Removed old unnecessary work.

1 files changed, 5 insertions, 63 deletions

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}