Counting Coxeter's friezes over a finite field via moduli spaces

We count the number of Coxeter's friezes over a finite field. Our method uses geometric realizations of the spaces of friezes in a certain completion of the classical moduli space $\mathcal{M}_{0,n}$ allowing repeated points in the configurations. Counting points in the completed moduli space over a finite field is related to the enumeration problem of counting partitions of cyclically ordered set of points into subsets containing no consecutive points. In Appendix we provide an elementary solution for this enumeration problem.


Introduction
The notion of friezes goes back to Coxeter in the early 70's [Cox71]. They are arrays of numbers in which every four adjacent values forming a square are related by the same arithmetic condition ad − bc = 1. The classification of friezes over positive integers is given by a beautiful result of Conway and Coxeter [CC73] establishing an unexpected bijection between friezes and triangulations of polygons. As a consequence the number of friezes over positive integers with a fixed number of rows is given by a Catalan number.
Friezes became popular in the last decade due to connections with the theory of cluster algebras of Fomin and Zelevinsky. Many variants of friezes have been recently introduced and studied. In particular Coxeter's friezes over other sets of numbers, or other types of friezes over positive integers, were investigated in various cases leading to new interesting combinatorics, e.g. [BM09], [BR10], [Fon14], [FP16], [Cun17], [Ovs18], [CH19], [HJ].
In the present paper we determine the number of Coxeter's friezes over an arbitrary finite field. Our main result gives explicit polynomial formulas according to the "width" of the friezes, i.e. the number of rows in the friezes. We express the formulas with the help of the q-integer [m] q 2 = q 2m −1 q 2 −1 and the q-binomial coefficient m 2 q = (q m −1)(q m−1 −1) (q−1)(q 2 −1) . Theorem 1. (i) The number of tame friezes of width 2m − 2 over F q is given by for all m ≥ 2; (ii) The total number of tame friezes of width 2m − 3 over F q is given by if char(F q ) = 2, and m even (q − 1) m 2 q + q m−1 , otherwise for all m ≥ 2.
We solve this problem by using geometric realizations of the spaces of friezes in moduli spaces of configurations of points in the projective line. The realizations are different according to the parity of the number of rows in the friezes leading to different formulas according to this parity.
Friezes over complex numbers with even number of rows containing no zero entries, form an algebraic variety isomorphic to the classical moduli space M 0,n , where n is the width plus three, [Cox71], [MGOT12]. Tame friezes are more general friezes allowing zero entries. When the number of rows is even, tame friezes form a bigger variety denoted M 0,n which is some completion of M 0,n [MGOST14], [MG15]. This is no longer true when the friezes have an odd number of rows, one has to consider a subvariety of M 0,n that we denote M + 0,n . Section 1 is a brief introduction to Coxeter's friezes. In particular we recall the definition of tame friezes and their main properties.
In Section 2 we explain the geometric realizations of the spaces of friezes. Cases of friezes with odd or even number of rows need to be distinguished. As a new result we get the geometric realization of the space of tame friezes with odd number of rows as M + 0,n , Theorem 2. In Section 3 we determine the number of points in the spaces M 0,n and M + 0,n when considered over F q , Theorem 3 and Theorem 4 respectively, and we prove Theorem 1. We end the section with examples of friezes over F q and open problems.
Appendix A can be read independently of the rest of the paper. It deals with an enumeration problem of restricted set partitions for which we provide an elementary solution. More precisely we establish an explicit formula for the number of partitions of n points in the circle into k subsets avoiding consecutive points, Proposition A.4. Our method is related to the counting of points in M 0,n over F q and requires only the formula of Lemma 3.1 established for the proof of Theorem 3. These numbers of restricted partitions are known and many information can be found about them on A261139 of OEIS.
1. Coxeter's friezes 1.1. Definition. Coxeter's friezes [Cox71] are arrays of numbers that can be described as follows: (i) The array has finitely many rows, all of them being infinite on the right and left; (ii) The first two top rows are a row of 0's followed by a row of 1's, and the last two bottom rows are a row 1's; (iii) Consecutive rows are displayed with a shift, and every four adjacent entries a, b, c, d forming a diamond b a d c satisfy the unimodular rule: ad − bc = 1.
The number of rows strictly between the border rows of 1's is called the width of the frieze. The following array (3) is an example of a frieze pattern of width w = 4.
Theorem 1.1 ( [CC73]). The number of friezes of width w over positive integers is given by the Catalan number C w+1 = 1 w+2 2w+2 w+1 . Several enumerative interpretations for the entries appearing in the friezes are known, see [MG15] for an overview.
1.2. Tame friezes. Following [BR10] a frieze is said to be tame if it satisfies the following extra condition: every nine adjacent entries in a diamond Friezes with no zero entries are all tame, but tame friezes may contain zeroes. Generically when e = 0 the determniant of the matrix (4) formed by the ajacent entries in a frieze vanishes by Desnanot-Jacobi identity. Hence the tameness condition has to be checked only for entries centered at the zeroes of the frieze.
Many properties of friezes established by Coxeter in the case of friezes over positive real numbers can be directly generalized to tame friezes with entries in any unital ring. Non-tame friezes are called "wild" and may have completely different behavior, see [Cun17]. In this paper we always assume that the friezes are tame.
Let us recollect some of the main properties of tame friezes. repeated horizontally under glide reflections. We will sometimes represent a fundamental domain instead of the whole frieze.
The entries in the first row of a tame frieze of width w will be denoted a 1 , . . . , a n , with n = w + 3.
). The South-East diagonal (∆ i ) 0≤i≤w+1 in a tame friezes of width w, starting at ∆ 0 = 1, ∆ 1 = a 1 , satisfies the recursion and similarly for the other diagonals up to shifts in the indices.
The above recursion gives an alternative way to compute the entries in the frieze from the first row without using the relation ad − bc = 1, which may cause trouble in the case where one of the entries is zero. A tame frieze starts as follows.
One can extend the friezes top and bottom with rows of 0's and −1's so that the rule ad−bc = 1 is still satisfied. The recursion (5) remains valid in the extended frieze and allows to immediately deduce the glide reflection property of Theorem 1.2. Furthermore, the recursion (5) in the extended frieze can be encoded by 2 × 2 matrices and lead to the following criterion.
Theorem 1.4 ( [Cox71]). The n-tuple (a 1 , . . . , a n ) determines the first row of a tame frieze if and only if (6) a n −1 The condition (6) is established in [Cox71] in an equivalent form, using continuants, in the case of friezes over positive real numbers. It was also recovered in this case in [BR10]. The arguments can be extended straightforward in the case of tame friezes over any unital ring. The condition (6) is also established in [CH19] for tame friezes over subsets of C and can also be extended straightforward in the case of tame friezes over arbitrary unital ring.
2. Friezes and he moduli space M 0,n 2.1. Definition of M 0,n . We fix a ground field K. The field will be later specify to the field of complex numbers C or to the finite field F q with q elements. A variety V (K) defined over K will be denoted simply V when no confusion occurs, e.g P 1 = P 1 (K), GL 2 = GL 2 (K). The classical moduli space of genus 0 curves with n distinct marked points is We consider a completion of this space by allowing repetitions of points as long as they are not cyclically next to each other We often consider the configuration space used in the definition of M 0,n ; we define where again the indices are considered modulo n.

2.2.
The subvariety of M + 0,2m . The varieties M 0,n have different properties according to the parity of n. In the case n = 2m we will need to consider the algebraic subvariety of M + 0,2m that we define below. We first introduce two subspaces of C 2m and then define These are algebraic subvarieties of M ± 0,2m of codimension 1.
2.3. Geometric realizations of tame friezes. Coxeter's friezes have deep connections with geometry [Cox71], [Cox91]. The link with the moduli space M 0,n was explained in [MGOT12] and then further developped in higher dimension with higher versions of friezes in [MGOST14].
). When n ≥ 3 is odd, tame friezes of width n − 3 over K = R or C form an algebraic subvariety of K n isomorphic to M 0,n (K).
Note that the subvariety of tame friezes in K n , of codimension 3, is given by the polynomial equations of (6) (since the involved matrices belong to SL 2 one gets three independent equations).
Originally the theorem was established in [MGOST14] for K = R but can be easily extended to the case K = C, see [MG15]. In Section 2.4 we will give a bijection between the set of tame friezes and the moduli space M 0,n , for odd n, in the case of an arbitrary field K.
In the case of even n the situation is different. We need to consider friezes up to rescaling of the first row. More precisely K * acts on the space of tame friezes with odd width by sending the frieze defined by the cycle of the first row (a 1 , . . . , a n ) to the frieze defined by (λa 1 , 1 λ a 2 , . . . , λa n−1 , 1 λ a n ) for λ ∈ K * . Also, some elements in M 0,n do not correspond to friezes. We need to restrict to M + 0,n . One gets the following.
Theorem 2. When n ≥ 3 is even, the variety of tame friezes of width n − 3 over K, modulo the action of K * , is isomorphic to M + 0,n (K).
We prove the above theorem in Section 2.5.

2.4.
Correspondence between tame friezes and M 0,n (K), case of odd n. For completeness we explain the bijection between the tame friezes of width w = n−3 with odd n over K and the set M 0,n (K) = C n (K)/PGL 2 (K). The construction is slightly adapted from [MGOST14] and [MG15] in order to work over an arbitrary field K.
Proof. Consider first an arbitrary lift (V ′ 1 , . . . , V ′ n ) of v in K 2 . We want to extend the sequence by antiperiodicity, V ′ i+n = −V ′ i , and rescale the points in order to have constant consecutive determinants. Introducing the unknown scalars c and λ i , 1 ≤ i ≤ n, we consider the system given by the n conditions det( This leads to the following The fact that n is odd is now crucial to solve the above system. Using λ 1 as a parameter one gets the expressions for all 1 ≤ i ≤ (n − 1)/2, using all but the last equation of (7). In order to have the full system satisfied one needs the final relation It is clear that there is a choice of c ∈ K * leading to a solution (λ 1 , . . . , λ n ) for the system. The sequence defined by V i = λ i V ′ i has the desired property. From a lift (V i ) of v given by Lemma 2.3 one gets n coefficients (a 1 , . . . , a n ) defined by The fact that the last coefficients in the expansion of V i in the basis (V i−1 , V i−2 ) are all equal to −1 comes form the fact that consecutive determinants are constant.
Lemma 2.4. The coefficients (a 1 , . . . , a n ) defined by (9) only depend on the class of v modulo PGL 2 (K) and characterize the class.
Proof. We need to check that the coefficients do not depend on the choice of the lift in Lemma 2.3 and that they are invariant under the action of PGL 2 (K). Let w be a configuration of C n equivalent to v modulo PGL 2 (K) (this of course includes the case w = v). Consider lifts (W 1 , . . . , W n ) and (V 1 , . . . , V n ) for w and v as in Lemma 2.3. There exist a matrix M ∈ GL 2 (K) and scalars µ 1 , . . . , µ n in K * such that Since the consecutive determinants are constant in both sequences one immediately gets that µ i = µ 1 for all 1 ≤ i ≤ n (here again we use the fact that n is odd). Thus W i = µ 1 M V i for all 1 ≤ i ≤ n and this leads to the same coefficients for the expansions (9). Now, if two lifts (W 1 , . . . , W n ) and (V 1 , . . . , V n ) for two configurations w and v lead to the same coefficients (9) then the matrix M defined by M V 1 = W 1 and M V 2 = W 2 will satisfy M V i = W i for all 1 ≤ i ≤ n. Thus w and v are equivalent modulo PGL 2 (K).
Proof. By writing the coordinates of the vectors V i and V i−1 in the rows of a 2 × 2 matrix one obtains from (9) Hence (6) holds.
The aboce lemma and Theorem 1.4 imply that the coefficients (a 1 , . . . , a n ) defined by (9) determine a unique tame frieze of width n − 3. Hence we have constructed an injective map from M 0,n (K) to the set of tame friezes of width n − 3.
Conversely, starting form a tame frieze of width n − 3, the corresponding point in M 0,n (K) is obtained from the sequence of n vectors −1 whose components are given by the first and second diagonal of the frieze.
2.5. Correspondence between tame friezes and M 0,n (K), case of even n. In this section we prove Theorem 2. When n is even there is no bijection between the tame friezes of width w = n − 3 and the set M 0,n (K). We need to adapt the strategy used in the case of odd n and use the subspace M + 0,n (K).
Lemma 2.6. When n is even Lemma 2.3 holds if and only if v ∈ C + n . Proof. We fix v ∈ C + n and proceed as in the proof of Lemma 2.3. We are led to the system (7). In general, for even n the system has no solution. But with the choice of v ∈ C + n , by definition, every arbitrary lift of v to (V ′ i ) 1≤i≤n will satisfy the relation . This makes the system consistent and solutions are given by (8) where λ 1 is any non zero parameter. Hence the rescaling V i = λ i V ′ i provides us with a antiperiodic lift with constant consecutive determinants.
Conversely, if such a lift exists the element v does belong to C + n . (Note that the choice of V n+1 = −V 1 makes a difference of sign in the equation defining C + n .) For v ∈ C + n and a lift (V i ) given by Lemma 2.3 one gets n coefficients (a 1 , . . . , a n ) defined by the recursion (9). Now these coefficients depend on the choice of the lift, but another choice of lift will lead to the coefficients (λa 1 , 1 λ a 2 , . . . , λa n−1 , 1 λ a n ) for some λ = 0. We immediately obtain the following analogs of Lemma 2.4 and Lemma 2.5.
Theorem 3. Let n ≥ 2. The number of points in M 0,n (F q ) is Proof. We first count points in the space C n .
Proof. One easily computes the first values c 2 = (q + 1)q, c 3 = (q + 1)q(q − 1), which are obtained as the choice of two, resp. three, distinct points among the q + 1 points of P 1 (F q ). Consider a configuration (v 1 , . . . , v n+1 , v n+2 ) in C n+2 . If v 1 = v n+1 then the configuration can be viewed as a configuration (v 1 , . . . , v n ) ∈ C n with an extra point v n+2 = v 1 . There are qc n configurations of this type. If v 1 = v n+1 then the configuration can be viewed as a configuration (v 1 , . . . , v n+1 ) ∈ C n+1 with an extra point v n+2 ∈ {v 1 , v n+1 }. There are (q−1)c n+1 configurations of this type. This establishes the recurrence relation This linear recurrence of order 2 can be easily solved and leads to the formula (11).
When a configuration contains three distinct points it will lead to q(q 2 − 1) other configurations modulo the action of PGL 2 (F q ). When n = 2m + 1 the configurations in C n always contain three distinct points, therefore When n = 2m, the (q + 1)q configurations of the form (v 1 , v 2 , v 1 , v 2 , . . . , v 1 , v 2 ), v 1 = v 2 are all the same modulo PGL 2 (F q ). All other configurations contain three distinct points. Therefore Lemma 3.1 is the only result needed in Appendix A.

3.2.
Counting points in M + 0,2m (F q ). We consider K = F q and want to count the points in the variety M + 0,2m defined in the previous section. Different cases appear according to the characteristic of the field is 2 or not and according to m is even or odd. Recall the definition of the q-binomial coefficient m 2 q = (q m −1)(q m−1 −1) (q−1)(q 2 −1) , which symplify to a polynomial in q. Proof. We introduce the cardinals c ± n = |C ± n (F q )|. Recall that c n = |C n (F q )| is given by Equation (11).
Lemma 3.2. For all n ≥ 4, even, one has c ± n = c n−1 + qc ∓ n−2 . Proof. Let us choose v = (v 1 , . . . , v n ) ∈ C + n . Assume first that v 1 = v n−1 . This means that v ′ = (v 1 , . . . , v n−1 ) defines an element of C n−1 . For such an element v ′ there is a unique choice of v n that makes (v ′ , v n ) belong to C + n . Indeed, choose any lift of v ′ to a tuple (V 1 , V 2 , . . . , V n−1 ) of points in K 2 and denote d i = det(V i , V i+1 ) for i = 1 . . . n − 2. We want to add V n such that the following condition d 1 d 3 · · · d n−3 det(V n−1 , V n ) = −d 2 d 4 · · · d n−2 det(V n , V 1 ) holds. Since v 1 = v n−1 , this leads to a unique choice of V n , up to scalar multiple. We conclude that there exist c n−1 elements v in C + n such that v 1 = v n−1 .
Assume now v 1 = v n−1 . The sequence v ′′ = (v 1 , . . . , v n−2 ) is an element in C n−2 . Choose any lift of v ′′ to (V 1 , V 2 , . . . , V n−2 ) and denote d i = det(V i , V i+1 ) for i = 1 . . . n − 3 and d n−2 = det(V n−2 , V 1 ). We want to add V n−1 = λV 1 and V n in order to have Substituting V n−1 = λV 1 and cancelling λ's we obtain the simplified condition which can hold if and only if d 1 d 3 · · · d n−3 = d 2 d 4 · · · d n−2 , i.e. if and only if v ′′ ∈ C − n−2 . And from this condition we see that if v ′′ ∈ C − n−2 any choice of v n in P 1 (F q )\{v 1 } will make (v ′′ , v 1 , v n ) belong to C + n . We conclude that there exist qc − n−2 elements v in C + n such that v 1 = v n−1 . We have established the recursion c + n = c n−1 + qc − n−2 . Same arguments hold when taking opposite signs so we also have c − n = c n−1 + qc + n−2 . From the above lemma one obtains where (−) m−1 = + or − if m is odd or even respectively. So we will need to separate the cases m odd or even. Also if char(F q ) = 2 one has c + n = c − n since C + n = C − n . So we will need to separate the cases char(F q ) = 2 and char(F q ) > 2.
if char(F q ) = 2 and m even, Proof. Assume char(F q ) > 2 and n = 2m with even m. We have C − 2 = ∅ so that c − 2 = 0 and we obtain from the recursion c + 2m = c 2m−1 + qc 2m−3 + q 2 c 2m−5 + . . . + q m−2 c 3 . The space C + 2m does not contain the elements of the form (v 1 , v 2 , v 1 , v 2 , . . . , v 1 , v 2 ). Thus elements of C + 2m always contain three distinct points and the orbits under PGL 2 will have same cardinality |PGL 2 | . One obtains Assume char(F q ) > 2 and n = 2m with odd m. We have C + 2 = C 2 so that c + 2 = c 2 = q(q + 1) and we obtain from the recursion The space C + 2m does contain the elements of the form (v 1 , v 2 , v 1 , v 2 , . . . , v 1 , v 2 ). These elements are all the same under the action of PGL 2 . The other elements of C + 2m always contain three distinct points and their orbits under PGL 2 have same cardinality q 3 − q. One obtains Assume char(F q ) = 2. In that case c + 2 = c − 2 = c 2 = q(q + 1). The computations are the same as in the previous case, independently of m is odd or even.
Finally one gives alternative expressions in the formula of Lemme 3.3. One first notices that In the sums one can group the terms for k = 2ℓ − 1 and k = 2ℓ. If m = 2m ′ one obtains and if m = 2m ′ + 1 one obtains One observes that in both cases Theorem 4 is proved.
To count all the tame friezes, or equivalently the solutions of (6), over F q , we use the geometric realizations of the previous sections.
Proof of Theorem 1 (i) Let n = 2m + 3. The construction in Section 2.4 gives a bijection between the set of tame friezes of width n − 3 with odd n and the set M 0,n (K). Theorem 3 gives the number of elements in the sets when K = F q . Hence we get (1).
Hence we have just established that the number of tame friezes over F q is 2q − 1 if the characteristic is 2 and q − 1 otherwise, which agrees with the value of f 1 given in Theorem 1.
For q = 4, let F 4 = {0, 1, α, β} with β = 1 + α = α −1 . The 17 friezes are the same as the one above for F 2 together with the friezes containing one of the following three fundamental domains 1 1 1 1 α 0 β 1 1 1 Note that the domain on the left contributes to 10 friezes under dihedral action.
When working over F p with p prime, the original construction of Conway-Coxeter [CC73] for friezes over Z ≥0 by means of triangulations of polygons provide solutions to this problem by reducing modulo p. But not all the solutions. A construction producing all the friezes over F 2 and F 3 has been given [Mab19].
Open problem: Find a combinatorial model to construct all friezes over F q .
Appendix A. Partitions of a circular set As an application of the computation in Section 3 and independently of Section 2 and Section 1 we solve an enumeration problem of restricted partitions of a circular set.
A.1. Definition. We consider the set {1, 2, . . . , n} as the labels of n cyclically ordered points on the circle, i.e. 1 and n are consecutive points. We denote by A k,n the number of partitions of the set into k parts such that no two consecutive points are in the same part.
Example A.1. One always has A 1,n = 0 and A n,n = 1. For n = 3, one has A 1,3 = A 2,3 = 0 and A 3,3 = 1. For n = 4 one has A 1,4 = 0, A 2,4 = 1, since the only allowed partition into two parts is exactly Our next goal is to deduce the exact value of A k,n from §3.1. These values are known, see A261139 and A105794 of [OEI19]. The corresponding generating function is given in [KL07]. They are also a particular case of graphical Stirling numbers [GT13].
A.2. Recounting points of C n (F q ). We consider the set of configurations C n (F q ) = {(v 1 , v 2 , . . . , v n ) ∈ (P 1 (F q )) n , v 1 = v n , v i = v i+1 , ∀i} introduced in §2.1. The cardinal c n = |C n (F q )| has been computed in Lemma 3.1. It can be related to the numbers A k,n .
Proof. Let us choose a partition S 1 ⊔ . . . ⊔ S k of cyclic {1, 2, . . . , n} such that no two consecutive integers are in the same subset S ℓ . Consider a configuration in C n satisfying v i = v j if and only if i and j belong to the same part S ℓ . There are exactly (q +1)q k−2 j=1 (q −j) such configurations, corresponding to a choice of k points in P 1 (F q ). Hence the results.
In the next section we use the exact value of c n given by (11) to deduce the value of A k,n .
A.3. Determining A k,n . The problem has now been translated into a simple problem of linear algebra consisting in expanding a given polynomial P (q) ∈ Q[q] into the basis of the falling factorials (q) k := q(q − 1) . . . (q − (k − 1)).
This exercise has been probably already solved in the literature. We write down a brief solution.
Let P be a polynomial of degree n and denote by B k its coefficients in the basis {(q) k } k , i.e. P (q) = (−1) j j!(k − j)! P (j).
Hence, one obtains Applying the above lemma for P = c n /q(q + 1) (note that this is indeed a polynomial in q since c n vanishes for q = 0 and q = −1), one immediately deduces the following formula.