Lots of small tweaks.

1 files changed, 98 insertions, 89 deletions

diff --git a/frsb_kac-rice.tex b/frsb_kac-rice.tex
index 6994a48..3cc1ba6 100644
--- a/frsb_kac-rice.tex
+++ b/frsb_kac-rice.tex
@@ -115,8 +115,9 @@ foothold for numerically computing the complexity everywhere else.
 \S\ref{sec:frsb} explains aspects of the solution specific to the case of full
 RSB, and derives the RS--FRSB transition line. \S\ref{sec:examples} details the
 landscape topology of two example models: a $3+16$ model with a 2RSB ground
-state, and a $2+4$ with a FRSB ground state. Finally \S\ref{sec:interpretation}
-provides some interpretation of our results.
+state and a 1RSB complexity, and a $2+4$ with a FRSB ground state and a FRSB
+complexity. Finally \S\ref{sec:interpretation} provides some interpretation of
+our results.
 
 \section{The model}
 \label{sec:model}
@@ -134,7 +135,7 @@ constants that define the particular model. For instance, the so-called `pure'
 models have $a_p=1$ for some $p$ and all others zero.
 
 The variance of the couplings implies that the covariance of the energy with
-itself depends only on the dot product (or overlap) between two configurations.
+itself depends on only the dot product (or overlap) between two configurations.
 In particular, one finds
 \begin{equation} \label{eq:covariance}
   \overline{H(\mathbf s_1)H(\mathbf s_2)}=Nf\left(\frac{\mathbf s_1\cdot\mathbf s_2}N\right)
@@ -148,7 +149,7 @@ One needn't start with a Hamiltonian like
 \eqref{eq:covariance} can be specified for arbitrary, non-polynomial $f$, as in
 the `toy model' of M\'ezard and Parisi \cite{Mezard_1992_Manifolds}.
 
-The family of mixed $p$-spin models may be thought of as the most general
+The family of mixed $p$-spin models may be considered as the most general
 models of generic Gaussian functions on the sphere.  To constrain the model to
 the sphere, we use a Lagrange multiplier $\mu$, with the total energy being
 \begin{equation}
@@ -165,9 +166,10 @@ where $\partial=\frac\partial{\partial\mathbf s}$ always. An important
 observation was made by Bray and Dean  \cite{Bray_2007_Statistics} that
 gradient and Hessian are independent for Gaussian random functions.  The
 average over disorder breaks into a product of two independent averages, one
-for any function of the gradient and one for any function of the Hessian, and
-in particular the number of negative eigenvalues, the index $\mathcal I$ of the
-saddle (see Fyodorov \cite{Fyodorov_2007_Replica} for a detailed discussion).
+for any function of the gradient and one for any function of the Hessian.  In
+particular, the number of negative eigenvalues at a stationary point, which
+sets the index $\mathcal I$ of the saddle, is a function of the Hessian alone
+(see Fyodorov \cite{Fyodorov_2007_Replica} for a detailed discussion).
 
 \section{Equilibrium}
 \label{sec:equilibrium}
@@ -242,7 +244,7 @@ correspond to the ground state energy $E_0$.  The zero temperature limit is
 most easily obtained by putting $x_i=\tilde x_ix_k$ and
 $x_k=\tilde\beta/\beta$, $q_k=1-z/\beta$, which ensures the $\tilde x_i$,
 $\tilde\beta$, and $z$ have nontrivial limits. Inserting the ansatz and taking
-the limit carefully treating the $k$th term in each sum separately from the
+the limit, carefully treating the $k$th term in each sum separately from the
 rest, one can show after some algebra that
 \begin{equation} \label{eq:ground.state.free.energy}
   \tilde\beta\langle E\rangle_0=\tilde\beta\lim_{\beta\to\infty}\frac{\partial(\beta F)}{\partial\beta}
@@ -251,7 +253,7 @@ rest, one can show after some algebra that
   \right)
 \end{equation}
 where $\tilde Q$ is a $(k-1)$RSB matrix with entries $\tilde q_1=\lim_{\beta\to\infty}q_1$, \dots, $\tilde q_{k-1}=\lim_{\beta\to\infty}q_{k-1}$ parameterized by $\tilde x_1,\ldots,\tilde x_{k-1}$.
-This is a $k-1$ RSB ansatz whose spectrum in the determinant is scaled by
+This is a ($k-1$)RSB ansatz whose spectrum in the determinant is scaled by
 $\tilde\beta z^{-1}$ and shifted by 1, with effective temperature
 $\tilde\beta$, and an extra term. In the continuum case, this is
 \begin{equation} \label{eq:ground.state.free.energy.cont}
@@ -260,15 +262,15 @@ $\tilde\beta$, and an extra term. In the continuum case, this is
     \tilde\beta^2f''(q)\tilde\chi(q)+\frac1{\tilde\chi(q)+\tilde\beta z^{-1}}
   \right)
 \end{equation}
-where $\tilde\chi$ is bound by the same constraints as before.
+where $\tilde\chi$ is bound by the same constraints as $\chi$.
 
 The zero temperature limit of the free energy loses one level of replica
-symmetry breaking. Physically, this is a result of the fact that in $k$-RSB,
-$q_k$ gives the overlap within a state, e.g., within the basin of a well inside
+symmetry breaking. Physically, this is a result of the fact that in $k$RSB,
+$q_k$ gives the overlap within a state, i.e., within the basin of a well inside
 the energy landscape. At zero temperature, the measure is completely localized
-on the bottom of the well, and therefore the overlap with each state becomes
+on the bottom of the well, and therefore the overlap within each state becomes
 one. We will see that the complexity of low-energy stationary points in
-Kac--Rice computation is also given by a $(k-1)$-RSB anstaz. Heuristically, this is because
+Kac--Rice computation is also given by a $(k-1)$RSB anstaz. Heuristically, this is because
 each stationary point also has no width and therefore overlap one with itself.
 
 \section{Landscape complexity}
@@ -278,12 +280,12 @@ The stationary points of a function can be counted using the Kac--Rice formula,
 which integrates 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}. It gives the number of stationary points $\mathcal N$ as
-\begin{equation}
+\begin{equation} \label{eq:kac-rice}
   \mathcal N
     =\int d\mathbf s\, d\mu\,\delta\big(\tfrac12(\|\mathbf s\|^2-N)\big)\,\delta\big(\nabla H(\mathbf s,\mu)\big)\,\big|\det\operatorname{Hess}H(\mathbf s,\mu)\big|
 \end{equation}
 It is more interesting to count stationary points which share certain
-properties, like energy density $E$ or index $\mathcal I$. These properties can
+properties, like energy density $E$ or index density $\mathcal I$. These properties can
 be fixed by inserting additional $\delta$-functions into the integral. Rather
 than fix the index directly, we fix the trace of the Hessian, which we'll soon
 show is equivalent to fixing the value $\mu$, and fixing $\mu$ fixes the index
@@ -317,7 +319,7 @@ model we study in detail later), the annealed complexity predicts that minima
 vanish well before the dominant saddles, a contradiction for any bounded
 function.
 
-A sometimes more illuminating quantity to consider is the Legendre transform $G$ of the complexity:
+A sometimes more illuminating quantity is the Legendre transform $G$ of the complexity, defined by
 \begin{equation}
     e^{NG(\hat \beta, \mu^*)} = \int dE \; e^{ -\hat \beta E +\Sigma(\hat \beta, \mu^*)}
 \end{equation}
@@ -345,7 +347,7 @@ product, which gives
   \end{aligned}
 \end{equation}
 As discussed in \S\ref{sec:model}, it has been shown that to the largest order
-in $N$, the Hessian of Gaussian random functions in independent from their
+in $N$, the Hessian of Gaussian random functions is independent from their
 gradient, once both are conditioned on certain properties. Here, they are only
 related by their shared value of $\mu$. Because of this statistical
 independence, we may write
@@ -400,7 +402,7 @@ factorizes into the product of averages, each of which is given by the same
 expression depending only on $\mu$ \cite{Ros_2019_Complex}. We therefore find
 \begin{equation}
   \overline{\prod_a^n |\det\operatorname{Hess}(\mathbf s_a,\mu_a)|\,\delta\big(N\mu^*-\operatorname{Tr}\operatorname{Hess}H(\mathbf s_a,\mu_a)\big)}
- \rightarrow \prod_a^ne^{N{\cal D}(\mu_a)}\delta(\mu_a-\mu^*)
+  \rightarrow \prod_a^ne^{N{\cal D}(\mu_a)}\delta\big(N(\mu^*-\mu_a)\big)
 \end{equation}
 where the function $\mathcal D$ is defined by
 \begin{equation}
@@ -432,7 +434,7 @@ the value of the stability $\mu$ in all replicas to the value $\mu^*$.
 
 The $\delta$-functions in the remaining factor are treated by writing them in
 the Fourier basis.  Introducing auxiliary fields $\hat{\mathbf s}_a$ and
-$\hat\beta$, for each replica replica one writes
+$\hat\beta$ for this purpose, for each replica replica one writes
 \begin{equation}
   \begin{aligned}
     &\delta\big(\tfrac12(\|\mathbf s_a\|^2-N)\big)\,\delta\big(\nabla H(\mathbf s_a,\mu^*)\big)\delta(NE-H(\mathbf s_a)) \\
@@ -440,7 +442,7 @@ $\hat\beta$, for each replica replica one writes
       e^{\frac12\hat\mu(\|\mathbf s_a\|^2-N)+\hat\beta(NE-H(\mathbf s_a))+i\hat{\mathbf s}_a\cdot(\partial H(\mathbf s_a)+\mu^*\mathbf s_a)}
   \end{aligned}
 \end{equation}
-Anticipating the Parisi-style solution, we don't label $\hat\mu$ or $\hat\beta$
+Anticipating a Parisi-style solution, we don't label $\hat\mu$ or $\hat\beta$
 with replica indices, since replica vectors won't be broken in the scheme.  The
 average over disorder can now be taken for the pieces which depend explicitly
 on the Hamiltonian, and since everything is Gaussian this gives
@@ -558,17 +560,18 @@ matrix products and Hadamard products. In particular, the determinant of the blo
 \begin{equation}
   \ln\det\begin{bmatrix}C&iR\\iR&D\end{bmatrix}=\ln\det(CD+R^2)
 \end{equation}
-The is straightforward to write down at $k$RSB, since the product and sum of
+This is straightforward to write down at $k$RSB, since the product and sum of
 the hierarchical matrices is still a hierarchical matrix. Recall the
-standard formulas for the product $C=AB$,
+standard formulas for the product $C=AB$ of two hierarchical matrices:
 \begin{align} \label{eq:replica.prod}
   c_d&=a_db_d-\sum_{j=0}^k(x_{j+1}-x_j)a_jb_j \\
   c_i&=b_da_i+a_db_i-\sum_{j=0}^{i-1}(x_{j+1}-x_j)a_jb_j+(2x_{i+1}-x_i)a_ib_i
   -\sum_{j=i+1}^k(x_{j+1}-x_j)(a_ib_j+a_jb_i)
 \end{align}
-and for the sum $C=A+B$, $c_d=a_d+b_d$ and $c_i=a_i+b_i$. Using these, one can
-write down the hierarchical matrix $CD+R^2$, and then compute the $\ln\det$
-using the previous formula \eqref{eq:replica.logdet}.
+and for the sum $C=A+B$ of two hierarchical matrices: $c_d=a_d+b_d$ and
+$c_i=a_i+b_i$. Using these, one can write down the hierarchical matrix
+$CD+R^2$, and then compute the $\ln\det$ using the previous formula
+\eqref{eq:replica.logdet}.
 
 The extremal conditions are given by differentiating the complexity with respect to its parameters, yielding
 \begin{align}
@@ -596,9 +599,9 @@ To these conditions must be added the addition condition that $\Sigma$ is extrem
 \begin{equation} \label{eq:cond.x}
   0=\frac{\partial\Sigma}{\partial x_i}\qquad 1\leq i\leq k
 \end{equation}
-These stationary conditions are the most numerically taxing, because when the
+The stationary conditions for the $x$s are the most numerically taxing, because when the
 formulas above are chained together and substituted into the complexity it
-results in a complicated formula.
+results in a complicated expression.
 
 In addition to these equations, we often want to maximize the complexity as a
 function of $\mu^*$, to find the most common type of stationary points. These
@@ -638,7 +641,7 @@ absolute value of the determinant is neglected and the trace of the Hessian is
 not fixed. This supersymmetry has been studied in great detail in the
 complexity of the Thouless--Anderson--Palmer (TAP) free energy
 \cite{Annibale_2003_The, Annibale_2003_Supersymmetric,
-Annibale_2004_Coexistence}. When the absolute value is dropped, the determinant can be
+Annibale_2004_Coexistence}. When the absolute value is dropped, the determinant in \eqref{eq:kac-rice} can be
 represented by an integral over Grassmann variables, which yields a complexity
 depending on `bosons' and `fermions' that share the supersymmetry. The Ward
 identities associated with the supersymmetry imply that $D=\hat\beta R$
@@ -676,7 +679,7 @@ which is precisely the condition \eqref{eq:mu.minima} for dominant minima. There
 supersymmetric solution counts the most common minima}
 \cite{Annibale_2004_Coexistence}. When minima are not the most common type of
 stationary point, the supersymmetric solution correctly counts minima that
-satisfy \eqref{eq:mu.minima}, but these do not have any special significance.
+satisfy \eqref{eq:mu.minima}, but these do not have any other special significance.
 
 Inserting the supersymmetric ansatz $D=\hat\beta R$ and $R=r_dI$, one gets for the complexity
 \begin{equation} \label{eq:diagonal.action}
@@ -702,7 +705,7 @@ $\Sigma(E_0,\mu^*)=0$, gives
   \right)
 \end{equation}
 which is precisely the ground state energy predicted by the equilibrium
-solution\eqref{eq:ground.state.free.energy} with $r_d=z$,
+solution \eqref{eq:ground.state.free.energy} with $r_d=z$,
 $\hat\beta=\tilde\beta$, and $C=\tilde Q$.
 
 {\em Therefore a $(k-1)$RSB ansatz in
@@ -724,16 +727,16 @@ complexity in the ground state are
   &&
   d_d=\hat\beta r_d
 \end{align}
-Unlike the case for the TAP complexity, this correspondence between complexity
+Unlike the case for the TAP complexity, this correspondence between landscape complexity
 and equilibrium solutions only exists at the ground state. We will see in our
 examples in \S\ref{sec:examples} that there appears to be little correspondence
 between these parameters away from the ground state.
 
 The supersymmetric solution produces the correct complexity for the ground
-state and for a class of minima. Moreover, it produces the correct parameters
+state and for a class of minima, including dominant ones. Moreover, it produces the correct parameters
 for the fields $C$, $R$, and $D$ at those points. This is an important foothold
 in the problem of computing the general complexity. The full saddle point
-equations at $k$-RSB are not very numerically stable, and a `good' saddle point
+equations at $k$RSB are not very numerically stable, and a `good' saddle point
 has a typically small radius of convergence under methods like Newton's
 algorithm. With the supersymmetric solution in hand, it is possible to take
 small steps in the parameter space to find non-supersymmetric numeric
@@ -753,9 +756,9 @@ each of its rows, with
   &&
   R\;\leftrightarrow\;[r_d, r(x)]
   &&
-  D\;\leftrightarrow \;d_d, d(x)]
+  D\;\leftrightarrow\;[d_d, d(x)]
 \end{align}
-the complexity becomes
+The complexity becomes
 \begin{equation}
   \begin{aligned}
     \Sigma(E,\mu^*)
@@ -772,7 +775,7 @@ Define for any function of $x$
 \begin{equation}
   \langle a\rangle=\int_0^1dx\,a(x)
 \end{equation}
-In general, the product of hierarchical matrices becomes
+In general, the product of hierarchical matrices gives
 \begin{align}
   (a\ast b)_d&=a_db_d-\langle ab\rangle \\
   (a\ast b)(x)&=(b_d-\langle b\rangle)a(x)+(a_d-\langle a\rangle)b(x)
@@ -787,9 +790,9 @@ and the $\ln\det$ becomes, for the hierarchical matrix $A=CD+R^2$,
   \lim_{n\to0}\frac1n\ln\det A
   =\ln(a_d-\langle a\rangle)
   +\frac{a(0)}{a_d-\langle a\rangle}
-  -\int_0^1\frac{dx}{x^2}\ln\left(\frac{a_d-\langle a\rangle+\int_0^xdy\,a(y)-xa(x)}{a_d-\langle a\rangle}\right)
+  -\int_0^1\frac{dx}{x^2}\ln\left(\frac{a_d-\langle a\rangle-xa(x)+\int_0^xdy\,a(y)}{a_d-\langle a\rangle}\right)
 \end{equation}
-The saddle point equations then take the form
+The saddle point equations take the form
 \begin{align}
   0&=\hat\mu c(x)+[(\hat\beta^2f'(c)+(2\hat\beta r-d)f''(c)+r^2f'''(c))\ast c](x)+(f'(c)\ast d)(x) \label{eq:extremum.c} \\
   0&=-\mu^* c(x)+[(\hat\beta f'(c)+rf''(c))\ast c](x)+(f'(c)\ast r)(x) \label{eq:extremum.r} \\
@@ -822,7 +825,7 @@ over all functions $\chi$. This gives
 \begin{equation}
   \chi_0(q\mid\hat\beta,r_d)=\frac1{\hat\beta}\left(f''(q)^{-1/2}-r_d\right)
 \end{equation}
-Satisfying the boundary condition $\chi(1)=0$ requires $r_d=f''(1)^{-1/2}$ and $\hat\beta=\frac12f'''(1)/f''(1)^{3/2}$.
+Satisfying the boundary conditions requires $r_d=f''(1)^{-1/2}$ and $\hat\beta=\frac12f'''(1)/f''(1)^{3/2}$.
 This in turn implies $\mu^*=\frac1{r_d}+f''(1)r_d=\sqrt{4f''(1)}=\mu_m$.
 Therefore, the FRSB ground state is always marginal, as excepted. It is
 straightforward to check that these conditions are indeed a saddle of the
@@ -856,12 +859,12 @@ related to $c(x)$ by $-\chi'(c(x))=x$, there is a corresponding $x_\textrm{max}$
 \subsection{Expansion near the transition}
 \label{subsec:expansion}
 
-Working with the general equations in their continuum form away from the
+Working with the continuum equations away from the
 supersymmetric solution is not generally tractable. However, there is another
 point where they can be treated analytically: near the onset of replica
 symmetry breaking. Here, the off-diagonal components of $C$, $R$, and $D$ are
 expected to be small. In particular, we expect the functions $c(x)$, $r(x)$, and $d(x)$
-to approach zero at the transition, and moreover take the form
+to approach zero at the transition, and moreover take the piecewise linear form
 \begin{align}
   c(x)=\begin{cases}\bar cx&x\leq x_\mathrm{max}\\\bar cx_\mathrm{max}&\text{otherwise}\end{cases}
                            &&
@@ -887,7 +890,7 @@ $\bar r$ can be found in terms of $\bar c$, yielding
     -\hat\beta-\frac1{f'(1)+f''(0)}\left(r_d(f''(0)+f''(1))-\mu^*\right)+O(x_\mathrm{max})
   \end{aligned}
 \end{equation}
-Likewise, \eqref{eq:extremum.d} depends linearly on $\bar d$ to all orders, and can be solved using this form for $\bar r$ to give
+Likewise, \eqref{eq:extremum.d} depends linearly on $\bar d$ to all orders, and can be solved to give
 \begin{equation}
   \begin{aligned}
     \frac{\bar d}{\bar c}
@@ -910,12 +913,11 @@ treat the determinant. To quadratic order, this gives
   \right]x_\textrm{max}^2
   \end{aligned}
 \end{equation}
-The spectrum of the Hessian of $\Sigma$ evaluated at the replica symmetric
-solution with $x_\text{max}=0$ gives the stability of the replica symmetric
-solution with respect to perturbations of the type described above. When the
-values of $\bar r$ and $\bar d$ above are substituted in and everything is
-evaluated at the replica symmetric solution, the  eigenvalue of interest takes
-the form
+The spectrum of the Hessian of $\Sigma$ with evaluated at the RS
+solution gives its stability with respect to these functional perturbations. When the
+values of $\bar r$ and $\bar d$ above are substituted into the Hessian and
+$\hat\beta$, $r_d$, and $d_d$ are evaluated at their RS values, the eigenvalue
+of interest takes the form
 \begin{equation}
   \lambda
   =-\bar c^2\frac{(f'(1)-2f(1))^2(f'(1)-f''(0))f''(0)}{2(f'(1)+f''(0))(f'(1)^2-f(1)(f'(1)+f''(1)))^2}(\mu^*-\mu^*_+(E))(\mu^*-\mu^*_-(E))
@@ -926,9 +928,9 @@ where
   =\pm\frac{(f'(1)+f''(0))(f'(1)^2-f(1)(f'(1)+f''(1)))}{(2f(1)-f'(1))f'(1)f''(0)^{-1/2}}
   -\frac{f''(1)-f'(1)}{f'(1)-2f(1)}E
 \end{equation}
-This eigenvalues changes sign when $\mu^*$ crosses $\mu^*_\pm(E)$.  We expect
-that this is the line of stability for the replica symmetric solution. The
-numerics in \S\ref{sec:examples} bear this out.
+This eigenvalue changes sign when $\mu^*$ crosses $\mu^*_\pm(E)$.  We expect
+that this is the line of stability for the replica symmetric solution when the
+transition is RS-FRSB. The numerics in \S\ref{sec:examples} bear this out.
 
 
 \section{General solution: examples}
@@ -943,7 +945,7 @@ to its parameters, since the equilibrium solution is equivalent to a
 variational problem. Second, the mapping \eqref{eq:equilibrium.complexity.map}
 is used to find the corresponding Kac--Rice saddle parameters in the ground
 state. With these parameters in hand, small steps are then made in energy $E$
-or stability $\mu$, after which known values are used as the initial condition
+or stability $\mu$, after which known these values are used as the initial condition
 for a saddle-finding problem. In this section, we use this basic numeric idea
 to map out the complexity for two representative examples: a model with a
 2RSB equilibrium ground state and therefore 1RSB complexity in its
@@ -961,7 +963,7 @@ to 1RSB in Kac--Rice. For this example, we take
 established to have a 2RSB ground state \cite{Crisanti_2011_Statistical}.
 With this covariance, the model sees a replica symmetric to 1RSB transition at
 $\beta_1=1.70615\ldots$ and a 1RSB to 2RSB transition at
-$\beta_2=6.02198\ldots$. At these transitions, the average energies are
+$\beta_2=6.02198\ldots$. At these transitions, the average energies in equilibrium are
 $\langle E\rangle_1=-0.906391\ldots$ and $\langle E\rangle_2=-1.19553\ldots$,
 respectively, and the ground state energy is $E_0=-1.287\,605\,530\ldots$.
 Besides these typical equilibrium energies, an energy of special interest for
@@ -984,7 +986,7 @@ In this model, the RS complexity gives an inconsistent answer for the
 complexity of the ground state, predicting that the complexity of minima
 vanishes at a higher energy than the complexity of saddles, with both at a
 lower energy than the equilibrium ground state. The 1RSB complexity resolves
-these problems, predicting the same ground state as equilibrium and that the
+these problems, predicting the same ground state as equilibrium and with a ground state stability $\mu_0=6.480\,764\ldots>\mu_m$. It predicts that the
 complexity of marginal minima (and therefore all saddles) vanishes at
 $E_m=-1.287\,605\,527\ldots$, which is very slightly greater than $E_0$. Saddles
 become dominant over minima at a higher energy $E_\mathrm{th}=-1.287\,575\,114\ldots$.
@@ -1002,7 +1004,7 @@ not inherit a 1RSB description until the energy is with in a close vicinity of
 the ground state. On the other hand, for high-index saddles the complexity
 becomes described by 1RSB at quite high energies. This suggests that when
 sampling a landscape at high energies, high index saddles may show a sign of
-replica symmetry breaking when minima do not.
+replica symmetry breaking when minima or inherent states do not.
 
 \begin{figure}
   \includegraphics{figs/316_complexity.pdf}
@@ -1012,7 +1014,7 @@ replica symmetry breaking when minima do not.
     dominant minima (green) of the $3+16$ model. Solid lines show the result of
     the 1RSB ansatz, while the dashed lines show that of a RS ansatz. The
     complexity of marginal minima is always below that of dominant critical
-    points except at the red dot, where they are dominant.
+    points except at the black dot, where they are dominant.
     The inset shows a region around the ground state and the fate of the RS solution.
   } \label{fig:2rsb.complexity}
 \end{figure}
@@ -1026,8 +1028,11 @@ replica symmetry breaking when minima do not.
   \raisebox{3em}{\includegraphics{figs/316_complexity_contour_leg.pdf}}
 
   \caption{
-    Complexity of the $3+16$ model in the energy $E$ and stability $\mu$
-    plane. The right shows a detail of the left.
+    Complexity of the $3+16$ model in the energy $E$ and stability $\mu^*$
+    plane. The right shows a detail of the left. Below the yellow marginal line
+    the complexity counts saddles of increasing index as $\mu^*$ decreases.
+    Above the yellow marginal line the complexity counts minima of increasing
+    stability as $\mu^*$ increases.
   } \label{fig:2rsb.contour}
 \end{figure}
 
@@ -1044,7 +1049,10 @@ below the place where most saddle transition to 1RSB is suggestive; we
 speculate that an analysis of the typical minima connected to these saddles by
 downward trajectories will coincide with the algorithmic limit. An analysis of
 the typical nearby minima or the typical downward trajectories from these
-saddles at 1RSB is warranted \cite{Ros_2019_Complex, Ros_2021_Dynamical}.
+saddles at 1RSB is warranted \cite{Ros_2019_Complex, Ros_2021_Dynamical}. Also
+notable is that $E_\mathrm{alg}$ is at a significantly higher energy than
+$E_\mathrm{th}$; according to the theory, optimal smooth algorithms in this
+model stall in a place where minima are exponentially subdominant.
 
 \begin{figure}
   \centering
@@ -1063,11 +1071,12 @@ saddles at 1RSB is warranted \cite{Ros_2019_Complex, Ros_2021_Dynamical}.
     $E_\mathrm{alg}$ that bounds the performance of smooth algorithms, and the
     average energies at the $2$RSB and $1$RSB equilibrium transitions $\langle
     E\rangle_2$ and $\langle E\rangle_1$, respectively. Though the figure is
-    suggestive, $E_\mathrm{alg}$ lies slightly below the termination of the RS
+    suggestive, $E_\mathrm{alg}$ lies at slightly lower energy than the termination of the RS
     -- 1RSB transition line.
   } \label{fig:2rsb.phases}
 \end{figure}
 
+
 Fig.~\ref{fig:2rsb.comparison} shows the saddle parameters for the $3+16$
 system for notable species of stationary points, notably the most common, the
 marginal ones, those with zero complexity, and those on the transition line.
@@ -1099,7 +1108,7 @@ point in the complexity of high-index saddles in worth further study.
   \caption{
     Comparison of the saddle point parameters for the $3+16$ model along
     different trajectories in the energy and stability space, and with the
-    equilibrium values (when they exist) at the same value of $\langle
+    equilibrium values (when they exist) at the same value of average energy $\langle
     E\rangle$.
   } \label{fig:2rsb.comparison}
 \end{figure}
@@ -1107,13 +1116,13 @@ point in the complexity of high-index saddles in worth further study.
 
 \subsection{Full RSB complexity}
 
-On the other hand, if the covariance $f$ is chosen to be concave, then one develops FRSB in equilibrium. To this purpose, we choose
+If the covariance $f$ is chosen to be concave, then one develops FRSB in equilibrium. To this purpose, we choose
 \begin{equation}
   f(q)=\frac12\left(q^2+\frac1{16}q^4\right)
 \end{equation}
 also studied before in equilibrium \cite{Crisanti_2004_Spherical, Crisanti_2006_Spherical}. Because the ground state is FRSB, for this model
 \begin{equation}
-  E_0=E_\mathrm{alg}=E_\mathrm{th}=-\int_0^1dq\sqrt{f''(q)}=-1.059\,384\,319\ldots
+  E_0=E_\mathrm{alg}=E_\mathrm{th}=-\int_0^1dq\,\sqrt{f''(q)}=-1.059\,384\,319\ldots
 \end{equation}
 In the equilibrium solution, the transition temperature from RS to FRSB is $\beta_\infty=1$, with corresponding average energy $\langle E\rangle_\infty=-0.53125\ldots$.
 
@@ -1127,8 +1136,8 @@ $k$RSB solution, but approximate the FRSB one, we drop the extremal condition
 \end{equation}
 and extremize over $x_\textrm{max}$ alone. This dramatically simplifies the
 equations that must be solved to find solutions. In the results that follow, a
-20RSB solution is used to trace the dominant saddles and marginal minima, while
-a 5RSB solution is used to trace the (much longer) boundaries of the
+20RSB approximation is used to trace the dominant saddles and marginal minima, while
+a 5RSB approximation is used to trace the (much longer) boundaries of the
 complexity.
 
 Fig.~\ref{fig:frsb.complexity} shows the complexity for this model as a
@@ -1196,7 +1205,8 @@ goes to zero with finite slopes.
 \section{Interpretation}
 \label{sec:interpretation}
 
-Let $\langle A\rangle$ be the average of $A$ over stationary points with given $E$ and $\mu^*$, i.e.,
+Let $\langle A\rangle$ be the average of any function $A$ over stationary
+points with given $E$ and $\mu^*$, i.e.,
 \begin{equation}
   \langle A\rangle
   =\frac1{\mathcal N}\sum_{\mathbf s\in\mathcal S}A(\mathbf s)
@@ -1216,8 +1226,7 @@ of a hierarchical matrix. The fields $C$, $R$, and $D$ defined in
 
 \subsection{\textit{C}: distribution of overlaps}
 
-First,
-consider $C$, which has an interpretation nearly identical to that of Parisi's
+First consider $C$, which has an interpretation nearly identical to that of Parisi's
 $Q$ matrix of overlaps in the equilibrium case. Its off-diagonal corresponds to
 the probability distribution $P(q)$ of the overlaps $q=(\mathbf s_1\cdot\mathbf
 s_2)/N$ between stationary points.  Let $\mathcal S$ be the set of all
@@ -1251,7 +1260,7 @@ disorder (again, for fixed energy and index) gives
     =\lim_{n\to0}{\int\overline{\prod_a^n d\nu(\mathbf s_a)}\,\left(\frac{\mathbf s_1\cdot\mathbf s_2}N\right)^p} \\
     &=\lim_{n\to0}{\int D[C,R,D] \,
     \left(C_{12}\right)^p\; } e^{nN\Sigma[C,R,D]}
-    =\frac1{n(n-1)}\lim_{n\to0}{\int D[C,R,D] \,\sum_{a\neq b}\left(C_{ab}\right)^p\; } e^{nN\Sigma[C,R,D]}
+    =\lim_{n\to0}{\int D[C,R,D] \,\frac1{n(n-1)}\sum_{a\neq b}\left(C_{ab}\right)^p\; } e^{nN\Sigma[C,R,D]}
   \end{aligned}
 \end{equation}
 In the last line, we have used that there is nothing special about replicas one and two.
@@ -1280,7 +1289,7 @@ sphere, one will still find zero average overlap. However, if the disconnected
 level sets are \emph{few}, i.e., less than order $N$, then it is possible to
 draw two stationary points from the same set with nonzero probability.
 Therefore, the picture in this case is of few, large basins each containing
-exponentially many stationary points.
+exponentially many stationary points. A cartoon of this picture is shown in Fig.~\ref{fig:cartoon}.
 
 \begin{figure}
   \centering
@@ -1307,9 +1316,9 @@ exponentially many stationary points.
 \subsubsection{A concrete example}
 
 One can construct a schematic 2RSB model from two 1RSB models.
-Consider two independent pure models with $p_1$ and $p_2$ couplings,
+Consider two independent pure models of size $N$ and with $p_1$-spin and $p_2$-spin couplings,
 respectively, with energies $H_{p_1}({\mathbf s})$ and
-$H_{p_2}({\boldsymbol\sigma})$ of sizes  $N$, and couple them weakly with
+$H_{p_2}({\boldsymbol\sigma})$, and couple them weakly with
 $\varepsilon \; {\boldsymbol \sigma} \cdot {\mathbf s}$. The landscape of the
 pure models is much simpler than that of the mixed because, in these models,
 fixing the stability $\mu$ is equivalent to fixing the energy: $\mu=pE$. This
@@ -1344,13 +1353,13 @@ this point, $\Sigma_1(E_1)=0$ and $E_1$ is the ground state energy. At an even
 higher value $\hat \beta=\hat \beta_f$, both systems will become frozen in
 their ground states.  For $\hat \beta_f > \hat \beta> \hat \beta_c$ one system
 is unfrozen, while the other is, because of coupling, frozen at inverse
-temperature $\hat \beta_c$.  The overlap between two solutions  is:
+temperature $\hat \beta_c$.  The overlap between two solutions in this intermediate phase is
 \begin{equation}
   \frac1 {2N}\Big\langle({\mathbf s^1},{\boldsymbol\sigma^1})\cdot ({\mathbf s^2},{\boldsymbol\sigma^2})\Big\rangle
   =\frac1 {2N}\Big[\langle{\mathbf s^1}\cdot {\mathbf s^2}\rangle+ \langle{\boldsymbol\sigma^1}\cdot {\boldsymbol\sigma^2}\rangle\Big]
   = \frac1 {2N}\langle{\mathbf s^1}\cdot {\mathbf s^2}\rangle>0
 \end{equation}
-nonzero because there are few low-energy stationary points in system one, and
+which is nonzero because there are only a few low-energy stationary points in system one, and
 there is a nonvanishing probability of selecting one of them twice.
 The distribution of this overlap is one-half the overlap distribution of a
 frozen spin-glass at temperature $\hat \beta$, a 1RSB system like the Random
@@ -1361,7 +1370,7 @@ global overlap between different states is at most $1/2$.  At $\hat \beta>\hat
 \beta_f$ there is a further transition.
 
 This schematic example provides a metaphor for considering what happens in
-ordinary models when replica symmetry is broken. At some point certain degree
+ordinary models when replica symmetry is broken. At some point certain degrees
 of freedom `freeze' onto a subextensive number of possible states, while the
 remainder are effectively unconstrained. The overlap measures something in the
 competition between the number of these unconstrained subregions and their
@@ -1371,9 +1380,9 @@ size.
 
 The matrix field $R$ is related to responses of the stationary points to
 perturbations of the tensors $J$. One adds to the Hamiltonian a random term
-$\varepsilon_p \tilde H_p = -\varepsilon_p \sum_{i_1,...,i_p} \tilde J_{i_1,...,i_p}
-s_{i_1}...s_{i_p}$, where the $\tilde J$ are random Gaussian uncorrelated with
-the $J$'s and having variance $\overline{(\tilde J)^2}=p!/2N^{p-1}$.  The response to these is:
+$\varepsilon_p \tilde H_p = -\frac1{p!}\varepsilon_p \sum_{i_1\cdots i_p} \tilde J_{i_1\cdots i_p}
+s_{i_1}\cdots s_{i_p}$, where the $\tilde J$ are random Gaussian uncorrelated with
+the $J$s and having variance $\overline{\tilde J^2}=p!/2N^{p-1}$.  The response to these is:
 \begin{equation}
 \frac1N\overline{\frac{\partial \langle \tilde H_p \rangle  }  {\partial \varepsilon_p}}
      =\lim_{n\to0}\int\left(\prod_a^nd\nu(\mathbf s_a)\right)\sum_b^n\left[
@@ -1409,7 +1418,7 @@ contribution due to the perturbation causing stationary points to move up or
 down in energy.
 
 The matrix field $D$ is related to the response of the complexity to
-perturbations to the variance of the tensors $J$. This can be found by taking
+perturbations of the variance of the tensors $J$. This can be found by taking
 the expression for the complexity and inserting the dependence of $f$ on the
 coefficients $a_p$, then differentiating:
 \begin{equation}
@@ -1437,28 +1446,28 @@ and in particular for $p=1$
   =\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.
+resulting magnetization of the stationary points for supersymmetric minima.
 
 \section{Conclusion}
 \label{se:conclusion}
 
 We have constructed a  replica solution for the general  problem of finding
 saddles of random mean-field landscapes, including systems with many steps of
-RSB.  For systems with full RSB, we find that minima are, at all energy
-densities above the ground state, exponentially subdominant with respect to
-saddles.  The solution contains valuable geometric information that has yet to
-be extracted in all detail.
+RSB.  For systems with full RSB, we find that minima are exponentially
+subdominant with respect to saddles at all energy densities above the ground
+state.  The solution contains valuable geometric information that has yet to be
+extracted in all detail.
 
 A first and very important application of the method here is to perform the
 calculation for high dimensional spheres, where it would give us a clear
-understanding of what happens in a low-temperature realistic jamming dynamics
+understanding of what happens in realistic low-temperature jamming dynamics
 \cite{Maimbourg_2016_Solution}. More simply, examining the landscape of a
 spherical model with a glass to glass transition from 1RSB to RS, like the
 $2+4$ model when $a_4$ is larger than we have taken it in our example, might
 give insight into the cases of interest for Gardner physics
 \cite{Crisanti_2004_Spherical, Crisanti_2006_Spherical}. In any case, our
 analysis of typical 1RSB and FRSB landscapes indicates that the highest energy
-signature of RSB phases at lower energies is in the overlap structure of the
+signature of RSB phases is in the overlap structure of the
 high-index saddle points. Though measuring the statistics of saddle points is
 difficult to imagine for experiments, this insight could find application in
 simulations of glass formers, where saddle-finding methods are possible.
@@ -1466,10 +1475,10 @@ simulations of glass formers, where saddle-finding methods are possible.
 A second application is to evaluate in more detail the landscape of these RSB
 systems. In particular, examining the complexity of stationary points with
 non-extensive indices (like rank-one saddles), the complexity of pairs of
-stationary points at fixed overlap, or the complexity of barriers
+stationary points at fixed overlap, or the complexity of energy barriers
 \cite{Auffinger_2012_Random, Ros_2019_Complexity}. These other properties of
 the landscape might shed light on the relationship between landscape RSB and
-dynamical features, like the algorithmic energy $E_\mathrm{alg}$. For our 1RSB
+dynamical features, like the algorithmic energy $E_\mathrm{alg}$, or the asymptotic level reached by physical dynamics. For our 1RSB
 example, because $E_\mathrm{alg}$ is just below the energy where
 dominant saddles transition to a RSB complexity, we speculate that
 $E_\mathrm{alg}$ may be related to the statistics of minima connected to the