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Diffstat (limited to '2-point.tex')
| -rw-r--r-- | 2-point.tex | 192 |
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} |