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?
Comments If we denote by $h(t;z,z)$ the difference of the heat kernels on the open subset $U$, then, since $\Delta_{A|U} = \Delta_{B|U}$, it is a solution of the heat equation $(\partial_t + \Delta)h(t;z) = 0$ with initial condition $h(0;z)=0$ for any $z\in U$; but I don't derive any boundary condition to make the solution of the problem above unique.
I ask this question because I remember having read something similar somewhere, and I would like it to be true; but the comment above seems to indicate that the case is hopeless, is it really the case?
Thank you in advance for your interest!
Note This question has originally been posted on Math.SE.