Classical Lie algebras and Yangians

Transcription

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

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

3 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

4 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.

5 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 ).

6 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.

7 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).

8 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.

9 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).

10 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.

11 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.

12 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.

13 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. Example. χ : E E NN λ λ N = l l N N(N 1)/2.

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

15 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.

16 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. The Lie algebra g N = o N is spanned by the elements F ij = E ij E j, i, n i, j n.

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

18 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.

19 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 n 1 1 n A B = B C = C A n

20 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.

21 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.

22 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.

23 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 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.

24 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 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. In the D case χ(z) is the sum of a symmetric polynomial in l 2 1,..., l 2 n and l 1... l n times a symmetric polynomial in l 2 1,..., l 2 n.

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

26 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 ) m<i<m F mi F im ) is the second degree Casimir element. Its Harish-Chandra image is l l 2 n.

27 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 C N, C i U(gl N ).

28 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.

29 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.

30 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.

31 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

32 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 E 12 E 21 + E 21 E 12 + E 2 22.

33 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 E 12 E 21 + E 21 E 12 + E These are Casimir elements and χ(tr E) = l 1 + l 2 1, χ(tr E 2 ) = l l l 1 + l 2.

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

35 A noncommutative analogue of the classical Newton formula: Theorem We have the equality of power series in u ( 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.

36 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.

37 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.

38 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.

39 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,

40 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,

41 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 ).

42 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 ).

43 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.

44 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. Then the pair (p 2, p N 2 ) is found as the image of (p 2, p N 1 ) under the above map, etc.

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

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

47 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)

48 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)

49 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.

50 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)

51 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) (N, N 1,..., 2, 1) (1,,...,, N 1, N)

52 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) (N, N 1,..., 2, 1) (1,,...,, N 1, N) (N, N 1,..., 2, 1) (1, 2,..., N 2, N 1, N) = id.

53 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.

54 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

55 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

56 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.

57 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 i=1

58 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,

59 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

60 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.

61 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,

62 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.

63 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 ).

64 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

65 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

66 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 ).

67 Theorem If N = 2n then 1 + 2u + 1 2u ( 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).

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

69 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 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.

70 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.

71 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.

72 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

73 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.

74 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 }.

75 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 }. Example. Φ (m) 1 = E mm m + 1 Φ (m) m 1 2 = (E mm m + 1) E m i E i m. i=1

76 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 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) m=1 1 l<m N E ml E lm.

77 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},

78 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 }.

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

80 Example. We have Φ (m) 2 = (F mm + ρ m ) Φ (m) 2 = (F mm + ρ m ) 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 ).

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

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

83 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).

84 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

85 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).

86 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:

87 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: 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.

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

89 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).

90 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.

91 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

92 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 = Hence S (2,1) = 1 3 tr E (E 1) (E + 1) (1 + P 12)(1 P 13 ).

93 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 ).

94 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}.

95 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).

96 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}.

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

98 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).

99 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). The s µ(λ) are certain symmetric polynomials in l 1,..., l N.

100 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.

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

102 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

103 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

104 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

105 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 These are symmetric polynomials in l 1,..., l N, l 1 = λ 1,..., l N = λ N N + 1.

106 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

107 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.

108 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. Set C k = ( 1) k Φ I Φ I, I = { i 2k,..., i 1 }. I

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

110 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

111 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 Corollary. C(u) n (u + ρ n )... (u + ρ n ) = 1 + k=1 C k (u 2 ρ 2 n k+1 )... (u2 ρ 2 n).

112 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.

113 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).

114 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 }.

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

116 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

117 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