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1 files changed, 312 insertions, 310 deletions
diff --git a/2-point.tex b/2-point.tex
index 9be13e9..ce2d1bc 100644
--- a/2-point.tex
+++ b/2-point.tex
@@ -612,314 +612,7 @@ $\pmb\sigma_1$ is special among the set of $\pmb\sigma$ replicas, since it alone
is constrained to lie a given overlap from the $\mathbf s$ replicas. This
replica asymmetry will be important later.
-\subsection{The Hessian factors}
-
-The double partial derivatives of the energy are Gaussian with the variance
-\begin{equation}
- \overline{(\partial_i\partial_jH(\mathbf s))^2}=\frac1Nf''(1)
-\end{equation}
-which means that the matrix of partial derivatives belongs to the GOE class. Its spectrum is given by the Wigner semicircle
-\begin{equation}
- \rho(\lambda)=\begin{cases}
- \frac2{\pi}\sqrt{1-\big(\frac{\lambda}{\mu_\text m}\big)^2} & \lambda^2\leq\mu_\text m^2 \\
- 0 & \text{otherwise}
- \end{cases}
-\end{equation}
-with radius $\mu_\text m=\sqrt{4f''(1)}$. Since the Hessian differs from the
-matrix of partial derivatives by adding the constant diagonal matrix $\omega
-I$, it follows that the spectrum of the Hessian is a Wigner semicircle shifted
-by $\omega$, or $\rho(\lambda+\omega)$.
-
-The average over factors depending on the Hessian alone can be made separately
-from those depending on the gradient or energy, since for random Gaussian
-fields the Hessian is independent of these \cite{Bray_2007_Statistics}. In
-principle the fact that we have conditioned the Hessian to belong to stationary
-points of certain energy, stability, and proximity to another stationary point
-will modify its statistics, but these changes will only appear at subleading
-order in $N$ \cite{Ros_2019_Complexity}. At leading order, the various expectations factorize, each yielding
-\begin{equation}
- \overline{\big|\det\operatorname{Hess}H(\mathbf s,\omega)\big|\,\delta\big(N\mu-\operatorname{Tr}\operatorname{Hess}H(\mathbf s,\omega)\big)}
- =e^{N\int d\lambda\,\rho(\lambda+\mu)\log|\lambda|}\delta(N\mu-N\omega)
-\end{equation}
-Therefore, all of the Lagrange multipliers are fixed to the stabilities $\mu$. We define the function
-\begin{equation}
- \begin{aligned}
- \mathcal D(\mu)
- &=\int d\lambda\,\rho(\lambda+\mu)\log|\lambda| \\
- &=\begin{cases}
- \frac12+\log\left(\frac12\mu_\text m\right)+\frac{\mu^2}{\mu_\text m^2}
- & \mu^2\leq\mu_\text m^2 \\
- \frac12+\log\left(\frac12\mu_\text m\right)+\frac{\mu^2}{\mu_\text m^2}
- -\left|\frac{\mu}{\mu_\text m}\right|\sqrt{\big(\frac\mu{\mu_\text m}\big)^2-1}
- -\log\left(\left|\frac{\mu}{\mu_\text m}\right|-\sqrt{\big(\frac\mu{\mu_\text m}\big)^2-1}\right) & \mu^2>\mu_\text m^2
- \end{cases}
- \end{aligned}
-\end{equation}
-and the full factor due to the Hessians is
-\begin{equation}
- e^{Nm\mathcal D(\mu_0)+Nn\mathcal D(\mu_1)}\left[\prod_a^m\delta(N\mu_0-N\varsigma_a)\right]\left[\prod_a^n\delta(N\mu_1-N\omega_a)\right]
-\end{equation}
-
-\subsection{The other factors}
-
-Having integrated over the Lagrange multipliers using the $\delta$-functions
-resulting from the average of the Hessians, any $\delta$-functions in the
-remaining integrand we Fourier transform into their integral representation
-over auxiliary fields. The resulting integrand has the form
-\begin{equation}
- e^{
- Nm\hat\beta_0E_0+Nn\hat\beta_1E_1
- -\sum_a^m\left[(\pmb\sigma_a\cdot\hat{\pmb\sigma}_a)\mu_0
- -\frac12\hat\mu_0(N-\pmb\sigma_a\cdot\pmb\sigma_a)
- \right]
- -\sum_a^n\left[(\mathbf s_a\cdot\hat{\mathbf s}_a)\mu_1
- -\frac12\hat\mu_1(N-\mathbf s_a\cdot\mathbf s_a)
- -\frac12\hat\mu_{12}(Nq-\pmb\sigma_1\cdot\mathbf s_a)
- \right]
- +\int d\mathbf t\,\mathcal O(\mathbf t)H(\mathbf t)
- }
-\end{equation}
-where we have introduced the linear operator
-\begin{equation}
- \mathcal O(\mathbf t)
- =\sum_a^m\delta(\mathbf t-\pmb\sigma_a)\left(
- i\hat{\pmb\sigma}_a\cdot\partial_{\mathbf t}-\hat\beta_0
- \right)
- +
- \sum_a^n\delta(\mathbf t-\mathbf s_a)\left(
- i\hat{\mathbf s}_a\cdot\partial_{\mathbf t}-\hat\beta_1
- \right)
-\end{equation}
-Here the $\hat\beta$s are the fields auxiliary to the energy constraints, the
-$\hat\mu$s are auxiliary to the spherical and overlap constraints, and the
-$\hat{\pmb\sigma}$s and $\hat{\mathbf s}$s are auxiliary to the constraint that
-the gradient be zero.
-We have written the $H$-dependent terms in this strange form for the ease of taking the average over $H$: since it is Gaussian-correlated, it follows that
-\begin{equation}
- \overline{e^{\int d\mathbf t\,\mathcal O(\mathbf t)H(\mathbf t)}}
- =e^{\frac12\int d\mathbf t\,d\mathbf t'\,\mathcal O(\mathbf t)\mathcal O(\mathbf t')\overline{H(\mathbf t)H(\mathbf t')}}
- =e^{N\frac12\int d\mathbf t\,d\mathbf t'\,\mathcal O(\mathbf t)\mathcal O(\mathbf t')f\big(\frac{\mathbf t\cdot\mathbf t'}N\big)}
-\end{equation}
-It remains only to apply the doubled operators to $f$ and then evaluate the simple integrals over the $\delta$ measures. We do not include these details, which are standard.
-
-\subsection{Hubbard--Stratonovich}
-
-Having expanded this expression, we are left with an argument in the exponential which is a function of scalar products between the fields $\mathbf s$, $\hat{\mathbf s}$, $\pmb\sigma$, and $\hat{\pmb\sigma}$. We will change integration coordinates from these fields to matrix fields given by their scalar products, defined as
-\begin{equation} \label{eq:fields}
- \begin{aligned}
- C^{00}_{ab}=\frac1N\pmb\sigma_a\cdot\pmb\sigma_b &&
- R^{00}_{ab}=-i\frac1N{\pmb\sigma}_a\cdot\hat{\pmb\sigma}_b &&
- D^{00}_{ab}=\frac1N\hat{\pmb\sigma}_a\cdot\hat{\pmb\sigma}_b \\
- C^{01}_{ab}=\frac1N\pmb\sigma_a\cdot\mathbf s_b &&
- R^{01}_{ab}=-i\frac1N{\pmb\sigma}_a\cdot\hat{\mathbf s}_b &&
- R^{10}_{ab}=-i\frac1N\hat{\pmb\sigma}_a\cdot{\mathbf s}_b &&
- D^{01}_{ab}=\frac1N\hat{\pmb\sigma}_a\cdot\hat{\mathbf s}_b \\
- C^{11}_{ab}=\frac1N\mathbf s_a\cdot\mathbf s_b &&
- R^{11}_{ab}=-i\frac1N{\mathbf s}_a\cdot\hat{\mathbf s}_b &&
- D^{11}_{ab}=\frac1N\hat{\mathbf s}_a\cdot\hat{\mathbf s}_b
- \end{aligned}
-\end{equation}
-We insert into the integral the product of $\delta$-functions enforcing these
-definitions, integrated over the new matrix fields, which is equivalent to
-multiplying by one. Once this is done, the many scalar products appearing
-throughout can be replaced by the matrix fields, and the original vector fields
-can be integrated over. Conjugate matrix field integrals created when the
-$\delta$-functions are promoted to exponentials can be evaluated by saddle
-point in the standard way, yielding an effective action depending on the above
-matrix fields alone.
-
-\subsection{Saddle point}
-
-We will always assume that the square matrices $C^{00}$, $R^{00}$, $D^{00}$,
-$C^{11}$, $R^{11}$, and $D^{11}$ are hierarchical matrices, with each set of
-three sharing the same hierarchical structure. In particular, we immediately
-define $c_\mathrm d^{00}$, $r_\mathrm d^{00}$, $d_\mathrm d^{00}$, $c_\mathrm d^{11}$, $r_\mathrm d^{11}$, and
-$d_\mathrm d^{11}$ as the value of the diagonal elements of these matrices,
-respectively. Note that $c_\mathrm d^{00}=c_\mathrm d^{11}=1$ due to the spherical constraint.
-
-Defining the `block' fields $\mathcal Q_{00}=(\hat\beta_0, \hat\mu_0, C^{00},
-R^{00}, D^{00})$, $\mathcal Q_{11}=(\hat\beta_1, \hat\mu_1, C^{11}, R^{11},
-D^{11})$, and $\mathcal Q_{01}=(\hat\mu_{01},C^{01},R^{01},R^{10},D^{01})$
-the resulting complexity is
-\begin{equation}
- \Sigma_{01}
- =\frac1N\lim_{n\to0}\lim_{m\to0}\frac\partial{\partial n}\int d\mathcal Q_{00}\,d\mathcal Q_{11}\,d\mathcal Q_{01}\,e^{Nm\mathcal S_0(\mathcal Q_{00})+Nn\mathcal S_1(\mathcal Q_{11},\mathcal Q_{01}\mid\mathcal Q_{00})}
-\end{equation}
-where
-\begin{equation} \label{eq:one-point.action}
- \begin{aligned}
- &\mathcal S_0(\mathcal Q_{00})
- =\hat\beta_0E_0-r^{00}_\mathrm d\mu_0-\frac12\hat\mu_0(1-c^{00}_\mathrm d)+\mathcal D(\mu_0)\\
- &\quad+\frac1m\bigg\{
- \frac12\sum_{ab}^m\left[
- \hat\beta_1^2f(C^{00}_{ab})+(2\hat\beta_1R^{00}_{ab}-D^{00}_{ab})f'(C^{00}_{ab})+(R_{ab}^{00})^2f''(C_{ab}^{00})
- \right]+\frac12\log\det\begin{bmatrix}C^{00}&R^{00}\\R^{00}&D^{00}\end{bmatrix}
-\bigg\}
-\end{aligned}
-\end{equation}
-is the action for the ordinary, one-point complexity, and remainder is given by
-\begin{equation} \label{eq:two-point.action}
- \begin{aligned}
- &\mathcal S(\mathcal Q_{11},\mathcal Q_{01}\mid\mathcal Q_{00})
- =\hat\beta_1E_1-r^{11}_\mathrm d\mu_1-\frac12\hat\mu_1(1-c^{11}_\mathrm d)+\mathcal D(\mu_1) \\
- &\quad+\frac1n\sum_b^n\left\{-\frac12\hat\mu_{12}(q-C^{01}_{1b})+\sum_a^m\left[
- \hat\beta_0\hat\beta_1f(C^{01}_{ab})+(\hat\beta_0R^{01}_{ab}+\hat\beta_1R^{10}_{ab}-D^{01}_{ab})f'(C^{01}_{ab})+R^{01}_{ab}R^{10}_{ab}f''(C^{01}_{ab})
- \right]\right\}
- \\
- &\quad+\frac1n\bigg\{
- \frac12\sum_{ab}^n\left[
- \hat\beta_1^2f(C^{11}_{ab})+(2\hat\beta_1R^{11}_{ab}-D^{11}_{ab})f'(C^{11}_{ab})+(R^{11}_{ab})^2f''(C^{11}_{ab})
- \right]\\
- &\quad+\frac12\log\det\left(
- \begin{bmatrix}
- C^{11}&iR^{11}\\iR^{11}&D^{11}
- \end{bmatrix}-
- \begin{bmatrix}
- C^{01}&iR^{01}\\iR^{10}&D^{01}
- \end{bmatrix}^T
- \begin{bmatrix}
- C^{00}&iR^{00}\\iR^{00}&D^{00}
- \end{bmatrix}^{-1}
- \begin{bmatrix}
- C^{01}&iR^{01}\\iR^{10}&D^{01}
- \end{bmatrix}
- \right)
- \bigg\}
- \end{aligned}
-\end{equation}
-Because of the structure of this problem in the twin limits of $m$ and $n$ to
-zero, the parameters $\mathcal Q_{00}$ can be evaluated at a saddle point of
-$\mathcal S_0$ alone. This means that these parameters will take the same value
-they take when the ordinary, 1-point complexity is calculated. For a replica
-symmetric complexity of the reference point, this results in
-\begin{align}
- \hat\beta_0
- &=-\frac{\mu_0f'(1)+E_0\big(f'(1)+f''(1)\big)}{u_f}\\
- r_\mathrm d^{00}
- &=\frac{\mu_0f(1)+E_0f'(1)}{u_f} \\
- d_\mathrm d^{00}
- &=\frac1{f'(1)}
- -\left(
- \frac{\mu_0f(1)+E_0f'(1)}{u_f}
- \right)^2
-\end{align}
-where we define for brevity (here and elsewhere) the constants
-\begin{align}
- u_f=f(1)\big(f'(1)+f''(1)\big)-f'(1)^2
- &&
- v_f=f'(1)\big(f''(1)+f'''(1)\big)-f''(1)^2
-\end{align}
-Note that because the coefficients of $f$ must be nonnegative for $f$ to
-be a sensible covariance, both $u_f$ and $v_f$ are strictly positive. Note also
-that $u_f=v_f=0$ if $f$ is a homogeneous polynomial as in the pure models.
-These expressions are invalid for the pure models because $\mu_0$ and $E_0$
-cannot be fixed independently; we would have done the equivalent of inserting
-two identical $\delta$-functions. For the pure models, the terms $\hat\beta_0$ and
-$\hat\beta_1$ must be set to zero in our prior formulae (as if the energy was
-not constrained) and then the saddle point taken.
-
-
-In general, we except the $m\times n$ matrices $C^{01}$, $R^{01}$, $R^{10}$,
-and $D^{01}$ to have constant \emph{rows} of length $n$, with blocks of rows
-corresponding to the \textsc{rsb} structure of the single-point complexity. For
-the scope of this paper, where we restrict ourselves to replica symmetric
-complexities, they have the following form at the saddle point:
-\begin{align} \label{eq:01.ansatz}
- C^{01}=
- \begin{subarray}{l}
- \hphantom{[}\begin{array}{ccc}\leftarrow&n&\rightarrow\end{array}\hphantom{\Bigg]}\\
- \left[
- \begin{array}{ccc}
- q&\cdots&q\\
- 0&\cdots&0\\
- \vdots&\ddots&\vdots\\
- 0&\cdots&0
- \end{array}
- \right]\begin{array}{c}
- \\\uparrow\\m-1\\\downarrow
- \end{array}
-\end{subarray}
- &&
- R^{01}
- =\begin{bmatrix}
- r_{01}&\cdots&r_{01}\\
- 0&\cdots&0\\
- \vdots&\ddots&\vdots\\
- 0&\cdots&0
- \end{bmatrix}
- &&
- R^{10}
- =\begin{bmatrix}
- r_{10}&\cdots&r_{10}\\
- 0&\cdots&0\\
- \vdots&\ddots&\vdots\\
- 0&\cdots&0
- \end{bmatrix}
- &&
- D^{01}
- =\begin{bmatrix}
- d_{01}&\cdots&d_{01}\\
- 0&\cdots&0\\
- \vdots&\ddots&\vdots\\
- 0&\cdots&0
- \end{bmatrix}
-\end{align}
-where only the first row is nonzero. The other entries, which correspond to the
-completely uncorrelated replicas in an \textsc{rsb} picture, are all zero
-because uncorrelated vectors on the sphere are orthogonal.
-
-The inverse of block hierarchical matrix is still a block hierarchical matrix, since
-\begin{equation}
- \begin{bmatrix}
- C^{00}&iR^{00}\\iR^{00}&D^{00}
- \end{bmatrix}^{-1}
- =
- \begin{bmatrix}
- (C^{00}D^{00}+R^{00}R^{00})^{-1}D^{00} & -i(C^{00}D^{00}+R^{00}R^{00})^{-1}R^{00} \\
- -i(C^{00}D^{00}+R^{00}R^{00})^{-1}R^{00} & (C^{00}D^{00}+R^{00}R^{00})^{-1}C^{00}
- \end{bmatrix}
-\end{equation}
-Because of the structure of the 01 matrices, the volume element will depend
-only on the diagonals of the matrices in this inverse block matrix. If we define
-\begin{align}
- \tilde c_\mathrm d^{00}&=[(C^{00}D^{00}+R^{00}R^{00})^{-1}C^{00}]_{\text d} \\
- \tilde r_\mathrm d^{00}&=[(C^{00}D^{00}+R^{00}R^{00})^{-1}R^{00}]_{\text d} \\
- \tilde d_\mathrm d^{00}&=[(C^{00}D^{00}+R^{00}R^{00})^{-1}D^{00}]_{\text d}
-\end{align}
-as the diagonals of the blocks of the inverse matrix, then the result of the product is
-\begin{equation}
- \begin{aligned}
- & \begin{bmatrix}
- C^{01}&iR^{01}\\iR^{10}&D^{01}
- \end{bmatrix}^T
- \begin{bmatrix}
- C^{00}&iR^{00}\\iR^{00}&D^{00}
- \end{bmatrix}^{-1}
- \begin{bmatrix}
- C^{01}&iR^{01}\\iR^{10}&D^{01}
- \end{bmatrix} \\
- &\qquad=\begin{bmatrix}
- q^2\tilde d_\mathrm d^{00}+2qr_{10}\tilde r^{00}_\mathrm d-r_{10}^2\tilde d^{00}_\mathrm d
- &
- i\left[d_{01}(r_{10}\tilde c^{00}_\mathrm d-q\tilde r^{00}_\mathrm d)+r_{01}(r_{10}\tilde r^{00}_\mathrm d+q\tilde d^{00}_\mathrm d)\right]
- \\
- i\left[d_{01}(r_{10}\tilde c^{00}_\mathrm d-q\tilde r^{00}_\mathrm d)+r_{01}(r_{10}\tilde r^{00}_\mathrm d+q\tilde d^{00}_\mathrm d)\right]
- &
- d_{01}^2\tilde c^{00}_\mathrm d+2r_{01}d_{01}\tilde r^{00}_\mathrm d-r_{01}^2\tilde d^{00}_\mathrm d
- \end{bmatrix}
- \end{aligned}
-\end{equation}
-where each block is a constant $n\times n$ matrix. Because the matrices
-$C^{00}$, $R^{00}$, and $D^{00}$ are diagonal in the replica symmetric case,
-the diagonals of the blocks above take a simple form:
-\begin{align}
- \tilde c_\mathrm d^{00}=f'(1) &&
- \tilde r_\mathrm d^{00}=r^{00}_\mathrm df'(1) &&
- \tilde d_\mathrm d^{00}=d^{00}_\mathrm df'(1)
-\end{align}
-Once these expressions are inserted into the complexity, the limits of $n$ and
-$m$ to zero can be taken, and the parameters from $D^{01}$ and $D^{11}$ can be
-extremized explicitly. The resulting expression for the complexity, which must
+The resulting expression for the complexity, which must
still be extremized over the parameters $\hat\beta_1$, $r^{01}$,
$r^{11}_\mathrm d$, $r^{11}_0$, and $q^{11}_0$, is
\begin{equation}
@@ -958,8 +651,6 @@ this paper was produced. Second, the complexity can be calculated in the near
neighborhood of a reference point by expanding in small $1-q$. This is what we
describe in the next subsection.
-\subsection{Expansion in the near neighborhood}
-
If there is no overlap gap between the reference point and its nearest
neighbors, their complexity can be calculated by an expansion in $1-q$. First,
we'll use this method to describe the most common type of stationary point in
@@ -1324,6 +1015,317 @@ INFN.
\appendix
+\section{Details of calculation for the two-point complexity}
+\label{sec:complexity-details}
+
+\subsection{The Hessian factors}
+
+The double partial derivatives of the energy are Gaussian with the variance
+\begin{equation}
+ \overline{(\partial_i\partial_jH(\mathbf s))^2}=\frac1Nf''(1)
+\end{equation}
+which means that the matrix of partial derivatives belongs to the GOE class. Its spectrum is given by the Wigner semicircle
+\begin{equation}
+ \rho(\lambda)=\begin{cases}
+ \frac2{\pi}\sqrt{1-\big(\frac{\lambda}{\mu_\text m}\big)^2} & \lambda^2\leq\mu_\text m^2 \\
+ 0 & \text{otherwise}
+ \end{cases}
+\end{equation}
+with radius $\mu_\text m=\sqrt{4f''(1)}$. Since the Hessian differs from the
+matrix of partial derivatives by adding the constant diagonal matrix $\omega
+I$, it follows that the spectrum of the Hessian is a Wigner semicircle shifted
+by $\omega$, or $\rho(\lambda+\omega)$.
+
+The average over factors depending on the Hessian alone can be made separately
+from those depending on the gradient or energy, since for random Gaussian
+fields the Hessian is independent of these \cite{Bray_2007_Statistics}. In
+principle the fact that we have conditioned the Hessian to belong to stationary
+points of certain energy, stability, and proximity to another stationary point
+will modify its statistics, but these changes will only appear at subleading
+order in $N$ \cite{Ros_2019_Complexity}. At leading order, the various expectations factorize, each yielding
+\begin{equation}
+ \overline{\big|\det\operatorname{Hess}H(\mathbf s,\omega)\big|\,\delta\big(N\mu-\operatorname{Tr}\operatorname{Hess}H(\mathbf s,\omega)\big)}
+ =e^{N\int d\lambda\,\rho(\lambda+\mu)\log|\lambda|}\delta(N\mu-N\omega)
+\end{equation}
+Therefore, all of the Lagrange multipliers are fixed to the stabilities $\mu$. We define the function
+\begin{equation}
+ \begin{aligned}
+ \mathcal D(\mu)
+ &=\int d\lambda\,\rho(\lambda+\mu)\log|\lambda| \\
+ &=\begin{cases}
+ \frac12+\log\left(\frac12\mu_\text m\right)+\frac{\mu^2}{\mu_\text m^2}
+ & \mu^2\leq\mu_\text m^2 \\
+ \frac12+\log\left(\frac12\mu_\text m\right)+\frac{\mu^2}{\mu_\text m^2}
+ -\left|\frac{\mu}{\mu_\text m}\right|\sqrt{\big(\frac\mu{\mu_\text m}\big)^2-1}
+ -\log\left(\left|\frac{\mu}{\mu_\text m}\right|-\sqrt{\big(\frac\mu{\mu_\text m}\big)^2-1}\right) & \mu^2>\mu_\text m^2
+ \end{cases}
+ \end{aligned}
+\end{equation}
+and the full factor due to the Hessians is
+\begin{equation}
+ e^{Nm\mathcal D(\mu_0)+Nn\mathcal D(\mu_1)}\left[\prod_a^m\delta(N\mu_0-N\varsigma_a)\right]\left[\prod_a^n\delta(N\mu_1-N\omega_a)\right]
+\end{equation}
+
+\subsection{The other factors}
+
+Having integrated over the Lagrange multipliers using the $\delta$-functions
+resulting from the average of the Hessians, any $\delta$-functions in the
+remaining integrand we Fourier transform into their integral representation
+over auxiliary fields. The resulting integrand has the form
+\begin{equation}
+ e^{
+ Nm\hat\beta_0E_0+Nn\hat\beta_1E_1
+ -\sum_a^m\left[(\pmb\sigma_a\cdot\hat{\pmb\sigma}_a)\mu_0
+ -\frac12\hat\mu_0(N-\pmb\sigma_a\cdot\pmb\sigma_a)
+ \right]
+ -\sum_a^n\left[(\mathbf s_a\cdot\hat{\mathbf s}_a)\mu_1
+ -\frac12\hat\mu_1(N-\mathbf s_a\cdot\mathbf s_a)
+ -\frac12\hat\mu_{12}(Nq-\pmb\sigma_1\cdot\mathbf s_a)
+ \right]
+ +\int d\mathbf t\,\mathcal O(\mathbf t)H(\mathbf t)
+ }
+\end{equation}
+where we have introduced the linear operator
+\begin{equation}
+ \mathcal O(\mathbf t)
+ =\sum_a^m\delta(\mathbf t-\pmb\sigma_a)\left(
+ i\hat{\pmb\sigma}_a\cdot\partial_{\mathbf t}-\hat\beta_0
+ \right)
+ +
+ \sum_a^n\delta(\mathbf t-\mathbf s_a)\left(
+ i\hat{\mathbf s}_a\cdot\partial_{\mathbf t}-\hat\beta_1
+ \right)
+\end{equation}
+Here the $\hat\beta$s are the fields auxiliary to the energy constraints, the
+$\hat\mu$s are auxiliary to the spherical and overlap constraints, and the
+$\hat{\pmb\sigma}$s and $\hat{\mathbf s}$s are auxiliary to the constraint that
+the gradient be zero.
+We have written the $H$-dependent terms in this strange form for the ease of taking the average over $H$: since it is Gaussian-correlated, it follows that
+\begin{equation}
+ \overline{e^{\int d\mathbf t\,\mathcal O(\mathbf t)H(\mathbf t)}}
+ =e^{\frac12\int d\mathbf t\,d\mathbf t'\,\mathcal O(\mathbf t)\mathcal O(\mathbf t')\overline{H(\mathbf t)H(\mathbf t')}}
+ =e^{N\frac12\int d\mathbf t\,d\mathbf t'\,\mathcal O(\mathbf t)\mathcal O(\mathbf t')f\big(\frac{\mathbf t\cdot\mathbf t'}N\big)}
+\end{equation}
+It remains only to apply the doubled operators to $f$ and then evaluate the simple integrals over the $\delta$ measures. We do not include these details, which are standard.
+
+\subsection{Hubbard--Stratonovich}
+
+Having expanded this expression, we are left with an argument in the exponential which is a function of scalar products between the fields $\mathbf s$, $\hat{\mathbf s}$, $\pmb\sigma$, and $\hat{\pmb\sigma}$. We will change integration coordinates from these fields to matrix fields given by their scalar products, defined as
+\begin{equation} \label{eq:fields}
+ \begin{aligned}
+ C^{00}_{ab}=\frac1N\pmb\sigma_a\cdot\pmb\sigma_b &&
+ R^{00}_{ab}=-i\frac1N{\pmb\sigma}_a\cdot\hat{\pmb\sigma}_b &&
+ D^{00}_{ab}=\frac1N\hat{\pmb\sigma}_a\cdot\hat{\pmb\sigma}_b \\
+ C^{01}_{ab}=\frac1N\pmb\sigma_a\cdot\mathbf s_b &&
+ R^{01}_{ab}=-i\frac1N{\pmb\sigma}_a\cdot\hat{\mathbf s}_b &&
+ R^{10}_{ab}=-i\frac1N\hat{\pmb\sigma}_a\cdot{\mathbf s}_b &&
+ D^{01}_{ab}=\frac1N\hat{\pmb\sigma}_a\cdot\hat{\mathbf s}_b \\
+ C^{11}_{ab}=\frac1N\mathbf s_a\cdot\mathbf s_b &&
+ R^{11}_{ab}=-i\frac1N{\mathbf s}_a\cdot\hat{\mathbf s}_b &&
+ D^{11}_{ab}=\frac1N\hat{\mathbf s}_a\cdot\hat{\mathbf s}_b
+ \end{aligned}
+\end{equation}
+We insert into the integral the product of $\delta$-functions enforcing these
+definitions, integrated over the new matrix fields, which is equivalent to
+multiplying by one. Once this is done, the many scalar products appearing
+throughout can be replaced by the matrix fields, and the original vector fields
+can be integrated over. Conjugate matrix field integrals created when the
+$\delta$-functions are promoted to exponentials can be evaluated by saddle
+point in the standard way, yielding an effective action depending on the above
+matrix fields alone.
+
+\subsection{Saddle point}
+
+We will always assume that the square matrices $C^{00}$, $R^{00}$, $D^{00}$,
+$C^{11}$, $R^{11}$, and $D^{11}$ are hierarchical matrices, with each set of
+three sharing the same hierarchical structure. In particular, we immediately
+define $c_\mathrm d^{00}$, $r_\mathrm d^{00}$, $d_\mathrm d^{00}$, $c_\mathrm d^{11}$, $r_\mathrm d^{11}$, and
+$d_\mathrm d^{11}$ as the value of the diagonal elements of these matrices,
+respectively. Note that $c_\mathrm d^{00}=c_\mathrm d^{11}=1$ due to the spherical constraint.
+
+Defining the `block' fields $\mathcal Q_{00}=(\hat\beta_0, \hat\mu_0, C^{00},
+R^{00}, D^{00})$, $\mathcal Q_{11}=(\hat\beta_1, \hat\mu_1, C^{11}, R^{11},
+D^{11})$, and $\mathcal Q_{01}=(\hat\mu_{01},C^{01},R^{01},R^{10},D^{01})$
+the resulting complexity is
+\begin{equation}
+ \Sigma_{01}
+ =\frac1N\lim_{n\to0}\lim_{m\to0}\frac\partial{\partial n}\int d\mathcal Q_{00}\,d\mathcal Q_{11}\,d\mathcal Q_{01}\,e^{Nm\mathcal S_0(\mathcal Q_{00})+Nn\mathcal S_1(\mathcal Q_{11},\mathcal Q_{01}\mid\mathcal Q_{00})}
+\end{equation}
+where
+\begin{equation} \label{eq:one-point.action}
+ \begin{aligned}
+ &\mathcal S_0(\mathcal Q_{00})
+ =\hat\beta_0E_0-r^{00}_\mathrm d\mu_0-\frac12\hat\mu_0(1-c^{00}_\mathrm d)+\mathcal D(\mu_0)\\
+ &\quad+\frac1m\bigg\{
+ \frac12\sum_{ab}^m\left[
+ \hat\beta_1^2f(C^{00}_{ab})+(2\hat\beta_1R^{00}_{ab}-D^{00}_{ab})f'(C^{00}_{ab})+(R_{ab}^{00})^2f''(C_{ab}^{00})
+ \right]+\frac12\log\det\begin{bmatrix}C^{00}&R^{00}\\R^{00}&D^{00}\end{bmatrix}
+\bigg\}
+\end{aligned}
+\end{equation}
+is the action for the ordinary, one-point complexity, and remainder is given by
+\begin{equation} \label{eq:two-point.action}
+ \begin{aligned}
+ &\mathcal S(\mathcal Q_{11},\mathcal Q_{01}\mid\mathcal Q_{00})
+ =\hat\beta_1E_1-r^{11}_\mathrm d\mu_1-\frac12\hat\mu_1(1-c^{11}_\mathrm d)+\mathcal D(\mu_1) \\
+ &\quad+\frac1n\sum_b^n\left\{-\frac12\hat\mu_{12}(q-C^{01}_{1b})+\sum_a^m\left[
+ \hat\beta_0\hat\beta_1f(C^{01}_{ab})+(\hat\beta_0R^{01}_{ab}+\hat\beta_1R^{10}_{ab}-D^{01}_{ab})f'(C^{01}_{ab})+R^{01}_{ab}R^{10}_{ab}f''(C^{01}_{ab})
+ \right]\right\}
+ \\
+ &\quad+\frac1n\bigg\{
+ \frac12\sum_{ab}^n\left[
+ \hat\beta_1^2f(C^{11}_{ab})+(2\hat\beta_1R^{11}_{ab}-D^{11}_{ab})f'(C^{11}_{ab})+(R^{11}_{ab})^2f''(C^{11}_{ab})
+ \right]\\
+ &\quad+\frac12\log\det\left(
+ \begin{bmatrix}
+ C^{11}&iR^{11}\\iR^{11}&D^{11}
+ \end{bmatrix}-
+ \begin{bmatrix}
+ C^{01}&iR^{01}\\iR^{10}&D^{01}
+ \end{bmatrix}^T
+ \begin{bmatrix}
+ C^{00}&iR^{00}\\iR^{00}&D^{00}
+ \end{bmatrix}^{-1}
+ \begin{bmatrix}
+ C^{01}&iR^{01}\\iR^{10}&D^{01}
+ \end{bmatrix}
+ \right)
+ \bigg\}
+ \end{aligned}
+\end{equation}
+Because of the structure of this problem in the twin limits of $m$ and $n$ to
+zero, the parameters $\mathcal Q_{00}$ can be evaluated at a saddle point of
+$\mathcal S_0$ alone. This means that these parameters will take the same value
+they take when the ordinary, 1-point complexity is calculated. For a replica
+symmetric complexity of the reference point, this results in
+\begin{align}
+ \hat\beta_0
+ &=-\frac{\mu_0f'(1)+E_0\big(f'(1)+f''(1)\big)}{u_f}\\
+ r_\mathrm d^{00}
+ &=\frac{\mu_0f(1)+E_0f'(1)}{u_f} \\
+ d_\mathrm d^{00}
+ &=\frac1{f'(1)}
+ -\left(
+ \frac{\mu_0f(1)+E_0f'(1)}{u_f}
+ \right)^2
+\end{align}
+where we define for brevity (here and elsewhere) the constants
+\begin{align}
+ u_f=f(1)\big(f'(1)+f''(1)\big)-f'(1)^2
+ &&
+ v_f=f'(1)\big(f''(1)+f'''(1)\big)-f''(1)^2
+\end{align}
+Note that because the coefficients of $f$ must be nonnegative for $f$ to
+be a sensible covariance, both $u_f$ and $v_f$ are strictly positive. Note also
+that $u_f=v_f=0$ if $f$ is a homogeneous polynomial as in the pure models.
+These expressions are invalid for the pure models because $\mu_0$ and $E_0$
+cannot be fixed independently; we would have done the equivalent of inserting
+two identical $\delta$-functions. For the pure models, the terms $\hat\beta_0$ and
+$\hat\beta_1$ must be set to zero in our prior formulae (as if the energy was
+not constrained) and then the saddle point taken.
+
+
+In general, we except the $m\times n$ matrices $C^{01}$, $R^{01}$, $R^{10}$,
+and $D^{01}$ to have constant \emph{rows} of length $n$, with blocks of rows
+corresponding to the \textsc{rsb} structure of the single-point complexity. For
+the scope of this paper, where we restrict ourselves to replica symmetric
+complexities, they have the following form at the saddle point:
+\begin{align} \label{eq:01.ansatz}
+ C^{01}=
+ \begin{subarray}{l}
+ \hphantom{[}\begin{array}{ccc}\leftarrow&n&\rightarrow\end{array}\hphantom{\Bigg]}\\
+ \left[
+ \begin{array}{ccc}
+ q&\cdots&q\\
+ 0&\cdots&0\\
+ \vdots&\ddots&\vdots\\
+ 0&\cdots&0
+ \end{array}
+ \right]\begin{array}{c}
+ \\\uparrow\\m-1\\\downarrow
+ \end{array}
+\end{subarray}
+ &&
+ R^{01}
+ =\begin{bmatrix}
+ r_{01}&\cdots&r_{01}\\
+ 0&\cdots&0\\
+ \vdots&\ddots&\vdots\\
+ 0&\cdots&0
+ \end{bmatrix}
+ &&
+ R^{10}
+ =\begin{bmatrix}
+ r_{10}&\cdots&r_{10}\\
+ 0&\cdots&0\\
+ \vdots&\ddots&\vdots\\
+ 0&\cdots&0
+ \end{bmatrix}
+ &&
+ D^{01}
+ =\begin{bmatrix}
+ d_{01}&\cdots&d_{01}\\
+ 0&\cdots&0\\
+ \vdots&\ddots&\vdots\\
+ 0&\cdots&0
+ \end{bmatrix}
+\end{align}
+where only the first row is nonzero. The other entries, which correspond to the
+completely uncorrelated replicas in an \textsc{rsb} picture, are all zero
+because uncorrelated vectors on the sphere are orthogonal.
+
+The inverse of block hierarchical matrix is still a block hierarchical matrix, since
+\begin{equation}
+ \begin{bmatrix}
+ C^{00}&iR^{00}\\iR^{00}&D^{00}
+ \end{bmatrix}^{-1}
+ =
+ \begin{bmatrix}
+ (C^{00}D^{00}+R^{00}R^{00})^{-1}D^{00} & -i(C^{00}D^{00}+R^{00}R^{00})^{-1}R^{00} \\
+ -i(C^{00}D^{00}+R^{00}R^{00})^{-1}R^{00} & (C^{00}D^{00}+R^{00}R^{00})^{-1}C^{00}
+ \end{bmatrix}
+\end{equation}
+Because of the structure of the 01 matrices, the volume element will depend
+only on the diagonals of the matrices in this inverse block matrix. If we define
+\begin{align}
+ \tilde c_\mathrm d^{00}&=[(C^{00}D^{00}+R^{00}R^{00})^{-1}C^{00}]_{\text d} \\
+ \tilde r_\mathrm d^{00}&=[(C^{00}D^{00}+R^{00}R^{00})^{-1}R^{00}]_{\text d} \\
+ \tilde d_\mathrm d^{00}&=[(C^{00}D^{00}+R^{00}R^{00})^{-1}D^{00}]_{\text d}
+\end{align}
+as the diagonals of the blocks of the inverse matrix, then the result of the product is
+\begin{equation}
+ \begin{aligned}
+ & \begin{bmatrix}
+ C^{01}&iR^{01}\\iR^{10}&D^{01}
+ \end{bmatrix}^T
+ \begin{bmatrix}
+ C^{00}&iR^{00}\\iR^{00}&D^{00}
+ \end{bmatrix}^{-1}
+ \begin{bmatrix}
+ C^{01}&iR^{01}\\iR^{10}&D^{01}
+ \end{bmatrix} \\
+ &\qquad=\begin{bmatrix}
+ q^2\tilde d_\mathrm d^{00}+2qr_{10}\tilde r^{00}_\mathrm d-r_{10}^2\tilde d^{00}_\mathrm d
+ &
+ i\left[d_{01}(r_{10}\tilde c^{00}_\mathrm d-q\tilde r^{00}_\mathrm d)+r_{01}(r_{10}\tilde r^{00}_\mathrm d+q\tilde d^{00}_\mathrm d)\right]
+ \\
+ i\left[d_{01}(r_{10}\tilde c^{00}_\mathrm d-q\tilde r^{00}_\mathrm d)+r_{01}(r_{10}\tilde r^{00}_\mathrm d+q\tilde d^{00}_\mathrm d)\right]
+ &
+ d_{01}^2\tilde c^{00}_\mathrm d+2r_{01}d_{01}\tilde r^{00}_\mathrm d-r_{01}^2\tilde d^{00}_\mathrm d
+ \end{bmatrix}
+ \end{aligned}
+\end{equation}
+where each block is a constant $n\times n$ matrix. Because the matrices
+$C^{00}$, $R^{00}$, and $D^{00}$ are diagonal in the replica symmetric case,
+the diagonals of the blocks above take a simple form:
+\begin{align}
+ \tilde c_\mathrm d^{00}=f'(1) &&
+ \tilde r_\mathrm d^{00}=r^{00}_\mathrm df'(1) &&
+ \tilde d_\mathrm d^{00}=d^{00}_\mathrm df'(1)
+\end{align}
+Once these expressions are inserted into the complexity, the limits of $n$ and
+$m$ to zero can be taken, and the parameters from $D^{01}$ and $D^{11}$ can be
+extremized explicitly.
\section{Details of calculation for the isolated eigenvalue}
\label{sec:eigenvalue-details}