This is probably a very silly question, but I'm curious to see if this has an interesting interpretation.
Let $M$ be a smooth $n$-dimensional parallelizable manifold. Since the tangent bundle is trivial, we get $TM \cong M \times \mathbb{R}^n$. So any vector field $X \in \Gamma(TM)$ can be expressed, via this isomorphism, as a map $X: M \to M \times \mathbb{R}^n$ sending $p \in M$ to, say, $(p, x(p))$ for a smooth function $x : M \to \mathbb{R}^n$. And WLOG, we might as well identify $X$ with $x$.
With this setup, we get $x = (x^1, \ldots, x^n)$ for $n$ functions $x^i \in C^\infty(M)$. Then my question is as follows.
Given $f \in C^\infty (M)$ and $X \in \Gamma(TM)$, how may we express $X(f)$ as "$x(f)$", that is, in terms of the functions $x^i$?
To clarify this confusing question, I wish to see if we can express the function $X(f)$ in terms of the component functions of the identified "vector field" $x$, in such a way that all the usual rules of vector fields follow, i.e. Leibniz rule and linearity.
So I would want a set of rules that defines $x(f)$ while satisfying
- $x(fg) = fx(g) + x(f)g$ for smooth functions $f,g$,
- $x(cf) = cx(f)$ for constants $c$.
My motivation is to see if there is a simpler way to take directional derivatives over parallelizable manifolds without using any local arguments.