Divergence Theorem on Manifolds

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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.
 

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  • #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.
 
  • #3
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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.
Thank you!

I guess I forgot about the isomorphism.
 

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