Timeline for What categorical property of monoidal categories picks out the ones with duals?
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12 events
| when toggle format | what | by | license | comment | |
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| Sep 12, 2016 at 13:56 | comment | added | HeinrichD | Can someone post a detailed answer based on Mike's answer? | |
| Dec 7, 2015 at 13:04 | comment | added | Chris Schommer-Pries | @PavelSafronov In D.1.3 of Appendix D in Gaitsgory's paper he says "... it is easy to show that O is rigid in the sense of Sect. D.1.1 if and only if every compact object of O admits both left and right monoidal duals". However right now I am only able to see that this implies "weak rigidity" as in Bakalov-Kirillov's Lectures on Tensor Cateogries and Modular Functors. Do you understand the argument here? | |
| Dec 7, 2015 at 1:57 | vote | accept | Theo Johnson-Freyd | ||
| Dec 7, 2015 at 1:41 | answer | added | Mike Shulman | timeline score: 8 | |
| Dec 6, 2015 at 16:42 | comment | added | Theo Johnson-Freyd | @PavelSafronov I had not looked there --- thanks! Gaitsgory's condition is probably just right (and certainly about how I like to think), although it doesn't make it clear that rigidity specializes to groupness. | |
| Dec 6, 2015 at 16:41 | comment | added | Theo Johnson-Freyd | @მამუკაჯიბლაძე In the Hopf case as well, the antipode map is a map to the "dual" (co)algebra, where "dual" here means in the Morita bicategory. Similarly, for a category $\mathcal C$, the opposite category $\mathcal C^\op$, which is the recipient of $X \mapsto X^*$, is the dual object in the "Morita" bicategory of categories, profunctors, and natural transformations. I don't know if that is a useful similarity or not. | |
| Dec 6, 2015 at 16:38 | comment | added | Theo Johnson-Freyd | @მამუკაჯიბლაძე Right, the problem is that sending an object of a rigid category to its dual is contravariant. It is a bit like the problem of defining "Hopf": a bialgebra is an algebra object among coalgebras, but the Hopf condition requires going down to the level of underlying vector spaces (existence of antipode is equivalent to the linear map $\Delta \circ m : H\otimes H \to H \otimes H$ being invertible; but this map is neither a map of algebras nor a map of coalgebras). | |
| Dec 6, 2015 at 12:23 | comment | added | Pavel Safronov | Have you looked at Appendix D in Gaitsgory's 1-affineness paper (arxiv.org/abs/1306.4304)? He discusses the setting of presentable categories. | |
| Dec 6, 2015 at 12:12 | answer | added | Michal R. Przybylek | timeline score: 2 | |
| Dec 6, 2015 at 7:59 | comment | added | მამუკა ჯიბლაძე | Probably $\cal C\times\cal C$ has to be replaced by some sort of Grothendieck construction. Reasoning might go like this: (1) for a monoid $G$ in a "plain" category $\cal S$, to say "for every element $x$ of $G$ multiplying by $x$ is an isomorphism" is to say this for the generic element (diagonal) $1_G\to G^*(G)$ of $G^*(G)$ in ${\cal S}/G$. (2) to repeat this for a monoid $\cal C$ in a 2- (as opposed to "plain") category like $\bf Cat$, it is not obvious anymore that one should use ${\bf Cat}/\cal C$; maybe more reasonable choice is ${\bf Cat}^{{\cal C}^\circ}$. | |
| Dec 6, 2015 at 6:46 | comment | added | Theo Johnson-Freyd | There are many close connections between groups and rigid monoidal categories; for example, a rigid monoidal category whose only morphisms are identities is a group. I learned from wikipedia about pregroups, which are monoidal categories whose underlying category is a partially ordered set; their sole use seems to be in linguistics and computer science. | |
| Dec 6, 2015 at 6:45 | history | asked | Theo Johnson-Freyd | CC BY-SA 3.0 |