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authorJaron Kent-Dobias <jaron@kent-dobias.com>2023-06-10 17:59:50 +0200
committerJaron Kent-Dobias <jaron@kent-dobias.com>2023-06-10 17:59:50 +0200
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Initial commit.
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-rw-r--r--when_annealed.tex51
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+*.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
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+\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}