}. 

\bibitem[Stan11]{stanley-ec1}Richard P. Stanley, \emph{Enumerative
Combinatorics, volume 1, 2nd edition},  no. 49 in Cambridge
Studies in Advanced Mathematics, Cambridge University Press, 2011.
Also available at \url{http://math.mit.edu/~rstan/ec/ec1/}. 

\bibitem[Stan86]{stanley-polytopes}Richard Stanley, \textit{Two Poset
Polytopes}, Disc. \& Comp. Geom., 1986, Volume 1, Issue 1, pp. 9--23.

\bibitem[StWi11]{striker-williams}Jessica Striker, Nathan Williams,
\textit{Promotion and Rowmotion}, Europ. J. of Combin. 33 (2012),
pp. 1919--1942, DOI 10.1016/j.ejc.2012.05.003.\newline%
\url{http://www.sciencedirect.com/science/article/pii/S0195669812000972}. 
Also available as \arxiv{1108.1172v3}.

\end{thebibliography}


\end{document}