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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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  • $\begingroup$ So the point must be to not assume local compactness; is there an interesting example? $\endgroup$
    – BCnrd
    Nov 11, 2010 at 15:49
  • $\begingroup$ 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? $\endgroup$
    – Gerald Edgar
    Nov 11, 2010 at 17:35
  • $\begingroup$ Gerry: Oops, sorry about that. Yes, the Hausdorffness would have to be checked as part of the example since it's not generally automatic. $\endgroup$
    – KConrad
    Nov 11, 2010 at 18:51
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    $\begingroup$ 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. $\endgroup$
    – KConrad
    Nov 11, 2010 at 18:58
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    $\begingroup$ @KConrad: The usual example of such a space (James space) was given in the linked question. Unfortunately this does not furnish an example for your question because the Pontryagin duality homomorphism is an isomorphism for all Banach spaces by a result of M. F. Smith: "The Pontrjagin duality theorem in linear spaces". Annals of Mathematics 56 (2): 248–253 $\endgroup$
    – Robert Furber
    May 8, 2016 at 23:03

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If I understand it well then Proposition A4.21 (along with Exercise EA4.10 containing hints for a proof) from the Hofmann-Morris book (Structure of compact groups) is an example.

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    $\begingroup$ Just to help the random reader: the indicated reference (page 835 of the 3rd edition) considers the finest vector space topology $\mathscr O'$ on a real vector space $E$ and the finest locally convex vector space topology $\mathscr O$. One has $\mathscr O\subset\mathscr O'$, and $(E,\mathscr O')''=(E,\mathscr O)''=(E,\mathscr O)$. However, if $\dim(E)$ is uncountable, then $\mathscr O'\neq\mathscr O$. $\endgroup$
    – ACL
    Oct 9, 2016 at 21:46
  • $\begingroup$ @ACL - I don't think that this is an answer to the question? $(E,\mathscr{O})$ is reflexive, and $(E,\mathscr{O}')$ is not isomorphic as a topological group to its bidual. Or am I missing the point? $\endgroup$
    – benblumsmith
    Apr 8, 2017 at 14:33

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