Timeline for A vector bundle analogy of the Nash embedding theorem
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| Dec 27, 2019 at 14:31 | history | edited | Ali Taghavi | CC BY-SA 4.0 |
added 10 characters in body
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| Dec 27, 2019 at 14:30 | comment | added | Ali Taghavi | @StevenLandsburg sorry for my delay. Thanks for your correction. I revise it. | |
| Dec 27, 2019 at 14:04 | comment | added | Sebastian Goette | Even more is true: if the vector bundle has a metric connection, then by a result of Narasimhan and Ramanan, there is a classifying map from $M$ to a sufficiently high-dimensional Grassmannian that classifies both metric and connection on $E$. This seems to me the correct "linearisation" of the Nash embedding theorem. | |
| Dec 26, 2019 at 23:42 | comment | added | mme | After applying Steven Landsburg's correction, this is a definition chase using the fact that there is a classifying map $f: M \to BO(k)$, where $BO(k)$ is the Grassmannian of $k$-planes in $\Bbb R^\infty$, and the fact that $f$ factors through some finite-dimensional Grassmannian $\text{Gr}(n,k)$. It is not analagous to Nash's theorem (which is much more difficult). | |
| Dec 26, 2019 at 22:00 | comment | added | Steven Landsburg | Should "trivial subbundle" be "subbundle of a trivial bundle"? | |
| Dec 26, 2019 at 15:49 | history | asked | Ali Taghavi | CC BY-SA 4.0 |