Friezes, weak friezes, and T-paths

Frieze patterns form a nexus between algebra, combinatorics, and geometry. T-paths with respect to triangulations of surfaces have been used to obtain expansion formulae for cluster variables. This paper will introduce the concepts of weak friezes and T-paths with respect to dissections of polygons. Our main result is that weak friezes are characterised by satisfying an expansion formula which we call the T-path formula. We also show that weak friezes can be glued together, and that the resulting weak frieze is a frieze if and only if so was each of the weak friezes being glued.


Introduction
This paper will introduce weak friezes and T -paths with respect to dissections of polygons, and show that weak friezes are characterised by satisfying an expansion formula which we call the T -path formula. See Definitions 0.1, 0.2 and Theorem A.
Weak friezes are strongly related to the generalised frieze patterns defined in [2, sec. 5] and to the frieze patterns of [4, sec. 1], which form a nexus between algebra, combinatorics, and geometry; see [11] for a recent survey. T -paths with respect to dissections of polygons are a generalisation of T -paths with respect to triangulations of polygons as defined in [14, sec. 1.2].
In preparation for the proof of Theorem A, we show in Theorem B that weak friezes can be glued together. There is also a notion of frieze, and Theorem C shows that when weak friezes are glued together, the result is a frieze if and only if each of the weak friezes being glued is a frieze.
Note that T -paths with respect to triangulations of polygons and general surfaces were used in [1, thm. 2.10], [7, thm. 3.8], [12, thm. 1.1], [14, thm. 1.2], [15, thm. 3.1], and [16, thm. 3.5] to obtain expansion formulae for cluster variables. This permitted the resolution of the positivity conjecture for cluster algebras arising from surfaces, stating that all Laurent polynomials for cluster variables have positive coefficients. Cluster algebras and the positivity conjecture were introduced in [ (0.i). Friezes and weak friezes. If P is a polygon, α = β vertices, then there is a diagonal {α, β}, and the set of diagonals of P is denoted diag(P ). A dissection D of P is a set of pairwise non-crossing diagonals between non-neighbouring vertices, and the empty dissection D = ∅ is allowed.
Throughout the paper, (K, +, ·) is a fixed semifield, see Definition 1.1. An important feature is that K has no "subtraction" in the sense of an inverse operation of +. Examples are (R >0 , +, ·), the positive real numbers with addition and multiplication, and (Z, max, +), the so-called tropical semifield. In the latter case, the operations denoted + and · in the abstract are indeed given by maximum and addition in Z.
(i) f is called a frieze if it satisfies the Ptolemy relation The relation between (weak) friezes and (generalised) frieze patterns will be explained in Section (0.iv). Note that friezes with values in the tropical semifield (Z, max, +) were defined in [8, sec. 2.1]; they are known as tropical friezes.
Any map is a weak frieze with respect to the empty dissection. For example, a trivial weak frieze is a map on diagonals with constant value 1 K . Gluing together three trivial weak friezes with values in R >0 , two of them on 4-gons, the third on a pentagon, gives the weak frieze f on a 9-gon shown in Figure 1. Weak friezes can be glued by Theorem B, and the edges along which we glue become internal diagonals which are adjoined to the dissection D.
The notion of T -path with respect to a triangulation of a polygon was defined in [14, sec. 1.2], and we generalise it to T -path with respect to a dissection as follows.
Definition 0.2 (T -paths and the T -path formula). Let P be a polygon, D a dissection of P . If π 1 = π p are vertices, then a T -path from π 1 to π p with respect to D is an ordered tuple π = (π 1 , . . . , π p ) of vertices satisfying the following.
The set of T -paths from π 1 to π p with respect to D is denoted P(D, π 1 , π p ). A map f : diag(P ) − → K is said to satisfy the T -path formula with respect to D if for all vertices α = β of P , where for a T -path π.  Figure 1, these are the T -paths from α to β with respect to the dissection D.
For example, in the situation of Figure 1, the T -paths from α to β with respect to D are shown in Figure 2.
(0.iii). Gluing. The proof of Theorem A relies on our second main result.
Theorem B (Gluing weak friezes). Let P be a polygon, let d 1 , . . . , d m be pairwise non-crossing internal diagonals dividing P into subpolygons P 1 , . . . , P m+1 , and let D i be a dissection of P i for each i.
Let f i : diag(P i ) − → K be a weak frieze with respect to D i for each i. Assume that if P i and P j share a diagonal d, then f i (d) = f j (d) (note that such a d must be an edge of both P i and P j ).
Then there is a unique weak frieze f : diag(P ) − → K with respect to D such that f Our third main result concerns friezes.
Theorem C (Gluing friezes). In Theorem B, the weak frieze f is a frieze if and only if each f i is a frieze.
(0.iv). Frieze patterns. We end the introduction by explaining how (weak) friezes are related to (generalised) frieze patterns.
Let n 3 be an integer and consider the coordinate system on a horizontal strip in Figure 3. The coordinates are elements of Z/n, and taking a step right adds 1 to each coordinate. The first coordinate is constant when ascending diagonally, then second when descending diagonally. A (weak) frieze f on an n-gon P is turned into a pattern on the strip by labelling the vertices of P by Z/n and placing the value f (i, j) at position (i, j) in the coordinate system. Note that since f (i, j) = f (j, i), the pattern has a glide symmetry.
First, it is classic that if f is a frieze with values in R >0 satisfying f (i, i+1) = 1 for each i ∈ Z/n, then the corresponding pattern is a frieze pattern as defined in [4, sec. 1]. Namely, if i, j ∈ Z/n are non-neighbouring vertices, then {i, j} and {i + 1, j + 1} are crossing diagonals and Equation The four values of f in the equation constitute a "diamond" in the corresponding frieze pattern as shown. Secondly, if f is any frieze with values in R >0 , then the corresponding pattern is a frieze pattern with coefficients as defined in [ Thirdly, if f is a weak frieze with values in N = {1, 2, . . .} satisfying f (i, i + 1) = 1 for each i ∈ Z/n, then the corresponding pattern is a generalised frieze pattern as defined in [2, sec. 5]. For instance, the weak frieze f in Figure 1 gives the generalised frieze pattern in Figure 5. The entries in the first and last row are 1 because they are the values f (i, i + 1). The entry f (α, β) = 4 is shown in red.
The paper is organised as follows: Section 1 collects some definitions, Section 2 proves Theorems B and C, Section 3 establishes some properties of T -paths, and Section 4 proves Theorem A.

Semifields and polygons
Recall that throughout the paper, (K, +, ·) is a fixed semifield in the following sense.
where K is a set and + and · are binary operations, satisfying the following.
• The operation + is associative and commutative.
• The operation · turns K into a commutative group. The unit element is denoted 1 K , the inverse of x by x −1 . • The operation · distributes over +.
We abbreviate (K, +, ·) to K and write xy := x · y and x y := x · y −1 . Definition 1.2. A polygon P is a finite set V of three or more vertices with a cyclic order.
The predecessor and successor of α ∈ V are denoted α − and α + .
A diagonal of P is a two-element subset of V . The set of diagonals is denoted diag(P ).
The diagonal {α, β} has end points α and β. Diagonals of the form {α, α + } are called edges. The remaining diagonals are called internal diagonals.
A subpolygon is a subset of V of three or more vertices equipped with the induced cyclic order.
A dissection of P is a set D of pairwise non-crossing internal diagonals. If D has m elements, then it divides P into m + 1 subpolygons. Note that D can be empty.
The notation is further extended by permitting the sign =, for instance,

Proofs of Theorems B and C
In the proofs of Theorems B and C we will assume m = 1, that is, there is one diagonal d 1 dividing P into subpolygons P 1 and P 2 . This implies the general case by an easy induction. Note that the proofs do not use subtraction, which is unavailable in the semifield K.
Let P have the set of vertices V , denote d 1 by d, and pick ζ, η ∈ V such that d = {ζ, η} while the sets of vertices of P 1 and P 2 are Figure 6. Set There are disjoint unions D = D 1 · ∪ {d} · ∪ D 2 and when {ζ, η} and {α, β} are crossing diagonals of P . If f also satisfies f and x := f 1 (ζ, η) = f 2 (ζ, η), then f (α, β) must be given by the entries of the following table, according to whether α and β are in U 1 , U 2 , or {ζ, η}. {ζ,η} This shows uniqueness of the weak frieze f claimed in Theorem B.
To show existence of f , let f (α, β) be defined by the table. We must show that f is a weak frieze with respect to D; that is, for crossing diagonals {α, β} and {γ, δ} with {γ, δ} ∈ D, Equation (0.1) holds. It is necessary to treat a number of cases. We leave most of them to the reader, but show the computation for the case ζ < α < η < γ < β < δ < ζ, see Figure 6 (left), in which The definition of f and Equations (2.1) and (2.2) give and analogous equations where β is replaced with γ or δ. Hence as desired, where each red factor is replaced in the subsequent step. For (a) and (c), use Equation Proof of Theorem C. "Only if" is clear.
"If": Assume that f 1 and f 2 are friezes. We must show that f is a frieze; that is, for crossing diagonals {α, β} and {γ, δ}, Equation (0.1) holds. It is necessary to treat a number of cases. We leave most of them to the reader, but show the computation for the case ζ < α < γ < η < β < δ < ζ, see Figure 6 (right), in which Equation (2.6) implies that {ζ, γ}, {η, α} are diagonals of P 1 . Since they cross while f 1 is a frieze, we get the first of the following equations, and the second follows by an analogous argument.
(ii): Equation (0.2) holds because the sum on the right hand side has only the term f (α, β) by part (i) of the lemma. Setup 3.3 (Ears). Let P be a polygon, V the set of vertices of P , and let D be a non-empty dissection of P . It is easy to see that we can accomplish the following setup, which is sketched in Figure 7 and will be assumed for the rest of Section 3: The T -path π in the proof of Lemma 3.6 starts as shown (dashed).
The diagonal d = {ζ, η} in D divides P into subpolygons P 1 and P 2 such that D = {d} · ∪ D 2 for a dissection D 2 of P 2 . The sets of vertices of P 1 and P 2 are Observe that no diagonal in D has an end point in U 1 . In particular, D contains no internal diagonals of P 1 . Accordingly, we say that P 1 is an ear.
Step 3: R is injective: This is clear from the formula defining R.
Step 4: Equation "If": Assume that f satisfies the T -path formula with respect to D. We will prove by induction on ℓ := |D| that f is a weak frieze with respect to D.
If ℓ = 0 then D = ∅ and it is immediate from Definition 0.1(ii) that f is a weak frieze with respect to D.
If ℓ 1, then we can assume to be in the situation of Setup 3.3 and use the notation introduced there. Setting m = 1, d 1 = d, D 1 = ∅ then puts us in the situation of Theorem B.
is a weak frieze with respect to D 1 = ∅.
is a weak frieze with respect to D 2 .
By Theorem B there exists a unique weak frieze f : diag(P ) − → K with respect to D such that for vertices α = β of P . There are four cases to consider.
"Only if": Assume that f is a weak frieze with respect to D. We will prove by induction on ℓ := |D| that f satisfies the T -path formula with respect to D.
If ℓ = 0 then D = ∅. It follows that if α = β are vertices of P , then {α, β} crosses no diagonal in D, so Equation (0.2) holds by Lemma 3.2(ii). That is, f satisfies the T -path formula with respect to D.
If ℓ 1, then we can assume to be in the situation of Setup 3.3 and use the notation introduced there. It is clear from Definition 0.1(ii) that f diag(P 2 ) is a weak frieze with respect to D 2 . Since satisfies the T -path formula with 16İLKE Ç ANAKÇ I AND PETER JØRGENSEN respect to D 2 . To show that f satisfies the T -path formula with respect to D, let α = β be vertices of P . We must show that Equation (0.2) holds, and there are four cases to consider. Case 4: α ∈ U 2 , β ∈ U 1 . This reduces to Case 3 because each side of Equation (0.2) is symmetric in α and β. The left hand side is symmetric since f is defined on diag(P ), and the right hand side is symmetric by Lemma 3.1(ii).