**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](https://math.stackexchange.com/questions/1046394/is-the-on-diagonal-heat-kernel-local-with-respect-to-the-metric).



Is the on-diagonal heat kernel “local” with respect to the metric?
The heat kernel is a non local object: $K\_A(t,z,z) \neq K\_B(t,z,z)$. 


There is a very intuitive probabilistic explanation: $K\_A(t,z,z)dz$ is the probability that a Brownian motion started at $z$ is, at time $t$, around $z$. A brownian path can go outside the domain $U$ and then come back inside the domain. This means that $K\_A(t,z,z)$ also depends on what is outside of $U$.


Observe that, in small times, $K\_A(t,z,z)$ is however very close to $K\_B(t,z,z)$ since we can prove that 


$ | K\_A(t,z,z)-K\_B(t,z,z) | \simeq e^{-C/t}$


