From 6bc7889a901eff2745d3531a357a44094b52cd2e Mon Sep 17 00:00:00 2001
From: Jaron Kent-Dobias <jaron@kent-dobias.com>
Date: Tue, 4 Jun 2024 10:45:35 -0700
Subject: Small tweaks.

---
 marginal.tex | 157 +----------------------------------------------------------
 1 file changed, 2 insertions(+), 155 deletions(-)

(limited to 'marginal.tex')

diff --git a/marginal.tex b/marginal.tex
index 309f433..ccfe579 100644
--- a/marginal.tex
+++ b/marginal.tex
@@ -667,9 +667,9 @@ of $M$ random functions $V_k:\mathbf S^{N-1}\to\mathbb R$ that are centered Gaus
 \begin{equation}
   \overline{V_i(\mathbf x)V_j(\mathbf x')}=\delta_{ij}f\left(\frac{\mathbf x^T\mathbf x'}N\right)
 \end{equation}
-The energy or cost function is the sum of squares of the $V_k$, or
+The energy or cost function is minus the sum of squares of the $V_k$, or
 \begin{equation}
-  H(\mathbf x)=\frac12\sum_{k=1}^MV_k(\mathbf x)^2
+  H(\mathbf x)=-\frac12\sum_{k=1}^MV_k(\mathbf x)^2
 \end{equation}
 The landscape complexity and large deviations of the ground state for this problem were recently studied in a linear context, with $f(q)=\sigma^2+aq$ \cite{Fyodorov_2020_Counting, Fyodorov_2022_Optimization}. Some results on the ground state of the general nonlinear problem can also be found in \cite{Tublin_2022_A}. In particular, that work indicates that the low-lying minima of the problem tend to be either replica symmetric or full replica symmetry breaking. This is not good news for our analysis or marginal states, because in the former case the problem is typically easy to solve, and in the latter the analysis becomes much more technically challenging.
 
@@ -701,159 +701,6 @@ Applying the Lagrange multiplier method detailed above to enforce the spherical
   \operatorname{Hess}H(\mathbf x,\omega)=\partial V_k(\mathbf x)\partial V_k(\mathbf x)+V_k(\mathbf x)\partial\partial V_k(\mathbf x)+\omega I
 \end{align}
 \begin{widetext}
-The number of stationary points in a circumstance where the determinants add constructively is
-\begin{equation}
-  \begin{aligned}
-    &\mathcal N(E,\mu)^n
-    =\int\prod_{a=1}^nd\mathbf x_a\frac{d\hat{\mathbf x}_a}{(2\pi)^N}d\omega_a\,d\hat\beta_a\,\hat\mu_a\,d\bar\eta_a\,d\eta_a\,\exp\bigg\{
-      i\hat{\mathbf x}_a^T(V^k(\mathbf x_a)\partial V^k(\mathbf x_a)+\omega\mathbf x_a)
-      +\hat\beta(NE-\frac12V^k(\mathbf x_a)V^k(\mathbf x_a)) \\
-    &  +\bar\eta_a^T(\partial V^k(\mathbf x_a)\partial V^k(\mathbf x_a)^T+V^k(\mathbf x_a)\partial\partial V^k(\mathbf x_a)+\omega_a I)\eta_a
-    +\hat\mu_a(N\mu-\partial V^k(\mathbf x_a)^T\partial V^k(\mathbf x_a)-V^k(\mathbf x_a)\operatorname{Tr}\partial\partial V^k(\mathbf x_a)-N\omega_a)
-    \bigg\}
-  \end{aligned}
-\end{equation}
-To linearize the argument of the exponential with respect to $V$, we define the following new fields: $w^k_a=V^k(\mathbf x_a)$ and $\mathbf v^k_a=\partial V^k(\mathbf x_1)$. Inserting these in $\delta$ functions, we have
-\begin{equation}
-  \begin{aligned}
-    &\mathcal N(E,\mu)^n
-    =\int\prod_{a=1}^nd\mathbf x_a\frac{d\hat{\mathbf x}_a}{(2\pi)^N}d\omega_a\,d\hat\beta_a\,\hat\mu_a\,d\bar\eta_a\,d\eta_a\,\exp\bigg\{
-      i\hat{\mathbf x}_a^T(w^k_a\mathbf v^k_a+\omega\mathbf x_a)
-      +\hat\beta(NE-\frac12w^k_aw^k_a) \\
-    &  +\bar\eta_a^T(\mathbf v^k_a(\mathbf v^k_a)^T+w^k_a\partial\partial V^k(\mathbf x_a)+\omega_a I)\eta_a
-    +\hat\mu_a(N\mu-(\mathbf v^k_a)^T\mathbf v^k_a-w^k_a\operatorname{Tr}\partial\partial V^k(\mathbf x_a)-N\omega_a) \\
-    & +i\hat w^k_a(w^k_a-V^k(\mathbf x_a))
-    +i(\hat{\mathbf v}^k_a)^T(\mathbf v^k_a-\partial V^k(\mathbf x_a))
-    \bigg\}
-  \end{aligned}
-\end{equation}
-which is now linear in $V$. Averaging over $V$ yields, from only the terms that depend on it and to highest order in $N$,
-\begin{equation}
-  -\frac12\left(
-    f(C_{ab})\hat w^k_a\hat w^k_b
-    +2f'(C_{ab})\hat w^k_a\frac{\mathbf x^T_a\hat{\mathbf v}^k_b}N
-    +f'(C_{ab})\frac{(\hat{\mathbf v}^k_a)^T\hat{\mathbf v}^k_b}N
-    +f''(C_{ab})\left(\frac{\mathbf x_a^T\hat{\mathbf v}^k_b}N\right)^2
-    +f''(C_{ab})w^k_aw^k_bG_{ab}^2
-  \right)
-\end{equation}
-The resulting integrand is Gaussian in the $w$, $\hat w$, $\mathbf y$, and $\hat{\mathbf y}$, with
-\begin{equation}
-  \exp\left\{
-    -\frac12\sum_{k=1}^M\sum_{ab}^n\begin{bmatrix}w_a^k\\\mathbf v_a^k\\\hat w_a^k\\\hat{\mathbf v}_a^k\end{bmatrix}^T
-    \begin{bmatrix}
-      \hat\beta_a\delta_{ab}+G_{ab}^2f''(C_{ab}) & -i\hat{\mathbf x}_a^T\delta_{ab} & -i\delta_{ab} & 0 \\
-      -i\hat{\mathbf x}_a\delta_{ab} & 2(\hat\mu_a I-\bar\eta_a\eta_a^T)\delta_{ab} & 0 & -i\delta_{ab}I\\
-      -i\delta_{ab} & 0 & f(C_{ab}) & \frac1Nf'(C_{ab})\mathbf x_a^T \\
-      0 & -i\delta_{ab}I & \frac1Nf'(C_{ab})\mathbf x_b & \frac1Nf'(C_{ab})I+\frac1{N^2}f''(C_{ab})\mathbf x_a\mathbf x_b^T
-    \end{bmatrix}
-    \begin{bmatrix}w_b^k\\\mathbf v_b^k\\\hat w_b^k\\\hat{\mathbf v}_b^k\end{bmatrix}
-  \right\}
-\end{equation}
-which produces
-\begin{equation}
-  \exp\left\{
-    \frac M2\log\det\left(
-      I+\begin{bmatrix}
-      \hat\beta_a\delta_{ac}+G_{ac}^2f''(C_{ac}) & -i\hat{\mathbf x}_a^T\delta_{ac} \\
-      -i\hat{\mathbf x}_a\delta_{ac} & 2(\hat\mu_a I-\bar\eta_a\eta_a^T)\delta_{ac}
-    \end{bmatrix}
-    \begin{bmatrix}
-      f(C_{cb})&\frac1Nf'(C_{cb})\mathbf x_c^T \\
-      \frac1Nf'(C_{cb})\mathbf x_b & \frac1Nf'(C_{cb})I+\frac1{N^2}f''(C_{cb})\mathbf x_c\mathbf x_b^T
-    \end{bmatrix}
-    \right)
-  \right\}
-\end{equation}
-\begin{equation}
-  \begin{bmatrix}
-    (\hat\beta_a\delta_{ac}+G_{ac}^2f''(C_{ac}))f(C_{cb}) + R_{ab}f'(C_{ab})
-    &
-    \frac1N\left[(\hat\beta_a\delta_{ac}+G_{ac}^2f''(C_{ac}))f'(C_{cb})+R_{ab}f''(C_{ab})\right]\mathbf x_b^T-\frac1Nif'(C_{ab})\hat{\mathbf x}_a^T
-    \\
-    -i\hat{\mathbf x}_af(C_{ab})+\frac1N\hat\mu f'(C_{ab})\mathbf x_b
-    &
-    -i\frac1Nf'(C_{ab})\hat{\mathbf x}_a\mathbf x_b^T
-    +2\frac1N(\hat\mu_aI-\bar{\pmb\eta}_a\pmb\eta_a^T)f'(C_{ab})
-    +\frac2{N^2}\hat\mu_af''(C_{ab})\mathbf x_a\mathbf x_b^T
-  \end{bmatrix}
-\end{equation}
-Here we already see that the terms dependent on $\hat\mu$ will be smaller by a factor of $N$ than those not. Therefore we can drop these terms safely at leading order in $N$.
-We treat this determinant by using block form, which gives two contributions
-\begin{equation}
-  \begin{aligned}
-    &\log\det\left[
-      \delta_{ab}+(\hat\beta_a\delta_{ac}+G_{ac}^2f''(C_{ac}))f(C_{cb}) + R_{ab}f'(C_{ab})
-    \right] \\
-    &\log\det\left(
-      I\delta_{ab}
-      -2\frac1N\bar{\pmb\eta}_a\pmb\eta_a^Tf'(C_{ab})
-      -\frac1Ni\hat{\mathbf x}_aB_{ab}\mathbf x_b^T-\frac1N\hat{\mathbf x}_af'(C_{ab})\hat{\mathbf x}_b^T
-    \right)
-  \end{aligned}
-\end{equation}
-\[
-  B=f'(C)+f(C)A^{-1}
-      \left[(\hat\beta I+G\odot G\odot f''(C))f'(C)+R\odot f''(C)\right]
-\]
-\[
-  \det B_{ab}\det\begin{bmatrix}
-    I&\frac1N\begin{bmatrix}\hat{\mathbf x}_a&\hat{\mathbf x}_a&\bar{\pmb\eta}_a\end{bmatrix} \\
-    \begin{bmatrix}i\mathbf x_b^T\\\hat{\mathbf x}_b^T\\\pmb\eta_b^T\end{bmatrix}
-     & \begin{bmatrix}
-       B_{ab} & 0 & 0\\ 0 & f'(C_{ab}) & 0 \\ 0 & 0 & f'(C_{ab})
-     \end{bmatrix}^{-1}
-  \end{bmatrix}
-\]
-\[
-  \det\left(
-    I-
-     \frac1N\begin{bmatrix}
-       B_{ab} & 0\\ 0 & f'(C_{ab})
-     \end{bmatrix}
-    \begin{bmatrix}i\mathbf x_b^T\\\hat{\mathbf x}_b^T\end{bmatrix}
-    \begin{bmatrix}\hat{\mathbf x}_a&\hat{\mathbf x}_a\end{bmatrix}
-  \right)
-  \det\left(
-    I-\begin{bmatrix}0&f'(C_{ab})\\f'(C_{ab})&0\end{bmatrix}\begin{bmatrix}\bar{\pmb\eta}_a^T&\pmb\eta_a^T\end{bmatrix}
-    \begin{bmatrix}\bar{\pmb\eta}_b\\\pmb\eta_b\end{bmatrix}
-  \right)^{-1}
-\]
-\[
-  \det\left(
-    I-
-     \begin{bmatrix}
-       B & 0\\ 0 & f'(C)
-     \end{bmatrix}
-     \begin{bmatrix}
-       -R&-R\\D&D
-     \end{bmatrix}
-  \right)
-  \det\left(
-    I-\begin{bmatrix}0&-f'(C)\\f'(C)&0\end{bmatrix}
-    \begin{bmatrix}0&-G\\G&0\end{bmatrix}
-  \right)^{-1}
-  =\det\left(
-     \begin{bmatrix}
-       1+B\odot R&B\odot R\\-f'(C)\odot D&1-f'(C)\odot D
-     \end{bmatrix}
-  \right)
-  \det\left(
-    \begin{bmatrix}1+f'(C)\odot G&0\\0&1+f'(C)\odot G\end{bmatrix}
-  \right)^{-1}
-\]
-\[
-  \det A\det\left[
-    I+B\odot R-f'(C)\odot D
-  \right]
-  =\det[
-  (I-f'(C)\odot D)A
-  +A(f'(C)\odot R)
-  +f(C)
-      \left[(\hat\beta I+G\odot G\odot f''(C))f'(C)+R\odot f''(C)\right]
-  ]
-\]
-
 \begin{equation}
   \begin{aligned}
     &\mathcal S
-- 
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