jcreed blog > Thinking About Opetopes

Thinking About Opetopes

Here's an idea that I found myself thinking about lately, which, if it works, might be a nice simple discrete-geometric definition of opetope. I haven't gotten much of the way into proving that it works, but it seems promising on small examples.

Preopetopes

Say an $n$-dimensional preopetope or simply $n$-preopetope is a tuple \[\O = (X, C, \sim)\] where $X$ is a set and $C = C_0, \ldots, C_n \subseteq X$ is a family of $n+1$ unary relations on $X$, and ${\sim} = {\sim_0}, \ldots, {\sim_n} \subseteq X \x X$ is a family of $n+1$ binary relations on $X$. Some intuitions motivating this data are:

Depicting Preopetopes

We can draw a preopetope by drawing it as a graph with colored edges, with colored annotations at the vertices. Here is an example 2-preopetope with twelve points, drawn as gray dots:
The edges of the graph represent $\sim_0$ (black) $\sim_1$ (light blue) and $\sim_2$ (red) holding between flags of the carrier set. The small colored dots near the flags represent them belonging to $C_0$ (black) $C_1$ (light blue) and/or $C_2$ (red).

Defining Opetopes

We inductively define the conditions for an $n$-preopetope to be an $n$-opetope. To make this definition make sense, we need to define $\btriv$, $\bleaf$, $\bnode$, and $\bglue$, which we do in the following section. We say:

Constructions on Preopetopes

Trivial

There is a trivial $(-1)$-preopetope $\btriv = (\{*\}, \cdot, \cdot)$ that has a one-element carrier set, and has no relations at all.

Leaf

Given an $n$-preopetope $\O = (X, \sim, C)$, we define an $(n+1)$-preopetope $\bleaf(\O) = (X', \sim', C')$ by saying \[X' = X \qquad C'_{i} = \begin{cases} X' & \hbox{if $i = n+1$}\\ C_i & \hbox{ otherwise.}\end{cases} \qquad { x_1 \sim_i x_2 \over x_1 \sim'_i x_2} \qquad { x_1 \sim_n x_2 \over x_1 \sim'_{n+1} x_2}\] What's going is that we are copying the (previously) top-dimension binary relation up one dimension higher, and saying everything belongs to the new top-dimension unary relation. For example:
We've duplicated every $\sim_1$ edge to become both a $\sim_1$ and a $\sim_2$ edge. We've said every flag belongs to the $2$-codomain; all of them are marked with red dots.

Node

Given an $n$-preopetope $\O = (X, \sim, C)$, we define an $(n+1)$-preopetope $\bnode(\O) = (X', \sim', C')$ by saying \[X' = 2 \x X \qquad C_i' = \begin{cases} \{(1,x) \in X' \} & \hbox{if $i = n+1$}\\ C_i & \hbox{ otherwise.}\end{cases} \]\[ { x_1 \sim_i x_2 \over (b, x_1) \sim'_i (b, x_2)} \qquad { b_1 \ne b_2 \over (b_1, x) \sim'_{n+1} (b_2, x)}\] What's going is that we take two copies of the entire flag set of $\O$, and join them along the newly top-dimensional relation $\sim_{n+1}$. We say one copy belongs to the codomain $C_{n+1}$, and the other one does not. For example:
We've made two copies of the ring-shaped 1-preopetope and related them with $\sigma_2$. We've said the outer copy belongs to $C_2$.

Glue

Suppose we're given two $n$-preopetopes $\O = (X,\sim,C)$ and $\O' = (X',\sim',C')$ Suppose there is a bijection $h : S \cong S'$ between some subsets $S \subseteq X$ and $S' \subseteq X'$. Suppose that $h$ preserves the relations $C_0,\ldots,C_{n-2}$ and $\sim_0,\ldots,\sim_{n-2}$, and symmetrically for $h^{-1}$. Write $s \approx s'$ for $h(s) = s'$. Then we define an $n$-preopetope $\bglue(\O, \O', h) = (X'', \sim'', C'')$ by \[X' = (X\setminus S) + (X' \setminus S') \qquad C''_i = C_i \cup C'_i\] \[{x \sim_i y \over x \sim''_i y} \qquad {x' \sim'_i y' \over x' \sim''_i y'} \qquad {x \sim_i s\qquad s \approx s' \qquad s' \sim'_i x' \over x \sim''_i x'}\] Here's an example:
What happens is that we delete all of the flags in the domain and codomain of $h$, and we "patch up" all of the binary relations as appropriate.

Understanding Low-Dimensional Cases

(-1)-Opetopes

We can see immediately that $\btriv$ is the unique (-1)-opetope.

0-Opetopes

We can see that $\bnode(\btriv)$ is the unique (0)-opetope. It looks like this:

1-Opetopes

For 1-opetopes, we must consider leaves constructed from $(-1)$-opetopes, (of which there is one) nodes constructed from $0$-opetopes, (of which there is one) and gluing different $1$-opetopes together. The leaf case is $\bleaf(\bnode(\btriv))$. It looks like this:
Then we have $\bnode(\bnode(\btriv))$. It looks like this:
There are now 4 ways we can glue these two known $1$-opetopes together:
The first three rows of this table yield preopetopes we've already constructed, but the fourth produces something novel. If we keep turning the crank, we find that there is exactly one $1$-opetope $P_m$ for every natural number $m$, which has $2m+2$ flags. Exactly half of its flags are in $C_0$, and exactly two of its flags are in $C_1$. The leaf case was $m=0$ and the node case was $m=1$. The relations $\sim_0$ and $\sim_1$ are both irreflexive and symmetric, and they relate every element of the flag set to exactly one other element. We also find there is a unique way to glue together $P_m$ and $P_{m'}$, which yields $P_{m+m'}$.

2-Opetopes

I'm going to be even less formal and more suggestive in this section. 2-opetopes correspond to planar trees. The leaf case is the construction $\bleaf(\bnode(\bnode(\btriv)))$. I'm going to depict it three ways: On the left is what it is as a preopetope. In the center is an interpretation of the same structure as a tree, thought of as a string diagram. On the right is it presented as a commutative diagram. All three are labelled suggestively with names for the cells involved.

There's one node construction available to us for each $1$-opetope. Here's a sketch of $\bnode(P_0), \bnode(P_1), \bnode(P_2), \bnode(P_3)$... with the same three columns showing preopetope, string diagram, and pasting diagram.

Let's consider an interestingly nontrivial example of gluing now. We'll glue a 0-ary node to a 3-ary node:

Future Work

The obvious thing to try to do here is prove this is equivalent to some well-known category of opetopes. Morphisms of the above structures would surely be maps of the carrier that preserve all relations.