Journal of Algebraic Combinatorics, 21, 111–130, 2005
c 2005 Springer Science + Business Media, Inc. Manufactured in The Netherlands.
A Formula for N-Row Macdonald Polynomials
ELLISON-ANNE WILLIAMS
North Carolina State University
kingpuck20@yahoo.com
Received February 13, 2003; Revised October 30, 2003; Accepted October 11, 2004
Abstract. We derive a formula for the n-row Macdonald polynomials with the coefficients presented both
combinatorically and in terms of very-well-poised hypergeometric series.
Keywords:
1.
Macdonald polynomials, symmetric functions, hypergeometric series
Introduction
Denote the ring of symmetric functions over the field F as F and let nF denote its
nth graded space. The space nF consists of all symmetric
functions of total degree n ∈ Z ,
indexed by the partitions λ = (λ1 , . . . , λk ) for which i λi = n. There are four Z -bases
and
one Q-basis of n . The Q-basis consists of the power sum symmetric functions pn = i xin ,
where pλ = pλ1 · · · pλk , and the four Z -bases are: The monomial symmetric functions
m λ = i1 <···<ik xiλ11 . . . xiλkk , the elementary symmetric functions en = i1 <···<in xi1 . . . xin ,
where eλ = eλ1 · · · eλk , the complete symmetric functions h λ = i1 ≤...≤ik xiλ11 . . . xiλkk , and
λ +k− j
k− j
the Schur functions sλ (x1 , . . . , xk ) = det(xi j
)1≤i, j≤k /det(xi ).
Let H = Q(q, t) be the field of rational functions in q and t. In 1988, Macdonald
introduced a new class of two-parameter symmetric functions Pλ (q, t), over the ring H ,
which generalize several classes of symmetric functions. In particular, taking q = t we
obtain the Schur functions, setting t = 1 we have the monomial symmetric functions, and
letting q = 0 gives the Hall-Littlewood functions.
We know from [4] that the (Pλ ) are a basis of nH . Further, with respect to the scalar
product:
pλ , pµ = δλ,µ
i mi m i !
l(λ)
1 − qλj
i
j=1
we have that
Pλ , Pµ = 0
if
λ = µ,
1 − tλj
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WILLIAMS
where m i denotes the multiplicity of i in λ and l(λ) denote the length of λ. We also know
that for each λ, there exists a unique Pλ (q, t) such that:
Pλ = m λ +
cλµ m µ
where cλµ ∈ Q(q, t).
µ<λ
Define:
Qλ =
Pλ
.
Pλ , Pλ Then, the bases (Pλ ) and (Q λ ) of nH are dual to each other, Q λ , Pµ = δλ,µ , and from [4],
for λ = (n):
Q (n) =
|λ|=n i
l(λ)
1 1 − tλj
pλ .
m
i i m i ! j=1 1 − q λ j
For partitions of the form λ = (λ1 , . . . , λk ), we will derive a formula for the Macdonald
polynomials Q λ with coefficients presented first combinatorically in Section 3 and then as
very-well-poised hypergeometric series in section 5. A formula for Q (λ1 ,λ2 ) can be found in
[2] and following the completion of this work, the author discovered the paper [3] which
gives a formula, without proof, for Q (λ1 ,λ2 ,λ3 ) .
2.
Preliminaries
Let λ, µ be partitions such that µ ⊂ λ; the diagram of µ being contained in the diagram of
λ. For compactness, we will denote the skew diagram λ − µ as λ\µ. Moreover, we will
view the diagram of a partition with row one being the largest and row n begin the smallest
(English notation).
The diagram of λ\µ is said to be a horizontal r-strip if the number of blocks contained
in λ\µ equals r and, of the remaining r blocks, there is at most one in each column of λ\µ.
For each block d found in the diagram of λ, let:

e(d) s(d)+1
)
 (1 − q t
bλ (d) = (1 − q e(d)+1 t s(d) )

1
if d ∈ λ
(1)
if d ∈
/ λ,
where e(d) denotes the number of blocks to the east of d and s(d) denotes the number of
blocks to the south of d in the diagram of λ [4].
Let Rλ\µ denote the union of the rows and let Cλ\µ denote the union of the columns which
intersect the diagram of λ\µ.
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N -ROW MACDONALD POLYNOMIALS
Using a Pieri-type formula [4], for λ, µ such that µ ⊂ λ is a horizontal r -strip, we have:
Q µ Qr =
bµ (d)
Qλ
bλ (d)
λ
d
(2)
for d ∈ Rλ\µ − Cλ\µ .
Let r = (r1 , . . . , rn ) and r = (r1 , . . . , rn−1 ) be partitions with the length of row i equal
to ri . In order to derive a formula for Q (r1 ,...,rn ) , we begin with Q (r1 ,...,rn−1 ) .
To utilize (2), we must obtain all partitions ω = (w1 , . . . , wn ) such that r ⊂ ω and ω\r
is a horizontal rn -strip. Denote this set as .
3.
Construction of Ω
Begin with the desired partition r = (r1 , . . . , rn−1 ). For all ω ∈ , ω\r is a horizontal
rn -strip and therefore we must add exactly rn blocks to r = (r1 , . . . , rn−1 ). To this end, we
strategically enlarge the length of the rows, in some cases creating an nth row, by adding
the appropriate number of blocks to row ri , and below row rn−1 , for all i.
Let ji0 denote the number of blocks added to row ri . We may add a maximum of rn blocks
to row r1 and thus:
j10 = 0, . . . , rn .
Consequently, the number of blocks added to all other rows of r is first dependent on
Second, beginning with row r1 and adding blocks to r in order of row succession, we
see that the number of blocks which can be added to row ri is depedent upon the number
of blocks added to rows rk for 1 ≤ k < i. Finally, since we may add at most one block
anywhere in r in order to create any new column which appears in ω, the number of blocks
added to row rm is dependent on the difference between its length and the length of the
preceeding row, (rm−1 − rm ).
Assimilating this, for m ∈ {1, . . . , (n − 1)}:

m−1
m−1

0
0

 0, . . . , rn −
if
rn −
js
js ≤ (rm−1 − rm )


s=1
s=1
0
jm =
m−1


0


if
rn −
js > (rm−1 − rm ).
 0, . . . , (rm−1 − rm )
j10 .
s=1
Lastly, for row wn ∈ ω, where 0 ≤ wn ≤ rn , set:
wn = rn −
(n−1)
jm0 .
m=1
Taking all possible combinations of jm0 , beginning with m = 1 and ending with m =
(n − 1), we generate = {ω = (w1 , . . . , wn )}.
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WILLIAMS
Example 3.1 Let r = (5, 4, 3) and r4 = 2. Then, j10 = {0, 1, 2} , j20 = {0, 1}, and
j30 = {0, 1}. Thus: = {(5, 4, 3, 2), (5, 4, 4, 1), (5, 5, 3, 1), (6, 4, 3, 1), (5, 5, 4), (6, 4, 4),
(6, 5, 3), (7, 4, 3)}.
4.
Construction of the coefficients
For the partition λ = (λ1 , . . . , λn ) , define:
vi = { p where p = max{1, . . . , n} such that λ p ≥ i}.
Proposition 4.1 Let λ = (λ1 , . . . , λn ) be a partition with row i, λi . For each row λm and
each block dk ∈ λm , 1 ≤ k ≤ λm , numbered left to right,
s(dk ) = (vk − m).
Proof: Let λ denote the conjugate partition of λ, the partition whose diagram is the
transpose of the diagram of λ. Let dk be the kth block of row λm . Then, dk is the mth block
of the kth row of λ . Since row λm of λ is equal to column m of λ , it follows that the number
of blocks in row λm is equal to p where:
p = max{1, . . . , n} such that
λ p ≥ m.
This implies that the number of blocks to the east of dk is the diagram of λ is equal to
p − m. Taking the transpose, it follows that the number of blocks south of dk in the diagram
of λ equals p − m.
Remark 4.1 As demonstrated by the proof of Proposition 4.1, for λ = (λ1 , . . . , λn ) , vi ≡
λi . As we shall see, the desirability of using vi rather than the conjugate partition is evidenced
by the considerable facility that it gives to the implimentation of the principal formula. For
example, when λ = (6, 4, 2, 2) and i = 3, we may simply observe from λ that vi = 2.
Proposition 4.2 Let λ = (λ1 , . . . , λn ) be a partition with row i, λi . For each row λm ,
let {dk }, 1 ≤ k ≤ λm , denote the set of blocks which compose λm , numbered left to right.
Then:
b λm = bλ ({dk }) =
λ
m −1
i=0
1 − q λm −(i+1) t v(i+1) −(m−1)
1 − q λm −i t v(i+1) −m
Proof: For the product limits {i = 0, . . . , (λm − 1)} , let block k , dk , for 1 ≤ k ≤ λm ,
correspond to i = (k − 1). Then, {i = 0, . . . , (λm − 1)} corresponds directly to {dk }, where
1 ≤ k ≤ λm .
It is easily seen that the number of blocks east of block dk in the diagram of λ is equal to
(λm − k). By Proposition 4.1, we know that the number of blocks south of dk in the diagram
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N -ROW MACDONALD POLYNOMIALS
of λ is equal to (vk − m). Therefore:
1 − q λm −k t v(k) −(m−1)
.
bλ (dk ) = 1 − q λm −k+1 t v(k) −m
Shifting block bk to block bi where i = (k − 1), we have:
bλ (dk ) = bλ (di ) =
1 − q λm −(i+1) t v(i+1) −(m−1)
.
1 − q λm −i t v(i+1) −m
Taking the product over {i = 0, . . . , (λm − 1)} to encompass {dk }, 1 ≤ k ≤ λm , yields the
desired result.
Let ω ∈ . In order to utilize the Pieri-type formula, we must construct the required
coefficent for each Q (ω) :
br (d)
bω (d)
d
d ∈ Rω\r − Cω\r .
For m ∈ {1, . . . , (n − 1)}, we compute these coefficients according to row, rm , and
according to the number of blocks added to rm , jm0 , where ωm = (rm + jm0 ).
Proposition 4.3 Let ω ∈ and r = (r1 , . . . , rn−1 ).
For row wm = (rm + jm0 ), 1 ≤ m ≤ (n − 1), such that d ∈ wm and d ∈ Rω\r − Cω\r ,
br (d)
m
= T jm0 ,0 T jm0 , jk0 ,0
b
(d)
w
m
d
m < k ≤ (n − 1)
(3)
where
T jm0 ,0 =



r
m −1
0
i=rn − n−1
s=1 js

0
1 − q rm −(i+1) t v(i+1) −(m−1) 1 − q rm + jm −i t v(i+1) −m
0
1 − q rm −i t v(i+1) −m 1 − q rm + jm −(i+1) t v(i+1) −(m−1)
1
jm0 = 0
jm0 = 0
(4)
and
T jm0 , jk0 ,0
r + j 0 −1
0
k
k
r −i k−m−1

)(1 − q rm + jm −(i+1) t k−m )
 (1 − q m t
0
=
(1 − q rm −(i+1) t k−m )(1 − q rm + jm −i t k−m−1 )

 i=rk
1
jm0 = 0 and jk0 = 0
otherwise
(5)
restricting v(i+1) to r ; v(i+1) = { p where p = max{1, . . . (n − 1)} such that r p ≥
(i + 1)}.
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WILLIAMS
Proof: First, note that within this proof, we use the expressions br m and bw m to identify
“incomplete stages” in the development of brm and bwm .
For row rm , we include only the d ∈ rm for which d ∈ Rω\r − Cω\r . Since row
0
wn = (rn − n−1
m=1 jm ), it follows that, for each row of ω and hence of r , the blocks
{d1 , . . . , d(rn − n−1 jm0 ) } are not included in Rω\r − Cω\r . Therefore, by Proposition 4.2:
m=1
br m
=
r
m −1
0
i = rn − n−1
m=1 jm
1 − q rm −(i+1) t v(i+1) −(m−1)
.
1 − q rm −i t v(i+1) −m
(6)
Similarly, for row wm = (rm + jm0 ):
bw m
=
r
m −1
0
i = rn − n−1
m=1 jm
1 − q rm + jm −(i+1) t v(i+1) −(m−1)
0
1 − q rm + jm −i t v(i+1) −m
0
(7)
where we restrict v(i+1) to r in order to exclude any blocks d ∈
/ r which lie beneath row
rm .
If jm0 = 0, we do not include blocks from rm in (3), and thus from (6) and (7):
br m
bw m
= T jm0 ,0
=



r
m −1
0
i=rn − n−1
s=1 js

1 − q rm −(i+1) t v(i+1) −(m−1) )(1 − q rm + jm −i t v(i+1) −m
0
1 − q rm −i t v(i+1) −m )(1 − q rm + jm −(i+1) t v(i+1) −(m−1)
0
jm0 = 0
jm0 = 0.
1
However, we do not include any d ∈ wm such that d ∈ Cω\r ; therefore, we must get rid
of any terms corresponding to these blocks in the quotient (4). These blocks, which generate
the unwanted terms, directly correspond to the jk0 , for (m + 1) ≤ k ≤ (n − 1) , which were
added below them on row rk to create row wk . In order for these terms to appear in (4),
we must have jm0 = 0 and jk0 = 0. Therefore, to correct this in order to include only the
d ∈ wm such that d ∈ Rω\r − Cω\r , we multiply (4) by:
T jm0 , jk0 ,0
r + j 0 −1 0
k
r −i k−m−1
k

)(1 − q rm + jm −(i+1) t k−m
 1−q m t
rm −(i+1) t k−m )(1 − q rm + jm0 −i t k−m−1
=
1
−
q
i=r
k


1
yeilding the desired result.
jm0 = 0 and jk0 = 0
otherwise,
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N -ROW MACDONALD POLYNOMIALS
5.
Construction of the principal formula
Let r = (r1 , . . . , rn ) be a partition. For l ∈ {1, . . . , (rn − 1)}, let
n−1 l−1
j1l = 0, . . . , rn −
jsc .
s=1 c=0
And, for m ∈ {2, . . . , (n − 1)}, let

n−1 m−1
l−1


c
l

js −
js
0, . . . , rn −




s=1 c=0
s=1



n−1 m−1
l−1
l−1

c


c
l
c

jm−1 − jm
js −
js ≤ rm−1 − rm +

 if rn −
s=1 c=0
s=1
c=0
l
jm =
l−1

c

c


0, . . . , rm−1 − rm +
jm−1 − jm



c=0




n−1 m−1
l−1
l−1

c

c
l
c

js −
js > rm−1 − rm +
jm−1 − jm .

 if rn −
s=1 c=0
s=1
c=0
Remark 5.1 The purpose of the jil , 1 ≤ i ≤ (n − 1), is to allow us to systematically
n−1 0
reduce row wn = (rn − i=1
ji ) to zero. To this end, we first must compute the ji0 in
n−1 l
n−1 l
order to generate . Then, restricting the jil such that i=1
ji = 0 , 0 <
i=1 ji ≤
n−1 l−1 c
(rn − s=1 c=0 js ) , we are able to achieve the desired result. Further, restrict the ji0 such
n−1 0
that i=1
ji = 0, excluding the original partition r = (r1 , . . . , rn ) from this reduction.
We must now add at least one block to the partition r , and to any subsequently created
partitions (i.e., ω ∈ ); thus, the maximum
of times that we may need to repeat
number
n−1 rn −1 l
this reduction process is rn , yielding (rn − i=1
l=0 ji ) = 0.
Example 5.1 Building upon Example 3.1, for the partition ω = (6, 4, 3, 1) , j10 =
1 , j20 = 0, and j30 = 0, we desire to reduce row w4 to zero. Using the restriction given in
3
1
Remark 5.1, we have j11 = {0, 1} , j21 = {0, 1} , j31 = {0, 1} where 0 <
m=1 jm ≤ 1.
0
0
0
1
Taking all possible combinations of the jm dictated by the j1 = 1 , j2 = 0, and j3 = 0 , 1 ≤
m ≤ 3, and applying them to ω, yields the set of three-row partitions {(7, 4, 3), (6, 5, 3),
(6, 4, 4)}.
Given jml , 1 ≤ m ≤ (n − 1), (4) becomes:
 l−1
rm + c=0 jmc −1








i=
rn −n−1 l j c s
l
s=1 r c=0+l−1 j c −(i+1) v −(m−1) T jml ,0 =
c
m
c=0 m

1
−
q
1 − q rm + c=0 jm −i t v(i+1) −m
t (i+1)


×
l−1 c
l


c

1 − q rm + c=0 jm −i t v(i+1) −m 1 − q rm + c=0 jm −(i+1) t v(i+1) −(m−1)



1
jml = 0
jml = 0
(8)
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WILLIAMS
and, (5) becomes:
T jml , jkl ,0
 l c
rk + c=0 jk −1







i = rk +l−1
jc
l
c=0 kr +l−1 j c −i k−m−1 c
=
1 − q rm + c=0 jm −(i+1) t k−m
1 − q m c=0 m t


×

jml = 0 and jkl = 0
c

rm + l−1
rm + lc=0 jmc −i t k−m−1

c=0 jm −(i+1) t k−m
1
−
q
1
−
q



1
otherwise
(9)
c
where for l = 0, l−1
c=0 jm = 0 and where v(i+1) is restricted to the partition (r 1 +
l−1 c
l−1 c
c=0 j1 , . . . , r n−1 +
c=0 jn−1 ); v(i+1) = { p where p = max{1, . . . , (n − 1)} such that
l−1 c
(r p + c=0 j p ) ≥ (i + 1)}.
Theorem 5.1 For the partition r = (r1 , . . . , rn ) , m ∈ {1, . . . , (n − 1)}, and l ∈ {1, . . . ,
(rn − 1)}, we have
Q (r1 ,...,rn ) = Q (r1 ,...,rn−1 ) Q (rn )
n−1
n−1
i−1
n −1
r
i
+
(−1)
T jml ,0
T jml , jkl ,0
jm0 , jml i=1
l=0
m=1
×Q i−1 l
i−1 l r1 + l=0
j1 ,...,rn−1 + l=0
jn−1
+
j0 , jl
n−1 m rnm−1
m=1 jm
×Q rn
(−1)
r
n −1
l=0
k=m+1
Q
i−1 l rn − n−1
m=1
l=0 jm
n−1
T jml ,0
m=1
n−1
T jml , jkl ,0
k=m+1
= 0
n −1 l
n −1 l r1 + rl=0
j1 ,...,rn−1 + rl=0
jn−1
signifies to sum over all possible combinations of the values of j10 and jm0
n−1 0
such that 0 < m=2 ( j1 + jm0 ) ≤ rn and all possible combinations of the values jml dictated
l
by the jm0 such that n−1
m=1 jm = 0.
where
Proof:
jm0 , jml
We want to show that:
Q (r1 ,...,rn ) − Q (r1 ,...,rn−1 ) Q (rn )
n−1
n−1
i−1
n −1
r
−
(−1)i
T jml ,0
T jml , jkl ,0
j10 , jm0 i=1
× Q
l=0
m=1
i−1 l
i−1 l
r1 + l=0
j1 ,...,rn−1 + l=0
jn−1
k=m+1
Q
i−1 l rn − n−1
m=1
l=0 jm
119
N -ROW MACDONALD POLYNOMIALS
−
(−1)
Using (2), for
l=0
j0, j0
n−1 1 rnm−1
m=1 jm
× Q r
r
n −1
rn
n−1
T jml ,0
m=1
n−1
T jml , jkl ,0
k=m+1
= 0
rn −1 l
rn −1 l 1 + l=0 j1 ,...,rn−1 + l=0 jn−1
n−1
0
m=1 jm
= 0.
(10)
= 0, we have:
Q (r1 ,...,rn−1 ) Q (rn ) = Q (r1 ,...,rn ) +
×Q r
j10 , jm0
n−1
T jm0 ,0
m=1
n−1
T jk0 , jk0 ,0
k=m+1
n−1
0
0
1 + j1 ,...,rn−1 + jn−1 ,rn −
0
m=1 jm
.
(11)
Substituting (11) into (10), we need only to show:
j10 , jm0
+
n−1
T jm0 ,0
m=1
k=m+1
n −1
r
i
(−1)
j10 , jm0 i=1
×Q r
T jk0 , jk0 ,0 Q r
i−1
l=0
i−1
1+
l
l=0 jn−1
rn −1
(−1)
r
n −1
rn −1
1+
l=0
n−1
T jml ,0
0
m=1 jm
T jml , jkl ,0
k=m+1
Q
i−1 l rn − n−1
m=1
l=0 jm
l=0
j0, j0
n−1 1 rmn −1
=0
m=1 jm
×Q r
n−1
i−1
n−1
0
0
1 + j1 ,...,rn−1 + jn−1 ,rn −
m=1
l
l=0 j1 ,...,rn−1 +
=
n−1
n−1
T jml ,0
m=1
n−1
T jml , jkl ,0
k=m+1
n −1 l .
j1l ,...,rn−1 + rl=0
jn−1
(12)
Note that each Pieri-formula expansion (via (2) and (11)) of
j10 , jm0
n−1
T jm0 ,0
m=1
n−1
k=m+1
T jk0 , jk0 ,0 Q r
n−1
0
0
1 + j1 ,...,rn−1 + jn−1 ,rn −
0
m=1 jm
(13)
yields two terms. To show our desired result, we will show that with each successive
“numbered” Pieri-formula expansion of (13), the term which contains the product of two
Macdonald polynomials cancells with the correspondingly numbered term i in
n −1
r
j10 , jm0
(−1)i
i=1
×Q r
i−1
l=0
i−1
1+
n−1
T jml ,0
m=1
i−1
l
l=0 j1 ,...,rn−1 +
n−1
l
l=0 jn−1
T jml , jkl ,0
k=m+1
Q
i−1 l rn − n−1
m=1
l=0 jm
(14)
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WILLIAMS
and that, in the final Pieri expansion of (13), we will be left only with
(−1)rn −1
j0, j0
n−1 1 rnm−1
m=1 jm
×Q r
n −1
n−1
l=0
T jml ,0
m=1
n−1
T jml , jkl ,0
k=m+1
= 0
n −1 l
n −1 l .
r1 + rl=0
j1 ,...,rn−1 + rl=0
jn−1
(15)
To this end, we will induct on i.
Consder i = 1. Completing the first Pieri expansion of (13), and computing i = 1 from
(14), (12) becomes:
j10 , jm0
n−1
T jm0 ,0
m=1
−
n−1
m=1
j10 , jm0 l=0
−
(−1)
=−
l=0
+
(−1)
j10 , jm0 i=2
×Q r
i−1
1+
Q r
T jm0 , jk0 ,0
n−1
T jml ,0
0
0
1 + j1 ,...,rn−1 + jn−1
Q
0
rn − n−1
m=1 jm
T jml , jkl ,0
k=m+1
i−1 l rn − n−1
m=1
l=0 jm
n−1
T jml , jkl ,0
k=m+1
n−1 1
1
i
l
l=0 jm
Q
T jml ,0
l
l=0 j1 ,...,rn−1 +
n −1
r
l
l=0 jn−1
m=1
1
1+
n−1
i−1
n−1
T jml , jkl ,0
m=1
l
l=0 j1 ,...,rn−1 +
j10 , jm0 l=0
×Q r
i−1
i−1
1
m=1
k=m+1
j10 , jm0 i=2
1+
n−1
T jm0 ,0
i
0
rn − n−1
m=1 jm
n−1 1
l
l=0 jn−1 ,rn −
m=1
Q
k=m+1
n−1
0
0
1 + j1 ,...,rn−1 + jn−1
n−1
T jml ,0
1
l
l=0 j1 ,...,rn−1 +
n −1
r
×Q r
1
1+
j10 , jm0
+
T jk0 , jk0 ,0 Q r
k=m+1
1
×Q r
n−1
l
l=0 jn−1 ,rn −
i−1
l=0
n−1
m=1
T jml ,0
m=1
i−1
l
l=0 j1 ,...,rn−1 +
l
l=0 jn−1
l
l=0 jm
n−1
T jml , jkl ,0
k=m+1
Q
i−1 l .
rn − n−1
m=1
l=0 jm
Therefore, in the first Pieri expansion of (13), the term containing the product of two
Macdonald polynomials cancelled with the correspondingly numbered i = 1 in (14), as
desired.
Assume that with each further “numbered” Pieri expansion of (13), up to (rn − 2), the
term containing a product of two Macdonald polynomials cancells with the correspondingly
numbered i in (14). We will show the result for (rn − 1).
121
N -ROW MACDONALD POLYNOMIALS
For i = (rn − 2), we have that (12) equals:
r
n −3
rn −3
(−1)
j10 , jm0
l=0
×Q r
+
n−1
T jml ,0
m=1
T jml , jkl ,0
k=m+1
n −3 l j1l ,...,rn−1 + rl=0
jn−1
rn −3
1+
n−1
l=0
(−1)rn −2
Q r −n−1 rn −3 j l n
m
m=1
l=0
r
n−1
n−1
n −2
T jml ,0
T jml , jkl ,0
j10 , jm0
l=0
r
n −3
m=1
×Q k=m+1
rn −2 l n −2 l
n −2 l
r1 + rl=0
j1 ,...,rn−1 + rl=0
jn−1 ,rn − n−1
m=1
l=0 jm
+
(−1)rn −2
j10 , jm0
×Q r
+
l=0
n−1
T jml ,0
m=1
l=0
(−1)rn −1
l=0
×Q (−1)rn −2
j10 , jm0
m=1
+
r
n −2
l=0
×Q r
rn −2
1+
l=0
k=m+1
×Q r
×Q r
rn −2
l=0
r
n −2
m=1
n−1
T jml ,0
T jml , jkl ,0
k=m+1
n−1
T jml ,0
m=1
n−1
T jml , jkl ,0
k=m+1
n −2 l j1l ,...,rn−1 + rl=0
jn−1
n−1 rn −2
n−
rn −2 l rn − n−1
m=1
l=0 jm
m=1
l=0
1+
n−1
Q
rn −2 l n −2 l
j1l ,...,rn−1 + rl=0
jn−1 ,rn − n−1
m=1
l=0 jm
(−1)rn −1
j10 , jm0
T jml , jkl ,0
k=m+1
n −2 l
n −2 l
r1 + rl=0
j1 ,...,rn−1 + rl=0
jn−1
=
Q r −n−1 rn −3 j l n
m
m=1
l=0
r
n−1
n−1
n −2
T jml ,0
T jml , jkl ,0
j10 , jm0
n−1
n −3 l j1l ,...,rn−1 + rl=0
jn−1
rn −3
1+
l=0
jml
.
Performing the Pieri expansion on (13) a final time, (12) is equal to:
j10 , jm0
r
n −2
(−1)rn −2
l=0
×Q r
+
j10 , jm0
×Q rn −2
1+
l=0
n−1
n−1
T jml ,0
m=1
T jml , jkl ,0
k=m+1
n −2 l j1l ,...,rn−1 + rl=0
jn−1
(−1)rn −1
Q r −n−1 rn −2 j l n
m
m=1
l=0
r
n−1
n−1
n −1
T jml ,0
T jml , jkl ,0
l=0
m=1
n −1 l
n −1 l
r1 + rl=0
j1 ,...,rn−1 + rl=0
jn−1
k=m+1
122
WILLIAMS
+
rn −1
(−1)
j10 , jm0
×Q r
rn −2
1+
j0, j0
n−1 1 rnm−1
m=1 jm
×Q r
l=0
l=0
=
r
n −2
n−1
T jml ,0
m=1
n−1
T jml , jkl ,0
k=m+1
rn −2 l n −2 l Q j1l ,...,rn−1 + rl=0
jn−1
rn − n−1
m=1
l=0 jm
rn −1
(−1)
r
n −1
l=0
n−1
T jml ,0
m=1
n−1
T jml , jkl ,0
k=m+1
= 0
rn −1 l
rn −1 l .
1 + l=0 j1 ,...,rn−1 + l=0 jn−1
Since, by definition of jml , we must have
rn −
n−1 r
n −1
jml = 0.
m=1 l=0
Example 5.2 We shall build upon Example 3.1 to calculate the expansion of the
Macdonald polynomial Q (5,4,3,2) .
Given the nature of the partition r = (5, 4, 3, 2), it is fitting and instructive to use the
following notation:
n−1
m=1
n−1
k=m+1
T jml ,0 ≡ T(lj l ,..., j l
(n−1) )
1
ll
T jll , j l ,0 ≡ T(0,...,
jl
m
l
jml , jm+i
= 0,
l
m+1 ,..., j(n−1) )
k
1 ≤ i ≤ (n − m − 1)
For 2 ≤ m ≤ 3, we have:
3 l−1
c
,
= 0, . . . , 2 −
js
j10
jm0
= {0, 1, 2}
=
j1l
s=1 c=0

m−1



js0
0,
.
.
.
,
2
−


if 2 −




0, . . . , (rm−1 − rm )
if 2 −
s=1
m−1
s=1
m−1
s=1
≤ (rm−1 − rm )
js0
js0
> (rm−1 − rm ),
123
N -ROW MACDONALD POLYNOMIALS
jml =

3 m−1
l−1


c

j
−
jsl
0,
.
.
.
,
2
−

s



s=1 c=0
s=1



3 m−1
l−1
l−1

c


c
l
c
 if 2 −
j
≤
r
j
−
j
−
r
+
−
j

m−1
m
s
s
m−1
m

s=1 c=0
s=1
c=0
s=1 c=0
s=1
c=0
l−1

c

c


0,
.
.
.
,
r
j
−
r
+
−
j
m−1
m

m−1
m


c=0




3 m−1
l−1
l−1

c

c
l
c

if
2
−
>
r
j
−
j
−
r
+
−
j
j
.

m−1
m
s
s
m−1
m

We begin with
jm0
since these values determine the subsequent choices for jml . We have:
j10 = {0, 1, 2}
j20 = {0, 1}
j11 = {0, 1}
j21 = {0, 1}
j30 = {0, 1}
j31 = {0, 1}.
Taking all possible combinations of the jml , l = {0, 1} , such that 0 <
0 < 3m=1 jm1 ≤ 1 , we have:
3
3
Q (5,4,3,2) = Q (5,4,3) Q (2) −
T jm0 ,0
T jm0 , jk0 ,0
×Q j10 , jm0
5+ j10 ,4+ j20 ,3+ j30
+
1
(
j10 , jm0 l=0
m=1
Q
3
T jml ,0
m=1
0
m=1 jm
≤ 2 and
k=m+1
2− 3m=1 jm0
3
3
k=m+1
T jml , jkl ,0
Q 5+1
1
l
l=0 j1 ,4+
1
l
l=0 j2 ,3+
l
l=0 j3
0
0
1
0
1
= Q (5,4,3) Q (2) + T(1,0,0)
T(1,0,0)
− T(2,0,0)
T(0,1,0)
Q (7,4,3) + T(1,0,0)
0
1
0
00
Q (6,5,3)
+ T(0,1,0)
T(1,0,0)
− T(1,1,0)
T(0,1,0)
0
1
0
1
0
00
Q (6,4,4)
+ T(1,0,0) T(0,0,1) + T(0,0,1) T(1,0,0) − T(1,0,1)
T(0,0,1)
0
1
0
1
0
00
+ T(0,1,0) T(0,0,1) + T(0,0,1) T(0,1,0) − T(0,1,1) T(0,0,1) Q (5,5,4)
0
0
0
− T(1,0,0)
Q (6,4,3) Q (1) − T(0,1,0)
Q (5,5,3) Q (1) − T(0,0,1)
Q (5,4,4) Q (1) .
6.
Very-well-poised hypergeometric series
We may express the coefficients T jml , for m ∈ {1, . . . , (n − 1)} , and l ∈ {0, . . . , rn−1 } ,
in terms of very-well-poised hypergeometric series.
We will use the following notations.
(a; q)0 = 1
n−1
(a; q)n =
(1 − aq i )
i=0
124
WILLIAMS
(a; q)∞ =
∞
(1 − aq i )
i=0
(a; q)∞
(aq n ; q)∞
(a1 , . . . , am ; q)n = (a1 ; q)n · · · (am ; q)n
(a1 , . . . , am ; q)∞ = (a1 ; q)∞ · · · (am ; q)∞
(a; q)n =
The basic hypergeometric series φr +1,r is of the form:
a1 , a2
b1 , b2
, . . . , ar +1
, . . . , br
q, z
where
φr +1,r =
∞
n=0
(a1 , . . . , ar +1 ; q)n
zn .
(b1 , . . . , br ; q)n (q; q)n
We say that φr +1,r is very-well-poised, denoted
Wr +1,r = (a1 ; a4 , . . . , ar +1 ; q, z),
if
a1 q = a2 b1 = . . . = ar +1 br
1
1
and a2 = qa12 , a3 = −qa12 .
From [1], we have:
, aq , aq ; q ∞
aq, aq
aq
bc bd cd
W6,5 a; b, c, d; q,
= aq aq aq aq .
bcd
, c , d , bcd ; q ∞
b
(16)
The Coefficients Tj0m ,0 and Tj0m ,j0k ,0
We first compute the coefficient T jm0 in terms of very-well-poised hypergeometric series.
Beginning with row rm , we identify the “indentions” which lie “under” it in the diagram
of r . In other words, we identify all rows ri , m < i ≤ (n − 1) , such that ri < ri+1 .
For k1 ∈ {1, . . . , (n − m)}, let rn−k1 be the largest indexed row of r such that rn−k1 ≤ rm .
Proposition 6.1 Suppose rn−k1 = rm . It follows that k1 = (n − m) and rm ≡ rn−1 . Then:
T jm0 ,0 =

n−1 0
n−1 0
0
0
W6,5 q rm −rn + s=1 js t n−m−1 ; qt −1 , q − jm , q rn−1 −rn + s=1 js ; q, q jm t n−m

1
jm0 = 0
jm0 = 0.
125
N -ROW MACDONALD POLYNOMIALS
Proof: Let rn−k1 = rm ; k1 = (n − m) and rm ≡ rn−1 . There are two cases: jm = 0 and
jm0 = 0.
For jm0 = 0, we have that T jm0 ,0 = 1 by Proposition 4.3.
For jm0 = 0, using properties of hypergeometric series and (16), (4) becomes:
T jm0 ,0 =
rn−1
−1
i=rn −
(1 − q rm −(i+1) t n−m )(1 − q rm + jm −i t n−m−1 )
0
(1 − q rm −i t n−m−1 )(1 − q rm + jm −(i+1) t n−m )
0
n−1
s=1
js0
(q rm −rn−1 t n−m , q rm −rn−1 + jm +1 t n−m−1 ; q)rn−1 −rn +n−1 js0
0
=
s=1
(q rm −rn−1 +1 t n−m−1 , q rm −rn−1 + jm t n−m ; q)rn−1 −rn +n−1 js0
0
s=1
(q rm −rn−1 t n−m , q rm + jm −rn−1 +1 t n−m−1 , q rm −rn +
0
=
(q rm −rn +
n−1
s=1
=W6,5 (q
rm −rn +
=W6,5 (q
rm −rn +
js0 t n−m , q rm + jm0 −rn +
n−1
s=1
n−1
s=1
js0 n−m−1
t
js0 n−m−1
t
; qt
; qt
n−1
s=1
js0 +1 n−m−1
t
n−1
, q rm + jm −rn +
0
n−1
s=1
js0 n−m
t
; q)∞
0
0
s=1 js +1 t n−m−1 , q rm −rn−1 +1 t n−m−1 , q rm + jm −rn−1 t n−m ; q)
∞
n−1 0
0
0
−1
− jm
rn−1 −rn + s=1 js
rm −rn−1 + jm n−m
,q
−1
,q
,q
− jm0
,q
0
rn−1 −rn + n−1
s=1 js
; q, q
t
jm0 n−m
; q, q t
)
).
Remark 6.1 Let rm = rn−k1 . It follows that for all m < k ≤ (n − 1), we have jk0 = 0,
and thus T jm0 , jk0 ,0 = 1.
Consider the case when rn−k1 = rm . Let rn−k2 be the largest indexed row of r =
(r1 , . . . , rn−1 ) such that rn−k1 = rn−k2 . If rn−k2 = rm , let rn−k3 be the largest indexed
row of r for which rn−k2 = rn−k3 , ect. Continue this process until one reaches row rn−k f for
which rn−k f = rm , creating the chain:
rn−k1 < rn−k2 < · · · < rn−k f = rm
for ki ∈ {1, . . . , (n − m)}, 1 ≤ i ≤ f .
Proposition 6.2 Suppose rn−k1 = rm . Then, (4) becomes:
T jm0 ,0 =

rm −rn +n−1 j 0 n−m−k1 −1 − j 0 rn−k −rn +n−1 j 0
s=1 s t
s=1 s ;

; qt , q m , q 1
q
W
6,5




f −1


 q, q rm −rn−k1 + jm0 t n−m−k1 +1 · W q rm −rn−k(h+1) + jm0 t n−k(h+1) −m ;








6,5
h=2
rm −rn−kh n−m−k(h+1) +1
qt −1 , q jm , q rn−kh −rn−k(h+1) ; q, q
0
1
t
jm0 = 0
jm0 = 0.
Proof: Using our chain, we are able to identify the indentions under row rm in the diagram
of r ; thus, we obtain all intervals of blocks d ∈ rm on which s(d) changes. Using properties
of hypergeometric series and (16), we decompose (4) as follows:
126
WILLIAMS
rn−k1 −1
T jm0 ,0 =
js0
0
1 − q rm −(i+1) t n−m−k1 +1 1 − q rm + jm −i t n−m−k1
(1 − q rm −i t n−m−k1 )(1 − q rm + jm −(i+1) t n−m−k1 +1 )


0
rn−k(h+1) −1
f −1
1 − q rm −(i+1) t n−m−k(h+1) +1 1 − q rm + jm −i t n−m−k(h+1)


·
0
1 − q rm −i t n−m−k(h+1) 1 − q rm + jm −(i+1) t n−m−k(h+1) +1
i=rn−kh
h=2
0
q rm −rn−k1 t n−m−k1 +1 , q rm + jm −rn−k1 +1 t n−m−k1 ; q
0
rn−k −rn + n−1
s=1 js
= r + j 0 −r
·
1
r
−r
+1
q m m n−k1 t n−m−k1 +1 , q m n−k1 t n−m−k1 ; q rn−k −rn +n−1 j 0
s=1 s
1
0
rm −rn−k(h+1) n−m−k(h+1) +1
t
, q rm + jm −rn−k(h+1) +1 t n−m−k(h+1) ; q
f −1 q
rn−k(h+1) −rn−kh
rm −rn−k(h+1) +1 n−m−k(h+1)
rm −rn−k(h+1) n−m−k(h+1) +1
q
t
,q
t
;q
h=2
i=rn − n−1
s=1
0
rn−k(h+1) −rn−kh
n−1 0
n−1 0
0
0
q rm −rn−k1 t n−m−k1 +1 , q rm + jm −rn−k1 +1 t n−m−k1 , q rm −rn + jm + s=1 js t n−m−k1 +1 , q rm −rn + s=1 js +1 t n−m−k1 ; q
∞
=
n−1 0
n−1 0 0
0
0
q rm + jm −rn−k1 t n−m−k1 +1 , q rm + jm −rn−k1 t n−m−k1 , q rm −rn + s=1 js t n−m−k1 +1 , q rm −rn + s=1 js + jm +1 t n−m−k1 ; q
∞
f −1
P
·
Q
h=2
n−1 0
n−1 0
0
0
= W6,5 q rm −rn + s=1 js t n−m−k1 ; qt −1 , q − jm , q rn−k1 −rn + s=1 js ; q, q rm + jm −rn−k1 t n−m−k1 +1
·
f −1
0
0
W6,5 q rm −rn−k(h+1) + jm t n−k(h+1) −m ; qt −1 , q jm , q rn−kh −rn−k(h+1) ; q, q rm −rn−kh t n−m−k(h+1) +1 .
h=2
Where:
0
P = q rm −rn−k(h+1) t n−m−k(h+1) +1 , q rm + jm −rn−k(h+1) +1 t n−m−k(h+1) , q rm −rn−kh +1 t n−m−k(h+1) ,
0
q rm −rn−kh + jm t n−m−k(h+1) +1 ; q ∞
0
Q = q rm −rn−k(h+1) +1 t n−m−k(h+1) , q rm + jm −rn−k(h+1) t n−m−k(h+1) +1 , q rm −rn−kh t n−m−k(h+1) +1 ,
0
q rm −rn−kh + jm +1 t n−m−k(h+1) ; q ∞ .
Proposition 6.3 For rn−k1 = rm , the coefficient T jm0 , jk0 ,0 becomes:
0
0
0
0
T jm0 , jk0 ,0 = W6,5 q rm −rk − jk t k−m−1 ; q − jk , qt −1 , q − jm ; q, q rm −rk + jm t k−m .
Proof:
Using properties of hypergeometric series and (16), we have:
0
(1 − q rm −i t k−m−1 ) 1 − q rm + jm −(i+1) t k−m
rm −(i+1) t k−m ) 1 − q rm + jm0 −i t k−m−1
i=rk (1 − q
r −r − j 0 −1 k−m−1 r −r + j 0 − j 0 k−m qm k k t
,q m k m kt
; q j0
= k
0
0
0
q rm −rk − jk t k−m , q rm −rk + jm − jk +1 t k−m−1 ; q j 0
T jm0 , jk0 ,0 =
rk +
jk0 −1
k
127
N -ROW MACDONALD POLYNOMIALS
r −r − j 0 +1 k−m−1 r −r + j 0 − j 0 k−m r −r k−m r −r + j 0 +1 k−m−1 qm k k t
,q m k m kt
,q m kt
,q m k m t
;q ∞
= 0
0 − j 0 +1 k−m−1
0 k−m
r
−r
−
j
r
−r
+
j
r
−r
+
j
k−m
r
−r
+1
k−m−1
q m k kt
,q m k m k t
,q m k t
,q m k mt
;q ∞
0
0
0
0
= W6,5 q rm −rk − jk t k−m−1 ; q − jk , qt −1 , q − jm ; q, q rm −rk + jm t k−m .
The Coefficients Tjlm ,0 and Tjlm ,jlk ,0 .
For l = {1, . . . , (rn − 1)}, we compute the coefficients T jml ,0 and T jml , jkl ,0 in terms of verywell-poised hypergeometric series. To this end, fix l and carry out the procedure previously
described to obtain the chain:
rn−k1 +
l−1
c
jn−k
1
< · · · < rn−k f +
c=0
l−1
c
jn−k
f
c=0
= rm +
l−1
jmc
.
c=0
Notation 6.1 Set:
l−1
jmc ≡ Jm
c=0
n−1 l
jsc ≡ Js
s=1 c=0
l−1
c
jn−k
≡ Jn−ki
i
c=0
Proposition 6.4 Suppose that (rn−k1 + Jn−k1 ) = (rm + Jm ). It follows that (rm + Jm ) ≡
(rn−1 + Jn−1 ) and (8) becomes:
T jml ,0 =

l

W6,5 q rm −rn +Jm +Js t n−m−k1 ; qt −1 , q − jm , q rn−k1 +Jn−k1 −rn Js ;




f −1



rm −rn−k1 −Jn−k1 +Jm + jml n−m−k1

q,
q
·
t
W6,5 q rm −rn−ki −Jn−ki +Jm t n−ki −m+1 ;











h=2
qt −1 , q − jm , q rn−ki+1 −rn−ki −Jn−ki +Jn−ki+1 ; q, q rm −rn−ki+1 −Jn−ki+1 +Jm + jm t n−m−ki
jml = 0
l
1
l
jml = 0.
Proof: Let (rn−k1 + Jn−k1 ) = (rm + Jm ) ⇒ (rm + Jm ) ≡ (rn−1 + Jn−1 ). There are two
cases: jml = 0 and jml = 0.
For jml = 0, by Proposition 4.3, we have that T jml ,0 = 1.
128
WILLIAMS
For jml = 0, using properties of hypergeometric series, we have:
rn−1 +J
n−1 −1 1 − q rm +Jm −(i+1) t n−m 1 − q rm +Jm + jml −i t n−m−1
T jml ,0 =
l
1 − q rm +Jm −i t n−m−1 1 − q rm +Jm + jm −(i+1) t n−m
i=rn −Js
r +J −r −J n−m r +J + j l −r −J +1 n−m−1 q m m n−1 n−1 t
, q m m m n−1 n−1 t
; q r +J −r +J
P
= r +J −J −r +1 n−m−1 r −r −J +J + j l n−m n−1 n−1 n s =
Q
q m m n−1 n−1 t
, q m n−1 n−1 m m t
; q rn−1 −rn +Js +Jn−1
l
= W6,5 q rm −rn +Js +Jm t n−m−1 ; qt −1 , q − jm , q rn−1 −rn +Js +Jn−1 ;
l
q, q rm −rn−1 +Jm −Jn−1 + jm t n−m .
Where:
l
P = q rm −rn−1 +Jm −Jn−1 t n−m , q rm −rn−1 +Jm −Jn−1 + jm +1 t n−m−1 , q rm −rn +Js +Jm +1 t n−m−1 ,
l
q rm −rn +Js +Jm + jm t n−m ; q ∞
l
Q = q rm −rn +Js +Jm t n−m , q rm −rn +Js +Jm + jm +1 t n−m−1 , q rm −rn−1 +Jm −Jn−1 t n−m+1 ,
l
q rm −rn−1 +Jm −Jn−1 + jm t n−m ; q ∞ .
Remark 6.2 Since (rm + Jm ) ≡ (rn−1 + Jn−1 ) ⇒ jkl = 0 for all m < k ≤ (n − 1), we
have that T jml , jkl ,0 = 1.
Proposition 6.5 Suppose that (rn−k1 + Jn−k1 ) = (rm + Jm ). Obtaining the chain
rn−k1 + Jn−k1 < · · · < rn−k f + Jn−k f = (rm + Jm ),
we have that:
T jml ,0 =

l
W6,5 q rm −rn +Js +Jm t n−m−k1 ; qt −1 , q − jm , q rn−k1 −rn +Jn−k1 +Js ;




f −1


l

rm −rn−k1 −Jn−k1 +Jm + jml n−m−k1 +1

q,
q
·
t
W6,5 q rm −rn−k(h+1) +Jm + jm −Jn−k(h+1)












t
n−k(h+1) −m
; qt
−1
jml
,q ,q
h=2
rn−kh −rn−k(h+1) +Jn−kh −Jn−k(h+1)
q, q rm −rn−kh −Jn−kh +Jm t n−m−k(h+1)
1
;
0
jml =
l
jm = 0.
Proof: For (rn−k1 + Jn−k1 ) = (rm + Jm ), using properties of hypergeometric series, (8)
becomes:
l
1 − q rm +Jm −(i+1) t n−m−k1 +1 1 − q rm +Jm + jm −i t n−m−k1
T jml ,0 =
l
1 − q rm +Jm −i t n−m−k1 1 − q rm +Jm + jm −(i+1) t n−m−k1 +1
i=rn −Js
f −1 rn−k(h+1) +J
n−k(h+1) −1 1 − q rm +Jm −(i+1) t n−m−k(h+1) +1 1 − q rm +Jm + jml −i t n−m−k(h+1)
·
l
1 − q rm +Jm −i t n−m−k(h+1) 1 − q rm +Jm + jm −(i+1) t n−m−k(h+1) +1
i=rn−k +Jn−k
h=2
rn−k1 +Jn−k1 −1
h
h
129
N -ROW MACDONALD POLYNOMIALS
r −r +J −J
l
q m n−k1 m n−k1 t n−m−k1 +1 , q rm −rn−k1 +Jm −Jn−k1 + jm +1 t n−m−k1 ; q rn−k −rn +Js +Jn−k
1
1
= r −r +J −J + j l
q m n−k1 m n−k1 m t n−m−k1 +1 , q rm −rn−k1 +Jm −Jn−k1 t n−m−k1 ; q rn−k −rn +Jn−k +Js
1
1
f −1
P1
·
Q1
h=2
f −1
P2 P3
·
Q 2 h=2 Q 3
l
=W6,5 q rm −rn +Jm +Js t n−m−k1 ; qt −1 , q − jm , q rn−k1 +Jn−k1 −rn +Js ;
l
q, q rm −rn−k1 −Jn−k1 +Jm + jm t n−m−k1 +1 .
=
·
f −1
l
W6,5 q rm −rn−k(h+1) +Jm + jm −Jn−k(h+1) t n−k(h+1) −m ;
i=2
l
qt −1 , q jm , q rn−kh −rn−k(h+1) +Jn−kh −Jn−k(h+1) ; q, q rm −rn−kh −Jn−kh +Jm t n−m−k(h+1) .
Where:
P1 = q rm −rn−k(h+1) +Jm −Jn−k(h+1) t n−m−k(h+1) +1 ,
q rm −rn−k(h+1) +Jm −Jn−k(h+1) + jm +1 t n−m−k(h+1) ; q
Q 1 = q rm −rn−k(h+1) +Jm −Jn−k(h+1) +1 t n−m−k(h+1) ,
l
P2
Q2
P3
Q3
rn−k(h+1) −rn−kh +Jn−k(h+1) −Jn−kh
l
q rm −rn−k(h+1) +Jm −Jn−k(h+1) + jm t n−m−k(h+1) +1 ; q rn−k +Jn−k −rn−k −Jn−k
h
h
(h+1)
(h+1)
rm +Jm −rn−k −Jn−k n−m−k1 +1 rm +Jm + j l −rn−k −Jn−k +1 n−m−k1
m
1
1t
1
1
= q
,q
t
,
rm −rn +Js +Jm + jml n−m−k1 +1
rm −rn +Js +Jm +1 n−m−k1
q
t
,q
t
;q ∞
rm −rn−k +Jm −Jn−k + j l n−m−k1 +1 rm −rn−k +Jm −Jn−k + j l n−m−k1
mt
mt
1
1
1
1
= q
,q
,
rm −rn +Js +Jm n−m−k1 +1
rm −rs +Jn +Jm + jml +1 n−m−k1
q
t
,q
t
;q ∞
rm −rn−k +Jm −Jn−k
r −rn−k(h+1) +Jm −Jn−k(h+1) + jml +1 n−m−k(h+1)
(h+1)
(h+1) t n−m−k(h+1) +1 , q m
= q
t
,
rm −rn−kh +Jn−kh +Jm +1 n−m−k(h+1)
rm −rn−kh −Jn−kh +Jm + jml n−m−k(h+1) +1
q
t
,q
t
;q ∞
rm −rn−k −Jn−k +Jm +1 n−m−k(h+1) rm +Jm + j l −rn−k −Jn−k
m
(h+1)
(h+1)
(h+1)
(h+1) t n−m−k(h+1) +1 ,
= q
t
,q
l
q rm −rn−kh +Jm −Jn−kh t n−m−k(h+1) +1 , q rm −rn−kh −Jn−kh +Jm + jm +1 t n−m−k(h+1) ; q ∞ .
Proposition 6.6 For T jml , jkl ,0 , m < k ≤ (n − 1), we have:
T jml , jkl ,0

r −r −J +J − j l k−m−1

;
W6,5 ql m k k ml k t
− jk
−1
− jm
rm −rk +Jm +Jk + jml k−m
q , qt , q ; q, q
t
=


1
jml = 0 and jkl = 0
otherwise.
130
WILLIAMS
Proof:
Using properties of hypergeometric series, (9) becomes:
T jml , jkl ,0
l
(1 − q rm +Jm −i t k−m−1 ) 1 − q rm +Jm + jm −(i+1) t k−m
=
l
1 − q rm +Jm −(i+1) t k−m 1 − q rm +Jm + jm −i t k−m−1
i=rk +Jk
r −r +J −J − j l +1 k−m−1 r −r +J −J + j l − j l k−m qm k m k k t
,q m k m k m kt
; q jl
= k
l
l
l
q rm −rk −Jk +Jm − jk t k−m , q rm −rk +Jm −Jk + jm − jk +1 t k−m−1 ; q j l
l
rk +J
k + jk −1
k
P
=
Q
l
l
l
l
= W6,5 q rm −rk −Jk +Jm − jk t k−m−1 ; q − jk , qt −1 , q − jm ; q, q rm −rk +Jm +Jk + jm t k−m .
Where:
l
l
l
P = q rm −rk −Jk +Jm − jk +1 t k−m−1 , q rm −rk +Jm −Jk + jm − jk t k−m ,
l
q rm −rk −Jk +Jm t k−m , q rm −rk +Jm −Jk + jm +1 t k−m−1 ; q ∞
l
l
l
Q = q rm −rk −Jk +Jm − jk t k−m , q rm −rk +Jm −Jk + jm − jk +1 t k−m−1 ,
l
q rm −rk −Jk +Jm +1 t k−m−1 , q rm −rk +Jm −Jk + jm t k−m ; q ∞ .
Acknowlegdments
I would like to thank Professor Ernest L. Stitzinger and Professor Naihuan Jing for their
professional support and encouragement during the course of this work and for their helpful comments and suggestions regarding this manuscript. I am grateful to Prof. Jing for
introducing me to Macdonald polynomials and for his suggestion of this problem.
I would also like to thank my husband, Michael Williams, for his comments and suggestions concerning this text.
References
1. G. Gasper and M. Rahman, Basic Hypergeometric Series, Cambridge University Press, Cambridge.
1990.
2. N. Jing and T. Jozefiak, “A Formula for two-row macdonald functions,” Duke Math. 67(2) (1992), 377–385.
3. M. Lassalle, “Explication des polynomes de Jack et de Macdonald en loungueur trois,” C.R. Acad. Sci. Paris.
333(I) (2001), 505–508.
4. I.G. Macdonald, Symmetric Functions and Hall Polynomials, second edition, Oxford University Press, Oxford,
1995.