Automorphisms and twisted forms of Lie conformal superalgebras

Transcription

1 Algebra Seminar Automorphisms and twisted forms of Lie conformal superalgebras Zhihua Chang University of Alberta April 04, Dept of Math and Stats, University of Alberta, Edmonton, Alberta, T6G 2G1, Canada. 1 / 40

2 2 / 40 Outlines Twisted forms of finite dimensional algebras. Differential conformal superalgebras and their twisted forms. The case of the N = 1, 2, 3 Lie conformal superalgebras. The case of the N = 4 Lie conformal superalgebras.

3 3 / 40 Twisted affine Kac-Moody algebras Let g a finite dimensional simple Lie algebra over k, σ an automorphism of g of order m. The twisted loop algebra L(g, σ) is the subalgebra L(g, σ) = i Z g i k kt i m g k k[t ± 1 m ], where g i = {x g σ(x) = ζ i mx}, ζ m is a m-th primitive root of unit. In particular, L(g, id) = g k k[t ±1 ].

4 4 / 40 Twisted affine Kac-Moody algebras (continued) Facts: Every affine Kac-Moody algebra is an extension of a twisted loop algebra. L(g, σ) is not only a Lie algebra over k, but also a Lie algebra over the ring k[t ±1 ]. L(g, σ) k[t ±1 ] k[t ± 1 m ] = L(g, id) k[t ±1 ] k[t ± 1 m ] = g k k[t ± 1 m ]. L(g, σ) is a twisted form of L(g, id) with respect to the ring extension k[t ±1 ] k[t ± 1 m ].

5 Twisted forms of algebras Setting: R is a commutative ring. A is an algebra over R. R S a ring homomorphism. Change of base ring from R to S: An S algebra structure on A R S given by (a r)(b s) = ab rs, a, b A, r, s S S/R-form of A: An R algebra B is called a S/R form of A if A R S = B R S as S algebras. 5 / 40

6 6 / 40 Classification of twisted forms Theorem Let A be an algebra over R and R S a faithfully flat ring extension, then the set of isomorphism classes of S/R forms of A one to one H 1 (S/R, Aut(A)).

7 7 / 40 The automorphism group functor Aut(A): a functor from the category of R algebras to the category of groups S Aut S (A R S). For a finite dimensional algebra A over k, Aut(A ) is a linear algebraic group. Aut(A ) is a representable functor: R Aut R (A k R). If A = A k R, then Aut(A) = Aut(A ) R.

8 8 / 40 Outlines Twisted forms of finite dimensional algebras. Differential conformal superalgebras and their twisted forms. The case of the N = 1, 2, 3 Lie conformal superalgebras. The case of the N = 4 Lie conformal superalgebras.

9 9 / 40 Lie conformal algebras over k Definition A k Lie conformal algebra is k[ ] module A equipped with a k bilinear product (n) : A A A for each n N satisfying: for a, b, c A, (C0) a (n) b = 0 for n 0, (C1) A (a) (n) b = na (n 1) b, (C2) a (n) b = j N ( 1)j+n (j) A (b (n+j)a), (C3) a (m) (b (n) c) = m ( m ) j=0 j (a(j) b) (m+n j) c + b (n) (a (m) c),

10 10 / 40 The λ bracket Let A be a k Lie conformal algebra. For a, b A, define where λ (n) = 1 n! λn. [a λ b] = n N λ (n) a (n) b, The axioms (C0)-(C3) are equivalent respectively to (C0) λ [a λ b] is a polynomial in λ, (C1) λ [( a) λ b] = λ[a λ b], (C2) λ [a λ b] = [b λ a], (C3) λ [a λ [b µ c]] = [[a λ b] λ+µ c] + [b µ [a λ c]].

11 11 / 40 Example The centreless Virasoro Lie algebra: basis: {l n, n Z}, [l m, l n ] = (m n)l m+n, m, n Z. Form a formal series: L(z) = n Z l n z n 2. The operator product expansion: [L(z), L(w)] := [l m, l n ]z m 2 w n 2 where m,n Z = ( w L(w))δ(z w) + 2L(w) w δ(z w) δ(z w) = 1 z n Z ( ) w n. z

12 12 / 40 Example (continued) The Virasoro Conformal algebra: the k[ ] module V = k[ ]L. the n-th product: L (0) L = L, L (1) L = 2L, L (n) L = 0, n 2. The λ bracket [L λ L] = ( + 2λ)L.

13 Differential rings To define conformal superalgebra over a ring, rings = differential rings A k differential ring is a pair R = (R, δ) consisting of a commutative associative k algebra R, and a k linear derivation δ : R R. k differential rings form a category k-drng, where a morphism f : R = (R, δ R ) S = (S, δ S ) is a homomorphism of k algebra f : R S such that f δ R = δ S f. 13 / 40

14 14 / 40 Differential Lie conformal superalgebras Definition Let R = (R, δ) be a k differential ring, an R Lie conformal superalgebra is a triple (A, A, ( (n) ) n N ) consisting of a Z/2Z graded R module A = A 0 A 1, A End k (A) stabilizing the even and odd parts of A, a k bilinear product (a, b) a (n) b, a, b A for each n N, satisfying the following axioms for r D, a, b, c A, m, n N: (CS0) a (n) b = 0 for n 0, (CS1) A (a) (n) b = na (n 1) b and a (n) A (b) = A (a (n) b) + na (n 1) b, (CS2) A (ra) = r A (a) + δ(r)a, (CS3) a (n) (rb) = r(a (n) b) and (ra) (n) b = j N δ(j) (r)(a (n+j) b), (CS4) a (n) b = p(a, b) j N ( 1)j+n (j) A (b (n+j)a), (CS5) a (m) (b (n) c) = m j=0 ( m j ) (a(j) b) (m+n j) c + p(a, b)b (n) (a (m) c),

15 Change the base ring Let R = (R, δ R ) be a k differential ring, A an R-Lie conformal superalgebra, R = (R, δ R ) S = (S, δ S ) an extension of k differential rings. Then we can form a S Lie conformal superalgebra A R S: The underlying S module: A R S, A R S = A id + id δ S, The n-th product for n N: (a f ) (n) (b g) = j N a (n+j) b δ (j) S (f )g, where δ (j) S = 1 j! δj S. 15 / 40

16 16 / 40 Twisted forms Definition (Kac, Lau, Pianzola, 2009) Let R S be an extension of k differential rings, A an R Lie conformal superalgebra. An S/R form of A is an R Lie conformal superalgebra B such that B R S = A R S as S Lie conformal superalgebras.

17 17 / 40 Automorphisms Let A be an R Lie conformal superalgebra, a map φ : A A is called an R automorphism of A if: φ is an isomorphism of Z/2Z graded R modules, φ A = A φ, φ preserves all n-th products. All R automorphisms of A form a group, denoted by Aut R-conf (A).

18 18 / 40 Automorphism group functor The automorphism group functor Aut(A): the category of R extensions the category of groups S Aut S-conf (A R S)

19 19 / 40 Classification of twisted forms Theorem (Kac, Lau, Pianzola, 2009) Let R S be a faithfully flat extension of k differential rings, A an R conformal superalgebra. Then the set of isomorphism classes of S/R forms of A (up to R automorphism) one to one H 1 (S/R, Aut(A))

20 20 / 40 Twisted loop conformal superalgebras Let A be a Lie conformal superalgebra over k, σ an automorphism of A of order m, D = (k[t ±1 ], d dt ), D m = (k[t ± 1 m ], d dt ), D = lim D m = (k[t q, q Q], d dt ). A twisted loop Lie conformal algebra L(A, σ) = i Z A i kt i m A k D m where A i = {a A σ(a) = ζ i ma}, i Z, ζ m is a primitive m-th root of unit.

21 21 / 40 Twisted loop conformal superalgebras (continued) Facts: L(A, σ) is a D Lie conformal superalgebra; L(A, σ) is a D m /D form of L(A, id) = A k D. L(A, σ) is a D/D form of L(A, id).

22 Finiteness condition Proposition Let A an D conformal superalgebras. If A satisfies: There exist a 1,, a n A such that { l A (ra i) r D, l 0} spans A. Then H 1 ( D/D, Aut(A)) = H 1 ct(π 1 (D), Aut(A)( D)), where π 1 (D) is the algebraic fundamental group of Spec(D) at the geometric point a = Spec(C(t)) and H 1 ct denotes the continuous non-abelian cohomology of the profinite group π 1 (D) = Ẑ acting (continuously) on the group Aut(A)( D). 22 / 40

23 23 / 40 Centroid trick method The centroid of A is Ctd R (A) = {χ End R-smod (A) χ(a (n) b) = a (n) χ(b), a, b A, n Z + }, Proposition Let A i, i = 1, 2 be two R conformal superalgebras. Assume that Aut k (R) = 1. If R Ctd k (A i ), i = 1, 2 are both k algebra isomorphisms. Then A 1 = A2 as k Lie conformal superalgebras if and only if A 1 = A2 as R Lie conformal superalgebras.

24 24 / 40 Outlines for completed works Automorphisms and twisted forms of N = 1, 2, 3 Lie conformal superalgebras Automorphisms of the N = 4 Lie conformal superalgebras

25 25 / 40 N = 1, 2, 3 Lie conformal superalgebras K N K N = k[ ] k Λ(N), where Λ(N) is the Grassmann superalgebra over k in N variables ξ 1,, ξ N. For n N, the n-th product on K N is given by ( ) 1 f (0) g = 2 f 1 fg ( 1) f ( ) 1 f (1) g = ( f + g ) 2 fg, 2 f (n) g = 0, n 2. N i=1 ( i f )( i g)

26 26 / 40 A subgroup functor GrAut(K N )(R) = {φ Aut(K N )(R) φ(k k Λ(N) k R) k k Λ(N) k R}. Theorem Let R = (R, δ) be an arbitrary k differential ring. For N = 1, 2, 3, GrAut(K N )(R) = O N (R), where O N is the group scheme of N N orthogonal matrices. The isomorphism is functorial in R.

27 27 / 40 The construction of the isomorphism. Given A O N (R), it defines an automorphism ϕ A of K N k R N = 1, A = a R, ϕ A (1) = 1, ϕ A (ξ 1 ) = ξ 1 a. N = 2, A = (a ij ), φ A (1) = 1 + ξ 1 ξ 2 r, φ A (ξ 1 ) = ξ 1 a 11 + ξ 2 a 21, φ A (ξ 1 ξ 2 ) = ξ 1 ξ 2 det(a), φ A (ξ 2 ) = ξ 1 a 12 + ξ 2 a 22, where ( ) 0 r = 2δ(A)A r 0 T.

28 28 / 40 The construction of the isomorphism (continued) For N = 3, 3 φ A (1) = 1 + ɛ mnl ξ m ξ n r l, φ A (ξ j ) = l=1 3 ξ l a lj + ξ 1 ξ 2 ξ 3 s j, l=1 φ A (ξ i ξ j ) = ɛ ijl 3 ɛ mnl ξ m ξ n A l l, l =1 φ A (ξ 1 ξ 2 ξ 3 ) = ξ 1 ξ 2 ξ 3 det(a), i, j = 1, 2, 3, i j, where A l l is the cofactor of a l l in A and 0 r 3 r 2 r 3 0 r 1 = 2δ(A)A T, 0 s 3 s 2 s 3 0 s 1 = 2(det A)A T δ(a). r 2 r 1 0 s 2 s 1 0

29 29 / 40 Theorem Let R = (R, δ) be a k differential ring. If R is an integral domain, then for N = 1, 2, 3, GrAut(K N )(R) = Aut(K N )(R).

30 30 / 40 Classification up to R isomorphism Theorem Let N = 1, 2, 3. There are exactly two D/D forms of K N k D (up to isomorphism of D Lie conformal superalgebras): L(K N, id) and L(K N, ω N ).

31 31 / 40 Classification up to k isomorphism Proposition (Chang, Pianzola, 2011) Let A = L(K N, σ) where σ is an automorphism of K N of order m, and N = 1, 2, 3. Then the canonical map D Ctd k (A) is a k algebra isomorphism. Theorem (Chang, Pianzola, 2011) There are exactly two twisted loop Lie conformal superalgebra (up to k isomorphism) based on each K N, N = 1, 2, 3: L(K N, id) and L(K N, ω N ).

32 32 / 40 Corresponding Lie superalgebras Let A be a k Lie conformal superalgebra, every twisted loop algebra L(A, σ) corresponds to a Lie superalgebra over k: L(A, σ) Alg(A, σ) = ( ), + d dt L(A, σ) where the super bracket comes from the 0-th product on L(A, σ). For K N, N = 1, 2, 3, L(K N, id) and L(K N, ω N ) yield Alg(K N, id) and Alg(K N, ω N ). Theorem (Chang, Pianzola, 2011) Alg(K N, id) and Alg(K N, ω N ) are non-isomorphic.

33 33 / 40 Outlines for completed works Automorphisms and twisted forms of N = 1, 2, 3 Lie conformal superalgebras Automorphisms of the N = 4 Lie conformal superalgebras

34 34 / 40 N = 4 Lie conformal superalgebra F where F = F 0 F 1 = (k[ ] V 0) (k[ ] V 1), V 0 = kl kt 1 kt 2 kt 3, V 1 = kg 1 kg 2 kg 1 kg 2. The λ bracket on L is given by [L λ L] = ( + 2λ)L, [L λ T i ] = ( + λ)t i, [L λ G p ] = ( λ) G p, [T i λt j ] = iɛ ijk T k, [ Lλ G p ] ( = λ) G p, [G p λg q ] = 0, [T i λg p ] = q=1 σi pqg q [ p, G λg q ] = 0, [ T i λ G p ] = q=1 σi qpg q, [ G p λ G q ] = 2δpq L 2( + 2λ) 3 i=1 σi pqt i,

35 35 / 40 A subgroup functor GrAut(F)(R) = {φ Aut(F)(R) φ(k k V k R) k k V k R}, where V = V 0 V 1 and R = (R, δ) is a k differential ring.

36 36 / 40 GrAut(F) Theorem For every k differential ring R = (R, δ), there is an exact sequence of groups: 1 µ 2 (R 0 ) SL 2 (R) SL 2 (R 0 ) ιr GrAut(F)(R), where R 0 = ker δ, SL 2 is the group scheme of matrices with determinant 1, µ 2 is the group scheme of square roots of 1. The exact sequence is functorial in R.

37 37 / 40 The surjectivity of ι R Proposition For every k differential ring R = (R, δ R ) with R an integral domain, and ϕ GrAut(F)(R), there is an étale extension S = (S, δ S ) of R, and an element (A, B) SL 2 (S) SL 2 (S 0 ) such that ι S (A, B) = ϕ S, where ϕ S is the image of ϕ under GrAut(F)(R) GrAut(F)(S). Proposition Let R = (R, δ) be a k differential ring. If R is an integral domain, and the étale cohomology set H 1 ét (R, µ 2) is trivial, then ι R is surjective.

38 38 / 40 Aut(F) Theorem Let R = (R, δ) be a k differential ring such that R is an integral domain. Then GrAut(F)(R) = Aut(F)(R).

39 39 / 40 References Z. Chang, and A. Pianzola, Automorphisms and twisted forms of the N = 1, 2, 3 Lie conformal superalgebras, Communications in Number Theory and Physics, 5(4), , (2011). V. Kac, M. Lau and A. Pianzola, Differential Conformal superalgebras and their forms, Advances in Mathematics 222 (2009), A. Schwimmer, N. Seiberg, Comments on the N = 2, 3, 4 superconformal algebras in two dimensions, Physics Letters B, 184, , (1987). W. Waterhouse, Introduction to Affine Group Schemes, Graduate Texts in Mathematics 66, Springer, (1979).

40 40 / 40 Thank You!