\input amstex
\documentstyle{amsppt}
\magnification=\magstep1
\hsize 6truein
\parskip=1pt plus 1pt
\define\s{\sigma}
\define\ch{\Cal H}
\define\z{\bold Z}
\define \cu{\Cal U}
\define\bt{\bold T}
\define\r{\bold R}
 
\leftheadtext\nofrills{{\rightline{P. M. Bleher and P. Major}}}
\rightheadtext\nofrills{\leftline{Phase transition in statistical
physics}}
 
\topmatter
\title PHASE-TRANSITION IN STATISTICAL PHYSICAL MODELS\\ WITH DISCRETE
AND CONTINUOUS SYMMETRIES\endtitle
\author P. M. BLEHER${}^{1}$ and P. MAJOR${}^{2}$ \endauthor
\affil ${}^{1}$Tel Aviv University, \\
${}^{2}$Mathematical Institute of the Hungarian
Academy of Sciences \endaffil
\abstract We discuss the problem of existence or non-existence of
phase-transition in statistical physical models. The main
technical difficulties connected with this problem are formulated. We
point out the difference between models with discrete and continouous
symmetry. A particular model, Dyson's hierarchical model is considered
in some more detail. \endabstract
\endtopmatter
\subheading{1. Introduction}
In statistical physics the investigation of the existence and
uniqueness of a random field, called equilibrium state, plays a most
important role. This random field takes its values on the
configurations $\s(j)$,~
$j\in \z$, where the so-called spin-variables $\s(\cdot)$ are in some
metric space $S$, generally $S$ is a subset of the Euclidean space
$\r^1$ or $\r^s$
with some $s\ge2$, $\z$ is  a parameter set, generally  the integer
lattice $\z^p$ in the
$p$-dimensional Euclidean space with some $p\ge 1$. We have to define
a probability measure on the space $S^{\z}$. This measure depends
on
a Hamiltonian function, a physical parameter, the temperature, and a
so-called free measure. For the sake of simpler notations we
restrict ourselves to the case when the model contains only
pair-interaction. In this case the Hamiltonian function is a formal
series,
$$
\ch(\s)=\sum_{j\in \z}\sum_{k\in \z}
\cu_{j,k}(\{\s(j),\s(k)\}),\quad\s=\{\s(j);\;j\in\z\},\tag1
$$
and
$$
\sum_{k\in
\z}\cu_{j,k}(\{\s(j),\s(k)\})<\infty,\quad\text{for all }j\in\z
\tag1$'$
 $$
where $\cu_{j,k}(\cdot,\cdot)$  are measurable functions on $S\times
S$.
 
Let us fix some measure $\nu$ on $S$ which is called the free measure
in the literature. Given some finite set $V\subset\z$  and a
configuration $\bar\s=\{\bar\s(k);\;k\in\z\setminus V\}$ and a
parameter $T>0$,
the temperature, we define the conditional Gibbs distribution
$\mu_V$ with
some density $f_V(\s(j);\;j\in V|\bar\s)$ with respect to the
product measure $\prod_{j\in V}d\nu(\s(j))$ by the formula
$$
\aligned
\frac{d\mu_V(\s(j);\;j\in
V|\bar\s)}{\prod_{j\in V}d\nu(\s(j))}
&=f_V(\s(j);\;j\in V),T|\bar\s) \\
&=\frac 1Z\exp\left\{-\frac1T\sum_{j\in
V, k\in\z\setminus V}\cu_{j,k}(\{\s(j),\bar\s(k)\})\right\},
\endaligned \tag2
$$
where  the norming factor $Z$ is defined as
$$
Z=Z(V,\bar\s)=\int\exp\left\{-\frac1T\sum_{j\in V,
k\in\z\setminus V}\cu_{j.k}(\{\s(j),\bar\s(k)\})\right\}\prod_{j\in
V}d\nu(\s(j)), \tag$2'$ $$
if formula (2) is meaningful.
 
We call a probability measure  $\mu$ on $S^{\z}$ an equilibrium state
with a Hamiltonian $\ch$  defined in (1) and free measure $\nu$ at
temperature $T$, if for any finite set $V$  the
conditional distribution of $\mu(\{\s(j);j\in V\})$ with respect to
the configurations $\bar\s$ in $\z\setminus V$, or more precisely an
appropriate version of it, is defined by the formula
$$
\mu(\{\s(j);\;j\in V\}\in A|\bar\s)=\int_A f_V(\{\s(j);\;j\in
V\}),T|\bar\s) \prod_{j\in V}\,d\nu(\s(j))  \tag3
$$
for all measurable $A\subset S^{V}$, where the function $f_V$ is given
by formula (2).
 
The first question to be clarified is whether such a measure $\mu$
exists
and whether it is unique. The classical results of probability theory
cannot be applied directly to answer this question. If there are
several equilibrium states $\mu$ with the same Hamiltonian and free
measure at the same temperature, then one speaks in the literature
about phase-transition. A natural way to construct equilibrium
states is to carry out the following procedure: Choose a sequence of
finite sets $V_n\subset\z$ such that $\lim_{n\to\infty} V_n=\z$ and
configurations
$\bar\s_n=\{\bar\s(j);\,j\in\z\setminus V_n\}$ on their complementary
sets. Define the measures $\mu_{V_n}$ on the sets $V_n$ with some
boundary
condition $\bar\s_n$ by formulas $(2)$ and $(2')$. It is natural to
expect that they are good approximations of the equilibrium states we
are looking for. Hence, we try to prove that the sequence of measures
$\mu_{V_n}$ has a convergent subsequence, i.e.\ the sequence
$\mu_{V_n}$ is compact, and the
limit of its convergent subsequences is an equilibrium state. The
compactness of the sequence
$\mu_{V_n}$ holds automatically, if  $S$ is a compact set. In more
general
cases, when $S$ can be e.g.\ the Euclidean space $\r^s$, the proof of
compactness may be a
really hard problem. It is true under very general conditions that the
limit of a subsequence of $\mu_{V_n}$ is an equilibrium state.
Moreover,
it is true that all equilibrium states are in the closure of the
convex linear combination of measures obtained in such a way, if the
boundary conditions $\bar\s_n$ can be  chosen  arbitrarily.
Hence, the really hard problem is the problem of uniqueness. This is
connected to the following question: Let us fix some $j\in \z$
together with a large neighbourhood of $V$ of $j$, and let us fix some
configuration $\bar\s=\{\bar\s(k);\;k\in \z\setminus V\}$ on
$\z\setminus
V$. Take the measure $\mu_V(\cdot|\bar\s)$ on the set $V$. Does the
distribution of $\s(j)$ with respect to this measure ``feel strongly''
the boundary condition $\bar\s$, i.e.\ can it have different limits
for different boundary conditions?  The investigation of this question
requires a more refined analysis.
 
\subheading{2. On translation invariant models}
In this section we discuss some basic results about the uniqueness
and non-uniqueness of models on the integer lattice $\z^p$ of the
Euclidean space $\r^p$ and such  that $\cu_{j,k}(x,y)=\cu_{0,k-j}(x,y)$
for all $j,\,k\in\z$ and $x,\,y\in S$. Such models are called
translation invariant.
 
Let us remark that the multiplying term $\frac1T$ in the exponent of
formula (2) plays an important role. Let us compare the density
$f_V(\cdot|\bar \s)$ of two configurations
$\s^{(1)}=\{\s^{(1)}(j);\;j\in V\}$ and
$\s^{(2)}=\{\s^{(2)}(j);\;j\in V\}$ in  the volume $V$ with respect to
the measure $\mu_V(\cdot|\bar\s)$. If one of them has less energy with
respect to the boundary condition $\bar \s$, then its density is much
greater for small $T$, but these configurations have  almost the same
density for large $T$. Hence, it is natural to expect that, since at
high temperatures the $\mu_V$ probability of a configuration in a
volume $V$ weakly depends on the boundary condition $\bar\s$
on $\z\setminus V$, there is a unique equilibrium state
$\mu_V=\mu_V(T)$ at  high temperatures. More precisely, at the
temperature $T=\infty$ the spins $\s(j)$ are independent random
variables with distribution $\nu$ with respect to the
$\mu=\mu(\infty)$ measure. This measure is stable in the following
sense: For large $T$ the (unique) measure $\mu(T)$ is a small
perturbation of the measure $\mu(\infty)$ under very general
conditions. (See e.g.\ [5] or [8] Chapter 1.)
 
On the other hand, the situation at low temperatures is more complex.
We shall call a configuration $\bar\s=\{\s(k);\;k\in\z\}$ a
configuration with (locally) minimal energy if for all configurations
$\bar\s'=\{\s'(k);\;k\in\z\}$ such that
the configurations $\bar\s$ and $\bar\s'$ differ only at finitely many
places $k\in\z$ the conditional energy
$$
\ch(\bar\s'|\bar\s)=\sum_{j\in \z}\sum_{k\in \z}
[\cu_{j,k}(\{\s'(j),\s'(k)\})- \cu_{j,k}(\{\s(j),\s(k)\})]\ge0
$$
is non-negative. (The sum in the last expression is finite because of
$(1')$.)
 
The previous argument would suggest that for small $T>0$ and a
configuration $\bar\s$ with minimal energy there is an equilibrium
state $\mu(T)$
which is essentially concentrated on configurations that are small
perturbations of $\bar\s$. Namely, if we  take a sequence of
volumes $V_n$  such that $\lim_{n\to\infty}V_n=\z$ and take the
measures $\mu_{V_n}$ defined by formula (2) in $V_n$ with the boundary
condition $\bar\s$ on$\z\setminus V_n$ then the limit of these
measures should be such an equilibrium state. This would mean in
particular, that if there exist  several configurations with minimal
energy, then there are several equilibrium states at low temperatures.
Nevertheless, this is not always the case. Let us call a
configuration with minimal energy stable if for small $T>0$ there is
some equilibrium state $\mu(T)$  which  is essentially concentrated on
configurations that are close to it.
Whether a configurations with minimal energy is stable or not that
depends on how big is the conditional energy of a configuration
differing of it on a large finite set with respect to this
configuration. We can expect that a configuration with
minimal energy is stable if this conditional energy is large,
otherwise we expect that it is not stable. However, it is very
hard to decide when this conditional energy is sufficiently large to
ensure stability, and only partial results are available about this
problem. An important result in this direction is the Pirogov-Sinai
theorem, (see e.g.\ ~[8], Chapter~2). Here the situation is
considered  when the potential $\cu$ has a finite range of
interaction, i.e.\ $\cu(j,k)=0$ if $|j-k|\le r$ with some $r>0$, and
the set $S$ where the spins $\s(\cdot)$ take their values is finite.
The question whether a periodic
configuration with minimal energy is stable is investigated.  The
situation is rather complex even in this particular case. Pirogov and
Sinai gave a satisfactory sufficient condition for the stability of
such configurations. We omit the exact formulation of their result,
because it requires the introduction of some new notions, and the
questions we are
interested in in this paper  are only loosely connected
with this result. We only remark that the most important conditions
they require are that the dimension of the parameter space $\z=\z^p$
must be at least $p\ge2$ and the model must satisfy the so-called
Peierls's condition, which says the following:
The conditonal energy of a configuration that  differs from a periodic
configuration with minimal energy on a finite set $B$ has a
conditional energy with respect to this periodic configuration which
is greater than $\rho$ times the cardinality  of the set $B$, where
$\rho>0$ is an appropriate fixed number. Peierls's condition
should guarantee that the conditional energy of a configuration
differing on a finite set from a periodic configuration with minimal
energy must be sufficiently large.
 
The case when the state-space $S$ is a connected set is  much less
known. The main problem is that in interesting cases there is no
natural candidate for the analogue of the Peierls's condition. In this
case only partial results are available, but there are some results
which indicate that the situation in more general state space can be
essentially different. Let us discuss some such models.
 
We consider models with the Hamiltonian function
$$
\ch(\s)=-\sum_{j\in \z}\sum_{k\in \z}
U_{j,k}\s(j)\s(k),\quad\s=\{\s(j);\;j\in\z\},
$$
where the numbers $U_{j,k}\ge0$ is such that $U_{j,k}=U(j-k)$. First
we want to compare the following two models.
 
\parindent=25pt
\item{1.a)}$\z=\z^p$ with some $p\ge2$, $U(j-k)=1$ if $j$ and $k$ are
neighbours in $\z^p$, $S=\{-1,\,1\}$, and the measure $\nu$ is defined
as $\nu\{-1\}=\nu\{1\}=1/2$.
\item{1.b)}$\z=\z^p$ with some $p\ge2$, $U(j-k)=1$ if $j$ and $k$ are
neighbours in $\z^p$, $S$ is the unit-sphere in the Euclidean space
$\r^s$ with some $s\ge2$. Here $\s(j)\s(k)$ denotes scalar product,
and $\nu$ is the Lebesgue measure on the unit sphere.
 
\parindent=16pt
Let us compare the following two models too:
 
\parindent=25pt
\item{2.a)} The same as model 1.a), only $\z=\z^1$, and
$U(n)=n^{-\alpha}$, with some $\alpha>1$.
\item{2.b)} The same as model 1.b), only $\z=\z^1$, and
$U(n)=n^{-\alpha}$, with some $\alpha>1$.
 
\parindent=16pt
Models 1.a) and 2.a) have the following symmetry property: If we
multiply all $\s(j)$ by $-1$ then both the Hamiltonian $\ch(\s)$ and
the  free measure $\nu$ remain the same. Hence, we say that these
models  are invariant with respect to the multiplication group
$\{-1,1\}$. In the same way models 1.b) and 2.b) are invariant with
respect to the group of rotations $U(s)$ of the $s$-dimensional space.
The first invariance is called a discrete and the second one a
continuous symmetry.
 
In models 1.a) and  2.a) the  configurations $\s(j)=1$ for all
$j\in\z$ or $\s(j)=-1$ for all $j\in\z$ are configurations with
minimal energies. It is proved that in model 1.a) for small $T>0$ at
any dimension $p\ge2$ there is a translation invariant equilibrium
state $\mu^+=\mu^+(T)$ such that $\mu^+\{\s(j)=1\}>\frac12$  for all
$j\in\z$. Similarly, there is an equilibrium  state $\mu^-$ such that
$\mu^-\{\s(j)=-1\}>\frac12$ for all $j\in\z$. This means a phase
transition which is connected  to a break-down of symmetry, i.e. to
the fact
that the measures $\mu^+$ and $\mu^-$ do not preserve the symmetry the
Hamiltonian $\ch$ and free measure $\nu$ have. They are in the
vicinity of a configuration with minimal energy instead. In model 2.a)
the same result holds if $\alpha\le2$. On the other hand, the
equilibrium state is unique for $\alpha>2$ at any temperature $T$.
In these results we should emphasize the phase-transition in the case
$\alpha=2$. This is a very delicate boundary case, and model 2.a) with
this parameter has certain peculiar properties.   (See ~[1].)
 
In models 1.b) and 2.b) the configurations $\s(j)=e;\;\forall j\in\z$
with some $e\in S$ are configurations with minimal energies. One is
interested in which of these models have a phase-transition  at low
temperatures and which have not. In model 1.b) there is a
phase-transition at low temperatures if the dimension of the lattice
$\z^p$ is $p\ge3$, (see ~[7])
and there is no phase-transition for $p=2$. More precisely, the result
for $p=2$ is proved completely only in the case when the state space
$S$ is the unit circle, (see [4]).  In the general case only the
weaker result
is proven that any equilibrium state is invariant with respect to
rotations [6]. This excludes the possibility of such equilibrium
states, where the configurations are
in the vicinity of a configuration with minimal energy with
probability almost ~one, and it is
believed that there is no phase-transition, if this result holds. In
model 2.b) there is
a phase-transition for $\alpha<2$, and there is no phase-transition
for $\alpha\ge2$. Models 1.a) and 2.a) behave differently in the case
$p=2$, and the difference between
the behaviour of models 1.b) and 2.b) appears in the case $\alpha=2$.
 
The above examples show that models with continuous and discrete
symmetries behave differently. The heuristic explanation of this
difference is clear.   Let us fix a  configuration with minimal energy
in a neighbourhood of infinity, and let us look at how much energy
is needed to change this configuration radically in a neighbourhood of
zero. It may happen that in models with continuous symmetry we can
achieve this change at the expense of less energy by rotating the
configuration in such a way that the relative rotation between
neighbour points is small. In models with discrete symmetry this
cannot be done. However, this heuristic argument is not strong enough
to give an orientation about what to expect in the general case.
Hence,
it may be interesting a model where  these questions can be solved
completely. We discuss a one-dimensional model, Dyson's
hierarchical model in detail. This is a version of models 1.b) and
2.b). The main difference is that the number $U(i,j)$
appearing in the Hamiltonian of this model depends not on the usual
distance $|i-j|$, but some  different distance  on $\z$. Hence this
model is not translation invariant, but it has some other symmetries
which makes it simpler to handle.
 
\subheading{3. Dyson's hierarchical model}
Dyson's hierarchical model is a model on the positive integers
$\z=\{1,2,\dots\}$ with Hamiltonian function
$$
\ch(\s)=-\sum_{i\in\z}\sum\Sb j\in \z\\j>i\endSb
\varphi(d(i,j))\s(i)\s(j)\,,               \tag4
$$
where the so-called hierarchical distance $d(\cdot,\cdot)$ is defined
by the formula $d(i,j)=2^{n(i,j)-1}$, and
$$
n(i,j)=\min n,\;\exists \;\text{some $k$ such that }(k-1)2^n<i,j\le
k2^n,
$$
$\varphi(\cdot)$ is a real function, and the free measure $\nu$ has
the density function $p_0(x)$
$$
p_0(x)=\frac {d\nu}{dx}(x) =C(t)\exp\left\{-
\frac{x^2}2-\frac t4|x|^4\right\},  \tag5
$$
where $t>0$ is some small number.
We consider both the scalar case when the spins $\s(\cdot)$ take
values on the real line $\r^1$ and  the vector case when they take
values in $\r^s$ with some $s\ge 2$. In the latter case, the product
$\s(j)\s(k)$ in formula (4) means scalar product. We are interested in
the question for which functions $\varphi(\cdot)$ the model has a
phase transition at low temperatures and for which one it has not.
 
The hierarchical distance $d(\cdot,\cdot)$ appearing in this model is
a version of the usual distance. It is not translation invariant,
but it has some other symmetries wich makes the model simpler. The
density function $p_0(x)$ is a small perturbation of  the normal
density, and the condition $t>0$ guarantees that all integrals we need
are convergent. The scalar case is a version of problem 2.a) and the
vector case of problem 2.b) of the previous section. The crucial step
in solving the
problem about phase transition consists of investigating the following
question:
 
Put $V_n=\{1,2,\dots,2^n\}$, and
$$
\ch_{V_n}(x_1,\dots,x_{2^n})=-\sum_{i\in V_n}\sum\Sb j\in
V_n\\j>i\endSb \varphi(d(i,j))x_ix_j.
$$
Define the probability measure $\mu_n=\mu_{n,T}$  on
$\r^{V_n}$ (on $(\r^s)^{V_n}$ if we have a model with
$s$-dimensional spins) with the density function
$p_n(x_1,\dots,x_{2^n})$ by the following formula:
$$
p_n(x_1,\dots,x_{2^n})= \frac{d\mu_n(x_1,\dots,x_{2^n})}
{dx_1\dots dx_{2^n}  }
=C_n\exp\left\{-\frac1T\ch_{V_n}(x_1,\dots,x_{2^n})\right\}
\prod_{j=1}^{2^n}p(x_j).
$$
Let $\left(\s(1), \s(2), \dots,\s(2^n)\right)$ be a $\mu_n$~distributed
random vector, and let $p_n(x)$ denote the density function
of the average $2^{-n}\sum_{i=1}^{2^n}\s(i)$. Give a good
asymptotic formula for $p_n(x)$.
 
This function $p_n(x)$ has the symmetry property $p_n(-x)=p_n(x)$ in
the scalar case, and it is rotation invariant in the vector case. As
some further analysis shows, there are two possibilities. Either the
function $p_n(\cdot)$ is essentially concentrated in a small
neighbourhood of the  origin for large $n$ or it is concentrated
around some points $\pm M$, \ $M>0$ in the scalar case and around the
sphere $|R|=M$ in the vector case. In the second case there is a phase
transition, and in the first case there is not. The last statement is
far from trivial, it requires the investigation of the approximating
measures $\mu_n$ with some boundary condition discussed in the first
section of this paper. The main technical difficulty is to have a
control on the finite dimensional projections (its dimension is
independent of $n$) of these measures $\mu_n$. This problem can be
translated to a purely analytical question, where the formulas contain
explicitly the above defined functions $p_n(\cdot)$. As a deeper
analysis
shows the existence or non-existence of phase transition depends on
the behaviour of these functions. This question is discussed for
instance in our paper [3], and here we omit the details.
 
The investigation of the function $p_n(x)$ also leads to a purely
analytic question: Observe that
$$
\align
\ch_{V_n}(x_1,\dots,x_{2^{n+1}})&=\ch_{V_n}(x_1,\dots,x_{2^n})+
\ch_{V_n}(x_{2^n+1},\dots,x_{2^{n+1}})\\
&\qquad-\varphi\left(2^n\right)
\left(\sum_{j=1}^{2^n}x_j\right)
\left(\sum_{j=2^n+1}^{2^{n+1}}x_j\right).
\endalign
$$
This relation leads to the following recursive formula:
$$
p_{n+1}(x)=C_n\int\exp\left\{\frac{4^n\varphi(2^n)}T(x^2-u^2)\right\}
p_n(x-u)p_n(x+u)\,du,    \tag6
$$
where $C_n$ is an appropriate norming constant, turning $p_{n+1}(x)$
into a density function. So we have to study the asymptotic behaviour
of the functions $p_n(x)$ defined by the recursion formula (6). To
complete the formulation of the problem we have to remark that the
starting function $p_0(x)$ is defined in (5). The problem can be
slightly simplified by introducing the following transformation of
$p_n(x)$.
 
Put
$$
A_n=1+\sum_{j=n+1}^\infty 2^j\frac{\varphi(2^{n+j})}{\varphi(2^n)},
$$
and
$$
q_n(x)=\exp\left\{\frac{ A_n}{2(1+A_n)}4^n\varphi(2^n) x^2\right\}
p_n\left(\sqrt{ \frac T{1+A_n} }x\right).
$$
Some calculation shows that the above defined functions $q_n(x)$
satisfy the relation
$$
\aligned
q_{n+1}(x,T)=\bar C_n(T)\int \exp\left\{-4^n\varphi(2^n)u^2\right\}
&q_{n}\left(\sqrt{\frac{1+A_{n}}{1+A_{n+1}}}x-u,T\right)\\
&\qquad q_{n}\left(\sqrt{\frac{1+A_{n}}{1+A_{n+1}}}x+u,T\right)\,du,
\endaligned \tag7
$$
and the starting function $q_0(x)$ is
$$
q_0(x)=q_0(x,T)=C_0(T)\exp\left\{ \frac{A_0-T}{1+A_0}\frac
{x^2}2-\frac{tT^2}{(1+A_0)^2}\frac{|x|^4}4\right\}.\tag$7'$
$$
We can study the functions $q_n(x)$ defined by formulas (7) and ($7'$)
instead of the functions $p_n(x)$. Observe that the recursive relation
(7) does not contain the parameter $T$, it appears only  in the
starting function $q_0(x)$. In formula (7$'$) the coefficient of
$|x|^4$ is always negative and the coefficient  of $x^2$ is negative
for $T>A_0$, and it is positive for $T<A_0$. As a consequence, the
function $q_0(x)$ has its minimum at zero in the first case and at
$M_0=\sqrt{\frac{A_0-T}{tT^2}(1+A_0)}$ in the second
case. The integral operator in (7) is very similar to the convolution,
the main difference between them is the appereance of the kernel
$\exp\left\{-4^n\varphi(2^n)u^2\right\}$  in our case. The appereance
of this kernel has far reaching consequences. It causes some
localization in the following sense: The main contribution to the
integral in (7) is given for small $u$. As a consequence, the original
form of the starting function may be preserved, and this is the reason
why the function $q_n(x)$ may have different behaviour for large and
small $T$. For large  $T$ the starting function  $q_0(x)$ is
essentially concentrated in a small neighbourhood of the origin, and
the integral (7) can be well approximated by convolution. For  small
$T$ the situation is different, and actually we are interested in the
following question: If the maximum $M_0>0$ around which point the
function $q_0(|x|)$ is concentrated is very large, will  this
property be preserved for large $n$ too? The answer  depends on how
large the coefficient $4^n\varphi(2^n)$ in the kernel is. Let us also
remark that in the vector valued case formulas (7) and (7$'$) imply
that the function $q_n(x)$ is in the space of rotation invariant
functions for all $n$. This constraint implies a different behaviour
in scalar and vector valued cases. We explain how to investigate these
models at low temperatures.
 
In the scalar valued case at small $T$ the function $q_0(x)$ is
concentrated
around some point $\pm M$, $M>0$, and we are interested in whether
this property will be preserved for all $n$. For this reason we make
an appropriate rescaling. We try to define some new function
$g_n(x)=A_nq_n(M_n+A_nx)$, where $M_n$ is the place of maximum
of the function $q_n(x)$, and the number $A_n$ is chosen in such a way
that the relation $g_{n+1}(x)=\bt g_n(x)$ hold with some operator
$\bt$ not depending on $n$. Then the stability properties of this
operator $\bt$ must be investigated, and if it is stable enough, then
the relation $\lim_{n\to \infty}g_n(x)=g^*(x)$ holds, where $g^*(x)$
is the solution of the fixed-point equation $g^*(x)=\bt g^*(x)$.
During these  calculations we also get a recursive relation for the
numerical sequence $M_n$, and  the question whether there is a phase
transition or not depends on whether $\lim_{n\to\infty}M_n>0$ or not.
It is relatively simple to carry out this program if
$4^n\varphi(n)>\alpha^n$ with some $\alpha>1$. In this case there is
a phase-transition. The question is much harder in the case when
although $4^n\varphi(n)\to\infty$, but it does not increase
exponentially fast. Actually, this is the case we are first of all
interested in.
 
In this case the kernel in the integral of (7)  has a smaller effect,
and to carry out the above sketched program one needs to study the
behaviour
of the function $q_n(x)$ in the interval $[-M_n,M_n]$ more carefully.
In this interval the localization property of the integral (7)
behaves differently. In the point zero the main contribution is given
by $u=\pm M_n$, in the point $M/2$ by  $u=\pm M_n/2$, etc. If the
kernel in the integral in (7) tends to infinity slowly, then these
contributions can be very essential. It may happen that  new peaks
appear in this interval which may be the new maximum. The above
sketched program can be carried out, but it is much more sophisticated
because of the intricate behaviour of the function $q_n(x)$ in the
interval $[-M_n,M_n]$. We cannot explain the main ideas  of this
argument in this short note, we only give  the recursive formula for
$M_n$ this procedure gives. (For some more discussion see ~[2].)
It is:
$$
M_{n+1}\sim M_n\left[1-\exp
\left\{-\frac1T4^n\varphi(2^n)M^2_n\right\}\right].\tag8
$$
 
Formula (8) shows that in the case
$4^n\varphi(2^n)>\text{const.}\,\log
n$ and $T>0$ is small, (hence $M_0>0$ is large)
$\lim_{n\to\infty}M_n>0$, therefore there is a phase-transition. On
the other hand, if $4^n\varphi(2^n)/\log n\to 0$ then $M_n\to 0$, and
there  is no phase transition. Moreover, the special case
$4^n\varphi(2^n)=\text{const.}\,\log n$ has the following remarkable
property: If $\frac {M_k^2}T<1$ for some $k$,  then
$M_{n+1}<(1-n^{-\alpha})M_n$ for all $n\ge k$ with some $\alpha<1$,
hence $\lim_{n\to\infty}M_n=0$. Define the function
$M(T)=\lim_{n\to\infty}M_n(T)$, which is called in the literature the
spontaneous magnetization. Since there is some $T_0>0$
such that $M(T)=0$ for $T>T_0$, the above property implies that either
$M^2(T)>T_0$ or $M(T)=0$, i.e.\ the  spontaneous magnetization as a
function of the temperature has a discontinuity. This is called the
Thouless effect in the literature. Model 2.a) of the previous section
has a similar property in the special case $\alpha=2$. (See [1].)
 
In the vector valued case the situations when $4^n\varphi(2^n)$
increases exponentially fast and when it increases slower can be
similarly investigated. Since the function $q_n(x)$ is rotation
invariant, it is useful to introduce the scalar valued function
$Q_n(x)=q_n(x,0,\dots,0)$, where $x\in \r^1$ and $(x,0,\dots,0)\in
\r^s$ and to study this function. For the sake of simpler notations
let us assume that $s=2$. The function $Q_n(x)$ is
essentially concentrated in a small neighbourhood of the points $\pm
M_n$, where $M_n$ is defined  as
$$
M_n=\int_0^\infty xQ_n(x)\,dx.
$$
We want to give a good asymptotic of the function $Q_n(x)$ in these
neighbourhoods, together with a recursive formula on the sequence
$M_n$. The main technical difficulty is that when rewriting formula
(7) for the function $Q_n(x)$    the arguments
$\sqrt{\frac{1+A_{n}}{1+A_{n+1}}}x\pm u$
turn into the upleasant expressions $\sqrt{(B_nx\pm u_1)^2+u_2^2}$,
with $B_n=\frac{1+A_{n}}{1+A_{n+1}}$. Since we are interested in a
good asymptotic only in the case $x\sim M_n$, and the main
contribution to the integral expressing $Q_n(x)$ for such $x$ is given
when $u_1$ and $u_2$ are small, we commit a small error by replacing
this argument by the expression $B_nx\pm
u_1+\frac{u_2^2}{2B_n^2M_n^2}$.
Then we can continue our argument similarly to the scalar valued case.
Let us introduce the functions $f_n(x)=4^{-n}\varphi(2^n)^{-1} Q_n
(M_n+4^{-n}\varphi(2^n)^{-1}x)$. We cannot express $f_{n+1}(x)$ as an
operator of $f_n(x)$ not depending on $n$,  but the following weaker
statement holds: We can write $f_{n+1}(x)=\bt
f_n(x)+\varepsilon_n(x)$, where $\varepsilon_n(x)$ is a small error
term. The operator $\bt$ is written down explicitly in [3], and its
stability is also studied there. Here we omit the details. What is
important for us is that this investigation gives us the relation
$$
M_{n+1}\sim M_n-\frac1{M_n4^n\varphi(2^n)},
$$
or because of the relation $M_n\sim M_{n+1}$ we commit negligible
error by rewriting the last  formula as
$$
M_{n+1}^2- M_n^2\sim \frac2{4^n\varphi(2^n)}.
$$
This relation shows that in Dyson's vector valued model there is a
phase-transition at low temperatures if the sum
$$
\sum_{n=1}^\infty \frac1{4^n\varphi(2^n)}
$$
is convergent, and there is no phase-transition if this sum is
divergent.
 
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\bye
 
