\section{The Main Result}
\label{small}


\subsection{Statement of the Result}
\label{main result}

Let $\omega=\exp(2 \pi i/8)$ be the usual primitive $8$th root of unity.
Let $P$ be the regular octagon whose vertices are powers of $\omega$.
Let $\Sigma_0$ be the pinwheel strip associated to $P$ as in \S \ref{octopix}.
Let $\Sigma_0^1 \subset \Sigma_0$ denote the half-strip that
lies to the right of the line $x=1+\sqrt 2$.   See Figure 5.1.

\begin{center}
\resizebox{!}{3.5in}{\includegraphics{Pix/halfstrip.eps}}
\newline
{\bf Figure 5.1:\/} The relevant sets
\end{center}

\S \ref{countproof} we prove

\begin{lemma}
\label{ocount}
Relative to the the pinwheel map for the regular
octagon, any periodic orbit 
that intersects $\Sigma_0^1$
nontrivially does so in exactly $3^k$ points
for some $k=0,1,2...$.
\end{lemma}

Given a periodic point $x \in \Sigma_0^1$, we define $|x|$
to be the exponent $k$ such that the pinwheel
orbit of $x$ intersects 
$\Sigma_0^1$ exactly $3^k$ times.
We call a sequence $\{x_n\} \in \Sigma_0^1$ a {\it good sequence\/} if
$|x_n| \to \infty$ as $n \to \infty$ and
he congruence of $|x_n|$ mod $4$ is odd, and independent of $n$.

Recall that $\pi_2$ and
$\pi_3$ are the projections from $\R^8$ to $\R^2$ defined
in the previous chapter.   Define
\begin{equation}
s_2=\sqrt 3; \hskip 30 pt
s_3=1+\sqrt 2.
\end{equation}
Let $\phi: I_3 \to I_2$ be the canonical surjection described above.

\begin{theorem}[Main Theorem]
\label{main}
Let $\{x_n\}$ be a good sequence. Also, let
$A_n \subset \R^8$ be the arithmetic graph of $x_n$.  
Then the following is true.
\begin{enumerate}
\item $\pi_2(A_n)$ and $\pi_3(A_n)$ are closed polygons for all $n$.
\item Let $S_{k,n}$ be dilation by
$s_k^{-|x_n|}$ about the origin.
Then the following limit exists.
$$\Gamma_k=\lim_{n \to \infty} S_{k,n} \circ  \pi_k (A_n).$$

\item $\Gamma_3=I_3$, where 
$I_3$ is scaled so that one of the $8$ isosceles triangles
in its seed has vertices 
$$
(0,0); \hskip 20 pt
(1/2,-s_3/2) \hskip 30 pt
(-s_3/2,-1/2).
$$

\item $\Gamma_2=I_2$, where $I_2$ is scaled
so that its center is 
$(3/2,3/2)$ and one of its main corners is $(-3/2,3/2)$.

\item Identifying points on $\pi_3(A_n)$ and
$\pi_2(A_n)$ if they come from the same point on $A_n$, we
get a polygonal surjectin
$$\phi_n:  S_{3,n}\pi_3(A_n) \to S_{2,n} \pi_2(A_n).$$
We have $\lim_{n \to \infty} \phi_n = F \circ \phi$.
Here $F$ is an isometry of $I_2$ and
$\phi: I_3 \to I_2$ is the canonical surjection from
the snowflake to the carpet.
\end{enumerate}
\end{theorem}

\noindent
{\bf Remarks:\/} \newline
(i)
Experimentally, we see that Theorem \ref{main} also holds when
we consider sequences of points in $\Sigma_0-\Sigma_0^1$, but
these cases lead to annoying complications.  We are interested
in establishing a nice robust version of the phenomenon, but not
necessarily the sharpest possible version. \newline
(ii) 
The limits are exactly the same when $|x_n| \equiv 1$ mod $4$ and
when $|x_n| \equiv 3$ mod $4$.    The only thing that changes
is the global isometry $F$.  In the former case, $F$ is
orientation preserving (and can be taken to be the identity)
and in the other case $F$ is orientation reversing.
\newline

The rest of the paper is devoted to proving Theorem \ref{main}.

\subsection{Restricting the Domain}
Let $P$ be the regular octagon, and let $\Sigma_0$ be the
pinwheel strip described in \S \ref{octo}.  Consider the octagons
\begin{equation}
O_k^{\pm}=P+k(\sqrt 2+1,\pm 1); \hskip 30 pt k=1,2,3 \ldots
\end{equation}

$O_k^+$ and $O_k^-$ share a common vertex and
fit inside $\Sigma_0$ as shown in Figure 5.2.   Figure 5.2
shows the cases $k=1,2$, but the remaining cases look the same.

\begin{center}
\resizebox{!}{3in}{\includegraphics{Pix/regions0.eps}}
\newline
{\bf Figure 5.2:\/} The region $R_1$.
\end{center}

We let $R_k$ be the region of $\Sigma_0$
bounded by $O_k^+ \cup O_k^-$ and
$O_{k+1}^+ \cup O_{k+1}^-$. Figure 5.2
shows the region $R_1$, which is the main
region of interest to us.  Note that $R_k$ is translation
equivalent to $R_1$ for all $k \geq 1$,  so the
picture in Figure 5.2 is typical.

The first purpose of this chapter is to prove the
Invariance Lemma.   This result implies that each
$R_k$ is
forward invariant under the pinwheel map.  
Following this, we will prove

\begin{lemma}
\label{main2}
Each $R_k$ is an invariant set for the pinwheel map.
Furthermore, if Theorem \ref{main} is true for
sequences in $R_1$ then it is true as originally stated.
\end{lemma}

\subsection{The Pattern of Octagons}
\label{id}
\label{qr}

let $P$ be as above. The octagons we described above are
part of the infinite pattern of octagons shown in Figure 5.3.
Figure 5.3 shows a part of an infinite
set of octagons that, as it turns out, is invariant under
the outer billiards map.  Figure 5.3 also shows the
strip $\Sigma_0$.

\begin{center}
\resizebox{!}{4in}{\includegraphics{Pix/octo1.ps}}
\newline
{\bf Figure 5.3:\/} Octagonal orbits
\end{center}

The $8$ dark rows of octagons are obtained by translating
the central one by a vector of the form
\begin{equation}
\label{translate}
k\omega^n(\sqrt 2 + 1,1); \hskip 30 pt
k,n \in \Z; \hskip 30 pt \omega=\exp(2 \pi i/8).
\end{equation}
The remaining octagons interpolate between these dark ones.
The octagons are situated in such a way that the outer billiards map
is entirely defined on the interior of each one.
The octagons come in an infinite family of rings, or
{\it necklaces\/} $N_1, N_2, N_3,...$.  The octagons
$O_k^{\pm}$ are precisely the two octagons of $\Sigma_0 \cap N_k$.
Compare  [{\bf GS\/}] and [{\bf K\/}].

\begin{lemma}
Each necklace is an outer billiards orbit.  Moreover, the
region between consecutive necklaces is invariant under the
outer billiards map.
\end{lemma}

\startproof
We change notation slightly.
Let $O(n,k;0)$ be the octagon we get by translating the central one by 
the vector in Equation \ref{translate}.
Let $O(n,k;i)$ denote the octagon that is $i$ ``clicks'' away from
$O(n,k;0)$ going counterclockwise around $N_k$. 

The outer billiard map has the same action on all the
octagons lying on the vertical line segment between
$O(-1;k;0)$ and $O(1,k;-1)$ because such octagons
all lie in the sector shown in Figure 5.4.
The outer billiards map reflects each of the octagons
under construction through the apex of the sector in
Figure 5.4.  We check easily that the action is as follows.
\begin{equation}
O(-1,k;0) \to O(4,k;1); \hskip 30 pt
O(1,k;-1) \to O(5,k;0).
\end{equation}

\begin{center}
\resizebox{!}{1.8in}{\includegraphics{Pix/sector.eps}}
\newline
{\bf Figure 5.4:\/} Octagons in a sector
\end{center}

The octagons $O(4,k;1)$ and $O(5,k;0)$ have their centers
on the same vertical line segment $L$, as shown for
$k=2$ in Figure 5.4.  By symmetry, the
octagons between $O(-1,k;0)$ and $O(1,k;-1)$ map to the
octagons of the grid that have their centers on $L$.
These octagons all belong to $N_k$.
For the remaining octagons in $N_k$, the result follows
from the $8$-fold symmetry of the picture.

Call two octagons $O_1 \in N_k$ and $O_2 \in N_{k+1}$ {\it close\/}
if they are as close together as possible.   Two close octagons determine
a polygonal region ``between them'' like the one that is shaded in
Figure 5.4.  Call such a region a {\it buffer\/}.
  If the two octagons belong to the sector from Figure 5.4,
then the buffer is mapped between $N_{k}$ and $N_{k+1}$.
But then, by symmetry, all buffers are mapped between
$N_k$ and $N_{k+1}$.   But every point between $N_k$ and
$N_{k+1}$ is contained in a buffer.
\endproof

\subsection{Invariant Domains in the Strip}

Let $\Sigma'_0$ denote those points in $\Sigma_0$
that lie to the central octagon $P$.   A point
$p$ lies in $\Sigma_0'$ if every vector pointing
from $P$ to $p$ has positive $x$ coordinate.
Let $\Phi': \Sigma_0' \to \Sigma_0'$ be the
first return map of the $\phi^2$, the square
outer billiards map.

\begin{corollary}
\label{neck1}
For $k \geq 2$ the set $R_k$ is invariant under $\Phi'$.
\end{corollary}

\startproof
For $k \geq 2$, the region between $N_k$ and $N_{k+1}$
intersects $\Sigma_0'$ exactly in the set $R_k$.
\endproof

\noindent
{\bf Remark:\/}
The region between $N_1$ and $N_2$ intersects $\Sigma_0$ in a
set that is somewhat larger than $R_1$., as one can see by
looking carefully at Figure 5.3. It is for this reason
that Corollary \ref{neck1} fails for $R_1$.
\newline


\begin{corollary}
For $k \geq 4$ the set $R_k$ is invariant under $\Phi$,
the pinwheel map.
\end{corollary}

\startproof
For $k \geq 4$ the argument in Lemma \ref{far2} goes through,
and shows that $\Phi=\Phi'$.
\endproof

Below we will prove the following result.

\begin{lemma}
\label{ag1}
Let $\mu=1,2,3...$
Let $p \in R_1$ and 
$$q=p+2\mu (\sqrt 2+1,1) \in R_{1+\lambda}.$$
Then the arithmetic graphs $\Gamma(p)$ and $\Gamma(q)$
coincide.
\end{lemma}

\begin{corollary}
\label{neck2}
The set $R_k$ is $\Phi$-invariant for all $k \geq 1$.
\end{corollary}

\startproof
By Lemma \ref{ag1}, translation by the
vector $4(\sqrt 2+1,1)$ conjugates the action
of $\Phi$ on $R_1$ to the action of $\Phi$ on
$R_5$.  We already know that $R_5$ is
invariant under $\Phi$ and the conjugation
property implies the same result for $R_1$.
A similar argument takes care of $R_2$ and $R_3$.
We have already handled the remaining cases.
\endproof

\noindent
{\bf Remark:\/}
Note that $\Phi$ and $\Phi'$ do not coincide on
$R_1$, because Corollary \ref{neck1} is false for
$R_1$ whereas Corollary \ref{neck2} is true.
These maps in fact coincide for $k=2,3,4...$.
\newline

\noindent
{\bf Proof of the Invariance Lemma:\/}
We want to show that
$\Sigma_0^k$ is $\Phi$ invariant.
Let $N_k^+$ be the set of half-octagons
obtained by taking the right halves of all the
octagons in $N_k$.   Since $\phi^2$ is
a piecewise translation, the set $N_k^+$
is invariant under $\phi^2$.  But then
the set
$$S_k=N_k^+ \cap \Sigma_0$$ is $\Phi'$ invariant
for $k \geq 4$.   Hence, by the argument in
Lemma \ref{far2}, the set $S_k$
is $\Phi$-invariant for $k \geq 4$.
As in the proof of Corollary \ref{neck2}, we now conclude
that $S_k$ is $\Phi$ invariant for all $k=1,2,3...$
Finally
$$\Sigma_0^k=S_k \cup R_{k+1} \cup S_{k+1} \cup R_{k+2} \ldots.$$
We have decomposed $\Sigma_0^k$ into
$\Phi$ invariant sets.  Hence $\Sigma_0^k$ is also
$\Phi$ invariant.
\endproof

\noindent
{\bf Proof of Lemma \ref{main2}:\/}
Suppose we want to prove Theorem \ref{main} for
a sequence of periodic points $\{x_n\}$ in $\Sigma_0^1$.
Since $\Phi$ is the identity on each of the orbits
$N_k$, we have $x_n \in R_{k_n}$ for large $n$.
Chopping off the initial portion of the sequence,
we can assume that $x_n \in R_{k_n}$ for all $n$.

By Lemma \ref{ag1}, we can find a point
$y_n \in R_1 \cup R_2$ such that $x_n$ and $y_n$
have the same arithmetic graph.
Hence, it suffices to prove Theorem \ref{main}
for the sequence $\{y_n\} \in R_1 \cup R_2$.

The intersection $R_1 \cup R_2$ is a single point,
namely $\zeta=(2 + 2 \sqrt 2,1)$.    We call
two points {\it specially related\/} if
reflection through $\zeta$ interchanges the two
points.    We check by hand the following property,
which we call {\it Property P\/}:
If $y$ and $z$ are specially related then
$\Phi(y)$ and $\Phi(z)$ are specially related and
$\widehat \Phi(y)+\widehat \Phi(z)=0$.

From Property P we see that the arithmetic graphs of
$y$ and $z$ are isometric to each other.
Using this basic principle, we can replace each
$y_n \in R_2$ by the specially related point
$z_n$ without changing the isometry type 
of the rescaled limits of the arithmetic graphs.
\endproof

\noindent
{\bf Remark:\/}  In \S \ref{pinwheeldyn}
we will explain in detail how
we actually check Property P.

\subsection{Equivalent Points in Strips}
\label{stripmap}

The rest of the chapter is devoted to the proof of
Lemma \ref{ag1}.
In this section we make a general
observation about strip maps.
In the section following this one, we will
apply our observation to the case of the regular octagon.


Let $\Sigma_1$ and $\Sigma_2$ be non-parallel strips.
Let $V$ and $W_1,W_2$ be the three vectors shown in Figure 5.3.
We say that $p,q \in \Sigma_1$ are {\it related\/}
if $p-q=kW_1$ for some integer $k$.  Likewise, we say that
$p,q \in \Sigma_2$ are {\it related\/} if $p-q=kW_2$
for some integer $k$.

\begin{center}
\resizebox{!}{3in}{\includegraphics{Pix/strips.eps}}
\newline
{\bf Figure 5.4:\/} A Strip map
\end{center}

\begin{lemma}
\label{propagate}
Let $f$ be the strip map associated to $(\Sigma_2,V)$.  If
$p,q \in \Sigma_1$ are related then $f(p),f(q) \in \Sigma_2$
are related.
\end{lemma}

\startproof
Our statement is an affinely invariant one.  So, we normalize
our strips so that they are perpendicular and each have width $1$.
In this case, $f$ translates the horizontal squares $A,B,C,...$ to
the vertical squares with the same labels.  If $p,q \in \Sigma_1$
are related then they are in the same positions relative to the
horizontal squares that contain them.  But then $f(p),f(q) \in \Sigma_2$
are in the same position relative to the vertical squares that
contain them.  Hence $f(p)$ and $f(q)$ are related.
\endproof


\subsection{Proof of Lemma \ref{ag1}}

We first prove Lemma \ref{ag1} in case that
$\lambda=2 \mu$ is even.

We define
\begin{equation}
W_{2n}=W_{2n+1}=2\omega^n (\sqrt 2 + 1,1).
\end{equation}
The vector $W_0=W_1$ is parallel to both
pointed strips $\Sigma_0$ and $\Sigma_1$.  The
vector $W_2=W_3$ is parallel to both
$\Sigma_2$ and $\Sigma_3$, and so on.

We say that two points $p,q \in \Sigma_{k-1}$ are {\it equivalent\/}
\begin{equation}
\label{equiv}
q-p=\mu W_{k-1}; \hskip 30 pt \mu \in \Z.
\end{equation}
We let $\mu_{k-1}(p,q)$ be the integer in Equation \ref{equiv}.

Let $f_k$ be the strip map associated to
$(\Sigma_k,V_k)$, as in \S \ref{ag0}.
Let $m_k(p)$ be as in Equation \ref{gr2}.  That is,
\begin{equation}
f_{k}(p)=p+m_{k}(p) V_k.
\end{equation}

\begin{lemma}
\label{odd}
Let $f=f_k$.
If $p,q \in \Sigma_{k-1}$ are equivalent and
$k$ is odd, then
$f(q),f(q)$ are equivalent in $\Sigma_k$ and
$$\mu_{k}(f(p),f(q))=\mu_{k-1}(p,q);
\hskip 30 pt m_k(p)=m_k(q).$$
\end{lemma}

\startproof
By rotational symmetry, it suffices to consider the
case when $k=1$.
Suppose that $k=1$.     Note that $\Sigma_0 \cap \Sigma_1$ is
a strip parallel to $W_0=W_1$. So, $p \in \Sigma_0 \cap \Sigma_1$
if and only if $q \in \Sigma_0 \cap \Sigma_1$.     
Suppose that $p,q \in \Sigma_0 \cap \Sigma_1$.
Then $f(p)=p$ and $f(q)=q$.  Also, $W_1=W_0$.
Hence $f(p)$ and $f(q)$ are equivalent in
$\Sigma_1$ and  
$$\mu_1(f(p),f(q))=\mu_0(p,q); \hskip 30 pt
f_1(p)=f_1(q)=0.$$
If $p,q \in \Sigma_0-\Sigma_1$ then
$p+V_1 \in \Sigma_1$ and likewise $q+V_1 \in \Sigma_1$.
Here $$f_1(q)-f_1(p)=q-p.$$  Hence
$$\mu_1(f(p),f(q))=\mu_0(p,q); \hskip 30 pt
f_1(p)=f_1(q)=1.$$
\endproof


\begin{lemma}
\label{even}
Let $f=f_k$.
If $p,q \in \Sigma_{k-1}$ are equivalent and
$k$ is even, then
$f(q),f(q)$ are equivalent in $\Sigma_k$ and
$$\mu_{k}(f(p),f(q))=\mu_{k-1}(p,q);
\hskip 30 pt m_k(q)=m_k(p)+\mu_{k-1}(p,q).$$
\end{lemma}

\startproof
By rotational symmetry, it suffices to consider the case $k=2$.
In this case, the strips $\Sigma_1$ and $\Sigma_2$, as well
as the vectors
$W_1$ and $W_2$ and $V=V_2$ are related precisely as
discussed in \S \ref{stripmap}.   Lemma
\ref{propagate} now implies that $p$ and $q$ are
equivalent in $\Sigma_2$ and
$\mu_2(p,q)=\mu_{1}(p,q)$.

Referring to Figure 5.4, which is an affine image
of the octagon picture, the number $\mu_1(p,q)$
counts the number of parallelograms one needs to
shift in order to move $p$ over to $q$.  For
instance, if $p$ were in parallelogram $A$ and
$q$ were in parallelogram $C$ in Figure 5.4, then
$\mu_0(p,q)=2$.    From Figure 5.4, we see
clearly that  $m_2(q)=m_k(p)+\mu_{1}(p,q).$
\endproof

Now we can finish the proof of Lemma \ref{ag1}.
Suppose that $p,q \in \Sigma_0$ are equivalent.
Let $\mu=\mu_0(p,q)$.  Then, referring
to Equation \ref{spectrum}, we have
\begin{equation}
\widehat \Phi(q) - \widehat \Phi(p)=
\sum_{i=1}^8 (m_k(q)-m_k(p)) \widetilde V_k=^1
\sum_{k\  {\rm even\/}} \mu \widetilde V_k=^20.
\end{equation}
\label{cancel}
Equality 1 comes from Lemmas \ref{odd} and \ref{even}, and
Equality 2 comes from the fact that
$\widetilde V_6-=\widetilde V_2$
and $\widetilde V_8=-\widetilde V_4$.

\newpage
{"config": "arxiv", "file": "1006.2782/5pinwheel.tex"}