Given $E,F$ vector bundles over a manifold $M$, I would like to know a proof of $\Gamma(E\otimes F) = \Gamma(E) \otimes_{C^\infty(M)} \Gamma(F)$. Where $\Gamma$ denotes the smooth sections over $M$. Thanks


1 Answer 1


The statement is trivially true if $E$ is a trivial vector bundle. Note also that if the statement is true for $E$, then it is also true for any direct summand of $E$. The conclusion follows from the fact that any vector bundle on a manifold is a subbundle of a trivial vector bundle.

  • $\begingroup$ you are using a deep fact to prove a trivial result! $\endgroup$
    – Juan Fran
    Jul 14, 2016 at 14:43
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    $\begingroup$ Well, "deep" is of course a relative notion. The fact that any vector bundle on a manifold is a direct summand of a trivial vector bundle does not seem so deep to me. It is quite easily proved using partitions of unity. Exactly the same technique can be used to show OP's statement. I thought it was kind of economical to use the former fact in order to prove the latter... $\endgroup$ Jul 14, 2016 at 19:31
  • $\begingroup$ That is true, but these modules are canonically isomorphic, and I don't quite see how your argument gives a canonical isomorphism (which I think the op desires, for he writes "=" instead of just isomorphic). A nice exposition (containing these canonical isomorphism) is "connections, curvature, and cohomology I" $\endgroup$
    – Juan Fran
    Jul 15, 2016 at 13:53
  • $\begingroup$ There is a canonical morphism, but it just needs to be checked that it is an isomorphism. This approach does work, as shown in the answer to this question (of which this is a duplicate). $\endgroup$
    – ಠ_ಠ
    Mar 8, 2017 at 6:19
  • $\begingroup$ BTW if the base space is not compact then a vector bundle is not necessarilly a summand of a trivial bundle. For example the tautological line bundle on \mathbb{R}P^{\infty}$. But I think that the result in the OP question is true for non compact base space. $\endgroup$ Feb 10, 2022 at 12:32

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