# On Frobenius reciprocity theorem

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

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 $G$.)