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diff --git a/frsb_kac_new.tex b/frsb_kac_new.tex index 3db21e4..4d938da 100644 --- a/frsb_kac_new.tex +++ b/frsb_kac_new.tex @@ -403,7 +403,7 @@ Inserting the diagonal ansatz \eqref{ansatz} one gets Using standard manipulations (Appendix B), one finds also a continuous version \begin{equation} \label{eq:functional.action} \begin{aligned} - S + \Sigma(\epsilon,\mu) =\mathcal D(\mu) + \hat\beta\epsilon-\mu R_d @@ -427,7 +427,7 @@ Note the close similarity of this action to the equilibrium replica one, at fini The dominant stationary points are given by maximizing the action with respect to $\mu$. This gives \begin{equation} \label{eq:mu.saddle} - 0=\frac{\partial S}{\partial\mu}=\mathcal D'(\mu)-R_d + 0=\frac{\partial\Sigma}{\partial\mu}=\mathcal D'(\mu)-R_d \end{equation} as expected. To take the derivative, we must resolve the real part inside the definition of @@ -476,20 +476,17 @@ Adding $2(D_d/R_d)$ times \eqref{eq:saddle.d} to \eqref{eq:saddle.r} and multipl \begin{equation} 0=-R_d\mu+1+R_d^2f''(1)+f'(1)(R_d\hat\beta-D_d) \end{equation} -There are two scenarios: one where the dominant stationary points -in the vicinity of the ground state are minima, and one where they are saddles. In the case where the dominant stationary points are minima, we can use the optimal $\mu$ from \S\ref{sec:counting.minima}, which gives +At the ground state, minima will always dominate (even if marginal). We can +therefore use the optimal $\mu$ from \S\ref{sec:counting.minima}, which gives \begin{equation} 0=f'(1)(R_d\hat\beta-D_d) \end{equation} -Therefore, in any situation where minima dominate, the optimal $\mu$ will have $R_d\hat\beta=D_d$. - -When the dominant stationary points are saddles, we can use the $\mu$ from \S\ref{sec:counting.saddles}, which implies $R_d=\mu/2f''(1)$ and -\begin{equation} - 0=1-\frac{\mu^2}{4f''(1)}+f'(1)(R_d\hat\beta-D_d) -\end{equation} -If saddles dominate all the way to the ground state, then they must become marginal minima at the ground state. Therefore at the ground state energy $\mu=\mu_m=\sqrt{4f''(1)}$, and once again $R_d\hat\beta-D_d=0$. +In order to satisfy this equation we must have +$D_d=R_d\hat\beta$. This relationship holds for the most common minima whenever +they dominante, including in the ground state. -In any case, at the ground state $D_d=R_d\hat\beta$. Substituting this into the action, and also substituting the optimal $\mu$ for saddles or minima, and taking $\Sigma(\epsilon_0,\mu^*)=0$, gives +Substituting this into the action, and also substituting the optimal $\mu$ for +minima, and taking $\Sigma(\epsilon_0,\mu^*)=0$, gives \begin{equation} \hat\beta\epsilon_0 =-\frac12R_d\hat\beta f'(1)-\frac12\lim_{n\to0}\frac1n\left( @@ -497,12 +494,60 @@ In any case, at the ground state $D_d=R_d\hat\beta$. Substituting this into the +\log\det(\hat\beta R_d^{-1} Q+I) \right) \end{equation} -which is precisely \eqref{eq:ground.state.free.energy} with $R_d=z$ and $\hat\beta=\tilde\beta$. +which is precisely \eqref{eq:ground.state.free.energy} with $R_d=z$, +$\hat\beta=\tilde\beta$, and $Q=\tilde Q$. -{\em We arrive at one of the main results of our paper: 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 We arrive at one of the main results of our paper: 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 } Moreover, there is an +exact correspondance between the saddle parameters of each. If the equilibrium +is given by a Parisi matrix with parameters $x_1,\ldots,x_k$ and +$q_1,\ldots,q_k$, then the parameters $\hat\beta$, $R_d$, $D_d$, $\tilde +x_1,\ldots,\tilde x_{k-1}$, and $\tilde q_1,\ldots,\tilde q_{k-1}$ for the +complexity in the ground state are +\begin{align} + \hat\beta=\lim_{\beta\to\infty}\beta x_k + && + \tilde x_i=\lim_{\beta\to\infty}\frac{x_i}{x_k} + && + \tilde q_i=\lim_{\beta\to\infty}q_i + && + R_d=\lim_{\beta\to\infty}\beta(1-q_k) + && + D_d=\hat\beta R_d +\end{align} \subsection{The continuum situation at a glance} +In the case where any FRSB is present, one must work with the functional form +of the complexity \eqref{eq:functional.action}, which must be extremized with +respect to $\chi$ under the conditions that $\chi$ is concave, monotonically +decreasing, and $\chi(1)=0$, $\chi'(1)=-1$. The annealed case is found by +taking $\chi(q)=1-q$, which satisfies all of these conditions. $k$-RSB is +produced by breaking $\chi$ into $k+1$ piecewise linear segments. + +Forget for the moment these tricky requirements. The function would then be +extremized by satisfying +\begin{equation} + 0=\frac{\delta\Sigma}{\delta\lambda(q)}=\frac12\hat\beta^2f''(q)-\frac12\frac1{(\lambda(q)+R_d^2/D_d)^2} +\end{equation} +which implies the solution +\begin{equation} + \lambda^*(q)=\frac1{\hat\beta}f''(q)^{-1/2}-\frac{R_d^2}{D_d} +\end{equation} +If $f''(q)^{-1/2}$ is not concave anywhere, there is little use of this +solution. However, if it is concave everywhere it may constitute a portion of +the full solution. + +We suppose that solutions are given by +\begin{equation} + \lambda(q)=\begin{cases} + \lambda^*(q) & q<q_\textrm{max} \\ + 1-q & q\geq q_\textrm{max} + \end{cases} +\end{equation} +Continuity requires that $1-q_\textrm{max}=\lambda^*(q_\textrm{max})$. + Here a picture of $\chi$ vs $C$ or $X$ vs $C$ showing limits $q_{max}$, $x_{max}$ for different energies and typical vs minima. @@ -662,130 +707,6 @@ Integrating by parts, &=\log[\hat\beta R_d^{-1}\lambda(q_{k-1})+1]+\frac{\hat\beta}{R_d}\int_{q_0^+}^{q_{k-1}}dq\,\frac1{\hat\beta R_d^{-1}\lambda(q)+1} \end{align*} -\begin{align*} - \Sigma - =-\epsilon\hat\beta+ - \frac12\hat\beta R_df'(1) - +\frac12\int_0^1dq\,\left[ - \hat\beta^2\lambda(q)f''(q) - +\frac1{\lambda(q)+R_d/\hat\beta} - \right] -\end{align*} -for $\lambda$ concave, monotonic, $\lambda(1)=0$, and $\lambda'(1)=-1$ -\[ - 0=\frac{\delta\Sigma}{\delta\lambda(q)}=\frac12\hat\beta^2f''(q)-\frac12\frac1{(\lambda(q)+R_d/\hat\beta)^2} -\] -\[ - \lambda^*(q)=\frac1{\hat\beta}\left[f''(q)^{-1/2}-R_d\right] -\] - -We suppose that solutions are given by -\begin{equation} - \lambda(q)=\begin{cases} - \lambda^*(q) & q<q^* \\ - 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. - -Inserting this into the complexity, we find -\begin{align*} - \Sigma - &=-\epsilon\hat\beta+\frac12\hat\beta R_df'(1) - +\frac12\int_0^{q^*}dq\left[ - \hat\beta(f''(q)^{-1/2}-R_d)f''(q)+\hat\beta f''(q)^{1/2} - \right] - +\frac12\int_{q^*}^1dq\left[ - \hat\beta^2(1-q)f''(q)+\frac1{q-1+R_d/\hat\beta} - \right] \\ - &=-\epsilon\hat\beta+\frac12\hat\beta R_d\left[f'(1)-f'(q^*)\right] - +\hat\beta\int_0^{q^*}dq\,f''(q)^{1/2} - +\frac12\hat\beta^2\int_{q^*}^1dq\, - (1-q)f''(q) - -\log\left[1-(1-q^*)\hat\beta/R_d\right] -\end{align*} -$R_d$ can be extremized now, with -\[ - R_d=\frac12\left( - (1-q^*)\hat\beta\pm\sqrt{ - (1-q^*)\left( - (1-q^*)\hat\beta^2+8/[f'(1)-f'(q^*)] - \right) - } - \right) -\] - -This all is for $\mu=\mu^*$, which counts the dominant saddles. We can also count by fixed macroscopic index $\mu$ by leaving it unoptimized in the complexity. This gives -\[ - F_d=\frac1{2f''(1)}\left[\mu\pm\sqrt{\mu^2-4f''(1)}\right] -\] -and -\begin{align*} - \Sigma - =-\epsilon\hat\beta+ - \frac12\hat\beta R_df'(1) - +\frac12\int_0^1dq\,\left[ - \hat\beta^2\lambda(q)f''(q) - +\frac1{\lambda(q)+R_d/\hat\beta} - \right]-\mu R_d+\frac12R_d^2f''(1)+\log R_d\\ - +\operatorname{Re}\left\{\frac12\left(1+\frac\mu{2f''(1)}\left(\mu\pm\sqrt{\mu^2-4f''(1)}\right)\right)-\log\left(\frac1{2f''(1)}\left(\mu\pm\sqrt{\mu^2-4f''(1)}\right)\right)\right\} -\end{align*} - - - -\section{Main result} - -\begin{equation} - \begin{aligned} - \overline{\Sigma(\epsilon,\mu)} - =\mathcal D(\mu) - +\operatorname*{extremum}_{\substack{R_d,D_d,\hat\beta\in\mathbb R\\\chi\in\Lambda}} - \left\{ - \hat\beta\epsilon+\mu R_d - +\frac12(2\hat\beta R_d-D_d)f'(1)+\frac12R_d^2f''(1) - +\frac12\log R_d^2 \right.\\\left. - +\frac12\int_0^1dq\,\left( - \hat\beta^2f''(q)\chi(q)+\frac1{\chi(q)+R_d^2/D_d} - \right) - \right\} - \end{aligned} -\end{equation} -where -\begin{equation} - \mathcal D(\mu) - =\operatorname{Re}\left\{ - \frac12\left(1+\frac\mu{2f''(1)}\left(\mu\pm\sqrt{\mu^2-4f''(1)}\right)\right) - -\log\left(\frac1{2f''(1)}\left(\mu\pm\sqrt{\mu^2-4f''(1)}\right)\right) - \right\} -\end{equation} -and $\Lambda$ is the space of functions $\chi:[0,1]\to[0,1]$ which are -monotonically decreasing, concave, and have $\chi(1)=0$ and $\chi'(1)=-1$. -If there is more than one extremum of this function, choose the one with the -smallest value of $\Sigma$. The sign of the root inside $\mathcal D(\mu)$ is -negative for $\mu>0$ and positive for $\mu<0$. - -The $k$-RSB ansatz is equivalent to piecewise linear $\chi$ with $k+1$ -pieces, with replica symmetric or 0-RSB giving $\chi(q)=1-q$. Our other major -result is that, if the equilibrium state in the vicinity of zero temperature is -given by a $k$-RSB ansatz, then the complexity is given by a $(k-1)$-RSB -ansatz. Moreover, there is an exact correspondence between the parameters of -the equilibrium saddle point in the limit of zero temperature and those of the -complexity saddle at the ground state. If the equilibrium is given by -$x_1,\ldots,x_k$ and $q_1,\ldots,q_k$, then the parameters $\tilde -x_1,\ldots,\tilde x_{k-1}$ and $\tilde q_1,\ldots,\tilde q_{k-1}$ for the -complexity in the ground state are -\begin{align} - \hat\beta=\lim_{\beta\to\infty}\beta x_k - && - \tilde x_i=\lim_{\beta\to\infty}\frac{x_i}{x_k} - && - \tilde q_i=\lim_{\beta\to\infty}q_i - && - R_d=\lim_{\beta\to\infty}\beta(1-q_k) - && - D_d=R_d\hat\beta -\end{align} - \section{ A motivation for the ansatz} We may encode the original variables in a superspace variable: |