From dcc2d3358d4be2a0eab883870898bdfbafe4eacd Mon Sep 17 00:00:00 2001
From: Jaron Kent-Dobias <jaron@kent-dobias.com>
Date: Tue, 12 Jul 2022 17:30:58 +0200
Subject: Some more small edits.

---
 frsb_kac-rice.tex | 64 +++++++++++++++++++++++++++++++++++--------------------
 1 file changed, 41 insertions(+), 23 deletions(-)

(limited to 'frsb_kac-rice.tex')

diff --git a/frsb_kac-rice.tex b/frsb_kac-rice.tex
index 680cc89..f5a259e 100644
--- a/frsb_kac-rice.tex
+++ b/frsb_kac-rice.tex
@@ -13,7 +13,11 @@
   filecolor=MidnightBlue,
   linkcolor=MidnightBlue
 ]{hyperref} % ref and cite links with pretty colors
-\usepackage[style=phys,eprint=true]{biblatex}
+\usepackage[
+  style=phys,
+  eprint=true,
+  maxnames = 100
+]{biblatex}
 
 \addbibresource{frsb_kac-rice.bib}
 
@@ -25,11 +29,11 @@
 \author{Jaron Kent-Dobias \& Jorge Kurchan}
 \maketitle
 \begin{abstract}
-    We derive the general solution for counting the stationary points of
-    mean-field complex landscapes. It incorporates Parisi's solution
-    for the ground state, as it should. Using this solution, we count the
-    stationary points of two models: one with multi-step replica symmetry
-    breaking, and one with full replica symmetry breaking.
+  We derive the general solution for counting the stationary points of
+  mean-field complex landscapes. It incorporates Parisi's solution
+  for the ground state, as it should. Using this solution, we count the
+  stationary points of two models: one with multi-step replica symmetry
+  breaking, and one with full replica symmetry breaking.
 \end{abstract}
 
 \section{Introduction}
@@ -166,7 +170,8 @@ saddle (see Fyodorov \cite{Fyodorov_2007_Replica} for a detailed discussion).
 
 Here we review the equilibrium solution, which has been studied in detail
 \cite{Crisanti_1992_The, Crisanti_1993_The, Crisanti_2004_Spherical,
-Crisanti_2006_Spherical}.  The free energy, averaged over disorder, is
+Crisanti_2006_Spherical}. For a succinct review, see \cite{Folena_2020_The}.
+The free energy, averaged over disorder, is
 \begin{equation}
   \beta F = - \overline{\ln \int d\mathbf s \;\delta(\|\mathbf s\|^2-N)\, e^{-\beta H(\mathbf s)}}
 \end{equation}
@@ -1322,7 +1327,9 @@ both $\Sigma_1(E_1)$ and $\Sigma_2(E_2)$ are non-zero. In this situation, two sy
 and, because many points contribute, the overlap between two global
 configurations is zero:
 \begin{equation}
-  \frac1{2N}({\mathbf s^1},{\boldsymbol\sigma^1})\cdot ({\mathbf s^2},{\boldsymbol\sigma^2})=\frac1 {2N}[ {\mathbf s^1}\cdot {\mathbf s^2}+ {\boldsymbol\sigma^1}\cdot {\boldsymbol\sigma^2}] =0
+  \frac1 {2N}\Big\langle({\mathbf s^1},{\boldsymbol\sigma^1})\cdot ({\mathbf s^2},{\boldsymbol\sigma^2})\Big\rangle
+  =\frac1 {2N}\Big[\langle{\mathbf s^1}\cdot {\mathbf s^2}\rangle+ \langle{\boldsymbol\sigma^1}\cdot {\boldsymbol\sigma^2}\rangle\Big]
+  =0
 \end{equation}
 This is the `annealed' phase of a Kac-Rice calculation.
 
@@ -1335,8 +1342,12 @@ their ground states.  For $\hat \beta_f > \hat \beta> \hat \beta_c$ one system
 is unfrozen, while the other is, because of coupling, frozen at inverse
 temperature $\hat \beta_c$.  The overlap between two solutions  is:
 \begin{equation}
-  \frac1 {2N}({\mathbf s^1},{\boldsymbol\sigma^1})\cdot ({\mathbf s^2},{\boldsymbol\sigma^2}) =\frac1 {2N}[ {\mathbf s^1}\cdot {\mathbf s^2}+ {\boldsymbol\sigma^1}\cdot {\boldsymbol\sigma^2}] = \frac1 {2N} {\mathbf s^1}\cdot {\mathbf s^2}\neq0
+  \frac1 {2N}\Big\langle({\mathbf s^1},{\boldsymbol\sigma^1})\cdot ({\mathbf s^2},{\boldsymbol\sigma^2})\Big\rangle
+  =\frac1 {2N}\Big[\langle{\mathbf s^1}\cdot {\mathbf s^2}\rangle+ \langle{\boldsymbol\sigma^1}\cdot {\boldsymbol\sigma^2}\rangle\Big]
+  = \frac1 {2N}\langle{\mathbf s^1}\cdot {\mathbf s^2}\rangle>0
 \end{equation}
+nonzero because there are few low-energy stationary points in system one, and
+there is a nonvanishing probability of selecting one of them twice.
 The distribution of this overlap is one-half the overlap distribution of a
 frozen spin-glass at temperature $\hat \beta$, a 1RSB system like the Random
 Energy Model. The value of $x$ corresponding to it depends on $\hat \beta$,
@@ -1345,35 +1356,44 @@ starting at $x=1$ at $\hat \beta_c$ and decreasing with increasing $\hat
 global overlap between different states is at most $1/2$.  At $\hat \beta>\hat
 \beta_f$ there is a further transition.
 
+This schematic example provides a metaphor for considering what happens in
+ordinary models when replica symmetry is broken. At some point certain degree
+of freedom `freeze' onto a subextensive number of possible states, while the
+remainder are effectively unconstrained. The overlap measures something in the
+competition between the number of these unconstrained subregions and their
+size.
+
 \subsection{\textit{R} and \textit{D}: response functions}
 
 The matrix field $R$ is related to responses of the stationary points to
 perturbations of the tensors $J$. One adds to the Hamiltonian a random term
-$\varepsilon \tilde H_p = \varepsilon \sum_{i_1,...,i_p} \tilde J_{i_1,...,i_p}
+$\varepsilon_p \tilde H_p = -\varepsilon_p \sum_{i_1,...,i_p} \tilde J_{i_1,...,i_p}
 s_{i_1}...s_{i_p}$, where the $\tilde J$ are random Gaussian uncorrelated with
-the $J$'s. ] The response to these is:
+the $J$'s and having variance $\overline{(\tilde J)^2}=p!/2N^{p-1}$.  The response to these is:
 \begin{equation}
-  \begin{aligned}
-  & \overline{  \frac{\partial \langle \tilde H_p \rangle_{\tilde J}  }  {\partial \varepsilon} } 
-  %  \frac1{N^p}\sum_{i_1\cdots i_p}\frac{\partial\langle s_{i_1}\cdots s_{i_p}\rangle}{\partial J^{(p)}_{i_1\cdots i_p}}
-  %  &=\lim_{n\to0}\frac1{N^p}\sum_{i_1\cdots i_p}\frac\partial{\partial J^{(p)}_{i_1\cdots i_p}}
-  %  \int\left(\prod_a^nd\nu(\mathbf s_a)\right)\,s^1_{i_1}\cdots s^1_{i_p} \\
-    & =\lim_{n\to0}\int\left(\prod_a^nd\nu(\mathbf s_a)\right)\sum_b^n\left[
+\frac1N\overline{\frac{\partial \langle \tilde H_p \rangle  }  {\partial \varepsilon_p}}
+     =\lim_{n\to0}\int\left(\prod_a^nd\nu(\mathbf s_a)\right)\sum_b^n\left[
       \hat\beta\left(\frac{\mathbf s_1\cdot\mathbf s_b}N\right)^p+
         p\left(-i\frac{\mathbf s_1\cdot\hat{\mathbf s}_b}N\right)\left(\frac{\mathbf s_1\cdot\mathbf s_b}N\right)^{p-1}
       \right]
-  \end{aligned}
 \end{equation}
 Taking the average of this expression over disorder and averaging over the equivalent replicas in the integral gives, similar to before,
 \begin{equation}
   \begin{aligned}
-   \overline{  \frac{\partial \langle \tilde H_p \rangle_{\tilde J}  }  {\partial \varepsilon} } 
+  \frac1N\overline{\frac{\partial\langle \tilde H_p \rangle  }  {\partial \varepsilon_p}}
   %  \overline{\frac1{N^p}\sum_{i_1\cdots i_p}\frac{\partial\langle s_{i_1}\cdots s_{i_p}\rangle}{\partial J^{(p)}_{i_1\cdots i_p}}}
     &=\lim_{n\to0}\int D[C,R,D]\,\frac1n\sum_{ab}^n(\hat\beta C_{ab}^p+pR_{ab}C_{ab}^{p-1})e^{nN\Sigma[C,R,D]}\\
     &=\hat\beta+pr_d-\int_0^1dx\,c^{p-1}(x)(\hat\beta c(x)+pr(x))
   \end{aligned}
 \end{equation}
-In particular, when the energy is unconstrained ($\hat\beta=0$) and there is replica symmetry,
+The responses as defined by this average perturbation in the pure $p$-spin
+energy can be directly related to responses in the tensor polarization of the
+stationary points:
+\begin{equation}
+  \frac1{N^p}\overline{\sum_{i_1\cdots i_p}\frac{\partial\langle s_{i_1}\cdots s_{i_p}\rangle}{\partial J^{(p)}_{i_1\cdots i_p}}}
+  =\frac1N\overline{\frac{\partial\langle \tilde H_p \rangle  }  {\partial \varepsilon_p}}
+\end{equation}
+In particular, when the energy is unconstrained ($\hat\beta=0$) and there is replica symmetry, the above formulas imply that
 \begin{equation}
   \frac1N\sum_i\frac{\partial\langle s_i\rangle}{\partial J_i^{(1)}}
   =r_d
@@ -1415,10 +1435,8 @@ and in particular for $p=1$
 i.e., the change in complexity due to a linear field is directly related to the
 resulting magnetization of the stationary points.
 
-
-
-
 \section{Conclusion}
+\label{se:conclusion}
 
 We have constructed a  replica solution for the general  problem of finding
 saddles of random mean-field landscapes, including systems with many steps of
-- 
cgit v1.2.3-70-g09d2

