Timeline for Are there two groups which are categorically Morita equivalent but only one of which is simple
Current License: CC BY-SA 2.5
21 events
| when toggle format | what | by | license | comment | |
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| Jan 5, 2010 at 6:11 | answer | added | Dmitri Nikshych | timeline score: 12 | |
| Nov 2, 2009 at 0:14 | vote | accept | Noah Snyder | ||
| Nov 2, 2009 at 0:14 | vote | accept | Noah Snyder | ||
| Nov 2, 2009 at 0:14 | |||||
| Oct 31, 2009 at 18:21 | comment | added | David Jordan | Perhaps functors should be replaced by (not-necessarily-invertible) bimodule categories, as that seems to be the more appropriate notion of morphism in this business? | |
| Oct 31, 2009 at 18:17 | comment | added | David Jordan | By the way, what's your definition of and subsequent objection with the notion of simple for fusion categories? I'm interested because Pavel once explained to me that "short exact sequences" would naturally be defined as sequences of functors Vect --> K --> C --> Q --> Vect where arrows are functors and K,C,Q are fusion categories. The problem, though is that this forces K necessarily to be representations of a Hopf algebra because the sequence gives a functor to Vect. So it is interesting to come up with the correct version of short exact sequences, and thus probably simplicity. | |
| Oct 31, 2009 at 15:42 | answer | added | David Jordan | timeline score: 12 | |
| Oct 29, 2009 at 23:08 | history | edited | Noah Snyder | CC BY-SA 2.5 |
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| Oct 29, 2009 at 19:23 | answer | added | pasquale zito | timeline score: 4 | |
| Oct 29, 2009 at 19:19 | answer | added | Noah Snyder | timeline score: 3 | |
| Oct 29, 2009 at 19:16 | comment | added | Noah Snyder | Another definition: A and B are morita equivalent if there exists an algebra object X in A such that B is equivalent as a tensor category to the category of X-X bimodule objects in A. | |
| Oct 29, 2009 at 19:14 | history | edited | Noah Snyder | CC BY-SA 2.5 |
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| Oct 29, 2009 at 19:08 | comment | added | Noah Snyder | Yeah, they should commute. | |
| Oct 29, 2009 at 19:07 | history | edited | Noah Snyder | CC BY-SA 2.5 |
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| Oct 29, 2009 at 19:06 | comment | added | S. Carnahan♦ | Don't the actions of A and B have to commute? | |
| Oct 29, 2009 at 18:58 | comment | added | Noah Snyder | If you like subfactor planar algebras, A and B are Morita equivalent if you can realize them as the shaded-shaded and unshaded-unshaded parts of a single shaded planar algebra. | |
| Oct 29, 2009 at 18:57 | comment | added | Noah Snyder | Invertible is a little trickier. Basically it means that ${}_A M_B \otimes_B {}_B M_A \cong {}_A A_A$, but for that you need to know what it means to take a tensor product of module categories over a tensor category. Alternately, you can say that two tensor cateogries A and B are Morita equivalent if there exists a module category M over A such that B=A* the category of monoidal endofunctors of M. | |
| Oct 29, 2009 at 18:46 | comment | added | David E Speyer | And what does "invertible" mean? | |
| Oct 29, 2009 at 18:44 | comment | added | Noah Snyder | The representation category C[G]-mod has a tensor product. Thus it makes sense to ask about module categories over it. A module category M over a monoidal category R is a category together with a functor (R,M)->M which satisfies associativity (possibly up to a coherent associator). A bimodule category is just a category which is a left module category for one monoidal category and a right module category for another monoidal category. | |
| Oct 29, 2009 at 18:40 | history | edited | Noah Snyder | CC BY-SA 2.5 |
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| Oct 29, 2009 at 18:36 | comment | added | Tom Leinster | Can you spell out what you mean by "categorically Morita equivalent", or "invertible bimodule category"? Regarding G and H as one-object categories, they are Morita equivalent as categories iff they are isomorphic as groups. (By definition, categories are Morita equivalent iff their presheaf categories are equivalent. The result on groups is a little theorem.) But it sounds like that's not what you mean. | |
| Oct 29, 2009 at 18:27 | history | asked | Noah Snyder | CC BY-SA 2.5 |