What does $X(f)$ mean (where $X$ is a vector field and $f$ is a function?
What I am referring to is for example the section Vector Field On Manifolds in the Wikipedia article.
If a vector field $X$ is a map from a manifold $M$ to its tangent bundle $TM$ and a funcrion $f$ is a map $C^\infty(M) \rightarrow \mathbb{R}$, how can we create $X(f)$?
Do I interpret right that the $f$ is like an "argument" of $X$ here? I thought that if $f$ "returns" a scalar from $\mathbb{R}$, but the $X$ takes elements from the manifold $M$ as arguments, how can it take $f$ as an argument and what does it do with it?
I have also seen $X$ defined as a (linear) map $X :C^\infty(M) \rightarrow C^\infty(M)$ , which confuses me even more. How does that correspond to the other definition that uses $X: M \rightarrow TM$? If $C^\infty(M)$ means just infinitely-differentiable maps from $M$ to $\mathbb{R}$, does that mean that what $X$ "returns" is still such a map?
Summary: I hope my question is clear. I am just confused about what $X$ takes as an argument (where does the argument live and what it is) and what $X$ returns and where does the outcome live. This confusion stems from seing an expression $X(f)$ which doesnt make sense to me.
Thank you.
Notice that $X$ must not only be a linear endomorphism on $C^\infty(M)$ but a derivation: for any two smooth functions $f,g$, one has that $X(fg) = X(f)g + f X(g)$. This captures the fact that if $X$ is a field, then $X(f)$ is the derivative of $f$ along $X$.
In case that $X$ is given as a map $X : M \longrightarrow TM$, you can define $X(f)$ as follows: for each $p\in M$, the tangent vector $X_p$ is just the datum of a vector (at the point $p$) so you may compute the directional derivative of $f$ at $p$ along $X_p$:
$$X(f)_p = df_p(X_p).$$
This is linear on scalars (because $df_p$ is linear), and we have that
$$X(fg)_p = d(fg)_p(X_p) = f(p) dg_p(X_p) + g(p) df_p(X_p),$$
so this gives an assignment that turns a field $X : M\longrightarrow TM$ into a derivation $X : C^\infty(M) \longrightarrow C^\infty(M)$.