From 2b03110226d0f5fed8a30c740a3a1da42b3b8e40 Mon Sep 17 00:00:00 2001
From: Jaron Kent-Dobias <jaron@kent-dobias.com>
Date: Tue, 8 Dec 2020 18:56:41 +0100
Subject: Added reference, reorganized existing content.

---
 bezout.tex | 27 +++++++++++++--------------
 1 file changed, 13 insertions(+), 14 deletions(-)

(limited to 'bezout.tex')

diff --git a/bezout.tex b/bezout.tex
index 9b72684..d39a2e9 100644
--- a/bezout.tex
+++ b/bezout.tex
@@ -23,7 +23,7 @@
 \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 all its $N$ variables are allowed to be complex. The problem becomes
+  We study the saddle-points of the $p$-spin mode -- the best understood example of `complex (rugged) landscape' -- in the space in which all its $N$ variables are allowed to be complex. The problem becomes
   a system of $N$ random equations of degree $p-1$. 
   We solve for quantities averaged over randomness in the $N \rightarrow \infty$ limit. 
   We show that the number of solutions saturates the Bézout bound $\ln {\cal{N}}\sim N \ln (p-1) $\cite{Bezout_1779_Theorie}.
@@ -78,7 +78,7 @@ 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}
-It is easy to see that {\em for a pure $p$ spin}, at  any critical point $\epsilon=H/N$, the average energy.
+It is easy to see that {\em for a pure $p$-spin}, at  any critical point $\epsilon=H/N$, the average energy.
 
 
 Since $H$ is holomorphic, a point is a critical point of its real part if and
@@ -195,7 +195,7 @@ is shifted by $p\epsilon$.
     Eigenvalue and singular value spectra of the matrix $\partial\partial H$
     for $p=3$, $a=\frac54$, $\kappa=\frac34e^{i3\pi/4}$, and
     $\epsilon=i|\epsilon|$ with various values of $|\epsilon|$. The shaded
-    regions of the lefthand plots show the support of the eigenvalue
+    region of the lefthand plots shows the support of the eigenvalue
     distribution. The solid line on the righthand plots shows the distribution
     of singular values, while the overlaid histograms show empirical
     distributions from $2^{10}\times2^{10}$ complex normal matrices with the
@@ -217,7 +217,14 @@ corresponds to the origin leaving the support of the eigenvalue spectrum.
 Weyl's theorem requires that the product over the norm of all eigenvalues must
 not be greater than the product over all singular values \cite{Weyl_1912_Das}.
 Therefore, the absence of zero eigenvalues implies the absence of zero singular
-values.
+values. The determination of the threshold energy is therefore reduced to a
+geometry problem, and yields
+\begin{equation} \label{eq:threshold.energy}
+  |\epsilon_{\mathrm{th}}|^2
+  =\frac{p-1}{2p}\frac{(1-|\delta|^2)^2a^{p-2}}
+  {1+|\delta|^2-2|\delta|\cos(\arg\kappa+2\arg\epsilon)}
+\end{equation}
+for $\delta=\kappa a^{-(p-2)}$.
 
 % This is kind of a boring definition...
 \begin{equation} \label{eq:count.def.marginal}
@@ -247,27 +254,19 @@ For $|\kappa|<1$,
   =(p-1)^{N/2}
 \end{equation}
 
-\begin{equation} \label{eq:threshold.energy}
-  |\epsilon_{\mathrm{th}}|^2
-  =\frac{p-1}{2p}\frac{(1-|\delta|^2)^2a^{p-2}}
-  {1+|\delta|^2-2|\delta|\cos(\arg\kappa+2\arg\epsilon)}
-\end{equation}
-for $\delta=\kappa a^{-(p-2)}$.
-
 
 {\color{teal} {\bf somewhere else}
 
 Another instrument we have to study this problem is to compute the following partition function:
-
 \begin{equation}
   \begin{aligned}
     Z(a,\beta)&=\int dx\, dy \, e^{-\mathop{\mathrm{Re}}(\beta H_0)}\\
-     &\qquad\delta(\sum_i z_i^2-N) \delta\left(\sum_i y_i^2 -N \frac{a-1}{2}\right) 
+     &\qquad\delta(\sum_i z_i^2-N) \delta\left(\sum_i y_i^2 -N \frac{a-1}{2}\right).
   \end{aligned}
 \end{equation}
 The energy $\Re H_0, \Im H_0$ are in a one-to one relation with the temperatures $\beta_R,\beta_I$. The entropy $S(a,H_0) = \ln Z+ +\beta_{R} \langle \Re H_0 \rangle +\beta_I \langle \Im H_0\rangle$
 is the logarithm of the number of configurations of a given $(a,H_0)$.
-This problem may be solved exactly with replicas, {\em but it may also be simulated}
+This problem may be solved exactly with replicas, {\em but it may also be simulated} \cite{Bray_1980_Metastable}.
 Consider for example the ground-state energy for given $a$, that is, the energy in the limit $\beta_R \rightarrow \infty$ taken adjusting  $\beta_I$ so that $\Im H_0=0$ . For $a=1$ this coincides with the ground-state of the real problem.
 
 \begin{center}
-- 
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