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Let $M=S^2$ then the hairy ball theorem, states that there is no non-vanishing smooth tangent vector field on even-dimensional $n$-spheres. Hence, we can multiply any smooth vector field $v\in\Gamma(TS^2)$ by a function $f\in \mathcal{C}^\infty(S^2)$, which is zero everywhere except where v is, obtaining $fv=0$ despite $f\neq 0$ and $v\neq 0$. Therefore, there is no set of linearly independent vector fields on $S^2$ over $C^{\infty}(S^2)$-module. I think the same argument goes for the vector space over an algebraic field $K$. Moreover we can not express all the vector field as $g = fv$ where $f \in K \text{ and }v,g \in \Gamma(TS^2)$. If I am right does there exist any basis for vector space over the sphere?

Swakshar Deb
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    "Hence, we can multiply any smooth vector field $v\in\Gamma(TS^2)$ by a function $f\in \mathcal{C}^\infty(S^2)$, obtaining $fv=0$" This is not true : if $v$ is zero on a finite set (eg two antipodal points) then $f$ would have to be zero everywhere in order to have $fv = 0$ and $f$ continuous – SolubleFish May 11 '21 at 14:09
  • yes, you're right. – Swakshar Deb May 11 '21 at 14:10

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