From de48114dc57c85bd9bec94400d232133f452f294 Mon Sep 17 00:00:00 2001
From: Jaron Kent-Dobias <jaron@kent-dobias.com>
Date: Fri, 24 Jun 2022 14:39:30 +0200
Subject: Moved file back to original

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-\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.\\
-or Full solution for the counting of saddles of mean-field glass models}
-\author{Jaron Kent-Dobias \& Jorge Kurchan}
-\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 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, 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 (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 reproduces  the Parisi result in the limit
-of small temperature for the lowest states, as it should.
-
-
-\section{The model}
-
-Here we consider, for definiteness, the  mixed $p$-spin model, itself a particular case 
-of the `Toy Model' of M\'ezard and Parisi \cite{Mezard_1992_Manifolds}
-\begin{equation}
-  H(s)=\sum_p\frac{a_p^{1/2}}{p!}\sum_{i_1\cdots i_p}J_{i_1\cdots i_p}s_{i_1}\cdots s_{i_p}
-\end{equation}
-for $\overline{J^2}=p!/2N^{p-1}$. Then
-\begin{equation}
-  \overline{H(s_1)H(s_2)}=Nf\left(\frac{s_1\cdot s_2}N\right)
-\end{equation}
-for
-\begin{equation}
-  f(q)=\frac12\sum_pa_pq^p
-\end{equation}
-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(s\cdot s-N)
-\end{equation}
-
-At any critical point, the hessian is
-\begin{equation}
-  \operatorname{Hess}H=\partial\partial H+\mu I
-\end{equation}
-$\partial\partial H$ is a GOE matrix with variance
-\begin{equation}
-  \overline{(\partial_i\partial_jH)^2}=\frac1Nf''(1)\delta_{ij}
-\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{4f''(1)-\lambda^2}
-\end{equation}
-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?}
-
-In order to try to visualize what one should expect, consider two pure p-spin models, with 
-\begin{equation}
-    H = H_1 + H_2=\alpha_1 \sum_{ijk} J^1_{ijk} s_i s_j s_k +
-    \alpha_2 \sum_{ijk} J^2_{ijk} \bar s_i \bar s_j \bar s_k +\epsilon \sum_i s_i \bar s_i
-\end{equation}
-The complexity of the first and second systems in terms of $H_1$ and of $H_2$ 
-have, in the absence of coupling,  the same dependence, but are stretched to one another:
-\begin{equation}
-    \Sigma_1(H_1)= \Sigma_o(H_1/\alpha_1) \qquad ; \qquad \Sigma_2(H_2)= \Sigma_o(H_2/\alpha_2) 
-\end{equation}
-Each system has a ground state energy $E_{gs}^{1,2}$, a threshold energy $E_{thres}^{1,2}$ (a well-defined notion, since we are considering pure p-spins),  the corresponding limit values $X^{1,2}_{gs}=\left. \frac{d \Sigma_1}{dE_{1,2}}\right|_{E^{gs}_{12}}$
-and $X^{1,2}_{thres}=\left. \frac{d \Sigma_1}{dE_{1,2}}\right|_{E^{thres}_{12}}$
-Considering the cartesian product of both systems, we have, in terms of the total energy
-$H=H_1+H_2$ three regimes:
-\begin{itemize}
-\item  {\bf Unfrozen}: 
-\begin{eqnarray}
-& & X_1 \equiv \frac{d \Sigma_1}{dE_1}=  X_2 \equiv \frac{d \Sigma_2}{dE_2}
- \end{eqnarray}
-\item  {\bf Semi-frozen}
-As we go down in energy, one of the systems (say, the first) reaches its frozen phase,
- the first system is thus concentrated in a few states of $O(1)$ energy, while the second is not, so that $X_1=X_1^{gs}> X_2$. The lowest energy is reached when  systems are frozen.
-\item {\bf Semi-threshold }  As we go up from the unfrozen upwards in energy,
-the second system reaches its threshold $X_2^{thres}$. At higher energies minima are extremely rare, so the minima of the second system remain stuck at its threshold for higher energies.
-
-\item{\bf Both systems reach their thresholds} There essentially no more minima above that.
-\end{itemize}
-Consider now two combined vectors $({\bf s},{\bf \hat s})$ and $({\bf s}',{\bf \hat s}')$
-chosen at the same energies.\\
-
-$\bullet$ Their normalized overlap is close to one when both subsystems are frozen,
-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}
-
-
-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}
-    &\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)
-    +
-      \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\beta^2f''(q)\chi(q)+\frac1{\chi(q)+R_d^2/D_d}
-      \right)
-  \end{aligned}
-\end{equation}
-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}
-  \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}
-
-\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\beta=\lim_{\beta\to\infty}\beta x_k
-  &&
-  \tilde x_i=\lim_{\beta\to\infty}\frac{x_i}{x_k}
-  &&
-  \tilde q_i=\lim_{\beta\to\infty}q_i
-  &&
-  R_d=\lim_{\beta\to\infty}\beta(1-q_k)
-  &&
-  D_d=\hat\beta R_d
-\end{align}
-
-\section{Examples}
-
-\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
-\end{equation}
-$\log S_\infty=1+\log2\pi$.
-\begin{align*}
-  \beta F=
-  -\frac12\log S_\infty
-  -\frac12\lim_{n\to0}\frac1n\left(\beta^2nf(1)+\beta^2\sum_{i=0}^kn(x_i-x_{i+1})f(q_i)
-  +\log\left[
-    \frac{
-      1+\sum_{i=0}^k(x_i-x_{i+1})q_i
-    }{
-      1+\sum_{i=1}^k(x_i-x_{i+1})q_i-x_1q_0
-    }
-  \right]\right.\\
-  +\frac n{x_1}\log\left[
-      1+\sum_{i=1}^k(x_i-x_{i+1})q_i-x_1q_0
-  \right]\\
-  \left.+\sum_{j=1}^kn(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{align*}
-
-\begin{align*}
-  \lim_{n\to0}\frac1n
-  \log\left[
-    \frac{
-      1+\sum_{i=0}^k(x_i-x_{i+1})q_i
-    }{
-      1+\sum_{i=1}^k(x_i-x_{i+1})q_i-x_1q_0
-    }
-  \right]
-  &=
-  \lim_{n\to0}\frac1n
-  \log\left[
-    \frac{
-      1+\sum_{i=0}^k(x_i-x_{i+1})q_i
-    }{
-      1+\sum_{i=0}^k(x_i-x_{i+1})q_i-nq_0
-    }
-  \right] \\
-  &=q_0\left(1+\sum_{i=0}^k(x_i-x_{i+1})q_i\right)^{-1}
-\end{align*}
-
-
-\begin{align*}
-  \beta F=
-  -\frac12\log S_\infty
-  -\frac12\left(\beta^2f(1)+\beta^2\sum_{i=0}^k(x_i-x_{i+1})f(q_i)
-  +q_0\left(1+\sum_{i=0}^k(x_i-x_{i+1})q_i\right)^{-1}\right. \\
-  +\frac1{x_1}\log\left[
-      1+\sum_{i=1}^{k}(x_i-x_{i+1})q_i-x_1q_0
-  \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{align*}
-$q_0=0$
-\begin{align*}
-  \beta F=
-  -\frac12\log S_\infty
-  -\frac12\left(\beta^2f(1)+\beta^2\sum_{i=0}^k(x_i-x_{i+1})f(q_i)
-  +\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{align*}
-$x_i=\tilde x_ix_k$, $x_k=y/\beta$, $q_k=1-z/\beta$
-\begin{align*}
-  \beta F=
-  -\frac12\log S_\infty-
-  \frac12\left(\beta^2f(1)+\beta^2(y\beta^{-1}-1)f(1-z\beta^{-1})+y\beta\sum_{i=0}^{k-1}(\tilde x_i-\tilde x_{i+1})f(q_i)\right. \\
-  +\frac\beta{\tilde x_1 y}\log\left[
-    y\sum_{i=1}^{k-1}(\tilde x_i-\tilde x_{i+1})q_i+y+z-yz/\beta
-  \right]\\
-  +\sum_{j=1}^{k-1}\frac\beta y(\tilde x_{j+1}^{-1}-\tilde x_j^{-1})\log\left[
-    y\sum_{i=j+1}^{k-1}(\tilde x_i-\tilde x_{i+1})q_i+y+z-yz/\beta-y\tilde x_{j+1}q_j
-\right]\\
-    \left.-\frac\beta{\tilde x_1 y}\log\beta-\sum_{j=1}^{k-1}\frac\beta y(\tilde x_{j+1}^{-1}-\tilde x_j^{-1})\log\beta+(1-\beta y^{-1})\log\left[
-      z/\beta
-  \right]
-\right)
-\end{align*}
-\begin{align*}
-  \lim_{\beta\to\infty}F=
-  -\frac12\left(yf(1)+zf'(1)+y\sum_{i=0}^{k-1}(\tilde x_i-\tilde x_{i+1})f(q_i)
-  +\frac1{\tilde x_1 y}\log\left[
-    y\sum_{i=1}^{k-1}(\tilde x_i-\tilde x_{i+1})q_i+y+z
-  \right]\right.\\
-  \left.+\sum_{j=1}^{k-1}\frac1 y(\tilde x_{j+1}^{-1}-\tilde x_j^{-1})\log\left[
-    y\sum_{i=j+1}^{k-1}(\tilde x_i-\tilde x_{i+1})q_i+y+z-y\tilde x_{j+1}q_j
-    \right]
-    -\frac1y\log z
-\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 $\tilde x_k=1$.
-
-
-
-\section{ RSB for the Kac--Rice integral}
-
-
-\subsection{Solution}
-
-
-
-\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*}
-
-\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}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) =
-Q_{ab}  -i\left[\bar\theta_1\theta_1+\bar\theta_2\theta_2\right]  R_{ab}
- +(\bar\theta_1\theta_2+\theta_1\bar\theta_2)F_{ab}
- + \bar\theta_1\theta_1 \bar \theta_2 \theta_2 D_{ab} \\ 
-&+  \text{odd terms in the $\bar \theta,\theta$}~.
-\end{aligned}
-\label{Q12}
-\end{equation}
-\begin{equation}
-  \overline{\Sigma(\epsilon,\mu)}
-  =\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\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}
-The odd and even fermion numbers decouple, so we can neglect all odd terms in $\theta,\bar{\theta}$.
-
-\cite{Annibale_2004_Coexistence}
-
-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}.
-Not surprisingly, and for the same reason as in the quantum case, this ansatz closes, as we shall see.For example, consider the convolution:
-
-\begin{equation}
-  \begin{aligned}
-  \int d3\,\mathbb Q_1(1,3)\mathbb Q_2(3,2)
-  =\int d3\,(
-  Q_1  -i(\bar\theta_1\theta_1+\bar\theta_3\theta_3)  R_1
-   +(\bar\theta_1\theta_3+\theta_1\bar\theta_3)F_1
- + \bar\theta_1\theta_1 \bar \theta_3 \theta_3 D_1
- ) \\ (
-  Q_2  -i(\bar\theta_3\theta_3+\bar\theta_2\theta_2)  R_2
-   +(\bar\theta_3\theta_2+\theta_3\bar\theta_2)F_2
- + \bar\theta_3\theta_3 \bar \theta_2 \theta_2 D_2
- ) \\
- =-i(Q_1R_2+R_1Q_2)
- +Q_1D_2\bar\theta_2\theta_2+D_1Q_2\bar\theta_1\theta_1
- -i\bar\theta_1\theta_1\bar\theta_2\theta_2R_1D_2
- -i\bar\theta_1\theta_1\bar\theta_2\theta_2D_1R_2
- \end{aligned}
-\end{equation}
-\end{appendix}
-
-
-\bibliographystyle{plain}
-\bibliography{frsb_kac-rice}
-
-
-
-
-\end{document}
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