Simple classical Lie algebras in characteristic 2 and their gradations, II.

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1 Simple classical Lie algebras in characteristic 2 and their gradations, II. A. N. Grishkov 1 M. Guerreiro 2 Rua do Matão, Butantã CEP São Paulo-SP - Brazil grishkov@ime.usp.br Departamento de Matemática - UFV CEP Viçosa-MG - Brazil marines@ufv.br 1 Introduction This paper is a continuation of [GG]. Here we prove Conjecture 5.1 [GG]. Recall some notations and definition of [GG]. Definition 1.1. Let I n = {1,..., n}. We call a P(I n ) = {σ σ I n } an even set if for all σ, τ a, we have σ τ 0 and σ τ 0 mod 2. We note that P(I n ) is an elementary abelian group with the operation σ τ = (σ \ τ) (τ \ σ). For a P(I n ), < a > denotes the group generated by a. Definition 1.2. A subset H of P(I n ) is connected if, for every partition I n = I J, there is σ H such that σ I and σ J. Definition 1.3. A subset σ I n is called a-even if µ τ 0 mod 2 for all τ a. A subset B P(I n ) is called an a-even set if all its elements are a-even. For an even set a P(I n ), in [GG] we defined a commutative algebra S = S(a) with basis { e i, h i, f i, h σ i I n, σ < a > \ } and multiplication given 1 Supported by FAPEMIG as visitor to the Depto. de Matemática - UFV in February Supported by FAPESP as visitor to IME-USP in February

2 by e i f i = h i, e i h σ = e i, f i h σ = f i, for i σ, (1) and zero for all other cases. We denoted h µ = i µ h i and defined h = 0, h = 0. The algebra S(a) contains a central ideal I generated by {h σ + h τ + h σ τ + h σ τ σ, τ < a >}. We denote S(a) = S = S(a)/I. For every a-even set σ, we also defined an S-module Λ σ whose basis is {(σ, µ) µ σ} and the S-action is given by (σ, µ) e i = (σ, µ i), i σ \ µ; (σ, µ) f i = (σ, µ \ i), i µ; (σ, µ) h i = (σ, µ), i σ; (2) ( ) σ ϕ (σ, µ) h ϕ = + ϕ µ (σ, µ), for ϕ a, 2 and for all other cases the action is zero. Now let = {0} a. Definition 1.4. An algebra A is called a -algebra if A = A α α and, for every α β a, we have A α A β A α β, A 2 0 A 0, A 0 A α A α, A α A α A 0 + A and A 0 A = 0. Define a commutative -graded algebra Λ as follows. As a k-space, Λ is Λ = Λ 0 Λ σ, where Λ 0 = S(a). (3) σ a Moreover, S = S(a) is a subalgebra of Λ and, by (2), each Λ σ is an S-module. For σ τ a, the multiplication is given by (σ, µ) (τ, ϕ) = (σ τ, (µ \ τ) (ϕ \ σ)), if µ ϕ =, µ ϕ σ τ. (4) 2

3 e i, µ ϕ = i, µ ϕ = σ, (σ, µ) (σ, ϕ) = f i, µ ϕ =, µ ϕ = σ \ i, h σ + h ϕ + (, ), µ ϕ =, µ ϕ = σ, and all other products are zero. (5) Recall the definition of the product of two -algebras. Let A = A α α and B = B α be two -algebras. Then A B = A α B α is a α α -algebra with multiplication [, ] given by [a α b α, a β b β ] = γ c γ d γ, if a α a β = γ c γ, b α b β = γ d γ. Proposition 1.1. Let a be an even set, Λ = Λ(a) and = {0} a. Let M = M 0 σ a M σ be a commutative -algebra. Then the algebra L = Λ M is a Lie algebra if and only if M satisfies a list of -identities given in Proposition 3.2 [GG]. We recall some of the -identities which will be used in this paper. a σ b τ c λ + b τ c λ a σ + c λ a σ b τ = 0, σ τ λ = 0, σ τ λ σ τ λ, (6) (a σ b σ ) c τ = 0, (a σ b σ ) 0 c τ = a σ c τ b σ, σ τ, σ τ = 2, (7) a σ c τ b σ = a σ c τ b σ + (a σ b σ ) c τ, σ τ = 0, (8) (a σ b τ c λ ) 0 = (a σ b τ c λ ) 0, λ = σ τ, (9) (ab) 0 c = (cb) 0 a, (ca) b = 0, a, b, c M τ, τ = 4, (10) (ab) 0 c + (bc) 0 a = (ac) b, a, b, c M τ, τ = 2, (11) (a σ b σ ) 0 c τ = 0, σ > 4, (12) (a b σ c σ ) + (b σ c σ ) a + (c σ a b σ ) = 0 (13) (a σ b σ ) 0 c 0 = (a σ c 0 b σ ) 0, σ (14) We observe that if M is simple, then L = Λ M is not necessarily a simple algebra, but L/Z(L) is simple, where Z(L) is the center of L. 3

4 Let a be an even connected set and = {0} a. Let M be the variety of -algebras satisfying the list of identities of Proposition 3.2 [GG]. Let M = M 0 σ a M σ M be a commutative -algebra in M. In [GG] (see Theorem 3.1 [GG]) we classified the simple -algebras of the variety M, for which M = 0. Now we consider the case when M is abelian. In the final section of [GG], we remarked that Theorem 3.1 [GG] is not true if we omit the condition / a and we formulated the following conjecture. Conjecture 1.1. Let M be an arbitrary simple finite dimensional -algebra which satisfies all the list of identities of Proposition 3.2 [GG] and M 2 = 0. Then the corresponding Lie algebra L = M Λ is a simple Lie algebra of type B 2l, C l, D 2l+1, E 7 or E 8. 2 Proof of Conjecture 1.1 In this section we prove Conjecture 1.1. For each σ a, define M 0 σ = {x M σ xm σ M } = {x M σ (xm σ ) 0 = 0}. Lemma 2.1. I = Mσ 0 σ a\ σ a\ (M σ Mσ) 0 is an ideal in M. Proof. (a) First we prove that M τ M 0 σ Mσ τ 0, for all σ τ a. Indeed, by (9), for a τ M τ, b σ M 0 σ, c λ M 0 σ τ, we have (a τ b σ c λ ) 0 = (b σ a τ c λ ) 0 = 0. (b) Now we prove that (M σ M 0 σ)m τ M 0 τ, for all τ a. We need to prove that (((b σ c σ ) a τ )d τ ) 0 = 0 for all a τ M τ, b σ M 0 σ, c σ M σ, d τ M τ. We have two cases: (b.1) σ τ. If σ τ = 2, we have by (7) that (((b σ c σ ) a τ )d τ ) 0 = 0. If σ τ = 0 then, by (8) and (9), (((b σ c σ ) a τ )d τ ) 0 = ((b σ a τ c σ )d τ ) 0 + ((c σ a τ b σ )d τ ) 0 = (b σ a τ c σ d τ ) 0 + (b σ (c σ a τ d τ )) 0 = (b σ a τ (c σ d τ )) 0 = 0. 4

5 (b.2) σ = τ. If σ = 2, then by (11) we have (((b σ c σ ) a σ )d σ ) 0 = ((b σ a σ ) 0 c σ )d σ ) 0 + ((c σ a σ ) 0 b σ )d σ ) 0 k (b σ d σ ) 0 = 0, as b σ Mσ 0. If σ = 4, then by (10) (((b σ c σ ) a σ )d σ ) 0 = 0. This proves the lemma. By Lemma 2.1, if M is simple then I = 0 and, for each σ, Mσ 0 = 0. Lemma 2.2. For a simple algebra M as defined above and σ a, we have 0. M 0 = k s, for s 2 = s. 1. If σ = 4 then M σ = k a σ where a 2 σ = s. 2. If σ = 2 then 2.1) M σ = k a σ, where (a 2 σ) 0 = s, or 2.2) M σ = k a σ k b σ, where (a 2 σ) 0 = (b 2 σ) 0 = s and (a σ b σ ) 0 = 0 or 2.3) M σ = k a σ k b σ, where (a σ b σ ) 0 = s and (a 2 σ) 0 = (b 2 σ) 0 = 0. Proof. The proof of item 0. is the same as in Lemma 3.1 of [GG]. Let σ = 4 and a σ M σ. By Lemma 2.1, there exists b σ M σ such that (a σ b σ ) 0 = s, then we have on the one hand (a σ b σ ) 0 a σ = a σ and on the other hand, by (10), (a σ b σ ) 0 a σ = (a σ a σ ) 0 b σ = α b σ, if (a σ a σ ) 0 = α s. Hence a σ = α b σ. If c M σ and (b c) 0 = γ s, then by (10), c = (a b) 0 c = (b c) 0 a = γ a. Hence dim M σ = 1 and so item 1. is proved. Now let σ = 2. (a) There exists a M σ such that (a 2 ) 0 = s. If dim M σ = 1, then we have case 2.1. Suppose that there exists b M σ \ k a. If (a b) 0 = αs 0 then we can replace b by b + αa = b and we get ( b a ) 0 = 0. Hence we can suppose that b satisfies (a b) 0 = 0. (a.1) Suppose that for all a M σ such that (a b) 0 = 0 we have (b 2 ) 0 = 0. By Lemma 2.1, there exists c M σ such that (c b) 0 = s. We can suppose that (c a) 0 = 0 (by replacing c by c + αa = c as before). Now, using identity (11), we get (ab) c = (cb) 0 a + (ac) 0 b = a, (bc) a = (ba) 0 c + (ca) 0 b = 0 and 5

6 (bc) c = (bc) 0 c + (cc) 0 b = c. Hence, [(ab), (bc) ] c = a 0, contradicting the fact that M is abelian. (a.2) There exists b M σ such that (b 2 ) 0 = s and (a b) 0 = 0. If dim M σ = 2 then item 2.2 is proved. Suppose that dim M σ > 2. By Lemma 2.1, there exists c M σ such that (a c) 0 = (b c) 0 = 0. (a.3) If (c 2 ) 0 = s then by (11), (ac) c = (cc) 0 a = a, (ab) c = 0 and (ab) a = b. Hence [(ac), (ab) ] c = b 0, contradicting the fact that M is abelian. (a.4) Suppose that for all c M σ such that (a c) 0 = (b c) 0 = 0 we have (c 2 ) 0 = 0. By Lemma 2.1, there exists d M σ such that (d 2 ) 0 = 0 and (c d) 0 = s. Then, by identity (11), (ab) a = b, (ab) d = 0 and (ac) d = a. Hence [(ab), (ac) ] d 0 and again the fact that M is abelian is contradicted. (b) For all a M σ, (a 2 ) 0 = 0. By Lemma 2.1, there exist a, b M σ such that (a 2 ) 0 = (b 2 ) 0 = 0 and (a b) 0 = s. If dim M σ = 2, then we have case 2.3. If dim M σ > 2, then by Lemma 2.1 there exist c, d M σ such that (a c) 0 = (a d) 0 = (b c) 0 = (b d) 0 = (c 2 ) 0 = (d 2 ) 0 = 0 and (c d) 0 = s. In this case, by (11), (ab) a = a and (ac) d = a. Hence [(ab), (ac) ] d = a 0, contradicting the fact that M is abelian. This proves the lemma. Lemma 2.3. Let a P(I n ) be an even set and = {0} a. Let M be the variety of -algebras satisfying the list of identities of Proposition 3.2 [GG]. If M M is a simple -algebra (containing no graded ideals), then M = {σ a M σ 0} is one of the following sets: (i) {(2i 1, 2i, 2j 1, 2j) 1 i < j l} = C 2l, (ii) {(2i 1, 2i, 2j 1, 2j), (2i 1, 2i) 1 i < j l} = B 2l, (iii) { (1234), (1256), (1357), (3456), (2457), (2367), (1467) } = E 7, (iv) E 7 { σ σ E 7, σ = I 8 \ σ } = E 8. 6

7 Proof. The proof of this lemma in [GG] is based on the following facts: (1) for all σ M, we have σ = 2 or 4. (2) If σ τ M and σ τ then σ τ M. The item (2) may be proved as in [GG]. Let us prove item (1). Suppose that σ M and σ > 4. Thus, by (12), (M σ M σ ) 0 M σ = 0, hence (M σ M σ ) 0 = 0 and M σ = Mσ 0. But by Lemma 2.1, Mσ 0 = 0. Theorem 2.1. Let M M be a simple -algebra such that M 0 and M 2 = 0. Then M = B 2l and M has a basis { s, d ij, a i, b i, λ 1 i < j l } with one of the following set of multiplication rules: d ij d jk = d ik, d ij a j = b i, d ij b j = a i, a i b j = d ij, (a i b i ) = λ, λ a i = b i, (15) λ b i = a i, (d 2 ij) 0 = s, (a 2 i ) 0 = (b 2 i ) 0 = s or d ij d jk = d ik, d ij a i = a j, d ij b j = b i, a i b j = d ij, a i b i = s + λ, (d 2 ij) 0 = s, (16) λ a i = a i, λ b i = b i, where M (2i 1,2i,2j 1,2j) = k d ij, M (2i 1,2i) = k a i k b i and M = k λ. Proof. If M = B 2l = {(2i 1, 2i, 2j 1, 2j), (2i 1, 2i) 1 i < j l}, then by Lemma 2.2 for σ = (2i 1, 2i) M we have three cases (a) M σ = k a σ, where (a 2 σ) 0 = s, or 7

8 (b) M σ = k a σ k b σ, where (a 2 σ) 0 = (b 2 σ) 0 = s and (a σ b σ ) 0 = 0 or (c) M σ = k a σ k b σ, where (a σ b σ ) 0 = s and (a 2 σ) 0 = (b 2 σ) 0 = 0. Let us consider each case. (a) For σ = 2, by identity (11), we have (a σ a σ ) a σ = 2 (a 2 σ) 0 a σ = 0 and by (7), for σ = τ = 2 with σ τ = (a σ a σ ) a τ = 2 a σ a τ a σ = 0. If σ = 2 and σ τ = 2 with σ τ, then by (7) (a σ a σ ) c τ = 0. Therefore, (a σ a σ ) Z(M) = 0. Let µ M such that M µ = k d and σ µ, τ = µ \ σ. Denote b τ = d a σ. If c M τ then, by (7), d c d = c(dd) 0 = c. But d c M σ = k a σ. Thus c = d c d = α a σ d = α b τ and M τ = k b τ. In this case, M is the algebra obtained in [GG]. (b) Let d = d 12 M (1234) and denote b 2 = d b 1, a 2 = d a 1, a 1 a 2 = α d, b 2 b 1 = β d By (7), d b 2 = d d b 1 = (d d) 0 b 1 = b 1 and d a 2 = d d a 1 = (d d) 0 a 1 = a 1. Now, by (9), we have (b 2 b 2 ) 0 = (d b 1 b 2 ) 0 = (d b 1 b 2 ) 0 = (d b 2 b 1 ) 0 = s. Hence, β = 1. Moreover, (a 2 a 2 ) 0 = (d a 1 a 2 ) 0 = (d a 2 a 1 ) 0 = (d a 1 a 2 ) 0 = s. Hence, α = 1. Again by (9), (b 2 a 2 ) 0 = (d b 1 a 2 ) 0 = (d b 1 a 2 ) 0 = (d a 2 b 1 ) 0 = 0. Hence, b 1 a 2 = 0. Analogously, a 1 b 2 = 0. Now denote τ = (a 1 a 1 ), ξ = (b 1 b 1 ), λ = (a 1 b 1 ). For c M (12), by identity (11), we have (a 1 a 1 ) c = 2(a 1 c) 0 a 1 = 0. By (8), for c M (2i 1,2i), i 1, we have (a 1 a 1 ) c = 2a 1 c a 1 = 0. Hence, (a 1 a 1 ) Z(M) = 0 and analogously (a i a i ) = (b i b i ) = 0. Moreover, by (11), λ a 1 = (a 1 b 1 ) a 1 = (a 1 a 1 ) 0 b 1 = b 1 and analogously λ b 1 = a 1. By (7), λ a 2 = (a 1 b 1 ) a 2 = b 1 a 2 a 1 = d a 1 = b 2 and in the same way λ b 2 = a 2. Now we denote b i = d i1 b 1 and a i = d i1 a 1. As above, we can prove that b i b j = a i a j = d ij, a i b j = 0 and (a i a i ) = (b i b i ) = 0 and (a i b i ) = λ. In 8

9 this case, we have the multiplication rules given by (15). (c) Let d = d 12 M (1234) be such that d 2 = s and denote a 2 = d a 1, b 2 = d b 1, a 1 b 2 = α d, a 1 a 2 = γ d, a 2 b 1 = β d, b 1 b 2 = τ d, As in case (b), by (7), we have d a 2 = d d a 1 = (d d) 0 a 1 = a 1 and d b 2 = d d b 1 = (d d) 0 b 1 = b 1. Now, by (9), we have (a 2 a 2 ) 0 = (d a 1 a 2 ) 0 = (d a 2 a 1 ) 0 = (a 1 a 1 ) 0 = 0, (b 2 b 2 ) 0 = 0. Again by (9), (a 2 b 2 ) 0 = (d a 1 b 2 ) 0 = (d b 2 a 1 ) 0 = (b 1 a 1 ) 0 = s. Now for c M (12), by identity (11), we have (a 1 a 1 ) c = 2(a 1 c) 0 a 1 = 0 and by (8), for c M (2i 1,2i), i 1, we have (a 1 a 1 ) c = 2a 1 c a 1 = 0. Hence, (a i a i ) = (b i b i ) Z(M) = 0. By (11) we have (a 1 b 1 ) a 1 = (a 1 b 1 ) 0 a 1 = a 1 and, by (7), (a 1 b 1 ) a 2 = a 2 b 1 a 1 = d a 1 = a 2. From this, analogously to the previous case, for λ = (a 1 b 1 ) we get λ a i = a i and λ b i = b i. Now for d ij M (2i 1,2i,2j 1,2j), it is clear, by (7), (8) and the fact that σ 4, that λ d ij = (a 1 b 1 ) d ij = 0 We will denote by D 2l+1 the -algebra M with multiplication rules given by (15) and by C 2l+1 the one with multiplication rules given by (16). 3 Irreducible Representations of -algebras In this section we study the action of the -algebras M of Theorem 2.1. Theorem 3.1. Let M be a -algebra as in Theorem 2.1 and V be an irreducible M-module. Then 1. M = D 2l+1 and 9

10 1.1. V = < v 1,..., v l, ξ, µ >, where v i V (2i 1,2i), ξ, µ V and v i d ij = v j, (v i a i ) = ξ, (v i b i ) = µ, ξ a i = v i, µ b i = v i, λ µ = ξ, λ ξ = µ (17) and all the other products are zero V is the adjoint module. 2. M = C 2l V = < v 1,..., v l, τ, µ >, where v i V (2i 1,2i), τ, µ V and v i d ij = v j, (v i a i ) = τ, (v i b i ) = µ, τ b i = µ a i = v i, λ τ = τ, λ µ = µ (18) and all the other products are zero V is the adjoint module. Proof. 1. Let V 0 0 and v 0 V 0. Define v ij = v 0 d ij, v i = v 0 a i, w i = v 0 b i, (v i b i ) = µ i (w i a i ) = ξ i. (19) By (14) we have (v i a i ) 0 = (v 0 a i a i ) 0 = v 0 (a i a i ) 0 = v 0. Thus, by (8) and (11), we have that µ = µ 1 = = µ l = ξ 1 = = ξ l and µ i a i = (v i b i ) a i = (v i a i ) 0 b i + (a i b i ) 0 v i = v 0 b i = w i and analogously µ b i = v i. Hence V has a basis {v 0, v ij, v i, w i, µ i i, j l} and V is the adjoint M-module. Now suppose that V 0 = 0 and take V µ, µ 0. As µ (2i 1, 2i) = 0 or 2, then (2i 1, 2i) µ or µ (2i 1, 2i) = for all 1 i l. Suppose (1, 2) µ. If (2i 1, 2i) µ, i > 2, then σ 1i = (1, 2, 2i 1, 2i) µ and, by (12), V µ = s V µ = (d 1i d 1i ) 0 V µ = 0, a contradiction. Hence, µ = (12). Let 0 v 1 V (12) and denote v i = v 1 d 1i, (v i b i ) = τ i (v i a i ) = µ i. (20) 10

11 Now by (6), we have v i d ij = v 1 d 1i d ij = v 1 d ij d 1i + v 1 d 1i d ij = v 1 d 1j = v j. By (11), µ i a i = (v i a i ) a i = (v i a i ) 0 a i + (a i a i ) 0 v i = v i and µ i b i = (v i a i ) b i = (v i b i ) 0 a i + (a i b i ) 0 v i = 0. Analogously, τ i a i = 0 and τ i b i = w i. Moreover, by (8), µ i a j = (v i a i ) a j = v i a j a i + v i a i a j = v i d ij = v j. Analogously, we prove that µ i b j = 0, τ i b j = w i, τ i a j = 0. Hence µ = µ 1 = = µ l and τ = τ 1 = = τ l. Furthermore, by (13), λµ = (a 1 b 1 ) µ = (a 1 µ b 1 ) + (a 1 µ b 1 ) = (v 1 b 1 ) = τ and, analogously, λτ = µ. Hence V = < v 1,..., v l, τ, µ >, is the standard M-module. 2. Suppose V 0 0. As in case 1, we can prove that V is the adjoint M- module. Thus, let V 0 = 0. Again as in case 1 we can prove that there exists µ = (12) such that V µ 0. Denote, as in the previous case, v i = v 1 d 1i, (v i a i ) = µ i (v i b i ) = τ i v i d ij = v j. (21) By (11), µ i a i = (v i a i ) a i = (v i a i ) 0 a i + (a i a i ) 0 v i = 0 and µ i b i = v i, τ i a i = v i, τ i b i = 0. By (8), µ i a j = (v i a i ) a j = v i a j a i + v i a i a j = 0, µ i b j = (v i a i ) b j = v i b j a i + v i a i b j = v i d ij = v j. Analogously, τ i a j = v i, τ i b j = 0. Hence µ = µ 1 = = µ l and τ = τ 1 = = τ l. Furthermore, by (13), λµ = (a 1 b 1 ) µ = (a 1 µ b 1 ) + (a 1 µ b 1 ) = (a 1 v 1 ) = µ and, analogously, λτ = τ. 4 Lie algebras from -algebras First recall some well known facts about quadratic forms over an algebraically closed field of characteristic 2 and its corresponding Lie algebras. Let V be a n-dimensional k-space and f : V V k be a non-degenerate symmetric bilinear form. This means that f(x, y) = f(y, x), for all x, y V and f(x, V ) = 0 implies x = 0. A non-degenerate symmetric bilinear form f is called symplectic if f(x, x) = 0 and orthogonal otherwise. A vector space V 11

12 has a unique orthogonal form f and in some basis {v 1,..., v n } the form can be written as n f(v, w) = x i y i n n where v = x i v i and w = y i v i. i=1 i=1 i=1 If dim V is odd, then the vector space V does not have a symplectic form and if dim V = 2l then it has a unique symplectic form. In this last case, the form can be written, in an appropriate basis {v 1,..., v l, w 1,..., w l }, as follows f(v, w) = l (x i t i + y i z i ) i=1 l l where v = (x i v i + y i w i ) and w = (z i v i + t i w i ). i=1 i=1 Let End(V ) be the associative algebra of all linear transformations of V. Consider the following sets S(f) = {a End(V ) f(va, w) = f(v, wa), v, w V }, O(f) = {a End(V ) f(va, v) = 0, v V }. It is clear that O(f) S(f). For f orthogonal, we denote D l = O(f) when dim V = 2l and B l = O(f) when dim V = 2l + 1. For f symplectic, C l = O(f). Theorem 4.1. In the notation above we have 1. [S(f), S(f)] = O(f). 2. Z(S(f)) = 0 if dim V = 2l + 1 and Z(O(f)) = 1 if dim V = 2l. 3. O(f)/Z(O(f)) is simple if dim V > 2 and dim V C l is a 2-algebra. 5. B l and D l are not 2-algebras and S(f) is the 2-envelope of B l (resp., D l ) in End(V ). 6. dim C l = 2l 2 l, dim B l = 2l 2 + l and dim D l = 2l 2 l. 12

13 Theorem 4.2. Let M be a simple -algebra in a -variety M as described above and L = M Λ be the corresponding Lie algebra. Then 1. L = C 2l if M has a basis {s, a ij 1 i < j l}, where a ij M (2i 1,2i,2j 1,2j). 2. L = B 2l if M has a basis {s, a ij, a i 1 i < j l}, where a ij M (2i 1,2i,2j 1,2j), a i M (2i 1,2i). 3. L = D 2l+1 or C 2l+1 if M has a basis {s, a ij, a i, b i, λ 1 i < j l}, where a ij M (2i 1,2i,2j 1,2j) and the multiplication rules are given by (15) or (16). 4. L is a Lie algebra of type E 7 or E 8, if M = {σ M σ 0} = E 7 or E 8. Proof. 1. By Theorem [GG], a -algebra M has a module V with a basis {v 1,..., v l }, v i V (2i 1,2i). This M-module admits an M-invariant bilinear form given by (v i, v j ) = δ ij. Note that if a M-module V = V 0 V σ admits an M-invariant symmetric bilinear form f, then the corresponding L-module W = V Λ admits a L-invariant symmetric bilinear form as follows: f(v x, w y) = f(v, w) (x, y), v, w V, x, y Λ Moreover, f is symplectic (orthogonal) if and only if the restriction of f to V 0 V is symplectic (orthogonal). In our case, V 0 V = 0 hence this form is non-degenerate and symplectic. As dim W = 4l and dim L = 8l 2 2l, we have that L = C 2l. 2. and 3. In all this cases M has a module V with a basis {v 1,..., v l, µ, τ} described in Theorem 3.1, with v i V (2i 1,2i). The M-module V admits an M- invariant bilinear form given by (v i, v j ) = δ ij and (λ, µ) = 0, (λ, λ) = (µ, µ) = 1, if M has multiplication rules defined by (15) or (v i, v j ) = δ ij and (λ, µ) = 1, (λ, λ) = (µ, µ) = 0, if M has multiplication rules defined by (16). In the first case, the corresponding L-invariant bilinear form on the L-module W = V Λ is orthogonal and, in the second case, it is symplectic. As dim L = 8l 2 + 6l + 1, then L = D 2l+1 in the first and L = C 2l+1, in the second case. 13

14 4. We prove this statement in the case E 8. The case E 7 is a corollary of this. By definition, a Lie algebra L over a field k of characteristic 2 is a Lie algebra of type E 8 if there exists a Z-form L Z of the Lie algebra L over the field C of all complex numbers such that L = L Z Z k. Let L be the Lie algebra of type E 8 over the complex field C constructed in [?] with a basis { e 1, f 1,..., e 8, f 8, h 1,..., h 8, (σ, µ), µ σ E 8 } and multiplication rules stated by Theorem 1 [?]. Let L Z be a Z- module with generators { e i, f i, h i, i = 1,..., 8, (σ, µ), h σ = 1 2 ( i σ h i ), µ σ E 8 }. Note that [L Z, L Z ] L Z, since for ϕ ψ =, ϕ ψ = σ we have, by Theorem 1 [?], that (σ, ϕ)(σ, ψ) = ( 1) ψ +1 ( h i h j )/2 = ( 1) ψ +1 (h σ h j ), i ψ j ϕ j ϕ (σ, µ)h τ = 1 ( µ τ µ τ )(σ, µ), where µ = σ \ µ. (22) 2 But ( µ τ µ τ ) = ( σ τ 2 µ τ ) σ τ 0(mod 2). Hence (σ, µ)h τ L Z. Now we prove that L = L Z Z k. Define ξ : L L Z Z k given by ξ(e i ) = e i, ξ(f i ) = f i, ξ(σ, µ) = (σ, µ), ξ(h i ) = h i ξ(h σ ) = h σ. (Note that although the notation for the elements is the same, they are in two diferent algebras.) To prove that ξ is an algebra isomorphism, it is enough to prove that ξ((σ, µ)h ϕ ) = [ξ(σ, µ), ξ(h ϕ )] ( ). By (2), ( σ ϕ ξ((σ, µ) h ϕ ) = 2 14 ) + ϕ µ ξ((σ, µ))

15 and by (22) ξ((σ, µ)) ξ(h ϕ ) = 1 ( µ ϕ µ ϕ )ξ((σ, µ)). 2 Now since 1( µ ϕ µ ϕ ) = 1 ( µ ϕ + µ ϕ ) + µ ϕ = σ ϕ ( σ ϕ + µ ϕ + ϕ µ, the equality (*) holds. 2 2 References [GG] Grishkov, A.N., Guerreiro, M., Simple classical Lie algebras in characteristic 2 and their gradations, I. International Journal of Mathematics, Game Theory, and Algebra,(to appear.). 15