Proof of a Conjecture of Philippe di Francesco and Paul Zinn-Justin related to the qKZ equations and to Dave Robbins’ Two Favorite Combinatorial Objects

Doron ZEILBERGER^{1}

For any Laurent polynomial P (z_{1}, …, z_{n}) in the variables (z_{1}, …, z_{n}), recall that the anti-symmetrizer, AS(P) is defined by:

AS(P)(z_{1}, …, z_{n}) = ∑_{π∈S_{n}}(−1)^{inv(π)}P (z_{π(1)}, …, z_{π(n)}) ,

where S_{n} is the group of permutations on {1, …, n} and inv(π) is the number of inversions of π (the number of pairs 1 ≤ i < j ≤ n such that π(i) > π(j)).

Also define the negative-part of P, N(P), to be the polynomial in 1/z_{1}, …, 1/z_{n} obtained by removing all monomials that have at least one positive exponent.

For example, AS(5+z_{1}^{−1} +z_{1}^{3}z_{2}^{−2}) = z_{1}^{−1} +z_{1}^{3}z_{2}^{−2} −z_{2}^{−1} −z_{2}^{3}z_{1}^{−2}, and N(5+z_{1}^{−1} +z_{1}^{3}z_{2}^{−2}) = 5+z_{1}^{−1}.

Let

F_{n}(z_{1}, …, z_{n}) = ∏_{i=1}^{n} z_{i}^{−2i+2} ∏_{1≤i≤j≤n}(1 − z_{i}z_{j}) ∏_{1≤i<j≤n}(1 + z_{i}z_{j} + T z_{j}) ,

B_{n}(z_{1}, …, z_{n}) = ∏_{1≤i<j≤n}(z_{j}^{−1} − z_{i}^{−1})(T + z_{i}^{−1} + z_{j}^{−1}) .

In [1] (Eq. (4.4)), the following conjecture was made:

N [AS [F_{n}(z_{1}, …, z_{n})]] = B_{n}(z_{1}, …, z_{n}) .  (diF JZ)

Let G_{n} = AS(F_{n}). By breaking-up the summation over S_{n} into those π for which π(n) = i (i = 1, …, n), G_{n} is seen to satisfy the recurrence

G_{n}(z_{1}, …, z_{n}) = ∑_{j=1}^{n}(−1)^{n−j} (z_{j}^{1−2n}(1 − z_{j}^{2}) ∏_{i=1,i≠j}^{n} [(1 + T z_{j} + Lz_{i}z_{j})(1 − z_{i}z_{j})]) ·

G_{n−1}(z_{1}, …, z_{j−1}, z_{j+1}, …, z_{n}) .

Let H_{n} = N(G_{n}) (our object of desire). Applying the N-operation to both sides, and noting that since the factor in front of G_{n−1}(z_{1}, …, z_{j−1}, z_{j+1}, …, z_{n}) only has positive exponents in (z_{1}, …, z_{j−1}, z_{j+1}, …, z_{n}), we have

1 Department of Mathematics, Rutgers University (New Brunswick), Hill Center-Busch Campus, 110 Frelinghuysen

Rd., Piscataway, NJ 08854-8019, USA. zeilberg at math dot rutgers dot edu , http://www.math.rutgers.edu/~zeilberg . March 21, 2007. Exclusively published in the Personal Journal of S.B.

Ekhad and D. Zeilberger http://www.math.rutgers.edu/~zeilberg/pj.html . Supported in part by the NSF.

H_{n}(z_{1}, …, z_{n}) = N{∑_{j=1}^{n}(−1)^{n−j}z_{j}^{1−2n}(1 − z_{j}^{2}) ∏_{i=1,i≠j}^{n} [(1 + T z_{j} + Lz_{i}z_{j})(1 − z_{i}z_{j})] ·

H_{n−1}(z_{1}, …, z_{j−1}, z_{j+1}, …, z_{n})} .

By the inductive hypotheis H_{n−1} = B_{n−1} so we have to prove that

B_{n}(z_{1}, …, z_{n}) = N{∑_{j=1}^{n}(−1)^{n−j}z_{j}^{1−2n}(1 − z_{j}^{2}) ∏_{i=1,i≠j}^{n} [(1 + T z_{j} + Lz_{i}z_{j})(1 − z_{i}z_{j})] ·

B_{n−1}(z_{1}, …, z_{j−1}, z_{j+1}, …, z_{n})} .

Now comes a:

Nice Surprise:

∑_{j=1}^{n} ((−1)^{n−j}z_{j}^{1−2n}(1 − z_{j}^{2}) ∏_{i=1,i≠j}^{n} [(1 + T z_{j} + Lz_{i}z_{j})(1 − z_{i}z_{j})])·B_{n−1}(z_{1}, …, z_{j−1}, z_{j+1}, …, z_{n}) =

B_{n}(z_{1}, …, z_{n})(1 − z_{1}^{2}z_{2}^{2} ⋯ z_{n}^{2}) .  (NiceSurprise)

By the pigeonhole principle, each and every monomial feauting in B_{n}(z_{1}, …, z_{n})z_{1}^{2}z_{2}^{2} ⋯ z_{n}^{2} has at least one positive exponent, hence

N(B_{n}(z_{1}, …, z_{n})(1 − z_{1}^{2}z_{2}^{2} ⋯ z_{n}^{2})) = B_{n}(z_{1}, …, z_{n}) ,

and (diF JZ) would follow. It remains to prove the Nice Surprise. By dividing both sides of (NiceSurprise) by B_{n}, and rearranging terms, we see that it is equivalent to the following Intriguing Identity:

∑_{j=1}^{n} \frac{(2z_{j} + T) ∏_{i=1}^{n}[(1 + z_{i}(T + z_{j}))(1 − z_{i}z_{j})]}{(1 + z_{j}T + z_{j}^{2}) ∏_{i=1,i≠j}^{n}(z_{j} − z_{i}) ∏_{i=1}^{n}(z_{j} + z_{i} + T)} = (−1)^{n−1}(1 − z_{1}^{2} ⋯ z_{n}^{2}) .

(IntriguingIdentity)

Let α = α(T) and β = β(T) be the two roots of 1 + zT + z^{2} = 0. Note that α + β = −T and αβ = 1.

Now it is obvious whom to call for help! Dear old Count Joe!, a.k.a. as Joseph-Louis Lagrange, Comte de l’Empire, whose humble (and a posteriori trivial) Lagrange Interpolation Formula is at least as useful as Euler-Lagrange and Lagrange multipliers and sum-of-four-squares combined! To wit, for any polynomial P (z) of degree ≤ N − 1 and any numbers w_{1}, …, w_{N}, we have

P (z) = ∑_{j=1}^{N} \frac{P (w_{j}) ∏_{i=1,i≠j}^{N}(z − w_{i})}{∏_{i=1,i≠j}^{N}(w_{j} − w_{i})} .  (LI)

(Proof: both sides are equal at the N numbers z = w_{1}, …, w_{N}, hence they always equal (being polynomials of degree ≤ N − 1).

In fact, we will only need the corollary obtained by comparing the coefficient of z^{N−1} on both sides

Coeff of z^{N−1} ofP (z) = ∑_{j=1}^{N} \frac{P (w_{j})}{∏_{i=1,i≠j}^{n}(w_{j} − w_{i})} .  (LI′)

Now Lagrange-Interpolate

P (z) := (2z + T) ∏_{i=1}^{n} [(1 + z_{i}(T + z))(1 − z_{i}z)] ,

with respect to the 2n + 2 numbers

{z_{1}, …, z_{n}, −z_{1} − T, …, −z_{n} − T, α, β} ,

in order to get

2(−1)^{n}z_{1}^{2} ⋯ z_{n}^{2} = ∑_{j=1}^{n} \frac{P (z_{j})}{(1 + z_{j}T + z_{j}^{2}) ∏_{i=1,i≠j}^{n}(z_{j} − z_{i}) ∏_{i=1}^{n}(z_{j} + z_{i} + T)}

+ ∑_{j=1}^{n} \frac{P (−T − z_{j})}{(1 + z_{j}T + z_{j}^{2}) ∏_{i=1,i≠j}^{n}(−z_{j} − T − z_{i}) ∏_{i=1}^{n}(−z_{j} − T + z_{i} + T)}

+\frac{P (α)}{(α − β) ∏_{i=1}^{n}(α − z_{i}) ∏_{i=1}^{n}(α + z_{i} + T)}

+\frac{P (β)}{(β − α) ∏_{i=1}^{n}(β − z_{i}) ∏_{i=1}^{n}(β + z_{i} + T)} .

Since P (−T − z_{j}) = −P (z_{j}) (check!) and using αβ = 1 and α + β = −T to simplify, we get that (IntriguingIdenity) is indeed true. .

Reference

1. P. di Francesco and P. Zinn-Justin, Quantum Knizhnik-Zamolodchikov equation, Totally Symmetric Self-Complementary Plane Partitions and Alternating Sign Matrices, http://www.arxiv.org/abs/math-

ph/0703015 .