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:

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

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.