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{\bf
A STEMBRIDGE-STANTON STYLE PROOF OF THE HABSIEGER-KADELL}
\centerline
{\bf q-MORRIS IDENTITY}
\medskip
\centerline
{\it by Doron ZEILBERGER}
{\bf Appeared in: Discrete Math. 79 (1989), 313-322}
{\bf 0. NOMENCLATURE}
q and $ x=( x_1 , ... , x_n ) $ are commuting indeterminates.
If $\alpha = ( \alpha_1 , ... , \alpha_n ) $ is any vector of
integers , then $ x^{\alpha} $ stands for
$ x^{ \alpha_1 } ... x^{\alpha_n } $. For example if
$\alpha$=(1,-2,5), then $x^{\alpha}$ =
$ x_1 x_2^{-2} x_3^5$.
A {\it Laurent polynomial} is a finite linear combination of monomials
$x^{\alpha} $, where the $\alpha$s have integer components. All our
Laurent polynomials will be with integer coefficients.
"C.T." stands for "the constant term of", with respect to
$x= ( x_1 , ... , x_n )$. For example
C.T.(1-qx)(1-q/x)= 1+q.
The symmetric group $S_n$ acts on vectors of integers by permuting
the coordinates, for example 321(-1,2,1)=(1,2,-1). A permutation $\pi$
acts on monomials $x^{\gamma} $ by
$ \pi ( x^{\gamma} ) = x^{ \pi ( \gamma ) } $ , and by linearity on
any Laurent polynomial. For example,
$$
(321)[ x_1^{-1} x_2^2x_3 + 4 + x_1^2x_2^3x_3^{-5} ] = x_1 x_2^2x_3^{-1} + 4 + x_1^{-5} x_2^3x_1^2.
$$
A Laurent polynomial P in $x=( x_1 , ... , x_n ) $ is
{\it symmetric} if $ \pi ( P)= P $ for all permutations $\pi$, and is
{\it antisymmetric} if $\pi ( P)= (sgn \,\,\pi )P $ for every permutation
$\pi$.
$(y;Q)_a $, the "q-analog of $ {(1-y)}^a $ to base Q " is
defined by
$$
(y;Q)_a = (1-y)(1-Qy)...(1- Q^{a-1} y ),
$$
and if the base Q is q then we often abbreviate $(y;q)_a $ to
$ (y)_a $ .
A vector of integers $\alpha$ is a {\it bad guy} if it has two or more
identical components, otherwise it is a {\it good guy}. For example
$(1,3,1)$ and $(-1,-1,-1)$ are bad guys while $(1,-1,0)$ and $(2,1,8)$ are good
guys.
$ \delta $ is the vector $(0,1,...,n-1)$ and $ \bar{\delta} $ is its reverse:
$ \bar{\delta}$ = $( n-1, ... , 0 )$.
Throughout this paper $t= q^a $, $s= q^b $, $u= q^c $ .
{\bf 1. THE HABSIEGER-KADELL q-MORRIS IDENTITY}
Let, for a,b,c,n nonnegative integers,
$$
{F'_{a,b,c}}^{(n)} (x)= \prod_{i=1}^{n} { {( x_i )}_b {(q/ x_i )}_c }
\prod_{ 1 \leq i < j \leq n } { {( x_i / x_j ) }_a {( q x_j / x_i ) }_a }
\eqno(1.1)
$$
$$
{H'_{a,b,c}}^{(n)} = C.T. {F'_{a,b,c}}^{(n)} .
\eqno(1.2)
$$
$$
{R'_{a,b,c}}^{(n)} = \prod_{j=0}^{n-1} { {{(q)_{b+c+ja} (q)_{(j+1)a} } \over { (q)_{b+ja} (q)_{c+ja} (q)_a }} }
\eqno(1.3)
$$
In this paper I give a new proof of
THEOREM'( The Habsieger-Kadell q-Morris identity )
$$
{H'_{a,b,c}}^{(n)} = {R'_{a,b,c}}^{(n)} .
\eqno(1.4)
$$
This result was conjectured by Morris[Mo] who proved the q=1 case. It
was recently proved independently by Habsieger[H] and Kadell[K]. Both
Habsieger and Kadell first proved a q-analog of Selberg's integral that
was conjectured by Askey[As] and then deduced from it the q-Morris
identity.
The q-Morris identity is a generalization of the so-called "A cases of
Macdonald's root system conjecture"([Ma]) , also known as " the equal
parameter case of the Zeilberger-Bressoud q-Dyson theorem" . The general
q-Dyson theorem was proved in [Z-B]. Indeed substituting
b=c=0 in the
q-Morris identity (1.4) gives the equal parameter case of q-Dyson.
John Stembridge[Ste], standing on the shoulders of Dennis Stanton[Sta],
has recently come up with a short, elegant and
elementary proof of the equal-parameter case of q-Dyson. In this paper
I adapt Stembridge's proof to give a relatively short, elegant and
{\it elementary} proof of the q-Morris identity.
The word "elementary" has at least two meanings. The first one is the
colloquial "Holmesian" one that means "easy". The second one is the
technical-philosophical "Kroneckerian" meaning of only using finite
algebraic operations on integers. The present proof is elementary in
both senses. The statement of the q-Morris identity (1.4) is completely
elementary and God-created and it was disturbing that so far one had to
resort to such artificial man-made analytical notions as limits and
q-integration to prove it.
{\bf 2. AN EQUIVALENT IDENTITY AND THE ROLE OF ANTISYMMETRY}
It turns out that instead of ${F'_{a,b,c}}^{(n)} $ of (1.1) it is much
easier to consider
$$
{F_{a,b,c}}^{(n)} (x)= \prod_{i=1}^{n} { {( x_i )}_b {(q/ x_i )}_c } \prod_{ 1 \leq i < j \leq n } { {( x_i / x_j ) }_a {( q x_j / x_i ) }_{a-1} }
\eqno(2.1)
$$
and to try and evaluate
$$
{H_{a,b,c}}^{(n)} = C.T. {F_{a,b,c}}^{(n)} ,
\eqno(2.2)
$$
that will turn out to be equal to
$$
{R_{a,b,c}}^{(n)} = \prod_{j=0}^{n-1}
{ {(q)_{b+c+ja} (q)_{(j+1)a-1} }
\over
{ (q)_{b+ja} (q)_{c+ja} (q)_{a-1} } }
\eqno(2.3)
$$
We will actually prove instead of the original statement (1.4) of the
q-Morris identity the identity (2.4) below that turns out to be
equivalent to it.
THEOREM
$$
{H_{a,b,c}}^{(n)} = {R_{a,b,c}}^{(n)}
\eqno(2.4)
$$
The reason that ${F_{a,b,c}}^{(n)} (x) $ is more congenial than
${F'_{a,b,c}}^{(n)} (x)$ is that the former is almost antisymmetric.
Indeed, peeling off the first layer of ${( x_i / x_j )}_a $
yields
$$
{F_{a,b,c}}^{(n)} (x)= \prod_{ 1 \leq i < j \leq n } { ( 1- x_i / x_j ) } \prod_{i=1}^{n} { {( x_i )}_b {(q/ x_i )}_c } \prod_{ 1 \leq i < j \leq n } { {( q x_i / x_j ) }_{a-1} {( q x_j / x_i ) }_{a-1} }
\eqno(2.5)
$$
$$
= x_2^{-1} x_3^{-2} ... x_n^{-(n-1)} \cdot \prod_{ 1 \leq i < j \leq n } {( x_j - x_i ) } \cdot ( something \quad symmetric).
$$
Define $ \delta $ = (0,1,2,..., n-1 ), and
$$
{G_{a,b,c}}^{(n)} (x) = x^{\delta} {F_{a,b,c}}^{(n),}
\eqno(2.6)
$$
then it follows from (2.5) that $ {G_{a,b,c}}^{(n)} $ is an
antisymmetric Laurent polynomial. In terms of $ {G_{a,b,c}}^{(n)} $,
the quantity of interest ${H_{a,b,c}}^{(n)} $ is expressed as
$$
{H_{a,b,c}}^{(n)} = C.T.( x^{ - \delta } {G_{a,b,c}}^{(n)} ) .
\eqno(2.7)
$$
The proof of the equivalence of the original q-Morris identity (1.4) and
its variant (2.4) is a pleasant exercise in antisymmetry. We will not
give it here since the proof in section 4 of [Ste] passes verbatim ( see
also section 3 of [Z]).
The reason antisymmetry is so important is the following
CRUCIAL LEMMA
Let $ G=G(x)= G( x_1 , ... , x_n ) $ be an antisymmetric Laurent
polynomial.
(i) For any vector of integers $ \gamma $ and any permutation $ \pi $ we
have
$$
C.T.[ x^{ \pi ( \gamma ) } G ] = sgn \pi C.T.[ x^{\gamma} G].
$$
(ii) If $ \gamma $=
$( \gamma_1 , ... , \gamma_n )$ is a bad guy ( i.e. there are i
and j , $ 1 \leq i < j \leq n $, such that $ \gamma_i $ =
$ \gamma_j $ ), then
$ C.T.[ x^{\gamma} G ]=0 $
PROOF:
(i) follows straight
from the definitions of antisymmetry and the action of a
permutation on a monomial, while (ii) follows from (i) by using the
transposition (ij) whose sign is -1.
{\bf 3. INDUCTION ON n}
Of course $ H_{a,b,c}^{(0)} \equiv 1 $ and
$ R_{a,b,c}^{(0)} \equiv 1 $, so (2.4) is true for n=0 and we have a
basis to start induction on n. From the definition (2.1) it follows
that
$$
F_{a,0,0}^{(n+1)} ( x_1 , ... , x_{n+1} ) = F_{a,a,a-1}^{(n)} ( x_1 / x_{n+1} , ... , x_n / x_{n+1} )
\eqno(3.1)
$$
Thus taking the constant term yields
$$
H_{a,0,0}^{(n+1)} = H_{a,a,a-1}^{(n)}
\eqno(3.2)
$$
From the definition (2.3) of $R_{a,b,c}^{(n)} $ we have
$$
R_{a,0,0}^{(n+1)} = R_{a,a,a-1}^{(n)}.
\eqno(3.3)
$$
So if we knew that (2.4) was true for n and {\it all} a,b,c then by
plugging b=a,c=a-1 we would have that it is true for n+1 with b=c=0.
This will take care of climbing up the n induction ladder. Now we have
to show that for a fixed n, the truth of (2.4) for b=c=0 implies its
truth for all b,c. So it seems that we have to climb first the c
induction ladder: showing the truth of (2.4) for b=0 and all c, and then
the b ladder :showing that (2.4) for b=0 implies it for all b. Luckily
we get the first ascent {\it gratis}. Indeed, since
$$
\prod_{ 1 \leq i < j \leq n } { { ( x_i / x_j ) }_a { ( q x_j / x_i ) }_{a-1} }
$$
is homogeneous, we obviously have
$$
H_{a,0,c}^{(n)} = C.T. F_{a,0,c}^{(n)} = C.T. F_{a,0,0}^{(n)} = H_{a,0,0}^{(n)} .
$$
Also from the definition (2.3) we have
$$
R_{a,0,c}^{(n)} = R_{a,0,0}^{(n)} .
$$
So we know that if (2.4) is true for b=c=0 then it is true for b=0 and
all c. It remains to climb the b induction ladder.
{\bf 4. INDUCTION ON b}
(2.4) would follow by induction on b once we show that
$$
{{H_{a,b+1,c}^{(n)}} \over {H_{a,b,c}^{(n)}}} = {{R_{a,b+1,c}^{(n)}} \over {R_{a,b,c}^{(n)}}} .
\eqno(4.1)
$$
A routine calculation using the definition (2.3) shows that
( $ t= q^a $)
$$
{{ R_{a,b+1,c} } \over { R_{a,b,c} }} = \prod_{j=0}^{n-1} { {{ (q)_{b+c+ja+1} (q)_{b+ja} } \over { (q)_{b+c+ja} (q)_{b+ja+1} }} } = \prod_{j=0}^{n-1}
{
{( 1- q^ {b+c+ja+1} )} \over {( 1- q^{b+ja+1} )}
}
=
{ {( q^{ b+c+1} ; t )_n }
\over
{( q^{b+1} ; t )_n }} \quad .
\eqno(4.2
$$
Now by the definitions (2.1) (2.2) , and by peeling off the last layer
out of $ {( x_i )}_{b+1} $,
$$
H_{a,b+1,c} = C.T. F_{a,b+1,c}^{(n)} = C.T. \prod_{i=1}^{n} { ( 1- q^b x_i ) } F_{a,b,c}^{(n)} = C.T.\{ x^{ - \delta } \prod_{i=1}^{n} {
( 1- q^b x_i )
}
{G_{a,b,c}}^{(n)}
\}
$$
Now let $s= q^b $ and by expanding the product we get
$$
H_{a,b+1,c}^{(n)} = \sum_{\beta} { (-s)^{ | \beta | } C.T.[ x^{ \beta - \delta } G_{a,b,c}^{(n)} ] }
\eqno(4.3)
$$
where the sum is over all (0-1) vectors
$ \beta = ( \beta_1 , ... , \beta_n ) $, and
$| \beta | = \beta_1 + ... + \beta_n $ = (the number of
ones in $ \beta $).
Now comes the gory Stembridge-Stanton {\it massacre of the bad guys}.
The only way $ \beta - \delta $ can be a good guy is if $ \beta $ has
the form (1,...1,0,...,0), where for some r between 0 and n there are r
1's followed by n-r 0's. The reason is, of course, that if $\beta$
had a zero followed by a one, say in the i and i+1 places:
$\beta_i = 0,\beta_{i+1} = 1 $ then the i and i+1 components
of $\beta - \delta $ are going to be equal to each other. By the
{\it crucial lemma} (ii) the terms in (4.3) that correspond to bad guys
vanish and (4.3) becomes
$$
H_{a,b+1,c}^{(n)} = \sum_{r=0}^{n} { (-s)^r C.T. [ x_1 ... x_r \cdot x^{ - \delta } G_{a,b,c}^{(n)} ] }
\eqno(4.4)
$$
The term corresponding to r=0 in the above sum is nothing but
$C.T.[ x^{ - \delta } G_{a,b,c}^{(n)} ] $ alias
$ H_{a,b,c}^{(n)} $. We have thus expressed
$ H_{a,b+1,c}^{(n)} $ in terms of $H_{a,b,c}^{(n)} $ {\it and}
(unfortunately) some of its "buddies". We would be done if we will be
able to express all the terms that feature in (4.4) in terms of
$H_{a,b,c}^{(n)} $. Luckily it is indeed possible and in the next
section we will prove
(set $s= q^b $, $ t= q^a $, $ u= q^c $)
$$
C.T. [ x_1 ... x_r \cdot x^{ - \delta } G_{a,b,c}^{(n)} ] = (-q)^r {{ (t;t)_n (u;t)_r (qs;t)_{n-r} } \over { (t;t)_r (t;t)_{n-r} (qs;t)_n }} H_{a,b,c}^{(n)}
\eqno(4.5)
$$
Substituting in (4.4) we get
$$
{{ H_{a,b+1,c}^{(n)} } \over { H_{a,b,c}^{(n)} }} = \sum_{r=0}^{n} { (qs)^r {{ (t;t)_n (u;t)_r (qs;t)_{n-r} } \over { (t;t)_r (t;t)_{n-r} (qs;t)_n }} }
\eqno(4.6)
$$
In order to conclude the proof of (4.1) (modulo (4.5)) we must show that
the right hand sides of (4.6) and (4.2) are the same, i.e. we have to show
( as before we set $ t= q^a ,s= q^b , u= q^c $)
$$
{{ (qsu;t)_n }\over{ (qs;t)_n }} = \sum_{r=0}^{n} { (qs)^r {{ (t;t)_n (u;t)_r (qs;t)_{n-r} } \over { (t;t)_r (t;t)_{n-r} (qs;t)_n }} }
\eqno(4.7)
$$
But (4.7) follows immediately by setting X=qs, Y=u, in the following
identity (4.8), taking the base to be t instead of the customary q
(i.e. ${()}_a = {( ;t)}_a $)
SIMPLE LEMMA (A variant of q-Vandermonde)
$$
(XY)_n = \sum_{r=0}^{n} { X^r {{(t)_n }\over{ (t)_r (t)_{n-r} }} (Y)_r (X)_{n-r} }
\eqno(4.8)
$$
PROOF OF THE SIMPLE LEMMA
Cauchy's famous q-analog of the binomial theorem ( e.g. [An] p.10,
(2.9)) says
$$
{ {(az;t)_{\infty} } \over {(z;t)_{\infty} }} =
\sum { {{(a;t)_n }\over{ (t;t)_n }} z^n }
\eqno(4.9)
$$
(Incidentally, the " $|z|<1,|t|<1$" that is added as a "condition of
validity " in [An] is completely superfluous, at least in {\it my}
book).
Of course
$$
{{(zXY;t)_{\infty} } \over { (z;t)_{\infty} }} = {{ (zXY;t)_{\infty} } \over {(zX;t)_{\infty} }} {{ (zX;t)_{\infty} } \over {(z;t)_{\infty} }}
\eqno(4.10)
$$
Now, using (4.9), we expand each of the three ratios in (4.10) as formal
power series in z, and compare coefficients of $ z^n $, which yields
the desired identity (4.8).
We have thus completed the proof of the theorem {\it modulo} the
identity (4.5). To get to where we are we have climbed two induction
ladders: the n ladder (section 3) and the b ladder (section 4). In
order to prove (4.5) we need to climb one more induction ladder: the
r-ladder.
{\bf 5. PROOF OF (4.5): INDUCTION ON r.}
In this section n,a,b,c are fixed throughout. As before $t= q^a $,
$s= q^b $, $u= q^c $.
Let
$$
C_r = C.T. \,\,
[x_1 ... x_r \cdot x^{ - \delta } G_{a,b,c}^{(n)} ] / H_{a,b,c}^{(n)} ,
\eqno(5.1a)
$$
$$
D_r = {(-q)}^r {{ (t;t)_n (u;t)_r (qs;t)_{n-r} } \over { (t;t)_r (t;t)_{n-r} (qs;t)_n }} ,
\eqno(5.1b)
$$
Then (4.5) can be rewritten as
$$
C_r = D_r
\eqno(5.2)
$$
Since $C_0 =1$ by definition and $D_0 =1 $ by plugging r=0 in
(5.1b), it follows that (5.2) is true for the base case r=0. The general
case would then follow by induction if we can prove that
$$
{{C_{r+1} }\over{ C_r }} = {{D_{r+1} }\over{ D_r }} .
\eqno(5.3)
$$
A routine calculation shows that
$$
{{D_{r+1}} \over {D_r}} =
-q \cdot {{(1- t^{n-r} )( 1- u t^r )} \over {( 1- t^{r+1} ) (1- qs t^{n-r-1} ) }} .
\eqno(5.4)
$$
Thus we have to prove that
$$
{{ C_{r+1} }\over{ C_r }} = -q \cdot {{(1- t^{n-r} )( 1- u t^r )} \over {( 1- t^{r+1} ) (1- qs t^{n-r-1} ) }} .
\eqno(5.5)
$$
It turns out that instead of $C_r $ of (5.1a) it is more convenient
to consider
$$
A_j = C.T.[ x_j ... x_n x^{ - \bar{\delta} } G_{a,b,c}^{(n)} ] ,
\eqno(5.6)
$$
where
$$
\bar{\delta} = ( n-1, ... , 0).
$$
But since $ \bar{\delta} = rev( \delta )$, where rev is the "reverse
permutation" $rev(i)=n-i+1$, whose sign is $(-1)^{ n(n-1)/2) } $,
$$
A_j = (-1)^{ n(n-1)/2 } C.T.[ x_1 ... x_{n-j+1} \cdot x^{ - \delta } G_{a,b,c}^{(n)} ] = (-1)^{n(n-1)/2} C_{n-j+1} .
\eqno(5.7)
$$
It is readily seen that in terms of the $ A_j $ (5.5) is equivalent
to ( take r= n-j+1),
$$
{{A_{j-1}}\over{ A_j }} = -q {{ (1- t^{j-1} )(1-u t^{n-j+1} )} \over {(1- t^{n-j+2} )(1-qs t^{j-2} )}} .
\eqno(5.9)
$$
We now go on and prove (5.9).
By using the definitions (2.1),(2.6) and by routine telescoping, we
obtain ( from now on $G= G_{a,b,c}^{(n)} $ , recall that
$t= q^a $,$s= q^b $, $ u= q^c $ ).
$$
{{G(q x_1 , ... , x_n )}\over{G( x_1 , ... , x_n )}} =
{ {(1- s x_1 ) \prod_{ j=2}^{n} { (1 - t x_1 / x_j )} }
\over {(u- x_1 ) \prod_{j=2}^{n} { ( q^{-1} t - x_1 / x_j )} }} .
\eqno(5.10)
$$
By cross multiplying we get,
$$
{(u- x_1 ) \prod_{j=2}^{n} { ( q^{-1} t - x_1 / x_j )} }{G(q x_1 , ... , x_n )} = {(1- s x_1 ) \prod_{j=2}^{n} { (1 - t x_1 / x_j )} } {G( x_1 , ... , x_n )} .
\eqno(5.11)
$$
Expanding the product, we get
$$
\sum_{\beta} { u {( q^{-1} t ) }^{ n-1- | \beta | } (-1)^{ | \beta | } x_1^{| \beta | } x^{ - \beta } {G(q x_1 , ... , x_n )} } -
\eqno(5.12)
$$
$$
\sum_{\beta} { {( q^{-1} t ) }^{ n-1- | \beta | } (-1)^{ | \beta | } x_1^{| \beta |+1 } x^{ - \beta } {G(q x_1 , ... , x_n )} }
$$
$$
= \sum_{\beta} { (-1)^{ | \beta |} t^{ | \beta |} x_1^{| \beta |} x^{ - \beta } G } - \sum_{\beta} { (-1)^{ | \beta |}s t^{ | \beta |} x_1^{| \beta |+1 } x^{ - \beta } G },
$$
where the sums are over all (0-1) vectors whose first component is zero:
$\beta = ( 0, \beta_2 , ... \beta_n )$, where $ \beta_i = 0$ or
$1$ for $i=2, \dots , n$.
Let
$$
\alpha^{(j)} = ( n-1, ... , n-j+1 , n-j-1, ... , -1 ) = \bar{\delta} - ( 0_0 , ... 0_{j-1} , 1_j , ... , 1_n )
\eqno(5.13)
$$
for $j=2,..., n+1$ .
Because of (5.6), we have
$$
A^{(j)} = C.T. [ x^{ - \alpha^{(j)} } G ].
\eqno(5.14)
$$
Multiplying both sides of (5.12) by $ x^{ - \alpha^{(j)} } $ and
taking the constant term, we get ( recall that $ e_1 $= (1,0,...0) )
$$
\sum_{\beta} { u {( q^{-1} t ) }^{ n-1- | \beta | } (-1)^{ | \beta | } C.T.[ x^{ -[ \alpha^{(j)} + \beta - | \beta | e_1 ]} {G(q x_1 , ... , x_n )}] } -
\eqno(5.15)
$$
$$
\sum_{\beta} { {( q^{-1} t ) }^{ n-1- | \beta | } (-1)^{ | \beta | } C.T.[ x^{ -[ \alpha^{(j)} + \beta - (| \beta |+1) e_1 ]} {G(q x_1 , ... , x_n )} }
$$
$$
= \sum_{\beta} { (-1)^{ | \beta |} t^{ | \beta |}
C.T.[ x^{ -[ \alpha^{(j)} + \beta - | \beta | e_1 ]} G] } -
\sum_{\beta} { (-1)^{ | \beta |} s t^{ | \beta |}
C.T.[ x^{ -[ \alpha^{(j)} + \beta - (| \beta |+1) e_1 ]} G] }.
$$
Note that the first component of
$ \alpha^{(j)} + \beta - | \beta | e_1 is n-1-| \beta | $ and the first
component of
$ \alpha^{(j)} + \beta - (| \beta |+1) e_1$ is $n-2-| \beta | $ .
Now we use the obvious relation
$$
C.T.[ x^{ - \gamma } G(q x_1 , ... , x_n ) ]= q^{ \gamma_1 } C.T.[ x^{ - \gamma } G ]
\eqno(5.16)
$$
in the left side of (5.15) and we get
$$
\sum_{\beta} { u t^{ n-1- | \beta | } (-1)^{ | \beta | } C.T.[ x^{ -[ \alpha^{(j)} + \beta - | \beta | e_1 ]} G] } -
\eqno(5.17)
$$
$$
\sum_{\beta} { q^{-1} t^{ n-1- | \beta | } (-1)^{ | \beta | }
C.T.[ x^{ -[ \alpha^{(j)} + \beta - (| \beta |+1) e_1 ]} G] }
$$
$$
= \sum_{\beta} { (-1)^{ | \beta |} t^{ | \beta |} C.T.[ x^{ -[ \alpha^{(j)} + \beta - | \beta | e_1 ]} G] } - \sum_{\beta} { (-1)^{ | \beta |} t^{ | \beta |} s C.T.[ x^{ -[ \alpha^{(j)} + \beta - (| \beta |+1) e_1 ]} G] }.
$$
We now need the following simple, but crucial, lemma whose proof is left
as a pleasant exercise to the reader.
LEMMA
(i) $ \alpha^{(j)} + \beta - | \beta | e_1 $ is a bad guy unless
$ \beta $ has the form $(0,1,...,1,0,...0)$, where the first component is $0$
and then for some $r$, $0 \leq r \leq j-2 $, there are $r$ $1$'s
followed by $n-r-1$
$0$'s. In this case $ \alpha^{(j)} + \beta - | \beta | e_1 $ is the
image of $ \alpha^{(j)} $ under the cycle $(1,2,...,r+1)$, whose sign is
$(-1)^r $ .
(ii)$ \alpha^{(j)} + \beta - (| \beta |+1) e_1 $ is a bad guy unless
$\beta$ has the form $(0,1,...1,0,...0)$, where for some $r$
satisfying
$ j-2 \leq r \leq n-1 $ you have a 0 followed by r 1's followed by n-r-1
0's. In this case $ \alpha^{(j)} + \beta - (| \beta |+1) e_1 $ is the
image of $ \alpha^{(j-1)} $ under the cycle (1,2,...,r+1) whose sign is
$(-1)^r $.
Discarding all the bad guys in (5.17) and using the above lemma and
the {\it crucial lemma} ,the equation (5.17) shrinks to (recall (5.14))
$$
\{ \sum_{r=0}^{j-2} { u t^{ n-1- r } (-1)^r (-1)^r } \} A^{(j)} - \{ \sum_{r=j-2}^{n-1} { q^{-1} t^{ n-1-r } (-1)^r (-1)^r }
\} A_{(j-1)}
$$
$$
= \{ \sum_{r=0}^{j-2} { (-1)^r t^r (-1)^r } \} A^{(j)} - \{ \sum_{r=j-2}^{n-1} { (-1)^r t^r s (-1)^r
}
\} A_{(j-1)}
\eqno(5.18)
$$
By summing all the geometric series and performing very routine and
simple ninth grade algebra we get (5.9).
tav vav shin lamed bet ayin
REFERENCES
[An] Andrews, George \it
"q-Series: Their Development and Applications
in Analysis, Number Theory, Combinatorics, Physics, and Computer Algebra",
CBMS regional conference series in mathematics, number 66,
American Mathematical Society, Providence, 1986.
[As] Askey, Richard
\it
Some basic hypergeometric extensions of
integrals of Selberg and Andrews , SIAM J.Math.Anal. {\bf 11}(1980),
938-951.
[H] Habsieger, Laurent
{\it Une q-integrale de Selberg-Askey}, SIAM J.Math.Anal., to appear.
[K]Kadell, Kevin , \it
A proof of Askey's conjectured q-analog of Selberg's integral
and a conjecture of Morris\rm
, SIAM J.Math.Anal., to appear.
\medskip
[Ma] Macdonald, Ian G.
{\it Some conjectures for root systems}, SIAM J.Math.Anal. {\bf 13}
(1982), 988-1007.
[Mo] Morris, Walter G,II
"Constant Term Identities for Finite and Affine Root Systems", Ph.D.
thesis, Univ. of Wisconsin-Madison, 1982.
[ available for 25.50 dollars from University
Microfilms, 300 Zeeb Road, Ann Arbor, Michigan 48106, (313)761-4700].
[Sta] Stanton, Dennis
{\it Sign variations of the Macdonald identities}, SIAM
J.Math.Anal. {\bf 17}(1986),1454-1460.
[Ste] Stembridge, John \it
A short proof of Macdonald's conjecture for the root systems of
type A\rm
, UCLA preprint(1986).
[Z] Zeilberger, Doron,
{\it A unified approach to Macdonald's root-system conjectures},
preprint.
[Z-B] Zeilberger, Doron and Bressoud, David,
{\it A proof of Andrews' q-Dyson conjecture}, Discrete
Math.{\bf 54}(1985), 201-224.
Department of Mathematics, Drexel University, Philadelphia, PA19104.
Current Address
(1990-):
Department of Mathematics, Temple University, Philadelphia, PA19122.
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