The set of all sequences on $\{0,1\}$ with Tychonoff topology is homemorphic to the Cantor set.

The map that replaces $1$ with $10$ is a homeomorphism.

Tail equivalence is dense on this subspace. i.e. every equivalence class is dense.

``Penrose tilings up to isometry" is a trivial topological space.

Connes' school of noncommutative geometry studied Penrose tilings via $\mathbb{C}^*$-algebras and $K$-theory.

Paul Smith saw Penrose tilings as points in a noncommutative space whose quasicoherent sheaves were given by graded modules over a noncommutative algebra.

Smith's ring was \begin{align*} P=\frac{\mathbb{C}\langle x,y \rangle}{(x^2)} \end{align*} where $x,y$ both lived in degree $1$ and he considered $\mathbb{N}$-graded modules over $P$ modulo tail equivalence.

Categories for Grassmannian cluster algebras of infinite rank. And also Cluster structures for the $A_\infty$- singularity.

They had the ring \begin{align*} R = \frac{\mathbb{C}[x,y]}{(x^2)} \end{align*} where $x$ lived in degree $1$ and $y$ in $-1$.

Could we get Penrose tilings involved?

Yes, we could.

Consider the category of $\mathbb{Z}$-graded maximal Cohen-Macaulay modules over $R$. Cluster tilting subcategories of this Frobenius category are in bijection with fountains of the infinitigon.

We will define two classes of arcs for a right fountain $R$.

The class $A_1(R)$ will contain the fountain arcs.

The class $A_2(R)$ will contain the arcs that connect the fountain arcs.

We will define two sequences.

\begin{align*} \boldsymbol{x}^R_n = \begin{cases} 1 \quad \text{if } (0,n+1) \in R \\ 0 \quad \text{else} \end{cases} \end{align*}

In the example, \begin{align*} 0,1,0,0,0,1,1,\ldots\end{align*}

And we define the sequence $\boldsymbol{y}^R$ by the rule that $\boldsymbol{y}^R_n$ is the $(n+1)$st integer $\ell$ such that $(0,\ell) \in R$.

In the example, \begin{align*} 3,7,8,\ldots\end{align*}

Every nonboundary arc in a right fountain is a diagonal of a unique quadrileteral.

Mutation at a nonboundary arc in a right fountain replaces this arc with the other diagonal.

Transfinite mutations in the completed infinitigon were studied by Baur-Gratz.

We want more mutation classes.

Lemma 1. Let $m$ be a positive integer and $\mu$ be the mutation at the fountain arc $(0, \boldsymbol{y}^R_m)$. Then, the sequences $\boldsymbol{x}^R$ and $\boldsymbol{x}^{\mu(R)}$ differ exactly at one place.

Lemma 2. Let $m$ be a positive integer and $\mu$ be the mutation at the arc $(\boldsymbol{y}^R_m, \boldsymbol{y}^R_{m+1})$. Then, the sequences $\boldsymbol{x}^R$ and $\boldsymbol{x}^{\mu(R)}$ differ exactly at one place.

Two right fountains are called equivalent if one can be mutated into the other one by using only finitely many mutations at arcs in $A_1$ or $A_2$.

Two right fountains $R_1$ and $R_2$ are equivalent if and only if $\boldsymbol{x}^{R_1}$ and $\boldsymbol{x}^{R_2}$ differ at finitely many places.

For an integer $n \geq 1$ \[ \boldsymbol{x}_n^R = \boldsymbol{x}_{n+1}^R = 1 \iff \boldsymbol{y}_{n+1}^R - \boldsymbol{y}_n^R = 1 \] which is equivalent to saying that $(\boldsymbol{y}^R_n, \boldsymbol{y}^R_{n+1})$ is a boundary arc.

We say that a right fountain is special if it is equivalent to a right fountain where $A_2$ does not contain any boundary arcs.

An infinite (right) half-frieze is a set of positive integers $m_{(a,b)}$ with $0 \leq a \leq b$ which satisfies the following rules:

• $m_{(a,a)} = 0$ for any $a \geq 0$.
• $m_{(a,a_{a+1})} = 1$ for any $a \geq 0$.
• $m_{(a,b)}m_{(a+1,b+1)} - m_{(a+1,b)}m_{(a,b+1)} = 1$ for any pair $(a,b)$ with $0 \leq a \leq b$.

The quiddity sequence is the sequence $m_{(a, a+2)}$.

The quiddity sequence determines the freeze.

Let $\C$ denote the category of $\mathbb{Z}$-graded MCM modules over $R=\mathbb{C}[x,y]/(x^2)$.

$R$ has countable Cohen-Macaulay type.

The module $(x,y^k)(j)$ corresponds to the arc $(-j-k, 1-j)$.

The modules $R(j)$ correspond to boundary arcs.

The modules $\dfrac{R}{x}(j)$ correspond to infinite arcs $(j, \infty)$.

The AR-quiver is

Infinite arcs do not appear because they are not generically free.

A cluster character on $\C$ with values in a commutative ring $R$ is a map $X$ from $\mathrm{obj}(\C)$ to $R$ which takes the same values on isomorphic objects and which satisfies $X(M \oplus N) = X(M) X(N)$ and also for any $M,N$ with \[\dim_\mathbb{C}\Ext_R^1(M,N) = 1 = \dim_\mathbb{C}\Ext_R^1(N,M)\]

we want an equality \[ X(M)X(N) = X(A) + X(B)\] where \[0 \to M \to A \to N\to 0 \\ 0 \to N \to B \to M \to 0 \]