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Can you find two finite groups G and H such that their representation categories are Morita equivalent (which is to say that there's an invertible bimodule category over these two monoidal categories) but where G is simple and H is not. The standard reference for module categories and related notions is this paper of Ostrik's

This is a much stronger condition than saying that C[G] and C[H] are Morita equivalent as rings (where C[A_7] and C[Z/9Z] gives an example, since they both have 9 matrix factors). It is weaker than asking whether a simple group can be isocategorical (i.e. have representation categories which are equivalent as tensor categories) with a non-simple group, which was shown to be impossible by Etingof and Gelaki.

Matt Emerton asked me this question when I was trying to explain to him why I was unhappy with any notion of "simple" for fusion categories. It's of interest to the study of fusion categories where the dual even Haagerup fusion category appears to be "simple" while the principal even Haagerup fusion category appears to be "not simple" yet the two are categorically Morita equivalent.

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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. –  Tom Leinster Oct 29 '09 at 18:36
    
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. –  Noah Snyder Oct 29 '09 at 18:44
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And what does "invertible" mean? –  David Speyer Oct 29 '09 at 18:46
    
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. –  Noah Snyder Oct 29 '09 at 18:57
    
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. –  Noah Snyder Oct 29 '09 at 18:58

4 Answers 4

up vote 7 down vote accepted

Noah,

I think an answer to your question is given in http://arxiv.org/pdf/0810.0032, theorem 1.1.

Subcategories of the double D(G) are given by pairs of normal subgroups K,N in G which centralize each other, together with the data of a bicharacter K\times N \to C^\times.

So in particular if G has no normal subgroups and H does, then you're going to find that D(G) has no nontrivial subcategories, while D(H) will (one can take, K=the normal subgroup in H, N={id}, and the bicharacter K\to C^* to be trivial I guess.

-david

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Hi Noah,

Categorically Morita equivalent groups were studied by Deepak Naidu in arXiv:math/0605530. He obtained there a complete description of Morita equivalent groups. It is also shown that simple groups are categorically Morita rigid.

Best, Dmitri

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Wouldn't it follow that the quantum doubles of the two groups are isomorphic? Would this help to set the question? (Sorry for posting this as an answer, didn't manage to leave it as a comment).

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Good point. Morita equivalence of fusion categories is exactly the same as saying that the doubles are equivalent as braided tensor categories. –  Noah Snyder Oct 29 '09 at 19:39
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On the other hand it's not at all clear to me that the doubles being equivalent as braided tensor categories implies that they're isomorphic as (quasi-triangular) Hopf algebras. A priori the Hopf algebra structures could come from two different fiber functors. –  Noah Snyder Oct 29 '09 at 19:54
    
I might be saying nonsense (it's late here), but it seems to me we are looking at Rep(D(H)) and Rep(D(G)) as already embedded in Hilb, and we also know they are braided equivalent, so this leaves space only for isomorphism between D(H) and D(G). Last 1/2 thought: the 3-d TFTs given by G and H are equivalent, I guess it's possible to give direct counterexamples considering the TFT associated to G, H and H/N (N a normal subgroup of H). –  pasquale zito Oct 29 '09 at 23:54
    
If the doubles give equivalent categories then one of them is a twist of the other (a result of Nikshych). For this result you don't need the braided equivalence, just equivalence as monoidal). A somehow similar and related question would be for what groups G some twisted group algebras $(kG)^R$ are simple as Hopf algebras? (in the sense of no normal Hopf subalgebras) –  Sebastian Burciu Dec 28 '09 at 20:56

The farthest I got thinking about this problem (and I haven't thought about it all that much) is that module categories over C[G] are classified in Section 3.4. They correspond to pairs K a subgroup of G and a choice of central extension of K (or equivalently, a certain cohomology class). In the case where there's no central extension, the dual category is some sort of Hecke algebra category C[K\G/K]-mod that I've never totally understood. Also I don't know how to modify that construction when you introduce the central extension. Anyway, modulo understanding those issues the question comes down to when a twisted Hecke algebra category C[K\G/K]-mod can be equivalent as a tensor category to C[H]-mod for some group H.

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This only works for finite G, I think. Are you looking for a pair of finite groups? or would you be happy with any pair? –  Tom Church Oct 29 '09 at 20:55
    
Yeah, sorry, finite groups. –  Noah Snyder Oct 29 '09 at 23:08

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