What is the degree of 0 in the context of differential forms? 0 can be considered a smooth function, which is a 0-form.
However, it can also be considered as $0\,dx$ (?) which is a 1-form?
I am curious what is the official degree of 0 as a $k$-form?
Is it the same as in polynomials, where it is either undefined or $-\infty$?
Thanks.
 A: Formally speaking, the zero one-form is not the same as the zero function (for example, because they are sections of different bundles). They are different objects and for each $k$ you have a zero $k$-form. To keep track of those different objects, let me introduce the notation $0_k$ for the zero $k$-form on $M$ (where $M$ is an $m$-dimensional manifold).
When you combine all forms into a graded algebra $\Omega^{*}(M) = \bigoplus_{i=0}^{m} \Omega^{i}(M)$, the usual convention is that a non-zero element $\alpha$ of $\Omega^k(M)$ (identified naturally with the element $(0_0, \dots, 0_{k-1},\alpha,0_{k+1},\dots,0_m)$ of $\Omega^{*}(M)$) is said to be of degree $k$ and the zero elements of $\Omega^{i}(M)$ (which are all identified with the zero element of the algebra $\Omega^{*}(M)$) don't get a degree or get the degree $-\infty$. 
Inside $\Omega^{*}(M)$, there is no "zero-degree one-form" and "zero-degree two-form" because all those objects have been identified with a single object $(0_0,\dots,0_m) = 0$ which is the zero of the algebra and this object doesn't get a degree. Looking at each piece $\Omega^k(M)$ separately, in each piece you have a zero $k$-form and it doesn't have a degree.
