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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.

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Omar Antolín-Camarena
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    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? – Sam Gunningham Apr 26 '13 at 21:13
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    Yes, I think that's right @SamGunningham. I've added a remark about this to the question. – Omar Antolín-Camarena Apr 27 '13 at 02:30
  • Omar it is not clear what the homotopy theory of (infinity,infinity)-categories should be. Let J denote the nerve of the contractible groupoid with 2 objects. Then inverting J and all of its suspensions yields one homotopy theory. There is a second, different homotopy theory in which a category with all adjoints is contractible. If one looks only at (infinity,n)-categories these two homotopy theories are the same, but I believe they are not the same at (infinity,infinity). I think that the category of bordisms is only contractible in the second sense – Jeremy Hahn Apr 27 '13 at 04:46
  • Did you mean $\infty$-groupoid instead of contractible, @JeremyHahn? I mean an ordinary groupoid has all adjoints, right? I certainly don't want it to be contractible as an $(\infty,\infty)$-category. (Or do I?) If this second homotopy theory of $(\infty,\infty)$-categories makes either (1) a all homotopy types contractible, or (2) homotopy inverses not count as adjoints, it doesn't sound like such a great idea. – Omar Antolín-Camarena Apr 27 '13 at 12:51
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    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? – Dmitri Pavlov Apr 27 '13 at 14:40
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    I meant infinity groupoid. Homotopy inverses should always count as adjoints, but only in the second theory is there a kind of converse for towers of adjoints going all the way up to infinity. – Jeremy Hahn Apr 27 '13 at 15:46
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    I think the cobordism categories are much more interesting in the first homotopy theory, the advantage of the second being that it has a nice "coinductive" definition ala the model structure on strict omega categories. Presumably strict omega categories have a second model structure that better represents the first theory. – Jeremy Hahn Apr 27 '13 at 15:50
  • I agree that your question looks like a duplicate, but I am hesitant to close it unilaterally. – S. Carnahan Apr 28 '13 at 06:17
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    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. – David Ben-Zvi Apr 30 '13 at 15:52
  • @Jeremy Hahn: Can you describe what an invertible morphism in a (weak) $\omega$-category would look like in the first homotopy theory you consider? – Ricardo Andrade Apr 30 '13 at 18:36
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    @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. – Jeremy Hahn May 02 '13 at 16:57
  • @Jeremy: Thank you very much for the explanation. I just want to make sure I understand correctly what you mean by "higher identity morphism". Would that be a degenerate simplex (or globule, etc) in some multi-simplicial or globular approach? Or maybe it is a thin simplex in a complicial approach? Is that correct? Is there a better way to think about it? – Ricardo Andrade May 02 '13 at 19:14
  • Yes it seems you have the idea. The higher identity morphisms should indeed be degenerate multi-simplices or globules. Even in the case of complicial sets the higher identity morphisms should be characterized by degeneracy conditions rather than thinness. – Jeremy Hahn May 02 '13 at 23:41
  • @Jeremy: Thank you for the clarification. – Ricardo Andrade May 08 '13 at 04:30

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