Timeline for Does every vector bundle allow a finite trivialization cover?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 20, 2012 at 1:37 | comment | added | Fiktor | @ Andy Putman : Thank you for your references. I already feel, that I should learn more topology: I didn't know these facts (that every manifold can be triangulated, every manifold is homeomorphic to a CW-complex), as I didn't know about notion of Lebesgue covering dimension, which was essentially used in the answer by Renato G Bettiol ("general results in topology" = "Lebesgue dimension of n-dimensional manifold is equal to $n$"). The fact you mentioned certainly solves my question. | |
| Apr 19, 2012 at 20:17 | comment | added | John Klein | @Andy Putnam: What Andreas is really talking about a relative notion: the condition of having a map $f\:X \to Y$ and a covering $U_\alpha$ of $X$ such that $f_{|U_\alpha}\: U_\alpha \to Y$ is null for each $\alpha$. In Andreas' situation, $Y = BO(n)$. | |
| Apr 19, 2012 at 3:28 | comment | added | Andy Putman | Actually, you specified that your manifolds are smooth, so there is no need to appeal to fancy things like Kirby-Siebenmann. Smooth manifolds can be triangulated, and thus are homeomorphic to simplicial complexes (better than just CW complexes!). | |
| Apr 19, 2012 at 3:26 | comment | added | Andy Putman | @Fiktor : I guess that's true, but Kirby-Siebenmann proved that all manifolds have topological handle decompositions, and thus are homeomorphic to CW complexes. | |
| Apr 19, 2012 at 3:16 | comment | added | David Roberts♦ | ALso, since the OP is asking for a local trivialisation of a given vector bundle (as opposed to all vector bundles), the connected components of the open sets don't need to be contractible, merely that the vector bundle restricts to be trivial over them. If he has proved that the manifold has a cover by (disjoint unions of) contractible opens, then this is even stronger than what the question requires. | |
| Apr 19, 2012 at 3:14 | comment | added | Fiktor | @Andry Putman : I do allow a cover by disconnected open sets (I've just added this clarification to the question), but I didn't noticed my proof, you are referencing to. I have some proof for the case of CW-complexes, but Wikipedia says only that manifolds are homotopy equivalent to CW-complexes. | |
| Apr 19, 2012 at 2:23 | comment | added | Andy Putman | @Andreas Blass : It's not necessary that the open sets be contractible, merely that each of their components are contractible. Once you allow disconnected components, then the OP has a correct proof that $(n+1)$ open sets are necessary for an $n$-dimensional manifold. | |
| Apr 19, 2012 at 1:33 | history | answered | Andreas Blass | CC BY-SA 3.0 |