Timeline for Do cotangent bundles have "bounded geometry"?
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| S Jul 22 at 18:10 | history | suggested | Ali Taghavi |
I add a tag metric geometry
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| Jul 22 at 11:04 | review | Suggested edits | |||
| S Jul 22 at 18:10 | |||||
| S May 6, 2018 at 18:51 | history | suggested | Ali Taghavi |
I add a tag.
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| May 6, 2018 at 16:20 | review | Suggested edits | |||
| S May 6, 2018 at 18:51 | |||||
| Oct 9, 2017 at 8:04 | comment | added | AlexE | One should note that there are several different versions of the definition of "bounded geometry", and they are not mutually equivalent to each other. Everytime an author uses this notion one has to check which of these definitions he is using (and since some authors do not write it down explicitly, one often has to guess the version which is used). | |
| Oct 6, 2017 at 13:57 | answer | added | Alex M. | timeline score: 4 | |
| Jul 31, 2015 at 13:43 | answer | added | Vidit Nanda | timeline score: 8 | |
| Jul 31, 2015 at 12:58 | answer | added | Igor Belegradek | timeline score: 13 | |
| Jul 31, 2015 at 12:49 | answer | added | Jaap Eldering | timeline score: 7 | |
| Jul 31, 2015 at 11:16 | comment | added | ss78 | OK, so does a cotangent bundle equipped with the standard metric (induced from a metric on the base) have bounded geomerty if the metric satifies these curvature properties? | |
| Jul 31, 2015 at 10:50 | comment | added | Paul Siegel | Bounded geometry is a property of a metric space, so your question doesn't make sense. A Riemannian manifold has bounded geometry if and only if the curvature tensor and all of its covariant derivatives are uniformly bounded. | |
| Jul 31, 2015 at 10:16 | history | edited | ss78 | CC BY-SA 3.0 |
added 27 characters in body
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| Jul 31, 2015 at 10:15 | review | First posts | |||
| Jul 31, 2015 at 11:12 | |||||
| Jul 31, 2015 at 10:06 | history | asked | ss78 | CC BY-SA 3.0 |