Let $G = \{(x; y) : x \in \mathbb{R}, y > 0\}$. With $(x, y)(u, v) = (x + yu, yv)$, $G$ is a group. If we topologize $G$ as a subset of $\mathbb{R}^2$, it is known that $G$ is a locally compact group that is not unimodular (see (15.17) of Hewitt-Ross). Is there another topological structure of the group $G$ such that $G$ is a locally compact group and the subgroup $K = \{(0, y) : y > 0\}$ is compact?
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3$\begingroup$ Have you noticed that $K$ is isomorphic (as a group) to the additive group of the real line? Can you put a compact group topology on $\mathbb{R}$? $\endgroup$– Alain ValetteCommented Jul 8, 2015 at 17:14
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2$\begingroup$ It is possible to put a compact group topology on $\mathbb{R}$ (Halmos, "Comment on the real line", 1944). But any such topology is weird, in the sense that it cannot look very much like the usual topology on $\mathbb{R}$. For example, $[0,1)$ is not Haar measurable in any compact group topology on $\mathbb{R}$ (if it were, then you get a contradiction by the same argument that shows Vitali sets are non-measurable). $\endgroup$– Will BrianCommented Jul 8, 2015 at 18:07
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5$\begingroup$ Indeed the Pontryagin dual of the discrete group $\mathbf{Q}$ is a compact group whose underlying discrete group is isomorphic to the underlying discrete group of $\mathbf{R}$. It's not clear if this can be extended to the semidirect structure, but the question is reasonable and even if one can argue about its interest, it's not ambiguous at all and the "put on hold as unclear" is not justified. $\endgroup$– YCorCommented Jul 8, 2015 at 19:48
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$\begingroup$ Here $\hat{\mathbf{Q}}$ cannot work because the its automorphism group is not transitive on nonzero elements (because it's isomorphic to the automorphism group of $\mathbf{Q}$, namely $\mathbf{Q}^*$, which is countable hence can't be transitive on an uncountable set). But I'm not sure about $\hat{\mathbf{Q}}^{\mathbf{N}}$, which is abstractly isomorphic to $\mathbf{R}$ and has an uncountable automorphism group. $\endgroup$– YCorCommented Jul 8, 2015 at 19:54
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2 Answers
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[I realize I had misread the question, as I understood that the real subgroup required to be compact is the normal one. Since asking the question with the normal subgroup ($X$ in Dave's post) required to be compact seems much less trivial than the original question (answered by Dave), I'll include the full answer to the modified question.]
To avoid ambiguity, let me denote by $R$ the underlying additive group structure on the reals, without topology. So it's just a $\mathbf{Q}$-vector space of continuum dimension, so a group is isomorphic to $R$ iff it's abelian, torsion-free, and divisible, of continuum cardinal.
As mentioned in the comments, there exist compact group topologies on $R$.
However, I claim that if $T$ is a compact topology on $R$, then the group of automorphisms of the topological group $(R,T)$ is not transitive on $R-\{0\}$. This implies that the underlying group of $\mathbf{R}\rtimes\mathbf{R}$ cannot be endowed with a group topology making the normal $\mathbf{R}$ compact.
To prove the claim, write $K=(R,T)$. This is a compact abelian group, with no continuous homomorphism onto any nontrivial finite group. Hence its Pontryagin dual is a torsion-free abelian group $A$. If $I$ is a maximal free family in $A$, then the embedding $\mathbf{Z}^{(I)}\subset A$ induces a (continuous) surjection $K\to (\mathbf{R}/\mathbf{Z})^I$. Since $K$ has continuum cardinal, this implies that $I$ has cardinal less than continuum. Hence, noting that $A$ has the same cardinal as $I$, the cardinal of $A$ is less than continuum.
On the other hand, for each $x\in\mathbf{R}/\mathbf{Z}$, $x$ belongs to the image of some homomorphism $A\to \mathbf{R}/\mathbf{Z}$: this comes from basic Pontryagin duality: just find an infinite cyclic subgroup $\langle m\rangle$ in $A$, consider the homomorphism $m\mapsto x$, and extend it to $A$. As a consequence (using that $A$ has cardinal less than continuum), the number of possible images of homomorphisms $A\to\mathbf{R}/\mathbf{Z}$ is continuum. Now the group $\mathrm{Aut}(K)$ acts on this set of homomorphisms, and this action does not change the image, so the action of $\mathrm{Aut}(K)$ on $\mathrm{Hom}(A,\mathbf{R}/\mathbf{Z})$ has continuum many orbits. By Pontryagin duality, the action of $\mathrm{Aut}(K)$ on $\mathrm{Hom}(A,\mathbf{R}/\mathbf{Z})\simeq K$ has continuum many orbits, in particular it is not transitive on $K\smallsetminus\{0\}$.
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If I understand the question, then the answer is no. Let $X = \{(x,1) : x \in \mathbb{R}\}$. Then the conjugation action of $K$ has only three orbits on $X$. All three orbits must be closed (since $K$ is compact). So the orbits are also open (in $X$), since there are only finitely many of them. One of these orbits is the singleton $\{(0,1)\}$, so the topology on $X$ must be the discrete topology. This contradicts the fact that the other two (infinite) orbits are compact.
answered Jul 9, 2015 at 11:46