From 12ced356d81e1285dfcbff33c1526f0af20eb2b3 Mon Sep 17 00:00:00 2001
From: Jaron Kent-Dobias <jaron@kent-dobias.com>
Date: Thu, 30 Jun 2022 18:51:24 +0200
Subject: Some cleanup.

---
 frsb_kac-rice.tex | 81 +++++++++++++++++++++++++++++++------------------------
 1 file changed, 46 insertions(+), 35 deletions(-)

(limited to 'frsb_kac-rice.tex')

diff --git a/frsb_kac-rice.tex b/frsb_kac-rice.tex
index 3c551d7..9ba027c 100644
--- a/frsb_kac-rice.tex
+++ b/frsb_kac-rice.tex
@@ -184,17 +184,17 @@ The stationary points of a function can be counted using the Kac--Rice formula,
 which integrates a over the function's domain a $\delta$-function containing
 the gradient multiplied by the absolute value of the determinant
 \cite{Rice_1939_The, Kac_1943_On}. In addition, we insert a $\delta$-function
-fixing the energy density $\epsilon$, giving the number of stationary points at
-energy $\epsilon$ and radial reaction $\mu$ as
+fixing the energy density $E$, giving the number of stationary points at
+energy $E$ and radial reaction $\mu$ as
 \begin{equation}
-  \mathcal N(\epsilon, \mu)
-  =\int ds\,\delta(N\epsilon-H(s))\delta(\partial H(s)+\mu s)|\det(\partial\partial H(s)+\mu I)|
+  \mathcal N(E, \mu)
+  =\int ds\,\delta(NE-H(s))\delta(\partial H(s)+\mu s)|\det(\partial\partial H(s)+\mu I)|
 \end{equation}
 This number will typically be exponential in $N$. In order to find typical
 counts when disorder is averaged, we will want to average its logarithm
 instead, which is known as the complexity:
 \begin{equation}
-  \Sigma(\epsilon,\mu)=\lim_{N\to\infty}\frac1N\overline{\log\mathcal N(\epsilon, \mu)}
+  \Sigma(E,\mu)=\lim_{N\to\infty}\frac1N\overline{\log\mathcal N(E, \mu)}
 \end{equation}
 The radial reaction $\mu$, which acts like a kind of `mass' term, takes a
 fixed value here, which means that the complexity is for a given energy
@@ -205,14 +205,14 @@ given by maximizing the complexity as a function of $\mu$.
 
 If one averages over $\mathcal N$ and afterward takes its logarithm, one arrives at the annealed complexity
 \begin{equation}
-  \Sigma_\mathrm a(\epsilon,\mu)
-  =\lim_{N\to\infty}\frac1N\log\overline{\mathcal N(\epsilon,\mu)}
+  \Sigma_\mathrm a(E,\mu)
+  =\lim_{N\to\infty}\frac1N\log\overline{\mathcal N(E,\mu)}
 \end{equation}
 This has been previously computed for the mixed $p$-spin models \cite{BenArous_2019_Geometry}, with the result
 \begin{equation}
   \begin{aligned}
-    \Sigma_\mathrm a(\epsilon,\mu)
-    =-\frac{\epsilon^2(f'(1)+f''(1))+2\epsilon\mu f'(1)+f(1)\mu^2}{2f(1)(f'(1)+f''(1))-2f'(1)^2}-\frac12\log f'(1)\\
+    \Sigma_\mathrm a(E,\mu)
+    =-\frac{E^2(f'(1)+f''(1))+2E\mu f'(1)+f(1)\mu^2}{2f(1)(f'(1)+f''(1))-2f'(1)^2}-\frac12\log f'(1)\\
     +\operatorname{Re}\left[\frac\mu{\mu+\sqrt{\mu^2-4f''(1)}}
       -\log\left(\frac1{2f''(1)}\left(\mu-\sqrt{\mu^2-4f''(1)}\right)\right)
     \right]
@@ -235,9 +235,9 @@ Folena et al.~\cite{Folena_2020_Rethinking}. Here we review the derivation.
 In order to average the complexity over disorder properly, the logarithm must be dealt with. We use the standard replica trick, writing
 \begin{equation}
   \begin{aligned}
-    \log\mathcal N(\epsilon,\mu)
-    &=\lim_{n\to0}\frac\partial{\partial n}\mathcal N^n(\epsilon,\mu) \\
-    &=\lim_{n\to0}\frac\partial{\partial n}\int\prod_a^n ds_a\,\delta(N\epsilon-H(s_a))\delta(\partial H(s_a)+\mu s_a)|\det(\partial\partial H(s_a)+\mu I)|
+    \log\mathcal N(E,\mu)
+    &=\lim_{n\to0}\frac\partial{\partial n}\mathcal N^n(E,\mu) \\
+    &=\lim_{n\to0}\frac\partial{\partial n}\int\prod_a^n ds_a\,\delta(NE-H(s_a))\delta(\partial H(s_a)+\mu s_a)|\det(\partial\partial H(s_a)+\mu I)|
   \end{aligned}
 \end{equation}
 
@@ -252,8 +252,8 @@ eigenvalues (the index $\mathcal I$ of the saddle), (see Fyodorov
 therefore able to write
 \begin{equation}
   \begin{aligned}
-    \Sigma(\epsilon, \mu)
-    &=\lim_{N\to\infty}\frac1N\lim_{n\to0}\frac\partial{\partial n}\int\left(\prod_a^nds_a\right)\,\overline{\prod_a^n \delta(N\epsilon-H(s_a))\delta(\partial H(s_a)+\mu s_a)}
+    \Sigma(E, \mu)
+    &=\lim_{N\to\infty}\frac1N\lim_{n\to0}\frac\partial{\partial n}\int\left(\prod_a^nds_a\right)\,\overline{\prod_a^n \delta(NE-H(s_a))\delta(\partial H(s_a)+\mu s_a)}
     \times
     \overline{\prod_a^n |\det(\partial\partial H(s_a)+\mu I)|}
   \end{aligned}
@@ -270,11 +270,11 @@ To largest order in $N$, the average over the product of determinants factorizes
   \right\}
   \end{aligned}
 \end{equation}
-The $\delta$-functions are treated by writing them in the Fourier basis, introducing auxiliary fields $\hat s_a$ and $\hat beta$,
+The $\delta$-functions are treated by writing them in the Fourier basis, introducing auxiliary fields $\hat s_a$ and $\hat\beta$,
 \begin{equation}
-  \prod_a^n\delta(N\epsilon-H(s_a))\delta(\partial H(s_a)+\mu s_a)
+  \prod_a^n\delta(NE-H(s_a))\delta(\partial H(s_a)+\mu s_a)
   =\int \frac{d\hat\beta}{2\pi}\prod_a^n\frac{d\hat s_a}{2\pi}
-    e^{\hat\beta(N\epsilon-H(s_a))+i\hat s_a\cdot(\partial H(s_a)+\mu s_a)}
+    e^{\hat\beta(NE-H(s_a))+i\hat s_a\cdot(\partial H(s_a)+\mu s_a)}
 \end{equation}
 $\hat \beta$ is a parameter conjugate to the state energies, i.e. playing the
 role of an inverse temperature for the metastable states. The average over disorder can now be taken, and since everything is Gaussian it gives
@@ -322,8 +322,8 @@ change of variables in the integration from $s_a$ and $\hat s_a$ to these three
 matrices, we arrive at the form for the complexity
 \begin{equation}
   \begin{aligned}
-    &\Sigma(\epsilon,\mu)
-    =\mathcal D(\mu)+\hat\beta\epsilon+\\
+    &\Sigma(E,\mu)
+    =\mathcal D(\mu)+\hat\beta E+\\
     &\lim_{n\to0}\frac1n\left(
       -\mu\operatorname{Tr}R
       +\frac12\sum_{ab}\left[
@@ -346,7 +346,7 @@ extremal conditions, since hierarchical matrices commute and are closed under
 matrix products and Hadamard products. The extremal conditions are
 \begin{align}
   0&=\frac{\partial\Sigma}{\partial\hat\beta}
-  =\epsilon+\sum_{ab}\left[\hat\beta f(C_{ab})+R_{ab}f'(C_{ab})\right] \label{eq:cond.b} \\
+  =E+\sum_{ab}\left[\hat\beta f(C_{ab})+R_{ab}f'(C_{ab})\right] \label{eq:cond.b} \\
   \frac{\tilde c}2I&=\frac{\partial\Sigma}{\partial C}
   =\frac12\left[
     \hat\beta^2f'(C)+(2\hat\beta R-D)\odot f''(C)+R\odot R\odot f'''(C)
@@ -423,10 +423,10 @@ supersymmetric solution only counts dominant minima.}
 Inserting the supersymmetric ansatz $D=\hat\beta R$ and $R=r_dI$, one gets
 \begin{equation} \label{eq:diagonal.action}
   \begin{aligned}
-    \Sigma(\epsilon,\mu)
+    \Sigma(E,\mu)
     =\mathcal D(\mu)
     +
-      \hat\beta\epsilon-\mu r_d
+      \hat\beta E-\mu r_d
       +\frac12\hat\beta r_df'(1)+\frac12r_d^2f''(1)+\frac12\log r_d^2
       \\
       +\frac12\lim_{n\to0}\frac1n\left(\hat\beta^2\sum_{ab}f(C_{ab})+\log\det((\hat\beta/r_d)C+I)\right)
@@ -435,9 +435,9 @@ Inserting the supersymmetric ansatz $D=\hat\beta R$ and $R=r_dI$, one gets
 From here, it is straightforward to see that the complexity vanishes at the
 ground state energy. First, in the ground state minima will dominate (even if
 they are marginal), so we may assume \eqref{eq:mu.minima}. Then, taking
-$\Sigma(\epsilon_0,\mu^*)=0$, gives
+$\Sigma(E_0,\mu^*)=0$, gives
 \begin{equation}
-  \hat\beta\epsilon_0
+  \hat\beta E_0
   =-\frac12r_d\hat\beta f'(1)-\frac12\lim_{n\to0}\frac1n\left(
       \hat\beta^2\sum_{ab}^nf(C_{ab})
       +\log\det(\hat\beta r_d^{-1} C+I)
@@ -543,10 +543,10 @@ E\rangle_2$.
 Using standard manipulations (Appendix B), one finds also a continuous version
 \begin{equation} \label{eq:functional.action}
   \begin{aligned}
-    \Sigma(\epsilon,\mu)
+    \Sigma(E,\mu)
     =\mathcal D(\mu)
     +
-      \hat\beta\epsilon-\mu R_d
+      \hat\beta E-\mu R_d
       +\frac12(2\hat\beta R_d-D_d)f'(1)+\frac12R_d^2f''(1)+\frac12\log R_d^2
       \\
       +\frac12\int_0^1dq\,\left(
@@ -561,7 +561,7 @@ where $\chi(q)$ is defined with respect to $Q$ exactly as in the equilibrium cas
   \includegraphics{figs/24_complexity.pdf}
   \caption{
     The complexity $\Sigma$ of the mixed $2+4$ spin model as a function of
-    distance $\Delta\epsilon=\epsilon-\epsilon_0$ of the ground state. The
+    distance $\Delta E=E-E_0$ of the ground state. The
     solid blue line shows the complexity of dominant saddles given by the FRSB
     ansatz, and the solid yellow line shows the complexity of marginal minima.
     The dashed lines show the same for the annealed complexity. The inset shows
@@ -637,7 +637,7 @@ for different energies and typical vs minima.
 
 \section{Interpretation}
 
-Let $\langle A\rangle$ be average over stationary points with given $\epsilon$ and $\mu$, i.e.,
+Let $\langle A\rangle$ be average over stationary points with given $E$ and $\mu$, i.e.,
 \begin{equation}
   \langle A\rangle
   =\frac1{\mathcal N}\sum_{\sigma}A(s_\sigma)
@@ -646,13 +646,13 @@ Let $\langle A\rangle$ be average over stationary points with given $\epsilon$ a
 \end{equation}
 with
 \begin{equation}
-  d\nu(s)=ds\,\delta(N\epsilon-H(s))\delta(\partial H(s)+\mu s)|\det(\partial\partial H(s)+\mu I)|
+  d\nu(s)=ds\,\delta(NE-H(s))\delta(\partial H(s)+\mu s)|\det(\partial\partial H(s)+\mu I)|
 \end{equation}
 Then
 \begin{equation}
   \begin{aligned}
     \overline{\frac1{N^p}\sum_{i_1\cdots i_p}\langle s_{i_1}\cdots s_{i_p}\rangle\langle s_{i_1}\cdots s_{i_p}\rangle}
-    =\lim_{n\to0}\overline{\int\prod_\alpha^nd\nu(s_\alpha)\,\frac1{n(n-1)}\sum_{a\neq b}\left(\frac{s_a^Ts_b}N\right)^p} \\
+    =\lim_{n\to0}\overline{\int\prod_\alpha^nd\nu(s_\alpha)\,\frac1{n(n-1)}\sum_{a\neq b}\left(\frac{s_a\cdot s_b}N\right)^p} \\
     =\lim_{n\to0}\frac1{n(n-1)}\sum_{a\neq b}^nC^p_{ab}
     =\int_0^1 dx\,c^p(x)
   \end{aligned}
@@ -662,8 +662,8 @@ Then
   \begin{aligned}
     \overline{\frac1{N^p}\sum_{i_1\cdots i_p}\frac{\partial\langle s_{i_1}\cdots s_{i_p}\rangle}{\partial J^{(p)}_{i_1\cdots i_p}}}
     =\lim_{n\to0}\overline{\int\prod_\alpha^nd\nu(s_\alpha)\,\frac1n\sum_{ab}\left[
-      \hat\beta\left(\frac{s_a^Ts_b}N\right)^p+
-      p\left(-i\frac{\hat s_a^Ts_b}N\right)\left(\frac{s_a^Ts_b}N\right)^{p-1}
+      \hat\beta\left(\frac{s_a\cdot s_b}N\right)^p+
+      p\left(-i\frac{\hat s_a\cdot s_b}N\right)\left(\frac{s_a\cdot s_b}N\right)^{p-1}
     \right]} \\
     =\lim_{n\to0}\frac1n\sum_{ab}(\hat\beta C_{ab}^p+pR_{ab}C_{ab}^{p-1})
     =\hat\beta+pr_d-\int_0^1dx\,c^{p-1}(x)(\hat\beta c(x)+pr(x))
@@ -684,12 +684,23 @@ i.e., adding a linear field causes a response in the average saddle location pro
     \right]
   \end{aligned}
 \end{equation}
-In particular, when the energy is unconstrained ($\hat\beta=0$),
+In particular, when the energy is unconstrained ($\hat\beta=0$) and there is no replica symmetry breaking,
 \begin{equation}
-  \frac{\partial\Sigma}{\partial a_1}=-\frac14\lim_{n\to0}\frac1n\sum_{ab}D_{ab}=-\frac14d_d-\frac14\int_0^1dx\,d(x)
+  \frac{\partial\Sigma}{\partial a_1}=-\frac14\lim_{n\to0}\frac1n\sum_{ab}D_{ab}=-\frac14d_d
 \end{equation}
-i.e., adding a linear field decreases the complexity of solutions by an amount proportional to $d_d$.
+i.e., adding a random linear field decreases the complexity of solutions by an amount proportional to $d_d$ in the variance of the field.
 
+When the saddle point of the Kac--Rice problem is supersymmetric,
+\begin{equation}
+  \frac{\partial\Sigma}{\partial a_p}
+  =\frac{\hat\beta}4\overline{\frac1{N^p}\sum_{i_1\cdots i_p}\frac{\partial\langle s_{i_1}\cdots s_{i_p}\rangle}{\partial J^{(p)}_{i_1\cdots i_p}}}+\lim_{n\to0}\frac1n\sum_{ab}^np(p-1)R_{ab}^2C_{ab}^{p-2}
+\end{equation}
+and in particular for $p=1$
+\begin{equation}
+  \frac{\partial\Sigma}{a_1}
+  =\frac{\hat\beta}4\overline{\frac1N\sum_i\frac{\partial\langle s_i\rangle}{\partial J_i^{(1)}}}
+\end{equation}
+i.e., the change in complexity due to a linear field is directly related to the resulting magnetization of the stationary points.
 
 \section{Ultrametricity rediscovered}
 
-- 
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