From 919c8556bf2c4a8bfbd46774590d398461d556d0 Mon Sep 17 00:00:00 2001
From: Jaron Kent-Dobias <jaron@kent-dobias.com>
Date: Fri, 3 Jun 2022 18:06:10 +0200
Subject: Initial commit.

---
 frsb_kac-rice.tex | 242 ++++++++++++++++++++++++++++++++++++++++++++++++++++++
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+\documentclass[fleqn]{article}
+
+\usepackage{fullpage,amsmath,amssymb,latexsym}
+
+\begin{document}
+
+\section{Equilibrium}
+
+\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}
+$\log S_\infty=1+\log2\pi$.
+\begin{align*}
+  \beta F=
+  -\frac12\log S_\infty+
+  \frac12\lim_{n\to0}\frac1n\left(\beta^2nf(1)+\beta^2\sum_{i=0}^kn(x_i-x_{i+1})f(q_i)
+  +\log\left[
+    \frac{
+      1+\sum_{i=0}^k(x_i-x_{i+1})q_i
+    }{
+      1+\sum_{i=1}^k(x_i-x_{i+1})q_i-x_1q_0
+    }
+  \right]\right.\\
+  +\frac n{x_1}\log\left[
+      1+\sum_{i=1}^k(x_i-x_{i+1})q_i-x_1q_0
+  \right]\\
+  \left.+\sum_{j=1}^kn(x_{j+1}^{-1}-x_j^{-1})\log\left[
+      1+\sum_{i=j+1}^k(x_i-x_{i+1})q_i-x_{j+1}q_j
+  \right]
+\right)
+\end{align*}
+
+\begin{align*}
+  \lim_{n\to0}\frac1n
+  \log\left[
+    \frac{
+      1+\sum_{i=0}^k(x_i-x_{i+1})q_i
+    }{
+      1+\sum_{i=1}^k(x_i-x_{i+1})q_i-x_1q_0
+    }
+  \right]
+  &=
+  \lim_{n\to0}\frac1n
+  \log\left[
+    \frac{
+      1+\sum_{i=0}^k(x_i-x_{i+1})q_i
+    }{
+      1+\sum_{i=0}^k(x_i-x_{i+1})q_i-nq_0
+    }
+  \right] \\
+  &=q_0\left(1+\sum_{i=0}^k(x_i-x_{i+1})q_i\right)^{-1}
+\end{align*}
+
+
+\begin{align*}
+  \beta F=
+  -\frac12\log S_\infty+
+  \frac12\left(\beta^2f(1)+\beta^2\sum_{i=0}^k(x_i-x_{i+1})f(q_i)
+  +q_0\left(1+\sum_{i=0}^k(x_i-x_{i+1})q_i\right)^{-1}\right. \\
+  +\frac1{x_1}\log\left[
+      1+\sum_{i=1}^{k}(x_i-x_{i+1})q_i-x_1q_0
+  \right]\\
+  \left.+\sum_{j=1}^k(x_{j+1}^{-1}-x_j^{-1})\log\left[
+      1+\sum_{i=j+1}^{k}(x_i-x_{i+1})q_i-x_{j+1}q_j
+  \right]
+\right)
+\end{align*}
+$q_0=0$
+\begin{align*}
+  \beta F=
+  -\frac12\log S_\infty+
+  \frac12\left(\beta^2f(1)+\beta^2\sum_{i=0}^k(x_i-x_{i+1})f(q_i)
+  +\frac1{x_1}\log\left[
+      1+\sum_{i=1}^{k}(x_i-x_{i+1})q_i
+  \right]\right.\\
+  \left.+\sum_{j=1}^k(x_{j+1}^{-1}-x_j^{-1})\log\left[
+      1+\sum_{i=j+1}^{k}(x_i-x_{i+1})q_i-x_{j+1}q_j
+  \right]
+\right)
+\end{align*}
+$x_i=\tilde x_ix_k$, $x_k=y/\beta$, $q_k=1-z/\beta$
+\begin{align*}
+  \beta F=
+  -\frac12\log S_\infty+
+  \frac12\left(\beta^2f(1)+\beta^2(y\beta^{-1}-1)f(1-z\beta^{-1})+y\beta\sum_{i=0}^{k-1}(\tilde x_i-\tilde x_{i+1})f(q_i)\right. \\
+  +\frac\beta{\tilde x_1 y}\log\left[
+    y\sum_{i=1}^{k-1}(\tilde x_i-\tilde x_{i+1})q_i+y+z-yz/\beta
+  \right]\\
+  +\sum_{j=1}^{k-1}\frac\beta y(\tilde x_{j+1}^{-1}-\tilde x_j^{-1})\log\left[
+    y\sum_{i=j+1}^{k-1}(\tilde x_i-\tilde x_{i+1})q_i+y+z-yz/\beta-y\tilde x_{j+1}q_j
+\right]\\
+    \left.-\frac\beta{\tilde x_1 y}\log\beta-\sum_{j=1}^{k-1}\frac\beta y(\tilde x_{j+1}^{-1}-\tilde x_j^{-1})\log\beta+(1-\beta y^{-1})\log\left[
+      z/\beta
+  \right]
+\right)
+\end{align*}
+\begin{align*}
+  \lim_{\beta\to\infty}F=
+  \frac12\left(yf(1)+zf'(1)+y\sum_{i=0}^{k-1}(\tilde x_i-\tilde x_{i+1})f(q_i)
+  +\frac1{\tilde x_1 y}\log\left[
+    y\sum_{i=1}^{k-1}(\tilde x_i-\tilde x_{i+1})q_i+y+z
+  \right]\right.\\
+  \left.+\sum_{j=1}^{k-1}\frac1 y(\tilde x_{j+1}^{-1}-\tilde x_j^{-1})\log\left[
+    y\sum_{i=j+1}^{k-1}(\tilde x_i-\tilde x_{i+1})q_i+y+z-y\tilde x_{j+1}q_j
+    \right]
+    -\frac1y\log z
+\right)
+\end{align*}
+$F$ is a $k-1$ RSB ansatz with all eigenvalues scaled by $y$ and shifted by $z$. $\tilde x_0=0$ and $\tilde x_k=1$.
+\begin{equation} \label{eq:ground.state.free.energy}
+  \lim_{\beta\to\infty}F=\lim_{n\to0}\frac1n\frac12\left(nzf'(1)+y\sum_{ab}f(\tilde Q_{ab})+\frac1y\log\det(yz^{-1}\tilde Q+I)
+  \right)
+\end{equation}
+
+\section{Kac-Rice}
+
+\begin{align*}
+  \Sigma
+  =-\epsilon\hat\epsilon+\lim_{n\to0}\frac1n\left(
+    \sum_a\mu(F_{aa}-R_{aa})
+    +\frac12\sum_{ab}\left[
+      \hat\epsilon^2f(Q_{ab})+2\hat\epsilon R_{ab}f'(Q_{ab})
+      +D_{ab}f'(Q_{ab})+R_{ab}^2f''(Q_{ab})-F_{ab}^2f''(Q_{ab})
+    \right]\right.\\\left.
+  +\frac12\log\det\begin{bmatrix}Q&-iR\\-iR&-D\end{bmatrix}
+  -\log\det F
+  \right)
+\end{align*}
+\[
+  0=\frac{\partial\Sigma}{\partial R_{ab}}
+  =-\mu\delta_{ab}+\hat\epsilon f'(Q_{ab})+R_{ab}f''(Q_{ab})+\sum_c(R^2-DQ)^{-1}_{ac}R_{cb}
+\]
+\[
+  0=\frac{\partial\Sigma}{\partial D_{ab}}
+  =\frac12 f'(Q_{ab})-\frac12\sum_c(R^2-DQ)^{-1}_{ac}Q_{cb}
+\]
+The second equation implies
+\[
+  (R^2-DQ)^{-1}=Q^{-1}f'(Q)
+\]
+Insert the diagonal ansatz $R=R_dI$, $D=D_dI$. Then
+\[
+  0=(R_df''(1)-\mu)I+\hat\epsilon f'(Q)+R_d(R_d^2I-D_dQ)^{-1}
+  =(R_df''(1)-\mu)I+\hat\epsilon f'(Q)+R_dQ^{-1}f'(Q)
+\]
+and
+\[
+  Q^{-1}f'(Q)=(I+D_df'(Q))/R_d^2
+\]
+Substituting the second into the first, we have
+\[
+  0=(R_df''(1)-\mu)I+\hat\epsilon f'(Q)+\frac1{R_d}(I+D_df'(Q))
+\]
+\[
+  0=(R_df''(1)-\mu+R_d^{-1})I+(\hat\epsilon+D_d/R_d)f'(Q)
+\]
+The only way for this equation to be satisfied off the diagonal for nontrivial $Q$ is for $D_d=-R_d\hat\epsilon$. We therefore have
+\begin{align*}
+  \Sigma
+  =-\epsilon\hat\epsilon+\lim_{n\to0}\frac1n\left(
+    n\mu(F_d-R_d)+\frac12n\left[
+      \hat\epsilon R_df'(1)+R_d^2f''(1)-F_d^2f''(1)
+    \right]
+    +\frac12\sum_{ab}
+      \hat\epsilon^2f(Q_{ab})
+    ]\right.\\\left.
+    +\frac12\log\det(\hat\epsilon R_d^{-1} Q+I)
+    +n\log R_d
+  -n\log F_d
+  \right)
+\end{align*}
+Taking the saddle with respect to $\mu$ and $F_d$ yields
+\[
+  F_d=R_d
+\]
+\[
+  \mu=R_d^{-1}(1+R_d^2f''(1))
+\]
+and gives
+\begin{align*}
+  \Sigma
+  =-\epsilon\hat\epsilon+\lim_{n\to0}\frac1n\frac12\left(
+    n
+      \hat\epsilon R_df'(1)
+    +\hat\epsilon^2\sum_{ab}
+      f(Q_{ab})
+    +\log\det(\hat\epsilon R_d^{-1} Q+I)
+  \right)
+\end{align*}
+Finally, setting $0=\Sigma$ gives
+\[
+  \epsilon
+  =\lim_{n\to0}\frac1n\frac12\left(nR_df'(1)+\hat\epsilon\sum_{ab}
+      f(Q_{ab})
+      +\frac1{\hat\epsilon}\log\det(\hat\epsilon R_d^{-1} Q+I)
+  \right)
+\]
+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.
+
+\section{Full}
+
+\begin{align*}
+  \lim_{n\to0}\frac1n\log\det(\hat\epsilon R_d^{-1} Q+I)
+  =x_1^{-1}\log\left(\hat\epsilon R_d^{-1}(1-\bar q_k)+1\right)+\int_{q_0^+}^{q_{k-1}}dq\,\mu(q)\log\left[\hat\epsilon R_d^{-1}\lambda(q)+1\right]
+\end{align*}
+where
+\[
+  \mu(q)=\frac{\partial x^{-1}(q)}{\partial q}
+\]
+Integrating by parts,
+\begin{align*}
+  \lim_{n\to0}\frac1n\log\det(\hat\epsilon R_d^{-1} Q+I)
+  &=x_1^{-1}\log\left(\hat\epsilon R_d^{-1}(1-\bar q_k)+1\right)+\left[x^{-1}(q)\log[\hat\epsilon R_d^{-1}\lambda(q)+1]\right]_{q=q_0^+}^{q=q_{k-1}}-\frac{\hat\epsilon}{R_d}\int_{q_0^+}^{q_{k-1}}dq\,\frac{\lambda'(q)}{x(q)}\frac1{\hat\epsilon R_d^{-1}\lambda(q)+1}\\
+  &=\log[\hat\epsilon R_d^{-1}\lambda(q_{k-1})+1]+\frac{\hat\epsilon}{R_d}\int_{q_0^+}^{q_{k-1}}dq\,\frac1{\hat\epsilon R_d^{-1}\lambda(q)+1}
+\end{align*}
+
+\begin{align*}
+  \Sigma
+  =-\epsilon\hat\epsilon+
+    \frac12\hat\epsilon R_df'(1)
+    +\frac12\int_0^1dq\,\left[
+      \hat\epsilon^2\lambda(q)f''(q)
+      +\frac1{\lambda(q)+R_d/\hat\epsilon}
+    \right]
+\end{align*}
+for $\lambda$ concave, monotonic, and $\lambda(1)=0$
+\[
+  0=\frac{\partial\Sigma}{\partial R_d}
+  =\frac12\hat\epsilon f'(1)-\frac12\frac1{\hat\epsilon}\int_0^1dq\,\frac{\lambda(q)}{[\lambda(q)+R_d/\hat\epsilon]^2}
+\]
+\[
+  0=\frac{\partial\Sigma}{\partial\hat\epsilon}
+  =-\epsilon+\frac12R_d f'(1)+\hat\epsilon\int_0^1dq\,\lambda(q)f''(q)+\frac12\frac{R_d}{\hat\epsilon^2}\int_0^1dq\,\frac{\lambda(q)}{[\lambda(q)+R_d/\hat\epsilon]^2}
+\]
+\[
+  0=\frac{\delta\Sigma}{\delta\lambda(q)}=\frac12\hat\epsilon^2f''(q)-\frac12\frac1{(\lambda(q)+R_d/\hat\epsilon)^2}
+\]
+\[
+  \lambda^*(q)=\frac1{\hat\epsilon}\left[f''(q)^{-1/2}-R_d\right]
+\]
+
+\end{document}
-- 
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