3
$\begingroup$

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

$\endgroup$
13
  • 6
    $\begingroup$ The question as currently stated ($\Gamma$ is not assumed to lie in $G$) doesn't match the title. $\endgroup$
    – YCor
    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. $\endgroup$ 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. $\endgroup$ Nov 27, 2019 at 3:00
  • 1
    $\begingroup$ The fact that all surfaces arise this way should be noted. $\endgroup$
    – Will Sawin
    Nov 27, 2019 at 3:39
  • 2
    $\begingroup$ 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. $\endgroup$ Nov 27, 2019 at 15:58

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.