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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.

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  • $\begingroup$ 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 $\endgroup$
    – Callum
    Commented 2 days ago

2 Answers 2

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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.

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    $\begingroup$ 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. $\endgroup$
    – Arthur
    Commented 2 days ago
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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.

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