\magnification 1200
\hsize 17truecm
\vsize 24truecm
\font\ten=cmbx10 at 13pt
\font\twelve=cmbx10
\font\eight=cmr8

\baselineskip=14pt
\parskip=5pt plus 1pt minus 1pt
\null
\def \b{\beta}
\def \sg{\sigma}
\def \ep{\varepsilon}
\def\t{\theta}
\def \sk{\sum_{j=1}^k}
\def \pr{\prod_{j=1}^k a(h_j)\nu_n(h_j)} \def \prv{\prod_{v\in
S(x)}
a(h_v)\,\nu_v(h_v)} \def \ff{f(h_1)+\dots+f(h_k)}
\def \di{\displaystyle}
\def \artanh{\,{\rm artanh}\,}
\def \vi{\vskip 2mm\noindent}

\font\eightrm=cmr8
\font\ninerm=cmr9
\font\eightbf=cmbx8
\font\ninebf=cmbx9
\font\eighti=cmti8

{
\centerline{{\twelve Centre de Physique
Th\'eorique}\footnote{$^{\star}$}{\eight Unit\'e Propre de Recherche
7061}{\twelve - CNRS - Luminy, Case 907}}
\centerline{{\twelve F--13288 Marseille Cedex 9 -
France}}

\vskip 2truecm

\centerline{\ten ON THE PURITY OF THE }
\centerline{\ten LIMITING GIBBS STATE FOR THE}
\centerline{\ten  
ISING MODEL ON THE BETHE LATTICE}
\bigskip

\centerline{  {\bf P.M. BLEHER}\footnote{ $^1$}{\eight Department 
of Mathematical Sciences, Indiana University, Purdue
University at Indianapolis,}\footnote{}{\eight 402 N. Blackford Street,
Indianapolis, IN 46202-3216 USA}\footnote{}{\eight E-mail address: 
bleher@math.iupui.edu}{\bf,  J. RUIZ}\footnote{ $^2$}
{\eight E-mail address: ruiz@cpt.univ-mrs.fr}{\bf and V.A.
ZAGREBNOV}\footnote{$^3$}{\eight and D\'epartement de Physique, 
Universit\'e d'Aix-Marseille II}\footnote{}{\eight E-mail address: 
zagrebnov@cpt.univ-mrs.fr}  }

\vskip 2truecm

\centerline{\bf Abstract} 

\medskip





We give a proof that for the Ising model on the Bethe
lattice, 
the limiting Gibbs state 
with zero effective field (disordered state) 
persists to be pure for temperature below ferromagnetic critical
temperature
$T_c^F$ till the critical temperature $T_c^{SG}$ 
of the corresponding spin-glass model.
This new proof revises the one proposed in [1].

\vskip 3truecm

\noindent Key-Words : Pure Gibbs state, Ising-Bethe model,
spin-glass.
\bigskip


\noindent September 1994

\noindent CPT-94/P.3074

\bigskip

\noindent anonymous ftp or gopher : cpt.univ-mrs.fr

\footline={} 

\vfill\eject    }

\noindent
{\bf 1.}
The question about the possibility of the disordered limiting Gibbs
state (i.e., with zero effective field) of the ferromagnetic Ising model
on the Bethe lattice, to be pure below  the ferromagnetic
critical temperature $T_c$ has been formulated
as an open problem in [2,3] (some non--homogeneous pure Gibbs
states were constructed in [4,5]). In [1] it was proved that the
disordered phase is pure up to the spin--glass critical temperature
$T_c^{SG}$, but as was later pointed out to the author by
Hans-Otto Georgii, the proof contained a gap. 
A corrected version of the proof [6] was based on an
inequality valid for the Bethe lattice $\tau^k$ of degree $k \leq 6$.
In the present note we give a new proof valid for all $k$.
\vskip 2mm
\noindent
{\bf 2.} Let $\tau^k$ be the Bethe lattice of degree $k \geq 2$ such that
exactly $(k+1)$ edges  come out of any of it vertex.
Then the Ising model is defined by the Hamiltonian
$$
H(\sg)=- \sum_{ x,y }J_{xy} \;  \sg (x) \sg (y)
\eqno(2.1)
$$
where the sum is over pairs of nearest neighbors $x,y$ and $\sg (x)
= \pm1$ The ferromagnetic model corresponds to $J_{xy} = J > 0$, 
while the spin-glass model
corresponds to $J_{xy} = \pm J$ $(J > 0)$, where $\{ J_{xy} \}$ are
independent random variables with $\Pr\{J_{xy} = +J\} =\Pr\{J_{xy}
= -J\}=1/2$ for any pair $xy$. Denote 
$$
\theta = \tanh \b,\qquad \b=J / k_B T,
\eqno(2.2)
$$
where $T$ is the temperature.
Then for the ferromagnetic model the critical value $\theta^F_c =
1/k$
(see e.g. [3]), while
for the spin-glass model $\theta^{SG}_c = 1/\sqrt{k}$, see e.g.
[7].
The main result of this note is the following theorem
\vskip 2mm
\noindent
{\bf Theorem.}
\it 
The limiting Gibbs state with zero
effective field
(disordered phase) is pure for $0 < \theta \leq \theta ^{SG}_c$.
\rm
\vskip 2mm
\noindent
{\bf Remark.} 
For 
$0 < \theta \leq \theta ^{F}_c$ the limiting Gibbs state is unique,
and so
trivially pure.
On the other hand, for $\theta > 1/\sqrt{k}$ the disordered phase
is not
pure [2,4]. Thus
our main result concerns the interval $[1/k, 1/\sqrt{k}]$.


\vskip 2mm
\noindent
{\bf 3.}
Let $\tau^k=(V,L,i)$ be the Caley tree of order $k$ with a root
vertex
$x_*\in V$. Here $V$ is
the set of vertices, $L$ is the set of edges and $i$ is the
incidence function which
corresponds to each edge $l\in L$ its end-points $x_1,x_2\in V$.
There is a
distance $d(x,y)$
on $V$ which is the length of the minimal path from $x$ to $y$,
assuming
that the length of
any edge is 1. Denote by
$$
W_n=\{x\in V: \; d(x_*,x)=n\}
$$
the sphere of radius $n$ on $V$, and by
$$
V_n=\{x\in V: \; d(x_*,x)\le n\}
$$
the ball of radius $n$, so that
$$
V_n=\cup_{m=0}^n W_m.
$$
For any $x\in W_n,\; n=0,1,2,\dots,$ denote by 
$$
S(x)=\{y\in W_{n+1}:\; d(x,y)=1\}.
$$

Let $\{h(x),\; x\in V\}$ be a set of real numbers satisfying for
each $x\in
V$ the recursive 
relation 
$$
h(x)=\sum_{y \in S(x)} f_{\t}(h(y)),
\quad f_\t(h)={\rm artanh}\,(\theta\tanh(h)). 
\eqno (3.1)
$$
Define the Gibbs probability distribution on the configurational
space
$$
\Sigma(V_n)=\{\sigma_n=\{\sigma(x)=\pm 1,\; x\in V_n\}\} 
$$
by the formula
$$
\mu_n(\sg_n)=Z_n^{-1}
\exp\left(\b\sum_{x,y\in V_n:\, d(x,y)=1}\sg(x)\sg(y) +\sum_{x\in
W_n}
h(x)\sg(x)\right).
\eqno (3.2)
$$
Then (3.1) implies the consistency of $\mu_n$ for different $n$ so
that 
there exists a Gibbs measure $\mu$ on the infinite configurational
space 
$$
\Sigma(V)=\{\sigma=\{\sigma(x)=\pm 1,\; x\in V\}\} 
$$
whose finite--dimensional distributions are $\mu_n$. 
The function $h(x),\; x\in V,$ which satisfies (3.1), is called an
effective field.
Remark, that $h(x)\equiv 0$ satisfies (3.1), so that
the finite--dimensional distributions
$$
\mu_n^\#(\sg_n)=Z_n^{-1}
\exp\left(\b\sum_{x,y\in V_n:\, d(x,y)=1}\sg(x)\sg(y) 
\right)
\eqno (3.3)
$$
are consistent and generate a Gibbs measure $\mu^\#$, the
disordered phase.

To prove that $\mu^\#$ is pure we will prove that the spins 
$\{\sg(x),\; x\in V_n\}$ and $\{\sg(x),\; x\in W_N\}$ are
asymptotically
independent with respect to $\mu^\#$, when $N\to\infty$ and $n$ is
fixed.
To that end we first fix $N>0$ and
define recursively for every $x\in V_{N-1}$ a random variable $h_x$
such
that
$$
h_x=\sum_{y\in S(x)} f_\t(h_y),\qquad \forall\, x\in V_{N-2}, 
\eqno (3.4)
$$
with the initial data
$$
h_x=\b\sum_{y\in S(x)} \sg(y),\qquad \forall\, x\in W_{N-1}. 
\eqno (3.5)
$$
Eqs. (3.4), (3.5) define $h_x$ as a function of $\sg(y)$, 
$y\in W_N\cap V_x$, where $V_x$ is the subtree growing from $x$.
It is to be noted that in general the random variables 
$\{h_x,\; x\in V_{N-1}\}$ depend on $N$.

\vskip 3mm

\noindent
{\bf Lemma 3.1.} 
{\it For every $n\le N-1$, the joint distribution of
$\sg_n=\{\sg(x),\;x\in
V_n\}$ and
$h^{(n)}=\{h_x,\; x\in W_n\}$ with respect to $\mu^\#$ is given by 
$$
\mu^\#(\sg_n,h^{(n)})=Z^{-1}\exp\left(-H_n(\sg_n) +
\sum_{x\in W_n}h_x\sg(x)\right)
\prod_{x\in W_n}\nu_x(h_x),
\eqno (3.6)
$$
where the probability distributions $\nu_x(h)$ are defined by the
recursive
equations
$$
\nu_{x}(h)={\di\sum_{h_y,\dots,h_z:\,
f(h_y)+\dots+f(h_z)=h}
\left(\prv\right)\over
\di\sum_{h_y,\dots,h_z} \left(\prv\right)}\,, \qquad \forall\,x\in
V_{N-2},
\eqno (3.7)
$$
where
$$
S(x)=(y,\dots,z),\qquad f(h)=f_\t(h),
\qquad a(h)=[1+(1-\t^2)\sinh^2h]^{1/2},
\eqno (3.8)
$$
and by the initial condition
$$
\nu_{x}(h)=Z^{-1}\sum_{\b\,[\sg(y)+\dots+\sg(z)]=h} 
1\,,
\qquad \forall\,x\in W_{N-1}.
\eqno (3.9)
$$}

\noindent
{\bf Remark.} Lemma 3.1 allows the following extension.
Let $\mu$ be a Gibbs measure generated,
according to (3.2),
by some set of real numbers $\{ h(x),\; x\in V\}$ which
satisfy the consistency equations (3.1), and let $\{ h_x,\;
x\in W_{N-1}\}$ be random variables defined through (3.4) and
(3.5). 
Then the joint distribution $\mu (\sg_n,h^{(n)})$ has the 
form (3.6) with probability distributions $\nu_x(h)$ defined by
the recursive equations (3.6) and by the initial condition
$$
\nu_{x}(h)=Z^{-1}\sum_{\b\,[\sg(y)+\dots+\sg(z)]=h} 
\exp [h(y)\sg(y)+\dots+h(z)\sg(z)]\,,
\qquad \forall\,x\in W_{N-1}.
\eqno (3.10)
$$
When $h(x)\equiv 0$, (3.10) reduces obviously to (3.9).
If, in addition, the coupling constant $J_{xy}$ depend on
$xy$ then again the formulas (3.2)--(3.8), (3.10) are valid with
$$
\b=J_{xy}/k_BT,\qquad 
\t=\tanh(J_{xy}/k_BT),
$$
and some natural modifications in these formulas.
The proof of this extension of
Lemma 3.1 is similar to the proof of Lemma 3.1.

\vskip 2mm
\noindent
{\it Proof of Lemma 3.1.} The proof of (3.7) is based on the
following identity:
$$
\sum_{\sg(y)=\pm 1}\exp [\b\sg(x)\sg(y)+h_y\sg(y)]
=Z\,a(h_y)\exp\left[f_\t(h_y)\,\sg(x)\right],\quad {\rm
if}\quad\sg(x)=\pm 1,
\eqno (3.11)
$$
where $Z=\cosh\b$.
To prove this indentity notice that according to $\sg(x)=\pm 1$, it
reduces to
two identities:
$$\eqalign{
&\exp(\b+h_y)+\exp(-\b-h_y)=Z\,a(h_y)\exp f_\t(h_y),\cr
&\exp(-\b+h_y)+\exp(\b-h_y)=Z\,a(h_y)\exp [-f_\t(h_y)].\cr}
$$
If we divide the first identity by the second one then we obtain
$$
{\exp(\b+h_y)+\exp(-\b-h_y)\over
\exp(-\b+h_y)+\exp(\b-h_y)}=\exp [2f_\t(h_y)],
$$
which follows easily from the formula $\exp[2\artanh
z]=(1+z)/(1-z)$.
If we multiply the first identity by the second one then we obtain
$$
\cosh^2\b+\sinh^2 h_y=Z^2a^2(h_y),
$$
which follows easily from the formula
$a^2(h)=1+\sinh^2h/\cosh^2\b$.
Thus (3.11) is proved. Multiplying (3.11) over $y\in S(x)$ we
obtain (3.7)
by induction.
(3.9) follows directly from (3.3). Lemma 3.1 is proved.
$\diamondsuit$


\vskip 3mm
\noindent
{\bf 4.}
By (3.9) and (3.7) the measures $\nu_x(h)$ coincide for all
$x\in W_n$ so we will denote them by $\nu_n(h)$. By (3.9) and (3.7)
 $\nu_n(h)$ is symmetric, hence
$$
\sum_h\sinh h\,\nu_n(h)=0.
$$
Let
$$
D_n=\sum_h\sinh^2 h\,\nu_n(h).
$$

\vskip 2mm
\noindent
{\bf Lemma 4.1.} {\it If $k\t^2\le 1$ then
$$
D_{n-1}\le k\t^2D_n.
\eqno (4.1)
$$} 

\vskip 2mm
\noindent
{\it Proof.} It is convenient to rewrite (3.4) and (3.7) in the
form
$$
h=\sk f(h_j),\qquad f(h)=f_\t(h),
$$
and
$$
\nu_{n-1}(h)=
{\di\sum_{h_1,\dots,h_k: 
f(h_1)+\dots+f(h_k)=h} \left(\pr\right)\over
\di\sum_{h_1,\dots,h_k}
\left(\pr\right)}. \eqno (4.2)
$$
Denote
$$
s=\sinh h,\qquad s_j=\sinh h_j,\qquad t_j=\tanh h_j. 
$$
We have
$$
s^2=\sinh^2h=\sinh^2(\ff)={\tanh^2(\ff)\over 1-\tanh^2(\ff)},
$$
and
$$
\tanh(x_1+\dots+x_k)=
{\sum_{{\rm odd}\; p}
\sum_{j_1<j_2<\dots<j_p}y_{j_1}\dots y_{j_p} \over
\sum_{{\rm even}\; p}
\sum_{j_1<j_2<\dots<j_p}y_{j_1}\dots y_{j_p}}\,, \qquad y_j=\tanh
x_j,
$$
so
$$
\tanh(\ff)=
{\sum_{{\rm odd}\; p}\t^p
\sum_{j_1<j_2<\dots<j_p}t_{j_1}\dots t_{j_p} \over
\sum_{{\rm even}\; p}\t^p
\sum_{j_1<j_2<\dots<j_p}t_{j_1}\dots t_{j_p}}, 
$$
and
$$\eqalign{
s^2=\sinh^2 h
&={\left(
{\sum_{{\rm odd}\; p}\t^p
\sum_{j_1<j_2<\dots<j_p}t_{j_1}\dots t_{j_p} \over
\sum_{{\rm even}\; p}\t^p
\sum_{j_1<j_2<\dots<j_p}t_{j_1}\dots t_{j_p}} \right)^2
\over
1-\left(
{\sum_{{\rm odd}\; p}\t^p
\sum_{j_1<j_2<\dots<j_p}t_{j_1}\dots t_{j_p} \over
\sum_{{\rm even}\; p}\t^p
\sum_{j_1<j_2<\dots<j_p}t_{j_1}\dots t_{j_p}} \right)^2}\cr
&={\left(
\sum_{{\rm odd}\; p}\t^p
\sum_{j_1<j_2<\dots<j_p}t_{j_1}\dots t_{j_p} \right)^2
\over
\di\prod_{j=1}^k (1-\t^2 t_j^2)}\,.\cr}
$$
Using the symmetry of $\nu_n(h)$ we obtain that 
$$
\eqalign{
D_{n-1}&=\sum_h\sinh^2 h\,\nu_{n-1}(h)\cr &=
{\di\sum_{h_1,\dots,h_k}
\left(
\sum_{{\rm odd}\; p}\t^{2p}
\sum_{j_1<j_2<\dots<j_p}t_{j_1}^2\dots t_{j_p}^2 \right)
\prod_{j=1}^k[(1-\t^2t_j^2)^{-1}a(h_j)\nu_n(h_j)] \over
\di\sum_{h_1,\dots,h_k}\left(\pr\right)}\,.\cr} 
$$
Now we substitute
$$
t_j^2={s_j^2\over 1+s_j^2}
$$
and obtain
$$
D_{n-1}=
{\di\sum_{h_1,\dots,h_k}
\left(
\sum_{{\rm odd}\; p}\t^{2p}
\sum_{j_1<j_2<\dots<j_p}s_{j_1}^2\dots s_{j_p}^2
\prod_{j\not=j_1,\dots,j_p}(1+s_j^2)
\right)
\prod_{j=1}^k a^{-1}(h_j)\nu_n(h_j)
\over
\di\sum_{h_1,\dots,h_k}\left(\prod_{j=1}^k
a(h_j)\nu_n(h_j)\right)}\,. 
\eqno (4.3)
$$
\vi
To estimate the RHS of (4.3) we use FKG inequality.

\vi
{\bf Lemma 4.2} (FKG inequality). {\it Assume that $b(x),\;
c(x)$ and $d(x)$ are functions on a finite set $X$ such that
$$
\eqalign{
&d(x)\ge 0\qquad \forall\, x\in X;\cr
&[b(x_1)-b(x_2)]\,[c(x_1)-c(x_2)]\ge 0\qquad
\forall\, x_1,x_2\in X;\cr
&\sum_{x}d(x)>0,\qquad \sum_{x} c(x)d(x)>0.}
$$
Then
$$
{\sum_{x} b(x)c(x)d(x)\over
\sum_{x} c(x)d(x)}
\ge
{\sum_{x} b(x)d(x)\over
\sum_{x} d(x)}.\eqno (4.4)
$$}

\vi
{\it Proof.} We have:
$$\eqalign{
0&\le \sum_{x_1,x_2}[b(x_1)-b(x_2)]\,[c(x_1)-c(x_2)]d(x_1)
d(x_2)\cr
&=2\sum_{x_1}b(x_1)c(x_1)d(x_1)\sum_{x_2}d(x_2)
-2\sum_{x_1} b(x_1)d(x_1)\sum_{x_2} c(x_2)d(x_2),\cr
}$$
which implies (4.4). $\diamondsuit$

\vskip 2mm

Applying FKG inequality to the RHS of (4.3), we obtain
$$
D_{n-1}\le
{\di\sum_{h_1,\dots,h_k}
\left(
\sum_{{\rm odd}\; p}\t^{2p}
\sum_{j_1<j_2<\dots<j_p}s_{j_1}^2\dots s_{j_p}^2
\prod_{j\not=j_1,\dots,j_p}(1+s_j^2)
\right)
\prod_{j=1}^k \nu_n(h_j)
\over
\di\sum_{h_1,\dots,h_k}\left(\prod_{j=1}^k a^2(h_j)\nu_n(h_j)
\right)}\,,
$$
which gives
$$
D_{n-1}\le
{\di\sum_{{\rm odd}\; p}C_k^p\t^{2p}D_n^p(1+D_n)^{k-p} \over
[1+(1-\t^2)D_n]^k}\,,
\qquad C_k^p={k!\over p!(k-p)!}\,.
\eqno (4.5)
$$

\vskip 3mm
\noindent
{\bf Lemma 4.3.} {\it If $k\t^2\le 1$ then 
$$
{\di\sum_{{\rm odd}\; p}C_k^p\t^{2p}D_n^p(1+D_n)^{k-p} \over
[1+(1-\t^2)D_n]^k}
\le k\t^2 D_n.
\eqno (4.6)
$$}




\noindent
{\it Proof.} Denote for a fixed $D_n>0$, 
$$
A(\t)=
{\di\sum_{{\rm odd}\; p}C_k^p\t^{2p}D_n^p(1+D_n)^{k-p} \over
k\t^2D_n[1+(1-\t^2)D_n]^k}
={\di\sum_{{\rm odd}\; p}C_k^p\t^{2(p-1)}D_n^{p-1} (1+D_n)^{k-p}
\over
k[1+(1-\t^2)D_n]^k}\,
$$
Then (4.6) is equivalent to
$$
A(\t)\le 1.
$$
Observe that $A(\t)$ is an increasing function of $\t$, so it is
sufficient
to prove that
$$
A(k^{-1/2})\le 1. 
\eqno (4.7)
$$
By the Newton binomial formula,
$$
\eqalign{
A(k^{-1/2})&=
{\di\sum_{{\rm odd}\; p}C_k^p k^{-p}D_n^p(1+D_n)^{k-p} \over
D_n[1+(1-k^{-1})D_n]^k}\cr
&={(1+D_n+k^{-1}D_n)^k
-(1+D_n-k^{-1}D_n)^k
\over
2D_n[1+(1-k^{-1})D_n]^k}\cr
&={\di\left(
{1+(1+k^{-1})D_n
\over
1+(1-k^{-1})D_n}\right)^k-1
\over
2D_n}\,.
\cr}
\eqno (4.8)
$$
Put
$$
z=
{1+(1+k^{-1})D_n
\over
1+(1-k^{-1})D_n}-1
={2D_n\over k+(k-1)D_n}\,.
$$
Then
$$
0<z<{2\over k-1}\,,\qquad {\rm when} \qquad 0<D_n<\infty. 
$$
In addition,
$$
D_n={kz\over 2-(k-1)z}\,,
$$
and (4.7) is equivalent (due to (4.8)) to 
$$
(1+z)^k-1\le 2D_n={kz\over 1-(k-1)z/2}\,. 
$$
Expanding the both sides of the
last inequality in the Taylor series in $z$ we reduce it to
$$
kz+{k(k-1)\over 2}z^2+{k(k-1)(k-2)\over 6}z^3
+\dots\le kz+{k(k-1)\over 2}z^2+{k(k-1)^2\over 4}z^3+\dots, 
\eqno (4.9)
$$
which is obvious. Thus (4.7) is proved and this finishes the proof
of Lemma
4.3.
$\diamondsuit$

{}From (4.5) and Lemma 4.3 we obtain (4.1). Lemma 4.1 is proved.
$\diamondsuit$

Now we observe that (4.9) can be strengthened to 
$$
kz+{k(k-1)\over 2}z^2+{k(k-1)(k-2)\over 6}z^3+\dots
\le { kz+\di{k(k-1)\over 2}z^2+{k(k-1)^2\over 4}z^3+\dots \over
1+\di{(k-1)^2\over 12}\,z^2}\,,
$$
which leads to
$$
(1+z)^k-1\le{2D_n\over
\di 1+{(k-1)^2\over 12}\,\left( {2D_n\over k+(k-1)D_n}\right)^2}
$$
and then we get the following

\vskip 3mm
\noindent
{\bf Lemma 4.4.} 
$$
D_{n-1}\le {k\t^2 D_n
\over
\di 1+{(k-1)^2\over 3}\,\left({D_n\over k+(k-1)D_n} \right)^2}\,.
\eqno (4.10)
$$

This is useful when $k\t^2=1$.

\vskip 3mm
\noindent
{\bf 5.} {\it Proof of the Theorem.}
The inequality (4.10) says that
$$
\lim_{N-n \to \infty} D_n =0
\eqno (5.1)
$$
for $ 0< \theta \leq 1/ \sqrt{k}$.
Following the line of reasoning of Section 3 from [1], one can now
check 
that (5.1) implies the extremality of the disordered state
$\mu^\#$.
This means that for any $\varepsilon >0$, $n>0$ and any
configuration 
$\sg_n  \subset \Sigma(V_n)$,
there exist  $N>n$ and a set $\Omega_N \subset \Sigma(W_N) $ such
that
\vskip 2mm
\begingroup
\item{(i)}
$\mu^\#(\Omega_N) > 1- \varepsilon$
\item{(ii)}
$\vert \mu^\#(\sg_n \mid \sg^{(N)}) -\mu^\#(\sg_n) \vert <
\varepsilon$, 
$\forall \sg^{(N)} \in \Omega_N$
\endgroup
\vskip 2mm
\noindent
This gives our main result. $\diamondsuit$

One of the open problems remains  about the purity of the limiting
Gibbs
states in the case of nonzero external field. 
Another open problem is the characterization of the limiting states
for
the random field model.
The ground states for this model for binary distribution were
examined in
[8].


\vskip 3mm
\noindent
\noindent {\bf Acknowledgements.}
One of us (P.M.B.) would like to express his sincere gratitude
to Hans-Otto Georgii for pointing out a gap in the original
proof [1] of the main theorem of the present paper. 
A corrected version of the proof for $k\le 6$ was given then
in the letter of P.M.B. to H.-O. Georgii [6] and the general case
is considered here. P.M.B. thanks 
the Centre de Physique
Th\'eorique--Marseille for
hospitality and the Universit\'e de Provence for financial support
during his visit to CPT--Marseille where this work was done.

\vfill\eject

\noindent {\bf References.}

\medskip

\begingroup

\item{1.} P.M. Bleher,
{\it Extremity of the disordered phase in the Ising model on the
Bethe
lattice}, 
Commun. Math. Phys. {\bf 128}, 411-419 (1990).
\smallskip
\item{2.} T. Moore, J.L. Snell,
{\it A branching process showing a phase transition}, 
J. Appl. Prob. {\bf 16}, 252-260 (1979).
\smallskip
\item{3.} H.-O. Georgii,
{\it Gibbs Measures and Phase Transitions} 
(De Gruyter Studies in Math., vol. 9)
Berlin-New York: De Gruyter 1998.
\smallskip
\item{4.}
Y. Higuchi,
{\it Remarks on the limiting Gibbs states on a ($d+1$)-tree},
Publ. RIMS Kyoto University
{\bf 13}, 335-348 (1977).
\smallskip
\item{5.} P.M. Bleher and N.N. Ganihodgaev,
{\it On pure phases of the Ising model on the Bethe lattice}, 
Theory Prob. Appl. {\bf 35}, 1-26 (1990).
\smallskip
\item{6.} P.M. Bleher, 
{\it Letter to H.-O. Georgii from January 21, 1992}.
\smallskip
\item{7.} J.M. Carlson, J.T. Chayes, L. Chayes, P. Shetna, D.J.
Thouless, 
{\it Critical behavior of the Bethe lattice spin-glass},
Europhys. Lett. {\bf 5}, 355-360 (1989).
\smallskip
\item{8.} R. Bruinsma,
{\it Random field Ising model on a Bethe lattice}, 
Phys. Rev. B {\bf 30}, 289-299 (1984).

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