From da6506e20a65ae62bdc03c9a999de35b406577e3 Mon Sep 17 00:00:00 2001
From: Jaron Kent-Dobias <jaron@kent-dobias.com>
Date: Sun, 3 Dec 2023 21:36:59 +0100
Subject: Lots more writing.

---
 2-point.tex | 99 ++++++++++++++++++++++++++++++-------------------------------
 1 file changed, 48 insertions(+), 51 deletions(-)

(limited to '2-point.tex')

diff --git a/2-point.tex b/2-point.tex
index f1c0832..a281119 100644
--- a/2-point.tex
+++ b/2-point.tex
@@ -680,7 +680,7 @@ Appendix~\ref{sec:complexity-details}.
 The resulting expression for the complexity, which must
 still be extremized over the order parameters $\hat\beta_1$, $r^{01}$,
 $r^{11}_\mathrm d$, $r^{11}_0$, and $q^{11}_0$, is
-\begin{equation}
+\begin{equation} \label{eq:complexity.full}
   \begin{aligned}
     &\Sigma_{12}(E_0,\mu_0,E_1,\mu_1,q)
     =\mathop{\mathrm{extremum}}_{\hat\beta_1,r^{11}_\mathrm d,r^{11}_0,r^{01},q^{11}_0}\Bigg\{
@@ -1321,14 +1321,6 @@ After the transformation of the previous section, the complexity has been
 brought to the form of an exponential integral over the matrix order parameters
 \eqref{eq:fields}, proportional to $N$. We are therefore in the position to
 evaluate this integral using a saddle point method.
-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. At most {\oldstylenums1}\textsc{rsb} in the equilibrium is guaranteed if the function
-$\chi(q)=f''(q)^{-1/2}$ is convex \cite{Crisanti_1992_The}. The complexity at the ground state must
-reflect the structure of equilibrium, and therefore be replica symmetric.
-Recent work has found that the complexity of saddle points can produce
-other \textsc{rsb} order even when the ground state is replica symmetric, but the $3+4$ model has a safely replica symmetric complexity everywhere \cite{Kent-Dobias_2023_When}.
-
 We will always assume that the square matrices $C^{00}$, $R^{00}$, $D^{00}$,
 $C^{11}$, $R^{11}$, and $D^{11}$ are hierarchical matrices, i.e., of the Parisi form, with each set of
 three sharing the same structure. In particular, we immediately
@@ -1336,6 +1328,14 @@ define $c_\mathrm d^{00}$, $r_\mathrm d^{00}$, $d_\mathrm d^{00}$, $c_\mathrm d^
 $d_\mathrm d^{11}$ as the value of the diagonal elements of these matrices,
 respectively. Note that $c_\mathrm d^{00}=c_\mathrm d^{11}=1$ due to the spherical constraint.
 
+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. At most {\oldstylenums1}\textsc{rsb} in the equilibrium is guaranteed if the function
+$\chi(q)=f''(q)^{-1/2}$ is convex \cite{Crisanti_1992_The}. The complexity at the ground state must
+reflect the structure of equilibrium, and therefore be replica symmetric.
+Recent work has found that the complexity of saddle points can produce
+other \textsc{rsb} orders even when the ground state is replica symmetric, but the $3+4$ model has a safely replica symmetric complexity everywhere \cite{Kent-Dobias_2023_When}.
+
 Defining the `block' fields $\mathcal Q_{00}=(\hat\beta_0, \hat\mu_0, C^{00},
 R^{00}, D^{00})$, $\mathcal Q_{11}=(\hat\beta_1, \hat\mu_1, C^{11}, R^{11},
 D^{11})$, and $\mathcal Q_{01}=(\hat\mu_{01},C^{01},R^{01},R^{10},D^{01})$
@@ -1356,10 +1356,10 @@ where
 \bigg\}
 \end{aligned}
 \end{equation}
-is the action for the ordinary, one-point complexity, and remainder is given by
+is the action for the ordinary, one-point complexity, and the remainder is given by
 \begin{equation} \label{eq:two-point.action}
   \begin{aligned}
-    &\mathcal S(\mathcal Q_{11},\mathcal Q_{01}\mid\mathcal Q_{00})
+    &\mathcal S_1(\mathcal Q_{11},\mathcal Q_{01}\mid\mathcal Q_{00})
     =\hat\beta_1E_1-r^{11}_\mathrm d\mu_1-\frac12\hat\mu_1(1-c^{11}_\mathrm d)+\mathcal D(\mu_1) \\
     &\quad+\frac1n\sum_b^n\left\{-\frac12\hat\mu_{12}(q-C^{01}_{1b})+\sum_a^m\left[
         \hat\beta_0\hat\beta_1f(C^{01}_{ab})+(\hat\beta_0R^{01}_{ab}+\hat\beta_1R^{10}_{ab}-D^{01}_{ab})f'(C^{01}_{ab})+R^{01}_{ab}R^{10}_{ab}f''(C^{01}_{ab})
@@ -1422,7 +1422,7 @@ not constrained) and then the saddle point taken.
 
 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
-corresponding to the \textsc{rsb} structure of the single-point complexity.
+corresponding to the \textsc{rsb} structure of the single-point complexity for the model.
 For
 the scope of this paper, where we restrict ourselves to replica symmetric
 complexities, they have the following form at the saddle point:
@@ -1470,7 +1470,7 @@ where only the first row is nonzero. The other entries, which correspond to the
 completely uncorrelated replicas in an \textsc{rsb} picture, are all zero
 because uncorrelated vectors on the sphere are orthogonal.
 
-The most challenging part of inserting our replica symmetry ansatz is the
+The most challenging part of inserting our replica symmetric ansatz is the
 volume element in the $\log\det$, which involves the product and inverse of
 block replica matrices. The inverse of block hierarchical matrix is still a
 block hierarchical matrix, since
@@ -1484,6 +1484,7 @@ block hierarchical matrix, since
     -i(C^{00}D^{00}+R^{00}R^{00})^{-1}R^{00} & (C^{00}D^{00}+R^{00}R^{00})^{-1}C^{00}
   \end{bmatrix}
 \end{equation}
+and hierarchical matrices are closed under inverses and products.
 Because of the structure of the 01 matrices, the volume element will depend
 only on the diagonals of the matrices in this inverse block matrix. If we define
 \begin{align}
@@ -1524,11 +1525,20 @@ the diagonals of the blocks above take a simple form:
 \end{align}
 Once these expressions are inserted into the complexity, the limits of $n$ and
 $m$ to zero can be taken, and the parameters from $D^{01}$ and $D^{11}$ can be
-extremized explicitly. 
+extremized explicitly. The result is \eqref{eq:complexity.full} from section \ref{sec:complexity} of the main text.
+
 \section{Details of calculation for the isolated eigenvalue}
 \label{sec:eigenvalue-details}
 
-Using the same methodology as above, the disorder-dependent terms are captured in the linear operator
+Many of the steps in the evaluation of the isolated eigenvalue are similar to
+those from the evaluation of the two-point complexity: the treatment of the
+average over disorder and the Hubbard--Stratonovich transformation follow the
+exact same reasoning. We will not repeat the details of techniques that were
+already reported in the previous appendix.
+
+The treatment of the factors in the average over disorder proceeds as it does
+for the complexity itself in \ref{subsec:other.factors}, now with the
+disorder-dependent terms captured in the linear operator
 \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)
@@ -1538,9 +1548,9 @@ Using the same methodology as above, the disorder-dependent terms are captured i
   \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 produces dependencies only on the
-scalar products in \eqref{eq:fields} and on the new scalar products involving
-the tangent plane vectors $\mathbf x$,
+resulting expression for the integrand produces dependencies 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
   &&
@@ -1552,6 +1562,7 @@ the tangent plane vectors $\mathbf x$,
   &&
   \hat X^1_{ab}=-i\frac1N\hat{\mathbf s}_a\cdot\mathbf x_b
 \end{align}
+Replacing the original variables using a Hubbard--Stratonovich transformation then proceeds as for the complexity in subsection \ref{subsec:hubbard.strat}.
 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
@@ -1563,11 +1574,11 @@ schematically as
   \int d\mathcal Q\,d\mathcal Q_x\,
   e^{N[
     m\mathcal S_0(\mathcal Q_{00})
-    +n\mathcal S(\mathcal Q_{11},\mathcal Q_{01}\mid\mathcal Q_{00})
+    +n\mathcal S_1(\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}
-where $\mathcal S_0$ is given by \eqref{eq:one-point.action}, $\mathcal S$ is
+where $\mathcal S_0$ is given by \eqref{eq:one-point.action}, $\mathcal S_1$ is
 given by \eqref{eq:two-point.action}, and the new action $\mathcal S_x$ is
 given by
 \begin{equation} \label{eq:action.eigenvalue}
@@ -1667,15 +1678,15 @@ constrained configurations have replica symmetric order, we expect
     \hat x_1^1&\cdots&\hat x_1^1
   \end{bmatrix}
 \end{align}
-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
+Here, the lower block of the 0 matrices is zero, because the replicas whose overlap they represent (that of the normalization of the reference configuration) have no
+correlation 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.
+\eqref{eq:action.eigenvalue}, but the term with $\log\det$ is again more complicated.
 We must invert the block matrix inside. We define
 \begin{equation}
     \begin{bmatrix}
@@ -1743,11 +1754,12 @@ where the blocks inside the inverse are given by
 Here, $M_{22}$ is the inverse of the matrix already analyzed 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 $n$.
+block constant matrices, things can be worked out from here using a computer
+algebra system. 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 can be written as
+The contribution from the product with the block inverse matrix can be written as
 \begin{equation} \label{eq:inverse.quadratic.form}
     \begin{bmatrix}
       X_0\\i\hat X_0
@@ -1775,7 +1787,7 @@ and without too much reasoning one can see that the result is an $\ell\times\ell
   \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 $x_{k+1}=1$.
-The basic form of the action is (for replica symmetric $A$)
+The basic form of the action is therefore (for replica symmetric $A$)
 \begin{equation}
   2\mathcal S_x(\mathcal Q_x\mid\mathcal Q)
   =-\beta\mu_1+\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
@@ -1877,35 +1889,20 @@ with associated saddle point conditions
   &&
   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_1-\sqrt{4f''(1)}=\mu_1-\mu_\mathrm m
-\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}
-  \|\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.
+as reported in the main text.
 
-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
+The solution described here also encodes information about the correlation
+between the eigenvector $\mathbf x_\text{min}$ associated with the minimum eigenvalue and the tangent
+direction connecting the two stationary points $\mathbf x_{0\leftarrow1}$.
+The overlap between these vectors is directly related to the value of the order parameter $x_0=\frac1N\pmb\sigma_1\cdot\mathbf x_a$. This tangent vector is $\mathbf x_{0\leftarrow
+1}=\frac1{1-q}\big(\pmb\sigma_1-q\mathbf s_a\big)$, which is normalized and
 lies strictly in the tangent plane of $\mathbf s_a$. Then
 \begin{equation}
   q_\textrm{min}=\frac{\mathbf x_{0\leftarrow 1}\cdot\mathbf x_\mathrm{min}}N
   =\frac{x_0}{1-q}
 \end{equation}
+where $\mathrm x_\text{min}\cdot\mathrm s_a=0$ because of the restriction of
+the $\mathrm x$ vectors to the tangent plane at $\mathrm s_a$.
 
 
 \section{Franz--Parisi potential}
-- 
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