Emily Gunawan (University of Connecticut)
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.
A (type $A$) frieze is an array such that
A Conway - Coxeter frieze consists of only positive integers.
% run attachment/friezedemo.py
friezepic = print_frieze(input_row=(1,2,2,3,1,2,4), width = 14, friezerow = 8)
view(friezepic)
The code produces a LaTeX source code which would produce the following picture
The corresponding frieze is below.
$%Answer: the number of triangles adjacent to each vertex.$
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.
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.
Q=ClusterQuiver([[1,0],[2,1],[2,0]])
Q.show(fig_size=0.1, circular=True)
Note: Python indexing starts at $0$, but our indexing starts at $1$.
A=ClusterAlgebra(Q)
A
A Cluster Algebra with cluster variables x0, x1, x2 and no coefficients over Integer Ring
Start with an initial cluster (a set of cluster variables) of size n.
S=A.initial_seed()
S.cluster_variables()
[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.
S=A.initial_seed()
S.mutate(0)
S.cluster_variables()
[(x1*x2 + 1)/x0, x1, x2]
We get a new quiver by reversing all arrows adjacent to vertex $0$.
ClusterQuiver(S.b_matrix()).show(fig_size=0.1, circular=True)
Continue mutating all clusters at all vertices.
Below, we mutate at $1$ after mutating at $0$.
S=A.initial_seed()
S.mutate([0,1])
S.cluster_variables()
[(x1*x2 + 1)/x0, (x1*x2^2 + x0 + x2)/(x0*x1), x2]
ClusterQuiver(S.b_matrix()).show(fig_size=0.1, circular=True)
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.
S=A.initial_seed()
S.mutate([0,1,2,0,1,2,1])
for x in S.cluster_variables():
print(x)
print('')
(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)
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.
For the rest of the talk, we will discuss Frieze vectors and unitary friezes, joint with R. Schiffler.
In general, a frieze (of type $A$) is an array of elements of an integral domain $R$ such that
satisfies the rule $ad-bc=1$.
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*}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*}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*}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*}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 $R=\mathbb{Z}$.
A positive integral frieze $F$ is called unitary if $F$ can be obtained by specializing every element in one cluster to $1$.
Let $Q$ be any quiver. The positive integral unitary friezes are in bijection with clusters.
Every positive integral frieze of type $\widetilde{\mathbb{A}}_{p,q}$ is unitary.
Type Dynkin affine $\widetilde{\mathbb{D}}$