Karin Baur

Eleonore Faber
email: e.m.faber@leeds.ac.uk

Sira Gratz
email: sira.gratz@glasgow.ac.uk

Khrystyna Serhiyenko
email: khrystyna.serhiyenko@berkeley.edu

Gordana Todorov
email: g.todorov@northeastern.edu

Michael J Cuntz

Pierre Guy Plamondon
email: pierre-guy.plamondon@math.u-psud.fr

FRIEZES SATISFYING HIGHER SL k -DETERMINANTS

Keywords: 2010 Mathematics Subject Classification. 05E10, 13F60, 16G20, 18D99, 14M15 frieze pattern, mesh frieze, unitary frieze, cluster category, Grassmannian, Iyama-Yoshino reduction

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Introduction 2

2. Background and notation 3 3. Integral tame SL k -friezes from Plücker friezes 10

Introduction

In this paper, we establish an explicit connection between SL k -friezes and Grassmannian cluster categories and algebras.

Integral SL k -friezes are certain arrays of integers consisting of finitely many rows of infinite length, see Example 2.4. Moreover, entries in an SL k frieze satisfy the so-called diamond rule, where for every k × k diamond formed by the neighboring entries, the determinant of the corresponding matrix equals 1. For example, when k = 2 each diamond a b c d satisfies the relation b a d c = 1. Moreover, we consider tame SL k frieze, that is friezes where the determinant of every (k + 1) × (k + 1) diamond is 0. Such friezes have horizontal period n, where n is determined by k and the number of rows. SL k -friezes were introduced in the seventies by Coxeter [START_REF] Harold | Frieze patterns[END_REF] in the case k = 2, while higher SL k -frieze patterns first appeared in work of Cordes-Roselle [START_REF] Cordes | Generalized frieze patterns[END_REF]. Conway and Coxeter further studied SL 2 -friezes in [START_REF] Conway | Triangulated polygons and frieze patterns[END_REF][START_REF] Conway | Triangulated polygons and frieze patterns[END_REF], where they showed that there exists a bijection between SL 2 friezes and triangulations of polygons. Interest in friezes and their various generalizations renewed after the introduction of cluster algebras in 2001, as cluster algebras coming from the Grassmannian of 2 planes in an n-dimensional space are also in bijection with triangulations of polygons. Moreover, it was shown later that SL 2 friezes can be obtained by specializing all cluster variables in a given cluster to one [START_REF] Caldero | Cluster algebras as Hall algebras of quiver representations[END_REF].

In this way, SL 2 friezes are well-understood and they are closely related to the combinatorics of cluster algebras. On the other hand, the classification of integral SL k -friezes remains elusive. Our paper makes a step in the direction of a complete classification. We show that in the finite type cases, all integral SL k -friezes can be obtained from the combinatorics of the cluster algebras on coordinate rings of Grassmannians, and Grassmannian cluster categories. We make use of the combinatorial tools that we call Plücker friezes, which arise from the cluster algebras, and mesh friezes, which arise from Grassmannian cluster categories.

Plücker friezes play a crucial role in our construction of SL k -friezes. Their entries are given by a constellation of Plücker coordinates in the homogeneous coordinate ring A(k, n) of the Grassmannian Gr(k, n) via the Plücker embedding. We show that the (specialized) Plücker frieze of type (k, n) deserves its name: it is indeed an SL k -frieze -all its k × k diamonds have determinant 1 (cf. Theorem 3.1). In particular, each map from A(k, n) to Z, given by a specialization of a cluster to one, yields a tame integral SL k -frieze (cf. Corollary 3.12). We call friezes obtained in this way unitary. This generalizes the results about friezes from the case k = 2, which have been well-known and studied for some time: Indeed, all SL 2 -friezes arise from the (specialized) Plücker frieze of type (2, n) in this way. However, for k > 2 not all friezes are unitary.

Next, we use categorification of cluster algebras to obtain all SL k friezes whenever A(k, n) is of finite type. A mesh frieze is an integral frieze on the Auslander-Reiten quiver of a Grassmannian cluster category such that entries coming from a mesh satisfy a certain frieze-like relation. In the case k = 2, a mesh frieze of the cluster category exactly agrees with the SL 2 -frieze. Thus, in light of the connection that we found, it makes sense to consider relationship between SL k -friezes and mesh friezes of Grassmannian cluster categories for k > 2. In finite type we show that there exists a bijection between the two. In particular, this implies that in finite type every SL k frieze is obtained by specializing the collection of cluster variables in a given cluster to some set of positive integers.

This also allows us to use tools from representation theory, such as Iyama-Yoshino reduction, to study SL k -friezes. It enables us to pass from a mesh frieze of a given "rank" (that is, a mesh frieze for a cluster category whose rank is given by the number of indecomposables in a cluster tilting object), to one of lower rank. Making use of known restrictions for smaller rank cases (in particular, of type A cases) we can then draw conclusions about the nature of our original mesh frieze. The natural bijection, in finite type cases, between mesh friezes and SL k -friezes means that Iyama-Yoshino reduction helps us better our understanding of integral SL k -friezes. In particular, we obtain new results on the number of such friezes and their possible entries.

An appendix by Cuntz and Plamondon determines the number of friezes in type (3, 7).

Acknowledgements. We want to thank Hugh Thomas for comments on an earlier version of the paper. We also want to thank the two anonymous referees for their helpful comments and suggestions.

Background and notation

2.1. SL k -friezes.

Definition 2.1. Let R be an integral domain. An SL k -frieze (over R) is a grid in the plane, consisting of a finite number of rows: 0 0 0 0 0 0 0 . The frieze is formed by k -1 rows of zeroes followed by a row of 1s from top and from bottom respectively and by w ≥ 1 rows of elements m ij ∈ R in between, such that every k × k-diamond of entries of the frieze has determinant 1, i.e. , whenever we consider a matrix having k successive entries of a row of an SL k -frieze on its diagonals and all other entries above and below the diagonal accordingly, the determinant of this matrix is 1. The integer w is called the width of the frieze.

Definition 2.2. Let F be an SL k -frieze over R.

(1) F is tame if all (k + 1) × (k + 1)-diamonds in F have determinant 0. In general, we define an s × s-diamond of the frieze, where 1 ≤ s ≤ k + 1, as the matrix having s successive entries of the frieze on its diagonals and all other entries above and below the diagonal accordingly.

(2) If all non-trivial entries m ij with 0 ≤ i, j ≤ w of F are positive integers, we call F an integral SL k -frieze.

Remark 2.3. One can prove that every tame SL k -frieze of width w over a field K has horizontal period n = w + k + 1, which was first proven for certain SL k -friezes in [START_REF] Cordes | Generalized frieze patterns[END_REF] and in general in [START_REF] Morier-Genoud | Linear difference equations, frieze patterns, and the combinatorial Gale transform[END_REF]Cor. 7 [START_REF] Cordes | Generalized frieze patterns[END_REF] (with extra conditions on the minors of the first k ×k-diamonds)

and seem only to have re-emerged with the introduction of cluster algebras: as 2-friezes [START_REF] Propp | The combinatorics of frieze patterns and Markoff numbers[END_REF], SL k -tilings in [START_REF] Bergeron | SL k -tilings of the plane[END_REF] and more systematically in [START_REF] Morier-Genoud | 2-frieze patterns and the cluster structure of the space of polygons[END_REF][START_REF] Morier-Genoud | Linear difference equations, frieze patterns, and the combinatorial Gale transform[END_REF][START_REF] Morier-Genoud | Arithmetics of 2-friezes[END_REF][START_REF] Cuntz | On wild frieze patterns[END_REF]. See also [START_REF] Morier-Genoud | Coxeter's frieze patterns at the crossroads of algebra, geometry and combinatorics[END_REF] for a survey on different types of frieze patterns.

Example 2.4. An SL 2 -frieze is simply called a frieze pattern. Frieze patterns have first been studied by Coxeter, [START_REF] Harold | Frieze patterns[END_REF] and by Conway and Coxeter in [START_REF] Conway | Triangulated polygons and frieze patterns[END_REF][START_REF] Conway | Triangulated polygons and frieze patterns[END_REF]. They showed that integral frieze patterns of width w are in bijection with triangulations of w + 3-gons. Their horizontal period is w + 3 or a divisor of w + 3. The following example

4 1 2 2 2 1 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 3 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 4 1 1 2 2 2 2 1 2 2 2 4 3 3 1 1 3 1 3 1
: : :

: : :

: : :

: : :

arises from a fan triangulation of a hexagon: The first non-trivial row of the frieze pattern (in bold face) is given by the number of triangles of the triangulation meeting at the vertices of the hexagon (counting these while going around the hexagon).

Plücker relations.

Throughout, we fix k, n ∈ Z >0 with 2 ≤ k ≤ n 2 and consider the Grassmannian Gr(k, n) as a projective variety via the Plücker embedding, with homogeneous coordinate ring

A(k, n) = C[Gr(k, n)].
This coordinate ring can be equipped with the structure of a cluster algebra as we briefly recall here. For details, we refer to [START_REF] Scott | Grassmannians and cluster algebras[END_REF].

We write [1, n] = {1, 2, . . . , n} for the closed interval of integers between 1 and n.

A k-subset of [1, n] is a k-element subset of [1, n]. Every k-subset I = {i 1 , . . . , i k } with i 1 < • • • < i k of [1, n]
gives rise to a Plücker coordinate p i 1 ,i 2 ,...,i k . We extend this definition to allow different ordering on the indices and repetition of indices: for an arbitrary ktuple I = (i 1 , . . . , i k ) we set p i 1 ,...,i k = 0 if there exists = m such that i = i m and p i 1 ,...,i k = sgn(π)p j 1 ,...,j k if (i 1 , . . . , i k ) = (j 1 , . . . , j k ), with j 1 < j 2 < • • • < j k and π is the permutation sending j m to i m for 1 ≤ m ≤ k. Sometimes, by abuse of notation, we will also call any tuple (i 1 , . . . , i k ) a k-subset. Scott proved in [START_REF] Scott | Grassmannians and cluster algebras[END_REF] that A(k, n) is a cluster algebra, where all the Plücker coordinates are cluster variables. For each (k, n), there exists a cluster consisting of Plücker coordinates and the exchange relations arise from the Plücker relations (see below).

Note that throughout the article we will consider sets I up to reduction modulo n, where we choose the representatives 1, . . . , n for the equivalence classes modulo n. If I consists of k consecutive elements, up to reducing modulo n, we call I a consecutive k-subset and p I a consecutive Plücker coordinate. These are frozen cluster variables or coefficients, and we will later set all of them equal to 1. Of importance for our purposes are the Plücker coordinates which arise from k-subsets of the form I = I 0 ∪ {m} where I 0 is a consecutive (k -1)-subset of [1, n] and m is any entry of [1, n] \ I 0 . We call such k-subsets almost consecutive. Note that each consecutive k-subset is almost consecutive. In particular the frozen cluster variables are consecutive and thus almost consecutive.

We follow the exposition of [Mar13, Section 9.2] for the presentation of the Plücker relations. In the cluster algebra A(k, n), the relations k =0 (-1) l p i 1 ,...,i k-1 ,j p j 0 ,..., j ,...,j k = 0 (1)

hold for arbitrary 1 ≤ i 1 < i 2 < • • • < i k-1 ≤ n and 1 ≤ j 0 < j 1 < • • • < j k ≤ n,
where • signifies omission. These relations are called the Plücker relations. For k = 2, the non-trivial Plücker relations are of the form

p a,c p b,d = p a,b p c,d + p a,d p b,c (2)
for arbitrary 1 ≤ a < b < c < d ≤ n. They are often called the three-term Plücker relations.

Obviously, the order of the elements in a tuple plays a significant role. We will often need to use ordered tuples, or partially ordered tuples. For this, we introduce the following notation. For a tuple J = {i 1 , . . . , i } of [1, n] and {a 1 , . . . , a

} = {i 1 , . . . , i } with 1 ≤ a 1 ≤ • • • ≤ a ≤ n we set o(J) = o(i 1 , . . . , i ) = (a 1 , . . . , a ).
With this notation, if 0 ≤ s, ≤ k, we will write p b 1 ,...,bs,o(i 1 ,...,i ),c 1 ,...,c k-+a for p b 1 ,...,bs,a 1 ,...,a ,c 1 ,...,c k-+a . For arbitrary I = {i 1 , . . . , i k-1 } and J = {j 0 , . . . , j k }, the Plücker relations (1) then take the form

k l=0 (-1) p o(I)j • p o(J\j ) = 0.

Plücker friezes, and special Plücker friezes.

To the homogeneous coordinate ring A(k, n) we now associate a certain frieze pattern which we will use later to obtain SL k -friezes. It uses the almost consecutive Plücker coordinates. For brevity, it is convenient to write [r] for the set {r, r + 1, . . . , r + -1}. For example, if k = 4 and n = 9, p [1] 3 ,6 is the short notation for p 1236 and p o([8] 3 ,4) is short for p 1489 .

Definition 2.5. The Plücker frieze of type (k, n) is a Z × {1, 2, . . . , n + k -1}-grid with entries given by the map

ϕ (k,n) : Z × {1, 2, . . . , n + k -1} → A(k, n) (r, m) → p o([r ] k-1 ,m ) ,
where r ∈ [1, n] is the reduction of r modulo n and m ∈ [1, n] is the reduction of m+r -1 modulo n. We denote the Plücker frieze by P (k,n) .

Example 2.6 and Figure 1 illustrate the map ϕ (k,n) .

Example 2.6. We draw P (2,5) as grid of Plücker coordinates with the positions in Z × {1, 2, . . . , 6} written in grey above them:

m=6 / / (-2, 6) (-1, 6) (0, 6) (1, 6) (2, 6) (3,
B B Remark 2.7. Notice that if m / ∈ {1, 2, . . . , k -1} then the image ϕ (k,n) (r, m) corresponds to a Plücker coordinate p o([r ] k-1 ,m ) =
0 and for m ∈ {1, 2, . . . , k -1}, the image is 0. In particular, the top and bottom k -1 rows of P (k,n) consist of solely 0 entries. Remark 2.8 (Specializing coefficients). In what follows, we will replace all the frozen variables by 1. For that, let J be the ideal of A(k, n) generated by {x -1 | x is a consecutive Plücker coordinate}.

p o([1] k-1 ,1) p o([2] k-1 ,2) p o([1] k-1 ,k-1) p o([2] k-1 ,k) p o([1] k-1 ,k) p o([2] k-1 ,k+1) p o([1] k-1 ,k+1) p o([2] k-1 ,k+2) p o([1] k-1 ,n-1) p o([2] k-1 ,n) p o([1] k-1 ,n) p o([2] k-1 ,1) p o([1] k-1 ,1) p o([2] k-1 ,2) p o([1] k-1 ,k-1) p o([2] k-1 ,k) . . .
p o([r] k-1 ,r) p o([r] k-1 ,r+k-2) p o([r] k-1 ,r+k-1) p o([r] k-1 ,r+k) p o([r] k-1 ,r+n-2) p o([r] k-1 ,r+n-1) p o([r] k-1 ,r+n) p o([r] k-1 ,r+k-2) . . . . . .
The consecutive Plücker coordinates are precisely the frozen variables in the cluster algebra A(k, n). The quotient A(k, n)/J is a coefficient-free cluster algebra, and therefore -as a subring of a field of rational functions over C -an integral domain. Let s : A(k, n) → A(k, n)/J be the map induced from the identity on non-consecutive Plücker coordinates and from replacing the consecutive Plücker coordinates by 1. This map is an algebra homomorphism A(k, n) → A(k, n)/J. We write sA(k, n) for A(k, n)/J. In this section, we prove the following result.

Theorem 3.1. The frieze sP (k,n) is a tame SL k -frieze over sA(k, n).

Before we provide a proof for Theorem 3.1, we make some observations on notation and set-up.

Notation 3.2. Throughout we calculate modulo n. More precisely, we will always reduce integers modulo n, and identify integers with their representatives in [1, n]. In addition, the following notation will be useful: For any a, b ∈ [1, n] we denote by [a, b] the closed interval between a and b modulo n, defined as Recall that we obtain the frieze sP (k,n) from the frieze P (k,n) by specialising the consecutive Plücker coordinates to 1. We will compute the determinants of a diamond in the frieze sP (k,n) via the corresponding diamond in P (k,n) . The k × k diamonds in the Plücker frieze P (k,n) are matrices of the form

[a, b] = {p ∈ [1, n] | a ≤ p ≤ b} if a ≤ b {p ∈ [1, n] | a ≤ p ≤ n or 1 ≤ p ≤ b} if b < a,
(a, b) = {p ∈ [1, n] | a < p < b} if a ≤ b {p ∈ [1, n] | a < p ≤ n or 1 ≤ p < b} if b < a.
A [m] k ;r := (p o([r+i-1] k-1 ,m+j-1) ) 1≤i,j≤k , with r ∈ [1, n] and m ∈ [r + k -1, r + n -1] = [r + k -1, r -1],
where as before we calculate modulo n, with representatives 1, . . . , n.

To inductively compute the determinants of the matrices A [m] k ;r , we need to consider other matrices of the following form: For r ∈ [1, n], 1 ≤ s ≤ n and m = (m 1 , . . . , m s ) with

m i ∈ [1, n] we define A m;r = (a ij ) 1≤i,j≤s
to be the (s × s)-matrix with entries a ij = p o([r+i-1] k-1 ,m j ) for 1 ≤ i, j ≤ s:

(3)

A m;r :=      p o([r] k-1 ,m 1 ) p o([r] k-1 ,m 2 ) . . . p o([r] k-1 ,ms) p o([r+1] k-1 ,m 1 ) p o([r+1] k-1 ,m 2 ) . . . p o([r+1] k-1 ,ms) . . . . . . . . . . . . p o([r+s-1] k-1 ,m 1 ) p o([r+s-1] k-1 ,m 2 ) . . . p o([r+s-1] k-1 ,ms)      .
Example 3.3. As an example, let k = 3 and n = 6 and consider the matrix A m;r in P (3,6) for r = 1, s = 3 and m = (3, 4, 5). It is given by

A (3,
m b < m b+1 < . . . < m s < m 1 < m 2 < . . . < m b-1 , if b ∈ {2, . . . , s} or m 1 < m 2 < . . . < m s if b = 1. (c2) We have r + k -2 / ∈ [m 1 , m s ).
Conditions (c1) and (c2) are technical conditions needed to ensure that we get the correct signs in our computations (cf. the proof of Proposition 3.5).

We now provide a formula for computing the determinants of the matrices A m;r with entries in A(k, n).

Proposition 3.5. Let r ∈ [1, n] and let 1 ≤ s ≤ k. Let m = (m 1 , . . . , m s ) with m i ∈ [1, n]
for all i, and assume that m satisfies conditions (c1) and (c2). Let b m;r be the determinant of the matrix A m;r from (3.3).

Then we have

b m;r = s-2 l=0 p o([r+l] k ) • p o([r+s-1] k-s ,m 1 ,...,ms) . (4)
The proof of this proposition can be found in Appendix A.

Remark 3.6. In particular, if k = s and m = [m] k in the statement of Proposition 3.5, then we have

b m;r = b m 1 ,m 2 ,...,m k ;r = k-2 l=0 p o([r+l] k ) • p o([m] k ) ;
a product of consecutive Plücker coordinates.

Proof of Theorem 3.1. All entries in the bottom k -1 rows and in the top k -1 rows of sP (k,n) are zero by definition and the entries in the kth row from top and from bottom are 1 since the consecutive Plücker coordinates are set to be 1, cf. Definition 2.9.

It remains to show that the k × k determinants in sP (k,n) are all 1 and that the (k + 1) × (k + 1)-determinants all vanish. We will use Proposition 3.5 for both claims. 3.3 is equal to k here) and the map s is the specialization of consecutive Plücker coordinates to 1 (cf. Definition 2.9). Recall that A m;r is of the form as in (3.3), where

Observe that each k × k-diamond in the special Plücker frieze sP (k,n) is of the form s(A m;r ) = (s(a ij )) 1≤i,j≤k , where A m;r = (a ij ) 1≤i,j≤k is a k × k-diamond in P (k,n) (the integer s from Equation
r ∈ [1, n] and m ∈ [r + k -1, r -1]
and m = (m, m + 1, . . . , m + k -1). Then the k-tuple m clearly satisfies Condition (c1). We are going to show that it also satisfies Condition (c2). Indeed, if we had

r + k -2 ∈ [m, m + k -1), then r + k -2 = m + j -1 for some 1 ≤ j < k. However, since m ∈ [r + k -1, r -1], this would imply r + k -2 = m + j -1 ∈ [r + k + j -2, r + j -2]. But for 1 ≤ j < k we have r + k -2 ∈ (r + j -2, r + k + j -2), and [r + k + j -2, r + j -2] ∩ (r + j -2, r + k + j -2) = ∅; a contradiction. So we must have r + k -2 / ∈ [m, m + k -1).
Therefore, both (c1) and (c2) are satisfied by m and we can apply Proposition 3.5 to obtain

det(A m;r ) = b m;r = k-2 l=0 p o([r+l] k ) • p o([m] k ) .

This yields det(s(A

m;r ) = s(b m;r ) = k-2 l=0 s(p o([r+l] k ) ) • s(p o([m] k ) ) = 1, for the k × k diamond s(A m;r ) of the special Plücker frieze sP (k,n) . Since this holds for every k × k diamond, sP (k,n) is indeed a SL k -frieze.
To show that it is tame, consider an arbitrary (k + 1) It follows that the determinant of the (k + 1)

× (k + 1) diamond of sP (k,n) . It must be of the form s(A [m] k+1 ;r ) = (s(a ij )) 1≤i,
det(A [m] k+1 ;r ) = k+1 l=1 (-1) k+1+l p o([r+k] k-1 ,m+l-1) • b ml ;r = k+1 l=1 (-1) k+1+l p o([r+k] k-1 ,m l ) • k-2 j=0 p o([r+j] k ) p o(m 1 ,..., m l ,...,m k+1 ) = k-2 j=0 p o([r+j] k ) • k+1 l=1 (-1) k+1+l p o([r+k] k-1 ,m l ) p o(m 1 ,..., m l ,...,m k+1 ) ) (
× (k + 1) diamond s(A [m] k+1 ;r ) vanishes, det(s(A [m] k+1 ;r )) = s(det(A [m] k+1 ;r )) = 0,
and thus sP (k,n) is tame.

Remark 3.7. In [MGOST14, Proposition 3.2.1], it is shown that the variety of SL k -friezes of width w = n -k -1, say over C, can be embedded into Gr(k, n) as the subvariety of points which can be represented by matrices whose consecutive k × k-minors all coincide (and are non-vanishing). In the strategy from [MGOST14, Proposition 3.2.1], we could choose the (k × n)-matrix P with entries in sA(k, n) given by 

   1 p o([n-k+2] k-1 2) . . . p o([n-k+2] k-1 n-k) 1 0 . . . 0 0 1 p o([n-k+3] k-1 3) . . . p o([n-k+3] k-1 n-k) p o([k-1] k-1 n-k+1) 1 . . . 0 . . . . . . . . . . . . . . . 0 0 . . . 1 p o([1] k-1 k+1) p o([1] k-1 k+2) . . . p o([1]n-1) 1     .
Each consecutive (k × k)-minor of the above matrix is 1 (as follows e.g. from Proposition 3.5). We can complete this to a unique tame SL k -frieze over sA(k, n). By Theorem 3.1 this must be precisely the specialised Plücker frieze of type (k, n).

Let now F be any tame SL k -frieze over C. As explained in [START_REF] Morier-Genoud | Linear difference equations, frieze patterns, and the combinatorial Gale transform[END_REF], it is uniquely determined by a k × n slice of the form

M =      1 m 12 m 13 . . . m 1,n-k 1 0 . . . 0 0 1 m 23 . . . m 2,n-k m 2,n-k+1 1 . . . 0 . . . . . . . . . . . . . . . 0 0 . . . 1 m k,k+1 . . . m k,n-1 1     
, representing a point in the cone over Gr(k, n) with respect to the Plücker embedding. The minors

M [r] k l of M , for 1 ≤ r ≤ k and l ∈ [r +k +1, r -2]
(with the interval taken cyclically modulo n), are precisely the entries of the matrix M , and we can view the matrix M as a pointwise evaluation of the matrix P at a point in the cone over the Grassmannian, and consequently, the frieze F as a pointwise evaluation of the specialised Plücker frieze sP (k,n) at a point in the cone over the Grassmannian Gr(k, n). This might justify considering the specialised Plücker frieze sA(k, n) the "universal" SL k -frieze of width w = n -k -1.

Integral tame SL k -friezes from Plücker friezes.

As an application of Theorem 3.1 we obtain that specialising a cluster to 1 yields an integral tame SL k -frieze as we will show now.

Definition 3.8. Let A be a cluster algebra of rank m, i.e., its clusters have cardinality m ∈ Z >0 . The specialization of a cluster x = (x 1 , . . . , x m ) in A to a tuple (a 1 , . . . , a m ) ∈ C m is the algebra homomorphism A → C determined by sending x i to a i for all 1 ≤ i ≤ m. If (a 1 , . . . , a m ) = (1, . . . , 1), we call this the specialization of x to 1.

Remark 3.9. Note that we consider clusters in a cluster algebra to be ordered tuples, rather than sets.

We observe here that specialising a cluster to a tuple (a 1 , . . . , a m ) determines values for all cluster variables, since the cluster algebra is generated by the cluster variables, each of which can be expressed as a Laurent polynomial in any given cluster. Since we are interested in integral SL k -friezes we consider specializations with respect to tuples in Z m >0 .

Remark 3.10. The image of the cluster variables in A under a specialization of a cluster to a tuple in (Z >0 ) m (or in (Q >0 ) m ) lies in Q >0 . This is due to the Laurent phenomenon ( [START_REF] Fomin | The Laurent phenomenon[END_REF]) and positivity (see [START_REF] Gross | Canonical bases for cluster algebras[END_REF][START_REF] Lee | Positivity for cluster algebras[END_REF][START_REF] Musiker | Positivity for cluster algebras from surfaces[END_REF]): Every non-initial cluster variable is a Laurent polynomial whose denominator is a monomial in the x i and whose numerator is a polynomial in the x i with positive coefficients . In particular, specialising a cluster to 1 sends every cluster variable in A to a positive integer.

Remark 3.11 (Tameness). Let F be a tame SL k -frieze over an integral domain R, and let ϕ : R → S be a unitary ring homomorphism from R to an integral domain S. Assume that the images of the entries of F lie in S for some subring S of S. Then the grid ϕ(F ) we obtain by evaluating ϕ entry-wise is a tame SL k -frieze over S .

Corollary 3.12. Let x be a cluster in A(k, n) and let ϕ

x : A(k, n) → C be the specialization of x to 1. Then ϕ x (sP (k,n) ) is a tame integral SL k -frieze of width w = n -k -1.
Proof. By Theorem 3.1 and Remark 3.10, ϕ x (sP (k,n) ) is an integral SL k -frieze; the tameness follows from Remark 3.11, since ϕ x is unitary. Its width follows from the definition of P (k,n) .

Remark 3.13. We see later (Lemma 4.8) that for k ≤ 3 and arbitrary n, as well as for k = 4 and n = 6 two different clusters x = x of a cluster algebra A produce different images ϕ x (sP (k,n) ) and ϕ x (sP (k,n) ) if and only if x is not a permutation of x .

Connection between the categories C(k, n) and SL k -friezes

In this section, we use indecomposable modules of the Grassmannian cluster categories C(k, n) to form friezes of the same shape as the (special) Plücker friezes. We describe the Ext-hammocks for entries in these friezes and characterize the cases where such a frieze of modules (see Def. 4.2) gives rise to cluster-tilting objects. With that, we will then give a bijection between friezes on Auslander-Reiten quivers of Grassmannian cluster categories which we will call mesh friezes (see Def. 4.9), and integral tame SL 3 -friezes.

4.1. The Grassmannian cluster categories.

We recall the definition of and results about the Grassmannian cluster categories from [START_REF] Bernt | A categorification of Grassmannian cluster algebras[END_REF].

Note that an alternative method for constructing an additive categorification of Grassmannian cluster algebras with coefficients has been provided by Demonet and Luo [START_REF] Demonet | Ice quivers with potential associated with triangulations and Cohen-Macaulay modules over orders[END_REF]; in the special case k = 2 they applied Amiot's construction of the generalized cluster category [START_REF] Amiot | Cluster categories for algebras of global dimension 2 and quivers with potential[END_REF] to an ice quiver with potential coming from a triangulation of a polygon. Let Q(n) be the cyclic quiver with vertices 1, . . . , n and 2n arrows

x i : i -1 → i, y i : i → i -1.
Let B = B k,n be the completion of the path algebra CQ(n)/ xy -yx, x k -y n-k , where xy -yx stands for the n relations x i y i -y i+1 x i+1 , i = 1, . . . , n and x k -y n-k stands for the n relations x i+k . . . x i+1 -y i+n-k+1 . . . y i (reducing indices modulo n).

Let t = x i y i . The centre Z = Z(B) is isomorphic to C[[t]].
Then we define the Grassmannian cluster category of type (k, n), denoted by C(k, n), as the category of maximal Cohen Macaulay modules for B. In particular, the objects of C(k, n) are (left) B-modules M , such that M is free over Z. This category is a Frobenius category, and it is stably 2-CY. It is an additive categorification of the cluster algebra structure of A(k, n), as proved in [START_REF] Bernt | A categorification of Grassmannian cluster algebras[END_REF]. Note that the stable category C(k, n) is triangulated, which follows from [Buc86, Theorem 4.4.1], since B is Iwanaga-Gorenstein, or from [Hap88, I, Theorem 2.6], since C(k, n) is a Frobenius category. Both C(k, n) and its stable version C(k, n) will be called Grassmannian cluster categories.

We recall from [START_REF] Bernt | A categorification of Grassmannian cluster algebras[END_REF] that the number of indecomposable non projective-injective summands in any cluster-tilting object of C(k, n) is (k -1)(n -k -1). This is called the rank of the cluster category C(k, n). We say that C(k, n) is of finite type, if if has finitely many isomorphism classes of indecomposable modules. Note also that (for k ≤ n/2) the category C(k, n) is of finite type if and only if either k = 2 and n arbitrary, or k = 3 and n ∈ {6, 7, 8}. These categories are of Dynkin type A n-3 , D 4 , E 6 and E 8 respectively. This means that one can find a cluster-tilting object in C(k, n) whose quiver is of the corresponding Dynkin type.

Remark 4.1 (Indecomposable objects of C(k, n)). The following results are from [START_REF] Bernt | A categorification of Grassmannian cluster algebras[END_REF]. The rank of an object M ∈ C(k, n) is defined to be the length of M ⊗ Z K for K the field of fractions of the centre Z ([JKS16, Definition 3.5]). There is a bijection between the rank one indecomposable objects of C(k, n) and k-subsets of [1, n]. We may thus write any rank one module as M I where I is a k-subset of [1, n]. In particular, the indecomposable projective-injective objects are indexed by the n consecutive k-subsets of [1, n] (note here that we consider cyclic consecutive k-subsets modulo n. The rank one indecomposables are thus in bijection with the cluster variables of A(k, n) which are Plücker coordinates, cf. Section 2.2. We denote the bijective association p I → M I by ψ k,n . Definition 4.2. Using the bijection between rank one modules and Plücker variables, we can form a frieze of rank 1 indecomposable modules through the composition

ψ k,n • ϕ (k,n) .
For the remainder of this section, we will write

M (k,n) to denote the image of ψ k,n • ϕ (k,n) .
We write 0s for the images of the p I where I has a repeated entry. As an example, we take (k, n) = (2, 5). The categories C(3, 9) and C(4, 8) are tame, these categories are known to be of tubular type, see [START_REF] Baur | Cluster categories from Grassmannians and root combinatorics[END_REF], where the non-homogenous tubes are described.

0 0 0 0 0 0 0 0 M 23 M 34 M 45 M 15 M 12 M 23 M 34 M 13 M 24 M 35 M 14 M 25 M 13 M 24 • • • • • • M 14 M 25 M 13 M 24 M 35 M 14 M 25 M 45 M 15 M 12 M 23 M 34 M 45 M 15 M 12 0 0 0 0 0 0 0 Remark 4.3

Description of Ext-hammocks.

We determine the shape of the Ext-hammocks in M (k,n) , that is, given I we describe the set of all J appearing in the frieze

M (k,n) such that Ext 1 C(k,n) (M I , M J ) is nonzero.
We prove that the Ext-hammocks consist of two maximal rectangles determined by I. We use the result in [START_REF] Bernt | A categorification of Grassmannian cluster algebras[END_REF], where the authors determine a precise combinatorial condition for two rank 1 modules to have a nonzero extension. For this section, we allow arbitrary 1 < k < n -1. Next, we define a maximal rectangle in the Plücker frieze starting (resp. ending) at I.

Let I = o([r] k-1 , m) be a non-consecutive k-subset.
Then we say that J belongs to the maximal rectangle starting at

I if J = o([s] k-1 ,
([r + 1] k-1 , m + 1) or the maximal rectangle ending at o([r -1] k-1 , m -1).
Proof. By Proposition 4.4 it suffices to show that I, J are crossing if and only if J belongs to the maximal rectangle starting at o([r + 1] k-1 , m + 1) or the maximal rectangle ending at o([r -1] k-1 , m -1). Recall that

I = o([r] k-1 , m) = {r, r + 1, . . . , r + k -2, m} and let J = o([s] k-1 , p) = {s, s + 1, . . . , s + k -2, p}
such that I and J cross. In the figure below we depict elements in I on a circle, and observe that I and J cross if and only if the following three conditions are satisfied.

m r r + 1 r + k -2 J ∩ (r + k -2, m) = ∅ J ∩ (m, r) = ∅ m ∈ J 1 1 1 • • • 1 1 o([m+1] k-1 ,m-1) o([r+1] k-1 ,r-1) o([m+1] k-1 ,r+k-1) o([r-1] k-1 ,m-1) I o([r+1] k-1 ,m+1) o([m-k+1] k-1 ,r-1) o([r-1] k-1 ,r+k-1) o([m-k+1] k-1 ,m+1) 1 1 1 • • • 1 1 Figure 2. Ext-hammock of M I where I = o([r] k-1 , m) Note that p ∈ [r, r + k -2],
because otherwise for the first two conditions above to be satisfied we must have m ∈ [s] k-1 ⊂ J. This clearly contradicts the third condition. Therefore, we have two cases p ∈ (m, r) or p ∈ (r + k -2, m).

Suppose p ∈ (m, r). Then I and J cross if and only if s + j ∈ (r + k -2, m) for some j ∈ {0, 1, . . . , k -2}, and also s + j = m for any such j. This implies that s + k -2 < m.

On the other hand, we must also have that r < s. All together, we obtain r < s < m -k + 2. Thus, we see from Figure 2 that J lies in the maximal rectangle starting at o([r + 1] k-1 , m + 1).

Similarly, suppose p ∈ (r + k -2, m). Then I and J cross if and only if s ∈ (m, r). From Figure 2 we see that this mean J lies in the maximal rectangle ending at o([r-1] k-1 , m-1). This shows the desired claim.

For arbitrary k and n, the category C(k, n) contains a cluster-tilting object consisting entirely of modules indexed by Plücker coordinates. This follows from the results in [START_REF] Scott | Grassmannians and cluster algebras[END_REF]. Next, we describe for which k and n there exists a cluster-tilting objected indexed by the almost consecutive Plücker coordinates.

Proposition 4.6. A Plücker frieze of type (k, n) contains a collection I of almost consecutive k-element subsets such that ⊕ I∈I M I is a cluster-tilting object of C(k, n) if and only if k = 2, 3 with n arbitrary, or k = 4 and n = 6.

Proof. In general, the cluster category C(k, n) contains a cluster-tilting object consisting of (k -1)(n -k -1) indecomposable summands. Therefore, we want to show that a Plücker frieze contains (k -1)(n -k -1) distinct k-element subsets that are pairwise non-crossing if and only if k = 2, 3 or k = 4 and n = 6.

If k = 2 then every object in C(k, n) is of the form M I where I is a 2 element subset of [1, n]. In particular, every such I lies in the associated Plücker frieze. This shows the claim for k = 2.

If k = 3 we want to find 2(n -4) pairwise non-crossing elements in the Plücker frieze.

Consider the following collection of 3-element subsets of [1, n] consisting of two disjoint families of size n -4.

{1, 2, m} m=4,5,...,n-1 and {s, s + 1, 1} s=3,4,...,n-2

Each of these corresponds to a diagonal in the Plücker frieze, thus by Proposition 4.5 we see that no two elements in the same family cross. Moreover, {1, 2, m} does not cross {s, s + 1, 1} by definition. This shows that the collection of 3-element subsets above are pairwise non-crossing.

If k ≥ 4 we want to show that the associated Plücker frieze does not contain (k-1)(n-k-1) pairwise non-crossing elements unless k = 4 and n = 6. We make use of the following key observation here: the Ext-hammocks in a frieze of width w have the same shape as the Ext-hammocks in the cluster category of type A with rank w. In particular, any region in the frieze that has the same shape as the fundamental region in the associated cluster category of type A, as depicted on the left in Figure 3, contains at most w pairwise non-crossing objects.

First we consider the case k = 4 separately. We see that the Plücker frieze admits two copies of the fundamental region followed by two columns, because we have w = n-k-1 = n -5 and the period of the frieze is n. See Figure 3 on the right. Now suppose that such frieze admits a collection I of 3w pairwise non-crossing objects. Because the frieze contains two such fundamental regions, each of which admits at most w non-crossing object, we conclude that the two remaining columns contain at least w element of I. However, a frieze is n-periodic, and the same argument as above implies that any two neighbouring columns contain at least w elements of I. Because there are w + 5 columns we conclude that I contains at least w + 5 2 w elements. We also know that the size of I is 3w which implies that w+5 2 ≤ 3. Therefore, there are two possibilities w = 1, k = 4, n = 6 or w = 2, k = 4, n = 7. In the first case, we see from the diagram below that we can take I = {1235, 1345, 1356}.

1 1 1 1 1 1 1 1 1235 2346 1345 2456 1356 1246 1235 1 1 1 1 1 1 1 1
In the second case, one can check that it is not possible to find I of size 6. This completes the proof for k = 4.

Finally, suppose k > 4 and let I be a collection of pairwise non-crossing elements in the Plücker frieze of size (k -1)(n -k -1) = (k -1)w. By the same reasoning as in the previous case we think of subdividing the frieze into fundamental regions. We begin by taking two copies of the frieze, so we have a rectangular array of elements of width w and length 2n. Next, we want to subdivide it into rectangles of width w and length w + 3, see Figure 3 Since k > 4 we can write

n > k -2 > (k -2)(k -2)
k -4 which in turn implies equations (6) and (5). This completes the proof in the case k > 4.

In terms of cluster algebras we get the following: We can apply this to determine when the specialization ϕ x of a cluster x to (1, . . . , 1) uniquely determines a specialized Plücker frieze.

Lemma 4.8. Let k ≤ 3 and n arbitrary or let k = 4 and n = 6. Let x and x be two clusters of Plücker variables. Then ϕ x (sP

(k,n) ) = ϕ x (sP (k,n) ) if and only if x is a permutation of x .
Proof. The assumptions on k and n guarantee that the Plücker frieze P (k,n) contains a cluster a of almost consecutive Plücker coordinates, see Prop. 4.6 and Cor. 4.7. If x is a permutation of x then ϕ x = ϕ x , which shows the reverse implication. On the other hand, assume that there are two clusters x = (x 1 , . . . , x N ) and x = (x 1 , . . . , x N ) of Plücker coordinates such that ϕ x (sP (k,n) ) = ϕ x (sP (k,n) ), where N = (k -1)(n -k -1) + n is the cardinality of a cluster in A(k, n). Since x and a are clusters, there exist Laurent polynomials Q

x,a i such that a i = Q

x,a i (x 1 , . . . , x N ). On the other hand, there also exist exchange Laurent polynomials Q a,x i such that x i = Q a,x i (a 1 , . . . , a N ) and Q a,x (Q x,a (x)) = x (where we understand Q x,a as the vector (Q x,a 1 , . . . , Q

x,a N )). Similarly we can find Q x ,x and Q x,x . By definition ϕ x (x) = (1, . . . , 1), and ϕ x (x ) = (1, . . . , 1), and in particular, the entries of both evaluations agree on the cluster a in sP (k,n) . This means:

(7) ϕ x (a) = ϕ x (Q x,a (x)) = Q x,a
(1, . . . , 1) .

On the other hand

(8) ϕ x (a) = ϕ x (Q x,a (Q x ,x (x ))) = Q x,a (Q x ,x (1, . . . , 1)) .
Since Q a,x (Q x,a (1, . . . , 1)) = (1, . . . , 1), applying Q a,x to (7) and (8) yields

(1, . . . , 1) = Q a,x (Q x,a (Q x ,x (1, . . . , 1))) = Q x ,x (1, . . . , 1) .
Note that, while we have defined the cluster algebra A(k, n) as a C-algebra, we have in fact

A(k, n) = C ⊗ C A Z (k, n),
where A Z (k, n) is a cluster algebra of geometric type, defined as the subring of Q(x 1 , . . . , x N ) generated by the cluster variables. Assume now as a contradiction that x is not a permutation of x . Then there is an 1 ≤ i ≤ N such that x i / ∈ {x 1 , . . . , x N } and

x i = Q x ,x i (x 1 , . . . , x N ) = f (x 1 , . . . , x N ) x d 1 1 • • • x d N N
, with x j f (x ) for all 1 ≤ j ≤ N . By positivity [START_REF] Gross | Canonical bases for cluster algebras[END_REF][START_REF] Lee | Positivity for cluster algebras[END_REF] this is a Laurent polynomial over Z ≥0 . Therefore, Q

x ,x i (x 1 , . . . , x N ) = 1 implies that f is the constant polynomial f = 1 and

x i = 1 x d 1 1 • • • x d N N .
However, by positivity of d-vectors, which was shown for skew-symmetric cluster algebras by [CL17, Thm. 1.2], we have d j ≥ 0 for all 1 ≤ j ≤ N and so

x i is a unit in A Z (k, n); contradicting [GLS13, Theorem 1.3].

Mesh friezes.

Assume k ≤ n/2. We recall that the Grassmannian cluster category C(k, n) is of finite type if and only if either k = 2, or k = 3 and n ∈ {6, 7, 8} (Section 4.1).

Definition 4.9. Let C(k, n) be of finite type. A mesh frieze F k,n for the Grassmannian cluster category C(k, n) is a collection of positive integers, one for each indecomposable object in C(k, n) such the F k,n (P ) = 1 for every indecomposable projective-injective P and such that all mesh relations on the AR-quiver of C(k, n) evaluate to 1 thus, whenever we have an AR-sequence

(9) 0 → A → I i=1 B i → C → 0,
where the B i are indecomposable, and I ≥ 1, we get Remark 4.12. Recall that C(k, n) is a categorification of the cluster algebra A(k, n). Furthermore, the mesh frieze has 1's at the projective-injective objects, i.e., the objects for consecutive k-subsets. In that sense, any mesh frieze F k,n for C(k, n) is also a mesh frieze associated to the cluster algebra sA(k, n) = A(k, n)/J with coefficients specialized to 1, cf. Remark 2.8. The requirement that mesh relations evaluate to 1 translates to requiring that certain exchange relations in the cluster algebra hold. We will use both points of view. The approach to friezes via C(k, n) is particularly useful when we want to check for the existence of cluster-tilting objects in the SL k -frieze. It will also be of importance in Section 5 as we want to use Iyama-Yoshino reductions on these categories to analyse frieze patterns.

F k,n (A)F k,n (C) = I i=1 F k,n (B i ) + 1. (Note that C(k, n) has
Remark 4.13. Note that the entries in a mesh frieze satisfy the exchange relations of the associated cluster algebra A(k, n). That is, denoting by x N the cluster variable in A(k, n) associated to an indecomposable object N in C(k, n) the following holds: Whenever we have an exchange relation

x A x C = m i=1 x Bi + m j=1 x B j in A(k, n), then F(A)F(B) = m i=1 F( Bi ) + m j=1 F( B j ).
Indeed, this follows from the fact that a mesh frieze is determined uniquely by the values of one cluster tilting object. Thus, we may take an acyclic cluster that forms a slice because k = 2 and n is arbitrary or k = 3 and n = 6, 7, 8, and then use sink/source mutations to determine all entries in the mesh frieze uniquely. Now this exact same cluster also determines values on the mesh frieze uniquely using arbitrary mutations of the cluster algebra. Thus, these two approaches should yield the exact same mesh frieze, which implies that the entries in the mesh frieze satisfy all relations of the cluster algebra, not just the ones coming from sink/source mutations.

4.4. Mesh friezes for C(k, n) yield SL k -friezes.

In this section, we show the correspondence between integral tame SL k -friezes and mesh friezes in the finite type cases. For k = 2, these two notions coincide (see Example 4.10). So for the rest of this section, we assume k = 3 and n ∈ {6, 7, 8}. The width of such an SL 3 -frieze is n -4, see Remark 2.3.

The following lemma shows that the entries in a tame SL 3 -frieze are determined by a cluster of A(k, n), which consists of consecutive Plücker coordinates.

Lemma 4.14. Let F 3,n with n ∈ {6, 7, 8} be a tame integral SL 3 -frieze. Then the entries of F 3,n are uniquely determined by the values on the 2w entries x 1 = m 00 , x 2 = m 01 , . . . , x w = m 0,n-5 and x w+1 = m 2,n-3 , . . . , x 2w = m n-3,n-3 . See below the picture for (3, 7).

0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1
x 3 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0
Proof. We show the assertion for F 3,7 . Here w = 3. Assume that the values of the entries (x 1 , . . . , x 6 ) are given. We first complete the frieze by repeating the entries along the diagonals (according to the (anti-)periodicity rule given in [START_REF] Morier-Genoud | Linear difference equations, frieze patterns, and the combinatorial Gale transform[END_REF][START_REF] Cuntz | On wild frieze patterns[END_REF]), that is, the frieze becomes infinite also in the vertical directions. See below the picture for F 3,7 :

. . . Then we can successively compute the remaining entries m ij of the frieze from the x i 's using the tameness condition, in particular the two rows of 0s for calculating the 4 × 4 determinants. First compute m 35 from the determinant of the 4

× 4 matrix A =     0 1 x 4 x 5 0 0 1 m 35 1 0 0 1 x 1 1 0 0     :
since by tameness det(A) = 0, this yields m 35 = x 1 +x 5 x 4 . Moreover from the determinant of the matrix

    0 0 1 m 35 1 0 0 1 x 1 1 0 0 x 2 m 11 1 0     we obtain m 11 = x 2 +m 35 x 1
.

Similarly, we obtain m 61 = x 2 +x 6

x 3 and m 33 = x 5 +m 61

1 -m 35 -x 4 (x 4 m 35 -x 5 ) B m 45 m 46 = -m 11 1 -x 6 .
This system has a unique solution (m 45 , m 46 ), since the determinant of the matrix B equals -x 5 = 0. Similarly, we may also uniquely determine m 50 and m 60 . The remaining entries may be calculated entirely by using determinants of the 3 × 3 diamonds.

The assertion for the cases (3, 6) and (3, 8) are proven by analogous tedious calculations.

Proposition 4.15. Let w ∈ {2, 3, 4} and n = w + 4. For every tame integral SL 3 -frieze of width w there exists a mesh frieze on C(3, n).

Proof. Let F = F 3,n be a SL 3 -frieze of horizontal period n = w + 4, for w ∈ {2, 3, 4}.

Consider the Plücker frieze P(k, n) of type (3, n). It has the same width and period as F , and we can overlay P(k, n) on top of F such that every p I for an almost consecutive 3-subset I of [1, n] comes to lie on top of an entry in F . (Note that a choice is involved here, the way to overlay F by P(k, n) is not unique.) This assigns to each entry p I in the Plücker frieze P(k, n) an integer f (p I ) lying underneath p I . By Proposition 4.6, the Plücker frieze P(k, n) contains a collection of entries {p

I | I ∈ I} such that I∈I M I is a cluster-tilting object of C(3, n). Its image under ψ -1 3,n is a cluster in A(3, n) (cf.
Corollary 4.7). Specializing the indecomposable object M I for I ∈ I to f (p I ) and calculating the remaining entries using the mesh relations yields a mesh frieze on the AR-quiver of C(3, n): By construction, the mesh relations are satisfied and as explained in Remark 3.10 all its entries are positive rationals. It remains to show that the entries are all integers. For this, we associate positive integers from F to their positions in the corresponding Auslander-Reiten quiver/mesh frieze and use the mesh relations to see that all the other entries are sums and differences of products of these given positive integers. That proves the claim. We include pictures of the Auslander-Reiten quivers of these three categories along the way.

(i) Let w = 2 and n = 6. The Auslander-Reiten quiver of C(3, 6) is shown in Figure 4. The indecomposables modules are indicated by black circles. Among them, the one corresponding to indecomposable rank 1 modules (or Plücker coordinates) appear with their label. The remaining two nodes correspond to rank 2 modules.

The entries of F 3,6 yield positive integers in the τ -orbits of the nodes 1, 3 and 4 (see Figure 7) of the Auslander-Reiten quiver of C(3, 6), as this is where the objects indexed by almost consecutive 3-subsets sit. To compute the entries of the τ -orbit of node 2, we use the mesh relations. Here, they are SL 2 -relations. They involve the τ -orbit of node 1 and the 1's above, if present. In particular, every entry in the τ -orbit of node 2 is of the form d = ab-1 1 = ab -1, for a and b in the τ -orbit of node 1. So d ∈ Z and we have a mesh frieze. The uniqueness follows since the entries in F 3,6 are uniquely determined by the choice of the entries corresponding to a cluster. (ii) The Auslander-Reiten quiver of C(3, 7) is in Figure 5. The unlabelled nodes correspond to rank 2 modules. Any integer SL 3 -frieze F = F 3,7 of width 3 yields positive integers in the τ -orbits of the nodes 1 (or 6) and 2, as the almost consecutive labels are exactly in these τ -orbits. (Note that the τ -orbits of node 1 and node 6 coincide). Using the frieze relation from Conway-Coxeter (see Equation ( 2) in [START_REF] Baur | Infinite friezes[END_REF]) one can write all entries of the mesh frieze as subtractions of multiples of the entries in the τ -orbit of node 1. So as in (i), all entries are integers and we have a mesh frieze.

(iii) Let F = F 3,8 be an integral SL 3 -frieze of width 4. Its entries yield positive integers in the τ -orbits of the nodes 1 and 8 of the Auslander-Reiten quiver of C(3, 8) (see Figure 6). The unlabelled nodes are rank 2 and rank 3 modules. Observe that only 32 of the entries in the mesh frieze arise from the SL 3 -frieze in this case: The entries in the first and fourth row of F form the τ -orbit of 8, the entries of the second and third row the τ -orbit of 1. Using the frieze relation from Conway-Coxeter one can write all entries in the τ -orbits of the nodes 3, 4, 5, 6, 7 of the mesh frieze of C(3, 8) as subtractions of multiples of entries in the τ -orbits above/below them (see Equation (2) in [START_REF] Baur | Infinite friezes[END_REF]). So again, all these entries are integers.

To see that the entries in the τ -orbit of node 2 are integers, use the Plücker relations. For example, to see that the entry for I = {2, 5, 7} is an integer, we use the Plücker relations (1) for (i 1 , i 2 ) = (5, 6) and (j 0 , . . . , j 3 ) = (2, 4, 5, 7): Proof. For n = 6, this can be found in [START_REF] Morier-Genoud | Coxeter's frieze patterns at the crossroads of algebra, geometry and combinatorics[END_REF]. We use Proposition 4.15 to see that there is a surjective map from the set of mesh friezes for C(3, n) to the set of integral tame SL 3friezes. This map is injective since by Lemma 4.14, in these cases, the SL 3 -frieze contains a cluster which uniquely determines it.

p

Mesh friezes via IY-reduction

In this section, we explain the effect of Iyama-Yoshino reduction on mesh friezes of cluster categories. We will consider cluster categories of not necessarily connected Dynkin type, i.e. the case where the Auslander-Reiten quiver consists of (possibly more than one) connected components, each of which is isomorphic to the Auslander-Reiten quiver of a cluster category D b (modH)/τ -1 Σ, where H is a hereditary algebra with quiver given by a disjoint union of Dynkin diagrams.

We say that such a cluster category is of finite type if it has only finitely many indecomposable objects. In this case, the rank is the number of summands in any cluster-tilting object.

Note that when the stable category of C(k, n) is of finite type (i.e. if k = 2 and n is arbitrary, or if k = 3 and n ∈ {6, 7, 8}), then its Auslander-Reiten quiver is isomorphic to the Auslander-Reiten quiver of the cluster category of the same type in the sense of

[BMR + 06].
Recall that for any A(k, n), specialising a cluster to 1 yields an integral tame SL k -frieze of width w = n -k -1 (Corollary 3.12). Mesh friezes for C(2, n) all arise from specialising a cluster to 1. It is known that there exist mesh friezes on finite type cluster categories which do not arise from specialising a cluster, the first example is a mesh frieze in type D 4 containing only 2s and 3s, see Example 5.9 below. It has appeared in [BM09, Appendix A], see also [START_REF] Fontaine | Counting friezes in type Dn[END_REF]Section 3], where all mesh friezes of type D are characterised. For C(3, 6) there are thus 51 different mesh friezes: 50 arising from specialising a cluster to 1 and the example above.

Mesh friezes and their reductions.

Let M be a rigid indecomposable object of a triangulated 2-Calabi-Yau category C. Let

M ⊥ := {Y ∈ C | Hom C (M, Y [1]) = 0} = {Y ∈ C | Hom C (Y, M [1]) = 0} and let C(M ) = M ⊥ /(M )
where (M ) denotes the ideal of the category M ⊥ consisting of morphisms that factor through add(M ). Then by [IY08, Theorem 4.2, Theorem 4.7], C(M ) is triangulated 2-Calabi-Yau. In particular, it has a Serre functor, and thus Auslander-Reiten triangles.

If C is a (stable) cluster category of finite type, every indecomposable object is rigid. Moreover, the reduction C(M ) is also of finite type. Let Γ C be the Dynkin diagram associated to C. See Figure 7. Then the Auslander-Reiten quiver of C has the shape of

A n : 1 2 n D n : 1 2 n n -1 E 6 : 1 3 4 5 6 2 E 7 : 1 3 4 5 6 2 7 E 8 : 1 3 4 5 6 2 7 8
Figure 7. The Dynkin diagrams Γ C of the finite type cluster categories a quotient of the AR-quiver of the bounded derived category of this type, i.e. of Γ C × Z (cf. [START_REF] Happel | Triangulated categories in the representation theory of finite-dimensional algebras[END_REF]). For M indecomposable, consider the indecomposable objects reached by sectional paths from M . The arrows of these paths form an oriented Dynkin diagram (see Figure 7) whose vertices are M and the indecomposable objects on the sectional paths. This is a slice in the Auslander-Reiten quiver. The vertex of M in the Dynkin diagram will be called the node of M .

By considering the Auslander-Reiten quiver of C(M ), it is straightforward to prove the following statement (note that C(M ) may be a product of cluster categories of finite type):

Proposition 5.1. Let C be a cluster category of type A,D or E. Then C(M ) is a cluster category. Its type is obtained by deleting the node of M of the underlying Dynkin diagram.

Notation 5.2. Let F be a mesh frieze on a finite type cluster category C. Let M be an indecomposable object in C. We write F(N ) for the (integer) value of an indecomposable N in F. Then F| C(M ) denotes the collection of numbers of F on the AR-quiver of C(M ).

Proposition 5.3. Let F be a mesh frieze on a finite type cluster category C, containing an entry 1. Let M ∈ ind C such that the frieze entry at M is 1. Then F| C(M ) is a mesh frieze for C(M ).

Proof. Denote by 1 suspension in C(M ) and let

(10) A → B → C → A 1
be an AR-triangle in C(M ). We need to show that the corresponding values of the mesh frieze satisfy the equation

F(A)F(C) = F(B) + 1, where F(B) = F(B 1 ) • • • F(B t ) when- ever B = ⊕ t i=1 B i .
Let T be a cluster tilting object in C(M ) consisting of all indecomposable objects N that admit a sectional path starting at A and ending at N . Note that T forms a slice in the AR-quiver of C(M ) with A being a source. Moreover, A, B ∈ add T while C ∈ add T , and the triangle (10) corresponds to a minimal left add (T /A)-approximation of A. By construction, the minimal right add (T /A)-approximation of A is zero.

(11) A -1 → C → 0 → A Next consider the object T = T ⊕M of C. Since, T ∈ C(M ) it follows that Ext 1 C (T, M ) = 0 Also, Ext 1 C (T, T ) = 0, because T is a cluster-tilting object in C(M ).
Finally, M is an indecomposable object of C, which means that it is rigid. This implies that T is rigid in C. Thus, T is a cluster-tilting object in C as it has the correct number of indecomposable summands.

The AR-triangles (10) and (11) in C(M ) lift to the corresponding triangles in C:

(12) A → B → C → A[1] A[-1] → C → B → A
where B ∼ = B ⊕ M i and B ∼ = M j for some non-negative integers i, j and [1] denotes suspension in C. Here, M 0 denotes the zero module. Moreover, we see that the two triangles in (12) are the left and right add T /A-approximations of A respectively.

Let A be the cluster algebra associated to C. For an object N ∈ C let x N denote the associated (product of) cluster variable(s) in A. Because C is the categorification of A it follows that the two triangles (12) give rise to the following exchange relation in the cluster algebra.

x A x C = x B + x B
Note that the entries of the associated mesh frieze F also satisfy the relations of the cluster algebra A. By Remark 4.13 we obtain the desired equation

F(A)F(C) = F( B) + F( B ) = F(B)F(M i ) + F(M j ) = F(B) + 1 because F(M 1 ) = F(M ) = 1 and F(M 0 ) = F(0) = 1.
Remark 5.4. We can use Proposition 5.3 to argue that a mesh frieze for C always has ≤ m entries 1, if m is the rank of C: Let F be a mesh frieze for C, a cluster category of rank m. Assume that F has > m entries 1. Take M 0 indecomposable such that its entry in the frieze is 1. Let C 1 := C(M 0 ). This has rank m -1. Let

F 1 = F | C 1 . Take M 1 ∈ ind C 1 such that the entry of M 1 in F 1 is 1. Let C 2 := C 1 (M 1 )
. This is a cluster category of rank m -2 and F 2 := F 1 | C 2 has ≥ m-1 entries equal to 1. Iterate until the resulting cluster category is a product of cluster categories of type A, whose rank is m (and m ≥ 3) and where the associated frieze F has ≥ m + 1 entries equal to 1s. Contradiction. Here, we also might need to justify that passing from F to F 1 we loose exactly one entry in the frieze that equals 1. This holds, because suppose on the contrary that F(M 0 ) = F(M 0 ) = 1, where M 0 is some other indecomposable object of C, and M 0 ∈ M ⊥ 0 . This implies that Ext 1 C (M 0 , M 0 ) = 0. In particular, there exist two triangles in

C M 0 → B → M 0 → M 0 [1] M 0 → B → M 0 → M 0 [1].
And by the same reasoning as in the proof of the above proposition, we obtain

F(M 0 )F(M 0 ) = F(B) + F(B ).
The right hand side is clearly ≥ 2, because F is an integral mesh frieze. This is a contradiction to F(M 0 ) = F(M 0 ) = 1.

Counting friezes.

In this section, we study the number of mesh friezes for the Grassmannian cluster categories C(3, n) in the finite types (types D 4 , E 6 and E 8 ) and for the cluster category of type E 7 . We recover the known number of mesh friezes for C(3, 7) and find all the known mesh friezes for C(3, 8) and for cluster categories of type E 7 . Recall that specialising a cluster to 1 gives a unique mesh frieze on C(k, n), cf. Corollary 3.12.

Definition 5.5. A unitary mesh frieze is a mesh frieze which arises from specialising a cluster to 1.

Not every frieze pattern arises this way. Some mesh friezes have less 1's than the rank of the cluster category. These are called non-unitary friezes. We will focus on these in what follows.

Remark 5.6. For k = 2, the notion of mesh frieze coincides with the notion of SL 2 -frieze and the numbers of them are well-known, they are counted by the Catalan numbers. In particular, they all arise from specialising a cluster to 1. From now on, we thus concentrate on k = 3 and n ∈ {6, 7, 8}. Recall first that the case (3, 6) corresponds to a cluster category of type D 4 , the case (3, 7) to a cluster category of type E 6 and the case (3,8) to a cluster category of type E 8 . The number of clusters in these types are 50, 833 and 25080 respectively, see e.g. [FZ03, Table 3].

It is known that the number of SL 3 -friezes of width 2 or, equivalently, the number of mesh friezes for C(3, 6) is 51 (this is proven via 2-friezes in [MGOT12, Prop. 6.2], and can be checked using a computer algebra system or counting the mesh friezes on a cluster category of type D 4 , see Example 5.9 below). The number of SL 3 -friezes (equivalently, the number of mesh friezes for C(3, 7)) of width 3 is 868 by Theorem B.1. Note that this number was already stated, citing a private communication by Cuntz, in [START_REF] Morier-Genoud | Coxeter's frieze patterns at the crossroads of algebra, geometry and combinatorics[END_REF]. Appendix B now provides a proof for this statement Fontaine and Plamondon found 26952 mesh friezes for C(3, 8), see the lists on [FP], see also [Cun17, Conjecture 2.1] and [MG15, Section 4.4].

We also recall that the number of clusters in a cluster category of type E 7 is 4160. Conjecturally, there are 4400 mesh friezes in type E 7 ( [FP]).

Corollary 5.7. Let F be a mesh frieze on a finite type cluster category C and M ∈ ind C an object whose frieze entry is 1. Then the following holds:

(1) If F := F| C(M ) is unitary, then F is unitary.

(2) If the Dynkin type of C(M ) is A or a product of A-types, then F is unitary.

Proof. Let n be the rank of C. By Proposition 5.3, F := F| C(M ) is a mesh frieze for C(M ).

If C(M ) is unitary, then F contains n -1 entries equal to 1. But then F contains n 1s and is unitary. The second statement then follows, since in type A, all mesh friezes are unitary.

An immediate consequence of Corollary 5.7 is the following.

Corollary 5.8. Let F be a mesh frieze on a cluster category of type D or E. If F contains an entry 1 at the branch node or at a node which is an immediate neighbour of the branch node, then F is unitary.

By Corollary 5.8, non-unitary friezes cannot exist in ranks ≤ 3: In terms of ranks, the smallest possible cluster category with non-unitary mesh friezes is of type D 4 . Indeed, we have one in this case, see Example 5.9.

Example 5.9. There exists a non-unitary mesh frieze in type D 4 ([BM09, Appendix A]. It is the only non-unitary mesh frieze in type D 4 , see [FP16, Section 3]).

2 2 2 2 3 3 3 3 • • • 2 2 2 2 • • • 2 2 2 2
5.2.1. Non-unitary mesh friezes in type E 6 . Let C be a cluster category of type E 6 . Using the result of Cuntz and Plamondon (B.1), one can deduce that there are 35 non-unitary mesh friezes for C. One can extract these from the lists available on [FP]. One can check that each of them contains two 1s.

Here, we explain how non-unitary mesh friezes arise in type E 6 and how any frieze with a 1 must contain at least two 1s. Observe that the AR-quiver of a cluster category of type E 6 consists of 7 slices of the Dynkin diagram E 6 :

• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •
(Note that the last slice with circles involves a twist, the top node is identified with the bottom node in the first slice, etc.).

By Corollary 5.7, if a non-unitary (mesh) frieze in type E 6 contains an entry 1, the node of this entry has to be 1 (or 6) (see Figure 7 for the labelling of nodes). So let F be a non-unitary frieze for C and M an indecomposable corresponding to an entry 1 in the top row. Then C(M ) is a cluster category of type D 5 . Furthermore, F| C(M ) is a non-unitary mesh frieze on C(M ) (since otherwise, F is unitary, by Corollary 5.7). There are exactly 5 non-unitary frieze on a cluster category of type D 5 , the five τ -translates of the following mesh frieze:

2 3 1 3 2 5 2 2 5 3 3 3 3 7 7 2 2 2 4 2 2
2 2 4 2 Since there are 7 positions for the first entry 1 we picked (respectively for M ), all in all there are exactly 35 different non-unitary friezes for C.

5.2.2.

Non-unitary mesh friezes in type E 7 . Let C be a cluster category of type E 7 . Conjecturally, there are 240 non-unitary mesh friezes for C. We now show how 240 non-unitary mesh friezes arise from non-unitary friezes for a cluster category of type D 6 or of type D 4 . First recall that the AR-quiver of C consists of 10 slices of a Dynkin diagram of type E 7 . If F is a non-unitary for C, it cannot contain entries equal to 1 in the nodes 2,3,4 or 5 (see Figure 7) for the labels of the nodes in the Dynkin diagram).

Remark 5.10. Let F be a non-unitary frieze for C.

(i) Assume that F contains an entry 1 in the τ -orbit of node 1, let M be the corresponding indecomposable object. Then C(M ) is of type D 6 . (a) There are 2 mesh friezes of type D 6 without any 1's, see Example 5.12. Hence we get 20 mesh friezes of type E 7 having an entry 1 in the τ -orbit of 1 and all other entries ≥ 2.

(b) So assume that F = F C(M ) contains an additional entry 1, with corresponding module N . Then doing a further reduction (C(M ))(N ) yields a cluster category of type D 5 or of type A 1 × D 4 . In both cases, F has two entries equal to 1. Analyzing the AR-quiver of C(M ) and the possible positions of these 1s yields another 220 nonunitary mesh friezes, as one can check. (i) The above mesh friezes cover are all known non-unitary mesh friezes in type E 7 having at least one entry 1, according to the lists of Fontaine and Plamondon.

Conjecture 5.11. There are no mesh friezes for C of type E 7 where all entries are ≥ 2.

Example 5.12. In type D 6 there are exactly two mesh friezes without any 1's, see [ 7. Non-unitary (mesh) friezes do not contain 1s in the τ -orbits of the nodes 2,3,4 or 5, cf. Corollary 5.8.

So in order to study non-unitary mesh friezes for C we can start considering friezes containing an entry 1 in the τ -orbit of node 6. There are 16 choices for this, as the Auslander-Reiten quiver of C has 16 slices. Let M be the corresponding indecomposable. Then C(M ) is a cluster category of type D 5 × A 2 . There must be two more 1s in the frieze in the factor of type A 2 and one in the factor of type D 5 . There are 5 choices of a cluster in type A 2 and there are five choices for an entry 1 in the type D 5 factor (cf. Subsection 5.2.1. All in all there are 16 × 5 × 5 = 400 possibilities for a non-unitary frieze with an entry 1 in the τ -orbit of node 6.

Continuing with similar arguments, one finds 1852 non-unitary friezes with four 1s and 16 friezes with two 1s. There are no known mesh friezes for C with only one entry = 1, giving additional evidence for conjecture 5.11:

Lemma 5.13. If Conjecture 5.11 is true, then every mesh frieze for C of type E 8 contains 0,2,4 or 8 entries equal to 1.

Proof. We only need to discuss non-unitary mesh friezes. If there is an entry 1 in the τ -orbit of node 8, the corresponding category C(M ) is of type E 7 and the claim follows by assumption (Conjecture 5.11) and by Remark 5.10 (i) (a) and (b). If there is an entry 1 in the τ -orbit of node 1, C(M ) is a cluster category of type D 7 . By the results in [FP16, Section 3], every non-unitary frieze of type D 7 contains either one or three entries = 1. All the other cases arise from two entries equal to 1 and a non-unitary mesh frieze for a cluster category of type E 6 , these are known to have two entries = 1. Appendix A. Proof of Proposition 3.5

In the following, we are mainly interested in Plücker coordinates p i 1 ,...,i k where a portion of the i l 's is consecutive. For brevity we will write [r] l for the set {r, r + 1, . . . , r + l -1}. We reduce integers modulo n. So if l = k -1, such a tuple forms an almost consecutive k-subset. As before, we will use the map o to indicate reordering by size, e.g. if k = 4, n = 9, p o([8] 3 ),4 stands for the Plücker coordinate p 1894 and p (o([8] 3 ,4) for p 1489 . We recall the proposition we are to prove here.

Proposition (Proposition 3.5). Let r ∈ [1, n] and let 1 ≤ s ≤ k. Let m = (m 1 , . . . , m s ) with m i ∈ [1, n] for all i, and assume that m satisfies conditions (c1) and (c2). Let b m;r be the determinant of the matrix A m;r from (3.3).

Then we have b

m;r = s-2 l=0 p o([r+l] k ) • p o([r+s-1] k-s ,m 1 ,...,ms) . ( 13
)
Proof of Proposition 3.5. If we have m i = m j for some 1 ≤ i < j ≤ s, then

b m;r = 0 = s-2 l=0 p o([r+l] k ) • p o([r+s-1] k-s ,m 1 ,...,ms) .
as the last term is 0. We can thus assume that m 1 , . . . , m s are mutually distinct. We prove the claim by induction over s. For s = 1 the claim holds, since b

m;r = det(p o([r] k-1 ,m 1 ) ) = p o([r] k-1 ,m 1 )
. Let now 2 ≤ s ≤ k and assume the statement is true for 1 ≤ l < s, that is for all b ( m1 ,..., ml );s , where the mj 's satisfy conditions (c1) and (c2). We show that it also holds for b m;r .

We have

(14) b m;r = s j=1 (-1) s+j p o([r+s-1] k-1 ,m j ) • b (m 1 ,..., m j ,...,ms);r = (-1) s • s-3 l=0 p o([r+l] k ) • • s j=1 (-1) j p o([r+s-1] k-1 ,m j ) • p o([r+s-2] k-(s-1) ,m 1 ,..., m j ...,ms) ,
where the first equality is by Laplace expansion along the last row and the second equality is by induction assumption and pulling the term X := (-1) s • s-3 l=0 p o([r+l] k ) out of the sum. Notice that the conditions (c1) and (c2) pass down from the set {m 1 , . . . , m s } to any subset {m 1 , . . . , m j , . . . , m s }, so we are justified in using the induction assumption on the minors of the form b (m 1 ,..., m j ,...,ms);r .

We distinguish the following cases. Case 1: Assume s = k and there exists a 1 ≤ l ≤ s with m l ∈ [r + s -2] k-(s-1) . Case 1a: Assume first that m l = r + s -2. Then m l ∈ [r + s -1] k-s and the following expressions all vanish:

p o([r+s-1] k-1 ,m l ) = p o([r+s-1] k-s ,m 1 ,...,ms) = 0
and p [r+s-2] k-(s-1) ,m 1 ,..., m j ...,ms = 0 for all j = l.

Combining this with Equation (14) it follows that b

m;r = 0 = s-2 l=0 p o([r+l] k ) • p o([r+s-1] k-s ,m 1 ,...,ms) .
This proves the claim in this case. Case 1b: Assume now that m l = r +s-2 and l < s. If we had m l+1 ∈ (r +k -2, m l ], then we would have r + k -2 ∈ (m l , m l+1 ) ⊆ (m 1 , m s ) contradicting condition (c2). Therefore, we must have m l+1 ∈ (m l , r + k -2] = [r + s -1] k-s and we are in case 1a with m l+1 taking on the role of m l . Case 1c: As a last subcase for case 1 assume that l = s and m s = r + s -2. Then all terms in (14) with factors p o([r+s-2] k-(s-1) ,m 1 ,..., m j ,...,ms) with j = s vanish and we obtain

b m;r = (-1) s • s-3 l=0 p o([r+l] k ) • • (-1) s p o([r+s-1] k-1 ,ms) • p o([r+s-2] k-(s-1) ,m 1 ,...,m s-1 ) , = s-2 l=0 p o([r+l] k ) • p o([r+s-1] k-s ,m 1 ,...,ms) ,
where the second equality follows directly from the equality m s = r + s -2. This proves the claim in this case. Case 3a: Assume that 1 = r + a ∈ [r + s -1] k-1 ∩ [r + s -2] k-(s-1) . Then I is ordered as follows: 1 = r + a < r + a + 1 < . . . < r + s + k -3 < r + s -1 < . . . < r + a -1 and J is ordered as follows: 1 = r + a < r + a + 1 < . . . < r + k -2 < m 1 < m 2 < . . . < m s < r + s -2 < r + s -1 < . . . < r + a -1 = n.

Notice that we must have m 1 < m l for all l = 1. Otherwise, if there exists a 1 < l ≤ s with m l < m 1 , then if we have m Our proof relies on a reduction to 2-friezes (whose definition we recall below); our strategy is to show that the entries in a 2-frieze of height 3 are bounded.

We have attempted to apply the methods used in this Appendix to types E 7 and E 8 , without success.

B.1. 2-friezes

We shall not be using the definition of a frieze from [START_REF] Assem | Friezes[END_REF], but rather a slightly different notion, that of a 2-frieze as defined in [START_REF] Morier-Genoud | 2-frieze patterns and the cluster structure of the space of polygons[END_REF].

Definition B.2. A 2-frieze of height h is an array of positive integers (a i,j ), where

• i ∈ Z and j ∈ {0, 1, . . . , h + 1}; • for all i ∈ Z, we have that a i,0 = a i,h+1 = 1; • for all i ∈ Z and all j ∈ {1, . . . , h}, we have that a i,j = a i-1,j a i+1,j -a i,j-1 a i,j+1 .

The reason we are interested in 2-friezes is the following result, which is a consequence of [MGOT12, Proposition 5.4] and of the fact that the quiver depicted in [START_REF] Morier-Genoud | 2-frieze patterns and the cluster structure of the space of polygons[END_REF](2.3)] is mutation-equivalent to a quiver of type E 6 .

Theorem B.2 ([MGOT12]

). Entries in a 2-frieze of height 3 are entries in a frieze of type E 6 which determine the frieze. This defines an injection from the set of friezes of type E 6 to the set of 2-friezes of height 3.

Our strategy to prove Theorem B.1 is thus to prove that the number of 2-friezes of height 3 is finite. We will do this by showing that there is a bound on the possible values appearing in a 2-frieze of height 3.

B.2. Two choices of initial variables for 2-friezes of height 3

If we fix the entries of the first two rows of a 2-frieze of height 3 to be s, t, u, v, w and x, then we get the following expressions for all its entries: If, instead, we fix the six first entries of the leftmost non-trivial column, then we get:

1 s t u 1 1 v w x

Figure 1 .

 1 Figure 1. The grid defined by ϕ (k,n) , reducing indices modulo n. The top and bottom k -1 rows (blue) are 0, the k-th row from top/bottom (red) consist of consecutive k-tuples, i.e. frozen Plücker coordinates.

  and by(a, b) the open interval between a and b modulo n, defined as

  Analogously, we define the half-open intervals [a, b) and (a, b].

  j≤k+1 , where A [m] k+1 ;r is the corresponding diamond in the Plücker frieze P (k,n) given by (3.3) for [m] k+1 = (m, m + 1, . . . , m + k) and m ∈ [r + k, r -2]. Similarly to the first part of the proof (for k × k diamonds), one can show that conditions (c1) and (c2) are satisfied for [m] k+1 and for any order-inheriting subtuple thereof. So we can apply Proposition 3.5 to compute the determinants b ml ;r for ml = (m, . . . , m + l, . . . , m + k), for 0 ≤ l ≤ k, and Laplace expansion for the determinant of A [m] k+1 ;r by the last row yields

  * ) Now since (*) is precisely the Plücker relation on I = [r + k] k-1 and J = [m 1 , m k+1 ], this term equals zero and so det(A [m] k+1 ;r ) = 0.

  Two k-subsets I, J of [1, n] are said to be non-crossing if there are no cyclically ordered a, b, c, d with a, c ∈ I \ J and b, d ∈ J \ I. If such a, b, c, d do occur, then I and J are crossing. Proposition 4.4. [JKS16, Proposition 5.6] Let I, J be k-subsets of [1, n]. Then Ext 1 C(k,n) (M I , M J ) = 0 if and only if I and J are non-crossing.

  on the right. Moreover, each such rectangular region contains at most 2w elements of I. But in the beginning we took two copies of the frieze, so dividing by two, we see that a Plücker frieze contains at most 2n w + 3 w pairwise non-crossing elements. To prove the claim it suffices to show that (k -1)w > 2n w in turn implies the equation above. Multiplying equation (5) by w + 3 and substituting w + 3 = n -k + 2 we obtain (6) (k -4)n > (k -2) 2

Figure 3 .

 3 Figure 3. Fundamental region (left) and Plücker frieze where k = 4 (right)

x 6 .

 6 Next we obtain m 45 , m 46 from the determinants of two determinants yield the system of linear equations:

Figure 4 .

 4 Figure 4. Auslander-Reiten quiver of C(3, 6). Projective-injectives are drawn as white circles.

Figure 6 .

 6 Figure 6. Auslander-Reiten quiver of C(3, 8).

Case 2 :

 2 Assume now s = k and there exists a 1 ≤ l ≤ k such that m l = r + k -2. This cannot happen if l = k: By condition (c2) we have r + k -2 / ∈ [m 1 , m k ), but by condition (c1), we have m l ∈ [m 1 , m k ) for 1 ≤ l < k. Therefore, m l = r + k -2 implies l = k, and thus in Case 2 we have m k = r + k -2. Then the same argument as in case 1c, replacing s by k, proves the claim. Case 3: We are now left with the case where m l / ∈ [r + s -2] k-(s-1) for all 1 ≤ l ≤ s. We are going to show thats j=1 (-1) j p o([r+s-1] k-1 ,m j ) • p o([r+s-2] k-(s-1) ,m 1 ,... m j ...,ms) = (-1) s • p o([r+s-2] k ) • p o([r+s-1] k-s ,m 1 ,...,ms) ,(15)which, when substituting into (14) yields the desired equality (13). We use the Plücker relations on I = [r + s -1] k-1 of cardinality |I| = k -1 and J = [r + s -2] k-(s-1) ∪ {m 1 , m 2 , . . . , m s } of cardinality |J| = k + 1.

  l ∈ [1, r + k -2] ⊆ [r + s -2, r + k -2]this contradicts the assumptions of case 3, and if we have m l ∈ (r + k -2, n], then it follows from m l < m 1 that r + k -2 ∈ [m 1 , m l ) ⊆ [m 1 , m s ), which contradicts condition (c2). The Plücker relation on I and J (cf. Equation (1)) thus reads as:0 = k-a-2 l=0 (-1) l • p o([r+s-1] k-1 ),r+a+l • p o(J\{r+a+l}) + s l=1 (-1) k-a-2+l • p o([r+s-1] k-1 ),m l • p o(J\m l ) k-a+s+l-1 • p o([r+s-1] k-1 ),r+s-2+l • p o(J\{r+s-2+l}) .Notice that p o([r+s-1] k-1 ),r+j = 0 for all j ∈ [s -1, k -2] and thus the whole first sum among those three vanishes, and all the summands, except for the first one (l = 0) in the last sum vanish as well. The Plücker relation thus reduces to0 = s l=1 (-1) k-a-2+l • p o([r+s-1] k-1 ),m l • p o(J\m l ) + (-1) k-a+s-1 • p o([r+s-1] k-1 ),r+s-2 • p o([r+s-1] k-s ,m 1 ,...,ms) .We need to get the remaining k-tuples in linear order. Observe thatp o([r+s-1] k-1 ,m l ) =p r+a,...,r+s+k-3,m l ,r+s-1,...,r+a-1 = sgn(π)p o([r+s-1] k-1 ),m l and p o([r+s-2] k ) =(r + a, . . . , r + s + k -3, r + s -2, r + s -1, . . . , r + a -1) = sgn(π)p o([r+s-1] k-1 ),r+s-2 where π = (s + k -a -1, s + k -a, . . . , k -1, k) ∈ S k .Substituting this into the Plücker relation on I and J (and multiplying by sign(π)• (-1) k-a l • p o([r+s-1] k-1 ,m l ) • p o(J\m l ) -(-1) s • p o([r+s-2] k ) • p o([r+s-1] k-s ,m 1 ,...,ms) , which, observing that p o(J\m l ) = p o([r+s-2] k-(s-1),m 1 ,..., m l ,...,ms) and pulling the second term to the left hand side yields the desired Equation (15). Case 3b: For the final case, assume that 1 / ∈[r + s -1] k-1 ∩ [r + s -2] k-(s-1). We first observe what the linear order on J looks like. Notice that we have r + s -2 < r + s -1 < . . . < r + k -2. Since m j / ∈ [r + s -2, r + k -2], the set J is ordered in one of the following three ways (16), (17), (18):m b < m b+1 < . . . < m s < r + s -2 < . . . < r + k -2 < m 1 < m 2 < . . . < m b-1 , (16)for some b ∈ {2, . . . , s},m 1 < m 2 < . . . < m s < r + s -2 < . . . < r + k -2, or (17) r + s -2 < . . . < r + k -2 < m 1 < . . . < m s . (18)If it is ordered as in (17), we set b = 1, and if it is ordered as in (18), we set b = s + 1. Then the Plücker relation on I and J reads as follows:l-b p o( [r+s-1] k-1 ),m l • p o(J\m l ) s-b+1+l • p o([r+s-1] k-1 ),r+s-2+l • p o(J\{r+s-2+l}) k-b+1+l • p o([r+s-1] k-1 ),m l • p o(J\m l ) .Notice again that p o([r+s-1] k-1 ),r+j = 0 for all j ∈ [s -1, k -2], and therefore the corresponding terms in the Plücker relation vanish. Again, we need to get the remaining k-tuples in linear order. Observe thatp o([r+s-1] k-1 ),m l = p o([r+s-1] k-1 ,m l ) if 1 ≤ l ≤ b -1 sgn(σ)p o([r+s-1] k-1 ,m l ) if b ≤ l ≤ s and that p o([r+s-1] k-1 ),r+s-2 = sgn(σ)p o([r+s-2] k ), where σ = (1, 2, . . . , k) ∈ S k with sgn(σ) = (-1) k-1 . Therefore, if J is ordered as in (17) or (18) the Plücker relation In this Appendix, we settle the case of type E 6 , and obtain the result for type F 4 as a corollary.Theorem B.1. The number of friezes of type E 6 is exactly 868.Since Dynkin type F 4 is a folding of type E 6 , it follows from the work of [FP16] that we then have:Corollary B.1. The number of friezes of type F 4 is exactly 112.

  .1.1]. SL 2 -friezes were first studied by Coxeter and Conway-Coxeter in the early 1970s, see Example 2.4 below. Higher SL k -frieze patterns made their first appearance 1972 in work of Cordes-Roselle

p 124 p 235 p 346 p 145 p 256 p 136 p 124

  Definition 2.9. The special Plücker frieze of type (k, n), denoted by sP (k,n) , is the grid we get from P (k,n) by substituting the consecutive Plücker coordinates with 1s. In other words, sP (k,n) is the grid with entries given by the map s • ϕ (k,n) with entries in sA(k, n). 24 p 35 p 14 p 25 p 13 p 24 • • • • • • p 14 p 25 p 13 p 24 p 35 p 14 p 25

	observed that, for k ≥ 3, cluster variables for A(k, n) appear as entries in the derived
	arrays of an SL k -frieze.
	3. Integral tame SL k -friezes from Plücker friezes
	3.1. The special Plücker frieze sP (k,n) is a tame SL k -frieze.
	Example 2.10. The special Plücker frieze of type (2, 5), sP (2,5) , (see Example 2.6) is as
	follows (writing 0 for the entries p ii ):
	0	0	0	0	0	0	0	0
		1	1	1	1	1	1	1
	p 13 p 1 1	1	1	1	1	1	1
		0	0	0	0	0	0	0
	And sP (3,6) :
	0	0	0		0	0	0	0
	0	0	0	0	0		0	0	0
	1	1	1		1	1	1	1
	p 146 p 125 p 236 p 134 p 245 p 356 p 146 • • •
	• • • 1	1	1	1	1		1	1	1
	0	0	0		0	0	0	0
	0	0	0	0	0		0	0	0
	Remark 2.11. For k = 3, [MGOT12, Proposition 3.3] describes an explicit correspondence
	between SL 3 -friezes and 2-friezes (for a definition of 2-friezes see Appendix B). It is stated
	in [MGOT12, Remark 5.8] that it can be checked that all the n(n -4) entries in a 2-
	frieze are Plücker coordinates for the Grassmannian Gr(3, n). Performing this check and
	combining the two observations yields an alternative way to obtain the Plücker frieze of
	type (k, n) for the special case k = 3. More generally, in [MG15, Remark 3.19] it is

  Notation 3.4. Fix r ∈ [1, n] and let 1 ≤ s ≤ k. Choose elements m j ∈ [1, n] for each 1 ≤ j ≤ s. We can impose the following conditions on the ordered tuple (m 1 , . . . , m s ) (c1) It is ordered cyclically modulo n, that is there exists a number b ∈ {1, . . . , s} such that

		 p 123 p 124 p 125		 p 123 p 124 p 125	
	4,5);1 :=	 p 233 p 234 p 235	 =	 0	p 234 p 235	
		p 334 p 344 p 345		0	0	p 345
	We set
		b m;r := det A m;r .
	In particular, for m = [m] k = (m, m + 1, . . . , m + k -1) the matrix A m;r is a k × k diamond
	in the frieze P (k,n) whose determinant it is our first goal to compute.
	We can compute the determinants b m;r provided the two conditions (c1) and (c2) are
	satisfied for the tuple m:

  . In the finite types, the indecomposable modules are well-known. For k = 2, they are exactly the rank 1 indecomposables. The Auslander-Reiten quivers of the categories C(3, n), for n = 6, 7, 8 are described in[START_REF] Bernt | A categorification of Grassmannian cluster algebras[END_REF]. We recall some of this information here.

(i) For (3, 6), there are 22 indecomposable objects. Among them, 20 are rank one modules. The additional two are rank two modules filtered by M 135 and M 246 (in both ways). Altogether, there are 6 projective-injective indecomposables and 16 other indecomposable objects. The Dynkin type of C(3, 6) is D 4 . (ii) For (3, 7), there are 14 rank 2 modules filtered by two rank 1 modules in addition to the 35 rank one modules. The category C(3, 7) has 7 projective-injectives objects and is a cluster category of type E 6 . (iii) In addition to the the 56 rank one modules, C(3, 8) has 56 rank two modules and 24 rank 3 modules, all filtered by rank 1 modules. It has 8 projective-injective objects. The Dynkin type of C(3, 8) is E 8 .

  p) where s ∈ {r, r + 1, . . . , m -k} and p ∈ {m, m + 1, . . . , r -2}. Similarly, we say that J belongs to the maximal rectangle ending at I whenever s ∈ {r, r -1, . . . m + 2} and p ∈ {m, m -1, . . . , r + k}. Note that the collection of such elements indeed forms a maximal rectangle in M (k,n) .

Proposition 4.5. Let I = o([r] k-1 , m) be a non-consecutive k-subset and let J be another element of

P (k,n) . Then Ext 1 C(k,n) (M I , M J ) is

nonzero if and only if J belongs to the maximal rectangle starting at o

  Example 4.10. In case k = 2, a mesh frieze F 2,n for C(2, n) is an integral frieze pattern (or an integral SL 2 -frieze). In that case, the notion of mesh friezes and of integral SL 2 -friezes coincide. Such friezes are in bijection with triangulations of an n-gon and hence arise from specialising a cluster to 1, see Example 2.4.Remark 4.11. Note that[START_REF] Assem | Friezes[END_REF] introduce the notion of 'friezes of Dynkin type'. It coincides with certain mesh friezes (on the stable version C(k, n)) in types A n , D 4 , E 6 , E 8 .

	Auslander-Reiten sequences and an Auslander-Reiten quiver, cf.
	for example [JKS16, Remark 3.3 ].)

  562 p 457 -p 564 p 257 +p 565 p 247 -p 567 p 245 = 0. This yields p 257 = p 256 p 457 -p 567 p 245 = p 256 p 457 -1 with p 256 ∈ Z, p 457 ∈ Z (as they are Plücker coordinates of almost consecutive 3-subsets). Let n ∈ {6, 7, 8}. There is a bijection between mesh friezes for C(3, n) and integral tame SL 3 -friezes of width n -4.

	123		456		127		345
	237	124	356	457	126	137	245	346	157
	256	347	125	367	145	236	147	256
	235 Figure 5. Auslander-Reiten quiver of C(3, 7). 156 157 237 167 234 567 134 467 823 356 681 134 467 123 456 781 234 567 123 124 457 782 235 568 124 824 357 682 135 468 725 258 583 836 361 Corollary 4.16. 267 346 345 245 346 578 678 671 246 571 147 472 237 156 481 734 267 512 845 378 562 815 348 673 126 451	712 623	812 713 614	813 784	245 156	246 345 147	346 237

  [START_REF] Fontaine | Counting friezes in type Dn[END_REF] Section 3]. One of them is here, the other one is the translate of it by τ : Non-unitary mesh friezes in type E 8 . Let C be a cluster category of type E 8 . Conjecturally, there are 1872 non-unitary mesh friezes for C. Recall the labelling of the nodes in the Dynkin diagram from Figure

	2	2	2	2	2	2
	3	3	3	3	3	3
		4	4	4	4	4	4
	• • •	5	5	5	5	5	5	• • •
			2	3	2	3	2	3
			3	2	3	2	3	2
	5.2.3.

  Conjecture 5.14. If F is a mesh frieze for C of type E 8 and if all entries of F are ≥ 2, then F is one of the four friezes from Example 5.15.Example 5.15. In type E 8 there are 4 known mesh friezes without any 1's, appearing in the lists by Fontaine and Plamondon, available at[FP]. They are the 4 translates of the following mesh frieze (a 4-periodic frieze, here, 8 slices are shown):

	3	2		2		3	3		2	2	3
		5	3		5	8		5	3	5	8
	13	7		7	13	13		7	7	13	• • •
	• • •	18	16		18	21		18	16	18	21
	29 6 41 7 41 6 29 5 29 6 41 7 41 6 29 5
		11	15		11	8		11	15	11	8
	3	4		4		3	3		4	4	3
	These mesh friezes arise from specialising the following cluster to (3, 3, 3, 3, 3, 3, 3, 3).
		p 235 O O	/ / p 236	o o	p 126 O O	/ / p 127
		p 356	o o	p 256		/ / p 267	o o	p 167

  + stwx + suvwx + suw 2 + t 2 vw + t 2 vx + t 2 w 2 + t 2 wx + tuvw + tuw 2 stuvwx ,andZ = stw + suvx + suw + t 2 w + tuw + tvx tvwx .

						1
		1		v+w s	vx+w t	w+x u	1
		1	svx+sw+tv+tw stv		X	tw+tx+uvx+uw tux	1
		1 suvx+suw+tuw+tvw+tvx tuvw	Y stw+suvx+suw+tvx+twx stwx	1
		1	sux+tw+tx+uw uwx		Z	suv+sw+tv+tw svw	1
		1		t+u x	su+t w	s+t v	1
		1		u		t	s	1
		1		x		w	v	1
	where
			X =	suvx + suw + tvw + tvx + tw 2 + twx stuw	,
	Y =	stuvx + stw 2

K.B. was supported by FWF grants P 30549-N26 and W1230. She is supported by a Royal Society Wolfson Research Merit Award. E.F. is a Marie Sk lodowska-Curie fellow at the University of Leeds (funded by the European Union's Horizon 2020 research and innovation programme under the Marie Sk lodowska-Curie grant agreement No 789580). K.S. was supported by NSF Postdoctoral Fellowship MSPRF -1502881.

on I and J can be rewritten as

Juggling the signs in a straight forward way, and pulling together the first and last sum we obtain 0 =(-1)

Pulling the second term on the left hand side and multiplying by (-1) Friezes (also called "friezes of type A" and "SL 2 -friezes") were introduced by H. Coxeter in [START_REF] Harold | Frieze patterns[END_REF] and later studied by H. Coxeter and J. Conway in [START_REF] Conway | Triangulated polygons and frieze patterns[END_REF][START_REF] Conway | Triangulated polygons and frieze patterns[END_REF]. It was observed by P. Caldero that the theory of cluster algebras of S. Fomin and A. Zelevinsky [START_REF] Fomin | Cluster algebras. I. Foundations[END_REF] allows for a far-reaching generalization of the original notion of frieze; this generalization was first studied in [START_REF] Assem | Friezes[END_REF]. Recall that in [START_REF] Assem | Friezes[END_REF], a frieze (of Dynkin type) is a mesh frieze on an associated Auslander-Reiten quiver, cf. Remark 4.11. Since then, many generalizations and variations on the notion of friezes have been introduced, as can be seen in the survey paper [START_REF] Morier-Genoud | Coxeter's frieze patterns at the crossroads of algebra, geometry and combinatorics[END_REF].

It is known that, for a given non-Dynkin type, there are infinitely many friezes of that type. However, it follows from [START_REF] Conway | Triangulated polygons and frieze patterns[END_REF][START_REF] Conway | Triangulated polygons and frieze patterns[END_REF] that friezes of Dynkin type A come in a finite number (given by Catalan numbers), and it was proved in [START_REF] Fontaine | Counting friezes in type Dn[END_REF] that friezes of Dynkin type B, D and G also come in a finite (explicit) number (the result for type D 4 was also proved in [START_REF] Morier-Genoud | 2-frieze patterns and the cluster structure of the space of polygons[END_REF]). It was conjectured in [Pro05, [START_REF] Morier-Genoud | 2-frieze patterns and the cluster structure of the space of polygons[END_REF][START_REF] Fontaine | Counting friezes in type Dn[END_REF] that the number of friezes in type E 6 is 868; in [START_REF] Fontaine | Counting friezes in type Dn[END_REF] the precise number of friezes in any Dynkin type is also conjectured.

where

B.3. Finiteness of the number of 2-friezes of height 3

We use the second choice of initial variables of the previous section. We can assume, without loss of generality, that the greatest entry in the first and third columns is u. Then

Hence (s -1)(w -1) < 2. This implies that s = 1 or w = 1 or s = w = 2.

B.3.1

The case s = 1 or w = 1. If s = 1 or w = 1, then the associated frieze of type E 6 contains a frieze of type D 5 , and it is known [START_REF] Fontaine | Counting friezes in type Dn[END_REF] that there are only 187 of these. Thus there is a finite number of cases where s = 1 or w = 1.

B.3.2

The case s = w = 2. If s = w = 2, then consider the following inequalities:

Thus we obtain (v -2)x < 4 which gives v < 6. For symmetry reasons, the same argument produces t < 6. But then u ≤ 9 since u ≤ 2t -1. Hence we have reduced the problem to a finite number of cases. In fact, an easy computation shows that the only solution is (s, t, u, v, w, x) = (2, 4, 5, 4, 2, 1) in which case the (transposed) 2-frieze is:

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 4 5 4 2 1 1 2 4 5 4 2 1 1 2 6 11 6 2 1 1 2 6 11 6 2 1 1 2 4 5 4 2 1 1 2 4 5 4 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1