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Exterior derivative problem

  1. Dec 5, 2011 #1
    I'm trying to work out that [itex]d^{2}=0[/itex] for any form, by applying it to a vierbein field [itex]e^{a}_{μ}[/itex], so for the first exterior derivative I get [itex](de)_{μ\nu}=∂_{μ}e_{\nu}-∂_{\nu}e_{μ}[/itex] (ignoring the differentials for now). Now here is where I get stuck. When you apply d again to this, I have no idea which indices to switch around, or moreover, what it even looks like. Could anyone help me on this?
  2. jcsd
  3. Dec 6, 2011 #2

    Ben Niehoff

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    It would be much easier to see what's going on, in fact, if you did not choose to "ignore the differentials". If you have some 1-form [itex]A = A_\mu \, dx^\mu[/itex], then

    [tex]dA = \partial_\nu A_\mu \, dx^\nu \wedge dx^\mu[/tex]

    [tex]ddA = \partial_\lambda \partial_\nu A_\mu \, dx^\lambda \wedge dx^\nu \wedge dx^\mu[/tex]
    Then it is easy to see that [itex]ddA = 0[/itex] because [itex]\partial_\lambda \partial_\nu[/itex] is symmetric, whereas [itex]dx^\lambda \wedge dx^\nu[/itex] is antisymmetric.
  4. Dec 6, 2011 #3


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    In short, you're saying you don't know how to write the exterior derivative of something with 2 lower indices?
  5. Dec 6, 2011 #4
    By the way, the fact that d^2 is 0 ought to be seen as a consequence of Stokes theorem. Just apply it twice. That gives you that the integral of d^2w over anything vanishes, since the boundary of a boundary is empty. So, d^2w itself must vanish.

    It's good to do the calculation, too.
  6. Dec 7, 2011 #5
    or maybe generalized stoke's theorem on manifolds should be seen as a consequence of this? to write the usual stoke's theorem on euclidean space you need to use divergence etc which you can do by composing this with hodge star operator etc
  7. Dec 7, 2011 #6
    Yes, the exterior derivative is sort of a generalization of curl and divergence to higher dimensions. But the general Stokes theorem, I would see as coming from the same sort of argument as the old Stokes theorem. Actually, differential forms make the actual proof much much better. The rough, heuristic idea is that the exterior derivative is the integral over an infinitesimal parallelepiped. So, if you want to integrate over a boundary, you break up the interior into little tiny parallelepipeds. So, the result is that you integrate the exterior derivative over the interior.
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