Sorry if this is a naive question-- I'm trying to learn this stuff (cross-posted from <https://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?



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> In the topological setting, are $\operatorname{Ind}^G\_H$ and $\operatorname{Res}$ in any sense adjoints?
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A slightly vague rider-- if (as I suspect) the answer is "no", can we be more precise about *why* the answer is no?



Induced representations of topological groups
If $G$ is compact this is the Frobenius Reciprocity Theorem, see e.g., Section 6.2 in Folland's A Course in Abstract Harmonic Analysis for a proof. When $G$ is not compact then this fails already for $H$ the trivial group and $U$, and $V$ trivial representations. Indeed, in this case Ind$\_H^G(1\_H) = L^2G$ the left regular representation, and since constant functions are not square integrable we have Hom$\_G( L^2G, 1\_G ) = \{ 0 \}$, while Hom$\_H(1\_H, 1\_H) \not= \{ 0 \}$.


