most recent 30 from http://mathoverflow.net 2013-05-23T22:44:10Z http://mathoverflow.net/feeds/question/45714 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/45714/are-there-non-reflexive-abelian-topological-groups-isomorphic-to-their-second-dua KConrad 2010-11-11T15:40:04Z 2010-11-11T19:02:40Z

<p>I posted the following question in a comment at <a href="http://mathoverflow.net/questions/43986/are-there-non-reflexive-vector-spaces-isomorphic-to-their-bi-dual" rel="nofollow">http://mathoverflow.net/questions/43986/are-there-non-reflexive-vector-spaces-isomorphic-to-their-bi-dual</a> and it got one upvote, but it didn't get an answer, so I'll post it as an independent question.</p> <p>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).</p> <p>[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.]</p>