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A Resolvent Approach to Metastability

C. Landim, D. Marcondes, I. Seo

Abstract

We provide a necessary and sufficient condition for the metastability of a Markov chain, expressed in terms of a property of the solutions of the resolvent equation. As an application of this result, we prove the metastability of reversible, critical zero-range processes starting from a configuration.

A Resolvent Approach to Metastability

Abstract

We provide a necessary and sufficient condition for the metastability of a Markov chain, expressed in terms of a property of the solutions of the resolvent equation. As an application of this result, we prove the metastability of reversible, critical zero-range processes starting from a configuration.

Paper Structure

This paper contains 30 sections, 48 theorems, 293 equations, 1 figure.

Table of Contents

  1. Introduction
  2. A Resolvent Approach to Metastability
  3. Applications
  4. Critical Zero-range Dynamics
  5. Condensation of particles
  6. Order process
  7. Main result
  8. Proof Theorem \ref{['1t_main2']}
  9. Condition ${\mathfrak R}_{{\mathscr L}}$ entails ${\mathfrak C}_{{\mathscr L}}$ and ${\mathfrak D}$
  10. Conditions ${\mathfrak C}_{{\mathscr L}}$ and ${\mathfrak D}$ imply ${\mathfrak R}_{{\mathscr L}}$
  11. Potential theory
  12. The solutions of the resolvent equation
  13. Visiting condition ${\mathfrak V}$
  14. Mixing condition ${\mathfrak M}$
  15. Local equilibration and condition ${\mathfrak D}$
  16. ...and 15 more sections

Key Result

Theorem \oldthetheorem

The process $\xi_N(\,\cdot\,)$ is ${\mathscr L}$-metastable if, and only if, condition ${\mathfrak R}_{{\mathscr L}}$ is fulfilled. In other words, Conditions ${\mathfrak D}$ and ${\mathfrak C}_{{\mathscr L}}$ hold if, and only if, condition ${\mathfrak R}_{{\mathscr L}}$ is in force.

Figures (1)

  • Figure 1: This picture illustrates the idea behind the statement of Lemma \ref{['l13']}. To simplify, we argue in a continuous setting, but the same idea applies to the discrete setting. Consider a diffusion on the potential field appearing in the picture. The valley ${\mathcal{E}}_N^y$ is a metastable set and ${\mathcal{E}}_N^z$ a stable one. As $\mu_N(\Delta_N)/\mu_N({\mathcal{E}}_N^y)$ does not converge to $0$, we decompose $\Delta_N$ as $\Delta'_N \cup {\mathcal{A}}_N$, so that $\mu_N (\Delta'_N) / \mu_N ({\mathcal{E}}_N^y) \rightarrow 0$. On the other hand, as ${\mathcal{A}}_N$ is a subset of the domain of attraction of the valley ${\mathcal{E}}_N^z$, we can expect \ref{['an_cond']} to hold.

Theorems & Definitions (105)

  • Definition \oldthetheorem
  • Remark \oldthetheorem
  • Theorem \oldthetheorem
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  • ...and 95 more