A manifold $M$ together with a transitive $G$-action is always diffeomorphic a quotient $G/H$ for $H < G$ Lie groups. On the other hand, there might be a proper subgroup of $G$ that also acts transitively on $M$, so this representation may not be unique.


If $H$ is compact, we may choose a metric on $G$ that descends into $G/H$ and makes it into a Riemannian homogeneous space, that is, we may choose a metric on $M \cong G/H$ such that the action is by isometries.


Is it always possible to choose a representation of a homogeneous space as a quotient of Lie groups $G/H$ such that $H$ is a compact Lie group? Or in other words, given a manifold $M$ with a transitive $G$-action, is there always a $G'$-action with $G' \leq G$ such that the isotropy group is compact?


Riemannian homogeneous spaces are in particular reductive, so this would imply that every homogeneous space has a representation as a reductive quotient. I have read that this is a non-trivial condition for homogeneous spaces, so this makes me think that this may not be true, but I have never seen an explicit counterexample either.


On matrix Lie groups one has that every connected Lie algebra closed under transposes is reductive, so I am assuming that the counterexamples may not be very nice-looking?



Is every homogeneous space Riemannian homogeneous?
You can take $M=\operatorname{SL}\_3 (\mathbb R)/ \mathbb R$, where we can choose any non-compact one-parameter subgroup of $\operatorname{SL}\_3$. Because the stabilizer is noncompact, $G'$ must be proper, but to act transitively, $G'$ must have codimension at most one in $\operatorname{SL}\_3(\mathbb R)$. But no codimension one subgroup of $\operatorname{SL}\_3(\mathbb R)$ exists.


