Effects of "weak" vs. "strict" categories in Eckmann-Hilton arguments

Question

A standard example for demonstrating the need for genuinely weak n-categories is that a weak 3-category with unique 0- and 1-cells amounts to the same thing as a braided monoidal category (by an Eckmann-Hilton argument), but were one to use a strict 3-category instead, this would automatically become a fortiori symmetric. In trying to get a better intuition for "the right notion" of weak categories (this is still unsettled, right?), I was wondering if anyone could give me a good intuition for what step in the argument for symmetry in the strict case we want to fail in the weak case and why. [I suppose this amounts to more generally giving a good intuition for what higher-dimensional coherence isomorphisms we should refrain from demanding to exist in the definition of weak categories, and why, but this example seems in particular like a usefully illustrative introductory context]

Answer 1

Here is a nice pictorial proof of the Eckmann-Hilton argument in a higher category (made by Eugenia Cheng). One way to read this is as a proof that in a weak 2-category with one 0-cell and one 1-cell, the composition of 2-cells is commutative: in this case each diagram is equal to the two diagrams next to it, either by a unit law or by an interchange law.

Alternately, we can read it as a proof that in a weak 3-category with unique 0- and 1-cells, the composition on 2-cells is braided. In this case each diagram is isomorphic to those next to it, via either a unit isomorphism or an interchange isomorphism. The braiding is the composite isomorphism along the top half (say), and the fact that it isn't a symmetry arises because the composite all the way around need not be the identity.

Of course, if both interchange and unit isomorphisms were to be identities, the braiding would be a symmetry. It is known in dimension 3 that it's actually good enough (i.e. things don't collapse and become too strict) if either the unit or interchange law is weak, and the other is strict. If interchange is weak but the unit is strict, we get Gray-categories as in the original Gordon-Power-Street paper on tricategories, while if the interchange is weak but the unit is strict, this is the subject of Simpson's conjecture as mentioned by David (for the n=3 case, see here). Note that associativity never enters the picture at all, and at least in dimension 3 it can also be made strict.

There isn't really a "one step" that we want to fail. In a "fully weak" higher category, all of the associativity, unit, and interchange laws would hold only weakly. It so happens that such fully weak categories can be "semi-strictified" to make some, but not (in general) all, of these laws hold strictly, while still being equivalent to the original category in a suitable sense. Such semistrictification can often be technically useful, but I don't regard it as really of foundational importance.

Regarding "intuition for what higher-dimensional coherence isomorphisms we should refrain from demanding to exist," I think it's misleading to view this example in that way. There isn't really a coherence isomorphism that we're refraining from demanding to exist---in a weak 3-category, it's still the case that all coherence isomorphisms exist and all "formally describable" diagrams commute. Rather, what's happening is that accidentally, if we happen to be considering cells whose source and target are both identities, then there are some structural isomorphisms that we happen to be able to compose in a way "unforeseen" by the general theory of weak n-categories. Therefore, in this particular case, the assertion of that general theory that "all diagrams commute" doesn't apply, since the diagram we're looking at is only well-defined by accident. This is kind of vague, but it can be made precise by the theory of contractible globular operads, where it is true that "all diagrams commute" in the formal, operadic, sense, but in some particular algebra over such an operad, there may be accidental "composites" which do not commute. See also this question.

Answer 2

One thought along these lines is Simpson's conjecture (proved for $n$-categories up to $n=3$). Essentially this states that a weak $n$-category is equivalent to one where everything is strict except for the identities. Since one of the assumptions in Eckmann-Hilton is that there is a unit for both multiplications, we lose the result that the product is abelian/symmetric.

Another approach is along the lines of weakening interchange (another assumption of Eckman-Hilton, namely that the products commute with each other), which is what happens in Gray-categories - categories enriched over 2Cat with the Gray tensor product (as opposed to the cartesian product). Gray-categories are known to model all weak 3-categories. Crans has taken this further and constructed an analogous tensor product on GrayCat, but I'm not certain if enriching using this tensor product gives weak 4-categories.