path: root/frsb_kac-rice.tex

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\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, including most notably the (pure) $p$-spin model ($p>2$) \cite{Rieger_1992_The, Crisanti_1995_Thouless-Anderson-Palmer}. 
In a parallel development, it 
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,  which we expect to include the Sherrington--Kirkpatrick model. It reproduces  the Parisi result in the limit
of small temperature for the lowest states, as it should.

To understand the importance of this computation, consider the following situation. When one solves the problem of spheres in large dimensions, one finds that there is
a transition at a given temperature to a one-step one step symmetry breaking (1RSB) phase at a Kauzmann temperature,
and, at a lower temperature,
another  transition to  a full RSB phase (see \cite{gross1985mean,gardner1985spin}, the `Gardner ' phase \cite{charbonneau2014fractal}. 
Now, this transition involves the lowest, equilibrium states. Because they are obviously unreachable at any reasonable timescale, an often addressed question  to ask is "what is the Gardner transition line for higher than equilibrium energy-densities"? (see, for a review \cite{berthier2019gardner})  For example, when studying `jamming' at zero temperature, the question is posed as to "on what side of the 1RSB-FRS transition
are the high energy (or low density) states reachable dynamically. 
Posed in this way, such a question does not have a clear definition.
In the present paper we give a concrete strategy to define unambiguously such an issue: we consider the local energy minima at a given energy and study their number and other properties: the solution involves a replica-symmetry breaking scheme that is well-defined, and corresponds directly to the topological characteristics of those minima.



\section{The model}

Here we consider, for definiteness, the  mixed $p$-spin model, 
\begin{equation}
  H(s)=-\sum_p\frac1{p!}\sum_{i_1\cdots i_p}J^{(p)}_{i_1\cdots i_p}s_{i_1}\cdots s_{i_p}
\end{equation}
for $\overline{(J^{(p)})^2}=a_pp!/2N^{p-1}$ 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} 
or, more generally, the `Toy Model' of M\'ezard and Parisi \cite{Mezard_1992_Manifolds} which involve non-polynomial forms for $f(q)$.

These may 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 gradient and Hessian are
\begin{equation}
  \operatorname{Grad}H=\partial H+\mu z \qquad ; \qquad  \operatorname{Hess}H=\partial\partial H+\mu I
\end{equation}
The important observation was made by Bray and Dean  \cite{Bray_2007_Statistics} that gradient and Hessian
are independent for  random Gaussian disorder.
The average over disorder
breaks into a product of two independent averages, one for the gradient factor
and one for any function of the Hessian, in particular its number of negative eigenvalues, the index $\mathcal I$ of the saddle (see Fyodorov
\cite{Fyodorov_2007_Replica} for a detailed discussion).. 






\section{Equilibrium}

Here we review the equilibrium solution. \cite{Crisanti_1992_The, Crisanti_1993_The, Crisanti_2004_Spherical, Crisanti_2006_Spherical}
The free energy averaged over disorder is 
\begin{equation}
\beta F = - \overline{\ln \int ds \; e^{-\beta H(s)}}
\end{equation}
Computing the logarithm as the limit of $n \rightarrow 0$ replicas is standard, and it 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 computation is also given by a $(k-1)$-RSB anstaz. Heuristically, this is because
each stationary point also has no width and therefore overlap one with itself.



\section{Kac--Rice calculation}

\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 $E$, giving the number of stationary points at
energy $E$ in terms of a Lagrange multiplier $\mu$ as:
\begin{equation}
  \mathcal N(E, {\cal I}_o)
  =\int ds d\mu\,\delta(NE-H(s))\delta(\partial H(s)+\mu s)|\det(\partial\partial H(s)+\mu I)| \Theta\left[{\cal I}(s,\mu)-{\cal I}_o)\right]
\end{equation}
where $\Theta$ is one in the region in which the Hessian the argument is vanishes, and zero elsewhere. 
This number will typically be exponential in $N$. In order to find typical
counts when disorder is averaged, we will want to average its logarithm
instead, which is known as the averaged complexity:
\begin{equation}
  \Sigma(E,\mu)=\lim_{N\to\infty}\frac1N\overline{\log\mathcal N(E, {\cal I}_o)}
\end{equation}
If one averages over $\mathcal N$ and afterward takes its logarithm, one arrives at the so-called {\em annealed} complexity
\begin{equation}
  \Sigma_\mathrm a(E,{\cal I}_o)
  =\lim_{N\to\infty}\frac1N\log\overline{\mathcal N(E,{\cal I}_o))}
\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(E,\mu)
    %=-\frac{E^2(f'(1)+f''(1))+2E\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}.

A sometimes more illuminating quantity to consider is the Legendre transform $G$ of the complexity:
\begin{equation}
    e^{NG(\hat \beta, {\cal I}_o)} = \int dE \; e^{ -\hat \beta E +\Sigma(\hat \beta, {\cal I}_o)}
\end{equation}
There will be a critical value $\hat \beta_c$ beyond  which the complexity is zero: above
this value the measure is split between the lowest $O(1)$ energy states. We shall not study here this  regime that interpolates between the dynamically relevant and the equilibrium states, but just mention that
it is an interesting object of study.






\subsection{The replicated problem}



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(E,{\cal I}_o)
    &=\lim_{n\to0}\frac\partial{\partial n}\mathcal N^n(E,{\cal I}_o) \\
    &=\lim_{n\to0}\frac\partial{\partial n}\int\prod_a^n ds_a\,\delta(NE-H(s_a))\delta(\partial H(s_a)+\mu s_a)|\det(\partial\partial H(s_a)+\mu I)| \Theta\left[{\cal I}(s_a,\mu)-{\cal I}_o)\right]
  \end{aligned}
\end{equation}
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 practice, we are
therefore able to write
\begin{equation}
  \begin{aligned}
    \Sigma(E, {\cal I}_o)
    &=\lim_{N\to\infty}\frac1N\lim_{n\to0}\frac\partial{\partial n}\int\left(\prod_a^nds_a\right)\,\overline{\prod_a^n \delta(NE-H(s_a))\delta(\partial H(s_a)+\mu s_a)}
    \times
    \overline{\prod_a^n |\det(\partial\partial H(s_a)+\mu I)| \Theta\left[{\cal I}(s_a,\mu)-{\cal I}_o)\right]}
  \end{aligned}
\end{equation}

\subsubsection{The Hessian factors}

The spectrum of the Hessian matrix $\partial\partial H$ is in the large $N$ limit 
for almost every point and realization of disorder a GOE matrix with variance
\begin{equation}
  \overline{(\partial_i\partial_jH)^2}=\frac1Nf''(1)\delta_{ij}
\end{equation}
and therefore its spectrum is given by the Wigner semicircle with radius $\sqrt{4f''(1)}$, or
\begin{equation}
  \rho(\lambda)=\frac1{2\pi 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 $\mu$ thus fixes the spectrum of the Hessian, when $\mu$ is taken to be within
the range $\pm\sqrt{4f''(1)}=\pm\mu_m$, the critical (or in fact, any) points  have
index density
\begin{equation}
  \mathcal I(\mu)=\int_0^\infty d\lambda\,\rho(\lambda+\mu)
  =N\left\{\frac12-\frac1\pi\left[
    \arctan\left(\frac\mu{\sqrt{\mu_m^2-\mu^2}}\right)
    +\frac\mu{\mu_m^2}\sqrt{\mu_m^2-\mu^2}
  \right]
  \right\}
\end{equation}
When $\mu>\mu_m$, the critical
points are minima whose sloppiest eigenvalue is $\mu-\mu_m$.
The factor $\Theta[({\cal I}(s_a,\mu)-{\cal I}_0]$ selects a domain of integration of $\mu,s_a$.


To largest order in $N$, the average over the product of determinants factorizes into the product of averages, each of which is given by the same expression depending only on $\mu$:
\begin{equation}
  \begin{aligned}
  & \overline{\prod_a^n |\det(\partial\partial H(s_a)+\mu I)| \Theta\left[{\cal I}(s_a,\mu)-{\cal I}_o)\right]}\\
    \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}

\subsubsection{The gradient factors}

The $\delta$-functions are treated by writing them in the Fourier basis, introducing auxiliary fields $\hat s_a$ and $\hat\beta$,
\begin{equation}
  \prod_a^n\delta(NE-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(NE-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} \label{eq:fields}
  C_{ab}=\frac1Ns_a\cdot s_b &&
  R_{ab}=-i\frac1N\hat s_a\cdot s_b &&
  D_{ab}=\frac1N\hat s_a\cdot\hat s_b
\end{align}
Their physical meaning is explained in \S\ref{sec:interpretation}.
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(E,\mu)
    =\mathcal D(\mu)+\hat\beta E+\\
    &\lim_{n\to0}\frac1n\left(
      -\mu\operatorname{Tr}R
      +\frac12\sum_{ab}\left[
        \hat\beta^2f(C_{ab})+(2\hat\beta R_{ab}-D_{ab})f'(C_{ab})
        +R_{ab}^2f''(C_{ab})
      \right]
    +\frac12\log\det\begin{bmatrix}C&iR\\iR&D\end{bmatrix}
    \right)
  \end{aligned}
\end{equation}
where $\hat\beta$, $C$, $R$ and $D$ must be evaluated at extrema of this
expression.


{\color{blue}



The complexity is defined as 
\begin{equation}
   \Sigma(E,\mu)=  \frac 1N \log\mathcal N(E,\mu)
\end{equation}
The same information is contained, and better expressed in its double Legendre
transform $J(\hat \beta, R_d)$:
\begin{equation}
   e^{N J(\hat \beta, R_d)} =\int \; d\mu de \; e^{N\Sigma(E,\mu)+R_d\mu -\hat \beta E +{\cal{D}}(\mu)}
\end{equation}
$R_d$ is conjugate to $\mu$ and through it to the Index density, while $\hat \beta$ plays the role of an inverse temperature conjugate to the complexity, that has been used since the beginning of the spin-glass field.


}



\section{Replica ansatz}

Based on previous work on the SK model and the equilibrium solution of the
spherical model, we expect $C$, and $R$ and $D$ to be hierarchical matrices,
i.e., to follow Parisi's scheme. This assumption immediately simplifies the
extremal conditions, since hierarchical matrices commute and are closed under
matrix products and Hadamard products. The extremal conditions are
\begin{align}
  0&=\frac{\partial\Sigma}{\partial\hat\beta}
  =E+\sum_{ab}\left[\hat\beta f(C_{ab})+R_{ab}f'(C_{ab})\right] \label{eq:cond.b} \\
  \frac{\tilde c}2I&=\frac{\partial\Sigma}{\partial C}
  =\frac12\left[
    \hat\beta^2f'(C)+(2\hat\beta R-D)\odot f''(C)+R\odot R\odot f'''(C)
    +(CD+R^2)^{-1}D
  \right] \label{eq:cond.q} \\
  0&=\frac{\partial\Sigma}{\partial R}
  =-\mu I+\hat\beta f'(C)+R\odot f''(C)
  +(CD+R^2)^{-1}R \label{eq:cond.r} \\
  0&=\frac{\partial\Sigma}{\partial D}
  =-\frac12f'(C)
    +\frac12(CD+R^2)^{-1}C \label{eq:cond.d}
\end{align}
where $\odot$ denotes the Hadamard product, or the componentwise product. The
equation for \eqref{eq:cond.q} would not be true on the diagonal save for the
arbitrary factor $\tilde c$.  Equation \eqref{eq:cond.d} implies that
\begin{equation} \label{eq:D.solution}
  D=f'(C)^{-1}-RC^{-1}R
\end{equation}

In addition to these equations, one is also often interested in maximizing the complexity as a function of $\mu$, to find the dominant or most common type of stationary points. These are given by the condition
\begin{equation} \label{eq:cond.mu}
  0=\frac{\partial\Sigma}{\partial\mu}
  =\mathcal D'(\mu)-r_d
\end{equation}
Since $\mathcal D(\mu)$ is effectively a piecewise function, with different forms for $\mu$ greater or less than $\mu_m$, there are two regimes. When $\mu>\mu_m$, the critical points are minima, and \eqref{eq:cond.mu} implies
\begin{equation} \label{eq:mu.minima}
  \mu=\frac1{r_d}+r_df''(1)
\end{equation}
When $\mu<\mu_m$, they are saddles, and
\begin{equation} \label{eq:mu.saddles}
  \mu=2f''(1)r_d
\end{equation}

\section{Interpretation}
\label{sec:interpretation}

Let $\langle A\rangle$ be the average of $A$ over stationary points with given $E$ and $\mu$, i.e.,
\begin{equation}
  \langle A\rangle
  =\frac1{\mathcal N}\sum_{\sigma}A(s_\sigma)
  =\frac1{\mathcal N}
  \int d\nu(s)\,A(s)
\end{equation}
with
\begin{equation}
  d\nu(s)=ds\,\delta(NE-H(s))\delta(\partial H(s)+\mu s)|\det(\partial\partial H(s)+\mu I)|
\end{equation}
the Kac--Rice measure. The fields $C$, $R$, and $D$ defined in
\eqref{eq:fields} can be related to certain averages of this type.

\subsection{\textit{C}: distribution of overlaps}

First,
consider $C$, which has an interpretation nearly identical to that of Parisi's
$Q$ matrix of overlaps. It can be shown that its off-diagonal corresponds to
the probability distribution of the overlaps between stationary points $P(q)$. First, define this distribution as
\begin{equation}
  P(q)=\frac1{\mathcal N^2}\sum_{\sigma,\sigma'}\delta\left(\frac{s_\sigma\cdot s_{\sigma'}}N-q\right)
\end{equation}
where the sum is twice over stationary points $\sigma$ and $\sigma'$.  It is
straightforward to show that moments of this distribution are related to
certain averages of the form
\begin{equation}
  \int dq\,q^p P(q)
  =q^{(p)}
  \equiv\frac1{N^p}\sum_{i_1\cdots i_p}\langle s_{i_1}\cdots s_{i_p}\rangle\langle s_{i_1}\cdots s_{i_p}\rangle
\end{equation}
The appeal of Parisi to properties of pure states is unnecessary here, since
the stationary points are points. These moments are related to our $C$ by
computing their average over disorder:
\begin{equation}
  \begin{aligned}
    \overline{q^{(p)}}
    =\overline{\frac1{N^p}\sum_{i_1\cdots i_p}\langle s_{i_1}\cdots s_{i_p}\rangle\langle s_{i_1}\cdots s_{i_p}\rangle}
    =\lim_{n\to0}\overline{\int\prod_\alpha^nd\nu(s_\alpha)\,\frac1{n(n-1)}\sum_{a\neq b}\left(\frac{s_a\cdot s_b}N\right)^p} \\
    =\lim_{n\to0}\frac1{n(n-1)}\sum_{a\neq b}^nC^p_{ab}
    =\int_0^1 dx\,c^p(x)
  \end{aligned}
\end{equation}
where $C$ is assumed to take its saddle point value. If we now change variables in the integral from $x$ to $c$, we find
\begin{equation}
  \overline{q^{(p)}}=\int dc\,c^p\frac{dx}{dc}
\end{equation}
from which we conclude $\overline{P(q)}=\frac{dx}{dc}\big|_{c=q}$.

With this
established, we now address what it means for $C$ to have a nontrivial
replica-symmetry broken structure. When $C$ is replica symmetric, drawing two
stationary points at random will always lead to the same overlap. In the case
when there is no linear field, they will always have overlap zero, because the
second point will almost certainly lie on the equator of the sphere with
respect to the first. Though other stationary points exist nearby the first
one, they are exponentially fewer and so will be picked with vanishing
probability in the thermodynamic limit.

When $C$ is replica-symmetry broken, there is a nonzero probability of picking
a second stationary point at some other overlap. This can be interpreted by
imagining the level sets of the Hamiltonian in this scenario. If the level sets
are disconnected but there are exponentially many of them distributed on the
sphere, one will still find zero average overlap. However, if the disconnected
level sets are \emph{few}, i.e., less than order $N$, then it is possible to
draw two stationary points from the same set. Therefore, the picture in this
case is of few, large basins each containing exponentially many stationary
points.

\subsection{\textit{R} and \textit{D}: response functions}

The matrix field $R$ is related to responses of the stationary points to perturbations of the tensors $J$:
\begin{equation}
  \begin{aligned}
    \overline{\frac1{N^p}\sum_{i_1\cdots i_p}\frac{\partial\langle s_{i_1}\cdots s_{i_p}\rangle}{\partial J^{(p)}_{i_1\cdots i_p}}}
    =\lim_{n\to0}\overline{\int\prod_\alpha^nd\nu(s_\alpha)\,\frac1n\sum_{ab}\left[
      \hat\beta\left(\frac{s_a\cdot s_b}N\right)^p+
      p\left(-i\frac{\hat s_a\cdot s_b}N\right)\left(\frac{s_a\cdot s_b}N\right)^{p-1}
    \right]} \\
    =\lim_{n\to0}\frac1n\sum_{ab}(\hat\beta C_{ab}^p+pR_{ab}C_{ab}^{p-1})
    =\hat\beta+pr_d-\int_0^1dx\,c^{p-1}(x)(\hat\beta c(x)+pr(x))
  \end{aligned}
\end{equation}
In particular, when the energy is unconstrained ($\hat\beta=0$) and there is replica symmetry,
\begin{equation}
  \frac1N\sum_i\frac{\partial\langle s_i\rangle}{\partial J_i^{(1)}}
  =r_d
\end{equation}
i.e., adding a linear field causes a response in the average stationary point
location proportional to $r_d$. If positive, for instance, stationary points
tend to align with a field. The energy constraint has a significant
contribution due to the perturbation causing stationary points to move up or
down in energy.

The matrix field $D$ is related to the response of the complexity to such perturbations:
\begin{equation}
  \begin{aligned}
    \frac{\partial\Sigma}{\partial a_p}
    =\frac14\lim_{n\to0}\frac1n\sum_{ab}^n\left[
      \hat\beta^2C_{ab}^p+p(2\hat\beta R_{ab}-D_{ab})C_{ab}^{p-1}+p(p-1)R_{ab}^2C_{ab}^{p-2}
    \right]
  \end{aligned}
\end{equation}
In particular, when the energy is unconstrained ($\hat\beta=0$) and there is no replica symmetry breaking,
\begin{equation}
  \frac{\partial\Sigma}{\partial a_1}=-\frac14\lim_{n\to0}\frac1n\sum_{ab}D_{ab}=-\frac14d_d
\end{equation}
i.e., adding a random linear field decreases the complexity of solutions by an amount proportional to $d_d$ in the variance of the field.

When the saddle point of the Kac--Rice problem is supersymmetric,
\begin{equation}
  \frac{\partial\Sigma}{\partial a_p}
  =\frac{\hat\beta}4\overline{\frac1{N^p}\sum_{i_1\cdots i_p}\frac{\partial\langle s_{i_1}\cdots s_{i_p}\rangle}{\partial J^{(p)}_{i_1\cdots i_p}}}+\lim_{n\to0}\frac1n\sum_{ab}^np(p-1)R_{ab}^2C_{ab}^{p-2}
\end{equation}
and in particular for $p=1$
\begin{equation}
  \frac{\partial\Sigma}{a_1}
  =\frac{\hat\beta}4\overline{\frac1N\sum_i\frac{\partial\langle s_i\rangle}{\partial J_i^{(1)}}}
\end{equation}
i.e., the change in complexity due to a linear field is directly related to the
resulting magnetization of the stationary points.


\section{Supersymmetric solution}

The Kac--Rice problem has an approximate supersymmetry, which is found when the
absolute value of the determinant is neglected. When this is done, the
determinant can be represented by an integral over Grassmann variables, which
yields a complexity depending on `bosons' and `fermions' that share the
supersymmetry. The Ward identities associated with the supersymmetry imply
that $D=\hat\beta R$ \cite{Annibale_2003_The}. Under which conditions can this relationship be expected to hold?

Any result of supersymmetry can only be valid when the symmetry itself is
valid, which means the determinant must be positive. This is only guaranteed
for minima, which have $\mu>\mu_m$. Moreover, this identity heavily constrains
the form that the rest of the solution can take. Assuming the supersymmetry holds, \eqref{eq:cond.q} implies
\begin{equation}
  \tilde cI=\hat\beta^2f'(C)+\hat\beta R\odot f''(C)+R\odot R\odot f'''(C)+\hat\beta(CD+R^2)^{-1}R
\end{equation}
Substituting \eqref{eq:cond.r} for the factor $(CD+R^2)^{-1}R$, we find substantial cancellation, and finally
\begin{equation} \label{eq:R.diagonal}
  (\tilde c-\mu)I=R\odot R\odot f'''(C)
\end{equation}
If $C$ has a nontrivial off-diagonal structure and supersymmetry holds, then
the off-diagonal of $R$ must vanish, and therefore $R=r_dI$. Therefore, a
supersymmetric ansatz is equivalent to a \emph{diagonal} ansatz.

Supersymmetry has further implications.
Equations \eqref{eq:cond.r} and \eqref{eq:cond.d} can be combined to find
\begin{equation}
  I=R\left[\mu I-R\odot f''(C)\right]+(D-\hat\beta R)f'(C)
\end{equation}
Assuming the supersymmetry holds implies that
\begin{equation}
  I=R\left[\mu I-R\odot f''(C)\right]
\end{equation}
Understanding that $R$ is diagonal, this implies
\begin{equation}
  \mu=\frac1{r_d}+r_df''(1)
\end{equation}
which is precisely the condition \eqref{eq:mu.minima}. Therefore, \emph{the
supersymmetric solution only counts dominant minima.}

Inserting the supersymmetric ansatz $D=\hat\beta R$ and $R=r_dI$, one gets
\begin{equation} \label{eq:diagonal.action}
  \begin{aligned}
    \Sigma(E,\mu)
    =\mathcal D(\mu)
    +
      \hat\beta E-\mu r_d
      +\frac12\hat\beta r_df'(1)+\frac12r_d^2f''(1)+\frac12\log r_d^2
      \\
      +\frac12\lim_{n\to0}\frac1n\left(\hat\beta^2\sum_{ab}f(C_{ab})+\log\det((\hat\beta/r_d)C+I)\right)
  \end{aligned}
\end{equation}
From here, it is straightforward to see that the complexity vanishes at the
ground state energy. First, in the ground state minima will dominate (even if
they are marginal), so we may assume \eqref{eq:mu.minima}. Then, taking
$\Sigma(E_0,\mu^*)=0$, gives
\begin{equation}
  \hat\beta E_0
  =-\frac12r_d\hat\beta f'(1)-\frac12\lim_{n\to0}\frac1n\left(
      \hat\beta^2\sum_{ab}^nf(C_{ab})
      +\log\det(\hat\beta r_d^{-1} C+I)
  \right)
\end{equation}
which is precisely \eqref{eq:ground.state.free.energy} with $r_d=z$,
$\hat\beta=\tilde\beta$, and $C=\tilde Q$.

{\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 $\tilde c_1,\ldots,\tilde c_{k-1}$ 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 c_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}

\subsection{Full RSB}

This reasoning applies equally well to FRSB systems.  Using standard
manipulations (Appendix B), one finds also a continuous version of the
supersymmetric complexity
\begin{equation} \label{eq:functional.action}
    \Sigma(E,\mu)
    =\mathcal D(\mu)
    +
      \hat\beta E-\mu r_d
      +\frac12\hat\beta r_df'(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/\hat\beta}
      \right)
\end{equation}
where $\chi(q)=\int_1^qdq'\int_0^{q'}dq''\,P(q)$, as in the equilibrium case.
Though the supersymmetric solution leads to a nice tractable expression, it
turn out to be useful only at one point of interest: the ground state. Indeed,
we know from the equilibrium that in the ground state $\chi$ is continuous in
the whole range of $q$. Therefore, the saddle solution found by extremizing
\begin{equation}
  0=\frac{\delta\Sigma}{\delta\chi(q)}=\frac12\hat\beta^2f''(q)-\frac12\frac1{(\chi(q)+r_d/\hat\beta)^2}
\end{equation}
given by
\begin{equation}
  \chi(q)=\frac1{\hat\beta}\left(f''(q)^{-1/2}-r_d\right)
\end{equation}
is correct. This is only correct if it satisfies the boundary condition
$\chi(1)=0$, which requires $r_d=f''(1)^{-1/2}$. This in turn implies
$\mu=\frac1{r_d}+f''(1)r_d=\sqrt{4f''(1)}=\mu_m$. Therefore, the FRSB ground state
is exactly marginal! It is straightforward to check that these conditions are
indeed a saddle of the complexity.

This has several implications. First, other than the ground state, there are
\emph{no} energies at which minima are most numerous; saddles always dominante.
As we will see, stable minima are numerous at energies above the ground state,
but these vanish at the ground state.

\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}
  \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
  \includegraphics{figs/316_complexity_contour_1.pdf}
  \hfill
  \includegraphics{figs/316_complexity_contour_2.pdf}

  \caption{
    Complexity of the $3+16$ model in the energy $E$ and radial reaction $\mu$
    plane. The right shows a detail of the left. The black line shows $\mu_m$,
    which separates minima above from saddles below. The white lines show the
    dominant stationary points at each energy, dashed when they are described
    by a RS solution and solid for 1RSB. The red line shows the transition
    between RS and 1RSB descriptions. The gray line shows where the RS
    description predicts zero 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}







\begin{figure}
  \begin{center}
    \includegraphics[width=12cm]{twosys.png}\end{center}
  \caption{ Weakly coupling two different pure p-spin models we may infer the  phases for the joint model. A sketch of the individual complexities versus energy of each model,
  the dotted part of the lines corresponds to saddles, and starts at the threshold level.
  (1) Both systems frozen, lowest energy density.
(2) One system unfreezes, $1RSB$ Kac-Rice computation. (3) Both systems unfreeze, the gradients are the same in the typical (maximal complexity) values, thus corresponding to the RS Kac-Rice phase.
The beginning of this phase may start when the system on the right is either
on the region with minima or with saddles, depending on the whether the gradient of the complexity in the threshold of the system is higher or lower to the one of the ground state of the other.
(4) and (5) Typical states are saddles, both gradients are the same and hence this is RS. (6) The highest possible energy for minima. Because the gradients at the thresholds of the two systems do not coincide,
this has subdominant complexity
  } \label{twosys}
\end{figure}


\begin{figure}
  \begin{center}
    \includegraphics[width=12cm]{phase.png}\end{center}
  \caption{The phase diagram corresponding to two coupled models.
  } \label{phase}
\end{figure}

















\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 $\Delta E=E-E_0$ 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}


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. 
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. 

A first and very important application of the method here is to perform the calculation for high dimensional spheres, where it would give us
a clear understanding of what happens in a low-temperature realistic jamming dynamics \cite{maimbourg2016solution}. 

\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{Diagonal}



\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*}


\subsection{Non-diagonal}

\end{appendix}


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