3

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

Ian Teixeira
  • 273
  • 6
    The question as currently stated ($\Gamma$ is not assumed to lie in $G$) doesn't match the title. – YCor Nov 26 '19 at 05:51
  • @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 02:56
  • 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 03:00
  • 1
    The fact that all surfaces arise this way should be noted. – Will Sawin Nov 27 '19 at 03:39
  • The question still doesn’t match the title. (Double coset space entails a specific way for $\Gamma$ to act on $G/H$.) – Francois Ziegler Nov 27 '19 at 04:30
  • @FrancoisZiegler I understand the current question with this specific way (although I'd have denoted $\Gamma\backslash G/H$) – YCor Nov 27 '19 at 05:21
  • @YCor Well, biquotients appear to have had several definitions, and then the OP seems to want “a bit more general”. (How exactly?) – Francois Ziegler Nov 27 '19 at 05:52
  • @Will Sawin I know all surfaces (both orientable and non orientable) arise this way, that’s part of why I asked the question. Is it true for all connected 2 manifolds (not necessarily compact)? – Ian Teixeira Nov 27 '19 at 14:21
  • @FrancoisZiegler The commit about “a bit more general” is leftover from before the first edit when the question was more general. I’ve deleted it. Now everything should match the title. – Ian Teixeira Nov 27 '19 at 14:23
  • 2
    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
  • I can't speak for Kapovich but my assumptions would lead me to think he is referring to the type of double-quotient construction used to make geometric manifolds. – Ryan Budney Nov 27 '19 at 18:45
  • I have a vague recollection that the answer is negative, and that this is an old theorem of Wu-Chung Hsiang's. I believe he shows exotic spheres are not of this form. – Ryan Budney Dec 09 '19 at 03:29
  • I'd be curious about the connected sum of two (or more) copies of $S^1\times S^2$ (or 3-tori). – YCor Dec 09 '19 at 10:50

0 Answers0