%diFrancesco.tex:
%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
%%a Plain TeX file by Doron Zeilberger (3 pages)

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\bf
\centerline
{
Proof of a Conjecture of Philippe di Francesco and Paul Zinn-Justin 
related to the qKZ equations 
}
\centerline
{and to
Dave Robbins' Two Favorite Combinatorial Objects
}
\rm
\bigskip
\centerline{ {\it
Doron 
ZEILBERGER}\footnote{$^1$}
{\eightrm  \raggedright
Department of Mathematics, Rutgers University (New Brunswick),
Hill Center-Busch Campus, 110 Frelinghuysen Rd., Piscataway,
NJ 08854-8019, USA.
{\eighttt zeilberg  at math dot rutgers dot edu} ,
\hfill \break
{\eighttt http://www.math.rutgers.edu/\Tilde zeilberg} .
March 21, 2007.
Exclusively published in the Personal Journal of S.B. Ekhad
and D. Zeilberger
{\eighttt http://www.math.rutgers.edu/\Tilde zeilberg/pj.html} .
Supported in part by the NSF.
}
}

For any Laurent polynomial $P(z_1, \dots, z_n)$ in the
variables $(z_1, \dots, z_n)$, recall that the
{\it anti-symmetrizer}, $AS(P)$ is defined by:
$$
AS(P)(z_1, \dots, z_n) =\sum_{\pi \in S_n} (-1)^{inv(\pi)} 
P(z_{\pi(1)}, \dots, z_{\pi(n)}) \quad ,
$$
where $S_n$ is the group of permutations on $\{1, \dots, n \}$
and $inv(\pi)$ is the number of inversions  of $\pi$
(the number of pairs $1 \leq i<j \leq n$ such that
$\pi(i)>\pi(j)$).

Also define the {\it negative-part} of $P$, $N(P)$, to be
the polynomial in $1/z_1, \dots, 1/z_n$ obtained by
removing all monomials that have at least one positive exponent.

For example, 
$AS(5+z_1^{-1}+z_1^3z_2^{-2})=z_1^{-1}+z_1^3z_2^{-2}-z_2^{-1}-z_2^3z_1^{-2}$, 
and 
$N(5+z_1^{-1}+z_1^3z_2^{-2})=5+z_1^{-1}$.

Let
$$
F_n(z_1, \dots, z_n)=
\prod_{i=1}^{n} z_i^{-2i+2}\prod_{1 \leq i \leq j \leq n}(1-z_i z_j)
\prod_{1 \leq i < j \leq n}(1+z_iz_j+Tz_j) \quad,
$$
$$
B_n(z_1, \dots, z_n)=
\prod_{1 \leq i < j \leq n}(z_j^{-1}-z_i^{-1}) (T+z_i^{-1}+z_j^{-1})
\quad .
$$
In [1] (Eq. (4.4)), the following conjecture was made:
$$
N \left [ AS \left [ F_n(z_1, \dots , z_n) \right ] \right ]=
B_n(z_1, \dots , z_n) \quad.
\eqno(diFJZ)
$$

Let $G_n=AS(F_n)$. By breaking-up  the summation over
$S_n$ into those $\pi$ for which $\pi(n)=i$
($i=1, \dots , n$), $G_n$ is seen to satisfy the recurrence
$$
G_n(z_1, \dots, z_n)=
\sum_{j=1}^{n} (-1)^{n-j} 
\left ( z_j^{1-2n} (1-z_j^2)
\prod_{i=1, i \neq j}^{n} \left [ (1+T z_j + Lz_i z_j) (1-z_i z_j)
\right ] \right ) \cdot
$$
$$
G_{n-1}(z_1, \dots, z_{j-1}, z_{j+1}, \dots, z_n) \quad .
$$
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 $G_{n-1}(z_1, \dots, z_{j-1}, z_{j+1}, \dots, z_n)$ only has
{\it positive} exponents in $(z_1, \dots, z_{j-1}, z_{j+1}, \dots, z_n)$,
\vfill\eject
we have
$$
H_n(z_1, \dots, z_n)=
N  \{
\sum_{j=1}^{n} 
(-1)^{n-j} z_j^{1-2n} (1-z_j^2)
\prod_{i=1, i \neq j}^{n} \left [ (1+T z_j + Lz_i z_j) (1-z_i z_j)
\right ]  \cdot
$$
$$
H_{n-1}(z_1, \dots, z_{j-1}, z_{j+1}, \dots, z_n) \} \quad.
$$

By the inductive hypotheis $H_{n-1}=B_{n-1}$ so we have to prove
that
$$
B_n(z_1, \dots, z_n)=
N  \{
\sum_{j=1}^{n} (-1)^{n-j} z_j^{1-2n} (1-z_j^2)
\prod_{i=1, i \neq j}^{n} \left [ (1+T z_j + Lz_i z_j) (1-z_i z_j)
\right ] \cdot
$$
$$
B_{n-1}(z_1, \dots, z_{j-1}, z_{j+1}, \dots, z_n)  \} \quad.
$$

Now comes a:

{\bf Nice Surprise:}
$$
\sum_{j=1}^{n} 
\left ( (-1)^{n-j} z_j^{1-2n} (1-z_j^2)
\prod_{i=1, i \neq j}^{n} \left [ (1+T z_j + Lz_i z_j) (1-z_i z_j)
\right ] \right ) \cdot
B_{n-1}(z_1, \dots, z_{j-1}, z_{j+1}, \dots, z_n)=
$$
$$
B_{n}(z_1, \dots, z_n) (1- z_1^2 z_2^2 \cdots z_n^2 ) \quad.
\eqno(NiceSurprise)
$$
By the {\it pigeonhole principle}, each and every monomial feauting in
$B_{n}(z_1, \dots, z_n) z_1^2 z_2^2 \cdots z_n^2 $
has at least one positive exponent, hence
$$
N(B_{n}(z_1, \dots, z_n) (1- z_1^2 z_2^2 \cdots z_n^2 ))=
B_{n}(z_1, \dots, z_n)  \quad ,
$$
and $(diFJZ)$ 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:
$$
\sum_{j=1}^{n}
{
{(2z_j+T) \prod_{i=1}^{n} [ (1+z_i(T+z_j))(1-z_iz_j)] }
\over
{ (1+z_jT+z_j^2) \prod_{i=1, i \neq j}^{n} (z_j-z_i) 
\prod_{i=1}^{n} (z_j+z_i+T) 
} }
=(-1)^{n-1} (1-z_1^2 \cdots z_n^2) \quad .
\eqno(IntriguingIdentity)
$$
Let $\alpha=\alpha(T)$ and $\beta=\beta(T)$ be the two roots of
$1+zT+z^2=0$. Note that $\alpha+\beta=-T$ and $\alpha \beta=1$.

Now it is obvious whom to call for help! Dear old Count Joe!,
a.k.a. as Joseph-Louis Lagrange, {\it Comte de l'Empire},
whose humble (and a posteriori trivial) {\it 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 $\leq N-1$ and
{\it any} numbers $w_1, \dots, w_{N}$, we have
$$
P(z)=\sum_{j=1}^{N}
{{P(w_j) \prod_{i=1, i \neq j}^{N} (z-w_i)} 
\over
{\prod_{i=1, i \neq j}^{N} (w_j-w_i)}} \quad .
\eqno(LI)
$$
(Proof: both sides are equal at the $N$ numbers $z=w_1, \dots, w_N$,
hence they {\it always} equal (being polynomials of degree $\leq N-1$).

In fact, we will only need the corollary obtained by
comparing the coefficient of $z^{N-1}$ on both sides
$$
Coeff \quad of \quad z^{N-1} \quad of P(z)=\sum_{j=1}^{N}
{{P(w_j)} 
\over
{\prod_{i=1, i \neq j}^{n} (w_j-w_i)}} \quad .
\eqno(LI')
$$
Now Lagrange-Interpolate
$$
P(z) :=
(2z+T) \prod_{i=1}^{n} \left [ (1+ z_i(T+z)) (1-z_iz) \right ] \quad ,
$$
with respect to the $2n+2$ numbers 
$$
\{ z_1, \dots, z_n, -z_1-T, \dots ,-z_n-T, \alpha, \beta \} \quad ,
$$
in order to get
$$
2(-1)^n z_1^2 \cdots z_n^2=
\sum_{j=1}^{n}
{ {P(z_j) }
\over
{ (1+z_jT+z_j^2) \prod_{i=1, i \neq j}^{n} (z_j-z_i) 
\prod_{i=1}^{n} (z_j+z_i+T) 
} }
$$
$$
+\sum_{j=1}^{n}
{
{P(-T-z_j) }
\over
{ (1+z_jT+z_j^2) \prod_{i=1, i \neq j}^{n} (-z_j-T-z_i) 
\prod_{i=1}^{n} (-z_j-T+z_i+T) 
} 
}
$$
$$
+
{
{P(\alpha) }
\over
{ (\alpha-\beta)  \prod_{i=1}^{n} (\alpha-z_i) 
\prod_{i=1}^{n}  (\alpha+z_i+T) } }
$$
$$
+
{
{P(\beta) }
\over
{ (\beta-\alpha)  \prod_{i=1}^{n} (\beta-z_i) 
\prod_{i=1}^{n}  (\beta+z_i+T) } } \quad .
$$
Since $P(-T-z_j)=-P(z_j)$ (check!) and using
$\alpha\beta=1$ and $\alpha+\beta=-T$ to simplify,
we get that $(IntriguingIdenity)$  is indeed true. \halmos.

{\bf Reference}

1. P. di Francesco and P. Zinn-Justin,
{\it Quantum Knizhnik-Zamolodchikov equation, Totally
Symmetric Self-Complementary Plane Partitions and Alternating
Sign Matrices}, 
{\eighttt http://www.arxiv.org/abs/math-ph/0703015} .
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