I was browsing through Todd Trimble's notes on tetracategories, and there are a number of details in the discussion of the tetracategorical data that I don't understand, related (IIRC) to the tricategorical coherence criteria, which I think of as a sort of warm-up for the tetracategorical coherators. My confusions are related to a bunch of apparently-missing details that I don't yet see how to fill in.
The first instance of this is at the type signature of the associator $\alpha$, on page 3, which is given as $$\alpha_{a, b, c, d} : \otimes_{a, c, d} (\otimes_{a, b, c} \times 1_{\mathcal C(c, d)}) \to \otimes_{a, b, d} (1_{\mathcal C(a, b)} \times \otimes_{b, c, d})$$ where $$1_A : A \to A$$ is the identity trifunctor and $$\otimes_{a, b, c} : \mathcal C(a, b) \times \mathcal C(b, c) \to \mathcal C(a, c)$$ is the compositor (also a trifunctor), and the notations $\to$ and $\times$ are overloaded.
Here, I assume that I'm supposed to see $\alpha_{a, b, c, d}$ as a natural transformation between functors between tricategories, where the source functor is $\otimes_{a, c, d} (\otimes_{a, b, c} \times 1)$ and the target functor is $\otimes_{a, b, d} (1 \times \otimes_{b, c, d})$. The trouble is, the first functor has type $((\mathcal C(a, b) \times \mathcal C(b, c)) \times \mathcal C(c, d)) \to \mathcal C(a, d)$ whereas the latter has type $(\mathcal C(a, b) \times (\mathcal C(b, c) \times \mathcal C(c, d))) \to \mathcal C(a, d)$ -- where, notably, in the former the source tuple is left-associated and in the latter it's right-associated. As such, it seems to me like it would be more correct to say $$\alpha_{a, b, c, d} : \otimes_{a, c, d} (\otimes_{a, b, c} \times 1_{\mathcal C(c, d)}) \to \rho ⨾ \otimes_{a, b, d} (1_{\mathcal C(a, b)} \times \otimes_{b, c, d}),$$ where $$\rho_{A, B, C} : ((A \times B) \times C) \to (A \times (B \times C))$$ is the obvious right associator, and I've suppressed some subscripts above for the sake of brevity.
This omission seems reasonable -- the right associator is pretty trivial (though, IIRC, not literally the identity), and perhaps the notes are just suppressing it. However, the fact that they don't mention this suppression causes me to think that I may instead be misunderstanding the associator. This worry is compounded given that fact that I get completely lost on page 3, for reasons that seem related.
In particular, on page 3 we see a pair of 2-cells that are supposed to act as boundaries of a 3-cell (which has the shape of the 3-dimensional associahedron $\mathcal K_5$). In the top-right corner of the source surface, there's a 2-cell running (in the notation of the notes) from $$(xy)((zu)v) \xrightarrow{(xy)\alpha} (xy)(z(uv)) \xrightarrow{\alpha} x(y(z(uv)))$$ to $$(xy)((zu)v) \xrightarrow{\alpha} x(y((zu)v)) \xrightarrow{x(y\alpha)} x(y(z(uv))),$$ with the 2-cell labeled $\alpha_1 = \alpha_{1, 1, \alpha}$. Long story short, I basically don't understand what this means. Which 2-cell fills that square?
More specifically: The notes say that words like $x(yz)$ are shorthand for $\otimes_{a, b, d} (x, \otimes_{b, c, d} (y, z))$, and understandably so. Furthermore, the diagram on page 3 occurs for every 6-tuple of cells, which are left unnamed in the notes, but which we might call $a, b, c, d, e, f$, with $x : \mathcal C(a, b), y : \mathcal C(b, c),$ and etc. We can then think of words like $(xy)((zu)v)$ as functors from 5-tuples to 1-tuples that apply the compositors in the given order. At this point, we see another instance of the confusion above: under this interpretation, the vertex $(xy)((zu)v)$ denotes a functor with a different source than the functor $(xy)(z(uv))$ -- there's an equivalence between the sources, sure, but they aren't exactly the same, and my best guess is that the notes are simply suppressing the tuple-reassociators. But under this interpretation, I'm struggling to see how the square (with endpoints given above) is filled.
Alltogether, I have two related questions here:
- What's up with all these natural transformations whose source and target functors have different sources? Is this detail being suppressed, or am I misunderstanding?
- How are the squares in the surfaces on page 3 filled? These are labeled with things like $\alpha_1 = \alpha_{1, 1, \alpha}$ and $\alpha_1^{-1} = \alpha_{1, 1, \alpha}^{-1}$ and $\alpha_1^{-1} = \alpha_{1, \alpha, 1}^{-1}$, and I don't understand what those labels mean (though I'd perhaps be able to work it out if I wasn't confused by the first question), and I also don't see how to fill those squares from just the pentagram & triangle diagrams that come before this 3-cell.
Also, more generally, while I'm interested in understanding Todd Trimble's notes, if there's a more modern treatment of the coherence conditions of higher categories, I'm also happy to be pointed towards those. In particular, I'm curious about the patterns of coherators in high-dimensional categories, where the question of how the squares in $\mathcal K_5$ get filled in (ie, are the square-fillers derived, or are they additional coherence criteria?) is a special case.
While Nate and I have been corresponding about this and I was able to answer his questions, I'll post something here as well in case anyone else is interested.
To answer question 1: the question is about the distinction between different ways of bracketing products of structures such as local hom-tricategories $(C(a, b) \times C(b, c)) \times C(c, d)$ and $C(a, b) \times (C(b, c) \times C(c, d))$. It's true that this distinction was unremarked on in my notes; informally, it was handled at a notational level by using 'unbiased products' instead of iterated binary products (see Tom Leinster's Higher Operads, Higher Categories). If pressed on the matter, I would say that weak $n$-categories (in such algebraic senses) are at bottom essentially algebraic structures modeled in $\mathbf{Set}$, and that we can if we like always pass to a monoidal strictification of the cartesian monoidal category $\mathbf{Set}$ where these distinctions disappear: my own preferred way of doing that is to use "cliques", for which a nice account is given in Geometry of Tensor Calculus by Joyal and Street. Long story short: it's nothing to worry about.
To answer question 2: as I explained to Nate privately, those "fillers" come out of the higher-dimensional data that is inherent when we stipulate that $\alpha$ is a tritransformation (as defined in the monograph Coherence for Tricategories by Gordan-Power-Street). Tritransformations are akin to (pseudo-natural) transformations (between homomorphisms between bicategories), but one dimension up.
Recall that if $B, C$ are bicategories and $F, G: B \to C$ are homomorphisms (that's Bénabou's term; other people say "pseudofunctors" or the like), a pseudo-natural transformation $\theta: F \to G$ assigns to each object or $0$-cell $b$ of $B$ a 1-cell $\theta b: Fb \to Gb$ of $C$, and to each $1$-cell $f: b \to c$ of $B$ an invertible 2-cell filler $\theta f$ of the form
$$\begin{array}{ccc} Fb & \stackrel{Ff}{\to} & Fc \\ \theta b \downarrow & \Leftarrow & \downarrow \theta c\\ Gb & \underset{Gf}{\to} & Gc \end{array}$$
going from $(\theta c)(Ff)$ to $(Gf)(\theta b)$, plus various compatibility/coherence conditions. It's the same deal with tritransformations, except that the fillers (which are one dimension up) are equivalences (and there is even more structure and higher-level coherence conditions which don't concern us here; see Gordan-Power-Street for the details). In any case, the square filler that Nate was asking about comes about where we have a tritransformation $\theta = \alpha$ and we apply it to an arrow $f$ of the form $(1_x, 1_y, \alpha_{z u v})$ to get a filler 2-cell (in a local tricategory).
So word to the wise: in order to read and make sense of my Notes, the Gordan-Power-Street monograph is required background. I should also say that those notes were originally written at Ross Street's request and known only to a few people (mostly committed experts) before John Baez took it upon himself to publicize them on his website, when I said I didn't mind. I say this to explain why they are not written in a more user-friendly form! But they have acquired, how shall I say, a certain notoriety.