\documentclass[fleqn]{article} \usepackage{fullpage,amsmath,amssymb,latexsym} \binoppenalty=10000 \relpenalty=10000 \usepackage{float} % Fix \cal and \mathcal characters look (so it's not the same as \mathscr) \DeclareSymbolFont{usualmathcal}{OMS}{cmsy}{m}{n} \DeclareSymbolFontAlphabet{\mathcal}{usualmathcal} \begin{document} \title{Full solution of the Kac-Rice problem for mean-field models} \maketitle \begin{abstract} We derive the general solution for the computation of saddle points of mean-field complex landscapes. The solution incorporates Parisi's solution for equilibrium, as it should. \end{abstract} \section{Introduction} Although the Bray-Moore computation for the SK model was the first application of some replica symmetry breaking scheme, it turned out that the problem has been open ever since. to this date the program has been only complete for a subset of models here we present what we believe is the general scheme \subsection{What to expect?} In order to try to visualize what one should expect, consider two pure p-spin models, with \begin{equation} H = H_1 + H_2=\alpha_1 \sum_{ijk} J^1_{ijk} s_i s_j s_k + \alpha_2 \sum_{ijk} J^2_{ijk} \bar s_i \bar s_j \bar s_k +\epsilon \sum_i s_i \bar s_i \end{equation} The complexity of the first and second systems in terms of $H_1$ and of $H_2$ have, in the absence of coupling, the same dependence, but are stretched to one another: \begin{equation} \Sigma_1(H_1)= \Sigma_o(H_1/\alpha_1) \qquad ; \qquad \Sigma_2(H_2)= \Sigma_o(H_2/\alpha_2) \end{equation} Each system has a ground state energy $E_{gs}^{1,2}$, a threshold energy $E_{thres}^{1,2}$ (a well-defined notion, since we are considering pure p-spins), abd the corresponding limit values $X^{1,2}_{gs}=\left. \frac{d \Sigma_1}{dE_{1,2}}\right|_{E^{gs}_{12}}$ and $X^{1,2}_{thres}=\left. \frac{d \Sigma_1}{dE_{1,2}}\right|_{E^{thres}_{12}}$ Considering the cartesian product of both systems, we have, in terms of the total energy $H=H_1+H_2$ three regimes: \begin{itemize} \item {\bf Unfrozen}: \begin{eqnarray} & & X_1 \equiv \frac{d \Sigma_1}{dE_1}= X_2 \equiv \frac{d \Sigma_2}{dE_2} \end{eqnarray} \item {\bf Semi-frozen} As we go down in energy, one of the systems (say, the first) reaches its ground state, At lower temperatures, the first system is thus frozen, while the second is not, so that $X_1=X_1^{gs}> X_2$. The lowest energy is such that both systems are frozen. \item {\bf Semi-threshold } As we go up from the unfrozen upwards in energy, the second system reaches its threshold $X_2^{thres}$. At higher energies minima are extremely rare, so the second system remains stuck at its threshold for higher energies. \item{\bf Both systems reach their thresholds} There essentially no more minima above that. \end{itemize} Consider now two combined vectors $({\bf s},{\bf \hat s})$ and $({\bf s}',{\bf \hat s}')$ chosen at the same energies.\\ $\bullet$ Their normalized overlap is close to one when both subsystems are frozen, close to a half in the semifrozen phase, and zero at all higher energies.\\ $\bullet$ In phases where one or both systems are stuck in their thresholds (and only in those), the minima are exponentially subdominant with respect to saddles. $\bullet$ {\bf note that the same reasoning leads us to the conclusion that minima of two total energies such that one of the systems is frozen have nonzero overlaps} \section{The model} Here we consider, for definiteness, the `toy' model introduced by M\'ezard and Parisi \section{Equilibrium} Here we review the equilibrium solution. \cite{Crisanti_1992_The, Crisanti_1993_The, Crisanti_2004_Spherical, Crisanti_2006_Spherical} \begin{equation} \beta F=\frac12\lim_{n\to0}\frac1n\left(\beta^2\sum_{ab}f(Q_{ab})+\log\det Q\right)-\frac12\log S_\infty \end{equation} $\log S_\infty=1+\log2\pi$. \begin{align*} \beta F= -\frac12\log S_\infty+ \frac12\lim_{n\to0}\frac1n\left(\beta^2nf(1)+\beta^2\sum_{i=0}^kn(x_i-x_{i+1})f(q_i) +\log\left[ \frac{ 1+\sum_{i=0}^k(x_i-x_{i+1})q_i }{ 1+\sum_{i=1}^k(x_i-x_{i+1})q_i-x_1q_0 } \right]\right.\\ +\frac n{x_1}\log\left[ 1+\sum_{i=1}^k(x_i-x_{i+1})q_i-x_1q_0 \right]\\ \left.+\sum_{j=1}^kn(x_{j+1}^{-1}-x_j^{-1})\log\left[ 1+\sum_{i=j+1}^k(x_i-x_{i+1})q_i-x_{j+1}q_j \right] \right) \end{align*} \begin{align*} \lim_{n\to0}\frac1n \log\left[ \frac{ 1+\sum_{i=0}^k(x_i-x_{i+1})q_i }{ 1+\sum_{i=1}^k(x_i-x_{i+1})q_i-x_1q_0 } \right] &= \lim_{n\to0}\frac1n \log\left[ \frac{ 1+\sum_{i=0}^k(x_i-x_{i+1})q_i }{ 1+\sum_{i=0}^k(x_i-x_{i+1})q_i-nq_0 } \right] \\ &=q_0\left(1+\sum_{i=0}^k(x_i-x_{i+1})q_i\right)^{-1} \end{align*} \begin{align*} \beta F= -\frac12\log S_\infty+ \frac12\left(\beta^2f(1)+\beta^2\sum_{i=0}^k(x_i-x_{i+1})f(q_i) +q_0\left(1+\sum_{i=0}^k(x_i-x_{i+1})q_i\right)^{-1}\right. \\ +\frac1{x_1}\log\left[ 1+\sum_{i=1}^{k}(x_i-x_{i+1})q_i-x_1q_0 \right]\\ \left.+\sum_{j=1}^k(x_{j+1}^{-1}-x_j^{-1})\log\left[ 1+\sum_{i=j+1}^{k}(x_i-x_{i+1})q_i-x_{j+1}q_j \right] \right) \end{align*} $q_0=0$ \begin{align*} \beta F= -\frac12\log S_\infty+ \frac12\left(\beta^2f(1)+\beta^2\sum_{i=0}^k(x_i-x_{i+1})f(q_i) +\frac1{x_1}\log\left[ 1+\sum_{i=1}^{k}(x_i-x_{i+1})q_i \right]\right.\\ \left.+\sum_{j=1}^k(x_{j+1}^{-1}-x_j^{-1})\log\left[ 1+\sum_{i=j+1}^{k}(x_i-x_{i+1})q_i-x_{j+1}q_j \right] \right) \end{align*} $x_i=\tilde x_ix_k$, $x_k=y/\beta$, $q_k=1-z/\beta$ \begin{align*} \beta F= -\frac12\log S_\infty+ \frac12\left(\beta^2f(1)+\beta^2(y\beta^{-1}-1)f(1-z\beta^{-1})+y\beta\sum_{i=0}^{k-1}(\tilde x_i-\tilde x_{i+1})f(q_i)\right. \\ +\frac\beta{\tilde x_1 y}\log\left[ y\sum_{i=1}^{k-1}(\tilde x_i-\tilde x_{i+1})q_i+y+z-yz/\beta \right]\\ +\sum_{j=1}^{k-1}\frac\beta y(\tilde x_{j+1}^{-1}-\tilde x_j^{-1})\log\left[ y\sum_{i=j+1}^{k-1}(\tilde x_i-\tilde x_{i+1})q_i+y+z-yz/\beta-y\tilde x_{j+1}q_j \right]\\ \left.-\frac\beta{\tilde x_1 y}\log\beta-\sum_{j=1}^{k-1}\frac\beta y(\tilde x_{j+1}^{-1}-\tilde x_j^{-1})\log\beta+(1-\beta y^{-1})\log\left[ z/\beta \right] \right) \end{align*} \begin{align*} \lim_{\beta\to\infty}F= \frac12\left(yf(1)+zf'(1)+y\sum_{i=0}^{k-1}(\tilde x_i-\tilde x_{i+1})f(q_i) +\frac1{\tilde x_1 y}\log\left[ y\sum_{i=1}^{k-1}(\tilde x_i-\tilde x_{i+1})q_i+y+z \right]\right.\\ \left.+\sum_{j=1}^{k-1}\frac1 y(\tilde x_{j+1}^{-1}-\tilde x_j^{-1})\log\left[ y\sum_{i=j+1}^{k-1}(\tilde x_i-\tilde x_{i+1})q_i+y+z-y\tilde x_{j+1}q_j \right] -\frac1y\log z \right) \end{align*} $F$ is a $k-1$ RSB ansatz with all eigenvalues scaled by $y$ and shifted by $z$. $\tilde x_0=0$ and $\tilde x_k=1$. \begin{equation} \label{eq:ground.state.free.energy} \lim_{\beta\to\infty}F=\lim_{n\to0}\frac1n\frac12\left(nzf'(1)+y\sum_{ab}f(\tilde Q_{ab})+\frac1y\log\det(yz^{-1}\tilde Q+I) \right) \end{equation} \section{Kac-Rice} \cite{Auffinger_2012_Random, BenArous_2019_Geometry} \begin{equation} \mathcal N(\epsilon, \mu) =\int ds\,\delta(N\epsilon-H(s))\delta(\partial H(s)-\mu s)|\det(\partial\partial H(s)-\mu I)| \end{equation} \begin{equation} \Sigma(\epsilon,\mu)=\frac1N\log\mathcal N(\epsilon, \mu) \end{equation} \subsection{The replicated problem} \cite{Ros_2019_Complex} \cite{Folena_2020_Rethinking} \begin{equation} \begin{aligned} \Sigma(\epsilon, \mu) &=\frac1N\lim_{n\to0}\frac\partial{\partial n}\mathcal N^n(\epsilon) \\ &=\frac1N\lim_{n\to0}\frac\partial{\partial n}\int\prod_a^n ds_a\,\delta(N\epsilon-H(s_a))\delta(\partial H(s_a)-\mu s_a)|\det(\partial\partial H(s_a)-\mu I)| \end{aligned} \end{equation} the question of independence \cite{Bray_2007_Statistics} \begin{equation} \begin{aligned} \overline{\Sigma(\epsilon, \mu)} &=\frac1N\lim_{n\to0}\frac\partial{\partial n}\int\left(\prod_a^nds_a\right)\,\overline{\prod_a^n \delta(N\epsilon-H(s_a))\delta(\partial H(s_a)-\mu s_a)} \times \overline{\prod_a^n |\det(\partial\partial H(s_a)-\mu I)|} \end{aligned} \end{equation} \begin{equation} \begin{aligned} \mathcal D(\mu) &=\frac1N\overline{\log|\det(\partial\partial H(s_a)-\mu I)|} =\int d\lambda\,\rho(\lambda-\mu)\log|\lambda| \\ &=\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{aligned} \end{equation} for $\rho$ a semicircle distribution with radius $\sqrt{4f''(1)}$. \begin{equation} \begin{aligned} \overline{\Sigma(\epsilon, \mu)} &=\frac1N\lim_{n\to0}\frac\partial{\partial n} e^{nN\mathcal D(\mu)} \int\left(\prod_a^nds_a\,d\hat s_a\right)\,d\hat\epsilon\,e^{nN\hat\epsilon\epsilon-\mu\sum_a^n\hat s_as_a}\overline{ \exp\left[ \sum_a^n (\hat s_a\partial_a-\hat\epsilon)H(s_a) \right] } \\ &=\frac1N\lim_{n\to0}\frac\partial{\partial n} e^{nN\mathcal D(\mu)} \int\left(\prod_a^nds_a\,d\hat s_a\right)\,d\hat\epsilon\,e^{nN\hat\epsilon\epsilon-\mu\sum_a^n\hat s_as_a} \exp\left[ \sum_{ab}^n (\hat s_a\partial_a-\hat\epsilon)(\hat s_b\partial_b-\hat\epsilon)f(s_as_b/N) \right] \\ &=\frac1N\lim_{n\to0}\frac\partial{\partial n} e^{nN\mathcal D(\mu)} \int\left(\prod_a^nds_a\,d\hat s_a\right)\,d\hat\epsilon\,e^{nN\hat\epsilon\epsilon-\mu\sum_a^n\hat s_as_a} \exp\left[ \sum_{ab}^n ( \hat\epsilon^2f(s_as_b/N)-2\hat\epsilon\hat s_as_bf'(s_as_b/N)+\hat s_a\hat s_bf'(s_as_b/N) +(\hat s_as_b)^2f''(s_as_b/N) ) \right] \end{aligned} \end{equation} all saddles versus only minima The parameters: \begin{equation} \begin{aligned} Q_{ab}=\frac1Ns_a\cdot s_b \\ R_{ab}=\frac1N\hat s_a\cdot s_b \\ D_{ab}=\frac1N\hat s_a\cdot\hat s_b \end{aligned} \end{equation} \begin{equation} S =\mathcal D(\mu)+\hat\epsilon\epsilon+\lim_{n\to0}\frac1n\left( -\mu\sum_a^nR_{aa} +\frac12\sum_{ab}\left[ \hat\epsilon^2f(Q_{ab})-2\hat\epsilon R_{ab}f'(Q_{ab}) +D_{ab}f'(Q_{ab})+R_{ab}^2f''(Q_{ab}) \right] +\frac12\log\det\begin{bmatrix}Q&R\\R&D\end{bmatrix} \right) \end{equation} \section{Replicated action} \begin{align*} \Sigma =-\epsilon\hat\epsilon+\lim_{n\to0}\frac1n\left( \sum_a\mu(F_{aa}-R_{aa}) +\frac12\sum_{ab}\left[ \hat\epsilon^2f(Q_{ab})+2\hat\epsilon R_{ab}f'(Q_{ab}) +D_{ab}f'(Q_{ab})+R_{ab}^2f''(Q_{ab})-F_{ab}^2f''(Q_{ab}) \right]\right.\\\left. +\frac12\log\det\begin{bmatrix}Q&-iR\\-iR&-D\end{bmatrix} -\log\det F \right) \end{align*} \[ 0=\frac{\partial\Sigma}{\partial R_{ab}} =-\mu\delta_{ab}+\hat\epsilon f'(Q_{ab})+R_{ab}f''(Q_{ab})+\sum_c(R^2-DQ)^{-1}_{ac}R_{cb} \] \[ 0=\frac{\partial\Sigma}{\partial D_{ab}} =\frac12 f'(Q_{ab})-\frac12\sum_c(R^2-DQ)^{-1}_{ac}Q_{cb} \] The second equation implies \[ (R^2-DQ)^{-1}=Q^{-1}f'(Q) \] \section{Replica ansatz} \subsection{Motivation} The reader who is happy with the ansatz may skip this section. We may encode the original variables in a superspace variable: \begin{equation} \phi_i(1)= q_i(t) + \bar \theta a_i + a_i^\dag \theta + p_i \bar \theta \theta~, \end{equation} \begin{equation} \begin{aligned} {\bf Q}(1,2)&=\frac 1 N \sum_i \phi_i(1) \phi_i (2) = Q_{ab} + (\bar \theta_2 - \bar \theta_1) \theta_2 R_{ab} + \bar \theta_1 \theta_1 R_{ab} + \bar\theta_1\theta_1 \bar \theta_2 \theta_2 D_{ab} \\ &+ \text{odd terms in the $\bar \theta,\theta$}~. \end{aligned} \label{Q12} \end{equation} Here $\theta_a$, $\bar \theta_a$ are Grassmann variables, and we denote the full set of coordinates in a compact form as $1= \theta_1 \overline\theta_1$, $d1= d\theta_1 d\overline\theta_1$, etc. The odd and even fermion numbers decouple, so we can neglect all odd terms in $\theta,\bar{\theta}$. The variables $\bar \theta \theta$ and $\bar \theta ' \theta'$ play the role of `times' in a superspace treatment. We have a long experience of making an ansatz for replicated quantum problems, which naturally involve a (Matsubara) time. The analogy strongly suggests that only the diagonal ${\bf Q}_{aa}$ depend on the $\theta$'s. This boils down to putting: \begin{eqnarray} Q_{ab}&=& {\mbox{ a Parisi matrix}}\nonumber\\ R_{ab}&=R \delta_{ab}&\nonumber\\ D_{ab}&=& D \delta_{ab} \end{eqnarray} Not surprisingly, this ansatz closes, as we shall see. \subsection{Solution} Insert the diagonal ansatz $R=R_dI$, $D=D_dI$. Then \[ 0=(R_df''(1)-\mu)I+\hat\epsilon f'(Q)+R_d(R_d^2I-D_dQ)^{-1} =(R_df''(1)-\mu)I+\hat\epsilon f'(Q)+R_dQ^{-1}f'(Q) \] and \[ Q^{-1}f'(Q)=(I+D_df'(Q))/R_d^2 \] Substituting the second into the first, we have \[ 0=(R_df''(1)-\mu)I+\hat\epsilon f'(Q)+\frac1{R_d}(I+D_df'(Q)) \] \[ 0=(R_df''(1)-\mu+R_d^{-1})I+(\hat\epsilon+D_d/R_d)f'(Q) \] The only way for this equation to be satisfied off the diagonal for nontrivial $Q$ is for $D_d=-R_d\hat\epsilon$. We therefore have \begin{align*} \Sigma =-\epsilon\hat\epsilon+\lim_{n\to0}\frac1n\left( n\mu(F_d-R_d)+\frac12n\left[ \hat\epsilon R_df'(1)+R_d^2f''(1)-F_d^2f''(1) \right] +\frac12\sum_{ab} \hat\epsilon^2f(Q_{ab}) ]\right.\\\left. +\frac12\log\det(\hat\epsilon R_d^{-1} Q+I) +n\log R_d -n\log F_d \right) \end{align*} Taking the saddle with respect to $\mu$ and $F_d$ yields \[ F_d=R_d \] \[ \mu=R_d^{-1}(1+R_d^2f''(1)) \] and gives \begin{align*} \Sigma =\epsilon\hat\epsilon+\hat\epsilon R_d f'(1)+\frac12D_df'(1)+\lim_{n\to0}\frac1n\frac12\left( \hat\epsilon^2\sum_{ab} f(Q_{ab}) +\log\det(-D_dR_d^{-2} Q+I) \right) \end{align*} Finally, setting $0=\Sigma$ gives \[ \epsilon =\lim_{n\to0}\frac1n\frac12\left(nR_df'(1)+\hat\epsilon\sum_{ab} f(Q_{ab}) +\frac1{\hat\epsilon}\log\det(\hat\epsilon R_d^{-1} Q+I) \right) \] 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.} \subsection{Full} \begin{align*} \lim_{n\to0}\frac1n\log\det(\hat\epsilon R_d^{-1} Q+I) =x_1^{-1}\log\left(\hat\epsilon R_d^{-1}(1-\bar q_k)+1\right)+\int_{q_0^+}^{q_{k-1}}dq\,\mu(q)\log\left[\hat\epsilon R_d^{-1}\lambda(q)+1\right] \end{align*} where \[ \mu(q)=\frac{\partial x^{-1}(q)}{\partial q} \] Integrating by parts, \begin{align*} \lim_{n\to0}\frac1n\log\det(\hat\epsilon R_d^{-1} Q+I) &=x_1^{-1}\log\left(\hat\epsilon R_d^{-1}(1-\bar q_k)+1\right)+\left[x^{-1}(q)\log[\hat\epsilon R_d^{-1}\lambda(q)+1]\right]_{q=q_0^+}^{q=q_{k-1}}-\frac{\hat\epsilon}{R_d}\int_{q_0^+}^{q_{k-1}}dq\,\frac{\lambda'(q)}{x(q)}\frac1{\hat\epsilon R_d^{-1}\lambda(q)+1}\\ &=\log[\hat\epsilon R_d^{-1}\lambda(q_{k-1})+1]+\frac{\hat\epsilon}{R_d}\int_{q_0^+}^{q_{k-1}}dq\,\frac1{\hat\epsilon R_d^{-1}\lambda(q)+1} \end{align*} \begin{align*} \Sigma =-\epsilon\hat\epsilon+ \frac12\hat\epsilon R_df'(1) +\frac12\int_0^1dq\,\left[ \hat\epsilon^2\lambda(q)f''(q) +\frac1{\lambda(q)+R_d/\hat\epsilon} \right] \end{align*} for $\lambda$ concave, monotonic, $\lambda(1)=0$, and $\lambda'(1)=-1$ \[ 0=\frac{\delta\Sigma}{\delta\lambda(q)}=\frac12\hat\epsilon^2f''(q)-\frac12\frac1{(\lambda(q)+R_d/\hat\epsilon)^2} \] \[ \lambda^*(q)=\frac1{\hat\epsilon}\left[f''(q)^{-1/2}-R_d\right] \] We suppose that solutions are given by \begin{equation} \lambda(q)=\begin{cases} \lambda^*(q) & q