Page de maths de Thierry Monteil

Tilings and connectedness

To avoid heavy formalism we will say that a tiling is a map $T:\mathbb{Z}^2\to \{0,1\}$. The set $\mathcal{T}$ of tilings is endowed with the usual product topology (two tilings are close if they coincide on a big square $[-n,n] \times [-n,n]$) on which $\mathbb{Z}^2$ acts continuously by translations (for $(u,v)\in\mathbb{Z}^2$ and $T\in\mathcal{T}$, the tiling $(u,v).T$ is the map which sends $(x,y)$ to $T(x-u,y-v)$).

A tiling $T$ is said to be connected if its geometric realization $G(T)=\bigcup_{T(x)=1} x + [0,1]\times[0,1]$ is a connected subset of $\mathbb{R}^2$. Connectedness is invariant under the translations.

Damien Jamet and Jean-Luc Toutant are interested in the connectedness of discrete planes (which can be described in our formalism), and ask (at least it was the case in 2006!) whether it depends on the translation parameter.

Since two discrete planes with the same slope and thickness are in the the orbit closure of eachother (and the orbit closure of one of them is the set of discrete planes sharing their slope and thickness, letting the translation parameter vary), we can look for a dynamical view on the problem.

Hence, we are interested in the behaviour of the connectedness of a tiling with respect to the orbit closure operation.

Of course, it is easy to push the problems far away when the patterns causing the (dis)connectedness do not occur uniformly:

Here is a disconnected tiling with a connected limit point.

$\longrightarrow$

Here is a connected tiling with a disconncted limit point.

$\longrightarrow$

But discrete planes (with irrational slope) are uniformly recurrent tilings (tilings leading to minimal systems), so we can ask whether connectedness is a property of the orbit closure for such tilings. The following example shows that it is not the case:

Let $T$ be the characteristic map of the set $\{(x,y) \in \mathbb{Z}^2 \mid y \mbox{ divides } 2^{d(x)}\}$, where $d(n)$ denotes the dyadic valuation of the integer $n$, that is the bigger interger $k$ such that $2^k$ divides $n$ (with the conventions that $d(0)=\infty$ and that $0$ divides any number including $\infty$).

Here is a picture of the tiling $T$, which is uniformly recurrent and connected.

Unfortunately, the tiling $T-\delta_{(0,0)}$ is in its orbit closure (and conversely by minimality) and is not connected:

Hence, there is a priori no dynamical reason for the connectedness of a discrete plane not to depend on the translation parameter.

If we require even more regularity for the tiling than uniform recurrence, we can remark that the previous example is substitutive, and therefore sofic, since it can be generated by the following substitution:

It turns out that "stable connectedness" (which is stable under orbit closure) can be characterised as a kind of "uniform connectedness" or "local connectedness", more precisely:

All the tilings of the orbit closure of a tiling $T$ are connected if, and only if, for any $r$, there exists some $f(r)$ such that between two points at distance at most $r$ in $G(T)$, there exists a path of length at most $f(r)$ joining them in $G(T)$.

Indeed, if a tiling $T$ is uniformly connected (with respect to some map $f$), let $T'$ be a tiling in the orbit closure of $T$, and let $(x,y)$ and $(x',y')$ be two elements of $G(T')$. Since $T'$ is in the orbit closure of $T$, there exists $(a,b)\in\mathbb{Z}^2$ such that $G((a,b).T)$ coincides with $G(T')$ on the big square $[-A,A] \times [-A,A]$, where $A= \max(|x|,|y|,|x'|,|y'|)+f(|(x,y)-(x',y')|)$. Since $G((a,b).T)$ is a translate of $G(T)$, there exists a path joining $(x,y)$ and $(x',y')$ in $G((a,b).T)\cap [-A,A] \times [-A,A] = G(T')\cap [-A,A] \times [-A,A]$. Hence $T'$ is connected, therefore $T$ is stably connected.

Conversely, assume by contradiction that $T$ is stably connected but not uniformly connected: there exists a number $r$ such that, for any $n$, there exists a pair of points $(x_n,y_n)$ and $(x'_n,y'_n)$ in $G(T)$ at distance at most $r$ to eachother such that there is no path of length at most $n$ joining them in $G(T)$. Since $G(T)$ is made of unit squares, any point $(x,y)\in G(T)$ can be joined to $([x],[y])$ by a short path, hence we can assume tha the numbers $x_n$,$y_n$,$x'_n$,$y'_n$ are all integers. Since $\mathcal{T}$ is compact, the sequence $(-x_n,-y_n).T$ admits a limit point $T'$. Let $t_n$ be an increasing integer sequence such that $(-x_{t_n},-y_{t_n}).T$ converges to $T'$ since $|(x_n,y_n)-(x'_n,y'_n)|$ is bounded by $r$, we can refine the subsequence and assume that $x'_n-x_n$ and $y'_n-y_n$ are constant, with values $x$ and $y$ respectively. Now, since $T'$ is assumed to be connected, there exists a path in $G(T')$ from $(0,0)$ to $(x,y)$. Let $L$ be its length. For $n$ large enough, $G((-x_{t_n},-y_{t_n}).T)$ coincides with $G(T')$ on the square $[-L,L]\times[-L,L]$, hence there is a path of length $L$ joining $(x_{t_n},y_{t_n})$ and $(x'_{t_n},y'_{t_n})$ in $G(T)$, leading to a contradiction when $t_n$ is bigger than $L$.