Suppose M is a smooth manifold, and we have two Riemannian metrics on M, say g and h, with g bigger than h (i.e. for every tangent vector at every point, the norm according to g is bigger than the norm according to h).
Suppose $H\_g$ and $H\_h$ are the heat kernels on (M, g) and (M, h) respectively.
Does ${\frac{H\_g(x, y, t)} {H\_g(x, x, t)} \leq \frac {H\_h(x, y, t)}{H\_h(x, x, t)}}$ necessarily hold for all x, y, and t?


This intuitively makes sense: if you have something diffusing on M, starting at the point x, it should diffuse "faster" with the metric h because distances are smaller, and so after time t it seems you should have a greater proportion of the diffusing substance at y compared to at x. On $\mathbb{R^n}$ this is trivially true, also I've tested it out numerically for a variety of metrics on the circle and it seems to hold. But a proof eludes me.


While I am curious about the general case, I'm especially interested when the manifolds in question are quotients of $\mathbb{R^n}$ by a lattice, with the corresponding flat metrics, because an affirmative answer in this special case can be used to solve [this problem](https://mathoverflow.net/questions/160507/inequality-regarding-sum-of-gaussian-on-lattices).


$\textbf{Update:}$ I've found a proof of this for the special case of translation-invariant metrics on the circle. For these spaces, the heat kernel $H(0, x ,t)$, as a function of x, is given by a wrapped Gaussian function. The variance of the Gaussian that gets wrapped is determined by t and the metric, and a smaller metric has larger variance for a fixed t. The problem then boils down to the following assertion:



> 
> If $a$, $b$, and $s$ are real numbers with $0 < a \leq b < 1$, then
>  $$\frac{\sum\limits\_{n = -\infty}^{\infty} a^{(n + s)^2}}{\sum\limits\_{n = -\infty}^{\infty} a^{n^2}} \leq \frac{\sum\limits\_{n = -\infty}^{\infty} b^{(n + s)^2}}{\sum\limits\_{n = -\infty}^{\infty} b^{n^2}} $$
> 
> 
> 


From the Poisson summation formula, the above is equivalent to:



> 
> If $\alpha$ and $\beta$ are real numbers with $1 > \alpha \geq \beta > 0$, and $z$ is a complex number of norm 1, then
>  $$\frac{\sum\limits\_{n = -\infty}^{\infty} {\alpha^{n^2} z^{n}}}{\sum\limits\_{n = -\infty}^{\infty} \alpha^{n^2}} \leq \frac{\sum\limits\_{n = -\infty}^{\infty} {\beta^{n^2} z^{n}}}{\sum\limits\_{n = -\infty}^{\infty} \beta^{n^2}} $$
> 
> 
> 


From the Jacobi triple product, the above is equivalent to:


$$\prod\limits\_{m=1}^{\infty}{\frac{(1 + \alpha^{2m - 1}z)(1 + \frac{\alpha^{2m - 1}}{z})}{(1 + \alpha^{2m - 1})^2}} \leq \prod\limits\_{m=1}^{\infty}{\frac{(1 + \beta^{2m - 1}z)(1 + \frac{\beta^{2m - 1}}{z})}{(1 + \beta^{2m - 1})^2}}$$


It's easy to verify that the expressions inside the infinite products are positive for each m (recall that $|z| = 1$, and so $\frac{1}{z} = \overline{z}$), and so it suffices to show that we have this for each $m \geq 1$:


$$\frac{(1 + \alpha^{2m - 1}z)(1 + \frac{\alpha^{2m - 1}}{z})}{(1 + \alpha^{2m - 1})^2} \leq \frac{(1 + \beta^{2m - 1}z)(1 + \frac{\beta^{2m - 1}}{z})}{(1 + \beta^{2m - 1})^2}$$


For the sake of concision, let $p = \alpha^{2m - 1}$ and $q = \beta^{2m - 1}$.
We are done if we can show the following:


$$\frac{(1 + pz)(1 + \frac{p}{z})}{(1 + p)^2} \leq \frac{(1 + qz)(1 + \frac{q}{z})}{(1 + q)^2}$$


Or equivalently:


$$0 \leq (1 + qz)(1 + \frac{q}{z})(1 + p)^2 - (1 + pz)(1 + \frac{p}{z})(1 + q)^2$$


By expanding and factoring the right side, it can be shown that the above is equivalent to:


$$0 \leq (1 - \Re(z))(p - q)(1 - pq)$$


It's easy to show that each of the three terms in the product is non-negative, and this proves the inequality, and so we are done.



Is the heat kernel more spread out with a smaller metric?
Oded Regev and I just posted a [paper](http://arxiv.org/abs/1502.04796) to the arXiv that shows that this does in fact hold over $\mathbb{R}^n/\mathcal{L}$ where $\mathcal{L}$ is a lattice. See Proposition 4.2. 


The general case does not hold. In particular, if the second-largest eigenvalue of the heat kernel is unique, then as $t$ approaches infinity, the point with the highest mass will be the point that maximizes the corresponding eigenfunction. I think Jeff Cheeger originally pointed this out to Oded.


