From 244d97f5cb05b89c7ffa742a5a2fcb21e6aadad6 Mon Sep 17 00:00:00 2001 From: Jaron Kent-Dobias 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(-) 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