Simple $\omega$-Lie Algebras and 4-Dimensional $\omega$-Lie Algebras Over ${\mathbb {C}}$

Abstract

In this paper, we study simple $\omega$-Lie algebras and 4-dimensional $\omega$-Lie algebras over the field of complex numbers. We provide an approach to classify all 4-dimensional non-Lie complex $\omega$-Lie algebras. We prove that any non-Lie finite-dimensional simple $\omega$-Lie algebra has dimension 3. A complete list of all non-Lie complex simple $\omega$-Lie algebras is also derived.

1 Introduction

Let ${\mathbb {C}}$ be the field of complex numbers and $L$ a finite dimensional vector space over ${\mathbb {C}}$. Let $[-,-]:L\times L \longrightarrow L$ be a skew-symmetric bracket and $\omega : L\times L \longrightarrow {\mathbb {C}}$ a bilinear form. We say that a triple $(L,[-,-],\omega )$ is an $\omega$ -Lie algebra if

$$\begin{aligned}{}[[x,y],z]+ [[y,z],x]+ [[z,x],y]=\omega (x,y)z+\omega (y,z)x+\omega (z,x)y \end{aligned}$$

(1.1)

for all $x,y,z\in L$. The Eq. (1.1) is said to be the $\omega$-Jacobi identity. Obviously, this notion generalizes the definition of Lie algebras. It follows from (1.1) that the bilinear form $\omega$ is skew-symmetric, and an $\omega$-Lie algebra is a Lie algebra if and only if $\omega \equiv 0$. Thus the usual Lie algebras are called trivial $\omega$ -Lie algebras and the $\omega$-Lie algebras with nonzero $\omega$ are called nontrivial $\omega$ -Lie algebras.

The notion of $\omega$-Lie algebras was introduced by Nurowski in the recent work [7] for some physical considerations. These new nonassociative algebras are very closed to the study of isoparametric hypersufaces in Riemannian geometry [4, 8]. As we know, an algebraic structure possessing “symmetry” always attracts the attention of pure mathematicians and physicists, such as groups, Lie algebras, left-symmetric algebras, and Lie superalgebras. Apparently, $\omega$-Lie algebras are also a kind of “symmetric” nonassociative algebras.

From the point of view of pure mathematics, the theoretical study of an algebraic object should contain its structures, representations, and classifications. A fundamental development of $\omega$-Lie algebras was made by Zusmanovich [11], in which a lot of basic concepts, such as modules, (quasi-) ideals and (generalized) derivations, were introduced and some properties of $\omega$-Lie algebras were investigated.

Like other nonassociative algebras (such as [1, 2, 9, 10]), low-dimensional $\omega$-Lie algebras provide a lot of interesting examples and a possible direction to more researches. One of Zusmanovich’s results asserts that finite-dimensional nontrivial $\omega$-Lie algebras are either low-dimensional or have an abelian subalgebra of small codimension with some restrictive conditions. In particular, three useful and surprising results are proved.

Proposition 1.1

([11], Lemma 8.1) If $L$ is a finite-dimensional $\omega$-Lie algebra with non-degenerate $\omega$, then dim $L=2$.

Proposition 1.2

([11], Lemma 8.2) A 4-dimensional $\omega$-Lie algebra over an algebraically closed field contains a 3-dimensional subalgebra.

Theorem 1.3

([11], Theorem 2) A finite-dimensional semisimple $\omega$-Lie algebra is either a Lie algebra, or has dimension $\leqslant 4$.

The above results show essentially the importance of classifying low-dimensional $\omega$-Lie algebras. By the definition of $\omega$-Lie algebras, it is easy to see that all $\omega$-Lie algebras are trivial in the cases of dimensions 1 and 2. The first example of nontrivial 3-dimensional real $\omega$-Lie algebra was given by Nurowski [7]. Actually, in that paper, the author gave a classification of 3-dimensional $\omega$-Lie algebras over the field of real numbers. Recently, we extended the method of classification of 3-dimensional complex Lie algebras appeared in [6] to obtain a classification of 3-dimensional complex (and real) $\omega$-Lie algebras [5]. Proposition 1.1 mentioned above is a key fact in our paper.

With the classification of 3-dimensional complex $\omega$-Lie algebras in hand, Proposition 1.2 and Theorem 1.3 lead us to consider the classification problem of 4-dimensional complex $\omega$-Lie algebras.

The purpose of this paper is to give a classification of all 4-dimensional complex $\omega$-Lie algebras. Let $L$ be a 4-dimensional complex $\omega$-Lie algebra. By Proposition 1.2, there exists a 3-dimensional subalgebra $W\subset L$, which must be in our classification list appeared in [5]. The following theorem is a classification of all 3-dimensional nontrivial $\omega$-Lie algebras.

Theorem 1.4

([5]) Let $L$ be a nontrivial (i.e., non-Lie) 3-dimensional $\omega$-Lie algebra with a basis $\{x,y,z\}$, then it must be isomorphic to one of the following algebras:

• (1) $L_{1}{:} [x,z]=0,[y,z]=z, [x,y]=y\text { and }\omega (y,z)=\omega (x,z)=0,\omega (x,y)=1.$

• (2) $L_{2}{:} [x,y]=0,[x,z]=y,[y,z]=z \text { and } \omega (x,y)=0, \omega (x,z)=1, \omega (y,z)=0.$

• (3) $A_{\alpha }{:} [x,y]=x,[x,z]=x+y, [y,z]=z+\alpha x\text { and } \omega (x,y)=\omega (x,z)=0,$

  $$\begin{aligned} \omega (y,z)=-1,\text { where }\alpha \in {\mathbb {C}}. \end{aligned}$$

• (4) $B{:} [x,y]=y, [x,z]=y+z, [y,z]=x\text { and } \omega (x,y)=\omega (x,z)=0,$ $\omega (y,z)=2.$

• (5) $C_{\alpha }{:} [x,y]=y, [x,z]=\alpha z, [y,z]=x\text { and } \omega (x,y)=\omega (x,z)=0,$

  $$\begin{aligned} \omega (y,z)=1+\alpha ,\text { where } 0,-1\ne \alpha \in {\mathbb {C}}. \end{aligned}$$

Now we assume that $\{x,y,z\}\subset L$ is a basis of $W$. Choose a nonzero vector $e\in L$ such that $e\notin W$. To obtain all 4-dimensional $\omega$-Lie algebras, we need only to determine six values:

$$\begin{aligned}{}[e,x],[e,y],[e,z]; \omega (e,x),\omega (e,y),\omega (e,z). \end{aligned}$$

Our main results are summarized as follows.

Theorem 1.5

Let $L$ be a nontrivial 4-dimensional $\omega$-Lie algebra and $W$ a 3-dimensional subalgebra of $L$.

• (1) If $W\cong L_{1}$, then $L$ must be isomorphic to one of the following algebras:

  $$\begin{aligned} L_{1,i} (i=1,2,\ldots ,8) ,E_{1,\alpha }(\alpha \ne 0,1) ,F_{1,\alpha } (\alpha \ne 0,1),G_{1,\alpha } ,H_{1,\alpha }. \end{aligned}$$

• (2) If $W\cong L_{2}$, then $L$ must be isomorphic to one of the following algebras: $L_{2,1}, L_{2,2}, L_{2,3}, L_{2,4}.$

• (3) If $W\cong A_{\alpha }$, then $L$ is isomorphic to $\widetilde{A}_{\alpha }$.

• (4) If $W\cong B$, then $L$ is isomorphic to $\widetilde{B}$.

• (5) If $W\cong C_{\alpha }$, then $L$ is isomorphic to $\widetilde{C}_{\alpha }(\alpha \ne 0,-1)$.

In all cases, the parameter $\alpha$ belongs to ${\mathbb {C}}$.

In particular, since every nontrivial 4-dimensional $\omega$-Lie algebra contains a 3-dimensional subalgebra (Proposition 1.2 above) and there is a classification of nontrivial 3-dimensional $\omega$-Lie algebras (Theorem 1.4), Theorem 1.5 gives rise to a classification of all nontrivial 4-dimensional $\omega$-Lie algebras.

Corollary 1.6

Any nontrivial 4-dimensional $\omega$-Lie algebra must be isomorphic to one of the following algebras:

$$\begin{aligned}&\bigg \{L_{1,1}, \ldots , L_{1,8}, L_{2,1}, L_{2,2}, L_{2,3}, L_{2,4}, \widetilde{B}, E_{1,\alpha } (\alpha \ne 0,1) ,F_{1,\alpha } (\alpha \ne 0,1),\\&\qquad G_{1,\alpha }, H_{1,\alpha },\widetilde{A}_{\alpha }, \widetilde{C}_{\alpha }(\alpha \ne 0,-1)\bigg \}. \end{aligned}$$

For all cases, the parameter $\alpha \in {\mathbb {C}}$.

Combining Theorems 1.3 and 1.4 with Corollary 1.6, we give a complete list consisting of all finite dimensional complex simple $\omega$-Lie algebras.

Theorem 1.7

Let $L$ be a nontrivial finite dimensional complex simple $\omega$-Lie algebra, then it must be 3-dimensional and isomorphic to one of the following algebras: $A_{\alpha }(\alpha \in {\mathbb {C}}) , B$ and $C_{\alpha } (0,-1\ne \alpha \in {\mathbb {C}})$.

Remark 1.8

We should notice that in [11] (page 1041), a claim states that there exists an example of 4-dimensional simple $\omega$-Lie algebra. By Theorem 1.7, however, this claim is not correct. See Sect. 8 for a detailed discussion.

The present paper is organized as follows. In Sects. 2–6, we discuss all 4-dimensional $\omega$-Lie algebras case by case according to the situation of their 3-dimensional subalgebras. Our classification methods in all cases are similar, so we just give a detailed computation in Sect. 2 and other sections only contain the main proof steps. All contents in Sects. 2–6 provide a proof of Theorem 1.5. Section 7 consists of the classification of finite-dimensional complex simple $\omega$-Lie algebras. In the last section, we give some discussions about infinite-dimensional $\omega$-Lie algebras and $\omega$-Jacobi-Jordan algebras. In particular, we explain the reason why the claim in [11] that there exists a 4-dimensional simple $\omega$-Lie algebra is not correct.

Throughout this paper, all vector spaces and algebras are finite dimensional over ${\mathbb {C}}$ unless stated otherwise. We use the notion $\circlearrowleft (x,y,z)$ to denote the $\omega$-Jacobi identity in elements $x,y,z$. Let $L$ be an $\omega$-Lie algebra with a basis $\{x,y,z,e\}$. We write $\wedge ^{2} L$ for the exterior power of $L$ and view the Lie bracket as a linear map $\varphi =[-,-]:\wedge ^{2} L\longrightarrow L$. We always assume that

$$\begin{aligned}{}[e,x]= & {} a_{1}e+b_{1}x+c_{1}y+d_{1}z;\nonumber \\ ~~[e,y]= & {} a_{2}e+b_{2}x+c_{2}y+d_{2}z;~~[e,z]= a_{3}e+b_{3}x+c_{3}y+d_{3}z. \end{aligned}$$

(1.2)

Recall that the $\omega$-Jacobi identities are

$$\begin{aligned}&\circlearrowleft (e,x,y){:}[[e,x],y]\,+\,[[x,y],e]\,+\, [[y,e],x]=\omega (e,x)y\,+\,\omega (x,y)e\,+\,\omega (y,e)x,\nonumber \\ \end{aligned}$$

(1.3)

$$\begin{aligned}&\circlearrowleft (e,x,z){:}[[e,x],z]+[[x,z],e]\,+\, [[z,e],x]=\omega (e,x)z\,+\,\omega (x,z)e\,+\,\omega (z,e)x,\nonumber \\ \end{aligned}$$

(1.4)

$$\begin{aligned}&\circlearrowleft (e,y,z){:}[[e,y],z]+[[y,z],e]\,+\, [[z,e],y]=\omega (e,y)z\,+\,\omega (y,z)e\,+\,\omega (z,e)y.\nonumber \\ \end{aligned}$$

(1.5)

2 $\omega$-Lie Algebras Containing $L_{1}$

Notice that in this case $[x,z]=0,[y,z]=z, [x,y]=y$, and $\omega (y,z)=\omega (x,z)=0,\omega (x,y)=1.$ The identity $\circlearrowleft (e,x,y)$ reads that $[a_{1}e+b_{1}x+c_{1}y+d_{1}z,y]-(a_{2}e+b_{2}x+c_{2}y+d_{2}z)+[x,a_{2}e+b_{2}x+c_{2}y+d_{2}z]=\omega (e,x)y+e+\omega (y,e)x.$ Thus $a_{1}(a_{2}e+b_{2}x+c_{2}y+d_{2}z)+b_{1}y-d_{1}z-a_{2}e-b_{2}x-c_{2}y-d_{2}z-a_{2}(a_{1}e+b_{1}x+c_{1}y+d_{1}z)+c_{2}y= \omega (e,x)y+e-\omega (e,y)x.$ By the linear independence of the vectors $e,x,y,z$, we have

$$\begin{aligned} a_{1}a_{2}-a_{2}- a_{2}a_{1} =1&\Longrightarrow&a_{2}=-1, \end{aligned}$$

(2.1)

$$\begin{aligned} a_{1}b_{2}-b_{2}-a_{2}b_{1}=-\omega (e,y)&\Longrightarrow&-a_{1}b_{2}+b_{2}-b_{1}=\omega (e,y), \end{aligned}$$

(2.2)

$$\begin{aligned} a_{1}c_{2}+b_{1}-c_{2}-a_{2}c_{1}+c_{2}=\omega (e,x)&\Longrightarrow&a_{1}c_{2}+b_{1}+c_{1}=\omega (e,x), \end{aligned}$$

(2.3)

$$\begin{aligned} a_{1}d_{2}-d_{1}-d_{2}-a_{2}d_{1} =0&\Longrightarrow&a_{1}d_{2}=d_{2}. \end{aligned}$$

(2.4)

The identity $\circlearrowleft (e,y,z)$ implies that $[-e+b_{2}x+c_{2}y+d_{2}z,z]-(a_{3}e+b_{3}x+c_{3}y+d_{3}z)+[y,a_{3}e+b_{3}x+c_{3}y+d_{3}z]=\omega (e,y)z-\omega (e,z)y$, i.e., $-(a_{3}e+b_{3}x+c_{3}y+d_{3}z)+c_{2}z-a_{3}e-b_{3}x-c_{3}y-d_{3}z+a_{3}(e-b_{2}x-c_{2}y-d_{2}z)-b_{3}y+d_{3}z=\omega (e,y)z-\omega (e,z)y.$ Thus,

$$\begin{aligned} -a_{3}-a_{3}+a_{3} =0&\Longrightarrow&a_{3} =0, \end{aligned}$$

(2.5)

$$\begin{aligned} -b_{3}-b_{3}=0&\Longrightarrow&b_{3}=0, \end{aligned}$$

(2.6)

$$\begin{aligned} -2c_{3}=-\omega (e,z)&\Longrightarrow&2c_{3}=\omega (e,z), \end{aligned}$$

(2.7)

$$\begin{aligned} -2d_{3}+c_{2}+d_{3} =\omega (e,y)&\Longrightarrow&c_{2}-d_{3}=\omega (e,y). \end{aligned}$$

(2.8)

The identity $\circlearrowleft (e,x,z)$ means that $[a_{1}e+b_{1}x+c_{1}y+d_{1}z,z]+[x,a_{3}e+b_{3}x+c_{3}y+d_{3}z]=\omega (e,x)z-\omega (e,z)x$, i.e., $a_{1}(a_{3}e+b_{3}x+c_{3}y+d_{3}z)+c_{1}z-a_{3}(a_{1}e+b_{1}x+c_{1}y+d_{1}z)+c_{3}y=\omega (e,x)z-\omega (e,z)x.$ Thus,

$$\begin{aligned} a_{1}a_{3}-a_{3}a_{1} =0,&\end{aligned}$$

(2.9)

$$\begin{aligned} a_{1}b_{3}-a_{3}b_{1}=-\omega (e,z)&\Longrightarrow&\omega (e,z)=0, \end{aligned}$$

(2.10)

$$\begin{aligned} a_{1}c_{3}-a_{3}c_{1}+c_{3}=0&\Longrightarrow&a_{1}c_{3}=-c_{3}, \end{aligned}$$

(2.11)

$$\begin{aligned} a_{1}d_{3}+c_{1}-a_{3}d_{1} =\omega (e,x)&\Longrightarrow&a_{1}d_{3}+c_{1}=\omega (e,x). \end{aligned}$$

(2.12)

It follows from Eqs. (2.7) and (2.10) that $c_{3}=0$. Equations (2.3) and (2.12) imply that

$$\begin{aligned} b_{1}=a_{1}(d_{3}-c_{2}). \end{aligned}$$

(2.13)

Combining Eqs. (2.2) and (2.8) with (2.13), we obtain

$$\begin{aligned} (1-a_{1})(b_{2}+d_{3}-c_{2})=0. \end{aligned}$$

(2.14)

Thus $[e,x]=a_{1}e+a_{1}(d_{3}-c_{2})x+c_{1}y+d_{1}z; ~~[e,y]=-e+b_{2}x+c_{2}y+d_{2}z;~~[e,z]=d_{3}z.$ Let

$$\begin{aligned} e'=e+(d_{3}-c_{2})x-d_{3}y,~~ c_{1}'=a_{1}d_{3}+c_{1}+d_{3},~~b_{2}'=b_{2}+d_{3}-c_{2}. \end{aligned}$$

Then $[e',x]=a_{1}e'+c_{1}'y+d_{1}z, [e',y]=-e'+b_{2}'x+d_{2}z$ and $[e',z]=0$. Notice that the image of $\varphi =[-,-]$ must contain $e',y,z$, so dim(im$\varphi$) is either 3 or 4.

Case 1. Assume that dim(im$\varphi$) is 3. Our arguments will be separated into three subcases: $a_{1}=0, a_{1}=1$ and $a_{1}\ne 0,1$.

Subcase 1. If $a_{1}=0$, then Eqs. (2.4) and (2.14) imply, respectively, that $d_{2}=0$ and $b_{2}'=b_{2}+d_{3}-c_{2}=0.$ Thus $[e',x]=c_{1}'y+d_{1}z, [e',y]=-e',[e',z]=0.$ If $c_{1}',d_{1}$ are both zero, then the $\omega$-Jacobi identities imply that $\omega (e',x)=\omega (e',y)=\omega (e',z)=0$. Therefore, we obtain a new nontrivial $\omega$-Lie algebra:

(2.15)

If $c_{1}'=0,d_{1}\ne 0,$ setting $z'=d_{1}z$, we have $[e',x]=z', [e',y]=-e',[e',z']=0, [x,z']=0,[y,z']=z', [x,y]=y$ and $\omega (e',x)=\omega (e',y)=\omega (e',z')=0.$ Thus we obtain

(2.16)

If $c_{1}'\ne 0,d_{1}=0,$ then $[c_{1}'^{-1}e',x]=y, [c_{1}'^{-1}e',y]=-c_{1}'^{-1}e',[c_{1}'^{-1}e',z]=0.$ This yields

(2.17)

If $c_{1}'\ne 0,d_{1}\ne 0,$ we can first use $c_{1}'^{-1}e'$ to replace $e'$ and then use $z'$ to replace $c_{1}'^{-1}d_{1}z$. We obtain a new $\omega$-Lie algebra:

(2.18)

Subcase 2. If $a_{1}=1$, as dim(im$\varphi$) is 3, so $b_{2}'=0$, again. Let $e''=e'+d_{1}z$, then $[e'',x]=e''+ay,[e'',y]=-e''+bz,[e'',z]=0$ for some $a,b\in {\mathbb {C}}.$ As in the previous subcase, we will obtain four new $\omega$-Lie algebras. More precisely, if $a=0,b=0$ then

$$\begin{aligned} L_{1,5} :&[e,x]=e, [e,y]=-e,[e,z]=0; \omega (e,x)=\omega (e,y)=\omega (e,z)=0;\qquad \qquad \\&[x,z]=0,[y,z]=z, [x,y]=y;\omega (y,z)=\omega (x,z)=0,\omega (x,y)=1.\nonumber \end{aligned}$$

(2.19)

If $a\ne 0,b=0$, we can multiply $[e'',x]=e''+ay$ with $a^{-1}$ and then use $a^{-1}e''$ to replace $e''$. Eventually, we obtain

$$\begin{aligned} L_{1,6} :&[e,x]\!=\!e\!+\!y, [e,y]\!=\!-e,[e,z]\!=\!0; \omega (e,x)\!=\!1,\omega (e,y)\!=\!\omega (e,z)\!=\!0;\qquad \quad \\&[x,z]=0,[y,z]=z, [x,y]=y;\omega (y,z)=\omega (x,z)=0,\omega (x,y)=1.\nonumber \end{aligned}$$

(2.20)

If $a=0,b\ne 0$, replacing $bz$ by $z$, we obtain

$$\begin{aligned} L_{1,7} :&[e,x]\!=\!e, [e,y]\!=\!-e\!+\!z,[e,z]\!=\!0; \omega (e,x)=\omega (e,y)=\omega (e,z)=0;\nonumber \\&[x,z]=0,[y,z]=z, [x,y]=y;\omega (y,z)=\omega (x,z)=0,\omega (x,y)=1.\nonumber \\ \end{aligned}$$

(2.21)

If $a\ne 0,b\ne 0$, set $e=a^{-1}e''$, then $[e,x]=e+y,[e,y]=-e+a^{-1}bz,[e,z]=0$. Replacing $a^{-1}bz$ by $z$, we obtain

$$\begin{aligned} L_{1,8} :&[e,x]\!=\!e\!+\!y, [e,y]\!=\!-e\!+\!z,[e,z]\!=\!0; \omega (e,x)=1,\omega (e,y)=\omega (e,z)=0;\nonumber \\&[x,z]=0,[y,z]=z, [x,y]=y;\omega (y,z)=\omega (x,z)=0,\omega (x,y)=1.\nonumber \\ \end{aligned}$$

(2.22)

Subcase 3. Assume that $a_{1}\ne 0,1$. It follows from Eqs. (2.4) and (2.14) that $b_{2}'=0$ and $d_{2}=0.$ Let $e''=e'+a_{1}^{-1}d_{1}z$, then $[e'',x]=a_{1}e''+c_{1}'y, [e'',y]=-e''$ and $[e'',z]=0$. If $c_{1}'=0$, then there exists a new family of $\omega$-Lie algebras:

$$\begin{aligned} E_{1,\alpha } :&[e,x]=\alpha e, [e,y]=-e,[e,z]=0; \omega (e,x)=\omega (e,y)=\omega (e,z)=0;\nonumber \\&[x,z]=0,[y,z]=z, [x,y]=y;\omega (y,z)=\omega (x,z)=0,\omega (x,y)=1,\nonumber \\&(\alpha \ne 0,1). \end{aligned}$$

(2.23)

If $c_{1}'\ne 0$, then the replacing $e''$ by $c_{1}'^{-1}e$ can yield a new family of $\omega$-Lie algebras:

$$\begin{aligned} F_{1,\alpha } :&[e,x]\!=\!\alpha e\!+\!y, [e,y]\!=\!-e,[e,z]\!=\!0; \omega (e,x)\!=\!1,\omega (e,y)\!=\!\omega (e,z)\!=\!0;\nonumber \\&[x,z]\!=\!0,[y,z]\!=\!z, [x,y]\!=\!y;\omega (y,z)=\omega (x,z)\!=\!0,\omega (x,y)\!=\!1, (\alpha \!\ne \! 0,1).\nonumber \\ \end{aligned}$$

(2.24)

Case 2. Assume that dim(im$\varphi$) is 4. In this case, $b_{2}'\ne 0$. (If $b_{2}'=0$, then dim(im$\varphi$) is 3, a contradiction.) It follows from Eq. (2.14) that $a_{1}=1$. Thus $[e',x]=e'+c_{1}'y+d_{1}z, [e',y]=-e'+b_{2}'x+d_{2}z$ and $[e',z]=0$. Let $e''=e'+d_{1}z$, then $[e'',x]=e''+c_{1}'y, [e'',y]=-e''+b_{2}'x+d_{2}z$ and $[e'',z]=0$. Since $b_{2}'\ne 0$, $e''$ can be replaced by $b_{2}'^{-1}e''$. If $d_{2}=0$, then we obtain

$$\begin{aligned} G_{1,\alpha } :&[e,x]\!=\!e\!+\!\alpha y, [e,y]\!=\!-e\!+\!x,[e,z]\!=\!0; \omega (e,x)\!=\!\alpha ,\omega (e,y)\!=\!\omega (e,z)\!=\!0;\nonumber \\&[x,z]=0,[y,z]=z, [x,y]=y;\omega (y,z)=\omega (x,z)=0,\omega (x,y)=1. \nonumber \\ \end{aligned}$$

(2.25)

If $d_{2}\ne 0$, we obtain

$$\begin{aligned} H_{1,\alpha } :&[e,x]=e+\alpha y, [e,y]=-e+x+z,[e,z]=0; \omega (e,x)=\alpha ,\qquad \\&\omega (e,y)=\omega (e,z)=0;[x,z]=0,[y,z]=z, [x,y]=y;\nonumber \\&\omega (y,z)=\omega (x,z)=0,\omega (x,y)=1.\nonumber \end{aligned}$$

(2.26)

3 $\omega$-Lie Algebras Containing $L_{2}$

In this case $[x,y]=0,[x,z]=y, [y,z]=z$ and $\omega (y,z)=\omega (x,y)=0,\omega (x,z)=1.$ A similar argument as in Sect. 2 will be used throughout this section. The identity $\circlearrowleft (e,x,y)$ implies that

$$\begin{aligned} a_{1}d_{2}-d_{1}-a_{2}d_{1}= & {} 0, \end{aligned}$$

(3.1)

$$\begin{aligned} a_{1}b_{2}-a_{2}b_{1}= & {} \omega (y,e),\end{aligned}$$

(3.2)

$$\begin{aligned} a_{1}c_{2}-a_{2}c_{1} +d_{2}= & {} \omega (e,x). \end{aligned}$$

(3.3)

The identity $\circlearrowleft (e,y,z)$ implies that

$$\begin{aligned} a_{3}= & {} 0,\end{aligned}$$

(3.4)

$$\begin{aligned} a_{2}b_{3}-b_{3}= & {} 0, \end{aligned}$$

(3.5)

$$\begin{aligned} a_{2}c_{3}+b_{2}-c_{3}= & {} \omega (z,e), \end{aligned}$$

(3.6)

$$\begin{aligned} a_{2}d_{3}+c_{2}= & {} \omega (e,y). \end{aligned}$$

(3.7)

Notice that $a_{3}=0$, so $\circlearrowleft (e,x,z)$ yields

$$\begin{aligned} a_{2}= & {} -1,\end{aligned}$$

(3.8)

$$\begin{aligned} a_{1}c_{3}+b_{1}-c_{2}+d_{3}= & {} 0, \end{aligned}$$

(3.9)

$$\begin{aligned} a_{1}b_{3}-b_{2}= & {} \omega (z,e), \end{aligned}$$

(3.10)

$$\begin{aligned} a_{1}d_{3}+c_{1}-d_{2}= & {} \omega (e,x). \end{aligned}$$

(3.11)

Since $a_{2}=-1$, it follows from Eq. (3.5) that $b_{3}=0$. Equations (3.6) and (3.10) imply that $b_{2}=c_{3}$. Combining Eqs. (3.2) and (3.7) with (3.9), we have

$$\begin{aligned} b_{1}=-a_{1}b_{2},~~ d_{3}=c_{2}. \end{aligned}$$

(3.12)

Equations (3.3) and (3.11) imply that $d_{2}=0.$

Our goal equation is as follows

$$\begin{aligned}{}[e,x]=a_{1}e-a_{1}b_{2}x+c_{1}y+d_{1}z; ~~[e,y]=-e+b_{2}x+c_{2}y;~~[e,z]=b_{2}y+c_{2}z. \end{aligned}$$

Let $e'=e-b_{2}x-c_{2}y+(c_{1}+a_{1}c_{2})z.$ The direct computation shows

$$\begin{aligned}{}[e',x]=a_{1}e'+d_{1}'z; ~~[e',y]=-e';~~[e',z]=0, \end{aligned}$$

(3.13)

where $d_{1}'=d_{1}-a_{1}(c_{1}+a_{1}c_{2})$. Our arguments can be separated into the following four cases.

Case 1. If $a_{1}=d_{1}'=0$, then we obtain

$$\begin{aligned} L_{2,1} :&[e,x]=0, [e,y]=-e,[e,z]=0; \omega (e,x)=\omega (e,y)=\omega (e,z)=0;\qquad \quad \\&[x,y]=0,[x,z]=y, [y,z]=z;\omega (y,z)=\omega (x,y)=0,\omega (x,z)=1.\nonumber \end{aligned}$$

(3.14)

Case 2. If $a_{1}=0,d_{1}'\ne 0$, set $e''=d_{1}'^{-1}e'$, then we get

$$\begin{aligned} L_{2,2} :&[e,x]=z, [e,y]=-e,[e,z]=0; \omega (e,x)=\omega (e,y)=\omega (e,z)=0;\qquad \quad \\&[x,y]=0,[x,z]=y, [y,z]=z;\omega (y,z)=\omega (x,y)=0,\omega (x,z)=1.\nonumber \end{aligned}$$

(3.15)

Case 3. If $a_{1}\ne 0,d_{1}'=0$, set $x'=a_{1}^{-1}x,z'=a_{1}z$, then we obtain

$$\begin{aligned} L_{2,3} :&[e,x]=e, [e,y]=-e,[e,z]=0; \omega (e,x)=\omega (e,y)=\omega (e,z)=0;\qquad \quad \\&[x,y]=0,[x,z]=y, [y,z]=z;\omega (y,z)=\omega (x,y)=0,\omega (x,z)=1.\nonumber \end{aligned}$$

(3.16)

Case 4. If $a_{1}\ne 0,d_{1}'\ne 0$, set $x'=a_{1}^{-1}x,z'=a_{1}z, e''=a_{1}^{2}d_{1}'^{-1}e'$, then we have

$$\begin{aligned} L_{2,4} :&[e,x]\!=\!e\!+\!z, [e,y]\!=\!-e,[e,z]\!=\!0; \omega (e,x)\!=\!\omega (e,y)\!=\!\omega (e,z)=0;\qquad \quad \\&[x,y]=0,[x,z]=y, [y,z]=z;\omega (y,z)=\omega (x,y)=0,\omega (x,z)=1.\nonumber \end{aligned}$$

(3.17)

4 $\omega$-Lie Algebras Containing $A_{\alpha }$

With the relations in Theorem 1.4, we compute the identity $\circlearrowleft (e,x,y)$ and obtain

$$\begin{aligned}&\displaystyle a_{1} = 0, \nonumber \\&\displaystyle d_{1}(a_{2}+2) = 0, \end{aligned}$$

(4.1)

$$\begin{aligned}&\displaystyle -a_{2}c_{1}-c_{1}+d_{2} = \omega (e,x),\end{aligned}$$

(4.2)

$$\begin{aligned}&\displaystyle -a_{2}b_{1}+c_{2}+d_{2}-\alpha d_{1} = \omega (y,e). \end{aligned}$$

(4.3)

The identity $\circlearrowleft (e,x,z)$ implies that

$$\begin{aligned}&\displaystyle a_{2} = 0, \nonumber \\&\displaystyle b_{1}-c_{1}-c_{2}+d_{3}-a_{3}c_{1} = 0, \end{aligned}$$

(4.4)

$$\begin{aligned}&\displaystyle c_{1}-d_{1}-d_{2}-a_{3}d_{1} = \omega (e,x),\end{aligned}$$

(4.5)

$$\begin{aligned}&\displaystyle \alpha c_{1}-b_{2}-a_{3}b_{1}+c_{3}+d_{3} = \omega (z,e). \end{aligned}$$

(4.6)

The identity $\circlearrowleft (e,y,z)$ implies that

$$\begin{aligned}&\displaystyle a_{3} = 1, \nonumber \\&\displaystyle \alpha c_{2}-2b_{3}-\alpha b_{1}+\alpha d_{3} = 0, \end{aligned}$$

(4.7)

$$\begin{aligned}&\displaystyle b_{2}-c_{3}-\alpha c_{1}-c_{2} = \omega (z,e),\end{aligned}$$

(4.8)

$$\begin{aligned}&\displaystyle c_{2}-\alpha d_{1}-d_{2} = \omega (e,y). \end{aligned}$$

(4.9)

Since $a_{2}=0$, it follows from Eq. (4.1) that $d_{1}=0$. Equations (4.3) and (4.9) imply that $c_{2}=0.$ By Eqs.(4.2) and (4.5), we have $c_{1}=d_{2}$. Combining Eqs. (4.6)–(4.8) with Eq. (4.4), we have

$$\begin{aligned} b_{3}=\alpha (c_{1}-b_{1}), c_{3}=b_{2}-\alpha c_{1}+b_{1}-c_{1}, d_{3}=2c_{1}-b_{1}. \end{aligned}$$

Thus $[e,x]=b_{1}x+c_{1}y; ~~[e,y]=b_{2}x+c_{1}z;~~[e,z]=e+b_{3}x+c_{3}y+d_{3}z.$

Let $e'=e+(\alpha c_{1}-b_{2})x+(b_{1}-c_{1})y+c_{1}z$, then $[e',x]=[e',y]=0, [e',z]=e'.$ Therefore, we obtain a new family of $\omega$-Lie algebras:

$$\begin{aligned} \widetilde{A}_{\alpha } :&[e,x]=0, [e,y]=0,[e,z]=e; \omega (e,x)=\omega (e,y)=\omega (e,z)=0;\qquad \quad \\&[x,y]=x,[x,z]=x+y, [y,z]=z+\alpha x;\omega (x,y)=\omega (x,z)=0,\nonumber \\&\omega (y,z)=-1 (\alpha \in {\mathbb {C}}).\nonumber \end{aligned}$$

(4.10)

5 $\omega$-Lie Algebras Containing $B$

Recall that in the algebra $B$, $[x,y]=y, [x,z]=y+z, [y,z]=x.$ The identity $\circlearrowleft (e,x,y)$ reads that

$$\begin{aligned}&\displaystyle a_{2} = 0, \nonumber \\&\displaystyle a_{1}d_{2} = 0, \end{aligned}$$

(5.1)

$$\begin{aligned}&\displaystyle a_{1}c_{2}+b_{1}+d_{2} = \omega (e,x),\end{aligned}$$

(5.2)

$$\begin{aligned}&\displaystyle a_{1}b_{2}-d_{1}-b_{2} = \omega (y,e). \end{aligned}$$

(5.3)

From $\circlearrowleft (e,x,z)$ we have

$$\begin{aligned}&\displaystyle a_{3} = 0, \nonumber \\&\displaystyle a_{1}c_{3}+b_{1}-c_{2}+d_{3} = 0, \end{aligned}$$

(5.4)

$$\begin{aligned}&\displaystyle a_{1}d_{3}+b_{1}-d_{2} = \omega (e,x),\end{aligned}$$

(5.5)

$$\begin{aligned}&\displaystyle a_{1}b_{3}+c_{1}-b_{2}-b_{3} = \omega (z,e). \end{aligned}$$

(5.6)

From $\circlearrowleft (e,y,z)$ we have

$$\begin{aligned}&\displaystyle a_{1} = -2, \nonumber \\&\displaystyle c_{2}-b_{1}+d_{3} = 0, \end{aligned}$$

(5.7)

$$\begin{aligned}&\displaystyle b_{2}-d_{1} = \omega (e,y),\end{aligned}$$

(5.8)

$$\begin{aligned}&\displaystyle b_{2}-b_{3}-c_{1} = \omega (z,e). \end{aligned}$$

(5.9)

By $a_{1}=-2$ and Eq. (5.1), we have $d_{2}=0$. As in Sect. 4, Eqs. (5.2) and (5.5) lead to a new equation, similarly, for Eqs. (5.3) and (5.8), Eqs. (5.6) and (5.9). Combining Eq. (5.4) with Eq. (5.7), we can write out all unknowns by linear combinations in $b_{2},b_{3}$, and $c_{2}$. More precisely,

$$\begin{aligned} d_{3}=c_{3}=c_{2}, b_{1}=2c_{2},c_{1}=b_{2}+b_{3},d_{1}=-b_{2}. \end{aligned}$$

This gives our goal equation,

$$\begin{aligned}{}[e,x]=-2e+2c_{2}x+(b_{2}+b_{3})y-b_{2}z; ~~[e,y]=b_{2}x+c_{2}y;~~[e,z]=b_{3}x+c_{2}y+c_{2}z. \end{aligned}$$

(5.10)

Let $e'=e-c_{2}x-b_{3}y+b_{2}z$, then

$$\begin{aligned}{}[e',x]=-2e'; ~~[e',y]=0;~~[e',z]=0. \end{aligned}$$

(5.11)

Therefore we obtain a new $\omega$-Lie algebra:

$$\begin{aligned} \widetilde{B} :&[e,x]=-2e, [e,y]=0,[e,z]=0; \omega (e,x)=\omega (e,y)=\omega (e,z)=0;\\&[x,y]=y,[x,z]=y+z, [y,z]=x;\omega (x,y)=\omega (x,z)=0,\omega (y,z)=2.\nonumber \end{aligned}$$

(5.12)

6 $\omega$-Lie Algebras Containing $C_{\alpha }$

Recall that in $C_{\alpha }$, $[x,y]=y, [x,z]=\alpha z$ and $[y,z]=x.$ The identity $\circlearrowleft (e,x,y)$ reads that

$$\begin{aligned}&\displaystyle a_{2} = 0, \nonumber \\&\displaystyle (a_{1}-1+\alpha )d_{2} = 0, \end{aligned}$$

(6.1)

$$\begin{aligned}&\displaystyle a_{1}c_{2}+b_{1} = \omega (e,x),\end{aligned}$$

(6.2)

$$\begin{aligned}&\displaystyle a_{1}b_{2}-d_{1}-b_{2} = \omega (y,e). \end{aligned}$$

(6.3)

From $\circlearrowleft (e,x,z)$ we have

$$\begin{aligned}&\displaystyle a_{3} = 0, \nonumber \\&\displaystyle (a_{1}+1-\alpha )c_{3} = 0, \end{aligned}$$

(6.4)

$$\begin{aligned}&\displaystyle a_{1}d_{3}+\alpha b_{1} = \omega (e,x),\end{aligned}$$

(6.5)

$$\begin{aligned}&\displaystyle a_{1}b_{3}+c_{1}-\alpha b_{3} = \omega (z,e). \end{aligned}$$

(6.6)

From $\circlearrowleft (e,y,z)$ we have

$$\begin{aligned}&\displaystyle a_{1} = -1-\alpha , \nonumber \\&\displaystyle c_{2}-b_{1}+d_{3} = 0, \end{aligned}$$

(6.7)

$$\begin{aligned}&\displaystyle \alpha b_{2}-d_{1} = \omega (e,y),\end{aligned}$$

(6.8)

$$\begin{aligned}&\displaystyle -b_{3}-c_{1} = \omega (z,e). \end{aligned}$$

(6.9)

We can write out all unknowns by linear combinations in $b_{2},b_{3}$, and $c_{2}$. More precisely,

$$\begin{aligned} d_{2}=c_{3}=0, d_{3}=\alpha c_{2}, b_{1}=(1+\alpha ) c_{2},c_{1}=\alpha b_{3},d_{1}=-b_{2}. \end{aligned}$$

This gives,

$$\begin{aligned}{}[e,x]=-(1+\alpha )e+(1+\alpha )c_{2}x+\alpha b_{3}y-b_{2}z; ~~[e,y]=b_{2}x+c_{2}y;~~[e,z]=b_{3}x+\alpha c_{2}z. \end{aligned}$$

(6.10)

Let $e'=e-c_{2}x-b_{3}y+b_{2}z$, then

$$\begin{aligned}{}[e',x]=-(1+\alpha )e'; ~~[e',y]=0;~~[e',z]=0. \end{aligned}$$

(6.11)

Therefore we obtain a new family of $\omega$-Lie algebras:

$$\begin{aligned} \widetilde{C}_{\alpha } :&[e,x]=-(1+\alpha )e, [e,y]=0,[e,z]=0; \omega (e,x)=\omega (e,y)=\omega (e,z)=0;\nonumber \\&[x,y]=y, [x,z]=\alpha z, [y,z]=x;\omega (x,y)=\omega (x,z)=0,\nonumber \\&\omega (y,z)=1+\alpha , (0,-1\ne \alpha \in {\mathbb {C}}). \end{aligned}$$

(6.12)

7 Simple $\omega$-Lie Algebras

In this section, we will discuss the complex simple $\omega$-Lie algebras in dimensions 3 and 4 we just have obtained in previous sections. First of all, we have the following 3-dimensional nontrivial (i.e., non-Lie) simple $\omega$-Lie algebras.

Proposition 7.1

The algebras $A_{\alpha }(\alpha \in {\mathbb {C}}) , B,$ and $C_{\alpha } (0,-1\ne \alpha \in {\mathbb {C}})$ are all non-Lie 3-dimensional complex simple $\omega$-Lie algebras.

Proof

Here we just prove that $A_{\alpha }$ is simple. The similar arguments can be applied to $B$ and $C_{\alpha }$. Let $I$ be a nonzero ideal of $A_{\alpha }$. Take a nonzero element $\xi \in I$. We write $\xi =ax+by+cz$ for some $a,b,c\in {\mathbb {C}}$ not all zero. Since $[x,\xi ]=[x,ax+by+cz]=(b+c)x+cy\in I$, it follows that $[x,[x,\xi ]]=cx\in I$.

Case 1. Assume that $c\ne 0$. Then $x\in I$. Notice that $[x,z]=x+y\in I$, so $y\in I$. The relation $[y,z]=z+\alpha x\in I$ implies that $z$ is also in $I$. Therefore $I=A_{\alpha }$.

Case 2. Assume that $c=0$. Then $\xi =ax+by$ and $[x,\xi ]=bx\in I$. If $b\ne 0$, then an argument similar to Case 1 will yield that $I$ contains $x,y,z$, so $I=A_{\alpha }$ again. If $b=0$, then $a\ne 0$ and $\xi =ax$. Thus $x\in I$ and finally $y,z$ are also contained in $I$. This means that there are not any nonzero proper ideals in $A_{\alpha }$, i.e., $A_{\alpha }$ is simple.

In our list of Theorem 1.4, $L_{1}$ and $L_{2}$ are not simple because they both have a proper ideal generated by $y,z$. This completes the proof. $\square$

We check all 4-dimensional $\omega$-Lie algebras obtained in Sects. 2–6 case by case and conclude that

Proposition 7.2

There do not exist non-Lie 4-dimensional complex simple $\omega$-Lie algebras.

Proof

It is easy to see that $e,y,z$ can generate a proper ideal for the algebras $L_{1,1},\ldots ,L_{1,8}, L_{2,1},\ldots ,L_{2,4}$ and $E_{1,\alpha }, F_{1,\alpha }$. The algebras $G_{1,\alpha }$ and $H_{1,\alpha }$ contain a proper ideal generated by $z$. For $\widetilde{A}_{\alpha }, \widetilde{B} ,\widetilde{C}_{\alpha }$, the element $e$ can generate a proper ideal.

8 Discussion

In the fundamental paper [11] (page 1041), there exists an example of 4-dimensional $\omega$-Lie algebra, obtained via appropriate generalized derivation:

$$\begin{aligned}{}[e_{1},e_{2}]\!=\!e_{2}, [e_{1},e_{3}]\!=\!e_{3}, [e_{2},e_{3}]\!=\!e_{1}, [e_{1},e_{4}]\!=\!-e_{3} \!+\!2e_{4}, [e_{2},e_{4}]\!=\!e_{1}, [e_{3},e_{4}]\!=\!0 \end{aligned}$$

with $\omega (e_{2},e_{3})=\omega (e_{2},e_{4})=2$ and others zero. We denote this $\omega$-Lie algebra by $D$. A claim states that $D$ is a simple $\omega$-Lie algebra. It seems that this claim is not correct. Actually, direct calculation shows that the ideal generated by $e_{4}-e_{3}$ is a proper ideal of $D$. Furthermore, if one replaces $e_{4}$ by $e_{4}'=e_{4}-e_{3}$, then it is not hard to see that $D$ is actually isomorphic to $\widetilde{C}_{1}$ in our list.

In the last section of [11], Zusmanovich asks “Are there ‘interesting’ examples of infinite-dimensional $\omega$-Lie algebras?” Here we present an example of infinite-dimensional $\omega$-Lie algebra.

Example 8.1

Let $R$ be a vector space over ${\mathbb {C}}$ with a basis $\big \{x,y,z,e_{i}\mid i\in \mathbb {N}\big \}$. We define

$$\begin{aligned}&[x,y]=y, [x,z]=y+z,[y,z]=x; \omega (x,y)=\omega (x,z)=0,\omega (y,z)=2;\\&[e_{i},x]=-2e_{i}, [e_{i},y]=[e_{i},z]=0;\omega (e_{i},x)=\omega (e_{i},y)=\omega (e_{i},z)=0;\\&[e_{i},e_{j}]=0; \omega (e_{i},e_{j})=0, i,j\in \mathbb {N}. \end{aligned}$$

This is an infinite-dimensional $\omega$-Lie algebra.

We close this paper with a natural idea. One might be interested in the analogs of $\omega$-Lie algebras for other classes of algebras, for example, “$\omega$-associative algebra,” “$\omega$-Leibniz algebra,” “$\omega$-pre-Lie algebra,” etc. (see [11], page 1043, Question 4). We notice that a recent paper Burde–Fialowski [3] concentrated on a kind of new algebraic structures, Jacobi–Jordan algebras, which are the algebras obtained by replacing the antisymmetric condition in Lie algebras by the symmetric condition and keeping the Jacobi identity. The paper [3] gave a classification of Jacobi–Jordan algebras of dimension less than 7. We are wondering if the algebras satisfying the symmetric condition and the $\omega$-Jacobi identity are “correct” for $\omega$-Jacobi–Jordan algebras.