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-rw-r--r--2-point.tex67
1 files changed, 22 insertions, 45 deletions
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
@@ -738,7 +745,7 @@ 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
@@ -746,51 +753,21 @@ which is \emph{not} a hierarchical matrix! But, all these new hat variables are
\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