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?
Best Answer
$\DeclareMathOperator\Hom{Hom}\DeclareMathOperator\Ext{Ext}\DeclareMathOperator\R{\mathbf{R}}\DeclareMathOperator\Z{\mathbf{Z}}\DeclareMathOperator\Q{\mathbf{Q}}\DeclareMathOperator\wQ{\widehat{\mathbf{Q}}}$
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.