# How to find basis of vector fields?

I'm figuring out definition of vector fields over a manifold as differentiations of algebra $$C^\infty(M)$$ of functions on $$M$$. How can we find their basis starting from this very definition? I know, in some chart $$(U, \varphi)$$, it must be some $$X|_U = \sum_{i=1}^n X_i \frac{\partial}{\partial x_i}$$ but how to obtain it?

All we know $$X$$ is some differentiation i.e. $$X(fg)=Xf\cdot g+f\cdot Xg$$ and that's it.

EDIT: I'm asking, of course, about a basis in some chart $$(U, \varphi)$$. I editted the question.

• What do you mean "how to obtain it"? As with a basis in a vector space you just choose one. You can extend any chosen basis of the tangent space at a point into vector fields locally although you can't guarantee globally that these vector spaces are non-vanishing let alone form a basis as @Lieven's answer mentions Commented 2 days ago

On the domain of an individual coordinate chart the vectors $$\frac{\partial}{\partial x^i}$$ ($$i=1,\ldots,n$$) form a basis in every fibre; but in a different coordinate system, on the intersection of two charts, these vectors will in general be linear combinations of the basis vectors of the other chart. The existence of a global basis of vector fields on a manifold is not guaranteed and depends on the global topology of the manifold.

As a commenter pointed out, the $$2$$-sphere cannot have a global nonzero vector field, so no single vector field can be part of a basis everywhere.

• Famously, the 2-sphere doesn't have a basis. Any continuous vector field is zero somewhere, so given a candidate basis with 2 fields there will be points where the span is one-dimensional, and if you have 3 fields there is bound to be a lot of linear dependence going on. Commented 2 days ago

For sake of simplicity let us explain in dimension 2, with two coordinate $$(x,y)$$, so in an open set of $$\bf R^2$$.

Let $$X$$ be a derivation, i.e. an operator on $$C^{\infty}$$ such that $$X(f.g)=fX(g)+gX(f)$$, $$X(\lambda f+g)=\lambda X.f+X.g$$. Let $$a$$ the function $$X.x$$, $$b$$ the function $$X.y$$ we want to show $$X= a.{\partial \over \partial x}+ b.{\partial \over \partial y}$$.

Let $$(x_0,y_0)$$ be a point, and write

$$f(x,y)=f(x_0,y_0)+(x-x_0) A(x,y)+(y-y_0)B(x,y)$$, with $$A,B$$ of class $$C^{\infty}$$. (Hadamard lemma)

then $$(X.f)(x_0,y_0)= [X.f(x_0,y_0)](x_0,y_0) +A(x_0,y_0)[ X. (x-x_0)](x_0,y_0) +B(x_0,y_0)[ X. (x-x_0)](x_0,y_0)$$.

We first note that $$X.c=0$$ if $$c$$ is a constant function : indeed $$X.c=cX.1=c X. 1^2=2cX.1=2. X.c$$, so the first term is $$0$$.

Remark that $$A(x_0,y_0)={\partial \over \partial x} f(x_0,y_0)$$, $$B(x_0,y_0)={\partial \over \partial y} f(x_0,y_0)$$, and that $$X(x-x_0)= X.x=a$$, $$X(y-y_0)= X.y=b$$.

Using linearity, we get

$$(X.f)(x_0,y_0)=(a.{\partial \over \partial x}+ b.{\partial \over \partial y}) f (x_0,y_0)$$.

This identity is true at every point whence the result.