Timeline for The category of subfactors extending the category of groups?
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Nov 3, 2014 at 21:10 | vote | accept | Sebastien Palcoux | ||
| Oct 29, 2014 at 2:20 | answer | added | Marcel Bischoff | timeline score: 2 | |
| Jan 31, 2014 at 20:11 | history | edited | Sebastien Palcoux |
I add the tag qa.quantum-algebra.
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| Jan 31, 2014 at 19:33 | comment | added | Jesse Peterson | You could define a ``morphism'' from $(N_1 \subset M_1)$ to $(N_2 \subset M_2)$ to be a group homomorphism from the normalizer group $\mathcal N_{M_1}(N_1) / \mathcal U(N_1)$ to $\mathcal N_{M_2}(N_2) / \mathcal U(N_2)$. But I don't think you'll get much insight from this perspective. | |
| Jan 31, 2014 at 18:54 | answer | added | Sebastien Palcoux | timeline score: 0 | |
| Jan 31, 2014 at 18:22 | comment | added | Sebastien Palcoux | @JessePeterson : because the isomorphic class of $R \subset R \rtimes G$ does not depend on the choice of the action (as above), I could say "something which holds for arbitrary actions", but because I'm looking for "other" subfactor-morphisms, the choice of specific action is perhaps relevant. | |
| Jan 31, 2014 at 18:08 | comment | added | Jesse Peterson | Are you taking specific actions or are you looking for something which holds for arbitrary actions? | |
| Jan 31, 2014 at 10:57 | comment | added | Sebastien Palcoux | @JessePeterson : the finite groups $G$ and $G'$ act as outer automorphisms of the hyperfinite II$_1$ factor $R$. | |
| Jan 31, 2014 at 3:47 | comment | added | Jesse Peterson | How are you having $G$ and $G'$ act on $R$? | |
| Jan 29, 2014 at 21:16 | comment | added | Sebastien Palcoux | The category of finite groups is also not a (natural) subcategory of the category of planar algebras because the subfactor planar algebras are simple (see here), so the subfactor planar algebra morphisms are also either injective or $0$... | |
| Jan 28, 2014 at 20:19 | history | asked | Sebastien Palcoux | CC BY-SA 3.0 |