\input amstex
\input amsppt.sty

\input texdraw

\magnification1200
\hsize14cm
\vsize19cm

\font\gigantic=cmr17

\TagsOnRight
\NoBlackBoxes

\def\CiKrAF{1}
\def\LascAZ{2}

%equation numbers
\def\AA{2.1}
\def\AB{2.2}
\def\ABa{2.4}
\def\ABb{2.6}
\def\AC{1.1}
\def\ACa{1.2}
\def\ACb{1.3}
\def\ACc{1.4}
\def\ACd{1.5}
\def\AD{5.1}
\def\AE{5.2}
\def\AF{5.3}
\def\AFa{5.5}
\def\AFb{5.4}
\def\AG{1}
\def\AGa{2}
\def\AGb{3}
\def\AGc{4}
\def\AGd{5}
\def\AH{6}
\def\AHa{7}
\def\AI{4.2}
\def\AJ{4.3}
\def\AK{4.4}
\def\AKa{2.3}
\def\AL{4.5}
\def\AMa{4.6}
\def\AM{4.7}
\def\AN{4.8}
\def\AO{4.9}
\def\AP{4.10}
\def\AQ{4.11}
\def\AR{2.7}
\def\AS{2.5}
\def\AT{6.1}
\def\AU{6.2}
\def\AV{6.3}
\def\AW{6.4}
\def\AX{6.5}
\def\AY{6.6}
\def\AZ{6.7}

\def\Pf{\operatorname{Pf}}



\topmatter
\title Two polynomial identities
\endtitle
\author C.~Krattenthaler and A. Lascoux
\endauthor
\affil Fakult\"at f\"ur Mathematik der Universit\"at Wien,\\
Nordbergstra{\ss}e 15, A-1090 Wien, Austria.\\
WWW: {\tt http://www.mat.univie.ac.at/\~{}kratt}\\\vskip6pt
Institut Gaspard Monge, Universit\'e de
Marne-la-Vall\'ee,\\
F-77454 Marne-la-Vall\'ee Cedex~2,
France\\
WWW: \tt http://www-igm.univ-mlv.fr/\~{}al
\endaffil
\address Fakult\"at f\"ur Mathematik der Universit\"at Wien,
Nordbergstra\ss e 15, A-1090 Wien, Austria.
\endaddress
\address{Institut Gaspard Monge, Universit\'e de Marne-la-Vall\'ee,
F-77454 Marne-la-Vall\'ee Cedex~2, France.}\endaddress

\thanks{The
research of the first author was partially supported by the Austrian
Science Foundation FWF, grants Z130-N13 and S9607-N13,
the latter in the framework of the National Research Network
``Analytic Combinatorics and Probabilistic Number Theory."}\endthanks
\subjclassyear{2000}
\subjclass Primary 05A19;
Secondary 05A15 05A17 05E15 11P81 20C15 20G05 
\endsubjclass
\keywords Vandermonde determinant, polynomial identities\endkeywords
\abstract
We establish two polynomial identities. One of which generalizes a key
identity in ``A factorization theorem for classical group characters, with
applications to plane partitions and rhombus tilings" by 
Mihai Ciucu and
the first author [preprint 2008]. The other shows that the two sides
in the identity can be expressed alternatively in form of a Pfaffian.
\endabstract
\endtopmatter
\document

\leftheadtext{C. Krattenthaler and A. Lascoux}
\rightheadtext{Two polynomial identities}

The purpose of this note is to establish two polynomial identities. 
The first (see Theorem~1 below) generalizes an identity in
\cite{\CiKrAF} which was crucial there to prove certain
factorization theorems for classical group characters, and for
numbers of rhombus tilings of certain regions. The second (see
Theorem~2) provides a surprising alternative expression for the two
sides in the identity involving a Pfaffian.

In the statement of our results, we use additive notation for sets of
variables (``alphabets") as in \cite{\LascAZ}. Further notations that
we use are:
the symbol $X_M$ denotes the alphabet $x_1+x_2+\dots+x_M$; given an
alphabet $A$, we write $V(A)$ for the ``Vandermonde product"
$$\underset{a<b}\to{\prod _{a,b\in A} ^{}}(a-b).$$

The following theorem generalizes Lemma~1 in \cite{\CiKrAF}.

\proclaim{Theorem 1}
Let $a,b$ be two variables. Then there holds the identity
$$
\underset{\vert A\vert=N}\to{\sum _{A+B=X_{2N}} ^{}}
V(a+A+A^{-1})V(b+B+B^{-1})
=
\sum _{A+B=X_{2N}} ^{}
V(a+A+B^{-1})V(b+B+A^{-1}),
\tag\AG$$
where $A^{-1}$ denotes the alphabet $\sum _{\alpha\in A} ^{}\alpha^{-1}$, 
with an analogous meaning for $B^{-1}$.
\endproclaim
\demo{Proof} 
We prove the assertion by induction on $N$. In fact, the induction
proof will prove {\it two} assertions at the same time, namely (\AG)
and
$$\multline
(-1)^N\underset{\vert A\vert=N}\to{\sum _{A+B=X_{2N-1}} ^{}}
V(A+A^{-1})V(a+b+B+B^{-1})\\
=
\sum _{A+B=X_{2N-1}} ^{}(-1)^{\vert A\vert}
V(a+A+B^{-1})V(b+B+A^{-1}).
\endmultline\tag\AGa$$
For $N=1$, identity~(\AG) reduces to
$$\align
(a-x_1)&(a-x_1^{-1})(x_1-x_1^{-1})(b-x_2)(b-x_2^{-1})(x_2-x_2^{-1})\\
&\kern1cm
+(a-x_2)(a-x_2^{-1})(x_2-x_2^{-1})(b-x_1)(b-x_1^{-1})(x_1-x_1^{-1})\\
&=
(a-x_1^{-1})(a-x_2^{-1})(x_1^{-1}-x_2^{-1})(b-x_1)(b-x_2)(x_1-x_2)\\
&\kern1cm
+(a-x_1)(a-x_2^{-1})(x_1-x_2^{-1})(b-x_2)(b-x_1^{-1})(x_2-x_1^{-1})\\
&\kern1cm
+(a-x_2)(a-x_1^{-1})(x_2-x_1^{-1})(b-x_1)(b-x_2^{-1})(x_1-x_2^{-1})\\
&\kern1cm
+(a-x_1)(a-x_2)(x_1-x_2)(b-x_1^{-1})(b-x_2^{-1})(x_1^{-1}-x_2^{-1}),
\endalign$$
while (\AGa) reduces to
$$
-(x_1-x_1^{-1})(a-b)
=-(a-x_1)(b-x_1^{-1})
+(a-x_1^{-1})(b-x_1),
$$
both of which can be readily verified.

Now let us suppose that we have proved (\AG) and (\AGa) with $N$ replaced by
$N-1$. Below, we shall show that this implies that identities (\AG)
and (\AGa) are
true if we specialize $x_1$ to $x_2$. Let us for the moment
suppose that this is already done. It is easy to
see that both sides of (\AG) are symmetric in the variables
$x_1,x_2,\dots,x_{2N}$, while both sides of (\AGa)
are symmetric in the variables $x_1,x_2,\dots,x_{2N-1}$. Moreover,
both sides of (\AG) and both sides of (\AGa) 
remain invariant up to an overall
multiplicative sign if we replace $x_1$ by $x_1^{-1}$
(on the left-hand sides of (\AG) respectively (\AGa), the latter assertion
already applies to each individual summand; 
on the right-hand sides, one has to group the summands in pairs:
the summands corresponding to a pair of alphabets $(A,B)$ and the pair of 
symmetric differences 
$(A\triangle \{x_1\},B\triangle \{x_1\})$ have to be considered
together).

This means
that identity (\AGa) holds for {\it all\/} specializations of $x_1$ to
one of $x_2,x_2^{-1},x_3,x_3^{-1},\mathbreak
\dots,x_{2N-1},x_{2N-1}^{-1}$. 
These are $4N-4$
specializations. The previously observed fact that the replacement of
$x_1$ by $x_1^{-1}$ changes the sign of both sides of (\AGa) implies
that both sides vanish for $x_1=\pm1$. Thus, we have $4N-2$
specializations of $x_1$ for which (\AGa) holds.

We now specialize $x_1=a$ on both sides of (\AGa). Under this
specialization, terms corresponding to alphabets $B$ which contain
$x_1$ vanish on the left-hand side of (\AGa) due to the
appearance of the Vandermonde product $V(a+b+B+B^{-1})$, while terms
corresponding to alphabets $A$ which contain $x_1$ vanish on the
right-hand side of (\AGa) due to the appearance of the Vandermonde
product $V(a+A+B^{-1})$.
Therefore, if $x_1=a$, identity (\AGa) reduces to
$$\multline 
(-1)^N(a-a^{-1})(a-b)
\bigg(\prod _{j=2}
^{2N-1}(a-x_j)(a-x_j^{-1})\bigg)\\
\times
\underset{\vert A\vert=N-1}\to{\sum _{A+B=X'} ^{}}
V(A+a^{-1}+A^{-1})V(b+B+B^{-1})\\
=
(a-a^{-1})(b-a)
\bigg(\prod _{j=2}
^{2N-1}(a-x_j)(a-x_j^{-1})\bigg)\kern4cm\\
\times
\sum _{A+B=X'} ^{}(-1)^{\vert A\vert}
V(A+a^{-1}+B^{-1})V(b+B+A^{-1}),
\endmultline\tag\AGb$$
where $X'=x_2+x_3+\dots+x_{2N-1}$.
After clearing the product which is common to both sides, and after
little rearrangement, we see that
the remaining identity is an instance of (\AG) with $N$ replaced by
$N-1$, the latter being true due to the induction hypothesis.

In summary, we have found $4N-1$ specializations of $x_1$ for which
both sides of (\AGa) agree.
On the other hand, as Laurent polynomials in $x_1$, 
the degree of both sides of (\AGa) is at most $2N-1$. (That is, the
maximal exponent $e$ of a power $x_1^e$ is $e=2N-1$, and the minimal
exponent is $e=-(2N-1)$.) Hence, both sides of (\AGa) must be
identical.

\smallskip
The earlier arguments show as well that 
identity (\AG) holds for {\it all\/} specializations of $x_1$ to
one of $x_2,x_2^{-1},x_3,x_3^{-1},\dots,x_{2N},x_{2N}^{-1}$ and $\pm1$. 
These are $4N$ specializations. 

We now specialize $x_1=a$ on both sides of (\AG). Under this
specialization, terms corresponding to alphabets $A$ which contain
$x_1$ vanish on both sides of (\AG) due to the
appearance of the Vandermonde products $V(a+A+A^{-1})$ respectively
$V(a+A+B^{-1})$. Therefore, if $x_1=a$, identity (\AG) reduces to
$$\multline 
(a-a^{-1})(b-a)
\bigg(\prod _{j=2}
^{2N}(a-x_j)(a-x_j^{-1})\bigg)\\
\times
\underset{\vert A\vert=N}\to{\sum _{A+B=X''} ^{}}
V(A+A^{-1})V(b+B+a^{-1}+B^{-1})\\
=
(a-a^{-1})(b-a)
\bigg(\prod _{j=2}
^{2N}(a-x_j)(a-x_j^{-1})\bigg)\kern4cm\\
\times
\sum _{A+B=X''} ^{}
V(A+a^{-1}+B^{-1})V(b+B+A^{-1}),
\endmultline\tag\AGc$$
where $X''=x_2+x_3+\dots+x_{2N}$.
After clearing the product which is common to both sides, and after
little rearrangement, we see that
the remaining identity is an instance of (\AGa) which has been just
established.

In summary, we have found $4N+1$ specializations of $x_1$ for which
both sides of (\AG) agree.
On the other hand, as Laurent polynomials in $x_1$, 
the degree of both sides of (\AG) is at most $2N$. (That is, the
maximal exponent $e$ of a power $x_1^e$ is $e=2N$, and the minimal
exponent is $e=-2N$.) Hence, both sides of (\AG) must be
identical.

\medskip
It remains to prove that, under the induction hypothesis,
Equations~(\AG) and (\AGa) hold for $x_1=x_2$. We provide the argument
establishing this fact for (\AG). The argument for (\AGa) is
completely analogous.

Indeed, under
this specialization, terms corresponding to alphabets $A$ which contain
{\it both\/} $x_1$ and $x_2$ and to those which contain {\it neither}  
$x_1$ {\it nor} $x_2$ vanish on both sides of (\AG), because of the
appearance of the Vandermonde products $V(A)$ respectively $V(B)$.
Therefore, if $x_1=x_2$, identity (\AG) reduces to
$$\multline 
2(a-x_1)(a-x_1^{-1})
(b-x_1)(b-x_1^{-1})
(x_1-x_1^{-1})^2\bigg(\prod _{j=3}
^{2N}(x_1-x_j)(x_1^{-1}-x_j^{-1})(x_1-x_j^{-1})(x_j-x_1^{-1})\bigg)\\
\times
\underset{\vert A\vert=N-1}\to{\sum _{A+B=X'''} ^{}}
V(a+A+A^{-1})V(b+B+B^{-1})\\
=2
(a-x_1)(a-x_1^{-1})
(b-x_1)(b-x_1^{-1})
(x_1-x_1^{-1})^2\bigg(\prod _{j=3}
^{2N}(x_1-x_j)(x_1^{-1}-x_j^{-1})(x_1-x_j^{-1})(x_j-x_1^{-1})\bigg)\\
\times
\sum _{A+B=X'''} ^{}
V(a+A+B^{-1})V(b+B+A^{-1}),
\endmultline\tag\AGd$$
where $X'''=x_3+x_4+\dots+x_{2N}$.
After clearing the product which is common to both sides, we see that
the remaining identity is equivalent to (\AG) with $N$ replaced by
$N-1$, the latter being true due to the induction hypothesis. This
finishes the proof of the theorem.\quad \quad \qed
\enddemo

\proclaim{Theorem 2}
Let $a,b$ be two variables, and let
$$\Phi(a,b;x_i,x_j)=\frac {(a-x_i)(a-x_i^{-1})(b-x_j)(b-x_j^{-1})+
(a-x_j)(a-x_j^{-1})(b-x_i)(b-x_i^{-1})} {(x_i+x_i^{-1}-x_j-x_j^{-1})}.$$ 
Then there holds the identity
$$\multline 
\prod _{i=1} ^{2N}(x_i-x_i^{-1})
\prod _{1\le i<j\le 2N} ^{}(x_i+x_i^{-1}-x_j-x_j^{-1})
\underset{1\le i<j\le 2N}\to\Pf
\left(\Phi(a,b;x_i,x_j)\right)
\\
=
\underset{\vert A\vert=N}\to{\sum _{A+B=X_{2N}} ^{}}
V(a+A+A^{-1})V(b+B+B^{-1}).
\endmultline\tag\AH$$
\endproclaim

\demo{Proof}
We prove the assertion by induction on $N$. Again, the induction
proof will prove {\it two} assertions at the same time, namely (\AH)
and
$$\multline
a^{2N-2}\prod _{i=1} ^{2N-1}(x_i-x_i^{-1})
\prod _{1\le i<j\le 2N-1} ^{}(x_i+x_i^{-1}-x_j-x_j^{-1})\\
\times
\underset{1\le i<j\le 2N}\to\Pf
\left(\matrix b-a&i=1\\  
\Phi(a^{-1},b;x_{i-1},x_{j-1})
&i\ge 2\endmatrix\right)
\\
=(-1)^N
\underset{\vert A\vert=N}\to{\sum _{A+B=X_{2N-1}} ^{}}
V(A+A^{-1})V(a+b+B+B^{-1}).
\endmultline\tag\AHa$$
For $N=1$, both (\AH) and (\AHa) reduce to tautologies.

Now let us suppose that we have proved (\AH) and (\AHa) with $N$ replaced by
$N-1$. Below, we shall show that this implies that identities (\AH)
and (\AHa) are
true if we specialize $x_1$ to $x_2$. Let us for the moment
suppose that this is already done. It is easy to
see that both sides of (\AH) are symmetric in the variables
$x_1,x_2,\dots,x_{2N}$, while both sides of (\AHa)
are symmetric in the variables $x_1,x_2,\dots,x_{2N-1}$. Moreover,
both sides of (\AH) and both sides of (\AHa) 
remain invariant up to an overall
multiplicative sign if we replace $x_1$ by $x_1^{-1}$.

This means
that identity (\AHa) holds for {\it all\/} specializations of $x_1$ to
one of $x_2,x_2^{-1},x_3,x_3^{-1},\mathbreak
\dots,x_{2N-1},x_{2N-1}^{-1}$. 
These are $4N-4$
specializations. The previously observed fact that the replacement of
$x_1$ by $x_1^{-1}$ changes the sign of both sides of (\AHa) implies
that both sides vanish for $x_1=\pm1$. Thus, we have $4N-2$
specializations of $x_1$ for which (\AHa) holds.

We now specialize $x_1=a$ on both sides of (\AHa). 
Since
$$\Phi(a^{-1},b;a,x_j)=a^{-1}(b-a)(b-a^{-1}),$$
under this
specialization, the left-hand side of (\AHa) becomes
$$\multline
a^{2N-2}(a-a^{-1})\prod _{i=2} ^{2N-1}(x_i-x_i^{-1})
\prod _{j=2} ^{2N-1}(a+a^{-1}-x_j-x_j^{-1})
\prod _{2\le i<j\le 2N-1} ^{}(x_i+x_i^{-1}-x_j-x_j^{-1})\\
\times
\underset{1\le i<j\le 2N}\to\Pf
\left(\matrix 
b-a&i=1\\  
a^{-1}(b-a)(b-a^{-1})&i=2\\  
\Phi(a^{-1},b;x_{i-1},x_{j-1})
&i\ge 3\endmatrix\right)\\
=a^{2N-2}(a-a^{-1})(b-a)\kern9cm\\
\times\prod _{i=2} ^{2N-1}(x_i-x_i^{-1})
\prod _{j=2} ^{2N-1}(a+a^{-1}-x_j-x_j^{-1})
\prod _{2\le i<j\le 2N-1} ^{}(x_i+x_i^{-1}-x_j-x_j^{-1})\\
\times
\underset{1\le i<j\le 2N}\to\Pf
\left(\matrix 
1&i=1,j=2\\  
b-a&i=1,j>2\\
a^{-1}(b-a^{-1})&i=2\\  
\Phi(a^{-1},b;x_{i-1},x_{j-1})
&i\ge 3\endmatrix\right).
\endmultline$$
Here, the Pfaffian can be simplified. To see this most transparently,
it is useful to display the skew symmetric matrix of which the Pfaffian is
taken:
$$\pmatrix 0&1&b-a&b-a&\dots&b-a\\
-1&0&a^{-1}(b-a^{-1})&a^{-1}(b-a^{-1})&\dots&a^{-1}(b-a^{-1})\\
-(b-a)&-a^{-1}(b-a^{-1})\\
\vdots&\vdots&&\text{\gigantic*}\\
-(b-a)&-a^{-1}(b-a^{-1})\endpmatrix.$$
If we subtract $a(b-a)/b-a^{-1}$ times the second row from the first row
(and subsequently do the same operation with the corresponding columns),
then this matrix is transformed to
$$\pmatrix 0&1&0&0&\dots&0\\
-1&0&a^{-1}(b-a^{-1})&a^{-1}(b-a^{-1})&\dots&a^{-1}(b-a^{-1})\\
0&-a^{-1}(b-a^{-1})\\
\vdots&\vdots&&\text{\gigantic*}\\
0&-a^{-1}(b-a^{-1})\endpmatrix,$$
without changing the Pfaffian. The Pfaffian of the last matrix however 
is $\Pf(\text{\raise-6pt\hbox{\gigantic*}})$. Therefore, under the specialization $x_1=a$,
the left-hand side of (\AHa) is equal to
$$\multline
=a^{2N-2}(a-a^{-1})(b-a)
\prod _{i=2} ^{2N-1}(x_i-x_i^{-1})
\prod _{j=2} ^{2N-1}(a+a^{-1}-x_j-x_j^{-1})\\
\times
\prod _{2\le i<j\le 2N-1} ^{}(x_i+x_i^{-1}-x_j-x_j^{-1})
\underset{1\le i<j\le 2N-2}\to\Pf
\left(\Phi(a^{-1},b;x_{i+1},x_{j+1})\right).
\endmultline$$
On the other hand, the specialization of the right-hand side of (\AHa)
to $x_1=a$ is exactly the left-hand side in (\AGb).
Hence, after clearing common factors, we see that for $x_1=a$ identity
(\AHa) reduces to
$$\multline 
\prod _{i=2} ^{2N-1}(x_i-x_i^{-1})
\prod _{2\le i<j\le 2N-1} ^{}(x_i+x_i^{-1}-x_j-x_j^{-1})
\underset{1\le i<j\le 2N-2}\to\Pf
\left(\Phi(a^{-1},b;x_{i+1},x_{j+1})\right)\\
=
\underset{\vert A\vert=N-1}\to{\sum _{A+B=X'} ^{}}
V(a^{-1}+A+A^{-1})V(b+B+B^{-1}),
\endmultline$$
where $X'=x_2+x_3+\dots+x_{2N-1}$.
Clearly, this is an instance of (\AH) with $N$ replaced by
$N-1$, the latter being true due to the induction hypothesis.

In summary, we have found $4N-1$ specializations of $x_1$ for which
both sides of (\AHa) agree.
On the other hand, as Laurent polynomials in $x_1$
(the left-hand side of (\AHa) is indeed a Laurent polynomial in the
$x_i$'s since the denominators in the Pfaffian cancel with the product
over $1\le i<j\le 2N-1$\,!), 
the degree of both sides of (\AHa) is at most $2N-1$. 
Hence, both sides of (\AHa) must be identical.
\smallskip

The earlier arguments show as well that 
identity (\AH) holds for {\it all\/} specializations of $x_1$ to
one of $x_2,x_2^{-1},x_3,x_3^{-1},\dots,x_{2N},x_{2N}^{-1}$ and $\pm1$. 
These are $4N$ specializations. 

We now specialize $x_1=a$ on both sides of (\AH). 
Since
$$\Phi(a,b;a,x_j)=a(b-a)(b-a^{-1}),$$
under this
specialization, the left-hand side of (\AH) becomes
$$\multline 
(a-a^{-1})a(b-a)\prod _{i=2} ^{2N}(x_i-x_i^{-1})
\prod _{j=2} ^{2N}(a+a^{-1}-x_j-x_j^{-1})\\
\times
\prod _{2\le i<j\le 2N} ^{}(x_i+x_i^{-1}-x_j-x_j^{-1})
\underset{1\le i<j\le 2N}\to\Pf
\left(\matrix b-a^{-1}&i=1\\
\Phi(a,b;x_i,x_j)&i\ge2
\endmatrix\right).
\endmultline$$
On the other hand, the specialization of the right-hand side of (\AH)
to $x_1=a$ is exactly the left-hand side in (\AGc).
Hence, after clearing common factors, we see that for $x_1=a$ identity
(\AH) reduces to
$$\multline 
a^{2-2N}\prod _{i=2} ^{2N}(x_i-x_i^{-1})
\prod _{2\le i<j\le 2N} ^{}(x_i+x_i^{-1}-x_j-x_j^{-1})
\underset{1\le i<j\le 2N}\to\Pf
\left(\matrix b-a^{-1}&i=1\\
\Phi(a,b;x_i,x_j)&i\ge2
\endmatrix\right)\\
=(-1)^N
\underset{\vert A\vert=N}\to{\sum _{A+B=X''} ^{}}
V(A+A^{-1})V(a^{-1}+b+B+B^{-1}),
\endmultline$$
where $X''=x_2+x_3+\dots+x_{2N}$.
This is an instance of (\AHa) which has been just
established.

In summary, we have found $4N+1$ specializations of $x_1$ for which
both sides of (\AH) agree.
On the other hand, as Laurent polynomials in $x_1$, 
the degree of both sides of (\AH) is at most $2N$. 
Hence, both sides of (\AH) must be identical.

\medskip
It remains to prove that, under the induction hypothesis,
Equations~(\AH) and (\AHa) hold for $x_1=x_2$. We provide the argument
establishing this fact for (\AH). The argument for (\AHa) is completely
analogous.

Indeed, under this specialization, due to the term
$$\prod _{1\le i<j\le 2N} ^{}(x_i+x_i^{-1}-x_j-x_j^{-1})
$$
on the left hand side of (\AH), in the expansion of the Pfaffian only
terms survive which contain the $(1,2)$-entry of the Pfaffian. Thus, 
for $x_1=x_2$, the left-hand side of (\AH) reduces to
$$\multline 
2(a-x_1)(a-x_1^{-1})
(b-x_1)(b-x_1^{-1})
(x_1-x_1^{-1})^2\prod _{i=3} ^{2N}(x_i-x_i^{-1})
\bigg(\prod _{j=3}
^{2N}(x_1+x_1^{-1}-x_j-x_j^{-1})^2\bigg)\\
\times
\prod _{3\le i<j\le 2N} ^{}(x_i+x_i^{-1}-x_j-x_j^{-1})
\underset{3\le i<j\le 2N}\to\Pf
\left(\Phi(a,b;x_i,x_j)\right).
\endmultline$$
On the other hand, the specialization of the right-hand side of (\AH) to
$x_1=a$ is exactly the left-hand side of (\AGd).
Hence, after clearing common factors, we see that for $x_1=a$ identity
(\AH) reduces to
$$\multline 
\prod _{i=3} ^{2N}(x_i-x_i^{-1})
\prod _{3\le i<j\le 2N} ^{}(x_i+x_i^{-1}-x_j-x_j^{-1})
\underset{3\le i<j\le 2N}\to\Pf
\left(\Phi(a,b;x_i,x_j)\right)\\
=
\underset{\vert A\vert=N-1}\to{\sum _{A+B=X'''} ^{}}
V(a+A+A^{-1})V(b+B+B^{-1}),
\endmultline$$
where $X'''=x_3+x_4+\dots+x_{2N}$.
This identity is equivalent to (\AH) with $N$ replaced by
$N-1$, the latter being true due to the induction hypothesis. This
finishes the proof of the theorem.\quad \quad \qed
\enddemo

\Refs

\ref\no \CiKrAF\by M. Ciucu annd C. Krattenthaler \yr
\paper A factorization theorem for classical group characters, with
applications to plane partitions and rhombus tilings
\inbook Advances in Combinatorial Mathematics: Proceedings of the
Waterloo Workshop in Computer Algebra 2008\eds
I.~Kotsireas, E.~Zima\publ Springer--Verlag\publaddr New York\toappear\pages
\endref     

\ref\no\LascAZ\by
A. Lascoux\book Symmetric functions and combinatorial operators on
polynomials\publ CBMS Regional Conference Series in Mathematics,
vol.~99, Amer.\ Math.\ Soc.\publaddr Providence, RI\yr 2003\endref


\endRefs

\enddocument
