diff options
Diffstat (limited to 'frsb_kac-rice.tex')
| -rw-r--r-- | frsb_kac-rice.tex | 106 |
1 files changed, 57 insertions, 49 deletions
diff --git a/frsb_kac-rice.tex b/frsb_kac-rice.tex index 1366988..8086c70 100644 --- a/frsb_kac-rice.tex +++ b/frsb_kac-rice.tex @@ -23,59 +23,18 @@ in fact the problem has been open ever since. Indeed, to this date the program of computing the number of saddles of a mean-field glass has been only carried out for a small subset of models. -These include most notable the p-spin model ($p>2$) \cite{rieger1992number,crisanti1995thouless} -In this paper we present what we believe is the general ansatz for the solution. -It includes the Parisi solution as the limit of lowest states, as it should. +These include most notably the p-spin model ($p>2$) \cite{rieger1992number,crisanti1995thouless}. +The problem of studying the critical points of these landscapes +has evolved into an active field in probability theory \cite{Auffinger_2012_Random,Auffinger_2013_Complexity,BenArous_2019_Geometry} +In this paper we present what we believe is the general ansatz for the +computation of saddles of generic mean-field models, including the Sherrington-Kirkpatrick model. It incorporates the Parisi solution as the limit of lowest states, as it should. -\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 (????? what) - -The mixed $p$-spin model +Here we consider, for definiteness, the mixed $p$-spin model, itself a particular case +of the `Toy Model' of M\'ezard and Parisi \cite{mezard1992manifolds} \begin{equation} H(s)=\sum_p\frac{a_p^{1/2}}{p!}\sum_{i_1\cdots i_p}J_{i_1\cdots i_p}s_{i_1}\cdots s_{i_p} \end{equation} @@ -116,6 +75,49 @@ points are minima whose sloppiest eigenvalue is $\mu-\mu_m$. Finally, when $\mu=\mu_m$, the critical points are marginal minima. +\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{Main result} \begin{equation} @@ -294,6 +296,11 @@ $F$ is a $k-1$ RSB ansatz with all eigenvalues scaled by $y$ and shifted by $z$. \Sigma(\epsilon,\mu)=\frac1N\log\mathcal N(\epsilon, \mu) \end{equation} +{\em The `mass' term $\mu$ may take a fixed value, or it may be an integration constant, +for example fixing the spherical constraint. +This will turn out to be important when we discriminate between counting all solutions, or selecting those of a given index, for example minima} + + \subsection{The replicated problem} \cite{Ros_2019_Complex} @@ -316,7 +323,8 @@ the question of independence \cite{Bray_2007_Statistics} \times \overline{\prod_a^n |\det(\partial\partial H(s_a)-\mu I)|} \end{aligned} -\end{equation} +\end{equation}{\bf The average breaks into a product of two independent averages, one for the gradient +and one for the determinant. The integration of all variables may be restricted to the domain such that the matrix $\partial\partial H(s_a)-\mu I$ has a specified number of negative eigenvalues Fyodorov? and } \begin{equation} \begin{aligned} |