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I already know that this is impossible for $(\Bbb Q, +, \mathcal T)$ to be a compact Hausdorff topological group (notice that the trivial topology does not work because it is not Hausdorff).

Indeed, this follows from Baire theorem: for if $(\Bbb Q, +, \mathcal T)$ were a compact Hausdorff topological group, then $\Bbb Q$ would be the union of the countable collection of closed sets $\{r\}$ (with $r \in \Bbb Q$). As we have a locally compact Hausdorff space, Baire theorem tells us that at least one of the closed set has non empty interior, i.e. some $\{r\}$ is open. Since we have a topological group, it follows that $(\Bbb Q, +, \mathcal T)$ is discrete and hence non compact.


But what about $(\Bbb R, +, \mathcal T)$ ? My first idea was to use the group (and even vector spaces) isomorphism $\Bbb R \cong \Bbb Q^{(\Bbb N)}$. Transporting the topology $\mathcal T$ on $\Bbb Q^{(\Bbb N)}$ preserves compactness, and we could try to use the projection $\Bbb Q^{(\Bbb N)} \to \Bbb Q$. But I was not sure what to do then.

Any comment would be appreciated!

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A good way to think about this is in terms of Pontryagin duality. Since we only care about the abelian group structure of $\mathbb{R}$, let's first get a nice characterization of this structure. As an abelian group, $\mathbb{R}$ is the unique $\mathbb{Q}$-vector space of its cardinality (up to isomorphism). An abelian group $A$ is a $\mathbb{Q}$-vector space iff for each nonzero $n\in \mathbb{Z}$, the multiplication by $n$ map $n:A\to A$ is an isomorphism.

Now the neat thing about this is that this condition is self-dual under Pontryagin duality. If $A$ is a locally compact abelian group, then the dual of the map $n:A\to A$ is just the map $n:\hat{A}\to\hat{A}$ on the dual group. So this says that a locally compact abelian group is a $\mathbb{Q}$-vector space iff its dual is a $\mathbb{Q}$-vector space.

In particular, let us use this to classify the compact abelian groups which are isomorphic (as groups) to $\mathbb{R}$. These are just the Pontryagin duals $\hat{V}$ of all $\mathbb{Q}$-vector spaces $V$ (with the discrete topology) for which $\hat{V}$ has cardinality $2^{\aleph_0}$. It is not hard to show that $\hat{\mathbb{Q}}$ has cardinality $2^{\aleph_0}$. If $V$ is a $\kappa$-dimensional $\mathbb{Q}$-vector space then $\hat{V}$ is a product of $\kappa$ copies of $\hat{\mathbb{Q}}$, which has cardinality $2^{\aleph_0\cdot\kappa}$.

So to sum up, there are indeed compact group topologies on $\mathbb{R}$. Up to continuous isomorphism, there is one such topology for each cardinal $\kappa$ such that $2^{\aleph_0\cdot\kappa}=2^{\aleph_0}$ (in particular, this includes all $\kappa$ such that $0<\kappa\leq\aleph_0$). The Pontryagin dual of this compact group is a $\mathbb{Q}$-vector space of dimension $\kappa$.

The case $\kappa=1$ gives $\hat{\mathbb{Q}}$, which is a solenoid. Explicitly, $\hat{\mathbb{Q}}$ is the inverse limit of the sequence $\dots\to S^1\stackrel{4}{\to}S^1\stackrel{3}{\to}S^1\stackrel{2}{\to}S^1$, since $\mathbb{Q}$ is the direct limit of the sequence $\mathbb{Z}\stackrel{2}{\to}\mathbb{Z}\stackrel{3}\to\mathbb{Z}\stackrel{4}{\to}\mathbb{Z}\to\dots$. For general $\kappa$, you just have a product of $\kappa$ copies of $\hat{\mathbb{Q}}$.

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  • $\begingroup$ This is a great answer, because I love Pontryagin duality! I will read it carefully. Thank you! $\endgroup$ – Watson Jul 21 '16 at 20:35
  • $\begingroup$ Just a little question: since $\widehat{\mathbb{Q}}$ is a compact topological group of cardinality $2^{\aleph_0}$, we can transport that structure on $\Bbb R$, but then the addition is not necessarily the usual addition on the reals. Am I right? $\endgroup$ – Watson Jul 21 '16 at 20:52
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    $\begingroup$ $\hat{\mathbb{Q}}$ is not just a compact topological group of cardinality $2^{\aleph_0}$ but also a $\mathbb{Q}$-vector space under its group structure, so it is isomorphic as a group to $\mathbb{R}$. $\endgroup$ – Eric Wofsey Jul 21 '16 at 21:09
  • $\begingroup$ Is there a way to construct such a topology on $\mathbb{R}$, though? $\endgroup$ – Cronus Aug 19 '16 at 1:28
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    $\begingroup$ @Cronus: Probably not without the axiom of choice. With the axiom of choice, it's "easy": just pick a basis for $\mathbb{R}$ over $\mathbb{Q}$ and a basis for $\hat{\mathbb{Q}}$ over $\mathbb{Q}$ and a bijection between them, and transport the topology along the induced bijection $\mathbb{R}\to\hat{\mathbb{Q}}$. $\endgroup$ – Eric Wofsey Aug 19 '16 at 1:33

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