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\begin{document}

\title{Complex complex landscapes: I: saturating the Bézout bound} % change me

\author{Jaron Kent-Dobias}
\author{Jorge Kurchan}

\affiliation{Laboratoire de Physique de l'Ecole Normale Supérieure, Paris, France}

\date\today

\begin{abstract}
  We study the saddle-points of the $p$-spin model, the best understood example of `complex (rugged) landscape' in the space in which
  saturates the Bézout bound \cite{Bezout_1779_Theorie}.
\end{abstract}

\maketitle

\begin{equation} \label{eq:bare.hamiltonian}
  H_0 = \frac1{p!}\sum_{i_1\cdots i_p}^NJ_{i_1\cdots i_p}z_{i_1}\cdots z_{i_p},
\end{equation}
where $z\in\mathbb C^N$ is constrained by $z^2=N$ and $J$ is a symmetric tensor
whose elements are complex normal with $\langle|J|^2\rangle=p!/2N^{p-1}$ and
$\langle J^2\rangle=\kappa\langle|J|^2\rangle$ for complex parameter
$|\kappa|<1$. The constraint is enforced using the method of Lagrange
multipliers: introducing the $\epsilon\in\mathbb C$, this gives
\begin{equation} \label{eq:constrained.hamiltonian}
  H = H_0+\frac p2\epsilon\left(N-\sum_i^Nz_i^2\right).
\end{equation}
At any critical point $\epsilon=H/N$, the average energy.

When compared with $z^*z=N$, the constraint $z^2=N$ may seem an unnatural
extension of the real $p$-spin spherical model. However, a model with this
nonholomorphic spherical constraint has a disturbing lack of critical points
nearly everywhere, since $0=\partial^* H=-p\epsilon z$ is only
satisfied for $\epsilon=0$, as $z=0$ is forbidden by the constraint.

Since $H$ is holomorphic, a point is a critical point of its real part if and
only if it is also a critical point of its imaginary part. The number of
critical points of $H$ is therefore the number of critical points of
$\mathop{\mathrm{Re}}H$. Writing $z=x+iy$, $\mathop{\mathrm{Re}}H$ can be
interpreted as a real function of $2N$ real variables. The number of critical
points it has is given by the usual Kac--Rice formula:
\begin{equation} \label{eq:real.kac-rice}
  \mathcal N(\kappa,\epsilon)
    = \int dx\,dy\,\delta(\partial_x\mathop{\mathrm{Re}}H)\delta(\partial_y\mathop{\mathrm{Re}}H)
      \left|\det\begin{bmatrix}
        \partial_x\partial_x\mathop{\mathrm{Re}}H & \partial_x\partial_y\mathop{\mathrm{Re}}H \\
        \partial_y\partial_x\mathop{\mathrm{Re}}H & \partial_y\partial_y\mathop{\mathrm{Re}}H
      \end{bmatrix}\right|.
\end{equation}
The Cauchy--Riemann relations can be used to simplify this. Using the Wirtinger
derivative $\partial=\partial_x-i\partial_y$, one can write
$\partial_x\mathop{\mathrm{Re}}H=\mathop{\mathrm{Re}}\partial H$ and
$\partial_y\mathop{\mathrm{Re}}H=-\mathop{\mathrm{Im}}\partial H$. With similar
transformations, the eigenvalue spectrum of the Hessian of
$\mathop{\mathrm{Re}}H$ can be shown to be equivalent to the singular value
spectrum of the Hessian $\partial\partial H$ of $H$, and as a result the
determinant that appears above is equivalent to $|\det\partial\partial H|^2$.
This allows us to write \eqref{eq:real.kac-rice} in the manifestly complex
form
\begin{equation} \label{eq:complex.kac-rice}
  \mathcal N(\kappa,\epsilon)
    = \int dx\,dy\,\delta(\mathop{\mathrm{Re}}\partial H)\delta(\mathop{\mathrm{Im}}\partial H)
      |\det\partial\partial H|^2.
\end{equation}

The Hessian of \eqref{eq:constrained.hamiltonian} is $\partial_i\partial_j
H=\partial_i\partial_j H_0-p\epsilon\delta_{ij}$, or the Hessian of
\eqref{eq:bare.hamiltonian} with a constant added to its diagonal. The
eigenvalue distribution $\rho$ of the constrained Hessian is therefore related
to the eigenvalue distribution $\rho_0$ of the unconstrained one by a similar
shift, or $\rho(\lambda)=\rho_0(\lambda+p\epsilon)$. The Hessian of
\eqref{eq:bare.hamiltonian} is
\begin{equation} \label{eq:bare.hessian}
  \partial_i\partial_jH_0
  =\frac{p(p-1)}{p!}\sum_{k_1\cdots k_{p-2}}^NJ_{ijk_1\cdots k_{p-2}}z_{k_1}\cdots z_{k_{p-2}},
\end{equation}
which makes its ensemble that of Gaussian complex symmetric matrices, whose
spectrum is constant inside the support of a certain ellipse and zero
everywhere else \cite{Nguyen_2014_The}. Given its variances
$\langle|\partial_i\partial_j H_0|^2\rangle=p(p-1)a^{p-2}/2N$ and
$\langle(\partial_i\partial_j H_0)^2\rangle=p(p-1)\kappa/2N$, $\rho_0(\lambda)$ is constant inside the ellipse
\begin{equation} \label{eq:ellipse}
  \left(\frac{\mathop{\mathrm{Re}}(\lambda e^{i\theta})}{1+|\kappa|/a^{p-2}}\right)^2+
  \left(\frac{\mathop{\mathrm{Im}}(\lambda e^{i\theta})}{1-|\kappa|/a^{p-2}}\right)^2
  <\frac12p(p-1)a^{p-2}
\end{equation}
where $\theta=\frac12\arg\kappa$.

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