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What is the dual group of the additive group of rational numbers equipped with the standard topology inherited from $\mathbb R$? As a group, this dual group is isomorphic to $\mathbb R$ (see the answer of Ekedahl given below), but it should be equipped with the topology of uniform convergence on compact subsets of $\mathbb Q$. What are the properties of this group? Is it locally compact? what are its connected components? does it have more natural structure?

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Something very interesting is the compact group obtained as the dual of $\mathbb{Q}$ with the discrete topology! – Gerald Edgar Aug 2 '10 at 12:33
Note that there is also a $p$-adic analogue of this question, for which the answer is also the analogous thing (i.e., replace $\mathbb{R}$ by $\mathbb{Q}_p$ in every instance). – Pete L. Clark Aug 3 '10 at 10:56
up vote 0 down vote accepted

In fact, uniform convergence on compact subsets of $\mathbb{Q}\subset\mathbb{R}$ induces the usual topology on its group of (continuous) characters $\mathbb{R}\simeq\{t\mapsto\exp(ixt)\}_{x\in\mathbb{R}}$.

Namely, consider $K=\{0\}\cup\{1/n,n\geq 1\}$. For $x\in\mathbb{R}$, the corresponding character is uniformly $\epsilon$-close on $K$ to the trivial character iff $$|exp(ix/n)−1|<\epsilon\;\;\;\; (*)$$ for all integers $n\geq1$. Then for $\epsilon<1/\sqrt{2}$, $x$ must be small : $|x|<2\epsilon/\pi$. Indeed, consider $k\in\mathbb{Z}$ such that $|x−k\pi|\leq\pi/2$ , and take $n=|k|$; if $k\neq 0$ we reach a contradiction in $(*)$. Hence $k=0$, and the claim follows easily.

This implies that uniform convergence on compact subsets of $\mathbb{Q}$ (in fact the one compact subset $K$) induces the usual topology on $\mathbb{R}\simeq\mathrm{Hom}(\mathbb{Q},S^1)$.

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Every continuous group homomorphism $\mathbb Q \rightarrow S^1$ extends to the completion of $\mathbb Q$ (cf., Bourbaki: General topology, Prop. III:4.8) which is $\mathbb R$ so the dual group of $\mathbb Q$ is the same as that of $\mathbb R$ which is $\mathbb R$. (There may be some question as to whether the topologies are the same but I am not even sure which topology to use for the dual group when the group is not locally compact.)

Addendum: Erased previous addendum as it was all wrong.

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A continuous group homomorphism is of course uniformly continuous, thus extends to the completions. – Gerald Edgar Aug 2 '10 at 12:32
Which is the argument of my reference (chosen because it is where uniformities are introduced)... – Torsten Ekedahl Aug 2 '10 at 12:58
Compact subsets of $\mathbb{Q}$ are finite. So uniform convergence on compact subsets of $\mathbb{Q}$ ist the same as convergence in all rational points. – Johannes Hahn Aug 3 '10 at 12:57
@Johannes : $\mathbb{Q}$ has infinite compact subsets, like a sequence converging to a rational and its limit. – BS. Aug 3 '10 at 14:24
@Johannes: Is that already an answer in the "common misconceptions" question? If not it should be. – Gerald Edgar Aug 3 '10 at 17:13

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