AFormula for N-Row Macdonald Polynomials

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1 Journa of Agebraic Combinatorics, 21, , 25 c 25 Springer Science + Business Media, Inc. Manufactured in The Netherands. AFormua for N-Row Macdonad Poynomias ELLISON-ANNE WILLIAMS North Caroina State University kingpuck2@yahoo.com Received February 13, 23; Revised October 3, 23; Accepted October 11, 24 Abstract. We derive a formua for the n-row Macdonad poynomias with the coefficients presented both combinatoricay and in terms of very-we-poised hypergeometric series. Keywords: Macdonad poynomias, symmetric functions, hypergeometric series 1. Introduction Denote the ring of symmetric functions over the fied F as F and et n F denote its nth graded space. The space n F consists of a symmetric functions of tota degree n Z, indexed by the partitions λ λ 1,...,λ k for which i λ i n. There are four Z-bases and one Q-basis of n. The Q-basis consists of the power sum symmetric functions p n i x i n, where p λ p λ1 p λk, and the four Z-bases are: The monomia symmetric functions m λ i 1 < <i k x λ 1 i 1...x λ k i k, the eementary symmetric functions e n i 1 < <i n x i1...x in, where e λ e λ1 e λk, the compete symmetric functions h λ i 1... i k x λ 1 i 1 the Schur functions s λ x 1,...,x k detx λ j +k j i 1 i, j k /detx k j i....x λ k i k, and Let H Q, t bethe fied of rationa functions in and t. In1988, Macdonad introduced a new cass of two-parameter symmetric functions P λ, t, over the ring H, which generaize severa casses of symmetric functions. In particuar, taking t we obtain the Schur functions, setting t 1wehave the monomia symmetric functions, and etting gives the Ha-Littewood functions. We know from [4] that the P λ are a basis of n H. Further, with respect to the scaar product: p λ, p µ δ λ,µ i λ i m i λ j m i! 1 t λ j j1 we have that P λ, P µ if λ µ,

2 112 WILLIAMS where m i denotes the mutipicity of i in λ and λ denote the ength of λ. Weaso know that for each λ, there exists a uniue P λ, t such that: P λ m λ + µ<λ c λµ m µ where c λµ Q, t. Define: Q λ P λ P λ, P λ. Then, the bases P λ and Q λ of n H are dua to each other, Q λ, P µ δ λ,µ, and from [4], for λ n: Q n 1 λ 1 t λ j i λ n i m i mi! λ p j λ. j1 For partitions of the form λ λ 1,...,λ k, we wi derive a formua for the Macdonad poynomias Q λ with coefficients presented first combinatoricay in Section 3 and then as very-we-poised hypergeometric series in section 5. A formua for Q λ1,λ 2 can be found in [2] and foowing the competion of this work, the author discovered the paper [3] which gives a formua, without proof, for Q λ1,λ 2,λ Preiminaries Let λ, µ be partitions such that µ λ; the diagram of µ being contained in the diagram of λ. For compactness, we wi denote the skew diagram λ µ as λ\µ. Moreover, we wi view the diagram of a partition with row one being the argest and row n begin the smaest Engish notation. The diagram of λ\µ is said to be a horizonta r-strip if the number of bocks contained in λ\µ euas r and, of the remaining r bocks, there is at most one in each coumn of λ\µ. For each bock d found in the diagram of λ, et: ed t sd+1 if d λ b λ d ed+1 t sd 1 if d / λ, 1 where ed denotes the number of bocks to the east of d and sd denotes the number of bocks to the south of d in the diagram of λ [4]. Let R λ\µ denote the union of the rows and et C λ\µ denote the union of the coumns which intersect the diagram of λ\µ.

3 N-ROW MACDONALD POLYNOMIALS 113 Using a Pieri-type formua [4], for λ, µ such that µ λ is a horizonta r-strip, we have: Q µ Q r λ d b µ d b λ d Q λ 2 for d R λ\µ C λ\µ. Let r r 1,...,r n and r r 1,...,r n 1 bepartitions with the ength of row i eua to r i.inorder to derive a formua for Q r1,...,r n,webegin with Q r1,...,r n 1. To utiize 2, we must obtain a partitions ω w 1,...,w n such that r ω and ω\r is a horizonta r n -strip. Denote this set as. 3. Construction of Ω Begin with the desired partition r r 1,...,r n 1. For a ω,ω\r is a horizonta r n -strip and therefore we must add exacty r n bocks to r r 1,...,r n 1. To this end, we strategicay enarge the ength of the rows, in some cases creating an nth row, by adding the appropriate number of bocks to row r i, and beow row r n 1, for a i. Let ji denote the number of bocks added to row r i.wemay add a maximum of r n bocks to row r 1 and thus: j 1,...,r n. Conseuenty, the number of bocks added to a other rows of r is first dependent on j 1. Second, beginning with row r 1 and adding bocks to r in order of row succession, we see that the number of bocks which can be added to row r i is depedent upon the number of bocks added to rows r k for 1 k < i. Finay, since we may add at most one bock anywhere in r in order to create any new coumn which appears in ω, the number of bocks added to row r m is dependent on the difference between its ength and the ength of the preceeding row, r m 1 r m. Assimiating this, for m {1,...,n 1}: m 1,..., r n jm j s,...,r m 1 r m if if m 1 r n r n Lasty, for row w n ω, where w n r n, set: n 1 w n r n j m. js m 1 js r m 1 r m > r m 1 r m. Taking a possibe combinations of j m,beginning with m 1 and ending with m n 1, we generate {ω w 1,...,w n }.

4 114 WILLIAMS Exampe 3.1 Let r 5, 4, 3 and r 4 2. Then, j1 {, 1, 2}, j 2 {, 1}, and j3 {, 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 λ 1,...,λ n, define: v i {p where p max{1,...,n} such that λ p i}. Proposition 4.1 Let λ λ 1,...,λ n be a partition with row i, λ i.foreach row λ m and each bock d k λ m, 1 k λ m, numbered eft to right, sd k v k m. Proof: Let λ denote the conjugate partition of λ, the partition whose diagram is the transpose of the diagram of λ. Let d k be the kth bock of row λ m. Then, d k is the mth bock of the kth row of λ. Since row λ m of λ is eua to coumn m of λ, it foows that the number of bocks in row λ m is eua to p where: p max{1,...,n} such that λ p m. This impies that the number of bocks to the east of d k is the diagram of λ is eua to p m.taking the transpose, it foows that the number of bocks south of d k in the diagram of λ euas p m. Remark 4.1 As demonstrated by the proof of Proposition 4.1, for λ λ 1,...,λ n,v i λ i.aswesha see, the desirabiity of using v i rather than the conjugate partition is evidenced by the considerabe faciity that it gives to the impimentation of the principa formua. For exampe, when λ 6, 4, 2, 2 and i 3, we may simpy observe from λ that v i 2. Proposition 4.2 Let λ λ 1,...,λ n be a partition with row i, λ i.foreach row λ m, et {d k }, 1 k λ m, denote the set of bocks which compose λ m, numbered eft to right. Then: b λm b λ {d k } λ m 1 i λ m i+1 t v i+1 m 1 λ m i t v i+1 m Proof: For the product imits {i,...,λ m 1}, et bock k, d k, for 1 k λ m, correspond to i k 1. Then, {i,...,λ m 1} corresponds directy to {d k }, where 1 k λ m. It is easiy seen that the number of bocks east of bock d k in the diagram of λ is eua to λ m k. By Proposition 4.1, we know that the number of bocks south of d k in the diagram

5 N-ROW MACDONALD POLYNOMIALS 115 of λ is eua to v k m. Therefore: λ m k t v k m 1 b λ d k λ m k+1 t. v k m Shifting bock b k to bock b i where i k 1, we have: λ m i+1 t v i+1 m 1 b λ d k b λ d i λ m i t. v i+1 m Taking the product over {i,...,λ m 1} to encompass {d k }, 1 k λ m, yieds the desired resut. Let ω. Inorder to utiize the Pieri-type formua, we must construct the reuired coefficent for each Q ω : d b r d b ω d d R ω\r C ω\r. For m {1,...,n 1}, wecompute these coefficients according to row, r m, and according to the number of bocks added to r m, j m, where ω m r m + j m. Proposition 4.3 Let ω and r r 1,...,r n 1. For row w m r m + j m, 1 m n 1, such that d w m and d R ω\r C ω\r, d b rm d b wm d T j m, T j m, j k, m < k n 1 3 where T j m, r m 1 ir n n 1 j s r m i+1 t v i+1 m 1 r m+ j m i t v i+1 m r m i t v i+1 m r m+ j m i+1 t v i+1 m 1 j m 1 j m 4 and r k + j k 1 rm i t k m 1 rm+ jm i+1 t k m j T j m, jk, ir k r m i+1 t k m r m+ jm i t k m 1 m and j k 1 otherwise 5 restricting v i+1 to r ; v i+1 {p where p max{1,...n 1} such that r p i + 1}.

6 116 WILLIAMS Proof: First, note that within this proof, we use the expressions b r m and b w m to identify incompete stages in the deveopment of b rm and b wm. For row r m,weincude ony the d r m for which d R ω\r C ω\r. Since row w n r n n 1, it foows that, for each row of ω and hence of r, the bocks {d 1,...,d rn n 1 } are not incuded in R ω\r C ω\r. Therefore, by Proposition 4.2: b r m r m 1 i r n n 1 j m r m i+1 t v i+1 m 1 r m i t v i+1 m. 6 Simiary, for row w m r m + j m : b w m r m 1 i r n n 1 j m r m + j m i+1 t v i+1 m 1 r m + j m i t v i+1 m 7 where we restrict v i+1 to r in order to excude any bocks d / r which ie beneath row r m. If j m, we do not incude bocks from r m in 3, and thus from 6 and 7: b r m T b w j m, m r m 1 r m i+1 t v i+1 m 1 r m+ jm i t v i+1 m ir n n 1 r m i t v i+1 m r m+ j j s m i+1 t j v i+1 m 1 m 1 jm. However, we do not incude any d w m such that d C ω\r ; therefore, we must get rid of any terms corresponding to these bocks in the uotient 4. These bocks, which generate the unwanted terms, directy correspond to the jk, for m + 1 k n 1, which were added beow them on row r k to create row w k.inorder for these terms to appear in 4, we must have jm and j k. Therefore, to correct this in order to incude ony the d w m such that d R ω\r C ω\r,wemutipy 4 by: r k + jk 1 r m i t k m 1 r m+ jm i+1 t k m T j m, jk, ir k r m i+1 t k m r m+ jm i t k m 1 and j k 1 otherwise, yeiding the desired resut.

7 N-ROW MACDONALD POLYNOMIALS Construction of the principa formua Let r r 1,...,r n beapartition. For {1,...,r n 1}, et j1 r,..., n 1 n. js c c And, for m {2,...,n 1}, et n 1,..., r n c n 1 1 m 1 js c m 1 if r n js c js r m 1 r m + jm c c,..., r m 1 r m + j c m 1 j c m c n 1 m 1 if r n js c js > r m 1 r m + c j s c j c m 1 jm c j c m 1 jm c. Remark 5.1 The purpose of the ji, 1 i n 1, is to aow us to systematicay reduce row w n r n n 1 i1 j i tozero. To this end, we first must compute the ji in order to generate. Then, restricting the ji such that n 1 i1 j i, < n 1 i1 j i r n n 1 1 c j s c, we are abe to achieve the desired resut. Further, restrict the ji such that n 1 i1 j i, excuding the origina partition r r 1,...,r n from this reduction. We must now add at east one bock to the partition r, and to any subseuenty created partitions i.e., ω ; thus, the maximum number of times that we may need to repeat this reduction process is r n, yieding r n n 1 rn 1 i1 ji. Exampe 5.1 Buiding upon Exampe 3.1, for the partition ω 6, 4, 3, 1, j1 1, j2, and j 3, we desire to reduce row w 4 to zero. Using the restriction given in Remark 5.1, we have j1 1 {, 1}, j 2 1 {, 1}, j 3 1 {, 1} where < Taking a possibe combinations of the jm 1 dictated by the j 1 1, j 2, and j 3, 1 m 3, and appying them to ω, yieds the set of three-row partitions {7, 4, 3, 6, 5, 3, 6, 4, 4}. Given jm, 1 m n 1, 4 becomes: r m + 1 c c 1 T j m, i r n n 1 c j s c r m + 1 c c i+1 t v i+1 m 1 r m+ c c i t v i+1 m r m + 1 c c i t v i+1 m r m+ c c i+1 t v i+1 m 1 j m 1 jm 8

8 118 WILLIAMS and, 5 becomes: r k + c j k c 1 i r k + 1 c T j m, jk, j k c r m + 1 c c i t k m 1 r m+ c c i+1 t k m r m + 1 c c i+1 t k m r m+ c c i t k m 1 j m and j k 1 otherwise 9 where for, 1 c c and where v i+1 is restricted to the partition r c j 1 c,...,r n c j n 1 c ; v i+1 {p where p max{1,...,n 1} such that r p + 1 c j p c i + 1}. Theorem 5.1 For the partition r r 1,...,r n, m {1,...,n 1}, and {1,..., r n 1},wehave Q r1,...,r n Q r1,...,r n 1 Q rn + r n 1 i 1 n 1 1 i jm, j i1 m Q r 1 + i 1 j 1,...,r n 1+ i 1 j n 1 + j m, j m n 1 1 jrn m 1 r n r n 1 T j m, Q n 1 Q r 1 + rn 1 j 1,...,r n 1+ rn 1 j n 1 n 1 r n n 1 T j m, km+1 i 1 j m T j m, j k, n 1 km+1 T j m, j k, where jm, signifies to sum over a possibe combinations of the vaues of j j m 1 and such that < n 1 m2 j 1 + r n and a possibe combinations of the vaues jm dictated by the jm such that n 1 j m. Proof: We want to show that: Q r1,...,r n Q r1,...,r n 1 Q rn r n 1 i 1 n 1 1 i j1, i1 Q r 1 + i 1 j 1,...,r n 1+ i 1 j n 1 T j m, Q n 1 km+1 r n n 1 i 1 j m T j m, j k,

9 N-ROW MACDONALD POLYNOMIALS 119 j 1, n 1 1 jrn m 1 r n r n 1 n 1 Q r 1 + rn 1 j 1,...,r n 1+ rn 1 j n 1 Using 2, for n 1 j m, we have: T j m, n 1 km+1 Q r1,...,r n 1 Q rn Q r1,...,r n + T j m, j1, Q r 1 + j1,...,r n 1+ jn 1,r n n 1 T j m, j k,. 1 n 1 Substituting 11 into 1, we need ony to show: j 1, j m n 1 T j m, n 1 km+1 r n 1 i 1 1 i + j1, i1 Q r 1 + i 1 j 1,...,r n 1+ i 1 j n 1 n 1 j 1, 1 jrn m 1 r n 1 n 1 km+1 T j k, j k, T j k, jk, Q r 1 + j1,...,r n 1+ jn 1,r n n 1 n 1 r n 1 Q r 1 + rn 1 j 1,...,r n 1+ rn 1 j n 1 T j m, Q n 1 n 1 r n n 1 T j m, km+1 i 1 j m. 11 T j m, j k, n 1 km+1 Note that each Pieri-formua expansion via 2 and 11 of j 1, j m n 1 T j m, n 1 km+1 T j m, j k,. 12 T j k, jk, Q r 1 + j1,...,r n 1+ jn 1,r n n 1 13 yieds two terms. To show our desired resut, we wi show that with each successive numbered Pieri-formua expansion of 13, the term which contains the product of two Macdonad poynomias cances with the correspondingy numbered term i in r n 1 i 1 1 i j1, i1 n 1 T j m, Q r 1 + i 1 j 1,...,r n 1+ i 1 j n 1 n 1 Q km+1 r n n 1 T j m, j k, i 1 14 j m

10 12 WILLIAMS and that, in the fina Pieri expansion of 13, we wi be eft ony with r n 1 1 r n 1 j 1, j m n 1 1 jrn m n 1 Q r 1 + rn 1 j 1,...,r n 1+ rn 1 j n 1 T j m, n 1 km+1 T j m, j k,. 15 To this end, we wi induct on i. Consder i 1. Competing the first Pieri expansion of 13, and computing i 1 from 14, 12 becomes: j 1, j m n 1 n 1 T j m, km+1 1 n 1 T jm, j1, Q r j 1, j m n 1 T j k, j k, Q r 1 + j1,...,r n 1+ j Q n 1 r n n 1 n 1 km+1 T j m, j k, j 1,...,r n 1+ 1 j n 1,r n n 1 1 j m T j m, r n 1 i 1 1 i + j1, i2 Q r 1 + i 1 1 n 1 j1, n 1 km+1 n 1 j 1,...,r n 1+ i 1 j n 1 T j m, T j m, T j m, j k, Q n 1 km+1 n 1 km+1 r n n 1 i 1 j m T j m, j k, Q r j 1,...,r n 1+ 1 j n 1,r n n 1 1 j m + r n 1 i 1 n 1 n 1 1 i T j m, j1, i2 Q r 1 + i 1 j 1,...,r n 1+ i 1 j n 1 Q Q r 1 + j1,...,r n 1+ j Q n 1 r n n 1 km+1 T j m, j k, r n n 1 i 1 j m T j m, j k, Therefore, in the first Pieri expansion of 13, the term containing the product of two Macdonad poynomias canceed with the correspondingy numbered i 1 in14, as desired. Assume that with each further numbered Pieri expansion of 13, up to r n 2, the term containing a product of two Macdonad poynomias cances with the correspondingy numbered i in 14. We wi show the resut for r n 1..

11 N-ROW MACDONALD POLYNOMIALS 121 For i r n 2, we have that 12 euas: j 1, j m 1 r n 3 r n 3 Q r 1 + rn r n 2 j1, Q r 1 + rn r n 2 j1, Q r 1 + rn r n 1 j1, Q r 1 + rn 2 1 r n 2 j1, Q r 1 + rn r n 1 j1, n 1 T j m, j1,...,r n 1+ rn 3 jn 1 r n 2 n 1 n 1 km+1 Q T j m, T j m, j k, r n n 1 n 1 km+1 j1,...,r n 1+ rn 2 jn 1,r n n 1 rn 2 jm r n 3 n 1 T j m, j1,...,r n 1+ rn 3 jn 1 r n 2 n 1 Q T j m, j1,...,r n 1+ rn 2 jn 1 r n 2 n 1 T j m, Q n 1 km+1 r n n 1 n 1 km+1 r n n 1 n 1 km+1 j1,...,r n 1+ rn 2 jn 1,r n n 1 rn 2 jm r n 2 n 1 T j m, Q r 1 + rn 2 j1,...,r n 1+ rn 2 jn 1 Q r n n 1 rn 2 j. m n 1 km+1 rn 3 jm T j m, j k, T j m, j k, rn 3 j m T j m, j k, rn 2 j m T j m, j k, T j m, j k, Performing the Pieri expansion on 13 a fina time, 12 is eua to: r n 2 1 r n 2 j1, Q r 1 + rn r n 1 j1, n 1 T j m, j1,...,r n 1+ rn 2 jn 1 r n 1 n 1 n 1 km+1 Q T j m, Q r 1 + rn 1 j 1,...,r n 1+ rn 1 j n 1 T j m, j k, r n n 1 n 1 km+1 rn 2 jm T j m, j k,

12 122 WILLIAMS + 1 r n 1 j1, Q r 1 + rn 2 j 1, n 1 1 jrn m r n 2 n 1 T j m, j1,...,r n 1+ rn 2 jn 1 1 r n 1 r n 1 Q n 1 Q r 1 + rn 1 j 1,...,r n 1+ rn 1 j n 1 n 1 km+1 r n n 1 T j m,. T j m, j k, rn 2 jm n 1 km+1 T j m, j k, Since, by definition of jm,wemust have n 1 r n 1 r n jm. Exampe 5.2 We sha buid upon Exampe 3.1 to cacuate the expansion of the Macdonad poynomia Q 5,4,3,2. Given the nature of the partition r 5, 4, 3, 2, it is fitting and instructive to use the foowing notation: n 1 n 1 km+1 T j m, T j 1,..., j n 1 T jm, T j k,,..., jm+1,..., j n 1 jm, j m+i, 1 i n m 1 For 2 m 3, we have: j1 {, 1, 2} j 1 {,..., 2 m 1,..., 2 jm j s,...,r m 1 r m 3 js c c m 1 if 2 if 2 js m 1 js }, r m 1 r m > r m 1 r m,

13 N-ROW MACDONALD POLYNOMIALS 123 jm,..., if 2 2,..., if 3 c 1 m 1 js c 3 m 1 js c c js 1 j s r m 1 r m + j c m 1 jm c 2 c 3 m 1 js c c j s r m 1 r m + c c j c m 1 jm c > r m 1 r m + j c m 1 jm c. We begin with jm since these vaues determine the subseuent choices for j m.wehave: j1 {, 1, 2} j 1 1 {, 1} j2 {, 1} j 2 1 {, 1} j3 {, 1} j 3 1 {, 1}. Taking a possibe combinations of the jm, {, 1}, such that < 3 2 and < 3 1 1, we have: Q 5,4,3,2 Q 5,4,3 Q T j m, T j m, jk, j1, km+1 Q 5+ j Q 1,4+ j 2,3+ j T j m, T j m, jk, Q 5+ 1 j1, km+1 j 1,4+ 1 j 2,3+ 1 j 3 Q 5,4,3 Q 2 + T1,, T 1,, 1 T 2,, Q7,4,3 + T1,, T,1, 1 + T,1, T 1,, 1 T 1,1, T,1, Q6,5,3 + T 1,, T 1,,1 + T,,1 T 1 1,, T 1,,1 T,,1 Q6,4,4 + T,1, T 1,,1 + T,,1 T 1,1, T,1,1 T,,1 Q5,5,4 T 1,, Q 6,4,3 Q 1 T,1, Q 5,5,3 Q 1 T,,1 Q 5,4,4 Q Very-we-poised hypergeometric series We may express the coefficients T j m, for m {1,...,n 1}, and {,...,r n 1 }, in terms of very-we-poised hypergeometric series. We wi use the foowing notations. a; 1 n 1 a; n 1 a i i

14 124 WILLIAMS a; 1 a i i a; n a; a n ; a 1,...,a m ; n a 1 ; n a m ; n a 1,...,a m ; a 1 ; a m ; The basic hypergeometric series φ r+1,r is of the form: [ ] a1, a 2,..., a r+1 b 1, b 2,..., b r, z where if φ r+1,r n a 1,...,a r+1 ; n b 1,...,b r ; n ; n z n. We say that φ r+1,r is very-we-poised, denoted W r+1,r a 1 ; a 4,...,a r+1 ;, z, a 1 a 2 b 1... a r+1 b r and a 2 a 1 2 1, a 3 a From [1], we have: W 6,5 a; b, c, d;, a bcd a, a bc, a bd, a a b, a c, a d, a bcd ; cd ;. 16 The Coefficients T j m, and T j m,j k, We first compute the coefficient T j m in terms of very-we-poised hypergeometric series. Beginning with row r m,weidentify the indentions which ie under it in the diagram of r.inother words, we identify a rows r i, m < i n 1, such that r i < r i+1. For k 1 {1,...,n m}, et r n k1 be the argest indexed row of r such that r n k1 r m. Proposition 6.1 Suppose r n k1 r m.itfoows that k 1 n m and r m r n 1. Then: W 6,5 r m r n + n 1 j s t n m 1 ; t 1,, r n 1 r n + n 1 j s ;, jm t n m jm T j m, 1 jm.

15 N-ROW MACDONALD POLYNOMIALS 125 Proof: Let r n k1 r m ; k 1 n m and r m r n 1. There are two cases: and jm. For jm, we have that T jm, 1byProposition 4.3. For jm, using properties of hypergeometric series and 16, 4 becomes: T j m, r n 1 1 ir n n 1 j s rm i+1 t n m rm+ j m i t n m 1 rm i t n m 1 rm+ j m i+1 t n m t n m, rm rn 1+ j rm rn 1 m +1 t n m 1 ; rn 1 r n 1 n+ j s rm rn 1+1 t n m 1, rm rn 1+ t n m ; rn 1 r n 1 n+ j s rm rn 1 t n m, rm+ rn 1+1 t n m 1, rm rn+ n 1 j s +1 t n m 1, rm+ rn+ n 1 j s t n m ; rm rn+ n 1 j s t n m, rm+ rn+ n 1 j s +1 t n m 1, rm rn 1+1 t n m 1, rm+ rn 1 t n m ; W 6,5 rm rn+ n 1 j s t n m 1 ; t 1,, r n 1 n 1 r n+ j s ;, r m r n 1+ jm t n m W 6,5 rm rn+ n 1 j s t n m 1 ; t 1,, r n 1 n 1 r n+ j s ;, jm t n m. Remark 6.1 Let r m r n k1.itfoows that for a m < k n 1, we have j k, and thus T j m, j k, 1. Consider the case when r n k1 r m. Let r n k2 be the argest indexed row of r r 1,...,r n 1 such that r n k1 r n k2.ifr n k2 r m, et r n k3 be the argest indexed row of r for which r n k2 r n k3, ect. Continue this process unti one reaches row r n k f for which r n k f r m, creating the chain: r n k1 < r n k2 < < r n k f r m for k i {1,...,n m}, 1 i f. Proposition 6.2 Suppose r n k1 r m. Then, 4 becomes: T j m, W 6,5 r m r n + n 1 j s t n m k 1 ; t 1, j m, r n k1 r n + n 1 j s ;, r m r n k1 + j m t n m k 1 +1 f 1 h2 W 6,5 r m r n kh+1 + j m t n k h+1 m ; t 1,, r n kh r n kh+1 ;, r m r n kh t n m k h+1 +1 jm 1 jm. Proof: Using our chain, we are abe to identify the indentions under row r m in the diagram of r ; thus, we obtain a intervas of bocks d r m on which sd changes. Using properties of hypergeometric series and 16, we decompose 4 as foows:

16 126 WILLIAMS T j m, rn k 1 1 irn n 1 f 1 h2 j s rn k h+1 1 irn k h rm i+1 t n m k1+1 rm + j m i t n m k1 rm i t n m k1 rm + j m i+1 t n m k1+1 rm i+1 t n m kh+1+1 rm + i t n m kh+1 rm i t n m kh+1 rm + jm i+1 t n m kh+1+1 rm rn k 1 t n m k1+1, rm + rn k 1 +1 t n m k1 ; rn k 1 rn+ n 1 j s rm + jm rn k 1 t n m k1+1, rm rn k 1 +1 t n m k1 ; rn k1 rn+ n 1 j s f 1 h2 rm rn k h+1 t n m kh+1+1, rm + rn k h+1 +1 t n m kh+1 ; rm rn k h+1 +1 t n m kh+1, rm rn k h+1 t n m kh+1+1 ; rn k h+1 rn k h rn k h+1 rn k h rm rn k 1 t n m k1+1, rm + rn k 1 +1 t n m k1, rm rn+ + n 1 j s t n m k1+1, rm rn+ n 1 j s +1 t n m k1 ; rm + rn k 1 t n m k1+1, rm + rn k 1 t n m k1, rm rn+ n 1 j s t n m k1+1, rm rn+ n 1 j s + +1 t n m k1 ; f 1 h2 P Q W 6,5 rm rn+ n 1 j s t n m k1 ; t 1, j m, rn k 1 rn+ n 1 j s ;, rm + j m rn k 1 t n m k1+1 f 1 W 6,5 rm rn k h+1 + j m t n kh+1 m ; t 1,, rn k h rn k h+1 ;, rm rn k h t n m kh+1+1. h2 Where: P r m r n kh+1 t n m k h+1 +1, r m+ jm r n k +1 h+1 t n m k h+1, r m r n kh +1 t n m k h+1, r m r n kh + jm t n m k h+1 +1 ; Q r m r n kh+1 +1 t n m k h+1, r m+ jm r n k h+1 t n m k h+1 +1, r m r n kh t n m k h+1 +1, r m r n kh + jm +1 t n m k h+1 ;. Proposition 6.3 For r n k1 r m, the coefficient T j m, j k, becomes: T j m, j k, W 6,5 r m r k j k t k m 1 ; j k, t 1, j m ;, r m r k + j m t k m. Proof: Using properties of hypergeometric series and 16, we have: T j m, j k, r k + j k 1 rm i t k m 1 r m + jm i+1 t k m ir k r m i+1 t k m r m+ jm i t k m 1 r m r k jk 1 t k m 1, r m r k + jm j k t k m ; jk r m r k jk t k m, r m r k + jm j k +1 t k m 1 ; jk

17 N-ROW MACDONALD POLYNOMIALS 127 r m r k j k +1 t k m 1, r m r k + j m j k t k m, r m r k t k m, r m r k + j m +1 t k m 1 ; r m r k j k t k m, r m r k + j m j k +1 t k m 1, r m r k +1 t k m 1, r m r k + j m t k m ; W 6,5 r m r k j k t k m 1 ; j k, t 1, j m ;, r m r k + j m t k m. The Coefficients T j m, and T j m,j k,. For {1,...,r n 1}, wecompute the coefficients T j m, and T j m, j k, in terms of verywe-poised hypergeometric series. To this end, fix and carry out the procedure previousy described to obtain the chain: r n k1 + jn k c 1 < < r n k f + jn k c f r m + c Notation 6.1 Set: c c j c m. c n 1 c jn k c i c j c m J m js c J s J n ki Proposition 6.4 Suppose that r n k1 + J n k1 r m + J m.itfoows that r m + J m r n 1 + J n 1 and 8 becomes: W 6,5 r m r n +J m +J s t n m k 1 ; t 1, j m, r n k1 +J n k1 r n J s ;, r m r n k1 J n k1 +J m + jm t n m k 1 W 6,5 r m r n ki J n ki +J m t n k i m+1 ; T j m, h2 t 1, j m, r n ki+1 r n ki J n ki +J n ki+1 ;, r m r n ki+1 J n ki+1 +J m + jm t n m k i jm 1 jm. Proof: Let r n k1 + J n k1 r m + J m r m + J m r n 1 + J n 1. There are two cases: jm and j m. For jm, by Proposition 4.3, we have that T jm, 1. f 1

18 128 WILLIAMS For jm, using properties of hypergeometric series, we have: r n 1 +J n 1 1 r m +J m i+1 t n m r m+j m + jm i t n m 1 T j m, r m +J m i t n m 1 r m+j m + jm i+1 t n m ir n J s r m +J m r n 1 J n 1 t n m, r m+j m + jm r n 1 J n 1 +1 t n m 1 ; r n 1+J n 1 r n +J s r m +J m J n 1 r n 1 +1 t n m 1, r m r n 1 J n 1 +J m + jm t n m ; r n 1 r n +J s +J n 1 P Q Where: W 6,5 r m r n +J s +J m t n m 1 ; t 1, j m, r n 1 r n +J s +J n 1 ;, r m r n 1 +J m J n 1 + jm t n m. P r m r n 1 +J m J n 1 t n m, r m r n 1 +J m J n 1 + jm +1 t n m 1, r m r n +J s +J m +1 t n m 1, r m r n +J s +J m + jm t n m ; Q r m r n +J s +J m t n m, r m r n +J s +J m + jm +1 t n m 1, r m r n 1 +J m J n 1 t n m+1, r m r n 1 +J m J n 1 + jm t n m ;. Remark 6.2 Since r m + J m r n 1 + J n 1 jk have that T j m, jk, 1. for a m < k n 1, we Proposition 6.5 Suppose that r n k1 + J n k1 r m + J m. Obtaining the chain rn k1 + J n k1 < < rn k f + J n k f rm + J m, we have that: W 6,5 r m r n +J s +J m t n m k 1 ; t 1, j m, r n k1 r n +J n k1 +J s ;, r m r n k1 J n k1 +J m + jm t n m k 1 +1 W 6,5 r m r n kh+1 +J m + jm J n k h+1 h2 T j m, t n k h+1 m ; t 1, j m, r n kh r n kh+1 +J n kh J n kh+1 ;, r m r n kh J n kh +J m t n m k h+1 jm 1 jm. Proof: For r n k1 + J n k1 r m + J m, using properties of hypergeometric series, 8 becomes: r n k1 +J n k1 1 T j m, ir n J s f 1 rn kh+1 +J n kh+1 1 h2 f 1 r m +J m i+1 t n m k 1+1 r m+j m + j m i t n m k 1 r m +J m i t n m k 1 r m +J m + j m i+1 t n m k 1+1 ir n kh +J n kh r m +J m i+1 t n m k h+1+1 r m+j m + jm i t h+1 n m k r m +J m i t n m k h+1 r m +J m + jm i+1 t n m k h+1+1

19 N-ROW MACDONALD POLYNOMIALS 129 r m r n k1 +J m J n k1 t n m k1+1, r m r n k1 +J m J n k1 + jm +1 t n m k 1 ; r n k1 r n+j s +J n k1 r m r n k1 +J m J n k1 + jm t n m k 1 +1, r m r n k1 +J m J n k1 t n m k 1 ; f 1 P 1 Q h2 1 P f 1 2 Q 2 h2 P 3 Q 3 W 6,5 r m r n +J m +J s t n m k 1 ; t 1, j m, r n k1 +J n k1 r n +J s ;, r m r n k1 J n k1 +J m + jm t n m k Where: f 1 i2 W 6,5 r m r n kh+1 +J m + j m J n k h+1 t n k h+1 m ; r n k1 r n +J n k1 +J s t 1, j m, r n kh r n kh+1 +J n kh J n kh+1 ;, r m r n kh J n kh +J m t n m k h+1. P 1 r m r n kh+1 +J m J n kh+1 t n m k h+1 +1, r m r n kh+1 +J m J n kh+1 + j m +1 t n m k h+1 ; r n kh+1 r n kh +J n kh+1 J n kh Q 1 r m r n kh+1 +J m J n kh+1 +1 t n m k h+1, r m r n kh+1 +J m J n kh+1 + j m t n m k h+1+1 ; r n kh+1 +J n kh+1 r n kh J n kh P 2 r m+j m r n k1 J n k1 t n m k 1 +1, r m+j m + jm r n k 1 J n k1 +1 t n m k 1, r m r n +J s +J m + jm t n m k 1 +1, r m r n +J s +J m +1 t n m k 1 ; Q 2 r m r n k1 +J m J n k1 + jm t n m k 1 +1, r m r n k1 +J m J n k1 + jm t n m k 1, r m r n +J s +J m t n m k1+1, r m r s +J n +J m + jm +1 t n m k 1 ; P 3 r m r n kh+1 +J m J n kh+1 t n m k h+1 +1, r m r n kh+1 +J m J n kh+1 + jm +1 t n m k h+1, r m r n kh +J n kh +J m +1 t n m k h+1, r m r n kh J n kh +J m + jm t n m k h+1 +1 ; Q 3 r m r n kh+1 J n kh+1 +J m +1 t n m k h+1, r m+j m + jm r n k h+1 J n kh+1 t n m k h+1 +1, r m r n kh +J m J n kh t n m k h+1 +1, r m r n kh J n kh +J m + jm +1 t n m k h+1 ;. Proposition 6.6 For T j m, jk,, m < k n 1, we have: W 6,5 r m r k J k +J m jk t k m 1 ; T j m, jk, j k, t 1, j m ;, r m r k +J m +J k + jm t k m jm and j k 1 otherwise.

20 13 WILLIAMS Proof: Using properties of hypergeometric series, 9 becomes: T j m, j k, Where: r k +J k + jk 1 rm+jm i t k m 1 r m +J m + jm i+1 t k m ir k +J k r m +J m i+1 t k m r m+j m + jm i t k m 1 r m r k +J m J k jk +1 t k m 1, r m r k +J m J k + jm j k t k m ; jk r m r k J k +J m jk t k m, r m r k +J m J k + jm j k +1 t k m 1 ; jk P Q W 6,5 r m r k J k +J m jk t k m 1 ; j k, t 1, j m ;, r m r k +J m +J k + jm t k m. P r m r k J k +J m jk +1 t k m 1, r m r k +J m J k + jm j k t k m, r m r k J k +J m t k m, r m r k +J m J k + jm +1 t k m 1 ; Q r m r k J k +J m jk t k m, r m r k +J m J k + jm j k +1 t k m 1, r m r k J k +J m +1 t k m 1, r m r k +J m J k + jm t k m ;. Acknowegdments I woud ike to thank Professor Ernest L. Stitzinger and Professor Naihuan Jing for their professiona support and encouragement during the course of this work and for their hepfu comments and suggestions regarding this manuscript. I am gratefu to Prof. Jing for introducing me to Macdonad poynomias and for his suggestion of this probem. I woud aso ike to thank my husband, Michae Wiiams, for his comments and suggestions concerning this text. References 1. G. Gasper and M. Rahman, Basic Hypergeometric Series, Cambridge University Press, Cambridge N. Jing and T. Jozefiak, A Formua for two-row macdonad functions, Duke Math , M. Lassae, Expication des poynomes de Jack et de Macdonad en oungueur trois, C.R. Acad. Sci. Paris. 333I 21, I.G. Macdonad, Symmetric Functions and Ha Poynomias, second edition, Oxford University Press, Oxford, 1995.