if $\omega$ is a $2$-form, and $\Bbb d \omega = 0$ what can we conclude? 
Let $f_1, f_2, f_3$ be smooth functions on an open subset $\Omega \subset \Bbb R ^3$, which contains the standard cube $I^3$. We define the differential $2$-form
$$ \omega = f_1 \Bbb d x^2 \wedge \Bbb d x^3 - f_2 \Bbb d x^1 \wedge \Bbb d x^3 + f_3 \Bbb d x^1 \wedge \Bbb d x^2 .$$
Assume that $\Bbb d \omega = 0$. What conclusions can we draw on the functions $f_i, \ i = 1, 2, 3$?

I have calculated $\Bbb d \omega = D_1f_1 + D_2f_2 + D_3f_3 $, but I am not sure what I can conclude if the sum is equal to $0$. Any help please.
 A: I suspect your conclusion is what they want; i.e. the divergence of f is zero.
The definition of $\omega$ above is just the hodge star applied to $d\eta$ where we define $\eta = f_1 dx^1 + f_2 dx^2 + f_3 dx^3$ then if follows that
$$\omega = *\eta$$ 
and so
$$ 0 = d\omega = d*\eta = (D_1 f_1 + D_2 f_2 + D_3 f_3)dx^1 \wedge dx^2 \wedge dx^3$$
Conclusion: the divergence of the vector field formed by $\left( \begin{array} {c} f_1 \\ f_2 \\ f_3 \end{array} \right)$ is zero .
Hope this helps!
A: I suspect that what the author wants you to think about is Poincaré's lemma, i.e. you are expected to conclude that there exist $\eta = \alpha \ \Bbb d x^1 + \beta \ \Bbb d x^2 + \gamma \ \Bbb d x^3$ such that $\omega = \Bbb d \eta$. Unfortunately, you may not draw this conclusion if $\Omega$ is not contractible, so this assumption should be added to your statement. (The hypothesis that $I^3 \subset \Omega$ remains mysterious to me.)
Once you apply Poincaré's lemma, you get that (remember that $\Bbb d x^i \wedge \Bbb d x^i = 0$ and that $\Bbb d x^i \wedge \Bbb d x^j = - \Bbb d x^j \wedge \Bbb d x^i$):
$$\omega = \Bbb d (\alpha \ \Bbb d x^1 + \beta \ \Bbb d x^2 + \gamma \ \Bbb d x^3) = \Bbb d \alpha \wedge \Bbb d x^1 + \Bbb d \beta \wedge \Bbb d x^2 + \Bbb d \gamma \wedge \Bbb d x^3 = \\ \left( \frac {\partial \gamma} {\partial x^2} - \frac {\partial \beta} {\partial x^3} \right) \Bbb d x^2 \wedge \Bbb d x^3 + \left( \frac {\partial \gamma} {\partial x^1} - \frac {\partial \alpha} {\partial x^3} \right) \Bbb d x^1 \wedge \Bbb d x^3 + \left( \frac {\partial \beta} {\partial x^1} - \frac {\partial \alpha} {\partial x^2} \right) \Bbb d x^1 \wedge \Bbb d x^2 .$$
One the other hand, $\omega = f_1 \Bbb d x^2 \wedge \Bbb d x^3 - f_2 \Bbb d x^1 \wedge \Bbb d x^3 + f_3 \Bbb d x^1 \wedge \Bbb d x^2$, so equating the two expressions for $\omega$ you get that
$$f_1 = \frac {\partial \gamma} {\partial x^2} - \frac {\partial \beta} {\partial x^3}, \quad f_2 = - \frac {\partial \gamma} {\partial x^1} + \frac {\partial \alpha} {\partial x^3}, \quad f_3 = \frac {\partial \beta} {\partial x^1} - \frac {\partial \alpha} {\partial x^2} .$$
In a more traditional notation, this means that $(f_1, f_2, f_3) = \text{rot} (\alpha, \beta, \gamma) = \nabla \times (\alpha, \beta, \gamma)$.
