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\begin{document}

\title{
  Arrangement of nearby minima and saddles in the mixed $p$-spin energy landscape
}

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

\maketitle
\begin{abstract}
\end{abstract}

\cite{Ros_2020_Distribution, Ros_2019_Complex, Ros_2019_Complexity}

The mixed $p$-spin 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.

We introduce the Kac--Rice \cite{Kac_1943_On, Rice_1944_Mathematical} measure
\begin{equation}
  d\nu_H(\mathbf s)
  =d\mathbf s\,\delta\big(\nabla H(\mathbf s)\big)\,
  \big|\det\operatorname{Hess}H(\mathbf s)\big|
\end{equation}
which counts stationary points of the function $H$. More interesting is the
measure conditioned on the energy density $E$ and stability $\mu$,
\begin{equation}
  d\nu_H(\mathbf s\mid E,\mu)
  =d\nu_H(\mathbf s)\,
  \delta\big(H(\mathbf s)-NE\big)\,
  \delta\big(N\mu-\operatorname{Tr}\operatorname{Hess}H(\mathbf s)\big)
\end{equation}


We want the typical number of stationary points with energy density
$E_2$ and stability $\mu_2$ that lie a fixed overlap $q$ from a reference
stationary point of energy density $E_1$ and stability $\mu_1$.
\begin{align*}
  \Sigma_{12}
  &=\frac1N\overline{\int\frac{d\nu_H(\mathbf s_0\mid E_0,\mu_0)}{\int d\nu_H(\mathbf s_0'\mid E_0,\mu_0)}\,
  \log\bigg(\int d\nu_H(\mathbf s_1\mid E_1,\mu_1)\,\delta(Nq-\mathbf s_0\cdot\mathbf s_1)\bigg)}
\end{align*}
\begin{align*}
  \Sigma_{12}
  &=\frac1N\lim_{n\to0}\lim_{m\to-1}\overline{\int d\nu_H(\mathbf s_0\mid E_0,\mu_0)\left(\int d\nu_H(\mathbf s_0'\mid E_0,\mu_0)\right)^m\,
  \frac\partial{\partial n}\bigg(\int d\nu_H(\mathbf s_1\mid E_1,\mu_1)\,\delta(Nq-\mathbf s_0\cdot \mathbf s_1)\bigg)^n}\\
  &=\frac1N\lim_{n\to0}\lim_{m\to0}\frac\partial{\partial n}\overline{\int\left(\prod_{b=1}^md\nu_H(\pmb\sigma_b\mid E_0,\mu_0)\right)\left(\prod_{a=1}^nd\nu_H(\mathbf s_a\mid E_1,\mu_1)\,\delta(Nq-\pmb \sigma_1\cdot \mathbf s_a)\right)}
\end{align*}
\begin{equation}
  \overline{\big|\det\operatorname{Hess}H(s)\big|\,\delta\big(N\mu-\operatorname{Tr}\operatorname{Hess}H(s)\big)}
  =e^{N\int d\lambda\,\rho(\lambda+\mu)\log|\lambda|}\delta(N\mu-s\cdot\partial H)
\end{equation}
\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}
\begin{equation}
  \begin{aligned}
    \mathcal D(\mu)
    &=\int d\lambda\,\rho(\lambda+\mu)\ln|\lambda| \\
    &=\begin{cases}
      \frac12+\log\left(\frac12\mu_\text m\right)+\frac\mu{\mu_\text m}\left(\frac\mu{\mu_\text m}-\sqrt{\big(\frac\mu{\mu_\text m}\big)^2-1}\right)
      -\log\left(\frac{\mu}{\mu_\text m}-\sqrt{\big(\frac\mu{\mu_\text m}\big)^2-1}\right) & \mu>\mu_\text m \\
      \frac12+\log\left(\frac12\mu_\text m\right)+\frac{\mu^2}{\mu_\text m^2}
       & -\mu_\text m\leq\mu\leq\mu_\text m \\
       \frac12+\log\left(\frac12\mu_\text m\right)+\frac\mu{\mu_\text m}\left(\frac\mu{\mu_\text m}+\sqrt{\big(\frac\mu{\mu_\text m}\big)^2-1}\right)
      -\log\left(\frac{\mu}{\mu_\text m}+\sqrt{\big(\frac\mu{\mu_\text m}\big)^2-1}\right) & \mu<-\mu_\text m
    \end{cases}
  \end{aligned}
\end{equation}

\begin{align}
  \mathcal Q_{00}=\begin{bmatrix}
    \hat\beta_0\\\hat\mu_0\\C^{00}\\R^{00}\\D^{00}
  \end{bmatrix}
  &&
  \mathcal Q_{11}=\begin{bmatrix}
    \hat\beta_1\\\hat\mu_1\\C^{11}\\R^{11}\\D^{11}
  \end{bmatrix}
  &&
  \mathcal Q_{01}=\begin{bmatrix}
    \hat\mu_{01}\\C^{01}\\R^{01}\\R_{10}\\D^{01}
  \end{bmatrix}
\end{align}
\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}
\begin{equation}
  \begin{aligned}
    &\mathcal S_0(\mathcal Q_{00})
    =-\hat\beta_0E_0-r^{00}_d\mu_0-\frac12\hat\mu_0(1-c^{00}_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}

\begin{equation}
  \begin{aligned}
    &\mathcal S(\mathcal Q_{00},\mathcal Q_{11},\mathcal Q_{01})
    =-\hat\beta_1E_1-\mu_1r^{11}_d-\frac12\hat\mu_1(1-c^{11}_d) \\
    &\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}

\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}\\
  \vphantom{\begin{array}{c}n\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}

The inverse of block hierarchical matrix is still a block hierarchical matrix, since (dropping the superscripts for clarity)
\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_d^{00}&=[(C^{00}D^{00}+R^{00}R^{00})^{-1}C^{00}]_{11} \\
  \tilde r_d^{00}&=[(C^{00}D^{00}+R^{00}R^{00})^{-1}R^{00}]_{11} \\
  \tilde d_d^{00}&=[(C^{00}D^{00}+R^{00}R^{00})^{-1}D^{00}]_{11}
\end{align}

In the replica symmetric case,
\begin{align}
  \tilde c_d^{00}=\frac1{(r^{00}_d)^2+d^{00}_d} &&
  \tilde r_d^{00}=\frac{r^{00}_d}{(r^{00}_d)^2+d^{00}_d} &&
  \tilde d_d^{00}=\frac{d^{00}_d}{(r^{00}_d)^2+d^{00}_d}
\end{align}

\begin{equation}
  \begin{bmatrix}
    q^2\tilde d_d^{00}+2qr_{10}\tilde r^{00}_d-r_{10}^2\tilde d^{00}_d
    &
    i\left[d_{01}(r_{10}\tilde c^{00}_d-q\tilde r^{00}_d)+r_{01}(r_{10}\tilde r^{00}_d+q\tilde d^{00}_d)\right]
    \\
    i\left[d_{01}(r_{10}\tilde c^{00}_d-q\tilde r^{00}_d)+r_{01}(r_{10}\tilde r^{00}_d+q\tilde d^{00}_d)\right]
    &
    d_{01}^2\tilde c^{00}_d+2r_{01}d_{01}\tilde r^{00}_d-r_{01}^2\tilde d^{00}_d
  \end{bmatrix}
\end{equation}
where each block is a constant $n\times n$ matrix.

In the twin limits of $m$ and $n$ to zero, the saddle point conditions for the variables involving only the reference critical point (those in $\mathcal Q_{00}$) reduce to the ordinary, 1-point conditions. With a replica-symmetric ansatz, these conditions are
\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_d^{00}
  &=\frac{\mu_0f(1)+\epsilon_0f'(1)}{f(1)\big(f'(1)+f''(1)\big)-f'(1)^2} \\
  d_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}

\begin{align*}
  &
  \begin{bmatrix}
    \tilde c&\tilde r\\\tilde r&\tilde d
  \end{bmatrix}
  =
  \begin{bmatrix}
    q&ir_{10}\\ir_{01}&d_{01}
  \end{bmatrix}
  \begin{bmatrix}
    1&ir_{0}\\
    ir_{0}&d_{0}
  \end{bmatrix}^{-1}
  \begin{bmatrix}
    q&ir_{01}\\ir_{10}&d_{01}
  \end{bmatrix}\\
  &=
  \frac1{r_{0}^2+d_{0}}\begin{bmatrix}
    q^2d_{0}+2qr_{0}r_{10}-r_{10}^2
    &
    i\left[d_{01}(r_{10}-r_0q)+r_{01}(r_0r_{10}+d_0q)\right]
    \\
    i\left[d_{01}(r_{10}-r_0q)+r_{01}(r_0r_{10}+d_0q)\right]
    &
    d_{01}^2+2r_{0}r_{01}d_{01}-d_{0}r_{01}^2
  \end{bmatrix}
\end{align*}
This matrix with modify the diagonal of the RS matrix for the second spin.

Define $\tilde C=C-\tilde c$, $\tilde R=R-\tilde r$, $\tilde D=D-\tilde d$. Then
\begin{align*}
  &\Sigma_{12}
  =\mathcal D(\mu_1)+\hat\beta_1E_1-\hat\mu_1
  +\hat\beta_0\hat\beta_1f(q)+(\hat\beta_0r_{01}+\hat\beta_1r_{10}-d_{01})f'(q)+r_{01}r_{10}f''(q)
  \\&
  +\lim_{n\to0}\frac1n\bigg\{
    \frac12\sum_{ab}\left[
      \hat\beta_1^2f(C_{ab})+(2\hat\beta_1R_{ab}-D_{ab})f'(C_{ab})+R_{ab}^2f''(C_{ab})
    \right]
  \\&
    +\hat\mu_1\operatorname{Tr}C-\mu_1\operatorname{Tr}R
    +\frac12\log\det((C-\tilde c)(D-\tilde d)+(R-\tilde r)^2)
  \bigg\}
\end{align*}
These equations for $D^*$ are the same as those for the unpinned case, or
\[
  0=-\frac12f'(C)+\frac12((C-\tilde c)(D-\tilde d)+(R-\tilde r)^2)^{-1}(C-\tilde c)
\]
Solving, we get
\[
  D=\tilde d+f'(C)^{-1}-(C-\tilde c)^{-1}(R-\tilde r)^2
\]
\begin{align*}
  &\Sigma_{12}
  =\mathcal D(\mu_1)+\hat\beta_1E_1-\frac12\hat\mu_1
  +\hat\beta_0\hat\beta_1f(q)+(\hat\beta_0r_{01}+\hat\beta_1r_{10}-d_{01})f'(q)+r_{01}r_{10}f''(q)
  \\&
  +\lim_{n\to0}\frac1n\bigg\{
    \frac12\sum_{ab}\left[
      \hat\beta_1^2f(C_{ab})+(2\hat\beta_1R_{ab}-(f'(C)^{-1}_{ab}-((C-\tilde c)^{-1}(R-\tilde r)^2)_{ab}-\tilde d))f'(C_{ab})+R_{ab}^2f''(C_{ab})
    \right]
  \\&
    +\frac12\hat\mu_1\operatorname{Tr}C-\mu_1\operatorname{Tr}R
    +\frac12\log\det(C-\tilde c)-\frac12\log\det f'(C)
  \bigg\}
\end{align*}



\begin{align*}
  0&=C^*-f'(C)(C^*D^*+R^*R^*) \\
  0&=\big[\hat\beta_1f'(C)-\mu_1I+R\odot f''(C)\big](C-\tilde c)+f'(C)(R-\tilde r)
\end{align*}

\begin{align*}
  0&=-f'(q)+\frac12(C^*D^*+R^*R^*)^{-1}_{ij}\left(
    C^*_{jk}\frac{D^*_{ki}}{d_{01}}
    +
    2R^*_{jk}\frac{R^*_{ki}}{d_{01}}
    \right)
\end{align*}

\begin{equation}
  \hat\beta_2E_2-r_{22}^{(0)}\mu_2\frac12\left\{
    \hat\beta_2^2\big(f(1)-f(q_{22}^{(0)})\big)
    +\left(
      r_{12}^2+2\hat\beta_2r_{22}-\frac{2q_{12}r_{12}(r_{22}-r_{22}^{(0)})}{1-q_{22}^{(0)}}
    \right)\big(f'(1)-f'(q_{22}^{(0)})\big)
  \right\}
\end{equation}

What about the average for the Hessian terms?

\[
  \overline{
    |\det\operatorname{Hess}H(s_0)|\delta(\mu_0-\operatorname{Tr}\operatorname{Hess}H(s_0))|\det\operatorname{Hess}H(s_a)|\delta(\mu_1-\operatorname{Tr}\operatorname{Hess}H(s_a))
  }
\]


\paragraph{Acknowledgements}

\paragraph{Funding information}

\printbibliography

\end{document}
