From d981124dd115fed96915350a510f30e4e94d72df Mon Sep 17 00:00:00 2001
From: Jaron Kent-Dobias <jaron@kent-dobias.com>
Date: Tue, 28 Jun 2022 15:30:02 +0200
Subject: Rewrote the solution section to make it honest.

---
 frsb_kac-rice.tex | 328 +++++++++++++++++++++---------------------------------
 1 file changed, 126 insertions(+), 202 deletions(-)

(limited to 'frsb_kac-rice.tex')

diff --git a/frsb_kac-rice.tex b/frsb_kac-rice.tex
index cd295e9..2015feb 100644
--- a/frsb_kac-rice.tex
+++ b/frsb_kac-rice.tex
@@ -1,7 +1,14 @@
 \documentclass[fleqn]{article}
 
 \usepackage{fullpage,amsmath,amssymb,latexsym,graphicx}
-\usepackage{appendix}
+\usepackage{appendix,xcolor}
+\usepackage[
+  colorlinks=true,
+  urlcolor=purple,
+  citecolor=purple,
+  filecolor=purple,
+  linkcolor=purple
+]{hyperref} % ref and cite links with pretty colors
 
 \begin{document}
 \title{Full solution of the Kac--Rice problem for mean-field models.\\
@@ -303,268 +310,159 @@ role of an inverse temperature for the metastable states. The average over disor
 
 We introduce new fields
 \begin{align}
-  Q_{ab}=\frac1Ns_a\cdot s_b &&
+  C_{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}
-$Q_{ab}$ is the overlap between spins belonging to different replicas. 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.
+$C_{ab}$ is the overlap between spins belonging to different replicas.
 
-By substituting these parameters into the expressions above and then making a change of variables in the integration from $s_a$ and $\hat s_a$ to these three matrices, we arrive at the form for the complexity
+By substituting these parameters into the expressions above and then making a
+change of variables in the integration from $s_a$ and $\hat s_a$ to these three
+matrices, we arrive at the form for the complexity
 \begin{equation}
   \begin{aligned}
     &\Sigma(\epsilon,\mu)
     =\mathcal D(\mu)+\hat\beta\epsilon+\\
     &\lim_{n\to0}\frac1n\left(
-      -\mu\sum_a^nR_{aa}
+      -\mu\operatorname{Tr}R
       +\frac12\sum_{ab}\left[
-        \hat\beta^2f(Q_{ab})+(2\hat\beta R_{ab}-D_{ab})f'(Q_{ab})
-        +R_{ab}^2f''(Q_{ab})
+        \hat\beta^2f(C_{ab})+(2\hat\beta R_{ab}-D_{ab})f'(C_{ab})
+        +R_{ab}^2f''(C_{ab})
       \right]
-    +\frac12\log\det\begin{bmatrix}Q&iR\\iR&D\end{bmatrix}
+    +\frac12\log\det\begin{bmatrix}C&iR\\iR&D\end{bmatrix}
     \right)
   \end{aligned}
 \end{equation}
-where $\hat\beta$, $Q$, $R$ and $D$ must be evaluated at extrema of this
-expression. With $Q$, $R$, and $D$ distinct replica matrices, this is
-potentially quite challenging.
-
-\begin{equation}
-  \begin{aligned}
-    &\Sigma(\epsilon,\mu)
-    =-\frac12+\mathcal D(\mu)+\hat\beta\epsilon+\\
-    &\lim_{n\to0}\frac1n\left(
-      -\mu\sum_a^nR_{aa}
-      +\frac12\sum_{ab}\left[
-        \hat\beta^2f(Q_{ab})+R_{ac}(2\hat\beta\delta_{cb}+Q^{-1}_{cd}R_{db})f'(Q_{ab})
-        +R_{ab}^2f''(Q_{ab})
-      \right]
-      +\frac12\log\det Q - \frac12\log\det f'(Q)
-    \right)
-  \end{aligned}
-\end{equation}
-
-{\bf  Inserting the expression for $D$ in the action, and writing $R=R_d+\tilde R$ ($\tilde R$ is zero in the diagonal) , the linear term in $\tilde R$ is 
-$Tr \left[\tilde R (\hat \beta -R_d Q^{-1} f')\right]$ }
+where $\hat\beta$, $C$, $R$ and $D$ must be evaluated at extrema of this
+expression.
 
 \section{Replica ansatz}
 
-We shall make the following ansatz for the saddle point:
-\begin{align}\label{ansatz}
-  Q_{ab}= \text{a Parisi matrix} &&
-  R_{ab}=R_d \delta_{ab} &&
-  D_{ab}= D_d \delta_{ab}
+Based on previous work on the SK model and the equilibrium solution of the
+spherical model, we expect $C$, and $R$ and $D$ to be hierarchical matrices,
+i.e., to follow Parisi's scheme. This assumption immediately simplifies the
+extremal conditions, since hierarchical matrices commute and are closed under
+matrix products and Hadamard products. The extremal conditions are
+\begin{align}
+  0&=\frac{\partial\Sigma}{\partial\hat\beta}
+  =\epsilon+\sum_{ab}\left[\hat\beta f(C_{ab})+R_{ab}f'(C_{ab})\right] \label{eq:cond.b} \\
+  \frac{\tilde c}2I&=\frac{\partial\Sigma}{\partial C}
+  =\frac12\left[
+    \hat\beta^2f'(C)+(2\hat\beta R-D)\odot f''(C)+R\odot R\odot f'''(C)
+    +(CD+R^2)^{-1}D
+  \right] \label{eq:cond.q} \\
+  0&=\frac{\partial\Sigma}{\partial R}
+  =-\mu I+\hat\beta f'(C)+R\odot f''(C)
+  +(CD+R^2)^{-1}R \label{eq:cond.r} \\
+  0&=\frac{\partial\Sigma}{\partial D}
+  =-\frac12f'(C)
+    +\frac12(CD+R^2)^{-1}C \label{eq:cond.d}
 \end{align}
-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.
+where $\odot$ denotes the Hadamard product, or the componentwise product. The
+equation for \eqref{eq:cond.q} would not be true on the diagonal save for the
+arbitrary factor $\tilde c$.  Equation \eqref{eq:cond.d} implies that
+\begin{equation} \label{eq:D.solution}
+  D=f'(C)^{-1}-RC^{-1}R
+\end{equation}
 
-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.
-\begin{equation}
- R_d =    \mathcal D'(\mu)
+In addition to these equations, one is also often interested in maximizing the complexity as a function of $\mu$, to find the dominant or most common type of stationary points. These are given by the condition
+\begin{equation} \label{eq:cond.mu}
+  0=\frac{\partial\Sigma}{\partial\mu}
+  =\mathcal D'(\mu)-r_d
 \end{equation}
-Similarly, ....  one shows that 
-\begin{equation}
-D_d = \hat \beta R_d
+Since $\mathcal D(\mu)$ is effectively a piecewise function, with different forms for $\mu$ greater or less than $\mu_m$, there are two regimes. When $\mu>\mu_m$, the critical points are minima, and \eqref{eq:cond.mu} implies
+\begin{equation} \label{eq:mu.minima}
+  \mu=\frac1{r_d}+r_df''(1)
+\end{equation}
+When $\mu<\mu_m$, they are saddles, and
+\begin{equation} \label{eq:mu.saddles}
+  \mu=2f''(1)r_d
 \end{equation}
 
+\section{Supersymmetric solution}
 
-Is it a saddle?
+The Kac--Rice problem has an approximate supersymmetry, which is found when the
+absolute value of the determinant is neglected. When this is done, the
+determinant can be represented by an integral over Grassmann variables, which
+yields a complexity depending on `bosons' and `fermions' that share the
+supersymmetry. The Ward identities associated with the supersymmetry imply
+that $D=\hat\beta R$ \cite{Annibale_2003_The}. Under which conditions can this relationship be expected to hold?
 
+Any result of supersymmetry can only be valid when the symmetry itself is
+valid, which means the determinant must be positive. This is only guaranteed
+for minima, which have $\mu>\mu_m$. Moreover, this identity heavily constrains
+the form that the rest of the solution can take. Assuming the supersymmetry holds, \eqref{eq:cond.q} implies
 \begin{equation}
-  0=\frac{\partial\Sigma}{\partial R}
-  =\lim_{n\to0}\frac1n\left(-\mu I+\hat\beta f'(Q)+R\odot f''(Q)
-    +(QD+R^2)^{-1}R
-  \right)
+  \tilde cI=\hat\beta^2f'(C)+\hat\beta R\odot f''(C)+R\odot R\odot f'''(C)+\hat\beta(CD+R^2)^{-1}R
 \end{equation}
-\begin{equation}
-  0=\frac{\partial\Sigma}{\partial D}
-  =\lim_{n\to0}\frac1n\left(-\frac12f'(Q)
-    +\frac12(QD+R^2)^{-1}Q
-  \right)
-\end{equation}
-\begin{equation}
-  D=f'(Q)^{-1}-RQ^{-1}R
-\end{equation}
-Diagonal anstaz requires that $f'(Q_{ab})^{-1}=R_d^2Q_{ab}^{-1}$ for $a\neq b$.
-\begin{equation}
-  0
-  =\lim_{n\to0}\frac1n\left(-\mu I+\hat\beta f'(Q)+R\odot f''(Q)
-    +Q^{-1}Rf'(Q)
-  \right)
-\end{equation}
-Diagonal ansatz requires that
-\begin{equation}
-  0=-\mu I+\hat\beta f'(Q)+R_df''(1)I+R_dQ^{-1}f'(Q)
-\end{equation}
-or $\hat\beta f'(Q_{ab})=-R_d[Q^{-1}f'(Q)]_{ab}$ for $a\neq b$.
-We also have from the first (before the diagonal ansatz) $Df'(Q)=I-RQ^{-1}Rf'(Q)$. Inserting the diagonal, we get $Q^{-1}f'(Q)=\frac1{R_d^2}(I-D_df'(Q))$, and
-\begin{equation}
-  0=-\mu I+\hat\beta f'(Q)+R_df''(1)I+\frac1{R_d}(I-D_df'(Q))
-  =(R_df''(1)+R_d^{-1}-\mu)I+(\hat\beta-D_d/R_d)f'(Q)
+Substituting \eqref{eq:cond.r} for the factor $(CD+R^2)^{-1}R$, we find substantial cancellation, and finally
+\begin{equation} \label{eq:R.diagonal}
+  (\tilde c-\mu)I=R\odot R\odot f'''(C)
 \end{equation}
+If $C$ has a nontrivial off-diagonal structure and supersymmetry holds, then
+the off-diagonal of $R$ must vanish, and therefore $R=r_dI$. Therefore, a
+supersymmetric ansatz is equivalent to a \emph{diagonal} ansatz.
 
-Always true:
-\begin{equation}
-  0=-\mu R+\hat\beta Rf'(Q)+R(R\odot f''(Q))+I-Df'(Q)
-\end{equation}
-Insert $D=D_dI+\delta D$, $R=R_dI+\delta R$, and $D_d=\hat\beta R_d$
+Supersymmetry has further implications.
+Equations \eqref{eq:cond.r} and \eqref{eq:cond.d} can be combined to find
 \begin{equation}
-  0=-\hat\beta \mu R_d I +\hat\beta R_df'(Q)+R_d^2f''(1)I+I-\hat\beta R_df'(Q)
-  -\mu\delta R
+  I=R\left[\mu I-R\odot f''(C)\right]+(D-\hat\beta R)f'(C)
 \end{equation}
-
+Assuming the supersymmetry holds implies that
 \begin{equation}
-  0=(\hat\beta Q+R) f'(Q)+(R\odot f''(Q)-\mu I)Q
+  I=R\left[\mu I-R\odot f''(C)\right]
 \end{equation}
-
+Understanding that $R$ is diagonal, this implies
 \begin{equation}
-  \begin{aligned}
-    &\Sigma(\epsilon,\mu)
-    =\frac12+\mathcal D(\mu)+\hat\beta\epsilon+\\
-    &\lim_{n\to0}\frac1n\left(
-      -\frac12\mu\sum_a^nR_{aa}
-      +\frac12\sum_{ab}\left[
-        \hat\beta^2f(Q_{ab})+\hat\beta R_{ab}f'(Q_{ab})
-      \right]
-      +\frac12\log\det(f'(Q)^{-1}Q)
-    \right)
-  \end{aligned}
+  \mu=\frac1{r_d}+r_df''(1)
 \end{equation}
+which is precisely the condition \eqref{eq:mu.minima}. Therefore, \emph{the
+supersymmetric solution only counts dominant minima.}
 
-
-\subsection{Solution}
-
-
-Inserting the diagonal ansatz \eqref{ansatz} one gets
+Inserting the supersymmetric ansatz $D=\hat\beta R$ and $R=r_dI$, one gets
 \begin{equation} \label{eq:diagonal.action}
   \begin{aligned}
     \Sigma(\epsilon,\mu)
     =\mathcal D(\mu)
     +
-      \hat\beta\epsilon-\mu R_d
-      +\frac12(2\hat\beta R_d-D_d)f'(1)+\frac12R_d^2f''(1)+\frac12\log R_d^2
-      \\
-      +\frac12\lim_{n\to0}\frac1n\left(\hat\beta^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 also a continuous version
-\begin{equation} \label{eq:functional.action}
-  \begin{aligned}
-    \Sigma(\epsilon,\mu)
-    =\mathcal D(\mu)
-    +
-      \hat\beta\epsilon-\mu R_d
-      +\frac12(2\hat\beta R_d-D_d)f'(1)+\frac12R_d^2f''(1)+\frac12\log R_d^2
+      \hat\beta\epsilon-\mu r_d
+      +\frac12\hat\beta r_df'(1)+\frac12r_d^2f''(1)+\frac12\log r_d^2
       \\
-      +\frac12\int_0^1dq\,\left(
-        \hat\beta^2f''(q)\chi(q)+\frac1{\chi(q)+R_d^2/D_d}
-      \right)
+      +\frac12\lim_{n\to0}\frac1n\left(\hat\beta^2\sum_{ab}f(C_{ab})+\log\det((\hat\beta/r_d)C+I)\right)
   \end{aligned}
 \end{equation}
-where $\chi(q)$ is defined with respect to $Q$ exactly as in the equilibrium case.
-Note the close similarity of this action to the equilibrium replica one, at finite temperature.
-\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}
-
-
-\subsubsection{Saddles}
-\label{sec:counting.saddles}
-
-The dominant stationary points are given by maximizing the action with respect
-to $\mu$. This gives
-\begin{equation} \label{eq:mu.saddle}
-  0=\frac{\partial\Sigma}{\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 $\mu=2f''(1)R_d$. Their index
-is thus $\mathcal I=\frac12N(1-R_d\sqrt{f''(1)})$. If
-$R_d\sqrt{f''(1)}>1$ then we were wrong to assume that saddles dominate, and
-the most numerous saddles will be found just above $\mu=\mu_m$.
-
-\subsubsection{Minima}
-\label{sec:counting.minima}
-
-When minima dominate, $\mu>\mu_m$ and all the roots inside $\mathcal D(\mu)$ are real. Therefore $\mathcal D(\mu)$ is given by its former expression with the real part dropped, and
-\begin{equation}
-  \mathcal D'(\mu)=\frac{2}{\mu+\sqrt{\mu^2-4f''(1)}}
-\end{equation}
-\begin{equation}
-  \mu=\frac1{R_d}+R_df''(1)
-\end{equation}
-
-\subsubsection{Recovering the equilibrium 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.
-
-Consider the extremum problem of \eqref{eq:diagonal.action} with respect to $R_d$ and $D_d$. This gives the equations
-\begin{align}
-  0
-  &=\frac{\partial\Sigma}{\partial D_d}
-  =-\frac12f'(1)+\frac12\frac1{R_d^2}\lim_{n\to0}\frac1n\operatorname{Tr}((D_d/R_d^2)Q+I)^{-1}Q \label{eq:saddle.d}\\
-  0
-  &=\frac{\partial\Sigma}{\partial R_d}
-  =-\mu+\hat\beta f'(1)+R_df''(1)+\frac1{R_d}-\frac{D_d}{R_d^3}\lim_{n\to0}\frac1n\operatorname{Tr}((D_d/R_d^2)Q+I)^{-1}Q \label{eq:saddle.r}
-\end{align}
-Adding $2(D_d/R_d)$ times \eqref{eq:saddle.d} to \eqref{eq:saddle.r} and multiplying by $R_d$ gives
-\begin{equation}
-  0=-R_d\mu+1+R_d^2f''(1)+f'(1)(R_d\hat\beta-D_d)
-\end{equation}
-At the ground state, minima will always dominate (even if marginal). We can
-therefore use the optimal $\mu$ from \S\ref{sec:counting.minima}, which gives
-\begin{equation}
-  0=f'(1)(R_d\hat\beta-D_d)
-\end{equation}
-In order to satisfy this equation we must have
-$D_d=R_d\hat\beta$. This relationship holds for the most common minima whenever
-they dominante, including in the ground state.
-
-Substituting this into the action, and also substituting the optimal $\mu$ for
-minima, and taking $\Sigma(\epsilon_0,\mu^*)=0$, gives
+From here, it is straightforward to see that the complexity vanishes at the
+ground state energy. First, in the ground state minima will dominate (even if
+they are marginal), so we may assume \eqref{eq:mu.minima}. Then, taking
+$\Sigma(\epsilon_0,\mu^*)=0$, gives
 \begin{equation}
   \hat\beta\epsilon_0
-  =-\frac12R_d\hat\beta f'(1)-\frac12\lim_{n\to0}\frac1n\left(
-      \hat\beta^2\sum_{ab}^nf(Q_{ab})
-      +\log\det(\hat\beta R_d^{-1} Q+I)
+  =-\frac12r_d\hat\beta f'(1)-\frac12\lim_{n\to0}\frac1n\left(
+      \hat\beta^2\sum_{ab}^nf(C_{ab})
+      +\log\det(\hat\beta r_d^{-1} C+I)
   \right)
 \end{equation}
-which is precisely \eqref{eq:ground.state.free.energy} with $R_d=z$,
-$\hat\beta=\tilde\beta$, and $Q=\tilde Q$.
+which is precisely \eqref{eq:ground.state.free.energy} with $r_d=z$,
+$\hat\beta=\tilde\beta$, and $C=\tilde Q$.
 
 {\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 } Moreover, there is an
 exact correspondance between the saddle parameters of each.  If the equilibrium
 is given by a Parisi matrix with parameters $x_1,\ldots,x_k$ and
-$q_1,\ldots,q_k$, then the parameters $\hat\beta$, $R_d$, $D_d$, $\tilde
-x_1,\ldots,\tilde x_{k-1}$, and $\tilde q_1,\ldots,\tilde q_{k-1}$ for the
+$q_1,\ldots,q_k$, then the parameters $\hat\beta$, $r_d$, $d_d$, $\tilde
+x_1,\ldots,\tilde x_{k-1}$, and $\tilde c_1,\ldots,\tilde c_{k-1}$ for the
 complexity in the ground state are
 \begin{align}
   \hat\beta=\lim_{\beta\to\infty}\beta x_k
   &&
   \tilde x_i=\lim_{\beta\to\infty}\frac{x_i}{x_k}
   &&
-  \tilde q_i=\lim_{\beta\to\infty}q_i
+  \tilde c_i=\lim_{\beta\to\infty}q_i
   &&
-  R_d=\lim_{\beta\to\infty}\beta(1-q_k)
+  r_d=\lim_{\beta\to\infty}\beta(1-q_k)
   &&
-  D_d=\hat\beta R_d
+  d_d=\hat\beta r_d
 \end{align}
 
 \section{Examples}
@@ -634,6 +532,21 @@ E\rangle_2$.
 \end{figure}
 
 \subsection{Full RSB complexity}
+Using standard manipulations (Appendix B), one finds also a continuous version
+\begin{equation} \label{eq:functional.action}
+  \begin{aligned}
+    \Sigma(\epsilon,\mu)
+    =\mathcal D(\mu)
+    +
+      \hat\beta\epsilon-\mu R_d
+      +\frac12(2\hat\beta R_d-D_d)f'(1)+\frac12R_d^2f''(1)+\frac12\log R_d^2
+      \\
+      +\frac12\int_0^1dq\,\left(
+        \hat\beta^2f''(q)\chi(q)+\frac1{\chi(q)+R_d^2/D_d}
+      \right)
+  \end{aligned}
+\end{equation}
+where $\chi(q)$ is defined with respect to $Q$ exactly as in the equilibrium case.
 
 \begin{figure}
   \centering
@@ -714,6 +627,17 @@ Continuity requires that $1-q_\textrm{max}=\lambda^*(q_\textrm{max})$.
 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{Interpretation}
+
+ 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.
+
+
 \section{Ultrametricity rediscovered}
 
 TENTATIVE BUT INTERESTING
-- 
cgit v1.2.3-70-g09d2

