Refined Catalan and Narayana cyclic sieving

We prove several new instances of the cyclic sieving phenomenon (CSP) on Catalan objects of type A and type B. Moreover, we refine many of the known instances of the CSP on Catalan objects. For example, we consider triangulations refined by the number of"ears", non-crossing matchings with a fixed number of short edges, and non-crossing configurations with a fixed number of loops and edges.


Introduction
The original inspiration for this paper is a natural interpolation between type A and type B Catalan numbers.For n ≥ 0 consider the expression For s = 0, we recover the n th Catalan number and for s = 1, we recover the (n + 1) th Catalan number.When s = n, we obtain the central binomial coefficient 2n n , which is known as the n th type B Catalan number, see [Arm09].There are several combinatorial families of objects which are counted by the expression in (1), certain standard Young tableaux and lattice paths to name a few.The expression in (1) has the q-analog given by the difference of q-binomials 2n n q − q s+1 2n n − s − 1 q . (2) For s ∈ {0, 1, n}, the polynomials in (2) appear in instances of the cyclic sieving phenomenon.Furthermore, it follows from [APRU20, Theorem 46] that there exist group actions such that the polynomials in (2) exhibit cyclic sieving for all s ∈ {0, 1, . . ., n}.
Definition 1 (Cyclic sieving, [RSW04]).Let X be a set and C n be the cyclic group of order n acting on X.Let f (q) ∈ N [q].We say that the triple (X, C n , f (q)) exhibits the cyclic sieving phenomenon (CSP) if for all d ∈ Z, |{x ∈ X : where ξ is a primitive n th root of unity.
Note that it follows immediately from the definition that |X| = f (1).In the study of cyclic sieving, it is mainly the case that the C n -action and the polynomial f (q) are natural in some sense.The group action could be some form of rotation or cyclic shift of the elements of X.The polynomial usually has a closed form and is also typically the generating polynomial for some combinatorial statistic defined on X. See B. Sagan's article [Sag11] for a survey of various types of CSP instances.
Many known instances of the cyclic sieving phenomenon involve a set X whose size is a Catalan number.Once such a CSP triple is obtained, one can ask if X can be partitioned X = j X j in such a way that the group action on X induces a group action on X j for all j, and, in that case, also ask if there is a refinement of the CSP triple in question.
Definition 2 (Refinement of cyclic sieving).The family {(X j , C n , f j (q))} j of CSP triples is said to refine the CSP triple (X, C n , f (q)) if • X j = X, • j f j (q) = f (q) and • the C n -action on X j coincides with the C n -action on X restricted to X j , for all j.
Typically, the sets X i are of the form X j = {x ∈ X : st(x) = j} for some statistic st : X → N that is preserved by the group action.Examples of such statistics are the number of cyclic descents of a word, the number of blocks of a partition, and the number of ears of a triangulation of an n-gon-all with the group action being (clockwise) cyclic rotation.Throughout the paper, we shall consistently use the order of the group (or group generator) as subscript.For example, rotation by 2π/n is denoted rot n .
For s ∈ {0, n}, the q-analog in (2) admits a natural refinement, so that the type A and type B q-Narayana polynomials are recovered.The q-Narayana polynomials can be used to refine the aforementioned instances of the CSP.It is therefore natural to ask if there is a q-analog of (1) for arbitrary s ∈ {0, 1, . . ., n} which also exhibits similar combinatorial properties as the type A and type B q-Narayana polynomials.We discuss partial results and motivations behind this problem in Section 3.
In the process of analyzing this intriguing question, we discovered several new instances of the cyclic sieving phenomenon.Some concern new q-analogs of Catalan numbers, while others refine known instances.In the tables in Section 2.5, we present a comprehensive (but most likely incomplete) overview of the current state-of-the-art regarding the cyclic sieving phenomenon involving Catalan and Narayana objects of type A and B.
1.1.Overview of our results.We only highlight some of the results in our paper; in addition we also prove several other results which fill gaps in the literature.In Section 4, the main result is the following theorem, which is a new refined CSP instance on Catalan objects.It can be stated either in terms of promotion (denoted ∂ 2n ) on two-row standard Young tableaux with k cyclic descents, SYT cdes (n 2 , k), or non-crossing perfect matchings with k short edges, NCM sh (n, k).
In Section 5, we study the set of so-called non-crossing (1,2)-configurations on n vertices, which we denote by NCC(n + 1).The cardinality of this set is the Catalan number Cat(n + 1) = 2n n − 2n n−2 .We define a simple "rotate-and-flip" action on NCC(n + 1) which has order 2n and is reminiscent of promotion.
Note that we use a quite non-standard q-analog of the Catalan numbers here, which has not appeared in the context of cyclic sieving before.Cyclic sieving on noncrossing (1,2)-configurations was studied earlier by M. Thiel [Thi17], with rotation as the group action.In Theorem 38 and Corollary 39, we refine Thiel's result.In particular, we obtain a new CSP instance involving the q-Narayana polynomial Nar(n + 1, k; q).
In Section 6, we study various instances of cyclic sieving involving the type B Catalan numbers, 2n n .Some results have more or less appeared in earlier works, but we make some of the results more explicit.One novel result is a type B version of Theorem 4, where we consider the twist action on type B non-crossing (1,2)configurations.Briefly, such objects are obtained from elements in NCC(n) by choosing to mark one edge.
As in type A, we also obtain a refined cyclic sieving result in Theorem 50 where we consider rotation instead.
In Section 7, we briefly consider two-column semistandard Young tableaux, and note in Theorem 54 that (SSYT(2 k , n), ∂n , Nar(n + 1, k + 1; q)) is a CSP triple, where ∂n denotes the so-called k-promotion and SSYT(2 k , n) is the set of semistandard Young tableaux of the rectangular shape 2 k whose maximal entry is at most n.
In Section 8, we refine the classical CSP triple on triangulations of an n-gon by taking ears into consideration.An ear in a triangulation is a triangle formed by three cyclically consecutive vertices.We let TRI ear (n, k) denote the set of triangulations of an n-gon with k ears.
In the last section, we consider another natural interpolation between type A and type B Catalan objects and prove a cyclic sieving result using standard methods.
Finally, a word about the proofs in this paper.There are traditionally two different approaches to proving instances of the cyclic sieving phenomenon -combinatorial1 or representation-theoretical (using vector spaces and diagonalization).In this paper we exclusively use the combinatorial approach, meaning that we need to explicitly evaluate the CSP-polynomials at roots of unity and also count the number fixed points of the sets under the group actions.It may also involve the use of equivariant bijections to derive new CSP triples from the previously known ones.

Preliminaries
We shall use standard notation in the area of combinatorics, see the go-to references [Sta01,Mac95].In particular, [n] := {1, 2, . . ., n} and it should not be confused with the q-analog [n] q defined further down.

Words and paths. Given a word w
We let the major index, denoted maj(w), be the sum of the descents of w.An inversion in w is a pair of indices i, j ∈ [n] such that i < j and w i > w j .We let inv(w) be the number of inversions of w.Let BW(n, k) denote the set of binary words of length n with exactly k ones.
Let PATH(n) be the set of paths from (0, 0) to (n, n) using north, (1, 0), and east, (0, 1), steps.A peak is a north step followed by an east step, and a valley is an east step followed by a north step.We have an obvious bijection PATH(n) ↔ BW(2n, n) where we identify north steps with zeros.Given P ∈ PATH(n), we let maj(P ) be defined as the sum of the positions of the valleys of the path P .Observe that this coincides with the major index of the corresponding binary word, as valleys correspond to descents.We shall also let pmaj(P ) denote the sum of the positions of the peaks.For a path P ∈ PATH(n), we let the depth, depth(P ) be the largest value of r ≥ 0 such that the path touches the line y = x − r.Let us define PATH s (n) ⊆ PATH(n) as the set of paths with depth(P ) ≤ s.We set DYCK(n) := PATH 0 (n).
2.2.q-analogs.Roughly, a q-analog of a certain expression is a rational function in the variable q from which we can obtain the original expression in the limit q → 1.
if n ≥ k ≥ 0, and n k q := 0 otherwise.Note that the q-binomial coefficients are polynomials in the variable q, see [Sta11] for more background.The q-multinomial coefficients are defined in a similar manner.
where Φ d (q) is the d th cyclotomic polynomial.In particular, we have When ξ is a root of unity, let o(ξ) denote the smallest positive integer with the property that ξ o(ξ) = 1.The following is a standard lemma that should not need a proof.
Lemma 10.Let n, k, d ∈ N and let ξ be a primitive n th root of unity.Then We will use Theorem 9 and Lemma 10 in later sections.
Lemma 11.Let ξ be a primitive n th root of unity, and suppose that Proof.In [AA19, Lem.2.2], it is proved that f (up to mod q n − 1) is a linear combination of It then suffices to verify that for all d | n, j ∈ Z, which is straightforward by using Lemma 10.
Hence, if we know that f (ξ d ) ∈ Z for all d ∈ Z, it suffices to verify (3) for all d | n.There is a related result about computing the number of fixed points.
Lemma 12. Suppose that C n = g acts on the set X.
Proof.Note that all elements of C n with order o generate the same subgroup S ⊆ C n .If h, h are both of order o, then h = h = S, and h Lemma 11 and Lemma 12 are useful facts and are used implicitly in many papers.We shall use them without further mention throughout the paper.

Catalan and Narayana numbers. The Catalan numbers
are indexed by natural numbers.These numbers occur frequently in combinatorics, see A000108 in the OEIS, and give the cardinalities of many families of combinatorial objects.For the purpose of this paper, we note that the following sets all have cardinality Cat(n).Throughout this paper, we use MacMahon's q-analog of the Catalan numbers.For any natural number n, the n th q-Catalan number is defined by A definition of maj on standard Young tableaux can be found in the next section.
The Narayana numbers Nar(n, k) := 1 n n k n k−1 , indexed by two natural numbers n and k such that 1 ≤ k ≤ n, are also well-known and have many applications, see the OEIS entry A001263.The Narayana numbers refine the Catalan numbers in the sense that k Nar(n, k) = Cat(n).For our purposes, it suffices to know that the following sets all have cardinality Nar(n, k).
• DYCK(n, k): the set of paths in DYCK(n) with exactly k peaks, • SYT(n 2 , k): the set of tableaux in SYT(n 2 ) with exactly k descents, • NCP(n, k): the set of partitions in NCP(n) with exactly k blocks, • NCC(n, k): the set of non-crossing (1, 2)-configurations in NCC(n) such that the numbers of proper edges plus the number of loops is equal to k − 1, • NCM(n, k − 1): the set of non-crossing matchings in NCM(n) with k − 1 even edges.The q-Narayana numbers are defined as the q-analog q maj(T )−n .
The q-Narayana numbers refine the q-Catalan numbers, that is, k Nar(n, k; q) is equal to Cat(n; q).We also mention that there is a bijection NCPtoDYCK from NCP(n, k) to DYCK(n, k) described in Bijection 8. Thus, For more background, see [Sim94] and [ZZ11].
2.4.Type B Catalan numbers.We shall now describe the type B analogs of the combinatorial objects we saw in the previous section.The type B Catalan numbers Cat B (n) are defined as The type B Narayana numbers Nar B (n, k) are defined as The type B Narayana numbers clearly refine the B Catalan numbers, as can be seen from ( 9).Among other things, they count the number of elements in PATH(n) with k valleys.For a more comprehensive list, see A008459 in the OEIS, and also the reference [Arm09] for more background.The q-analogs of the type B Catalan numbers and the type B q-Narayana numbers are defined as It is straightforward to verify that Cat B (n; q) = n k=0 Nar B (n, k; q).Moreover, one can show that t valleys(P ) q maj(P ) , (13) see [Sul98,Sul02].
The following combinatorial interpretation of the type B q-Narayana numbers is mentioned in I. Macdonald's book [Mac95,p. 400].Let V be a 2n-dimensional vector space over F q , and let U be an n-dimensional subspace of V .Then Nar B (n, k; q) is the number of n-dimensional subspaces U of V such that dim(U ∩ U ) = n − k.The bijections in Figure 1 respect a Narayana refinement and so, for example, SYT(n 2 , k), NCM(n, k − 1) and NCP(n, k) are all equinumerous.Furthermore, composing the natural bijections Bijection 6 and the inverse of Bijection 7, we get that promotion on SYT corresponds to rotation on non-crossing matchings.Theorem 27
However, promotion on standard Young tableaux does not preserve the number of descents, but rotation preserves the number of even edges of matchings.It follows that the cyclic sieving phenomenon on non-crossing matchings with a specified number of even edges does not correspond to one on SYT(n 2 ) with a fixed number of descents with promotion as the action.In general, a specific Narayana refinement might be incompatible with a cyclic group action.By Bijection 6, here the compatible statistic one should use for SYT(n 2 ) is the number of even entries in the first row.
A note on the general philosophy of the paper.Having many different sets of objects and well-behaved bijections between these sets turns out to be a very fruitful approach to proving instances of the CSP.In this context, well-behaved often times means that the bijection is equivariant.If a group action looks complicated on a certain set, it can perhaps be made easier if one first applies an equivariant bijection and then studies the image.For example, promotion on SYT(n 2 ) is complicated while rotation on NCM(n) is easier.There is a type of converse of the above.If one has two different CSP triples with identical CSP-polynomials and whose cyclic groups have the same order, then there exists an equivariant bijection between these two sets (by sending orbits to orbits of the same size).

Type A/B-Narayana numbers and a quest for a q-analog
We shall now discuss a natural interpolation between type A and type B Catalan numbers.The following observation illustrates this interpolation.For any s ≥ 0, the sets below are equinumerous: (1) the set of skew standard Young tableaux SYT((n + s, n)/(s)), (2) the set of lattice paths, PATH s (n), (3) the set of order ideals in the type B root poset with at most s elements on the top diagonal.
Note that for s = 0, we recover sets of cardinality Cat(n), and for s = n, we recover sets of cardinality Cat B (n). Bijective arguments are given below in Proposition 13 and Proposition 15.
Let T ∈ SYT(λ/µ) where the diagram of λ/µ has n boxes.A descent of T is an integer j ∈ {1, . . ., n − 1} such that j + 1 appears in a row below j.The major index of T is the sum of the descents.The major-index generating function for skew standard Young tableaux is defined as when λ/µ is a skew shape.Our motivation for studying this polynomial is [APRU20, Thm.46] which states that for any skew shape λ/µ where each row contains a multiple of m boxes, there must exist some cyclic group action C m of order m such that is a CSP triple.We do not know how such a group action looks like except in the case m = 2.In that case one can use evacuation, defined by Schützenberger [Sch63].
3.1.Skew standard Young tableaux with two rows.We now describe a bijection between skew SYT with two rows and certain lattice paths.
Proposition 13.Given s ∈ {0, . . ., n}, there is a bijection which sends descents in the tableau to peaks in the path.
Proof.A natural generalization of the standard bijection works: an i in the upper or lower row corresponds to the i th step in the path being north or east, respectively.Evidently, a descent in the tableau is sent to a peak in the path.
Recall that for a SYT T the statistic maj(T ) is the sum of the position of the descents, which is then sent to pmaj(P ) which is the sum of the positions of the peaks in the corresponding path P .Let X n,s (q) := P ∈PATHs(n) q pmaj(P ) and Y n,s (q) := P ∈PATHs(n) q maj(P ) .

Root lattices in type A/B.
The following illustrate the root ideals of B n where n = 3.There are in total 2•3 3 = 20 such ideals.A root ideal is simply a lower set in the root poset-marked as shaded boxes in the diagrams below.Root ideals are also called non-nesting partitions of type W , where W is the Weyl group of some root system.
An explicit bijection from the set of skew standard Young tableaux SYT((n + s, n)/(s)) to the root ideals of B n with at most s elements on the top diagonal is described below.First, let OI(n, s) be the set of root ideals with at most s elements on the top diagonal.
Bijection 1.Let a 1 , a 2 , . . ., a n be the top row of the skew tableau.We identify this top row using the bijection in Proposition 13 with a path α ∈ PATH s (n) and get that depth(α) = max i {a i − 2i + 1}.Let j be the smallest value for which the maximum is obtained, so depth(α) = a j − 2j + 1.We then define the map φ as changing the step a j − 1, just before reaching maximal depth for the first time, from an east step to a north step.That is, φ(α) = a 1 , . . ., a j−1 , a j − 1, a j , . . ., a n .This new path ends at (n − 1, n + 1) and has depth one less than α.We repeat depth(α) times and get φ depth(α) (α) which ends in (n − depth(α), n + depth(α)) and has depth zero.This path always starts with a north step, and the boxes below and to the right of it make up a root ideal o in B n with depth(α) elements in the top diagonal.Since depth(α) ≤ s this gives o ∈ OI(n, s) and the desired map.See Figure 2 for an example.The inverse φ −1 is easily obtained as follows.Given a root ideal o, let β(o) be the north-east path along its boundary, starting with an extra north step.Now, change the north step of β(o) after the last time the path has reached maximum depth to an east step.The inverse of the bijection is obtained by iterating φ −1 until the path ends in (n, n).The map φ has been used many times before, see e.g.[ALP19].

We naturally define maj(o) := maj(β(o)) and pmaj(o) := pmaj(β(o)) for o ∈ OI(n, s).
Proposition 15.The map in Bijection 1 is a bijection so Proof.The map is clearly a bijection and the first formula follows.For the second statement note that the map φ does not change the peaks and thus not pmaj, but it changes the position of one valley and decreases maj by 1. Thus the q-polynomial for pmaj is identical to X n,s (q) in Proposition 14.The maj-generating polynomial for paths with depth at most s is Y n,s (q) by Proposition 14.Thus, for paths having depth exactly d, it is q d 2n n−d q − q d+1 2n n−d−1 q .A path with depth d is mapped by φ d to a root ideal with exactly d elements in the top diagonal.This gives the sum o∈OI(n,s) which simplifies to the formula given.
Remark 16.There is a notion of rowmotion as an action on order ideals.Unfortunately, this action does not seem to have a nice order when restricted to OI(n, s) for general s, see [SW12].For example, for n = 3 and s = 1 we have the following orbit of length 4, implying that the action does not have the order we are looking for (which is n = 3).
Let ξ be a primitive (2n) th root of unity.For all integers n > s ≥ 0 and d | 2n, we have .
where χ(a, b) is equal to 1 if b divides a and 0 otherwise.
Proof.The evaluation follows from the q-Lucas theorem, Theorem 9.
In light of (15), it would be of great interest to explicitly describe a group action C n so that X n,s (q) or Y n,s (q) is the corresponding polynomial in a CSP triple (which must exist due to (15)).In Section 5 we give an explicit action in the case s = 1, which gives a new CSP triple involving the Catalan numbers.

Narayana connection.
We discuss an open problem regarding the interpolation between type A and type B q-Narayana numbers.This problem is part of a broader set of questions regarding the interplay of cyclic sieving and characters in the symmetric group, see [APRU20].We argue that the small special case discussed below is interesting in its own right.
Recall that where the sum ranges over Dyck paths of size n.Hence, n k=1 Nar(n, k; q) = X n,0 (q).Note that the set of Dyck paths with k peaks is in bijection with the set of noncrossing set partitions of [n] with k blocks.

Problem 18 (Main Narayana problem). Refine the expression
for all s ≥ 0 in the same way as the q-Narayana numbers Nar(n, k; q) refine the case s = 0.This problem is not really interesting unless we impose some additional requirements.In Problem 18, we are hoping to find a family of polynomials, N (s, n, k; q) ∈ N[q] with some of the following properties: Specializes to Nar(n, k; q): For s = 0, we have Refines the Y n,s (q) in (16): We want that for all s ≥ 0, we have the identity Is given by some generalization of the the peak statistic: We hope for some statistic peaks s (P ) such that peaks 0 (P ) is the usual number of peaks of a Dyck path and We can alternatively consider some other family of combinatorial objects mentioned in (3), such as type B root ideals with at most s elements on the top diagonal, or standard Young tableaux in SYT((n + s, n)/(s)) with some type of generalized descents.
Refines Cat B (n) at s = n: For s = n, we have a natural candidate Note that N (n, n, k; q) is not equal to Nar B (n, k; q) that appear in Section 2.4.The combinatorial interpretation in this case is as follows: where a modified peak is any occurrence of 01 (north-east) in the path, plus 1 if the path ends with a north step.

Palindromicity:
The Narayana numbers have quite nice properties.First of all, is a palindromic polynomial (in t).For example, for n = 5, this sum is t+10t 2 +20t 3 + 10t 4 + t 5 .One would therefore hope that for fixed s the sum n k=1 t k N (s, n, k; q) is palindromic.The s = n candidate given by [n + 1] q Nar(n, k; q) is also palindromic.Palindromicity II: Each N (s, n, k; q) is a palindromic polynomial (in q).This is true for Nar(n, k; q) and the expression in (18).Gamma-positivity: The sum n k=1 t k N (s, n, k; 0) is γ(t)-positive (see the survey [Ath18] for the definition).The corresponding statement seems to hold for the expression in (18).One might hope that the general expression

Values at roots of unity and cyclic sieving:
We require that N (s, n, k; ξ) is a non-negative integer whenever ξ is an n th root of unity.This resonates well with the palindromicity properties, and cyclic sieving for (15).Taking (17) into account, we would like that for every k ≥ 0, is a CSP triple for some action β n of order n.Note that such a refinement is known in the case s = 0, as shown in the table below.Note also that there cannot be a cyclic group action of order 2n that fits together with N (0, n, k; q) in a CSP triple: for example, at n = 4, k = 2 this is not an integer at a primitive 2n th root of unity.

Set
Group action q-statistic peak-statistic Dyck paths -maj peaks Non-crossing partitions † Rotation maj blocks Table 3.We only have the full Narayana refinement picture for the non-crossing partition family.That is, there is a "peak"-statistic and a group action of order n preserving the peak-statistic.Note that promotion on Dyck paths does not preserve the number of peaks.† For these sets maj is computed via a bijection to paths.
Example 19.For n = 2, s = 1, we have that Y 2,1 (q) = q 4 + q 3 + q 2 + q + 1.We want to refine this into two polynomials corresponding to k = 1, 2. The criteria to have non-negative evaluations at roots at unity, here −1, tells us that q 3 and q must be together with at least one other term each.By palindromicity II there are five possibilities for N (1, 2, 2; q): q 4 + q 3 + q + 1, q 4 + q 3 + q 2 + q, q 4 + q 3 + q 2 , q 4 + q 3 and q 4 .4. Case s = 0 and non-crossing matchings The goal of this section is to prove two Narayana-refinements of cyclic sieving on non-crossing perfect matchings by considering the number of even edges and short edges.The second result corresponds to a refinement of the CSP on SYT(n 2 ) under promotion, where we refine the set by the number of cyclic descents.4.1.Even edge refinement.Given a non-crossing perfect matching, let even(M ) denote the number of edges {i, j} where i < j and i is even.We refer to them as even edges, and all non-even edges are called odd.Let NCM(n, k) be the set of M ∈ NCM(n) such that even(M ) = k.
Note that for parity reasons an edge {i, j} must have i + j odd.Thus the set of non-crossing perfect matchings on 2n vertices with k even edges is invariant under rotation by rot n since • any odd edge (i, 2n) is mapped to the even edge (2, i + 2); • any even edge (j, 2n − 1) is mapped to the odd edge (1, j + 2).The first result is essentially just a restatement of [RSW04, Thm.7.2].
Proposition 20.For 0 ≤ k ≤ n, the triple exhibits the cyclic sieving phenomenon.
Proof.Mapping non-crossing matchings to non-crossing partitions via the inverse of NCPtoNCM takes matchings with k even edges to partitions with k + 1 blocks, see Bijection 7.This CSP result was proven already in [RSW04, Thm.7.2].

Short edge refinement.
Definition 21.We define promotion ∂ 2n : SYT(n 2 ) → SYT(n 2 ) as the following composition of bijections: Promotion is originally defined for Young tableaux of all shapes using the so-called jeu-de-taquin.The notion has been generalized to arbitrary posets by R. Stanley, see [Sta09].The above definition of cyclic descent set can be generalized in a straightforward manner to all rectangular standard Young tableaux-that is, tableaux of shape λ = (a b ).In [Hua20], an explicit construction is given, where it is shown that all shapes which are not connected ribbons admit a type of cyclic descent statistic.It follows that one can define the set SYT cdes (λ, k) for all such shapes λ as well.
The set SYT cdes (n 2 , k) is in bijection with a certain subset of DYCK(n) which we shall now describe.We first recall the standard bijection SYTtoDYCK between SYT(n 2 ) and DYCK(n): given a T ∈ SYT(n 2 ), let SYTtoDYCK(T ) = w 1 w 2 • • • w 2n be the Dyck path where w i = 0 if i is in the top row and w i = 1 otherwise.
Call a Dyck path These numbers are a shifted variant of the OEIS entry A108838.Define the following q-analog of these numbers.For any two natural numbers n and k, let where the second sum is taken over all D ∈ DYCK(n) that are either non-elevated with k peaks or elevated with k − 1 peaks.For integers k, n ≥ 1 we claim that To see this, consider the restriction of SYTtoDYCK to SYT cdes (n 2 , k).If D is an elevated Dyck path of size n with k peaks and D is the corresponding Dyck path of size n − 1, then maj(D) − maj(D ) = k − 1, as each of the k − 1 valleys contribute one less to the major index in D compared to in D.
The polynomials Syt(n, k; q) refine the q-Catalan numbers, which is easily seen by comparing their definition with (6).
Proposition 24.For all integers n, k Syt(n, k; q) = Cat(n; q).It is easy to see that Syt(0, 0; q) = Syt(1, 1; q) = 1 and Syt(n, k; q) = 0 for all other pairs of natural numbers n, k such that either n ≤ 1 or k ≤ 1.For larger n and k, we have the following closed form for Syt(n, k; q).Lemma 25.For all integers k, n ≥ 2, Proof.We may restrict ourselves to the case when n ≥ k as both sides of (23) are identically zero otherwise.We write Syt(n, k; q) using the expression in (22) and expand the q-Narayana numbers to obtain The expression in the parentheses is then rewritten as We must now show that this is equal to 1 + q n or, equivalently, that the following identity holds: If n = k, the identity is clearly true, so we may assume that n > k.By using [j] q = (1 − q j )/(1 − q) for integers j ≥ 1, we get the equivalent equation Clearing denominators and expanding the products gives which evidently holds for all integers n, k.
The edge xy in a non-crossing perfect matching is said to be short if either x = i and y = i + 1 for some i or if x = 1 and y = 2n.If M ∈ NCM(n), then we denote short(M ) its number of short edges and NCM sh (n, k) the set of M ∈ NCM(n) such that short(M ) = k.The set SYT cdes (n 2 , k) is in a natural bijection with NCM sh (n, k).To see this, we use the standard bijection SYTtoNCM between SYT(n 2 ) and NCM(n), see Bijection 6 in Appendix B.1.It follows from our definition of promotion that SYTtoNCM is an equivariant bijection in the sense that From the definition of SYTtoNCM and (25), one can prove the following lemma.
Lemma 26.Let T ∈ SYT(n 2 ).Then x ∈ cDes(T ) if and only if xy, where x < y, is a short edge in SYTtoNCM(T ).
Theorem 27.Let n, k be natural numbers.The triple exhibits the cyclic sieving phenomenon.
Proof.Let ξ be a primitive (2n Then, by dividing into cases and applying Theorem 9 (the q-Lucas theorem) twice, we get We prove that these evaluations agree with the number of fixed points in NCM sh (n, k) under rot d n on a case-by-case basis.Case d = 2n: Trivial.Case o(ξ d ) | n and k 0 = 0: By using Bijection 5, we see that such rotationally symmetric perfect matchings are in bijection with the set BW k1 (n/ o(ξ d )).To see that this set has the desired cardinality, we equate the two expressions in (32) and (33) and then take q = 1.Case o(ξ d ) = 2, n odd and 2 | k: It is easy to check that the assertion holds in the case n = 3 and k = 2.It thus remains to show the assertion for n > 3.Such a non-crossing perfect matching must have a diagonal (an edge that connects two vertices i and i + n (mod 2n)) that divides the matching into two halves.The diagonal can be chosen in n ways.The matching is now determined uniquely by one of its two halves.To choose one half, we choose a non-crossing matching on (n − 1)/2 vertices with k/2 short edges, not including a potential short edge between the vertices closest to the diagonal.Such a matching is either i) an element of NCM sh ((n − 1)/2, k/2) which does not have an edge between the two vertices that lie closest to the diagonal or ii) an element of NCM sh ((n − 1)/2, k/2 + 1) which has a short edge between the two vertices that lie closest to the diagonal.
Let us note that, in general, the fraction of elements in NCM sh (n, k) that have a short edge adjacent to a given side is equal to k/2n.This is easily seen by considering rotations of such a non-crossing perfect matching.Hence, in our case the number of matchings fixed by rot d 2n is equal to Substituting the values from (20) (recall that |NCM sh (a, b)| = SYT cdes a 2 , b ), it remains to show that this expression is identical to the one given by Syt(n, k; −1).This can now be verified with a computer algebra system, such as Sage [Sag20].
The remaining cases: We need to show that, in all the remaining cases, there are no rotationally symmetric non-crossing perfect matchings.Suppose first that o(ξ d ) = 2, n is odd and 2 k.It is clear that such a non-crossing perfect matching must have a diagonal dividing the matching into two halves.The two halves are identical up to a rotation of π radians and so, in particular, they must have the same number of short edges.In other words, the number of short edges must be even, contradicting 2 k.We now use the claim and note that there cannot be any non-crossing perfect matchings that are fixed under rotation by an odd number of steps, except in the case when o(ξ d ) = 2 and n is odd, i.e. when the matching has a diagonal.This exhausts all possibilities and thus the proof is complete.
Theorem 27 can be stated in an alternative way as follows.Since SYTtoNCM maps cyclic descents to short edges, we see that SYT cdes (λ, k) is closed under promotion for all rectangular λ.Recall that (SYT(λ), ∂ , Cat(n; q)) exhibits the cyclic sieving phenomenon, for rectangular λ.In the case when λ = (n, n), we have the following refinement with regards to the number of cyclic descents.
A related result is alluded to by C. Ahlbach, B. Rhoades and J. Swanson in the presentation slides [Swa18].They claim to have proven a refinement of the cyclic sieving phenomenon on standard Young tableaux with the group action being promotion in the Catalan case.It is not clear from the slides if they refine by cyclic descents or in some other way.Therefore, we cannot tell if their result is identical to Theorem 33 or not.
It follows from [Rho10, Lemma 3.3] that the number of cyclic descents remains fixed under promotion of rectangular standard Young tableaux.Experiments suggests that Corollary 28 generalizes to all rectangular standard Young tableaux.This would be a refinement of the famous CSP result on rectangular tableaux, see [Rho10, Theorem 1.3].More precisely, we denote f λ k (q) := T q maj(T ) where the sum is taken over all standard Young tableaux of shape λ with exactly k cyclic descents.

Case s = 1 and non-crossing (1,2)-configurations
For s = 1, there is a nice Catalan family, given by non-crossing (1, 2)-configurations described in [Sta15,Family 60].In the first subsection, we introduce a twisted rotation action on such configurations, and prove a new instance of Catalan CSP.In the second subsection, we refine a CSP result of Thiel, where the group action is given by rotation.

A new Catalan CSP under twisted rotation.
A non-crossing (1, 2)configuration of size n is constructed by placing vertices 1, . . ., n − 1 around a circle, and then drawing some non-intersecting edges between the vertices.Here, we allow vertices to have a loop, which is counted as an edge.There are Cat(n) elements in this family.Let NCC(n) be the set of such objects of size n, and let NCC(n, k) be the subset of those with k − 1 edges, loops included.See Appendix A for a figure when n = 3.
Bijection 2 (Laser construction).See Figure 3 for an example.Let P ∈ DP(n).Define the non-crossing (1, 2)-configuration DYCKtoNCC(P ) as follows.First, number the east-steps with 1, 2, . . ., n − 1.Secondly, if there is a valley at (i, j), draw a line (a laser) from (i, j) to (i + ∆, j + ∆), where ∆ is the smallest positive integer such that (i + ∆, j + ∆) lies on P .Now, consider an east-step ending in (i 1 , j 1 ) on P .If there is a laser drawn from (i 1 , j 1 ), then let (i 2 , j 2 ) be the vertex of P where this laser ends.Then there is an edge between j 1 and j 2 − 1 in DYCKtoNCC(P ) (this can be a loop).The remaining vertices in DYCKtoNCC(P ) will be unmarked, that is, unpaired and without a loop.

Proposition 30. The map DYCKtoNCC is a bijection DP(n) → NCC(n).
A proof of Proposition 30 can essentially be found in [Bod19, Prop.6.5].M. Bodnar studies so called n + 1, n-Dyck paths and shows that these are in bijection with NCC(n).It is not hard, however, to see that the set of n + 1, n-Dyck paths is in bijection with DYCK(n) by removing the first north-step.
Note that there is a natural correspondence between Dyck paths of size n and paths of size n−1 that stay weakly above the diagonal y = x−1.Let rot n denote rotation by one step, acting on NCC(n + 1).Furthermore, let γ denote the the action of removing the mark on vertex 1 if it is marked, and marking it if it is unmarked.It does not do anything if 1 is connected to an edge.We refer to this as a flip.
Let the twist action be defined as twist 2n := rot n • γ.It is straightforward to see that twist 2n generates a cyclic group of order 2n acting on NCC(n + 1).Alternatively, we can act by (rot n • γ) n−1 , which closely resembles promotion.Recall that promotion on SYT may be defined as a sequence of swaps, for i = 1, 2, . . ., n − 1, where swap i interchanges the labels i and i + 1 if possible.Theorem 33 (A new cyclic sieving on Catalan objects).The triple NCC(n + 1), twist 2n , 2n n q − q 2 2n n − 2 q exhibits the cyclic sieving phenomenon.Note that Proof.We compute the number of fixed points under twist m 2n , where we may without loss of generality assume m | 2n.There are two cases to consider, m odd and m even.In the first, we must, according to Lemma 17, show that the number of fixed points under twist m 2n is For the first expression we reason as follows.Since m is odd, any fixed point for such m must consist of a diagonal (an edge from i to i + m) and two rotationally symmetrical halves, both consisting of m − 1 vertices.In such a non-crossing configuration, no vertex can be isolated.To see this, note that if vertex j is isolated, then so are j + km (mod n), k ∈ Z, but the j + km (mod n) would need to be both marked and unmarked, a contradiction.Thus, all vertices in the non-crossing configuration are incident to an edge and it is in fact a non-crossing matching.The diagonal can be chosen in m ways and a non-crossing matching on one of the two halves can be chosen in Cat((m − 1)/2) ways, so there are mCat((m − 1)/2) = m (m−1)/2 fixed points.In the second case above, if n = m then at least one vertex has to be isolated since m is odd, which implies there can be no fixed points.For n/m > 2, we use a "parity" argument.Since any isolated vertices among S = {n − m + 2, n − m + 3, . . ., n, 1} change from unmarked to marked and vice versa under twist m 2n , the number of isolated vertices has to be even.Since m is odd, this implies there must be an odd number of edges from S to [n] \ S in a fixed point.However, note that S and the edges out of S completely determine the configuration.Hence the edges must have their other endpoints in the two neighboring intervals of length m.But this violates being rotationally symmetric under rotations of m steps since the number of edges is odd.
In the case 2 | m, according to Lemma 17 we must show that the number of fixed points under twist m 2n is equal to otherwise.
Counting fixpoints for the first case is trivial.For the second expression, note that twist n 2n is simply the action of flipping the markings on all vertices.A fixed point can therefore not have any isolated vertices.What remains are non-crossing matchings, which there are Cat(n/2) many of.
It remains to prove the third expression.Let m = 2d.
Case n and n/d are even.In this case, the only possible invariant configurations are non-crossing matchings that are rotationally symmetrical when rotating 2d steps.Recall from Bijection 5 that such matchings are in bijection with BW(2d, d) and this set clearly has cardinality 2d d .Case n/d is odd.Here we can have fixed points under twist m 2n with unpaired vertices, for examples see Figure 5.The orbit of a vertex j under the operation twist m 2n is {j + dk (mod n)} k∈Z and if 1 ≤ j ≤ d is an unpaired, unmarked (that is, without a loop) vertex, the vertices {j + 2dk (mod n)} 0≤k<n/2d must be unmarked whereas {j + 2dk (mod n)} n/2d<k<n/d will be marked.Note that in the latter case j + 2dk (mod n) ≡ j + d + 2dr for r = k − (n/d + 1)/2.Thus it suffices to understand the vertices from 1 to d.We claim that for every 0 ≤ i ≤ d/2 we get a valid fixed point by choosing i left vertices and i right vertices and matching them in a non-crossing manner as in the previous case.Then we can choose to put a loop at any subset of the remaining d − 2i unpaired vertices.Every fixed point is now constructed exactly once.This gives a total of Finally we need to prove that this sum is equal to 2d d .We will use a bijection to all possible subsets A of size d from two rows with numbers 1 to d, the numbers in the top row being blue and the bottom row red.For a given i we choose i numbers and let both the red and the blue belong to A and then i numbers such that neither blue nor red belong to A. Finally the term 2 d−2i corresponds to choosing any subset of the remaining d − 2i numbers such that the red numbers in that subset belong to A and the blue in the complement are in A. Problem 34.It would be nice to refine the CSP triple in Theorem 33 to the hypothetical Narayana case discussed in Section 3.That is, we want the following equality to hold: where N (1, n, k; 1) is the number of NCC on n vertices with (k − 1) edges.This is a natural consideration, as indicated by Lemma 31.

A refinement of Thiel's result.
Recall that rot n acts on non-crossing (1, 2)configurations of size (n + 1) via a 2π/n-rotation.M. Thiel proved the following.
Proof.Let ξ be a primitive n th root of unity and let d | n.Write e = e 1 (n/d)+e 0 and l = l 1 (n/d) + l 0 for the unique natural numbers e 1 , e 0 , l 1 , l 0 such that 0 ≤ e 0 < n/d and 0 ≤ l 0 < n/d.Using Theorem 9 (the q-Lucas theorem) repeatedly, we get We prove that these evaluations agree with the number of fixed points in NCC(n + 1, e, l) under rot d n on a case-by-case basis.Case d = n: Trivial.Case e 0 = 0 and l 0 = 0: A (1, 2)-configuration that is fixed by rot d n is completely determined by its first d vertices.Among these d vertices, there must 2e(d/n) = 2e 1 vertices that are incident to an edge and l(d/n) = l 1 loops.There are d 2e1 ways to choose 2e 1 from the first d vertices.The number of ways to arrange these edges in an admissible way is equal to the number of perfect matchings that are invariant when rotating 2e 1 steps.By Bijection 5, we know that there are 2e1 e1 such matchings.Lastly, choose l 1 loops among the remaining d − 2e 1 vertices in d−2e1 l1 ways.These choices are all independent and the desired result follows.Case d = n/2, e 0 = 1 and l 0 = 0: Such a (1, 2)-configuration must have a diagonal (an edge from i to i + d) that splits the (1, 2)-configuration into two halves.The diagonal can be chosen in d ways.The (1, 2)-configuration is now determined uniquely by one of its halves.Such a half must have d − 1 vertices with (e − 1)/2 = e 1 edges and l 1 loops.Choose the 2e 1 vertices that are incident to an edge from the d − 1 vertices in d−1 2e1 ways.The number of the ways to arrange these edges in an admissible way is equal to the number of non-crossing perfect matchings on 2e 1 vertices, namely Cat(e 1 ).Finally, choose l 1 loops from the remaining d − e vertices in d−e l1 ways.Since these choices are independent, the number of fixed points is given by d − e l 1 where equality follows from some simple manipulations of binomial coefficients.The remaining cases: Suppose that P ∈ NCC(n + 1, e, l) is invariant under rot d n , where d = n.There are l/(n/d) loops among the first d vertices, so if l 0 = 0, there cannot be such a P .Hence assume that l 0 = 0.If d = n/2 and e 0 = 0, then for each edge ij in P there must be edges (i+d)(j +d), (i+2d)(j +2d), . . ., (i+n−d)(j +n−d) in P (where addition is taken modulo n).Hence the number of edges must be a multiple of n/d which cannot be the case if e 0 = 0.This exhausts all possibilities and thus the proof is complete.
Recall that NCC(n + 1, k) is the set of non-crossing (1, 2)-configurations P on n vertices such that the number of loops plus proper edges of P is equal to k − 1.In other words, By applying Lemma 37, we obtain the following result.

exhibits the cyclic sieving phenomenon.
There is already a known instance of the cyclic sieving phenomenon with the q-Narayana numbers as the polynomial, namely that of non-crossing partitions with a fixed number of blocks and where the group action is rotation [RSW04, Thm 7.2].Note, however, that in Corollary 39 the cyclic group has a different order than the one with non-crossing partitions.
Remark 40.We cannot hope to find a refinement of the above CSP result involving the Kreweras numbers as in [RS18].For example, consider n = 4 and k = 2.There are two partitions of n into k parts, namely (3, 1) and (2, 2).There are 4 non-crossing partitions with parts given by (3, 1) and 2 non-crossing partitions with parts given by (2, 2).But NCC(4, 2) has two orbits under rotation, both of size 3.

Case s = n and type B Catalan numbers
In this section, we prove several instances of the CSP, related to type B Catalan numbers.We first consider a q-Narayana refinement on non-crossing matchings.In the subsequent subsection, we consider a cyclic descent refinement on binary words.Finally, in the last subsection we prove a type B analog of Theorem 38.

Type B Narayana CSP.
A type B non-crossing partition of size n is a noncrossing partition of {1, . . ., n, n + 1, . . ., 2n} which is preserved under a half-turn rotation.These were first defined by Reiner in [Rei97].We let this set be denoted NCP B (n) and let rotB n denote the action on NCP B (n) by rotation of π/n.Note that we only need to make a half-turn before arriving at the initial position.
Proposition 41.The triple Proof.There are many ways to prove this.For example, NCP B (n) can first be put in bijection with type B non-crossing matchings, which are non-crossing matchings on 4n vertices that are symmetric under a half-turn, by using Bijection 7.
We then consider the first 2n new vertices, and for each vertex u, we record a 1 if the edge u → v is oriented clockwise, and 0 otherwise.This is a binary word of length 2n with n ones.Furthermore, rotB n of the non-crossing partition corresponds to shift 2 2n on the binary word.The triple BW(2n, n), shift 2n , 2n n q exhibits the CSP (see [RSW04,Prop. 4.4]), so it is direct from the definition of cyclic sieving that replacing shift 2n by shift 2 2n also gives a CSP triple.We shall now consider Narayana refinements of type B non-crossing partitions and non-crossing matchings.First, we introduce the following polynomial: Note that by using the q-Pascal identity, Π n (q; 1) = j q j 2 n j 2 q = 2n n q , so the sum of the polynomials refines the type B q-Catalan numbers.With the polynomial formulated, cyclic sieving is easy to prove by following the proof of [RSW04, Thm 7.2].As a side note, the coefficients of the polynomial at q = 1 are given by the OEIS entry A088855.
Proposition 42.Let n, k ≥ 0 be integers.Then and exhibit the cyclic sieving phenomenon.Moreover, for every n ≥ 1 and k, 0 and exhibit the cyclic sieving phenomenon.
Proof.Everything is trivial unless 1 ≤ k ≤ n, so we assume this holds.Using Bijection 7 the first two statements are equivalent to the last two, so we only need to prove the former.We now evaluate [t k ]Π n (q; t) at a primitive d th root of unity.The case d = 1 is trivial.If k = 0, 1 (mod d), d ≥ 2, then it is clearly zero.For k = 2dr, we get [t 2dr ]Π n (q; t) = q (dr) 2 +n−dr n dr q n − 1 dr − 1 q which by Theorem 9 (the q-Lucas theorem) becomes , which is what we want.A similar calculation gives the case k = 2dr + 1.The expression q k 2 n k 2 q in (30) is just the sum of the two cases.
A cyclic sieving result involving type B Kreweras numbers (and thus type B Catalan numbers) was proven in [RS18, Thm.1.7].The downside is that the Kreweras numbers in type B are not indexed by usual partitions, but partitions of 2n + 1, where each even part has even multiplicity.
At q = 1, this is A335340 in the OEIS and two times A103371.Note that we have Bw(0, 0; q) = 1 and Bw(n, k; q) = 0 if k > n.
Lemma 43.For all integers 1 ≤ k ≤ n, Proof.The set BW k (n) is in bijection with a certain subset of PATH(n) which we shall now describe.Call binary words of the form b = 0b 2 b 3 • • • b 2n−1 1 elevated and call binary words that are not elevated non-elevated (so a binary word is elevated if cdes(b) = des(b) + 1).Elevated binary words in PATH(n) are in natural bijection with paths in PATH(n − 1) by letting the elevated binary word It follows that a word in BW k (n) corresponds either to a non-elevated path in PATH(n) with k valleys or to an elevated path in PATH(n) with k − 1 valleys.Using this correspondence and (12), one gets that Here, the factors q k and q k−1 appear since by translating a binary word c ∈ BW(2(n − 1), n − 1) with k descents into its corresponding elevated binary word c in BW(2n, n), we have maj(P ) − maj(P ) = k as each descent of c contributes one more to maj than in c.
It remains to show that the expression in (34) coincides with the one in (33).To do this, we rewrite It is therefore sufficient to show that the expression inside the parentheses is equal to 1 + q n or, equivalently, that the following equation holds: This equation can be derived from (24) by adding side of the equation.This concludes the proof.
The number of cyclic descents of a binary word is clearly invariant under cyclic shifts of the word so one has a group action of rot 2n on BW k (n).The following proposition follows from [AS18, Cor.1.6], although they do not compute the closedform expression of Equation (32).
Proposition 44.For all n, k ∈ N such that 1 ≤ k ≤ n, the triple BW k (n), shift 2n , Bw(n, k; q) exhibits the cyclic sieving phenomenon.6.3.Type B non-crossing configurations with a twist.Recall that NCC(n+1) denotes the set of non-crossing (1, 2)-configurations on n vertices.We shall now modify this family slightly.Definition 45.Let NCC B (n) be the set of non-crossing (1, 2)-configurations on n − 1 vertices, with the extra option that one of the proper edges may be marked.We let NCC B (n, e, l) ⊂ NCC B (n) be the subset with exactly e proper edges, and l loops.Finally, let NCC B (n, k) be the subset of NCC B (n) with k edges and loops, i.e.NCC B (n, k) := e+l=k NCC B (n, e, l).
It follows directly from the definition that |NCC B (n, e, l)| = (e + 1)|NCC(n, e, l)| and it is not difficult to sum over all possible e, l to prove that Theorem 46.We let twist 2n act on NCC B (n + 1) as before (the marked edge is also rotated), which gives an action of order 2n.Then is a CSP triple.
Proof.We compute the number of fixed points under (twist 2 2n ) d where we can without loss of generality assume d | n.Write n = md.By Theorem 9, 2n n q = 2d d at a primitive m th root of unity.The claim follows from Theorem 33 except in the cases where a marked edge can appear in a fixed point.Note that in the case 4 | n and 2d = n/2 or 3n/2 a marked edge would have to split the configuration into two non-crossing matchings on an odd number of vertices.Hence there cannot be a marked edge in a fixed point in this case.
The only case left is n | 2d.First, 2d = 2n is trivial.Second, if 2d = n, no fixed point can have marked vertices, as is noted in the proof of Theorem 33.Hence we only have non-crossing matchings on 2d vertices with one edge possibly marked, the number of which is (d + 1)Cat(d) = 2d d .It should be possible to prove Theorem 46 bijectively.Problem 47. Find an equivariant bijection between NCC B (n + 1) and BW(2n, n) sending twist 2 2n to shift 2 2n .Note that the triple in Theorem 46 exhibits the so-called Lyndon-like cyclic sieving [ALP19], which is not intuitively clear (as it is for BW(2n, n)).
Remark 48.Theorem 46 does not hold when only considering twist 2n .For n = 2, 2n n q evaluated at a primitive 4 th root of unity gives 0. However, there are 6 elements in NCC B (3), two of which are fixed under twist 4 ; consider an edge between vertices 1 and 2, which may or may not be unmarked.Since there are no loops or isolated vertices, these two elements are fixed.Can one modify the q-analog of 2n n so that it is compatible with twist 2n ?Problem 49.Is it possible to define a refinement P (n, e, l; q) of 2n n q so that  Unfortunately the polynomials Ncc B (n, e, l; q) := q e 2 +nl [e + 1] q n 2e q Cat(e; q) n − 2e l q do not serve this purpose even though they do satisfy the identities (proof omitted) Ncc B (n, e, l; q).
Proof.We can split NCC B (n + 1, e, l) into two sets, the first set A being the case without a marked edge, and the second set B the case with a marked edge.Then it suffices to prove that (A, rot n , Ncc(n, e, l; q)) and (B, rot n , [e] q Ncc(n, e, l; q)) are CSP triples.The first one is already proved in Theorem 38.
For the second, consider rot d n , and without loss of generality write n = kd.A single marked edge can only appear in a fixed point if d = n or d = n/2.The former is trivial.Now, rewrite the polynomial as .
By summing over the cases when e + l = k, we get the following corollary: Corollary 51.We have a q-analog of the type B Narayana numbers, which admits the CSP triple where U n,k (q) = k e=0 q e(e+1)+(n+1)(k−e) (1 + [e] q ) n 2e q Cat(e; q) n−2e k−e q .
Proof.As in the discussion before Theorem 46 it is not difficult to prove that when q = 1, we do indeed obtain n k 2 , so this is a q-Narayana refinement.Now, summing over all k gives cyclic sieving on NCC B (n + 1) under rotation.We leave it as an open problem to find a nice expression for k U n,k (q).

Two-column semistandard Young tableaux
The Schur polynomial s λ (x 1 , . . ., x n ) is defined as the sum where SSYT(λ, n) is the set of semi-standard Young tableaux of shape λ with maximal entry at most n.The product is taken over all labels in T .
P. Brändén gave the following interpretation of q-Narayana numbers.
Proof.We first note that s 2 k 1 k /1 k = (s 1 k ) 2 .To compute s 1 k , we simply sum over all k-subsets of [n].This gives immediately that s 1 k (q, q 2 , . . ., q n ) = q k(k+1)/2 n k q , and the theorem above follows.
It is then reasonable to interpret s 2 k 1 s /1 s (q, q 2 , . . ., q n ) (39) for 0 ≤ s ≤ k as an interpolation between type A (s = 0) and type B (s = k) q-Narayana polynomials.Note that this approach is different from what is sought after in Section 3. The expression in (39) can easily be computed by the dual Jacobi-Trudi identity, see [Mac95].We find that (39) is equal to The first part of the theorem below follows from combining [Rho10, Thm.1.4] and Theorem 52.
Proof.We can define the action ϕ as follows.Given T ∈ SSYT(2 k , n − 1), define This can be seen to be an action with the desired properties by referring to Bijection 3 from SSYT(2 k , n − 1) to NCP(n, k + 1).Then rotation one step of the latter set corresponds to ϕ where P are the other elements in the same block as n, b is the smallest element in the block of n − 1 (if n − 1 / ∈ P ), and is the largest element in the block of n other than n itself.Clearly, b + 1 must be removed from the first column since it will be the smallest element in the block of n in ϕ(T ), and + 1 must be added to the second since it will be the largest element in the block containing 1 in ϕ(T ).
Find the largest y ∈ T * ,1 , y ≤ x i , which is not in and set y i = y.Let p i = {z ∈ N : y i ≤ z ≤ x i , z ∈ P i−1 }.Repeat for i < k + 1.Finally, let p k+1 = [n] \ P k .Then the blocks p 1 , . . ., p k+1 form a non-crossing partition in NCP(n, k + 1) by construction.Note that exactly one element from each of T * ,1 and T * ,2 is contained in p i .Note also that this is in fact the unique non-crossing partition in NCP(n, k + 1) having parts whose smallest and largest elements are y i and x i respectively.
The inverse of the above bijection can be described as follows.Let p 1 |p 2 | • • • |p k+1 ∈ NCP(n, k + 1) and assume n ∈ p k+1 .Let the first column of T consist of the smallest elements from p i , 1 ≤ i ≤ k, in increasing order from top to bottom, and the second column of T of the largest elements from the same blocks, also in increasing order.To see why T ∈ SSYT(2 k , n − 1), it suffices to suppose that we on some row i have T i,1 > T i,2 , so the smallest element in the block containing T i,2 must be T j,1 for some j < i.Then the smallest element in the block containing T j,2 has to be T k,1 for some j < k < i, the smallest element in the block containing T j+1,2 some T k ,1 for j < k < i, and so on.We match all elements T j,2 , . . ., T i−1,2 to elements in the first column between T j,1 and T i,1 , but this yields a contradiction.Note that T is the unique element of SSYT(2 k , n − 1) whose first column consists of the smallest elements of p 1 , . . ., p k and whose second column consists of the largest elements of these parts.Hence it is clear that this indeed is the inverse.
are the sizes of the blocks with T 1,2 , T 2,2 , . . ., T k,2 as the largest elements.
There is a type B version of Theorem 54.
Proof.We describe a bijection from SSYT(2 exhibits the cyclic sieving phenomenon.It is not hard to see that the two different CSP-polynomials agree at n th roots of unity.This completes the proof. It is natural to ask if the first part of Theorem 54 and Theorem 56 generalize to the intermediate skew shapes.We would then hope that k-promotion acting on SSYT(2 k 1 s /1 s , n), for 1 ≤ s ≤ k − 1, has order n.However, this is not the case, as for n = 4, k = 2, s = 1, the tableaux Perhaps some other group action gives a CSP triple with (39) as the CSPpolynomial.In a recent preprint, Y.-T.Oh and E. Park [OP20] the authors show some closely related results, regarding cyclic sieving on SSYT.

Triangulations of n-gons with k-ears
We shall now consider type A triangulations of an n-gon.The main result in this section is a refinement of the CSP instance on triangulations of n-gons which are counted by Cat(n − 2), see [RSW04,Thm. 7.1].We also mention a cyclic sieving result on type B triangulations.
We now introduce the following q-analog of the expression in (40).For integers n and k satisfying 2 At a first glance, one might hope that there is an easier expression for Tri(n, k; q).However, note that Tri(n, k; q) is not palindromic in general.As an example, consider Tri(6, 2; q) = 1 + q 4 + q 5 + q 8 .This means that one cannot hope to find a formula for Tri(n, k; q) which only involves products of palindromic polynomials.In particular, it cannot be a product of q-binomial coefficients.
The choice of Tri(n, k; q) is motivated by the following theorem.
But this follows from substituting a = q 2 q n and c = q 3 in the following claim and then expanding the q-hypergeometric series, together with using the fact that (q 5 ; q) 2n = (q 5 ; q) n (q n q 5 ; q) n .Claim: For non-negative integers n, we have the identity Applying the first q-Chu-Vandermonde identity, the left-hand side becomes Now, using the identity (aq/q k ; q) k = q −( k 2 ) (−a) k (1/a; q) k we see that the left-hand side of ( 46) is equal to This is now a special case of the second q-Chu-Vandermonde identity and we are done.
We let rot n act on TRI(n) by rotating a triangulation one step clockwise.As rot n also preserves the set TRI ear (n, k), we have a group action of rot n on TRI ear (n, k).
exhibits the cyclic sieving phenomenon.
The polynomial [n] q Tri(n + 1, k; q) does not seem to give a refinement of Proposition 61.

Marked non-crossing matchings
A marked non-crossing matching is a non-crossing matching where some of the regions have been marked.Let NCM (r) (n) denote the set of marked non-crossing matchings with exactly r marked regions.Since every non-crossing matching in NCM(n) has n + 1 regions, it follows that |NCM (r) (n)| = n+1 r |NCM(n)|.In particular, for r = 1 we can think of our objects as non-crossing matchings of vertices on the outer boundary of an annulus rather than on a disk.This model is reminiscent of the non-crossing permutations considered in [Kim13], where points on the boundary of an annulus are matched in a non-crossing fashion, but with some other technicalities imposed.
The following generalizes Proposition 20.Here we recall several bijections on Catalan objects which have appeared earlier in the literature.We have tried to find the earliest reference for each.B.1.NCM and binary words.Suppose xy is an edge in a non-crossing perfect matching, with x < y.We say that x is the starting vertex and y is the end vertex.Further, denote NCM sh (n, k) the subset of all N ∈ NCM(n) such that short(N ) = k.
We now describe a well-known bijection NCMtoDYCK from NCM(n) to DYCK(n).Let d be a natural number such that d | n.If a matching M ∈ NCM(n) has d-fold rotational symmetry, it is sufficient to understand how the vertices 1, 2, . . ., d are matched up.Here, we restrict ourselves to the case when M does not have a diagonal-for the case when M has a diagonal, see the third case in the proof of Theorem 27.In this case, there is a bijection BWtoNCM between BW(2d, d) and rotationally symmetric non-crossing perfect matchings.Bijection 5. Let c = c 1 c 2 • • • c 2d ∈ BW(2d, d).We show how to construct the corresponding BWtoNCM(c) ∈ NCM(n).Think of the vertices 1, 2, . . ., 2d being arranged on a line.For all vertices i = 1, 2, . . .2d, we let vertex i be a "left" vertex if c i = 0 and a "right" vertex if c i = 1.Then we match every left vertex with the closest available right vertex to its right without creating a crossing of edges.There is a unique way of doing this so that no left vertex remains unpaired between a matched pair of vertices.However, there might be some left vertices which are not matched because there are not sufficiently many right vertices to their right.There must also be equally many unpaired right vertices because there are not enough left vertices to their left.Since this will be the same in every interval of vertices [2dk + 1, 2d(k + 1)] there is a unique way to pair the remaining left vertices with the remaining right vertices of the interval to the right and vice versa.
We can prove a stronger statement about rotationally symmetric NCMs.We study BWtoNCM restricted to NCM sh (n, k).Once again, such a matching is completely determined by how the vertices 1, 2, . . ., 2d are paired up.Among these 2d vertices, there are two possible cases to consider.
Case 1: Exactly dk/n short edges where either vertex 1 is a left vertex or vertex 2d is a right vertex.
Case 2: Exactly dk/n − 1 short edges where vertex 1 is a right vertex and vertex 2d is a left vertex.
If one applies BWtoNCM −1 (that is, left vertex corresponds to 0 and right vertex corresponds to 1) to the non-crossing perfect matchings in the two above cases, one gets the image BW dk/n (d).
B.2. NCM and SYT.We recall the definition of this bijection.Bijection 6.Let T ∈ SYT(n 2 ) and construct SYTtoNCM(T ) as follows.For i ∈ {1, 2, . . ., 2n}, let vertex i be a starting vertex if i is in the first row and an end vertex otherwise.It is not hard to show that determining the starting and ending vertices uniquely determines a non-crossing matching.

B.3. NCP and NCM.
There is a bijection between non-crossing partitions and non-crossing matchings, NCPtoNCM : NCP(n) → NCM(n), that directly restricts to a bijection between NCP B and NCM B .This bijection has another nice property; it is equivariant with regards to the Kreweras complement on the non-crossing perfect matchings and rotation on the non-crossing perfect matchings [Hei07, Thm.1].
Bijection 7. Consider π ∈ NCP(n), and for every vertex j ∈ {1, . . ., n}, insert a new vertex (2j − 1) immediately after j, and (2j − 2) immediately before j, where we insert (2n) immediately before 1.There we match (2j) to the closest point (2k − 1) , j < k, such that the edge between those vertices does not cross any of the blocks in π.If no such k exists, we match to the smallest number k, 0 < k ≤ j.Since j = k is always possible the map is well-defined.This gives a perfect matching σ on 1 , 2 , . . ., (2n) which is non-crossing, and in addition, does not cross any of the blocks of π.We can get the inverse of the map NCPtoNCM by putting back the vertices 1, 2, . . ., n and letting all vertices that can have an edge between them without crossing any edge of the perfect matching σ belong to the same block.The following is an example of this bijection.One can also illustrate the bijection as follows.For each singleton block, add an edge between two copies of the vertex.For each block of size two, split the edge into two non-crossing edges and hence each vertex into two copies.Finally, for each block with m ≥ 3 elements, push apart the edges at the vertices so that we have m non-crossing edges on 2m vertices.See Figure 7.
Note that, by definition, every block except the one whose minimum element is 1 corresponds to an even edge under the bijection.If j > 1 is the smallest element of a block and k is the largest, then (2j − 2) is matched to (2k − 1) .The remaining even vertices are matched to smaller odd vertices.Hence Bijection 7 in fact gives a bijection between the sets NCP(n, ) and NCM(n, − 1).B.4. NCP and Dyck paths.We briefly describe a bijection between non-crossing partitions and Dyck paths, NCPtoDYCK, with the property that the number of parts is sent to the number of peaks.This bijection is often attributed to P. Biane [Bia97].
Bijection 8. Let π = B 1 |B 2 | • • • |B k be a non-crossing partition of size n, where the blocks are ordered increasingly according to maximal member.By convention, we let B 0 be a block of size 0, where the maximal member is also 0. We construct a Dyck path D ∈ DYCK(n) from π as follows.For each j = 1, . . ., k, we have a sequence of max(B j ) − max(B j−1 ) north steps, immediately followed by |B j | east steps.

•
DYCK(n): the set of Dyck paths of size n, that is, the subset of paths in PATH(n) which never touch the line y = x − 1, • SYT(n 2 ): the set of standard Young tableaux with two rows of length n, • NCP(n): the set of non-crossing partitions on n vertices, • NCM(n): the set of non-crossing matchings on 2n vertices, • TRI(n): the set of triangulations of an (n + 2)-gon, • NCC(n): the set of non-crossing (1, 2)-configurations on n − 1 vertices.Examples of such objects are listed in Appendix A.
Suppose next that o(ξ d ) | n and k 0 = 0.Such a matching is completely determined by how the vertices 1, 2, . . ., d are paired up.It follows that the number of short edges must be a multiple of 2n/d, contradicting k 0 = 0. Suppose lastly that o(ξ d ) | 2n but o(ξ d ) n.To analyze this case, we first prove the following.Claim: If o(ξ d ) | 2n and o(ξ d ) n, then d is odd.Proof of Claim.The hypothesis implies that the number of 2's in the prime factorization of 2n is equal to the number of 2's in the prime factorization of o(ξ d ).Hence, the number 2n/o(ξ d ) is odd.Combining this with the fact that o(ξ d ) = 2n/ gcd(2n, d) yields that gcd(2n, d) is odd.But this implies that d is odd, so we are done.

Figure 3 .
Figure 3.An example of the bijection DYCKtoNCC in Bijection 2.

Figure 4 .
Figure 4.The result of rot n • γ on an element in NCC(13).
6.2.A second refinement of the type B Catalan numbers.Let BW(2n, n) be the set of binary words of length 2n with exactly n ones.Define a cyclic descent of a binary word b = b 1 b 2 • • • b 2n as an index i such that b i > b i+1 , where the indices are taken modulo 2n.The number of cyclic descents of b is denoted cdes(b).As an example, if b = 0110010111, then cdes(b) = 3.For any two natural numbers n and k, let BW k

and apply Theorem 9 ..
At a primitive k th root of unity, this evaluates to 0 unless k | 2e, k | e − 1 and k | l.The second implies gcd(k, e) = 1, so by the first k| 2. If k = 2, that is d = n/2,we get that the number of fixed points should be This is indeed the case.The marked edge has to split the configuration into two symmetric parts, and connects i to i + n/2 for some 1 ≤ i ≤ n/2.The symmetric configurations are on n/2 -1 vertices, and have (e − 1)/2 edges and l/2 marked vertices each.The number of fixed points is hence and let P be the subset of Q containing numbers not occurring as entries in T .Further, define = max P and b = max Q.If P = ∅, let = 0. Now ϕ(T ) is defined as follows.If = n − 1, then we just add one to every entry in T and are done.If = n − 1, then we add one to every entry in T and• in the first column: remove b + 1, add 1 in increasing order; • in the second column: remove n, add + 1 in increasing order.

Remark 55 .
Below is an example of Bijection 3 and ϕ, when n = 8, k = 4. {{1, 4}, {2}, {3}, {7}, {5, 6, 8}}, ϕ(T ) The bijection in the proof of [Brä04, Thm.5] together with Bijection 8 provides a different bijection between SSYT(2 k , n − 1) and NCP(n, k + 1) where, if The corresponding pair of binary words (b 1 , b 2 ) is constructed as follows.Write b 1 = b 11 . . .b 1n and let b 1i = 1 if T has an i in the left column, and otherwise, let b 1i = 0.In an analogous way, let b 2 be determined by the entries in the right column of T .It is easy to see that this bijection is equivariant with the corresponding group action being rot n on the pair of binary words.Here, rot n acts by cyclically shifting both of the words one step.It follows from [RSW04, Thm.1.1] that BW(n, k) × BW(n, k), rot n , n k q 2 under k-promotion, but we want a group action of order 4.

8. 1 .
Refined CSP on triangulations by considering ears.Let TRI(n) denote the set of triangulations of a regular n-gon.If the vertices i, i + 1, i + 2 (mod n) are pairwise adjacent for T ∈ TRI(n), we say they form an ear of T .Let TRI ear (n, k) denote the set of triangulations of a regular n-gon with exactly k ears, and let Tri(n, k) be the cardinality of this set.It was shown by F. Hurtado and M. Noy [HN96, Thm.1] that Tri

8. 2 .
Triangulations of type B. Let us define type B triangulations TRI B (n), as the set of elements in TRI(2n) which are invariant under rotation by a half-turn.In such a triangulation, there is always an edge through the center.There are n choices of such an edge, and then we need to choose a triangulation on one half of the 2n-gon.This gives n • type B triangulations.The following result is straightforward to prove but also follows from [EF08, Thm.4.1].Proposition 61 (See [EF08, Thm.4.1].).The triple

Figure 6 .
Figure 6.Partitioning a non-crossing perfect matching of size 2n = 18 with d-fold rotational symmetry, d = 3, into segments of length 2d.Each of the three edges from [2d] to vertices with bigger labels has a unique region to its left.

Bijection 4 .
Take M ∈ NCM(n) and construct the Dyck pathNCMtoDYCK(M ) = b 1 b 2 • • • b 2n as follows.For vertices i = 1, 2, . . ., 2n in M , let b i = 0 if i is a starting vertex and let b i = 1 if i is an end vertex.It is not hard to see that this procedure ensures that the resulting binary word is a Dyck path.

Figure 7 .
Figure7.A non-crossing partition given by the dashed edges and the corresponding non-crossing matching given by the solid edges.
Table 1 and Table 2 list the state-of-the-art of the CSP on Catalan-type objects of type A and B respectively, including the results proven in the present paper.Examples of such objects can be found in Appendix A. We use several bijections (described in Appendix B) between Catalan and Narayana objects, see Figure 1.

Table 2 .
The state-of-the-art regarding cyclic sieving on type B Catalan objects, including the new results presented in this article.
1 is also a Dyck path.A Dyck path which is not elevated is called non-elevated.Elevated Dyck paths of size n are in natural bijection with Dyck paths of size n − 1.The next lemma now easily follows.Let T ∈ SYT cdes (n 2 , k).Then 2n ∈ cDes(T ) if and only if the Dyck path SYTtoDYCK(T ) is elevated.It follows from Lemma 23 that the restriction of SYTtoDYCK to SYT cdes (n 2 , k) is a bijection to the set of D ∈ DYCK(n) such that D either is non-elevated and has k peaks or is elevated and has k − 1 peaks.Hence, we get (using an argument by T. Došlić [Doš10, Prop.2.1]),