q-selberg INTEGRALS AND MACDONALD POLYNOMIALS

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1 q-selberg INTEGRALS AND MACDONALD POLYNOMIALS S OLE WARNAAR Dedicated to Richard Askey on the occasion of his 70th birthday Abstract Using the theory of Macdonald polynomials a number of q-integrals of Selberg type are proved Introduction and summary In [3] Richard Askey conjectured several q-integrals extending the famous Selberg integral The third of his now ex-conjectures reads ( [0 n = x α ( cx i q α+β+2(n k ; q i ( cx i ; q x 2k i Γ q (α + (i kγ q (β + (i kγ q (ik + Γ q (α + β + (n + i 2kΓ q (k + ( xj q k x i ; q d q x 2k ϑ( cq α+2(i k ; q ϑ( c; q with k a positive integer Re(α > 0 Re(β > 0 and 0 < q < Here (a; q m = m j=0 ( aqj and (a; q = j=0 ( aqj are q-shifted factorials (2 Γ q (z = ( q z (q; q (q z ; q is the q-gamma function (3 ϑ(z; q = (z; q (q/z; q is a theta function and for x = (x x n and measure d q x = d q x d q x n (4 f(x d q x = ( q n f(q k q kn q k+ +kn [0 n k k n= is a multiple q-integral To see that ( generalizes a special case of the Selberg integral [24] (5 [0] n x α i ( x i β = x i x j 2γ dx Γ(α + (i γγ(β + (i γγ(iγ + Γ(α + β + (n + i 2γΓ(γ Mathematics Subject Classification 33D05 33D52 33D60

2 2 S OLE WARNAAR true for Re(α > 0 Re(β > 0 and Re(γ > min{/n Re(α/(n Re(β/(n } assume that c > 0 and let q tend to one in ( using the formal limits lim q Γ q (z = Γ(z and lim (a; q /(aq z ; q = ( a z q lim q [0 n f(x d q x = [0 n f(x dx with dx = dx dx n After rescaling cx i x i this yields the integral [ Exercise 84; γ = k] (6 [0 n x α i ( + x i α+β+2(n k = (x i x j 2k dx Γ(α + (i kγ(β + (i kγ(ik + Γ(α + β + (n + i 2kΓ(k + for Re(α > 0 and Re(β > 0 By the variable change x i x i /( x i this becomes the Selberg integral (5 for γ = k Askey s conjecture ( was proved for k = (and up to a symmetrization of the integrand by Milne [22 Theorem 48] and by Aomoto [2] for general k Subsequently Kaneko [8] found a proof based on a Ψ summation for Macdonald polynomials We will show that Kaneko s Ψ sum together with a symmetry property of the Macdonald polynomials implies an integral evaluation more general than ( Theorem Let = ( n be a partition and k a positive integer Then P (x; q q k x α ( cx i q α+β+2(n k ; q ( i x 2k xj q k i ; q d q x ( cx [0 n i ; q x i 2k = P (q kδ ; q q k ; q q k(2n 3i+i ϑ( cqα+2(i k+i ϑ( c; q Γ q(α + (n ik + i Γ q (β + (i k i Γ q (ik + Γ q (α + β + (n + i 2kΓ q (k + = q αk(n 2+2k 2 ( n 3 P (q kδ ; q q k ϑ( cq α+(2n i k+i ; q ϑ( cq (n ik ; q for Re(α > n Re(β > and 0 < q < Γ q(α + (n ik + i Γ q (β + (i k i Γ q (ik + Γ q (α + β + (n + i 2kΓ q (k + Here P (x; q t is a Macdonald polynomial (see Section 2 and P (q kδ ; q q k denotes the specialized Macdonald polynomial P ( q k q (n k ; q q k which can be expressed in closed form as (7 P (q kδ ; q q k = q kn( (q (j ik+i j ; q k (q (j ik ; q k

3 q-selberg INTEGRALS AND MACDONALD POLYNOMIALS 3 with n( = n (i i The equality of the two expressions on the right of the theorem simply follows from the (multiplicative quasi-periodicity of the theta functions: ϑ(aq j ; q = ( a j q (j 2 ϑ(a; q Denoting the empty partition by 0 we have P 0 (x; q t = Hence the = 0 case of the theorem reduces to ( Also since (8 P (x; q t = (x x n n P ( n 2 n0(x; q t it is easily seen by making the variable changes α α n and β β + n that the n > 0 case of the theorem reduces to the n = 0 case We also mention that Theorem remains correct for k = 0 However here and elsewhere in the paper we exclude k = 0 for being trivial but requiring special attention For example for k = 0 we have P (x; q = m (x with m the monomial symmetric function given by m (x = x ν xνn n where the sum is over all distinct permutations ν of Hence (7 for k = 0 is incorrect unless = (a a = (a n However with the correct interpretation of P ( ; q 0 = m ( the integral of Theorem with k = 0 trivializes to a sum over the one-dimensional integral (with α α + µ i and β β µ i 0 x α ( cxqα+β ; q ( cx; q d q x = Γ q(αγ q (β ϑ( cq α ; q Γ q (α + β ϑ( c; q This result which is equivalent to Ramanujan s ψ summation [3 5] corresponds to ( with n = A more interesting special case of the theorem is obtained in the q limit for c > 0 By rescaling x to eliminate c this yields the following generalization of the integration formula (6 Corollary 2 With the same conditions as in Theorem there holds [0 n P (/k (x = P (/k ( x α i ( + x i α+β+2(n k Here P (α (x is a Jack polynomial and P (/k ( = (x i x j 2k dx Γ(α + (n ik + i Γ(β + (i k i Γ(ik + Γ(α + β + (n + i 2kΓ(k + ((j ik + i j k ((j ik k with (a n = n j=0 (a + j By replacing q β q α 2(n k /c and then letting c q β in Theorem and by noting that (x i q; q = 0 for x i = q ki with k i a negative integer we obtain as second corollary another generalized q-selberg integral This integral was first conjectured by Kadell [ Conjecture 8] and proved for k = by Kadell [] and for general k by Kaneko [6 Proposition 52] and Macdonald [20 Example VI93] See also Kadell [4 Theorem ] for the q limit (To obtain the formulae of [] and [20] one has to first carry out a symmetrization of the integrand

4 4 S OLE WARNAAR Corollary 3 Let be a partition and k a positive integer Then P (x; q q k x α (x i q; q ( i (x [0] n i q β x 2k xj q k i ; q d q x ; q x i 2k = q αk(n 2+2k 2 ( n 3 P (q kδ ; q q k Γ q (α + (n ik + i Γ q (β + (i kγ q (ik + Γ q (α + β + (2n i k + i Γ q (k + for Re(α > n Re(β > 0 and 0 < q < For = 0 this is the well-known Askey Habsieger Kadell integral conjectured by Askey [3 Conjecture ] and proved independently by Habsieger [9] and Kadell [2 Theorem 2; l = m = 0] In the above f(x d q x = ( q n f(q k q kn q k+ +kn [0] n k k n=0 The second main result of this paper entails a generalization of the β = k case of Corollary 3 Theorem 4 For = ( n and µ = (µ µ n partitions and k a positive integer there holds (9 [0] n P (x; q q k P µ (x; q q k x α i (x i q; q k x 2k i ( xj q k = q αk(n 2+2k 2 ( n 3 ( q kn 2 P (q kδ ; q q k P µ (q kδ ; q q k for Re(α > n µ n and 0 < q < Γ q (ikγ q (ik + Γ q (k + ij= x i ; q d q x 2k (q α+(2n i jk+i+µj ; q k In the limit when q tends to this reproduces the Hua Kadell integration formula for Jack polynomials [0 Theorem 52] [3 Theorem 2] Corollary 5 With the same conditions as in Theorem 4 there holds µ (x x α i ( x i k (x i x j 2k dx [0] n P where (/k (xp (/k f k = = n!(γ(k n f k f k µ ij= ((j ik + i j k (α + (2n i jk + i + µ j k As another corollary of Theorem 4 we will prove the following integral representation of a terminating n+ Φ n multivariable basic hypergeometric series based on Macdonald polynomials (see definition (26

5 q-selberg INTEGRALS AND MACDONALD POLYNOMIALS 5 Corollary 6 Let y = (y y m For = ( n a partition and k a positive integer there holds (0 P (x; q q k ( x i y j [0] n i n j m = q αk(n 2+2k 2 ( n 3 P (q kδ ; q q k for Re(α > n and 0 < q < x α i (x i q; q k x 2k i ( xj q k x i ; q d qx 2k Γ q (α + (n ik + i Γ q (ikγ q (ik + Γ q (α + (2n ik + i Γ q (k + [ q nk ] q α 2(n k q α (n k n n+ Φ n ; q k q y q α (2n k q α nk n The generic numerator and denominator parameters in the above n+ Φ n are q α (2n i k i and q α (2n ik i Equation (0 extends the β = k case of the following integral of Kaneko [7 Equation (3]: ( x i y j x α (x i q; q ( i (x i q β x 2k xj q k i ; q d q x ; q 2k [0] n i n j m = q αk(n 2+2k 2 ( n 3 n Γ q (α + (i kγ q (β + (i kγ q (ik + Γ q (α + β + (n + i 2kΓ q (k + x i [ q nk q α (n k ] 2 Φ q α β 2(n k ; qk q q k β y true for Re(α > n Re(β > 0 and 0 < q < Here q k β y is shorthand for (q k β y q k β y m When m = the above two r+ Φ r series simplify to the ordinary one-variable basic hypergeometric series r+ φ r We also note the remarkable fact that m-dimensional hypergeometric series are expressed in terms of n-dimensional q-integrals There is a well-known connection between constant term identities and q-selberg integrals of the type given in Corollary 3 and Theorem 4 In the case of Corollary 3 the corresponding constant term identity was obtained by Kaneko Correcting his expression for the exponent D n in [7 Equation (35] Kaneko s identity asserts that [7 Theorem 4] [ ( q CT P (x; q q k (x i ; q a ; q x i = ( (qk ; q k n ( q k n b ( xi qx j ; q x j x i k ] q ( i + 2 (q; qa+b+(i k (q; q a+(n ik+i (q; q b+(i k i (q (j ik+i j ; q k

6 6 S OLE WARNAAR Here a b 0 = n i (a a r ; q m = r j= (a j; q m /(q; q m = 0 for m a positive integer and CT[f(x] denotes the constant term of the Laurent polynomial f(x Kaneko s constant term identity has a rich history For = 0 it was conjectured by Morris [23 Conjecture (42] who proved his conjecture for q = [23 Theorem (43] q-proofs for = 0 were first obtained by Kadell [2 Theorem 3; l = m = 0] and Habsieger [9 p 487] based upon their Askey Kadell Habsieger integral evaluation For general but q = the above identity was first established by Kadell [4 Equation (A] (see also [3 Theorem 4] Taking Theorem 4 as starting point we will prove the following generalization of the b = k a case of Kaneko s constant term identity Theorem 7 Let = ( n and µ = (µ µ n be partitions and a and k integers such that a {0 k } Then [ ( q ( CT P (x; q q k P µ (x ; q q k xi (x i ; q a ; q qx ] j ; q x i k a x j x i k = ( + µ (qk ; q k n [ ] ( q k n q (µ i i 2 k a + i µ i (q (j ik+i j ; q k (q (j ik+µi µj ; q k (q +a k+(j ik+i µj ; q k (q a+(j ik+µi j ; q k Here f(x = f(x x n and [ ] r (q; q r a {0 r} = (q; q a (q; q r a a q 0 otherwise is a q-binomial coefficient The correct interpretation of the statement of the theorem is that the constant term is zero whenever the q-binomial coefficient on the right vanishes ie whenever a + i µ i {0 k } for some i { n} This convention makes the right-hand side well-defined for all partitions and µ as we shall now show The double product on the right has a zero factor in the denominator if ( a + (j ik + i µ j {0 k } for i j { n} i j Now assume that for given i and j ( holds By adding and subtracting j this implies (j ik + ( i j + (a + j µ j {0 k } For i < j we have (j ik + i j k so that a + j µ j must be negative If i > j we have (j ik + i j k so that a + j µ j must be greater or equal to k We thus conclude that zero factors in the denominator cannot occur when the q-binomial coefficient on the right is non-vanishing Hence the constant term is either zero or a well-defined rational function We further note that the theorem is invariant under the simultaneous replacements a k a and µ On the left side this requires the change t i q/t n i+ and the symmetry and homogeneity of the Macdonald polynomials see Section 2 In the proof of the theorem we will in fact show the stronger statement that the non-zero values of a follow from a = 0 By letting q tend to in Theorem 7 we obtain a constant term identity of Kadell [3 Theorem 5] q

7 q-selberg INTEGRALS AND MACDONALD POLYNOMIALS 7 Corollary 8 With the same conditions as in Theorem 7 there holds [ CT P (/k (tp µ (/k (t ( t i a( k a ( t k ( i t ] k j t i t j t i n ( k = n!( + µ a + i µ i ((j ik + i j k ((j ik + µ i µ j k ( + a k + (j ik + i µ j k ( a + (j ik + µ i j k In Section 9 of Chapter VI of his book [20] Macdonald defines a scalar product on L n the Q(q t-algebra of Laurent polynomials in x When t = q k this scalar product reads (2 f g = n! CT [f(xg(x ( xi x j x j x i ; q for f g L n A straightforward symmetrization (see Lemma 32 shows that Theorem 7 is equivalent to n [ ] (3 P h ak P µ = ( + µ q (µ i i 2 k a + i µ i for (4 h ak (x = (q (j ik+i j ; q k (q (j ik+µi µj ; q k (q +a k+(j ik+i µj ; q k (q a+(j ik+µi j ; q k ( q (x i ; q a ; q x i k a Following Kadell s treatment of the case q = [3] we will show that this can be viewed as a generalization of Macdonald s orthogonality relation and norm evaluation for Macdonald polynomials with respect to the scalar product (2 which is given by [20 Example VI9(d] (5 P P µ (q (j ik+i j ; q k = δ µ (q k+(j ik+i j ; q k Still following Kadell [3 Theorem 8] we will then show that Theorem 7 implies the following q-binomial theorem for Macdonald polynomials Corollary 9 Let µ = (µ µ n be a partition and k a positive integer Then (6 c k µ(qp (x; q q k = P µ (x; q q k (x i ; q k where (7 c k µ(q = ( + µ n [ ] q ( i µ i 2 k i µ i q q k ] (q (j ik+µi µj ; q k (q k+(j ik+i j ; q k (q (j ik+µi j ; q k (q k+(j ik+i µj ; q k

8 8 S OLE WARNAAR Here we follow the same convention as in Theorem 7 in that c k µ (q = 0 unless i µ i {0 k } for all i { n} A slightly more general result is given in Corollary 33 Finally using an identity for the structure constants of the Macdonald polynomials we generalize Corollary 9 from t = q k to all t Theorem 0 If µ = (µ µ n is a partition then (qt n P (x (qt j i ; q i µ j c (qt j i = t (n µ P µ(x ; q i µ j P µ (t δ ij= for q < and max{ x i } n < Here (a = (a; q t = t n( n (at i ; q i is a q-shifted factorial labelled by the partition and c = c (q t = t n( (qt n (qt j i ; q i j (qt j i ; q i j (x i q/t; q (x i ; q is a generalized hook polynomial Since the summand on the left vanishes when µ ie when the partition µ is not contained in the partition ( i µ i for all i the sum over may be replaced by a sum over subject to µ When µ = 0 the double product in the summand on the left becomes (q/t /(qt n and Theorem 0 reduces to a special case of the q-binomial theorem for Macdonald polynomials given by (27 In the next section we present some necessary background material on Macdonald polynomials and in Sections 3 36 we prove all of the claims made in the introduction Finally in Section 4 we conclude with an (ex-conjecture of Askey closely related to the Askey Habsieger Kadell integral 2 Preliminaries on Macdonald polynomials We review some of the theory of Macdonald polynomials needed for our subsequent proofs For more details we refer to [20 Chapter VI] Let = ( 2 be a partition ie 2 with finitely many i unequal to zero The length and weight of denoted by l( and are the number and sum of the non-zero i respectively We shall not employ the frequency or multiplicity notation for partitions except for partitions of rectangular shape in which case we abbreviate (N N with n repeated Ns as (N n For a partition and a an integer we write (2 ± a = ( ± a 2 ± a For two partitions and µ we define the usual addition as + µ = ( + µ 2 + µ 2 and for x = (x x 2 a sequence (possibly a partition and z a scalar we set zx = (zx zx 2 The dominance partial order on the set of partitions is defined as follows: µ if and only if = µ and + + i µ + + µ i for all i As usual we identify a partition with its diagram or Ferrers graph defined by the set of points in (i j Z 2 such that j i The conjugate of is the

9 q-selberg INTEGRALS AND MACDONALD POLYNOMIALS 9 partition obtained by reflecting the diagram of in the main diagonal If and µ are partitions then µ means that (the diagram of µ is contained in (the diagram of ie i µ i for all i Let s = (i j be a square in the diagram of Then a(s a (s l(s and l (s are the arm-length arm-colength leg-length and leg-colength of s defined by (22a (22b a(s = i j a (s = j l(s = j i l (s = i An often used statistic on partitions is n( = i (i i = i ( i 2 In the following we will almost exclusively work with partitions = ( n with l( n For such a partition the monomial symmetric function m (x in the variables x = (x x n is defined as (23 m (x = x α where the sum is over all distinct permutations α of and x α = x α xαn n The monomial symmetric functions form a Z-basis of Λ n the ring of symmetric polynomials in x x n with integer coefficients The r-th power sum p r is given by p r (x = m (r (x = n xn i and more generally the power-sum products are defined as p (x = p (x p n (x Following Macdonald we now introduce the scalar product on Λ n by p p µ = δ µ z n q i t i with z = i m i! i mi m i being the number of parts of equal to i Denote the ring of symmetric functions in n variables over the field F = Q(q t of rational functions in q and t by Λ nf Then the Macdonald polynomial P (x; q t is the unique symmetric polynomial in Λ nf such that [20 Equation (VI47]: (24 P = µ u µ m µ where u µ F u = and P P µ = 0 if µ The Macdonald polynomials form an F -basis of Λ nf For t = and t = q they reduce to the monomial symmetric functions and Schur functions; P (x; q = m (x and P (x; q q = s (x More generally lim P (x; t α t = P (α t (x with P (α Jack s symmetric function Below we have listed several known results for Macdonald polynomials that will be needed subsequently P is homogeneous of degree ie for z a scalar (25 P (zx = z P (x This is of course immediate from (23 and (24 Also [20 Equation (VI47] (26 P +a (x = (x x n a P (x

10 0 S OLE WARNAAR provided (for the time being that n +a is a nonnegative integer so that both sides are well-defined Recalling (22 we introduce two generalized hook polynomials as [20 Equation (VI8] (27a (27b c = c (q t = s ( q a(s t l(s+ c = c (q t = s ( q a(s+ t l(s and set (28 b = b (q t = c (q t c (q t The quadratic norm evaluation of the Macdonald polynomials can be very simply expressed in terms of this last function as [20 Equation (VI69] Using a generalized q-shifted factorial (29 (a = (a; q t = s we explicitly have [6 Proposition 32] (20a (20b c = t n( (t n c = t n( (qt n P P = b n (t l (s aq a (s = t n( (at i ; q i (t j i ; q i j (t j i+ ; q i j (qt j i ; q i j (qt j i ; q i j The b also features in the Cauchy identity for Macdonald polynomials Defining y = (y y n in addition to the usual x = (x x n this identity states that [20 Equation (VI43] (2 b P (xp (y = ij= (tx i y j ; q (x i y j ; q This may be alternatively put in terms of y = (y y m as [20 Equation (VI54] (22 P (x; q tp (y; t q = ( + x i y j m i n j m Next we need the evaluation homomorphism u : Λ nf F defined by (23 u (f = f(q t δ for f Λ nf Here q t δ = (q t δ q 2 t δ2 q n t δn with δ denoting the special partition δ = (n n 2 0 Two very important results that will be frequently used are the specialization [20 Example VI65] u 0 (P = P (t δ = P ( t t n = t n( s q a (s t n l (s q a(s t l(s+

11 q-selberg INTEGRALS AND MACDONALD POLYNOMIALS or by (27a (29 and (20a [20 Example VI65] (24a (24b P (t δ = (tn c = t n( and the symmetry [20 Equation (VI66] (t j i+ ; q i j (t j i ; q i j (25 u (P µ u 0 (P = u µ (P u 0 (P µ Finally we follow Kaneko [6] and Macdonald [2] in defining multivariable basic hypergeometric series and multivariable bilateral basic hypergeometric series with Macdonald polynomial argument Extending the condensed notation for q-shifted factorials to the multivariable setting by r r (a a r = (a a r ; q t = (a j ; q t = (a j we define [ ] a a r+ (26 r+φ r ; q t x b b r = j= j= (a a r+ (b b r c P (x Often we employ the one-line notation r+ Φ r (a a r+ ; b b r ; q t x for the above series For n = we have P (k (x = x k and c (k = (q; q k so that the r+ Φ r series simplifies to the standard one-variable basic hypergeometric series r+ φ r [8] An important result needed a number of times is the following multivariable version of the q-binomial theorem due to Kaneko [6 Theorem 35] and Macdonald [2 Equation (22]: (27 Φ 0 (a; ; q t x = (ax i ; q (x i ; q for q < and max{ x i } n < In the Jack polynomial case this was also found by Yan [26 Proposition 3] and in the Schur polynomial case (Jack with α = by Milne [22 Theorem 8] When a = q N with N a nonnegative integer the series on the left of (27 terminates leading to (28 Φ 0 (q N ; ; q t x = (x i q N ; q N We also need the generalization of r+ Φ r to bilateral series Let us denote the set of weakly decreasing integer sequences of length n by P The relation (8 (which follows from (26 by the substitutions ( n n a n can be used to define the Macdonald polynomial for all P By extending the definition of the q-shifted factorials to all integers n by (29 (a; q n = (a; q (aq n ; q we can also give meaning to (a of (29 and c of (20b for P Observe that /(q; q n = 0 for n a positive integer and hence that /(qt n = 0 if is not a partition However (qt n /c is perfectly well-defined for all P We

12 2 S OLE WARNAAR now use the above definitions to define the bilateral multiple basic hypergeometric series [ ] a a r (220 rψ r ; q t x = (qt n a a r b b r (b b r c P (x P Again we also employ the one-line notation r Ψ r (a a r ; b b r ; q t x We remark that we have chosen a slightly different (more symmetrical form for r Ψ r compared to earlier definitions of Milne for t = q [22] and of Kaneko [8] for general t In view of our earlier comments the series (220 simplifies to the unilateral series (26 with r r if b r = qt n For n = the series (220 reduces to the classical bilateral basic hypergeometric series r ψ r [8] Kaneko [8 Equation (] proved the following multiple series analogue of Ramanujan s celebrated ψ summation formula: (22 Ψ (a; b; q t x = (qt n i bt i /a ax i q/ax i ; q (bt i qt n i /a x i bt n /ax i ; q for bt n /a < x i < for all i { n} Kaneko s proof utilizes the q-binomial theorem (27 (obtained for b = qt n and Ismail s analytic argument A different proof based on a Gauss summation for Macdonald polynomials was subsequently found by Baker and Forrester [6 Theorem 6] We should also mention that the Schur case of the theorem corresponding to t = q was first obtained by Milne using a determinant evaluation [22 Theorem 4] A last comment concerning the Macdonald polynomials P for P is that the relations (25 (26 and (25 still hold In the case of (26 we may now in fact assume that n + a is any integer In the case of (25 P (zx = (z n x x n n P n (zx = z (x x n n P n (x = z P (x by (8 then (25 and then (8 For (26 the above claim is evident as both sides may be rewritten by (8 as (x x n n+a P n (x Finally in the case of (25 u (P µ u 0 (P = q µn t (n+µn(n 2 u (P µ µn u 0 (P n = q µn +n µ nnµn t (n+µn(n 2 u n (P µ µn u 0 (P n = q µn +n µ nnµn t (n+µn(n 2 uµ µn (P n u 0 (P µ µn = = u µ (P u 0 (P µ by (8 then (25 then (25 et cetera 3 Proofs 3 Proof of Theorem We take the Ψ sum (22 replace x zx with z a scalar and use (25 Then we apply the evaluation homomorphism u µ defined in (23 and utilize the symmetry (25 to eliminate u µ (P with the summation

13 q-selberg INTEGRALS AND MACDONALD POLYNOMIALS 3 index of the Ψ series Hence (qt n a z P (b c u (P µ u 0 (P u 0 (P µ = (qt n i bt i /a azq µi t n i q µi t i n /az; q (bt i qt n i /a zq µi t n i bq µi t i 2n+ /az; q To transform this into Theorem we need to specialize t = q k z = q α a = cq (n k b = cq α+β+3(n k Using (7 and (20b in the form (3 c (q q k = q kn( (q +(n k ; q q k and noting that (32 (a; q k (aq k ; q n = (a; q k+n yields (33 q 2kn( (q +(j i k ; q k (q +(j i k+i j ; q k u 0 (P c (q qk = (q +(n k ; q q k q (j ik+i j (q +(j i k+i j ; q 2k q (j ik (q +(j i k ; q 2k q 2kn( q ik (q; q k = (q +(n k ; q q k q k (q; q (i k (q; q ik ( q (j ik+i j (q +(j i k+i j ; q 2k where here and in the following P (x = P (x; q q k Employing the definitions (2 and (3 of the q-gamma function and theta function we thus obtain the identity (34 P = q α +2kn( u (P µ u 0 (P µ ( cq α+β+(3n i 2k+i ; q ( cq (n ik+i ; q ( q (j ik+i j (q +(j i k+i j ; q 2k q k Γ q (α + (n ik + µ i Γ q (β + (i k µ i Γ q (ik + q ik Γ q (α + β + (n + i 2kΓ q (k + ϑ( cqα+(2n i k+µi ; q ϑ( cq (n ik ; q The next step of the proof relies on the following lemma Let S n be the symmetric group with elements w given by the permutations of ( n S n acts on f(x as w(f(x = f(w(x = f(x w x wn Lemma 3 For k a positive integer and w S n let (35a h = h w(

14 4 S OLE WARNAAR and (35b h = 0 if i j {0 k } for all i < j n Then formally ( tn (36 h +kδ = (t; t n P n= h q i tq j q i q j Note that because of the symmetry (35a not all of the inequalities in (35b are independent For example (35b follows from the weaker h = 0 if i i+ {0 k } for all i { n } Before giving a proof let us first show how Lemma 3 can be applied to transform (34 into the statement of Theorem To this end we identify h +kδ with the summand on the left of (34 Hence by δ = ( ( n 2 n(δ = n 3 and n( = i<j j h = q αk(n 2 2k 2 ( n P 3+α µ (q P µ (q kδ ( cq α+β+2(n k+i ; q ( cq i ; q q (2k j (q j q i (q k+i j ; q 2k Now clearly the first line on the right is symmetric in To see that also the second line is symmetric note that by (37 (a; q n = (q n /a; q n ( a n q (n 2 both i<j (qj q i and i<j q(2k j (q k+i j ; q 2k are skew-symmetric functions (of making their product a symmetric function Next observe that thanks to the factor (q k+i j ; q 2k also the inequalities (35b hold putting us in a position to apply Lemma 3 Choosing t = q k in the lemma recalling definition (4 and using the above-discussed fact that the second line in the expression for h is symmetric in so that we may interchange i and j we obtain the second integral evaluation of Theorem with replaced by µ The conditions on α and β imposed by the theorem follow from the fact that taking into account all the variable changes we have effectively applied (27 with x i q α+(n ik+i and b/a q α+β+2(n k Since 0 < q < the inequality bt n /a < x i < (with t = q k therefore translates into Re(α + β + (n k > Re(α + (n ik + i > 0 Since is a partition this yields Re(α > n and Re(β > Proof of Lemma 3 We take the left-hand side of the identity of the lemma and replace kδ ie i i k(n i This leads to the sum P h where the prime denotes the added restriction i i+ k Since h = 0 for i i+ {0 k } we may drop the prime to simply get P h Now by the symmetry (35a and the fact that h = 0 for i = j we may further replace this by (/n! h n= Next we need the summation identity [20 Equation (III4] (38 w S n w ( x i tx j = x i x j t i t (This follows by viewing the symmetric function on the left as the ratio of two skew-symmetric polynomials the denominator polynomial being the Vandermonde

15 q-selberg INTEGRALS AND MACDONALD POLYNOMIALS 5 determinant Since both polynomials are of degree n and vanish for x i = x j the left must be independent of x and may be evaluated by taking x = t δ For this specialization the only non-vanishing contribution comes from the (unique permutation of maximum length leading to the desired right-side Hence n= h = = ( tn (t; t n ( tn (t; t n ( tn = n! (t; t n h w n= w S n w n= w S n n= h ( h ( q i tq j q i q j q i tq j q i q j q i tq j q i q j 32 Proof of Theorem 4 Let the structure constants fµ ν = f µ ν (q t be given by [20 Equation (VI7 ] (39 P (xp µ (x = ν f ν µp ν (x Substituting this in the left-hand side of (9 and performing the resulting integral by the β = k case of Corollary 3 we get (30 LHS(9 = q αk(n 2+2k 2 ( n 3 n ( q kn 2 Γ q (ikγ q (ik + Γ q (k + fµ(q ν q k P ν (q kδ ; q q k (q α+(n ik+νi ; q ν nk where we have used (3 Γ q (a + n Γ q (a = (qa ; q n ( q n To perform the sum over ν we need the following proposition Proposition 3 Let = ( n and µ = (µ µ n be partitions Then fµp ν ν (t δ (zt n ν = P (t δ P µ (t δ (zt i j ; q i+µ j (z ν (zt 2 i j ; q i+µ j ν ij= Note that for given and µ this result is a rational function identity Indeed by the homogeneity of the Macdonald polynomials fµ ν = 0 if ν = + µ so that only a finite number of terms contribute to the sum on the left In view of this it clearly suffices to prove the proposition for z < min{ t n+i } n (ie z < t n if t and z < t 2n if t This restriction on z in the proof given below arises from the use of the q-binomial theorem (27 Proof of Proposition 3 By (29 we first rewrite the above identity as (32 fµp ν ν (t δ (zt i q νi ; q (zt n i = P (t δ P µ (t δ (zt 2 i j q i+µj ; q q νi ; q (zt i j q i+µj ; q ν ij=

16 6 S OLE WARNAAR For the sake of compactness of subsequent equations we set z = yt 2n and then use the Φ 0 sum (27 to expand the product over i in the summand on the left Hence LHS(32 = f ν (t n η y η µ c u ν (P η u 0 (P ν νη η where we have used (25 and P ν (t δ = u 0 (P ν By the symmetry (25 this yields LHS(32 = νη fµ ν (t n η y η c u η (P ν u 0 (P η η The sum over ν can now be carried out by (39 leading to LHS(32 = η By two more applications of (25 this gives (t n η y η u η (P u η (P µ u 0 (P η c η LHS(32 = u 0 (P u 0 (P µ η (t n η y η c η u (P η u µ (P η u 0 (P η From (28 and (24a it thus follows that LHS(32 = u 0 (P u 0 (P µ η b η y η u (P η u µ (P η This nearly completes the proof All that is needed now is the homogeneity of the Macdonald polynomials and the Cauchy identity (2 By virtue of these (yt 2n+ i j q i+µj ; q LHS(32 = u 0 (P u 0 (P µ (yt 2n i j q i+µj ; q ij= Recalling that z = yq 2n this is in accordance with the right-hand side of (32 We remark here that a dual version of Proposition 3 without the free parameter z is given by Proposition 32 of Section 36 We now return to the proof of Theorem 4 Replacing z q α t 2n and then choosing t = q k the identity of Proposition 3 can be written as fµ(q ν q k P ν (q kδ ; q q k (q α+(n ik+νi ; q nk ν = P (q kδ ; q q k P µ (q kδ ; q q k ij= (q α+(2n i jk+i+µj ; q k Using this to perform the sum over ν in (30 results in the right-hand side of (9 In the above we have not yet addressed to condition Re(α > n µ n imposed on the theorem To see how this arises note that we first applied Corollary 3 with β = k and ν This step is justified if Re(α > ν n The next step was to perform the sum over ν in (30 According to [20 Equation (74] f ν µ = 0 unless ν and ν µ Hence we may without loss of generality assume that ν n µ n Recalling the notation (2 let µ and ν be the partitions µ = µ µ n and ν = ν µ n From (39 and (8 we have P P µ = ν f ν µ P ν Since also

17 q-selberg INTEGRALS AND MACDONALD POLYNOMIALS 7 P P µ = ν f µ ν P ν we conclude that fµ ν = f µ ν But f µ ν = 0 if ν Hence fµ ν = 0 if ν n < n + µ n This last result implies that by summing over ν in (30 the condition Re(α > ν n on the summand translates into Re(α > n µ n 33 Proof of Corollary 6 To obtain this from Theorem 4 is rather straightforward Multiplying boths sides of (9 by ( µ P µ (y; q k q and summing over µ such that µ = l(µ m we find by the use of the Cauchy identity (22 that LHS(0 = q αk(n 2+2k 2 ( n 3 ( q kn 2 P (q kδ ; q q k ( µ P µ (y; q k qp µ (q kδ ; q q k µ µ m Next we eliminate P µ (q kδ ; q q k by (24a use ij= ij= Γ q (ikγ q (ik + Γ q (k + (q α+(2n i jk+i+µj ; q k (q α+(2n i jk+i+µj ; q k Γ q (α + (n ik + i (q α+(2n i k+i ; q q k µ = ( q kn2 Γ q (α + (2n ik + i (q α+(2n ik+i ; q q k µ as follows from (33 (aq n ; q k = (a; q k(aq k ; q n (a; q n (29 and (3 and replace the summation index µ by µ As a result LHS(0 = q αk(n 2+2k 2 ( n 3 P (q kδ ; q q k Γ q (α + (n ik + i Γ q (ikγ q (ik + Γ q (α + (2n ik + i Γ q (k + ( µ (qnk ; q q k µ P µ (y; q k q (q α+(2n i k+i ; q q k µ µ c µ (q q k (q α+(2n ik+i ; q q k µ µ n with µ = (µ µ m Comparing this with the right-hand side of (0 it remains to show that the last line can be written as the n+ Φ n series claimed in the corollary For this we need (a; q t = ( (a ; t q and c (q t = c (t q both of which follow immediately from the definitions of the respective functions in terms of the arm-(colength and leg-(colength and the fact that under transposition arms become legs and and legs become arms The sum over µ may thus be written as µ µ n (q nk ; q k q µ P µ (y; q k q c µ(q k q (q α (2n i k i ; q k q µ (q α (2n ik i ; q k q µ By definition (26 this is seen to be in accordance with the n+ Φ n series on the right of (0

18 8 S OLE WARNAAR 34 Proof of Theorem 7 The proof presented below is a streamlined q-version of the proof of Corollary 8 given by Kadell [3] First we show that it suffices to proof the theorem for a = 0 or equivalently for a = k To this end we assume a > 0 and replace µ µ + a Next we twice apply (26 (with µ and use both (32 and (37 to write (x i ; q a (q/x i ; q k a = ( x i a q (a 2 (q a /x i ; q k Comparing the resulting constant term identity with Theorem 7 for a = 0 we conclude that the a > 0 cases follow from the a = 0 case if [ ( q CT P (x; q q k P µ (x ; q q k a ( xi ; q qx ] j ; q x i k x j x i k [ ( q ( = gµ(q a CT P (x; q q k P µ (x ; q q k xi ; q qx ] j ; q x i k x j x i k with g a µ (q = qa( µ This is readily established by the substitution x i x i q a on the left Although this changes the expression within the square brackets it does not alter the constant term The homogeneity (25 of the Macdonald polynomials now does the rest Below we present a proof for a = k By the comment following Theorem 7 this is equivalent to proving the case a = 0 Let us denote the a = k instance of the expression within the square brackets in Theorem 7 by S k µ (q For N a fixed integer such that µ N denote ˆµ the complement of µ with respect to (N n ie ˆµ i = N µ n i+ for i { n} Then by (34 P µ (x; q t = (x x n N Pˆµ (x ; q t (see eg [6] (37 and (32 Sµ(q k = ( k(n 2 q ( n 2( k 2 P (x; q q k Pˆµ (x; q q k x (n k N i (x i ; q k x 2k i ( xj q k x i ; q 2k Since we need the constant term of Sµ k (q we may replace x qx and use the homogeneity of the Macdonald polynomials and nn ˆµ = µ to get (35 CT [ Sµ(q k ] [ = ( k(n 2 q ( n 2( k 2+ µ CT P (x; q q k Pˆµ (x; q q k x (n k N i (x i q; q k x 2k i ( xj q k x i ] ; q 2k Our next step is to use the well-known and easy to prove fact (see eg [3 9 7] that if f(x is a Laurent polynomial in x then ( q ε n ( n CT[f(x] = lim x ε ε 0 i f(xd q x q [0] n

19 q-selberg INTEGRALS AND MACDONALD POLYNOMIALS 9 According to the above we may compute the right-hand side of (35 by the α ε N (n k and µ ˆµ instance of Theorem 4 (where the integral in (9 is analytically continued using its evaluation Replacing q ε by ω this yields CT [ S k µ(q ] = ( k(n 2 q ( n 2( k 2 k 2 ( n+ 3 + µ Nk( n 2 P (q kδ ; q q k Pˆµ (q kδ ; q q k lim ω ( ω n ij= (q; q ik (q; q ik (q; q k (ωq (n i j+k N+i+ˆµj ; q k where we have also used (2 to eliminate some q-gamma functions We now eliminate ˆµ in favour of µ To achieve this we replace j n + j in the double product on the right and use (24b to obtain Hence Pˆµ (q δ = t N(n 2 (n µ P µ (q δ (36 CT [ S k µ(q ] = ( k(n 2 q ( n 2( k 2 k 2 ( n+ 3 + µ k(n µ P (q kδ ; q q k P µ (q kδ ; q q k (q; q ik (q; q ik (q; q k lim ω ( ω n ij= (ωq (j ik+i µj ; q k It remains to deal with the limit of the double product For fixed i consider (37 (ωq (j ik+i µj ; q k j= It has a pole at ω = if there is a j { n} such that (j ik + i µ j { k 0} Clearly there can at most be one such j since ((j + ik + i µ j+ ((j ik + i µ j = k + µ j µ j+ k Hence (37 has at most one simple pole at ω = Since ( ω n has a zero of order n at ω = it follows that the CT [ S k µ (q] = 0 unless for each i { n} the product (37 has a pole Hence for the constant term to be non-zero there must for each i be a (unique j say j i such that (38 (j i ik + i µ ji { k 0} Therefore ((j i ik + i µ ji ((j i+ i k + i+ µ ji+ { k k } Since i i+ 0 this implies that (j i j i+ k + µ ji+ µ ji < 0 If j i j i+ then µ ji+ µ ji 0 so that (j i j i+ k + µ ji+ µ ji 0 Since this would lead to a contradiction we infer that j i < j i+ for all i which implies that j i = i Substituting this back into (38 we conclude that CT [ S k µ (q] = 0 unless (39 µ i i {0 k } for all i { n} This is in accordance with the comments made following Theorem 7

20 20 S OLE WARNAAR In the remainder we assume (39 to hold Then lim ω ( ωn = ij= i j (ωq (j ik+i µj ; q ij= k lim (q (j ik+i µj ; q k ω = ( + µ (q; q n k [ ] q ( i µ i 2 k µ i i q ω (ωq i µi ; q k ij= i j (q (j ik+i µj ; q k = ( + µ +k(n 2 q k 2 ( n+ 3 ( n 2( k 2 k{n(+n(µ (n µ } [ ] (q; q n q ( i µ i 2 k k µ i i q (q (j ik+i µj ; q k (q k+(j ik+µi j ; q k where we have used the easily proved lim ω a+ 2 ω (ωq a = ( a q ( ; q k (q; q k [ ] k and (37 Substituting the above in (36 and using (7 to eliminate the specialized Macdonald polynomials completes the proof of the a = k instance of Theorem 7 a q 35 Proof of Corollary 9 We first invoke the following simple lemma which is essentially the content of parts (a (c of [20 Example VI9] to show that Theorem 7 is equivalent to (3 Lemma 32 Let f(x be a symmetric Laurent polynomial Then [ CT f(x x j ; q x j x i ( xi k ] = n! ( qk n [ (q k ; q k CT f(x n ( xi qx j ; q x j x i Since P (xp µ (x h ak (x with h ak given by (4 is a symmetric Laurent polynomial this clearly establishes the equivalence of Theorem 7 and the scalar product evaluation of (3 Proof of Lemma 32 Suppressing their k-dependence let us denote the expressions within the square brackets on the left and right by L(x and R(x respectively Clearly L(x = w(l(x CT[R(x] = CT[w(R(x] and R(x = L(x x i q k x j x i x j k ]

21 q-selberg INTEGRALS AND MACDONALD POLYNOMIALS 2 Combining these facts gives CT[R(x] = CT[w(R(x] n! w S n = [ n! CT L(x ( w S n w = (q k ; q k n n! ( q k n CT[L(x] x i q k ] x j x i x j where in the final step we have used (38 Next we show that (3 implies (5 Since P (xp µ (x is homogeneous of degree µ i<j (x i/x j x j /x i ; q k is homogeneous of degree zero and h k k (x is a polynomial with constant term it follows that P P µ = P P µ δ µ = P h k k P µ δ µ From (3 we know that P h k k P µ = 0 unless µ i i {0 k } for all i { n} Combined with the fact that = µ this yields that = µ Hence P P µ = P h k k P µ δ µ Using (3 with = µ and a = k to evaluate the scalar product on the right leads to (5 Having shown that Theorem 7 implies (3 and (5 has prepared us to show that the same theorem also implies Corollary 9 Since the right-hand side of (6 is a symmetric polynomial in x it may be expanded in terms of the Macdonald polynomials as in the left-hand side The exercise is thus to determine the coefficients c k µ (q Using (2 we obtain from (6 the identity c k µ(q P P ν = P µ h k k P ν By the orthogonality (5 this yields c k µ(q = P µh k k P P P where we have replaced ν by Thanks to (3 and (5 both scalar products can be explicitly computed resulting in the expression for c k µ (q as claimed in (7 Somewhat more general but easily derivable from Corollary 9 is the following q-binomial theorem Corollary 33 Let µ = (µ µ n be a partition and a and k integers such that a {0 k } Then P c ak µ (qp (x; q q k = P µ (x; q q k ( q (x i ; q a ; q x i k a

22 22 S OLE WARNAAR where c ak µ (q = ( + µ [ ] q ( i µ i 2 k a + µ i i q (q (j ik+µi µj ; q k (q k+(j ik+i j ; q k (q +a k+(j ik+µi j ; q k (q a+(j ik+i µj ; q k Here we again adopt the convention of Theorem 7 that c ak µ (q = 0 unless a + µ i i {0 k } for all i { n} Proof Replace (k a and x q k a x Using (25 (26 and (37 it then follows that we reduce to the case a = k 36 Proof of Theorem 0 Key in our proof is the following sum over the structure constants of Macdonald polynomials Proposition 32 If = ( n and µ = (µ µ n are partitions then (320 fµν (q/t ν c = t ( n µ P µ (t δ (qtn (qt j i ; q i µ j ν c (qt j i ; q i µ j ν ij= Below we present two proofs of this result The first and longer proof explains the origin of the above identity The second and shorter proof uses Corollary 9 but obscures what is really going on First however we show how (320 leads to Theorem 0 Multiplying both sides of (320 by P (x and summing over yields (qt n P (x c ij= (qt j i ; q i µ j (qt j i ; q i µ j = t (n µ P µ(x P µ (t δ Φ 0 (q/t; ; q t x where we have used (39 and definition (26 Summing the Φ 0 series by the q-binomial theorem (27 completes the proof First proof of Proposition 32 The proof hinges on the simple identity (32 f ν µ = f ˆ µˆν P P P ν P ν where as before the partitions ˆ and ˆν are the complements of and ν with respect to (N n with ν and not exceeding N The identity (32 may seem to lack the necessary symmetry but we note (see also the proof below that P P = Pˆ Pˆ The following proof of (32 is correct for all t provided we take the more general definition of the scalar product (2 in terms of an n-dimensional integral as given in [20 Equation (VI90] However since (32 is a rational function identity it certainly suffices to establish its truth for just t = q k We leave it to the reader to either take t = q k in all of the remainder or to assume the generalized definition of the scalar product From the orthogonality of the Macdonald polynomials with respect to the scalar product it follows that f ν µ = P P µ P ν P ν P ν

23 q-selberg INTEGRALS AND MACDONALD POLYNOMIALS 23 Let f(x = f(x x n and e n (x = x x n Then by f = f e n ē n = and (34 f ˆ µˆν = P µpˆν Pˆ Pˆ Pˆ = en n P µ Pν e N P n e N P n e N P = P P µ P ν n P P = fµ ν P ν P ν P P We now take z = t n q N in the identity of Proposition 3 and make the substitution (32 Also using [20 Equation (VI96] then gives P P P ν P ν = (qtn ν c P (t δ (qt n c ν P ν (t δ ν f ˆ µˆν (qt n q N ν (q N t n ν c ν = P µ (t δ (qtn c ij= (q N t n i j+ ; q i+µ j (q N t n i j+2 ; q i+µ j Next we replace the summation index ν by ˆν and eliminate ˆν and in favour of ν and ˆ using [6 Equation (4] and [6 Equation (45] (aˆ = ( q/a t n( q n( N (a (N n (q N t n /a Therefore c ˆ = ( t n( N(n 2 q n( N (qt n (N n (qt n q N c ˆν f ˆ µν (q/t ν c ν = t ( n µ P µ (t δ (qtn ˆ c ˆ ij= (qt j i ; qˆi µ j (qt j i ; qˆi µ j where we have also used that f ˆ µν = 0 unless µ + ν = ˆ and (a; q n k = (a; q n (q n /b; q ( k b k (b; q n k (b; q n (q n /b; q k a Next note that summing over the partition ˆν is equivalent to summing over ν Indeed (N n ˆν implies that also (N n ν But since f ˆ µν = 0 if ν > ˆ = N we may simply sum over all partitions ν = (ν ν n Finally renaming ˆ as completes the proof Second proof of Proposition 32 Since f µν = 0 unless = µ + µ only a finite number of terms contribute to the sum on the left Hence for fixed and µ equation (320 is a rational function identity It thus suffices to give a proof for sufficiently many values of t and in the following we will show that (320 is true for t = q k with k a positive integer

24 24 S OLE WARNAAR We take (6 replace x by q k x and then expand the product on the right using (28 Hence RHS(6 x xq k = P µ (xq k ; q q k Φ 0 (q k ; ; q q k x = q ( k µ P µ (x; q q k ν (q k ; q q k ν c ν(q q k P ν (x; q q k = q ( k µ ν fµν(q q k (q k ; q q k ν c ν(q q k P (x; q q k Equating coefficients of P (x; q q k with the left-hand side of (6 (with x xq k yields ν f µν(q q k (q k ; q q k ν c ν(q q k = q (k ( µ c k µ(q By (7 (3 (37 and (33 this is in accordance with equation (320 for t = q k 4 Back to Askey It seems appropriate to end this paper with another one of Askey s conjectures By examining a special limit of the Askey Habsieger Kadell integral (Conjecture 3 with = 0 Askey was led to conjecture the sum [3 Conjecture 7] (4 N n=0 ( ( i + a N i + b i = N i ( k + j i 2k (a (i k (b (i k (a + b N+(i k (ik! (a + b (n+i 2k (N (i k!k! This is contained in Askey s last q-integral [3 Conjecture 8] proved by Evans [7] Kadell [5] and Tarasov and Varchenko [25 Theorem E8] Although we are at present unable to use the Macdonald machinery to deal with Askey s last conjecture we will show that the multivariable basic hypergeometric series studied in this paper can certainly deal with (4 Let us begin by introducing the series [ Φ a a r+ r+ r b b r = ] ; q z n=0 (a a r+ ; q i z i (q b b r ; q i q 2ki (q k+j i ; q 2k for k a nonnegative integer and n a positive integer When k = 0 this of course trivializes to the n-th power of an ordinary r+ φ r series We now claim the following n-dimensional version of the q-pfaff Saalschütz summation

25 q-selberg INTEGRALS AND MACDONALD POLYNOMIALS 25 Theorem 4 For N a nonnegative integer there holds [ Φ a b q N ] 3 2 c abq N+2(n k /c ; q q = q 2k2 ( n 3+k( n 2 n ( (q; qik (a b q N ; q (n ik (q; q k (c abq N+2(n k /c; q (n ik (cq( ik /a cq ( ik /b; q N (n k (cq (n ik cq (2 n ik /ab; q N (n k Note that for N < (n k the right-hand side vanishes as it should Indeed because the summand on the left is zero for i j {0 k} or max{ i } n > N there are no non-zero contributions to the sum when N < (n k To see how the theorem relates to (4 we let b tend to zero and then replace a q a and c q N b By the usual manipulations involving q-shifted factorials this leads to (42 N n=0 [ ] i + a q bi i q = q 2k2 ( n 3+bk( n 2 n [ ] N i + b N i q q 2ki (q k+j i ; q 2k (q a q b ; q (i k (q a+b ; q N+(i k (q; q ik (q a+b ; q (n+i 2k (q; q N (i k (q; q k which is the obvious q-analogue of (4 For k = (42 is equivalent to [9 Theorem 6] of Krattenthaler Proof of Theorem 4 The theorem is really nothing but a rewriting of the Pfaff Saalschütz sum for Macdonald polynomials due to Baker and Forrester [6 Equation 47] [ a b q N ] 3Φ 2 c abt n q N ; q t qtδ = (c/a b/a (N n /c (c c/ab (N n Taking t = q k and using (33 this can be written as (a b q N ; q q k q 2kn(+ (q +(n k c abq N+(n k /c; q q k ( q (j ik+i j (q +(j i k+i j ; q 2k Multiplying both sides by = (c/a b/a; q qk (N n (c b/ab; q q k (N n q k q ik (q; q (i k (q; q ik (q; q k (43 (aq ( nk bq ( nk q N+( nk ; q (n ik (q cq ( nk abq N /c; q (n ik

26 26 S OLE WARNAAR and using (29 and (32 this becomes (aq ( nk bq ( nk q N+( nk ; q i+(n ik q i (q cq ( nk abq N /c; q i+(n ik q 2kj ( q (j ik+i j (q +(j i k+i j ; q 2k = (c/a b/a; q qk (N n (c c/ab; q q k (N n q k (q; q ik (aq ( nk bq ( nk q N+( nk ; q (n ik q ik (q; q k (cq ( nk abq N /c; q (n ik Because (43 is zero for N { (n k } the above sum is true for all integers N such that N +(n k 0 Next we replace a aq (n k b bq (n k c cq (n k and N N (n k This yields (44 (a b q N ; q i+(n ik q i (q c abq N+2(n k /c; q i+(n ik q 2ki ( q (j ik+i j (q +(j i k+i j ; q 2k = ( q k (q; q ik (a b q N ; q (n ik q ik (q; q k (c abq N+2(n k /c; q (n ik where N should now be a nonnegative integer Defining h = q 2k2 ( n 3 k( n 2 n (a b q N ; q i q i (cq( ik /a bq ( ik /a; q N (n k (cq (n ik cq (2 n ik /ab; q N (n k (q c abq N+2(n k /c; q i q 2ki ( q i j (q k+i j ; q 2k the left-hand side of (44 is nothing but h +kδ Since h satisfies (35a and (35b we may utilize the unilateral version of Lemma 3 obtained by replacing the sum on the left of (36 by and the sum on the right of (36 by n=0 Provided we take t = q k in the unilateral form of (36 and use the symmetry of h we obtain Theorem 4 Acknowledgements I thank Hjalmar Rosengren for helpful discussions References G E Andrews R Askey and R Roy Special functions Encyclopedia of Mathematics and its Applications Vol 7 (Cambridge University Press Cambridge K Aomoto Connection formulas of the q-analog de Rham cohomology in Functional Analysis on the Eve of the 2st Century Vol pp 2 S Gindikin et al eds Prog in Math 3 (Birkhauser Boston MA R Askey Some basic hypergeometric extensions of integrals of Selberg and Andrews SIAM J Math Anal (

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