# Is every homogeneous space Riemannian homogeneous?

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?

• You should write $G'\le G$ instead of $G'<G$ since you have no reason to exclude $G'=G$.
– YCor
Nov 19, 2019 at 0:34
• $SL_2(\mathbf{R})/L$, 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.
– YCor
Nov 19, 2019 at 0:39
• Corrected, and thanks for the answer! Nov 19, 2019 at 1:48

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.

• How does one know that there is no codimension-1 $G'$ in $\operatorname{SL}_3(\mathbb R)$? Nov 18, 2019 at 23:38
• @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.
– YCor
Nov 19, 2019 at 0:31

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 $$N$$.

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 $$H$$.