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diff --git a/frsb_kac-rice.tex b/frsb_kac-rice.tex index 11d5822..7c09dca 100644 --- a/frsb_kac-rice.tex +++ b/frsb_kac-rice.tex @@ -184,38 +184,14 @@ The free energy, averaged over disorder, is Once $n$ replicas are introduced to treat the logarithm, the fields $\mathbf s_a$ can be replaced with the new $n\times n$ matrix field $Q_{ab}\equiv(\mathbf s_a\cdot\mathbf s_b)/N$. This yields for the free energy -\begin{equation} +\begin{equation} \label{eq:eq.free.energy} \beta F=-1-\ln2\pi-\frac12\lim_{n\to0}\frac1n\left(\beta^2\sum_{ab}^nf(Q_{ab})+\ln\det Q\right) \end{equation} which must be evaluated at the $Q$ which maximizes this expression and whose diagonal is one. The solution is generally a hierarchical matrix \emph{à la} -Parisi. Each row of the matrix is the same up to permutation of their elements. -The so-called $k$RSB solution has $k+2$ different values in each row: $n-x_1$ -of those entries are $q_0$, $x_1-x_2$ of those entries are $q_1$, and so on -until $x_k-1$ entries of $q_k$, and of course one entry of $1$ (the diagonal). -Given such a matrix, there are standard ways of producing the sum and -determinant that appear in the free energy. These formulas are, for an -arbitrary $k$RSB matrix $A$ with $a_d$ on its diagonal (recall $q_d=1$), -\begin{equation} \label{eq:replica.sum} - \lim_{n\to0}\frac1n\sum_{ab}^nA_{ab} - =a_d-\sum_{i=0}^k(x_{i+1}-x_i)a_i -\end{equation} -\begin{equation} \label{eq:replica.logdet} - \begin{aligned} - \lim_{n\to0}\frac1n\ln\det A - &= - \frac{a_0}{a_d-\sum_{i=0}^k(x_{i+1}-x_i)a_i} - +\frac1{x_1}\log\left[ - a_d-\sum_{i=0}^{k}(x_{i+1}-x_i)a_i - \right]\\ - &\hspace{10pc}-\sum_{j=1}^k(x_j^{-1}-x_{j+1}^{-1})\log\left[ - a_d-\sum_{i=j}^{k}(x_{i+1}-x_i)a_i-x_ja_j - \right] - \end{aligned} -\end{equation} -where $x_0=0$ and $x_{k+1}=1$. Once these are substituted into the free energy, -optimizing the anstaz is equivalent to maximizing $F$ with respect to the -$q_0,\ldots,q_k$ and $x_1,\ldots,x_k$. +Parisi. The properties of these matrices is reviewed in \S\ref{sec:dict}, +including how to write down \eqref{eq:eq.free.energy} in terms of their +parameters. The free energy can also be written in a functional form, which is necessary for working with the solution in the limit $k\to\infty$, the so-called full @@ -561,19 +537,13 @@ matrix products and Hadamard products. In particular, the determinant of the blo \ln\det\begin{bmatrix}C&iR\\iR&D\end{bmatrix}=\ln\det(CD+R^2) \end{equation} This is straightforward to write down at $k$RSB, since the product and sum of -the hierarchical matrices is still a hierarchical matrix. Recall the -standard formulas for the product $C=AB$ of two hierarchical matrices: -\begin{align} \label{eq:replica.prod} - c_d&=a_db_d-\sum_{j=0}^k(x_{j+1}-x_j)a_jb_j \\ - c_i&=b_da_i+a_db_i-\sum_{j=0}^{i-1}(x_{j+1}-x_j)a_jb_j+(2x_{i+1}-x_i)a_ib_i - -\sum_{j=i+1}^k(x_{j+1}-x_j)(a_ib_j+a_jb_i) -\end{align} -and for the sum $C=A+B$ of two hierarchical matrices: $c_d=a_d+b_d$ and -$c_i=a_i+b_i$. Using these, one can write down the hierarchical matrix -$CD+R^2$, and then compute the $\ln\det$ using the previous formula -\eqref{eq:replica.logdet}. +the hierarchical matrices is still a hierarchical matrix. The algebra of +hierarchical matrices is reviewed in \S\ref{sec:dict}. Using the product formula +\eqref{eq:replica.prod}, one can write down the hierarchical matrix $CD+R^2$, +and then compute the $\ln\det$ using the formula \eqref{eq:replica.logdet}. -The extremal conditions are given by differentiating the complexity with respect to its parameters, yielding +The extremal conditions are given by differentiating the complexity with +respect to its parameters, yielding \begin{align} 0&=\frac{\partial\Sigma}{\partial\hat\mu} =\frac12(c_d-1) \\ @@ -758,7 +728,8 @@ each of its rows, with && D\;\leftrightarrow\;[d_d, d(x)] \end{align} -The complexity becomes +The algebra of hierarchical matrices under this continuous parameterization is +review in \S\ref{sec:dict}. The complexity becomes \begin{equation} \begin{aligned} \Sigma(E,\mu^*) @@ -770,34 +741,18 @@ The complexity becomes +\frac12\lim_{n\to0}\frac1n\ln\det(CD+R^2) \end{aligned} \end{equation} -The determinant takes more work to treat. -Define for any function of $x$ -\begin{equation} - \langle a\rangle=\int_0^1dx\,a(x) -\end{equation} -In general, the product of hierarchical matrices gives -\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 the $\ln\det$ becomes, for the hierarchical matrix $A=CD+R^2$, -\begin{equation} \label{eq:replica.det.cont} - \lim_{n\to0}\frac1n\ln\det A - =\ln(a_d-\langle a\rangle) - +\frac{a(0)}{a_d-\langle a\rangle} - -\int_0^1\frac{dx}{x^2}\ln\left(\frac{a_d-\langle a\rangle-xa(x)+\int_0^xdy\,a(y)}{a_d-\langle a\rangle}\right) -\end{equation} +The formula for the determinant is complicated, and can be found by using the +product formula \eqref{eq:cont.replica.prod} to write $CD$ and $R^2$, summing +them, and finally using the $\ln\det$ formula \eqref{eq:replica.det.cont}. The saddle point equations take the form \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 $(a\ast b)(x)$ denotes the functional parameterization of the diagonal of +the product of hierarchical matrices $AB$, defined in +\eqref{eq:cont.replica.prod}. \subsection{Supersymmetric complexity} @@ -1484,6 +1439,80 @@ dominant saddles transition to a RSB complexity, we speculate that $E_\mathrm{alg}$ may be related to the statistics of minima connected to the saddles at this transition point. +\begin{appendix} + + \section{Hierarchical matrix dictionary} + \label{sec:dict} + + Each row of a hierarchical matrix is the same up to permutation of their + elements. The so-called $k$RSB ansatz has $k+2$ different values in each + row. If $A$ is an $n\times n$ hierarchical matrix, then $n-x_1$ of those + entries are $a_0$, $x_1-x_2$ of those entries are $a_1$, and so on until + $x_a-1$ entries of $a_k$, and one entry of $a_d$, corresponding to the + diagonal. Given such a matrix, there are standard ways of producing the sum + and determinant that appear in the free energy. These formulas are, for an + arbitrary $k$RSB matrix $A$ with $a_d$ on its diagonal (recall $q_d=1$), + \begin{equation} \label{eq:replica.sum} + \lim_{n\to0}\frac1n\sum_{ab}^nA_{ab} + =a_d-\sum_{i=0}^k(x_{i+1}-x_i)a_i + \end{equation} + \begin{equation} \label{eq:replica.logdet} + \begin{aligned} + \lim_{n\to0}\frac1n\ln\det A + &= + \frac{a_0}{a_d-\sum_{i=0}^k(x_{i+1}-x_i)a_i} + +\frac1{x_1}\log\left[ + a_d-\sum_{i=0}^{k}(x_{i+1}-x_i)a_i + \right]\\ + &\hspace{10pc}-\sum_{j=1}^k(x_j^{-1}-x_{j+1}^{-1})\log\left[ + a_d-\sum_{i=j}^{k}(x_{i+1}-x_i)a_i-x_ja_j + \right] + \end{aligned} + \end{equation} + where $x_0=0$ and $x_{k+1}=1$. The sum of two hierarchical matrices results + in the sum of each of their elements: $(a+b)_d=a_d+b_d$ and + $(a+b)_i=a_i+b_i$. The product $A\ast B$ of two hierarchical matrices $A$ + and $B$ is given by + \begin{align} \label{eq:replica.prod} + (a\ast b)_d&=a_db_d-\sum_{j=0}^k(x_{j+1}-x_j)a_jb_j \\ + (a\ast b)_i&=b_da_i+a_db_i-\sum_{j=0}^{i-1}(x_{j+1}-x_j)a_jb_j+(2x_{i+1}-x_i)a_ib_i + -\sum_{j=i+1}^k(x_{j+1}-x_j)(a_ib_j+a_jb_i) + \end{align} + + There is a canonical mapping between the parameterization of a hierarchical + matrix described above and a functional parameterization that is particularly + convenient in the twin limit $n\to0$ and $k\to\infty$ + \cite{Parisi_1980_Magnetic, Mezard_1991_Replica}. The distribution of + diagonal elements of a matrix $A$ is parameterized by a continuous function + $a(x)$ on the interval $[0,1]$, while its diagonal is still called $a_d$. + Define for any function $g$ the average + \begin{equation} + \langle g\rangle=\int_0^1dx\,g(x) + \end{equation} + The sum of two hierarchical matrices so parameterized results in the sum of + these functions. The product $AB$ of hierarchical matrices $A$ and $B$ gives + \begin{align} \label{eq:cont.replica.prod} + (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} + The sum over all elements of a hierarchical matrix $A$ gives + \begin{equation} + \lim_{n\to0}\frac1n\sum_{ab}A_{ab}=a_d-\langle a\rangle + \end{equation} + The $\ln\det=\operatorname{Tr}\ln$ becomes + \begin{equation} \label{eq:replica.det.cont} + \lim_{n\to0}\frac1n\ln\det A + =\ln(a_d-\langle a\rangle) + +\frac{a(0)}{a_d-\langle a\rangle} + -\int_0^1\frac{dx}{x^2}\ln\left(\frac{a_d-\langle a\rangle-xa(x)+\int_0^xdy\,a(y)}{a_d-\langle a\rangle}\right) + \end{equation} +\end{appendix} + \paragraph{Acknowledgements} The authors would like to thank Valentina Ros for helpful discussions. |