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Endow the rational numbers (or any global field) with the discrete topology, what will be the (compact) Pontryagin dual of the additive group and of the multiplicative group?

I am suprised nobody mentioned this: but the part of the question of the additive group of the rational is answered here already: Representation theory of the additive group of the rationals?

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    late_learner: I had mentioned the link in your 2nd paragraph as a comment to my answer below.2011-07-02

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The dual of the additive group is A_Q/Q. See http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/characterQ.pdf

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    Nice. Is it too much to hope for that the answer for the multiplicative group is $A^\times /Q^\times$ is the answer for the multiplicative group?2011-06-28
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    If you add a `http://` in front of a link, it becomes clickable.2011-06-28
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    Okay, I see you use that $Q$ is a lattice in $A$. So the same argument works for the finite adeles $A_f^\times / Q^{times}$ for the multiplicative group, nice.2011-06-28
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    Nice answser, so it seems I was a bit too quick with mine (partly ashamed, partly amused). Now I wonder if there's a way to fix my earlier reasoning : To choose a character, you need to chose the image of $1$ in $\mathbb{R}/\mathbb{Z}$ (that's the part I forgot earlier), then for all $n \in \mathbb{N}^*$ choose an $n^{\textrm{th}}$ root of it (the image of $\frac{1}{n}$) in a compatible way yielding an element of $\hat{\mathbb{Z}}$. In the end you'd get $\hat{\mathbb{Z}} \times \mathbb{R}/\mathbb{Z}$, right ?2011-06-28
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    @Joel, yes I think strong approximation shows you what you were missing in your first guess. When commenting your previous answer, I was confusing for a second the Pontryagin dual with characters defined over $\mathbb{Q}$, sorry about that2011-06-28
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    See also the question http://math.stackexchange.com/questions/10603/representation-theory-of-the-additive-group-of-the-rationals2011-06-29
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    @Joel: Dear Joel, $\mathbb A/\mathbb Q$ is not the same as $\hat{\mathbb Z}\times \mathbb R/\mathbb Z$; e.g. the former is a $\mathbb Q$-vector space, while the latter contains torsion elements. There is a surjection from $\mathbb A/\mathbb Q$ to $\mathbb R/\mathbb Z$, with kernel equal to $\hat{\mathbb Z}$, but this surjection does not split. Regards,2011-06-29
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    @Matt : Ah yes, now I understand why it doesn't work (the bijection I had in mind fails to be a group morphism). Thank you very for clearing my misunderstanding.2011-06-29
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    @plusepsilon.de - the Pontryagin dual of $\Bbb Q^\times$ is a countable product of circle groups. So are you saying that $\Bbb A_f^\times / \Bbb Q^\times$ is a countable product of circle groups? If so, how?2015-11-10
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For those who don't know what is $A_Q$...

Hewitt and Ross, Abstract Harmonic Analysis, p. 404. The dual of the discrete rationals is described as an $\mathbf{a}$-adic solenoid. An inverse limit of a sequence of circles, $T_n$, say, where the map of $T_{n+1}$ onto $T_n$ wraps around $n$ times.

Their notes say this is due to Makoto Abe (1940) and independently to Anzai and Kakutani (1943).