From 7a8b694e888b65bcb3e80155e82d9bfcbd27c620 Mon Sep 17 00:00:00 2001
From: Jaron Kent-Dobias <jaron@kent-dobias.com>
Date: Mon, 7 Jun 2021 15:15:59 +0200
Subject: More writing.

---
 stokes.tex | 44 +++++++++++++++++++++++++++++++++-----------
 1 file changed, 33 insertions(+), 11 deletions(-)

diff --git a/stokes.tex b/stokes.tex
index c64f934..9982848 100644
--- a/stokes.tex
+++ b/stokes.tex
@@ -81,8 +81,9 @@ 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}.
+landscapes, which are typically constructed from the limits of series or
+integrals of analytic functions which are not themselves analytic
+\cite{Cavagna_1999_Energy}.
 
 \section{Dynamics}
 
@@ -230,26 +231,35 @@ 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{p-spin spherical models}
 
-\section{2-spin}
+For $p$-spin spherical models, one is constrained to the manifold $M=\{z\mid
+z^2=N\}$. The normal to this manifold at any point $z\in M$ is always in the
+direction $z$. The projection operator onto the tangent space of this manifold
+is given by
+\begin{equation}
+  P=I-\frac{zz^\dagger}{|z|^2},
+\end{equation}
+where indeed $Pz=z-z|z|^2/|z|^2=0$ and $Pz'=z'$ for any $z'$ orthogonal to $z$.
+
+\subsection{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
+$J$ is generically diagonalizable by a complex orthogonal matrix. In a diagonal basis, $J_{ij}=\lambda_i\delta_{ij}$. Then $\partial_i H=\lambda_iz_i$. 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
+  0=\partial_iH=(\lambda_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,
+which is only possible for $z_i=0$ or $\epsilon=\frac12\lambda_i$. Generically the $\lambda_i$ will all differ, so this can only be satisfied for one $\lambda_i$ 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,
+  \frac1N\frac12\lambda_iz_i^2=\epsilon=\frac12\lambda_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.
 
@@ -279,11 +289,23 @@ 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.
+Since they sit at the corners of a simplex, 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}
+\subsection{Pure \textit{p}-spin}
 
-\section{(2 + 4)-spin}
+\begin{equation}
+  H_p=\frac1{p!}\sum_{i_1\cdots i_p}J_{i_1\cdots i_p}z_{i_1}\cdots z_{i_p}
+\end{equation}
+
+\subsection{(2 + 4)-spin}
+
+\begin{equation}
+  H_2+H_4
+\end{equation}
 
 \section{Numerics}
 
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