The dual group of $\mathbb Q$ - MathOverflow

The dual group of $\mathbb Q$

2010-08-02T12:13:34Z

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 by Torsten Ekedahl for The dual group of $\mathbb Q$

2010-08-02T12:25:21Z

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.

Answer by BS for The dual group of $\mathbb Q$

2010-08-10T12:22:24Z

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