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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?

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    I'm not sure I understand what the confusion is here: What is the "opposite direction"? What you've outlined is that a smooth vector field on $M$ actions on smooth functions via a particular coordinate formula, but that it also acts on 1 forms via $X \cdot \omega = \omega(X)$ (just the natural pairing), and that these two actions are related by $Xf = df(X)$. (In fact, this latter formula shows that our expression for $Xf$ does not depend on a choice of coordinate chart.) – 2017-02-28
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    These two actions are different in a key way, however: The action of $X$ on $f \in C^{\infty}$ yields a function $Xf$, and the value of $Xf$ at a point depends on the first derivative of $f$. On the other hand, the action of $X$ on a $1$-form $\omega$ produces a function $\omega(X)$, but the value of this function at a point depends only on the value of $\omega$ at that point, i.e., unlike the action of $X$ on $C^{\infty}(M)$ does not depend on the derivatives of the object on which it is acting. – 2017-02-28
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    I understand how the field acts it just seems strange to me that a vector field can be defined as $X : C^{\infty}(M) \rightarrow C^{\infty}(M)$ and $X : \Omega^1(M) \rightarrow C^{\infty}(M)$. Is that not a problem? – 2017-02-28
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    Not every map $C^{\infty}(M) \to C^{\infty}(M)$ is a vector field! Among these only the maps that are also derivations, that is, satisfy $X(fg) = (Xf)g + f(Xg)$ are vector fields (this is sometimes taken to be the definition of vector field). Likewise, not all maps $\Omega^1(M) \to C^{\infty}(M)$ are vector fields: Only the maps that are linear on each fiber $T^*_x M$ of the bundle $\Omega^1(M) \to M$ (and vary smoothly in $x$) are vector fields. – 2017-02-28

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