# Is There a Basis for a Smooth Vector Fields on the 2-Sphere?

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.

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.