My undergraduate dissertation was on inverse semigroups and the key text I used for it was Lawson's, "Inverse Semigroups: The Theory of Partial Symmetries". In said book, Lawson describes tiling semigroups. Upon hearing about the einstein "hat" tile (and "spectre"), I began to wonder what the tiling semigroup of "hat" is.
What follows is some definitions, taken from Lawson's book and paraphrased slightly.
Definition: Let $\mathcal{T}$ be a tiling of $\Bbb R^2$. A finite subset of a tiling (as a union of tiles) is a pattern. Two patterns $P,Q$ are equivalent if there is a translation $\tau$ of $\Bbb R^2$ with $\tau(P)=Q$. (This need not be a symmetry of the tiling.) A doubly pointed pattern is a triple $(p_2,P, p_1)$, where $P$ is a pattern, and $p_1$, the in-tile, and $p_2$, the out-tile, are distinguished tiles from $P$. Define
$$[p_2, P, p_1]=\{(q_2,Q,q_1)\mid \exists \text{ translation }\tau, \tau(P)=Q, \tau(p_1)=q_1, \tau(p_2)=q_2\}$$
to be a doubly pointed pattern class; denote by $S(\mathcal{T})$ the set of all doubly pointed pattern classes, together with a new symbol, $0$. Given $[p_2, P, p_1], [q_2, Q, q_1]$, suppose there are translations $\tau_1,\tau_2$ of $\Bbb R^2$ such that $\tau_1(P),\tau_2(Q)$ are patterns and $\tau_1(p_1)=\tau_2(q_2)$. Define
$$[p_2, P, p_1]\cdot [q_2, Q, q_1]=[\tau_1(p_1), \tau_1(P)\cup\tau_2(Q), \tau_2(q_1)],$$
where $\tau_1(P)\cup\tau_2(Q)$ is the union of the patterns $\tau_1(P),\tau_2(Q)$; and if no such translations exist, the product is defined as $0$, and the product is $0$ whenever one of the classes is $0$. The semigroup $(S(\mathcal{T}), \cdot)$ is called the tiling semigroup of $\mathcal{T}$.
The Question:
Given a "hat" tiling $\mathcal{T}$, what is its tiling semigroup?
Thoughts:
My intuition is that it's either trivial or infinite.
I don't know whether "hat" uniquely defines a tiling (hence 'given a "hat" tiling'), but given any tiling, the tiling semigroup is unique.
I can't offer much more than that.
