I have searched the web for an answer to this question but couldn't quite find what I was looking for.
Given a Manifold $M$ and a choice of local coordinates $\lbrace{x}^{i}\rbrace$ one can give the expression for a tangent vector $V \in {T}_{p}M$ in these local coordinates as
$$V={V}^{i} {\partial}_{i}$$
Now a vector can be defined as a linear operator from smooth functions on $M$ to the reals, i.e. ${C}^{\infty} \left( M \right) \rightarrow \mathbb{R}$ and this is a concept I am completely fine with, however my trouble comes when considering the action of Vectors on one forms.
Take a one form in the same coordinate system $$\omega = {\omega}_{i} d{x}^{i}$$ where ${\omega}_{i} \in {C}^{\infty} \left( M \right)$ then what is the action of the vector on this one form? Is it defined? I know that from the usual construction of one forms as elements of the dual space of ${T}^{*}_{p}M$ allows us to think of one forms as $\mathbb{R}$-linear maps ${T}_{p}M \rightarrow \mathbb{R}$ with the following relation between the dual basis elements $$ d{x}^{i} \left( {\partial}_{j} \right) = \frac{\partial {x}^{i}}{\partial {x}^{j}}= {{\delta}^{i}}_{j}$$ Where the action of the basis element on an arbitrary vector $$d{x}^{i} \left( V \right) = {V}^{j} \frac{\partial {x}^{i}}{\partial {x}^{j}}={V}^{i}$$ Which is just the directional derivative of the $i$th coordinate function along $V$ and so this gives us our action
$$ \omega \left( V \right) = {\omega}_{i} d{x}^{i} \left( V \right) = {\omega}_{i}{V}^{i}$$ Is there a well defined action for $V \left( \omega \right)$? If not is there a reason why this action is not well defined and thus can't be constructed? My apprehension is derived from the fact the basis elements of $T_{p}M$ are differential operators and naturally obey a Leibniz law.
Thanks for reading and any responses will be greatly appreciated!
