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Inverse of exterior derivative

  1. Mar 24, 2014 #1
    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##
     
  2. jcsd
  3. Mar 24, 2014 #2

    Ben Niehoff

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    Generally what you want to do is solve

    [tex]d \alpha = \beta[/tex]
    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

    [tex]d^2 \alpha = d \beta = 0[/tex]
    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.
     
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