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The classical Frobenius reciprocity theorem asserts the following:

If $W$ is a representation of $H$, and $U$ a representation of $G$, then $$(\chi_{Ind W},\chi_{U})_{G}=(\chi_{W},\chi_{Res U})_{H}.$$

The proof in the standard textbook (Fulton&Harris, Dummit&Foote,etc) is easy to understand. What puzzled me is this Frobenius theorem that appears in Raoul Bott's paper:

"Proposition 2.1. Let $W$ be a $G$-module, let $M$ be an $H$-module and denote by $i^{*}W$, the restriction of $W$ to $H$. Then, $$Hom_{G}(W,\Gamma MG)\cong Hom_{H}(i^{*}W, M).$$

In here the $\Gamma MG$ is defined to be the section of the bundle $G\times_{H} M\rightarrow G/H$, with $G\times_{H}M$ defined to be $G\times M/(g,m)\approx (gh,h^{-1}m)$.

Bott claimed in his paper (Homogeneous differential operators) that this isomorphism is quite canoical, yet not only I could not understand his proof, but also I could not see how the isomorphism is anything but canonical. There should be some kind of relationship between this and the classical theorem, but I could not get it as well. After several hours pondering I decided to ask in here as the matter is purely technical.

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We have $\Gamma MG=Ind\ M$ and $i^*W=Res\ W$. – Pierre-Yves Gaillard Sep 11 '11 at 6:24
This should be true, but $\Gamma MG$ is not simply $Ind M$ as $\displaystyle Ind M=\oplus_{\o\in G/H} o*M$. It is the set of sections from $G/H$ to $G\times_{H}M$, with $G\times_{H}M\rightarrow G/H$ being the projection map. Also the classical theorem applies to characters, while in here we are dealing with module homomorphisms $\phi$ and $\theta$ such that $\phi g=g \phi$ and $\theta g=g \theta$. Thanks for your comment though. – Kerry Sep 11 '11 at 6:31
You're welcome! I should have said that this is one of the main examples of adjoint functors. It holds for representations, and a fortiori for characters. I'm sure you'll get nice answers soon. Many people on MSE know this stuff much better than I. (Suggestion: use the @ sign to notify users.) – Pierre-Yves Gaillard Sep 11 '11 at 6:55
@ Pierre-Yves Gaillard: I am very surprised category theory is lurking in this seemingly trivial problem. Obviously my knowledge in this subject is unsatisfying. Thank you for pointing this out as I am not aware of this myself. – Kerry Sep 11 '11 at 7:10
You're welcome again! It's funny to note that the statement in terms of characters is an adjunction in the original sense. - Suggestion: Spell out your assumptions. I suppose you're talking about complex representations of finite groups. [If you edit your question, please correct the typo "canoical".] – Pierre-Yves Gaillard Sep 11 '11 at 7:44
up vote 5 down vote accepted

Sections of the bundle are the same as maps $\phi:G \to M$ such that $\phi(gh) = h^{-1}\phi(g)$ for all $g \in G, h \in H$. This is one way to define the induction of $M$ from $H$ to $G$ in the context of not-necessarily finite groups.

Now the isomorphism you ask about is the adjunction $$Hom_G(W,Ind_H^G M) \cong Hom_H(W,M).$$ The map is easy to define: map $Ind_H^G M \to M$ via evaluation at $1 \in G$ (in Bott's geometric terms, look at the value of the section over $g =1$) --- this is $H$-equivariant, and induces a corresponding map of Hom spaces. To check it is an isomorphism is not much harder. (If sections/maps are understood to be smooth, then you will have to use the fact that $W$ is a smooth representation of $G$.)

Aside: you shouldn't be surprised at the appearance of the term "adjunction" here; that is what Bott's formula is, and all the basic facts from character theory of finite groups are manifestations of underlying facts about the categories of representations. Also, it is easier for many people (including me!) to prove the categorical facts first and then to deduce the character-theoretic facts directly, since there is more structure to work with.

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Dear Matt E, I hope I edited your answer correctly. (I like it very much and voted for it!) – Pierre-Yves Gaillard Sep 29 '11 at 10:36
@Matt E:Hi! I want to ask where to find appropriate to learn about categories of representations, the books I used (Humphreys, Fulton&Harris, Segal&MacDonald) do not have anything on this. Thank you for the proof, which I need sometime to digest properly. – Kerry Oct 10 '11 at 7:14
@Matt E: I get it. This feels totally trivial now. Thanks for the explanation in here. – Kerry Oct 10 '11 at 7:32

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