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| -rw-r--r-- | frsb_kac-rice.tex | 72 |
1 files changed, 27 insertions, 45 deletions
diff --git a/frsb_kac-rice.tex b/frsb_kac-rice.tex index 90bf081..cd295e9 100644 --- a/frsb_kac-rice.tex +++ b/frsb_kac-rice.tex @@ -61,7 +61,7 @@ $\partial\partial H$ is a GOE matrix with variance \end{equation} and therefore its spectrum is given by the Wigner semicircle with radius $\sqrt{4f''(1)}$, or \begin{equation} - \rho(\lambda)=\frac1{\pi\sqrt{f''(1)}}\sqrt{4f''(1)-\lambda^2} + \rho(\lambda)=\frac1{2\pi f''(1)}\sqrt{4f''(1)-\lambda^2} \end{equation} and the spectrum of $\operatorname{Hess}H$ is this shifted by $\mu$, or $\rho(\lambda+\mu)$. @@ -69,7 +69,16 @@ The parameter $\mu$ fixes the spectrum of the hessian. By manipulating it, one can decide to find the complexity of saddles of a certain macroscopic index, or of minima with a certain harmonic stiffness. When $\mu$ is taken to be within the range $\pm\sqrt{4f''(1)}=\pm\mu_m$, the critical points are constrained to have -index $\mathcal I=\frac12N(1-\mu/\mu_m)$. When $\mu>\mu_m$, the critical +index +\begin{equation} + \mathcal I(\mu)=N\int_0^\infty d\lambda\,\rho(\lambda+\mu) + =N\left\{\frac12-\frac1\pi\left[ + \arctan\left(\frac\mu{\sqrt{\mu_m^2-\mu^2}}\right) + +\frac\mu{\mu_m^2}\sqrt{\mu_m^2-\mu^2} + \right] + \right\} +\end{equation} +When $\mu>\mu_m$, the critical points are minima whose sloppiest eigenvalue is $\mu-\mu_m$. Finally, when $\mu=\mu_m$, the critical points are marginal minima. @@ -79,48 +88,6 @@ and one restricts the domain of all integrations to compute saddles of a certain of minima with a certain harmonic stiffness, its value is the `softest' mode that adapts to change the Hessian \cite{Fyodorov_2007_Replica}. When it is fixed, then the restriction of the index of saddles is `payed' by the realization of the eigenvalues of the Hessian, usually a `harder' mode. -{\tiny NOT SURE WORTHWHILE - -\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), 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 frozen phase, - the first system is thus concentrated in a few states of $O(1)$ energy, while the second is not, so that $X_1=X_1^{gs}> X_2$. The lowest energy is reached when 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 minima of the second system remain 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, -between zero and one 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, because a saddle is found by releasing the constraint of staying on the threshold. - -} - \section{Equilibrium} Here we review the equilibrium solution. \cite{Crisanti_1992_The, Crisanti_1993_The, Crisanti_2004_Spherical, Crisanti_2006_Spherical} @@ -196,7 +163,7 @@ $q_k$ gives the overlap within a state, e.g., within the basin of a well inside the energy landscape. At zero temperature, the measure is completely localized on the bottom of the well, and therefore the overlap with each state becomes one. We will see that the complexity of low-energy stationary points in -Kac--Rice is also given by a $(k-1)$-RSB anstaz. Heuristicall, this is because +Kac--Rice is also given by a $(k-1)$-RSB anstaz. Heuristically, this is because each stationary point also has no width and therefore overlap one with itself. @@ -367,6 +334,21 @@ 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 $Tr \left[\tilde R (\hat \beta -R_d Q^{-1} f')\right]$ } |