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%%%%% Auteur
\author{\firstname{Alexander} \lastname{Garver}}
\address{Department of Mathematics\\
University of Michigan}
\email{alexander.garver@gmail.com}
\author{\firstname{Thomas} \lastname{McConville}}
\address{Department of Mathematics\\
Kennesaw State University}
\email{thomasmcconvillea@gmail.com}
%%%%% Sujet
\subjclass{05A19, 06A07, 06B10}
\keywords{lattice, Catalan number, Cambrian lattice, noncrossing partition, nonkissing complex}
%%%%% Gestion
\DOI{10.5802/alco.142}
\datereceived{2020-01-03}
\dateaccepted{2020-08-02}
%%%%% Titre et résumé
\title{Chapoton triangles for nonkissing complexes}
\begin{abstract}
We continue the study of the nonkissing complex that was introduced by Petersen, Pylyavskyy, and Speyer and was studied lattice-theoretically by the second author. We introduce a theory of Grid--Catalan combinatorics, given the initial data of a nonkissing complex, and show how this theory parallels the well-known Coxeter--Catalan combinatorics. In particular, we present analogues of Chapoton's $F$-triangle, $H$-triangle, and $M$-triangle and give both combinatorial and lattice-theoretic interpretations of the objects defining these polynomials. In our Grid--Catalan setting, we prove that our analogue of Chapoton's $F$-triangle and $H$-triangle identity holds, and we conjecture that our analogue of Chapoton's $F$-triangle and $M$-triangle identity also holds.
\end{abstract}
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\begin{document}
\maketitle
\section{Introduction}
The Catalan numbers famously enumerate many naturally occuring classes of objects in mathematics~\cite{stanley2015catalan}. Among the most significant of these classes is the set of maximal faces of the \emph{simplicial associahedron}. This is a simplicial complex whose vertices correspond to $2$-element subsets of $[n]:=\{1,\ldots,n\}$ that are not cyclically adjacent modulo $n$. Its faces are collections of noncrossing collections of $2$-element subsets of $[n]$. Here, two subsets $\{i,k\}, \{j,l\} \subseteq [n]$ are \emph{noncrossing} unless $i\bigvee B$ for all proper subsets $B\subsetneq A$. We observe that an element $x$ is \emph{join-irreducible} if and only if the only irredundant join-representation of $x$ is $\{x\}$. We partially order irredundant join-representations of $x$, where $A\leq B$ means that for all $a\in A$ there exists $b\in B$ with $a\leq b$. If the set of irredundant join-representations of $x$ has a minimum element, this minimum representation is called the \emph{canonical join-representation} of $x$. The elements in a canonical join-representation are necessarily join-irreducible.
A lattice $L$ is \emph{semidistributive} if
\begin{align*}
x\vee z &=y\vee z\ \ \mbox{implies}\ \ x\vee z=(x\wedge y)\vee z\ \ \mbox{and}\\
x\wedge z &= y\wedge z \ \ \mbox{implies}\ \ x\wedge z=(x\vee y)\wedge z
\end{align*}
for all $x,y,z\in L$. Equivalently~\cite[Theorem 2.24]{freese.jezek.nation:1995free}, a lattice is semidistributive if and only if every element admits a canonical join-representation and a canonical meet-representation, defined dually.
If a set $A$ is a canonical join-representation of some element, then so is any subset of $A$. Hence, the set of canonical join-representations is the set of faces of a simplicial complex, known as the \emph{canonical join complex}. In~\cite{barnard:2016canonical}, it was shown that for any finite semidistributive lattice $L$, the canonical join complex is flag. For example, the faces of the canonical join complex of the lattice of order ideals of a finite poset $P$ is the set of antichains of $P$. The canonical join complex of the weak order of type A is a simplicial complex of noncrossing arc diagrams~\cite{reading:2015noncrossing}. This complex contains the canonical join complex of the Tamari lattice.
Let $L$ be a semidistributive lattice. If $x=\bigvee A$ is the canonical join-representation of some element $x\in L$, then there is a bijection between lower covers of $x$ and the elements of $A$~\cite{barnard:2016canonical}. Conversely, if $x=\bigwedge B$ is a canonical meet-representation, then the upper covers of $x$ are in bijection with the elements of $B$.
An equivalence relation $\Theta$ on a lattice $L$ is a \emph{lattice congruence} if whenever $x\equiv y\mod\Theta$, we have $x\wedge z\equiv y\wedge z\mod\Theta$ and $x\vee z\equiv y\vee z\mod\Theta$. For $x\in L$, we let $[x]=[x]_{\Theta}$ be the $\Theta$-equivalence class of $x$. The \emph{quotient lattice} $L/\Theta$ where $\Theta$ is a lattice congruence of $L$ is the lattice of $\Theta$-equivalence classes where $[x]\vee[y]=[x\vee y]$ and $[x]\wedge[y]=[x\wedge y]$ for $x,y\in L$. The following characterization of lattice congruences of finite lattices is well-known; see \eg~\cite[Proposition 9-5.2]{reading:2016lattice}.
\begin{lemma}\label{lem_lattice_congruence}
An equivalence relation $\Theta$ on a lattice $L$ is a lattice congruence if and only if
\begin{itemize}
\item the equivalence classes of $\Theta$ are all closed intervals of $L$, and
\item the maps $\pi^{\uparrow}: L \to L$ and $\pi_{\downarrow}: L \to L$ taking an element of $L$ to the largest and smallest elements of its $\Theta$-equivalence class are both order-preserving.
\end{itemize}
\end{lemma}
In particular, if $x$ covers $y$ in $L$, then either $[x]=[y]$ or $[x]$ covers $[y]$ in $L/\Theta$. The following stronger result holds, which we will use to determine the lattice congruences of the Grid--Tamari order.
\begin{lemma}\label{lem_lattice_quotient_cover}~\cite[Proposition 2.2]{reading:lattice_congruence}
Let $L$ be a finite lattice with lattice congruence $\Theta$. If $x$ is the minimum element in its $\Theta$-equivalence class, then for each $y\in L$ covered by $x$, the class $[y]$ is covered by $[x]$ in $L/\Theta$. Furthermore, this is a bijection between lower covers of $x$ and lower covers of $[x]$. Dually, if $x$ is the maximum element in its $\Theta$-equivalence class, then there is a similar bijection between the upper covers of $x$ and the upper covers of $[x]$.
\end{lemma}
Given a lattice $L$, its set of lattice congruences $\Con(L)$ forms a distributive lattice under refinement order. Hence when $L$ is finite, $\Con(L)$ is isomorphic to the lattice of order ideals of $\JI(\Con(L))$. If $y$ covers $x$, we write $\con(x,y)$ for the minimal lattice congruence in which $x\equiv y\ (\con(x,y))$ holds.
For any finite lattice $L$ with lattice congruence $\Theta$, we have
\[\Theta=\bigvee_{\substack{j\in \JI(L)\\ j\equiv j_*\mod\Theta}}\con(j_*,j).\]
Hence, the join-irreducible congruences are always of the form $\con(j_*,j)$ for some $j\in \JI(L)$. A finite lattice $L$ is \emph{congruence-uniform} (or \emph{bounded}) if
\begin{itemize}
\item the map $j\mapsto\con(j_*,j)$ is a bijection from $\JI(L)$ to $\JI(\Con(L))$, and
\item the map $m\mapsto\con(m,m^*)$ is a bijection from $\MI(L)$ to $\MI(\Con(L))$.
\end{itemize}
Alternatively, finite congruence-uniform lattices may be characterized as homomorphic images of free lattices with bounded fibers or as lattices constructible from the one-element lattice by a sequence of interval doublings~\cite{day:congruence}.
For $x\in L$, let $\psi(x)$ be the set
\[\left\{\con(w,z):\ \bigwedge_{i=1}^l y_i\leq w\lessdot z\leq x\right\},\]
where $y_1,\ldots,y_l$ are the elements covered by $x$. The \emph{core label order} $\CLO(L)$ is the collection of sets $\psi(x)$, ordered by inclusion. For a congruence-uniform lattice $L$, the map $x\mapsto\psi(x)$ is a bijection between $L$ and $\CLO(L)$.
This formulation in lattice-theoretic terms was given by Reading~\cite{reading:2016lattice} and was used in~\cite{garver.mcconville:2016oriented} to give a correspondence between noncrossing tree partitions and some of the partial triangulations of a polygon.
\subsection{The Grid--Tamari order}\label{subsec_GT_order}
If $\Delta$ is a pure simplicial complex, its \emph{dual graph} is the graph whose vertices are the facets of $\Delta$ with edges $\{F,F^{\pr}\}$ whenever $F\cap F^{\pr}$ is a ridge.
When $F\cap F^{\pr}$ is a ridge, we simply say that $F$ and $F^{\pr}$ are \emph{adjacent} facets. In~\cite{mcconville:2015lattice}, it was shown that one may associate a segment to adjacent facets of the nonkissing complex by using the following lemma.
\begin{lemma}\label{lem_GT_labeling}
Let $\lambda_{\textsf{M}}$ be a marked shape, and let $F$ and $F^{\pr}$ be adjacent facets of $\Delta^{NK}(\lambda_{\textsf{M}})$. Let $p\in F$ and $q\in F^{\pr}$ such that $F\setm\{p\}=F^{\pr}\setm\{q\}$. Then there is a unique segment $s$ such that $p$ and $q$ kiss along $s$.
\end{lemma}
\begin{proof}
This was proven for unmarked shapes in~\cite[Theorem 3.2.3]{mcconville:2015lattice}. Using Corollary~\ref{passing_to_lambda}, we may identify $F$ and $F^\prime$ with facets of $\Delta^{NK}(\lambda)$. Via this identification, we obtain the desired result by again applying~\cite[Theorem 3.2.3]{mcconville:2015lattice}.
\end{proof}
\begin{defi}
For any two adjacent facets $F,F^{\pr}\in\Delta^{NK}(\lambda_{\textsf{M}})$ with $p,q,s$ defined as in Lemma~\ref{lem_GT_labeling}, we orient the edge $F\ra F^{\pr}$ if $p$ enters $s$ from the West and leaves to the South and $q$ enters $s$ from the North and leaves to the East. We say that $F^{\prime}$ may be obtained from $F$ by a (directed) \emph{flip}.
The \emph{Grid--Tamari order}, denoted $\GT(\lambda_{\textsf{M}})$, is defined as the reflexive, transitive closure of the flip relation on facets of $\Delta^{NK}(\lambda_{\textsf{M}})$. That is, $F\le F^{\prime}$ holds if $F^{\prime}$ may be obtained from $F$ by a sequence of flips.
\end{defi}
\begin{exam}
In Figure~\ref{walls_ex}, we show the Grid--Tamari order for the $2\times3$ rectangle shape. To simplify the figure, we omit the vertical and horizontal routes from each facet of $\Delta^{NK}(\lambda)$.\end{exam}
\begin{figure}[htb]\centering
\includegraphics[scale=1]{Figures/A2_GT_v3.pdf}
\caption{{The Grid--Tamari order $\GT(\lambda)$ (with the vertical and horizontal routes omitted) where $\lambda$ is a $2\times 3$ rectangle. In $\lambda$, $s_1 = (\textsf{v}_1)$ and $s_2 = (\textsf{v}_2)$. }}
\label{walls_ex}
\end{figure}
In~\cite{mcconville:2015lattice}, $\GT(\lambda)$ was proved to be a lattice by identifying it with a lattice quotient of a different lattice. We will recall the lattice quotient description of $\GT(\lambda)$ and extend it to marked shapes in the following section.
\subsection{Biclosed sets}\label{subsec_gt_biclosed}
Recall from Section~\ref{subsec_transitive} that a collection $X \subseteq \Seg(\lambda_{\M})$ of segments is \emph{closed} if whenever $s,t\in X$ and $s\circ t$ is well-defined then $s\circ t\in X$. We say $X$ is \emph{biclosed} if it is closed and its complement $X^c:=\Seg(\lambda_{\textsf{M}})\setm X$ is closed. The set $\Bic(\lambda_{\textsf{M}})$ of biclosed subsets of $\Seg(\lambda_{\M})$ forms a poset under inclusion. That is, if $X,Y\in\Bic(\lambda_{\textsf{M}})$, we set $X\leq Y$ if $X\subseteq Y$.
In~\cite{mcconville:2015lattice}, the poset $\Bic(\lambda)$ was proved to be a congruence-uniform lattice for an unmarked shape $\lambda$. More precisely, the following was shown.
\begin{theorem}\label{thm_bic_prop}\cite[Theorem 5.2, Theorem 5.5]{mcconville:2015lattice}
The poset of biclosed sets has the following properties. These three properties together imply that $\Bic(\lambda)$ is a congruence-uniform lattice.
\begin{enumerate}
\item The poset $\Bic(\lambda)$ is graded, with rank function $X\mapsto|X|$.
\item\label{thm_bic_prop_2} The poset $\Bic(\lambda)$ is a lattice where
\[
X,Y,W\in\Bic(\lambda),\ W\subseteq X\cap Y\ \ \mbox{implies}\ \ W\cup\ov{(X\cup Y)\setm W}\in\Bic(\lambda).
\]
\item If $u\in\ov{\{s,t\}}\setm\{s,t\}$, then $u=s\circ t$.
\end{enumerate}
\end{theorem}
We note that taking $W=\emptyset$ in~\eqref{thm_bic_prop_2}, we have $\ov{X\cup Y}$ is biclosed whenever $X$ and $Y$ are biclosed. Since $\ov{X\cup Y}$ is the smallest closed set containing both $X$ and $Y$, this set is the join of $X$ and $Y$.
In order to extend the proof of congruence-uniformity to biclosed sets on marked shapes, we make use of the following lemma.
\begin{lemma}\label{lem_bic_marked}
Let $\lambda_\textsf{M}$ be any marked shape.
\begin{enumerate}
\item\label{lem_bic_marked_1} The inclusion $\Seg(\lambda_{\textsf{M}})\subseteq\Seg(\lambda)$ induces an inclusion of lattices $\Bic(\lambda_{\textsf{M}})\hookra\Bic(\lambda)$.
\item\label{lem_bic_marked_2} The map $X\mapsto X\cap\Seg(\lambda_{\textsf{M}})$ from subsets of $\Seg(\lambda)$ to subsets of $\Seg(\lambda_{\textsf{M}})$ induces a lattice quotient map $\Bic(\lambda)\twoheadrightarrow\Bic(\lambda_{\textsf{M}})$.
\end{enumerate}
\end{lemma}
\begin{proof}
\eqref{lem_bic_marked_1} Let $X$ be a set of segments of $\Seg(\lambda_{\M})$. We claim that $X$ is biclosed as a subset of $\Seg(\lambda_{\textsf{M}})$ if and only if it is biclosed as a subset of $\Seg(\lambda)$. It is clear that if $X$ is in $\Bic(\lambda)$ then $X$ is in $\Bic(\lambda_{\textsf{M}})$ since being biclosed relative to the marked shape $\lambda_{\textsf{M}}$ imposes fewer conditions on $X$.
Suppose $X\in\Bic(\lambda_{\textsf{M}})$, and let $s,t,u\in\Seg(\lambda)$ such that $s\circ t = u$. If $s,t\in X$, then $u$ is in $\Seg(\lambda_{\textsf{M}})$, so $u\in X$ holds. If $u\in X$, then both $s$ and $t$ are in $\Seg(\lambda_{\textsf{M}})$, so $s\in X$ or $t\in X$.
We have shown that the inclusion $\Seg(\lambda_{\textsf{M}})\subseteq\Seg(\lambda)$ induces an inclusion of sets $\Bic(\lambda_{\textsf{M}})\hookra\Bic(\lambda)$. Moreover, the latter inclusion identifies $\Bic(\lambda_{\textsf{M}})$ as a closed interval of $\Bic(\lambda)$, so it inherits the lattice structure from $\Bic(\lambda)$.
\eqref{lem_bic_marked_2} Set $S=\Seg(\lambda_{\textsf{M}})$. We observe that $X\cap S$ is biclosed whenever $X\in\Bic(\lambda)$ by a similar argument as in the proof of~\eqref{lem_bic_marked_1}. So the map $\Bic(\lambda)\ra\Bic(\lambda_{\textsf{M}})$ is well-defined, and its surjectivity follows from~\eqref{lem_bic_marked_1}.
To show that the map preserves joins, we let $X,Y\in\Bic(\lambda)$ and claim that $\ov{(X\cup Y)\cap S}=\ov{X\cup Y}\cap S$ holds. Indeed, the elements of $\ov{(X\cup Y)\cap S}$ are concatenations of segments in $X$ and $Y$ that are each in $\Seg(\lambda_{\textsf{M}})$. On the other hand, the elements of $\ov{X\cup Y}\cap S$ are those segments in $\Seg(\lambda_{\textsf{M}})$ that may be formed by concatenating segments in $X$ and $Y$. But if $s=s_1\circ\cdots\circ s_k$, then $s\in\Seg(\lambda_{\textsf{M}})$ if and only if $s_i\in\Seg(\lambda_{\textsf{M}})$ for all $i$. Hence, the claim is established, and we have
\begin{align*}
(X\cap S)\vee(Y\cap S) &= \ov{(X\cap S)\cup(Y\cap S)}\\
&= \ov{(X\cup Y)\cap S}\\
&= \ov{X\cup Y}\cap S\\
&= (X\vee Y)\cap S.
\end{align*}
As complementation commutes with the map $\Bic(\lambda)\twoheadrightarrow\Bic(\lambda_{\textsf{M}})$, we deduce that meets are preserved as well.
\end{proof}
\begin{coro}\label{latt_str_marked_bic}
{Let $\lambda_{\M}$ be any marked shape. Then $\Bic(\lambda_{\M})$ is a congruence-uniform lattice with the following properties.}
\begin{enumerate}
\item The poset $\Bic(\lambda_{\M})$ is graded, with rank function $X \mapsto |X|$.
\item\label{latt_str_marked_bic_2} For any $X,Y \in \Bic(\lambda_{\M})$, one has $X\vee Y = \overline{X \cup Y}$.
\item\label{latt_str_marked_bic_3} For any $X,Y \in \Bic(\lambda_{\M})$, one has $X \wedge Y = (X^c\vee Y^c)^c$.
\end{enumerate}
\end{coro}
\begin{proof}
That $\Bic(\lambda_{\M})$ is congruence-uniform follows from Theorem~\ref{thm_bic_prop} and from Lemma~\ref{lem_bic_marked}~\eqref{lem_bic_marked_2}. That $\Bic(\lambda_{\M})$ is graded with the given rank function Theorem~\ref{thm_bic_prop} and Lemma~\ref{lem_bic_marked}~\eqref{lem_bic_marked_1}. Property~\eqref{latt_str_marked_bic_2} is a consequence of the discussion following Theorem~\ref{thm_bic_prop} and Lemma~\ref{lem_bic_marked}~\eqref{lem_bic_marked_1}.
It remains to prove property~\eqref{latt_str_marked_bic_3}. Assume that $s \in X^c\vee Y^c$. We prove that $s \in (X\wedge Y)^c$ under the assumption that all segments $s^\prime \in X^c\vee Y^c$ with strictly smaller length than $s$ belong to $(X\wedge Y)^c$. Now write $s = s_1\circ \cdots \circ s_k$ for some segments $s_i \in X^c\cup Y^c$. If $k = 1$, then $s \in X^c \cup Y^c$. Therefore, $s \in (X \cap Y)^c$ so $s \in (X\wedge Y)^c$.
Suppose $k \ge 2$. By induction, $s_1, s_2\circ \cdots \circ s_k \in (X\wedge Y)^c$. Since $(X\wedge Y)^c$ is closed, we have that $s \in (X\wedge Y)^c$. We obtain that $X\wedge Y \subseteq (X^c \vee Y^c)^c$.
To prove the opposite inclusion, observe that
\[
X \wedge Y = \displaystyle \bigvee_{\substack{Z \in \Bic (\lambda_{\M})\\ Z \subseteq X, Z\subseteq Y}} Z.
\] Now notice that if $s \in (X^c \vee Y^c)^c$, then $s \not \in X^c$ and $s \not \in Y^c$. This implies that $s \in X$ and $s \in Y$. Thus $(X^c \vee Y^c)^c \subseteq X$ and $(X^c \vee Y^c)^c \subseteq Y$. Since $(X^c \vee Y^c)^c \in \Bic (\lambda_{\M})$, it follows that $(X^c \vee Y^c)^c$ is a joinand in the join-representation of $X \wedge Y$ shown above. We obtain that $(X^c \vee Y^c)^c \subseteq X \wedge Y$.
\end{proof}
In~\cite[Section 8]{mcconville:2015lattice}, the Grid--Tamari order $\GT(\lambda)$ was shown to be isomorphic to both a quotient lattice and a sublattice of $\Bic(\lambda)$. We recall the quotient lattice and sublattice maps below in order to extend them to marked shapes.
We recall the definition of the map $\eta$ from $\Bic(\lambda)$ to $\GT(\lambda)$ as follows. If $X \subseteq \Seg(\lambda)$ is a set of segments, let $\widehat{\eta}(X)$ be the set of routes
\[\widehat{\eta}(X)=\{p_e:\ e \mbox{ is a vertical edge of }\lambda\}\]
where $p_e=(\textsf{v}_0,\ldots, \textsf{v}_l)$ is the unique route such that $e=(\textsf{v}_{j-1}, \textsf{v}_j)$ for some $j$, and the following two conditions hold.
\begin{itemize}
\item For $1\leq i\leq j-1$, the vertex $\textsf{v}_{i-1}$ is North of $\textsf{v}_i$ if $p_e[\textsf{v}_i,\textsf{v}_{j-1}] := (\textsf{v}_i,\ldots, \textsf{v}_{j-1})\in X$ and $\textsf{v}_{i-1}$ is West of $\textsf{v}_i$ if $(\textsf{v}_i,\ldots, \textsf{v}_{j-1})\notin X$.
\item For $j\leq k1$ where each $u_i^{\pr}$ is in $\SW(u_j)$ for some $j$ (depending on $i$). There exist indices $i\leq j$ such that if $u^{\pr}=u_i^{\pr}\circ u_{i+1}^{\pr}\circ\cdots\circ u_j^{\pr}$:
\begin{itemize}
\item $s\neq u^{\pr}$,
\item either $i=1$ or $s$ enters $u^{\pr}$ from the West, and
\item either $j=m$ or $s$ leaves $u^{\pr}$ to the South.
\end{itemize}
Since $u^{\pr}\in \SW(s)\vee \SW(t)$, we have $u^{\pr}=t_1\circ\cdots\circ t_k$ where each $t_i$ is a SW-subsegment of $s$ or $t$. By the hypotheses on $u^{\pr}$, either $t_1\in \SW(t)$ or $t_k\in \SW(t)$ must hold. Suppose $t_1\in \SW(t)$. Choose $j$ maximal such that $t_1\circ\cdots\circ t_j$ is a SW-subsegment of $t$. If $j=k$, then $u^{\pr}$ is a SW-subsegment of $t$, and $s$ and $t$ are friendly along $u^{\pr}$. If $j