- #1

- 2

- 0

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 anyone help me understand why? Of course, this proves the theorem.

Thank you.