16
$\begingroup$

As is usual, let's say an (n, k)-category is something with objects, morphisms, 2-morphisms, ..., n-morphisms, such that all j-morphisms for j > k are invertible, everything meant in the weak sense. We can also take n = ∞ or n = k = ∞. In this terminology the weak ω-categories in the title question are (∞,∞)-categories.

I think the only examples I know of weak ω-categories that are not (∞, k)-categories for some finite k are the ∞-category of all ∞-categories and the ∞-category Cob whose n-morphisms are n-dimensional manifolds (with corners) thought of as cobordisms between some specified (n-1)-dimensional manifolds (with corners). (I saw Dominic Verity give a very nice talk about his construction of a PL-version of this as a weak complicial set.) Of course, Cob has many variants, and we could also look at constructions such as functor categories, coproducts, products, etc., starting from these.

I'd be very interested in hearing about other examples of (∞,∞)-categories, even if they haven't really been constructed in the literature yet. Specially examples like Cob which are not internal to the theory of (∞,∞)-categories.

EDIT: I think that Sam Gunningham is right and I forgot (again) that the difference between having duals and having inverses is supposed to fall of the edge of the world when you go all the way out to ∞, so that Cob is an ∞-groupoid (specifically, it should be the well-known space classifying whatever kind of cobordism you used to build Cob). This means that I actually don't know any examples of genuinely (∞,∞)-categories that come from outside higher category theory.

EDIT 2: I somehow missed this earlier question. Maybe my question should be closed as a duplicate.

EDIT 3: Jeremy Hahn has convinced me that Sam's comment is true or false depending on how you define the equivalences of (∞,∞)-categories, and that it is not clear whether you really want every (∞,∞)-category with all adjoints to be an ∞-groupoid.

Improve this question
$\endgroup$
14
  • 6
    $\begingroup$ I thought the $(\infty, \infty)$-category of bordisms should in fact be an $(\infty, 0)$-category. Naively, wouldn't being ``$\infty$-dualizable'' mean that every morphism is invertible? $\endgroup$
    – Sam Gunningham
    Commented Apr 26, 2013 at 21:13
  • 1
    $\begingroup$ Yes, I think that's right @SamGunningham. I've added a remark about this to the question. $\endgroup$
    – Omar Antolín-Camarena
    Commented Apr 27, 2013 at 2:30
  • 3
    $\begingroup$ This might be too naïve, but E_n-algebras can be organized into an (∞,n+1)-category, so perhaps E_∞-algebras can be organized into an (∞,∞)-category? $\endgroup$
    – Dmitri Pavlov
    Commented Apr 27, 2013 at 14:40
  • 3
    $\begingroup$ An example I think of Dmitri Pavlov's comment: spaces are $E_\infty$ algebras in the category of spaces with correspondences, and the corresponding Morita $(\infty,\infty)$ category should have objects spaces, morphisms correspondences, 2-morphisms correspondences of correspondences, and so on all the way up. OTOH I think any space (or $E_\infty$ algebra) is $n$-dualizable for any $n$, so not sure how this agrees with Sam's comment. $\endgroup$
    – David Ben-Zvi
    Commented Apr 30, 2013 at 15:52
  • 2
    $\begingroup$ @Ricardo I can try... it's hard without pictures. Basically, in the first homotopy theory a morphism is invertible if it is part of a tower of adjoints that terminates at a higher identity morphism. This is in contrast to the second theory where a morphism is invertible simply if it is part of a tower of adjoints, with no restriction that the tower eventually degenerate into higher identity morphisms. $\endgroup$
    – Jeremy Hahn
    Commented May 2, 2013 at 16:57

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .