Differential Forms, Exterior Derivative

Question

I have a question regarding differential forms.

Let $\omega = dx_1\wedge dx_2$. What would $d\omega$ equal? Would it be 0?

Answer 1

Yes. The same holds true for any differential form whose coefficients are constant functions. For example, if $\omega = 3(dx\land dy) + 5(dx\land dz) + 7 (dy\land dz)$, then $d\omega = 0$.

Edit: In general, the exterior derivative is defined by $$ d\bigl(f\, dx_{i_1}\land\cdots\land dx_{i_n}\bigr) \;=\; df\land dx_{i_1}\land\cdots\land dx_{i_n} $$ where $$ df \;=\; \frac{\partial f}{\partial x_1}dx_1 + \cdots + \frac{\partial f}{\partial x_n}dx_n $$ For example, in three dimensions $$\begin{align*} d\bigl(x^3y^2 z^4 dy\bigr) \;&=\; (3x^2y^2z^4 dx + 2x^3yz^4 dy + 4x^3y^2z^3 dz)\land dy \\ &=\; 3x^2y^2z^4 dx\land dy \,-\, 4x^3y^2z^3 dy\land dz \end{align*}$$ Note that the $2x^3yz^4$ term goes away since $dy\land dy = 0$.

Also, the exterior derivative of a sum of forms is the sum of the exterior derivatives of the forms, i.e. $$ d(z\,dx + x^2\,dy) \;=\; 2x\, dx\land dy \,-\, dx\land dz $$

Answer 2

Recall that $$d(\alpha \wedge \beta) = d\alpha \wedge \beta + (-1)^{\deg \alpha} \alpha \wedge d\beta.$$

Then \begin{align*} d\omega & = d(dx_1 \wedge dx_2) \\ & = ddx_1 \wedge dx_2 - dx_1 \wedge ddx_2 \\ & = 0 - 0 \\ & = 0, \end{align*} since $d^2 = 0$.

Answer 3

The differential form $\omega = dx_1 \wedge dx_2$ is constant hence we have $$ d\omega = d(dx_1 \wedge dx_2) = d(1) \wedge dx_1 \wedge dx_2 \pm 1 \, ddx_1 \wedge dx_2 \pm 1 \, dx_1 \wedge ddx_2$$ and because $d^2 = 0$, we have $$ d \omega = 0.$$