From 5963d605d3646b787e5557904b1330b3eda66c17 Mon Sep 17 00:00:00 2001
From: Jaron Kent-Dobias <jaron@kent-dobias.com>
Date: Thu, 10 Dec 2020 08:15:31 +0100
Subject: Added more explanation of replica calculation.

---
 bezout.tex | 33 +++++++++++++++++++++++----------
 1 file changed, 23 insertions(+), 10 deletions(-)

(limited to 'bezout.tex')

diff --git a/bezout.tex b/bezout.tex
index c586908..98e058b 100644
--- a/bezout.tex
+++ b/bezout.tex
@@ -195,7 +195,6 @@ and their conjugates. The result is, to leading order in $N$,
 \end{equation}
 where
 \begin{widetext}
-  \textcolor{red}{\textbf{[appendix?? I'm putting too much right now so as to trim later...]}}
   \begin{equation}
     f=2+\frac12\log\det\frac12\begin{bmatrix}
       1 & a & d & b \\
@@ -232,9 +231,6 @@ shift, or $\rho(\lambda)=\rho_0(\lambda+p\epsilon)$. The Hessian of
   \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}
-
-{\color{red} \bf here I would explain the question of the det and also of the appearance of the gap, would draw a picture of ellipse etc, and would send the reader to an appendix for most of this part of the calculation}
-
 which makes its ensemble that of Gaussian complex symmetric matrices. Given its variances
 $\overline{|\partial_i\partial_j H_0|^2}=p(p-1)a^{p-2}/2N$ and
 $\overline{(\partial_i\partial_j H_0)^2}=p(p-1)\kappa/2N$, $\rho_0(\lambda)$ is constant inside the ellipse
@@ -276,14 +272,31 @@ elements of $J$ are standard complex normal, this corresponds to a complex
 Wishart distribution. For $\kappa\neq0$ the problem changes, and to our
 knowledge a closed form is not known.  We have worked out an implicit form for
 this spectrum using the saddle point of a replica symmetric calculation for the
-Green function. {\color{red} the calculation is standard, we outline it in appendix xx} The result is
+Green function. Introducing replicas to bring the partition function to
+the numerator gives
 \begin{widetext}
   \begin{equation}
-    G(\sigma)=\lim_{n\to0}\int d\alpha\,d\chi\,d\chi^*\frac\alpha2
-    \exp nN\left\{
-      1+\frac{p(p-1)}{16}a^{p-2}\alpha^2-\frac{\alpha\sigma}2+\frac12\log(\alpha^2-|\chi|^2)
-      +\frac p4\mathop{\mathrm{Re}}\left(\frac{(p-1)}8\kappa^*\chi^2-\epsilon^*\chi\right)
-      \right\}
+    G(\sigma)=\frac1N\lim_{n\to0}\int d\zeta\,d\zeta^*\,(\zeta_i^{(0)})^*\zeta_i^{(0)}
+      \exp\left\{
+      \frac12\sum_\alpha^n\left[(\zeta_i^{(\alpha)})^*\zeta_i^{(\alpha)}\sigma
+        -\Re\left(\zeta_i^{(\alpha)}\zeta_j^{(\alpha)}\partial_i\partial_jH\right)
+      \right]
+    \right\}
+  \end{equation}
+  with sums taken over repeated latin indices.
+  The average can then be made over $J$ and Hubbard--Stratonovich used to change
+  variables to replica matrices
+  $N\alpha_{\alpha\beta}=(\zeta^{(\alpha)})^*\cdot\zeta^{(\beta)}$ and
+  $N\chi_{\alpha\beta}=\zeta^{(\alpha)}\cdot\zeta^{(\beta)}$ and a series of replica
+  vectors. Taking the replica-symmetric ansatz leaves all off-diagonal elements
+  and vectors zero, and $\alpha_{\alpha\beta}=\alpha_0\delta_{\alpha\beta}$,
+  $\chi_{\alpha\beta}=\chi_0\delta_{\alpha\beta}$. The result is
+  \begin{equation}
+    \overline G(\sigma)=\lim_{n\to0}\int d\alpha_0\,d\chi_0\,d\chi_0^*\,\alpha_0
+    \exp\left\{nN\left[
+      1+\frac{p(p-1)}{16}a^{p-2}\alpha_0^2-\frac{\alpha_0\sigma}2+\frac12\log(\alpha_0^2-|\chi_0|^2)
+      +\frac p4\mathop{\mathrm{Re}}\left(\frac{(p-1)}8\kappa^*\chi_0^2-\epsilon^*\chi_0\right)
+    \right]\right\}
   \end{equation}
 \end{widetext}
 The argument of the exponential has several saddles, but the one with the smallest value of $\mathop{\mathrm{Re}}\alpha$ gives the correct solution \textcolor{red}{\textbf{we have checked this, but a detailed analysis of the saddle-point integration is still needed to justify it.}}.
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