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\begin{document}

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\section*{tileorder}

\hypertarget{contents}{}\section*{{Contents}}\label{contents}

\noindent\hyperlink{idea}{Idea}\dotfill \pageref*{idea} \linebreak
\noindent\hyperlink{definition}{Definition}\dotfill \pageref*{definition} \linebreak
\noindent\hyperlink{characterizations}{Characterizations}\dotfill \pageref*{characterizations} \linebreak
\noindent\hyperlink{using_maximal_chain_properties}{Using maximal chain properties}\dotfill \pageref*{using_maximal_chain_properties} \linebreak
\noindent\hyperlink{using_local_configuration_structure}{Using local configuration structure}\dotfill \pageref*{using_local_configuration_structure} \linebreak
\noindent\hyperlink{references}{References}\dotfill \pageref*{references} \linebreak


\hypertarget{idea}{}\subsection*{{Idea}}\label{idea}

A \textbf{tileorder} is a \emph{double order}, i.e. a set $T$ equipped with two [[partial orders]] $\le$ and $\sqsubseteq$, which can be realized by a dissection of a rectangle into finitely many subrectangles, the subrectangles being the elements of $T$, in such a way that $a\sqsubseteq b$ iff $a$ ``lies below'' $b$, while $a\le b$ iff $a$ ``lies to the left of'' $b$. Interestingly and importantly, such orders can also be characterized in purely combinatorial, and effectively verifiable, terms.

Such orders are of interest in [[double category]] theory, since these are the arrangements of 2-cells in a double category which could potentially be composed (although in a general double category, not all of them can actually be composed).

\hypertarget{definition}{}\subsection*{{Definition}}\label{definition}

Suppose given a dissection of a rectangle into finitely many subrectangles. Define $T$ to be the set of subrectangles, and for $a,b\in T$ write $a\le b$ if there exists a finite list $a= x_0, x_1, \dots, x_n = b$ such that for each $i$, the right edge of $x_i$ intersects the left edge of $x_{i+1}$ in more than one point. Define $a\sqsubseteq b$ similarly using top and bottom edges instead of left and right. Clearly $\le$ and $\sqsubseteq$ are [[partial orders]].

A set $T$ equipped with two partial orders $\le$ and $\sqsubseteq$ (a priori, unrelated) is called a \textbf{double order}. A double order which arises in this way from a rectangle dissection is called a \textbf{tileorder}.

[[Todd Trimble|Todd]]: Maybe I'm missing what you mean by ``dissection'', but off the bat it looks like you are allowing the ``pinwheel'' as a possible dissection (I hope it is clear what the pinwheel is; Dawson discusses it in one of his papers), but this is the kind of configuration not interpretable in double categories.

[[Mike Shulman]]: Yes, we are allowing the pinwheel, even though it is not always composable in a double category. That's what I meant to imply above by

\begin{quote}%
arrangements of 2-cells in a double category which could potentially be composed (although in a general double category, not all of them can actually be composed)


\end{quote}
but maybe that isn't sufficiently clear. The notion of tileorder is purely geometric/combinatorial, and you then have to ask which tileorders are composable in a double category (essentially, all that don't contain a pinwheel). (BTW, a picture of the pinwheel can be found \href{http://ncatlab.org/nlab/show/double+category#unbiased_composition_and_associativity_9}{here}.)

\hypertarget{characterizations}{}\subsection*{{Characterizations}}\label{characterizations}

As proven by Dawson and Par\'e{}, tileorders can be characterized in purely combinatorial terms in a couple of interesting ways.

\hypertarget{using_maximal_chain_properties}{}\subsubsection*{{Using maximal chain properties}}\label{using_maximal_chain_properties}

A double order $T$ is said to have the \textbf{$\sqsubset$-parallel maximal chain property} if whenever $K$ and $L$ are maximal $\sqsubseteq$-chains in $T$, and we have $k_1,k_2\in K$ and $l_1,l_2\in L$ with $k_1 \sqsubset k_2$, $k_1 \le l_1$, and $l_2 \le k_2$, then there exists an $e\in K\cap L$ such that $k_1\sqsubset e$, $l_1\sqsubset e$, $e\sqsubset k_2$, and $e\sqsubset l_2$. In other words, two maximal $\sqsubseteq$-chains cannot ``swap places'' in the $\le$-order without intersecting.

Dually, we define the \textbf{$\lt$-parallel maximal chain property}.

A double order $T$ is said to have the \textbf{orthogonal maximal chain property} if every maximal $\le$-chain intersects every maximal $\sqsubseteq$-chain exactly once.

\begin{utheorem}
A double order is a tileorder iff it has both parallel maximal chain properties and the orthogonal maximal chain property.

\end{utheorem}
\hypertarget{using_local_configuration_structure}{}\subsubsection*{{Using local configuration structure}}\label{using_local_configuration_structure}

A double order is \textbf{strongly antisymmetric} if two unequal elements are never related (in either direction) by both $\lt$ and $\sqsubset$. That is, if $a\neq b$, then at most one of $a\lt b$, $b\lt a$, $a\sqsubset b$, and $b\sqsubset a$ holds.

A double order is \textbf{rectangular} if $\exists c.(a\sqsubseteq c \le b)$ iff $\exists d.(a\le d \sqsubseteq b)$, and similarly $\exists c.(a\sqsubseteq c \ge b)$ iff $\exists d.(a\ge d \sqsubseteq b)$.

A double order is \textbf{total} if for any $a,b$, there exists a $c$ such that either $a\sqsubseteq c \le b$, or $a\sqsubseteq c \ge b$, or $b \le c \sqsubseteq a$, or $b\ge c \sqsubseteq a$.

A double order has the \textbf{first $\sqsubset$-orthogonal butterfly factorization property} if given $a,b,c,d$ with $c\sqsubseteq a$, $c\sqsubseteq b$, $d\sqsubseteq b$, and $d\le a$, there exists an $e$ with $c\sqsubseteq e \sqsubseteq b$ and $d\le e \le a$. The \textbf{second} such property is defined by replacing $\le$ by $\ge$, but not changing $\sqsubseteq$. The \textbf{$\lt$-orthogonal} such properties are defined by switching the roles of $\sqsubseteq$ and $\le$.

A double order has the \textbf{$\sqsubset$-parallel butterfly factorization property} if given $a,b,c,d$ with $c\sqsubseteq a$, $b\le c$, $d\sqsubseteq b$, and $d\le a$, there exists an $e$ with $b\le e \le c$ and $d\le e \le a$. The \textbf{$\lt$-parallel} such property is defined by switching the roles of $\sqsubseteq$ and $\le$.

\begin{utheorem}
A double order is a tileorder iff it is strongly antisymmetric, rectangular, total, and has all the orthogonal and parallel butterfly factorization properties.

\end{utheorem}
\hypertarget{references}{}\subsection*{{References}}\label{references}

\begin{itemize}%
\item Dawson and Par\'e{}, ``Characterizing tileorders''

\end{itemize}
[[!redirects tileorders]] [[!redirects double order]] [[!redirects double orders]]



\end{document}