Am I correct with the following reasoning:
$\phantom{f}$
We defined tangent vectors on a manifold $M$ at point $p$ as equivalence classes of paths $\gamma: (-\epsilon, \epsilon) \rightarrow M$ with $\gamma(0) = p$. Two paths $\gamma_{1/2}$ are identified with as same tangent vectors if for all local coordinates x: $(x \circ \gamma_1)'(0) = (x \circ \gamma_2)'(0)$.
I think the following should be the definition for a vector $X$ as a differential operator (in coordinate independent form):
$$ X(f) := Tf(X) $$
Why?:
Let $M,N$ be two n-dim manifolds, $f:M \rightarrow N$. Then by our definition:
$$ Tf(X) = [f\circ \gamma] $$
After choosing local coordinates $(U,x)$ on $M$ and $(V,y)$ on $N$, we define the coordinate function Edit (in bold): $f_C:=\mathbf{y} \circ f \circ x^{-1} : \mathbb{R}^n \rightarrow \mathbb{R}^n$ and coordinate vector $X_C = (x\circ \gamma)'(0) = \begin{pmatrix} X^1 \\ … \\ X^n \end{pmatrix}$ and $p_C = x\circ p$. Then:
$$ Tf_C(X) = (f_C \circ (x \circ \gamma) )'(0) = Df_C((x \circ \gamma)(0)) (x \circ \gamma)'(0) = Df_C(p_C)(X_C) $$
$$= \underbrace{(\frac{\partial}{\partial x_1}f_C(p_C) , …, \frac{\partial}{\partial x_n} f_C(p_C))}_{\text{Jacobi-Matrix}} \begin{pmatrix} X^1 \\ … \\ X^n \end{pmatrix} = X^i \frac{\partial}{\partial x^i}f_C(p_C)$$
Where $\frac{\partial}{\partial x^i}f_C(p_C)$ lives in $y$-Coordinates of $N$.
That's exactly the formula that our professor introduced for $X(f)$ when he said we regard $X$ as a differential operator:
$$ X(f) = X^i \frac{\partial}{\partial x^i} f $$
Only that no distinction between the actual tangent vector/function and the respective coordinate forms is made.
Is the above reasoning correct? Is that a valid definition?
Best Answer
For the map derivative/"Tangent Map"/exterior derivative of a function we have the formula expressing it as a directional derivative: $df(X)=X(f)$, where $X$ is an arbitrary tangent vector.
$df$ is linear so is the $X$ as an operator. So in a coordinate chart, with coordinate curves, and the corresponding basis of the tangent space we have $$df(X)=X(f)=\left(X^i \frac{\partial}{\partial x^i}\right) f=X^i \frac{\partial f}{\partial x^i}$$
Where in the last equality there's some implicit identification reusing the same symbol $f$ for $f(x^1,\dots, x^n)=f\circ p(x^1,\dots, x^n)$, and the directional derivative is given by the usual partial derivative, it can be shown by properly considering the coordinate curves (e.g. here)