I read this statement a few times, for example in the answer to this question. Does anyone have a reference to this statement?
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What’s wrong with the proof given by Ivo? – peek-a-boo Aug 31 '23 at 10:59
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well he relies on "Stokes' theorem says that the integral of a divergence (hence of a Laplacian) over a compact manifold without boundary vanishes" which he does not proof. also he assumes that M is oriented as pointed out by "Quarto Bendir". So I would like to see a formal proof listing all assumptions etc. – Butters Aug 31 '23 at 11:04
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I have two comments to that. First, the divergence theorem does not require an orientation. People often formulate it as a corollary of Stokes’ theorem (which does require an orientation), but you don’t need to. For a reference, see Loomis and Sternberg’s Advanced Calculus, chapter 10.6. Second, on a compact boundaryless manifold, the boundary terms vanish so that part is trivial, so you should really just review the usual Stokes/divergence theorems. – peek-a-boo Aug 31 '23 at 11:15
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Ok I see. Thanks for your comments. I suppose what was confusing me then is that we can simply apply the divergence theorem for a manifold without a boundary although the divergence theorem is always stated for a manifold with a boundary. In that sense, a manifold without boundary is a manifold with the empty set as boundary and thus the integral vanishes? – Butters Aug 31 '23 at 11:24
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Yes that’s correct. But really if you analyze the proof of the theorem, you see that it is often split into two cases, one where the function/form has support not intersecting the boundary (which includes the case of empty boundary) and the other case where it does intersect the boundary. You can see that here (but note that my proof there requires one extra degree of differentiability than what I stated, but the theorem statement is still correct lol… but you’re dealing with $C^{\infty}$ so you can ignore these remarks). – peek-a-boo Aug 31 '23 at 11:33
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That makes sense. Thank you! Feel free to post your reference to Loomis and Sterberg as an asnwer. – Butters Aug 31 '23 at 11:46
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1You can answer your own question. – peek-a-boo Aug 31 '23 at 11:47