From 69bdd429373050bb5aafdf394b2844a037aca9cb Mon Sep 17 00:00:00 2001
From: "kurchan.jorge" <kurchan.jorge@gmail.com>
Date: Wed, 8 Jun 2022 10:06:19 +0000
Subject: Update on Overleaf.

---
 frsb_kac_new.tex | 60 ++++++++++++++++++++++++++++++++++++++++++++++++--------
 1 file changed, 52 insertions(+), 8 deletions(-)

(limited to 'frsb_kac_new.tex')

diff --git a/frsb_kac_new.tex b/frsb_kac_new.tex
index e1c5768..ec0083c 100644
--- a/frsb_kac_new.tex
+++ b/frsb_kac_new.tex
@@ -70,7 +70,7 @@ and one restricts the domain of all integrations to compute saddles of a certain
 of minima with a certain harmonic stiffness, its value is the `softest' mode that adapts to change the Hessian \cite{Fyodorov_2007_Replica}. When it is fixed, then the restriction of the index of saddles is `payed' by the realization of the eigenvalues of the Hessian, usually a
 `harder' mode.
 
-
+{\tiny  NOT SURE WORTHWHILE
 
 \subsection{What to expect?}
 
@@ -110,7 +110,7 @@ between zero and one  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, because a saddle is found by releasing the constraint of staying on the threshold.
 
-
+}
 
 \section{Equilibrium}
 
@@ -174,6 +174,7 @@ The zero temperature limit is most easily obtained by putting $x_i=\tilde x_ix_k
 $F$ is a $k-1$ RSB ansatz with all eigenvalues scaled by $y$ and shifted by $z$. $\tilde x_0=0$ and $\tilde x_k=1$.
 {\em We have lost one level of RSB because at zero temperature the states become points.}
 
+{\bf Jaron: should'nt we put here the continuum solution?}
 
 
 
@@ -270,13 +271,20 @@ $\hat \beta$ is a parameter conjugate to the state energies, i.e. playing the ro
   \end{aligned}
 \end{equation}
 
-Introducing the parameters:
+We introduce the parameters:
 \begin{align}
   Q_{ab}=\frac1Ns_a\cdot s_b &&
   R_{ab}=-i\frac1N\hat s_a\cdot s_b &&
   D_{ab}=\frac1N\hat s_a\cdot\hat s_b
 \end{align}
+The meaning of $R_{ab}$ is that of a response of replica $a$ to a linear
+field in replica $b$:
+\begin{equation}
+    R_{ab} = \frac 1 N \sum_i \overline{\frac{\delta s_i^a}{\delta h_i^b}}
+\end{equation}
+The $D$ may similarly be seen as the variation of the complexity with respect to a random field.
 
+In terms of these parameters, we have
 \begin{equation}
   \begin{aligned}
     S
@@ -330,14 +338,34 @@ We shall make the following ansatz:
   R_{ab}&=&R_d \; \delta_{ab} \nonumber\\
   D_{ab}&=& D_d \; \delta_{ab}
   \label{diagonal}\end{eqnarray}
-This ansatz closes under the operations that are involved in the replicated action.
-The reader who is uneasy about   this ansatz will find another  motivation in Appendix C.
+From what we have seen above, this means that replica $a$ is insensitive to
+a small field applied to replica $b$ if $a \neq b$, a property related to ultrametricity. A similar situation happens in quantum replicated systems, 
+with time appearing only on the diagonal terms: see Appendix C for details.
+
+From its very definition, it is easy to see just perturbing the equations 
+with a field that $R_d$ is the trace of the inverse Hessian, as one expect indeed of a response. Putting: 
+\begin{equation}
+   \mathcal D(\mu)
+    =\frac1N\overline{\log|\det(\partial\partial H(s_a)-\mu I)|}
+    =\int d\lambda\,\rho(\lambda-\mu)\log|\lambda| 
+\end{equation}
+this means that:
+\begin{equation}
+ R_d =    \mathcal D(\mu)'
+\end{equation}
 
 \subsection{Solution}
 
 
 Insert the diagonal ansatz \cite{diagonal} one gets
-
+REINSTATED THIS---------
+\begin{equation}
+  \begin{aligned}
+    \mathcal D(\mu)
+    &=\operatorname{Re}\left\{\frac12\left(1+\frac\mu{2f''(1)}\left(\mu\pm\sqrt{\mu^2-4f''(1)}\right)\right)-\log\left(\frac1{2f''(1)}\left(\mu\pm\sqrt{\mu^2-4f''(1)}\right)\right)\right\}
+  \end{aligned}
+\end{equation}
+--------------------------------------------------
 \begin{equation} \label{eq:diagonal.action}
   \begin{aligned}
     S
@@ -349,7 +377,15 @@ Insert the diagonal ansatz \cite{diagonal} one gets
       +\frac12\lim_{n\to0}\frac1n\left(\hat\epsilon^2\sum_{ab}f(Q_{ab})+\log\det((D_d/R_d^2)Q+I)\right)
   \end{aligned}
 \end{equation}
+
+
+
+
 Using standard manipulations (Appendix B), one finds
+
+
+
+
 \begin{equation} \label{eq:functional.action}
   \begin{aligned}
     S
@@ -367,7 +403,7 @@ Note the close similarity of this action to the equilibrium replica one, at fini
 \begin{equation}
   \beta F=-\frac12\lim_{n\to0}\frac1n\left(\beta^2\sum_{ab}f(Q_{ab})+\log\det Q\right)-1-\log2\pi
 \end{equation}
-I WOULD LOVE TO FIND A DIRECT BRIDGE AT THIS LEVEL
+
 
 \subsubsection{Saddles}
 \label{sec:counting.saddles}
@@ -377,12 +413,15 @@ to $\mu$. This gives
 \begin{equation} \label{eq:mu.saddle}
   0=\frac{\partial S}{\partial\mu}=\mathcal D'(\mu)-R_d
 \end{equation}
+as expected.
 To take the derivative, we must resolve the real part inside the definition of
 $\mathcal D$. When saddles dominate,, $\mu<\mu_m$, and
 \begin{equation}
   \mathcal D(\mu)=\frac12+\frac12\log f''(1)+\frac{\mu^2}{4f''(1)}
 \end{equation}
-It follows that the dominant saddles have $\mu=2f''(1)R_d$.
+It follows that the dominant saddles have $\mu=2f''(1)R_d$. Their index 
+is thus ${\cal{I}}=  $  THIS NEEDS MATHEMATICA
+
 
 \subsubsection{Minima}
 \label{sec:counting.minima}
@@ -400,6 +439,7 @@ When minima dominate, $\mu>\mu_m$ and all the roots inside $\mathcal D(\mu)$ are
 \includegraphics[width=13cm]{frsb_complexity-2.pdf}
 \end{center}
 \end{figure}
+
 \subsubsection{Recovering the replica ground state}
 
 The ground state energy corresponds to that where the complexity of dominant stationary points becomes zero. If the most common stationary points vanish, then there cannot be any stationary points. In this section, we will show that it reproduces the ground state produced by taking the zero-temperature limit in the equilibrium case.
@@ -443,6 +483,10 @@ which is precisely \eqref{eq:ground.state.free.energy} with $R_d=z$ and $\hat\ep
 
 {\em We arrive at one of the main results of our paper: 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 }
 
+\subsection{The continuum situation at a glance}
+
+Here a picture of $\chi$ vs $C$ or $X$ vs $C$ showing limits $q_{max}$, $x_{max}$
+for different energies and typical vs minima.
 
 \section{Ultrametricity rediscovered}
 
-- 
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