# 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?

Thank you in advance for your interest!

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

## 1 Answer

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

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