Differential Forms, Exterior Derivative I have a question regarding differential forms.
Let $\omega = dx_1\wedge dx_2$.  What would $d\omega$ equal?  Would it be 0?
 A: 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 $d^2 = 0$.
A: 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.
$$
d(z\,dx + x^2\,dy) \;=\; 2x\, dx\land dy \,-\, dx\land dz
$$
A: 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 $$ d \omega = 0.$$
