JACK-LAURENT SYMMETRIC FUNCTIONS

JACK-LAURENT SYMMETRIC FUNCTIONS

arXiv:1310.2462v1 [math-ph] 9 Oct 2013

A.N. SERGEEV AND A.P. VESELOV

Abstract. We develop the general theory of Jack-Laurent symmetric functions, which are certain generalisations of the Jack symmetric functions, depending on an additional parameter p0 .

Contents 1. Introduction 2. Laurent version of CMS operators in infinite dimension 3. Jack-Laurent symmetric polynomials 4. Jack–Laurent symmetric functions 5. Harish-Chandra homomorphism and Polychronakos operator 6. Pieri formula for Jack–Laurent symmetric functions 7. Evaluation theorem 8. Symmetric bilinear form 9. Special case k = −1 : Schur-Laurent symmetric functions 10. Some conjectures and open questions 11. Acknowledgements References

1 4 8 12 14 21 24 28 30 31 32 32

1. Introduction In the late 1960s Henry Jack [9, 10] introduced certain symmetric polynomials Z(λ, α) depending on a partition λ and an additional parameter α, which are now known as Jack polynomials. When α = 1 they reduce to the classical Schur polynomials, so the Jack polynomials can be considered as a one-parameter generalisation of Schur polynomials, whose theory goes back to Jacobi and Frobenius. When α = 2 they are naturally related to zonal spherical functions on the symmetric spaces U (n)/O(n), which was the main initial motivation for Jack. The theory of Jack polynomials was further developed by Stanley [33] and by Macdonald, who also extended them to the symmetric polynomials depending on two parameters, nowadays named after him [12]. Approximately at the same time Calogero [2] and Sutherland [3] initiated the theory of quantum integrable models, describing the interaction particles on the line, which in the classical case were studied by Moser [17]. 1

Although it was not recognised at the time Jack polynomials can be defined as symmetric polynomial eigenfunctions of (properly gauged) version Lk,N of the Calogero–Moser–Sutherland (CMS) operator (N )

Lk

=−

N N X X ∂2 2k(k + 1) + , 2 ∂zi sinh2 (zi − zj ) i=1 i0

a>0

a,b>0

where ∂a = a ∂p∂ a . Some explicit formulas for the higher order CMS integrals at infinity were recently found by Nazarov and Sklyanin in [18, 19]. In the present paper we define and study a Laurent version of Jack symmetric functions - Jack-Laurent symmetric functions and the corresponding infinite-dimensional Laurent analogue of the CMS operator acting on the algebra Λ± freely generated by pa with a ∈ Z \ {0} being both positive and negative. The variable p0 plays a special role and will be considered as an additional parameter. The idea to consider the Laurent polynomial eigenfunctions of CMS operator (1) is quite natural and was proposed already by Sutherland in [32]. The corresponding Laurent polynomials were later discussed in more details by Sogo [29, 30, 31]. However, as it was pointed out by Forrester in his MathSciNet review of the paper [29], in finite dimension it does not have 2

much sense since the corresponding Laurent polynomials always can be reduced to the usual Jack polynomials simply by multiplication by a suitable power of the determinant ∆ = x1 . . . xN . In the infinite-dimensional case one can not do this since the infinite product x1 x2 . . . does not belong to Λ. Moreover, in the Laurent case there is no stability (at least in the same sense as above, since one can not set xi to zero), so the corresponding Jack–Laurent symmetric functions essentially depend on both k and additional parameter p0 , which can be viewed as ”dimension”. Such a parameter appeared already in Jack’s paper [10] as S0 (see page 9 there) and Sogo’s papers, but its importance was probably first became clear after the work of Rains [22], who considered BC-case (see also [25] and [27]). Our main motivation for studying the Jack-Laurent symmetric functions came from the representation theory of Lie superalgebra gl(m, n) and related spherical functions, where these functions play an important role. We will discuss this in a separate publication. The structure of the paper is as follows. In section 2 we introduce the infinite-dimensional Laurent version of the CMS operator X X X Lk,p0 = pa+b ∂a ∂b − k[ pa pb ∂a+b − pa pb ∂a+b ] a,b∈Z

a,b>0

a,b0

a0 . The usual Jack symmetric functions are particular cases corresponding to (k,p ) (k) empty second partition µ: Pλ,∅ 0 = Pλ ∈ Λ. The simplest Laurent example corresponding to two one-box Young diagrams is given by (k,p0 )

P1,1

= p1 p−1 − 3

p0 . 1 + k − kp0

In sections 5-8 we study the Laurent analogues of Harish-Chandra homomorphism, Pieri and evaluation formulas and compute the square norms of (k,p ) Pα 0 for the corresponding symmetric bilinear form on Λ± . Section 9 is devoted to an important special case k = −1, corresponding to Schur-Laurent symmetric functions. We show that the limit Sλ,µ of Jack– (k,p ) Laurent symmetric functions Pλ,µ 0 when k → −1 for generic p0 does exist, does not depend on p0 and can be given by an analogue of Jacobi-Trudy formula. The related symmetric Laurent polynomials (called sometimes symmetric Schur polynomials indexed by a composite partition sµ¯,λ (x)) and their supersymmetric versions play an important role in representation theory of Lie superalgebra gl(m, n) (see [6, 16, 5]). In the last section we discuss some conjectures and open problems. 2. Laurent version of CMS operators in infinite dimension The finite dimensional CMS operators (1) preserve the algebra of symmetric Laurent polynomials ±1 ±1 SN , Λ± N = C[x1 , . . . , xN ]

generated (not freely) by pj (x) = xj1 + · · · + xjN , j ∈ Z. Let us define its infinite-dimensional version - the algebra of Laurent symmetric functions Λ± as the commutative algebra with the free generators pi , i ∈ Z \ {0}. The dimension p0 = 1 + 1 + · · · + 1 = N does not make sense in infinite-dimensional case, so we will add it as an additional formal parameter, which will play a very essential role in what will follow. Λ± has a natural Z-grading, where the degree of pi is i. There is a natural involution ∗ : Λ± → Λ± defined by p∗i = p−i ,

i ∈ Z.

(4)

This algebra can be also represented as Λ± = Λ+ ⊗Λ− , where Λ+ is generated by pi with positive i and Λ− by pi with negative i. Note that the involution ∗ swaps Λ+ and Λ− . For every natural N there is a homomorphism ϕN : Λ± → Λ± N : ϕN (pj ) = xj1 + · · · + xjN , j ∈ Z.

(5)

The involution ∗ under this homomorphism goes to the natural involution −1 on Λ± N mapping xi to xi . Define also the following algebra homomorphism θ : Λ± → Λ± by a ∈ Z \ {0}

θ(pa ) = kpa ,

(cf. [12], formula (10.6)). If we change also k → k −1 , p0 → kp0 then this map becomes an involution. 4

(6)

Now we are going to construct explicitly the infinite dimensional version of CMS operator and higher integrals. Our main tool is an infinite dimensional version of the Dunkl-Heckman operator [8]. Let us remind that the Dunkl-Heckman operator for the root system of the type An has the form k X xi + xj ∂ Di,N = ∂i − (1 − sij ), ∂i = xi , (7) 2 xi − xj ∂xi j6=i

where sij is a transposition, acting on the functions by permuting the coordinates xi and xj . Heckman proved [8] that the differential operators (r)

r r Lk,N = Res (D1,N + · · · + DN,N ),

(8)

where Res means the operation of restriction on the space of symmetric polynomials, commute and give the integrals for the quantum CMS system. We have the following simple, but important Lemma. ±1 Lemma 2.1. The operator Di,N maps the algebra Λ± N [xi ] into itself. ∂ Proof. For the operator ∂i = xi ∂x this is obvious since ∂i (pl ) = lxli . The i operator X xi + xj (1 − sij ) (9) ∆i,N = xi − xj j6=i

acts trivially on the algebra

Λ± N

and has the property

∆i,N (f ∗ ) = −(∆i,N (f ))∗ . ±1 Therefore it is enough to prove that ∆i,N (xli ) ∈ Λ± N [xi ] for l > 0, which follows from the identity X xi + xj X xi + xj ∆i,N (xli ) = (1 − sij )(xli ) = (xl − xlj ) xi − xj xi − xj i j6=i

j6=i

l = xli N + 2xl−1 i p1 + · · · + 2xi pl−1 + pl − 2lxi .

(10) 

Let Λ± [x, x−1 ] be the algebra of Laurent polynomials in x over the algebra Define the differentiation ∂ in Λ± [x, x−1 ] by the formulae

Λ± .

∂(x) = x, ∂(pl ) = lxl , and the operator ∆p0 : Λ± [x, x−1 ] → Λ± [x, x−1 ] by ∆p0 (xl f ) = ∆p0 (xl )f, ∆p0 (1) = 0, f ∈ Λ± , l ∈ Z and ∆p0 (xl ) = xl p0 + 2xl−1 p1 + · · · + 2xpl−1 + pl − 2lxl , l > 0, ∆p0 (xl ) = −(∆p0 (x−l ))∗ , l < 0, where we set x∗ = x−1 . 5

Define the infinite dimensional analogue of the Dunkl-Heckman operator Dk,p0 : Λ± [x, x−1 ] −→ Λ± [x, x−1 ] by 1 Dk,p0 = ∂ − k∆p0 . (11) 2 Introduce also a linear operator Ep0 : Λ± [x, x−1 ] −→ Λ± by the formula Ep0 (xl f ) = pl f, f ∈ Λ, l ∈ Z

(12)

(r)

and the operators Lk,p0 : Λ± −→ Λ± , r ∈ Z+ by (r)

r , Lk,p0 = Ep0 ◦ Dk,p 0

(13)

where the action of the right hand side is restricted to Λ± . We claim that these operators give a Laurent version of quantum CMS integrals at infinity. More precisely, we have the following result. (r)

Theorem 2.2. The operator Lk,p0 is a differential operator of order r with polynomial dependance on p0 and the following properties: (r)

(r)

θ−1 ◦ Lk,p0 ◦ θ = k r−1 Lk−1 ,kp0 , (r)

(14)

(r)

(Lk,p0 )∗ = (−1)r Lk,p0

(15)

where θ is defined by (6). (2) The operator Lk,p0 has the following explicit form: X X X (2) Lk,p0 = pa+b ∂a ∂b − k[ pa pb ∂a+b − pa pb ∂a+b ] a,b∈Z

−kp0 [

a,b>0

X a>0

p a ∂a −

X

pa ∂a ] + (1 + k)

a N , then ϕN (Pµ ) = (k) (ϕN (Pµ ))∗ = 0, so we proved the theorem when |λ| = 0. Let α be a bipartition. Denote by X(α) and Y (α) the sets of bipartitions, which can be obtained from α by adding one box to λ and by deleting one box from µ respectively and define Z(α) = X(α) ∪ Y (α). Similarly to the previous section define for any bipartition α X X X X ep0 (α) = λ2i + µ2j +k (2i−1)λi +k (2j−1)µj −kp0 (|λ|+|µ|) (32) and consider the following polynomial in t, depending rationally on p0 and on an additional parameter s Y t − ep (γ) 0 , Rα (p0 , t, s) := s − ep0 (γ) where the product is over all bipartitions γ ∈ Z(α) such that ep0 (γ) 6= s. 12

Now suppose that theorem is true for all α = (λ, µ) with |λ| ≤ M . Let β = (ν, τ ) be a bipartition such that |ν| = M + 1. Let α be a bipartition obtained from β by removing one box (i, νi ) from ν. Set V (α, β) = Vi (λ), where Vi is defined by formula (27) with i corresponding to the removed box. One can check that if k is not rational and p0 6= n + k −1 m with natural m, n the coefficient V (α, β) is well defined and nonzero. Therefore we can define (k,p ) Pβ 0 = V (α, β)−1 Rα (p0 , L2 , s)(p1 Pα(k,p0 ) ) with s = ep0 (β). Let γ ∈ X(α) with added box (j, λj + 1), then ep0 (β) − ep0 (γ) = 2(λi − λj ) + 2k(i − j). Similarly for γ ∈ Y (α) with deleted box (j, µj ) ep0 (β) − ep0 (γ) = 2(λi + µj ) + 2k(i + j − 1) − 2kp0 . From the previous formula we see that Pβ is well defined if p0 6= n + k −1 m, m, n ∈ Z>0 and (k,p0 )

ϕN (Pβ

) = V (α, β)−1 Rα (N, Lk,N , s)(ϕN (p1 )ϕN (Pα(k,p0 ) )).

Now we are going to compare two polynomials Rα (N, t, s) and RχN (α) (t, s). If l(α) < N then l(β) ≤ N and Rα (N, t, s) = RχN (α) (t, s), Vi (χN (α)) = V (α, β). By induction assumption and proposition 3.5 we have (k,p0 )

ϕN (Pβ

(k)

) = PχN (β) (x1 , . . . , xN ).

If l(α) = l(β) = N then there exists γ ∈ X(α) with the added box (l(λ)+1, 1) and it is easy to check that Rα (N, t, s) = Rχ(α) (t, s)

t − eN (γ) , s − eN (γ)

Lk,N − cN (γ) (k) (k) Pχ(β) = Pχ(β) , s − cN (γ) (k,p )

(k)

which by proposition 3.5 imply that ϕN (Pβ 0 ) = Pχ(β) (x1 , . . . , xN ). Suppose now that l(β) > N and consider two cases: l(α) > N and l(α) = (k,p ) N . In the first case by induction ϕN (Pα ) = 0, therefore ϕN (Pβ 0 ) = 0. In the second case we have again equality Rα (N, t, s) = RχN (α) (t, s), but this time s = eN (β) 6= eN (χ(α) + εj ), 1 ≤ j ≤ N and according to proposition 3.5 RχN (α) (N, Lk,N , s)(ϕN (p1 )ϕN (Pα(k,p0 ) )) = 0. This proves the existence. The uniqueness follows from the same arguments as in the proof of lemma 2.3.  We will show in the next section that the Jack-Laurent symmetric func(k,p ) (r) tions Pα 0 are the eigenfunctions of the CMS operators Lk,p0 . 13

Remark 4.2. The usual definition of Jack symmetric functions uses the basis of monomial symmetric functions. The problem in the Laurent case is that the monomial symmetric functions (corresponding to k = 0) also depend on the additional parameter p0 and the very existence of them (which can be proven in a similar way) is not quite obvious. Having this basis one can define the Jack-Laurent symmetric functions using the CMS operator and show that they have a triangular decomposition in the monomial basis with coefficients polynomially depending on p0 . This implies also that the Jack-Laurent symmetric functions form a basis in Λ± (p0 ). Here is the explicit form of the Jack-Laurent symmetric functions in the simplest cases: p0 (k,p ) P1,1 0 = p1 p−1 − , (33) 1 + k − kp0 2(p0 − 1) 1 (k,p ) p1 , (34) P12 ,1 0 = (p21 − p2 )p−1 − 2 2 + 4k − 2kp0 where 12 denotes partition λ = (1, 1). Define the involution w on the bipartitions by w(α) = w(λ, µ) := (µ, λ). Corollary 4.3. For any bipartition α the corresponding Jack-Laurent symmetric functions satisfy the property (k,p )

Pα(k,p0 )∗ = Pw(α)0 . Proof. Choose N ≥ l(α), then we have (k)∗

(k)

(k)

(k,p )

ϕN (Pα(k,p0 )∗ ) = PχN (α) = Pw(χN (α)) = PχN (w(α)) = ϕN (Pw(α)0 ), which implies the claim.



5. Harish-Chandra homomorphism and Polychronakos operator Recall that the usual Harish-Chandra homomorphism ψN : DN (k) −→ Λ∗N maps this algebra onto the algebra of shifted symmetric polynomials Λ∗N and there are natural homomorphims ϕN,N −1 : DN (k) −→ DN −1 (k). Consider the inverse limit D(k) = lim DN (k), ←

which we will call the algebra of stable CMS integrals, and the inverse limit of Harish-Chandra homomorphisms ψN ψ : D(k) −→ Λ∗ , where Λ∗ = lim← Λ∗N is the algebra of shifted symmetric functions [20]. 14

Algebra D(k) can be naturally considered as a subalgebra of the algebra of differential operators acting on Λ. It has a natural extension D(k)[p0 ] acting as differential operators on Λ polynomially depending on p0 , with Harish-Chandra homomorphism ψ acting trivially on p0 : ψ(p0 ) = p0 . We claim that D(k, p0 )[p0 ] = D(k)[p0 ]. To prove this we will need the following version of the Dunkl operator, which was introduced by Polychronakos [21]: X xi ∂ ˜ i,N , πi,N := xi −k (1 − sij ) = ∂i − k ∆ (35) ∂xi xi − xj j6=i

where ˜ i,N := ∆

X j6=i

xi (1 − sij ). xi − xj

Note the difference with Dunkl-Heckman operator: 1 X (1 − sij ). Di,N = πi,N + k 2

(36)

j6=i

The action of the operator πi,N on

± Λ± N [xi ]

is described by

∂i (xli ) = lxli , ∂i (pl ) = lxli ,  l−1 l  xi (N − l) + xi + · · · + xi pl−1 , l > 0 ˜ i,N (xli ) = 0, l = 0 ∆   l lxi + xl+1 i p−1 + · · · + pl , l < 0.

(37)

Polychronakos used this to give an alternative way to construct the CMS integrals. Theorem 5.1. [21] The operators (r)

Ik,N =

N X

r πi,N , r = 1, 2, . . .

(38)

i=1

commute with each other. Their restrictions to Λ± N are the quantum integrals (r) (r) Ik,N = Res Ik,N of the corresponding CMS system. r : The proof is simple and based on the commutation relations between πi,N

[πi,N , πj,N ] = k(πi,N − πj,N )sij .

(39) (r)

(r)

Now we would like to express Heckman’s CMS integrals Lk,N via Ik,N . Proposition 5.2. We have r Res Di,N = Res

r X a=0

15

r−a (r) πi,N fa ,

where Res as before is the restriction on symmetric polynomials Λ± N. The (r) operator coefficients fa satisfy the following recurrent relations 1 (r) fa(r+1) = fa(r) + kN fa−1 , a 6= r + 1, 2 (r+1)

fr+1

(40)

r

1 1 X r−a (r) = kN fr(r) − k Ik,N fa 2 2

(41)

a=0

(0)

with initial data f0

(r)

= 1 and fa = 0 when a < 0 or a > r. (b)

b b −I Proof. It is easy to see that Res Si πi,N = N Res πi,N k,N , where X Si = (1 − sij ). j6=i

By inductive assumption we have r Res Di,N = Res

r X

r−a (r) πi,N fa .

a=0

Therefore r+1 = Res Res Di,N

r X a=0

r X 1 r−a+1 (r) fa πi,N (πi,N + kSi )πir−a fa(r) = Res 2 a=0

r r r X 1 X 1 X (r−a) r−a+1 (r) r−a (r) r−a +Res k πi,N fa +Res k −Ik,N )fa(r) Si πi,N fa = Res (N πi,N 2 2 a=0

a=0

= Res

r X

a=0

r

r−a+1 (r) πi,N (fa

a=0

X (r−a) 1 1 (r) + kN fa−1 ) + Res k(N fr(r) − Ik,N fa(r) ), 2 2 a=0

which implies (40),(41).



Corollary 5.3. Two series of CMS integrals are related as follows (r)

Lk,N =

r X

(r−a)

Ik,N fa(r) =

a=0

r−1 X

(r−a)

Ik,N fa(r−1) .

(42)

a=0

Proof. Indeed, using (40),(41) we have (r) Lk,N

=

r X

(r−a) Ik,N fa(r)

=

a=0

r−1 X a=0

1 (r−a) (r−1) Ik,N (fa(r−1) + kN fa−1 ) 2

r−1

r−1

a=0

a=0

X (r−a−1) X (r−a) 1 1 (r−1) + kN 2 fr−1 − kN Ik,N fa(r−1) = Ik,N fa(r−1) 2 2 since

(0) Ik,N

= N.

 16

Let us define now the infinite dimensional analogue of the Polychronakos operator on Λ± [x] by the formulas ˜ p , ∂(xl ) = lxl , ∂(pl ) = lxl , l ∈ Z, πk,p0 = ∂ − k ∆ 0  l l−1  x (p0 − l) + x p1 + · · · + xpl−1 , l > 0 ˜ p (xl ) = 0, l = 0 ∆ 0   l lx + xl+1 p−1 + · · · + pl , l < 0. Let Ep0 : Λ± [x, x−1 ] −→ Λ± be the operator defined by formula (12) above and consider the following set of infinite dimensional CMS integrals (r)

r . Ik,p0 = Ep0 ◦ πk,p 0

(43)

Their commutativity follows from the same arguments as in Theorem 2.2. Theorem 5.4. Let D1 and D2 be the algebras generated by two sets of CMS (1) (2) (1) (2) integrals Lk,p0 , Lk,p0 , . . . and Ik,p0 , Ik,p0 , . . . respectively. Then D1 [p0 ] = D2 [p0 ] (r)

Proof. Define the operators fa

recursively by

1 (r) fa(r+1) = fa(r) + kp0 fa−1 , a 6= r + 1, 2 (r+1) fr+1

(44)

r

1 1 X (r−a) (r) = kp0 fr(r) − k Ik,p0 fa 2 2

(45)

a=0

(0)

(r)

with initial data f0 = 1 and fa = 0 when a < 0 or a > r. One can check that for such operators we have the relation r Dk,p = 0

r X

r πk,p f (r) 0 a

a=0

and hence the relation between the corresponding CMS integrals (r) Lk,p0

=

r X

(r−a)

Ik,p0 fa(r) .

(46)

a=0 (r)

Since f0

(r)

= 1 we can reverse these formulas to express polynomially Ik,p0

(a)

via Lk,p0 with a ≤ r.



It turns out that all three algebras D(k), D1 , D2 of CMS integrals generate the same algebra if we allow the coefficients to depend polynomially on p0 . Theorem 5.5. D(k)[p0 ] = D2 [p0 ]. 17

Proof. We claim first that on the algebra Λ[x] (π + kp0 )r ∂ = lim(πi,N + kN )r ∂i,N . ←−

(47)

Indeed, the operator ∂i,N maps algebra ΛN [xi ] into the ideal J generated by xi , while the operator πi,N + kN maps the ideal J into itself (see formula (37)). Therefore the operator (πi,N +kN )r ∂i,N maps the algebra ΛN [xi ] into the ideal J and does not depend on p0 . Consider the operators (r)

Hi,N := (πi,N + kN )r−1 πi,N : ΛN −→ ΛN [xi ]. From (47) it follows that they are stable and the inverse limit can be naturally identified with (r)

Hk := (π + kp0 )r−1 π : Λ −→ Λ[x]. Thus the integral (r) Hk

=E◦

(r) Hk

 r−1  X r−1 (j) = (kp0 )r−j Ik,p0 j j=1

(r)

is stable and therefore belongs to D(k). Since Hk D(k) the theorem follows.

generate the algebra 

Let p∗r,a = lim p∗r,a,N ←

be the inverse limit of shifted power sums (24). Theorem 5.6. The Harish-Chandra homomorphism ψ : D(k)[p0 ] −→ Λ∗ [p0 ] transforms the involution ∗ on the algebra D(k)[p0 ] to the involution w on Λ∗ [p0 ] defined by w(p∗r,a ) = (−1)r p∗r,k−kp0 −a . (48) More precisely, the following diagram is commutative: ∗

D(k)[p0 ] −→ D(k)[p0 ] ↓ψ ↓ψ w ∗ ∗ Λ [p0 ] −→ Λ [p0 ]. Proof. Let ϕN be defined by (17) and use the same notation for its action on D(k)[p0 ]. Define also ϕ∗N : Λ∗ [p0 ] → Λ∗N by setting p0 = N and ϕ∗N (p∗r,0 ) = p∗r,0,N . We have the following commutative diagram ψ

D(k)[p0 ] −→ Λ∗ [p0 ] ↓ ϕN ↓ ϕ∗N DN (k)

ψ

N −→

18

Λ∗N

so ϕ∗N (ψ(D∗ ) = ψN (ϕN (D∗ )) = wN (ψN (ϕN (D))). Therefore we only need to prove that ϕ∗N (w(p)) = wN (ϕ∗N (p)) for any p ∈ Λ∗ [p0 ]. It is enough to prove this for shifted power sums, when this reduces to the identity p∗r,a,N (w(χ)) = (−1)r p∗r,k−kN −a,N (χ), which is easy to check. Theorem is proved.



Λ∗ [p

We can consider the elements from 0 ] as functions on bipartitions. Namely for any bipartition α = (λ, µ) define a homomorphism fα : Λ∗ [p0 ] −→ C by fα (p∗r ) = p∗r (λ) + w(p∗r )(µ), (49) ∗ ∗ ∗ where pr = pr,0 . Define for any p ∈ Λ [p0 ] p(α) := fα (p). Corollary 5.7. Jack-Laurent symmetric functions are the eigenfunctions of the algebra D(k)[p0 ] and for any D ∈ D(k)[p0 ] we have DPα(k,p0 ) = ψ(D)(α)Pα(k,p0 ) .

(50)

Proof. Let us apply ϕN to both sides of (50). We have for N ≥ l(λ) + l(µ) (k)

ϕN (DPα(k,p0 ) ) = ϕN (D)ϕN (Pα(k,p0 ) ) = ϕN (D)PχN (α) (k)

(k)

= ψN (ϕN (D))(χN (α))PχN (α) = ϕ∗N (ψ(D))(χN (α))PχN (α) . On the other hand we have (k)

ϕN (ψ(D)(α)Pα(k,p0 ) ) = ϕN (ψ(D)(α))ϕN (Pα(k,p0 ) ) = ϕN (ψ(D)(α))PχN (α) , so we only need to prove that ϕN (ψ(D)(α)) = ϕ∗N (ψ(D))(χN (α)). It is enough to check this for ψ(D) = p∗r . We have ϕN (p∗r (α)) = ϕN (p∗r (λ)) + ϕN (w(p∗r )(µ)) = ϕN (p∗r (λ)) + wN (ϕN (p∗r (µ)) = p∗r,N (χN (λ)) + p∗r,N (wN (χN (µ))). Since N ≥ l(λ) + l(µ) we have χN (α) = χN (λ) + wN (χN (µ)). It is easy to see that since for any 1 ≤ i ≤ N , λi wN (µ)i = 0 we have p∗r,N (χN (λ)) + p∗r,N (wN (χN (µ))) = p∗r,N (χN (α)) = ϕ∗N (pr )(χN (α)), which completes the proof.



Now we can use this to prove the following important result. Theorem 5.8. If k ∈ / Q and kp0 6= m + nk for all m, n ∈ Z>0 then the spectrum of the algebra D(k)[p0 ] of quantum CMS integrals on Λ± is simple. 19

Proof. Consider the following shifted symmetric functions ∞ X bl (k, a)(x) = [Bl (xi + k(i − 1) + a) − Bl (k(i − 1) + a)] ,

(51)

i≥1

where Bl (z) are the classical Bernoulli polynomials [34] and a is a parameter. They generate the algebra of shifted symmetric functions Λ∗ [p0 ]. Lemma 5.9. We have the following formula X X bl (k, a)(α) = s c(, 0)s−1 + (−1)s s c(, 1 + k − kp0 )s−1 , ∈µ

∈λ

where for  = (ij) we define c(, a) := (j − 1) + k(i − 1) + a. Proof. We have Bl (z) =

l X s=0

  l bls z , bls = Bl−s , s s

where Bj are the Bernoulli numbers. This implies bl (k, a) = thus by (48) X w(bl (k, a)) = bls (−1)s p∗s,k−kp0 −a .

P

∗ s bls ps,a

and

s

Using the standard property of Bernoulli polynomials Bl (−x) = (−1)l Bl (1+ x), (see [34]) we have X X bls (−1)s p∗s,k−kp0 −a = bls p∗s,1+k−kp0 −a . s

s

Therefore w(bl (k, a)) = (−1)l bk,1+k−kp0 and lemma follows from the equality X bl (λ, k, a) = l [(j − 1) + k(i − 1) + a]l−1 .

(52)

(i,j)∈λ

 Let us assume now that bs+1 (α) = bs+1 (˜ α). Then we have X X X X c(x, 0)s +(−1)s c(y, 1+k−kp0 )s = c(x, 0)s +(−1)s c(y, 1+k−kp0 )s . x∈λ

˜ x∈λ

y∈˜ µ

y∈µ

If this is true for all s, then the sequences (c(x, 0), −c(y, 1 + k − kp0 ))x∈λ,y∈˜µ ,

(c(x, 0), −c(y, 1 + k − kp0 ))x∈λ,y∈µ ˜

coincide up to a permutation. Therefore we have two possibilities: c(x, 0) = c(˜ x, 0) where x ∈ λ, x ˜ ∈ ν, or c(x, 0) = −c(˜ y , 1+k−kp0 ) where x ∈ λ, y˜ ∈ µ. In the first case we have for x = (ij), x ˜ = (˜i, ˜j) that j − ˜j + k(i − ˜i) = 0, so ˜ ˜ j = j, i = i since k is not rational. 20

In the second case we have for y˜ = (˜i, ˜j) that kp0 = j + ˜j − 1 + k(i + ˜i − 1), which contradicts to our assumption, since both j + ˜j − 1 and i + ˜i − 1 are positive integers.  Corollary 5.10. Jack–Laurent symmetric functions obey the following θduality property (k−1 ,kp0 ) θ−1 (Pα(k,p0 ) ) = dα Pα0 (53) 0 0 0 with some constants dα = dα (k, p0 ) and α = (λ , µ ). Proof. Indeed, because of the symmetry property (14) of quantum inte(q) (k,p ) grals Lk,p0 the function θ−1 (Pα 0 ) is also an eigenfunction of the operator (q)

Lk−1 ,kp0 with the same eigenvalue up to a constant. Now the claim follows from the lemma (5.9), the duality property bl (λ, k, a) = k l−1 bl (λ0 , k −1 , k −1 a) and the simplicity of the spectrum for generic k and p0 . An explicit form of the constants dα (k, p0 ) is given below by (66).  6. Pieri formula for Jack–Laurent symmetric functions Let α be a bipartition represented by a pair of Young diagrams λ and µ. Define for any positive integers i, j the following functions cλ (ji, a) = λi − j − k(λ0j − i) + a, cα (ji, a) = λi + j + k(µ0j + i) + a, where λ0 as before is the Young diagram conjugated (transposed) to λ. Let x = (ij) be a box such that the union λ + x := λ ∪ x ∈ P is also a Young diagram and similarly to the Pieri formula for Jack polynomials [12] define i−1 Y cλ (jr, 1)cλ (jr, −2k) . (54) V (x, α) = cλ (jr, −k)cλ (jr, 1 − k) r=1

If x can not be added to λ we assume that the corresponding V (x, α) = 0. Similarly, if the box y = (ij) can be removed from the Young diagram µ in the sense that µ − y := µ \ y ∈ P we define l(µ)

U (y, α) =

Y cµ (jr, 1 + k)cµ (jr, −k) cµ (jr, 1)cµ (jr, 0)

r=i+1 l(λ)

×

Y r=1

×

cα (jr, −1 − k(p0 + 2))cα (jr, −kp0 ) cα (jr, −1 − k(p0 + 1))cα (jr, −k(p0 + 1))

(j − 1 + k(l(λ) + µ0j − p0 − 1))(j + k(µ0j − l(µ))) , (j + k(l(λ) + µ0j − p0 ))(j − 1 + k(µ0j − l(µ) − 1))

(55)

where l(λ) is the length, which is the number of non-zero parts in partition λ. If y = (ij) can not be removed from µ we define U (y, α) = 0. 21

i

Πλ,µ

j

Figure 1. Diagrammatic representation of a pair of partitions The following theorem follows from the Pieri formula for Jack-Laurent polynomials (28). (k,p0 )

Theorem 6.1. The Jack-Laurent symmetric functions Pλ,µ = Pα α = (λ, µ) satisfy the following Pieri formula: X X p1 Pλ,µ = V (x, α)Pλ+x,µ + U (y, α)Pλ,µ−y . x

with (56)

y

One can rewrite the formula in terms of the following diagrammatic representation of a bipartition α = (λ, µ) (cf. [14]). Consider the following geometric figure Y = Yλ,µ = Yλ ∪ Y−µ ∪ Πλ,µ , where Yλ = {(ji) | j, i ∈ Z, 1 ≤ i ≤ l(λ), 1 ≤ j ≤ λi }, Y−µ = {(ji) | j, i ∈ Z, −l(µ) ≤ i ≤ −1, −µi ≤ j ≤ −1} and Πλ,µ = {(ji) | j, i ∈ Z, 1 ≤ i ≤ l(λ), −l(µ0 ) ≤ j ≤ −1}. On Fig. 1 we have the corresponding representation for λ = (6, 5, 4, 2, 1) and µ = (7, 3, 2, 1, 1). Note that for λ we follow the French way of drawing Young diagram, for µ it is rotated by 180 degrees. Define the following analogues of rows ( λi , 1 1≤ i ≤ l(λ) yi = −µ−i , −l(µ) ≤ i ≤ −1 22

i

π3 π1 j

π2

Figure 2. Summation sets for the Pieri formula and columns yj0

( λ0j , 1 ≤ j ≤ l(λ0 ) = −µ0−j , −l(µ0 ) ≤ j ≤ −1

with all other yi , yj0 being zero. For every box  with integer coordinates (j, i) define the function cY (, a) = yi − j − k(yj0 − i) + a. Define for the added box  = (j, i) the following subset in Yλ π1 = {(j, r) | 1 ≤ r < i} and for deleted box  = (ji) the subsets in Y π2 = {(j, r) | −l(µ) ≤ r < −µ0−j }, π3 = {(j, r) | 1 ≤ r ≤ l(λ)}. The meaning of these subsets is clear from Fig. 2, where the deleted box is black and the added box is crosshatched. In these terms the Pieri formula (56) can be written as X X V (, α)Pλ∪,µ + U (, α)Pλ,µ\ (57) p1 Pλ,µ = 



with V (, α) =

Y ∈π1

cY (, −2k)cY (, 1) , cY (, −k)cY (, 1 − k)

1 Y cY (, −1 − k)cY (, k) U (, α) = cY (, −1)cY (, 0) ∈π2

23

(58)

µ

λ

Figure 3. Complementary Young diagrams λ and µ ×

Y ∈π3

cY (, −1 − k(p0 + 2))cY (, −kp0 ) cY (, −1 − k(p0 + 1))cY (, −k(p0 + 1))

(j + 1 + k(yj0 − l(λ) + p0 + 1))(j + k(yj0 + l(µ)) × (j + k(yj0 − l(λ) + p0 ))(j + 1 + k(yj0 + l(µ) + 1)

(59)

with the convention that the product over empty set is equal to 1. A non-symmetry between λ and µ is due to the choice of p1 in the left hand side of the Pieri formula (56). By applying ∗-involution to formula (56) one has the corresponding formula for p−1 , where the roles of λ and µ are interchanged. Another remark is that in the Pieri formula (57) one can replace the rectangle containing the figure Y by any bigger rectangle with −M ≤ i ≤ L by changing in formula (59) the lengths l(λ) and l(µ) to L and M respectively. 7. Evaluation theorem Consider a pair of Young diagrams λ and µ which can be joint together to form a × b rectangle (see Fig. 3): λi + µb−i+1 = a, λ0j + µ0a−j+1 = b. We will call such two diagrams complementary. Define the following function on Young diagrams depending on two parameters p and x: Y j − 1 + k(i − 1 − p) + x (60) ϕp (λ, x) = λi − j + k(i − 1 − λ0j ) + x (i,j)∈λ

=

Y (i,j)∈λ

λi − j + k(i − 1 − p) + x λi − j + k(i − 1 − λ0j ) + x 24

(61)

with the assumption that for empty Young diagram ϕp (∅, x) = 1. Such a function was first introduced by Stanley [33] in the theory of Jack polynomials. Lemma 7.1. For any pair of complementary Young diagrams λ and µ forming a × b rectangle ϕb (λ, x) = ϕb (µ, x). Proof. By induction in a + b. If a + b = 2 then we have λ = (1), µ = ∅ or the other way around. Therefore −k + x = 1 = ϕ1 (µ, x) −k + x Let now a + b > 2. There are two cases: first when λ1 = a or µ1 = a and the second when λ01 = b or µ01 = b. Let us consider the first case. By symmetry we can assume that λ1 = a and set ν = λ \ λ1 . Then we have Y j − 1 + k(i − 1 − b) + x Y νi − j + k(i − 1 − νj0 ) + x ϕb (λ) = ϕb−1 (ν) λi − j + k(i − 1 − λ0j ) + x j − 1 + k(i − b) + x ϕ1 (λ, x) =

(i,j)∈λ

(i,j)∈ν

= ϕb ((λ1 ), x)ϕb−1 (ν, x)ϕb−1 (ν, x)−1 = ϕb ((λ1 ), x) =

a Y j − 1 − kb + x . a − j − kλ0j + x

j=1

Now Y j − 1 + k(i − 1 − b) + x Y µi − j + k(i − 1 − µ0j ) + x ϕb (µ) = ϕb−1 (µ) µi − j + k(i − 1 − µ0j ) + x j − 1 + k(i − b) + x (i,j)∈µ

(i,j)∈µ µ0

j a Y Y j − 1 + k(i − 1 − b) + x Y j − 1 + k(i − 1 − b) + x = = j − 1 + k(i − b) + x j − 1 + k(i − b) + x

j=1 i=1

(i,j)∈µ

=

a Y j=1

a Y j − 1 − kb + x j − 1 − kb + x = . 0 j − 1 + k(µj − b) + x a − j − kλ0j + x j=1

where in the last row we have made the change j → a − j + 1 and use the equality λ0j + µ0a−j+1 = b in the denominator. Thus we see that ϕb (λ) ϕb (µ) = . ϕb−1 (ν) ϕb−1 (µ) Since by induction ϕb−1 (ν) = ϕb−1 (µ) we have ϕb (λ) = ϕb (µ) in that case. Consider now the second case. Set ν = λ \ λ01 . As before we can assume that λ01 = b. By inductive hypothesis ϕb (µ, x) = ϕb (ν, x). The equality ϕb (ν, x) = ϕb (λ, x) is clear from the second expression (61) for ϕb (λ, x).  Lemma 7.2. Let λ, µ be two partitions, a ≥ µ1 and N ≥ l(λ) + l(µ) and ν = (λ1 + a, . . . , λr + a, a, . . . , a, a − µs , . . . , a − µ1 ) ∈ P. Then ϕN (ν, x) = ϕN (λ, x)ϕN (µ, x)ϕN (λ, µ, x) 25

(62)

ν1 = λ ν2

ν3 µ

Figure 4. Three parts of the Young diagram ν. where ϕp (λ, µ, x) is given by the formula l(λ) l(µ0 )

ϕp =

Y Y λi + j − 1 + k(i − 1 − p) + x j − 1 + k(i − 1 + µ0j − p) + x . j − 1 + k(i − 1 − p) + x λi + j − 1 + k(i − 1 + µ0j − p) + x i=1 j=1

(63) Proof. Let τ ⊂ λ be some subset of λ ∈ P. Define ψp (λ, τ, x) similarly to ϕp (λ, x) as Y λi − j + k(i − 1 − p) + x ψp (λ, τ, x) = . (64) λi − j + k(i − 1 − λ0j ) + x (i,j)∈τ

Split the Young diagram ν into three parts as follows: ϕN (ν, x) = ψN (ν, ν1 , x)ψN (ν, ν2 , x)ψN (ν, ν3 , x) where ν1 = {(ij) | 1 ≤ i ≤ l(λ), a + 1 ≤ j ≤ a + λi } ν2 = {(ij) | 1 ≤ i ≤ l(λ), 1 ≤ j ≤ a} ν3 = {(ij) | 1 ≤ j ≤ a, l(λ) + 1 ≤ i ≤ N − µ0a−j+1 } (see Fig. 4). We have (using the second formula (61) for ϕp (λ, x)) ϕN (ν, x) = ψN (ν, ν1 , x)ψN (ν, ν2 , x)ψN (ν, ν3 , x). It is easy to see that ψN (ν, ν1 , x) = ϕN (λ, x), ϕN (ν2 ∪ ν3 , x) ψN (ν, ν3 , x) = ψN (ν2 ∪ ν3 , ν3 , x) = ψN (ν2 ∪ ν3 , ν2 , x) 26

1

But according to lemma 7.1 we have ϕN (ν2 ∪ ν3 , x) = ϕN (µ, x) since ν2 ∪ ν3 is complementary to µ. We have also Y νi − j + k(i − 1 − N ) + x ψN (ν, ν2 , x) = = νi − j + k(i − 1 − νj0 ) + x (ij)∈ν2

l(λ) a YY i=1 j=1

λi + a − j + k(i − 1 − N ) + x = λi + a − j + k(i − 1 − N + µ0a−j+1 ) + x

l(λ) a YY i=1 j=1

λi + j − 1 + k(i − 1 − N ) + x λi + j − 1 + k(i − 1 − N + µ0j ) + x

where we made the change j → a − j + 1. Now we only need to compute ψN (ν2 ∪ ν3 , ν2 , x). But this product we can get from the previous one by setting λi = 0 to have ψN (ν2 ∪ ν3 , ν2 , x) =

l(λ) a YY i=1 j=1

j − 1 + k(i − 1 − N ) + x . j − 1 + k(i − 1 − N + µj ) + x

Taking a = l(µ0 ) we have the claim.



Now we are ready to prove the main result of this section. Theorem 7.3. (Evaluation Theorem) Let εp0 be the homomorphism Λ± → C defined by ε(pi ) = p0 , i ∈ Z. Then the evaluation of the Jack-Laurent (k,p ) symmetric function Pλ,µ 0 can be given by (k,p0 )

εp0 (Pλ,µ

) = ϕp0 (λ, 0)ϕp0 (µ, 0)ϕp0 (λ, µ, 0),

(65)

where the functions ϕp (λ, x), ϕp (λ, µ, x) are defined by formulae (60), (63). Proof. Denote by r(λ, µ, k)(p0 ) the right hand side of the formula (65). Ac(k) cording to Stanley [33] for the usual Jack polynomials Pλ we have (k)

εp0 (Pλ ) = ϕp0 (λ, 0). (k,p0 )

For fixed λ, µ and k (assumed to be generic) the evaluation εp0 (Pλ,µ

) is a

(k,p ) εp0 (Pλ,µ 0 )

rational function of p0 . We need to prove that = r(λ, µ, k)(p0 ). Since both sides are rational functions we only need to verify this for large enough integers p0 = N ∈ Z>0 . But in that case we have (k,N )

εN (Pλ,µ ) = εN (Pν (k, N )) = ϕN (ν, 0) = r(λ, µ, k)(N ) according to lemma 7.2.

 (k,p0 )

Corollary 7.4. Jack–Laurent symmetric functions Pα duality (k,p0 )

θ−1 (Pλ,µ

(k−1 ,kp0 )

) = dλ,µ (k, p0 )Pλ0 ,µ0 27

,

satisfy the θ-

where as before θ is defined by θ(pa ) = kpa and (k,p0 )

εp0 (Pλ,µ

dλ,µ (k, p0 ) =

)

(k−1 ,kp0 ) λ0 ,µ0

εkp0 (P

.

(66)

)

(k,p )

(k−1 ,kp )

Proof. We know from Corollary 5.10 that θ−1 (Pλ,µ 0 ) = dλ,µ Pλ0 ,µ0 0 for some constants dλ,µ . Applying to both sides the evaluation homomorphism εkp0 and using θ−1 (pi ) = k −1 pi we have (k,p0 )

εkp0 ◦ θ−1 (Pλ,µ and thus (k,p0 )

εp0 (Pλ,µ

(k,p0 )

) = εp0 (Pλ,µ

),

(k−1 ,kp0 )

) = dλ,µ εkp0 (Pλ0 ,µ0

),

which implies (66).

 8. Symmetric bilinear form

Let us fix the parameter k, which we assume in this section to be negative real. We start with the finite-dimensional case. The original CMS operator is clearly formally self-adjoint with respect to the standard scalar product Z (ψ1 , ψ2 ) = ψ1 (z)ψ¯2 (z)dz with the standard Lebesgue measure dz = dz1 . . . dzN on RN . After the gauge ψ = f Ψ0 and change xj = e2zj we naturally come to the following symmetric bilinear form for the Laurent polynomials f, g ∈ Λ± N Z (f, g) := cN (k) f (x)g ∗ (x)∆N (x, k)dx (67) TN

TN

where is the complex torus with |xj | = 1, i = 1, . . . , N , dx is the Haar −1 measure on T N , g ∗ (x1 , . . . , xN ) = g(x−1 1 , . . . , xN ), and ∆N (x, k) =

N Y

(1 − xi /xj )−k

(68)

i,j:j6=i

(cf. Macdonald [12], p.383, who is using parameter α = −1/k). The normalisation constant cN (k) is chosen in such a way that (1, 1)N = 1: Z −1 cN (k) = ∆N (x, k)dx. TN

Note that for negative real k the integral (67) is clearly convergent for all Laurent polynomials f, g and that on the Laurent polynomials with real coefficients (in particular, for Jack-Laurent polynomials with real k) the product (67) coincides with the Hermitian scalar product Z (f, g) := cN (k) f (x)¯ g (x)∆N (x, k)dx. TN

28

Since the eigenfunctions of a self-adjoint operator are orthogonal the Jack (k) polynomials Pλ (x1 , . . . , xN ) are orthogonal with respect to the product (67). Using formulae (10.37), (10.22) from [12] we have (k)

(k)

(Pλ , Pλ ) =

Y λi − j − k(λ0j − i) + 1 j − 1 + k(i − 1) − kN , λi − j − k(λ0j − i) − k j + ki − kN

(i,j)∈λ

which can be rewritten in our notations as (k)

(k)

(Pλ , Pλ ) =

ϕN (λ, 0) . ϕN (λ, 1 + k)

(69) (k)

We can extend now this formula to the Jack-Laurent polynomials Pχ for any integer non-decreasing sequence χ = (χ1 , . . . , χN ) by adding a large a to all its parts to make them positive. Note that both ϕN (λ, 0) Rand ϕN (λ,∗1 + k) do not change under this operation and thata the integral T N f (x)f (x)∆(x, k)dx is invariant under f (x) → (x1 . . . xN ) f (x), so this procedure is well-defined. Now let’s look at the infinite-dimensional case. Theorem 8.1. There exists a unique symmetric bilinear form ( , )p0 on Λ± rationally dependent on p0 such that Jack-Laurent symmetric functions (k,p ) Pα 0 are orthogonal and (ϕN (Pα(k,N ) ), ϕN (Pα(k,N ) )) = (Pα(k,N ) , Pα(k,N ) )N

(70)

for all sufficiently large N, where the product in the left hand side is defined by (67). The corresponding square norm of the Jack-Laurent symmetric (k,p ) function Pα 0 with bipartition α = (λ, µ) is equal to (Pα(k,p0 ) , Pα(k,p0 ) )p0 =

ϕp0 (λ, 0)ϕp0 (µ, 0)ϕp0 (λ, µ, 0) . ϕp0 (λ, 1 + k)ϕp0 (µ, 1 + k)ϕp0 (λ, µ, 1 + k)

(71)

Proof. The uniqueness is obvious since the rational function is determined by its values at sufficiently large integers. To prove the existence we simply check that the formula (71) defines the symmetric bilinear form satisfying (70). We have according to (31) that ϕN (Pα(k,N ) ) = (x1 . . . xN )−a Pν(k) (x1 , . . . , xN ), so that (ϕN (Pα(k,N ) ), ϕN (Pα(k,p0 ) )) = (Pν(k) , Pν(k) ). By (69) we have (Pν(k) , Pν(k) ) =

ϕN (ν, 0) , ϕN (ν, 1 + k)

which by formula (62) from lemma 7.2 coincides with the right hand side of (71) for p0 = N.  29

Note that in contrast to the usual Jack case [12] the bilinear form ( , )p0 is not positive definite on real Laurent symmetric functions, as it follows from (71). In order to have positive definite form one should send p0 to infinity, see the last section. 9. Special case k = −1 : Schur-Laurent symmetric functions The case k = −1 is very important for representation theory of Lie superalgebra gl(n, m) (see [15, 5]). In this case the corresponding Jack-Laurent symmetric functions (whose existence is not obvious) do not depend on p0 , as one can see already in the simplest case p0 (k,p ) P1,1 0 = p1 p−1 − . 1 + k − kp0 Proposition 9.1. The limit (k,p0 )

Sλ,µ := lim Pλ,µ k→−1

does exist for generic p0 and does not depend on p0 . We will call Sλ,µ the Schur-Laurent symmetric functions. The image of these functions under the homomorphism ϕN coincide with the symmetric Schur polynomials sµ¯,λ (x) indexed by a composite partition µ ¯; λ (see [16] for a brief history of these polynomials and their role in representation theory). Here are the two simplest examples of Schur-Laurent symmetric functions 1 S1,1 = p1 p−1 − 1, S12 ,1 = (p21 − p2 )p−1 − p1 . 2 (k,p0 )

Proof. By induction on |λ|. When λ = ∅ then P∅,µ

(k)∗

= Pµ

(k)∗

, where Pµ (−1)

is the usual Jack symmetric function. It is well known that Pµ is welldefined (recall that k = −1 corresponds to α = 1 in Jack’s notations) and coincide with Schur symmetric function Sµ (see e.g [12]). To prove the induction step one can use the Pieri formula (56). The left hand side is well defined at k = −1 by induction assumption. Restrict the CMS operator Lk,p0 onto the invariant subspace generated by the linear combinations of the Jack–Laurent symmetric functions in the right hand side of Pieri formula for generic values of the parameters. One can check analysing proof of Theorem 3.1 that for k = −1 and generic p0 the corresponding eigenvalues E1 , . . . , Ek are distinct. This means that the component V (x, α)Pλ+x,µ = Q(Lk,p0 )(p1 Pλ,µ ) with polynomial Q(E) = C

k Y

(E − Ej ), C −1 =

j=2

k Y

(E1 − Ej ),

j=2

where E1 is the eigenvalue corresponding to Pλ+x,µ . Since Lk,p0 is polynomial in parameters and E1 6= Ej the product V (x, α)Pλ+x,µ is well defined for k = −1 and generic p0 . Since the coefficients V (x, α) tend to 1 when k → −1 this means that Pλ+x,µ is well-defined as well. This proves the existence of 30

Schur–Laurent symmetric functions for generic p0 ; their independence on p0 follows from the Laurent version of Jacobi–Trudy formula below.  Let hi ∈ Λ ⊂ Λ± , i ∈ Z be the complete symmetric functions [12] for i ≥ 0 and hi = 0 for i < 0. Define h∗i as the image of hi under the ∗-involution in Λ± . Theorem 9.2. The Schur-Laurent symmetric functions Sα , α = (λ, µ) can be given by the following Jacobi–Trudy formula as (r + s) × (r + s) determinant, where r = l(λ), s = l(µ) are the number of parts in λ and µ: ∗ ∗ ∗ h h . . . h µ µs −1 µs −s−r+1 s .. .. .. .. . . . . ∗ ∗ h∗ hµ1 +s−2 . . . hµ1 −r µ +s−1 1 Sα = (72) hλ1 −s+1 . . . hλ1 +r−1 hλ1 −s .. .. .. .. . . . . hλr −s−r+1 hλr −s−r+2 . . . h λr Proof. Applying the homomorphisms ϕN : Λ± → Λ± N we have in the left hand side by definition ϕN (Sλ,µ ) = (x1 . . . xN )a Sν (x1 , . . . , xN ), where νi = χN (α)i + a with any integer a ≥ µ1 and χN (α) defined by (30). Now the proof follows from the results of Cummins and King (see formulae (3.7), (3.8) in [6] or (1.21),(1.23) in [16]), who used the language of composite Young diagrams.  10. Some conjectures and open questions The usual Jack symmetric functions can be defined using the following scalar product in Λ defined in the standard basis pλ = pλ1 pλ2 . . . by Y < pλ , pµ >= (−k)−l(λ) j mj mj !δλ,µ , j≥1

where mj is the number of parts of λ equal to j (see [12], p. 305). It is known (see e.g. [12], p. 383) that this scalar product is the limit of the scalar product (71) restricted on Λ when p0 → ∞. An interesting question is what happens on Λ± . We believe that the limit ( , )∞ of the indefinite bilinear form (71) does exist and is positive definite on real Laurent symmetric functions for real negative k. More precisely, we conjecture that the limits of the Jack-Laurent symmetric functions Pα(k,∞) := lim Pα(k,p0 ) p0 →∞

exist for all k ∈ / Q. Then by (71) they would provide an orthogonal basis in ± Λ with (Pα(k,∞) , Pα(k,∞) )∞ = Φ(λ, k)Φ(µ, k), (73) 31

where

Y λi − j + 1 + k(i − λ0j ) Φ(λ, k) = , λi − j + k(i − 1 − λ0j )

(74)

(i,j)∈λ

which can be checked to be positive for k < 0 and all λ. Note that in contrast to the Jack polynomial case the Laurent polynomials (k) (k)∗ pλ,µ = pλ p∗µ , as well as the products of Jack symmetric functions Pλ Pµ , are not orthogonal with respect to ( , )∞ . What are the transition matrices (k,∞) between these bases and the Jack-Laurent basis Pλ,µ ? In the theory of Jack symmetric functions [12] it is known that the product (k) A(λ)Pλ with Y A(λ) = (λi − j + k(i − 1 − λ0j )) (ij)∈λ

depends on k polynomially. We conjecture that a similar fact is true for Jack-Laurent symmetric functions, namely that the product (k,p )

(k,p0 )

Jλ,µ 0 := A(λ, µ)A(λ)A(µ)Pλ,µ

(75)

is polynomial in k and p0 , where l(λ) l(µ0 )

A(λ, µ) =

YY

(j − 1 + k(i − 1 − p0 ))(λi + j − 1 + k(i − 1 + µ0j − p0 )).

i=1 j=1 (k,p )

A weaker version of this conjecture is that the product A(λ, µ)Pλ,µ 0 is polynomial in p0 . The case of special parameters k and p0 with p0 = n + mk −1 is very important for the representation theory of Lie superalgebras and will be discussed in our forthcoming paper. 11. Acknowledgements This work was partly supported by the EPSRC (grant EP/J00488X/1). ANS is grateful to Loughborough University for the hospitality during the autumn semesters 2012 and 2013. References [1] H. Awata Hidden algebraic structure of the Calogero-Sutherland model, integral formula for Jack polynomial and their relativistic analog. Calogero-Moser-Sutherland models (Montreal, QC, 1997), 23–35, CRM Ser. Math. Phys., Springer, New York, 2000. [2] F.Calogero Solution of the one-dimensional N-body problems with quadratic and/or inversely quadratic pair potentials. J. Math. Phys. 12, 1971, 419-436. [3] B. Sutherland Exact results for a quantum many-body problem in one dimension. Phys. Rev. A 4, 1971, 2019-2021. [4] Calogero-Moser-Sutherland models (Montreal, QC, 1997), 23–35, CRM Ser. Math. Phys., Springer, New York, 2000. [5] J. Comes, B. Wilson Deligne’s category Rep(GLδ ) and representations of general linear supergroups. arXiv:1108.0652 32

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[31] K. Sogo Excited states of Calogero-Sutherland-Moser model. II. Complete set of solutions. J. Phys. Soc. Japan 63 (1994), no. 3, 879–884. [32] B. Sutherland Exact results for a quantum many-body problem in one dimension. II Phys. Rev. A 5 (1972), 1372 - 1376. [33] R. Stanley Some combinatorial properties of Jack symmetric functions. Adv. Math. 77 (1989), no. 1, 76–115. [34] E. T. Whittaker and G. N. Watson A Course of Modern Analysis, Cambridge University Press, 1963. Department of Mathematics, Saratov State University, Astrakhanskaya 83, Saratov, 410012, Russia and Department of Mathematical Sciences, Loughborough University, Loughborough LE11 3TU, UK E-mail address: [email protected] Department of Mathematical Sciences, Loughborough University, Loughborough LE11 3TU, UK and Moscow State University, Moscow 119899, Russia E-mail address: [email protected]

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