If $G$ is a locally compact group, we can define its dual group $\hat G$. That is set of continuous homomorphism from $G$ to circle group $\mathbb T$. My question is how to define dual group $\hat G$ when $G$ is a noncommutative group?
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4There will be no dual group, which is a nicely structured collection of one-dimensional representations, since there will be representations that are more-dimensional. See the "noncommutative theory" section on Wikipedia's Pontryagin duality article. (Also relevant.) – anon Apr 01 '13 at 19:00
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3One meaning of $\hat{G}$ in the non-commutative case is the unitary dual, i.e., the set of unitary equivalence classes of irreducible representations of $G$ endowed with the Fell topology. It is the spectrum of the group $C^\ast$-algebra $C^\ast(G)$. – Martin Apr 01 '13 at 19:59
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As mentioned in the comments, the "correct" generalization of the dual group to the noncommutative case is the unitary representation theory of $G$ (studying this reduces to studying homomorphisms into the circle group, or $\text{U}(1)$, when $G$ is commutative). At least when $G$ is compact, it is possible to recover $G$ from its unitary representation theory using some version of Tannaka reconstruction, e.g. the Doplicher-Roberts theorem; this is the "correct" generalization of Pontrjagin duality to this case.
There are other possible generalizations; see this MathOverflow question for some.
Qiaochu Yuan
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It helps a lot. But I still confused what is the dual group, do you mean the dual group is set of unitary representation? Can you provide more information? – owen Apr 03 '13 at 05:30
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1@owen: the dual group isn't a group. You can think of the correct "dual thing" as being the "moduli space" of unitary representations, but I prefer to work with the category of unitary representations (at least when $G$ is compact). – Qiaochu Yuan Apr 03 '13 at 05:54