K-Forms of 2-Step Splitting Trivectors

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1 International Journal of Algebra, Vol. 2, 2008, no. 8, K-Forms of 2-Step Splitting Trivectors Lemnouar Noui Department of Mathematics, Faculty of Science Batna University, Algeria N. Midoune Department of Mathematics, Faculty of Science Msila University, Algeria Abstract. In this paper, we describe the K-forms of 2-step splitting trivectors of rank 8 where K is any field of characteristic other than 2 and 3. Mathematics Subject Classification: Primary 0C38, 1A1; Secondary 0A1, 1A18 Keywords: complexity, trivector, length 1. Introduction First we recall some definitions. Let E be an n-dimensional vector space over a field K and let Λ m E denote the exterior power of degree m over E, the classification of trivectors is the study of the action of General linear group GL(E) on the space Λ 3 E defined by f ω =(Λ 3 f)(ω). The equivalence classes are the GL(E)-orbits under this action. As Λ 3 E (Λ 3 E ), there is no difference between trilinear alternating forms and trivectors. The support of the trivector ω is the least subspace F of E such that ω Λ 3 F. Denote this subspace by S ω, its dimension is the rank of ω and will be denoted by r(ω). This classification could help to solve some open problems in the theory of self dual codes, see[10]. The relationship between trivectors and nilpotent lie algebras can be seen in [6]. A general classification of trivectors exists in dimension 7 and less [2,7,12]. For n=8 Gurevitch [4], D. Djokovic [3] and L.Noui [9] give an answer to the classification problem with K = C, K = R and for K algebraically closed field of arbitrary characteristic respectively.

2 370 L. Noui and N. Midoune If ω is a trivector defined over the field K, ak-form of ω is another trivector, of the same type as that of ω, defined over K which is isomorphic to ω over K, the algebraic closure of K. In [6], the K-forms of a trivector of rank 8 in the open dense orbit are determined when K has characteristic zero. In this paper, K is a field of characteristic other than 2 and 3, we are interested in describing the K-forms of 2-step splitting trivectors of rank Invariants and trivectors. First we talk about arithmetical invariant d 1 (ω) that one may assign to a trivector ω. For every subspace α of E let ω α the canonical image of ω in Λ 3 (E/α). We put d k (ω) = inf α r(ω α ) where α runs over all k-dimensional subspaces of E. The numbers d k (ω) form a chain d 0 (ω) =r(ω) >d 1 (ω) >d 2 (ω) > >d r (ω) =0 where r is less than or equal to dim E. Note that d 1 (ω) = 0 if and only if ω is divisible by a vector [8]. The element ω of Λ 3 E is called splitting if there exists a decomposition E = E 1 E 2 such that ω E 1 Λ 2 E 2. If dim E 1 = r, ω is called r step splitting. In particular ω is 1-step splitting means that ω is divisible: ω = e 1 u where e 1 E 1 and u Λ 2 E 2. We refer to Aut (ω) the automorphisms group of ω as the stabilizer (in GL(E) )ofω, other invariants as Derivation and Commutant are defined in [7, 8, 13]. Theorem 1. [11] There exists a system of divided powers in the ideal 2k E k>0 of the exterior power of E. This result is well known in the case of fields and it has a generalization in the ring case. If u 2 M, u is nondegenerate if and only if γ m (u) generates the A-module 2m M.Ifu = r x 2i 1 x 2i where {x 1,...,x 2r,...,x n } is a basis i=1 of M, γ k (u) = x 2i1 1x 2i1 x 2ik 1 i 1 i k r such that rank (u) =2r is equivalent to γ k (u) 0 and γ k+1 (u) = Trivectors of rank 8. Theorem 2. [9] Let E be a vector space of dimension 8 over an algebraically closed field K of arbitrary characteristic. Then any trivector of rank 8 in Λ 3 E is equivalent to one of the trivectors ω 8,i of Table 1.

3 K-forms of 2-step splitting trivectors 371 Table 1 ω 8,i ω 8,1 ω 8,2 ω 8,3 ω 8,4 ω 8, ω 8,6 ω 8,7 ω 8,8 ω 8,9 ω 8,10 ω 8,11 ω 8,12 ω 8,13 Notation of Gurevitch XV I XI XV II XIII XIX XIV XII XV III XXI XX XV XXII XXIII Trivector d 1 (ω 8,i ) e 1 (e 2 e 3 + e 4 e )+e 6 e 7 e 8 e 1 (e 2 e 3 + e 4 e + e 6 e 7 )+e e 6 e 8 e 1 (e 3 e 4 + e e 6 )+e 2 (e 3 e + e 7 e 8 ) e 1 (e 2 e 3 + e 4 e )+e 6 (e 2 e 7 + e 4 e 8 ) e 1 (e 2 e 3 + e 4 e )+e 6 (e 2 e 3 + e 7 e 8 ) e 1 (e 2 e 3 + e 4 e + e 6 e 7 )+e 8 (e 4 e 3 + e e 6 ) e 1 (e 2 e 3 + e 4 e + e 6 e 7 )+e 2 (e e 6 + e 7 e 8 ) e 1 (e 2 e 8 + e 3 e 6 + e 4 e 7 )+e 6 e 7 e 8 + e 3 e 4 e e 1 [e 2 (e 3 + e 4 )+e e 6 ]+e 3 e e 7 + e 4 e 6 e 8 e 1 (e 2 e 8 + e 6 e 7 )+e 2 e 3 e + e 3 e 4 e 6 + e 4 e e 7 e 1 (e 3 e 7 + e e 4 + e 8 e 2 )+e 8 (e 4 e 3 + e 6 e 7 )+e 2 e 4 e 6 e 1 [(e 4 e 7 )(e 3 e 8 )+e e 7 ]+e 2 (e 3 e 4 + e e 6 )+e 6 e 7 e 8 e 1 [e (e 3 + e 7 )+e 8 e 4 ]+e 2 (e 3 e 4 + e e 6 )+e 6 e 7 e Remark 1. The trivectors ω 8,i, i =1,...,6. are 2 step-splitting. 2. Automorphisms group and Galois Cohomology. In this section, K is an arbitrary field of characteristic other than 2 and 3, we will compute the automorphisms group A i of each trivector ω 8,i,1 i 6. If K is the algebraic closure of K and G is Gal(K/K), the classification of trivectors using Galois descent will involve the cohomology H 1 (G, A i ) for not necessarily abelian coefficients A i, H 1 (G, A i ) classifies trivectors over K that become isomorphic over K to ω 8,i. Indeed if A i = Aut K (ω 8,i ), there is a bijection from H 1 (G, A i ) to the set of K-isomorphisms classes of trivectors ω such that ω K ω 8,i ([12, 14]). To establish the main result for K-forms we need to introduce a few more concepts Etale algebras. Following [] by etale algebra we mean a finite dimensional commutative K-algebra L such that L K 1... K r, with the K i s finite separable field extentions of K. In other words, if we denote by K s a separable closure of K then a finite-dimensional K-algebra L is called etale if L K K s is isomorphic to a split K s -algebra K s... K s. For the set X = {1, 2,..., n}, let S n be the symmetric group of X, i.e the group of all permutations of X. It is well-known that the Galois cohomology set H 1 (G, S n )is in canonical one -to-one correspondence with the isomorphism classes of etale K-algebras of dimension n, see[ ], that is, the pointed set H 1 (G, S n ) classifies etale K-algebras of degree n. In particular, for n = 3 we obtain the cubic etale

4 372 L. Noui and N. Midoune algebras as commutative associative K-algebras L of dimension 3 over K such that L K s K s K s. They are of four possible types: L 1 K K K, the split algebra. L 2 K K, where K is a Galois quadratic field extension of K. L 3 a cyclic cubic field extension of K, and L 4 a general separable field extension of degree 3 over K Invariant domains. Let E be a vector space of dimension n over the field K and if ω is a trivector in Λ 3 E, we define two domains which are invariant under the group of automorphisms of the trivector ω. First, the domain R i (ω) ={x E / ω x is of type ω i } is invariant under Aut (ω), indeed, let x R i and put E = Kx E, then ω = xu + ω where ω is a trivector in Λ 3 E. Since x E, ω x = ω x = ω which is of type ω i. If f Aut (ω) then Λ 3 f(ω) =ω = f(x)λ 2 f(u)+λ 3 f(ω ), this yields ω f (x) = Λ 3 f(ω ) f (x),asf is a bijective linear map, Λ 3 f(ω )is equivalent to ω. f(x) f(e ) imply that ω f (x) =Λ 3 f(ω ) which is of type ω i,sof(x) R i, that is, f(r i ) R i. For the second domain, suppose x E and let f be a nonzero trilinear alternating form on E. Consider the bilinear alternating form f x on E defined by f x (y, z) =f(x, y, z) (y, z E). Its rank rk(f x ) is an even number, then the domain R i (f) ={x E/ rk(f x )= 2i} (0 2i n) is invariant under Aut (f) [ 2]. In order to compute the automorphisms group A i we use the previous invariant domains Automorphisms groups. Proposition 1. The automorphisms group A 1 = Aut (ω 8,1 ) verify the following exact sequences : where ω = e 1 (e 2 e 3 + e 4 e ) 1 SL 3 (K) A 1 Aut (ω ) 1 1 A 0 Aut (ω ) K 1 1 K 4 A 0 SP 4 (K) 1 Proof. Let ω 8,1 = e 1 (e 2 e 3 + e 4 e )+e 6 e 7 e 8, first we determine the domain R 61 (ω 8,1 )=R 0 (ω 8,1 ) R 1 (ω 8,1 ). Direct computation shows that R 61 (ω 8,1 )=<e 2,e 3,e 4,e > <e 6,e 7,e 8 >= V 1 V 2.

5 K-forms of 2-step splitting trivectors 373 As V 1 V 2 = {0} then if f A 1 = Aut (ω 8,1 ) we have f(r 61 (ω 8,1 )) R 61 (ω 8,1 ) so f either stabilizes each of the two linear spaces V 1,V 2 or interchanges them, that is, {f(v 1 ) V 1 and f(v 2 ) V 2 } or {f(v 1 ) V 2 and f(v 2 ) V 1 }. The second case is impossible and the matrix of f is of the shape: α 1 o 1 4 o 1 3. A o 4 3 α 8 o 3 4 B Let f A 1,Λ 3 f(ω 8,1 )=ω 8,1 implies B SL 3 (K) and α 2 = = α 8 = 0, in this case f (e 1 ) = α 1 e 1 and 3 f [e 1 (e 2 e 3 + e 4 e )] = e 1 (e 2 e 3 + e 4 e )= ω, consequently f < e 1,...,e > Aut (ω ), this allows us to define a homomorphism of groups A 1 f ϕ(f) =f <e1,...,e > Aut (ω ) ϕ is obviously surjective, we have ker ϕ = {f /f A 1 and α 1 =1,A = id 4 }. Λ 3 f(ω 8,1 )=ω 8,1 implies that ker ϕ SL 3 (K), so the sequence 1 SL 3 (K) A 1 (ω ) 1 is exact, as Aut (ω ) verify the following exact sequences [8] 1 A 0 (ω ) K 1 1 K A 0 SP 4 (K) 1 so we obtain the desired result. Proposition 2. The automorphisms group A 2 = Aut (ω 8,2 ) verify the following exact sequences : 1 A 8 2 A 2 K 1 1 A 88 2 A8 2 SL 2(K) 1 1 A A 88 2 K 1 1 U(16) A SL 2 (K) 1 Proof. Choosing a suitable basis of E, we can write ω 8,2 = e 1 (e e 6 + e 8 e 2 + e 3 e 7 )+e 2 e 3 e 4. { } First determine the domain R 3 (ω 8,2 )= x E/ ω 8,2 x has rank 3. Notice that e 1 R 3 (ω 8,2 ). Direct computation yields R 3 (ω 8,2 )=<e 1 > {0}, then we can define a groups homomorphism ϕ : A 2 K by ϕ(f )=α where f(e 1 )=αe 1, A 8 2 =kerϕ. ϕ is surjective so the first sequence is exact: 1 A 8 2 A 2 K 1.

6 374 L. Noui and N. Midoune Let f A 8 2, hence α = 1 and Λ3 f(ω 8,2 )=ω 8,2 imply that f stabilizes the linear space <e 1,e 2,e 3,e 4 >. In this case the matrix of f has the form : 1 a b c 0 B 0 A, A GL 0 3 (K), C,E GL 2 (K). C D O 4 4 O 2 2 E By simple computation we deduce that C SL 2 (K). This allows us to define a groups homomorphism π : A 2 SL 2 (K). π is surjective, therefore the following sequence is exact 1 A 88 2 A 8 2 SL 2 (K) 1 where A 88 2 =kerπ. As Λ 3 f(ω 8,2 )=ω 8,2, then by simple computation, the matrix A gives the last sequences with U(16) is unipotent group of dimension 16. Proposition 3. The automorphisms group A 3 = Aut (ω 8,3 ) verify the following exact sequences : 1 A 8 3 A 3 K K 1 1 A 88 3 A 8 3 SL 2 (K) 1 1 K 9 A 88 3 SL 2 (K) 1 { Proof. Let ω 8,3 = e 1 (e 3 e 4 + } e e 6 )+e 2 (e 3 e + e 7 e 8 ), the invariant domain R (ω 8,3 )= x E/ ω 8,3 x has rank is equal to {V 1 V 2 } {0} = {< e 1 > <e 2 >} {0}. Let f A 3, then f(r (ω 8,3 )) R (ω 8,3 )sof either stabilizes each of the two linear spaces V 1,V 2 or interchanges them, the case f(v 1 ) V 2 and f(v 2 ) V 1 is impossible because if f(e 1 )=λe 2 and f(e 2 )=βe 1, a computation of the coefficient of e 2 e 7 e 8 in the equality Λ 3 f(ω 8,3 ) = ω 8,3 leads to 1 = 0, a contradiction. So f(v 1 ) V 1 and f(v 2 ) V 2, hence f(e 1 )=αe 1 and f(e 2 )= βe 2, then we can define a groups homomorphism ϕ : A 3 K K by ϕ(f) =(α, β), this allows us to write the first exact sequence 1 A 8 3 A 3 K K 1 where A 8 3 =kerϕ. Let f A 8 3, then α = β = 1, from the equality Λ 3 f(ω 8,3 )= ω 8,3, we get: 1 0 A M(f) = 0 1 D B B GL 4(K) C GL 2 (K) O 2 2 O 2 4 C

7 K-forms of 2-step splitting trivectors 37 By a computation using the equality det M(f ) = det B det C 0, we deduce that C SL 2 (K), then we can define a groups homomorphism Ψ:A 8 3 SL 2 (K) byϕ(f) =C, consequently we obtain the second exact sequence. Let f A 88 3 = kerψ then C = Id 2. The requirement 3 f(ω 8,3 )=ω 8,3 gives the last sequence. Proposition 4. The automorphisms group A 4 = Aut (ω 8,4 ) verify the following exact sequences :: 1 A 8 4 A 4 Z/2Z 1 1 A 88 4 A 8 4 K K 1 1 K 12 A 88 4 SL 2(K) SL 2 (K) 1 Proof. Choosing a suitable basis of E, we can write ω 8,4 = e 1 (e 3 e 6 + e 4 e 7 )+ e 2 (e 3 e + e 4 e 8 ). The invariant domain R (ω 8,4 ) is equal to {< e 1,e 2 > <e 3,e 4 >} {0} = {V 1 V 2 } {0}, obviously V 1 V 2 = {0}, then if f A 4, f(r (ω 8,4 )) R (ω 8,4 ) consequently f either stabilizes each of the two linear spaces V 1,V 2 or interchanges them, this allows us to define a groups homomorphism { 0 if f(v1 ) V ϕ : A 4 Z/2Z, f ϕ (f) = 1 and f(v 2 ) V 2 1 otherwise ϕ is surjective, indeed ϕ (id E ) = 0 and ϕ (f 0 ) = 1 with f 0 is defined by f 0 (e 1 )= e 3,f 0 (e 2 )=e 4,f 0 (e 3 )=e 1,f 0 (e 4 )=e 2,f 0 (e )= e 7,f 0 (e 6 )= e 6,f 0 (e 7 )= e,f 0 (e 8 )= e 8. Thus we obtain the following sequence : 1 A 8 4 A 4 Z/2Z 1 where A 8 4 =kerϕ. Let f A 8 4, then f(v 1) V 1 and f(v 2 ) V 2, consequently the matrix of f has the form: A O 2 2 M(f) = B C A, B GL 2 (K) O 2 2 O 4 4 this allows us to define a groups homomorphism ψ : A 8 4 K K,ψ(f) = (det A, det B). ψ is surjective, hence we obtain the exact sequence 1 A 88 4 A 8 4 K K 1 where A 88 4 =kerψ. Let f A 88 4, then det A = det B =1,soA,B SL 2 (K), thus we can define a groups homomorphism π : A 88 4 SL 2 (K) SL 2 (K) by π(f) =(A, B), π is surjective, determine its kernel ker π. Let f ker π, then A = B = id 2 and 3 f(ω) =ω imply ker π K 12, as wanted.

8 376 L. Noui and N. Midoune Proposition. The automorphisms group A = Aut (ω 8, ) verify the following exact sequences : 1 A 8 A S A 88 A 8 K 1 1 A 888 A 88 SL 2 (K) 1 1 K 6 A 888 SL 2(K) SL 2 (K) 1 Proof. Choosing a suitable basis of E, we can write ω 8, = e 1 (e 3 e 4 + e e 6 )+ e 2 (e 3 e 4 + e 7 e 8 ). The invariant domain R (ω 8, ) is equal to {< e 1 > <e 2 > <e 1 + e 2 >} {0} = {V 1 V 2 V 3 } {0}. If f A,f(R (ω 8, )) R (ω 8, ) so that f(v i ) V σ(i) with σ S 3, this allows us to define a groups homomorphism ϕ : A S 3, ϕ(f) =σ where σ = ( ) V1 V 2 V 3. V σ(1) V σ(2) V σ(3) It readily follows that ϕ is surjective, so that the first sequence with A 8 = ker ϕ is exact. Let f A 8 then f(v 1 ) V 1 and f(v 2 ) V 2, f(v 3 ) V 3, consequently f(e 1 )=αe 1, f(e 2 )=βe 2, f(e 1 + e 2 )=δ(e 1 + e 2 ), hence α = β. This allows us to define a groups homomorphism ψ : A 8 K by ψ(f )=α. We readily find that ψ is surjective, so the second sequence with A 88 =kerψ is exact. Let f A 88, then α = 1 and from the equality Λ 3 f(ω 8, )=ω 8,, the matrix of f must have the form M(f )= 1 0 A D 0 1 B C GL 2(K) B GL O 6 2 O 2 4 C 4 (K) Since det M(f ) = det B det C 0, det C 0. Moreover 3 f(ω) =ω implies det C =1soC SL 2 (K), hence we can define a groups homomorphism π : A 88 SL 2 (K) byπ(f) =C, π is surjective then the third sequence is exact 1 A 888 A 88 SL 2 (K) 1 with A 888 =kerπ. Let f ker ( π, we get C = ) id 2. But Λ 3 f(ω) =ω shows that X O2 2 the matrix B has the form : with det X = det Y = 1, then O 2 2 Y X, Y SL 2 (K). This allows us to define a groups homomorphism h : A 888 SL 2 (K) SL 2 (K) byh(f) =(X, Y ), it readily follows that h is surjective and its kernel ker h is isomorphic to K 6.

9 K-forms of 2-step splitting trivectors 377 Proposition 6. The automorphisms group A 6 = Aut (ω 8,6 ) verify the following exact sequences : 1 A 8 6 A 6 K K 1 1 U(13) A 8 6 SL 2(K) 1 Proof. Choosing a suitable basis of E, we can write ω 8,6 = e 1 (e 8 e 3 + e 6 e + e 4 e 7 )+e 2 (e 6 e 3 + e e 4 ). A direct computation leads to the relations R (ω 8,6 )=<e 1 > {0} and R 7,1 (ω 8,6 ) = {x E/ω 8,6 x ) is of type ω 7,1 [8, 13]} = {x E/x= α 1 e 1 + α 2 e 2 and α 2 0}. Let f A 6, from previous invariant domains the matrix of f must have the form β α 1 0 α 2 A, O 6 2 thus we can define a groups homomorphism ϕ : A 6 K K by ϕ(f) = (β,α 2 ). Put A 8 6 =kerϕ, it readily follows that ϕ is surjective and the first sequence is exact. 1 A 8 6 A 6 K K 1 Let f A 8 6, f(e 1 )=e 2 + α 1 e 1 and Λ 3 f(ω 8,6 )=ω 8,6 implies det B = 1 where B is a 2 2 submatrix of the matrix A such that det M(f) = det B det C ; C GL 2 (K). Thus we can construct a homomorphism of groups ψ : A 8 6 SL 2 (K) byψ(f )=B, ψ is surjective and its kernel is isomorphic to U(13). Hence we obtain the exact sequence 1 U(13) A 8 6 SL 2 (K) 1. This ends the proof of the proposition. 3. K-forms of 2-step splitting trivectors of rank 8. We are now ready to formulate the main results. The following theorem gives us a description of K-forms of 2-step splitting trivectors of rank 8. Theorem 3. Let E be a vector space of dimension 8 over a field K of characteristic other than 2 and 3. Then any K-form of 2-step splitting trivector of rank 8 in Λ 3 E is equivalent to one of the trivectors of Table 2.

10 378 L. Noui and N. Midoune Table 2 ω 8,i Trivector d 1 (ω 8,i ) ω 8,1 ω 8,2 ω 8,3 ω 8,4 ω 8,4,d ω 8, ω 8,,d ω 8,,a ω 8,,b,c ω 8,6 e 1 (e 2 e 3 + e 4 e )+e 6 e 7 e 8 e 1 (e 2 e 3 + e 4 e + e 6 e 7 )+e e 6 e 8 e 1 (e 3 e 4 + e e 6 )+e 2 (e 3 e + e 7 e 8 ) e 1 (e 2 e 3 + e 4 e )+e 6 (e 2 e 7 + e 4 e 8 ) e (e 1 e 2 + e 3 e 4 )+e 6 (e 1 e 3 + de 2 e 4 )+e 7 (e 1 e 4 )+e 8 (e 2 e 3 ) e 1 (e 2 e 3 + e 4 e )+e 6 (e 2 e 3 + e 7 e 8 ) e 2 (e e 3 de 4 e 6 )+e 1 (e e 4 + e 7 e 8 + e 6 e 3 ) e 1 (ae 3 e 4 + ae e 6 + e 7 e 8 )+e 2 (e 3 e + e 4 e 7 + e 6 e 8 ) e 7 (e 1 e 2 + e 3 e 4 + e e 6 )+e 8 [e 1 (e 4 + be )+e 2 e 6 + ce 3 e ] e 1 (e 2 e 3 + e 4 e + e 6 e 7 )+e 8 (e 4 e 3 + e e 6 ) Here, d K K 2, a K K 3, and bc 0. Denote by C i the set of K-isomorphism classes of the K-forms of ω 8,i, 1 i 6. In order to prove Theorem 3, the following lemmas are necessary. Lemma 1. The trivector ω 8,4 has two K-forms ω 8,4 ω 8,4,d = e (e 1 e 2 + e 3 e 4 )+e 6 (e 1 e 3 + de 2 e 4 )+e 7 (e 1 e 4 )+e 8 (e 2 e 3 ) where d K K 2. Proof. According to [14] H 1 (G, X) = 0 whenever X is one of the groups K m, K m, SL m (K), by proposition 4 and using the exact sequences of Galois cohomology sets, we get C 4 = H 1 (G, 2Z Z). Since H1 (G, nz Z ) classifies cyclic K-algebras of degree n, then C 4 = H 1 (G, 2Z Z K ) =. We deduce that if K 2 L = K( d) is quadratic extension of K, d / K 2, there exists a trivector ω L 3 E such that ω L ω 8,4 and ω L L 3 (E K L)isL-isomorphic to ω 8,4. We construct ω L as follows. ω 8,4 = e 1 (e 2 e + e 3 e 6 )+e 4 (e 7 e 2 + e 8 e 3 ) is 4-step splitting because ω 8,4 = e u 1 + e 6 u 2 + e 7 u 3 + e 8 u 4 where u 1 = e 1 e 2,u 2 = e 1 e 3,u 3 = e 2 e 4 and u 4 = e 3 e 4,thusV = vect {u 1,u 2,u 3,u 4 } is a subspace of dimension 4 of 2 K 4. As in [ 12] we define the quadratic form γ 2 on V and we put ω L = ω 8,4,d = e v 1 + e 6 v 2 + e 7 v 3 + e 8 v 4 with v 1 = e 1 e 2 + e 3 e 4,v 2 = e 1 e 3 + de 2 e 4,v 3 = e 1 e 4,v 4 = e 2 e 3, where d/ K 2. To each of the forms ω 8,4, ω 8,4,d we associate a quadratic form γ 2 (xu 1 + yu 2 + zu 3 + tu 4 ), then we get γ 2 (xu 1 + yu 2 + zu 3 + tu 4 )= yz, γ 2 (xv 1 + yv 2 + zv 3 + tv 4 )=x 2 dy 2

11 K-forms of 2-step splitting trivectors 379 respectively. The two forms are not equivalent over K but they may become equivalent over the algebraic closure K. We can also prove that ω 8,4 is not equivalent to ω 8,4,d by using the arithmetical invariant d 1 (ω). For the K-forms of ω 8, we have the following Lemma 2. ω 8, has four K-forms. ω 8, ; ω 8,,d = e 2 (e e 3 de 4 e 6 )+e 1 (e 7 e 8 + e e 4 + e 6 e 3 ), d K K 2 ω 8,,a = e 1 (ae 3 e 4 + ae e 6 + e 7 e 8 )+e 2 (e 3 e + e 4 e 7 + e 6 e 8 ), a/ (F q ) 3. ω 8,,b,c = e 7 (e 1 e 2 + e 3 e 4 + e e 6 )+e 8 (e 1 [e 4 + be )+e 2 e 6 + ce 3 e )] where bc 0. Proof. For the K-forms of ω 8,, using the exact sequences of proposition we deduce that C = H 1 (G, S 3 ), on the other hand we know that there exists a bijective correspondence from H 1 (G, S 3 ) to the isomorphism classes of cubic etale algebras. By according 2-1 we get four types of cubic algebras so ω 8, has four K-forms. Let L be a cubic extension of K and F a L space of dimension 3. Choose a basis {e 1,e 2,e 3 } of F and let the determinant form ϕ : 3 F L such that ϕ(e 1,e 2,e 3 )=1. and Tr : L K the trace forme hence ω L = Tr L ϕ : F F F K: is an alternating trilinear form of rank 9 on E = F viewed as vector space over K. If cara(l) 3 we take L = K(t) with t 3 = a in this case {e 1,e 2,e 3,e 4 = te 1,e = te 2,e 6 = te 3,e 7 = t 2 e 1,e 8 = t 2 e 2,e 9 = t 2 e 3 } is a basis of E. Let (e i ), 1 i 9; its dual basis; Hence it readily follows that: ω L (e 1,e 2,e 3 )=3,ω L (e 4,e,e 6 )=3a, ω L (e 7,e 8,e 9 )=3a 2, ω L (e 1,e,e 9 )=3a, ω L (e 1,e 6,e 8 )= 3a, ω L (e 2,e 4,e 9 )= 3a, ω L (e 2,e 6,e 7 )=3a, ω L (e 9,e 4,e 8 )= 3a, ω L (e 3,e,e 7 )= 3a and ω L (e i,e j,e k ) = 0 otherwise. Hence 1 ω 3 L = e 1e 2e 3 + ae 4e e 6 + a 2 e 7e 8e 9 + ae 1e e 9 ae 1e 6e 8 + ae 2e 4e 9 ae 2e 6e 7 + ae 3e 4e 8 ae 3e e 7 and 1 3 ω e 9 ) = e 3(e 1e 2 + ae 4e 8 + ae 7e )+ae 6(e 4e + e 1e 8 + e 2e 7), then the trivector e 3 (e 1 e 2 +ae 4 e 8 +ae 7 e )+ae 6 (e 4 e +e 1 e 8 +e 2 e 7 ) is equivalent to trivector ω 8,,a ( it suffices to consider linear map f defined by f(e 1 )=e 3,f(e 2 )=ae 6, f(e 3 )=e 4, f(e 4 )=e 8, f(e )=e,f(e 6 )= e 7,f(e 7 )= e 1, f(e 8 )= e 2 ). If we take L = K 8 K with K 8 = K(α) and α 2 = d (d / K 2 ), in this case, {e 1,e 2,e 3,e 4 = αe 1,e = αe 2,e 6 = αe 3,e 7,e 8,e 9 } is a basis of E. Let (e i ),1 i 9; its dual basis then ω L (e 1,e 2,e 3 )=3, ω L (e 7,e 8,e 9 )=3,ω L (e 1,e,e 6 )=3d, ω L (e 2,e 4,e 6 )= 3d, ω L (e 3,e 4,e )=3d, and

12 380 L. Noui and N. Midoune ω L (e i,e j,e k ) = 0 otherwise. Then 1 3 ω L = e 1 e 2 e 3 + de 1 e e 6 + de 2 e 4 e 6 + de 3 e 4 e + e 7 e 8 e 9 and it follows that ω = e 1 e 2 e 3 + de 1 e e 6 + de 2 e 4 e 6 + de 3 e 4 e + e 7 e 8 e 9 = ω 6,1,d + e 7 e 8 e 9 because ω 6,1,d = e 1 (e 3 e 4 + e e 6 )+e 2 (e 3 e 6 de 4 e ) ([7]). We put x = e 9 e 1, then e 9 = x+e 1 and ω x = e 1 (e 7 e 8 +e 3 e 4 +e e 6 )+e 2 (e 3 e 6 de 4 e ) is equivalent to ω 8,,d. If L = K K K = K 3 then F = L 3 = K 9, by the same method we obtain ω 8,. In order to find the last trivector ω 8,,b,c, take L = K(α) a general separable field extension of degree 3 over K where α 3 + bα + c = 0 with b, c K/bc 0, Since the trivectors ω 8,,ω 8,,d, ω 8,,a and ω 8,,b,c are 2-step splitting, using the invariant γ 3 (xu 1 + yu 2 ) which is a cubic forme; u 1,u 2 2 K 6 we get : γ 3 (xu 1 + yu 2 )=xy(x + y), x ( x 2 dy 2),a 2 x 3 + y 3 and x 3 + bxy 2 + cy 3 for ω 8,,ω 8,,d, ω 8,,a and ω 8,,b,c respectively. Clearly, the three last forms may be equivalent to the first form if the field is algebraically closed. Proof of Theorem 3. Proof. As H 1 (G, X) = 0 whenever X is one of the groups K m, SP 2m(K), GL m (K), SL m (K) [14 ], by propositions 1, 2, 3 and 6, and using the exact sequences of Galois cohomology sets, we get H 1 (G, A i )=0 if i {1, 2, 3, 6}. Then C i = 1, consequently ω 8,i is the only K-form of ω 8,i for i {1, 2, 3, 6}. For ω 8,4 by Lemma 1 we obtain two K-forms, ω 8,4 and ω 8,4,d = e (e 1 e 2 + e 3 e 4 )+e 6 (e 1 e 3 + de 2 e 4 )+e 7 (e 1 e 4 )+e 8 (e 2 e 3 ) where d / K 2. For the K-forms of ω 8,, Lemma 2 gives us four K-forms ω 8,,ω 8,,d,ω 8,,a and ω 8,,b,c where bc 0. Notice that if b =0,ω 8,,b,c must be equivalent to ω 8,,a and for c =0,ω 8,,b,c is equivalent to ω 8,,d. This establishes Theorem 3. Corollary 1. There are trivectors of rank 8 over K that are not 2-step splitting but they may become 2-step splitting by extension of scalars. Proof. If the field K is not quadratically closed, it is enough to consider trivectors ω 8,4,d which are not 2-step splitting and becomes equivalent to ω 8,4 over the algebraic closure K. Thus(ω 8,4,d ) K is 2-step splitting. In order to deduce other results, the following definitions are necessary. An ordered field K is said to be real closed field if any algebraic extension of K which is ordered must be equal to K. For example R the field of real numbers is a real closed field.

13 K-forms of 2-step splitting trivectors 381 A field which is complete with respect to a discrete valuation is called a local field if its field of residue classes is finite. For example Q p the field of p adic Numbers for some prime p is a local field. As an immediate consequence of Theorem 3, we deduce nine F q -forms of 2 stepsplitting of rank eight over a finite field F q, indeed, we have H 1 (G, S 3 ) =3 because the last etale algebra L 4 does not exist [1] and H 1 (G, 2Z Z ) = 2. But over the real closed field we get eight K-forms of rank 8 of 2-step splitting trivectors, indeed, H 1 (G, S 3 ) = 2 because every polynomial of odd degree has arootink so we obtain only two etale algebras L 1 and L 2. H 1 (G, 2Z Z ) =2, hence we deduce the known result [3] about real forms of rank 8 of 2-step splitting trivectors. If K is local field such that its field of residue classes has a characteristic other than 2 and 3, then C 4 = K = K 2Z Z 2Z Z = 4 [8]. Hence we deduce the 2 following Corollary 2. a) There are nine F q -forms of 2 step-splitting trivectors of rank 8 over a finite field F q : ω 8,1, ω 8,2, ω 8,3, ω 8,4, ω 8,4,d, ω 8,, ω 8,,d, ω 8,,a, ω 8,6 where d/ (F q ) 2, a/ (F q ) 3. b) If K is real closed field (in particular if K = R), then the K-forms of 2-step splitting trivectors of rank 8 are ω 8,1, ω 8,2, ω 8,3, ω 8,4, ω 8,4, 1, ω 8,, ω 8,, 1, ω 8,6. c) If K is local field such that its field of residue classes has a characteristic different from 2 and 3. Then there are 12 K-forms of 2 step-splitting trivectors of rank 8. ω 8,1, ω 8,2, ω 8,3, ω 8,4, ω 8,4,α i β j, ω 8,, ω 8,,d, ω 8,,a, ω 8,,b,c, ω 8,6 where d/ K 2, a/ K 3, bc 0, and αβ 1 / K 2, i, j {0, 1}. References [1] N. Bourbaki, Algèbre, chapitres 4 à 7, Masson, paris [2] A. Cohen & A. Helminck., Trilinear alternating forms on a vector space of dimension 7, Comm. in Algebra. 16,(1), 1988, p [3] D. Djokovic s, Classification of trivectors of an eight dimensional real vector space, Lin.and multilin. Algebra. 13(3), 1983, p [4] G.B. Gurevitch, Theory of Algebraic invariants, P.Noordhof LTD, Groningen, the netherland, [] M.A. Knus, A. Merkurjev, M. Rost, J.P. Tignol, The Book of involutions, American Mathematical society colloquim publications, V.44 Providence, RI, [6] L. NOUI & Ph. Revoy, Algèbres de Lie orthogonales et formes trilinéaires alternées, communications in Algebra, 2(2), (1997). [7] L. NOUI & Ph. Revoy., Formes multilinèaires alternées, Ann. Math. Blaise Pascal, Vol1, n 0 2, p (1994). [8] L. NOUI, Classification des trivecteurs par l action du groupe linèaire, Thèse de Doctorat, Université de MontpellierII, France, 199. [9] L. NOUI., Transvecteur de rang 8 sur un corps algébriquement clos, C. R. Acad. Sci. Paris, t. 324, Serie i, Algèbre, p (1997). [10] Rains E. M., Sloane J. A. Self -dual Codes, Handbook of Coding Theory, Pless V. S. & Huffman W. C(editors), Elsevier, Amsterdam, 1998, pp [11] Ph. Revoy, Formes alternées et puissances divisées, sem.p. Dubreil 26 eme annee, ,n=8, 10p.

14 382 L. Noui and N. Midoune [12] Ph. Revoy, Trivecteurs de rang 6, in coll. Sur les formes quadratiques, Memoire SMF9., 1979, p [13] Ph. Revoy, Formes trilinéaires alternées de rang 7, Bull. Sc. Math. 112, 1988, p [14] J.P.Serre Corps locaux, Hermann, paris Received: August 29, 2007