Sorry if this is a naive question-- I'm trying to learn this stuff (cross-posted from http://math.stackexchange.com/questions/89248/induced-representations-of-topological-groups)

If $G$ is a group with subgroup $H$, then we have the restriction functor $\operatorname{Res}$ from $G-\operatorname{mod}$ to $H-\operatorname{mod}$. We also have this idea of induction, a functor $\operatorname{Ind}^G_H$ from $H-\operatorname{mod}$ to $G-\operatorname{mod}$. These are adjoints, which means (I think) that $\operatorname{Hom}_G(\operatorname{Ind}^G_H(V), U) \cong \operatorname{Hom}_H(V,\operatorname{Res}(U))$ naturally, for $G$-modules $U$ and $H$-modules $V$.

For locally compact groups, there is a theory worked out by MacKey and others. Actually, I have only read Rieffel's work on the subject (as I come from a functional Analysis background). For a locally compact $G$ and closed subgroup $H$, there is a very satisfactory notion of the functor $\operatorname{Ind}^G_H$ (where we consider "Hermitian modules", i.e. unitary representations on Hilbert spaces). What I don't see is how (or even if) this relates to the restriction functor?

In the topological setting, are $\operatorname{Ind}^G_H$ and $\operatorname{Res}$ in any sense adjoints?

A slightly vague rider-- if (as I suspect) the answer is "no", can we be more precise about *why* the answer is no?

rightadjoint-- I think it's too much to hope that $\operatorname{Hom}_G(U,\operatorname{Ind}^G_H(V)) \cong \operatorname{Hom}_H(\operatorname{Res}(U),V)$ I think. So, yes, I agree! – Matthew Daws Dec 13 '11 at 20:28leftadjoint to restriction; the induced representation is the one "freely generated" by a given representation of $H$. The isomorphism in your comment would be saying that induction isrightadjoint to restriction. – Yemon Choi Dec 13 '11 at 23:52