Some consistency fixes, including a lot in the GOE example.

1 files changed, 108 insertions, 83 deletions

diff --git a/marginal.tex b/marginal.tex
index 26a88fe..45d65dc 100644
--- a/marginal.tex
+++ b/marginal.tex
@@ -202,12 +202,14 @@ 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{min}(B+\mu I)=\lambda^*}
-  =\overline{\delta\big(N\lambda^*-N\lambda_\mathrm{min}(B+\mu I)\big)}
+  \begin{aligned}
+    e^{NG_{\lambda^*}(\mu)}
+    &=P\big(\lambda_\mathrm{min}(B+\mu I)=\lambda^*\big) \\
+    &=\overline{\delta\big(N\lambda^*-N\lambda_\mathrm{min}(B+\mu I)\big)}
+  \end{aligned}
 \end{equation}
 where the overline is the average over $B$, and we have defined the large
-deviation function $G_\sigma(\mu)$.
+deviation function $G_{\lambda^*}(\mu)$.
 Using the representation of $\lambda_\mathrm{min}$ defined in \eqref{eq:λmin}, we have
 \begin{widetext}
 \begin{equation}
@@ -225,10 +227,10 @@ representation, we have
   =\overline{\lim_{\beta\to\infty}\lim_{m\to0}\int d\hat\lambda\prod_{\alpha=1}^m\left[d\mathbf s^\alpha\,\delta(N-\|\mathbf s^\alpha\|^2)\right]
   \exp\left\{-\beta\sum_{\alpha=1}^m(\mathbf s^\alpha)^T(B+\mu I)\mathbf s^\alpha+\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
+having introduced the auxiliary parameter $\hat\lambda$ in the Fourier representation of
+the $\delta$-function. The whole expression, so transformed, is an
 exponential integral linear in the matrix $B$. Taking the average over $B$, we
-have
+find
 \begin{equation}
   \begin{aligned}
   &e^{NG_{\lambda^*}(\mu)}
@@ -241,27 +243,27 @@ have
 \end{equation}
 \end{widetext}
 We make the Hubbard--Stratonovich transformation to the matrix field
-$Q_{ab}=\frac1N\mathbf s_a^T\mathbf s_b$. This gives
+$Q^{\alpha\beta}=\frac1N\mathbf s^\alpha\cdot\mathbf s^\beta$. This gives
 \begin{equation}
   e^{NG_{\lambda^*}(\mu)}
   =\lim_{\beta\to\infty}\lim_{m\to0}\int d\hat\lambda\,dQ\,
-  e^{N\mathcal U_\mathrm{GOE}(\hat\lambda,Q\mid\mu,\lambda^*,\beta)}
+  e^{N\mathcal U_\mathrm{GOE}(\hat\lambda,Q\mid\beta,\lambda^*,\mu)}
 \end{equation}
 where the effective action is given by
 \begin{equation}
   \begin{aligned}
-    &\mathcal U_\textrm{GOE}(\hat\lambda, Q\mid\lambda^*,\mu,\beta)
+    &\mathcal U_\textrm{GOE}(\hat\lambda, Q\mid\beta,\lambda^*,\mu)
     =\hat\lambda(\lambda^*-\mu)-m\beta\mu \\
-    &+\sigma^2\left[\beta^2\sum_{ab}^mQ_{ab}^2
-        +2\beta\hat\lambda\sum_a^mQ_{1a}^2
+    &+\sigma^2\left[\beta^2\sum_{\alpha\gamma}^m(Q^{\alpha\gamma})^2
+        +2\beta\hat\lambda\sum_\alpha^m(Q^{1\alpha})^2
       +\hat\lambda^2
     \right]+\frac12\log\det Q
   \end{aligned}
 \end{equation}
-where $Q_{aa}=1$ because of the spherical constraint. We can evaluate this
+and $Q^{\alpha\alpha}=1$ 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
+of its unique coupling with $\hat\lambda$. The resulting matrix has the form
 \begin{equation} \label{eq:Q.structure}
   Q=\begin{bmatrix}
     1&\tilde q_0&\tilde q_0&\cdots&\tilde q_0\\
@@ -271,16 +273,17 @@ of its unique coupling with $\hat\lambda$. This gives
     \tilde q_0&q_0&q_0&\cdots&1
   \end{bmatrix}
 \end{equation}
-with $\sum_{ab}Q_{ab}^2=m+2(m-1)\tilde q_0^2+(m-1)(m-2)q_0^2$, $\sum_aQ_{1a}^2=1+(m-1)\tilde q_0^2$,
+The relevant expressions in the effective action produce $\sum_{\alpha\beta}(Q^{\alpha\beta})^2=m+2(m-1)\tilde q_0^2+(m-1)(m-2)q_0^2$, $\sum_\alpha(Q^{1\alpha})^2=1+(m-1)\tilde q_0^2$,
 and
 \begin{equation}
   \log\det Q=(m-2)\log(1-q_0)+\log(1+(m-2)q_0-(m-1)\tilde q_0^2)
 \end{equation}
-Inserting these expressions and taking the limit of $m$ to zero, we find
+Inserting these expressions into the effective action and taking the limit of
+$m$ to zero, we arrive at
 \begin{equation}
   e^{NG_{\lambda^*}(\mu)}
   =\lim_{\beta\to\infty}\int d\hat\lambda\,dq_0\,d\tilde q_0\,
-  e^{N\mathcal U_\textrm{GOE}(\hat\lambda,q_0,\tilde q_0\mid\mu,\lambda^*,\beta)}
+  e^{N\mathcal U_\textrm{GOE}(\hat\lambda,q_0,\tilde q_0\mid\beta,\lambda^*,\mu)}
 \end{equation}
 with the effective action
 \begin{equation}
@@ -305,32 +308,37 @@ 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 U_\textrm{GOE}(\hat\lambda,y,\Delta z\mid\mu,\lambda^*,\infty)
-    &=\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)
+    &\mathcal U_\textrm{GOE}(\hat\lambda,y,\Delta z\mid\infty,\lambda^*,\mu)
+    =\hat\lambda(\lambda^*-\mu) \\
+    &\qquad+\sigma^2\big[
+      \hat\lambda^2+4(y+\Delta z)
+    \big]
+    +\frac12\log\left(1-\frac{2\Delta z}{y^2}\right)
   \end{aligned}
 \end{equation}
 Extremizing this action over the new parameters $y$, $\Delta z$, and $\hat\lambda$, we have
 \begin{align}
-  \hat\lambda=-\frac1\sigma\sqrt{\frac{(\mu+\lambda^*)^2}{(2\sigma)^2}-1}
+  \hat\lambda&=\frac1\sigma\sqrt{\left(\frac{\mu-\lambda^*}{2\sigma}\right)^2-1}
   \\
-  y=\frac1{2\sigma}\left(\frac{\mu+\lambda^*}{2\sigma}-\sqrt{\frac{(\mu+\lambda^*)^2}{(2\sigma)^2}-1}\right)
-  &\\
-  \Delta z=\frac1{4\sigma^2}\left(1-\frac{\mu+\lambda^*}{2\sigma}\left(\frac{\mu+\lambda^*}{2\sigma}-\sqrt{\frac{(\mu+\lambda^*)^2}{(2\sigma)^2}-1}\right)\right)
+  y&=\frac1{2\sigma}\left[
+    \frac{\mu-\lambda^*}{2\sigma}+\sqrt{\left(\frac{\mu-\lambda^*}{2\sigma}\right)^2-1}
+  \right]^{-1}
+  \\
+  \Delta z&=\frac1{4\sigma^2}\left[
+    \left(\frac{\mu-\lambda^*}{2\sigma}\right)^2-1
+    -\frac{\mu-\lambda^*}{2\sigma}\sqrt{\left(\frac{\mu-\lambda^*}{2\sigma}\right)^2-1}
+  \right]
 \end{align}
 Inserting this solution into $\mathcal S_\infty$ we find
 \begin{equation} \label{eq:goe.large.dev}
   \begin{aligned}
     &G_{\lambda^*}(\mu)
-    =\mathop{\textrm{extremum}}_{y,\Delta z,\hat\lambda}
-    \mathcal U_\mathrm{GOE}(y,\Delta z,\hat\lambda\mid\mu,\lambda^*,\infty) \\
-    &=-\tfrac{\mu+\lambda^*}{2\sigma}\sqrt{\Big(\tfrac{\mu+\lambda^*}{2\sigma}\Big)^2-1}
-    +\log\left(
-      \tfrac{\mu+\lambda^*}{2\sigma}+\sqrt{\Big(\tfrac{\mu+\lambda^*}{2\sigma}\Big)^2-1}
-    \right)
+    =\mathop{\textrm{extremum}}_{\hat\lambda,y,\Delta z}
+    \mathcal U_\mathrm{GOE}(\hat\lambda,y,\Delta z\mid\infty,\lambda^*,\mu) \\
+    &=-\frac{\mu-\lambda^*}{2\sigma}\sqrt{\left(\frac{\mu-\lambda^*}{2\sigma}\right)^2-1} \\
+    &\hspace{5em}-\log\left[
+      \frac{\mu-\lambda^*}{2\sigma}-\sqrt{\left(\frac{\mu-\lambda^*}{2\sigma}\right)^2-1}
+    \right]
   \end{aligned}
 \end{equation}
 This function is plotted in Fig.~\ref{fig:large.dev} for $\lambda^*=0$. For $\mu<2\sigma$ $G_{0}(\mu)$ has an
@@ -1376,7 +1384,7 @@ taking the zero-temperature limit, we find
     for $\alpha=\frac32$ and $f(q)=q^2+q^3$. The ground state energy
     $E_\mathrm{gs}$ and the threshold energy $E_\mathrm{th}$ are marked on the
     plot.
-  }
+  } \label{fig:ls.complexity}
 \end{figure}
 
 \section{Conclusions}
@@ -1572,24 +1580,28 @@ fixing the trace of the Hessian, this gives
 \end{align}
 so that the differential form of the symmetry is
 \begin{equation}
-  \mathcal D=\bar{\pmb\eta}\frac\partial{\partial\mathbf x}
-  -i\hat\beta\bar{\pmb\eta}\frac\partial{\partial\hat{\mathbf x}}
-  -i\hat{\mathbf x}\frac\partial{\partial\pmb\eta}
+  \mathcal D=\bar{\pmb\eta}\cdot\frac\partial{\partial\mathbf x}
+  -i\hat\beta\bar{\pmb\eta}\cdot\frac\partial{\partial\hat{\mathbf x}}
+  -i\hat{\mathbf x}\cdot\frac\partial{\partial\pmb\eta}
 \end{equation}
 The Ward identities associated with this symmetry give rise to relationships among the order parameters. These identities are
 \begin{align}
-  0=\frac1N\mathcal D\langle\mathbf x_a^T\pmb\eta_b\rangle
-  =\frac1N\left[
-    \langle\bar{\pmb\eta}_a^T\pmb\eta_b\rangle-
-    i\langle\mathbf x_a^T\hat{\mathbf x}_b\rangle
-  \right]
-  =G_{ab}+R_{ab} \\
-  0=\frac iN\mathcal D\langle\hat{\mathbf x}_a^T\pmb\eta_b\rangle
-  =\frac1N\left[
-    \hat\beta\langle\bar{\pmb\eta}_a^T\pmb\eta_b\rangle
-    +\langle\hat{\mathbf x}_a^T\hat{\mathbf x}_b\rangle
-  \right]
-  =\hat\beta G_{ab}+D_{ab}
+  \begin{aligned}
+    0&=\frac1N\mathcal D\langle\mathbf x_a\cdot\pmb\eta_b\rangle
+    =\frac1N\left[
+      \langle\bar{\pmb\eta}_a\cdot\pmb\eta_b\rangle-
+      i\langle\mathbf x_a\cdot\hat{\mathbf x}_b\rangle
+    \right] \\
+     &=G_{ab}+R_{ab}
+  \end{aligned} \\
+  \begin{aligned}
+    0&=\frac iN\mathcal D\langle\hat{\mathbf x}_a\cdot\pmb\eta_b\rangle
+    =\frac1N\left[
+      \hat\beta\langle\bar{\pmb\eta}_a\cdot\pmb\eta_b\rangle
+      +\langle\hat{\mathbf x}_a\cdot\hat{\mathbf x}_b\rangle
+    \right] \\
+     &=\hat\beta G_{ab}+D_{ab}
+  \end{aligned}
 \end{align}
 These identities establish $G_{ab}=-R_{ab}$ and $D_{ab}=\hat\beta R_{ab}$,
 allowing elimination of the matrices $G$ and $D$ in favor of $R$. Fixing the
@@ -1600,74 +1612,85 @@ trace to $\mu$ explicitly breaks this symmetry, and the simplification is lost.
 
 In this appendix we derive an expression for the asymptotic spectral density in
 the two-sphere multispherical spin glass that we describe in Section
-\ref{sec:multispherical}. \cite{Livan_2018_Introduction}
+\ref{sec:multispherical}. We use a typical approach of employing replicas to
+compute the resolvent \cite{Livan_2018_Introduction}. The resolvent for the
+Hessian of the multispherical model is given by an integral over $\mathbf
+y=[\mathbf y^{(1)},\mathbf y^{(2)}]\in\mathbb R^{2N}$ as
+\begin{widetext}
 \begin{equation}
   \begin{aligned}
-    &G(\lambda)
-    =\lim_{n\to0}\int\|\mathbf y_1\|^2\,\prod_{a=1}^nd\mathbf y_a\,
+    G(\lambda)
+    &=\lim_{n\to0}\int\|\mathbf y_1\|^2\,\prod_{a=1}^nd\mathbf y_a\,
     \exp\left\{
-      -\frac12\mathbf y_a^T(\operatorname{Hess}H(\mathbf x,\pmb\omega)+\lambda I)\mathbf y_a
+      -\frac12\mathbf y_a^T(\operatorname{Hess}H(\mathbf x,\pmb\omega)-\lambda I)\mathbf y_a
     \right\} \\
     &
-    =\lim_{n\to0}\int\big(\|\mathbf y_1^{(1)}\|^2+\|\mathbf y_1^{(2)}\|^2\big)\,\prod_{a=1}^nd\mathbf y_a\, \\
-    &\times\exp\left\{
+    =\lim_{n\to0}\int\big(\|\mathbf y_1^{(1)}\|^2+\|\mathbf y_1^{(2)}\|^2\big)\,\prod_{a=1}^nd\mathbf y_a\,
+    \exp\left\{
       -\frac12\begin{bmatrix}\mathbf y_a^{(1)}\\\mathbf y_a^{(2)}\end{bmatrix}^T
       \left(
         \begin{bmatrix}
           \operatorname{Hess}H_1(\mathbf x^{(1)},\omega_1) & -\epsilon \\
           -\epsilon & \operatorname{Hess}H_2(\mathbf x^{(2)},\omega_2)
         \end{bmatrix}
-        +\lambda I
+        -\lambda I
       \right)\begin{bmatrix}\mathbf y_a^{(1)}\\\mathbf y_a^{(2)}\end{bmatrix}
-    \right\} \\
+    \right\}
   \end{aligned}
 \end{equation}
-If $Y_{ab}^{(ik)}=\frac1N\mathbf y_a^{(i)}\cdot\mathbf y_b^{(j)}$ is the matrix
-of overlaps of the $\mathbf y$, then a short and standard calculation yields
+If $Y_{ab}^{(ij)}=\frac1N\mathbf y_a^{(i)}\cdot\mathbf y_b^{(j)}$ is the matrix
+of overlaps of the vectors $\mathbf y$, then a short and standard calculation involving the average over $H$ and the change of variables from $\mathbf y$ to $Y$ yields
 \begin{equation}
-  G(\lambda)=N\lim_{n\to0}\int dY\,(Y_{11}^{(11)}+Y_{11}^{(22)})\,
+  \overline{G(\lambda)}=N\lim_{n\to0}\int dY\,\big(Y_{11}^{(11)}+Y_{11}^{(22)}\big)\,
   e^{nN\mathcal S(Y)}
 \end{equation}
-for
+where the effective action $\mathcal S$ is given by
 \begin{equation}
   \begin{aligned}
     &\mathcal S(Y)
-    =\frac1n\sum_{ab}\left[
+    =\lim_{n\to0}\frac1n\left\{
+      \frac14\sum_{ab}^n\left[
       \sigma_1^2(Y_{ab}^{(11)})^2
       +\sigma_2^2(Y_{ab}^{(22)})^2
-    \right]+\frac12\log\det\begin{bmatrix}
-      Y^{(11)}&Y^{(12)}\\Y^{(12)}&Y^{(22)}
-    \end{bmatrix}\\
-    &+\frac1n\sum_a^n\left[
+    \right]
+    +\frac12\sum_a^n\left[
       2\epsilon Y_{aa}^{(12)}
-      -\omega_1Y_{aa}^{(11)}
-      -\omega_2Y_{aa}^{(22)}
-      +\lambda(Y_{aa}^{(11)}
-      +Y_{aa}^{(22)})
+      +(\lambda-\omega_1)Y_{aa}^{(11)}
+      +(\lambda-\omega_2)Y_{aa}^{(22)}
     \right]
+    +\frac12\log\det\begin{bmatrix}
+      Y^{(11)}&Y^{(12)}\\Y^{(12)}&Y^{(22)}
+    \end{bmatrix}
+    \right\}
   \end{aligned}
 \end{equation}
-Making the replica symmetric ansatz $Y_{ab}^{(ij)}=y^{(ij)}\delta_{ab}$ yields
+\end{widetext}
+Making the replica symmetric ansatz $Y_{ab}^{(ij)}=y^{(ij)}\delta_{ab}$ for
+each of the matrices $Y^{(ij)}$ yields
 \begin{equation}
   \begin{aligned}
-    &\mathcal S(y)
-    =
-      \sigma_1^2(y^{(11)})^2
-      +\sigma_2^2(y^{(22)})^2
-    +\frac12\log(
+    \mathcal S(y)
+    &=
+      \frac14\left[\sigma_1^2(y^{(11)})^2
+      +\sigma_2^2(y^{(22)})^2\right]+\epsilon y^{(12)}
+      \\
+    &
+    \qquad+\frac12\left[(\lambda-\omega_1)y^{(11)}
+    +(\lambda-\omega_2)y^{(22)}\right] \\
+    &
+    \qquad+\frac12\log(
       y^{(11)}y^{(22)}-y^{(12)}y^{(12)}
-      )\\
-    &+2\epsilon y^{(12)}
-      -\omega_1y^{(11)}
-      -\omega_2y^{(22)}
-      +\lambda(y^{(11)}
-      +y^{(22)})
+      )
   \end{aligned}
 \end{equation}
+while the average resolvent becomes
 \begin{equation}
   \overline{G(\lambda)}
   =N(y^{(11)}+y^{(22)})
 \end{equation}
+for $y^{(11)}$ and $y^{(22)}$ evaluated at a saddle point of $\mathcal S$. The
+spectral density at large $N$ is then given by the discontinuity in its
+imaginary point on the real axis, or
 \begin{equation}
   \rho(\lambda)
   =\frac1{i\pi N}
@@ -1840,7 +1863,9 @@ $\hat\beta$, $c_0$, $r$, and $r_0$ is
   \end{aligned}
 \end{equation}
 When $f(0)=0$ as in the cases directly studied in this work, this further
-simplifies as $c_0=r_0=0$.
+simplifies as $c_0=r_0=0$. Extremizing this expression with respect to the
+order parameters $\hat\beta$ and $r$ produces the red line of dominant minima
+shown in Fig.~\ref{fig:ls.complexity}.
 
 \end{widetext}