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The integral of the unit normal field over a closed surface vanishes. There's a simple proof based on the divergence theorem. But is there a proof that "stays within the surface" (in other words, doesn't refer to objects in the ambient space) that would reveal the deeper geometric reason behind this fact?

Clarification in response to the comments: I'm not looking for an "intrinsic" proof, but a proof restricted to objects defined on the surface. For example, the proof of same statement for the mean curvature normal is of this kind.

Fana Voyler
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    The "normal field" is - as far as I remember - an extrinsic property using the ambient space itself. – martini Aug 07 '21 at 17:04
  • Yes, but restricted to the surface. – Fana Voyler Aug 07 '21 at 17:10
  • How would you define the normal vector without reference to the ambient space? How could you even state the theorem? There doesn't seem to be any hope of giving an intrinsic proof of the theorem unless you can state it intrinsically. – saulspatz Aug 07 '21 at 17:19
  • In view of the proof, the deep geometric reason is that the surface is the boundary of a solid region in the ambient space. You could perhaps reformulate the question as how to recognize intrinsically that a manifold is the boundary of a solid region in a higher dimensional space. – Gribouillis Aug 07 '21 at 17:20
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    That reformulation should be that the $2$-cycle represented by the closed surface is trivial in homology. But ultimately we're still applying Stokes's Theorem (abstractly) to deduce that a $2$-form on the surface has zero integral. – Ted Shifrin Aug 07 '21 at 17:24
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    Presumably the "intrinsic" proof you reference proceeds by applying Stokes's Theorem in the other dimension, i.e., by showing that the $2$-form is exact (rather than by using $S=\partial \Omega$). We can do that in this case, too. Finding the flux of $\vec i$ across $S$ is integrating the $2$-form $dy\wedge dz$ over the surface, and this $2$-form is plainly exact. – Ted Shifrin Aug 07 '21 at 17:40

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