More work.

1 files changed, 148 insertions, 44 deletions

diff --git a/2-point.tex b/2-point.tex
index d4bff09..7e3c818 100644
--- a/2-point.tex
+++ b/2-point.tex
@@ -440,6 +440,9 @@ Because the matrices $C^{00}$, $R^{00}$, and $D^{00}$ are diagonal in this case,
 \begin{equation}
   \mu_1=\mu_0-\frac{f'(1)(f'''(1)+f''(1))-f''(1)^2}{f(1)(f'(1)+f''(1))-f'(1)^2}(E_0-E_0^*)(1-q)+O\big((1-q)^2\big)
 \end{equation}
+\begin{equation}
+  E_1=E_0+\frac12\frac{f'(1)(f'''(1)+f''(1))-f''(1)^2}{f(1)(f'(1)+f''(1))-f'(1)^2}(E_0-E_0^*)(1-q)^2+O\big((1-q)^3\big)
+\end{equation}
 
 \section{Isolated eigenvalue}
 
@@ -465,19 +468,20 @@ Because the matrices $C^{00}$, $R^{00}$, and $D^{00}$ are diagonal in this case,
     &=\lim_{m\to0}\int\left[\prod_{a=1}^m d\nu(\pmb\sigma_a\mid E_0,\mu_0)\right]\,F(\beta\mid E_1,\mu_1,q,\pmb\sigma_1)
   \end{aligned}
 \]
-\[
-  \sum_a^m(i\hat{\pmb\sigma}_{\pmb\sigma_a}\cdot\partial_a-\hat\beta_0)H(\pmb\sigma_a)
-  +
-  \sum_b^n(i\hat{\mathbf s}_{\mathbf s_b}\cdot\partial_b-\hat\beta_1)H(\mathbf s_b)
+\begin{equation}
+  \mathcal O(\mathbf t)=
+  \sum_a^m\delta(\mathbf t-\pmb\sigma_a)(i\hat{\pmb\sigma}_a\cdot\partial_\mathbf t-\hat\beta_0)
   +
-  \sum_c^\ell(\mathbf x_c\cdot\partial_{\mathbf s_1})^2H(\mathbf s_1)
-\]
+  \sum_b^n\delta(\mathbf t-\mathbf s_b)(i\hat{\mathbf s}_b\cdot\partial_\mathbf t-\hat\beta_1)
+  -\frac12
+  \delta(\mathbf t-\mathbf s_1)\beta\sum_c^\ell(\mathbf x_c\cdot\partial_{\mathbf t})^2
+\end{equation}
 \begin{align*}
   &\sum_{ab}^\ell(\mathbf x_a\cdot\partial_{\mathbf s_1})^2(\mathbf x_b\cdot\partial_{\mathbf s_1'})^2\overline{H(\mathbf s_1)H(\mathbf s_1')}\\
   &=(\mathbf x_a\cdot\mathbf s_1)^2(\mathbf x_b\cdot\mathbf s_1)^2f''''(1)
   +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) \\
-  &=f''(1)\sum_{ab}^\ell A_{ab}
+  &=\frac14\beta^2f''(1)\sum_{ab}^\ell A_{ab}^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)}\\
@@ -485,13 +489,13 @@ Because the matrices $C^{00}$, $R^{00}$, and $D^{00}$ are diagonal in this case,
   +i(\hat{\mathbf s}_b\cdot\mathbf s_1)(\mathbf x_a\cdot\mathbf s_b)^2f'''(C^{11}_{1b})
   +2i(\hat{\mathbf s}_b\cdot\mathbf x_a)(\mathbf x_a\cdot\mathbf s_b)f''(C^{11}_{1b}) \\
   &=
-  \sum_{a=1}^\ell\sum_{b=2}^n\left[
-    -\hat\beta_1(X^1_{ab})^2f''(C^{11}_{1b})-(X^1_{ab})^2R^{11}_{1b}f'''(C^{11}_{1b})
-    -2X^1_{ab}\hat X_{ab}f''(C^{11}_{1b})
+  \frac12\beta\sum_{a=1}^\ell\sum_{b=2}^n\left[
+    \hat\beta_1(X^1_{ab})^2f''(C^{11}_{1b})+(X^1_{ab})^2R^{11}_{1b}f'''(C^{11}_{1b})
+    +2X^1_{ab}\hat X_{ab}f''(C^{11}_{1b})
   \right] \\
-  &=\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})
-    -x_1\hat x_1\sum_{b=2}^nf''(C^{11}_{1b})
+  &=\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]
 \end{align*}
 \begin{align*}
@@ -500,13 +504,14 @@ Because the matrices $C^{00}$, $R^{00}$, and $D^{00}$ are diagonal in this case,
   +i(\hat{\pmb \sigma}_b\cdot\pmb \sigma_1)(\mathbf x_a\cdot\pmb \sigma_b)^2f'''(C^{01}_{1b})
   +2i(\hat{\pmb \sigma}_b\cdot\mathbf x_a)(\mathbf x_a\cdot\pmb \sigma_b)f''(C^{01}_{1b}) \\
   &=
-  \sum_{a=1}^\ell\sum_{b=1}^m\left[
-    -\hat\beta_0(X^0_{ab})^2f''(C^{01}_{1b})-(X^0_{ab})^2R^{01}_{1b}f'''(C^{01}_{1b})
-    -2X^0_{ab}\hat X^0_{ab}f''(C^{01}_{1b})
+  \frac12\beta\sum_{a=1}^\ell\sum_{b=1}^m\left[
+    \hat\beta_0(X^0_{ab})^2f''(C^{01}_{b1})
+    +(X^0_{ab})^2R^{10}_{b1}f'''(C^{01}_{b1})
+    +2X^0_{ab}\hat X^0_{ab}f''(C^{01}_{b1})
   \right] \\
-  &=\ell\left[-\hat\beta_0x_0^2f''(q)
-    -x_0^2r_{10}f'''(q)
-    -x_0\hat x_0f''(q)
+  &=\frac12\beta\ell\left[\hat\beta_0x_0^2f''(q)
+    +x_0^2r_{10}f'''(q)
+    +2x_0\hat x_0f''(q)
   \right]
 \end{align*}
 
@@ -543,28 +548,70 @@ Because the matrices $C^{00}$, $R^{00}$, and $D^{00}$ are diagonal in this case,
 \end{equation}
 \begin{equation}
     A=
+    \left(
       \begin{bmatrix}
         C^{00}&iR^{00}\\iR^{00}&D^{00}
-      \end{bmatrix}^{-1}
-      +
-      \begin{bmatrix}
-        C^{00}&iR^{00}\\iR^{00}&D^{00}
-      \end{bmatrix}^{-1}
+      \end{bmatrix}
+      -
       \begin{bmatrix}
         C^{01}&iR^{01}\\
         iR^{10}&D^{01}
       \end{bmatrix}
-      D
+      \begin{bmatrix}
+        C^{11}&iR^{11}\\iR^{11}&D^{11}
+      \end{bmatrix}^{-1}
       \begin{bmatrix}
         C^{01}&iR^{01}\\
         iR^{10}&D^{01}
       \end{bmatrix}^T
+    \right)^{-1}
+\end{equation}
+\begin{align}
+  \hat c^{11}=\sum_{ij}^n[(C^{11}D^{11}+R^{11}R^{11})^{-1}D^{11}]_{ij}\\
+  \hat r^{11}=\sum_{ij}^n[(C^{11}D^{11}+R^{11}R^{11})^{-1}R^{11}]_{ij}\\
+  \hat d^{11}=\sum_{ij}^n[(C^{11}D^{11}+R^{11}R^{11})^{-1}D^{11}]_{ij}
+\end{align}
+Based on the structure of the 01 matrices established above, the second term inside the inverse is
+\begin{equation}
+  \begin{aligned}
+     &   \begin{bmatrix}
+          C^{01}&iR^{01}\\iR^{10}&D^{01}
+        \end{bmatrix}
+        \begin{bmatrix}
+          C^{11}&iR^{11}\\iR^{11}&D^{11}
+        \end{bmatrix}^{-1}
+        \begin{bmatrix}
+          C^{01}&iR^{01}\\iR^{10}&D^{01}
+        \end{bmatrix}^T \\
+    &\qquad=\begin{bmatrix}
+      q^2\hat d^{11}+2qr_{10}\hat r^{11}-r_{10}^2\hat d^{11}
+      &
+      i\left[d_{01}(r_{10}\hat c^{11}-q\hat r^{11})+r_{01}(r_{10}\hat r^{11}+q\hat d^{11})\right]
+      \\
+      i\left[d_{01}(r_{10}\hat c^{11}-q\hat r^{11})+r_{01}(r_{10}\hat r^{11}+q\hat d^{11})\right]
+      &
+      d_{01}^2\hat c^{11}+2r_{01}d_{01}\hat r^{11}-r_{01}^2\hat d^{11}
+    \end{bmatrix}
+  \end{aligned}
+\end{equation}
+where each block is proportional to the $m\times m$ matrix
+\begin{equation}
+  \begin{bmatrix}
+    1&0&\cdots&0\\
+    0&0&\cdots&0\\
+    \vdots&\vdots&\ddots&\vdots\\
+    0&0&\cdots&0
+  \end{bmatrix}
+\end{equation}
+which is \emph{not} a hierarchical matrix! But, all these new hat variables are proportional to $n$, and will vanish when the eventual limit is taken. So, the whole contribution is zero, and
+\begin{equation}
+    A=
       \begin{bmatrix}
         C^{00}&iR^{00}\\iR^{00}&D^{00}
       \end{bmatrix}^{-1}
 \end{equation}
 \begin{equation}
-    B=-
+  B=-
       \begin{bmatrix}
         C^{00}&iR^{00}\\iR^{00}&D^{00}
       \end{bmatrix}^{-1}
@@ -572,18 +619,22 @@ Because the matrices $C^{00}$, $R^{00}$, and $D^{00}$ are diagonal in this case,
         C^{01}&iR^{01}\\
         iR^{10}&D^{01}
       \end{bmatrix}
-      D
+      \begin{bmatrix}
+        C^{11}&iR^{11}\\iR^{11}&D^{11}
+      \end{bmatrix}^{-1}
 \end{equation}
 \begin{equation}
     C=-
-      D
+      \begin{bmatrix}
+        C^{11}&iR^{11}\\iR^{11}&D^{11}
+      \end{bmatrix}^{-1}
       \begin{bmatrix}
         C^{01}&iR^{01}\\
         iR^{10}&D^{01}
       \end{bmatrix}^T
       \begin{bmatrix}
         C^{00}&iR^{00}\\iR^{00}&D^{00}
-      \end{bmatrix}^{-1}
+      \end{bmatrix}^{-1}=B^T
 \end{equation}
 \begin{equation}
     D=
@@ -608,31 +659,24 @@ Because the matrices $C^{00}$, $R^{00}$, and $D^{00}$ are diagonal in this case,
 
 \begin{equation}
     \begin{bmatrix}
-      X_0\\\hat X_0
+      X_0\\i\hat X_0
     \end{bmatrix}^TA
     \begin{bmatrix}
-      X_0\\\hat X_0
-    \end{bmatrix}
-    +
-    \begin{bmatrix}
-      X_1\\\hat X_1
-    \end{bmatrix}^TC
-    \begin{bmatrix}
-      X_0\\\hat X_0
+      X_0\\i\hat X_0
     \end{bmatrix}
     +
-    \begin{bmatrix}
-      X_0\\\hat X_0
+    2\begin{bmatrix}
+      X_0\\i\hat X_0
     \end{bmatrix}^TB
     \begin{bmatrix}
-      X_1\\\hat X_1
+      X_1\\i\hat X_1
     \end{bmatrix}
     +
     \begin{bmatrix}
-      X_1\\\hat X_1
+      X_1\\i\hat X_1
     \end{bmatrix}^TD
     \begin{bmatrix}
-      X_1\\\hat X_1
+      X_1\\i\hat X_1
     \end{bmatrix}
 \end{equation}
 
@@ -698,7 +742,67 @@ Because the matrices $C^{00}$, $R^{00}$, and $D^{00}$ are diagonal in this case,
 \end{align}
 
 \[
-  2(A-X^TC^{-1}X)^{-1}X^TC^{-1}
+  x_0^2\tilde d^{00}_\mathrm d-\hat x_0^2\tilde c^{00}_\mathrm d+2x\hat x\tilde r^{00}_\mathrm d
+  -2
+      \begin{bmatrix}
+        x_0q\tilde d^{00}_\mathrm d+\hat x_0q\tilde r^{00}_\mathrm d+x_0r_{10}\tilde r^{00}_\mathrm d-\hat x_0r_{10}\tilde c^{00}_\mathrm d
+        \\
+        i(x_0(r_{01}\tilde d^{00}_\mathrm d-d_{01}\tilde d^{00}_\mathrm d)
+        +\hat x_0(d_{01}\tilde c^{00}_\mathrm d+r_{01}\tilde r^{00}_\mathrm d))
+      \end{bmatrix}^T
+      \begin{bmatrix}
+        C^{11}&iR^{11}\\iR^{11}&D^{11}
+      \end{bmatrix}^{-1}
+    \begin{bmatrix}
+      X_1\\i\hat X_1
+    \end{bmatrix}
+\]
+\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 $A$)
+\[
+  \beta_x x_1+\frac14\beta^2f''(1)(1-a_0^2)+\frac12\log(1-a_0)+\frac12\frac{a_0}{1-a_0}+\frac12\beta\begin{bmatrix}x_0\\\hat x_0\\x_1\\\hat x_1\end{bmatrix}^TB\begin{bmatrix}x_0\\\hat x_0\\x_1\\\hat x_1\end{bmatrix}-\frac1{1-a_0}\frac12\begin{bmatrix}x_0\\\hat x_0\\x_1\\\hat x_1\end{bmatrix}^TC\begin{bmatrix}x_0\\\hat x_0\\x_1\\\hat x_1\end{bmatrix}
+\]
+Use $X$ for the big vector. Then
+\[
+  0=-\frac12\beta^2f''(1)a_0+\frac12\frac{a_0-X^TCX}{(1-a_0)^2}
+\]
+\[
+  0=\beta BX-\frac1{1-a_0}CX+\beta_X
+\]
+\[
+  X=\left(\frac1{1-a_0}C-\beta B\right)^{-1}\beta_X
+\]
+Suppose $a_0=1-y/\beta$
+\[
+  X=\beta^{-1}\left(\frac1y C-B\right)^{-1}\beta_X
+\]
+\[
+  0=-\frac12\beta f''(1)(\beta-y)+\beta\frac12\frac{\beta-y-\beta X^TCX}{y^2}
+\]
+For large $\beta$
+\[
+  0=-\frac12\beta^2 f''(1)+\beta^2\frac12\frac{1-X^TCX}{y^2}
+\]
+\[
+  y^2=\frac{1-X^TCX}{f''(1)}
 \]
 
 \paragraph{Acknowledgements}