Timeline for Example of a manifold which is not a homogeneous space of any Lie group
Current License: CC BY-SA 3.0
19 events
| when toggle format | what | by | license | comment | |
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| S Mar 29, 2016 at 13:18 | history | suggested | Ali Taghavi |
I add a tag
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| Mar 29, 2016 at 13:15 | review | Suggested edits | |||
| S Mar 29, 2016 at 13:18 | |||||
| Feb 26, 2012 at 5:56 | vote | accept | MTS | ||
| Feb 25, 2012 at 23:01 | history | edited | MTS | CC BY-SA 3.0 |
Removed part of question as per Vitali's comment
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| Feb 25, 2012 at 0:06 | comment | added | Vitali Kapovitch | @MTS you should probably remove the assumption of compactness of G from the last line of your question. you ask it several times but only mention compactness once. and it's clear from your comments that you are interested in the general case of a possibly noncompact G. Assuming that G is compact is more restrictive and makes the question less interesting. In particular it rules out quotients of Lie groups by cocompact lattices such as nilmanifolds. Also note that if G is compact then the quotient G/H admits a homogeneous G-invariant metric coming form a biinvariant metric on G. | |
| Feb 24, 2012 at 22:03 | comment | added | MTS | Wow, these answers are all great so far! It's interesting that the examples are coming from so many different perspectives. | |
| Feb 24, 2012 at 20:05 | answer | added | Lee Mosher | timeline score: 11 | |
| Feb 24, 2012 at 16:50 | answer | added | Igor Belegradek | timeline score: 8 | |
| Feb 24, 2012 at 13:03 | comment | added | Marc Palm | As you mention, locally symmetric spaces give a huge class. The example you describe with the genus at least two is actually $M = \Gamma \backslash PSL_2( \mathbb{R}) /PSO(2)$, where $\Gamma \cong \pi_1(M)$. | |
| Feb 24, 2012 at 6:54 | answer | added | Dylan Wilson | timeline score: 28 | |
| Feb 24, 2012 at 5:11 | comment | added | Ryan Budney | @MTS: Sometimes people underestimate a question, oversimplifying it. I suspect in this case it's just a mistake on the part of the people who cast votes to close. But if they don't say anything it's hard to tell. I don't think you have to worry about this thread being closed. | |
| Feb 24, 2012 at 5:04 | comment | added | MTS | I must say I am surprised that this question has garnered two votes to close as "not a real question." Would anybody care to explain? The answers below seem to imply (to me, at least) that the question isn't so trivial. | |
| Feb 24, 2012 at 3:03 | answer | added | Vitali Kapovitch | timeline score: 48 | |
| Feb 24, 2012 at 3:02 | answer | added | Paul Siegel | timeline score: 34 | |
| Feb 24, 2012 at 1:21 | answer | added | Tom Goodwillie | timeline score: 73 | |
| Feb 24, 2012 at 0:52 | comment | added | Tom Goodwillie | Well a compact orientable surface of genus two or more has no transitive action of a compact Lie group because such an action would necessarily preserve a Riemann metric and therefore a conformal structure. The group of conformal automorphisms of such a surface is finite. | |
| Feb 24, 2012 at 0:48 | comment | added | Qiaochu Yuan | Well, a compact surface of genus at least two is the quotient of $\mathbb{H}$ by a group action, and $\mathbb{H}$ is a homogeneous space for $\text{SL}_2(\mathbb{R})$, so in some sense these examples still come from Lie groups. | |
| Feb 24, 2012 at 0:29 | answer | added | Roberto Frigerio | timeline score: 9 | |
| Feb 24, 2012 at 0:11 | history | asked | MTS | CC BY-SA 3.0 |