27
$\begingroup$

This title probably seems strange, so let me explain.

Out of the several different ways of modeling $(\infty, n)$-categories, complicial sets and comical sets allow $n = \infty$, providing mathematical definitions of $(\infty, \infty)$-categories. I've asked people a few times for interesting examples of $(\infty, \infty)$-categories that could fit into these definitions, and I've always gotten the answer: the $(\infty, \infty)$-category of (small) $(\infty, \infty)$-categories.

This is not a bad example, and I think it's cool, but I would like to know what kinds of examples are out there other than just categories of categories. For example, for $(\infty, n)$-categories with $n$ finite, "non-categorical" examples include $(\infty, n)$-categories of bordisms as well as the Morita $(\infty, n)$-category of $E_{n-1}$-algebras in an $(\infty, 1)$-category: people care about bordisms and $E_{n-1}$-algebras before learning that they have this higher-categorical structure.

I'm interested in hearing about examples like these for $(\infty, \infty)$-categories. It doesn't matter a lot to me whether something's been rigorously shown to be an example of one of these models or not; and maybe your favorite example is a different kind of $(\infty, \infty)$-category, such as the ones discussed in Theo's question from several years ago; that's also welcome.

What would be really neat is an example of a new phenomenon at the $n = \infty$ level, so an example of an $(\infty, \infty)$-category that's not similar to an $(\infty, n)$-category example for any $n$, but that seems like a lot to ask for.

In addition to Theo's question that I linked above, this question by Alec Rhea and this question by Giorgio Mossa are also relevant, asking similar questions for $n$ finite.

$\endgroup$
12
  • 10
    $\begingroup$ This is obviously not relevant to the main point but I would somewhat dispute the claim that "the $(\infty,\infty)$-category of all $(\infty,\infty)$-categories" is not a bad example. Some would even say that it is an excellent example of a bad example... $\endgroup$
    – Will Sawin
    Jan 29, 2021 at 0:38
  • 8
    $\begingroup$ So there are two non-equivalent definition of $(\infty,\infty)$-categories as explained here mathoverflow.net/a/134099/22131 I will refer to these as inductive and coinductive $(\infty,\infty)$-categories. If you are using the inductive definition, there is an $(\infty,\infty)$-category of cobordisms and an $(\infty,\infty)$-category of Higher spans. These becam trivial using the coinductive definition however. $\endgroup$ Jan 29, 2021 at 1:41
  • 5
    $\begingroup$ I appreciate the mononymy, but of course there are multiple "Theo"s who do mathematics :) In any case, @SimonHenry got to it before me, but spans are naturally an $(\infty,\infty)$-category. This is true also for spans-with-structure. An important example is the $(\infty,\infty)$-category of (shifted) symplectic manifolds and Lagrangian correspondences, which I believe is carefully defined in upcoming work by Calaque, Haugseng, and Scheimbauer. $\endgroup$ Jan 30, 2021 at 0:40
  • 1
    $\begingroup$ Actually, the category of Lagrangian correspondences can also be extended "down", where you form a sort of "loop spectrum" of $(\infty,\infty)$-categories — what Scheimbauer termed a tower in her PhD thesis. $\endgroup$ Jan 30, 2021 at 0:42
  • 2
    $\begingroup$ @TimCampion Not obviously, at least: an (inductive) $(\infty,\infty)$-category is a sequence $(X_n)$ where $X_n$ is an $(\infty,n)$-category with underlying $(\infty,n-1)$-category $X_{n-1}$. But for the Morita $(\infty,n)$-categories of $E_n$-algebras in a symmetric monoidal $\infty$-category $V$ then the objects are different for each $n$. But maybe there should be a Morita $(\infty,\infty)$-category of $V$-$(\infty,\infty)$-categories? (I don't think you can view $E_\infty$-algebras as special $(\infty,\infty)$-categories, as you can for $E_n$, since you are extending "down" not "up".) $\endgroup$ Feb 4, 2021 at 8:20

1 Answer 1

11
$\begingroup$

As mentioned by Simon Henry: The $(\infty,\infty)$-category of cobordisms.

(Not constructed, but if you did it you could presumably have any of the usual bells and whistles you might want.)

To clarify Simon Henry's comment: The statement is that that $(\infty,\infty)$-category of cobordisms in the coinductive setting is an $\infty$-groupoid by Cheng's theorem (so it's whatever Thom spectrum you expect by GMTW). In the inductive setting, Cheng's theorem doesn't hold. So non-invertible $(\infty,\infty)$-TFT's should be a thing. I think nobody's formally written down this $(\infty,\infty)$-category -- I assume because $(\infty,n)$-TFTs are hard enough so there's not much demand for it. Please challenge that assumption!

One nice thing about complicial sets (and I guess also comical sets) is that they (ought to) naturally put you in the (more general) inductive setting, and you might hope they'd be a good place to construct these (∞,∞)-categories.

Anyway, this ticks a few boxes:

  1. The inductive / coinductive distinction is arguably a "new phenomenon" (though maybe it's just a "new complication"), and this example already illustrates how it works.

  2. It's a super-canonical example, and should be super-interesting for all the reasons its lower brethren are.

  3. It's not a category of categories.

$\endgroup$
3
  • 2
    $\begingroup$ I think Dominic Verity has given a description of this $(\infty,\infty)$-category as a complicial sets. (well, given that the precise connection between complicial sets and $(\infty,\infty)$-categories is still unclear one cannot prove that it really model this object, but it supposed too) I don't know if it is written out in his paper, I only heard gave him talk about this. $\endgroup$ Feb 3, 2021 at 22:58
  • 2
    $\begingroup$ As of a year ago my understanding from talking to Dom was that he had thought seriously about what would go into writing down such a definition, but had not written it up. But it's possible that's partly because he has high standards for what counts as writing and is very modest. $\endgroup$
    – Tim Campion
    Feb 3, 2021 at 23:02
  • $\begingroup$ @TimCampion Thanks, this is a great answer! $\endgroup$ Feb 11, 2021 at 20:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.