Let $(M,g)$ be a Riemannian $n$-manifold with $Ric_g\ge -Kg$, $\Omega\subset M$ be an open subset. We can define Dirichlet heat kernel on $\Omega$, $p_{\Omega}(y,t,y',t')$ as the minimal fundamental solution to heat equation with Laplace-Beltrami operator on $\Omega\times(t',+\infty)$. The following paper


gives a 2-side Gaussian bound, especially a lower Gaussian bound in Theorem 6.1 in terms of Riemannian distance. I will copy down the lower bound part, for simplicity I only state this theorem for Laplace-Beltrami operator, the author did it in much more general context. In that paper's notation $\beta=0, \alpha=1,\mu=1$ for Laplace-Beltrami operator.

$\bf Theorem$ Let $0<\delta<1$ be fixed and suppose $B(x,r)\subset \Omega$. Then for all $y,y'\in B(x,\delta r)$, $t'<t<t'+r^2$ we have $$e^{-C_1(1+K\tau)}V(y,\sqrt{t})^{-1/2}V(y',\sqrt{t})^{-1/2}\exp(-C_2\rho^2(y,y')/t)\le p_{\Omega}(y,t,y',t')$$ where $\tau=t-t'$, $\rho$ is the Riemannian distance on $M$,and $V(x,r)$ is the volume of $B(x,r)$ the constants depend on dimension $n$, $\delta$, possibly $K$, but (seem to) do not depend on $\Omega$.

The author claims that this is a consequence of the Harnack inequality, i.e. Theorem 5.3 in the same paper. I did not understand the proof and I am a bit surprised by the fact that this lower Gaussian estimate is in terms of the Riemannian distance but not the intrinsic distance on $\Omega$. In general the intrinsic distance can be quite different from the Riemmanian distance on $M$.

For example take $M=S^1$ with large radius, and $\Omega=S^1\setminus[0,\varepsilon]$ be a major arc that is almost the full circle(when $\varepsilon$ is small). Then we can take $B=\Omega$ since arcs are geodesic balls in $S^1$, the heat kernel on $\Omega$ behaves like a Gaussian with intrinsic distance on the arc. If we take $y,y'\in \Omega$ to be close to the 2 endpoints, then $d_{S^1}(y,y')$ is close to $\epsilon$ but $d_\Omega(y,y')$ can be arbitrarily large, it is hard to imagine that the heat kernel on $\Omega$ can be bounded below by the Gaussian with distance on $S^1$.

So could anybody tell me why and how this kind of Gaussian bound is possible? Any help is appreciated.

  • $\begingroup$ Aren't $y$ and $y'$ assumed to be in a smaller ball $B(x, \delta r)$ rather than in $\Omega = B(x, \delta)$? $\endgroup$ Mar 5, 2021 at 20:37
  • $\begingroup$ @MateuszKwaśnicki I think the $\delta$ can be close to 1, so really no much restriction is put on $y,y'$? Or I misunderstood. $\endgroup$
    – WhiteDwarf
    Mar 5, 2021 at 20:38
  • $\begingroup$ The constants depend on $\delta$, do they not? $\endgroup$ Mar 5, 2021 at 20:57

1 Answer 1


Maybe, the book

N. Th. Varopoulos, L. Saloff-Coste, T. Coulhon, Analysis and Geometry on Groups, volume 100, Cambridge university press, 2008.

would be of help, especially, Theorem IV.4.3 and its proof in page 49. There are also other lower bounds similar to the theorem you mentioned. They use a Harnack inequality given in Proposition IX.1.1, page 125.


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