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$\DeclareMathOperator\Rep{Rep}$Related to this question I also had some troubles to understand the classification of module categories over $\Rep(G)$. Specifically, on page 12 of Ostrik's paper what is the category $\mathrm{\Rep}^1(\tilde{H})$? $k^*$ acting as "identity character on V" means $a.v=av$ for all $a \in k^*$ and $v \in V$? Then what is the structure of module category over $\Rep(G)$? Tensor product should be after restricting representations of $G$ to $H$ and then inducing back to $\tilde{H}$?

Concretely, I was thinking about the following example. Let $H$ be a subgroup of $G$. Then $\Rep(H)$ is a module category over $\Rep(G)$ via tensor product as $H$-modules. What is the decomposition of $\Rep(H)$ in indecomposable module categories and what are the corresponding subgroups $H$ and cocyles $\omega \in H^2(H,\;k^*)$ for each indecomposable subcategory?

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  • $\begingroup$ 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? $\endgroup$
    – t3suji
    Commented Jan 5, 2010 at 17:09
  • $\begingroup$ 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, $\mathcal{O}$. $\endgroup$ Commented Jan 5, 2010 at 17:13
  • $\begingroup$ @ Leonid: It gives a classification of all modules categories over $Rep(G)$. Maybe I should have said this in the post. $\endgroup$ Commented Jan 5, 2010 at 17:17
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    $\begingroup$ @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? $\endgroup$
    – t3suji
    Commented Jan 5, 2010 at 17:19
  • $\begingroup$ @ t3suji : Yes, it's something wrong. I will rewrite the second part, shortly. One can also take $N=Z(G)$ to get a contradiction. $\endgroup$ Commented Jan 5, 2010 at 17:30

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Sebastian: your definition of $Rep^1(\tilde H)$ is absolutely correct. If you have a representation of $G$ you can restrict it to $H$ and consider it as a representation of $\tilde H$ (this operation is called inflation). Now you can tensor it with any representation of $\tilde H$; this tensoring preserves $Rep^1(\tilde H)$; this is a module category structure (same thing was explained above by t3suji).

The category $Rep(H)$ considered as a module category over $Rep(G)$ is indecomposable. It corresponds to subgroup $H$ and trivial cocycle $\omega$ (so $\tilde H$ is a direct product of $H$ and multiplicative group $G_m$).

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