The dual group of $\mathbb Q$ - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T14:39:39Z http://mathoverflow.net/feeds/question/34251 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/34251/the-dual-group-of-mathbb-q The dual group of $\mathbb Q$ Hany 2010-08-02T12:13:34Z 2010-08-20T09:49:13Z <p>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? </p> http://mathoverflow.net/questions/34251/the-dual-group-of-mathbb-q/34254#34254 Answer by Torsten Ekedahl for The dual group of $\mathbb Q$ Torsten Ekedahl 2010-08-02T12:25:21Z 2010-08-03T15:24:05Z <p>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.)</p> <p><b>Addendum</b>: Erased previous addendum as it was all wrong.</p> http://mathoverflow.net/questions/34251/the-dual-group-of-mathbb-q/35105#35105 Answer by BS for The dual group of $\mathbb Q$ BS 2010-08-10T12:22:24Z 2010-08-20T09:49:13Z <p>In fact, uniform convergence on compact subsets of $\mathbb{Q}\subset\mathbb{R}$ induces the usual topology on its group of (continuous) characters <code>$\mathbb{R}\simeq\{t\mapsto\exp(ixt)\}_{x\in\mathbb{R}}$</code>.</p> <p>Namely, consider <code>$K=\{0\}\cup\{1/n,n\geq 1\}$</code>. For $x\in\mathbb{R}$, the corresponding character is uniformly $\epsilon$-close on $K$ to the trivial character iff <code>$$|exp(ix/n)−1|&lt;\epsilon\;\;\;\; (*)$$</code> for all integers $n\geq1$. Then for $\epsilon&lt;1/\sqrt{2}$, $x$ must be small : $|x|&lt;2\epsilon/\pi$. 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. </p> <p>This implies that uniform convergence on compact subsets of $\mathbb{Q}$ (in fact the one compact subset $K$) induces the usual topology on $\mathbb{R}\simeq\mathrm{Hom}(\mathbb{Q},S^1)$.</p>