Coxeter frieze patterns

1 Field-valued patterns

Throughout this document, we will study frieze patterns of finite height. This terminology (as compared to finite width or finite order) is unconventional but more convenient for formalisation. For us, the height of a frieze pattern corresponds to the number of rows, including the rows of ones but excluding the rows of zeros. Throughout this section, we fix an arbitrary field \(F\).

Definition 1.1
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Fix \(n \in \mathbb {N}^*\). A map \(f : \{ 0,1,\ldots , n,n+1\} \times \mathbb {Z} \longrightarrow F\) is called an \(F\)-valued pattern of height n if,

1) for all \(m \in \mathbb {Z}\), \( f(0,m) = f (n+1,m) = 0\),

2) for all \(m \in \mathbb {Z}\), \( f(1,m) = f (n,m) = 1\), and

3) for all \((i,m) \in \{ 1,2,\ldots , n\} \times \mathbb {Z}\), we have

\begin{equation} \label{eq:diamond} f(i,m) f(i,m+1) = 1 + f(i+1,m) f(i-1, m+1). \end{equation}
1

An \(F\)-valued pattern \(f\) of height \(n\) is said to be nowhere zero if \(f(i,m) \neq 0\), for all \(i \in \{ 1,\ldots , n\} \) and for all \(m \in \mathbb {Z}\).

Lemma 1.2

Let \(f\) be a nowhere-zero \(F\)-valued pattern of height \(n\). For all \(m\), we have

\begin{align*} f(i+2,m) & = f(2,m+i) f(i+1,m) - f(i,m), \qquad i \in \{ 0, \ldots , n-1\} \\ f (i,m) & = f (n-1,m) f (i+1,m-1) - f(i+2,m-2), \qquad i \in \{ 0,n-1\} . \end{align*}
Proof

We begin by proving the first statement. That is, we prove

\[ P_i : \forall m \in \mathbb {Z}, f(i+2,m) = f(2,m+i+1) f(i+1,m) - f(i,m), \]

for \( i \in \{ 0, \ldots , n-1\} \). We do so by induction on \(i\).

Base case \(P_0\): We have that for all \(m \in \mathbb {Z}\), \(f(2,m) f(1,m) - f(0,m) = f (2,m+1)*1 - 0 = f (2,m)\).

Inductive hypothesis. Suppose that our claim holds for some \(i \in \{ 0,\ldots , n-2\} \) fixed. Then,

\begin{align*} f (i+3,m) f (i+1,m+1) & = f (i+2,m) f (i+2,m+1)-1 \\ & = f (i+2,m) (f(2,m+i+1) f(i+1,m+1) - f(i,m+1)) -1 \\ & = f (i+2,m) f (2,m+i+1) f (i+1,m+1) - (f (i+2,m)f (i,m+1) + 1) \\ & = f (i+2,m) f (2,m+i+1) f (i+1,m+1) - f(i+1,m) f (i+1,m+1). \end{align*}

Since \(f\) is nowhere-zero, we may divide both sides of the equation by \(f (i+1,m+1)\) to obtain the desired equality.

The second statement is proved almost identically. Namely, we prove

\[ Q_i: \forall m \in \mathbb {Z}, f (i,m) = f (n-1,m) f (i+1,m-1) - f(i+2,m-2), \]

by induction on \(i\), starting with \(i = n-1\) and proving the inductive step \(Q_i \Rightarrow Q_{i-1}\).

Base case \(Q_{n-1}\): for all \(m \in \mathbb {Z}\), \( f (n-1,m) f (n,m-1) - f(n+1,m-2) = f (n-1,m)*1 - 0 = f (n-1,m)\).

Inductive hypothesis. Suppose that \(Q_{i+1}\) holds for some fixed \(i \in \{ 0,\ldots , n-2\} \). Then,

\begin{align*} f (i,m) f (i+2,m-1) & = f (i+1,m-1) f (i+1,m) - 1\\ & = f (i+1,m-1) (f (i+2,m-1) f (n-1,m) - f(i+3,m-2)) -1 \\ & = f (i+1,m-1)f (n-1,m) f (i+2,m-1) - (f (i+1,m-1) f (i+3,m-2) + 1) \\ & = f (i+1,m-1)f (n-1,m) f (i+2,m-1) - f (i+2,m-2) f (i+2,m-1). \end{align*}

Again since \(f\) is nowhere-zero, dividing by \(f (i+2,m-1)\) on both sides we obtain \(Q_i\).

Proposition 1.3
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Let \(f\) be a nowhere-zero \(F\)-valued pattern of height \(n\). Then, for all \(m \in \mathbb {Z}\) and all \(i \in \{ 0,\ldots , n+1\} \), we have

\[ f(i,m) = f(i,m+n+1). \]
Proof

We prove a stronger statement, called the glide symmetry of frieze patterns. First, consider the map \(\rho _n: \{ 0,1,\ldots , n+1\} \times \mathbb {Z} \longrightarrow \{ 0,1,\ldots , n+1\} \times \mathbb {Z}\) given by

\begin{equation} \label{def:glide} \rho _n(i,m) = (n+1-i, m+i). \end{equation}
2

We show that every nowhere-zero \(F\)-valued pattern of height \(n\) is \(\rho _n\)-invariant, i.e. satisfies

\[ f(\rho _n(i,m)) = f(i,m), \qquad \forall (i,m) \in \{ 0,1,\ldots , n+1\} \times \mathbb {Z}. \]

The proposition will then follow by observing that \(\rho _n^2 : (i,m) \mapsto (i,m+n+1)\). Thus, consider the statement

\[ P_i: \forall m \in \mathbb {Z}, f (i,m) = f (n+1-i,m+i), \]

where \(i \in \{ 0, \ldots , n+1\} \). To prove that \(P_i\) holds for all \(i\), it is sufficient to prove that \(P_0, P_1\) hold, and that \(P_i \wedge P_{i+1} \Rightarrow P_{i+2}\).

\(P_0:\) for all \(m \in \mathbb {Z}, f(0,m) = 0 = f(n+1,m)\).

\(P_1:\) for all \(m \in \mathbb {Z}, f(1,m) = 1 = f(n,m+1)\).

Now suppose we are given \(i \in \{ 0,1,\ldots , n-1\} \) such that \(P_i\) and \(P_{i+1}\) hold. Then, for any fixed \(m \in \mathbb {Z}\), we have

\begin{align*} f(i+2,m) & = f(2,m+i) f(i+1,m) - f(i,m)\\ & = f (n-1,m+i+2) f (n-i,m+i+1) - f (n+1- i, m + i) \\ & = f (n-i-1,m + i + 2). \end{align*}
Corollary 1.4
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Let \(f\) be a nowhere-zero \(F\)-valued pattern of height \(n\). Then, \({\rm Im}(f) := \{ f (i,m) : i \in \{ 1,\ldots , n\} , m \in \mathbb {Z}\} \) is a finite set.

Proof

Consider the finite set \(\mathcal{D} = \{ (i,m) : i \in \{ 1,\ldots , n\} , m \in \{ 0,\ldots , n\} \} \). By Proposition 1.3,

\[ {\rm Im}(f) = \{ f (i,m) : (i,m) \in \mathcal{D}\} , \]

and the right-hand side is obviously finite.