Moved interpretation section up.

1 files changed, 126 insertions, 125 deletions

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