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%% define your title in the usual way
%\title[Optional shorter title]{Something to do with math}
\title[Chainlink Polytopes]{Chainlink Polytopes}

%% define your authors in the usual way
%% use \addressmark{1}, \addressmark{2} etc for the institutions, and use \thanks{} for contact details
%\author[Optional shorter names]{Longer names me\thanks{\href{mailto:hello@world.c}{hello@world.c}. Longer names me was partially supported by Grant 2017.11.14.$\partial$\;supp.}\addressmark{1}, \and you\addressmark{2}}
\author{Ezgi Kantarcı Oğuz\thanks{\href{mailto:ezgikantarcioguz@gmail.com}{ezgikantarcioguz@gmail.com}. EKO was partially supported by Tübitak BİDEP 2218-121C385.}\addressmark{1}, Cem Yalım Özel\thanks{\href{mailto:yalim98@gmail.com}{yalim98@gmail.com}.}\addressmark{2}, \and Mohan Ravichandran\thanks{\href{mailto:mohan.ravichandran@gmail.com}{mohan.ravichandran@gmail.com}. MR gratefully acknowledges financial support from the Bogazici Solidarity fund.}\addressmark{2}}

%% then use \addressmark to match authors to institutions here
\address{\addressmark{1}Department of Mathematics, Galatasaray University, İstanbul \\ \addressmark{2}Department of Mathematics, Bogazici University, İstanbul}

%% put the date of submission here
\received{\today}

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%\revised{}

%% put your English abstract here, or comment this out if you don't have one yet

%% please don't use custom commands in your abstract / resume, as these will be displayed online
%% likewise for citations -- please don't use \cite, and instead write out your citation as something like (author year)
\abstract{We introduce a class of polytopes that we call chainlink polytopes and which allow us to construct infinite families of pairs of non isomorphic rational polytopes with the same Ehrhart quasi-polynomial. Our construction is inspired by a non-obvious and non-trivial symmetry in the rank sequences of  circular fence posets. We show that this symmetry can be lifted to yield an analogous symmetry  at the level of polytopes.  We show this symmetry property of Chainlink polytopes by introducing the related class of chainlink posets and show that they exhibit the same symmetry properties using linear algebraic techniques. We further prove an outstanding conjecture on the unimodality of circular rank polynomials. }

%% put your French abstract here, or comment this out if you don't have one

%% put your keywords here, or comment this out if you don't have them yet
\keywords{polytope, fence poset, symmetry, Ehrhart theory}

%% you can include your bibliography however you want, but using an external .bib file is STRONGLY RECOMMENDED and will make the editor's life much easier
%% regardless of how you do it, please use numerical citations; i.e., [xx, yy] in the text

%% this sample uses biblatex, which (among other things) takes care of URLs in a more flexible way than bibtex
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\usepackage[backend=bibtex]{biblatex}
\addbibresource{chainlink.bib}
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\begin{document}

\maketitle
%% note that you DO NOT have to put your abstract here -- it is generated by \maketitle and the \abstract and \resume commands above


This paper is about a class of polytopes, that naturally arise in poset theory, specifically in the study of fence posets and related objects. They are easy to describe, pliable of study, possess certain unexpected properties and throw up several puzzles. These polytopes will be indexed by compositions; let $\bar{a} = (a_1, \ldots, a_s)$ be a composition of $n$ and let $l$ be a non-negative integer. The \definition{chainlink polytope} $\CL(\bar{a},l)$ with \definition{chain composition} $\bar{a}$ and \definition{link number} $l$ is defined to be the polytope:
\[\CL(\bar{a},l)= \{x \in \mathbb{R}^s \mid 0 \leq x_i \leq a_i,  \, x_{i} - x_{i+1 \,(\operatorname{mod} s)} \leq a_{i}-l, \, i \in [s]\}.\]
This is a polytope that naturally lies in $\mathbb{R}^s$ and has a maximum of $3s$ facets. When the link number $l$ is equal to zero, the second set of constraints become redundant and the polytope becomes a cuboid, $\CL(\bar{a},0)= [0, a_1] \times [0, a_2] \times \ldots \times [0, a_s].$
When the link number is larger, new facets emerge. For an example, see \cref{fig:ChainlinkFirst}.


We will also work with certain special sections of these chainlink polytopes. For a positive real number $t$, we define \[\HL^t(\bar{a},l)=\HL(\bar{a},l) \cap \{x_1 + \ldots + x_s = t\}.\]

The polytopes $CL^t(\bar{a}, l)$ are rational, see \cref{prop:vertices}, are non-empty for $t \in [0, n]$, where $n = a_1 + \ldots + a_s$. One of the main results in this paper is the following (unexpected) symmetry property of these sections of chainlink polytopes. 


\begin{theorem0} \label{thm:vol} Let $\bar{a} = (a_1, \ldots, a_s)$ be a composition of $n$, let $l$ be a positive integer such that $2l \leq \operatorname{min} \{a_i\}_{i \in [s]}$ and let $t$ be a positive integer. Then complementary sections of the chainlink polytope have the same volume,
 \[|\HL^t(\bar{a},l)| = |\HL^{n-t}(\bar{a},l)|,\]
 where $|P|$ for a polytope denotes the relative volume. 
\end{theorem0}
There will be no ambiguity in the definition of the relative volume for us. All our polytopes will lie on hyperplanes of the form $\{x_1 + \ldots + x_s = t\}$ and we will work with the volume form that assigns volume $1$ to the polytope $\mathcal{P} = \operatorname{conv}\{0, e_1 - e_2, e_1 - e_3, \ldots, e_1 - e_s\}$. This theorem is a special case of the following more general theorem. The terms used will be formally defined in the next section. 

\begin{thmn}  Let $\bar{a} = (a_1, \ldots, a_s)$ be a composition of $n$, let $l$ be a non-negative integer such that $2l \leq \operatorname{min} \{a_i\}_{i \in [s]}$ and let $t$ be a positive integer. Then complementary sections of the chainlink polytope have the same Ehrhart quasipolynomial,
 \[\operatorname{Ehr}\, \HL^t(\bar{a},l) = \,\operatorname{Ehr}\,\,\HL^{n-t}(\bar{a},l).\]
\end{thmn}

To see why this is unexpected, consider the chainlink polytope $CL((6, 4, 5), 2)$ as above. The sections at $t = 4$ and $t = 11$ have the same volume, but are non-isomorphic. We plot the first one on the triangular lattice, the natural choice given that these lie on the hyperplanes $x + y + z = \operatorname{const}$. 
\begin{center}
\begin{figure}[ht]
\usetikzlibrary{calc}\newcommand*\rows{4}

\qquad \includegraphics[width = 0.25\linewidth]{Hl-Example.png}\qquad \qquad \begin{tikzpicture}[scale=0.7]
    \foreach \row in {0, 1, ...,4} {
        \draw[gray] ($(\row*.5, {\row*0.5*sqrt(3)})$) -- ($(4-\row*.5,{ \row*0.5*sqrt(3)})$);
        \draw[gray] ($\row*(1, 0)$) -- ($(\rows/2,{\rows/2*sqrt(3)})+\row*(0.5,{-0.5*sqrt(3)})$);
        \draw[gray] ($\row*(1, 0)$) -- ($(0,0)+\row*(0.5,{0.5*sqrt(3)})$);
    }
        \fill[red] (1,0)  circle[radius=2pt];
    \fill[red] (3,0)  circle[radius=2pt];
    \fill[red] (3.75,0.43)  circle[radius=2pt];
    \fill[red] (2,3.46)  circle[radius=2pt];
    \fill[red] (1,1.73)  circle[radius=2pt];

    \draw[red, very thick] (1,0) -- (3, 0) -- (3.75, 0.43) -- (2, 3.46) -- (1, 1.73) -- (1, 0);
\end{tikzpicture} \qquad \qquad 
\begin{tikzpicture}[scale=0.7]
    \foreach \row in {0, 1, ...,4} {
        \draw[gray] ($(2.75*2.18181818-.88+\row*.5, -{\row*0.5*sqrt(3)})$) -- ($(2.75*2.18181818-.88+4-\row*.5,-{ \row*0.5*sqrt(3)})$);
        \draw[gray] ($(\row+2.75*2.18181818-.88, 0)$) -- ($(+2.75*2.18181818-.88+\rows/2,-{\rows/2*sqrt(3)})+\row*(0.5,-{-0.5*sqrt(3)})$);
        \draw[gray] ($(\row+2.75*2.18181818-.88, 0)$) -- ($(2.75*2.18181818-.88,0)+\row*(0.5,-{0.5*sqrt(3)})$);
    }
    \fill[blue] ($(2.75*2.18181818+1.12, -{2*sqrt(3)})$) circle[radius=2pt];%bottom
    \fill[blue] ($(2.75*2.72727273+1.12,{2.75*1.57272727-1.71-2*sqrt(3)})$) circle[radius=2pt];
    \fill[blue] ($(2.75*1.63636364+1.12, {2.75*1.57272727-1.71-2*sqrt(3)})$) circle[radius=2pt];
    \fill[blue] ($(2.75*1.63636364+1.12, {2.75*1.88727273-1.71-2*sqrt(3)})$) circle[radius=2pt];
    \fill[blue] ($(2.7*2.18181818+1.12, {2.75*1.88727273-1.71-2*sqrt(3)})$) circle[radius=2pt];

    \draw[blue, very thick] ($(2.75*2.18181818+1.12, -{2*sqrt(3)})$)--
   ($(2.75*2.72727273+1.12,{2.75*1.57272727-1.71-2*sqrt(3)})$)-- 
 ($(2.7*2.18181818+1.12, {2.75*1.88727273-1.71-2*sqrt(3)})$)--
 ($(2.75*1.63636364+1.12, {2.75*1.88727273-1.71-2*sqrt(3)})$)--
($(2.75*1.63636364+1.12, {2.75*1.57272727-1.71-2*sqrt(3)})$)--
($(2.75*2.18181818+1.12, -{2*sqrt(3)})$);
\end{tikzpicture}
\caption{The polytope $\CL(\bar{a}= (6, 4, 5), \, l = 2)$ with sections \color{red}$CL^4(\bar{a}, l)$ \color{black} 
 and \color{blue} $CL^{11}(\bar{a}, l)$\color{black}.}\label{fig:ChainlinkFirst}
\end{figure}
\end{center}

%Through out this paper, given a polytope $P$, we will use the notation $\#P$ to denote the number of lattice points in $P$.
%\begin{rmk} The theorem may fail when $2l > \operatorname{min} \{a_i\}_{i \in [s]}$. For instance, take our running example $\bar{a} = (6, 4, 5)$, but with $l = 3$ rather than $2$. Let $t = 7$ with complementary section corresponding to $n - t = 15 - 7 = 8$. In this case, we have that even the number of lattice points (the Ehrhart quasi polynomials evaluated at $1$) are different, 
%\[\# \CL^t(\bar{a}, l) = 9, \qquad \# \CL^{n - t}(\bar{a}, l) = 10,\]
%as are the volumes. We have that 
%\[| \CL^t(\bar{a}, l)| = \dfrac{71}{12} \sqrt{3}, \qquad |\CL^{n - t}(\bar{a}, l)| = 6\sqrt{3}.\]

%\end{rmk}

%We do not have a conceptual explanation for why we may lose symmetry when $2l > \operatorname{min} \{a_i\}_{i \in [s]}$. However, in \cref{sec:2}, we will closely relate these polytopes to a natural class of posets, the so called \emph{circular fence posets} and this perspective will provide some insight into this phenomenon. 
%In the third section of this paper, we will explicate why this happens by relating the lattice point enumeration to the structure of certain related posets. \ezgi{Do we actually do that? + we should not say "third" section, and actually cite where it happens}
   
   
\vspace{-1.5cm}
\section{Background}
Where do these chainlink polytopes come from? At first sight, they (might perhaps) seem unmotivated, if (again perhaps) natural. We were led to these following the paper by the first and third authors \cite{main} on fence posets and in particular, a tricky problem that they had been unable to solve. We first recall the definition of fence posets. 
\begin{defn}
Given a composition $\bar{c} = (c_1, \ldots, c_k)$, the \definition{fence poset} is the poset on $n+1$ nodes, where $n = c_1 + \ldots + c_k + 1$ defined by the cover relations 
\[x_1 \prec x_2 \ldots \prec x_{c_1+1} \succ x_{c_1 + 2} \succ \ldots \succ x_{c_1 + c_2 + 1} \prec x_{c_1 + c_2 + 2} \prec \ldots.\]
\end{defn}
For an example, see \cref{fig:2112}. These posets arise in cluster algebras, quiver representation theory and combinatorics. They also appeared in recent work of Morier-Genoud and Ovsienko \cite{morier2020continued}, where they introduced a $q-$deformation of the rational numbers. In this same paper, the authors conjectured the following, which was proved in \cite{main}.
\begin{thm}\label{thm:1}
The rank polynomials of fence posets are unimodal. 
\end{thm}
The main step in the proof of this theorem involved the introduction of the ancillary class of \emph{circular fence posets}, and an unexpected property of these posets. 
\begin{defn}
Given an even length composition $\bar{c} = (c_1, \ldots, c_{2s})$, the circular fence poset $\bar{\mathcal{F}}(\bar{c})$ is the poset on $n$ nodes where $n = c_1 + \ldots + c_{2s}$, defined by the cover relations 
\begin{multline*}
x_1 \prec  \ldots \prec x_{c_1+1} \succ x_{c_1 + 2} \succ \ldots \\ \succ x_{c_1 + c_2 + 1} \prec x_{c_1 + c_2 + 2} \prec \ldots \prec x_{1+\sum_{1}^{2s-1} c_i}  \succ  \ldots \succ x_{\sum_{1}^{2s} c_i} \succ x_1.
\end{multline*}
In other words, this is what we get by identifying the two end points of a regular fence poset. 
\end{defn}
\begin{figure}[ht]
\begin{tikzpicture}[scale=.7]
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\fill(6,0) circle(.1);
\draw (0,0)--(2,2)--(3,1)--(4,2)--(6,0);
\draw (0,-.5) node{$x_1$};
\draw (1,0.5) node{$x_2$};
\draw (2,1.5) node{$x_3$};
\draw (3,0.5) node{$x_4$};
\draw (4,1.5) node{$x_5$};
\draw (5,0.5) node{$x_6$};
\draw (6,-0.5) node{$x_7$};
\end{tikzpicture}\qquad 
\begin{tikzpicture}[scale=.7]
\fill[pea](-.2,0)--(0,-.2)--(.2,0)--(0,.2)--(-.2,0);
\fill[pea](1,1) circle(.1);
\fill(2,2) circle(.1);
\fill(3,1) circle(.1);
\fill(4,2) circle(.1);
\fill(5,1) circle(.1);
\fill[pea] (5.8,0)--(6,-.2)--(6.2,0)--(6,.2)--(5.8,0);
\draw (0,0)--(2,2)--(3,1)--(4,2)--(6,0);
\draw (0,-.5) node{$x_1$};
\draw (1,0.5) node{$x_2$};
\draw (2,1.5) node{$x_3$};
\draw (3,0.5) node{$x_4$};
\draw (4,1.5) node{$x_5$};
\draw (5,0.5) node{$x_6$};
\draw (6,-0.5) node{$x_1$};
\end{tikzpicture} 
\quad 
\begin{tikzpicture}[scale=.7]
\fill(1,1) circle(.1);
\fill(2,2) circle(.1);
\fill(3,1) circle(.1);
\fill(4,2) circle(.1);
\fill(5,1) circle(.1);
\fill(3,0) circle(.1);

\draw (3, 0)--(1, 1)--(2,2)--(3,1)--(4,2)--(5, 1)--(3, 0);
\draw (0.7,0.5) node{$x_2$};
\draw (2,2.5) node{$x_3$};
\draw (3,1.5) node{$x_4$};
\draw (4,2.5) node{$x_5$};
\draw (5.3,0.5) node{$x_6$};
\draw (3,-0.5) node{$x_1$};

\end{tikzpicture} 
\caption{The fence poset $F(2,1,1,2)$ (left) and two depictions of the circular fence poset $\bar{F}(2, 1, 1, 2)$(center, right). In the middle one, the nodes marked $x_1$ are identified.}\label{fig:2112}
\end{figure}
In \cite{main}, the authors showed that circular fence posets satisfy an unexpected property. 
\begin{thm}[\cite{main}]\label{thm:sym}
Rank polynomials of circular fence posets are symmetric. 
\end{thm}

Let us make a comment on why this result is unexpected. Given a composition $\bar{c} = (c_1, \ldots, c_{2s})$, let $\bar{c}_{\operatorname{shift}}$ be the composition that is the cyclical shift of $\bar{c}$, that is $\bar{c}_{\operatorname{shift}} = (c_{2s}, c_1, c_2, \ldots, c_{2s-1})$. A calculation shows that the symmetry of the rank polynomial of $\bar{F}(\bar{c})$ is equivalent to the statement that the posets $\bar{F}(\bar{c})$ and $\bar{F}(\bar{c}_{\operatorname{shift}})$ have the same rank polynomial. It is also possible to see that this same rank symmetry may also be expressed as saying that the poset of lower ideals (our $\bar{F}(\bar{a})$) and the poset of upper ideals of the same fence poset have the same rank polynomial. However, except in very special cases, the two posets \emph{are not isomorphic}. For instance, take $\bar{c} = (2, 1, 1, 2)$ as in \cref{fig:2112} and compare the Hasse diagrams of $\bar{F}(\bar{c})$ and $\bar{F}(\bar{c}_{\operatorname{shift}})$. The second fence can be seen as the vertical reflection of the first and we have labeled the elements appropriately. 

\begin{table}
\begin{tabular}{|c|c|c|}
\hline
Composition&Fence Poset & Hasse Diagram of Lattice of Lower Ideals\\
%\begin{figure}[ht]
\hline 
\raisebox{1.4cm}{\begin{tabular}{c} $\bar{c}$\\ \\$(2, 1, 1, 2)$\end{tabular}} & \raisebox{.5cm}{\begin{tikzpicture}[scale=.7]
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\draw (0,-.5) node{$1$};
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\fill[pea](-.2,0)--(0,-.2)--(.2,0)--(0,.2)--(-.2,0);
\end{tikzpicture} }&
\begin{tikzpicture}[scale=1.1]
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\end{tikzpicture}\\
\hline
\raisebox{1.4cm}{
\begin{tabular}{c} $\bar{c}_{\operatorname{shift}}$\\ \\$(2, 2, 1, 1)$\end{tabular} }
 & \raisebox{.5cm}{
 \begin{tikzpicture}[scale=0.7]
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\draw (2,0)--(3,0);
\draw (2,-1)--(3,0);
\draw (2,-1)--(3,-1);
\draw (3,1)--(4,1);
\draw (3,0)--(4,0);
\draw (3,-1)--(4,-1);

\draw (3,-1)--(4,0);
\draw (3,1)--(4,-1);


\draw (4,1)--(5,1);
\draw (4,0)--(5,1);
\draw (4,-1)--(5,-1);
\draw (4,-1)--(5,1);


\draw (5, 1)--(6, 0);
\draw (5, -1)--(6, 0);


\end{tikzpicture}\\
\hline
\end{tabular}
\caption{Example showing that lattices of upper and lower ideals can be non-isomorphic (the inclusion in the Hasse diagrams given is in the direction left to right)}
\end{table}
%\caption{The fence poset $F(2,1,1,2)$ (left) and two equivalent depictions of the circular fence poset $\bar{F}(2, 1, 1, 2)$(center and right).}\label{fig:2112}
%\end{figure}

A second, this time bijective, proof of \cref{thm:sym} was given by Elizalde and Sagan in \cite{es22}. Both proofs of this result are highly intricate and it is natural to seek a transparent proof of this basic result. We present such a proof in this paper, see \cref{cor:main}.
%\ezgi{Again, too informal. Let us actually cite, and not say "we". Yalım is also in this paper, and any statement should include him.}

As mentioned above, in \cite{main},  the symmetry of the rank polynomials of circular fence posets was used to prove \cref{thm:1}, that rank polynomials of (regular) fence posets are unimodal. Generically, rank polynomials of circular fence posets seemed to be unimodal as well, though there are certain exceptions; a calculation shows that 
\[\overline{R}((1, 1, 1, 1);q) = 1 + 2q + q^2 + 2q^3 + q^4.\]
Extensive computer calculations however suggested the following conjecture. 
\begin{conj}[\cite{main}] \label{conj:unimodality}
The rank polynomial $\overline{R}(\bar{a};q)$ of a circular fence poset $\overline{F}(\bar{c})$ is unimodal except when $\bar{c} = (a, 1, a, 1)$ or $(1, a, 1, a)$ for some positive integer $a$. 
\end{conj}
In this same paper, the authors showed that if $\bar{R}(\bar{a})$ is not unimodal, then the composition $\bar{a}$ may not have any two adjacent entries larger than $1$. For the purposes of this paper, we will focus on the more concrete case, where our compositions have form:
\begin{align}\label{comp2}
\bar{a} = (a_1, 1, a_2, 1, \ldots, a_s, 1).
\end{align}
Such compositions play an important role in the bijective proof of symmetry of Elizalde and Sagan in \cite{es22}, where the authors refer to such circular fence posets as \definition{gate posets}. 

%It turns out that ideals of circular fence posets coming from gate posets are precisely the lattice points in the chainlink polytope $\CL(\bar{a}, 1)$, where $\bar{a} = (a_1, a_2, \ldots, a_s)$. Further, the number of rank $k$ ideals is precisely the number of lattice points in $\CL^k(\bar{a}, 1)$. The symmetry of the rank polynomials of gates may then be written as 
 %\[\#\CL^k(\bar{a}, 1) = \#\CL^{n-k}(\bar{a},1),\]
 %where $n = a_1 + a_2 + \ldots + a_s + s$ and $\#P$ for a polytope $P$ is the number of lattice points contained in $P$. We will see in \cref{prop:vertices} that rank the polytopes $\CL^k(\bar{a}, 1)$ while not necessarily integral, are always half integral, in particular, rational. 
 
 %We were naturally led to investigate whether the syntactic generalization 
 
 It turns out that the number of rank $k$ ideals of gate posets coming from a composition $(a_1, 1, \ldots, a_s, 1)$ equal the number of lattice points in $\CL^k(\bar{a}, 1)$ where $\bar{a} = (a_1, a_2, \ldots, a_s)$. We were naturally led to investigate whether the generalization \begin{align}\label{stmt:Ehr}\operatorname{Ehr}\, \CL^k(\bar{a}, 1) = \operatorname{Ehr}\, \CL^{n-k}(\bar{a},1),
 \end{align}
 where $n = a_1 + \ldots + a_s$ is true as well. Well, it is! And this is the content of the main theorem in this paper. The equality of Ehrhart quasipolynomials yields as a corollary the equality of volumes of these polytopes, see \cref{thm:main}. 
\begin{rmk}
Given a composition $\bar{c} = (a_1, 1, \ldots, a_s, 1)$ of $n$, it is also true that the order polytope   $\mathcal{O}(\bar{F}(\bar{c}))$ \cite{stanley1986two}  of the composition is such that for each integer,
\[\#\mathcal{O}^k(\bar{F}(\bar{c})) = \#\mathcal{O}^{n-k}(\bar{F}(\bar{c})),\]
where $\mathcal{O}^k(\bar{F}(\bar{c})) = \mathcal{O}^k(\bar{F}(\bar{c})) \cap \{x_1 + \ldots + x_k = n\}$. However the Ehrhart polynomials of these polytopes need not be equal, see \cite{korarxiv}.
\end{rmk}
Proving \cref{thm:main} needed several new ideas. Denote the Ehrhart polynomial $\operatorname{Ehr}\, \CL^k(\bar{a}, 1)$ by $f_k$. This polynomial evaluated at $1$ counts the number of ideals of size $k$ in the associated circular fence poset, that corresponding to $\bar{c} =(a_1, 1, \ldots, a_s, 1)$. When evaluated at other integers, say $f_k(m)$, we will show that the value can again be interpreted as the number of lower ideals of size $mk$ in a certain poset, which we call a \definition{chainlink poset}.  These posets share a familial resemblance to circular fence posets; they are introduced in \cref{sec:2} (see \cref{fig:posetexample} for an example). 

We will show that all chainlink posets have symmetric rank polynomials. The strate-
gies for showing symmetry for circular fence posets in \cite{main, es22} do not carry over and we needed to approach the problem differently. The new ingredient is a linear algebraic approach coming from the theory of oriented posets (see \cite{ops}) that we believe yields a transparent proof. We note that this yields a new (third) proof of rank symmetry for circular fences as well (See \cref{cor:main}). %This linear algebraic formulation was also used by the first author in \cite{ops} to resolve a conjecture of Morier-Genoud and Leclerc from \cite{leclere2021q} on traces of matrices in a $q$ deformation of the modular group. 

This approach has as its starting point the following basic feature of fence posets: They can be built up by gluing chains in an iterative manner. We review in \cref{MatFor} how the rank polynomials of fence (and chainlink) posets can be computed by multiplying certain $2 \times 2$ matrices: The entries of these matrices are certain polynomials that  encode refined order relations. 

This approach has another felicitous consequence : We discovered new recurrences, that lead to a proof of  \cref{conj:unimodality}. We include a proof of this in \cref{Sec:Unimodality}. 

\section{Chainlink Polytopes and Chainlink Posets}\label{sec:2}
We note here a basic integrality property of chainlink polytopes. 
\begin{prop}\label{prop:vertices}
 Let $\bar{a} = (a_1, \ldots, a_s)$ be a composition of $n$, let $l, t \in \mathbb{Z}_{\geq 0}$. Then 
 \begin{itemize}
 \item The vertices of $\CL(\bar{a}, l)$ are integral.
\item Assume that $2l \leq \operatorname{min} \bar{a}$. Then the vertices of $\CL^t(\bar{a}, l)$ are either integral or half integral. 
 \end{itemize}
\end{prop}

%Some remarks are in order. 
%\begin{enumerate}
%\item Half-integral vertices can occur. The polytope $\CL^2((2, 2), 1)$, is given by 
%\[0 \leq x_1 \leq 2, \quad 0 \leq x_2 \leq 2, \quad x_1 - x_2 \leq  1, \quad x_1 + x_2 = 2,\]
% has dimension $1$ and has as its' two vertices $(3/2, 1/2)$ and $(1/2, 3/2)$.
%\item The condition on $l$ is necessary. Take for instance the polytope $\CL^5((3, 3, 3), 2)$. One may check that the point $(8/3, 5/3, 2/3)$ is a vertex of the polytope. 
%\item It is possible to show that if the vertices of $\CL^t(\bar{a}, l)$ are all integral or half integral for every integer $l$, then we must have that $2l \leq \operatorname{min} \{a_i\}_{i \in [s]}$. 
%\end{enumerate}

Let $\HL(\bar{a},l)$ be a chainlink polytope with  $2l\leq \mathrm{min}_i(a_i)$. Consider the integer points that lie inside the polytope.  When $l=1$, these points correspond to ideals of the circular fence poset $F(a_1-1,1,a_2-1,1,\ldots,a_s-1,1)$, where the rank of the ideal corresponds to the sum of the coordinates of the point. For general $l$, the integer points can be interpreted as ideals of a poset $F_l(\bar{a})$ formed by adding extra edges to the Hasse diagram of $F(a_1-1,1,a_2-1,1,\ldots,a_s-1,1)$ as shown in \cref{fig:posetexample}. More precisely, we can define chainlink posets as follows:
\begin{defn} Let $\bar{a}$  be a composition, with $l(\bar{a})$ parts and $l$ be a positive integer satisfying $2l\leq \operatorname{min}_i{a_i}$ as in the case for chainlink polytopes. The \definition{chainlink poset} $\CLP(\bar{a},l)$ is given by points $x_{i,j}$ for $1\leq i \leq \ell(\bar{a})$ and $0\leq j\leq a_i$  with the generating relations $x_{i,0}\prec x_{i,1} \prec \cdots \prec x_{i,a_i}$ and $x_{i,a_i-l+1+k}\succ x_{i+1,k}$ for $0 \leq k < l$ and for each $i$, where $i+1$ is calculated cyclically.
\end{defn}  

\begin{figure}
    \centering
    \begin{tikzpicture}[scale=.4]\setlength{\tabcolsep}{0pt}
		\node  (1) at (-16.5, 1) {};
		\node  (2) at (-16, 2) {};
		\node  (3) at (-17, 0) {};
		\node  (4) at (-15.5, 3) {};
		\node  (5) at (-15, 4) {};
		\node  (6) at (-14.5, 5) {};
		\node  (7) at (-13, 1) {};
		\node  (8) at (-12.5, 2) {};
		\node  (9) at (-12, 3) {};
		\node  (10) at (-11.5, 4) {};
		\node  (11) at (-10, 0) {};
		\node  (12) at (-9.5, 1) {};
		\node  (13) at (-9, 2) {};
		\node  (14) at (-8.5, 3) {};
		\node  (15) at (-8, 4) {};
		 \draw[->,red,dashed,ultra thick] (15.center) to (-7,3);
		\draw[>-,red,dashed,ultra thick] (-17.5,2)--(-16.5,1);
		\draw[->,blue,dotted, ultra thick] (14.center) to (-7.5,2);
		\draw[>-,blue,dotted,ultra thick] (-18,1)--(-17,0);
		\fill(-16.5,1) circle(.2); 
	    \fill(-16,2) circle(.2); 
		\fill(-17,0) circle(.2); 
		\fill(-15.5,3) circle(.2); 
	    \fill(-15,4) circle(.2); 
		\fill(-14.5,5) circle(.2); 
		\fill(-13,1) circle(.2); 
	    \fill(-12.5,2) circle(.2); 
		\fill(-12,3) circle(.2); 
		\fill(-11.5,4) circle(.2); 
	    \fill(-10,0) circle(.2); 
		\fill(-8,4) circle(.2); 
		\fill(-9.5,1) circle(.2); 
	    \fill(-9,2) circle(.2); 
		\fill(-8.5,3) circle(.2); 
		\draw (3.center) to (6.center);
		\draw (6.center) to (8.center);
		\draw (5.center) to (7.center);
		\draw (7.center) to (10.center);
		\draw (10.center) to (12.center);
		\draw (9.center) to (11.center);
		\draw (11.center) to (15.center);
    \end{tikzpicture}
    \caption{The chainlink poset with $\bar{a}=(6,4,5)$ and $l=2$, the two top right nodes are connected to the two bottom left nodes.}
    \label{fig:posetexample}
\end{figure}

Recall that we use $\HL^t(\bar{a},l)$ to denote the slice of the polytope with respect to the hyperplane $x_1 + \ldots + x_s = t$. Note that this slice can be non-empty only when $t\in[0,n]$. Furthermore, the number of integer points in $\HL^t(\bar{a},l)$ is given by the coefficient of $q^t$ in the rank polynomial of $\CLP(\bar{a},l)$.% The fact that 
%\[\# \HL^t(\bar{a},1) = \# \HL^{n-t}(\bar{a},1)\]
% is equivalent to the symmetry of the rank polynomial of  $\CLP(a,1)$ which was recently proved using an inductive argument by Kantarcı Oğuz and Ravichandran and then bijectively by Elizalde and Sagan. 

The connection between integer points of the polytope and rank polynomial of the corresponding poset still holds if we scale the polytope by a number $k$. This allows us to describe the coefficients of the Ehrhart quasi polynomial of slices of the chainlink polytope in terms of coefficients of rank polynomials of some chainlink posets. That means a general statement about the symmetry of rank polynomials of all chainlink posets can be used to prove the main theorem stated in the introduction, which is precisely what we do in the next few sections.

\subsection{Matrix Formulation}\label{MatFor}


An oriented poset $ {P\!\nearrow}=(P,x_L,x_R)$ consists of a poset $P$ with two specialized vertices $x_L$ and $x_R$ which can be thought as the target (left) vertex $\target\,$  and the source (right) vertex ${\color{blue}\rightarrow}$. One can think of an oriented poset as a poset with an upwards arrow coming out of the source vertex $x_R$. One can combine oriented posets by linking the arrow of one poset with target of another via $x_R \preceq y_L$ ($x_R\nearrow y_L$) to get $ ({P}\!\nearrow {Q})\!\nearrow$\footnote{Linking via  $x_R \succeq y_L$ is also an option, see \cite{ops}.  }. 

The effect of this operation on the rank polynomial can be calculated easily by $2\times2$ matrices. A \definition{rank matrix} of an oriented poset ${P\!\nearrow}$ is defined as follows:

    $$	\displaystyle \drm( {P}\!\nearrow):=\begin{bmatrix} \rank( {P};w)|_{x_R\in I} & \rank( {P};w)|_{x_R\notin I}\\ \rank( {P};w)|_{\substack{x_R\in I\\x_L \notin I}} & \rank( {P};w)|_{\substack{x_R\notin I\\x_L \notin I}}
			\end{bmatrix}$$

The entries are partial rank polynomials, which correspond to restricting to ideals of the poset $P$ that satisfy the given constraints. We also use the notation $\close( {P}\searrow)$ (resp. $\close( {P}\nearrow)$ to denote the structure obtained by adding the relation $x_R \succeq x_L$ (resp. $x_R \preceq x_L$). On the rank matrix level, this corresponds to taking the trace. See \cref{table:tableofmoves} for precise formulas and examples of these operations.  


\begin{table}
    \centering
    \begin{tabular}[t]{|c|c|c|}
\hline
         & Formula & Example \\
         \hline 

    $ {P}\!\nearrow {Q}$
        &\begin{tabular}{c}

 {\footnotesize$\drm(( {P}\nearrow {Q})\!\nearrow)=\drm( {P}\!\nearrow) \cdot \drm( {Q}\!\nearrow)$}
 \end{tabular}&\begin{tabular}{c}
 \begin{tikzpicture}[scale=.45]
\draw (0,0)--(1,1)--(3,-1);
\fill[white] (0,0) circle(.2) ;
\fill[red] (0,0) circle(.1) ;
\draw[red] (0,0) circle(.2);
\fill (1,1) circle(.1)  ;
\fill (2,0) circle(.1)  ;
\fill[blue] (3,-1) circle(.15)  ;
\draw[->, blue,dotted, thick] (3,-1)--(3.5,-.5);
\node at (1.5,-1.5) {$\scriptstyle{ {P}\!\nearrow}$\quad,};
\end{tikzpicture}\quad \begin{tikzpicture}[scale=.45]
\fill[white] (0,1) circle(.2) ;
\draw (0,-.5)--(1,.5);
\fill[white] (0,-.5) circle(.2) ;
\fill[red] (0,-.5) circle(.1) ;
\draw[red] (0,-.5) circle(.2);
\fill[blue] (1,.5) circle(.15)  ;
\draw[->, blue,dotted,thick] (1,.5)--(1.5,1);
\node at (.5,-1.5) {$\scriptstyle{ {Q}\!\nearrow}$};
\end{tikzpicture}$\,$ \raisebox{.6cm}{$\rightarrow$} $\,$\begin{tikzpicture}[scale=.45]
\draw (0,0)--(1,1)--(3,-1)--(5,1);
\fill[white] (0,0) circle(.2) ;
\fill[red] (0,0) circle(.1) ;
\draw[red] (0,0) circle(.2);
\fill (1,1) circle(.1)  ;
\fill(2,0) circle(.1);
\fill(3,-1) circle(.1);
\fill (4,0) circle(.1) ;
\fill[blue] (5,1) circle(.15)  ;
\draw[->, blue,dotted, thick] (5,1)--(5.5,1.5);
\node at (3,-1.5) {$\scriptstyle{( {P}\nearrow {Q})\!\nearrow}$};
\end{tikzpicture} \end{tabular}\\
\hline 
 $\close\,( {P}\!\nearrow)$ &$\rank( \close( {P}\nearrow);w)=\operatorname{tr}(\rmm({ {P}\!\nearrow}))$ &\begin{tabular}{c}
        \begin{tikzpicture}[scale=.45]
\draw (0,0)--(1,1)--(3,-1);
\fill[white] (0,0) circle(.2) ;
\fill[red] (0,0) circle(.1) ;
\draw[red] (0,0) circle(.2);
\fill (1,1) circle(.1)  ;
\fill (2,0) circle(.1)  ;
\fill[blue] (3,-1) circle(.15)  ;
\draw[->, blue,dotted, thick] (3,-1)--(3.5,-.5);
\node at (1.5,-1.5) {$\scriptstyle{ {P}\!\nearrow}$};
\end{tikzpicture} $\quad$ \raisebox{.6cm}{$\rightarrow$} $\quad$   \begin{tikzpicture}[scale=.45]
\draw (0,0)--(1,1)--(3,-1)--(0,0);
\fill (0,0) circle(.1) ;
\fill (1,1) circle(.1)  ;
\fill (2,0) circle(.1)  ;
\fill (3,-1) circle(.1);
\node at (1.5,-1.5) {$\scriptstyle{\close\,( {P}\!\nearrow)}$};
\end{tikzpicture}\end{tabular} \\
\hline
    \end{tabular}
    \caption{The moves are shown through examples}
    \label{table:tableofmoves}
\end{table}


In particular, consider when $P$ is formed of a single node equal to both $x_R$ and $x_L$. We call this oriented poset an \definition{up step} and denote the corresponding matrix by $U$.
    $$	\displaystyle U:=\drm(\bullet\!\nearrow):=\begin{bmatrix}q&1\\0&1
			\end{bmatrix}, \qquad \qquad \drm(C_k\!\nearrow)=U^{k-1}.$$


\begin{wrapfigure}{r}{6cm}
\centering
     \begin{tikzpicture}[scale=.6]
		\node  (2) at (-16, 2) {};
		\node  (4) at (-15.5, 3) {};
		\node  (5) at (-15, 4) {};
		\node  (6) at (-14.5, 5) {};
		\node  (7) at (-13, 1) {};
		\node  (8) at (-12.5, 2) {};
		\node  (9) at (-12, 3) {};
		\node  (10) at (-11.5, 4) {};
		\draw (2.center) to (6.center);
		\draw (4.center) to (8.center);
		\draw (2.center) to (7.center);
		\draw (7.center) to (10.center);
		\draw (10.center) to (6.center);
		\draw (9.center) to (5.center);
		\draw (-13,4.5) to (-14.5,1.5);
		\fill(-15.5,3) circle(.15); 
	    \fill(-15,4) circle(.15); 
		\fill(-14.5,5) circle(.15); 
		\fill(-13,1) circle(.15); 
	    \fill(-12.5,2) circle(.15); 
		\fill(-12,3) circle(.15); 
		\fill[blue](-11.5,4) circle(.22); 
	    \fill(-14.5,1.5) circle(.15); 
		\fill(-14,2.5) circle(.15); 
		\fill(-13.5,3.5) circle(.15); 
	    \fill(-13,4.5) circle(.15); 
	\fill[white] (-16,2) circle(.3) ;
\fill[red] (-16,2) circle(.15) ;
\draw[red] (-16,2) circle(.3);
\draw[->, blue, thick] (-11.5, 4)--(-11, 5);
    \end{tikzpicture}
    \caption{$B_{3\times4}\!\nearrow$}
        \label{fig:box}

    \end{wrapfigure}

The second part comes from the fact that combining $k+1$ such posets gives us a chain $C_k$ of length $k$ with $x_L$ corresponding to the minimal element, and $x_R$ to the maximal. 


    
Let $ {B_{a\times b}}\!\nearrow$ denote the $ab$-element oriented box poset given by the direct product of two chains $C_{a-1}$ and $C_{b-1}$ with the left vertex given by $(a-1,0)$ and the right vertex is given by $(0,b-1)$. The directed poset $B_{3\times4}\!\nearrow$ is shown in \cref{fig:box}. 
  %  \caption{The directed poset $B_{3\times4}\!\nearrow$}
   % \label{fig:my_label}
%\end{figure}
\begin{comment}
\begin{tabular}{c c}
\begin{tabular}{p{110mm}}
Let $ {B_{a\times b}}\!\nearrow$ denote the $ab$-element oriented box poset given by the direct product of two chains $C_{a-1}$ and $C_{b-1}$ with the left vertex given by $(a-1,0)$ and the right vertex is given by $(0,b-1)$ ($B_{3\times4}\!\nearrow$ is shown on the right). 
\end{tabular} &
\begin{tabular}{c}

      \rotatebox{-30}{
    \begin{tikzpicture}[scale=.6]
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		\node  (4) at (-15.5, 3) {};
		\node  (5) at (-15, 4) {};
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		\node  (9) at (-12, 3) {};
		\node  (10) at (-11.5, 4) {};
		\draw (2.center) to (6.center);
		\draw (4.center) to (8.center);
		\draw (2.center) to (7.center);
		\draw (7.center) to (10.center);
		\draw (10.center) to (6.center);
		\draw (9.center) to (5.center);
		\draw (-13,4.5) to (-14.5,1.5);
		\fill(-15.5,3) circle(.15); 
	    \fill(-15,4) circle(.15); 
		\fill(-14.5,5) circle(.15); 
		\fill(-13,1) circle(.15); 
	    \fill(-12.5,2) circle(.15); 
		\fill(-12,3) circle(.15); 
		\fill[blue](-11.5,4) circle(.22); 
	    \fill(-14.5,1.5) circle(.15); 
		\fill(-14,2.5) circle(.15); 
		\fill(-13.5,3.5) circle(.15); 
	    \fill(-13,4.5) circle(.15); 
	\fill[white] (-16,2) circle(.3) ;
\fill[red] (-16,2) circle(.15) ;
\draw[red] (-16,2) circle(.3);
\draw[->, blue, thick] (-11.5, 4)--(-11, 5);
    \end{tikzpicture}}
\end{tabular}    
\end{tabular}
\end{comment}

The rank matrix of an oriented box poset is given as follows:
    $$	\displaystyle \drm(B_{a\times b}\!\nearrow)=\begin{bmatrix} q^b\qbinom{a+b-1}{b}_q & \qbinom{a+b-1}{b-1}_q  \\  q^b\qbinom{a+b-2}{b}_q  & \qbinom{a+b-2}{b-1}_q \end{bmatrix} = \begin{bmatrix} \qbinom{a+b}{b}_q -\qbinom{a+b-1}{b-1}_q & \qbinom{a+b}{b}_q- q^b\qbinom{a+b-1}{b}_q  \\  q^b\qbinom{a+b-2}{b}_q  & \qbinom{a+b-2}{b-1}_q 
			\end{bmatrix}$$

Next we will see that any given chainlink poset can be realized by combining copies of the up step and the box poset $B_{2\times l}\!\nearrow$.

\begin{prop}  Consider the chainlink poset $\CLP(\bar{a},l)$ with $2l\leq \operatorname{min}_i(a_i)$. Let $\bar{d}$ be the weak composition formed by taking $d_i=a_i-2l$.  The rank polynomial of $\CLP(\bar{a},l)$ is given by:

\begin{equation} \label{eq:matrixconnection}
    \rank (\CLP(\bar{a},l);q)= \operatorname{tr}(U^{d_1}\cdot B \cdot U^{d_2} \cdot B \cdot \,\cdots\,\cdot U^{d_1} \cdot B)
\end{equation} where $B$ denotes the rank matrix $\drm(B_{2\times l}\!\nearrow)$.

\end{prop}



\begin{example} The chainlink poset given in \cref{fig:posetexample} with $\bar{a}=(6,4,5)$ and $l=2$ can be formed by combining $2\times 2$ boxes with up steps and then taking the closure: $\close(B_{2\times 2}\!\nearrow \cdot \bullet\!\nearrow \cdot  \bullet\!\nearrow \cdot B_{2\times 2}\! \nearrow \cdot B_{2\times 2}\!\nearrow \cdot \! \bullet\!\nearrow)$. The corresponding rank polynomial is given by:

\begin{eqnarray*}
  &&\operatorname{tr} \left( \begin{bmatrix} q^2 [3]_q & [3]_q\\ q^2 & [2]_q\end{bmatrix} \cdot \begin{bmatrix} q&1\\0&1 \end{bmatrix} \cdot \begin{bmatrix} q&1\\0&1 \end{bmatrix} \cdot \begin{bmatrix} q^2 [3]_q & [3]_q\\ q^2 & [2]_q\end{bmatrix} \cdot \begin{bmatrix} q^2 [3]_q & [3]_q\\ q^2 & [2]_q\end{bmatrix} \cdot \begin{bmatrix} q&1\\0&1 \end{bmatrix} \right)\\
  && =1+3q+6q^2+9q^3+12q^4+14q^5+16q^6+17q^7\\
  && \hspace{4cm} +17q^8+16q^9+14q^{10}+12q^{11}+9q^{12} +6q^{13}+3q^{14}+q^{15}.
\end{eqnarray*}

\end{example}
Note that the rank polynomial given in this instance is symmetric. Next, we will show that this is always the case.

\section{Recurrence Relations and Rank Symmetry}\label{sec:sym}

One advantage of building posets via matrices is that the characteristic equations of matrices give us recurrence relations for the rank polynomial. For example, consider the characteristic polynomial of $U$. Plugging $U$ in the place of $x$ gives us
\begin{equation*}
 U^2 =(q+1) U + q,
\end{equation*}
 Note that the coefficient of $U$ is symmetric around $q^{1/2}$ and $q$ is trivially symmetric around $q$.


\begin{lemma}\label{lem:characteristic}
Let $B= \drm(B_{a\times b}\!\nearrow)$ for some fixed $a,b$. The characteristic polynomials of the matrices $B$ and $B\cdot U$ have coefficients that are symmetric polynomials in $q$. In particular, the trace and determinant of $B$ and $B\cdot U$ are symmetric about $ab/2$, $ab$, $(ab+1)/2$ and $ab+1$ respectively.
\end{lemma}

\begin{comment}
\begin{lemma} \label{lem:charpol} Let $X$ be a matrix whose trace is given by a symmetric polynomial on $q$ with center $C$.
Then we have the following:
\begin{enumerate}
    \item If $\mathrm{tr}(BX)$ is symmetric around $C+(ab)/2$, then for all $k$, $\mathrm{tr}(B^k X)$ is symmetric with center $C+k(ab)/2$.
    \item If $\mathrm{tr}(U X)$ is symmetric around $C+1/2$, then for all $k$, $\mathrm{tr}(U ^kX)$ is symmetric with center $C+k/2$.
    \item If $\mathrm{tr}(B U X)$ is symmetric around $C+(ab+1)/2$, then for all $k$, $\mathrm{tr}((B U)^kX)$ is symmetric with center $C+k(ab+1)/2$.
\end{enumerate}
\end{lemma}\end{comment}


\begin{thm}\label{prop:symmetry} Let $(d_1,d_2,\ldots,d_s)$ be a weak composition and 
$B_{a\times b}\!\nearrow$ be an oriented box poset where $B$ denotes the rank matrix $\drm(B_{a\times b}\!\nearrow)$. Then the following polynomial is symmetric.
\begin{equation*}
    \operatorname{tr}(U^{d_1} \cdot B \cdot U^{d_2} \cdot B \cdots U^{d_1} \cdot B).
\end{equation*}
\end{thm}


Note that when $l=1$, we recover the rank symmetry of gate posets. 

\begin{corollary} \label{cor:main} The rank polynomial of any chainlink poset is symmetric.
\end{corollary}

We use this machinery to prove our main theorem. 
\begin{thm}\label{thm:main} Let $\bar{a}$ be a composition of $n$, let $l$ be a positive integer such that $2l \leq \operatorname{min} \{a_i\}_{i \in [s]}$ and let $t$ be a non-negative integer. Then complementary sections of the chainlink polytope have the same Ehrhart quasi-polynomial,
 \[\operatorname{Ehr}\, \HL^t(\bar{a},l) = \operatorname{Ehr}\,\HL^{n-t}(\bar{a},l).\]
 \end{thm}
 Using a scaling argument, we can deduce the following. 
 \begin{corollary}
     Let $a_i$ be positive real numbers such that $a_1 + \ldots + a_s = n$, let $l$ be a positive real number such that $2l \leq \operatorname{min} \{a_i\}_{i \in [s]}$ and let $t$ be a non-negative real number. Then complementary sections of the chainlink polytope have the same volume,
 \[|\HL^t(\bar{a},l)| = |\HL^{n-t}(\bar{a},l)|.\]
 \end{corollary}

 %The most natural polytope such that the lattice points in any $k$ section equals the rank $k$ ideals in a gate poset $\mathcal{P}$, is the order polytope $\mathcal{O}(\mathcal{P})$, see \cite{stanley1986two}. However, order polytopes need not obey rank symmetry, see \cite{korarxiv}. 
 \begin{comment}
 \begin{rmk}\label{rem:order}
 Consider the gate poset for the composition $\bar{c} = (3,1,2,1,1,1)$, let $P = \mathcal{O}(\bar{\mathcal{F}}(\bar{c}))$ be the order polytope and let $P^k = P \cap \{x_1 + \ldots + x_n = k\}$. Then the  lattice points in the second dilates of $P^{4}$ and the complementary section $P^{5}$ number respectively $84$ and $83$, i.e. we do not have equality of Ehrhart polynomials. This indicates the subtlety involved in this problem. 
 \end{rmk}
 \end{comment}

\section{Unimodality}\label{Sec:Unimodality}


The recurrence relations deduced from characteristic matrices have other applications as well. In this subsection, we prove the following result.

\begin{thm}\label{unimodality} Rank polynomials of circular fence posets $\bar{F}(\bar{c})$ are unimodal except when $\bar{c} = (a, 1, a, 1)$ or $(1, a, 1, a)$ for some positive integer $a$. 
\end{thm}

We define the matrix for a \definition{down step} denoted by $D$ as follows:

    $$	\displaystyle D:=\begin{bmatrix}1+q&-q\\1&0
			\end{bmatrix}.$$

The following lemma is an easy consequence of the work in \cite{ops}. The interested reader is referred there to learn about how down steps fit into the framework of oriented posets.

\begin{lemma} Let $DC_n$ denote a decreasing chain, an $n$-element chain poset oriented by taking the maximal vertex as the target and the minimal vertex as the source. Then we have,
$$ \drm(DC_n\!\nearrow)=D^{n-1}\cdot U.$$
\end{lemma}

That means the above theorem may be restated as follows:
\begin{reptheorem}{unimodality}
For any composition $\bar{a}$ with an even number of parts the following polynomial is unimodal except when $\bar{a} = (a, 1, a, 1)$ or $(1, a, 1, a)$ for some positive integer $a$:
$$\overline{R}(\bar{a},q)=\operatorname{tr}(D^{a_1}U^{a_2}D^{a_3}U^{a_4}\cdots D^{a_{s-1}}U^{a_s})=\operatorname{tr}(U^{a_1}D^{a_2}U^{a_3}D^{a_4}\cdots U^{a_{s-1}}D^{a_s}).$$
\end{reptheorem}

We prove this theorem by  using recurrence identities similar to those described in the previous section. A particular one we use is the following matrix identity:
\begin{equation} \tag{Id 1}
    DUD = DU + UD - U + D^3 - D^2 .
\end{equation}

On the rank polynomial level, (Id 1) translates to:
\begin{align*} \tag{Id 1$'$}
\overline{R}((a, 1, b, X);q)=& \overline{R}((a-1, 1, b, X);q) + \overline{R}((a, 1, b-1, X);q) \\ -&  \overline{R}((a-1, 1, b-1, X);q) +  \overline{R}((a+b+1,X);q) \\-& \overline{R}((a+b, X);q).
\end{align*}

That allows us to prove the theorem inductively, using the following  to reduce the problem to compositions formed of $2$'s and $1$'s, and resolve those case by case.

\begin{prop}\label{prop:simplify}
For an odd-length sequence $X=(x_1,x_2,\ldots,x_k)$ of positive integers suppose that  $a,b \geq 1$ and $\bar{R}(a-1, 1, b-1, X)$ is unimodal. If $a>1$ or $\ell(X)>1$ with $x_1>1$ or  $b\geq x_2$  then $\bar{R}(a, 1, b, X)$ is also unimodal.
\end{prop}


\section{Properties of Chainlink polytopes}

In this section, we investigate properties of our chainlink polytopes.
\begin{lemma}
Let $\bar{a} \in \mathbb{N}^s$ be a composition of $n$ and let $l \in \mathbb{R}$. The chainlink polytope $\CL(\bar{a}, l)$ is full dimensional when $l < \operatorname{min} (\bar{a})$.
\end{lemma}

Determining exactly when these polytopes are non-empty is a tricky problem and does not seem to have a nice solution. We note though that a routine application of LP duality shows that the condition $l \leq (a_1 + \ldots + a_s)/s$ is necessary. 
\begin{lemma}
 Let $\bar{a} \in \mathbb{R}_{>0}^s$ and $ l \in \mathbb{R}_{\geq 0}$. Suppose that $ 0<l <\min_{i \in [s]} a_i$. Then the polytope $\CL ( \bar{a},l)$ has exactly $3s$ facets, defined by the equalities $ x_i =0$, $ x_i = a_i$ and $ x_i -x_{i+1} = a_i -l$. 
\end{lemma}
\begin{comment}
\begin{lemma}
\label{vertexlemma} Let $\bar{a} \in \mathbb{R}_{>0}^s$ and $ l \in \mathbb{R}_{\geq 0}$. Suppose that $ 2l \leq \min_{i \in [s]} a_i$. 
The vertices $v$  of $ \CL ( \bar{a} , l )$ have the form $v_i \in \lbrace 0, l , a_i -l, a_i \rbrace $. Moreover each edge must be parallel either to the standard base vectors $e_i$ or to $e_{i}+e_{i+1} $ for some $i \in [s].$
\end{lemma}
\end{comment}
\begin{prop}
If we have the strict inequality $ 2l < \min_{i \in [s]} a_i$, then the polytope
$ \CL ( \bar{a}, l )$ is simple and the combinatorial structure is independent of $\bar{a}$ and $l$.
\end{prop}
\begin{prop}
Let $\bar{a} \in \mathbb{R}_{>0}^s$ and $ l \in \mathbb{R}_{\geq 0}$. Suppose that $ 2l \leq \min_{i \in [s]} a_i$. The number of vertices of the chainlink polytope $ \CL ( \bar{a}, l) $ is given by 
\begin{align*}
   \textbf{Vert} ( \CL ( \bar{a}, l))= \operatorname{tr} ( A_1 \cdots A_s)
\end{align*}
where each $ A_i= A$ if $ a_i > 2l$ or $A_i =B$ if $ a_i = 2l$, where 
\begin{align*}
    A= \begin{bmatrix} 1 & 1 &1 \\
    1 & 0 & 0 \\
    1 & 1 & 1
    \end{bmatrix}, \quad  B = \begin{bmatrix} 1 & 1 &1 \\
    1 & 0 & 0 \\
    1 & 0 & 1
    \end{bmatrix} .
\end{align*}
\end{prop}


\begin{corollary}\label{cor:vertices}
In particular if $ 2l < \min_{ i \in [s] } a_i$, then 
\begin{align*}
    \textbf{Vert} ( \CL ( \bar{a}, l)) = \operatorname{tr} (A^s)
\end{align*}
which satisfies the linear recurrence $\operatorname{tr} (A^s)= 2 \operatorname{tr} (A^{s-1}) + \operatorname{tr} (A^{s-2})$
for $ s \geq 3$. The number of vertices is the "Companion Pell Numbers" in OEIS A002203. %$ \operatorname{tr} (A) =2$ and $ \operatorname{tr} (A^2) =6$. These are dubbed the "Companion Pell Numbers" in A002203.
\end{corollary}

\begin{comment}
\begin{corollary}
If $ a_i =2l$ for each $ i \in [s]$, then 
\begin{align*}
    \textbf{Vert} ( \CL ( \bar{a}, l)) = \operatorname{tr} (B^s)
\end{align*}
which satisfies the linear recurrence $\operatorname{tr} (B^s)=2  \operatorname{tr} (B^{s-1}) +\operatorname{tr} (B^{s-2}) - \operatorname{tr} (B^{s-3})$
for $ s \geq 3$. It's easy to see that $ \operatorname{tr} (B^{0}) =3 $, $ \operatorname{tr} (B)=2 $ and $ \operatorname{tr} (B^{2})=6 $. This is the sequence
A033304. Note that the matrix $B$ shows up but it is conjugated by a symmetric matrix. 
\end{corollary}
\end{comment}
%We'll consider $ \bar{a } \in \mathbb{R}^s_{>0}$ and $ l \in \mathbb{R}$ so that $ 2l \leq \min \lbrace a_i: i \in [s] \rbrace$. The combinatorial structure of the polytope is the same for all $\bar{a},l$ for which $ 2l < \min \lbrace a_i: i \in [s] \rbrace$ so we pay special attention to the indices where $a_i =2l$. In this case we call the $i$th index \definition{singular}. If there is no singular index we call $ \bar{a}$ and the resulting chainlink polytope \definition{generic}. 

The calculation of the volume of a chainlink polytope has quite a straightforward formula in the case $ 2l \leq \min_{i \in [s]} a_i$. 
\begin{prop}\label{prop:volume}
Let $\bar{a} \in \mathbb{R}_{>0}^s$ and $ l \in \mathbb{R}_{\geq 0}$. Suppose that $ 2l \leq \min_{i \in [s]} a_i$.  The volume of the chainlink polytope $ \HL( \bar{a},l )$ is given by the following trace formula.
 \begin{align*}
    \Vol ( \HL( \bar{a}, l ))= \operatorname{tr} \left(\begin{bmatrix}
    a_1 & \frac{-l^2}{2} \\
    1 & 0   
    \end{bmatrix}\cdot \begin{bmatrix}
    a_2 & \frac{-l^2}{2} \\
    1 & 0   
    \end{bmatrix}\cdot \begin{bmatrix}
    a_3 & \frac{-l^2}{2} \\
    1 & 0   
    \end{bmatrix}\cdots\begin{bmatrix}
    a_s & \frac{-l^2}{2} \\
    1 & 0   
    \end{bmatrix}\right).
\end{align*}

%\begin{align*}
 %   \Vol ( \HL( \bar{a}, l )) =  \sum_{k=0}^{{\lfloor \frac{s}{2}} \rfloor} \big( \sum_{m \in  \mathcal{M}_k(s)} \prod_{i \not \in m} a_i \big) \frac{l^{2k}}{2^k} (-1)^k .
%\end{align*}
\end{prop}

%The decomposition used in the above proposition also gives rise to a different formula for the rank polynomial of the chainlink polytope. By the rank polynomial, I mean 
%\begin{align*}
   %R(q)= \sum_{k=0}^{n} | \CL^k( \bar{a},l) \cap \mathbb{Z}^s | q^{k}.
%\end{align*}
%The formula in question is
%\begin{align*}
  %  R(q) &= tr \bigg( \prod_{i=1}^s \begin{pmatrix}
 %   1 + q + \cdots + q^{a_i} & - q^{a_i} T_l (q) \\
%    1 & 0   
   % \end{pmatrix} \bigg) \\
  %  &= tr \bigg( \prod_{i=1}^s \begin{pmatrix}
 %   q & 1 \\
%    0 & 1   
   % \end{pmatrix}^{a_i}  \begin{pmatrix}
  %  1  & - T_l (q) \\
  %  1 & 0   
 %\end{align*}
%where 
%\begin{align*}
 %   T_l (q) = \sum_{k=0}^{l-1} (k+1) q^{-k}.
%\end{align*}

\section{Remarks and further work}
There are several questions about these chainlink polytopes that naturally arise. 
\begin{itemize}
\item \textbf{Ehrhart-Equivalence:} Two  rational polytopes $P, Q \in \mathbb{R}^d$ are said to be Ehrhart-Equivalent if they have the same  Ehrhart quasi-polynomial.  They are said to be $GL$ equidecomposable if we may partition $P = U_1 \cup \ldots \cup U_n$ and $Q = V_1 \cup \ldots \cup V_n$ into relatively open simplices such that for each $i$, $U_i$ and $V_i$ are $GL_d(\mathbb{Z})$ equivalent. In \cite{haase2008quasi}, it was conjectured that Ehrhart-equivalent polytopes are $GL$ equidecomposable. This is known to be true in dimensions $2$ \cite{greenberg1993piecewise} and $3$ \cite{erbe2019ehrhart}. Sections of chainlink polytopes provide us with a large class of examples to test this conjecture. 
\item \textbf{Multimodality:} \cref{unimodality} can be expressed in the following way: Let $\bar{a}$ be a composition of $n$. Then the function from $\{0, \ldots, n\}$ to $\mathbb{N}$ given by $k \rightarrow \# CL^k(\bar{a}, 1)$
is unimodal save when $\bar{a} = (a, 1, a, 1)$ or $(1, a, 1, a)$  and is bimodal in these cases. If we instead fix a positive integer $l$ such that $2l \leq \operatorname{min} \{a_i\}$ and look at $k \rightarrow \# CL^k(\bar{a}, l)$, the function may be multimodal. Indeed, we have that when $\bar{a} = (2k, 2k)$ and $l = k$, we seem to have $k+1$ peaks. Can one describe the maximal number of modes that may arise for fixed $l$ and when these are attained? 
\item \textbf{Polytopal models for general circular fences:} In this paper, we developed a polytopal model for \emph{gates}, which are circular fences coming from compositions of the form $(a_1, 1, \ldots, a_s, 1)$, i.e. where all the down steps have size $1$.% As noted in the paper in Remark \ref{rem:order}, the order polytope for the gate poset \emph{does not} obey symmetry, which is why we worked with chainlink polytopes instead.  Now, our proof works equally well when the down steps all have the same size, say $k$. 
What about general circular fences, those coming from compositions of the form $(a_1, b_1, \ldots, a_s, b_s)$ with differing lengths of down steps $b_1, \ldots, b_s$? 

A natural proposal is as follows. The polytope \cite{korarxiv} will consist of all real tuples $(x_1, y_1, \ldots, x_s, y_s)$ such that 
\[0 \leq x_i \leq a_i + 1, \quad 0 \leq y_i \leq b_i - 1, \quad (b_i-1)(x_i - a_i) \leq y_i , \quad y_i \leq (b_i- 1)x_{i+1}.\]
where the indices are taken cyclically. %When all the down steps are $1$, all save the first set of inequalities become trivial, but if agree to combine the last two set of inequalities, we get $x_i - a_i \leq x_{i+1}$, which are the defining equations in the chainlink polytope with link number $1$. 
Unfortunately, these polytopes do not have the symmetry that chainlink polytopes have. Is there a way of defining polyhedral models for general circular fence posets so that this symmetry does hold?
\item \textbf{General Chainlink Polytopes:} In the case $ 2l > \min_{i \in [s]} a_i$, the polytopes $\CL(\bar{a}, l)$ begin to exhibit wild behaviour. \Cref{prop:vertices,prop:volume}, and \cref{cor:vertices} no longer hold. Given that the chainlink polytope $ \CL ( \bar{a}, l)$ is full-dimensional when $ l < \min_{i \in [s]} a_i$, this leaves a lot to be investigated, both combinatorially and geometrically.
\end{itemize}
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\end{document}
