How Can You Find the Inverse of the Exterior Derivative?

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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##
 
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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.
 
Hello! There is a simple line in the textbook. If ##S## is a manifold, an injectively immersed submanifold ##M## of ##S## is embedded if and only if ##M## is locally closed in ##S##. Recall the definition. M is locally closed if for each point ##x\in M## there open ##U\subset S## such that ##M\cap U## is closed in ##U##. Embedding to injective immesion is simple. The opposite direction is hard. Suppose I have ##N## as source manifold and ##f:N\rightarrow S## is the injective...

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