%% if you are submitting an initial manuscript then you should have submission as an option here
%% if you are submitting a revised manuscript then you should have revision as an option here
%% otherwise options taken by the article class will be accepted
\documentclass[finalversion]{FPSAC2018}
\articlenumber{2}
%% but DO NOT pass any options (or change anything else anywhere) which alters page size / layout / font size etc

%% note that the class file already loads {amsmath, amsthm, amssymb}

\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{example}[theorem]{Example}

\newcommand*{\vcenteredhbox}[1]{\begingroup
\setbox0=\hbox{#1}\parbox{\wd0}{\box0}\endgroup}


\usepackage{lipsum}

%% define your title in the usual way
%\title[The canonical join complex for biclosed sets]{The canonical join complex for biclosed sets}
\title{The canonical join complex for biclosed sets}

%% define your authors in the usual way
%% use \addressmark{1}, \addressmark{2} etc for the institutions, and use \thanks{} for contact details
\author{Alexander
Clifton\thanks{\href{mailto:aclift2@emory.edu}{aclift2@emory.edu}}\addressmark{1},
Peter
Dillery\thanks{\href{mailto:dillery@umich.edu}{dillery@umich.edu}}\addressmark{2},
\and Alexander
Garver\thanks{\href{mailto:alexander.garver@lacim.ca}{alexander.garver@lacim.ca}}\addressmark{3}}

%% then use \addressmark to match authors to institutions here
\address{\addressmark{1}Department of Mathematics and Computer Science, Emory University \\ \addressmark{2}Department of Mathematics, University of Michigan\\ \addressmark{3}Laboratoire de Combinatoire et d'Informatique Math\'ematique, Universit\'e du Qu\'ebec \`a Montr\'eal}

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

%% leave this blank until submitting a revised version
%\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{The canonical join complex of a semidistributive lattice is a simplicial complex whose faces are canonical join representations of elements of the semidistributive lattice. We give a combinatorial classification of the faces of the canonical join complex of the lattice of biclosed sets of segments supported by a tree, as introduced by the third author and McConville. We also use our classification to describe the elements of the shard intersection order of the lattice of biclosed sets. As a consequence, we prove that this shard intersection order is a lattice.}

%% put your French abstract here, or comment this out if you don't have one
\resume{Le complexe sup-canonique d'un treillis semi-distributif est un complexe simplicial dont les faces sont des repr\'esentations sup-canoniques d'\'el\'e-ments du treillis semi-distributif. Nous donnons une classification combinatoire des faces du complexe sup-canonique du treillis des ensembles bi-ferm\'es de segments sur un arbre, qui ont \'et\'e introduits par le troisi\`eme auteur et McConville. Nous utilisons notre classification pour 
d\'ecrire les \'el\'ements de l'ordre d'intersection des tessons du treillis des ensembles bi-ferm\'es. En cons\'equence, nous prouvons que cet ordre d'intersection des tessons est un treillis.}

%{Le complexe de jointure canonique d'un r\'eseau semidistributif est un complexe simplicial dont les visages sont des repr\'esentations d'articulation canoniques d'\'el\'ements du r\'eseau semidistributif. Nous donnons une classification combinatoire des visages du complexe de jointure canonique du r\'eseau d'ensembles de segments biferm\'e soutenus par un arbre, tel qu'introduit par le troisi\`eme auteur et McConville. Nous utilisons \'egalement notre classification pour d\'ecrire les \'el\'ements de l'ordre d'intersection du morceau du r\'eseau d'ensembles biferm\'e. En cons\'equence, nous prouvons que cet ordre d'intersection du morceau est un r\'eseau.}

%% put your keywords here, or comment this out if you don't have them yet
\keywords{lattice, weak order, canonical join representation, shard intersection order}

%% 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, ie. [xx, yy] in the text

%% this sample uses biblatex, which (among other things) takes care of URLs in a more flexible way than bibtex
%% but you can use bibtex if you want
\usepackage[backend=bibtex]{biblatex}
\addbibresource{02-Clifton-Dillery-Garver.bib}
%% note the \printbibliography command at the end of the file which goes with these biblatex commands

\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


\section{Introduction}

In \cite{mcconville2017lattice}, McConville introduced a lattice of biclosed sets as a tool for studying the lattice structure of Grid-Tamari orders. The class of lattices of biclosed sets includes the weak order on permutations. As the weak order on permutations appears in many mathematical contexts, including (and certainly not limited to) geometric combinatorics \cite{reading:cambrian_lattices, hohlweg2011permutahedra} and representation theory of preprojective algebras \cite{mizuno2014classifying, thomas2017stability}, it is natural to study the lattice-theoretic aspects of biclosed sets.

In subsequent work by McConville and the third author \cite{garver2015lattice, garver2016oriented, garver2017enumerative}, biclosed sets were used to understand lattice properties of Grid-Tamari orders and oriented flip graphs. Futhermore, in \cite{garver2016oriented, garver2017enumerative}, they describe the shard intersection order, in the sense of \cite{ReadingPAB}, of Grid-Tamari orders and of oriented flip graphs. Neither one of the classes of lattices of biclosed sets used in these two settings contains the other, but both contain the weak order on permutations. However, both classes of lattices belong to the class of lattices of biclosed sets studied in \cite{palu2017non}.  The goal of this paper is to gain a combinatorial description of this shard intersection order of the lattice of biclosed sets appearing in \cite{garver2016oriented}. 

The lattice of biclosed sets is a congruence-uniform lattice and therefore a semidistributive lattice. As a semidistributive lattice, the lattice of biclosed sets has a well-defined canonical join complex. The canonical join complex of a semidistributive lattice $L$ is the simplicial complex whose faces are canonical join representations of elements of $L$. As a congruence-uniform lattice, the lattice of biclosed sets $L$ has a well-defined shard intersection order, denoted $\Psi(L).$ The poset $\Psi(L)$ is an alternative partial order on the elements of $L$ that is constructed using the data of canonical join representations of elements of $L$. If a finite lattice $L$ is not congruence-uniform, then $\Psi(L)$ may not be a partial order (see \cite[Exercise 9.73]{ReadingPAB}). We are not aware of a characterization of lattices whose shard intersection order is a partial order.

%To understand this shard intersection order, it is very useful to understand the canonical join complex of the lattice of biclosed sets. The canonical join complex is defined for any semidistributive lattice $L$.  The lattices of biclosed sets that we consider in this paper are all congruence-uniform, which implies that they are semidistributive. The canonical join complex is the simplicial complex whose faces are canonical join representations of elements of $L$. The shard intersection order of $L$, denoted $\Psi(L)$, is an alternative partial order on the elements of $L$ that is constructed using the data of canonical join representations of elements of $L$. We remark that if $L$ is not congruence-uniform, then $\Psi(L)$ may not be partially ordered.

Our approach is to, first, describe the join-irreducible biclosed sets (see Proposition~\ref{Prop_join_irr_des}). After that, we use this description to classify the faces of the canonical join complex of biclosed sets (see Theorem~\ref{Thm_CJC}). We are then in a position to describe the elements of the shard intersection order (see Theorem~\ref{psib}) and prove that the shard intersection order is a lattice (see Theorem~\ref{Thm_lattice_prop}). The latter was conjectured by the first and second author in \cite[Conjecture 6.3]{cliftondillery}. The full version of our paper can be found at \cite{clifton2017canonical}.

%Our approach is to, first, describe the join-irreducible biclosed sets (see Proposition~\ref{Prop_join_irr_des}). After that, we use this description to classify the faces of the canonical join complex of biclosed sets (see Theorem~\ref{Thm_CJC}). We are then in a position describe the elements of the shard intersection order (see Theorem~\ref{psib}) and prove that this shard intersection order is a lattice (see Theorem~\ref{Thm_lattice_prop}). The latter was conjectured by the first and second author in \cite[Conjecture 6.3]{cliftondillery}. The full version of our paper can be found at \cite{clifton2017canonical}.

Generally speaking, it is an open problem in \cite[Problem 9.5]{ReadingPAB} to determine which congruence-uniform lattices $L$ have the property that $\Psi(L)$ is a lattice. Theorem~\ref{Thm_lattice_prop} provides many new examples of congruence-uniform lattices with this property. When $L$ is the weak order on permutations, it was already shown by Reading in \cite{reading:noncrossing} that $\Psi(L)$ is a lattice. As the weak order on permutations is an example of a lattice of biclosed sets, our Theorem~\ref{Thm_lattice_prop} recovers this result.  We also remark that in \cite[Theorem 1.1]{muhle}, M\"uhle found a necessary condition on $L$ in order for  $\Psi(L)$ to be a lattice: he showed that the M\"obius function on $L$, denoted $\mu_L(-,-)$, must satisfy $\mu_L(\hat{0}, \hat{1}) \neq 0 $ if $\Psi(L)$ is a lattice. 

The paper is organized as follows. We remind the reader of the lattice theory that we will use throughout the paper in Section~\ref{Sec_lattice}. We describe the lattices of biclosed sets we will work with in Section~\ref{Sec_biclosed}. In Section~\ref{cul}, we construct a special labeling of the covering relations in lattices of biclosed sets. In Section~\ref{Sec_cjc}, we use this labeling to describe the join- and meet-irreducible biclosed sets and the faces of the canonical join complex of biclosed sets. We study the shard intersection order of biclosed sets in Section~\ref{shard_section}.



\section{Lattices}\label{Sec_lattice}


Let $(L,\le_L)$ be a finite lattice. For $x,y \in L$, if $x < y$ and there does not exist $z \in L$ such that $x < z <y$, we write $x \lessdot y$. Let $\text{Cov}(L) := \{(x,y) \in L^{2} \mid x \lessdot y\}$ be the set of \textbf{covering relations} of $L$. We let $\hat{0}, \hat{1} \in L$ denote the unique minimal and unique maximal elements of $L$, respectively. 

A set map $\lambda : \text{Cov}(L) \to Q$, where $(Q, \leq_{Q})$ is some poset is called an \textbf{edge labeling}. We review the concepts of join- and meet-irreducibility in order to discuss an important type of edge labeling. 

We say that an element $j \in L$ is \textbf{join-irreducible} if $j \neq \hat{0}$ and whenever $j = x \vee y$, one has that $j = x$ or $j = y$. \textbf{Meet-irreducible} elements $m \in L$ are defined dually. We denote the subset of join-irreducible (resp., meet-irreducible) elements by $\text{JI}(L)$ (resp., $\text{MI}(L)$). For $j$ (resp., $m$) in $\text{JI}(L)$ (resp., $\text{MI}(L)$), we let $j_{*}$ (resp., $m^{*}$) denote the unique element of $L$ covered by (resp., that covers) $j$ (resp., $m$). 

For $A\subseteq L$, the expression $\bigvee A := \bigvee_{a \in A} a$ is \textbf{irredundant} if there does not exist a proper subset $A^{\prime}\subsetneq A$ such that $\bigvee A^{\prime}=\bigvee A$. Given $A,B\subseteq \text{JI}(L)$ such that $\bigvee A$ and $\bigvee B$ are irredundant and $\bigvee A=\bigvee B$, we set $A\preceq B$ if for any $a\in A$ there exists $b\in B$ with $a\leq b$. If $x\in L$ and $A\subseteq\text{JI}(L)$ such that $x=\bigvee A$ is irredundant, we say $\bigvee A$ is a \textbf{canonical join representation} of $x$ if $A\preceq B$ for any other irredundant join representation $x=\bigvee B,\ B\subseteq\text{JI}(L)$. Dually, one defines \textbf{canonical meet representations}.

We define the \textbf{canonical join complex} of $L$, denoted $\Delta^{CJ}(L)$, to be the abstract simplicial complex whose vertex set is $\text{JI}(L)$ and whose faces are sets of join-irreducibles whose join is a canonical join representation of some  element of $L$. 

Now we assume that $L$ is a \textbf{semidistributive} lattice. This means that for any three elements $x, y, z \in L$, the following properties hold:

$\bullet$ if $x \wedge z = y \wedge z$, then $(x \vee y) \wedge z = x \wedge z$, and

$\bullet$ if $x \vee z = y \vee z$, then $(x \wedge y) \vee z = x \vee z$.

\noindent It is known that a lattice $L$ is semidistributive if and only if each element of $L$ has a canonical join representation and a canonical meet representation \cite[Theorem 2.24]{freese1995free}. In this case, there is a canonical bijection $L \to \Delta^{CJ}(L)$ sending $x \mapsto A$ where $\bigvee A$ is the canonical join representation of $x$.

With these notions in hand, we arrive at the notions of \textbf{CN-} and \textbf{CU-labeling}, the latter of which plays a prominent role in this paper.

\begin{definition}\label{defn_cu} A labeling  $\lambda: \text{Cov}(L) \to Q$ is a \textbf{CN-labeling} if $L$ and its dual $L^{*}$ satisfy the following: given $x,y,z \in L$ with $(z,x), (z,y) \in \text{Cov}(L)$ and maximal chains $C_{1}$ and $C_{2}$ in $[z, x \vee y]$ with $x \in C_{1}$ and $y \in C_{2}$, 

(CN1) {the elements $x' \in C_{1}, y' \in C_{2}$ such that $(x', x \vee y), (y', x \vee y) \in \text{Cov}(L)$ satisfy $$\lambda(x', x \vee y) = \lambda(z,y), \lambda(y', x \vee y) = \lambda(z, x);$$}

%\vspace{-.25in}
(CN2) {if $(u,v) \in \text{Cov}(C_{1})$ with $z < u, v < x \vee y$, then $\lambda(z,x), \lambda(z,y) <_{Q} \lambda(u,v)$;}

(CN3) {the labels on $\text{Cov}(C_{1})$ are pairwise distinct.}


\noindent We say that $\lambda$ is a \textbf{CU-labeling} if, in addition, it satisfies 

(CU1) {$\lambda(j_{*}, j) \neq \lambda(j_{*}', j')$ for $j, j' \in \text{JI}(L)$, $j \neq j'$, and}

(CU2) {$\lambda(m, m^{*}) \neq \lambda(m', m'^{*})$ for $m, m' \in \text{MI}(L)$, $m \neq m'$.}


\noindent If $L$ admits a CU-labeling, it is said to be \textbf{congruence-uniform}. 
\end{definition}


%\begin{remark}\label{Remark_def_of_CU_lattice}
%For completeness, we include the more standard definition of a congruence-uniform lattice. 
%
%Given $(x, y) \in \text{Cov}(L)$, we let $\text{con}(x,y)$ denote the most refined lattice congruence for which $x \equiv y$. Such congruences are join-irreducible elements of the lattice of lattice congruences of $L$, denoted $\text{Con}(L)$. When $L$ is a finite lattice, the join-irreducibles (resp., meet-irreducibles) of $\text{Con}(L)$ are the congruences of the form $\text{con}(j_*,j)$ (resp., $\text{con}(m,m^*)$). We thus obtain surjections 
%$$\begin{array}{rclccrcl}
%\text{JI}(L) & \rightarrow & \text{JI}(\text{Con}(L)) & & &  \text{MI}(L) & \rightarrow & \text{MI}(\text{Con}(L)) \\
%j & \mapsto & \text{con}(j_*,j) & & &  m & \mapsto & \text{con}(m,m^*).
%\end{array}$$
%If these maps are bijections, we say that $L$ is \textbf{congruence-uniform}. It follows from \cite[Proposition 2.6]{garver2016oriented} that this definition and the one given in Definition~\ref{defn_cu} are equivalent.
%\end{remark}

It was proved by Day in \cite{day:congruence} that any finite congruence-uniform lattice is also semidistributive.

We now mention some general properties of CU-labelings and the definition of the shard intersection order of $L$. Given an edge labeling $\lambda: \text{Cov}(L) \to Q$, one defines $$\lambda_{\downarrow}(x) := \{\lambda(y,x): \ y\lessdot x\}, \ \ \ \lambda^{\uparrow}(x):= \{\lambda(x,z): \ x \lessdot z\}.$$ 

\begin{lemma}[{\cite[Lemma 2.6]{garver2016oriented}}]\label{cu-label_ji_mi}
Let $L$ be a congruence-uniform lattice with CU-labeling $\lambda:\text{Cov}(L)\rightarrow P$. For any $s\in P$, there is a unique join-irreducible $j \in \text{JI}(L)$ (resp., meet-irreducible $m \in \text{MI}(L)$) such that $\lambda(j_*,j) = s$ (resp., $\lambda(m,m^*) = s$). Moreover, this join-irreducible $j$ (resp., meet-irreducible $m$) is the minimal (resp., maximal) element of $L$ such that $s \in \lambda_\downarrow(j)$ (resp., $s \in \lambda^\uparrow(m)$).
\end{lemma}

Later, in Propositions~\ref{Prop_join_irr_des} and \ref{Prop_meet_irr_des}, we use Lemma~\ref{cu-label_ji_mi} to characterize join- and meet-irreducible elements of $\text{Bic}(T),$ the lattice of biclosed sets defined in the next section. 


One can also use CU-labelings to determine canonical join representations and canonical meet representations of elements of a congruence-uniform lattice. We state this precisely as follows.

\begin{lemma}[{\cite[Proposition 2.9]{garver2016oriented}}]\label{Lemma_cjr_cmr}
Let $L$ be a congruence-uniform lattice with CU-labeling $\lambda$. For any $x\in L$, the canonical join representation of $x$ is $\bigvee D$, where $D = \{j \in \text{JI}(L): \ \lambda(j_*,j) \in \lambda_\downarrow(x)\}$. Dually, for any $x\in L$, the canonical meet representation of $x$ is $\bigwedge U$, where $U = \{m \in \text{MI}(L): \ \lambda(m,m^*) \in \lambda^\uparrow(x)\}$.
\end{lemma}


Given a lattice $L$ with a CU-labeling, one can define a new partial order on the elements of $L$ known as the \textbf{shard intersection order} of $L$. Reading introduced this concept in \cite{ReadingPAB}.

\begin{definition} Let $L$ be a congruence-uniform lattice with CU-labeling $\lambda : \text{Cov}(L) \to P$. Let $x \in L$ and let $y_{1}, \dots y_{k}$ be the elements of $L$ satisfying $(y_{i}, x) \in \text{Cov}(L)$. Define the \textbf{shard intersection order} of $L$, denoted $\Psi(L)$, to be the collection of sets of the form $$\psi(x) := \{ \lambda(w,z) \mid \bigwedge_{i}^{k} y_{i} \leq w < z \leq x, (w,z) \in \text{Cov}(L)\}$$ partially ordered by inclusion. We may refer to the interval $[\bigwedge_{i}^{k} y_{i},x]$ as a \textbf{facial interval}.
\end{definition}

%\begin{definition} Let $L$ be a congruence-uniform lattice with CU-labeling $\lambda : \text{Cov}(L) \to P$. Let $x \in L$ and let $y_{1}, \dots y_{k}$ be the elements of $L$ satisfying $(y_{i}, x) \in \text{Cov}(L)$. We denote the set $\{\lambda(y_{i}, x)\}$ by $\lambda_{\downarrow}(x)$. Define the \textbf{shard intersection order} of $L$, denoted $\Psi(L)$, to be the collection of sets of the form $$\psi(x) := \{ \lambda(w,z) \mid \bigwedge_{i}^{k} y_{i} \leq w < z \leq x, (w,z) \in \text{Cov}(L)\}$$ partially ordered by inclusion. We may refer to the interval $[\bigwedge_{i}^{k} y_{i},x]$ as a \textbf{facial interval}.
%\end{definition}

\begin{remark}\label{rel_to_hyperplanes}
The shard intersection order was originally defined by Reading in \cite{reading:noncrossing} given the data of $\text{Pos}(\mathcal{A}, B)$, the poset of regions of a simplicial hyperplane arrangement $\mathcal{A}$ with base region $B$. Such posets are congruence-uniform lattices, as shown by Reading in \cite[Proposition 9-3.33 and Corollary 9-7.22]{ReadingPAB}. It follows from \cite[Proposition 9-7.13]{ReadingPAB} that Reading's original definition of the shard intersection order of $\text{Pos}(\mathcal{A}, B)$ coincides with the definition of $\Psi(\text{Pos}(\mathcal{A}, B))$. The definition of $\Psi(L)$ where $L$ is any congruence-uniform lattice is therefore a generalization of Reading's original definition of the shard intersection order.
\end{remark}

%The shard intersection order was originally defined by Reading in \cite{reading:noncrossing} on $\text{Pos}(\mathcal{A}, B)$, the poset of regions of a simplicial hyperplane arrangement $\mathcal{A}$ with base region $B$.

%If, in \textcolor{blue}{addition} \textcolor{red}{CUT addtion}, $L$ is congruence-uniform, it follows from \cite[Proposition 9-7.13]{ReadingPAB} that these two definitions of the shard intersection order produce the same partial order.

\section{Biclosed sets}\label{Sec_biclosed}

A \textbf{tree} is a finite connected acyclic graph. The degree-one vertices of a tree are called \textbf{leaves}. We can always embed a tree $T$ into the disk $D^{2}$ so that exactly the leaves lie on the boundary. Unless stated otherwise, a tree is assumed to be equipped with such an embedding. Non-leaf vertices of $T$ are thus in the interior of $D^{2}$, and we call these \textbf{interior vertices}. We also assume that the interior vertices of $T$ have degree at least 3.

An \textbf{acyclic path} is a sequence of pairwise distinct vertices $(v_{i_{1}}, \dots, v_{i_{n}})$ of $T$ such that there is an edge connecting $v_{i_{j}}$ and $v_{i_{j+1}}$ for all $1 \leq j \leq n-1$. Since an acyclic path is uniquely determined by its endpoints, we can denote $(v_{i_{1}}, \dots, v_{i_{n}})$ by $[v_{i_{1}}, v_{i_{n}}]$. 


Observe that the embedding of $T$ in $D^2$ determines a cyclic ordering of the edges of $T$ that are incident to a given vertex. An acyclic path $(v_{i_{1}}, \dots, v_{i_{n}})$ is called a \textbf{segment} if, for each $1 \leq j \leq n-2$, $[v_{i_{j+1}}, v_{i_{j+2}}]$ is immediately clockwise or counterclockwise from $[v_{i_{j}}, v_{i_{j+1}}]$ with respect to the cyclic ordering on the edges incident to $v_{i_{j+1}}$. The set of all segments supported by a tree $T$ is denoted by $\text{Seg}(T)$. Figure~\ref{seg_fig} shows some examples and non-examples of segments.



%\begin{figure}
%$$\includegraphics[scale=1.5]{SegEx1.pdf}$$
%\caption{Acyclic paths $[1,3], [5,4],$ and $[7,10]$ are segments, but $[4,10]$ is not.}
%\label{seg_fig}
%\end{figure}

\begin{figure}
\captionsetup{width=0.45\textwidth}
\centering
\begin{minipage}{.5\textwidth}
  \centering
  \def\svgwidth{140pt}
\includegraphics[scale=1.32]{SegEx1.pdf}
\caption{Acyclic paths $[1,3], [5,4],$ $[7,10]$ are segments, but $[4,10]$ is not.}
\label{seg_fig}
 % \centering
 % \includegraphics[width=.4\linewidth]{image1}
 % \captionof{figure}{A figure}
 % \label{fig:test1}
\end{minipage}%
\begin{minipage}{.5\textwidth}
\centering
\includegraphics[scale=1]{Sn_bic.pdf}
\caption{A tree whose biclosed sets are permutations in $\mathfrak{S}_n$.}
\label{perm_bic_fig}
  %\centering
  %\includegraphics[width=.4\linewidth]{image1}
  %\captionof{figure}{Another figure}
  %\label{fig:test2}
\end{minipage}
\end{figure}

%\begin{figure}
%\captionsetup{width=0.45\textwidth}
%\centering
%\begin{minipage}{.5\textwidth}
%  \centering
%  \def\svgwidth{140pt}
%  \input{elem_lamination.pdf_tex}
%  \caption{Lamination corresponding to the given set of arcs.}
%   \label{fig:elem_lamination}
% % \centering
% % \includegraphics[width=.4\linewidth]{image1}
% % \captionof{figure}{A figure}
% % \label{fig:test1}
%\end{minipage}%
%\begin{minipage}{.5\textwidth}
%\centering
%\includegraphics[scale=1]{lamin.pdf}
%\caption{The shear coordinate of $\gamma$ with respect to $\ell_1$ is $+1$, and the shear coordinate of $\delta$ with respect to $\ell_2$ is $-1$.}
%\label{lamin_convention}
%  %\centering
%  %\includegraphics[width=.4\linewidth]{image1}
%  %\captionof{figure}{Another figure}
%  %\label{fig:test2}
%\end{minipage}
%\end{figure}






Given two segments $s_{1} = (v_{i_1}, v_{i_2}, \ldots, v_{i_k}), s_{2} = (v_{i_k}, v_{i_{k+1}}, \ldots, v_{i_{n}}) \in \text{Seg}(T)$ that share an endpoint $v_{i_k}$ but differ at all other vertices, we define their \textbf{composition} to be the acyclic path $s_{1} \circ s_{2} := (v_{i_1}, \ldots, v_{i_k}, \ldots, v_{i_n})$. We say that two segments $s_1$ and $s_2$ are \textbf{composable} if $s_1 \circ s_2 \in \text{Seg}(T)$. A subset $B\subset \text{Seg}(T)$ is \textbf{closed} if for all composable $s_1, s_2 \in B, s_1 \circ s_2 \in B$. The set $B$ is \textbf{biclosed} if both $B$ and its complement, $B^c := \text{Seg}(T)\backslash B$, are closed. Additionally, if $B \subset \text{Seg}(T)$, we let $\overline{B}$ denote the smallest closed set containing $B$. We let $\text{Bic}(T)$ denote the poset of biclosed subsets of $\text{Seg}(T)$ ordered by inclusion.

%Given two segments $s_{1} = (v_{i_1}, v_{i_2}, \ldots, v_{i_k}), s_{2} = (v_{i_k}, v_{i_{k+1}}, \ldots, v_{i_{n}}) \in \text{Seg}(T)$ that share an endpoint $v_{i_k}$ but differ at all other vertices, we define their \textbf{composition} to be the acyclic path $s_{1} \circ s_{2} := (v_{i_1}, \ldots, v_{i_k}, \ldots, v_{i_n})$. We say that two segments $s_1$ and $s_2$ are \textbf{composable} if $s_1 \circ s_2 \in \text{Seg}(T)$. A subset $B\subset \text{Seg}(T)$ is \textbf{closed} if for all composable $s_1, s_2 \in B, s_1 \circ s_2 \in B$. $B$ is \textbf{biclosed} if both $B$ and its complement, $B^c := \text{Seg}(T)\backslash B$, are closed. We will also often say that $B$ is \textbf{coclosed} when $B^c$ is closed. Additionally, if $B \subset \text{Seg}(T)$, we define $\overline{B}$ to be the smallest closed set containing $B$.


We remark that although the poset $\text{Bic}(T)$ depends on the embedding of $T$, each tree $T$ is equipped with an embedding in $D^2$ so it makes sense to write $\text{Bic}(T)$. The poset structure of $\text{Bic}(T)$ is studied in \cite{garver2016oriented}, where the following result is proved:

\begin{theorem}[{\cite[Theorem 4.1]{garver2016oriented}}]\label{bic_structure} The poset $\text{Bic}(T)$ is a congruence-uniform lattice. Moreover, the poset $\text{Bic}(T)$ has the following properties:

1. for any $X, Y \in \text{Bic}(T)$, if $X < Y$, there is a segment $y \in Y$ such that $X \sqcup \{y\} \in \text{Bic}(T);$

2. for any $W, X, Y \in \text{Bic}(T)$ with $W \le X\cap Y$, the set $W \cup \overline{(X\cup Y)\backslash W}$ is biclosed;

3. the edge-labeling $\lambda: \text{Cov}(\text{Bic}(T)) \to \text{Seg}(T)$ defined by $\lambda(X,Y) = s$ if $Y\backslash X = \{s\}$ is a CN-labeling.
\end{theorem}

\label{joinmeet_bijection} Theorem~\ref{bic_structure} implies that the map $(-)^{c} \colon \text{Bic}(T) \to \text{Bic}(T)$, $B \mapsto B^{c}$, gives rise to a bijection $\text{JI}(\text{Bic}(T)) \to \text{MI}(\text{Bic}(T))$. We frequently use the next lemma, which follows from property (2) in Theorem~\ref{bic_structure} with $X = B_1,$ $Y = B_2$, and $W = \emptyset$.

\begin{lemma}\label{Join_Description}
For any $B_{1}, B_{2} \in \text{Bic}(T)$, we have $B_{1} \vee B_{2} = \overline{B_{1} \cup B_{2}}.$
\end{lemma}

\begin{example}
Let $T$ be the tree shown in Figure~\ref{perm_bic_fig} with the indicated labeling of the interior vertices. Define a map that sends a segment $s$ to $(i,j) \in \mathbb{N}^2$ with $i < j$ where $i$ and $j$ are the vertex labels of the endpoints of $s$. This induces a map on biclosed sets that sends each biclosed set to the inversion set of a permutation in $\mathfrak{S}_n$. Moreover, it induces a poset isomorphism $\text{Bic}(T) \to \text{Weak}({\mathfrak{S}_n})$ where the latter denotes the weak order on permutations.

Additionally, it follows from \cite[Theorem 10-3.1]{ReadingPAB} that
$\text{Weak}(\mathfrak{S}_n)$ is isomorphic to $\text{Pos}(\mathcal{A},
B)$ where $\mathcal{A}$ is Coxeter arrangement of $\mathfrak{S}_n$ and
$B$ is the region of the Coxeter arrangement containing the identity permutation. Now it follows from Remark~\ref{rel_to_hyperplanes} that the class of lattices of the form $\Psi(\text{Bic}(T))$ includes the shard intersection orders of type $A$ Coxeter arrangements.
\end{example}

%\begin{figure}
%\centering
%\includegraphics[scale=1]{Sn_bic.pdf}
%\caption{A tree whose biclosed sets are identified with permutations in $\mathfrak{S}_n$.}
%\label{perm_bic_fig}
%\end{figure}



%\begin{lemma}\label{Lemma_meet_in_bic}
%For $B_1, B_2 \in \text{Bic}(T)$, one has $B_1 \wedge B_2 = (B_1^c \vee B_2^c)^c.$
%\end{lemma}
%\begin{proof}
%Let $s \in B_1 \wedge B_2$. It follows that $s \in B_1 \cap B_2$. Now $s \not \in B_1^c\cup B_2^c \subset {B_1^c \vee B_2^c}.$ This implies that $s \in (B_1^c \vee B_2^c)^c.$
%
%To prove the opposite inclusion, observe that $$B_1 \wedge B_2 = \displaystyle \bigvee_{\begin{array}{c}\small\text{$B \in \text{Bic}(T)$}\\ \small \text{$B \subset B_1, B\subset B_2$} \end{array}} B.$$ Now notice that if $s \in (B_1^c \vee B_2^c)^c$, then $s \not \in B_1^c$ and $s \not \in B_2^c$. This implies that $s \in B_1$ and $s \in B_2$. Thus $(B_1^c \vee B_2^c)^c \subset B_1$ and $(B_1^c \vee B_2^c)^c \subset B_2$. Since $(B_1^c \vee B_2^c)^c \in \text{Bic}(T)$, it follows that $(B_1^c \vee B_2^c)^c$ is a joinand in the join representation of $B_1 \wedge B_2$ shown above. We obtain that $(B_1^c \vee B_2^c)^c \subset B_1 \wedge B_2$.\end{proof}

%Using the following lemma, one obtains an explicit description of the facial intervals of $\text{Bic}(T)$. One can find a proof of the following lemma, in \cite[Lemma 4.14]{garver2017enumerative}.
%
%\begin{lemma}\label{meet_of_B_is}
%Given $B, B\backslash\{s_1\}, \ldots, B\backslash\{s_k\} \in \text{Bic}(T)$, we have that $\bigwedge_{i = 1}^k B_i = B\backslash\overline{\{s_1,\ldots, s_k\}}.$ \end{lemma}

\section{A CU-labeling of $\text{Bic}(T)$}\label{cul}
In \cite{garver2016oriented}, the authors prove that $\text{Bic}(T)$ is congruence-uniform, and thus it admits a CU-labeling. In this section, we explicitly construct such a labeling.







We say a segment $s \in \text{Seg}(T)$ is a \textbf{split} of a segment $t$ if $s$ is a proper subsegment of $t$, and $s$ and $t$ share an endpoint. A \textbf{break} of a segment $[a,c]$ is a pair of splits of $[a,c]$, denoted $\{[a,b],[b,c]\}$, for some vertex $b$ of segment $[a,c]$ where $b \neq a$ and $b \neq c$. We say that $b$ is the \textbf{faultline} of the break $\{[a,b],[b,c]\}$.


Define a poset $\mathcal{S}_T$ whose elements are of the form $(s,{\{s_1,s_2,\ldots,s_m\}}) \in \text{Seg}(T)\times 2^{\text{Seg}(T)}$ with the following properties:

$\bullet$ $s = (v_0, v_1,\ldots, v_{m+1})$ has $m$ breaks,

$\bullet$ each $s_i$ is a split of $s$, and

$\bullet$ two distinct splits $s_i$ and $s_j$ do not appear in the same break of $s$. 

We will typically denote $(s,{\{s_1,s_2,\ldots,s_m\}}) \in \mathcal{S}_T$ by $s_\mathcal{D} = s_{\{s_1,s_2,\ldots,s_m\}}$. The elements of $\mathcal{S}_T$ are partially ordered as follows: given $s_{\{s_1,s_2,\ldots,s_k\}}, {s^\prime_{\{t_1,t_2,\ldots,t_l\}}} \in \mathcal{S}_T$, one has ${s^\prime_{\{t_1,t_2,\ldots,t_l\}}} \le s_{\{s_1,s_2,\ldots,s_k\}}$ if $s^\prime$ is a proper subsegment of $s$. At times, we will also write $s^\prime \subseteq s$ (resp., $s^\prime \not \subseteq s$) to indicate that $s^\prime$ is a subsegment (resp., is not a subsegment) of $s$. We will refer to elements of $\mathcal{S}_T$ as \textbf{labels}.


For an example of this poset of labels, let $T$ be the tree shown in Figure~\ref{poset_of_labels_fig}. There we also show the poset of labels $\mathcal{S}_T$.% of the covering relations of $\text{Bic}(T)$.


%\begin{example}
%Let $T$ be the tree shown in Figure~\ref{poset_of_labels_fig}. In this same figure, we show the poset of labels $\mathcal{S}_T$ of the covering relations of $\text{Bic}(T)$.
%\end{example}

\begin{figure}
$$\includegraphics[scale=.8]{poset_of_labels_v2.pdf}$$
\caption{A tree $T$ and the corresponding poset of labels $\mathcal{S}_T$. The shortest segments of $T$ are labeled $a, b,$ and $c$.}
\label{poset_of_labels_fig}
\end{figure}

\begin{definition}
Define a map $\widetilde{\lambda}:\text{Cov}(\text{Bic}(T))\rightarrow \mathcal{S}_T$ by
 $\widetilde{\lambda}(B,B\sqcup\{{s}\})=s_{\{s_1,s_2,\ldots,s_k\}}$ where $s_1,s_2,\ldots,s_k$ are the splits of $s$ which are contained in $B$. It is clear that $\widetilde{\lambda}$ is an edge-labeling of $\text{Bic}(T)$. If we let $\lambda: \text{Cov}(\text{Bic}(T)) \to \text{Seg}(T)$ denote the first coordinate function of $\widetilde{\lambda}$, we have that $\lambda$ is the CN-labeling of $\text{Bic}(T)$ from Theorem~\ref{bic_structure} (3) (see Figure~\ref{label_fig}).
\end{definition}

%\begin{example}
%Let $B$ denote the biclosed set shown in Figure~\ref{label_fig}. One checks that $$\widetilde{\lambda}(B\backslash\{[9,8]\}, B) = [9,8]_{\{[9,2], [9,4], [6,8], [3,8], [7,8]\}}.$$
%\end{example}

%\vspace{-2in}

\begin{figure}[t]
$$\raisebox{-.8in}{\includegraphics[scale=1.25]{SegEx2.pdf} }\ \ \ \ \ \ \ \ \ \begin{array}{rcl}
s & = & [9,8] \\
\mathcal{D} & = & \{[9,2], [9,4], [6,8], [3,8], [7,8]\} \\
S([9,2]) & = & \{[6,2]\} \\
S([9,4]) & = & \{[6,4], [3,4]\} \\
S([6,8]) & = & \{[6,2], [6,4]\} \\
S([3,8]) & = & \{[3,4]\}\\
S([7,8]) & = & \emptyset.
\end{array}$$
\caption{The join irreducible biclosed set $B = J([9,8]_{\{[9,2], [9,4], [6,8], [3,8], [7,8]\}})$. One checks that $\widetilde{\lambda}(B\backslash\{[9,8]\}, B) = [9,8]_{\{[9,2], [9,4], [6,8], [3,8], [7,8]\}}$}% where the tree shown here is the same tree from Figure~\ref{seg_fig}.}
\label{label_fig}
\end{figure}

\begin{theorem}\label{cuproof}
The edge-labeling $\widetilde{\lambda}:\text{Cov}(\text{Bic}(T))\rightarrow \mathcal{S}_T$ is a CU-labeling of $\text{Bic}(T)$.
\end{theorem}

%\begin{proof}[Proof of Theorem \ref{cuproof}]
%We first verify axioms (CN1), (CN2), and (CN3). Note that it is enough verify these axioms for $\text{Bic}(T)$ since $\text{Bic}(T)$ is self-dual. 
%
%Let $B_1, B_2 \in \text{Bic}(T)$ and assume that they both cover some biclosed set $B$. By Theorem~\ref{bic_structure}, there exists $s, t \in \text{Seg}(T)$ such that $B_1 = B \sqcup \{s\}$ and $B_2 = B\sqcup \{t\}$. The interval $[B, B_1\vee B_2]$ has one of the two forms shown in Figure~\ref{polygons_fig} where the left (resp., right) figure occurs when $s$ and $t$ are not composable (resp., when $s$ and $t$ are composable). One deduces axioms (CN1), (CN2), and (CN3) from Figure~\ref{polygons_fig}.
%	
%We now verify axioms (CU1) and (CU2).
%	
%	(CU2): Consider two meet-irreducibles $M_1, M_2 \in \text{JI}(\text{Bic}(T))$ which are covered by $M_1^*$ and $M_2^*$, respectively. Assume for the sake of contradiction that $\widetilde{\lambda}(M_1, M_1^*)=\widetilde{\lambda}(M_2, M_2^*),$ and denote this label by $s_\mathcal{D}$. Thus $M_1^*=M_1\sqcup\{{s}\}$ and $M_2^*=M_2\sqcup\{{s}\}$. Note that $s \in M_1\vee M_2$ so there exists $s_1, \ldots, s_k \in M_1 \cup M_2$ such that $s = s_1\circ \cdots \circ s_k$. Without loss of generality, assume that $s_1 \in M_1,$ and that $s_{i} \in M_2$ if $s_{i-1} \in M_1$ and $s_i \in M_1$ if $s_{i-1} \in M_2$ for each $i = 1, \ldots, k$.
%
%Next, since $\widetilde{\lambda}(M_1, M_1^*)=\widetilde{\lambda}(M_2, M_2^*)$, sets $M_1$ and $M_2$ both contain the same split $s^\prime$ or $s^{\prime\prime}$ from a given break $\{s^\prime, s^{\prime\prime}\}$. We know that $s_1$ is a split of $s$ so $s_1 \in M_1 \cap M_2$. Since $s_2 \in M_2$, we know $s_1 \circ s_2 \in M_2$. Now $s_1\circ s_2$ is a split of $s$ so it follows that $s_1\circ s_2 \in M_1\cap M_2$. By continuing this argument, we obtain that $s \in M_1$, a contradiction.
%	
%	(CU1): Consider elements $J^1$ and $J^2$ of Bic($T$) which cover $J^1_*$ and $J^2_*$, respectively. Notice that $(J^1)^c$ and $(J^2)^c$ are meet-irreducibles of $\text{Bic}(T)$ which are covered by $(J^1_*)^c$ and $(J^2_*)^c$, respectively. If $\widetilde{\lambda}(J^1_*, J^1)=\widetilde{\lambda}(J^2_*, J^2)$ and we denote this label by $s_\mathcal{D}$, then $\widetilde{\lambda}((J^1)^c, (J^1_*)^c)=\widetilde{\lambda}((J^2)^c, (J^2_*)^c)$ and the common label is $s_{\mathcal{D}^\prime}$ where $\mathcal{D}\cap \mathcal{D}^\prime = \emptyset.$ This contradicts property (CU1).
%\end{proof}

%\begin{figure}
%$$\vcenteredhbox{\includegraphics[scale=1.2]{polygon1.pdf}} \ \ \ \ \ \ \vcenteredhbox{\includegraphics[scale=1.2]{polygon2.pdf}}$$
%\caption{{The two forms of the interval $[B,B_1\vee B_2]$ of $\text{Bic}(T)$. The labels on the covering relations as defined by the labeling $\widetilde{\lambda}: \text{Cov}(\text{Bic}(T)) \to \mathcal{S}_T$ are in blue. The set $\mathcal{D}^1$ (resp., $\mathcal{D}^2$) consists of all splits of $s$ (resp., $t$) belonging to $B$. Similarly, the set $\mathcal{D}$ consists of all splits of $s\circ t$ that belong to $B$.}}
%\label{polygons_fig}
%\end{figure}

\section{Canonical join complex of $\text{Bic}(T)$}\label{Sec_cjc}


In this section, we classify the join- and meet-irreducible biclosed sets. We do this by choosing a label $s_\mathcal{D} \in \mathcal{S}_T$ and constructing the minimal biclosed set $B$ where $\widetilde{\lambda}_\downarrow(B) = \{s_\mathcal{D}\}$ and by constructing the maximal biclosed set where $\widetilde{\lambda}^\uparrow(B) = \{s_\mathcal{D}\}.$

Given $s_\mathcal{D} \in \mathcal{S}_T$, define $$J(s_\mathcal{D}) := \{s\} \sqcup \mathcal{D} \sqcup \bigcup_{t \in \mathcal{D}} S(t)$$ where $S(t) = S(t,\mathcal{D}) \subset \text{Seg}(T)$ is defined to be the set of all splits $s^\prime$ of $t$ satisfying the following:

i) segment $s^\prime$ is not a split of $s$, and

ii) segment $s^\prime$ is not composable with any segment in $\mathcal{D}$ (see Figure~\ref{label_fig}).
 



\begin{proposition}\label{Prop_join_irr_des}
The set $J(s_\mathcal{D})$ satisfies $\widetilde{\lambda}_\downarrow(J(s_\mathcal{D})) = \{s_\mathcal{D}\}.$ Moreover, any biclosed set $B$ with $s_\mathcal{D} \in \widetilde{\lambda}_\downarrow(B)$ satisfies $J(s_\mathcal{D}) \le B$, and the reverse containment holds if and only if $\widetilde{\lambda}_{\downarrow}(B) = \{s_\mathcal{D}\}$. Consequently, the map $J: \mathcal{S}_T \to \text{JI}(\text{Bic}(T))$ is a bijection.
\end{proposition}


Next, we classify the meet-irreducible biclosed sets. Given $s_\mathcal{D} \in \mathcal{S}_T$, define $$M(s_\mathcal{D}) := J(s_\mathcal{D})\backslash\{s\} \sqcup \{t \in \text{Seg}(T): \ \text{$t \not \subseteq s$}\}\sqcup \bigcup_{t \in \mathcal{D}}R(t)$$ where $R(t) = R(t,\mathcal{D}) \subset \text{Seg}(T)$ is defined to be the set of all splits $s^\prime$ of $t$ satisfying the following:

i) segment $s^\prime$ is not a split of $s$, and 

ii) segment $s^\prime$ is composable with some element of $\mathcal{D}$. 

Observe that if $s^\prime \in R(t)$ for some $t \in \mathcal{D}$, then there is necessarily a unique element $t^\prime \in \mathcal{D}$ where $t^\prime \subseteq t$, $t^\prime$ is composable with $s^\prime$, and $s^\prime\circ t^\prime = t$.

\begin{proposition}\label{Prop_meet_irr_des}
The set $M(s_\mathcal{D})$ satisfies $\widetilde{\lambda}^\uparrow(M(s_\mathcal{D})) = \{s_\mathcal{D}\}.$ Moreover, any biclosed set $B$ with $s_\mathcal{D} \in \widetilde{\lambda}^\uparrow(B)$ satisfies $B \le M(s_\mathcal{D})$, and the reverse containment holds if and only if $\widetilde{\lambda}^{\uparrow}(B) = \{s_\mathcal{D}\}$. Consequently, the map $M: \mathcal{S}_T \to \text{MI}(\text{Bic}(T))$ is a bijection.
\end{proposition}


We now describe the faces of the canonical join complex of $\text{Bic}(T)$. In \cite[Theorem 1.1]{barnard2016canonical}, it is shown that the canonical join complex of a finite semidistributive lattice $L$ is a \textbf{flag complex}. That is, the minimal nonfaces of $\Delta^{CJ}(L)$ have size two. Thus, it is enough to determine the pairs of elements of $\text{JI}(\text{Bic}(T))$ that join canonically.

\begin{theorem}\label{Thm_CJC}
A collection $\{J(s^1_{\mathcal{D}^1}), \ldots, J(s^k_{\mathcal{D}^k})\} \subset \text{JI}(\text{Bic}(T))$ is a face of $\Delta^{CJ}(\text{Bic}(T))$ if and only if labels $s^i_{\mathcal{D}^i}$ and $s^j_{\mathcal{D}^j}$ satisfy the following:

1) segments $s^i$ and $s^j$ are distinct,

2) neither $s^i$ nor $s^j$ is a composition of at least two segments in $J(s^i_{\mathcal{D}^i})\cup J(s^j_{\mathcal{D}^j})$, and

3) neither $J(s^i_{\mathcal{D}^i}) \le J(s^{j}_{\mathcal{D}^j})$ nor $J(s^{j}_{\mathcal{D}^j}) \le J(s^i_{\mathcal{D}^i})$.

\noindent for any distinct $i,j \in \{1,\ldots, k\}$.
\end{theorem}


\begin{figure}
$$\includegraphics[scale=.8]{arcEx2prime_v2.pdf}$$
\caption{A tree $T$ and the canonical join complex $\Delta^{CJ}(\text{Bic}(T))$. The shortest segments of $T$ are labeled $a$, $b$, and $c$.}
\label{CJ_complex_figure}
\end{figure}

Let $T$ be the tree shown in Figure~\ref{CJ_complex_figure}. In this same figure, we show the canonical join complex of the biclosed sets of $T$. Each join-irreducible of $\text{Bic}(T)$ is written next to its corresponding vertex of $\Delta^{CJ}(\text{Bic}(T))$.



\section{The shard intersection order of $\text{Bic}(T)$}\label{shard_section} Let $B \in \text{Bic}(T)$ be a biclosed set that covers exactly the following biclosed sets: $B_1,$ $B_2,\ldots,$ $B_k$. Let ${s_i}_{\mathcal{D}_i} = \widetilde{\lambda}(B_i,B)$ for $i=1, \ldots, k$ and $\lambda_\downarrow(B) =\{s_1, s_2,\ldots,s_k\}$ where $s_i = \lambda(B_i,B)$ for $i = 1, \ldots, k$. Now fix a segment $s \in\overline{\lambda_{\downarrow}(B)}$ expressed as $s = s_{i_1}\circ s_{i_2}\circ \ldots\circ s_{i_\ell}$ with each $s_{i_j}\in \lambda_{\downarrow}(B)$. If $t \in \text{Seg}(T)$ is a split of $s$ that can be expressed as either $t = s_{i_1}\circ \cdots \circ s_{i_j}$ for some $j = 1, \ldots, \ell-1$ or $t = s_{i_j}\circ \cdots \circ s_{i_\ell}$ for some $j = 2,\ldots, \ell$, we say that $t$ is a \textbf{faultline split} of $s$. Otherwise, we say that $t$ is a \textbf{non-faultline split} of $t$. %Additionally, we refer to $\{s_{i_1}\circ\cdots \circ s_{i_j}, s_{i_{j+1}}\circ \cdots \circ s_{i_\ell}\}$ as a \textbf{faultine break}.


\begin{theorem}\label{psib}
	Given a biclosed set $B \in \text{Bic}(T)$, we have that $\psi(B)$ is the set of all labels of the form $(s_{i_1}\circ s_{i_2}\circ \cdots\circ s_{i_\ell})_{\mathcal{D}}$ with $s_{i_j} \in \lambda_{\downarrow}(B)$ where $s_{i_1}\circ s_{i_2}\circ \cdots\circ s_{i_\ell}$ is any element of $\overline{\lambda_{\downarrow}(B)}$ and where $\mathcal{D}$ is any set of segments that satisfies the following properties: 

(i) $|\mathcal{D}| = |\{\text{breaks of } s_{i_1}\circ s_{i_2}\circ \cdots\circ s_{i_\ell}\}|,$

(ii) each segment $t \in \mathcal{D}$ is a split of $s_{i_1}\circ \cdots \circ s_{i_\ell}$,

(iii) no two distinct splits $t_1, t_2 \in \mathcal{D}$ appear in the same break of $s_{i_1}\circ \cdots \circ s_{i_\ell}$, and

(iv) {whenever $t \in \mathcal{D}$ is a non-faultline split of $s_{i_1}\circ \cdots \circ s_{i_\ell}$, we have that $t = s_{i_1} \circ \cdots \circ s_{i_{j-1}}\circ t_j$ for some $j = 1,\ldots, \ell$ and some $t_j \in \mathcal{D}_{i_j}$ or $t = t_j \circ s_{i_{j+1}}\circ \cdots \circ s_{i_{\ell}}$ for some $j = 1,\ldots, \ell$ and some $t_j \in \mathcal{D}_{i_j}$. In the former case if $j = 1$, we mean $t = t_1$, and in the latter case, if $j = \ell$, we mean $t = t_\ell$.}
\end{theorem}

\begin{figure}
\centering
\includegraphics[scale=1]{cycle_trees.pdf}
\caption{The trees from Example~\ref{shard_calc_ex}.}
\label{cycle_trees}
\end{figure}


\begin{example}\label{shard_calc_ex}
Let $T$ be the tree on the left in Figure~\ref{cycle_trees}, and let $T^\prime$ be the tree on the right. In Figure~\ref{shard_ex_a3}, we show the shard intersection order of $\text{Bic}(T)$. The atoms in this lattice are the 9 labels in $\mathcal{S}_T$. The presence of a dashed segment $s$ indicates that both labels $s_\mathcal{D}$ and $s_{\mathcal{D}^\prime}$ belong to the corresponding set $\psi(B)$. This indicates that given $B \in \text{Bic}(T)$, one has $$|\psi(B)| = |\{\text{dark red segments in $\psi(B)$}\}| + 2|\{\text{dashed segments in $\psi(B)$}\}|.$$

The shard intersection order of biclosed sets fails to be graded in general, although $\Psi(\text{Bic}(T))$ is graded of rank 3. For more information, we refer the reader to \cite[Example 5.2]{clifton2017canonical}. The trees $T$ and $T^\prime$ in this example belong to the one parameter family of trees that are completely determined by the choice of degree on their central vertex. In \cite[Conjecture 6.13]{cliftondillery}, the first and second authors conjectured that for this one parameter family of trees $\Psi(\text{Bic}(T))$ is graded if and only if $n$ is odd.\end{example}


\begin{figure}
\centering
\rotatebox{270}{\includegraphics[scale=1.35]{redgreenarcs_v3.pdf}}
\caption{The shard intersection order $\Psi(\text{Bic}(T))$ when $T$ is the tree on the left in Figure~\ref{cycle_trees}.}
\label{shard_ex_a3}
\end{figure}





%We use the following lemma in the proof of Theorem~\ref{psib}.
%
%\begin{lemma}\label{Lemma_facial_int_atoms}
%Given a biclosed $B \in \text{Bic}(T)$ where $B_1, \ldots, B_k \in \text{Bic}(T)$ are all of the biclosed sets covered by $B$ and where $s_j = \lambda(B_j, B)$ for $j = 1, \ldots, k$, we have that $B^\prime \in \left[\bigwedge_{i = 1}^k B_i, B\right]$ if and only if $B^\prime = (\bigwedge_{i = 1}^k B_i) \sqcup S^\prime$ where $S^\prime$ and $\overline{\{s_1,\ldots, s_k\}}\backslash S^\prime$ are closed subsets of $\overline{\{s_1,\ldots, s_k\}}$. 
%\end{lemma}
%\begin{proof}
%By \cite[Proposition 4.2]{garver2016oriented}, it is enough to show that $(\bigwedge_{i = 1}^k B_i)\sqcup \{s_j\}$ is a biclosed set for any $j = 1, \ldots, k.$ Note that $B = (\bigwedge_{i = 1}^k B_i) \sqcup \overline{\{s_1, \ldots, s_k\}}$ by Lemma~\ref{meet_of_B_is}.
%
%
%
%To show that $(\bigwedge_{i = 1}^k B_i)\sqcup \{s_j\}$ is closed, it is enough to show that if $s \in \bigwedge_{i = 1}^k B_i$ and segments $s$ and $s_j$ are composable, then $s \circ s_j \in (\bigwedge_{i = 1}^k B_i) \sqcup \{s_j\}$. Since $B$ is closed and $B = (\bigwedge_{i = 1}^k B_i) \sqcup \overline{\{s_1, \ldots, s_k\}}$, we have that $s \circ s_j \in B$. Therefore, we can assume that $s \circ s_j \not \in \bigwedge_{i = 1}^k B_i$. Write $s\circ s_j = s_{i_1}\circ \cdots \circ s_{i_\ell}$ where $s_{i_1}, \ldots, s_{i_\ell} \in \{s_1, \ldots, s_k\}$, and we can assume that $\ell \ge 2$. By Lemma~\ref{nosplit}, we see that $s_j = s_{i_1}$ or $s_j = s_{i_\ell}$. We conclude that $s \in \overline{\{s_1, \ldots, s_k\}}$, a contradiction.
%
%Next, we show that $((\bigwedge_{i = 1}^k B_i)\sqcup \{s_j\})^c$ is closed. Note that $((\bigwedge_{i = 1}^k B_i)\sqcup \{s_j\})^c = (\bigvee_{i = 1}^k B_i^c)\backslash\{s_j\}$ by Lemma~\ref{Lemma_meet_in_bic}. It is clear that if $s, t \in (\bigvee_{i = 1}^k B_i^c)\backslash\{s_j\}$ are composable, then $s\circ t \in (\bigvee_{i = 1}^k B_i^c)$. Thus, it remains to show that $s\circ t \neq s_j$. 
%
%Suppose $s\circ t = s_j$. If $s, t \in  (\bigvee_{i = 1}^k B_i^c)\backslash\overline{\{s_1,\ldots, s_k\}}$, then since $(\bigvee_{i = 1}^k B_i^c)\backslash\overline{\{s_1,\ldots, s_k\}}$ is a closed set, we know that $s\circ t \in (\bigvee_{i = 1}^k B_i^c)\backslash\overline{\{s_1,\ldots, s_k\}}$. However, this contradicts that $s\circ t = s_j$. We conclude that $s \in \overline{\{s_1,\ldots, s_k\}}$ or $t \in \overline{\{s_1,\ldots, s_k\}}$. We assume $s \in \overline{\{s_1,\ldots, s_k\}}$, without loss of generality. Write $s = s_{i_1}\circ \cdots \circ s_{i_\ell}$ for some $s_{i_1}, \ldots, s_{i_\ell} \in \{s_1,\ldots, s_k\}$. Since $s\circ t = s_j$, we have $s_{i_1}\circ \cdots \circ s_{i_\ell} \circ t = s_j$. This implies that $s_{i_1}$ is a split of $s_j$, and this contradicts Lemma~\ref{nosplit}.
%\end{proof}

%\begin{proof}[Proof of Theorem~\ref{psib}]
%
%
%We know that the set $\psi(B)$ consists of labels of the form $(s_{i_1}\circ \cdots \circ s_{i_\ell})_{\mathcal{D}}$ where $\mathcal{D}$ is some subset of $\text{Seg}(T)$ and $s_{i_1},\ldots, s_{i_\ell} \in \lambda_\downarrow(B)$. Since any biclosed set $B^\prime \in \left[\bigwedge_{i=1}^k B_i, B\right]$ is of the form $\bigwedge_{i=1}^k B_i \sqcup S^\prime$ where $S^\prime$ is some biclosed subset of $\overline{\{s_1,\ldots, s_k\}}$, the segment $s_{i_1}\circ \cdots \circ s_{i_\ell}$ may be any element of $\overline{\{s_1,\ldots, s_k\}}$. Moreover, we know that any two sets $\mathcal{D}$ and $\mathcal{D}^\prime$ appearing on labels $(s_{i_1}\circ \cdots \circ s_{i_\ell})_{\mathcal{D}}$ and $(s_{i_1}\circ \cdots \circ s_{i_\ell})_{\mathcal{D}^\prime}$ contain the same non-faultline splits. 
%
%It remains to show that given a segment $s_{i_1}\circ \cdots \circ s_{i_\ell}$, the subsets $\mathcal{D}$ appearing on labels $(s_{i_1}\circ \cdots \circ s_{i_\ell})_{\mathcal{D}}$ are exactly those sets of splits of $s_{i_1}\circ \cdots \circ s_{i_\ell}$ with the properties appearing in the statement of the theorem. 
%
%First, given a label $(s_{i_1}\circ \cdots \circ s_{i_\ell})_{\mathcal{D}} \in \psi(B)$, we show that $\mathcal{D}$ has the desired properties. It is clear from the definition of $\widetilde{\lambda}$ that $\mathcal{D}$ satisfies properties (i), (ii), and (iii). Thus, we proceed by induction on $\ell$ to prove property (iv). Assume that $\ell = 1$ so we consider labels of the form $(s_{i_1})_\mathcal{D}$. Observe that there are no nontrivial faultline splits of $s_{i_1}$ so $\mathcal{D}$ contains no faultline splits. If $({s_{i_1}})_{\mathcal{D}_{i_1}}$ is another label appearing on a covering relation in $\left[\bigwedge_{i=1}^k B_i, B\right]$, we have $|\mathcal{D}| = |\mathcal{D}_{i_1}|$. Thus $\mathcal{D} = \mathcal{D}_{i_1}$.
%
%Next, suppose that for any label $(s_{j_1}\circ \cdots \circ s_{j_{\ell^\prime}})_{\mathcal{D}^\prime} \in \psi(B)$ where $s_{j_1},  \ldots, s_{j_{\ell^\prime}} \in \{s_{i_1}, \ldots, s_{i_\ell}\}$ and $\ell^\prime < \ell$ the subset $\mathcal{D}^\prime$ satisfies property (iv). Now consider $(s_{i_1}\circ \cdots \circ s_{i_\ell})_\mathcal{D} \in \psi(B)$ and let $t \in \mathcal{D}$ be a non-faultline split of $s_{i_1}\circ \cdots \circ s_{i_\ell}$. Without loss of generality, $t = s_{i_1}\circ \cdots \circ s_{i_{j-1}}\circ t_j$. Since $t_j$ is a non-faultline split of $s_{i_j}$, by induction, $t_j \in \mathcal{D}_{i_j}$. 
%
%To prove the converse, it is enough to show that any family $\mathcal{F} \subset \overline{\{s_{i_{1}}, \ldots, s_{i_\ell}\}}$ of $\ell-1$ faultline splits of $s_{i_1}\circ \cdots \circ s_{i_\ell}$ satisfying the properties in the statement of the theorem belongs to a set $\mathcal{D}$ that appears on some label $(s_{i_1}\circ \cdots \circ s_{i_\ell})_\mathcal{D}$. Given $f = s_{i_1}\circ \cdots \circ s_{i_j} \in \mathcal{F}$, define $$D(f) := \{f =s_{i_1}\circ \cdots \circ s_{i_j}, s_{i_2}\circ \cdots \circ s_{i_j}, \ldots, s_{i_{j-1}} \circ s_{i_j}, s_{i_j} \}.$$ The set $D(f)$ is defined analogously when $f = s_{i_j}\circ \cdots \circ s_{i_\ell}$. By construction, $D(f)$ and $\overline{\{s_{i_{1}}, \ldots, s_{i_\ell}\}}\backslash D(f)$ are closed subsets of $\overline{\{s_{i_{1}}, \ldots, s_{i_\ell}\}}$. 
%
%Now observe that $\overline{\bigcup_{f \in \mathcal{F}} D(f)}$ and $\overline{\{s_{i_{1}}, \ldots, s_{i_\ell}\}}\backslash \overline{\bigcup_{f \in \mathcal{F}} D(f)}$ are closed subsets of $\overline{\{s_{i_{1}}, \ldots, s_{i_\ell}\}}$. This implies that $\overline{\bigcup_{f \in \mathcal{F}} D(f)} \sqcup \bigwedge_{i=1}^k B_i \in \left[\bigwedge_{i=1}^k B_i, B\right]$ is a biclosed set. Moreover, the only (necessarily) faultline splits of $s_{i_1}\circ \cdots \circ s_{i_\ell}$ in $\overline{\bigcup_{f \in \mathcal{F}} D(f)}$ are the elements of $\mathcal{F}$. Thus, setting $B^{1} := \overline{\bigcup_{f \in \mathcal{F}} D(f)} \sqcup \bigwedge_{i=1}^k B_i$ and $B^{2} := B^{1} \sqcup \{s_{i_1}\circ \cdots \circ s_{i_\ell}\}$, we obtain that $\widetilde{\lambda}(B^{1}, B^{2}) = (s_{i_1}\circ \cdots \circ s_{i_\ell})_{\mathcal{F} \sqcup \mathcal{N}}$ where $\mathcal{N}$ is the set of non-faultline splits of $s_{i_1}\circ \cdots \circ s_{i_\ell}$ that appears in every label of the form $(s_{i_1}\circ \cdots \circ s_{i_\ell})_\mathcal{D}$.\end{proof}


%As a consequence of Theorem~\ref{psib}, we now show that the segment $s$ appearing in a label $s_\mathcal{D} \in \psi(B)$ may be expressed in a unique way as a composition of segments in $\{s_1, \ldots, s_k\}$. 

%\begin{lemma}\label{Lemma_ker_coker}
%Given $B \in \text{Bic}(T)$ and any $s_{\mathcal{D}} \in \psi(B)$, there is a unique way to express $s$ as a composition of some subset of the segments $s_i := \lambda(B_i, B)$ with $i = 1, \ldots, k$ where $B_1, \ldots, B_k$ are all of the biclosed sets covered by $B$. In particular, for any $s_{\mathcal{D}}, t_{\mathcal{D}^\prime} \in \psi(B)$ where $s = t\circ t^\prime$, we have that $t^\prime_{\mathcal{D}^{\prime\prime}} \in \psi(B)$ for some $\mathcal{D}^{\prime\prime} \subset \text{Seg}(T)$.
%\end{lemma}
%\begin{proof}
%We know from Theorem~\ref{psib} that $s$ may be expressed as a composition of the elements $s_i := \lambda(B_i, B)$. Now suppose $s = s_{i_1}\circ \cdots \circ s_{i_l}$ and $s = s_{j_1} \circ \cdots \circ s_{j_m}$ for some $s_{i_1},\ldots, s_{i_1}, s_{j_1}, \ldots, s_{j_m} \in \{s_1, \ldots, s_k\}$. If $s_{i_l} \subseteq s_{j_m}$, then $s_{j_m} = s_{i_l} \circ t^\prime$ for some segment $t^\prime \in \text{Seg}(T)$. However, such an equation contradicts Lemma~\ref{nosplit}. The analogous argument shows $s_{j_m}$ is not properly contained in $s_{i_l}$. We conclude that $s_{j_m} = s_{i_l}$. By repeating this argument and removing pairs of equal segments $s_{j_n} = s_{i_r}$ with $n \le m$ and $r \le l$, we either obtain an equation $s_{j_1}\circ \cdots \circ s_{j_{n-1}} = s_{i_1}$ or $s_{j_1} = s_{i_1}\circ \cdots \circ s_{i_{m-1}}$. In either case, we reach a contradiction.
%\end{proof}


%\begin{remark}\label{remark_same_N}
%As is mentioned in the proof of Theorem~\ref{psib}, given any two labels $s_\mathcal{D}, s_{\mathcal{D}^\prime} \in \psi(B)$, the sets $\mathcal{D}$ and $\mathcal{D}^\prime$ contain exactly the same non-faultline splits of $s$. In addition, for any biclosed set $B \in \text{Bic}(T)$, the set $\widetilde{\lambda}_{\downarrow}(B)$ consists of exactly the elements $s_\mathcal{D} \in \psi(B)$ with the property that there does not exist $s^\prime_{\mathcal{D}^\prime} \in \psi(B)$ where $s = s^\prime \circ t^\prime$ for some $t^\prime \in \text{Seg}(T)$.
%\end{remark}


%\section{Lattice structure of the shard intersection order}\label{shard_section_2} 

We arrive at our main theorem.

\begin{theorem}\label{Thm_lattice_prop}
The shard intersection order $\Psi(\text{Bic}(T))$ is a lattice.
\end{theorem}

Since $\Psi(\text{Bic}(T))$ is a finite poset whose unique maximal element is $\psi(\text{Seg}(T)) = \mathcal{S}_T$, it is enough to show that $\psi(B) \cap \psi(B^\prime) \in \Psi(\text{Bic}(T))$ for any $B, B^\prime \in \text{Bic}(T)$. We sketch our approach below.

Let $\psi(B), \psi(B^\prime) \in \Psi(\text{Bic}(T))$ and let $B_1, \ldots, B_k \in \text{Bic}(T)$ (resp., $B^\prime_1, \ldots, B^\prime_l \in \text{Bic}(T)$) be all of the biclosed sets covered by $B$ (resp., $B^\prime$). As above, set $s_j := \lambda(B_j,B)$ for $j = 1, \ldots, k$ and $t_j := \lambda(B^\prime_j, B^\prime)$ for $j = 1, \ldots, l$. To show that $\psi(B)\cap \psi(B^\prime) \in \Psi(\text{Bic}(T))$ we construct a biclosed set $B^{\prime\prime}$ that satisfies $\psi(B^{\prime\prime}) = \psi(B) \cap \psi(B^\prime)$. With this goal in mind, we let $\{s^{(i)}_{\mathcal{D}^{(i)}}\}_{i = 1}^\ell$ denote the elements of $\psi(B)\cap \psi(B^\prime)$ where $s^{(i)}$ appears in exactly one label in $\psi(B)\cap \psi(B^\prime)$. We can prove that such a collection of labels exists, and thus it is unique. After that, we show that $\bigvee_{i=1}^\ell J(s^{(i)}_{\mathcal{D}^{(i)}})$ is a canonical join representation using Theorem~\ref{Thm_CJC}. We then use Theorem~\ref{psib} to prove that $\psi\left(\bigvee_{i=1}^\ell J(s^{(i)}_{\mathcal{D}^{(i)}})\right) = \psi(B)\cap \psi(B^\prime)$.

%Consider $\psi(B) \cap \psi(B^\prime)$, and let $\text{Seg}(\psi(B)\cap \psi(B^\prime))$ denote the set of segments that appear in some label in the set $\psi(B)\cap \psi(B^\prime)$. 

%We will prove that $\Psi(\text{Bic}(T))$ is a lattice by showing that $\psi(B) \cap \psi(B^\prime)$ is the meet of $\psi(B)$ and $\psi(B^\prime)$. This also shows that $\Psi(\text{Bic}(T))$ is a meet-subsemilattice of the Boolean lattice on the elements of $\mathcal{S}_T$. %We first use Lemma~\ref{Lemma_ker_coker} and Remark~\ref{remark_same_N} to make the following important observation.

%\begin{lemma}\label{intersection_ker_coker}
%If $s_\mathcal{D}, s^\prime_{\mathcal{D}^\prime} \in \psi(B) \cap \psi(B^\prime)$ with $s = s^\prime \circ t^{\prime\prime}$ for some $t^{\prime\prime} \in \text{Seg}(T)$, then there exists $\mathcal{D}^{\prime\prime} \subset \text{Seg}(T)$ such that $t^{\prime\prime}_{\mathcal{D}^{\prime\prime}} \in \psi(B)\cap \psi(B^\prime)$.
%\end{lemma}
%\begin{proof}
%By Lemma~\ref{Lemma_ker_coker}, we know that there exists $\mathcal{D}^1, \mathcal{D}^2 \subset \text{Seg}(T)$ such that $t^{\prime\prime}_{\mathcal{D}^1} \in \psi(B)$ and $t^{\prime\prime}_{\mathcal{D}^2} \in \psi(B^\prime).$ By Remark~\ref{remark_same_N}, this implies that $s^\prime$ and $t^{\prime\prime}$ are faultline splits of $s = s_{i_1}\circ \cdots \circ s_{i_n}$ and $s = t_{j_1}\circ \cdots \circ t_{j_m}$. 
%
%Now define $$\mathcal{D}^{\prime\prime} := \left\{\begin{array}{lcl} t^* & : & \text{if $t^* \in \mathcal{D}$ and $t^*$ is a proper subsegment of $t^{\prime\prime}$,}\\
%t^* & : & \text{if $s^\prime \circ t^{*} \in \mathcal{D}$ where $t^*$ is a split of $t^{\prime\prime}$.} \end{array}\right.$$ The splits in $\mathcal{D}^{\prime\prime}$ satisfy the properties in the statement of Theorem~\ref{psib} with respect to both $\psi(B)$ and $\psi(B^\prime)$, since the splits in $\mathcal{D}$ satisfy these. Thus $t^{\prime\prime}_{\mathcal{D}^{\prime\prime}} \in \psi(B) \cap \psi(B^\prime)$.
%\end{proof}

%To show that $\psi(B)\cap \psi(B^\prime) \in \Psi(\text{Bic}(T))$ we must construct a biclosed set $B^{\prime\prime}$ that satisfies $\psi(B^{\prime\prime}) = \psi(B) \cap \psi(B^\prime)$. With this goal in mind, we let $\{s^{(i)}_{\mathcal{D}^{(i)}}\}_{i = 1}^\ell$ denote the elements of $\psi(B)\cap \psi(B^\prime)$ where $s^{(i)}$ appears in exactly one label in $\psi(B)\cap \psi(B^\prime)$. If such a collection of labels exists, it is unique. We prove that such a collection exists in the following lemma. After that, we work toward showing that $\psi\left(\bigvee_{i=1}^\ell J(s^{(i)}_{\mathcal{D}^{(i)}})\right) = \psi(B)\cap \psi(B^\prime)$, which is the content of Lemma~\ref{meet_in_Psi}.

%\begin{lemma}\label{nonempty_s^i}
%The collection of elements $\{s^{(i)}_{\mathcal{D}^{(i)}}\}_{i = 1}^\ell \subset \psi(B) \cap \psi(B^\prime)$ is nonempty. In particular, the segments $s^{(1)}, \ldots, s^{(\ell)}$ are pairwise distinct.
%\end{lemma}
%\begin{proof}
%Suppose that any segment in $\text{Seg}(\psi(B)\cap \psi(B^\prime))$ belongs to at least two labels of $\psi(B) \cap \psi(B^\prime)$. Let $s_\mathcal{D}$ and $s_{\mathcal{D}^\prime}$ be minimal elements of the subposet $\psi(B)\cap \psi(B^\prime)$ of $\mathcal{S}_T$. Since $\mathcal{D} \neq \mathcal{D}^\prime$, there exists $t \in \mathcal{D}$ and $t^\prime \in \mathcal{D}^\prime$ such that $s = t \circ t^\prime$. From Remark~\ref{remark_same_N}, segments $t$ and $t^\prime$ are faultline splits of $s$ in terms of its expression as a composition of segments in $\{s_1,\ldots, s_k\}$. Segments $t$ and $t^\prime$ are also faultline splits of $s$ in terms of its expression as a composition of segments in $\{t_1, \ldots, t_l\}$. In particular, $t, t^\prime \in \overline{\{s_1, \ldots, s_k\}}\cap \overline{\{t_1, \ldots, t_l\}}.$
%
%We now show that $t_{\mathcal{D}(t)} \in \psi(B)\cap \psi(B^\prime)$ where $\mathcal{D}(t) := \{s^\prime \in J(s_\mathcal{D}): \text{$s^\prime$ is a split of $t$}\}.$ This contradicts that $s_\mathcal{D}$ is a minimal element of the subposet $\psi(B)\cap \psi(B^\prime)$ and therefore completes the proof. We show that $t_{\mathcal{D}(t)} \in \psi(B)$ and an analogous argument shows that $t_{\mathcal{D}(t)} \in \psi(B^\prime)$ so we omit the latter.
%
%First, consider the biclosed sets $$B^1 := \left(\bigwedge_{i=1}^k B_i\right) \vee (J(s_\mathcal{D})\backslash \{s\}), \ B^2 := \left(\bigwedge_{i=1}^k B_i\right) \vee (J(t_{\mathcal{D}(t)})\backslash \{t\}) \in \left[\bigwedge_{i=1}^k B_i, B\right].$$ We know that $B^2 \in \left[\bigwedge_{i=1}^k B_i, B\right]$ since $J(t_{\mathcal{D}(t)})\backslash \{t\} < J(s_\mathcal{D})\backslash \{s\}$ by Lemma~\ref{Lemma_join_irr_containment}. Note that $t \not \in B^2$ since $t \in \overline{\{s_1, \ldots, s_k\}}\cap \overline{\{t_1, \ldots, t_l\}}.$  We show that $\widetilde{\lambda}(B^2, B^2 \sqcup \{t\}) = t_{\mathcal{D}(t)}$ by showing that $B^2 \sqcup \{t\}$ is biclosed. From this it follows that $t_{\mathcal{D}(t)} \in \psi(B)$.
%
%By construction, $B^2\sqcup \{t\}$ is coclosed. Since $B^2$ is closed, we show that $B^2\sqcup \{t\}$ is closed by showing that, without loss of generality, for any $t^1 \in \bigwedge_{i = 1}^k B_i$ that is composable with $t$, we have $t\circ t^1 \in \bigwedge_{i = 1}^k B_i$. We know that $t, t^1 \in B$ so $t \circ t^1 \in B.$ 
%
%Now, suppose that $t\circ t^1 \in \overline{\{s_1,\ldots, s_k\}}$. This means there exists $\mathcal{D}^{*} \subset \text{Seg}(T)$ such that $(t \circ t^1)_{\mathcal{D}^*} \in \psi(B)$. By Lemma~\ref{Lemma_ker_coker}, we know that there is a unique way to express $t,$ and $t\circ t^1$ as a composition of elements of $\overline{\{s_1,\ldots, s_k\}}$. It follows that $t^1 \in \overline{\{s_1, \ldots, s_k\}}$, but this contradicts that $t^1 \in \bigwedge_{i = 1}^k B_i$. \end{proof}


%The following proposition is a crucial step in proving the lattice property of $\Psi(\text{Bic}(T))$. The reader should compare the statement with the characterization of elements of $\widetilde{\lambda}_{\downarrow}(B)$ in Remark~\ref{remark_same_N}.


%\begin{proposition}\label{Lemma_property_of_s_i}
%The labels $\{s^{(i)}_{\mathcal{D}^{(i)}}\}_{i = 1}^\ell$ are precisely the elements $s_\mathcal{D} \in \psi(B)\cap \psi(B^\prime)$ with the property that there does not exist $s^\prime_{\mathcal{D}^\prime} \in \psi(B) \cap \psi(B^\prime)$ where $s = s^\prime \circ t^\prime$ for some $t^\prime \in \text{Seg}(T)$.
%\end{proposition}
%\begin{proof}
%Suppose there exists a label $s^\prime_{\mathcal{D}^\prime} \in \psi(B) \cap \psi(B^\prime)$ such that $s^{(i)} = s^\prime \circ t^\prime$ for some $t^\prime \in \text{Seg}(T)$. By Lemma~\ref{intersection_ker_coker}, there exists $\mathcal{D}^{\prime\prime} \subset \text{Seg}(T)$ such that $t^{\prime}_{\mathcal{D}^{\prime\prime}} \in \psi(B)\cap \psi(B^\prime)$. By Remark~\ref{remark_same_N}, $s^\prime$ and $t^\prime$ are faultline splits of $s^{(i)}$ regardless of whether this segment is expressed in terms of $s_1, \ldots, s_k$ or $t_1, \ldots, t_l$. Without loss of generality, we assume that $s^\prime \in \mathcal{D}^{(i)}$. Thus $s^{(i)}_{\mathcal{D}^{(i)}}, s^{(i)}_{(\mathcal{D}^{(i)}\backslash\{s^\prime\}) \sqcup \{t^\prime\}} \in \psi(B)\cap \psi(B^\prime)$. However, this contradicts the definition of $s^{(i)}_{\mathcal{D}^{(i)}}$.
%
%Let $s_{\mathcal{D}} \in \psi(B)\cap \psi(B^\prime)$ be a label with the stated property. Suppose there are two distinct labels $s_{\mathcal{D}}, s_{\mathcal{D}^\prime} \in \psi(B)\cap \psi(B^\prime)$. Then there exist $t \in \mathcal{D}$ and $t^\prime \in \mathcal{D}^\prime$ such that $s = t \circ t^\prime$. We claim that $t_{\mathcal{D}(t)}, t^\prime_{\mathcal{D}^\prime(t^\prime)} \in  \psi(B)\cap \psi(B^\prime)$ where $\mathcal{D}(t) := \{s^\prime \in J(s_\mathcal{D}): \text{$s^\prime$ is a split of $t$}\}$ and $\mathcal{D}^\prime(t^\prime) := \{s^\prime \in J(s_{\mathcal{D}^\prime}): \text{$s^\prime$ is a split of $t^\prime$}\}.$ This claim contradicts that $s_{\mathcal{D}}$ has the property in the statement of the lemma. Thus $s$ appears in exactly one label in $\psi(B)\cap \psi(B^\prime)$. Therefore, $s = s^{(i)}$ and $\mathcal{D}= \mathcal{D}^{(i)}$ for some $i = 1, \ldots, \ell$.
%
%One proves our claim by adapting the proof of Lemma~\ref{nonempty_s^i}.
%\end{proof}


%\begin{lemma}\label{Lemma_seg_of_int}
%$\text{Seg}(\psi(B)\cap \psi(B^\prime)) = \overline{\{s^{(1)}, \ldots, s^{(\ell)}\}}$. 
%\end{lemma}
%\begin{proof}
%To show that $\overline{\{s^{(1)}, \ldots, s^{(\ell)}\}} \subset \text{Seg}(\psi(B)\cap \psi(B^\prime))$, it is enough to show that the composition of any two composable segments $s^{(i)}$ and $s^{(j)}$ belongs to $\text{Seg}(\psi(B)\cap \psi(B^\prime))$. We know that $s^{(i)}\circ s^{(j)} \in \overline{\{s_1, \ldots, s_k\}} \cap \overline{\{t_1, \ldots, t_l\}}$ since this is true for $s^{(i)}$ and $s^{(j)}$. Now from the description of $\psi(B)$ and $\psi(B^\prime)$ obtained in Theorem~\ref{psib}, we see that $s^{(i)}\circ s^{(j)} \in \text{Seg}(\psi(B)\cap \psi(B^\prime))$.
%
%To prove the opposite containment, let $s \in \text{Seg}(\psi(B) \cap \psi(B^\prime))$, and assume that any segment in $\text{Seg}(\psi(B)\cap \psi(B^\prime))$ that is shorter than $s$ belongs to $\overline{\{s^{(1)}, \ldots, s^{(\ell)}\}}$. Since $s \in \text{Seg}(\psi(B) \cap \psi(B^\prime))$, there exists $\mathcal{D} \subset \text{Seg}(T)$ such that $s_{\mathcal{D}} \in \psi(B) \cap \psi(B^\prime)$.
%
%We can further assume that $s \neq s^{(i)}$ for any $i = 1, \ldots, \ell$. By Proposition~\ref{Lemma_property_of_s_i}, this implies that there exists $s^{\prime}_{\mathcal{D}^{\prime}} \in \psi(B)\cap \psi(B^\prime)$ where $s = s^{\prime}\circ t^{\prime\prime}$ for some $t^{\prime\prime} \in \text{Seg}(T)$. By Lemma~\ref{intersection_ker_coker}, there exists $\mathcal{D}^{\prime\prime} \subset \text{Seg}(T)$ such that $t^{\prime\prime}_{\mathcal{D}^{\prime\prime}} \in \psi(B)\cap \psi(B^\prime)$. Since $s^\prime$ and $t^{\prime\prime}$ are both shorter than $s$, we have that $s^\prime, t^{\prime\prime} \in \overline{\{s^{(1)}, \ldots, s^{(\ell)}\}}$ so $s = s^\prime\circ t^{\prime\prime} \in \overline{\{s^{(1)}, \ldots, s^{(\ell)}\}}$. \end{proof}


%\begin{lemma}\label{Lemma_cjr_for_meet}
%{The expression $\bigvee_{i=1}^\ell J(s^{(i)}_{\mathcal{D}^{(i)}})$ is a canonical join representation.}
%\end{lemma}
%\begin{proof}
%By Theorem~\ref{Thm_CJC}, it is enough to show that for any $s^{(i)}_{\mathcal{D}^{(i)}}, s^{(j)}_{\mathcal{D}^{(j)}} \in \psi(B) \cap \psi(B^\prime)$ with $i \neq j$ the expression $J(s^{(i)}_{\mathcal{D}^{(i)}})\vee J(s^{(j)}_{\mathcal{D}^{(j)}})$ is a canonical join representation. It follows from Lemma~\ref{nonempty_s^i} that $s^{(i)} \neq s^{(j)}$ so we need to show that $s^{(i)}_{\mathcal{D}^{(i)}}$ and $s^{(j)}_{\mathcal{D}^{(j)}}$ satisfy properties 2) and 3) from Theorem~\ref{Thm_CJC}.
%
%
%
%
%
%First, we show $s^{(i)}_{\mathcal{D}^{(i)}}$ and $s^{(j)}_{\mathcal{D}^{(j)}}$ satisfy property 2) by showing that, without loss of generality, $s^{(i)}$ is not expressible as a composition of at least two segments from $J(s^{(i)}_{\mathcal{D}^{(i)}})\cup J(s^{(j)}_{\mathcal{D}^{(j)}})$. Suppose we can write $s^{(i)} = t^1\circ \cdots \circ  t^r$ where $t^1, \ldots, t^r \in J(s^{(i)}_{\mathcal{D}^{(i)}})\cup J(s^{(j)}_{\mathcal{D}^{(j)}})$ with $r \ge 2$. Clearly, $t^1, t^2\circ \cdots \circ t^r \in J(s^{(i)}_{\mathcal{D}^{(i)}}) \vee J(s^{(j)}_{\mathcal{D}^{(j)}})$. 
%
%We also know that $J(s^{(i)}_{\mathcal{D}^{(i)}}) \vee J(s^{(j)}_{\mathcal{D}^{(j)}}) \le B$ and $J(s^{(i)}_{\mathcal{D}^{(i)}}) \vee J(s^{(j)}_{\mathcal{D}^{(j)}}) \le B^\prime.$ To see this, note that $s^{(i)}_{\mathcal{D}^{(i)}}, s^{(j)}_{\mathcal{D}^{(j)}} \in \psi(B)$ implies that there exists $(B^1, B^2), (B^3, B^4) \in \text{Cov}([\bigwedge_{i =1}^k, B_i, B])$ such that $\widetilde{\lambda}(B^1, B^2) = s^{(i)}_{\mathcal{D}^{(i)}}$ and $\widetilde{\lambda}(B^3, B^4) = s^{(j)}_{\mathcal{D}^{(j)}}$. By Proposition~\ref{Prop_join_irr_des}, we have that $J(s^{(i)}_{\mathcal{D}^{(i)}}) \le B^2$ and $J(s^{(j)}_{\mathcal{D}^{(j)}}) \le B^4.$ Thus $J(s^{(i)}_{\mathcal{D}^{(i)}}) \vee J(s^{(j)}_{\mathcal{D}^{(j)}}) \le B$. The proof that $J(s^{(i)}_{\mathcal{D}^{(i)}}) \vee J(s^{(j)}_{\mathcal{D}^{(j)}}) \le B^\prime$ is similar.
%
%
%By Remark~\ref{remark_same_N}, we obtain that $t^1$ and $t^2\circ \cdots \circ t^r$ are faultline splits of $s^{(i)} = s_{i_1}\circ \cdots \circ s_{i_n}$ and $s^{(i)} = t_{j_1} \circ \cdots \circ t_{j_m}$. Therefore, $t^1, t^2\circ \cdots \circ t^r \in \text{Seg}(\psi(B)\cap \psi(B^\prime)).$ We can assume, without loss of generality, that $t^1 \in \mathcal{D}^{(i)}$. We obtain that $s^{(i)}_{\mathcal{D}^{(i)}}, s^{(i)}_{(\mathcal{D}^{(i)}\backslash\{t^1\}) \sqcup \{t^2\circ \cdots \circ t^r\}} \in \psi(B)\cap \psi(B^\prime)$. However, this contradicts that $s^{(i)}$ is a segment in exactly one label in $\psi(B) \cap \psi(B^\prime)$. 
%
%Lastly, we show $s^{(i)}_{\mathcal{D}^{(i)}}$ and $s^{(j)}_{\mathcal{D}^{(j)}}$ satisfy property 3). Assume that we have $s^{(i)} \subseteq s^{(j)}$. Write $s^{(j)} = s_{i_1}\circ \cdots \circ s_{i_n}$ for some $s_{i_1}, \ldots, s_{i_n} \in \{s_1, \ldots, s_k\}$ and $s^{(j)} = t_{j_1} \circ \cdots \circ t_{j_m}$ for some $t_{j_1}, \ldots, t_{j_m} \in \{t_1, \ldots, t_l\}$. One checks that there are the following three cases describing how $s^{(i)}$ may be expressed in terms of $s_{i_1}, \ldots, s_{i_n}$ and $t_{j_1}, \ldots, t_{j_m}$:
%\begin{itemize}
%\item[i)] $s^{(i)} = s_{i_\ell}\circ \cdots \circ s_{i_{\ell^\prime}}$ and $s^{(i)} = t^r\circ t_{j_{r+1}}\circ \cdots \circ t_{j_{r^\prime-1}}\circ t^{r^\prime},$
%\item[ii)] $s^{(i)} = t_{j_r}\circ \cdots \circ t_{j_{r^\prime}}$ and $s^{(i)} = t^\ell \circ s_{i_{\ell+1}}\circ \cdots \circ s_{i_{\ell^\prime-1}}\circ t^{\ell^\prime},$
%\item[iii)] $s^{(i)} = t^\ell \circ s_{i_{\ell+1}}\circ \cdots \circ s_{i_{\ell^\prime-1}}\circ t^{\ell^\prime}$ and $s^{(i)} = t^r\circ t_{j_{r+1}}\circ \cdots \circ t_{j_{r^\prime-1}}\circ t^{r^\prime}.$
%\end{itemize}
%In these cases, $t^r$ is a split of $t_{j_r}$, $t^{r^\prime}$ is a split of $t_{j_{r^\prime}}$, $t^\ell$ is a split of $s_{i_\ell}$, and $t^{\ell^\prime}$ is a split of $s_{i_{\ell^\prime}}$. In the degenerate case when $r = r^\prime$ (resp., $\ell = \ell^\prime$), we mean that $s^{(i)}$ is proper subsegment of $t_{j_r}$ (resp., $s_{i_\ell}$) that is not a split of $t_{j_r}$ (resp., $s_{i_\ell}$).
%
%Now write $s^{(i)} \subseteq s_{i_\ell}\circ \cdots \circ s_{i_{\ell^\prime}}$ with $\ell^\prime - \ell$ as small as possible. We can assume without loss of generality that we are in case ii) or iii). This means we can express $s^{(i)}$ in terms of $\{s_1, \ldots, s_k\}$ as $s^{(i)} = s_{k_1}\circ \cdots \circ s_{k_r}$ where $s_{k_1}$ is not a split of $s_{i_\ell}$ or $s_{i_{\ell^\prime}}$ and no segment in $\{s_{i_\ell}, \ldots, s_{i_{\ell^\prime}}\}$ is a split of $s_{k_1}$. Similarly, segment $s_{k_r}$ will have these same properties. It follows that $s_{k_1}, s_{k_r} \not \in \{s_{i_1}, \ldots, s_{i_n}\}$. 
%
%We claim that $\{s_{i_1}, \ldots, s_{i_n}\}\cap \{s_{k_1},\ldots, s_{k_r}\} = \emptyset$. If this were not true, then from the properties of $s_{k_1}$ and $s_{k_r}$, there must exist $s_{k_j} \in \{s_{k_1},\ldots, s_{k_r}\}$ that is either a split of an element in $\{s_{i_\ell},\ldots, s_{i_{\ell^\prime}}\}$ or an element of this set is a split of $s_{k_j}$. The existence of such an element contradicts Lemma~\ref{nosplit}. 
%
%We obtain that $J({s_{k_j}}_{\mathcal{D}_{k_j}}) \not \le \bigvee_{r = 1}^n J({s_{i_r}}_{\mathcal{D}_{i_r}})$ for any $s_{k_j} \in \{s_{k_1}, \ldots, s_{k_r}\}$. If this containment did hold, then $\bigvee_{i = 1}^k J({s_i}_{\mathcal{D}_i})$ would not be an irredundant join representation. This would contradict that $\bigvee_{i = 1}^k J({s_i}_{\mathcal{D}_i})$ is a canonical join representation. 
%
%
%
%
%Next, by Lemma~\ref{Lemma_special_subseg}, there exists $s_{k^*} \in \{s_{k_1}, \ldots, s_{k_r}\}$ such that $s_{k^*} \in J(s^{(i)}_{\mathcal{D}^{(i)}}).$ Observe that $J({s_{k^*}}_{\mathcal{D}_{k^*}}) \le J(s^{(i)}_{\mathcal{D}^{(i)}})$ by Lemma~\ref{Lemma_join_irr_containment}. This implies that $J(s^{(i)}_{\mathcal{D}^{(i)}}) \not \le \bigvee_{r = 1}^n J({s_{i_r}}_{\mathcal{D}_{i_r}})$. {Since $J(s^{(j)}_{\mathcal{D}^{(j)}}) \le \bigvee_{r = 1}^n J({s_{i_r}}_{\mathcal{D}_{i_r}})$,} we see that $J(s^{(i)}_{\mathcal{D}^{(i)}}) \not \le J(s^{(j)}_{\mathcal{D}^{(j)}})$. 
%
%Since $s^{(i)}\subseteq s^{(j)}$, it must be a proper subsegment of $s^{(j)}$, and it is not a split of $s^{(j)}$. Thus $J(s^{(j)}_{\mathcal{D}^{(j)}}) \not \le J(s^{(i)}_{\mathcal{D}^{(i)}}),$ since $s^{(j)} \not \in J(s^{(i)}_{\mathcal{D}^{(i)}})$. We obtain that $s^{(i)}_{\mathcal{D}^{(i)}}$ and $s^{(j)}_{\mathcal{D}^{(j)}}$ satisfy property 3). We conclude that the expression $J(s^{(i)}_{\mathcal{D}^{(i)}}) \vee J(s^{(j)}_{\mathcal{D}^{(j)}})$ is a canonical join representation.\end{proof}



%\begin{lemma}\label{Lemma_special_subseg}
%If $J(s_\mathcal{D}) \in \text{JI}(\text{Bic}(T))$ and $s = s_{k_1}\circ \cdots \circ s_{k_r}$, then there exists $s_{k^*} \in \{s_{k_1}, \ldots, s_{k_r}\}$ such that $s_{k^*} \in J(s_{\mathcal{D}})$.
%\end{lemma}
%\begin{proof}
%If $s_{k_r} \not \in J(s_\mathcal{D})$, then $s_{k_1}\circ \cdots \circ s_{k_{r-1}} \in J(s_{\mathcal{D}})$. By Lemma~\ref{Lemma_join_irr_containment}, we have that $$J((s_{k_1}\circ \cdots \circ s_{k_{r-1}})_{\mathcal{D}(s_{k_1}\circ \cdots \circ s_{k_{r-1}})}) \le J(s_\mathcal{D})$$ where ${\mathcal{D}(s_{k_1}\circ \cdots \circ s_{k_{r-1}})} = \{s^\prime \in J(s_\mathcal{D}): \ \text{$s^\prime$ is a split of $s_{k_1}\circ \cdots \circ s_{k_{r-1}}$}\}.$ If $s_{k_{r-1}} \not \in J(s_{\mathcal{D}})$, then $s_{k_{r-1}} \not \in J((s_{k_1}\circ \cdots \circ s_{k_{r-1}})_{\mathcal{D}(s_{k_1}\circ \cdots \circ s_{k_{r-1}})})$ so $s_{k_1}\circ \cdots \circ s_{k_{r-2}} \in J((s_{k_1}\circ \cdots \circ s_{k_{r-1}})_{\mathcal{D}(s_{k_1}\circ \cdots \circ s_{k_{r-1}})})$. Now by Lemma~\ref{Lemma_join_irr_containment}, we have that $$J((s_{k_1}\circ \cdots \circ s_{k_{r-2}})_{\mathcal{D}(s_{k_1}\circ \cdots \circ s_{k_{r-2}})}) \le J(s_\mathcal{D})$$ where ${\mathcal{D}(s_{k_1}\circ \cdots \circ s_{k_{r-2}})} = \{s^\prime \in J(s_\mathcal{D}): \ \text{$s^\prime$ is a split of $s_{k_1}\circ \cdots \circ s_{k_{r-2}}$}\}.$ Continuing this process, we obtain that there exists $s_{k^*} \in  J(s_{\mathcal{D}})$.
%\end{proof}

%\begin{lemma}\label{meet_in_Psi}
%{We have that $\psi\left(\bigvee_{i=1}^\ell J(s^{(i)}_{\mathcal{D}^{(i)}})\right) = \psi(B)\cap \psi(B^\prime).$}
%\end{lemma}
%\begin{proof}
%By Lemma~\ref{Lemma_cjr_for_meet}, we know that $\widetilde{\lambda}_{\downarrow}\left(\bigvee_{i=1}^\ell J(s^{(i)}_{\mathcal{D}^{(i)}})\right) = \{s^{(i)}_{\mathcal{D}^{(i)}}\}_{i = 1}^\ell.$ Thus, by Theorem~\ref{psib}, any element of $\psi\left(\bigvee_{i=1}^\ell J(s^{(i)}_{\mathcal{D}^{(i)}})\right)$ is of the form $(s^{(i_1)}\circ\cdots \circ s^{(i_r)})_\mathcal{D}$ for some $i_1, \ldots, i_r \in \{1,\ldots, \ell\}$. From Lemma~\ref{Lemma_seg_of_int}, we know that $s^{(i_1)}\circ\cdots \circ s^{(i_r)} \in \text{Seg}(\psi(B)\cap \psi(B^\prime))$. 
%
%We now show that $(s^{(i_1)}\circ\cdots \circ s^{(i_r)})_\mathcal{D} \in \psi(B)\cap \psi(B^\prime)$. We show $(s^{(i_1)}\circ\cdots \circ s^{(i_r)})_\mathcal{D} \in \psi(B)$ and the analogous argument shows that $(s^{(i_1)}\circ\cdots \circ s^{(i_r)})_\mathcal{D} \in \psi(B^\prime)$. Consider the biclosed set $$B^* := \left(\bigwedge_{i=1}^kB_i\right) \vee \left(J((s^{(i_1)}\circ \cdots \circ s^{(i_r)})_\mathcal{D})\backslash \{s^{(i_1)}\circ \cdots \circ s^{(i_r)}\}\right).$$ Clearly, $\bigwedge_{i=1}^kB_i \le B^*$ and $B^* < B$ as $s^{(i_1)}\circ \cdots \circ s^{(i_r)} \in B\backslash B^*$. 
%
%If we assume that $B^*\sqcup \{s^{(i_1)}\circ \cdots \circ s^{(i_r)}\}$ is biclosed, we obtain that $\widetilde{\lambda}(B^*, B^*\sqcup \{s^{(i_1)}\circ \cdots \circ s^{(i_r)}\}) = (s^{(i_1)}\circ\cdots \circ s^{(i_r)})_\mathcal{D}$ and $B^*\sqcup \{s^{(i_1)}\circ \cdots \circ s^{(i_r)}\} \le B$. From this it follows that $(s^{(i_1)}\circ\cdots \circ s^{(i_r)})_\mathcal{D} \in \psi(B)$. One shows that $B^*\sqcup \{s^{(i_1)}\circ \cdots \circ s^{(i_r)}\}$ is biclosed by adapting the argument that we used in Lemma~\ref{nonempty_s^i}.
%
%Conversely, suppose $(s^{(i_1)}\circ \cdots \circ s^{(i_r)})_{\mathcal{D}} \in \psi(B)\cap \psi(B^\prime)$ where this label can be expressed this way by Lemma~\ref{Lemma_seg_of_int}. By Theorem~\ref{psib}, we know that there exist labels $\widetilde{\lambda}(B^1, B^2) \in \psi\left(\bigvee_{i=1}^\ell J(s^{(i)}_{\mathcal{D}^{(i)}})\right)$ such that $\lambda(B^1, B^2) = s^{(i_1)}\circ \cdots \circ s^{(i_r)}.$ Thus it is enough to show that $\mathcal{D}$ satisfies the properties appearing in the statement of Theorem~\ref{psib} with respect to the labels $\{s^{(i)}_{\mathcal{D}^{(i)}}\}_{i = 1}^\ell$. In fact, it is enough to check that the non-faultline splits of $\mathcal{D}$ satisfy (iv) in the statement of Theorem~\ref{psib}. If $r = 1$, the label under consideration is $s^{(i_1)}_{\mathcal{D}^{(i_1)}}$ and the result follows.
%
%
%
%Now suppose that $s^{(i_1)}\circ \cdots \circ s^{(i_{j-1})}\circ t^{(i_j)} \in \mathcal{D}$ is a non-faultline split of $s^{(i_1)}\circ \cdots \circ s^{(i_{r})}$ where $t^{(i_j)}$ is a split of $s^{(i_j)}$. We assume that the result holds for labels $(s^{(j_1)}\circ \cdots \circ s^{(j_{r^\prime})})_{\mathcal{D}} \in \psi(B)\cap \psi(B^\prime)$ with $r^\prime < r$. Using the approach in the proof of Lemma~\ref{intersection_ker_coker}, we have that the label $(s^{(i_j)}\circ \cdots \circ s^{(i_r)})_{\mathcal{D}^{\prime\prime}} \in \psi(B)\cap \psi(B^\prime)$ where $$\mathcal{D}^{\prime\prime} := \left\{\begin{array}{lcl} t^* & : & \text{if $t^* \in \mathcal{D}$ and $t^*$ is a proper subsegment of $s^{(i_j)}\circ \cdots \circ s^{(i_r)}$,}\\
%t^* & : & \text{if  $s^{(i_1)}\circ \cdots \circ s^{(i_{j-1})} \circ t^{*} \in \mathcal{D}$ where $t^*$ is a split of $s^{(i_j)}\circ \cdots \circ s^{(i_r)}$.} \end{array}\right.$$
%
%We observe that $t^{(i_j)} \in \mathcal{D}^{\prime\prime}$. By induction, $t^{(i_j)} \in \mathcal{D}^{(i_j)}$. We conclude that $(s^{(i_1)}\circ \cdots \circ s^{(i_r)})_{\mathcal{D}} \in \psi\left(\bigvee_{i=1}^\ell J(s^{(i)}_{\mathcal{D}^{(i)}})\right)$. \end{proof}

%\begin{theorem}\label{Thm_lattice_prop}
%The shard intersection order $\Psi(\text{Bic}(T))$ is a lattice.
%\end{theorem}
%\begin{proof}
%The shard intersection order $\Psi(\text{Bic}(T))$ is finite. So, to prove that it is a lattice, we show that it has a unique maximal element and that for any $\psi(B), \psi(B^\prime) \in \Psi(\text{Bic}(T))$ one has $\psi(B)\cap \psi(B^\prime) \in \Psi(\text{Bic}(T))$. The latter is a consequence of Lemma~\ref{meet_in_Psi}.
%
%Next, the shard intersection order of $\text{Bic}(T)$ has a unique maximal element. The set $\text{Seg}(T)$ is the top element of $\text{Bic}(T)$, and the coatoms of $\text{Bic}(T)$ are the sets of the form $\text{Seg}(T)\backslash\{[a,b]\}$ where $a$ and $b$ are interior vertices connected by an edge of T. Thus the meet of all coatoms of $\text{Bic}(T)$ is the empty set. Now, using the definition of $\Psi(\text{Bic}(T))$ and Proposition~\ref{Prop_join_irr_des}, we see that $\psi(\text{Seg}(T)) = \widetilde{\lambda}(\text{Cov}(\text{Bic}(T))) = \mathcal{S}_T.$ This implies that $\psi(\text{Seg}(T))$ is the unique maximal element of $\Psi(\text{Bic}(T))$.
%\end{proof}








\acknowledgements{This project began at the 2016 combinatorics REU at the University of Minnesota, Twin Cities, and was supported by NSF RTG grant DMS-1148634. The authors thank Craig Corsi, Thomas McConville, Henri M\"uhle, and Vic Reiner for their valuable advice and for useful conversations. The authors are also grateful to two anonymous referees who provided helpful feedback on an earlier version of the paper. Alexander Garver also received support from NSERC and the Canada Research Chairs program.}

%Alexander Garver also received support from NSERC and the Canada Research Chairs program during various parts of this project.

%% if you use biblatex then this generates the bibliography
%% if you use some other method then remove this and do it your own way
\printbibliography



\end{document}
