SageDays@ICERM: Cluster algebras and unitary friezes

Emily Gunawan (University of Connecticut)


  • I. Background: Friezes
  • II. Background: Cluster algebras
  • III. New results: Frieze vectors and unitary friezes

Part I. Friezes

A frieze is an image that repeats itself along one direction. The name comes from architecture, where a frieze is a decoration running horizontally below a ceiling or roof. From M. Ascher, Ethnomathematics, p. 162.

Drawing Drawing Drawing

Conway - Coxeter frieze (1970s)

A (type $A$) frieze is an array such that

  • it is bounded above and below by a row of $1$s
  • every diamond $$ \begin{array}{ccccccc} &b&\\[-1pt] a&&d\\[-1pt] &c& \end{array}$$ satisfies the diamond rule $ad-bc=1$.

A Conway - Coxeter frieze consists of only positive integers.

Example (a Conway - Coxeter frieze)

\begin{equation*} \begin{array}{lcccccccccccccccccccccccc} &&&1&&1&& 1&&1&&1&&1&&1&& \cdots\\[4pt] \text{Row $2$} &&\cdots&&\mathbf{{3}}&&\mathbf{1}&&\mathbf{2}&&\mathbf{2}&&\mathbf{1}&&3&&1&& \\[4pt] &&&2&&2&&1&&3&&1&&2&&2&&\cdots&\ \\[4pt] &&\cdots&&1&&1&&1&&1&&1&&1&&1&& \end{array} \end{equation*}
In [75]:
% run attachment/
friezepic = print_frieze(input_row=(1,2,2,3,1,2,4), width = 14, friezerow = 8)

The code produces a LaTeX source code which would produce the following picture Drawing

Children practicing arithmetic

Note: every frieze is completely determined by the 2nd row.


Children practicing arithmetic: Answer Key

Drawing Drawing

What do the numbers around the integers count?


The corresponding frieze is below.


$%Answer: the number of triangles adjacent to each vertex.$

Theorem (Conway and Coxeter, 1970s)

A Conway - Coxeter frieze with $n$ nontrivial rows $\large\longleftrightarrow$ a triangulation of an $(n+3)$-gon

Note: Hence Conway - Coxeter friezes are Catalan objects.

Part II. Cluster Algebra (Fomin - Zelevinsky, 2001)

Let $Q$ be a quiver (a directed graph) on $n$ vertices with no loop and no 2-cycle.
E.g. Below is a quiver of affine Dynkin type $\widetilde{\mathbb{A}}_{1,2}$, which has 1 arrow pointing counterclockwise and 2 arrows pointing clockwise.

In [76]:
Q=ClusterQuiver([[1,0],[2,1],[2,0]]), circular=True)
  • A cluster algebra from $Q$ is a subalgebra $\mathcal{A}$ of the field of rational functions in $n$ variables.
  • The generators of $\mathcal{A}$ are called cluster variables, which are computed from $Q$ as we explain below.

Note: Python indexing starts at $0$, but our indexing starts at $1$.

In [77]:
A Cluster Algebra with cluster variables x0, x1, x2 and no coefficients over Integer Ring

Initial Seed (cluster + quiver)

Start with an initial cluster (a set of cluster variables) of size n.

In [78]:
[x0, x1, x2]


We can mutate a cluster at each of the vertices 1, 2, $\dots$, n.
Below, we mutate the initial cluster at vertex $0$ to get a new cluster variable.

In [79]:
[(x1*x2 + 1)/x0, x1, x2]

We get a new quiver by reversing all arrows adjacent to vertex $0$.

In [80]:
ClusterQuiver(S.b_matrix()).show(fig_size=0.1, circular=True)

Mutating all clusters

Continue mutating all clusters at all vertices.
Below, we mutate at $1$ after mutating at $0$.

In [81]:
[(x1*x2 + 1)/x0, (x1*x2^2 + x0 + x2)/(x0*x1), x2]
In [82]:
ClusterQuiver(S.b_matrix()).show(fig_size=0.1, circular=True)

Laurent Phenomenon and Positivity

As we mutate many times, the cluster variables (which we are dividing by) get more and more complicated, but we keep producing positive Laurent polynomials.

In [83]:
for x in S.cluster_variables():
(x1^2*x2^4 + x0^3*x1 + x0^2*x1*x2 + 2*x0*x1*x2^2 + 2*x1*x2^3 + x0^2 + 2*x0*x2 + x2^2)/(x0^2*x1^2*x2)

(x0 + x2)/x1

(x1^3*x2^6 + x0^5*x1^2 + x0^4*x1^2*x2 + 2*x0^3*x1^2*x2^2 + 2*x0^2*x1^2*x2^3 + 3*x0*x1^2*x2^4 + 3*x1^2*x2^5 + 2*x0^4*x1 + 4*x0^3*x1*x2 + 5*x0^2*x1*x2^2 + 6*x0*x1*x2^3 + 3*x1*x2^4 + x0^3 + 3*x0^2*x2 + 3*x0*x2^2 + x2^3)/(x0^3*x1^3*x2^2)

Theorem (Fomin - Zelevinsky, Gross - Hacking, Sean Keel - Kontsevich, Lee - Schiffler):

Every cluster variable is a Laurent polynomial with positive coefficients in the initial cluster variables, that is, every cluster variable $x$ is $$x=\frac{g(x_1,\dots,x_n)}{x_1^{d_1} \dots x_n^{d_n}}$$ where $g(x_1,\dots,x_n)\in \mathbb{Z}_{>0}[x_1,\dots,x_n]$, that is, a polynomial with positive coefficients.

Part III. Frieze vectors and Unitary friezes

For the rest of the talk, we will discuss Frieze vectors and unitary friezes, joint with R. Schiffler.

  • Comments are welcome.

Type $A$ Frieze (over an integral domain)

In general, a frieze (of type $A$) is an array of elements of an integral domain $R$ such that

  • it is bounded above and below by a row of $1$s
  • every diamond
$$ \begin{array}{ccccccc} &b&\\[-1pt] a&&d\\[-1pt] &c& \end{array} $$

satisfies the rule $ad-bc=1$.

Example: a frieze over the cluster algebra

Note: I omit the rows of $1$s.

\begin{align*} &\hspace{10pt} x_3\hspace{10pt} && \frac{x_1x_3+1+x_2}{x_2x_3} && \frac{x_2+1}{x_1} && \hspace{10pt} x_1 \hspace{10pt} \\ \hspace{5pt} x_2\hspace{5pt} && \frac{x_1x_3+1}{x_2} && \hspace{-5pt} \hspace{-5pt} \frac{x_2^2+2x_2+1+x_1x_3}{x_1x_2x_3}\hspace{-5pt}\hspace{-5pt} && \hspace{5pt} x_2\hspace{5pt} && \\ & \hspace{10pt} x_1 && \frac{x_1x_3+1+x_2}{x_1x_2} && \frac{x_2+1}{x_3} && \hspace{10pt} x_3 \end{align*}

Friezes over the integers

Specializing $x_1=x_2=x_3=1$ gives a Conway - Coxeter (positive integer) frieze

\begin{equation*} \begin{array}{lcccccccccccccccccccccccc} & 1 && 3 && 2 && 1 && \\ 1 && 2 && 5 && 1 && \\ & 1 && 3 && 2 && 1 && \end{array} \end{equation*}

Specializing $x_1=x_2=1$ and $x_3=-1$ gives

\begin{equation*} \begin{array}{lcccccccccccccccccccccccc} & -1 && -1 && 2 && 1 && \\ 1 && 0 && -3 && 1 && \\ & 1 && 1 && -2 && -1 && \end{array} \end{equation*}

Frieze over the Gaussian integers $\mathbb{Z}[i]$

Specializing $x_1=1$, $x_2=i$, and $x_3=i$ gives

\begin{equation*} \begin{array}{lcccccccccccccccccccccccc} & \hspace{5pt} i \hspace{5pt} && \hspace{-5pt}-1-2i \hspace{-5pt} && 1+i && \hspace{5pt}1 \hspace{5pt} && \\ i && 1-i && -3i && i && \\ & 1 && 2-i && 1-i && i && \\ 1 && 1 && 1 && 1 && \end{array} \end{equation*}

Frieze over the quadratic integer ring $\mathbb{Z}\left[\sqrt{-3}\right]$

Specializing $x_1=1$, $x_2=\frac{1+\sqrt{-3}}{2}$, $x_3=1$ gives

\begin{equation*} \begin{array}{lcccccccccccccccccccccccc} & 1 && \scriptstyle 2-\sqrt{-3} && \frac{3+\sqrt{-3}}{2} && 1 && \\ \frac{1+\sqrt{-3}}{2} && \scriptstyle 1-\sqrt{-3} && \frac{7-\sqrt{-3}}{2} && \frac{1+\sqrt{-3}}{2} && \\ & 1 && \scriptstyle 2-\sqrt{-3} && \frac{3+\sqrt{-3}}{2} && 1 && \end{array} \end{equation*}

Friezes as ring homomorphisms

Given any quiver $Q$, let a frieze be a ring homomorphism $$F:\mathcal{A}(Q) \to R$$ where $R$ is an integral domain, for example $R=\mathbb{Z}$.

Examples of friezes

  • $Id: \mathcal{A}(Q) \to \mathcal{A}(Q)$
  • $F: \mathcal{A}(Q) \to \mathbb{Z}$ defined by
    $(x_1,\dots,x_n) \mapsto (1,\dots,1)$ for a cluster $(x_1,\dots,x_n)$

Unitary friezes

A positive integral frieze $F$ is called unitary if $F$ can be obtained by specializing every element in one cluster to $1$.

Theorem 1 (G, Schiffler)

Let $Q$ be any quiver. The positive integral unitary friezes are in bijection with clusters.

Theorem 2 (G, Schiffler)

Every positive integral frieze of type $\widetilde{\mathbb{A}}_{p,q}$ is unitary.


  • An acyclic (no oriented cycles) quiver of type $\widetilde{\mathbb{A}}_{p,q}$ is a cyle with $p+q$ vertices with $p$ arrows oriented clockwise and $q$ arrows oriented counterclockwise.


  • For type $\mathbb{A}$ and $\widetilde{\mathbb{A}}$, every positive integral frieze is unitary.
  • For type $\mathbb{D}$, $\mathbb{E}$, $\widetilde{\mathbb{D}}$, and $\widetilde{\mathbb{E}}$, there are non-unitary friezes.

Future directions

Type Dynkin affine $\widetilde{\mathbb{D}}$

  • Classify the non-unitary friezes
  • Conjecture (based on Sage experiments): Up to cluster automorphisms (symmetry of the quiver), there are finitely many friezes.