Exterior derivative of a coordinate function

Question

I'm starting to learn about differential forms. From what I understand the coordinate differential forms $dx^1, \dots, dx^n$ are actually the exterior derivatives of the coordinate functions $x^1, \dots, x^n$ (not necessarily Cartesian coordinates). How is this shown?

The only way I know how to take the exterior derivative of a function ($0$-form) $f$ is to use the formula $df = \frac {\partial f}{\partial x^1}dx^1 + \cdots + \frac {\partial f}{\partial x^n}dx^n$, but applying this to a coordinate function $x^1$ would give $$d(x^1) = \frac{\partial x^1}{\partial x^1}dx^1 + \cdots + \frac{\partial x^1}{\partial x^n}dx^n = dx^1$$ which is wholly unconvincing to me. If someone showed me this proof I'd think that they'd come up with the method just because they knew what the answer should be.

I think the problem is that I need a definition of the exterior derivative that doesn't already contain the coordinate functions and then I can confirm that for any arbitrary smooth $f$ the exterior derivative $df$ does in fact equal $\frac {\partial f}{\partial x^1}dx^1 + \cdots + \frac {\partial f}{\partial x^n}dx^n$ where $dx^i$ is the exterior derivative of the coordinate function $x^i$.

So what is the rigorous way to show that the things that we expand the $1$-form $df$ in ($dx^1, \dots, dx^n$ in the usual notation) are the same things that we get be taking the exterior derivative of each coordinate function ($x^1, \dots, x^n$)?

Answer 1

Well, $df$ doesn't make any sense outside exterior derivatives (in this context).

You can see that if $(U,x)$ is a local chart, and $\partial/\partial x^\mu$ are the coordinate basis vectors, then the $dx^\mu$ one-forms are their dual basis. After all, the definition of $d$ on functions is that for an arbitrary vector field $X$, $$ X(f)=df(X). $$ Then, $$dx^\mu\left(\frac{\partial}{\partial x^\nu}\right)=\frac{\partial}{\partial x^\nu}(x^\mu)=\frac{\partial x^\mu}{\partial x^\nu}=\delta^\mu_\nu, $$ thus our proposition is proven. Since we want to work with dual bases, this is why we use the $dx^\mu$s.

Answer 2

If $f\in C^{\infty}(M)$ and $X$ is a vector field on $M^n$, then

$$X(f)=df(X),$$

where in a coordinate chart $(U,x^i)$, $df$ is given by

$$df=\sum \frac{df}{dx^i}dx^i.$$

Note that $\{dx^1,\cdots,dx^n\}$ is the dual basis to the basis of coordinate vectors $\{\frac{\partial}{\partial x^1},\cdots,\frac{\partial}{\partial x^n}\}.$

Advice: I recommend two excellent readings of about differential forms, in my point-view.

1: Differential Forms, by Henri Cartan. This book is ideal for understand differential forms in various contexts, for example, Cartan develops the theory of forms in space of finite and infinite dimension.

2: Differential Forms, by Manfredo do Carmo. This book is ideal to learn the concepts of differential forms to apply in Differential Geometry.

So a quick reading of Manfredo's book will be great for you and your doubts, but I recommend, principally for geometry study, a reading in Cartan's book.

Answer 3

If you have an open subset $O \subset \mathbb{R}^n$ and a smooth function $f: O \rightarrow \mathbb{R}$, then the exterior derivative of $f$ at $x \in O$ is defined to be the linear function $df(x): \mathbb{R}^n \rightarrow \mathbb{R}$ such that, given any $v \in \mathbb{R}^n$, $\langle df(x), v\rangle$ is the directional derivative of $f$ in the direction $v$. More precisely, if $c: (-\delta, \delta) \rightarrow M$ is any smooth curve such that $c(0) = x$ and $c'(0) = v$, then $$ \langle df(x), v\rangle = \left.\frac{d}{dt}\right|_{t=0}f(c(t)) $$ If you apply this definition to the functions $x^i: O \rightarrow \mathbb{R}$, then it is easy to check that given any smooth function $f$, the formula $$ df = \frac{\partial f}{\partial x^1}\,dx^1 + \cdots + \frac{\partial f}{\partial x^n}\,dx^n $$ So this formula is not a definition but a consequence of the definitions of $x^1, \dots, x^n$ as functions and the definition of the exterior derivative of a function I wrote above.