15

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?

Community
  • 1
Marc Palm
  • 4,641

2 Answers2

22

The dual of the additive group $\mathbf Q$ with the discrete topology is $A_\mathbf Q/\mathbf Q$, where $A_\mathbf Q$ is the adele ring of $\mathbf Q$ (viewed as an additive group). See a proof here.

The standard topology on $A_\mathbf Q$ makes it locally compact and $\mathbf Q$ (embedded diagonally) is a discrete subgroup for which the quotient topology on $A_\mathbf Q/\mathbf Q$ is compact, as it needs to be if it's going to be the Pontryagin dual of a discrete abelian group.

KCd
  • 46,062
  • 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? – Marc Palm Jun 28 '11 at 10:19
  • 2
    If you add a http:// in front of a link, it becomes clickable. – t.b. Jun 28 '11 at 10:19
  • 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. – Marc Palm Jun 28 '11 at 10:25
  • 1
    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 ? – Joel Cohen Jun 28 '11 at 10:44
  • @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 that – Marc Palm Jun 28 '11 at 10:57
  • See also the question http://math.stackexchange.com/questions/10603/representation-theory-of-the-additive-group-of-the-rationals – KCd Jun 29 '11 at 00:37
  • 2
    @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, – Matt E Jun 29 '11 at 21:08
  • @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. – Joel Cohen Jun 29 '11 at 21:41
  • @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? – Mike Battaglia Nov 10 '15 at 23:58
16

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).

GEdgar
  • 111,679