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Given any manifold $ M $ does there exist $ G $ a Lie group and $ H,\Gamma $ closed subgroups of $ G $ such that $$ M \cong \Gamma \backslash G/H $$

I was inspired to ask by this question: Example of a manifold which is not a homogeneous space of any Lie group

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    $\begingroup$ The question as currently stated ($\Gamma$ is not assumed to lie in $G$) doesn't match the title. $\endgroup$
    – YCor
    Commented Nov 26, 2019 at 5:51
  • $\begingroup$ @Ycor Ok you are right. I edited the question so it aligns with the title. I expect the answer now to be "no". The question I'm more interested in is if $ \Gamma $ is not necessarily a subgroup. In particular I'm curious if there is a general enough structure such that every manifold "come from a lie group/ comes from a homogeneous space" using only algebraic data (e.g. quotienting by group actions). Although I should probably just ask that in a different question with a different title. $\endgroup$
    – Ian Gershon Teixeira
    Commented Nov 27, 2019 at 2:56
  • $\begingroup$ Similarly I'd be interested in a large class of manifolds all of which come from homogeneous spaces by quotienting out by a group action. I'd be interested in results of the sort "manifolds with X geometric structure always arise as the quotient by a free and proper action of their fundamental group on their universal cover and their universal cover is always homogeneous." Again I should probably make another question. $\endgroup$
    – Ian Gershon Teixeira
    Commented Nov 27, 2019 at 3:00
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    $\begingroup$ The fact that all surfaces arise this way should be noted. $\endgroup$
    – Will Sawin
    Commented Nov 27, 2019 at 3:39
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    $\begingroup$ it is true for all 2-manifolds, (I think this requires assuming that your definition of manifold requires second countability) in the sense that noncompact 2-manifolds with nontrivial fundamental group can be shown to have the hyperbolic plane as their universal cover. $\endgroup$
    – Robbie Lyman
    Commented Nov 27, 2019 at 15:58

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