Skip to main content
MathOverflow

Return to Answer

Added name of paper; removed opening and salutation; TeXed while I was here
Source Link
LSpice
LSpice
  • 12.9k
  • 4
  • 45
  • 69

Noah,

I think an answer to your question is given in http://arxiv.org/pdf/0810.0032Naidu, Nikshych, and Witherspoon - Fusion subcategories of representation categories of twisted quantum doubles of finite groups, theorem 1.1.

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

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

-david

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

I think an answer to your question is given in Naidu, Nikshych, and Witherspoon - Fusion subcategories of representation categories of twisted quantum doubles of finite groups, 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 datum of a bicharacter $K\times N \to \mathbb C^\times$.

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 $K\to \mathbb C^\times$ to be trivial, I guess).

Post Undeleted by David Jordan
deleted 416 characters in body; added 2 characters in body
Source Link
David Jordan
David Jordan
  • 6.1k
  • 31
  • 43

Noah,

Maybe I'm goingI think an answer to make a dumb mistake hereyour question is given in http://arxiv.org/pdf/0810.0032, since I'm writing this down rather quickly. Tell me what you thinktheorem 1. I think I claim this can happen all the time1.

What about taking G to be a finite simple group, letting C=RepSubcategories of the double D(G) are given by pairs of normal subgroups K, and letting D=C^* VN in G which centralize each other, where C^**V means taketogether with the dual category relative to V=categorydata of Vector spaces, regarded as a module category over C via the fiber functor. Now, D will be pointed, as it will be equal to representations of the dual hopf algebra to C[G], which is commutativebicharacter K\times N \to C^\times. A pointed fusion category is necessarily equal to Rep(H,w), where H is a finite abelian group

So in particular if G has no normal subgroups and w is a cocycle. HoweverH does, this is actuallythen you're going to be trivial, because you already realizedfind that D as Rep(C[G]^*G). So that means has no nontrivial subcategories, while D is just Rep[H] for H abelian(H) will (one can take, andK=the normal subgroup in particular not simple. By definition, these share an invertible bimodule categoryH, namely VecN={id}, soand the bicharacter K\to C^* to be trivial I think this answers your questionguess.

-david

Noah,

Maybe I'm going to make a dumb mistake here, since I'm writing this down rather quickly. Tell me what you think. I think I claim this can happen all the time.

What about taking G to be a finite simple group, letting C=Rep(G), and letting D=C^* V, where C^**V means take the dual category relative to V=category of Vector spaces, regarded as a module category over C via the fiber functor. Now, D will be pointed, as it will be equal to representations of the dual hopf algebra to C[G], which is commutative. A pointed fusion category is necessarily equal to Rep(H,w), where H is a finite abelian group and w is a cocycle. However, this is actually going to be trivial, because you already realized D as Rep(C[G]^*). So that means D is just Rep[H] for H abelian, and in particular not simple. By definition, these share an invertible bimodule category, namely Vec, so I think this answers your question.

-david

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

Post Deleted by David Jordan
Source Link
David Jordan
David Jordan
  • 6.1k
  • 31
  • 43

Noah,

Maybe I'm going to make a dumb mistake here, since I'm writing this down rather quickly. Tell me what you think. I think I claim this can happen all the time.

What about taking G to be a finite simple group, letting C=Rep(G), and letting D=C^* V, where C^**V means take the dual category relative to V=category of Vector spaces, regarded as a module category over C via the fiber functor. Now, D will be pointed, as it will be equal to representations of the dual hopf algebra to C[G], which is commutative. A pointed fusion category is necessarily equal to Rep(H,w), where H is a finite abelian group and w is a cocycle. However, this is actually going to be trivial, because you already realized D as Rep(C[G]^*). So that means D is just Rep[H] for H abelian, and in particular not simple. By definition, these share an invertible bimodule category, namely Vec, so I think this answers your question.

-david