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1 files changed, 667 insertions, 376 deletions

diff --git a/frsb_kac-rice.tex b/frsb_kac-rice.tex
index e02cab8..90bf081 100644
--- a/frsb_kac-rice.tex
+++ b/frsb_kac-rice.tex
@@ -1,6 +1,7 @@
 \documentclass[fleqn]{article}
 
 \usepackage{fullpage,amsmath,amssymb,latexsym,graphicx}
+\usepackage{appendix}
 
 \begin{document}
 \title{Full solution of the Kac--Rice problem for mean-field models.\\
@@ -9,27 +10,24 @@ or Full solution for the counting of saddles of mean-field glass models}
 \maketitle
 \begin{abstract}
     We derive the general solution for the computation of saddle points
-    of  mean-field complex landscapes. The solution incorporates Parisi's solution
-    for equilibrium, as it should.
+    of  mean-field complex landscapes. The solution incorporates Parisi's solution for the ground state, as it should.
 \end{abstract}
 \section{Introduction}
 
 The computation of the number of metastable states of mean field spin glasses
 goes back to the beginning of the field. Over forty years ago,
 Bray and Moore \cite{Bray_1980_Metastable} attempted the first calculation for 
- the Sherrington--Kirkpatrick model, a paper remarkable for being the first practical application of a replica symmetry breaking scheme. As became clear when the actual
- ground-state of the model was computed by Parisi \cite{Parisi_1979_Infinite}, the Bray--Moore result
- was not exact, and
-in fact  the problem has been open 
-ever since.
-Indeed, to this date the program of computing the number of saddles of a mean-field 
+ the Sherrington--Kirkpatrick model, in a paper remarkable for being one of the first applications of a replica symmetry breaking scheme. As became clear when the actual ground-state of the model was computed by Parisi \cite{Parisi_1979_Infinite} with a different scheme, the Bray--Moore result
+ was not exact, and in fact  the problem has been open ever since.
+To this date the program of computing the number of saddles of a mean-field 
 glass has been only  carried out for a small subset of models.
-These include most notably the $p$-spin model ($p>2$) \cite{Rieger_1992_The, Crisanti_1995_Thouless-Anderson-Palmer}. 
+These include most notably the (pure) $p$-spin model ($p>2$) \cite{Rieger_1992_The, Crisanti_1995_Thouless-Anderson-Palmer}. 
 The problem of studying the critical points of these landscapes
 has evolved into an active field in probability theory \cite{Auffinger_2012_Random, Auffinger_2013_Complexity, BenArous_2019_Geometry}
 
 In this paper we present what we argue is the general replica ansatz for the 
-computation of the number of saddles of generic mean-field models,  including the Sherrington--Kirkpatrick model. It incorporates  the Parisi solution as the limit of lowest states, as it should.
+computation of the number of saddles of generic mean-field models,  including the Sherrington--Kirkpatrick model. It reproduces  the Parisi result in the limit
+of small temperature for the lowest states, as it should.
 
 
 \section{The model}
@@ -50,12 +48,12 @@ for
 Can be thought of as a model of generic gaussian functions on the sphere.
 To constrain the model to the sphere, we use a Lagrange multiplier $\mu$, with the total energy being
 \begin{equation}
-  H(s)+\frac\mu2(N-s\cdot s)
+  H(s)+\frac\mu2(s\cdot s-N)
 \end{equation}
 
 At any critical point, the hessian is
 \begin{equation}
-  \operatorname{Hess}H=\partial\partial H-\mu I
+  \operatorname{Hess}H=\partial\partial H+\mu I
 \end{equation}
 $\partial\partial H$ is a GOE matrix with variance
 \begin{equation}
@@ -63,16 +61,25 @@ $\partial\partial H$ is a GOE matrix with variance
 \end{equation}
 and therefore its spectrum is given by the Wigner semicircle with radius $\sqrt{4f''(1)}$, or
 \begin{equation}
-  \rho(\lambda)=\frac1{\pi\sqrt{f''(1)}}\sqrt{\lambda^2-4f''(1)}
+  \rho(\lambda)=\frac1{\pi\sqrt{f''(1)}}\sqrt{4f''(1)-\lambda^2}
 \end{equation}
-and the spectrum of $\operatorname{Hess}H$ is this shifted by $\mu$, or $\rho(\lambda-\mu)$.
+and the spectrum of $\operatorname{Hess}H$ is this shifted by $\mu$, or $\rho(\lambda+\mu)$.
+
+The parameter $\mu$ fixes the spectrum of the hessian. By manipulating it, one
+can decide to find the complexity of saddles of a certain macroscopic index, or
+of minima with a certain harmonic stiffness. When $\mu$ is taken to be within
+the range $\pm\sqrt{4f''(1)}=\pm\mu_m$, the critical points are constrained to have
+index $\mathcal I=\frac12N(1-\mu/\mu_m)$. When $\mu>\mu_m$, the critical
+points are minima whose sloppiest eigenvalue is $\mu-\mu_m$. Finally,
+when $\mu=\mu_m$, the critical points are marginal minima.
+
 
 The parameter $\mu$ fixes the spectrum of the hessian. When it is an integration variable,
 and one restricts the domain of all integrations to compute saddles of a certain macroscopic index, or
 of minima with a certain harmonic stiffness, its value is the `softest' mode that adapts to change the Hessian \cite{Fyodorov_2007_Replica}. When it is fixed, then the restriction of the index of saddles is `payed' by the realization of the eigenvalues of the Hessian, usually a
 `harder' mode.
 
-
+{\tiny  NOT SURE WORTHWHILE
 
 \subsection{What to expect?}
 
@@ -112,51 +119,462 @@ between zero and one  in the semifrozen phase, and zero at all higher energies.\
 $\bullet$ In phases where one or both systems are stuck in their thresholds (and only in those), the
 minima are exponentially subdominant with respect to saddles, because a saddle is found by releasing the constraint of staying on the threshold.
 
+}
+
+\section{Equilibrium}
+
+Here we review the equilibrium solution. \cite{Crisanti_1992_The, Crisanti_1993_The, Crisanti_2004_Spherical, Crisanti_2006_Spherical}
+The free energy is well known to take the form
+\begin{equation}
+  \beta F=-1-\log2\pi-\frac12\lim_{n\to0}\frac1n\left(\beta^2\sum_{ab}^nf(Q_{ab})+\log\det Q\right)
+\end{equation}
+which must be extremized over the matrix $Q$. When the solution is a Parisi matrix, this can also be written in a functional form. If $P(q)$ is the probability distribution for elements $q$ in a row of the matrix, then
+\begin{equation}
+  \chi(q)=\int_1^qdq'\,\int_0^{q'}dq''\,P(q'')
+\end{equation}
+Since it is the double integral of a probability distribution, $\chi$ must be concave, monotonically decreasing and have $\chi(1)=0$, $\chi'(1)=1$. The free energy can be written as a functional over $\chi$ as
+\begin{equation}
+  \beta F=-1-\log2\pi-\frac12\int dq\,\left(\beta^2f''(q)\chi(q)+\frac1{\chi(q)}\right)
+\end{equation}
+
+
+We insert a  $k$ step RSB ansatz (the steps are standard, and are reviewed in
+Appendix A), and obtain $q_0=0$
+\begin{equation}
+  \begin{aligned}
+    \beta F=
+    -1-\log2\pi
+    -\frac12\left\{\beta^2\left(f(1)+\sum_{i=0}^k(x_i-x_{i+1})f(q_i)\right)
+    +\frac1{x_1}\log\left[
+        1+\sum_{i=1}^{k}(x_i-x_{i+1})q_i
+    \right]\right.\\
+    \left.+\sum_{j=1}^k(x_{j+1}^{-1}-x_j^{-1})\log\left[
+        1+\sum_{i=j+1}^{k}(x_i-x_{i+1})q_i-x_{j+1}q_j
+    \right]
+  \right\}
+  \end{aligned}
+\end{equation}
+The zero temperature limit is most easily obtained by putting $x_i=\tilde
+x_ix_k$ and  $x_k=\tilde\beta/\beta$, $q_k=1-z/\beta$, which ensures the $\tilde x_i$,
+$\tilde\beta$, and $z$ have nontrivial limits. Inserting the ansatz and taking the limit
+carefully treating the $k$th term in each sum separately from the rest, we get
+\begin{equation}
+  \begin{aligned}
+    \lim_{\beta\to\infty}\tilde\beta F=
+    -\frac12z\tilde\beta f'(1)-\frac12\left\{\tilde\beta^2\left(f(1)+\sum_{i=0}^{k-1}(\tilde x_i-\tilde x_{i+1})f(q_i)\right)
+    +\frac1{\tilde x_1}\log\left[
+      \tilde\beta z^{-1}\left(1+\sum_{i=1}^{k-1}(\tilde x_i-\tilde x_{i+1})q_i\right)+1
+    \right]\right.\\
+    \left.+\sum_{j=1}^{k-1}(\tilde x_{j+1}^{-1}-\tilde x_j^{-1})\log\left[
+        \tilde\beta z^{-1}\left(1+\sum_{i=j+1}^{k-1}(\tilde x_i-\tilde x_{i+1})q_i-\tilde x_{j+1}q_j\right)+1
+      \right]
+    \right\}
+  \end{aligned}
+\end{equation}
+This is a $k-1$ RSB ansatz with all eigenvalues scaled by $\tilde\beta z^{-1}$ and shifted by
+1, with effective temperature $\tilde\beta$, and an extra term. This can be seen
+more clearly by rewriting the result in terms of the matrix $\tilde Q$, a
+$(k-1)$-RSB Parisi matrix built from the $\tilde x_1,\ldots,\tilde x_{k-1}$ and
+$q_1,\ldots,q_{k-1}$, which gives
+\begin{equation} \label{eq:ground.state.free.energy}
+  \lim_{\beta\to\infty}\tilde\beta F
+  =-\frac12z\tilde\beta f'(1)-\frac12\lim_{n\to0}\frac1n\left(
+    \tilde\beta^2\sum_{ab}^nf(\tilde Q_{ab})+\log\det(\tilde\beta z^{-1}\tilde Q+I)
+  \right)
+\end{equation}
+In the continuum case, this is
+\begin{equation} \label{eq:ground.state.free.energy.cont}
+  \lim_{\beta\to\infty}\tilde\beta F
+  =-\frac12z\tilde\beta f'(1)-\frac12\int dq\left(
+    \tilde\beta^2f''(q)\tilde\chi(q)+\frac1{\tilde\chi(q)+\tilde\beta z^{-1}}
+  \right)
+\end{equation}
+
+The zero temperature limit of the free energy loses one level of replica
+symmetry breaking. Physically, this is a result of the fact that in $k$-RSB,
+$q_k$ gives the overlap within a state, e.g., within the basin of a well inside
+the energy landscape. At zero temperature, the measure is completely localized
+on the bottom of the well, and therefore the overlap with each state becomes
+one. We will see that the complexity of low-energy stationary points in
+Kac--Rice is also given by a $(k-1)$-RSB anstaz. Heuristicall, this is because
+each stationary point also has no width and therefore overlap one with itself.
+
+
+
+\section{Kac--Rice}
+
+\cite{Auffinger_2012_Random, BenArous_2019_Geometry}
+
+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 $\epsilon$, giving the number of stationary points at
+energy $\epsilon$ and radial reaction $\mu$ as
+\begin{equation}
+  \mathcal N(\epsilon, \mu)
+  =\int ds\,\delta(N\epsilon-H(s))\delta(\partial H(s)+\mu s)|\det(\partial\partial H(s)+\mu I)|
+\end{equation}
+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 complexity:
+\begin{equation}
+  \Sigma(\epsilon,\mu)=\lim_{N\to\infty}\frac1N\overline{\log\mathcal N(\epsilon, \mu)}
+\end{equation}
+The radial reaction $\mu$, which acts like a kind of `mass' term, takes a
+fixed value here, which means that the complexity is for a given energy
+density and hessian spectrum. This will turn out to be important when we
+discriminate between counting all solutions, or selecting those of a given
+index, for example minima. The complexity of solutions without fixing $\mu$ is
+given by maximizing the complexity as a function of $\mu$.
+
+If one averages over $\mathcal N$ and afterward takes its logarithm, one arrives at the annealed complexity
+\begin{equation}
+  \Sigma_\mathrm a(\epsilon,\mu)
+  =\lim_{N\to\infty}\frac1N\log\overline{\mathcal N(\epsilon,\mu)}
+\end{equation}
+This has been previously computed for the mixed $p$-spin models \cite{BenArous_2019_Geometry}, with the result
+\begin{equation}
+  \begin{aligned}
+    \Sigma_\mathrm a(\epsilon,\mu)
+    =-\frac{\epsilon^2(f'(1)+f''(1))+2\epsilon\mu f'(1)+f(1)\mu^2}{2f(1)(f'(1)+f''(1))-2f'(1)^2}-\frac12\log f'(1)\\
+    +\operatorname{Re}\left[\frac\mu{\mu+\sqrt{\mu^2-4f''(1)}}
+      -\log\left(\frac1{2f''(1)}\left(\mu-\sqrt{\mu^2-4f''(1)}\right)\right)
+    \right]
+  \end{aligned}
+\end{equation}
+The annealed complexity is known to equal the actual (quenched) complexity in
+circumstances where there is at most one level of RSB. This is the case for the
+pure $p$-spin models, or for mixed models where $1/\sqrt{f''(q)}$ is a convex
+function. However, it fails dramatically for models with higher replica
+symmetry breaking. For instance, when $f(q)=\frac12(q^2+\frac1{16}q^4)$, the
+anneal complexity predicts that minima vanish well before the dominant saddles,
+a contradiction for any bounded function, as seen in Fig.~\ref{fig:frsb.complexity}.
+
+\subsection{The replicated problem}
+
+The replicated Kac--Rice formula was introduced by Ros et al.~\cite{Ros_2019_Complex}, and its
+effective action for the mixed $p$-spin model has previously been computed by
+Folena et al.~\cite{Folena_2020_Rethinking}. Here we review the derivation.
+
+In order to average the complexity over disorder properly, the logarithm must be dealt with. We use the standard replica trick, writing
+\begin{equation}
+  \begin{aligned}
+    \log\mathcal N(\epsilon,\mu)
+    &=\lim_{n\to0}\frac\partial{\partial n}\mathcal N^n(\epsilon,\mu) \\
+    &=\lim_{n\to0}\frac\partial{\partial n}\int\prod_a^n ds_a\,\delta(N\epsilon-H(s_a))\delta(\partial H(s_a)+\mu s_a)|\det(\partial\partial H(s_a)+\mu I)|
+  \end{aligned}
+\end{equation}
+
+As noted by Bray and Dean  \cite{Bray_2007_Statistics}, gradient and Hessian
+are independent for a random Gaussian function, and the average over disorder
+breaks into a product of two independent averages, one for the gradient factor
+and one for the determinant. The integration of all variables, including the
+disorder in the last factor, may be restricted to the domain such that the
+matrix $\partial\partial H(s_a)-\mu I$ has  a specified number of negative
+eigenvalues (the index $\mathcal I$ of the saddle), (see Fyodorov
+\cite{Fyodorov_2007_Replica} for a detailed discussion). In practice, we are
+therefore able to write
+\begin{equation}
+  \begin{aligned}
+    \Sigma(\epsilon, \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(N\epsilon-H(s_a))\delta(\partial H(s_a)+\mu s_a)}
+    \times
+    \overline{\prod_a^n |\det(\partial\partial H(s_a)+\mu I)|}
+  \end{aligned}
+\end{equation}
+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}
+    \mathcal D(\mu)
+    &=\frac1N\overline{\log|\det(\partial\partial H(s_a)+\mu I)|}
+    =\int d\lambda\,\rho(\lambda+\mu)\log|\lambda| \\
+    &=\operatorname{Re}\left\{
+    \frac12\left(1+\frac\mu{2f''(1)}\left(\mu-\sqrt{\mu^2-4f''(1)}\right)\right)
+    -\log\left(\frac1{2f''(1)}\left(\mu-\sqrt{\mu^2-4f''(1)}\right)\right)
+  \right\}
+  \end{aligned}
+\end{equation}
+The $\delta$-functions are treated by writing them in the Fourier basis, introducing auxiliary fields $\hat s_a$ and $\hat beta$,
+\begin{equation}
+  \prod_a^n\delta(N\epsilon-H(s_a))\delta(\partial H(s_a)+\mu s_a)
+  =\int \frac{d\hat\beta}{2\pi}\prod_a^n\frac{d\hat s_a}{2\pi}
+    e^{\hat\beta(N\epsilon-H(s_a))+i\hat s_a\cdot(\partial H(s_a)+\mu s_a)}
+\end{equation}
+$\hat \beta$ is a parameter conjugate to the state energies, i.e. playing the
+role of an inverse temperature for the metastable states. The average over disorder can now be taken, and since everything is Gaussian it gives
+\begin{equation}
+  \begin{aligned}
+    \overline{
+      \exp\left\{
+        \sum_a^n(i\hat s_a\cdot\partial_a-\hat\beta)H(s_a)
+      \right\}
+    }
+    &=\exp\left\{
+        \frac12\sum_{ab}^n
+        (i\hat s_a\cdot\partial_a-\hat\beta)
+        (i\hat s_b\cdot\partial_b-\hat\beta)
+        \overline{H(s_a)H(s_b)}
+      \right\} \\
+    &=\exp\left\{
+        \frac N2\sum_{ab}^n
+        (i\hat s_a\cdot\partial_a-\hat\beta)
+        (i\hat s_b\cdot\partial_b-\hat\beta)
+        f\left(\frac{s_a\cdot s_b}N\right)
+      \right\} \\
+    &\hspace{-14em}=\exp\left\{
+        \frac N2\sum_{ab}^n
+        \left[
+          \hat\beta^2f\left(\frac{s_a\cdot s_b}N\right)
+          -2i\hat\beta\frac{\hat s_a\cdot s_b}Nf'\left(\frac{s_a\cdot s_b}N\right)
+          -\frac{\hat s_a\cdot \hat s_b}Nf'\left(\frac{s_a\cdot s_b}N\right)
+          +\left(i\frac{\hat s_a\cdot s_b}N\right)^2f''\left(\frac{s_a\cdot s_b}N\right)
+        \right]
+      \right\}
+  \end{aligned}
+\end{equation}
+
+We introduce new fields
+\begin{align}
+  Q_{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
+\end{align}
+$Q_{ab}$ is the overlap between spins belonging to different replicas. The
+meaning of $R_{ab}$ is that of a response of replica $a$ to a linear field in
+replica $b$:
+\begin{equation}
+    R_{ab} = \frac 1 N \sum_i \overline{\frac{\delta s_i^a}{\delta h_i^b}}
+\end{equation}
+The $D$ may similarly be seen as the variation of the complexity with respect to a random field.
+
+By substituting these parameters into the expressions above and then making a change of variables in the integration from $s_a$ and $\hat s_a$ to these three matrices, we arrive at the form for the complexity
+\begin{equation}
+  \begin{aligned}
+    &\Sigma(\epsilon,\mu)
+    =\mathcal D(\mu)+\hat\beta\epsilon+\\
+    &\lim_{n\to0}\frac1n\left(
+      -\mu\sum_a^nR_{aa}
+      +\frac12\sum_{ab}\left[
+        \hat\beta^2f(Q_{ab})+(2\hat\beta R_{ab}-D_{ab})f'(Q_{ab})
+        +R_{ab}^2f''(Q_{ab})
+      \right]
+    +\frac12\log\det\begin{bmatrix}Q&iR\\iR&D\end{bmatrix}
+    \right)
+  \end{aligned}
+\end{equation}
+where $\hat\beta$, $Q$, $R$ and $D$ must be evaluated at extrema of this
+expression. With $Q$, $R$, and $D$ distinct replica matrices, this is
+potentially quite challenging.
+
+{\bf  Inserting the expression for $D$ in the action, and writing $R=R_d+\tilde R$ ($\tilde R$ is zero in the diagonal) , the linear term in $\tilde R$ is 
+$Tr \left[\tilde R (\hat \beta -R_d Q^{-1} f')\right]$ }
+
+\section{Replica ansatz}
+
+We shall make the following ansatz for the saddle point:
+\begin{align}\label{ansatz}
+  Q_{ab}= \text{a Parisi matrix} &&
+  R_{ab}=R_d \delta_{ab} &&
+  D_{ab}= D_d \delta_{ab}
+\end{align}
+From what we have seen above, this means that replica $a$ is insensitive to
+a small field applied to replica $b$ if $a \neq b$, a property related to ultrametricity. A similar situation happens in quantum replicated systems,
+with time appearing only on the diagonal terms: see Appendix C for details.
 
+From its very definition, it is easy to see just perturbing the equations
+with a field that $R_d$ is the trace of the inverse Hessian, as one expect indeed of a response.
+\begin{equation}
+ R_d =    \mathcal D'(\mu)
+\end{equation}
+Similarly, ....  one shows that 
+\begin{equation}
+D_d = \hat \beta R_d
+\end{equation}
 
-\section{Main result}
+
+Is it a saddle?
+
+\begin{equation}
+  0=\frac{\partial\Sigma}{\partial R}
+  =\lim_{n\to0}\frac1n\left(-\mu I+\hat\beta f'(Q)+R\odot f''(Q)
+    +(QD+R^2)^{-1}R
+  \right)
+\end{equation}
+\begin{equation}
+  0=\frac{\partial\Sigma}{\partial D}
+  =\lim_{n\to0}\frac1n\left(-\frac12f'(Q)
+    +\frac12(QD+R^2)^{-1}Q
+  \right)
+\end{equation}
+\begin{equation}
+  D=f'(Q)^{-1}-RQ^{-1}R
+\end{equation}
+Diagonal anstaz requires that $f'(Q_{ab})^{-1}=R_d^2Q_{ab}^{-1}$ for $a\neq b$.
+\begin{equation}
+  0
+  =\lim_{n\to0}\frac1n\left(-\mu I+\hat\beta f'(Q)+R\odot f''(Q)
+    +Q^{-1}Rf'(Q)
+  \right)
+\end{equation}
+Diagonal ansatz requires that
+\begin{equation}
+  0=-\mu I+\hat\beta f'(Q)+R_df''(1)I+R_dQ^{-1}f'(Q)
+\end{equation}
+or $\hat\beta f'(Q_{ab})=-R_d[Q^{-1}f'(Q)]_{ab}$ for $a\neq b$.
+We also have from the first (before the diagonal ansatz) $Df'(Q)=I-RQ^{-1}Rf'(Q)$. Inserting the diagonal, we get $Q^{-1}f'(Q)=\frac1{R_d^2}(I-D_df'(Q))$, and
+\begin{equation}
+  0=-\mu I+\hat\beta f'(Q)+R_df''(1)I+\frac1{R_d}(I-D_df'(Q))
+  =(R_df''(1)+R_d^{-1}-\mu)I+(\hat\beta-D_d/R_d)f'(Q)
+\end{equation}
+
+Always true:
+\begin{equation}
+  0=-\mu R+\hat\beta Rf'(Q)+R(R\odot f''(Q))+I-Df'(Q)
+\end{equation}
+Insert $D=D_dI+\delta D$, $R=R_dI+\delta R$, and $D_d=\hat\beta R_d$
+\begin{equation}
+  0=-\hat\beta \mu R_d I +\hat\beta R_df'(Q)+R_d^2f''(1)I+I-\hat\beta R_df'(Q)
+  -\mu\delta R
+\end{equation}
+
+\begin{equation}
+  0=(\hat\beta Q+R) f'(Q)+(R\odot f''(Q)-\mu I)Q
+\end{equation}
 
 \begin{equation}
   \begin{aligned}
-    \overline{\Sigma(\epsilon,\mu)}
+    &\Sigma(\epsilon,\mu)
+    =\frac12+\mathcal D(\mu)+\hat\beta\epsilon+\\
+    &\lim_{n\to0}\frac1n\left(
+      -\frac12\mu\sum_a^nR_{aa}
+      +\frac12\sum_{ab}\left[
+        \hat\beta^2f(Q_{ab})+\hat\beta R_{ab}f'(Q_{ab})
+      \right]
+      +\frac12\log\det(f'(Q)^{-1}Q)
+    \right)
+  \end{aligned}
+\end{equation}
+
+
+\subsection{Solution}
+
+
+Inserting the diagonal ansatz \eqref{ansatz} one gets
+\begin{equation} \label{eq:diagonal.action}
+  \begin{aligned}
+    \Sigma(\epsilon,\mu)
     =\mathcal D(\mu)
-    +\operatorname*{extremum}_{\substack{R_d,D_d,\hat\epsilon\in\mathbb R\\\chi\in\Lambda}}
-    \left\{
-      \hat\epsilon\epsilon+\mu R_d
-      +\frac12(2\hat\epsilon R_d-D_d)f'(1)+\frac12R_d^2f''(1)
-      +\frac12\log R_d^2 \right.\\\left.
+    +
+      \hat\beta\epsilon-\mu R_d
+      +\frac12(2\hat\beta R_d-D_d)f'(1)+\frac12R_d^2f''(1)+\frac12\log R_d^2
+      \\
+      +\frac12\lim_{n\to0}\frac1n\left(\hat\beta^2\sum_{ab}f(Q_{ab})+\log\det((D_d/R_d^2)Q+I)\right)
+  \end{aligned}
+\end{equation}
+Using standard manipulations (Appendix B), one finds also a continuous version
+\begin{equation} \label{eq:functional.action}
+  \begin{aligned}
+    \Sigma(\epsilon,\mu)
+    =\mathcal D(\mu)
+    +
+      \hat\beta\epsilon-\mu R_d
+      +\frac12(2\hat\beta R_d-D_d)f'(1)+\frac12R_d^2f''(1)+\frac12\log R_d^2
+      \\
       +\frac12\int_0^1dq\,\left(
-        \hat\epsilon^2f''(q)\chi(q)+\frac1{\chi(q)+R_d^2/D_d}
+        \hat\beta^2f''(q)\chi(q)+\frac1{\chi(q)+R_d^2/D_d}
       \right)
-    \right\}
   \end{aligned}
 \end{equation}
-where
+where $\chi(q)$ is defined with respect to $Q$ exactly as in the equilibrium case.
+Note the close similarity of this action to the equilibrium replica one, at finite temperature.
 \begin{equation}
-  \mathcal D(\mu)
-  =\operatorname{Re}\left\{
-    \frac12\left(1+\frac\mu{2f''(1)}\left(\mu\pm\sqrt{\mu^2-4f''(1)}\right)\right)
-    -\log\left(\frac1{2f''(1)}\left(\mu\pm\sqrt{\mu^2-4f''(1)}\right)\right)
-  \right\}
+  \beta F=-\frac12\lim_{n\to0}\frac1n\left(\beta^2\sum_{ab}f(Q_{ab})+\log\det Q\right)-1-\log2\pi
+\end{equation}
+
+
+\subsubsection{Saddles}
+\label{sec:counting.saddles}
+
+The dominant stationary points are given by maximizing the action with respect
+to $\mu$. This gives
+\begin{equation} \label{eq:mu.saddle}
+  0=\frac{\partial\Sigma}{\partial\mu}=\mathcal D'(\mu)-R_d
+\end{equation}
+as expected.
+To take the derivative, we must resolve the real part inside the definition of
+$\mathcal D$. When saddles dominate, $\mu<\mu_m$, and
+\begin{equation}
+  \mathcal D(\mu)=\frac12+\frac12\log f''(1)+\frac{\mu^2}{4f''(1)}
+\end{equation}
+It follows that the dominant saddles have $\mu=2f''(1)R_d$. Their index
+is thus $\mathcal I=\frac12N(1-R_d\sqrt{f''(1)})$. If
+$R_d\sqrt{f''(1)}>1$ then we were wrong to assume that saddles dominate, and
+the most numerous saddles will be found just above $\mu=\mu_m$.
+
+\subsubsection{Minima}
+\label{sec:counting.minima}
+
+When minima dominate, $\mu>\mu_m$ and all the roots inside $\mathcal D(\mu)$ are real. Therefore $\mathcal D(\mu)$ is given by its former expression with the real part dropped, and
+\begin{equation}
+  \mathcal D'(\mu)=\frac{2}{\mu+\sqrt{\mu^2-4f''(1)}}
+\end{equation}
+\begin{equation}
+  \mu=\frac1{R_d}+R_df''(1)
 \end{equation}
-and $\Lambda$ is the space of functions $\chi:[0,1]\to[0,1]$ which are
-monotonically decreasing, convex, and have $\chi(1)=0$ and $\chi'(1)=-1$.
-If there is more than one extremum of this function, choose the one with the
-smallest value of $\Sigma$. The sign of the root inside $\mathcal D(\mu)$ is
-negative for $\mu>0$ and positive for $\mu<0$.
-
-The $k$-RSB ansatz is equivalent to piecewise linear $\chi$ with $k+1$
-pieces, with replica symmetric or 0-RSB giving $\chi(q)=1-q$. Our other major
-result is that, if the equilibrium state in the vicinity of zero temperature is
-given by a $k$-RSB ansatz, then the complexity is given by a $(k-1)$-RSB
-ansatz. Moreover, there is an exact correspondence between the parameters of
-the equilibrium saddle point in the limit of zero temperature and those of the
-complexity saddle at the ground state. If the equilibrium is given by
-$x_1,\ldots,x_k$ and $q_1,\ldots,q_k$, then the parameters $\tilde
-x_1,\ldots,\tilde x_{k-1}$ and $\tilde q_1,\ldots,\tilde q_{k-1}$ for the
+
+\subsubsection{Recovering the equilibrium ground state}
+
+The ground state energy corresponds to that where the complexity of dominant
+stationary points becomes zero. If the most common stationary points vanish,
+then there cannot be any stationary points. In this section, we will show that
+it reproduces the ground state produced by taking the zero-temperature limit in
+the equilibrium case.
+
+Consider the extremum problem of \eqref{eq:diagonal.action} with respect to $R_d$ and $D_d$. This gives the equations
+\begin{align}
+  0
+  &=\frac{\partial\Sigma}{\partial D_d}
+  =-\frac12f'(1)+\frac12\frac1{R_d^2}\lim_{n\to0}\frac1n\operatorname{Tr}((D_d/R_d^2)Q+I)^{-1}Q \label{eq:saddle.d}\\
+  0
+  &=\frac{\partial\Sigma}{\partial R_d}
+  =-\mu+\hat\beta f'(1)+R_df''(1)+\frac1{R_d}-\frac{D_d}{R_d^3}\lim_{n\to0}\frac1n\operatorname{Tr}((D_d/R_d^2)Q+I)^{-1}Q \label{eq:saddle.r}
+\end{align}
+Adding $2(D_d/R_d)$ times \eqref{eq:saddle.d} to \eqref{eq:saddle.r} and multiplying by $R_d$ gives
+\begin{equation}
+  0=-R_d\mu+1+R_d^2f''(1)+f'(1)(R_d\hat\beta-D_d)
+\end{equation}
+At the ground state, minima will always dominate (even if marginal). We can
+therefore use the optimal $\mu$ from \S\ref{sec:counting.minima}, which gives
+\begin{equation}
+  0=f'(1)(R_d\hat\beta-D_d)
+\end{equation}
+In order to satisfy this equation we must have
+$D_d=R_d\hat\beta$. This relationship holds for the most common minima whenever
+they dominante, including in the ground state.
+
+Substituting this into the action, and also substituting the optimal $\mu$ for
+minima, and taking $\Sigma(\epsilon_0,\mu^*)=0$, gives
+\begin{equation}
+  \hat\beta\epsilon_0
+  =-\frac12R_d\hat\beta f'(1)-\frac12\lim_{n\to0}\frac1n\left(
+      \hat\beta^2\sum_{ab}^nf(Q_{ab})
+      +\log\det(\hat\beta R_d^{-1} Q+I)
+  \right)
+\end{equation}
+which is precisely \eqref{eq:ground.state.free.energy} with $R_d=z$,
+$\hat\beta=\tilde\beta$, and $Q=\tilde Q$.
+
+{\em We arrive at one of the main results of our paper: a $(k-1)$-RSB ansatz in
+Kac--Rice will predict the correct ground state energy for a model whose
+equilibrium state at small temperatures is $k$-RSB } Moreover, there is an
+exact correspondance between the saddle parameters of each.  If the equilibrium
+is given by a Parisi matrix with parameters $x_1,\ldots,x_k$ and
+$q_1,\ldots,q_k$, then the parameters $\hat\beta$, $R_d$, $D_d$, $\tilde
+x_1,\ldots,\tilde x_{k-1}$, and $\tilde q_1,\ldots,\tilde q_{k-1}$ for the
 complexity in the ground state are
 \begin{align}
-  \hat\epsilon=\lim_{\beta\to\infty}\beta x_k
+  \hat\beta=\lim_{\beta\to\infty}\beta x_k
   &&
   \tilde x_i=\lim_{\beta\to\infty}\frac{x_i}{x_k}
   &&
@@ -164,12 +582,185 @@ complexity in the ground state are
   &&
   R_d=\lim_{\beta\to\infty}\beta(1-q_k)
   &&
-  D_d=R_d\hat\epsilon
+  D_d=\hat\beta R_d
 \end{align}
 
-\section{Equilibrium}
+\section{Examples}
 
-Here we review the equilibrium solution. \cite{Crisanti_1992_The, Crisanti_1993_The, Crisanti_2004_Spherical, Crisanti_2006_Spherical}
+\subsection{1RSB complexity}
+
+It is known that by choosing a covariance $f$ as the sum of polynomials with
+well-separated powers, one develops 2RSB in equilibrium. This should correspond
+to 1RSB in Kac--Rice. For this example, we take
+\begin{equation}
+  f(q)=\frac12\left(q^3+\frac1{16}q^{16}\right)
+\end{equation}
+With this covariance, the model sees a RS to 1RSB transition at
+$\beta_1=1.70615\ldots$ and a 1RSB to 2RSB transition at $\beta_2=6.02198\ldots$. At these points, the average energies are $\langle E\rangle_1=-0.906391\ldots$ and $\langle E\rangle_2=-1.19553\ldots$, and the ground state energy is $E_0=-1.2876055305\ldots$.
+
+In this model, the RS complexity gives an inconsistent answer for the
+complexity of the ground state, predicting that the complexity of minima
+vanishes at a higher energy than the complexity of saddles, with both at a
+lower energy than the equilibrium ground state. The 1RSB complexity resolves
+these problems, predicting the same ground state as equilibrium and that the
+complexity of marginal minima (and therefore all saddles) vanishes at
+$E_m=-1.2876055265\ldots$, which is very slightly greater than $E_0$. Saddles
+become dominant over minima at a higher energy $E_s=-1.287605716\ldots$.
+Finally, the 1RSB complexity transitions to a RS description at an energy
+$E_1=-1.27135996\ldots$. All these complexities can be seen plotted in
+Fig.~\ref{fig:2rsb.complexity}.
+
+All of the landmark energies associated with the complexity are a great deal
+smaller than their equilibrium counterparts, e.g., comparing $E_1$ and $\langle
+E\rangle_2$.
+
+\begin{figure}
+  \centering
+  \includegraphics{figs/316_complexity.pdf}
+
+  \caption{
+    Complexity of dominant saddles (blue), marginal minima (yellow), and
+    dominant minima (green) of the $3+16$ model. Solid lines show the result of
+    the 1RSB ansatz, while the dashed lines show that of a RS ansatz. The
+    complexity of marginal minima is always below that of dominant critical
+    points except at the red dot, where they are dominant.
+    The inset shows a region around the ground state and the fate of the RS solution.
+  } \label{fig:2rsb.complexity}
+\end{figure}
+
+\begin{figure}
+  \centering
+  \begin{minipage}{0.7\textwidth}
+    \includegraphics{figs/316_comparison_q.pdf}
+    \hspace{1em}
+    \includegraphics{figs/316_comparison_x.pdf} \vspace{1em}\\
+    \includegraphics{figs/316_comparison_b.pdf}
+    \hspace{1em}
+    \includegraphics{figs/316_comparison_R.pdf}
+  \end{minipage}
+  \includegraphics{figs/316_comparison_legend.pdf}
+
+  \caption{
+    Comparisons between the saddle parameters of the equilibrium solution to
+    the $3+16$ model (blue) and those of the complexity (yellow). Equilibrium
+    parameters are plotted as functions of the average energy $\langle
+    E\rangle=\partial_\beta(\beta F)$ and complexity parameters as functions of
+    fixed energy $E$. Solid lines show the result of a 2RSB ansatz, dashed
+    lines that of a 1RSB ansatz, and dotted lines that of a RS ansatz. All
+    paired parameters coincide at the ground state energy, as expected.
+  } \label{fig:2rsb.comparison}
+\end{figure}
+
+\subsection{Full RSB complexity}
+
+\begin{figure}
+  \centering
+  \includegraphics{figs/24_complexity.pdf}
+  \caption{
+    The complexity $\Sigma$ of the mixed $2+4$ spin model as a function of
+    distance $\Delta\epsilon=\epsilon-\epsilon_0$ of the ground state. The
+    solid blue line shows the complexity of dominant saddles given by the FRSB
+    ansatz, and the solid yellow line shows the complexity of marginal minima.
+    The dashed lines show the same for the annealed complexity. The inset shows
+    more detail around the ground state.
+  } \label{fig:frsb.complexity}
+\end{figure}
+
+\begin{figure}
+  \centering
+  \includegraphics{figs/24_func.pdf}
+  \hspace{1em}
+  \includegraphics{figs/24_qmax.pdf}
+
+  \caption{
+    \textbf{Left:} The spectrum $\chi$ of the replica matrix in the complexity
+    of dominant saddles for the $2+4$ model at several energies.
+    \textbf{Right:} The cutoff $q_{\mathrm{max}}$ for the nonlinear part of the
+    spectrum as a function of energy $E$ for both dominant saddles and marginal
+    minima. The colored vertical lines show the energies that correspond to the
+    curves on the left.
+  } \label{fig:24.func}
+\end{figure}
+
+\begin{figure}
+  \centering
+  \includegraphics{figs/24_comparison_b.pdf}
+  \hspace{1em}
+  \includegraphics{figs/24_comparison_Rd.pdf}
+  \raisebox{3em}{\includegraphics{figs/24_comparison_legend.pdf}}
+
+  \caption{
+    Comparisons between the saddle parameters of the equilibrium solution to
+    the $3+4$ model (black) and those of the complexity (blue and yellow). Equilibrium
+    parameters are plotted as functions of the average energy $\langle
+    E\rangle=\partial_\beta(\beta F)$ and complexity parameters as functions of
+    fixed energy $E$. Solid lines show the result of a FRSB ansatz and dashed
+    lines that of a RS ansatz. All paired parameters coincide at the ground
+    state energy, as expected.
+  } \label{fig:2rsb.comparison}
+\end{figure}
+
+In the case where any FRSB is present, one must work with the functional form
+of the complexity \eqref{eq:functional.action}, which must be extremized with
+respect to $\chi$ under the conditions that $\chi$ is concave, monotonically
+decreasing, and $\chi(1)=0$, $\chi'(1)=-1$. The annealed case is found by
+taking $\chi(q)=1-q$, which satisfies all of these conditions. $k$-RSB is
+produced by breaking $\chi$ into $k+1$ piecewise linear segments.
+
+Forget for the moment these tricky requirements. The function would then be
+extremized by satisfying
+\begin{equation}
+  0=\frac{\delta\Sigma}{\delta\lambda(q)}=\frac12\hat\beta^2f''(q)-\frac12\frac1{(\lambda(q)+R_d^2/D_d)^2}
+\end{equation}
+which implies the solution
+\begin{equation}
+  \lambda^*(q)=\frac1{\hat\beta}f''(q)^{-1/2}-\frac{R_d^2}{D_d}
+\end{equation}
+If $f''(q)^{-1/2}$ is not concave anywhere, there is little use of this
+solution. However, if it is concave everywhere it may constitute a portion of
+the full solution.
+
+We suppose that solutions are given by
+\begin{equation}
+  \lambda(q)=\begin{cases}
+    \lambda^*(q) & q<q_\textrm{max} \\
+    1-q          & q\geq q_\textrm{max}
+  \end{cases}
+\end{equation}
+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{Ultrametricity rediscovered}
+
+TENTATIVE BUT INTERESTING
+
+The frozen phase for a given index ${\cal{I}}$ is the one for values of $\hat \beta> \hat \beta_{freeze}^{\cal{I}}$. 
+
+[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. 
+
+
+
+\section{Conclusion}
+We have constructed a  replica solution for the general  problem of finding saddles of random mean-field landscapes, including systems
+with many steps of RSB. 
+The main results of this paper are the ansatz (\ref{ansatz}) and the check that the lowest energy is the correct one obtained with the usual Parisi ansatz. 
+For systems with full RSB, we find that minima are, at all energy densities above the ground state, exponentially subdominant with respect to saddles.
+The solution contains valuable geometric information that has yet to be
+extracted in all detail.
+
+\paragraph{Funding information}
+J K-D and J K are supported by the Simons Foundation Grant No. 454943.
+
+\begin{appendix}
+
+\section{RSB for the Gibbs-Boltzmann measure}
 
 \begin{equation}
   \beta F=-\frac12\lim_{n\to0}\frac1n\left(\beta^2\sum_{ab}f(Q_{ab})+\log\det Q\right)-\frac12\log S_\infty
@@ -272,175 +863,39 @@ $x_i=\tilde x_ix_k$, $x_k=y/\beta$, $q_k=1-z/\beta$
 \right)
 \end{align*}
 $F$ is a $k-1$ RSB ansatz with all eigenvalues scaled by $y$ and shifted by $z$. $\tilde x_0=0$ and $\tilde x_k=1$.
-\begin{equation} \label{eq:ground.state.free.energy}
-  \lim_{\beta\to\infty}F=-\lim_{n\to0}\frac1n\frac12\left(nzf'(1)+y\sum_{ab}f(\tilde Q_{ab})+\frac1y\log\det(yz^{-1}\tilde Q+I)
-  \right)
-\end{equation}
 
 
 
+\section{ RSB for the Kac--Rice integral}
 
-\section{Kac-Rice}
-
-\cite{Auffinger_2012_Random, BenArous_2019_Geometry}
-
-\begin{equation}
-  \mathcal N(\epsilon, \mu)
-  =\int ds\,\delta(N\epsilon-H(s))\delta(\partial H(s)-\mu s)|\det(\partial\partial H(s)-\mu I)|
-\end{equation}
-\begin{equation}
-  \Sigma(\epsilon,\mu)=\frac1N\log\mathcal N(\epsilon, \mu)
-\end{equation}
-
-{\em The `mass' term $\mu$ may take a fixed value, or it may be an integration constant,
-for example fixing the spherical constraint.
-This will turn out to be important when we discriminate between counting all solutions, or selecting those of a given index, for example minima}
-
-
-\subsection{The replicated  problem}
-
-\cite{Ros_2019_Complex}
-\cite{Folena_2020_Rethinking}
-
-\begin{equation}
-  \begin{aligned}
-    \Sigma(\epsilon, \mu)
-    &=\frac1N\lim_{n\to0}\frac\partial{\partial n}\mathcal N^n(\epsilon) \\
-    &=\frac1N\lim_{n\to0}\frac\partial{\partial n}\int\prod_a^n ds_a\,\delta(N\epsilon-H(s_a))\delta(\partial H(s_a)-\mu s_a)|\det(\partial\partial H(s_a)-\mu I)|
-  \end{aligned}
-\end{equation}
-
-
-
-\begin{equation}
-  \begin{aligned}
-    \overline{\Sigma(\epsilon, \mu)}
-    &=\frac1N\lim_{n\to0}\frac\partial{\partial n}\int\left(\prod_a^nds_a\right)\,\overline{\prod_a^n \delta(N\epsilon-H(s_a))\delta(\partial H(s_a)-\mu s_a)}
-    \times
-    \overline{\prod_a^n |\det(\partial\partial H(s_a)-\mu I)|}
-  \end{aligned}
-\end{equation}{\bf
-As noted by Bray and Dean  \cite{Bray_2007_Statistics}, gradient and Hessian are independent
-for a Gaussian distribution, and
-the average over disorder breaks into a product of two independent averages, one for the gradient factor and one for the determinant. The integration of all variables, including the disorder in the last factor, may be restricted to the domain such that the matrix $\partial\partial H(s_a)-\mu I$ has  a specified number of negative eigenvalues (the index {\cal{I}} of the saddle), 
-(see Fyodorov \cite{Fyodorov_2007_Replica} for a detailed discussion) }
-
-\begin{equation}
-  \begin{aligned}
-    \mathcal D(\mu)
-    &=\frac1N\overline{\log|\det(\partial\partial H(s_a)-\mu I)|}
-    =\int d\lambda\,\rho(\lambda-\mu)\log|\lambda| \\
-    &=\operatorname{Re}\left\{\frac12\left(1+\frac\mu{2f''(1)}\left(\mu\pm\sqrt{\mu^2-4f''(1)}\right)\right)-\log\left(\frac1{2f''(1)}\left(\mu\pm\sqrt{\mu^2-4f''(1)}\right)\right)\right\}
-  \end{aligned}
-\end{equation}
-for $\rho$ a semicircle distribution with radius $\sqrt{4f''(1)}$.
-
-all saddles versus only minima
-
-\begin{equation}
-  \prod_a^n\delta(N\epsilon-H(s_a))\delta(\partial H(s_a)-\mu s_a)
-  =\int \frac{d\hat\epsilon}{2\pi}\prod_a^n\frac{d\hat s_a}{2\pi}
-    e^{\hat\epsilon(N\epsilon-H(s_a))+i\hat s_a\cdot(\partial H(s_a)-\mu s_a)}
-\end{equation}
-
-\begin{equation}
-  \begin{aligned}
-    \overline{
-      \exp\left\{
-        \sum_a^n(i\hat s_a\cdot\partial_a-\hat\epsilon)H(s_a)
-      \right\}
-    }
-    &=\exp\left\{
-        \frac12\sum_{ab}^n
-        (i\hat s_a\cdot\partial_a-\hat\epsilon)
-        (i\hat s_b\cdot\partial_b-\hat\epsilon)
-        \overline{H(s_a)H(s_b)}
-      \right\} \\
-    &=\exp\left\{
-        \frac N2\sum_{ab}^n
-        (i\hat s_a\cdot\partial_a-\hat\epsilon)
-        (i\hat s_b\cdot\partial_b-\hat\epsilon)
-        f\left(\frac{s_a\cdot s_b}N\right)
-      \right\} \\
-    &\hspace{-13em}\exp\left\{
-        \frac N2\sum_{ab}^n
-        \left[
-          \hat\epsilon^2f\left(\frac{s_a\cdot s_b}N\right)
-          -2i\hat\epsilon\frac{\hat s_a\cdot s_b}Nf'\left(\frac{s_a\cdot s_b}N\right)
-          -\frac{\hat s_a\cdot \hat s_b}Nf'\left(\frac{s_a\cdot s_b}N\right)
-          +\left(i\frac{\hat s_a\cdot s_b}N\right)^2f''\left(\frac{s_a\cdot s_b}N\right)
-        \right]
-      \right\}
-  \end{aligned}
-\end{equation}
-
-The parameters:
-\begin{align}
-  Q_{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
-\end{align}
-
-\begin{equation}
-  \begin{aligned}
-    S
-    =\mathcal D(\mu)+\hat\epsilon\epsilon+\lim_{n\to0}\frac1n\left(
-      \mu\sum_a^nR_{aa}
-      +\frac12\sum_{ab}\left[
-        \hat\epsilon^2f(Q_{ab})+2\hat\epsilon R_{ab}f'(Q_{ab})
-        -D_{ab}f'(Q_{ab})+R_{ab}^2f''(Q_{ab})
-      \right] \right. \\ \left.
-    +\frac12\log\det\begin{bmatrix}Q&iR\\iR&D\end{bmatrix}
-    \right)
-  \end{aligned}
-\end{equation}
 
+\subsection{Solution}
 
-\section{Replicated action}
 
 
-\begin{equation}
-  \begin{aligned}
-    S
-    =\mathcal D(\mu)+\hat\epsilon\epsilon+\lim_{n\to0}\frac1n\left(
-      \mu\sum_a^nR_{aa}
-      +\frac12\sum_{ab}\left[
-        \hat\epsilon^2f(Q_{ab})+2\hat\epsilon R_{ab}f'(Q_{ab})
-        -D_{ab}f'(Q_{ab})+R_{ab}^2f''(Q_{ab})
-      \right] \right. \\ \left.
-    +\frac12\log\det\begin{bmatrix}Q&iR\\iR&D\end{bmatrix}
-    \right)
-  \end{aligned}
-\end{equation}
-\begin{align}
-  0&=\frac{\partial S}{\partial R_{ab}}
-  =\mu\delta_{ab}+\hat\epsilon f'(Q_{ab})+R_{ab}f''(Q_{ab})+\sum_c(DQ+R^2)^{-1}_{ac}R_{cb} \\
-  0&=\frac{\partial S}{\partial D_{ab}}
-  =-\frac12 f'(Q_{ab})+\frac12\sum_c(DQ+R^2)^{-1}_{ac}Q_{cb}
-\end{align}
-The second equation implies
-\begin{equation}
-  (DQ+R^2)^{-1}=Q^{-1}f'(Q)
-\end{equation}
-
-\section{Replica ansatz}
-
-\subsection{Motivation}
+\begin{align*}
+  \lim_{n\to0}\frac1n\log\det(\hat\beta R_d^{-1} Q+I)
+  =x_1^{-1}\log\left(\hat\beta R_d^{-1}(1-\bar q_k)+1\right)+\int_{q_0^+}^{q_{k-1}}dq\,\mu(q)\log\left[\hat\beta R_d^{-1}\lambda(q)+1\right]
+\end{align*}
+where
+\[
+  \mu(q)=\frac{\partial x^{-1}(q)}{\partial q}
+\]
+Integrating by parts,
+\begin{align*}
+  \lim_{n\to0}\frac1n\log\det(\hat\beta R_d^{-1} Q+I)
+  &=x_1^{-1}\log\left(\hat\beta R_d^{-1}(1-\bar q_k)+1\right)+\left[x^{-1}(q)\log[\hat\beta R_d^{-1}\lambda(q)+1]\right]_{q=q_0^+}^{q=q_{k-1}}-\frac{\hat\beta}{R_d}\int_{q_0^+}^{q_{k-1}}dq\,\frac{\lambda'(q)}{x(q)}\frac1{\hat\beta R_d^{-1}\lambda(q)+1}\\
+  &=\log[\hat\beta R_d^{-1}\lambda(q_{k-1})+1]+\frac{\hat\beta}{R_d}\int_{q_0^+}^{q_{k-1}}dq\,\frac1{\hat\beta R_d^{-1}\lambda(q)+1}
+\end{align*}
 
-We shall make the following ansatz
-to putting:
-\begin{align}\label{ansatz}
-  Q_{ab}= \text{a Parisi matrix} &&
-  R_{ab}=R_d \delta_{ab} &&
-  D_{ab}= D_d \delta_{ab}
-\end{align}
-This ansatz closes under the operations that are involved in the replicated action.
-The reader who is happy with the ansatz may skip the rest of this section.
+\section{ A motivation for the ansatz}
 
  We may encode the original variables in a superspace variable:
 \begin{equation}
   \phi_a(1)= s_a + \bar\eta_a\theta_1+\bar\theta_1\eta_a +  \hat s_a \bar \theta_1 \theta_1
-\end{equation}
+\end{equation}Here $\theta_a$, $\bar \theta_a$ are Grassmann variables, and we denote the full set of coordinates 
+in a  compact form as $1=  \theta_1 \overline\theta_1$, $d1=   d\theta_1 d\overline\theta_1$, etc.
+The correlations are encoded in
 \begin{equation}
 \begin{aligned}
   \mathbb Q_{a,b}(1,2)&=\frac 1 N \phi_a(1)\cdot\phi_b (2) =
@@ -453,22 +908,18 @@ Q_{ab}  -i\left[\bar\theta_1\theta_1+\bar\theta_2\theta_2\right]  R_{ab}
 \end{equation}
 \begin{equation}
   \overline{\Sigma(\epsilon,\mu)}
-  =\hat\epsilon\epsilon\lim_{n\to0}\frac1n\left[
+  =\hat\beta\epsilon\lim_{n\to0}\frac1n\left[
     \mu\int d1\sum_a^n\mathbb Q_{aa}(1,1)
-    +\int d2\,d1\,\frac12\sum_{ab}^n(1+\hat\epsilon\bar\theta_1\theta_1)f(\mathbb Q_{ab}(1,2))(1+\hat\epsilon\bar\theta_2\theta_2)
+    +\int d2\,d1\,\frac12\sum_{ab}^n(1+\hat\beta\bar\theta_1\theta_1)f(\mathbb Q_{ab}(1,2))(1+\hat\beta\bar\theta_2\theta_2)
     +\frac12\operatorname{sdet}\mathbb Q
   \right]
 \end{equation}
-
-Here $\theta_a$, $\bar \theta_a$ are Grassmann variables, and we denote the full set of coordinates 
-in a  compact form as
-$1=  \theta_1 \overline\theta_1$, $d1=   d\theta_1 d\overline\theta_1$, etc.
 The odd and even fermion numbers decouple, so we can neglect all odd terms in $\theta,\bar{\theta}$.
 
 \cite{Annibale_2004_Coexistence}
 
-
-The variables $\bar \theta \theta$ and $\bar \theta ' \theta'$ play
+This encoding also works for dynamics, where the coordinates then read 
+$1= (\bar \theta,  \theta, t)$, etc.   The variables $\bar \theta \theta$ and $\bar \theta ' \theta'$ play
 the role of `times' in a superspace treatment. We have a long experience of
 making an ansatz for replicated quantum problems, which naturally involve a (Matsubara) time. The dependence on this time only holds for diagonal replica elements, a consequence of ultrametricity.  The analogy strongly
 suggests that only the diagonal ${\bf Q}_{aa}$ depend on the $\theta$'s. This boils down the ansatz \ref{ansatz}.
@@ -492,173 +943,13 @@ Not surprisingly, and for the same reason as in the quantum case, this ansatz cl
  -i\bar\theta_1\theta_1\bar\theta_2\theta_2D_1R_2
  \end{aligned}
 \end{equation}
-
-\subsection{Solution}
-
-
-Insert the diagonal ansatz $R=R_dI$, $D=D_dI$. Then
-\[
-  0=(R_df''(1)-\mu)I+\hat\epsilon f'(Q)+R_d(R_d^2I-D_dQ)^{-1}
-  =(R_df''(1)-\mu)I+\hat\epsilon f'(Q)+R_dQ^{-1}f'(Q)
-\]
-and
-\[
-  Q^{-1}f'(Q)=(I+D_df'(Q))/R_d^2
-\]
-Substituting the second into the first, we have
-\[
-  0=(R_df''(1)-\mu)I+\hat\epsilon f'(Q)+\frac1{R_d}(I+D_df'(Q))
-\]
-\[
-  0=(R_df''(1)-\mu+R_d^{-1})I+(\hat\epsilon+D_d/R_d)f'(Q)
-\]
-The only way for this equation to be satisfied off the diagonal for nontrivial $Q$ is for $D_d=-R_d\hat\epsilon$. We therefore have
-\begin{align*}
-  \Sigma
-  =-\epsilon\hat\epsilon+\lim_{n\to0}\frac1n\left(
-    n\mu(F_d-R_d)+\frac12n\left[
-      \hat\epsilon R_df'(1)+R_d^2f''(1)-F_d^2f''(1)
-    \right]
-    +\frac12\sum_{ab}
-      \hat\epsilon^2f(Q_{ab})
-    ]\right.\\\left.
-    +\frac12\log\det(\hat\epsilon R_d^{-1} Q+I)
-    +n\log R_d
-  -n\log F_d
-  \right)
-\end{align*}
-Taking the saddle with respect to $\mu$ and $F_d$ yields
-\[
-  F_d=R_d
-\]
-\[
-  \mu=R_d^{-1}(1+R_d^2f''(1))
-\]
-and gives
-\begin{align*}
-  \Sigma
-  =\epsilon\hat\epsilon+\hat\epsilon R_d f'(1)+\frac12D_df'(1)+\lim_{n\to0}\frac1n\frac12\left(
-    \hat\epsilon^2\sum_{ab}
-      f(Q_{ab})
-    +\log\det(-D_dR_d^{-2} Q+I)
-  \right)
-\end{align*}
-Finally, setting $0=\Sigma$ gives
-\[
-  \epsilon
-  =\lim_{n\to0}\frac1n\frac12\left(nR_df'(1)+\hat\epsilon\sum_{ab}
-      f(Q_{ab})
-      +\frac1{\hat\epsilon}\log\det(\hat\epsilon R_d^{-1} Q+I)
-  \right)
-\]
-which is precisely \eqref{eq:ground.state.free.energy} with $R_d=z$ and $\hat\epsilon=y$. 
-
-{\em Therefore, a $(k-1)$-RSB ansatz in Kac--Rice will predict the correct ground state energy for a model whose equilibrium state at small temperatures is $k$-RSB.}
-
-\subsection{Full}
-
-\begin{align*}
-  \lim_{n\to0}\frac1n\log\det(\hat\epsilon R_d^{-1} Q+I)
-  =x_1^{-1}\log\left(\hat\epsilon R_d^{-1}(1-\bar q_k)+1\right)+\int_{q_0^+}^{q_{k-1}}dq\,\mu(q)\log\left[\hat\epsilon R_d^{-1}\lambda(q)+1\right]
-\end{align*}
-where
-\[
-  \mu(q)=\frac{\partial x^{-1}(q)}{\partial q}
-\]
-Integrating by parts,
-\begin{align*}
-  \lim_{n\to0}\frac1n\log\det(\hat\epsilon R_d^{-1} Q+I)
-  &=x_1^{-1}\log\left(\hat\epsilon R_d^{-1}(1-\bar q_k)+1\right)+\left[x^{-1}(q)\log[\hat\epsilon R_d^{-1}\lambda(q)+1]\right]_{q=q_0^+}^{q=q_{k-1}}-\frac{\hat\epsilon}{R_d}\int_{q_0^+}^{q_{k-1}}dq\,\frac{\lambda'(q)}{x(q)}\frac1{\hat\epsilon R_d^{-1}\lambda(q)+1}\\
-  &=\log[\hat\epsilon R_d^{-1}\lambda(q_{k-1})+1]+\frac{\hat\epsilon}{R_d}\int_{q_0^+}^{q_{k-1}}dq\,\frac1{\hat\epsilon R_d^{-1}\lambda(q)+1}
-\end{align*}
-
-\begin{align*}
-  \Sigma
-  =-\epsilon\hat\epsilon+
-    \frac12\hat\epsilon R_df'(1)
-    +\frac12\int_0^1dq\,\left[
-      \hat\epsilon^2\lambda(q)f''(q)
-      +\frac1{\lambda(q)+R_d/\hat\epsilon}
-    \right]
-\end{align*}
-for $\lambda$ concave, monotonic,  $\lambda(1)=0$, and $\lambda'(1)=-1$
-\[
-  0=\frac{\delta\Sigma}{\delta\lambda(q)}=\frac12\hat\epsilon^2f''(q)-\frac12\frac1{(\lambda(q)+R_d/\hat\epsilon)^2}
-\]
-\[
-  \lambda^*(q)=\frac1{\hat\epsilon}\left[f''(q)^{-1/2}-R_d\right]
-\]
-
-We suppose that solutions are given by
-\begin{equation}
-  \lambda(q)=\begin{cases}
-    \lambda^*(q) & q<q^* \\
-    1-q          & q\geq q^*
-  \end{cases}
-\end{equation}
-where $1-q$ guarantees the boundary conditions at $q=1$, and corresponds to the 0RSB or annealed solutions (annealed Kac--Rice is recovered by substituting in $1-q$ for $\lambda$). We will need to require that $1-q^*=\lambda^*(q^*)$, i.e., continuity.
-
-Inserting this into the complexity, we find
-\begin{align*}
-  \Sigma
-  &=-\epsilon\hat\epsilon+\frac12\hat\epsilon R_df'(1)
-  +\frac12\int_0^{q^*}dq\left[
-    \hat\epsilon(f''(q)^{-1/2}-R_d)f''(q)+\hat\epsilon f''(q)^{1/2}
-  \right]
-  +\frac12\int_{q^*}^1dq\left[
-    \hat\epsilon^2(1-q)f''(q)+\frac1{q-1+R_d/\hat\epsilon}
-  \right] \\
-  &=-\epsilon\hat\epsilon+\frac12\hat\epsilon R_d\left[f'(1)-f'(q^*)\right]
-  +\hat\epsilon\int_0^{q^*}dq\,f''(q)^{1/2}
-  +\frac12\hat\epsilon^2\int_{q^*}^1dq\,
-    (1-q)f''(q)
-    -\log\left[1-(1-q^*)\hat\epsilon/R_d\right]
-\end{align*}
-$R_d$ can be extremized now, with
-\[
-  R_d=\frac12\left(
-    (1-q^*)\hat\epsilon\pm\sqrt{
-      (1-q^*)\left(
-        (1-q^*)\hat\epsilon^2+8/[f'(1)-f'(q^*)]
-      \right)
-    }
-  \right)
-\]
-
-This all is for $\mu=\mu^*$, which counts the dominant saddles. We can also count by fixed macroscopic index $\mu$ by leaving it unoptimized in the complexity. This gives
-\[
-  F_d=\frac1{2f''(1)}\left[\mu\pm\sqrt{\mu^2-4f''(1)}\right]
-\]
-and
-\begin{align*}
-  \Sigma
-  =-\epsilon\hat\epsilon+
-    \frac12\hat\epsilon R_df'(1)
-    +\frac12\int_0^1dq\,\left[
-      \hat\epsilon^2\lambda(q)f''(q)
-      +\frac1{\lambda(q)+R_d/\hat\epsilon}
-    \right]-\mu R_d+\frac12R_d^2f''(1)+\log R_d\\
-    +\operatorname{Re}\left\{\frac12\left(1+\frac\mu{2f''(1)}\left(\mu\pm\sqrt{\mu^2-4f''(1)}\right)\right)-\log\left(\frac1{2f''(1)}\left(\mu\pm\sqrt{\mu^2-4f''(1)}\right)\right)\right\}
-\end{align*}
-
-
-\section{Ultrametricity rediscovered}
-
-Three states chosen at the same energy share some common information if there is some `frozen' element common to all. Suppose we choose randomly 
-these states but restrict to those whose overlaps
-take values $Q_{12}$ and $Q_{13}$. Unlike an equilibrium situation, where the Gibbs measure allows us to find such pairs (in a FRSB case) the cost in probability of this in the present case will be exponential.
-Once conditioned this way, we compute $Q_{23}= \min(Q_{12},Q_{13})$
+\end{appendix}
 
 
+\bibliographystyle{plain}
+\bibliography{frsb_kac-rice}
 
-{\tiny wild guess:
-Consider two states chosen at  different energies, corresponding to maximal values $q^{max}_1$, $q^{max}_2$. Their mutual overlap 
-should be (I am guessing this) $Q_{12}=\min(q^{max}_1, q^{max}_2)$
 
-}
-\section{Conclusion}
 
-\bibliographystyle{plain}
-\bibliography{frsb_kac-rice}
 
 \end{document}