REFINED CYCLIC SIEVING ON WORDS AND THRALL S PROBLEM (DRAFT)

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1 REFINED CYCLIC SIEVING ON WORDS AND THRALL S PROBLEM (DRAFT CONNOR AHLBACH AND JOSH SWANSON Abstract. Thrall [Th] famously considered the computation of certain Schur characters L λ of GL(V -modules L λ coming from the study of free Lie algebras. Kraskiewicz-Weyman [KW] computed the Schur expansion of L (n using major indices of standard Young tableaux. On the other hand, numerous authors have related the L λ to certain group actions involving cyclic groups. Later, Reiner-Stanton-White [RSW] defined the cyclic sieving phenomenon associated to a cyclic group action. Motivated by Kraskiewicz-Weyman s result, we formulate and prove a refined cyclic sieving result for the natural cyclic action on words using the major index generating function and cyclic descents. As an application, this yields a new, nearly bijective proof of Kraskiewicz- Weyman s characterization, as well as Schocker s generalization for higher Lie characters. 1. Introduction We begin by summarizing our main results. Most definitions are deferred to Section 2. Let W n denote the set of words in the alphabet P = {1, 2, 3,...} of length n and let C n := σ denote the cyclic group of order n, which acts on W n via (1 σ w 1 w n 1 w n := w n w 1 w n 1. Let W (α, δ denote the subset of W n consisting of words with content α and cyclic descent type δ. Finally, let W (α, δ maj (q denote the corresponding major index generating function. The Cyclic Sieving Phenomenon was introduced by Reiner, Stanton, and White [RSW]. We show: Theorem 1. The triple (W (α, δ, C n, W (α, δ maj exhibits the cyclic sieving phenomenon. One may deduce the less refined version of Theorem 1 with W (α, δ replaced by δ W (α, δ from Theorem 1.1 in [RSW]; see Theorem 33. We recover Kraskiewicz-Weyman s result from Theorem 1 in Theorem 33, as well as Schocker s generalization in 36. Each step of our argument is bijective, except for the application of Theorem 1. A bijective proof of the following Date: May 19,

2 2 CONNOR AHLBACH AND JOSH SWANSON reformulation of Theorem 1 would make the argument in Theorem 33 fully bijective: the modular major index and the frequency-lex statistics on W (α, δ are equidistributed. Our proof of Theorem 1 is combinatorial and involves an analysis of the sequences of runs and falls of elements in W (α, δ using a version of a standard insertion lemma (Lemma 7. We organize the relevant combinatorial data into certain trees T (α, δ whose edges are labeled by sets and multisets. A key step is the following product formula: Theorem 2. Suppose α = n with α 1 0. Set g := gcd(α 1,... α m, δ 1,... δ m, β x = α α x, and k = k k x. Then W (α, δ maj (q n m ( ( βx 1 k x 1 kx + (α x δ x 1 (mod q g 1 α 1 δ x α x δ x The reduction mod q g 1 is essential. The remainder of the proof of Theorem 1 analyzes cyclic sieving on sets and multisets and essentially builds up Theorem 1 from Theorem 2. In Section 7 we discuss the application of Theorem 1 to Thrall s problem. The paper is organized as follows. In Section 2 we give combinatorial background on cyclic sieving and words. Section 3 summarizes the relevant representation theory and Lie representations for use in Section 7. Section 4 analyzes T (α, δ and proves Theorem 2, while Section 5 deduces Theorem 1. Section 6 discusses a naturally arising instance of cyclic sieving in terms of a method inspired by Wagon-Wilf [WW]. Finally, Section 7 applies the preceding results to Thrall s problem. Section 8 discusses generalizations to higher Lie characters. Section 9 discusses another generalization of Section 7 to arbitrary cyclic sugroups of S n. Section 10 concludes with a few remarks on symmetric and quasisymmetric functions and to the represenation of S n acting on its own conjugacy classes by conjugation. q q 2. Combinatorial Background In this section, we briefly recall or introduce combinatorial notions on words and fix our notation. We use the alphabet of positive integers P throughout. A word w of length n =: w is a sequence w = w 1 w 2... w n of letters w i P. The descent set of w is Des(w := {1 i < n : w i > w i+1 } and the number of descents is des(w := # Des(w. The major index of w is maj(w := i Des(w i. The cyclic descent set of w is CDes(w := {1 i n : w i > w i+1 }, where now the subscripts are taken mod n, and we write cdes(w := # CDes(w. Any position that is not a cyclic descent is a cyclic weak ascent, though we often drop the word weak. The cyclic major

3 REFINED CYCLIC SIEVING ON WORDS AND THRALL S PROBLEM (DRAFT 3 index is cmaj(w := i CDes(w i. Often the distinction between cyclic major index and major index will be irrelevant as cmaj(w n maj(w. We use periods to indicate cyclic descents throughout the paper. For example, if w = , then Des(w = {3, 4, 7}, des(w = 3, CDes(w = {3, 4, 7, 8}, cdes(w = 4, maj(w = 14, and cmaj(w = 22. A necklace is an orbit of the set W n of words of length n under the action of the cyclic group C n by rotation as in (1. The product of two words is their concatenation. A word is primitive if it is not a power of a smaller word. Any word w may be written uniquely as w = v f for f 1 maximal, in which case v is primitive. We call v the period of w, written period(w, and f the frequency of w, written freq(w. The period of a word is the size of its necklace, and freq(w period(w = w. Primitivity, period, frequency, and cdes are all well-defined (constant on necklaces. Example 3. The necklace of w = = ( is [w] := { , , , }, which is not primitive, has period 4, frequency 2. Since the cyclic action on words moves each cyclic descent forward one place modulo n, cmaj forms an arithmetic sequence mod n on [w] with differences k := cdes(w. Hence, cmaj of a necklace is well-defined (constant modulo d := gcd(n, k. The content of a word w, denoted cont(w, is the sequence α whose j-th part is the number of j s in w. Let w (i denote the subsequence of w with all elements larger than i removed. For w W n, cont(w is a weak composition of n. Thus, if m := max(w, we have a filtration 1 α 1 = w (1 w (2 w (m 1 w (m = w, where u v means that u is a subsequence of v. We think of this filtration as building up w by recursively adding all of the copies of the next largest letter where they fit. The cyclic descent type of a word w is the sequence which tracks the number of new cyclic descents at each stage of the filtration. Formally, CDT(w := (cdes w (1, cdes w (2 cdes w (1,..., cdes w (m cdes w (m 1. Note CDT is well-defined (constant on necklaces since rotating w rotates each w (j. We write W (α := {w W n : cont(w = α}, W (α, δ := {w W n : cont(w = α, CDT(w = δ}. We also let N n, N(α, N(α, δ denote the set of necklaces (orbits that appear in W n, W (α, W (α, δ, respectively. Note that we could define W (α, δ more symmetrically by replacing cont with an analogous cyclic weak ascent type

4 4 CONNOR AHLBACH AND JOSH SWANSON which would be the point-wise difference of cont and CDT. However, cont is ubiquitous, so we use it instead. Example 4. Suppose w = We find w (1 = 1111 cdes w (1 = 0, w (2 = cdes w (2 = 2, w (3 = cdes w (3 = 3, w (4 = cdes w (4 = 5. Hence cont(w = (4, 3, 2, 3 and CDT(w = (0, 2 0, 3 2, 5 3 = (0, 2, 1, 2. Given a necklace, consider ordering its elements lexicographically, and assign indexes to these elements starting from 0. The lex statistic lex(w of a word w is the index so assigned to w when lexicographically ordering the necklace of w. For instance, the necklace in Example 3 has lex statistics 0, 3, 2, 1, respectively, so that lex( = 3. The flex statistic of a word w is the product flex(w := freq(w lex(w, so flex( = 2 3 = 6. We briefly recall the cyclic sieving phenomenon (CSP. For more, see [RSW] or the excellent survey in [SaCSP]. Suppose the cyclic group C n of order n acts on a set X, and f(q N[q] is a polynomial in q. In this paper, f(q will be some statistic generating function on X. The triple (X, C n, f(q is said to exhibit the cyclic sieving phenomenon (CSP if for any τ C n, letting X τ := {x X : τ x = x}, #X τ = f(ω p, where p is the order of τ in C n, and ω p is a primitive p-th root of unity. Given a function stat: X Z on a set X, the corresponding generating function is written as X stat (q := x X q stat(x N[q, q 1 ]. We sometimes use the natural multivariable analogue of this notation. For instance, viewing cont: W n Z n, flex: W n Z, we have (cont wn Wn cont,flex (x 1,..., x n, q := x (cont w 1 1 x n q flex w. x X Finally, we assume familiarity with the elementary combinatorics and notation associated to representations of the symmetric group, such as partitions, compositions, (semistandard Young tableaux, descent sets of tableaux, the RSK algorithm, and Schur functions. For details, see [Fu] and/or [Sa].

5 REFINED CYCLIC SIEVING ON WORDS AND THRALL S PROBLEM (DRAFT 5 3. Representation Theory Background We next recall key aspects of Thrall s problem. We assume some familiarity with the representation theory of symmetric and general linear groups over C, though we recall some key points. See [Fu] for more details. For simplicity, all representations are over C, so G-modules are CG-modules. The complex irreducible inequivalent representations of S n are the Specht modules S λ for λ n. The Frobenius character ch is the additive map which sends S λ to the Schur function s λ (x 1, x 2,... := T SSYT(λ xt. Given a subgroup H of S n and a (left H-module M, the induced S n -representation is M Sn H := CS n CH M = Hom CH (CS n, M. Let V be a complex vector space of dimension m. Endow V n with the natural left GL(V -action by linear substitutions and the natural right S n -action by permutation of indexes. Given an S n -module N, define a corresponding GL(V -module E(N := V n CSn N, which we call the Schur module of N. The irreducible inequivalent polynomial representations of GL(V are precisely the Schur modules E λ := E(S λ associated to Specht modules S λ where λ has at most dim(v parts. Pick a finite-dimensional, polynomial representation E of GL(V and a basis, v 1,..., v m for V, and consider the action of a diagonal matrix diag(x 1,..., x m GL(V as a linear endomorphism of E. The Schur character of E, ch(e, is the function which sends (x 1,..., x m to the trace of diag(x 1,..., x m. Indeed, ch(e N[x 1,..., x m ] and ch E(S λ = s λ (x 1,..., x m, 0, 0,.... Given an S n -module N, we then have in the m limit equality of Schur and Frobenius characters, ch E(N = ch N. In light of this, we often leave dependence on m or V implicit. The Littlewood- Richardson rule gives a combinatorial description for the coefficients c λ µν N in ch(e µ E ν = λ c λ µνe λ. We turn to Thrall s problem. As motivation, the Schur module associated to the regular representation is simply V n. The corresponding Schur character is then (2 ch V n = ch E(CS n = s λ = f λ s λ λ n T SYT λ λ n

6 6 CONNOR AHLBACH AND JOSH SWANSON where f λ := # SYT(λ. Our version of Thrall s problem is to find the Schur expansion of certain GL(V -modules arising from the study of free Lie algebras, which we next describe. The tensor algebra of V is T (V := n=0 V n, which is naturally a graded GL(V -representation. Let L(V denote the free Lie algebra on V over C, which is the Lie subalgebra of T (V generated by V. The universal enveloping algebra U(L(V is T (V itself, which is a standard consequence of the PBW theorem. Consider choosing an ordered basis for L(V by successively picking ordered bases for the homogeneous components L n (V := L(V V n. By the PBW theorem, weakly increasing monomials in this basis yield a basis for U(L(V = T (V. It is then straightforward to see that we have a degree-preserving vector space isomorphism (3 U(L(V = Sym m 1 (L 1 (V Sym m 2 (L 2 (V, λ=1 m 1 2 m 2 where the sum is over all partitions of all non-negative integers and Sym m (E denotes the mth symmetric power of a GL(V -representation E. A more careful argument involving Hall bases yields an isomorphism as in (3, except of GL(V -modules. For details, see [Re] (Lemma (Indeed, a comparison of characters gives an abstract isomorphism of GL(V -modules directly. Thus, given λ = 1 m 1 2 m2, we define the higher Lie module To summarize, (4 L λ (V := Sym m 1 (L 1 (V Sym m 2 (L 2 (V. T (V = λ L λ (V = λ ( i L (i m i where L (a b = Sym b (L a (V, L (n (V = L n (V = V n L(V T (V as GL(V -modules. Since our aim is the consideration of the coefficients of s µ in ch L λ, it follows from (4 and the Littlewood-Richardson rule that we may restrict attention to rectangular partitions λ = (a b. The one-row case of this problem was solved by Kraskiewicz-Weyman: Theorem 5 ([KW]. We have ch L (n = λ n a λ,1 s λ. where a λ,1 = #{T SYT(λ : maj(t n 1}. Our primary motivation is giving a proof of Theorem 5 which is as bijective as possible; see Section 7. We next summarize the representation theory of generalized permutation groups. For a finite group G, the group G S n is the semidirect product G n S n where S n acts on G n by permuting the entries, as in σ (g 1,..., g n = (g σ 1 (1,... g σ 1 (n.

7 REFINED CYCLIC SIEVING ON WORDS AND THRALL S PROBLEM (DRAFT 7 More specifically, we multiply two elements in G S n as follows: (g 1,..., g n, σ (h 1,..., h n, τ = (g 1 h σ 1 (1,..., g n h σ 1 (n, στ for all g 1, h 1,... g n, h n G and σ, τ S n. In particular, C a S b is more concretly the subgroup of S ab generated by the cycles c i for i = 1,..., b, and ˆσ for σ S b, where c i = ( ((i 1a + 1 ((i 1a (ia ˆσ( (i 1a + j = (σ(i 1a + j for all i = 1,..., b, and j = 0,... (a 1. That is, c i cycles the interval Ia i := [(i 1a + 1, ia], and ˆσ permutes these b intervals Ia i according to σ. The abelianization of C a S b is C a (S b /A b. For example, at b = 2 we find that #(C a S 2 /#(C a C 2 = 2!a2 2a = a, so 1/a-th of the dimensions of the regular representation of C a S 2 are accounted for by one-dimensional representations. 4. Major Index, Runs, and Falls Throughout this section, w W n, and k = cdes(w. Our definitions may differ slightly from other literature because we are interested in cyclic phenomena. A run of a word w is a maximal sequence of consecutive, weakly increasing entries, allowing wrapping. That is, w i w i+1... w j is a run of w if and only if w i 1 > w i w i+1 w j > w j+1, where the indices are taken modulo n. Similarly, a fall of a word w is a maximal sequence of consecutive, strictly decreasing entries, allowing wrapping. Observe that the runs are separated by the cyclic descents, and the falls are separated by cyclic weak ascents. Example 6. If w = = , then the runs of w are 1126, 5, 346, and the falls are 2, 653, 4, 61,1. As each position is either a cyclic descent or ascent, w has k runs and n k falls. Note the constant word w = a n has n has 0 runs and n falls. (This is consistent with definition since no weakly increasing sequence is maximal as we can wrap around forever. Our next task is to characterize what happens to maj when we insert another element to w. Suppose v is obtained from w by inserting an x in some position. Then, cdes(v = cdes(w, or cdes(v = cdes(w + 1. If cdes(v = cdes(w, we say x is added to a NDA (non-descent-adding position. In this case, the number of runs is unchanged, so x must fit into one of the runs of w - that is, x increases the length of one of the runs

8 8 CONNOR AHLBACH AND JOSH SWANSON of w by 1. Else, cdes(v = cdes(w + 1, and we say x is added to a DA (descent-adding position. In this case, the number of falls is unchanged because v cdes(v = w + 1 (cdes(w + 1 = w cdes(w. Thus, x must fit into one of the falls of w - that is, x increases the length of one of the falls of w by 1. See the example after the proof of Lemma 7. We label the runs of w from left to right with 0 to (k 1 and the falls from 0 to (n k 1 starting from whichever run, fall involves the first letter of the word. We describe a few variations on how maj changes when we add a letter to a word. Lemma 7. Let v be the word w with an x inserted in any position, and let k = cdes(w. Then, x fits into some run or some fall of w, and 0 if x fits into 0-th run of w at end, k i if x fits into i-th run of w, cmaj(v cmaj(w = n + 1 if x fits into 0-th fall of w at end. k j if x fits into j-th fall of w Proof. (a If x fits into the 0-th run of w at the end, then the cyclic descent positions remain the same, so cmaj(w = cmaj(v. (b If x fits into the i-th run of w (at the start if i = 0, then x pushes the k i cyclic descents after this run forward 1 index, so cmaj(v cmaj(w = k i. (c Finally, if x fits into the 0-th fall of w at the end, then we add a cyclic descent at the end, with the rest of the cyclic descent positions unchanged. So, cmaj(v cmaj(w = n + 1. (d If x fits into the j-th fall of w, (at the start if j = 0, x pushes the n k j cyclic ascents after this fall forward 1 index. Consider the comaj, given by comaj(w := i [n]\cdes(w (i = ( n+1 2 cmaj(w. Thus, comaj(v comaj(w = n k j, which we can rewrite as cmaj(v cmaj(w = k j. Example 8. Consider w = , which has cmaj(w = = 21, n = 8, and k = 4. We look at an example of each of the four cases of Lemma 7, underlining the element we add. (a If v = , then cmaj(v = 21 = cmaj(w. (b If v = , then 6 fits into run 1, and cmaj(v = 24 = cmaj(w+ k 1. (c If v = then 8 fits into fall 3, and cmaj(v = 29 = cmaj(w+ k (d If v = , then cmaj(v = 30 = maj(w + n + 1.

9 REFINED CYCLIC SIEVING ON WORDS AND THRALL S PROBLEM (DRAFT 9 Given w and x, there are (n + 1 positions in which we can add x, which correspond bijectively with the possibilities listed in Lemma 7, with (a removed if x > w 1, and (c removed if x w 1. Lemma 7 is the engine of Sections 4 to 6. For our purposes, the distinction between DA versus NDA positions is the critical piece we need to demonstrate cyclic sieving for the triple (W (α, δ, C n, W (α, δ maj (q. Weaker results that characterize the change in major index upon insertion of new element are well-known, such as the following Corollary, see [No] or [Gu], which we deduce from our stronger result. Corollary 9. Suppose w is a word of length n, and x is an element that does not appear in w. Letting w i (x denote the word formed by inserting x after the i-th position of w, and at the start if i = 0. Then, (maj(w 0 (x, maj(w 1 (x,..., maj(w n (x maj(w is a permutation of {0, 1,... n}. Proof. Suppose we always treat position n as a cyclic ascent (whether or not it is according to the definition, which replaces cmaj by maj and excludes (c in Lemma 7. The other cases of Lemma 7 correspond bijectively to positions of w in which to add x, so (maj(w 0 (x, maj(w 1 (x,..., maj(w n (x maj(w is a permutation of {0, 1,... n}. Next, we deduce a special case of Lemma 7 when w ends in a 1, and we forbid x from being added to the end of the word. The motivation for these restrictions is that we can rotate a word with a 1 so it ends in a 1, and more importantly, we need there to be 1 position between any two adjacent elements, as there is in a necklace. Without this cyclic symmetry of 1 position between any two elements, many of our future arguments would fall apart. Conveniently, ending in a 1 means position n is a cyclic ascent, so cmaj reduces to maj. Lemma 10. Suppose w is a word ending in 1. Let v be obtained from w by inserting x in any position but the end. Then, { k i if x fits into the i-th run of w maj(v maj(w = k j if x fits into the j-th fall of w Proof. We can never add to 0th run or 0th fall at the end of the word, so (a and (c of Lemma 7 disappear. We are left with the remaining cases of Lemma 7, and we can replace cmaj by maj, as stated above. Finally, we consider the insertion of multiple x s into w. First, we will need to break down this insertion as inserting x s into certain falls, then

10 10 CONNOR AHLBACH AND JOSH SWANSON inserting x s into certain runs. Let [a, b] denote the set of integers between a and b inclusive, and [a] := [1, a]. We let X Y denote that X is a multisubset of Y, so for example, {1, 1, 1, 2, 2, 4} [4]. Lemma 11. Let w end in a 1. Forbidding insertion at the end, any other insertion of α x x s into w can be viewed uniquely as the insertion of x s into some subset {a 1, a 2,..., a δx } [0, n k 1] of falls of w to obtain u, followed by the insertion of x s into a multisubset {b 1, b 2,..., b αx δ x } [0, k+δ x 1] of the runs of u. Proof. Consider the descent-adding (DA positions of w that at least one x gets added to, and say there are δ x such positions. This accounts for all of the DA x s, say which fit into some subset {a 1, a 2,..., a δx } [0, n k 1] of the falls of w, as no two x s can fit into the same fall, to obtain u. Now, the rest of the x s to be inserted into u are NDA, and since multiple x s could fit into a run, the runs inserted into give a multisubset {b 1, b 2,..., b αx δ x } [0, k + δ x 1] of the runs of u, as u has δ x more runs than w. On the other hand, given any subset {a 1, a 2,..., a δx } [0, n k 1] and multisubset {b 1, b 2,..., b αx δx } [0, k + δ x 1], there is a unique way to insert x s to fit into falls a 1, a 2,..., a δx of w to get u, and then insert x s to fit into runs b 1, b 2,..., b αx δx of u. Example 12. Let w = and v = be w with some 4 s added to it. If we first add DA 4 s, we get u = , which is w with 4 s added to falls {2, 6} [0, 6]. Adding NDA 4 s to u gives v = , which is u with 4 s added to runs {0, 1, 2, 2} [0, 4]. So, if w ends in a 1, each step of the filtration from w (x 1 to w (x consists of the insertion of x s into some subset {a 1, a 2,..., a δx } [0, n k 1] of falls of w (x 1 to obtain v, followed by the insertion of x s into a multisubset {b 1, b 2,..., b αx δ x } [0, k + δ x 1] of runs of v to obtain w (x. A clearer way to present this building up process is in the form of a rooted tree. Let W 1 (α, δ = {w W (α, δ : w ends in a 1}. Definition 13. Form a rooted, vertex- and edge-labeled, directed, tree graph T recursively as follows. Begin at the 0th step with root. At the x-th step, at each terminal vertex w in T, pick a subset A of the falls of w, compute the result w of inserting x into falls A of w, and then pick a multiset B on the runs of w. Let w be the result of inserting x into runs B of w. Add an edge from w to w labeled by (x, A, B. For each word w that ends in a 1, let T w be the subgraph of T consisting of paths to terminal vertices that

11 REFINED CYCLIC SIEVING ON WORDS AND THRALL S PROBLEM (DRAFT 11 are rotations of w ending in a 1. Finally, let T α,δ be the union (in T of T w as w ranges over W 1 (α, δ. The uniqueness in Lemma 11 guarantees that T is a tree. Each w W 1 (α, δ occurs once as a terminal vertex in T (α, δ, there is a unique path from the root to w in T (α, δ. Example 14. If w = , then the graph T w, a subgraph of T ((4, 2, 3, (0, 2, 3 is (1,, {0, 0, 0, 0} 1111 (2, {0, 2}, (2, {1, 3}, (3, {0}, {0, 1} (3, {2}, {1, 2} (3, {1}, {0, 1} (3, {3}, {1, 2} Corollary 15. Let w be a word ending in a 1. If we insert x s into the {a 1, a 2,..., a δx } [0, n k 1] falls of w to obtain v, followed by the insertion of x s into a multisubset {b 1, b 2,..., b αx δx } [0, k + δ x 1] of runs of u to obtain v, forbidding insertion at the end, then maj(v maj(w = (k + 1δ x + δ x i=1 a i + ( δx 2 + α x δ x (k + δ x b j. Proof. We add the DA x s from left to right. Note that after adding (i 1 of these DA x s, the number of runs is now k + (i 1, so when we add to the a i -th fall, the maj is increased by k + (i a i. Hence, δ x δ x ( δx maj(u maj(w = (k + (i a i = (k + 1δ x + a i + 2 i=1 We then add the NDA x s from left to right. Since v has k + δ x runs, when we add an x to run b j, we increase maj by k + δ x b j. Hence, maj(v maj(u = (k + δ x b j. α x δ x i=1

12 12 CONNOR AHLBACH AND JOSH SWANSON Putting these together gives Corollary 15. Suppose we fix content α = (α 1,..., α m and circular descent type δ = (δ 1,..., δ m. We assume δ 1 = 0 as otherwise, W (α, δ =. Also, without loss of generality, we assume α has no parts of size 0. Else, let ˆα denote α with all parts of size 0 removed, and consider the natural bijection Flatten : W (α W (ˆα which replaces all copies of the i-th smallest element by i. Clearly, Flatten preserves CDes, maj, period, and frequency. Define the reduced CDT (RCDT of w W n by RCDT(w := CDT(Flatten(w. Equivalently, for w W (α, we obtain RCDT(w by removing all indices j in CDT(w where α j = 0. Note CDT(w j = 0 at these indices as well. Example 16. If w = , then Flatten(w = , w has content (1, 0, 2, 1, 0, 0, 2, 0, 1, and CDT(w = (0, 0, 1, 1, 0, 0, 1, 0, 0, RCDT(w = (0, 1, 1, 1, 0. Thus, if ˆδ is δ with all parts δ j where α j = 0 removed, then Flatten restricts to a bijecton Flatten : W (α, δ W (ˆα, ˆδ. In summary, if α has a part of size 0, replace α by ˆα, CDT by RCDT, and δ by ˆδ. So, assume α j 0 for j = 1,... m. Let (β 1,..., β m, (k 1,..., k m be the partial sums β x := α 1 + α α x, k x := δ 1 + δ δ x, so that β x = w (x and k x = cdes(w (x. Let k := cdes(w = k m. Next, we calculate maj after adding x s to falls in A x := {a x,1, a x,2,... a x,δx } [0, β x 1 k x 1 1] and runs in B x := {b x,1, b x,2,... b x,αx δx } [0, k x 1] for x = 2,..., m starting with 1 α 1. From Corollary 15, for all x 2, maj(w (x maj(w (x 1 = (k x 1 + 1δ x + Since maj(w (1 = 0, m maj(w = (k x 1 + 1δ x + = m δ x k x m δ x + δ x i=1 δ x i=1 a i + (a i + ( δx 2 + α x r x ( α δx x δ x + (k x b j 2 ( m δx a i + m (α x δ x + m i=1 α x δ x m ( δx 2 (k x 1 b j. (k x b j

13 REFINED CYCLIC SIEVING ON WORDS AND THRALL S PROBLEM (DRAFT 13 Now, let B x := k x 1 B x [0, k x 1], and define the sum function on sets or multisets by sum(a = a A a. Then, ( m m ( δx m m m maj(w = δ x k x α x + sum(a x + sum(b 2 x ( k m = + n α 1 + (sum(a x + sum(b 2 x. where we used that k x = x j=2 δ j to deduce that ( m m δ x k x 1 + ( δx 2 = ( k. 2 We also recall the formulas ( [0, n 1] sum ( n (q = e k (1, q, q 2,... q n 1 = q (k 2 k k q ( ( [0, n 1] n + k 1 sum (q = h k (1, q, q 2,... q n 1 = k k Therefore, by uniqueness of paths in T (α, δ to each terminal vertex in W 1 (α, δ, the maj generating function on W 1 (α, δ is given by m ( W 1 (α, δ maj (q = q (k 2+n α 1 [0, βx 1 k x 1 1] sum ( [0, kx 1] (q sum (q δ x α x δ x m ( ( = q (k 2+n α 1 q (δx 2 βx 1 k x 1 kx + (α x δ x 1 δ x α x δ x This gives the following theorem. Theorem 17. The maj generating function on W 1 (α, δ is given by W 1 (α, δ maj (q = q (k 2+ m ( m ( ( δx 2 +n α 1 βx 1 k x 1 kx + (α x δ x 1 δ x α x δ x Next, each necklace in N(α, δ has α 1 n of its elements in W 1 (α, δ. Since maj on each necklace is well-defined modulo d := gcd(k, n, Theorem 17 tells us the following Theorem about the maj generating function on W (α, δ. Theorem 18. The generating function for maj on the set W (α, δ satisfies W (α, δ maj (q = n q (k 2+ m ( m ( ( δx 2 α 1 βx 1 k x 1 kx + (α x δ x 1 α 1 δ x α x δ x modulo q d 1. q q q. q. q q q

14 14 CONNOR AHLBACH AND JOSH SWANSON Let g := gcd(α, δ := gcd(α 1,... α m, δ 1,... δ m, and notice ( k m ( δx m m + = δ i δ j δ x 0 (mod g. 2 2 i j 2 So, reducing modulo q g 1, we get the following Theorem. Theorem 19. The generating function for maj on the set W (α, δ satisfies W (α, δ maj (q n m ( ( βx 1 k x 1 kx + (α x δ x 1 (mod q g 1 α 1 δ x α x δ x 5. Refining the CSP to fixed content and Circular Descent Type Recall, the triple (X, C n, f(q is said to exhibit the CSP, if for all τ C n, or equivalently, if f(q Orbits R X q #X τ = f(ω τ, R 1 i=0 q i n R (mod q n 1, where the orbits are from the action of C n on X. Such an equivalence makes sense since prescribing values of f(q at the n-th roots of unity determines f(q modulo q n 1. In our case, we can rephrase the CSP as saying that two statistics are equidistributed on W (α, δ. Letting maj n (w = maj(w (mod n we can restate the CSP on (W (α, δ, C n, W (α, δ maj (q as W (α, δ maj n(q = N N(α,δ N 1 j=0 q j freq(n = W (α, δ flex (q The second equality comes from the fact that flex has values j freq(n for j = 0,..., N 1 on necklace N. Hence, the CSP on (W (α, δ, C n, W (α, δ maj (q is equivalent to maj modulo n and flex being equidistributed on W (α, δ. To prove the CSP on (W (α, δ, C n, W (α, δ maj (q, we first show the polynomial W (α, δ maj (q has period gcd(α 1,..., α m, δ 1,... δ m modulo n (Definition 20. Then, we show that the CSP holds for (W (α, δ, C g, W (α, δ maj (q using well-known instances of the CSP on subsets and multisubsets. Putting these results together demonstrates the CSP for (W (α, δ, C n, W (α, δ maj (q. Definition 20. We say a statistic stat : X N has period a modulo b on X if for all i, #{x X : stat(x b i} = #{x X : stat(x b i + a}. Similarly, we say a polynomial f(q has period a modulo b if for all i [0, b 1], h(q q i= h(q q (i+a (mod b. q

15 REFINED CYCLIC SIEVING ON WORDS AND THRALL S PROBLEM (DRAFT 15 where h(q q j is the coefficient of q j in q, and h(q is the unique polynomial of degree at most (b 1 that f(q is equivalent to modulo q b 1. By definition, stat has period a modulo b on X if and only if X stat (q has period a modulo b. Here are some useful facts we will be using frequently: (a If stat has period a modulo b and period b modulo c, then stat has period a modulo c. (b If stat has period a modulo c and period b modulo c, then stat has period gcd(a, b modulo c. The same facts hold if stat is replaced by a polynomial f(q. Theorem 21. maj has period g := gcd(α, δ modulo n on W (α, δ. In order to transfer between W (α, δ and W 1 (α, δ, the following lemma will be useful. Lemma 22. maj has period a modulo n on W (α, δ if and only if maj has period a modulo d on W 1 (α, δ. Proof. By the action of rotation, maj has period d modulo n on W (α, δ. Thus, maj has period a modulo n on W (α, δ if and only if maj has period a modulo d on W (α, δ. As α 1 n of each necklace in N(α, δ shows up in W 1 (α, δ, and maj is well-defined modulo d on necklaces, maj has period a modulo d on W (α, δ if and only if maj has period a modulo d on W 1 (α, δ. We set g := gcd(α 1,... α m, δ 1,... δ m. We prove the result by induction on m, the m = 1 being clear as g = n. For m 2, by Theorem 18, (5 W (α, δ maj (q n α 1 q C G(α, δ, q (mod q d 1 where C = ( k 2 + m ( δx2 α1, and m ( ( βx 1 k x 1 kx + (α x δ x 1 G(α, δ, q = δ x q α x δ x q m ( = q (δx 2 [0, βx 1 k x 1 1] sum ( [0, kx 1] (q δ x α x δ x Let γ x := α x δ x. Notice β x = β x 1 + (k x k x 1 + γ x = β x 1 k x 1 = β x k x γ x sum (q By the action of rotation on multisets in ( ( kx α x δ x and sets in ( [0,β x 1 k x 1 1] δ x, which affects the sum by the size of the multiset or set modulo the size of the interval, we deduce that for all x = 2,..., m,

16 16 CONNOR AHLBACH AND JOSH SWANSON (a G(α, δ, q has period γ x = α x δ x modulo k x. (b G(α, δ, q has period δ x modulo β x 1 k x 1 = β x k x γ x. But by equation 5, if G(α, δ, q has period a modulo b, then maj has period gcd(a, d modulo b on W (α, δ. In particular, for x = m, G(α, δ, q has period γ m modulo k. Because maj has periods d, k modulo n on W (α, δ, maj has period gcd(γ m, d mod n. Also, for x = m, we find that G(α, δ, q has period δ m modulo β m k m γ m = n k γ m. Therefore, W (α, q has period gcd(δ m, d modulo gcd(n k γ m, d on W (α, δ, which reduces to maj having period gcd(δ m, d modulo gcd(γ m, d on W (α, δ. Combining with maj having period gcd(γ m, d mod n on W (α, δ, maj has period gcd(δ m, γ m, d on W (α, δ. Now, inductively assume maj has period g := gcd(α 1,..., α m 1, δ 1,... δ m 1 on W (α, δ, where α = (α 1,... α m 1 and δ = (δ 1,..., δ m 1. By Lemma 22, maj has period g modulo gcd(n α m, k m 1 on W 1 (α, δ. This periodicity extends to the set W 1 (α, δ because a choices of falls then runs to add to and a word in W 1 (α, δ uniquely determines a word in W 1 (α, δ, and the difference of their major indices is independent of the word chosen. Again, by Lemma 22, maj has periods δ m, γ m, d, k modulo d on W 1 (α, δ, so maj has period gcd(n α m, k m 1 modulo d on W 1 (α, δ because (6 n α m = n (δ m + γ m, k m 1 = k δ m. Combining this with maj having period g modulo gcd(n α m, k m 1, we deduce that maj has period g modulo d on W 1 (α, δ. So, by Lemma 22, maj has period g modulo n on W (α, δ. Finally, we already have shown maj has periods α m = δ m + γ m and δ m modulo n on W (α, δ. Therefore, maj has period g modulo n on W (α, δ, proving Theorem 21. We acknowledge the previous proof is the least elegant part of our argument, and we would appreciate improvements toward a more elegant proof. Now, we will prove following Theorem, bringing us closer to proving Theorem 1. Theorem 23. (W (α, δ, C g, W (α, δ maj (q exhibits the CSP. If C n = σ n, then C g = σn n/g C n, and let C g act on W (α, δ as this subgroup of C n. Also, assuming g a, C g acts on ( [a] ( b and [a] b by rotation of values by a/g. Recall that if (X, C n, f(q exhibits the CSP, then so does (X, C g, f(q when g n. And, if (X, C n, f(q, (Y, C n, h(q exhibit the CSP, then so does (X Y, C n, f(qh(q, where C n acts on X Y by τ (x, y = (τ x, τ y [RSW]. It is well-known that ( ([n] k, C n, ( n k q, ( ([n] k, C n, ( n + k 1, k q

17 REFINED CYCLIC SIEVING ON WORDS AND THRALL S PROBLEM (DRAFT 17 each exhibit the CSP from action of rotation on sets and multisets [RSW], so ( ([n] ( ( ([n] ( n n + k 1, C g,,, C g,, k k k k q exhibit the CSP. Because g (β x 1 k x 1, δ x, k x, α x δ x for all x, ( m ( [0, βx 1 k x 1 1] m ( [0, kx 1], C g, G(α, δ, q δ x α x δ x exhibits the CSP. So, letting D x = ( [0,β x 1 k x 1 1] δ x and Ex = m #(Dx#(E τ x τ = G(α, δ, ω τ. for all τ C g. Therefore, by Theorem 19, n α 1 m #(Dx#(E τ x τ = W (α, δ maj (ω τ. q ( ([0,kx 1] α x δ x, Next, we need to relate the cyclic action on sets and multisets above to that on words in W (α, δ. For a subset A [0, r 1] or multisubset A [0, r 1], define period(a to be the number of distinct elements in the orbit of A under rotation of values. The frequency freq(a = r/ period(a counts the maximum number of times A can be written as the same set or multiset repeated one after another. Note the concept of frequency depends on what we view it as a subset of. For example, [0, 3] [0, 5] has frequency 2, but [0, 3] [0, 6] has frequency 1. We relate the concepts of frequency on words, subsets, and multisubsets using the tree decomposition in section 4. Recall each w W 1 (α, δ can be uniquely broken down into a choice of falls A x = {a 1,... a δx } D x and runs B x E x for the x s to fit into. Lemma 24. The frequency of a word w W 1 (α, δ is given by freq(w = gcd(freq(a 2, freq(b 2,..., freq(a m, freq(b m, where A 2, B 2,... A m, B m represent the choices of falls and runs to insert into in order to obtain w. Proof. Since w is a word repeated freq(w times, each A x, B x is a subset or multisubset repeated freq(w times. Conversely, since each A x, B x chosen is a subset repeated h := gcd(freq(a 2, freq(b 2,..., freq(a m, freq(b m, times, w is a word repeated h times. Hence, freq(w = h. Lemma 24 tells us τ C g fixes w W 1 (α, δ if and only if τ fixes the corresponding A 2,... A m, B 2,... B m, so #W 1 (α, δ τ = m #(Dτ x#(e τ x.

18 18 CONNOR AHLBACH AND JOSH SWANSON As α 1 n of each necklace in N(α, δ shows up in W 1 (α, δ, #W (α, δ τ = n m #(D α x#(e τ x τ = W (α, δ maj (ω τ. 1 As this holds for all τ C g, we have proven Theorem 23. Finally, we extend the CSP from C g to C n. By Theorem 23, W (α, δ maj (q N N(α,δ N 1 j=0 q j freq(n (mod q g 1. For any N N(α, δ, freq(n g. Thus, noting freq(n = n/ N, W (α, δ maj (q n g N N(α,δ g N n 1 j=0 q j freq(n (mod q g 1 But now, using that W (α, δ maj (q has period g modulo n, W (α, δ maj (q N N(α,δ N 1 j=0 q j freq(n (mod q n 1, so (W (α, δ, C n, W (α, δ maj (q exhibits the CSP, proving Theorem An Explanation of the CSP for subsets and Multisubsets Fix X = ( [0,n 1] ( k, and f(q = n k q. Though it is known the triple (X, C n, f(q exhibits the CSP, the literature is lacking a more insightful proof that links the elements of X with terms in f(q through a statistic, the subset sum function on X. The idea for our argument comes from Wagon and Wilf in [WW], who studied the distribution of subset sums modulo m, and characterize when the subset sum statistic is equidistributed modulo m. We apply the same rotation within intervals they use, but this time in order to obtain cyclic sieving. Let C n = σ act on X by, for A = {a 1, a 2,..., a k } X, σ A = {a (mod n, a (mod n,... a n + 1 (mod n}. Let d = gcd(k, n. For all x n, and j [0, n x 1], let Ij x := [jx, (j + 1x 1], which we call x-intervals. Let C(a, b = {A X : gcd(a, #(A Ia, 0 #(A Ia, 1..., #(A Ia n/a 1 = b}. In particular, C(a, b is empty unless b a. Define the map Int : X {x : x n} by Int(A is the largest r n such that each r-interval in A is full or empty. Also, let S r = {A X : Int(A = r}.

19 REFINED CYCLIC SIEVING ON WORDS AND THRALL S PROBLEM (DRAFT 19 Example 25. If n = 8, k = 4, Int(0134 = 1, Int(0145 = 2, Int(0123 = 4. We first restrict attention to a particular set of subsets satisfying certain gcd requirements and obtain a result that looks remarkably like cyclic sieving. Theorem 26. Let C := C(n, d 0 C(d 0, d 1 C(d p 1, g C(g, x X. Then, the distribution of sum on C satisfies f C (q := q (k 2 C sum (q r x Proof. We proceed by induction on x. Let Then, (7 #(C S r r g/r 1 q ir (mod q g 1 g B(g, a 0,..., a n/g 1 := {A X : #(A I j g = a j for all j}, C = (a 0,...,a n/g 1 i=0 B(g, a 0,..., a n/g 1, where the union is over all (a 0,..., a n/g 1 satisfying gcd ( d r, a a dr/g 1,..., a (n/g dr/g + + a n/g 1 = dr+1 for all r = 1,..., p, where d p = g, d p+1 = x. In particular, gcd(g, a 0,... a n/g 1 = x, so there exists integers c 0, c 1... c n/g 1 such that c 0 a 0 + c 1 a c n/g 1 a n/g 1 x (mod g Thus, the action on B(g, a 0,..., a n/g 1 of rotating the elements in I j g forward c j times in I j g for all j increases sum statistic by a total of x modulo g. This action extends to C, as C by equation 7. Hence, the polynomial f C (q has period x modulo g. Lemma 27. If C S r, then r x. Proof. Fix A C. First, r n. As A C(n, d 0, and each r-interval is full or empty in A, r d 0. Inductively assuming r d i, as each d i -interval is union of r-intervals, r gcd(d i, #(A Id 0 i, #(A Id 2 i,..., #(A I n/d i 1 d i = d i+1. Hence, r x. For the base case, when x = 1, f C has period 1 modulo g, so f C (q #C 1 g 1 q i = #(C S 1 1 g 1 g g i=0 i=0 (mod q g 1,

20 20 CONNOR AHLBACH AND JOSH SWANSON as C S 1 by Lemma 27. Now, Suppose x 2. By conditioning on the gcd of x and the sizes of the intersections with the x-intervals, C = y x (C C(x, y. First, if y < x, by our induction hypothesis, (8 q (k 2 (C C(x, y sum (q r y #(C C(x, y S r r x/r 1 q jr (mod q x 1 x We claim the same identity holds if y = x as well. First, C C(x, x S x by Lemma 27 and definition of C(x, x. Therefore, the right side of equation (8 with y = x becomes #(C C(x, xq 0. But on the other hand, if A S x, then by shifting x-intervals by multiples of x, we can form the set [0, (k 1], which preserves the sum modulo x. Hence, sum(a ( k 2 x 0. Therefore, j=0 q (k 2 (C C(x, x sum (q #(C C(x, xq 0 (mod q x 1, proving equation (8 when y = x. Summing equation (8 over y x, q (k 2 C sum (q = r x #(C S r r x/r 1 q jr (mod q x 1 x j=0 because C = y x (C C(x, y. However, as f C (q has period x modulo g, f C (q = q (k 2 C sum (q r x proving Theorem 26. #(C S r r g/r 1 q ir (mod q g 1, g i=0 Example 28. Suppose C = C(4, 2 C(2, 1 = {02, 03, 12, 13} S 1. Then, f C (q = q 1 + 2q 2 + q 3 2(q 0 + q 1 (mod q 2 1. Note f C (q is not equivalent to q 0 + q 1 + q 2 + q 3 modulo q 4 1, for Theorem 26 only gives equivalence modulo q 2 1 in this case. Also, if C = C(4, 2 = {01, 02, 03, 12, 13, 23}, then #(C S 1 = 4 and #(C S 2 = 2. Then, f C (q (q 0 + q 1 + q 2 + q 3 + (q 0 + q 2 (mod q 4 1. Both of these calculations agree with Theorem 26. In particular, when C = C(n, d = X, f(q = f X (q r n #S r r n/r 1 q ir (mod q n 1 n Define F r := {A X : freq(a = r}, recalling the the frequency of A is n over the size of the orbit of A under C n. First, x, r x n S x is the subset i=0

21 REFINED CYCLIC SIEVING ON WORDS AND THRALL S PROBLEM (DRAFT 21 of X where the r-intervals are full or empty, so choosing which of the k/r r-intervals are full, ( n/r #S x =. k/r x,r x n Also, x,r x n F x is the subset of X where each n/r-interval is shifted version of the first n/r-interval, which must contain k/r elements. So, ( n/r #F x = k/r x,r x n By Mobius inversion on the divisibility lattice of n, #S r = #F r for all r. Therefore, f(q r n #(F r r n/r 1 q ir = n i=0 Orbits R X R 1 i=0 q i n R (mod q n 1, which demonstrates (X, C n, f(q exhibits the CSP. Furthermore, our proof shows that the terms where the orbits have size n/r correspond to the elements of X in S r under the sum statistic., and h(q = ( n+k 1 k Next, let Y = ( [0,n 1] k multisubsets by, if A = {a 1, a 2... a k } Y, then q. Let C n = σ act on σ A = {a (mod n, a (mod n,... a n + 1 (mod n}. Recall the bijection ϕ : Y Z := ( [0,n+k 2] k given by ϕ({a1, a 2,..., a k } = {a 1, a 2 + 1,..., a k + (k 1}, where a 1 a k. This means sum(ϕ(a = sum(a + ( k 2, so ( n + k 1 h(q = = q (k 2 Z sum (q = Y sum (q. k q Just as before, consider the x-intervals Ix j = [jx, (j + 1x 1], but this time for j = 1, 2,..., ( n+k x 1 so the x-intervals cover [0, n + k 1 x]. Note [n + k x, n + k 2], which we call the x-remainder, is leftover after we remove the x-intervals. In this case, for B Z, define Int(B = max{x : x d, I x j B is full or empty for all j}, which differs from Int defined earlier in that we force the output to divide d = gcd(k, n. Also, let S r = {B Z : Int(B = r}. Thus, S r = unless r d. Then, from the same definitions and reasoning for the proof of Theorem 26, we deduce the following Theorem. Theorem 29. Let C := C(d, d 1 C(d 1, d 2 C(d p 1, g C(g, x Z.

22 22 CONNOR AHLBACH AND JOSH SWANSON Then, the distribution of sum on C is given by f C (q := q (k 2 C sum (q r x In particular, for Z = d 1 d C(d, d 1, h(q = q (k 2 Z sum (q r d #(C S r r g/r 1 q ir (mod q g 1 g i=0 #(S r r d/r 1 q ir (mod q d 1 d The action of rotation of multisubsets shows h(q has period k and thus period d modulo n, so h(q r d i=0 #(S r r n/r 1 q ir (mod q n 1 n i=0 Now, set F r = {A Y : freq(a = r}, recalling the the frequency of A is n over the size of the orbit of A under C n. Note F r = unless r d. Then, r x F x is the set of multisubsets where each n/r-interval is shifted version of the first n/r-interval, which must contain k/r elements. Thus, for all r d, ( n/r + k/r 1 x,r x d F x = ( n/r k/r = But on the other hand, r x S x is the set of subsets where each r-interval is full or empty. For B S x, the condition r d k forces the r-remainder to contain no elements of B. Thus, for all r d, by choosing which k/r of the (n/r + k/r 1 r-intervals is full, ( n/r + k/r 1 S x =. k/r x,r x d By Mobius inversion on the divisibility lattice of d, #S r = #F r. Therefore, h(q r n #(F r r n/r 1 q ir = n i=0 Orbits R Y k/r R 1 i=0. q i n R (mod q n 1 This demonstrates that (Y, C n, h(q exhibits the CSP. Furthermore, our proof shows that the terms where the orbits have size n/r correspond to the inverse image of ϕ of the subsets in S r under the sum statistic. 7. Application to Thrall s Problem In this section, we use the work of Kraskiewicz-Weyman to motivate the cyclic sieving phenomenon on words of a fixed content. Reversing the argument, we also deduce Theorem 5 using Theorem 1 to bridge the cyclic

23 REFINED CYCLIC SIEVING ON WORDS AND THRALL S PROBLEM (DRAFT 23 sieving gap. Every step of the argument except our proof of Theorem 1 is bijective. Since Theorem 1 is significantly more refined than necessary for this application, there is hope for a fully bijective proof of Theorem 5 as well as an underlying representation-theoretic explanation of Theorem 1. We begin by summarizing a classical generating function for ch L n and Klyachko s observation connecting L n to the Schur modules of induced representations of irreducible representations of cyclic groups. Theorem 30 ([Ha], Lemma ; [Re], Theorem 7.5. There is a basis for L(V, called a Hall basis, whose elements are certain iterated bracketings of v 1,..., v m. Moreover, there is a natural, weight-preserving bijection between the Hall basis and the set of primitive necklaces on [m] of length n. Let P N n denote the set of primitive necklaces of length n. Corollary 31. ch L n is the content-generating function of primitive necklaces of length n, i.e. ch L n = P N cont n. Proof. The action of diag(x 1,..., x m on an iterated bracketing of v 1,..., v m multiplies the bracketing by x a 1 1 xam m, where a i is the number of v i s. The trace of the endomorphism induced by diag(x 1,..., x m is the sum of these monomials, so the result follows directly from the theorem. The argument in 31 generalizes to say that if a GL(V -representation M has a basis with simple tensors, then the Schur character of M is the content generating function for the indices which show up in a basis. On the other hand, let χ r be the one-dimensional representation of C n := σ given by letting σ act on the left as multiplication by ωn r := e 2πir/n. Note that CC n = r 1 i=0 χ r. We now summarize and slightly generalize observations due to Klyachko [Kl] (equation (6. Theorem 32. There is a basis for E(χ r Sn C n indexed by necklaces of length n on [m] with frequency dividing r, or equivalently indexed by words of length n on [m] with flex r. Hence n 1 (ch χ r Sn C n q r = Wn cont,flex (x 1, x 2,... ; q. r=0 Proof. The suggested necklace basis comes from a sequence of natural identifications, which we now describe. We may restrict the right S n -action on V n to a C n -action where σ := (12 n S n as needed. Now E(χ r Sn C n = V n CSn χ r Sn C n = V n CCn χ r.

24 24 CONNOR AHLBACH AND JOSH SWANSON We may give χ r a right C n -module structure in addition to its left C n -module structure by declaring that σ acts on the right as multiplication by ωn. r In this fashion, n 1 r=0 χ r = CC n as a (C n, C n -bimodule. Hence we have natural right CC n -module isomorphisms V n CCn CC n = V n = HomCCn (CC n, V n. Applying CC n = n 1 r=0 χ r and linearity gives n 1 n 1 V n CCn χ r = Hom CCn (χ r, V n r=0 i=0 which descend to right CC n -module isomorphisms V n CCn χ r = HomCCn (χ r, V n. Moreover, these isomorphisms respect the (left GL(V -module structures. Since χ r is one-dimensional, we may identify Hom CCn (χ r, V n with the subset of V n of images of 1 C, from which we find E(χ r Sn C n = {v V n : v σ = ω r nv}. = Span C {v i1 v i2 v in + ω r nv in v i1 v in 1 + : i 1 i n W n, 1 i j m}. In the final span, we need only include one word in any fixed necklace. If the necklace has period p, we may factor out a constant 1 + ω rp n + ω 2rp n + + ωn r(n p = ωn n 1 ωn rp 1 which is nonzero if and only if ωn rp = 1, so if and only if rp n 0. Since p = n/f where f is the frequency of the necklace, it follows that rp n 0 if and only if f r. Using this basis, the Schur character E(χ r Sn C n is now clearly the weight generating function of the suggested necklaces. Finally, given a necklace N of frequency f r, flex N = {0, f, 2f,..., n f}, so N contains a unique word of flex r, which completes the proof. At r = 1, we require f = 1, resulting in primitive necklaces. Thus we have arrived at Klyachko s observation [Kl] (Proposition 1, ch E(χ 1 Sn C n = ch L (n. Moreover, the argument used a series of natural bijections. The corresponding Schur expansion is well-known: Theorem 33 ([KW]. We have ch χ r Sn C n = λ n a λ,r s λ where a λ,r = #{Q SYT(λ : maj(q n r}.

25 REFINED CYCLIC SIEVING ON WORDS AND THRALL S PROBLEM (DRAFT 25 We now show that Theorem 33 is equivalent to (W (α, C n, W (α maj(q exhibiting the CSP for all contents α. First, let M n,r = {w W n : maj(w n r}, NF D n,r = {w N n : freq(n r}, and let M n,r (α, NF D n,r (α denote the elements in M n,r, NF D n,r content α, respectively. with Proof. We can restate (W (α, C n, W (α maj(q exhibiting the CSP for all contents α as W (α maj (q N N(α N 1 q j freq(n (mod q n 1, or, by comparing the coefficients of q r, #M n,r (α = #NF D n,r (α for all α, or just Mn,r cont = NF Dn,r cont. First, assuming that (W (α, C n, W (α maj(q exhibits the CSP for all α, and using that under RSK, Des(Q(w = Des(w, ch χ r Sn C n = NF Dn,r cont = Mn,r cont = SSYT(λ cont = a λ,r s λ, λ n λ n Q SYT(λ maj(q nr which proves Theorem 33. Here, {Q SYT(λ : maj(q n r} is the set of possible Q (recording-tableau of a given shape for w M n,r, and for a given Q-tableau, all P (insertion-tableau of the same shape are possible. Secondly, assuming Theorem 33, NF D cont n,r = ch χ r Sn C n = a λ,r s λ = Mn,r cont, λ n so (W (α, C n, W (α maj(q exhibits the CSP for all α. The formula NF Dn,r cont = λ n a λ,rs λ also explains the following symmetry among the a λ,r coefficients. Corollary 34. For all n 1 and λ n, a λ,r := #{T SYT(λ : maj(t n r} only depends on gcd(r, n. Proof. For N N n, freq(n n, so freq(n r if and only if freq(n gcd(n, r. Thus, NF Dn,r cont only depends on gcd(n, r. Hence, the coefficients a λ,r in the Schur expansion of NF Dn,r cont must depend only on gcd(n, r.

26 26 CONNOR AHLBACH AND JOSH SWANSON 8. Application to Higher Lie Modules The one-dimensional irreducible representations of C a S b for b > 1 are as follows. Pick 0 r < a, ɛ {0, 1}. Define χ r,ɛ : C a S b C (i 1,..., i b, τ ω r(i 1+ +i b a ( 1 ɛ sgn τ. Given w W ab, define the standardization of w with size-a intervals, written std b (w, as follows. Put the subwords coming from the size-a intervals of w into lexicographic order, numbered 1, 2,.... Replace each size-a interval with the position of its subword in this order, and then apply the usual standardization map to get an element of S b. For example, if w = with b = 3, then the subwords are 12 < 35, so we compute std b (w by standardizing 121, which yields std 3 ( = 132 S 3. Theorem 35. There is a basis for E(χ r,ɛ S ab C a S b indexed by words w of length ab with the following properties: Each length-a sub-interval of w has flex statistic r. If ɛ = 0, std b (w = id. If ɛ = 1, std b (w = w 0 and the length-a intervals of w are all distinct. Proof. Much of the proof of Theorem 32 may be carried out by replacing C n, S n, χ r with C a S b, S ab, χ r,ɛ. Doing so gives E(χ r,ɛ S ab C a S b = {φ(1 V ab : φ Hom C(Ca S b (χ r,ɛ, V ab }. To obtain a generating set for the right-hand side, first consider the natural action of C a S b on W ab. For convenience, we identify words with their image under the natural map W ab V ab, which is (C a S b -equivariant by definition. We have (C a S b -submodules of V ab spanned by each (C a S b - orbit of W ab, so we may break φ(1 into components arising from these orbits. Now consider w W ab and suppose For g C a S b we have φ(1 = v w c v v v w c v (v g = φ(1 g = φ(1 g = (1 gφ(1 = u w (1 gc u u so we must define c w =: (1 gc w g. This may be inconsistent if the same w g appears multiple times on the right-hand side. Assuming the coefficients are non-zero, consistency is equivalent to requiring Stab(w Stab(1. To describe Stab(w more conveniently, we may assume that any two a-intervals of w with equal necklaces are themselves equal. Letting e i be the ith standard basis vector of Ca, b and letting p i be the period of the ith a-interval