Question

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?

Answer 1

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 $\mathbb{R}\simeq\mathrm{Hom}(\mathbb{Q},S^1)$.

Answer 2

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.