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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 $0\leq t0$ 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 $\epsilon 0$ 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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\end{document}
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