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If $U$ is a representation of $G$ and $W$ is a representation of $H$, then why is $$ U \otimes \operatorname{Ind}(W) = \operatorname{Ind}(\operatorname{Res}(U) \otimes W)$$

I've tried to simply use the definitions to prove they are equal, but I've hit a road block.

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up vote 2 down vote accepted

I take it $H \leq G$. These reps are not equal, but they are isomorphic. It will be more natural to prove that $ (W \otimes U|_H) \uparrow^G \cong (W \uparrow ^G) \otimes U $, but this is equivalent as $A\otimes B \cong B \otimes A$ for modules over group algebras.

We want a module map $kG \otimes_H (W \otimes U|_H) \to (kG\otimes_H W) \otimes U$. Take $\beta: x\otimes (w \otimes u) \mapsto (x\otimes w) \otimes x u $. This is easily verified to be a module map (and well-defined: there is an issue because of the $\otimes_H$).

Now define $\alpha : (kG\otimes_H W) \otimes U \to kG \otimes_H (W \otimes U|_H)$ by $(x \otimes w) \otimes u \mapsto x \otimes ( w \otimes x^{-1} u )$. This is inverse to $\beta$, and is well-defined and a module map: thus the two modules are isomorphic.

Surprisingly, all this generalises to arbitrary Hopf algebras (even non-cocommutative ones). The maps are slightly more messy to write down though.

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Is there a proof of this using Frobenius reciprocity? I think there is, and it'd be nice to know (nicer than having to explicitly define an isomorphism, at least) –  Bey Nov 12 '12 at 22:19
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