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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?

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  • $\begingroup$ You should write $G'\le G$ instead of $G'<G$ since you have no reason to exclude $G'=G$. $\endgroup$
    – YCor
    Nov 19, 2019 at 0:34
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    $\begingroup$ $SL_2(\mathbf{R})/L$, where $L$ is non-compact closed 1-dimensional (up to conjugation, upper unipotent or diagonal) is an example. Indeed it's homeomorphic to a plane minus a point. The only candidates for being transitive are of codimension 1, hence conjugates of the upper triangular group; since stabilizers are Zariski-closed and 0-dimensional, they have to be finite, so it doesn't fit with the topology. $\endgroup$
    – YCor
    Nov 19, 2019 at 0:39
  • $\begingroup$ Corrected, and thanks for the answer! $\endgroup$
    – Lezkus
    Nov 19, 2019 at 1:48

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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.

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    $\begingroup$ How does one know that there is no codimension-1 $G'$ in $\operatorname{SL}_3(\mathbb R)$? $\endgroup$
    – LSpice
    Nov 18, 2019 at 23:38
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    $\begingroup$ @LSpice in many ways. It can be by checking algebraically in the Lie algebra. Or showing that $SL_3(\mathbf{R})$ has no faithful transitive action on $\mathbf{R}$ or the circle. Both are not hard. $\endgroup$
    – YCor
    Nov 19, 2019 at 0:31
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It immediately follows from the long exact homotopy sequence of the bundle $H\to G\to G/H$ that $\pi_1$ of any Riemannian homogeneous space $M$ is virtually abelian. So if you have a homogeneous space $G'/H'$ with non virtually abelian fundamental group then it can not possibly admit a Riemannian homogeneous metric even if you don't assume any relationship between $Iso(M)$ and $G'$. There are many such manifolds, for example all compact $N/\Gamma$ where $N$ is simply connected nilpotent non abelian and $\Gamma$ is a cocompact lattice in $N$.

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There are many counter examples in the case when $G$ are simply connected solvable Lie groups (so $G/H$ are solvmanifolds).

Some series of "concrete" examples:

If $\dim H =1$ then $H$ is diffeomorphic to $\bf R$ (so noncompact). But in general there are no $G^\prime \subset G$ which is complementary to such $H$.

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