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diff --git a/2-point.tex b/2-point.tex
index 8c2107d..4ce640a 100644
--- a/2-point.tex
+++ b/2-point.tex
@@ -101,7 +101,7 @@
structure not present in the pure models, we find nothing that distinguishes
the points that do attract the dynamics. Instead, we find new geometric
significance of the old threshold energy, and invalidate pictures of
- the arrangement most of marginal inherent states into a continuous `manifold.'
+ the arrangement most of marginal inherent states into a continuous manifold.
\end{abstract}
\tableofcontents
@@ -133,7 +133,7 @@ energy} $E_\mathrm{th}$, slices of the landscape at fixed energy undergo a
percolation transition. In fact, this threshold energy is significant in other
ways: it attracts the long-time dynamics after quenches in temperature to below
the dynamical transition from any starting temperature
-\cite{Biroli_1999_Dynamical}. All of this can be understood in terms of the
+\cite{Biroli_1999_Dynamical, Sellke_2023_The}. All of this can be understood in terms of the
landscape structure.
In slightly less simple models, the mixed spherical models, the story changes.
@@ -231,7 +231,7 @@ The gradient and Hessian at a stationary point are then
&&
\operatorname{Hess}H(\mathbf s,\omega)=\partial\partial H(\mathbf s)+\omega I
\end{align}
-where $\partial=\frac\partial{\partial\mathbf s}$ will always denote the derivative with respect to $\mathbf s$.
+where $\partial=\frac\partial{\partial\mathbf s}$ denotes the derivative with respect to $\mathbf s$.
\begin{figure}
\includegraphics{figs/spectrum_saddle.pdf}
@@ -296,7 +296,7 @@ points are found. We define the threshold energy $E_\mathrm{th}$ as the energy
at which most stationary points are marginal. Note that crucially this is
\emph{not} the energy that has the most marginal stationary points: this energy
level with the largest number of marginal points has even more saddles of
-extensive index! So $E_\mathrm{th}$ contains a \emph{minority} of the
+extensive index. So $E_\mathrm{th}$ contains a \emph{minority} of the
marginal points, even if those marginal points are the \emph{majority} of
stationary points with energy $E_\mathrm{th}$.
@@ -1013,7 +1013,7 @@ When $\mu=\mu_\mathrm m$, the linear term above vanishes. Under these conditions
\begin{equation}
\Sigma_{12}
=\frac12\frac{f'''(1)v_f}{f''(1)^{3/2}u_f}
- \left(\sqrt{2+\frac{2f''(1)(f''(1)-f'(1))}{f'''(1)f'(1)}}-1\right)\big(E_0-E_\textrm{th}\big)(1-q)^2+O\big((1-q)^3\big)
+ \left(\sqrt{\frac{2\big[f'(1)(f'''(1)-f''(1))+f''(1)^2\big]}{f'(1)f'''(1)}}-1\right)\big(E_0-E_\textrm{th}\big)(1-q)^2+O\big((1-q)^3\big)
\end{equation}
Note that this expression is only true for $\mu=\mu_\mathrm m$. Therefore,
among marginal minima, when $E_0$ is greater than the threshold one finds
@@ -1061,11 +1061,11 @@ by the coefficient in \eqref{eq:expansion.E.1} and
v_f
}{3u_f}
\big[
- f'(1)(2f''(1)-(2-(2-\delta q_0)\delta q_0)f'''(1))
+ f'(1)(2f''(1)-(2-(2-\delta q_0)\delta q_0)f'''(1))-2f''(1)^2
\big]
\\
- &\hspace{15pc}\times
- \big[f'(1)\big(6f''(1)+(18-(6-\delta q_0)\delta q_0)f'''(1)\big)
+ &\hspace{12pc}\times
+ \big[f'(1)\big(6f''(1)+(18-(6-\delta q_0)\delta q_0)f'''(1)\big)-6f''(1)^2
\big]
\bigg)^\frac12
\end{aligned}
@@ -1685,11 +1685,11 @@ limit of $\beta\to\infty$. There are two possibilities. First, in the replica
symmetric case $x=1$, and in the limit of large $\beta$ $q_0$ will scale like
$q_0=1-(y_0\beta)^{-1}$. Inserting this, the limit is
\begin{equation}
- V_\infty^{\textsc{rs}}=-\hat\beta_0 f(q)-r^{11}_\mathrm df'(q)q-\frac12\left(y_0(1-q^2)+\frac{f'(1)^2-f'(q)^2}{y_0f'(1)}\right)
+ V_\infty^{\textsc{rs}}=-\hat\beta_0 f(q)-r^{00}_\mathrm df'(q)q-\frac12\left(y_0(1-q^2)+\frac{f'(1)^2-f'(q)^2}{y_0f'(1)}\right)
\end{equation}
The saddle point in $y_0$ can now be taken, taking care to choose the solution for $y_0>0$. This gives
\begin{equation}
- V_\infty^{\textsc{rs}}(q\mid E_0,\mu_0)=-\hat\beta_0f(q)-r^{11}_\mathrm df'(q)q-\sqrt{(1-q^2)\left(1-\frac{f'(q)^2}{f'(1)^2}\right)}
+ V_\infty^{\textsc{rs}}(q\mid E_0,\mu_0)=-\hat\beta_0f(q)-r^{00}_\mathrm df'(q)q-\sqrt{(1-q^2)\left(1-\frac{f'(q)^2}{f'(1)^2}\right)}
\end{equation}
The second case is when the inner statistical mechanics problem has replica
symmetry breaking. Here, $q_0$ approaches a nontrivial limit, but
@@ -1697,7 +1697,7 @@ $x=z\beta^{-1}$ approaches zero and $q_1=1-(y_1\beta)^{-1}$ approaches one. The
\begin{equation}
\begin{aligned}
V_\infty^{\oldstylenums{1}\textsc{rsb}}(q\mid E_0,\mu_0)
- &=-\hat\beta_0f(q)-r^{11}_\mathrm df'(q)q-\frac12\bigg(
+ &=-\hat\beta_0f(q)-r^{00}_\mathrm df'(q)q-\frac12\bigg(
z(f(1)-f(q_0))+\frac{f'(1)}{y_1}-\frac{y_1(q^2-q_0)}{1+y_1z(1-q_0)} \\
&\hspace{8pc}-(1+y_1z(1-q_0))\frac{f'(q)^2}{y_1f'(1)}+\frac1z\log\left(1+zy_1(1-q_0)\right)
\bigg)
@@ -1735,6 +1735,17 @@ approached. To understand something better about why certain states attract the
dynamics in certain situations, nonlocal information, like the
structure of their entire basin of attraction, seems vital.
+It is possible that replica symmetry breaking among the constrained stationary
+points could change the details of the two-point complexity of very nearby
+states. Indeed, it is difficult to rule out \textsc{rsb} in complexity
+calculations. However, such corrections would not change the overarching
+conclusions of this paper, namely that most marginal minima are separated from
+each other by a macroscopic overlap gap and high barriers. This is because the
+replica symmetric complexity bounds any \textsc{rsb} complexities from above,
+and so \textsc{rsb} corrections can only decrease the complexity. Therefore,
+the overlap gaps, which correspond to regions of negative complexity, cannot be
+removed by a more detailed saddle point ansatz.
+
The methods developed in this paper are straightforwardly (if not easily)
generalized to landscapes with replica symmetry broken complexities
\cite{Kent-Dobias_2023_How}. We suspect that many of the qualitative features