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\topmatter
\title
Shapes of growing droplets --- a model of escape from a
metastable phase
\endtitle
%
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\leftheadtext{}
\rightheadtext{}
%
%
\author
R. Koteck\'y and E. Olivieri
\endauthor
\affil
Charles University in Prague and
Universit\`a di Roma ``Tor Vergata''
\endaffil
\address
Roman Koteck\'y
\hfill\newline
Center for Theoretical Study, Charles
University,\hfill\newline
T\'aboritsk\'a 23, 130 87 Praha 3,
Czech Republic
\hfill\newline
\phantom{18.}and
\hfill\newline
Department of
Theoretical Physics, Charles University,\hfill\newline
V~Hole\v sovi\v
ck\'ach~2, 180~00~Praha~8, Czech Republic
\endaddress
\email
kotecky\@cspuni12.bitnet
\endemail
\address
Enzo Olivieri
\hfill\newline
Dipartimento di Matematica, Universit\`a di Roma ``Tor
Vergata'', \hfill\newline
Via della Ricerca Scientifica, 00133 Roma, Italy
\endaddress
\email
olivieri\@irmtvm51.bitnet
\endemail
\keywords
Stochastic dynamics, Ising model,
next nearest neighbour interaction,
metastability, crystal growth, first excursion
\endkeywords
\abstract
Nucleation from a metastable state is studied for an Ising
ferromagnet with nearest and next
nearest neighbour interaction and at very
low temperatures.
The typical escape path is shown to follow a
sequence of configurations with a growing droplet of stable
phase whose shape is determined by
dynamical considerations and differs significantly from
equilibrium shape corresponding to the instanantenous volume.
\endabstract
\subjclass Primary 82A05; secondary 82A25
\endsubjclass
\thanks
The research has been partially supported by CNR -- GNFM.
\endthanks
\endtopmatter
% last update 31.5.
\head 1. Introduction \endhead
Relaxation to equilibrium of a system close to a first order
transition is a problem whose rigorous treatment is recently
receiving a lot of attention \cite{S 1, 2, NS 1, 2, MOS}. In
particular, an interesting question is to grasp the escape
pattern of a system relaxing from a metastable starting
configuration toward a stable equilibrium state. The escape is
through a formation of small droplets or crystals of stable
phase that are stabilized once they reach certain critical size.
Here, we are interested in the shape of such crystal during the
process of growth. It seems that the mechanism of growth depends
on the size of a crystal. While for large supercritical crystals
one has to take into account the transport of matter and heat
around the crystal, the growth of tiny subcritical
crystals should be governed in a more direct way by the
instanantenous microscopic dynamics.
Our aim in the present paper is to discuss the shape of a growing
crystal modeled by the Ising model. Namely, we are considering
the Ising model with ferromagnetic nearest neighbour and next
nearest neighbour interactions in the presence of a small
positive external field. Starting from the configuration
$-\underline 1$, minus spins at all lattice sites in a fixed
finite volume under periodic boundary conditions, we study the
relaxation pattern of the stochastic process yielded by a
standard Glauber dynamics. In particular, we are interested in
the typical configurations during the first excursion from the
configuration $-\underline 1$
to the configuration $+\underline 1$
with all sites occupied by plus spins. We present a detailed
description of the escape pattern in the asymptotic region of
vanishing temperatures. It turns out that in this asymptotics
(and in a finite volume) one can consider a single droplet
of plus spins, playing the role of the crystal, that grows in a
very particular manner to the critical droplet.
On the first
sight one could suppose that growth is through crystals that
minimize the surface tension under fixed instanantenous volume
\cite{Z}.
The shape of such crystals is
the equilibrium shape yielded by the
Wulff construction \cite{W, RW, DKS}. For our model at low
temperatures the Wulff construction can be easily shown to lead
to a shape that closely follows an octagon with coordinate and
oblique sides proportional to the nearest neighbour and next
nearest
neighbour interactions, respectively. The main result of the
present paper (Theorem 3) asserts that the typical growth of
subcritical crystals is through a sequence of particular shapes
that significantly differ from the equilibrium Wulff octagons.
This is in agreement with a similar result [KO 1] concerning
Ising model with anisotropic nearest neighbour interaction.
Notice that the fact that we are considering the asymptotics of
vanishing temperature with fixed (small) external field is
technically crucial for our proofs. It would be interesting
(and difficult) to extend these type of results also to the
region of vanishing external field under a fixed (small)
temperature or to the region where both external field and
temperature vanish in such a way that their ratio is fixed.
The escape time and asymptotics of the metastable state in the
latter region has been recently discussed \cite{S 2} in the case
of the nearest neighbour $d$-dimensional Ising model.
We believe that a difference between dynamical and equilibrium
shapes would occur alrady for the simplest nearest neighbour
Ising model.
But while it would reveal itself only as a higher order effect
at low temperatures, for the model with an additional next
nearest neighbour interaction discussed here it appears already
in the first order of the low temperature asymptotics.
The main effect is thus captured in a situation that is
sufficiently simple to be studied in a rigorous way.
The paper is organized in the following way. In Section~2 we
introduce the model, the dynamics and the notation concerning
octagonal shapes that yield local minima of considered
interaction (Lemma~2.1). Then we summarize our results in Theorem
1 and 2 (asymptotics of the hitting time to the configuration
$+\underline 1$)
and Theorem 3 (describing the sequence of droplet shapes on the
escape path).
It turns out that the evolution of the crystal can be evaluated
by discussing ``the movement in the energy landscape''.
Section 3
is devoted to a detailed investigation of passages between
neighbouring (octagonal) local minima. An important role is
played by a characterization of different basins of attraction
and saddle points on their boundaries.
A closely linked fact is the existence of two different
important time scales in the problem.
The shorter one is a typical time needed for a passage from a
particular octagonal local minimum to a close octagon with
identical circumscribed rectangle.
The height of the corresponding saddle point is proportional to
the value of the next nearest neighbour coupling. Only when
passing to an octagon with larger circumscribed rectangle one
has to overcome a saddle point whose height is proportional to
the nearest neighbour coupling and the typical passage time is
correspondingly longer.
(We are supposing that the nearest neighbour coupling is
stronger than the next nearest neighbour one.)
In terms of this longer time scale we observe a
growing octagonal shape whose oblique sides are ``breathing''
around ``equilibrium positions'' when observed on
a shorter time scale.
The global saddle point, a
configuration minimizing the maximal energies on pathes from
$-\underline 1$ to $+\underline 1$, is discussed in Section 4.
For its investigation it is useful to introduce a ``global basin
of attraction of the configuration $-\underline 1$'' --- resp\.
its subset $\Cal A$ characterized by the fact that starting from
a configuration in $\Cal A$, a typical path first hits the
configuration $-\underline 1$ before reaching the global minimum
$+\underline 1$. The goal is to choose the set $\Cal A$ small
enough to satisfy this condition but, on the other side, large
enough to make the minimum on its boundary to coincide with the
global saddle point.
The results of Section~3 and 4 are than merged in Section~5 to
the proofs of Theorems 1-3.
The basic estimate in getting a typical escape time is the
lower bound on the probability to reach a global saddle point.
A crucial observation is that when passing through an octagonal
local minimum configuration, one has a freedom to stay in its
local basin of attraction for a random time in a time interval
characterized by the minimal saddle point on its boundary. The
resulting estimate can thus be multiplied by corresponding
``resistance times''. It leads to a correct lower bound
only if the path enters the local basin of attraction through
lowest saddle point. This criterion determines the optimal
escape path and is responsible for a particular dynamically
optimal sequence of shapes mentioned above.
The
results of the present paper, as well as those from \cite{KO
1}, were announced in \cite{KO~2}.
\vfill \newpage
% last update 31.5.
\head 2. Setting and results \endhead
We will consider a {\it discrete time Metropolis dynamics}
for a~totally
ferromagnetic two dimensional {\it Ising model} with
{\it nearest neighbours} and
{\it next nearest neighbours interaction}.
The choice of~a~discrete instead of~a~continuous time
evolution is
made only for the~sake of~simplicity of~the~exposition.
It will
appear clear to the~reader that, with some minor changes,
all our
results can be extended to~the~continuous case.
Our dynamics will be
given by~a~Markov chain whose space of~states is
$\Gamma=\{-1,1\}^{\Lambda}$ where $\Lambda$ is a~two
dimensional torus
--- namely, $\Lambda$ is the~square $\{1,\dots,M\}^2$
with periodic
boundary conditions\footnote{For instance $x=(M+1,b)$
is, for every $b\leq M$, identified with $x'=(1,b)$ and
so on.}.
A~configuration~$\sigma$ is a~function
$$
\sigma\:\Lambda\to\{-1,1\},
\tag{2.1}
$$
i.e. $\sigma\in\Gamma=\{-1,1\}^{\Lambda}$. The value
$\sigma(x)$ is the~{\it spin}
at~ the~site $x$.
The {\it energy} of~a~configuration $\sigma$ is
$$
H(\sigma)=-\frac{\tilde J}2\sum_{\subset
\Lambda}\sigma(x)\sigma(y)-\frac
K2\sum_{<\!\!\!\!>\subset\Lambda}
\sigma(x)\sigma(y)-\frac
h2\sum_{x\in\Lambda}\sigma(x),
\tag{2.2}
$$
where we suppose that $K$, $\tilde J$, $h>0$.
Here $<\!\!x,y\!\!>$
denotes a~pair
of~nearest neighbours in~$\Lambda\:|x-y|=1$ and
$<\!\!<\!x,y\!>\!\!>$ denotes a~pair of~next nearest
neighbours
in~$\Lambda\: |x-y|=\sqrt2$.
Our dynamics is defined with the help of the following
{\it updating rule} :
\flushpar
Given the~configuration $\sigma$ at time $t$ we
first choose
at random (with uniform probability) a~site $x\in\Lambda$.
Then we flip the
spin at the site $x$ with probability
$$
\exp(-\beta(\Delta_x H(\sigma))^{+}),
\tag{2.3}
$$
where
$$
\Delta_x H(\sigma)= H\bigl(\sigma^{(x)}\bigr)-H(\sigma)
\tag{2.4}
$$
with
$$
\sigma^{(x)}(y)=
\cases\phantom{-} \sigma(y)&\text{ whenever
}y\ne x,\\-\sigma(x)&\text{ whenever } y=x.
\endcases
\tag{2.5}
$$
Here, for every $c\in\Bbb R$ we denote $(c)^{+}=\min(c,0)$;
$\beta$
is the~
inverse temperature.
The {\it transition probabilities} are then given by
$$
P(\sigma\to\eta)=\cases\frac1{|\Lambda|}\exp(-\beta
(\Delta_x
H(\sigma))^{+}) &\text{ if } \eta=\sigma^{(x)}
\text{ for some }x,\\
0&\text{ otherwise.}\endcases
\tag{2.6}
$$
The {\it space of~the~trajectories} of~our process is
$$
\Omega=\Gamma^{\Bbb N}\equiv(\{-1,1\}^{\Lambda})^{\Bbb
N}.
\tag{2.7}
$$
An~element in $\Omega$ is denoted by $\omega$, it is a~function
$$
\omega:\quad \Bbb N\to\Gamma.
$$
If
$$
\omega=\sigma_0,\sigma_1,\dots,\sigma_t,\dots,
$$
we set
$$
\omega_t\equiv\omega\restriction_t=\sigma_t.
$$
The dynamics is {\it reversible} with regard to the~Gibbs
measure in
$\Lambda$ in the~sense that
$$
P(\sigma\to\eta)e^{-\beta H(\sigma)}\equiv P(\eta\to\sigma)
e^{-\beta
H(\eta)}.
\tag{2.8}
$$
We will discuss a~behaviour at very low temperatures.
Thus it is natural
to~describe configurations in terms of~contours.
Namely, for every
$\sigma\in\Gamma$ consider the union $C(\sigma)$ of~all
closed unit
squares centered at lattice sites $x$ for which
$\sigma(x)=+1$.
The~
boundary $\partial C$ of $C$ is a~polygon with vertices on
the~dual
lattice $\Bbb Z^2+(\frac12,\frac12)$ such that in any vertex
of~the~
dual lattice an~even number ($0$, $2$, or $4$) of~unit
segments belonging to~this polygon meet. Any connected
component $\gamma$ of~the~boundary
$\partial C$ is called a {\it contour} of~the~configuration
$\sigma$.
It is easy to see that if a configuration $\sigma$ has
the boundary
$\partial C(\sigma)$ consisting of a single contour
$\gamma$,
the energy
of $\sigma$ is
$$
H(\sigma)-H(-\underline 1)=J|\gamma|-
K|A(\gamma)|-h|I(\gamma)|.
\tag{2.9}
$$
Here
$$
J=\tilde J+2K,
\tag{2.10}
$$
$|\gamma|$ is the length of $\gamma$, $|I(\gamma)|$ is the
cardinality
(area) of the interior $I(\gamma)\equiv C(\sigma)$.
Finally, $|A(\gamma)|$ is the number of corners
(right angles) of
$\gamma$. Notice that in the situation like that on
Fig\. 2.1
we count 4
corners.
%
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\midinsert
\centerline{\scaledpicture 2.1in by 1.42in (one scaled 800)}
\botcaption{Fig\. 2.1}
\endcaption
\endinsert
%
%
A contour $\gamma$ is said to be {\it isolated} if it lies
at a distance at
least $\sqrt2$ from other contours.
A relevant role will be played by a particular class of
contours that
we call octagons. An {\it octagon} is a closed contour
inscribed in a
rectangle $R$ with edges parallel to the lattice axes.
Call
$P_1,P_2,P_3,P_4$ the vertices of $R$. The octagon
contains four
straight edges with extremes $x_i,y_i,\ i=1,\dots,4:
(x_1,y_1)
\subseteq P_1P_2$, $(x_2,y_2)\subseteq P_2P_3$,
$(x_3,y_3)\subseteq P_3P_4$, and $(x_4,y_4)
\subseteq P_4P_1$,
called
{\it coordinate edges}, and four {\it oblique edges} that
have a local
staircase structure with extremes
$y_1x_2$, $y_2x_3$, $y_3x_4$,
$y_4x_1$. (See
Fig\. 2.2)
%
%
\midinsert
\centerline{\scaledpicture 3.42in by 1.97in (two scaled 900)}
\botcaption{Fig\. 2.2}
\endcaption
\endinsert
%
%
When referring to a {\it stable octagon} we suppose that
$|x_i-y_i|\geq
2$, $|y_i-x_{i+1}|\geq \sqrt2$, $i=1,\dots,4$.
When discussing a particular octagon $Q$, it will be
useful to introduce
$$
L_i=|x_i-y_i|,\quad
i=1,\dots,4,
\tag2.11
$$
for the lengthes of its coordinate edges,
$$
l_i=1+\frac1{\sqrt2}
|y_i-x_{i+1}|,
\quad
i=1,\dots,4,\, (x_5\equiv x_1),
\tag2.12
$$
for the lengths of its oblique edges,
$$
\aligned
D_1 &=\overline{P_1P_2}= L_1 + l_4 -1 +l_1-1
=\overline{P_4P_3}= L_3 + l_3 -1 +l_2 -1,\\
D_2 &=\overline{P_2P_3}= L_2 + l_1 -1 +l_2-1
=\overline{P_4P_1}= L_4 + l_3 -1 +l_4 -1,
\endaligned
\tag2.13
$$
for the lengthes of the sides of its {\it rectangular
envelope} $R(Q)=P_1P_2P_3P_4$, and, finally,
$$
\aligned
{d}_1 &=L_1+L_4 +2(l_4-1) = L_2+L_3 +2(l_2-1),\\
{d}_2 &=L_1+L_2 +2(l_1-1) = L_4+L_3 +2(l_3-1)
\endaligned
\tag2.14
$$
for the distances (in units of $1/\sqrt 2$) between pairs of
opposite parallel oblique edges.
We will use
$$
Q(D_1,D_2,l_1,l_2,l_3,l_4)
\tag{2.15}
$$
to denote the corresponding octagon. Another way of
characterize it, is by specifying, say,
$d_1$, $d_2$, $L_1$, $L_2$, $L_3$, $L_4$.
The previous definitions made no
reference to the {\it
location of the octagon}. Sometimes,
in the sequel, we will consider
a canonical location.
We say that an octagon $Q$ is {\it centered}
if the upper left corner
$x_1$ of $ Q$ is the point $(-1/2,1/2)$ of
the dual lattice
(namely, the uppermost left + spin, the first in
lexicographic order, is the origin).
Sometimes, when not specifying a location of an octagon
$Q$, we tacitly assume it to be centered.
We will often use the same
symbol
$Q$ also to denote the set of all spin
configurations $\sigma$
giving
rise to a unique closed contour
(full of pluses) consisting, up to a
translation on the torus, of the octagon $Q$;
sometimes we write
$\sigma\in Q$. We set $$
H(Q)\equiv H(\sigma)\ ,\quad\text{for }\sigma\in Q\ .
$$
Clearly, for $Q=Q(D_1,D_2,(l_i)_{i=1,\dots,4})$
we have
$$
H(Q)=2J(D_1+D_2)-hD_1D_2-\sum_{j=1}^4[K(2l_j-1)-
\frac12hl_j(l_j-1)].
\tag{2.16}
$$
Let us define
$$
\gathered l^{*}=\left[\frac{2K}h\right]+1\ ,\quad\text{i.e.
}h(l^{*}-1)<2K \bar\tau_{-\underline 1}
\: \sigma_t \in \Cal P\}.
\tag2.26
$$
\proclaim{Theorem 1}
$$
\lim_{\beta\to\infty}P_{-\underline 1}(\bar\tau_{\Cal P}
<\tau_{+\underline 1})=1.
\tag2.27
$$
\endproclaim
\proclaim{Theorem 2}
$$
\lim_{\beta\to\infty}P_{-\underline 1}
(\exp[\beta(E^*-\varepsilon)]
<\tau_{+\underline 1}<\exp[\beta(E^*+\varepsilon)])=1
\tag2.28
$$
for every $\varepsilon > 0$.
\endproclaim
\bigskip
In addition, we are getting a much more detailed information about
a typical path followed by our process $\sigma_t$ during its first
excursion from $-\underline 1$ to $+\underline 1$, or, in other
words, between the moments
$\bar\tau_{-\underline 1}$ and $\tau_{+\underline 1}$.
Theorem 3 below states,
roughly speaking, that with high probability for large
$\beta$, the path $\sigma_t$ sticks to a certain {\it tube }
of trajectories. A precise definition involves a
lot of other preliminary definitions and notions and
we will be able to present it only at the
end of Section 5.
There, we will introduce the concept of an {\it $\epsilon$-typical
path} that wil be determined in terms of its
geometrical properties, but also with a specified times of
passing through certain configurations. For the moment we only say
that, roughly speaking, the typical trajectories during the first
excursion begin by following a sequence of almost regular octagons
up to an edge $l^*$; after that the oblique edge stay almost
constant at the value $l^*$ while the coordinate edges grow,
keeping the rectangular envelope almost squared, from the value
$l^*$ up to the value $L^*$ corresponding to the critical nucleus.
This first part of the first excursion can be viewed as a
nucleation phenomenon and it involves, in average,
``ascending'' transitions with growing energy.
Finally, the oblique edges stay further almost constant
at the value $l^*$ whereas, the coordinate edges continue to grow,
with larger fluctuations, still preserving, however, the average
squared shape of the rectangular envelope until the whole volume is
invaded by pluses. This ``supercritical growth'' is, in
average, a descent in energy.
\proclaim {Theorem 3}
For every $\epsilon >0$ one has
$$
\lim_{\beta\to\infty} P_{-\underline
1}(\{\sigma_t\}_{t\in [\bar \tau_{-\underline 1},
\tau_{+\underline 1}]}\text { is an $\epsilon$-typical
path})=1.
\tag2.29
$$
\endproclaim
\vfill
\newpage
% last update 31.5.
\head 3. Passage between neighbouring local minima \endhead
To be able to discuss in detail the growth or shrinking of
a droplet we
first introduce some ``elementary events'' --- namely,
certain particular
spin-flips.
Notice first that the energy increment, when flipping the
spin of a
configuration $\sigma$ at a site $x$, can be expressed in
the form
$$
\Delta_xH=H\bigl(\sigma^{(x)}\bigr)-H(\sigma)=\bigl(\tilde
JM_{\tilde J}^{(x)}(\sigma)+KM_K^{(x)}(\sigma)+h\bigr)\sigma(x),
\tag{3.1}
$$
where
$$
M_{\tilde J}^{(x)}(\sigma)=\sum_{y:|x-y|=1}\sigma(y)\ ,\
M_K^{(x)}(\sigma)=\sum_{y:|y-x|=\sqrt2}\sigma(y).
$$
We are interested in the region of phase diagram where $K$
is small with
respect to $\tilde J$ (next nearest neighbour interaction is a
perturbation of nearest neighbour interaction). Recall
that we are assuming that
$$
0<7 hx_0,
y>y_0\}$, $\{(x,y)\in R;
x>x_0, yy_0\}$,
and $\{(x,y)\in R;
x0$ one has
a~selfavoiding curve, we can consider (as will
be useful later) a~path
winding around along all the boundary $\partial_{\text{out}}C$.
Our aim now
is to prove first that $\partial_{\text{out}}C\equiv\partial Q$
(see Fig\. 3.3).
%
%
\midinsert
\centerline{\picture 1.92in by 1.61in (threethree)}
\botcaption{Fig\. 3.3}
The set $C$ is shaded, thickened line denotes
$\partial_{\text{out}}C$.
\endcaption
\endinsert
%
%
To this end we inspect a catalogue of locally stable
configurations (with
respect to spin flip at $x$).
We say that a configuration
$\sigma$ is {\it stable at} $x$ if the spin flip
$\sigma(x)\to-\sigma(x)$
increases the energy.
Consider thus a configuration $\sigma$ and a site
$x$ with $\sigma(x)=-1$.
Whether the configuration $\sigma$ is stable
at $x$ depends only
on its value at the nearest and next nearest neighbour
sites of $x$.
Namely, it is stable at $x$
whenever either
$M_{\tilde J}^{(x)}<0$ or $M_{\tilde J}^{(x)}=0$
and $M_K^{(x)}<0$. As
a~result we get the following catalogue
(up to rotations and reflections) of
stable situations around $x$ (dots stand for an~arbitrary spin):
$$\gather
\vbox{%
\halign{\strut\ $#$\ &\ $#$\ &\ $#$\ &#\qquad&\ $#$\
&\ $#$\ &\ $#$\ &#\qquad&
\ $#$\ &\ $#$\ &\ $#$\ &#\qquad&\ $#$\ &\ $#$\ &\ $#$\ \cr
\multispan{3}\hfil a)\hfil&&\multispan{3}\hfil b)
\hfil&&\multispan{3}\hfil c)\hfil
&&\multispan{3}\hfil d)\hfil\cr
\noalign{\vskip 1pt}
\bullet&\bullet&\bullet&&+&+&-&&-&+&-&&-&+&+\cr
-&-&-&&-&-&-&&-&-&-&&-&-&+\cr
\bullet&-&\bullet&&-&+&-&&-&+&-&&-&-&-\cr}
\vskip 3pt}\\
\vbox{%
\halign{\strut\ $#$\ &\ $#$\ &\ $#$\ &#\qquad&\ $#$\
&\ $#$\ &\ $#$\ &#\qquad&
\ $#$\ &\ $#$\ &\ $#$\ \cr
\multispan{3}\hfil e)\hfil&&\multispan{3}\hfil f)
\hfil&&\multispan{3}\hfil g)\hfil\cr
\noalign{\vskip 1pt}
+&+&-&&-&+&-&&-&+&-\cr
-&-&+&&-&-&+&&-&-&+\cr
-&-&-&&+&-&-&&-&-&-\cr}}\endgather$$
Suppose now that $\partial_{\text{out}}C\ne\partial Q$.
Then, considering
a path along $\partial Q$ oriented in the same sense as
that along
$\partial_{\text{out}}C$, there exist two points
$A,B\in\partial_{\text{out}}C\cap\partial Q$ such that
the paths along
$\partial_{\text{out}}C$ and $\partial Q$ between these
two points do not
have a common unit segment and such that either
\roster
\item"a)" the points $A$, $B$ belong to the same side of $Q$
(see Fig\. 3.4A) or
\item"b)" they belong to two neighbouring sides of $Q$ (see Fig\.
3.4B).
\endroster
\vfill\newpage
%
%
\midinsert
\centerline{\scaledpicture 4.79in by 2.07in (threefour scaled 800)}
\botcaption{Fig\. 3.4a\phantom{xxxxxxxxxxxxxxxxxxxxxxxx}Fig\. 3.4b}
\endcaption
\endinsert
%
%
The easiest case to tackle is A) with the side
in question being, say,
horizontal. Namely, consider the path $\gamma$
between $A$ and $B$ and the lowermost
horizontal line $l$ touching $\gamma$. (See Fig\. 3.5)
%
%
\midinsert
\centerline{\scaledpicture 2.40in by 0.89in
(threefive scaled 800)}
\botcaption{Fig\. 3.5}
\endcaption
\endinsert
%
%
At the point where $\gamma$ for the first time (going from $A$ to $B$)
touches $l$, we have the configuration \!\!\!\!
\raise 6pt \hbox{$\quad\vtop{\offinterlineskip
\halign{\strut#&\quad\quad#&\ $#$\ &#&\ $#$\ &#\quad\quad\cr
&&+&\vrule width 1pt&-&\cr
\omit&&&\multispan{2}\leaders\hrule height 0.5pt
depth 0.5pt\hfill&\cr
\omit&\multispan{5}\leaders\hrule height 0.05pt depth
0.05pt\hfill\cr $l$ &&&&+&\cr}}\quad$}\!\!.
For the spin $-$ to be stable it must be \!\!
\raise 6pt \hbox{$\quad\vtop{\offinterlineskip
\halign{\strut#&\quad\quad\ $#$\ &\ $#$\ &\ $#$\
&#\quad\quad\cr
&+&-&-&\cr
\omit&\multispan{4}\leaders\hrule height 0.05pt depth
0.05pt\hfill\cr
$l$ &&+&&\cr}}\quad$}\!\!, and, $l$ being the
lowermost line touching
$\gamma$, we must have
\raise 6pt \hbox{$\quad\vtop{\offinterlineskip
\halign{\strut#&\quad\quad\
$#$\ &\ $#$\ &\ $#$\ &#\quad\quad\cr
&+&-&-&\cr
\omit&\multispan{4}\leaders\hrule height 0.05pt depth
0.05pt\hfill\cr
$l$ &&+&+&\cr}}\quad$}. In view of e) from our catalogue we
necessarily have
\raise 12pt \hbox{$\quad\vtop{\offinterlineskip
\halign{\strut#&\quad\quad\ $#$\ &\ $#$\ &\ $#$\ &#\quad\quad\cr
&-&-&-&\cr
&+&-&-&\cr
\omit&\multispan{4}\leaders\hrule height 0.05pt depth
0.05pt\hfill\cr
$l$ &-&+&+&\cr}}\quad$} --- otherwise the spin $-$
in the centre
would not be stable.
The spin $+$ above the line $l$ is, according to a)
from our catalogue,
always unstable (it has three $-$ nearest neighbours) and we get a
contradiction. In the remaining
cases shown in Fig\. 3.4 we get a
contradiction by the same reasoning, once the lowermost
horizontal line $l$ touching $\gamma$ does not pass
through the~point $B$.
The argument above can be also
interpreted in the following way: whenever we encounter
a concave corner \raise 6pt \hbox{$\quad\vtop{\offinterlineskip
\halign{\strut\ $#$\ &#&\ $#$\ \cr +&\vrule
width 1pt&-\cr
\omit&\multispan{2}\leaders\hrule height 0.5pt
depth 0.5pt\hfill\cr
&&+\cr}}\quad$},
then, for the spin $-$ to be stable, we must have
$\matrix
+&-&-\\
&+&\\
\endmatrix$
and hence
$\matrix
+&-&-\\
&+&-\\
\endmatrix$, because supposing the configuration
$\matrix
+&-&-\\
&+&+\\
\endmatrix$ leads to a contradiction. Discussing in the
same fashion also the value of the spin above the upper
$+$, we conclude that the configuration necessarily is
\raise 16pt \hbox{$\quad\vtop{\offinterlineskip
\halign{\strut\ $#$\ &#&\ $#$\ &#&\ $#$\ \cr
-&&-&&\cr
\multispan{2}\leaders\hrule height 0.5pt
depth 0.5pt\hfill&&&\cr
+&\vrule width 1pt&-&&-\cr
\omit&\multispan{3}\leaders\hrule height 0.5pt
depth 0.5pt\hfill&\cr
&&+&\vrule width 1pt&-\cr}}\quad$}
and our concave corner is surrounded by two convex
corners. We can use this observation in the following way.
Consider the horizontal
line $l$ passing through $B$ and the point $B'$
where $\gamma$ first hits $l$.
Knowing already that $\gamma$ does not pass below $l$,
it proceeds from $B'$
horizontaly so that a concave corner is formed:
\raise 6pt \hbox{$\quad\vtop{\offinterlineskip
\halign{\strut\ $#$\ &#&\ $#$\ \cr
+&\vrule width 1pt&-\cr
\omit&\multispan{2}\leaders\hrule height 0.5pt
depth 0.5pt\hfill\cr
B'&&+\cr}}\quad$}.
According to the above observation there must be two
convex corner attached:
\raise 12pt \hbox{$\quad\vtop{\offinterlineskip
\halign{\strut#&\qquad#&\ $#$\ &#&\ $#$\ &#\quad\quad\cr
\omit&&\multispan{2}\leaders\hrule height 0.5pt
depth 0.5pt\hfill&&\cr
&&&\vrule width 1pt&&\cr
\omit&&&\multispan{2}\leaders\hrule height 0.5pt
depth 0.5pt\hfill&\cr
\omit&\multispan{5}\leaders\hrule height 0.05pt depth
0.05pt\hfill\cr $l$ &&B'&&\quad&\vrule width
1pt\cr}}\quad$}. But since the line $\gamma$ does not,
before reaching $B$, pass below of $l$, the point $B'$
must coincide with $B$.
Consider now the point $\overline{B}$ on $\gamma$,
two units backward from
$B$ and repeat the above argument with the
curve $\overline{\gamma}$ joining
$A$ with $\overline{B}$ and the horizontal
line $\overline{l}$ passing through
$\overline{B}$ (see Fig\. 3.6).
%
%
\midinsert
\centerline{\scaledpicture 2.78in by 0.92in
(threesix scaled 800)}
\botcaption{Fig\. 3.6}
\endcaption
\endinsert
%
%
If $\overline{l}$ also passes through $A$,
the curve $\overline{\gamma}$
has to be just a horizontal segment joining
$A$ and $\overline{B}$.
If not
we can repeat the argument and get a point
$\overline{\overline{B}}$.
Iterating this process we eventually get a
line passing through $A$. As
a result the curve $\gamma$ actually follows
the boundary $\partial Q$
and we can conclude that $\partial_{\text{out}}\equiv\partial Q$.
To show, finally, that $\partial C=\partial_{\text{out}}C$
(there are no holes
in C)
consider the set $\partial C\setminus\partial_{\text{out}}C$ and
the lowermost
horizontal line touching it. Taking the most left touching point,
we
nec\-es\-sa\-ri\-ly have a concave corner \!\!\!
\raise 6pt \hbox{$\quad\vtop{\offinterlineskip
\halign{\strut#&\quad\quad#&\ $#$\ &#&\ $#$\ &#\quad\quad\cr
&&+&\vrule width 1pt&-&\cr
\omit&&&\multispan{2}\leaders\hrule height 0.5pt
depth 0.5pt\hfill&\cr
\omit&\multispan{5}\leaders\hrule height 0.05pt depth
0.05pt\hfill\cr $l$ &&&&+&\cr}}\quad$} and thus also
\raise 6pt \hbox{$\quad\vtop{\offinterlineskip
\halign{\strut#&\qquad\ $#$\ &#&\ $#$\ &#&\ $#$\ &#\qquad\cr
&+&\vrule width 1pt&-&&-&\cr
\omit&&\multispan{3}\leaders\hrule height 0.5pt
depth 0.5pt\hfill&&\cr
\omit&\multispan{6}\leaders\hrule height 0.05pt depth
0.05pt\hfill\cr $l$ &&&+&\vrule width 1pt&-&\cr}}\quad$}\!\!\!.
The newly attached bond lies below $l$ and
in the same time does not belong
to $\partial_{\text{out}}C$; this would
be possible only if the upper right
spin were $+$ which is not the case.
Hence we are getting a contradiction
with the fact that the set $\partial
C\setminus\partial_{\text{out}}C$ does
not
reach below $l$.
The fact that the octagon $C$ has to be stable
$(L_i, l_i \geq2)$ is
obvious. Finally, if the configuration $\sigma$
contained two
octagonal components of mutual distance one,
there would always exist, as it is
easy to convince oneself by inspection of possible cases,
a minus spin between
them, whose flipping would lead to a decrease of energy.
\qed
\enddemo
Among octagons with the same circumscribed rectangle,
$Q=Q(D_1,D_2,l_1,\dots,l_4)$, the standard ones,
or at least those
with oblique sides as close as possible to $l^{\ast}$,
minimize the energy.
This is stated, in a slightly more general form,
in the following
lemma.
\vfill
\newpage
\proclaim{Lemma 3.2} Let~$R$ be a~rectangle with
sides~$D_1\geq D_2$ and
consider the~set~$\Cal M(D_1,D_2)$ of~all monotone
droplets with connected
interior%
\footnote{Notice that even though a monotone envelope
is connected,
its interior
might split into disjoint components in a situation like
that illustrated
on Fig. 2.1. Cf\. Lemma 4.3.}
whose circumscribed rectangle is~$R$. Let~$\sigma_0$ be
the~configuration corresponding to \roster
\item"a)" the~octagon $Q(D_1,D_2,l^{*})$ if $D_2\geq2l^{*}-1$,
\item"b)" the~octagon $Q(D_1,D_2,l_1=l_2=l_3=l_4=\frac12(D_2+1))$
if $D_2<2l^{*}-1$ and it is odd, and
\item"c)" the octagon $Q(D_1,D_2,l_1=l_2=
\frac{D_2}2,l_3=l_4=\frac{D_2}2+1)$
if $D_2<2l^{*}-1$ and it is even.
\endroster
Then
$$
\min_{\sigma\in\Cal M(D_1,D_2)}H(\sigma)=H(\sigma_0).
$$
\endproclaim
\demo{Proof} Let~$\sigma\in\Cal M(D_1,D_2)$ and
consider its
circumscribed octagon $Q(D_1,D_2,l_1,\dots,l_4)$.
The~configuration
$\overline{\sigma}$ represented by~$Q$ has clearly
lower energy than $\sigma$
because it occupies larger area then $C(\sigma)$ and,
taking into account that in every site of the dual
lattice at most two its edges meet, its boundary
necessarily has at~least that number of~corners as
$C(\sigma)$ (this is
clearly true for every its oblique side separately). The~energy
of~$\overline{\sigma}$ is
$$
H(\overline{\sigma})=H(- \underline 1)+2J(D_1+D_2)-hD_1D_2+\sum_{a=1}^4F(l_a).
$$ Notice that
the~function
$$
F(l)\equiv-K(2l-1)+\frac12hl(l-1)
\tag3.6
$$
is minimized
for~$l=l^{*}$. Indeed, for~$l\in\Bbb R$,
the~parabola~$F$ has minimum at~
$l=\frac{2K}h+\frac12$ and this point is closer
to~$l^{*}$ than to
either~$l^{*}-1$ or~$l^{*}+1$. Hence,
whenever~$D_2\geq2l^*-1$, the~energy
of~$\overline{\sigma}$ is larger or equal to~that of~$\sigma_0$
minimizing every
term~$F(l_a)$ separately.
If $D_2<2l^*-1$, we first fix~$\sum l_a$
and under this condition minimize
$\sum F(l_a)$. The~minimum is achieved
for a~maximally symetric quadruple~
$l_1',\dots,l_4'$, for which $\max_al_a'-\min_al_a'\leq1$,
such that
$\sum l_a'=\sum l_a$.
If $\sum l_a$ is not divisible by four, there is some
freedom in~the~choice
of~$l_1',\dots,l_4'$ and, in~particular, the~conditions
$$|l_1'+l_4'-(l_2'+l_3')|\leq1
\text{ and }|l_1'+l_2'-(l_3'+l_4')|\leq1$$
can be satisfied. Given that~$\sum l_a\leq2(D_2+1)$,
we have also
$$
\max(l_1'+l_4',l_2'+l_3')\leq D_2+1\text{ and }
\max(l_1'+l_2',l_3'+l_4')\leq
D_2+1\leq D_1+1
$$
and thus the~octagon~$Q(D_1,D_2,l_1',l_2',l_3',l_4')$
exists. We can further
decrease its energy by~increasing one by~one~$l_1',
\dots,l_4'$ maintaining
maximal possible symmetry. The~resulting octagon
depends on the parity of~$D_2$.
\qed
\enddemo
With every octagon~$Q$ we associate two {\it basins
of attraction} --- a narrow one
$$
\Cal B(Q)=\{\sigma:S\sigma=Q\text{ for every
standard }S\},
\tag3.7
$$
and a wide one
$$
\widehat{\Cal B}(Q)=\{\sigma:\text{ there
exists a standard }S\text{ such that }S\sigma=Q\}.
\tag3.8
$$
Clearly,
$$
\widehat{\Cal B}(Q)\supset\Cal B(Q)
$$
and the sets $\Cal B(Q)$ are disjoint for
different $Q$'s.
When discussing a growth of a droplet, we are
studying a passage between close
standard octagons with different rectangular envelops.
It is possible to pass between different octagons
with the same rectangular
envelope at the cost of overcomming energetical
barriers between them.
A crucial circumstance here is that the barriers
are much lower than those one
has to pass when changing the rectangular envelope ---
the corresponding time
scales are much shorter and on the scale relevant for
the passage to a
``neighbouring'' standard octagon,
these barriers can be overcome with a nonvanishing probability.
If we choose to disregard the lower barriers,
we can introduce ``domains of
attraction'' associated
with every standard octagon $Q=Q(D_1,D_2)$ specified
by its rectangular envelope $R(D_1,D_2)$. Namely, we combine
different octagons with the same rectangular
envelope and introduce a broad
{\it domain of attraction} $\widehat {\Cal D}(D_1,D_2)$
and a naturally
restricted one ${\Cal D}(D_1,D_2)$.
We thus define
$$
\widehat{\Cal D}(D_1,D_2)=
\bigcup\limits_{Q:(D_1(Q),D_2(Q))=(D_1,D_2)}\widehat
{\Cal B}(Q)
\tag3.9
$$
for every~$(D_1,D_2)\in\Bbb Z_{+}^2$.
For $(D_1,D_2)$ such that $\min (D_1,D_2)\geq 3l^{\ast}-2$
we define
$$
\multline
\Cal D(D_1,D_2)=\{\sigma :\text{ for every
standard }S, S\sigma=Q\text{ with
} Q \text{ such that }\\ (D_1(Q),D_2(Q))=(D_1,D_2)\text{ and }
\min (L_1(Q),L_2(Q))\geq l^{\ast}\}.
\endmultline
\tag3.10
$$
Notice that the condition $\min (D_1,D_2)\geq 3l^{\ast}-2$
assures
that the standard octagon $Q(D_1,D_2)\in\Cal D(D_1,D_2)$.
Let us introduce also the {\it boundaries}
$$
\partial\Cal D=\{\sigma\notin\Cal D\text{ such that
there exists }x\text{
such that }\sigma^{(x)}\in\Cal D\}
\tag3.11
$$
and
$$
\partial\widehat{\Cal
D}=\{\sigma\in\widehat{\Cal D}\text{ such that there exists }x
\text{ such
that }\sigma^{(x)}\notin\widehat{\Cal D}\}.
\tag3.12
$$
Notice that, even though we
take~$\partial\Cal D$ outside~$\Cal D$, clearly~$\partial\Cal
D\subset\widehat{\Cal D}$ and thus all configurations in~$\partial\Cal D$
and~$\partial\widehat{\Cal D}$ can ``fall down''
to a minimum ``inside'' $R(D_1
,D_2)$.
To see
that~$\partial\Cal D\subset\widehat{\Cal D}$ we
observe that if~$\xi\in\Cal
D$ and~$\sigma=\xi^{(x)}\in\partial\Cal D$,
then $H(\xi)
\min_{\sigma\in\partial\widehat{\Cal
D}(D_1,D_2)}H(\sigma)=H(Q(D_1,D_2))+\widehat E(D_1,D_2),
\tag 3.15
$$
where the minimum is over all paths starting at any
octagon $Q=Q(D_1,D_2,l_1,l_2,l_3,l_4)$ and leaving the
set $\Cal M$ of all monotonous configurations.
If, moreover, $\min(D_1,D_2)\geq 3 l^{\ast}-2$, then also
$$
\min_{\omega
:\sigma_0\to\Cal D^c}\sup_{\sigma\in\omega}H(\sigma)=
\min_{\sigma\in\partial\Cal
D(D_1,D_2)}H(\sigma)=H(Q(D_1,D_2))+ E(D_1,D_2),
\tag 3.16
$$
where the minimum is
over all paths starting at the standard octagon $Q(D_1,D_2)$ and
leaving the set $\Cal D(D_1,D_2)$.
\endproclaim
\demo{Proof} Let~$\sigma\in\partial\widehat{\Cal
D}$,~$\xi=\sigma^{(x)}\notin \widehat{ \Cal D}$.
Since~$\sigma\in\widehat{\Cal D}$, there exists a standard~$S$
such that~$S\sigma=Q$ and~$(D_1(Q),D_2(Q))=(D_1,D_2)$. Consider
the sequence of configurations obtained by applying~$S$
on~$\sigma$. Taking it in the opposite order, we get a
path~$\omega:Q\to\sigma$ such that the energy increases in every
step.
With~$Q=Q(D_1,D_2,l_1,l_2,l_3,l_4)$, $D_1\geq D_2$, and using the
function
$F$ defined in (3.6), we shall prove that
$$
H(\sigma)\geq H(Q)+\min\bigl[h(D_2-1)-4K,
2J-4K-h\bigr]-\sum_{a=1}^4[F(l_a)-F(l^{*})].
\tag{3.17}
$$
On the other side, particular cases of
$\sigma\in\partial\widehat{\Cal D}$ yielding equality in
\thetag{3.17} can be displayed. Namely, if $D_2\leq D^{\ast}$, we
take the standard octagon $Q(D_1,D_2)$ and cut, except one spin,
all row of~$(D_2-1)$ spins; the resulting droplet corresponds to
$\sigma\in\partial\widehat{\Cal D}$ since cutting the last spin
we get~$\xi\notin\widehat{\Cal D}$. If $D_2> D^{\ast}$, we get
$\sigma\in\partial\widehat{\Cal D}$ just by attaching one plus
spin to the coordinate side.
Taking into account that
$$
H(Q)=H(- \underline 1)+2J(D_1+D_2)-hD_1D_2+\sum_{a=1}^4F(l_a),
\tag{3.18}
$$
we get from (3.17) the sought minimum for
$\partial\widehat{\Cal D}(D_1,D_2)$,
$$
\min_{\sigma\in\partial\widehat{\Cal
D}(D_1,D_2)}H(\sigma)=H(Q(D_1,D_2))+\widehat E(D_1,D_2).
$$
To prove \thetag{3.17}, notice first that,
without a loss of generality, we
can suppose that the path~$\omega$ consists of at
most~$D_2-1-\max(l_1+l_2-2, 2l^{*}-2)$ steps. (For concreteness
we suppose that $\max(l_1+l_2,l_2+l_3,l_3+l_4,l_4+l_1)=l_1+l_2$.)
Indeed, in every step the energy increases by at least~$h$ and $$
h(D_2-(2l^{*}-1))>h(D_2-1)-4K $$
since~$h(l^{*}-1)<2K$. Hence, the number of steps can be taken to be at most
$D_2-1-2(l^{*}-1)$. If~$l_1+l_2>2l^{*}$, we have even
stronger restriction on the number of steps. Indeed, it suffices
to prove a lower bound on the increase of energy, $$
h[D_2-1-(l_1-1)-(l_2-1)]\geq
h(D_2-1)-4K-\sum_{a=1}^4[F(l_a)-F(l^{*})].
\tag{3.19}
$$ This is true, once we verify that
$$
F(l)-F(l^{*})\geq h(l-1)-2K
\tag3.20
$$
for any $l$.
To see it, we observe that the line
(as function of~$l$) on the right
hand side above
touches the parabola on the left hand side
in the point~$l=l^{*}+1$ and
is below of it
for~$l=l^{*}$ and~$l=l^{*}+2$ :
$$
\align l=l^{*}:\quad&F(l)-F(l^{*})=0>h(l^{*}-1)-2K,\\
l=l^{*}+1:\quad&F(l)-F(l^{*})=-2K+\frac12hl^{*}2=-2K+hl^{*},\\
l=l^{*}+2:\quad&F(l)-F(l^{*})=-4K+h(2l^{*}+1)>-2K+h(l^{*}+1).
\endalign
$$
(Notice that equality in \thetag{3.19} is attained only
for~$l_1=l_2=l^{*}+1$.)
%
%
\midinsert
\centerline{\scaledpicture 4.83in by 2.83in (proofone scaled 600)}
\botcaption {Fig\. 3.7}
\endcaption
\endinsert
%
%
Next, we can suppose that all droplets~$\zeta\in\omega$
(including the~last one,~$\sigma$) are monotone with connected
interior. Indeed, starting from the~octagon~$Q$ and supposing
that after~$n\leq D_2+1-l_1-l_2$ steps we still have such a
monotone droplet, minimal energy flips leading to~a~non-monotone
droplet are those shown in Fig\. 3.7.
(Given number of
steps is clearly not sufficient to allow any configuration like
%
%
\midinsert
\centerline{\scaledpicture 0.42in by 0.2in (prooftwo scaled 600)}
\endinsert
%
%
\flushpar
with subsequent splitting into disjoint components etc.) The
increase of~energy in~the~case a), b), and c) is $2J-4K+h$,
$2J-2K+h$, and
$2J-4K-h$, respectively. In all three cases this value is
at least as large as $\min[h(D_2-1)-4K, 2J-4K-h]$
and already this single step
would suffice
for our claim.
This remark proves, in particular, the first inequality in (3.15).
Observe now that the spin flip $\sigma\to\sigma^{(x)}=\xi$
necessarily decreases
energy (otherwise one would have $\xi\in\widehat{\Cal D}$).
Taking into account that $\sigma$ is monotone with connected
interior,
the droplet $\xi$ is also monotone with interior consisting
of at most two
components.
Suppose first that $\xi$ has connected interior.
For such $\xi$, there clearly exists a standard sequence
leading to
its octagonal envelope $Q(\xi)$
and, since $\xi\notin \widehat{\Cal D}$, one has
$(D_1(\xi),D_2(\xi))\ne(D_1,D_2)$.
As a consequence
either~$(D_1(\sigma),D_2(\sigma))\ne(D_1,D_2)$
(actually~$D_1(\sigma)>D_1$ or~$D_2(\sigma)>D_2$)
or~$(D_1(\sigma),D_2(\sigma))=(D_1,D_2)$ and~$\bar d_a=1$ for
some~$a=1,\dots,4$
\hfill\newline
\phantom{xxxxxxx}(we use here
$\bar d_a(\sigma)$
to denote the lengths of segments along which
the set $C(\sigma)$
\hfill\newline
\phantom{xxxxxxx}intersects the sides of the
circumscribed rectangle
$R(\sigma)$).
\flushpar
Consider the~first configuration $\zeta$ in~$\omega$
with this property and its
predecessor~$\overline{\zeta}$. If~$D_1(\zeta)>D_1$
or~$D_2(\zeta)>D_2$, then
$$
H(\sigma)-H(Q)\geq H(\zeta)-H(\overline{\zeta})\geq2J-4K-h\geq
\min[h(D_2-1)-4K, 2J-4K-h].
\tag 3.21
$$
If~$(D_1(\zeta),D_2(\zeta))=(D_1,D_2)$ and in
the~same time~$\bar d_a=1$, one had
to~cut at least~$L_a-\bar d_a=L_a-1$ spins touching
the~side of~$R(Q)$
to reach this configuration, and thus
$$
H(\zeta)-H(Q)\geq h(L_a-1)\geq h(D_2-1-(l_1-1)-(l_2-1)).
\tag 3.22
$$
Hence, using again
\thetag{3.19} we get \thetag{3.17}.
If the interior of $\xi$ consists of two components,
than the path
$\omega$ reaching the configuration $\sigma$
should consist of at least $L_a-1$ steps.
Indeed, consider the horizontal (or vertical) line
passing through the point
in which the closures of these two components intersect.
In the configuration $Q$ there is at least $2L_a$ plus
spins at sites of
distance
$1/2$ from this line, while in the configuration $\xi$
at lest $L_a$
of them is missing (``two quadrants
filled with minuses touch at the considered
intersection point'').
As a result, the inequality (3.22) holds for
$\sigma$ and we get (3.17).
To get the bound for
$\partial\Cal D(D_1,D_2)$,
consider first $ \sigma\in\partial
\Cal D(D_1,D_2)\setminus\partial\widehat{\Cal D} (D_1,D_2)$.
Notice that then there exists a standard sequence $S$
mapping $\sigma$ onto
an octagon $Q$, $S\sigma=Q$, such that
$\min(L_1(Q),L_2(Q))< l^{\ast}$.
Indeed, by the same argument as above we can show
that all $\zeta$ on the (reversed)
path from $Q$ to $\sigma$ are monotone,
$(D_1(\sigma),D_2(\sigma))=
(D_1,D_2)$, and $\bar d_a(\sigma)\geq 2$, $a=1, \dots , 4$
(otherwise we would have $\sigma\in\partial\widehat{\Cal D}(D_1,D_2)$).
Hence, there does not exist any standard sequence $S$ mapping $\sigma$ onto an
octagon $Q$ such that
$(D_1(Q),D_2(Q))\ne(D_1,D_2)$
Moreover, for at least one $\zeta$ on our path it is
$Q(\zeta)\supset Q$
and $Q(\zeta)\neq Q$. There exists a standard sequence mapping
$\zeta$ into $Q$ and
thus
$Q(\zeta)\supset Q$. If one had $Q(\zeta)= Q$
for every $\zeta$ including $\sigma$,
then one could use the fact that, since $\sigma \in \partial \Cal D$, there exists
a standard sequence $\bar S$ mapping $\sigma$ into $\bar Q$ such that $(D_1(\bar
Q),D_2(\bar Q))= (D_1,D_2)$ and $\min(L_1(\bar Q),L_2(\bar Q))\geq l^{\ast}$
to get a contradiction.
Indeed, for such $\bar Q$ one cannot have
$\bar Q\subset Q(\sigma)=Q$
and in the same time
$(D_1(\bar Q),D_2(\bar Q))=(D_1( Q),D_2( Q))$.
Thus, on the uphill path from $Q$ to $\sigma$ a
$K$-proturberance appears at least once
and thus
$$
H(\sigma)\geq H(Q)+2K-h.
\tag3.23
$$
We can label the sides of the octagon $Q$ in such a way that
$L_2= D_2-l_1-l_2 +2\leq l^{\ast}-1$,
which is equivalent to
$$
l_1+l_2 -2\geq D_2-1-(l^{\ast}-2).
\tag3.24
$$
The difference of the energy of the octagon $Q$ and
the standard octagon $Q(D_1,D_2)$ is at least
$$
H(Q)-H(Q(D_1,D_2))\geq\sum_{a=1}^4 (F(l_a)-F(l^{\ast})).
\tag 3.25
$$
If $\min(D_1,D_2)\geq 3 l^{\ast}-1$, we use the bound (3.20)
to evaluate,
using (3.24), the right hand side
$$
\multline
\sum_{a=1}^4 (F(l_a)-F(l^{\ast}))\geq
F(l_1)-F(l^{\ast})+F(l_2)-F(l^{\ast})\geq\\
\geq h(l_1+l_2 -2)-4K\geq (D_2-1)h-4K -(l^{\ast}-2)h.
\endmultline
\tag3.26
$$
With the help of the equality (2.18),
$\eta h = 2K - h - (l^{\ast}-2)h$,
we get from (3.23) and the above inequality
the sought lower bound with $1$ in
place of $3 l^{\ast}-\min(D_1,D_2)$ in the definition (3.14).
If $\min(D_1,D_2)= D_2= 3 l^{\ast}-2$,
the bound (3.24) asserts that
$$
l_1+l_2 -2\geq 2(l^{\ast}-1)+1.
\tag3.27
$$
Supposing, say, $\max(l_1,l_2)= l_1$, we can infer that
$$
l_1\geq l^{\ast}+1
\tag 3.28
$$
and thus
$$
\sum_{a=1}^4 (F(l_a)-F(l^{\ast}))\geq
F(l_1)-F(l^{\ast})
\geq h l^{\ast}-2K.
\tag3.29
$$
Using now the bound (3.23) and then the bound (2.17)
and the definition
(2.18), we get
$$
\multline
H(\sigma) \geq H(Q(D_1,D_2)) + h l^{\ast} - 2K +2K-h =
H(Q(D_1,D_2)) + h (l^{\ast} - 1)
= \\
= H(Q(D_1,D_2)) + h (3l^{\ast}-3) - 4K +2\eta h.
\endmultline
\tag 3.30
$$
Thus, we finished the proof for $ \sigma\in\partial\Cal
D(D_1,D_2)\setminus\partial\widehat{\Cal D} (D_1,D_2)$.
Let us now turn to the case $ \sigma\in\partial\Cal
D(D_1,D_2)\cap\partial\widehat{\Cal D} (D_1,D_2)$.
If $\min(D_1,D_2)\geq 3 l^{\ast}$, the needed bound
is the already proved inequality
(3.15).
In the case ($D_2=$) $\min(D_1,D_2)= 3 l^{\ast}-1$,
one can consider a
path leading to $Q$ with $L_2(Q)\geq l^{\ast}$.
In the case leading to (3.21) the needed bound is
amply satisfied since
$$
D_2 h -4K+2h < 2J -4K -h.
\tag 3.31
$$
In the case leading to (3.22) we get
$$
H(\sigma) - H(Q)\geq h(L_2-1) \geq h(l^{\ast}-1)
\geq h(3l^{\ast}-2)-4K+\eta h
\tag 3.32
$$
using (2.17).
If ($D_2=$) $\min(D_1,D_2)= 3 l^{\ast}-2$, we get
again either the sufficient
bound (3.21) or
$$
H(\sigma) - H(Q)\geq h(L_2-1) \geq h(l^{\ast}-1) =
h(3l^{\ast}-3)-4K+2\eta h.
\tag 3.33
$$
Finally, to prove that
$$
\min_{\omega :\sigma_0\to\Cal D^c}
\sup_{\sigma\in\omega}H(\sigma)
=H(Q(D_1,D_2))+ E(D_1,D_2),
\tag 3.34
$$
we have to find a path $\omega$ from the standard
octagon $Q(D_1,D_2)$ to a
saddle configuration $\sigma$ with
$$
H(\sigma)=H(Q(D_1,D_2))+ E(D_1,D_2),
$$
such that for every $\zeta\in\omega$ one has
$$
H(\zeta) \leq H(Q(D_1,D_2))+ E(D_1,D_2).
$$
If $\min(D_1, D_2)\geq 3 l^{\ast}$, the saddle
point $\sigma$ is actually on
the boundary
$\partial\widehat{\Cal D}(D_1,D_2)$.
A path satisfying the above condition can be taken,
for example, by first
cutting one layer along two oblique sides of the standard octagon
and then along the coordinate side between them.
The highest point on this path,
before the final steady growth when cutting
coordinate side, is
$$
H(Q(D_1,D_2)) + 2(l^{\ast}-1)h-(2K-h)0$ and all $\sigma\ne\eta \in \Gamma$,
$H(\sigma) < H(\eta) $,
one has
$$
P_{\sigma}(\tau_{\eta}\leq\exp\{\beta(H(\eta)-H(\sigma)-
\epsilon)\})@>>{\beta\to\infty}>0.
$$
\endproclaim
\demo{Proof}
Given~$T\in\Bbb N$, one has
$$
P_{\sigma}(\tau_{\eta} v_n : \sigma_t \not = \sigma_{t-1}\},\cr &v_n =
\inf\{ t\geq u_n:
\sigma_t \in
\partial B \cup Q\},
& (3.36)\cr}
$$
we set
$$ \xi_n = \sigma_{v_n},\qquad
\sigma_0\in Q\cup\partial B,
\tag{3.37}
$$
and
$$
\aligned
\theta&=\inf\{ n:\xi_n=Q\}\\
\nu &= \inf
\{n\geq\theta:\xi_n\in \partial B\}.
\endaligned
\tag3.38
$$
For every $s\in \Bbb N$ one has
$$
P_Q(\tau_{\partial B}>
s)\geq P_Q(\nu>s)=P(Q\rightarrow Q)^s =
(1-P(Q\rightarrow \partial B))^s,
$$
where
$$
\aligned
&P(Q\rightarrow Q)=P(\xi_1=Q|\xi_0=Q),\\
&P(Q\rightarrow\partial B) = \sum_{\eta \in
\partial B} P(\xi_1=\eta
|\xi_0=Q)=P(\xi_1\in\partial B|\xi_0=Q).
\endaligned
\tag3.39
$$
We have
$$
\multline
P(\xi_1\in\partial B|\xi_0=Q)=\\
=\sum^{\infty}_{s_0=0}\sum^{\infty}_{s=1}
\sum_{\bar\sigma_1,\dots,\bar\sigma_{s-1}\in B\setminus\{Q\}}
\sum_{\bar\sigma_s\in\partial B}
P(\sigma_0\!=Q,\dots ,
\sigma_{s_0}\!=Q,
\sigma_{s_0+1}\!=\bar\sigma_1,\dots,
\sigma_{s_0+s}\! = \bar\sigma_s) =\\
\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\!\!\!\!\!\!\!\!\!
\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\!\!\!\!\!\!\!\!\!\!\!\!
=\sum^{\infty}_{s_0=1}\sum^{\infty}_{s=1}
\sum_{\bar\sigma_1,\dots,
\bar\sigma_{s-1}\in B\setminus\{Q\}}
\sum _{\bar\sigma_s\in \partial
B} e^{-\beta \lbrack H(\bar\sigma_s)-H(Q)\rbrack} \times\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,
\,\,\,\,\,\,\,\,\,\,
\times P(\sigma_0 \!= \bar\sigma_s,
\sigma_1\!=\bar\sigma_{s-1},\dots, \sigma_{s-1}\!=
\bar\sigma_1, \sigma_s\!=Q,\dots, \sigma_{s+s_0}\!=Q)\leq\\
\leq e^{-\beta \min_{\sigma\in\partial B}\lbrack
H(\sigma)-H(Q)\rbrack}
\sum_{\bar\sigma\in\partial B}\sum ^{\infty}_{s_0=1}
\sum^{\infty}_{s=1}
P_{\bar\sigma}( \sigma_t\in B\setminus Q ,
\forall t~~0$ one has
$$
P(Q\rightarrow
\partial B)i$),
the process $\xi^j$ is defined considering the portion of the
trajectory between the moment it first enters $\partial B_j$
until, after visiting $Q_j$, it first enters $\partial B_{j+1}$
(visiting, possibly, other $B_i$'s with $i \tau_{Q_j} :
\sigma_t \in \partial B_j \},
\tag{3.55}
$$
and putting
$$ \bar \Cal F = (\bar
\Cal F^{1,u};\bar \Cal F^{1,d}_{\bar t^1_d};\bar\Cal
F^{2,u};\dots;\bar \Cal F^{N-1,d}_{\bar t^{N-1}_d}; \bar \Cal
F^{N,u}),
\tag{3.56}
$$
we get
$$
\eqalignno{ (\bar \Cal F^{1,u}_{\bar t^1_u};
\dots;\bar \Cal
F^{N-1,d}_{\bar t^{N-1}_d}; \bar
\Cal F^{N,u}_{\bar t^N_u})&=(\bar \Cal
F^{1,u};\bar \Cal F^{1,d}_{ \bar t^1_d};\bar\Cal F^{2,u};
\dots;\bar \Cal F^{N-1,d}_{\bar
t^{N-1}_d};\bar \Cal F^{N,u})\cap
\Cal G=\cr &=\lbrack
\cup^{\infty}_{t^1_u=1}\cup^{\bar t^1_d}_{t^1_d=1}
\cup^{\infty}_{t^2_u=1}
\cdots \cup^{\bar
t^{N-1}_d}_{t^{N-1}_d=1} \cup^{\infty}_{t^N_u=1}
(\Cal F^{1,u}_{ t^1_u};\dots;
\Cal F^{N,u}_{t^N_u})\rbrack \cap \Cal G\equiv \cr
&\equiv \bar \Cal F\cap
\Cal G.
&(3.57)\cr}
$$
To get a lower bound on the probability of
$\bar \Cal F\cap \Cal G$ one can use the lower bound
$$
P(\bar \Cal F\cap \Cal
G)\geq P(\bar \Cal F) - P(\Cal G^c).
\tag{3.58}
$$
Thus, if we got a
bound of the form
$$
P(\Cal G^c) < a \leq {1\over 2}P(\bar \Cal F),
\tag{3.59}
$$
we would have
$$
P(\bar \Cal F\cap \Cal G) > {1\over 2}P(\bar \Cal F).
\tag{3.60}
$$
To evaluate the probability of the event $\bar\Cal F$, we can
use the following general strategy. First, we
evaluate the probability $P(\xi^j_{\nu _j}=S_{j+1})$ relative to
the Markov chain $\xi ^j$ (see the equations (3.37) and
(3.38)). We have $$
P_{Q_j}(\xi^j_{\nu_j}=S_j) = \sum^{\infty}_{n=0} [1-
P(Q_j\rightarrow \partial B_j)]^n P(Q_j\rightarrow S_{j+1}).
\tag{3.61}
$$
>From now on we suppose that the
configurations $ S_{i+1}\in \partial B_i\ \cap\ \partial
B_{i+1}$,
$i=1,\dots,N-1$, $S_1\in \partial B_1$, and $ S_{N+1}\in \partial B_N $
are chosen so
that
$$
\min_{\sigma\in\partial B_j} H(\sigma ) = H(S_j),
\quad j=1,\dots ,N.
\tag{3.62}
$$
Using (3.41) we can thus infer that, for every $\epsilon > 0$
and $\beta$ sufficiently
large,
$$
P(Q_j\rightarrow \partial B_j) \leq \exp
\bigl\{ -\beta [ H(S_j)-H(Q_j)-\epsilon]
\bigr\}.
\tag{3.63}
$$
Let us suppose further that
$$
H(S_j) > H(S_{j-1}) ,
\tag{3.64}
$$
and that we have the lower bounds
$$
P(Q_j\rightarrow S_{j+1} ) \geq \exp\bigl\{ -\beta
[ H(S_{j+1})-H(Q_j) + \epsilon]\bigr\}
\tag{3.65}
$$
and
$$
P_{S_{j+1}}(\theta _{j+1} \leq \bar t^j_d) \geq \exp(-\epsilon \beta)
\tag{3.66}
$$
for every $\epsilon > 0$ and $\beta$
sufficiently large.
Combining the bounds (3.63) -- (3.65), we get
a lower bound (3.61).
Using now this bound together with the definitions
(3.50), (3.53), and (3.56), the equation (3.51), and the bound
(3.63),
we conclude that
$$
P(\bar \Cal F) \geq \exp\bigl\{ -\beta [H(S_N) - H(S_1) + \epsilon
]\bigr\}
\tag{3.67}
$$
for
every $\epsilon > 0$ and $\beta$ sufficiently large.
In our particular case, the equation (3.62) and the inequalities
(3.64) and (3.65) will be
satisfied and we will get
estimates of the form (3.66). On the other hand, we get upper
bounds on $P(\Cal G^c)$ that are superexponentially small in
$\beta$ (see the condition (C1) in the proof of Proposition 1) yielding (3.59)
for $\beta$ sufficiently large. Thus we will get
$$
P(\bar \Cal F\cap \Cal G) > \exp \bigl\{-\beta [H(S_N) - H(S_1) +
\epsilon ]\bigr\}
\tag{3.68}
$$
for every $\epsilon > 0$ and $\beta$ sufficiently large.
Finally, to state Proposition 1, we
need the definititon of yet another stopping time.
Namely, considering standard octagons
$Q(D_1,D_2)$ with $D_i \geq 3l^* -2$, $i=1,2$, we take $\bar
\tau $ to be the
first hitting time
to a standard octagon different from $Q(D_1,D_2)$,
$$
\bar \tau = \tau_{\text{\seveneufm S}_Q\setminus Q(D_1,D_2)}.
\tag{3.69}
$$
Here
$$
\text{\teneufm S}_Q = \bigcup _{D_1,D_2} Q(D_1,D_2)
\tag{3.70}
$$
is the set of all standard octagons.
Supposing that $D_2\leq D_1$ with $3l^*-2l^*+1$ and
$$
E(l^*+1)=2h(l^*-1)-(2K-h).
\tag3.72
$$
for $L_2=l^*+1$.
\proclaim{Proposition 1} Consider a~standard
octagon~$Q(D_1,D_2)$
with~$D_2\leq D_1$ and let
$$
3l^*-20$, one has
$$
\sup_{\sigma\in
Q(D_1,D_2)}P_{\sigma}(\tau<\exp\bigl\{(\beta(E(L_2)
+\epsilon) \bigr\}\text { and } \tau = \bar
\tau)@>>{\beta\to\infty}>1.
\tag3.76
$$
\endproclaim
\demo{Proof} We suppose that~$D_1>D_2$. The~case~$D_1=D_2$
does not present
any supplementary difficulty and is left to the~reader.
First, let us consider the case $L_2 > l^{\ast} + 1$.
The main step in the proof will be to show that
$$
P_{\sigma }(\tau_{\partial \Cal D(D_1,D_2)} >
\exp \bigl\{\beta(E (L_2) +\epsilon)\bigr\})
@>>{\beta\to\infty}> 0
\tag3.77
$$
and
$$
P_{\sigma }(\tau_Q >
\exp \bigl\{\beta(2K-h +\epsilon)\bigr\})
@>>{\beta\to\infty}> 0
\tag3.78
$$
for every $\epsilon >0$
and $\sigma
\in \Cal D(D_1,D_2) $.
Once (3.77) and (3.78) will be proven, we can reason in the following way.
>From (3.77) we know, in particular, that starting from $Q$ one reaches with high
probability the set $\partial\Cal D(D_1,D_2) $ before the time
$\exp \bigl\{\beta(E (L_2) +\epsilon)\bigr\}$.
Moreover, it is unprobable to reach $\partial\Cal D(D_1,D_2) $
outside the set
$\Cal S(D_1,D_2)$ of configurations yielding minima of $H$ on
$\partial \Cal D(D_1,D_2)$
(see the definitions (3.10) and (3.11)).
Indeed, one clearly has
$$
P_Q(\tau_{\partial \Cal D \setminus \Cal S} <
\tau_{\partial \Cal D})
\leq P_Q(\tau_{\partial \Cal D
\setminus \Cal S} <
\exp \bigl\{\beta(E (L_2) +\epsilon)\bigr\})
+ P_Q(\tau_{\partial \Cal D} \geq
\exp \bigl\{\beta(E (L_2) +\epsilon)\bigr\})
\tag3.79
$$
for every $\epsilon > 0$
(we have dropped the explicit dependence on $D_1,D_2$).
The first term on the right hand side can be bounded with the help of Lemma
3.4, taking into account that according to
Lemma 3.3,
for every $\bar \sigma \in \Cal S(D_1,D_2)$ one has
$$
H(\bar \sigma)
= \min _{ \sigma \in \partial \Cal D(D_1,D_2) } H(\sigma) =
H(Q(D_1,D_2)) +
E (L_2).
\tag3.80
$$
Once reaching $\Cal S(D_1,D_2)$, with high probability one of the two
possibilities occurs. Either we descent to
$Q(D_1-1,D_2)$ with a fixed nonvanishing probability,
or we return back to $\Cal D(D_1,D_2)$.
The saddles in $\Cal S$ consist just from a contracted
octagon united with a unit
square protuberance and, with probability
approaching $1$ as $\beta \to
\infty$, this protuberance is either cut off,
or it is made stable by a flip of a
minus spin
adjacent to it.
Actually,
following the argument of the proof of Lemma 3.3
we can show that
for every $\bar \sigma \in \Cal S(D_1,D_2) $ one has
$$
P_{\bar \sigma} (\tau = 1) > \frac1{|\Lambda |} \text{ and }
P_{\bar \sigma} (\tau_{\Cal D(D_1,D_2)}<
\tau_{\Cal D ^{c}(D_1,D_2)\setminus
\{\bar \sigma \} }\mid \tau > 1)@>>{\beta\to\infty}> 1.
\tag3.81
$$
After returning to $\Cal D(D_1,D_2)$ and reaching $Q$, according to
(3.78), we can repeat the attempt and prove finally,
with the help of the strong Markov
property, that for every $\sigma \in \Cal D(D_1,D_2) $ one has
$$
P_{\sigma} (\tau < \exp \bigl\{\beta(E (L_2) +\epsilon)\bigr\})
@>>{\beta\to\infty}> 1.
\tag3.82
$$
To prove that
$$
P_Q(\tau = \bar \tau) @>>{\beta\to\infty}> 1,
\tag3.83
$$
for
$\bar\tau$ defined in (3.69), or, in other terms, that
$$
P_Q(\tau < \bar \tau) @>>{\beta\to\infty}> 0,
\tag3.84
$$
we observe that, since $L_2 < L^*$, one has
$$
\min_{\omega :Q(D_1,D_2)\to\text{\seveneufm S}_Q\setminus \{
Q(D_1,D_2), Q(D_1 - 1,D_2)\} }\max_{\sigma\in\omega}
[H(\sigma) - H(Q(D_1,D_2))]>
E(L_2) .
\tag3.85
$$
Thus, for all sufficiently small $\epsilon > 0 $,
the set
$$
\bigcup _{ D_{1}^{'} , D_{2}^{'} \not= (D_1,D_2) ,
(D_1 -1 ,D_2) } \Cal D
( D_{1}^{'}, D_{2}^{'} )
$$
cannot be reached in a time
$ \exp \bigl\{\beta(E (L_2) +\epsilon)\bigr\} $
with probability approaching
one as $\beta \to \infty $.
For $L_2 = l^* + 1$, similar reasoning is slightly more complicated.
Instead of (3.81) we introduce the set
$$
\Cal C = \Cal D (D_1 -1,D_2) \cup
\{Q: (D_1(Q),D_2(Q)) =
(D_1,D_2) \text{ and }
\min ( L_1(Q), L_2(Q)) < l^*\},
\tag3.86
$$
for which, for every $\bar \sigma \in \Cal S(D_1,D_2) $, one has
$$
P_{\bar \sigma} (\tau_{\Cal D(D_1,D_2)}<
\tau_{\Cal D ^{c}(D_1,D_2)\setminus
\{\bar \sigma \} }\mid \tau_{\Cal C} > 1)@>>{\beta\to\infty}> 1.
\tag3.87
$$
Moreover, (still $ L_2=l^* + 1$)
we will show that
$$
P_{\bar \sigma }(\tau <
\exp \bigl\{\beta h (l^* - 2 + \epsilon)\bigr\}\mid\tau_{\Cal C}=
1) @>>{\beta\to\infty} >1
\tag3.88
$$
for any
$\bar \sigma \in \Cal
S(D_1,D_2)$ and $\epsilon > 0 $.
Thus, noticing that
$$
h(l^* -2) < E (l^* +1).
$$
we can conclude that (3.82) holds true.
Hence, to prove the Proposition we are left
with the proof of (3.77), (3.78),
and for $ L_2=l^* + 1$ also (3.88).
Actually, the crucial point is to get (3.77).
The claim (3.78) will be a
byproduct of the proof of (3.77) (see the condition (C3) below).
To derive the bound (3.77), we will look more carefuly on a possible way
of reaching the set $\partial \Cal D$ from a configuration
$\sigma\in \Cal D$.
The trajectory will be characterized by visiting
several special
octagons (see Fig\. 3.8):
First, we use
$\widetilde Q_0 \in Q(D_1,D_2,l^*,l^*,l^*,l^*)$ to
denote the standard
octagon representing our initial condition. We can suppose,
without loss of
generality, that it is centered --- its upper left corner
is the point $(-1/2,1/2)$ of
the dual lattice.
As usually, we use~$R(D_1,D_2)$ to denote the rectangle
circumscribed to $\widetilde Q_0$.
The octagon $\widetilde Q_1$
is contained in $\widetilde Q_0$; it has the same rectangular
envelope $R(D_1,D_2)$ and it differs from
$\widetilde Q_0$ only by the length of
one of its oblique
edges --- namely, the first one in the clockwise
numeration $l_i$, $i = 1,2,3,4$ starting with the uppermost
right one is of length $l^* +
1 $, $l_1=l^* +1 $.
The length of all the remaining oblique edges is again $l^*$,
$$
\widetilde Q_1 \in Q(D_1,D_2,(l_i)_{i=1,\dots ,4},
l_1 = l^* +1,
l_2 =l_3 =l_4 =
l^* ).
$$
Further, the octagon $\widetilde Q_2$ is
obtained from $\widetilde Q_1$ by replacing
the oblique edge $l_2$ (of length $l^*$) adjacent to the
shortest coordinate edge by
an edge of length $ l^* + 1$.
The octagon
$\widetilde Q_2$ is thus again centered and has two oblique
edges of length
$l^* + 1 $ adjacent to the shortest coordinate edge. Finally,
the octagon $\widetilde Q_3$ is the element of $Q(D_1,D_2
-1,l^*,l^*,l^*,l^*)$ obtained
from $\widetilde Q_2$ by erasing the
shortest coordinate edge (the vertical one on
the right hand side) adjacent to the oblique
edges of length $l^* + 1$.
The saddle
configurations $\widetilde S_i $, $i=1,2,3 $, are obtained
from $\widetilde Q_i$ by adding a unit square protuberance
(the first one in
lexicographic order --- denoted by a dot in Fig\. 3.8) to
\flushpar
1) the first
(uppermost left) oblique edge in the case
of $\widetilde Q_1$;
\flushpar
2) the second oblique edge
(down on the right) in the
case of $\widetilde Q_2$, and
\flushpar
3) one of the shortest
coordinate edges --- namely, to the vertical one on the right
hand side ---
in the case of
$\widetilde Q_3$.
%
%
\midinsert
\centerline{\picture 4.40in by 6.89in (threeeight)}
\botcaption {Fig\. 3.8}
\endcaption
\endinsert
%
%
Let us first consider the case $L_2 > l^* +1$.
To prove (3.77), we follow [KO1] and introduce a
event $\Cal E_{\sigma}^s$ (of shrinking) starting from an arbitrary $\sigma$ in
$\Cal D(D_1,D_2)$, taking place in an interval of time
$T_1=T_1(\delta _1)$,
$$
T_1(\delta _1)=\exp\bigl\{\beta (2K-h+\delta_1)\bigr\},
\tag3.89
$$
with sufficiently small $ \delta _1 > 0$ to be
specified later, and such that:
\roster
\item
If $\Cal E_{\sigma}^s$ takes place
then necessarily the set $\partial \Cal
D$ is reached (in a particular manner) before the time
$T_1(\delta_1)$.
\item
The probability $P(\Cal E_{\sigma}^s)$ satisfies
the uniform lower bound
$$
\inf _{\sigma \in \Cal D (D_1,D_2)} P(\Cal E_{\sigma}^s) \geq
\alpha
\tag3.90
$$
with $\alpha$ chosen large enough to satisfy
$$
\lim _{\beta \to \infty} (1- \alpha)^{{T_1\over T_2}} = 0.
\tag3.91
$$
\endroster
Here $T_2 = T_2(\delta _2)$ (again, the constant $\delta _2 > 0 $
will be specified later) is given by
$$
T_2(\delta _2)=\exp\bigl\{\beta(E (L_2) +\delta _2)\bigr\}.
\tag3.92
$$
Splitting now the interval $T_2$ into $T_2/T_1$ intervals of length
$T_1$, we can argue that the probability that, during any such interval,
one has not reached $\partial \Cal D$ is at most $1-\alpha$, since not
reaching $\partial \Cal D$
in time $T_1$ means that $\Cal E_{\sigma}^s$ certainly did not take place.
Using now the strong Markov
property and taking into account (3.91), we see that an attempt to
realize the event $\Cal E_{\sigma}^s$ not later than $T_2$ will be
succesful with high probability for large $\beta$ --- the
equality (3.77) follows.
Now, let us first describe the event $\Cal
E_{\sigma}^s$ of shrinking in words --- a formal definition will follow.
Starting from any $\sigma$ in $\Cal D(D_1,D_2)$, we first
descent to $\widetilde Q_0$. Then we stay for a time of the order
$\exp\{ \beta(2K - h) \}$ inside $\Cal B(\widetilde Q_0)$, the basin
of attraction of $\widetilde Q_0$, leaving it afterwards through
the saddle $\widetilde S_1$. After reaching $\widetilde S_1$ we
descent to $\widetilde Q_1$ in a unique step. Staying
inside $\Cal B(\widetilde Q_1)$, for a time
of order $\exp\{ \beta(2K - h) \}$,
we leave through $\widetilde S_2$ from which, again in one step,
we descent to $\widetilde Q_2$. Finally, again staying in $\Cal
B(\widetilde Q_2)$ for a time of order $\exp\{ \beta(2K - h) \}$,
we ascent to $\widetilde S_3 \in \Cal
S(D_1,D_2)$.
Let us turn to a formal definition of $\Cal
E_{\sigma}^s$.
For every $\sigma$ in $\Cal D(D_1,D_2)$
and every $t_0 \in \Bbb N$, let
$$
\Cal E_{\sigma,t_0}^0 =\{ \sigma _0 = \sigma, \,
\tau _{\widetilde Q_0} = t_0\}.
\tag3.93
$$
Namely, $\Cal E_{\sigma,t_0}^0$
describes the event, after starting from $\sigma$, of hitting in the moment $t_0$,
for the first time, the octagon $\widetilde Q_0$. Now, consider the events
$\Cal F^{1,u}_{t^1_u}$,
$\Cal F^{1,d}_{t^1_d}$,
$\Cal F^{2,u}_{t^2_u}$,
$\Cal F^{2,d}_{t^2_d}$, and
$\Cal F^{3,u}_{t^3_u}$,
defined by (3.47) -- (3.50)
with
$N = 3$, $Q_1, Q_2, Q_3 = \widetilde Q_0,\widetilde Q_1,
\widetilde Q_2$, $B_1,B_2,B_3
= \Cal B(\widetilde Q_0),\Cal B(\widetilde Q_1),
\Cal B(\widetilde Q_2)$, and
$S_2,S_3,S_4 =\widetilde S_1,\widetilde S_2,\widetilde S_3$.
Moreover, we take $\bar t^1_u = \bar t^2_u =\bar t^3_u =
\exp\{\beta (2K - h + \delta)\}$ and
$\bar t^1_d = \bar t^2_d = 1 $.
Further, we consider the time
$\bar t_0 = \exp\{\beta (2K - h + \delta)\}\equiv
T_1(\delta)$, ($\delta > 0$,
sufficiently small, to be specified later)
for the first descent from $\partial \Cal D(D_1,D_2)$
to $\widetilde Q_0$.
Finally, the configuration $ S_1$ is
one of the minimal saddles in $\partial \Cal B (
\widetilde Q_0)$ satisfying the equality
$$
\min_{\sigma\in\partial \Cal B(\widetilde Q_0)} H(\sigma ) =
H( S_1).
$$
It is obtained by adding a unit square
protuberance to one of the oblique edges of $\widetilde Q_0$.
Clearly,
$$
H( S_1) = H(\widetilde Q_0 ) + 2K - h.
\tag 3.94
$$
Introducing now
$$
\Cal E_{\sigma}^s = (\bar \Cal E_{\sigma,\bar t^0}^0;\bar
\Cal F^{1,u};\Cal F^{1,d}_1;\bar \Cal F^{2,u};\Cal F^{2,d}_1
\bar \Cal F^{3,u})\cap \Cal G.
\tag3.95
$$
(cf\. (3.57)),
we will suppose that the following conditions are satisfied:
\roster
\item "(C1)" $P(\Cal G^c
) \leq \exp (-e^{C\beta })$ for a constant $C > 0$.
\item "(C2)"
$P(\bar \Cal F_1^{j,d}) >\lambda $, $j=1,2$,
for a constant $\lambda > 0$ independent of $\beta$.
\item "(C3)" $P(\bar \Cal E _{\sigma,\bar t_0}^0) =
\sum _{t_0 =1}^{\bar t_0} P( \Cal E _{\sigma,t_0}^0 )
\geq \exp(-\epsilon \beta)$
for every $\epsilon > 0$ and all $\beta $ sufficiently
large.
\endroster
Let us notice that the equality (3.62) and the
inequality (3.64)
are satisfied in our case,
that the validity of (3.66) immediately follows from an
explicit computation, based on the definition of Metropolis
dynamics, taking into account that the transitions $ \widetilde
S_{j+1} \to \widetilde Q_j$, $ j=1,2,3$, are just single spin flip
events, and, finally, that the validity of (3.65) is immediate
by considering a particular event leading from $Q_j$ to
$S_{j+1}$ consisting of a set of subsequent $h$-erosions in
the sites adjacent from the interior to the concerned oblique
or vertical edge of $\widetilde Q_0$, $\widetilde Q_1$, and
$\widetilde Q_2$ (see Fig\. 3.8). (The same argument actually
proves also the stronger condition (C2).)
>From the strong Markov property we get
$$
P(\Cal E_{\sigma}^s ) = P(\bar \Cal E_{\sigma,\bar t^0}^0) P[(\bar
\Cal F^{1,u};\Cal F^{1,d}_1;\Cal F^{2,u};\Cal F^{2,d}_1
\bar \Cal F^{3,u})\cap \Cal G].
\tag3.96
$$
Using (3.66) together with (3.65),
we can imply, as we have seen, the lower bound (3.67). From (C1)
and (3.67), for $\beta$ sufficiently large, we get (3.59), (3.60),
and thus also (3.68). Finally from (C3), (3.68), (3.94), and (3.96) we
deduce that
$$
\inf_{\sigma \in \Cal D(D_1,D_2)}P(\Cal E_{\sigma}^s )
\geq \alpha \equiv
\exp \{ - \beta( E (L_2) - 2K + h + \epsilon) \}
\tag3.97
$$
for every $\epsilon >0$
and $\beta$ sufficiently large.
The validity of (3.91) then follows from the bound (3.97) --- for
any fixed $\delta_2$ (see the equation (3.92))
once we take, say, $\delta = \frac {\delta_1}{2} =
\frac {\delta_2}{4} $ for $\delta$ from the definition of times $\bar t_u^i$.
Hence, to conclude the proof of (3.77), we only need to verify
the conditions
(C1), (C2), and (C3) above.
To get the bound (C1), we suppose that for every $j = 0,1,2$ there
exist a time $\tilde t_u^j$ and, for every $\sigma \in \Cal B(\widetilde
Q_j)$,
an event
$\widehat{\Cal E}^j_{\sigma}$ of escape
from $\Cal B ( \widetilde Q_j)$
such that:
\roster
\item"(i)" The occurrence of $\widehat{\Cal E}^j_{\sigma}$ implies
that
$$
\tau_{\partial \Cal B ( \widetilde Q_j)} < \tilde t_u^j
\tag3.98
$$
\item"(ii)" For every $\epsilon_0 > 0 $ sufficiently small and for
every
$\beta $ sufficiently large,
$$
\inf _{\sigma \in B_j} P(\widehat{\Cal E}^j_{\sigma})
\geq \frac{ \tilde t_u^j}{\bar t_u^j} \exp (\epsilon_0 \beta).
\tag3.99
$$
\endroster
The superexponential estimate
(C1) then directly follows
from (3.98) and (3.99) with the help of the strong Markov
property.
To construct the event $\widehat{\Cal E}^j_{\sigma}$, we
repeat, in the present simpler situation of escape
from $ \Cal B =\Cal B ( \widetilde Q_j)$,
a construction similar to the one discussed above in the
case of escape from $\Cal D$. Namely, we chose time $ \tilde
t_u^j = T_0 + 1 $
with
$$
T_0=(8\widetilde{J}+8K+2h){|\Lambda|} / h,
\tag{3.100}
$$
the time that suffices,
starting from any configuration, to reach a local minimum
(see the equation (3.5)),
and set
$$
\widehat{\Cal E}_{\sigma }^j = (\widehat {\Cal E}_{\sigma, T_0}^0;\widehat
{\Cal F}^{j,u}),
\tag3.101
$$
where
$$
\widehat {\Cal F} ^{j,u} = \{\xi_0^j=\widetilde Q_j,\quad \xi^j_{\nu_j}=
\widetilde S_j,\quad
v_{\nu_j} = 1 \}
\tag3.102
$$
and
$$
\widehat {\Cal E}_{\sigma, T_0}^0 = \{ \sigma_0 = \sigma,\,
\tau_{\widetilde Q_j} < T_0 ,\, \sigma_t = \widetilde Q_j
\text{ for all }
t \in [\tau_{\widetilde Q_j} , T_0] \}.
\tag3.103
$$
The saddle configurations
$\widetilde S _j$ for $j=1,2$ (and $3$) have already been
defined. The configuration $\widetilde S_0$ is
identical to $S_1$ introduced above and
it is obtained by adding a unit square
protuberance to one of the oblique edges of $\widetilde Q_0$.
The event $\widehat{\Cal E}_{\sigma }^j$ thus describes, starting
from a generic $\sigma \in \Cal B (\widetilde Q_j)$,
a descent to
$\widetilde Q_j$ in a time shorter than $T_0$ and staying in $\widetilde Q_j$
up to $T_0$. The event $\widehat {\Cal F} ^{j,u}$ simply consists of
creating of a $K$-protuberance on an oblique edge of
$\widetilde Q_j$.
It is easy see that, given $\delta$ (see definition of $\bar t^j_u$),
the bound (3.99) is satisfied provided $\epsilon _0 < \delta /4$.
As we already mentioned, the condition (C2) follows from the
definition of $\widetilde S_1$ and $\widetilde S_2$ with $\lambda
= \frac1{|\Lambda|}$. With this probability the spin chosen for updating
is the one on the unit square protuberance of
$\widetilde S_1,\widetilde S_2$, respectively.
The idea of the proof of the condition (C3) is as follows.
Starting from any $\sigma \in \Cal D(D_1,D_2)$,
after a time of order $T_0$ (3.100),
we descend to octagon $Q$ in $ \Cal
D(D_1,D_2)$ with high
probability. Then, in a time of order $\exp \bigl\{\beta(2K-h+
\tilde \epsilon )\bigr\}$, with $\tilde \epsilon > 0$ such that
$$
2K-h+ \tilde
\epsilon < h(l^* - 1) - \tilde \epsilon,
\tag3.104
$$
many $K$-protuberances occur with high probability,
but no total erosion of any oblique edge of length
$l \geq l^*$ takes place.
To provide a~formal proof we introduce again a~Markov chain ---
this time the chain $\{\eta_n\}$ obtained by looking
at our process~$\sigma_t$ at
the~times of passing through octagons. The space of
states of our chain is the set $Y$ of all
octagons $Q$
in $\Lambda$,
$$
Y = \{ Q; Q\subset \Lambda\}.
$$
Let
$$
\gathered \overline v_0=0,\quad\overline u_n=\inf\{t>\overline
v_n,\overline{\sigma}_t\notin Y\}, \text { and }\\
\overline v_{n+1}=
\inf\{t>\overline
u_n:\overline{\sigma}_t\in Y\}.\endgathered
$$
We set
$$
\eta_n=\sigma_{\overline v_n}
$$
and, for every $ A \subset Y $, introduce
$$
\bar \nu _A = \inf \{ n ; \eta_n \in A \} .
\tag3.105
$$
Let
$$
\bar Y = \{Q : (D_1(Q),D_2(Q) ) = (D_1,D_2)\}
\supset \{ Q \in \Cal D(D_1,D_2) \}
\tag3.106
$$
and
$$
Y^<=\{Q\equiv Q(D_1,D_2,(l_i)_{i=1,\dots,4})\in
\bar Y : l_i\leq l^*,\,
i=1,\dots,4\}.
\tag3.107
$$
We will prove that for all $\epsilon > 0$ one has
$$
\sup _{Q\in Y} P_Q(\bar \nu _{Y^<} >
\exp\bigl\{\beta (2K-h + \epsilon ) \bigr\})
@>>{\beta\to\infty}> 0.
\tag3.108
$$
Postponing the proof of (3.108), let us first show that
for every $\sigma \in \Cal D(D_1,D_2) , \epsilon > 0$,
and $\beta$ sufficiently large,
$$
P_{\sigma} (\tau_Y < \exp(\epsilon \beta)) > 1/2.
\tag3.109
$$
It implies, with the help of the strong Markov property,
that for every $\epsilon
> 0$ , $\sigma \in \Cal D (D_1,D_2)$, and $\beta$ sufficiently
large, one has
$$
P_{\sigma} (\tau_Y > \exp(2\epsilon \beta)) <
(1/2)^{\exp (\epsilon \beta)}.
\tag3.110
$$
To get the lower bound (3.109), one again uses
the property of time $T_0$ ---
with a strictly positive
probability independent of $\beta$,
starting from any configuration one reaches a local minimum
in a time at most $T_0$.
>From (3.110) one gets, for every $A\subset \bar Y$,
$$
P_Q (\tau_A <\exp (\epsilon \beta) \nu_A) @>>{\beta\to\infty}>
1.
\tag{3.111}
$$
Hence, by (3.111) and (3.108) we get
$$
P_Q(\tau_{Y^<} <
\exp\bigl\{\beta (2K-h +
\epsilon ) \bigr\}
) @>>{\beta\to\infty}> 1
\tag{3.112}
$$
for every $\epsilon
> 0$.
On the other hand, with the same reasoning as that one leading to (3.110), we
obtain
$$
\inf_{\sigma \in \Cal D(D_1,D_2)} P_{\sigma}
(\tau_{\bar Y} < T_0) > \exp(
-\epsilon \beta)
\tag{3.113}
$$
for every $\epsilon > 0$ and $\beta$ sufficiently large.
Now, let us prove (3.108).
This can be obtained very easily, once we take $\epsilon$ in
(3.108) sufficiently small --- namely, $\epsilon = \tilde
\epsilon$ satisfying the inequality (3.104). For every
$Q=Q(D_1,D_2, (l_i)_{i=1,\dots ,4})$ such that for a nonempty
subset of indices, $J\subset\{1,\dots , 4\}$, one has $l_j \geq
l^*$, $ j \in J$, let $Y^< (Q)$ be the set $$
Y^< (Q) = \{Q'\equiv Q'(D_1,D_2, (l_i')_{i=1,\dots ,4}):l_j'
\leq l_j \text{ for all } j
\in J \}.
\tag3.114
$$
We have
$$
P(Q\rightarrow Y\setminus (Y^<(Q) \cup \{ Q\}))=
P_Q(\sigma_{\bar \upsilon_1} \notin Y^<(Q)) \leq
\exp\bigl\{-\beta (h(l^* - 1)
-\epsilon ) \bigr\}
\tag3.115
$$
for every $\epsilon >0$ and $\beta$ sufficiently large.
Indeed, the bound (3.115) follows easily by reversibility, in a similar way as
the equation (3.41), since for every $Q \in \Cal D(D_1,D_2)$ one has
$$
\min _{\omega : Q \rightarrow Y\setminus (Y^<(Q) \cup \{ Q\})}
\max _{\sigma
\in \omega} H(\sigma) \geq H(Q) + h(l^* - 1) \beta.
\tag3.116
$$
On the other hand, one easily sees that for every
$\epsilon >0$ and $\beta$ sufficiently large,
$$
P(Q\rightarrow Y\setminus\{ Q\}) \equiv
P_Q(\sigma_{\bar \upsilon _1} \neq Q)
\geq
\exp\bigl\{-\beta (2K - h + \epsilon) \bigr\}
$$
and, moreover,
$$
P(Q\rightarrow Y^<(Q)\setminus\{ Q\})
\geq \exp -(2K - h + \epsilon)\beta.
$$
It is easy to see that, starting from any configuration
$\sigma \in Y^<$
we reach the minimum $\widetilde Q _0$,
with high probability, in a time of order $\exp\{ \beta h (l^* - 2 )\}$.
Indeed, in a time of the order
$$
T_1=
\exp\bigl\{\beta (h(l^*-2)+\epsilon_1) \bigr\},
\tag3.117
$$
if $\epsilon_1 $ is sufficiently
small,
no $K$-protuberances (or $K$-erosions) or {\it a fortiori}
elementary events involving
bigger increments of energy,
take place with probability larger than $\exp(-\epsilon_1\beta)$
for $\beta$
sufficiently large. Hence, with the same probability, the only
possible elementary events are $h$-erosions and recoveries.
Taking this fact into account, for all $\sigma \in Y^<$, we get
$$
P_{\sigma}( \tau_{\widetilde Q_0}e^{-\beta\epsilon}.
\tag3.118
$$
The~proof of
\thetag{3.118} can be obtained by adapting
the argument of proof of Theorem 1 a) of \cite{NS}.
We only present here the~main idea leading to~\thetag{3.118}. Starting
from any~$\sigma\in\Cal B(\widetilde Q_0)$, there is~a~probability
larger than
$\exp\bigl\{-\beta (h (l^*-2)-\epsilon) \bigr\}$
to completely
erode, in a time $l^* - 1$, any~oblique edge of~$\widetilde Q_0$
of the length~$ \leq l^*-1$. Hence, in a~time of order~$T_1$,
we certainly have to reach~$\widetilde Q_0$ before~$T_1$
since
the~circumscribed octagon can~not grow and since {\it a~fortiori} we know from
the~inequality
$$
h(l^*-1)>(2K-h)
$$
that no oblique edge~$l\geq l^*$ can be completely
eroded.
>From \thetag{3.112}, \thetag{3.113}, and\thetag{3.118} we get
(C3). Namely, for every $\epsilon > 0$ one has
$$
P(\bar \Cal E _{\sigma,\bar t_0}^0) =
\sum _{t_0 =1}^{\bar t_0} P( \Cal E _{\sigma,t_0}^0 )
\geq \exp(-\epsilon \beta).
\tag3.119
$$
This concludes the proof of Proposition 1 when $L_2 > l^* +1$.
Consider now the case $L_2 = l^* + 1$ (supposing always $L_1 > L_2$ ).
The proof of Proposition 1, in this case, is similar to the one
in the case $L_2 > l^* + 1$. For any $\sigma \in \Cal
D(D_1,D_2)$ we will again introduce
an event $ \Cal E_{\sigma}^s$. The main
differences can be summarized in the
following 2 points.
1) It follows from Lemma 3.3 above that the minimum of H in
$\partial \Cal D(D_1,D_2)$ is not achieved in $\widetilde S_3 $
but in $\widetilde S_2 $ (cf\. Fig\. 3.8) and in $\hat S_2 $
defined as the saddle
configuration obtained from $\widetilde Q_1$ by eroding
the last $ L_2 - 2 = l^* - 1 $ unit squares adjacent to
the coordinate edge of
length $L_2 - 1 = l^* $ (in other words, $\hat S_2 $
is obtained by adding
a unit square
protuberance to the coordinate edge of length $l^*+2 $
of an
octagon in $Q(D_1 - 1, D_2,l^* , l^* - 1,l^* ,l^*)$ --- see
Fig\. 3.9).
%
%
\midinsert
\centerline{\picture 1.40in by 1.44in (threenine)}
\botcaption {Fig\. 3.9}
\endcaption
\endinsert
%
%
2) The path starting from $\widetilde S_2$ and reaching
$\widetilde Q_3$ is not a pure descent anymore, but it involves
tunnelling.
We have
$$
H(\widetilde S_2) - H(\widetilde Q_0) = H(\hat S_2) -
H(\widetilde Q_0) =
2h(l^* - 1 ) - ( 2K - h) = E( l^* + 1 ),
$$
and
$$
H(\widetilde S_2) = H(\hat S_2) > H(\widetilde S_3) =H(\widetilde Q_0)+
2h(l^* - 1 ) - ( 2K - h) + h(l^* - 2 ) - ( 2K - h).
$$
Let $D_1 > D_2 = l^* + 1 + 2 (l^* - 1) $, i.e. $L_2 = l^* + 1 $.
The configurations
$ \widetilde Q_0,\widetilde Q_1, \widetilde Q_2$;
$\widetilde S_1,\widetilde S_2, \widetilde S_3$ are defined in
exactly the same way as in the case $ L_2 > l^* + 1 $.
Again, we define for every $\sigma \in \Cal D (D_1,D_2)$,
the event $\Cal
E_{\sigma , t_0}^0$ (cf\. (3.93)) in the same manner like before.
Now, let
$$
Q_1,Q_2 = \widetilde Q_0,\widetilde Q_1 ;\quad
S_1, S_2 = \widetilde S_1,\widetilde S_2
$$
and let $\bar t_0 = \exp\{\beta (2K - h + \delta)\}$,
$\bar t^1_u = \bar t^2_u = \exp\{\beta (2K - h + \delta)\}$,
and $t^1_d = 1 $.
We define
$$
\Cal E_{\sigma}^s = (\bar \Cal E_{\sigma,\bar t^0}^s ;\bar
\Cal F^{1,u};\Cal F^{1,d}_1;\bar\Cal F^{2,u})\cap \Cal G.
\tag3.120
$$
(cf\. (3.57) and (3.94)).
We get
$$
\inf_{\sigma \in \Cal D(D_1,D_2)}P(\Cal E_{\sigma}^s )
\geq \alpha \equiv
\exp \{ - \beta( E (l^*+1) - 2K + h + \epsilon) \}.
\tag3.121
$$
One verifies (3.121) in the exactly same way as in
the case $L_2 >
l^*+1$. We only have to check, in our case when $L_2=l^*+1$, the
validity of the bound (3.119).
However, the proof of (3.119) is even simpler since now
the number of possible cases to be considered, namely the number
of $ Q$'s in $\bar Y$, is much smaller; in fact $E (l^*+1)
3l^* -2$. Now we turn to the case $D_1>D_2=3l^* -2$.
It means that $L_2= l^{\ast}$ and
$$
E(L_2)=E(l^*)=h(l^*-1)
\tag{3.122}
$$
according to (3.14) and (2.18).
\proclaim{Proposition 2} Consider a~standard
octagon~$Q(D_1,D_2)$ with
$$
D_1>D_2=3 l^*-2.
$$
Then, for all $\epsilon>0$, one has
$$
P_{Q(D_1,D_2)}(\tau_{Q(D_1-1,D_2)}>
\exp\bigl\{\beta (E(l^*)+\epsilon) \bigr\})@>>
{\beta\to\infty}>0.
\tag{3.123}
$$
\endproclaim
\demo{Proof}
The proof is obtained along the same lines as in the case $L_2 = l^* +1$
above. The
main difference (and simplification) is that typical paths to the saddle
configurations in $\Cal S(D_1,D_2)$ are now purely uphill, while the paths
from $\Cal S(D_1,D_2)$ to $Q(D_1 -1,D_2)$ involve two, instead of one,
tunnelling phenomena.
Let the octagons $ \widetilde Q_0,\widetilde Q_1,\widetilde Q_3$, and the saddle
configuration $\widetilde S_1$
be defined as for $L_2 \geq l^*+1$. In particular $\widetilde S_1$ is
obtained from $ \widetilde Q_1 \in Q(D_1,D_2,l^*+1,l^*,l^*,l^*)$ by adding a~unit
square to~the~oblique edge of length~$l^*+1$. Moreover,
let $S^*_1$ be the~saddle configuration
obtained from the centered octagon $\widehat Q _1$ in
$Q(D_1-1,D_2,l^*-1,l^*-1,l^*,l^*)$ by
adding a~unit square to its right hand vertical edge of length~$l^*+2$.
We have
$H(\widetilde S_1)=H(S^*_1)=H(\widetilde Q_0)+h(l^*-1)$.
The minimum of $H$ in
$\partial \Cal D (D_1,D_2)$ is now reached,
again acording to Lemma 3.3,
in $\widetilde S_1$ and $S^*_1$.
We define $\Cal C$ as in (3.86) and the shrinking event $\Cal E^s_{\sigma}$
simply by
$$
\Cal E^s_{\sigma} =( \bar \Cal E^0_{\sigma, \bar t_0}; \bar \Cal F^{1,u})
\bigcap \Cal G,
$$
where the event $\bar \Cal E^0_{\sigma, \bar t_0}$
is, for all $\sigma \in \Cal D (D_1,D_2)$,
defined in terms of events $\Cal E^0_{\sigma, t_0}$, as in (3.44), while
$\Cal E^0_{\sigma, t_0}$ is
defined like in (3.93); for $\bar t^0$ we take $\bar t^0 = \exp (h(l^*-2)
+\delta)$ and
$$
\Cal F^{1,u}=
\{\xi^1_0=\widetilde Q_0,\, \xi^1_{\nu_1}=\widetilde S_1\},
$$
$$
\Cal G=\{
\tilde \tau_{\partial \Cal B (\widetilde Q_0)}\leq \bar t_u \}, \text { and }
$$
$$
\bar t_u =
\exp\bigl\{\beta (2K-h+\delta ) \bigr\}.
$$
Proposition 2 follows once we prove, in the present
case, the bounds (3.88) and (3.119).
To get these two bounds we again argue as in the proof of
(3.118) for the case $L_2=l^*+1$.
In particular, we notice that now, entering into $\Cal C$ in one step
actually means to descent
to $\widetilde Q _1$. The octagon $\widetilde Q _1$ itself now has
a coordinate edge (the vertical one on the righ hand side) of length $l^*-1$.
Thus, in a time $t= \exp\bigl\{\beta (h(l^*-2)+\delta) \bigr\}$,
with $\delta$ so small that
$$
h(l^*-2)+\delta < (2K-h) -\delta,
$$
no $K$-protuberance (or elementary events involving
even higher increment in energy) takes place with high probability and, as
a conseqence, the
vertical edge of length
$l^* -1$ will be completely eroded.
In this way we reach the octagon
$$
Q^*_2 \in Q(D_1-1,D_2,l^*,l^*-1,l^*,l^*),
$$
from which, in time $t=\exp \{\beta(h(l^*-2)+\delta)\}$, with the help of the
same mechanism, we descent to $\widetilde Q_3$.
We leave the details to the
reader.
\qed\enddemo
Propositions 3 A--C describe subsequent shrinking, once
we reached a regular octagon $Q(l^*)$. In the first step it shrinks by cutting
an arbitrary one, coordinate or oblique, edge.
\proclaim{Proposition 3 A}
Consider a~regular
octagon~$Q(l^*)$. Let~$ G_1$ be the~set of (not
standard) octagons
$$
Q(D_1,D_2,l^*-1,l^*,l^*,l^*-1),\quad D_1=3l^*-2,\quad D_2=3l^*-3,
$$
i.e.
$$
L_1=l^*+2,\quad L_2=L_3=L_4=l^* \text{ (modulo rotations) },
$$
and $ G_2$ be the~set of octagons
$$
\gather Q(D_1,D_2,l^*+1,l^*,l^*,l^*),\quad
D_1=D_2=3l^*-2,\\ L_1=l^*-1=L_2,\quad L_3=L_4=l^* \text{ (modulo rotations). }
\endgather
$$
Then, for any~$\epsilon>0$, one has
$$
P_{Q(l^*)}(\tau_{ G_1\cup
G_2}>\exp\bigl\{\beta(E(l^*)+\epsilon)\bigr\})@>>{\beta\to\infty}>0.
\tag{3.124}
$$
\endproclaim
\demo{Proof}
The proof is a straightforward adaptation of the proof of Proposition 2.
\qed\enddemo
Thus, starting from $Q(l^\ast)$ we reach an octagon with at
least one side shorter than $l^\ast$. This condition is
maintained also during the subsequent shrinking and we get:
\proclaim{Proposition 3 B}
Consider an octagon $Q$ with the values
$d_1,d_2,D_1,D_2$ such that
$$
\aligned
\text { either } \min (d_1,d_2) < &4l^*-2,\\
\text { or } \min (D_1,D_2) < &3l^*-2.
\endaligned
\tag3.125
$$
Then, for every $\epsilon >0$,
$$
P_Q(\tau_{-\underline
1}>\exp\{\beta (h(l^*-2)+\epsilon)\})@>>{\beta\to\infty}>0.
\tag3.126
$$
\endproclaim
\demo{Proof}
We give only a sketch of the proof leaving the details to the
reader. (See also the discussion of the growth event $\Cal
E^{(r)}$ introduced in Section 5.) We first
observe that by the hypothesis, one has initialy at least one
edge of length $l2$. Suppose, we apply to
$Q$
sequentially a series of arbitrary canonical cuts (i.e., we always
arbitrarily choose the edge to cut among the ones of minimal
length). Then, for every $l$ and for any such sequence
\roster
\item"i)"
After exactly 14 canonical cuts we always reach an
element of $Q(l-1)$. Namely, a regular octagon with the edge length
decreased by one. We use $\Cal M(l)$ to denote the set of all sequences of
canonical contractions from $Q(l)$ to $Q(l-1)$ .
\item"ii)"
Any sequence $M=\{Q(l), Q^{(1)},\dots,Q^{(13)},Q^{(14)}
\in Q(l-1) \} \in \Cal M(l)$ contains only ``almost regular''
octagons in the sense that the differences in the lengths of
any two oblique or coordinate edges in any $Q \in M $ is, for every $M \in
\Cal M(l)$, at most 3, $\max \Cal L_i - \min \Cal L_i \leq 3$.
Moreover, the minimal possible length of an edge during any
canonical sequence $M$ is $l-2$, while the maximal one is $l+2$.
\item"iii)"
In any sequence $M \in \Cal M(l)$ we perform exactly 4 oblique
cuts in the NE-SW direction, 4 cuts in NW-SE direction, 3 horizontal cuts, and 3
vertical cuts. There is 1 cut of an edge of length $l$, 9 cuts of edges of length
$l-1$ and 4 cuts of edges of length $l-2$.
\endroster
\endproclaim
\demo{Proof}
The proof is a straightforward exercise.
\qed\enddemo
We will now introduce some notions that will be useful to describe
the time dependence during a typical shrinking of a subcritical
droplet.
Let $\tau_0,\tau_1,\dots , \tau_n ,\dots $ be random times in
which our process $\sigma_t$ visits (after a change) the set $\Cal Q $ of
configurations containing a unique octagon:
$$
\tau _0 = \inf \{ t \geq 0 : \;\sigma_t \in \Cal Q\},
$$
$$
\tau _{n+1} = \inf \{ t > \tau _n : \;\sigma_t \in \Cal Q\},
\ \ n=0,1,2,\dots.
$$
Given $l ,\; 2\leq l \leq l^*$, and $\epsilon > 0$, we say that
$\sigma_t$ is an {\it $\epsilon$-canonical contraction path}
from $Q(l)$ to $Q(l-1)$ if
$$
\tau _0 = 0, \; \sigma_0 \in Q(l),\; \sigma_{\tau_1}\in
Q^{(1)},\dots,\sigma_{\tau_{14}}\in Q^{(14)},
$$
where
\roster
\item"i)" $( Q^{(1)},\dots ,Q^{(14)} \equiv Q(l-1) ) $ is an
element of the set $\Cal M(l)$ of canonical contractions,
\item"ii)" $ \exp\{
\beta (h (\hat \Cal L ^{(i)} -1)- \epsilon )\} < \tau _{i+1} - \tau
_i < \exp\{ \beta (h (\hat \Cal L ^{(i)}-1) + \epsilon)\},\ \
i=1,\dots , 14$, where $\hat \Cal L ^{(i)} = \min_{j=i,\dots , 8}
\hat \Cal L ^{(i)}_j$ and $ \hat \Cal L ^{(i)}_j$ are the lengths
of the edges of $Q^{(i)}$.
\endroster
\proclaim{Proposition 3 C}
Let $G_1,G_2$ be the set of octagons defined in Proposition 3 A
and let $\epsilon >0$. Then
$$
P_{G_1\cup
G_2}(\tau_{-\underline
1}>\exp\{\beta(h(l^*-2)+\epsilon)\}@>>{\beta\to\infty}>0.
\tag3.127
$$
Moreover,
with probability tending to one as $\beta \to \infty$,
\roster
\item"i)" starting from $G_1 \cup G_2$ our process will follow
the remaining 13 steps of an
$\epsilon$-canonical contraction path up to $Q(l^*-1)$,
\item"ii)" starting from $Q(l^*-1)$ it will follow an
$\epsilon$-canonical contraction path up to $Q(l^*-2)$ and so on
up to $Q(2)$,
\item"iii)" finally, it will persist in $Q(2)$ for a time $t< \exp
\{\beta (h + \epsilon)\}$ and then, after an $h$-erosion, it will proceed
downhill to $ - \underline 1$.
\endroster
\endproclaim
\demo{Proof}
The validity of (3.127) is a corollary of Proposition 3 A.
The statements i) and ii) follow from Lemma 3.5 and the fact that in a
time less than $\exp \{\beta ( h(l^*-2) +\epsilon)\}$, for
$\epsilon$ sufficiently small, the only elementary events are, with
high probability for $\beta $ large, $h$-erosions and recoveries.
The statement iii) is immediate.
\qed\enddemo
\remark {Remark}
By Lemma 3.5 we get a lot of additional geometrical information about an
$\epsilon$-canonical contraction path that could have been, actually, included in
the statement of Proposition 3 C.
\endremark
Finally, we consider the much simpler case of growth of supercritical octagons.
\proclaim{Proposition 4} Let $Q(D_1,D_2)$ be a~standard octagon
with~$D_2\leq D_1$ such that
$$
L_2=D_2-2(l^*-1) \geq L^*.
$$
Let
$$
\hat \tau=\tau _{Q(D_1,D_2+1)\cup Q(D_1+1,D_2)}.
\tag3.128
$$
Then, for every $\epsilon>0$, one has
$$
\sup_{\sigma\in
Q(D_1,D_2)}P_{\sigma}(\hat \tau<\exp\bigl\{\beta(2J-4K-h
+\epsilon)\bigr\} \text { and }\hat \tau = \bar \tau)@>>{\beta\to\infty}>1,
\tag3.129
$$
where $\bar \tau $ has been defined in (3.69).
\endproclaim
\demo{Proof}
According to Lemma 3.3, we have
$$
\inf_{\omega :\sigma_0\to\Cal D^{c}(D_1,D_2)}\sup_{\sigma\in\omega}H(\sigma)=
\min_{\sigma\in\partial\Cal
D(D_1,D_2)}H(\sigma)=E(D_1,D_2) =H(Q) +2J-4K-h.
\tag 3.130
$$
In the present case, $L_2>L^*$, the configurations in
$\Cal S (D_1,D_2)$, minimizing $H$ on $\partial \Cal D(D_1,
D_2)$, are obtained by adding a unit square protuberance
to one of the coordinate edges of an octagon in $Q(D_1,D_2)$.
Consider the time
$t(\delta) =
\exp\bigl\{\beta (2J-4K-h+\delta) \bigr\}$.
It follows from (3.130) and Lemma 3.4 that if
$\delta$ is sufficiently small, any $\sigma \in \partial \Cal D \setminus \Cal
S$ cannot be reached before $t(\delta)$ with a probability approaching one as
$\beta \to \infty$.
Let us use $\widetilde Q_0$ to denote the octagon in $Q(D_1,D_2)$, corresponding
to our initial condition and $\hat S_1$, the saddle in $\Cal S(D_1,D_2)$ obtained
by adding to $\widetilde Q_0$ the first unit square protuberance to its vertical
right side.
For every $\sigma \in \Cal D(D_1,D_2)$, let $\Cal E^0_{\sigma,t_0}$ and $\bar
\Cal E^0_{\sigma,\bar t_0} $ be defined as in (3.93) and (3.44) with $\bar t_0 =
\exp\bigl\{\beta ( 2K-h+\bar \delta) \bigr\}$ with $\bar \delta$ sufficiently
small. We notice that, by Lemma 3.4, before the time $t(\delta)$ one cannot see,
with probability
approaching one as $\beta \to \infty$,
any $J$-protuberance occuring on a configuration $\bar \sigma \in \Cal
D(D_1,D_2) $ that would differ from $ \widetilde Q_0$. Otherwise we would
touch the configuration
$$
\sigma' = \bar \sigma + J\text{-protuberance}
$$
with $ H(\sigma') - H(\widetilde Q_0) \geq 2J -4K -h +\Delta$, for some
positive $\Delta$.
Thus we can apply the same argument as that one
used in the case $L_2 < L^*$. In
this way it is easy to get (3.119).
Now, consider the set up for introducing our auxiliary Markov
chains with $ N=1$, $Q_1 = \widetilde Q_0$, $B_1 = \Cal
B(\widetilde Q_0)$, $S_2 = \hat S _1$, and $S_1$ --- the saddle
obtained from $\widetilde Q_0$ by adding a $K$-protuberance
to one of its oblique edge,
$$
\min_{\sigma\in\partial \Cal B(\widetilde Q_0)} H(\sigma ) = H(S_1).
$$
For every $\sigma \in \Cal D(D_1,D_2)$ consider the
event (of growth)
$$
\Cal E^g_{\sigma} =( \bar \Cal E^0_{\sigma, \bar t_0}; \bar \Cal F^{1,u})
\bigcap \Cal G,
$$
where
$$
\Cal F^{1,u}=
\{\xi^1_0=\widetilde Q_0\quad \xi^1_{\nu_1}=\hat S_1\},
$$
$$
\Cal G=\{
\tilde \tau_{\partial B (\widetilde Q_0)}\leq \bar t_u \},
$$
and
$$
\bar t_u =
\exp\bigl\{\beta (2K-h+\tilde \delta ) \bigr\}.
$$
As in the case $L_2 0 $ and $ \beta$ sufficiently large,
$$
\inf_{\sigma \in \Cal D(D_1,D_2)}P(\Cal E_{\sigma}^g ) \geq \alpha \equiv
\exp \{ - \beta[ (2J -4K -h) -(2K-h) + \epsilon] \} .
\tag3.131
$$
Hence, by a recurrence argument similar to the one given in estimates
(3.90), (3.91), with the help of the strong Markov property, we get
$$
P_{\sigma }(\tau_{\partial \Cal D} >
\exp\bigl\{\beta (2J-4K-h + \epsilon) \bigr\})
@>>{\beta\to\infty}> 0
\tag3.132
$$
for
every $\epsilon >0$ and any $\sigma \in \Cal D (D_1,D_2)$.
Then, again with the help of the strong Markov property, we get the desired results
in the same way as in the proof of
Proposition 1 since, again,
$$
P_{\bar \sigma} (\hat \tau = 1) > \frac1{|\Lambda |},\quad
P_{\bar \sigma} (\tau_{\Cal D(D_1,D_2)}<\tau_{\Cal D ^{c}(D_1,D_2)\setminus
\{\bar \sigma \} }| \hat \tau > 1)@>>{\beta\to\infty}> 1.
\tag3.133
$$
Once more, we leave the details to the reader.
\qed
\enddemo
\vfill
\newpage
\head 4. Global saddle point \endhead
Similarly as in \cite{KO 1}, the proof of our Theorems about ``escape time
and optimal route'' are based on the existence of a~set~$\Cal A$ with
the~following properties:
\roster \item"i)" For every~$\sigma\in\Cal A$ and
any~$\epsilon>0$ one has $$
\lim_{\beta\to \infty}P_{\sigma}(\tau_{-\underline 1}<\tau_{+\underline 1})=1
$$
and
$$
\lim_{\beta\to\infty}P_{\sigma}(\tau_{-\underline
1}<\exp\{\beta(h(D^*-1)+\epsilon)\})=1\ .
$$
\item"ii)" Every path $\omega$ starting in $-\underline 1$ and ending
in $+\underline1$ has to~pass through~the~boundary
$\partial\Cal A$ of~$\Cal
A$ defined by
$$
\partial\Cal A=\{\sigma\notin\Cal A,\text{ there exists } x\text{
such that }\sigma^{(x)}\in\Cal A\}\ .
$$
\item"iii)" The~minimal energy in~$\partial\Cal A$ is attained for
``protocritical'' (global saddle) con\-fi\-gu\-rations~$\sigma \in \Cal P$ ---
a~single
unit square attached to~the~longer coordinate side of~a~standard
octagon~$Q(D^*-1,D^*)$ --- with the~energy
$E^*\equiv H(Q(D^{\ast}-1,D^{\ast})) +2J-4K-h$. All
configurations in~$\partial\Cal A$ that are not of~this form have
the~energy by at least~$h$ higher.
\endroster
As a first step toward the~construction of~the~set~$\Cal A$ we
construct for~every configuration~$\sigma$ (from certain class)
a~configuration~$\overline{\sigma}$ such that
$$
S \sigma \prec\overline{\sigma}
$$
for every standard~$S$. Here we use the~natural order on the~set of
configurations:
$$
\sigma_1\prec\sigma_2\text{ iff }
\{x:\sigma_1(x)=+1\}\subset\{x:\sigma_2(x)=+1\}\ .
$$
In other words, the~configuration~$\overline{\sigma}$, to be called
{\it the blown up envelope of} $\sigma$, is the
``maximal'' one
(and
actually even larger) to~which we can arrive applying a~standard
sequence.
We begin by~constructing~$\overline{\sigma}$ in~the~case when~$\sigma$
corresponds to~a~single droplet $C=C(\sigma)$ (i.e.
the~set~$C(\sigma)$ is connected). Suppose also that the~rectangular
envelope~$R(C)$ does not wind around the~torus and consider
the~monotone envelope~$M=M(C)$. The~set~$M$ consists of~monotonous
blocks connected in~corners, of~the~form shown on Fig\. 2.1, where four
edges meet. We say that such point is a {\it bottleneck\/} of the~set~$M$,
once the~intersection of~the~boundary~$\partial M$ with each of~
the~two~outside right angles touching in~this point has at~least along
one side the~length one (Fig\. 4.1).
%
%
\midinsert
\centerline{\scaledpicture 4.92in by 3.29in (fourone scaled 800)}
\botcaption{Fig\. 4.1}
A set $M$ with bottlenecks denoted by a thick dot.
The points denoted by~an~arrow cease to~be bottlenecks after ``enveloping
the~components''.
\endcaption
\endinsert
%
%
Disconnecting now the~set~$M$ in~bottlenecks,
consider the~union~$M^{(1)}$ of~octagonal envelopes of~corresponding
components. Some of~the~resulting points that were bottlenecks for~$M$
may not be such any more for~the~resulting set~$M^{(1)}$. Considering only
its bottlenecks we repeat the~procedure and iterate until the~set
of~bottlenecks does not change. We use~$N(C)$ to denote
the~droplet constructed above from the~set~$C$ and call it a~{\it
string\/} (i.e. a~string is a~monotonous droplet whose components, after
disconnecting its bottlenecks, are octagons with oblique sides $l_i\geq 1$
--- not necessarily larger or equal $2$).
The configuration~$\overline{\sigma}$ corresponding
to~the~droplet~$N(C(\sigma))$ is the~sought blown up envelope
with the~desired properties.
\proclaim{Lemma 4.1a} Let~$\sigma$ be a~configuration with a~single
droplet~$C(\sigma)$ contained in~a~rectangle that is not wrapped around
the~torus. Then
\roster
\item"i)" $H(\overline{\sigma})\leq H(\sigma)$,
\item"ii)" $\sigma\prec\overline{\sigma}$ and $S\sigma\prec\overline{\sigma}$
for every standard sequence.
\endroster
\endproclaim
\demo{Proof} Since taking octagonal envelopes of components decreases
the energy,
for our proof of (i) it is sufficient to~show that
$$
H(\sigma)\geq H(M)
$$
(we identify the~droplet~$M$ with the~corresponding configuration).
We will show that there exists a~sequence of~configurations with
decreasing energy and leading from~$\sigma$ to~$M$.
First, we show that the~holes inside~$C=C(\sigma)$ (minus spins
inside~$\partial_{\text{out}}C$ --- see Fig\. 3.2) can be filled up. Consider
the~boundary of the holes, $\partial C\setminus\partial_{\text{out}}C$, and
consider the lowermost horizontal line touching it. It touches it along a
certain number of segments, with~the~spins around each of~them
necessarily taking the~values shown in Fig\. 4.2.
$$
\vbox{%
\centerline{\vbox{%
\offinterlineskip
\halign{\strut#&\ \ \ $#$\ &#&\ $#$\ &\ $#$\ &\ $#$\ &\ $#$\ &\ $#$\ &\ $#$\
&#&\ $#$\ \ \ \cr
&\bullet&&\bullet&\bullet&\cdots&\bullet&\bullet&\bullet&&\bullet\cr
&+&\vrule width 1pt&-&-&\cdots&-&-&-&\vrule width 1pt&+\cr
\omit&&\vrule height 1pt width 1pt&
\multispan{2}\leaders\hrule height 0.5pt depth 0.5pt\hfill&\dots&
\multispan{4}\leaders\hrule height 0.5pt depth 0.5pt\hfill
\vrule height 1pt width 1pt\cr
\omit&\multispan{4}\leaders\hrule height 0.05pt depth
0.05pt\hfill&
&\multispan{5}\leaders\hrule height 0.05pt depth
0.05pt\hfill\cr
$l$&\bullet&&+&+&\cdots&+&+&+&&\bullet\cr}}}
\vskip 10pt
\centerline{Fig\. 4.2}}
$$
Notice, in particular, the~$+$ spins below the~line. If any of them
were replaced by~$-$, the~line~$l$ would be pushed lower. The~spins
denoted by dots are arbitrary. Labelling the~spins in~the~first row
as~~$\sigma_{\roman I},\sigma_1,\dots,\sigma_k,\sigma_{\roman{II}}$ and those in
the~third line that are not fixed
as~$\sigma_{\roman{III}},\sigma_{\roman{IV}}$, we
get an~energy decrease after flipping simultaneously all minus
spins in~the~second row. Namely,
$$
\multline
-\Delta H=2\widetilde J+\widetilde
J\sum_{i=1}^{k}(1+\sigma_i)+K(1+\sigma_{\roman I}+
\sigma_{\roman{III}}+\sigma_2)+\\
+K(1+\sigma_{\roman{II}}+\sigma_{\roman{IV}}+\sigma_{k-1})+K\sum_{i=2}^{k-1}(2+\
+\sigma_{i-1}+\sigma_{i+1})+hk
\endmultline
$$
whenever $k\geq2$ and
$$
-\Delta H=2\widetilde J+\widetilde J(1+\sigma_1)+K(\sigma_{\roman I}
+\sigma_{\roman{II}}+\sigma_{\roman{III}}+\sigma_{\roman{IV}})+h
$$
for $k=1$. In both cases its minimum is attained
if all spins denoted by dots are minus,
$\sigma_i=-1,i=1,\dots,k,\ \sigma_{\roman
I}=\sigma_{\roman{II}}= \sigma_{\roman{III}}=
\sigma_{\roman{IV}}=-1$,
and we get
$$
-\Delta H\geq2\widetilde J-4K+h>0
$$
acording to our assumption~$K\leq\frac12(\widetilde J-h)$ (cf\. Lemma~
3.1).
Flipping thus all considered minus spins above the~line~$l$, it will be
pushed higher. Iterating the~process we finally erase all holes in~$C$,
decreasing in~the~same time the~energy of~the~configuration. Hence, we
can suppose that~$\partial_{\text{out}}C=\partial C$.
In a similar way we can prove that filling the~droplet up to~$M$ we
further decrease the~energy. Namely, consider a~right
angle, say~$\{(x,y)\in\Bbb R^2;x\geq x_0,y\geq y_0\}$, such that~$C$
does not intersect its interior and such that it touches~$\partial C$
in at~least two distinct points. Consider those two such points~$A$
and~$B$ whose distance is maximal (see Fig\.~4.3 --- even though we show
here~$A$ and~$B$ belonging one to the~horizontal and one to
the~vertical side of~the~angle, nothing prevents them to belong
to~the~same side).
%
%
\midinsert
\centerline{\scaledpicture 4.90in by 2.78in (fourthree scaled 800)}
\botcaption{Fig\. 4.3}
\endcaption
\endinsert
%
%
Considering now an~order on~the~path winding around~$\partial C$
(as~in the proof of Lemma~2.1), take the~portion~$\gamma$ between~$A$
and~$B$ and the~lowermost horizontal line~$l$ touching~$\gamma$. If~$l$
does not contain the~point~$(x_0,y_0)$, we can decrease the~energy
by~flipping intervals of~minuses just above the~line in~the~same way
as~we did when erasing the~holes of~$C$. Iterating the~procedure
unless~$l$ passes through~$(x_0,y_0)$ and repeating then all with
leftmost vertical line, we finally replace~$\gamma$ by~a~corresponding
segment of~the~boundary of~the~angle. Repeating the~same for all
relevant angles, we finally obtain the~configuration~$M$ and prove thus
(i).
To prove (ii) we assume the~contrary and consider a~standard
sequence~$S$ for which~$S \sigma\prec\overline{\sigma}$ does not hold.
Take the~first site~$x\in S$
outside~$C(\overline{\sigma})$, $x\notin C(\overline{\sigma})$. Considering
an~arbitrary configuration~$\xi$ inside~$C(\overline{\sigma})$
and~$\xi(y)=-1$ for all~$y\notin C(\overline{\sigma})$, we will show that
the spin flip~$\xi(x)=-1\to+1$ always increases energy. This is clear
if~$x$ is not attached to~the~boundary of~$C(\overline{\sigma})$.
Taking into account that $C(\overline{\sigma})$ is a string, the local
situation around $x$ attached to $\partial C(\overline{\sigma})$ is among
the~following ones (with the site $x$ in the center of the square and up
to~rotations and reflections): $$\matrix \bullet&\bullet&\bullet\\
-&-&-\\
-&-&-\\
\noalign{\vskip 2 pt}
&{\roman A)}&\endmatrix\qquad
\matrix
-&\bullet&-\\
-&-&\bullet\\
-&-&\bullet\\
\noalign{\vskip 2pt}
&{\roman B)}&\endmatrix\qquad
\matrix
-&\bullet&\bullet\\
-&-&\bullet\\
-&-&-\\
\noalign{\vskip 2pt}
&{\roman C)}&\endmatrix$$
(Notice that the case A) covers for example the situations
$$\matrix
\bullet&-&-\\
-&-&-\\
-&-&-\endmatrix\quad,\quad
\matrix
-&\bullet&-\\
-&-&-\\
-&-&-\endmatrix\quad,\text{ etc.},$$
while B) e.g.
$$\matrix
-&\bullet&-\\
-&-&\bullet\\
-&-&-\endmatrix\quad,\quad\matrix
-&\bullet&-\\
-&-&-\\
-&-&\bullet\endmatrix\quad\text{ etc.})$$
A glance into the catalogue of~stable situations (page 12) insures us that
in all three cases the minus spin in the~center is stable irrespectively
of~the~values of~spins at~sites denoted by dots, in ~contradiction
with the~assumption that the~spin flip~$\xi(x)=-1\to+1$ decreases
the~energy. The~stability in~the~case A) follows from a)
of~the~catalogue, B) from a), e), and g), and C) from a), d), and
g).\hfil\qed
\enddemo
Notice that the string~$\overline{\sigma}$ might still contain unstable
lattice sites --- namely those plus spins that are surrounded by~4
nearest neighbour minuses. When applying a~standard sequence~$S$, one
will eventually erase them obtaining a~union of~octagons contained
in~$C(\overline{\sigma})$.
Next we proceed to~a~case of~a~configuration~$\sigma$ consisting
of~several components. To treat this case we shall repeatedly use
the~following lemma to evaluate the sum of~energies of~two
configurations by~the~energy of~the~configuration whose area of~pluses
is~the~union of~areas of~pluses of~the~two concerned configurations.
For configurations~$\sigma_1$ and~$\sigma_2$ we define
the~minimum~$\sigma_1\wedge\sigma_2$ and
the~maximum~$\sigma_1\vee\sigma_2$ with respect to~the~order~$\prec$.
Namely, these configurations are given by~taking minimum and
maximum, respectively, of~$\sigma_1$ and~$\sigma_2$ site by site. Then
we have the standard inequality (a base of the FKG inequality):
\proclaim{Lemma 4.2} For any~$\sigma_1,\sigma_2$ and the~Hamiltonian
(2.2) one has
$$
H(\sigma_1)+H(\sigma_2)\geq
H(\sigma_1\vee\sigma_2)+H(\sigma_1\wedge\sigma_2).
\tag{4.1}
$$
\endproclaim
\demo{Proof} Using the~equality
$$
a+b=\max(a,b)+\min(a,b),
\tag{4.2}
$$
we get
$$
h\sum\sigma_1(x)+h\sum\sigma_2(x)=h\sum(\sigma_1\vee\sigma_2)(x)+
h\sum(\sigma_1\wedge\sigma_2)(x).
$$
The remaining two sums in the~definition (2.2) of~the~Hamiltonian
consist of terms of the form $\sigma(x)\sigma(y)$ and the inequality
\thetag{4.1} will be verified once we show that
$$
\sigma_1(x)\sigma_1(y)+\sigma_2(x)\sigma_2(y)\leq(\sigma_1\vee\sigma_2)(x)
(\sigma_1\vee\sigma_2)(y)+
(\sigma_1\wedge\sigma_2)(x)(\sigma_1\wedge\sigma_2)(y).
\tag{4.3}
$$
If~$\sigma_1(x)=\sigma_2(x)$ we have an~equality in~\thetag{4.3}
by~\thetag{4.2}.\newline
If~$\sigma_1(x)\ne\sigma_2(x)$, say ~$\sigma_1(x)=+1$, $\sigma_2(x)=-1$,
the~inequality \thetag{4.3} follows from
$$
\sigma_1(y)-\sigma_2(y)\leq\max(\sigma_1(y),\sigma_2(y))-
\min(\sigma_1(y),\sigma_2(y))\ .
\eqno{\lower 12pt \hbox{\text{\qed}}}
$$
\enddemo
Consider now a~configuration~$\sigma$ whose set~$C(\sigma)$ has several
components $C_1,\dots,C_k$. Let us suppose that
the~rectangles~$R(C_1),\dots,R(C_k)$ are not wrapped around the~torus
and let~$\overline{\sigma}_1,\dots,\overline{\sigma}_k$ be the~strings obtained
as blown up envelopes of the~configurations\break $\sigma_1,\dots,\sigma_k$
corresponding to~droplets~$C_1,\dots,C_k$.
Using $\overline H$ to denote the relative energy with respect to the energy
of $-\underline 1$, i.e. $\overline H(\sigma)= H(\sigma) - H(-\underline 1)$,
by Lemma~4.1 we have
$$
\overline H(\sigma)\geq \overline H(\overline{\sigma}_1)+\cdots+
\overline H(\overline{\sigma}_k).
\tag{4.2}
$$
If the~droplets~$N_1,\dots,N_k$ corresponding
to~configurations~$\overline{\sigma}_1,\dots,\overline{\sigma}_k$ were isolated,
we would define~$\overline{\sigma}$ by
taking~$C(\overline{\sigma})=\cup_{i=1}^kN_i$.
If the~droplets~$N_1,\dots,N_k$ are not isolated, we first glue them
together and repeat the~procedure. Namely, if two droplets~$N$, $N'$, have
a~nonempty intersection, consider the~droplet~$N\cup N'$. By Lemma~4.2
we have
$$
H(N)+H(N')\geq H(N\cup N')+H(N\cap N').
$$
Let us suppose that the~rectangular envelopes~$R(N)$ and~$R(N')$ are
subcritical (i.e. not winding around the~torus and with minimal sides
not exceeding~$D^*$). Then~$\overline H(N\cap N')\geq0$. Indeed, applying any
standard sequence to~$N\cap N'$, we get a~union of subcritical octagons
of~even lower energy. And the~energy of~a~subcritical octagon with respect to
the energy of the configuration $- \underline 1$ is positive. Thus
$$
\overline H(N)+\overline H(N')\geq \overline H(N\cup N').
\tag{4.3}
$$
Using this observation, we can consider the~components $\widetilde
N_1,\dots,\widetilde N_k$ of $N_1\cup N_2\cup\dots\cup N_k$ and show
that
$$
\overline H(\sigma)\geq \overline H(\widetilde N_1)+\cdots+
\overline H(\widetilde N_l),
\tag{4.4}
$$
once we suppose that the circumscribed rectangles of~$\widetilde
N_1,\dots,\widetilde N_l$ are subcritical.
Further, we say that two droplets~$\widetilde N$ and~$\widetilde N'$
{\it stick\/} together if there exists a~site~$x\notin \widetilde
N\cup\widetilde N'$ (necessarily touching~$\partial\widetilde N$ as
well as~$\partial\widetilde N'$) and a~configuration\footnote{
It is easy to~notice that if such configuration exists, one can take the
configuration with pluses in~$\widetilde N\cup\widetilde N'$, but we do
not need this fact.} inside~$\widetilde N$ and~$\widetilde N'$ such
that, flipping~$-1\to+1$ in~$x$, we decrease the~energy of
the~configuration corresponding to~$\widetilde N\cup\widetilde N'$ and
obtain a~connected set~$\widetilde N\cup\widetilde N'\cup q(x)$
(here~$q(x)$ is the~unit square around~$x$).
To avoid ambiguities, we choose the~pairs of~droplets sticking together
as well as the~particular site~$x$ to be flipped using some canonical
order --- say, lexicographic order of~the~uppermost left corner
of~the~droplet and the~same order for~the~site~$x$. Flipping the concerned site
$x$ and iterating the procedure we obtain a~set ~$C_1',\dots,C'_m$ of~disjoint
droplets such that neither pair of~them sticks together. Clearly,
$$
C'_1\cup\dots\cup C_m'\supset N_1\cup\dots\cup N_k
$$
and
$$
H(\sigma)\geq H(C_1')+\cdots+H(C_m').
$$
Supposing again that the~circumsribed
rectangles~$R(C_1'),\dots,R(C_m')$ are subcritical (and in~particular
are not winding around the~torus), we repeat the~same procedure as when
we started from~$\sigma$. Iterating it we finally get a~set of~disjoint
strings~$A_1,\dots,A_n$ such that no pair of~them sticks together. If
their circumscribed rectangles are not wrapped around the~torus and are
subcritical, we say that the~original configuration $\sigma$ is {\it
acceptable\/} and define~$\overline{\sigma}$ so
that~$C(\overline{\sigma})=A_1\cup\dots\cup A_n$. Clearly,
$$
H(\sigma)\geq H(\overline{\sigma}).
$$
Thus we get the~definitive
\proclaim{Lemma 4.1} Let~$\sigma$ be an~acceptable configuration. Then
\roster
\item"(i)" $H(\overline{\sigma})\leq H(\sigma)$,
\item"(ii)" $S\sigma\prec\overline{\sigma}$ for every standard sequence,
\item"(iii)" $\sigma\prec\overline{\sigma}$ and the~summary length of
boundaries of~all droplets of~$\sigma$ inside a~component~$A_i$
of~$C(\overline{\sigma})$ is at~least as large as the~length of the~boundary
of~the~circumscribed rectangle~$R(A_i)$.
\endroster
\endproclaim
\demo{Proof} The~statement (i) was already proven during
the~construction.
To prove (ii) we use the~same argument as when proving (ii) in
Lemma~4.1a. One has only to~observe that, considering a~site~$x$
attached to a~component~$A_i$ of~$C(\overline{\sigma})$ and supposing that
another component~$A_j$ is touching the~unit square~$q(x)$, the~spin
flip~$\sigma(x)=-1\to+1$ necessarily increases energy irrespectively
of~the~configuration inside~$A_i$ and~$A_j$. Indeed, if a~configuration
inside~$A_i\cup A_j$ existed so that the~energy decreases,
the~strings~$A_i$ and~$A_j$ would stick together which is not the~case.
To prove (iii) we first notice that in~the~first stages
of~the~construction of~$\overline{\sigma}$ we only decreased the~number of
bounds --- the~length of~the~boundary of~$\cup \widetilde N_i$ is not
larger than the~boundary of~the~original~$C(\sigma)$. Further, whenever
gluing two sticking components we do not change the~length of
the~boundary.\hfil\qed\enddemo
We say that a~string~$A$ is {\it ephemere\/} if
$\min(D_1(R(A)),D_2(R(A)))<3l^*-3$. Strings that are not ephemere have
energy larger or equal the~energy of the standard octagon inscribed
in~their rectangular envelope.
\proclaim{Lemma 4.3} Let~$A$ be a~string that is not ephemere and, $R$
be its rectangular envelope, $R=R(A)$, and~$Q$ be the standard octagon
with~$R(Q)=R$. Then
$$
H(A)\geq H(Q).
\tag{4.5}
$$
\endproclaim
\demo{Proof}
The string $A$ consists of certain number of octagonal blocks touching in
bottlenecks.
Actually, the inside blocks --- those not touching the boundary of the
rectangle $R(A)$ ---
are in general ``hexagons'', while the two corner blocks --- those touching the
boundary of $R(A)$ --- are in general ``heptagons'' (cf\. Fig\. 4.1).
The energy of a string does not change if we reorder the inside blocks.
In the case when any corner block contains also the vertex of the rectangle
$R(A)$, (it is a ``hexagon''), it can be interchanged with any inside
block without changing the energy of the configuration.
After a reordering some bottlenecks may disapear and one can decrease the
energy of the corresponding configuration by replacing a group of blocks,
attached at points that are not bottlenecks any more, by an octagonal envelope.
Using this observation in an iterative manner, one can finally replace the
string $A$ by a string $\bar A$ such that $R(A)=R(\bar A)$,
$H(A)\geq H(\bar A)$, and all the inside blocks of $\bar A$ are unit squares.
Indeed, starting from an arbitrary string $A$,
one easily gets a string whose all inside blocks are rectangles such that, say,
the vertical side of all of them is of length one.
We decrease further the energy and get $\bar A$ by shrinking all inside blocks
to unit squares and expanding in the corresponding manner one of the corner
blocks (see Fig\. 4.4). This last step does not change the number of corners
while increasing the area of the droplet.
%
%
\midinsert
\centerline{\scaledpicture 5.17in by 1.29in (fourfour scaled 800)}
\botcaption{Fig\. 4.4}
\endcaption
\endinsert
%
%
To evaluate now the energy of such a droplet $\bar A$ with $d$ inside
unit square blocks, we will consider two cases:
\roster
\item"$\bullet$"
$d\geq \frac{1}{2}(3+\sqrt{5})l^{\ast}+3$, or
\item"$\bullet$"
$d < 3 l^{\ast}-2$.
\endroster
These two cases cover all values of $d$ once $(3-\sqrt{5})l^{\ast}
>10$, i.e. if $K>7 h$.
In the first case, the energy decreases if we replace all inside (unit square)
blocks by a single hexagon with two oblique sides of length $l^{\ast}$.
Indeed, to prove this we have to show that (cf\. (3.18) and (3.6))
$$
-hd -(d-1)6K-2K>-hd^2+2F(l^{\ast}),
$$
or equivalently, using (3.6) and (2.18),
$$
P(d)\equiv d^2-d(3l^{\ast}-2+3\eta)+2(l^{\ast}-1+\eta)+[(l^{\ast}-1)^2
+\eta(2l^{\ast}-1)] >0.
$$
Taking into account that $\eta\in (0,1)$, the discriminant of the quadratic
expression $P(d)$ is bounded from above by $5(l^{\ast}+2)^2$.
Observing further that the term $d^2$ in $P(d)$ has a positive coefficient,
the sought inequality is fulfilled once $d$ exceeds the larger one from the
solutions of the equation $P(d)=0$. This solution is bounded from above by
$$
\frac{1}{2}[3l^{\ast}+1+\sqrt{5}(l^{\ast}+2)]\leq
\frac{3+\sqrt{5}}{2}l^{\ast}
+3.
$$
The resulting string contains a hexagon that is attached to (if any)
two corner blocks in points that are not bottlenecks and thus the energy
decreases if we replace this string by its octagonal envelope.
Its energy can be further lowered by replacing it by the standard octagon $Q$
with the same circumscribed rectangle $R(A)$ (cf\. Lemma 3.2).
On the other side, if $d < 3 l^{\ast}-2$ we consider the monotonous set
$\overline M$ constructed from $\overline A$ as the union of
the $d\times d$ square $\overline Q$ circumscribed to the union of inside
(unit square) blocks with the corner blocks and the union of all their shifts,
up to the distance $d$, along those sides of the rectangle
$R(A)$ they are touching (see Fig\. 4.5).
%
%
\midinsert
\centerline{\scaledpicture 3.04in by 1.92in (fourfive scaled 800)}
\botcaption{Fig\. 4.5}
\endcaption
\endinsert
%
%
Using $D$ to denote $\min(D_1(R(A)),D_2(R(A)))$, the area covered by $\overline
M\setminus \overline A$ is at least
$$
2Dd-d^2+d.
$$
On the other side, the droplet $\overline M$ has by $6(d-1)$ less corners and
the replacement of $\overline A$ by $\overline M$ is favourable once
$$
[(2D+1)d-d^2]h>6(d-1)K.
$$
This inequality is fulfilled if
$$
[2D+1-d]h > 6K=3(l^{\ast}-1+\eta)h.
$$
Thus, it is enough to take $d$ such that
$$
d<2D+4-3l^{\ast}.
$$
Since $D\geq 3l^{\ast}-3$, the inequality is fulfilled once
$d<3l^{\ast}-2$. The energy of the droplet $\overline M$ is lower than that
of its octagonal envelope and this, in turn, is lower than the energy of
standard octagon $Q$.
\hfil\qed\enddemo
Now, we are ready to define the set $\Cal A$.
Consider an acceptable configuration $\sigma$ and the corresponding
$\bar \sigma$ with nonsticking components $A_1, \dots, A_k$.
Consider further the family of sets $R_1, \dots, R_k$, where
$R_i = A_i$ if $A_i$ is ephemere; otherwise $R_i$ is the rectangular envelope
of $A_i$. Two strings $A_i, A_j$ are said to {\it interact} if at least one of
them is not ephemere and the sets $R_i$ and $R_j$ (but not necessarily $A_i$ and
$A_j$) stick together (or intersect)%
\footnote{However, if $A_i$ and $A_j$ stick together, they necessarily
interact.}.
A family of strings $A_1, \dots, A_k$ is said to form a {\it chain} $\Cal C$
if every pair $(A_i, A_j)$ of them
can be linked by a sequence $\{A_{i_1},\dots, A_{i_n}\}$
of pairwise interacting strings from $\Cal C$; $A_{i_1}=A_i$,
$A_{i_n}=A_j$, and $A_{i_{l}}$
and $A_{i_{l+1}}$ are interacting for all $l=1,\dots, n-1$.
Given a collection of chains $\Cal C_1,\dots,\Cal C_n$ corresponding
to the family $A_1, \dots, A_k$, we start the
following iterative procedure:
\roster
\item
The chains $\Cal C^{(1)}_j$ of the ``first generation'' are
identical to $\Cal C_j$,
$j=1,\dots,n$.
\item
Having defined $\Cal C^{(r)}_j$, we construct the sets
$R^{(r)}_j$ as the rectangular envelopes of the unions
$$
\bigcup_{R\in\Cal C^{(r)}_j}R
$$
whenever $\Cal C^{(r)}_j$ contains at least two sets;
we put $R^{(r)}_j=R$ if $\Cal C^{(r)}_j=\{R\}$.
\endroster
We define $\Cal C^{(r+1)}_j$ as the maximal chains of the family of sets
$\{R^{(r)}_j\}$.
Iterating this procedure we reach a family of chains, each consisting of a
single set. We call them {\it complete sets of the configuration $\sigma$}.
It is easy to observe that every complete set
from the resulting family
$\bar R_1,\dots,\bar R_s$ is either
a rectangle containing certain amount of original strings
$A_1, \dots, A_k$, or it is one of ephemere strings contained
in the family $A_1, \dots, A_k$.
We introduce $\Cal A$ as the set of all those acceptable configurations for
which all complete sets $\bar R_1,\dots,\bar R_s$ are subcritical ---
they can be placed in a rectangle that is not wrapped around the torus and
whose minimal side does not exceed $D^{\ast}-1$.
In the remaining part of the present Section we shall verify the
properties i) -- iii) of the set $\Cal A$.
It is easy to see that the property ii) of $\Cal A$ is obvious from
the definition, while the property i) follows from Propositions 1--3.
To prove the property iii) we first consider the configuration $\hat \sigma$
obtained by placing plus spins at all lattice sites in $\bar R_i$ for
ephemere $\bar R_i=\bar A_i$ and in the standard octagon $Q_i$
inscribed in $\bar R_i$ for $\bar R_i$ that is not ephemere.
We will repeatedly use the bound
$$
H(\sigma) \geq H(\hat \sigma).
\tag 4.6
$$
This follows, iterating with the help of Lemma 4.3, from the following Lemma.
\proclaim{Lemma 4.4}
Let $A'$ and $A''$ be a pair of interacting strings and let $\xi$ be the
configuration with the set of pluses coinciding with $A' \cup A''$.
Suppose that the rectangular envelope $R$ of the set $A' \cup A''$
is subcritical.
Then
$$
H(\xi) \geq H(\hat \xi),
\tag 4.7
$$
where $\hat \xi$ is the configuration with pluses in the standard octagon $Q$
inscribed into $R$.
\endproclaim
\demo{Proof}
Suppose first that both $A'$ and $A''$ are not ephemere and consider the
standard octagons $Q'$ and $Q''$ inscribed into the rectangular envelopes $R'$
and $R''$ of $A'$ and $A''$, respectively.
Clearly, by lemma 4.3, $\bar H(\xi)\geq \bar H(Q') +
\bar H(Q'') \geq \bar H(Q' \cup Q'')$.
If $Q'$ and $Q''$ stick together (i.e. $Q'$ and $Q''$ intersect each other or
there exists a site $x$ such that flipping the spin at $x$ decreases the energy)
the energy of the single droplet $Q' \cup Q'' \cup q(x)$ (resp\. $Q' \cup Q''$
if $Q'$ and $Q''$ intersect) is smaller than $H(\xi)$. It can be further
lowered by taking its monotonous envelope and, finally, by replacing it with
the octagon $Q$.
If $Q'$ and $Q''$ do not stick together, there does not exist a site $x$ such
that the unit square $q(x)$ intersects the boundaries of both $Q'$ and $Q''$
along a bond and we can see that the area of $Q$ is by at least
$$
\multline
\min_{y\in [0,l^{\ast}-1]}[(2(l^{\ast}-1)-y)(2(l^{\ast}-1)+y)+(2(l^{\ast}-1)+y)
(2(l^{\ast}-1)-y)]\geq \\
\geq \min_{y\in [0,l^{\ast}-1]}[8(l^{\ast}-1)^2-2y^2]\geq 6(l^{\ast}-1)^2
\endmultline
$$
larger than the area of $Q' \cup Q''$ (see Fig\. 4.6).
%
%
\midinsert
\centerline{\scaledpicture 2.79in by 2.54in (foursix scaled 800)}
\botcaption{Fig\. 4.6}
\endcaption
\endinsert
%
%
Taking into account that $4F(l^{\ast})\geq -2 (l^{\ast}-1)h$, we get
$$
6(l^{\ast}-1)h > -4 F(l^{\ast}),
$$
and thus the energy decreases once we replace the configuration $\xi$ by $Q$.
(We are even not using the fact that the overall number of bonds in the
boundaries of $Q'$ and $Q''$
might be larger than the number of bonds in the boundary of $Q$ leading to an
even larger decrease of energy.)
It remains to consider the case when one of the strings, say $A''$,
is ephemere. (See Fig\. 4.7.)
The only nontrivial situation is when $Q'$ and $A''$ do not stick together.
When replacing the configuration $\xi$ (with energy not lower than that of the
union of $Q'$ and $A''$) by $Q$, we have to compensate the loss of all corners
in $\partial A''$ by the surplus area $|Q|-|Q'\cup A''|$ and surplus length of
the boundary $|\partial Q'|+|\partial A''| - |\partial Q|$.
Consider now the strips $H$ and $V$ obtained as the union of all horizontal
and vertical shifts of $Q'$, respectively.
It is easy to see that the surplus length is at least the number $N_h$ of
horizontal bonds of $\partial A''$ in $V$ plus the number $N_v$ of vertical
bonds of $\partial A''$ in $H$.
Each corner of $\partial A'' $ in $V$ is linked with at least one
horizontal bond of $\partial A'' $ in $V$.
The number of corners attached to $N_h$ horizontal bonds
in $V$ is at most $6\frac{N_h}{2}+6$.
Their loss is compensated by the surplus of those $N_h$ bonds since
$$
(3N_h+6)K \leq 6 N_h K \leq N_h J.
$$
Similarly for the vertical bonds in $H$.
%
%
\midinsert
\centerline{\scaledpicture 4.40in by 2.57in (fourseven scaled 800)}
\botcaption{Fig\. 4.7}
\endcaption
\endinsert
%
%
Thus it remains to compensate for corners not contained in the closed set
$H\cup V$.
Consider the four quadrants --- components of $(H\cup V)^c$.
The set $A''$ intersects at most two of them. We use $U_1$, $U_2$, to
denote these two quadrants and $M_i = U_i\cap A'' $, $i=1$, $2$.
Moreover, there is at most one quadrant of $(H\cup V)^c$, say $U_1$,
whose both sides intersect $A''$.
Without loss of generality we can suppose that $M_2$ does not intersect
$H$ and the circumscribed rectangle $R(M_2)$ to $M_2$ is $m_2\times n_2$.
Then there is at least
$\min (m_2, n_2) (3 l^{\ast} -2)$
of surplus area of $|Q| - |Q'\cup A'| $ in the component of $H\setminus
Q'$ that touches $U_2$. On the other side, at most $\min (m_2, n_2) l^{\ast}
$ of the area may be lost from the portion of $M_2$ sticking out of $Q$.
Using $b_2$ to denote the number of bottlenecks in $M_2$, there is also at
least $b_2(3 l^{\ast} -2)$ of surplus area of $|Q| - |Q'\cup A'| $
in the component of $V\setminus
Q'$ that touches $U_2$.
In the same time, the number of corners in $M_2$
(that are not at the boundary of $U_2$ and thus were not accounted for
before) is at most $6b_2 +4(\min (m_2, n_2)-1-b_2)$ and we have
$$
6b_2 K+4(\min (m_2, n_2)-1-b_2)K\leq [(3 l^{\ast} -2) \min (m_2,
n_2)-\min (m_2, n_2) l^{\ast}+b_2(3 l^{\ast} -2)] h.
$$
To analyze the quadrant $U_1$, we distinguish two possibilities.
First, suppose that $M_1$ contains at least one bottleneck, consider the
quadrant $U'$, with vertex in this bottleneck, that contains the octagon
$Q'$, and use $n_h$ and $n_v$ to denote the number of horizontal and
vertical bonds in $M_1$, respectively.
The number of corners of
$M_1$ is at most $3\min(n_h,n_v)$.
In the same time the surplus of the area $|(Q\setminus(Q'\cup A''))\cap
U'\cap (H\cup V)|$ is at least $\frac{n_h +n_v}{2}(3l^{\ast}-2)$.
To get the sought bound we just notice that
$$
3\min (n_h,n_v)K=\frac32\min (n_h,n_v)(l^{\ast}-1+\eta)h <
\min (n_h,n_v)\frac32 l^{\ast} h\leq \frac{n_h +n_v}{2}(3l^{\ast}-2)h.
$$
If there is no bottleneck of $A''$ in $U_1$,
the portion of $A''$ between two
bottlenecks containing the set $M_1$ is clearly contained
in a halfplane $P$, the oblique boundary $\partial P$ of which passes
through the vertex of of $U_1$ and is orthogonal to the axis of symmetry of
$U_1$. The number of corners of
$M_1$, not accounted for before, is in this case at most
$2\min(n_h,n_v)$.
On the other side, the area $|(Q\setminus(Q'\cup A''))\cap
P^c\cap(H\cup V)|$ is at least
$
\frac{\min(n_h,n_v)}{2}(3l^{\ast}-2)
$
and
$$
2\min(n_h,n_v)K
\leq
\frac{\min(n_h,n_v)}{2}(3l^{\ast}-2)h.
$$
\hfil\rightline{\raise 6pt
\hbox{\qed}}
\enddemo
To verify the property iii) of $\Cal A$,
let $\xi\in\partial\Cal A$ and
$\sigma\in\Cal A$ be such that $\xi=\sigma^{(x)}\notin\Cal A$.
The mapping, assigning to a configuration $\zeta$ its complete sets, is
monotonous in the ordering $\prec$ on configurations and the ordering by
inclusion on complete sets. As a consequence, the value $\sigma(x)$ is
necessarily $-1$; otherwise $\sigma \in\Cal A$ would imply also $\xi \in\Cal A$.
Moreover, the site $x$ lies outside of all complete sets $\bar R_1, \dots ,
\bar R_s$ of $\sigma$. Among the complete sets of $\xi$ there exists a
rectangle $\bar R(D_1,D_2)$ with the following properties:
\roster
\item"i)" $\bar R$ is supercritical, $\min(D_1,D_2)\geq D^{\ast}$,
\item"ii)" it contains the site $x$ and several complete sets $\bar R_i$, say,
$\bar R_1, \dots , \bar R_k$, of $\sigma$; the remaining
complete sets $\bar R_{k+1}, \dots , \bar R_s$ of $\sigma$
are also complete sets
of $\xi$.
\endroster
Our aim now is to prove that
$$
H(\xi)\geq H(Q(D^{\ast},D^{\ast}))+h(D^{\ast}-1)-4K=H(Q(D^{\ast}-1,D^{\ast}))
+2J-4K-h.
\tag4.8
$$
If $\bar R$ is winding around the torus, referring briefly to (4.6),
the
inequality (4.8) is clearly satisfied.
Thus, let us suppose that $\bar R$ is not wrapped around the torus.
Consider first
the configuration $\tilde\xi$, $H(\xi) \geq H(\tilde\xi)$, obtained by
restricting the configuration $\xi$ to the union $\cup_{i=1}^{k}\bar R_i\cup
q(x)$, where $q(x)$ is the unit square centered at the site $x$ (i.e.
considering $ \xi$ to be plus on
$\cup_{i=1}^{k}\bar R_i\cup q(x)$ and taking minuses outside; we will see in a
moment (Eq. (4.9)) that the energy decreases when skipping the subcritical sets
$\bar
R_{k+1}, \dots , \bar R_s$). Further, consider the set $C^{(0)}$ consisting of
the union of the unit square $q(x)$ and those sets among
$\bar R_1, \dots , \bar R_k$ that have a common edge with $q(x)$.
Let us take the rectangular envelope $\tilde R^{(1)}$ of $C^{(0)}$ if
$C^{(0)}$ is not ephemere and $\tilde R^{(1)}=C^{(0)}$ for ephemere
$C^{(0)}$,
and distinguish two cases, either the set $\tilde R^{(1)}$ is
supercritical or not. (Notice that both, $C^{(0)}$ and $\tilde R^{(1)}$
may actually coincide with $q(x)$.)
If $\tilde R^{(1)}$ is supercritical, we decrease the energy of $\tilde\xi$
further by erasing all the remaining sets from $\bar R_1, \dots , \bar R_k$
that were not contributing to the set $C^{(0)}$ and considering the
configuration $\xi^{(0)}$ yielded by the restriction of $\xi$ to the set
$C^{(0)}$. To see this, notice first that for any subcritical $\bar R_i$,
the energy of the restriction $\xi_i$ of $\xi$ to $\bar R_i$,
$\xi_i=\xi\restriction \bar R_i$, can be bounded from below as
$$
\bar H(\xi_i)\geq 2K.
\tag4.9
$$
Indeed, if $\bar R_i$ is subcritical and nonephemere and denoting $\bar
D_2=D_2(\bar R_i)\leq D_1=D_1(\bar R_i)$, we have
$$
\multline
\bar H(\xi_i)\geq 2J(\bar D_1+\bar D_2)-h\bar D_1 \bar D_2 - 2h(l^\ast-1)^2
=\\
= 2J\bar D_1 - h\bar D_1 \bar D_2 + 2J \bar D_2
-2h(l^\ast-1)^2\geq\\ \geq
h(l^\ast-1)\bigl[(D^\ast-1)3-2(l^\ast-1)\bigr]\geq 2K. \endmultline
\tag4.10
$$
If $\bar R_i$ is ephemere, we have
$$
\bar H(\xi_i)\geq 2J(\bar D_1+\bar D_2)-6K \bar D_2 - h\bar D_1(3l^\ast-3)+2K
\geq 2K
\tag4.11
$$
once $K\leq J/3$. Thus, the energy decreases if we erase the configuration in
those $\bar R_i$ that are not contributing to $C^{(0)}$ --- the bound $2K$
on the right hand side of (4.9) is needed in case $\bar R_i$ is touching
$q(x)$ in a corner.
Observing that $\bar R_1, \dots , \bar R_k$ are not interacting,
there can be at most two sets, say $\bar R_1,\bar R_2$,
contributing to $C^{(0)}$ (i.e. intersecting $q(x)$
along its edge).
Let us suppose first that $\bar R_1$ is the only set from
$\bar R_1, \dots ,$
contributing to $C^{(0)}$. Recalling that $\bar
R_1$ is subcritical and $\tilde R^{(1)}$ is supercritical in the same time, we
infer that the set $\bar R_1$ must be nonephemere and, moreover, it is
necessarily the rectangle $(D^{\ast}-1)\times D$ (or $ D\times
(D^{\ast}-1)$) with $D\geq D^{\ast} $ and with the unit square $q(x)$ attached
to its longer side. Consider the standard octagon $\bar Q_1$ inscribed into
$\bar
R_1$. The configuration $\xi^{(0)}$ restricted to $\bar R_1$ is actually the
configuration $\sigma^{(0)}$ obtained as the restriction of $\sigma$ to $\bar
R_1$ with the energy not lower than the energy of $\bar Q_1$
(cf\. (4.6)), $H(\sigma^{(0)})\geq H(\bar Q_1)$.
If $q(x)$ does not touch $C(\sigma^{(0)})$,
then
$$
\bar H(\xi^{(0)})=\bar H(q(x))+\bar H(\sigma^{(0)})
\geq \bar H(q(x))+ \bar H(\bar Q_1)
\tag4.12
$$
and the bound (4.8) follows since $\bar H(q(x))=4J-4K-h>2J-4K-h$
and $H(\bar Q_1)\geq H(Q(D^{\ast}, D^{\ast}-1))$.
If $q(x)$ touches $C(\sigma^{(0)})$, then the monotone envelope
of $\xi^{(0)}$ is a union of a monotone configuration $M$ in $\bar R_1$ with
the unit square $q(x)$ sharing a common edge with $M$.
Hence
$$
H(\xi^{(0)})\geq H(M)+2J-4K-h.
\tag4.13
$$
Using the bound $H(M)\geq H(\bar Q_1)$ according to Lemma 3.2, we get the sought
inequality.
If there are two sets, $\bar R_1$ and $\bar R_2$,
contributing to $C^{(0)}$, then at least one of them must be ephemere
(otherwise $\bar R_1$ and $\bar R_2$ would interact). Suppose first that
$\bar R_1$ is nonephemere and $\bar R_2$ is ephemere.
Considering the halfplane $p$ containing $\bar R_1$,
whose boundary contains the edge separating $q$ from $\bar R_1$,
then, necessarily, $\bar R_2\subset \Bbb R^2\setminus p$.
Moreover, the first row along the
boundary of the halfplane $p$ (the difference $p'\setminus p$,
where $p'$ is the halfplane $p$ shifted by the unit vector orthogonal to
its boundary) does not intersect the interior of $\bar R_2$ (it
contains only the square $q(x)$), the square $q(x)$ is attached to
$\bar R_1$ in a corner, and the second row contains a single square, to
be denoted $\bar q$, of $\bar R_2$ (cf\. Fig. 4.8).
%
%
\midinsert
\centerline{\scaledpicture 2.67in by 2.04in (foureight scaled 800)}
\botcaption{Fig\. 4.8}
\endcaption
\endinsert
%
%
Since $\tilde R^{(1)}$ is supercritical, there exists a
$D\times D^{\ast}$
(or $D^{\ast}\times D$)
rectangle $R^{\ast}$ with $D\geq D^{\ast}$
such that
\roster
\item"i)" $R^{\ast}$ contains $q\cup\bar R_2$,
\item"ii)" $R^{\ast}$ is the circumbscribed rectangle of
$R^{\ast}\cap (\bar R_1\cup q \cup \bar R_2) $,
\item"iii)" $R^{\ast}\cap\bar R_1$ is not ephemere (here, we use the
assumption $10K (2(D^{\ast}-2)+l^{\ast}-1)h \geq 2D^{\ast}h >4J
\tag 4.20
$$
satisfied once $l^{\ast}\geq 3$ (cf\. (3.2)),
we get
$$
H(\xi^{(0)}) - H(Q(D^{\ast}-1,D))\geq (D^{\ast}-2)h-2K
>2J -4K-h.
\tag4.21
$$
To get the sought inequality it
remains to realize that
$H(Q(D^{\ast}-1,D))\geq H(Q(D^{\ast}-1,D^{\ast}))$
since
$(D^{\ast}-1)h <2J$.
Consider now the case when both $\bar R_1$ and $\bar R_2$ are ephemere.
For the strings not to interact, at least one of them, say $\bar R_2$, must be
attached to the rectangle $q$ through a rectangle $\bar q$, in the similar
manner as above.
Using the assumption $10K < J$, one can show that
$$
10(l^{\ast}-1)< D^{\ast}.
\tag 4.22
$$
Observing now that any ephemere string is contained in a strip of thicknes
$3l^{\ast}-3$, we can conclude that, to combine to a supercritical
rectangle $\tilde R^{(1)}$, the strips
containing $\bar R_1$ and $\bar R_2\cup q$ are necessarily orthogonal to each
other. Further, there exists a $D\times D^{\ast}$
(or $D^{\ast}\times D$)
rectangle $R^{\ast}$ with $D\geq D^{\ast}$
such that
$R^{\ast}$ is the circumbscribed rectangle of
$R^{\ast}\cap (\bar R_1\cup q \cup \bar R_2) $.
The surplus area in $R^{\ast}$ with respect to $\bar R_1\cup q \cup \bar R_2$
is at least $(D^{\ast}- 3l^{\ast}+3)^2$, while the number of corners in the
strings $\bar R_1$ and $\bar R_2$ is at most $2\times 6\times (3l^{\ast}-3)$.
Taking into account that $2K < l^{\ast} h $,
$2(D^{\ast}- 3 l^{\ast})\geq
D^{\ast}$, and $-4F(l^{\ast}) \geq 2(l^{\ast}-1)^2 h$,
we get the bound
$$
\multline
H(\xi^{(0)})-H(Q(D^{\ast}-1,D))\geq -2J - 12(3l^{\ast}-3) K +
(D^{\ast}- 3l^{\ast}+3)^2 h -4F(l^{\ast})\geq\\
\geq -2J - 6(3l^{\ast}-3) l^{\ast} h
+ 3D^{\ast} h +(D^{\ast}- 3l^{\ast})^2 h +2(l^{\ast}-3)^2 h.
\endmultline
\tag 4.23
$$
The sought inequality then follows from
$$
- 6(3l^{\ast}-3) l^{\ast} h +
3D^{\ast} h +(D^{\ast}- 3l^{\ast})^2 h +2(l^{\ast}-3)^2 h
> 4J-4K-h
\tag 4.24
$$
since
$$
(D^{\ast}- 3l^{\ast})^2 +2(l^{\ast}-3)^2 + 2(l^{\ast}-1)
> (6 l^{\ast})^2 + (l^{\ast})^2 > 18 (l^{\ast})^2 > 6(3 l^{\ast} - 3) l^{\ast}.
\tag 4.25
$$
Next, consider the case when $\tilde R^{(1)}$ is subcritical. Let us introduce
$\bar R^{(1)}=\tilde R^{(1)}$
if $\tilde R^{(1)}$ is nonephemere. Otherwise we take for
$\bar R^{(1)}$ the string $N(C^{(0)})$ obtained as the blown up envelope of
the configuration corresponding to $C^{(0)}$.
Consider now $\bar R^{(1)}$
and all complete sets among
$\bar R_1, \dots , \bar R_k$ that were not used for $C^{(0)}$ and
construct from them the set of chains $\Cal C_j^{(1)}$
of the first generation.
A sequence $\Cal C_j^{(r)}, r=1,\dots , m$ of chains of following
generations is obtained from it by iteration.
Since the sets $\bar R_1, \dots , \bar R_k$ are mutually
noninteracting, for every generation $r$ we get a chain, say
$\Cal C_1^{(r)}$, consisting of a set $\bar R^{(r)}$ containing
the site $x$ and certain subset of $\bar R_1, \dots , \bar R_k$,
each of them interacting directly with $\bar R^{(r)}$.
The remaining chains $\Cal C_j^{(r)}, j=2,\dots $ contain each just one
complete set from those among $\bar R_1, \dots , \bar R_k$ that have not
appeared
in $\Cal C_1^{(p)}$, $ p\leq r$, in the preceding steps.
Clearly, there is only one chain in the last generation,
$\Cal C_1^{(m)}=\{\bar R^{(m)}\}\equiv \{\bar R \}$.
Let us consider the last set $\bar R^{(p)}$ among $\bar R^{(r)}$,
$r=1,\dots,m$, that is subcritical and take the chain $\Cal C_1^{(p)}$
with the complete sets in $\Cal C_1^{(p)}\setminus \bar R^{(p)}$
ordered in a particular way, say in lexicographic order of their
left upper corner.
Let us unite them, one by one in the given order, with the set $\bar
R^{(p)}$ until the circumscribed rectangle is supercritical.
Cutting off the remaining complete sets from the chain $\Cal C_1^{(p)}$
we get the chain
$\tilde\Cal C_1^{(p)}\subset \Cal C_1^{(p)}$.
Let us use $\tilde R'$ to denote the last complete set that was attached to
form the chain
$\tilde\Cal C_1^{(p)}$ and $\tilde R$ either the blown up envelope or the
circumscribed rectangle of the union of complete sets from $\tilde\Cal
C_1^{(p)}\setminus \{\tilde R'\}$, in dependence on whether it is ephemere or
not. Clearly, $\tilde R$ and $\tilde R'$ are subcritical interacting sets
with a supercritical rectangular envelope of their union.
The energy of the configuration $\tilde \xi$ can be evaluated by the sum of the
energies of its restriction to $\tilde R$ and $\tilde R'$, respectively, as
$$
\bar H(\tilde \xi)\geq \bar H(\tilde \xi\restriction \tilde R)
+\bar H(\tilde \xi\restriction \tilde R') -2K.
\tag 4.27
$$
Indeed, the pluses of the original configuration $\tilde \xi$
are inside of the noninteracting sets $\bar R_1, \dots , \bar R_k$, and $q$, and
thus only when $\tilde R'$ is touching by its corner the square $q$ (included in
$\tilde R$), the additional $2K$ may appear when joining the configurations
$\tilde \xi\restriction \tilde R$ and $\tilde \xi\restriction \tilde R'$.
Consider further, depending on whether $\tilde R$ ($\tilde R'$) is ephemere or
not, the set $\tilde Q$ ($\tilde Q'$) defined as the
set itself or its inscribed standard octagons.
According to (4.6) one has
$$
\bar H(\tilde \xi\restriction \tilde R)
+\bar H(\tilde \xi\restriction \tilde R')\geq
\bar H( \tilde Q)
+\bar H( \tilde Q').
\tag 4.28
$$
Let us observe now that there
exists a $D^{\ast} \times D^{\ast}$ square $R^{\ast}$
such that
\roster
\item"i)" it intersects both sets $\tilde R$ and $\tilde R'$ in
nondegenerated sets $R$ and $R'$, $R=\tilde R\cap R^{\ast}$ and $R'=\tilde
R'\cap R^{\ast}$,
\item"ii)" it is the rectangular envelope of $R\cup R'$,
\item"iii)" it contains the intersection $\tilde R\cap\tilde R'$ (if it is
nonempty), and
\item"iv)" if
$\tilde R$ ($\tilde R'$) is nonephemere, the same is true for $R$ ($R'$).
\endroster
Since both sets $\tilde R$ and $\tilde R'$ are subcritical,
we decrease the energy of $\tilde Q$ ($\tilde Q'$),
to the energy of
$Q$ ($Q'$) defined as $Q\cap R^{\ast}$ ($Q'\cap R^{\ast}$)
for ephemere $\tilde Q$ ($\tilde Q'$) and as the inscribed standard
octagon for nonephemere $\tilde Q$ ($\tilde Q'$).
Hence
$$
\bar H(\tilde\xi)\geq \bar H(Q) +\bar H(Q') -2K.
\tag 4.29
$$
Thus our task is to evaluate the sum of energies of two interacting
sets $Q$ and $Q'$
(nonephemere standard octagons or ephemere string) whose union has
$R^{\ast}$ as the rectangular envelope.
Consider first the case when both $Q$ and $Q'$ are nonephemere.
Replacing $Q$ and $Q'$ by the standard octagon $Q^{\ast}$ inscribed
in $R^{\ast}$, the sum on the right hand side of (4.29) decreases by at least
$h(D^{\ast}-1)$, yielding thus (4.8).
Indeed, if the rectangles $R$ and $R'$ just touch
in the corner, the boundary has the same number of
bonds as in $Q^{\ast}$, while the surplus area of $R^{\ast}$,
as compared with that of $R\cup R'$, is at least
$D^{\ast}+2(3l^{\ast} -4)^2$.
The sought bound follows, once we realize that
$$
4F(l^{\ast})\geq -2(l^{\ast} -1)^2 h -2\eta (2 l^*-1)\geq -2(3l^{\ast}
-4)^2h.
\tag 4.30
$$
If the rectangles $R$ and $R'$ are intersecting, the surplus area shrinks.
However, each line of at most $D^{\ast}$ surplus sites lost in this way is
compensated by surplus $2$ bonds in the joint boundary of $Q$ and $Q'$,
as compared with the boundary of $Q^{\ast}$.
If, on the other side, the distance of the rectangles $R$ and $R'$, say in the
vertical direction, is $1$, there must exist at least two surplus horizontal
bonds for them to interact, compensating thus the lack of two vertical bonds.
It remains to consider the case with, say, nonephemere $R$
and ephemere $R'$. If $R$ and $R'$ just touch in the corner, the
reasoning is the same as when discussing the case of supercritical
$\tilde R^{(1)}$ above (cf. the bound (4.21)).
The cases of intersecting $R$ and $R'$ or of $R$ and $R'$ whose
distance is 1 are then treated with the same modifications as when
both $R$ and $R'$ are nonephemere.
It is easy to observe that, except the case when a single $\bar
R_1$ is contributing to $C^{(0)}$, the lower bounds are always
sharp --- at least once during the process of getting a lower
bound we use a sharp bound. See e.g. the lower bounds (4.21),
(4.24), (4.25). As a result, we can conclude that the only
configurations from $\partial\Cal A$ on which the bound can be
achieved are those from $\Cal P$, the remaining ones having strictly
higher energy by at least the minimal amount $h$. \hfil\qed
\vfill\newpage
\NoBlackBoxes
\magnification=1200 \font\arm=cmr10 at
10truept \font\ait=cmti10 at
10truept
\TagsOnRight \CenteredTagsOnSplits
\head 5. Proofs of Theorems \endhead
Similarly as when proving the statement
of Proposition 1 from the bound (3.77),
we will get Theorems 1 and 2 from
$$
\lim{_{\beta\rightarrow\infty}} P_{-
\underline 1}\bigl(\tau_{\partial \Cal A}\geq
T(\tilde\varepsilon)\bigr)
= 0
\tag{5.1}
$$
that will be shown to be valid for
all $\tilde\varepsilon > 0$
with $T(\tilde\epsilon)=e^{\beta(E^*+\tilde\epsilon)}
$ (cf\. (2.24), (2.28)).
To prove (5.1), we follow \cite{KO 1} and define, in
a similar manner as in the proof
of Proposition 3.1, an event $\Cal
E_\sigma$ starting from
an arbitrary $\sigma$ in $\Cal A$, taking
place over an interval of time
$T_1= \exp\{\beta(E(l^*-1)+\delta)\}$
(cf\. (3.71)) and such that:
\item{(1)}
If $\Cal E_\sigma$ takes place then
necessarily the set $\partial\Cal A$
is reached (in a particular
manner) before the time $T_1$.
\item{(2)}
For the probability $P(\Cal E_\sigma)$ the
uniform lower bound
$$
\inf_{\sigma \in \Cal A} P(\Cal E_{\sigma})\ge \alpha
\tag{5.2}
$$
holds with $\alpha$ such that
$$
\lim{_{\beta\rightarrow\infty}}(1-\alpha)^{T_2\over T_1} = 0,
\tag{5.3}
$$
where $T_2 = T(\tilde\epsilon)$.
Using the equations (5.2) and (5.3),
it follows by the strong Markov
property that an attempt to reach
$\partial\Cal A$ not later then
$T_2$ will be successful with high
probability for large $\beta$.
Indeed, it is enough to observe that
after splitting $T_2$ into $T_2/T_1$
intervals of length $T_1$,
staying inside of $\Cal A$ for
any of these intervals
means that the event
$\Cal E_\sigma$ did not take place.
Once we have (5.1),
we use the reversibility (Lemma 3.4) and refer to the
property iii) of the set $\Cal A$, to get an upper
bound on the probability of reaching $\partial\Cal A$
in a configuration outside $\Cal P$.
Noticing that, after starting from $\Cal P$, there is
a finite probability to go to $+\underline 1$
before returning to $\Cal A$ (cf\. Proposition 4),
we get Theorems 1 and 2.
Thus, our aim is to construct an event
$\Cal E_\sigma$ such that (5.2) and (5.3) hold true.
First, we present the idea for
the construction of $\Cal E_\sigma$;
formal definitions will follow.
We begin by recalling that $\Cal A$
is defined in such a way that
for all $\sigma \in \Cal A$ one descents
(energy decreases)
to a set of non-interacting
subcritical
octagons in time of order $T_0$ (see the
equation (3.100)). Then, with
high probability, in a time shorter
than $T_1=\exp \{\beta
E(L^{\ast} - 1)\}$ we reach
the configuration $-\underline 1$.
The first part of $\Cal E_\sigma$
refers to this
shrinking phenomenon.
This stage of $\Cal E_\sigma$ is called
{\it contraction} and is denoted by $\Cal E^{(c)}$.
The subsequent stage consists just
in staying in $-\underline 1$ for a
time of order $\exp \{\beta E(L^{\ast} - 1)\}$;
it is called {\it
waiting} and is denoted by $\Cal E^{(w)}$.
This (random) time spent in
$-\underline 1$ before the growth of
a droplet up to the critical
nucleus can be considered as a
``global resistence time'' as it will
be clear later; its introduction will
lead to a gain of a factor $\exp
\{\beta E(L^{\ast} - 1)\}$, due to the
``temporal entropy'', in our
lower bound for the probability
$P(\Cal E_\sigma)$. (Remember that
$E(L^{\ast} - 1) < 2 \tilde J = 2J - 4K$
and the energy for creating
a plus spin in a sea of minuses is
$4\tilde J + 4K = 4J - 4K$; hence,
the time for the creation of
a plus spin is, with high probabiltity,
much longer than $\exp \{\beta E(L^* - 1)\}$).
%
%
\midinsert
\centerline{\scaledpicture 4.17in by 1.97in (fiveone scaled 800)}
\botcaption{Fig\. 5.1}
\endcaption
\endinsert
%
%
Next, during the third stage,
that we call {\it embryonal} and denote
by $\Cal E^{(e)}$, we create, sequentially,
the first stable regular octagon%
\footnote{Recall that
we identify an octagon $Q$ with
the spin configuration where the pluses
are precisely the spins inside $Q$.}
(of edge 2) $Q(l=2)$ in 12
elementary (single spin flip)
steps with increase of energy
during the first 11, and with a loss of
energy during the 12th one.
At each step the configuration will
consist of a unique closed contour
$\bar\gamma_i$, $i=1,\dots,12$ as
indicated in Fig.~5.1. Notice that
in $\Cal E_\sigma$, from
$\bar\gamma_1$ up to the protocritical
droplet, all the octagons will
be centered --- their upper left corner
belonging to the dual lattice
is located in the point $({-{1\over 2}},
{1\over 2})$. The dot in
Fig.~5.1 represents the origin.
%
%
\midinsert
\centerline{\scaledpicture 3.24in by 2.54in
(fivetwo scaled 800)}
\botcaption{Fig\. 5.2}
\endcaption
\endinsert
%
%
Subsequently, there is a stage called
{\it regular}, $\Cal E^{(r)}$,
during which one passes through
regular octagons $Q(l)$.
The passage from an octagon $Q(l)$
to the following one, $Q(l+1)$,
is by a sequence
of canonical growth
(reversed to sequence of canonical
cuts (cf\. Lemma 3.5)).
Fig\. 5.2 shows one particular sequence
of octagons on the path from $Q(l)$
to $Q(l+1)$ (here $l=3$).
Namely, the octagonal droplet
$ \gamma_l^{(i)}$,
$i=1,\dots, 14$, is obtained
from the preceeding one, $\gamma_l^{(i-
1)}$, by adding the $i$-th edge
as indicated. Notice that
$\gamma_l^{(14)}\equiv
\gamma _{l+1}^{(0)}\in Q(l+1)$. Further, use
$S(l)$ to denote the saddle obtained
by erasing from $Q(l+1)$ the last
(in lexicographic order) $l$ contiguous
unit squares adjacent from the
interior to its upper right oblique edge.
More generally we denote by
$S_l^{(i)}$ the sadle configuration obtained by adding to
$\gamma_l^{(i-1)}$ the first unit square of the $i$-th edge.
%
%As we
%will see in more detail later in the
%formal definitions, the event
%$\Cal E^{(r)}$ will consist of a sequence
% of elementary steps that
%can be represented as in Fig. 5.3 with
%$\eta_1=\gamma_l^{(i-1)}$, $
%\zeta_1=S_l^{(i)}$, $ \eta_2 = \gamma_l^{(i)}$ and so on.
%
The path, visiting 14 octagons
indicated in Fig.~5.2, obtained in
this way is {\it almost monotonous} path ---
it consists of a series of elementary
transitions of the
following form:
\roster
\item"$\bullet$"
first, starting from an octagon $Q^{(1)}$, say
$Q^{(1)}=\gamma_l^{(i-1)}$, a monotonous ascent to \hglue 5mm
$S^{(1)}$ (a saddle between $Q^{(1)}$ and $Q^{(2)}$);
\item"$\bullet$"
then a descent to a
configuration $Q^{(2)}$ (again an octagon), higher then
$Q^{(1)}$: \hglue 5mm $ H(Q^{(2)})>H(Q^{(1)})$;
\item"$\bullet$"
another ascent to
$S^{(2)}$ with $H(S^{(2)})>H(S^{(1)})$ and so on (see
Fig.~5.3).
\endroster
%
%
\midinsert
\centerline{\scaledpicture 3.04in by
1.83in (fivethree scaled 900)}
\botcaption{Fig\. 5.3}
\endcaption
\endinsert
%
%
\comment
To be more specific let us introduce for
$l$ such that $$ 2\leq l \leq
l^{\ast} -1 $$ (see the equation (2.14))
an $l${\it -canonical sequence}
of 14 octagons starting from a regular
octagon with edge $l$ (the zeroth
one) and ending with the 14th one which
is again regular with edge
$l+1$. They have all the same fixed left
upper corner in the point
$(-{1\over 2}, {1\over 2})$ belonging to
the dual lattice. We denote
by $\gamma ^0_l$ the element of $Q(l)$\
($\equiv$ the set of all th e
spin configuration where the plus spins
are precisely those inside a
{\it regular octagon} of size $l$ (see
definition after the equation
(2.17)) with upper left corner in $(-{1\over 2}, {1\over 2})$.
\endcomment
To get a lower bound on probability of
$\Cal E^{(r)}$, we suppose that a path in
$\Cal E^{(r)}$
stays in the basin of attraction
$\Cal B(\gamma_l^{(i-
1)})$ (see (3.7)) up to a random
time shorter then $\exp \{\beta [
H(S_l^{(i-1)})- H(\gamma_l^{(i-1)})
+\varepsilon]\}$ with $\varepsilon
> 0$ chosen sufficiently small;
then it ascents to $S_l^{(i)}$ and
afterwards descents to $\gamma _l^{(i)}$;
in the next step it stays in
$\Cal B(\gamma_l^{(i)})$ up to
a time of order at most $\exp \{\beta[
H(S_l^{(i)})- H(\gamma_l^{(i)}) +\varepsilon]\}$
then it ascents to
$S_l^{(i)}$ and so on.
The above times are called {\it resistance times}.
Their
introduction in definition of
$\Cal E^{(r)} $ (and in the subsequent
stages) is motivated by
the necessity of exploiting the ``temporal
entropy'' to get a correct lower bound.
It turns out that, by
the above choice of the resistance times,
one gets exactly the needed
factors. Notice that since the minimum of
$H$ on $\partial \Cal B
(\gamma_l^{(i)})$ is
reached in $S_l^{(i)}$, the probability that,
during the time interval of the order $\exp\{
\beta[H(S_l^{(i)})-H(\gamma_l^{(i)})]\}$,
the process does not leave
$\Cal B(\gamma_l^{(i)})$ is almost one
(here we are using the
reversibility of the process
(see \cite{S 1} and Lemma 3.4 above)).
On the other side, by
the particular choice of a sequence of elementary
transitions in the definition of
$\Cal E^{(r)}$, the process enters
the basin of attraction
$\Cal B(\gamma_l^{(i)})$ through the saddle
point $S_l^{(i)}$. The combination of these two facts
is crucial to get a lower bound
that is sufficient for the event $\Cal
E$ to satisfy the condition (5.3).
Aa already mentioned in the Introduction,
a ``local'' criterium can be
formulated that allows to choose a
sequence of elementary transitions
to yield a correct probability
estimate for the event $\Cal E^{(r)}$:
the passage to a successive
minimum of $H$ has to occur
through a saddle whose height is exactly
the one of the minimum of $H$ on
the boundary of the basin of
attraction of the successive minimum.
We stay in the basin of
attraction of this new minimum for
a ``typical'' resistance time and
then pass to the next one.
As we will see later,
such a criterion can be generalized to
non-almost-monotonous sequences of steps
appearing in the subsequent
stages of $\Cal E$, provided
we substitute the basin of attraction
$\Cal B(Q)$ of a single octagon $Q$
by its generalization $\Cal
D(D_1, D_2)$ (see (3.10)).
The ``resistance'' inside
$\Cal D(D_1, D_2)$ will be against a mechanism
of escape that is not monotonous
(in energy) any more. Actually, it
is exactly the escape described
in the shrinking event $ \Cal
E^s_{\sigma}$ introduced in
the proof of Proposition 1 (see the
definition (3.95)). In addition,
the descent from a saddle (in
$\partial \Cal D(D_1,D_2)$) to
the corresponding minimum will
involve some tunneling phenomena
(passing through some local saddle
points).
The stage $\Cal E^{(r)}$ ends once
we reach the octagon $Q(l^*)$.
After it, there is a stage that
we call {\it transient} and denote
by $\Cal E^{(t)}$ during which,
following a very special (not almost
monotone) sequence of octagons,
we reach a standard octagon with $L_1
= L_2 = l^{\ast} +2$. Namely,
the first one with $\min (L_1, L_2) >
l^{\ast} + 1$. To describe
the ``transient'', which turns out to be a
somehow complicated mechanism,
we need to define several particular
octagons. To this end we
will make a repeated use of drawings.
The droplet
$Q^{(0)}_{l^{\ast},l^{\ast} }$
is the regular
(and simultaneously the
smallest standard) centered octagon
with $l = l^{\ast}$.
The configurations $\bar
Q^{(i)}_{l^{\ast},l^{\ast} }$,
$i=1, 2, 3$ are the
octagons obtained
by adding, sequentially, to
$Q^{(0)}_{l^{\ast},l^{\ast} }$ the
edges $1$, $2$, and $ 3$, as indicated in
Fig.~5.4. We notice here that
$\bar Q^{(1)}_{l^{\ast},l^{\ast} }$,
$\bar Q^{(2)}_{l^{\ast},l^{\ast}}$,
respectively, correspond to $Q^*_2$,
$\widetilde Q _1 $ introduced in the
proof of Proposition~2.
%
%
\midinsert
\centerline{\scaledpicture 3.00in
by 2.56in (fivefour scaled
700)}
\botcaption{Fig\. 5.4}
\endcaption
\endinsert
%
%
The configurations
$\bar S^{(i)}_{l^{\ast},l^{\ast} }$,
$i=1, 2, 3$ are the saddles
obtained by adding to
$Q^{(i-1)}_{l^{\ast},l^{\ast} }$ the first unit
square of the $i$-th edge.
Further, let $Q^{(0)}_{{l^{\ast}+1},
l^{\ast}}
\equiv \bar Q^{(3)}_{l^{\ast},l^{\ast} }$ (= standard
octagon with $L_1 = l^{\ast} + 1$,
$L_2=l^\ast$) and let
$Q^{(i)}_{{l^{\ast}+1}, l^{\ast}}$,
$i=1, 2, 3$ be the octagons
obtained by adding, sequentially,
to $Q^{(i-1)}_{{l^{\ast}+1},
l^{\ast}}$ the edges $1, 2$ and
$ 3$ as indicated in Fig. 5.5.
Again, the configurations
$S^{(i)}_{{l^{\ast}+1}, l^{\ast}}$ are the
saddles obtained by adding to
$Q^{(i-1)}_{l^{\ast}+1, l^{\ast}}$ the
first unit square of the
$i$-th edge. Similarly, let
$Q^{(0)}_{l^{\ast}+1,
l^{\ast}+1}\equiv Q^{(3)}_{l^{\ast}+1, l^{\ast}}$
(= standard octagon with
$L_1=L_2=l^{\ast}+1$) and
$Q^{(i)}_{l^{\ast}+1, l^{\ast}+1}$,
$i=1,2,3$, be the octagons
obtained by adding, sequentially,
to $Q^{(i- 1)}_{l^{\ast}+1,
l^{\ast}+1}$ the edges
$1, 2$ and $ 3$ (cf\. Fig. 5.5). Again, the
configurations
$S^{(i)}_{l^{\ast}+1, l^{\ast}+1}$ are the saddles
defined in the same way as $S^{(i)}_{l^{\ast}+1, l^{\ast}}$.
%
%
\midinsert
\centerline{\scaledpicture 6.54in by 3.68in (fivefive
scaled 700)}
\botcaption{Fig\. 5.5}
\endcaption
\endinsert
%
%
In the
first part of the event
$\Cal E^{(t)}$ our process will stay in $\Cal
D(D_1=D_2=l^{\ast})$ ---
the generalized basin of attraction of the
standard (and regular) octagon
$Q(D_1=D_2=l^{\ast})$ that has been
defined in the equation (3.10) ---
for a time of order $ \exp \{\beta
h(l^{\ast}-1)\}$. Then, after visiting for the last time
$Q^{(0)}_{l^{\ast},l^{\ast} }$,
it will pass to $\bar
Q^{(1)}_{l^{\ast},l^{\ast} }$
through $\bar
S^{(1)}_{l^{\ast},l^{\ast} }$,
staying subsequently in $\Cal
B(\bar Q^{(1)}_{l^{\ast},l^{\ast}})$
(the true basin of
attraction of
$\bar Q^{(1)}_{l^{\ast},l^{\ast} }$) for a time of
order $\exp \{\beta h(l^{\ast}-2)\}$.
Then it jumps to $\bar
Q^{(2)}_{l^{\ast},l^{\ast} } $,
passing through $\bar
S^{(2)}_{l^{\ast},l^{\ast} }$;
it stays in $\Cal B(\bar
Q^{(2)}_{l^{\ast},l^{\ast} })$
for a time of order $\exp \{\beta
h(l^{\ast}-2)\}$ and then
it passes again to
$Q^{(3)}_{l^{\ast},l^{\ast} }$
through $\bar
S^{(3)}_{l^{\ast},l^{\ast} }$.
(Notice that when we say ``the
process stays in certain set of
configurations for a time of order
$\exp\{\beta\Delta\}$'', we actually
mean ``it stays there for a
random time shorter than
$\exp \{\beta(\Delta+\varepsilon)\}$ with a
suitable, sufficiently small,
positive $\varepsilon$'').
This first part of $\Cal E^{(t)}$
can be better understood by observing
the landscape of the energy depicted in Fig.~5.6.
%
%
\midinsert
\centerline{\scaledpicture 4.75in by 3.39in
(fivesix scaled 700)}
\botcaption{Fig\. 5.6}
\endcaption
\endinsert
%
%
The second and third parts of $\Cal E^{(t)}$,
consisting of the
transitions from the octagons
$Q^{(0)}_{l^{\ast}+1, l^{\ast}}$ to
$Q^{(0)}_{l^{\ast}+1, l^{\ast}+1}$
and from $Q^{(0)}_{l^{\ast}+1, l^{\ast}+1}$
to $Q^{(0)}_{l^{\ast}+2,
l^{\ast}+1}$, respectively, are very similar.
Namely, we stay in
$\Cal B(Q^{(0)}_{l^{\ast}+1, l^{\ast}})$
for a time
of order $\exp\{\beta h(l^{\ast}-1)\}$
then pass to
$Q^{(1)}_{l^{\ast}+1, l^{\ast}}$ through
$S^{(1)}_{l^{\ast}+1, l^{\ast}}$
and stay in $\Cal
B(Q^{(1)}_{l^{\ast}+1, l^{\ast}})$
for a time $\exp \{\beta h ( l^*
-2)\}$; after that we go to
$Q^{(2)}_{l^{\ast}+1, l^{\ast}}$ passing
through $ S^{(2)}_{l^{\ast}+1, l^{\ast}}$
and after staying in $\Cal
B(Q^{(2)}_{l^{\ast}+1, l^{\ast}})$
for a time $\exp \{\beta (2K -
h)\}$ we pass through
$S^{(3)}_{l^{\ast}+1, l^{\ast}} $ to
$Q^{(3)}_{l^{\ast}+1, l^{\ast}+1}$.
After reaching the octagon
$Q^{(0)}_{l^{\ast}+2, l^{\ast}+1}$, the
last part of $\Cal E^{(t)} $ starts;
it consists of a transition to
$Q^{(0)}_{l^{\ast}+2, l^{\ast}+2}$
via a mechanism that will be
repeated several times during
the subsequent stage of $\Cal E$ that we
call {\it standard} and denote by $\Cal E^{(s)}$.
In other words,
the last part of $\Cal E^{(t)}$
can be considered also as the first
part of $\Cal E^{(s)}$.\par
%
%
\midinsert
\centerline{\scaledpicture 5.54in by 3.10in
(fiveseven scaled 700)}
\botcaption{Fig\. 5.7}
\endcaption
\endinsert
%
%
During $\Cal E^{(s)}$ our process
will visit a sequence of growing
standard octagons inscribed
in squares or almost squares of the form:
\flushpar $L_1, L_2\equiv L,
L+1\rightarrow L+1, L\rightarrow L+1,
L+1\rightarrow L+2, L+1,\dots$
from $L_1=l^{\ast}+2$, $L_2=l^{\ast}+1$
up to $L_1=L^{\ast}$, $ L_2=L^{\ast}- 1$
(cf\. (2.17)).
These transitions are performed
via a canonical (not almost monotonous)
sequence of (not standard) octagons.
Consider a standard octagon with
$L_1=L_2=L$, $ l^{\ast}+2\leq L\leq L^{\ast}-1$
and with upper left
corner in $(-{1\over 2}, {1\over 2})$;
call it $Q^0_{L, L}$. Let
$Q^{(i)}_{L, L}$, $ i=1, 2, 3$
be the octagons obtained by adding,
sequentially, to $Q^{(i-1)}_{L, L}$
the edges $1, 2 $ and $3$ as
indicated in Fig\. 5.7.
Similarly we call $Q^{(0)}_{L+1, L}\equiv
Q^{(3)}_{L,L}$ the standard octagon
with $L_1=L+1, L_2=L,\quad
l^{\ast}+1\leq L\leq L^{\ast}-1,$
and with left upper corner in
$(-{1\over 2}, {1\over 2})$;
$Q^{(i)}_{L+1, L}$, $i=1, 2, 3,$ are the
octagons obtained by adding,
sequentially to $Q^{(i-1)}_{L+1, L}$
the edges $1 , 2, 3$ as indicated in Fig\.~5.7.
The configurations $\bar S^{(i)}_{L_1,
L_2}$, $i=1, 2, 3,$ are the saddles
obtained from $Q^{(i-1)}_{L,
L_2}$ by adding the first
unit square to the $i$-th edge.
%
%
\midinsert
\centerline{\scaledpicture 6.65in
by 2.39in (fiveeight scaled 800)}
\botcaption{Fig\. 5.8}
\endcaption
\endinsert
%
%
The canonical mechanism to pass
from a standard octagon, say
$Q^{(0)}_{L,L}$ to the
following one in the sequence,
$Q^{(0)}_{L+1,L}$, is as follows.
We stay in $\Cal
D(D_1, D_2)$, $D_1=L+2(l^{\ast}-1)$,
$D_2= L+2(l^{\ast}-1)$, for the
time $\exp\{\beta E(L)\}$
(see the equations (3.71) and (3.72)) then
we jump to $S^{(1)}_{L,L}$
and then in a time $\sim \exp
\{(2K-h)\beta\}$ we pass to
the next standard octagon $Q^{(0)}_{L+1,
L}$. This transition from
$S^{(1)}_{L, L}$ to $Q^{(0)}_{L+1, L}$ is
not a purely downhill path
but involves two tunnelings. The case of
the transition
$Q^{(0)}_{L+1, L}\rightarrow Q^{(0)}_{L+1, L+1}$ is
completely analogous:
in general the resistance time in $\Cal
D(D_1,D_2)$ with $D_1=L_1+2(l^{\ast}-1)$
and $ D_2=L_2 +
2(l^{\ast}-1)$ is
$\exp \{\beta E(L_1\wedge L_2)\}$. A transition
from $Q_{L,L}$ to $Q_{L+1, L}$ for
$L\geq l^{\ast}+2$ is represented
in Fig.~5.8.
%
%
\midinsert \centerline{\scaledpicture
6.92in by 4.06in (fivenine
scaled 800)}
\botcaption{Fig\. 5.9}
\endcaption
\endinsert
%
%
A similar picture describes the
transition $Q^{(0)}_{L+1, L}\rightarrow
Q^{(0)}_{L+1, L+1}$. In Fig.~5.9
we represent the transition from
$Q^{(0)}_{l^{\ast}+2, l^{\ast}+1}$
to $Q^{(0)}_{l^{\ast}+2,
l^{\ast}+2}$, namely the
last part of the transient event.
We summarize the description of
the transient event $\Cal E^{(t)}$ in
Figs.~5.10, 5.11.
%
%
\midinsert
\centerline{\scaledpicture 3.35in by 3.25in
(fiveten scaled 700)}
\botcaption{Fig\. 5.10}
\endcaption \endinsert
%
%
%
%
\midinsert
\centerline{\scaledpicture 6.38in by 13.0in
(fiveeleven scaled 600)}
\botcaption{Fig\. 5.11}
\endcaption
\endinsert
%
%
Before passing to a formal
description of the event $\Cal E$, we
would like to make several additional
observations concerning the events
$\Cal E^{(t)}$ and $\Cal E^{(s)} $.
1) The transition
$\gamma^{(13)}_{l^{\ast}-1} \rightarrow
S(l^{\ast})\rightarrow
Q^{(0)}_{l^{\ast},l^{\ast} } \rightarrow
\bar S^{(1)}_{l^{\ast},l^{\ast} }
\rightarrow\dots$ is not almost
monotonous.
2) The sequence $\bar
Q^{(1)}_{l^{\ast},l^{\ast} }$,
$\bar Q^{(2)}_{l^{\ast},l^{\ast}
}$, $ \bar Q^{(3)}_{l^{\ast},l^{\ast} }$
is not analogue of $
Q^{(1)}_{l^{\ast}+1,l^{\ast}}$,
$ Q^{(2)}_{l^{\ast}+1,l^{\ast}}$ , $
Q^{(3)}_{l^{\ast}+1,l^{\ast}}$ or,
more generally of $Q^{(1)}_{L_1,
L_2}$, $ Q^{(2)}_{L_1, L_2}$,
$ Q^{(3)}_{L_1, L_2}$ for the
subsequent values of $L_1, L_2$.
This part of the event $\Cal
E^{(t)}$ provides a good example
to clarify the local criterion we
referred to before. In the sequence
$Q^{(0)}_{l^{\ast},l^{\ast}
}$, $ \bar S^{(1)}_{l^{\ast},l^{\ast} } $, $ \bar
Q^{(1)}_{l^{\ast},l^{\ast} } $,
and $ \bar
S^{(2)}_{l^{\ast},l^{\ast} }$,
we always respect the condition of
passing through a saddle $S$ such
that $H(S)$ is the minimum in the
boundary of the (possibly generalized)
subsequent basin of
attraction.
To clarify better this point
let us suppose that we defined the
first part of $\Cal E^{(t)}$
concerning the transition
$Q^{(0)}_{l^{\ast},l^{\ast} }\rightarrow
Q^{(0)}_{l^{\ast}+1,l^{\ast}}$
in a different way. Namely, instead of
$\bar Q^{(i)}_{l^{\ast},l^{\ast} }$
we take the sequence
of octagons $Q^{(i)}_{l^{\ast},l^{\ast}}$,
$i=1,2$ and $3$ defined,
in the usual way, according to
Fig\.~5.12. Notice that
$\bar Q^{(1)}_{l^{\ast},l^{\ast} }\not =
Q^{(1)}_{l^{\ast},l^{\ast} }$,
whereas $\bar
Q^{(2)}_{l^{\ast},l^{\ast} }
= Q^{(2)}_{l^{\ast},l^{\ast} }
$, $ \bar Q^{(3)}_{l^{\ast},l^{\ast} }=
Q^{(3)}_{l^{\ast},l^{\ast}
}\equiv Q^{(0)}_{l^{\ast}+1,l^{\ast}}$,
and $ Q^{(i)}_{l^{\ast},l^{\ast} }$
would be the analogue for
$L_1=L_2=l^{\ast}$ of our
$Q^{(i)}_{L_1,L_2}$ as they are defined
for the subsequent $L_1,L_2$'s.
%
%
\midinsert
\centerline{\scaledpicture 2.79in
by 2.54in (fivetwelf scaled 800)}
\botcaption{Fig\. 5.12} \endcaption
\endinsert
%
%
If we define
$S^{(i)}_{l^{\ast},l^{\ast} }$ as
the saddle obtained by adding to
$Q^{(i-1)}_{l^{\ast},l^{\ast} }$
the first unit square in the
$i$-th edge, we can represent
the landscape of energy for this
different transition as indicated in Fig\.~5.12.
%
%
\midinsert
\centerline{\scaledpicture 3.83in by 2.93in (fivethirteen
scaled 800)}
\botcaption{Fig\. 5.13}
\endcaption
\endinsert
%
%
We observe that in the mechanism
described in Fig\.~5.13 what
gets wrong with respect to our criterion
is the transition
$Q^{(1)}_{l^{\ast},l^{\ast} }\rightarrow
S^{(2)}_{l^{\ast},l^{\ast} }
\rightarrow Q^{(2)}_{l^{\ast},l^{\ast}
}$, since the minimum of $H$ in $\partial \Cal
D(Q^{(2)}_{l^{\ast},l^{\ast} })$ is reached in
$\bar S^{(2)}_{l^{\ast},l^{\ast} }$ and not in $
S^{(2)}_{l^{\ast},l^{\ast}
}$.
3) In the sequence
$Q^{(0)}_{l^{\ast}+1,l^{\ast}}\rightarrow
Q^{(0)}_{l^{\ast}+1,l^{\ast}+1} \rightarrow
Q^{(0)}_{l^{\ast}+2,l^{\ast}+2}$
we could have followed another path
without violating our local criterion.
This is a consequence of a
degeneracy of the minimum of $H$ in
$\Cal D(D_1=3l^{\ast} - 1,
D_2=3l^{\ast} - 1)$ and in
$\Cal D(D_1=3l^{\ast} , D_2=3l^{\ast} - 1)$;
as we already noticed in the
proof of Proposition 2,
this minimum is
reached for $L_1\wedge L_2=l^{\ast}+1$,
both in $\widetilde S_2$ and in
$\hat S_2$ where $\widetilde S_2$ and
$ \hat S_2$ are saddles defined
during the proof of Proposition~1
and represented in Fig.~3.8, 3.9,
respectively. To be more precise,
consider the sequence
$\bar Q^{(1)}_{l^{\ast}+1,l^{\ast}}$,
$ \bar
Q^{(2)}_{l^{\ast}+1,l^{\ast}}\equiv
Q^{(2)}_{l^{\ast}+1,l^{\ast}}$, $
\bar Q^{(3)}_{l^{\ast}+1,l^{\ast}} =
Q^{(3)}_{l^{\ast}+1,l^{\ast}}$, $
\bar S^{(1)}_{l^{\ast}+1,l^{\ast}}$,
$ \bar
S^{(2)}_{l^{\ast}+1,l^{\ast}}$,
$ \bar S^{(3)}_{l^{\ast}+1,l^{\ast}}$
similarly to $\bar Q^{(1)}_{l^{\ast},
l^{\ast}}$, $ \bar
Q^{(2)}_{l^{\ast},l^{\ast}}$,
$ \bar Q^{(3)}_{l^{\ast},l^{\ast}}$,
$ \bar
S^{(1)}_{l^{\ast},l^{\ast}}$,
$ \bar S^{(2)}_{l^{\ast},l^{\ast}}$,
and $ \bar
S^{(3)}_{l^{\ast},l^{\ast}}$
(see Fig\. 5.4). Then, for $D_1 =
3l^{\ast}- 1=D_2$, the configuration
$ \bar
S^{(2)}_{l^{\ast},l^{\ast}}$
coincides with $\hat S _2$ (see Fig\.
3.9) and
$$
\min_{\sigma\in\partial B (D_1, D_2)}
H(\sigma)= H(\bar
S^{(2)}_{l^{\ast}+1,l^{\ast}})=
H(\hat S^{(2)}_{l^{\ast}+1,l^{\ast}}).
\tag{5.4}
$$
A similar statement is true for $D_1=3l^{\ast},
D_2=3l^{\ast}-1$.
4) Our mechanism of growth is,
in all the stages,
exactly the reverse of
the best mechanism of shrinking described in
the proof of Propositions 1, 2, 3.
To clarify the relation between
the notation we used in
Propositions 1, 2, and 3, to describe the
shrinking phenomenon
(following the drift) and the present
construction for describing
the growth up to the critical size
(against the drift), consider,
for example, a standard octagon
$\widetilde Q_0 $ with
$L_2=L+1$, $ L_1=L$. We recall that $\widetilde
Q_3$ is again a standard octagon with $L_1=L_2=L$. We have
$$
Q^{(3)}_{L,L-1} \equiv Q^{(0)}_{L,L}= \widetilde Q_3
$$
$$
Q^{(1)}_{L,L} = \widetilde Q_2,\quad Q^{(2)}_{L,L}
= \widetilde Q_1
,\quad Q^{(3)}_{L,L} = \widetilde Q_{0}.
$$
5) Looking at
Fig.~5.10 one easily realizes that
the relative heights in energy of
\break $S^{(1)}_{L_1, L_2}$,
$ S^{(2)}_{L_1, L_2}$, and $
S^{(3)}_{L_1, L_2}$, in
the transitions from $L_1, L_2\rightarrow
L'_1, L'_2$ in $\Cal E^{(t)}$,
namely, in the transitions
$l^{\ast},l^{\ast}
\rightarrow l^{\ast}+1, l^{\ast}\rightarrow
l^{\ast}+1, l^{\ast}+1\rightarrow l^{\ast}+2,
l^{\ast}+1\rightarrow
l^{\ast}+2, l^{\ast}+2$,
change according to the value of $L'_1\wedge
L'_2$.
For $L'_1\wedge L'_2=l^{\ast}$,
the saddle $S^{(3)} $ is the highest
one.
For $L'_1\wedge L'_2=l^{\ast}+1$,
the saddle $S^{(2)}$ is the
highest one.
For $L'_1\wedge L'_2=l^{\ast}+2 $
(or larger) the saddle $S^{(1)}$ is
the highest one.
In any case the differences
in height between the saddles (that
change sign passing from
$L'_1\wedge L'_2=l^{\ast}$ to $L'_1\wedge
L'_2=l^{\ast}+2$) are, in absolute value,
of order $\eta h$ with $\eta <
1$ defined in the equation (2.18).
6) The introduction of the
generalized basin $\Cal D(D_1, D_2)$
in place of the usual $\Cal B(Q)$
(that was needed in $\Cal E^{(t)}$
and $\Cal E^{(s)}$)
is strictly related
to the presence of a non almost monotonous path.
The next to the last transition in
the standard stage ends with the
formation of a standard octagon with
$L_1=L^{\ast}$ and $
L_2=L^{\ast} -1$.
The very last transition is then
just a flip of a minus spin adjacent
to a long coordinate edge
in the standard octagon with
$L_1=L^{\ast}$ and $ L_2=L^{\ast} -1$
(namely, the creation of a unit
square protuberance adjacent to that edge).
In this way we form a
global saddle
(``protocritical'') droplet in $\Cal P$ (see [NS]) and
enter into $\partial \Cal A$ (see Fig\.~5.14).
%
%
\midinsert
\centerline{\scaledpicture 3.78in
by 3.56in (fivefourteen scaled 700)}
\botcaption{Fig\. 5.14}
\endcaption
\endinsert
%
%
To conclude our preliminary discussion we
want to say that
the property of $Q(D^{\ast},D^{\ast})$ of being
the minimal supercritical
standard octagon can be expressed by the fact
that $L^{\ast}$ is
the minimal value of $L$ for which
$$
H(S^{(1)}_{L,
L})-H(Q_{L+1, L}) = E(L) > 2\tilde J-h.
$$
Remember that during the
growth up to the protocritical droplet,
described by $\Cal E^{(s)}$ we
had
$$
H(S^{(1)}_{L, L})-H(Q_{L+1, L}) < 2\tilde J-h
$$
as indicated in
Fig\.~5.8.
Let us now start with the formal
definition of $\Cal E$. Given
$\sigma\in \Cal A$ and
$ t_c\in \Bbb N$ to be fixed later, we
define
$$
\Cal E^{(c)}_{\sigma, t_c} =
\{\sigma_0 = \sigma,\tau_{
-\underline 1}=t_c\}.
\tag{5.5}
$$
Given $t_w\in \Bbb N$ we set
$$
\Cal
E^{(w)}_{t_w}=\{\sigma_t=
- \underline 1, 0\leq t\leq t_w\}.
\tag{5.6}
$$
Now,
consider the sequence of clusters
$\bar\gamma_1, \dots \bar\gamma_{12}$
specified in Fig\.~5.1. We define
$$
\Cal E^{(e)}=\{
\sigma_0=-\underline 1, \sigma_1=\bar\gamma_1, \dots ,
\sigma_{12}=\bar\gamma_{12}\}.
\tag{5.7}
$$
We recall that the energy
is strictly increasing from $-1$
to $\bar\gamma_{12}$; indeed, we have
$$
\eqalign{
&H(\bar\gamma_1)- H(-\underline 1)=4J-4K-h,\cr
&H(\bar\gamma_3)-H(\bar\gamma_2)=2J - h-2K,\cr
&H(\bar\gamma_5)-H(\bar\gamma_4)=2J -2K- h,\cr
&H(\bar\gamma_7)-H(\bar\gamma_6)=2K - h,\cr
&H(\bar\gamma_9)-H(\bar\gamma_8)=2J -4K- h,\cr
&H(\bar\gamma_{11})-H(\bar\gamma_{10})=2 J-4K - h\cr}
\eqalign{
&\qquad H(\bar\gamma_2)-H(\bar\gamma_1)=2J - h,\cr
&\qquad H(\bar\gamma_4)-H(\bar\gamma_3)= - h+2K,\cr
&\qquad H(\bar\gamma_6)-H(\bar\gamma_5)=2J -4K- h,\cr
&\qquad H(\bar\gamma_8)-H(\bar\gamma_7)=2K- h,\cr
&\qquad H(\bar\gamma_{10})-H(\bar\gamma_9)=2K - h,\cr
&{}\cr
}\tag{5.8}
$$
To present explicit
definitions for the subsequent stages
$\Cal E^{(r)},\Cal E^{(t)},
\Cal E^{(s)}$ of the event $\Cal
E_{\sigma}$, we would like
to use the general set-up,
described in Section 3, based
on the introduction of a set of
auxiliary Markov chains.
Thus, we have to specify
an integer $N$, a sequence of octagons
$Q_1,\dots ,Q_N$,
connected sets $B_1,\dots ,B_N$, saddles
$S_2,\dots ,S_{N +1}$,
with $S_i \in \partial B_{i} \cap B_{i+1}
$, $ i=1,\dots ,N$, $ S_{N+1}
\in \partial B_{N}$, as well as
resistance times $\bar t^i_u$ and
``descent'' times $\bar t^i_d$.
We begin by saying that part
of the octagons $Q_1,\dots ,Q_N$ entering
in our construction will
be standard (see the equation (2.17)). The
other ones will be non-standard
(some $l_i$'s will differ from
$l^*$); further, the sets $ B_i$,
$ i=1, \dots ,N$ will be either
\flushpar
1) the basin of attraction
$\Cal B(Q_i)$ of $Q_i$ (see the equation
(3.7)), if $Q_i$ is not standard or
\flushpar
2) the domain of attraction
$\Cal D(D_1, D_2)$
(see the equation (3.10))
if $ Q=Q(D_1,D_2)$ is a standard octagon.
At the end of our construction
we will consider the set $\bar \Cal F
\cap \Cal G$ (cf\. the equations (3.57), (3.54), (3.56),
(3.48), and (3.53). It will contain the regular,
the transient and the
standard stages. We have, with obvious meaning of the symbols,
$$
\aligned
N = &N_r +N_t +N_s \\ N_r = 14 (l^* -2) ,\; &N_t
= 7 ,\; N_r = 2[L^* - (l^* +2)],
\endaligned
\tag5.9
$$
and
$$
\multline
Q_1,S_2,Q_2,S_3, \dots ,Q_N,S_{N+1}
\equiv \gamma^0_2,S^1_2,\dots,
\gamma^{13}_2,S^{14}_2,\gamma^0_3,\\
\dots ,\gamma^{13}_{l^*-1},
S^{14}_{l^*-1}, Q^{(0)}_{l^*,l^*},\bar
S^{(2)}_{l^*,l^*},
Q^{(2)}_{l^*,l^*},S^{(3)}_{l^*,l^*},Q^{(0)}_{l^*+1,l^*},
S^{(1)}_{l^*+1,l^*},Q^{(1)}_{l^*+1,l^*},S^{(2)}_{l^*,l^*}, \\
Q^{(0)}_{l^*+1,l^*+1},S^{(1)}_{l^*+1,l^*+1},
Q^{(1)}_{l^*+1,l^*+1},S^{(2)}_{l^*+1,l^*+1},
Q^{(0)}_{l^*+2,l^*+1},S^{(1)}_{l^*+2,l^*+1}, \\
Q^{(0)}_{l^*+2,l^*+2},S^{(1)}_{l^*+2,l^*+2},
\dots, S^{(1)}_{L^*-1,L^*-
1},Q^{(0)}_{L^*,L^*-1},S^{(1)}_{L^*,L^*-1}.
\endmultline
\tag{5.10}
$$
Now we specify the times $\bar t^i_u,\bar t^i_d $
corresponding to the
different stages: regular,
transient and standard.
The times $\bar t^i_u,\bar t^i_d$
for the regular stages are denoted by
$\bar t^{j,l}_u , \bar t^{j,l}_d$.
Thy correspond, respectively, to the
transitions
$$
\gamma^{(j)}_l \to S^{j+1}_l , \; j=0, \dots , 13,
S^{j+1}_l\to \gamma^{(j+1)}_l, j=0,\dots,12, S^{14}_l \to
\gamma^{(0)}_{l+1},
$$
for
$l : 2 \leq l \leq l^*-1$.
They are given by
$$
\aligned
\bar t^{j,l}_u = \exp \{\beta [h(&l-1)+\delta]\} \;
\text{ for } j \neq 1,2,4,6,
\\
\bar t^{j,l}_u = \exp \{\beta [h(&l-
2)+\delta] \}\; \text{ for } j= 1,2,4,6,
\\
&\bar t^{j,l}_d = \exp
\{\beta\delta\}
,\endaligned
\tag{5.11}
$$
with $\delta$ to be chosen later.
For the transient stage we have
$$
\bar t^j_u
= \exp\{ \beta(h(l^*-2)+\delta) \}
$$
for the transitions
$$
Q^{(2)}_{l^*,l^*} \to S^{(3)}_{l^*,l^*} ,
Q^{(1)}_{l^*+1,l^*} \to
S^{(2)}_{l^*+1,l^*}, Q^{(1)}_{l^*+1,l^*+1}
\to S^{(2)}_{l^*+1,l^*+1} ,
$$
further,
$$
\bar t^j_u = \exp \{\beta(h(l^*-1)+\delta)\}
$$
for the transitions
$$
Q^{(0)}_{l^*,l^*} \to \bar S^{(2)}_{l^*,l^*},
Q^{(0)}_{l^*+1,l^*} \to
S^{(1)}_{l^*+1,l^*} ,
$$
as well as
$$
\bar t^j_u = \exp \{\beta(2h(l^*-1)-2(K-
h))+\delta) \}
$$
for
$$
Q^{(0)}_{l^*+1,l^*+1} \to S^{(1)}_{l^*+1,l^*+1},
Q^{(0)}_{l^*+2,l^*+1} \to S^{(1)}_{l^*+2,l^*+1} ,
$$
and
$$
\bar t^j_u =
\exp \{\beta (E(l^*+2)+\delta)\}
$$
for
$$
Q^{(0)}_{l^*+2,l^*+1} \to
S^{(1)}_{l^*+2,l^*+2} .
$$
Moreover, we take
$$
\bar t^j_d = \exp\{
\beta \delta \}
$$
for the transitions
$$
S(l^*) \to Q^{(0)}_{l^*,l^*},
\bar S^{(2)}_{l^*,l^*}
\to Q^{(2)}_{l^*,l^*}, S^{(3)}_{l^*,l^*} \to
Q^{(0)}_{l^*+1,l^*}, S^{(1)}_{l^*,l^*} \to Q^{(1)}_{l^*+1,l^*},
S^{(1)}_{l^*,l^*+1} \to Q^{(1)}_{l^{*}+1,l^{*}+1} .
$$
For all the other
transitions of the transient stage we take
$$
\bar t^j_d = \exp \{\beta(2K-
h+\delta)\}
$$
and for any transition
$$
Q^{(0)}_{L_1,L_2} \to
S^{(1)}_{L_1,L_2}
$$
of the standard stage,
$$ \bar t^j_u =
\exp \{\beta ( E(L_1 \wedge L_2) +\delta)\}.
$$
Finally, for every {\it descent}
time of the standard stage we take
$$
\bar t^j_d = \exp
\{\beta(2K-h+\delta)\}.
$$
Now we are ready to present the definition of our
event $\Cal E _{\sigma}$,
$$
\multline
\Cal E_{\sigma} = (\bar \Cal E_{\bar
t_c}^{(c)};\bar \Cal E^{(w)}_{\bar t _w};
\Cal E^{(e)} ; (\bar \Cal
F\bigcap \Cal G) = \\
= \bigcup^{\bar t_c}_{t_c=1} \bigcup^{\bar
t_w}_{t_w = 1} \Cal E_{t_c}^{(c)}
\bigcap T_{t_c}\Cal E^{(w)}_{
t_w}\bigcap T_{t_c+t_w}\Cal E^{(e)}
\bigcap T_{t_c+t_w+12}(\bar \Cal F
\bigcap \Cal G).
\endmultline
\tag{5.12}
$$
We repeat that $\bar \Cal F \bigcap \Cal
G$ is the event, corresponding to the regular,
transient and standard
stages defined like in the equation (3.57)
using the previously defined $Q$'s, $B$'s,
$S$'s (see the equation
(5.10) ) and the corresponding times
$\bar t^u_i , \bar t^d_i$ .
It is immediate to verify inequality (3.62) in our case.
Now, to get the basic estimate
given by inequality (3.67), namely, in
our case,
$$
P(\bar \Cal F) \geq \exp\{ -\beta (H(\Cal P) - H(S_1) +
\epsilon )\},
\tag{5;13}
$$
we first need to verify (3.66).
For every {\it downhill}
transition, namely for every transition of the regular
case as well as
for the {\it descent} transitions of
the transient case corresponding
to $\bar t^j_d =\exp\{\beta\delta\}$
this is an immediate consequence of
the inequality
$$
\Bigl({1\over |\Lambda|}\Bigr)^{T_o} > \exp(-\epsilon \beta),
$$
valid for every $\epsilon >0$ and $\beta$
sufficiently large (see (3.100)).
For the transitions
$$
S^{(2)}_{l^*+1,l^*} \to Q^{(0)}_{l^*+1,l^*+1},
S^{(2)}_{l^*+1,l^*+1} \to Q^{(0)}_{l^*+2,l^*+1},
S^{(2)}_{l^*+2,l^*+1}
\to Q^{(0)}_{l^*+2,l^*+2}
$$
of the transient stage,
as well as for every
transition of the standard case,
we notice that from $S_j$ in one step,
with probability larger that
$1\over |\Lambda|$ we go to $B_{j+1}$.
Then the equation (3.66) is an immediate
consequence of the argument of proof of
Proposition 1.
The inequality (3.65) is very easy
to deduce by remarking that $Q_j
\to S_{j+1}$ is always an uphill
single spin flip transition with the
exception of the first part of
the transient stage, $Q^{(0)}_{l^*,l^*}
\to \bar S^{(2)}_{l^*,l^*}$.
In this last case it is easy to prove
the equation (3.65) by using a trial
event with the resistance time of
the order $\exp \{\beta( h
(l^*-2) + \delta)\} $ in $\bar Q
^{(1)}_{l^*,l^*}$ exploited
in the usual way. We leave the details
to the reader.
For all the other cases that,
we repeat, are single spin flip uphill
transitions, the lower bound
given by (3.65) is immediate.
To get the equation of the form
(3.59) we proceed like in the proof of
Proposition 1. Namely, we prove condition (C1)
$$
\text{ there exists } C > 0 \text{ such that }
\Cal P (\Cal G^c) < \exp(-e^{C\beta}),
\tag {5;14}
$$
by introducing,
for every $j = 1,\dots,N$ certain
events $\widehat {\Cal E}^j_{\sigma}$
and times $\tilde t^j_u$ satisfying
conditions (3.98) and (3.99).
To this end, we observe that, for
every non-standard (subcritical)
octagon $Q_j$ appearing in $\Cal G$,
the corresponding $B_j$ is just the
usual basin of attraction and we can
introduce the event $ \widehat
{\Cal E}^j_{\sigma}$ in the following way:
\flushpar
(1) First, we pass to $Q_j$ via a
descent path in a time of
order $T_0$. (For every $\epsilon > 0 $
and $\beta$ sufficiently
large the corresponding
probability is larger than $ \exp (-\epsilon
\beta)$.)
\flushpar (2) Then we pass to
the minimal saddle $S$ in $\partial B_j$
(corresponding to the shrinking mechanism
since $Q_j$ is subcritical)
by a sequence of corner erosions.
The corresponding probability
estimate is like (3.65) with a proper
choice of $\epsilon$.
Otherwise, for the cases of standard
octagons (appearing both in the
transient and in the standard stages)
we take $ B_j = \Cal D ( L_1
+2(l^*-1) , L_2 +2(l^*-1))$ for
the domain of attraction. The time
$\tilde t^j_u$ can be taken as
$ \exp \{\beta( E(L_1 \wedge L_2)
+\epsilon)\}$ and the escape event
$ \widehat {\Cal E}^j_{\sigma}$ is
just the shrinking event
$ \Cal E^s_{\sigma}$ constructed in the proof
of Proposition 1.
Now, it is easy to see that,
for every sufficiently small $\epsilon > 0
$ and $\beta$ sufficiently large, if
$$
\bar t_w = \exp \{\beta[ E(L^*-1 )
+\delta]\} , \quad \delta >0,
$$
then
$$
P(\bar {\Cal E} ^{(w)}_{\bar
t_w} ; \Cal E^e) \geq \exp \{\beta[ E(L^*-1 )-H(\bar
\gamma_{11}) +H(-
\underline 1) -\epsilon] \}.
\tag {5;15}
$$
Moreover, from
Propositions 1, 2, and 3, it follows
that for every sufficiently small
$\epsilon $ and $\beta$ sufficiently large, once
$$
\bar t_c = \bar t_w = \exp
\{\beta[ E(L^*-1 ) +\delta]\},
$$
then
$$
P(\bar {\Cal E}_{\bar t_c})
\geq \exp \{-\beta\epsilon \}
\tag {5;16}
$$
From
(5.12),(5;13), (5;14), (3.67), (5;15),
and (5;16) it follows that
$$
P( \Cal
E_{\sigma}) \geq \exp \{\beta[ E(L^*-1 )+
H(\Cal P)- H(-\underline 1) -
\epsilon]\}
$$
for all sufficiently small $\epsilon $ and $\beta$
sufficiently large.
Finally from (5.2),(5.3), and (5;15) we get (5.1).
Describing the event $\Cal E$, we actually defined an
{\it$\epsilon$-typical path} appearing
in the statement of Theorem 3.
The only difference is that
while in the definition of $\Cal
E_\sigma$ we considered
a very particular sequence of configurations
(for example, all concerned
octagons are centered), defining the set
$\Cal U_{\epsilon}$
of all $\epsilon$-typical
paths we can be slightly more flexible and
allow also droplets of different
positions and orientations.
Namely, an
$\epsilon$-typical path describes
the typical way followed by our
process starting from $-\underline 1$
to form a critical nucleus and
then to go to $+\underline 1$.
It contains, in particular, the stages
that we have described when
defining our trial event $\Cal
E_{\sigma}$ except for the initial
{\it contraction} and {\it waiting}
stages. In other words an
$\epsilon$-typical path will pass,
initially, through the {\it embryonal,
regular, transient,} and {\it
standard} stages spending,
in the appropriate basins or domains of
attraction, suitable intervals
of time ({\it resistance times}).
Then, after getting the set
of global saddles (protocritical
droplets in $\Cal P$)
a new stage starts that we call {\it
supercritical}: we pass from
$\Cal P$ to $+\underline 1$ through a
suitable sequence of growing
standard octagons with proper
resistance times.
The embryonal stage is uphill;
the regular, transient, and standard
are uphill in average and, finally,
the supercritical is downhill
in average.
The first (subcritical) portions
(embryonal, regular, transient,
and standard) of an $\epsilon$-typical
path will involve notions
generalizing the ones already seen
in the definition of the event
$\Cal E_{\sigma}$. Embryonal, regular,
transient, and standard
stages in $\Cal E_{\sigma}$
are strictly related to a particular
example, with a particular
choice of locations and orientations, of
the corresponding ones in
$\Cal U_{\epsilon}$. Namely, the generalization
in $\Cal U_{\epsilon}$ with respect
to $\Cal E_{\sigma}$ is only related
to geometrical trasformations
like translations, rotations or
reflections with respect to
some lattice axes of the corresponding
clusters, octagons, standard
octagons involved in $\Cal
E_{\sigma}$.
The family of sequences of eleven
clusters taking part to the
embryonal stage of an
$\epsilon$-typical path (sequentially
visited in eleven steps)
will be specified in great detail.
Then, assigning the regular,
transient and standard stages,
similarly to what we did in
the definition of $\Cal E_{\sigma}$,
will consist in specifying a set of
sequences of {\it octagons} $
Q_1,\dots,Q_{N+1}$ with $Q_1 = Q(2)$,
$ Q_{N+1} = Q(D^*,D^*) \equiv
$ critical nucleus, connected regions
$B_1,\dots,B_{N+1}$ with $
B_i \ni Q_i$, saddles $S_2,\dots,S_{N+1}$ with
$ S_{i+1} \in B_i \cup
B _ {i+1} $, resistance times (see below) .
Further, $N = N_r +N_t +N_s$ with
$ N_r$, $N_t$ and $N_s $ given in
the equation (5.9) (the numbers
of octagons in {\it any} sequence, in
the concerned stage are exactly
the same as the numbers of the corresponding ones
in $\Cal E_{\sigma}$).
Then, after $Q(D^*,D^*)$, comes
the supercritical stage
$Q_{N+1},\dots,Q_{N+2(M-D^*)},
+ \underline 1$, with the
corresponding $B_{N+1},\dots,B_{N+2(M-D^*)}$,
$S_{N+2},\dots,S_{N+2(M-D^*)}$ and resistance times.
Regular and part of the transient
stage will involve non-standard
octagons. Part of transient,
standard and supercritical stages will
involve standard octagons.
In the definitions introducing
$\Cal U_{\epsilon}$, during the evolution
along a typical path we are
less and less specific in the following
sense: at the very beginning
we assign a class of sequences of
clusters (no resistance times
up to $Q(2)$), then, in the regular
and part of transient stage,
we specify a class of non-standard
octagons (without specifying the sequences of
non-octagonal clusters in between).
Subsequently (in part of
transient stage and in the standard
stage) we will be able to specify only a class of sequences of
standard octagons (without specifying the sequences of
non-standard octagons in between).
Finally, in the supercritical stage,
we will not even be able to
specify a precise class of sequences
of standard octagons and
much larger fluctuations have to be allowed.
We can say that, as the time goes on our tube of trajectories
becomes less and less narrow:
it will correspond to the maximum
possible specification compatible
with an almost full probability
estimate.
Now let us start with the detailed definitions.
The first part of the $\Cal U_{\epsilon}$,
called {\it embryonal}, is
given by the set of all paths
($\sigma_1=\gamma_1,\dots ,\sigma_{11}=
\gamma_{11})$ where
$(\gamma_1,\dots ,\gamma_{11})$ is a generic
sequence
of connected clusters with
\roster
\item"i)" $\gamma_1$ given by a unit square;
$\gamma_{11}$ given
by the saddle configuration
$\bar \gamma_{11}$ of Fig\.~5.1
arbitrarily located and oriented
(the droplet $\bar \gamma_{11}$ can go downhill to $Q(2)$
by a single spin-flip);
\item"ii)" $\gamma_j$ obtained
from $\gamma_{j-1}$ by adding a unit
square touching it and, in this
way, increasing the energy
without bypassing the energy
level of $\bar \gamma_{11}$ :$ H(\gamma _j) >
H(\gamma _{j-1})$,\hfill\newline
$ H(\gamma _j) < H(\bar \gamma_{11})$, $j=2,\dots,11$.
\endroster
It is clear by inspection that,
with the rule ii), starting from $\gamma_1$,
after 11 steps we always
end up in $\gamma_{11}$.
An example of a sequence implementing
the characteristics of the
embryonal stage is given in Figure 5.1.
Now, for the definition of
the subsequent stages we have
to define, preliminarily,
the $Q_i$'s, $B_i$'s, and $S_i$'s.
Like in $\Cal E_{\sigma}$, for all $i=1,\dots,
N+2(M-D^*)$ we take $ B_i=
\Cal B (Q_i) =$ basin of attraction of $Q_i$
if $Q_i$ is a non-standard octagon
and $B_i=\Cal D (D_1, D_2) =$
extended domain of attraction of $Q_i$,
if $Q_i$ is a standard octagon.
The saddles $S_i \in \partial B_i
\cap B_{i+1},\; i=1,\dots,N
+2(M-D^*)$ and, for $i=2,\dots,N$,
$S_i \in \{ \text {set of minimal saddles in }
\partial B_i \}$;
whereas
for $i=N+1,\dots,N+2( M-D^*)$,
$S_i \in \{ \text {set of minimal saddles in }
\partial B_{i+1} \}$.
Then the $B_i$'s and the $S_i$'s
are determined once the $Q_i$'s are
given.
Any sequence $Q_1,\dots, Q_{N+2(M-D^*)}$
corresponding to
a typical path will be called
{\it typical sequence of octagons}
(it will follow the embryonal stage).
The set of all typical sequences
will be called {\it typical tube}
and will be denoted by $\Cal T_{\epsilon}$.
Now let $V_i \equiv V(Q_i)$ be given by:
\roster
\item"i)"
$V_i = h \hat \Cal L_j $ with $\hat
\Cal L_j\equiv \min_{i=1,\dots,8} \Cal L_i$
(see definitions before Lemma 3.5), whenever $Q_i$
is a non-standard octagon,
\item"ii)"
$V_i = E(L)$, whenever $Q_i$ is a standard
octagon with $\min \{L_1(Q_i), L_2(Q_i) \} =
L$ , with $l^*
\leq L \leq L^* -1$,
\item"iii)"
$V_i = 2J-4K-h$, whenever $Q_i$ is
standard and supercritical,\hfill\newline
$L =\min
\{ L_1(Q_i),L_2(Q_i) \} \geq L^*$.
\endroster
{\sl A path $\sigma_t$ will be an
$\epsilon$-typical path if, after
the embryonal stage it will
visit sequentially $Q_1 \in Q(2),\dots,
Q_{N+2(M-D^*)}$ in the following way:
\flushpar
Starting from $Q_i$ it will
spend some time $t_i$ inside $B_i$.
\flushpar
Then, after passing through $S_{i+1}$, it will reach
$Q_{i+1}$ for the first time.
Calling $t_i$ the time interval
between first arrivals in $Q_i$ and $Q_{i+1} $ we have:}
$$
\exp \beta (V_i -\epsilon) < t_i < \exp \beta (V_i +\epsilon).
$$
So, to conclude
the definition of $\Cal U_{\epsilon}$ we
only need to assign the typical tube of octagons.
Now we start this definition by distinguishing the regular,
transient, standard and supercritical portions.
The {\it regular portion }
of the typical
tube is denoted by
$\Cal T^{(r)}_{\epsilon}$ .
Starting from $Q(2)$,
$\Cal T^{(r)}_{\epsilon}$ contains the set
of all the reverse of a
sequence of canonical
contractions (see Section 3) from $Q(3)$ to $Q(2)$
and so on up to $Q(l^*)$.
The transient and standard
portions $\Cal T^{(t)}_{\epsilon}$ and
$\Cal T^{(s)}_{\epsilon}$
of the typical tube are simply
given by the set of sequences of
octagons obtained by the set
of sequences appearing in the
definitions given above when
introducing $\Cal E_{\sigma}$, modulo
translations and the obvious
rotations and reflections (we do not
enter into a detailed
classification leaving the easy exercise to
the interested reader).
After getting $S_{N+1} \in \Cal P$
the supercritical stage
$\Cal T^{(sc)}_{\epsilon}$ starts.
It will consist of the set of
sequences of standard octagons $
Q(D^{(j)}_1,D^{(j)}_2)$,\hfill\newline
$j=1, \dots,2(M-D^*)$ with the following
monotonicity property
$$
\text {either } (D^{(j+1)}_1,D^{(j+1)}) =(D^{(j)}_1+1,D^{(j)}) \text
{ or } (D^{(j+1)}_1,D^{(j+1)})=(D^{(j)}_1,D^{(j)}+1).
$$
\demo{Proof of Theorem 3}
Theorem 3 follows directly
by Propositions 1--4 and
the easy observation that
the embryonal path is the reverse of a
downhill shrinking starting
from $\gamma _{11}$, via a
straightforward adaptation of
the results of [S 1] based on
reversibility of the process
(Lemmas 2, 3, 4 and Theorem 1 therein).
\enddemo
\vfill
\newpage
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ENDBODY
~~