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tilings, mixing, substitution, subshifts, conjugacy
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\begin{document}

\title{When size matters: subshifts and their related tiling spaces}

\author{Alex Clark and Lorenzo Sadun}

\address{Alex Clark: Department of Mathematics, University of North Texas,
             Denton, Texas 76203}
\email{AlexC@unt.edu}

\address{Lorenzo Sadun: Department of Mathematics, The University of Texas at Austin,
 Austin, TX 78712-1082 U.S.A.}
\email{sadun@math.utexas.edu}   

\subjclass{52C23, 37A25, 37A30, 37A10, 37B10}


%\dedicatory{}

\begin{abstract}
We investigate the dynamics of substitution subshifts and their 
associated tiling spaces.  For a given subshift, the associated tiling
spaces are all homeomorphic, but their dynamical properties may differ.
We give criteria for such a tiling space to be weakly mixing, and for
the dynamics of two such spaces to be topologically conjugate. 
\end{abstract}

\maketitle

\markboth{Clark-Sadun}{When size matters: subshifts and tiling spaces}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Introduction}

We consider the dynamics of 1-dimensional minimal substitutions subshifts
(with a natural $\mathbb{Z}$ action) and the associated 1-dimensional tiling
spaces (with natural $\mathbb{R}$ actions). Given an alphabet $\mathcal{A}$
of $n$ symbols $\left\{ a_{1},...,a_{n}\right\} $, a \emph{substitution }on $
\mathcal{A}$ is a function $\sigma $ from $\mathcal{A}$ into the non-empty,
finite words of $\mathcal{A}$. Associated with such a substitution is the $
n\times n$ matrix $M$ which has as its $\left( i,j\right) $ entry the number
of occurrences of $a_{j}$ in $\sigma \left( a_{i}\right) $. A substitution
is \emph{primitive }if some positive power of $M$ has strictly positive
entries. The substitution $\sigma $ induces the map $\sigma :\mathcal{A}^{
\mathbb{Z}}\rightarrow \mathcal{A}^{\mathbb{Z}}$ given by $\left\langle
\dots u_{-1}.u_{0}u_{1}\dots \right\rangle \overset{\sigma }{\mapsto }
\left\langle \dots \sigma \left( u_{-1}\right) .\sigma \left( u_{0}\right)
\sigma \left( u_{1}\right) \dots \right\rangle .$ For any primitive
substitution $\sigma $ there is at least one point $u\in \mathcal{A}^{
\mathbb{Z}}$ which is periodic under $\sigma ,$ and the closure of the orbit
of any such $u$ under the left shift map $s$ of $\mathcal{A}^{\mathbb{Z}}$
forms a minimal subshift $\mathcal{S}$. This subshift $\mathcal{S}$ is
uniquely determined by $\sigma ,$ and the restriction $\sigma |_{\mathcal{S}}
$ is a homeomorphism onto $\mathcal{S}.$ To avoid trivialities, we shall
only consider \emph{aperiodic} substitutions $\sigma ,$ those for which the
subshift $\mathcal{S}$ is not periodic.

Given a collection of intervals $\mathcal{I}=\left\{ I_{1},...,I_{n}\right\} 
$, a \emph{tiling} $T$ of $\mathbb{R}$ by $\mathcal{I}$ is a collection of
closed intervals $\left\{ T_{i}\right\} _{i\in \mathbb{Z}}$ satisfying

\begin{enumerate}
\item $\cup _{i\in \mathbb{Z}}T_{i}=\mathbb{R},$

\item For each $i\in \mathbb{Z},$ $T_{i}$ is the translate of some $I_{\tau
\left( i\right) }\in \mathcal{I},$ and

\item $T_{i}\cap T_{i+1}$ is a singleton for each $i\in \mathbb{Z}.$
\end{enumerate}

\noindent If the function $\tau :\mathbb{Z\rightarrow }\left\{
1,...,n\right\} $ is an element of a minimal substitution subshift
$\mathcal{ S}$ of $\left\{ 1,...,n\right\} ^{\mathbb{Z}}$, then $T$ is
called a \emph{substitution tiling.} There is a natural topology on
the space $\mathfrak{T}$ of tilings of $\mathbb{R}$ that is induced by
a metric which measures as close any two tilings $T$ and $T^{\prime }$
that agree on a large neighborhood of $0$ up to an $\varepsilon $
translation. There is then the continuous translation action
$\mathbf{T}$ of $\mathbb{R}$ on $\mathfrak{T:}$ for $t\in \mathbb{R}$
and the tiling $T=\left\{ T_{i}\right\} _{i\in \mathbb{ Z}}$,
$\mathbf{T}:\left( T,t\right) \mapsto \mathbf{T}_{t}T=\left\{
T_{i}-t\right\} _{i\in \mathbb{Z}}.$ (A positive element of $\red$
moves the origin to the right, or equivalently moves tiles to the
left).  The closure of the translation orbit of any substitution
tiling $T$ in $\mathfrak{T}$ is then a minimal set of the action, the
(\emph{substitution})\emph{\ tiling space }$\mathcal{T}$ of the tiling
$T$. As the tiling space for any iterate of a substitution $\sigma $
is the same as the tiling space of $\sigma $ and since any
substitution has a point $u$ whose right half $\left\langle
u_{0}u_{1}\dots \right\rangle $ is periodic under substitution, for
ease of discussion we shall only consider substitutions with a point
$u$ whose right half $\left\langle u_{0}u_{1}\dots \right\rangle $ is
fixed under substitution.

Flows under a function provide an alternative description of tiling spaces
convenient for our purposes. Given a minimal subshift $\left( \mathcal{S}
,s\right) $ of the shift on $\mathcal{A}^{\mathbb{Z}}$ and $f:\mathcal{A}
\rightarrow \left( 0,\infty \right) ,$ there is the flow under $f$ given by
the natural $\mathbb{R}$ action on $\calT_{f}=\calS\times \mathbb{R}/\sim $,
where $(u,f(u_{0}))\sim (s(u),0)$. If we then associate with each $a_{i}\in 
\mathcal{A}$ a closed interval $I_{i}$ of length $f\left( a_{i}\right) $ and
form the tiling on $T$ of $\mathbb{R}$ by $\left\{ I_{1},\dots
,I_{n}\right\} $ with associated function $\tau =u\in \mathcal{A}^{\mathbb{Z}
}$, which has the left endpoint of the interval corresponding to $u_{0}$ at $
0\in \mathbb{R}$, then the function sending $T$ to the class of $\left(
u,0\right) $ in $\calT_{f}$ extends uniquely to a homeomorphism $\mathcal{T}
\rightarrow \mathcal{T}_{f}$ which conjugates the respective $\mathbb{R}$
actions.

The primary focus of this paper is the extent to which the dynamical systems 
$\mathcal{T}_{f}$ depend on the function $f$. If all we care about is the
topological space, they don't:

\begin{thm}
\label{homeo}Let $\calS$ be a subshift, and let $f,g$ be two positive
functions on the alphabet of $\calS$. Then $\calT_{f}$ and $\calT_{g}$ are
homeomorphic.
\end{thm}

\medskip \noindent \textbf{Proof.\enspace} Every class $x\in \calT_{f}$ has
a (unique) representative of the form $(u,t)$, with $0\leq t<f(u_{0})$. Let $
h(x)$ be the class of $(u,g(u_{0})t/f(u_{0}))$ in $\calT_{g}$. As the
function on quotient spaces induced by a continuous function $\mathcal{S}
\times \mathbb{R\rightarrow }\mathcal{S}\times \mathbb{R}$, $h$ is
continuous, and it has a well-defined continuous inverse.\hfill $\square $

But what about dynamics? Dekking and Keane \cite{DK} showed that, for the
substitution $a\rightarrow abab$, $b\rightarrow bbba$, the associated tiling
space with $f(a)=2,f(b)=1$ has point spectrum only at multiples of $2\pi $
. (That is, the associated discrete dynamical system is weakly mixing.)
Further, Berend and Radin \cite{BR} showed that a tiling space 
based on this substitution has trivial
point spectrum if $f(a)/f(b)$ is irrational. Radin and Sadun \cite{RS}
obtained very different results for the Fibonacci substitution $a\rightarrow
b$, $b\rightarrow ab$. For the Fibonacci substitution, $\calT_{f}$ is
topologically conjugate to $\calT_{g}$ whenever $f(a)+\tau f(b)=g(a)+\tau
g(b)$, where $\tau =(1+\sqrt{5})/2$ is the golden mean. (The conjugacy is
not the map $h$, described above, but is homotopic to $h$. A generalization
of this construction appears below in Section 3.) In particular, all
Fibonacci tiling spaces have pure point spectrum, regardless of the sizes
of the tiles.

In Section 2 we examine the general ergodic properties of the flows $\calT
_{f},$ showing that these flows are never mixing. We also provide general
criteria for the existence of point spectrum and constraints on the form
of that spectrum, in terms of the eigenvalues of the substitution matrix and
the ratios of lengths of the tiles. The Dekking-Keane and Berend-Radin
results are then easily understood. In Section 3 we provide criteria for $
\calT_{f}$ and $\calT_{g}$ to be topologically conjugate.

Sadun and Williams \cite{SW} have shown that every finite dimensional
tiling space meeting mild conditions (consisting of a finite number of
polygonal tiles, appearing in a finite number of orientations, and
meeting edge-to-edge) can be deformed into a tiling space that is
topologically conjugate to the $d$-fold suspension of a ${}
\mathbb{Z}^{d}$ subshift. This deformation changes the shapes and
sizes of the tiles, but not the combinatorics of which tiles touch
which others.  The difference between the dynamics of general tiling
spaces and of subshifts is therefore not combinatorial in nature, but
depends rather on the geometry of the individual tiles.  
This paper is an exploration of this phenomenon in one dimension, where
the geometry of a tile is simply its length. 
Moreover, our results have implications for the large class of higher
dimensional substitution tiling spaces that can be realized by taking
the product of one-dimensional substitution tiling spaces.

\section{Ergodic Properties}

For a substitution tiling space $\mathcal{T}_{f},$ let
$L=(f(a_{1}),\ldots ,f(a_{n}))$ be the (row) vector that gives the
lengths of the various tiles.  Let $L_0$ be the normalized
positive left Perron-Frobenius eigenvector with eigenvalue $\lambda
_{PF}$ of the substitution matrix $M$, and let $\calT_{0}$ be the tiling
space associated with the length vector $L_{0}$. Every such tiling space
$\mathcal{T}_{0}$ is known to admit a self-homeomorphism derived from
$ \sigma $ (also denoted $\sigma )$ which inflates all the tiles by a
factor of $\lambda _{PF}$ and which maps the class of a point $\left(
x,0\right) $ to the class of $\left( \sigma \left( x\right) ,0\right)
,$ see, e.g., \cite {BD}. The more general substitution tiling space
$\mathcal{T}_{f}$ then admits a substitution homeomorphism, also
denoted $\sigma ,$ conjugate to the substitution homeomorphism of
$\mathcal{T}_{0}$ via the homeomorphism $h$ of theorem \ref{homeo},
but this substitution homeomorphism is not generally affine. Solomyak
\cite{S} has investigated the ergodic properties of the self-affine
$\mathcal{T}_{0}$ in the case that the matrix $M$ is
diagonalizable. Now we proceed to investigate the ergodic properties
of the more general tiling spaces $\mathcal{T}_{f},$ requiring only
that the substitution be primitive and aperiodic. As most proofs in
this section are similar to those for analogous known results for
substitutions, the proofs shall be abbreviated.

It should be noted that most of the literature is devoted to the ergodic
properties of one-sided substitutions, while we are dealing with two-sided
substitutions. However, as there are at most $n^{2}$ $s-$orbits of words in
a one-sided substitution that admit more than one left extension \cite{BDH},
measure theoretic considerations are not altered. For example, by placing
all two-sided words with the same right halves into the same partition
element, partitions of one-sided substitutions lead to partitions of
two-sided substitutions with corresponding partition elements having the
same measure.

\medskip \noindent \textbf{Definitions.\enspace }For a word $w=w_{0}\cdots
w_{k}$ from the alphabet $\mathcal{A}$ of the substitution subshift $
\mathcal{S}$ derived from the substitution $\sigma ,$ 
\begin{equation}
\left[ w\right] \overset{\text{def}}{=}\left\{ u\in \mathcal{S}
\,|\,u_{0}\cdots u_{k}=w\right\} .
\end{equation}
A \textit{cylinder set }$\left[ w\right] \times I$ in $\mathcal{T}_{f}$ for
a word $w=w_{0}\cdots w_{k}$ and interval $I\subseteq \left[ 0,f\left( w_{0}\right) \right) $ 
is the set of all $x\sim \left( u,t\right)
\in \mathcal{T}_{f}$ with $u\in \left[ w\right] $ and $t\in I.$

The \textit{population vector }$v=\left( v_{1},...,v_{n}\right) ^{T}$ of the
finite word $w$ gives the number of occurrences $v_{i}$ of the letter $a_{i}$
in $w.$  Note that $v$ is a column vector, not a row, and that the
substitution matrix acts from the left.

For $u\in \mathcal{A}^{\mathbb{Z}}$, a \textit{recurrence word }is a
finite word $w$ in $u,$ $w=u_{r}u_{r+1}\cdots u_{s}$ satisfying the
condition that $ u_{s+1}=u_{r}$, and a \textit{recurrence vector} is
the population vector for a recurrence word. For example, for $\ldots
abbcba\ldots $ in $\left\{ a,b,c\right\} ^{\mathbb{Z}}$, $abbcb$ is a
recurrence word for which $ (1,3,1)^{T}$ is the recurrence vector.
The vectors $(0,1,0)^{T}$, $(0,1,1)^{T}$, and $(0,2,1)^{T}$ are
recurrence vectors for the successive $b$'s.

A recurrence vector $v$ is called \textit{full} if the vectors
$\{M^kv\}$, with $k$ ranging from 0 to $n-1$, are linearly
independent.  If $M$ is diagonalizable, this is equivalent to $v$
pairing nontrivially with every left-eigenvector of $M$.  Note that
every substitution on two letters admits a full recurrence vector,
namely $(0,1)^T$ or $(1,0)^T$.

Primitive aperiodic substitution tilings are recurrent, meaning that
any finite patch of one tiling appears somewhere in every other
tiling. As a result, the set of recurrence vectors is the same for
every tiling in the space, and so we can speak of the recurrence
vectors of a tiling \textit{space}.
\medskip
 
The following lemma and its implication for mixing are proved in much the
same way as for substitutions \cite[Theorem 2]{DK}.

\begin{lem} Let $v$ be the recurrence
vector of a recurrence word $r$ and let $t_{m}\overset{\text{def}}{=}
LM^{m}v. $ For any cylinder set\textit{\ }$\left[ w\right] \times I$ in $
\mathcal{T}_{f}$ and for any $m$, let $S_{m}\overset{\text{def}}{=}\left( 
\left[ w\right] \times I\right) \cap \left( \mathbf{T}_{-t_{m}}\left( \left[
w\right] \times I\right) \right) .$ Then 
\begin{equation}
\underset{m\rightarrow \infty }{\lim \inf }\;\mu \left( S_{m}\right) \geq
\gamma \cdot \mu \left( \left[ w\right] \times I\right) ,
\end{equation}
where $\gamma $ is a positive constant independent of $w$ and $I,$ depending
only on the recurrence word $r$ of $v.$
\label{olap} \end{lem}

\demo{Proof.}   With $t_{m}^{\prime }\overset{
\text{def}}{=}\left( 1,\dots ,1\right) M^{m}v,$ it follows from \cite{DK}
that the measures $\mu ^{\prime }$ in the substitution subshift of the sets $
S_{m}^{\prime }\overset{\text{def}}{=}\left[ w\right] \cap s^{-t_{m}^{\prime
}}\left[ w\right] $ satisfy
\begin{equation}
\underset{m\rightarrow \infty }{\lim \inf }\;\mu ^{\prime }\left(
S_{m}^{\prime }\right) \geq \gamma ^{\prime }\cdot \mu ^{\prime }\left( 
\left[ w\right] \right) ,
\end{equation}
for some positive constant $\gamma ^{\prime }$ independent of $w.$ Observe
that $\mu ^{\prime }\left( \left[ w\right] \right) $ is the limit as $
k\rightarrow \infty $ of the average number of iterates $\left\{ s^{i}\left(
u\right) |\,i=0,...,k-1\right\} $ in $\left[ w\right] $ $,$ for any given $
u\in \mathcal{A}^{\mathbb{Z}}$. And the measure $\mu \left( \left[ w\right]
\times I\right) $ in $\mathcal{T}_{f}$ is the limit as $T\rightarrow \infty $
of the average amount of time that $\left\{ \mathbf{T}^{t}\left( x\right)
|\,t\in \lbrack 0,T]\right\} $ spends in $\mu \left( \left[ w\right] \times
I\right) $ for any given $x\in \mathcal{T}_{f}.$ Hence, $\mu \left( \left[ w
\right] \times I\right) =\mu ^{\prime }\left( \left[ w\right] \right) \times 
\dfrac{\text{length}\left( I\right) }{\mu ^{\prime }\left( \left[ a_{1}
\right] \right) f\left( a_{1}\right) +\cdots +\mu ^{\prime }\left( \left[
a_{n}\right] \right) f\left( a_{n}\right) }.$ Now, if $u\in \left[ w\right] $
and $t\in I,$ then $s^{t_{m}^{\prime }}\left( u\right) \in \left[ w\right] $
if and only if $\mathbf{T}_{t_{m}}\left( \left( u,t\right) \right) \in \left[
w\right] \times \left\{ t\right\} , $and the result follows.

\hfill $\square $\medskip

By choosing the cylinder set smaller in measure than a $\gamma $ for a fixed
recurrence word, this leads to the following conclusion.

\begin{thm} None of the substitution tiling space
flows on $\mathcal{T}_{f}$ are (strongly) mixing.
\end{thm}

\begin{thm}

The number $k$ is in the point spectrum of $\mathcal{T}_{f}\ $if and only
if $\ $for any recurrence vector $v$, 
\begin{equation}
\dfrac{k}{2\pi }LM^{m}v\rightarrow 0 \text{(mod 1) as } m\rightarrow \infty .
\end{equation}
Moreover, when this condition is met there is a constant $C$ and a $\rho <1$
satisfying
\begin{equation}
\left| \exp \left( ikLM^{m}v\right) -1\right| <C\rho ^{m}
\end{equation}
for any given $v.$

\label{zeromod1}\ \end{thm} \medskip

\demo{Proof.}  



Essentially the same arguments as found in \cite{FMN} using Rokhlin stacks
and columns apply to show that the condition is both necessary and
sufficient for \emph{return words, }finite words $w$ in $u$,  $
w=u_{r}u_{r+1}\cdots u_{s}$ satisfying for sufficiently large $m$ the
conditions $\sigma ^{m}u_{r}=\sigma ^{m}u_{s+1}$ and $\sigma ^{m}u_{r}\neq
\sigma ^{m}u_{j}$ for $j=r+1,...,s$. Since the eigenfunction
constructed as in  \cite{FMN} using the geometric convergence 
and the condition for return words is continuous,
one need only consider continuous eigenfunctions.

We now show that for continuous eigenfunctions the condition for the general 
\emph{recurrence }word is necessary, which is all that need be shown since
the condition with return words is already sufficient. Consider
then a continuous eigenfunction $f$ with eigenvalue $k$. Let $x\sim \left(
u,0\right) \in \mathcal{T}_{f}$, where $u\in \mathcal{A}^{\mathbb{Z}}$ is \
a point of the substitution $\sigma $ with fixed right half $\left\langle
u_{0}u_{1}\dots \right\rangle $, and let $v$ be a recurrence vector for a
recurrence word between $u_{0}$ and some $u_{s}$, and let $t_{m}\overset{
\text{def}}{=}LM^{m}v$. Note that $\mathbf{T}_{t_{0}}x$ agrees with $x$ on
the interval covered by the single tile of $x$ represented by $u_{0}$. Then $
\mathbf{T}_{t_{m}}x=\sigma ^{m}(\mathbf{T}_{t_{0}}x)$ agrees with $x$ on
forward time intervals of increasing unbounded length, and so $\exp (ikt_{m})
$ must converge to 1, and $kt_{m}/(2\pi )$ must converge to 0 (mod 1). Now
let $v_{1}$ and $v_{2}$ be two such recurrence vectors. Since $
kLM^{m}v_{1}/(2\pi )$ and $kLM^{m}v_{2}/(2\pi )$ both converge to zero (mod
1), so does $kLM^{m}(v_{1}-v_{2})/(2\pi )$. In other words, we can use any
recurrence vector for the type of letter (say, $a$) that occurs at $u_{0}$.
Finally, suppose that $v_{3}$ is a recurrence vector for letter $b$. Since $
\sigma $ is primitive, suppose $a$ appears somewhere in the $\ell $-th
substitution of $b$, for some integer $\ell $. Then $v_{4}=M^{\ell }v_{3}$
is a recurrence vector for $a$, so $kLM^{m}v_{3}/(2\pi )=kLM^{m-\ell
}v_{4}/(2\pi )$ converges to $0($mod $1)$. To obtain the geometric convergence 
as in the statement of the theorem, one applies the argument of \cite[Lemme 1]{H} with
the lattice $f\left( a_{1}\right) \mathbb{Z\oplus \cdots }\mathbb{\oplus }
f\left( a_{n}\right) \mathbb{Z}$ in place of the lattice $\mathbb{Z}^{n}.$
\hfill $\square $\medskip 


The application of this criterion depends on the eigenvalues of $M$ and on
the possible forms of the recurrence vectors.

\begin{thm}
Suppose that all the eigenvalues of $M$ are of magnitude 1 or greater,
and that there exists a full recurrence vector. If the ratio of any
two tile lengths is irrational, then there is trivial point
spectrum. If the ratios of tile lengths are all rational, then the
point spectrum is contained in $2\pi\qed/L_{1}$.
\label{allbig1}
\end{thm}

\medskip \noindent \textbf{Proof.\enspace} Let $k$ be in the point
spectrum, and consider the sequence of real numbers
$t_{m}=kLM^{m}v/(2\pi )$, where $v$ is a fixed full recurrence
vector. Let $p(\lambda )=\lambda ^{n}+a_{n-1}\lambda ^{n-1}+\cdots
+a_{0}$ be the characteristic polynomial of $M$. Note that the $a_{i}$'s are
all integers, since $M$ is an integer matrix. Since $p(M)=0$, the
$t_{m}$'s satisfy a recurrence relation:
\begin{equation}
t_{m+n}=-\sum_{k=0}^{n-1}a_{k}t_{m+k}.  \label{intrecur}
\end{equation}
By Theorem \ref{zeromod1}, the $t_{m}$'s converge to zero mod 1. That is, we
can write 
\begin{equation}
t_{m}=i_{m}+\epsilon _{m}  \label{splits}
\end{equation}
where the $i_{m}$'s are integers, and the $\epsilon _{m}$'s converge to zero
as real numbers. By substituting the division (\ref{splits}) into the
recursion (\ref{intrecur}), we see that both the $i$'s and the $\epsilon $'s
must separately satisfy the recursion (\ref{intrecur}), once $m$ is
sufficiently large (e.g., large enough that the $\epsilon $'s are bounded by 
$1/\sum |a_{i}|$). However, any solution to this recursion relation is a
linear combination of powers of the eigenvalues of $M$ (or polynomials in $m$
times eigenvalues to the $m$-th power, if $M$ is not diagonalizable). 
Since the eigenvalues are all of magnitude one or greater, such a linear
combination converges to zero only if it is identically zero. Therefore $
\epsilon _{m}$ must be identically zero for all sufficiently large values of 
$m$.

Each $t_{m}$ is an integer linear combination of the elements of the vector $
kL/(2\pi )$. From a sequence of $n$ consecutive $s$'s, one can recover all
the elements of $kL$ by inverting an integer matrix whose columns are powers
of $M$ times $v$. (The invertibility of this matrix depends on the fact that 
$v$ is full and that zero is not an eigenvalue.) By Cramer's rule,
this inversion involves integer multiplication and addition, followed by
division by the determinant of this matrix. Since the $t_{m}$'s are integers
(for $m$ large enough), the components of $kL/(2\pi )$ must then all be
rational. If $L_{i}/L_{j}$ is irrational, this implies that $k=0$. If all
the $L$'s are rationally related, it implies that $k$ lies in $2\pi\qed/L_{1}$
.\hfill $\square $\medskip

Theorem \ref{allbig1} gives partial spectral information. When the alphabet
has only two letters and the substitution has constant length, we can say
considerably more:

\begin{thm}
\label{constantlength} Suppose that we have a primitive substitution on two
letters of constant length $n$, where $\sigma(a)$ contains $n_a$ $a$'s and $
n-n_a$ $b$'s, while $\sigma(b)$ contains $n_b$ $a$'s and $n-n_b$ $b$'s.
Suppose further that $1 \le n_a,n_b \le n-1$ and $n_a \ne n_b$. Let $z$ be
the greatest common factor of $n$ and $n_a-n_b$. Then the point spectrum $
\sigma_{pp}$ depends as follows on the ratio $L_1/L_2$:

\begin{enumerate}
\item If $L_{1}=L_{2}$, then there is a positive integer $N$ such that 
\begin{equation}
N{}\mathbb{Z}[1/n]\subset N{}L_{1}\sigma _{pp}/2\pi \subset \mathbb{Z}
{}[1/n].
\end{equation}

\item If $L_{1}/L_{2}\in \mathbb{Q}-\{1\}$, then there exist positive
integers $N_{1}$ and $N_{2}$ such that 
\begin{equation}
N_{1}{}\mathbb{Z}[1/z]\subset N_{2}{}L_{1}{}\sigma _{pp}/2\pi \subset {}
\mathbb{Z}[1/z].
\end{equation}

\item If $L_{1}/L_{2}\notin \mathbb{Q}$, then $\sigma _{pp}=\{0\}$.
\end{enumerate}
\end{thm}

\medskip \noindent \textbf{Proof.\enspace} The $L_{1}=L_{2}$ case is not new
-- Coven and Keane \cite{CK} proved an even stronger result over 30 years
ago. We prove all three cases together, since the proof neatly illustrates
the significance of the ratio $L_{1}/L_{2}$.

The substitution matrix 
\begin{equation}
M = 
\begin{pmatrix}
n_a & n_b \cr n-n_a & n-n_b
\end{pmatrix}
\end{equation}
has eigenvalues $n$ and $n_a-n_b$, with left-eigenvectors $(1,1)$ and $
(n-n_a, -n_b)$ and right-eigenvectors $(n_b, n-n_a)^T$ and $(1, -1)^T$.
Since $n_a \ne n_b$, both eigenvalues have magnitude one or greater.

Since the system is aperiodic and there are only two letters, either $
(1,0)^{T}$ or $(0,1)^{T}$ is a (full) recurrence vector. Suppose that
$(1,0)^{T}$ is (the other case is similar). Then we take

\begin{equation}
v = 
\begin{pmatrix}
1 \\ 
0
\end{pmatrix}
= \frac{1}{n+n_b-n_a} 
\begin{pmatrix}
n_b \\ 
n-n_a
\end{pmatrix}
+ \frac{n-n_a}{n+n_b-n_a} 
\begin{pmatrix}
1 \\ 
-1
\end{pmatrix}
.
\end{equation}

As in the proof of Theorem \ref{allbig1}, if $k\in \sigma _{pp}$ then the
numbers $t_{m}=kLM^{m}v/2\pi $ must eventually be integers. Note that 
\begin{equation}
\begin{split}
2\pi t_{m}=& k 
\begin{matrix}
LM^{m}
\end{matrix}
\begin{pmatrix}
1 \\ 
0
\end{pmatrix}
\\
2\pi t_{m+1}=& k 
\begin{matrix}
LM^{m+1}
\end{matrix}
\begin{pmatrix}
1 \\ 
0
\end{pmatrix}
=k 
\begin{matrix}
LM^{m}
\end{matrix}
\begin{pmatrix}
n_{a} \\ 
n-n_{a}
\end{pmatrix}
,
\end{split}
\end{equation}
so 
\begin{equation}
2\pi (t_{m},t_{m+1})=kLM^{m} 
\begin{pmatrix}
1 & n_{a} \\ 
0 & n-n_{a}
\end{pmatrix}
,
\end{equation}
and 
\begin{equation}
kLM^{m}=2\pi (t_{m},t_{m+1}){\ 
\begin{pmatrix}
1 & n_{a} \\ 
0 & n-n_{a}
\end{pmatrix}
}^{-1}=\frac{1}{n-n_{a}}\left( (n-n_{a})t_{m},t_{m+1}-n_{a}t_{m}\right) .
\end{equation}
Multiplying on the right by the right-eigenvectors of $M$ gives two scalar
equations for $k$: 
\begin{equation}
\begin{split}
k(L_{1}-L_{2})(n_{a}-n_{b})^{m}=& \frac{2\pi }{n-n_{a}}(nt_{m}-t_{m+1}) \\
k(n_{b}L_{1}+(n-n_{a})L_{2})n^{m}=& 2\pi ((n_{b}-n_{a})t_{m}+t_{m+1}).
\end{split}
\label{kequations}
\end{equation}

If $L_{1}=L_{2}$, the first equation reads $0=0$, but the second equation
says that $NkL_{1}n^{m}/2\pi $ is an integer for $m$ sufficiently large,
where $N=n+n_{b}-n_{a}$. Thus $NL_{1}\sigma _{pp}/2\pi \subset \mathbb{Z}
[1/n]$.

If $L_{1}/L_{2}$ is rational but not one, then the first equation says that $
\sigma _{pp}L_{1}/2\pi $ is contained in a fixed rational multiple of $
\mathbb{Z}[1/(n_{a}-n_{b})]$, while the second says that $\sigma
_{pp}L_{1}/2\pi $ is contained in a fixed rational multiple of $\mathbb{Z}
[1/n]$. Thus an integer multiple of $\sigma _{pp}L_{1}/2\pi $ lies in ${}
\mathbb{Z}[1/n]\cap \mathbb{Z}[1/(n_{a}-n_{b})]=\mathbb{Z}[1/z]$.

If $L_{1}/L_{2}$ is irrational, then the only simultaneous solution to both
equations is $k=0$. For if $k\neq 0$ is a solution, then the ratio of the
two equations would imply that $(L_{1}-L_{2})/(n_{b}L_{1}+(n-n_{a})L_{2})$
would be rational, which is inconsistent with the irrationality of $
L_{1}/L_{2}$. Thus $\sigma _{pp}\subset \{0\}$ and so $\sigma _{pp}=\{0\}.$
All that remains is to show that $L_{1}\sigma _{pp}/2\pi $ contains ${}
\mathbb{Z}[1/n]$ if $L_{1}=L_{2}$, and that $L_{1}\sigma _{pp}/2\pi $
contains a rational multiple of ${}\mathbb{Z}[1/z]$ if $L_{1}/L_{2}$ is
rational but not one.

We call the result of applying the substitution $\ell $ times to a single
tile (or a single letter) a \textit{supertile} of order $\ell $. If $
L_{1}=L_{2}$, then all supertiles of order $m$ have length $n^{m}L_{1}$.
Thus the coordinates of the endpoints of such supertiles agree modulo $
n^{m}L_{1}$. For each pair of integers $(j,m)$ we define a function on our
tiling space. If $x$ is a tiling, let 
\begin{equation}
\psi _{j,m}(x)=\exp (2\pi ijp/n^{m}L_{1}),
\end{equation}
where $p$ is the endpoint of \textit{any} supertile of order $m$ in $x$.
This is manifestly an eigenfunction of translation with eigenvalue 
$2 \pi j/ n^{m} L_1$.

If $L_{1}/L_{2}$ is rational but not integral, then the length of each
supertile of order $m$ is a fixed rational linear combination of $
L_{1}n^{m} $ and $L_{1}(n_{a}-n_{b})^{m}$. Thus there is a constant $c$, a
rational multiple of $L_{1}$, such that every supertile of order $m$ has
length divisible by $cz^{m}$. For each pair $(j,m)$ we then define the
eigenfunction 
\begin{equation}
\phi _{j,m}(x)=\exp (2\pi ijp/cz^{m}),
\end{equation}
which shows that some rational multiple of $\mathbb{Z}[1/z]$ is contained in 
$L_{1}\sigma _{pp}/2\pi $.\hfill $\square $

\medskip

As a special case, we recover the results of Dekking and Keane, and then of
Berend and Radin, since their substitution $\sigma $ has $
n=4,n_{a}=2,n_{b}=1 $, so $z=1$. When $L_{1}=2$ and $L_{2}=1$, the point
spectrum is $2\pi \mathbb{Z}[1/z]=2\pi \mathbb{Z}$. This also illustrates
the sharp contrast between substitution tiling spaces admitting a flow with
point spectrum and spaces admitting equicontinuous flows. If $X$ admits
an equicontinuous $\mathbb{R}$ action $\phi $ with point spectrum $\sigma
_{pp}$ and if $\phi ^{\prime }$ is any $\mathbb{R}$ action on $X$ obtained
by a time change of $\phi ,$ then the point spectrum of $\phi ^{\prime }$
is $c\cdot \sigma _{pp}$ for some possibly $0$ constant $c,$ see Egawa \cite
{E}. In the substitution $\sigma $ currently under consideration, the $
\left( 1,1\right) $ flow has purely point spectrum $\sigma _{pp}$ but
admits a time change represented by $\left( 2,1\right) $ which has discrete
spectrum which can only be represented as a proper subset of a multiple of $
\sigma _{pp}.$

What about the Fibonacci tiling? In that case the smaller eigenvalue is $
1-\tau$, whose magnitude is less than 1, so Theorem \ref{allbig1} does not
apply directly. However, there is still something that can be said about
such cases:

\begin{thm}
\label{somesmall1} Suppose that there exists a full recurrence vector. 
Let $S$ be
the span of the (generalized left-) eigenspaces of $M$ with eigenvalue
of magnitude strictly less than 1. If $k$ is in the point spectrum,
then $kL/ {2 \pi}$ is the sum of a rational vector and an element of
$S$.
\end{thm}

\medskip\noindent\textbf{Proof.\enspace} As before, $t_m=LM^m v$ converges
to zero (mod 1), so we can write $t_m = i_m + \epsilon_m$ with $i_m$
integral and $\epsilon_m$ converging to zero, and with both the $i_m$'s and
the $\epsilon_m$'s eventually satisfying the recursion relation (\ref
{intrecur}). Since the $\epsilon_m$'s converge to zero, they must eventually
be linear combinations of $m$-th powers of the small eigenvalues of $M$
(possibly times polynomials in $m$, if $M$ is not diagonalizable). 
By adding an element of $S$ to $L$ (and hence to $kL/2\pi$), we can then get
all the $\epsilon_m$'s to be identically zero beyond a certain point. 


If $M$ is invertible, the resulting value of $kL/2\pi$ must then be
rational by the same argument as in the proof of Theorem
\ref{allbig1}. If $M$ is not invertible, then we need an extra step,
since the vectors $M^m v, M^{m+1} v, \ldots, M^{m+n-1}v$ are no longer
linearly independent.  Let $S_0 \subset S$ be the generalized
left-eigenspace with eigenvalue zero, and suppose that this space has
dimension $\ell$. The vectors $M^mv, \ldots, M^{m+n-\ell-1}$ are
linearly independent and span the annihilator of $S_0$, which can be
identified with the dual space of ${\red^n}^*/S_0$.  From the
integers $t_m, \ldots, t_{m+n=\ell-1}$ we can reconstruct, by rational
operations, a representative of $kL/2\pi$ in ${\red^n}^*/S_0$.  But that
implies that $kL/2\pi$ is a rational vector plus an element of $S_0$. 
\hfill $\square$
\medskip

Another way of stating the same result is to say that $kL/2\pi $, projected
onto the span of the large eigenvectors, equals the projection of a rational
vector onto this span.

This theorem can be used in two different ways. First, it constrains the set
of vectors $L$ for which the system admits point spectrum. Let $d_b$
be the number of large eigenvalues, counted with (algebraic) multiplicity.
There are only a countable number of possible values for the
projection of $kL/2\pi$ onto the span of the large (generalized) eigenvectors. 
If the projection of $L$ does not lie in the ray generated by one of these
points, there is no point spectrum. In other words, one
must tune $d_b-1$ parameters to a countable number of possible values in
order to achieve a nontrivial point spectrum.

A second usage is to constrain the spectrum for fixed $L$.  The
rational points in $\red^n$, projected onto the span of the large
eigenvalues, and then intersected with the ray defined by a fixed $L$,
form a vector space over $\qed$ of dimension at most $n+1-d_b$. As a
result, the point spectrum tensored with $\mathbb{Q}$ is a vector
space over $\qed$ whose dimension is bounded by one plus the number of
small eigenvalues.  Below we derive an even stronger result, in which
only the small eigenvalues that are conjugate to the Perron-Frobenius
eigenvalue contribute to the complexity of the spectrum.

So far we have assumed that there exists a full recurrence vector $v$.
This may fail, either because there is something peculiar about the
available recurrence vectors, or because the matrix $M$ has repeated
eigenvalues.  In
such cases we can repeat our analysis on a proper subspace of
${}\mathbb{R}^{n}$. There are necessarily fewer constraints on the
vector $L$, but for fixed $L$ the constraints on $k$ are 
actually stronger than before.

\begin{thm}
\label{notfull} Let $b_{PF}$ be the number of large eigenvalues that
are algebraically conjugate to the Perron-Frobenius eigenvalue
$\lambda_{PF}$ (including
$\lambda_{PF}$ itself), and let
$s_{PF}$ be the number of small eigenvalues conjugate to $\lambda_{PF}$. 
For the system to have nontrivial point spectrum, $L$ must lie in a 
countable union of subspaces of $\red^n$, each of codimension $b_{PF}-1$
or greater.  For fixed $L$, the dimension over $\qed$ of the point
spectrum tensored with $\qed$ is at most $s_{PF}+1$.  
\end{thm}

\medskip\noindent\textbf{Proof.\enspace} 
As a first step we diagonalize $M$ over the rationals as far as
possible.  By rational operations we can always put $M$ in
block-diagonal form, where the characteristic polynomial of each block
is a power of an irreducible polynomial.  Since the Perron eigenvalue
$\lambda _{PF}$ has algebraic multiplicity one, every eigenvalue
algebraically conjugate to $\lambda _{PF}$ also has multiplicity
one. Thus there is a unique block whose characteristic polynomial has
$\lambda _{PF}$ for a root. Every recurrence vector is integral, and
pairs nontrivially with $L_0$, and so pairs nontrivially with every
left-eigenvector of this block.  We then consider the constraints on
the spectrum that can be obtained from this block alone.

We repeat the argument of the previous two theorems.  As before, we
modify $L$ by adding a linear combination of small eigenvectors so as
to make $t_m$ eventually integral.  Since the block diagonalization was
over the rationals, the vector $\vec t= (t_m, \ldots, t_{m+n-1})^T$, expressed
in the new basis, is still rational. From this vector we deduce, as
before, the projection of $kL/(2\pi)$ on any large eigenvector with
which our recurrence vector $v$ pairs nontrivially.  This gives at
least $b_{PF}$ independent constraints on the pair ($L,k$), hence
$b_{PF}-1$ constraints on $L$ if the spectrum is to be nontrivial.

Next we consider the projection of $kL/(2\pi)$ onto the large
eigenspaces of the Perron-Frobenius
block.  Since only $b_{PF}+s_{PF}$ components of $\vec t$ (expressed
in the new basis) contribute, this is the projection of $\qed^{b_{PF}+s_{PF}}$
onto $\red^{b_{PF}}$, whose real span is all of $\red^{b_{PF}}$. 
Intersected with the ray defined by a fixed $L$, this gives a vector space 
of dimension at most $s_{PF}+1$ in which $k$ can live. 
\hfill $\square$ 

\section{Topological conjugacies}

In Section 2 we studied the point spectrum of a tiling space. This is a
conjugacy invariant; systems with different discrete spectra cannot be
conjugate. However, we found many different values of $L$ that gave exactly
the same point spectrum -- namely $\{0\}$. In this section we give
criteria for determining when two values of $L$ give rise to conjugate
tiling spaces. We supply sufficient conditions for conjugacy in terms of the
matrix $M$. We also supply necessary conditions; these require knowing
something about the actual substitution, and not just its matrix $M$.
Finally, we exhibit examples that demonstrate the need for such details.
Different substitutions with the same matrix may result in different
criteria for conjugacy.

Throughout this section, $\calS$ is a substitution subshift with
substitution map $\sigma $ and substitution matrix $M$ and we consider
two possible length vectors $L=(f(a_{1}),f(a_{2}),\ldots )$ and
$L^{\prime }=(g(a_{1}),g(a_{2}),\ldots )$.  The substitution applied
$\ell $ times to a single tile (or letter) is called a
\textit{supertile} of order $\ell$.

\begin{thm}
If, for some integer $k$, 
\begin{equation}
\lim_{m \to \infty} \left ( L M^{m+k} - L^{\prime}M^m \right ) = 0,
\label{samesize}
\end{equation}
then $\calT_f$ is conjugate to $\calT_g$. \label{sufficient}
\end{thm}

\medskip\noindent\textbf{Proof.\enspace} We prove the theorem in two steps.
First we prove it in the special case that $k=0$, and then in the case that $
L = L^{\prime}M$. Combining these results then gives the theorem for all $k
\le 0$. The situation for $k >0$ is the same, only with $\calT_f$ and $\calT
_g$ reversed.

If (\ref{samesize}) holds with $k=0$, then the supertiles in the $\calT_{f}$
system asymptotically have the same size as the corresponding supertiles in
the $\calT_{g}$ system, and the convergence is exponential. We then adapt
the argument that Radin and Sadun \cite{RS} applied to the Fibonacci tiling.
If $x$ is a tiling in $\calT_{f}$ with tiles in a sequence $u=\ldots
u_{-1}u_{0}u_{1},\ldots $, then we require $\phi _{\ell }(x)\in \calT_{g}$
to be a tiling with the exact same sequence of tiles. The only question is
where to place the origin. Let $d_{\ell }$ be the coordinate of the right
edge of the order-$\ell $ supertile of $x$ that contains the origin, and let 
$-e_{\ell }$ be the coordinate of the left edge. Place the origin in $\phi
_{\ell }(x)$ a fraction $e_{\ell }/(d_{\ell }+e_{\ell })$ of the way across
the corresponding supertile of order $\ell $ in the $\calT_{g}$ system.
(This is essentially the map $h$ of Theorem \ref{homeo}, only applied to
supertiles of order $\ell $.) Since the sizes of the supertiles converge
exponentially, the location of the origin in $\phi _{\ell }(x)$ converges,
and we can define $\phi (x)=\lim_{\ell \rightarrow \infty }\phi _{\ell }(x)$
. All that remains is to show that $\phi $ is a conjugacy. The approximate
map $\phi _{\ell }$ is not a conjugacy; translations in $x$ the keep the
origin in the same supertile of order $\ell $ get magnified (in $\phi (x)$)
by the ratio of the sizes of the supertiles of order $\ell $ containing the
origin in $x$ and $\phi (x)$. However, this ratio goes to one as $\ell
\rightarrow \infty $, while the range of translations to which this ratio
applies grows exponentially. In the $\ell \rightarrow \infty $ limit, $\phi $
commutes with translation by any $s\in (-e_{\infty },d_{\infty })$.
Typically this is everything. The case where $x$ contains two
infinite-order supertiles is only slightly trickier -- once
one sees that $\phi $ preserves the boundary between these infinite-order
supertiles, it is clear that $\phi $ commutes with all translations.

Now suppose that $L=L^{\prime }M$. Then the tiles in the $\calT_{f}$ system
have exactly the same size as the corresponding supertiles of order 1 in the 
$\calT_{g}$ system. We define our conjugacy $\phi $ as follows: If $x$ is a
tiling in $\calT_{f}$ with tiles in a sequence $u=\ldots
u_{-1}u_{0}u_{1}\ldots $, then $\phi (x)$ to a tiling in $\calT_{g}$ with
sequence $\sigma (u)$, with each tile in $x$ aligned with the corresponding
order-1 supertile in $\phi (x)$.

Combining the cases, we obtain conjugacy whenever (\ref{samesize})
applies.\hfill $\square $\medskip

Theorem \ref{sufficient} gives sufficient conditions for two tiling spaces
to be conjugate. To derive necessary conditions we must understand the
extent to which words in our subshift $\calS$ (and its associated tiling
spaces) can repeat themselves. Moss\'{e} shows in \cite{M1} that in every
substitution subshift there exists an integer $N$ so that no word is ever
repeated $N$ or more times. We will need to refine these results by
considering words that are repeated a fractional number of times. Counting
fractional periods depends on the length vector: the word $ababa$ has its
basic period $ab$ repeated $2+f(a)/(f(a)+f(b))$ times.

\noindent \textbf{Definition.} A recurrence vector $v$ is a \textit{
repetition vector of degree }$p$\textit{\ }of a tiling space $\calT$ if
there are finite words $w$ (with population vector $v$) and $w^{\prime }$,
such that: 1) $w^{\prime }$ contains $w$, 2) $w^{\prime }$ is periodic with
period equal to the length of $w$, 3) the length of $w^{\prime }$ is at
least $p$ times the length of $w$, and 4) $w^{\prime }$ appears in some (and
therefore every) tiling in $\calT$.\smallskip

Note that every repetition vector of degree $p>p^{\prime}$ is also a
repetition vector of degree $p^{\prime}$. Note also that the word $w$ is
typically not uniquely defined by $v$. A cyclic permutation of the tiles in $
w$ typically yields a word that works as well. For instance, if $
w^{\prime}=ababa$ appears in a tiling, then $v=(1,1)^T$ is a repetition
vector with period $2+ f(a)/(f(a)+f(b))$, and we may take either $w=ab$ or $
w=ba$.

We denote the length $L_{0}w$ of a word $w$ in $\calT_{0}$ as described
above by $|w|$. If $v$ is a repetition vector of
degree $p$ for $\calT_{0}$, representing the word $w$ sitting inside $
w^{\prime }$ in $u\in \mathcal{A}^{\mathbb{Z}}$, then $Mv$ is a repetition
vector of degree $p$, representing the word $\sigma (w)$ sitting inside $
\sigma (w^{\prime })$ in $\sigma (u)\in \mathcal{A}^{\mathbb{Z}}$. The
degree is the same, since in $\calT_{0}$ the substitution $\sigma $
stretches each word by exactly the same factor, namely $\lambda _{PF}$.
Thus, every repetition vector $v$ gives rise to an infinite family of
repetition vectors $M^{k}v$. The following theorem limits the number of such
families.

\begin{thm}
Let $p>1$. There is a finite collection of vectors $\left\{ v_{1},\ldots
,v_{N}\right\} $ such that every repetition vector of degree $p$ for $\calT
_{0}$ is of the form $M^{k}v_{i}$ for some pair $(k,i)$. \label{families}
\end{thm}

\medskip \noindent \textbf{Proof.\enspace} There is a recognition length $D_{
\mathcal{S}}$ of the subshift $\calS$ such that knowing a letter, its $D_{_{
\mathcal{S}}}$ immediate predecessors and its $D_{_{\mathcal{S}}}$ immediate
successors determines the supertile of order 1 containing that letter, and
the position within that supertile of the letter. Thus for every
substitution tiling space $\calT$ there is a recognition length $D_{\mathcal{
T}}$ such that the neighborhood of radius $D_{_{\mathcal{T}}}$ about a point
determines the supertile of order 1 containing that point, and the position
of that point within the supertile. (E.g., one can take $D_{_{\mathcal{T}}}$
to be $1+D_{_{\mathcal{S}}}$ times the size of the largest tile.) Let $D=D_{
\mathcal{T}_{0}}$.

Suppose $v$ is a repetition vector of degree $p$ for $\mathcal{T}_{0}$,
corresponding to a word $w$ that is repeated $p$ times in a word $w^{\prime
} $, whose length is much greater than $D$. Then there is an interval of
size $|w^{\prime }|-2D$ in which the supertiles of order 1 are periodic with
period $|w|$. Thus $v$ is the population vector of a collection of
supertiles of order 1, so there exists a population vector $v_{1}$ of a word 
$w_{1}$, such that $v=Mv_{1}$, and such that $v_{1}$ is a repetition vector
of degree $(|w^{\prime }|-2D)/|w|=p-2D/\lambda _{PF}|w_{1}|$. Note that $w$
itself does not have to equal $\sigma (w_{1})$ --- it may happen that $w$ is
a cyclic permutation of $\sigma (w_{1})$.

Repeating the process, we find words $w_i$ such that $\sigma(w_i)$ equals $
w_{i-1}$ (up to cyclic permutation of tiles), and such that the population
vector of $w_i$ is a repetition vector of degree 
\begin{equation}
p - \frac{2 D}{|w_i|} (\lambda_{PF}^{-1} + \lambda_{PF}^{-2} + \cdots +
\lambda_{PF}^{-i}) \ge p - \frac{2 D}{|w_i|} \sum_{\ell=1}^\infty
\lambda_{PF}^{-\ell} = p - \frac{2D}{|w_i| (\lambda_{PF}-1)}.
\end{equation}

Now pick $\epsilon <p-1$. We have shown that every repetition vector of
degree $p$ is of the form $M^{k}v_{i}$, where $v_{i}$ is a repetition vector
of degree $p-\epsilon $, and where $L_{0}v_{i}$ is bounded by $\frac{
2D\lambda _{PF}}{\epsilon (\lambda _{PF}-1)}$. However, there are only a
finite number of words of this length, hence only a finite number of
possible repetition vectors $v_{i}$.\hfill $\square $

\medskip

As defined, the degree of a repetition vector depends on the length vector
of the tiling space. This difference between tiling spaces disappears in the
limit of long repetition vectors:

\begin{lem}
Let $L$ be decomposed as $L= cL_0 + L_r$, where $L_r$ is a linear
combination of (generalized) eigenvectors with eigenvalue less than $
\lambda_{PF}$. For each word $w$, let $r_w$ be the ratio of the length of $w$
in $\calT_f$ to the length of $w$ in $\calT_0$. There exist constants $
c_{1,2} > 0$ such that $|r_w - c| < c_1 / |w|^{c_2}$.
\end{lem}

\medskip \noindent \textbf{\enspace} The proof is trivial.\hfill $\square $
\medskip

\begin{lem}
Let $\epsilon >0$, and let $L$ and $L^{\prime}$ be fixed. There is a length $
R$ such that, if $v$ is a repetition vector of degree $p$ in $\calT_f$,
representing a word $w$ of length greater than $R$, then $v$ is a repetition
vector of degree $p-\epsilon$ in $\calT_g$. \label{samerepvecs}
\end{lem}

\medskip \noindent \textbf{\enspace} Again, trivial.\hfill $\square $\medskip

Pick $p_0>1$ such that there exist repetition vectors of degree greater than 
$p_0$ in $\calT_0$. By Theorem \ref{families}, there are a finite number of
vectors $v_i$ such that every repetition vector of degree $p_0$ is of the
form $M^k v_i$. Now, for each $k,i$, let $p_{k,i}$ be the maximal degree of $
M^k v_i$, and let $p_i = \lim_{k \to \infty} p_{k,i}$. Since the $p_i$'s are
a finite set of real numbers, all greater than $p_0$, there are real numbers 
$p$ and $\epsilon$ such that $p > p_0 + \epsilon$ and such that each $p_i$
is greater than $p+ \epsilon$. Let $L$, $L^{\prime}$ and $L_0$ be given. By
lemma \ref{samerepvecs}, there is a length $R$ such that the set of
repetition vectors $v$ of degree $p$ and with $L_0 v>R$ is precisely the
same for the three tiling systems $\calT_f$, $\calT_g$ and $\calT_0$. Pick
generators $v_1, \ldots, v_N$ of the finite families of these repetition
vectors, with $L_0 v_i \le L_0 v_{i+1}$, and with $L_0 v_N < \lambda_{PF}
L_0 v_1$.

The asymptotic lengths of these repetition vectors are conjugacy invariants:

\begin{thm}
Suppose that $\calT_{f}$ and $\calT_{g}$ are conjugate and that $\left\{
v_{1},\dots v_{N}\right\} $ is a collection of repetition vectors of degree $
p$ that generates all repetition vectors of degree $p.$ Then, given $\delta
>0$, for every $i\in \{1,\ldots ,N\}$ and for any sufficiently large integer 
$m$, there exist $j,m^{\prime }$ such that $|LM^{m}v_{i}-L^{\prime
}M^{m^{\prime }}v_{j}|<\delta $. \label{necessary}
\end{thm}

\medskip \noindent \textbf{Proof.\enspace} Let $\phi :\calT_{f}\rightarrow 
\calT_{g}$ be the conjugacy. Since $\calT_{f}$ and $\calT_{g}$ are compact
metric spaces, $\phi $ is uniformly continuous. Now suppose $v_{i}$ is a
repetition vector of degree $p$. Let $t_{m,i}=LM^{m}v_{i}$. Then for every
tiling $x\in \calT_{f}$ there is a range of real numbers $r$ of size roughly 
$(p-1)t_{m,i}$, such that $T_{r}(x)$ is very close to $T_{r+t_{m,i}}(x)$,
where $T_{r}$ denotes translation by $r$. Specifically, as $m\rightarrow
\infty $, the distance from $T_{r+t_{m,i}}(x)$ to $T_{r}(x)$ can be made
arbitrarily small, and the size of the range of $r$'s, divided by $t_{m,i}$,
can be made arbitrarily close to $p-1$. This implies that $\phi (T_{r}(x))$
and $\phi (T_{r+t_{m,i}}(x))$ are extremely close. But $\phi
(T_{r}(x))=T_{r}(\phi (x))$ and $\phi (T_{r+t_{m,i}}(x))=T_{r+t_{m,i}}(\phi
(x))$, so $\calT_{g}$ admits a repetition vector whose pairing with $
L^{\prime }$ is extremely close to $t_{m,i}$, and whose degree is extremely
close to $p$. However, for $m$ large, all such repetition vectors are of the
form $M^{m^{\prime }}v_{j}$. \hfill $\square $\medskip

As an illustration of the power of this theorem, consider the substitution $
a\rightarrow aaaabb$, $b\rightarrow babbba$. The substitution matrix is $ 
\left(\begin{smallmatrix}
4 & 2\cr2 & 4
\end{smallmatrix}\right)
$, whose eigenvalues, namely 6 and 2, are both greater than one. It is easy
to see that $v=(1,0)^{T}$ is a repetition vector of degree 5, and this
vector generates the only family of repetition vectors with $p=5$. If $\calT
_{f}$ and $\calT_{g}$ are conjugate, for each large $m$ there must exist $
m^{\prime }$ such that $(LM^{m}-L^{\prime }M^{m^{\prime }})v$ is small.
Since $v$ is full, this implies there
is an integer $k$ such that $L^{\prime }=LM^{k}$. Those are the only
possible conjugacies.

As a second example, consider the substitution $a\rightarrow aaaabb$, $
b\rightarrow bbbbaa$. This has exactly the same substitution matrix, but the
periodicities are different. $v_{1}=(1,0)^{T}$ and $v_{2}=(0,1)^{T}$ are
both repetition vectors with degree 6. The possibility of having $j\neq i$
in Theorem \ref{necessary} allows for additional conjugacies. Indeed, it is
not hard to see that $\calT_{f}$ and $\calT_{g}$ are conjugate if (and only
if) either $L^{\prime }=(L_{1},L_{2})M^{k}$ or $L^{\prime
}=(L_{2},L_{1})M^{k}$ for some (possibly negative) integer $k$.

The difference between these two examples illustrates that one cannot obtain
sharp conditions on conjugacy from the substitution matrix alone. One needs
some details about the substitution itself.

Another class of examples illustrating the importance of the actual
substitution as opposed to its matrix can be constructed using powers of
matrices. For instance, the substitution $a\rightarrow aabaabbba,$ $
b\rightarrow bbabbaaab$ has matrix $M=
\left(\begin{smallmatrix}
5 & 4\cr4 & 5
\end{smallmatrix}\right)
=
\left(\begin{smallmatrix}
2 & 1\cr1 & 2
\end{smallmatrix}\right)^{2}$ 
and is the square of the substitution $a\rightarrow aab,$ $
b\rightarrow bba,$ and so it can be seen that the associated tiling space admits
(among others) conjugacies with $L^{\prime }=L
\left(\begin{smallmatrix}
2 & 1\cr1 & 2
\end{smallmatrix}\right)^m$, 
where $m$ can be any integer. 
But the substitution $a\rightarrow aaaaabbbb,$ $b\rightarrow
abbbbbaaa $, with the same matrix $M$, has only one family of repetition
vectors of degree $8$, namely those generated by $\left( 1,0\right) ^{T}$. 
In this case all conjugacies are of the form $L^{\prime }=LM^{k}
=L \left(\begin{smallmatrix} 2 & 1 \cr 1 & 2 \end{smallmatrix}\right)^{2k}.$

\bigskip
We thank Felipe Voloch, Bob Williams and Charles Radin for
helpful discussions.  This work was partially supported by Texas ARP
grant 003658-152 and by a Faculty Research Grant from the University of 
North Texas.

\bigskip
 
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