path: root/frsb_kac-rice.tex

\documentclass[fleqn]{article}

\usepackage{fullpage,amsmath,amssymb,latexsym}

\begin{document}
\title{Full solution of the Kac-Rice problem for mean-field models}
\maketitle
\begin{abstract}
    We derive the general solution for the computation of saddle points
    of complex mean-field landscapes. The solution incorporates Parisi's solution
    for equilibrium, as it should.
\end{abstract}
\section{Introduction}


Although the Bray-Moore computation for the SK model was the first application of 
some replica symmetry breaking scheme, it turned out that the problem has been open 
ever since.



to this date the program has been only complete for a subset of models 

here we present what we believe is the general scheme


\section{The model}

Here we consider, for definiteness, the `toy' model introduced by M\'ezard and Parisi



\section{Equilibrium}

Here we review the equilibrium solution.

\begin{equation}
  \beta F=\frac12\lim_{n\to0}\frac1n\left(\beta^2\sum_{ab}f(Q_{ab})+\log\det Q\right)-\frac12\log S_\infty
\end{equation}
$\log S_\infty=1+\log2\pi$.
\begin{align*}
  \beta F=
  -\frac12\log S_\infty+
  \frac12\lim_{n\to0}\frac1n\left(\beta^2nf(1)+\beta^2\sum_{i=0}^kn(x_i-x_{i+1})f(q_i)
  +\log\left[
    \frac{
      1+\sum_{i=0}^k(x_i-x_{i+1})q_i
    }{
      1+\sum_{i=1}^k(x_i-x_{i+1})q_i-x_1q_0
    }
  \right]\right.\\
  +\frac n{x_1}\log\left[
      1+\sum_{i=1}^k(x_i-x_{i+1})q_i-x_1q_0
  \right]\\
  \left.+\sum_{j=1}^kn(x_{j+1}^{-1}-x_j^{-1})\log\left[
      1+\sum_{i=j+1}^k(x_i-x_{i+1})q_i-x_{j+1}q_j
  \right]
\right)
\end{align*}

\begin{align*}
  \lim_{n\to0}\frac1n
  \log\left[
    \frac{
      1+\sum_{i=0}^k(x_i-x_{i+1})q_i
    }{
      1+\sum_{i=1}^k(x_i-x_{i+1})q_i-x_1q_0
    }
  \right]
  &=
  \lim_{n\to0}\frac1n
  \log\left[
    \frac{
      1+\sum_{i=0}^k(x_i-x_{i+1})q_i
    }{
      1+\sum_{i=0}^k(x_i-x_{i+1})q_i-nq_0
    }
  \right] \\
  &=q_0\left(1+\sum_{i=0}^k(x_i-x_{i+1})q_i\right)^{-1}
\end{align*}


\begin{align*}
  \beta F=
  -\frac12\log S_\infty+
  \frac12\left(\beta^2f(1)+\beta^2\sum_{i=0}^k(x_i-x_{i+1})f(q_i)
  +q_0\left(1+\sum_{i=0}^k(x_i-x_{i+1})q_i\right)^{-1}\right. \\
  +\frac1{x_1}\log\left[
      1+\sum_{i=1}^{k}(x_i-x_{i+1})q_i-x_1q_0
  \right]\\
  \left.+\sum_{j=1}^k(x_{j+1}^{-1}-x_j^{-1})\log\left[
      1+\sum_{i=j+1}^{k}(x_i-x_{i+1})q_i-x_{j+1}q_j
  \right]
\right)
\end{align*}
$q_0=0$
\begin{align*}
  \beta F=
  -\frac12\log S_\infty+
  \frac12\left(\beta^2f(1)+\beta^2\sum_{i=0}^k(x_i-x_{i+1})f(q_i)
  +\frac1{x_1}\log\left[
      1+\sum_{i=1}^{k}(x_i-x_{i+1})q_i
  \right]\right.\\
  \left.+\sum_{j=1}^k(x_{j+1}^{-1}-x_j^{-1})\log\left[
      1+\sum_{i=j+1}^{k}(x_i-x_{i+1})q_i-x_{j+1}q_j
  \right]
\right)
\end{align*}
$x_i=\tilde x_ix_k$, $x_k=y/\beta$, $q_k=1-z/\beta$
\begin{align*}
  \beta F=
  -\frac12\log S_\infty+
  \frac12\left(\beta^2f(1)+\beta^2(y\beta^{-1}-1)f(1-z\beta^{-1})+y\beta\sum_{i=0}^{k-1}(\tilde x_i-\tilde x_{i+1})f(q_i)\right. \\
  +\frac\beta{\tilde x_1 y}\log\left[
    y\sum_{i=1}^{k-1}(\tilde x_i-\tilde x_{i+1})q_i+y+z-yz/\beta
  \right]\\
  +\sum_{j=1}^{k-1}\frac\beta y(\tilde x_{j+1}^{-1}-\tilde x_j^{-1})\log\left[
    y\sum_{i=j+1}^{k-1}(\tilde x_i-\tilde x_{i+1})q_i+y+z-yz/\beta-y\tilde x_{j+1}q_j
\right]\\
    \left.-\frac\beta{\tilde x_1 y}\log\beta-\sum_{j=1}^{k-1}\frac\beta y(\tilde x_{j+1}^{-1}-\tilde x_j^{-1})\log\beta+(1-\beta y^{-1})\log\left[
      z/\beta
  \right]
\right)
\end{align*}
\begin{align*}
  \lim_{\beta\to\infty}F=
  \frac12\left(yf(1)+zf'(1)+y\sum_{i=0}^{k-1}(\tilde x_i-\tilde x_{i+1})f(q_i)
  +\frac1{\tilde x_1 y}\log\left[
    y\sum_{i=1}^{k-1}(\tilde x_i-\tilde x_{i+1})q_i+y+z
  \right]\right.\\
  \left.+\sum_{j=1}^{k-1}\frac1 y(\tilde x_{j+1}^{-1}-\tilde x_j^{-1})\log\left[
    y\sum_{i=j+1}^{k-1}(\tilde x_i-\tilde x_{i+1})q_i+y+z-y\tilde x_{j+1}q_j
    \right]
    -\frac1y\log z
\right)
\end{align*}
$F$ is a $k-1$ RSB ansatz with all eigenvalues scaled by $y$ and shifted by $z$. $\tilde x_0=0$ and $\tilde x_k=1$.
\begin{equation} \label{eq:ground.state.free.energy}
  \lim_{\beta\to\infty}F=\lim_{n\to0}\frac1n\frac12\left(nzf'(1)+y\sum_{ab}f(\tilde Q_{ab})+\frac1y\log\det(yz^{-1}\tilde Q+I)
  \right)
\end{equation}




\section{Kac-Rice}

\subsection{The replicated Kac-Rice problem}

\begin{eqnarray}
&=& \Pi_a \delta(Eq_a)  \;  \Pi_a \left| \det_a( )\right| \delta(E_a-E(s_a))\nonumber\\
&\rightarrow& \overline{\Pi_a \delta(Eq_a)}  \; \overline{ \Pi_a \left| \det_a( )\right|\delta(E_a-E(s_a))}\nonumber\\
\end{eqnarray}

the question of independence

all saddles versus only minima

The parameters:
\begin{eqnarray}
Q_{ab}&=&\nonumber\\
R_{ab}&=&\nonumber\\
D_{ab}&=&
\end{eqnarray}




\section{Replicated action}
\begin{align*}
  \Sigma
  =-\epsilon\hat\epsilon+\lim_{n\to0}\frac1n\left(
    \sum_a\mu(F_{aa}-R_{aa})
    +\frac12\sum_{ab}\left[
      \hat\epsilon^2f(Q_{ab})+2\hat\epsilon R_{ab}f'(Q_{ab})
      +D_{ab}f'(Q_{ab})+R_{ab}^2f''(Q_{ab})-F_{ab}^2f''(Q_{ab})
    \right]\right.\\\left.
  +\frac12\log\det\begin{bmatrix}Q&-iR\\-iR&-D\end{bmatrix}
  -\log\det F
  \right)
\end{align*}
\[
  0=\frac{\partial\Sigma}{\partial R_{ab}}
  =-\mu\delta_{ab}+\hat\epsilon f'(Q_{ab})+R_{ab}f''(Q_{ab})+\sum_c(R^2-DQ)^{-1}_{ac}R_{cb}
\]
\[
  0=\frac{\partial\Sigma}{\partial D_{ab}}
  =\frac12 f'(Q_{ab})-\frac12\sum_c(R^2-DQ)^{-1}_{ac}Q_{cb}
\]
The second equation implies
\[
  (R^2-DQ)^{-1}=Q^{-1}f'(Q)
\]

\section{Replica ansatz}

\subsection{Motivation}

One may write 
\begin{equation}
    {\bf Q}_{ab}(\bar \theta \theta, \bar \theta ' \theta')=
Q_{ab} + \bar \theta ... R_{ab} +\bar \theta \theta \bar \theta ' \theta' D_{ab}
\end{equation}

The variables $\bar \theta \theta$ and $\bar \theta ' \theta'$ play
the role of `times' in a superspace treatment. We have a long experience of
making an ansatz for replicated quantum problems. The analogy strongly
suggests that only the diagonal ${\bf Q}_{aa}$ depend on the $\theta$'s. This boils down
to putting:
\begin{eqnarray}
Q_{ab}&=& {\mbox{ a Parisi matrix}}\nonumber\\
R_{ab}&=R \delta_{ab}&\nonumber\\
D_{ab}&=& D \delta_{ab}
\end{eqnarray}
Not surprisingly, this ansatz closes, as we shall see.

\subsection{Solution}


Insert the diagonal ansatz $R=R_dI$, $D=D_dI$. Then
\[
  0=(R_df''(1)-\mu)I+\hat\epsilon f'(Q)+R_d(R_d^2I-D_dQ)^{-1}
  =(R_df''(1)-\mu)I+\hat\epsilon f'(Q)+R_dQ^{-1}f'(Q)
\]
and
\[
  Q^{-1}f'(Q)=(I+D_df'(Q))/R_d^2
\]
Substituting the second into the first, we have
\[
  0=(R_df''(1)-\mu)I+\hat\epsilon f'(Q)+\frac1{R_d}(I+D_df'(Q))
\]
\[
  0=(R_df''(1)-\mu+R_d^{-1})I+(\hat\epsilon+D_d/R_d)f'(Q)
\]
The only way for this equation to be satisfied off the diagonal for nontrivial $Q$ is for $D_d=-R_d\hat\epsilon$. We therefore have
\begin{align*}
  \Sigma
  =-\epsilon\hat\epsilon+\lim_{n\to0}\frac1n\left(
    n\mu(F_d-R_d)+\frac12n\left[
      \hat\epsilon R_df'(1)+R_d^2f''(1)-F_d^2f''(1)
    \right]
    +\frac12\sum_{ab}
      \hat\epsilon^2f(Q_{ab})
    ]\right.\\\left.
    +\frac12\log\det(\hat\epsilon R_d^{-1} Q+I)
    +n\log R_d
  -n\log F_d
  \right)
\end{align*}
Taking the saddle with respect to $\mu$ and $F_d$ yields
\[
  F_d=R_d
\]
\[
  \mu=R_d^{-1}(1+R_d^2f''(1))
\]
and gives
\begin{align*}
  \Sigma
  =-\epsilon\hat\epsilon+\lim_{n\to0}\frac1n\frac12\left(
    n
      \hat\epsilon R_df'(1)
    +\hat\epsilon^2\sum_{ab}
      f(Q_{ab})
    +\log\det(\hat\epsilon R_d^{-1} Q+I)
  \right)
\end{align*}
Finally, setting $0=\Sigma$ gives
\[
  \epsilon
  =\lim_{n\to0}\frac1n\frac12\left(nR_df'(1)+\hat\epsilon\sum_{ab}
      f(Q_{ab})
      +\frac1{\hat\epsilon}\log\det(\hat\epsilon R_d^{-1} Q+I)
  \right)
\]
which is precisely \eqref{eq:ground.state.free.energy} with $R_d=z$ and $\hat\epsilon=y$. 

{\em Therefore, a $(k-1)$-RSB ansatz in Kac-Rice will predict the correct ground state energy for a model whose equilibrium state at small temperatures is $k$-RSB.}

\subsection{Full}

\begin{align*}
  \lim_{n\to0}\frac1n\log\det(\hat\epsilon R_d^{-1} Q+I)
  =x_1^{-1}\log\left(\hat\epsilon R_d^{-1}(1-\bar q_k)+1\right)+\int_{q_0^+}^{q_{k-1}}dq\,\mu(q)\log\left[\hat\epsilon R_d^{-1}\lambda(q)+1\right]
\end{align*}
where
\[
  \mu(q)=\frac{\partial x^{-1}(q)}{\partial q}
\]
Integrating by parts,
\begin{align*}
  \lim_{n\to0}\frac1n\log\det(\hat\epsilon R_d^{-1} Q+I)
  &=x_1^{-1}\log\left(\hat\epsilon R_d^{-1}(1-\bar q_k)+1\right)+\left[x^{-1}(q)\log[\hat\epsilon R_d^{-1}\lambda(q)+1]\right]_{q=q_0^+}^{q=q_{k-1}}-\frac{\hat\epsilon}{R_d}\int_{q_0^+}^{q_{k-1}}dq\,\frac{\lambda'(q)}{x(q)}\frac1{\hat\epsilon R_d^{-1}\lambda(q)+1}\\
  &=\log[\hat\epsilon R_d^{-1}\lambda(q_{k-1})+1]+\frac{\hat\epsilon}{R_d}\int_{q_0^+}^{q_{k-1}}dq\,\frac1{\hat\epsilon R_d^{-1}\lambda(q)+1}
\end{align*}

\begin{align*}
  \Sigma
  =-\epsilon\hat\epsilon+
    \frac12\hat\epsilon R_df'(1)
    +\frac12\int_0^1dq\,\left[
      \hat\epsilon^2\lambda(q)f''(q)
      +\frac1{\lambda(q)+R_d/\hat\epsilon}
    \right]
\end{align*}
for $\lambda$ concave, monotonic,  $\lambda(1)=0$, and $\lambda'(1)=-1$
\[
  0=\frac{\delta\Sigma}{\delta\lambda(q)}=\frac12\hat\epsilon^2f''(q)-\frac12\frac1{(\lambda(q)+R_d/\hat\epsilon)^2}
\]
\[
  \lambda^*(q)=\frac1{\hat\epsilon}\left[f''(q)^{-1/2}-R_d\right]
\]

We suppose that solutions are given by
\begin{equation}
  \lambda(q)=\begin{cases}
    \lambda^*(q) & q<q^* \\
    1-q          & q\geq q^*
  \end{cases}
\end{equation}
where $1-q$ guarantees the boundary conditions at $q=1$, and corresponds to the 0RSB or annealed solutions (annealed Kac-Rice is recovered by substituting in $1-q$ for $\lambda$). We will need to require that $1-q^*=\lambda^*(q^*)$, i.e., continuity.

Inserting this into the complexity, we find
\begin{align*}
  \Sigma
  &=-\epsilon\hat\epsilon+\frac12\hat\epsilon R_df'(1)
  +\frac12\int_0^{q^*}dq\left[
    \hat\epsilon(f''(q)^{-1/2}-R_d)f''(q)+\hat\epsilon f''(q)^{1/2}
  \right]
  +\frac12\int_{q^*}^1dq\left[
    \hat\epsilon^2(1-q)f''(q)+\frac1{q-1+R_d/\hat\epsilon}
  \right] \\
  &=-\epsilon\hat\epsilon+\frac12\hat\epsilon R_d\left[f'(1)-f'(q^*)\right]
  +\hat\epsilon\int_0^{q^*}dq\,f''(q)^{1/2}
  +\frac12\hat\epsilon^2\int_{q^*}^1dq\,
    (1-q)f''(q)
    -\log\left[1-(1-q^*)\hat\epsilon/R_d\right]
\end{align*}
$R_d$ can be extremized now, with
\[
  R_d=\frac12\left(
    (1-q^*)\hat\epsilon\pm\sqrt{
      (1-q^*)\left(
        (1-q^*)\hat\epsilon^2+8/[f'(1)-f'(q^*)]
      \right)
    }
  \right)
\]

This all is for $\mu=\mu^*$, which counts the dominant saddles. We can also count by fixed macroscopic index $\mu$ by leaving it unoptimized in the complexity. This gives
\[
  F_d=\frac1{2f''(1)}\left[\mu\pm\sqrt{\mu^2-4f''(1)}\right]
\]
and
\begin{align*}
  \Sigma
  =-\epsilon\hat\epsilon+
    \frac12\hat\epsilon R_df'(1)
    +\frac12\int_0^1dq\,\left[
      \hat\epsilon^2\lambda(q)f''(q)
      +\frac1{\lambda(q)+R_d/\hat\epsilon}
    \right]-\mu R_d+\frac12R_d^2f''(1)+\log R_d\\
    +\operatorname{Re}\left\{\frac12\left(1+\frac\mu{2f''(1)}\left(\mu\pm\sqrt{\mu^2-4f''(1)}\right)\right)-\log\left(\frac1{2f''(1)}\left(\mu\pm\sqrt{\mu^2-4f''(1)}\right)\right)\right\}
\end{align*}


\section{Overview of the landscape}
For the full model minima are always dominated exponentially by saddles, whose
index density  goes smoothly down to zero with energy density. (I hope) 
The meaning of the parameter $Q_{ab}$ is  ????????

\section{Conclusion}


\end{document}