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diff --git a/frsb_kac-rice.tex b/frsb_kac-rice.tex index 61d283f..42c5696 100644 --- a/frsb_kac-rice.tex +++ b/frsb_kac-rice.tex @@ -379,6 +379,132 @@ When $\mu<\mu_m$, they are saddles, and \mu=2f''(1)r_d \end{equation} +\section{Interpretation} +\label{sec:interpretation} + +Let $\langle A\rangle$ be the average of $A$ over stationary points with given $E$ and $\mu$, i.e., +\begin{equation} + \langle A\rangle + =\frac1{\mathcal N}\sum_{\sigma}A(s_\sigma) + =\frac1{\mathcal N} + \int d\nu(s)\,A(s) +\end{equation} +with +\begin{equation} + d\nu(s)=ds\,\delta(NE-H(s))\delta(\partial H(s)+\mu s)|\det(\partial\partial H(s)+\mu I)| +\end{equation} +the Kac--Rice measure. The fields $C$, $R$, and $D$ defined in +\eqref{eq:fields} can be related to certain averages of this type. + +\subsection{\textit{C}: distribution of overlaps} + +First, +consider $C$, which has an interpretation nearly identical to that of Parisi's +$Q$ matrix of overlaps. It can be shown that its off-diagonal corresponds to +the probability distribution of the overlaps between stationary points $P(q)$. First, define this distribution as +\begin{equation} + P(q)=\frac1{\mathcal N^2}\sum_{\sigma,\sigma'}\delta\left(\frac{s_\sigma\cdot s_{\sigma'}}N-q\right) +\end{equation} +where the sum is twice over stationary points $\sigma$ and $\sigma'$. It is +straightforward to show that moments of this distribution are related to +certain averages of the form +\begin{equation} + \int dq\,q^p P(q) + =q^{(p)} + \equiv\frac1{N^p}\sum_{i_1\cdots i_p}\langle s_{i_1}\cdots s_{i_p}\rangle\langle s_{i_1}\cdots s_{i_p}\rangle +\end{equation} +The appeal of Parisi to properties of pure states is unnecessary here, since +the stationary points are points. These moments are related to our $C$ by +computing their average over disorder: +\begin{equation} + \begin{aligned} + \overline{q^{(p)}} + =\overline{\frac1{N^p}\sum_{i_1\cdots i_p}\langle s_{i_1}\cdots s_{i_p}\rangle\langle s_{i_1}\cdots s_{i_p}\rangle} + =\lim_{n\to0}\overline{\int\prod_\alpha^nd\nu(s_\alpha)\,\frac1{n(n-1)}\sum_{a\neq b}\left(\frac{s_a\cdot s_b}N\right)^p} \\ + =\lim_{n\to0}\frac1{n(n-1)}\sum_{a\neq b}^nC^p_{ab} + =\int_0^1 dx\,c^p(x) + \end{aligned} +\end{equation} +where $C$ is assumed to take its saddle point value. If we now change variables in the integral from $x$ to $c$, we find +\begin{equation} + \overline{q^{(p)}}=\int dc\,c^p\frac{dx}{dc} +\end{equation} +from which we conclude $\overline{P(q)}=\frac{dx}{dc}\big|_{c=q}$. + +With this +established, we now address what it means for $C$ to have a nontrivial +replica-symmetry broken structure. When $C$ is replica symmetric, drawing two +stationary points at random will always lead to the same overlap. In the case +when there is no linear field, they will always have overlap zero, because the +second point will almost certainly lie on the equator of the sphere with +respect to the first. Though other stationary points exist nearby the first +one, they are exponentially fewer and so will be picked with vanishing +probability in the thermodynamic limit. + +When $C$ is replica-symmetry broken, there is a nonzero probability of picking +a second stationary point at some other overlap. This can be interpreted by +imagining the level sets of the Hamiltonian in this scenario. If the level sets +are disconnected but there are exponentially many of them distributed on the +sphere, one will still find zero average overlap. However, if the disconnected +level sets are \emph{few}, i.e., less than order $N$, then it is possible to +draw two stationary points from the same set. Therefore, the picture in this +case is of few, large basins each containing exponentially many stationary +points. + +\subsection{\textit{R} and \textit{D}: response functions} + +The matrix field $R$ is related to responses of the stationary points to perturbations of the tensors $J$: +\begin{equation} + \begin{aligned} + \overline{\frac1{N^p}\sum_{i_1\cdots i_p}\frac{\partial\langle s_{i_1}\cdots s_{i_p}\rangle}{\partial J^{(p)}_{i_1\cdots i_p}}} + =\lim_{n\to0}\overline{\int\prod_\alpha^nd\nu(s_\alpha)\,\frac1n\sum_{ab}\left[ + \hat\beta\left(\frac{s_a\cdot s_b}N\right)^p+ + p\left(-i\frac{\hat s_a\cdot s_b}N\right)\left(\frac{s_a\cdot s_b}N\right)^{p-1} + \right]} \\ + =\lim_{n\to0}\frac1n\sum_{ab}(\hat\beta C_{ab}^p+pR_{ab}C_{ab}^{p-1}) + =\hat\beta+pr_d-\int_0^1dx\,c^{p-1}(x)(\hat\beta c(x)+pr(x)) + \end{aligned} +\end{equation} +In particular, when the energy is unconstrained ($\hat\beta=0$) and there is replica symmetry, +\begin{equation} + \frac1N\sum_i\frac{\partial\langle s_i\rangle}{\partial J_i^{(1)}} + =r_d +\end{equation} +i.e., adding a linear field causes a response in the average stationary point +location proportional to $r_d$. If positive, for instance, stationary points +tend to align with a field. The energy constraint has a significant +contribution due to the perturbation causing stationary points to move up or +down in energy. + +The matrix field $D$ is related to the response of the complexity to such perturbations: +\begin{equation} + \begin{aligned} + \frac{\partial\Sigma}{\partial a_p} + =\frac14\lim_{n\to0}\frac1n\sum_{ab}^n\left[ + \hat\beta^2C_{ab}^p+p(2\hat\beta R_{ab}-D_{ab})C_{ab}^{p-1}+p(p-1)R_{ab}^2C_{ab}^{p-2} + \right] + \end{aligned} +\end{equation} +In particular, when the energy is unconstrained ($\hat\beta=0$) and there is no replica symmetry breaking, +\begin{equation} + \frac{\partial\Sigma}{\partial a_1}=-\frac14\lim_{n\to0}\frac1n\sum_{ab}D_{ab}=-\frac14d_d +\end{equation} +i.e., adding a random linear field decreases the complexity of solutions by an amount proportional to $d_d$ in the variance of the field. + +When the saddle point of the Kac--Rice problem is supersymmetric, +\begin{equation} + \frac{\partial\Sigma}{\partial a_p} + =\frac{\hat\beta}4\overline{\frac1{N^p}\sum_{i_1\cdots i_p}\frac{\partial\langle s_{i_1}\cdots s_{i_p}\rangle}{\partial J^{(p)}_{i_1\cdots i_p}}}+\lim_{n\to0}\frac1n\sum_{ab}^np(p-1)R_{ab}^2C_{ab}^{p-2} +\end{equation} +and in particular for $p=1$ +\begin{equation} + \frac{\partial\Sigma}{a_1} + =\frac{\hat\beta}4\overline{\frac1N\sum_i\frac{\partial\langle s_i\rangle}{\partial J_i^{(1)}}} +\end{equation} +i.e., the change in complexity due to a linear field is directly related to the +resulting magnetization of the stationary points. + + \section{Supersymmetric solution} The Kac--Rice problem has an approximate supersymmetry, which is found when the @@ -644,131 +770,6 @@ Continuity requires that $1-q_\textrm{max}=\lambda^*(q_\textrm{max})$. Here a picture of $\chi$ vs $C$ or $X$ vs $C$ showing limits $q_{max}$, $x_{max}$ for different energies and typical vs minima. -\section{Interpretation} -\label{sec:interpretation} - -Let $\langle A\rangle$ be the average of $A$ over stationary points with given $E$ and $\mu$, i.e., -\begin{equation} - \langle A\rangle - =\frac1{\mathcal N}\sum_{\sigma}A(s_\sigma) - =\frac1{\mathcal N} - \int d\nu(s)\,A(s) -\end{equation} -with -\begin{equation} - d\nu(s)=ds\,\delta(NE-H(s))\delta(\partial H(s)+\mu s)|\det(\partial\partial H(s)+\mu I)| -\end{equation} -the Kac--Rice measure. The fields $C$, $R$, and $D$ defined in -\eqref{eq:fields} can be related to certain averages of this type. - -\subsection{\textit{C}: distribution of overlaps} - -First, -consider $C$, which has an interpretation nearly identical to that of Parisi's -$Q$ matrix of overlaps. It can be shown that its off-diagonal corresponds to -the probability distribution of the overlaps between stationary points $P(q)$. First, define this distribution as -\begin{equation} - P(q)=\frac1{\mathcal N^2}\sum_{\sigma,\sigma'}\delta\left(\frac{s_\sigma\cdot s_{\sigma'}}N-q\right) -\end{equation} -where the sum is twice over stationary points $\sigma$ and $\sigma'$. It is -straightforward to show that moments of this distribution are related to -certain averages of the form -\begin{equation} - \int dq\,q^p P(q) - =q^{(p)} - \equiv\frac1{N^p}\sum_{i_1\cdots i_p}\langle s_{i_1}\cdots s_{i_p}\rangle\langle s_{i_1}\cdots s_{i_p}\rangle -\end{equation} -The appeal of Parisi to properties of pure states is unnecessary here, since -the stationary points are points. These moments are related to our $C$ by -computing their average over disorder: -\begin{equation} - \begin{aligned} - \overline{q^{(p)}} - =\overline{\frac1{N^p}\sum_{i_1\cdots i_p}\langle s_{i_1}\cdots s_{i_p}\rangle\langle s_{i_1}\cdots s_{i_p}\rangle} - =\lim_{n\to0}\overline{\int\prod_\alpha^nd\nu(s_\alpha)\,\frac1{n(n-1)}\sum_{a\neq b}\left(\frac{s_a\cdot s_b}N\right)^p} \\ - =\lim_{n\to0}\frac1{n(n-1)}\sum_{a\neq b}^nC^p_{ab} - =\int_0^1 dx\,c^p(x) - \end{aligned} -\end{equation} -where $C$ is assumed to take its saddle point value. If we now change variables in the integral from $x$ to $c$, we find -\begin{equation} - \overline{q^{(p)}}=\int dc\,c^p\frac{dx}{dc} -\end{equation} -from which we conclude $\overline{P(q)}=\frac{dx}{dc}\big|_{c=q}$. - -With this -established, we now address what it means for $C$ to have a nontrivial -replica-symmetry broken structure. When $C$ is replica symmetric, drawing two -stationary points at random will always lead to the same overlap. In the case -when there is no linear field, they will always have overlap zero, because the -second point will almost certainly lie on the equator of the sphere with -respect to the first. Though other stationary points exist nearby the first -one, they are exponentially fewer and so will be picked with vanishing -probability in the thermodynamic limit. - -When $C$ is replica-symmetry broken, there is a nonzero probability of picking -a second stationary point at some other overlap. This can be interpreted by -imagining the level sets of the Hamiltonian in this scenario. If the level sets -are disconnected but there are exponentially many of them distributed on the -sphere, one will still find zero average overlap. However, if the disconnected -level sets are \emph{few}, i.e., less than order $N$, then it is possible to -draw two stationary points from the same set. Therefore, the picture in this -case is of few, large basins each containing exponentially many stationary -points. - -\subsection{\textit{R} and \textit{D}: response functions} - -The matrix field $R$ is related to responses of the stationary points to perturbations of the tensors $J$: -\begin{equation} - \begin{aligned} - \overline{\frac1{N^p}\sum_{i_1\cdots i_p}\frac{\partial\langle s_{i_1}\cdots s_{i_p}\rangle}{\partial J^{(p)}_{i_1\cdots i_p}}} - =\lim_{n\to0}\overline{\int\prod_\alpha^nd\nu(s_\alpha)\,\frac1n\sum_{ab}\left[ - \hat\beta\left(\frac{s_a\cdot s_b}N\right)^p+ - p\left(-i\frac{\hat s_a\cdot s_b}N\right)\left(\frac{s_a\cdot s_b}N\right)^{p-1} - \right]} \\ - =\lim_{n\to0}\frac1n\sum_{ab}(\hat\beta C_{ab}^p+pR_{ab}C_{ab}^{p-1}) - =\hat\beta+pr_d-\int_0^1dx\,c^{p-1}(x)(\hat\beta c(x)+pr(x)) - \end{aligned} -\end{equation} -In particular, when the energy is unconstrained ($\hat\beta=0$) and there is replica symmetry, -\begin{equation} - \frac1N\sum_i\frac{\partial\langle s_i\rangle}{\partial J_i^{(1)}} - =r_d -\end{equation} -i.e., adding a linear field causes a response in the average stationary point -location proportional to $r_d$. If positive, for instance, stationary points -tend to align with a field. The energy constraint has a significant -contribution due to the perturbation causing stationary points to move up or -down in energy. - -The matrix field $D$ is related to the response of the complexity to such perturbations: -\begin{equation} - \begin{aligned} - \frac{\partial\Sigma}{\partial a_p} - =\frac14\lim_{n\to0}\frac1n\sum_{ab}^n\left[ - \hat\beta^2C_{ab}^p+p(2\hat\beta R_{ab}-D_{ab})C_{ab}^{p-1}+p(p-1)R_{ab}^2C_{ab}^{p-2} - \right] - \end{aligned} -\end{equation} -In particular, when the energy is unconstrained ($\hat\beta=0$) and there is no replica symmetry breaking, -\begin{equation} - \frac{\partial\Sigma}{\partial a_1}=-\frac14\lim_{n\to0}\frac1n\sum_{ab}D_{ab}=-\frac14d_d -\end{equation} -i.e., adding a random linear field decreases the complexity of solutions by an amount proportional to $d_d$ in the variance of the field. - -When the saddle point of the Kac--Rice problem is supersymmetric, -\begin{equation} - \frac{\partial\Sigma}{\partial a_p} - =\frac{\hat\beta}4\overline{\frac1{N^p}\sum_{i_1\cdots i_p}\frac{\partial\langle s_{i_1}\cdots s_{i_p}\rangle}{\partial J^{(p)}_{i_1\cdots i_p}}}+\lim_{n\to0}\frac1n\sum_{ab}^np(p-1)R_{ab}^2C_{ab}^{p-2} -\end{equation} -and in particular for $p=1$ -\begin{equation} - \frac{\partial\Sigma}{a_1} - =\frac{\hat\beta}4\overline{\frac1N\sum_i\frac{\partial\langle s_i\rangle}{\partial J_i^{(1)}}} -\end{equation} -i.e., the change in complexity due to a linear field is directly related to the -resulting magnetization of the stationary points. - \section{Ultrametricity rediscovered} TENTATIVE BUT INTERESTING |