# Is the on-diagonal heat kernel “local” with respect to the metric?

Question Let $X$ be a manifold, and $\mu_A$, $\mu_B$ two Riemannian metric on it which agree on an open subset $U\subset X$, i.e. $\mu_{A\,|U} = \mu_{B\,|U}$. Let $K_A(t;z,w)$ resp. $K_B(t;z,w)$ be the heat kernel associated to the metric $\mu_A$ resp. $\mu_B$, is then the formula $$K_A(t;z,z) = K_B(t;z,z) \qquad \forall z \in U$$ true?

My Progresses We want to prove that $$h(t;z) := K_A(t;z,z) - K_B(t;z,z) = 0 \qquad \forall z \in U.$$

The Laplacian is clearly local in the metric, i.e. $\Delta_{A\,|U} = \Delta_{B\,|U}$. Denoting by $\mu$ any such metric and by $\Delta$ any such Laplacian, by the properties of the heat kernel we deduce that $h(t;z)$ can be continuously extended to $t=0$ and is a solution of the problem $$\begin{cases} \left( \partial_t + \Delta\right)h(t;z) = 0& \quad\forall t>0,\, z\in U,\\ h(0;z) = 0& \quad \forall z \in U. \end{cases}$$

And, by hypoellipticity of the heat equation, a solution of this problem must be $\mathcal{C}^\infty$ in $U$.

Now I would like to apply a uniqueness-result such as Cauchy-Kowalevski theorem, but the heat equation doesn't satisfies the hypothesis. Still it looks to me that "it doesn't satisfy the hypothesis needed for existence, rather than those needed for uniqueness". Somehow I'm a bit lost on this point.

• I don't think that the initial value problem you posed has a unique solution. My heuristic reasoning is that one could consider $g(t,x)$, a solution to the heat equation on all of $X$, and even if $g(0,z)=0$ for $z \in U$, $g(t,z)$ need not be zero for $z \in U$ at later times $t$ (heuristically, heat must flow into U). Note that Dirichlet boundary conditions ($h = 0$ on $\partial U$), for example, would give that $h \equiv 0$ is the unique solution to your problem, but I'm not sure those boundary conditions are relevant here. – Phillip Andreae Dec 1 '14 at 16:03
• Thank you for your interest Phillip! I agree that Dirichlet condition would provide uniqueness, and also that they wouldn't be relevant for the problem. I was now thinking to the strong maximum principle; if $B$ is a closed ball of small radius contained in $U$, then the maximum of $h$ for any time on $B$ is obtained for $t=0$; and the minimum as well. This forces $h\equiv 0$ for any time and any $z\in B$. It looks a bit too easy to be true though, and I want to check if there's a flaw in the argument. – Giovanni De Gaetano Dec 2 '14 at 15:39
• Indeed I was obviously ignoring the fact that the $\sup$ can also be achieved for $t>0$ and $z \in \partial B$; which invalidates the argument. – Giovanni De Gaetano Dec 2 '14 at 16:18