3-Lie Algebras Realized by Cubic Matrices∗

2014-06-04RuipuBAIHuiLIUMengZHANG

Chinese Annals of Mathematics,Series B 2014年2期

Ruipu BAI Hui LIU Meng ZHANG

1 Introduction

Kerner in[1]presented the survey which scattered in a series of papers whose common denominator is the use of cubic relations in various algebraic structures,and constructed the non-associative ternary algebra of cubic matrices(the object with elements having three indices).For cubic matricesA=Kerner defined the ternary multiplication of three cubic matricesand(A∗B∗=in[1].The symmetry properties of the ternary algebras were studied,and theZ3-graded generalization of Grassmann algebras,a ternary generalization of Clifford algebras,also the description of quark fields were discussed.Awata,Li,Minicc and Yoneyad in[2]presented several nontrivial examples of the 3-dimensional quantum Nambu bracket with cubic matrices.For cubic matricesA=(),B=(),C=(),1≤i,j,k≤N,they defined the traces functions:andABC=which satisfy

defined the multiplication of three cubic matrices(ABC)by(ABC)=then obtain a skew-symmetric quantum Nambu bracket

[A,B,C]=(ABC)+(BCA)+(CAB)−(ACB)−(BAC)−(CBA),

which satisfies the generalized Jacobi identity(3.1).

The multi-index constants are often used to describe structures of algebras.For example,ifLis a 3-Lie algebra(see[3])with a basisx1,···,xN,the entire multiplication table can be described by the structure constantswhich occur in the expressionThen we obtain a four indices matrixA=which satisfiesand=0,whereσis arbitrary 3-ary permutation.In[4],the authors studied structures of 2-step nilpotent metric 3-Lie algebras by means of four indices matrices.By the properties of four indices matrices,it is proved that there do not exist 2,4,6 and 10-dimensional 2-step nilpotent metric 3-Lie algebras with corank zero,there exist 8,or greater than 10-dimensional 2-step nilpotent metric 3-Lie algebras with corank zero.And up to isomorphisms only one 8-dimensional 2-step nilpotent metric 3-Lie algebras with corank zero.

In this paper,we discuss the multiplications of two cubic matrices which satisfy the associative law,and define thesth-determinant,and “traces” of cubic matrix.By means of cubic matrices,we construct 3-Lie algebras.

Throughout this paper,we assume that the cubic matrices are over a fieldFof characteristic zero,and the symbol=for positive integersiandj.

2 Multiplications of Cubic Matrix

AnN-order cubic matrixA=over a fieldFis an ordered object which the elements with 3 indicesi,j,kand 1≤i,j,k≤N.The element in the position(i,j,k)is denoted by(A1≤i,j,k≤N.

Denote the set of all cubic matrices over a fieldFby Ω.Then Ω is anN3-dimensional vector space overFwith

∀A=B=∈Ω,λ∈F,that is,(A+

Denotea cubic matrix which the element in the position(i,j,k)is 1 and elsewhere are zero,that is,1≤l,m,n≤N;Ei=1≤j≤N.Then{1≤i,j,k≤N}is a basis of Ω,and for everyA=)∈Ω,A=

Every cubic matrixAcan be written as the following three types of blocking forms:

whereare usual(N×N)-order matrices with the elements at the position of thejth-row and thekth-column,respectively,1≤i≤N.

Define the multiplicationsof cubic matrices are as follows:∀A=∈Ω,AB=,1≤i,j,k≤N,where

Then the multiplications of two cubic matrices defined as above in the blocking form are as follows:

whereis the product of two(N×N)-order matrices

Define the linear isomorphismτ:Ω→Ω,that is,∀A=()∈Ω,

Thenτsatisfiesτ3=IdΩ,and

Theorem 2.1are non-isomorphic associative algebras.And there exists the unit element U(1)=where1≤i,j,k≤N,that is,for every A∈(Ω,U(1)=U(1)∗11A=A.And the multiplication tables of the associative algebras in the basis{1≤i,j,k≤N}are as follows,respectively,

ProofThe result follows from the direct computations.

The determinant|A|of a cubic matrixAis defined as

Then we have|AB|=|BA|=|A||B|,|U(1)|=1.If|A|0,thenAis called a non-degenerate cubic matrix,the inverse cubic matrix ofAis denoted byA−1,that is,

And for arbitrary non-degenerate cubic matricesAandB,we have

For constructing 3-Lie algebras by cubic matrices according to the multiplicationswe define the “sth-trace” linear functions:Ω→F,s=1,2,3,4 as follows:∀A=()∈Ω,

Theorem 2.2For arbitrary cubic matrices A,B∈Ω,we have

ProofThe result follows from the direct computations.

3 Construction of 3-Lie Algebras by Cubic Matrix

In this section we construct 3-Lie algebras by cubic matrices.First we introduce some notions onn-Lie algebras(see[3]).

Ann-Lie algebraJover a fieldFis a vector space endowed with ann-ary multilinear skewsymmetric multiplication which satisfies then-Jacobi identity:∈J,

Then-ary skew-symmetry of the operationmeans that

for any permutationσ∈Sn.A subspaceWofJis called a subalgebra if[W,...,W]⊆W.In particular,the subalgebra generated by the vectorsfor any∈Jis called the derived algebra ofJ,which is denoted byJ1.IfJ1=0,thenJis called an abeliann-Lie algebra.

An ideal of ann-Lie algebraJis a subspaceIsuch that[I,J,···,J]⊆I.

An idealIof ann-Lie algebraJis called nilpotent,ifIs=0 for somes≥0,whereI0=IandIsis defined asIs=fors≥1.IfI=J,thenJis nilpotentn-Lie algebra.If

thenJis called a 2-step nilpotentn-Lie algebra.

The subsetZ(J)={x∈L|=0,∈L}is called the center ofJ.

Now we define the 3-ary linear multiplications on Ω as follows according to the multiplications∀A,B,C∈Ω,

Theorem 3.1The3-ary algebras(Ω,[,,and(Ω,[,,are3-Lie algebras,which are denoted byrespectively.

ProofThanks to Proposition 2.2,∀A,B∈Ω,

Following from Theorems 2.1 and 3.1 in[5],(Ω,[,,),(Ω,[,,),(Ω,[,,),(Ω,[,,)and(Ω,[,,]31)are 3-Lie algebras.

Define the linear isomorphismω:Ω→Ω,that is,∀A=where(ω(A=(A,1≤i,j,k≤N.Then by Theorems 2.1–2.2,for arbitraryA,B∈Ω,ω(AB)=ω(B)Then

Therefore,3-Lie algebrais anti-isomorphic to 3-Lie algebrain the isomorphismω:that is,∀A,B,C∈Ω,ω([A,B,C)=−[ω(A),ω(B),ω(C).

Following from the multiplicationsthe multiplication tables of the 3-Lie algebras in the basis1≤i,j,k≤N}are as follows:

where 1≤i,j,k,l,m,n,p,q,r≤N.

4 Structures of 3-Lie Algebras Jrs

Now we study the structures of the 3-Lie algebrasFirst,we study the structure of 3-Lie algebra.Denote

For everyA∈A=(,0,···,0)+(0,,0,···,0)+···+(0,···,0,),then

as the direct sum of subspaces.From the multiplication(3.3),we have

Then,1≤l≤Nare subalgebras and satisfy

And the center ofis

Summarizing above discussions,we obtain the following result.

Theorem 4.1(1)The3-Lie algebra can be decomposed into the direct sum of subalgebras

and the derived algebraare minimal ideals of,1≤l≤N.

(2)For arbitrary1≤<<···<≤N,the subspaceis a subalgebra of

(3)is the semidirect productwhere I=is the maximal ideal of J11with codimension one.

ProofThe result follows from identities(4.1)–(4.4).

Next we study the structure of the 3-Lie algebra.For everyA∈Ω,

Denote

Then Ω=Φ˙+Ψ as the direct sum of subspaces and for everyA∈Ψ,==0.

Since identity(3.4)and=0 forA∈Ψ,Ψ is an(N3−N2)-dimensional abelian ideal ofJ21,that is,[Ψ,Ψ,=0,and Φ is a subalgebra with

Summarizing the above discussions,we obtain the following result.

Theorem 4.2(1)The3-Lie algebrawhere

(2)The3-Lie algebrais the semidirect product=Ψ˙+Φ,andΨ=is an(N3−N2)-dimensional abelian ideal of ,andΦis a subalgebra with

(3)The abelian idealΨhas a decompositionΨwhere

are(N−1)-dimensional minimal ideals of.Therefore,,i=2,···,N,n=1,···,N,are irreducible modules of

ProofThe results(1)and(2)follow from the identity(3.4).Denote=−),i=2,···,N,n=1,···,N,then Ψ =is the direct sum of subspaces.Since the identity(2.2)and Proposition 2.2,

Therefore,,i=2,···,N,n=1,···,N,are minimal ideals of.

Now we study the structure of the 3-Lie algebra.Denote

Since identity(3.5),

Then,1≤p≤N,are subalgebras andSince

=0.Therefore,are abelian subalgebras,1≤i,p≤N.And

Theorem 4.3The3-Lie algebrais an indecomposable3-Lie algebra,and can be decomposed into the direct sum of subalgebras

For arbitrary1≤,···,≤N,andare subalgebrasof

ProofThe result follows from identities(3.5)and(4.6).

Lastly,we study the structure of the 3-Lie algebra.Denote

Theorem 4.4(1)The3-Lie algebra is a2-step-nilpotent3-Lie algebra,

(2)The3-Lie algebracan be decomposed into the direct sum of abelian subalgebras

ProofThe identity(4.7)follows from the direct computation according to the multiplication(3.6).By the identities(2.2)and(2.4),we haveand=1,1≤i,j,k,l,m,n≤N.Then for arbitrary