.EQ
delim $$
.EN
.nr Pi 10
.nr Pt 1
.SP1
.DS
0.NOMENCLATURE
.DE
.SP1
.LP
q and $ x=( x sub 1 , ... , x sub n ) $ are commuting indeterminates.
If $alpha = ( alpha sub 1 , ... , alpha sub n ) $ is any vector of
integers , then $ x sup alpha $ stands for
$ x sup { alpha sub 1 } ... x sup { alpha sub n } $. For example if
$alpha$=(1,2,5), then $x sup alpha$ =
$ x sub 1 x sub 2 sup 2 x sub 3 sup 5 $.
.SP1
.LP
A \fILaurent polynomial\fR is a finite linear combination of monomials
$x sup alpha $, where the $alpha$s have integer components. All our
Laurent polynomials will be with integer coefficients.
.SP1
.LP
"C.T." stands for "the constant term of", with respect to
$x= ( x sub 1 , ... , x sub n )$. For example
C.T.(1qx)(1q/x)= 1+q.
.SP1
.LP
The symmetric group $S sub 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 sup gamma $ by
$ pi ( x sup gamma ) = x sup { pi ( gamma ) } $ , and by linearity on
any Laurent polynomial. For example,
.SP1
.EQ
(321)[ x sub 1 sup 1 x sub 2 sup 2 x sub 3 + 4 +
x sub 1 sup 2 x sub 2 sup 3 x sub 3 sup 5 ] ~=~
x sub 1 x sub 2 sup 2 x sub 3 sup 1 + 4 + x sub 1 sup 5 x sub 2 sup 3
x sub 1 sup 2 ~~.~
.EN
.SP1
.LP
A Laurent polynomial P in $x=( x sub 1 , ... , x sub n ) $ is
\fIsymmetric\fR if $ pi ( P)= P $ for all permutations $pi$, and is
\fIantisymmetric\fR if $pi ( P)= (sgn pi )P $ for every permutation
$pi$.
.SP1
.LP
$(y;Q) sub a $, the "qanalog of $ {(1y)} sup a $ to base Q " is
defined by
.SP1
.EQ
(y;Q) sub a = (1y)(1Qy)...(1 Q sup a1 y )~~~,
.EN
.SP1
.LP
and if the base Q is q then we often abbreviate $(y;q) sub a $ to
$ (y) sub a $ .
.SP1
.LP
A vector of integers $alpha$ is a \fIbad guy\fR if it has two or more
identical components, otherwise it is a \fIgood guy\fR. For example
(1,3,1) and (1,1,1) are bad guys while (1,1,0) and (2,1,8) are good
guys.
.LP
$ delta $ is the vector (0,1,...,n1) and $ delta bar $ is its reverse:
$ delta bar$ = ( n1, ... , 0 ).
.LP
Throughout this paper $t= q sup a $, $s= q sup b $, $u= q sup c $ .
.SP3
.DS
1.THE HABSIEGERKADELL qMORRIS IDENTITY
.DE
.SP1
.P
Let, for a,b,c,n nonnegative integers,
.SP2
.EQ(1.1)
F' sub a,b,c sup (n) (x)=
prod from {i=^1} to n
{
{( x sub i )} sub b {(q/ x sub i )} sub c
}
prod from { 1 <= i < j <= n }
{
{( x sub i / x sub j ) } sub a
{( q x sub j / x sub i ) } sub a
}
.EN
.SP2
.EQ(1.2)
H' sub a,b,c sup (n) = C.T. F' sub a,b,c sup (n) ~~.
.EN
.SP3
.EQ(1.3)
R' sub a,b,c sup (n) =
prod from {j=^0} to n1
{
{(q) sub b+c+ja (q) sub (j+1)a }
over
{ (q) sub b+ja (q) sub c+ja (q) sub a }
}
.EN
.SP2
.LP
In this paper I give a new proof of
.SP1
.DS
THEOREM'( The HabsiegerKadell qMorris identity )
.DE
.SP1
.EQ(1.4)
H' sub a,b,c sup (n) = R' sub a,b,c sup (n) ~.~
.EN
.SP1
.LP
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 qanalog of Selberg's integral that
was conjectured by Askey[As] and then deduced from it the qMorris
identity.
.P
The qMorris identity is a generalization of the socalled "A cases of
Macdonald's root system conjecture"([Ma]) , also known as " the equal
parameter case of the ZeilbergerBressoud qDyson theorem" . The general
qDyson theorem was proved in [ZB]. Indeed substituting
b=c=0 in the
qMorris identity (1.4) gives the equal parameter case of qDyson.
.P
John Stembridge[Ste], standing on the shoulders of Dennis Stanton[Sta],
has recently come up with a short, elegant and
elementary proof of the equalparameter case of qDyson. In this paper
I adapt Stembridge's proof to give a relatively short, elegant and
\fIelementary\fR proof of the qMorris identity.
.P
The word "elementary" has at least two meanings. The first one is the
colloquial "Holmesian" one that means "easy". The second one is the
technicalphilosophical "Kroneckerian" meaning of only using finite
algebraic operations on integers. The present proof is elementary in
both senses. The statement of the qMorris identity (1.4) is completely
elementary and Godcreated and it was disturbing that so far one had to
resort to such artificial manmade analytical notions as limits and
qintegration to prove it.
.SP3
.DS
2.AN EQUIVALENT IDENTITY AND THE ROLE OF ANTISYMMETRY
.DE
.SP2
.P
It turns out that instead of $F' sub a,b,c sup (n) $ of (1.1) it is much
easier to consider
.SP2
.EQ(2.1)
F sub a,b,c sup (n) (x)=
prod from {i=^1} to n
{
{( x sub i )} sub b {(q/ x sub i )} sub c
}
prod from { 1 <= i < j <= n }
{
{( x sub i / x sub j ) } sub a
{( q x sub j / x sub i ) } sub a1
}
.EN
.SP2
.LP
and to try and evaluate
.SP2
.EQ(2.2)
H sub a,b,c sup (n) = C.T. F sub a,b,c sup (n) ~~,
.EN
.SP1
.LP
that will turn out to be equal to
.SP3
.EQ(2.3)
R sub a,b,c sup (n) =
prod from j=0 to n1
{
{(q) sub b+c+ja (q) sub (j+1)a1 }
over
{ (q) sub b+ja (q) sub c+ja (q) sub a1 }
}
.EN
.SP2
.P
We will actually prove instead of the original statement (1.4) of the
qMorris identity the identity (2.4) below that turns out to be
equivalent to it.
.SP1
.DS
THEOREM
.DE
.SP1
.EQ(2.4)
H sub a,b,c sup (n) = R sub a,b,c sup (n)
.EN
.SP1
.P
The reason that $F sub a,b,c sup (n) (x) $ is more congenial than
$F' sub a,b,c sup (n) (x)$ is that the former is almost antisymmetric.
Indeed, peeling off the first layer of ${( x sub i / x sub j )} sub a $
yields
.SP2
.EQ(2.5)
F sub a,b,c sup (n) (x)=
prod from { 1 <= i < j <= n }
{ ( 1 x sub i / x sub j ) }
prod from {i=^1} to n
{
{( x sub i )} sub b {(q/ x sub i )} sub c
}
prod from { 1 <= i < j <= n }
{
{( q x sub i / x sub j ) } sub a1
{( q x sub j / x sub i ) } sub a1
}
.EN
.SP2
.EQ
= x sub 2 sup 1 x sub 3 sup 2 ... x sub n sup (n1)
cdot
prod from { 1 <= i < j <= n }
{( x sub j  x sub i ) }
cdot
( ~something~symmetric~)~~.
.EN
.SP2
.P
Define $ delta $ = (0,1,2,..., n1 ), and
.SP1
.EQ(2.6)
G sub a,b,c sup (n) (x) = x sup delta F sub a,b,c sup (n)~~,
.EN
.SP1
.LP
then it follows from (2.5) that $ G sub a,b,c sup (n) $ is an
antisymmetric Laurent polynomial. In terms of $ G sub a,b,c sup (n) $,
the quantity of interest $H sub a,b,c sup (n) $ is expressed as
.SP1
.EQ(2.7)
H sub a,b,c sup (n) = C.T.( x sup {  delta } G sub a,b,c sup (n) )
~~~~~~~~~.
.EN
.SP1
.P
The proof of the equivalence of the original qMorris 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]).
.P
The reason antisymmetry is so important is the following
.SP1
.DS
CRUCIAL LEMMA
.DE
.SP1
.LP
Let $ G=G(x)= G( x sub 1 , ... , x sub n ) $ be an antisymmetric Laurent
polynomial.
.SP1
.LP
(i) For any vector of integers $ gamma $ and any permutation $ pi $ we
have
.SP1
.EQ
C.T.[ x sup { pi ( gamma ) } G ] =
sgn pi C.T.[ x sup gamma G]~~.
.EN
.SP1
.LP
(ii) If $ gamma $=
$( gamma sub 1 , ... , gamma sub n )$ is a bad guy ( i.e. there are i
and j , $ 1 <= i < j <=n $, such that $ gamma sub i $ =
$ gamma sub j $ ), then
$ C.T.[ x sup gamma G ]=0 $
.SP1
.LP
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.
.SP3
.DS
3.INDUCTION ON n
.DE
.SP1
.P
Of course $ H sub a,b,c sup (0) == 1 $ and
$ R sub a,b,c sup (0) == 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
.SP1
.EQ(3.1)
F sub a,0,0 sup (n+1) ( x sub 1 , ... , x sub n+1 ) =
F sub a,a,a1 sup (n) ( x sub 1 / x sub n+1 , ... , x sub n / x sub n+1
)
.EN
.SP1
.LP
Thus taking the constant term yields
.SP1
.EQ(3.2)
H sub a,0,0 sup (n+1) = H sub a,a,a1 sup (n)
.EN
.SP1
.LP
>From the definition (2.3) of $R sub a,b,c sup (n) $ we have
.SP1
.EQ(3.3)
R sub a,0,0 sup (n+1) = R sub a,a,a1 sup (n)~~~~~~~~.
.EN
.SP1
.LP
So if we knew that (2.4) was true for n and \fIall\fR a,b,c then by
plugging b=a,c=a1 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 \fIgratis\fR. Indeed, since
.SP1
.EQ
prod from { 1 <= i < j <= n }
{
{ ( x sub i / x sub j ) } sub a
{ ( q x sub j / x sub i ) } sub a1
}
.EN
.SP1
.LP
is homogeneous, we obviously have
.SP1
.EQ
H sub a,0,c sup (n) =
C.T. F sub a,0,c sup (n) =
C.T. F sub a,0,0 sup (n) =
H sub a,0,0 sup (n) ~~~~~~~~.
.EN
.SP1
.LP
Also from the definition (2.3) we have
.SP1
.EQ
R sub a,0,c sup (n) =
R sub a,0,0 sup (n) ~~~~~~~.
.EN
.SP1
.LP
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.
.SP3
.DS
4.INDUCTION ON b
.DE
.SP1
.P
(2.4) would follow by induction on b once we show that
.SP2
.EQ(4.1)
{H sub a,b+1,c sup (n)}
over
{H sub a,b,c sup (n)}
=
{R sub a,b+1,c sup (n)}
over
{R sub a,b,c sup (n)} ~~~~~.
.EN
.SP2
.LP
A routine calculation using the definition (2.3) shows that
( $ t= q sup a $)
.SP3
.EQ(4.2)
{ R sub a,b+1,c }
over
{ R sub a,b,c }
=
prod from {j=^0} to n1
{
{ (q) sub b+c+ja+1 (q) sub b+ja }
over
{ (q) sub b+c+ja (q) sub b+ja+1 }
}
=
~prod from {j=^0} to n1
{
{( 1 q sup b+c+ja+1 )} over {( 1 q sup b+ja+1 )}
}
=
{ {( q sup b+c+1 ;^ t ) } sub n }
over
{ {( q sup b+1 ;^ t ) } sub n } ~~.
.EN
.SP2
.P
Now by the definitions (2.1) (2.2) , and by peeling off the last layer
out of $ {( x sub i )} sub b+1 $,
.SP2
.EQ
H sub a,b+1,c = C.T. F sub a,b+1,c sup (n) =
C.T.
prod from i=1 to n
{
( 1 q sup b x sub i )
}
F sub a,b,c sup (n)
=
C.T."{"
x sup {  delta }
prod from i=1 to n
{
( 1 q sup b x sub i )
}
G sub a,b,c sup (n)
"}"
.EN
.SP2
.P
Now let $s= q sup b $ and by expanding the product we get
.SP2
.EQ(4.3)
H sub a,b+1,c sup (n)
=
sum from beta
{
(s) sup {  beta  }
C.T.[ x sup { beta  delta } G sub a,b,c sup (n) ]
}
.EN
.SP1
.LP
where the sum is over all (01) vectors
$ beta = ( beta sub 1 , ... , beta sub n ) $, and
$ beta  = beta sub 1 + ... + beta sub n $ = (the number of
ones in $ beta $).
.P
Now comes the gory StembridgeStanton \fImassacre of the bad guys\fR.
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 nr 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 sub i = 0,~~beta sub i+1 = 1 $ then the i and i+1 components
of $beta  delta $ are going to be equal to each other. By the
\fIcrucial lemma\fR (ii) the terms in (4.3) that correspond to bad guys
vanish and (4.3) becomes
.SP2
.EQ(4.4)
H sub a,b+1,c sup (n) =
sum from {r=^0} to n
{
(s) sup r C.T.
[
x sub 1 ... x sub r ~ cdot ~ x sup {  delta }
G sub a,b,c sup (n)
]
}
.EN
.SP2
.LP
The term corresponding to r=0 in the above sum is nothing but
$C.T.[ x sup {  delta } G sub a,b,c sup (n) ] $ alias
$ H sub a,b,c sup (n) $. We have thus expressed
$ H sub a,b+1,c sup (n) $ in terms of $H sub a,b,c sup (n) $ \fIand\fR
(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 sub a,b,c sup (n) $. Luckily it is indeed possible and in the next
section we will prove
(set $s= q sup b $, $ t= q sup a $, $ u= q sup c $)
.SP2
.EQ(4.5)
C.T.
[
x sub 1 ... x sub r ~ cdot ~ x sup {  delta } G sub a,b,c sup (n)
]
=
(q) sup r
{ (t;^t) sub n (u;^t) sub r (qs;^t) sub nr }
over
{ (t;^t) sub r (t;^t) sub nr (qs;^t) sub n }
H sub a,b,c sup (n)
.EN
.SP2
.P
Substituting in (4.4) we get
.SP2
.EQ(4.6)
{ H sub a,b+1,c sup (n) }
over
{ H sub a,b,c sup (n) }
=
sum from r=0 to n
{
(qs) sup r
{ (t;^t) sub n (u;^t) sub r (qs;^t) sub nr }
over
{ (t;^t) sub r (t;^t) sub nr (qs;^t) sub n }
}
.EN
.SP2
.P
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 sup a ~,~s= q sup b ~, u= q sup c $)
.SP2
.EQ(4.7)
{ (qsu;^t) sub n } over { (qs;^t) sub n }
=
sum from r=0 to n
{
(qs) sup r
{ (t;^t) sub n (u;^t) sub r (qs;^t) sub nr }
over
{ (t;^t) sub r (t;^t) sub nr (qs;^t) sub n }
}
.EN
.SP2
.LP
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. ${(~~)} sub a = {( ~~;^t)} sub a $)
.SP1
.DS
SIMPLE LEMMA (A variant of qVandermonde)
.DE
.SP2
.EQ(4.8)
(XY) sub n = sum from r=0 to n
{
X sup r
{(t) sub n } over { (t) sub r (t) sub nr }
(Y) sub r
(X) sub nr
}
.EN
.SP2
.DS
PROOF OF THE SIMPLE LEMMA
.DE
.SP1
Cauchy's famous qanalog of the binomial theorem ( e.g. [An] p.10,
(2.9)) says
.SP2
.EQ(4.9)
{(az;^t) sub inf }over {(z;^t) sub inf }
=
sum
{
{(a;^t) sub n } over { (t;^t) sub n } z sup n
}
.EN
.SP2
.LP
(Incidentally, the " z<1,t<1" that is added as a "condition of
validity " in [An] is completely superfluous, at least in \fImy\fR
book).
.P
Of course
.SP2
.EQ(4.10)
{(zXY;^t) sub inf } over { (z;^t) sub inf } =
{ (zXY;^t) sub inf } over {(zX;^t) sub inf }
{ (zX;^t) sub inf } over {(z;^t) sub inf }
.EN
.SP2
.LP
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 sup n $, which yields
the desired identity (4.8).
.P
We have thus completed the proof of the theorem \fImodulo\fR 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
rladder.
.SP3
.DS
5.PROOF OF (4.5): INDUCTION ON r.
.DE
.SP1
.P
In this section n,a,b,c are fixed throughout. As before $t= q sup a $,
$s= q sup b $, $u= q sup c $.
.P
Let
.SP1
.EQ(5.1a)
C sub r =
C.T.[ x sub 1 ... x sub r ^ cdot ^ x sup {  delta } G sub a,b,c sup (n) ]
/
H sub a,b,c sup (n) ~~~~~,
.EN
.SP2
.EQ(5.1b)
D sub r
=
{(q)} sup r
{ (t;^t) sub n (u;^t) sub r (qs^;t) sub nr }
over
{ (t;^t) sub r (t;^t) sub nr (qs;^t) sub n } ~~~,
.EN
.SP2
.LP
Then (4.5) can be rewritten as
.SP1
.EQ(5.2)
C sub r = D sub r
.EN
.SP1
.LP
Since $C sub 0 =1$ by definition and $D sub 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
.SP2
.EQ(5.3)
{C sub r+1 } over { C sub r }
=
{D sub r+1 } over { D sub r } ~~~~.
.EN
.SP2
.LP
A routine calculation shows that
.SP2
.EQ(5.4)
{D sub r+1} over D sub r
=
q ^ cdot ^
{(1 t sup nr )( 1 u t sup r )}
over
{( 1 t sup r+1 ) (1 qs t sup nr1 ) } ~~~~.
.EN
.SP2
.LP
Thus we have to prove that
.SP2
.EQ(5.5)
{ C sub r+1 } over { C sub r }
=
q ^ cdot ^
{(1 t sup nr )( 1 u t sup r )}
over
{( 1 t sup r+1 ) (1 qs t sup nr1 ) } ~~~.
.EN
.SP2
.LP
It turns out that instead of $C sub r $ of (5.1a) it is more convenient
to consider
.SP1
.EQ(5.6)
A sub j = C.T.[ x sub j ... x sub n x sup {  delta bar } G sub a,b,c
sup (n) ] ~~~,
.EN
.SP1
.LP
where
.SP1
.EQ
delta bar = ( n1, ... , 0)~.
.EN
.SP1
.LP
But since $ delta bar = rev( delta )$, where rev is the "reverse
permutation" rev(i)=ni+1, whose sign is $(1) sup { n(n1)/2) } $,
.SP2
.EQ(5.7)
A sub j =
(1) sup { n(n1)/2 }
C.T.[ x sub 1 ... x sub nj+1 ^ cdot ^
x sup {  delta } G sub a,b,c sup (n) ]
^=^ (1) sup n(n1)/2 C sub nj+1 ~~~.
.EN
.SP2
.LP
It is readily seen that in terms of the $ A sub j $ (5.5) is equivalent
to ( take r= nj+1),
.SP2
.EQ(5.9)
{A sub j1} over { A sub j }
=
q
{ (1 t sup j1 )(1u t sup nj+1 )}
over
{(1 t sup nj+2 )(1qs t sup j2 )} ~~~.
.EN
.SP2
.P
We now go on and prove (5.9).
.SP1
.P
By using the definitions (2.1),(2.6) and by routine telescoping, we
obtain ( from now on $G= G sub a,b,c sup (n) $ , recall that
$t= q sup a $,$s= q sup b $, $ u= q sup c $ ).
.SP3
.EQ(5.10)
{G(q x sub 1 , ... , x sub n )}
over
{G( x sub 1 , ... , x sub n )}
=
{(1 s x sub 1 ) prod from { j=^2} to n { (1  t x sub 1 / x sub j )} }
over
{(u x sub 1 ) prod from {j=^2} to n { ( q sup 1 t  x sub 1 / x sub j
)} } ~~~~.
.EN
.SP2
.LP
By cross multiplying we get,
.SP2
.EQ(5.11)
{(u x sub 1 ) prod from {j=^2} to n { ( q sup 1 t  x sub 1 / x sub j
)} }{G(q x sub 1 , ... , x sub n )}
=
{(1 s x sub 1 ) prod from {j=^2} to n { (1  t x sub 1 / x sub j )} }
{G( x sub 1 , ... , x sub n )} ~~.
.EN
.SP2
.LP
Expanding the product, we get
.SP2
.EQ(5.12)
sum from beta
{~
u {( q sup 1 t ) } sup { n1  beta  } (1) sup {  beta  }
x sub 1 sup {  beta  } x sup {  beta }
{G(q x sub 1 , ... , x sub n )}
}

.EN
.SP1
.EQ
sum from beta
{
{( q sup 1 t ) } sup { n1  beta  } (1) sup {  beta  }
x sub 1 sup {  beta +1 } x sup {  beta }
{G(q x sub 1 , ... , x sub n )}
}
.EN
.SP1
.EQ
=
sum from beta
{
(1) sup {  beta } t sup {  beta }
x sub 1 sup {  beta } x sup {  beta } G
}~
~
sum from beta
{
(1) sup {  beta }^s^ t sup {  beta }
x sub 1 sup {  beta +1 } x sup {  beta } G
}~~~,
.EN
.SP2
.LP
where the sums are over all (01) vectors whose first component is zero:
$beta = ( 0, beta sub 2 , ... beta sub n )$, where $ beta sub i $ = 0 or
1 for i=2,...n.
.P
Let
.SP2
.EQ(5.13)
alpha sup (j) = ( n1, ... , nj+1 , nj1, ... , 1 ) =
delta bar  ( 0 sub 0 , ... 0 sub j1 , 1 sub j , ... , 1 sub n )
.EN
.SP1
.LP
for j=2,..., n+1 .
.SP1
.LP
Because of (5.6), we have
.SP2
.EQ(5.14)
A sup (j) = C.T. [ x sup {  alpha sup (j) } G ]~~~.
.EN
.SP1
.LP
Multiplying both sides of (5.12) by $ x sup {  alpha sup (j) } $ and
taking the constant term, we get ( recall that $ e sub 1 $= (1,0,...0) )
.SP2
.EQ(5.15)
sum from beta
{~
u {( q sup 1 t ) } sup { n1  beta  } (1) sup {  beta  }
C.T.[ x sup { [ alpha sup (j) + beta   beta  e sub 1 ]}
{G(q x sub 1 , ... , x sub n )}]
}

.EN
.SP1
.EQ
sum from beta
{~
{( q sup 1 t ) } sup { n1  beta  } (1) sup {  beta  }
C.T.[ x sup { [ alpha sup (j) + beta  ( beta +1) e sub 1 ]}
{G(q x sub 1 , ... , x sub n )}
}
.EN
.SP1
.EQ
=
sum from beta
{
(1) sup {  beta } t sup {  beta }
C.T.[ x sup { [ alpha sup (j) + beta   beta  e sub 1 ]} G]
}

sum from beta
{
(1) sup {  beta } s t sup {  beta }
C.T.[ x sup { [ alpha sup (j) + beta  ( beta +1) e sub 1 ]} G]~
}~~.
.EN
.SP2
.P
Note that the first component of
$ alpha sup (j) + beta   beta  e sub 1 ~is ~n1 beta  $ and the first
component of
$ alpha sup (j) + beta  ( beta +1) e sub 1 ~is ~n2 beta  $ .
Now we use the obvious relation
.SP2
.EQ(5.16)
C.T.[ x sup {  gamma } G(q x sub 1 , ... , x sub n ) ]=
q sup { gamma sub 1 } C.T.[ x sup {  gamma } G ]
.EN
.SP2
.LP
in the left side of (5.15) and we get
.SP2
.EQ(5.17)
sum from beta
{~
u t sup { n1  beta  } (1) sup {  beta  }
C.T.[ x sup { [ alpha sup (j) + beta   beta  e sub 1 ]}
G]
}

.EN
.SP1
.EQ
sum from beta
{~
q sup 1 t sup { n1  beta  } (1) sup {  beta  }
C.T.[ x sup { [ alpha sup (j) + beta  ( beta +1) e sub 1 ]}
G]
}
.EN
.SP1
.EQ
=
sum from beta
{
(1) sup {  beta } t sup {  beta }
C.T.[ x sup { [ alpha sup (j) + beta   beta  e sub 1 ]} G]
}

sum from beta
{
(1) sup {  beta } t sup {  beta } ^s^
C.T.[ x sup { [ alpha sup (j) + beta  ( beta +1) e sub 1 ]} G]
}~~.
.EN
.SP2
.P
We now need the following simple, but crucial, lemma whose proof is left
as a pleasant exercise to the reader.
.SP1
.DS
LEMMA
.DE
.LP
(i) $ alpha sup (j) + beta   beta  e sub 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 <= r <= j2 $, there are r 1's followed by nr1
0's. In this case $ alpha sup (j) + beta   beta  e sub 1 $ is the
image of $ alpha sup (j) $ under the cycle (1,2,...,r+1), whose sign is
$(1) sup r $ .
.SP1
.LP
(ii)$ alpha sup (j) + beta  ( beta +1) e sub 1 $ is a bad guy unless
$beta$ has the form (0,1,...1,0,...0), where for some r satisfying
$ j2 <= r <= n1 $ you have a 0 followed by r 1's followed by nr1
0's. In this case $ alpha sup (j) + beta  ( beta +1) e sub 1 $ is the
image of $ alpha sup (j1) $ under the cycle (1,2,...,r+1) whose sign is
$(1) sup r $.
.P
Discarding all the bad guys in (5.17) and using the above lemma and
the \fIcrucial lemma\fR ,the equation (5.17) shrinks to (recall (5.14))
.SP2
.EQ(5.18)
"{"
sum from r=0 to j2
{
u t sup { n1 r } (1) sup r (1) sup r
}
"}" A sup (j)

"{"
sum from r=j2 to n1
{
q sup 1 t sup { n1r } (1) sup r (1) sup r
}
"}" A sup (j1)
.EN
.SP1
.EQ
=
"{"
sum from r=0 to j2
{
(1) sup r t sup r (1) sup r
}
"}" A sup (j)

"{"
sum from r=j2 to n1
{
(1) sup r t sup r s (1) sup r
}
"}" A sup (j1)
.EN
.SP2
.P
By summing all the geometric series and performing very routine and
simple ninth grade algebra we get (5.9). \(sq
tav vav shin lamed bet ayin
.SP3
.DS
REFERENCES
.DE
.LP
Andrews, George
.br
.LP
[An]\fI"qSeries: Their Development and Applications
in Analysis, Number Theory, Combinatorics, Physics, and Computer Algebra",
\fPCBMS regional conference series in mathematics, number 66,
American Mathematical Society, Providence, 1986.
.SP1
.LP
Askey, Richard
.br
.LP
[As]\fI Some basic hypergeometric extensions of
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938951.
.SP1
.LP
Cohen, Daniel I.A.
.br
.LP
"Basic Techniques of Combinatorial Theory", JOhn Wiley, New York, 1978,
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.SP1
.LP
Habsieger, Laurent
.br
.LP
[H]\fIUne qintegrale de SelbergAskey\fP, SIAM J.Math.Anal., to appear.
.SP1
.LP
Kadell, Kevin
.br
.LP
[K]\fIA proof of Askey's conjectured qanalog of Selberg's integral
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.SK
.LP
Macdonald, Ian G.
.br
.LP
[Ma]\fISome conjectures for root systems\fR, SIAM J.Math.Anal. \fB13\fR
(1982), 9881007.
.SP1
.LP
Morris, Walter G,II
.br
.LP
[Mo]"Constant Term Identities for Finite and Affine Root Systems", Ph.D.
thesis, Univ. of WisconsinMadison, 1982.
[ available for 25.50 dollars from University
Microfilms, 300 Zeeb Road, Ann Arbor, Michigan 48106, (313)7614700].
.SP1
.LP
Stanton, Dennis
.br
.LP
[Sta]\fISign variations of the Macdonald identities\fR, SIAM
J.Math.Anal. \fB17\fR(1986),14541460.
.SP1
.LP
Stembridge, John
.br
.LP
[Ste]\fIA short proof of Macdonald's conjecture for the root systems of
type A\fR, UCLA preprint(1986).
.SP1
.LP
Zeilberger, Doron
.br
.LP
[Z]\fIA unified approach to Macdonald's rootsystem conjectures\fR,
preprint.
.SP1
.LP
Zeilberger, Doron and Bressoud, David
.br
.LP
[ZB]\fIA proof of Andrews' qDyson conjecture\fP, Discrete
Math.\fB54\fP(1985), 201224.
.SP3
.LP
Department of Mathematics, Drexel University, Philadelphia, PA19104.
.SP1
.LP
March 30, 1987.