# Is every manifold a double coset space?

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

• The question as currently stated ($\Gamma$ is not assumed to lie in $G$) doesn't match the title. – YCor Nov 26 '19 at 5:51
• @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. – Ian Teixeira Nov 27 '19 at 2:56
• 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. – Ian Teixeira Nov 27 '19 at 3:00
• The fact that all surfaces arise this way should be noted. – Will Sawin Nov 27 '19 at 3:39
• 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. – Rylee Lyman Nov 27 '19 at 15:58