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According to the periodic table of k-tuply monoidal n-categories, it should be the case that a tetracategory (= weak 4-category) with one object, one 1-morphism and one 2-morphism is effectively equivalent to a symmetric monoidal category. I understand geometrically why one should expect this, but I cannot see how it arises directly from an algebraic definition of tetracategory.

I do know how we get a braided monoidal category: if $X$ and $Y$ are 3-morphisms, then the braiding $\gamma_{X,Y}$ is defined as the interchanger of $X$ and $Y$. What one then needs to show is that $\gamma_{X,Y} \circ \gamma_{Y,X} = \text{id}$. My understanding is that one would not expect an algebraic definition of tetracategory to carry this equation explicitly as an axiom; it would have to be derived.

I am aware that, of course, there does not exist a stable easily-referenced definition of tetracategory. (Edit: John Baez links below to an explicit definition due to Todd Trimble, written up by Alex Hoffnung.) I would be happy if someone could give me an answer along the lines of ''this is the sort of way a tetracategory should be defined, and one would reasonably expect it to have such-and-such algebraic data associated to it, and hence $\gamma_{X,Y} \circ \gamma _{Y,X} = \text{id}$ would follow in such-and-such a way."

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2 Answers 2

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It will come from a compatibility between different ways of composing interchangers.

(I'm going to use = to mean iso/homotopy in a HoTT-like way throughout, for ease of notation. I will also confuse proofs and homotopies throughout.)

To get a better intuition, let's first think about the case of higher groupoids. As a warmup let's think about the braided case first. So there we have a homotopy 3-type which is 1-connected and we want to understand why there's a braiding on 2-loops. (Note that universally it's enough to just do this for the sphere $S^2$, so secretly we're trying to understand $\pi_3(S^2)$.) But, as you know, this is just the usual Eckmann-Hilton argument where we just braid the maps around each other. Translating this into compositions turns into the interchange law because the first half of the argument is: $$x \circ_1 y = (x \circ_2 1) \circ_1 (1 \circ_2 y) = (x \circ_1 1) \circ_2 (1 \circ_1 y) = x \circ_2 y.$$ Note carefully that there are actually two proofs of Eckmann-Hilton: one where you rotate clockwise and one where you rotate counterclockwise. These proofs are different and they correspond to the two generators of $\pi_3(S^2) = \mathbb{Z}$. It is the fact that these two proofs of Eckmann-Hilton are different that makes these categories braided and not symmetric.

Alright, now let's move up one dimension higher. Now we're looking at 2-connected 4-groupoids. Now the algebraic topology question is the well-known fact that the generator of $\pi_3(S^2) = \mathbb{Z}$ has only order 2 when you stabilize it to put it in $\pi_4(S^3)$. Where does this come from? Well, it's clearly the same thing as asking why the two proofs of Eckmann-Hilton become the same when applied to 3-loops. Here the topological intuition is clear: take your clockwise proof and rotate it around the x-axis until it comes back into the plane becoming the counterclockwise proof.

Let's think about this in more detail. You can think of Eckmann-Hilton as taking place on a big square made up of four little squares with two tiles labelled x and y and the blanks labelled by identities and then sliding them around. Now we're looking in 3-dimensions at cubes and we want to slide one proof into the other by moving into the third dimension.

So we start with the proof: $$x \circ_1 y = (x \circ_2 1) \circ_1 (1 \circ_2 y) = (x \circ_1 1) \circ_2 (1 \circ_1 y) = x \circ_2 y = \ldots = y \circ_1 x.$$ Next we need to introduce a third dimension to give ourselves more room to work. So maybe: $$x \circ_1 y = (x \circ_2 1) \circ_1 (1 \circ_2 y) = (x \circ_1 1) \circ_2 (1 \circ_1 y) = ((x \circ_1 1) \circ_2 (1 \circ_1 y)) \circ_3 ((1 \circ_1 1) \circ_2 (1 \circ_1 1)) = \ldots$$ Then one gradually moves the clockwise proof around into the third dimension (so the interesting part is in the $\circ_2$ $\circ_3$ plane instead of the $\circ_1$ $\circ_2$ plane) and then back into the plane until it becomes the counterclockwise proof. I'm not going to write down the rest of this proof because it would take a while to get it entirely right and it would be hard to fit nicely on the screen anyway. But I'm confident that given a couple days I could write such a proof in Agda, and hopefully the idea is clear.

Just like the key lemma in proving Eckmann-Hilton is the interchange law, in order to prove this result we're going to need a higher interchanger relating the three compositions and their pairwise interchangers. Once you have such a higher interchanger the above proof should work for any doubly monoidal 4-category without groupoid-ness appearing anywhere.

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    $\begingroup$ Hi @Noah, thanks for this answer, which is extremely clarifying. The key is precisely how the different interchangers interact, as you mention in your last paragraph. It would be good to know what these interactions are, and how they follow from the tetracategory axioms. For example, here is an attempt at describing the interaction between a 3-interchanger and a 4-interchanger: dropbox.com/s/a6i1cogyrk28fpq/2014-12-24%2017.37.18.jpg?dl=0 $\endgroup$ Dec 24, 2014 at 17:47
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    $\begingroup$ (In that picture $\sigma$ and $\tau$ are 3-cells, $j_{\sigma,\tau}$ and $k_{\sigma,\tau}$ are different 4-interchangers that exchange the heights of $\sigma$ and $\tau$, the crossing present in each subdiagram is a 3-interchanger, and the fill-in 5-cell is a putative homotopy between the upper and lower paths that ought to follow from the tetracategory axioms.) $\endgroup$ Dec 24, 2014 at 17:50
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    $\begingroup$ "...ought to follow from the tetracategory axioms" that phrase delights me in a way I wouldn't have expected. I didn't imagine people using Todd's definition, when I first learned of it. $\endgroup$
    – David Roberts
    Dec 25, 2014 at 22:52
  • $\begingroup$ Alex Hoffnung is the first one who really used it, as far as I know. $\endgroup$
    – John Baez
    Jan 4, 2015 at 0:09
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The most easily referenced definition of a tetracategory - due to Todd Trimble - is in this paper by a former student of mine:

Tetracategories with just one object and one morphism almost surely won't be quite the same as symmetric monoidal categories: there will be some extra 'fluff' generated by the coherence 2-morphisms, 3-morphisms and 4-morphisms of the one object and the one morphism. This 'fluff' has been carefully analyzed in the tricategorical case here:

Mike Shulman has proposed one solution to this subtle problem, which I quite like, and the authors above have proposed another:

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  • $\begingroup$ Thanks for these great references. I hadn't seen that nice paper by Alex Hoffnung. If someone could answer my question based on the definition that Alex gives, that would be perfect. There is lots of subtlety here (the 'fluff'), which I'm very happy to sidestep --- I 'just' want to see why the double braid $\gamma_{X,Y} \circ \gamma_{Y,X} = \text{id}$. $\endgroup$ Dec 24, 2014 at 16:11
  • $\begingroup$ Below Noah Snyder has given the 'moral' proof that the braiding is a symmetry - it's a refinement of the Eckmann-Hilton argument. This may provide the insight you're seeking, though someone someday should still turn it into a proof based on Todd Trimble's definition of tetracategory as written up by Hoffnung. (Somehow I mistakenly thought you wanted an actual proof of an actual theorem, rather than 'the idea'.) $\endgroup$
    – John Baez
    Dec 24, 2014 at 16:47
  • $\begingroup$ I would be happy with a 'moral' proof, together with the 'hooks' that show how it can be executed with a real-world algebraic definition. $\endgroup$ Dec 24, 2014 at 17:01
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    $\begingroup$ What's Mike's proposed solution? $\endgroup$
    – David Roberts
    Dec 25, 2014 at 22:48
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    $\begingroup$ In a nutshell, Mike's proposed solution is that when you talk about an n-category "with one object, one 1-morphism, ... one k-morphism", you should treat all these things as extra structure rather than extra properties. We have had extensive online conversations about this but I'm too lazy to find them. There is probably a bit about it in the appendix to "Lectures on n-Categories and Cohomology". $\endgroup$
    – John Baez
    Jan 4, 2015 at 0:05

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