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\begin{document}
%\font\twelverm=cmr12
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\openup 1.5\jot
\centerline{A random walk on the permutation group, some formal long-
time asymptotic relations}
\vspace{1in}
\centerline{Paul Federbush}
\centerline{Department of Mathematics}
\centerline{University of Michigan}
\centerline{Ann Arbor, MI 48109-1109}
\centerline{(pfed@umich.edu)}
\vspace{1in}
\centerline{\underline{Abstract}}
We consider the group of permutations of the vertices of a lattice.
A random walk is generated by unit steps that each interchange two
nearest neighbor vertices of the lattice. We study the heat equation
on the permutation group, using the Laplacian associated to the
random walk. At $t=0$ we take as initial conditions a probability
distribution concentrated at the identity. A natural conjecture for
the probability distribution at long times is that it is
'approximately' a product of Gaussian distributions for each vertex.
That is, each vertex diffuses independently of the others. We obtain
some formal asymptotic results in this direction. The problem arises
in certain ways of treating the Heisenberg model in statistical
mechanics.
\vfill\eject
This paper represents work in progress, and is written in a casual
manner, as notes for a seminar or perhaps a physics article. But the
work is interesting and non-trivial, and perhaps will inspire
research, many natural questions will appear. We of course will be
clear about what is proved and what is conjectured, some proofs will
be sketched. As will be seen this work is somewhat discouraging
about the application sought to the Heinsenberg model problem, but
opens some pleasant mathematical vistas.
We study a d-dimensional periodic lattice cube, $\Lambda$ of edge
size $L$. The number of its vertices, $\cal V$, is given as
\be |\Lambda| \equiv N \equiv L^d \equiv \# \{ {\cal V} \}. \ee
We set ${\cal G}$ to be the permutation group on $\cal V$. $H$ is
the element in the group algebra given as
\be H = - \sum_{i \sim j } (I_{ij} - I). \ee
here $i\sim j$ indicates that $i$ and $j$ are neighboring vertices in
the lattice. $I$ is the identity element in $\cal G$, and $I_{ij}$ is
the group element that interchanges vertices $i$ and $j$ leaving the
other vertices alone. $H$ is thus the Heisenberg model Hamiltonian
``promoted" from an operator in the Hilbert space to an element of
the group algebra. (One need not be familiar with the Heisenberg
model for purposes of this paper.)
We then consider the group algebra element $e^{-Ht}$. This can be
expanded as a linear combination of group elements, the $g_p$
\be e^{-Ht} = \sum_p \ f(g_p, t)g_p \ee
where $p$ labels the group elements. Each $g_p$ represents a mapping
of the vertices,
\be g_p : i \rightarrow p(i) \ee
so $p$ is specified by
\be \Big( p(1), p(2), ..... \Big) \ee
Equation (5) is a point in $(Z^d)^N$, in fact in $(\Lambda)^N$. The
cardinality of the set of such points is $N!$. Such points in $
(\Lambda)^N$ are restricted by the condition that all the $p(i)$ in
(5) are distinct. This subset we label $(\Lambda)^{N*}$. The
respective cardinalities of $(\Lambda)^{N*}$ and $(\Lambda)^{N}$ are
$N!$ and $N^N$. The latter space is simpler, being a periodic
lattice cube.
We have a time dependent probability function $f(g,t)$ on the
permutation group (or on $(\Lambda)^{N*}$). We now extend $f(g,t)$
to $f^e(\vec{x}, t), \vec{x}$ in $(\Lambda)^{N}$. Of course such
extensions are not unique. The motivation for extending $f$ will
soon be clear, largely depending on the simplicity of $(\Lambda)^{N}$
over $(\Lambda)^{N*}$. The extended function will no longer be a
probability function.
The relation between the Heisenberg model of ferromagnetism and our
random walk on the permutation group was beautifully developed by R.
Powers in [1]. Inspired by this work, the author presented a
possible avenue towards proving the phase transition of the
Heisenberg model in [2], a development of ideas in [1]. The central
relation needed in the proof envisioned in [2] is of the form
\be f^e(\vec{x}, t) \cong C_N \prod_{i\in {\cal V}} \ (e^{\Delta
t})_{i,x_i} \ \ {\rm for} \ t \ {\rm large}. \ee
$\Delta$ is the lattice Laplacian on $\Lambda$. (Of course, equation
(6) need only hold on $f(g,t)$, but we presume the extension
satisfies (6).) We do not now make explicit the degree of
approximation implied by $\cong$. The right side of (6) is a product
of gaussians (associated to independent random walks). It also is a
solution of the heat equation on $(\Lambda)^{N}$ with its natural
lattice Laplacian!
At the very least we would want (6) to imply
\be \lim_{t \ra \infty} f^e(\vec{x}, t) = C_N \lim_{t \ra \infty}
\prod_{i\in {\cal V}} \ (e^{\Delta t})_{i,x_i}. \ee
We note
\be \lim_{t \ra \infty} f(g,t) = \frac 1{N!} \ee
and
\be \lim_{t \ra \infty} (e^{\Delta t})_{i,j} = \frac 1{N}. \ee
Restricting to points $\vec x$ in $(\La)^{N*}$ equation (7) becomes
\be \frac 1{N!} = C_N\left(\frac 1 N\right)^N \ee
From (10) and Stirling's formula we have determined the $C_N$ in (6)
satisfy
\be \lim_{N \ra \infty} (C_N)^{\frac 1 N} = e \ee
We will find an extension $f^e$ of $f$ satisfying a differential
equation
\be \frac {\pa f^e}{\pa t} = \Delta \ f^e + V \ f^e \ee
$\Delta$ is the usual lattice Laplacian on $(\La)^N$ a periodic
version of the lattice Laplacian on $Z^{dN}$. Here we will only
consider $V$ built up with ``two-particle" interactions. Thus
\be V = \sum \ V_{i,j} \ee
where $V_{i,j}$ describes the interaction of the two vertices that at
time $t=0$ were at $i$ and $j$ respectively. The $V$ must be such
that the solution of (12) (with initial conditions, the identity at
$t=0$) restricted to $(\La)^{N*}$ agrees with $f$ as defined in
equation (3). It indeed is possible to find such $V$ that achieve this.
A form for $V_{i,j}$ that works is given as follows. We apply this $V
$ to a product function
\be \phi(\vec x) = \prod_i \ \phi_i(x_i) \ee
($V$ applied to product functions determines $V$ uniquely.)
\[ \left(V_{i,j} \phi \right) (\vec x) = - \; \prod_{k\not= i,j}
\phi_k(x_k) \cdot \sum_{y \in {\cal V}} \cdot \sum_{{\vec i}} \cdot \]
\[ \cdot \left[ \de_{x_i,y} \; \de_{x_j,y+{\vec i}} + \de_{x_j,y} \de_
{x_i,y+{\vec i}} + r\; \de_{x_i,y}\de_{x_j,y} + r \;
\de_{x_i,y+{\vec i}} \de_{x_j,y+{\vec i}} \right] \cdot \]
\be \cdot \left[\phi_i(y) - \phi_i(y+{\vec i})\right] \cd \left
[ \phi_j(y) - \phi_j(y + \vec i)\right] \ee
$r$ is arbitrary. There are $d$ orthonormal unit vectors, $\vec i$.
So the paths interact only when the vertices are in nearest neighbor
position. The expression (15) certainly is uniquely determined by
the conditions above. This is straightforward to show....though I
labored weeks on it. Herein we work with $r$ set equal to zero.
(But we believe working with an $r \not = 0$ at least if $|r| < 1$,
leads to no essential changes in the form of our calculations and
results. We have studied this $r\not= 0$ situation a little, one
need consider more diagrams than in the $r=0$ case. We will briefly
comment on this again later.)
We return to equation (6) for some further deliberations. We
consider summing $\vec {x}$ over $(\La)^N \equiv A$ and over $(\La)^
{N*} \equiv B$.
\be
\sum_{\vec {x} \in A} f^e(\vec{x}, t) \cong C_N \sum_{\vec {x} \in A}
\ \prod_{i\in {\cal V}} (e^{\Delta t})_{i,x_i} = C_N \ {\rm for} \
t \ {\rm large}
\ee
\be
\sum_{\vec {x} \in B} f^e(\vec{x}, t) = \sum_{\vec {x} \in B} f(\vec
{x}, t) = 1 \cong C_N \sum_{\vec {x} \in B} \ \prod_{i\in {\cal V}}
(e^{\Delta t})_{i,x_i} \ {\rm for} \ t \ {\rm large}
\ee
We will want the content of (16) and (17) to be given by the
following conjectures.
\bigskip
\centerline{--------------------}
\underline{Conjecture 1}
\be \lim_{t\ra \infty} \left( \sum_{\vec{x} \in A} f^e(\vec{x}, t)
\right)^{1/N} = \left(C_N\right)^{1/N} \ee
where $C_N$ is given by (10) and the limit is uniform in $N$.
\bigskip
\centerline{--------------------}
It is the uniformity requirement that makes the conjecture most
difficult.
\bigskip
\centerline{--------------------}
\underline{Conjecture 2}
\be \lim_{t\ra \infty} \left( \sum_{\vec{x} \in B} \ \prod_{i\in
{\cal V}} (e^{\Delta t})_{i,x_i}\right)^{1/N} = \left(C_N\right)^{-
\; \frac1 N} \ee
where $C_N$ is given by (10) and the limit is uniform in $N$.
\centerline{--------------------}
We believe Conjecture 2 is not very difficult to prove, and we plan
to turn to it soon.
We now consider the solution of (12), treating $V$ as a perturbation
in the form of a Rayleigh-Schrodinger expansion.
\be
f^e(\vec x, t) = \sum^\infty_{n=0} \int^t_0 dt_n \int^{t_n}_0 dt_
{n-1} \cdots \int^{t_2}_0 dt_1 \ e^{\Del(t-t_n)} Ve^{\Del(t_n-t_
{n-1})} \cdots Ve^{\Del t_1}. \ee
The right side of (20) is naturally represented as a sum of
contributions of diagrams. In some more detail as a sum of products
of the contributions of connected diagrams. All the computations of
this paper deal with results for the sum over final states (in (20)
over $\vec x$ in $(\La)^N)$ for connected tree-graph diagrams. If in
equation (6) we sum over final states, over $\vec x$, on both sides
we get
\be \sum_{{\vec x}} f^e(\vec x, t) \cong C_N \ \ \ {\rm for} \ \ t \
\ {\rm large}. \ee
(See Conjecture 1.)
Yes, in this paper we study the truth of (21), a much weakened form
of (6). But we expect that if we can get on top of (21) we are well
on our way to treating (6). Studying the decomposition of (20) into
connected diagram contributions puts us in the ``cluster expansion"
framework familiar in statistical mechanics. A very complete
treatment is in [3], but the level of sophistication of an
undergraduate course in statistical mechanics is more than adequate.
\bigskip
\centerline{--------------------}
\begin{theorem}
We consider a two-particle (vertex) connected tree-graph contribution
to (20). The diagram is an $H$ shaped figure. The bottom legs end
at vertices $i$ and $j$ and the upper arms end at $x_i$ and ${x_j}$.
The cross segment represents an interaction at time $t_1$. The
contribution of this diagram, with $x_i$ and $x_j$ summed over $
{\La^2}$, but before integrating over $t_1$ is
\be \sum_y \frac d{dt} \left( \phi_1(y,t) \ \phi_2(y,t)\right)\bigg|_
{t=t_1} \ee
where $\phi_1$ satisfies
\begin{eqnarray}
\frac \pa{\pa t} \phi_1 &=& \Del \phi_1 \\
\phi_1(x,0) &=& \de_{x,i}
\end{eqnarray}
with $\Del$ in (23) the Laplacian in $\La$. Similarly for $\phi_2$.
\end{theorem}
\centerline{--------------------}
This is a simple computation patterned on the continuum or lattice
version of
\be
-2 \int \vec \nabla \phi_1 \vec \nabla \phi_2 = \int \left( \phi_1
\Del \phi_2 + \phi_2 \Del \phi_1 \right) =
\int \left( \dot \phi_1 \phi_2 + \phi_2 \dot \phi_1 \right) = \int
\frac d{dt} (\phi_1 \phi_2)
\ee
We will always be dealing with tree-graph diagrams. Non tree-graph
diagrams have contributions that fall off with $t$ (for $r = 0)$, and
we are considering the $t \ra \infty$ limit. We do not control the
uniformity of this limit for the sum over all such diagrams, one
reason for the formal nature of our computations. Our final comment
on the $r \not= 0$ case is that corresponding to Theorem 1, in this
case, one must consider all ``ladder" diagrams to get the same formal
estimate, and not just the single tree graph. Choosing $r \not= 0$
leads to much more work and no gain. To do better than our
"results", if that is possible, one must consider potentials $V$ with
other than two-body forces.
\bigskip
\centerline{-----------------------}
Theorem 2 will be the analog of Theorem 1 for connected diagrams
involving $n$ particles, $n$ not necessarily 2.
\bigskip
\centerline{-----------------------}
\begin{theorem}
We consider all $n$-particle (vertex) connected tree-graph
contributions to (20) involving vertices $z_1,z_2,...,z_n$ at $t=0$.
We sum over final positions, over $\La^n$. We do not integrate over
$t_1$, and over the other times in the order $t_n, t_{n-1}, ..., t_2
$. Using Theorem 1, and Theorem 2 inductively on $n$, each of these
integrals will be of an explicit time derivative. In evaluating
these integrals \underline{we keep only the lower limit}, as if in
\be \int^b_a f'(t)dt = f(b) - f(a) \ee
we keep only the $-f(a)$ term. We will collect contributions of the
upper limits later. The ``contribution of lower limits" is
\be (-1)^N (n-1)! \frac d{dt} \sum_y \ \prod^n_{i=1} \ \phi_i(y)
\bigg|_{t=t_1} \ee
where
\begin{eqnarray}
\frac \pa{\pa t} \; \phi_i &=& \Delta \phi_i \\
\phi_i(y,0) &=& \delta_{y,z_i}
\end{eqnarray}
\end{theorem}
\bigskip
\centerline{---------------------}
\bigskip
\underline{Proof.} The proof is by induction on $n$. Let the
earliest interaction corresponding to $t_1$ be between vertices $n-1$
and $n$. (That is, one of these vertices at $z_n$ at $t=0$ and the
other at $z_{n-1}$.) Upon integrating over the later times and
keeping only lower limits we have terms with vertices $1,2,...,j$ at
$y$, and vertices $j+1, ..., n-2$ at $y + \vec i$, where vertices $n$
and $n-1$ are at these two points at $t_1$. One sums over the value
of $j$, the points $y$, the unit vectors $\vec i$, and the
permutation of different possibilities for the vertices attached to $y
$ and $y+\vec i$. We let $S$ stand for the sum of the $(n-1)!$
permutations of vertices $1,...,n-1$. The following telescopic sum
relation is the heart of the proof.
\[ \sum_y \sum_{\vec i} S \left\{ \left[ \sum^{n-2}_{j=0} \phi_1
(y) .... \phi_j(y) \phi_{j+1}(y+\vec i) ... \phi_{n-2}(y+ \vec i)
\right] \cd \left(\phi_{n-1}(y) - \phi_{n-1}(y+\vec i)\right) \right
\} \left(\phi_n(y) - \phi_n(y+\vec i)\right) \]
\[ = \sum_y \sum_{\vec i} S \left[ \phi_1(y) ... \phi_{n-1}(y) -
\phi_1(y+\vec i) ... \phi_{n-1}(y + \vec i)\right]
\left(\phi_n(y) - \phi_n(y+\vec i)\right) \]
\be = \sum_y \sum_{\vec i} (n-1)! \left[ \phi_1(y) ... \phi_{n-1}(y)
- \phi_1(y+\vec i) ... \phi_{n-1}(y + \vec i)\right]
\left(\phi_n(y) - \phi_n(y+\vec i)\right)
\ee
This will represent the contribution from terms where the first
interaction involves vertex $n$. One then sums over the $n$
possibilities for this first vertex coupled. (There is a factor of 2
in the first, $t=t_1$, interaction we have absorbed against the fact
that we are counting double since either end of the interaction at
$t=t_1$, could have been called $n$.) Thus the proof is short. Even
writing out the details which we have raced over. But the proof is
tricky enough, so that it's hard to be sure you're right. Counting
is hard.
\centerline{--------------}
Theorem 1 may be included in the statement of Theorem 2 as the $n=2$
case. In the next theorem we take the same contributions as in
Theorem 2 but in addition integrate over $t_1$ from $0$ to $t$, and
sum over $z_2,...,z_n$ but requiring that $z_1,...,z_n$ be a point in
$(\La)^{n*}$. We first define
\be a_i \equiv - (-1)^i. \ee
\centerline{--------------}
\begin{theorem} Let $T_n(t)$ be all the contributions considered in
Theorem 2 for given $n$ in addition integrated over $t_1$ from 0 to $t
$, and summed over $z_2,...,z_n$ with the requirement that
$z_1,...,z_n$ be a point in $(\La)^{n*}$.
\be \lim_{t\ra \infty} T_n(t) = a_{n-1} \ee
\end{theorem}
\centerline{--------------}
\underline{Proof.} We detail the proof for $n=2$ which contains all
the essential points. From (27) to (29) we have upon integrating
(27) from 0 to $t$ and summing over $z_2$
\be \sum_{z_2 \not= z_1} \left( \sum_y \ \phi_1(y,t) \phi_2(y,t) -
\sum_y \ \phi_1(0,t) \phi_2(0,t) \right) \ee
with $\phi_1(y,0) = \de_{y,z_1}, \ \ \phi_2(y,0) = \de_{y,z_2}$.
\noindent
The second term in (33) therefore vanishes, leaving
\be \sum_{z_2 \not= z_1} \sum_y \phi_1(y,t) \phi_2(y,t) \ee
which equals
\be \sum_{z_2} \sum_y \phi_1(y,t) \phi_2(y,t) - \sum_y \ \phi^2_1
(y,t) . \ee
The first term is 1 by
\begin{eqnarray}
\sum_{z_2} \phi_2(y,t) &=& 1 \\
\sum_y \phi_1(y,t) &=& 1.
\end{eqnarray}
and the second term in (35) gets to zero with $t$. This concludes
the proof for $n=2$. For $n=3$, say, the initial points $z_1, z_2,
z_3$ and $z_1, z_3, z_2$ lead to the same set of diagram
contributions explaining the lost $(n-1)!$ factors when pursued.
We now define quantities $A_i$ defined recursively from the $a_i$ of
(31). We set
\be P \equiv \sum_{j=0} \ A_j \ t^j \ee
a formal power series in $t$. Then the $A_i$
are defined by
\be A_0 = 1 \ee
\be A_i = a_i + \sum^{i-1}_{k=1} a_k \ {\rm coef} \ \left(P^{2k+1}, t^
{i-k} \right), \ i > 0 \ee
where using Maple notation, coef $(f,t^s)$ picks out the coefficient
of $t^s$ in the formal power series $f$. This is the procedure by
which we found the $A_i$. Actually, with $A_1 = 1, A_2 = 2, A_3 =5,
A_4 =14....$, the $A_i$ are the Catalan numbers, given as
\be A_i = \frac 1{i+1} \left( \begin{array}{c}
2i \\
\ \\
i
\end{array} \right). \ee
\centerline{--------------}
\begin{theorem} Let $\tilde{T} _ n(t)$ be the analog of $T_n(t)$ of
Theorem 3 but now \underline{including upper limits}, the whole
megillah. Again integrating over all times, summing final states
over $(\La)^n$, keeping one initial vertex fixed and summing the
other vertices at $t=0$ over points lying in $(\La)^{n*}$. Then
\be \lim_{t\ra \infty} \tilde{T} _ n(t) = A_{n-1} \ee
\end{theorem}
\centerline{--------------}
\underline{On the Proof}
The computation of the right side of (42), arising as a solution of
(40), was perhaps the most difficulty and tricky business I have ever
been associated with. Also I would find it extremely difficult to
write a presentable proof. Perhaps someone can come up with a
reasonable proof. (Skeptics may prefer to call Theorem 4 a
conjecture, but it is certainly true.) I content myself here with
some points on the computation of $A_2$.
Equation (40), for $i=2$ becomes
\be A_2 + a_2 + 3a_1. \ee
The contributions of all contributing diagrams,when only the lower
limit of the $t_2$ integration is kept, is $a_2 = -1$, the first term
in (43). This by Theorem 3.
Keeping the upper limit on the $t_2$ integration involves us with
three cases.
1) Case 1, associated to $t_1$ is $V_{z_1,z_2}$, associated to $t_2$
is $V_{z_1, z_3}$.
2) Case 2, associated to $t_1$ is $V_{z_1,z_2}$, associated to $t_2$
is $V_{z_2, z_3}$.
3) Case 3, associated to $t_1$ is $V_{z_2,z_3}$, associated to $t_2$
is $V_{z_1, z_2}$.
With this notation there is a sum over $z_2$ and $z_3$ with the
restriction $z_2 \not= z_3 \not= z_1 \not= z_2$. Here the
contribution of $z_1, z_2, z_3$ does not equal the contribution of
$z_1, z_3, z_2$.
Each of these three cases contributes a factor $a_1 = 1$ to equation
(43). Case 3 is the most interesting, and we will deal with this one
case.
The contribution of Case 3 may be represented as
\be \sum_{x_1,x_2,x_3} \ \sum_{z_2,z_3} \int^t_0 \; dt_1 \; \int^t_
{t_1} dt_2 \; K(x_1,x_2,x_3,z_1,z_2,z_3,t_1,t_2) \ee
$x_1,x_2,x_3$ lie in $\La^3$ and $z_1, z_2, z_3$ are restricted to $
\La^{3*}$. $x_1,x_2,x_3$ are positions of the vertices at $t=t$ and
$z_1, z_2, z_3$, the positions at $t=0$. Recall we are keeping only
the upper limit in the integral over $t_2$, getting
\be \sum_{x_1,x_3} \ \sum_{z_2,z_3} \int^t_0 \; dt_1 \; k
(x_1,x_3,z_2,z_3,t_1) \left( e^{\Delta t}\right)_{z_1,x_1} \ee
$k$ is the kernel of $a$ two-vertex diagram with a single interaction
at $t=t_1$. We rewrite this as
\be \sum_{x_1} \left( e^{\Delta t} \right)_{z_1,x_1} \left( \sum_
{x_3} \; \sum_{z_2,z_3} \int^t_0 \; dt_1 \; k\left
(x_1,x_3,z_2,z_3,t_1 \right) \right). \ee
We wish to compare this expression to
\be \sum_{x_1} \left( e^{\Delta t} \right)_{z_1,x_1} \left( \sum_
{\bar x_1, x_3} \; \sum_{z_3} \int^t_0 \; dt_1 \; k\left(\bar
x_1,x_3,z_2,z_3,t_1 \right) \right). \ee
\be \ \ \ \ \ \ \ \ \ \ \cong \ \ 1 \ \ \ \ \cdot \ \ \ \ 1. \ee
by Theorem 3. Using translation invariance of the kernel $k$, (46)
and (47) differ by an error that goes to zero with $t$ (from the
different restrictions on the $z$'s in the two expressions). In
(47) we neglect the restriction that $z_2$ and $z_3$ may not equal
$z_1$.
\bigskip
\centerline{--------------------}
\bigskip
We turn to the relation of Conjecture 1, equation (18)
\be \lim_{t\ra \infty} \left( \sum_{\vec{x} \in A} f^e(\vec{x}, t)
\right)^{1/N} = \left(C_N\right)^{1/N}. \ee
We work in the limit $N$ large, and $t \ra \infty$ (before $N \ra
\infty)$. We thus want
\be \lim_{t \ra \infty} \left( \sum_{{\vec x} \in A} f^e(x,t) \right)^
{1/N} = e \ee
and eschew considering uniformity of $t$ limit with respect to $N$.
We view $f^e(x, t)$ expressed a sum of products of connected
diagrams. For $N$ large we expect this sum to be dominated by terms
with some fixed number of connectivity patterns. That is, in terms
kept in the product there are $x_1N$ two-connected terms in the
product, $x_2N$ three-connected terms in the product, $x_3N$ four-
connected terms in the product, and so on. The expression for $
\displaystyle{\sum_{{\vec x} \in A}} f^e(x,t)$ in this limit is
\vfill\eject
\[ \prod_i \left( \frac{A_i \; i!}{N^i} \right)^{x_iN} \ \cdot \ \frac
{N\;!}{(\sum_i(i+1)Nx_i)!(N - \sum (i+1)Nx_i)!} \ \cdot \]
\be \cdot \ \frac{(\sum(i+1)Nx_i)!}{\Pi(x_iN)!\;\Pi\left( (i+1)!
\right)^{x_iN}} . \ee
The first set of parentheses includes the contributions of the
diagrams $t \ra \infty$ limit, from Theorem 4. The next factor, a
ratio of factorials, sums over which set of initial vertices are
included in the set that are connected to other vertices. The final
ratio of factorials sums over the connectivities of the vertices
(which vertices are connected with which vertices).
Maximizing (51) over the choice of $x_i$ one finds
\be \lim_{t \ra \infty} \ \left(\sum_{{\vec x} \in A} \; f^e(x,t)
\right)^{1/N} \ \cong e^q \ee
with
\be q = -1 + \sum_{i=0} \ A_i \frac{p^{i+1}}{i+1} - \ell n \ p \ee
where
\be 1 = \sum_{i=0} \ A_i p^{i+1} . \ee
One wants $q=1$ but (53) and (54) do not yield $q=1$. (It is not
clear how to define a solution of (54) for $p$, but no reasonable
definition works.) At this point we have reached complete frustration!
Sometime after arriving at this impasse, we decided to consider a
random walk not involving all the lattice vertices, but rather a
fraction $\rho$ of the vertices, ``uniformly distributed". This is
achieved by ``integrating out" a fraction $(1-\rho)$ of the vertices
in the probability function $f(x,t)$. This is done before the
extension to $f^e\;!$. It is easy to make the corresponding changes
in all the computations of this paper, a matter of a day or two.
Replacing (52), (53) and (54) one finds
\be \lim_{t \ra \infty} \left(\Sigma \; f^e(x,t)\right)^{1/N} \cong e^
{\tilde q} \ee
with $f^e$ here depending on $\rho N$ vertices and with
\be \tilde q = - 1 + \sum_{i=0} \ A_i \; p^{i+1} \ \frac{\rho^i}{i
+1} - \ell n \; p \ee
and
\be 1 = \sum_{i=0} \ A_i \; p^{i+1} \rho^i . \ee
Where before one wanted $q=1$, here we desire
\be \tilde q = 1 + \frac{1 - \rho}{\rho} \ \ell n (1 - \rho). \ee
The miracle that happens is as follows. For $\rho < 1/2$, equation
(57) is satisfied with
\be p = 1 - \rho \ee
and substituting this expression for $p$ into (56) one finds (56) and
(58) yield the same formal expansion in powers of $\rho$, valid for $
\rho < 1/2$. It is interesting to note that (56), (57) and (58)
determines both (59) and the $A_i$ (expanding $p$ and $\tilde q$ in
powers of $\rho$).
If we set
\be f(\rho\, p) \equiv \sum_{i=1} \ A_i \; p^i \rho^i \ee
we can solve, from (57) and (59)
\be p + pf(\rho\, p) = 1 \ee
\be p = 1 - \rho \ee
to get
\be f(z) = \frac{1 - \sqrt{1 - 4z}}{1 + \sqrt{1 - 4z}} \ee
and see the singularity at $z = 1/4$, or $\rho = \frac 1 2$.
So we are led to believe that Equation (6) may still be true in some
suitable sense, but that a perturbation expansion development as
undertaken in this paper is not promising. Presumably as $\rho$
increases to value $1/2$ one must consider diagrams of arbitrarily
high connectivity.
Many interesting questions suggest themselves, of which we choose
two. For $\rho < 1/2$, control the perturbation expansion, and
obtain uniformity in $N$ of the $t \ra \infty$ limits. Find some way
of treating $\rho > 1/2$. The first question is likely a problem
about which to develop several Ph.D. theses. The second still
requires some further ideas to gauge its difficulty.
\vfill\eject
\centerline{\underline{References}}
\begin{itemize}
\item[[1]] Robert T. Powers, ``Heisenberg Model and a Random Walk on
the Permutation Group", {\it Lett. in Math. Phys.} {\bf 1}, 125-130
(1976).
\item[[2]] P. Federbush, ``For the Quantum Heisenberg Ferromagnet,
Tao to the Proof of a Phase Transition", math-ph/0202044.
\item[[3]] David C. Brydges, ``A Short Course in Cluster Expansions,
Phenomenes critiques, systems aleatoires, theories de gauge, Part I,
II" (Les Houches, 1984), 129-183, North Holland, Amsterdam, 1986.
\end{itemize}
\end{document}