Update on Overleaf.

1 files changed, 90 insertions, 10 deletions

diff --git a/frsb_kac-rice.tex b/frsb_kac-rice.tex
index 600c81d..2be5afc 100644
--- a/frsb_kac-rice.tex
+++ b/frsb_kac-rice.tex
@@ -1,13 +1,24 @@
 \documentclass[fleqn]{article}
 
 \usepackage{fullpage,amsmath,amssymb,latexsym}
+\binoppenalty=10000
+\relpenalty=10000
+
+
+\usepackage{float}
+
+% Fix \cal and \mathcal characters look (so it's not the same as \mathscr)
+\DeclareSymbolFont{usualmathcal}{OMS}{cmsy}{m}{n}
+\DeclareSymbolFontAlphabet{\mathcal}{usualmathcal}
+
+
 
 \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
+    of  mean-field complex landscapes. The solution incorporates Parisi's solution
     for equilibrium, as it should.
 \end{abstract}
 \section{Introduction}
@@ -24,6 +35,49 @@ to this date the program has been only complete for a subset of models
 here we present what we believe is the general scheme
 
 
+\subsection{What to expect?}
+
+In order to try to visualize what one should expect, consider two pure p-spin models, with 
+\begin{equation}
+    H = H_1 + H_2=\alpha_1 \sum_{ijk} J^1_{ijk} s_i s_j s_k +
+    \alpha_2 \sum_{ijk} J^2_{ijk} \bar s_i \bar s_j \bar s_k +\epsilon \sum_i s_i \bar s_i
+\end{equation}
+The complexity of the first and second systems in terms of $H_1$ and of $H_2$ 
+have, in the absence of coupling,  the same dependence, but are stretched to one another:
+\begin{equation}
+    \Sigma_1(H_1)= \Sigma_o(H_1/\alpha_1) \qquad ; \qquad \Sigma_2(H_2)= \Sigma_o(H_2/\alpha_2) 
+\end{equation}
+Each system has a ground state energy $E_{gs}^{1,2}$, a threshold energy $E_{thres}^{1,2}$ (a well-defined notion, since we are considering pure p-spins), abd the corresponding limit values $X^{1,2}_{gs}=\left. \frac{d \Sigma_1}{dE_{1,2}}\right|_{E^{gs}_{12}}$
+and $X^{1,2}_{thres}=\left. \frac{d \Sigma_1}{dE_{1,2}}\right|_{E^{thres}_{12}}$
+Considering the cartesian product of both systems, we have, in terms of the total energy
+$H=H_1+H_2$ three regimes:
+\begin{itemize}
+\item  {\bf Unfrozen}: 
+\begin{eqnarray}
+& & X_1 \equiv \frac{d \Sigma_1}{dE_1}=  X_2 \equiv \frac{d \Sigma_2}{dE_2}
+ \end{eqnarray}
+\item  {\bf Semi-frozen}
+As we go down in energy, one of the systems (say, the first) reaches its ground state,
+At lower temperatures, the first system is thus frozen, while the second is not,
+so that $X_1=X_1^{gs}> X_2$. The lowest energy is such that both systems are frozen.
+\item {\bf Semi-threshold }  As we go up from the unfrozen upwards in energy,
+the second system reaches its threshold $X_2^{thres}$. At higher energies minima are extremely rare,
+so the second system remains stuck at its threshold for higher energies.
+
+\item{\bf Both systems reach their thresholds} There essentially no more minima above that.
+\end{itemize}
+Consider now two combined vectors $({\bf s},{\bf \hat s})$ and $({\bf s}',{\bf \hat s}')$
+chosen at the same energies.\\
+
+$\bullet$ Their normalized overlap is close to one when both subsystems are frozen,
+close to a half in the semifrozen phase, and zero at all higher energies.\\
+
+$\bullet$ In phases where one or both systems are stuck in their thresholds (and only in those), the
+minima are exponentially subdominant with respect to saddles.
+
+$\bullet$ {\bf note that the same reasoning leads us to the conclusion that
+minima of two total energies such that one of the systems is frozen have nonzero overlaps}
+
 \section{The model}
 
 Here we consider, for definiteness, the `toy' model introduced by M\'ezard and Parisi
@@ -145,7 +199,7 @@ $F$ is a $k-1$ RSB ansatz with all eigenvalues scaled by $y$ and shifted by $z$.
 
 \section{Kac-Rice}
 
-\subsection{The replicated Kac-Rice problem}
+\subsection{The replicated  problem}
 
 \begin{eqnarray}
 &=& \Pi_a \delta(Eq_a)  \;  \Pi_a \left| \det_a( )\right| \delta(E_a-E(s_a))\nonumber\\
@@ -196,15 +250,32 @@ The second equation implies
 
 \subsection{Motivation}
 
-One may write 
+The reader who is happy with the ansatz may skip this section. 
+ We may encode the original variables in a superspace variable:
+  \begin{equation}
+  \phi_i(1)= q_i(t) + \bar \theta a_i + a_i^\dag \theta + p_i \bar \theta \theta~,
+  \end{equation} 
 \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}
+\begin{aligned}
+{\bf Q}(1,2)&=\frac 1 N \sum_i \phi_i(1) \phi_i (2) =
+Q_{ab} + (\bar \theta_2 - \bar \theta_1)
+\theta_2  R_{ab}
++ \bar \theta_1 \theta_1 R_{ab} + \bar\theta_1\theta_1 \bar \theta_2 \theta_2 D_{ab} \\ 
+&+  \text{odd terms in the $\bar \theta,\theta$}~.
+\end{aligned}
+\label{Q12}
 \end{equation}
+Here $\theta_a$, $\bar \theta_a$ are Grassmann variables, and we denote the full set of coordinates 
+in a  compact form as
+$1=  \theta_1 \overline\theta_1$, $d1=   d\theta_1 d\overline\theta_1$, etc.
+The odd and even fermion numbers decouple, so we can neglect all odd terms in $\theta,\bar{\theta}$.
+
+
+
 
 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
+making an ansatz for replicated quantum problems, which naturally involve a (Matsubara) time. The analogy strongly
 suggests that only the diagonal ${\bf Q}_{aa}$ depend on the $\theta$'s. This boils down
 to putting:
 \begin{eqnarray}
@@ -375,11 +446,20 @@ All that changes is now
 \]
 
 
-\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{Ultrametricity rediscovered}
+
+Three states chosen at the same energy share some common information if there is some `frozen' element common to all. Suppose we choose randomly 
+these states but restrict to those whose overlaps
+take values $Q_{12}$ and $Q_{13}$. Unlike an equilibrium situation, where the Gibbs measure allows us to find such pairs (in a FRSB case) the cost in probability of this in the present case will be exponential.
+Once conditioned this way, we compute $Q_{23}= \min(Q_{12},Q_{13})$
+
+
+
+{\tiny wild guess:
+Consider two states chosen at  different energies, corresponding to maximal values $q^{max}_1$, $q^{max}_2$. Their mutual overlap 
+should be (I am guessing this) $Q_{12}=\min(q^{max}_1, q^{max}_2)$
 
+}
 \section{Conclusion}