\documentclass[fleqn,a4paper]{article}

\usepackage[utf8]{inputenc} % why not type "Bézout" with unicode?
\usepackage[T1]{fontenc} % vector fonts plz
\usepackage{fullpage,amsmath,amssymb,latexsym,graphicx}
\usepackage{newtxtext,newtxmath} % Times for PR
\usepackage{appendix}
\usepackage[dvipsnames]{xcolor}
\usepackage[
  colorlinks=true,
  urlcolor=MidnightBlue,
  citecolor=MidnightBlue,
  filecolor=MidnightBlue,
  linkcolor=MidnightBlue
]{hyperref} % ref and cite links with pretty colors
\usepackage[
  style=phys,
  eprint=true,
  maxnames = 100
]{biblatex}
\usepackage{anyfontsize,authblk}

\usepackage{tikz}
\usetikzlibrary{calc,fadings,decorations.pathreplacing,calligraphy}

\newcommand\pgfmathsinandcos[3]{%
  \pgfmathsetmacro#1{sin(#3)}%
  \pgfmathsetmacro#2{cos(#3)}%
}
\newcommand\LongitudePlane[3][current plane]{%
  \pgfmathsinandcos\sinEl\cosEl{#2} % elevation
  \pgfmathsinandcos\sint\cost{#3} % azimuth
  \tikzset{#1/.style={cm={\cost,\sint*\sinEl,0,\cosEl,(0,0)}}}
}

\newcommand\LatitudePlane[3][current plane]{%
  \pgfmathsinandcos\sinEl\cosEl{#2} % elevation
  \pgfmathsinandcos\sint\cost{#3} % latitude
  \pgfmathsetmacro\yshift{\cosEl*\sint}
  \tikzset{#1/.style={cm={\cost,0,0,\cost*\sinEl,(0,\yshift)}}} %
}
\newcommand\NewLatitudePlane[4][current plane]{%
  \pgfmathsinandcos\sinEl\cosEl{#3} % elevation
  \pgfmathsinandcos\sint\cost{#4} % latitude
  \pgfmathsetmacro\yshift{#2*\cosEl*\sint}
  \tikzset{#1/.style={cm={\cost,0,0,\cost*\sinEl,(0,\yshift)}}} %
}
\newcommand\TangentPlane[5][current plane]{%
  \pgfmathsinandcos\sint\cost{#3} % elevation
  \pgfmathsinandcos\sinb\cosb{-#4} % latitude
  \pgfmathsinandcos\sina\cosa{#5+90} % azimuth
  \pgfmathsetmacro\xshift{\cosb*\sina}
  \pgfmathsetmacro\yshift{-\cost*\sinb-\cosa*\cosb*\sint}
  \tikzset{#1/.style={cm={-\sina*\sinb,\cosa*\sinb*\sint-\cost*\cosb,\cosa,\sina*\sint,(#2*\xshift,#2*\yshift)}}} %
}
\newcommand\DrawLongitudeCircle[2][1]{
  \LongitudePlane{\angEl}{#2}
  \tikzset{current plane/.prefix style={scale=#1}}
   % angle of "visibility"
  \pgfmathsetmacro\angVis{atan(sin(#2)*cos(\angEl)/sin(\angEl))} %
  \draw[current plane] (\angVis:1) arc (\angVis:\angVis+180:1);
  \draw[current plane,dashed] (\angVis-180:1) arc (\angVis-180:\angVis:1);
}
\newcommand\DrawLatitudeCircle[2][1]{
  \LatitudePlane{\angEl}{#2}
  \tikzset{current plane/.prefix style={scale=#1}}
  \pgfmathsetmacro\sinVis{sin(#2)/cos(#2)*sin(\angEl)/cos(\angEl)}
  % angle of "visibility"
  \pgfmathsetmacro\angVis{asin(min(1,max(\sinVis,-1)))}
  \draw[current plane] (\angVis:1) arc (\angVis:-\angVis-180:1);
  \draw[current plane,dashed] (180-\angVis:1) arc (180-\angVis:\angVis:1);
}

\tikzset{%
  >=latex, % option for nice arrows
  inner sep=0pt,%
  outer sep=2pt%
}

\addbibresource{2-point.bib}

\begin{document}

\title{
  Arrangement of nearby minima and saddles in the mixed spherical energy landscapes
}

\author{Jaron Kent-Dobias}
\affil{\textsc{DynSysMath}, Istituto Nazionale di Fisica Nucleare, Sezione di Roma}

\maketitle
\begin{abstract}
  The mixed spherical models were recently found to violate long-held
  assumptions about mean-field glassy dynamics. In particular, the threshold
  energy, where most stationary points are marginal and which in the simpler
  pure models attracts long-time dynamics, seems to lose significance. Here,
  we compute the typical distribution of stationary points relative to each
  other in mixed models with a replica symmetric complexity. We examine the
  stability of nearby points, accounting for the presence of an isolated
  eigenvalue in their spectrum due to their proximity. Despite finding rich
  structure not present in the pure models, we find nothing that distinguishes
  the points that do attract the dynamics. Instead, we find new geometric
  significance of the old threshold energy.
\end{abstract}

\tableofcontents

\section{Introduction}

\cite{Biroli_1999_Dynamical}
\cite{Folena_2020_Rethinking, Folena_2021_Gradient}

\cite{Ros_2020_Distribution, Ros_2019_Complex, Ros_2019_Complexity}

\section{Model}

\cite{Crisanti_1992_The, Crisanti_1993_The}

The mixed spherical models are defined by the Hamiltonian
\begin{equation} \label{eq:hamiltonian}
  H(\mathbf s)=-\sum_p\frac1{p!}\sum_{i_1\cdots i_p}^NJ^{(p)}_{i_1\cdots i_p}s_{i_1}\cdots s_{i_p}
\end{equation}
where the vectors $\mathbf s\in\mathbb R^N$ are confined to the sphere
$\|\mathbf s\|^2=N$.  The coupling coefficients $J$ are fully-connected and random, with
zero mean and variance $\overline{(J^{(p)})^2}=a_pp!/2N^{p-1}$ scaled so that
the energy is typically extensive. The overbar denotes an average
over the coefficients $J$. The factors $a_p$ in the variances are freely chosen
constants that define the particular model. For instance, the `pure'
$p$-spin model has $a_{p'}=\delta_{p'p}$. This class of models encompasses all
statistically isotropic Gaussian random Hamiltonians defined on the
hypersphere.

The covariance between the energy at two different points is a function of the overlap, or dot product, between those points, or
\begin{equation} \label{eq:covariance}
  \overline{H(\mathbf s_1)H(\mathbf s_2)}=Nf\left(\frac{\mathbf s_1\cdot\mathbf s_2}N\right)
\end{equation}
where the function $f$ is defined from the coefficients $a_p$ by
\begin{equation}
  f(q)=\frac12\sum_pa_pq^p
\end{equation}
In this paper, we will focus on models with a replica symmetric complexity, but
many of the intermediate formulae are valid for arbitrary replica symmetry
breakings. At most {\oldstylenums1}\textsc{rsb} in the equilibrium is guaranteed if the function
$\chi(q)=f''(q)^{-1/2}$ is convex. The complexity at the ground state must
reflect the structure of equilibrium, and therefore be replica symmetric. We
are not aware of any result guaranteeing this for the complexity away from the
ground state, but we check that our replica-symmetric solutions satisfy the
saddle point equations at 1RSB.

To enforce the spherical constraint at stationary points, we make use of a Lagrange multiplier $\omega$. This results in the extremal problem
\begin{equation}
  H(\mathbf s)+\frac\omega2(\|\mathbf s\|^2-N)
\end{equation}
The gradient and Hessian at a stationary point are then
\begin{align}
  \nabla H(\mathbf s,\omega)=\partial H(\mathbf s)+\omega\mathbf s
  &&
  \operatorname{Hess}H(\mathbf s,\omega)=\partial\partial H(\mathbf s)+\omega I
\end{align}
where $\partial=\frac\partial{\partial\mathbf s}$ will always denote the derivative with respect to $\mathbf s$.

\section{Results}

\begin{figure}
  \includegraphics{figs/gapped_min_energy.pdf}
  \raisebox{5em}{\includegraphics{figs/gapped_min_energy_legend.pdf}}
  \hfill
  \includegraphics{figs/gapped_min_stability.pdf}
  \raisebox{5em}{\includegraphics{figs/gapped_min_stability_legend.pdf}}

  \caption{
    The neighborhood of a reference minimum with $E_0=-1.71865<E_\mathrm{th}$
    and $\mu_0=6.1>\mu_\mathrm m$. \textbf{Left:} The most common type of
    stationary point lying at fixed overlap $q$ and energy $E_1$ from the
    reference minimum. The black line gives the smallest or largest energies
    where neighbors can be found at a given overlap. \textbf{Right:} The most
    common type of stationary point lying at fixed overlap $q$ and stability
    $\mu_1$ from the reference minimum. Note that this describes a different
    set of stationary points than shown in the left plot. On both plots, the
    shading of the righthand part depicts the state of an isolated eigenvalue
    in the spectrum of the Hessian of the neighboring points. Those more
    lightly shaded are minima with an isolated eigenvalue that does not change
    their stability, i.e., $\lambda_\mathrm{min}>0$. Those more darkly shaded
    are saddles with an isolated eigenvalue, either with many unstable
    directions ($\mu_1<\mu_\mathrm m$) or with only one, corresponding to minima
    destabilized by the isolated eigenvalue ($\mu_1>\mu_\mathrm m$). The
    dot-dashed lines on both plots depict the trajectory of the solid line on
    the other plot. In this case, the points lying nearest to the reference
    minimum are saddles with $\mu<\mu_\mathrm m$, but with energies smaller than
    the threshold energy.
  } \label{fig:min.neighborhood}
\end{figure}

\begin{figure}
  \centering
  \includegraphics{figs/franz_parisi.pdf}

  \caption{
    Comparison of the lowest-energy stationary points at overlap $q$ with a
    reference minimum of $E_0=-1.71865<E_\mathrm{th}$ and
    $\mu_0=6.1>\mu_\mathrm m$ (yellow, top), and the zero-temperature Franz--Parisi potential
    with respect to the same reference minimum (blue, bottom). The two curves
    coincide precisely at their minimum $q=0$ and at the local maximum $q\simeq0.5909$.
  } \label{fig:franz-parisi}
\end{figure}

\section{Complexity}

We introduce the Kac--Rice \cite{Kac_1943_On, Rice_1944_Mathematical} measure
\begin{equation}
  d\nu_H(\mathbf s,\omega)
  =2\,d\mathbf s\,d\omega\,\delta(\|\mathbf s\|^2-N)\,
  \delta\big(\nabla H(\mathbf s,\omega)\big)\,
  \big|\det\operatorname{Hess}H(\mathbf s,\omega)\big|
\end{equation}
which counts stationary points of the function $H$. More interesting is the
measure conditioned on the energy density $E$ and stability $\mu$ of the points,
\begin{equation}
  d\nu_H(\mathbf s,\omega\mid E,\mu)
  =d\nu_H(\mathbf s,\omega)\,
  \delta\big(NE-H(\mathbf s)\big)\,
  \delta\big(N\mu-\operatorname{Tr}\operatorname{Hess}H(\mathbf s,\omega)\big)
\end{equation}
While $\mu$ is strictly the trace of the Hessian, we call it the stability
because in this family of models all stationary points have a bulk spectrum of
the same shape, shifted by different constants. The stability $\mu$ sets this
shift, and therefore determines if the spectrum has bulk support on zero.

We want the typical number of stationary points with energy density
$E_1$ and stability $\mu_1$ that lie a fixed overlap $q$ from a reference
stationary point of energy density $E_0$ and stability $\mu_0$.
\begin{equation} \label{eq:complexity.definition}
  \Sigma_{12}
  =\frac1N\overline{\int\frac{d\nu_H(\pmb\sigma,\varsigma\mid E_0,\mu_0)}{\int d\nu_H(\pmb\sigma',\varsigma'\mid E_0,\mu_0)}\,
  \log\bigg(\int d\nu_H(\mathbf s,\omega\mid E_1,\mu_1)\,\delta(Nq-\pmb\sigma\cdot\mathbf s)\bigg)}
\end{equation}
Both the denominator and the logarithm are treated using the replica trick, which yields
\begin{equation}
  \Sigma_{12}
  =\frac1N\lim_{n\to0}\lim_{m\to0}\frac\partial{\partial n}\overline{\int\left(\prod_{b=1}^md\nu_H(\pmb\sigma_b,\varsigma_b\mid E_0,\mu_0)\right)\left(\prod_{a=1}^nd\nu_H(\mathbf s_a,\omega_a\mid E_1,\mu_1)\,\delta(Nq-\pmb \sigma_1\cdot \mathbf s_a)\right)}
\end{equation}
Note that because of the structure of \eqref{eq:complexity.definition},
$\pmb\sigma_1$ is special among the set of $\pmb\sigma$ replicas, since only it
is constrained to lie a given overlap from the $\mathbf s$ replicas. This
replica asymmetry will be important later.

\subsection{The Hessian factors}

The double partial derivatives of the energy are Gaussian with the variance
\begin{equation}
  \overline{(\partial_i\partial_jH(\mathbf s))^2}=\frac1Nf''(1)
\end{equation}
which means that the matrix of partial derivatives belongs to the GOE class. Its spectrum is given by the Wigner semicircle
\begin{equation}
  \rho(\lambda)=\begin{cases}
    \frac2{\pi}\sqrt{1-\big(\frac{\lambda}{\mu_\text m}\big)^2} & \lambda^2\leq\mu_\text m^2 \\
    0 & \text{otherwise}
  \end{cases}
\end{equation}
with radius $\mu_\text m=\sqrt{4f''(1)}$. Since the Hessian differs from the
matrix of partial derivatives by adding the constant diagonal matrix $\omega
I$, it follows that the spectrum of the Hessian is a Winger semicircle shifted
by $\omega$, or $\rho(\lambda+\omega)$.

The average over factors depending on the Hessian alone can be made separately
from those depending on the gradient or energy, since for random Gaussian
fields the Hessian is independent of these \cite{Bray_2007_Statistics}. In
principle the fact that we have conditioned the Hessian to belong to stationary
points of certain energy, stability, and proximity to another stationary point
will modify its statistics, but these changes will only appear at subleading
order in $N$ \cite{Ros_2019_Complexity}. At leading order, the various expectations factorize, each yielding
\begin{equation}
  \overline{\big|\det\operatorname{Hess}H(\mathbf s,\omega)\big|\,\delta\big(N\mu-\operatorname{Tr}\operatorname{Hess}H(\mathbf s,\omega)\big)}
  =e^{N\int d\lambda\,\rho(\lambda+\mu)\log|\lambda|}\delta(N\mu-N\omega)
\end{equation}
Therefore, all of the Lagrange multipliers are fixed identically to the stabilities $\mu$. We define the function
\begin{equation}
  \begin{aligned}
    \mathcal D(\mu)
    &=\int d\lambda\,\rho(\lambda+\mu)\log|\lambda| \\
    &=\begin{cases}
      \frac12+\log\left(\frac12\mu_\text m\right)+\frac{\mu^2}{\mu_\text m^2}
       & \mu^2\leq\mu_\text m^2 \\
      \frac12+\log\left(\frac12\mu_\text m\right)+\frac{\mu^2}{\mu_\text m^2}
      -\left|\frac{\mu}{\mu_\text m}\right|\sqrt{\big(\frac\mu{\mu_\text m}\big)^2-1}
      -\log\left(\left|\frac{\mu}{\mu_\text m}\right|-\sqrt{\big(\frac\mu{\mu_\text m}\big)^2-1}\right) & \mu^2>\mu_\text m^2
    \end{cases}
  \end{aligned}
\end{equation}
and the full factor due to the Hessians is
\begin{equation}
  e^{Nm\mathcal D(\mu_0)+Nn\mathcal D(\mu_1)}\left[\prod_a^m\delta(N\mu_0-N\varsigma_a)\right]\left[\prod_a^n\delta(N\mu_1-N\omega_a)\right]
\end{equation}

\subsection{The other factors}

Having integrated over the Lagrange multipliers using the $\delta$ functions
resulting from the average of the Hessians, the remaining part of the integrand
has the form
\begin{equation}
  e^{
    Nm\hat\beta_0E_0+Nn\hat\beta_1E_1
    -\sum_a^m\left[(\pmb\sigma_a\cdot\hat{\pmb\sigma}_a)\mu_0
      -\frac12\hat\mu_0(N-\pmb\sigma_a\cdot\pmb\sigma_a)
    \right]
      -\sum_a^n\left[(\mathbf s_a\cdot\hat{\mathbf s}_a)\mu_1
      -\frac12\hat\mu_1(N-\mathbf s_a\cdot\mathbf s_a)
      -\frac12\hat\mu_{12}(Nq-\pmb\sigma_1\cdot\mathbf s_a)
    \right]
    +\int d\mathbf t\,\mathcal O(\mathbf t)H(\mathbf t)
  }
\end{equation}
where we have introduced the linear operator
\begin{equation}
  \mathcal O(\mathbf t)
  =\sum_a^m\delta(\mathbf t-\pmb\sigma_a)\left(
    i\hat{\pmb\sigma}_a\cdot\partial_{\mathbf t}-\hat\beta_0
  \right)
  +
  \sum_a^n\delta(\mathbf t-\mathbf s_a)\left(
    i\hat{\mathbf s}_a\cdot\partial_{\mathbf t}-\hat\beta_1
  \right)
\end{equation}
We have written the $H$-dependant terms in this strange form for the ease of taking the average over $H$: since it is Gaussian-correlated, it follows that
\begin{equation}
  \overline{e^{\int d\mathbf t\,\mathcal O(\mathbf t)H(\mathbf t)}}
  =e^{\frac12\int d\mathbf t\,d\mathbf t'\,\mathcal O(\mathbf t)\mathcal O(\mathbf t')\overline{H(\mathbf t)H(\mathbf t')}}
  =e^{N\frac12\int d\mathbf t\,d\mathbf t'\,\mathcal O(\mathbf t)\mathcal O(\mathbf t')f\big(\frac{\mathbf t\cdot\mathbf t'}N\big)}
\end{equation}
It remains only to apply the doubled operators to $f$ and then evaluate the simple integrals over the $\delta$ measures. We do not include these details, which are standard.

\subsection{Hubbard--Stratonovich}

Having expanded this expression, we are left with an argument in the exponential which is a function of scalar products between the fields $\mathbf s$, $\hat{\mathbf s}$, $\pmb\sigma$, and $\hat{\pmb\sigma}$. We will change integration coordinates from these fields to matrix fields given by the scalar products, defined as
\begin{align} \label{eq:fields}
  C^{00}_{ab}=\frac1N\pmb\sigma_a\cdot\pmb\sigma_b &&
  R^{00}_{ab}=-i\frac1N{\pmb\sigma}_a\cdot\hat{\pmb\sigma}_b &&
  D^{00}_{ab}=\frac1N\hat{\pmb\sigma}_a\cdot\hat{\pmb\sigma}_b \\
  C^{01}_{ab}=\frac1N\pmb\sigma_a\cdot\mathbf s_b &&
  R^{01}_{ab}=-i\frac1N{\pmb\sigma}_a\cdot\hat{\mathbf s}_b &&
  R^{10}_{ab}=-i\frac1N\hat{\pmb\sigma}_a\cdot{\mathbf s}_b &&
  D^{01}_{ab}=\frac1N\hat{\pmb\sigma}_a\cdot\hat{\mathbf s}_b \\
  C^{11}_{ab}=\frac1N\mathbf s_a\cdot\mathbf s_b &&
  R^{11}_{ab}=-i\frac1N{\mathbf s}_a\cdot\hat{\mathbf s}_b &&
  D^{11}_{ab}=\frac1N\hat{\mathbf s}_a\cdot\hat{\mathbf s}_b
\end{align}
We insert into the integral the product of $\delta$ functions enforcing these
definitions, integrated over the new matrix fields, which is equivalent to
multiplying by one. Once this is done, the many scalar products appearing
throughout can be replaced by the matrix fields, and the original vector fields
can be integrated over. Conjugate matrix field integrals created when the
$\delta$ functions are promoted to exponentials can be evaluated by saddle
point in the standard way, yielding an effective action depending on the above
matrix fields alone.

We will always assume that the square matrices $C^{00}$, $R^{00}$, $D^{00}$,
$C^{11}$, $R^{11}$, and $D^{11}$ are hierarchical matrices, with each set of
three sharing the same hierarchical structure. In particular, we immediately
define $c_\mathrm d^{00}$, $r_\mathrm d^{00}$, $d_\mathrm d^{00}$, $c_\mathrm d^{11}$, $r_\mathrm d^{11}$, and
$d_\mathrm d^{11}$ as the value of the diagonal elements of these matrices,
respectively. Note that $c_\mathrm d^{00}=c_\mathrm d^{11}=1$ due to the spherical constraint.

Defining the `block' fields $\mathcal Q_{00}=(\hat\beta_0, \hat\mu_0, C^{00},
R^{00}, D^{00})$, $\mathcal Q_{11}=(\hat\beta_1, \hat\mu_1, C^{11}, R^{11},
D^{11})$, and $\mathcal Q_{01}=(\hat\mu_{01},C^{01},R^{01},R^{10},D^{01})$
the resulting complexity is
\begin{equation}
  \Sigma_{01}
  =\frac1N\lim_{n\to0}\lim_{m\to0}\frac\partial{\partial n}\int d\mathcal Q_{00}\,d\mathcal Q_{11}\,d\mathcal Q_{01}\,e^{Nm\mathcal S_0(\mathcal Q_{00})+Nn\mathcal S_1(\mathcal Q_{00},\mathcal Q_{11},\mathcal Q_{01})}
\end{equation}
where
\begin{equation}
  \begin{aligned}
    &\mathcal S_0(\mathcal Q_{00})
    =\hat\beta_0E_0-r^{00}_\mathrm d\mu_0-\frac12\hat\mu_0(1-c^{00}_\mathrm d)+\mathcal D(\mu_0)\\
    &\quad+\frac1m\bigg\{
      \frac12\sum_{ab}^m\left[
        \hat\beta_1^2f(C^{00}_{ab})+(2\hat\beta_1R^{00}_{ab}-D^{00}_{ab})f'(C^{00}_{ab})+(R_{ab}^{00})^2f''(C_{ab}^{00})
  \right]+\frac12\log\det\begin{bmatrix}C^{00}&R^{00}\\R^{00}&D^{00}\end{bmatrix}
\bigg\}
\end{aligned}
\end{equation}
is the action for the ordinary, one-point complexity, and remainder is given by
\begin{equation}
  \begin{aligned}
    &\mathcal S(\mathcal Q_{00},\mathcal Q_{11},\mathcal Q_{01})
    =\hat\beta_1E_1-r^{11}_\mathrm d\mu_1-\frac12\hat\mu_1(1-c^{11}_\mathrm d)+\mathcal D(\mu_1) \\
    &\quad+\frac1n\sum_b^n\left\{-\frac12\hat\mu_{12}(q-C^{01}_{1b})+\sum_a^m\left[
        \hat\beta_0\hat\beta_1f(C^{01}_{ab})+(\hat\beta_0R^{01}_{ab}+\hat\beta_1R^{10}_{ab}-D^{01}_{ab})f'(C^{01}_{ab})+R^{01}_{ab}R^{10}_{ab}f''(C^{01}_{ab})
    \right]\right\}
    \\
    &\quad+\frac1n\bigg\{
      \frac12\sum_{ab}^n\left[
        \hat\beta_1^2f(C^{11}_{ab})+(2\hat\beta_1R^{11}_{ab}-D^{11}_{ab})f'(C^{11}_{ab})+(R^{11}_{ab})^2f''(C^{11}_{ab})
      \right]\\
    &\quad+\frac12\log\det\left(
      \begin{bmatrix}
        C^{11}&iR^{11}\\iR^{11}&D^{11}
      \end{bmatrix}-
      \begin{bmatrix}
        C^{01}&iR^{01}\\iR^{10}&D^{01}
      \end{bmatrix}^T
      \begin{bmatrix}
        C^{00}&iR^{00}\\iR^{00}&D^{00}
      \end{bmatrix}^{-1}
      \begin{bmatrix}
        C^{01}&iR^{01}\\iR^{10}&D^{01}
      \end{bmatrix}
    \right)
  \bigg\}
  \end{aligned}
\end{equation}
Because of the structure of this problem in the twin limits of $m$ and $n$ to
zero, the parameters $\mathcal Q_{00}$ can be evaluated at a saddle point of
$\mathcal S_0$ alone. This means that these parameters will take the same value
they take when the ordinary, 1-point complexity is calculated. For a replica
symmetric complexity of the reference point, this results in
\begin{align}
  \hat\beta_0
  &=-\frac{(\epsilon_0+\mu_0)f'(1)+\epsilon_0f''(1)}{f(1)\big(f'(1)+f''(1)\big)-f'(1)^2}\\
  r_\mathrm d^{00}
  &=\frac{\mu_0f(1)+\epsilon_0f'(1)}{f(1)\big(f'(1)+f''(1)\big)-f'(1)^2} \\
  d_\mathrm d^{00}
  &=\frac1{f'(1)}
  -\left(
    \frac{\mu_0f(1)+\epsilon_0f'(1)}{f(1)\big(f'(1)+f''(1)\big)-f'(1)^2}
  \right)^2
\end{align}


In general, we except the $m\times n$ matrices $C^{01}$, $R^{01}$, $R^{10}$,
and $D^{01}$ to have constant \emph{rows} of length $n$, with blocks of rows
corresponding to the \textsc{rsb} structure of the single-point complexity. For
the scope of this paper, where we restrict ourselves to replica symmetric
complexities, they have the following form at the saddle point:
\begin{align}
  C^{01}=
  \begin{subarray}{l}
  \hphantom{[}\begin{array}{ccc}\leftarrow&n&\rightarrow\end{array}\hphantom{\Bigg]}\\
  \left[
    \begin{array}{ccc}
      q&\cdots&q\\
      0&\cdots&0\\
      \vdots&\ddots&\vdots\\
      0&\cdots&0
    \end{array}
  \right]\begin{array}{c}
    \\\uparrow\\m-1\\\downarrow
  \end{array}
\end{subarray}
  &&
  R^{01}
  =\begin{bmatrix}
    r_{01}&\cdots&r_{01}\\
    0&\cdots&0\\
    \vdots&\ddots&\vdots\\
    0&\cdots&0
  \end{bmatrix}
  &&
  R^{10}
  =\begin{bmatrix}
    r_{10}&\cdots&r_{10}\\
    0&\cdots&0\\
    \vdots&\ddots&\vdots\\
    0&\cdots&0
  \end{bmatrix}
  &&
  D^{01}
  =\begin{bmatrix}
    d_{01}&\cdots&d_{01}\\
    0&\cdots&0\\
    \vdots&\ddots&\vdots\\
    0&\cdots&0
  \end{bmatrix}
\end{align}
where only the first row is nonzero as a result of the sole linear term
proportional to $C_{1b}^{01}$ in the action.

The inverse of block hierarchical matrix is still a block hierarchical matrix, since
\begin{equation}
  \begin{bmatrix}
    C^{00}&iR^{00}\\iR^{00}&D^{00}
  \end{bmatrix}^{-1}
  =
  \begin{bmatrix}
    (C^{00}D^{00}+R^{00}R^{00})^{-1}D^{00} & -i(C^{00}D^{00}+R^{00}R^{00})^{-1}R^{00} \\
    -i(C^{00}D^{00}+R^{00}R^{00})^{-1}R^{00} & (C^{00}D^{00}+R^{00}R^{00})^{-1}C^{00}
  \end{bmatrix}
\end{equation}
Because of the structure of the 01 matrices, the volume element will depend only on the diagonal if this matrix. If we write
\begin{align}
  \tilde c_\mathrm d^{00}&=[(C^{00}D^{00}+R^{00}R^{00})^{-1}C^{00}]_{\text d} \\
  \tilde r_\mathrm d^{00}&=[(C^{00}D^{00}+R^{00}R^{00})^{-1}R^{00}]_{\text d} \\
  \tilde d_\mathrm d^{00}&=[(C^{00}D^{00}+R^{00}R^{00})^{-1}D^{00}]_{\text d}
\end{align}
then the result is
\begin{equation}
  \begin{aligned}
     &   \begin{bmatrix}
       C^{01}&iR^{01}\\iR^{10}&D^{01}
     \end{bmatrix}^T
     \begin{bmatrix}
       C^{00}&iR^{00}\\iR^{00}&D^{00}
     \end{bmatrix}^{-1}
     \begin{bmatrix}
       C^{01}&iR^{01}\\iR^{10}&D^{01}
     \end{bmatrix} \\
     &\qquad=\begin{bmatrix}
       q^2\tilde d_\mathrm d^{00}+2qr_{10}\tilde r^{00}_\mathrm d-r_{10}^2\tilde d^{00}_\mathrm d
      &
      i\left[d_{01}(r_{10}\tilde c^{00}_\mathrm d-q\tilde r^{00}_\mathrm d)+r_{01}(r_{10}\tilde r^{00}_\mathrm d+q\tilde d^{00}_\mathrm d)\right]
      \\
      i\left[d_{01}(r_{10}\tilde c^{00}_\mathrm d-q\tilde r^{00}_\mathrm d)+r_{01}(r_{10}\tilde r^{00}_\mathrm d+q\tilde d^{00}_\mathrm d)\right]
      &
      d_{01}^2\tilde c^{00}_\mathrm d+2r_{01}d_{01}\tilde r^{00}_\mathrm d-r_{01}^2\tilde d^{00}_\mathrm d
     \end{bmatrix}
  \end{aligned}
\end{equation}
where each block is a constant $n\times n$ matrix.


Because the matrices $C^{00}$, $R^{00}$, and $D^{00}$ are diagonal in this case, the diagonals of the inverse block matrix from above are simple expressions:
\begin{align}
  \tilde c_\mathrm d^{00}=f'(1) &&
  \tilde r_\mathrm d^{00}=r^{00}_\mathrm df'(1) &&
  \tilde d_\mathrm d^{00}=d^{00}_\mathrm df'(1)
\end{align}


\begin{equation}
  \begin{aligned}
    &\Sigma_{12}=\mathcal D(\mu_1)-\frac12+\hat\beta_1E_1-r^{11}_\mathrm d\mu_1
    +\hat\beta_1\big(r^{11}_\mathrm df'(1)-r^{11}_0f'(q^{11}_0)\big)
    +\hat\beta_0\hat\beta_1f(q)+(\hat\beta_0r^{01}+\hat\beta_1r^{10}+r^{00}_\mathrm d r^{01})f'(q)
    \\
    &+\frac{r^{11}_\mathrm d-r^{11}_0}{1-q^{11}_0}(r^{10}-qr^{00}_\mathrm d)f'(q)+
    \frac12\Bigg\{
      \hat\beta_1^2\big(f(1)-f(q^{11}_{0})\big)
      +(r^{11}_\mathrm d)^2f''(1)+2r^{01}r^{10}f''(q)-(r^{11}_0)^2f''(q^{11}_0)
      \\
    &+\left(
      (r^{01})^2-\frac{r^{11}_\mathrm d-r^{11}_0}{1-q^{11}_0}\left(2qr^{01}-\frac{(1-q^2)r^{11}_0-(q^{11}_0-q^2)r^{11}_\mathrm d}{1-q^{11}_0}\right)
    \right)\big(f'(1)-f'(q_{22}^{(0)})\big) \\
    &+\frac{1-q^2}{1-q^{11}_0}+\frac{(r^{10}-qr^{00}_\mathrm d)^2}{1-q^{11}_0}f'(1)
    -\frac1{f'(1)}\frac{f'(1)^2-f'(q)^2}{f'(1)-f'(q^{11}_0)}
    +\frac{r^{11}_\mathrm d-r^{11}_0}{1-q^{11}_0}\big(r^{11}_\mathrm df'(1)-r^{11}_0f'(q^{11}_0)\big)
    \\
    &+\log\left(\frac{1-q_{11}^0}{f'(1)-f'(q_{11}^0)}\right)
  \Bigg\}
  \end{aligned}
\end{equation}

\subsection{Most common neighbors with given overlap}

\begin{align}
  \hat\beta_1=0 &&
  \mu_1=2r^{11}_\mathrm df''(1)
\end{align}

\begin{equation}
  \Sigma_{12}=\frac{f'''(1)}{8f''(1)^2}(\mu_\mathrm m^2-\mu_0^2)\left(\sqrt{2+\frac{2f''(1)(f''(1)-f'(1))}{f'''(1)f'(1)}}-1\right)(1-q)
  +O\big((1-q)^2\big)
\end{equation}
\begin{equation}
  E_0^*=-\frac{f'(1)^2+f(1)\big(f''(1)-f'(1)\big)}{2f''(1)f'(1)}\mu_0
\end{equation}
\begin{equation}
  \mu_1=\mu_0-\frac{f'(1)(f'''(1)+f''(1))-f''(1)^2}{f(1)(f'(1)+f''(1))-f'(1)^2}(E_0-E_0^*)(1-q)+O\big((1-q)^2\big)
\end{equation}
\begin{equation}
  E_1=E_0+\frac12\frac{f'(1)(f'''(1)+f''(1))-f''(1)^2}{f(1)(f'(1)+f''(1))-f'(1)^2}(E_0-E_0^*)(1-q)^2+O\big((1-q)^3\big)
\end{equation}

\section{Franz--Parisi potential}

\cite{Franz_1995_Recipes}

\begin{equation} \label{eq:franz-parisi.definition}
  \beta V_\beta(q\mid E_0,\mu_0)
  =-\frac1N\overline{\int\frac{d\nu_H(\pmb\sigma,\varsigma\mid E_0,\mu_0)}{\int d\nu_H(\pmb\sigma',\varsigma'\mid E_0,\mu_0)}\,
  \log\bigg(\int d\mathbf s\,\delta\big(\|\mathbf s\|^2-N\big)\,\delta(\pmb\sigma\cdot\mathbf s-Nq)\,e^{-\beta H(\mathbf s)}\bigg)}
\end{equation}
Both the denominator and the logarithm are treated using the replica trick, which yields
\begin{equation}
  \beta V_\beta(q\mid E_0,\mu_0)
  =-\frac1N\lim_{\substack{m\to0\\n\to0}}\frac\partial{\partial n}\overline{\int\left(\prod_{b=1}^md\nu_H(\pmb\sigma_b,\varsigma_b\mid E_0,\mu_0)\right)\left(\prod_{a=1}^nd\mathbf s_a\,\delta(\|\mathbf s_a\|^2-N)\,\delta(\pmb \sigma_1\cdot \mathbf s_a-Nq)\,e^{-\beta H(\mathbf s_a)}\right)}
\end{equation}
\begin{equation}
  \mathcal O(\mathbf t)
  =\sum_a^m\delta(\mathbf t-\pmb\sigma_a)\left(
    i\hat{\pmb\sigma}_a\cdot\partial_{\mathbf t}-\hat\beta_0
    \right)
    -\beta
    \sum_a^n\delta(\mathbf t-\mathbf s_a)
\end{equation}
\begin{equation}
  \beta V_\beta(q\mid E_0,\mu_0)=-\frac1N\lim_{\substack{m\to0\\n\to0}}\frac\partial{\partial n}\int d\mathcal Q_0\,d\mathcal Q_1\,e^{Nm\mathcal S_0(\mathcal Q_0)+Nn\mathcal S_\mathrm{FP}(\mathcal Q_1)}
\end{equation}

\begin{equation}
  n\mathcal S_{\mathrm{FP}}
  =\frac12\beta^2\sum_{ab}^nf(Q_{ab})
  +\beta\sum_a^m\sum_b^n\left[
    \hat\beta_0f(C^{01}_{ab})
    +R^{10}_{ab}f'(C^{01}_{ab})
  \right]
  +\frac12\log\det\left(
    Q-\begin{bmatrix}C^{01}\\iR^{10}\end{bmatrix}^T\begin{bmatrix}C^{00}&iR^{00}\\iR^{00}&D^{00}\end{bmatrix}^{-1}\begin{bmatrix}C^{01}\\iR^{10}\end{bmatrix}
    \right)
\end{equation}

\begin{equation}
  \begin{aligned}
    \beta V_\beta&=\frac12\beta^2\big[f(1)-(1-x)f(q_1)-xf(q_0)\big]
    +\beta\hat\beta_0f(q)+\beta r^{10}f'(q)-\frac{1-x}x\log(1-q_1)
    +\frac1x\log(1-(1-x)q_1-xq_0) \\
     &+\frac{q_0-d^{00}_df'(1)q^2-2r^{00}_df'(1)r^{10}q+(r^{10})^2f'(1)}{
      1-(1-x)q_1-xq_0
    }
  \end{aligned}
\end{equation}
The saddle point for $r^{10}$ can be taken explicitly. After this, we take the
limit of $\beta\to\infty$. There are two possibilities. First, in the replica
symmetric case $x=1$, and in the limit of large $\beta$ $q_0$ will scale like
$q_0=1-(y_0\beta)^{-1}$. Inserting this, the limit is
\begin{equation}
  V_\infty^{\textsc{rs}}=-\hat\beta_0 f(q)-r^{11}_\mathrm df'(q)q-\frac12\left(y_0(1-q^2)+\frac{f'(1)^2-f'(q)^2}{y_0f'(1)}\right)
\end{equation}
The saddle point in $y_0$ can now be taken, taking care to choose the solution for $y_0>0$. This gives
\begin{equation}
  V_\infty^{\textsc{rs}}=-\hat\beta_0f(q)-r^{11}_\mathrm df'(q)q-\sqrt{(1-q^2)\left(1-\frac{f'(q)^2}{f'(1)^2}\right)}
\end{equation}
The second case is when the inner statistical mechanics problem has replica symmetry breaking. Here, $q_0$ approaches a nontrivial limit, but $x=z\beta^{-1}$ approaches zero and $q_1=1-(y_1\beta)^{-1}$ approaches one.
\begin{equation}
  \begin{aligned}
    V_\infty^{\oldstylenums{1}\textsc{rsb}}(q\mid E_0,\mu_0)
    &=-\hat\beta_0f(q)-r^{11}_\mathrm df'(q)q-\frac12\bigg(
      z(f(1)-f(q_0))+\frac{f'(1)}{y_1}-\frac{y_1(q^2-q_0)}{1+y_1z(1-q_0)} \\
    &\hspace{8pc}-(1+y_1z(1-q_0))\frac{f'(q)^2}{y_1f'(1)}+\frac1z\log\left(1+zy_1(1-q_0)\right)
    \bigg)
  \end{aligned}
\end{equation}
Though the saddle point in $y_1$ can be evaluated in this expression, it
delivers no insight. The final potential is found by taking the saddle over
$z$, $y_1$, and $q_0$.

\section{Isolated eigenvalue}

The two-point complexity depends on the spectrum at both stationary points
through the determinant of their Hessians, but only on the bulk of the
distribution. As we saw, this bulk is unaffected by the conditions of energy
and proximity. However, these conditions give rise to small-rank perturbations
to the Hessian, which can lead a subextensive number of eigenvalues leaving the
bulk. We study the possibility of \emph{one} stray eigenvalue.

We use a technique recently developed to find the smallest eigenvalue of a
random matrix \cite{Ikeda_2023_Bose-Einstein-like}. One defines a quadratic
statistical mechanics model with configurations defined on the sphere, whose
interaction tensor is given by the matrix of interest. By construction, the
ground state is located in the direction of the eigenvector associated with the
smallest eigenvalue, and the ground state energy is proportional to that
eigenvalue.

Our matrix of interest is the Hessian evaluated at a stationary point of the mixed spherical
model, conditioned on the relative position, energies, and stabilities
discussed above. We must restrict the artificial spherical model to lie in the
tangent plane of the `real' spherical configuration space at the point of
interest, to avoid our eigenvector pointing in a direction that violates the
spherical constraint. The free energy of this model given a point $\mathbf s$
and a specific realization of the disordered Hamiltonian is
\begin{equation}
  \begin{aligned}
    \beta F_H(\beta\mid\mathbf s,\omega)
    &=-\frac1N\log\left(\int d\mathbf x\,\delta(\mathbf x\cdot\mathbf s)\delta(\|\mathbf x\|^2-N)\exp\left\{
        -\beta\frac12\mathbf x^T\operatorname{Hess}H(\mathbf s,\omega)\mathbf x
    \right\}\right) \\
    &=-\lim_{\ell\to0}\frac1N\frac\partial{\partial\ell}\int\left[\prod_{\alpha=1}^\ell d\mathbf x_\alpha\,\delta(\mathbf x_\alpha^T\mathbf s)\delta(N-\mathbf x_\alpha^T\mathbf x_\alpha)\exp\left\{
      -\beta\frac12\mathbf x^T_\alpha\big(\partial\partial H(\mathbf s)+\omega I\big)\mathbf x_\alpha
    \right\}\right]
  \end{aligned}
\end{equation}
where the first $\delta$-function keeps the configurations in the tangent
plane, and the second enforces the spherical constraint. We have anticipated
treating the logarithm with replicas. We are of course interested in
points $\mathbf s$ that have certain properties: they are stationary points of
$H$ with given energy density and stability, and fixed overlap from a reference
configuration $\pmb\sigma$. We therefore average the free energy above over
such points, giving
\begin{equation}
  \begin{aligned}
    F_H(\beta\mid E_1,\mu_1,q,\pmb\sigma)
    &=\int\frac{d\nu_H(\mathbf s,\omega\mid E_1,\mu_1)\delta(Nq-\pmb\sigma\cdot\mathbf s)}{\int d\nu_H(\mathbf s',\omega'\mid E_1,\mu_1)\delta(Nq-\pmb\sigma\cdot\mathbf s')}F_H(\beta\mid\mathbf s,\omega) \\
    &=\lim_{n\to0}\int\left[\prod_{a=1}^nd\nu_H(\mathbf s_a,\omega_a\mid E_1,\mu_1)\,\delta(Nq-\pmb\sigma\cdot\mathbf s_a)\right]F_H(\beta\mid\mathbf s_1,\omega_1)
  \end{aligned}
\end{equation}
again anticipating the use of replicas. Finally, the reference configuration $\pmb\sigma$ should itself be a stationary point of $H$ with its own energy density and stability. Averaging over these conditions gives
\begin{equation}
  \begin{aligned}
    F_H(\beta\mid\epsilon_1,\mu_1,\epsilon_2,\mu_2,q)
    &=\int\frac{d\nu_H(\pmb\sigma,\varsigma\mid E_0,\mu_0)}{\int d\nu_H(\pmb\sigma',\varsigma'\mid E_0,\mu_0)}\,F_H(\beta\mid E_1,\mu_1,q,\pmb\sigma) \\
    &=\lim_{m\to0}\int\left[\prod_{a=1}^m d\nu_H(\pmb\sigma_a,\varsigma_a\mid E_0,\mu_0)\right]\,F_H(\beta\mid E_1,\mu_1,q,\pmb\sigma_1)
  \end{aligned}
\end{equation}
This formidable expression is now ready to be averaged over the disordered Hamiltonians $H$. Once averaged,
the minimum eigenvalue of the conditioned Hessian is then given by twice the ground state energy, or
\begin{equation}
  \lambda_\text{min}=2\lim_{\beta\to\infty}\overline{F_H(\beta\mid\epsilon_1,\mu_1,\epsilon_2,\mu_2,q)}
\end{equation}
For this calculation, there are three different sets of replicated variables.
Note that, as for the computation of the complexity, the $\pmb\sigma_1$ and
$\mathbf s_1$ replicas are \emph{special}. The first again is the only of the
$\sigma$ replicas constrained to lie at fixed overlap with \emph{all} the
$\mathbf s$ replicas, and the second is the only of the $\mathbf s$ replicas at
which the Hessian is evaluated.

\begin{figure}
  \centering
  \begin{tikzpicture}
    \def\R{4 } % sphere radius
    \def\Rt{2 } % tangent plane radius
    \def\angEl{15} % elevation angle
    \def\angsa{-160} % azimuth of s_1
    \def\angq{40} % elevation of constraint circle
    \filldraw[ball color=white] (0,0) circle (\R);
  %  \filldraw[fill=white] (0,0) circle (\R);

    \foreach \t in {0,\angq} { \DrawLatitudeCircle[\R]{\t} }
    %\foreach \t in {\angsa} { \DrawLongitudeCircle[\R]{\t} }

    \pgfmathsetmacro\H{\R*cos(\angEl)} % distance to north pole
    \coordinate (O) at (0,0);
    \node[circle,draw,black,scale=0.3] at (0,0) {};
    \coordinate (N) at (0,\H);
    \draw node[right=10,below] at (0,\H){$\pmb\sigma_1$};
    \draw[thick, ->](O)--(N);

    \NewLatitudePlane[planeP]{\R}{\angEl}{\angq};
    \path[planeP] (\angsa:\R) coordinate (P);
    \path[planeP] (0:1.5*\R) coordinate (Q);
    \path[planeP] (0:\R) coordinate (Q2);
    \draw[left] node at (P){$\mathbf s_1$};

    \NewLatitudePlane[equator]{\R}{\angEl}{00};
    \path[equator] (-30:\R) coordinate (Pprime);
    \path[equator] (0:{1.5*cos(\angq)*\R}) coordinate (Qe);
    \path[equator] (0:\R) coordinate (Qe2);
    \draw node[right=5,below] at (Pprime){$\pmb\sigma_c$};

    \NewLatitudePlane[sbplane]{\R}{\angEl}{\angq};
    \path[sbplane] (20:\R) coordinate (sb);
    \draw node[right=3,above=1] at (sb){$\mathbf s_b$};

    \TangentPlane[tplane]{\R}{\angEl}{\angq}{\angsa};
    \draw[tplane,fill=gray,fill opacity=0.3] circle (\Rt);
    \draw[tplane,->,thick] (0,0) -> ({\Rt*cos(160)},{\Rt*sin(160)}) node[above=1.5,right] {$\mathbf x_a$};
    \draw[tplane,->,thick] (0,0) -> ({\Rt*cos(250)},{\Rt*sin(250)}) node[above=1.5,left=0.1] {$\mathbf x_b$};

    \draw[thick, ->] (O)->(P);
    \draw[thick, ->] (O)->(Pprime);
    \draw[thick, ->] (O)->(sb);

    \draw[dotted] (Qe) -- (Qe2);
    \draw[dotted] (Q2) -- (Q);
    \draw[decorate, decoration = {brace,raise=3}] (Q) -- (Qe) node[pos=0.5,right=7]{$q$};
  \end{tikzpicture}
  \caption{
    A sketch of the vectors involved in the calculation of the isolated
    eigenvalue. All replicas $\mathbf x$ sit in an $N-2$ sphere corresponding
    with the tangent plane (not to scale) of the first $\mathbf s$ replica. All of the
    $\mathbf s$ replicas lie on the sphere, constrained to be at fixed overlap
    $q$ with the first of the $\pmb\sigma$ replicas, the reference
    configuration. All of the $\pmb\sigma$ replicas lie on the sphere.
  }
\end{figure}

Using the same methodology as above, the disorder-dependant terms are captured in the linear operator
\begin{equation}
  \mathcal O(\mathbf t)=
  \sum_a^m\delta(\mathbf t-\pmb\sigma_a)(i\hat{\pmb\sigma}_a\cdot\partial_\mathbf t-\hat\beta_0)
  +
  \sum_b^n\delta(\mathbf t-\mathbf s_b)(i\hat{\mathbf s}_b\cdot\partial_\mathbf t-\hat\beta_1)
  -\frac12
  \delta(\mathbf t-\mathbf s_1)\beta\sum_c^\ell(\mathbf x_c\cdot\partial_{\mathbf t})^2
\end{equation}
\begin{align}
  A_{ab}=\frac1N\mathbf x_a\cdot\mathbf x_b
  &&
  X^0_{ab}=\frac1N\pmb\sigma_a\cdot\mathbf x_b
  &&
  \hat X^0_{ab}=-i\frac1N\hat{\pmb\sigma}_a\cdot\mathbf x_b
  &&
  X^1_{ab}=\frac1N\mathbf s_a\cdot\mathbf x_b
  &&
  \hat X^1_{ab}=-i\frac1N\hat{\mathbf s}_a\cdot\mathbf x_b
\end{align}

\begin{equation}
  \lambda_\mathrm{min}
  =-2\lim_{\beta\to\infty}
  \lim_{\substack{\ell\to0\\m\to0\\n\to0}}\frac\partial{\partial\ell}\frac1{\beta N}
  \int d\mathcal Q\,d\mathcal X\,
  e^{N[
    m\mathcal S_0(\mathcal Q_{00})
    +n\mathcal S_1(\mathcal Q_{11},\mathcal Q_{01}\mid\mathcal Q_{00})
    +\ell\mathcal S_x(\mathcal X\mid\mathcal Q_{00},\mathcal Q_{01},\mathcal Q_{11})
  ]}
\end{equation}

\begin{equation}
  \begin{aligned}
    \ell\mathcal S_x(\mathcal X\mid\mathcal Q)
    =-\frac12\beta\mu+
    \frac12\beta\sum_b^\ell\bigg\{
      \frac12\beta&f''(1)\sum_a^lA_{ab}^2\\
    &+\sum_a^m\left[
        \big(\hat\beta_0f''(C^{01}_{a1})+R^{10}_{a1}f'''(C^{01}_{a1})\big)(X^0_{ab})^2
        +2f''(C^{01}_{a1})X^0_{ab}\hat X^0_{ab}
      \right] \\
    &+\sum_a^n\left[
        \big(\hat\beta_1f''(C^{11}_{a1})+R^{11}_{a1}f'''(C^{11}_{a1})\big)(X^1_{ab})^2
        +2f''(C^{11}_{a1})X^1_{ab}\hat X^1_{ab}
      \right]
    \bigg\}\\
    &+\frac12\log\det\left(
      A-
      \begin{bmatrix}
        X^0\\\hat X^0\\X^1\\\hat X^1
      \end{bmatrix}^T
      \begin{bmatrix}
        C^{00}&iR^{00}&C^{01}&iR^{01}\\
        iR^{00}&D^{00}&iR^{10}&D^{01}\\
        (C^{01})^T&(iR^{10})^T&C^{11}&iR^{11}\\
        (iR^{01})^T&(D^{01})^T&iR^{11}&D^{11}\\
      \end{bmatrix}^{-1}
      \begin{bmatrix}
        X^0\\\hat X^0\\X^1\\\hat X^1
      \end{bmatrix}
    \right)
  \end{aligned}
\end{equation}

\begin{align}
  X^0
  =
  \begin{subarray}{l}
    \hphantom{[}\begin{array}{ccc}\leftarrow&\ell&\rightarrow\end{array}\hphantom{\Bigg]}\\
  \left[
    \begin{array}{ccc}
      x_0&\cdots&x_0\\
      0&\cdots&0\\
      \vdots&\ddots&\vdots\\
      0&\cdots&0
    \end{array}
  \right]\begin{array}{c}
    \\\uparrow\\m-1\\\downarrow
  \end{array}\\
  \vphantom{\begin{array}{c}n\end{array}}
  \end{subarray}
  &&
  \hat X^0
  =
  \left[
    \begin{array}{ccc}
      \hat x_0&\cdots&\hat x_0\\
      0&\cdots&0\\
      \vdots&\ddots&\vdots\\
      0&\cdots&0
    \end{array}
  \right]
  &&
  X^1
  =
  \begin{subarray}{l}
    \hphantom{[}\begin{array}{ccc}\leftarrow&\ell&\rightarrow\end{array}\hphantom{\Bigg]}\\
  \left[
    \begin{array}{ccc}
      0&\cdots&0\\
      x_1&\cdots&x_1\\
      \vdots&\ddots&\vdots\\
      x_1&\cdots&x_1
    \end{array}
  \right]\begin{array}{c}
    \\\uparrow\\n-1\\\downarrow
  \end{array}\\
  \vphantom{\begin{array}{c}n\end{array}}
  \end{subarray}
  &&
  \hat X^1
  =\begin{bmatrix}
    \hat x_1^0&\cdots&\hat x_1^0\\
    \hat x_1^1&\cdots&\hat x_1^1\\
    \vdots&\ddots&\vdots\\
    \hat x_1^1&\cdots&\hat x_1^1
  \end{bmatrix}
\end{align}
\begin{equation}
  \begin{aligned}
    \frac2\beta\lim_{\ell\to0}\mathcal S_x(\mathcal X\mid\mathcal Q)
    &=
      -\mu+\frac12\beta f''(1)(1-a_0^2)
    +\big(\hat\beta_0f''(q)+r_{10}f'''(q)\big)x_0^2
      +2f''(q)x_0\hat x_0 \\
    &-
      \big(\hat\beta_1f''(q^{11}_0)+r^{11}_0f'''(q^{11}_0)\big)x_1^2
      -2f''(q^{11}_0)x_1\hat x_1^1 \\
    &+\lim_{\ell\to0}\frac1\ell\frac1\beta\log\det\left(
      A-
      \begin{bmatrix}
        X^0\\\hat X^0\\X^1\\\hat X^1
      \end{bmatrix}^T
      \begin{bmatrix}
        C^{00}&iR^{00}&C^{01}&iR^{01}\\
        iR^{00}&D^{00}&iR^{10}&D^{01}\\
        (C^{01})^T&(iR^{10})^T&C^{11}&iR^{11}\\
        (iR^{01})^T&(D^{01})^T&iR^{11}&D^{11}\\
      \end{bmatrix}^{-1}
      \begin{bmatrix}
        X^0\\\hat X^0\\X^1\\\hat X^1
      \end{bmatrix}
    \right)
  \end{aligned}
\end{equation}

\begin{equation}
    \begin{bmatrix}
      C^{00}&iR^{00}&C^{01}&iR^{01}\\
      iR^{00}&D^{00}&iR^{10}&D^{01}\\
      (C^{01})^T&(iR^{10})^T&C^{11}&iR^{11}\\
      (iR^{01})^T&(D^{10})^T&iR^{11}&D^{11}\\
    \end{bmatrix}^{-1}
    =
  \begin{bmatrix}
    A & B \\
    C & D
  \end{bmatrix}
\end{equation}
\begin{equation}
    A=
    \left(
      \begin{bmatrix}
        C^{00}&iR^{00}\\iR^{00}&D^{00}
      \end{bmatrix}
      -
      \begin{bmatrix}
        C^{01}&iR^{01}\\
        iR^{10}&D^{01}
      \end{bmatrix}
      \begin{bmatrix}
        C^{11}&iR^{11}\\iR^{11}&D^{11}
      \end{bmatrix}^{-1}
      \begin{bmatrix}
        C^{01}&iR^{01}\\
        iR^{10}&D^{01}
      \end{bmatrix}^T
    \right)^{-1}
\end{equation}
\begin{align}
  \hat c^{11}=\sum_{ij}^n[(C^{11}D^{11}+R^{11}R^{11})^{-1}D^{11}]_{ij}\\
  \hat r^{11}=\sum_{ij}^n[(C^{11}D^{11}+R^{11}R^{11})^{-1}R^{11}]_{ij}\\
  \hat d^{11}=\sum_{ij}^n[(C^{11}D^{11}+R^{11}R^{11})^{-1}D^{11}]_{ij}
\end{align}
Based on the structure of the 01 matrices established above, the second term inside the inverse is
\begin{equation}
  \begin{aligned}
     &   \begin{bmatrix}
          C^{01}&iR^{01}\\iR^{10}&D^{01}
        \end{bmatrix}
        \begin{bmatrix}
          C^{11}&iR^{11}\\iR^{11}&D^{11}
        \end{bmatrix}^{-1}
        \begin{bmatrix}
          C^{01}&iR^{01}\\iR^{10}&D^{01}
        \end{bmatrix}^T \\
    &\qquad=\begin{bmatrix}
      q^2\hat d^{11}+2qr_{10}\hat r^{11}-r_{10}^2\hat d^{11}
      &
      i\left[d_{01}(r_{10}\hat c^{11}-q\hat r^{11})+r_{01}(r_{10}\hat r^{11}+q\hat d^{11})\right]
      \\
      i\left[d_{01}(r_{10}\hat c^{11}-q\hat r^{11})+r_{01}(r_{10}\hat r^{11}+q\hat d^{11})\right]
      &
      d_{01}^2\hat c^{11}+2r_{01}d_{01}\hat r^{11}-r_{01}^2\hat d^{11}
    \end{bmatrix}
  \end{aligned}
\end{equation}
where each block is proportional to the $m\times m$ matrix
\begin{equation}
  \begin{bmatrix}
    1&0&\cdots&0\\
    0&0&\cdots&0\\
    \vdots&\vdots&\ddots&\vdots\\
    0&0&\cdots&0
  \end{bmatrix}
\end{equation}
which is \emph{not} a hierarchical matrix! But, all these new hat variables are proportional to $n$, and will vanish when the eventual limit is taken. So, the whole contribution is zero, and
\begin{equation}
    A=
      \begin{bmatrix}
        C^{00}&iR^{00}\\iR^{00}&D^{00}
      \end{bmatrix}^{-1}
\end{equation}
\begin{equation}
  B=-
      \begin{bmatrix}
        C^{00}&iR^{00}\\iR^{00}&D^{00}
      \end{bmatrix}^{-1}
      \begin{bmatrix}
        C^{01}&iR^{01}\\
        iR^{10}&D^{01}
      \end{bmatrix}
      \begin{bmatrix}
        C^{11}&iR^{11}\\iR^{11}&D^{11}
      \end{bmatrix}^{-1}
\end{equation}
\begin{equation}
    C=-
      \begin{bmatrix}
        C^{11}&iR^{11}\\iR^{11}&D^{11}
      \end{bmatrix}^{-1}
      \begin{bmatrix}
        C^{01}&iR^{01}\\
        iR^{10}&D^{01}
      \end{bmatrix}^T
      \begin{bmatrix}
        C^{00}&iR^{00}\\iR^{00}&D^{00}
      \end{bmatrix}^{-1}=B^T
\end{equation}
\begin{equation}
    D=
    \left(
      \begin{bmatrix}
        C^{11}&iR^{11}\\iR^{11}&D^{11}
      \end{bmatrix}
      -
      \begin{bmatrix}
        C^{01}&iR^{01}\\
        iR^{10}&D^{01}
      \end{bmatrix}^T
      \begin{bmatrix}
        C^{00}&iR^{00}\\iR^{00}&D^{00}
      \end{bmatrix}^{-1}
      \begin{bmatrix}
        C^{01}&iR^{01}\\
        iR^{10}&D^{01}
      \end{bmatrix}
    \right)^{-1}
\end{equation}

\begin{equation}
    \begin{bmatrix}
      X_0\\i\hat X_0
    \end{bmatrix}^TA
    \begin{bmatrix}
      X_0\\i\hat X_0
    \end{bmatrix}
    +
    2\begin{bmatrix}
      X_0\\i\hat X_0
    \end{bmatrix}^TB
    \begin{bmatrix}
      X_1\\i\hat X_1
    \end{bmatrix}
    +
    \begin{bmatrix}
      X_1\\i\hat X_1
    \end{bmatrix}^TD
    \begin{bmatrix}
      X_1\\i\hat X_1
    \end{bmatrix}
\end{equation}


\[
  \log\det(A-c)=\log\det A-\frac{c}{\sum_{i=0}^k(a_{i+1}-a_i)x_{i+1}}
\]
where $a_{k+1}=1$ and $x_{k+1}=1$.
So the basic form of the action is (for replica symmetric $A$)
\[
  -\mu+\frac14\beta^2f''(1)(1-a_0^2)+\frac12\log(1-a_0)+\frac12\frac{a_0}{1-a_0}+\frac12\begin{bmatrix}x_0\\\hat x_0\\x_1\\\hat x_1^1\\\hat x_1^0\end{bmatrix}^T\left(\beta B-\frac1{1-a_0}C\right)\begin{bmatrix}x_0\\\hat x_0\\x_1\\\hat x_1^1\\\hat x_1^0\end{bmatrix}
\]
for
\[
  B=\begin{bmatrix}
    \hat\beta_0f''(q)+r_{10}f'''(q)&f''(q)&0&0&0\\
    f''(q)&0&0&0&0\\
    0&0&-\hat\beta_1f''(q^{11}_0)-r^{11}_0f'''(q^{11}_0)&-f''(q_0^{11})&0\\
    0&0&-f''(q_0^{11})&0&0\\
    0&0&0&0&0
  \end{bmatrix}
\]
\begin{align}
  &
  C_{11}=d^{00}_\mathrm df'(1)
  \quad
  C_{12}=r^{00}_\mathrm df'(1)
  \quad
  C_{22}=-f'(1)
  \\
  &
  C_{13}
  =\frac1{1-q_0}\left(
    (r^{11}_d-r^{11}_0)\left(r^{01}-q\frac{r^{11}_d-r^{11}_0}{1-q_0}\right)(f'(1)-f'(q_0))+qf'(1)d^{00}_d+r^{00}_d(r^{10}f'(1)+(r^{11}_d-r^{11}_0)f'(q))
    \right)
  \\
  &
  C_{15}=r^{00}_df'(q)+\left(r^{01}-q\frac{r^{11}_d-r^{11}_0}{1-q_0}\right)(f'(1)-f'(q_0))
  \quad
  C_{14}=-C_{15}
  \\
  &
  C_{23}=\frac1{1-q_0}\left((qr^{00}_d-r^{10})f'(1)-(r^{11}_d-r^{11}_0)f'(q)\right)
  \quad
  C_{24}=f'(q)
  \quad
  C_{25}=-C_{24}
  \\
  &
  C_{33}
  =
    -\frac{r^{11}_d-r^{11}_0}{1-q_0}\left[
      \frac{r^{11}_d-r^{11}_0}{1-q_0}f'(1)
      -2\left(
        \frac{qr^{01}-r^{11}_0}{1-q_0}+\frac{1-q^2}{1-q_0}\frac{r^{11}_d-r^{11}_0}{1-q_0}
        \right)(f'(1)-f'(q_0))
        -2\frac{qr^{00}-r^{10}}{1-q_0}f'(q)
    \right]\\
  &\qquad-\frac{1-q^2}{(1-q_0)^2}-\frac{(r^{10}-qr^{00}_d)^2}{(1-q_0)^2}f'(1)
    \\
  &
  C_{34}
  =-(qr^{01}-r^{11}_0)\frac{f'(1)-f'(q_0)}{1-q_0}-\frac{r^{11}_d-r^{11}_0}{1-q_0}\left(
    \frac{1-q^2}{1-q_0}(f'(1)-f'(q_0))-f'(q_0)
    \right)-f'(q)\frac{qr^{00}_d-r^{10}}{1-q_0}
    \\
  &
    C_{35}=-C_{34}-\frac{r^{11}_d-r^{11}_0}{1-q_0}(f'(1)-f'(q_0))
  \quad
    C_{44}=f'(1)-2f'(q_0)
    \quad
    C_{45}=f'(q_0)
    \quad
    C_{55}=-f'(1)
\end{align}
Use $X$ for the big vector. Then
\[
  0=-\beta^2f''(1)a_0+\frac{a_0-X^TCX}{(1-a_0)^2}
\]
\[
  0=\bigg(\beta B-\frac1{1-a_0}C\bigg)X
\]
For a non-trivial solution, we require the following: change of basis in $X$ to diagonalize the matrix $\beta B-C/(1-a_0)$. Then, all of the new basis vectors of $X$ are zero except one, and the second equation is satisfied by tuning $a_0$ to make the coefficient zero. Finally, the first equation is satisfied by choice of the magnitude of this basis vector.
Suppose $a_0=1-1/(y\beta)$, $X\sim X+O(1/\beta)$. Then
\[
  0=-f''(1)(1-(y\beta)^{-1})+y^2\left(1-(y\beta)^{-1}-X^TCX\right)
\]
For large $\beta$
\[
  0=-f''(1)+y^2(1-X^TCX)
\]
\[
  0=(B-yC)X
\]
which gives
\[
  \mathcal S=-\frac12\beta\mu+\frac14\beta^2f''(1)\big(2(y\beta)^{-1}-(y\beta)^{-2}\big)-\frac12\log(y\beta)+\frac12y\beta(1-(y\beta)^{-1})+\frac12\beta X^TBX-\frac12y\beta X^TCX
\]
so
\[
  \lambda_\mathrm{min}=-2\lim_{\beta\to\infty}\frac{\partial\mathcal S}{\partial\beta}
  =\mu-\left(y+\frac1yf''(1)\right)-X^T(B-yC)X
  =\mu-\left(y+\frac1yf''(1)\right)
\]
assuming the last equation is satisfied. The trivial solution, which gives the bottom of the semicircle, is for $X=0$, so the first equation is $y^2=f''(1)$, and
\[
  \lambda_\mathrm{min}=\mu-\sqrt{4f''(1)}=\mu-\mu_\mathrm m
\]
as expected. We need to first the nontrivial solutions with nonzero $X$, but
because the coefficients are so nasty this will be a numeric problem...
Specifically, we are good for $y$ where one of the eigenvalues of $B-yC$ is
zero. In this case, if the associated normalized eigenvector is $\hat X$, its magnitude is set by
\begin{equation}
  \|X\|^2=\frac1{\hat X^TC\hat X}\left(1-\frac{f''(1)}{y^2}\right)
\end{equation}
In practice, we find that $\hat X^TC\hat X$ is positive. Therefore, for the
solution to make sense we must have $y^2>f''(1)$. In practice, there is at most
\emph{one} $y$ which produces a zero eigenvalue of $B-yC$ and satisfies this
inequality, so the solution seems to be unique.

\section{Conclusion}

The methods developed in this paper are straightforwardly (if not easily)
generalized to landscapes with replica symmetry broken complexities.

\paragraph{Acknowledgements}

\paragraph{Funding information}

\printbibliography

\end{document}
