Timeline for Does every manifold have a flat connection?
Current License: CC BY-SA 3.0
18 events
| when toggle format | what | by | license | comment | |
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| Aug 25, 2015 at 1:55 | answer | added | Qiaochu Yuan | timeline score: 14 | |
| S Aug 23, 2015 at 15:49 | history | suggested | Ali Taghavi |
I add two tags
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| Aug 23, 2015 at 15:36 | review | Suggested edits | |||
| S Aug 23, 2015 at 15:49 | |||||
| Oct 3, 2013 at 4:20 | vote | accept | pavpanchekha | ||
| Oct 3, 2013 at 2:35 | comment | added | Misha | @PaulReynolds: Yes, there is a notion of connections and curvature for vector bundles, but it is no longer Riemann's (and is no longer a tensor) and should not be referred to by this name, as Riemann curvature tensor is something much more specific. | |
| Oct 2, 2013 at 21:42 | comment | added | Peter Crooks | Yes, this is true. Indeed, you will find that I only referred to the exterior derivative connection on a trivial bundle. | |
| Oct 2, 2013 at 21:20 | comment | added | Urs Schreiber | Peter, on a general non-trivial bundle "the exterior derivative connection" may not exist, not even make sense. This is what Paul is referring to. It exists locally after a choice of local trivialization, but not in general globally. | |
| Oct 2, 2013 at 20:54 | answer | added | valeri | timeline score: 17 | |
| Oct 2, 2013 at 20:42 | comment | added | Paul Reynolds | @Misha, why not? There is a well-defined two-form with values in the adjoint bundle. | |
| Oct 2, 2013 at 20:35 | comment | added | Misha | Also, Riemann curvature tensor makes no sense for arbitrary vector bundles. | |
| Oct 2, 2013 at 20:16 | comment | added | Peter Crooks | Sorry, I was referring to the "exterior derivative" connection on a trivial bundle over the manifold. When studying connections on trivial bundles, one often takes it to be the origin of the space of connections. This connection is definitely flat, though. | |
| Oct 2, 2013 at 20:15 | comment | added | Ryan Budney | The answer is no, as Liviu points out. The lowest-dimensional example is the tangent bundle of $S^2$, by Gauss-Bonnet. | |
| Oct 2, 2013 at 20:11 | comment | added | Paul Reynolds | @Peter, there isn't a zero connection. Unless you mean something perculiar? Either way, your conclusion isn't right. | |
| Oct 2, 2013 at 20:06 | comment | added | Paul Reynolds | Very close question: mathoverflow.net/questions/91852/… | |
| Oct 2, 2013 at 19:59 | history | edited | pavpanchekha | CC BY-SA 3.0 |
We have a vector bundle.
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| Oct 2, 2013 at 19:57 | comment | added | Peter Crooks | Since you are working without a metric, you can take the zero-connection on any vector bundle over your manifold. Its curvature is certainly zero (ie. the connection is flat). | |
| Oct 2, 2013 at 19:56 | comment | added | Liviu Nicolaescu | Your question needs some edition because it is very imprecise. (E.g.a connection is defined on a vector bundle, not on a manifold.) In any case, if the manifold is compact, oriented and the Euler characteristic is $\neq 0$, then there cannot exist any metric on the tangent bundle and connection compatible with it whose curvature is zero. This is a consequence of the Gauss-Bonnet-Chern theorem. | |
| Oct 2, 2013 at 19:45 | history | asked | pavpanchekha | CC BY-SA 3.0 |