A classification of low dimensional multiplicative Hom-Lie superalgebras

Chunyue Wang; Qingcheng Zhang; Zhu Wei

doi:10.1515/math-2016-0056

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A classification of low dimensional multiplicative Hom-Lie superalgebras

Chunyue Wang , Qingcheng Zhang and Zhu Wei

Published/Copyright: September 20, 2016

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$Open Mathematics$

From the journal Open Mathematics Volume 14 Issue 1

Abstract

We study a twisted generalization of Lie superalgebras, called Hom-Lie superalgebras. It is obtained by twisting the graded Jacobi identity by an even linear map. We give a complete classification of the complex multiplicative Hom-Lie superalgebras of low dimensions.

Keywords: Hom-Lie superalgebras; Multiplicative; Classification

MSC 2010: 17B99; 55U15

1 Introduction

In recent years, Hom-Lie algebras and other Hom-algebras have been extensively studied. They first arised in quasideformation of Lie algebras of vector fields. The Hom-Lie and quasi-Hom-Lie structures are obtained from twisted derivations of discrete modifications of vector fields. For those algebras, the Jacobi condition is twisted [1]. Homalgebras were studied in [2–10]. Furthermore, Hom-Lie algebras on some q-deformations of Witt and Virasoro algebras were considered in [11–13].

Hom-Lie algebras were generalized to Hom-Lie superalgebras by Ammar and Makhlouf [14] and to Hom-Lie color algebras by Yuan [15]. In particular, Cao and Luo [16] proved that there were only the trivial Hom-Lie superalgebras on the finite-dimensional simple Lie superalgebras. The representation theory and the cohomology theory of Hom-Lie superalgebras were studied in [17]. In [18], some infinite-dimensional Hom-Lie superalgebras were constructed, induced by affinizations of the Hom-Balinskii-Novikov superalgebras and Hom-Novikov superalgebras. Furthermore, the central extensions of the infinite-dimensional Hom-Lie superalgebras were studied.

During the past few years, the theories of Hom-Lie superalgebras make great advances. But, only few people study the classification of low dimensional Hom-Lie superalgebras. There are few results about it. Zhang [19] and Li [20] studied the classifications of low dimensional Lie superalgebras and low dimensional multiplicative Hom-Lie algebras, respectively. In this paper, we will discuss the classification of low dimensional multiplicative Hom-Lie superalgebras. For the pure even Hom-Lie superalgebras, they are Hom-Lie algebras. Their classifications have been obtained. It can be found in [20], so we only discuss the odd part, which is nontrivial. We know that derived algebras of non-abelian Hom-Lie superalgebras are non-zero. We note that the properties of the derived algebras can be hold up to isomorphisms. Thus, it is crucial to determine the multiplication tables by using the dimension and properties of derived algebras. Finally, we can obtain the classifications of the low dimensional multiplicative Hom-Lie superalgebras.

The paper is organized as follows: In Section 2, we give some notations and some basic properties of Hom-Lie superalgebras. We give a method to determine whether two different multiplicative Hom-Lie superalgebras are isomorphism or not, and then we obtain a complete classification of multiplicative Hom-Lie superalgebras up to isomorphism. In Section 3 and 4, we get the classifications of the low dimensional multiplicative Hom-Lie superalgebras.

In this paper, we discuss all algebras and vector spaces over a complex field ℂ. The parity of the homogeneous element x is denoted by |x|. hg(L) denotes the set of all homogeneous elements of Hom-Lie superalgebras (L, [, ], σ). L0¯ and L1¯ denote the even part and the odd part of Hom-Lie superalgebras (L, [, ], σ), respectively.

2 Preliminaries

In this section, we recall the notions of Hom-Lie superalgebras. Then we can get some basic properties of Hom-Lie superalgebras. We aim to give a method to determine whether two different multiplicative Hom-Lie superalgebras are isomorphism or not.

Definition 2.1 ([14])

A Hom-Lie superalgebra is a triple (L, [, ], σ) consisting of a super space L, an even bilinear [, ] : L × L → Land an even linear mapσ : L → L satisfying

[x, y]=−(−1)|x||y|[y, x], (1)

(−1)|x||z|[σ(x), [y, z]]+(−1)|z||y|[σ(z), [x, y]]+(−1)|y||x|[σ(y), [z, x]]=0.(2)

for all homogeneous elements x, y, z in L.

Let (L, [, ], σ) and (L′, [, ]′, σ′) be two Hom-Lie superalgebras. An even linear map ϕ : L → L′ is said to be a homomorphism of Hom-Lie superalgebras if the following conditions are satisfied:

ϕ([x, y])=[ϕ(x), ϕ(y)]′, (3)

ϕoσ=σ′oϕ, (4)

for any x, y ∈ hg(L). If ϕ is a bijection, then ϕ is called an isomorphism of Hom-Lie superalgebras.

Definition 2.2

A Hom-Lie superalgebra (L, [, ], σ) is called a multiplicative Hom-Lie superalgebra if the even linear map σ satisfiesσ[x, y] = [σ(x, σ(y)] for any x, y ∈ hg(L). A Hom-Lie superalgebra (L, [, ], σ) is called a classical Hom-Lie superalgebra if (L, [, ], σ) is a Lie superalgebra.

Remark 2.3

In the following paper, Hom-Lie superalgebras always denote to be the multiplicative Hom-Lie superalgebras.

Proposition 2.4

Let (L, [, ], σ) be a Hom-Lie superalgebra and σ be an even invertible linear map. Define [, ]₁ : L × L → L by [x, y]₁ = [σ^-1(x), σ^-1(y)], for any x, y ∈ hg(L). Then (L, [, ]₁) is a Lie superalgebra and σ is a Lie superalgebra endomorphism.

Proof

It is clear that [, ]₁ is an even bilinear map and is super skew-symmetric. Since

σ−1[x, y]=σ−1(σ[σ−1(x), σ−1(y)])=[σ−1(x), σ−1(y)]=[x, y]1,

we have, for any x, y, z ∈ hg(L),

(−1)|x||z|[[x, y]1, z]1+(−1)|x||y|[[y, z]1, x]1+(−1)|y||z|[[z, x]1, y]1(−1)|x||z|[σ−1[x, y], z]1+(−1)|x||y|[σ−1[y, z], x]1+(−1)|y||z|[σ−1[z, x], y]1+(−1)|x||z|σ−1[σ−1[x, y], z]+(−1)|x||y|σ−1[σ−1[y, z], x]+(−1)|y||z|σ−1[σ−1[z, x], y]=σ−2((−1)|x||z|[[x, y], σ(z)]+(−1)|x||y|[[y, z], σ(x)]+(−1)|y||z|[[z, x], σ(y)])= 0.

Obviously, σ is an even endomorphism of the Lie superalgebra (L, [, ]₁).

Let L be a linear super space and [, ] : L × L → L be an even bilinear map. Mor(L, [, ]) denotes the set of all even linear maps σ such that (L, [, ], σ) is a multiplicative Hom-Lie superalgebra.

Lemma 2.5

Let L be a linear super space, [, ]₁and [, ]₂be two even bilinear maps on L. If there exists an even invertible linear map ϕ of L satisfying

ϕ[x, y]1=[ϕ(x), ϕ(y)]2,

then

Mor(L, [, ]2)={ϕσϕ−1|σ∈Mor(L, [, ]1)}.

Proof

Since ϕ is invertible, we have, for any σ₁∈Mor(L, [, ]₁), x, y, z ∈ hg(L)

ϕσ1ϕ−1[x, y]2=ϕσ1[ϕ−1(x), ϕ−1(y)]1=[ϕσ1ϕ−1(x), ϕσ1ϕ−1(y)]2,

and

(−1)|x||z|[[x,y]2,ϕσ1ϕ−1(z)]2+(−1)|x||y|[[y,z]2,ϕσ1ϕ−1(x)]2+(−1)|y||z|[[z,x]2,ϕσ1ϕ−1(y)]2=(−1)|x||z|[ϕ[ϕ−1(x),ϕ−1(y)]1,ϕσ1ϕ−1(z)]2+(−1)|x||y|[ϕ[ϕ−1(y),ϕ−1(z)]1,ϕσ1ϕ−1(x)]2+(−1)|y||z|[ϕ[ϕ−1(z),ϕ−1(x)]1,ϕσ1ϕ−1(y)]2=(−1)|x||z|ϕ[[ϕ−1(x),ϕ−1(y)]1,σ1ϕ−1(z)]1+(−1)|x||y|ϕ[[ϕ−1(y),ϕ−1(z)]1,σ1ϕ−1(x)]1+(−1)|y||z|ϕ[[ϕ−1(z),ϕ−1(x)]1,σ1ϕ−1(y)]1= 0.

So ϕσ1ϕ−1∈Mor(L, [, ]2).

Conversely, for any σ₂ ∈ Mor(L, [, ]₂), similar to above, we have ϕ^-1σ₂ϕ ∈ Mor(L, [, ]₁). Set ϕ^-1σ₂ϕ = σ₁. Then σ₂ = ϕσ₁ϕ^-1. Thus, we have Mor(L, [, ]2)={ϕσϕ−1|σ∈Mor(L, [, ]1)}.

Remark 2.6

By Lemma 2.5, we know that there exists a one-to-one correspondence between the multiplicative Hom-Lie superalgebras (L, [, ]₁, σ₁) and the multiplicative Hom-Lie superalgebras (L, [, ]₂, σ₂). In order to simplify the classification of multiplicative Hom-Lie superalgebras, we can simplify [, ]₁to [, ]₂by an even invertible linear map ϕ. Then we determine the multiplicative Hom-Lie superalgebra structure by means of [, ]₂.

Remark 2.7

Let (L, [, ]₁, σ₁) and (L, [, ]₂, σ₂) be two Hom-Lie superalgebras with the same underlying vector super space L. (L, [, ]₁, σ₁) is isomorphic to (L, [, ]₂, σ₂) if and only if there exists an invertible matrix T such that T[e_i, e_j]₁ = [T(e_i), T(e_j)]₂andA₂ = TA₁T^-1, where A₁, A₂andTdenote matrices corresponding to σ₁, σ₂and the homomorphism ϕ : L → Lwith respect to the basis {e₁, e₂, ..., e_n}, respectively.

We denote the derived algebra of Hom-Lie superalgebras (L, [, ], σ) by L′ = [L, L]. It consists of all linear combinations of commutators [x, y] for x, y ∈ hg(L). The Hom-Lie superalgebra (L, [, ], σ) is called an abelian Hom-Lie superalgebra if [L, L] = 0.

Lemma 2.8

Let (L₁, [, ]1, σ₁) and (L₂, [, ]2, σ₂) be two abelian Hom-Lie superalgebras. If (L₁, [, ]1, σ₁) is isomorphic to (L₂, [, ]2, σ₂), thendimℂL10¯=dimℂL20¯anddimℂL11¯=dimℂL21¯.

Proof

An isomorphism from (L₁, [, ]1, σ₁) to (L₂, [, ]2, σ₂) is an isomorphism of their underlying vector super space. So there exists an even invertible linear map ϕ : L10¯→L20¯→L11¯→L21¯.

Lemma 2.9

Let (L, [, ]1, σ₁) and (L, [, ]2, σ₂) be two abelian Hom-Lie superalgebras with the same underlying vector super space L. (L, [, ]1, σ₁) is isomorphic to (L, [, ]2, σ₂) if and only if the matrices corresponding to σ₁ and σ₂ have the same elementary divisors.

Proof

It is straightforward.

Lemma 2.10

Let (L₁, [, ]1, σ₁) and (L₂, [, ]2, σ₂) be two Hom-Lie superalgebras. Ifϕ:L1→L2is a surjective homomorphism of Hom-Lie superalgebras, thenϕ(L′1)=L′2.

Proof

For any x, y ∈ hg(L₁), we have ϕ[x, y]1=[ϕ(x), ϕ(y)]2. So by linearity ϕ(L′1)⊆L′2. Now L′₂ is spanned by commutators [y₁, y₂] with y₁, y₂ ∈ hg(L₂). Set ϕ(x₁) = y₁, ϕ(x₂) = y₂ It follows that ϕ[x₁, x₂]₁ = ϕ[y₁, y₂]₂.

3 A classification of 2-dimensional Hom-Lie superalgebras

In this section, we are going to give the classification of 2-dimensional Hom-Lie superalgebras.

Theorem 3.1

Every 2-dimensional Hom-Lie superalgebra is isomorphic to one of the following nonisomorphic Hom-Lie superalgebra: each algebra is denoted byLi, jk, where i is the dimension ofL0¯, j is is the dimension ofL1¯, k is the number.

(1) L0, 21is an abelian Hom-Lie superalgebra.

(2) L1, 12is an abelian Hom-Lie superalgebra.

(3) L1, 13, b:[e0, e1]=e1, [e1, e1]=0, σ=(100b).L1, 13, bis isomorphic toL1, 13, b′if and only if b = b′.

(4) L1, 14, a:[e0, e1]=e1, [e1, e1]=0, σ=(a000)(a≠0, 1). L1, 14, ais isomorphic toL1, 14, a′if and only if a = a′.

(5) L1, 15, b:[e0, e1]=0, [e1, e1]=e0, σ=(b2000)(b≠0). L1, 15, bis isomorphic toL1, 15, b′if and only if b = b′.

In order to prove the above theorem, we give some lemmas.

Let (L, [, ]1, σ) be a 2-dimensional Hom-Lie superalgebra and {e₀, e₁} be a basis of (L, [, ], σ).

If e₀, e₁ ∈ L1¯ and L0¯ = 0, then (L, [, ], σ) is an abelian Hom-Lie superalgebra.

If e₀ ∈ L0¯, e₁ ∈ L1¯ and [L, L]=0, then (L, [, ], σ) is also an abelian Hom-Lie superalgebra.

In the following, we will discuss the situation: e0∈L0¯, e1∈L1¯ and [L, L] ≠ 0. Obviously, (L, [, ], σ) is a non-abelian Hom-Lie superalgebra.

Let [e0, e1]=αe1, [e1, e1]=βe0, α2+β2≠0.

If α ≠ 0, β = 0, by Lemma 2.5, then there exists an even invertible linear mapϕ=(α001α) such that the bracket can be simplified as

[e0, e1]=e1, [e1, e1]=0.(5)

Similarly, if α = 0, β ≠ 0, the bracket can be simplified as

[e0, e1]=0, [e1, e1]=e0.(6)

If α ≠ 0, β ≠ 0, the bracket can be simplified as

[e0, e1]=e1, [e1, e1]=e0.(7)

However, there does not exist an even invertible linear map such that formulas (5), (6) and (7) are equivalent.

Lemma 3.2

For the bracket (5), Hom-Lie superalgebras are (L, [, ], σ_i)(i = 1, 2) up to isomorphism, whereσ1=(100b), σ2(a000)(a≠0, 1).

Proof

Let σ : L → L be an even linear map of Hom-Lie superalgebras and σ(e₀) = ae₀, σ(e₁) = be₁. The Hom-Jacobi identity is satisfied for any a, b ∈ ℂ. Since σ is multiplicative, we have a = 1 or b = 0.

Lemma 3.3

For the bracket (6), the Hom-Lie superalgebra is (L, [, ], σ) up to isomorphism, whereσ=(b200b)(b≠0).

Proof

It is similar to Lemma 3.2.

For the bracket (7), there is not a non-zero even linear map σ such that it is multiplicative and satisfies Hom-Jacobi identity.

Proof of Theorem 3.1

From Lemmas 3.2 and 3.3, we obtain Theorem 3.1.

Remark 3.4

(1) All 2-dimensional Hom-Lie superalgebras are classical Lie superalgebras.

(2) From Lemma 2.8, Hom-Lie superalgebrasL0, 21andL1, 12are not isomorphism.

4 A classification of 3-dimensional Hom-Lie superalgebras

By Lemma 2.10, we note that it can keep the structures of the derived algebras with the isomorphism. Thus, we can use the properties of the derived algebras to determine the structures of 3-dimensional Hom-Lie superalgebras.

Theorem 4.1

Every 3-dimensional Hom-Lie superalgebra is isomorphic to one of the following non-isomorphic Hom-Lie superalgebra: each algebra is denoted byLi, jk, where i is the dimension ofL0¯, j is is the dimension ofL1¯, k is the number.

(1) L0, 31is an abelian Hom-Lie superalgebra.

(2) L2, 12is an abelian Hom-Lie superalgebra.

(3) L2, 13, a, b, c:[e1, e2]=0, [e1, e3]=0, [e2, e3]=0, [e3, e3]=e1, σ=(c2a00b000c)(a2+b2+c2≠0). L2, 13, a, b, cis isomorphic toL2, 13, a′, b′, c′if and only ifb = b′, c = c′ anda = x(c² - b) + ya′(y ≠ 0).

(4) L2, 14, a, b, c:[e1, e2]=e1, [e1, e3]=0, [e2, e3]=0, [e3, e3]=0, σ=(0a00b000c)(a2+b2+c2≠0). L2, 14, a, b, cis isomorphic toL2, 14, a′, b′, c′if and only ifb = b′, c = c′ and a′ = xa + yb(x ≠ 0).

(5) L2, 15, a, b, c:[e1, e2]=e1, [e1, e3]=0, [e2, e3]=0, [e3, e3]=0, σ=(a1a2001000b)(a1≠0). L2, 15, a1, a2, bis isomorphic toL2, 15, a′1, a′2, b′if and only ifb = b′, a₁ = a′₁anda₂x + y = ya′₁ + a₂(x ≠ 0).

(6) L2, 16:[e1, e2]=e1, [e1, e3]=0, [e2, e3]=0, [e3, e3]=e1, σ=(010000000)

(7) L2, 17, a, b, c:[e1, e2]=0[e1, e3]=0, [e2, e3]=e3, [e3, e3]=0, σ=(ab00c0000) (a2+b2+c2≠0). L2, 17, a, b, cis isomorphic to L2, 17, a′, b′, c′if and only if a = a^′, c = c^′andb^′ = xb + yc – yc^′ (x ≠ 0).

(8) L2, 18, a, b, c:[e1, e2]=0, [e1, e3]=0, [e2, e3]=e3, [e3, e3]=0, σ=(ab001000c) (c≠0). L2, 18, a, b, cisomorphic to L2, 18, a′, b′, c′if and only if a = a^′, c = c^′andb^′ = xb - ya^′ + yc(x ≠ 0).

(9) L2, 19, a:[e1, e2]=e1, [e1, e3]=0, [e2, e3]=0, [e3, e3]=e2, σ=(0000a2000a)(a≠0). L2, 19, a is isomorphic to L2, 19, a′if and only if a = a^′.

(10) L2, 110, a:[e1, e2]=e1, [e1, e3]=e3, [e2, e3]=0, [e3, e3]=0, σ=(00001000a) .

(11) L2, 111, a:[e1, e2]=e1, [e1, e3]=e3, [e2, e3]=0, [e3, e3]=e2, σ=(0000a000a) (a≠1). L2, 111, ais isomorphic toL2, 111, a′if and only if a^′.

(12) L2, 112, a:[e1, e2]=e1, [e1, e3]=e3, [e2, e3]=0, [e3, e3]=0, σ=(a0001000a) (a≠1). L2, 112, ais isomorphic toL2, 112, a′if and only if a = a^′.

(13) L2, 113, a:[e1, e2]=0, [e1, e3]=0, [e2, e3]=e3, [e3, e3]=e1, σ=(0a0010000) . L2, 113, ais isomorphic toL2, 113, a′if and only if a^′x²a + y(x ≠ 0).

(14) L2, 114, a1, a2:[e1, e2]=0, [e1, e3]=0, [e2, e3]=e3, [e3, e3]=e1, σ=(0a100a20000) (a≠1). L2, 114, a1, a2is isomorphic toL2, 114, a′1, a′2if and only ifa2=a′2, a′1=x2a1+ya2(x≠0).

(15) L2, 115, a, a:[e1, e2]=0, [e1, e3]=0, [e2, e3]=e3, [e3, e3]=e1, σ=(a2b0010000) (a≠1). L2, 115, a, bis isomorphic toL2, 115, a′, b′if and only ifa=a′, b′=x2b−ya2+y(x≠0).

(16) L2, 116, a:[e1, e2]=αe1, [e1, e3]=0, [e2, e3]=λe3, [e3, e3]=0, σ=(00001000a) (a≠0). L2, 116, ais isomorphic toL2, 116, a′if and only ifa = a^′.

(17) L2, 117:[e1, e2]=αe1, [e1, e3]=0, [e2, e3]=λe3, [e3, e3]=0, σ=(00001000a).

(18) L2, 118, a, b:[e1, e2]=αe1, [e1, e3]=0, [e2, e3]=λe3, [e3, e3]=0, σ=(0a00b0000) (b≠1). L2, 118, a, bis isomorphic toL2, 118, a′, b′if and only ifb=b′, a′=xa+yb(x≠0).

(19) L2, 119, a, b:[e1, e2]=αe1, [e1, e3]=0, [e2, e3]=λe3, [e3, e3]=0, σ=(ab0010000) (a≠0). L2, 119, a, bis isomorphic toL2, 119, a′, b′if and only ifa=a′, b′=xb−ya′+y(x≠0).

(20) L2, 120, a, b, c:[e1, e2]=αe1, [e1, e3]=0, [e2, e3]=λe3, [e3, e3]=0, σ5=(ab0010000) (a≠0, c≠0). L2, 120, a, b, cis isomorphic toL2, 120, a′, b′, c′if and only ifa=a′, c=c′, b′=xb−ya′+y(x≠0).

(21) L2, 121, a:[e1, e2]=0, [e1, e3]=e3, [e2, e3]=0, [e3, e3]=e1, σ=(0000a0000) (a≠0). L2, 121, ais isomorphic toL2, 121, a′if and only if a = a^′.

(22) L2, 122, a:[e1, e2]=0, [e1, e3]=e3, [e2, e3]=0, [e3, e3]=e1, σ=(1000a000±1). L2, 122, ais isomorphic toL2, 122, a′if and only if a = a^′.

(23) L2, 123:[e1, e2]=αe1, [e1, e3]=0, [e2, e3]=λe3, [e3, e3]=μe1, σ=(010000000)

(24) L1, 224:(L, [, ], σ)is an abelian Hom-Lie superalgebra.

(25) L2, 125, a, b:[e2, e2]=e1, [e2, e3]=0, [e3, e3]=0, σ=(0000000ab)(a2+b2≠0).L1, 225, a, bis isomorphic toL1, 225, a′, b′if and only ifb=b′, a=xa′+yb′(x≠0).

(26) L2, 126, a, b, c:[e2, e2]=e1, [e2, e3]=0, [e3, e3]=0, σ=(a2000a00bc)(a≠0). L1, 226, a, b, cis isomorphic toL1, 226, a′, b′, c′if and only ifa=a′, b′=xa+yb−xc′(x≠0).

(27) L2, 127, a:[e2, e2]=0, [e2, e3]=e1, [e3, e3]=0, σ=(00000a000)(a≠0). L1, 227, ais isomorphic toL1, 227, a′if and only ifa′=xa(x≠0).

(28) L2, 128, a:[e2, e2]=0, [e2, e3]=e1, [e3, e3]=0, σ=(00000000a)(a≠0). L1, 228, ais isomorphic toL1, 228, a′if and only if a^′ = a.

(29) L2, 129, a, b:[e2, e2]=0, [e2, e3]=e1, [e3, e3]=0, σ=(0000000ab)(a≠0, b≠0). L1, 229, a, bis isomorphic toL1, 229, a′, b′if and only ifb′=b, a′=xa(x≠0).

(30) L2, 130, a, b:[e2, e2]=0, [e2, e3]=e1, [e3, e3]=0, σ=(ab0000b0a0)(a≠0, b≠0). L1, 230, a, bis isomorphic toL1, 230, a′, b′if and only ifa=xa, b′=1xb(x≠0).

(31) L2, 131, a, b:[e2, e2]=0, [e2, e3]=e1, [e3, e3]=0, σ=(ab0000b0a0)(a≠0, b≠0). L1, 231, a, bis isomorphic toL1, 231, a′, b′if and only if a^′ = a, b = b^′.

(32) L1, 232, a:[e1, e2]=e2, [e1, e3]=0, σ=(10000000a).L1, 232, ais isomorphic toL1, 232, a′if and only if a′ = a.

(33) L1, 233, a:[e1, e2]=e2, [e1, e3]=0, σ=(00000000a).(a≠0).L1, 233, ais isomorphic toL1, 233, a′if and only if a′ = a.

(34) L1, 234, a, b:[e1, e2]=e2, [e1, e3]=0, σ=(00000a00b).(a≠0).L1, 234, a, bis isomorphic toL1, 234, a′, b′if and only if b′ = b, a′ = |a.

(35) L1, 235, a, b:[e1, e2]=e2, [e1, e3]=0, σ=(a0000000b).(a≠0, 1).L1, 235, a, bis isomorphic toL1, 235, a′, b′if and only if b′ = b, a′ = |a.

(36) L1, 236, a, b:[e1, e2]=e2, [e1, e3]=0, σ=(1000a000b)(a≠0).L1, 236, a, bis isomorphic toL1, 236, a′, b′if and only if b′ = b, a′ = a.

(37) L1, 237, a, b:[e1, e2]=0, [e1, e3]=e2, σ=(00000a00b)(a2+b2≠|0).L1, 237, a, bis isomorphic toL1, 237, a′, b′if and only if b′ = b, a′ = xa + yb(x ≠ 0).

(38) L1, 238, a, b, c:[e1, e2]=0, [e1, e3]=e2, σ=(a000acb00c)(a≠0, c≠|0).L1, 238, a, b, cis isomorphic toL1, 238, a′, b′, c′if and only if a′ = a, c = c′, b′ = xb + yc - ya′c′(x≠0).

(39) L1, 239, a1, a2, b1, b2:[e1, e2]=αe2, [e1, e3]=μe3, σ=(1000a1b10a2b2).L1, 239, a1, a2, b1, b2is isomorphic toL1, 239, a′1, a′2, b′1, b′2if and only ifa′1=a1, b2=b′2, b′1=xb1, a′2=|1xa2(x≠0).

(40) L1, 240, a, b:[e1, e2]=αe2, [e1, e3]=μe3, σ=(1000a000b).L1, 240, a, bis isomorphic toL1, 240, a′, b′if and only if a = a′, b = b′.

(41) L1, 241, a:[e1, e2]=αe2, [e1, e3]=μe3, σ=(a00000000)(a≠0, 1).L1, 241, ais isomorphic toL1, 241, a′if and only if a = a′.

(42) L1, 242, a:[e1, e2]=αe2, [e1, e3]=μe3, σ=(αμ000000a0)(a≠0).L1, 242, ais isomorphic toL1, 242, a′if and only if a′ = xa(x ≠ 0).

(43) L1, 243, a:[e1, e2]=0, [e1, e3]=βe2, [e2, e2]=0, [e2, e3]=0, [e3, e3]=γe1, σ=(00000a0a0)(a≠0).L1, 243, ais isomorphic toL1, 243, a′if and only if a′ = a.

(44) L1, 244, a, b:[e1, e2]=0, [e1, e3]=βe2, [e2, e2]=0, [e2, e3]=0, [e3, e3]=γe1, σ=(a2000a3b00a)(a≠0).L1, 244, a, bis isomorphic toL1, 244, a′, b′if and only if a′ = a, b′ = xb + ya − ya′³(x ≠ 0).

L1, 245, a: [e1, e2] = 0, [e1, e3] = βe2, [e2, e2] = 0, [e2, e3] = μe1, [e3, e3] = γe1, σ = (00000a000) (a ≠ 0). L1, 245, ais isomorphic toL1, 245, a′if and only ifa^′ = a.

(46) L1, 246, a, b: [e1, e2] = 0, [e1, e3] = βe2, [e2, e2] = 0, [e2, e3] = μe1, [e3, e3] = 0, σ = (00000a00b) (a2 = b2 ≠ 0). L1, 246, a, bis isomorphic toL1, 246, a′, b′if and only ifb′ = b, a′ = xa(x ≠ 0).

(47) L1, 247, a: [e1, e2] = 0, [e1, e3] = βe2, [e2, e2] = 0, [e2, e3] = μe1, [e3, e3] = 0, σ = (a000±a000±1) (a ≠ 0). L1, 247, ais isomorphic toL1, 247, a′if and only ifa = a′.

(48) L1, 248, a: [e1, e2] = αe2, [e1, e3] = 0, [e2, e2] = λe1, [e2, e3] = 0, [e3, e3] = 0, σ = (00000000a) (a ≠ 0). L1, 248, ais isomorphic toL1, 248, a′if and only ifa = a′.

(49) L1, 249, a: [e1, e2] = αe2, [e1, e3] = 0, [e2, e2] = 0, [e2, e3] = 0, [e3, e3] = γe1, σ = (a20000000a) (a ≠ 0). L1, 249, ais isomorphic toL1, 249, a′if and only ifa = a′.

(50) L1, 250, a: [e1, e2] = αe2, [e1, e3] = βe2, [e2, e2] = 0, [e2, e3] = 0, [e3, e3] = γe1, σ = (a20000−βαa00a) (a ≠ 0). L1, 250, ais isomorphic toL1, 250, a′if and only ifa = a′.

In order to prove the above theorem, we give some lemmas.

Let (L, [, ], σ) be a 3-dimensional Hom-Lie superalgebra and {e₁e₂e₃} be a basis of (L, [, ], σ).

If e1, e2, e3 ∈ L1¯ and L0¯ = 0, then (L, [, ], σ) is an abelian Hom-Lie superalgebra.

Now we consider the 3-dimensional Hom-Lie superalgebra (L, [, ], σ), where dim L0¯ = 2 and dim L1¯ = 1. Set e1, e2 ∈ L0¯, e3 ∈ L1¯.

(1) If [L, L] = 0, then (L, [, ], σ) is an abelian Hom-Lie superalgebra.

(2) If dim [L, L] = 1, we consider two cases as follows.

Case I: [L, L] = ℂe1. Let [e1, e2] = αe1, [e1, e3] = 0, [e2, e3] = 0, [e3, e3] = βe1, where α² + β² ≠ 0.

If α = 0, β ≠ 0, by Lemma 2.5, there exists an even invertible linear map ϕ = (1β10010001) such that the bracket can be simplified as

[e1, e2]=0, [e1, e3]=0, [e2, e3]=0, [e3, e3]=e1.(8)

Similarly, if α ≠ 0, β = 0, the bracket can be simplified as

[e1, e2]=e1, [e1, e3]=0, [e2, e3]=0, [e3, e3]=0.(9)

If α ≠ 0, β = 0, the bracket can be simplified as

[e1, e2]=e1, [e1, e3]=0, [e2, e3]=0, [e3, e3]=e1.(10)

However, there is not an even invertible linear map such that formulas (8), (9) and (10) are equivalent.

Lemma 4.2

For the bracket (8), the Hom-Lie superalgebra is (L, [, ], σ) up to isomorphism, where σ = (c2a00b000c) (a2 + b2 c2≠ 0).

Proof

Let σ : L → L be an even linear map of the Hom-Lie superalgebras and σ(e₁) = a₁e₁ + a₂e₂, σ(e₂) = b₁e₁ + b₂e₂, σ(e₃) = ce₃. For any a1, a2, b1, b2, c ∈ ℂ Hom-Jacobi identity is satisfied. Since σ is multiplicative, we have a₁ = c₂ and a₂ = 0. Set b₁ = a, b₂ = b.

Lemma 4.3

For the bracket (9), the Hom-Lie superalgebra is (L, [, ], σ_i)(i = 1, 2) up to isomorphism, where σ1 = (0a00b000c) (a2 + b2 c2≠ 0), σ2 = (a1a2001000b) (a1≠ 0).

Proof

It is similar to Lemma 4.2.

Lemma 4.4

For the bracket (10), the Hom-Lie superalgebra is (L, [, ], σ_i up to isomorphism, where σ = (010000000).

Proof

Let σ : L → L be an even linear map of the Hom-Lie superalgebras and σ(e₁) = a₁e₁ + a₂e₂, σ(e₂) = b₁e₁ + b₂e₂, σ(e₃) = ce₃. Since σ is multiplicative, we have a₁ = c² = a₁b₂ - a₂b₁ and a₂ = 0. By Hom-Jacobi identity, we obtain b₂ = 0, then c = 0. Set σ′ = (0b10000000) (b1≠ 0). There exists an even invertible linear map ϕ = (1b11001000±1b1) satisfying

[ϕ(e1), ϕ(e2)]=ϕ(e1), [ϕ(e1), ϕ(e3)]=0, [ϕ(e2), ϕ(e3)]=0, [ϕ(e3), ϕ(e3)]=ϕ(e1)

such that ϕσ′ϕ−1 = σ = (010000000).

Case II: [L, L] = ℂe3 Let [e1, e2] = 0, [e1, e3] = αe3, [e2, e3] = βe3, [e3, e3] = 0, where α² + β² ≠ 0.

If α = 0, β ≠ 0, by Lemma 2.5, there exists an even invertible linear map ϕ = (1100β0001) such that the bracket can be simplified as

[e1, e2]=0, [e1, e3]=0, [e2, e3]=e3, [e3, e3]=0.(11)

Lemma 4.5

For the bracket (11), the Hom-Lie superalgebras are (L, [, ], σ_i)(i = 1, 2) up to isomorphism, where σ1 = (ab00c0000) (a2 + b2 c2≠ 0), σ2 = (ab001000c) (c≠ 0).

Proof

It is similar to Lemma 4.4.

If α ≠ 0, β = 0 or α ≠ 0, β ≠, 0 by Lemma 2.5, we know that Hom-Lie superalgebras are isomorphic to those Hom-Lie superalgebras in Lemma 4.5.

Remark 4.6

All Hom-Lie superalgebras in Theorem 4.2-4.5 are classical Lie superalgebras.

(3) If dim [L, L] = 2, we will consider two cases as follows.

Case I: [L, L]=ℂe1+ℂe2=L0¯. Let [e₁, e₂] = α₁e₁+α₂e₂, [e₁, e₃] = 0, [e₂, e₃] = 0, [e₃, e₃] = β₁e₁+α₂e₂.

The matrix A=(α1β1α2β2) is non-degenerate, otherwise, it is contradictory to dim [L, L] = 2.

If β₁ = α₂ = 0, we have α₁ ≠ 0, β₂ ≠ 0. By Lemma 2.5, there exists an even invertible linear map ϕ=(1000α1000±α1β2) such that the bracket can be simplified as

[e1, e2]=e1, [e1, e3]=0, [e2, e3]=0, [e3, e3]=e2.(12)

Other cases can be simplified as (12) by Lemma 2.5.

Lemma 4.7

For the bracket (12), the Hom-Lie superalgebra is (L, [, ], σ) up to isomorphism, where σ = (1000a2000a)(a≠0).

Proof

It is similar to Lemma 4.4.

Case II: [L, L] = ℂe₁ + ℂe₃. Let [e₁, e₂] = αe₁, [e₁, e₃] = βe₃, [e₂, e₃] = λe₃, [e₃, e₃] = μe₁. We will consider situations as follows.

If α ≠ 0, β ≠ 0, λ = 0, μ = 0, by Lemma 2.5, there exists an even invertible linear map ϕ=(β000α0001) such that the bracket can be simplified as

[e1, e2]=e1, [e1, e3]=e3, [e2, e3]=0, [e3, e3]=0.(13)

Moreover, for α ≠ 0, β ≠ 0, λ ≠ 0, μ = 0, the bracket can be simplified as (13).

Lemma 4.8

For the bracket (13), Hom-Lie superalgebras are (L, [, ], σ_i) (1 ≥ i ≥ 3) up to isomorphism, whereσ1=(000010000), σ2=(0000a0000)(a≠1), σ3=(a00010000)(a≠1).

Proof

It is similar to Lemma 4.4.

If α = 0, β = 0, λ ≠ 0, μ ≠ 0, then the bracket can be simplified as

[e1, e2]=0, [e1, e3]=0, [e2, e3]=e3, [e3, e3]=e1.(14)

Lemma 4.9

For the bracket (14), Hom-Lie superalgebras are (L, [, ], σ_i) (1 ≥ i ≥ 3) up to isomorphism, whereσ1=(0a0010000), σ2=(0a100a20000)(a2≠1), σ3=(a2b001000a)(a≠0).

Proof

It is similar to Lemma 4.4.

If α ≠ 0, β = 0, λ ≠ 0, μ = 0, then

[e1, e2]=αe1, [e1, e3]=0, [e2, e3]=λe3, [e3, e3]=0.(15)

Lemma 4.10

For the bracket (15), Hom-Lie superalgebras are (L, [, ], σ_i) (1 ≥ i ≥ 5) up to isomorphism, whereσ1=(000010000)(a≠0), σ2=(000010000), σ3=(0a00b0000)(b≠1), σ4=(ab0010000)(a≠0), σ5=(ab001000c)(a≠0, c≠0).

Proof

It is similar to Lemma 4.4.

If α = 0, β ≠ 0, λ = 0, μ ≠ 0, then the bracket can be simplified as

[e1, e2]=0, [e1, e3]=e3, [e2, e3]=0, [e3, e3]=e1.(16)

Moreover, for α = 0, β ≠ 0, λ ≠ 0, μ ≠ 0, the bracket can be simplified as (16).

Lemma 4.11

For the bracket (16), Hom-Lie superalgebras are (L, [, ], σ_i) (1 ≥ i ≥ 2) up to isomorphism, whereσ1=(0000a0000)(a≠0), σ2=(1000a000±1).

Proof

It is similar to Lemma 4.4.

If α ≠ 0, β ≠ 0, λ = 0, μ ≠ 0, then the bracket can be simplified as

[e1, e2]=e1, [e1, e3]=e3, [e2, e3]=0, [e3, e3]=e1.(17)

Moreover, for α ≠ 0, β ≠ 0, λ ≠ 0, μ ≠ 0, the bracket can be simplified as (17). However, for formula (17), there are not non-zero even linear maps σ satisfying σ[e_i, e_j] = [σ(e_i), σ(e_j)] such that (L, [, ], σ) is a Hom-Lie superalgebra.

If α ≠ 0, β = 0, λ ≠ 0, μ ≠ 0, then

[e1, e2]=αe1, [e1, e3]=0, [e2, e3]=λe3, [e3, e3]=μe1.(18)

Lemma 4.12

For the bracket (17), the Hom-Lie superalgebra is (L, [, ], σ) up to isomorphism, where σ = (010000000)

Proof

It is similar to Lemma 4.4.

(4) If dim [L, L] = 3, then [L, L] = L. Let [e₁, e₂] = α₁e₁ + α₂e₂, [e₁, e₃] = μe₃, [e₃, e₃] = β₁e₁ + β₂e₂. From dim [L, L] = 3, it follows that λ² + μ² ≠ 0 and the matrix A=(α1β1α2β2) is non-degenerate. So we only consider three cases as follows, the other cases are equivalent to those.

If μ ≠ 0, α₁ ≠ 0, β₂ ≠ 0, λ = β₁ = α₂ = 0, then [e₁, e₂] = α₁e₁, [e₁, e₃] = 0, [e₂, e₃] = μe₃, [e₃, e₃] = β₂e₂.

If λ ≠ 0, α₁ ≠ 0, β₂ ≠ 0, μ = β₁ = α₂ = 0, then [e₁, e₂] = α₁e₁, [e₁, e₃] = λe₃, [e₂, e₃] = 0, [e₃, e₃] = β₂e₂.

If λ ≠ 0, α₁ ≠ 0, β₂ ≠ 0, μ ≠ 0, β₁ = α₂ = 0, then [e₁, e₂] = α₁e₁, [e₁, e₃] = λe₃, [e₂, e₃] = μe₃, [e₃, e₃] = β₂e₂.

For the above formulas, there are not non-zero even linear maps σ satisfying σ[e_i, e_j] = [σ(e_i), σ(e_j)] such that (L, [, ], σ) is a Hom-Lie superalgebra.

In the following paragraph, we will consider the 3-dimensional Hom-Lie superalgebra (L, [, ], σ), where dim L0¯=1, dim L1¯=2. Set e₁ ∈ L0¯, e₂, e₃ ∈ L1¯.

(1) If dim [L, L] = 0, then (L, [, ], σ) is an abelian Hom-Lie superalgebra.

(2) If dim [L, L] = 1, we consider the two cases as follows.

Case I: [L, L] = ℂe₁ = L0¯. Let [e₁, e₂] = 0, [e₁, e₃] = 0, [e₂, e₂] = αe₁, [e₂, e₃] = βe₁, [e₃, e₃] = γe₁, where α² + β² + γ² ≠ 0.

Similar to above, we have

[e2, e2]=e1, [e2, e3]=0, [e3, e3]=0.(19)

[e2, e2]=0, [e2, e3]=e1, [e3, e3]=0.(20)

Lemma 4.13

For the bracket (19), Hom-Lie superalgebras are (L, [, ], σ_i) (1 ≥ i ≥ 2) up to isomorphism, whereσ1=(0000000ab)(a2+b2≠0), σ2=(a2000a00bc)(a≠0).

Proof

It is similar to Lemma 4.4.

Lemma 4.14

For the bracket (20), Hom-Lie superalgebras are (L, [, ], σ_i)(1 ≥ i ≥ 5) up to isomorphism, whereσ1=(00000a000)(a≠0), σ2=(00000000a)(a≠0), σ3=(0000000ab)(a≠0, b≠0), σ4=(ab0000b0a0)(a≠0, b≠0), σ5=(ab000a000b)(a≠0, b≠0).

Proof

It is similar to Lemma 4.4.

Case II: [L, L] = ℂe₂ ⊆ L1¯. Let [e₁, e₂] = αe₂, [e₁, e₃] = βe₂, where α² + β² ≠ 0.

Similar to above, we have

[e1, e2]=e2, [e1, e3]=0.(21)

[e1, e2]=0, [e1, e3]=e2(22)

Lemma 4.15

For the bracket (21), Hom-Lie superalgebras are (L, [, ], σ_i)(1 ≥ i ≥ 5) up to isomorphism, whereσ1=(10000000a), σ2=(00000000a)(a≠0), σ3=(00000a00b)(a≠0), σ4=(a0000000b)(a≠0, 1), σ5=(1000a000b)(a≠0).

Proof

It is similar to Lemma 4.4.

Lemma 4.16

For the bracket (22), Hom-Lie superalgebras are (L, [, ], σ_i)(1 ≥ i ≥ 2) up to isomorphism, whereσ1=(00000a00b)(a2+b2≠0), σ3=(a000acb00c)(a≠0, c≠0).

Proof

It is similar to Lemma 4.4.

Remark 4.17

Hom-Lie superalgebras in Lemma 4.15 and Lemma 4.16 are classical Lie superalgebras.

(3) If dim [L, L] = 2, we consider the two cases as follows.

Case I: L, L=Ce2+Ce3=L1¯⋅Lete1, e2=αe2+βe3, e1, e3=λe2+μe3 , where the matrix A=αλβμ is non-degenerate. Then we only consider α ≠ 0, μ ≠ 0, β = λ = 0, other cases are equivalent to this one. By the direct calculation, we can get

e1, e2=αe2, e1, e3=μe3(23)

Lemma 4.18

For the bracket (23), Hom-Lie superalgebras are (L, [, ], σ_i) (1 ≤ i ≤ 4) up to isomorphism, whereσ1=1000a1b10a2b2, σ2=1000a000b, σ3=a00000000a≠0, 1, σ4=αμ000000a0a≠0.

Proof

It is similar to Lemma 4.4.

Case II: L, L=Ce1+Ce2⋅Lete1, e2=αe2, e1, e3=βe2, e2, e2=λe1, e2, e3=μe1, e3, e3=γe1,

where α² + β² ≠ 0, λ² + μ² + γ² ≠ 0.

Similar to above, we have

e1, e2=0, e1, e3=βe2, e2, e2=0, e2, e3=0, e3, e3=γe1.(24)

e1, e2=0, e1, e3=βe2, e2, e2=0, e2, e3=μe1, e3, e3=γe1.(25)

e1, e2=0, e1, e3=βe2, e2, e2=0, e2, e3=μe1, e3, e3=0.(26)

e1, e2=αe2, e1, e3=0, e2, e2=λe1, e2, e3=0, e3, e3=0.(27)

e1, e2=αe2, e1, e3=0, e2, e2=0, e2, e3=0, e3, e3=γe1.(28)

e1, e2=αe2, e1, e3=βe2, e2, e2=0, e2, e3=0, e3, e3=γe1.(29)

Lemma 4.19

For the bracket (24), Hom-Lie superalgebras are (L, [, ], σ_i)(1 ≤ i ≤ 2) up to isomorphism, whereσ1=00000a000a≠0, σ2=a2000a3b00aa≠0.

Proof

It is similar to Lemma 4.4.

Lemma 4.20

For the bracket (25), the Hom-Lie superalgebra is (L, [, ]σ) up to isomorphism, where σ = σ=00000a000a≠0.

Proof

It is similar to Lemma 4.4.

Lemma 4.21

For the bracket (26), Hom-Lie superalgebras are (L, [, ], σ_i)(1 ≤ i ≤ 2) up to isomorphism, whereσ1=00000a00ba2+b2≠0, σ2=a000±a000±1a≠0.

Proof

It is similar to Lemma 4.4.

Lemma 4.22

For the bracket (27), the Hom-Lie superalgebra is (L, [, ], σ) up to isomorphism, where σ = σ=00000000aa≠0.

Proof

It is similar to Lemma 4.4.

Lemma 4.23

For the bracket (28), the Hom-Lie superalgebra is (L, [, ], σ) up to isomorphism, where σ = σ=a20000000aa≠0.

Proof

It is similar to Lemma 4.4.

Lemma 4.24

For the bracket (29), Hom-Lie superalgebras are (L, [, ], σ) up to isomorphism,, whereσ=a20000−βαa00aa≠0.

Proof

It is similar to Lemma 4.4.

(4) If dim [L, L] = 3, then [L, L] = L. Let [e₁, e₂] = α₁e₂ + β₁e₃, [e₁, e₃] = α₂e₂ + β₂e₃, [e₂, e₂] = λe₁, [e₂, e₃] = μe₁, [e₃, e₃] = γe₁. From dim[L, L] = 3, it follows that λ² + μ² + γ² ≠ 0 and the matrix A=α1α2β1β2 is non-degenerate. So we only consider three cases as follows, other cases are equivalent to those.

e1, e2=α1e2, e1, e3=β2e3, e2, e2=e1, e2, e3=0, e3, e3=0.

e1, e2=α1e2, e1, e3=β2e3, e2, e2=e1, e2, e3=0, e3, e3=e1.

e1, e2=α1e2, e1, e3=β2e3, e2, e2=0, e2, e3=e1, e3, e3=0.

For the above formulas, there are not non-zero even linear maps σ satisfying σ [e_i, e_j] = [σ(e_i), σ(ej)] such that (L, [, ], σ) is a Hom-Lie superalgebra.

Proof of Theorem 4.1

From Lemmas 4.2-4.24, we obtain Theorem 4.1.

Remark 4.25

By Lemma 2.8, Hom-Lie superalgebrasL0;31 , L2, 12andL1, 224are not isomorphism.

Acknowledgement

This work was supported by National Natural Science Foundation of China(11471090) and Science and Technology Research Fund of Jilin Province (2016111).

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Received: 2015-5-3

Accepted: 2016-8-1

Published Online: 2016-9-20

Published in Print: 2016-1-1

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