Semi-Associative $3$-Algebras

TL;DR

This paper introduces semi-associative 3-algebras, explores their structures, modules, and extensions, and establishes their connection to 3-Lie algebras, expanding the understanding of non-associative algebraic systems.

Contribution

It defines semi-associative 3-algebras, constructs their modules and extensions, and links them to 3-Lie algebras, providing new frameworks and tools for non-associative algebra research.

Findings

Defined semi-associative 3-algebras and their properties.

Constructed double modules and extensions using cocycles.

Established the relationship between semi-associative 3-algebras and 3-Lie algebras.

Abstract

A new 3-ary non-associative algebra, which is called a semi-associative $3$-algebra, is introduced, and the double modules and double extensions by cocycles are provided. Every semi-associative $3$-algebra $(A, \{ , , \})$ has an adjacent 3-Lie algebra $(A, [ , , ]_c)$. From a semi-associative $3$-algebra $(A, \{, , \})$, a double module $(\phi, \psi, M)$ and a cocycle $\theta$, a semi-direct product semi-associative $3$-algebra $A\ltimes_{\phi\psi} M $ and a double extension $(A\dot+A^*, \{ , , \}_{\theta})$ are constructed, and structures are studied.

Equations

$$\{A,B,C\}=\frac{\partial (A,B,C)}{(\partial x,\partial y,\partial z)},$$

$$\{A,B,\{C,D,E\}\}=\{\{A,B,C\},D,E\}+\{C,\{A,B,D\},E\}+\{C,D,\{A,B,E\}\}.$$

$$[A,B,C]=A[B,C]+B[C,A]+C[A,B]=[B,C]A+[C,A]B+[A,B]C.$$

$$⟨A,B,C⟩=[A,B]Tr(C)+[C,A]Tr(B)+[B,C]Tr(A).$$

$$\{x_1,x_2,x_3\}=-\{x_2,x_1,x_3\},$$

$$\{x_1,\{x_2,x_3,x_4\},x_5\}=\{x_1,x_2,\{x_3,x_4,x_5\}\},$$

$$\{x_1,\{x_2,x_3,x_4\},x_5\}=\{x_5,\{x_2,x_3,x_4\},x_1\}+\{x_1,\{x_5,x_3,x_4\},x_2\}.$$

$$\{v_1,v_2,v_2\}=-\{v_2,v_1,v_2\}=v_3,\text{mbox}andothersarezero.$$

$$Z_A(V)=\{x|x\in A,\{x,V,A\}=\{V,A,x\}=0\},$$

$$f\{x,y,z\}=\{f(x),f(y),f(z)\},\forall x,y,z\in A,$$

$$\{x+I,y+I,z+I\}=\{x,y,z\}+I,\forall x,y,z\in A,$$

$$\{e_1,\lambda e_1+\mu e_2,\frac{1}{\mu }{e}_3\}=\lambda e_1+\mu e_2\ne 0.$$

$$\{e_i,e_j,e_k\}\ne 0.$$

$$\{e_i,e_j,e_k\}=e_l.$$

$$\{e_i,e_j,e_k\}=a{e}_i+b{e}_j+c{e}_k,a,b,c\in \mathbb{F}.$$

$$\frac{c}{a}\{e_j,e_k,e_k\}=\frac{1}{a}\{e_j,c{e}_k,e_k\}=\frac{1}{a}\{e_j,\{e_i,e_j,e_k\}-a{e}_i-b{e}_j,e_k\}=e_l.$$

$$\{B,e_4,B\}=\{B,B,e_4\}=0,\{C,e_4,C\}=\{C,C,e_4\}=0,$$

$$\{e_5,e_4,e_6\}=a_3{e}_1+b_3{e}_2+c_3{e}_3+d_3{e}_4+{\lambda }_3{e}_5+{\mu }_3{e}_6,a_3,b_3,c_3,{\lambda }_3,{\mu }_3\in \mathbb{F}.$$

$$D\{x_1,x_2,x_3\}=\{D{x}_1,x_2,x_3\}+\{x_1,D{x}_2,x_3\}+\{x_1,x_2,D{x}_3\},\forall x_1,x_2,x_3\in A,$$

$$L(x_1,x_2)(x)=\{x_1,x_2,x\},R(x_1,x_2)(x)=\{x,x_1,x_2\},\forall x\in A,$$

$$S(x_1,x_2)=L(x_1,x_2)-R(x_1,x_2).$$

$$L(x_1,x_2)=-L(x_2,x_1),$$

$$L(x_1,x_2)L(x_3,x_4)=L(x_1,\{x_2,x_3,x_4\}),$$

$$L(x_1,\{x_2,x_3,x_4\})=R(\{x_2,x_3,x_4\},x_1)-R(x_1,x_2)R(x_3,x_4),$$

$$L(x_1,x_2)L(x_3,x_4)=L(x_3,x_1)L(x_2,x_4)=L(x_4,x_2)R(x_3,x_1)+L(x_3,x_1)R(x_4,x_2),$$

$$R(x_3,x_4)(R(x_1,x_2)+R(x_2,x_1))=0.$$

$$L(A)=⟨L(x_1,x_2)\mid \forall x_1,x_2\in A⟩,R(A)=⟨R(x,x_2)\mid \forall x_1,x_2\in A⟩,$$

$$S(A)=⟨S(x_1,x_2)\mid \forall x_1,x_2\in A⟩,T(A)=L(A)+R(A).$$

$$\Gamma (A)=\{\phi \in End(A)|\phi \{x_1,x_2,x_3\}=\{\phi (x_1),x_2,x_3\}=\{x_1,x_2,\phi (x_3)\},\forall x_1,x_2,x_3\in A\}$$

$$\phi \{x_1,x_2,x_3\}=\{\phi (x_1),x_2,x_3\}=\{x_1,\phi (x_2),x_3\}=\{x_1,x_2,\phi (x_3)\}.$$

Full text

††2010 Mathematics Subject Classification: 15A69; 17D99.††Key words and phrases: semi-associative 3-algebra, 3-Lie algebra, double module, extension

1. Introduction

3-ary algebraic systems are applied in mathematics and mathematical physics. For example, 3-Lie algebras provided by Filippov in 1985 [1, 2], have close relation with the theory of integrable-systems and Nambu mechanical-systems [3, 4]. Bagger and Lambert [5, 6, 7] proposed a supersymmetric fields theory model for multiple $M_{2}$-branes based on the metric $3$-Lie algebras [8].

C. Bai and coauthors [9], defined pre-3-Lie algebras and local cocycle 3-Lie bialgebras and generalized Yang-Baxter equation in 3-Lie algebras, and proved that tensor form solutions of 3-Lie Yang-Baxter equation can be constructed by local cocycle 3-Lie bialgebras. Three classes of 3-algebras are constructed in [10]:

(1) The classical Nambu bracket based on three variables ( say $x,y,z$ ) is the simplest to compute, in most situations. It involves the Jacobian-like determinant of partial derivatives of three functions $A,B,C$,

[TABLE]

and it satisfies Filippove identity:

[TABLE]

(2) The sum of single operators producted with commutators of the remaining two, or as anticommutators acting on the commutators,

[TABLE]

The multiplication does not satisfy Filippov identity, and the 3-algebra can by realized by operators [10].

(3) The re-packaging commutators of any Lie algebra to define

[TABLE]

This is again a totally antisymmetric trilinear combination, but it is a singular construction for finite dimensional realizations in the sense that $Tr\langle A,B,C\rangle=0,$ and authors in [11] proved that all the $m$-dimensional 3-Lie algebras with $m\leq 5$, except for the unique simple 3-Lie algebra, can be constructed by the method from Lie algebras.

Likewise, in [9], the paper provides an algebra which is called the $3$-pre-Lie algebra, and induces a sub-adjacent $3$-Lie algebra. However, the $3$-pre-Lie algebra does not satisfy associative property. Motivated by this, in this paper, we define a new 3-algebra, the 3-ary multiplication is none completely antisymmetry, but it has close relation with 3-Lie algebras.

In the following we assume that $\mathbb{Z}$ is the set of integers, $\mathbb{F}$ is an algebraic closed field of characteristic zero, and for a subset $S$ of a linear space $V$, $\langle S\rangle$ denotes the subspace spanned by $S$. And we usually omit zero products of basis vectors when we list the multiplication of $3$-algebras.

2. Natures of semi-associative 3-algebras

Definition 2.1**.**

A semi-associative 3-algebra $(A,\{,,\})$ is a vector space $A$ with a 3-ary linear multiplication $\{,,\}:A\otimes A\otimes A\rightarrow A$ satisfying that for all $x_{i}\in A,1\leq i\leq 5$,

[TABLE]

[TABLE]

[TABLE]

For subspaces $B_{1},B_{2},B_{3}$ of $A$, denotes $\{B_{1},B_{2},B_{3}\}$ the subspace spanned by vectors $\{x_{1},x_{2},x_{3}\}$, $\forall x_{i}\in B_{i}$, $i=1,2,3$; $\{A,A,A\}$ is denoted by $A^{1}$, which is called the derived algebra of $A$. If $A^{1}=0$, then $A$ is called abelian.

For example, let $A$ be a $3$-dimensional vector space with a basis $v_{1},v_{2},v_{3}$. Then $A$ is a semi-associative 3-algebra with the multiplication

[TABLE]

Definition 2.2**.**

A subalgebra of a semi-associative 3-algebra $A$ is a subspace $B$ satisfying $\{B,B,B\}\subseteq B$. If $B$ satisfies that $\{A,A,B\}\subseteq B$ and $(A,B,A)\subseteq B$, then $B$ is called an ideal of $A$.

Obviously, $\{0\}$ and $A$ are ideals of $A$, which are called trivial ideals. If $A$ has non proper ideals, then $A$ is called a simple semi-associative 3-algebra.

For a given subspace $V\leq A$, the subalgebra

[TABLE]

is called the centralizer of $V$ in $A$. $Z_{A}(A)$ is called the center of $A$, and is simply denoted by $Z(A)$, that is, $Z(A)=\{~{}x\mid x\in A,\{x,A,A\}=\{A,A,x\}=0~{}\}$. It is clear that $Z(A)$ is an ideal.

Definition 2.3**.**

Let $A$ and $A_{1}$ be semi-associative 3-algebras. If a linear mapping (linear isomorphism) $f:A\rightarrow A_{1}$ satisfies

[TABLE]

then $f$ is called an algebra homomorphism (an algebra isomorphism).

Let $I$ be an ideal of $A$. Then quotient space $A/I=\{x+I\mid~{}~{}x\in A\}$ is a semi-associative 3-algebra in the multiplication

[TABLE]

which is called the quotient algebra of $A$ by $I$, and $\pi:A\rightarrow A/I$, $\pi(x)=x+I$, $\forall x\in A$, is a surjective algebra homomorphism.

Proposition 2.4**.**

Let $A$ be a semi-associative algebra, $I_{1},I_{2},I_{3}$ be ideals of $A$. Then

1) $I_{1}+I_{2}$, $I_{1}\cap I_{2}$ and $\{I_{1},I_{2},I_{3}\}$ are ideals of $A$.

2) If $I_{1}\subseteq I_{2}$, then $I_{2}/I_{1}$ is an ideal of $A/I_{1}$.

Proof.

It can be verified by a direct computation according to Definition 2.2. ∎

Proposition 2.5**.**

Let $A$ and $A_{1}$ be semi-associative algebras, $f:A\rightarrow A_{1}$ be an algebra homomorphism. Then

1) $K=Kerf=\{x\in A\mid f(x)=0\}$ is an ideal of $A$, and $f(A)$ is a subalgebra of $A_{1}$.

2) If $f(A)=A_{1}$, then linear mapping $\bar{f}:A/K\rightarrow A_{1}$, $\bar{f}(x+K)=f(x)$, for all $x\in A$, is an algebra isomorphism, and there is one to one correspondence between subalgebras of $A$ containing $K$, with the subalgebras of $A_{1}$, and ideals correspond to ideals.

Proof.

The result is easily verified by a direct computation. ∎

Theorem 2.6**.**

Let $A$ be a semi-associative 3-algebra.

1) If there exist nonzero vectors $e_{1},e_{2},e_{3},e_{4}\in A$ such that $\{e_{1},e_{2},e_{3}\}=e_{4}$, then $e_{4}\neq\lambda e_{s}$, where $\lambda\in\mathbb{F},\lambda\neq 0,$ $s=1,2,3$.

2) If there exist nonzero vectors $e_{1},e_{2},e_{3}\in A$ such that $\{e_{1},e_{2},e_{3}\}\neq 0$, then $\{e_{1},e_{2},e_{3}\}\neq\lambda e_{1}+\mu e_{2},\forall\lambda,\mu\in\mathbb{F}$. Therefore, if $\{e_{1},e_{2},e_{3}\}=ae_{1}+be_{2}+ce_{4}\neq 0$, then $c\neq 0$, and $e_{1},e_{2},e_{4}$ are linearly independent.

Proof.

If there exist nonzero vectors $e_{1},e_{2},e_{3},e_{4}\in A$ such that $\{e_{1},e_{2},e_{3}\}=e_{4}=\lambda e_{1}$, $\lambda\in\mathbb{F},\lambda\neq 0$, then by Eqs (1) and (2)

$e_{4}=\{e_{1},e_{2},e_{3}\}=\frac{1}{\lambda}\{e_{2},\{e_{2},e_{1},e_{3}\},e_{3}\}=\frac{1}{\lambda}\{e_{2},e_{2},\{e_{1},e_{3},e_{3}\}\}=0$. Contradiction. Therefore, $e_{4}\neq\lambda e_{1}$. By the similar discussion to the above, $e_{4}\neq\mu e_{2}$, $\mu\in\mathbb{F}$.

If $e_{4}=\alpha e_{3}$, $\alpha\in\mathbb{F},\alpha\neq 0$, then

$e_{4}=\{e_{1},e_{2},e_{3}\}=\frac{1}{\alpha}\{e_{1},e_{2},\{e_{1},e_{2},e_{3}\}\}=-\frac{1}{\alpha}\{e_{1},e_{1},\{e_{2},e_{2},e_{3}\}\}=0$. Contradiction. The result 1) follows.

If there exist nonzero vectors $e_{1},e_{2},e_{3}\in A$ such that $\{e_{1},e_{2},e_{3}\}=\lambda e_{1}+\mu e_{2}\neq 0,\lambda,\mu\in\mathbb{F}$. Without loss of generality, we can suppose $\mu\neq 0$. Then

[TABLE]

Contradiction. Therefore, $\{e_{1},e_{2},e_{3}\}\neq\lambda e_{1}+\mu e_{2},\forall\lambda,\mu\in\mathbb{F}$. The proof is complete. ∎

Theorem 2.7**.**

Let $A$ be a non-abelian semi-associative 3-algebra with $\dim A=m\geq 3$. Then there exist linearly independent vectors $e_{i},e_{j},e_{k}\in A$ such that

[TABLE]

Proof.

Since $A$ is non-abelian, by (1), there are linearly independent vectors $e_{1},e_{2}\in A$, such that $\{e_{1},e_{2},A\}\neq 0.$ Suppose $\{e_{1},e_{2},\cdots,e_{m}\}$ is a basis of $A$. If $\{e_{1},e_{2},e_{l}\}=0$, for all $l\geq 3$, then there exist $a,b\in\mathbb{F}$, such that $\{e_{1},e_{2},ae_{1}+be_{2}\}\neq 0$, we get $\{e_{1},e_{2},ae_{1}+be_{2}+e_{3}\}\neq 0$, and vectors $e_{1},e_{2},ae_{1}+be_{2}+e_{3}$ are linearly independent. The proof is complete. ∎

Theorem 2.8**.**

Let $A$ be an $s$-dimensional semi-associative 3-algebra with $s\leq 6$. Then $A^{1}\subseteq Z(A)$.

Proof.

If $A$ is abelian, then the result is trivial.

If $A^{1}\neq 0$, then we will prove $\{A,A^{1},A\}=\{A,A,A^{1}\}=0$.

Since $A^{1}\neq 0$, then there are nonzero vectors $e_{i},e_{j},e_{k},e_{l}\in A$ such that

[TABLE]

$\bullet$ $\{e_{i},e_{j},e_{k}\}$ are linearly dependent.

Then we can suppose $e_{k}=ae_{i}+be_{j}$, $a,b\in\mathbb{F}$, and $e_{l}=a\{e_{i},e_{j},e_{i}\}+b\{e_{i},e_{j},e_{j}\}$.

By (2) and (3), for all $e_{m},e_{n}\in A$, we have

$\{e_{m},e_{l},e_{n}\}=a\{e_{m},\{e_{i},e_{j},e_{i}\},e_{n}\}+b\{e_{m},\{e_{i},e_{j},e_{j}\},e_{n}\}$

$=-a\{e_{m},e_{j},\{e_{i},e_{i},e_{n}\}\}+b\{e_{m},e_{i},\{e_{j},e_{j},e_{n}\}\}=0$,

$\{e_{m},e_{n},e_{l}\}=a\{e_{m},e_{n},\{e_{i},e_{j},e_{i}\}\}+b\{e_{m},e_{n},\{e_{i},e_{j},e_{j}\}\}$

$=a\{e_{i},e_{i},\{e_{n},e_{j},e_{m}\}\}+b\{e_{j},e_{j},\{e_{n},e_{i},e_{m}\}=0.$

Therefore, $\{A,e_{l},A\}=\{A,A,e_{l}\}=0$.

$\bullet\bullet$ $\{e_{i},e_{j},e_{k}\}$ are linearly independent.

$\bullet\bullet_{1}$ If there $a,b,c\in\mathbb{F},$ such that $e_{l}=ae_{i}+be_{j}+ce_{k}$, then

[TABLE]

By Theorem 2.6, $ac\neq 0$, or $bc\neq 0$. Without loss of generality, suppose $ac\neq 0$, then

[TABLE]

Therefore, for all $e_{m},e_{n}\in A$,

$\{e_{m},e_{l},e_{n}\}=\{e_{m},\frac{c}{a}\{e_{j},e_{k},e_{k}\},e_{n}\}=\{e_{m},\frac{c}{a}e_{j},\{e_{k},e_{k},e_{n}\}\}=0$, and

$\{e_{m},e_{n},e_{l}\}=\{e_{m},e_{n},\frac{c}{a}\{e_{j},e_{k},e_{k}\}\}=\frac{c}{a}\{e_{k},\{e_{k},e_{n},e_{j}\},e_{m}\}=0$.

It follows $\{A,A,e_{l}\}=\{A,e_{l},A\}=0.$

$\bullet\bullet_{2}$ $\{e_{i},e_{j},e_{k},e_{l}\}$ are linearly independent.

Then in the case $\dim A=4$, $\{e_{i},e_{j},e_{k},e_{l}\}$ is a basis of $A$. For all $e_{m},e_{n}\in A$, set

$e_{m}=a_{1}e_{i}+b_{1}e_{j}+c_{1}e_{k}+d_{1}e_{l},$ $e_{n}=a_{2}e_{i}+b_{2}e_{j}+c_{2}e_{k}+d_{2}e_{l},$ $a_{i},b_{i},c_{i},d_{i}\in\mathbb{F},i=1,2$. Then $\{e_{m},e_{l},e_{n}\}=\{a_{1}e_{i}+b_{1}e_{j}+c_{1}e_{k}+d_{1}e_{l},e_{l},e_{n}\}$

$=a_{1}\{e_{i},e_{i},\{e_{j},e_{k},e_{n}\}\}-b_{1}\{e_{j},e_{j},\{e_{i},e_{k},e_{n}\}\}-c_{1}\{e_{k},e_{i},\{e_{k},e_{j},e_{n}\}\}$

$=c_{1}\{e_{k},e_{k},\{e_{i},e_{j},e_{n}\}\}=0.$

By the similar to the above, $\{e_{m},e_{n},e_{l}\}=0$.

In the case $\dim A=5$, then we can suppose that $\{e_{i},e_{j},e_{k},e_{l},e_{t}\}$ is a basis of $A$.

From the discussion of the case $\dim A=4$, we only need to prove that $\{e_{t},e_{l},e_{t}\}=0$. Thanks to (1), (2) and (3),

$\{e_{t},e_{l},e_{t}\}=\{e_{t},\{e_{i},e_{j},e_{k}\},e_{t}\}=-\{e_{t},e_{t},\{e_{j},e_{k},e_{i}\}\}=0$.

In the case $\dim A=6$, then we can suppose $e_{1},e_{2},e_{3},e_{4},e_{5},e_{6}$ is a basis of $A$, where $e_{1}=e_{i}$, $e_{2}=e_{j}$, $e_{3}=e_{k}$, and $e_{4}=e_{l}$. Then $\{e_{1},e_{2},e_{3}\}=e_{4}.$

By the above discussion,

[TABLE]

where $B=\langle e_{1},e_{2},e_{3},e_{4},e_{5}\rangle$, $C=\langle e_{1},e_{2},e_{3},e_{4},e_{6}\rangle$. Therefore, we only need to discuss the products $\{e_{5},e_{4},e_{6}\}$, $\{e_{6},e_{4},e_{5}\}$ and $\{e_{5},e_{6},e_{4}\}$. Suppose

[TABLE]

Thanks to (4),

$\lambda_{3}\{e_{5},e_{4},e_{6}\}=\{\{e_{5},e_{4},e_{6}\}-a_{3}e_{1}-b_{3}e_{2}-c_{3}e_{3}-d_{3}e_{4}-\mu_{3}e_{6},e_{4},e_{6}\}=\{\{e_{5},e_{4},e_{6}\},e_{4},e_{6}\}$

$=\{e_{4},\{e_{4},e_{5},e_{6}\},e_{6}\}=\{e_{4},e_{4},\{e_{5},e_{6},e_{6}\}\}=0.$

Therefore, $\lambda_{3}\{e_{5},e_{4},e_{6}\}=\lambda_{3}ae_{1}+\lambda_{3}b_{3}e_{2}+\lambda_{3}c_{3}e_{3}+\lambda_{3}d_{3}e_{4}+\lambda_{3}^{2}e_{5}+\lambda_{3}\mu-3\lambda_{3}e_{6}=0$, $\lambda_{3}=0$.

By the complete similar discussion, we have $a_{3}=b_{3}=c_{3}=d_{3}=\mu_{3}=0$.

Therefore, $\{e_{5},e_{4},e_{6}\}=\{e_{5},e_{6},e_{4}\}$ $=\{e_{6},e_{4},e_{5}\}=0$, and $A^{1}\subseteq Z(A)$. ∎

3. Derivations and centroid of semi-associative 3-algebras

3.1. Derivations of semi-associative 3-algebras

Definition 3.1**.**

Let $A$ be a semi-associative 3-algebra, and $D:A\rightarrow A$ be a linear mapping. If $D$ satisfies that

[TABLE]

then $D$ is called a derivation of $A$. $Der(A)$ denotes the set of all derivations of $A$.

Proposition 3.2**.**

$Der(A)$* is a subalgebra of the general linear algebra $gl(A)$.*

Proof.

The result follows from a direct computation. ∎

For $D\in Der(A)$, if $D$ satisfies $D(A)\subseteq Z(A)$ and $D(A^{1})=0$, then $D$ is called a central derivation of $A$. And $Der_{C}(A)$ denotes the set of central derivations. It is clear that $Der_{C}(A)$ is a subalgebra of $Der(A)$.

For all $x_{1},x_{2}\in A$, define linear mappings,

$L(x_{1},x_{2}),$  $R(x_{1},x_{2}),~{}~{}S(x_{1},x_{2}):A\times A\longrightarrow A$, by

[TABLE]

[TABLE]

$L(x_{1},x_{2})$ and $R(x_{1},x_{2})$ are called the left multiplication and the right multiplication, respectively.

Lemma 3.3**.**

Let $A$ be a semi-associative 3-algebra. Then for all $x_{1},x_{2},x_{3},x_{4}\in A$,

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

Proof.

Eqs (8), (9) and (10) follow from (1), (2) and (3), directly.

For $\forall x_{i}\in A$, $1\leq i\leq 5$, by (2) and (3), we have

$R(x_{3},x_{4})R(x_{1},x_{2})(x_{5})+R(x_{3},x_{4})R(x_{2},x_{1})(x_{5})$

$=\{\{x_{5},x_{1},x_{2}\},x_{3},x_{4}\}+\{\{x_{5},x_{2},x_{1}\},x_{3},x_{4}\}$

$=-\{x_{3},\{x_{5},x_{1},x_{2}\},x_{4}\}-\{x_{3},\{x_{5},x_{2},x_{1}\},x_{4}\}$

$=-\{x_{3},\{x_{5},x_{1},x_{2}\},x_{4}\}+\{x_{3},\{x_{5},x_{1},x_{2}\},x_{4}\}=0,$

Similarly, we have

$L(x_{1},x_{2})L(x_{3},x_{4})(x_{5})-L(x_{3},x_{1})L(x_{2},x_{4})(x_{5})$

$=\{x_{1},x_{2},\{x_{3},x_{4},x_{5}\}\}-\{x_{3},x_{1},\{x_{2},x_{4},x_{5}\}\}=0,$ and

$L(x_{1},x_{2})L(x_{3},x_{4})(x_{5})-(L(x_{4},x_{2})R(x_{3},x_{1})(x_{5})+L(x_{3},x_{1})R(x_{4},x_{2})(x_{5}))=0.$

It follows (11) and (12). ∎

Let $L(A)$, $R(A)$ and $S(A)$ be subspaces of $End(A)$ spanned by linear mappings $L(x_{1},x_{2})$, $R(x,x_{2})$ and $S(x_{1},x_{2})$, respectively, $\forall x_{1},x_{2}\in A,$ that is,

[TABLE]

[TABLE]

Then $L(A)\subseteq T(A)$ and $R(A)\subseteq T(A).$

Theorem 3.4**.**

Let $A$ be a semi-associative 3-algebra. Then $T(A)$ is an abelian subalgebra of $gl(A)$.

Proof.

By (1) and (2), for all $x_{i}\in A$, $1\leq i\leq 5,$

$[L(x_{1},x_{2}),L(x_{3},x_{4})](x_{5})=(L(x_{1},x_{2})L(x_{3},x_{4})-L(x_{3},x_{4})L(x_{1},x_{2}))(x_{5})$

$=\{x_{1},\{x_{2},x_{3},x_{4}\},x_{5}\}-\{x_{3},\{x_{4},x_{1},x_{2}\},x_{5}\}$

$=\{x_{1},x_{3},\{x_{4},x_{2},x_{5}\}\}-\{x_{3},\{x_{4},x_{1},x_{2}\},x_{5}\}$

$=(\{x_{3},\{x_{4},x_{1},x_{2}\},x_{5}\}-\{x_{3},\{x_{4},x_{1},x_{2}\},x_{5}\}=0.$

Similarly, $[L(x_{1},x_{2}),R(x_{3},x_{4})]x_{5}=0.$ Thanks to (3),

$[R(x_{1},x_{2}),R(x_{3},x_{4})](x_{5})=(R(x_{1},x_{2})R(x_{3},x_{4})-R(x_{3},x_{4})R(x_{1},x_{2}))(x_{5})$

$=\{\{x_{5},x_{3},x_{4}\},x_{1},x_{2}\}-\{\{x_{5},x_{1},x_{2}\},x_{3},x_{4}\}$

$=-\{x_{1},\{x_{5},x_{3},x_{4}\},x_{2}\}+\{x_{3},\{x_{5},x_{1},x_{2}\},x_{4}\}$

$=-\{x_{1},\{x_{5},x_{3},x_{4}\},x_{2}\}+\{x_{5},\{x_{1},x_{3},x_{2}\},x_{4}\}$

$=-\{x_{2},\{x_{5},x_{3},x_{4}\},x_{1}\}-\{x_{1},\{x_{2},x_{3},x_{4}\},x_{5}\}$ $+\{x_{4},\{x_{1},x_{3},x_{2}\},x_{5}\}$

$+\{x_{5},\{x_{4},x_{3},x_{2}\},x_{1}\}=0.$

Therefore, $[L(A),T(A)]=[R(A),T(A)]=0,[T(A),T(A)]=0.$ The proof is complete. ∎

Theorem 3.5**.**

Let $A$ be a semi-associative 3-algebra. Then

1) for any $x_{1},x_{2}\in A$, $S(x_{1},x_{2})$ is a derivation of $A$;

2) $S(A)$ is an ideal of $Der(A)$, and $[S(A),S(A)]=0$.

Proof.

For $\forall x_{1},x_{2},x,y,z\in A$, by (6) and (7),

S(x_{1},x_{2})\{x,y,z\}$$=\{x_{1},x_{2},\{x,y,z\}\}-\{\{x,y,z\},x_{1},x_{2}\}

$=\{x_{1},\{x_{2},x,y\},z\}+\{x_{1},\{x,y,z\},x_{2}\},$

$\{S(x_{1},x_{2})(x),y,z\}+\{x,S(x_{1},x_{2})(y),z\}+\{x,y,S(x_{1},x_{2})(z)\}$

$=\{\{x_{1},x_{2},x\},y,z\}-\{\{x,x_{1},x_{2}\},y,z\}+\{x,\{x_{1},x_{2},y\},z\}$

$-\{x,\{y,x_{1},x_{2}\},z\}+\{x,y,\{x_{1},x_{2},z\}\}-\{x,y,\{z,x_{1},x_{2}\}\}$

$=\{x_{1},\{x_{2},x,y\},z\}+\{x_{1},\{x,y,z\},x_{2}\}.$ Therefore,

$S(x_{1},x_{2})\{x,y,z\}=\{S(x_{1},x_{2})(x),y,z\}+\{x,S(x_{1},x_{2})(y),z\}+\{x,y,S(x_{1},x_{2})(z)\}.$

The result 1) follows.

For any $S(x_{1},x_{2})\in S(A),D\in Der(A)$, and $x\in A$, since

$[S(x_{1},x_{2}),D](x)$ $=S(x_{1},x_{2})D(x)-DS(x_{1},x_{2})(x)$

$=\{x_{1},x_{2},D(x)\}-\{D(x),x_{1},x_{2}\}-D\{x_{1},x_{2},x\}+D\{x,x_{1},x_{2}\}$

$=\{x_{1},x_{2},D(x)\}-\{D(x),x_{1},x_{2}\}-\{D(x_{1}),x_{2},x\}-\{x_{1},D(x_{2}),x\}$

$-\{x_{1},x_{2},D(x)\}+\{D(x),x_{1},x_{2}\}+\{x,D(x_{1}),x_{2}\}+\{x,x_{1},D(x_{2})\}$

$=(S\{x_{1},D(x_{2})\}-S\{D(x_{1}),x_{2}\})(x)$.

Then $[S(A),Der(A)]\subseteq S(A)$, that is, $S(A)$ is an ideal of $Der(A)$. Thanks to Theorem 3.4, $[S(A),S(A)]=[L(A)-R(A),L(A)-R(A)]=0.$ It follows 2). ∎

For any $x_{1},x_{2}\in A$, $S(x_{1},x_{2})$ is called an inner derivation of $A$.

3.2. The centroid of semi-associative 3-algebras

Definition 3.6**.**

Let $A$ be a semi-associative 3-algebra. The vector space

[TABLE]

is called the centroid of $A$.

For any $\varphi\in End(A)$, by (1), $\varphi\in\Gamma(A)$ if and only if for all $x_{1},x_{2},x_{3}\in A$,

[TABLE]

Theorem 3.7**.**

Let $A$ be a semi-associative 3-algebra. Then

1) $\Gamma(A)$ is a subalgebra of the general linear Lie algebra $gl(A)$.

2) For any $\varphi\in\Gamma(A)$, if $\varphi(A)\subseteq Z(A)$ and $\varphi(A^{1})=0$, then $\varphi$ is a central derivation.

3) For any $\varphi\in\Gamma(A)$, and $D\in Der(A)$. Then $\varphi D\in Der(A)$.

4) $Der_{C}(A)=\Gamma(A)\cap Der(A)$.

Proof.

For all $\varphi_{1},\varphi_{2}\in\Gamma(A)$, and $x_{1},x_{2},x_{3}\in A$, by Definition 3.6,

$[\varphi_{1},\varphi_{2}]\{x_{1},x_{2},x_{3}\}=(\varphi_{1}\varphi_{2}-\varphi_{2}\varphi_{1})\{x_{1},x_{2},x_{3}\}$

$=\{\varphi_{1}\varphi_{2}(x_{1}),x_{2},x_{3}\}-\{\varphi_{2}\varphi_{1}(x_{1}),x_{2},x_{3}\}$ $=\{[\varphi_{1},\varphi_{2}](x_{1}),x_{2},x_{3}\}$,

$[\varphi_{1},\varphi_{2}]\{x_{1},x_{2},x_{3}\}=(\varphi_{1}\varphi_{2}-\varphi_{2}\varphi_{1})\{x_{1},x_{2},x_{3}\}$

$=\{x_{1},x_{2},\varphi_{1}\varphi_{2}(x_{3})\}-\{x_{1},x_{2},\varphi_{2}\varphi_{1}(x_{3})\}$ $=\{x_{1},x_{2},[\varphi_{1},\varphi_{2}](x_{3})\}$.

Therefore, $[\varphi_{1},\varphi_{2}]\in gl(A)$, the result 1) follows.

For any $\varphi\in\Gamma(A)$, if $\varphi(A)\subseteq Z(A)$ and $\varphi(A^{1})=0$, then by Definition 3.6 and (5), $\varphi\in DerA$. It follows 2).

For any $\varphi\in\Gamma(A)$, $D\in DerA$, and $x_{1},x_{2},x_{3}\in A$,

$\varphi D\{x_{1},x_{2},x_{3}\}=\varphi\{D(x_{1}),x_{2},x_{3}\}+\varphi\{x_{1},D(x_{2}),x_{3}\}+\varphi\{x_{1},x_{2},D(x_{3})\}$

$=\{\varphi D(x_{1}),x_{2},x_{3}\}+\{x_{1},\varphi D(x_{2}),x_{3}\}+\{x_{1},x_{2},\varphi D(x_{3})\}$,

$\varphi D\in DerA$. It follows 3).

For any $\varphi\in\Gamma(A)\cap Der(A)$, by Definition 3.6, and (5), $\forall x_{1},x_{2},x_{3}\in A$, we have

$\varphi\{x_{1},x_{2},x_{3}\}=\{\varphi(x_{1}),x_{2},x_{3}\}+\{x_{1},\varphi(x_{2}),x_{3}\}+\{x_{1},x_{2},\varphi(x_{3})\}=3\varphi\{x_{1},x_{2},x_{3}\}$.

Therefore, $\varphi(A^{1})=0$. By $\varphi\{x_{1},x_{2},x_{3}\}=\{\varphi(x_{1}),x_{2},x_{3}\}=\{x_{1},x_{2},\varphi(x_{3})\}=0$, we have $\varphi(A)\subseteq Z(A)$, it follows $\Gamma(A)\cap Der(A)\subseteq Der_{C}(A)$.

It is clear that if $\varphi\in Der_{C}(A)$, then $\varphi\in\Gamma(A)$. This implies $Der_{C}(A)=\Gamma(A)\cap Der(A)$. ∎

Theorem 3.8**.**

Let $A$ be a semi-associative 3-algebra. Then $\forall D\in Der(A),\varphi\in\Gamma(A)$

1) $[D,\varphi]\subseteq\Gamma(A)$,   2) $D\varphi\in\Gamma(A)$ if and only if $\varphi D\in Der_{C}(A)$,

3) $D\varphi\in Der(A)$ if and only if $[D,\varphi]\in Der_{C}(A)$.

Proof For any $D\in Der(A),\varphi\in\Gamma(A)$, and $x_{1},x_{2},x_{3}\in A$, since

$D\varphi\{x_{1},x_{2},x_{3}\}=D\{\varphi(x_{1}),x_{2},x_{3}\}$

$=\{D\varphi(x_{1}),x_{2},x_{3}\}+\{\varphi(x_{1}),D(x_{2}),x_{3}\}+\{\varphi(x_{1}),x_{2},D(x_{3})\}$

$=\{D\varphi(x_{1}),x_{2},x_{3}\}+\varphi D\{x_{1},x_{2},x_{3}\}-\{\varphi D(x_{1}),x_{2},x_{3}\}$, and

$D\varphi\{x_{1},x_{2},x_{3}\}=D\{x_{1},x_{2},\varphi(x_{3})\}$

$=\{D(x_{1}),x_{2},\varphi(x_{3})\}+\{x_{1},D(x_{2}),\varphi(x_{3})\}+\{x_{1},x_{2},D\varphi(x_{3})\}$

$=\varphi D\{x_{1},x_{2},x_{3}\}-\{x_{1},x_{2},\varphi D(x_{3})\}+\{x_{1},x_{2},D\varphi(x_{3})\}$,

we have $(D\varphi-\varphi D)\{x_{1},x_{2},x_{3}\}=\{(D\varphi-\varphi D)(x_{1}),x_{2},x_{3}\}$, and

$(D\varphi-\varphi D)\{x_{1},x_{2},x_{3}\}=\{x_{1},x_{2},(D\varphi-\varphi D\}(x_{3})\}$. Therefore, $[D,\varphi]\in\Gamma(A)$.

Thanks to Theorem 3.7, $\varphi D\in DerA$. If $D\varphi\in\Gamma(A)$, then by $[D,\varphi]\in\Gamma(A)$, we have $\varphi D\in\Gamma(A)$. Therefore, $\varphi D\in Der(A)\cap\Gamma(A)$. Conversely, if $\varphi D\in\Gamma(A)$, then by $[D,\varphi]\in\Gamma(A)$, we have $D\varphi\in\Gamma(A)$. We get 2). The result 3) follows from 1) and 2).

4. Sub-adjacent $3$-Lie algebras and double modules of semi-associative 3-algebras

4.1. Sub-adjacent $3$-Lie algebras of semi-associative 3-algebras

3-Lie algebras have close relationships with Lie algebras, pre-Lie algebras, associative algebras, commutative associative algebras and et al. Now we discuss the relation between 3-Lie algebras with semi-associative 3-algebras.

A $3$-Lie algebra $(L,[,,])$ is a vector space $L$ with a linear multiplication $[,,]:L\wedge L\wedge L\rightarrow L$ which satisfies that for all $x_{i}\in L$, $1\leq i\leq 5,$

[TABLE]

Theorem 4.1**.**

Let $A$ be a semi-association $3$-algebra. Then $(A,[,,]_{c})$ is a $3$-Lie algebra, where for all $x_{1},x_{2},x_{3}\in A,$

[TABLE]

Proof.

By (1), (2) and (3), the multiplication $[,,]_{C}$ given by (14) is skew-symmetric, and for all $x_{i}\in A,1\leq i\leq 5$,

$[[x_{1},x_{2},x_{3}]_{c},x_{4},x_{5}]_{c}-[[x_{1},x_{4},x_{5}]_{c},x_{2},x_{3}]_{c}-[x_{1},[x_{2},x_{4},x_{5}]_{c},x_{3}]_{c}$

$-[x_{1},x_{2},[x_{3},x_{4},x_{5}]_{c}]_{c}$

$=\{\{x_{1},x_{2},x_{3}\},x_{4},x_{5}\}+\{\{x_{2},x_{3},x_{1}\},x_{4},x_{5}\}+\{\{x_{3},x_{1},x_{2}\},x_{4},x_{5}\}$ $+\{x_{4},x_{5},\{x_{1},x_{2},x_{3}\}\}$

$+\{x_{4},x_{5},\{x_{2},x_{3},x_{1}\}\}+\{x_{4},x_{5},\{x_{3},x_{1},x_{2}\}\}$ $+\{x_{5},\{x_{1},x_{2},x_{3}\},x_{4}\}$ $+\{x_{5},\{x_{2},x_{3},x_{1}\},x_{4}\}$

$+\{x_{5},x_{3},x_{1},x_{2}\},x_{4}\}$ $-\{\{x_{1},x_{4},x_{5}\},x_{2},x_{3}\}-\{\{x_{4},x_{5},x_{1}\},x_{2},x_{3}\}$ $-\{\{x_{5},x_{1},x_{4}\},x_{2},x_{3}\}$

$-\{x_{2},x_{3},\{x_{1},x_{4},x_{5}\}\}$ $-\{x_{2},x_{3},\{x_{4},x_{5},x_{1}\}\}-\{x_{2},x_{3},\{x_{5},x_{1},x_{4}\}\}$ $-\{x_{3},\{x_{1},x_{4},x_{5}\},x_{2}\}$

$-\{x_{3},\{x_{4},x_{5},x_{1}\},x_{2}\}$ $-\{x_{3},\{x_{5},x_{1},x_{4}\},x_{2}\}$ $-\{x_{1},\{x_{2},x_{4},x_{5}\},x_{3}\}$ $-\{x_{1},\{x_{4},x_{5},x_{2}\},x_{3}\}$

$-\{x_{1},\{x_{5},x_{2},x_{4}\},x_{3}\}$ $-\{\{x_{2},x_{4},x_{5}\},x_{3},x_{1}\}-\{\{x_{4},x_{5},x_{2}\},x_{3},x_{1}\}-\{\{x_{5},x_{2},x_{4}\},x_{3},x_{1}\}$

$-\{x_{3},x_{1},\{x_{2},x_{4},x_{5}\}\}-\{x_{3},x_{1},\{x_{4},x_{5},x_{2}\}\}-\{x_{3},x_{1},\{x_{5},x_{2},x_{4}\}\}$ $-\{x_{1},x_{2},\{x_{3},x_{4},x_{5}\}\}$

$-\{x_{1},x_{2},\{x_{4},x_{5},x_{3}\}\}-\{x_{1},x_{2},\{x_{5},x_{3},x_{4}\}\}$ $-\{x_{2},\{x_{3},x_{4},x_{5}\},x_{1}\}-\{x_{2},\{x_{4},x_{5},x_{3}\},x_{1}\}$

$-\{x_{2},\{x_{5},x_{3},x_{4}\},x_{1}\}$ $-\{\{x_{3},x_{4},x_{5}\},x_{1},x_{2}\}-\{\{x_{4},x_{5},x_{3}\},x_{1},x_{2}\}-\{\{x_{5},x_{3},x_{4}\},x_{1},x_{2}\}$

$=\{x_{4},x_{5},\{x_{1},x_{2},x_{3}\}\}+\{x_{5},\{x_{4},x_{2},x_{3}\},x_{1}\}+\{x_{4},x_{5},\{x_{2},x_{3},x_{1}\}\}$ $+\{x_{5},\{x_{4},x_{3},x_{1}\},x_{2}\}$

$+\{x_{4},x_{5},\{x_{3},x_{1},x_{2}\}\}+\{x_{5},\{x_{4},x_{1},x_{2}\},x_{3}\}$ $-\{x_{2},x_{3},\{x_{1},x_{4},x_{5}\}\}-\{x_{3},\{x_{2},x_{4},x_{5}\},x_{1}\}$

$-\{x_{2},x_{3},\{x_{4},x_{5},x_{1}\}\}$ $-\{x_{3},\{x_{2},x_{5},x_{1}\},x_{4}\}-\{x_{2},x_{3},\{x_{5},x_{1},x_{4}\}\}-\{x_{3},\{x_{2},x_{1},x_{4}\},x_{5}\}$

$-\{x_{3},x_{1},\{x_{2},x_{4},x_{5}\}\}-\{x_{1},\{x_{3},x_{4},x_{5}\},x_{2}\}-\{x_{3},x_{1},\{x_{4},x_{5},x_{2}\}\}$ $-\{x_{1},\{x_{3},x_{5},x_{2}\},x_{4}\}$

$-\{x_{3},x_{1},\{x_{5},x_{2},x_{4}\}\}-\{x_{1},\{x_{3},x_{2},x_{4}\},x_{5}\}$ $-\{x_{1},x_{2},\{x_{3},x_{4},x_{5}\}\}-\{x_{2},\{x_{1},x_{4},x_{5}\},x_{3}\}$

$-\{x_{1},x_{2},\{x_{4},x_{5},x_{3}\}\}$ $-\{x_{2},\{x_{1},x_{5},x_{3}\},x_{4}\}-\{x_{1},x_{2},\{x_{5},x_{3},x_{4}\}\}-\{x_{2},\{x_{1},x_{3},x_{4}\},x_{5}\}$ $=0$.

Therefore, the multiplication $[~{},~{},~{}]_{c}$ satisfies (13), and $A_{c}$ is a 3-Lie algebra. The proof is complete. ∎

The 3-Lie algebra $(A,[,,]_{c})$ is called the sub-adjacent $3$-Lie algebra of the semi-association 3-algebra $A$, and is simply denoted by $A_{c}$. $Der(A_{c})$ denotes the derivation algebras of the sub-adjacent $3$-Lie algebra $A_{c}$, and

Theorem 4.2**.**

Let $A$ be a semi-associative $3$-algebra. Then

1) derivation algebra $Der(A)$ is a subalgebra of $Der(A_{c})$;

2) for all $x_{i}\in A$, $1\leq i\leq 4,$ and $S(x_{1},x_{2})$, $S(x_{3},x_{4})\in Der(A)$,

[TABLE]

3) if $I$ is an ideal ( a subalgebra ) of semi-associative algebra $A$, then $I$ is also an ideal ( a subalgebra ) of the $3$-Lie algebra $A_{c}$.

Proof.

For any $D\in Der(A)$, and $x,y,z\in A$,

$D[x,y,z]_{c}=D(\{x,y,z\}+\{y,z,x\}+\{z,x,y\})$ $=\{Dx,y,z\}+\{x,Dy,z\}+\{x,y,Dz\}$

$+\{Dy,z,x\}+\{y,Dz,x\}+\{y,z,Dx\}$ $+\{Dz,x,y\}+\{z,Dx,y\}$

$+\{z,x,Dy\}$ $=[Dx,y,z]_{C}+[x,Dy,z]_{C}+[x,y,Dz]_{c},$

therefore, $D\in Der(A_{c})$. It follows 1). By the similar discussion to the above, we get 2) and 3). ∎

Theorem 4.3**.**

Let $A$ be a semi-associative $3$-algebra. Then $L(A),R(A)\subseteq$ $Der(A_{c})$.

Proof.

By (1)-(3), (6) and (14), for any $x,y,u,v,w\in A$,

[TABLE]

[TABLE]

Therefore, $L(x,y)[u,v,w]_{c}=[L(x,y)(u),v,w]_{c}+[u,L(x,y)(v),w]_{c}+[u,v,L(x,y)(w)]_{c}.$

By the similar discussion to the above, we have

[TABLE]

The proof is complete. ∎

4.2. Double modules of the semi-associative 3-algebras

In this section, we discuss double modules of semi-associative $3$-algebras.

A representation (or a module) $(V,\rho)$ of a $3$-Lie algebra $(G,[,,])$ is a vector space $V$ and a linear transformation $\rho:$ $G\wedge G\longrightarrow End(V)$, satisfying, $\forall x_{1},x_{2},x_{3},x_{4}\in G$,

[TABLE]

[TABLE]

We know that $(V,\rho)$ is a 3-Lie algebra $(G,[,,])$-module if and only if $G\dot{+}V$ is a 3-Lie algebra in the multiplication $[,,]_{\rho}$, where $\forall x_{1},x_{2},x_{3}\in A$ and $v_{1},v_{2},v_{3}\in V,$

[TABLE]

Definition 4.4**.**

Let $A$ be a semi-associative 3-algebra, $V$ be a vector space, and $\phi,\psi$: $A\times A\longrightarrow End(V)$ be linear mappings. If $\phi$ and $\psi$ satisfy the following properties, $\forall x_{1},x_{2},x_{3},x_{4}\in A$,

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

then $(\phi,\psi,V)$ is called a double representation of $A$, or is simply called a double module of $A$.

Theorem 4.5**.**

Let $A$ be a semi-associative 3-algebra, and $l,r$: $A\times A\longrightarrow End(V)$ be linear mappings. Then triple $(\phi,\psi,V)$ is a double module of $A$ if and only if $(A\dot{+}V,\{,,\}_{\phi\psi})$ is a semi-associative 3-algebra, where $\forall x_{i}\in A,v_{i}\in V,i=1,2,3$,

[TABLE]

Proof.

If $(\phi,\psi,V)$ is a double mode of the semi-associative algebra $A$, we will prove that $(A\dot{+}V,\{,,\}_{\phi\psi})$ is a semi-associative 3-algebra.

Thanks to (24), for all $x_{i}\in A,v_{i}\in V,1\leq i\leq 5$,

$\{x_{1}+v_{1},x_{2}+v_{2},x_{3}+v_{3}\}_{\phi\psi}=-\{x_{2}+v_{2},x_{1}+v_{1},x_{3}+v_{3}\}_{\phi\psi}$, (1) holds.

By (24), (19), (20) and (23),

$\{x_{1}+v_{1},\{x_{2}+v_{2},x_{3}+v_{3},x_{4}+v_{4}\}_{\phi\psi},x_{5}+v_{5}\}_{\phi\psi}$

$=\{x_{1},x_{2},\{x_{3},x_{4},x_{5}\})+\psi(x_{2},\{x_{3},x_{4},x_{5}\})v_{1}-\psi(x_{1},\{x_{3},x_{4},x_{5}\})v_{2}$

$+\phi(x_{1},x_{2})\psi(x_{4},x_{5})v_{3}-\phi(x_{1},x_{2})\psi(x_{3},x_{5})v_{4}+\phi(x_{1},x_{2})\phi(x_{3},x_{4})v_{5}$

$=\{x_{1}+v_{1},x_{2}+v_{2},\{x_{3}+v_{3},x_{4}+v_{4},x_{5}+v_{5}\}_{\phi\psi}\}_{\phi\psi}$, (2) holds.

Thanks to (24), (20)- (22),

$\{x_{1}+v_{1},\{x_{2}+v_{2},x_{3}+v_{3},x_{4}+v_{4}\}_{\phi\psi},x_{5}+v_{5}\}_{\phi\psi}$

$=\{x_{5},\{x_{2},x_{3},x_{4}\},x_{1}\}+\{x_{1},\{x_{5},x_{3},x_{4}\},x_{2}\}+(\phi(x_{5},\{x_{2},x_{3},x_{4}\})$

$+\psi(\{x_{5},x_{3},x_{4}\},x_{2}))v_{1}+(\phi(x_{1},\{x_{5},x_{3},x_{4}\})-\psi(x_{5},x_{1})\psi(x_{3},x_{4}))v_{2}$

$+(\psi(x_{5},x_{1})\psi(x_{2},x_{4})+\psi(x_{1},x_{2})\psi(x_{5},x_{4}))v_{3}-(\psi(x_{5},x_{1})\phi(x_{2},x_{3})$

$+\psi(x_{1},x_{2})\phi(x_{5},x_{3}))v_{4}+(\psi(\{x_{2},x_{3},x_{4}\},x_{1})-\psi(x_{1},x_{2})\psi(x_{3},x_{4}))v_{5}$

$=\{x_{5}+v_{5},\{x_{2}+v_{2},x_{3}+v_{3},x_{4}+v_{4}\}_{\phi\psi},x_{1}+v_{1}\}\}_{\phi\psi}$

$+\{x_{1}+v_{1},\{x_{5}+v_{5},x_{3}+v_{3},x_{4}+v_{4}\}_{\phi\psi},x_{2}+v_{2}\}_{\phi\psi}$, (3) holds.

Therefore, $(A\dot{+}V,\{,,\}_{\phi\psi})$ is a semi-associative 3-algebra.

Conversely, if $(A\dot{+}V,\{,,\}_{\phi\psi})$ is a semi-associative algebra. Then by (1), $\forall x_{i}\in A,v_{i}\in V$, $1\leq i\leq 3$,

[TABLE]

Thanks to (24),

$\{x_{1},x_{2},x_{3}+v_{3}\}_{\phi\psi}=\{x_{1},x_{2},x_{3}\}+\phi(x_{1},x_{2})v_{3}$, $\{x_{2},x_{1},x_{3}+v_{3}\}=\{x_{2},x_{1},x_{3}\}+\phi(x_{2},x_{1})v_{3}$,

therefore, $\phi(x_{1},x_{2})=-\phi(x_{2},x_{1})$, (18) holds.

By (2), and (24),

$\{x_{1}+v_{1},\{x_{2}+v_{2},x_{3}+v_{3},x_{4}+v_{4}\}_{\phi\psi},x_{5}+v_{5}\}_{\phi\psi}$

$=\{x_{1}+v_{1},x_{2}+v_{2},\{x_{3}+v_{3},x_{4}+v_{4},x_{5}+v_{5}\}_{\phi\psi}\}_{\phi\psi}$,

$\{x_{1}+v_{1},\{x_{2}+v_{2},x_{3}+v_{3},x_{4}+v_{4}\}_{\phi\psi},x_{5}+v_{5}\}_{\phi\psi}$

$=\{x_{1},\{x_{2},x_{3},x_{4}\},x_{5}\}+\psi(\{x_{2},x_{3},x_{4}\},x_{5})v_{1}-\psi(x_{1},x_{5})\psi(x_{3},x_{4})v_{2}$

$+\psi(x_{1},x_{5})\psi(x_{2},x_{4})v_{3}-\psi(x_{1},x_{5})\phi(x_{2},x_{3})v_{4}+\phi(x_{1},\{x_{2},x_{3},x_{4}\})v_{5}$,

$\{x_{1}+v_{1},x_{2}+v_{2},\{x_{3}+v_{3},x_{4}+v_{4},x_{5}+v_{5}\}_{\phi\psi}\}_{\phi\psi}$

$=\{x_{1},x_{2},\{x_{3},x_{4},x_{5}\}\}+\psi(x_{2},\{x_{3},x_{4},x_{5}\})v_{1}-\psi(x_{1},\{x_{3},x_{4},x_{5}\})v_{2}$

$+\phi(x_{1},x_{2})\psi(x_{4},x_{5})v_{3}-\phi(x_{1},x_{2})\psi(x_{3},x_{5})v_{4}+\phi(x_{1},x_{2})\phi(x_{3},x_{4})v_{5}$.

If we suppose $v_{i}\neq 0,v_{j}=0$, for $1\leq i\neq j\leq 5$, then we get

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

Thanks to (3), and (24),

$\{x_{5}+v_{5},\{x_{2}+v_{2},x_{3}+v_{3},x_{4}+v_{4}\}_{\phi\psi},x_{1}+v_{1}\}_{\phi\psi}+\{x_{1}+v_{1},\{x_{5}+v_{5},x_{3}+v_{3},x_{4}+v_{4}\}_{\phi\psi},x_{2}+v_{2}\}_{\phi\psi}$

=\{x_{5},\{x_{2},x_{3},x_{4}\},x_{1}\}+\{x_{1},\{x_{5},x_{3},x_{4}\},x_{2}\}+(\phi(x_{5},\{x_{2},x_{3},x_{4}\})$$+\psi(\{x_{5},x_{3},x_{4}\},x_{2}))v_{1}

+(\phi(x_{1},\{x_{5},x_{3},x_{4}\})-\psi(x_{5},x_{1})\psi(x_{3},x_{4}))v_{2}$$+(\psi(x_{5},x_{1})\psi(x_{2},x_{4})+\psi(x_{1},x_{2})\psi(x_{5},x_{4}))v_{3}

-(\psi(x_{5},x_{1})\phi(x_{2},x_{3})$$+\psi(x_{1},x_{2})\phi(x_{5},x_{3}))v_{4}+(\psi(\{x_{2},x_{3},x_{4}\},x_{1})-\psi(x_{1},x_{2})\psi(x_{3},x_{4}))v_{5},

similarly, for $v_{i}\neq 0,v_{j}=0$, for $1\leq i\neq j\leq 5$, we have

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

Now we analyze the above identities. (19) follows from (26); (22) follows from (33); and it is clear that (30) is equivalent to (28), (29) and (31); and (34) is equivalent to (25), (27) and (33); (20) follows from (25) and (26); (21) follows from (28), (29) and (31); and (23) follows from (28), (29), (31) and (32). Therefore, $(l,r,V)$ is a double mode of the semi-associative algebra $A$. The proof is complete. ∎

Let $\tau:$ $V\otimes V\rightarrow V\otimes V$ be the exchange mapping, that is,

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Theorem 4.6**.**

Let $A$ be a semi-associative $3$-algebra, and $(l,r,V)$ be the double module of $A$. Then $(V,\rho)$ is a 3-Lie algebra $A_{c}$-module, where

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Proof.

Thanks to Theorem 4.5, $A\dot{+}V$ is a semi-associative $3$-algebra in the multiplication (24). Therefore, the multiplication of the subjacent 3-Lie algebra $(A\dot{+}V)_{c}$ of $A\dot{+}V$ is (14), that is, $\forall x_{1},x_{2},x_{3}\in A,$ $v_{1},v_{2},v_{3}\in V$,

$[x_{1}+v_{1},x_{2}+v_{2},x_{3}+v_{3}]_{c}$

$=\{x_{1}+v_{1},x_{2}+v_{2},x_{3}+v_{3}\}_{\phi\psi}+\{x_{2}+v_{2},x_{3}+v_{3},x_{1}+v_{1}\}_{\phi\psi}+\{x_{3}+v_{3},x_{1}+v_{1},x_{2}+v_{2}\}_{\phi\psi}$

$=\{x_{1},x_{2},x_{3}\}+\phi(x_{1},x_{2})v_{3}-\psi(x_{1},x_{3})v_{2}+\psi(x_{2},x_{3})v_{1}+\{x_{2},x_{3},x_{1}\}+\psi(x_{2},x_{3})v_{1}$

$-\psi(x_{2},x_{1})v_{3}+\psi(x_{3},x_{1})v_{2}+\{x_{3},x_{1},x_{2}\}+\phi(x_{3},x_{1})v_{2}-\psi(x_{3},x_{2})v_{1}+\psi(x_{1},x_{2})v_{3}$

$=[x_{1},x_{2},x_{3}]_{c}+(\phi-\psi\tau+\psi)(x_{1},x_{2})v_{3}+(\phi-\psi\tau+\psi)(x_{2},x_{3})v_{1}+(\phi-\psi\tau+\psi)(x_{3},x_{1})v_{2}.$

Follows from (17), $(V,\rho)$ is a 3-Lie algebra $A_{c}$, where $\rho=\phi-\psi\tau+\psi$. ∎

Theorem 4.7**.**

Let $A$ be a semi-associative 3-algebra, and $L(x,y),R(x,y):A\times A\rightarrow A$ be left and right multiplications defined as (6), $\forall x,y\in A$. Then $(L,R,A)$ is a double module of $A$.

Proof.

Apply Lemma 12, Theorem 4.5 and Definition 4.4.∎

The double module $(L,R,A)$ of the semi-associative 3-algebra $A$ is called the regular representation of $A$, or the adjoint module of $A$.

Theorem 4.8**.**

Let $A$ be a semi-associative 3-algebra, and $(l,r,V)$ be a double module of $A$. Then $(\phi^{*},\psi^{*},V^{*})$ is also a double module of $A$, where $V^{*}$ is the dual space of $V$, and $\phi^{*},\psi^{*}:A\times A\longrightarrow End(V^{*})$ are defined by $\forall x,y\in A,v\in V,\xi\in V^{*}$,

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Proof.

Apply Lemma 12, Definition 4.4 and Theorem 4.5.∎

Corollary 4.9**.**

Let $A$ be a semi-associative 3-algebra. Then $(L^{*},R^{*},A^{*})$ is a double module of $A$.

Proof.

The result follows from Theorem 4.7 and Theorem 35, directly.

∎

4.3. Double extensions of semi-associative 3-algebras by cocycles

Definition 4.10**.**

Let $A$ be a semi-associative 3-algebra, if linear mapping $\theta:A\otimes A\otimes A\rightarrow A^{*}$ satisfies that for all $x,y,z,w,u\in A$,

[TABLE]

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$=\theta\{x,\{u,z,w\},y\}+R^{*}(x,u)\theta\{y,z,w\}$**

$-R^{*}(u,x)\theta\{y,z,w\}-R^{*}(x,y)\theta\{u,z,w\}$*,

then $\theta$ is called a cocycle of the semi-associative 3-algebra $A$, where $L^{*},R^{*}$ are defined by (35).*

Theorem 4.11**.**

Let $A$ be a semi-associative 3-algebra, and $\theta:A\otimes A\otimes A\rightarrow A^{*}$ be a cocycle of $A$. Then $(A\dot{+}A^{*},\{,,\}_{\theta})$ is a semi-associative 3-algebra, where $\forall x_{i}\in A,\xi_{i}\in A^{*},1\leq i\leq 3,$

[TABLE]

$-R^{*}(x_{1},x_{3})\xi_{2}+R^{*}(x_{2},x_{3})\xi_{1}.$**

Proof.

Thanks to (18) - (23), $\forall x_{i}\in A,\xi_{i}\in A^{*},1\leq i\leq 3,$

\{x_{1}+\xi_{1},x_{2}+\xi_{2},x_{3}+\xi_{3}\}_{\theta}$$=-\{x_{2}+\xi_{2},x_{1}+\xi_{1},x_{3}+\xi_{3}\}_{\theta}, (1) holds.

By Definition 4.10,

$\{x_{1}+\xi_{1},\{x_{2}+h_{2},x_{3}+h_{3},x_{4}+\xi_{4}\}_{\theta},x_{5}+\xi_{5}\}_{\theta}$

$=\{x_{1},x_{2},\{x_{3},x_{4},x_{5}\}\}+\theta\{x_{1},x_{2},\{x_{3},x_{4},x_{5}\}\}+L^{*}(x_{1},x_{2})L^{*}(x_{3},x_{4})\xi_{5}$

$+L^{*}(x_{1},x_{2})\theta\{x_{3},x_{4},x_{5}\}-L^{*}(x_{1},x_{2})R^{*}(x_{3},x_{5})\xi_{4}+L^{*}(x_{1},x_{2})R^{*}(x_{4},x_{5})\xi_{3}$

$-R^{*}(x_{1},\{x_{3},x_{4},x_{5}\})\xi_{2}+R^{*}(x_{2},\{x_{3},x_{4},x_{5}\})\xi_{1}$

$=\{x_{1}+\xi_{1},x_{2}+\xi_{2},\{x_{3}+\xi_{3},x_{4}+\xi_{4},x_{5}+\xi_{5}\}_{\theta}\}_{\theta}$, (2) holds.

Thanks to Theorem 35,

$\{x_{1},\{x_{2},x_{3},x_{4}\},x_{5}\}+\theta\{x_{1},\{x_{2},x_{3},x_{4}\},x_{5}\}+L^{*}(x_{1},\{x_{2},x_{3},x_{4}\})\xi_{5}$

$-R^{*}(x_{1},x_{5})\theta\{x_{2},x_{3},x_{4}\}-R^{*}(x_{1},x_{5})L^{*}(x_{2},x_{3})\xi_{4}+R^{*}(x_{1},x_{5})R^{*}(x_{2},x_{4})\xi_{3}$

$-R^{*}(x_{1},x_{5})R^{*}(x_{3},x_{4})\xi_{2}+R^{*}(\{x_{2},x_{3},x_{4}\},x_{5})\xi_{1}$

$=\{x_{5}+\xi_{5},\{x_{2}+\xi_{2},x_{3}+\xi_{3},x_{4}+\xi_{4}\}_{\theta},x_{1}+\xi_{1}\}_{\theta}+\{x_{1}+\xi_{1},\{x_{5}+\xi_{5},x_{3}+\xi_{3},x_{4}+\xi_{4}\}_{\theta},x_{2}+\xi_{2}\}_{\theta}$,

we get

$\{x_{1}+\xi_{1},\{x_{2}+\xi_{2},x_{3}+\xi_{3},x_{4}+\xi_{4}\}_{\theta},x_{5}+\xi_{5}\}_{\theta}$

$=\{x_{5}+\xi_{5},\{x_{2}+\xi_{2},x_{3}+\xi_{3},x_{4}+\xi_{4}\}_{\theta},x_{1}+\xi_{1}\}_{\theta}+\{x_{1}+\xi_{1},\{x_{5}+\xi_{5},x_{3}+\xi_{3},x_{4}+\xi_{4}\}_{\theta},x_{2}+\xi_{2}\}_{\theta}$, that is, (3) holds. Therefore, $(A\dot{+}A^{*},\{,,\}_{\theta})$ is a semi-associative 3-algebra. ∎

The semi-associative 3-algebra $(A\dot{+}A^{*},\{,,\}_{\theta})$ is called a double extension of $3$-semi-associative algebra by $\theta$.