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diff --git a/frsb_kac-rice.tex b/frsb_kac-rice.tex index cd295e9..2015feb 100644 --- a/frsb_kac-rice.tex +++ b/frsb_kac-rice.tex @@ -1,7 +1,14 @@ \documentclass[fleqn]{article} \usepackage{fullpage,amsmath,amssymb,latexsym,graphicx} -\usepackage{appendix} +\usepackage{appendix,xcolor} +\usepackage[ + colorlinks=true, + urlcolor=purple, + citecolor=purple, + filecolor=purple, + linkcolor=purple +]{hyperref} % ref and cite links with pretty colors \begin{document} \title{Full solution of the Kac--Rice problem for mean-field models.\\ @@ -303,268 +310,159 @@ role of an inverse temperature for the metastable states. The average over disor We introduce new fields \begin{align} - Q_{ab}=\frac1Ns_a\cdot s_b && + C_{ab}=\frac1Ns_a\cdot s_b && R_{ab}=-i\frac1N\hat s_a\cdot s_b && D_{ab}=\frac1N\hat s_a\cdot\hat s_b \end{align} -$Q_{ab}$ is the overlap between spins belonging to different replicas. The -meaning of $R_{ab}$ is that of a response of replica $a$ to a linear field in -replica $b$: -\begin{equation} - R_{ab} = \frac 1 N \sum_i \overline{\frac{\delta s_i^a}{\delta h_i^b}} -\end{equation} -The $D$ may similarly be seen as the variation of the complexity with respect to a random field. +$C_{ab}$ is the overlap between spins belonging to different replicas. -By substituting these parameters into the expressions above and then making a change of variables in the integration from $s_a$ and $\hat s_a$ to these three matrices, we arrive at the form for the complexity +By substituting these parameters into the expressions above and then making a +change of variables in the integration from $s_a$ and $\hat s_a$ to these three +matrices, we arrive at the form for the complexity \begin{equation} \begin{aligned} &\Sigma(\epsilon,\mu) =\mathcal D(\mu)+\hat\beta\epsilon+\\ &\lim_{n\to0}\frac1n\left( - -\mu\sum_a^nR_{aa} + -\mu\operatorname{Tr}R +\frac12\sum_{ab}\left[ - \hat\beta^2f(Q_{ab})+(2\hat\beta R_{ab}-D_{ab})f'(Q_{ab}) - +R_{ab}^2f''(Q_{ab}) + \hat\beta^2f(C_{ab})+(2\hat\beta R_{ab}-D_{ab})f'(C_{ab}) + +R_{ab}^2f''(C_{ab}) \right] - +\frac12\log\det\begin{bmatrix}Q&iR\\iR&D\end{bmatrix} + +\frac12\log\det\begin{bmatrix}C&iR\\iR&D\end{bmatrix} \right) \end{aligned} \end{equation} -where $\hat\beta$, $Q$, $R$ and $D$ must be evaluated at extrema of this -expression. With $Q$, $R$, and $D$ distinct replica matrices, this is -potentially quite challenging. - -\begin{equation} - \begin{aligned} - &\Sigma(\epsilon,\mu) - =-\frac12+\mathcal D(\mu)+\hat\beta\epsilon+\\ - &\lim_{n\to0}\frac1n\left( - -\mu\sum_a^nR_{aa} - +\frac12\sum_{ab}\left[ - \hat\beta^2f(Q_{ab})+R_{ac}(2\hat\beta\delta_{cb}+Q^{-1}_{cd}R_{db})f'(Q_{ab}) - +R_{ab}^2f''(Q_{ab}) - \right] - +\frac12\log\det Q - \frac12\log\det f'(Q) - \right) - \end{aligned} -\end{equation} - -{\bf Inserting the expression for $D$ in the action, and writing $R=R_d+\tilde R$ ($\tilde R$ is zero in the diagonal) , the linear term in $\tilde R$ is -$Tr \left[\tilde R (\hat \beta -R_d Q^{-1} f')\right]$ } +where $\hat\beta$, $C$, $R$ and $D$ must be evaluated at extrema of this +expression. \section{Replica ansatz} -We shall make the following ansatz for the saddle point: -\begin{align}\label{ansatz} - Q_{ab}= \text{a Parisi matrix} && - R_{ab}=R_d \delta_{ab} && - D_{ab}= D_d \delta_{ab} +Based on previous work on the SK model and the equilibrium solution of the +spherical model, we expect $C$, and $R$ and $D$ to be hierarchical matrices, +i.e., to follow Parisi's scheme. This assumption immediately simplifies the +extremal conditions, since hierarchical matrices commute and are closed under +matrix products and Hadamard products. The extremal conditions are +\begin{align} + 0&=\frac{\partial\Sigma}{\partial\hat\beta} + =\epsilon+\sum_{ab}\left[\hat\beta f(C_{ab})+R_{ab}f'(C_{ab})\right] \label{eq:cond.b} \\ + \frac{\tilde c}2I&=\frac{\partial\Sigma}{\partial C} + =\frac12\left[ + \hat\beta^2f'(C)+(2\hat\beta R-D)\odot f''(C)+R\odot R\odot f'''(C) + +(CD+R^2)^{-1}D + \right] \label{eq:cond.q} \\ + 0&=\frac{\partial\Sigma}{\partial R} + =-\mu I+\hat\beta f'(C)+R\odot f''(C) + +(CD+R^2)^{-1}R \label{eq:cond.r} \\ + 0&=\frac{\partial\Sigma}{\partial D} + =-\frac12f'(C) + +\frac12(CD+R^2)^{-1}C \label{eq:cond.d} \end{align} -From what we have seen above, this means that replica $a$ is insensitive to -a small field applied to replica $b$ if $a \neq b$, a property related to ultrametricity. A similar situation happens in quantum replicated systems, -with time appearing only on the diagonal terms: see Appendix C for details. +where $\odot$ denotes the Hadamard product, or the componentwise product. The +equation for \eqref{eq:cond.q} would not be true on the diagonal save for the +arbitrary factor $\tilde c$. Equation \eqref{eq:cond.d} implies that +\begin{equation} \label{eq:D.solution} + D=f'(C)^{-1}-RC^{-1}R +\end{equation} -From its very definition, it is easy to see just perturbing the equations -with a field that $R_d$ is the trace of the inverse Hessian, as one expect indeed of a response. -\begin{equation} - R_d = \mathcal D'(\mu) +In addition to these equations, one is also often interested in maximizing the complexity as a function of $\mu$, to find the dominant or most common type of stationary points. These are given by the condition +\begin{equation} \label{eq:cond.mu} + 0=\frac{\partial\Sigma}{\partial\mu} + =\mathcal D'(\mu)-r_d \end{equation} -Similarly, .... one shows that -\begin{equation} -D_d = \hat \beta R_d +Since $\mathcal D(\mu)$ is effectively a piecewise function, with different forms for $\mu$ greater or less than $\mu_m$, there are two regimes. When $\mu>\mu_m$, the critical points are minima, and \eqref{eq:cond.mu} implies +\begin{equation} \label{eq:mu.minima} + \mu=\frac1{r_d}+r_df''(1) +\end{equation} +When $\mu<\mu_m$, they are saddles, and +\begin{equation} \label{eq:mu.saddles} + \mu=2f''(1)r_d \end{equation} +\section{Supersymmetric solution} -Is it a saddle? +The Kac--Rice problem has an approximate supersymmetry, which is found when the +absolute value of the determinant is neglected. When this is done, the +determinant can be represented by an integral over Grassmann variables, which +yields a complexity depending on `bosons' and `fermions' that share the +supersymmetry. The Ward identities associated with the supersymmetry imply +that $D=\hat\beta R$ \cite{Annibale_2003_The}. Under which conditions can this relationship be expected to hold? +Any result of supersymmetry can only be valid when the symmetry itself is +valid, which means the determinant must be positive. This is only guaranteed +for minima, which have $\mu>\mu_m$. Moreover, this identity heavily constrains +the form that the rest of the solution can take. Assuming the supersymmetry holds, \eqref{eq:cond.q} implies \begin{equation} - 0=\frac{\partial\Sigma}{\partial R} - =\lim_{n\to0}\frac1n\left(-\mu I+\hat\beta f'(Q)+R\odot f''(Q) - +(QD+R^2)^{-1}R - \right) + \tilde cI=\hat\beta^2f'(C)+\hat\beta R\odot f''(C)+R\odot R\odot f'''(C)+\hat\beta(CD+R^2)^{-1}R \end{equation} -\begin{equation} - 0=\frac{\partial\Sigma}{\partial D} - =\lim_{n\to0}\frac1n\left(-\frac12f'(Q) - +\frac12(QD+R^2)^{-1}Q - \right) -\end{equation} -\begin{equation} - D=f'(Q)^{-1}-RQ^{-1}R -\end{equation} -Diagonal anstaz requires that $f'(Q_{ab})^{-1}=R_d^2Q_{ab}^{-1}$ for $a\neq b$. -\begin{equation} - 0 - =\lim_{n\to0}\frac1n\left(-\mu I+\hat\beta f'(Q)+R\odot f''(Q) - +Q^{-1}Rf'(Q) - \right) -\end{equation} -Diagonal ansatz requires that -\begin{equation} - 0=-\mu I+\hat\beta f'(Q)+R_df''(1)I+R_dQ^{-1}f'(Q) -\end{equation} -or $\hat\beta f'(Q_{ab})=-R_d[Q^{-1}f'(Q)]_{ab}$ for $a\neq b$. -We also have from the first (before the diagonal ansatz) $Df'(Q)=I-RQ^{-1}Rf'(Q)$. Inserting the diagonal, we get $Q^{-1}f'(Q)=\frac1{R_d^2}(I-D_df'(Q))$, and -\begin{equation} - 0=-\mu I+\hat\beta f'(Q)+R_df''(1)I+\frac1{R_d}(I-D_df'(Q)) - =(R_df''(1)+R_d^{-1}-\mu)I+(\hat\beta-D_d/R_d)f'(Q) +Substituting \eqref{eq:cond.r} for the factor $(CD+R^2)^{-1}R$, we find substantial cancellation, and finally +\begin{equation} \label{eq:R.diagonal} + (\tilde c-\mu)I=R\odot R\odot f'''(C) \end{equation} +If $C$ has a nontrivial off-diagonal structure and supersymmetry holds, then +the off-diagonal of $R$ must vanish, and therefore $R=r_dI$. Therefore, a +supersymmetric ansatz is equivalent to a \emph{diagonal} ansatz. -Always true: -\begin{equation} - 0=-\mu R+\hat\beta Rf'(Q)+R(R\odot f''(Q))+I-Df'(Q) -\end{equation} -Insert $D=D_dI+\delta D$, $R=R_dI+\delta R$, and $D_d=\hat\beta R_d$ +Supersymmetry has further implications. +Equations \eqref{eq:cond.r} and \eqref{eq:cond.d} can be combined to find \begin{equation} - 0=-\hat\beta \mu R_d I +\hat\beta R_df'(Q)+R_d^2f''(1)I+I-\hat\beta R_df'(Q) - -\mu\delta R + I=R\left[\mu I-R\odot f''(C)\right]+(D-\hat\beta R)f'(C) \end{equation} - +Assuming the supersymmetry holds implies that \begin{equation} - 0=(\hat\beta Q+R) f'(Q)+(R\odot f''(Q)-\mu I)Q + I=R\left[\mu I-R\odot f''(C)\right] \end{equation} - +Understanding that $R$ is diagonal, this implies \begin{equation} - \begin{aligned} - &\Sigma(\epsilon,\mu) - =\frac12+\mathcal D(\mu)+\hat\beta\epsilon+\\ - &\lim_{n\to0}\frac1n\left( - -\frac12\mu\sum_a^nR_{aa} - +\frac12\sum_{ab}\left[ - \hat\beta^2f(Q_{ab})+\hat\beta R_{ab}f'(Q_{ab}) - \right] - +\frac12\log\det(f'(Q)^{-1}Q) - \right) - \end{aligned} + \mu=\frac1{r_d}+r_df''(1) \end{equation} +which is precisely the condition \eqref{eq:mu.minima}. Therefore, \emph{the +supersymmetric solution only counts dominant minima.} - -\subsection{Solution} - - -Inserting the diagonal ansatz \eqref{ansatz} one gets +Inserting the supersymmetric ansatz $D=\hat\beta R$ and $R=r_dI$, one gets \begin{equation} \label{eq:diagonal.action} \begin{aligned} \Sigma(\epsilon,\mu) =\mathcal D(\mu) + - \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 - \\ - +\frac12\lim_{n\to0}\frac1n\left(\hat\beta^2\sum_{ab}f(Q_{ab})+\log\det((D_d/R_d^2)Q+I)\right) - \end{aligned} -\end{equation} -Using standard manipulations (Appendix B), one finds also a continuous version -\begin{equation} \label{eq:functional.action} - \begin{aligned} - \Sigma(\epsilon,\mu) - =\mathcal D(\mu) - + - \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 + \hat\beta\epsilon-\mu r_d + +\frac12\hat\beta r_df'(1)+\frac12r_d^2f''(1)+\frac12\log r_d^2 \\ - +\frac12\int_0^1dq\,\left( - \hat\beta^2f''(q)\chi(q)+\frac1{\chi(q)+R_d^2/D_d} - \right) + +\frac12\lim_{n\to0}\frac1n\left(\hat\beta^2\sum_{ab}f(C_{ab})+\log\det((\hat\beta/r_d)C+I)\right) \end{aligned} \end{equation} -where $\chi(q)$ is defined with respect to $Q$ exactly as in the equilibrium case. -Note the close similarity of this action to the equilibrium replica one, at finite temperature. -\begin{equation} - \beta F=-\frac12\lim_{n\to0}\frac1n\left(\beta^2\sum_{ab}f(Q_{ab})+\log\det Q\right)-1-\log2\pi -\end{equation} - - -\subsubsection{Saddles} -\label{sec:counting.saddles} - -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\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 -$\mathcal D$. When saddles dominate, $\mu<\mu_m$, and -\begin{equation} - \mathcal D(\mu)=\frac12+\frac12\log f''(1)+\frac{\mu^2}{4f''(1)} -\end{equation} -It follows that the dominant saddles have $\mu=2f''(1)R_d$. Their index -is thus $\mathcal I=\frac12N(1-R_d\sqrt{f''(1)})$. If -$R_d\sqrt{f''(1)}>1$ then we were wrong to assume that saddles dominate, and -the most numerous saddles will be found just above $\mu=\mu_m$. - -\subsubsection{Minima} -\label{sec:counting.minima} - -When minima dominate, $\mu>\mu_m$ and all the roots inside $\mathcal D(\mu)$ are real. Therefore $\mathcal D(\mu)$ is given by its former expression with the real part dropped, and -\begin{equation} - \mathcal D'(\mu)=\frac{2}{\mu+\sqrt{\mu^2-4f''(1)}} -\end{equation} -\begin{equation} - \mu=\frac1{R_d}+R_df''(1) -\end{equation} - -\subsubsection{Recovering the equilibrium ground state} - -The ground state energy corresponds to that where the complexity of dominant -stationary points becomes zero. If the most common stationary points vanish, -then there cannot be any stationary points. In this section, we will show that -it reproduces the ground state produced by taking the zero-temperature limit in -the equilibrium case. - -Consider the extremum problem of \eqref{eq:diagonal.action} with respect to $R_d$ and $D_d$. This gives the equations -\begin{align} - 0 - &=\frac{\partial\Sigma}{\partial D_d} - =-\frac12f'(1)+\frac12\frac1{R_d^2}\lim_{n\to0}\frac1n\operatorname{Tr}((D_d/R_d^2)Q+I)^{-1}Q \label{eq:saddle.d}\\ - 0 - &=\frac{\partial\Sigma}{\partial R_d} - =-\mu+\hat\beta f'(1)+R_df''(1)+\frac1{R_d}-\frac{D_d}{R_d^3}\lim_{n\to0}\frac1n\operatorname{Tr}((D_d/R_d^2)Q+I)^{-1}Q \label{eq:saddle.r} -\end{align} -Adding $2(D_d/R_d)$ times \eqref{eq:saddle.d} to \eqref{eq:saddle.r} and multiplying by $R_d$ gives -\begin{equation} - 0=-R_d\mu+1+R_d^2f''(1)+f'(1)(R_d\hat\beta-D_d) -\end{equation} -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} -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. - -Substituting this into the action, and also substituting the optimal $\mu$ for -minima, and taking $\Sigma(\epsilon_0,\mu^*)=0$, gives +From here, it is straightforward to see that the complexity vanishes at the +ground state energy. First, in the ground state minima will dominate (even if +they are marginal), so we may assume \eqref{eq:mu.minima}. Then, 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( - \hat\beta^2\sum_{ab}^nf(Q_{ab}) - +\log\det(\hat\beta R_d^{-1} Q+I) + =-\frac12r_d\hat\beta f'(1)-\frac12\lim_{n\to0}\frac1n\left( + \hat\beta^2\sum_{ab}^nf(C_{ab}) + +\log\det(\hat\beta r_d^{-1} C+I) \right) \end{equation} -which is precisely \eqref{eq:ground.state.free.energy} with $R_d=z$, -$\hat\beta=\tilde\beta$, and $Q=\tilde Q$. +which is precisely \eqref{eq:ground.state.free.energy} with $r_d=z$, +$\hat\beta=\tilde\beta$, and $C=\tilde Q$. {\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 } 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 +$q_1,\ldots,q_k$, then the parameters $\hat\beta$, $r_d$, $d_d$, $\tilde +x_1,\ldots,\tilde x_{k-1}$, and $\tilde c_1,\ldots,\tilde c_{k-1}$ 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 + \tilde c_i=\lim_{\beta\to\infty}q_i && - R_d=\lim_{\beta\to\infty}\beta(1-q_k) + r_d=\lim_{\beta\to\infty}\beta(1-q_k) && - D_d=\hat\beta R_d + d_d=\hat\beta r_d \end{align} \section{Examples} @@ -634,6 +532,21 @@ E\rangle_2$. \end{figure} \subsection{Full RSB complexity} +Using standard manipulations (Appendix B), one finds also a continuous version +\begin{equation} \label{eq:functional.action} + \begin{aligned} + \Sigma(\epsilon,\mu) + =\mathcal D(\mu) + + + \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 + \\ + +\frac12\int_0^1dq\,\left( + \hat\beta^2f''(q)\chi(q)+\frac1{\chi(q)+R_d^2/D_d} + \right) + \end{aligned} +\end{equation} +where $\chi(q)$ is defined with respect to $Q$ exactly as in the equilibrium case. \begin{figure} \centering @@ -714,6 +627,17 @@ 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. +\section{Interpretation} + + The +meaning of $R_{ab}$ is that of a response of replica $a$ to a linear field in +replica $b$: +\begin{equation} + R_{ab} = \frac 1 N \sum_i \overline{\frac{\delta s_i^a}{\delta h_i^b}} +\end{equation} +The $D$ may similarly be seen as the variation of the complexity with respect to a random field. + + \section{Ultrametricity rediscovered} TENTATIVE BUT INTERESTING |