Why Does the Contraction Term Vanish in the Divergence Theorem on Manifolds?

[mathmeat]

Hi,

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

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

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

I understand that you can do the following:

<br /> \[ <br /> \int_{M} \mathcal{L}_X(\Omega) &amp;= \int_{M} d(\iota_{X}(\Omega))) \\<br /> &amp;= \int_{\partial M} \iota_{X}(\Omega) <br /> \]<br />

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

\hat N(x) \wedge \omega = \Omega.

By the formula for the contraction, we know that

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

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

Thank you.

 

[quasar987]

If I understand you correctly, \hat{N} is the image of N by the musical isomorphism; that is, the 1-form \hat{N}_x=g_x(N(x),\cdot). Clearly this vanishes on \partial M (that is, \hat{N}_x|_{T_x(\partial M)}=0 for all x in dM) since N is, by definition, normal to dM.

 

[mathmeat]

If I understand you correctly, \hat{N} is the image of N by the musical isomorphism; that is, the 1-form \hat{N}_x=g_x(N(x),\cdot). Clearly this vanishes on \partial M (that is, \hat{N}_x|_{T_x(\partial M)}=0 for all x in dM) since N is, by definition, normal to dM.

Thank you!

I guess I forgot about the isomorphism.