# 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?

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?

Note This question has originally been posted on Math.SE.

• As an easy way to see this cannot be true, consider $X$ compact and connected. For fixed $z$, as $t \to \infty$ we have $K_A(t; z,y) \to 1/\mathrm{Vol}_A(X)$ uniformly in $y$. (This is because $K_A(t;z,\cdot)$ must converge to a constant, and it also must integrate to 1 with respect to $d\mathrm{Vol}_A$.) Now let $\mu_B$ be any metric which agrees with $\mu_A$ on $U$ but gives different total volume to $X$. For sufficiently large $t$, $K_A(t;z, y)$ and $K_B(t;z,y)$ must differ for all $y$. – Nate Eldredge Dec 10 '14 at 2:06

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
$| K_A(t,z,z)-K_B(t,z,z) | \simeq e^{-C/t}$