Some writing and small fixes.

1 files changed, 32 insertions, 31 deletions

diff --git a/2-point.tex b/2-point.tex
index 3ab4b07..b6f13c4 100644
--- a/2-point.tex
+++ b/2-point.tex
@@ -140,7 +140,12 @@ where the function $f$ is defined from the coefficients $a_p$ by
 \end{equation}
 In this paper, we will focus on models with a replica symmetric complexity, but
 many of the intermediate formulae are valid for arbitrary replica symmetry
-breakings.
+breakings. At most {\oldstylenums1}\textsc{rsb} in the equilibrium is guaranteed if the function
+$\chi(q)=f''(q)^{-1/2}$ is convex. The complexity at the ground state must
+reflect the structure of equilibrium, and therefore be replica symmetric. We
+are not aware of any result guaranteeing this for the complexity away from the
+ground state, but we check that our replica-symmetric solutions satisfy the
+saddle point equations at 1RSB.
 
 To enforce the spherical constraint at stationary points, we make use of a Lagrange multiplier $\omega$. This results in the extremal problem
 \begin{equation}
@@ -408,7 +413,20 @@ is the action for the ordinary, one-point complexity, and remainder is given by
 Because of the structure of this problem in the twin limits of $m$ and $n$ to
 zero, the parameters $\mathcal Q_{00}$ can be evaluated at a saddle point of
 $\mathcal S_0$ alone. This means that these parameters will take the same value
-they take when the ordinary, 1-point complexity is calculated.
+they take when the ordinary, 1-point complexity is calculated. For a replica
+symmetric complexity of the reference point, this results in
+\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}\\
+  r_\mathrm d^{00}
+  &=\frac{\mu_0f(1)+\epsilon_0f'(1)}{f(1)\big(f'(1)+f''(1)\big)-f'(1)^2} \\
+  d_\mathrm d^{00}
+  &=\frac1{f'(1)}
+  -\left(
+    \frac{\mu_0f(1)+\epsilon_0f'(1)}{f(1)\big(f'(1)+f''(1)\big)-f'(1)^2}
+  \right)^2
+\end{align}
+
 
 In general, we except the $m\times n$ matrices $C^{01}$, $R^{01}$, $R^{10}$,
 and $D^{01}$ to have constant \emph{rows} of length $n$, with blocks of rows
@@ -500,23 +518,6 @@ then the result is
 \end{equation}
 where each block is a constant $n\times n$ matrix.
 
-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}\\
-  r_\mathrm d^{00}
-  &=\frac{\mu_0f(1)+\epsilon_0f'(1)}{f(1)\big(f'(1)+f''(1)\big)-f'(1)^2} \\
-  d_\mathrm d^{00}
-  &=\frac1{f'(1)}
-  -\left(
-    \frac{\mu_0f(1)+\epsilon_0f'(1)}{f(1)\big(f'(1)+f''(1)\big)-f'(1)^2}
-  \right)^2
-\end{align}
-
-$(r^{00}_\mathrm d)^2+d^{00}_\mathrm d=1/f'(1)$
 
 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}
@@ -669,12 +670,12 @@ spherical constraint. The free energy of this model given a point $\mathbf s$
 and a specific realization of the disordered Hamiltonian is
 \begin{equation}
   \begin{aligned}
-    \beta F_H(\beta\mid\mathbf s)
+    \beta F_H(\beta\mid\mathbf s,\omega)
     &=-\frac1N\log\left(\int d\mathbf x\,\delta(\mathbf x\cdot\mathbf s)\delta(\|\mathbf x\|^2-N)\exp\left\{
-      -\beta\frac12\mathbf x^T\partial\partial H(\mathbf s)\mathbf x
+        -\beta\frac12\mathbf x^T\operatorname{Hess}H(\mathbf s,\omega)\mathbf x
     \right\}\right) \\
     &=-\lim_{\ell\to0}\frac1N\frac\partial{\partial\ell}\int\left[\prod_{\alpha=1}^\ell d\mathbf x_\alpha\,\delta(\mathbf x_\alpha^T\mathbf s)\delta(N-\mathbf x_\alpha^T\mathbf x_\alpha)\exp\left\{
-      -\beta\frac12\mathbf x^T_\alpha\partial\partial H(\mathbf s)\mathbf x_\alpha
+      -\beta\frac12\mathbf x^T_\alpha\big(\partial\partial H(\mathbf s)+\omega I\big)\mathbf x_\alpha
     \right\}\right]
   \end{aligned}
 \end{equation}
@@ -688,8 +689,8 @@ such points, giving
 \begin{equation}
   \begin{aligned}
     F_H(\beta\mid E_1,\mu_1,q,\pmb\sigma)
-    &=\int\frac{d\nu_H(\mathbf s,\omega\mid E_1,\mu_1)\delta(Nq-\pmb\sigma\cdot\mathbf s)}{\int d\nu_H(\mathbf s',\omega'\mid E_1,\mu_1)\delta(Nq-\pmb\sigma\cdot\mathbf s')}F_H(\beta\mid\mathbf s) \\
-    &=\lim_{n\to0}\int\left[\prod_{a=1}^nd\nu_H(\mathbf s_a,\omega_a\mid E_1,\mu_1)\,\delta(Nq-\pmb\sigma\cdot\mathbf s_a)\right]F_H(\beta\mid\mathbf s_1)
+    &=\int\frac{d\nu_H(\mathbf s,\omega\mid E_1,\mu_1)\delta(Nq-\pmb\sigma\cdot\mathbf s)}{\int d\nu_H(\mathbf s',\omega'\mid E_1,\mu_1)\delta(Nq-\pmb\sigma\cdot\mathbf s')}F_H(\beta\mid\mathbf s,\omega) \\
+    &=\lim_{n\to0}\int\left[\prod_{a=1}^nd\nu_H(\mathbf s_a,\omega_a\mid E_1,\mu_1)\,\delta(Nq-\pmb\sigma\cdot\mathbf s_a)\right]F_H(\beta\mid\mathbf s_1,\omega_1)
   \end{aligned}
 \end{equation}
 again anticipating the use of replicas. Finally, the reference configuration $\pmb\sigma$ should itself be a stationary point of $H$ with its own energy density and stability. Averaging over these conditions gives
@@ -808,7 +809,7 @@ Using the same methodology as above, the disorder-dependant terms are captured i
 \begin{equation}
   \begin{aligned}
     \ell\mathcal S_x(\mathcal X\mid\mathcal Q)
-    =
+    =-\frac12\beta\mu+
     \frac12\beta\sum_b^\ell\bigg\{
       \frac12\beta&f''(1)\sum_a^lA_{ab}^2\\
     &+\sum_a^m\left[
@@ -896,7 +897,7 @@ Using the same methodology as above, the disorder-dependant terms are captured i
   \begin{aligned}
     \frac2\beta\lim_{\ell\to0}\mathcal S_x(\mathcal X\mid\mathcal Q)
     &=
-      \frac12\beta f''(1)(1-a_0^2)
+      -\mu+\frac12\beta f''(1)(1-a_0^2)
     +\big(\hat\beta_0f''(q)+r_{10}f'''(q)\big)x_0^2
       +2f''(q)x_0\hat x_0 \\
     &-
@@ -1074,7 +1075,7 @@ which is \emph{not} a hierarchical matrix! But, all these new hat variables are
 where $a_{k+1}=1$ and $x_{k+1}=1$.
 So the basic form of the action is (for replica symmetric $A$)
 \[
-  \frac14\beta^2f''(1)(1-a_0^2)+\frac12\log(1-a_0)+\frac12\frac{a_0}{1-a_0}+\frac12\begin{bmatrix}x_0\\\hat x_0\\x_1\\\hat x_1^1\\\hat x_1^0\end{bmatrix}^T\left(\beta B-\frac1{1-a_0}C\right)\begin{bmatrix}x_0\\\hat x_0\\x_1\\\hat x_1^1\\\hat x_1^0\end{bmatrix}
+  -\mu+\frac14\beta^2f''(1)(1-a_0^2)+\frac12\log(1-a_0)+\frac12\frac{a_0}{1-a_0}+\frac12\begin{bmatrix}x_0\\\hat x_0\\x_1\\\hat x_1^1\\\hat x_1^0\end{bmatrix}^T\left(\beta B-\frac1{1-a_0}C\right)\begin{bmatrix}x_0\\\hat x_0\\x_1\\\hat x_1^1\\\hat x_1^0\end{bmatrix}
 \]
 for
 \[
@@ -1160,17 +1161,17 @@ For large $\beta$
 \]
 which gives
 \[
-  \mathcal S=\frac14\beta^2f''(1)\big(2(y\beta)^{-1}-(y\beta)^{-2}\big)-\frac12\log(y\beta)+\frac12y\beta(1-(y\beta)^{-1})+\frac12\beta X^TBX-\frac12y\beta X^TCX
+  \mathcal S=-\frac12\beta\mu+\frac14\beta^2f''(1)\big(2(y\beta)^{-1}-(y\beta)^{-2}\big)-\frac12\log(y\beta)+\frac12y\beta(1-(y\beta)^{-1})+\frac12\beta X^TBX-\frac12y\beta X^TCX
 \]
 so
 \[
   \lambda_\mathrm{min}=-2\lim_{\beta\to\infty}\frac{\partial\mathcal S}{\partial\beta}
-  =-\left(y+\frac1yf''(1)\right)-X^T(B-yC)X
-  =-\left(y+\frac1yf''(1)\right)
+  =\mu-\left(y+\frac1yf''(1)\right)-X^T(B-yC)X
+  =\mu-\left(y+\frac1yf''(1)\right)
 \]
 assuming the last equation is satisfied. The trivial solution, which gives the bottom of the semicircle, is for $X=0$, so the first equation is $y^2=f''(1)$, and
 \[
-  \lambda_\mathrm{min}=-\sqrt{4f''(1)}
+  \lambda_\mathrm{min}=\mu-\sqrt{4f''(1)}=\mu-\mu_\mathrm m
 \]
 as expected. We need to first the nontrivial solutions with nonzero $X$, but because the coefficients are so nasty this will be a numeric problem... Specifically, we are good for $y$ where one of the eigenvalues of $B-yC$ is zero.