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author | Jaron Kent-Dobias <jaron@kent-dobias.com> | 2022-06-04 23:44:30 +0200 |
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committer | Jaron Kent-Dobias <jaron@kent-dobias.com> | 2022-06-04 23:44:30 +0200 |
commit | 8fa28ca67e971c653e720db69d0d992c74252dbd (patch) | |
tree | d97c50af077a31d9c3cb8c90dae09b3d5173f86c /frsb_kac-rice.tex | |
parent | b5fbe51398322260a93e2e33cde5d29b4855374e (diff) | |
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En dashes for Kac-Rice.
Diffstat (limited to 'frsb_kac-rice.tex')
-rw-r--r-- | frsb_kac-rice.tex | 8 |
1 files changed, 4 insertions, 4 deletions
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*} |