5
$\begingroup$

Recently, I am reading Sturm's paper "Analysis on local Dirichlet form III: X is a locally compact separable Hausdorff space and m is a positive Radon measure with supp[m]=X.

$\varepsilon$ is a symmetric and strongly local Dirichlet form with domain F on a real Hilbert space $L^2(X,m)$. Any such form can be written as $$ \varepsilon(u,v)=\int_X d\Gamma(u,v) $$ where $\Gamma$ is a postive semidefinite, symmetric bilinear form on F with values in the signed Radon measures on X. Let $$ F_{loc}(X)=\{u \in L^2_{loc}(X,m):\Gamma(u,u)\mbox{ is a Radon measure}\} $$ The energy measure defines in an intrinsic way a pseudo metric $\rho$ on X by $$ \rho(x,y)=sup\{u(x)-u(y):u \in F_{loc}(X) \cap C(X),\Gamma(u,u) \leq \mbox{ m on X}\} $$ The Dirichlet form is called strongly regular if it's regular and if $\rho$ is a metric on X whose topology coincides with the original one.

Question 1: On which kind of spaces, $\rho$ is a metric on X whose topology coincides with the original one? Riemannian manifold (Alexandrov spaces) with Ricci curvature bounded below, $CD(K,N)$ and $CD(K, \infty)$ satisfy?

The paper says: If "completeness property", "doubling" and "Poincare inequality" hold globally on X, then there are both Gaussian lower and upper bounds for the heat kernel,i.e. $$ 1/C (\mu(B_{\sqrt(t)}(x))^{-1/2}(\mu(B_{\sqrt(t)}(y))^{-1/2}e^{-\frac{\rho(x,y)^2}{C_1t}} \leq p_t(x,y) $$ $$\leq C(\mu(B_{\sqrt(t)}(x))^{-1/2}(\mu(B_{\sqrt(t)}(y))^{-1/2}e^{-\frac{\rho(x,y)^2}{C_2t}} $$ But Yau's book "differential geometry" just give a lower bound for nonnegative curved Riemannian manifold, why?

Question 2: Does the heat kernel of Riemannian manifolds (possibly noncompact and with boundary) with $Ric \ge -(n-1)k(k \ge 0)$ have both Guassian upper and lower bounds?

$\endgroup$
1

4 Answers 4

4
$\begingroup$

There are already some excellent answers, and I'll just list some known facts in the setting of manifolds: (1) On a complete (weighted) manifold, parabolic Harnack inequality is equivalent to Gaussian two-sided bounds, which is in turn equivalent to volume doubling property plus a (scale invariant) Poincare inequality. This is the work of Grigor'yan and Saloff-Coste (indepently). You may have a look at Grigor'yan http://www.math.uni-bielefeld.de/~grigor/harsbor.pdf and Saloff-Coste http://imrn.oxfordjournals.org/content/1992/2/27.full.pdf+html. The book ``Aspects of Sobolev-Type Inequalities" of Saloff-Coste contains a detailed exposition of this result. (2) Volume doubling property and Poincare inequality both hold for a complete manifold with nonnegative Ricci curvature, so the Gaussian two-sided estimates follow. Indeed, in this case Li and Yau's gradient estimates for the heat equation is stronger than Gaussian estimates. (3) The volume doubling property may fail for a manifold with Ricci bounded from below, as can be seen from the example of hyperbolic upper half space.

I'm not familiar enough with the intrinsic distance to answer your first question. I only heard that the intrinsic distance defined using Carre du Champ often breaks down for fractals. For a complete Riemannian manifold, it should coincide with the geodesic distance. Please correct me if I'm wrong.

$\endgroup$
4
  • $\begingroup$ I should have said that Buser proved the Poincare inequality for manifolds with Ricci curvature bounded from blow. In the case of nonnegative Ricci curvature, it gives a scale invariant Poincare inequality. $\endgroup$
    – Xueping
    Nov 6, 2013 at 9:07
  • $\begingroup$ :You mean scale invariant doubling plus scale invariant Poincare equivalent to Gaussian two-sided bounds? $\endgroup$ Nov 6, 2013 at 11:49
  • $\begingroup$ :And for manifold with Ricci bounded from below, we can get neither a scale invariant doubling nor a scale invariant Poincare? $\endgroup$ Nov 6, 2013 at 12:06
  • $\begingroup$ @jiangsaiyin Yes, and I think so. I assume by "scale invariant doubling" you mean the usual ``volume doubling" property (i.e. space of homogeneous type). $\endgroup$
    – Xueping
    Nov 6, 2013 at 19:37
3
$\begingroup$

I do not know much about the case of metric spaces, but for the heat kernel on a (noncompact) manifold without boundary, these estimates exist. There are various papers and book by Grigor`yan on that subject. Which estimates hold exactly, depends on the question you are asking: Are you looking for global or near-diagonal estimates, long-time or short-time estimates?

The case where one chooses the constants $C_1$ and $C_2$ equal to $4$ in the exponent has been dealt with e.g. by Davies and Pang ("Sharp Heat kernel bounds for some Laplace operators", 1988) and Molchanov ("Diffusion Processes and Riemannian Geometry", 1975).

Indeed, a lower bound on the Ricci curvature is important for global Gaussian bounds (the precise bound is irrelevant though). However, there are versions which yield bounds even without these assumptions (where the constant in front has to depend in some way on $x$ and $y$).

I do not really know anything about the case with boundary though. For sets whose closure is contained in the interior of a manifold with boundary, the same results hold as before. If this does not hold, things get complicated.

$\endgroup$
1
$\begingroup$

I'll try to say something about each question raised one by one. As @Kofi points out, there has been a lot of work on heat kernel estimates, it would be impossible to sum it up in a few paragraphs (I will second the recommendation to look at the work of Grigor'yan).

As far as which spaces have $\rho$ induce the original topology, this is a non-trivial assumption in general, but they agree on a Riemannian manifolds. It is useful to know that $\rho$ often agrees with other notions of intrinsic metrics, say from a Carre du Champ or upper gradients or spectral triples.

CD(K,n) holding for Riemannian manifolds and Alexandrov spaces, I believe that is covered in Sturm's "On the geometry of metric measure spaces." I think it comes from the Bochner identity for Riemannian manifolds.

On a non-negative Ricci curved Riemannian manifolds, I thought those estimates went back to Li and Yau's 1986 paper. I don't know why, in that particular book, Yau did not bring up both sides of the estimate, I assume it makes sense in context... a similar upper bound can be proven in the case where Ricci curvature is bounded below by a negative number.

I remember reading something about getting Li-Yau gradient inequalities for Riemannian manifolds with boundary for the Neumann Laplacian (I think there was also a convexity assumption), but I'm having trouble finding the reference.

$\endgroup$
1
$\begingroup$

An addition to the first question:

If $(X,d,m)$ is an $\mathsf{RCD}(K,\infty)$ space (see [1] for the definiton and equivalent characterizations) and $\epsilon$ the Cheeger energy $\epsilon(u)=\int_X |Du|_w^2\,dm$ (some authors call $\frac 1 2\epsilon$ the Cheeger energy instead), then the intrinisic metric $\rho$ induced by $\epsilon$ coincides with $d$, see [1], Thm. 7.4, [2], Thm. 6.10.

Some remarks:

  • Instead of the curvature-dimension condition $\mathsf{CD}(K,\infty)$ one needs the infinitesimally Hilbertian version $\mathsf{RCD}(K,\infty)$ to ensure that $\epsilon$ is a Dirichlet form. If $\epsilon$ is a Dirichlet form, then $\mathsf{CD}(K,\infty)$ and $\mathsf{RCD}(K,\infty)$ are equivalent.
  • Riemannian manifolds with Ricci curvature bounded below by $K$ satisfy $\mathsf{RCD}(K,\infty)$, see [3]. The Dirichlet form $\epsilon$ coincides with the standard energy form generated by the Laplacian.
  • $n$-dimensional Alexandrov spaces with sectional curvature bounded below by $K$ equipped with the Hausdorff measure satisfy $\mathsf{CD}(K(n-1),N)$, see [4].

References:

[1] Ambrosio, Gigli, Savaré. Metric measure spaces with Riemannian Ricci curvature bounded from below

[2] Ambrosio, Gigli, Mondino, Savaré. Riemannian Ricci curvature lower bounds in metric measure spaces with $ \sigma$-finite measure

[3] Sturm, von Renesse. Transport Inequalities, Gradient Estimates, Entropy, and Ricci Curvature

[4] Petrunin. Alexandrov meets Lott-Villani-Sturm

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.