Lots of interpretation writing.

1 files 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}