Explicit Computations with the Divided Symmetrization Operator

Given a multi-variable polynomial, there is an associated divided symmetrization (in particular turning it into a symmetric function). Postinkov has found the volume of a permutohedron as a divided symmetrization (DS) of the power of a certain linear form. The main task in this paper is to exhibit and prove closed form DS-formulas for a variety of polynomials. We hope the results to be valuable and available to the research practitioner in these areas. Also, the methods of proof utilized here are simple and amenable to many more analogous computations. We conclude the paper with a list of such formulas. Throughout, let Sn denote the symmetric group of permutations of the n-element set {1, 2, . . . , n}. Given a function f(λ1, . . . , λn), the divided symmetrization (DS) of f , is defined by 〈f〉 = ∑ σ∈Sn f(λσ(1), . . . , λσ(n)) n−1 ∏ k=1 1 λσ(k) − λσ(k+1) . Our motivation comes from a beautiful work [1] of Alexander Postnikov who has found the volume of a permutohedron in terms of divided symmetrization of a certain expression, in addition he offers a combinatorial interpretation of the resulting coefficients. To put in context, we recall the following: Proposition 0 [Postnikov]. If f is a polynomial of degree n− 1 in the variables λ1, . . . , λn, then its divided symmetrization 〈f〉 is a constant. If deg f < n− 1, then 〈f〉 = 0. Proof. Write 〈f〉 = g/∆ where ∆ = ∏ i<j(λi−λj) is the antisymmetric Vandermonde determinant and g is a some antisymmetric polynomial. Since 〈f〉 is symmetric, g is divisible by ∆. Because deg g = deg ∆ = ( n 2 ) , their quotient must be a constant. A similar argument shows if deg g < n− 1 then deg g < deg ∆ and, hence, g = 0. The proof is complete. This proposition is applicable to (most) identities discussed in this note. But then, we ask: is it possible to obtain the actual values of these constants? If so, then: do these constants have a closed form? In this paper, we answer both questions for a variety of polynomial functions. The selection of the results is not meant to be representative, rather reflects the author’s personal taste. For starters, we invite the reader to ponder on proving the following three numerical identities before continuing further on. Typeset by AMS-TEX 1 2 TEWODROS AMDEBERHAN TULANE UNIVERSITY [email protected] Involving products: ∑ σ∈Sn+1 n ∏ k=1 σ(1) + · · ·+ σ(k) σ(k)− σ(k + 1) = n!. ∑ σ∈Sn+1 σ(1) n ∏ k=1 σ(1)− σ(k + 1) σ(k)− σ(k + 1) = 1 + 2 + · · ·+ (n+ 1). Involving sums: ∑ σ∈Sn+1 n ∑ k=1 σ(1) + · · ·+ σ(k) σ(k)− σ(k + 1) = ( n+ 1 2 ) n!, which equals the numbers of edges in the Hasse diagram of weak Bruhat order of Sn+1. From here on, as promised, we state and prove numerous divided difference formulas. Lemma 1. If λ1, λ2, . . . are variables, then ∑ σ∈Sn+1 n ∑ k=1 λσ(k) λσ(k) − λσ(k+1) = ( n+ 1 2 ) n!. Proof. Break up the sum according to σ(k) = i, σ(k + 1) = j so that ∑ σ∈Sn+1 n ∑ k=1 λσ(k) λσ(k) − λσ(k+1) = n ∑ k=1 n+1 ∑ i,j=1 i 6=j λi λi − λj ∑ σ′∈Sn−1 1 = (n− 1)! n ∑ k=1 n+1 ∑ i,j=1 i<j λi − λj λi − λj = (n− 1)! · ( n+ 1 2 ) n ∑ k=1 1 = ( n+ 1 2 ) n!. Corollary 2. If λ1, λ2, . . . are variables, then ∑ σ∈Sn+1 n ∑ k=1 λσ(1) + · · ·+ λσ(k) λσ(k) − λσ(k+1) = ( n+ 1 2 ) n!. Proof. If j 6= k, k + 1 then ∑ σ∈Sn+1 λσ(j) λσ(k)−λσ(k+1) = 0. Thus by Lemma 1, ∑ σ∈Sn+1 n ∑ k=1 λσ(1) + · · ·+ λσ(k) λσ(k) − λσ(k+1) = ∑ σ∈Sn+1 n ∑ k=1 λσ(k) λσ(k) − λσ(k+1) = ( n+ 1 2 ) n!. Lemma 3. If λ1, λ2, . . . are variables, then ∑ σ∈Sn+1 n ∑ k=1 λσ(1) + · · ·+ λ 2 σ(k) λσ(k) − λσ(k+1) = n(λ1 + · · ·+ λn+1)n!. Proof. Since the terms λσ(`) contribute to zero, whenever ` < k, we restrict to ` = k and proceed as in the proof of Lemma 1 with σ(k) = i, σ(k + 1) = j. That means, ∑ σ∈Sn+1 n ∑ k=1 λσ(k) λσ(k) − λσ(k+1) = n ∑ k=1 n+1 ∑ i,j=1 i<j λi − λj λi − λj ∑ σ′∈Sn−1 1 = (n− 1)! n+1 ∑ i,j=1 i<j (λi + λj) n ∑