From 244d97f5cb05b89c7ffa742a5a2fcb21e6aadad6 Mon Sep 17 00:00:00 2001
From: Jaron Kent-Dobias <jaron@kent-dobias.com>
Date: Wed, 8 Jun 2022 20:22:50 +0200
Subject: Added more discussion of equilibrium case.

---
 frsb_kac_new.tex | 10 +++++++++-
 1 file changed, 9 insertions(+), 1 deletion(-)

(limited to 'frsb_kac_new.tex')

diff --git a/frsb_kac_new.tex b/frsb_kac_new.tex
index c2e5eb9..86e38ba 100644
--- a/frsb_kac_new.tex
+++ b/frsb_kac_new.tex
@@ -173,7 +173,6 @@ $q_1,\ldots,q_{k-1}$, which gives
     \tilde\beta^2\sum_{ab}^nf(\tilde Q_{ab})+\log\det(\tilde\beta z^{-1}\tilde Q+I)
   \right)
 \end{equation}
-{\em We have lost one level of RSB because at zero temperature the states become points.}
 In the continuum case, this is
 \begin{equation} \label{eq:ground.state.free.energy}
   \lim_{\beta\to\infty}\tilde\beta F
@@ -182,6 +181,15 @@ In the continuum case, this is
   \right)
 \end{equation}
 
+The zero temperature limit of the free energy loses one level of replica
+symmetry breaking. Physically, this is a result of the fact that in $k$-RSB,
+$q_k$ gives the overlap within a state, e.g., within the basin of a well inside
+the energy landscape. At zero temperature, the measure is completely localized
+on the bottom of the well, and therefore the overlap with each state becomes
+one. We will see that the complexity of low-energy stationary points in
+Kac--Rice is also given by a $(k-1)$-RSB anstaz. Heuristicall, this is because
+each stationary point also has no width and therefore overlap one with itself.
+
 
 
 \section{Kac-Rice}
-- 
cgit v1.2.3-70-g09d2

