- #1
mathmeat
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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 anyone help me understand why? Of course, this proves the theorem.
Thank you.
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.