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%%%%% Auteur
\author{\firstname{Emily} \lastname{Barnard}}%%%1
\address{Department of Mathematical Sciences\\
DePaul University\\
2320 N. Kenmore Ave.\\
Suite 502\\
Chicago\\
IL 60614, USA}
\email{e.barnard@depaul.edu}
%%%%
\author{\firstname{Andrew} \lastname{Carroll}} %%%2
\address{3778 Keating St. San Diego\\
CA 92110, USA }
\email{antcarro@gmail.com}
%%%
\author{\firstname{Shijie} \lastname{Zhu}} %%%3
\address{Mathematics Department\\
University of Iowa\\
14 MacLean Hall\\
Iowa City\\
IA 52242, USA}
\email{shijie-zhu@uiowa.edu}
%%%%% Sujet
\subjclass{05E10, 06B15}
\keywords{lattice theory, torsion classes, canonical join representations}
%%%%% Gestion
\DOI{10.5802/alco.72}
\datereceived{2018-05-18}
\daterevised{2019-01-14}
\dateaccepted{2019-03-06}
%%%%% Titre et résumé
\title[Minimal inclusions of torsion classes]{Minimal inclusions of torsion classes}
\begin{abstract}
Let $\Lambda$ be a finite-dimensional associative algebra.
The torsion classes of $\mathrm{mod}\Lambda$ form a lattice under containment, denoted by $\mathrm{tors} \Lambda$.
In this paper, we characterize the cover relations in $\mathrm{tors} \Lambda$ by certain indecomposable modules.
We consider three applications:
First, we show that the completely join-irreducible torsion classes
(torsion classes which cover precisely one element) are in bijection
with bricks.
Second, we characterize faces of the canonical join complex of
$\mathrm{tors} \Lambda$ in terms of representation theory.
Finally, we show that, in general, the algebra $\Lambda$ is not characterized by its lattice $\mathrm{tors} \Lambda$.
In particular, we study the torsion theory of a quotient of the preprojective algebra of type $A_n$.
We show that its torsion class lattice is isomorphic to the weak order on~$A_n$.
\end{abstract}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{document}
\maketitle
\section{Introduction}
\looseness-1
Let $\Lambda$ be a
finite-dimensional associative algebra over a field~$k$, and write
$\tors \Lambda$ for the set of torsion classes of finitely generated
modules over $\Lambda$, partially ordered by containment.
The poset $\tors \Lambda$ is a complete lattice in which the \newword{meet} (or greatest lower bound) $\Meet \{\calT, \calT'\}$ coincides with the intersection $\calT \cap \calT'$, and the \newword{join} (or smallest upper bound) $\Join \{\calT, \calT'\}$ coincides with the iterative extension closure of the union $\calT\cup \calT'$.
In this paper, we study the cover relations $\calT'\covers \calT$ in $\tors \Lambda$.
Recall that a torsion class $\calT'$ \newword{covers} $\calT$ if $\calT\subsetneq \calT'$ and for each $\calY\in \tors\Lambda$, if $\calT\subsetneq \calY \subseteq \calT'$ then $\calY= \calT'$.
In~\cite{AIR}, the authors describe the lattice of functorially-finite torsion classes by way of $\tau$-tilting pairs.
They show the existence of a unique module which encodes each cover relation as follows:
When $\calT$ and $\calT'$ are functorially finite torsion classes, with $\calT'\covers \calT$,
there exists a unique module $M$ with the property that $\calT'$ is the closure of $\add(\calT \cup\{M\})$ under taking quotients.
In our complimentary approach, we show that for each cover relation $\calT'\covers \calT$ in $\tors\Lambda$, there exists a unique ``minimal'' module $M$ with the property that $\calT'$ is the closure of $\calT\cup \{M\}$ under taking \emph{iterative extensions}.
Below, we make the notation of ``minimal'' precise.
\begin{definition}\label{def:min-ext-module}
A module $M$ is a \newword{minimal extending module for $\calT$} if it satisfies the following three properties:
\begin{enumerate}[label={(P\arabic*)}]
\item Every proper factor of \label{propertyone}
$M$ is in $\calT$;
\item If $0\rightarrow M\rightarrow X
\rightarrow T\rightarrow 0$ is a non-split exact sequence with
$T\in \calT$, then $X\in \calT$; \label{propertytwo}
\item $\Hom(\calT, M)=0$. \label{propertythree}
\end{enumerate}
\end{definition}
In general, a minimal extending module $M$ is a quotient of the module studied in~\cite{AIR}.
For a more precise description of this relationship see~\cite[Proposition~1.9]{Asai},~\cite[Theorem~4.1]{DIJ}, and~\cite[Theorem~3.11]{DIRRT}.
The following remarks will be useful.
Assuming Property~\ref{propertyone}, Property
\ref{propertythree} is equivalent to the fact that $M\notin \calT$.
Observe that if there is a non-trivial homomorphism from a module in
$\calT$ to $M$, then both the image and cokernel are in $\calT$, and
$M$ is an extension of the cokernel by the image.
Thus, any minimal extending module is indecomposable.
Indeed, direct summands are proper factors, and torsion classes are closed under direct sums.
In the statement of the following theorem, and throughout the paper we have the following notation:
We write $[M]$ for the isoclass of the module $M$; $\ME(\calT$) for the set of isoclasses~$[M]$ such that $M$ is a minimal extending module for $\calT$; and $\filt(\calT \cup \{M\})$ for the iterative extension closure of $\calT \cup \{M\}$.
\begin{theorem}\label{main: covers}
Let $\calT$ be a torsion class over $\Lambda$.
Then the map $$\eta_\calT: [M]\mapsto \filt(\calT\cup \{M\})$$ is a bijection from the set $\ME(\calT)$ to the set of $\calT' \in \tors \Lambda$ such that $\calT \covered \calT'$.
\end{theorem}
Recall that a module $M$ is called a \newword{brick} if the
endomorphism ring of $M$ is a division ring. (That is, the non-trivial
endomorphisms are invertible.)
\begin{theorem}\label{main: cor schur}
Let $\Lambda$ be a finite-dimensional associative algebra and $M\in\module\Lambda$.
Then $M$ is a minimal extending module for some torsion class if and
only if $M$ is a brick.
\end{theorem}
As an immediate consequence of Theorem~\ref{main: cor schur} we
obtain a labeling of cover relations for $\tors
\Lambda$ (when they exist) by bricks.
Independently and concurrently with our work, the authors of~\cite{DIRRT} also give a labeling of the cover relations in the lattice of torsion classes by brick modules which coincides with the labeling proposed here.
We illustrate in Example~\ref{kronecker} that in $\tors
\Lambda$ there exists pairs $\calT'> \calT$ such that there is no
torsion class~$\calY$ satisfying $\calT' \covers \calY \ge \calT$.
This paper fits into a larger body of research which studies the combinatorial structure of the lattice of (functorially finite) torsion classes.
Connections between the combinatorics of a finite simply laced Weyl group $W$ and the corresponding preprojective algebra $\Pi W$ are of particular interest.
In~\cite{Miz}, Mizuno showed the lattice of functorially finite torsion classes $\ftors \Pi W$ is isomorphic to the weak order on $W$.
Building on this work, the authors of~\cite{IRRT} have shown that the lattice of (functorially finite) torsion classes of quotients of $\Pi W$ are \emph{lattice quotients} of the weak order on~$W$.
They also obtain an analogous labeling of $\ftors \Pi W$ by certain modules called \emph{layer modules}.
(See~\cite[Theorem~1.3]{IRRT}.)
Cover relations in the lattice of torsion classes are also closely related to maximal green sequences~\cite{BST}.
In a related direction, the authors of~\cite{G-M} study certain lattice properties of $\ftors \Lambda$ when $\Lambda$ arises from a quiver that is mutation-equivalent to a path quiver or oriented cycle, including a certain minimal ``factorization'' called the \emph{canonical join representation}.
Inspired by these results, we will give three applications of Theorems~\ref{main: covers} and~\ref{main: cor schur}.
Before describing them, we recall some terminology in lattice theory.
In a (not necessarily finite) lattice $L$, a \newword{join representation} for an element $w\in L$ is an expression $\Join A = w$, where $A$ is a (possibly infinite) subset of $L$.
An element $w$ is \newword{join-irreducible}, if $w\in A$ for any join representation $w=\Join A$, where $A$ is a finite set.
An element $w$ is \newword{completely join-irreducible}, if $w\in A$ for any join representation $w=\Join A$, where $A$ is any subset of $L$.
(By convention, the smallest element in $L$ is not join-irreducible or completely join-irreducible.)
% because it can be written as $\Join A$ where $A$ is the empty set.)
Equivalently, $w$ is completely join-irreducible if and only if $w$ covers only one element $v$ and any element $u] (A.east)--(B.west);
\draw[->] (A.south)--(C.north);
\draw[->] (C.east)--(D.west);
\draw[right hook->] (B.south)--(D.north);
\end{tikzpicture}
\]
Then $M=M_{l^\prime} \supsetneq M_{l^\prime-1}\supsetneq \cdots
M_0=N=N_l \supsetneq N_{l-1} \supsetneq \cdots
\supsetneq N_0 = 0$ is an $\calS$-filtration of $M$.
\end{proof}
\begin{lemma}\label{lem-epi-closure}
Suppose that $\calS$ is a class of indecomposable modules that is
closed under taking indecomposable summands of factors.
Then $\filt(\calS)$ is closed under factors.
Dually, if $\calS$ is closed under taking indecomposable summands of submodules, then $\filt(\calS)$ is closed under submodules.
\end{lemma}
\begin{proof}
Suppose that $N\in \filt^{(l)}(\calS)$ and $\phi: N\rightarrow U$ is an epimorphism.
We assume, without loss of generality, that $U$ is indecomposable and proceed by induction on $l$.
If $l=1$, then $N\in \calS$.
% so $U$ is an indecomposable factor of $N$.
Since $\calS$ is closed under epimorphisms, we have $U\in \calS$.
Now suppose that all factors of modules in $\filt^{(l^\prime)}(\calS)$
are in $\filt(\calS)$ for $l'1$, and let $i$ to be the smallest index such that $N_i/N_{i-1} \cong M$ (one must exist since $N\notin
\calT$).
If $i=1$, then there is a short exact sequence
\[0 \rightarrow M \rightarrow N \rightarrow N/N_1 \rightarrow 0.\]
\looseness-1
Since $N$ was assumed indecomposable (and $l>1$), the sequence does
not split.
If $N/N_1 \in \calT$, then by Property~\ref{propertytwo}, $N\in
\calT$, a contradiction.
Otherwise, $N/N_1 \not\in \calT$ is in $\filt^{(l-1)}(\calT\cup \{M\})$,
so by induction, $M$ is a factor of $N/N_1$, so it is also a factor of
$N$.
If $i>1$, consider the short exact sequence
\[0 \rightarrow N_{i-1} \rightarrow N \rightarrow N/N_{i-1} \rightarrow 0\]
which again is non-split by assumption.
Note that since $N_{i-1}$ has a filtration by modules in $\calT$, $N_{i-1}\in
\calT$.
If $N/N_{i-1}\in \calT$, then so is $N$, since torsion classes are
closed under extensions.
Therefore, $N/N_{i-1} \notin \calT$, and it has a
$\calT\cup\{M\}$-filtration of length $l-i+11$, so that $N_1\in \calT$.
Then we have a nonzero homomorphism $N_1\hookrightarrow N \hookrightarrow X$ from a module in $\calT$ to $X$.
That is a contradiction.
Thus, $M\cong N_1 \hookrightarrow N \hookrightarrow X$ as desired.
\end{proof}
We close this section by completing the proof of Theorem~\ref{main: cor schur}.
Recall that in Lemma~\ref{lem: torsion and schur}, we showed that if
$M$ is a minimal extending module, then it is a brick (the forward implication in Theorem~\ref{main: cor schur}).
By Proposition~\ref{prop:torsion-free-to-torsion-relation} it is enough to show:
If $M$ is a brick, then there exists a torsion-free class $\calF$ such that $M$ is a minimal co-extending module for $\calF$. Also recall that $\filt(\Gen M)$ is always a torsion class.
\begin{proposition}\label{if part}
If $M$ is a brick over $\Lambda$, then $M$ is a minimal co-extending module for $\filt(\Gen M)^{\perp}$.
\end{proposition}
\begin{proof}[Proof of Proposition~\ref{if part} and Theorem~\ref{main: cor schur}]
Suppose that $M$ is a brick over $\Lambda$. We will argue that $M$ is a minimal co-extending module for the torsion-free class $\calF=\filt(\Gen M)^\perp$.
First, we claim that $\filt(\Gen M)^\perp= M^\perp:=\{X:\,\Hom_\Lambda(M, X)=0\}$.
It is immediate that $\filt(\Gen M)^\perp \subseteq M^\perp$. Conversely, let $X\in M^\perp$. Since $\Hom(-,X)$ is left exact, it follows that $\Hom(L,X)\hookrightarrow \Hom(\oplus M,X)=0$ for each epimorphism $\oplus M\rightarrow L$. Hence $ X\in \Gen(M)^\perp $. Then using induction on the length of a filtration of $N$, for all $N\in\filt(\Gen(M))$, it is easy to see $\Hom(N,X)=0$. Hence $X$ is in $\filt(\Gen(M))^\perp$.
Therefore Property~\ref{copropertythree} holds.
%Now suppose that $X$ is an indecomposable module satisfying:
%$\Hom_\Lambda(M,X)=0$ and $\Hom_\Lambda(N, X)\neq 0$ for some
%indecomposable $N\in \filt(\Gen M)$.
%Write $f: N\rightarrow X$ for such a non-zero homomorphism.
%Let $N_l \supsetneq \dotsc \supsetneq N_0=0$ be a filtration of $N$ such that $N_i/N_{i-1}$ is a factor of $M$.
%Choose $i$ to be the smallest index for which $f(N_i)\neq 0$, so that $f(N_i)/f(N_{i-1})=f(N_i)\subseteq X$.
%Since is $N_i/N_{i-1}$ is a factor of $M$, we have the following nonzero composition:
%\[M \twoheadrightarrow N_i/N_{i-1} \twoheadrightarrow f(N_i) \subseteq X.\]
%That is a contradiction, because $\Hom_\Lambda(M,X) = 0$.
To verify ~\ref{copropertyone}, let $M'$ be a proper submodule of
$M$. If $\Hom_\Lambda(M,M')\neq 0$, then $M$ is not a brick, a contradiction. Thus, $M'\in\{X:\,\Hom(M,X)=0\}=\filt(\Gen M)^\perp$.
To verify~\ref{copropertytwo}, suppose that there is a non-split short exact sequence \[ 0\rightarrow F \xrightarrow{f} X
\xrightarrow{g} M \rightarrow 0\] with $F\in \calF$.
If $X\notin \calF$, then there is a nonzero homomorphism $\pi: M\rightarrow X$.
Since $M$ is a brick, and $g\circ \pi$ is a endomorphism of $M$ (which is not an isomorphism since the sequence is non-split), $g\circ \pi=0$.
Hence, $\pi$ factors through $f$.
Since $\Hom_\Lambda(M,F)=0$, we have a contradiction.
Therefore, $M$ is a minimal co-extending module for $\calF$.
The statement follows from Proposition~\ref{prop:torsion-free-to-torsion-relation}.
\end{proof}
\section{Applications}
We consider two applications of Theorems~\ref{main: covers} and~\ref{main: cor schur}.
First, in Proposition~\ref{ji}, we prove that there is a bijective
correspondence between isoclasses $[M]$ of bricks over $\Lambda$ and
completely join-irreducible torsion classes in $\tors \Lambda$ via the
map sending $[M]$ to $\filt(\Gen(M)).$
Second, we consider the canonical join complex of $\tors \Lambda$,
proving the forward implication of Theorem~\ref{main: cjr} in Proposition~\ref{cjr only if}, and we completing the proof in Proposition~\ref{cjr end}.
\label{sec: ji}
\subsection{Completely join-irreducible torsion classes}
In this section, we prove Theorem~\ref{schur modules to join-irreducibles} as Proposition~\ref{ji} below.
Recall that $\calT$ is \newword{completely join-irreducible} if for each (possibly infinite) subset $A\subseteq \tors \Lambda$, $\calT = \Join A $ implies that $\calT\in A$.
Equivalently, $\calT$ is completely join-irreducible if and only if it covers precisely one torsion class $\calS$, and $\calG\subseteq \calS$ for any torsion class $\calG\subsetneq \calT$.
In the statement below, recall that $\Gen M$ denotes the factor-closure of $\add M$.
\begin{proposition}\label{ji}
The torsion class $\calT$ is completely join-irreducible if and only
if there exists a brick $M$ such that $\calT$ is equal to $\filt(\Gen M)$.
In this case, the brick $M$ is unique up to isomorphism.
In particular, the map $\zeta: [M]\mapsto \filt(\Gen M)$ from
Theorem~\ref{schur modules to join-irreducibles} is a bijection.
\end{proposition}
\begin{proof}[Proof Proposition~\ref{ji} and Theorem~\ref{schur modules to join-irreducibles}]
Suppose that the torsion class $\calT$ is completely join-irreducible
and write $\calS$ for the unique torsion class covered by $\calT$.
Theorem~\ref{main: cor schur} implies that there exists a unique (up to isomorphism) brick $M$ such that ${\calT = \filt(\calS \cup \{M\})}$.
We claim that $\calT = \filt(\Gen M)$.
Observe that $\filt(\Gen M)$ is contained in $\calT$, and $\filt(\Gen M) \not\subseteq\calS$ (because $M\not\in \calS$).
The claim follows.
Conversely, let $M$ be a brick over $\Lambda$.
In the proof of Proposition~\ref{if part}, we showed that $M$ is a minimal co-extending module for $\filt(\Gen M)^\perp$.
By Proposition~\ref{prop:torsion-free-to-torsion-relation}, there exists a torsion class $\calS$ such that $\filt(\Gen M) \covers \calS$ and $M$ is a minimal extending module for $\calS$.
(That is $\filt(\Gen M) = \filt(\calS \cup \{M\})$.)
Suppose that $\calG\subseteq \filt(\Gen M)$.
If $\calG\not \subseteq \calS$ then there exists some module $N\in \calG\setminus \calS$.
In particular, $N\in\filt(\Gen M) \setminus \calS$.
Proposition~\ref{prop: epis and covers} says that $M$ is a factor of $N$, hence $M\in \calG$.
We conclude that $\calG= \filt(\Gen M)$.
Thus any torsion class $\calG\subsetneq\filt(\Gen M)$ also satisfies
$\calG\subseteq\calS$.
We conclude that $\filt(\Gen M)$ is completely join-irreducible.
\end{proof}
\begin{remark}\label{kronecker}
\normalfont
There exist join-irreducible torsion classes which are not \emph{completely} join-irreducible.
Consider $\module kQ$ where $Q$ is the Kronecker quiver and $k$ is
algebraically closed (see Figure~\ref{fig: kronecker}).
\begin{figure}[htb]
\begin{tikzpicture}
\node (A) at (0,0) {1};
\node (B) at (2,0) {2};
\draw[->] ([yshift=-.2em] A.east)--([yshift=-.2em]B.west) node[midway,below] {$b$};
\draw[->] ([yshift=.2em] A.east)--([yshift=.2em] B.west) node[midway,above]
{$a$};
\end{tikzpicture}
\caption{The Kronecker quiver.}
\label{fig: kronecker}
\end{figure}
Let $n$ be a non-negative integer, and write $V_n$ for the representation defined as follows:
$V_n(1) = k^{n+1}$, $V_n(2) = k^n$; $V_n(a)=\begin{bmatrix} I_n \,\, \mathbf{0}\end{bmatrix}$ and $V_n(b)=\begin{bmatrix} \mathbf{0}\,\, I_n\end{bmatrix}$ where $I_n$ is the $n\times n$ identity matrix, and $\mathbf{0}$ is a column of zeros.
(The module corresponding to each $V_n$ is indecomposable and preinjective; see e.g.~\cite[VIII.1]{ARS}.)
Let $\calI_n$ denote the additive closure of $\{V_0,V_1,\dotsc, V_n\}$, and $\calI_\infty=\bigcup\limits_{n\geq 0} \calI_n$.
It is an easy exercise to verify that both $\calI_n$ and $\calI_\infty$ are torsion classes and $\calI_n \covered \calI_{n+1}<\calI_\infty$ for each $n$.
Observe that $\calI_\infty$ is join-irreducible, but not \emph{completely} join-irreducible.
In particular, it does not cover any elements in $\tors kQ$.
Each brick in $\calI_\infty$ is isomorphic to $V_n$ for some $n\geq
0$, and it can be shown that $\calI_\infty$ cannot be expressed as $\calI_n$ for
any such $n$.
\end{remark}
\subsection{The canonical join complex of \texorpdfstring{$\tors \Lambda$}{tors Lambda}}\label{canonical join complex}
In this section, we characterize certain faces of the canonical join complex of $\tors \Lambda$.
Before we begin, we review the necessary lattice-theoretic terminology.
Recall that a lattice $L$ is a poset such that, for each finite subset $A\subseteq L$, the \newword{join} or least upper bound $\Join A$ exists and, dually, the \newword{meet} or greatest lower bound $\Meet A$ exists.
The lattice $\tors \Lambda$ is a \newword{complete lattice}, meaning that $\Join A$ and $\Meet A$ exist for arbitrary subsets $A$ of torsion classes.
A subset $A\subseteq L$ is an \newword{antichain} if the elements in $A$ are not comparable.
The \newword{order ideal} generated by $A$ is the set of $w\in L$ such that $w\le a$ for some element $a\in A$.
Recall that the canonical join representation of an element $w$ is the unique ``lowest'' way to write $w$ as the join of smaller elements.
Below, we make these notions precise.
A \newword{join representation} of an element $w$ in a complete lattice is an expression $w=\Join A$, where $A$ is a (possibly infinite) subset of $L$.
We say that $\Join A$ is \newword{irredundant} if $\Join A' < \Join A$, for each proper subset $A'\subsetneq A$.
Observe that if $\Join A$ is irredundant, then $A$ is an antichain.
Consider the set of all irredundant join representation for $w$.
(Note that $\Join \{w\}$ is an irredundant join representation of $w$.)
We partially order the irredundant join representations of $w$ as follows:
Say $A~\refines B$ if the order ideal generated by $A$ is contained in the order ideal generated by $B$.
Equivalently, $A\refines B$ if, for each $a\in A$, there exists some $b\in B$ such that $a\le b$.
Informally, we say that $A$ is ``lower'' than~$B$.
When $A$ is strictly ``lower'' than $B$ (e.g. $A\refines B$ and $A\ne B$) then we write $A\strictrefines B$.
The \newword{canonical join representation of $w$} is the unique lowest irredundant join representation $\Join A$ of $w$, when such a representation exists.
In this case, we also say that the set $A$ is a canonical join representation.
%We write $\Join \can(w)$ or simply $\can(w)$, for the canonical join representation of $w$.
The elements of $A$ are called \newword{canonical joinands} of $w$.
If $w$ is join-irreducible then $\Join \{w\}$ is the canonical join representation of $w$.
Conversely, each canonical joinand is join-irreducible.
\begin{figure}[htb]
\centering
{\scalebox{.8}{ \includegraphics{168_Figures/M3.pdf}}}
\caption{The top element does not have a canonical join representation.}
\label{fig:diamond}
\end{figure}
\begin{remark}\label{cjr dne}
\normalfont
\looseness-1
In general, the canonical join representation of an element may not exist.
For example, see Figure~\ref{fig:diamond}.
Observe that the join of each pair of atoms is a minimal, irredundant join representation of the top element.
Since there is no \emph{unique} such join representation, we conclude that the canonical join representation does not exist.
\end{remark}
The \newword{canonical join complex of $L$}, denoted $\Gamma(L)$, is the collection of subsets $A\subseteq L$ such that $A$ is a canonical join representation.
In the following proposition (essentially~\cite[Proposition~2.2]{arcs}) we show that $\Gamma(L)$ is closed under taking subsets.
In the statement of~\cite[Proposition~2.2]{arcs}, the lattice $L$ is finite.
A standard argument from lattice theory shows that the proposition also holds when $L$ is a complete meet-semilattice.
\begin{proposition}\label{complex}
Suppose that $\Join A$ is a canonical join representation in a complete meet-semilattice $L$.
Then, for each $A'\subseteq A$, the join $\Join A'$ is also canonical join representation.
\end{proposition}
%\begin{proof}
%Write $w$ for the element $\Join A$.
%Observe that $\Join A'$ exists for any $A'\subseteq A$.
%Indeed that the set $S= \{v \in L: \text{$v\ge a$ for each $a \in A'$}\}$ is nonempty.
%(In particular, $w$ belongs to this set.)
%Taking the meet $\Meet S$ yields the desired element in $L$.
%
%The remainder of the argument is identical to the original proof of~\cite[Proposition~2.2]{arcs}.
%Write $w'$ for $\Join A'$.
%Suppose that $\Join B$ is another join representation of $w'$.
%Observe that $\Join (A\setminus A') \cup B = w$.
%Thus, for each element $a \in A$, there exists some element $b \in (A\setminus A')\cup B$ satisfying $a \le b$.
%If $a \in A'$, then the corresponding element $b \in B$ (because $A$ is an antichain).
%Hence, $\Join A'$ is the unique lowest join representation for $w'$.
%\end{proof}
With Proposition~\ref{complex}, we conclude that the canonical join complex $\Gamma(\tors \Lambda)$ is indeed a simplicial complex.
We are finally prepared to prove Theorem~\ref{main: cjr}.
In the next proposition, we tackle the easier direction of the proof.
\begin{proposition}\label{cjr only if}
Suppose that $\calE$ is a collection of bricks over $\Lambda$.
If the set $\{\filt(\Gen M):\, M\in \calE\}$ is a canonical join representation, then each pair of modules $M$ and $M'$ is hom-orthogonal.
\end{proposition}
\begin{proof}
By Proposition~\ref{complex}, it is enough to show that the statement holds when $\calE$ contains two elements, say $M$ and $M'$.
We write $\calT$ for $\filt(\Gen M) \join \filt(\Gen M')$.
Suppose that $f:M'\to M$ is a nonzero homomorphism, and write $N$ for~$\image f$.
Since $\filt(\Gen M) \join \filt(\Gen M')$ is irredundant, it follows that $N$ is not isomorphic to $M$.
Hence, $M/N$ is a proper factor of $M$, and the torsion class $\filt(\Gen(M/N))$ is strictly contained in $\filt(\Gen M)$.
(Indeed, if $M$ belongs to $\filt(\Gen(M/N))$, then there exists a submodule $Y\subseteq M$ that is also a proper factor of $M$.
That is a contradiction to the fact that $M$ is a brick.)
Finally, we observe that $\filt(\Gen M/N) \join \filt(\Gen M')$ is an irredundant join representation for $\calT$, and because $\filt(\Gen M/N) \subsetneq \filt(\Gen M)$,
$$\{\filt(\Gen M/N), \filt(\Gen M')\} \strictrefines \{\filt(\Gen M), \filt(\Gen M')\}.$$
That is a contradiction to the fact that $\filt(\Gen M)\join \filt(\Gen M')$ is the canonical join representation for $\calT$.
\end{proof}
Next, we consider the reverse implication of Theorem~\ref{main: cjr}.
As above, let $\calE$ be a collection of hom-orthogonal bricks.
We write $\calT$ for the torsion class $\Join \{\filt(\Gen M):\, M\in
\calE\}$, and argue that $\Join \{\filt(\Gen M):\, M\in \calE\}$ is the
canonical join representation of $\calT$.
This will require the following lemma.
For context, recall that a torsion class need not have any upper or lower cover relations.
(For example, the torsion class $\calI_\infty$ in the Kronecker quiver
from Example~\ref{kronecker} does not cover any other torsion class.)
\begin{lemma}\label{lem:join-has-lower-faces}
Suppose that $\calE$ is a collection of hom-orthogonal bricks over
$\Lambda$ and let $\calT = \Join \{\filt(\Gen M):\,M\in \calE\}$.
Then for each module $M\in \calE$:
\begin{enumerate}
\item there is a torsion class $\calS\covered \calT$ such that $\calT = \filt(\calS \cup M)$;
\item $M$ is a minimal extending module for $\calS$; and
\item $\calS$ contains $\calE \setminus \{M\}$.
\end{enumerate}
\end{lemma}
\begin{proof}
The torsion class $\calT=\Join \{\filt(\Gen(M)):\, M\in \calE\}$ is the smallest torsion class in $\tors \Lambda$ that contains the modules in $\calE$.
Thus, a module $N$ belongs to $\calT$ if and only if $N$ admits a filtration $N=N_l \supsetneq \cdots
\supsetneq N_0=0$ such that each $N_i/N_{i-1}$ is a factor of some indecomposable module in $\calE$.
For each $M\in \calE$, we claim that $M$ is a minimal co-extending module for the torsion-free class $\calT^\perp$.
First we check Property~\ref{copropertyone}.
Suppose that $Y$ is an indecomposable proper submodule of $M$ and that $Y\notin \calT^\perp$.
So, there exists a module $N\in \calT$ and a nonzero homomorphism $f: N\rightarrow Y$.
We may assume, without loss of generality, that $f$ is injective (if not, we take the map $f:N/\ker f\rightarrow Y$, noting that $N/\ker f$ is in $\calT$ by closure under factors).
From the filtration of $N$ described above, observe that the submodule $N_1$ is a factor of $M'$, for some $M'\in \calE$.
Also $f(N_1)$ is a non-trivial submodule of $Y$.
So, we have the following sequence of homomorphisms \[ M' \onto N_1 \onto f(N_1) \subseteq Y\subseteq M.\]
The composition of these homomorphisms is non-zero, contradicting our hypothesis that $\dim \Hom_\Lambda (M', M) = 0$.
Therefore, $Y\in \calT^\perp$.
Futhermore, since $M\in \calT$, Property~\ref{copropertythree} holds immediately.
To verify Property~\ref{copropertytwo}, suppose that $0 \rightarrow F \xrightarrow{i} X \xrightarrow{\pi} M \rightarrow 0$ is a non-split exact sequence with $F\in \calT^\perp$, and that $X\notin \calT^\perp$.
As above, let $f: N\rightarrow X$ be a non-zero homomorphism, where $N\in \calT$ indecomposable.
We may again assume that $f$ is injective, and in particular the restriction of the map to $N_1$ is injective.
As above, $N_1$ is a factor of some module $M'\in\calE$.
Thus, we have the nonzero composite homomorphism:
\[\tilde{f}: M' \onto N_1 \onto f(N_1) \subseteq X.\]
Since $\dim \Hom_{\Lambda}(M', M) =0$ the composition $\pi\circ \tilde{f}$ is zero.
Therefore, $\tilde{f}$ factors through the module $F$, and we have a nonzero homomorphism from $M'$ to $F$.
Since $F\in \calT^\perp$, that is a contradiction.
We conclude that $X\in \calT^\perp$, and $M$ a minimal co-extending module for~$\calT^\perp$.
Proposition~\ref{prop:torsion-free-to-torsion-relation} implies that there exists a torsion class $\calS\covered \calT$ such that $\calT= \filt(\calS \cup \{M\})$ and $M$ a minimal extending module for $\calS$.
To prove the third statement, suppose that $M'\in \calE\setminus \{M\}$ does not belong to~$\calS$.
Proposition~\ref{prop: epis and covers} implies that $M$ is a factor of $M'$, and that is a contradiction.
The statement follows.
\end{proof}
The following proposition completes our proof of Theorem~\ref{main: cjr}.
\begin{proposition}\label{cjr end}
Suppose that $\calE$ is a collection of hom-orthogonal $\Lambda$ brick modules and write $\calT$ for the torsion class $\Join \{\filt( \Gen M):\,M\in \calE\}$.
Then the expression $\Join \{\filt( \Gen M):\,M\in \calE\}$ is the canonical join representation of $\calT$.
\end{proposition}
\begin{proof}
We assume that $\calE$ has at least two elements (otherwise the statement follows from Theorem~\ref{schur modules to join-irreducibles}).
First we show that $\Join \{\filt(\Gen M):\,M\in \calE\}$ is irredundant.
Let $M\in \calE$, and consider the torsion class $\calT'= \Join \{\filt(\Gen M'):\,M'\in \calE\setminus \{M\}\}$.
By Lemma~\ref{lem:join-has-lower-faces}, there exists $\calS\covered \calT$ such that each module in $\calE\setminus \{M\}$ lies in $\calS$.
Thus $\calT' \le \calS \covered \calT$.
Next, we show the expression $\Join \{\filt(\Gen M):\,M\in \calE\}$ is the unique lowest irredundant join representation for $\calT$.
Suppose that $\Join A$ is another irredundant join representation.
We claim that for each torsion class $\filt(\Gen M)$ such that $M\in \calE$, there exists some $\calG\in A$ such that $\filt(\Gen M) \le \calG$.
(Thus, $\{\filt(\Gen M):\,M\in \calE\}$ is ``lower'' than $A$.)
%It suffices to show that for each $M\in \calE$, there is a torsion class $\calG\in A$ such that $M\in \calG$.
Fix a module $M\in \calE$, let $\calS$ be the torsion class covered by $\calT$ satisfying $\calT = \filt(\calS \cup \{M\})$ and $M$ is a minimal extending module for $\calS$.
(Such a torsion class $\calS$ exists by Lemma~\ref{lem:join-has-lower-faces}.)
Observe that there exists $\calG\in A$ such that $\calG\not \subseteq \calS$.
Indeed, if each $\calG$ is contained in $\calS$, then $\Join A \subseteq \calS$.
Let $N\in \calG \setminus \calS$.
Proposition~\ref{prop: epis and covers} implies that $M$ is a factor of $N$, hence $M\in \calG$.
We conclude that $\filt(\Gen M) \le \calG$.
We have proved the claim and the proposition.
\end{proof}
\begin{remark}\label{finite case}
\normalfont
When $\tors \Lambda$ is finite (equivalently, when $\Lambda$ is $\tau$-rigid finite) each join-irreducible torsion class is completely join-irreducible.
Thus, the canonical join complex $\Gamma(\tors \Lambda)$ is isomorphic to the complex of hom-orthogonal bricks.
Moreover, the proof Proposition~\ref{cjr end} implies that the number of canonical joinands of $\calT\in \tors \Lambda$ is equal to the number of torsion classes $\calS$ covered by $\calT$.
More precisely:
\begin{corollary}\label{cor: torsion labeling}
Suppose that $\calT$ is a torsion class over $\Lambda$ with the
following property: for every torsion class $\calS$ with
$\calS<\calT$, there is a torsion class $\calT'$ such that $\calS \leq
\calT' \lessdot \calT$.
Then the canonical join representation of $\calT$ is equal to \[\Join \{\filt(\Gen(M)):\,\text{$M$ is a minimal co-extending module of $\calT^\perp$}\}.\]
In particular, if $\Lambda$ is $\tau$-rigid finite, each torsion class has a canonical join representation.
\end{corollary}
\begin{proof}
The statement follows immediately form Corollary~\ref{hom orthogonality}.
\end{proof}
Indeed, the canonical join representation ``sees'' the geometry of the Hasse diagram for \emph{any} finite lattice.
We summarize this useful fact below (see~\cite[Proposition~2.2]{flag}).
\begin{proposition}\label{prop: labeling}
Suppose that $L$ is a finite lattice, and for each element $w\in L$ the canonical join representation of $w$ exists.
Then, for each $w\in L$, the number of canonical joinands of $w$ is equal to the number of elements covered by $w$.
\end{proposition}
\end{remark}
\begin{figure}[htb]
\centering
\scalebox{.7}{
\begin{tikzpicture}
%distributive lattice
%\draw[help lines] (-3,0) grid (3,3);
\draw[fill] (-2,0) circle [radius=.1];
\draw[fill] (-2,1) circle [radius=.1];
\draw[fill] (-3,2) circle [radius=.1];
\draw[fill] (-1,2) circle [radius=.1];
\draw[fill] (-2,3) circle [radius=.1];
\draw [black, line width=.65 mm] (-2,0) to (-2,1);
\draw[black, line width=.65 mm] (-2,1) to (-3,2);
\draw [black, line width=.65 mm] (-2,1) to (-1,2);
\draw [black, line width=.65 mm] (-2,3) to (-3,2);
\draw[black, line width=.65 mm] (-2,3) to (-1,2);
%nondistributive
\draw[fill] (2,0) circle [radius=.1];
\draw[fill] (1,1) circle [radius=.1];
\draw[fill] (3,1) circle [radius=.1];
\draw[fill] (1,2) circle [radius=.1];
\draw[fill] (2,3) circle [radius=.1];
\draw [black, line width=.65 mm] (2,0) to (1,1);
\draw [black, line width=.65 mm] (2,0) to (3,1);
\draw [black, line width=.65 mm] (1,1) to (1,2);
\draw [black, line width=.65 mm] (2,3) to (1,2);
\draw [black, line width=.65 mm] (2,3) to (3,1);
\end{tikzpicture}}
\caption{Two nonisomorphic lattices with isomorphic canonical join complexes.}
\label{fig: noniso}
\end{figure}
\begin{remark}\label{isomorphism}
\normalfont
It is natural to ask if the canonical join complex $\Gamma(\tors \Lambda)$ characterizes the underlying algebra $\Lambda$ or the torsion theory of $\Lambda$.
In fact, non-isomorphic algebras may have the same torsion theory.
We explore such an example in the following section when we show that the algebra $RA_n$ (an algebra of finite representation type for all $n$) has the same torsion theory as the preprojective algebra $\Pi A_n$ (which is \emph{not} finite representation type for $n\ge 4$).
Furthermore, nonisomorphic lattices $L$ and $L'$ may have isomorphic canonical join complexes.
For example, consider the two (nonisomorphic) lattices shown in Figure~\ref{fig: noniso}.
It is an easy exercise to check that canonical join complex of both lattices consists of an edge and an isolated vertex.
(See Example~\ref{Tamari example}.)
\end{remark}
\section{\texorpdfstring{$\tors RA_n$}{tors RAn} and the weak order on \texorpdfstring{$A_n$}{An}}
\label{sec:example:-shards}
Mizuno showed (in~\cite[Theorem~2.3]{Miz}) that the lattice of torsion classes for the preprojective algebra of Dynkin type $W$ is isomorphic to the weak order on the associated Weyl group, when $W$ is simply laced.
In this last section, we construct a different algebra, which we
refer to as $RA_n$, and show that $\tors RA_n$ is isomorphic to the weak
order on $A_n$.
This is carried out in two steps:
First, in Theorem~\ref{thm: cjc isomorphism}, we show that the canonical join complex of $\tors RA_n$ is isomorphic to the canonical join complex of the weak order on $A_n$.
Second, in Proposition~\ref{inversion and cover}, we map each cover relation $\calT\covered \calT'$ in $\tors RA_n$ bijectively to a cover relation of permutations in weak order.
%We will show in Section~\ref{RA_n background} that each indecomposable module corresponds to an orientation of some (possibly smaller) type A Dynkin Diagram.
%The benefit of working with $RA_n$ in lieu of $\Pi A_n$ is that $RA_n$ is
%finite type for all $n$.
%Our main tool will be an isomorphism $\Gamma$ from the canonical join complex of $\tors(RA_n)$ to the canonical join complex of the weak order on~$A_n$.
Before we begin, we establish some useful notation.
Throughout we write $[n]$ for the set $\{1,2,\ldots, n\}$, $[i,k]$ for
the set $\{i, i+1, \ldots, k\}$ and $(i,k)$ for $\{i+1, \ldots,
k-1\}$.
\subsection{The algebra \texorpdfstring{$RA_n$}{RAn}}\label{RA_n background}
Let $Q$ be the quiver with vertex set $Q_0=[n]$ and arrows $Q_1=\{a_i:
i\rightarrow i+1, a_i^*: i+1\rightarrow i\}_{i\in [n-1]}$. Define $I$ to be the two-sided ideal in the
path algebra $\kk
Q$ generated by all two-cycles, $I=\langle a_ia_i^*, a_i^*a_i \mid
i\in [n-1]\rangle$, and define $RA_n$ to be the algebra $\kk Q/I$.
Recall that a representation of $(Q,I)$ is a collection of vector spaces $M(x)$, one for each
vertex $i\in Q_0$, and linear maps $M(a_i): M(i)\rightarrow M(i+1)$,
$M(a_i^*):M(i+1)\rightarrow M(i)$, for each of the arrows in $Q$, that satisfy the
relations given by $I$. We will make generous use of the equivalence between the category of
modules over $RA_n$ and that of the representations of the bound
quiver $(Q, I)$, generally referring to the objects of interest as
modules, while describing them as representations.
The \newword{support} of a representation is the set of vertices $i$ for which $M(i)\neq 0$.
\begin{proposition}
There are finitely many isoclasses of indecomposable representations
of $RA_n$ for each $n$.
\end{proposition}
\begin{proof}
The algebra $RA_n$ is gentle (see~\cite{AS}) with no band modules since any cycle contains a 2-cycle,
each of which lies in $I$. By the work of Butler--Ringel~\cite{Butler-Ringel},
then, there are finitely many isoclasses of indecomposable modules.
\end{proof}
As a quiver representation, each indecomposable module $M$ (up to
isomorphism) over $RA_n$ corresponds to a connected subquiver $Q_M$ of
$Q$ satisfying the condition that at most one of either $a_i$ or $a_i^*$ belongs to $(Q_M)_1$.
Thus, each indecomposable module can be identified with an orientation of a type-A Dynkin diagram with rank less than or equal to $n$.
More precisely, the quiver representation corresponding to $Q_M$ satisfies:
\begin{itemize}
\item $M(i)=k$ for all $i\in (Q_M)_0$ and $M(i)=0$ for $i\notin (Q_M)_0$;
\item $M(a)\neq 0$ if and only if $a\in (Q_M)_1$.
\end{itemize}
%Because these algebras are representation directed, we obtain the following proposition.
%What's more, we can parameterize these indecomposables fairly
%easily. In particular, there is a bijection between indecomposable
%$RA_n$ modules and connected subquivers $Q'\subseteq Q$ such that for
%each $i\in [n]$, at most one of $a_i, a_i^*$ in $Q'_1$. The module $M$
%corresponding to $Q'$ has $\dim M(x) = \begin{cases} 1 &
% x\in Q'_0 \\ 0 & \textrm{otherwise}\end{cases}$, and $M(a)$ acts
%non-trivially if and only if $a\in Q'$. Under this bijection, we
%denote by $Q_M$ the subquiver of $Q$ corresponding to the
%indecomposable $M$. Since each representation of
%$RA_n$ can be identified as a representation of an orientation of a
%type-A Dynkin diagram, and these algebras are representation directed, we obtain the following proposition.
\begin{proposition}\label{cor: RA_n schur}
Each indecomposable module over $RA_n$ is a brick.
In particular, the canonical join complex of $\tors RA_n$ is isomorphic to the complex of hom-orthogonal indecomposable modules over $RA_n$.
\end{proposition}
\begin{proof}
Let $M$ be an indecomposable module over $RA_n$ and let $Q_M$ be the corresponding quiver.
An endomorphism $f:M\rightarrow M$ is a set of maps $f=(f_i)_{i\in Q_1}$ where $f_i:M(i)\rightarrow~M(i)$ and for every arrow $a:i\rightarrow j\in Q_1$, the composition $M(a)\circ f_i$ is equal to ${f_j\circ M(a)}$. Since $M(i)=k$ for all $i\in (Q_M)_0$, the map $M(a):k\rightarrow k$ is just a scalar multiplication. Therefore $f_i=f_j$ for each $i,j\in (Q_M)_0$, and hence $f$ is a scalar multiple of the identity map.
\end{proof}
%
%\begin{figure}[h]
%\centering
%\scalebox{.8}{
%\begin{tikzpicture}
%%\draw[help lines] (-7,-3) grid (7,3);
%%more help lines
%\draw [black] (0,-3) to (0,3);
%\draw [black] (-4,-3) to (-4,3);
%\draw [black] (4,-3) to (4,3);
%\draw[black] (-7,0) to (7,0);
%
%%top row
%%first
%\node at (-6.5,1) {$1$};
%\node at (-6,1.5) {$2$};
%\node at (-5.5,1) {$3$};
%\node at (-5,1.5) {$4$};
%
%%second
%\node at (-2.5,1.5) {$1$};
%\node at (-2,1) {$2$};
%\node at (-1.5,1.5) {$3$};
%\node at (-1,1) {$4$};
%
%%third
%\node at (2.5,1.5) {$4$};
%\node at (2,2) {$3$};
%\node at (1.5,1.5) {$2$};
%\node at (1,1) {$1$};
%
%%fourth
%\node at (6.5,1.5) {$4$};
%\node at (6,1) {$3$};
%\node at (5.5,1.5) {$2$};
%\node at (5,2) {$1$};
%
%%bottom row
%%first
%\node at (-6.5,-1.5) {$1$};
%\node at (-6,-1) {$2$};
%\node at (-5.5,-1.5) {$3$};
%\node at (-5,-2) {$4$};
%
%%second
%\node at (-2.5,-1.5) {$1$};
%\node at (-2,-2) {$2$};
%\node at (-1.5,-1.5) {$3$};
%\node at (-1,-1) {$4$};
%
%%third
%\node at (2.5,-2.5) {$4$};
%\node at (2,-2) {$3$};
%\node at (1.5,-1.5) {$2$};
%\node at (1,-1) {$1$};
%
%%fourth
%\node at (6.5,-1) {$4$};
%\node at (6,-1.5) {$3$};
%\node at (5.5,-2) {$2$};
%\node at (5,-2.5) {$1$};
%
%\end{tikzpicture}}
%\caption{A visualization of the indecomposable modules with full
% support over $RA_4$, where one vertex $i$ is above its neighbor $j$ if the
% arrow $i\rightarrow j$ acts non-trivially on $M$.}
%\label{fig: quivers for A4}
%\end{figure}
From now on, we abuse notation and refer to $\Gamma(\tors RA_n)$ as the complex of hom-orthogonal indecomposable modules over $RA_n$.
(Although, more precisely, $\Gamma(\tors RA_n)$ is a simplicial complex on the set of join-irreducible torsion classes $\filt(\Gen(M))$ in $\tors RA_n$, not the set of indecomposable modules.)
We close this section with a technical lemma that will be useful in Section~\ref{sec:shards}.
\begin{lemma}\label{lem: ext}
Suppose that $M$ is an indecomposable module over $RA_n$, and
$S$ is an interval in $[n]$ containing $\supp(M)$. Then:
\begin{enumerate}
\item $M$ is a submodule of some indecomposable $M'$ with
$\supp(M')=S$ and
\item there is an indecomposable $M''$ with $\supp(M'')=S$, of which
$M$ is a quotient.
\end{enumerate}
\end{lemma}
\begin{proof}
We prove only the first statement, since the second is similar by
Proposition~\ref{cor: factors and subs}. Let $Q_M$ be the quiver
associated with $M$, and write $[p,q]=\supp(M)$. Let $Q_{M'}$ be any
quiver with support equal to $S$ satisfying the following:
the orientation of $Q_{M'}$ on the interval $[p,q]$ coincides with that of $Q_M$, and
$Q_{M'}$ contains the the arrows $a_{p-1}$ if $p-1\in S$ and $a_{q}^*$ if $q+1\in S$.
Since $Q_M$ is a connected successor closed subquiver of $Q_{M'}$, we
obtain the desired result.
\end{proof}
%The weak order on any finite Coxeter group is a finite semidistributive lattice~\cite[Lemma~9]{weak order}.
%Thus, each permutation has a canonical join representation.
%The \newword{transitive closure} of $S$, denoted $\tran(S)$, is the set of all pair $(i,k)$ satisfying:
%There is a finite subset $\{(i_0, j_0), (j_0, j_1), \ldots, (j_r, k)\}\subseteq S$.
\subsection{Noncrossing arc diagrams and canonical join representations}
\label{sec:arc-diagrams}
In this section, we construct a model for the canonical join complex of $\tors RA_n$ called the noncrossing arc complex.
The noncrossing arc complex was first defined in~\cite{arcs} where it was used to study certain aspects of the symmetric group.
(We will make use of this connection in the following section.)
Informally, the noncrossing arc complex is a simplicial complex whose faces are collections of non-intersecting curves called arcs.
We will see that the ``noncrossing'' criteria that defines such a face also encodes the hom-orthogonality of indecomposable modules in $\module RA_n$.
%For further background, we direct the reader to~\cite{arcs}.
A \newword{noncrossing arc diagram} on $n+1$ nodes consists of a vertical column of nodes, labeled $0,\ldots, n$ in increasing order from bottom to top, together with a (possibly empty) collection of curves called arcs.
Each arc $\alpha$ has two endpoints, and travels monotonically up from its bottom endpoint $b(\alpha)$ to its top endpoint~$t(\alpha)$.
For each node in between, $\alpha$ passes either to the left or to the right.
Each pair of arcs $\alpha$ and $\beta$ in a diagram satisfies two compatibility conditions:
\begin{enumerate}[label={(C\arabic*)}]
\item\label{C1} $\alpha$ and $\beta$ do not share a bottom endpoint or a top endpoint;
%\item\label{C2} $\alpha$ and $\beta$ do not share a common top endpoint;
\item\label{C2} $\alpha$ and $\beta$ do not cross in their interiors.
\end{enumerate}
%Noncrossing diagrams were first defined by Reading in~\cite{arcs}, where it was shown that the set of noncrossing diagrams on $n+1$ vertices are in bijection with the set of permutations in $A_n$
\begin{figure}[htb]
\centering
\scalebox{1}{ \includegraphics{168_Figures/Sym3.pdf}}
\caption{The noncrossing arc diagrams on $3$ nodes.}
\label{S3 arcs}
\end{figure}
Each arc is considered only up to combinatorial
equivalence.
That is, each arc $\alpha$ is characterized by its endpoints and which
side each node the arc passes (either left or right) as it travels from
$b(\alpha)$ up to $t(\alpha)$.
Furthermore, a collection of arcs is drawn so as to have the smallest
number of intersections.
The \newword{support of an arc} $\alpha$, written $\supp(\alpha)$, is the set $[b(\alpha), t(\alpha)]$.
We write $\suppo(\alpha)$ for the set $(b(\alpha), t(\alpha))$.
We say that $\alpha$ has full support if $\supp(\alpha) = [0,n]$.
We say that $\beta$ is a \newword{subarc} of $\alpha$ if both of the following conditions are satisfied:
\begin{itemize}
\item $\supp(\beta)\subseteq \supp(\alpha)$;
\item $\alpha$ and $\beta$ pass on the same side of each node in $\suppo(\beta)$.
\end{itemize}
\begin{figure}[htb]
\centering
\scalebox{.45}{
\begin{tikzpicture}
%\draw[help lines] (-4,-2) grid (4,2);
%\draw[black] (1,-2) to (1,2);
%\draw[black] (-1.5,-2) to (-1.5,2);
%\draw[fill] (0,0) circle [radius=.04];
%diagram 1
\draw[fill] (-5,1) circle [radius=.07];
\draw[fill] (-5,-2) circle [radius=.07];
\draw[fill] (-5,0) circle [radius=.07];
\draw[fill] (-5,2) circle [radius=.07];
\draw[fill] (-5,-1) circle [radius=.07];
\draw [black, thick] (-5,-2) to [out= 150, in =-150 ] (-5,2);
\draw [red, ultra thick, dashed] (-5,-1) to [out= 30, in =-30 ] (-5,1);
%diagram 2
\draw[fill] (-2,1) circle [radius=.07];
\draw[fill] (-2,-2) circle [radius=.07];
\draw[fill] (-2,0) circle [radius=.07];
\draw[fill] (-2,2) circle [radius=.07];
\draw[fill] (-2,-1) circle [radius=.07];
\draw [black, thick] (-2,-2) to [out= 150, in =-150 ] (-2,2);
\draw [red, ultra thick, dashed] (-2,-1) to [out= 150, in =-150 ] (-2,1);
%diagram 3
\draw[fill] (0,1) circle [radius=.07];
\draw[fill] (0,-2) circle [radius=.07];
\draw[fill] (0,0) circle [radius=.07];
\draw[fill] (0,2) circle [radius=.07];
\draw[fill] (0,-1) circle [radius=.07];
\draw [black, thick] (0,-2) to [out= 150, in =210 ] (0,-.5);
\draw [black, thick] (0,-.5) to [out = 30, in = -30] (0,.5);
\draw [black, thick] (0,.5) to [out= 150, in =-150 ] (0,2);
\draw [red, ultra thick, dashed] (0,-1) to [out= 30, in =-30 ] (0,1);
%\draw [black, thick] (-2,1.5) to [out= 30, in =-30 ] (-2,2);
%diagram 4
\draw[fill] (2,1) circle [radius=.07];
\draw[fill] (2,-2) circle [radius=.07];
\draw[fill] (2,0) circle [radius=.07];
\draw[fill] (2,2) circle [radius=.07];
\draw[fill] (2,-1) circle [radius=.07];
\draw [black, thick] (2,-1) to [out= 150, in =210 ] (2,.5);
\draw [black, thick] (2, .5) to [out= 30, in =-30 ] (2,2);
\draw [red, ultra thick, dashed] (2,-2) to [out= 30, in =-30 ] (2,0);
%%%diagram 5
\draw[fill] (4.5,1) circle [radius=.07];
\draw[fill] (4.5,-2) circle [radius=.07];
\draw[fill] (4.5,0) circle [radius=.07];
\draw[fill] (4.5,2) circle [radius=.07];
\draw[fill] (4.5,-1) circle [radius=.07];
%arcs
\draw [black, thick] (4.5,-2) to [out= 150, in =-150 ] (4.5,0);
\draw [red, ultra thick, dashed] (4.5,-1) to [out= 30, in =-30 ] (4.5,2);
\end{tikzpicture}
}
\caption{Some pairs of compatible arcs.}
\label{fig: left and right position}
\end{figure}
A set of arcs are \newword{compatible} if there is a noncrossing arc diagram that contains them.
We define the \newword{noncrossing arc complex} on $n+1$ nodes to be the simplicial complex whose vertex set is the set of arcs and whose face set is the collection of all sets of compatible arcs.
(We view each collection of compatible arcs as a noncrossing arc diagram.
When we refer to ``the set of arcs'' we mean ``the set of noncrossing arc diagrams, each of which contains precisely one arc''.)
The next proposition is a combination of~\cite[Proposition~3.2 and Corollary~3.6]{arcs}.
Recall that a simplicial complex is \newword{flag} if its minimal non-faces have size equal to 2.
Equivalently, a subset~$F$ of vertices is a face if and only if each pair of vertices is a face.
\begin{proposition}\label{arc flag}
A collection of arcs can be drawn together in a noncrossing arc diagram if and only if each pair of arcs is compatible.
Thus, the noncrossing arc complex is flag.
\end{proposition}
Our goal is to prove:
\begin{theorem}\label{thm: cjc isomorphism}
The canonical join complex of the lattice $\tors RA_n $ is isomorphic to the noncrossing arc complex on $n+1$ nodes.
\end{theorem}
We begin the proof of Theorem~\ref{thm: cjc isomorphism} by mapping vertices to vertices.
More precisely, we define a bijection $\sigma$ from the set of
indecomposable modules over $RA_n$ to the set of arcs on $n+1$ nodes.
For an arc $\alpha$ with support $[p-1,q]$ we define:
\begin{align*}
R(\alpha)&=\{i\in [p,q-1]:\, \text{$\alpha$ passes on the right side of $i$}\};\\
L(\alpha)&=\{i\in [p,q-1]:\, \text{$\alpha$ passes on the left side of $i$}\}.
\end{align*}
For an indecomposable $RA_n$ module with support $[p,q] \subseteq [n]$,
we define:
\begin{align*}
R(M)&=\{i\in [p,q-1]:\, \text{$a_{i}$ acts nontrivially on $M$}\};\\
L(M)&= \{i\in [p,q-1]:\, \text{$a_{i}^*$ acts nontrivially on $M$}\}.
\end{align*}
%Define $\sigma(M)$ to be the
%arc with support $[p-1,q]$ which passes to the right (resp. left) of the node $i\in
%(p-1,q)$ if $a_i$ (resp. $a_i^*$) acts non-trivially on $M$.
Just as an arc is determined by its endpoints and the binary Left-Right
data, so too is an indecomposable module over $RA_n$ determined by the
binary data of the action of its Lowering-Raising arrows ($a_i^*$ and
$a_i$, respectively). Therefore, we have the following:
\begin{proposition}\label{prop: arc bij}
Let $\sigma$ be the map which sends an indecomposable~$RA_n$ module~$M$ with support $[p,q-1]$ to the arc $\sigma(M)=\alpha$ satisfying
\begin{itemize}
\item $b(\alpha)=p-1$ and $t(\alpha)=q$;
\item $L(\alpha)=L(M)$;
\item $R(\alpha)=R(M).$
\end{itemize}
Then $\sigma$ is a bijection from the set of indecomposable modules
over $RA_n$ the set of arcs on $n+1$ nodes. (See Figure~\ref{fig:
arcmap example}.)
\end{proposition}
\begin{figure}[htb] \centering
\scalebox{.8}{%
%\begin{subfigure}[b]{0.4\linewidth}
\begin{tikzpicture}
\node (A) at (0,0.5) {1};
\node (B) at (0,1.5) {2};
\node (C) at (0,2.5) {3};
\node (D) at (0,3.5) {4};
\draw[->] (B)--(A) node[midway,right] {$a_1^*$};
\draw[->] (B)--(C) node[midway,left] {$a_2$};
\draw[->] (D)--(C) node[midway,right] {$a_3^*$};
\draw (-3,-0.5);
\end{tikzpicture}}
\put(-58,-5){(A)~$Q_M$ for $RA_4$}%
\put(80,-5){(B)~The arc $\sigma(M)$}
%\caption{$Q_M$ for $RA_4$}
%\end{subfigure}
\qquad\qquad\qquad\qquad
%\begin{subfigure}[b]{0.3\textwidth}
\scalebox{.8}{
\begin{tikzpicture}
\foreach \n in {0,1,2,3,4}{
\node[draw] at (0,\n) {\n};
}
\draw[thick] plot [smooth,tension=1] coordinates
{(-0.25,0) (-0.5,1) (0.5,2) (-0.5,3) (-0.25,4)};
\draw (-1.5,-0.5);
\end{tikzpicture}}
%\caption{The arc $\sigma(M)$} %HERVE
%\end{subfigure}
%}
\caption{Visualization of $\sigma(M)$ and $Q_M$ for a module over $RA_4$. }
\label{fig: arcmap example}
\end{figure}
For the remainder of the section, we let $M$ and $M'$ be indecomposable modules, and write $\alpha$ for $\sigma(M)$ and $\alpha'$ for $\sigma(M')$.
We wish is to reinterpret the hom-orthogonality of $M$ and $M'$ in terms of certain subarcs of $\alpha$ and $\alpha'$.
Recall that a quiver $Q'$ is called a \newword{predecessor closed subquiver of $Q$} if
$i\rightarrow j$ with $j\in Q'$ implies $i\in Q'$. \newword{Successor closed
subquivers} are defined similarly.
The following result is well-known (and an easy exercise).
See~\cite[Section~2]{Crawley-Boevey}.
\begin{proposition}\label{cor: factors and subs}
Suppose that $M$ and $M'$ are indecomposable modules over $RA_n$
and let $Q_M$ and $Q_{M'}$ be the corresponding quivers.
Then:
\begin{enumerate}
\item $M'$ is a quotient of $M$ if and only if $Q_{M'}$ is a connected predecessor closed subquiver of $Q_M$.
\item $M'$ is a submodule of $M$ if and only if $Q_{M'}$ is a connected successor closed subquiver of $Q_M$.
\end{enumerate}
\end{proposition}
We define an analogous notion for arcs.
We say that $\beta$ is a \newword{predecessor closed subarc} of $\alpha$ if $\beta$ is a subarc of $\alpha$, and $\alpha$ does not pass to the right of $b(\beta)$, nor to the left of $t(\beta)$.
Similarly, $\beta$ is a \newword{successor closed subarc} if $\alpha$ does not pass to the left of $b(\beta)$ nor to the right of $t(\beta)$.
Observe that each predecessor closed subarc of $\alpha$ corresponds (via the map $\sigma$) to an indecomposable quotient of~$M$--that is, a \emph{predecessor closed} subquiver of $Q_M$.
(The analogous statement holds for successor closed subarcs of $\alpha$.)
The next result is from Crawley-Boevey~\cite[Section~2]{Crawley-Boevey}, rephrased for our context.
\begin{proposition}\label{prop:homs-between-shards}
Let $\alpha$ and $\alpha'$ be arcs on $n+1$ nodes, and write $M$ and $M'$ for the corresponding indecomposable modules over $RA_n$.
Then the vector space $\Hom_{RA_n} (M, M')$ has dimension equal to the number of predecessor closed subarcs of $\alpha$ which are also successor closed subarcs of $\alpha'$.
\end{proposition}
%By Theorem~\ref{main: canonical join complex}, the torsion classes $\calT$ and $\calT'$ are compatible if and only if there is no arc $\beta$ is that is a predecessor closed subarc of one arc, either $\alpha$ or $\alpha'$, and also a successor closed subarc of the other arc.
%Thus, to prove Theorem~\ref{thm: iso to arc complex}, we argue that $\alpha$ and $\alpha'$ are compatible if and only if there exists no arc $\beta$ that is a predecessor closed subarc of one, either $\alpha$ or $\alpha'$ and successor closed subarc of the other.
We argue that $\alpha$ and $\alpha'$ are compatible if and only if
there exists no arc~$\beta$ that is a predecessor closed subarc of one
which is simultaneously a successor closed subarc of the other.
\begin{figure}[htb]
\scalebox{.8}{
\begin{tikzpicture}
%\draw[help lines] (-1,-3) grid (1,3);
%\draw[fill] (0,0) circle [radius=.04];
%nodes of the diagram
\draw[fill] (0,1) circle [radius=.02];
\draw[fill] (0,.5) circle [radius=.02];
%\draw[fill] (0,.25) circle [radius=.02];
\draw[fill] (0,.75) circle [radius=.02];
\draw[fill] (0,1.25) circle [radius=.02];
\draw[fill] (0,1.75) circle [radius=.02];
\draw[fill] (0,1.5) circle [radius=.02];
\draw[fill] (0,-2) circle [radius=.07];
\draw[fill] (0,2) circle [radius=.07];
\draw[fill] (0,-1) circle [radius=.02];
\draw[fill] (0,-.5) circle [radius=.02];
\draw[fill] (0,-1.25) circle [radius=.02];
\draw[fill] (0,-1.75) circle [radius=.02];
\draw[fill] (0,-1.5) circle [radius=.02];
%\draw[fill] (0,-.25) circle [radius=.02];
\draw[fill] (0,-.75) circle [radius=.02];
%arcs
\draw [red, ultra thick, dashed] (1,-2.5) to [out= 90, in =-20 ] (0,0);
\draw [red, ultra thick, dashed] (0,0) to [out= 160, in =-90 ] (-1,2.5);
\draw [black, thick] (-1,-2.5) to [out= 90, in =200 ] (0,0);
\draw [black, thick] (0,0) to [out = 20, in = -90] (1,2.5);
%labels
\node at (-1.5, -2) {$\alpha$};
\node at (0,-2.5) {$b(\beta)$};
\node at (0, 2.5) {$t(\beta)$};
\node at (1.5, -2) {$\alpha'$};
\end{tikzpicture}
}
\caption{Suppose that $\beta$ is a predecessor closed subarc of $\alpha$ and a successor closed subarc of $\alpha'$.
Then the endpoints of~$\beta$ lie between these two arcs.
The two arcs switch from left side to right side (and vice versa) as they travel from $b(\beta)$ to~$t(\beta)$. }
\label{fig: subarc}
\end{figure}
The following two lemmas give one direction of that argument.
Figure~\ref{fig: subarc} may help build some intuition.
At times it will be convenient to consider the relative position of a pair of ``overlapping'' arcs.
We say that $\alpha$ and $\alpha'$ \newword{overlap} if the set $$\left(\supp(\alpha)\cap\suppo(\alpha')\right)\cup \left(\supp(\alpha')\cap \suppo(\alpha)\right)$$ is nonempty.
For two arcs $\alpha$ and $\alpha'$ that overlap, we say that $\alpha'$ is right of $\alpha$ if both of the following hold:
\begin{itemize}
\item $\left(L(\alpha') \cup \{t(\alpha'),b(\alpha')\}\right) \cap \left(\supp(\alpha) \right)\subseteq L(\alpha)$
\item $\{t(\alpha), b(\alpha)\} \cap \left(\supp(\alpha')\right) \subseteq R(\alpha')$.
\end{itemize}
%When $\alpha$ and $\alpha'$ are compatible, we say that $\alpha$ is
%left of $\alpha'$ if, whenever $\alpha$ passes to the right of a node, either $\alpha'$ also passes to the right of the same node or $\alpha'$ ends at that node.
For example, in Figure~\ref{fig: left and right position}, the dashed arc is \emph{right} of the solid arc.
Observe that if $\alpha'$ is right of $\alpha$ then the two arcs are compatible.
\begin{lemma}\label{special subarc shared endpoint}
Suppose that $\alpha$ and $\alpha'$ are distinct arcs that share a bottom endpoint or a top endpoint.
Then there exists an arc $\beta$ satisfying:
$\beta$ is a predecessor closed subarc of one arc, either $\alpha$ or $\alpha'$, and a successor closed subarc of the other arc.
\end{lemma}
\begin{proof}
By symmetry, we assume that $\alpha$ and $\alpha'$ share a bottom node, and, without loss of generality, this bottom endpoint is equal to 0.
Let $q$ be the smallest number such that either of two conditions below is satisfied:
\begin{itemize}
\item $\alpha$ and $\alpha'$ pass on opposite sides of $q$;
\item $q= \min\{t(\alpha), t(\alpha')\}$.
\end{itemize}
Let $\beta$ be the arc with endpoints $b(\beta)=0$ and $t(\beta)=q$ such that $\beta$ is a subarc of $\alpha$.
(Note, if $q= t(\alpha)$ then $\beta$ coincides with $\alpha$.
If $q\ne t(\alpha)$, we can visualize $\beta$ by cutting $\alpha$ where it passes beside the node $q$, and anchoring the resulting segment at $q$.)
Since $\alpha$ and $\alpha'$ pass on the same side of each node in the
set $[0,q-1]$, we conclude that $\beta$ is also a subarc of $\alpha'$.
At least one of the two arcs, $\alpha$ or $\alpha'$, passes $q$.
By symmetry, assume that this is the arc $\alpha$, and assume that $\alpha$ passes on the right side of $q$.
Then $\beta$ is a predecessor closed subarc of $\alpha$ and also a
successor closed subarc of $\alpha'$.
%The other choices yield similar results.
%If $\alpha$ passes to the left, then $\beta$ is a successor closed subarc of $\alpha$ and also a predecessor closed subarc of $\alpha'$.
\end{proof}
\begin{lemma}\label{crossing}
Suppose that $\alpha$ and $\alpha'$ are distinct arcs that neither
share a bottom nor top endpoint.
If $\alpha$ and $\alpha'$ intersect in
their interiors, then there exists an arc $\beta$ which is a predecessor closed subarc of one arc and a successor closed subarc of the other.
\end{lemma}
\begin{proof}
\looseness-1
First, we consider the case in which one of the two arcs is a subarc of the other.
By symmetry, we assume that $\alpha'$ is a subarc of $\alpha$.
Observe that $b(\alpha')$ and $t(\alpha')$ must be on opposite sides of $\alpha$.
(Otherwise, $\alpha$ and $\alpha'$ can be drawn so they do not cross.)
Therefore, by definition, either $\alpha'$ is a predecessor closed or
successor closed subarc of~$\alpha$.
Next, suppose that $\alpha'$ and $\alpha$ pass on the same side of each node in the set $\suppo(\alpha)\cap\suppo(\alpha')$, but neither arc is a subarc of the other.
By symmetry, assume that $b(\alpha')**w_j$ and $ii+1$
(resp. $j***w_{i+1}$.
% (That is, whenever the pair $w_i$ and $w_{i+1}$ form a descent.)
% Finally, we move all of the points into a vertical line, bending the straight line segments so that they become the arcs in our diagram.
Since descents become arcs, it follows that $\delta$ restricts to a bijection from the set of join-irreducible permutations (on $[0,n]$) to the set of arcs (on $n+1$ nodes).
\begin{figure}[htb]
\centering
\scalebox{.8}{
\begin{tikzpicture}
%\draw [black, thick] (0.1,1.45) -- (.68,2.1);
\draw [black, thick] (0,0) -- (-1,.9);
\draw [black, thick] (0,0) -- (1,.9);
\draw [black, thick] (1,1) -- (1,1.9);
\draw [black, thick] (-1,1) -- (-1,1.9);
\draw [black, thick] (1,2) -- (.1,2.9);
\draw [black, thick] (-1,2) -- (-.1,2.9);
\node at (0,-.2) {$012$};
\node at (1.4,1) {$021$};
\node at (-1.4,1) {$102$};
\node at (1.4,2) {$201$};
\node at (-1.4,2) {$120$};
\node at (0,3.2) {$210$};
\end{tikzpicture}
\qquad
\qquad
\qquad
%
\begin{tikzpicture}
%nodes
\draw[fill] (0,2) circle [radius=.07];
\draw[fill] (0,1) circle [radius = .07];
\draw[fill] (0,0) circle [radius=.07];
%arcs
\draw [black, thick] (0,0) to (0,1);
\draw [black, thick] (0,2) to (0,1);
%\draw [black, thick] (0,0) to [out = 20, in = -90] (1,2.5);
%labels
\node at (0.25,0) {$0$};
\node at (.25, 2) {$2$};
\node at (.25, 1) {$1$};
\end{tikzpicture}
}
\caption{The weak order on the symmetric group $A_2$ and the noncrossing arc diagram corresponding to $210$.}
\label{cayley}
\end{figure}
\begin{example}\label{delta example}
\normalfont
Consider the permutation $w= 210$, the top element in the weak order on $A_2$.
The noncrossing arc diagram $\delta(w)$ consists of two arcs, one connecting 0 to 1, and the second connecting 1 to 2.
(See Figure~\ref{cayley}.)
Each arc corresponds (via $\delta$) to a join-irreducible permutation.
In this example, the arc connecting 0 to 1 corresponds to $102$, and the arc which connects 1 to 2 corresponds to $021$.
Observe that that $210 = \Join \{102, 021\}$.
In fact, $\Join \{102, 021\}$ is the canonical join representation of $210$.
This fact is no coincidence.
\end{example}
In general, the weak order on $A_n$ is a lattice in which each permutation has a canonical join representation~\cite{weak_order}.
The next theorem is a combination of~\cite[Theorem~3.1 and
Corollary~3.4]{arcs}.
\begin{theorem}\label{prop: A_n CJC}
The bijection~$\delta$ induces an isomorphism from the canonical join complex of the weak order on $A_n$ to the noncrossing arc complex on $n+1$ nodes.
\end{theorem}
We immediately obtain the following corollary.
\begin{corollary}\label{cor: RA_n cjc}
The canonical join complex of $RA_n$ is isomorphic to the canonical join complex of the weak order on $A_n$.
\end{corollary}
Let us a consider the following composition:
\[
\Gamma(\tors RA_n) \xrightarrow{\sigma} \{\text{Noncrossing arc diagrams on $n+1$ nodes}\} \xrightarrow{\delta^{-1}} \Gamma(A_n).
\]
%We claim that $\delta^{-1} \circ \sigma$ induces a bijection from the set of torsion classes in $\tors RA_n$ to the set of permutations in $A_n$.
Recall that the map $\sigma$ sends a collection of hom-orthogonal modules $\calE$ to a collection of compatible arcs $\sigma(\calE)= A$.
By Theorem~\ref{prop: A_n CJC}, $\delta^{-1}$ sends this collection of compatible arcs to the permutation $w = \Join \{\delta^{-1}(\alpha):\, \alpha\in A\}$, where this join $\Join \{\delta^{-1}(\alpha):\, \alpha\in A\}$ is the canonical join representation of $w$.
Define a map ${\phi: \tors RA_n \to A_n}$ as follows:
\[\phi: \Join \{\filt(\Gen(M)):\, M\in \calE\} \mapsto \Join \{\delta^{-1}(\sigma(M)):\, M\in \calE\}.\]
%Because each torsion class in $\tors RA_n$ has a canonical join representation (by Corollary~\ref{cor: torsion labeling}) and because the canonical join representation is unique, we conclude that $\phi$ is a bijection.
Because each torsion class in $\tors RA_n$ and each permutation in $A_n$ has a canonical join representation (see Corollary~\ref{cor: torsion labeling}), and because the canonical join representation is unique, we conclude that $\phi$ is a bijection.
\begin{example}\label{phi example}
\looseness-1
Consider the join of torsion classes $\calT = \Join \{\{1\}, \{2\}\}$ in $\tors RA_2$.
(We write $\{i\}$ for the torsion class consisting of the simple module at vertex $i$.)
Observe that $\calT$ is equal to the entire module category over $RA_2$.
Since the simple modules $1$ and $2$ are hom-orthogonal, the join $\Join \{\{1\}, \{2\}\} $ is the canonical join representation of $\calT$.
The map $\sigma$ sends $\{i\}$ to the arc with endpoints $i-1$ and $i$.
Thus, $\sigma(\calT)$ is the noncrossing arc diagram with two arcs: one arc that connects 0 to 1; and another arc that connects~1 to~2.
Recall from Example~\ref{delta example} that the permutation associated to this diagram is $210$.
Thus, $\phi(\calT)= 210$.
\end{example}
\looseness-1
Our goal is to show that $\phi$ is actually a lattice isomorphism.
To that end, we now give an alternative description of $\phi$ in terms of inversion sets.
Recall that permutations in $A_n$ are ordered by containment of inversion sets.
Each inversion set $I = \inv(w)$ is \newword{transitively closed}, meaning that whenever $\{(p,q), (q,r)\}$ is a subset of $I$ then $(p,r)\in I$.
Consider the set $S$ of all pairs $(p,q)$ such that $p*r$ such that $(r,s)\in \inv(\calT)$ then $(q,s)$ also belongs to $\inv(\calT)$.
We conclude that $\inv(\calT') = \inv(\calT)\cup \{(q,r)\}$, as desired.
In particular, $\phi(\calT')\covers \phi(\calT)$.
For any finite lattice, define $\covdown(L)$ to be the set of pairs $(w',w)$ such that $w'\covers w$.
As is the case for $\tors RA_n$ and $A_n$, suppose that each element $w'\in L$ has a canonical join representation.
Recall that the number of canonical joinands of $w'$ is equal to the number of elements covered by $w'$.
(This is Proposition~\ref{prop: labeling}.)
Thus, the number of pairs $(w',w)$ in $\covdown(L)$ is equal to a weighted sum of the faces in $\Gamma(L)$, where each face is weighted by its size.
In particular, the sets $\covdown(\tors RA_n)$ and $\covdown(A_n)$ are equinumerous.
Since $\phi$ is a bijection, we conclude that the mapping $(\calT', \calT) \mapsto (\phi(\calT),\phi(\calT'))$ is a bijection from $\covdown(\tors RA_n)$ to $\covdown(A_n)$.
Hence, the lattice $\tors RA_n$ is isomorphic to the weak order on $A_n$.
\end{proof}
\longthanks{The authors thank David Speyer and Gordana Todorov, who helped start this project.
We also thank Alexander Garver and Thomas McConville, who led us to the correct statement of Theorem~\ref{main: cjr}.}
%\nocite{*}
\bibliographystyle{amsplain-ac}
\bibliography{ALCO_Barnard_168}
\end{document}