Timeline for Does the tensor bundle of a compact manifold have a bounded geometry?
Current License: CC BY-SA 3.0
17 events
| when toggle format | what | by | license | comment | |
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| Feb 18, 2016 at 13:51 | comment | added | valeri | @Kaveh - welcome! | |
| Feb 18, 2016 at 13:37 | comment | added | Kaveh | @valeri- many thanks, I think I am getting the answer. | |
| Feb 18, 2016 at 13:14 | comment | added | valeri | @Kaveh - I am not sure to whom this result was attributed (you may check referred posts), but to construct bounded metric on a finite handle body - take such metrics on handles and (accurately) glue them together (then average over the action of the structure group) - smth like this. To construct handle decomposition you may consider (the non-degenerate deformation of) the distance to the zero section as a Morse function. | |
| Feb 18, 2016 at 13:00 | comment | added | Kaveh | Thanks valeri, I just need to know about the existence of a metric on the fiber bundle $S^2_+T^*M$ that has a bounded geometry. By what you said there is such a metric. Can you give me any some references about that argment. | |
| Feb 18, 2016 at 12:43 | comment | added | valeri | Kaveh - if I understand your pairing correctly, two Cheeger-Gromoll $T^*M pair T^*M$ is unbounded. But every open manifold which is finite handle-body - like any bundle over compact $M$ with fibres vector spaces; has a metric of bounded geometry. May be even invariant under the structure group action. The point is - to have fibers look like a cylinder - check the original Cheeger-Gromoll example. | |
| Feb 18, 2016 at 12:17 | comment | added | Kaveh | so it seems to me that the question is that does Cheeger-Gromoll induces, by pairing, a metric to the cotangent bundle with bounded geometry. Or is there a metric unrelated to the metrics on tangent bundle which has a bounded geometry. | |
| Feb 18, 2016 at 11:58 | comment | added | Anton Petrunin | see also mathoverflow.net/q/94322/1441 | |
| Feb 18, 2016 at 11:51 | comment | added | valeri | in the referred post $TM$ was considered with two types of metrics: Sasaki with flat fibers - not bounded geometry, and Cheeger-Gromoll, coming from submersion $GM \to TM$ of bounded geometry. So, the choice is yours. | |
| Feb 18, 2016 at 11:41 | comment | added | Kaveh | About Riemannian metric, I think the natural metric is the tensor product of a Riemannian metric on the cotangent bundle. I dont not know yet what kind of metrics exist on cotangent bundles. | |
| Feb 18, 2016 at 11:30 | comment | added | Kaveh | I read mathoverflow.net/questions/212713/…. But I can not see the conclusion. $ s_+^2T^*M$ is a subbundle of $T^*M \otimes T^*M$, so the question is that is finite tensor product of cotangent bundle has bounded geometry. | |
| Feb 18, 2016 at 11:25 | comment | added | Ali Taghavi | @Kaveh May you more explain on the structure of the metric on the bundle? | |
| S Feb 18, 2016 at 11:18 | history | suggested | Ali Taghavi | CC BY-SA 3.0 |
I add a few words
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| Feb 18, 2016 at 11:11 | review | Suggested edits | |||
| S Feb 18, 2016 at 11:18 | |||||
| Feb 18, 2016 at 11:05 | comment | added | valeri | see mathoverflow.net/questions/212713/… | |
| Feb 18, 2016 at 10:15 | history | edited | Ben McKay | CC BY-SA 3.0 |
spelling, grammar, formatting
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| Feb 18, 2016 at 10:11 | review | First posts | |||
| Feb 18, 2016 at 10:14 | |||||
| Feb 18, 2016 at 10:10 | history | asked | Kaveh | CC BY-SA 3.0 |