I am trying reach an understanding of precisely how the space of differential forms is related to the space of vector fields. These are the definitions that I understand and am using for these objects: Let $X$ be an open subset of $\mathbb{R}^n$. A vector field $v$ on $X$ is a mapping that assigns to each point $p \in X$ to an element $v(p) \in T_pX$ of the (geometric) tangent space $T_pX = \{(p, x) | x \in \mathbb{R}^n\}$. A one form $\omega$ on $X$ is a mapping that assigns to each point $p \in X$ an element $\omega(p) \in T^*_pX$, the cotangent space of $X$ at $p$ which is defined as the algebraic dual of the tangent space at $p$. If $C^1(X)$ denotes the set of all continuously differentiable functions on $X$ we can form the ring of $C^1$ functions using the standard pointwise addition and multiplication of functions. We can then form the two $C^1(X)$-modules $\mathcal{V}^1(X)$ and $\Omega^1(X)$ whose underlying groups are the set of all $C^1$ vector fields on $X$ and all differential forms on $X$, respectively.
Finally, now, I can ask my question: what is the precise relation between the modules $\mathcal{V}^1(X)$ and $\Omega^1(X)$?
My thoughts: I notice that a differential form $\omega$ carries a point $p$ to the cotangent space at $p$ while a vector field $v$ will carry the same point $p$ to the tangent space. We can't really form the composition, per-se of $\omega$ and $v$ since the both have the same domain $X$ but for $v(p) \in T_pX$ we can evaluate $\omega(p)$ at $v(p)$ since the domain of $\omega(p)$ is $T_pX$. This looks a little like a dual pairing; can it be that $\Omega^1(X)$ and $\mathcal{V}^1(X)$ are dual modules in some sense?