What exactly is a 0-form?

Solution 1:

In general, a $k$-form on a vector space $\mathbb{V}$ over a field $\mathbb{F}$ is a multilinear, alternating map that takes $k$ vectors and returns a scalar (an element of $\Bbb F$). So a $0$-form is a map that takes no vectors at all and returns a scalar: We can concretely think of a $0$-form as a map $f : \Bbb F \to \Bbb F$, and for many purposes we may as well just identify this map with the scalar $f(1)$ itself.

On an open set $U \subset \mathbb{R}^n$, a smooth $k$-form is a smoothly varying choice of $k$-form on each tangent space $T_p U$, and so we may identify a smooth $0$-form $f$ with the smooth function given by mapping each $p$ to the scalar corresponding to the $0$-form $f_p$.

(To make clear the distinction between these two kinds of objects, we occasionally call the latter a $k$-form field on $U$.)

Solution 2:

A zero form is a smooth function defined on a manifold. Here smoothness means that the map $U\rightarrow \mathbb{R}$ is smooth by considering the smoothness of chart maps $\phi:\mathbb{R}^{n}\rightarrow U\rightarrow \mathbb{R}$. In general an $n$-form is a section of the anti-symmetric cotangent tensor bundle $\wedge^{n}(TM^{*})$.

Solution 3:

The notion of a '0-form' is just a convenient device. Basically k-form means for a fixed point $x$ you pick k-vectors starting at this point $x$ and pointing elsewhere. That is, you pick k number of vectors from the tangent space of some point $x$

$$f(x)\{(x;v_1), ... , \underbrace{(x; v_k)}_{\text{we pick k vectors starting at point x and ending at point $v_k$ of some $\mathcal{R}^n$}}\}$$, which is a real-valued function. Zero means we pick no vectors from the tangent space, we assign to x some scalar (=real) value.