Is there a topology such that $(\Bbb R, +, \mathcal T)$ is a compact Hausdorff topological group? 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!
 A: 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 $\hat{\mathbb{Q}}$.
