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.
Thank you in advance for your interest!
Note I hope the level of the question is appropriate for this website, I thought to post it on Math.OF instead.
This question has been answered on MathOverflow. The answer is no, this is not true.