Timeline for Ricci flow descending from an universal cover
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Nov 23, 2012 at 21:58 | comment | added | Renato G. Bettiol | @Deane: Yes, we need uniqueness for the statement to work properly; see also Section 4.1.2.1 here: books.google.com/… | |
| Nov 23, 2012 at 21:56 | comment | added | Renato G. Bettiol | @Deane: I actually had closed manifolds in mind (and just realized the OP talks about open manifolds instead), but it seems to me that isometries are preserved under any Ricci flow. Quoting from Hamilton [JDG, 1982], "degeneracies are there because the equation is invariant under the full diffeomorphism group. This has the interesting consequence that any isometries which exist in the metric to begin with are preserved as the metric evolves". Although he was talking about closed 3-manifolds in this context, in my understanding these observations about the evolution equation are general. | |
| Nov 23, 2012 at 16:51 | history | edited | malik | CC BY-SA 3.0 |
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| Nov 23, 2012 at 14:14 | comment | added | Robert Haslhofer | @malik: If the isometries are preserved, then in particular the deck transformations remain isometries. Thus it doesn't matter which lift you pick and you get a well defined metric on the quotient. | |
| Nov 23, 2012 at 11:21 | history | edited | malik | CC BY-SA 3.0 |
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| Nov 23, 2012 at 9:37 | history | edited | malik | CC BY-SA 3.0 |
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| Nov 23, 2012 at 5:29 | answer | added | Otis Chodosh | timeline score: 6 | |
| Nov 22, 2012 at 17:43 | comment | added | Deane Yang | Renato, don't you need a uniqueness theorem for this to be true? | |
| Nov 22, 2012 at 17:33 | history | edited | malik | CC BY-SA 3.0 |
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| Nov 22, 2012 at 17:33 | comment | added | malik | Thank you, somehow I overlooked that fact. So there is no need for the last part about Kotschwar's result and I will delete it. | |
| Nov 22, 2012 at 17:17 | comment | added | Renato G. Bettiol | @malik: Regarding the invariance under the isometry group, since both the Ricci tensor and the metric are invariant under isometries (and these constitute the equation for the Ricci flow), any isometries of the original metric $g_0$ are also isometries of $g_t$, regardless of any curvature conditions. Of course, $g_t$ has to exist and be a smooth metric for this statement to make sense, but as long as it exists the isometry group is preserved. | |
| Nov 22, 2012 at 14:56 | history | asked | malik | CC BY-SA 3.0 |