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Let $M$ be a complete Riemannian manifold and $-\Delta$ denote the Laplace-Beltrami operator on $M$. We can prove that $(-\Delta f, g) = (\nabla f, \nabla g) = (f, -\Delta g)$, when $f, g \in C^\infty_0(M)$. My question is, when one extends the Laplace-Beltrami operator as a self-adjoint operator, what is the domain of the extension?

Edit: As Jack Lee points out, here we are thinking of $-\Delta$ as an unbounded operator on $L^2(M)$.

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  • $\begingroup$ I think it some how up to you. What domain do you want? $\endgroup$ – spatially Jun 6 '15 at 14:55
  • $\begingroup$ @HomegrownTomato: If OP is looking for the domain as a unbounded operator on $L^2$, it has to be $H^2(M)$ (or at least contained in it), by elliptic regularity. $\endgroup$ – Jack Lee Jun 6 '15 at 17:06
  • $\begingroup$ See also math.stackexchange.com/questions/63251 $\endgroup$ – punctured dusk Jul 15 '18 at 11:25
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Take a look at this reference from Strichartz http://www.sciencedirect.com/science/article/pii/0022123683900903

The operator is essentially selfadjoint on its natural domain. The domain is, as the one commenter noted, $H^{2}(M)$.

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  • $\begingroup$ Could you give a reference that the domain is $H^2$? $\endgroup$ – punctured dusk Jul 15 '18 at 11:53

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