Firstly, I have to apologize if my question is very trivial.
My question is about differential forms; what does it mean to say that a differential form on a smooth manifold can be viewed as a smooth section of the exterior power of the cotangent bundle and that this is the geometric way to view it. Does this mean that there is an isomorphism between the space of differential forms on a smooth manifold $M$ and the space of smooth sections of the exterior power of the cotangent bundle $\Gamma(M, \wedge T^*(M))$, if so my thought is that this map associates to a differential form $\alpha \in \mathcal{A}^k(M)$, the smooth section $\tilde{\alpha}$ which is given by $$(\tilde{\alpha}(m))(v_1,...,v_k):= \alpha_m(v_1,...,v_k),$$ where $m \in M$ and $ v_1,...,v_k$ are tangent vectors in $T_mM$ ?
- What it means to say that this is the geometric way to view differential forms?
Edit(Clarification of notations): $\mathcal{A}^k(M)$ denotes the space of differential forms on $M$: we say that $\alpha \in \mathcal{A}^k(M)$ if it is a multilinear, antisymmetric function which takes as input $k$ smooth vector fields on $M$, and outputs a scalar function on $M$.
On the other hand $\Gamma(M, \wedge^kT^*(M)$ denotes the space of smooth sections on the exterior power of the cotangent bundle: We say that $s\in \Gamma(M, \wedge^kT^*(M)$ if $s$ is a smooth map which takes as input an element of $ m \in M$ and outputs a $k$ multilinear antisymmetric on $s_m :T_mM \times ...\times T_mM \rightarrow \mathbb{R}$.