Can vector fields be thought of as rank-1 contravariant tensor fields?

Question

I have some confusion regarding the definition of vector fields and how they act on smooth functions and 1-forms. Let $M$ be a smooth manifold. Let $X$ be a smooth vector field on $M$ and let $f \in C^{\infty}(M)$. The action of $X$ on $f$ is given by $$ Xf = X^i \frac{df}{dx^i} \in C^{\infty}(M) $$ Thus, applying $X$ to a $C^{\infty}$ function (0-form) produces another $C^{\infty}$ function. However, it was my understanding that a vector field may be thought of as a rank 1 contravariant tensor field. Thus, I would expect that a vector field would be defined by $X : \Omega^1(M) \rightarrow C^{\infty}(M)$, i.e. applying $X$ to a 1-form would produce another $C^{\infty}$ function. It is certainly the case that applying a 1-form $df$ to a vector field $X$ produces a $C^{\infty}$ function via $$ df(X) = X^i \frac{df}{dx^i} $$ so why does it not seem to be the case in the opposite direction? Am I misunderstanding something about vector fields?

Answer 1

Question: "..so why does it not seem to be the case in the opposite direction? Am I misunderstanding something about vector fields?"

Answer: If $M$ is connected the Serre-Swan theorem says (wikipedia)

"Another way of stating the above is that for any connected smooth manifold M, the section functor Γ from the category of smooth vector bundles over M to the category of finitely generated, projective C∞(M)-modules is full, faithful, and essentially surjective. Therefore the category of smooth vector bundles on M is equivalent to the category of finitely generated, projective C∞(M)-modules. Details may be found in (Nestruev 2003)."

You may use the Serre-Swan theorem to prove the following:

Example: If $\Omega^1:=\Omega^1(M)$ is the module of smooth sections of the cotangent bundle on $M$ and $R:=C^{\infty}(M)$ the ring of smooth real-valued functions on $M$, $T$ the module of sections of the tangent bundle $T(M)$ of $M$ it follows $\Omega^1, T$ are finite rank projective $R$-modules and there is an isomorphisn

$$\rho: T\cong Hom_R(\Omega^1, R),$$

hence for any global vector field $X: R \rightarrow R$ there is an $R$-linear map $\phi_X: \Omega^1 \rightarrow R$ with $\phi_X \circ d =X$. Here $d: R \rightarrow \Omega^1$ maps a smooth function $f$ to its differential form $df$. You must write down the map $\rho$ and understand why it is an isomorphism. For a connected manifold $M$, the module $\Omega^1$ is finite rank and projective and it follows

$$(\Omega^1)^{**}\cong T^* \cong \Omega^1.$$