From 1e2aea428aeb56acc378d01e014b59b8ae6fff42 Mon Sep 17 00:00:00 2001
From: Jaron Kent-Dobias <jaron@kent-dobias.com>
Date: Wed, 8 Jun 2022 23:23:29 +0200
Subject: Several changes.
---
frsb_kac_new.tex | 197 +++++++++++++++++--------------------------------------
1 file changed, 59 insertions(+), 138 deletions(-)
diff --git a/frsb_kac_new.tex b/frsb_kac_new.tex
index 3db21e4..4d938da 100644
--- a/frsb_kac_new.tex
+++ b/frsb_kac_new.tex
@@ -403,7 +403,7 @@ Inserting the diagonal ansatz \eqref{ansatz} one gets
Using standard manipulations (Appendix B), one finds also a continuous version
\begin{equation} \label{eq:functional.action}
\begin{aligned}
- S
+ \Sigma(\epsilon,\mu)
=\mathcal D(\mu)
+
\hat\beta\epsilon-\mu R_d
@@ -427,7 +427,7 @@ Note the close similarity of this action to the equilibrium replica one, at fini
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 S}{\partial\mu}=\mathcal D'(\mu)-R_d
+ 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
@@ -476,20 +476,17 @@ Adding $2(D_d/R_d)$ times \eqref{eq:saddle.d} to \eqref{eq:saddle.r} and multipl
\begin{equation}
0=-R_d\mu+1+R_d^2f''(1)+f'(1)(R_d\hat\beta-D_d)
\end{equation}
-There are two scenarios: one where the dominant stationary points
-in the vicinity of the ground state are minima, and one where they are saddles. In the case where the dominant stationary points are minima, we can use the optimal $\mu$ from \S\ref{sec:counting.minima}, which gives
+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}
-Therefore, in any situation where minima dominate, the optimal $\mu$ will have $R_d\hat\beta=D_d$.
-
-When the dominant stationary points are saddles, we can use the $\mu$ from \S\ref{sec:counting.saddles}, which implies $R_d=\mu/2f''(1)$ and
-\begin{equation}
- 0=1-\frac{\mu^2}{4f''(1)}+f'(1)(R_d\hat\beta-D_d)
-\end{equation}
-If saddles dominate all the way to the ground state, then they must become marginal minima at the ground state. Therefore at the ground state energy $\mu=\mu_m=\sqrt{4f''(1)}$, and once again $R_d\hat\beta-D_d=0$.
+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.
-In any case, at the ground state $D_d=R_d\hat\beta$. Substituting this into the action, and also substituting the optimal $\mu$ for saddles or minima, and taking $\Sigma(\epsilon_0,\mu^*)=0$, gives
+Substituting this into the action, and also substituting the optimal $\mu$ for
+minima, and 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(
@@ -497,12 +494,60 @@ In any case, at the ground state $D_d=R_d\hat\beta$. Substituting this into the
+\log\det(\hat\beta R_d^{-1} Q+I)
\right)
\end{equation}
-which is precisely \eqref{eq:ground.state.free.energy} with $R_d=z$ and $\hat\beta=\tilde\beta$.
+which is precisely \eqref{eq:ground.state.free.energy} with $R_d=z$,
+$\hat\beta=\tilde\beta$, and $Q=\tilde Q$.
-{\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 }
+{\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
+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
+ &&
+ R_d=\lim_{\beta\to\infty}\beta(1-q_k)
+ &&
+ D_d=\hat\beta R_d
+\end{align}
\subsection{The continuum situation at a glance}
+In the case where any FRSB is present, one must work with the functional form
+of the complexity \eqref{eq:functional.action}, which must be extremized with
+respect to $\chi$ under the conditions that $\chi$ is concave, monotonically
+decreasing, and $\chi(1)=0$, $\chi'(1)=-1$. The annealed case is found by
+taking $\chi(q)=1-q$, which satisfies all of these conditions. $k$-RSB is
+produced by breaking $\chi$ into $k+1$ piecewise linear segments.
+
+Forget for the moment these tricky requirements. The function would then be
+extremized by satisfying
+\begin{equation}
+ 0=\frac{\delta\Sigma}{\delta\lambda(q)}=\frac12\hat\beta^2f''(q)-\frac12\frac1{(\lambda(q)+R_d^2/D_d)^2}
+\end{equation}
+which implies the solution
+\begin{equation}
+ \lambda^*(q)=\frac1{\hat\beta}f''(q)^{-1/2}-\frac{R_d^2}{D_d}
+\end{equation}
+If $f''(q)^{-1/2}$ is not concave anywhere, there is little use of this
+solution. However, if it is concave everywhere it may constitute a portion of
+the full solution.
+
+We suppose that solutions are given by
+\begin{equation}
+ \lambda(q)=\begin{cases}
+ \lambda^*(q) & q<q_\textrm{max} \\
+ 1-q & q\geq q_\textrm{max}
+ \end{cases}
+\end{equation}
+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.
@@ -662,130 +707,6 @@ Integrating by parts,
&=\log[\hat\beta R_d^{-1}\lambda(q_{k-1})+1]+\frac{\hat\beta}{R_d}\int_{q_0^+}^{q_{k-1}}dq\,\frac1{\hat\beta R_d^{-1}\lambda(q)+1}
\end{align*}
-\begin{align*}
- \Sigma
- =-\epsilon\hat\beta+
- \frac12\hat\beta R_df'(1)
- +\frac12\int_0^1dq\,\left[
- \hat\beta^2\lambda(q)f''(q)
- +\frac1{\lambda(q)+R_d/\hat\beta}
- \right]
-\end{align*}
-for $\lambda$ concave, monotonic, $\lambda(1)=0$, and $\lambda'(1)=-1$
-\[
- 0=\frac{\delta\Sigma}{\delta\lambda(q)}=\frac12\hat\beta^2f''(q)-\frac12\frac1{(\lambda(q)+R_d/\hat\beta)^2}
-\]
-\[
- \lambda^*(q)=\frac1{\hat\beta}\left[f''(q)^{-1/2}-R_d\right]
-\]
-
-We suppose that solutions are given by
-\begin{equation}
- \lambda(q)=\begin{cases}
- \lambda^*(q) & q<q^* \\
- 1-q & q\geq q^*
- \end{cases}
-\end{equation}
-where $1-q$ guarantees the boundary conditions at $q=1$, and corresponds to the 0RSB or annealed solutions (annealed Kac--Rice is recovered by substituting in $1-q$ for $\lambda$). We will need to require that $1-q^*=\lambda^*(q^*)$, i.e., continuity.
-
-Inserting this into the complexity, we find
-\begin{align*}
- \Sigma
- &=-\epsilon\hat\beta+\frac12\hat\beta R_df'(1)
- +\frac12\int_0^{q^*}dq\left[
- \hat\beta(f''(q)^{-1/2}-R_d)f''(q)+\hat\beta f''(q)^{1/2}
- \right]
- +\frac12\int_{q^*}^1dq\left[
- \hat\beta^2(1-q)f''(q)+\frac1{q-1+R_d/\hat\beta}
- \right] \\
- &=-\epsilon\hat\beta+\frac12\hat\beta R_d\left[f'(1)-f'(q^*)\right]
- +\hat\beta\int_0^{q^*}dq\,f''(q)^{1/2}
- +\frac12\hat\beta^2\int_{q^*}^1dq\,
- (1-q)f''(q)
- -\log\left[1-(1-q^*)\hat\beta/R_d\right]
-\end{align*}
-$R_d$ can be extremized now, with
-\[
- R_d=\frac12\left(
- (1-q^*)\hat\beta\pm\sqrt{
- (1-q^*)\left(
- (1-q^*)\hat\beta^2+8/[f'(1)-f'(q^*)]
- \right)
- }
- \right)
-\]
-
-This all is for $\mu=\mu^*$, which counts the dominant saddles. We can also count by fixed macroscopic index $\mu$ by leaving it unoptimized in the complexity. This gives
-\[
- F_d=\frac1{2f''(1)}\left[\mu\pm\sqrt{\mu^2-4f''(1)}\right]
-\]
-and
-\begin{align*}
- \Sigma
- =-\epsilon\hat\beta+
- \frac12\hat\beta R_df'(1)
- +\frac12\int_0^1dq\,\left[
- \hat\beta^2\lambda(q)f''(q)
- +\frac1{\lambda(q)+R_d/\hat\beta}
- \right]-\mu R_d+\frac12R_d^2f''(1)+\log R_d\\
- +\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{align*}
-
-
-
-\section{Main result}
-
-\begin{equation}
- \begin{aligned}
- \overline{\Sigma(\epsilon,\mu)}
- =\mathcal D(\mu)
- +\operatorname*{extremum}_{\substack{R_d,D_d,\hat\beta\in\mathbb R\\\chi\in\Lambda}}
- \left\{
- \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 \right.\\\left.
- +\frac12\int_0^1dq\,\left(
- \hat\beta^2f''(q)\chi(q)+\frac1{\chi(q)+R_d^2/D_d}
- \right)
- \right\}
- \end{aligned}
-\end{equation}
-where
-\begin{equation}
- \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{equation}
-and $\Lambda$ is the space of functions $\chi:[0,1]\to[0,1]$ which are
-monotonically decreasing, concave, and have $\chi(1)=0$ and $\chi'(1)=-1$.
-If there is more than one extremum of this function, choose the one with the
-smallest value of $\Sigma$. The sign of the root inside $\mathcal D(\mu)$ is
-negative for $\mu>0$ and positive for $\mu<0$.
-
-The $k$-RSB ansatz is equivalent to piecewise linear $\chi$ with $k+1$
-pieces, with replica symmetric or 0-RSB giving $\chi(q)=1-q$. Our other major
-result is that, if the equilibrium state in the vicinity of zero temperature is
-given by a $k$-RSB ansatz, then the complexity is given by a $(k-1)$-RSB
-ansatz. Moreover, there is an exact correspondence between the parameters of
-the equilibrium saddle point in the limit of zero temperature and those of the
-complexity saddle at the ground state. If the equilibrium is given by
-$x_1,\ldots,x_k$ and $q_1,\ldots,q_k$, then the parameters $\tilde
-x_1,\ldots,\tilde x_{k-1}$ and $\tilde q_1,\ldots,\tilde q_{k-1}$ 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
- &&
- R_d=\lim_{\beta\to\infty}\beta(1-q_k)
- &&
- D_d=R_d\hat\beta
-\end{align}
-
\section{ A motivation for the ansatz}
We may encode the original variables in a superspace variable:
--
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