Formula of 1 form in R3

Question

I want to solve the following problem: Given $d ◦ d = 0$, d acting on $0$-form and $1$-form, I want to show that both $1$-form and $0$-form yield $0$.

I finished in the case of $0$-form using that the following equation; $df =(∂f/∂x)dx + (∂f/∂y)dy + (∂f/∂z)dz$.

But I have no clue whatsoever in the case of 1$$-form.

Does 1$$-form looks like $fdx\land dy\land dz$?

Answer 1

A generic 1-form on $\Bbb R^3$ has the form $$ \omega = f_1(x, y, z)~dx + f_2(x, y, z) ~dy + f_3 (x, y, z) ~dz, $$ where one typically assumes that the $f_i$ are differentiable functions.

Answer 2

One the other properties of $d$ is that it is a derivation. Suppose $\omega = f(x,y,z)\,dx$. Then \begin{align*} d\omega &= df \wedge dx = \left(\frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy + \frac{\partial f}{\partial z}dz\right)\wedge dx \\ &= \frac{\partial f}{\partial y}dy\wedge dx + \frac{\partial f}{\partial z}dz \wedge dx \end{align*} So \begin{align*} d^2 \omega &= d\left(\frac{\partial f}{\partial y}\right) \wedge dy \wedge dx +d\left(\frac{\partial f}{\partial z}\right) \wedge dz \wedge dx \\ &= \left(\frac{\partial^2 f}{\partial x\,\partial y}dx +\frac{\partial^2 f}{\partial y^2}dy + \frac{\partial^2 f}{\partial z\,\partial y}dz\right) \wedge dy \wedge dx +\left(\frac{\partial^2 f}{\partial x\,\partial z}dx + \frac{\partial^2 f}{\partial y\,\partial z}dy + \frac{\partial^2 f}{\partial z^2}dz\right) \wedge dz \wedge dx \\ &= \frac{\partial^2 f}{\partial z\,\partial y}dz\wedge dy \wedge dx +\frac{\partial^2 f}{\partial y\,\partial z}dy\wedge dz \wedge dx \end{align*} If $f$ is $C^2$, then the two partial derivatives above are equal. Since \begin{align*} dz \wedge dy \wedge dx &= - dx\wedge dy \wedge dz \\ dy \wedge dz \wedge dx &= dx\wedge dy \wedge dz \end{align*} the two terms sum to zero.