For the integral kernel of the Laplacian $\Delta$ on $\mathbb{R}^n$, consider the resolvent $R(\lambda) := (\lambda - \Delta)^{-1}$ and let $R(\lambda; x, y)$ be its kernel, which is a smooth function away from the diagonal $\{x=y\}$. If I calculated correctly, we have estimates of the form 
$$|R(\lambda;x, y)| \leq C e^{-\delta(\lambda)|x-y|},$$
whenever $|x-y| \geq 1$, where $\delta(\lambda)$ is a certain positive constant, depending (quite explicitly) on the distance of $\lambda$ to the spectrum of $\Delta$.


**Q:** On general Riemannian manifolds, do we still have estimates like this on the resolvent kernel of the Laplacian (or more general operators)?


More specifically, if $M$ is a complete Riemannian manifold, say with bounded geometry, and $\Delta$ is the Laplace-Beltrami operator, is it true that for any $\varepsilon>0$, there exist constants $C, \delta>0$ such that we have
$$|R(\lambda;x, y)| \leq C e^{-\delta\,d(x, y)}$$
whenever $d(x, y) \geq 1$ and $\lambda$ has distance at least $\varepsilon$ from the spectrum of $\Delta$? Here $d(x, y)$ is the Riemannian distance function.



Exponential decay of resolvent kernel
The local statement is correct for any positive self-adjoint extension of the Laplace operator on a Riemannian manifold (not necessarily complete). It is an old result by Hoermander. See the proof of Prop. 4.8 in
"L. HOERMANDER, On the Riesz means of spectral functions and eigenfunction expansions
for elliptic differential operators, in “Belfer Graduate School of Science Annual Science
Conference Proceedings” (A. Gelbart, Ed.), pp. 155-202, Vol. 2, 1969," where this is proved even much more generally for any differential operator. If you want uniform constants you need to assume bounded geometry and adapt his proof, using the appropriate calculus.


