Let $M = G/H$ be a homogeneous space, where $G$ is a Lie group and $H$ is a closed Lie subgroup. Can it happen that $M$ does not admit an invariant Riemannian metric?
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$\begingroup$ Could you define "homogeneous" and "equivariant"? $\endgroup$– YCorCommented Apr 8 at 17:26

7$\begingroup$ If the stabilizer of a point is noncompact, there is no invariant metric because isometries act properly. For example, the group of conformal automorphisms of the round sphere acts transitively on the sphere but its point stabilizers are noncompact. $\endgroup$– Igor BelegradekCommented Apr 8 at 17:28

3$\begingroup$ There is no Riemannian metric on the circle invariant under all selfdiffeomorphisms of the circle. Not even of we restrict ourselves to the (finitedimensional, but non compact) subgroup $PGL_2$ (acting on the circle as the projective line). $\endgroup$– GroTsenCommented Apr 8 at 17:35

2$\begingroup$ Perhaps the OP meant to ask whether there is a homogeneous manifold that does not admit a homogeneous Riemannian metric. A simple example is $TS^2$, the tangent bundle of the $2$sphere. It can be written as a homogeneous space $G/H$ where $G$ has dimension $6$ and $H$ is a closed (but not compact) subgroup of dimension $2$, but there does not exist a Riemannian metric on $TS^2$ whose isometry group acts transitively on $TS^2$. $\endgroup$– Robert BryantCommented Apr 13 at 18:50

$\begingroup$ It is possible to have an invariant Riemannian metric, even with noncompact stabilizer, but then the group $G$ is only a subgroup of the isometry group acting transitively on $M$. As Igor Belegradek points out, the isometry group of any homogeneous Riemannian manifold acts properly, but some transitive subgroups don't for some homogeneous Riemannian metrics. $\endgroup$– Ben McKayCommented Apr 24 at 20:32
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