From 1a2dc62a5e75c94e3cfc7da8334061743ba70d05 Mon Sep 17 00:00:00 2001
From: Jaron Kent-Dobias <jaron@kent-dobias.com>
Date: Wed, 12 Jun 2024 18:15:51 +0200
Subject: More writing.

---
 marginal.tex | 169 +++++++++++++++++++++++++++++++++++++++++------------------
 1 file changed, 119 insertions(+), 50 deletions(-)

(limited to 'marginal.tex')

diff --git a/marginal.tex b/marginal.tex
index 39adc40..0eb6e5b 100644
--- a/marginal.tex
+++ b/marginal.tex
@@ -103,24 +103,24 @@ expressed as
 \begin{equation} \label{eq:λmin}
   g(\lambda_\textrm{min}(A))
   =\lim_{\beta\to\infty}\int
-    \frac{d\mathbf s\,\delta(N-\mathbf s^T\mathbf s)e^{-\beta\mathbf s^TA\mathbf s}}
-      {\int d\mathbf s'\,\delta(N-\mathbf s'^T\mathbf s')e^{-\beta\mathbf s'^TA\mathbf s'}}
+    \frac{d\mathbf s\,\delta(N-\|\mathbf s\|^2)e^{-\beta\mathbf s^TA\mathbf s}}
+      {\int d\mathbf s'\,\delta(N-\|\mathbf s'\|^2)e^{-\beta\mathbf s'^TA\mathbf s'}}
     g\left(\frac{\mathbf s^TA\mathbf s}N\right)
 \end{equation}
 Assuming
 \begin{equation}
   \begin{aligned}
     &\lim_{\beta\to\infty}\int\frac{
-      d\mathbf s\,\delta(N-\mathbf s^T\mathbf s)e^{-\beta\mathbf s^TA\mathbf s}
+      d\mathbf s\,\delta(N-\|\mathbf s\|^2)e^{-\beta\mathbf s^TA\mathbf s}
     }{
-      \int d\mathbf s'\,\delta(N-\mathbf s'^T\mathbf s')e^{-\beta\mathbf s'^TA\mathbf s'}
+      \int d\mathbf s'\,\delta(N-\|\mathbf s'\|^2)e^{-\beta\mathbf s'^TA\mathbf s'}
     }g\left(\frac{\mathbf s^TA\mathbf s}N\right) \\
     &=\int\frac{
-      d\mathbf s\,\delta(N-\mathbf s^T\mathbf s)\mathbb 1_{\operatorname{ker}(A-\lambda_\mathrm{min}(A)I)}(\mathbf s)
+      d\mathbf s\,\delta(N-\|\mathbf s\|^2)\mathbb 1_{\operatorname{ker}(A-\lambda_\mathrm{min}(A)I)}(\mathbf s)
     }{
-      \int d\mathbf s'\,\delta(N-\mathbf s'^T\mathbf s')\mathbb 1_{\operatorname{ker}(A-\lambda_\mathrm{min}(A)I)}(\mathbf s')}g\left(\frac{\mathbf s^TA\mathbf s}N\right) \\
+      \int d\mathbf s'\,\delta(N-\|\mathbf s'\|^2)\mathbb 1_{\operatorname{ker}(A-\lambda_\mathrm{min}(A)I)}(\mathbf s')}g\left(\frac{\mathbf s^TA\mathbf s}N\right) \\
     &=g(\lambda_\mathrm{min}(A))
-    \frac{\int d\mathbf s\,\delta(N-\mathbf s^T\mathbf s)\mathbb 1_{\operatorname{ker}(A-\lambda_\mathrm{min}(A)I)}(\mathbf s)}{\int d\mathbf s'\,\delta(N-\mathbf s'^T\mathbf s')\mathbb 1_{\operatorname{ker}(A-\lambda_\mathrm{min}(A)I)}(\mathbf s')} \\
+    \frac{\int d\mathbf s\,\delta(N-\|\mathbf s\|^2)\mathbb 1_{\operatorname{ker}(A-\lambda_\mathrm{min}(A)I)}(\mathbf s)}{\int d\mathbf s'\,\delta(N-\|\mathbf s'\|^2)\mathbb 1_{\operatorname{ker}(A-\lambda_\mathrm{min}(A)I)}(\mathbf s')} \\
     &=g(\lambda_\mathrm{min}(A))
   \end{aligned}
 \end{equation}
@@ -174,17 +174,17 @@ Using the representation of $\lambda_\mathrm{min}$ defined in \eqref{eq:λmin},
 \begin{equation}
   e^{NG_{\lambda^*}(\mu)}
   =\overline{
-    \lim_{\beta\to\infty}\int\frac{d\mathbf s\,\delta(N-\mathbf s^T\mathbf s)e^{-\beta\mathbf s^T(B+\mu I)\mathbf s}}
-    {\int d\mathbf s'\,\delta(N-\mathbf s'^T\mathbf s')e^{-\beta\mathbf s'^T(B+\mu I)\mathbf s'}}\,\delta\big(N\lambda^*-\mathbf s^T(B+\mu I)\mathbf s\big)
+    \lim_{\beta\to\infty}\int\frac{d\mathbf s\,\delta(N-\|\mathbf s\|^2)e^{-\beta\mathbf s^T(B+\mu I)\mathbf s}}
+    {\int d\mathbf s'\,\delta(N-\|\mathbf s'\|^2)e^{-\beta\mathbf s'^T(B+\mu I)\mathbf s'}}\,\delta\big(N\lambda^*-\mathbf s^T(B+\mu I)\mathbf s\big)
   }
 \end{equation}
-Using replicas to treat the denominator ($x^{-1}=\lim_{n\to0}x^{n-1}$)
+Using replicas to treat the denominator ($x^{-1}=\lim_{m\to0}x^{m-1}$)
 and transforming the $\delta$-function to its Fourier
 representation, we have
 \begin{equation}
   e^{NG_{\lambda^*}(\mu)}
-  =\overline{\lim_{\beta\to\infty}\lim_{n\to0}\int d\hat\lambda\prod_{a=1}^n\left[d\mathbf s_a\,\delta(N-\mathbf s_a^T\mathbf s_a)\right]
-  \exp\left\{-\beta\sum_{a=1}^n\mathbf s_a^T(B+\mu I)\mathbf s_a+\hat\lambda\left[N\lambda^*-\mathbf s_1^T(B+\mu I)\mathbf s_1\right]\right\}}
+  =\overline{\lim_{\beta\to\infty}\lim_{m\to0}\int d\hat\lambda\prod_{\alpha=1}^m\left[d\mathbf s^\alpha\,\delta(N-\|\mathbf s^\alpha\|^2)\right]
+  \exp\left\{-\beta\sum_{\alpha=1}^m(\mathbf s^\alpha)^T(B+\mu I)\mathbf s^\alpha+\hat\lambda\left[N\lambda^*-(\mathbf s^1)^T(B+\mu I)\mathbf s^1\right]\right\}}
 \end{equation}
 having introduced the parameter $\hat\lambda$ in the Fourier representation of
 the $\delta$-function. The whole expression, so transformed, is a simple
@@ -193,9 +193,9 @@ have
 \begin{equation}
   \begin{aligned}
   &e^{NG_{\lambda^*}(\mu)}
-  =\lim_{\beta\to\infty}\lim_{n\to0}\int d\hat\lambda\prod_{a=1}^n\left[d\mathbf s_a\,\delta(N-\mathbf s_a^T\mathbf s_a)\right] \\
-  &\hspace{10em}\exp\left\{N\left[\hat\lambda(\lambda^*-\mu)-n\beta\mu\right]+\frac{\sigma^2}{N}\left[\beta^2\sum_{ab}^n(\mathbf s_a^T\mathbf s_b)^2
-    +2\beta\hat\lambda\sum_a^n(\mathbf s_a^T\mathbf s_1)^2
+  =\lim_{\beta\to\infty}\lim_{m\to0}\int d\hat\lambda\prod_{\alpha=1}^m\left[d\mathbf s^\alpha\,\delta(N-\|\mathbf s^\alpha\|^2)\right] \\
+  &\hspace{10em}\exp\left\{N\left[\hat\lambda(\lambda^*-\mu)-m\beta\mu\right]+\frac{\sigma^2}{N}\left[\beta^2\sum_{\alpha\gamma}^m(\mathbf s^\alpha\cdot\mathbf s^\gamma)^2
+    +2\beta\hat\lambda\sum_\alpha^m(\mathbf s^\alpha\cdot\mathbf s^1)^2
     +\hat\lambda^2N^2
   \right]\right\}
   \end{aligned}
@@ -231,12 +231,14 @@ and
 \end{equation}
 Inserting these expressions and taking the limit of $n$ to zero, we find
 \begin{equation}
-  e^{NG_{\lambda^*}(\mu)}=\lim_{\beta\to\infty}\int d\hat\lambda\,dq_0\,d\tilde q_0\,e^{N\mathcal S_\beta(q_0,\tilde q_0,\hat\lambda)}
+  e^{NG_{\lambda^*}(\mu)}
+  =\lim_{\beta\to\infty}\int d\hat\lambda\,dq_0\,d\tilde q_0\,
+  e^{N\mathcal U_\textrm{GOE}(q_0,\tilde q_0,\hat\lambda\mid\beta)}
 \end{equation}
 with the effective action
 \begin{equation}
   \begin{aligned}
-    &\mathcal S_\beta(q_0,\tilde q_0,\hat\lambda) \\
+    &\mathcal U_\mathrm{GOE}(q_0,\tilde q_0,\hat\lambda\mid\beta) \\
     &\quad=\hat\lambda(\lambda^*-\mu)+\sigma^2\left[
       2\beta^2(q_0^2-\tilde q_0^2)+2\beta\hat\lambda(1-\tilde q_0^2)+\hat\lambda^2
     \right] \\
@@ -256,7 +258,7 @@ However, taking the limit with $y\neq\tilde y$ results in an expression for the
 action that diverges with $\beta$. To cure this, we must take $\tilde y=y$. The result is
 \begin{equation}
   \begin{aligned}
-    \mathcal S_\infty(y,\Delta z,\hat\lambda)
+    \mathcal U_\textrm{GOE}(y,\Delta z,\hat\lambda\mid\infty)
     &=\hat\lambda(\lambda^*-\mu)
     +\sigma^2\big[
       \hat\lambda^2-4(y+\Delta z)
@@ -276,7 +278,8 @@ Inserting this solution into $\mathcal S_\infty$ we find
 \begin{equation} \label{eq:goe.large.dev}
   \begin{aligned}
     &G_{\lambda^*}(\mu)
-    =\mathop{\textrm{extremum}}_{y,\Delta z,\hat\lambda}\mathcal S_\infty(y,\Delta z,\hat\lambda) \\
+    =\mathop{\textrm{extremum}}_{y,\Delta z,\hat\lambda}
+    \mathcal U_\mathrm{GOE}(y,\Delta z,\hat\lambda\mid\infty) \\
     &=-\tfrac{\mu+\lambda^*}{2\sigma}\sqrt{\Big(\tfrac{\mu+\lambda^*}{2\sigma}\Big)^2-1}
     +\log\left(
       \tfrac{\mu+\lambda^*}{2\sigma}+\sqrt{\Big(\tfrac{\mu+\lambda^*}{2\sigma}\Big)^2-1}
@@ -411,8 +414,8 @@ We further want to control the value of the minimum eigenvalue of the Hessian at
     &\mathcal N_H(E,\mu,\lambda^*)
     =\int d\nu_H(\mathbf x,\pmb\omega\mid E,\mu)\,\delta\big(N\lambda^*-\lambda_\mathrm{min}(\operatorname{Hess}H(\mathbf x,\pmb\omega))\big) \\
     &=\lim_{\beta\to\infty}\int d\nu_H(\mathbf x,\pmb\omega\mid E,\mu)
-    \frac{d\mathbf s\,\delta(N-\mathbf s^T\mathbf s)\delta(\mathbf s^T\partial\mathbf g(\mathbf x))e^{-\beta\mathbf s^T\operatorname{Hess}H(\mathbf x,\pmb\omega)\mathbf s}}
-    {\int d\mathbf s'\,\delta(N-\mathbf s'^T\mathbf s')\delta(\mathbf s'^T\partial\mathbf g(\mathbf x))e^{-\beta\mathbf s'^T\operatorname{Hess}H(\mathbf x,\pmb\omega)\mathbf s'}}
+    \frac{d\mathbf s\,\delta(N-\|\mathbf s\|^2)\delta(\mathbf s^T\partial\mathbf g(\mathbf x))e^{-\beta\mathbf s^T\operatorname{Hess}H(\mathbf x,\pmb\omega)\mathbf s}}
+    {\int d\mathbf s'\,\delta(N-\|\mathbf s'\|^2)\delta(\mathbf s'^T\partial\mathbf g(\mathbf x))e^{-\beta\mathbf s'^T\operatorname{Hess}H(\mathbf x,\pmb\omega)\mathbf s'}}
     \delta\big(N\lambda^*-\mathbf s^T\operatorname{Hess}H(\mathbf x,\pmb\omega)\mathbf s\big)
   \end{aligned}
 \end{equation}
@@ -437,7 +440,7 @@ again to treat each of the normalizations in the numerator. This leads to the ex
     &=\lim_{\beta\to\infty}\lim_{n\to0}\frac1N\frac\partial{\partial n}\int\prod_{a=1}^n\Bigg[d\nu_H(\mathbf x_a,\pmb\omega_a\mid E,\mu)\,\delta\big(N\lambda^*-(\mathbf s_a^1)^T\operatorname{Hess}H(\mathbf x_a,\pmb\omega_a)\mathbf s_a^1\big)\\
     &\hspace{12em}\times\lim_{m_a\to0}
       \left(\prod_{\alpha=1}^{m_a} d\mathbf s_a^\alpha
-      \,\delta\big(N-(\mathbf s_a^\alpha)^T\mathbf s_a^\alpha\big)
+      \,\delta\big(N-\|\mathbf s_a^\alpha\|^2\big)
       \,\delta\big((\mathbf s_a^\alpha)^T\partial\mathbf g(\mathbf x_a)\big)
       \,e^{-\beta(\mathbf s_a^\alpha)^T\operatorname{Hess}H(\mathbf x_a,\pmb\omega_a)\mathbf s_a^\alpha}\right)
     \Bigg]
@@ -651,7 +654,7 @@ have introduced through their scalar products with each other. We therefore make
   &D_{ab}=\frac1N\hat{\mathbf x}_a\cdot\hat{\mathbf x}_b
   &G_{ab}=\frac1N\bar{\pmb\eta}_a^T\pmb\eta_b&
   \\
-  &A_{ab}^{\alpha\gamma}=\frac1N\mathbf s_a^\alpha\cdot\mathbf s_b^\gamma
+  &Q_{ab}^{\alpha\gamma}=\frac1N\mathbf s_a^\alpha\cdot\mathbf s_b^\gamma
   &X^\alpha_{ab}=\frac1N\mathbf x_a\cdot\mathbf s_b^\alpha&
   \\
   &\hat X^\alpha_{ab}=-i\frac1N\hat{\mathbf x}_a\cdot\mathbf s_b^\alpha&&
@@ -663,34 +666,57 @@ This transformation changes the measure of the integral, with
 \begin{equation}
   \begin{aligned}
     &\prod_{a=1}^nd\mathbf x_a\,\frac{d\hat{\mathbf x}_a}{(2\pi)^N}\,d\bar{\pmb\eta}_a\,d\pmb\eta\,\prod_{\alpha=1}^{m_a}d\mathbf s_a^\alpha \\
-    &\quad=dC\,dR\,dD\,dG\,dA\,dX\,d\hat X\,(\det J)^{N/2}(\det G)^{-N/2}
+    &\quad=dC\,dR\,dD\,dG\,dQ\,dX\,d\hat X\,(\det J)^{N/2}(\det G)^{-N}
   \end{aligned}
 \end{equation}
 where $J$ is the Jacobian of the transformation and takes the form
 \begin{equation} \label{eq:coordinate.jacobian}
   J=\begin{bmatrix}
-    C&iR&X^1&\cdots&X^n \\
-    iR&D&i\hat X^1&\cdots&i\hat X^m\\
-    (X^1)^T&i(\hat X^1)^T&A^{11}&\cdots&A^{1n}\\
+    C&iR&X_1&\cdots&X_n \\
+    iR&D&i\hat X_1&\cdots&i\hat X_n\\
+    X_1^T&i\hat X_1^T&Q_{11}&\cdots&Q_{1n}\\
     \vdots&\vdots&\vdots&\ddots&\vdots\\
-    (X^n)^T&i(\hat X^n)^T&A^{n1}&\cdots&A^{nn}
+    X_n^T&i\hat X_n^T&Q_{n1}&\cdots&Q_{nn}
   \end{bmatrix}
 \end{equation}
 and the contribution of the Grassmann integrals produces its own inverted
-Jacobian.
+Jacobian. The block matrices indicated above are such that $A_{ab}$ is an
+$m_a\times m_b$ matrix indexed by the upper indices, while $X_a$ is an $n\times
+m_a$ matrix with one lower and one upper index.
 
 After these steps, which follow identically to those more carefully outlined in
 the cited papers \cite{Folena_2020_Rethinking, Kent-Dobias_2023_How}, we arrive at a form of the integral as over an effective action
+\begin{widetext}
 \begin{equation}
   \begin{aligned}
     &\Sigma_{\lambda^*}(E,\mu)
-    =\lim_{\beta\to\infty}\lim_{n\to0}\frac1N\frac\partial{\partial n}
-    \int dC\,dR\,dD\,dG \\
-    &dA\,dX\,d\hat X\,
-    d\hat\beta\,d\hat\lambda\,e^{N
-    n\mathcal S_\mathrm{KR}(\hat\beta,\omega,C,R,D,G)
-    +N\mathcal S_\beta(\omega,\hat\lambda,A,X,\hat X)
-  }
+    =\lim_{\beta\to\infty}\lim_{n\to0}\lim_{m_1\cdots m_n\to0}
+    \frac1N\frac\partial{\partial n}
+    \int dC\,dR\,dD\,dG\,dQ\,dX\,d\hat X\,d\hat\beta\,d\hat\lambda\,
+    \exp\Bigg\{
+      nN\mathcal S_\mathrm{SSG}(\hat\beta,C,R,D,G\mid E,\mu) \\
+      &\qquad
+      +nN\mathcal U_\mathrm{SSG}(\hat\lambda,C,Q,X,\hat X\mid\beta)
+      +\frac N2\log\det\left[
+        I+\begin{bmatrix}
+          Q_{11}&\cdots&Q_{1n}\\
+          \vdots&\ddots&\vdots\\
+          Q_{n1}&\cdots&Q_{nn}
+        \end{bmatrix}^{-1}
+        \begin{bmatrix}
+          X_1^T&i\hat X_1^T\\
+          \vdots&\vdots\\
+          X_n^T&i\hat X_n^T
+        \end{bmatrix}
+        \begin{bmatrix}
+          C&iR\\iR&D
+        \end{bmatrix}^{-1}
+        \begin{bmatrix}
+          X_1\cdots X_n\\
+          i\hat X_1\cdots i\hat X_n
+        \end{bmatrix}
+      \right]
+    \Bigg\}
   \end{aligned}
 \end{equation}
 where the matrix $J$ is the Jacobian associated with the change of variables
@@ -700,20 +726,52 @@ terms which only share a dependence on the Lagrange multiplier $\omega$ that
 enforces the constraint, is generic to Gaussian problems. This is the
 appearance in practice of the fact mentioned before that conditions on the
 Hessian do not mostly effect the rest of the complexity problem.
-\begin{widetext}
-  \begin{equation}
-    \mathcal S_\mathrm{KR}
-    =\frac12\sum_{ab}\left(
-      \hat\beta_a\hat\beta_bf(C_{ab})
-      +\big(2\hat\beta_a(R_{ab}-F_{ab})-D_{ab}\big)f'(C_{ab})
-      +(R_{ab}^2-F_{ab}^2)f''(C_{ab})
+
+The effective action $\mathcal S_\mathrm{SSG}$ is precisely that for the
+ordinary complexity of stationary points, or
+\begin{equation}
+  \begin{aligned}
+    &\mathcal S_\mathrm{SSG}(\hat\beta,C,R,D,G\mid E,\mu)
+    =\hat\beta E-(r_d+g_d)\mu \\
+    &+\frac1n\left\{\frac12\sum_{ab}\left(
+      \hat\beta^2f(C_{ab})
+      +\big(2\hat\beta R_{ab}-D_{ab}\big)f'(C_{ab})
+      +(R_{ab}^2-G_{ab}^2)f''(C_{ab})
     \right)
-    -\log\det F
-  \end{equation}
-  \begin{equation}
-    \mathcal S_\beta
-    =\sum_{ab}^n\left[\beta\omega A_{aa}^{bb}+\hat x\omega A_{aa}^{11}+\beta^2f''(1)\sum_{cd}^m(A_{ab}^{cd})^2+\hat x^2f''(1)(A_{ab}^{11})^2+\beta\hat xf''(1)\sum_c^m A_{ab}^{1c}\right]
-  \end{equation}
+    +\frac12\log\det\begin{bmatrix}C&iR\\iR^T&D\end{bmatrix}
+    -\log\det G\right\}
+  \end{aligned}
+\end{equation}
+where $r_d$ and $g_d$ are the diagonal elements of $R$ and $G$, respectively.
+\begin{equation}
+  \begin{aligned}
+    &\mathcal U_\mathrm{SSG}(\hat\lambda,Q,X,\hat X\mid\lambda^*,\mu,C)
+    =\hat\lambda\lambda^*
+    +\frac1n\Bigg\{
+    \frac12\log\det Q+
+      \sum_{a=1}^n\bigg(
+        \sum_{\alpha=1}^{m_a}\beta\mu Q_{aa}^{\alpha\alpha}
+        +\hat\lambda\mu Q_{aa}^{11}
+      \bigg)
+    +2\sum_{ab}^nf''(C_{ab})
+    \\
+     &\qquad\times\Bigg[\beta\sum_\alpha^{m_a}\left(
+        \sum_\gamma^{m_b}(Q_{ab}^{\alpha\gamma})^2
+        -\hat\beta(X_{ab}^\alpha)^2
+        -2X_{ab}^\alpha\hat X_{ab}^\alpha
+      \right)
+    +\hat\lambda\left(
+      \hat\lambda(Q_{ab}^{11})^2
+      -\hat\beta(X_{ab}^1)^2
+      -2X_{ab}^1\hat X_{ab}^1
+    \right)
+    +\beta\hat\lambda\left(
+        \sum_\alpha^{m_a} Q_{ab}^{\alpha1}
+        +\sum_\alpha^{m_b} Q_{ab}^{1\alpha}
+    \right)\Bigg]
+    \Bigg\}
+  \end{aligned}
+\end{equation}
 \end{widetext}
 There are some dramatic simplifications that emerge from the structure of this
 particular problem. First, notice that (outside of the `volume' term due to
@@ -730,7 +788,18 @@ external field, the preferred direction can polarize both the direction of
 typical stationary points \emph{and} their soft eigenvectors. Therefore, in
 these instances one must account for solutions with nonzero $X$ and $\hat X$.
 
-When the $X$ and $\hat X$ order parameters are zero, as they are here, the term associated with the Jacobian separates into two terms, one dependent only on the order parameters of the traditional complexity problem $C$, $R$, and $D$, and one dependent only on the overlap of the minimum eigenvector, $A$. Now we see that, outside of the Lagrange multiplier $\omega$, the Kac--Rice complexity problem and the problem of fixing the smallest eigenvalue completely decouple.
+
+
+When we take $X=\hat X=0$, $Q^{\alpha\beta}_{ab}=\delta_{ab}Q^{\alpha\beta}$
+independent, and $Q$ to have the planted replica symmetric form of
+\eqref{eq:Q.structure}, we find that
+\begin{equation}
+  \mathcal U_\mathrm{SSG}(\hat\lambda,Q,0,0\mid\beta,\lambda^*,\mu,C)
+  =\mathcal U_\mathrm{GOE}(\hat\lambda,q_0,\tilde q_0\mid\beta)
+\end{equation}
+with $\sigma=f''(1)$. That is, the effective action for the terms related to
+fixing the eigenvalue in the spherical Kac--Rice problem is exactly the same as
+that for the \textrm{GOE} problem.
 
 \begin{equation}
   \Sigma_{\lambda^*}(E,\mu)
-- 
cgit v1.2.3-70-g09d2

