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I was recently trying to think of a simple example that demonstrates that the natural inclusion of an abelian group $G$ into $$\widehat{\widehat{G}}=\text{Hom}_{\mathsf{Ab}}(\text{Hom}_{\mathsf{Ab}}(G,\mathbb{C}^\times),\mathbb{C}^\times)$$ is not necessarily an isomorphism. Note that I'm looking at all homomorphisms; no topology on the group is involved (or, if you prefer, they are all discrete).

Obviously, $G$ has to be infinite. However, I was having a bit of trouble finding an example satisfactorily simple - in fact, the only one I could prove worked was $G=\mathbb{Z}$, in which case $\widehat{G}=\text{Hom}_{\mathsf{Ab}}(\mathbb{Z},\mathbb{C}^\times)\cong\mathbb{C}^\times$, so that $$\widehat{\widehat{G}}\cong\text{Hom}_{\mathsf{Ab}}(\mathbb{C}^\times,\mathbb{C}^\times)$$ which is uncountable due to the existence of uncountably many automorphisms of the field $\mathbb{C}$. However, that requires the axiom of choice, which seems like it ought not to be necessary. I'm sure I'm missing an obvious one - could someone provide a $G$ which doesn't require the axiom of choice to prove the map isn't surjective?

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The axiom of choice is butting in because you're thinking of C* as a discrete group, which isn't natural; you should be working in the topological category and looking for G which isn't locally compact. –  Qiaochu Yuan Feb 27 '11 at 22:09
    
restricting to continuous homomorphisms gives $\mathbb{Z}$ double dual equal to $\mathbb{Z}$ –  yoyo Feb 27 '11 at 22:13
    
Yes, I'm aware that one usually looks at continuous homomorphisms (though my actual knowledge about topological groups is sorely lacking). But I think it's as natural a question as asking for an explicit example of a vector space that isn't isomorphic to its double dual. –  Zev Chonoles Feb 27 '11 at 22:27
    
Well, not quite; in the category of vector spaces the only reasonable object you can try to get a theory of duality out of is the one-dimensional vector space. In the category of discrete groups there are other choices more natural than C*; for example there was a recent MO (math.SE?) question about using Q/Z instead. –  Qiaochu Yuan Feb 27 '11 at 22:55

1 Answer 1

up vote 1 down vote accepted

The question was answered on MO: http://mathoverflow.net/questions/56955.

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