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\begin{document}

\title{Analytic continuation of theories with complex landscapes}

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

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

\date\today

\begin{abstract}
In this paper we follow up the study of `complex-complex landscapes'  \cite{Kent-Dobias_2021_Complex}, rugged landscapes in spaces of many complex variables.
Contrary to the real case, the index of saddles is not itself relevant, as it is always half the total dimension. The relevant topological objects here are Stokes trajectories, gradient lines joining two saddles, where Lefschetz thimbles merge. The well-studied `threshold level', separating regions with minima from regions with saddles in the real case, here separates regions where Stokes lines are rare, from region where they proliferate.
Likewise, when a real landscape is prolonged to complex variables, the distinction between "one step replica-symmetry breaking" and "many step replica symmetry breaking"  is that in the former case the saddles are at first free of Stokes lines, while in the latter these immediately proliferate.
\end{abstract}

\maketitle

\cite{Witten_2011_Analytic}

Consider a thermodynamic calculation involving the (real) $p$-spin model for a
particular instantiation of the coupling tensor $J$
\begin{equation} \label{eq:partition.function}
  Z_J(\beta)=\int_{S^{N-1}}ds\,e^{-\beta H_J(s)}
\end{equation}
where $S^{N-1}$ is the $N-1$-sphere of radius $N$. Quantities of interest are
usually related to the quenched free energy, produced by averaging over the
$J$s the sample free energy $F_J$
\begin{equation}
  \overline F(\beta,\kappa)=-\beta^{-1}\int dJ\,p_\kappa(J)\log Z_J
\end{equation}
which can depend in general on the inverse temperature $\beta$ and on some
parameter $\kappa$ which governs the distribution of $J$s. For most
applications, $\beta$ is taken to be real and positive, and the distribution
$p_\kappa$ is taken to be Gaussian or discrete on $\pm1$.

We are interested in analytically continuing expressions like $\overline F$
into the region of complex $\beta$ or distributions $p_\kappa$ involving
complex $J$. 

When the argument of the exponential integrand in \eqref{eq:partition.function}
acquires an imaginary component, various numeric and perturbative schemes for
approximating its value can face immediate difficulties due to the emergence of
a sign problem, resulting from rapid oscillations coinciding with saddles.

Unfortunately the study is not so relevant for low-dimensional `rugged'
landscapes, which are typically series or integrals of analytic functions whose
limit are not themselves analytic \cite{Cavagna_1999_Energy}.


\section{Dynamics}

For a holomorphic Hamiltonian $H$, dynamics are defined by gradient descent on $\operatorname{Re}H$, or
\begin{equation} \label{eq:flow}
  \dot z=-\operatorname{grad}\operatorname{Re}H
  =-\frac12(\partial H)^\dagger\left(I-\frac{zz^\dagger}{|z|^2}\right)
  =-\tfrac12(\partial H)^\dagger P,
\end{equation}
where $P=I-\hat z\hat z^\dagger$ is the projection matrix on to the tangent space of $M$.
Gradient descent on $\operatorname{Re}H$ is equivalent to Hamiltonian dynamics
with the Hamiltonian $\operatorname{Im}H$. This is because $(M, h)$ is Kähler
and therefore admits a symplectic structure, but that the flow conserves
$\operatorname{Im}H$ can be seen from the Cauchy--Riemann equations and
\eqref{eq:flow}:
\begin{equation}
  \begin{aligned}
    \frac d{dt}\operatorname{Im}H
    &=\dot z_i\partial_i\operatorname{Im}H+\dot{\bar z}_i\bar\partial_i\operatorname{Im}H \\
    &=\frac i4\left(
      \bar\partial_j\bar HP_{ji}\partial_i H-\partial_j H\bar P_{ji}\bar\partial_i\bar H
    \right)
    =0
  \end{aligned}
\end{equation}
since $P$ is a Hermitian operator. This conservation law indicates that surfaces of constant $\operatorname{Im}H$ will be important when evaluting the possible endpoints of dynamic trajectories.

Let us consider the generic case, where the critical points of $H$ have
distinct energies. What is the topology of the $C=\operatorname{Im}H$ level
set? We shall argue its form by construction. Consider initially the situation
in the absence of any critical point. In this case the level set consists of a
single simply connected surface, locally diffeomorphic to $\mathbb R^{2(N-1)-1}$. Now, `place' a generic
(nondegenerate) critical point in the function at $z_0$. In the vicinity of the critical
point, the flow is locally
\begin{equation}
  \begin{aligned}
    \dot z_i
    &\simeq-\frac12\left[
      \partial_j\left(
        H(z_0)+\frac12\partial_k\partial_\ell H(z_0)\Delta z_k\Delta z_\ell
      \right)
    \right]^* P_{ji} \\
    &=-\frac12\left(
      \partial_j\partial_kH(z_0)\Delta z_k
    \right)^* P_{ji} \\
    &=-\frac12\Delta z_k^*(\partial_k\partial_jH(z_0))^*P_{ji}
  \end{aligned}
\end{equation}
The matrix $(\partial\partial H)^\dagger P$ has a spectrum identical to that of
$\partial\partial H$ save a single marginal direction corresponding to $z_0$,
the normal to the constraint surface. Assuming we are working in a diagonal basis, we find
\begin{equation}
  \dot z_i=-\frac12\lambda_i\Delta z_i^*+O(\Delta z^2)
\end{equation}
Breaking into real and imaginary parts gives
\begin{equation}
  \begin{aligned}
    \frac{d\Delta x_i}{dt}&=-\frac12\left(
      \operatorname{Re}\lambda_i\Delta x_i+\operatorname{Im}\lambda_i\Delta y_i
    \right) \\
    \frac{d\Delta y_i}{dt}&=-\frac12\left(
      \operatorname{Im}\lambda_i\Delta x_i-\operatorname{Re}\lambda_i\Delta y_i
    \right)
  \end{aligned}
\end{equation}
Therefore, in the complex plane defined by each eigenvector of $(\partial\partial H)^\dagger P$ there is a separatrix flow of the form in Figure \ref{fig:local_flow}. The effect of these separatrices in each complex direction of the tangent space $T_{z_0}M$ is to separate that space into four quadrants: two disconnected pieces with greater imaginary part than the critical point, and two with lesser imaginary part. This partitioning implies that the level set of $\operatorname{Im} H=C$ for $C\neq\operatorname{Im}H(z_0)$ is splint into two disconnected pieces, one lying in each of two quadrants corresponding with its value relative to that at the critical point.

\begin{figure}
  \includegraphics[width=\columnwidth]{figs/local_flow.pdf}
  \caption{
    Gradient descent in the vicinity of a critical point, in the $z$--$z*$
    plane for an eigenvector $z$ of $(\partial\partial H)^\dagger P$. The flow
    lines are colored by the value of $\operatorname{Im}H$.
  } \label{fig:local_flow}
\end{figure}

Continuing to `insert' critical points whose imaginary energy differs from $C$,
one repeatedly partitions the space this way with each insertion. Therefore,
for the generic case with $\mathcal N$ critical points, with $C$ differing in
value from all critical points, the level set $\operatorname{Im}H=C$ has
$\mathcal N+1$ connected components, each of which is simply connected,
diffeomorphic to $\mathbb R^{2(N-1)-1}$ and connects two sectors of infinity to
each other.

When $C$ is brought to the same value as the imaginary part of some critical
point, two of these disconnected surfaces pinch grow nearer and pinch together
at the critical point when $C=\operatorname{Im}H$, as in the black lines of
Figure \ref{fig:local_flow}. The unstable manifold of the critical point, which
corresponds with the portion of this surface that flows away, is known as a
\emph{Lefshetz thimble}.

Stokes lines are trajectories that approach distinct critical points as time
goes to $\pm\infty$. From the perspective of dynamics, these correspond to
\emph{heteroclinic orbits}. What are the conditions under which Stokes lines
appear? Because the dynamics conserves imaginary energy, two critical points
must have the same imaginary energy if they are to be connected by a Stokes
line. This is not a generic phenomena, but will happen often as one model
parameter is continuously varied. When two critical points do have the same
imaginary energy and $C$ is brought to that value, the level set
$C=\operatorname{Im}H$ sees formally disconnected surfaces pinch together in
two places. We shall argue that generically, a Stokes line will exist whenever
the two critical points in question lie on the same connected piece of this
surface.

What are the ramifications of this for disordered Hamiltonians? When some process brings two critical points to the same imaginary energy, whether a Stokes line connects them depends on whether the points are separated from each other by the separatrices of one or more intervening critical points. Therefore, we expect that in regions where critical points with the same value of $\operatorname{Im}H$ tend to be nearby, Stokes lines will proliferate, while in regions where critical points with the same value of $\operatorname{Im}H$ tend to be distant compared to those with different $\operatorname{Im}H$, Stokes lines will be rare.


\section{2-spin}

The Hamiltonian of the $2$-spin model is defined for $z\in\mathbb C^N$ by
\begin{equation}
  H_0=\frac12z^TJz.
\end{equation}
$J$ is generically diagonalizable by a complex orthogonal matrix $P$. With
$z\mapsto Pz$, $J\mapsto D$ for diagonal $D$. Then $\partial_i H=d_iz_i$. We will henceforth assume to be working in this basis. To find the critical points, we use the method of Lagrange multipliers. Introducing $\epsilon$, the constrained Hamiltonian is
\begin{equation}
  H=H_0+\epsilon(N-z^2)
\end{equation}
As usual, $\epsilon$ is equivalent to the energy per spin at any critical point.
Critical points must satisfy
\begin{equation}
  0=\partial_iH=(d_i-2\epsilon)z_i
\end{equation}
which is only possible for $z_i=0$ or $\epsilon=\frac12 d_i$. Generically the $d_i$ will all differ, so this can only be satisfied for one $d_i$ at a time, and to be a critical point all other $z_j$ must be zero. In the direction in question,
\begin{equation}
  \epsilon=\frac1N\frac12d_iz_i^2=\frac12 d_i,
\end{equation}
whence $z_i=\pm\sqrt{N}$. Thus there are $2N$ critical points, each corresponding to $\pm$ the cardinal directions in the diagonalized basis.

Suppose that two critical points have the same imaginary energy; without loss
of generality, assume these are associated with the first and second
cardinal directions. Since the gradient is proportional to $z$, any components that are
zero at some time will be zero at all times. The dynamics for the components of
interest assuming all others are zero are
\begin{equation}
  \begin{aligned}
    \dot z_1
    &=-z_1^*\left(d_1^*-\frac{d_1^*z_1^*z_1+d_2^*z_2^*z_2}{|z_1|^2+|z_2|^2}\right) \\
    &=-(d_1-d_2)^*z_1^*\frac{|z_2|^2}{|z_1|^2+|z_2|^2}
  \end{aligned}
\end{equation}
and the same for $z_2$ with all indices swapped.  Since $\Delta=d_1-d_2$ is
real, if $z_1$ begins real it remains real, with the same for $z_2$. Since the
critical points are at real $z$, we make this restriction, and find
\begin{equation}
  \begin{aligned}
    \frac d{dt}(z_1^2+z_2^2)=0\\
    \frac d{dt}\frac{z_2}{z_1}=\Delta\frac{z_2}{z_1}
  \end{aligned}
\end{equation}
Therefore $z_2/z_1=e^{\Delta t}$, with $z_1^2+z_2^2=N$ as necessary.  Depending on the sign of $\Delta$, $z$ flows
from one critical point to the other over infinite time. This is a Stokes line,
and establishes that any two critical points in the 2-spin model with the same
imaginary energy will possess one.

The critical points of the 2-spin model are all adjacent: no critical point is separated from another by the separatrix of a third. This means that when the imaginary energies of two critical points are brought to the same value, their surfaces of constant imaginary energy join.

\section{p-spin}

\section{(2 + 4)-spin}

\section{Numerics}

To confirm the presence of Stokes lines under certain processes in the $p$-spin, we studied the problem numerically.
\begin{equation}
  \mathcal L(z(t), z'(t))
  = 1-\frac{\operatorname{Re}(\dot z^*_iz'_i)}{|\dot z||z'|}
\end{equation}
\begin{equation}
  \mathcal C[\mathcal L]=\int dt\,\mathcal L(z(t), z'(t))
\end{equation}
$\mathcal C$ has minimum of zero, which is reached only by functions $z(t)$ whose tangent is everywhere parallel to the direction of the dynamics. Therefore, finding functions which reach this minimum will find time-reparameterized Stokes lines. For the sake of numeric stability, we look for functions whose tangent has constant norm.



\begin{acknowledgments}
  MIT mathematicians have been no help
\end{acknowledgments}

\bibliography{stokes}

\appendix

\section{Geometry}

The surface $M\subset\mathbb C^N$ defined by $N=f(z)=z^2$ is an $N-1$ dimensional
\emph{Stein manifold}, a type of complex manifold defined by the level set of a
holomorphic function \cite{Forstneric_2017_Stein}. One can define a Hermitian
metric $h$ on $M$ by taking the restriction of the standard metric of $\mathbb
C^N$ to vectors tangent along $M$. For any smooth function $\phi:M\to\mathbb
R$, its gradient is a holomorphic vector field given by
\begin{equation}
  \operatorname{grad}\phi=\bar\partial^\sharp\phi
\end{equation}
Suppose $u:M\to\mathbb C^{N-1}$ is a coordinate system. Then
\begin{equation}
  \operatorname{grad}\phi=h^{\bar\beta\alpha}\bar\partial_{\bar\beta}\phi\frac{\partial}{\partial u^\alpha}
\end{equation}
Let $z=u^{-1}$.
\begin{equation}
  \frac\partial{\partial u^\alpha}=\frac{\partial z^i}{\partial u^\alpha}\frac\partial{\partial z^i}
\end{equation}
\begin{equation}
  \bar\partial_{\bar\beta}\phi=\frac{\partial\bar z^i}{\partial\bar u^{\bar\beta}}\frac{\partial\phi}{\partial\bar z^i}
\end{equation}
\begin{equation}
  \operatorname{grad}\phi
  =\frac{\partial\bar z^{\bar\jmath}}{\partial\bar u^{\bar\beta}}h^{\bar\beta\alpha}\frac{\partial z^i}{\partial u^\alpha}\frac{\partial\phi}{\partial\bar z^{\bar\jmath}}\frac\partial{\partial z^i}
\end{equation}
At any point $z_0\in M$, $z_0$ is parallel to the normal to $M$. Without loss of generality, take $z_0=e^N$. In a neighborhood of $z_0$, the map $u^\alpha=z^\alpha$ is a coordinate system.
its inverse is $z^i=u^i$ for $1\leq i\leq N-1$ and
\begin{equation}
  z^N=\sqrt{N-u^2}.
\end{equation}
The Jacobian is
\begin{equation}
  \frac{\partial z^i}{\partial u^\alpha}=\delta^i_\alpha-\delta_N^{i}\frac{u^\alpha}{\sqrt{N-u^2}}
\end{equation}
and therefore the Hermitian metric induced by the map is
\begin{equation}
  h_{\alpha\bar\beta}=\frac{\partial\bar z^i}{\partial\bar u^\alpha}\frac{\partial z^{\bar\jmath}}{\partial u^{\bar\beta}}\delta_{i\bar\jmath}
  =\delta_{\alpha\bar\beta}+\frac{\bar u^{\alpha}u^{\bar\beta}}{|N-u^2|}
\end{equation}
The metric can be inverted explicitly:
\begin{equation}
  h^{\bar\beta\alpha}
  =\delta^{\bar\beta\alpha}-\frac{\bar u^{\bar\beta}u^\alpha}{|N-u^2|+|u|^2}.
\end{equation}
Putting these pieces together, we find
\begin{equation}
  \frac{\partial\bar z^{\bar\jmath}}{\partial\bar u^{\bar\beta}}h^{\bar\beta\alpha}\frac{\partial z^i}{\partial u^\alpha}
  =\delta^{\bar\jmath i}-\frac{z^{\bar\jmath}\bar z^i}{|z|^2}
\end{equation}
\begin{equation}
  \operatorname{grad}\phi
  =\left(\delta^{\bar\jmath i}-\frac{z^{\bar\jmath}\bar z^i}{|z|^2}\right)
  \frac{\partial\phi}{\partial\bar z^{\bar\jmath}}\frac\partial{\partial z^i}
\end{equation}

\end{document}