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Integration of closed forms

  1. Aug 10, 2011 #1
    Hi guys!
    I am reading a paper which uses closed forms [tex] \omega [/tex] on a p-dimensional closed submanifold [tex] \Sigma[/tex] of a larger manifold [tex]M[/tex]. When we integrate [tex]\omega[/tex] we get a number
    [tex] Q(\Sigma) =\int _{\Sigma}\omega [/tex] which, in principle, depends on the choice of [tex]\Sigma [/tex] but because [tex] \omega [/tex]is closed, [tex] Q(\Sigma)[/tex] is said to be unchanged by continuous deformations of [tex]\Sigma[/tex]. The converse is supposed to be true as well.

    Now, I understand this as a generalization of Gauss' law in electrostatics (three dimensions) but I only remember that for this particular case, the demonstration I saw was purely geometrical. I know that this has to use the fact that [tex] \Sigma [/tex] does not have a boundary (closed submanifold). How exactly is this proved?
     
    Last edited: Aug 10, 2011
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  3. Aug 10, 2011 #2

    Ben Niehoff

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    If [itex]\Sigma[/itex] is continuously deformed to [itex]\Sigma'[/itex], then we have that

    [tex]\Sigma - \Sigma' = \partial \mathcal{R}[/tex]

    where [itex]\mathcal{R}[/itex] is the (p+1)-dimensional region enclosed by [itex]\Sigma[/itex] and [itex]\Sigma'[/itex], and [itex]\partial[/itex] is the boundary operator. Applying the boundary operator again and using [itex]\partial^2 = 0[/itex], we get

    [tex]\partial \Sigma - \partial \Sigma' = 0[/tex]

    which means that the continuous deformations of [itex]\Sigma[/itex] must leave its boundary unchanged (if it has a boundary). This is analogous to Ampere's law: the B flux through a loop is proportional to the current enclosed, and the B flux can be evaluated on any surface having the loop as its boundary.

    Then we can apply the generalized Stokes' theorem:

    [tex]\int_{\Sigma - \Sigma'} \omega = \int_{\partial \mathcal{R}} \omega = \int_{\mathcal{R}} d\omega = 0[/tex]

    Finally, we have, by linearity,

    [tex]\int_{\Sigma} \omega = \int_{\Sigma'} \omega[/tex]
     
  4. Aug 12, 2011 #3
    hi thanks! that is partly what I wanted. However, this argument still doesn't use the fact that [tex]\Sigma [/tex] is a closed submanifold, that is, its boundary is the empty set. I think it should be important.
     
  5. Aug 13, 2011 #4

    Ben Niehoff

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    It's not important, as you can see in my argument above. It is simpler to assume that [itex]\Sigma[/itex] has no boundary, but it is not necessary, as long as [itex]\Sigma'[/itex] has the same boundary. That is, any continuous deformation which leaves the boundary unchanged is allowed.
     
  6. Aug 13, 2011 #5

    lavinia

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    The general theorem is Stoke's theorem if the deformation is smooth. The two bounding manifolds of the deformation are homologous so a cocycle will have the same value on each. The closed form is a cocycle in the de Rham complex of the larger manifold.
     
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