From 9bef8900ec13b6f38c96987fd8e2cffe55c7a22e Mon Sep 17 00:00:00 2001
From: Jaron Kent-Dobias <jaron@kent-dobias.com>
Date: Sat, 3 Jun 2023 19:40:22 +0200
Subject: Lots of writing and clarification.

---
 2-point.tex | 362 ++++++++++++++++++++++++++++--------------------------------
 1 file changed, 166 insertions(+), 196 deletions(-)

(limited to '2-point.tex')

diff --git a/2-point.tex b/2-point.tex
index 2c61253..2fc35d4 100644
--- a/2-point.tex
+++ b/2-point.tex
@@ -429,18 +429,20 @@ It remains only to apply the doubled operators to $f$ and then evaluate the simp
 \subsection{Hubbard--Stratonovich}
 
 Having expanded this expression, we are left with an argument in the exponential which is a function of scalar products between the fields $\mathbf s$, $\hat{\mathbf s}$, $\pmb\sigma$, and $\hat{\pmb\sigma}$. We will change integration coordinates from these fields to matrix fields given by the scalar products, defined as
-\begin{align} \label{eq:fields}
-  C^{00}_{ab}=\frac1N\pmb\sigma_a\cdot\pmb\sigma_b &&
-  R^{00}_{ab}=-i\frac1N{\pmb\sigma}_a\cdot\hat{\pmb\sigma}_b &&
-  D^{00}_{ab}=\frac1N\hat{\pmb\sigma}_a\cdot\hat{\pmb\sigma}_b \\
-  C^{01}_{ab}=\frac1N\pmb\sigma_a\cdot\mathbf s_b &&
-  R^{01}_{ab}=-i\frac1N{\pmb\sigma}_a\cdot\hat{\mathbf s}_b &&
-  R^{10}_{ab}=-i\frac1N\hat{\pmb\sigma}_a\cdot{\mathbf s}_b &&
-  D^{01}_{ab}=\frac1N\hat{\pmb\sigma}_a\cdot\hat{\mathbf s}_b \\
-  C^{11}_{ab}=\frac1N\mathbf s_a\cdot\mathbf s_b &&
-  R^{11}_{ab}=-i\frac1N{\mathbf s}_a\cdot\hat{\mathbf s}_b &&
-  D^{11}_{ab}=\frac1N\hat{\mathbf s}_a\cdot\hat{\mathbf s}_b
-\end{align}
+\begin{equation} \label{eq:fields}
+  \begin{aligned}
+    C^{00}_{ab}=\frac1N\pmb\sigma_a\cdot\pmb\sigma_b &&
+    R^{00}_{ab}=-i\frac1N{\pmb\sigma}_a\cdot\hat{\pmb\sigma}_b &&
+    D^{00}_{ab}=\frac1N\hat{\pmb\sigma}_a\cdot\hat{\pmb\sigma}_b \\
+    C^{01}_{ab}=\frac1N\pmb\sigma_a\cdot\mathbf s_b &&
+    R^{01}_{ab}=-i\frac1N{\pmb\sigma}_a\cdot\hat{\mathbf s}_b &&
+    R^{10}_{ab}=-i\frac1N\hat{\pmb\sigma}_a\cdot{\mathbf s}_b &&
+    D^{01}_{ab}=\frac1N\hat{\pmb\sigma}_a\cdot\hat{\mathbf s}_b \\
+    C^{11}_{ab}=\frac1N\mathbf s_a\cdot\mathbf s_b &&
+    R^{11}_{ab}=-i\frac1N{\mathbf s}_a\cdot\hat{\mathbf s}_b &&
+    D^{11}_{ab}=\frac1N\hat{\mathbf s}_a\cdot\hat{\mathbf s}_b
+  \end{aligned}
+\end{equation}
 We insert into the integral the product of $\delta$ functions enforcing these
 definitions, integrated over the new matrix fields, which is equivalent to
 multiplying by one. Once this is done, the many scalar products appearing
@@ -466,7 +468,7 @@ the resulting complexity is
   =\frac1N\lim_{n\to0}\lim_{m\to0}\frac\partial{\partial n}\int d\mathcal Q_{00}\,d\mathcal Q_{11}\,d\mathcal Q_{01}\,e^{Nm\mathcal S_0(\mathcal Q_{00})+Nn\mathcal S_1(\mathcal Q_{00},\mathcal Q_{11},\mathcal Q_{01})}
 \end{equation}
 where
-\begin{equation}
+\begin{equation} \label{eq:one-point.action}
   \begin{aligned}
     &\mathcal S_0(\mathcal Q_{00})
     =\hat\beta_0E_0-r^{00}_\mathrm d\mu_0-\frac12\hat\mu_0(1-c^{00}_\mathrm d)+\mathcal D(\mu_0)\\
@@ -478,7 +480,7 @@ where
 \end{aligned}
 \end{equation}
 is the action for the ordinary, one-point complexity, and remainder is given by
-\begin{equation}
+\begin{equation} \label{eq:two-point.action}
   \begin{aligned}
     &\mathcal S(\mathcal Q_{00},\mathcal Q_{11},\mathcal Q_{01})
     =\hat\beta_1E_1-r^{11}_\mathrm d\mu_1-\frac12\hat\mu_1(1-c^{11}_\mathrm d)+\mathcal D(\mu_1) \\
@@ -514,13 +516,13 @@ 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}\\
+  &=-\frac{(E_0+\mu_0)f'(1)+E_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} \\
+  &=\frac{\mu_0f(1)+E_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}
+    \frac{\mu_0f(1)+E_0f'(1)}{f(1)\big(f'(1)+f''(1)\big)-f'(1)^2}
   \right)^2
 \end{align}
 
@@ -744,7 +746,7 @@ giving
 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
 \begin{equation}
   \begin{aligned}
-    F_H(\beta\mid\epsilon_1,\mu_1,\epsilon_2,\mu_2,q)
+    F_H(\beta\mid E_1,\mu_1,E_2,\mu_2,q)
     &=\int\frac{d\nu_H(\pmb\sigma,\varsigma\mid E_0,\mu_0)}{\int d\nu_H(\pmb\sigma',\varsigma'\mid E_0,\mu_0)}\,F_H(\beta\mid E_1,\mu_1,q,\pmb\sigma) \\
     &=\lim_{m\to0}\int\left[\prod_{a=1}^m d\nu_H(\pmb\sigma_a,\varsigma_a\mid E_0,\mu_0)\right]\,F_H(\beta\mid E_1,\mu_1,q,\pmb\sigma_1)
   \end{aligned}
@@ -752,7 +754,7 @@ again anticipating the use of replicas. Finally, the reference configuration $\p
 This formidable expression is now ready to be averaged over the disordered Hamiltonians $H$. Once averaged,
 the minimum eigenvalue of the conditioned Hessian is then given by twice the ground state energy, or
 \begin{equation}
-  \lambda_\text{min}=2\lim_{\beta\to\infty}\overline{F_H(\beta\mid\epsilon_1,\mu_1,\epsilon_2,\mu_2,q)}
+  \lambda_\text{min}=2\lim_{\beta\to\infty}\overline{F_H(\beta\mid E_1,\mu_1,E_2,\mu_2,q)}
 \end{equation}
 For this calculation, there are three different sets of replicated variables.
 Note that, as for the computation of the complexity, the $\pmb\sigma_1$ and
@@ -813,11 +815,13 @@ which the Hessian is evaluated.
   \end{tikzpicture}
   \caption{
     A sketch of the vectors involved in the calculation of the isolated
-    eigenvalue. All replicas $\mathbf x$ sit in an $N-2$ sphere corresponding
-    with the tangent plane (not to scale) of the first $\mathbf s$ replica. All of the
-    $\mathbf s$ replicas lie on the sphere, constrained to be at fixed overlap
-    $q$ with the first of the $\pmb\sigma$ replicas, the reference
-    configuration. All of the $\pmb\sigma$ replicas lie on the sphere.
+    eigenvalue. All replicas $\mathbf x$, which correspond with candidate
+    eigenvectors of the Hessian evaluated at $\mathbf s_1$, sit in an $N-2$
+    sphere corresponding with the tangent plane (not to scale) of the first
+    $\mathbf s$ replica. All of the $\mathbf s$ replicas lie on the sphere,
+    constrained to be at fixed overlap $q$ with the first of the $\pmb\sigma$
+    replicas, the reference configuration. All of the $\pmb\sigma$ replicas lie
+    on the sphere.
   }
 \end{figure}
 
@@ -830,6 +834,10 @@ Using the same methodology as above, the disorder-dependant terms are captured i
   -\frac12
   \delta(\mathbf t-\mathbf s_1)\beta\sum_c^\ell(\mathbf x_c\cdot\partial_{\mathbf t})^2
 \end{equation}
+that is applied to $H$ by integrating over $\mathbf t\in\mathbb R^N$. The
+resulting expression for the integrand is produces dependencies only on the
+scalar products in \eqref{eq:fields} and on the new scalar products involving
+the tangent plane vectors $\mathbf x$,
 \begin{align}
   A_{ab}=\frac1N\mathbf x_a\cdot\mathbf x_b
   &&
@@ -841,23 +849,28 @@ Using the same methodology as above, the disorder-dependant terms are captured i
   &&
   \hat X^1_{ab}=-i\frac1N\hat{\mathbf s}_a\cdot\mathbf x_b
 \end{align}
-
+Defining as before a block variable $\mathcal Q_x=(A,X^0,\hat X^0,X^1,\hat X^1)$
+and consolidating the previous block variables $\mathcal Q=(\mathcal Q_{00},
+\mathcal Q_{01},\mathcal Q_{11})$, we can write the minimum eigenvalue
+schematically as
 \begin{equation}
   \lambda_\mathrm{min}
   =-2\lim_{\beta\to\infty}
   \lim_{\substack{\ell\to0\\m\to0\\n\to0}}\frac\partial{\partial\ell}\frac1{\beta N}
-  \int d\mathcal Q\,d\mathcal X\,
+  \int d\mathcal Q\,d\mathcal Q_x\,
   e^{N[
     m\mathcal S_0(\mathcal Q_{00})
-    +n\mathcal S_1(\mathcal Q_{11},\mathcal Q_{01}\mid\mathcal Q_{00})
-    +\ell\mathcal S_x(\mathcal X\mid\mathcal Q_{00},\mathcal Q_{01},\mathcal Q_{11})
+    +n\mathcal S(\mathcal Q_{11},\mathcal Q_{01}\mid\mathcal Q_{00})
+    +\ell\mathcal S_x(\mathcal Q_x\mid\mathcal Q_{00},\mathcal Q_{01},\mathcal Q_{11})
   ]}
 \end{equation}
-
-\begin{equation}
+where $\mathcal S_0$ is given by \eqref{eq:one-point.action}, $\mathcal S$ is
+given by \eqref{eq:two-point.action}, and the new action $\mathcal S_x$ is
+given by
+\begin{equation} \label{eq:action.eigenvalue}
   \begin{aligned}
-    \ell\mathcal S_x(\mathcal X\mid\mathcal Q)
-    =-\frac12\beta\mu+
+    \ell\mathcal S_x(\mathcal Q_x\mid\mathcal Q)
+    =-\frac12\ell\beta\mu+
     \frac12\beta\sum_b^\ell\bigg\{
       \frac12\beta&f''(1)\sum_a^lA_{ab}^2\\
     &+\sum_a^m\left[
@@ -886,7 +899,9 @@ Using the same methodology as above, the disorder-dependant terms are captured i
     \right)
   \end{aligned}
 \end{equation}
+As usual in these quenched Franz--Parisi style computations, the saddle point expressions for the variables $\mathcal Q$ in the joint limits of $m$, $n$, and $\ell$ to zero are independent of $\mathcal Q_x$, and so these quantities take the same value they do for the two-point complexity that we computed above. The saddle point conditions for the variables $\mathcal Q_x$ are then fixed by extremizing with respect to the final action.
 
+To evaluate this expression, we need a sensible ansatz for the variables $\mathcal Q_x$. The matrix $A$ we expect to be an ordinary hierarchical matrix, and since the model is a spherical 2-spin the finite but low temperature order will be {\oldstylenums1}\textsc{rsb}. The expected form of the $X$ matrices follows our reasoning for the 01 matrices of the previous section: namely, they should have constant rows and a column structure which matches that of the level of \textsc{rsb} order associated with the degrees of freedom that parameterize the columns. Since both the reference configurations and the constrained configurations have replica symmetric order, we expect
 \begin{align}
   X^0
   =
@@ -941,34 +956,16 @@ Using the same methodology as above, the disorder-dependant terms are captured i
     \hat x_1^1&\cdots&\hat x_1^1
   \end{bmatrix}
 \end{align}
-\begin{equation}
-  \begin{aligned}
-    \frac2\beta\lim_{\ell\to0}\mathcal S_x(\mathcal X\mid\mathcal Q)
-    &=
-      -\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 \\
-    &-
-      \big(\hat\beta_1f''(q^{11}_0)+r^{11}_0f'''(q^{11}_0)\big)x_1^2
-      -2f''(q^{11}_0)x_1\hat x_1^1 \\
-    &+\lim_{\ell\to0}\frac1\ell\frac1\beta\log\det\left(
-      A-
-      \begin{bmatrix}
-        X^0\\\hat X^0\\X^1\\\hat X^1
-      \end{bmatrix}^T
-      \begin{bmatrix}
-        C^{00}&iR^{00}&C^{01}&iR^{01}\\
-        iR^{00}&D^{00}&iR^{10}&D^{01}\\
-        (C^{01})^T&(iR^{10})^T&C^{11}&iR^{11}\\
-        (iR^{01})^T&(D^{01})^T&iR^{11}&D^{11}\\
-      \end{bmatrix}^{-1}
-      \begin{bmatrix}
-        X^0\\\hat X^0\\X^1\\\hat X^1
-      \end{bmatrix}
-    \right)
-  \end{aligned}
-\end{equation}
-
+Here, the lower block of the 0 matrices is zero, because these replicas have no
+overlap with the reference or anything else. The first row of the $X^1$ matrix
+needs to be zero because of the constraint that the tangent space vectors lie
+in the tangent plane to the sphere, and therefore have $\mathbf x_a\cdot\mathbf
+s_1=0$ for any $a$. This produces five parameters to deal with, which we
+compile in the vector $\mathcal X=(x_0,\hat x_0,x_1\hat x_1^1,\hat x_1^0)$.
+
+Inserting this ansatz is straightforward in the first part of
+\eqref{eq:action.eigenvalue}, but the term with $\log\det$ is more complicated.
+We must invert the block matrix inside. Defining
 \begin{equation}
     \begin{bmatrix}
       C^{00}&iR^{00}&C^{01}&iR^{01}\\
@@ -978,12 +975,13 @@ Using the same methodology as above, the disorder-dependant terms are captured i
     \end{bmatrix}^{-1}
     =
   \begin{bmatrix}
-    A & B \\
-    C & D
+    M_{11} & M_{12} \\
+    M_{12}^T & M_{22}
   \end{bmatrix}
 \end{equation}
-\begin{equation}
-    A=
+where the blocks inside the inverse are given by
+\begin{align}
+  M_{11}&=
     \left(
       \begin{bmatrix}
         C^{00}&iR^{00}\\iR^{00}&D^{00}
@@ -1001,56 +999,9 @@ Using the same methodology as above, the disorder-dependant terms are captured i
         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=-
-      \begin{bmatrix}
-        C^{00}&iR^{00}\\iR^{00}&D^{00}
-      \end{bmatrix}^{-1}
+    \\
+  M_{12}&=-
+  M_{11}
       \begin{bmatrix}
         C^{01}&iR^{01}\\
         iR^{10}&D^{01}
@@ -1058,22 +1009,8 @@ which is \emph{not} a hierarchical matrix! But, all these new hat variables are
       \begin{bmatrix}
         C^{11}&iR^{11}\\iR^{11}&D^{11}
       \end{bmatrix}^{-1}
-\end{equation}
-\begin{equation}
-    C=-
-      \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}=B^T
-\end{equation}
-\begin{equation}
-    D=
+      \\
+  M_{22}&=
     \left(
       \begin{bmatrix}
         C^{11}&iR^{11}\\iR^{11}&D^{11}
@@ -1091,42 +1028,50 @@ which is \emph{not} a hierarchical matrix! But, all these new hat variables are
         iR^{10}&D^{01}
       \end{bmatrix}
     \right)^{-1}
-\end{equation}
-
-\begin{equation}
+\end{align}
+Here, $M_{22}$ is the inverse of the matrix already analyzed in as part of
+\eqref{eq:two-point.action}. Following our discussion of the inverses of block
+replica matrices above, and reasoning about their products with the rectangular
+block constant matrices, things can be worked out from here. For instance, the
+second term in $M_{11}$ contributes nothing once the appropriate limits are
+taken, because each contribution is proportional to $n$.
+
+The contribution con be written as
+\begin{equation} \label{eq:inverse.quadratic.form}
     \begin{bmatrix}
       X_0\\i\hat X_0
-    \end{bmatrix}^TA
+    \end{bmatrix}^TM_{11}
     \begin{bmatrix}
       X_0\\i\hat X_0
     \end{bmatrix}
     +
     2\begin{bmatrix}
       X_0\\i\hat X_0
-    \end{bmatrix}^TB
+    \end{bmatrix}^TM_{12}
     \begin{bmatrix}
       X_1\\i\hat X_1
     \end{bmatrix}
     +
     \begin{bmatrix}
       X_1\\i\hat X_1
-    \end{bmatrix}^TD
+    \end{bmatrix}^TM_{22}
     \begin{bmatrix}
       X_1\\i\hat X_1
     \end{bmatrix}
 \end{equation}
-
-
-\[
+and without too much reasoning one can see that the result is an $\ell\times\ell$ constant matrix. If $A$ is a replica matrix and $c$ is a constant, then
+\begin{equation}
   \log\det(A-c)=\log\det A-\frac{c}{\sum_{i=0}^k(a_{i+1}-a_i)x_{i+1}}
-\]
+\end{equation}
 where $a_{k+1}=1$ and $x_{k+1}=1$.
-So the basic form of the action is (for replica symmetric $A$)
-\[
-  -\mu+\frac14\beta^2f''(1)(1-a_0^2)+\frac12\log(1-a_0)+\frac12\frac{a_0}{1-a_0}+\frac12\begin{bmatrix}x_0\\\hat x_0\\x_1\\\hat x_1^1\\\hat x_1^0\end{bmatrix}^T\left(\beta B-\frac1{1-a_0}C\right)\begin{bmatrix}x_0\\\hat x_0\\x_1\\\hat x_1^1\\\hat x_1^0\end{bmatrix}
-\]
-for
-\[
+The basic form of the action is (for replica symmetric $A$)
+\begin{equation}
+  2\mathcal S_x(\mathcal Q_x\mid\mathcal Q)
+  =-\beta\mu+\frac12\beta^2f''(1)(1-a_0^2)+\log(1-a_0)+\frac{a_0}{1-a_0}+\mathcal X^T\left(\beta B-\frac1{1-a_0}C\right)\mathcal X
+\end{equation}
+where the matrix $B$ comes from the $\mathcal X$-dependant parts of the first
+lines of \eqref{eq:action.eigenvalue} and is given by
+\begin{equation}
   B=\begin{bmatrix}
     \hat\beta_0f''(q)+r_{10}f'''(q)&f''(q)&0&0&0\\
     f''(q)&0&0&0&0\\
@@ -1134,33 +1079,39 @@ for
     0&0&-f''(q_0^{11})&0&0\\
     0&0&0&0&0
   \end{bmatrix}
-\]
+\end{equation}
+and where the matrix $C$ encodes the coefficients of the quadratic form
+\eqref{eq:inverse.quadratic.form}, and is given element-wise by
 \begin{align}
+  \notag
   &
   C_{11}=d^{00}_\mathrm df'(1)
-  \quad
+  \qquad
   C_{12}=r^{00}_\mathrm df'(1)
-  \quad
+  \qquad
   C_{22}=-f'(1)
   \\
+  \notag
   &
   C_{13}
   =\frac1{1-q_0}\left(
     (r^{11}_d-r^{11}_0)\left(r^{01}-q\frac{r^{11}_d-r^{11}_0}{1-q_0}\right)(f'(1)-f'(q_0))+qf'(1)d^{00}_d+r^{00}_d(r^{10}f'(1)+(r^{11}_d-r^{11}_0)f'(q))
     \right)
   \\
+  \notag
   &
   C_{15}=r^{00}_df'(q)+\left(r^{01}-q\frac{r^{11}_d-r^{11}_0}{1-q_0}\right)(f'(1)-f'(q_0))
-  \quad
+  \qquad
   C_{14}=-C_{15}
   \\
   &
   C_{23}=\frac1{1-q_0}\left((qr^{00}_d-r^{10})f'(1)-(r^{11}_d-r^{11}_0)f'(q)\right)
-  \quad
+  \qquad
   C_{24}=f'(q)
-  \quad
+  \qquad
   C_{25}=-C_{24}
   \\
+  \notag
   &
   C_{33}
   =
@@ -1171,72 +1122,91 @@ for
         \right)(f'(1)-f'(q_0))
         -2\frac{qr^{00}-r^{10}}{1-q_0}f'(q)
     \right]\\
+  \notag
   &\qquad-\frac{1-q^2}{(1-q_0)^2}-\frac{(r^{10}-qr^{00}_d)^2}{(1-q_0)^2}f'(1)
     \\
+  \notag
   &
   C_{34}
   =-(qr^{01}-r^{11}_0)\frac{f'(1)-f'(q_0)}{1-q_0}-\frac{r^{11}_d-r^{11}_0}{1-q_0}\left(
     \frac{1-q^2}{1-q_0}(f'(1)-f'(q_0))-f'(q_0)
     \right)-f'(q)\frac{qr^{00}_d-r^{10}}{1-q_0}
     \\
+  \notag
   &
     C_{35}=-C_{34}-\frac{r^{11}_d-r^{11}_0}{1-q_0}(f'(1)-f'(q_0))
-  \quad
+  \qquad
     C_{44}=f'(1)-2f'(q_0)
-    \quad
+    \qquad
     C_{45}=f'(q_0)
-    \quad
+    \qquad
     C_{55}=-f'(1)
+  \notag
+\end{align}
+The saddle point conditions read
+\begin{align}
+  0=-\beta^2f''(1)a_0+\frac{a_0-\mathcal X^TC\mathcal X}{(1-a_0)^2}
+  &&
+  0=\bigg(\beta B-\frac1{1-a_0}C\bigg)\mathcal X
 \end{align}
-Use $X$ for the big vector. Then
-\[
-  0=-\beta^2f''(1)a_0+\frac{a_0-X^TCX}{(1-a_0)^2}
-\]
-\[
-  0=\bigg(\beta B-\frac1{1-a_0}C\bigg)X
-\]
-For a non-trivial solution, we require the following: change of basis in $X$ to diagonalize the matrix $\beta B-C/(1-a_0)$. Then, all of the new basis vectors of $X$ are zero except one, and the second equation is satisfied by tuning $a_0$ to make the coefficient zero. Finally, the first equation is satisfied by choice of the magnitude of this basis vector.
-Suppose $a_0=1-1/(y\beta)$, $X\sim X+O(1/\beta)$. Then
-\[
-  0=-f''(1)(1-(y\beta)^{-1})+y^2\left(1-(y\beta)^{-1}-X^TCX\right)
-\]
-For large $\beta$
-\[
-  0=-f''(1)+y^2(1-X^TCX)
-\]
-\[
-  0=(B-yC)X
-\]
-which gives
-\[
-  \mathcal S=-\frac12\beta\mu+\frac14\beta^2f''(1)\big(2(y\beta)^{-1}-(y\beta)^{-2}\big)-\frac12\log(y\beta)+\frac12y\beta(1-(y\beta)^{-1})+\frac12\beta X^TBX-\frac12y\beta X^TCX
-\]
-so
-\[
-  \lambda_\mathrm{min}=-2\lim_{\beta\to\infty}\frac{\partial\mathcal S}{\partial\beta}
-  =\mu-\left(y+\frac1yf''(1)\right)-X^T(B-yC)X
+Note that the second of these conditions implies that the quadratic form in
+$\mathcal X$ in the action vanishes at the saddle.
+
+We would like to take the limit of $\beta\to\infty$. As is usual in the
+two-spin model, the appropriate limits of the order parameter is
+$a_0=1-(y\beta)^{-1}$. Upon inserting this scaling and taking the limit, we
+finally find
+\begin{equation}
+  \lambda_\mathrm{min}=-2\lim_{\beta\to\infty}\frac1\beta\mathcal S_x
   =\mu-\left(y+\frac1yf''(1)\right)
-\]
-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
-\[
+\end{equation}
+with associated saddle point conditions
+\begin{align}
+  0=-f''(1)+y^2(1-\mathcal X^TC\mathcal X)
+  &&
+  0=(B-yC)\mathcal X
+\end{align}
+The trivial solution, which gives the bottom of the semicircle, is for
+$\mathcal X=0$. When this is satisfied, the first equation gives $y^2=f''(1)$,
+and
+\begin{equation}
   \lambda_\mathrm{min}=\mu-\sqrt{4f''(1)}=\mu-\mu_\mathrm m
-\]
-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. In this case, if the associated normalized eigenvector is $\hat X$, its magnitude is set by
+\end{equation}
+as expected. The nontrivial solutions have nonzero $\mathcal X$. The only way
+to satisfy this with the saddle conditions is for $y$ such that one of the
+eigenvalues of $B-yC$ is zero. In this case, if the normalized eigenvector
+associated with the zero eigenvector is $\hat{\mathcal X}_0$, $\mathcal
+X=\|\mathcal X_0\|\hat{\mathcal X}_0$ is a solution. The magnitude of the solution
+is set by the other saddle point condition, namely
 \begin{equation}
-  \|X\|^2=\frac1{\hat X^TC\hat X}\left(1-\frac{f''(1)}{y^2}\right)
+  \|\mathcal X_0\|^2=\frac1{\hat{\mathcal X}_0^TC\hat{\mathcal X}_0}\left(1-\frac{f''(1)}{y^2}\right)
+\end{equation}
+In practice, we find that $\hat{\mathcal X}_0^TC\hat{\mathcal X}_0$ is positive
+at the saddle point. Therefore, for the solution to make sense we must have
+$y^2\geq f''(1)$. In practice, there is at most \emph{one} $y$ which produces a
+zero eigenvalue of $B-yC$ and satisfies this inequality, so the solution seems
+to be unique.
+
+In this solution, we simultaneously find the smallest eigenvalue and information
+about the orientation of its associated eigenvector: namely, its overlap with
+the tangent vector that points directly toward the reference spin. This is
+directly related to $x_0$. This tangent vector is $\mathbf x_{0\leftarrow
+1}=\frac1{1-q}\big(\pmb\sigma_0-q\mathbf s_a\big)$, which is normalized and
+lies strictly in the tangent plane of $\mathbf s_a$. Then
+\begin{equation}
+  \frac{\mathbf x_{0\leftarrow 1}\cdot\mathbf x_\mathrm{min}}N
+  =\frac{x_0}{1-q}
 \end{equation}
-In practice, we find that $\hat X^TC\hat X$ is positive. Therefore, for the
-solution to make sense we must have $y^2>f''(1)$. In practice, there is at most
-\emph{one} $y$ which produces a zero eigenvalue of $B-yC$ and satisfies this
-inequality, so the solution seems to be unique.
 
 \section{Franz--Parisi potential}
 \label{sec:franz-parisi}
 
-\cite{Franz_1995_Recipes}
+Here, we compute the Franz--Parisi potential for this model at zero
+temperature, with respect to a reference configuration fixed to be a stationary
+point of energy $E_0$ and stability $\mu_0$ as before.
+
+The Franz--Parisi 
+\cite{Franz_1995_Recipes, Franz_1998_EffectivePotential}
 
 \begin{equation} \label{eq:franz-parisi.definition}
   \beta V_\beta(q\mid E_0,\mu_0)
-- 
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