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I posted the following question in a comment at Are there non-reflexive vector spaces isomorphic to their bi-dual? and it got one upvote, but it didn't get an answer, so I'll post it as an independent question.

Is there an example of an abelian Hausdorff topological group G such that G and its second dual G^^ are isomorphic as topological groups but the natural map G ---> G^^ is not a topological group isomorphism? The dual group of an abelian Hausdorff topological group is given the compact-open topology, which makes the dual group an abelian topological group, although [edit] a priori it is not clear that G^ separates points in G, so the Hausdorfness of G^^ is part of the conditions that would need to be checked in an example (rather than being automatic).

[edit: Since G^ need not be Hausdorff, maybe I'm even willing to drop that condition. If G is an abelian top. group then G^ with the compact-open topology is an abelian top. group and G^^ is as well. Is there such G isomorphic to G^^ but not by the natural map? If a non-Hausdorff example turns out to be silly then maybe I'll stick the Hausdorff condition back in.]

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So the point must be to not assume local compactness; is there an interesting example? – BCnrd Nov 11 '10 at 15:49
You say "natural map", but in general we cannot expect $\widehat{G}$ separates points of $G$, right? So is there a convenient reference that discusses the setting you want here? – Gerald Edgar Nov 11 '10 at 17:35
Gerry: Oops, sorry about that. Yes, the Hausdorffness would have to be checked as part of the example since it's not generally automatic. – KConrad Nov 11 '10 at 18:51
BCnrd: I assume by "interesting example" you mean is there an example where Pontryagin duality works for a non-locally compact group. In S. Kaplan "Extensions of the Pontryagin Duality I: Infinite Products" Duke 15 (1948), 649--658, Theorem 4 says that any product of abelian topological groups which each satisfy Pontryagin duality also satisfies Pontryagin duality (i.e., the natural map G ---> G^^ is a top. group isomorphism). So the product of R and each Q_p (the "fake adeles") is an example. Can one say this example is interesting? Not for anything else, but that it works is curious. – KConrad Nov 11 '10 at 18:58
I cannot locate the reference at the moment, but I seem to recall that there do exist Banach spaces that are not reflexive (so are not naturally isomorphic to their second duals), but are isomorphic to their second duals. – paul garrett Jul 24 '11 at 20:38

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