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As the title saying,

Why $C^{\infty}(M)$ module of sections of a vector bundle $E\rightarrow M$ is a reflexive module?

Here we are considering vector bundles with finite-dimensional fibers.

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    That is true, since the global section is a finitely generated projective module.2015-07-31

2 Answers 2

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As explained in a comment to this question, we have $\Gamma(E)^\vee\cong\Gamma(E^\vee)$, and hence $\Gamma(E)^{\vee\vee}\cong\Gamma(E^{\vee\vee})\cong\Gamma(E)$.

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This follows from the Serre-Swan Theorem, which states that modules of section are characterized as projective finitely generated $C^\infty(M)$-modules.

Hence for any vector bundle $E\to M$ the natural map $\Gamma(E)\to\Gamma(E)^{\vee\vee}$ is

  • injective, because $\Gamma(E)$ is finitely generated,
  • surjective, because $\Gamma(E)$ is projective.

Fun fact: Any manifold $X$ admits a finite atlas consisting of $\dim X+1.$ (You can fing this fact in the book below on page 5) Applaying this to $X=E$ we have that $\Gamma(E)$ can be generated by $m+e+1$ elements where $m=\dim M$ and $e$ is the dimension of the fibres of $E.$

Kolář, Ivan; Michor, Peter W.; Slovák, Jan, Natural operations in differential geometry, Berlin: Springer-Verlag. vi, 434 p. (corrected electronic version) (1993). ZBL0782.53013.