From 1c306ecbae3231022f704d72593980a3788396a4 Mon Sep 17 00:00:00 2001
From: Jaron Kent-Dobias <jaron@kent-dobias.com>
Date: Tue, 4 Jun 2024 08:24:38 -0700
Subject: Lots of writing and changes.

---
 marginal.tex | 365 +++++++++++++++++++++++++++++++++++++++++++++--------------
 1 file changed, 281 insertions(+), 84 deletions(-)

(limited to 'marginal.tex')

diff --git a/marginal.tex b/marginal.tex
index 745b12d..309f433 100644
--- a/marginal.tex
+++ b/marginal.tex
@@ -93,21 +93,28 @@ more useful.
 
 An arbitrary function $g$ of the minimum eigenvalue of a matrix $A$ can be
 expressed as
-\begin{equation} \label{eq:λmax}
-  g(\lambda_\textrm{max}(A))
+\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^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'}}
     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}}{\int d\mathbf s'\,\delta(N-\mathbf s'^T\mathbf s')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{max}(A)I)}(\mathbf s)}{\int d\mathbf s'\,\delta(N-\mathbf s'^T\mathbf s')\mathbb 1_{\operatorname{ker}(A-\lambda_\mathrm{max}(A)I)}(\mathbf s')}g\left(\frac{\mathbf s^TA\mathbf s}N\right) \\
-    &=g(\lambda_\mathrm{max}(A))
-    \frac{\int d\mathbf s\,\delta(N-\mathbf s^T\mathbf s)\mathbb 1_{\operatorname{ker}(A-\lambda_\mathrm{max}(A)I)}(\mathbf s)}{\int d\mathbf s'\,\delta(N-\mathbf s'^T\mathbf s')\mathbb 1_{\operatorname{ker}(A-\lambda_\mathrm{max}(A)I)}(\mathbf s')} \\
-    &=g(\lambda_\mathrm{max}(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'}
+    }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)
+    }{
+      \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) \\
+    &=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')} \\
+    &=g(\lambda_\mathrm{min}(A))
   \end{aligned}
 \end{equation}
 The first relation extends a technique first introduced in
@@ -149,18 +156,18 @@ as the emergence of an imaginary part in the function.
 As an example, we compute
 \begin{equation} \label{eq:large.dev}
   e^{NG_\lambda^*(\mu)}
-  =P_{\lambda_\mathrm{max}(B-\mu I)=\lambda^*}
-  =\overline{\delta\big(N\lambda^*-N\lambda_\mathrm{max}(B-\mu I)\big)}
+  =P_{\lambda_\mathrm{min}(B+\mu I)=\lambda^*}
+  =\overline{\delta\big(N\lambda^*-N\lambda_\mathrm{min}(B+\mu I)\big)}
 \end{equation}
 where the overline is the average over $B$, and we have defined the large
 deviation function $G_\sigma(\mu)$.
-Using the representation of $\lambda_\mathrm{max}$ defined in \eqref{eq:λmax}, we have
+Using the representation of $\lambda_\mathrm{min}$ defined in \eqref{eq:λmin}, we have
 \begin{widetext}
 \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^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)
   }
 \end{equation}
 Using replicas to treat the denominator ($x^{-1}=\lim_{n\to0}x^{n-1}$)
@@ -169,17 +176,18 @@ 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\}}
+  \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\}}
 \end{equation}
-having introduced the parameter $\hat\lambda$ in the Fourier representation of the $\delta$-function.
-The whole expression, so transformed, is a simple exponential integral linear in the matrix $B$.
-Taking the average over $B$, we have
+having introduced the parameter $\hat\lambda$ in the Fourier representation of
+the $\delta$-function. The whole expression, so transformed, is a simple
+exponential integral linear in the matrix $B$. Taking the average over $B$, we
+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(\mu+\lambda^*)-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
+  &\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
     +\hat\lambda^2N^2
   \right]\right\}
   \end{aligned}
@@ -189,8 +197,8 @@ We make the Hubbard--Stratonovich transformation to the matrix field $Q_{ab}=\fr
   e^{NG_{\lambda^*}(\mu)}
   =\lim_{\beta\to\infty}\lim_{n\to0}\int d\hat\lambda\,dQ\,
   \exp N\left\{
-    \hat\lambda(\mu+\lambda^*)-n\beta\mu+\sigma^2\left[\beta^2\sum_{ab}^nQ_{ab}^2
-      -2\beta\hat\lambda\sum_a^nQ_{1a}^2
+    \hat\lambda(\lambda^*-\mu)-n\beta\mu+\sigma^2\left[\beta^2\sum_{ab}^nQ_{ab}^2
+      +2\beta\hat\lambda\sum_a^nQ_{1a}^2
     +\hat\lambda^2
   \right]+\frac12\log\det Q\right\}
 \end{equation}
@@ -199,7 +207,7 @@ where $Q_{aa}=1$ because of the spherical constraint. We can evaluate this
 integral using the saddle point method. We make a replica symmetric ansatz for
 $Q$, because this is a 2-spin model, but with the first row singled out because
 of its unique coupling with $\hat\lambda$. This gives
-\begin{equation}
+\begin{equation} \label{eq:Q.structure}
   Q=\begin{bmatrix}
     1&\tilde q_0&\tilde q_0&\cdots&\tilde q_0\\
     \tilde q_0&1&q_0&\cdots&q_0\\
@@ -221,8 +229,8 @@ with the effective action
 \begin{equation}
   \begin{aligned}
     &\mathcal S_\beta(q_0,\tilde q_0,\hat\lambda) \\
-    &\quad=\hat\lambda(\mu+\lambda^*)+\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
+    &\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] \\
     &\qquad-\log(1-q_0)+\frac12\log(1-2q_0+\tilde q_0^2)
   \end{aligned}
@@ -232,21 +240,21 @@ We expect the overlaps to concentrate on one as $\beta$ goes to infinity. We the
 \begin{align}
   q_0&=1-y\beta^{-1}-z\beta^{-2}+O(\beta^{-3})
   \\
-  \tilde q_0&=1-\tilde y\beta^{-1}-(z-\Delta z)\beta^{-2}+O(\beta^{-3})
+  \tilde q_0&=1-\tilde y\beta^{-1}-(z+\Delta z)\beta^{-2}+O(\beta^{-3})
 \end{align}
 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)
-    &=\hat\lambda(\mu+\lambda^*)
+    &=\hat\lambda(\lambda^*-\mu)
     +\sigma^2\big[
       \hat\lambda^2-4(y+\Delta z)
     \big] \\
     &\qquad+\frac12\log\left(1+\frac{2\Delta z}{y^2}\right)
   \end{aligned}
 \end{equation}
-Extremizing this action over the new parameters $y$, $\Delta z=z-\tilde z$, and $\hat\lambda$, we have
+Extremizing this action over the new parameters $y$, $\Delta z$, and $\hat\lambda$, we have
 \begin{align}
   \hat\lambda=-\frac1\sigma\sqrt{\frac{(\mu+\lambda^*)^2}{(2\sigma)^2}-1}
   \\
@@ -265,9 +273,8 @@ Inserting this solution into $\mathcal S_\infty$ we find
     \right)
   \end{aligned}
 \end{equation}
-This function is plotted in Fig.~\ref{fig:large.dev}. For $\mu<2\sigma$ $G_{\lambda^*}(\mu)$ has an
-imaginary part, which makes any additional integral over $\mu$ highly
-oscillatory. This indicates that the existence of a marginal minimum for this
+This function is plotted in Fig.~\ref{fig:large.dev} for $\lambda^*=0$. For $\mu<2\sigma$ $G_{0}(\mu)$ has an
+imaginary part. This indicates that the existence of a marginal minimum for this
 parameter value corresponds with a large deviation that grows faster than $N$,
 rather like $N^2$, since in this regime the bulk of the typical spectrum is
 over zero and therefore extensively many eigenvalues have to have large
@@ -318,7 +325,7 @@ breakdown of the large-deviation principle at order $N$.
 In isotropic or zero-signal landscapes, there is another way to condition on a
 pseudogap. In such landscapes, the typical spectrum does not have an isolated
 eigenvalue. Therefore, the condition associated with the bulk of the spectrum
-teaching zero, i.e., the pseudogap, will always correspond to the most common
+touching zero, i.e., the pseudogap, will always correspond to the most common
 configuration. We can therefore choose $\mu=\mu_\textrm m$ such that
 \begin{equation}
   0=\frac\partial{\partial\lambda^*}G_{\lambda^*}(\mu_\mathrm m)\bigg|_{\lambda^*=0}
@@ -355,10 +362,10 @@ The number of stationary points in a landscape for a particular realization $H$
   d\mu_H(\mathbf x,\pmb\omega)=d\mathbf x\,d\pmb\omega\,\delta\big(\nabla H(\mathbf x,\pmb\omega)\big)\,\delta\big(\mathbf g(\mathbf x)\big)\,\big|\det\operatorname{Hess}H(\mathbf x,\pmb\omega)\big|
 \end{equation}
 with a $\delta$-function of the gradient and the constraints ensuring that we
-count valid stationary points, and the Hessian entering in the determinant of
+count valid stationary points, and the Hessian entering in the determinant as
 the Jacobian of the argument to the $\delta$-function. It is usually more
 interesting to condition the count on interesting properties of the stationary
-points, like the energy,
+points, like the energy and spectrum trace,
 \begin{equation}
   \begin{aligned}
     &d\mu_H(\mathbf x,\pmb\omega\mid E,\mu) \\
@@ -367,17 +374,15 @@ points, like the energy,
     \,\delta\big(N\mu-\operatorname{Tr}\operatorname{Hess}H(\mathbf x,\pmb\omega)\big)
   \end{aligned}
 \end{equation}
-In this paper we in particular want to exploit our method to condition
-complexity on the marginality of stationary points. We therefore define the
-number of marginal points in a particular instantiation $H$ as
+We further want to control the value of the minimum eigenvalue of the Hessian at the stationary points. Using the method introduced above, we can write the number of stationary points with energy $E$, Hessian trace $\mu$, and smallest eigenvalue $\lambda^*$ as
 \begin{widetext}
 \begin{equation}
   \begin{aligned}
     &\mathcal N_H(E,\mu,\lambda^*)
-    =\int d\mu_H(\mathbf x,\pmb\omega\mid E,\mu)\,\delta\big(N\lambda^*-\lambda_\mathrm{max}(\operatorname{Hess}H(\mathbf x,\pmb\omega))\big) \\
+    =\int d\mu_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\mu_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^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'}}
     \delta\big(N\lambda^*-\mathbf s^T\operatorname{Hess}H(\mathbf x,\pmb\omega)\mathbf s\big)
   \end{aligned}
 \end{equation}
@@ -388,7 +393,7 @@ where the $\delta$-functions
 \end{equation}
 ensure that the integrals are constrained to the tangent space of the
 configuration manifold at the point $\mathbf x$. This likewise allows us to
-define the complexity of points with a specific energy, stability, and maximum eigenvalue as
+define the complexity of points with a specific energy, stability, and minimum eigenvalue as
 \begin{equation}
   \Sigma_{\lambda^*}(E,\mu)
   =\frac1N\overline{\log\mathcal N_H(E,\mu,\lambda^*)}
@@ -396,7 +401,7 @@ define the complexity of points with a specific energy, stability, and maximum e
 In practice, this can be computed by introducing replicas to treat the
 logarithm ($\log x=\lim_{n\to0}\frac\partial{\partial n}x^n$) and replicating
 again to treat each of the normalizations in the numerator. This leads to the expression
-\begin{equation} \label{eq:max.complexity.expanded}
+\begin{equation} \label{eq:min.complexity.expanded}
   \begin{aligned}
     \Sigma_{\lambda^*}(E,\mu)
     &=\lim_{\beta\to\infty}\lim_{n\to0}\frac1N\frac\partial{\partial n}\int\prod_{a=1}^n\Bigg[d\mu_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)\\
@@ -404,16 +409,18 @@ again to treat each of the normalizations in the numerator. This leads to the ex
       \left(\prod_{b=1}^{m_a} d\mathbf s_a^b
       \,\delta\big(N-(\mathbf s_a^b)^T\mathbf s_a^b\big)
       \,\delta\big((\mathbf s_a^b)^T\partial\mathbf g(\mathbf x_a)\big)
-      \,e^{\beta(\mathbf s_a^b)^T\operatorname{Hess}H(\mathbf x_a,\pmb\omega_a)\mathbf s_a^b}\right)
+      \,e^{-\beta(\mathbf s_a^b)^T\operatorname{Hess}H(\mathbf x_a,\pmb\omega_a)\mathbf s_a^b}\right)
     \Bigg]
   \end{aligned}
 \end{equation}
 \end{widetext}
-Finally, the \emph{marginal} complexity is given by fixing $\mu=\mu_\text{m}$ so that the complexity is stationary with respect to changes in the value of the maximum eigenvalue, or
+for the complexity of stationary points of a given energy, trace, and smallest eigenvalue.
+
+Finally, the \emph{marginal} complexity is given by fixing $\mu=\mu_\text{m}$ so that the complexity is stationary with respect to changes in the value of the minimum eigenvalue, or
 \begin{equation}
   0=\frac\partial{\partial\lambda^*}\Sigma_{\lambda^*}(E,\mu_\text{m}(E))\bigg|_{\lambda^*=0}
 \end{equation}
-Finally, the marginal complexity is defined by evaluating the complexity conditioned on $\lambda_{\text{min}}=0$ at $\mu_\text{m}$,
+Finally, the marginal complexity is defined by evaluating the complexity conditioned on $\lambda_{\text{min}}=0$ at $\mu=\mu_\text{m}(E)$,
 \begin{equation}
   \Sigma_\text{m}(E)
   =\Sigma_0(E,\mu_\text m(E))
@@ -467,13 +474,13 @@ signs around the determinant in the Kac--Rice measure. This can potentially
 lead to severe problems with the complexity. However, it is a justified step
 when the parameters of the problem, i.e., $E$, $\mu$, and $\lambda^*$ put us in
 a regime where the exponential majority of stationary points have the same
-index. This is true for maxima and minima, and for saddle points with a single
-outlier. Dropping the absolute value sign allows us to write
+index. This is true for maxima and minima, and for saddle points whose spectra have a strictly positive bulk with a fixed number of negative
+outliers. Dropping the absolute value sign allows us to write
 \begin{equation}
   \det\operatorname{Hess}H(\mathbf x_a, \pmb\omega_a)
   =\int d\pmb\eta_a\,d\bar{\pmb\eta}_a\,e^{\bar{\pmb\eta}_a^T\operatorname{Hess}H(\mathbf x_a,\pmb\omega)\pmb\eta_a}
 \end{equation}
-for $N$-dimensional Grassmann variables $\bar{\pmb\eta}$ and $\pmb\eta$. For
+for $N$-dimensional Grassmann variables $\bar{\pmb\eta}_a$ and $\pmb\eta_a$. For
 the spherical models this step is unnecessary, since there are other ways to
 treat the determinant keeping the absolute value signs, as in previous works
 \cite{Folena_2020_Rethinking, Kent-Dobias_2023_How}. However, since other of
@@ -481,7 +488,7 @@ our examples are for models where the same techniques are impossible, it is
 useful to see the fermionic method in action in this simple case.
 
 Once these substitutions have been made, the entire expression
-\eqref{eq:max.complexity.expanded} is an exponential integral whose argument is
+\eqref{eq:min.complexity.expanded} is an exponential integral whose argument is
 a linear functional of $H$. This allows for the average to be taken over the
 disorder. If we gather all the $H$-dependant pieces into the linear functional
 $\mathcal O$ then the average gives
@@ -525,27 +532,14 @@ the cited papers \cite{Folena_2020_Rethinking, Kent-Dobias_2023_How}, we arrive
     d\hat\beta\,d\hat\lambda\,e^{N
     n\mathcal S_\mathrm{KR}(\hat\beta,\omega,C,R,D,F)
     +N\mathcal S_\beta(\omega,\hat\lambda,A,X,\hat X)
+    +\frac12N\log\det J
   }
   \end{aligned}
 \end{equation}
-The structure of the integrand, with the effective action split between two
-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}
-  \begin{aligned}
-    &\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]\\
-    &+\hat\beta^2f(C_{ab})+(2\hat\beta(R_{ab}-F_{ab})-D_{ab})f'(C_{ab})+(R_{ab}^2-F_{ab}^2)f''(C_{ab})
-    +\log\det\begin{bmatrix}C&iR\\iR&D\end{bmatrix}-\log\det F
-  \end{aligned}
-\end{equation}
-\end{widetext}
-
-$X^a$ is $n\times m_a$, and $A^{ab}$ is $m_a\times m_b$.
-\begin{equation}
-  \begin{bmatrix}
+where the matrix $J$ is the Jacobian associated with the change of variables
+from the $\mathbf x$, $\hat{\mathbf x}$, and $\mathbf s$, and has 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}\\
@@ -553,14 +547,42 @@ $X^a$ is $n\times m_a$, and $A^{ab}$ is $m_a\times m_b$.
     (X^n)^T&i(\hat X^n)^T&A^{n1}&\cdots&A^{nn}
   \end{bmatrix}
 \end{equation}
-$X_{ab}^c$ will be nonzero if the lowest eigenvector of the hessian at the
-point $\mathbf s_c$ are correlated with the direction of the point $\mathbf
-s_a$. Since the eigenvector problem is always expected to be replica symmetric,
-we expect no $b$-dependence of this matrix. $A^{aa}$ is the usual
-replica-symmetric overlap matrix of the spherical two-spin problem. $A^{ab}$
-describes overlaps between eigenvectors at different stationary points and should be a constant $m_a\times m_b$ matrix.
-
-We will discuss at the end of this paper when these order parameters can be expected to be nonzero, but in this and most isotropic problems all of the $X$s, $\hat X$s, and $A^{ab}$ for $a\neq b$ are zero.
+The structure of the integrand, with the effective action split between two
+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})
+    \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}
+\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
+$J$) the dependence on the parameters $X$ and $\hat X$ are purely quadratic.
+Therefore, there will always be a saddle point condition where they are both
+zero. In this case, we except this solution to be correct. We can reason about
+why this is so: $X$, for instance, quantifies the correlation between the
+typical position of stationary points and the direction of their typical
+eigenvectors. In an isotropic landscape, where no direction is any more
+important than any other, we don't expect such correlations to be nonzero:
+where a state is location does not give any information as to the orientation
+of its soft directions. On the other hand, in the spiked case, or with an
+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.
 
 \begin{equation}
   \Sigma_{\lambda^*}(E,\mu)
@@ -723,7 +745,7 @@ The resulting integrand is Gaussian in the $w$, $\hat w$, $\mathbf y$, and $\hat
       \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})+\frac1{N^2}f''(C_{ab})\mathbf x_a\mathbf x_b^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\}
@@ -733,12 +755,12 @@ which produces
   \exp\left\{
     \frac M2\log\det\left(
       I+\begin{bmatrix}
-      \hat\beta_a\delta_{ab}+G_{ab}^2f''(C_{ab}) & -i\hat{\mathbf x}_a^T\delta_{ab} \\
-      -i\hat{\mathbf x}_a\delta_{ab} & 2(\hat\mu_a I-\bar\eta_a\eta_a^T)\delta_{ab}
+      \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_{ab})&\frac1Nf'(C_{ab})\mathbf x_a^T \\
-      \frac1Nf'(C_{ab})\mathbf x_b & \frac1Nf'(C_{ab})+\frac1{N^2}f''(C_{ab})\mathbf x_a\mathbf x_b^T
+      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\}
@@ -747,11 +769,90 @@ which produces
   \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-\frac1Nf'(C_{ab})\hat{\mathbf x}_a^T
+    \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}
@@ -790,11 +891,107 @@ In the case where $\mu$ is not specified, in which the model is supersymmetric,
   +\frac12\log\frac{\hat\beta+r_d}{r_d}
 \end{equation}
 
-The condition fixing the maximum eigenvalue adds to the integrand
+\cite{DeWitt_1992_Supermanifolds}
+Consider supervectors in the $\mathbb R^{N|4}$ superspace of the form
+\begin{equation}
+  \pmb\phi_{a\alpha}(1,2)
+  =\mathbf x_a
+  +\bar\theta_1\pmb\eta_a+\bar{\pmb\eta}_a\theta_1
+  +i\hat{\mathbf x}_a\bar\theta_1\theta_1
+  +\mathbf s_{a\alpha}(\bar\theta_1\theta_2+\bar\theta_2\theta_1)
+\end{equation}
+The Kac--Rice measure with the eigenvalue-fixing term included is
+\begin{equation}
+  \begin{aligned}
+    \mathcal N(E,\mu,\lambda^*)^n
+    &=\int\prod_{a=1}^n\prod_{\alpha=1}^{m_a}d\pmb\phi_{a\alpha}
+    \exp\left\{
+      \delta_{\alpha1}N(\hat\beta_aE+\hat\lambda_a\lambda^*)
+      +\int d1\,d2\,B_{a\alpha}(1,2)\left[H(\pmb\phi_{a\alpha})+\frac12\mu(\|\pmb\phi_{a\alpha}\|^2-N)\right]
+    \right\}
+  \end{aligned}
+\end{equation}
+\begin{equation}
+  B_{a\alpha}(1,2)=\delta_{\alpha1}\bar\theta_2\theta_2
+        (1-\hat\beta_a\bar\theta_1\theta_1)
+        -\delta_{\alpha1}\hat\lambda_a-\beta
+\end{equation}
+\begin{align}
+  d\pmb\phi_{a\alpha}
+  =d\mathbf x_a\,\delta(\|\mathbf x_a\|^2-N)\,\frac{d\hat{\mathbf x}_a}{(2\pi)^N}\,d\pmb\eta_a\,d\bar{\pmb\eta}_a\,
+  d\mathbf s_{a\alpha}\,\delta(\|\mathbf s_{a\alpha}\|^2-N)\,
+  \delta(\mathbf x_a^T\mathbf s_{a\alpha})
+\end{align}
+
+\begin{equation}
+  i\int d1\,d2\,\hat v_{a\alpha}^k(1,2)(V^k(\pmb\phi_{a\alpha}(1,2))-v_{a\alpha}^k(1,2))
+\end{equation}
+\begin{equation}
+  -\sum_{ab}\sum_{\alpha\gamma}\sum_k\frac12\int d1\,d2\,d3\,d4\,
+  \hat v_{a\alpha}^kf\big(\pmb\phi_{a\alpha}(1,2)^T\pmb\phi_{b\gamma}(3,4)\big)\hat v_{b\gamma}^k
+\end{equation}
+We're now quadratic in the $v$ and $\hat v$ with the kernel
+\begin{equation}
+  \begin{bmatrix}
+    B_{a\alpha}(1,2)\delta(1,3)\delta(2,4)\delta_{ab}\delta_{\alpha\gamma} & i\delta(1,3)\,\delta(2,4) \delta_{ab}\delta_{\alpha\gamma}\\
+    i\delta(1,3)\,\delta(2,4) \delta_{ab}\delta_{\alpha\gamma}& f\big(\pmb\phi_{a\alpha}(1,2)^T\pmb\phi_{b\gamma}(3,4)\big)
+  \end{bmatrix}
+\end{equation}
+Upon integration, this results in a term in the effective action of the form
+\begin{equation}
+  -\frac M2\log\operatorname{sdet}\left(
+    \delta(1,3)\,\delta(2,4) \delta_{ab}\delta_{\alpha\gamma}
+    +B_{a\alpha}(1,2)f\big(\pmb\phi_{a\alpha}(1,2)^T\pmb\phi_{b\gamma}(3,4)\big)
+  \right)
+\end{equation}
+When expanded, this supermatrix is constructed of the scalar products of the
+real and Grassmann vectors that make up $\pmb\phi$. The change of variables to
+these order parameters again results in the Jacobian of \eqref{eq:coordinate.jacobian}, contributing
+\begin{equation}
+  \frac N2\log\det J(C,R,D,G,Q,X,\hat X)
+\end{equation}
+Up to this point, the expressions above are general and independent of a given
+ansatz. However, we expect that the order parameters $X$ and $\hat X$ are zero,
+since this case is isotropic. Applying this ansatz here avoids a dramatically
+more complicated expression for the effective action found in the case with
+arbitrary $X$ and $\hat X$. We also will apply the ansatz that $Q_{a\alpha
+b\gamma}$ is zero for $a\neq b$, which is equivalent to assuming that the soft
+directions of typical pairs of stationary points are uncorrelated, and further
+that $Q_{\alpha\gamma}=Q_{a\alpha a\gamma}$ independently of the index $a$,
+implying that correlations in the tangent space of typical stationary points
+are the same.
+
+Given these simplifying forms of the ansatz, taking the superdeterminant yields
 \begin{equation}
-  \frac12\beta\sum_b^{m_a}\mathbf s^T_b(\mathbf v^k_a(\mathbf v^k_a)^T+w^k_a\partial\partial V^k(\mathbf x_a)+\omega I)\mathbf s_b
-  +\frac12\hat\lambda\mathbf s_1^T(\mathbf v^k_a(\mathbf v^k_a)^T+w^k_a\partial\partial V^k(\mathbf x_a)+\omega I)\mathbf s_1
+  \begin{aligned}
+    \log\det\left\{
+      \left[
+        f'(C)\odot D-\hat\beta I+\left(R^{\circ2}-G^{\circ2}+I\sum_{\alpha\gamma}2(\delta_{\alpha1}\hat\lambda+\beta)(\delta_{\gamma1}\hat\lambda+\beta)Q_{\alpha\gamma}^2\right)\odot f''(C)
+      \right]f(C)
+      +(I-R\odot f'(C))^2
+    \right\} \\
+    +n\log\det_{\alpha\gamma}(\delta_{\alpha\gamma}-2(\delta_{\alpha1}\hat\lambda+\beta)Q_{\alpha\gamma})
+    -2\log\det(I+G\odot f'(C))
+  \end{aligned}
 \end{equation}
+where once again $\odot$ is the Hadamard product and $A^{\circ n}$ gives the
+Hadamard power of $A$. We can already see one substantive difference between
+the structure of this problem and that of the spherical models: the effective
+action in this case mixes the order parameters $G$ due to the fermions with the
+ones $C$, $R$, and $D$ due to the other variables. This is the realization of
+the fact that the Hessian properties are no longer independent of the energy
+and gradient.
+
+Now we have reduced the problem to an extremal one over the order parameters
+$\hat\beta$, $\hat\lambda$, $C$, $R$, $D$, $G$, and $Q$, it is time to make an
+ansatz for the form of order we expect to find. We will focus on a regime where
+the structure of stationary points is replica symmetric, and further where
+typical pairs of stationary points have no overlap. This gives
+\begin{align}
+  C=I && R=r_dI && D = d_dI && G = g_dI
+\end{align}
+We further take a planted replica symmetric structure for the matrix $Q$,
+identical to that in \eqref{eq:Q.structure}.
 \end{widetext}
 
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