Split off hierarchical matrix algebra to an appendix.

1 files changed, 92 insertions, 63 deletions

diff --git a/frsb_kac-rice.tex b/frsb_kac-rice.tex
index 11d5822..7c09dca 100644
--- a/frsb_kac-rice.tex
+++ b/frsb_kac-rice.tex
@@ -184,38 +184,14 @@ The free energy, averaged over disorder, is
 Once $n$ replicas are introduced to treat the logarithm, the fields $\mathbf
 s_a$ can be replaced with the new $n\times n$ matrix field $Q_{ab}\equiv(\mathbf
 s_a\cdot\mathbf s_b)/N$. This yields for the free energy
-\begin{equation}
+\begin{equation} \label{eq:eq.free.energy}
   \beta F=-1-\ln2\pi-\frac12\lim_{n\to0}\frac1n\left(\beta^2\sum_{ab}^nf(Q_{ab})+\ln\det Q\right)
 \end{equation}
 which must be evaluated at the $Q$ which maximizes this expression and whose
 diagonal is one. The solution is generally a hierarchical matrix \emph{à la}
-Parisi. Each row of the matrix is the same up to permutation of their elements.
-The so-called $k$RSB solution has $k+2$ different values in each row: $n-x_1$
-of those entries are $q_0$, $x_1-x_2$ of those entries are $q_1$, and so on
-until $x_k-1$ entries of $q_k$, and of course one entry of $1$ (the diagonal).
-Given such a matrix, there are standard ways of producing the sum and
-determinant that appear in the free energy. These formulas are, for an
-arbitrary $k$RSB matrix $A$ with $a_d$ on its diagonal (recall $q_d=1$),
-\begin{equation} \label{eq:replica.sum}
-  \lim_{n\to0}\frac1n\sum_{ab}^nA_{ab}
-  =a_d-\sum_{i=0}^k(x_{i+1}-x_i)a_i
-\end{equation}
-\begin{equation} \label{eq:replica.logdet}
-  \begin{aligned}
-    \lim_{n\to0}\frac1n\ln\det A
-    &=
-    \frac{a_0}{a_d-\sum_{i=0}^k(x_{i+1}-x_i)a_i}
-    +\frac1{x_1}\log\left[
-      a_d-\sum_{i=0}^{k}(x_{i+1}-x_i)a_i
-    \right]\\
-    &\hspace{10pc}-\sum_{j=1}^k(x_j^{-1}-x_{j+1}^{-1})\log\left[
-        a_d-\sum_{i=j}^{k}(x_{i+1}-x_i)a_i-x_ja_j
-    \right]
-  \end{aligned}
-\end{equation}
-where $x_0=0$ and $x_{k+1}=1$. Once these are substituted into the free energy,
-optimizing the anstaz is equivalent to maximizing $F$ with respect to the
-$q_0,\ldots,q_k$ and $x_1,\ldots,x_k$.
+Parisi. The properties of these matrices is reviewed in \S\ref{sec:dict},
+including how to write down \eqref{eq:eq.free.energy} in terms of their
+parameters.
 
 The free energy can also be written in a functional form, which is necessary
 for working with the solution in the limit $k\to\infty$, the so-called full
@@ -561,19 +537,13 @@ matrix products and Hadamard products. In particular, the determinant of the blo
   \ln\det\begin{bmatrix}C&iR\\iR&D\end{bmatrix}=\ln\det(CD+R^2)
 \end{equation}
 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 $C=AB$ 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 $C=A+B$ 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 hierarchical matrices is still a hierarchical matrix. The algebra of
+hierarchical matrices is reviewed in \S\ref{sec:dict}. Using the product formula
+\eqref{eq:replica.prod}, one can write down the hierarchical matrix $CD+R^2$,
+and then compute the $\ln\det$ using the formula \eqref{eq:replica.logdet}.
 
-The extremal conditions are given by differentiating the complexity with respect to its parameters, yielding
+The extremal conditions are given by differentiating the complexity with
+respect to its parameters, yielding
 \begin{align}
   0&=\frac{\partial\Sigma}{\partial\hat\mu}
   =\frac12(c_d-1) \\
@@ -758,7 +728,8 @@ each of its rows, with
   &&
   D\;\leftrightarrow\;[d_d, d(x)]
 \end{align}
-The complexity becomes
+The algebra of hierarchical matrices under this continuous parameterization is
+review in \S\ref{sec:dict}.  The complexity becomes
 \begin{equation}
   \begin{aligned}
     \Sigma(E,\mu^*)
@@ -770,34 +741,18 @@ The complexity becomes
       +\frac12\lim_{n\to0}\frac1n\ln\det(CD+R^2)
   \end{aligned}
 \end{equation}
-The determinant takes more work to treat.
-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 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)
-  -\int_0^xdy\,\big(
-    a(x)-a(y)
-    \big)\big(
-    b(x)-b(y)
-  \big)
-\end{align}
-and the $\ln\det$ becomes, for the hierarchical matrix $A=CD+R^2$,
-\begin{equation} \label{eq:replica.det.cont}
-  \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-xa(x)+\int_0^xdy\,a(y)}{a_d-\langle a\rangle}\right)
-\end{equation}
+The formula for the determinant is complicated, and can be found by using the
+product formula \eqref{eq:cont.replica.prod} to write $CD$ and $R^2$, summing
+them, and finally using the $\ln\det$ formula \eqref{eq:replica.det.cont}.
 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} \\
   0&=c(x)-\big(f'(c)\ast(c\ast d+r\ast r)\big)(x) \label{eq:extremum.d}
 \end{align}
+where $(a\ast b)(x)$ denotes the functional parameterization of the diagonal of
+the product of hierarchical matrices $AB$, defined in
+\eqref{eq:cont.replica.prod}.
 
 \subsection{Supersymmetric complexity}
 
@@ -1484,6 +1439,80 @@ dominant saddles transition to a RSB complexity, we speculate that
 $E_\mathrm{alg}$ may be related to the statistics of minima connected to the
 saddles at this transition point.
 
+\begin{appendix}
+
+  \section{Hierarchical matrix dictionary}
+  \label{sec:dict}
+
+  Each row of a hierarchical matrix is the same up to permutation of their
+  elements.  The so-called $k$RSB ansatz has $k+2$ different values in each
+  row. If $A$ is an $n\times n$ hierarchical matrix, then $n-x_1$ of those
+  entries are $a_0$, $x_1-x_2$ of those entries are $a_1$, and so on until
+  $x_a-1$ entries of $a_k$, and one entry of $a_d$, corresponding to the
+  diagonal.  Given such a matrix, there are standard ways of producing the sum
+  and determinant that appear in the free energy. These formulas are, for an
+  arbitrary $k$RSB matrix $A$ with $a_d$ on its diagonal (recall $q_d=1$),
+  \begin{equation} \label{eq:replica.sum}
+    \lim_{n\to0}\frac1n\sum_{ab}^nA_{ab}
+    =a_d-\sum_{i=0}^k(x_{i+1}-x_i)a_i
+  \end{equation}
+  \begin{equation} \label{eq:replica.logdet}
+    \begin{aligned}
+      \lim_{n\to0}\frac1n\ln\det A
+      &=
+      \frac{a_0}{a_d-\sum_{i=0}^k(x_{i+1}-x_i)a_i}
+      +\frac1{x_1}\log\left[
+        a_d-\sum_{i=0}^{k}(x_{i+1}-x_i)a_i
+      \right]\\
+      &\hspace{10pc}-\sum_{j=1}^k(x_j^{-1}-x_{j+1}^{-1})\log\left[
+          a_d-\sum_{i=j}^{k}(x_{i+1}-x_i)a_i-x_ja_j
+      \right]
+    \end{aligned}
+  \end{equation}
+  where $x_0=0$ and $x_{k+1}=1$. The sum of two hierarchical matrices results
+  in the sum of each of their elements: $(a+b)_d=a_d+b_d$ and
+  $(a+b)_i=a_i+b_i$.  The product $A\ast B$ of two hierarchical matrices $A$
+  and $B$ is given by
+  \begin{align} \label{eq:replica.prod}
+    (a\ast b)_d&=a_db_d-\sum_{j=0}^k(x_{j+1}-x_j)a_jb_j \\
+    (a\ast b)_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}
+
+  There is a canonical mapping between the parameterization of a hierarchical
+  matrix described above and a functional parameterization that is particularly
+  convenient in the twin limit $n\to0$ and $k\to\infty$
+  \cite{Parisi_1980_Magnetic, Mezard_1991_Replica}. The distribution of
+  diagonal elements of a matrix $A$ is parameterized by a continuous function
+  $a(x)$ on the interval $[0,1]$, while its diagonal is still called $a_d$.
+  Define for any function $g$ the average
+  \begin{equation}
+    \langle g\rangle=\int_0^1dx\,g(x)
+  \end{equation}
+  The sum of two hierarchical matrices so parameterized results in the  sum of
+  these functions. The product $AB$ of hierarchical matrices $A$ and $B$ gives
+  \begin{align} \label{eq:cont.replica.prod}
+    (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)
+    -\int_0^xdy\,\big(
+      a(x)-a(y)
+      \big)\big(
+      b(x)-b(y)
+    \big)
+  \end{align}
+  The sum over all elements of a hierarchical matrix $A$ gives
+  \begin{equation}
+    \lim_{n\to0}\frac1n\sum_{ab}A_{ab}=a_d-\langle a\rangle
+  \end{equation}
+  The $\ln\det=\operatorname{Tr}\ln$ becomes
+  \begin{equation} \label{eq:replica.det.cont}
+    \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-xa(x)+\int_0^xdy\,a(y)}{a_d-\langle a\rangle}\right)
+  \end{equation}
+\end{appendix}
+
 \paragraph{Acknowledgements}
 The authors would like to thank Valentina Ros for helpful discussions.