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Let $M$ be a $n$ dimensional manifold and $p:TM\to M$ be the projection map. Then $\ker Dp$ is a $n$ dimensional vector bundle on $TM$, as a sub bundle of $TT(M)$.

For what type of manifolds, $\ker Dp$ is a trivial bundle? Or at least it admits a global non vanishing section? In particular put $M= S^{n}$. Does $\ker Dp$ admit a nonvanishing global section?

I need to the answer to this question for the "Note" in the final part of this question

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Ali Taghavi
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    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. – Ryan Budney Nov 06 '14 at 19:42
  • @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 ? – Ali Taghavi Nov 06 '14 at 21:00
  • @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 $S^{2}$. – Ali Taghavi Nov 06 '14 at 21:04
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    My differential geometry notes (couched in this iterated tangent bundle language and automorphisms of iterated tangent bundles) are here: http://rybu.org/DGNotes This material is in the 1st chapter, but the 3rd also covers some relevant material. – Ryan Budney Nov 06 '14 at 21:11
  • 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. – Ryan Budney Nov 06 '14 at 21:29
  • @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. – Ali Taghavi Nov 07 '14 at 17:25
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    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 $TS^2 \to S^2$. – Ryan Budney Nov 07 '14 at 17:45
  • @RyanBudney thanks a lot for your help. – Ali Taghavi Nov 07 '14 at 19:26

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