Does the simplex map to the cube?

Question

For every $n \geq 0$ there is an inclusion of the ordered set $\{0<1<\dots<n\}$ into the product $\{0<1\}^{\times n}$ sending $i$ to the increasing sequence $(0 < \dots<0<1<\dots<1)$, in which $1$ appears $i$ many times. Is there a higher analogue of this fact in n-category theory?

Answer 1

Yes, though I don’t know a clean description. Several relevant constructions are discussed in §8 of Aitchison 1986/2010, The geometry of iterated cubes, especially §8.1–8.3; the following description is from my notes when reading that paper a few years back, and I think it was a working-out from the constructions there, but I can’t quite see now how it comes from them.

I’ll describe the map in terms of parity-complexes — each cell of the oriented $n$-simplex is sent to a composable set of cells of the lax $n$-cube. I follow Aitchison’s notation: cells of the $n$-simplex are subsets of $[n]$, written without punctuation as e.g. $014 \subseteq [5]$; cells of the $n$-cube are words of length $n$ from $\{{-},0,{+}\}$.

The general case is as follows: Given a cell $\sigma$ in the $n$-simplex, read it as splitting the $n$ positions of a potential cube cell into blocks, with each $i \in \sigma$ meaning “split after the $i$th position”: so $125 \subseteq [5]$ divides $\newcommand{\w}[1]{(#1\,)} \newcommand{\u}{\,\_}\newcommand{\s}{\,/} \w{\u\u\u\u\u}$ into $\w{\u \s \u \s \u \u \u \s}$. (Note the first+last blocks may be empty, but interior blocks never are.) Now take $F(\sigma)$ to be the set of all words of the following form: their first block (under the splitting specified by $\sigma$) must be all $\newcommand{\p}{\mathord{+}}\newcommand{\m}{\mathord{-}} +$; their last block must be all $-$; and each interior block must be of the form $\m^i\;\!0\;\!\p^j$. So in the example above, $F(125) = \{\p0\m\m0,\p0\m0\p,\p00\p\p\}$.

Some examples, in the case $n=3$ (of course, drawing some pictures is recommended here): $$\begin{align*} F(0) &= \{\m\m\m\} \\ F(1) &= \{\p\m\m\} \\ F(2) &= \{\p\p\m\} \\ F(01) &= \{0\m\m\} \\ F(12) &= \{\p0\m\} \\ F(02) &= \{\m0\m,0\p\m \} \\ F(03) &= \{\m\m0,\m0\p,0\p\p \} \\ F(012) &= \{00- \} \\ F(013) &= \{0\m0,00\p \} \\ \end{align*}$$

So $F$ gives a map from simplex cells to sets of cube cells. $F$ respects dimension: the splitting induced by a $k$-cell $\sigma$ has $k$ interior blocks, so each cube cell in $F\sigma$ contains $k$ zeroes, i.e. is a $k$-cell. And the sets $F\sigma$ are pairwise disjoint, are each a segment in the adjacency ordering (i.e. form a composable configuration), and satisfy the condition $s(F\sigma) \sqcup F(t \sigma) = F(s \sigma) \sqcup t(F \sigma)$. They thus induce the desired map of strict $n$-categories from the oriented $n$-simplex to the lax $n$-cube.

I’m sure there must be a better way to present these — perhaps e.g. following abstractly from the presentation of the oriented simplices as iterated coning, compared to the presentation of the lax cubes as iterated lax product with the interval — but I haven’t seen it given anywhere.

(Incidentally, I strongly encourage avoiding the term orientals — outside maths, its main meaning as a noun in English is as a rather dated racist slur. By the standards of the 1980s, when Street introduced the term mathematically, it was a piece of mildly questionable-taste wordplay; today I think many more people would agree on finding it pretty distasteful.)

Edit — some clarifications in answer to OP’s comments: To see why the data above generates a map of $\omega$-categories $\mathcal{O}_n \to {\boxtimes}^n$, we rely on two facts:

• The universal property of $\mathcal{O}_n$ as “freely generated” by its atomic cells, i.e. the cells of the parity complex: roughly to give a map from $\newcommand{\O}{\mathcal{O}}\O_n$ to any $\omega$-category, it suffices to define the map just on atomic cells, preserving the source and target. This universal property is defined and proved for $\O_n$ in Street 1986, The algebra of oriented simplices, §4 (unpaywalled scan), and generalised to arbitrary parity complexes in Street 1991 Parity complexes, §4. Simon Forest showed (summarised here, with link to full paper) that Street’s general version of the theorem is wrong, but notes that it holds for most well-known examples, in particular for $\O_n$.

• The descriptions of the cells of the $\omega$-category $\boxtimes^n$ as suitable finite sets of atomic cells, and the calculations of their sources, targets, and composition. This construction is presented for arbitrary parity complexes in Street 1991 §3, and the specific parity complex for $\boxtimes^n$ is described in Aitchison 1986/2010.