Why is $\cos(x)dx$ a differential form

Question

I am trying to understand the concept of the differential

A differential $d$ is a map that sends functions to 1-forms

A preliminary example is

$$d\sin(x) = \cos(x) dx$$

So the operator $d$ sends a function $\sin(x)$ to a differential form $\cos(x)dx$

My question is that while $dx$ is obviously a differential form, but $\cos(x)$ is just a function.

What makes a function multiplying a differential form a differential form? i.e why would $\cos(x)dx$ be a differential form when the only identifiable differential form is the quantity $dx$

Answer 1

Differential forms can be added, multiplied by scalars, and given a product, namely the wedge product. Let $E^k (M)$ denote the set of all smooth $k$-forms on $M$, a manifold.

Proposition: $E^0 (M)$, the set of all smooth 0-forms, can be identified with $C^\infty (M)$.

Forms can be given a multiplication, namely the wedge product. In your case, you have $\cos (x) \in E^0 (M)$ and $dx \in E^1 (M)$. Writing $\cos (x) dx$ is actually shorthand for writing $\cos (x) \wedge dx \in E^{0 + 1} (M)$ and hence is a 1-form.

Answer 2

(Just learning all this myself, so if something seems wrong it probably is.) A $k$-form is a function that takes $k$ vectors and returns a scalar.

$$\cos(x)\,dx$$

is a $1$-form -- its value evaluated on the vector $(a, b, c)$ is

$$a\cdot\cos(a)$$

A one-form, let's say on $\mathbb{R}^3$:

$$f_1(x, y, z)\,dx + f_2(x, y, z)\,dy + f_3(x, y, z)\,dz$$

is called differential if $f_1, f_2, f_3$ are all differentiable. So $\cos(x)\,dx$ is a differential form.

It's true that $dx$ by itself is a differential form, and it's true that $\cos(x)\,dx$ is also a differential form. It just has a scalar in front of it.