From b353ec93c1161e1619e068a8830f3db7ec5ba520 Mon Sep 17 00:00:00 2001 From: Jaron Kent-Dobias Date: Thu, 30 Jun 2022 20:19:30 +0200 Subject: Lots of interpretation writing. --- frsb_kac-rice.tex | 65 ++++++++++++++++++++++++++++++++++++++++++++++++------- 1 file changed, 57 insertions(+), 8 deletions(-) diff --git a/frsb_kac-rice.tex b/frsb_kac-rice.tex index 9ba027c..130738d 100644 --- a/frsb_kac-rice.tex +++ b/frsb_kac-rice.tex @@ -310,7 +310,7 @@ role of an inverse temperature for the metastable states. The average over disor \end{equation} We introduce new fields -\begin{align} +\begin{align} \label{eq:fields} C_{ab}=\frac1Ns_a\cdot s_b && R_{ab}=-i\frac1N\hat s_a\cdot s_b && D_{ab}=\frac1N\hat s_a\cdot\hat s_b @@ -637,7 +637,7 @@ for different energies and typical vs minima. \section{Interpretation} -Let $\langle A\rangle$ be average over stationary points with given $E$ and $\mu$, i.e., +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) @@ -648,16 +648,59 @@ 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} -Then +the Kac--Rice measure. The fields $C$, $R$, and $D$ defined in +\eqref{eq:fields} can be related to certain averages of this type. 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} +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} +In particular, 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{\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} + \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 $\overline{P(q)}=\frac{dx}{dc}|_{c=q}$. + +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}}} @@ -669,13 +712,18 @@ Then =\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 ($\hat\beta=0$), +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-\int_0^1dx\,r(x) + =r_d \end{equation} -i.e., adding a linear field causes a response in the average saddle location proportional to $r_d$. +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} @@ -700,7 +748,8 @@ and in particular for $p=1$ \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. +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} -- cgit v1.2.3-70-g09d2