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-rw-r--r--2-point.tex202
1 files changed, 128 insertions, 74 deletions
diff --git a/2-point.tex b/2-point.tex
index d39f12c..6d3c437 100644
--- a/2-point.tex
+++ b/2-point.tex
@@ -107,12 +107,61 @@
\section{Introduction}
+Many systems exhibit ``glassiness,'' characterized by rapid slowing of dynamics
+over a short parameter interval. These include actual (structural) glasses,
+spin glasses, certain inference and optimization problems, and more
+\cite{lots}. Glassiness is qualitatively understood to arise from structure of
+an energy or cost landscape, whether due to the proliferation of metastable
+states, to the raising of barriers which cause effective dynamic constraints.
+However, in most models there is no quantitative correspondence between these
+properties and the behavior.
+
+In perhaps the simplest mean-field model of glasses, such a correspondence
+exists and is well understood. In the pure spherical models, the dynamic
+transition corresponds precisely with the energy level at which all marginal
+minima are concentrated. At that level, called the \emph{threshold energy}
+$E_\mathrm{th}$, slices of the landscape at fixed energy undergo a percolation
+transition. In fact, this threshold energy is significant in other ways: it
+attracts the long-time dynamics after quenches in temperature to below the
+dynamical transition from any starting temperature. All of this can be understood in terms of the landscape structure.
\cite{Biroli_1999_Dynamical}
-\cite{Folena_2020_Rethinking, Folena_2021_Gradient}
-\cite{Ros_2020_Distribution, Ros_2019_Complex, Ros_2019_Complexity}
+In slightly less simple models, the mixed spherical models, the story changes.
+There are now a range of energies with exponentially many marginal minima. It
+was believed that the energy level at which these marginal minima are the most
+common type of stationary point would play the same role as the threshold
+energy in the pure models. However, recent work has shown that this is
+incorrect. Quenches from different starting temperatures above the dynamical
+transition temperature result in dynamics that approach different energy
+levels, and the purported threshold does not attract the long-time dynamics in
+most cases \cite{Folena_2020_Rethinking, Folena_2021_Gradient}.
+
+This paper studies the two-point structure of stationary points in the mixed
+spherical models, or their arrangement relative to each other, previously
+studied only for the pure models \cite{Ros_2019_Complexity}. This gives various
+kinds of information. When one point is a minimum, we see what other kinds of
+minima are nearby, and what kind of saddle points (barriers) separate them.
+When both points are saddles, we see the arrangement of barriers relative to
+each other, perhaps learning something about the geometry of the basins of
+attraction that they surround.
+
+In order to address the open problem of what attracts the long-time dynamics,
+we focus on the neighborhoods of the marginal minima, to see if there is
+anything interesting to differentiate sets of them from each other. Though we
+find rich structure in this population, their properties pivot around the
+debunked threshold energy, and the apparent attractors of long-time dynamics
+are not distinguished by this measure. Therefore, with respect to the problem
+of dynamics this paper merely deepens the outstanding problems.
+
+In \S\ref{sec:model}, we introduce the mixed spherical models and discuss their
+properties. In \S\ref{sec:results}, we share the main results of the paper. In
+\S\ref{sec:complexity} we detail the calculation of the two-point complexity,
+and in \S\ref{sec:eigenvalue} and \S\ref{sec:franz-parisi} we do the same for
+the properties of the isolated eigenvalue and for the zero-temperature
+Franz--Parisi potential.
\section{Model}
+\label{sec:model}
\cite{Crisanti_1992_The, Crisanti_1993_The}
@@ -252,6 +301,7 @@ and a $3+8$ model tuned to maximize the ``interesting'' region of the dynamics r
\end{figure}
\section{Complexity}
+\label{sec:complexity}
We introduce the Kac--Rice \cite{Kac_1943_On, Rice_1944_Mathematical} measure
\begin{equation}
@@ -618,80 +668,9 @@ Because the matrices $C^{00}$, $R^{00}$, and $D^{00}$ are diagonal in this case,
E_1=E_0+\frac12\frac{f'(1)(f'''(1)+f''(1))-f''(1)^2}{f(1)(f'(1)+f''(1))-f'(1)^2}(E_0-E_0^*)(1-q)^2+O\big((1-q)^3\big)
\end{equation}
-\section{Franz--Parisi potential}
-
-\cite{Franz_1995_Recipes}
-
-\begin{equation} \label{eq:franz-parisi.definition}
- \beta V_\beta(q\mid E_0,\mu_0)
- =-\frac1N\overline{\int\frac{d\nu_H(\pmb\sigma,\varsigma\mid E_0,\mu_0)}{\int d\nu_H(\pmb\sigma',\varsigma'\mid E_0,\mu_0)}\,
- \log\bigg(\int d\mathbf s\,\delta\big(\|\mathbf s\|^2-N\big)\,\delta(\pmb\sigma\cdot\mathbf s-Nq)\,e^{-\beta H(\mathbf s)}\bigg)}
-\end{equation}
-Both the denominator and the logarithm are treated using the replica trick, which yields
-\begin{equation}
- \beta V_\beta(q\mid E_0,\mu_0)
- =-\frac1N\lim_{\substack{m\to0\\n\to0}}\frac\partial{\partial n}\overline{\int\left(\prod_{b=1}^md\nu_H(\pmb\sigma_b,\varsigma_b\mid E_0,\mu_0)\right)\left(\prod_{a=1}^nd\mathbf s_a\,\delta(\|\mathbf s_a\|^2-N)\,\delta(\pmb \sigma_1\cdot \mathbf s_a-Nq)\,e^{-\beta H(\mathbf s_a)}\right)}
-\end{equation}
-\begin{equation}
- \mathcal O(\mathbf t)
- =\sum_a^m\delta(\mathbf t-\pmb\sigma_a)\left(
- i\hat{\pmb\sigma}_a\cdot\partial_{\mathbf t}-\hat\beta_0
- \right)
- -\beta
- \sum_a^n\delta(\mathbf t-\mathbf s_a)
-\end{equation}
-\begin{equation}
- \beta V_\beta(q\mid E_0,\mu_0)=-\frac1N\lim_{\substack{m\to0\\n\to0}}\frac\partial{\partial n}\int d\mathcal Q_0\,d\mathcal Q_1\,e^{Nm\mathcal S_0(\mathcal Q_0)+Nn\mathcal S_\mathrm{FP}(\mathcal Q_1)}
-\end{equation}
-
-\begin{equation}
- n\mathcal S_{\mathrm{FP}}
- =\frac12\beta^2\sum_{ab}^nf(Q_{ab})
- +\beta\sum_a^m\sum_b^n\left[
- \hat\beta_0f(C^{01}_{ab})
- +R^{10}_{ab}f'(C^{01}_{ab})
- \right]
- +\frac12\log\det\left(
- Q-\begin{bmatrix}C^{01}\\iR^{10}\end{bmatrix}^T\begin{bmatrix}C^{00}&iR^{00}\\iR^{00}&D^{00}\end{bmatrix}^{-1}\begin{bmatrix}C^{01}\\iR^{10}\end{bmatrix}
- \right)
-\end{equation}
-
-\begin{equation}
- \begin{aligned}
- \beta V_\beta&=\frac12\beta^2\big[f(1)-(1-x)f(q_1)-xf(q_0)\big]
- +\beta\hat\beta_0f(q)+\beta r^{10}f'(q)-\frac{1-x}x\log(1-q_1)
- +\frac1x\log(1-(1-x)q_1-xq_0) \\
- &+\frac{q_0-d^{00}_df'(1)q^2-2r^{00}_df'(1)r^{10}q+(r^{10})^2f'(1)}{
- 1-(1-x)q_1-xq_0
- }
- \end{aligned}
-\end{equation}
-The saddle point for $r^{10}$ can be taken explicitly. After this, we take the
-limit of $\beta\to\infty$. There are two possibilities. First, in the replica
-symmetric case $x=1$, and in the limit of large $\beta$ $q_0$ will scale like
-$q_0=1-(y_0\beta)^{-1}$. Inserting this, the limit is
-\begin{equation}
- V_\infty^{\textsc{rs}}=-\hat\beta_0 f(q)-r^{11}_\mathrm df'(q)q-\frac12\left(y_0(1-q^2)+\frac{f'(1)^2-f'(q)^2}{y_0f'(1)}\right)
-\end{equation}
-The saddle point in $y_0$ can now be taken, taking care to choose the solution for $y_0>0$. This gives
-\begin{equation}
- V_\infty^{\textsc{rs}}=-\hat\beta_0f(q)-r^{11}_\mathrm df'(q)q-\sqrt{(1-q^2)\left(1-\frac{f'(q)^2}{f'(1)^2}\right)}
-\end{equation}
-The second case is when the inner statistical mechanics problem has replica symmetry breaking. Here, $q_0$ approaches a nontrivial limit, but $x=z\beta^{-1}$ approaches zero and $q_1=1-(y_1\beta)^{-1}$ approaches one.
-\begin{equation}
- \begin{aligned}
- V_\infty^{\oldstylenums{1}\textsc{rsb}}(q\mid E_0,\mu_0)
- &=-\hat\beta_0f(q)-r^{11}_\mathrm df'(q)q-\frac12\bigg(
- z(f(1)-f(q_0))+\frac{f'(1)}{y_1}-\frac{y_1(q^2-q_0)}{1+y_1z(1-q_0)} \\
- &\hspace{8pc}-(1+y_1z(1-q_0))\frac{f'(q)^2}{y_1f'(1)}+\frac1z\log\left(1+zy_1(1-q_0)\right)
- \bigg)
- \end{aligned}
-\end{equation}
-Though the saddle point in $y_1$ can be evaluated in this expression, it
-delivers no insight. The final potential is found by taking the saddle over
-$z$, $y_1$, and $q_0$.
\section{Isolated eigenvalue}
+\label{sec:eigenvalue}
The two-point complexity depends on the spectrum at both stationary points
through the determinant of their Hessians, but only on the bulk of the
@@ -1232,7 +1211,82 @@ solution to make sense we must have $y^2>f''(1)$. In practice, there is at most
\emph{one} $y$ which produces a zero eigenvalue of $B-yC$ and satisfies this
inequality, so the solution seems to be unique.
+\section{Franz--Parisi potential}
+\label{sec:franz-parisi}
+
+\cite{Franz_1995_Recipes}
+
+\begin{equation} \label{eq:franz-parisi.definition}
+ \beta V_\beta(q\mid E_0,\mu_0)
+ =-\frac1N\overline{\int\frac{d\nu_H(\pmb\sigma,\varsigma\mid E_0,\mu_0)}{\int d\nu_H(\pmb\sigma',\varsigma'\mid E_0,\mu_0)}\,
+ \log\bigg(\int d\mathbf s\,\delta\big(\|\mathbf s\|^2-N\big)\,\delta(\pmb\sigma\cdot\mathbf s-Nq)\,e^{-\beta H(\mathbf s)}\bigg)}
+\end{equation}
+Both the denominator and the logarithm are treated using the replica trick, which yields
+\begin{equation}
+ \beta V_\beta(q\mid E_0,\mu_0)
+ =-\frac1N\lim_{\substack{m\to0\\n\to0}}\frac\partial{\partial n}\overline{\int\left(\prod_{b=1}^md\nu_H(\pmb\sigma_b,\varsigma_b\mid E_0,\mu_0)\right)\left(\prod_{a=1}^nd\mathbf s_a\,\delta(\|\mathbf s_a\|^2-N)\,\delta(\pmb \sigma_1\cdot \mathbf s_a-Nq)\,e^{-\beta H(\mathbf s_a)}\right)}
+\end{equation}
+\begin{equation}
+ \mathcal O(\mathbf t)
+ =\sum_a^m\delta(\mathbf t-\pmb\sigma_a)\left(
+ i\hat{\pmb\sigma}_a\cdot\partial_{\mathbf t}-\hat\beta_0
+ \right)
+ -\beta
+ \sum_a^n\delta(\mathbf t-\mathbf s_a)
+\end{equation}
+\begin{equation}
+ \beta V_\beta(q\mid E_0,\mu_0)=-\frac1N\lim_{\substack{m\to0\\n\to0}}\frac\partial{\partial n}\int d\mathcal Q_0\,d\mathcal Q_1\,e^{Nm\mathcal S_0(\mathcal Q_0)+Nn\mathcal S_\mathrm{FP}(\mathcal Q_1)}
+\end{equation}
+
+\begin{equation}
+ n\mathcal S_{\mathrm{FP}}
+ =\frac12\beta^2\sum_{ab}^nf(Q_{ab})
+ +\beta\sum_a^m\sum_b^n\left[
+ \hat\beta_0f(C^{01}_{ab})
+ +R^{10}_{ab}f'(C^{01}_{ab})
+ \right]
+ +\frac12\log\det\left(
+ Q-\begin{bmatrix}C^{01}\\iR^{10}\end{bmatrix}^T\begin{bmatrix}C^{00}&iR^{00}\\iR^{00}&D^{00}\end{bmatrix}^{-1}\begin{bmatrix}C^{01}\\iR^{10}\end{bmatrix}
+ \right)
+\end{equation}
+
+\begin{equation}
+ \begin{aligned}
+ \beta V_\beta&=\frac12\beta^2\big[f(1)-(1-x)f(q_1)-xf(q_0)\big]
+ +\beta\hat\beta_0f(q)+\beta r^{10}f'(q)-\frac{1-x}x\log(1-q_1)
+ +\frac1x\log(1-(1-x)q_1-xq_0) \\
+ &+\frac{q_0-d^{00}_df'(1)q^2-2r^{00}_df'(1)r^{10}q+(r^{10})^2f'(1)}{
+ 1-(1-x)q_1-xq_0
+ }
+ \end{aligned}
+\end{equation}
+The saddle point for $r^{10}$ can be taken explicitly. After this, we take the
+limit of $\beta\to\infty$. There are two possibilities. First, in the replica
+symmetric case $x=1$, and in the limit of large $\beta$ $q_0$ will scale like
+$q_0=1-(y_0\beta)^{-1}$. Inserting this, the limit is
+\begin{equation}
+ V_\infty^{\textsc{rs}}=-\hat\beta_0 f(q)-r^{11}_\mathrm df'(q)q-\frac12\left(y_0(1-q^2)+\frac{f'(1)^2-f'(q)^2}{y_0f'(1)}\right)
+\end{equation}
+The saddle point in $y_0$ can now be taken, taking care to choose the solution for $y_0>0$. This gives
+\begin{equation}
+ V_\infty^{\textsc{rs}}=-\hat\beta_0f(q)-r^{11}_\mathrm df'(q)q-\sqrt{(1-q^2)\left(1-\frac{f'(q)^2}{f'(1)^2}\right)}
+\end{equation}
+The second case is when the inner statistical mechanics problem has replica symmetry breaking. Here, $q_0$ approaches a nontrivial limit, but $x=z\beta^{-1}$ approaches zero and $q_1=1-(y_1\beta)^{-1}$ approaches one.
+\begin{equation}
+ \begin{aligned}
+ V_\infty^{\oldstylenums{1}\textsc{rsb}}(q\mid E_0,\mu_0)
+ &=-\hat\beta_0f(q)-r^{11}_\mathrm df'(q)q-\frac12\bigg(
+ z(f(1)-f(q_0))+\frac{f'(1)}{y_1}-\frac{y_1(q^2-q_0)}{1+y_1z(1-q_0)} \\
+ &\hspace{8pc}-(1+y_1z(1-q_0))\frac{f'(q)^2}{y_1f'(1)}+\frac1z\log\left(1+zy_1(1-q_0)\right)
+ \bigg)
+ \end{aligned}
+\end{equation}
+Though the saddle point in $y_1$ can be evaluated in this expression, it
+delivers no insight. The final potential is found by taking the saddle over
+$z$, $y_1$, and $q_0$.
+
\section{Conclusion}
+\label{sec:conclusion}
The methods developed in this paper are straightforwardly (if not easily)
generalized to landscapes with replica symmetry broken complexities.