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diff --git a/frsb_kac-rice.tex b/frsb_kac-rice.tex index 04cac67..d513e48 100644 --- a/frsb_kac-rice.tex +++ b/frsb_kac-rice.tex @@ -433,7 +433,17 @@ $K(\hat \beta,r_d)$ contains all the information about saddle densities. 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 +i.e., to follow Parisi's scheme. In the end, when the limit of $n\to0$ is +taken, each can be represented in the canonical way by its diagonal and a +continuous function on the domain $[0,1]$ which parameterizes each of its rows, with +\begin{align} + C\;\leftrightarrow\;[c_d, c(x)] + && + R\;\leftrightarrow\;[r_d, r(x)] + && + D\;\leftrightarrow \;d_d, d(x)] +\end{align} +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} @@ -472,6 +482,17 @@ When $\mu^*<\mu_m$, they are saddles, and \mu^*=2f''(1)r_d \end{equation} +We will find it often useful to have the extremal conditions in a form without matrix inverses, both for numerics at finite $k$-RSB and for expanding in the continuous case. By simple manipulations, the matrix equations can be written as +\begin{align} + 0&=\left[\hat\beta^2f'(C)+(2\hat\beta R-D)\odot f''(C)+R\odot R\odot f'''(C)+\hat\mu I\right]C+f'(C)D \\ + 0&=\left[\hat\beta f'(C)+R\odot f''(C)-\mu^*I\right]C+f'(C)R \\ + 0&=C-f'(C)(CD+R^2) +\end{align} +The right-hand side of each of these equations is also a hierarchical matrix, +since products, Hadamard products, and sums of hierarchical matrices are such. +The equations for the continuous case are found by using the mappings to +functions $c(x)$, $r(x)$ and $d(x)$, then carrying through the appropriate +operations. \section{Supersymmetric solution} @@ -480,7 +501,8 @@ 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? +that $D=\hat\beta R$ \cite{Annibale_2003_The}. Under which conditions can this +relationship be expected to hold? We find that their applicability is limited. 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 @@ -559,16 +581,39 @@ complexity in the ground state are d_d=\hat\beta r_d \end{align} -\subsection{Full RSB} +\section{Full replica symmetry breaking} + +This reasoning applies equally well to FRSB systems. +\begin{align} + 0&=\hat\mu c(x)+[(\hat\beta^2f'(c)+(2\hat\beta r-d)f''(c)+r^2f'''(c))\ast c](x)+(f'(c)\ast d)(x) \label{eq:extremum.c} \\ + 0&=-\mu^* c(x)+[(\hat\beta f'(c)+rf''(c))\ast c](x)+(f'(c)\ast r)(x) \label{eq:extremum.r} \\ + 0&=c(x)-\big(f'(c)\ast(c\ast d+r\ast r)\big)(x) \label{eq:extremum.d} +\end{align} +where the product of two hierarchical matrix results in +\begin{align} + (a\ast b)_d&=a_db_d-\langle ab\rangle \\ + (a\ast b)(x)&=(b_d-\langle b\rangle)a(x)+(a_d-\langle a\rangle)b(x) + -\int_0^xdy\,\big( + a(x)-a(y) + \big)\big( + b(x)-b(y) + \big) +\end{align} +and +\begin{equation} + \langle a\rangle=\int_0^1dx\,a(x) +\end{equation} -This reasoning applies equally well to FRSB systems. Using standard +\subsection{Supersymmetric complexity} + +Using standard manipulations (Appendix B), one finds also a continuous version of the supersymmetric complexity \begin{equation} \label{eq:functional.action} - \Sigma(E,\mu) - =\mathcal D(\mu) + \Sigma(E,\mu^*) + =\mathcal D(\mu^*) + - \hat\beta E-\mu r_d + \hat\beta E-\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/\hat\beta} @@ -588,7 +633,7 @@ given by \end{equation} is correct. This is only correct if it satisfies the boundary condition $\chi(1)=0$, which requires $r_d=f''(1)^{-1/2}$. This in turn implies -$\mu=\frac1{r_d}+f''(1)r_d=\sqrt{4f''(1)}=\mu_m$. Therefore, the FRSB ground state +$\mu^*=\frac1{r_d}+f''(1)r_d=\sqrt{4f''(1)}=\mu_m$. Therefore, the FRSB ground state is exactly marginal! It is straightforward to check that these conditions are indeed a saddle of the complexity. @@ -597,6 +642,32 @@ This has several implications. First, other than the ground state, there are As we will see, stable minima are numerous at energies above the ground state, but these vanish at the ground state. +\subsection{Expansion near the transition} + +Working with the general equations in their continuum form away from the +supersymmetric solution is not generally tractable. However, there is another +point where they can be treated analytically: near the onset of replica +symmetry breaking. Here, the off-diagonal components of $C$, $R$, and $D$ are +expected to be small. In particular, we expect the function $c$, $r$, and $d$ +to approach zero at the transition, and moreover take the form +\begin{equation} + c(x)=\begin{cases}\bar cx&x\leq x_\mathrm{max}\\\bar cx_\mathrm{max}&\text{otherwise}\end{cases} +\end{equation} +with $x_\mathrm{max}$ vanishing at the transition, with the slopes $\bar c$, +$\bar r$, and $\bar d$ remaining nonzero. This ansatz is informed both by the +experience of the equilibrium solution, and by empirical observation within the +numerics. + +Given this ansatz, we take the equations \eqref{eq:extremum.c}, +\eqref{eq:extremum.r}, and \eqref{eq:extremum.d}, which are true for any $x$, and +integrate them over $x$. We then expand the result about small +$x_\mathrm{max}$. The result is +\begin{align} + 0&=(-\mu^*\bar c+\bar rf'(1)+\hat\beta\bar cf'(1)+\bar rf''(0)+\hat\beta\bar c f''(0)+\bar cr_d(f''(1)+f''(0)) +\end{align} + + + \section{General solution: examples} \subsection{1RSB complexity} |