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| -rw-r--r-- | .gitignore | 14 | ||||
| -rw-r--r-- | when_annealed.tex | 51 |
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diff --git a/.gitignore b/.gitignore new file mode 100644 index 0000000..9a36fc7 --- /dev/null +++ b/.gitignore @@ -0,0 +1,14 @@ +*.aux +*.fdb_latexmk +*.fls +*.log +/*.pdf +*.synctex.gz +*.bbl +*.blg +*.out +*.bcf +*.run.xml +*.synctex(busy) +*.toc +*Notes.bib diff --git a/when_annealed.tex b/when_annealed.tex new file mode 100644 index 0000000..b91c9ee --- /dev/null +++ b/when_annealed.tex @@ -0,0 +1,51 @@ +\documentclass[fleqn,a4paper]{article} + +\usepackage[utf8]{inputenc} % why not type "Bézout" with unicode? +\usepackage[T1]{fontenc} % vector fonts plz +\usepackage{fullpage,amsmath,amssymb,latexsym,graphicx} +\usepackage{newtxtext,newtxmath} % Times for PR +\usepackage{appendix} +\usepackage[dvipsnames]{xcolor} +\usepackage[ + colorlinks=true, + urlcolor=MidnightBlue, + citecolor=MidnightBlue, + filecolor=MidnightBlue, + linkcolor=MidnightBlue +]{hyperref} % ref and cite links with pretty colors +\usepackage[ + style=phys, + eprint=true, + maxnames = 100 +]{biblatex} +\usepackage{anyfontsize,authblk} + +\begin{document} + +\title{ + When is the annealed complexity correct? +} + +\author{Jaron Kent-Dobias} +\affil{\textsc{DynSysMath}, Istituto Nazionale di Fisica Nucleare, Sezione di Roma} + +\maketitle +\begin{abstract} + The difference between quenched and annealed averages is crucial for + disordered systems. In isotropic mean-field systems, they differ when replica + symmetry is broken. When computing the average free energy in equilibrium, + there are robust conditions to understand when {\oldstylenums1}\textsc{rsb} + is sufficient. When computing the average complexity, or the number of + stationary points of the energy, there is only robust reasoning at the ground + state, where a {\oldstylenums1}\textsc{rsb} equilibrium implies that the + annealed complexity \emph{at the ground state} is correct. Here, we + demonstrate that in the mixed spherical models, the annealed complexity can + be wrong away from the ground state even when the equilibrium free energy is + guaranteed to be at most {\oldstylenums1}\textsc{rsb} everywhere. Therefore, + simple equilibrium order cannot be used to assume a simple landscape + geometry. +\end{abstract} + +Parisi construction, $f''(q)^{-1/2}$ is concave + +\end{document} |