Timeline for Is a fibre bundle over a vector bundle trivializable on each fibre?
Current License: CC BY-SA 2.5
5 events
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| Dec 1, 2010 at 21:43 | comment | added | mathbekunkus | Thanks everyone for your comments, thanks Sean for your answer. @David Roberts: by homogeneous I mean that the trivialization of the bundle is given by the following diffeomorphisms of the base: $x\mapsto x+y$, that is, homogeneous under the action of the vector space as an additive group on the total space given by translations. Apparently what it takes for such a "meta-bundle" to be homogeneous in this sense is that it is diffeomorphic to the fibred product of the two bundles over $M$. | |
| Dec 1, 2010 at 21:39 | vote | accept | mathbekunkus | ||
| Nov 18, 2010 at 12:44 | comment | added | Sean Tilson | You are right, that is even better. My first thoughts are typically "is there an LES we can use?" Not to mention you then need to apply Whiteheads Thm. | |
| Nov 18, 2010 at 8:52 | comment | added | Johannes Ebert | @Sean: you are killing flies with a sledgehammer. For any vector bundle, the zero section is a homotopy inverse to the bundle projection. | |
| Nov 18, 2010 at 5:34 | history | answered | Sean Tilson | CC BY-SA 2.5 |