From 557887f782cf28547e52f2212b4b18ea3501efc6 Mon Sep 17 00:00:00 2001
From: Jaron Kent-Dobias <jaron@kent-dobias.com>
Date: Mon, 14 Feb 2022 18:25:20 +0100
Subject: Updated things relating to the two-replica bound.

---
 figs/bound.pdf         | Bin 30251 -> 32466 bytes
 figs/example_bound.pdf | Bin 0 -> 126951 bytes
 stokes.tex             |  43 +++++++++++++++++++++++++++----------------
 3 files changed, 27 insertions(+), 16 deletions(-)
 create mode 100644 figs/example_bound.pdf

diff --git a/figs/bound.pdf b/figs/bound.pdf
index 0e6814f..aa7f4d7 100644
Binary files a/figs/bound.pdf and b/figs/bound.pdf differ
diff --git a/figs/example_bound.pdf b/figs/example_bound.pdf
new file mode 100644
index 0000000..1cbbe79
Binary files /dev/null and b/figs/example_bound.pdf differ
diff --git a/stokes.tex b/stokes.tex
index fa65c10..5674fdd 100644
--- a/stokes.tex
+++ b/stokes.tex
@@ -921,8 +921,8 @@ Green function \cite{Livan_2018_Introduction} gives
 \end{equation}
 The average is then made over
 $J$ and Hubbard--Stratonovich is used to change variables to the replica matrices
-$N\alpha_{\alpha\beta}=(\zeta^{(\alpha)})^\dagger\zeta^{(\beta)}$ and
-$N\chi_{\alpha\beta}=(\zeta^{(\alpha)})^T\zeta^{(\beta)}$, and a series of
+$N\alpha_{\alpha\beta}=(z^{(\alpha)})^\dagger z^{(\beta)}$ and
+$N\chi_{\alpha\beta}=(z^{(\alpha)})^Tz^{(\beta)}$, and a series of
 replica vectors. The replica-symmetric ansatz leaves all replica vectors
 zero, and $\alpha_{\alpha\beta}=\alpha_0\delta_{\alpha\beta}$,
 $\chi_{\alpha\beta}=\chi_0\delta_{\alpha\beta}$. The result is
@@ -943,7 +943,7 @@ $\chi_{\alpha\beta}=\chi_0\delta_{\alpha\beta}$. The result is
   \includegraphics{figs/spectra_1.5.pdf}
 
   \caption{
-    Eigenvalue and singular value spectra of a random matrix $A+\lambda_0I$, where the entries of $A$ are complex-normal distributed with $\overline{|A|^2}=A_0=\sqrt{5/4}$ and $\overline{A^2}=C_0=\frac34e^{-i3\pi/4}$.
+    Eigenvalue and singular value spectra of a random matrix $A+\lambda_0I$, where the entries of $A$ are complex-normal distributed with $\overline{|A|^2}=A_0=5/4$ and $\overline{A^2}=C_0=\frac34e^{-i3\pi/4}$.
     The diaginal shifts differ in each plot, with (a) $\lambda_0=0$, (b)
     $\lambda_0=\frac12|\lambda_{\mathrm{th}}|$, (c)
     $\lambda_0=|\lambda_{\mathrm{th}}|$, and (d)
@@ -1440,23 +1440,29 @@ 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. Making this ansatz, the equations
-can be solved for the remaining 5 bilinear products, eliminating all the
-fermionic fields.
+replicas, like $\eta_x\cdot\eta_z$, 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
+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
 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
-$\Delta=\Delta z^\dagger\Delta z$ and $\delta=\Delta z^T\Delta z$. Once again
+$\Delta=\Delta z^\dagger\Delta z=\|\Delta z\|^2$ and $\delta=\frac{\Delta z^T\Delta z}{|\Delta z^\dagger\Delta z|}$. Once again
 we have one real (and strictly positive) variable $\Delta$ and one complex
 variable $\delta$.
 
-Though the value of $\delta$ is bounded by $\Delta$ by $|\delta|\leq\Delta$, in
-reality this bound is not the relevant one, because we are confined on the
-manifold $N=z^2$. This bound is most easily established by returning to a
-$2N$-dimensional real problem, with $x=x_1$ and $z=x_2+iy_2$. The constraint gives $x_2^Ty_2=0$, $x_1^Tx_1=1$, and $x_2^Tx_2=1+y_2^Ty_2$. Then
+Though the value of $\delta$ is bounded  by $|\delta|\leq1$ as a result of the
+inequality $\Delta z^T\Delta z\leq\|\Delta z\|^2$, in reality this bound is not
+the relevant one, because we are confined on the manifold $N=z^2$. The relevant
+bound is most easily established by returning to a $2N$-dimensional real
+problem, with $x=x_1$ and $z=x_2+iy_2$. The constraint gives $x_2^Ty_2=0$,
+$x_1^Tx_1=1$, and $x_2^Tx_2=1+y_2^Ty_2$. Then, by their definitions,
 \begin{equation}
   \Delta=1+x_2^Tx_2+y_2^Ty_2-2x_1^Tx_2=2(1+y_2^Ty_2-x_1^Tx_2)
 \end{equation}
@@ -1464,19 +1470,24 @@ $2N$-dimensional real problem, with $x=x_1$ and $z=x_2+iy_2$. The constraint giv
   \Delta=2(1+|y_2|^2-\sqrt{1-|y_2|^2}\cos\theta_{xx})
 \end{equation}
 \begin{eqnarray}
-  \delta&=2-2x_1^Tx_2-2ix_1^Ty_2=2(1-|x_2|\cos\theta_{xx}-i|y_2|\cos\theta_{xy}) \\
+  \delta\Delta&=2-2x_1^Tx_2-2ix_1^Ty_2=2(1-|x_2|\cos\theta_{xx}-i|y_2|\cos\theta_{xy}) \\
         &=2(1-\sqrt{1-|y_2|^2}\cos\theta_{xx}-i|y_2|\cos\theta_{xy})
 \end{eqnarray}
-$\cos^2\theta_{xy}\leq1-\cos^2\theta_{xx}$
-These equations along with the inequality produce the required bound on $|\delta|$ as a function of $\Delta$ and $\arg\delta$.
+There is also an inequality between the angles $\theta_{xx}$ and $\theta_{xy}$
+between $x_1$ and $x_2$ and $y_2$, respectively, which takes that form
+$\cos^2\theta_{xy}+\cos^2\theta_{xx}\leq1$. This results from the fact that
+$x_2$ and $y_2$ are orthogonal, a result of the constraint.  These equations
+along with the inequality produce the required bound on $|\delta|$ as a
+function of $\Delta$ and $\arg\delta$.
 
 \begin{figure}
   \includegraphics{figs/bound.pdf}
+  \includegraphics{figs/example_bound.pdf}
 
   \caption{
     The line bounding $\delta$ in the complex plane as a function of
-    $\Delta=1,2,\ldots,6$ (inner to outer). Notice that for $\Delta\leq4$,
-    $|\delta|=\Delta$ is saturated for positive real $\delta$, but is not for
+    $\Delta=0,1,2,\ldots,6$ (outer to inner). Notice that for $\Delta\leq4$,
+    $|\delta|=1$ is saturated for positive real $\delta$, but is not for
     $\Delta>4$, and $\Delta=4$ has a cusp in the boundary. This is due to
     $\Delta=4$ corresponding to the maximum distance between any two points on
     the real sphere.
-- 
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