%%%AMSTeX 2.1
%%Rotational Dynamics - Long Version

%\documentstyle{amsppt} 
\documentstyle{amsjcm}  %Use to print on sun3 with Publisher at UAB
\magnification=\magstep1
\NoBlackBoxes

%%Definitions
\define\lb{\Lambda}
\define\a{\alpha}
\define\e{\epsilon}
\define\g{\gamma}
\define\Int#1{\operatorname{Int}(#1)}
\define\Ext#1{\operatorname{Ext}(#1)}
\define\Cl#1{\overline{#1}}
\define\Bd#1{\partial#1}
\define\dist{\operatorname{d}}
\define\id{\operatorname{id}}
\define\real{\bold R}
\define\zed{\bold Z}
\define\complex{\bold C}
\define\disk{\bold D}
\define\wt{\widetilde}
\define\preprint{{\rm (preprint)}}
\define\bru{basic repetitive unit}

%%Definitions for placement of figures.
\define\figone{\midspace{3.25 true in}\caption{Figure 1. Prime end compactification of
the domain $\Int\lb$.}}

\define\figtwo{\midspace{6.25 true in}\caption{Figure 2. Construction of $\lb\in\Cal H$
as an intersection of annuli $A_k$.}}

\define\figthree{\midspace{3.25 true in}\caption{Figure 3. Inducing a pseudorotation by
a sequence of rational rotations of the defining annuli $A_k$.}}

\define\figfour{\midspace{4.25 true in}\caption{Figure 4. The $(d-c_{k+1})q_k$
fractional part of $A_{k+1}$ meets $D_k$, the $d$ fractional part of $A_k$.}}

\define\figfive{\midspace{4.25 true in}\caption{Figure 5.  ``Straightening-out" $A_2$
in $A_1$, and $A_3$ in $A_2$, to show that $\lb$ is a simple closed curve when the
$s_k$'s are bounded above.}}

%%Definitions for proofs and defining terms
\define\pr{\demo{Proof}}
\define\qd{\qed\enddemo}
\define\df{\it} %Font for defining new terms

%\NoBlackBoxes

\topmatter

\title The Rotational Dynamics of Cofrontiers \endtitle

\author B. L. Brechner, M. D. Guay, and J. C. Mayer \endauthor

\address Department of Mathematics, University of Florida,
Gainesville, FL 32611 \endaddress
\email brechner\@math.ufl.edu \endemail
\address Department of Mathematics, University of Southern Maine,
Portland, ME 04103 \endaddress
\address Department of Mathematics, University of Alabama at
Birmingham, Birmingham, AL 35294-2060 \endaddress
\email mayer\@math.uab.edu \endemail

%\date December 10, 1991 \enddate

\keywords almost periodicity, cofrontier, dynamics, equicontinuity,
indecomposable continuum, constant type, minimal
set, minimally invariant, plane homeomorphism, prime ends,
recurrence, rotation, rotation number, pseudorotation, Siegel disk
\endkeywords

\subjclass 54F20, 54F50, 54G15, 54G20, 54H20 \endsubjclass

\abstract
A {\df pseudorotation}\/ is defined to be an extendable
homeomorphism of a cofrontier $\lb\subset\real^2$ whose induced
prime end homeomorphisms from both sides are conjugate to a single
rotation $R_\alpha$ on $S^1$.  A class $\Cal H$ of cofrontiers in the
plane is constructed following the recipe suggested by M.
Handel [Proc.~AMS {\bf 86}(1982), 163-168].  The authors
study periodicity, almost periodicity, and recurrence
properties of pseudorotations of cofrontiers $\lb\in\Cal
H$, as well as cofrontiers in general.  They also study
the relationship between each of these properties on $\lb$
and the corresponding property on the induced prime end
circles and disks. \endabstract

\thanks Supported in part by NSF-Alabama EPSCoR grant
number RII-8610669.  A summary version of this paper without proofs
appeared in {\df Contemporary Mathematics\/} {\bf 117} (1991).
\endthanks
 
\endtopmatter

\document

\heading 1. Introduction \endheading
\nopagebreak

We construct a class $\Cal H$ of cofrontiers in the plane, together
with a class of automorphisms which we call pseudorotations.  These
examples are modeled on Handel's construction in \cite{Ha}.  We are
interested in studying both the topological and  the
embedding-dependent topological properties of these cofrontiers as
described by their prime end structure, as well as the dynamical
properties of the pseudorotations.

Initially, some (corollaries of) theorems of
Hemmingsen \cite{He} and Cartwright \cite{Ca} are observed,
characterizing (1) invariant cofrontiers under irrational rotations
of the plane and (2) a minimal cofrontier under an almost periodic
homeomorphism of the cofrontier, respectively (Theorems 2.7 and
2.9), and relating almost periodicity on an arbitrary
cofrontier-plus-interior with almost periodicity on its prime end
circle (Theorem 2.8).

Theorem 2.11, and Theorem 2.12 with Corollary 2.13, are prime end
theorems, about periodic pseudorotations, and about the prime end
structure of indecomposable cofrontiers admitting irrational
pseudorotations, respectively. Part of our motivation in studying
pseudorotations comes from the well-known Siegel disk problem in
complex analytic dynamics.  For a discussion of this classical
problem see Rogers' paper \cite{Ro2} (these Proceedings).  In this independent, and
earlier, work Rogers proves theorems (originally announced in \cite{Ro1}) about
``extendable intrinsic rotations" of a bounded domain similar to our Theorems
2.12 and 2.13.

Next, we construct the family $\Cal H$ carefully, and obtain some
topological, dynamical, and number-theoretic results about
cofrontiers $\lb\in\Cal H$ and pseudorotations $f_\a$ of $\lb$. 
($f_\a$ is a pseudorotation on $\lb$ if and only if the induced
map on the circle of prime ends on each side of $\lb$ is conjugate
to the rotation of angle $2\pi\a$ for some $\a\in [0,1)$.)  In
Theorem 5.1, we prove that {\df every} $\lb\in\Cal H$ admits an
irrational pseudorotation $f_\a$, for some $\a$.  The main theorems
of this paper are Theorems 2.12 and 5.8.  Theorem 2.12 describes the
prime end structure of indecomposable cofrontiers admitting
irrational pseudorotations, including members of our family $\Cal H$,
while Theorem 5.8 describes the dynamical properties and a
number-theoretic property of certain 
indecomposable members of $\Cal H$.  In Theorem 5.8, we put
restrictions on several of the technical conditions in the
construction of $\lb\in\Cal H$ to show that such restricted $\lb$
are indecomposable, and if $f_\a$ is a \lq\lq standard"
pseudorotation on such a restricted $\lb\in\Cal H$, then $\a$ is
irrational and not of constant type, and $f_\a$ is recurrent,
minimal, and not almost periodic on $\lb$. 


In \cite{BG}, Barge and Gillette show that if $\lb$ is a cofrontier left invariant by
a homeomorphism $f$ of the plane onto itself, and if the set of rotation numbers of
$f$ associated with $\lb$ is nondegenerate, the $\lb$ is indecomposable.  This
improves a result of Charpentier \cite{C}, which says that if the interior and
exterior rotation numbers are distinct, then $\lb$ is indecomposable.

 \heading 2. Cofrontiers, prime ends, and pseudorotations \endheading
\nopagebreak

\subheading{ 2.1. Cofrontier}
A {\df continuum\/} is a compact, connected metric space.  A
continuum is {\df  indecomposable\/} iff it cannot be written as the
union of two of its proper subcontinua.
A {\df cofrontier}\/ $\lb$ is a continuum in the plane $\real^2$
which irreducibly separates $\real^2$ into exactly two components. 
That $\lb$ separates $\real^2$ into exactly two components is a
topological property, as is the property that a cofrontier is the
common boundary of its complementary domains.  Two cofrontiers
$\lb_1$ and $\lb_2$ are said to be {\df equivalent}\/ iff there is a
homeomorphism $h:(\real^2,\lb_1)\rightarrow (\real^2,\lb_2)$.  A
cofrontier may well have inequivalent embeddings in $\real^2$
(though $S^1$ does not, of course).  We usually denote a cofrontier
together with its embedding into $\real^2$, by $\lb$.  We denote the
unbounded component of $\real^2-\lb$ by $\Ext\lb$ and the bounded
component by $\Int \lb$.


\subheading{2.2. Prime end compactification and prime end rotation
number} Suppose $U$ is a bounded, simply connected domain (connected
planar open set) with nondegenerate boundary and $f:\Cl U\to\Cl U$
is a homeomorphism.  The {\df prime end compactification}\/
$\phi:U\to \disk$ (where $\disk$ is the open unit disk), and the
homeomorphism $f|U$ induce a homeomorphism $F:\Cl\disk\to\Cl\disk$
as the extension of $\phi f\phi^{-1}$.  Note that $\phi$ may be
taken to be a conformal homeomorphism, and exists by the Riemann
Mapping Theorem.  We call $F|\Bd\disk$ the {\df induced
homeomorphism\/} on the circle $\Bd\disk$ of prime ends.  (For more
on prime ends, see Section 2.10.)  Always, $F|\Bd\disk$ has a
well-defined {\df  rotation number\/} given by
$$\rho(F)=\lim_{n\to\infty}\left(\frac{\wt F^n(\wt
x)}{n}\right)(\text{mod\ } 1)$$ where $\wt F$ is any lift of
$F|\Bd\disk$ to the universal cover $\real$, and $\wt x$ is any
point in $\real$.  We call this number the {\df prime end rotation
number\/} of $f$ on $U$.

Now suppose $\lb$ is a cofrontier and
$f:(\real^2,\lb)\to (\real^2,\lb)$ is a homeomorphism. The {\df
prime end compactification}\/ $\phi_i:\Int \lb\to \disk$, and the
homeomorphism $f|\Int \lb$ induce a homeomorphism
$F_i:\Cl\disk\to\Cl\disk$ as the extension of $\phi_i
f\phi_i^{-1}$.  We call $F_i|\Bd\disk$ the {\df induced
homeomorphism\/} on the circle $\Bd\disk$ of prime ends.    See
Figure 1.  Similarly, the prime end compactification
$\phi_e:\Ext\lb\cup\{\infty\}\to \disk$, and the homeomorphism
$f|\Ext\lb$ induce a homeomorphism $F_e:\Cl\disk\to\Cl\disk$ as the
extension of $\phi_e f\phi_e^{-1}$.  In general, neither
$F_i|\Bd\disk$ nor $F_e|\Bd\disk$ need be conjugate to any rotation
of $\Bd\disk=S^1$, let alone the same rotation.

\figone

\subheading{2.3. Pseudorotation}
A {\df pseudorotation}\/ of a cofrontier $\lb$ is a
homeomorphism  $f:\lb\to\lb$, extendable to a homeomorphism of 
$\real^2$, such that each of the induced homeomorphisms on the
circles of prime ends, corresponding respectively to  $\Int \lb$ and 
$\Ext\lb$, is conjugate to the same rotation $R_\a$ of $S^1$
through angle $2\pi\a$ for some $\a\in [0,1)$.  Since this $\a$ is
uniquely determined, we write $f_\a$ when $f$ is a pseudorotation and
$R_\a$ is the induced rotation on the circle of prime ends.  Since
$f_\a$ is extendable, we often assume a particular extension 
$f_\a:(\real^2,\lb)\to(\real^2,\lb)$.  A given pseudorotation will
have many extensions to homeomorphisms.

\subheading{2.4. Generalized rotation number}
Though a pseudorotation is associated with a single prime
end rotation number $\a$ via the prime end compactification, it is
not necessarily the case that every point of $\lb$ has $\a$ as its
rotation number.  We define the {\df  generalized
rotation number} for a cofrontier $\lb$ under a homeomorphism  $f$,
when it exists, as follows:

Suppose $0\in \Int\lb$ and $f:(\real^2,\lb,0)\rightarrow
(\real^2,\lb,0)$ is a homeomorphism.  Let
$\g:\real^2-\{0\}\rightarrow S^1$ be radial projection, and let $\wt
\g:\real^1\times (0,1)\rightarrow \real^1$ be a lift of $\g$ to the
 universal cover $\real\times(0,\infty)$ (of $\real^2-\{0\}\cong
S^1\times(0,\infty)$) with covering projection $\pi\times\id$.  Let
$\wt f:(\real^1\times (0,\infty),\wt\lb)\rightarrow (\real^1\times
(0,\infty),\wt\lb)$ be a lift of $f$.  Let $x \in \lb$ and let $\wt x
\in \wt\lb$ be such that $(\pi\times\id)(\wt x)=x$. Define
$$\rho(x):=\left\{ \text{accumulation points of}
\left\{\frac{\wt\g\wt f^k(\wt x)}{k}\right\}_{k=1}^\infty \right\}
(\text{mod\ } 1)$$  \noindent Equivalently,
$$\rho(x)=\left\{ \text{accumulation points of}
\left\{\frac{\wt\g\wt f^k(\wt x)-\wt\g(\wt
x)}{k}\right\}_{k=1}^\infty \right\} (\text{mod\ } 1)$$

In general, though $\rho(x)$ is independent of the choice of lifts
$\wt f$ of $f$ and $\wt \g$ of $\g$, and of the point $\wt
x\in \pi^{-1}(x)$, it may not be a single real number, but
rather a set, and $\rho(x)$ may vary with $x\in\lb$.  We define
the {\df rotation set}\/ of $\lb$ by
$\rho(\lb):=\bigcup_{x\in\lb}\rho(x)$.
We say the {\df generalized rotation number}\/ of
$x\in \lb$ is $\a$ iff there is a real number $\a$ such
that $\rho(x)=\a$.  Roughly, $\a$ is the average angular displacement
of $x$ under successive iterates of $f$.  We say the {\df
generalized rotation number}\/ of $\lb$ is $\a$ iff for all $x\in
\lb$, $\rho(x)=\a$.  We write $\rho(\lb)=\a$.  In case $\lb$ is
homeomorphic to $S^1$ (by a homeomorphism
$h:(\real^2,\lb)\rightarrow (\real^2,S^1)$), the generalized
rotation number of $f$ on $\lb$ is just the rotation number of the
conjugate $hfh^{-1}$.

One does have the following well-known relationship between the prime
end rotation number and the generalized rotation number:

\proclaim{2.5. Theorem}  Suppose
$f:(\real^2,\lb)\to(\real^2,\lb)$ is a homeomorphism of a cofrontier
$\lb$.  Then the prime end rotation number of $f$ on $\Int\lb$
(respectively, $\Ext\lb$) is the same as the generalized rotation
number of any point of $\lb$ accessible from $\Int\lb$
(respectively, $\Ext\lb$).\endproclaim

\subheading{2.6. Recurrent, almost periodic, and minimal
homeomorphisms}

A homeomorphism $h:X\to X$ is said to be {\df recurrent}\/ iff
for every $\e>0$, there is an $n\in \zed$ such that
$\dist(h^n,\id_X)<\e$.  Equivalently, $h$ is recurrent iff
$\{h^n|n\in \zed\}$ has a subsequence converging uniformly to the
identity.

A homeomorphism $h:X\to X$ is said to be {\df almost periodic}\/ iff
for every $\e>0$, there is a subsequence $\{h^{n_i}|i\in \zed\}$ of
$\{h^n|n\in \zed\}$ with bounded gaps such that for all $i$,
$\dist(h^{n_i},\id_X)<\e$.  On a compact metric space, $h$ is almost
periodic iff $\{h^n|n\in \zed\}$ is an equicontinuous family of
iterates.  Clearly, $h$ periodic implies $h$ almost periodic implies $h$ recurrent.

Recall that the recurrent homeomorphisms of $S^1$ are almost
periodic, and the almost periodic homeomorphisms of $S^1$ are
conjugate to rotations, with the rational rotations being periodic
and the irrational rotations being nonperiodic, almost periodic.  In
particular, recurrent is equivalent to almost periodic on $S^1$.

A homeomorphism $h:X\to X$ is {\df minimal}\/ iff $X$ is the only
nonempty  closed invariant subset of $X$ under $h$.  It follows that
$h:X\to X$ is  minimal iff every orbit is dense.  Recall that an
irrational rotation of $S^1$ is minimal.

In \cite{He}, Hemmingsen shows that topological disks and annuli are
the only plane continua with interior points which (continua) admit
nonperiodic, almost periodic homeomorphisms (i.e. homeomorphisms
whose families of iterates are equicontinuous), and that any such
homeomorphism must be conjugate to an irrational rotation on the
disk or annulus. 
The first two theorems below may be seen as corollaries of
Hemmingsen's theorem, although the first one is very easy to prove
directly.  The third is an observation of Cartwright \cite{Ca}
based on Hemmingsen's results.

\proclaim{2.7. Theorem}  Let
$h:\real^2\to\real^2$ be conjugate to an irrational rotation, and
let $\lb$ be a cofrontier.  If $h(\lb)=\lb$, then $\lb$ is
homeomorphic to $S^1$. \endproclaim

\proclaim{2.8. Theorem}  Let $h$ be an almost
periodic, nonperiodic homeomorphism of $\lb\cup \Int \lb$ onto
itself, where $\lb$ is a cofrontier in $\real^2$.  Then $\lb$ is a
simple closed curve and $h|\lb$ is conjugate to an irrational
rotation.  Thus, the induced homeomorphism on the prime end disk
will also be conjugate to an irrational rotation, and therefore
almost periodic on the circle of prime ends. \endproclaim

\proclaim{2.9. Theorem}  Let
$h:(\real^2,\lb)\to (\real^2,\lb)$ be a homeomorphism, where $\lb$ is
a nondegenerate continuum.  If $h$ is both almost periodic and
minimal on $\lb$, then $\lb$ is a simple closed curve and $h|\lb$ is
conjugate to an irrational rotation. \endproclaim 

\subheading{2.10. Prime ends}  Basic definitions and theory of prime
ends, and further references, may be found in [Br2,P].  Suppose
$U$ is a bounded domain with nondegenerate boundary.  For
convenience, we identify $\real/\zed=\Bd\disk$, and by $\eta$ we
denote both the point $\eta\in\Bd\disk$ and the prime end $\eta$ of
$U$ corresponding to that point.  Suppose $\{Q_i\}_{i=1}^\infty$ is
a chain of crosscuts of $U$ defining $\eta$, and for each $i$, $U_i$
is the complementary subdomain of $U$ cut off by $Q_i$ that contains
$Q_{i+1}$.  We denote the {\df impression\/} of a prime end $\eta$
by $I(\eta)$ and the {\df principal continuum\/} of $\eta$ by
$P(\eta)$. 
  By definition
$$\align
P(\eta)=&\{w\in\real^2\mid \text{$\exists$ chain
$\{Q_i\}_{i=1}^\infty$ of crosscuts of $U$ defining $\eta$ with
$Q_i@>>i> w$}\}\\ I(\eta)=&\bigcap_{i=1}^\infty
\Cl{U_i}=\bigcap_{i=1}^\infty \Bd{U_i},\endalign$$
Note that
$P(\eta)\subset I(\eta)$.

Consider the conditions \roster\item $P(\eta)$ is degenerate. \item
$I(\eta)=P(\eta)$.\endroster  A prime end $\eta$ is of the {\it first,
second, third, or fourth kind}\/ accordingly, as $\eta$ satisfies both
(1) and (2), (1) only, (2) only, or neither (1) nor (2).  A prime end
of the third kind whose impression is all of $\lb$ is said to be a
{\df simple dense canal\/}.

A point $b\in\Bd U$ is called {\it accessible with respect to the domain\/} $U$ iff
there is an endcut $[a,b)$ such that $[a,b)\subset U$ and $\Cl{[a,b)}=[a,b]$.  We
call both the half-open arc $[a,b)$ and the arc $[a,b]$, {\it an arc of
accessibility to\/} $b$.  The point $b\in\Bd U$ is said to be {\it accessible with
respect to the prime end $\eta$ of $U$}\/  iff there exist both a chain of crosscuts
$\{Q_i\}_{i=1}^\infty$ defining $\eta$ and an arc of accessibility $[a,b]$ to $b$
such that $Q_i\cap(a,b)$ is a (necessarily different) singleton for each $i$.  Since
endcuts go to endcuts under a C-map, it follows that the map $\phi$ may be extended
(though not generally continuously) to include the acessible points of $U$.  Thus,
our definition of arc of accessibility is not ambiguous.

We note that if $\eta$ is a prime end of the first or second kind, there is exactly
one principal point.  In each case, this point is both acessible from $U$ and
accessible with respect to $\eta$.  Simple examples show that a point $p$ may be
accessible from $U$, yet be the principal point of two (or more) distinct prime
ends.  For example, let $U$ be the interior of the unit circle in the plane minus the
arc $[0,1]\times\{0\}$.  Then the point $({1\over2},0)$ is accessible from $U$, and is
also accessible with respect to two distinct prime ends, but evidently not by the
same endcut in $U$.

We obtain the following additional theorems about 
pseudorotations in general.

\proclaim{2.11. Theorem  (Periodic pseudorotations)} Suppose $\lb$ is
a cofrontier, $h:(\real^2,\lb)\to(\real^2,\lb)$
is a homeomorphism, and $H:\Cl \disk\to\Cl \disk$ denotes
the induced homeomorphism on either the interior or exterior prime
end compactification.  Then $h|\lb$ is periodic of period $n$ iff $H|\Bd \disk$ is
periodic of period $n$. \endproclaim

\pr Fix $U$, one of the complementary domains of $\lb$.  Assume $h|\lb$ is periodic
of period $n>1$.  Let $b$ be a point of $\lb$ such that $b$ is accessible from $U$
and is of period $n>1$.  Let $K=[a,b)$ be an endcut to $b$ in $U$.  Then $h^n(K)$ is
also an endcut to $b$, since $h^n(b)=b$.  Thus, $K$ and $h^n(K)$ are endcuts to the
same acessible point $b$ in $\Bd U$.  By the proof of Lemma 3.11 of \cite{Br2}, since
$\lb$ is an irreducible separator of $\real^2$, each accessible point of $\Bd U$ is
associated with exactly one prime end of $U$, so that $K$ and $h^n(K)$ define the
same prime end $\eta$ of $U$.  It follows that $\phi(K)$ and $\phi(h^n(K))$ are
endcuts to the same point $b'\in\Bd \disk$.  But
$\phi(h^n(K))=\phi(h^n(\phi^{-1}(\phi(K))))=H^n(\phi(K))$.  Therefore, both
$\phi(K)$ and $H^n(\phi(K))$ are endcuts to the same point $b'$ in $\Bd\disk$.  Thus,
$H^n(b')=b'$.  Since distinct accessible points of $\Bd U$ map to distinct points of
$\Bd\disk$, $b'$ is a periodic point of period $n$ of $\Bd\disk$.  By Theorem 3.7 and
Corollary 3.8 of \cite{Br2}, the set $A$ of points of $\lb$ accessible from $U$ is
dense in $\lb$, while the set $A'$ of points of $\Bd\disk$ corresponding to the
points of $A$ is dense in $\Bd\disk$.  Thus, $H|\Bd\disk$ is periodic of period $n$.

Conversely, assume that $H|\Bd\disk$ is of period $n$, and let $b'\in\Bd\disk$
correspond to an accessible (from $U$) point $b\in\Bd U$.  Let $K$ be an endcut to
$b$.  Then $\phi(K)$ is an endcut to $b'\in\Bd\disk$, and therefore, $H^n(\phi(K))$
is an endcut to $H^n(b')\in\Bd\disk$.  But $H^n(b')=b'$ by hypothesis.  So 
$H^n(\phi(K))$ is also an endcut to $b'$.  Now $h^n(K)\subset\Cl U$ is an endcut to
$h^n(b)$ in $\Bd U$.  But $\phi(h^n(K))=\phi h^n\phi^{-1}(\phi(K))=H^n(\phi(K))$ is
an endcut to $b'$.  So $h^n(K)$ is an endcut to $b$.  That is, $h^n(b)=b$.  Since the
accessible points are dense in $\lb=\Bd U$, $h|\lb$ is periodic of period $n$.
\qd

The following theorem and corollary, which follow from theorems of
Rutt \cite{Ru} on prime ends, show that the fine structure of an
indecomposable cofrontier admitting an irrational pseudorotation
must be quite complicated.  Rogers \cite{Ro1-2} has obtained
theorems similar to Theorem 2.12 and Corollary 2.13 for extendable
intrinsic rotations of a bounded domain.  Related theorems are
proved for quadratic Julia sets in \cite{MR}.

\proclaim{2.12. Theorem (Prime end structure)}  Suppose  $\lb$ is a
cofrontier which admits an irrational pseudorotation.  Then the
following are equivalent:
\roster
\item  $\lb$ is indecomposable.
\item Some prime end of $\Int \lb$ (respectively, $\Ext\lb$) has
as its impression all of $\lb$.
\item There exists a dense subset $\Cal D$ of the
circle of prime ends of $\Int \lb$ (respectively, $\Ext\lb$) such
that for every $\eta\in\Cal D$ the impression of $\eta$ is all of
$\lb$. \item  Every prime end of $\Int \lb$ (respectively,
$\Ext\lb$) has as its impression all of $\lb$.
\item  The set of prime ends of $\Int\lb$ (respectively,
$\Ext\lb$) corresponding to simple dense canals is second category in
the circle of prime ends.
\item Some prime end of $\Int \lb$ (respectively, $\Ext\lb$) 
is a simple dense canal.
\endroster\endproclaim

\pr  We refer to $\Int\lb$ in the proofs below, but the statements
apply to $\Ext\lb$ as well.  Throughout we assume
$\phi:\Int\lb\to\disk$ is the prime end compactification of
$\Int\lb$, $f_\a:(\real^2,\lb)\to(\real^2,\lb)$ is the
irrational pseudorotation, and $F$ the induced homeomorphism on the
circle $S^1$ of prime ends which is conjugate to $R_\a$.

$(1)~\Longrightarrow~(2)$.  Rutt's proofs, and hence theorems
\cite{Ru}, hold for cofrontiers, as well as for the
nonseparating plane continua for which he originally stated the
theorems.  By Rutt's Theorem 1, if $\lb$ is indecomposable, then
some prime end of $\Int\lb$ has as its impression all of $\lb$.

$(2)~\Longrightarrow~(3)$.  Suppose prime end $\eta$ has
$I(\eta)=\lb$.  For all $n\in\zed$, $I(F^n(\eta))=\lb$, since
$f^n$ is a homeomorphism of $\real^2$, and therefore of
$\Cl{\Int{\lb}}$.  Let $\Cal D=\{F^n(\eta) \mid n\in\zed\}$.  Since
$F$ is conjugate to an irrational rotation, $\Cal D$ is dense in
$S^1$.

$(3)~\Longrightarrow~(4)$.  Suppose $\Cal D$ is a set of prime ends
dense in $S^1$ such that $I(\eta)=\lb$ for all $\eta\in\Cal D$. 
Let $\mu$ be a prime end of $\Int\lb$ and $\{Q_i\}_{i=1}^\infty$ a
chain of crosscuts of $\Int\lb$ defining $\mu$.  Fix $i$, and
consider the crosscut $\phi(Q_i)$ of $\disk$.  Since $\Cal D$ is
dense in $\Bd\disk=S^1$, there is an element $\eta_i\in\Cal D$ such
that $\eta_i$ is in the subdomain $\phi(U_i)$ of $\disk$ cut
off by $\phi(Q_i)$ that contains $\phi(Q_{i+1})$.  Consequently, the
corresponding subdomain $U_i$ of $\Int\lb$ cut off by $Q_i$ has
$I(\eta_i)=\lb$ in its boundary.  Since this holds for all $i$,
$I(\mu)=\lb$.

$(4)~\Longrightarrow~(5)$.  Suppose every prime end of $\Int\lb$
has $\lb$ as its impression.  Then $\Int\lb$ contains no prime ends
of the first kind.  By a theorem of Cartwright \cite{P}, the prime
ends of the first and third kinds together form a set of second
category in $S^1$.  Consequently, the set of prime ends
corresponding to simple dense canals is second category in $S^1$.

$(5)~\Longrightarrow~(6)$.  Trivial.

$(6)~\Longrightarrow~(1)$.  By \cite{Ru, Theorem 3}, the
existence of a simple dense canal implies $\lb$ is indecomposable.
\qd 

\proclaim{2.13. Corollary} Suppose $\lb$ is an indecomposable
cofrontier admitting an irrational pseudorotation.  Then each
composant of $\lb$ contains the principal continuum of at most one
prime end of $\Int\lb$ (respectively, $\Ext\lb$).  Further, no composant
can contain the principal continuum of both a prime end, $\eta$, of $\Int\lb$ and
another prime end, $\gamma$, of $\Ext\lb$. \endproclaim

\pr  Let $U$ be one of the complementary domains of $\lb$, and suppose $K$ is a
composant of $\lb$ such that $K\supset X\cup Y$, where $X$ and $Y$ are the
principal continua of the distinct prime ends $\eta$ and $\gamma$, respectively, of
$U$.  Let $Z$ be the irreducible continuum containing $X\cup Y$; note that $Z\subset
K$.  Let $\pi:(\real^2,\lb,Z)\to(\real^2,\lb,p)$ be the quotient map of $\real^2$
taking $Z$ to the point $p$.

We first note that by Theorem 2.12 above and by Iliadis' theorem (see \cite{Br1}), no
composant contains more than one accessible point; for otherwise, there would exist
prime ends whose impresions were proper in $\lb$.  Thus, it follows that any
crosscut in a chain of crosscuts defining $\eta$ (or $\gamma$) must have its
endpoints outside $K$.  It follows from this, that chains of crosscuts defining
different prime ends must go to chains of crosscuts defining different prime ends
under $\pi$.

Let $U=\Int\lb$. Since $\pi$ is monotone,
$\pi(\lb)$ is an irreducible separator of $\real^2$, so $\pi(U)=\Int{\lb/Z}$.  Let $A$
and $B$ be rays defining the prime ends $\eta$ and $\gamma$, respectively, such that
$A$ converges to $X$ and $B$ converges to $Y$.  Then $\pi(A)$ and $\pi(B)$ are each
endcuts to $p$, with $\pi(A)\cap\pi(B)=\emptyset$.  Further, $\pi(A)$ and $\pi(B)$
represent different prime ends of $\pi(U)$, since $A$ and $B$ represent different
prime ends of $U$.  This follows from the remarks in the above paragraph.  However,
$\pi(A)$ and $\pi(B)$ are endcuts to the same point, $p$, of $\pi(\lb)$.  As before,
it follows from the proof of Lemma 3.11 of \cite{Br2} that each accessible point of
$\pi(\lb)$ corresponds to exactly one prime end of $\pi(U)$.  This is a contradiction,
and the first part of the theorem is proved.

For the second part, let $U$ and $V$ be the complementary domains of $\lb$, and
suppose $K$ is a composant of $\lb$ such that $K\supset X\cup Y$, where $X$ and $Y$
are the principal continua of the distinct prime ends $\eta$ of $U$ and $\gamma$ of
$V$, respectively.  Let $Z$ be the irreducible continuum containing $X\cup Y$; note
that $Z\subset K$.  As before, let $\pi:(\real^2,\lb,Z)\to(\real^2,\lb,p)$ be the
quotient map of $\real^2$ taking $Z$ to the point $p$.  Let $A$ and $B$ be rays
defining the prime ends $\eta$ and $\gamma$, respectively, such that $A$ converges to
$X$ and $B$ converges to $Y$.  Then $\pi(A)$ and $\pi(B)$ are each endcuts to $p$
from the disjoint domains $\pi(U)$ and $\pi(V)$ in $\real^2$.  Thus, $p$
is a local cut point of $\pi(\lb)$.

Now, $\lb$ admits an irrational pseudorotation, $h$.  Thus, under $h$, no prime end
of either complementary domain is fixed.  Therefore, $H(\eta)$ and $H(\gamma)$ are
distinct from $\eta$ and $\gamma$, respectively.  It follows from the first part of
the corollary that $X$ and $h(X)$ (respectively, $Y$ and $h(Y)$) lie in different
composants of $\lb$.  Thus, $h(Z)$ is a proper subcontinuum of $\lb$, lying in a
different composant from the one containing $Z$.

We now let $\theta$ be a monotone map from $(\real^2,\lb,p,h(Z))\to(\real^2,\lb,p,q)$
taking $h(Z)$ to the point $q$.  Then $\theta\circ\pi$ is one-to-one everywhere, except
that it takes $Z$ to $p$ and $h(Z)$ to $q$.  Since $p$ is a local cut point of
$\pi(\lb)$, both $p$ and $q$ are local cut points of $W=\theta(\pi(\lb))$.  But $W$
must be an indecomposable continuum, since it is the monotone image of an
indecomposable continuum.  However, $W$ contains two local cut points, each leading
from $\theta(\pi(U))$ to $\theta(\pi(V))$.  Thus, $W$ is decomposable.  This
contradiction proves the second part of the corrolary.
\qd  

\heading 3. Construction of the family $\Cal H$ of cofrontiers
\endheading \nopagebreak

We are interested in looking at a class $\Cal H$ of cofrontiers,
including indecomposable cofrontiers, which are constructed in a
regular fashion, and which are a potential source of examples and
counterexamples.

\subheading{3.1. Outline}  A cofrontier $\lb\in \Cal H$ is constructed as the
intersection $\lb=\bigcap_{k=1}^\infty A_k$ of annuli $A_k$, where
$A_{k+1}$ is embedded essentially in $A_k$ and is invariant under
certain topological rotations of $A_1,\dots,A_k$.  Refer to Figure 2 for an 
example of
the construction outlined below.
\roster
\item  $A_k$ is divided into $q_k$  {\it links\/} (closed topological disks) of
diameter $\le\e_k$.
\item  The embedding of $A_{k+1}$ in $A_k$ is induced by lifting an
embedding of an annulus $\bar A_{k+1}$ in a \lq\lq single link"
annulus $A$ (where $A$ is the image of $A_k$ under a $q_k$--to--one 
covering map).
\item  Each segment of $A_{k+1}$ that {\df spans} (that is,
extends across) a link of $A_k$ is divided into $r_{k+1}>1$ links.  This can be 
done
via the lift.  (In the example, $r_{k+1}=2$.)
\item  A {\df \bru} of $A_{k+1}$ spans some number, say
$s_{k+1}$, of links of $A_k$.  (In the example, $s_{k+1}=2$, and a \bru\  is
high-lighted.) \item  The \lq\lq fractional part" of $A_k$ that a \bru\ of
$A_{k+1}$ spans is $c_{k+1}=\frac{s_{k+1}}{q_k}$. (In the example,
$c_{k+1}=\frac28=\frac14$.) \endroster
See the first paragraph of Section 4 for the significance of the $s_k$'s and the
$c_k$'s.

\figtwo

\subheading{3.2. Details} We will want to draw some conclusions about how 
rapidly $q_k$
grows relative to choices for $r_j$, $s_j$, and $c_j$ for $j<k$.  We also want to
make sure that $A_{k+1}$ is invariant under certain topological rotations of
$A_1,\dots,A_k$.  To this
end, we need to describe the above construction more precisely.  We will 
impose certain
restrictions on $r_k$, $s_k$, and $c_k$, for all $k$.

We suppose each annulus $A_k$ is divided into a number of {\df
links}\/ so that it is a circular chain of closed topological disks, which meet
adjacent links in a separating arc of their union (as in Figure 2, where 
$q_k=8$). We
use $A_k$ somewhat ambiguously to denote both the annulus and the 
collection of links
whose union is the annulus.

Let $q_1$ denote
the number of links of $A_1$.  We suppose that $A_1$ is $S^1\times [0,1]$ 
embedded as
the annulus between the circle of diameter 1/2 and the circle of diameter 1
centered at the origin, and that $q_1>1$.  Let $g_1$ be a periodic homeomorphism of
$A_1$ of period $q_1$ that carries links to links, moving forward
one link.  (At this stage we may take $g_1$ to be a rigid rotation
of $A_1$.)  Let $A$ denote the annulus $S^1\times [0,1]$ with $T$ the arc
$\{0\}\times [0,1]$ in $A$.  We suppose $A_1$ is mapped onto $A$ by a
$q_1$-to-one covering map $h_1$  that takes each component of the intrinsic
boundary of each link of $A_1$ onto the arc $T$, and maps the outer boundary
component of $A_1$ to the  outer boundary component of $A$.  We require that
$h_1=h_1g_1$.  (At this stage we may take $h_1$ to be 
$\exp(q_1z)\times\id$.)

The embedding of $A_2$ in $A_1$ is induced by lifting an
embedding of an annulus $\bar A_2$ in $A$ whose algebraic
intersection number with $T$ is 1. Choose consistent orientations
on $\bar A_2$ and $A$.  For simplicity, we suppose $\bar A_2$
wraps forward some number of times around the hole in $A$ spiraling
in toward the inner boundary of $A$, and wraps backward one less
time spiraling toward the outer boundary of $A$, with no extra
bending.  (In other constructions, we may want to allow more
bending in order to permit, for example, the pseudocircle to be one
of our cofrontiers.)  Since the algebraic intersection number of
$\bar A_2$ with $T$ is 1, $\bar A_2$ is a retract of $A$ by
a map which is homotopic to the identity.

Let $c_2$ denote the
ratio of the number $s_2$ of forward wraps of $\bar A_2$ in
$A$ to the number $q_1$  of links of $A_1$.  If it were the case
that $s_2=c_2q_1=1$, then $\bar A_2$ would go once around
$A$ with no bending.  In our construction, we require that
$s_2>1$ and that $c_2\le 1$.  We lift $\bar
A_2$ via covering map $h_1$ to define $A_2$ contained in $A_1$.  Thus, 
$A_2$ is symmetric with respect to the previously designated topological rotations of
$A_1$ generated by $g_1$.

Observe that a \lq\lq
stretched out copy" of $\bar A_2$ will have been repeated $q_1$ times, each 
link of $A_1$ containing the first link of such a copy, which we call a {\df
\bru\/} of $A_2$, and each \bru\ will itself {\it span\/} (that is, extend
through) $s_2=c_2q_1$  links of
$A_1$, turning around in the next link, and returning to the other boundary 
of the link of $A_1$ in which it started, turning around again to begin the
next  basic repetitive unit.  A \bru\ is outlined more heavily
in Figure 2.  Thus, $c_2$ is the {\df fractional part\/} of
$A_1$ that a \bru\ of $A_2$ {\df spans}.

We subdivide $A_2$ into links by first subdividing $\bar A_2$
and lifting this subdivision.  Thus, the links of $A_2$ are invariant
under $g_1$.  Let $r_2$ denote the number of links into which each full wrap
(forward or backward) of $\bar A_2$ is subdivided. We suppose the two bends
where the forward and backward wraps of $\bar A_2$ join consist of just one
link each.  In our construction, we require that $r_2>1$.
We show below how to compute $q_2$ from
$c_2$, $r_2$, and $q_1$.  For now, suppose $q_2=n_2q_1$.

The distinction that arises in embedding $A_3$ in $A_2$, and which we will 
have to carry forward inductively, is that we want $A_3$ to be invariant, not
only  under certain topological rotations of $A_2$ that carry links to links,
but also under  the topological rotations of $A_1$ generated by $g_1$.  To
this end, let $g_2$ be  a periodic homeomorphism of $A_2$ carrying links to
links, moving forward one link, and which generates a group (of
cardinality $q_2$) of topological rotations of $A_2$ that commutes with
$g_1$ on $A_2$. This can be done by the Lemma below.   As a result,
$g_1|A_2$ will be an element of the group of topological rotations of $A_2$,
namely $g_2^{n_2}$.  Choose a $q_2$-to-one  covering map $h_2:A_2\to A$ so
that it commutes with the previously designated  topological rotation
$g_2$ of $A_2$, and, therefore, with the topological rotation $g_1$ of $A_1$.  Then
$A_3\subset A_2$  may be defined by lifting, via $h_2$, an embedding of an annulus
$\bar A_3\subset A$ whose algebraic  intersection number with $T$ is 1; simultaneously,
$s_3$, $c_3$, and $r_3$ are defined in a similar manner to the above.  In particular,
$A_3$ will be  invariant  under the designated topological rotations of
$A_2$ and $A_1$.

In general, given $A_k$ invariant under the topological rotations $g_1, \dots 
, g_{k-1}$, of ${A_1,\dots,} A_{k-1}$, with $q_1,\dots,q_k$, $s_2,\dots,s_k$,
$c_2,\dots,c_k$,   $r_2,\dots,r_k$, and covering maps $h_1,\dots,h_{k-1}$
defined as above, we  choose a periodic generator $g_k$ of a group (of
cardinality $q_k$) of topological  rotations of $A_k$ carrying links to
links, moving forward one link, so as to commute with the
topological rotations $g_1, \dots , g_{k-1}$ of $A_1,\dots,A_{k-1}$,
respectively.  We can  do this by the Lemma below.  Then we choose a
$q_k$-to-one covering map $h_k:A_k\to A$, so as to commute with the
topological rotations $g_1, \dots ,  g_k$ of $A_1,\dots,A_{k-1},A_k$.  The
embedding of $A_{k+1}$ in $A_k$ is induced  by lifting, via $h_k$, an
embedding of an annulus $\bar A_{k+1}$ in $A$ whose  algebraic intersection
number with $T$ is 1. Let $c_{k+1}$ denote the ratio of the number 
$s_{k+1}$ of forward wraps of $\bar A_{k+1}$ in $A$ to the number $q_k$  of
links of  $A_k$.    We lift $\bar A_{k+1}$ via the covering map $h_k$ to
define $A_{k+1}$  contained in $A_k$.  Thus, $A_{k+1}$ is symmetric with
respect to the topological  rotations $g_1, \dots , g_k$ of $A_1,\dots,A_k$,
respectively.  A \bru\  of  $A_{k+1}$  will have been repeated $q_k$ times,
each link of $A_k$ containing the first link of  a \bru\ which will span
$s_{k+1}=c_{k+1}q_k$ links of $A_k$, turning around in the  $s_{k+1}+1$
link, and returning to the other boundary of the link of $A_k$ in which it 
started, turning around again to begin the next basic repetitive unit.  We
subdivide  $A_{k+1}$ into links by first subdividing $\bar A_{k+1}$ and
lifting this subdivision.   Let $r_{k+1}$ denote the number of links into
which each full wrap (forward or  backward) of $\bar A_{k+1}$ is subdivided. 
In our construction, we require that  $r_{k+1}>1$,
$s_{k+1}>1$, and $c_{k+1}\le 1$, for all k. 

Since $s_{k+1}=c_{k+1}q_k>1$, and since there are a total of
$2c_{k+1}q_k-1$ full wraps in $\bar A_{k+1}$, each wrap having been
subdivided into $r_{k+1}$ links, plus two bend links, the number of
links of $A_{k+1}$ is given by $$\align q_{k+1} 
&=(2r_{k+1}c_{k+1}q_k-r_{k+1}+2)q_k \\ 
&=(r_{k+1}+2)c_{k+1}q^2_k+(r_{k+1}-2)q_k(c_{k+1}q_k-1)
\endalign$$ Since $c_{k+1}q_k$ is a positive integer, and
$r_{k+1}>1$, the second term in the line above is
nonnegative.  Therefore, we have for all k, $$q_{k+1}\ge
(r_{k+1}+2)c_{k+1}q^2_k \tag1$$ Note that $q_k$
divides $q_{k+1}$, and that it takes $r_{k+1}$ links of $A_{k+1}$ to
{\df span\/} a link (that is, go from one boundary component to the
opposite boundary component of the link) of $A_k$.

Let $\e_k$ denote the maximum diameter of a link of $A_k$. 
We may assume that $\e_1\le\frac{2\pi}{q_1}$, and that $\e_k$
is on the close order of $$\e_1(\prod_{i=2}^k r_i)^{-1}$$
Since $r_k>1$, we may assume that the
$\e_k$'s are summable, with sum at most 1.

The intersection 
$$\lb=\bigcap_{k=1}^\infty A_k$$
of the annuli described above is a cofrontier.  Depending upon the
choices we make for $c_{k+1}$ and $r_{k+1}$, $\lb$ may be
homeomorphic to the circle $S^1$ (as we show in Theorem 4.2),
indecomposable (as we show in Theorem 4.1), or possibly something
in between (as we ask in Question 6.4).  If we allow more bending,
$\lb$ can be made hereditarily indecomposable, i.e. the pseudocircle.  

The necessary lemma that we use above is the following. 

\proclaim{3.2.1. Lemma}

\roster

\item  Suppose $A_1$, $A_2$, and $g_1$ are given as above, with $A_1$ and 
$A_2$ invariant under the period $q_1$ homeomorphism $g_1$ of $A_1$ onto
itself, taking  outer boundary  of $A_1$ to outer boundary of $A_1$,
intrinsic boundary segments to such  segments, and  links of $A_i$ to links
of $A_i$ for $i=1,2$.  Assume that each \bru\ of $A_2$ has  $n_2$ links, where
$q_2=n_2q_1$.  Then there exists a period $q_2$  homeomorphism $g_2$ of  $A_2$
carrying links of $A_2$ to links of $A_2$, moving forward one link, such that $g_2$
generates a  cyclic group of topological rotations of $A_2$ which commute with the
topological rotations  of $A_1$ generated by $g_1$.

\item   Suppose $A_1, \dots , A_k$, and $g_1, \dots , g_{k-1}$ are given as 
above, with $A_k$ invariant under the topological rotations $g_1, \dots ,
g_{k-1}$ of  periods $q_1, \dots , q_{k-1}$, respectively, respecting the
boundaries appropriately.   Then there exists a period $q_k=n_kq_{k-1}$
homeomorphism $g_k$ of  $A_k$ carrying links of $A_k$ to links of $A_k$, moving
forward one link, such that $g_k$ generates a  cyclic group of topological rotations of
$A_k$ which commute with the topological rotations  of $A_i$ generated by $g_i$, for
$i = 1, \dots, k-1$.

\endroster\endproclaim

{\it Proof of (1).\quad} Let $R_1$ be the period $q_1$ clockwise rotation of $A$
onto itself through angle ${{2\pi}\over{q_1}}$.  Let $\phi_2 : A_2 \to A$ be a
homeomorphism of $A_2$  onto $A$, taking outer boundary onto outer boundary,
inner boundary onto inner  boundary, and  intrinsic boundary segments of
links onto the iterates of ${0} \times [0,1]$  under the period $q_2$
clockwise rotation $R_2$ of $A$ onto itself through angle
${{2\pi}\over{q_2}}$. We may choose $\phi_2$ so that $R_1\phi_2= \phi_2g_1$
on $A_2$. Name the links of $A_2$, in clockwise order and in chunks
consisting of the  basic  repetitive units, as $a_{2,1,1}, \dots ,
a_{2,q_1,n_2}$, where $a_{i,j,k}$ is the  $k$th  link of the $j$th basic
repetitive unit of $A_i$.

Define $g_2$ on the $j$th \bru\ of $A_2$, $1 \leq j \leq q_1$, as follows:
on link $a_{2,j,k}$, for $1 \leq
k \leq n_2 - 1$, let $g_2 = \phi_2^{-1}R_2\phi_2$; on link $a_{2,j,n_2}$, let
$g_2 = \phi_2^{-1}R_1R_2^{-(n_2- 1)}\phi_2 =
g_1\phi_2^{-1}R_2^{-(n_2- 1)}\phi_2$.  The last equality follows because
$\phi_2$ conjugates $g_1|A_2$ to $R_1$ on $A$.
In this way, $g_2$ is not only periodic of period $q_2$ on $A_2$, but 
$g_2^{n_2} = g_1|A_2$ and $g_2$ commutes with $g_1$ on $A_2$.

{\it Proof of (2).\quad}  At the next stage, we need only insure that
$g_3^{n_3} =  g_2|A_3$ and that $g_3$ commutes with $g_2$ on $A_3$ in our
construction.  It will  automatically follow that $g_3$ also commutes with
$g_1$. We leave it to the reader to complete the induction. 
\qed

We use the following version of a well-known theorem of Cook and
Ingram \cite{CI} to detect indecomposability in our examples:

\proclaim{3.3. Theorem} 
Let $\lb=\bigcap{}_{k=1}^\infty A_k$ be a cofrontier constructed as
above.  Then $\lb$ is indecomposable iff there is a subsequence
$\{A_{k_j}\}_{j=1}^\infty$ such that \roster\item"{(*)}" for each
pair of links $a,b \in A_{k_{j+1}}$, at least one of the maximally
coherent subcollections of $A_{k_{j+1}}-\{a,b\}$ meets every link of
$A_{k_j}$.\endroster  \endproclaim

If $A_j$ in $A_i$ satisfies condition (*) of Theorem 3.3, we say that
$A_j$ is {\df folded\/} in $A_i$.

\subheading{3.4. Inducing pseudorotations on $\lb\in\Cal H$}
We induce a pseudorotation $f_\a$ on $\lb\in\Cal H$ as the uniform limit of
topological rotations $f_k$ of the annuli $A_k$.  The existence of the 
required
topological rotations has been established in our construction of $\lb$
in  Section
3.2.  Refer to Figure 3 for an example.

\roster
\item  Homeomorphism $f_1=g_1^{m_1}$ rotates $A_1$ onto itself by the
fractional amount $\frac{m_1}{q_1}$, taking links to links, and
leaving $A_2,A_3,\dots$ invariant, so $f_1|\lb$ is a homeomorphism. \item       
Homeomorphism $f_2$ first does $f_1$, then rotates $A_2$ onto itself by the 
amount $\frac{m_2}{q_2}$ within its own annular structure, taking links to
links, and leaving $A_3,A_4,\dots$ invariant.  Thus, $f_2=g_2^{m_2}f_1=
g_2^{m_2}g_1^{m_1}$ rotates $A_2$ the total amount
$\frac{m_1}{q_1}+\frac{m_2}{q_2}$, and is a homeomorphism  on $\lb$.
\item  In general, homeomorphism $f_k$ first does $f_{k-1}$, then
rotates $A_k$ an additional amount $\frac{m_k}{q_k}$ within its own
annular structure leaving $A_{k+1},A_{k+2},\dots$ invariant.
Thus, $f_k=g_k^{m_k}f_{k-1}=g_k^{m_k}g_{k-1}^{m_{k-1}}\dots g_1^{m_1}$ rotates
$A_k$ the total amount $$\frac{p_k}{q_k}=\sum_{j=1}^k \frac{m_j}{q_j}$$ and
is a homeomorphism on $\lb$. \item  Assume that the $m_k$'s are not too
large,  i.e.
$m_k<r_k$, and recall that the $\e_k$'s are summable.  In the limit,
$$\frac{p_k}{q_k}\to\a=\sum_{j=1}^\infty \frac{m_j}{q_j}$$
and the homeomorphisms $f_k$ converge uniformly to the pseudorotation
$f_\a$ on $\lb$. \endroster
We call any homeomorphism on $\lb$, induced via uniform limits of the
designated (in Section 3.2) topological rotations of the $A_k$'s as above, a 
{\df
standard}\/ pseudorotation.

It is easy to extend $f_\alpha$ to a homeomorphism of $\real^2$.   We
provide only a sketch (for more details see \cite{Ha}).  Since $f_1$ is a
rigid rotation of $A_1$, we may extend it to the corresponding rotation of
$\real^2-\Int{A_1}$.  On $A_1-\Int{A_2}$, we may continuously deform
$f_1|\Bd{A_1}$ to $f_2|\Bd{A_2}$ by using an embedding of an isotopy
between appropriate rigid rotations of $S^1$.  Proceeding inward, we deform
$f_2$ to $f_3$ as we move in toward the boundary of $A_3$.  This procedure
can be continued inductively so as to define a homeomorphism on $\real^2$
whose restriction to $\lb$ is $f_\alpha$.  Note that this extension of
$f_\alpha$ is not topologically conjugate to a rotation because of the
differing amounts of rotation on the boundaries of the $A_k$'s.  However,
this extension does induce a topological rotation on the inner and outer
circles of prime ends.  (To see this, one may construct the prime end
uniformizations stage-by-stage in tandem with the
extension of $f_\alpha$ to $\real^2$ so that the topological rotations on
the circles of prime ends are each a limit of topological
rotations on circles converging to the boundary of the unit disk.  We recall for the
reader that a prime end uniformization need not be conformal, but only a C-map
\cite{UY}.)

\figthree

\heading 4. Properties of cofrontiers $\lb\in\Cal H$ \endheading
\nopagebreak

The \lq\lq length" of {\bru}s determines whether $\lb$ is
indecomposable or homeomorphic to $S^1$.  There are two measures of
this length: $s_{k+1}$, the absolute number of links of $A_k$ that a
\bru\ of $A_{k+1}$ spans, and $c_{k+1}$, the fractional part of $A_k$
that a \bru\ of $A_{k+1}$ spans.  We made certain assumptions about
various numbers in the construction, of which we remind the reader:
$q_1>1$, $c_{k+1} \le 1$, $c_{k+1}q_k=s_{k+1}>1$, and $r_{k+1}>1$.

\proclaim{4.1. Theorem  ($\lb$ indecomposable)} 
If $\sum_{k=2}^\infty c_k=\infty$,
then $\lb$ is indecomposable.\endproclaim

\pr  It suffices to show that for all $i$, there is a $j$ such that
$A_j$ is folded in $A_i$.  Without loss of generality, suppose
$i=1$.  For all $k\ge 1$, $c_{k+1}q_k>1$.
Let $D_k$ be a chain of $dq_k$ links of $A_k$, where $\frac1{q_k}\le
d\le 1$, where the fraction $d$ is an integral multiple of
$\frac1{q_k}$.  We call $D_k$ the {\df d (fractional) part}\/ of
$A_k$.

\figfour

{\bf Claim:}  We need at least one \bru, and at most the
first $(d-c_{k+1})q_k$, {\bru}s of $A_{k+1}$ to
meet the $d$ part of $A_k$.

{\bf Proof of claim:}  Refer to Figure 4, where we illustrate the
following argument with $d=\frac{3}{4}$ and $c_{k+1}=\frac{1}{4}$. 
Since $c_{k+1}q_k>1$, each \bru\ of $A_{k+1}$ meets $c_{k+1}q_k+1$
links of $A_k$ (one more than it spans).  Hence, the \bru\ of
$A_{k+1}$ which begins in link number $(d-c_{k+1})q_k$ of $D_k$
meets the last $c_{k+1}q_k+1$ links of $D_k$.  The \bru\ which
begins in link $m$ of $D_k$ (for all $m<(d-c_{k+1})q_k$) certainly
meets the link in which it starts.  Therefore, the first
$(d-c_{k+1})q_k$, basic repetitive units of $A_{k+1}$ meet the
entire $d$ part of $A_k$.

Suppose $d=1$ and $k=1$; it follows that we need at least one, and at
most $(1-c_2)q_1$ consecutive {\bru}s of $A_2$ in
order to meet all the links of $A_1$.  It follows by induction with
$d=1-c_2-c_3-\dots -c_k$ that for each $k>1$, we need at least one,
and at most $(1-c_2-c_3-\dots -c_{k+1})q_k$, consecutive basic
repetitive units of $A_{k+1}$ in order to meet $(1-c_2-c_3-\dots
-c_k)q_k$ consecutive links of $A_k$.  Consequently, we need at most
$(1-c_2-c_3-\dots -c_{k+1})q_k$ consecutive {\bru}s
of $A_{k+1}$ in order to meet all of $A_1$.  Since the $c_k$'s are
not summable, there is a $j$ such that $\sum_{i=2}^j c_i\ge
\frac12$.  Therefore, we need at most half of the {\bru}s of
$A_j$ to meet every link of $A_1$.  It follows by the symmetry of
$A_j$ in $A_1$, that $A_j$ is folded in $A_1$. \qd

\figfive

\proclaim{4.2. Theorem ($\lb\cong S^1$)}  If the\/ $s_k$'s are
uniformly bounded above, then $\lb$ is homeomorphic to
$S^1$.\endproclaim

\pr  Refer to Figure 5, which illustrates the proof below.  In that diagram, two
basic repetitive units of $A_2$ in $A_1$ are shown, as are three basic repetitive
units of $A_3$ in $A_2$.  Note that a basic repetitive unit of $A_3$ spans two links
of $A_2$, but meets three links of $A_2$.  Note also that a basic repetitive unit of
$A_2$ contains 8 links.

Let $B_i$ be a sequence of concentric, circular chains, with $B_{i+1}$ contained in
$B_i$ as $B_2$ is contained in $B_1$ in Figure 5.  In particular, if there are
$t_{i+1}$ links in a basic repetitive unit of $A_{i+1}$, then each link of $B_i$
contains $t_{i+1}$ links of $B_{i+1}$.

Let $g_1:A_1\to B_1$ be a one-to-one, order-preserving correspondence between the
links.  Let $g_{i+1}:A_{i+1}\to B_{i+1}$ be a one-to-one, order-preserving function
from the links of $A_{i+1}$ to the links of $B_{i+1}$, taking the links of a basic
repetitive unit of $A_{i+1}$ to the links of $B_{i+1}$ which refine a link of $B_i$. 
Since basic repetitive units begin at one side of a link of the previous stage and
end at the other side (see Figure 5 again), we may preserve containment with respect
to the first and last links of each basic repetitive unit, and straigten out the
intermediate links so that they correspond to a straight chain refinement of $B_{i+1}$
in $B_i$.

In the standard manner, every $x\in \lb$ can be written as the intersection of a tower
of links (or consolidations of links) of the $A_i$'s with diameters tending to zero. 
Then let $g(x)$ be defined as the intersection of the tower of links of the $B_i$'s
corresponding under the $g_i$'s.  Clearly, $g$ is one-to-one.

To see that $g$ is also continuous (and thus a homeomorphism), let $m$ be an upper
bound for the $s_k$'s.  Thus for all $k$, a basic repetitive unit of $A_{k+1}$ spans no
more than $m$ links of $A_k$.  Recall that each $r_k\ge 2$, so that each segment of
$A_{k+1}$ that spans a link of $A_k$ is divided into at least two links.  There is an
integer k such that $2^k>(m+1)$.  Thus, a basic repetitive unit of $A_{2^k+1}$
refines a single link of $A_1$ (and perhaps meets one neighbor), so that all
subsequent ``straightening-out" of basic repetitive units is done to sets of diameter
less than twice the diameter of links of $A_1$.

Let $A_{t_1}=A_1$, and let $A_{t_2}=A_{2^k+1}$.  Similarly, we can find $A_{t_3}$
such that a basic repetitive unit of $A_{t_3}$ refines a single link of $A_{t_2}$. 
As before, a basic repetitive unit of $A_{t_3}$ refines a single link of $A_{t_2}$
(and perhaps meets one neighbor), so that all subsequent ``straightening-out" of basic
repetitive units is done to sets of diameter less than twice the diameter of links of
$A_{t_2}$.  This process may be continued inductively to see that, for any
$\epsilon>0$, there is a stage $A_{t_j}$, so that from that point on,
``straightening-out" is done only to sets of diameter less than $\epsilon$.  Thus,
$g$ is continuous, and the theorem follows.
\qd

\subhead 4.3. Remark \endsubhead We note that in Theorem 5.1 below,
we prove that every $\lb\in\Cal H$ admits an irrational
pseudorotation.  Thus it follows that {\df for every indecomposable
member of $\Cal H$, the conclusions of Theorem 2.12 hold.}

\heading 5. Properties of standard pseudorotations \endheading
\nopagebreak

Assume $f_\a$ is a standard pseudorotation induced on $\lb\in\Cal H$.  We made 
assumptions about various numbers in the construction, of which we
remind the reader: $q_1>1$, $c_k \le 1$, $c_{k+1}q_k=s_{k+1}>1$,
$r_k>1$, the $\e_k$'s are summable, and $0\le m_k<r_k$.   We obtain
the following theorems detailing some of the dynamical properties of
these standard pseudorotations.

\proclaim{5.1. Theorem ($\a$ irrational)}  Assume $f_\a$ is a 
standard pseudorotation on $\lb\in\Cal H$. If $\lb$ is not
homeomorphic to $S^1$ (e.g., $\lb$ is indecomposable) and for
infinitely many values of $k$, $m_k>0$, then $\a$ is
irrational.  Thus every $\lb\in\Cal H$ admits an irrational
pseudorotation.\endproclaim

\pr  Recall that by construction, we have inequality 3.2(1)
$q_{k+1}\ge (r_{k+1}+2)c_{k+1}q^2_k$; hence, $q_k$ increases to
$\infty$.  Suppose by way of contradiction that
$$\a=\lim \frac{p_k}{q_k}=\sum_{k=1}^\infty
\frac{m_k}{q_k}=\frac{p}{q}$$
is rational. 
Then, we have for each $k$,
$$\frac1{qq_k}\le
\frac{p}{q}-\frac{p_k}{q_k}=\sum_{j=k+1}^\infty
\frac{m_j}{q_j}\le\sum_{j=k+1}^\infty\frac{m_j}{(r_j+2)c_jq_{j-1}^2}$$
Since $0\le m_j<r_j$ for all $j$, this becomes
$$\frac1{qq_k}\le\sum_{j=k+1}^\infty\frac1{c_jq_{j-1}^2}$$

Since for infinitely many values of $k$, $m_k>0$, the above
inequality holds for all k.  We proceed to expand the
right-hand-side of the inequality as follows: $$\align
\frac1{qq_k}
&\le\frac1{c_{k+1}q_k^2}+\frac1{c_{k+2}q_{k+1}^2}+
\frac1{c_{k+3}q_{k+2}^2}+\dots \\
&\le\frac1{(c_{k+1}q_k)q_k}+\frac1{(c_{k+2}q_{k+1})q_{k+1}}
+\frac1{(c_{k+3}q_{k+2})q_{k+2}}+\dots \\
&\le\frac1{(c_{k+1}q_k)q_k}+\frac1{(c_{k+2}q_{k+1})
(r_{k+1}+2)(c_{k+1}q_k)q_k} \\
&\quad\quad+\frac1{(c_{k+3}q_{k+2})(r_{k+2}+2)(c_{k+2}q_{k+1})
(r_{k+1}+2)(c_{k+1}q_k)q_k}+\dots
\endalign$$
Therefore, multiplying by $q_k$, factoring out $\frac1{c_{k+1}q_k}$,
and simplifying (using $c_jq_{j-1}>1$ and $r_j>1$, for all $j$),
we have $$\frac1{q} \le \frac1{c_{k+1}q_k}\left
(1+\frac12+\frac14+\dots\right)
\le \frac2{c_{k+1}q_k}$$
Since $c_{k+1}q_k=s_{k+1}$, and since $\lb$ not homeomorphic to
$S^1$ implies $\limsup s_{k+1}=\infty$ by Theorem 4.2, it
follows that this cannot hold for all $k$.  Therefore, $\a$ is
irrational. \qd

If the number of subdivisions of a spanning segment is
large, and the incremental rotations ($\frac{m_k}{q_k}$) are small,
then $f_\a$ is recurrent on $\lb$, and $\lb$ has a well-defined
generalized rotation number, for any $\lb$ in $\Cal H$ (whether $\lb$
is indecomposable or not).  More precisely, 

\proclaim{5.2. Theorem ($f_\a$ recurrent)}  Assume $f_\a$ is a 
standard pseudorotation on $\lb\in\Cal H$. If for some $M>0$ and for
all k, $r_{k+1}\ge q_k$ and $m_k\le M$, then $f_\a|\lb$ is
recurrent.  Moreover, $\rho(\lb)=\a$. \endproclaim

\pr  We first prove a useful lemma.  

\proclaim{5.2.1. Lemma}  The hypotheses of Theorem 5.2 imply that
\roster
\item"{(1)}" there is a
constant $C>0$ such that $d(f^i_{k+1},f^i_k)<C\e_k$ for all $i$ with
$0\le i \le q_k$.
\endroster
Consequently, we have
\roster
\item "{(2)}"
$d(f_\a^i,f_k^i)\le C\sum_{j=k}^\infty \e_j$
for all $i$ with $0\le i \le q_k$.
\endroster
\endproclaim

\pr   First we show condition (1)
implies (2).  By the uniform convergence of $f^i_k$ to $f^i_\a$,
and the triangle inequality, for
each $\beta >0$, there is an $m>0$ such that
$$d(f_\a^i,f_k^i)<\beta+\sum_{j=k}^{k+m} C\e_j
<\beta+\sum_{j=k}^\infty C\e_j$$
Thus, in the limit as $\beta$ goes to $0$,
$$d(f_\a^i,f_k^i)\le\sum_{j=k}^\infty
C\e_j\le C\sum_{j=k}^\infty \e_j$$

That the hypotheses of Theorem 5.2 imply (1) follows from the
construction:  Choose $j$ so that $M<q_j$.  Choose
$C=\max\{M,\e_j^{-1}\}$.  If $k<j$ the conclusion follows
trivially.  So suppose $k\ge j$.  Each iteration of $f_{k+1}$ moves a
link of $A_{k+1}$, $m_{k+1}\le M$ links of $A_{k+1}$ away from what
$f_k$ would.  Since $r_{k+1}\ge q_k$ and $M<q_k$, $f_{k+1}$ moves a
link of $A_{k+1}$ at most 1 link of $A_k$ away in the annular
structure of $A_k$ from what $f_k$ would. In general, $f_{k+1}^i$
moves a link of $A_{k+1}$ at most $\lceil\frac{Mi}{q_k}\rceil$ links
of $A_k$ away in the annular structure of $A_k$.  Since the diameter
of a link of $A_k$ is at most $\e_k$, it follows that for all $i\le
q_k$, $d(f_{k+1}^i,f_k^i)<M\e_k$.  \qd

Now we will show that for all $\e>0$, there is a
positive integer $i$ such that for every $x\in \lb$,
$d(f_\a^i(x),x)<\e$.  Let $C$ be
the constant of Lemma 5.2.1. Since the $\e_k$'s are summable, we
may choose k such that $C\sum_{j=k}^\infty\e_j<\e$.  Let $x\in \lb$. 
Then there is a link $a\in A_k$ such that $x\in a$.  By
construction, $f_k^{q_k}$ is the identity on $A_k$, so
$f_k^{q_k}(x)=x$.  By Lemma 5.2.1, 
$$d(f_\a^{q_k}(x),f_k^{q_k}(x))\le
C\sum_{j=k}^\infty \e_j<\e$$ Consequently, $$d(f_\a^{q_k}(x),x)<\e$$
Hence, $f_\a|\lb$ is recurrent.

We want to show that the generalized rotation number for $\lb$ is
well-defined.  For each k, let $\g_k$ denote the projection of $A_k$
to its core circle $S^1$.  Let $\wt \g_k:\wt A_k@>>>\real^1$ denote
the lift of $\g_k$ to the respective universal covers (with $\pi_k$
and $\theta$ the respective covering projections).  Then
$\wt\lb=\bigcap{}_{k=1}^\infty \wt A_k$ covers $\lb$ under each
projection $\pi_k$.  Let $\wt f_k$ and $\wt f_\a$ denote the lifts of
$f_k$ and $f_\a$, respectively.  Since two lifts of the same map may
differ by an integer, we choose the \lq\lq simplest" lift in each
case.  We want to show that the generalized rotation number at each
point of $\lb$ is $\a$.  It suffices to show that for all $\wt
x\in\wt\lb$ $$\frac{\wt\g_1\wt f_\a^i(\wt x)-\wt\g_1(\wt
x)}{i}@>>i>\a$$

Since $c_k\le 1$ for all $k$, it follows that
$|\wt\g_{k+1}-\wt\g_k|\le 2$.  Consequently,
$$|\wt\g_k-\wt\g_1|\le 2k \tag1$$
Since $f_k$ is, by definition, rotation of $A_k$ by the fractional
amount $\a_k=\frac{p_k}{q_k}$, it follows that
$$\wt\g_k\wt f_k^i(\wt x)-\wt\g_k(\wt x)=i\a_k \tag2$$
Since $q_k\ge 2^k$, it follows by Lemma 5.2.1 that for all
sufficiently large $k$, for all $i$ with $0\le i\le 2^k$,
$$|\wt\g_1\wt f_\a^i-\wt\g_1\wt f_k^i|<1 \tag3$$

{\bf Claim:}  For all sufficiently large i, there exists $k=k(i)$
with $2^{k-1}\le i <2^k$ such that
$$\left| \frac{\wt\g_1\wt f_\a^i(\wt x)-\wt\g_1(\wt x)}{i}-\a\right|
\le |\a-\a_k|+\frac{4k+1}{i} \tag4$$

{\bf Proof of Claim:}  It follows from (2) that
$$|\wt\g_k\,\wt f_k^i(\wt x)-\wt\g_k(\wt x)-i\a|=|i\a_k-i\a|$$
Therefore, by (1) applied twice, and the triangle inequality
$$|\wt\g_1\wt f_k^i(\wt x)-\wt\g_k(\wt x)-i\a|\le |i\a_k-i\a|+2k$$
$$|\wt\g_1\wt f_k^i(\wt x)-\wt\g_1(\wt x)-i\a|\le |i\a_k-i\a|+4k$$
So by (3) and the triangle inequality,
$$|\wt\g_1\wt f_a^i(\wt x)-\wt\g_1(\wt x)-i\a|\le |i\a_k-i\a|+4k+1$$
Then (4) follows by dividing by $i$.

Since $|\a-\a_{k(i)}|@>>i>0$ and
$$\frac{4k+1}{2^k}<\frac{4k+1}{i}\le \frac{4k+1}{2^{k-1}}$$
it follows that the right-hand-side of (4) goes to $0$ as
$i@>>>\infty$.
\qd

\subheading {5.3. Irrationals of constant type}
The irrational\  $\a$ is of {\df constant type}\/ iff
there is an $\e>0$ such that for all rationals $\frac{p}{q}$
in lowest terms, $$|\a-\frac{p}{q}|\ge \frac{\e}{q^2}$$
Thus, in an informal sense, $\a$ is poorly approximated by rationals.

\proclaim{5.4. Theorem ($\a$ not of constant type)}  Assume $f_\a$ is
a  standard pseudorotation on $\lb\in\Cal H$.  If
\roster\item $\lb$ is not homeomorphic to $S^1$, \item for
infinitely many values of $k$, $m_k>0$, \item for some $M>0$ and for
all k, $m_k\le M$, and \item for all $k$, $r_{k+1}\ge
q_k$, \endroster then the irrational $\a$ is not of constant
type; that is, in an informal sense, $\a$ is
well-approximated by rationals.\endproclaim

\pr  We have shown in Theorem 5.1 that $\a$ is
irrational.  Now suppose by way of contradiction that $\a$ is of
constant type.  Then there exists an $\e>0$ such that for every
rational $\frac{p}{q}$, $|\a-\frac{p}{q}|\ge \frac{\e}{q^2}$.  By
hypotheses (2) and (3) and inequality 3.2(1), we have for every k,
$$\frac{\e}{q_k^2}\le \a-\frac{p_k}{q_k}=\sum_{j=k+1}^\infty
\frac{m_j}{q_j}\le M\sum_{j=k+1}^\infty\frac1{q_j}
\le M\sum_{j=k+1}^\infty\frac1{(r_j+2)c_jq_{j-1}^2}$$

Expanding the sum, applying 3.2(1) and hypothesis
(4), and simplifying, we have
$$\align
\frac{\e}{q_k^2}
&\le M\left( \frac1{(r_{k+1}+2)(c_{k+1}q_k^2)}+
\frac1{(r_{k+2}+2)(c_{k+2}q_{k+1}^2)}+
\frac1{(r_{k+3}+2)(c_{k+3}q_{k+2}^2)}+\dots \right) \\
&\le M\biggl(\frac1{(r_{k+1}+2)(c_{k+1}q_k^2)}+
\frac1{(r_{k+2}+2)(c_{k+2}q_{k+1})(r_{k+1}+2)(c_{k+1}q_k^2)} \\
&\quad\quad+\frac1{(r_{k+3}+2)(c_{k+3}q_{k+2})(r_{k+2}+2)
(c_{k+2}q_{k+1})(r_{k+1}+2)(c_{k+1}q_k^2)}+
\dots \biggr) \\
&\le M\left( \frac1{c_{k+1}q_k^3}+
\frac1{2(c_{k+1}q_k^3)}+\frac1{2^2(c_{k+1}q_k^3)}+\dots
\right) \\
&\le \frac{M}{c_{k+1}q_k^3}\left( 1+\frac12+\frac14+\dots \right) \\
\frac{\e}{q_k^2}&\le \frac{2M}{c_{k+1}q_k^3}
\endalign$$

Consequently, multiplying by $q_k^2$ and dividing by $2M$, we have
$$\frac{\e}{2M}\le \frac1{c_{k+1}q_k}$$
Since $c_{k+1}q_k=s_{k+1}$, and since $\limsup s_{k+1}=\infty$ by
hypothesis (1) and Theorem 4.2, it follows that this cannot hold for
all $k$.  Therefore, $\a$ is not of constant type. \qd

The following example serves two purposes: it shows that the
equivalence of almost periodic and recurrent does not hold for all
cofrontiers, and it shows that Theorem 2.11 cannot be extended to
almost periodic.

\proclaim{5.5. Example ($f_\a$ not almost periodic)}  follow the
construction recipe:
\roster \item $q_1=4$, \item $c_{k+1}=\frac12$ for all $k\ge 1$,
and \item $r_{k+1}=q_k$ for all $k\ge 1$, \endroster with minimal
bending.\endproclaim

The cofrontier $\lb\in\Cal H$ obtained will be indecomposable by (2) and Theorem 4.1. 
Induce a pseudorotation $f_\a$ on $\lb$ by choosing $m_k=1$ for all
$k$.  Then by Theorems 5.1-5.3, $f_\a$ is recurrent on $\lb$ and
$\a$ is an irrational not of constant type. By Theorem 5.6 below,
$f_\a$ is minimal on $\lb$.  It then follows by Theorem 2.9 that
$f_\a$ is {\bf not} almost periodic on $\lb$.

\proclaim{5.6. Theorem  ($\lb$ minimal under $f_\a$)}  Suppose
$f_\a$ is a standard pseudorotation of $\lb\in\Cal H$.  If 
$r_{k+1}\ge q_k$ and  $0<m_k\le M$ for all $k$, then $f_\a$ is
minimal on $\lb$.\endproclaim

\pr  
The first condition in Lemma 5.2.1 is the analog of one
of the conditions used in [Ha] to deduce the minimality of
$f_\a$ on $\lb$.  The hypotheses of Theorem 5.6 seem to be the most
direct way of realizing that condition in a specific construction. 
(The other hypothesis in \cite{Ha} used to obtain minimality, that
$f_k$ is transitive on the links of $A_k$, is superfluous, as we
show below in Lemma 5.6.1.)

We will show that $\lb$
is the closure of the orbit of each of its points.  First we need the
following lemma:

\proclaim{5.6.1. Lemma}  The hypotheses of Theorem 5.6 imply
that for sufficiently large $k$, for any link $a\in A_{k+1}$, the
sequence of iterates
$\{(f_{k+1}^{q_k})^s(a)\}_{s=1}^t$, where $t=\frac{q_{k+1}}{q_k}$,
meets every link of $A_{k-1}$. \endproclaim

\pr   Since $q_k@>>>\infty$, we may choose k such that
$M<q_{k-1}$.  Let $a\in A_{k+1}$.  Then $f_{k+1}^{q_k}(a)$ is
at least $q_k$ and at most $Mq_k$ links of $A_{k+1}$ away from its
starting link in the annular structure of $A_k$.  It takes at least
$q_{k-1}$ links of $A_k$ to span a link of $A_{k-1}$, and at least
$q_k$ links of $A_{k+1}$ to span a link of $A_k$.  Since
$Mq_k<q_{k-1}q_k$, $f_{k+1}^{q_k}(a)$ is within one link of its
starting link in $A_{k-1}$.  Recall that $f_{k+1}^{q_{k+1}}$ is the
identity on $A_{k+1}$.   Therefore, $f_{k+1}^{sq_k}(a)$ steps
through the links of $A_{k-1}$ at most one link at a time, with
increasing $s$, meeting every link of $A_{k-1}$ before
$sq_k=q_{k+1}$. \qd

Let $x \in \lb$
and let $U$ be an open set in the plane such that
$U\cap\lb\ne\emptyset$.  Let $C$ be the constant of Lemma 5.2.1.
Since the $\e_k$'s are summable, we may choose $k-1$ such that there
is a link $a \in A_{k-1}$ and $\e>0$ such that $B(a,\e)\subset U$ and
$C\sum_{j={k+1}}^\infty \e_j<\e$.  By Lemma 5.6.1, there is a
positive integer $i<q_{k+1}$ and a link $b\in A_{k+1}$ such that
$x\in b$ and $f_{k+1}^i(b)\subset a$.  By Lemma 5.2.1,
$$d(f_\a^i(x),f_{k+1}^i(x))\le C\sum_{j={k+1}}^\infty \e_j<\e$$
Since $f_{k+1}^i(x)\in a$, it follows that $$f_\a^i(x)\in
B(a,\e)\subset U\ \ \qed$$ \enddemo

We have shown that if the number of subdivisions of a spanning
segment is large, and the incremental rotations are small, then
every orbit of $f_\a|\lb$ is dense in $\lb$.  We leave the proof of
the following theorem, showing that Example 5.5 is far from unique
in $\Cal H$, to the reader.

\proclaim{5.7. Theorem  ($f_\a$ recurrent, minimal, and not almost
periodic)}  Suppose $f_\a$ is a standard pseudorotation of
$\lb\in\Cal H$.  If there are constants $c$ and $M$ such that
$c_{k+1}\ge c>0$, $r_{k+1}\ge q_k$, and $0<m_k\le M$ for all $k$,
then $f_\a$ is recurrent and minimal, but not almost periodic, on
$\lb$.\endproclaim

The following theorem summarizes the preceding results concerning cofrontiers in $\Cal
H$:

\proclaim {5.8. Theorem}  Suppose $f_\a$ is a standard
pseudorotation on $\lb\in\Cal H$.  Suppose further that
\roster
\item"{(a)}"  there exists $c>0$ such that for all $k$,
$c_{k+1}\ge c$,
\item"{(b)}"  for all $k$, $r_{k+1}\ge q_k$, 
\item"{(c)}"  for some $M>0$ and for all $k$, $0<m_k\le M$.
\endroster
Then 
\roster
\item $\lb$ is indecomposable, by (a) and Theorem 4.1, 
\item $\lb$ is a minimal set under $f_\a$, by (b), (c), and Theorem
5.6,
\item $f_\a$ is recurrent, by (b), (c), and Theorem 5.2,
\item $f_\a$ is not almost periodic, by (a), (b), (c), and Theorem
5.7, or by conclusions (1) and (2) and Theorem 2.9,
\item $\a$ is irrational, by (a), (c), and Theorem 5.1, and 
\item $\a$ is not of constant type, by (a), (b), (c), and Theorem
5.4. \endroster
\endproclaim

\heading 6. Questions \endheading
\nopagebreak

We conclude with some questions about cofrontiers
and pseudorotations, both arbitrary ones and those in our class $\Cal
H$.

\subheading{6.1. Question}  Does there exist an indecomposable
co\-frontier $\lb$ admitting an irrational pseudorotation $h_\a$ for
which $\a$ is of constant type?

\subheading{6.2. Question}  Suppose $\lb$ is a cofrontier and $h_\a$ is
an irrational pseudorotation of $\lb$.  Must $h_\a|\lb$ be minimal? 
Recurrent?

Note that M. Barge has asked the \lq\lq minimal" part.  The
example which motivated Barge's question is constructed in
\cite{Wa}: the  \lq\lq split hairy circle" is a decomposable
cofrontier $X$ invariant under a homeomorphism $g$ of the plane on
which the induced map on one circle of prime ends is a
rigid rotation, but on the other circle is a Denjoy map. 
Thus, $g$ is a ``one-sided pseudorotation."  However, $g$ is neither
minimal nor recurrent on $X$.  So rotation-like from both sides is
necessary in 6.2.  In \cite{MO} (these Proceedings), an
indecomposable cofrontier $Y$ is constructed invariant under a
homeomorphism $h$ of the plane with the induced prime end maps being
a rotation on one side and a Denjoy map on the other.  In this
case, $h$ is minimal on $Y$, though $h$ does not appear to be
recurrent on $Y$; still, it is only a one-sided pseudorotation.

\subheading{6.3. Question}  Let $\Cal A$ be the set of
all irrationals $\a$ such that there exists an indecomposable
cofrontier $\lb$ (in $\Cal H$ or arbitrary) and a pseudorotation
$h_\a$ on $\lb$. Is $\Cal A$ a proper subset of the irrationals? 
Does there exist a number-theoretic characterization of $\Cal A$?

\subheading{6.4. Question}  Are there any cofrontiers in $\Cal H$
that are neither indecomposable, nor homeomorphic to $S^1$?

In an attempt to answer Question 6.4, one might try to construct a
$\lb\in\Cal H$ for which $\limsup s_k=\infty$ and
$\sum_{k=2}^\infty c_k<\infty$ (so neither of Theorems 4.1 or 4.2
applies).

\subheading{6.5 Recurrent and analytic extensions}
We have briefly outlined (see Section 3.4, last paragraph) one way in which one may
extend the induced pseudorotations we obtain to $\real^2$.  There are many possible
extensions.  Another such procedure is outlined in Handel's construction of an
area-preserving diffeomorphism of $\real^2$ leaving a cofrontier $\lb$ ($\cong$
pseudocircle) invariant \cite{Ha}.   Similar procedures can be applied in the authors'
constructions.  However, neither Handel's nor the authors'
extensions are recurrent on $\Int \lb$, nor conjugate on
$\Int\lb$ to the rotation $R_\a$ on the open unit disk $\disk$.

\subheading{6.6. Question}  Suppose $f_\a$ is a standard pseudorotation on
$\lb\in\Cal H$, and that $f_\a|\lb$ is recurrent. 
Can $f_\a|\lb$ be extended so as to be recurrent on $\Int\lb$? 
Can $f_\a|\lb$ be extended to $\Int\lb$ so as to be conjugate to
the rotation $R_\a$ on the open disk $\disk$?

M. R. Herman \cite{Hr} obtains pseudorotations of some cofrontiers
in $S^2$ that are complex analytically conjugate on $\Int\lb$ to
$R_\a$ on the complex unit disk {\bf D} (and $C^\infty$ conjugate on
$\Ext\lb$).

\subheading{6.7. Question}  In general, when can a pseudorotation of a cofrontier
$\lb\subset S^2$ be extended so as to be recurrent on $\Int\lb$ (and $\Ext\lb$)? 
(Complex analytically) conjugate on $\Int\lb$ to $R_\a$ on {\bf D}?

\Refs
\widestnumber\key{BGM}

\ref \key BG \by M. Barge and R. M. Gillette \paper Rotation and periodicity in plane
separating continua \jour Ergodic Theory and Dynamical Systems \vol 11 \yr 1991
\pages 619--631 \endref

\ref \key Br1	\by B. L. Brechner \paper On stable
homeomorphisms and imbeddings of the pseudo arc \jour Illinois
J. of Math. \vol 22 \yr 1978 \pages 630--661 \endref

\ref \key Br2 \bysame \paper Extendable periodic
homeomorphisms on chainable continua \jour Houston J. of Math. \vol
7 \yr 1981 \pages 327-344 \endref

\ref \key BGM \by B. L. Brechner, J. C. Mayer, and M. D. Guay \paper
Rotational dynamics on cofrontiers \jour Contemporary Mathematics \yr
1991 \vol 117 \pages 39--48 \endref

\ref \key Ca  \by M. L. Cartwright \paper Equicontinuous mappings of plane
minimal sets \jour Proc. London Math. Soc. (3) \vol 14A \yr 1965 \pages 51--54 \endref

\ref\key C \by M. Charpentier \paper Sur quelques proprites des courbes de M. Birkoff
\jour Bull. Soc. Math. France \vol 62 \yr 1934 \pages 193--224\endref

\ref\key CI \by H. Cook and W. T. Ingram \paper A characterization
of indecomposable compact continua \jour Proceedings of Topology
Conference \yr held at Arizona State University, Tempe, AZ, March,
1967 \pages 168--169 \endref

\ref \key  Ha  \by M. Handel \paper A pathological area-preserving
diffeomorphism of the plane \jour Proceedings AMS \vol 86
\yr 1982 \pages 163--168 \endref

\ref \key He \by E. Hemmingsen \paper Plane continua admitting non-periodic
autohomeomorphisms with equicontinuous iterates \jour Math. Scand. \vol 2 \yr 1954
\pages 119--141 \endref

\ref \key Hr  \by M. R. Herman \paper Construction of some curious
diffeomorphisms of the Riemann sphere \jour J. London Math. Soc. (2) \vol 34 \yr 1986
\pages 375--384 \endref

\ref\key MO \by J. C. Mayer and L. G. Oversteegen \paper
Denjoy meets rotation on an indecomposable cofrontier 
\jour these Proceedings \endref

\ref\key MR \by J. C. Mayer and J. T. Rogers, Jr. \paper
Indecomposable continua and the Julia sets of polynomials \preprint
\endref

\ref \key P \by G. Piranian \paper The boundary of a simply
connected domain \jour Bull. Amer. Math. Soc. \vol 64 \yr 1958
\pages 45--55 \endref

\ref \key Ro1 \by J. T. Rogers, Jr. \paper  Rotations of simply
connected regions and circle-like continua \jour Contemporary Math.
\vol 117 \yr 1991 \pages 139--148 \endref

\ref\key Ro2 \bysame \paper Indecomposable continua, prime ends, and
Julia sets \jour these Proceedings \endref

\ref \key  Ru  \by N. E. Rutt \paper Prime ends and
indecomposability \jour Bull. A. M. S. \vol 41 \yr 1935 \pages
265--273 \endref

\ref\key UY \by H. D. Ursell and L. C. Young \paper Remarks on the theory of prime
ends \jour Mem. Amer. Math. Soc. \vol 3 \yr 1951 \pages 1--29 \endref

\ref \key Wa  \by R. Walker \paper Basin boundaries with irrational
rotations on prime ends \jour Transactions AMS \vol 324 \yr 1991
\pages 303--317 \endref

\endRefs

\enddocument

\figone

\newpage
\figtwo

\newpage
\figthree
\figfour

\newpage
\figfive