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Let $M$ be a closed Riemannian manifold, and let $h(t, x, y)$ denote the heat kernel on $M$. We know that there exists short time upper Gaussian heat kernel bounds of the following kind: $$ h(t, x, y) \leq Ct^{-n/2}e^{-\frac{c\text{ dist}(x, y)^2}{t}}. $$

My question is, is it known how the constant $c$ on the exponent behaves with respect to curvature conditions on $M$? For example, do we have, let's say, estimates of the following kind: if $g_1, g_2$ are two metrics on $M$, and $\text{Ric}$ denotes Ricci curvature, then $\text{Ric}_1 \leq \text{Ric}_2$, then $h_1(t, x, y) \leq h_2(t, x, y)$? I am guessing that if the volume grows faster, the heat kernel will decay faster (as then the "heat" will have more "area" to cover). It seems intuitive that such things should be true, but I would really appreciate a reference if they are actually true.

Edit: Now I have realized that the previous version of my question is a little vague, after parsing through a bit of Fabrice Baudoin's blog, and the comments and answers below. Now, I would like to ask the following specific question: in addition to short time, if we are also looking at near-diagonal ($x$ and $y$ are close), then what happens to the constant $C$? Are there estimates in terms of curvature? From this MO post, it is clear that one can take $c = 1/4$.

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    $\begingroup$ Possible duplicate of Quick question on the constants involved in heat kernel upper bounds $\endgroup$ – Raziel Mar 4 at 7:05
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    $\begingroup$ It is not true that $h_1 (t,x,y) \le h_2(t,x,y)$ because both kernels integrate to 1, so we would have $h_1=h_2$. A somewhat explicit dependency on a lower bound of the Ricci curvature is given in fabricebaudoin.wordpress.com/2013/10/05/… $\endgroup$ – Fabrice Baudoin Mar 4 at 10:05
  • $\begingroup$ You maybe mean c = 1/4 ? If you are happy with c = 1/(4+\varepsilon) then I think that he Li-Yau heat kernel estimate below provides a complete characterization of C (for any fixed epsilon, small as you like) in terms of ricci curvature lower bound, and volume of balls. The latter can be controlled if you have also a sectional curvature upper bound, but for times smaller than the injectivity radius (and this is the best you can hope for). $\endgroup$ – Raziel Mar 18 at 11:23
  • $\begingroup$ @Raziel Yeah sorry, I wanted to mean $c = 1/4$. If we take sectional curvature upper bounds (and below also if required), can we say that $c = 1/4$, and $C$ depends only on the sectional curvature bounds? $\endgroup$ – user136537 Mar 18 at 18:51
  • $\begingroup$ In general, I do not know. What I know is that if t is smaller than the injectivity radius and EPs>0, then from the Li-Yau estimate follows that C depends only on the curvature bounds. If t is larger, it is unlikely that you can control C only with curvature bounds. $\endgroup$ – Raziel Mar 18 at 19:00
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This is related to Li-Yau gradient estimates. I think the usual set up is for a complete manifold with $\textrm{Ric}(M)>-k$. You probably need some Harnark type inequalities for parabolic equations. If you can do this, I think you can also get comparison formulas on the spectral gap associated to the manifold (there is recent work by Xiangjin Xu on this). But all of above things may be very technical, and I am not sure how does it answer your original question.

One elementary thing that may or may not be helpful is that you may expand the trace of the heat kernel locally using the curvature tensors. If the inequalities you suggest does exist, then it would extend over to the heat trace. So this gives some zeroth approximation whether your suggestion is true (if I recall correctly, the first term involves the scalar curvature, etc).

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I think that the closest statement to an answer is contained in the famous paper:

Li, Peter; Yau, Shing Tung, On the parabolic kernel of the Schrödinger operator, Acta Math. 156, 154-201 (1986). ZBL0611.58045.

In particular, in Corollary 3.1, they prove the following:

enter image description here

The estimate is global in time, and the constants are explicit in the proof. Furthermore, $V$ represents the volume of balls of radius $\sqrt{t}$. If you want to make them even more explicit in terms of $t$, you need to add an upper bound on the sectional curvature in order to bound from below the volume of balls with standard comparison theorems, at least for sufficiently small times.

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