Is df a 1-form or a "1-form field"?

Question

Derivatives can be defined as $df/dx=f'$ iff $df=f' dx$ where $df$ is the exterior derivative of the scalar field $f$. The exterior derivative is supposed to map differential $k$-forms to differential $(k+1)$-forms but $df$ isn't a differential 1-form because a 1-form maps a vector to a scalar but $df$ maps a vector to a scalar field. Furthermore, $df(x)$ and $dx$ would be true 1-forms and $df$ would be a one-form field, right? Or is it true that $df$ maps a vector to a scalar?

Answer 1

The exterior derivative $\mathrm df$ indeed is a field. Note however that it maps a vector field to a scalar field. At each point $x$, $\mathrm df(x)$ maps the vector $v(x)$ living in the tangential vector space at $x$ to the scalar $\phi(x)=\mathrm df(x)(v(x))$. Different points have different (but isomorphic) tangential vector spaces.

Answer 2

I guess it's both. If one sees $df$ as a function of point $p$ and displacement $dx$, then we can use the symbol to denote both the function $df$ (which is a field) and the dependent variable $df_p$ (which is the '1-form' as you call it).

Actually 1-form is a covariant vector field, a section of alternating covariant 1-tensor space. I guess what you mean by saying '1-form' is an element (instead of section) of alternating covariant 1-tensor space.