Higher Lie Characters and Cyclic Descent Extension on Conjugacy Classes

. A now-classical cyclic extension of the descent set of a permutation has been introduced by Klyachko and Cellini. Following a recent axiomatic approach to this notion, it is natural to ask which sets of permutations admit such a (not necessarily classical) extension. The main result of this paper is a complete answer in the case of conjugacy classes of permutations. It is shown that the conjugacy class of cycle type λ has such an extension if and only if λ is not of the form ( r s ) for some square-free r . The proof involves a detailed study of hook constituents in higher Lie characters.

Background and main result.The study of descent sets for permutations may be traced back to Euler.A cyclic extension of this classical concept was introduced in the study of Lie algebras [20] and descent algebras [8].Surprising connections of the cyclic descent notion to a variety of mathematical areas were found later.
There exists a well-established notion of descent set for standard Young tableaux (SYT), but it has no obvious cyclic analogue.In a breakthrough work, Rhoades [27] defined a notion of cyclic descent set for standard Young tableaux of rectangular shape.The properties common to Cellini's definition (for permutations) and Rhoades' construction (for SYT) appeared in other combinatorial settings as well [24,23,11,3].This led to an abstract definition [2], as follows.
For connections of cyclic descents to Kazhdan-Lusztig theory see [27]; for topological aspects and connections to the Steinberg torus see [10]; for twisted Schützenberger promotion see [27,17]; for cyclic quasisymmetric functions and Schur-positivity see [1]; for Postnikov's toric Schur functions see [2].The goal of this paper is to determine which conjugacy classes of the symmetric group carry a cyclic descent extension.
Cellini's cyclic descent set, denoted CDes, is a special case of a cyclic descent extension, denoted in general cDes, as attested by the following observation.
Observation 1.2.Let Des and CDes denote the classical descent set and Cellini's cyclic descent set on permutations, respectively.Let p : S n → S n be the rotation Then the pair (CDes, p) is a cyclic descent extension of Des on S n in the sense of Definition 1.1.
Unlike the full symmetric group, for many conjugacy classes, Cellini's definition does not provide a cyclic extension, see Section 7.2 below.The goal of this paper is to show that most conjugacy classes in S n carry a cyclic descent extension.In fact, we obtain a full characterization.
Recall that an integer is square-free if no prime square divides it; in particular, 1 is square-free.Our main result is Theorem 1.4.Let λ be a partition of n, and let C λ ⊆ S n be the corresponding conjugacy class.The descent map Des on C λ has a cyclic extension (cDes, p) if and only if λ is not of the form (r s ) for some square-free r.
1.2.Proof method.The proof of Theorem 1.4 is non-constructive and involves a detailed study of the hook constituents in higher Lie characters.1.2.1.Higher Lie characters.Definition 1.5.For a partition λ of n, let C λ be the conjugacy class consisting of all the permutations in S n of cycle type λ, and let χ λ denote the irreducible S n -character corresponding to λ.Let Z λ be the centralizer of a permutation in C λ (defined up to conjugacy).If k i denotes the number of parts of λ equal to i, then Z λ is isomorphic to the direct product Here and in the rest of the paper Z i denotes the cyclic group of order i.
For each i, let ω i be the linear character on Z i ≀ S k i indexed by the i-tuple of partitions (∅, (k i ), ∅, . . ., ∅).In other words, let ζ i be a primitive irreducible character on the cyclic group Z i , and extend it to the wreath product Z i ≀ S k i so that it is homogeneous on the base subgroup Z k i i and trivial on the wreathing subgroup S k i .Denote this extension by ω i .Now let a linear character on Z λ .Define the corresponding higher Lie character to be the induced character The study of higher Lie characters can be traced back to Schur [30].An old problem of Thrall [39] is to provide an explicit combinatorial interpretation of the multiplicities of the irreducible characters in the higher Lie character, see also [33,Exercise 7.89(i)].Only partial results are known: the case λ = (n) was solved by Kraskiewicz and Weyman [21]; Désarménien and Wachs [9] resolved a coarser version of Thrall's problem for the sum of higher Lie characters over all derangements, see also [26].The best result so far is Schocker's expansion [29,Theorem 3.1], which however involves signs and rational coefficients.For recent discussions see, e.g., [25,5,36].
A remarkable theorem of Gessel and Reutenauer [15, Theorem 2.1] applies higher Lie characters to describe the fiber sizes of the descent set map on conjugacy classes.Their proof applies an interpretation of higher Lie character ψ λ in terms of quasisymmetric functions (Theorem 2.6 below).It follows that higher Lie characters can be used to prove the existence of cyclic descent extensions as explained below.For an integer 0 ≤ k < n and a Schur-positive subset A ⊆ S n denote where s (n−k,1 k ) is the Schur function indexed by the hook partition First we prove the following key lemma, which provides an algebraic criterion for the existence of a cyclic descent extension.

Next we prove
Proposition 1.7.The hook-multiplicity generating function of the higher Lie character is divisible by 1 + x if and only if λ is not of the form (r s ) for a square-free integer r.
This divisibility condition is proved using an explicit evaluation of the higher Lie character on n-cycles.;see Section 3.3 below.

Non-negativity.
In order to prove Theorem 1.4, it remains to show that the coefficients of the quotient M λ (x)/(1 + x) are nonnegative, whenever λ is not of the form (r s ) for a square-free r.It turns out that partitions λ with more than one cycle length are the easiest to handle.In that case, a factorization of the associated higher Lie character ψ λ is applied to prove Lemma 1.8.Let λ = µ ⊔ ν be a disjoint union of nonempty partitions with no common part.Then and its coefficients are thus non-negative.
The core of the proof of Theorem 1.4 is the case of λ = (r s ).For a fixed positive integer r, consider the formal power series where m i,(r s ) is the multiplicity of the hook character χ (rs−i,1 i ) in the higher Lie character ψ (r s ) .The following theorem completes the proof of Theorem 1.4.Theorem 1.9.If r is not square-free then the formal power series M r (x, y) 1 + x has non-negative integer coefficients.

1.3.
Outline.The rest of the paper is organized as follows.
Necessary background, including a cyclic analogue of the Gessel-Reutenauer Theorem, is given in Section 2.
The role of hooks in the study of cyclic descent extensions is explained in Section 3. In particular, a necessary and sufficient criterion for Schur-positive sets to carry a cyclic descent extension (Lemma 1.6) is proved in Subsection 3.2.Using this criterion, the proof of Theorem 1.4 is reduced to Proposition 1.7 (divisibility) and Theorem 1.9 (non-negativity).Proposition 1.7 is proved in Subsection 3.3.
The proof of Theorem 1.9, stating the non-negativity of the coefficients of M λ (x)/(1 + x) whenever this quotient is a polynomial, spans Sections 4, 5 and 6: the case of more than one cycle length is considered in Section 4; the case of n-cycles is considered in Section 5; and the case of cycle type (r s ) with s > 1 is considered in Section 6.In the case of more than one cycle length, non-negativity is proved using a factorization of the associated higher Lie character (Lemma 1.8).In the case of n-cycles, combining a combinatorial formula for inner products with a variant of the Witt transform proves unimodality of the sequence of hook-multiplicities (Proposition 5.2).This, in turn, implies the desired non-negativity of the quotient.
In Section 6 we lift the n-cycle result to the case of cycle type (r s ) with s > 1.In Subsection 6.1 we provide an explicit expression for the coefficients of (1 + x)M r (x, y), see Theorem 6.3.This expression is used to obtain a product formula for the bivariate polynomial 1 + (1 + x)M r (x, y) in Subsection 6.2, and to deduce Theorem 1.9 in Subsection 6.3.
Additional results are presented in Section 7. In Subsection 7.1 it is shown that Lemma 3.5 and Theorem 6.3 imply well-known combinatorial identities.In Subsection 7.2 it is shown that the natural approach does not provide a cyclic descent extension for conjugacy classes in S n .Palindromicity of the hook-multiplicity generating function M (r s ) (x) is studied in Subsection 7.3.
Section 8 concludes the paper with final remarks and open problems.
Acknowledgements.The second author is grateful for the hospitality of Bar-Ilan University during his Sabbatical leave.Special thanks are due to Jan Saxl, whose personality influenced the development of this paper in many ways; in particular, the paper is inspired by the ingenious use of a higher Lie character in [18].Thanks also to Eli Bagno, Jonathan Bloom, Sergi Elizalde, Darij Grinberg, Vic Reiner, Richard Stanley, Sheila Sundaram and Josh Swanson for their comments.

Preliminaries
The role of quasisymmetric in the study of the distribution of descent sets is discussed in Subsection 2.1.The results presented here are used in Section 3 to establish the reduction of existence of cyclic descent extension on conjugacy classes to the study of hook-multiplicities in higher Lie characters.In Subsection 2.2 we present cyclic analogues of a few classical results.These analogues are used for certain enumerative applications; the reader may skip this subsection.
The following theorem is due to Gessel.Denote the set of standard Young tableau of skew shape λ/µ by SYT(λ/µ).There is an established notion of descent set for SYT(λ/µ) Given any subset A ⊆ S n , define the quasisymmetric function Finding subsets of permutations A ⊆ S n , for which Q(A) is symmetric (Schur-positive), is a long-standing problem, see [15].
We write λ ⊢ n to denote that λ is a partition of a positive integer n.
Lemma 2.3.For every subset A ⊆ S n and a family {c λ } λ⊢n of coefficients, the equality x Des(T ) .
Next recall from [33,Ch. 7] that the fundamental quasisymmetric functions in x 1 , . . ., x n form a basis of the vector space QSym n of quasisymmetric functions in n variables.Finally, apply the vector space isomorphism from QSym n to the space of square-free polynomials in x 1 , . . ., x n , which maps F n,D to x D .x Des(T ) .
Corollary 2.4 will be combined with the following Theorem 2.5 to provide criteria for the existence of cyclic descent extensions for Schur-positive sets, see proof of Lemma 3.2 and Remark 4.2 below.
A ribbon is a skew shape "of width 1", namely: containing no 2 × 2 rectangle.The following theorem was proved in [2] using Postnikov's toric symmetric functions.
A constructive proof was recently given by Huang [17].
Let λ ⊢ n be a partition of n and let ψ λ be the higher Lie character indexed by λ (see Definition 1.5).
The following result is proved in [15].

Cyclic analogues.
The following theorem is due to Gessel.
For a subset defines a connected ribbon having the entries of α(J, n) as row lengths, from bottom to top.Let s α(J,n) be the associated (skew) ribbon Schur function.
Theorem 2.7.[14, an immediate consequence of Theorem 3] Let A be a finite set, equipped with a descent map Des : is symmetric then For a subset ∅ = J = {j 1 < j 2 < . . .< j t } ⊆ [n] define the corresponding cyclic composition of n as α cyc (J,n) := (j 2 − j 1 , . . ., j t − j t−1 , j 1 + n − j t ), with α cyc (J,n) := (n) when J = {j 1 }; note that α cyc (∅,n) is not defined.The corresponding affine (cyclic) ribbon Schur function was defined in [2] as Theorem 2.8.[1,Cor. 4.13] Let A be a finite set, equipped with a descent map Des : is symmetric then the fiber sizes of (any) cyclic descent map satisfy Proposition 2.9.[2, Lemma 2.2] If A ⊆ S n carries a cyclic descent extension, then the cyclic descent set generating function is uniquely determined.
Furthermore, Corollary 2.10.Let A ⊆ S n be a symmetric set which carries a cyclic descent extension and S be a finite set of skew shapes of size n which are not connected ribbons.Then for every cyclic descent extension the following equations are equivalent: a∈A x cDes(T ) .
For the opposite direction, let x n = 1 in Equation ( 7) and apply Corollary 2.4 to deduce Equation ( 6).
The following cyclic analogue of Theorem 2.11 results from Theorem 2.6 and Theorem 2.8.
Theorem 2.12.For every conjugacy class C λ , which carries a cyclic descent set extension, all cyclic extensions of the descent set map Des have fiber sizes given by

The role of hooks
3.1.Hooks and near-hooks.It turns out that hooks and near-hooks play a crucial role in the study of cyclic descent extensions.
A hook is a partition with at most one part larger than 1.Explicitly, it has the form A near-hook of size n is a hook of size n + 1 with its (northwestern) corner cell removed; see [3] for a somewhat more inclusive definition of this notion.Equivalently, recall the direct sum operation on shapes (partitions), denoted λ ⊕ µ, yielding a skew shape having the diagram of λ strictly southwest of the diagram of µ, with no rows or columns in common.A near-hook of size n is the direct sum of a one-column partition (1 k ) and a one-row partition (n − k), for some 0 ≤ k ≤ n.For example, When φ is understood from the context, we use the abbreviated notations m k := m k,φ and e k := e k,φ .By Pieri's rule [33,Theorem 7.5.17]combined with the (inverse) Frobenius characteristic map, Equivalently, Thus the sequences {m k } n−1 k=0 and {e k } n k=0 determine each other via the relations ( 9) where m k := 0 for k = −1 and k = n.Note that, in particular, n i=0 (−1) n−i e i = 0.

Cyclic descent extension and hook-multiplicities. Let
For λ = (n − k, 1 k ) we use the abbreviation The hook-multiplicity generating function is defined as The function M A (x), where A is a conjugacy class, was studied and applied to the enumeration of unimodal permutations with a given cycle type by Thibon [38].Indeed, Proof.For every 0 ≤ k < n there exists a unique standard Young tableau T of size n with Des(T ) = [k] (where [0] := ∅).The shape of T is (n − k, 1 k ).Comparing the coefficients of x [k] on both sides of Equation ( 5) completes the proof.
We now restate and prove Lemma 1.6.Lemma 3.2.A Schur-positive set A ⊆ S n carries a cyclic descent extension if and only if the hook-multiplicity generating function M A (x) is divisible by 1 + x and the quotient M A (x)/(1 + x) has non-negative coefficients; equivalently, if and only if there exist nonnegative integers , Proof.If A carries a cyclic descent extension then, by Observation 3.1 and the equivariance of cDes, for every 0 ≤ k ≤ n − 1: The numbers satisfy the required conditions, which imply the corresponding properties of M A (x).
For the opposite direction, assume that there exist non-negative integers d k (with Since By Corollary 2.4, this is equivalent to x Des(T ) .By Theorem 2.5, the set SYT(λ) carries a cyclic descent extension if and only if λ ⊢ n is not a hook; and each of the sets SYT(( carries a cyclic descent extension.Hence A also carries a cyclic descent extension, completing the proof. x cDes(T ) , where m λ and d k are the non-negative integers defined above.
Proof.By Corollary 2.10 together with Equation ( 10), the generating function for the corresponding cyclic descent set is uniquely determined and satisfies Equation (11).

Divisibility of the hook-multiplicity generating function. Recall the notation
for an S n -character φ.
Lemma 3.4.For every S n -character φ, the hook-multiplicity generating function is divisible by 1 + x if and only if the value of φ on an n-cycle is zero, i.e., φ (n) = 0.
Proof.By [28, Lemma 4.10.3],for every partition λ ⊢ n Thus which equals zero if and only if 1 + x divides M φ (x), completing the proof.
Letting φ = ψ λ , the higher Lie character indexed by a partition λ, reduces Proposition 1.7 to the following character evaluation.
Recall the Möbius function µ(n), the sum of all primitive (complex) n-th roots of 1.If n has a prime square divisor then µ(n) = 0; otherwise, n is a product of k distinct primes and µ(n) = (−1) k .The following lemma is equivalent to a combinatorial identity due to Garsia, as shown in Proposition 7.2 below.We give here an independent direct algebraic proof.
otherwise, where µ is the Möbius function.
Proof.Let c be an n-cycle in S n , and let Z λ = Z Sn (g) be the centralizer in S n of a specific element g ∈ C λ .An explicit formula for the induced character [19, (5.1)] is An n-cycle commutes only with its own powers.Thus, if λ is not of the form (r s ) for some r and s, then there is no n-cycle in Z λ ; equivalently, x −1 cx ∈ Z λ for every x ∈ S n .It follows that, for such partitions λ ⊢ n, ψ λ (n) = 0. Assume now that λ = (r s ), and let let g = g 1 g 2 • • • g s ∈ C λ be a fixed product of s disjoint r-cycles.The order of the centralizer for some integer k with gcd(k, n) = s; equivalently, u s = g j for some 0 < j < r with gcd(j, r) = 1.Conversely, if u ∈ S n is an n-cycle satisfying u s = g j for some 0 < j < r with gcd(j, r) = 1, then g is a power of u and therefore u ∈ Z λ .Thus the number of n-cycles in Z λ , namely the number of elements of where ϕ is Euler's totient function.
Viewing Z λ as the group of s × s monomial ("generalized permutation") matrices whose nonzero entries are complex r-th roots of unity, an element u ∈ Z λ ∩ C (n) corresponds to a matrix whose underlying permutation is a full s-cycle and the product of its nonzero entries is a primitive r-th root of unity.This product is equal to ω λ (u), so it is a primitive r-th root of unity.For u, v ∈ Z λ ∩ C (n) , write u ∼ v if v = u i for some integer i (necessarily coprime to n).This clearly defines an equivalence relation on Z λ ∩ C (n) .On each equivalence class, all primitive r-th roots of unity appear with the same frequency as values of ω λ .This property thus holds for all of Z λ ∩ C (n) , where this frequency is (s − 1)!r s−1 .Denoting by ξ any specific primitive r-th root of unity, the sum of all values of ω λ on Z λ ∩ C (n) is therefore Given any c, u ∈ C (n) , there are exactly n = rs permutations x ∈ S n which satisfy u = x −1 cx.Thus Proof of Proposition 1.7.By Lemma 3.4, 1 + x divides the hook-multiplicity generating function of the higher Lie character ψ λ if and only if ψ λ (n) = 0. Lemma 3.5 completes the proof.
1.If λ = (r s ) for some square-free integer r and positive integer s, then 1 + x does not divide the hook-multiplicity generating function M λ (x), and the descent set map on the conjugacy class C λ does not have a cyclic extension.2. If λ is not equal to (r s ) for any square-free r, then 1 + x divides M λ (x).In this case, the descent set map on C λ has a cyclic extension if and only if the quotient M λ (x)/(1 + x) has non-negative coefficients.
Proof.By the Gessel-Reutenauer theorem (Theorem 2.6), for every λ ⊢ n the conjugacy class C λ is Schur-positive, with Q(C λ ) = ch(ψ λ ).Combining this with Lemma 3.2 and Proposition 1.7 completes the proof of both parts.
In the following sections we will prove the non-negativity of the coefficients of the quotient M λ (x)/(1 + x) for partitions (cycle types) which are not equal to (r s ) for a square-free r: cycle types with more than one cycle length will be considered in Section 4, n-cycles will be considered in Section 5, and cycle types λ = (r s ) with non square-free r and s > 1 will be considered in Section 6 (this is the most difficult case).

Non-negativity: the case of more than one cycle length
Consider, first, the case of a conjugacy class with more than one cycle length.This is the easiest case to handle.
Proof of Lemma 1.8.The centralizer Z λ of a permutation in C λ is isomorphic, in this case, to the direct product Z µ × Z ν .By Definition 1.5, ω λ := ω µ ⊗ ω ν and ( 12) We deduce Corollary 4.1.If λ is a partition with more than one cycle length then M λ (x) is divisible by 1 + x and the quotient has non-negative coefficients.
Remark 4.2.In this case, the existence of a cyclic descent extension may be proved directly as follows.By Equation ( 12), ψ λ is a sum of characters indexed by disconnected shapes.Thus, by Corollary 2.4, the distribution of the descent set over C λ is equal to a sum of distributions over the sets of SYT of various disconnected shapes.By Theorem 2.5, each of these sets carries a cyclic descent extension, hence so does C λ .

Non-negativity: the single cycle case
Consider now the case of a conjugacy class with a single cycle.By Corollary 3.6, if r is not square-free then 1 + x divides M (r) (x).The main result of this section is Proposition 5.1.If r is not square-free then the coefficients of M (r) (x)/(1 + x) are nonnegative It follows from Lemma 5.12 below that, in order to prove Proposition 5.1, it suffices to show the unimodality (to be defined) of M (r) (x).This is the content of the following statement.
In Subsection 5.1 we use a variant of the Witt transform to produce explicit formulas for the coefficients m j,(r) (Lemma 5.11).Then, in Subsection 5.2, we prove their unimodality.

5.1.
A variant of the Witt transform.In this subsection we present a variant of the Witt transform, which will be used to prove non-negativity in Sections 5.2 and 6.
In particular, f j (r) is a non-negative integer.
Remark 5.6.Proposition 5.5 will not be proved here, since it is the special case s = 1 of Theorem Equation ( 9) and Proposition 5.5 imply Observation 5.9.
Proof.Use Definition 5.3, and write j = kd if d|(r, j).Then Remark 5.10.Recall from [22] that the r-th Witt transform of a polynomial p(x) is defined by The proof of Theorem 4 and Lemma 1 in [22] could have been used to prove that the coefficients of F r (x) are nonnegative integers.This non-obvious property of the numbers f j (r) also follows, of course, from their interpretation in Proposition 5.5 as inner products of two characters.What we really need, in Proposition 5.1, is the nonnegativity of the coefficients of F r (x)/(1 + x) 2 .
We now produce an explicit formula for each coefficient m j,(r) .For a combinatorial interpretation of these numbers, see Lemma 7.5 below.Proof.By Corollary 5.8, Using Definition 5.3 and Observation 5.9, we can write

Unimodality.
A sequence (a 0 , . . ., a n ) of real numbers is called unimodal if there exists an index 0 ≤ i 0 ≤ n such that the sequence is weakly increasing up to position i 0 and weakly decreasing afterwards: Lemma 5.12.Let a(x) = a 0 + a 1 x + . . .+ a n x n be a polynomial with real, nonnegative and unimodal coefficients.Assume that 1 + x divides a(x), and let b(x) := a(x)/(1 + x).Then the coefficients of b(x) are nonnegative.
Of course, divisibility of a(x) by 1 + x implies that Inverting (13) we get ( 14) and, similarly, By assumption, the sequence (a 0 , . . ., a n ) is nonnegative and unimodal, namely: there exists an index 0 ≤ i 0 ≤ n such that It follows from ( 14) that, for odd indices By ( 15), a similar argument holds for indices greater or equal to i 0 , and the proof is complete.
Lemma 5.12 shows that non-negativity of a sequence can be proved by showing unimodality of a related sequence.In particular, Proposition 5.1 would follow once we show the unimodality of the polynomial M (n) (x).In order to do that, we need the following technical lemma.Lemma 5.13.Assume that r > 7 and 1 < j < r/2; if j = (r − 1)/2 assume also that r > 11.Let d > 1 be a divisor of r, and denote Let ℓ := ⌊j/d⌋.Then ℓd is the largest multiple of d not exceeding j, hence The quotient A r,j,d can therefore be written in the form By assumption j < r/2, thus i/(r − i) < 1 for all 1 ≤ i ≤ j.It follows that A r,j,d is a decreasing function of j, with A r,1,d = 1/(r − 1).
It remains to consider the case j = (r − 1)/2, namely r = 2j + 1, for d ≥ 2. Note that in this case we assumed that r > 11, namely j > 5.
For 1 ≤ r ≤ 7, computing the polynomial M r (x) := F r (x)/(1 + x) explicitly, using Observation 5.9, gives The claim clearly holds in these cases.Assume from now on that r > 7.
Informally, the explicit formula for m j,(r) in Lemma 5.11 has a dominant term corresponding to d = 1, i.e., rm j,(r) is approximately equal to r−1 j .We will show that this approximation is good enough to make the sequence m 0,(r) , . . ., m r−1,(r) unimodal, like the sequence of binomial coefficients.Note that, unlike the binomial coefficients, this sequence is not always palindromic; see Proposition 7.13 below.

We conclude
Corollary 5.15.The conjugacy class of n-cycles C (n) carries a cyclic descent extension if and only if n is not square-free.
Proof.Combine Lemma 3.2 with Corollary 3.6.2and Proposition 5.1 6. Non-negativity: the case of cycle type (r s ) In this section we consider the case λ = (r s ).We fix r, while s and hence n = rs vary.The arguments below also work for the trivial case r = 1.
As in the previous section, instead of the hook multiplicities we prefer to work with their consecutive sums, Here is the structure of the current section.In Subsection 6.1 we obtain an explicit description of e i,(r s ) (Theorem 6.3).The proof involves a detailed computation of character values and inner products of characters.In Subsection 6.2 we transform this description into a product formula (Corollary 6.9) for the formal power series The product formula is a substantial merit of working with e i,(r s ) , and it facilitates the extension of the case s = 1 to s > 1.This is done in Subsection 6.3, where the result for s = 1 is used to obtain the general case.
6.1.Formulas for inner products.In this subsection we obtain explicit formulas for the inner products e k,(r s ) := ψ (r s ) , χ Recall that, by Equation ( 9), the sequences {m k } n−1 k=0 and {e k } n k=0 determine each other, via the relations where m k := 0 for k = −1 and k = n.Nota bene, these multiplicities depend on r and s but this dependence is suppressed in the notation.Definition 6.1.For given non-negative integers i, r and s, let P r,s (i) := γ = (γ 1 , . . ., γ s ) : The partition γ = (5, 3, 3, 2, 0) ∈ P 6,5 (13) denote the set of all partitions of i into at most s parts, each of size at most r.Denote the multiplicity of j in γ ∈ P r,s (i) by k j (γ Example 6.2.Let r = 6, s = 5 and i = 13.Then γ = (5, 3, 3, 2, 0) ∈ P 6,5 (13) is a partition of 13 with at most 5 parts, each of size at most 6; see Figure 1.The multiplicities of the parts are k 0 Recall f j (r) from Definition 5.3.The main result of this subsection is the following formula.Theorem 6.3.For every s ≥ 1 and i ≥ 0 we have In particular, for s = 1 we have e i,(r) = f i (r).
Remark 6.4.The special case s = 1 was stated (but not proved) in Proposition 5.5.The result in that case is not new, as noted in Remark 5.6.This case shows that f i (r) = e i,(r) is an inner product of two characters, and is therefore always a non-negative integer.The factor is therefore also a non-negative integer, and is zero if and only if either j is odd and k j > f j , or j is even and k j > 0 = f j .If k j = 0, this factor is equal to 1 and may be ignored.
In order to prove Theorem 6.3 we need a formula for a certain inner product of characters (Lemma 6.6).
First recall some notations from Definition 1.5: the centralizer Z (r s ) ∼ = Z r ≀ S s of an element of cycle type (r s ), the linear character ω (r s ) on Z (r s ) , and the higher Lie character ψ (r s ) := ω (r s ) ↑ Sn Z (r s ) .
Embed Z (r s ) ∼ = Z r ≀ S s into K r,s ∼ = S r ≀ S s ≤ S n , where Z r ≤ S r is generated by a full cycle.Denote φ r,s := ω (r s ) ↑ Kr,s Z (r s ) , so that ψ (r s ) = φ r,s ↑ Sn Kr,s .Observation 6.5.If s = s 1 + s 2 then Z (r s 1 ) × Z (r s 2 ) ≤ Z (r s ) , K r,s 1 × K r,s 2 ≤ K r,s and also for the characters In the lemma below we express the multiplicity of a certain linear character in a restriction of φ r,s .This expression will be used, in the proof of Theorem 6.3, to compute e i,(r s ) .
For every 0 ≤ j ≤ r, let R r,j := S j × S r−j ≤ S r , in the natural embedding.Then R r,j ≀ S s = K r,s ∩ S js × S (r−j)s .
Denote by 1 Sn the trivial character and by ε Sn the sign character of S n , so that .
It is a linear character on R r,j ≀ S s .
Lemma 6.6.For every 0 ≤ j ≤ r, f j (r) is a non-negative integer and , if j is even.
Remark 6.7.As a byproduct, Lemma 6.6 provides a new proof of the non-negativity of f j (r).Indeed, if s = 1 then φ r,1 R r,j , ν r,j,1 = f j (r) is clearly a non-negative integer.The rest of this subsection consists of the proofs of Lemma 6.6 and Theorem 6.3.In these proofs r, s and j are fixed, unless specified otherwise.For convenience, we omit the indices and write Z := Z (r s ) , ω := ω (r s ) , ψ := ψ (r s ) , K := K r,s , φ := φ r,s , R := R r,j , and ν := ν r,j,s , Proof of Lemma 6.6.The proof consists of two parts.First we determine the character values of the induced character φ = ω ↑ K Z on the wreath product K = S r ≀ S s ; the resulting formula is Equation (17).In the second part we apply this formula to compute the inner product.
Let ζ : Z r → C be the primitive linear character used to define ω; see Definition 1.5.Recall the explicit formula for an induced character [19, (5.1)]:For a subgroup H ≤ G and a character χ of H, define χ 0 : G → C by χ 0 (g) = χ(g) if g ∈ H and χ 0 (g) = 0 otherwise.Then (16) χ where T is a full set of right coset representatives of H in G.
An element of K = S r ≀ S s can be represented by an s-tuple of elements of S r and a wreathing permutation from S s , so that K = {(x 1 , . . ., x s ; σ) | x 1 , . . ., x s ∈ S r , σ ∈ S s } with the product (x 1 , . . ., x s ; σ)(y 1 , . . ., y s ; τ ) = (x 1 y σ −1 (1) , . . ., x s y σ −1 (s) ; στ ).A full set of right coset representatives of Z r in S r is S r−1 (in the natural embedding).Hence, a full set of right coset representatives of For any z 1 , . . ., z s ∈ Z r , the shifted set (z 1 , . . ., z s ; 1)T is also a full set of right coset representatives.Instead of taking the sum over T in (16), we will take it over the union of all the shifted sets, namely {(x 1 , . . ., x s ; 1)| (∀i) x i ∈ S r }, and divide by r s .Since (x 1 , . . ., x s ; 1) −1 (y 1 , . . ., y s ; σ)(x 1 , . . ., x s ; 1) = (x −1 1 y 1 x σ −1 (1) , . . ., x −1 s y s x σ −1 (s) ; σ), we conclude that, for any y = (y 1 , . . ., y s ; σ) ∈ K, The factors in the product over i are complex numbers, thus commute.We can therefore rearrange them in an order fitting the decomposition of σ −1 ∈ S s into disjoint cycles: if Since ζ is a linear character, cancellation gives Hence if φ(y) = 0 then, necessarily, each cycle-product c k ∈ S r must be conjugate to an element of Z r .Since Z r ≤ S r is generated by a full cycle, a necessary and sufficient condition for c k to have a conjugate in Z r is that it is a product of disjoint cycles of the same length.
For any divisor There are r ℓ k −1 such choices, where ℓ k is the length of the cycle C k .We conclude that, for any y ∈ K for which c k ∈ S r is a product of disjoint If g is a generator of Z r , then o(g m ) = d if and only if m = jr/d for some integer j coprime to d.It follows that Since t k=1 (ℓ k − 1) = s − t, we now have an explicit formula for the values of φ: To determine the inner product φ ↓ K R≀Ss , ν we evaluate the linear character where sgn(σ) denotes the sign of σ ∈ S s .We obtain For each nonzero summand and each cycle C k of σ −1 , by Equation ( 17), the cycle-product , and its restrictions to S j and S r−j have cycle types For any common divisor d of r and j, let n d denote the number of elements of R which are products of disjoint d-cycles.If C k has length ℓ k , then the cycle-product c k can be a product of disjoint d k -cycles in exactly (j!(r − j)!) ℓ k −1 n d k ways.For different cycles C k the choices of the respective d k are independent; the only restriction is d k |(r, j).
Of course, Putting everything together, we obtain In particular, if s = 1 then f = φ ↓ K R , ν is a non-negative integer.Finally, by [32,Proposition 1.3.4],for any s ≥ 0 and indeterminate x, Substituting −x for x and noting that sgn(σ) = (−1) s−cyc(σ) , we get This yields the desired formula, depending on the parity of j, for φ ↓ K R≀Ss , ν .To prove Theorem 6.3 we need a final ingredient, a combinatorial parametrization of (S r ≀ S s , S i × S rs−i ) double cosets of S rs by partitions.
Recall Definition 6.1.The idea and definition of P r,s (i) actually appear already in [16], in a similar context but without explicit reference to double cosets or Mackey's formula; see Definitions 2.8-2.10 and Proposition 2.11 there.
An example of a partition of 13 representing a certain (S 6 ≀ S 5 , S 13 × S 17 ) double coset appears in Figure 1.Lemma 6.8.Let n = rs, K = K r,s ∼ = S r ≀ S s ≤ S n .There is a bijection between the (K, S i × S n−i ) double cosets of S n and P r,s (i), the set of partitions of i into at most s parts, all of size at most r.
Now fix a decomposition for the action of K ∼ = S r ≀ S s : {1, . . ., n} = B 1 ∪ . . .∪ B s , where B j := {(j − 1)r + 1, (j − 1)r + 2, . . ., (j − 1)r + r} (j = 1, . . ., s).The elements of S s permute these blocks, and each of the s copies of S r acts on one of the blocks.Given g ∈ S n , we map the double coset Kg(S i ×S n−i ) to the partition γ which is the non-increasing rearrangement of the sequence This sequence consists of s non-negative integers, each at most r, which sum up to i. Thus γ ∈ P r,s (i).We will show that this map is a bijection.For arbitrary x ∈ K and y ∈ S i × S n−i we have The element x −1 ∈ K = S r ≀ S s permutes the blocks and permutes the elements of each block.This shows that the mapping Kg(S i × S n−i ) → γ is well defined.The mapping from K\S n /(S i ×S n−i ) to P r,s (i) is clearly onto: for each γ = (a 1 , . . ., a s ) ∈ P r,s (i) there exists a permutation g ∈ S n such that |B j ∩ g({1, . . ., i})| = a j (1 ≤ j ≤ s).
Finally, if Kg 1 (S i × S n−i ) and Kg 2 (S i × S n−i ) are mapped to the same partition γ, then there exists a permutation π ∈ S s satisfying Therefore there exist an element x ∈ K = S r ≀ S s such that It follows that g 1 ({1, . . ., i}) = xg 2 ({1, . . ., i}), and therefore g 1 = xg 2 y for a suitable permutation y ∈ S i × S n−i .
We are now ready to prove Theorem 6.3.The proof applies Lemma 6.8 and the explicit bijection described in its proof, combined with Lemma 6.6.

6.2.
A product formula.Now we derive a restatement of Theorem 6.3 as a product formula for formal power series.Corollary 6.9.For a positive integer r, define the formal power series E r (x, y) := i,s≥0 e i,(r s ) x i y s .
Remark 6.10.For small r and arbitrary s, Corollary 6.9 enables us to determine explicitly the hook-multiplicities m i,(r s ) .This is done by recalling Equation ( 9) and the fact that, by definition, e i,(r s ) is the coefficient of x i y s in E r (x, y).For example, by Corollary 6.9 and Observation 5.4, E 2 (x, y) = E 3 (x, y) = (1 + xy)(1 − x 2 y) −1 .Thus, for r ∈ {2, 3} and any s ≥ 1, the value of e i,(r s ) is 1 for i ∈ {2s − 1, 2s} and zero otherwise.Combining this with Equation ( 9), it follows that, for r ∈ {2, 3} and s ≥ 1, the hook multiplicity m i,(r s ) is 1 for i = 2s − 1 and zero otherwise.
Proof of Theorem 1.9.Assume that r is not square-free.Recall, from Definition 5.7, the notation F r (x) := r j=0 f j (r)x j .By Proposition 5.1 and Corollary 5.8 we may write , where G r (x) = r−2 j=0 g j (r)x j is a polynomial with non-negative integer coefficients.Let g j (r) := 0 for j < 0 or j > r − 2. Then f j (r) = g j (r) + 2g j−1 (r) + g j−2 (r) (∀j).
We claim that each factor in this product has the form 1 + (1 + x) 2 p j (x, y), with p j (x, y) a formal power series with non-negative integer coefficients.This implies that (E r (x, y) − 1)/(1 + x) 2 is itself a formal power series with non-negative integer coefficients, completing the proof of Theorem 1.9.Indeed, if j is odd then the corresponding factor is , where x j y (1 − x j+1 y) 2 = x j y • i≥0 (i + 1)(x j+1 y) i is a formal power series with non-negative integer coefficients.
Finally, if j is even then the corresponding factor is , where x j y is a formal power series with non-negative integer coefficients.

Additional results
7.1.Combinatorial identities.In this subsection it will be shown that Lemma 3.5 and Theorem 6.3 imply well-known combinatorial identities.
The following identity is due to Garsia [13].A purely combinatorial proof was given by Wachs [40].
Proposition 7.1.[13, Equation 5.8] For every partition λ ⊢ n, where ζ is a primitive n-th root of unity and µ is the Möbius function.
The following lemma follows from [34].There is a combinatorial description, due to M. Schocker, of the multiplicity of an arbitrary irreducible character of S n in the higher Lie character.In its full generality it is too complicated to be presented here, see [29] for details.The special case of a full cycle, λ = (n) (for which ψ Zn is the Lie character), is due to Kraśkiewicz and Weyman [21].
Corollary 7.5.For every 0 ≤ k ≤ n, the multiplicity m k,(n) := ψ (n) , χ (n−k,1 k ) is equal to the cardinality of the set k is a bijection.The major index of a tableau is the sum of the elements of its descent set.
Consider the following combinatorial identity.Proposition 7.6.For every 0 ≤ k ≤ n, c k,t x k q t mod (q n − 1), the cardinality we are interested in is the coefficient c k,1 of x k q.For any d|n and η a primitive d-th root of unity (we write o where the last equality holds since both sides are polynomials of the same degree, with exactly the same roots and the same constant term.
Let ω be a primitive n-th root of unity.Then as required.
In light of Corollary 7.5 one observes Observation 7.7.Proposition 5.5 is equivalent to Proposition 7.6.
Proof.Comparing Corollary 7.5 with Proposition 5.5 yields The opposite direction is similar.
Remark 7.8.Noting that Proposition 5.5 is the special case s = 1 of Theorem 6.3, one concludes that Proposition 7.6 is a consequence of the latter.
It remains a challenge to find such a direct link between Schocker's general description of the multiplicity and our version in Theorem 6.3.Remark 7.9.Proposition 5.5 is not new.For example, by the Gessel-Reutenauer Theorem (Theorem 2.11) together with Observation 3.1, Theorem 6.3 at s = 1 is equivalent to the following equation which is an immediate consequence of a recent result of Elizalde and Troyka [12,Theorem 3.1].An older proof was presented to us by Sheila Sundaram [37], deducing Proposition 5.5 from [35,Lemma 2.7].The reader is referred to [12] for further discussion and relations to the enumeration of Lyndon words.
7.2.Cellini's cyclic descents.In this subsection it is shown that the natural approach does not provide a cyclic descent extension for conjugacy classes in S n .
Recall the notion of cyclic descent set on permutations which was defined by Cellini [8] CDes(π) := {1 ≤ i ≤ n : π i > π i+1 }, with the convention π n+1 := π 1 .Special subsets of S n , for which Cellini's cyclic descent set map is closed under cyclic rotation, thus determines a cyclic extension of Des, are presented in [6].However, it seems that there are only two conjugacy classes (2-cycles in S 3 and 3-cycles in S 4 ) for which Cellini's cyclic descent map is closed under cyclic rotation modulo n.Hence this cyclic extension does not serve our purposes.
It is unsettled whether there are other conjugacy classes, for which Cellini's CDes map is closed under cyclic rotation.Below are some partial results.Proposition 7.10.For every n > 1 and conjugacy class of cycle type (r s ), (n = rs), Cellini's cyclic descent map is not is closed under cyclic rotation modulo n.
Proof.Recall the notation C λ from Section 2. For r = 1, C (1 n ) consists of the identity permutation only, whose Cellini's cyclic descent set is {n}.Thus not closed under rotation.For r > 1 let σ = [s + 1, s + 2, . . ., n, 1, 2, . . ., s], in other words σ is the permutation in S n defined by ).Then σ is a product of s disjoint cycles of length r and Cellini's CDes of σ is the singleton {n−s}.By equivariance property there must be a permutation of cycle type (r s ) with cyclic descent set {n}.The only permutation in S n with this property is the identity permutation, contradiction.namely, for every 3 < i ≤ x + 1, π(i) = i − 2 and for every x + 1 < i ≤ n, π(i) = i − 1.Thus π has no fixed points unless x = 2, in this case π has cycle type (n − 1, 1).One deduces that statement holds for k < n − 1.

Final remarks and open problems
Recall the notation m k,λ := ψ λ , χ (n−k,1 k ) .By Proposition 5.2, the hook-multiplicity sequence m 0,(n) , m 1,(n) , . . ., m n−1,(n) is unimodal; We conjecture that it is unimodal for all partitions λ ⊢ n.The conjecture has been verified for all partitions of size n ≤ 15 and for all partitions of rectangular shape (r s ) with r ≤ 40 and s ≤ 5.
Note that, by Lemma 5.12, Conjecture 8.1 can provide an alternative proof of Theorem 1.9.
A sequence a 0 , . . ., a n of real numbers is log-concave if a i−1 a i+1 ≤ a 2 i for all 0 < i < n.It is not hard to show that a log-concave sequence with no internal zeros is unimodal, see e.g.[7,31].Since log-concavity implies unimodality, it is tempting to check whether the hook-multiplicity sequence is log-concave.Conjecture 8.2.For every partition λ = (r s ) with even r = 6, the hook-multiplicity sequence m 0,λ , m 1,λ , . . ., m n−1,λ is log-concave.Conjecture 8.2 was checked for all r ≤ 40 and s ≤ 5.
Our proof of Theorem 1.4 is not constructive.conclude the paper with the following challenging problem.Problem 8.3.Find an explicit combinatorial description of the cyclic descent extension on the conjugacy class of cycle type µ, not equal to (r s ) for a square-free r.
A solution of this problem for the conjugacy classes of involutions is presented in [4].The analogous problem for standard Young tableaux of fixed non-ribbon shape was solved by Huang [17].

1. 2 . 2 .
Hook multiplicities and cyclic descent extensions.Recall the standard notation s λ for the Schur function indexed by a partition λ, as well as F n,D for the fundamental quasisymmetric function indexed by a subset D ⊆ [n − 1]; see Definition 2.1.A subset A ⊆ S n is Schur-positive if the associated quasisymmetric function Q(A) := a∈A F n,Des(a) , is symmetric and Schur-positive.

Lemma 1 . 6 .
A Schur-positive set A ⊆ S n has a cyclic descent extension if and only if the following two conditions hold: (divisibility) the polynomial n−1 k=0 m k,A x k is divisible by 1 + x; (non-negativity) the quotient has nonnegative coefficients.See Lemma 3.2 below.1.2.3.Divisibility.By the Gessel-Reutenauer theorem, for every conjugacy class C λ the quasisymmetric function Q(C λ ) is the Frobenius image of the higher Lie character ψ λ , thus C λ is Schurpositive; see Theorem 2.6 below.For a partition λ ⊢ n denote(1)

2. 1 .Definition 2 . 1 .
Quasisymmetric functions and descents.A symmetric function is called Schurpositive if all the coefficients in its expansion in the basis of Schur functions are non-negative.Recall the notation s λ/µ for the Schur function indexed by a skew shape λ/µ.For each subset D ⊆ [n − 1] define the fundamental quasisymmetric function F n,D (x) :

Theorem 2 . 6 and
Theorem 2.7 imply Theorem 2.11.[15, Theorem 2.1] For every conjugacy class C λ of cycle type λ ⊢ n the descent set map Des has fiber sizes given by |{π ∈ C λ : Des
February 9, 2023.PH was partially supported by Hungarian National Research, Development and Innovation Office (NK-FIH) Grant No. K115799 and by Bar-Ilan University visiting grant.The project leading to this application has received funding from the European Research Council (ERC) under the European Union's Horizon 2020 research and innovation programme, Grant agreement No. 741420.RMA and YR were partially supported by the Israel Science Foundation, grant no.1970/18 and by an MIT-Israel MISTI grant. Date: appears in a lower row of T than i}.
|µ| ×S |ν| .By the Littlewood-Richardson rule [33, Theorem A1.3.3], the outer product of two irreducible characters χ α ⊗ χ β ↑ Sn Sm×S n−m contains irreducible representations indexed by hooks if and only if both α and β are hooks; in the latter case, 6.3 below.It also follows from a well known result of Kraśkiewicz and Weyman [21] (Lemma 7.5 below).A symmetric functions proof which applies [35, Lemma 2.7] was presented by Sheila Sundaram [37].Another proof follows from [12, Theorem 3.1].See Subsection 7.1 below for a discussion.Definition 5.7.For a fixed positive integer r, collect the multiplicities f j th root of unity.By varying the conjugating element x a k , each element of Z r of order d is obtained with the same multiplicity |Z (d r/d ) | = (r/d)!dr/d .The other x i 's, for i ∈ C k \ {a k }, are arbitrary, as long as x −1 i