tileorder

A **tileorder** is a *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).

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 **double order**. A double order which arises in this way from a rectangle dissection is called a **tileorder**.

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

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)

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 here.)

As proven by Dawson and Paré, tileorders can be characterized in purely combinatorial terms in a couple of interesting ways.

A double order $T$ is said to have the **$\sqsubset$-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 **$\lt$-parallel maximal chain property**.

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

A double order is a tileorder iff it has both parallel maximal chain properties and the orthogonal maximal chain property.

A double order is **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 **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 **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 **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 **second** such property is defined by replacing $\le$ by $\ge$, but not changing $\sqsubseteq$. The **$\lt$-orthogonal** such properties are defined by switching the roles of $\sqsubseteq$ and $\le$.

A double order has the **$\sqsubset$-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 **$\lt$-parallel** such property is defined by switching the roles of $\sqsubseteq$ and $\le$.

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

- Dawson and Paré, “Characterizing tileorders”

Last revised on April 8, 2010 at 05:07:53. See the history of this page for a list of all contributions to it.