Update on Overleaf.

1 files changed, 92 insertions, 2 deletions

diff --git a/frsb_kac-rice.tex b/frsb_kac-rice.tex
index 3d06e20..600c81d 100644
--- a/frsb_kac-rice.tex
+++ b/frsb_kac-rice.tex
@@ -3,9 +3,37 @@
 \usepackage{fullpage,amsmath,amssymb,latexsym}
 
 \begin{document}
+\title{Full solution of the Kac-Rice problem for mean-field models}
+\maketitle
+\begin{abstract}
+    We derive the general solution for the computation of saddle points
+    of complex mean-field landscapes. The solution incorporates Parisi's solution
+    for equilibrium, as it should.
+\end{abstract}
+\section{Introduction}
+
+
+Although the Bray-Moore computation for the SK model was the first application of 
+some replica symmetry breaking scheme, it turned out that the problem has been open 
+ever since.
+
+
+
+to this date the program has been only complete for a subset of models 
+
+here we present what we believe is the general scheme
+
+
+\section{The model}
+
+Here we consider, for definiteness, the `toy' model introduced by M\'ezard and Parisi
+
+
 
 \section{Equilibrium}
 
+Here we review the equilibrium solution.
+
 \begin{equation}
   \beta F=\frac12\lim_{n\to0}\frac1n\left(\beta^2\sum_{ab}f(Q_{ab})+\log\det Q\right)-\frac12\log S_\infty
 \end{equation}
@@ -112,8 +140,33 @@ $F$ is a $k-1$ RSB ansatz with all eigenvalues scaled by $y$ and shifted by $z$.
   \right)
 \end{equation}
 
+
+
+
 \section{Kac-Rice}
 
+\subsection{The replicated Kac-Rice problem}
+
+\begin{eqnarray}
+&=& \Pi_a \delta(Eq_a)  \;  \Pi_a \left| \det_a( )\right| \delta(E_a-E(s_a))\nonumber\\
+&\rightarrow& \overline{\Pi_a \delta(Eq_a)}  \; \overline{ \Pi_a \left| \det_a( )\right|\delta(E_a-E(s_a))}\nonumber\\
+\end{eqnarray}
+
+the question of independence
+
+all saddles versus only minima
+
+The parameters:
+\begin{eqnarray}
+Q_{ab}&=&\nonumber\\
+R_{ab}&=&\nonumber\\
+D_{ab}&=&
+\end{eqnarray}
+
+
+
+
+\section{Replicated action}
 \begin{align*}
   \Sigma
   =-\epsilon\hat\epsilon+\lim_{n\to0}\frac1n\left(
@@ -138,6 +191,32 @@ The second equation implies
 \[
   (R^2-DQ)^{-1}=Q^{-1}f'(Q)
 \]
+
+\section{Replica ansatz}
+
+\subsection{Motivation}
+
+One may write 
+\begin{equation}
+    {\bf Q}_{ab}(\bar \theta \theta, \bar \theta ' \theta')=
+Q_{ab} + \bar \theta ... R_{ab} +\bar \theta \theta \bar \theta ' \theta' D_{ab}
+\end{equation}
+
+The variables $\bar \theta \theta$ and $\bar \theta ' \theta'$ play
+the role of `times' in a superspace treatment. We have a long experience of
+making an ansatz for replicated quantum problems. The analogy strongly
+suggests that only the diagonal ${\bf Q}_{aa}$ depend on the $\theta$'s. This boils down
+to putting:
+\begin{eqnarray}
+Q_{ab}&=& {\mbox{ a Parisi matrix}}\nonumber\\
+R_{ab}&=R \delta_{ab}&\nonumber\\
+D_{ab}&=& D \delta_{ab}
+\end{eqnarray}
+Not surprisingly, this ansatz closes, as we shall see.
+
+\subsection{Solution}
+
+
 Insert the diagonal ansatz $R=R_dI$, $D=D_dI$. Then
 \[
   0=(R_df''(1)-\mu)I+\hat\epsilon f'(Q)+R_d(R_d^2I-D_dQ)^{-1}
@@ -195,9 +274,11 @@ Finally, setting $0=\Sigma$ gives
       +\frac1{\hat\epsilon}\log\det(\hat\epsilon R_d^{-1} Q+I)
   \right)
 \]
-which is precisely \eqref{eq:ground.state.free.energy} with $R_d=z$ and $\hat\epsilon=y$. Therefore, a $(k-1)$-RSB ansatz in Kac-Rice will predict the correct ground state energy for a model whose equilibrium state at small temperatures is $k$-RSB.
+which is precisely \eqref{eq:ground.state.free.energy} with $R_d=z$ and $\hat\epsilon=y$. 
+
+{\em Therefore, a $(k-1)$-RSB ansatz in Kac-Rice will predict the correct ground state energy for a model whose equilibrium state at small temperatures is $k$-RSB.}
 
-\section{Full}
+\subsection{Full}
 
 \begin{align*}
   \lim_{n\to0}\frac1n\log\det(\hat\epsilon R_d^{-1} Q+I)
@@ -293,4 +374,13 @@ All that changes is now
   \right)
 \]
 
+
+\section{Overview of the landscape}
+For the full model minima are always dominated exponentially by saddles, whose
+index density  goes smoothly down to zero with energy density. (I hope) 
+The meaning of the parameter $Q_{ab}$ is  ????????
+
+\section{Conclusion}
+
+
 \end{document}