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Divergence Theorem on Manifolds

  1. Mar 17, 2010 #1
    Hi,

    I'm having some trouble understanding this theorem in Lang's book, (pp. 497) "Fundamentals of Differential Geometry." It goes as follows:

    [tex] \int_{M} \mathcal{L}_X(\Omega)= \int_{\partial M} \langle X, N \rangle \omega [/tex]

    where [tex] N [/tex] is the unit outward normal vector to [tex] \partial M [/tex], [tex] X [/tex] is a vector field on [tex] M [/tex], [tex] \Omega [/tex] is the volume element on [tex] M [/tex], [tex] \omega [/tex] is the volume element on the boundary [tex]\partial M[/tex], and [tex]\mathcal{L}_X[/tex] is the lie derivative along $X$.

    I understand that you can do the following:

    [tex]
    \[
    \int_{M} \mathcal{L}_X(\Omega) &= \int_{M} d(\iota_{X}(\Omega))) \\
    &= \int_{\partial M} \iota_{X}(\Omega)
    \]
    [/tex]

    by Stokes' theorem. Now, we can take [tex] N(x)[/tex] with an appropriate sign so that if [tex]\hat N(x)[/tex] is the dual of $N$, then

    [tex] \hat N(x) \wedge \omega = \Omega [/tex].

    By the formula for the contraction, we know that

    [tex] \iota_X (\Omega) = \langle X, N \rangle \omega - \hat{N(x)} \wedge \iota_X(\omega) [/tex]

    Lang claims that [tex] \hat{N(x)} \wedge \iota_X(\omega) [/tex] vanishes on the boundary at this point, and doesn't give an explanation. Can any one help me understand why? Of course, this proves the theorem.

    Thank you.
     
  2. jcsd
  3. Mar 17, 2010 #2

    quasar987

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    If I understand you correctly, [tex]\hat{N}[/tex] is the image of N by the musical isomorphism; that is, the 1-form [tex]\hat{N}_x=g_x(N(x),\cdot)[/tex]. Clearly this vanishes on [itex]\partial M[/itex] (that is, [tex]\hat{N}_x|_{T_x(\partial M)}=0[/tex] for all x in dM) since N is, by definition, normal to dM.
     
  4. Mar 17, 2010 #3
    Thank you!

    I guess I forgot about the isomorphism.
     
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