From f69b610ac37d25f21f32c0b9a6c0b1942249cb00 Mon Sep 17 00:00:00 2001
From: Jaron Kent-Dobias <jaron@kent-dobias.com>
Date: Sat, 10 Jun 2023 17:59:50 +0200
Subject: Initial commit.

---
 when_annealed.tex | 51 +++++++++++++++++++++++++++++++++++++++++++++++++++
 1 file changed, 51 insertions(+)
 create mode 100644 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}
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