Fixed mistake surrounding the relationship between μ and Lagrange multipliers in some cases.

1 files changed, 45 insertions, 25 deletions

diff --git a/marginal.tex b/marginal.tex
index b9fabc9..442e7e2 100644
--- a/marginal.tex
+++ b/marginal.tex
@@ -641,16 +641,31 @@ treat the determinant keeping the absolute value signs, as in previous works
 \cite{Folena_2020_Rethinking, Kent-Dobias_2023_How}. However, other of
 our examples are for models where the same techniques are impossible.
 
-For the cases studied here, fixing the trace results in a relationship
-between $\mu$ and the Lagrange multipliers enforcing the constraints. This is
-because the trace of $\partial\partial H$ is typically an order of $N$ smaller
-than the trace of $\partial\partial g_i$. The result is that
+Finally, the $\delta$-function fixing the trace of the Hessian to $\mu$ in
+\eqref{eq:kac-rice.measure.2} must be addressed. One could treat it using a
+Fourier representation as in (\ref{eq:delta.grad}--\ref{eq:delta.eigen}), but
+this is inconvenient because a term of the form
+$\operatorname{Tr}\partial\partial H(\mathbf x)$ in the exponential integrand
+cannot be neatly captured in superspace representation introduced in the next
+section. However, in the cases we study in this paper a simplification can be made: the trace of $\partial\partial H$ can be separated into two pieces, one
+that is spatially independent and one that is typically small, or
+\begin{equation} \label{eq:mu.star}
+  \operatorname{Tr}\partial\partial H(\mathbf x)=N\mu^*_H+\Delta_H(\mathbf x)
+\end{equation}
+where $\overline{\mu^*_H}=\mu^*$ and $\overline{\Delta_H(\mathbf x)}=O(N^0)$.
+Then fixing the trace of the Hessian to $\mu$ implies that
 \begin{equation}
-  \mu
-  =\frac1N\operatorname{Tr}\operatorname{Hess}H(\mathbf x)
-  =\frac1N\sum_{i=1}^r\omega_i\operatorname{Tr}\partial\partial g_i(\mathbf x)
-  +O(N^{-1})
+  \begin{aligned}
+    \mu
+    &=\frac1N\operatorname{Tr}\operatorname{Hess}H(\mathbf x)
+    =\frac1N\left(\partial\partial H(\mathbf x)+
+    \sum_{i=1}^r\omega_i\operatorname{Tr}\partial\partial g_i(\mathbf x)\right)
+    \\
+    &=\mu^*+\frac1N\sum_{i=1}^r\omega_i\operatorname{Tr}\partial\partial g_i(\mathbf x)
+    +O(N^{-1})
+  \end{aligned}
 \end{equation}
+for typical samples $H$.
 In particular, here we study only cases with quadratic $g_i$, which results in
 a linear expression relating $\mu$ and the $\omega_i$ that is independent of $\mathbf
 x$. Since $H$ contains the disorder of the problem, this simplification means
@@ -720,7 +735,7 @@ functions and the determinant made. The new measures
   \delta\big((\mathbf s_a^\alpha)^T\partial\mathbf g(\mathbf x_a)\big)
   \\
   d\pmb\omega&=\bigg(\prod_{i=1}^rd\omega_i\bigg)
-  \,\delta\bigg(N\mu-\sum_i^r\omega_i\operatorname{Tr}\partial\partial g_i\bigg)
+  \,\delta\bigg(N\mu-\mu^*-\sum_i^r\omega_i\operatorname{Tr}\partial\partial g_i\bigg)
 \end{align}
 collect the individual measures of the various fields embedded in the superfield, along with their constraints.
 \end{widetext}
@@ -765,7 +780,8 @@ unit normal distribution \cite{Crisanti_1993_The}. We focus on marginal minima
 in models with $f'(0)=0$, which corresponds to models without a random external
 field. Such a random field would correspond in each individual sample $H$ to a
 signal, and therefore complicate the analysis by correlating the positions of
-stationary points and the eigenvectors of their Hessians.
+stationary points and the eigenvectors of their Hessians. Here, $\mu^*$ of
+\eqref{eq:mu.star} is zero.
 
 The marginal optima of these models can be studied without the methods
 introduced in this paper, and have been in the past \cite{Folena_2020_Rethinking,
@@ -1030,7 +1046,7 @@ for $\mathbf x,\mathbf x'\in\mathbb R^N$ by
   \overline{H_i(\mathbf x)H_j(\mathbf x')}
   =N\delta_{ij}f_i\left(\frac{\mathbf x\cdot\mathbf x'}N\right)
 \end{equation}
-with the functions $f_1$ and $f_2$ not necessarily the same.
+with the functions $f_1$ and $f_2$ not necessarily the same. As for the spherical spin glasses, $\mu^*$ of \eqref{eq:mu.star} is zero.
 
 In this problem, there is an energetic competition between the independent spin
 glass energies on each sphere and their tendency to align or anti-align through
@@ -1374,9 +1390,12 @@ As in the previous sections, we used the method of Lagrange multipliers to analy
     -V_k(\mathbf x)\partial\partial V_k(\mathbf x)\right]+\omega I
   \end{aligned}
 \end{align}
-As in the spherical and multispherical spin glasses, fixing the trace of the Hessian
-is equivalent to constraining the value of the Lagrange
-multiplier $\omega=\mu$.
+Unlike in the spherical and multispherical spin glasses, the value $\mu^*$
+defined in \eqref{eq:mu.star} giving the typical value of
+$\frac1N\operatorname{Tr}\partial\partial H$ is not always zero. Instead
+$\mu^*=-f'(0)$, nonzero where there is a linear term in $V$. Fixing the trace
+of the Hessian is therefore equivalent to setting $\omega=\mu+f'(0)$.
+
 
 The derivation of the marginal complexity for this model is complicated, but
 can be made schematically like that of the derivation of the equilibrium free
@@ -1389,10 +1408,11 @@ $\lambda^*$ is given by
   \begin{aligned}
     \mathcal N(E,\mu,\lambda^*)^n
     &=\int d\hat\beta\,d\hat\lambda\prod_{a=1}^n\lim_{m_a\to0}\prod_{\alpha=1}^{m_a}d\pmb\phi_a^\alpha
-    \exp\left\{
+    \\
+    &\qquad\times\exp\left\{
       \delta^{\alpha1}N(\hat\beta E+\hat\lambda\lambda^*)
       -\frac12\int d1\,d2\,\left[B^\alpha(1,2)\sum_{k=1}^MV_k(\pmb\phi_a^\alpha(1,2))^2
-      -\mu\|\pmb\phi_a^\alpha(1,2)\|^2\right]
+      -\big(\mu+f'(0)\big)\|\pmb\phi_a^\alpha(1,2)\|^2\right]
     \right\}
   \end{aligned}
 \end{equation}
@@ -1509,7 +1529,7 @@ with an effective action
 \begin{equation}
   \begin{aligned}
     &\mathcal S_\mathrm{RSS}(\hat\beta,\hat\lambda,r,d,g,q_0,\tilde q_0\mid\lambda^*,E,\mu,\beta)
-    =\hat\beta E-\mu(r+g+\hat\lambda)
+    =\hat\beta E-\big(\mu+f'(0)\big)(r+g+\hat\lambda)
     +\hat\lambda\lambda^*
     +\frac12\log\left(\frac{d+r^2}{g^2}
     \times\frac{1-2q_0+\tilde q_0^2}{(1-q_0)^2}\right) \\
@@ -1532,7 +1552,7 @@ taking the zero-temperature limit, we find
 \begin{equation}
   \begin{aligned}
     &\mathcal S_\mathrm{RSS}(\hat\beta,\hat\lambda,r,d,g,y,\Delta z\mid\lambda^*,E,\mu,\infty)
-    =\hat\beta E-\mu(r+g+\hat\lambda)
+    =\hat\beta E-\big(\mu+f'(0)\big)(r+g+\hat\lambda)
     +\hat\lambda\lambda^*
     +\frac12\log\left(\frac{d+r^2}{g^2}\times\frac{y^2-2\Delta z}{y^2}\right)
     \\
@@ -2012,9 +2032,9 @@ the replicated count of stationary points can be written
     =\int d\hat\beta\prod_{a=1}^n\,d\pmb\phi_a\,
     \exp\bigg[
       N\hat\beta E \\
-    &\qquad-\frac12\int d1\,\left(
+    &-\frac12\int d1\,\left(
         B(1)\sum_{k=1}^MV_k(\pmb\phi_a(1))^2
-        -\mu\|\pmb\phi_a(1)\|^2
+        -\big(\mu+f'(0)\big)\|\pmb\phi_a(1)\|^2
       \right)
     \bigg]
   \end{aligned}
@@ -2057,9 +2077,9 @@ Making the $M$ independent Gaussian integrals, we find
 \begin{equation}
   \begin{aligned}
     &\mathcal N(E,\mu)^n
-    =\int d\hat\beta\left(\prod_{a=1}^nd\pmb\phi_a\right)
-    \exp\bigg[
-      nN\hat\beta E+\frac\mu2\sum_a^n\int d1\,\|\pmb\phi_a\|^2 \\
+    =\int d\hat\beta\left(\prod_{a=1}^nd\pmb\phi_a\right) \\
+    &\times\exp\bigg[
+      nN\hat\beta E+\frac{\mu+f'(0)}2\sum_a^n\int d1\,\|\pmb\phi_a\|^2 \\
     &\quad-\frac M2\log\operatorname{sdet}\left(
         \delta_{ab}\delta(1,2)+B(1)f\left(\frac{\pmb\phi_a(1)\cdot\pmb\phi_b(2)}N\right)
       \right)
@@ -2080,7 +2100,7 @@ We therefore have
     &\mathcal N(E,\mu)^n
     =\int d\hat\beta\,d\mathbb Q\,
     \exp\bigg\{
-      nN\hat\beta E+N\frac\mu2\operatorname{sTr}\mathbb Q
+      nN\hat\beta E+N\frac{\mu+f'(0)}2\operatorname{sTr}\mathbb Q
     +\frac N2\log\operatorname{sdet}\mathbb Q
     -\frac M2\log\operatorname{sdet}\left[
       \delta_{ab}\delta(1,2)+B(1)f(\mathbb Q_{ab}(1,2))
@@ -2110,7 +2130,7 @@ where the effective action is given by
 \begin{equation}
   \begin{aligned}
     \mathcal S_\mathrm{KR}(\hat\beta,C,R,D,G)
-    &=\hat\beta E+\lim_{n\to0}\frac1n\Bigg(-\mu\operatorname{Tr}(G+R)
+    &=\hat\beta E+\lim_{n\to0}\frac1n\Bigg(-\big(\mu+f'(0)\big)\operatorname{Tr}(G+R)
     +\frac12\log\det\big[G^{-2}(CD+R^2)\big]
     +\alpha\log\det\big[I+G\odot f'(C)\big]
     \\