From a9fc182c93f564d19a3d983e6696bd201b9670aa Mon Sep 17 00:00:00 2001
From: Jaron Kent-Dobias <jaron@kent-dobias.com>
Date: Tue, 16 May 2023 11:44:26 +0200
Subject: Some fixes in light of new information, and removal of inaccurate old
 information.

---
 2-point.tex | 67 ++++++++++++++++++++-----------------------------------------
 1 file changed, 22 insertions(+), 45 deletions(-)

(limited to '2-point.tex')

diff --git a/2-point.tex b/2-point.tex
index 5c4f58b..7b9e9e1 100644
--- a/2-point.tex
+++ b/2-point.tex
@@ -522,17 +522,17 @@ Because the matrices $C^{00}$, $R^{00}$, and $D^{00}$ are diagonal in this case,
 \begin{align}
   &\log\det
     \begin{bmatrix}
-      C^{00}&iR^{00}&C^{01}&iR^{01}&X_1\\
-      iR^{00}&D^{00}&iR^{10}&D^{01}&\hat X_1\\
-      C^{01})^T&iR^{10})^T&C^{11}&iR^{11}&X_2\\
-      iR^{01})^T&D^{10})^T&iR^{11}&D^{11}&\hat X_2\\
-      X_1)^T&\hat X_1)^T&X_2)^T&\hat X_2)^T&A
+      C^{00}&iR^{00}&C^{01}&iR^{01}&X_0\\
+      iR^{00}&D^{00}&iR^{10}&D^{01}&\hat X_0\\
+      (C^{01})^T&(iR^{10})^T&C^{11}&iR^{11}&X_1\\
+      (iR^{01})^T&(D^{10})^T&iR^{11}&D^{11}&\hat X_1\\
+      X_0^T&\hat X_0^T&X_1^T&\hat X_1^T&A
     \end{bmatrix}\\
   &=\log\det\left(
     A-
     \begin{bmatrix}
-      X_1\\\hat X_1\\X_2\\\hat X_2
-    \end{bmatrix})^T
+      X_0\\\hat X_0\\X_1\\\hat X_1
+    \end{bmatrix}^T
     \begin{bmatrix}
       C^{00}&iR^{00}&C^{01}&iR^{01}\\
       iR^{00}&D^{00}&iR^{10}&D^{01}\\
@@ -540,11 +540,18 @@ Because the matrices $C^{00}$, $R^{00}$, and $D^{00}$ are diagonal in this case,
       (iR^{01})^T&(D^{10})^T&iR^{11}&D^{11}\\
     \end{bmatrix}^{-1}
     \begin{bmatrix}
-      X_1\\\hat X_1\\X_2\\\hat X_2
+      X_0\\\hat X_0\\X_1\\\hat X_1
     \end{bmatrix}
   \right)
 \end{align}
 \begin{equation}
+    \begin{bmatrix}
+      C^{00}&iR^{00}&C^{01}&iR^{01}\\
+      iR^{00}&D^{00}&iR^{10}&D^{01}\\
+      (C^{01})^T&(iR^{10})^T&C^{11}&iR^{11}\\
+      (iR^{01})^T&(D^{10})^T&iR^{11}&D^{11}\\
+    \end{bmatrix}^{-1}
+    =
   \begin{bmatrix}
     A & B \\
     C & D
@@ -738,59 +745,29 @@ which is \emph{not} a hierarchical matrix! But, all these new hat variables are
   &&
   \hat X_1
   =\begin{bmatrix}
-    0&\cdots&0\\
+    \hat x_1^0&\cdots&\hat x_1^0\\
     \hat x_1^1&\cdots&\hat x_1^1\\
     \vdots&\ddots&\vdots\\
     \hat x_1^1&\cdots&\hat x_1^1
   \end{bmatrix}
 \end{align}
 
-\[
-  x_0^2\tilde d^{00}_\mathrm d-\hat x_0^2\tilde c^{00}_\mathrm d+2x\hat x\tilde r^{00}_\mathrm d
-  -2
-      \begin{bmatrix}
-        x_0q\tilde d^{00}_\mathrm d+\hat x_0q\tilde r^{00}_\mathrm d+x_0r_{10}\tilde r^{00}_\mathrm d-\hat x_0r_{10}\tilde c^{00}_\mathrm d
-        \\
-        i(x_0(r_{01}\tilde d^{00}_\mathrm d-d_{01}\tilde d^{00}_\mathrm d)
-        +\hat x_0(d_{01}\tilde c^{00}_\mathrm d+r_{01}\tilde r^{00}_\mathrm d))
-      \end{bmatrix}^T
-      \begin{bmatrix}
-        C^{11}&iR^{11}\\iR^{11}&D^{11}
-      \end{bmatrix}^{-1}
-    \begin{bmatrix}
-      X_1\\i\hat X_1
-    \end{bmatrix}
-\]
-\begin{align}
-  \hat c^{11}=\sum_{i}^n\sum_{j=2}^n[(C^{11}D^{11}+R^{11}R^{11})^{-1}D^{11}]_{ij}\\
-  \hat r^{11}=\sum_{i}^n\sum_{j=2}^n[(C^{11}D^{11}+R^{11}R^{11})^{-1}R^{11}]_{ij}\\
-  \hat d^{11}=\sum_{i}^n\sum_{j=2}^n[(C^{11}D^{11}+R^{11}R^{11})^{-1}D^{11}]_{ij}
-\end{align}
-\[
-  \begin{aligned}
-    &(\tilde d^{00}_\mathrm d\hat d^{11}q+\tilde r^{00}_\mathrm d\hat d^{11}r_{10}-\tilde r^{00}_\mathrm d\hat r^{11}d_{01}+\tilde d^{00}_\mathrm d\hat r^{11}r_{01})x_0x_1
-    +(\tilde r^{00}_\mathrm d\hat d^{11}q-\tilde c^{00}_\mathrm d\hat d^{11}r_{10}+\tilde c^{00}_\mathrm d\hat r^{11}d_{01}+\tilde r^{00}_\mathrm d\hat r^{11}r_{01})\hat x_0x_1 \\
-    &+(\tilde r^{00}_\mathrm d\hat c^{11}d_{01}-\tilde d^{00}_\mathrm d\hat c^{11}r_{01}+\tilde d^{00}_\mathrm d\hat r^{11}q+\tilde r^{00}_\mathrm d\hat r^{11}r_{10})x_0\hat x_1
-    -(\tilde c^{00}_\mathrm d\hat c^{11}d_{01}+\tilde r^{00}_\mathrm d\hat c^{11}r_{01}-\tilde r^{00}_\mathrm d\hat c^{11}q+\tilde c^{00}_\mathrm d\hat r^{11}r_{10})\hat x_0\hat x_1
-  \end{aligned}
-\]
-all a constant $\ell\times\ell$ matrix.
-
 \[
   \log\det(A-c)=\log\det A-\frac{c}{\sum_{i=0}^k(a_{i+1}-a_i)x_{i+1}}
 \]
 where $a_{k+1}=1$ and $x_{k+1}=1$.
 So the basic form of the action is (for replica symmetric $A$)
 \[
-  \frac14\beta^2f''(1)(1-a_0^2)+\frac12\log(1-a_0)+\frac12\frac{a_0}{1-a_0}+\frac12\beta\begin{bmatrix}x_0\\\hat x_0\\x_1\\\hat x_1\end{bmatrix}^TB\begin{bmatrix}x_0\\\hat x_0\\x_1\\\hat x_1\end{bmatrix}-\frac1{1-a_0}\frac12\begin{bmatrix}x_0\\\hat x_0\\x_1\\\hat x_1\end{bmatrix}^TC\begin{bmatrix}x_0\\\hat x_0\\x_1\\\hat x_1\end{bmatrix}
+  \frac14\beta^2f''(1)(1-a_0^2)+\frac12\log(1-a_0)+\frac12\frac{a_0}{1-a_0}+\frac12\begin{bmatrix}x_0\\\hat x_0\\x_1\\\hat x_1^1\\\hat x_1^0\end{bmatrix}^T\left(\beta B-\frac1{1-a_0}C\right)\begin{bmatrix}x_0\\\hat x_0\\x_1\\\hat x_1^1\\\hat x_1^0\end{bmatrix}
 \]
 for
 \[
   B=\begin{bmatrix}
-    \hat\beta_0f''(q)+r_{10}f'''(q)&f''(q)&0&0\\
-    f''(q)&0&0&0\\
-    0&0&-\hat\beta_1f''(q^{11}_0)-r^{11}_0f'''(q^{11}_0)&-f''(q_0^{11})\\
-    0&0&-f''(q_0^{11})&0
+    \hat\beta_0f''(q)+r_{10}f'''(q)&f''(q)&0&0&0\\
+    f''(q)&0&0&0&0\\
+    0&0&-\hat\beta_1f''(q^{11}_0)-r^{11}_0f'''(q^{11}_0)&-f''(q_0^{11})&0\\
+    0&0&-f''(q_0^{11})&0&0\\
+    0&0&0&0&0
   \end{bmatrix}
 \]
 Use $X$ for the big vector. Then
-- 
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