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.
$\textbf{Update:}$ 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.
