From 892a57ae59ef90474a5be664d52f369a0bc1f3a9 Mon Sep 17 00:00:00 2001
From: Jaron Kent-Dobias <jaron@kent-dobias.com>
Date: Thu, 24 Feb 2022 21:47:59 +0100
Subject: More futzing with superfield language and notation.

---
 stokes.tex | 154 +++++++++++++++++++++++++++++--------------------------------
 1 file changed, 72 insertions(+), 82 deletions(-)

(limited to 'stokes.tex')

diff --git a/stokes.tex b/stokes.tex
index b6f3e36..54b0b64 100644
--- a/stokes.tex
+++ b/stokes.tex
@@ -1184,37 +1184,30 @@ trick. Based on the experience from similar problems \cite{Castellani_2005_Spin-
 expected to be exact wherever the complexity is positive.
 
 As in \S\ref{sec:stationary.hessian}, these can be bright into a manifestly complex form using Cauchy--Riemann relations. This gives
-\begin{eqnarray} \label{eq:real.kac-rice}
+\begin{equation} \label{eq:real.kac-rice}
   \mathcal N
-  &=\int dz^*dz\,d\hat z^*d\hat z\,d\eta^*d\eta\,d\gamma^*d\gamma\exp\left\{
-    -\operatorname{Re}\left(
-      \hat z^T\partial\tilde\mathcal S_p+\eta^T\partial\partial\mathcal S_p\gamma
-    \right)
-  \right\} \\
-  &=\int dz^*dz\,d\hat z^*d\hat z\,d\eta^*d\eta\,d\gamma^*\,d\gamma\exp\left\{
-    -\frac12\left(
-      \hat z^T\partial\tilde\mathcal S_p+\hat z^\dagger(\partial\tilde\mathcal S_p)^*+
-      \eta^T\partial\partial\mathcal S_p\gamma+
-      \eta^\dagger(\partial\partial\mathcal S_p)^*\gamma^*
+   =\int dz^*dz\,d\hat z^*d\hat z\,d\eta^*d\eta\,d\gamma^*d\gamma\exp\left\{
+    \operatorname{Re}\left(
+      \hat z^T\partial\tilde\mathcal S_p+\eta^T\partial\partial\tilde\mathcal S_p\gamma
     \right)
   \right\}
-\end{eqnarray}
+\end{equation}
 where $\eta$ and $\gamma$ are Grassmann variables. This can be more
 conveniently studied using the method of superfields. Introducing the
 one-component Grassman variables $\theta$ and $\bar\theta$, define the
 superfield
-\begin{eqnarray}
-  \zeta(i)&=z+\bar\theta(i)\eta^*+\gamma\theta(i)+\hat z\bar\theta(i)\theta(i)
-\end{eqnarray}
+\begin{equation}
+  \phi(1)=z+\bar\theta(1)\eta+\gamma\theta(1)+\hat z\bar\theta(1)\theta(1)
+\end{equation}
+and its measure $d\phi=dz\,d\hat z\,d\eta\,d\gamma$.
 Then the expression for the number of stationary points can be written in a very compact form, as
 \begin{equation}
-  \mathcal N=\int d\zeta^*d\zeta\,\exp\left\{
-    -\frac12\int d1\,\left(
-      \tilde\mathcal S_p(\zeta(1))+\tilde\mathcal S_p(\zeta^*(1))
-    \right)
+  \mathcal N=\int d\phi^*d\phi\,\exp\left\{
+    \int d1\,\operatorname{Re}
+      \tilde\mathcal S_p(\phi(1))
   \right\}
 \end{equation}
-where $d1=d\bar\theta(1)\,d\theta(1)$ is a shorthand for integration over the auxiliary grassman variables.
+where $d1=d\bar\theta(1)\,d\theta(1)$ denotes the integration over the grasssman variables.
 This can be related to the previous expression by expansion with respect to
 the Grassman variables, recognizing that $\theta^2=\bar\theta^2=0$ restricts
 the series to two derivatives.
@@ -1223,46 +1216,54 @@ From here the process can be treated as usual, averaging over the couplings and
 replacing bilinear combinations of the fields with their own variables via a
 Hubbard--Stratonovich transformation. Defining the supermatrix
 \begin{equation}
-  A(i,j)=\frac1N\{\zeta(i),\zeta^*(i)\}\otimes\{\zeta(j),\zeta^*(j)\}
+  Q(1,2)=\frac1N\left[\matrix{
+      \phi(1)^T\phi(2)&\phi(1)^T\phi(2)^*\cr
+      \phi(1)^\dagger\phi(2)&\phi(1)^\dagger\phi(2)^*
+  }\right]
+\end{equation}
+the result can be written, neglecting constant factors,
+\begin{equation}
+  \overline\mathcal N\simeq\int dQ\,e^{NS_\mathrm{eff}(Q)}
 \end{equation}
-the result can be written
+for an effective action functional of the supermatrix $Q$
 \begin{equation}
-  \Sigma=
+  S_{\mathrm{eff}}=
   \int d1\,d2\,\operatorname{Tr}\left(
-      \frac1{16}\left[
-        \matrix{1&1\cr1&1}
-      \right]A^{(p)}(1,2)+\frac p4\left[
-        \matrix{\epsilon&0\cr0&\epsilon^*}
-      \right](I-A(1,1))\delta(1,2)
+      \frac14\left[
+        \matrix{\frac14&\frac14\cr\frac14&\frac14}
+      \right]Q^{(p)}(1,2)-\frac p2\left[
+        \matrix{\frac\epsilon2&0\cr0&\frac{\epsilon^*}2}
+      \right](Q(1,1)-I)\delta(1,2)
     \right)
-    +\frac12\log\det A
+    +\frac12\log\det Q
 \end{equation}
 where the exponent in parentheses denotes element-wise exponentiation, and
 \begin{equation}
-  \delta(i,j)=(\theta(i)-\theta(j))(\bar\theta(i)-\bar\theta(j))
+  \delta(1,2)=(\bar\theta(1)-\bar\theta(2))(\theta(1)-\theta(2))
 \end{equation}
-is the super Dirac-$\delta$, and the determinant is a superdeterminant. This leads to the condition for a saddle point of
+is the superspace $\delta$-function, and the determinant is a superdeterminant.
+This leads to the condition for a saddle point of
 \begin{equation}
   0
-  =\frac{\partial\Sigma}{\partial A(1,2)}
-  =\frac p{16}A^{(p-1)}(1,2)-\frac p4\left[
-    \matrix{\epsilon&0\cr0&\epsilon^*}
+  =\frac{\partial S_\mathrm{eff}}{\partial Q(1,2)}
+  =\frac p{16}Q^{(p-1)}(1,2)-\frac p2\left[
+    \matrix{\frac\epsilon2&0\cr0&\frac{\epsilon^*}2}
   \right]\delta(1,2)
-  +\frac12A^{-1}(1,2)
+  +\frac12Q^{-1}(1,2)
 \end{equation}
-where $\odot$ denotes element-wise multiplication and the inverse superfield is defined by
+where the inverse supermatrix is defined by
 \begin{equation}
-  I\delta(1,2)=\int d3\,A^{-1}(1,3)A(3,2)
+  I\delta(1,2)=\int d3\,Q^{-1}(1,3)Q(3,2)
 \end{equation}
 Making such a transformation, we arrive at the saddle point equations
 \begin{eqnarray}
   0
-  &=\int d3\,\frac{\partial\Sigma}{\partial A(1,3)}A(3,2) \\
-  &=\frac p{16}\int d3A^{(p-1)}(1,3)A(3,2)-\frac p4\left[
-    \matrix{\epsilon&0\cr0&\epsilon^*}
-  \right]A(1,2)+\frac12I\delta(1,2)
+  &=\int d3\,\frac{\partial S_\mathrm{eff}}{\partial Q(1,3)}Q(3,2) \\
+  &=\frac p{16}\int d3\,Q^{(p-1)}(1,3)Q(3,2)-\frac p2\left[
+  \matrix{\frac\epsilon2&0\cr0&\frac{\epsilon^*}2}
+  \right]Q(1,2)+\frac12I\delta(1,2)
 \end{eqnarray}
-When expanded, the supermatrix $A$ contains nine independent bilinear
+When expanded, the supermatrix $Q$ contains nine independent bilinear
 combinations of the original variables: $z^\dagger z$, $\hat z^T z$, $\hat
 z^\dagger z$, $\hat z^T\hat z$, $\hat z^\dagger\hat z$, $\eta^\dagger\eta$,
 $\gamma^\dagger\gamma$, $\eta^\dagger\gamma$, and $\eta^T\gamma$. The saddle
@@ -1288,69 +1289,59 @@ up the bulk of its adjacent points in the sense of being susceptible to Stokes
 points. The distribution of these near neighbors in the complex plane therefore
 gives a sense of whether many Stokes lines should be expected, and when.
 
-To determine this, we perform the same Kac--Rice produce as in the previous
+To determine this, we perform the same Kac--Rice procedure as in the previous
 section, but now with two probe points, or replicas of the system. The number of
 stationary points with given energies $\epsilon_1$ and $\epsilon_2$ are
-\begin{equation}
-  \mathcal N(\epsilon_1,\epsilon_2)
-  =\int dx\,dz\,dz^*\,\delta(\partial H(x))\,\delta(\operatorname{Re}\partial H(z))\delta(\operatorname{Im}\partial H(z))|\det\operatorname{Hess}H(z)|^2|\det\operatorname{Hess}H(x)|
-\end{equation}
-
 \begin{eqnarray}
-  \mathcal N(\epsilon_1,\epsilon_2)
-  &=\int d\phi\,d\zeta^*d\zeta\exp\left\{
-    \int d\bar\theta\,d\theta \left[
-      H(\phi)+\operatorname{Re}H(\zeta)
-    \right]
-  \right\} \\
-  &=\int d\phi\,d\zeta^*d\zeta\exp\left\{
-    \int d\bar\theta\,d\theta \left[
-      H(\phi)+\frac12H(\zeta)+\frac12H^*(\zeta^*)
+  \mathcal N
+  &=\int d\phi_1\,d\phi_2^*\,d\phi_2\,\exp\left\{
+    \int d1 \left[
+      \tilde\mathcal S_p(\phi_1(1))+\operatorname{Re}\tilde\mathcal S_p(\phi_2(1))
     \right]
   \right\}
 \end{eqnarray}
-\begin{eqnarray}
-  \phi(i)&=x+\bar\theta(i)\eta_x^*+\eta_x^*\theta(i)+\hat x\bar\theta(i)\theta(i) \\
-  \zeta(i)&=z+\bar\theta(i)\eta_z^*+\eta_z\theta(i)+\hat z\bar\theta(i)\theta(i) \\
-  \zeta^*(i)&=z^*+\bar\theta(i)\eta_{z^*}^*+\eta_{z^*}\theta(i)+\hat z^*\bar\theta(i)\theta(i)
-\end{eqnarray}
 \begin{equation}
-  A(i,j)=\{\phi(i),\zeta(i),\zeta^*(i)\}\otimes\{\phi(j),\zeta(j),\zeta^*(j)\}
+  Q(1,2)
+  =\left[
+    \matrix{
+      \phi_1(1)^T\phi_1(2) & \phi_1(1)^T\phi_2(2) & \phi_1(1)^T\phi_2(2)^* \cr
+      \phi_2(1)^T\phi_1(2) & \phi_2(1)^T\phi_2(2) & \phi_2(1)^T\phi_2(2)^* \cr
+      \phi_2(1)^\dagger\phi_1(2) & \phi_2(1)^\dagger\phi_2(2) & \phi_2(1)^\dagger\phi_2(2)^*
+    }
+  \right]
 \end{equation}
 \begin{equation}
-  S
+  S_\mathrm{eff}
   =\int d1\,d2\,\operatorname{Tr}\left\{
-    \frac14\left[\matrix{1&\frac12&\frac12\cr\frac12&\frac14&\frac14\cr\frac12&\frac14&\frac14}\right]A^{(p)}(1,2)
-    +\frac p2\left[
+    \frac14\left[\matrix{1&\frac12&\frac12\cr\frac12&\frac14&\frac14\cr\frac12&\frac14&\frac14}\right]Q^{(p)}(1,2)
+    -\frac p2\left[
     \matrix{\epsilon_1&0&0\cr0&\frac12\epsilon_2&0\cr0&0&\frac12\epsilon_2^*}
       \right]\left(
-      I-A(1,1)
+      Q(1,1)-I
     \right)\delta(1,2)
-  \right\}+\frac12\det A
+  \right\}+\frac12\det Q
 \end{equation}
-where the exponent in parentheses denotes the element-wise power.
-
 \begin{equation}
-  0=\frac{\partial S}{\partial A(1,2)}
+  0=\frac{\partial S_\mathrm{eff}}{\partial Q(1,2)}
   =
-    \frac p4\left[\matrix{1&\frac12&\frac12\cr\frac12&\frac14&\frac14\cr\frac12&\frac14&\frac14}\right]\odot A^{(p-1)}(1,2)
+    \frac p4\left[\matrix{1&\frac12&\frac12\cr\frac12&\frac14&\frac14\cr\frac12&\frac14&\frac14}\right]\odot Q^{(p-1)}(1,2)
     -\frac p2\left[
   \matrix{\epsilon_1&0&0\cr0&\frac12\epsilon_2&0\cr0&0&\frac12\epsilon_2^*}\right]\delta(1,2)
-  +\frac12A^{-1}(1,2)
+  +\frac12Q^{-1}(1,2)
 \end{equation}
 where $\odot$ denotes element-wise multiplication.
 \begin{eqnarray}
   0
-  &=\int d3\,\frac{\partial S}{\partial A(1,3)}A(3,2) \\
+  &=\int d3\,\frac{\partial S_\mathrm{eff}}{\partial Q(1,3)}Q(3,2) \\
   &=\frac p4\int d3\,
-    \left\{\left[\matrix{1&\frac12&\frac12\cr\frac12&\frac14&\frac14\cr\frac12&\frac14&\frac14}\right]\odot A^{(p-1)}(1,3)\right\}A(3,2)
+    \left\{\left[\matrix{1&\frac12&\frac12\cr\frac12&\frac14&\frac14\cr\frac12&\frac14&\frac14}\right]\odot Q^{(p-1)}(1,3)\right\}Q(3,2)
     -\frac p2 \left[
-  \matrix{\epsilon_1&0&0\cr0&\frac12\epsilon_2&0\cr0&0&\frac12\epsilon_2^*}\right]A(1,2)
+  \matrix{\epsilon_1&0&0\cr0&\frac12\epsilon_2&0\cr0&0&\frac12\epsilon_2^*}\right]Q(1,2)
   +\frac12I\delta(1,2)
 \end{eqnarray}
 Despite being able to pose the saddle point problem in a compact way, a great
-deal of complexity lies within. The supermatrix $A$ depends on 35 independent
-bilinear products, and when the superfields are expanded produces 48 equations.
+deal of complexity lies within. The supermatrix $Q$ depends on 35 independent
+bilinear products, and when the superfields are expanded produces 48 (not entirely independent) equations.
 These equations can be split into 30 involving bilinear products of the
 fermionic fields and 18 without them. The 18 equations without fermionic
 bilinear products can be solved with a computer algebra package to eliminate 17
@@ -1359,7 +1350,7 @@ unfortunately more complicated.
 
 They can be simplified somewhat by examination of the real two-replica problem.
 There, all bilinear products involving fermionic fields from different
-replicas, like $\eta_x\cdot\eta_z$, vanish. This is related to the influence of
+replicas, like $\eta_1^T\eta_2$, vanish. This is related to the influence of
 the relative position of the two replicas to their spectra, with the vanishing
 being equivalent to having no influence, e.g., the value of the determinant at
 each stationary point is exactly what it would be in the one-replica problem
@@ -1367,9 +1358,8 @@ with the same invariants, e.g., energy and radius. Making this ansatz, the
 equations can be solved for the remaining 5 bilinear products, eliminating all
 the fermionic fields.
 
-This leaves three bilinear products: $z^\dagger z$, $z^\dagger x$, and
-$(z^\dagger x)^*$, or one real and one complex number. The first is the radius
-of the complex saddle, while the others are a generalization of the overlap in
+This leaves two bilinear products: $z_2^\dagger z_2$ and $z_2^\dagger x_1$, or one real and one complex number. The first is the radius
+of the complex saddle, while the other is a generalization of the overlap in
 the real case. For us, it will be more convenient to work in terms of the
 difference $\Delta z=z-x$ and the constants which characterize it, which are
 $\Delta=\Delta z^\dagger\Delta z=\|\Delta z\|^2$ and $\delta=\frac{\Delta z^T\Delta z}{|\Delta z^\dagger\Delta z|}$. Once again
-- 
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