Confused about differential forms and operators. I am looking for concrete examples of m-forms, which are really hard to find on the internet for some reason.
This is what confuses me specifically: a 0-form is defined to be a function. Therefore, in $\mathbb{R}^3$, a 0-form is, for example, $f(x, y, z) = x^2 + y^2 + z^2$. But, a 1-form on the same set could be $\omega(\langle dx, dy, dz \rangle) = f dx + f dy + f dz$. Or, at least, that is how it's commonly written. Now, this doesn't make much sense to me, as $f$ should take some inputs in. What is inputted into $f$ in this case?
Also, consider the 0-form $g(x, y, z) = x + y + z$ and the 1-form $\omega(\langle dx, dy, dz \rangle) = dx + dy + dz$. Both forms represent the exact same thing, no? How can we differentiate between 0-forms and 1-forms if every 0-form can be written equivalently as a 1-form? I suppose I've really misunderstood something here, but I've been trying to figure it out for days.
On a slightly related note, there are these functions $x^i$ defined in this way: $x^i(\langle  x_1, ..., x_i, ..., x_n \rangle) = x_i$. After the operator $d$ acts on this function, we get a function $dx^i$ defined as: $dx^i(\langle x_1, ..., x_i, ..., x_n \rangle) = x_i$. But this is the exact same thing, is it not? I don't really understand how this operator even acts on 0-forms, probably because I don't understand 0-forms.
Any explanation of these things would be greatly appreciated.
 A: First of all, you should note that there is a difference between a covector and a covector field (although, in the literature you'll see the two used interchangeably when the context is clear). A covector is just an element of the dual space. A covector field is a covector-valued function esssentially. This is analogous to how there's a difference between a real number and a real valued function. A more precise definition is given as below.
A function is a map from the manifold to the reals $f:M \to \mathbb{R}$
A covector field (or a one-form) is a map (more specifically, a section) from the manifold to the cotangent bundle:
$$w: M \to T^*M$$
So in your case, the one form should read $$w(p) = f(p)dx|_p+f(p)dy|_p+f(p)dz|_p$$
Where the $dx|_p$ refers to the covector basis element in $T^*_pM$ (as do the others as well). So notice that the input is a point $p$ on the manifold $M$, and the output is a covector, which is an element of $T^*M$. The fact that such a map should be a section means $w(p)$ shouldn't be an arbitrary element of $T^{*}M$ but rather should specifically be in $T^*_pM$. The intuition should be similar to how you visualize a vector field in the plane, say: Given a point $(x,y)$ in the plane, you draw the vector emanating from that same point and not from any other point. Above every point you should visualize there exist a tangent plane/space (and a cotangent space), a covector field is an assignment to each point in your space/manifold, an element of the cotangent space directly associated with that particular point, i.e. $w(p) \in T^*_pM$.
So as you can see the one forms and zero forms are quite different objects; essentially a zero form is a real valued function on the manifold, while covector fields are covector-valued functions on the manifold.
Edit:
The operator $d$ acts on functions and gives you a covector field. More generally, $d$ is called the exterior derivative and acts on n-forms to give you n+1 forms; but to answer your question taking n$=0$ will suffice.
Fix a point $p$ on the m-dimensional manifold; then as you know the tangent space $T_pM$ is spanned by the basis vectors $\dfrac{\partial}{\partial x_i}$ with $1 \le  i \le m$.
Then define the dual basis to be $w_i$ such that $w_i (\dfrac{\partial}{\partial x_j})=\delta_{ij}$
Note that I'm not going to bother with contravariant vs covariant indices since they're beside the point.
These dual basis elements turn out to be exactly the differentials of the coordinate functions (i.e. $w_i = dx_i$ where $dx_i$ is the differential of the coordinate function $x_i$).
this is shown as follows:
Note that the differential of a function $f$ on a manifold looks like the following in coordinates: $$df = \dfrac{\partial f}{\partial x_i}w_i$$ Notice if we apply the above definition for the particular case where $f=x_j$ we get:
$$dx_j = \dfrac{\partial x_j}{\partial x_i}w_i = \delta_{ji}w_i=w_j$$
Hence $$w_j=dx_j$$
That is, the differential of the i-th coordinate function is the i-th dual basis covector. Hence why we just write the dual basis as $dx_i$
