The module structure of the vector field over a smooth manifold

Asked Jan 03 '24 at 12:33

Active Jan 03 '24 at 12:33

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Let $M$ be a manifold of dimension m and $X:M \rightarrow TM$ a smooth vectorial field. The set of all vectorial fields,( $\mathcal{X}(M)$,+,.) forms a $C^{\infty}(M)$ module.

Is this module free?

Also, locally by chosing a local chart, I can prove that this module is finite generated. But is it globally finite generated?

asked Jan 03 '24 at 12:33

mmm

1

Not necessarily free (or see this), but is finitely generated. – peek-a-boo Jan 03 '24 at 12:38
If $M$ is compact, by Swan's theorem the module of vector fields is projective. This is slightly weaker than saying that it is free. – Federico T. Jan 10 '24 at 10:43

The module structure of the vector field over a smooth manifold

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