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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.

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    $\begingroup$ Are you aware of the asymptotic expansion for the heat kernel in terms of $t:$ $H_t(p,q)= h_t(p,q)(\theta_0(p,q)+t\theta_1(p,q)+...),$ where $h_t(p,q)=\frac{1}{(4 \pi t)^{dim M/2}}exp(-d(p,q)^2/4t)$ and $\theta_0(p,p)=1.$ At least for small $t$ and nearby points $p,q$ this should give your inequality. $\endgroup$ – Sebastian Nov 7 '14 at 8:12
  • $\begingroup$ @Sebastian No I wasn't aware, thanks for pointing that out. I see that the $h_t$ functions will satisfy the inequality, but I don't see how that is necessarily preserved after multiplying by the series on the right without more information about these $\theta_n$. $\endgroup$ – Tom Price Nov 7 '14 at 16:54
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    $\begingroup$ The only thing I know about the $\theta$'s is that they can be expressed in terms of connection data along the diagonal, see for example the book of John Roe: Elliptic operators,... . But if I remember correctly, the series $(\theta_0(p,q)+t\theta_1(p,q)+...)$ converges for small t and nearby $p,q$ to a smooth map which is the identity for $t=0$ and $p=q.$ Therefore, you can get the inequality in a neighbourhood of the diagonal times the $t=0$ slice. $\endgroup$ – Sebastian Nov 8 '14 at 9:04
  • $\begingroup$ The heat operator has the "finite propagation speed" property which allows one to estimate the support of $e^{t\Delta}f$ in terms of the support of $f$. There are estimates of the propagation speed involving metric data, but I don't know off the top of my head if they are sharp enough to get what you want. $\endgroup$ – Paul Siegel Dec 12 '14 at 5:01
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    $\begingroup$ @Paul Siegel: I am surprised by your assertion. The heat semigroup typically has infinite speed of propagation, that is, as soon as the initial data are positive but not identically 0, the solution of the heat equation is instanteneously strictly larger than 0 at any point of the domain/manifold. This can be e.g. proved by irreducibility of the semigroup, see e.g. Ouhabaz' 2005 book. Am I missing something? $\endgroup$ – Delio Mugnolo Dec 12 '14 at 9:17
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Oded Regev and I just posted a paper 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.

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