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Jan 5, 2010 at 19:36 comment added Sebastian Burciu @t3suji: Here is what I had in mind before. Take the orbit $\mathcal{O}$ and induce it up to $G$. All the irreducible constituents that are obtained in this way generate a module category over $Rep(G/N)$. This is also indecomposable.
Jan 5, 2010 at 18:18 vote accept Sebastian Burciu
Jan 5, 2010 at 18:15 answer added Victor Ostrik timeline score: 10
Jan 5, 2010 at 17:37 history edited Sebastian Burciu CC BY-SA 2.5
reformulated the second part; it was an incorrect statement,sorry!
Jan 5, 2010 at 17:30 comment added Sebastian Burciu @ t3suji : Yes, it's something wrong. I will rewrite the second part, shortly. One can also take $N=Z(G)$ to get a contradiction.
Jan 5, 2010 at 17:19 comment added t3suji @Sebastian Burciu: Are they? Take $G=H$ and $\mathcal O$ to be the orbit of the one-point orbit of the trivial representation of $G$, for instance. Aren't you saying that the tensor product of the trivial representation and any representation is trivial?
Jan 5, 2010 at 17:17 comment added Sebastian Burciu @ Leonid: It gives a classification of all modules categories over $Rep(G)$. Maybe I should have said this in the post.
Jan 5, 2010 at 17:13 comment added Sebastian Burciu One rstricts a rep. of $G$ to $H$ and then it tensors over $H$ with some rep. from $\mathcal{O}$. The irreducible constituents of the tensor product are in the same orbit, $\mathcal{O}$.
Jan 5, 2010 at 17:09 comment added t3suji I'd think tensor product is by pulling back a representation from $G$ to $\tilde H$ and then tensoring over $\tilde H$. How is $\mathcal M$ a module category?
Jan 5, 2010 at 16:56 history edited Mariano Suárez-Álvarez CC BY-SA 2.5
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Jan 5, 2010 at 16:56 history edited Sebastian Burciu CC BY-SA 2.5
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Jan 5, 2010 at 16:50 history asked Sebastian Burciu CC BY-SA 2.5