From e3ed82dd20340b32281986d90de6cac54ced38ba Mon Sep 17 00:00:00 2001
From: Jaron Kent-Dobias <jaron@kent-dobias.com>
Date: Thu, 11 May 2023 16:52:53 +0200
Subject: Removed old unnecessary work.
---
2-point.tex | 68 +++++--------------------------------------------------------
1 file changed, 5 insertions(+), 63 deletions(-)
(limited to '2-point.tex')
diff --git a/2-point.tex b/2-point.tex
index 380d677..c5e404e 100644
--- a/2-point.tex
+++ b/2-point.tex
@@ -372,63 +372,12 @@ then the result is
\end{equation}
where each block is a constant $n\times n$ matrix.
-
-Define $\tilde C=C-\tilde c$, $\tilde R=R-\tilde r$, $\tilde D=D-\tilde d$. Then
-\begin{align*}
- &\Sigma_{12}
- =\mathcal D(\mu_1)+\hat\beta_1E_1-\hat\mu_1
- +\hat\beta_0\hat\beta_1f(q)+(\hat\beta_0r_{01}+\hat\beta_1r_{10}-d_{01})f'(q)+r_{01}r_{10}f''(q)
- \\&
- +\lim_{n\to0}\frac1n\bigg\{
- \frac12\sum_{ab}\left[
- \hat\beta_1^2f(C_{ab})+(2\hat\beta_1R_{ab}-D_{ab})f'(C_{ab})+R_{ab}^2f''(C_{ab})
- \right]
- \\&
- +\hat\mu_1\operatorname{Tr}C-\mu_1\operatorname{Tr}R
- +\frac12\log\det((C-\tilde c)(D-\tilde d)+(R-\tilde r)^2)
- \bigg\}
-\end{align*}
-These equations for $D^*$ are the same as those for the unpinned case, or
-\[
- 0=-\frac12f'(C)+\frac12((C-\tilde c)(D-\tilde d)+(R-\tilde r)^2)^{-1}(C-\tilde c)
-\]
-Solving, we get
-\[
- D=\tilde d+f'(C)^{-1}-(C-\tilde c)^{-1}(R-\tilde r)^2
-\]
-\begin{align*}
- &\Sigma_{12}
- =\mathcal D(\mu_1)+\hat\beta_1E_1-\frac12\hat\mu_1
- +\hat\beta_0\hat\beta_1f(q)+(\hat\beta_0r_{01}+\hat\beta_1r_{10}-d_{01})f'(q)+r_{01}r_{10}f''(q)
- \\&
- +\lim_{n\to0}\frac1n\bigg\{
- \frac12\sum_{ab}\left[
- \hat\beta_1^2f(C_{ab})+(2\hat\beta_1R_{ab}-(f'(C)^{-1}_{ab}-((C-\tilde c)^{-1}(R-\tilde r)^2)_{ab}-\tilde d))f'(C_{ab})+R_{ab}^2f''(C_{ab})
- \right]
- \\&
- +\frac12\hat\mu_1\operatorname{Tr}C-\mu_1\operatorname{Tr}R
- +\frac12\log\det(C-\tilde c)-\frac12\log\det f'(C)
- \bigg\}
-\end{align*}
-
-
-
-\begin{align*}
- 0&=C^*-f'(C)(C^*D^*+R^*R^*) \\
- 0&=\big[\hat\beta_1f'(C)-\mu_1I+R\odot f''(C)\big](C-\tilde c)+f'(C)(R-\tilde r)
-\end{align*}
-
-\begin{align*}
- 0&=-f'(q)+\frac12(C^*D^*+R^*R^*)^{-1}_{ij}\left(
- C^*_{jk}\frac{D^*_{ki}}{d_{01}}
- +
- 2R^*_{jk}\frac{R^*_{ki}}{d_{01}}
- \right)
-\end{align*}
-
\section{Replica symmetric case}
-With a replica-symmetric ansatz, these conditions are
+We focus now on models whose equilibrium has at most one level of replica
+symmetry breaking, which corresponds to a replica symmetric complexity.
+For these models, the saddle point parameters for the reference stationary
+point are well known, and take the values.
\begin{align}
\hat\beta_0
&=\frac{(\epsilon_0+\mu_0)f'(1)+\epsilon_0f''(1)}{f(1)\big(f'(1)+f''(1)\big)-f'(1)^2}\\
@@ -441,7 +390,7 @@ With a replica-symmetric ansatz, these conditions are
\right)^2
\end{align}
-In the replica symmetric case,
+Because the matrices $C^{00}$, $R^{00}$, and $D^{00}$ are diagonal in this case, the diagonals of the inverse block matrix from above are simple expressions:
\begin{align}
\tilde c_\mathrm d^{00}=\frac1{(r^{00}_\mathrm d)^2+d^{00}_\mathrm d} &&
\tilde r_\mathrm d^{00}=\frac{r^{00}_\mathrm d}{(r^{00}_\mathrm d)^2+d^{00}_\mathrm d} &&
@@ -458,13 +407,6 @@ In the replica symmetric case,
\right\}
\end{equation}
-What about the average for the Hessian terms?
-
-\[
- \overline{
- |\det\operatorname{Hess}H(s_0)|\delta(\mu_0-\operatorname{Tr}\operatorname{Hess}H(s_0))|\det\operatorname{Hess}H(s_a)|\delta(\mu_1-\operatorname{Tr}\operatorname{Hess}H(s_a))
- }
-\]
\section{Isolated eigenvalue}
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