SYMMETRIC FUNCTIONS AND MACDONALD POLYNOMIALS ...

SYMMETRIC FUNCTIONS AND MACDONALDPOLYNOMIALSROBIN LANGERSubmitted in total fulfilment of the requirements of the degree ofMaster of Mathematics by Research at the University of MelbourneDate: September 15 2008.1

2 ROBIN LANGERAbstract. The ring of symmetric functions Λ, with natural basisgiven by the Schur functions, arise in many different areas of mathematics.For example, as the cohomology ring of the grassmanian,and as the representation ring of the symmetric group. One maydefine a coproduct on Λ by the plethystic addition on alphabets. Inthis way the ring of symmetric functions becomes a Hopf algebra.The Littlewood–Richardson numbers may be viewed as the structureconstants for the co-product in the Schur basis. The first partof this thesis, inspired by the umbral calculus of Gian-Carlo Rota,is a study of the co-algebra maps of Λ. The Macdonald polynomialsare a somewhat mysterious qt-deformation of the Schur functions.The second part of this thesis contains a proof a generating functionidentity for the Macdonald polynomials which was originallyconjectured by Kawanaka.

SYMMETRIC FUNCTIONS AND MACDONALD POLYNOMIALS 3Declaration. The first part of this thesis is entirely the author’s ownwork. The idea for the second part was suggested by Ole Warnaarwho also carried out the derivation of equation 15 on pages 59–60 andverified proposition 2.1. The proof of lemma 2.1 was suggested by PaulZinn-Justin. Steps two and three in the proof of the second part of thethesis are the author’s own work.

4 ROBIN LANGERAcknowledgements. The author would like to thank Ole Warnaar forsuggesting the subject of Macdonald polynomials, Peter Forrester, AlainLascoux, Paul Zinn-Justin and the two anonymous referees for usefulcomments and suggestions on the manuscript as well as the Universityof Montreal for hosting an interesting lecture series and workshop onMacdonald Polynomials and Combinatorial Hopf Algebras.

SYMMETRIC FUNCTIONS AND MACDONALD POLYNOMIALS 5for Mrs Mclaren

6 ROBIN LANGERContents1. Symmetric functions of Littlewood–Richardson type 71.1. Symmetric Functions 91.1.1. Partitions 91.1.2. Monomial symmetric functions 151.1.3. Plethystic notation 161.1.4. Schur functions 171.2. The Umbral Calculus 181.2.1. Coalgebras 181.2.2. Sequences of Binomial Type 221.3. The Hall inner-product 261.3.1. Preliminaries 261.3.2. Column operators 281.3.3. Duality 301.4. Littlewood–Richardson Bases 331.4.1. Generalized complete symmetric functions 341.4.2. Umbral operators 361.4.3. Column operators 371.4.4. Generalized elementary symmetric functions 401.5. Examples 412. A generating function identity for Macdonald polynomials 472.1. Macdonald Polynomials 472.1.1. Notation 472.1.2. Operator definition 482.1.3. Characterization using the inner product 492.1.4. Arms and legs 492.1.5. Duality 502.1.6. Kawanaka conjecture 502.2. Resultants 522.2.1. Residue calculations 532.3. Pieri formula and recurrence 562.3.1. Arms and legs again 572.3.2. Pieri formula 582.3.3. Recurrence 592.4. The Proof 592.4.1. The Schur case 592.4.2. Step one 622.4.3. Step two 652.4.4. Step three 70References 75

SYMMETRIC FUNCTIONS AND MACDONALD POLYNOMIALS 71. Symmetric functions of Littlewood–Richardson typeThe Schur polynomials {s λ (X)} arise in both geometry and representationtheory as a natural basis for the ring of symmetric polynomialswhich we denote by Λ. The Littlewood–Richardson numbers {c λ µν} aredefined by:s µ (X)s ν (X) = ∑ c λ µνs λ (X)λThe plethystic addition of alphabets gives a coproduct structure onΛ, which induces, via the Hall inner product, the natural product structureon the dual space Λ ∗ - which may be thought of as the ring ofsymmetric “formal power series” in some dual alphabet Y .Since the Schur polynomials are self-dual with respect to the Hallinner product, the Littlewood–Richardson numbers may also be definedby:s λ (X + X ′ ) = ∑ c λ µνs µ (X)s ν (X ′ )µ,νA sequence of binomial type is a basis p n (x) for the one variablepolynomial ring Q(x) with the property that:n∑( np n (x + x ′ ) = p k (x)p n−k (xk)′ )k=0Sequences of binomial type arise as images, under co-algebra maps, ofthe standard basis {x n }. Furthermore, each such co-algebra map, orumbral operator is associated to an a formal power series f(z) which isinvertible with respect to function composition.The homomorphism from the one variable polynomial ring Q[x] tothe ring of symmetric functions Λ given by:x nn! ↦→ h n(X)preserves the coproduct structure. Similarly, the homomorphism fromthe ring of formal power series in one variable Q[[y]] to Λ ∗ given by:y n ↦→ m (n) (Y )preserves the product structure.In the special case where the alphabet X contains a single variable,both these maps are in fact isomorphisms, and the Hall inner productbecomes the bracket used in the umbral calculus of Gian-Carlo Rota:〈f(y),p(x)〉 = Lf(D)[p(x)]Here L denotes the constant term operator, while D is the differentialoperator.

8 ROBIN LANGERThe Schur functions may be expressed as either a determinant in thecomplete symmetric functions:s λ (X) = det(h λi +j−i(X))or as a determinant in the elementary symmetric functions:s λ (X) = det(e λ ′i+j−i(X))The main result of the first part of this thesis is the following. Let{p n (x)} be the sequence of binomial type arising as the image of {x n }under the umbral operator U f which is associated to the invertibleformal power series f(z). Let {r n (X)} be the image of { pn(x) } undern!the embedding of Q[x] into Λ just described. Then the vector spacebasis for Λ defined by:P λ (X) = det(r λi +j−i(X))has the Littlewood–Richardson property:P λ (X + X ′ ) = ∑ µ,νc λ µνP µ (X)P ν (X ′ )Furthermore, suppose that g(z) is the compositional inverse of f(z)and that {q n (x)} is the sequence of binomial type arising as the imageof the standard basis {x n } under the umbral operator U g . Let {˜r n (X)}denote the image of { qn(x) } under the embedding of Q[x] into Λ andn!let:c n (X) = ω[˜r n (X)]where ω is the natural involution on Λ which maps the complete symmetricfunctions to the elementary symmetric functions.The aforementioned basis {P λ (X)} has an alternate expression ofthe form:P λ (X) = det(c λi +j−i(X))Every basis {P λ (X)} which has the Littlewood–Richardson property,and which also admits an expansion of the form:P λ (X) = ∑ µ⊆λ□ λµ s µ (X)arises in this way.In section 1.1 we review some classical results from the theory ofsymmetric functions. In section 1.2 we review classical results fromthe umbral calculus, in section 1.3 we prove the main theorem and insection 1.4 we give some examples.

SYMMETRIC FUNCTIONS AND MACDONALD POLYNOMIALS 91.1. Symmetric Functions. The main references for this section areMacdonald [29] and Lascoux [24]. Another good reference is Stanley[42]. Also, see Bergeron’s online notes (in French) [3]. For connectionswith the geometry of grassmanians see Fulton [9] and Manivel [30].For connections with the representation theory of the symmetric andgeneral linear groups see Fulton [9] and Sagan [41]. For more aboutλ-rings see Knutson [22]. For more about the q-exponential functionsand q-binomial coefficients see the book by Kac and Cheung [20] orthe online notes by Foata and Han [8]. The standard reference forbasic hypergeometric series is Gasper and Rahmen [15]. For an interestingapproach to obtaining basic hypergeometric series identities byspecialization of symmetric functions see Bowman [4]1.1.1. Partitions. For our purposes, a composition is just a list of nonnegativeintegers, finitely many of which are greater than zero. Formost purposes we will identify lists of different lengths which differonly by a tail of zeros.A partition is a composition whose parts are weakly decreasing. Ifthe sum of the parts of λ is n then we say that λ is a partition of nand write λ ⊢ n.Some authors do not allow partitions to contain zero parts, howeverthere are some circumstances in which it is convenient to envision thepartition as sitting inside some rectangle of pre-determined dimensions,in which case the tail of zeros becomes significant.The set of all partitions of n with k parts (any number of which maybe zero) is denoted by Pk n . The generating series for partitions is thefirst q-exponential function:E q (z) =1(z;q) ∞= ∑ n,k|P n k |q n z k = ∑ kz k(q;q) kHere we are making use of the hypergeometric notation:n−1∏(a;q) n = (1 − aq k )k=0It is common practice to represent a partition pictorially as rowsof boxes. The conjugate of a partition is obtained by reflecting inthe main diagonal. The prime symbol is used to indicate conjugation.Conjugating a partition interchanges the role of rows and columns. Theparts of λ ′ correspond to the columns of λ.

10 ROBIN LANGERBelow, on the left, is the diagram for the partition λ = (4, 2, 1) andon the right, its conjugate λ ′ = (3, 2, 1, 1)We are using the English convention.We define the length of a partition, denoted by l(λ), to be the numberof nonzero parts. With this convention, the length of a partition isalways less than or equal to its number of parts. For example, λ =(3, 2, 2, 1, 0, 0) is a partition of 8 of length 4 with 6 parts.There are two natural additions on the set of partitions. The firstone corresponds to the concatenation of the individual parts:λ + µ = (λ 1 + µ 1 ,λ 2 + µ 2 ,...,λ n + µ n )The second corresponds to taking the union of the two sets of parts,and then re-ordering them as appropriate. For example:One may check that:(5, 3, 1) ∪ (7, 3, 2, 2) = (7, 5, 3, 3, 2, 2, 1)λ ′ + µ ′ = (λ ∪ µ) ′Of particular interest are partitions that contain no repeated parts,including no repeated zero parts. The smallest such partition with kdistinct parts is the staircase partition:δ k = (k − 1,k − 2,...,2, 1, 0)Observe that δ k is a partition of k(k − 1)/2. If λ is a partition of ninto k parts then λ + δ k is a partition of n + k(k − 1)/2 into k distinctparts. Conversely, every partition of n+k(k−1)/2 into k distinct partsis equal to λ + δ k for some λ ⊢ n.We denote by Dk n the set of all partitions of n + k(k − 1)/2 into kdistinct parts (one of which may be zero). The generating series forpartitions with distinct parts is the second q-exponential function:e q (z) = (−z;q) ∞ = ∑ n,k|D n k |q n z k = ∑ kq (k 2) z k(q;q) kNote that the conjugate of a partition with distinct parts will not,in general, be another partition with distinct parts. In fact, the only

SYMMETRIC FUNCTIONS AND MACDONALD POLYNOMIALS 11partition with distinct parts whose conjugate also has distinct parts isthe staircase partition δ k .It is perhaps worth pointing out here, though we won’t need it untilmuch later, that if π q is the function which maps q to 1/q then:and so:π q [(q;q) n ] = (−1) n q − n(n+1)2 (q;q) nπ q [E q (z)] = e q (−qz)If the diagram for a partition µ sits properly inside the diagramfor the partition λ, that is, if each part of µ is less than or equal tothe corresponding part of λ then we write µ ⊆ λ. For such a pair ofpartitions, we may form the skew partition λ/µ which is simply thecollection of boxes in λ which are not also boxes of µ.A skew partiton λ/µ is said to be a horizontal strip if:λ 1 ≥ µ 1 ≥ λ 2 ≥ ... ≥ λ n ≥ µ nThat is, no column contains more than a single box.For an arbitrary partition µ we let U(µ) denote the set of partitionsλ such that λ/µ is a horizontal strip, and let D(µ) denote the set ofpartitions ν such that µ/ν is a horizontal strip.More specifically, we let U r (µ) denote the set of partitions λ such thatλ/µ is a horizontal strip with exactly r boxes, and let D r (µ) denotethe set of partitions ν such that µ/ν is a horizontal strip with exactlyr boxes.Similarly a skew partition λ/µ is a vertical strip if:λ ′ 1 ≥ µ ′ 1 ≥ λ ′ 2 ≥ ... ≥ λ ′ n ≥ µ ′ nThat is, no row contains more than a single box.

12 ROBIN LANGERFor an arbitrary partition µ we let Ũ(µ) denote the set of partitionsλ such that λ/µ is a vertical strip, and let ˜D(µ) denote the set ofpartitions ν such that µ/ν is a vertical strip.More specifically, we let Ũr(µ) denote the set of partitions λ suchthat λ/µ is a vertical strip with exactly r boxes, and let ˜D r (µ) denotethe set of partitions ν such that µ/ν is a vertical strip with exactly rboxes.Clearly, if λ/µ is a horizontal strip, then the conjugate λ ′ /µ ′ is avertical strip.An (n,m)-binomial path is a binary string containing exactly n zerosand m ones. Pictorially, we may represent an (n,m)-binomial path bydrawing an n by m grid and tracing a path from the bottom left handcorner to the top right hand corner by reading the binary string fromleft to right and taking a step up each time we read a one, and a stepacross each time we read a zero.For example, if s = 00110101000 we have the following path:s =Observe that the boxes lying “above” a binomial path form a Youngdiagram. We shall call this diagram α(s). Likewise, after a rotation,the boxes lying “below” a binomial path form a Young diagram. Weshall call this diagram β(s). In our example, we have:α(s) = β(s) =If s is an (n,m) binomial path then the dual of s is the (n,m)-binomial path ˜s which would be obtained by reading the binary stringof s from right to left, rather than from left to right. In our example˜s = 00010101100˜s =It is not hard to see that the upper diagram associated with a binomialpath s is equal to the lower diagram associated with the dual

SYMMETRIC FUNCTIONS AND MACDONALD POLYNOMIALS 13path ˜s. Likewise the lower diagram of s is equal to the upper diagramof ˜s.α(˜s) = = β(s) β(˜s) = = α(s)Now let s ′ denote the binary string obtained from s by swapping allthe zeros for ones and vice versa. In our case s ′ = 11001010111. Notethat s ′ is now an (m,n) binomial path.s ′ =Observe that (α(s ′ ),β(s ′ )) = (β(s) ′ ,α(s) ′ ), and similarly (α(˜s ′ ),β(˜s ′ )) =(β(˜s) ′ ,α(˜s) ′ ) = (α(s) ′ ,β(s) ′ ). In other words, reversing the string andinterchanging the role of zeros and ones is equivalent to conjugatingthe partition.α(s ′ ) = = β(s) ′ β(s ′ ) = = α(s) ′There is a also a way that we may associate a pair of diagrams(ν(s),γ(s)) with distinct parts to any given binomial path. Supposethat we label the steps of s from left to right with the integers from 0to n + m − 1.s =32Then ν(s) is the diagram which has a row of length k if and onlyif there is some upstep of s labelled with k. We have of course thatν(s) = α(s) + δ n57

14 ROBIN LANGERν(s) =The other partition γ(s) is the diagram which has a row of length kif and only if there is some across step which is labelled with a k.γ(s) =Of course ν(s) ∪ γ(s) = δ m+n .Performing the same procedure with the dual path s ′ we get:This time:s ′ =10461098ν(s ′ ) == γ(s)and:γ(s ′ ) == ν(s)Observe that, a little surprisingly, we have γ(s) = ν(s ′ ) and ν(s) =γ(s ′ ), which gives us the following proposition:

SYMMETRIC FUNCTIONS AND MACDONALD POLYNOMIALS 15Proposition 1.1. Let λ be any partition contained within an n by mbox, and let µ be its compliment with respect to this box, then we have:(λ + δ n ) ∪ (µ ′ + δ m ) = δ n+mProof. Suppose that λ = α(s), so that λ + δ n = ν(s). Now µ ′ = α(s ′ )and µ ′ + δ m = ν(s ′ ) = γ(s). The result now follows from the fact thatν(s) ∪ γ(s) = δ n+m .□1.1.2. Monomial symmetric functions. Consider now the multivariatepolynomial ring Q[x 1 ,...,x k ]. Each monomial in Q[x 1 ,...,x k ] correspondsto a composition. For example, the monomial x 1 x 2 3 in Q[x 1 ,x 2 ,x 3 ]corresponds to the list (1, 0, 2).For notational convenience, if η = (η 1 ,η 2 ,...,η k ) is a composition,then by X η we mean the monomial x η 11 x η 22 ...x η kk. There is a naturaladdition on the space of compositions which corresponds to multiplicationin the polynomial ring. if η = (η 1 ,...,η k ) and γ = (γ 1 ,...,γ k )then η + γ = (η 1 + γ 1 ,...,η k + γ k ), and X (η+γ) = X η X γ .The symmetric group S k acts on the set of compositions with k partsby permuting the parts. Each orbit of S k contains a unique partition.For any composition λ let r(λ,i) denote the number of parts of λ equalto i. Let us define:r λ = ∏ r(λ,i)!i≥0Then r λ is the order of the subgroup of S k that stabilizes the compositionλ. If λ is a partition with distinct parts then r λ = 1.For each partition λ we define the monomial symmetric polynomialto be:m λ = 1 ∑X σ(λ)r λσ∈S nSince r λ is the number of permutations in S k that will stabilize themonomial x λ , the monomial symmetric function is the sum of all distinctpermutations of this monomial.As λ runs over the set of all permutations of n with at most k parts,the set {m λ |λ ∈ Pn} k forms a basis for the vector space Λ n k consistingof all symmetric polynomials of homogeneous degree k in n variables.We shall write:Λ n = ⊕ kΛ k nto denote the ring of symmetric functions in n variables. The map:π n : Λ n → Λ n−1

16 ROBIN LANGERwhich sets the nth variable equal to zero is a homomorphism whosekernel is generated by the monomial symmetric functions indexed bypartitions with exactly n nonzero parts.1.1.3. Plethystic notation. It is often more convenient to work in infinitelymany variables. It is also convenient to make use of the plethysticnotation. In the plethystic notation one writes: X = x 1 +x 2 +x 3 + · · ·to denote the set of indeterminates: X = {x 1 ,x 2 ,x 3 ,...} and Y =y 1 +y 2 +y 3 +· · · to denote the set of indeterminates Y = {y 1 ,y 2 ,y 3 ,...}Furthermore by X + Y we denote the disjoint union of the variablesin the sets X and YX + Y = x 1 + x 2 + · · · + y 1 + y 2 + · · ·= {x 1 ,x 2 ,...,y 1 ,y 2 ,...}= X ∪ Yand by XY we denote the cartesian product:XY = (x 1 + x 2 + · · · )(y 1 + y 2 + · · · )= x 1 y 1 + x 1 y 2 + · · · + x 2 y 1 + x 2 y 2 + · · ·= {x 1 y 1 ,x 1 y 2 ,...x 2 y 1 ,x 2 y 2 ,...}= X × YThe complete and elementary symmetric functions may be defined interms of their generating functions:Ω z (X) = Ω(Xz) = ∏ 11 − xz = ∑ ∞h n (X)z nx∈Xn=0˜Ω z (X) = ˜Ω(Xz) =x∈X(1 ∏ ∞∑+ xz) = e n (X)z nNote that the first q-exponential funtion may be expressed as:( ) zE q (z) = Ω1 − q1where is the alphabet 1 + q + 1−q q2 + .... Similarly the second q-exponential function may be expressed as:( )e q (z) = ˜Ω z1 − qThe involution ω is defined by:ω[h n (X)] = e n (X)n=0

SYMMETRIC FUNCTIONS AND MACDONALD POLYNOMIALS 17(extended multiplicatively). The “forgotten” symmetric functionsare defined by:ω[m λ (X)] = f λ (X)For each variable x let us define an anti-variable x such that:{x} ∪ {x} = {}Alternatively one can think in terms of anti-sets:{x} ∪ {x} = {}Now let −X = x 1 + x 2 + x 3 + · · · denote the set of anti-variables{x 1 ,x 2 ,x 3 ,...} or, the anti-set of variables {x 1 ,x 2 ,x 3 ,...}.For an arbitrary symmetric function f we define:f(−X) = ωf(ǫX)where ǫX is the more obvious but less natural form of negation:ǫX = {−x 1 , −x 2 , −x 3 ,...}We may now express the elementary symmetric functions in terms ofthe complete symmetric functions:˜Ω(X) = Ω(−ǫX)1.1.4. Schur functions. The most important basis for the ring of symmetricfunctions, from the perspective of geometry or representationtheory, is the Schur functions.Let H(X) denote the infinite Toepliz matrix (h i−j (X)), where we areassuming that h −k (X) = 0 for all k ≥ 1. In other words the matrix isupper triangular with ones on the diagonal. For λ and µ partitions withn parts (any number of which may be zero), the skew Schur functions λ/µ (X) may be defined by minors of this matrix as follows:s λ/µ (X) = det(H I,J (X)) = det(h λi −µ j +j−i(X))where I = µ+δ n and J = λ+δ n . Note that we indexing both the rowsand the columns from zero rather than one. In particular the regularSchur functions are given by:s λ (X) = det(H [n] , λ+δn ) = det(h λi +j−i(X))where [n] = {0, 1, 2,...,n − 1}. Note that n does not appear explicitlyin this equation, and if n is taken to be larger than the number ofnonzero parts of λ then the resulting matrix is block diagonal, withthe first block of size l(λ) independent of n, and the remaining blockscontaining ones, which do not affect the determinant.

18 ROBIN LANGERThe fact that the generating function for the complete symmetricfunctions has the property Ω z (X + X ′ ) = Ω z (X)Ω z (X ′ ) lifts to thefact that: H(X + X ′ ) = H(X)H(X ′ ), and so, by the Cauchy–Binettheorem we have:s λ (X + X ′ ) = ∑ µ⊆λs λ/µ (X)s µ (X ′ )Consider now the infinite matrix E(X) = (e i−j (X)) The fact thatΩ(X)˜Ω(ǫX) = 1 now lifts to the fact that H(X)E(ǫX) = I. Thus byJacobi’s formula for the minors of the inverse matrix, we have:det(H I,J (X)) = (−1) P rk=1 j k−i kdet(E J ′ ,I ′(ǫX))= det(E J ′ ,I ′(X))Now by proposition 1.1, if these are (n + m) by (n + m) matrices andI = λ + δ n then I ′ = λ ′ + δ m . In other words we have an alternativeexpression for the Schur functions given by:We also get from this that:s λ/µ (X) = det(e λ ′i −µ ′ j +j−i (X))ω[s λ (X)] = s λ ′(X)1.2. The Umbral Calculus. A nice reference for finite dimensionallinear algebra is Hoffman and Kunze [17]. For more about algebras,co-algebras, Hopf algebras and quantum groups see the online notesby Arun Ram [36]. The book by Roman [38] discusses the infinitedimensionalsubtleties more carefully than is done here.1.2.1. Coalgebras. An algebra is a vector space V over some field Fequipped with a multiplication:m : V ⊗ V → Vand a unit:e : F → Vsuch that the following two diagrams commute:V ⊗ V ⊗ V m ⊗ id ✲ V ⊗ Vid ⊗ m❄ m ✲ VmV ⊗ V❄This is just the associative law.

SYMMETRIC FUNCTIONS AND MACDONALD POLYNOMIALS 19Ve ⊗ id ✲ V ⊗ Vid ⊗ eidm✲❄ m ❄V ⊗ V ✲ VThis says that multiplication by the identity has no effect.A commutative algebra is an algebra with the additional propertythat:V ⊗ Vτ✲ V ⊗ Vmm✲V✛where τ : V ⊗ V → V ⊗ V is the map τ(x ⊗ y) = y ⊗ x.The ring Λ of symmetric functions, and the ring Q[x] of polynomialsin one variable are both commutative algebras.An algebra morphism is a vector space morphism: ψ : V → Vsatisfying the two properties:V ⊗ V ψ ⊗ ψ ✲ W ⊗ Wm❄VThe multiplication is preserved.V✛ψψm❄✲ W✲ W✲eeThe identity is preserved.F

20 ROBIN LANGERThe map which sends the complete symmetric function h n (X) to x nis an algebra morphism from Λ onto Q[x].A coalgebra is a vector space V equipped with a comultiplication:and a counit:∆ : V → V ⊗ Vc : V → Fsuch that the following diagrams commute:V ⊗ V ⊗ V ✛∆ ⊗ id V ⊗ V✻✻id ⊗ ∆∆V ⊗ VThis is the co-associative law.✛ ∆ VV ✛ c ⊗ id V ⊗ V✻ ✻id ⊗ c∆id∆V ⊗ V ✛ VThis says that co-multiplying by the co-unit has no affect.A co-commutative co-algebra is a co-algebra with the additionalproperty that:V ⊗ V ✛✛τ✲✲V ⊗ V∆∆VBoth the ring of symmetric functions and the ring of polynomials areco-commutative co-algebras. In the case of Q[x] we have:and the coproduct is given by:Q[x] ⊗ Q[x] ∼ = Q[x,x ′ ]∆[p(x)] = p(x + x ′ )

SYMMETRIC FUNCTIONS AND MACDONALD POLYNOMIALS 21In the symmetric function case we have, when working in an infinitealphabet, that:Λ ⊗ Λ ∼ = ΛThe coproduct is given by the plethystic addition of alphabets.A co-algebra morphism is a vector space morphism: φ : V → Vsatisfying the two properties:V ⊗ V ✛φ ⊗ φ W ⊗ W✻ ✻∆∆V ✛ φ WThe comultiplication is preserved.V ✛ φWcc✲✛ FThe counit is preserved.The map which sends x n /n! to the complete symmetric functionh n (X) is a co-algebra morphism from Q[x] into Λ. We shall make useof this homomorphism to lift properties of Q[x] to Λ.Categorically, the notions of algebra and co-algebra are dual. Combinatoriallythe product describes how the elements of an algebra canbe combined, while the coproduct describes how the elements of a coalgebramay be decomposed [19, 39].Every vector space V has a dual V ∗ which is the space of linearfunctionals from V into the field F. There is a natural map:given by:(−, −) : V ∗ ⊗ V → F(w,v) = w[v]In finitely many dimensions, every vector space is isomorphic to itsdual, and once an isomorphism φ : V → V ∗ has been fixed we maydefine a non-degenerate bilinear form:〈−, −〉 : V ⊗ V → F

22 ROBIN LANGERby:〈v 1 ,v 2 〉 = (φ(v 1 ),v 2 )Still in finite dimensions, if V is an algebra then V ∗ is a coalgebrawith coproduct:and counit:(∆w,v 1 ⊗ v 2 ) = (w,m(v 1 ,v 2 ))c(w) = (w,e(1))Furthermore, if ψ : V → W is an algebra morphism, then its adjointψ ∗ : W ∗ → V ∗ is a coalgebra morphism.Similarly, if V is a co-algebra then V ∗ is an algebra with multiplication:and unit:(m(w 1 ,w 2 ),v) = (w 1 ⊗ w 2 , ∆(v))(e(x),v) = xc(v)If φ : V → V is an co-algebra morphism, then its adjoint φ ∗ : V ∗ → V ∗is an algebra morphism.Unfortunately, both Λ and Q[x] are infinite dimensional. In infinitelymany dimensions, a vector space V need not be isomorphic to its dualV ∗ , and instead of equality we have:V ∗ ⊗ V ∗ ⊆ (V ⊗ V ) ∗Since ∆(w) ∈ V ∗ ⊗ V ∗ whenever w ∈ V ∗ we are still able, in infinitedimensions, to induce a product structure on the dual from an existingproduct structure on the original space, and we still have that algebramorphisms of V ∗ correspond to co-algebra morphisms of V . Theconverse is however no longer true.1.2.2. Sequences of Binomial Type. The classification of the co-algebraisomorphisms of the polynomial ring Q[x] by consideration of the algebramorphisms of the dual space goes back to Gian-Carlo Rota [33, 40].Since then there have been many other nice expositions such as Gessel[16] and Roman [38], as well as attempts to find q-analogs [1, 18, 37],and attempts to generalize the umbral calculus to symmetric functions:Loeb [26], Mendez [31, 32] and Chen [5, 6]. We follow here, in particular,the presentation of Garsia [10, 12, 13]. See the following onlineresources [7, 25] for a more extensive bibliography.

SYMMETRIC FUNCTIONS AND MACDONALD POLYNOMIALS 23The ring of polynomials in one indeterminate V = Q[x] has a distinguishedHamel basis given by:{1,x, x22! , x33! ,...}The dual space V ∗ = Q[[y]] has a distinguished Schauder basis:{1,y,y 2 ,y 3 ,...}We may think of y k as the kth coefficient extraction operator (actingon exponential generating series):[ ] x(y k k,p(x)) = p(x)k!Note that coefficient extraction and differentiation are related by:[ ] xkf(x) = LD k f(x)k!where D denotes the ordinary differential operator d/dx, and L is theconstant term operator. Thus we may also write the pairing betweenV and V ∗ as:(f(y),p(x)) = Lf(D)[p(x)]By the Leibniz rule, the co-multiplication on V :induces the multiplication on V ∗ :∆p(x) = p(x + x ′ )m(y r ,y s ) = y r+sDefine a delta series to be a formal power series f(y) which has acompositional inverse. For every delta series f(y) the map:φ f (y) = f(y)induces an algebra morphism of V ∗ with inverse:φ g (y) = g(y)where g(y) is the compositional inverse of f(y), that is:f(g(y)) = g(f(y)) = yA sequence of polynomials {p n (x)} is said to be of binomial type if:p n (x + x ′ ) = ∑ ( np k (x)p n−k (xk)′ )kA sequence of polynomials {p n (x)} is said to be of convolution type if:p n (x + x ′ ) = ∑ kp k (x)p n−k (x ′ )

24 ROBIN LANGERClearly, if p n (x) is of binomial type, then p n (x)/n! is of convolutiontype.Sequences of convolution type arise as images, under co-algebra maps,of the standard basis {x n /n!} while the corresponding sequences of binomialtype arise as images, under co-algebra maps, of the basis {x n }.By duality, the co-algebra maps of V are each adjoint to an algebramap of V ∗ . We shall write U f = φ ∗ f to denote the umbral operatordefined by:[ ] xnU f = p n (x)n!where {p n (x)} is the convolution type basis of V which is dual to thebasis {1,g(y),g(y) 2 ,...} of V ∗ . As before g(y) is the compositionalinverse of f(y).Let f 1 (y) = ∑ k a ky k and f 2 (y) = ∑ k b ky k be delta series. Now let:α nk = [y n ]f 2 (y) kβ nk = [y n ]f 1 (y) kγ nk = [y n ]f 1 (f 2 (y)) kand let A = (α nk ) and B = (β nk ) and C = (γ nk ) be infinite matriceswith n,k ≥ 1. Since:We have that:f 1 (f 2 (y)) n = ∑ i= ∑ kβ in f 2 (y) i( ∑iAB = CThese are called Jabotinsky matrices [13].α ki β in)Proposition 1.2. The umbral operator has the explicit expression:U f = ∑ x nn! Lfn (D)nAlternatively, the umbral operator may be characterized by the two properties:(i)D ◦ U f = U f ◦ g(D)y k(ii)L ◦ U f = L

SYMMETRIC FUNCTIONS AND MACDONALD POLYNOMIALS 25Proof. For the first part, note that:[ ] x[y n n]f(y) = f(D)n!Let A = {a nk } be the matrix:a nk = [y k ]f(y) nand let B = {b nk } be the inverse matrix:We have:b nk = [y k ]g(y) n〈g(y) k ,U f [x n /n!]〉 = Lg(D) k ∑ jx jj! [yn ]f(y) j= ∑ j[y j ]g(y) k [y n ]f(y) j= ∑ jb kj a jn= δ nkFor the second part, we have:〈g(y) k ,g(D)[p n (x)]〉 = Lg(D) k+1 [p n (x)]= δ k+1,n= δ k,n−1= 〈g(y) k ,p n−1 (x)〉Which shows that the umbral operator satisfies the first property. Thesecond property is obvious. To see that these properties characterizethe umbral operator, recall Taylor’s formula:I = ∑ nx nn! LDn

= U f□26 ROBIN LANGERSuppose that U is any operator satisfying the above two properties,then we have:U = ∑ x nn! LDn Un= ∑ x nn! LU ◦ fn (D)n= ∑ x nn! L ◦ fn (D)n1.3. The Hall inner-product. The results in this section are takenfrom the book of Lascoux [24] and the two papers by Garsia, Haimanand Tesler [11, 14]. See also the paper by Zabrocki [43].1.3.1. Preliminaries. The Hall inner product may be defined by:〈m λ (Y ),h µ (X)〉 = δ λµNormally this is thought of as a map:〈−, −〉 : Λ ⊗ Λ → Qbut we want to think of it as a map:〈−, −〉 : Λ ∗ ⊗ Λ → QIn analogy with the one variable case, Λ is symmetric polynomialswhile Λ ∗ is symmetric “formal power series”. In lieu of the fact that apair of bases {P λ (X)} and {Q λ (Y )} are dual with respect to the Hallinner product if and only if:∑λP λ (X)Q λ (Y ) = Ω(XY ) = ∏ x∈Xx∈Y11 − xyWe shall always write elements of Λ as symmetric functions in thealphabet X and elements of Λ ∗ as symmetric functions in the alphabetY .Any symmetric function f(X) may be thought of as the operator“multiplication by f(X)”, and as such there is an adjoint operator,which we denote by ∂ f . By definition:〈∂ f [g(Y )],h(X)〉 = 〈g(Y ),f(X)h(X)〉

Q µ (Y )Q ν (Y ) = ∑ λP λ (X + X ′ ) = ∑ µ,νQ µ (Y )Q ν (Y ) = ∑ λΩ((X + X ′ )Y ) = ∑ λ= ∑ γ,µ= ∑ γ,µ= ∑ λ= ∑ λQ λ (Y ) ∑ γ,µP γ (X)P µ (X ′ ) ∑ λQ λ (Y ) ∑ γ,µSYMMETRIC FUNCTIONS AND MACDONALD POLYNOMIALS 27The next proposition says that, via the Hall inner product, the coproducton Λ (given by plethystic addition of alphabets) induces thenatural product on the dual. In fact, this property characterizes theHall inner product.Proposition 1.3. If {P λ (X)} and {Q λ (Y )} are a pair of dual bases(with respect to the Hall inner product), and:P λ (X + X ′ ) = ∑ µ,νd λ µνP ν (X)P µ (X ′ )then:d λ µνQ λ (Y )Proof. Let:d λ µνP ν (X)P µ (X ′ )and let:˜d λ µνQ λ (Y )We have:P λ (X + X ′ )Q λ (Y )d λ γ,µP γ (X)P µ (X ′ )while:)( ∑γΩ(XY )Ω(X ′ Y ) =)( ∑P γ (X)Q γ (Y )µP µ (X ′ )Q γ (Y )P γ (X)P µ (X ′ )Q γ (Y )Q µ (Y )˜d λ µγQ λ (Y )˜d λ γµP γ (X)P µ (X ′ )The result follows now since Ω(XY + X ′ Y ) = Ω(XY )Ω(X ′ Y ).□

28 ROBIN LANGERCorollary 1.1 (Taylor’s theorem). If {P λ (X)} and {Q λ (Y )} are a pairof dual bases then for any symmetric function f(X) we have:f(X + X ′ ) = ∑ γ∂ Qγ [f(X)]P γ (X ′ )Proof. By linearity, it is sufficient to check on the basis {P λ (X)}, fromwhich the proposition follows by observation that:〈Q µ (Y ),∂ Qγ [P λ (X)]〉 = 〈Q µ (Y )Q γ (Y ),P λ (X)〉= d λ µγNote that as a consequence of Taylor’s theorem, and the self-duality ofSchur functions, the fact that:implies:s λ (X + X ′ ) = ∑ µs λ/µ (X)s µ (X ′ )s λ/µ (X) = ∂ sµ [s λ (X)]1.3.2. Column operators. Let us introduce now the “translation” and“multiplication” operators of Garsia, Haiman and Tesler [11, 14]:(1) T (z)[f(X)] = f (X + z)(2) P(z)[f(X)] = Ω(Xz)f(X)Working with the pair of dual bases {m λ (Y )} and {h λ (X)}, andnoting that unless λ is a row partition, m λ (X) will vanish when Xis an alphabet containing a single letter, we get, by Taylor’s theorem,that:□Similarly:T (z) = ∑ nP(z) = ∑ nz n ∂ hnz n h n (X)Clearly we have:(3) T (z) = P ∗ (z)Similarly, working with the dual bases {f λ (Y )} and {e λ (X)} andrecalling that f λ (−Y ) = m λ (ǫY ) we get that:T (−z) = ∑ n(ǫz) n ∂ en

SYMMETRIC FUNCTIONS AND MACDONALD POLYNOMIALS 29and:P(−z) = ∑ n(ǫz) n e n (X)and:In other words:P(−z) = ω ◦ P(ǫz) ◦ ωT (−z) = ω ∗ ◦ T (ǫz) ◦ ω ∗The following lemma tells us how the translation and multiplicationoperators commute:Lemma 1.1.T (v)P(w) = Ω(vw)P(w)T (v)Proof.T (v)P(w)[h(X)] = T (v)[Ω(Xw)h(X)]= Ω((X + v)w)h(X + v)= Ω(vw)Ω(Xw)T (v)[h(X)]= Ω(vw)P(w)T (v)[h(X)]□We can now define the column operator:(4) C(z) = ∑ m(−1) m C m z m = P(−z)T (1/z)and the row operator:(5) R(z) = ∑ nz m (−1) m R m = P(z)T (−1/z) = C(−z)Proposition 1.4. The Schur function has a natural expression interms of row operators:or in terms of column operators:s λ (X) = R λ1 ◦ R λ2 ◦ · · · ◦ R λn [1]s λ (X) = C λ ′1◦ C λ ′2◦ · · · ◦ C λ ′ m[1]Proof. Starting from the definition of the Schur function as:s λ (X) = det(e λ ′i+j−i(X))

30 ROBIN LANGERand expanding along the first row we get that:∑n−1s λ (X) = (−1) k e λ ′1+k(X)s µ/(k) (X)k=0∑n−1= (−1) k e λ ′1+k(X)∂ hk (X)[s µ (X)]k=0= C λ ′1[s µ (X)]where µ is the partition obtained from λ by removing the first column.Similarly, from the definition of the Schur function as:s λ (X) = det(h λi +j−i(X))we get by expanding along the first row that:s λ (X) = R λ1 [s µ (X)]where this time µ is the partition obtained from λ by removing thefirst row.□Corollary 1.2. In the row and column cases we have:s (n) (X) = h n (X)s (1 n )(X) = e n (X)1.3.3. Duality. The Schur functions are, up to scaling, the unique homogeneousbasis which is self dual with respect to the Hall inner product.We give here a slightly non-standard proof of this fact, which willbe generalized in the next section.Proposition 1.5. The Schur functions are self-dual with respect to theHall inner productProof. By proposition 1.3 and corollary 1.2 it suffices to demonstratethat:〈Ω z (Y )s µ (Y ),s λ (X)〉 = 〈s µ (Y ),s λ (X + z)〉Begin by observing that:which tells us that:T (v)C(z) = T (v)P(−z)T (1/z)= Ω(−vz)P(−z)T (v)T (1/z)= (1 − vz)C(z)T (v)

SYMMETRIC FUNCTIONS AND MACDONALD POLYNOMIALS 31[C(z), T (v)] = vz C(z)T (v)Equating coefficients of (−1) m v k z m on both sides we get that:[C m ,∂ hk ] = −C m−1 ∂ hk−1Rewriting this in the form:we obtain:∂ hm C k = C k ∂ hm + C k−1 ∂ hm−1∂ hk [s λ (X)] = ∂ hk ◦ C λ ′1◦ C λ ′2◦ · · ·C λ ′ n[1]= ∑ I⊆[n]|I|=kC λ ′1 −a 1◦ C λ ′2 −a 2◦ · · · ◦ C λ ′ n −a n[1]where a i = 1 if i ∈ I and zero otherwise.Of course it is possible that for some k we have λ k = λ k+1 whilsta k ∈ I and a k−1 ∉ I, in which case the sequence of integers (λ 1 −a 1 ,λ 2 −a 2 ,...,λ n −a n ) will not form a partition. In this case, however, we havethat the determinant: det(e λ ′i −a i +j−i(X)) contains a repeated column,and thus vanishes.We conclude that:(6) T (z)[s λ (X)] = s λ (X + z) = ∑µ∈D(λ)s µ (X)z |λ|−|µ|where D(λ) denotes all the partitions which can be obtained from λby removing a horizontal strip. This is the first form of the recurrencefor the Schur functions.Dually:which tells us that:P(v)C(z) = P(v)P(−z)T (1/z)= P(−z)P(v)T (1/z)= P(−z)Ω(−v/z)T (1/z)P(−v)= (1 − v/z)C(z)P(v)[C(z), P(v)] = v/z C(z)P(v)By equating coefficients of (−1) m v k z m on both sides we find that:[C m ,h k (X)] = −C m+1 h k−1 (X)

32 ROBIN LANGERRewriting this in the form:h m (X)C k = C k h m (X) + C k+1 h m−1a similar argument to the previous reveals:(7) P(z)[s µ (Y )] = s µ (Y )Ω z (Y ) = ∑λ∈U(µ)s λ (Y )z |λ|−|µ|where U(µ) denotes the set of all partitions which can be obtainedfrom µ by adding a horizontal strip. This is the first form of the Pieriformula for the Schur functions.Combining these two facts, we see that:〈s µ (Y )Ω(Y z),s λ (X)〉 = 〈s µ (Y ),s λ (X + z)]〉as claimed.□Note that identical arguments using the row operator rather thanthe column operator can be used to show that:(8) T (−z)[s λ (X)] = s λ (X − z) = ∑s µ (X)(ǫz) |λ|−|µ|µ∈ ˜D(λ)where ˜D(λ) denotes all the partitions which may be obtained from λby removing a vertical strip. This is the second form of the recurrencefor the Schur functions.Also we have:(9) P(−z)[s µ (Y )] = s µ (Y )˜Ω z (Y ) = ∑λ∈Ũ(µ) s λ (Y )(ǫz) |λ|−|µ|where Ũ(µ) denotes the set of all partitions which can be obtainedfrom µ by adding a vertical strip. This is the second form of the Pieriformula for the Schur functions.Putting these two facts together we see that we also have:〈s µ (Y )˜Ω(Y z),s λ (X)〉 = 〈s µ (Y ),s λ (X − z)〉

SYMMETRIC FUNCTIONS AND MACDONALD POLYNOMIALS 331.4. Littlewood–Richardson Bases. The Littlewood–Richardson coefficientsc λ µν are defined by:s λ (X + X ′ ) = ∑ µ,νc λ µνs µ (X)s ν (X ′ )These numbers may be described nicely by a combinatorial objectknown as puzzles [34]. A very nice proof of the Littlewood-Richardsonrule using ideas from Quantum Integrability can be found in [44]. Inthis section, we describe all the other bases {P λ (X)} of the ring ofsymmetric functions Λ with the property:P λ (X + X ′ ) = ∑ µ,νc λ µνP µ (X)P ν (X ′ )We shall say that such a basis is of Littlewood–Richardson type.Note that we are only interested in bases for Λ (the ring of symmetricfunctions in infinitely many variables) which are stable in the sensethat if λ is a partition with at most k parts, its expansion in terms ofSchur functions is unchanged under the projection onto Λ k (the ring ofsymmetric functions in just k variables).We saw in the proof of proposition 1.5 that:s λ (X + z) = ∑s µ (X)s (|λ|−|µ|) (z)µ∈D(λ)To prove that a given basis {P λ (X)} is of Littlewood–Richardsontype it suffices to demonstrate that:P λ (X + z) = ∑P µ (X)P (|λ|−|µ|) (z)µ∈D(λ)Since the Schur functions are self-dual we have, by proposition 1.3,that:s µ (Y )s ν (Y ) = ∑ c λ µνs λ (Y )λIf {P λ (X)} is a basis of Littlewood–Richardson type, and {Q λ (Y )} isits dual basis, then we must have:Q µ (Y )Q ν (Y ) = ∑ λc λ µνQ λ (Y )We also saw in the proof of proposition 1.3 that:s µ (Y ) ∑ s (n) (Y )s (n) (z) = ∑s λ (Y )s (|λ|−|µ|) (z)nλ∈U(µ)

34 ROBIN LANGERThis will generalize to:Q µ (Y ) ∑ nQ (n) (Y )Q (n) (z) = ∑λ∈U(µ)Q λ (Y )Q (|λ|−|µ|) (z)1.4.1. Generalized complete symmetric functions. In the row case, theSchur functions reduce to the complete symmetric functions, whichhave generating function:Ω(Xz) = ∑ nh n (X)z nThe complete symmetric functions have the property that:h n (X + X ′ ) = ∑ kh k (X)h n−k (X ′ )Let f(z) be an arbitrary delta series, with compositional inverse g(z).Define generalized complete symmetric functions by:(10) Ω f (Xz) = ∏ 11 − f(z)x = ∑ r n (X)z nxnAlso define, in the dual space:(11) Φ g (Y z) = ∏ 11 − zg(y) = ∑ ρ n (Y )z nynSince:we have that:Ω f (X + X ′ ) = Ω f (X)Ω f (X ′ )r n (X + X ′ ) = ∑ kr k (X)r n−k (X ′ )Furthermore, we have that:r n (X) = ∑ kα nk h k (X)where:α nk = [y n ]f(y) kIn other words, r n (X) is the image of the convolution type sequenceassociated to f(y) under co-algebra embedding of Q[x] into Λ given by:x nn! ↦→ h n(X)

SYMMETRIC FUNCTIONS AND MACDONALD POLYNOMIALS 35Note that there are plenty of other collections of ring generators {p n (X)}for Λ with the property that:p n (X + X ′ ) = ∑ kp k (X)p n−k (X ′ )For any pair of sequences {f n (x)} and {g n (y)} such:∑f n (x)g n (y) = 11 − xynthe ring generators {p n (X)} defined by:∑p n (X)z n = ∏ f n (x)z n = ∏ x n g n (z)nxxwill have this property. However, unless they are of the special formabove, it will not be possible to write them as:p n (X) = ∑ □ nk h k (X)kinstead they must have more general expansions of the form:p n (X) = ∑ λTheir image under the projection:□ λ h λ (X)h n (X) ↦→ xnn!will still be a sequence of convolution type associated to some f(y),however.The idea now is to show that the same determinantal constructionwhich produces the Schur functions from the complete symmetric functionscan be used to produce more general bases with the Littlewood–Richardson property.In the special case where the ring bases are images of convolutiontype sequences, we shall see that, as a direct consequence of the multilinearityof the determinant, the associated Littlewood–Richardsonbases have expansions of the form:P λ (X) = ∑ ν⊆λ□ λν s ν (X)while the dual basis will have an expansion of the form:Q µ (Y ) = ∑ µ⊆ν□ νµ s ν (Y )

36 ROBIN LANGER1.4.2. Umbral operators. Let U f be the operator which is defined onthe complete symmetric functions by:U f [Ω(Xz)] = Ω f (Xz)extendeded multiplicatively to the whole of Λ. Clearly we have:U −1f= U gIt turns out that if P λ (X) = U f [s λ (X)] then {P λ (X)} is a basis ofLittlewood–Richardson type with dual basis {Q λ (Y )} where Q λ (Y ) =U ∗ g [s λ (Y )].Proposition 1.6.U ∗ g [Ω(Y z)] = Φ g (Y z)Proof. Let {n λ (Y )} denote the dual basis to {r λ (X)}. We have:〈U ∗ f[n µ (Y )],h λ (X)〉 = 〈n µ (Y ),U f [h λ (X)]〉= 〈n µ (Y ),r λ (X)〉and so:That is:Now:= δ λµUf[n ∗ µ (Y )] = m µ (Y )n µ (Y ) = Ug ∗ [m µ (Y )]Ω(XY ) = ∏ y= ∏ yΩ(Xy)∑r n (X)g(y) nnwhich gives us the following explicit expression for n λ (Y ):n λ (Y ) = ∑ ∏g(y k ) λ σ(k)σ kwhere the sum is over all distinct permutations of λ. Finally:Ug ∗ [Ω(Y z)] = ∑ Ug ∗ [m λ (Y )]z |λ|n= ∑ n= ∏ xn λ (Y )z |λ|11 − zg(y)

SYMMETRIC FUNCTIONS AND MACDONALD POLYNOMIALS 37□Proposition 1.7. • ∂ rn = Ug ∗ ◦ ∂ hn ◦ Uf∗• ∂ ρn = U g ◦ ∂ hn ◦ U fProof. For the first part:〈∂ rn [n µ (Y )],r λ (X)〉 = 〈n µ (Y ),r λ (X)r n (X)〉= δ µ,n+λ= 〈m µ (Y ),h λ (X)h n (X)〉= 〈∂ hn [m µ (Y )],h λ (X)〉= 〈∂ hn ◦ U ∗ f[n µ (X)],U g [r λ (X)]〉= 〈U ∗ g ◦ ∂ hn ◦ U ∗ f[n µ (Y )],r λ (X)〉For the second part, let {η λ (X)} denote the dual basis to {ρ λ (Y )}.As before we have:〈h µ (Y ),U f [η λ (X)]〉 = 〈U ∗ f[h µ (Y )],η λ (X), 〉= 〈ρ µ (Y ),η λ (X)〉and so:That is:Now:= δ λµU f [η λ (X)] = m λ (X)η λ (X) = U g [m λ (X)]〈ρ µ (Y ),∂ ρn [η λ (X)]〉 = 〈ρ µ (Y )ρ (n) (Y ),η λ (X)〉= δ µ+(n),λ= 〈h µ (Y )h (n) (Y ),m λ (X)〉= 〈h µ (Y ),∂ hn [m λ (X)]〉= 〈U ∗ g [ρ µ (Y )],∂ hn ◦ U f [η λ (X)]= 〈ρ µ (Y ),U g ◦ ∂ hn ◦ U f [η λ (X)]1.4.3. Column operators. The (generalized) column operator for theLittlewood–Richardson basis {P λ (X)} is:□C f (z) = ∑ m(−1) m C f mz m = U f ◦ C(z) ◦ U gwhere C(z) is the column operator for the Schur functions.

38 ROBIN LANGERThe generalized column operator for the dual basis {Q λ (Y )} is:Ĉ f (z) = ∑ m(−1) m Ĉ f mz m = C ∗ f(z) = U ∗ g ◦ C ∗ (z) ◦ U ∗ fSince these operators are adjoint, its clear that the resulting bases:P λ (X) = C f λ◦ C f ′1 λ◦ · · · ◦ C f ′ λ [1]2′ nQ λ (Y ) = Ĉf λ ′ 1are dual. More explicitly:◦ Ĉf λ ′ 2◦ · · · ◦ Ĉf λ ′ n [1]〈Q µ (Y ),P λ (X)〉 = 〈Ug ∗ [s µ (Y )],U f [s λ (X)]〉= 〈s µ (Y ),U g ◦ U f [s λ (X)〉= 〈s µ (Y ),s λ (X)〉= δ λµNote that by the same argument used in proposition 1.4 we have:P λ (X) = det(r λi +j−i(X))Q λ (Y ) = det(ρ λi +j−i(Y ))Now define the generalized multiplication operator:P f (z) = ∑ k= ∑ kz k r k (X)f(z) k h k (X)= U f ◦ P(z) ◦ U gas well as the generalized translation operator:T g (z) = ∑ kz k ∂ ρk= ∑ kr k (z)∂ hk= U f ◦ T (z) ◦ U g

SYMMETRIC FUNCTIONS AND MACDONALD POLYNOMIALS 39Also let us define, in the dual space:ˆP g (z) = T ∗g (z)= U ∗ g ◦ P(z) ◦ U ∗ f= ∑ k= ∑ kz k ρ k (Y )r k (z)h k (X)as well as:ˆT f (z) = P ∗ f(z)= U ∗ g ◦ T (z) ◦ U ∗ f= ∑ kz k ∂ rk= ∑ kf(z) k ∂ hkWe have now that:while:C f (z) = P −f (z) ◦ T 1/g (z)Ĉ f (z) = ˆP −f (z) ◦ ˆT 1/g (z)Proposition 1.8. The (non-homogeneous) basis for the ring of symmetricfunctions defined by:is of Littlewood–Richardson type.Proof. We must show that:P λ (X) = C f λ◦ C f ′1 λ◦ · · · ◦ C f ′ λ [1]2′ m〈Q λ (Y )Φ g (Y z),P µ (X)〉 = 〈Q λ (Y ),P µ (X + z)〉By conjugating Lemma 1.1 by U f on the left and U g on the right, wefind that:T g (v) ◦ P f (w) = Ω(vw) ◦ P f (w) ◦ T g (v)This in turn implies that:[ ˜C(z), T g (v)] = vz ◦ ˜C(z) ◦ T g (v)By arguments which are essentially identical to those in the proof ofproposition 1.5 we obtain the following generalized recurrence:∂ ρk [P λ (X)] = ∑P µ (X)µ∈D k (λ)

40 ROBIN LANGERNext by using the fact that:T (z) = ∑ kr k (z)∂ ρkwe can re-write the above as:T (z)[P λ (X)] = ∑µ∈D(λ)P µ (X)P (|λ|−|µ|) (z)Similarly, by conjugating Lemma 1.1 on the left by Ug ∗ and the rightby Uf ∗ we find that:This tells us that:ˆT g (v) ◦ ˆP f (w) = Ω(vw) ◦ ˆP f (w) ◦ ˆT g (v)[Ĉ(z), ˆP f (v)] = v/z ◦ Ĉ(z) ◦ ˆP f (v)and so we have the following Pieri formula :Q µ (Y )ρ k (Y ) = ∑Q λ (Y )That is:λ∈U k (µ)ˆP f (z)[Q µ (Y )] = Q µ (Y )Φ g (Y z) = ∑λ∈U(µ)Putting these two facts together we find that:as claimed.Q λ (Y )z |λ|−|µ|〈Q λ (Y )Φ g (Y z),P µ (X)〉 = 〈Q λ (Y ),P µ (X + z)〉1.4.4. Generalized elementary symmetric functions. Now define generalizedelementary symmetric functions:˜Ω f (Xz) = ∏ xand in the dual space:(1 − f(−z)x) = ∑ nc n (X)z n□˜Φ g (Y z) = ∏ y(1 + zg(y)) = ∑ nγ n (Y )z nA minor modification of the argument at the end of section 1.1.4 canbe used to show that:det(r λi +j−i(X)) = det(c λ ′i+j−i(ǫX))det(ρ λi +j−i(Y )) = det(γ λ ′i +j−i(ǫY ))

SYMMETRIC FUNCTIONS AND MACDONALD POLYNOMIALS 41The generalized row operators are:One can show that:which implies that:R f (z) = C f (−z)ˆR f (z) = Ĉf(−z)P λ (X) = R f λ 1◦ R f λ 2◦ · · · ◦ R f λ n[1]Q λ (Y ) = ˆR f λ 1◦ ˆR f λ 2◦ · · · ◦ ˆR f λ n[1]T (−z)[P λ (X)] = P λ (X − z) = ∑µ∈ ˜D λP µ (X)P (|λ|−|µ|) (ǫz)ˆP −f (z)[Q µ (Y )] = Q µ (Y )˜Φ f (Y z) = ∑λ∈Ũ(µ) Q λ (Y )Q (|λ|−|µ|) (ǫz)In other words:〈Q λ (Y )˜Φ f (Y z),P µ (X)〉 = 〈Q λ (Y ),P µ (X − z)〉1.5. Examples. In this section we shall give symmetric function analoguesof the rising and falling factorials, as well as the rook polynomials,leading to a generalized family of Stirling and Lah numbers whichare indexed by partitions rather than integers.For any formal power series f(z) let ¯f(z) = f(−z).Let us use the following notation for the falling factorial:and the rising factorial:(x) n = x(x − 1)(x − 2) · · · (x − n + 1)(x) n = x(x + 1)(x + 2) · · · (x + n − 1)The Stirling numbers of the first kind are defined by:(x) n = ∑ ks(n,k)x kWe have:(x) n = ∑ k|s(n,k)|x kOne may check that { (x)n } is the sequence of convolution type associatedto f(z) = (exp(z) − 1) while { (x)n } is the sequence of convolutionn!n!type associated to − ¯f(z) = (1 − exp(−z)).

42 ROBIN LANGERNow let {A λ (X)} denote the Littlewood–Richardson basis for thering of symmetric functions Λ that is associated to f(z) = exp(z) − 1.The generating function for the row case is:Ω f (Xz) = ∏ x∈X11 − x log(1 + z)while the generating function for the column case is:˜Ω f (Xz) = ∏ x∈X(1 − x log(1 − z))Similarly let {B λ (X)} denote the Littlewood–Richardson basis associatedto − ¯f(z) = 1 − exp(−z). The generating function for the rowcase is:Ω − ¯f(Xz) = ∏ 11 − x log ( )1x∈X1−z= ∏ 11 + x log(1 − z)x∈X= ˜Ω f (−ǫXz)= ω[˜Ω f (Xz)]The generating function for the column case is:˜Ω − ¯f(Xz) =x∈X(1 ∏ ( ) 1− x log )1 + z= ∏ x∈X(1 + x log(1 + z))= Ω f (−ǫXz)= ω[Ω f (Xz)]Let A = (a λµ ) denote the transition matrix from the basis {A λ (X)}to the usual Schur basis {s λ (X)}, and let B = (b λµ ) denote the transitionmatrix from {B λ (X)} to {s λ (X)}. On the next page we give thecorner of these matrices corresponding to partitions with at most fiveparts.

SYMMETRIC FUNCTIONS AND MACDONALD POLYNOMIALS 43The order which we are using on partitions is:{{1}}{{2}, {1, 1}}{{3}, {2, 1}, {1, 1, 1}}{{4}, {3, 1}, {2, 2}, {2, 1, 1}, {1, 1, 1, 1}}{{5}, {4, 1}, {3, 2}, {3, 1, 1}, {2, 2, 1}, {2, 1, 1, 1}, {1, 1, 1, 1, 1}}Here is a portion of the matrix A:⎛1 − 1 1 1− 1 12 2 3 3 3 −1 10 − 1 1 1− 1 10 0 − 1 1 ⎞4 4 4 4 5 5 5 5 50 1 0 −1 1 0 11− 7 − 1 10 − 5 7 1− 5 − 1 102 12 12 12 3 6 12 12 12 12 4 . . . . 1 0 − 1 1 0 1− 1 − 7 110 − 1 1 5− 1 − 7 52 3 12 12 12 4 12 12 12 12 61 0 0 − 3 170 0 0 − 5 − 1 10 0 02 2 4 6 12 3 111 0 0 −1 0 1 0 0 − 1 12 12 −13 − 1 11012 12 12 1 0 0 0 − 1 310 0 0 − 1 2 2 3 12 −5 76 41 0 0 0 0 −2 1 0 0 0 0 02 1 0 0 0 0 − 3 0 1 0 0 02 1 0 0 0 0 −1 0 1 0 031 0 0 0 0 −1 0 02 1 0 0 0 0 0 − 1 2 21 0 0 0 0 0 01 0 0 0 0 01 0 0 0 01 0 0 01 0 0⎜⎟⎝.... 1 0⎠0 · · · · · · 0 1The columns are the images of the Schur functions under the transformationh k ↦→ ∑ i≤k c k,ih i where c k,i are the coefficients of the powers oflog(1 + z).

44 ROBIN LANGERHere is a portion of the matrix B:⎛1 1 − 1 12 2 3 −1 1 1− 1 10 − 1 13 3 4 4 4 4 5 −1 10 0 − 1 1 ⎞5 5 5 50 1 0 1 − 1 0 11− 7 − 1 10 5 − 7 − 1 5 1− 1 02 12 12 12 3 6 12 12 12 12 4 .... 1 0 1 1−1 0 − 1 − 7 110 1− 1 − 5 1 72 3 12 12 12 4 12 12 12 12 −5 631 0 0 − 1 0 0 0 7 2 2 4 −5 − 1 10 0 06 12 3 1 0 0 1 0 −1 0 0 11 − 112 12 −13 − 1 11012 12 1211 0 0 0 − 3 0 0 0 1− 12 2 3 12 −5 76 41 0 0 0 0 2 − 1 0 0 0 0 02 31 0 0 0 0 0 −1 0 0 02 1 0 0 0 0 1 0 −1 0 01 0 0 0 0 1 0 − 3 02 11 0 0 0 0 0 −22 1 0 0 0 0 0 01 0 0 0 0 01 0 0 0 01 0 0 01 0 0⎜⎟⎝.... 1 0⎠0 · · · · · · 0 1A similar remark applies to B with log(1+z) replaced with − log(1−z).Observe that b λµ = a λ ′ µ ′, and b λµ = (−1) |λ|−|µ| a λµ .If one looks at just entries of A corresponding to row diagrams (scaledby n!/k!), or alternatively those of B corresponding to column diagrams,one recovers the Stirling numbers of the first kind:⎛⎜⎝1 −1 2 −6 240 1 −3 11 −500 0 1 −6 350 0 0 1 −100 0 0 0 1Switching the roles of A and B, one obtains the unsigned Stirling numbersof the first kind:⎛⎞1 1 2 6 240 1 3 11 50⎜ 0 0 1 6 35⎟⎝ 0 0 0 1 10 ⎠0 0 0 0 1⎞⎟⎠

SYMMETRIC FUNCTIONS AND MACDONALD POLYNOMIALS 45Let us define the Lah numbers by:(x) n = ∑ kL(n,k)(x) kLet h(z) = z and observe that 1+z h−1 (z) = zWe have:1−z( )1−ḡ(h(z)) = log1 − zz+1= log(1 + z)= g(z)In other words:U − ¯f ◦ U h −1 = U fSuppose now that the rook polynomials have the expansion:then we have:[ ] xnl n (x) = U h −1 = ∑ n!ka nkx kk!U − ¯f [l n (x)] = ∑ ka nk(x) kk!while:[ ] xnU f = (x) nn! n!Thus we must have:l n (x) = ∑ k|L(n,k)|x k

46 ROBIN LANGERNow consider the Littlewood–Richardson basis {L λ (X)} associated toh(z) = z . Here is a portion of the transition matrix L = (l 1−z λµ):⎛⎞1 1 −1 1 −1 1 1 −1 0 1 −1 1 −1 0 1 0 −1 10 1 0 2 −1 0 3 −2 0 1 0 4 −3 0 2 0 −1 0.... 1 0 1 −2 0 1 0 −2 3 0 1 0 −2 0 3 −41 0 0 3 −1 0 0 0 6 −3 0 1 0 0 01 0 0 2 0 −2 0 0 3 0 −4 0 3 01 0 0 0 1 −3 0 0 0 1 0 −3 61 0 0 0 0 4 −1 0 0 0 0 01 0 0 0 0 3 0 −2 0 0 01 0 0 0 0 2 0 −2 0 01 0 0 0 0 2 0 −3 01 0 0 0 0 0 1 −41 0 0 0 0 0 01 0 0 0 0 01 0 0 0 01 0 0 01 0 0⎜⎟⎝ .... 1 0 ⎠0 · · · · · · 0 1The columns are the images under the Schur functions under the transformationh k ↦→ ∑ i≤k( ki)hi .Looking at just the rows (scaled by n!/k!), we recover the Lah numbers:⎛⎞1 2 6 24 1200 1 6 36 240⎜ 0 0 1 12 120⎟⎝ 0 0 0 1 20 ⎠0 0 0 0 1One may check that AL = B.

SYMMETRIC FUNCTIONS AND MACDONALD POLYNOMIALS 472. A generating function identity for MacdonaldpolynomialsThe Macdonald polynomials are a qt-deformation of the Schur functions.The main result of the second part of this thesis is the proof ofa generating function identity for Macdonald polynomials which wasoriginally conjectured by Kawanaka [21].In section 2.1 we define the Macdonald polynomials and state theKawanaka conjecture. Section 2.2 contains some technical lemmaswhich are not needed until the final step of the proof. In section 2.3 wediscuss the Pieri formula and recurrence for the Macdonald formulas,which generalize those for the Schur functions which were discussed indetail in the first part of this thesis. The proof itself is contained insection 2.4.2.1. Macdonald Polynomials.2.1.1. Notation. Recall from section 1.1.3 the generating series for thecomplete symmetric functions:Ω(zX) = ∏ 11 − z xx∈XHere we again use the plethystic notation. Inside a symmetric functionthe expression 1−t denotes the alphabet:1−q1 − t1 − q = {1,q,q2 ,... − t, −tq, −tq 2 ,...}Recall also the important distinction between plethystic negation:Ω(−X) = ∏ x∈X(1 − x)and formal negation:Ω(ǫX) = ∏ 11 + xx∈XWe remark that the operator Ω acting on alphabets plays a role insymmetric function theory that is in many ways analogous to that ofthe exponential function in the theory of functions of a single variable.In particular:Ω(X + Y ) = Ω(X)Ω(Y )Also note that:( ) zΩ =1 − q1(z;q) ∞= ∑ kz k(q;q) k

48 ROBIN LANGERis the first q-exponential function, mentioned in section 1.1.1, whilst:( ) −zΩ = (−z;q) ∞ = ∑ q (k 2) z k1 − q(q;q) kkis the second q-exponential function.2.1.2. Operator definition. In this section we shall be workng with afinite alphabet X n with exactly n letters:X n = x 1 + x 2 + · · · + x nLet ∆(X n ) denote the Vandemonde determinant:1 1 ... 1∆(X n ) =x 1 x 2 ... x n... ... ... ...= ∏ (x j − x i )∥ x n−11 x2 n−1 ... xnn−1 i

SYMMETRIC FUNCTIONS AND MACDONALD POLYNOMIALS 492.1.3. Characterization using the inner product. Macdonald showed in[28] that his operator D is self-adjoint (and thus its eigenfunctions areorthogonal) with respect to the deformed Hall inner product:〈f(Y ),g(X)〉 q,t =〈f(Y 1 − t1 − q),g(X)Dual bases {P λ (X)} and {Q λ (Y )} with respect to the Macdonald innerproduct are characterized by the property that:∑(P λ (X)Q λ (Y ) = Ω XY 1 − t )= ∏ (txy,q) ∞1 − q (xy,q) ∞λx∈Xy∈YNote the important fact that:(Ω (X + Y ) 1 − t ) (= Ω X 1 − t ) (Ω Y 1 − t )1 − q 1 − q 1 − qThe Macdonald polynomials may be characterized as the unique basisfor the ring of symmetric functions over Q(q,t) which is both orthogonal(but not orthonormal) with respect to the Macdonald innerproduct, and whose expansion in terms of the monomial symmetricfunctions is strictly upper triangular with respect to the dominanceorder [28] on partitions:P λ (X) = ∑ µ⊳λ□ λµ m µ (X)In particular in the column case the Macdonald polynomials correspondto the elementary symmetric functions:P (1 n )(X) = e n (X)2.1.4. Arms and legs. For s ∈ λ some box in a partition λ the armlength a λ (s) is defined to be the number of boxes in λ lying directly tothe right of the box s, while the leg length is defined to be the numberof boxes in the partition λ lying directly below the box s.〉In other words, if s = (i,j) then a λ (s) = λ i − j and l λ (s) = λ ′ j − i. Ifthe box s lies outside the partition λ then we defined a λ (s) = l λ (s) = 0.Let n(λ) = ∑ s∈λ a λ(s) and let ñ(λ) = ∑ s∈λ l λ(s) = n(λ ′ ).

50 ROBIN LANGERLet us define:B λ (q,t) = ∑ λq a λ(s) t l λ(s)We have, of course, that:B λ ′(t,q) = B λ (q,t)It is a surprising fact [29] (pages 338,339) that:∏〈P λ (Y ),P λ (X)〉 q,t = ∏ s∈λ(1 − qaλ(s)+1 t lλ(s) )s∈λ (1 − qa λ(s)t l λ(s)+1) = Ω((t − q)B λ(q,t))2.1.5. Duality. The Macdonald Q-functions are defined to be dual tothe Macdonald P-functions. Since the Macdonald P-functions are orthogonalwith respect to the Hall inner product we have:Q λ (X) =The operator ω q,t given by:ω q,t [f(X)] = ωfP λ (X)〈P λ (Y )P λ (X)〉 q,t( 1 − q1 − t X )is self-adjoint with respect to the Macdonald inner product.It is another surprising fact [29] (page 327) that:ω q,t [P λ (X;q,t)] = Q λ ′(X;t,q)Using this, one can show that in the one row case, the Macdonaldpolynomials correspond to the modified version of the complete symmetricfunctions [29] (page 311) defined by:Where:∑nP (n) (X) = (q;q) n(t;q) ng n (X)(g n (X)z n = Ω z X 1 − t )1 − q2.1.6. Kawanaka conjecture. In part II of this thesis we shall prove thefollowing generating function identity for the Macdonald polynomials,which was originally conjectured by Kawanaka [21] and proved in thecase of Hall-Littlewood Polynomials (q = 0):(∑ ∏s∈λλ1 + q a λ(s) t l λ(s)+11 − q a λ(s)+1t l λ(s))P λ (X;q 2 ,t 2 ) = ∏ i(−tx i ;q) ∞(x i ;q) ∞∏i

SYMMETRIC FUNCTIONS AND MACDONALD POLYNOMIALS 51By P λ (X;q 2 ,t 2 ) we mean the Macdonald polynomial P λ (X) for whichevery occurance of the variable q has been replaced by q 2 and similarlyevery occurence of the variable t has been replaced by t 2 .This identity complements the following two generating functionidentities for Macdonald polynomials which can be found in the casewhere b = 0 or b = 1 b on page 349 of Macdonald [29]:∑λ∑λb c(λ)b r(λ)∏s∈λl λ (s) even∏s∈λl λ (s) even1 − q aλ(s) t l λ(s)+11 − q a λ(s)+1t P λ(X;q,t) = ∏ ∏ (btx i ;q) ∞ (tx i x j ;q) ∞l λ(s)(bxi≤j i i ) ∞ (x i x j ;q) ∞1 − q aλ(s) t l λ(s)+11 − q a λ(s)+1t P λ(X;q,t) = ∏ ∏ (btx i ;q) ∞ (tx i x j ;q) ∞l λ(s)(bxi≤j i i ) ∞ (x i x j ;q) ∞Here c(λ) and r(λ) are the number of columns and rows of odd length,respectively.More recently some extension of the Hall-Littlewood version of theKawanaka identity have been proved in [27] but no proof of the moregeneral identity has yet appeared in the litterature. The results in thesecond part of this thesis will eventually be published in [35].We may rewrite the identity to be proved in the plethystic notationas:∑( 1 − ǫtΩ((q − ǫt))B λ (q,t)P λ (X;q 2 ,t 2 ) = Ω1 − q X + 1 − )t21 − q 2e 2(X)λBy e 2 (X) we mean the alphabet:e 2 (X) = ∑ i

52 ROBIN LANGERthe identity to a rational function in the variables a k = q µ k t m−k . Thefinal step is an induction on the residues at the poles.2.2. Resultants. The resultant is defined by:R(Z : A) = ∏ ∏(z − a)z∈Z a∈ABy the fundamental theorem of algebra, up to a scalar multiple, anypolynomial p(z) may be written in the form:p(z) = R(z : A)where A is the alphabet of zeros. More generally, any rational functionr(z) = p(z) may be written, up to a scalar multiple, in the form:q(z)r(z) =R(z : A)R(z : B)where A is the alphabet of zeros B is the set of poles.We shall need to make use of an extended version of the resultantdefined by:R(X − Y : A − B) =Be warned that:R(X : A − B)R(Y : A − B)Ω(X − Y ) = R(1 : Y − X)=R(X : A)R(Y : B)R(Y : A)R(X : B)In the previous lemma we assumed that the alphabets A and B weredistinct. In what follows we shall need to consider the case where onealphabet is a scalar multiple of the other.Let us define:Observe that:W(X : Y ) q,t = R(X : (q/t − 1)Y ) = ∏ x∈Xy∈YV (X : Y ) q,t = R(X : (t/q − 1)Y ) = ∏ x∈Xy∈Y(x − qy/t)(x − y)(x − ty/q)(x − y)V (X : Y ) q,t = W(X : Y ) 1/q,1/t = W(X : Y ) t,q

SYMMETRIC FUNCTIONS AND MACDONALD POLYNOMIALS 53We shall also need to make use of the functions:v(X : Y ) q,t = V (X : qY ) q,t = R(X : (t − q)Y ) = ∏ x∈Xy∈Y(x − ty)(x − qy)w(X : Y ) q,t = W(X : Y/q) q,t = R(X : (1/t − 1/q)Y ) = ∏ x∈Xy∈YObserve that:v(X : Y ) −1q,t = W(X : tY )w(X : Y ) −1q,t = V (X : Y/t)(x − y/t)(x − y/q)Now let us define:as well as:Θ(X : Y ) q,t = v(X : Y ) q,t W(X : Y ) q,t= R(X : ((t − q) + (q/t − 1))Y )= ∏ x∈Xy∈Y(x − ty)(x − qy/t)(x − qy)(x − y)Observe that:Φ(X : Y ) q,t = V (X : Y ) q,t w(X : Y ) q,t= R(X : ((t/q − 1) + (1/t − 1/q))Y )= ∏ x∈Xy∈Y(x − ty/q)(x − y/t)(x − y)(x − y/q)Φ(X : Y ) q,t = Θ(Y : X) q,t = Θ(X : Y ) 1/q,1/t = Θ(qX : Y ) q,tThese rational functions will play a key role in the proof of the Kawanakaconjecture. We shall suppress the subscripts when no ambiguity canresult.2.2.1. Residue calculations.Lemma 2.1. For any k ≥ 1 we have:∑(Φ(X ′ : X ′′ ) − Φ(X ′′ : X ′ )) = 0X ′ +X ′′ =X|X ′ |=k

54 ROBIN LANGERProof. We wish to show that for all 1 ≤ k < |A| we have:∑ ∏ (x − ty/q)(x − y/t)= ∑ ∏(x − y)(x − y/q)I+J=A|I|=kI+J=A|I|=kx∈Iy∈Jx∈Iy∈JI+J=A|I|=kx∈Iy∈J(x − qy/t)(x − ty)(x − y)(x − qy)Firstly make the substitution α = t/q and β = 1/t to rewrite this as:∑ ∏ (x − αy)(x − βy)(x − y)(x − αβy) = ∑ ∏ (αx − y)(αx − y)(x − y)(αβx − y)i=1j≠iI+J=A|I|=kx∈Iy∈JWrite A = x 1 + x 2 + x 3 + · · · + x n . The k = 1 case reduces to:n∑ ∏ (x i − αx j )(x i − βx j )n∑(x i − x j )(x i − αβx j ) = ∏ (αx i − x j )(βx i − x j )(x i − x j )(αβx i − x j )i=1j≠iNow consider the following contour integral:∮ ∮(1r∏ ∏ n· · ·r!(2πi) k C 1dw 1 · · ·dw rC rw 1 · · ·w ri=1k=1)(w i − αx k )(w i − βx k )(w i − x k )(w i − αβx k ) − 1 ×∏i≠j(w i − αβw j )(w i − w j )(w i − αw j )(w i − βw j )where the contours of integration corresponding to the variables w 1 ,w 2 ,...w rare taken such that the only poles of the integrand inside these contoursare the points x 1 ,...x n . Note that the integrand has no poles atzero or at infinity.When r = 1 the above integral reduces to:∮ ( n)1 dw ∏ (w − αx k )(w − βx k )2πi w (w − x k )(w − αβx k ) − 1Ck=1By Cauchy’s theorem this is equal to the sum over the residues at thepoles inside C 1 which occur precicely when w = x i for some i:( )(1 − α)(1 − β) ∑ ∏ (x i − αx j )(x i − βx j )(1 − αβ) (xi i − x j )(x i − αβx j )j≠iOne can on the other hand consider the poles outside the contour:they are of the form w = αβx i for some i. We obtain:( )(αβ − α)(αβ − β) ∑ ∏ (αβx i − αx j )(αβx i − βx j )(αβ − 1) αβ(αβxii − x j )(αβx i − αβx j )j≠i

SYMMETRIC FUNCTIONS AND MACDONALD POLYNOMIALS 55which simplifies to:(1 − α)(1 − β)−(1 − αβ)( ∑i∏i≠j)(αx i − x j )(βx i − x j )(x i − x j )(αβx i − x j )Noting that the result should be the opposite, we recover the k = 1version of our identity.Consider now the case of general r. We compare once again theresidues inside and outside the contours. The integration over eachvariable w i results in a sum over poles of the form w i = x ai for somea i . It is easy to check that due to the factors w i − w j , the a i must bedistinct. Furthermore, due to the symmetry of the integral by exchangeof the w i , each permutation of the a i produces the same contributionand compensates the factor r!. The result is a sum over subsets I =a 1 + · · · + a r and we obtain∑ 1r∏( (xai − αx ai )(x ai − βx ai )I⊂{1,...,n}|I|=r×x a1 · · ·x ari=1(x ai − αβx ai )n∏ (x ai − αx k )(x ai − βx k )(x ai − x k )(x ai − αβx k )k=1k≠a i) ∏i≠j(x ai − x aj )(x ai − αβx aj )(x ai − αx aj )(x ai − βx aj )which coincides with the left hand side of our identity after obviouscancellations, up to the factor((1 − α)(1 − β)/(1 − αβ)) rA careful analysis of the poles outside the contours shows that onlythe poles of the form w i = αβx ai contribute, so that we find a similarsum:∑ 1r∏( (αβxai − αx ai )(αβx ai − βx ai )I⊂{1,...,n}|I|=r(αβ) r x a1 · · ·x ari=1n∏ (αβx ai − αx k )(αβx ai − βx k )×(αβx ai − x k )(αβx ai − αβx k )k=1k≠a i) ∏i≠j(αβx ai − x ai )(αβx ai − x aj )(αβx ai − αβx aj )(αβx ai − αx aj )(αβx ai − βx aj )which coincides with the right hand side up to the factor( ) r1 (αβ − α)(αβ − β)(αβ) r 1 − αβ□

56 ROBIN LANGERLemma 2.2.Φ(z : X)−1 = 1 − t1 − q∑(w(z : x)Φ(X − x : x) − W(z : x)Φ(x : X − x))x∈XProof. By the k = 1 case of the previous lemma, both sides vanishas z goes to infinity, thus one need only compare the residues at thepoles.□Proposition 2.1.k∑s=0(q;t) s(t;t) s∑X ′ +X ′′ =X|X ′ |=k−s(w(z : X ′′ )V (z : t s−1 X ′′ )W(z : t s X ′ )Φ(X ′ ,X ′′ ))− w(z : X ′ )Φ(X ′′ ,X ′ ) = 0Proof. When k = 1 we have only two terms. When s = 1 we must haveX ′ = ∅ and X ′′ = X thus the s = 1 term reduces to:1 − q1 − t(w(z : X)V (z : X) − 1) =1 − q1 − t(Φ(z : X) − 1)while when s = 0 we must have that |X ′ | = 1. Since we have:w(z : X ′′ )V (z : X ′′ /t) = w(z : X ′′ )w(z : X ′′ ) −1 = 1The s = 0 term becomes:∑(W(z : x)Φ(x : X − x) − w(z : x)Φ(X − x : x))x∈XThus the k = 1 case of the identity is equivalent to the previous lemma.More generally, the left hand side is a rational function in z of degreezero, which vanishes in the limit as z goes to infinity. The poles arelocated at:z ∈ {a k t s ,a k /q : k = 1...m, s = 0...k}By considering separately the terms for which a k ∈ X ′ and those forwhich a k ∈ X ′′ one may check that the residue at the pole z = a k t svanishes for all s, while the residue at the pole a k /q is equivalent tothe k − 1 form of the identity in m − 1 variables.□2.3. Pieri formula and recurrence.

SYMMETRIC FUNCTIONS AND MACDONALD POLYNOMIALS 572.3.1. Arms and legs again. For µsubseteqλ, let C λ/µ denote the set ofboxes of λ in columns which are longer than the corresponding columnsof µ and let R λ/µ denotes the set of boxes of λ which are in rows longerthan the corresponding rows of µ. For example, if λ = (8, 6, 5, 3, 2) andµ = (8, 5, 5, 2, 2),C λ/µ : R λ/µ :Let ˜C λ/µ denote the set of boxes of λ in columns which are the samelength as the corresponding columns of µ, and let ˜R λ/µ denote the setof boxes of λ in rows which are the same length as the correspondingrows of µ. With the same example,˜C λ/µ : ˜Rλ/µ :Now let us define:C λ/µ (q,t) = ∑q aλ(s) t lλ(s) −∑q aµ(s) t lµ(s)s∈C λ/µ s∈µ∩C λ/µR λ/µ (q,t) = ∑q aλ(s) t lλ(s) −∑q aµ(s) t lµ(s)s∈R λ/µ s∈µ∩R λ/µ˜C λ/µ (q,t) = ∑It is clear that we have:as well as that:s∈ ˜C λ/µ(q a λ(s) t l λ(s) − q aµ(s) t lµ(s) )˜R λ/µ (q,t) = ∑s∈ ˜R λ/µ(q a λ(s) t l λ(s) − q aµ(s) t lµ(s) )C λ ′ /µ ′(t,q) = R λ/µ(q,t)˜C λ ′ /µ ′(t,q) = ˜R λ/µ (q,t)B λ (q,t) − B µ (q,t) = C λ/µ (q,t) + ˜C λ/µ (q,t)= R λ/µ (q,t) + ˜R λ/µ (q,t)

58 ROBIN LANGER2.3.2. Pieri formula. The Macdonald polynomials are, in fact, the eigenfunctionsof the more general family of operators [29] (page 315):1 ∑∆ n (X + (1 − t)X I )f(X + (1 − q)X I )∆ n (X)I⊆[n]|I|=rwhere X I is the alphabet:X I = ∑ i∈Ix iwith eigenvalues:e r( n∑i=1q λ it n−i )Using this fact, Macdonald was also able to show that his polynomialssatisfy the following generalization of the Pieri formulae [29] (page340):(P µ (X)Ω z X 1 − t )= ∑ϕ λ/µ (q,t)P λ (q,t)z |λ|−|µ|1 − qλ∈U(µ)Q µ (X)˜Ω z (X) = ∑ϕ ′ λ/µ(q,t)Q λ (X)z |λ|−|µ|λ∈Ũ(µ)where:ϕ λ/µ (q,t) = Ω((q − t)C λ/µ (q,t))ϕ ′ λ/µ(q,t) = Ω((t − q)R λ/µ (q,t))As before U(µ) denotes the set of partitions obtained from µ by addinga horizontal strip while Ũ(µ) denotes the set of partitions obtained fromµ by adding a vertical strip. Note that the latter may be obtained fromthe former by applying the operator ω q,t to both sides and then formallyreplacing λ by λ ′ and µ by µ ′ . The two additional Pieri formulae:(Q µ (X)Ω z X 1 − t )= ∑ψ λ/µ (q,t)Q λ (X)z |λ|−|µ|1 − qλ∈U(µ)P µ (X)˜Ω z (X) = ∑ψ λ/µ(q,t)P ′ λ (X)z |λ|−|µ|λ∈Ũ(µ)

SYMMETRIC FUNCTIONS AND MACDONALD POLYNOMIALS 59where:ψ λ/µ (q,t) = Ω((t − q) ˜C λ/µ (q,t))ψ ′ λ/µ(q,t) = Ω((q − t) ˜R λ/µ (q,t))are obtained from by multiplying, or dividing by Ω((q − t)(B λ (q,t) −B µ (q,t))) as appropriate.2.3.3. Recurrence. Dual to the Pieri formulae, we have the followingrecurrences [29] (pages 346, 348):P λ (X + z) = ∑ψ λ/µ (q,t)P µ (X)z |λ|−|µ|µ∈D(λ)Q λ (X − z) = ∑µ∈ ˜D(λ)Q λ (X + z) = ∑µ∈D(λ)P λ (X − z) = ∑µ∈ ˜D(λ)ψ ′ λ/µ(q,t)Q µ (X)(ǫz) |λ|−|µ|ϕ λ/µ (q,t)Q µ (X)z |λ|−|µ|ϕ ′ λ/µ(q,t)P µ (X)(ǫz) |λ|−|µ|Here D(λ) denotes the set of partitions obtained from λ by removing ahorizontal strip, while ˜D(λ) denotes the set of partitions obtained fromλ by removing a vertical strip.Note that it is also possible to take any one of the recurrences or thePieri formulae, together with the definition of the Macdonald polynomialof a row, or a column accordingly as the definition of the Macdonaldpolynomials.2.4. The Proof.2.4.1. The Schur case. To aid the reader in not getting lost in a thicketof q’s and t’s we first give a proof of the classical Schur identity, whichmirrors, in simplified form, the main steps of the proof of the morecomplicated Kawanaka identity.Proposition 2.2.∑s λ (x 1 ,...,x n+1 ) =λn∏i=11(1 − x i )∏1≤i

60 ROBIN LANGERProof. The proof is by induction on the number of variables. The firststep makes use of the Pieri formula and recurrence for Schur functions.The second step is a bijection between two families of partitions.When n = 1 we get the expansion of the geometric series:∑x k 1 = 11 − x 1k≥0By the recurrence for the Schur functions we have:LHS(n + 1) = ∑ λ= ∑ λ= ∑ µBy the induction hypothesis:RHS(n + 1) ==n+1∏i=1s λ (x 1 ,...,x n+1 )∑µ∈D(λ)1 ∏1 − x is µ (x 1 ,...,x n )x |λ|−|µ|n+1s µ (x 1 ,...,x n ) ∑1≤i

SYMMETRIC FUNCTIONS AND MACDONALD POLYNOMIALS 61By considering the coefficient of s µ (X) on both sides, it suffices todemonstrate that:∑x |λ|−|µ| 1 ∑(12)n+1 =x |µ|−|γ|n+11 − x n+1λ∈U(µ)x |λ|−|µ|n+1 =λ∈Ũ(µ)γ∈D(µ)Conjugating all the partitions involved, this is equivalent to:∑ 1 ∑(13)x |µ|−|γ|n+11 − x n+1γ∈ ˜D(µ)Suppose now that µ is a partition with distinct m distinct nonzeroparts. For each α ⊆ [m] such that |α| = k let µ − (α) denote thepartition obtained from µ by removing a box from the end of each rowindexed by an element of α. We have:˜D k (µ) = {µ − (α),α ⊆ [m], |α| = k}Similarly, let µ + (α) denote the partition obtained from µ by addinga box to the end of each row indexed by α. Let:SŨk(µ) = {µ + (α),α ⊆ [m], |α| = k}There is a natural bijection between ˜D k (µ) and SŨk(µ) which sendµ − (α) to µ + (α).If µ contains repeated parts, then it is no longer true that µ + (α) andµ − (α) will be valid partitions for all α. For example, if µ i = µ i+1 andi ∈ α but (i + 1) ∉ α then µ − (α) will not be a valid partition.Conversely, if i ∉ α while (i + 1) ∈ α then µ + (α) will not be a validpartition. There is still, however a natural bijection between ˜D k (µ)and SŨk(µ), though it is slightly more cumbersome to describe.More generally for each α ⊆ [m] such that |α| = k and for each p ≥ 0let µ + (α,p) denote the partition obtained from µ by adding a box tothe end of each row indexed by an element of α, and then adding pnew rows each of length one. Let:SŨk(µ,p) = {µ + (α,p),α ⊆ [m], |α| = k}

62 ROBIN LANGERUpon consideration of the coefficient of x k n in equation (13), to completethe proof we must show that:(14) |Ũk(µ)| = ∑ s| ˜D k−s (µ)|This follows immediately from the fact that:k⋃Ũ k (µ) = SŨk−p(µ,p)p=0and that for each p we have:|SŨk(µ,p)| = |SŨk(µ, 0)| = | ˜D k (µ)|□2.4.2. Step one. For the Kawanaka identity, let us define:LHS(n) = ∑ Ω ((q − ǫt)B λ (q,t))P λ (X n ;q 2 ,t 2 )λ( 1 − ǫtRHS(n) = Ω1 − q X n + 1 − )t21 − q 2e 2(X n )The case n = 1 is a special case of the q-binomial formula:( ) 1 − ǫtΩ1 − q x 1 = ∑ ( )(q − ǫt)(1 − q n )Ωx n 1(1 − q)n= ∑ nΩ ( (q − ǫt)B (n) (q,ǫt) ) x n 1We have now, by the recurrence for the Macdonald P-functions:LHS(n + 1)= ∑ λ= ∑ λ= ∑ µΩ ((q − ǫt)B λ (q,t))P λ (X n + x n+1 ;q 2 ,t 2 )Ω ((q − ǫt)B λ (q,t)) ∑P µ (X n ,q 2 ,t 2 ) ∑ kµ∈D(λ)x k n+1λ∈U k (µ)Ω((t 2 − q 2 ) ˜C λ/µ (q 2 ,t 2 ))P µ (X n ;q 2 ,t 2 )x |λ|−|µ|n+1∑Ω ((q − ǫt)B λ (q,t)) Ω((t 2 − q 2 ) ˜C λ/µ (q 2 ,t 2 ))

SYMMETRIC FUNCTIONS AND MACDONALD POLYNOMIALS 63while, by the induction hypothesis:( ) 1 − ǫtRHS(n + 1) = Ω1 − q x n+1( 1 − ǫt= Ω1 − q x n+1( 1 − t2RHS(n)Ω)LHS(n)Ω)1 − q 2X nx n+1( )1 − t21 − q 2X nx n+1But, by the Pieri formula for the Macdonald P functions, we have:( )1 − t2LHS(n)Ω1 − q 2X nx n+1= ∑ ( )1 − tΩ ((q − ǫt)B γ (q,t))P γ (X n ;q 2 ,t 2 2) Ω1 − q 2X nx n+1γ= ∑ Ω ((q − ǫt)B γ (q,t)) ∑Ω((q 2 − t 2 )C µ/γ (q 2 ,t 2 ))P µ (X n ;q 2 ,t 2 )x |µ|−|γ|n+1γµ∈U(γ)= ∑ P µ (X n ;q 2 ,t 2 ) ∑ ∑x k n+1 Ω ((q − ǫt)B γ (q,t)) Ω((q 2 − t 2 )C µ/γ (q 2 ,t 2 ))µkγ∈D k (µ)By considering the coefficient of P µ (X;q 2 ,t 2 ) on both sides, and thendividing by Ω((q − ǫt)B µ (q,t)), we need to show that, for any µ wehave:∑ ∑Ω ((q − ǫt)(B λ (q,t) − B µ (q,t))) Ω((t 2 − q 2 ) ˜C λ/µ (q 2 ,t 2 ))kx k n+1λ∈U k (µ)( ) 1 − ǫt= Ω1 − q x n+1 ×∑ ∑Ω ((q − ǫt)(B γ (q,t) − B µ (q,t))) Ω((q 2 − t 2 )C µ/γ (q 2 ,t 2 ))kx k n+1γ∈D k (µ)Observe that when q = ǫt this reduces to equation (12).By conjugating all the partitions involved, and interchanging the rolesof q and t we obtain the dual form:∑ ∑x k n+1 Ω((t − ǫq)(B λ (q,t) − B µ (q,t)))Ω((q 2 − t 2 ) ˜R λ/µ (q 2 ,t 2 ))k( λ∈Ũr(µ) ) 1 − ǫq= Ω1 − t x n+1 ×∑ ∑Ω((t − ǫq)(B γ (q,t) − B µ (q,t)))Ω((t 2 − q 2 )R λ/µ (q 2 ,t 2 ))kx k n+1γ∈ ˜D z(µ)

64 ROBIN LANGERHere we have made use of the fact that B λ ′(t,q) = B λ (q,t) as wellas ˜C λ ′ /µ ′(t,q) = ˜R λ/µ (q,t) and C λ ′ /µ ′(t,q) = R λ/µ(q,t).Note that when q = ǫt this reduces to equation (13).Finally, making use of the q-binomial formula once again, on considerationof the coefficient of x r n+1 on both sides, we must show that forany µ and any r we have:(15)∑Ω ((t − ǫq)(B λ (q,t) − B µ (q,t))) Ω((q 2 − t 2 )R λ/µ (q 2 ,t 2 ))λ∈Ũr(µ)= ∑ Ω((t − ǫq)B (1 r−s )(ǫq,t))×s∑γ∈ ˜D s(µ)Ω ((t − ǫq)(B γ (q,t) − B µ (q,t))) Ω((t 2 − q 2 ) ˜R λ/µ (q 2 ,t 2 ))Let us define:H λ (s) = Ω((q − ǫt)q a λ(s) t l λ(s) ) = 1 + qa λ(s) t l λ(s)+11 − q a λ(s)+1t l λ(s)˜H λ (s) = Ω((t − ǫq)q a λ(s) t l λ(s) ) = 1 + qa λ(s)+1 t l λ(s)1 − q a λ(s)t l λ(s)+1G λ (s) = Ω((t − q)q 2a λ(s) t 2l λ(s) ) = 1 − q2(a λ(s)+1) t 2l λ(s)1 − q 2a λ(s)t 2(l λ(s)+1)We have, by the difference of perfect squares Ω(a 2 ) = Ω(a + ǫa) that:G λ (s)H λ (s) = Ω((t 2 − q 2 )q 2a λ(s) t 2l λ(s) + (q − ǫt)q a λ(s) t l λ(s) )= Ω((t − ǫq)q a λ(s) t l λ(s) )= ˜H λ (s)

SYMMETRIC FUNCTIONS AND MACDONALD POLYNOMIALS 65Now we may simplify equation (15) by writing:and:L(λ,µ) = Ω ((t − ǫq)(B λ (q,t) − B µ (q,t)) Ω((q 2 − t 2 ) ˜R λ/µ (q 2 ,t 2 ))(= Ω (q − ǫt) ˜R)λ/µ (q,t) + (t − ǫq)R λ/µ (q,t)= ∏ H λ (s)Hs∈ ˜R µ (s)λ/µ∏s∈R λ/µ∏ H λ (s)=Hs∈ ˜R µ (s)λ/µ ∩C λ/µ˜Hλ (s)˜H µ (s)∏s∈R λ/µ˜Hλ (s)˜H µ (s)R(µ,γ) = Ω ((t − ǫq)(B γ (q,t) − B µ (q,t))) Ω((t 2 − q 2 ) ˜R µ/γ (q 2 ,t 2 ))(= Ω −(q − ǫt)R µ/γ (q,t) − (t − ǫq) ˜R)µ/γ (q,t)= ∏ H γ (s)H µ (s)s∈R µ/γ= ∏ H γ (s)H µ (s)s∈R µ/γ∏s∈ ˜R µ/γ∏˜Hγ (s)˜H µ (s)s∈ ˜R µ/γ ∩C µ/γ˜Hγ (s)˜H µ (s)To complete the proof we must show that for any partition µ and forany positive integer k we have:∑L(λ,µ) = ∑ (−q;t) s∑(16)R(µ,γ)(t;t)λ∈Ũk(µ) s sγ∈ ˜D k−s (µ)Note that when q = ǫt we have L(λ,µ) = R(µ,γ) = 1 and the abovereduces to equation (14).2.4.3. Step two. Suppose that µ is a partition with m distinct nonzeroparts. For each k ∈ [m] = {1, 2,...m} let a k = q µ k t m−k . For anyK ⊆ [m] let A K denote the alphabet:A K = ∑ k∈Ka kThe next step is to obtain several related combinatorial descriptionsfor the rational function first introduced in section 2.2:Φ(A I ,A J ) ǫq,t = ∏ (a i + t/q a j )(a i − a j /t)(a i − a j )(a i + a j /q)i∈Ij∈J

66 ROBIN LANGERNote that q has been replaced by ǫq, that is every occurence of q hasbeen replaced by its formal (as opposed to plethystic) negative.Let α and ˜α be such that α ∪ ˜α = [m]. Let γ = µ − (α) denote thepartition obtained from µ by removing a box from the end of each rowindexed by α and let ˜γ = µ − (˜α) denote the partition obtained from µby removing a box from the end of each row indexed by ˜α.If s ∈ ˜R µ/γ ∩C µ/γ then s must be of the form s = (i,µ j ) for some j ∈ αand i ∉ α with i < j. In this case we have that a γ (s) = a µ (s) = µ i −µ j ,while l γ (s) + 1 = l µ (s) = j − i.For example:µ = (8, 6, 5, 3, 2) and α = {2, 4}, ˜R µ/γ ∩C µ/γ = {(1,µ 4 ), (1,µ 2 ), (3,µ 4 )} ={(1, 3), (1, 6), (3, 3)}.In this case, one may check that:so that:(17)˜H µ (i,µ j ) = 1 + q a ia −1j1 − t a i a −1j˜H µ− (α)(i,µ j ) = 1 + q/t a ia −1j1 − a i a −1j˜H µ− (α)(i,µ j )˜H µ (i,µ j )= v(a j : a i ) −1ǫq,t= W(a j : a i ) ǫq,t= Θ(a j ,a i ) ǫq,t = Φ(a i ,a j ) ǫq,tSimilarly if s ∈ ˜R µ/˜γ ∩ C µ/˜γ then s must be of the form s = (j,µ i ) forsome j ∉ ˜α and i ∈ ˜α with j < i. now a˜γ (s) = a µ (s) = µ j − µ i , whilel˜γ (s) + 1 = l µ (s) = i − j.For example:

SYMMETRIC FUNCTIONS AND MACDONALD POLYNOMIALS 67µ = (8, 6, 5, 3, 2) and ˜α = {1, 3, 5}, ˜R µ/˜γ ∩C µ/˜γ = {(2,µ 5 ), (2,µ 3 ), (4,µ 5 )} ={(2, 2), (2, 5), (4, 2)}.In this case we have:˜H µ (j,µ i ) = v(a i : a j ) −1ǫq,t˜H µ− (˜α)(j,µ i ) = W(a i : a j ) ǫq,tso that:(18)˜H µ− (˜α)(j,µ i )˜H µ (j,µ i )= Θ(a i ,a j ) ǫq,t = Φ(a j ,a i ) ǫq,tNow let λ = µ + (α) and let ˜λ = µ + (˜α).If s ∈ ˜R λ/µ ∩C λ/µ then we must have s = (i,µ j +1) for some j ∈ α andi ∉ α with i < j. This time we have that a µ (s) = a λ (s) = µ i − µ j − 1,while l µ (s) + 1 = l λ (s) = j − i.For example:If µ = (8, 6, 5, 3, 2) and α = {2, 4} then ˜R λ/µ ∩ C λ/µ = {(1,µ 4 +1), (1,µ 2 + 1), (3,µ 4 + 1)} = {(1, 4), (1, 7), (3, 4)}.Now:so that:(19)H µ+ (α)(i,µ j + 1) = 1 + t/q a ia −1j1 − a i a −1jH µ (i,µ j + 1) = 1 + 1/q a ia −1j1 − 1/t a i a −1jH µ+ (α)(i,µ j + 1)H µ (i,µ j + 1)= V (a j ,a i ) ǫq,t= w(a j : a i ) −1ǫq,t= Φ(a j ,a i ) ǫq,tSimilarly, if s ∈ ˜R˜λ/µ ∩C˜λ/µthen we must have s = (j,µ i +1) for somej ∉ ˜α and i ∈ ˜α with j < i. This time a µ (s) = a λ (s) = µ j − µ i − 1,while l µ (s) + 1 = l λ (s) = i − j

68 ROBIN LANGERFor example:If µ = (8, 6, 5, 3, 2) and ˜α = {1, 3, 5} then ˜R˜λ/µ ∩ C˜λ/µ= {(2,µ 5 +1), (2,µ 3 + 1), (4,µ 5 + 1)} = {(2, 3), (2, 6), (4, 3)}We have:so that:(20)H µ+ (˜α)(j,µ i + 1) = V (a i ,a j ) ǫq,tH µ (j,µ i + 1) = w(a i ,a j ) −1ǫq,tH µ+ (˜α)(j,µ i + 1)H µ (j,µ i + 1)= Φ(a i ,a j ) ǫq,tNote that in equations (17) and (19) we have the restriction that i < jwhile in equations (18) and (20) we have the restriction that j < i.We may combine all these observations into the following:Proposition 2.3. We have:Φ(A I ,A J ) ǫq,t = ∏ i∉αj∈αi

SYMMETRIC FUNCTIONS AND MACDONALD POLYNOMIALS 69Now let µ be the partition obtained from µ by removing a box to theend of every row (or equivalently removing the first column), and let µbe the partition obtained from µ by adding a box to every row.µ =µ = µ =Although the highlighted box has the same arm and leg length inevery partition, it has different co-ordinates with respect to the top lefthand corner, giving us:and similarly:H µ (i − 1,µ j ) = H µ (i,µ j ) = H µ (i + 1,µ j + 1)˜H µ (i − 1,µ j ) = ˜H µ (i,µ j ) = ˜H µ (i + 1,µ j + 1)Observe further that µ − (˜α) = µ + (α) while µ +(˜α) = µ − (α). Usingthese facts we may rewrite the previous proposition as:Proposition 2.4. We have:Φ(A I ,A J ) ǫq,t = ∏ i∉αj∈αi

70 ROBIN LANGERH µ− (α)(j,µ i ) = H µ+ (˜α)(j,µ i+ 1) = H µ+ (˜α)(j,µ i + 1)and in the second case that:˜H µ+ (α)(j,µ i + 1) = ˜H µ− (˜α)(j,µ i ) = ˜H µ− (˜α)(j,µ i )2.4.4. Step three. Using proposition 2.4 we may now give an alternativedescription for the expressions R(µ,γ) and L(µ,γ) arising in equation(16).Proposition 2.5. If γ = µ − (α) then:R(µ,γ) = w(1/t : A J ) ǫq,t Φ(A I ,A J ) ǫq,t□Proof. By equation (17) we can write:∏s∈ ˜R µ/γ ∩C µ/γ˜Hγ (s)˜H µ (s) = ∏ i∉αj∈αi

SYMMETRIC FUNCTIONS AND MACDONALD POLYNOMIALS 71For j ∈ α, we have that H µ (j,k + 1) = H µ− (α)(j,k) unless k = µ i forsome i ∉ α with i > j.In the above diagram µ = (8, 5, 4, 2, 1) and α = {2, 4}. The grey boxescorrespond to the terms in the expression which do not get cancelled.It follows then, by the first part of proposition 2.4, that:∏µ j −1∏j∈α k=1H µ− (α)(j,k)H µ (j,k + 1) = ∏ H µ− (α)(j,µ i )Hj∈α µ (j,µ i + 1)i∉αi>j= ∏ j∈αi∉αi>jΦ(a i ,a j ) ǫq,tIt remains to observe that in the first column we have:∏ 1Hj∈α µ (j, 1) = ∏ 1 − a j1 + taj∈α j /q = w(1/t : A J) ǫq,t□Proposition 2.6. If λ = µ + (α,p) then:L(λ,µ) = (−q;t) p(t;t) p×w(1/t : A I ) ǫq,t V (1/t : t p+1 A I ) ǫq,t W(1/t : t p A J ) ǫq,t Φ(A J ,A I ) ǫq,tProof. The proof is similar to the previous one, only we must be slightlymore careful with the first column. Begin by splitting, into two pieces,the expression:∏ H λ (s)Hs∈ ˜R µ (s)λ/µ ∩C λ/µThe first piece is the contribution from the first column:n∏i=1i∉αH µ+ (α,p)(i, 1)H µ (i, 1)

72 ROBIN LANGERThe second piece is everything else:∏ H µ+ (α,p)(i,µ j + 1)H µ (i,µ j + 1)i∉αj∈αi

SYMMETRIC FUNCTIONS AND MACDONALD POLYNOMIALS 73In the above diagram µ = (8, 5, 4, 2, 1) and α = {2, 4}. Again the greyboxes correspond to the terms which do not get cancelled.We have, by the second part of proposition 2.4, that:µ j∏ ∏˜H µ+ (α)(j,k + 1)˜H µ+ (α)(j,µ i + 1)j∈α k=1˜H µ (j,k)= ∏ j∈αi∉αi>j˜H µ (j,µ i )= ∏ j∈αi∉αi>jΦ(a j ,a i ) ǫq,tIt remains to find suitable expressions for the four remaining factorscoming from the first column:p∏l=1˜H µ+ (α,p)(m + l, 1) = Ω((t − ǫq)(1 + t + · · · + t p−1 )) = (−q;t) p(t;t) p∏˜H µ+ (α,p)(j, 1) = ∏ 1 + qt p a j= W(1/t : t p A1 − t p+1 J ) ǫq,taj∈αj∈α jm∏H µ+ (α,p)(i, 1) =i=1i∉αm∏i=1i∉αm∏i=1i∉αm1H µ (i, 1) = ∏i=1i∉α1 + t/q a i t p1 − a i t p = V (1/t : t p−1 A I ) ǫq,t1 − a i1 + t/q a i= w(1/t : A I ) ǫq,t□Proposition 2.7. The Kawanaka identity is equivalent to the following:k∑s=0(q;t) s(t;t) s∑I∪J=[m]|J|=k−s(w(1/t : A I )V (1/t : t s−1 A I )W(1/t : t s A J )Φ(A J ,A I ))− w(1/t : A J )Φ(A I ,A J ) = 0

74 ROBIN LANGERProof. Equation (16) may be rewritten in the form:∑k∑ (−q;t) s∑L(λ,µ) −R(µ,γ) = 0(t;t) sλ∈Ũk(µ)s=0γ∈ ˜D k−s (µ)If µ is a partition with distinct parts, then, as remarked at the endof the proof of proposition 2.2, we may replace the sum over verticalstrips with a sum over subsets of the rows, to obtain the equivalentexpression:k∑∑s=0 |α|=k−s(L(µ + (α,s),µ) − (−q;t) )sR(µ,µ − (α)) = 0(t;t) sThe result now follows, in this special case, by applying propositions2.5 and 2.6 and then replacing ǫq with q (formal negative).If µ contains a repeated part, and α is such that µ − (α) is not a partitionthen we have a i−1 = ta i for some j = i − 1 ∈ α and i ∉ α and so theexpression:vanishes.Φ(A I ,A J ) = (a i + a j t/q)(a i − a j /t)(a i − a j )(a i + a j /q)Similarly if µ contains a repeated part, and α is such that µ + (α,p) is nota partition then we have 1/ta i = a i+1 for some i ∉ α and j = i + 1 ∈ αand the expression:vanishes.Φ(A J ,A I ) = ∏ i∉αj∈α(a j + a i t/q)(a j − a i /t)(a j − a i )(a j + a i /q)Thus, even when µ contains a repeated part, we may still replace thesum over vertical strips with a sum over subsets of the rows, since theterms which do not correspond to partitions all vanish. The resultfollows.□The above identity is none other than a special case of proposition2.1 with z = 1/t, which completes the proof of the Kawanaka identity:∑ ∏( 1 + qa λ)(s) t l λ(s)+1P1 − q a λ(s)+1t l λ (X;q 2 ,t 2 ) = ∏ (−tx i ;q) ∞∏ (t 2 x i x j ;q 2 ) ∞λ(s)(xi≥1 i ;q) ∞ (xi

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