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How to prove $U \otimes Ind W = Ind(Res(U) \otimes W)$? where U is a representation of G and W is a rep of H, a subgroup of G. $Ind(W)$ is the induced rep and $Res(U)$ is the restrict rep.

I got the answer, by both approaches: groups algebra and constructing isomorphic map. Thanks for all the very helpful comments and answer.

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Tell us your background and what books you have read on representation theory. – KConrad Aug 22 '10 at 22:24
This is an exercise in Fulton and Harris, so without more information one might be tempted to think that this is "homework". – José Figueroa-O'Farrill Aug 22 '10 at 22:31
Just me being ornery, but I strongly recommend you edit the question so that the title is a complete sentence, ending with a question mark, and the body doesn't begin mid-thought. As written, the question looks unprofessional. – Theo Johnson-Freyd Aug 22 '10 at 23:00
The two are canonically isomorphic. (This is what the equality means.) Both $\otimes$ and Ind satisfy universal mapping properties; try combining them to figure out the universal property of each side of the purported isomorphism, and check that both satisfy the same universal mapping property, and hence are naturally isomorphic. – Emerton Aug 23 '10 at 0:08
If the above comments aren't enough to help you, I recommend asking this on as I think you'll get an answer more at the right level there. – Loop Space Aug 23 '10 at 10:28
up vote 4 down vote accepted

This is really a comment which got too long.

Personally, I always find this one rather confusing. If you think in terms of modules over group rings, we want to show that $U \otimes (\mathbb{C}[G]\otimes_{\mathbb{C}[H]} W) \cong \mathbb{C}[G] \otimes_{\mathbb{C}[H]} (U\otimes W)$. The $G$-equivariant isomorphism is not given by sending $u\otimes (x\otimes w)$ to $x \otimes (u\otimes w)$. There's a lot wrong with this formula, but that's not the point.

The point is that to get the right formula, one really needs to remember exactly how the universal property of induction works. I don't have Fulton and Harris in front of me to see what they say, but Serre's book has a good discussion of induction which will lead one right to the answer.

Also, unless I'm confused, this really seems to depend on the structure of $\mathbb{C}[G]$ as both a ring and as a $\mathbb{C}[H]$-module. One needs to know that it's a free $\mathbb{C}[H]$-module, and that it has a decomposition as a $C[H]$-module into summands isomorphic to $C[H]$ that are permuted by the units of the ring $C[G]$. One could ask, for morphisms of rings $C\to R\to S$, when it's true that for an R-module M and an S-module N we have the formula $N \otimes_C (S\otimes_R M) \cong S \otimes_R (N \otimes_C M)$ (as S-modules). (Above I wrote $\otimes$ instead of $\otimes_{\mathbb{C}}$; now C is the ground ring). I don't know how to prove this without assuming S has the sort of structure mentioned above (free as an R-module, etc.).

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Beware that the approach to induced representations in Serre's book is in the spirit of what is called "compact induction" from a broader point of view (beyond finite groups), as is seen from how the Frobenius reciprocity as in Serre's book is "opposite" from how it appears in books which allow infinite-dimensional repns of more general groups (where the distinction between induction and compact induction becomes more significant). In the finite case the two notions are canonically isomorphic, but nonetheless it can be a source of some confusion when comparing references. – BCnrd Aug 23 '10 at 13:14
Good to know. I guess I've never thought about the situations in which this distinction arises. – Dan Ramras Aug 24 '10 at 2:22

If we are talking about complex representations of finite groups, module facts can usually be proved easily using characters. The module isomorphism in question corresponds to the following character formula. If $\chi$ is a character of $G$ and $\psi$ is a character of $H$, the formula is $\chi \psi^G = (\chi_H \psi)^G$. This follows directly from the formula defining an induced character. (See, for example, Chapter 5 of my character theory book.)

Alternatively, (and amusingly) the formula follows by Frobenius reciprocity. It suffices to show that the multiplicities of an arbitrary irreducible character $\alpha$ of $G$ in each of $\chi\psi^G$ and $(\chi_H\psi)^G$ are equal. We have $$ [\chi\psi^G,\alpha] = [\psi^G,\alpha\overline\chi] = [\psi,\alpha_H\overline \chi_H] = [\chi_H\psi,\alpha_H] = [(\chi_H\psi)^G,\alpha], $$ as wanted. The second and last equalities follow by reciprocity and the first and third follow from the definition of the character inner product.

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