Proof of a Conjecture of Philippe Di Francesco and Paul Zinn-Justin related to the qKZ equation 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}), the anti-symmetrizer, A(P), is defined by:

A(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)).

Define the negativizer of P, N (P), to be the polynomial in z_{1}^{−1}, …, z_{n}^{−1} obtained by removing all monomials that have at least one positive exponent.

For example, A(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 [A [F_{n}(z_{1}, …, z_{n})]] = B_{n}(z_{1}, …, z_{n}) .  (DiF ZJ)

Let G_{n} = A(F_{n}). By breaking-up the summation over S_{n} into those π for which π(n) = j (j = 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 + z_{i}z_{j} + T 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}). We want to prove that for every n, H_{n} = B_{n}. We will do it by induction on n. The case n = 1 is obviously true (check!). 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 . First Version: March 21, 2007. This Version: March 27, 2007, [incorporating remarks of referee Philippe Di Francesco]. 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 + z_{i}z_{j} + T z_{j})(1 − z_{i}z_{j})] ·

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

By the inductive hypothesis 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 + z_{i}z_{j} + T z_{j})(1 − z_{i}z_{j})]) ·

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

It turns out that the negativizand on the right side is nice:

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

We now need:

A Simple Fact: Each and every monomial in B_{n}(z_{1}, …, z_{n})z_{1}^{2}z_{2}^{2} ⋯ z_{n}^{2} has at least one positive exponent, hence its negativizer is 0.

Proof of the Simple Fact: (This simple argument of referee Philippe Di Francesco replaces a previous more complicated argument).

B_{n}(z_{1}, …, z_{n}) = ∏_{1≤i<j≤n}(T z_{j}^{−1} + z_{j}^{−2} − T z_{i}^{−1} − z_{i}^{−2}) = ∏_{1≤i<j≤n}[(T/2 + z_{j}^{−1})^{2} − (T/2 + z_{i}^{−1})^{2}] .

So each term in the last product can’t have all the variables, and the same is true when it is fully expanded, and it follows that each monomial 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 ZJ) would follow. It remains to prove (NiceSurprise). By dividing both sides by B_{n}, letting z_{i} → z_{i}^{−1} (1 ≤ i ≤ n), and rearranging terms, we see that it is equivalent to the following:

∑_{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.

Recall the Lagrange Interpolation Formula. 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 are 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

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

Now let

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

and pick the w’s to be 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}^{n}(−z_{j} − T − z_{i}) ∏_{i=1,i≠j}^{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)} .

Using P (−T − z_{j}) = −P (z_{j}) (check!), αβ = 1 and α + β = −T, and manipulating the obvious cancellations, and observing that the first two terms are identical, while the sum of the last two terms simplifies to 2(−1)^{n}, we get that (IntriguingIdentity) 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 .