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Most textbooks define a vector field on a smooth manifold $M$ as a section of the tangent bundle of $M$. My question is: why is it even necessary to talk about bundles when defining vector fields on $M$?

To be clear, when I refer to the tangent bundle, I refer to the triple $(M, TM, \pi)$, where $\pi: TM \to M$ is a smooth surjection. The problem is, vector fields can be defined without ever referring to $\pi$ at all. We can just define a smooth vector field as a smooth map $f: M \to TM$ satisfying $f (p) \in T_pM$ for all points $p \in M$ (that is, once we've given $TM$ a topology and a smooth atlas).

I'm likely missing something about the importance of bundles. Any comments would be most welcome. Thanks!

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    $\begingroup$ The definition you gave for $f$ is exactly what is meant by "a vector field is a smooth section of the tangent bundle of $M$". So, I'm not really sure what the issue is $\endgroup$
    – peek-a-boo
    Commented May 22, 2020 at 22:38
  • $\begingroup$ I've edited the post to make it clearer what I mean by "bundle". $\endgroup$
    – Adithya Chakravarthy
    Commented May 22, 2020 at 22:56
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    $\begingroup$ How do you define $TM$, give it a topology and a smooth atlas? Don't you need $\pi$ for properly introducing $TM$? $\endgroup$
    – Paul Frost
    Commented May 22, 2020 at 22:57
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    $\begingroup$ If $M$ is given as a submanifold of Euclidean space, then you can in fact avoid understanding the tangent bundle. Otherwise, ... $\endgroup$
    – Ted Shifrin
    Commented May 22, 2020 at 23:15
  • $\begingroup$ I agree with everyone above: if it makes sense to say $f: M\to TM$ is smooth, you have a smooth manifold structure on $TM;$ and if it makes sense to say $f(p) \in T_p M,$ you have the projection $\pi;$ so your definition necessitates already having the triple $(M, TM, \pi).$ You could define vector fields abstractly as derivations of $C^\infty(M),$ though you're missing a load of intuition if you don't then identify these as sections of a bundle. $\endgroup$
    – Anthony Carapetis
    Commented May 23, 2020 at 1:56

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For each $p$, $f(p)\in T_pM$ happens if and only if $\pi(f(p))=p$. This is the same as saying "$f$ is section for $\pi$", so indeed you said the same without ever using the word "section". Why use it? Well, it's just convenient because the same concept applies for maps which are not neccesarilly bundles. E.g., sheaf theory, algebraic topology, category theory.

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