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It is well-known that the Pontryagin dual of a discrete locally compact abelian group must be compact, which means that $\operatorname{Hom}(\mathbb{Q}, \mathbb{R}/\mathbb{Z})$ must be compact with the compact-open topology (where $\mathbb{Q}$ is taken with the discrete topology). However, one can prove that $\operatorname{Hom}(\mathbb{Q}, \mathbb{R}/\mathbb{Z}) = \mathbb{R}^2$ (or at least I think so, the below proof could have a mistake I have overlooked). This leaves me wondering what topology does $\mathbb{R}^2$ have to make it compact.

We start with the short exact sequence $0 \to \mathbb{Z} \to \mathbb{R} \to \mathbb{R}/\mathbb{Z} \to 0$. Since $\mathbb{R}$ is divisible, $\operatorname{RHom} (\mathbb{Q}, \mathbb{R}) = \mathbb{R}$. Also, $\operatorname{RHom}(\mathbb{Q}, \mathbb{Z}) = \mathbb{R}[-1]$ (see this reference). Also, $\mathbb{R}/\mathbb{Z}$ is divisible so $\operatorname{Ext}(\mathbb{Q}, \mathbb{R}/\mathbb{Z}) = 0$. We now use the induced long exact sequence

$$ 0 \to \mathbb{R} \to \operatorname{Hom}(\mathbb{Q}, \mathbb{R}/\mathbb{Z}) \to \mathbb{R} \to 0$$

However, $\mathbb{R}$ is injective so $\operatorname{Ext}(\mathbb{R}, \mathbb{R}) = 0$ and the extension must be trivial. Hence, $\operatorname{Hom}(\mathbb{Q}, \mathbb{R}/\mathbb{Z}) \cong \mathbb{R}^2$.

Due to the indirect computation, I find it hard to see how the compact-open topology applies to this group. What topology does $\mathbb{R}^2$ have in these circumstances?

Sofía Marlasca Aparicio
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  • The fact that Hom(Q, R/Z) = R2 sounds unlikely to me. I didn't check the proof in detail but one thing you need to take into account is that you can consider Q and R either with their discrete topology or their Euclidean topology, and it coulo well be that Hom's and Ext's are different depending on which you pick. So you'll want to make sure you don't mix them up in your comparison maps. – Jeroen van der Meer Feb 28 '21 at 10:05
  • @JeroenvanderMeer I am considering group homs, which is the same thing as considering groups maps that are continuous with the discrete topology, so I don't see a mistake there. – Sofía Marlasca Aparicio Feb 28 '21 at 10:10
  • @user10354138 discrete – Sofía Marlasca Aparicio Feb 28 '21 at 10:17
  • Anyway, this question has been asked before – user10354138 Feb 28 '21 at 10:23
  • @user10354138 yes but I want to see explicitly what topology this gives to $\mathbb{R}^2$ – Sofía Marlasca Aparicio Feb 28 '21 at 10:26
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    It's hard to answer your question if you don't say what you mean by "$=R^2$", or $\simeq R^2$". This is definitely not an equality, and if you mean isomorphism, you should specify isomorphism of what. – YCor Feb 28 '21 at 15:23
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    Related: https://math.stackexchange.com/questions/1866861/is-there-a-topology-such-that-bbb-r-mathcal-t-is-a-compact-hausdorff-t – Eric Wofsey Feb 28 '21 at 16:18

1 Answers1

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$\DeclareMathOperator\Hom{Hom}\DeclareMathOperator\Ext{Ext}\DeclareMathOperator\R{\mathbf{R}}\DeclareMathOperator\Z{\mathbf{Z}}\DeclareMathOperator\Q{\mathbf{Q}}\DeclareMathOperator\wQ{\widehat{\mathbf{Q}}}$

The exact sequence obtained from $0\to\Z\to\R\to\R/\Z\to 0$ by applying $\Hom(\Q,-)$ yields: $$0=\Hom(\Q,\Z)\to\Hom(\Q,\R)\to\Hom(\Q,\R/\Z)\to \Ext^1(\Q,\Z)\to\Ext^1(\Q,\R)=0,$$ so $$0\to \R\to\wQ\to \Ext^1(\Q,\Z)\to 0.$$

I think this embedding of $\R$ has a dense image in the torsion-free compact connected group $\wQ:=\Hom(\Q,\Q/\Z)$; the quotient being divisible, torsion-free, of the right cardinal, it's indeed, as an abstract group, isomorphic to $\R$.

So, yes, as an abstract group, $\wQ$ is isomorphic to $\R^2$, which is itself isomorphic to $\R$, to $\R^{2021}$, etc. Of course this is not the most useful way to think of $\wQ$, as it's especially interesting as (compact) topological group.

YCor
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  • Thank you! This is really helpful. Mostly, I think I got hung up on the idea that it is isomorphic to $\mathbb{R}^2$ but I guess that, as you said, that is not the most useful way to think about it. I was doing a bit of research and couldn't find anything specific. Do you know where I could find more information (and an explicit description) of the solenoid? – Sofía Marlasca Aparicio Feb 28 '21 at 18:00
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    @marlasca23 for $p$-adic solenoids see http://www-users.math.umn.edu/~garrett/m/mfms/notes_2013-14/12_1_adeles.pdf from the last line of page 4 ("To warm up to this situation...") and all of page 5. Note circles $\mathbf R/N\mathbf Z$ are written there as $N\mathbf Z\backslash,\mathbf R$ since it's an analogue of a more technical situation treated there with nonabelian groups where left and right cosets are not the same thing. For a longer account of $p$-adic solenoids, see the appendix to chapter 1 of Alain Robert's book "A course in $p$-adic analysis". – KCd Feb 28 '21 at 18:32
  • For an "explicit" description, you can view $\mathbf{Q}$ as inductive limit of copies of $Z$ where the $n$th map is $m\mapsto n!m$. This yields the Pontryagin dual as projective limit of copies of $\mathbf{R}/\mathbf{Z}$ where the $n$th map is the surjective map $x\mapsto n!x$. – YCor Feb 28 '21 at 18:53