Timeline for Is the Moebius strip Riemannian homogeneous?
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Jan 17, 2022 at 17:52 | comment | added | Ian Gershon Teixeira | Can this proof also be used to show that the only noncompact connected surfaces that can be a linear group orbit are the plane and the cylinder? It seems that you make essential use of the fact that the stabilizer must be compact since it is an action by isometries. But maybe you see an obvious reworking of the proof that I do not? | |
| Dec 8, 2021 at 17:20 | comment | added | YCor | @IanGershonTeixeira not really. Possibly it would better be separated indeed. | |
| Dec 8, 2021 at 16:35 | comment | added | Ian Gershon Teixeira | Do you have any thoughts about the general vector bundle case? Or do you think I should make that a separate question? | |
| Dec 8, 2021 at 16:34 | vote | accept | Ian Gershon Teixeira | ||
| Dec 8, 2021 at 16:20 | comment | added | YCor | You're right, I assumed noncompact. I've corrected. In the compact case, one gets $K'/K$ with $K'$ compact, and either $K'$ is 2-dimensional and we get the torus, and otherwise the only possibility is indeed $K'$ locally isomorphic to $\mathrm{SU}(2)$, and we get the 2-sphere or $\mathbf{P}^2_\mathbf{R}$. | |
| Dec 8, 2021 at 16:20 | history | edited | YCor | CC BY-SA 4.0 |
added 11 characters in body
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| Dec 8, 2021 at 16:06 | comment | added | Ian Gershon Teixeira | What about the sphere? | |
| Dec 8, 2021 at 15:00 | history | answered | YCor | CC BY-SA 4.0 |