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ON THE GROUP OF ALTERNATING COLORED PERMUTATIONS 4 1 ELI BAGNO, DAVID GARBER AND TOUFIK MANSOUR 0 2 Abstract. The group of alternating colored permutations is the n naturalanalogueoftheclassicalalternatinggroup,insidethewreath a J product Zr ≀Sn. We present a ’Coxeter-like’ presentation for this group and compute the length function with respect to that pre- 2 2 sentation. Then, we present this group as a covering of Zr ≀Sn 2 andusethispointofviewtogiveanotherexpressionforthelength ] function. Wealsousethiscoveringtoliftseveralknownparameters R of Zr ≀Sn to the group of alternating colored permutations. G 2 . h t a m 1. Introduction [ The group of colored permutations, G , is a natural generalization r,n 1 of the Coxeter groups of types A (the symmetric group) and B (the v hyperoctahedral group). Extensive research has been devoted to ex- 8 6 tending the enumerative combinatorics aspects and methods from the 5 symmetric group to the group of colored permutations (see for example 5 . [2, 3, 7, 13, 15, 16], and many more). 1 Itiswell-known thatthesymmetric groupS hasasystemofCoxeter 0 n 4 generators which consists of the adjacent transpositions: 1 : {(i,i+1) | 1 ≤ i ≤ n−1}. v i X The alternating subgroup, A , which is the kernel of the sign homo- n r morphism, is awell-known subgroup ofthesymmetric groupofindex 2. a A pioneering work, expanding one of the fascinating branches of enumerative combinatorics, namely, the study of permutation statistics to A , has been done by Roichman and Regev in [14]. They defined some n natural statistics which equidistribute over A and yielded identities n for their generating functions. Brenti, Reiner, andRoichman[6]dealtwiththealternatingsubgroup of an arbitrary Coxeter group. They started by exploring Bourbaki’s presentation [4, Chap. IV, Sec. 1, Exer. 9] and elaborated on a huge spectrum of extensions of the permutation statistics of S to the n (general) alternating group. Date: January 23, 2014. 1 2 ELI BAGNO,DAVIDGARBER AND TOUFIK MANSOUR In this paper, we study the subgroup of G , consisting of what r,n we call alternating colored permutations, which is the analogue of the usualalternatinggroupA inthecoloredpermutationgroup. Forevery n n ∈ Nandevenr, themappingwhichsendsall’Coxeter-like’ generators of G (see the definition in Section 2) to −1 is a Z -character, whose r,n 2 kernel is what we call here the group of alternating colored permutations, denoted by A . We present here a generalization of Bourbaki’s r,n presentation, for r = 4k + 2, equipped with a set of canonical words, an algorithm to find a canonical presentation for each element of the group, and a combinatorial length function. For the study of permutation statistics of A , Regev and Roichman n [14] used a covering map from A to S , which enabled them to pass n+1 n parameters from S to the alternating group A . In this paper, we n n+1 use a similar idea, where in this time we consider the group of alternating colored permutations as a 2n−1-cover of the group of colored permutations of half the number of colors. We use this technique to shed a combinatorial flavor on our length function and to pass some statistics and their generating function to the group of alternating colored permutations. Note that there are two additional candidates for the group of alternating colored permutations. Namely, every Z -character of G 2 r,n provides a kernel which deserves to be called a group of alternating colored permutations. A work in this direction which gives a profound treatment to the other two non-trivial kernels, and points out the con- nections between the three groups, and some interesting properties of each group separately is in progress. Thispaperisorganizedasfollows. InSection2, wegathertheneeded definitions on the colored permutation group, as well as some notations which we usein thesequel. ACoxeter-like presentation forthegroupof colored permutations, G , is presented at the end of this section. The r,n notion of alternating colored permutations is introduced in Section 3. Wepresentitssetofgenerators, andshowtheircorrespondingrelations. In Section 4, we present an algorithm for writing each element as a product of the generators. A detailed analysis of that algorithm yields a set of canonical words, as well as a length function. Section 5 is devoted to some technical proofs, as well as to the generating function of the length function. In Section 6, we present the covering map and study the structure of the cosets, thereby providing a way to decompose the length function via the quotient group. The part of the length which varies over each coset (fiber) is called the fibral length and is studied here in a GROUP OF ALTERNATING COLORED PERMUTATIONS 3 combinatorial way. Then we provide a generating function for this pa- rameter. In Section 7, we give some examples for using the covering map for lifting parameters from the colored permutations group of half the number of colors to the group of alternating colored permutations. 2. Preliminaries and notations In this section, we gathered some notations as well as preliminary notions which will be needed for the rest of the paper. 2.1. The group of colored permutations. Definition 2.1. Let r and n be positive integers. The group of colored permutations of n digits with r colors is the wreath product G = Z ≀S = Zn ⋊S , r,n r n r n consisting of all pairs (~z,τ), where ~z is an n-tuple of integers between 0 and r−1 and τ ∈ S . The multiplication is defined by the following n rule: for ~z = (z ,...,z ) and z~′ = (z′,...,z′), 1 n 1 n (1) (~z,τ)·(z~′,τ′) = ((z +z′ ,...,z +z′ ),τ ◦τ′) 1 τ−1(1) n τ−1(n) (here + is taken modulo r). Here is another way to present G : Consider the alphabet r,n Σ = {1,...,n,¯1,...,n¯,...,1[r−1],...,n[r−1]} as the set [n] colored by the colors 0,...,r − 1. Then, an element of G is a colored permutation, i.e., a bijection π : Σ → Σ satisfying r,n the following condition: if π i[α] = j[β], then π i[α+1] = j[β+1] (the addition in the exponents is taken modulo r). Using this approach, the (cid:0) (cid:1) (cid:0) (cid:1) element π = ((z ,...,z ),τ) ∈ G is the permutation on Σ, satisfying 1 n r,n π(i) = π(i[0]) = τ(i)[zτ(i)] for each 1 ≤ i ≤ n. For example, the element 1 2 3 4 π = (2,1,0,3), ∈ G satisfies: π(1) = 2[1],π(2) = 2 1 4 3 3,4 (cid:18) (cid:18) (cid:19)(cid:19) 1[2],π(3) = 4[3],π(4) = 3[0]. For an element π = (~z,τ) ∈ G with ~z = (z ,...,z ), we write r,n 1 n z (π) = z , and denote |π| = (~0,τ). We define also c (π) = r −z (π−1) i i i i and ~c(π) = ~c = (c ,...,c ). Using this notation, the element π = 1 n 1 2 3 4 (~z,τ) = (2,1,0,3), satisfies ~c = (1,2,3,0)). 2 1 4 3 (cid:18) (cid:18) (cid:19)(cid:19) We usually write π in its window notation (or one line notation): π = a[c1]···a[cn] , where a = τ(i), so in our example we have: π = 1 n i (2[1]1(cid:16)[2]4[3]3[0]) or j(cid:17)ust ¯2¯¯1¯¯¯43 . (cid:16) (cid:17) 4 ELI BAGNO,DAVIDGARBER AND TOUFIK MANSOUR Note that z is the color of the digit i (i is taken from the window i notation), while c is the color of the digit τ(j). Here, j stands for the j place, whence i stands for the value. ThegroupG isgeneratedbythesetofgeneratorsS = {s ,s ,...,s }, r,n 0 1 n−1 defined by their action on the set {1,...,n} as follows: i+1 j = i s (j) = i j = i+1 i  j otherwise,  whereas the generator s is defined by 0  ¯1 j = 1 s (j) = 0 j otherwise. (cid:26) It is easy to see that the group G has the following ’Coxeter-like’ r,n presentation with respect to the set of generators S: Presentation 2.2. • sr = 1, 0 • s2 = 1 for 1 ≤ i ≤ n−1, i • s s s = s s s for 1 ≤ i < n, i i+1 i i+1 i i+1 • s s = s s for 1 ≤ i < j < n, j −i > 1, i j j i • (s s )2r = 1. 0 1 2.2. Some permutation statistics. For π ∈ G , define the length r,n of π with respect to the set of generators S to be the minimal number of generators whose product is π. Formally: ℓ(π) = min{r ∈ N : π = s ···s ,for i ,...,i ∈ {0,...,n−1} }. i1 ir 1 r Definition 2.3. The length order on the alphabet Σ = {1,...,n,¯1,...,n¯,...,1[r−1],...,n[r−1]} is defined as follows: (2) n[r−1] < ··· < n¯ < ··· < 1[r−1] < ··· < ¯1 < 1 < ··· < n Let σ ∈ G . We define: r,n n n csum(σ) = c (σ) = z (σ). i i i=1 i=1 X X For π ∈ G , the inversion number, inv(π), is defined as follows: r,n inv(π) = |{(i,j) | i < j,π(i) > π(j)}|. where the partial order is the length order defined above. GROUP OF ALTERNATING COLORED PERMUTATIONS 5 For any a,n ∈ N, let R (a) be the representative of [a] ∈ Z satisfy- n n ing 0 ≤ a < n. In the sequel, we will use the following operator: Definition 2.4. Let a ∈ N. Rr(a) a ≡ 0(mod 2) (3) a⊘2 = 2a+2r ( Rr( 2) a 6≡ 0(mod 2) 2 2 It is easy to see that the operator ⊘ commutes with the addition operation in Zr, i.e. 2 r (4) ((a+b)⊘2) ≡ ((a⊘2)+(b⊘2)) mod 2 (cid:16) (cid:17) 3. The group of alternating colored permutations The main target of this paper is the group of alternating colored permutations. We proceed now to its definition. Let ϕ be the function defined on the set S by ϕ(s ) = −1 for any 0 ≤ i ≤ n−1. It is easy to i see that for even r, ϕ can be uniquely extended to a homomorphism from G to Z , so the following is well-defined: r,n 2 Definition 3.1. Let r be an even positive number. Define: A = ker(ϕ). r,n The group A is called the alternating subgroup of G . r,n r,n Since A is a subgroup of index 2, we have: |A | = rnn!. r,n r,n 2 In this paper, we concentrate on the case r = 4k+2. The other case will be treated in a subsequent paper. We start by presenting a set of generators for A (we prove that r,n they indeed generate the group in the next section). Define: A = {a ,a ,a−1,a ,...,a }, 0 1 1 2 n−1 where: r a = s2s for 1 ≤ i ≤ n−1 i 0 i a = s2. 0 0 It is easy to see that the following translation relations hold in G : r,n (1) s s = a a for i,j ∈ {2,...,n−1}, i j i j (2) s s = a−1a for i ∈ {2,...,n−1}, 1 i 1 i (3) s s = a a for i ≥ 2, i 1 i 1 r+2 (4) s s = a 4 a , 0 1 0 1 r+2 (5) s s = a−1a 4 , 1 0 1 0 6 ELI BAGNO,DAVIDGARBER AND TOUFIK MANSOUR r (6) a2 = 1, 0 r+2 (7) s s = a 4 a . 0 i 0 i 4. The Combinatorial algorithm In this section, we introduce an algorithm which presents each element of A as a product of the set of generators A of A in a r,n r,n canonical way. Let π ∈ A . We first refer to π as an element of G and apply the r,n r,n known algorithm on π to write it as a product of elements in S. In the second step, we translate that presentation into the set of generators A of A . r,n The algorithm for writing π as a product of elements in S consists of two parts: the coloring part and the ordering part. In the coloring part, we start from the identity element and color all the digits i having z 6= 0. This part terminates with an ordered i permutation σ with respect to the length order. In the second part, we use only generators of the set S −{s } to arrive at π from the ordered 0 permutation σ. 4.1. The coloring part. Define: Col(π) = {1 ≤ i ≤ n | z (π) 6= 0}, i and col(π) = |Col(π)|. Note that the set Col(π) contains the colored digits in the image of π, (i.e. those appearing in the window notation), and not their places. We order Col(π) as follows: Col(π) = {i < i < ··· < i }. 1 2 col(π) We start with the identity element and color each digit i ∈ Col(π) by z colors. This process is done according to the order of the elements i in Col(π). We use s s ···s sz to color the digit i by z colors. ik−1 ik−2 1 0 k Example 4.1. Let π = 12[2]45[1]3[3] ∈ G . 6,5 (12345) s→1s20 2[2]1345 s2(cid:0)→s1s30 3[3]2[2]1(cid:1)45 s4s3→s2s1s0 5[1]3[3]2[2]14 = σ. Thepermutat(cid:0)ionσ isa(cid:1)nordere(cid:0)dpermuta(cid:1)tionwithre(cid:0)specttothe(cid:1)length order. 4.2. The ordering part. Forsimplifyingthepresentation, inthispart we start with π and arrive at the ordered permutation σ, instead of continuing the algorithm from the point we have left it at the end of the coloring part. GROUP OF ALTERNATING COLORED PERMUTATIONS 7 We start by pushing the element i = |σ|(1) in the window notation 1 of π to its correct place. Let p = |π|−1(i ). The pushing is done by 1 multiplying π (from the right) by the element u = s s ···s . 1 p−1 p−2 1 Now, we continue to push the other digits of π: For each 1 < k ≤ n−2, assuming that i = |σ|(k) is now located at position p, we use k the element s s ···s in order to push the digit i to its correct p−1 p−2 k k place. Example 4.2. We continue the previous example. Again, let π = 12[2]45[1]3[3] . The coloring part ends with the following ordered permutation: (cid:0) (cid:1) σ = 5[1]3[3]2[2]14 . Now, we go the other way ar(cid:0)ound: we sta(cid:1)rt with π and order it until we reach σ: π = 12[2]45[1]3[3] s3→s2s1 5[1]12[2]43[3] s4→s3s2 (cid:0) (cid:1) (cid:0) (cid:1) → 5[1]3[3]12[2]4 →s3 5[1]3[3]2[2]14 = σ. Therefore, we ha(cid:0)ve: (cid:1) (cid:0) (cid:1) π = s s2 ·s s s3 ·s s s s s ·s ·s s s ·s s s . 1 0 2 1 0 4 3 2 1 0 3 2 3 4 1 2 3 coloring part ordering part The algorithm| described{zabove give}s a|reduced{zword re}presenting π in the generators of G . This fact was proved in [2, Theorem 4.3]. r,n The same algorithm can also be found in [16]; see also [15]. The word which was obtained in this way is called the canonical decomposition of π. 4.3. Translation. Now, we translate the word obtained by the algorithm described above into a word in the generators in A: Let π ∈ A . r,n Usetheabovealgorithmtowriteareducedexpression ofπ (intheusual generators of G ) in the form: s s ···s . Divide the elements of r,n i1 i2 i2k the reduced expression into pairs: (s s )···(s s ). Now, insert i1 i2 i2k−1 i2k r r s2s2 inside each pair, as follows: 0 0 r r r r r r r r s s2s2s ··· s s2s2s = s s2 s2s ··· s s2 s2s = i1 0 0 i2 i2k−1 0 0 i2k i1 0 0 i2 i2k−1 0 0 i2k (cid:16) (cid:17) (cid:16) (cid:17) = a(cid:16)εi1 ···(cid:17)a(cid:16)εi2k, (cid:17) (cid:16) (cid:17)(cid:16) (cid:17) i1 i2k where ε = 1 if i > 1, ε ∈ {±1} if i = 1, and ε ∈ 1,..., r −1 if ij j ij j ij 2 i = 0. j (cid:8) (cid:9) 8 ELI BAGNO,DAVIDGARBER AND TOUFIK MANSOUR Example 4.3. We continue with π = 12[2]45[1]3[3] ∈ A from the 6,5 previous examples. As we saw, π = s s2s s s3s s s s s s s s s s s s . 1 0(cid:0)2 1 0 4 3 2 (cid:1)1 0 3 2 3 4 1 2 3 Now, we perform the translation: π = (s s )(s s )(s s )(s s )(s s )(s s )(s s )(s s )(s s )(s s ) = 1 0 0 2 1 0 0 0 4 3 2 1 0 3 2 3 4 1 2 3 = (s s3)(s3s )(s s3)(s3s )(s s3)(s3s )(s s3)(s3s )(s s3)(s3s )· 1 0 0 0 0 0 0 2 1 0 0 0 0 0 0 0 4 0 0 3 ·(s s3)(s3s )(s s3)(s3s )(s s3)(s3s )(s s3)(s3s )(s s3)(s3s ) = 2 0 0 1 0 0 0 3 2 0 0 3 4 0 0 1 2 0 0 3 = a−1a2a2a a−1a2a2a2a a a a a2a a a a a a a = 1 0 0 2 1 0 0 0 4 3 2 1 0 3 2 3 4 1 2 3 = a−1a a a−1a a a a a2a a a a a a a . 1 0 2 1 4 3 2 1 0 3 2 3 4 1 2 3 In the last equality, we cancelled some appearances of the bold-faced generator a , since in A , a3 = 1. 0 6,5 0 4.4. Analysis of the algorithm. For analyzing the algorithm described above, we define the following sets of elements of G and A . r,n r,n 4.4.1. The coloring part. Let r−2 C = 1,s2,s4...,sr−2 = 1,a ,...,a 2 . 1 0 0 0 0 0 n o For each 1 < i ≤(cid:8)n, define for odd(cid:9)i−1: C0 = s s ···s ,s s ···s s2,s s ···s s4,...,s s ···s sr−2 i 0 i−1 1 0 i−1 1 0 0 i−1 1 0 0 i−1 1 0 C1 = 1,s ···s s ,s ···s s3,...,s s ···s sr−1 , i (cid:8) i−1 1 0 i−1 1 0 i i−1 1 0 (cid:9) or, in the language of the set A of generators of A : (cid:8) r,n (cid:9) r+2 r+2 r+2 a 4 a ···a ,a 4 a ···a a ,a 4 a ···a a2,..., C0 = 0 i−1 1 0 i−1 1 0 0 i−1 1 0 i r+2 r−1 ( a 4 a ···a a2 ) 0 i−1 1 0 C1 = 1,a ···a−1,a ···a a−1a ,...,a ···a−1a2r−1 . i i−1 1 i−1 2 1 0 i−1 1 0 n o For even i−1, we define: C0 = s s ···s s ,s s ···s s3,...,s s ···s sr−1 i 0 i−1 1 0 0 i−1 1 0 0 i−1 1 0 C1 = 1,s ···s ,s ···s s2,s ···s s4,...,s ···s sr−2 , i (cid:8) i−1 1 i−1 1 0 i−1 1 0 i−1 (cid:9)1 0 or, in the language of the set A of generators of A : (cid:8) r,n (cid:9) r+2 r+2 r+2 a 4 a ···a−1,a 4 a ···a−1a ,a 4 a ···a−1a2,..., C0 = 0 i−1 1 0 i−1 1 0 0 i−1 1 0 i ( ar+42a ···a−1a2r−1 ) 0 i−1 1 0 C1 = 1,a ···a ,a ···a a ,a ···a a2,...,a ···a a2r−1 . i i−1 1 i−1 1 0 i−1 1 0 i−1 1 0 n o Define also: r+2 C = C0 ∪C1 and C = 1,a 4 . i i i n+1 0 n o GROUP OF ALTERNATING COLORED PERMUTATIONS 9 Let π ∈ A . Write π as a product of the generators of G in the r,n r,n canonical form described above. If there is no coloring part, then π ∈ A (the classical alternating n group in S ), so its expression contains an even number of generators n from the set {s ,...,s }. We can easily make the pairing by the 1 n−1 relations mentioned above. The length of such an expression is clearly inv(π) (note that in this case, it does not matter whether we use the length order or the usual order). Otherwise, we start with the coloring part. Denote by i = i the 1 smallest colored digit in the window notation of π, and by z = z its i1 color. We divide our treatment into four cases: (1) i−1 and z are both even: In thiscase, we translates ···s i−1 1 z to a ···a and sz to a2, so the contribution of this sub- i−1 1 0 0 expression is i − 1 + Rr(z) = i − 1 + Rr(z ⊘ 2). We have 2 2 2 z used a ···a a2 ∈ C . i−1 1 0 i (2) i−1 isevenand z isodd: Inthiscase, wetranslates ···s sz−1 i−1 1 0 z−1 to a ···a a 2 ∈ C and leave an additional generator s i−1 1 0 i 0 which will be treated during the coloring of the next digit, or just before the ordering part. Note that since π ∈ A , r,n there must be some s , j 6= 0, appearing right after the sub- j expression s ···s sz. In calculating the contribution of color- i−1 1 0 ing the current digit (including the missing generator s which 0 will be paired later), consider the sub-expression szs = s2rsz+r2s = az+22r s2rs = az+2r2a = az⊘2a . 0 j 0 0 j 0 0 j 0 j 0 j Hence, i contributes i−1+Rr(z ⊘2). We have used 2 z−1 a ···a a 2 ∈ C , i−1 1 0 i andnotethat inthenext colored digit, we complete theremain- r+2 ing a 4 since z ⊘2 = z−1 + r+2. If i is the last colored digit, 0 2 4 r+2 then the term a 4 ∈ C will be chosen from the set C . 0 n+1 n+1 (3) i − 1 and z are both odd: In this case, the sub-expression s ···s will be translated to a ···a and s sz will be writ- i−1 2 i−1 2 1 0 ten as s s2rsz+r2 = a−1az⊘2. This expression contributes i−1+ 1 0 0 1 0 Rr(z ⊘2) to the length of π, and we have used 2 a ···a a−1az⊘2 ∈ C . i−1 2 1 0 i (4) i −1 is odd and z is even: Here, again, the sub-expression s ···s will be translated to a ···a and s sz−1 will be i−1 2 i−1 2 1 0 10 ELI BAGNO,DAVIDGARBER AND TOUFIK MANSOUR translated to a−1az−12+2r = a−1a(z−1)⊘2, so we use: 1 0 1 0 a ···a a−1a(z−1)⊘2 ∈ C , i−1 2 1 0 i and leave an additional generator s which will be paired with 0 some s during the coloring of the next digit or just before the j ordering part. In order to calculate the contribution of coloring this digit to the length of π (including the missing generator s which will be paired later), we borrow the generator s ap- 0 j pearing just after the coloring expression of the current digit: s szs = s s2rsz+2rs . Since we wrote: s sr2 = a−1, we are left 1 0 j 1 0 0 j 1 0 1 with sz+2rs = az⊘2. The contribution in this case is again 0 j 0 i−1+Rr(z ⊘2). Now, since 2 r +2 r z ⊘2 ≡ (z −1)⊘2+ mod , 4 2 (cid:18) (cid:19) (cid:16) (cid:17) we take a ···a−1a(z−1)⊘2 ∈ C i−1 1 0 i r+2 andtheremaining a 4 will betaken fromthenext coloreddigit 0 or from C (as in case (2)). n+1 Now, we apply the same procedure to the next colored digits, but note that there might be a situation in which the expression coloring thedigitj iss s s ···, duetoourdebt ofthegenerators fromthe 0 j−1 j−2 0 preceding colored digit, so the cases might be switched after converting r+2 s s to a 4 a . 0 j 0 j−1 The following example will illuminate the situation. Example 4.4. Let π = 12[2]45[1]3[3] ∈ A . Then the ordered per- 6,5 mutation is: σ = 5[1]3[3]2[2]14 . We perform the coloring part: (cid:0) (cid:1) (12345) (cid:13)1−→s1s0 2[(cid:0)1]1354 (cid:13)2 −s0→(cid:1)s2s1s30 3[3]2[2]145 (cid:13)3 −s4→s3s2s1 53[3]2[2]14 = σ. Step (cid:13)1 : The(cid:0)smallest(cid:1)colored digi(cid:0)t is 2, whi(cid:1)ch has to b(cid:0)e colored b(cid:1)y two colors, so we are in case (4). We choose s s = a−1a2 from C1. 1 0 1 0 2 The additional generator s will be treated in the next step. Note that 0 in the calculation of the contribution of this step to the length of π we borrow the generator s from the next colored digit: 2 s s2s = s s3+2+3s = s s3 s2 s3s = a−1a a . 1 0 2 1 0 2 1 0 0 0 2 1 0 2 This expression contributes only(cid:0)2 to(cid:1)the(cid:0)lengt(cid:1)h of π. The generator a will be counted in the next step. 2