Excedance Numbers for Permutations in Complex Reflection Groups

Home , Cyclic permutation

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We use some conventions along this paper. For an element σ = (z,τ) ∈ Gr,n with z = n (z1, ..., zn) we write zi(σ)= zi. For σ = (z,τ), we denote |σ| = (0,τ), (0 ∈ Zr ).

A much more natural way to present Gr,n is the following: Consider the alphabet Σ = {1[0],...,n[0], 1[1],...,n[1],..., 1[r−1],...,n[r−1]} as the set [n] = {1,...,n} colored by the colors 0, . . . , r−1. Then, an element of Gr,n is a colored permutation, i.e. a bijection σ :Σ → Σ such that if σ(i) = k[t] then σ(i[j]) = k[t+j] where 0 ≤ j ≤ r − 1 and the addition is taken modulo r. Occasionally, we write j bars over a digit i instead of i[j]. For example, an element ¯ ¯ (z,τ) = ((1, 2, 1, 2), (3, 1, 2, 4)) ∈ G3,4 will be written as (3¯1¯2¯4).¯ For each s|r we deﬁne the complex reﬂection group:

Gr,s,n := {σ ∈ Gr,n | csum(σ) ≡ 0 mod s}, n where csum(σ)= zi(σ). i=1 One can define theP following well-known statistics on Sn. For any permutation σ ∈ Sn, i ∈ [n] is an excedance of σ if and only if σ(i) > i. We denote the number of excedances by exc(σ). Another natural statistic on Sn is the number of fixed points, denoted by fix(σ). We can similarly define some statistics on Gr,n. The complex reflection group Gr,s,n inherits all of them. Given any ordered alphabet Σ′, we recall the definition of the excedance set of a permutation σ on Σ′: Exc(σ)= {i ∈ Σ′ | σ(i) > i}, and the excedance number is defined to be exc(σ)= |Exc(σ)|. We define the color order on the set Σ = {1,...,n, 1¯,..., n,...,¯ 1[r−1],...,n[r−1]} for 0 ≤ j < i < r by 1[i] < 2[i] < · · · < n[i] < 1[j] < 2[j] < · · · < n[j]. We note that there are some other possible ways of defining orders on Σ, some of them lead to other versions of the excedance number, see for example [1]. For example, given the color order 1¯ < 2¯ < 3¯ < 1¯ < 2¯ < 3¯ < ¯ 1 < 2 < 3, we write σ = (21¯3)¯ ∈ G3,3 in an extended form 1¯ 2¯ 3¯ 1¯ 2¯ 31¯ 23 (⋆) 2¯ 1 3¯ 2¯ 1¯ 3 2 1¯ 3¯ which implies that Exc(σ)= {1¯, 2¯, 3¯, 1¯, 3¯, 1} and exc(σ) = 6.

Deﬁne ExcA(σ) = {i ∈ [n − 1] | σ(i) > i}, where the comparison is with respect to the ¯ ¯ color order, and denote excA(σ) = |ExcA(σ)|. For instance, if σ = (1¯32¯ 4)¯ ∈ G3,4, then csum(σ) = 5, ExcA(σ)= {3} and hence excA(σ) = 1. Clr Now we can deﬁne the colored excedance number for Gr,n by exc (σ)= r·excA(σ)+csum(σ). Let Σ ordered by the color order, then we can state that exc(σ) = excClr(σ) obtained by Bagno and Garber [1] for any σ ∈ Gr,n.

For σ = (z,τ) ∈ Gr,n, |σ| is the permutation of [n] satisfying |σ|(i) = τ(i). We say that i ∈ [n] is an absolute fixed point of σ ∈ Gr,n if |σ|(i) = i. We denote the number of absolute fixed point of σ ∈ Gr,n by fix(σ).

3. Main results and proofs

m m Our main result can be formulated as follows. Recall that Gr,s,n = {σ ∈ Gr,s,n|σ = 1}, deﬁne

(m) ﬁx(σ) excA(σ) csum(σ) Hr,s,n(u, v, w)= u v w , σ m ∈GXr,s,n EXCEDANCE NUMBERS FOR PERMUTATIONS IN COMPLEX REFLECTION GROUPS 3 and xn xn H(m)(x; u, v, w)= H(m) (u, v, w) = uﬁx(σ)vexcA(σ)wcsum(σ) . r,s r,s,n n! n! n n σ m X≥0 X≥0 ∈GXr,s,n Theorem 3.1. For positive integer r, m ≥ 1, there holds (m) Hr,1 (x; u, v, w) d−1 k t xd k k−i (i) t = exp xuw + d! Ad−1,k i v Ud−k,tw , {t|0≤t

If d ≥ 2, we can claim that there is Ad−1,k cyclic permutations of length d in Sd with k excedances for 1 ≤ k ≤ d − 1, where Ad−1,k are the Eulerian numbers which are also the number of permutations on [d − 1] with k − 1 excedances. This claim can be proved by induction on d and k by noting that they obey the recurrence relation

(3.1) Ad,k = kAd−1,k + (d − k + 1)Ad−1,k−1, with the initial values A0,0 = A1,1 = 1 and Ad,k = 0 if d < k or k < 1 ≤ d. Note that for any cyclic permutation Cd = (i1, i2,...,id) ∈ Sd with k excedances, inserting d + 1 into the positions wherever ij < ij+1, there are k ways to obtain a cyclic permutation Cd+1 ∈ Sd+1 with k excedances, and for any cyclic permutation Cd = (i1, i2,...,id) ∈ Sd with k − 1 excedances, inserting d + 1 into the positions wherever ij > ij+1, there are d − k + 1 ways to obtain a cyclic permutation Cd+1 ∈ Sd+1 with k excedances. Conversely, any cyclic permutation Cd+1 ∈ Sd+1 with k excedances can reduce to the two cases above by deleting the symbol d + 1. This analysis makes us get the recurrence relation (3.1).

For any cyclic permutation C of length d in Sd with Exc(C) = {j ∈ [d − 1]|C(j) > j} such that exc(C) = k, we can color the symbols in C with the color set {[0], [1], · · · , [r − 1]} and ′ ′ obtain the colored cyclic permutation C . Suppose that excA(C ) = k − i, we know that ′ ExcA(C ) ⊆ Exc(C), which means that exc(C) − excA(C) = i, in other words, there are i k number of symbols in Exc(C) with color numbers ranging from [1] to [r − 1], so there are i ways to do this. ′ Let t = csum(C ) and tℓ be the color number of ℓ ∈ [d], then we have the equation t = t1 + t2 + · · · + td with 0 ≤ t1,t2, · · · ,td ≤ r − 1 such that

• tj = 0 for j ∈ Exc(C) and j has a color number [0], and ′ • 1 ≤ tj ≤ r − 1 for all j ∈ Exc(C) − ExcA(C ), so there are i number of such j’s. 4 EXCEDANCE NUMBERS FOR PERMUTATIONS IN COMPLEX REFLECTION GROUPS

(i) Therefore there are Ud−k,t number of solutions of the above equation, totally, there are k (i) ′ ′ i Ud−k,t ways to color the symbols in C such that csum(C )= t and excA(C )= k −i, where (i) t 2 r−1 i r−1 d−k Ud−k,t is the coefficient of x in (x + x + · · · + x ) (1 + x + · · · + x ) , which can be expressed as (i) t 2 r−1 i r−1 d−k Ud−k,t = [x ](x + x + · · · + x ) (1 + x + · · · + x ) 1 − xr i 1 − xr d−k = [xt] − 1 1 − x 1 − x i i 1 − xr d+j−k = [xt] (−1)i−j j 1 − x j X=0 i i d + j − k d + j + t − k − ℓr − 1 = (−1)i−j (−1)ℓ . j ℓ t − ℓr j X=0 Xℓ≥0 ′ [t1] [t2] [td] ′d [t] [t] [t] Let C = (i1 , i2 ,...,id ), then C = (i1 , i2 ,...,id ) with t = t1 + t2 + · · · + td, hence tm tm tm ′m [ d ] [ d ] [ d ] tm m C = (i1 , i2 ,...,id ) = 1 implies that r| d . For any π ∈ Gr,1,n such that the symbol n ′ n−1 lies in a cycle C of length d ≥ 2 with d|m (note that there are d−1 ways to choose the digits of such a cycle), define π′′ ∈ Gm in the following way: write π in its complete notation, r,1,n−d i.e., as a matrix of two rows, see (⋆). The first row of π′′ is (1, 2, · · · ,n − d) while the second row is obtained from the second row of π by ignoring the digits in C′ and the other digits are placed with the numbers 1, 2,...,n − d in an order preserving way with respect to the second row of π. The parameters satisfy fix(π) = fix(π′′), ′′ ′ excA(π) = excA(π ) + excA(C ), csum(π) = csum(π′′) + csum(C′). The above consideration gives the following recurrence (m) Hr,1,n(u, v, w) (m) t (m) n−1 = Hr,1,n−1(u, v, w) {t|0≤t

(m) Thus, the generating function Hr,1 (x; u, v, w) satisﬁes

∂ (m) d−1 ∂x Hr,1 (x; u, v, w) t x = uw + Am,d(v, w) . H(m)(x; u, v, w) (d − 1)! r,1 {t|0≤t

Specially, if m = p is a prime, then we have (p) Corollary 3.2. Let r ≥ 1 and p be a prime. The generating function Hr,1 (x; u, v, w) is given by

p p−1 k x k k−i (i) jr exp uxλr,p(w)+ A v U w ,  p! p−1,k i p−k,jr  i j  Xk=1 X=0 X≥0  (i) jr where Ap−1,k is the Eulerian number, Up−k,jr is the coeﬃcient of x in  (x + x2 + · · · + xr−1)i(1 + x + · · · + xr−1)p−k, ir p−1 p λr,p(w)= i=0 w for p|r, and λr,p(w) = 1 for p 6 |r. For the sakeP of comparison, the cases p = 2 and p = 3 in Corollary 3.2 generate the explicit (2) (3) formulas for Hr,1 (x; u, v, w) and Hr,1 (x; u, v, w), that is 2 (2) x r H (x; u, v, w) = exp(uxλr, (w)+ (v + (r − 1)w )), r,1 2 2 3 (3) x H (x; u, v, w) = exp(uxλr, (w)+ B , (v, w)), r,1 3 6 3 3 2 r 2 r 2r where B3,3(v, w)= v + v(1 + 3(r − 1)w ) + (r − 1)w + (r − 1)(r − 2)w . (m) Now let us compute the exponential generating function Hr,s (x; u, v, w) for the sequence (m) m {Hr,s,n(u, v, w)}n≥0. For any σ ∈ Gr,s,n, we have csum(σ) ≡ 0 (mod s), so we should collect (m) all the terms in which the exponent of w in Hr,1 (u, v, w) is a multiplication of s. This observation can make us get the following

(m) n Theorem 3.3. Let r,m,s ≥ 1, deﬁne Hr,1 (x; u, v, yw)= n≥0 Gm,r,n(x; u, v, w)y . Then (m) Hr,s (x; u, v, w) = Gm,r,sk(xP; u, v, w). k X≥0 Now let us focus on the case m = 2. Recall that 1 ux+ x2(v+(r−1)wr ) (2) e 2 , if r odd, r Hr,1 (x; u, v, w)= 2 1 2 r ( eux(1+w )+ 2 x (v+(r−1)w ), if r even . (2) Then by Theorem 3.3, we can compute the explicit formula for Hr,s (x; u, v, w). Since s | r, we have two cases either r odd or r even. • If r is an odd number, then it is clear that the exponent of y in each term of the (2) expansions of Hr,1 (x; u, v, yw) is always a multiplication of s. Hence, (2) (2) Hr,s (x; u, v, w)= Hr,1 (x; u, v, w). 6 EXCEDANCE NUMBERS FOR PERMUTATIONS IN COMPLEX REFLECTION GROUPS

r • Similarly, if r is an even number and s| 2 , we have that (2) (2) Hr,s (x; u, v, w)= Hr,1 (x; u, v, w). r r k r ux(1+(yw) 2 ) ux (ux(yw) 2 ) • Let r be any even number such that s ∤ 2 . Since e = e k≥0 k! 1 2 r 1 2 2 r k 2 x (v+(r−1)(yw) ) 2 x v ((r−1)x (yw) ) and e = e k≥0 2kk! , then by collectingP the coeﬃcients (2) of y in Hr,1 (x; u, v, w) such thatP the exponent y is a multiplication of s, we get that r r 2k kr uxw 2 −uxw 2 1 2 r (ux) (yw) 1 2 r e + e e 2 x (v+(r−1)(yw) ) = eux+ 2 x (v+(r−1)(yw) ) . (2k)! 2 Xk≥0 Therefore, the above cases gives the following result. Proposition 3.4. We have 1 2 r eux+ 2 x (v+(r−1)w ), if r odd, r 2 1 2 r (2) ux(1+w )+ 2 x (v+(r−1)w ) r Hr,s (x; u, v, w) =  e , if r even and s ∤ , r r 2  1 2 r uxw 2 −uxw 2  ux+ 2 x (v+(r−1)w ) e +e r e 2 , if r even and s ∤ 2 . (2)  (2) Note that Hr,s (x; u, v, w) is the generating function for the number of involutions in Gr,s,n. By expanding the generating functions, Bagno, Garber and Mansour [2] obtained the explicit (2) formulas for the number of involutions in Gr,s,n. But the expression in Proposition 5.7 [2] (2) should be corrected by the third case of Hr,s (x; u, v, w) and hence Corollary 5.8, 5.9 and 5.10 therein should be the following three corollaries, respectively. (2) Corollary 3.5. The polynomial Hr,s,n(u, v, w) is given by n! uk1+2k2 wrk2 (v + (r − 1)wr)k3 · . k3 k1!(2k2)!k3! 2 k1+2kX2+2k3=n (2) r Corollary 3.6. Let r ≥ 1. The number of colored involutions in Gr,s,n (r is even, s ∤ 2 ) with exactly k absolute ﬁxed points and excA(π)= ℓ is given by k n! (r − 1)k3−ℓ 3 · · . k3 ℓ k1!(2k2)!k3! 2 k+2k3=n,kX1+2k2=k (2) r Clr Corollary 3.7. The number of involutions π ∈ Gr,s,n (r is even, s ∤ 2 ) with exc (π)= k is given by k n! r r · . k1!(2k2)!k3! 2 k1+2k2+2k3=Xn, r(k2+k3)=k Acknowledgment The authors would like to thank Eli Bagno and David Garber for reading previous version of the present paper and for a number of helpful discussions.

References

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