Kac-Moody algebras, vertex operators and applications

Transcription

1 , vertex operators and applications Vyacheslav Futorny University of São Paulo, Brazil AIMS Capetown, July 19-31, 2015

2 Outline Examples of Lie algebras - Virasoro algebra - Heisenberg/Weyl algebras - - Krichever-Novikov algebras - DJKM (Date-Jimbo-Kashiwara-Miwa) algebra

3 Sophus Lie, 1870 M manifold, G = Diff(M) 1-parameter subgroup {ϕ t, t R} G; γ m (t) := ϕ t (m), m M; v(m) = γ m(0) Vect(M) = {v : m v(m), m M} Example: M = S 1, Vect C (S 1 ), A n = e inϕ d dϕ, [A n, A m ] = (m n)a n+m (Witt algebra) [A n, A m ] = (m n)a n+m + n3 n 12 δ n, m c (Virasoro algebra)

4 Sophus Lie, 1870 M manifold, G = Diff(M) 1-parameter subgroup {ϕ t, t R} G; γ m (t) := ϕ t (m), m M; v(m) = γ m(0) Vect(M) = {v : m v(m), m M} Example: M = S 1, Vect C (S 1 ), A n = e inϕ d dϕ, [A n, A m ] = (m n)a n+m (Witt algebra) [A n, A m ] = (m n)a n+m + n3 n 12 δ n, m c (Virasoro algebra)

5 Sophus Lie, 1870 M manifold, G = Diff(M) 1-parameter subgroup {ϕ t, t R} G; γ m (t) := ϕ t (m), m M; v(m) = γ m(0) Vect(M) = {v : m v(m), m M} Example: M = S 1, Vect C (S 1 ), A n = e inϕ d dϕ, [A n, A m ] = (m n)a n+m (Witt algebra) [A n, A m ] = (m n)a n+m + n3 n 12 δ n, m c (Virasoro algebra)

6 A Lie algebra g over field k: a vector space over k + a bilinear map [, ] : g g g such that: ( anticommutativiity) (Jacobi identity) [x, x] = 0, x g [x, [y, z]] + [z, [x, y]] + [y, [z, x]] = 0

7 1. Let A be an associative algebra. Define on a vector space A a Lie structure by: [a, b] = ab ba. hence every associative algebra gives rise to a Lie algebra. Reciprocally, if g is Lie algebra then one associates the universal enveloping algebra U(g), certain universal associative algebra which contains g 2. Let V be a vector space over k with basis e 1,..., e n. The symmetric algebra S(V ) is spanned over field k by the elements e k ekn n with commutative multiplication.

8 1. Let A be an associative algebra. Define on a vector space A a Lie structure by: [a, b] = ab ba. hence every associative algebra gives rise to a Lie algebra. Reciprocally, if g is Lie algebra then one associates the universal enveloping algebra U(g), certain universal associative algebra which contains g 2. Let V be a vector space over k with basis e 1,..., e n. The symmetric algebra S(V ) is spanned over field k by the elements e k ekn n with commutative multiplication.

9 1. Let A be an associative algebra. Define on a vector space A a Lie structure by: [a, b] = ab ba. hence every associative algebra gives rise to a Lie algebra. Reciprocally, if g is Lie algebra then one associates the universal enveloping algebra U(g), certain universal associative algebra which contains g 2. Let V be a vector space over k with basis e 1,..., e n. The symmetric algebra S(V ) is spanned over field k by the elements e k ekn n with commutative multiplication.

10 Let g be a simple Lie algebra with basis g 1,..., g n and the structure constants c k ij : [g i, g j ] = ik c k ij g k Then S(g) has a structure of a Lie algebra with the Lie Poisson bracket {g i, g j } = k=1 c k ij g k + product rule (Hamiltonian equation of the motion)

11 3. V, dimv = 2n, <, > symplectic form on V (nondegenerate, skew-symmetric), x i, y i, i = 1,..., n symplectic coordinates. C (V, R) smooth functions on V, Poisson bracket: (F.Berezin, 1967) {P, Q} = P y j Q x i P x i Q y j

12 4. Heisenberg algebra H n, n : [x i, x j ] = δ i, j z, [z, x i ] = 0. H n = kz n i= n,i 0 kx i, Take A n = U(H n )/(z a) for any a k., a 0. Then A n is n-th Weyl algebra (a = 1). Origin. System of quantum mechanics: Hilbert space + Hamiltonian operator, e.g. L 2 (R), T = d 2 dt 2 x 2 (quantum harmonic oscil.); Problem: find eigenvalues of T. To solve use p(f )(x) = df /dx, q(f )(x) = ixf (x), [p, q] = i.

13 4. Heisenberg algebra H n, n : [x i, x j ] = δ i, j z, [z, x i ] = 0. H n = kz n i= n,i 0 kx i, Take A n = U(H n )/(z a) for any a k., a 0. Then A n is n-th Weyl algebra (a = 1). Origin. System of quantum mechanics: Hilbert space + Hamiltonian operator, e.g. L 2 (R), T = d 2 dt 2 x 2 (quantum harmonic oscil.); Problem: find eigenvalues of T. To solve use p(f )(x) = df /dx, q(f )(x) = ixf (x), [p, q] = i.

14 Weyl algebras Examples of Lie algebras The n-th Weyl algebra A n = A n (k): generators: x i, i, i = 1,..., n relations x i x j = x j x i, i j = j i, i x j x j i = δ ij, i, j = 1,..., n. A n is a simple Noetherian domain;

15 Weyl algebras Examples of Lie algebras The n-th Weyl algebra A n = A n (k): generators: x i, i, i = 1,..., n relations x i x j = x j x i, i j = j i, i x j x j i = δ ij, i, j = 1,..., n. A n is a simple Noetherian domain;

16 Dixmier problem: End(A n ) Aut(A n )? e.g. x, y A 1, s.t. xy yx = 1 < x, y >= A 1?

17 Dixmier problem: End(A n ) Aut(A n )? e.g. x, y A 1, s.t. xy yx = 1 < x, y >= A 1?

18 Let P n = C[x 1,..., x n ]. φ End(P n ) define φ i = φ(x i ). Then φ defines a polynomial function F : C n C n, F(z 1,..., z n ) = (φ 1 (z 1 ),..., φ n (z n )). Let J F ( x) = ( f i / x j ), (F) = det(j F ). If F is invertible then (F) C.

19 Let P n = C[x 1,..., x n ]. φ End(P n ) define φ i = φ(x i ). Then φ defines a polynomial function F : C n C n, F (z 1,..., z n ) = (φ 1 (z 1 ),..., φ n (z n )). Let J F ( x) = ( f i / x j ), (F) = det(j F ). If F is invertible then (F) C.

20 Let P n = C[x 1,..., x n ]. φ End(P n ) define φ i = φ(x i ). Then φ defines a polynomial function F : C n C n, F (z 1,..., z n ) = (φ 1 (z 1 ),..., φ n (z n )). Let J F ( x) = ( f i / x j ), (F) = det(j F ). If F is invertible then (F) C.

21 Let P n = C[x 1,..., x n ]. φ End(P n ) define φ i = φ(x i ). Then φ defines a polynomial function F : C n C n, F (z 1,..., z n ) = (φ 1 (z 1 ),..., φ n (z n )). Let J F ( x) = ( f i / x j ), (F) = det(j F ). If F is invertible then (F) C.

22 Jacobian Conjecture: Let F : C n C n be a polynomial function such that (F ) C. Then F is invertible and the inverse is polynomial (Keller, 1939). Open for n > 2 True for n = 1 Dixmier Problem Jacobian Conjecture (Bass, McConnel, Wright, 1982; Bavula, 2001) Jac(2n) Dix(n) Jac(n) (Tsuchimoto, 2005; Belov-Kanel, Kontsevich, 2007)

23 Jacobian Conjecture: Let F : C n C n be a polynomial function such that (F ) C. Then F is invertible and the inverse is polynomial (Keller, 1939). Open for n > 2 True for n = 1 Dixmier Problem Jacobian Conjecture (Bass, McConnel, Wright, 1982; Bavula, 2001) Jac(2n) Dix(n) Jac(n) (Tsuchimoto, 2005; Belov-Kanel, Kontsevich, 2007)

24 Case n = 1: F(x) is polynomial and df dx F(x) = ax + b = a 0. Hence and the inverse is G = x b a F : C C, F(z) = e z. Then J F (z) = e z 0; F is not injective, e z is periodic with period 2πi. F : R R, F(x) = x 3 + x. Then J F (x) = 3x , F is bijective but the inverse is not polynomial.

25 Case n = 1: F(x) is polynomial and df dx F(x) = ax + b = a 0. Hence and the inverse is G = x b a F : C C, F(z) = e z. Then J F (z) = e z 0; F is not injective, e z is periodic with period 2πi. F : R R, F(x) = x 3 + x. Then J F (x) = 3x , F is bijective but the inverse is not polynomial.

26 Let g be a simple finite dimensional Lie algebra over C, A a commutative associative algebra. Consider a tensor product of vector spaces g A. Then g A becomes a Lie algebra with [x a, y b] = [x, y] ab, x, y g, a, b A. Denote Ω 1 A the space of Kähler 1-forms - the quotient of the free A-module with basis da, a A, modulo the relations: (1) d(ab) = (da)b + a(db), (2) d(a + b) = da + db and (3) dk = 0 for a, b A, k C.

27 Let g be a simple finite dimensional Lie algebra over C, A a commutative associative algebra. Consider a tensor product of vector spaces g A. Then g A becomes a Lie algebra with [x a, y b] = [x, y] ab, x, y g, a, b A. Denote Ω 1 A the space of Kähler 1-forms - the quotient of the free A-module with basis da, a A, modulo the relations: (1) d(ab) = (da)b + a(db), (2) d(a + b) = da + db and (3) dk = 0 for a, b A, k C.

28 C.Kassel (mid 1990 s): the universal central extension ĝ of L(g) = g A is the vector space L(g) Ω 1 A /da, where 0 Ω 1 A/dA ĝ L(g) 0, [x f, y g] := [xy] fg + (x, y)fdg, [x f, ω] = 0 for x, y g, f, g A, ω Ω 1 A /da, and (, ) denotes the Killing form on g.

29 Examples of Lie algebras Cartan matrix A = (a ij ), a ij 0, i j, a ii = 2, a ij = 0 a ji = 0 for all i, j, A is symmetrizable (there exists a diagonal invertible matrix D such that DA is symmetric) and DA is positive definite generators + Serre relations simple complex finite-dimensional Lie algebras. V.Kac, R.Moody (1967): without "positive definite" ; If A is positive semidefinite (det(a ij ) = 0, with positive principal minors) Affine Lie algebras

30 Examples of Lie algebras Cartan matrix A = (a ij ), a ij 0, i j, a ii = 2, a ij = 0 a ji = 0 for all i, j, A is symmetrizable (there exists a diagonal invertible matrix D such that DA is symmetric) and DA is positive definite generators + Serre relations simple complex finite-dimensional Lie algebras. V.Kac, R.Moody (1967): without "positive definite" ; If A is positive semidefinite (det(a ij ) = 0, with positive principal minors) Affine Lie algebras

31 Examples of Lie algebras Cartan matrix A = (a ij ), a ij 0, i j, a ii = 2, a ij = 0 a ji = 0 for all i, j, A is symmetrizable (there exists a diagonal invertible matrix D such that DA is symmetric) and DA is positive definite generators + Serre relations simple complex finite-dimensional Lie algebras. V.Kac, R.Moody (1967): without "positive definite" ; If A is positive semidefinite (det(a ij ) = 0, with positive principal minors) Affine Lie algebras

32 Examples of Lie algebras Cartan matrix A = (a ij ), a ij 0, i j, a ii = 2, a ij = 0 a ji = 0 for all i, j, A is symmetrizable (there exists a diagonal invertible matrix D such that DA is symmetric) and DA is positive definite generators + Serre relations simple complex finite-dimensional Lie algebras. V.Kac, R.Moody (1967): without "positive definite" ; If A is positive semidefinite (det(a ij ) = 0, with positive principal minors) Affine Lie algebras

33 Example Examples of Lie algebras A = C[t, t 1 ], ĝ = g C[t, t 1 ], with the Lie bracket [x t n, y t m ] = [x, y] t m+n (loop algebra). the universal central extension g = ĝ Cc Cd, [x t n, y t m ] = [x, y] t m+n + n(x, y)δ n+m,0 c, d : ĝ ĝ degree derivation, d(x t n ) = n(x t n ), d(c) = 0 (non-twisted affine Kac-Moody algebra).

34 Example Examples of Lie algebras A = C[t, t 1 ], ĝ = g C[t, t 1 ], with the Lie bracket [x t n, y t m ] = [x, y] t m+n (loop algebra). the universal central extension g = ĝ Cc Cd, [x t n, y t m ] = [x, y] t m+n + n(x, y)δ n+m,0 c, d : ĝ ĝ degree derivation, d(x t n ) = n(x t n ), d(c) = 0 (non-twisted affine Kac-Moody algebra).

35 Geometric interpretation: Examples of Lie algebras View C[t, t 1 ] as algebra of meromorphic functions on the Riemann sphere, holomorphic outside of t = 0 and t =. Then ĝ can be viewed as g-valued meromorphic functions on the Riemann sphere with possible poles only at t = 0 and t =. Generalise: take any compact Riemann surface X of arbitrary genus g and arbitrary set of points P where the poles are allowed. Then define g(x, P) = g R, where R the ring of meromorphic on X functions with poles in P: [x f, y g] = [x, y] fg. (Krichever-Novikov, Schlichenmaier, Sheinman, Bremner).

36 Geometric interpretation: Examples of Lie algebras View C[t, t 1 ] as algebra of meromorphic functions on the Riemann sphere, holomorphic outside of t = 0 and t =. Then ĝ can be viewed as g-valued meromorphic functions on the Riemann sphere with possible poles only at t = 0 and t =. Generalise: take any compact Riemann surface X of arbitrary genus g and arbitrary set of points P where the poles are allowed. Then define g(x, P) = g R, where R the ring of meromorphic on X functions with poles in P: [x f, y g] = [x, y] fg. (Krichever-Novikov, Schlichenmaier, Sheinman, Bremner).

37 Elliptic curve case: Examples of Lie algebras Consider a compact complex algebraic curve. (It can be represented as a quotient of the complex plane C by the lattice Λ = Z Zλ with positive imaginary part of λ. ) Take the ring A of meromorphic functions on Σ which are holomorphic outside of the set {0, 1 (1 + λ)}. 2 Let be the Weierstrass function: (z) := z 2 + ( 1 (z ω) 2 1 ) ω 2 Then 0 ω Λ

38 Elliptic curve case: Examples of Lie algebras Consider a compact complex algebraic curve. (It can be represented as a quotient of the complex plane C by the lattice Λ = Z Zλ with positive imaginary part of λ. ) Take the ring A of meromorphic functions on Σ which are holomorphic outside of the set {0, 1 (1 + λ)}. 2 Let be the Weierstrass function: (z) := z 2 + ( 1 (z ω) 2 1 ) ω 2 Then 0 ω Λ

39 Elliptic curve case: Examples of Lie algebras Consider a compact complex algebraic curve. (It can be represented as a quotient of the complex plane C by the lattice Λ = Z Zλ with positive imaginary part of λ. ) Take the ring A of meromorphic functions on Σ which are holomorphic outside of the set {0, 1 (1 + λ)}. 2 Let be the Weierstrass function: (z) := z 2 + ( 1 (z ω) 2 1 ) ω 2 Then 0 ω Λ

40 Elliptic curve case: Examples of Lie algebras Consider a compact complex algebraic curve. (It can be represented as a quotient of the complex plane C by the lattice Λ = Z Zλ with positive imaginary part of λ. ) Take the ring A of meromorphic functions on Σ which are holomorphic outside of the set {0, 1 (1 + λ)}. 2 Let be the Weierstrass function: (z) := z 2 + ( 1 (z ω) 2 1 ) ω 2 Then 0 ω Λ

41 (z) 2 = 4 (z) 3 g 2 (z) g 3, g 2 = 60 ξ 4, g 3 = 140 ξ 6. 0 ξ Λ 0 ξ Λ Theorem (Bremner) A = C[t, t 1, u u 2 = t 3 2bt 2 + t] ( ) 1 where b is some constant determined by m = (1 + λ). 2 In our example: Ω 1 A /da has a basis ω 0 := t 1 dt, ω := t 2 u dt, ω + := t 1 u dt.

42 (z) 2 = 4 (z) 3 g 2 (z) g 3, g 2 = 60 ξ 4, g 3 = 140 ξ 6. 0 ξ Λ 0 ξ Λ Theorem (Bremner) A = C[t, t 1, u u 2 = t 3 2bt 2 + t] ( ) 1 where b is some constant determined by m = (1 + λ). 2 In our example: Ω 1 A /da has a basis ω 0 := t 1 dt, ω := t 2 u dt, ω + := t 1 u dt.

43 (z) 2 = 4 (z) 3 g 2 (z) g 3, g 2 = 60 ξ 4, g 3 = 140 ξ 6. 0 ξ Λ 0 ξ Λ Theorem (Bremner) A = C[t, t 1, u u 2 = t 3 2bt 2 + t] ( ) 1 where b is some constant determined by m = (1 + λ). 2 In our example: Ω 1 A /da has a basis ω 0 := t 1 dt, ω := t 2 u dt, ω + := t 1 u dt.

44 (z) 2 = 4 (z) 3 g 2 (z) g 3, g 2 = 60 ξ 4, g 3 = 140 ξ 6. 0 ξ Λ 0 ξ Λ Theorem (Bremner) A = C[t, t 1, u u 2 = t 3 2bt 2 + t] ( ) 1 where b is some constant determined by m = (1 + λ). 2 In our example: Ω 1 A /da has a basis ω 0 := t 1 dt, ω := t 2 u dt, ω + := t 1 u dt.

45 Consider the sequence of polynomials p k (b), q k (b) determined by: t k 2 u dt = p k (b)t 1 u dt + q k (b)t 2 u dt. These are Pollaczek polynomials that satisfy the following recurrent relation: (k+γ)p k (b) = 2[(k+λ+α+γ 1)b+β]p k 1 (b) (k+2λ+γ 2)p k 2 (b) for the parameters λ = 1/2, α = 0, β = 1, γ = 1/2 together with the initial conditions p 0 (b) = 0, p 1 (b) = 1.

46 Consider the sequence of polynomials p k (b), q k (b) determined by: t k 2 u dt = p k (b)t 1 u dt + q k (b)t 2 u dt. These are Pollaczek polynomials that satisfy the following recurrent relation: (k+γ)p k (b) = 2[(k+λ+α+γ 1)b+β]p k 1 (b) (k+2λ+γ 2)p k 2 (b) for the parameters λ = 1/2, α = 0, β = 1, γ = 1/2 together with the initial conditions p 0 (b) = 0, p 1 (b) = 1.

47 Examples of Lie algebras Let p(t) = (t 2 a 2 )(t 2 b 2 ) = t 4 (a 2 + b 2 )t 2 + (ab) 2 with a ±b and neither a nor b is zero. Set A = C[t, t 1, u u 2 = (t 2 a 2 )(t 2 b 2 )].

48 Examples of Lie algebras Let p(t) = (t 2 a 2 )(t 2 b 2 ) = t 4 (a 2 + b 2 )t 2 + (ab) 2 with a ±b and neither a nor b is zero. Set A = C[t, t 1, u u 2 = (t 2 a 2 )(t 2 b 2 )].

49 (B.Cox, V.F., 2010): The universal central extension of the is the Z 2 -graded Lie algebra ĝ = ĝ 0 ĝ 1, ( ) ĝ 0 = g C[t, t 1 ] Cω 0, ( ) ĝ 1 = g C[t, t 1 ]u Cω 4 Cω 3 Cω 2 Cω 1, ω k = t k udt, k = 1, 2, 3, 4, ω 0 = t 1 dt. Then the recursion relation looks like

50 (B.Cox, V.F., 2010): The universal central extension of the is the Z 2 -graded Lie algebra ĝ = ĝ 0 ĝ 1, ( ) ĝ 0 = g C[t, t 1 ] Cω 0, ( ) ĝ 1 = g C[t, t 1 ]u Cω 4 Cω 3 Cω 2 Cω 1, ω k = t k udt, k = 1, 2, 3, 4, ω 0 = t 1 dt. Then the recursion relation looks like

51 (B.Cox, V.F., 2010): The universal central extension of the is the Z 2 -graded Lie algebra ĝ = ĝ 0 ĝ 1, ( ) ĝ 0 = g C[t, t 1 ] Cω 0, ( ) ĝ 1 = g C[t, t 1 ]u Cω 4 Cω 3 Cω 2 Cω 1, ω k = t k udt, k = 1, 2, 3, 4, ω 0 = t 1 dt. Then the recursion relation looks like

52 (6 + 2k)t k u dt = 2(k 3)t k 4 u dt + 4kct k 2 u dt, after setting c = (a 2 + b 2 )/2, so that p(t) = t 4 2ct Let P k := P k (c) be the polynomial in c satisfying the recursion relation for k 0. (6 + 2k)P k (c) = 4kcP k 2 (c) 2(k 3)P k 4 (c)

53 (6 + 2k)t k u dt = 2(k 3)t k 4 u dt + 4kct k 2 u dt, after setting c = (a 2 + b 2 )/2, so that p(t) = t 4 2ct Let P k := P k (c) be the polynomial in c satisfying the recursion relation for k 0. (6 + 2k)P k (c) = 4kcP k 2 (c) 2(k 3)P k 4 (c)

54 Set P(c, z) := P k (c)z k+4 = P k 4 (c)z k. k 4 k 0 Then P(c, z) satisfies the differential equation d dz P(c, z) 3z4 4cz z 5 2cz 3 P(c, z) = + z 2 (P 1 + cp 3 ) z 3 + P 2 z 2 + (4cz 2 1)P 4 z 5 2cz 3 + z

55 Set P(c, z) := P k (c)z k+4 = P k 4 (c)z k. k 4 k 0 Then P(c, z) satisfies the differential equation d dz P(c, z) 3z4 4cz z 5 2cz 3 P(c, z) = + z 2 (P 1 + cp 3 ) z 3 + P 2 z 2 + (4cz 2 1)P 4 z 5 2cz 3 + z

56 Take initial conditions P 3 (c) = P 2 (c) = P 1 (c) = 0 and P 4 (c) = 1 then the generating function P 4 (c, z) := k 4 P 4,k (c)z k+4 = k 0 is defined in terms of an elliptic integral P 4 (c, z) = z 1 2cz 2 + z 4 P 4,k 4 (c)z k, 4cz 2 1 z 2 (z 4 2cz 2 dz. + 1) 3/2

57 Take initial conditions P 3 (c) = P 2 (c) = P 1 (c) = 0 and P 4 (c) = 1 then the generating function P 4 (c, z) := k 4 P 4,k (c)z k+4 = k 0 is defined in terms of an elliptic integral P 4 (c, z) = z 1 2cz 2 + z 4 P 4,k 4 (c)z k, 4cz 2 1 z 2 (z 4 2cz 2 dz. + 1) 3/2

58 One way to interpret the right hand integral is to expand (z 4 2cz 2 + 1) 3/2 as a Taylor series about z = 0 and then formally integrate term by term and multiply the result by the Taylor series of z 1 2cz 2 + z 4. More precisely one integrates formally with zero constant term = n=0 (4c z 2 ) 4cQ (3/2) n n=0 (c) z 2n+1 2n + 1 Q (3/2) n (c)z 2n dz n=0 Q (3/2) n (c) 2n 1 z2n 1 where Q (λ) n (c) is the n-th Gegenbauer polynomial.

59 One way to interpret the right hand integral is to expand (z 4 2cz 2 + 1) 3/2 as a Taylor series about z = 0 and then formally integrate term by term and multiply the result by the Taylor series of z 1 2cz 2 + z 4. More precisely one integrates formally with zero constant term = n=0 (4c z 2 ) 4cQ (3/2) n n=0 (c) z 2n+1 2n + 1 Q (3/2) n (c)z 2n dz n=0 Q (3/2) n (c) 2n 1 z2n 1 where Q (λ) n (c) is the n-th Gegenbauer polynomial.

60 After multiplying this by z 1 2cz 2 + z 4 = n=0 one arrives at the series P 4 (c, z): z 1 2cz 2 + z 4 Q ( 1/2) n (c)z 2n+1 4cz 2 1 z 2 (z 4 2cz 2 + 1) 3/2 dz = P 4,n (c)z n n=0 = 1 + z 4 + 4c 5 z6 + 1 ( ) 32c 2 5 z ( ) c 8c 2 3 ( 2048c c ) 1155 z 12 + O(z 14 ) z 10

61 The first nonzero polynomials in c are P 4,4 (c) = 1, P 4,6 = 4c 5, P 4,8 = 32c whose coefficients satisfy the following differential equation:

62 16(c 2 1) 2 P (4) n + 160c(c 2 1)P (3) n a(n, c)p n 24c(n 2 4n 2)P n + (n 6)(n 2) 2 (n + 2)P n = 0

63 Vertex realizations Examples of Lie algebras Let dim g <, b a Borel subalgebra, g = LieG, b = LieB G acts transitively on the flag variety X = G/B (smooth algebraic variety) Take B = B. Action of N + on X leads to the decomposition of X into Shubert cells: X = w W C(w), C(w) = B + wb /B, W = N(T )/T the Weyl group, T = B + /N +.

64 Vertex realizations Examples of Lie algebras Let dim g <, b a Borel subalgebra, g = LieG, b = LieB G acts transitively on the flag variety X = G/B (smooth algebraic variety) Take B = B. Action of N + on X leads to the decomposition of X into Shubert cells: X = w W C(w), C(w) = B + wb /B, W = N(T )/T the Weyl group, T = B + /N +.

65 Vertex realizations Examples of Lie algebras Let dim g <, b a Borel subalgebra, g = LieG, b = LieB G acts transitively on the flag variety X = G/B (smooth algebraic variety) Take B = B. Action of N + on X leads to the decomposition of X into Shubert cells: X = w W C(w), C(w) = B + wb /B, W = N(T )/T the Weyl group, T = B + /N +.

66 Vertex realizations Examples of Lie algebras Let dim g <, b a Borel subalgebra, g = LieG, b = LieB G acts transitively on the flag variety X = G/B (smooth algebraic variety) Take B = B. Action of N + on X leads to the decomposition of X into Shubert cells: X = w W C(w), C(w) = B + wb /B, W = N(T )/T the Weyl group, T = B + /N +.

67 The largest Shubert cell U = N + (w = 1) can be identified with some affine space the ring of regular functions O U on U is a polynomial ring and g has an embedding into some Weyl algebra (acts by certain differential operators). Example Let g = sl(2), standard basis e, f, h, [e, f ] = h, [h, e] = 2e, [h, f ] = 2f, b = span{f, h} G = SL 2 (C) and X = G/B can be identified with the projective line P 1, U = A 1. Let x be a coordinate in U, O U = C[x] e d/dx, h 2xd/dx, f x 2 d/dx

68 The largest Shubert cell U = N + (w = 1) can be identified with some affine space the ring of regular functions O U on U is a polynomial ring and g has an embedding into some Weyl algebra (acts by certain differential operators). Example Let g = sl(2), standard basis e, f, h, [e, f ] = h, [h, e] = 2e, [h, f ] = 2f, b = span{f, h} G = SL 2 (C) and X = G/B can be identified with the projective line P 1, U = A 1. Let x be a coordinate in U, O U = C[x] e d/dx, h 2xd/dx, f x 2 d/dx

69 First free field realization for Affine Lie algebras Let g = sl(2), e n = e t n, h n = h t n, f n = f t n, n Z. Using natural Borel subalgebra (which leads to "semi-infinite variety" by Feigin, Frenkel, Voronov) we obtain e n x n, h n 2 m Z x m x n+m, f n m,k Z x m x k x n+m+k (not well-defined on V = C[x m, m Z]) apply the anti-involutions: e n f n, h n h n ; x n x n, n Z

70 First free field realization for Affine Lie algebras Let g = sl(2), e n = e t n, h n = h t n, f n = f t n, n Z. Using natural Borel subalgebra (which leads to "semi-infinite variety" by Feigin, Frenkel, Voronov) we obtain e n x n, h n 2 m Z x m x n+m, f n m,k Z x m x k x n+m+k (not well-defined on V = C[x m, m Z]) apply the anti-involutions: e n f n, h n h n ; x n x n, n Z

71 First free field realization for Affine Lie algebras Let g = sl(2), e n = e t n, h n = h t n, f n = f t n, n Z. Using natural Borel subalgebra (which leads to "semi-infinite variety" by Feigin, Frenkel, Voronov) we obtain e n x n, h n 2 m Z x m x n+m, f n m,k Z x m x k x n+m+k (not well-defined on V = C[x m, m Z]) apply the anti-involutions: e n f n, h n h n ; x n x n, n Z

72 First free field realization for Affine Lie algebras Let g = sl(2), e n = e t n, h n = h t n, f n = f t n, n Z. Using natural Borel subalgebra (which leads to "semi-infinite variety" by Feigin, Frenkel, Voronov) we obtain e n x n, h n 2 m Z x m x n+m, f n m,k Z x m x k x n+m+k (not well-defined on V = C[x m, m Z]) apply the anti-involutions: e n f n, h n h n ; x n x n, n Z

73 First free field realization for Affine Lie algebras Let g = sl(2), e n = e t n, h n = h t n, f n = f t n, n Z. Using natural Borel subalgebra (which leads to "semi-infinite variety" by Feigin, Frenkel, Voronov) we obtain e n x n, h n 2 m Z x m x n+m, f n m,k Z x m x k x n+m+k (not well-defined on V = C[x m, m Z]) apply the anti-involutions: e n f n, h n h n ; x n x n, n Z

74 First free field realization for Affine Lie algebras Let g = sl(2), e n = e t n, h n = h t n, f n = f t n, n Z. Using natural Borel subalgebra (which leads to "semi-infinite variety" by Feigin, Frenkel, Voronov) we obtain e n x n, h n 2 m Z x m x n+m, f n m,k Z x m x k x n+m+k (not well-defined on V = C[x m, m Z]) apply the anti-involutions: e n f n, h n h n ; x n x n, n Z

75 Theorem (Jakobsen-Kac, 85; Bernard-Felder, 90) Formulas e n ( m f n x n, h n (λ n + 2 m Z x n+m x m ), λ m+n x m + m,k Z x n+m+k x m x k ) define a representation of ĝ on a Fock space C[x m, m Z] with c = 0, where λ m = S 1 z m dµ for any finite measure µ on the unit circle

76 Second free field realization Examples of Lie algebras a(z) = n<0 a n z n 1, a(z) + = n 0 a n z n 1 normal ordering : a(z)b(z) := a(z) b(z) + b(z)a + (z) Let a n = { x n, n < 0 x n, n 0, a n = { x n, n 0 x n, n > 0, [a n, a m] = δ n+m,0

77 Second free field realization Examples of Lie algebras a(z) = n<0 a n z n 1, a(z) + = n 0 a n z n 1 normal ordering : a(z)b(z) := a(z) b(z) + b(z)a + (z) Let a n = { x n, n < 0 x n, n 0, a n = { x n, n 0 x n, n > 0, [a n, a m] = δ n+m,0

78 Second free field realization Examples of Lie algebras a(z) = n<0 a n z n 1, a(z) + = n 0 a n z n 1 normal ordering : a(z)b(z) := a(z) b(z) + b(z)a + (z) Let a n = { x n, n < 0 x n, n 0, a n = { x n, n 0 x n, n > 0, [a n, a m] = δ n+m,0

79 Theorem (Wakimoto, 86) The formulas c K, e(z) a(z), h(z) 2 : a (z)a(z) : f (z) : a (z) 2 a(z) : +K z a (z) define the second free field realization of the affine sl(2) acting on the Fock space C[x n, n Z]. B.Feigin, E.Frenkel (1990) - for any Affine Lie algebra. Theorem (Cox-V.F., 06) Boson type realizations exist for any Borel subalgebra of ŝl(n)

80 Theorem (Wakimoto, 86) The formulas c K, e(z) a(z), h(z) 2 : a (z)a(z) : f (z) : a (z) 2 a(z) : +K z a (z) define the second free field realization of the affine sl(2) acting on the Fock space C[x n, n Z]. B.Feigin, E.Frenkel (1990) - for any Affine Lie algebra. Theorem (Cox-V.F., 06) Boson type realizations exist for any Borel subalgebra of ŝl(n)

81 Theorem (Wakimoto, 86) The formulas c K, e(z) a(z), h(z) 2 : a (z)a(z) : f (z) : a (z) 2 a(z) : +K z a (z) define the second free field realization of the affine sl(2) acting on the Fock space C[x n, n Z]. B.Feigin, E.Frenkel (1990) - for any Affine Lie algebra. Theorem (Cox-V.F., 06) Boson type realizations exist for any Borel subalgebra of ŝl(n)

82 Vertex algebras Examples of Lie algebras McKay-Thompson conjecture: M and a Z-graded M-module V s.t. (dimv n )q n = J(q), n 1 J(q) = q q q R.Griess, 1980, largest sporadic group - Monster ("friendly giant"), M 10 54, M < AutB, dimb = , commutative, non-associative I.Frenkel-Lepowsky-Meurman: constructed V ("moonshine module"). It has a structure of "vertex operator algebra", M = AutV

83 Vertex algebras Examples of Lie algebras McKay-Thompson conjecture: M and a Z-graded M-module V s.t. (dimv n )q n = J(q), n 1 J(q) = q q q R.Griess, 1980, largest sporadic group - Monster ("friendly giant"), M 10 54, M < AutB, dimb = , commutative, non-associative I.Frenkel-Lepowsky-Meurman: constructed V ("moonshine module"). It has a structure of "vertex operator algebra", M = AutV

84 Vertex algebras Examples of Lie algebras McKay-Thompson conjecture: M and a Z-graded M-module V s.t. (dimv n )q n = J(q), n 1 J(q) = q q q R.Griess, 1980, largest sporadic group - Monster ("friendly giant"), M 10 54, M < AutB, dimb = , commutative, non-associative I.Frenkel-Lepowsky-Meurman: constructed V ("moonshine module"). It has a structure of "vertex operator algebra", M = AutV

85 Example(Lepowsky-Wilson). W = C[x 1/2, x 3/2..., ], Y = exp( n=1/2,3/2,... y n n x n )exp( 2 n=1/2,3/2,... y n x n ) < 1, y n, / y n, Y i, n, i 1/2Z, n > 0 > ŝl(2). This is principal realization of ŝl(2). Homogeneous realization is due to I.Frenkel-Kac and Seagal. Integrable representations character formula Macdonald identities for Dedekind η-function (Kac, Lepowsky-Wilson, Dyson,...), e.g. Jacobi identity: Π n 1 (1 x n y n )(1 x n 1 y n )(1 x n y n 1 ) = j Z ( 1) j x 1/2j(j+1) y 1/2j(j 1)

86 Example(Lepowsky-Wilson). W = C[x 1/2, x 3/2..., ], Y = exp( n=1/2,3/2,... y n n x n )exp( 2 n=1/2,3/2,... y n x n ) < 1, y n, / y n, Y i, n, i 1/2Z, n > 0 > ŝl(2). This is principal realization of ŝl(2). Homogeneous realization is due to I.Frenkel-Kac and Seagal. Integrable representations character formula Macdonald identities for Dedekind η-function (Kac, Lepowsky-Wilson, Dyson,...), e.g. Jacobi identity: Π n 1 (1 x n y n )(1 x n 1 y n )(1 x n y n 1 ) = j Z ( 1) j x 1/2j(j+1) y 1/2j(j 1)

87 Example(Lepowsky-Wilson). W = C[x 1/2, x 3/2..., ], Y = exp( n=1/2,3/2,... y n n x n )exp( 2 n=1/2,3/2,... y n x n ) < 1, y n, / y n, Y i, n, i 1/2Z, n > 0 > ŝl(2). This is principal realization of ŝl(2). Homogeneous realization is due to I.Frenkel-Kac and Seagal. Integrable representations character formula Macdonald identities for Dedekind η-function (Kac, Lepowsky-Wilson, Dyson,...), e.g. Jacobi identity: Π n 1 (1 x n y n )(1 x n 1 y n )(1 x n y n 1 ) = j Z ( 1) j x 1/2j(j+1) y 1/2j(j 1)

88 Vertex algebras Examples of Lie algebras Chiral algebras in CFT (Belavin, Polyakov, Zamolodchikov); modular invarience (Witten); vertex algebras (Borcherds). A vertex algebra is a vector space V with a distinguished vector 1l (vacuum vector) in V, an operator D (translation) on the space V, and a linear map Y (state-field correspondence) Y (, z) : V (EndV )[[z, z 1 ]], where a (n) EndV. a Y (a, z) = n Z a (n) z n 1

89 Vertex algebras Examples of Lie algebras Chiral algebras in CFT (Belavin, Polyakov, Zamolodchikov); modular invarience (Witten); vertex algebras (Borcherds). A vertex algebra is a vector space V with a distinguished vector 1l (vacuum vector) in V, an operator D (translation) on the space V, and a linear map Y (state-field correspondence) Y (, z) : V (EndV )[[z, z 1 ]], where a (n) EndV. a Y (a, z) = n Z a (n) z n 1

90 following axioms hold: (V1) For any a, b V, a (n) b = 0 for n sufficiently large; (V2) [D, Y (a, z)] = Y (D(a), z) = d dz Y (a, z) for any a V ; (V3) Y (1l, z) = Id V z 0 ; (V4) Y (a, z)1l V [[z]] and Y (a, z)1l z=0 = a for any a V (self-replication); (V5) Locality: (z w) n [Y (a, z), Y (b, w)] = 0, for n sufficiently large.

91 A vertex algebra V is called a vertex operator algebra (VOA) if, in addition, V contains a vector ω (Virasoro element) such that (V6) The components L n = ω (n+1) of the field Y (ω, z) = n Z ω (n) z n 1 = n Z L n z n 2 satisfy the Virasoro algebra relations: [L n, L m ] = (n m)l n+m + δ n, m n 3 n 12 C vir, where C vir acts on V by scalar, called the rank of V. (V7) D = L 1 ; (V8) Operator L 0 is diagonalizable on V.

92 Affine gl N VOA. Examples of Lie algebras. Let with the standard Lie bracket. ĝl N = C[t, t 1 ] gl N Cc Let ĝl N = ĝl N ĝl0 N ĝl+ N with ĝl0 N = 1 gl N Cc and ĝl ±1 N = t±1 C[t ±1 ] gl N.

93 Let C1l be a 1-dimensional module for ĝl0 N ĝl+ N with the following action: (C[t] gl N ) 1l = 0, c1l = 1l. The space of the affine ĝl N vertex algebra at level 1 is the induced ĝl N -module V gl = Ind ĝl N C1l = ĝl 0 U(ĝl N ) 1l. N ĝl+ N