From d3b287ae1a886850724cb5d55139b50a393af5a2 Mon Sep 17 00:00:00 2001
From: Jaron Kent-Dobias <jaron@kent-dobias.com>
Date: Tue, 16 May 2023 18:47:31 +0200
Subject: Some writing and citations.

---
 2-point.tex | 22 ++++++++++++++++++++++
 1 file changed, 22 insertions(+)

(limited to '2-point.tex')

diff --git a/2-point.tex b/2-point.tex
index f0fec59..3cd8422 100644
--- a/2-point.tex
+++ b/2-point.tex
@@ -446,6 +446,15 @@ Because the matrices $C^{00}$, $R^{00}$, and $D^{00}$ are diagonal in this case,
 
 \section{Isolated eigenvalue}
 
+The two-point complexity depends on the spectrum at both stationary points
+through the determinant of their Hessians, but only on the bulk of the
+distribution. As we saw, this bulk is unaffected by the conditions of energy
+and proximity. However, these conditions give rise to small-rank perturbations
+to the Hessian, which can lead a subextensive number of eigenvalues leaving the
+bulk (we will study \emph{one}).
+
+\cite{Ikeda_2023_Bose-Einstein-like}
+
 \begin{equation}
   \begin{aligned}
     \beta F(\beta\mid\mathbf s)
@@ -471,6 +480,17 @@ Because the matrices $C^{00}$, $R^{00}$, and $D^{00}$ are diagonal in this case,
     &=\lim_{m\to0}\int\left[\prod_{a=1}^m d\nu(\pmb\sigma_a,\varsigma_a\mid E_0,\mu_0)\right]\,F(\beta\mid E_1,\mu_1,q,\pmb\sigma_1)
   \end{aligned}
 \end{equation}
+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}\frac\partial{\partial\beta}\overline{F(\beta\mid\epsilon_1,\mu_1,\epsilon_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
+$\mathbf s_1$ replicas are \emph{special}. The first again is the only of the
+$\sigma$ replicas constrained to lie at fixed overlap with \emph{all} the
+$\mathbf s$ replicas, and the second is the only of the $\mathbf s$ replicas at
+which the Hessian is evaluated.
+
 \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)
@@ -479,6 +499,8 @@ Because the matrices $C^{00}$, $R^{00}$, and $D^{00}$ are diagonal in this case,
   -\frac12
   \delta(\mathbf t-\mathbf s_1)\beta\sum_c^\ell(\mathbf x_c\cdot\partial_{\mathbf t})^2
 \end{equation}
+
+
 \begin{equation}
   \begin{aligned}
     &\frac18\beta^2\int d\mathbf t\,d\mathbf t'\,\delta(\mathbf t-\mathbf s_1)\delta(\mathbf t'-\mathbf s_1)\sum_{ab}^\ell(\mathbf x_a\cdot\partial_{\mathbf t})^2(\mathbf x_b\cdot\partial_{\mathbf t'})^2f\left(\frac{\mathbf t\cdot\mathbf t'}N\right)\\
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