Determinantal formulas with major indices
aa r X i v : . [ m a t h . C O ] F e b DETERMINANTAL FORMULAS WITH MAJOR INDICES
THOMAS MCCONVILLE, DONALD ROBERTSON, CLIFFORD SMYTH
Abstract.
We give a simple proof of a major index determinant formula in the symmet-ric group discovered by Krattenthaler and first proved by Thibon using noncommutativesymmetric functions. We do so by proving a factorization of an element in the groupring of the symmetric group. By applying similar methods to the groups of signed per-mutations and colored permutations, we prove determinant formulas in these groups asconjectured by Krattenthaler. Introduction
Let S n be the group of permutations of [ n ] := { , . . . , n } . We often consider permutationsin one–line notation w = w · · · w n where w i = w ( i ). An integer i ∈ [ n −
1] is a descent of apermutation w if w i > w i +1 . The major index maj( w ) is the sum of the descents of w . Forexample, maj(314652) = 1 + 4 + 5 = 10. The major index matrix is ( q maj( uv − ) ) u,v ∈ S n . Inhis survey of determinantal formulas, Krattenthaler discovered and communicated a proofby Thibon of the following identity. Theorem 1.1 (Theorem 56, [Kra01]) . For all n ≥ (cid:16) q maj( uv − ) (cid:17) u,v ∈ S n = n Y k =2 (1 − q k ) n ! · ( k − /k . Example 1.2.
The identity is trivial if n = 1. For n = 2, Theorem 1.1 givesdet (cid:18) qq (cid:19) = (1 − q ) . For n = 3, if we index the rows and columns by (123 , , , , , q q q q q q q q q q q q q q qq q q q q q q q q qq q q q q = (1 − q ) (1 − q ) . To prove Theorem 1.1, Thibon explicitly determined the eigenvalues of the major indexmatrix with multiplicity using the theory of noncommutative symmetric functions developedin [KLT97]. Stanley also determined the eigenvalues with multiplicity in [Sta01, Theorem2.2] by applying a theorem of Bidigare, Hanlon, and Rockmore [BHR +
99, Theorem 1.2].In this paper, we present a new, simpler proof of Theorem 1.1. Our proof relies on aclever interpretation of the major index of a permutation given by Adin and Roichman in[AR01], which we recall here. Let t k = ( k, k − , . . . ,
1) be a k -cycle for 2 ≤ k ≤ n . Each w in S n can be uniquely expressed in the form t c n n t c n − n − · · · t c where 0 ≤ c k < k , and the majorindex of w is c n + c n − + · · · + c . This means that the sequence ( t n , t n − , . . . , t ) is a perfectbasis of S n , the definition of which we recall in Section 3. This perfect basis determinesa factorization of the major index matrix, which we use to evaluate its determinant inSection 4. Date : February 26, 2021. ur proof of Theorem 1.1 was motivated by Zagier’s proof [Zag92] of the identitydet (cid:16) q inv( uv − ) (cid:17) u,v ∈ S n = n Y k =2 (1 − q k − k ) n ! · ( n − k +1) / ( k − k ) , where inv( w ) is the number of inversions of a permutation w . Zagier considered an elementof the group algebra C ( q ) S n whose image under the regular representation is the matrix( q inv( uv − ) ). By factoring this element of the group algebra, he obtained a correspondingfactorization of the matrix ( q inv( uv − ) ) u,v ∈ S n for which the determinants of the factors couldbe readily evaluated.A colored permutation ( w, x ) consists of a permutation w ∈ S n and x ∈ ( Z /m Z ) n . Werecall the group structure on colored permutations in Section 5. When m = 2, this group isisomorphic to the group of signed permutations, the real reflection group of type B n .Based on extensive computational evidence, Krattenthaler conjectured in [Kra05] severalanalogues of Theorem 1.1 for colored permutations using variations on the major index givenin [ABR01, AR01, Rei93]. We prove all of his conjectured formulas and more in Sections 5, 6,and 7. For each case, we construct a basis such that the relevant statistics can be read fromthe exponent vectors.Theorem 1.1 and the various extensions we consider for colored or signed permutationsare all specializations of group determinants. For a finite group G , its group determinant is det( r gh − ) g,h ∈ G where { r g | g ∈ G } is a set of elements of a commutative ring. Inhis pioneering work on the representation theory of finite groups, Frobenius proved that if { r g | g ∈ G } is a set of indeterminates in a polynomial ring over C , then the irreduciblefactors of the group determinant naturally correspond to irreducible representations of G ;see [Haw78].We consider examples of the group determinant of the form det( q stat( gh − ) ) g,h ∈ G forsome statistic on G . We are especially intrigued by examples for which this determinant isa product of binomials. This behavior was proved for the length statistic on finite Coxetergroups in [Var93], extending the aforementioned result from [Zag92].The rest of this paper is structured as follows. Preliminary results on group determinantsand perfect bases are given in Sections 2 and 3. Theorem 1.1 is proved in Section 4.In Section 5, we recall the flag major index on colored permutations introduced by Adinand Roichman in [AR01] and prove a formula for the corresponding group determinant,answering [Kra05, Problem 49]. In Section 6, we consider another statistic on coloredpermutations that we call the absolute flag major index, and we generalize and prove [Kra05,Conjecture 48]. Finally, in Section 7 we prove an identity from which we can derive proofsof Conjectures 46, 47, and 50 in [Kra05].2. Group determinants
Let G be a finite group and let R be a commutative ring with 1. The group ring RG is the free R -module with a distinguished basis that we identify with the elements of G .Multiplication of basis elements is the same as in G and is extended linearly to RG .Given a complex vector space V ∼ = C m , a representation is a group homomorphism G → GL( V ), which we may extend to a ring homomorphism RG → End( V ). For anyelement α = P g ∈ G r g g ∈ RG and any representation φ , we set∆ φ ( α ) := det φ ( α ) = det X g ∈ G r g φ ( g ) . Example 2.1. (1) If φ triv : G → GL( C ) is the trivial representation, then ∆ φ triv ( α ) = P g ∈ G r g .(2) If G acts on a finite set X , then the permutation representation φ X : G → C X assigns to an element g the transformation x g · x for x ∈ X . So φ X ( g ) is a ermutation matrix, i.e. it is a 0 , regular representation φ reg is the permutation representation induced by theaction of G on itself by left multiplication. The group determinant of G is∆ φ reg ( α ) = det ( r uv − ) u,v ∈ G where α = P r g g .Two representations φ, η are equivalent if there exists an invertible matrix U such that η ( g ) = U φ ( g ) U − for all g ∈ G . It is clear that ∆ φ ( α ) = ∆ η ( α ) whenever φ and η are equivalent representations. The direct sum φ ⊕ η of two representations satisfies∆ φ ⊕ η ( α ) = ∆ φ ( α )∆ η ( α ). As any representation is equivalent to a direct sum of irreduciblerepresentations, we can always factor the determinant ∆ φ ( α ) as a product over the irre-ducible direct summands of φ .For an m × m matrix M , let θ M ( q ) = det( I − qM ). For permutation representations, wehave the following result. Proposition 2.2.
Let G act on a finite set X , and let φ = φ X be the correspondingpermutation representation. Fix g ∈ G , and let O , . . . , O N be the orbits of the cyclicsubgroup h g i . Then θ φ ( g ) ( q ) = N Y k =1 (1 − q |O k | ) . Proof.
For k ∈ [ N ], let T k be the |O k | × |O k | permutation matrix with ( i, j )-entry equal to1 if i = j + 1 mod |O k | . Up to a simultaneous permutation of rows and columns, the matrix φ ( g ) is the direct sum T ⊕ · · · ⊕ T N . Hence, θ φ ( g ) ( q ) = det( I − qφ ( g )) = N Y k =1 det( I − qT k ) = N Y k =1 (1 − q |O k | ) . (cid:3) Let o ( g ) denote the order of an element g . In the regular representation, every orbit of h g i has size equal to o ( g ). We immediately deduce the following corollary of Proposition 2.2. Corollary 2.3. If φ = φ reg is the regular representation of G , then θ φ ( g ) ( q ) = (1 − q o ( g ) ) | G | /o ( g ) . Perfect bases
We follow the terminology of [ARS07] for bases of groups.Let G be a finite group. A sequence ( g , . . . , g n ) of elements of G is a basis if there existpositive integers m , . . . , m n such that every element g ∈ G may be uniquely expressed inthe form g = g c · · · g c n n where 0 ≤ c i < m i for i ∈ [ n ]. This basis is perfect if m i = o ( g i )for all i . A group admits a perfect basis if and only if it may be identified as a set with aCartesian product of cyclic groups by the following map. h g i × · · · × h g n i ∼ → G ( g c , . . . , g c n n ) g c · · · g c n n This map is not necessarily a group isomorphism.
Example 3.1.
Let G = S . Take g = (123) , g = (12)(3), written in cycle notation.Then ( g , g ) is a perfect basis since every element of S is uniquely expressible in the form g c g c for 0 ≤ c < , ≤ c < g g = (1)(2)(3) g g = (12)(3) g g = (123) g g = (13)(2) g g = (132) g g = (1)(23) xample 3.2. We extend Example 3.1 to any dihedral group of order 2 n for n ≥ G of isometries of a regular n –gon whose vertices are labeled1 , , . . . , n in clockwise order. As a group of permutations of [ n ], G is generated by a rotation g = (123 · · · n ) and a reflection g = (1 , n − , n − · · · ( ⌊ n − ⌋ , ⌊ n +22 ⌋ )( n ). Every rotationsymmetry is of the form g k for some 0 ≤ k < n , and every reflection symmetry is of theform g k g for some 0 ≤ k < n . Hence, ( g , g ) is a perfect basis.For the remainder of this section, we let R = Q [ x , . . . , x n ] and we consider the groupring RG . Lemma 3.3.
If ( g , . . . , g n ) is a basis of G , then we have the identity X g = g c ··· g cnn ≤ c i The first statement immediately follows by expanding the right hand side. The latterstatement follows from the assumption g m i i = 1. (cid:3) Theorem 3.4. Suppose G has a perfect basis ( g , . . . , g n ), and set α = X g = g c ··· g cnn ≤ c i By Lemma 3.3, we have α n Y i =1 (1 − x i g i ) = n Y i =1 (1 − x m i i ) · . Hence, det( φ ( α )) = det φ ( Q ni =1 (1 − x m i i )1)det φ ( Q ni =1 − x i g i )= n Y i =1 det((1 − x m i i ) I r )det( I r − x i φ ( g i ))= n Y i =1 (1 − x m i i ) r θ φ ( g i ) ( x i ) . (cid:3) Corollary 3.5. Suppose G has a perfect basis ( g , . . . , g n ), and set α = X g = g c ··· g cnn ≤ c i Let G be the dihedral group in Example 3.2. For h ∈ G , set rot( h ) = h ( n ) if h ( n ) = n , and rot( h ) = 0 if h ( n ) = n . Let refl( h ) = 0 if h is a rotation and refl( h ) = 1 if h isa reflection. Then for 0 ≤ c < n, ≤ c < 2, we have rot( g c g c ) = c and refl( g c g c ) = c .Let α = P h ∈ G x rot( h )1 x refl( h )2 · h , and let φ be a representation of G of dimension r . ByTheorem 3.4, we have ∆ φ ( α ) = (1 − x n ) r θ φ ( g ) ( x ) (1 − x ) r θ φ ( g ) ( x ) . If φ = φ reg is the regular representation, then r = 2 n, θ φ ( g ) ( q ) = (1 − q n ) , and θ φ ( g ) ( q ) =(1 − q ) n . Hence, ∆ φ reg ( α ) = (1 − x n ) n − (1 − x ) n . We end this example by evaluating ∆ φ ( α ) for all irreducible representations over Q . Let C n = h g i . • For the trivial representation φ triv we have∆ φ triv ( α ) = (1 − x n )(1 − x )(1 − x )(1 − x ) . • There is a one-dimensional representation φ sign ( g ) = ( − c ( g ) coming from theaction on the cosets of C n . We have∆ φ sign ( α ) = (1 − x n )(1 − x )(1 − x )(1 + x ) . • For each divisor d of n , there is an irreducible representation ρ d of C n . WritingΦ d ( q ) = a + a q + · · · + q ℓ for the d th cyclotomic polynomial, we can realize ρ d on V = Q ℓ via ρ d ( g ) = · · · − a · · · − a · · · − a · · · − a ... ... ... . . . ... ... ...0 0 0 · · · − a ℓ − · · · − a ℓ − Let A be the anti-diagonal matrix with all anti-diagonal entries equal to 1. Then A is the identity and Aρ d ( g ) = ρ d ( g − ) A . Hence, we have an irreducible representa-tion φ d of the dihedral group via φ d ( h ) = ρ d ( g ) rot( h ) A refl( h ) . In this representation,one has θ φ d ( g ) ( q ) = q ℓ Φ d (1 /q ) and θ φ d ( g ) ( q ) = ( (1 − q ) ℓ ℓ even(1 − q ) ℓ − (1 − q ) ℓ oddfrom which ∆ φ d ( α ) can be written down. • Lastly, for any divisor d of n we can tensor the representation φ d with the represen-tation ψ to obtain another irreducible representation. . Major index matrix For k ∈ [ n ], let t k = ( k, k − , . . . , 1) be a k -cycle in S n . In [AR01], Adin and Roichmangave an alternative interpretation of the major index of a permutation. We recall thisstatement and its proof. Lemma 4.1 (Claim 2.1, [AR01]) . The ( n − t n , t n − , . . . , t ) is a perfect basisof S n . Moreover, if w = t c n n t c n − n − · · · t c for some c i with 0 ≤ c i < i , then maj( w ) = c n + c n − + · · · + c . Proof. Let w ∈ S n be given. Since t n is the n –cycle ( n, n − , . . . , c n with 0 ≤ c n < n such that t − c n n w fixes n . By similar reasoning, there is a unique c n − with 0 ≤ c n − < n − t − c n − n − t − c n n w fixes n − 1. Since t n − fixes n , the element t − c n − n − t − c n n w also fixes n . Continuing in this manner, we find unique c i with 0 ≤ c i < i suchthat t − c · · · t − c n n w is the identity permutation. Hence, the factorization w = t c n n · · · t c isunique.As there are n ! = | S n | choices for the exponents ( c n , . . . , c ), we conclude that ( t n , . . . , t )is a basis. In fact, it is a perfect basis since the order of t i is i and the exponent c i can beany value in the range 0 ≤ c i < i .For each k , let g k = t c n n · · · t c k k . We prove that the major index of g k is c n + · · · + c k .Taking k = 2 gives maj( w ) = c n + · · · + c .We observe that the values in g n are cyclically ordered , i.e. there exists a unique j ∈ Z /n Z such that g n ( | j | + 1) < · · · < g n ( n ) < g n (1) < · · · < g n ( | j | ). Moreover, | j | = c n , so themajor index of g n is c n .Now let k ≥ k +1 values of g k +1 are cyclically ordered. Multiplying g k +1 on the right by t k rotates the first k values of g k +1 . Hence, the first k values of g k arecyclically ordered.If the first k values of g k +1 are in increasing order, then by the same argument as in thebase case, maj( g k ) = maj( g k +1 t c k k ) = maj( g k +1 ) + c k .Otherwise, there exists j ∈ Z /k Z , j = 0 such that g k +1 ( | j | + 1) < · · · < g k +1 ( k + 1) 1. All higher descents of g k are sharedwith g k +1 . Hence, maj( g k ) = maj( g k +1 ) + c k , as desired. (cid:3) For the remainder of the section, we consider the element α ∈ C ( q ) S n where α = X w ∈ S n q maj( w ) · w. Lemma 4.1 together with Theorem 3.4 immediately implies the following. Corollary 4.2. Let n ≥ V ∼ = C r and φ : S n → GL( V ), then∆ φ ( α ) = n Y k =2 (1 − q k ) r θ φ ( t k ) ( q ) . If φ reg is the regular representation of S n , then φ reg (cid:16)X q maj( w ) · w (cid:17) = (cid:16) q maj( uv − ) (cid:17) u,v ∈ S n . Theorem 1.1 now follows immediately from Corollary 3.5. Example 4.3. The symmetric group S n naturally acts on [ n ]. The corresponding per-mutation representation φ def is called the defining representation . Explicitly, the matrixrepresenting α is φ def ( α ) = X w ( i )= j q maj( w ) i,j ∈ [ n ] . e verify that det X w ∈ S n w ( i )= j q maj( w ) i,j ∈ [ n ] = (1 − q )( n )([ n ]! q ) n − . The left-hand side is ∆ φ def ( α ), sodet X w ∈ S n w ( i )= j q maj( w ) i,j ∈ [ n ] = n Y k =2 (1 − q k ) n θ φ def ( t k ) ( q ) . The element t k has one orbit of size k and n − k orbits of size 1. Hence, θ φ def ( t k ) ( q ) = (1 − q k )(1 − q ) n − k . Hence, the determinant is equal to n Y k =2 (1 − q k ) n θ φ def ( t k ) ( q ) = n Y k =2 (1 − q k ) n (1 − q k )(1 − q ) n − k = n Y k =2 ([ k ] q ) n − (1 − q ) k − = (1 − q )( n )([ n ]! q ) n − . Example 4.4. One can consider the action of S n on tuples. For example, here we cal-culate ∆ φ ( α ) where φ is the representation determined by the action of S n on (cid:0) n (cid:1) . FromCorollary 2.3 and Theorem 3.4 it suffices to determine the orbit decomposition of the cycles t k acting on (cid:0) n (cid:1) .For the t k orbit of ( i, j ) ∈ (cid:0) n (cid:1) with i < j there are three possibilities: • k < i in which case t k fixes ( i, j ); • i ≤ k < j in which case the j is fixed by t k and the orbit has size k ; • j ≤ k in which case the orbit is the same as the orbit of ( i, j ) ∈ (cid:0) k (cid:1) .It therefore suffices to determine the orbit structure of the action of t k on (cid:0) k (cid:1) . When k isodd all orbits have size k . When k is even the possibility 2( j − i ) = k gives the unique orbitof size k . We therefore have, for the three possibilities above: • (cid:0) n − k (cid:1) orbits of size 1; • n − k orbits of size k ; • k − orbits of size k when k is odd or k − orbits of size k and one orbit of size k when k is even;and can calculate that θ φ ( t k ) ( q ) = ( (1 − q )( n − k )(1 − q k ) n − k (1 − q k ) k − k odd(1 − q )( n − k )(1 − q k ) n − k (1 − q k ) k − (1 − q k ) k evenwith the determinant formula following from Theorem 3.4.We are interested in the extent to which ∆ φ λ ( α ) = det φ λ ( α ) can be calculated where λ is any partition of n ∈ N and φ λ is the corresponding irreducible representation of S n . Thefactorization n Y i =2 (1 + qt i + · · · + q i − t i − i )(1 − qt i ) = n Y i =2 (1 − q i ) · x i = q and g i = t i in (1), together with the fact that t n − , . . . , t is a perfect basisof S n − , suggests an inductive approach. Indeed, if λ is a partition of n , the restriction φ λ | S n − of φ λ to S n − is known to be a direct sum φ λ | S n − = M η ≺ λ φ η f those φ η where η immediately precedes λ in the Young lattice. Thus, for 2 ≤ i ≤ n − φ λ (1 − qt i ) = Y η ≺ λ det φ η (1 − qt i )and it remains to calculate θ φ λ ( t n ) ( q ) = det φ λ (1 − qt n ). For this calculation it suffices todetermine the eigenvalues of φ λ ( t n ). These can be found using work of Stembridge [Ste89,Theorem 3.3] which we recall here.Fix a partition λ of n and g ∈ S n of order m . The eigenvalues of φ λ ( g ) are of the form ω e , . . . , ω e r where ω = e πi/m . The exponents e , . . . , e r are called the cyclic exponents of g and are defined modulo m .A standard tableau over λ is any filling of λ by { , . . . , n } with rows and columns strictlyincreasing. One calls 1 ≤ k ≤ n a descent of a standard tableau if k + 1 appears in a rowstrictly below that of k .Let µ = ( µ , µ , . . . , µ ℓ ) be the cycle type of our element g ∈ S n . Form b µ = (cid:18) mµ , mµ , . . . , m, mµ , mµ , . . . , m, . . . (cid:19) which is a tuple of length µ + · · · + µ ℓ . For example b (4 , , , = (3 , , , , , , , , , , , , g is an n -cycle we have b ( n ) = (1 , , . . . , n ). For any standard tableau T over λ its µ index is ind µ ( T ) = X k ∈ D ( T ) b µ ( k ) mod m where D ( T ) is the set of descents of T . The content of [Ste89, Theorem 3.3] is that q e + · · · + q e r = X T | λ q ind µ ( T ) modulo 1 − q m . Example 4.5 (The standard representation) . The standard representation corresponds tothe partition λ = [ n − , φ λ ( t n ). The standardtableaux over λ are indexed by the entry 2 , . . . , n on the second row. Each has a singledescent of 1 , . . . , n − φ λ ( t n ) has eigenvalues ω, ω , . . . , ω n − and that its characteristic polynomial is [ n ] q . Then θ φ λ ( t n ) ( q ) = q n − det φ λ ( q − t n ) = [ n ] q as well. For all 2 ≤ i ≤ n − θ φ λ ( t i ) ( q ) = (1 − q ) n − i θ φ [ i − , ( t i ) ( q ) = (1 − q ) n − i [ i ] q from repeated application of (2). We conclude that∆ λ ( α ) = n Y i =2 (1 − q i ) n − n Y i =2 (1 − q ) n − i [ i ] q = 1[ n ] q ! n Y i =2 (1 − q i ) n − n Y i =2 (1 − q ) n − n Y i =2 (1 − q ) i − = ([ n ] q !) n − (1 − q )( n )which can also be obtained from dividing the result of Example 4.3 by [ n ] q !. Example 4.6 (The [2 , 2] representation) . Fix λ = [2 , λ are T = 1 23 4 S = 1 32 4with descent sets { } and { , } respectively. The element t has eigenvalues − q − 1. Thus θ φ λ ( t ) ( q ) = q det φ λ ( q − t ) = 1 − q nd θ φ λ ( t ) ( q ) = θ φ [2 , ( t ) ( q ) = [3] q θ φ λ ( t ) ( q ) = θ φ [2 , ( t ) ( q ) = (1 − q )[2] q from the previous example. Finally∆ λ ( α ) = (1 − q ) (1 − q ) (1 − q ) (1 − q ) · (1 + q + q ) · (1 − q )(1 + q ) = (1 − q )(1 − q )(1 − q ) Flag major index matrix Let H, N be groups such that H acts on N on the right. The semidirect product H ⋉ N is the group whose elements are ( g, x ) for g ∈ H, x ∈ N , where( g, x )( h, y ) = ( gh, ( x · h ) y ) . The symmetric group S n acts on ( Z /m Z ) n by permuting coordinates. That is, if w ∈ S n and x ∈ ( Z /m Z ) n , then x · w ∈ ( Z /m Z ) n where ( x · w ) i = x w ( i ) . The group ofcolored permutations is the semidirect product S mn = S n ⋉ ( Z /m Z ) n . We express a coloredpermutation ( w, x ) by writing w in one–line notation with x k bars above w ( k ). For example,the colored permutation (1342 , (1 , , , b = (1 , , , . . . , ∈ ( Z /m Z ) n . As in Section 4, we set t k = ( k, k − , . . . , ∈ S n .In particular, we let t be the identity permutation.For k ∈ [ n ], let ˜ t k = ( t k , b ) ∈ S mn . Then the order of ˜ t k is mk . For example, if m = n = k = 3, then h ˜ t k i = { , ¯312 , ¯2¯31 , ¯1¯2¯3 , ¯¯3¯1¯2 , ¯¯2¯¯3¯1 , ¯¯1¯¯2¯¯3 , , } . Colored letters are totally ordered as n > ( n − > · · · > > ¯ n > · · · > ¯1 > ¯¯ n > · · · giving rise to a major index for colored permutations. For example, the colored permutation¯13¯¯4¯2 only has a descent at 2 since 3 > ¯¯4 but ¯1 < < ¯2. So, the major index ismaj(¯13¯¯4¯2) = 2.The flag major index of a colored permutation g ∈ S mn is fmaj( g ) = m maj( g ) + col( g ).For example, if m = 3, then fmaj(¯13¯¯4¯2) = 3 · Lemma 5.1. The n –tuple (˜ t n , . . . , ˜ t ) is a perfect basis of S mn . Moreover, if g = ˜ t c n n · · · ˜ t c for some 0 ≤ c k < mk , then fmaj( g ) = c n + · · · + c . Example 5.2. Consider the colored permutation g = ¯13¯¯4¯2. The only power of ˜ t with ¯2 inthe last position is (˜ t ) = ¯¯3¯¯4¯1¯2. This permutation has no descents, so its flag major indexis fmaj(˜ t ) = 3 maj(˜ t ) + col(˜ t ) = 0 + 6 = 6 . Next, we rotate the first three entries once to put ¯¯4 into the third position, i.e. ˜ t ˜ t = ¯¯1¯¯3¯¯4¯2.There are still no descents, and its flag major index is 7. Rotating the first two entries twicewill put 3 into the second position, i.e. ˜ t ˜ t ˜ t = 13¯¯4¯2. The colors are removed from the firsttwo values, but a descent at 2 is created, sofmaj(˜ t ˜ t ˜ t ) = 3 · . Finally, we change the color of the first entry to find g = ˜ t ˜ t ˜ t ˜ t and fmaj( g ) = 10 is thesum of the exponents of this factorization. heorem 5.3. Let n, m ≥ (cid:16) q fmaj( gh − ) (cid:17) g,h ∈ S mn = n Y k =1 (1 − q mk ) n ! m n (1 − / ( mk )) Proof. Set α = P q fmaj( g ) · g . If φ reg is the regular representation of S mn , then φ reg ( α ) = (cid:16) q fmaj( gh − ) (cid:17) g,h ∈ S mn . By Lemma 5.1 and Corollary 3.5,∆ φ reg ( α ) = n Y k =1 (1 − q o (˜ t k ) ) | S mn | (1 − /o (˜ t k )) = n Y k =1 (1 − q mk ) n ! m n (1 − / ( mk )) (cid:3) We identify S n with the subgroup of colored permutations { ( , w ) ∈ S mn | w ∈ S n } .Observe that for g ∈ S mn , we have g ∈ S n if and only if col( g ) = 0. If h is any coloredpermutation, there is a unique ordering of the colored values of h with no descents. That is,there is a unique g ∈ S n such that maj( hg ) = 0. Hence, the set T = { h ∈ S mn | maj( h ) = 0 } is a left transversal to S n in S mn . Lemma 5.4. Let T be the transversal to S n in S mn defined above. Then X p maj( g ) q col( g ) · g = X h ∈ T q col( h ) · h ! X w ∈ S n p maj( w ) · w ! . Proof. We first expand the right–hand side of the equation. Then X h ∈ T q col( h ) · h ! X w ∈ S n p maj( w ) · w ! = X g ∈ S mn p maj( w ) q col( h ) · g, where in the latter sum, g = hw, h ∈ T , and w ∈ S n . Fix g ∈ S mn , and decompose g = hw accordingly. Since multiplication by w on the right rearranges colors without changing theirvalues, it is clear that col( g ) = col( h ). On the other hand, since the colored values of h are inincreasing order, it follows that w and g have the same descents. Hence, maj( g ) = maj( w ).We conclude that p maj( w ) q col( h ) = p maj( g ) q col( g ) , as desired. (cid:3) Let H be a subgroup of G . For g ∈ G , the subgroup H acts on the right as the regularrepresentation on the vector space Q [ gH ]. Hence, the restriction of the regular representa-tion of G is isomorphic to a direct sum of [ G : H ] copies of the regular representation of H . Theorem 5.5. For m, n ≥ (cid:16) p maj( gh − ) q col( gh − ) (cid:17) g,h ∈ S mn = n Y k =2 (1 − p k ) n ! m n ( k − /k n Y k =1 (1 − q mk ) n ! m n − ( m − /k . Proof. Let α = P q fmaj( g ) · g as in the proof of Theorem 5.3. Let β = P p maj( g ) q col( g ) · g in Q ( p, q ) S mn . Then φ reg ( β ) = (cid:16) p maj( gh − ) q col( gh − ) (cid:17) g,h ∈ S mn , so we seek to prove that the right–hand side of the theorem statement is equal to ∆ φ reg ( β ).By Lemma 5.4, we have β = (cid:0)P h ∈ T q col( h ) · h (cid:1) (cid:0)P w ∈ S n p maj( w ) · w (cid:1) . Hence, there existspolynomials A ( p ) , B ( q ) such that ∆ φ reg ( β ) = A ( p ) B ( q ). Since α = β | p = q m , we have∆ φ reg ( α ) = A ( q m ) B ( q ), so A ( q m ) B ( q ) = n Y k =1 (1 − q mk ) n ! m n (1 − / ( mk )) . ince S n is a subgroup of S mn of index m n , the restriction of the regular representationof S mn to S n is isomorphic to a direct sum of m n copies of the regular representation of S n .Combined with Theorem 1.1, we have A ( p ) = ∆ φ reg X w ∈ S n p maj( w ) · w ! = n Y k =2 (1 − p k ) n ! m n ( k − /k . Therefore, B ( q ) = ∆ φ reg ( β ) A ( q m )= Q nk =1 (1 − q mk ) n ! m n (1 − / ( mk )) Q nk =2 (1 − q mk ) n ! m n ( k − /k = n Y k =1 (1 − q mk ) n ! m n − ( m − /k . The theorem now follows by multiplying the formulas for A ( p ) and B ( q ). (cid:3) Absolute flag major index matrix In contrast with Section 5, here we consider a simpler statistic that takes the descentsof ( w, x ) to be those of w . The absolute major index of ( w, x ) is amaj( w, x ) = maj( w ).The absolute flag major index of ( w, x ) is amaj( w, x ) + col( w, x ). We prove the followingidentity. Krattenthaler conjectured the m = 2 case in [Kra05, Conjecture 48]. Theorem 6.1. For all m, n ≥ 1, we havedet (cid:16) p amaj( gh − ) q col( gh − ) (cid:17) g,h ∈ S mn = (1 − q m ) n ! m n − ( m − n n Y k =2 (1 − p k ) n ! m n ( k − /k . To prove Theorem 6.1, we produce a different perfect basis of S mn than the one consideredin Section 5. The construction of this perfect basis can be formulated more generally asfollows.Let H, N be groups such that H acts on N on the right. Consider the semidirectproduct G = H ⋉ N . We identify H and N with the subgroups { ( h, ∈ G | h ∈ H } and { (1 , x ) ∈ G | x ∈ N } , respectively. If ( h , . . . , h k ) is a perfect basis of H and ( x , . . . , x ℓ ) isa perfect basis of N , then ( h , . . . , h k , x , . . . , x ℓ ) is a perfect basis of G .For S mn we combine our perfect basis for S n with one for Z /m Z n . For each i ∈ [ n ], let y ( i ) ∈ ( Z /m Z ) n where ( y ( i ) ) j = ( i = j . It is clear that ( y (1) , . . . , y ( n ) ) is a perfect basis of ( Z /m Z ) n since x = P i | x i | y ( i ) for all x ∈ ( Z /m Z ) n . Hence, ( t n , . . . , t , y (1) , . . . , y ( n ) ) is a perfect basis of S mn . Moreover, if g = t c n n · · · t c ( y (1) ) d · · · ( y ( n ) ) d n is the factorization of g , then amaj( g ) = c n + · · · + c andcol( g ) = d + · · · + d n . Proof of Theorem 6.1. Let φ reg be the regular representation of S mn . The restriction of φ reg to S n is isomorphic to a direct sum of [ S mn : S n ] = m n copies of the regular representationof S n . The restriction to ( Z /m Z ) n is isomorphic to a direct sum of [ S mn : ( Z /m Z ) n ] = n !copies of the regular representation of ( Z /m Z ) n . We deduce the following sequence ofidentities. et (cid:16) p amaj( gh − ) q col( gh − ) (cid:17) g,h ∈ S mn = ∆ φ reg X g ∈ S mn p amaj( g ) q col( g ) · g = ∆ φ reg X w ∈ S n p maj( w ) · w ! ∆ φ reg X x ∈ ( Z /m Z ) n q col( x ) · x = n Y k =2 (1 − p k ) n ! m n ( k − /k n Y i =1 (1 − q m ) n ! m n − ( m − (cid:3) Signed permutations A signed permutation is a pair ( ε, w ) where w ∈ S n and ε ∈ {− , } n . We refer to ε asthe sign vector of the signed permutation ( ε, w ). The symmetric group acts on the set ofsign vectors on the left such that for w ∈ S n , ε ∈ {− , } n , ( w · ε ) i = ε w − ( i ) for all i . Let B n be the group of signed permutations, i.e. the semidirect product { , − } n ⋊ S n where( ε, u )( ε ′ , v ) = ( ε ( u · ε ′ ) , uv ). This is also known in the literature as the hyperoctahedralgroup since it is isomorphic to the group of symmetries of a hyperoctahedron.We may write a signed permutation in one–line notation with a bar above a value i if ε i = − 1. For example, the signed permutation (cid:18) (1 , − , − , , (cid:18) (cid:19)(cid:19) would be written as ¯214¯3. We refer to elements of { , , , . . . , ¯1 , ¯2 , ¯3 , . . . } as signed letters.We consider two total orderings on signed letters. The first ordering is the naturalordering on integers, · · · < A ¯ n < A · · · < A ¯1 < A < A < A · · · < A n < A · · · . The second ordering is0 < B < B · · · < B n < B · · · < B ¯ n < B · · · < B ¯2 < B ¯1 . For i ∈ [ n − i is A–descent of a signed permutation ( ε, w ) = w · · · w n if w i > A w i +1 . Furthermore, 0 is an A–descent if w is negative. Similarly, i is a B–descent of w = w · · · w n if w i > B w i +1 . Furthermore, n is a B–descent if w n is negative. Let maj A ( ε, w )(respectively, maj B ( ε, w )) be the sum of the A–descents (respectively, B–descents) of ( ε, w ).The negative set is Neg( ε, w ) = { i | ε i = − } . Let neg( ε, w ) = | Neg( ε, w ) | , and letsneg( ε, w ) be the sum of elements in Neg( ε, w ).For example, { , } is the set of A–descents of ¯214¯3, so maj A (¯214¯3) = 3. The set ofB–descents of ¯214¯3 is { , } , so maj B (¯214¯3) = 5. The negative set is Neg(¯214¯3) = { , } , soneg( ε, w ) = 2 and sneg( ε, w ) = 5.The statistic maj B was introduced by Reiner in [Rei93]. The statistics maj A and snegwere used by Adin, Brenti, and Roichman in [ABR01] to prove a Carlitz-type formula for ajoint Euler–Mahonian distribution in Type B n .The statistics maj A , maj B , and neg are related as follows. Lemma 7.1. For any signed permutation ( ε, w ),maj B ( ε, w ) = maj A ( ε, w ) + neg( ε, w ) . Proof. Let ( ε, w ) = w · · · w n in one–line notation. Set w = 0 and w n +1 = n + 1. Then for i ∈ { , , . . . , n } , i is an A–descent if w i > A w i +1 and i is a B–descent if w i > B w i +1 . Let X be the set of A–descents and Y be the set of B–descents of ( ε, w ). There is a bijection φ : X → Y where φ ( i ) = i if w i and w i +1 have the same sign, and φ ( i ) = min { j | j >i, w j +1 > } if w i and w i +1 have different signs. We observe the identity X i ∈ X φ ( i ) − i = neg( ε, w ) , rom which the lemma follows. (cid:3) For X ⊆ [ n ], let q X = Q i ∈ X q i . We prove the following identity. Theorem 7.2. For all n ≥ (cid:16) p maj A ( gh − ) q Neg( gh − ) (cid:17) g,h ∈ B n = n Y k =1 (1 − q kk ) n !2 n − /k n Y k =2 (1 − p k ) n !2 n ( k − /k . Specializing Theorem 7.2 gives the following identities, which are Conjectures 46, 47,and 50 in [Kra05]. Corollary 7.3. If n ≥ 1, thendet (cid:16) p maj A ( gh − ) q neg( gh − ) (cid:17) g,h ∈ B n = n Y k =1 (1 − q k ) n !2 n − /k n Y k =2 (1 − p k ) n !2 n ( k − /k , det (cid:16) q maj B ( gh − ) (cid:17) g,h ∈ B n = n Y k =1 (1 − q k ) n !2 n − /k n Y k =2 (1 − q k ) n !2 n ( k − /k , anddet (cid:16) p maj A ( gh − ) q sneg( gh − ) (cid:17) g,h ∈ B n = n Y k =1 (1 − q k ) n !2 n − /k n Y k =2 (1 − p k ) n !2 n ( k − /k . Proof. For g ∈ B n , q Neg( g ) specializes to q neg( g ) by setting q i = q for all i . This gives thefirst identity. The second follows from the first by setting p = q . For the third identity, weobserve that q Neg( g ) specializes to q sneg( g ) by setting q i = q i for all i . (cid:3) To prove Theorem 7.2, we construct a certain basis for B n . This basis is not perfect for n ≥ 2, but it is “close enough” for our purposes.Let ε ( k ) ∈ { , − } n where ( ε ( k ) ) j = − k = j and ( ε ( k ) ) j = 1 if k = j . We againlet t k = ( k, k − , . . . , 1) be a k -cycle. Let s k = ( ε ( k ) , t k ) and u k = ( , t k ) be signed andunsigned versions of t k , respectively. Lemma 7.4. Let n ≥ s , . . . , s n , u n , u ) is a basis of B n .In particular, every signed permutation g may be uniquely expressed in the form g = s d · · · s d n n u c n n · · · u c where 0 ≤ d k < ≤ c k < k for all k . Moreover, maj A ( g ) =maj( t c n n · · · t c ) and Neg( g ) = { i | d i = 1 } . Proof. Let g = ( ε, w ) ∈ B n . For v ∈ S n , we have ( ε, w )( , v ) = ( ε, wv ). That is, rightmultiplication by an element ( , v ) rearranges the positions of the signed integers in g without changing the set of signed integers present.Let h = ( ε, u ) be the rearrangement of signed integers in g in increasing order relative to < A . Then h is the unique element in the left coset g S n such that maj A ( h ) = 0. Furthermore,for v ∈ S n , we have maj A ( ε, uv ) = maj( v ). In particular, maj A ( ε, w ) = maj( u − w ). ByLemma 4.1, there is a unique factorization u − w = t c n n · · · t c where 0 ≤ c k < k for all k ,and maj( u − w ) = c n + · · · + c .We have seen that the set T = { h ∈ B n | maj A ( h ) = 0 } is a left transversal to S n in B n . Since [ B n : S n ] = 2 n , we have | T | = 2 n . To complete the proof, we show that eachelement h ∈ T is uniquely expressible in the form h = s d · · · s d n n with 0 ≤ d i < i ,and Neg( h ) = { i | d i = 1 } .Let 0 ≤ d i < i , and let ( ε, u ) = s d · · · s d n − n − . Then u fixes n , and by induction,we may assume maj A ( ε, u ) = 0 and Neg( ε, u ) = { i | d i = 1 , i < n } . But s d · · · s d n n =( ε, u )( ε ( n ) , t n ) = ( ε + ε ( n ) , ut n ). Multiplying u on the right by t n rotates the values of u andputs n at the beginning. Since n ∈ Neg( s d · · · s d n n ), the signed values of s d · · · s d n n are stillin increasing order, i.e. maj A ( s d · · · s d n n ) = 0.Hence, { s d · · · s d n n | ∀ i, ≤ d i < } ⊆ T . Since both sets contain 2 n elements, theymust be equal. (cid:3) roof of Theorem 7.2. Let α = P g ∈ B n p maj A ( g ) q Neg( g ) · g . If φ reg is the regular representa-tion of B n , then ∆ φ reg ( α ) = det (cid:16) p maj A ( gh − ) q Neg( gh − ) (cid:17) g,h ∈ B n . By Lemma 7.4, we obtain a factorization α = (1 + q s ) · · · (1 + q n s n )(1 + pu n + · · · + p n − u n ) · · · (1 + pu ) . Therefore, ∆ φ reg ( α ) = n Y k =1 ∆ φ reg (1 + q s ) n Y k =2 ∆ φ reg (1 + pu k + · · · + p k − u k )= n Y k =1 θ φ reg ( s k ) ( − q k ) n Y k =2 (1 − p k ) | B n | θ φ reg ( u k ) ( p )The order of u k = ( , t k ) is k and the order of s k = ( ε ( k ) , t k ) is 2 k . Hence,∆ φ reg ( α ) = n Y k =1 (1 − ( − q k ) k ) n !2 n / (2 k ) n Y k =2 (1 − p k ) n !2 n (1 − /k ) . (cid:3) Acknowledgements The authors thank Victor Reiner and Volkmar Welker for initial discussions on thisproject. 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