How Can You Find the Inverse of the Exterior Derivative?

[Jhenrique]

If given an one-form like: $\omega = u dx + v dy$, dω is $d\omega = \left ( \frac{\partial v}{\partial x} - \frac{\partial u}{\partial y}\right )dxdy$. So, is possible to make the inverse path?

Given: $d\omega = Kdxdy$ , which is the expression for ω ?

$\omega = ? dx + ?dy$

 

[Ben Niehoff]

Generally what you want to do is solve

d \alpha = \beta
for a $p$-form $\alpha$ and a $(p+1)$-form $\beta$. In order for this equation to have a solution, it must be consistent with $d^2 = 0$. So you must have

d^2 \alpha = d \beta = 0
If that is true, then you can always find a local solution. However, you might not find a global solution if your manifold has nontrivial topology.