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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}