TITLE: Relationship Between Differential Forms and Vector Fields
QUESTION [7 upvotes]: 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?

REPLY [2 votes]: If you need to calculate in the local coordinates with a metric, then you need to use the musical isomorphism. You can treat it purely algebraically if you like, but if you want to do calculation in an actual manifold this is needed.