Hankel Hyperdeterminants, Rectangular Jack Polynomials and Even Powers of the Vandermonde Hacene Belbachir, Adrien Boussicault, Jean-Gabriel Luque

Hankel hyperdeterminants, rectangular Jack polynomials and even powers of the Vandermonde Hacene Belbachir, Adrien Boussicault, Jean-Gabriel Luque

To cite this version:

Hacene Belbachir, Adrien Boussicault, Jean-Gabriel Luque. Hankel hyperdeterminants, rectangular Jack polynomials and even powers of the Vandermonde. Journal of Algebra, Elsevier, 2008, 320 (11), pp.3911-3925. hal-00173319

HAL Id: hal-00173319 https://hal.archives-ouvertes.fr/hal-00173319 Submitted on 19 Sep 2007

HAL is a multi-disciplinary open access L’archive ouverte pluridisciplinaire HAL, est archive for the deposit and dissemination of sci- destinée au dépôt et à la diffusion de documents entific research documents, whether they are pub- scientifiques de niveau recherche, publiés ou non, lished or not. The documents may come from émanant des établissements d’enseignement et de teaching and research institutions in France or recherche français ou étrangers, des laboratoires abroad, or from public or private research centers. publics ou privés. hal-00173319, version 1 - 19 Sep 2007 ipetoe endfra for defined one, simplest a-zoa le,Agei.eal hacenebelbachir@gma Algérie. email: Alger, Bab-Ezzouar n rpsdsvrletnin ohge iesoa rasu arrays dimensional determinant higher for to name notation extensions modern several the proposed introduced he after Few Introduction 1 apr-og 75 an-aVleCdx2 mi:Adrien email: 2. Cedex Marne-la-Valle 77454 Gaspard-Monge apr-og 75 an-aVleCdx2 mi:Jean-G email: 2. Cedex Marne-la-Valle 77454 Gaspard-Monge dmninlsaeby space -dimensional aklhpreemnns etnua Jack rectangular hyperdeterminants, Hankel ∗ † ‡ nvri´ e cecse el ehooi oaiBoumed Houari Technologie la de et Université Sciences des nvri´ ePrsEtMrel-al,Isiu d’ Institut Marne-la-Valle, Paris-Est de Université nvri´ ePrsEtMrel-al,Isiu d’ Institut Marne-la-Valle, Paris-Est de Université fteee oe fteVnemnei emo akpolynomia Jack of term in expr Vandermonde the an of give power we even application the an of As hyperdeterminants. Hankel hyperdeterminant eivsiaeteln ewe etnua akpolynomia Jack rectangular between link the investigate We oyoil n vnpwr fthe of powers even and polynomials .Belbachir H. Det M 3 ] h oincniee eei paetythe apparently is here considered notion The 4]. [3, .Boussicault A. , ∗ etme 9 2007 19, September = Vandermonde k hodrtensor order th n 1 ! σ =( Abstract σ 1 , X ··· 1 ,σ k ) ∈ S n k † M sign( n .G Luque J.-G. and ( = lcrnqee d’Informatique et Electronique lcrnqee d’Informatique et Electronique ´ ´ il.com σ ) M M [email protected] [email protected] ee P3 SH 16111 USTHB 32 BP iene. i 1 σ ··· , i k ) 1 ≤ drtesame the nder i 1 ,...,i 2,Cayley [2], s sand ls ession ‡ k ≤ ls. n nan on where sign(σ) = sign(σ1) ··· sign(σk) is the product of the signs of the per- σ S mutations, M = Mσ1(1)...σk(1) ··· Mσ1(n)...σk(n) and n denotes the symmetric group. Note that others hyperdeterminants are found in literature. For example, those considered by Gelfand, Kapranov and Zelevinsky [7] are bigger polynomials with geometric properties. Our hyperdeterminant is a special case of the riciens [26, 21] with only alternant indices. For an hypermatrice with an even number of indices, it generates the space of the polynomial in- variants of lowest degree. Easily, one obtains the nullity of Det when k is odd ×k and its invariance under the action of the group SLn . Few references exists on the topics see [26, 29, 30, 11, 1, 8] and before [22] the multidimensional analogues of Hankel determinant do not seem to have been investigated. In this paper, we discuss about the links between Hankel hyperdeter- minsants and Jack’s symmetric functions indexed by rectangular partitions. Jack’s symmetric functions are a one parameter (denoted by α in this paper) generalization of Schur functions. They were defined by Henry Jack in 1969 in the aim to interpolate between Schur fonctions (α = 1) and zonal polynomials (α = 2) [15, 16]. The story of Jack’s polynomials is closely related to the generalizations of the Selberg integral [1, 12, 14, 19, 27, 22]. The relation between Jack’s polynomials and hyperdeterminants appeared implicitly in this context in [22], when one of the author with J.-Y. Thibon gave an expression of the Kaneko integral [12] in terms of Hankel hyperdeterminant. More recently, Matsumoto computed [25] an hyperdeterminantal Jacobi-Trudi type formula for rectangular Jack polynomials. The paper is organized as follow. In Section 2, after we recall defini- tions of Hankel and Toeplitz hyperdeterminants, we explain that an Hankel hyperdeterminant can be viewed as the umber of an even power of the Van- n n n dermonde via the substitution Y : x → Λ (Y) where Λ (Y) denotes the nth elementary symmetric functions on the alphabet Y. Section 3 is devoted R to the generalization of the Matsumoto formula [25] to almost rectangular Jack polynomials. In Section 4, we give an equality involving the substitution Y and skew Jack polynomials. As an application, we give in Section 5 expressions of even powers of the Vandermonde determinant in terms of R Schur functions and Jack polynomials.

2 2 Hankel and Toeplitz Hyperdeterminants of symmetric functions

2.1 Symmetric functions Symmetric functions over an alphabet X are functions which are invariant under permutation of the variables. The C-space of the symmetric functions over X is an algebra which will be denoted by Sym(X). Let us consider the complete symmetric functions whose generating series is 1 σ (X) := Si(X)ti = , t 1 − xt i X X xY∈ the elementary symmetric functions

i i −1 λt(X) := Λ (X)t = (1 + xt)= σ−1(X) , i X X xY∈ and power sum symmetric functions

ti ψ (X) := Ψi(X) = log(σ (X)). t i t i X When there is no algebraic relation between the letters of X, Sym(X)isa free (associative, commutative) algebra over complete, elementary or power sum symmetric functions

Sym = C[S1,S2, ··· ]= C[Λ1, Λ2, ··· ]= C[Ψ1, Ψ2, ··· ].

As a consequence, the algebra Sym(X)(X being inﬁnite or not) is spanned by the set of the decreasing products of the generators

Sλ = Sλn ...Sλ1 , Λλ =Λλn . . . Λλ1 , Ψλ =Ψλn . . . Ψλ1 , where λ =(λ1 ≥ λ2 ≥···≥ λn) is a (decreasing) partition. The algebra Sym(X) admits also non multiplicative basis. For example, the monomial functions deﬁned by

m X xλ1 xλn , λ( )= i1 ··· in X 3 xλ1 xλn x ,...,x where the sum is over all the distinct monomials i1 ··· in with i1 in ∈ X, and the Schur functions deﬁned via the Jacobi-Trudi formula

λi−i+j Sλ(X) = det(S (X)). (1) Note that the Schur basis admits other alternative deﬁnitions. For example, it is the only basis such that 1. It is orthogonal for the scalar product deﬁned on power sums by

hΨλ, Ψµi = δλ,µzλ (2)

where δµ,ν is the Kronecker symbol (equal to 1 if µ = ν and 0 otherwise) mi(λ) and zλ = i i mi(λ)! if mi(λ) denotes the multiplicity of i as a part of λ. Q 2. The coeﬃcient of the dominant term in the expansion in the monomial basis is 1,

Sλ = mλ + uλµmµ. Xµ<λ

When X = {x1,...,xn} is ﬁnite, a Schur function has another determi- nantal expression det(xλj +n−j) S (X)= i , λ ∆(X) X where ∆( )= i

Hk (X) = Det Λi1+···+i2k (X) , (3) n 0≤i1,...,i2k≤n−1 where Λm(X) is the mth elementary function on the alphabet X. Let us consider a shifted version of Hankel hyperdeterminants

k i +···+i +v H (X) = Det Λ 1 2k i1 (X) . (4) v 0≤i1,...,i2k≤n−1 4 n where v =(v0,...,vn−1) ∈ Z . Note that (3) implies M0,...,0 =Λ0(X) = 1 by convention. But if M0,...,0 =06 , 1, this property can be recovered using a suit- able normalization and, if M0...0 = 0, by using the shifted version (4) of the Hankel hyperdeterminant. As in [25], one deﬁnes To¨eplitz hyperdeterminant by giving directly the shifting version

k i +···+i −(i +···+i )+v T (X) = Det Λ 1 k k+1 2k i1 (X) . (5) v 0≤i1,...,i2k≤n−1 To¨eplitz hyperdeterminants are related to Hankel hyperde terminants by the following formulae.

− k kn(n 1) k H X 2 T X Proposition 2.1 1. v ( )=(−1) v+(k(n−1))n ( )

− k kn(n 1) k T X 2 H X 2. v( )=(−1) v+(k(1−n))n ( ). Proof The equalities (1) and (2) are equivalent and are direct consequences of the deﬁnitions (4) and (5),

k X i1+···+ik+(n−1−ik+1+···+(n−1−i2k−k(n−1)) X Tv( ) = Det Λ ( ) kn(n−1) 2 H n X = (−1) v+(k(1−n)) ( ). .

2.3 The substitution xn → Λn(Y)

Let X = {x1,...,xn} be a ﬁnite alphabet and Y be another (potentially inﬁnite) alphabet. For simplicity we will denote by Y the substitution R xp =Λp(Y), Y Z for each x ∈ X and each p ∈ Z. The main tool of this paper is the following proposition.

Proposition 2.2 For any integer k ∈ N −{0}, one has

1 X 2k k Y ∆( ) = Hn( ) n! Y Z X where ∆( )= i

2k X 2k j−1 σ1(i)+···+σ2k(i)−2k ∆( ) = det xi = sign(σ1 ··· σ2k) xi . σ1,··· ,σ ∈Sn i X2k Y Hence, applying the substitution, one obtains 1 1 2k σ1(i)+···+σ2k(i)−2k ∆(X) = sign(σ1 ··· σ2k) Λ (Y) n! Y n! S Z σ1,··· ,σ2k∈ n i k Y X Y = Hn( ).

More generally, the Jacobi-Trudi formula (1) implies the following result.

Proposition 2.3 One has 1 X X 2k Hk Y Sλ( )∆( ) = reversen(λ)( ), n! Y Z where reversen(v)=(vn,...,v1) if v = (v1,...,vp) is a composition with p ≤ n and vp+i =0 for 1 ≤ i ≤ n − p. Proof It suﬃces to remark that

2k−1 X X 2k λn−j+1+j−1 j−1 Sλ( )∆( ) = det(xi ) det xi , and apply the same computation than in the proof of Propositi on 2.2.

Example 2.4 If k =1 then using the second Jacobi-Trudi formula

′ λ − −i+j Sλ = det(Λ n i ) (6) where λ′ denotes the conjugate partition of λ, Proposition 2.3 implies

− 1 2 n(n 1) Sλ(X)∆(X) =(−1) 2 S(λ+(n−1)n)′ (Y). n! Y Z

6 3 Jack Polynomials and Hyperdeterminants

In this section, we will consider the symmetric functions as a λ-ring endowed with the operator Si, and we will use the definition of addition and multiplication of alphabets in this context (see e.g. [18]). Let X and Y be two alphabets, the symmetric functions over the alphabet X + Y are generated by the com- i X Y X Y X Y i X Y i plete functions S ( + ) defined by σt( + )= σt( )σt( )= i S ( + )t . 2 If X = Y, one has σt(2X) := σt(X + X) = σt(X) . Similarly one defines X X α X P X σt(α )= σt( ) . In particular, the equality σt(− )= x(1 − xt)= λ−t( ) gives Si(−X)=(−1)iΛi(X). The product of two alphabet X and Y is de- XY i XY i 1 Q XY fined by σt( ) = S ( )t = x∈X y∈Y 1−xyt . Note that σ1( ) = K(X, Y)= S (X)S (Y) is the Cauchy Kernel. λ λ Pλ Q Q P 3.1 Jack polynomials One considers a one parameter generalization of the scalar product (2) defined l(λ) by hΨλ, Ψµiα = δλ,µzλα , where l(λ)= n denotes the length of the partition (α) λ = (λ1 ≥ ··· ≥ λn) with λn > 0. The Jack polynomials Pλ are the (α) unique symmetric functions orthogonal for h , iα and such that Pλ = mλ + (α) µ<λ uλµ mµ. Note that in the case when α = 1, one recovers the definition P (1) S Q(α) P (α) Pof Schur functions, i.e. λ = λ. Let( λ ) be the dual basis of ( λ ). The (α) (α) polynomials Pλ and Qλ are equal up to a scalar factor and the coefficient of proportionality is computed explicitly in [24] VI. 10:

P (α) α(λ − j)+ λ′ − i +1 b(α) λ P (α),P (α) −1 i j . λ := (α) = h λ λ i = ′ (7) Q α(λi − j +1)+ λj − i λ (i,jY)∈λ ∨ Let X = {x1, ··· , xn} be a ﬁnite alphabet and denote by X the alphabet of −1 −1 the inverse {x1 ,...,xn }. Let us introduce the second scalar product by

′ 1 ∨ −1 1 hf,gi = C.T.{f(X)g(X ) (1 − x x ) α }, n,α n! i j Yi6=j (α) (α) see [24] VI. 10. The polynomials Pλ and Qλ are also orthogonal for this scalar product. For simplicity, we will consider also another normalisation deﬁned by (α),n Y (1/α) (1/α) ′ (α) Y Rλ ( ) := hPλ′ , Qλ′ i 1 Qλ ( ). n, α

7 (α),n Note that the polynomial Rλ is not zero only when l(λ) ≤ n and in this (1/α) (1/α) ′ case the value of the coeﬃcient hPλ′ , Qλ′ i 1 is known to be n, α

1 (1/α) (1/α) ′ 1 n + α (j − 1) − i +1 −1 α P ′ , Q ′ . . x x h λ λ in, 1 = j C T 1 − i j α n! ′ n + α − i ( ) (i,jY)∈λ Yi6=j (8) see [24] VI 10.

3.2 The operator and almost rectangle Jack poly- Y nomials Z

Suppose that X = {x1,...,xn} is a ﬁnite alphabet. Let Y = {y1, ···} be another (potentially inﬁnite) alphabet and consider the integral

1 I (Y)= C.T.{Λn(X∨)p+k(n−1)Λl(X∨) (1 + x y )∆(X)2k}. (9) n,k n! i j Y Note that,

kn(n−1) (−1) 2 n ∨ p l ∨ xi k In,k(Y)= C.T.{Λ (X ) Λ (X ) (1 + xiyj) (1 − ) }. n! xj i6=j Y Y (10) But n X∨ p l X∨ (1/k) X∨ Λ ( ) Λ ( )= P(p+1)lpn−l ( ), (11) and (1/k) X (k) Y (1 + xiyj)= Qλ ( )Qλ′ ( ). (12) Y Xλ (α) (α) Hence, from the orthogonality of Pλ and Qλ , equalities (10), (11) and (12) imply kn(n−1) (k),n Y 2 Y In,k( )=(−1) Rnpl ( ). (13) On the other hand, one has the equality

m −m x =Λm(Y) = C.T.{x (1 + xyi)}. Y i Z Y 8 n(n−1) ∆(X) X∨ 2 n X m l X X Remarking that ∆( )=(−1) Λn(X)n−1 , and Λ ( ) Λ ( )= S(mn)+(1l)( ) for each m ∈ Z, Equality (9) can be written as

− 1 2k kn(n 1) k Y n l X X 2 T X In,k( ) = S((p−k(n−1)) )+(1 )( )∆( ) = (−1) pn−l(p+1)l ( ). n! Y Z (14) One deduces an hyperdeterminantal expression for a Jack polynomial indexed by the partition npl. Proposition 3.1 For any positive integers n, p, l and k, one has. (k),n k Rnpl = Tpn−l(p+1)l . The constant term appearing in (8) is a special case of the the Dyson Con- jecture [6]. The conjecture of Dyson has been proved the same year indepen- dently by Gunson [10] and Wilson [31] ( in 1970 I. J. Good [9] have shown an elegant elementary proof involving Lagrange interpolation),

−1 ai a1 + ··· + an C.T. 1 − xixj = , a1,...,an i6=j Y for a1,...,an ∈ N. Hence, one has −1 (k) kn k Q p (Y)= n! κ(n, p, l; k)T n−l l (Y) n l k, ··· ,k p (p+1) where p n j + k(i − 1) l p +1+ k(n − i) κ(n, p, l; k)= . j − 1+ ki p + k(n − i + 1) i=1 j=1 i=1 Y Y Y In particular, when l = 0, one recovers a theorem by Matsumoto. Corollary 3.2 (Matsumoto [25]) −1 (k) kn k P p (Y)= n! T n (Y). n k,...,k p Proof From equalities (8) and (7), one has

(1/k) (1/k) ′ 1 kn (k) (k) hP n , Q n i = hP p ,P p i . p p 1/k,n n! k,...,k n l n l k Applying Proposition 3.1, one ﬁnds the result. Setting p = k(n − 1), one obtains the expression of an Hankel hyperdeterminant as a Jack polynomials.

9 Corollary 3.3

kn(n−1) k (−1) 2 kn (k) H (Y)= P − (Y). n n! k,...,k nk(n 1) 1 3.3 Jack polynomials with parameter α = k

Let Y = {y1,y2, ···} be a (potentially inﬁnite alphabet). Consider the endomorphism deﬁned on the power sums symmetric functions Ψp(Y) by p−1 ωα(Ψp(Y)) := Ψp(−αY)=(−1) αΨp(Y) (see [24] VI 10), where Y = {−y1, −y2, ···} . This map is known to satisfy the identities

( 1 ) (α) Y α Y ωαPλ ( )= Qλ′ ( ) and n n n ωαΛ (Y)= g 1 (Y):=Λ (−αY). ( α )

Applying ωk on Proposition 3.1, one obtains the expression of a Jack polyno- 1 l n−l mial with parameter α = k for an almost rectangular shape λ =(p +1) p as a shifted Toeplitz hyperdeterminant whose entries are

− i1+···+ik−ik+1−···−i2k+λn i1+1 Mi1...i2k = g 1 . k

Proposition 3.4 One has

−1 1 ( k ) kn (k) P − (Y)= n! κ(n, p, l; k)T − (−kY). (p+1)lpn l k,...,k pn l(p+1)l Proof It suﬃces to apply Proposition 3.1 with the alphabet −αY to ﬁnd

−1 (k) kn (k) Q p (−kY)= n! κ(n, p, l; k)T − (−kY). n l k,...,k pn l(p+1)l The result follows from

( 1 ) (k) Y (k) Y k Y Qnpl(−k )= ωkQnpl( )= P(p+1)lpn−l ( ).

10 4 Skew Jack polynomials and Hankel hyperdeterminants

4.1 Skew Jack polynomials Let us deﬁne as in [24] VI 10, the skew Q functions by

(α) (α) (α) (α) (α) hQλ/µ,Pν i := hQλ ,Pµ Pν i. Straightforwardly, one has

(α) (α) (α) (α) (α) Qλ/µ = hQλ ,Pν Pµ iQν . (15) ν X Classically, the skew Jack polynomials appear when one expands a Jack polynomial on a sum of alphabet.

Proposition 4.1 Let X and Y be two alphabets, one has

(α) X Y (α) X (α) Y Qλ ( + )= Qµ ( )Qλ/µ( ), µ X or equivalently (α) X Y (α) X (α) Y Pλ ( + )= Pµ ( )Pλ/µ( ). µ X Proof See [24] VI.7 for a short proof of this identity.

An other important normalisation is given by

(α) (α) ′ (α) Jλ = cλ(α)Pλ = cλ(α)Qλ , where ′ cλ(α)= (α(λi − j)+ λj − i + 1), (i,jY)∈λ and ′ ′ cλ(α)= (α(λi − j +1)+ λj − i), (i,jY)∈λ if λ′ denotes the conjugate partition of λ.

11 If one deﬁnes skew J function by

hJ (α),J (α)J (α)i J (α) λ µ ν α J (α) λ/µ := (α) (α) ν ν hJν ,Jν iα X (α) (α) (α) then Jλ/µ is again proportional to Pλ/µ and Qλ/µ :

(α) ′ (α) ′ (α) Jλ/µ = cλ(α)cµ(α)Pλ/µ = cλ(α)cµ(α)Qλ/µ. (16)

4.2 The operator Y and the skew Jack symmetric functions R Let X, Y and Z be three alphabets such that ♯X = n< ∞ and ♯Z = m< ∞. Consider the polynomial

Y Z 1 −m 2k X In,k( , )= xi (xi + zj)∆ ( ). n! Y i i,j Z Y Y Remarking that

m p−m p+i−m m−i x (x + zi) = Λ (Y)Λ (Z) Y Z i i=0 Y =X Λp(Y + Z) = xp, Y Z Z + one obtains Y Z k Y Z In,k( , ) = Hn( + ) kn(n−1) (−1) 2 nk (k) (17) P − Y Z . = n! k,...,k nk(n 1) ( + )

1 ( k ) X∨ X 2k Hence, the image of Qλ ( )∆( ) by is a Jack polynomial. Y Z Corollary 4.2 One has,

1 kn(n−1) ( k ) nk (k) (k) X∨ X 2k 2 −1 Y Qλ ( )∆( ) =(−1) (bnk(n−1) ) Qnk(n−1)/λ′ ( ), Y k,...,k Z (k) (2(n−1))!(nk)!((n−1)k)! b − . where nk(n 1) = kn!(n−1)!((2n−1)k−1)!

12 Proof The equality follows from

1 −m zj (k) ( k ) ∨ x (x + z )= (1 + )= Q ′ (Z)Q (X ). i i j x λ λ i i,j i,j i Y Y Y Xλ Indeed, one has

1 1 Y Z (k) (k) Z ( k ) X∨ X 2k In,k( , )= bλ′ Pλ′ ( ) Qλ ( )∆( ) . n! Y Xλ Z And in the other hand, by (17) one obtains

kn(n−1) (−1) 2 nk (k) (k) I (Y, Z)= P (Z)P − (Y). n,k n! k,...,k λ nk(n 1)/λ Xλ (k) Z Identifying the coeﬃcient of Pλ′ ( ) in the two expressions, one ﬁnds,

1 kn(n−1) ( k ) nk (k) (k) X∨ X 2k 2 −1 Y Qλ ( )∆( ) =(−1) (bλ ) Pnk(n−1)/λ′ ( ), Y k,...,k Z (k) (k) Pλ (α) where the value of bλ := (k) is given by equality (7). But, from 16, Pλ/µ = Qλ (α) bλ (α) (α) Qλ/µ. Hence, bµ

1 kn(n−1) ( k ) nk (k) (k) X∨ X 2k 2 −1 Y Qλ ( )∆( ) =(−1) (bnk(n−1) ) Qnk(n−1)/λ′ ( ). Y k,...,k Z (k) b − The value of nk(n 1) is obtained from Equality (7) after simpliﬁcation.

5 Even powers of the Vandermonde determinant

5.1 Expansion of the even power of the Vandermonde on the Schur basis The expansion of even power of the Vandermonde polynomial on the Schur functions is an open problem related the fractional quantum Hall eﬀect as

13 described by Laughlin’s wave function [20]. In particular is of considerable in- terest to determine for what partitions the coeﬃcients of the Schur functions in the expansion of the square of Vandermonde vanishe [5, 32, 33, 28, 17]. The aim of this subsection is to give an hyperdeterminantal expression for 2k the coeﬃcient of Sλ(X) in ∆(X) . n Let us denote by A0 the alphabet verifying Λ (A0) = 0 for each n =6 0 0 (and by convention Λ (A0) = 1). The second orthogonality of Jack polynomials can be written as

1 1 ′ X X∨ −1 α hf,gin,α = f( )g( ) (1 − xi xj) . n! A0 Z Yi6=j In the case when α = 1, it coincides with the ﬁrst scalar product. In particular, X X ′ hSλ( ),Sµ( )in,1 = δλµ. 2k Hence, the coeﬃcient of Sλ(X) in the expansion of ∆(X) is

X X 2k ′ 1 X X∨ 2k −1 hSλ( ), ∆( ) in,1 = Sλ( )∆( ) (1 − xi xj). n! A0 Z Yi6=j One has

n(n−1) 1 X X 2k ′ 2 X n X (2k+1)(1−n) X 2(k+1) hSλ( ), ∆( ) in,1 = (−1) Sλ( )Λ ( ) ∆( ) n! A0 − Z n(n 1) 1 2(k+1) = (−1) 2 Sλ+((2k+1)(1−n))n (X)∆(X) . n! A Z 0 By Proposition 2.3, one obtains an hyperdeterminantal expression for the coeﬃcients of the Schur functions in the expansion of the even power of the Vandermonde determinant.

2k Corollary 5.1 The coeﬃcient of Sλ(X) in the expansion of ∆(X) is the hyperdeterminant

n(n−1) 2k ′ 2 k+1 S X , X H n A0 . h λ( ) ∆( ) in,1 =(−1) reversen(λ)−((2k+1)(n−1)) ( ) It should be interesting to study the link between the notion of admissible partitions introduced by Di Francesco and al [5] and such an hyperdeterminantal expression.

14 5.2 Jack polynomials over the alphabet −X

In this paragraph, we work with Laurent polynomials in X = {x1, ··· , xn}. The space of symmetric Laurent polynomials is spanned by the family in-

˜ X n dexed by decreasing vectors (Sλ( ))(λ1≥···≥λn)∈Z and defined by det(xλj +n−j) S˜ (X)= i . λ ∆(X) Indeed, each symmetric Laurent polynomial f can be written as f(X)=Λn(X)−mg(X) where g(X) is a symmetric polynomial in X. As, g(x) is a linear combination of Schur functions, it follows that f(X) is a linear combination of S˜λ’s. Let X = {−x1,..., −xn} be the alphabet of the inverse of the letters of X. We p X consider the operation −X (i.e. the substitution sending each x for x ∈ and p ∈ Z to the complete symmetric function Sn(X)). R Consider the alternant σλ aλ(X) := ǫ(σ)x σ X λj = det(xi ) = S˜λ−δ(X)∆(X), where δ = (n − 1, n − 2,..., 1, 0). From this definition, one obtains that 1 the operator n! −X sends the product of 2k alternants aλ, aµ,..., aρ is an hyperdeterminant R 1 aλ(X)aµ(X) ··· aρ(X)= n! −X (18) Z λ +µ +···+ρ i1 i2 i2ki X Det(S ( ))1≤i1,...,i2k≤n. Consider the linear operator Ω+ defined by

+ λi+i−j Ω S˜λ(X) := Sλ(X) := det(S (X)). (19) In particular, the operator Ω+ lets invariant the symmetric polynomials.

Furthermore, it admits an expression involving −X. Lemma 5.2 One has R 1 Ω+ = a (X)a (X). n! δ −δ Z −X

15 Proof It suﬃces to show that 1 a (X)a (X)S˜ (X)= S (X). n! δ −δ λ λ Z −X But aδ(X)a−δ(X)S˜λ(X)= aλ+δ(X)a−δ(X), and by (18), one obtains the result. .

Proposition 5.3 Let X = {x1, ··· , xn} be a ﬁnite alphabet and 0 ≤ l ≤ p ∈ N.

− − (k),n (k 1)n(n 1) +np+l 2(k−1) R − − X 2 S l n−l X X . np+(k 1)(n 1)l(− )=(−1) (p+1) p ( )∆( ) (20) Proof By Lemma 5.2, one has

2(k−1) + (2k−1) Sλ(X)∆(X) = Ω Sλ(X)∆(X) 1 2(k−1) = aλ+δ(X)aδ(X) (X)a−δ(X). n! X Z − By Equality (18), one obtains

2(k−1) λi +n−i1+···+n−i2k−1+i2k−n Sλ(X)∆(X) = Det S 1 (X) 1≤i1,...,i2k≤n λ − +i +···+i − −i = Det S n i1 1 2k 1 2k (X) 0≤i1,...,i2k≤n−1 n(n−1) λ − +1−n+i +···+i2k = (−1) 2 Det S n i1 1 (X) 0≤i1,...,i2k≤n−1 n(n−1) 2 H n X = (−1) reverse n(λ)−[(n−1) ](− ) n(n−1)(k−1) 2 n X = (−1) Treversen(λ)−[((k−1)(n−1)) ](− ). In particular, from Proposition 3.1

(k),n (k) R − − X T − X [np+(k 1)(n 1)l](− ) = [((k−1)(n−1))n]+[pn l(p+1)l](− ) n(n−1)(k−1) − X X 2(k−1) = (−1) 2 S[(p+1)lpn l]( )∆( ) .

(α),n But, the Jack polynomial Rλ being homogeneous, one has (α),n X |λ| (α),n X Rλ (− )=(−1) Rλ (− ). The result follows.

16 Remark 5.4 1. Note that a special case of Proposition 5.3 appeared in [22].

2. Proposition 5.3 can be reformulated as

(k) X (k) X X 2(k−1) The polynomials Pnp+(k−1)(n−1)l(− ) and P(pn)+(1l)( )∆( ) are proportional.

This kind of identities relying Jack polynomials in X and in −X can be deduced from more general ones involving Macdonald polynomials when t is specialized to a power of q. This will be investigated in a forthcoming paper.

As a special case of Proposition 5.3, the even powers of the Vandermonde determinants ∆(X)2k are Jack polynomials on the alphabet −X.

Corollary 5.5 Setting l = p =0 in Equality (20), one obtains

(kn(n−1) 2k (−1) 2 (k + 1)n (k+1) ∆(X) = P − (−X). n! k +1,...,k +1 n(n 1)k In the same way, using Corollary 4.2, one ﬁnds a surprising identity relying Jack polynomials in the alphabets −X and X∨.

Proposition 5.6 One has

+ (1/k) X∨ X 2(k−1) n X n−1 Ω Qλ ( )∆( ) Λ ( ) = − − n(n 1)(k 1) +|λ| (−1) 2 nk (k) −1 (k) (b − ) Q − ′ (−X). n! k,...,k nk(n 1) nk(n 1)/λ Proof The result is a straightforward consequence of Lemma 5.2 and Corol- lary 4.2.

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