From ea3953cf944cfec604e4a941d53f531c6e7323a6 Mon Sep 17 00:00:00 2001
From: "kurchan.jorge" <kurchan.jorge@gmail.com>
Date: Wed, 6 Jul 2022 09:40:32 +0000
Subject: Update on Overleaf.

---
 frsb_kac-rice.tex | 389 +++++++++++++++++++++++++++++-------------------------
 1 file changed, 208 insertions(+), 181 deletions(-)

(limited to 'frsb_kac-rice.tex')

diff --git a/frsb_kac-rice.tex b/frsb_kac-rice.tex
index bc6647e..fc2d911 100644
--- a/frsb_kac-rice.tex
+++ b/frsb_kac-rice.tex
@@ -177,23 +177,23 @@ The stationary points of a function can be counted using the Kac--Rice formula,
 which integrates a over the function's domain a $\delta$-function containing
 the gradient multiplied by the absolute value of the determinant
 \cite{Rice_1939_The, Kac_1943_On}. In addition, we insert a $\delta$-function
-fixing the energy density $E$, giving the number of stationary points at
-energy $E$ in terms of a Lagrange multiplier $\mu$ as:
+fixing the energy density $E$, and another delta function to count the number of 
+saddles with trace of the Hessian $=\mu^*$. The latter will give us everything we need
+to characterize the saddles, as we shall see later
 \begin{equation}
-  \mathcal N(E, {\cal I}_o)
-  =\int ds d\mu\,\delta(NE-H(s))\delta(\partial H(s)+\mu s)|\det(\partial\partial H(s)+\mu I)| \Theta\left[{\cal I}(s,\mu)-{\cal I}_o)\right]
+  \mathcal N(E, \mu^*)
+  =\int ds d\mu\,\delta(NE-H(s))\delta\big(\partial H(s)+\mu s\big)\Big|\det(\partial\partial H(s)+\mu I) \Big| \delta\Big({\mbox{Tr}} [\partial\partial H(s)+\mu I)-N\mu^*] \Big)
 \end{equation}
-where $\Theta$ is one in the region in which the Hessian the argument is vanishes, and zero elsewhere. 
 This number will typically be exponential in $N$. In order to find typical
 counts when disorder is averaged, we will want to average its logarithm
 instead, which is known as the averaged complexity:
 \begin{equation}
-  \Sigma(E,\mu)=\lim_{N\to\infty}\frac1N\overline{\log\mathcal N(E, {\cal I}_o)}
+  \Sigma(E,\mu^*)=\lim_{N\to\infty}\frac1N\overline{\log\mathcal N(E, \mu^*})
 \end{equation}
 If one averages over $\mathcal N$ and afterward takes its logarithm, one arrives at the so-called {\em annealed} complexity
 \begin{equation}
-  \Sigma_\mathrm a(E,{\cal I}_o)
-  =\lim_{N\to\infty}\frac1N\log\overline{\mathcal N(E,{\cal I}_o))}
+  \Sigma_\mathrm a(E,\mu^*)
+  =\lim_{N\to\infty}\frac1N\log\overline{\mathcal N(E,\mu^*)}
 \end{equation}
 This has been previously computed for the mixed $p$-spin models \cite{BenArous_2019_Geometry}, with the result
 %\begin{equation}
@@ -215,7 +215,7 @@ a contradiction for any bounded function, as seen in Fig.~\ref{fig:frsb.complexi
 
 A sometimes more illuminating quantity to consider is the Legendre transform $G$ of the complexity:
 \begin{equation}
-    e^{NG(\hat \beta, {\cal I}_o)} = \int dE \; e^{ -\hat \beta E +\Sigma(\hat \beta, {\cal I}_o)}
+    e^{NG(\hat \beta, \mu^*)} = \int dE \; e^{ -\hat \beta E +\Sigma(\hat \beta, \mu^*)}
 \end{equation}
 There will be a critical value $\hat \beta_c$ beyond  which the complexity is zero: above
 this value the measure is split between the lowest $O(1)$ energy states. We shall not study here this  regime that interpolates between the dynamically relevant and the equilibrium states, but just mention that
@@ -245,16 +245,17 @@ Folena et al.~\cite{Folena_2020_Rethinking}. Here we review the derivation.
 therefore able to write
 \begin{equation}
   \begin{aligned}
-    \Sigma(E, {\cal I}_o)
-    &=\lim_{N\to\infty}\frac1N\lim_{n\to0}\frac\partial{\partial n}\int\left(\prod_a^nds_a\right)\,\overline{\prod_a^n \delta(NE-H(s_a))\delta(\partial H(s_a)+\mu s_a)}
+    \Sigma(E, \mu^*)
+    &=\lim_{N\to\infty}\frac1N\lim_{n\to0}\frac\partial{\partial n}\int\left(\prod_a^nds_a\right)\,\overline{\prod_a^n \delta(NE-H(s_a))\delta(\partial H(s_a)+\mu s_a)}\nonumber\\
+    & 
     \times
-    \overline{\prod_a^n |\det(\partial\partial H(s_a)+\mu I)| \Theta\left[{\cal I}(s_a,\mu)-{\cal I}_o)\right]}
+    \overline{\prod_a^n |\det(\partial\partial H(s_a)+\mu I)|\delta({\mbox{Tr}} [\partial\partial H(s_a)+\mu I)-N\mu^*] )}
   \end{aligned}
 \end{equation}
 
 \subsubsection{The Hessian factors}
 
-The spectrum of the Hessian matrix $\partial\partial H$ is in the large $N$ limit 
+The spectrum of the Hessian matrix $\partial\partial H$ is uncorrelated from the gradient. In the large $N$ limit 
 for almost every point and realization of disorder a GOE matrix with variance
 \begin{equation}
   \overline{(\partial_i\partial_jH)^2}=\frac1Nf''(1)\delta_{ij}
@@ -283,7 +284,9 @@ The factor $\Theta[({\cal I}(s_a,\mu)-{\cal I}_0]$ selects a domain of integrati
 To largest order in $N$, the average over the product of determinants factorizes into the product of averages, each of which is given by the same expression depending only on $\mu$:
 \begin{equation}
   \begin{aligned}
-  & \overline{\prod_a^n |\det(\partial\partial H(s_a)+\mu I)| \Theta\left[{\cal I}(s_a,\mu)-{\cal I}_o)\right]}\\
+  & \prod_a^n \overline{|\det(\partial\partial H(s_a)+\mu I)| 
+  \delta({\mbox{Tr}} [\partial\partial H(s_a)+\mu I)-N\mu^*] )
+  } \rightarrow e^{n{\cal D}(\mu^*)}\delta(\mu-\mu^*)  \quad {\mbox{with}}\\
     \mathcal D(\mu)
     &=\frac1N\overline{\log|\det(\partial\partial H(s_a)+\mu I)|}
     =\int d\lambda\,\rho(\lambda+\mu)\log|\lambda| \\
@@ -294,6 +297,11 @@ To largest order in $N$, the average over the product of determinants factorizes
   \end{aligned}
 \end{equation}
 
+
+What we have described is the {\em typical} spectrum for given $\mu$. What about the deviations of the spectrum -- we are particularly interested in the number of negative eigenvalues -- at given $\mu$. The result is well known qualitatively: there are two possibilities:\\
+$\bullet$ For $|\mu|>\mu_m$ there is the possibility of a finite number of eigenvalues of
+the second derivative matrix 
+
 \subsubsection{The gradient factors}
 
 The $\delta$-functions are treated by writing them in the Fourier basis, introducing auxiliary fields $\hat s_a$ and $\hat\beta$,
@@ -363,23 +371,48 @@ where $\hat\beta$, $C$, $R$ and $D$ must be evaluated at extrema of this
 expression.
 
 
-{\color{blue}
 
 
 
-The complexity is defined as 
+
+The same information is contained, and better expressed in its 
+Legendre
+transform
+\begin{equation}
+  \begin{aligned}
+    &G(\hat \beta,\mu)
+    =\mathcal D(\mu)+\\
+    &\lim_{n\to0}\frac1n\left(
+      -\mu\operatorname{Tr}R
+      +\frac12\sum_{ab}\left[
+        \hat\beta^2f(C_{ab})+(2\hat\beta R_{ab}-D_{ab})f'(C_{ab})
+        +R_{ab}^2f''(C_{ab})
+      \right]
+    +\frac12\log\det\begin{bmatrix}C&iR\\iR&D\end{bmatrix}
+    \right)
+  \end{aligned}
+\end{equation}
+Denoting $R_d \equiv \frac 1 n {\mbox Tr} R$, we have the double Legendre transform $K(\hat \beta, R_d)$:
 \begin{equation}
-   \Sigma(E,\mu)=  \frac 1N \log\mathcal N(E,\mu)
+   e^{N K(\hat \beta, R_d)} =\int \; d\mu de \; e^{N\Sigma(E,\mu)+R_d\mu -\hat \beta E +{\cal{D}}(\mu)}
 \end{equation}
-The same information is contained, and better expressed in its double Legendre
-transform $J(\hat \beta, R_d)$:
+given by
 \begin{equation}
-   e^{N J(\hat \beta, R_d)} =\int \; d\mu de \; e^{N\Sigma(E,\mu)+R_d\mu -\hat \beta E +{\cal{D}}(\mu)}
+  \begin{aligned}
+    &K(\hat \beta,R_d)
+   = \lim_{n\to0}\frac1n\left(
+      +\frac12\sum_{ab}\left[
+        \hat\beta^2f(C_{ab})+(2\hat\beta R_{ab}-D_{ab})f'(C_{ab})
+        +R_{ab}^2f''(C_{ab})
+      \right]
+    +\frac12\log\det\begin{bmatrix}C&iR\\iR&D\end{bmatrix}
+    \right)
+  \end{aligned}
 \end{equation}
-$R_d$ is conjugate to $\mu$ and through it to the Index density, while $\hat \beta$ plays the role of an inverse temperature conjugate to the complexity, that has been used since the beginning of the spin-glass field.
+$R_d$ is conjugate to $\mu$ and through it to the Index density, while $\hat \beta$ plays the role of an inverse temperature conjugate to the complexity, that has been used since the beginning of the spin-glass field. In this way $K(R_d,\hat \beta)$ contains all the information about saddle densities.
+
 
 
-}
 
 
 
@@ -426,131 +459,6 @@ 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}
 
@@ -676,7 +584,7 @@ This has several implications. First, other than the ground state, there are
 As we will see, stable minima are numerous at energies above the ground state,
 but these vanish at the ground state.
 
-\section{Examples}
+\section{General solution: examples}
 
 \subsection{1RSB complexity}
 
@@ -765,30 +673,6 @@ E\rangle_2$.
 
 
 
-\begin{figure}
-  \begin{center}
-    \includegraphics[width=12cm]{twosys.png}\end{center}
-  \caption{ Weakly coupling two different pure p-spin models we may infer the  phases for the joint model. A sketch of the individual complexities versus energy of each model,
-  the dotted part of the lines corresponds to saddles, and starts at the threshold level.
-  (1) Both systems frozen, lowest energy density.
-(2) One system unfreezes, $1RSB$ Kac-Rice computation. (3) Both systems unfreeze, the gradients are the same in the typical (maximal complexity) values, thus corresponding to the RS Kac-Rice phase.
-The beginning of this phase may start when the system on the right is either
-on the region with minima or with saddles, depending on the whether the gradient of the complexity in the threshold of the system is higher or lower to the one of the ground state of the other.
-(4) and (5) Typical states are saddles, both gradients are the same and hence this is RS. (6) The highest possible energy for minima. Because the gradients at the thresholds of the two systems do not coincide,
-this has subdominant complexity
-  } \label{twosys}
-\end{figure}
-
-
-\begin{figure}
-  \begin{center}
-    \includegraphics[width=12cm]{phase.png}\end{center}
-  \caption{The phase diagram corresponding to two coupled models.
-  } \label{phase}
-\end{figure}
-
-
-
 
 
 
@@ -857,18 +741,161 @@ this has subdominant complexity
 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{Ultrametricity rediscovered}
 
-TENTATIVE BUT INTERESTING
+\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. These are evaluated for a given energy, index, etc, but
+we shall omit these subindices for simplicity.
+
+\begin{eqnarray}
+q^{(p)}
+  &\equiv& \frac1{N^p}\sum_{i_1\cdots i_p}\langle s^1_{i_1}\cdots s^1_{i_p}\rangle\langle s^2_{i_1}\cdots s^2_{i_p}\rangle
+  \nonumber \\
+  &=&\frac1{N^p} \; \frac{1}{{\cal{N}}^2} \left\{ \sum_{{\mathbf s}^1,{\mathbf s}^2}\; \sum_{i_1\cdots i_p} s^1_{i_1}\cdots s^1_{i_p} s^2_{i_1}\cdots s^2_{i_p}\right\}
+  =\frac1{N^p}  \lim_{n\to0} \left\{ \sum_{{\mathbf s}^1,{\mathbf s}^2...{\mathbf s}^n}\; \sum_{i_1\cdots i_p} s^1_{i_1}\cdots s^1_{i_p} s^2_{i_1}\cdots s^2_{i_p}\right\}
+\end{eqnarray}
+The $(n-2)$ extra replicas providing the normalization.
+The average over disorder, and again, for given enegrgy, index,etc reads:
+\begin{eqnarray}
+    \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}{\int\prod_\alpha^n d\nu(s_\alpha)\,\left(\frac{s_1 s_2}N\right)^p} \times  \Big[ {\mbox {REPLICATED AVERAGED MEASURE}}  \Big]\nonumber \\
+    &=&\lim_{n\to0}{\int D[C,R,D] \,
+    \left(C_{12}\right)^p\; } e^{N\Sigma[C,R,D]}
+    =\frac1{n(n-1)}\lim_{n\to0}{\int D[C,R,D] \,\sum_{a\neq b}\left(C_{ab}\right)^p\; } e^{N\Sigma[C,R,D]}
+\end{eqnarray}
+In the last line, we have used that there is nothing special about replicas one and two.
+Using the Parisi ansatz, evaluating by saddle point  {\em summing over all the $n(n-1)$ saddles related by permutation} we then have
+\begin{equation}
+ \overline{q^{(p)}}=\int_0^1 dx\,c^p(x) = \int dq\,q^p P(q) \qquad ; \qquad {\mbox{with}} \qquad P(q)=\frac{dx}{dq}
+\end{equation}
+The appeal of Parisi to properties of pure states is unnecessary here, since
+the stationary points are points.
+
+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{A concrete example}
+
+Consider two independent pure $p$ spin models $H_{p_1}({\mathbf s})$ and $H_{p_2}({\mathbf \sigma})$ of sizes  $N$, and couple them weakly with $\varepsilon \;
+{\mathbf \sigma} \cdot {\mathbf s}$.
+The complexities are
+\begin{eqnarray}
+    e^{N\Sigma(e)}&=&\int de_1 de_2 \; e^{N[ \Sigma_1(e_1) + \Sigma_2(e_2) + O(\varepsilon) -\lambda N [(e_1+e_2)-e]}\nonumber \\
+    e^{-G(\hat \beta)}&=&\int de de_1 de_2 \; e^{N[-\hat \beta e+ \Sigma_1(e_1) + \Sigma_2(e_2) + O(\varepsilon) -\lambda N [(e_1+e_2)-e]}
+\end{eqnarray}
+The maximum is given by $\Sigma_1'=\Sigma_2'=\hat \beta$, provided it occurs in the phase
+in which both $\Sigma_1$ and $\Sigma_2$ are non-zero. The two systems are `thermalized',
+and it is easy to see that, because many points contribute, the overlap between two 
+global configurations $$\frac1 {2N}({\mathbf s^1},{\mathbf \sigma^1})\cdot ({\mathbf s^2},{\mathbf \sigma^2})=0$$ This is the `annealed' phase of a Kac-Rice calculation.
+
+Now start going down in energy, or up in $\hat \beta$: there will be a point $e_c$, $\hat \beta_c$
+at which one of the subsystems freezes at its lower energy density, say it is system one,
+while system two is not yet frozen. At an even higher $\hat \beta=\hat \beta_f$, both systems are frozen. 
+For $\hat \beta_f > \hat \beta> \hat \beta_c$ one system is unfrozen, while the other is,
+because of coupling, frozen at inverse temperature $\hat \beta_c$.
+The overlap between two solutions  is:
+$$\frac1 {2N}({\mathbf s^1},{\mathbf \sigma^1})\cdot ({\mathbf s^2},{\mathbf \sigma^2}) =\frac1 {2N}[ {\mathbf s^1}\cdot {\mathbf s^2}+ {\mathbf \sigma^1}\cdot {\mathbf \sigma^2}] = \frac1 {2N} {\mathbf s^1}\cdot {\mathbf s^2}   $$
+The distribution of this overlap is the one of a frozen spin-glass at temperature $\hat \beta$,  a $1RSB$ system like the Random Energy Model. The value of $x$
+corresponding to it depends on $\hat \beta$, starting at $x=1$ at $\hat \beta_c$ and decreasing wuth increasing $\hat \beta$. 
+Globally, the joint Kac-Rice system is  $1RSB$, but note that the global overlap between different states is at most $1/2$. 
+At $\hat \beta>\hat \beta_f$ there is a further transition.
+
+\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.
 
-The frozen phase for a given index ${\cal{I}}$ is the one for values of $\hat \beta> \hat \beta_{freeze}^{\cal{I}}$. 
+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.
 
-[Jaron: does $\hat \beta^I_{freeze}$ have a relation to the largest $x$  of the ansatz? If so, it would give an interesting interpretation for everything]
- 
- The complexity of that index is zero, and we are looking at the lowest saddles 
-in the problem, a question that to the best of our knowledge has not been discussed
-in the Kac--Rice context -- for good reason, since the complexity - the original motivation - is zero.
-However, our ansatz tells us something of the actual organization of  the lowest saddles of each index in phase space. 
 
 
 
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