Classical Lie algebras and Yangians

Classical Lie algebras and Yangians Alexander Molev University of Sydney Advanced Summer School Integrable Systems and Quantum Symmetries Prague 2007

Lecture 1. Casimir elements for classical Lie algebras Lecture 2. Yangians: algebraic structure Lecture 3. Yangians: representations

Classical Lie algebras over C A type: special linear Lie algebra sl N general linear Lie algebra gl N B type: orthogonal Lie algebra o 2n+1 C type: D type: symplectic Lie algebra sp 2n orthogonal Lie algebra o 2n

General linear Lie algebra The Lie algebra gl N has the basis of the standard matrix units E ij with 1 i, j N so that dim gl N = N 2. The commutation relations are [E ij, E kl ] = δ kj E il δ il E kj. The universal enveloping algebra U(gl N ) is the associative algebra with generators E ij and the defining relations E ij E kl E kl E ij = δ kj E il δ il E kj.

By the Poincaré Birkhoff Witt theorem, given any ordering on the set of generators {E ij }, any element of U(gl N ) can be uniquely written as a linear combination of the ordered monomials in the E ij. The center Z(gl N ) of U(gl N ) is Z(gl N ) = { z U(gl N ) z x = x z for all x U(gl N ) }. The Casimir elements for gl N are elements of Z(gl N ).

Given an N-tuple of complex numbers λ = (λ 1,..., λ N ) the Verma module M(λ) for gl N is the quotient of U(gl N ) by the left ideal generated by the elements E ij, i < j, and E ii λ i, i = 1,..., N. The Verma module has a unique maximal submodule K. Set L(λ) = M(λ)/K, the unique irreducible quotient of M(λ).

Equivalently, L(λ) is an irreducible module generated by a nonzero vector ζ such that E ij ζ = 0 for 1 i < j N, and E ii ζ = λ i ζ for 1 i N.

Equivalently, L(λ) is an irreducible module generated by a nonzero vector ζ such that E ij ζ = 0 for 1 i < j N, and E ii ζ = λ i ζ for 1 i N. Then ζ is the highest vector and λ = (λ 1,..., λ N ) is the highest weight of L(λ). The representation L(λ) is finite-dimensional if and only if λ i λ i+1 Z + for all i = 1,..., N 1.

Any Casimir element z Z(gl N ) acts as a multiplication by a scalar χ(z) in L(λ). This scalar is a polynomial in λ 1,..., λ N ; this polynomial is symmetric in the shifted variables l 1 = λ 1, l 2 = λ 2 1,..., l N = λ N N + 1. The map χ : Z(gl N ) C[l 1,..., l N ] S N in an algebra isomorphism called the Harish-Chandra isomorphism.

Orthogonal and symplectic Lie algebras For N = 2n or N = 2n + 1, respectively, set g N = o 2n+1, sp 2n, o 2n.

Orthogonal and symplectic Lie algebras For N = 2n or N = 2n + 1, respectively, set g N = o 2n+1, sp 2n, o 2n. We will number the rows and columns of N N matrices by the indices { n,..., 1, 0, 1,..., n} if N = 2n + 1, and by { n,..., 1, 1,..., n} if N = 2n.

g N = o 2n+1 g N = o 2n n 1 0 1 n n 1 1 n ṇ ṇ.. 1 0 1 A = A 1 1 A = A.. n n Skew-symmetric matrices with respect to the second diagonal.

The Lie algebra g N = sp N with N = 2n is spanned by the elements F ij = E ij sgn i sgn j E j, i, n i, j n.

The Lie algebra g N = sp N with N = 2n is spanned by the elements F ij = E ij sgn i sgn j E j, i, n i, j n. ṇ. 1 1. n 1 1 n A B = B C = C A n

Commutation relations in g N : F j, i = θ ij F ij and [F ij, F kl ] = δ kj F il δ il F kj θ k, j δ i, k F j,l + θ i, l δ l,j F k, i, where 1 in the orthogonal case, θ ij = sgn i sgn j in the symplectic case.

For any n-tuple of complex numbers λ = (λ 1,..., λ n ) the corresponding irreducible highest weight representation V (λ) of g N is generated by a nonzero vector ξ such that F ij ξ = 0 for n i < j n, and F ii ξ = λ i ξ for 1 i n. The representation V (λ) is finite-dimensional if and only if λ i λ i+1 Z + for i = 1,..., n 1 and λ 1 λ 2 Z + if g N = o 2n, λ 1 Z + if g N = sp 2n, 2 λ 1 Z + if g N = o 2n+1.

Any element z Z(g N ) of the center of U(g N ) acts as a multiplication by a scalar χ(z) in V (λ). This scalar is a polynomial in λ 1,..., λ n. In the B and C cases, this polynomial is symmetric in the variables l1 2,..., l n 2, where l i = λ i + ρ i and i + 1 for g N = o 2n, ρ i = ρ i = i + 1 2 for g N = o 2n+1, i for g N = sp 2n, for i = 1,..., n. Also, ρ 0 = 1/2 in the case g N = o 2n+1.

The map χ : Z(g N ) algebra of polynomials is an algebra isomorphism called the Harish-Chandra isomorphism.

The map χ : Z(g N ) algebra of polynomials is an algebra isomorphism called the Harish-Chandra isomorphism. Example. For g N = o N n m=1 ( (F mm + ρ m ) 2 + 2 m<i<m F mi F im ) is the second degree Casimir element. Its Harish-Chandra image is l 2 1 + + l 2 n.

Newton s formulas Denote by E the N N matrix whose ij-th entry if E ij. Denote by C(u) the Capelli determinant C(u) = sgn p (u + E) p(1),1... (u + E N + 1) p(n),n. p S N This is a polynomial in u with coefficients in the universal enveloping algebra U(gl N ), C(u) = u N + C 1 u N 1 + + C N, C i U(gl N ).

Example. For N = 2 we have C(u) = (u + E 11 ) (u + E 22 1) E 21 E 12 = u 2 + (E 11 + E 22 1) u + E 11 (E 22 1) E 21 E 12.

Example. For N = 2 we have C(u) = (u + E 11 ) (u + E 22 1) E 21 E 12 = u 2 + (E 11 + E 22 1) u + E 11 (E 22 1) E 21 E 12. Note that C 1 = E 11 + E 22 1, C 2 = E 11 (E 22 1) E 21 E 12 are Casimir elements for gl 2 and χ(c 1 ) = l 1 + l 2, χ(c 2 ) = l 1 l 2.

Theorem The coefficients C 1,..., C N belong to Z(gl N ). The image of C(u) under the Harish-Chandra isomorphism is given by χ : C(u) (u + l 1 )... (u + l N ), so that χ(c k ) is the elementary symmetric polynomial of degree k in l 1,..., l N, χ(c k ) = i 1 < <i k l i1... l ik. Moreover, the algebra Z(gl N ) is generated by C 1,..., C N.

Gelfand invariants are the elements of U(gl N ) defined by N tr E k = E i1 i 2 E i2 i 3... E ik i 1, k = 0, 1,.... i 1,i 2,...,i k =1

Gelfand invariants are the elements of U(gl N ) defined by N tr E k = E i1 i 2 E i2 i 3... E ik i 1, k = 0, 1,.... i 1,i 2,...,i k =1 Example. For N = 2 we have tr E = E 11 + E 22, tr E 2 = E 2 11 + E 12 E 21 + E 21 E 12 + E 2 22.

A noncommutative analogue of the classical Newton formula: Theorem We have the equality of power series in u 1 1 + k=0 ( 1) k tr E k C(u + 1) =. (u N + 1) k+1 C(u)

A noncommutative analogue of the classical Newton formula: Theorem We have the equality of power series in u 1 1 + ( 1) k tr E k C(u + 1) =. (u N + 1) k+1 C(u) k=0 Proof. This is equivalent to the Perelomov Popov formulas 1 + k=0 ( 1) k χ(tr E k ) N (u N + 1) k+1 = i=1 u + l i + 1 u + l i.

Capelli-type determinant for g N Introduce a special map ϕ N : S N S N, p p from the symmetric group S N into itself.

Capelli-type determinant for g N Introduce a special map ϕ N : S N S N, p p from the symmetric group S N into itself. If N = 2 we define ϕ 2 as the map S 2 S 2 whose image is the identity permutation. Given a set of positive integers a 1 < < a N we regard S N as the group of their permutations.

For N > 2 define a map from the set of ordered pairs (a k, a l ) with k l into itself by the rule (a k, a l ) (a l, a k ), k, l < N,

For N > 2 define a map from the set of ordered pairs (a k, a l ) with k l into itself by the rule (a k, a l ) (a l, a k ), k, l < N, (a k, a N ) (a N 1, a k ), k < N 1, (a N, a k ) (a k, a N 1 ), k < N 1,

For N > 2 define a map from the set of ordered pairs (a k, a l ) with k l into itself by the rule (a k, a l ) (a l, a k ), k, l < N, (a k, a N ) (a N 1, a k ), k < N 1, (a N, a k ) (a k, a N 1 ), k < N 1, (a N 1, a N ) (a N 1, a N 2 ), (a N, a N 1 ) (a N 1, a N 2 ).

Let p = (p 1,..., p N ) be a permutation of the indices a 1,..., a N. Its image under the map ϕ N is the permutation of the form p = (p 1,..., p N 1, a N ).

Let p = (p 1,..., p N ) be a permutation of the indices a 1,..., a N. Its image under the map ϕ N is the permutation of the form p = (p 1,..., p N 1, a N ). The pair (p 1, p N 1 ) is the image of the ordered pair (p 1, p N ) under the above map.

Examples. p = (3, 5, 7, 6, 1, 2, 4).

Examples. p = (3, 5, 7, 6, 1, 2, 4). (3, 5, 7, 6, 1, 2, 4) (,,,,,, 7)

Examples. p = (3, 5, 7, 6, 1, 2, 4). (3, 5, 7, 6, 1, 2, 4) (,,,,,, 7) (3, 5, 7, 6, 1, 2, 4) (4,,,,, 3, 7)

Examples. p = (3, 5, 7, 6, 1, 2, 4). (3, 5, 7, 6, 1, 2, 4) (,,,,,, 7) (3, 5, 7, 6, 1, 2, 4) (4,,,,, 3, 7) (3, 5, 7, 6, 1, 2, 4) (4, 2,,, 5, 3, 7)

Examples. p = (3, 5, 7, 6, 1, 2, 4). (3, 5, 7, 6, 1, 2, 4) (,,,,,, 7) (3, 5, 7, 6, 1, 2, 4) (4,,,,, 3, 7) (3, 5, 7, 6, 1, 2, 4) (4, 2,,, 5, 3, 7) (3, 5, 7, 6, 1, 2, 4) (4, 2, 1, 6, 5, 3, 7) = p.

Examples. p = (3, 5, 7, 6, 1, 2, 4). (3, 5, 7, 6, 1, 2, 4) (,,,,,, 7) (3, 5, 7, 6, 1, 2, 4) (4,,,,, 3, 7) (3, 5, 7, 6, 1, 2, 4) (4, 2,,, 5, 3, 7) (3, 5, 7, 6, 1, 2, 4) (4, 2, 1, 6, 5, 3, 7) = p. (N, N 1,..., 2, 1) (,,...,,, N)

Each fiber of the map ϕ N is an interval in S N with respect to the Bruhat order, and this interval is isomorphic to a Boolean poset.

Each fiber of the map ϕ N is an interval in S N with respect to the Bruhat order, and this interval is isomorphic to a Boolean poset. 321 231 312 213 132 123

Each fiber of the map ϕ N is an interval in S N with respect to the Bruhat order, and this interval is isomorphic to a Boolean poset. 321 231 312 213 132 123 213 312 132 123 213 321 231 123

Denote by F the N N matrix whose ij-th entry is F ij. Introduce the Capelli-type determinant C(u) = ( 1) n sgn p p (u + ρ n + F ) bp(1), b p (1) p S N (u + ρ n + F ) bp(n), b, p (N) where (b 1,..., b N ) is a fixed permutation of the indices ( n,..., n) and p is the image of p under the map ϕ N.

Theorem The polynomial C(u) does not depend on the choice of the permutation (b 1,..., b N ). All coefficients of C(u) belong to Z(g N ). Moreover, the image of C(u) under the Harish-Chandra isomorphism is given by n χ : C(u) (u 2 li 2 ), if N = 2n, i=1 and χ : C(u) ( u + 1 ) n (u 2 li 2 ), if N = 2n + 1. 2 i=1

Examples. For g N = sp 2 take (b 1, b 2 ) = ( 1, 1). We have ρ 1 = ρ 1 = 1, l 1 = λ 1 1,

Examples. For g N = sp 2 take (b 1, b 2 ) = ( 1, 1). We have ρ 1 = ρ 1 = 1, l 1 = λ 1 1, C(u) = (u + F 1, 1 + 1) (u + F 11 1) F 1, 1 F 1,1 = u 2 (F 11 1) 2 F 1, 1 F 1,1

Examples. For g N = sp 2 take (b 1, b 2 ) = ( 1, 1). We have ρ 1 = ρ 1 = 1, l 1 = λ 1 1, C(u) = (u + F 1, 1 + 1) (u + F 11 1) F 1, 1 F 1,1 = u 2 (F 11 1) 2 F 1, 1 F 1,1 and χ : C(u) u 2 l 2 1.

For g N = o 3 take (b 1, b 2, b 3 ) = ( 1, 0, 1). Here ρ 1 = ρ 0 = ρ 1 = 1/2, l 1 = λ 1 1/2,

For g N = o 3 take (b 1, b 2, b 3 ) = ( 1, 0, 1). Here ρ 1 = ρ 0 = ρ 1 = 1/2, l 1 = λ 1 1/2, C(u) = (u + F 1, 1 + 1/2) (u + 1/2) (u + F 11 1/2) F 0, 1 F 1,0 (u + F 11 1/2) F 10 (u + F 1, 1 + 1/2) F 01.

For g N = o 3 take (b 1, b 2, b 3 ) = ( 1, 0, 1). Here ρ 1 = ρ 0 = ρ 1 = 1/2, l 1 = λ 1 1/2, C(u) = (u + F 1, 1 + 1/2) (u + 1/2) (u + F 11 1/2) F 0, 1 F 1,0 (u + F 11 1/2) F 10 (u + F 1, 1 + 1/2) F 01. Hence C(u) = (u + 1/2) (u 2 (F 11 1/2) 2 2F 10 F 01 ) and χ : C(u) (u + 1/2) (u 2 l 2 1 ).

Gelfand invariants are the elements of U(g N ) defined by n tr F k = F i1 i 2 F i2 i 3... F ik i 1, k = 0, 1,.... i 1,i 2,...,i k = n

Gelfand invariants are the elements of U(g N ) defined by n tr F k = F i1 i 2 F i2 i 3... F ik i 1, k = 0, 1,.... i 1,i 2,...,i k = n We have tr F = 0, n tr F 2 = F ij F ji i,j= n

Gelfand invariants are the elements of U(g N ) defined by n tr F k = F i1 i 2 F i2 i 3... F ik i 1, k = 0, 1,.... i 1,i 2,...,i k = n We have tr F = 0, and tr F 2 = χ(tr F 2 ) = 2 n i,j= n n i=1 F ij F ji (l 2 i ρ 2 i ).

Theorem If N = 2n then 1 + 2u + 1 2u + 1 1 ( 1) k tr F k C(u + 1) =, (u + ρ n ) k+1 C(u) k=0 where the upper sign is taken in the orthogonal case and the lower sign in the symplectic case. If N = 2n + 1 then 1 + 2u + 1 2u ( 1) k tr F k C(u + 1) =, (u + ρ n ) k+1 C(u) k=0 where C(u) = 2u 2u + 1 C(u).

All Gelfand invariants tr F k belong to Z(g N ).

All Gelfand invariants tr F k belong to Z(g N ). Their images under the Harish-Chandra isomorphism are found by the Perelomov Popov formulas 1 + 2u + 1 2u + 1 1 k=0 ( 1) k χ(tr F k ) (u + ρ n ) k+1 = n i= n u + l i + 1 u + l i, where the zero index is skipped in the product if N = 2n, while for N = 2n + 1 one should set l 0 = 0.

Noncommutative Cayley Hamilton theorem C(u) denotes the Capelli determinant for gl N or the Capelli-type determinant for g N = o 2n, sp 2n, o 2n+1.

Noncommutative Cayley Hamilton theorem C(u) denotes the Capelli determinant for gl N or the Capelli-type determinant for g N = o 2n, sp 2n, o 2n+1. Theorem (i) For gl N we have C( E + N 1) = 0 and C( E t ) = 0. (ii) For g N we have C( F ρ n ) = 0.

Corollary (Characteristic identities of Bracken and Green) (i) The image of the matrix E in the representation L(λ) of gl N satisfies N (E l i N + 1) = 0 and i=1 N (E t l i ) = 0. i=1

Corollary (Characteristic identities of Bracken and Green) (i) The image of the matrix E in the representation L(λ) of gl N satisfies N (E l i N + 1) = 0 and i=1 N (E t l i ) = 0. i=1 (ii) The image of the matrix F in the representation V (λ) of g N satisfies n (F l i + ρ n ) = 0, i= n The zero index is skipped in the product if N = 2n, while for N = 2n + 1 one should set l 0 = 1/2.

Noncommutative power sums Casimir elements For 1 m N and any positive integer k set Φ (m) k = k α(i ) + 1 E m i 1 E i1 i 2... E ik 1 m, summed over i 1,..., i k 1 {1,..., m}, where E ij = E ij δ ij (m 1) and α(i ) is the multiplicity of m in the multiset I = {i 1,..., i k 1 }.

Theorem. For any k 1 the element Φ k = Φ (1) k + + Φ (N) k belongs to Z(gl N ). Moreover, χ(φ k ) = l k 1 + + l k N. (Gelfand, Krob, Lascoux, Leclerc, Retakh and Thibon, 95). Φ 1 = Φ 2 = N (E mm m + 1), m=1 N (E mm m + 1) 2 + 2 m=1 1 l<m N E ml E lm.

Orthogonal and symplectic case For 1 m n and any positive integer k set Φ (m) 2k = 2k α(i ) + 1 F m i 1 F i1 i 2... F i2k 1 m, summed over i 1,..., i 2k 1 { m,..., m},

Orthogonal and symplectic case For 1 m n and any positive integer k set Φ (m) 2k = 2k α(i ) + 1 F m i 1 F i1 i 2... F i2k 1 m, summed over i 1,..., i 2k 1 { m,..., m}, and Φ (m) 2k = 2k α(i ) + 1 F m i 1 F i1 i 2... F i2k 1 m, summed over i 1,..., i 2k 1 { m + 1,..., m}, where F ij = F ij + δ ij ρ m and α(i ) is the multiplicity of m in the multiset I = {i 1,..., i 2k 1 }.

Example. We have Φ (m) 2 = (F mm + ρ m ) 2 + 2 Φ (m) 2 = (F mm + ρ m ) 2 + 2 m i<m m<i<m F m i F i m, F m i F i m. and

Example. We have Φ (m) 2 = (F mm + ρ m ) 2 + 2 Φ (m) 2 = (F mm + ρ m ) 2 + 2 m i<m m<i<m F m i F i m, F m i F i m. and Theorem. For any k 1 the element Φ 2k = Φ (1) 2k (1) + Φ 2k + + Φ(n) (n) 2k + Φ 2k belongs to Z(g N ). Moreover, χ(φ 2k ) = 2 ( l1 2k + + ln 2k ).

If g N = o N then the second order Casimir element is Φ 2 = 2 n m=1 ( (F mm + ρ m ) 2 + 2 m<i<m F m i F i m ).

If g N = o N then the second order Casimir element is Φ 2 = 2 n m=1 ( (F mm + ρ m ) 2 + 2 m<i<m F m i F i m ). If g N = sp 2n then Φ 2 = 2 n m=1 ( (F mm m) 2 + F m, m F m,m + 2 m<i<m F m i F i m ).

Quantum immanants for gl N A diagram (or partition) is a sequence µ = (µ 1,..., µ N ) of integers µ i such that µ 1 µ N 0, depicted as an array of unit cells (or boxes).

Quantum immanants for gl N A diagram (or partition) is a sequence µ = (µ 1,..., µ N ) of integers µ i such that µ 1 µ N 0, depicted as an array of unit cells (or boxes). Example. The diagram µ = (5, 5, 3) is µ = 13 l(µ) = 3 The number of cells is the weight of the diagram, denoted µ. The number of nonzero rows is its length, denoted l(µ).

For a diagram µ with l(µ) N and µ = k consider the row tableau T 0 obtained by filling in the cells by the numbers 1,..., k from left to right in successive rows: 1 6 9 2 7 3 4 5 8

For a diagram µ with l(µ) N and µ = k consider the row tableau T 0 obtained by filling in the cells by the numbers 1,..., k from left to right in successive rows: 1 6 9 2 7 3 4 5 8 Let R µ and C µ denote the row symmetrizer and column antisymmetrizer of T 0 respectively: R µ = σ σ, C µ = τ sgn τ τ.

Set c µ (r) = j i if the cell (i, j) of T 0 is occupied by r.

Set c µ (r) = j i if the cell (i, j) of T 0 is occupied by r. Consider the matrix E as the element E = N e ij E ij End (C N ) U(gl N ) i,j=1 and define the quantum immanant S µ by S µ = 1 h(µ) tr (E c µ(1)) (E c µ (k)) R µ C µ, where h(µ) is the product of the hooks of µ (Okounkov, 96).

The symmetric group S k acts in a natural way in the tensor space (C N ) k. We identify elements of S k and hence R µ and C µ with the corresponding operators.

The symmetric group S k acts in a natural way in the tensor space (C N ) k. We identify elements of S k and hence R µ and C µ with the corresponding operators. Example For µ = (2, 1) we have T 0 = 1 2 3

Explicitly, E (E 1) (E + 1) = e i1 j 1 e i2 j 2 e i3 j 3 E i1 j 1 (E i2 j 2 δ i2 j 2 ) (E i3 j 3 + δ i3 j 3 ).

Explicitly, E (E 1) (E + 1) = e i1 j 1 e i2 j 2 e i3 j 3 E i1 j 1 (E i2 j 2 δ i2 j 2 ) (E i3 j 3 + δ i3 j 3 ). Hence S (2,1) = 1 3 i 1,i 2,i 3 ( E i1 i 1 (E i2 i 2 1) (E i3 i 3 + 1) + E i1 i 2 (E i2 i 1 δ i2 i 1 ) (E i3 i 3 + 1) E i1 i 3 (E i2 i 2 1) (E i3 i 1 + δ i3 i 1 ) ) E i1 i 2 (E i2 i 3 δ i2 i 3 ) (E i3 i 1 + δ i3 i 1 ), summed over the indices i 1, i 2, i 3 {1,..., N}.

Examples. Capelli elements (quantum minors) S (1 k ) = i 1 > >i k p S k sgn p E i1,i p(1)... (E + k 1) ik,i p(k).

Examples. Capelli elements (quantum minors) S (1 k ) = i 1 > >i k p S k sgn p E i1,i p(1)... (E + k 1) ik,i p(k). Quantum permanents S (k) = 1 α 1!... α n! i 1 i k p S k E i1,i p(1)... (E k + 1) ik,i p(k), where α i is the multiplicity of i in i 1,..., i k, each i r {1,..., N}.

Theorem The quantum immanants S µ with l(µ) N form a basis of the center of the universal enveloping algebra U(gl N ).

Theorem The quantum immanants S µ with l(µ) N form a basis of the center of the universal enveloping algebra U(gl N ). Moreover, χ(s µ ) = sµ, where sµ = sµ(λ) is the shifted symmetric polynomial. (Okounkov, 96).

Explicit formula: sµ(λ) = ( λt (α) c(α) ), sh(t )=µ α µ summed over all reverse µ-tableaux T with entries in {1,..., N} such that the entries of T weakly decrease along the rows and strictly decrease down the columns. Here c(α) = j i for α = (i, j) and T (α) is the entry of T in the cell α.

Example. For µ = (2, 1) the reverse tableaux are i j with i j and i > k k

Example. For µ = (2, 1) the reverse tableaux are i j with i j and i > k k Hence s(2,1) (λ) = λ i (λ j 1) (λ k + 1). i j, i>k

Examples. Harish-Chandra images of the Capelli elements χ(s (1 k )) = λ i1 (λ i2 + 1)... (λ ik + k 1). i 1 > >i k

Examples. Harish-Chandra images of the Capelli elements χ(s (1 k )) = λ i1 (λ i2 + 1)... (λ ik + k 1). i 1 > >i k Harish-Chandra images of the quantum permanents χ(s (k) ) = λ i1 (λ i2 1)... (λ ik k + 1). i 1 i k

Noncommutative Pfaffians and Hafnians The Pfaffian Pf A of a 2k 2k matrix A = [A ij ] is defined by Pf A = 1 2 k sgn σ A k! σ(1),σ(2)... A σ(2k 1),σ(2k). σ S 2k

Noncommutative Pfaffians and Hafnians The Pfaffian Pf A of a 2k 2k matrix A = [A ij ] is defined by Pf A = 1 2 k k! σ S 2k sgn σ A σ(1),σ(2)... A σ(2k 1),σ(2k). Let g N = o N. For any subset I of { n,..., n} containing 2k elements i 1 < < i 2k, the 2k 2k matrix [F ip, i q ] is skew-symmetric. We denote its Pfaffian by Φ I = Pf [F ip, i q ], p, q = 1,..., 2k.

If N = 2n then C n = ( 1) n (Pf F ) 2.

If N = 2n then C n = ( 1) n (Pf F ) 2. Theorem For all k = 1,..., n the C k are Casimir elements for o N. Moreover, the image of C k under the Harish-Chandra isomorphism is given by χ : C k ( 1) k (li 2 1 ρ 2 i 1 )... (li 2 k ρ 2 i k k+1 ). 1 i 1 < <i k n

For any k 1 let I = {i 1,..., i 2k } be a multiset whose elements belong to { n,..., n}. Denote by A I the 2k 2k matrix whose (a, b) entry is A iai b.

For any k 1 let I = {i 1,..., i 2k } be a multiset whose elements belong to { n,..., n}. Denote by A I the 2k 2k matrix whose (a, b) entry is A iai b. The Hafnian Hf A I of the matrix A I is defined by (Caianiello, 56). Hf A I = 1 2 k k! σ S 2k A iσ(1),i σ(2)... A iσ(2k 1),i σ(2k).

Let g N = sp 2n. Set F ij = sgn i F ij. Then we have F i, j = F j, i. Let I be any sequence of length 2k of elements from the set { n,..., n}. Denote the multiplicity of an element i in I by α i. Denote the Hafnian of the symmetric matrix [ F ip, iq ] by Ψ I = Hf [ F ip, i q ], p, q = 1,..., 2k. Set D k = I sgn (i 1... i 2k ) α n!... α n! Ψ I Ψ I, I = { i 2k,..., i 1 }.

Theorem For all k 1 the D k are Casimir elements for sp 2n.

Theorem For all k 1 the D k are Casimir elements for sp 2n. Moreover, the image of D k under the Harish-Chandra isomorphism is given by χ : D k ( 1) k (li 2 1 i1 2 )... (li 2 k (i k + k 1) 2 ). 1 i 1 i k n Corollary. ( C(u) (u + ρ n )... (u + ρ n ) ) 1 ( 1) k D k = 1 + (u 2 (n + 1) 2 )... (u 2 (n + k) 2 ). k=1