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Let $G$ is a compact group. We can form the Pontriagin dual $\widehat{G}$ of $G$: it is then discrete space. One can consider the Bohr compactification $b\widehat{G}$ of $\widehat{G}$ which is compact and once again we can go to the dual $\widehat{b\widehat{G}}$ which is again discrete. I heard that this coincides with $G$ equipped with the discrete topology. Can anyone give me some references where I can find the proof of this fact/or whether it is elementary, some explanation of the proof?

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This is basically immediate from the definition and Pontryagin duality. Pontryagin duality gives a contravariant involution on the category $LCAb$ of locally compact abelian groups which sends the subcategory $CAb$ of compact groups to the subcategory $Ab$ of discrete groups and conversely. The Bohr compactification $b:LCAb\to CAb$ is defined as the left adjoint to the inclusion functor $CAb\to LCAb$. It follows immediately that the functor $G\mapsto \widehat{b\widehat{G}}$ (i.e., $b$ conjugated by Pontryagin duality) is right adjoint to the inclusion functor $Ab\to LCAb$. It is easy to see that this right adjoint is the functor which takes a locally compact group to its underlying discrete group.

More explicitly, if $H$ is a discrete group, we have natural bijections $$\operatorname{Hom}(H,\widehat{b\widehat{G}})=\operatorname{Hom}(b\widehat{G},\widehat{H})=\operatorname{Hom}(\widehat{G},\widehat{H})=\operatorname{Hom}(H,G)=\operatorname{Hom}(H,G_d),$$ where $G_d$ is $G$ with the discrete topology. By the Yoneda lemma, it follows that $\widehat{b\widehat{G}}\cong G_d$.

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  • $\begingroup$ Thank you! So the assumption of $G$ being compact was superflous, am I right? $\endgroup$
    – truebaran
    Jul 26, 2015 at 20:26
  • $\begingroup$ Yes, that's right. $\endgroup$ Jul 26, 2015 at 20:48

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