Timeline for Do cotangent bundles have "bounded geometry"?
Current License: CC BY-SA 3.0
8 events
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| Aug 1, 2015 at 9:28 | comment | added | valeri | Yes, if "standard metric" means Sasaki. Although with this correction the question seems quite different. Btw, for TS^n the Cheeger-Gromoll metric for me much more "standard" - which I believe has bounded geometry. | |
| Aug 1, 2015 at 1:50 | comment | added | Jaap Eldering | @valeri: well, the OP clarified in the comments to the question that he was using the "standard metric induced from a metric on the base", which I interpreted as the natural Sasaki metric. So I think my answer as it stands correctly answers the question. | |
| Jul 31, 2015 at 21:32 | comment | added | valeri | all these claims are about very special Sasaki metric on TM, and do not imply non existence of bounded geometry. Actually, as was already answered above - every manifold admits bounded geometry [Greene], which means that your answer is wrong. To "believe" in Greene's result you might just represent a manidold as a handlebody and endow every handle with bounded geometry which is product near the boundary - then any union of handles is provided with bounded geometry. | |
| Jul 31, 2015 at 19:57 | comment | added | Jaap Eldering | @valeri: Sorry, I wasn't careful and reference the wrong thing. Now corrected it to Theorem 7.8. If $TM$ would have bounded curvature, then also its sectional curvature would be bounded, and Theorem 7.8 then implies that $TM$ is flat, which is equivalent to $M$ flat using Theorem 7.6. But can also look at the explicit formulas for the curvature of $TM$ expressed in terms of the curvature of $M$. | |
| Jul 31, 2015 at 19:52 | history | edited | Jaap Eldering | CC BY-SA 3.0 |
Fix reference to correct theorem.
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| Jul 31, 2015 at 18:57 | comment | added | valeri | The corollary 7.11 you mention (called Theorem 7.11) claims "Let (M, g) be a Riemannian manifold and the tangent bundle TM be equipped with the Sasaki metric g. Then (TM, g) has constant scalar curvature if and only if (M, g) is fiat." How you infer that TM admits no bounded geometry unless M is flat? | |
| Jul 31, 2015 at 12:55 | history | edited | Jaap Eldering | CC BY-SA 3.0 |
added 263 characters in body
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| Jul 31, 2015 at 12:49 | history | answered | Jaap Eldering | CC BY-SA 3.0 |