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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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    $\begingroup$ 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. $\endgroup$
    – Ryan Budney
    Commented Nov 6, 2014 at 19:42
  • $\begingroup$ @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 ? $\endgroup$
    – Ali Taghavi
    Commented Nov 6, 2014 at 21:00
  • $\begingroup$ @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 $S^{2}$. $\endgroup$
    – Ali Taghavi
    Commented Nov 6, 2014 at 21:04
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    $\begingroup$ 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. $\endgroup$
    – Ryan Budney
    Commented Nov 6, 2014 at 21:11
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    $\begingroup$ 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 $TS^2 \to S^2$. $\endgroup$
    – Ryan Budney
    Commented Nov 7, 2014 at 17:45

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