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I've been watching Frederic Schuller's course on "The Mathematics and Physics of Gravity and Light" and at a moment during Lecture 6 he claims there is no basis for $\Gamma(TS^2)$ (the space of smooth sections $\sigma \colon S^2 \to TS^2$) when considered as a $C^\infty(S^2)$-module and gives an example through the claim that since every vector field must vanish somewhere, you can't get a basis. During his example, he uses a vector field which vanishes at points $(\pm 1, 0, 0)$ and $(0, 0, \pm 1)$ and says that since on, say, $(0, 0, \pm 1)$ you have only one nonvanishing vector field, and thus a linear combination can't point towards a different direction.

My first question is: adding a new vector field which vanishes at $(0, \pm 1, 0)$ wouldn't solve the problem? At every point you have at least two non-parallel vectors, so it seems to me it would work, at least in principle. Furthermore, every vector space admits a basis, so $\Gamma(TM)$ (for some smooth manifold $M$) when considered as a real vector space has a basis. Since every real number can be regarded as a constant (and hence smooth) function, wouldn't it allow us to obtain a basis for $\Gamma(TM)$ when considered as a $C^\infty(M)$-module?

I believe it is worth mentioning I do not have much background in Differential Geometry nor Module Theory.

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The problem is if you cannot find one vector field that does not vanish at a point you can never make a basis since the set will be linearly dependent at some point (since it contains the $0$ vector). You are correct that $\Gamma(TM)$ can be considered a vector space however it would be an infinite dimensional one. As an example we can think about $\Gamma(T\mathbb{R}^3)$ as being a 3 dimensional module with generating set $\{ \partial_x, \partial_y ,\partial_z\}$, however that set under linear combination by scalars does not generate every vector field on $\mathbb{R}^3$, take as an example the vector field $x\partial_x$. If viewed as a vector space then this space is infinite dimensional and having a basis for that space does not give us a basis for $\Gamma(T\mathbb{R}^3)$ as a module. Hope this helps.

Also I love Frederich Schuller's lectures they are very good, I would recommend them to any (Mathematical) Physics undergraduate.

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    $\begingroup$ Helped a lot! Thank you! $\endgroup$ – Níckolas Alves Apr 19 '20 at 2:53

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