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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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  • $\begingroup$ late_learner: I had mentioned the link in your 2nd paragraph as a comment to my answer below. $\endgroup$ – KCd Jul 2 '11 at 3:15
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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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  • $\begingroup$ 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? $\endgroup$ – Marc Palm Jun 28 '11 at 10:19
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    $\begingroup$ If you add a http:// in front of a link, it becomes clickable. $\endgroup$ – t.b. Jun 28 '11 at 10:19
  • $\begingroup$ 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. $\endgroup$ – Marc Palm Jun 28 '11 at 10:25
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    $\begingroup$ 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 ? $\endgroup$ – Joel Cohen Jun 28 '11 at 10:44
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    $\begingroup$ @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, $\endgroup$ – Matt E Jun 29 '11 at 21:08
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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).

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