From 8fa28ca67e971c653e720db69d0d992c74252dbd Mon Sep 17 00:00:00 2001
From: Jaron Kent-Dobias <jaron@kent-dobias.com>
Date: Sat, 4 Jun 2022 23:44:30 +0200
Subject: En dashes for Kac-Rice.

---
 frsb_kac-rice.tex | 8 ++++----
 1 file changed, 4 insertions(+), 4 deletions(-)

(limited to 'frsb_kac-rice.tex')

diff --git a/frsb_kac-rice.tex b/frsb_kac-rice.tex
index 1b06dd0..8881f10 100644
--- a/frsb_kac-rice.tex
+++ b/frsb_kac-rice.tex
@@ -3,7 +3,7 @@
 \usepackage{fullpage,amsmath,amssymb,latexsym,graphicx}
 
 \begin{document}
-\title{Full solution of the Kac-Rice problem for mean-field models}
+\title{Full solution of the Kac--Rice problem for mean-field models}
 \author{Jaron Kent-Dobias \& Jorge Kurchan}
 \maketitle
 \begin{abstract}
@@ -203,7 +203,7 @@ $F$ is a $k-1$ RSB ansatz with all eigenvalues scaled by $y$ and shifted by $z$.
 
 
 
-\section{Kac-Rice}
+\section{Kac--Rice}
 
 \cite{Auffinger_2012_Random, BenArous_2019_Geometry}
 
@@ -447,7 +447,7 @@ Finally, setting $0=\Sigma$ gives
 \]
 which is precisely \eqref{eq:ground.state.free.energy} with $R_d=z$ and $\hat\epsilon=y$. 
 
-{\em 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.}
+{\em 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.}
 
 \subsection{Full}
 
@@ -490,7 +490,7 @@ We suppose that solutions are given by
     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.
+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*}
-- 
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