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 $h_U$.