From 393c6c861cdd1548ee059632fffd0a410e2a269f Mon Sep 17 00:00:00 2001
From: Jaron Kent-Dobias <jaron@kent-dobias.com>
Date: Mon, 15 May 2023 18:43:10 +0200
Subject: Lots of writing.

---
 2-point.tex | 54 ++++++++++++++++++++++++++++++++++++++----------------
 1 file changed, 38 insertions(+), 16 deletions(-)

(limited to '2-point.tex')

diff --git a/2-point.tex b/2-point.tex
index 7e3c818..5c4f58b 100644
--- a/2-point.tex
+++ b/2-point.tex
@@ -482,6 +482,7 @@ Because the matrices $C^{00}$, $R^{00}$, and $D^{00}$ are diagonal in this case,
   +2(\mathbf x_a\cdot\mathbf s_1)(\mathbf x_b\cdot\mathbf s_1)(\mathbf x_a\cdot\mathbf x_b)f'''(1)
   +(\mathbf x_a\cdot\mathbf x_b)^2f''(1) \\
   &=\frac14\beta^2f''(1)\sum_{ab}^\ell A_{ab}^2
+  =\frac14\beta^2f''(1)(1-a_0^2)
 \end{align*}
 \begin{align*}
   &\sum_{a}^\ell\sum_b^n(i\hat{\mathbf s}_{\mathbf s_b}\cdot\partial_b-\hat\beta_1)(\mathbf x_a\cdot\partial_{\mathbf s_1})^2\overline{H(\mathbf s_1)H(\mathbf s_b)}\\
@@ -496,6 +497,9 @@ Because the matrices $C^{00}$, $R^{00}$, and $D^{00}$ are diagonal in this case,
   &=\frac12\beta\ell\left[\hat\beta_1x_1^2\sum_{b=2}^nf''(C^{11}_{1b})
     +x_1^2\sum_{b=2}^nR^{11}_{1b}f'''(C^{11}_{1b})
     +2x_1\hat x_1\sum_{b=2}^nf''(C^{11}_{1b})
+  \right] \\
+  &=-\frac12\beta\left[
+    (\hat\beta_1f''(q^{11}_0)+r^{11}_0f'''(q^{11}_0))x_1^2+2f''(q^{11}_0)x_1\hat x_1
   \right]
 \end{align*}
 \begin{align*}
@@ -509,9 +513,9 @@ Because the matrices $C^{00}$, $R^{00}$, and $D^{00}$ are diagonal in this case,
     +(X^0_{ab})^2R^{10}_{b1}f'''(C^{01}_{b1})
     +2X^0_{ab}\hat X^0_{ab}f''(C^{01}_{b1})
   \right] \\
-  &=\frac12\beta\ell\left[\hat\beta_0x_0^2f''(q)
-    +x_0^2r_{10}f'''(q)
-    +2x_0\hat x_0f''(q)
+  &=\frac12\beta\left[\big(\hat\beta_0f''(q)
+    +r_{10}f'''(q)\big)x_0^2
+    +2f''(q)x_0\hat x_0
   \right]
 \end{align*}
 
@@ -718,7 +722,7 @@ which is \emph{not} a hierarchical matrix! But, all these new hat variables are
   X_1
   =
   \begin{subarray}{l}
-    \hphantom{[}\begin{array}{ccc}\leftarrow&n&\rightarrow\end{array}\hphantom{\Bigg]}\\
+    \hphantom{[}\begin{array}{ccc}\leftarrow&\ell&\rightarrow\end{array}\hphantom{\Bigg]}\\
   \left[
     \begin{array}{ccc}
       0&\cdots&0\\
@@ -727,14 +731,14 @@ which is \emph{not} a hierarchical matrix! But, all these new hat variables are
       x_1&\cdots&x_1
     \end{array}
   \right]\begin{array}{c}
-    \\\uparrow\\m-1\\\downarrow
+    \\\uparrow\\n-1\\\downarrow
   \end{array}\\
   \vphantom{\begin{array}{c}n\end{array}}
   \end{subarray}
   &&
   \hat X_1
   =\begin{bmatrix}
-    \hat x_1^0&\cdots&\hat x_1^0\\
+    0&\cdots&0\\
     \hat x_1^1&\cdots&\hat x_1^1\\
     \vdots&\ddots&\vdots\\
     \hat x_1^1&\cdots&\hat x_1^1
@@ -778,32 +782,50 @@ all a constant $\ell\times\ell$ matrix.
 where $a_{k+1}=1$ and $x_{k+1}=1$.
 So the basic form of the action is (for replica symmetric $A$)
 \[
-  \beta_x x_1+\frac14\beta^2f''(1)(1-a_0^2)+\frac12\log(1-a_0)+\frac12\frac{a_0}{1-a_0}+\frac12\beta\begin{bmatrix}x_0\\\hat x_0\\x_1\\\hat x_1\end{bmatrix}^TB\begin{bmatrix}x_0\\\hat x_0\\x_1\\\hat x_1\end{bmatrix}-\frac1{1-a_0}\frac12\begin{bmatrix}x_0\\\hat x_0\\x_1\\\hat x_1\end{bmatrix}^TC\begin{bmatrix}x_0\\\hat x_0\\x_1\\\hat x_1\end{bmatrix}
+  \frac14\beta^2f''(1)(1-a_0^2)+\frac12\log(1-a_0)+\frac12\frac{a_0}{1-a_0}+\frac12\beta\begin{bmatrix}x_0\\\hat x_0\\x_1\\\hat x_1\end{bmatrix}^TB\begin{bmatrix}x_0\\\hat x_0\\x_1\\\hat x_1\end{bmatrix}-\frac1{1-a_0}\frac12\begin{bmatrix}x_0\\\hat x_0\\x_1\\\hat x_1\end{bmatrix}^TC\begin{bmatrix}x_0\\\hat x_0\\x_1\\\hat x_1\end{bmatrix}
+\]
+for
+\[
+  B=\begin{bmatrix}
+    \hat\beta_0f''(q)+r_{10}f'''(q)&f''(q)&0&0\\
+    f''(q)&0&0&0\\
+    0&0&-\hat\beta_1f''(q^{11}_0)-r^{11}_0f'''(q^{11}_0)&-f''(q_0^{11})\\
+    0&0&-f''(q_0^{11})&0
+  \end{bmatrix}
 \]
 Use $X$ for the big vector. Then
 \[
-  0=-\frac12\beta^2f''(1)a_0+\frac12\frac{a_0-X^TCX}{(1-a_0)^2}
+  0=-\beta^2f''(1)a_0+\frac{a_0-X^TCX}{(1-a_0)^2}
+\]
+\[
+  0=\bigg(\beta B-\frac1{1-a_0}C\bigg)X
 \]
+For a non-trivial solution, we require the following: change of basis in $X$ to diagonalize the matrix $\beta B-C/(1-a_0)$. Then, all of the new basis vectors of $X$ are zero except one, and the second equation is satisfied by tuning $a_0$ to make the coefficient zero. Finally, the first equation is satisfied by choice of the magnitude of this basis vector.
+Suppose $a_0=1-1/(y\beta)$, $X\sim X+O(1/\beta)$. Then
 \[
-  0=\beta BX-\frac1{1-a_0}CX+\beta_X
+  0=-f''(1)(1-(y\beta)^{-1})+y^2\left(1-(y\beta)^{-1}-X^TCX\right)
 \]
+For large $\beta$
 \[
-  X=\left(\frac1{1-a_0}C-\beta B\right)^{-1}\beta_X
+  0=-f''(1)+y^2(1-X^TCX)
 \]
-Suppose $a_0=1-y/\beta$
 \[
-  X=\beta^{-1}\left(\frac1y C-B\right)^{-1}\beta_X
+  0=(B-yC)X
 \]
+which gives
 \[
-  0=-\frac12\beta f''(1)(\beta-y)+\beta\frac12\frac{\beta-y-\beta X^TCX}{y^2}
+  \mathcal S=\frac14\beta^2f''(1)\big(2(y\beta)^{-1}-(y\beta)^{-2}\big)-\frac12\log(y\beta)+\frac12y\beta(1-(y\beta)^{-1})+\frac12\beta X^TBX-\frac12y\beta X^TCX
 \]
-For large $\beta$
+so
 \[
-  0=-\frac12\beta^2 f''(1)+\beta^2\frac12\frac{1-X^TCX}{y^2}
+  E_{gs}=\lim_{\beta\to\infty}(\mathcal S/\beta)=\frac12\left(y+\frac1yf''(1)\right)+\frac12X^T(B-yC)X
+  =\frac12\left(y+\frac1yf''(1)\right)
 \]
+assuming the last equation is satisfied. The trivial solution, which gives the bottom of the semicircle, is for $X=0$, so the first equation is $y^2=f''(1)$, and
 \[
-  y^2=\frac{1-X^TCX}{f''(1)}
+  E_{gs}=-\sqrt{f''(1)}
 \]
+as expected. We need to first the nontrivial solutions with nonzero $X$, but because the coefficients are so nasty this will be a numeric problem... Specifically, we are good for $y$ where one of the eigenvalues of $B-yC$ is zero.
 
 \paragraph{Acknowledgements}
 
-- 
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