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We have the $C^{\infty}(M)$-module $\mathcal{D}^1(M)$ of vector fields over a $C^{\infty}$ manifold $M$. Is being parallelizable equivalent to this module being free, of dimension $n$?

I have the impression it is, since we can take as a basis the $n$ vector fields given by the assumption that $M$ is parallelizable and decompose every vector field in their components in each tangent space, and reciprocally the basis of the module would give the vector fields needed to parallelizability. Am I overlooking something?

If this is true, is it ever possible for $\mathcal{D}^1(M)$ to be a free module of a different dimension, other than $n$?

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  • $\begingroup$ see math.stackexchange.com/q/1071283/36530 $\endgroup$ Jan 15, 2016 at 1:02
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    $\begingroup$ @JamesS.Cook Sorry, I don't see how that link is relevant to my question. Could you clarify? $\endgroup$
    – Aloizio Macedo
    Jan 15, 2016 at 15:28
  • $\begingroup$ Particularly, math.stackexchange.com/a/1071338/36530 shows how parallelizable and free are connected. I don't think your question is a duplicate or anything, I just thought that Q and A might be helpful as you're thinking about these things. $\endgroup$ Jan 15, 2016 at 15:47

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That's correct, although you need to say a few words to justify the claim that if $n$ vector fields form a basis for $\mathcal{D}^1(M)$ over $C^\infty(M)$, then they are linearly independent at every point (those words are, if they weren't linearly independent at a point, then some tangent vector at that point would be linearly independent of them, and then a vector field that takes that value at the point could not be in their span over $C^\infty(M)$).

It is impossible for $\mathcal{D}^1(M)$ to be free of any other rank. One way to show this is to note that for any $p\in M$, the ideal $I_p\subset C^\infty(M)$ of functions vanishing at $p$ is a maximal ideal with quotient $\mathbb{R}$, and that $I_p\mathcal{D}^1(M)$ is exactly the vector fields vanishing at $p$ (this is not entirely obvious, but is easy locally and follows globally using bump functions). Thus $\mathcal{D}^1(M)/I_p\mathcal{D}^1(M)$ can be naturally identified with the tangent space at $p$, and thus must be $n$-dimensional over $C^\infty(M)/I_p\cong\mathbb{R}$. So if $\mathcal{D}^1(M)$ is free, it can only be of rank $n$.

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