diff options
| -rw-r--r-- | frsb_kac_new.tex | 185 |
1 files changed, 96 insertions, 89 deletions
diff --git a/frsb_kac_new.tex b/frsb_kac_new.tex index c89abb3..99455de 100644 --- a/frsb_kac_new.tex +++ b/frsb_kac_new.tex @@ -51,12 +51,12 @@ for Can be thought of as a model of generic gaussian functions on the sphere. To constrain the model to the sphere, we use a Lagrange multiplier $\mu$, with the total energy being \begin{equation} - H(s)+\frac\mu2(N-s\cdot s) + H(s)+\frac\mu2(s\cdot s-N) \end{equation} At any critical point, the hessian is \begin{equation} - \operatorname{Hess}H=\partial\partial H-\mu I + \operatorname{Hess}H=\partial\partial H+\mu I \end{equation} $\partial\partial H$ is a GOE matrix with variance \begin{equation} @@ -66,7 +66,7 @@ and therefore its spectrum is given by the Wigner semicircle with radius $\sqrt{ \begin{equation} \rho(\lambda)=\frac1{\pi\sqrt{f''(1)}}\sqrt{\lambda^2-4f''(1)} \end{equation} -and the spectrum of $\operatorname{Hess}H$ is this shifted by $\mu$, or $\rho(\lambda-\mu)$. +and the spectrum of $\operatorname{Hess}H$ is this shifted by $\mu$, or $\rho(\lambda+\mu)$. The parameter $\mu$ fixes the spectrum of the hessian. When it is an integration variable, and one restricts the domain of all integrations to compute saddles of a certain macroscopic index, or @@ -176,7 +176,7 @@ $F$ is a $k-1$ RSB ansatz with all eigenvalues scaled by $y$ and shifted by $z$. \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)| + =\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) @@ -196,7 +196,7 @@ This will turn out to be important when we discriminate between counting all sol \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)| + &=\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} @@ -212,9 +212,9 @@ $\hat \beta$ is a parameter conjugate to the state energies, i.e. playing the ro \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)} + &=\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)|} + \overline{\prod_a^n |\det(\partial\partial H(s_a)+\mu I)|} \end{aligned} \end{equation} @@ -227,9 +227,9 @@ $\hat \beta$ is a parameter conjugate to the state energies, i.e. playing the ro \begin{equation} - \prod_a^n\delta(N\epsilon-H(s_a))\delta(\partial H(s_a)-\mu s_a) + \prod_a^n\delta(N\epsilon-H(s_a))\delta(\partial H(s_a)+\mu s_a) =\int \frac{d\hat\epsilon}{2\pi}\prod_a^n\frac{d\hat s_a}{2\pi} - e^{\hat\epsilon(N\epsilon-H(s_a))+i\hat s_a\cdot(\partial H(s_a)-\mu s_a)} + e^{\hat\epsilon(N\epsilon-H(s_a))+i\hat s_a\cdot(\partial H(s_a)+\mu s_a)} \end{equation} \begin{equation} @@ -274,7 +274,7 @@ Introducing the parameters: \begin{aligned} S =\mathcal D(\mu)+\hat\epsilon\epsilon+\lim_{n\to0}\frac1n\left( - \mu\sum_a^nR_{aa} + -\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}) @@ -287,8 +287,8 @@ Introducing the parameters: \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| + &=\frac1N\overline{\log|\det(\partial\partial H(s_a)+\mu I)|} + =\int d\lambda\,\rho(\lambda+\mu)\log|\lambda| \end{aligned} \end{equation} @@ -300,7 +300,7 @@ Following the usual steps (Appendix B) we arrive at the replicated action: \begin{aligned} S =\mathcal D(\mu)+\hat\epsilon\epsilon+\lim_{n\to0}\frac1n\left( - \mu\sum_a^nR_{aa} + -\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}) @@ -325,26 +325,64 @@ The reader who is uneasy about this ansatz will find another motivation in Ap \subsection{Solution} -Insert the diagonal ansatz \cite{diagonal} one gets, using standard manipulations (Appendix B) +Insert the diagonal ansatz \cite{diagonal} one gets -\begin{eqnarray} - \Sigma - &=&\epsilon\hat\epsilon+\hat\epsilon R_d f'(1)+\frac12D_df'(1)+ \log(-D_dR_d^{-2}) \nonumber\\ - &+& \lim_{n\to0}\frac1n\frac12\left( - \hat\epsilon^2\sum_{ab} - f(Q_{ab}) - +\log\det( Q-D_d^{-1}R_d^{2}) - \right) -\end{eqnarray} +\begin{equation} \label{eq:diagonal.action} + \begin{aligned} + S + =\mathcal D(\mu) + + + \hat\epsilon\epsilon-\mu R_d + +\frac12(2\hat\epsilon R_d-D_d)f'(1)+\frac12R_d^2f''(1)+\frac12\log R_d^2 + \\ + +\frac12\lim_{n\to0}\frac1n\left(\hat\epsilon^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 +\begin{equation} \label{eq:functional.action} + \begin{aligned} + S + =\mathcal D(\mu) + + + \hat\epsilon\epsilon-\mu R_d + +\frac12(2\hat\epsilon R_d-D_d)f'(1)+\frac12R_d^2f''(1)+\frac12\log R_d^2 + \\ + +\frac12\int_0^1dq\,\left( + \hat\epsilon^2f''(q)\chi(q)+\frac1{\chi(q)+R_d^2/D_d} + \right) + \end{aligned} +\end{equation} 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} I WOULD LOVE TO FIND A DIRECT BRIDGE AT THIS LEVEL -\subsubsection{All saddles} +\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 S}{\partial\mu}=\mathcal D'(\mu)-R_d +\end{equation} +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 $\mu=2f''(1)R_d$. \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} \begin{figure} \begin{center} @@ -353,14 +391,43 @@ I WOULD LOVE TO FIND A DIRECT BRIDGE AT THIS LEVEL \end{figure} \subsubsection{Recovering the replica ground state} -The lowest states are obtained by setting $0=\Sigma$, which gives -\[ - \epsilon - =\lim_{n\to0}\frac1n\frac12\left(nR_df'(1)+\hat\epsilon\sum_{ab} +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 S}{\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 S}{\partial R_d} + =-\mu+\hat\epsilon f'(1)+R_df''(1)+\frac1{R_d}+\frac{D_d}{R_d}\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)$ of \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\epsilon-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 +\begin{equation} + 0=f'(1)(R_d\hat\epsilon-D_d) +\end{equation} +Therefore, in any situation where minima dominate, the optimal $\mu$ will have $R_d\hat\epsilon=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\epsilon-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\epsilon-D_d=0$. + +In any case, at the ground state $D_d=R_d\hat\epsilon$. Substuting this into the action, and also substituting the optimal $\mu$ for saddles or minima, gives +\begin{equation} + \epsilon_0 + =-\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) -\] +\end{equation} + which is precisely \eqref{eq:ground.state.free.energy} with $R_d=z$ and $\hat\epsilon=y$. {\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 } @@ -504,66 +571,6 @@ $F$ is a $k-1$ RSB ansatz with all eigenvalues scaled by $y$ and shifted by $z$. \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.} - - \begin{align*} \lim_{n\to0}\frac1n\log\det(\hat\epsilon R_d^{-1} Q+I) |