Timeline for A question on tangent bundle (and second tangent bundle)
Current License: CC BY-SA 3.0
10 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
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| Nov 7, 2014 at 19:26 | comment | added | Ali Taghavi | @RyanBudney thanks a lot for your help. | |
| Nov 7, 2014 at 17:45 | comment | added | Ryan Budney | Okay, yes, the answer to that question is yes. This is because $TS^2 \oplus TS^2$ as a bundle over $S^2$ has an everywhere non-zero vector field (just take two vector fields on $S^2$ that have no common zeros). You could then translate this vector field fibrewise to be sitting over any vector field of $TS^2 \to S^2$. | |
| Nov 7, 2014 at 17:25 | comment | added | Ali Taghavi | @RyanBudney I mean: let $P:T^{1}S^{2}\to S^{2}$ is the natural projection from the unit tangent bundle. Is it possible to lift a vec. field on $S^{2}$ to a nonvanishing vec. field on $T^{1}S^{2}$? The Motivation: As I said here, I would like to run away from $S^{2}$ in any possible way. | |
| Nov 6, 2014 at 21:29 | comment | added | Ryan Budney | Could you re-state your last comment, perhaps using things like the maps $p$, $Dp$, and the bundle projection $\pi : T^2 M \to TM$ ? I'm not quite sure which objects you're referring to. | |
| Nov 6, 2014 at 21:11 | comment | added | Ryan Budney | My differential geometry notes (couched in this iterated tangent bundle language and automorphisms of iterated tangent bundles) are here: rybu.org/DGNotes This material is in the 1st chapter, but the 3rd also covers some relevant material. | |
| Nov 6, 2014 at 21:04 | comment | added | Ali Taghavi | @RyanBudney So should I understand from your comment that every vector field on $S^{2}$ can be lifted to a non vanishing vec. field on the unit tangent bundle of $S^{2}$? By lifting I mean a vector field on total space whose solution are mapped to the solution of our initial vec. field on $S^{2}$. | |
| Nov 6, 2014 at 21:00 | comment | added | Ali Taghavi | @RyanBudney thanks for the comment. It remind me of your comment to one of my question about equivalency of two different structure of vector bundle on $T^{2}M\to TM$. you attached a pdf file of your lecture for a proof of this some thing related to this. your comment is not available now. may you resend that file ? | |
| Nov 6, 2014 at 19:42 | comment | added | Ryan Budney | The kernel of $Dp$ is canonically identified with the kernel of the bundle map $T^2 M \to TM$, thinking of $T^2M$ as the tangent bundle of $TM$. The identification is via the involution of the double tangent bundle. So your question is the same question as asking if the manifold $TM$ is parallelizable, which is the same question of if $TM \oplus TM \to M$ is a trivial vector bundle. I'm not sure if there's any cute answers to that question, but likely there's some cute sufficient conditions. | |
| Nov 6, 2014 at 19:14 | history | asked | Ali Taghavi | CC BY-SA 3.0 |