If $M$ is a complete Riemannian manifold, it possesses a unique self-adjoint positive operator $-\Delta$ on $L^2(M)$. If $M$ is not complete, though, it is known that the Laplace-Beltrami operator $-\Delta$ has several self-adjoint extensions in $L^2(M)$. Nevertheless, the heat kernel is uniquely defined (by the minimality property), regardless of whether $M$ is complete or not.

If $M$ is not complete, what is the self-adjoint extension of $-\Delta$ that corresponds to the heat kernel (i.e. what is the self-adjoint extension that generates the heat semigroup)? Is it the Friedrichs extension? If so, where can I find a proof of this?

I would be satisfied with an answer to the following, simpler question: if $M$ is complete and $U \subset M$ is an open subset with negligible complementary, and $h_U$ and $h_M$ their corresponding heat kernels, what is the relationship between $h_U$ and $h_M \big| _U$? By minimality, $h_U \le h_M \big| _U$; I expect them to be equal, though, at least as integral operators on $L^2 (U) = L^2 (M)$ - but I do not know whether this is true or how to prove it, because I do not know what extension of $-\Delta_U$ corresponds to $h_U$.

  • $\begingroup$ Could you state precisely how you define "the" heat kernel? I'm guessing it's going to turn out to be the Dirichlet heat kernel, which I believe should in fact correspond to the Friedrichs extension. $\endgroup$ – Nate Eldredge Jun 22 '18 at 15:18
  • $\begingroup$ Notice that the Laplace Beltrami can be essentially self adjoint (and then have a unique heat kernel) even if the manifold is not complete. For example this is the case for R^n minus a point if n>4. But there are also more complicated cases. $\endgroup$ – Raziel Jun 22 '18 at 15:40
  • $\begingroup$ @Raziel: I chose a well-known sufficient condition, I don't claim that it is also necessary. $\endgroup$ – Alex M. Jun 22 '18 at 15:54
  • $\begingroup$ @NateEldredge: The heat kernel of a Riemannian manifold is the (unique, a posteriori) positive, minimal fundamental solution of the heat equation. Smoothness is a consequence of the definition. Notice that the definition only uses the laplacian acting on smooth functions, not in $L^2$. $\endgroup$ – Alex M. Jun 22 '18 at 15:56

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