4
$\begingroup$

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?

  • 0
    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.2011-02-27

1 Answers 1

1

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