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Hom-Lie superalgebra structures on exceptional simple Lie superalgebras of vector fields

  • Liping Sun and Wende Liu EMAIL logo
Published/Copyright: November 13, 2017
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Open Mathematics
From the journal Open Mathematics Volume 15 Issue 1

Abstract

According to the classification by Kac, there are eight Cartan series and five exceptional Lie superalgebras in infinite-dimensional simple linearly compact Lie superalgebras of vector fields. In this paper, the Hom-Lie superalgebra structures on the five exceptional Lie superalgebras of vector fields are studied. By making use of the ℤ-grading structures and the transitivity, we prove that there is only the trivial Hom-Lie superalgebra structures on exceptional simple Lie superalgebras. This is achieved by studying the Hom-Lie superalgebra structures only on their 0-th and (−1)-th ℤ-components.

Keywords: Exceptional Lie superalgebra; Hom-Lie superalgebra structure; Gradation; Transitivity
MSC 2010: 17B40; 17B65; 17B66

1 Introduction

The motivations to study Hom-Lie algebras and related algebraic structures come from physics and deformations of Lie algebras, in particular, Lie algebras of vector fields. Recently, this kind of investigation has become rather popular [1, 2, 3, 4, 5, 6, 7, 8, 9, 10], partially due to the prospect of having a general framework in which one can produce many types of natural deformations of Lie algebras, in particular, q-deformations. The notion of Hom-Lie algebras was introduced by J. T. Hartwig, D. Larsson and S. D. Silvestrov in [1] to describe the structures on certain deformations of Witt algebras and Virasoro algebras, which are applied widely in the theoretical physics [11, 12, 13, 14]. Later, W. J. Xie and Q. Q. Jin gave a description of Hom-Lie algebra structures on semi-simple Lie algebras [9, 10]. In 2010, F. Ammar and A. Makhlouf generalized Hom-Lie algebras to Hom-Lie superalgebras [15]. In 2013 and 2015, B. T. Cao, L. Luo, J. X. Yuan and W. D. Liu studied the Hom-structures on finite dimensional simple Lie superalgebras [16, 17].

In the well-known paper [18], Kac classified the infinite-dimensional simple linearly compact Lie superalgebras of vector fields, including eight Cartan series and five exceptional simple Lie superalgebras. In 2014, the Hom-Lie superalgebra structures on the eight infinite-dimensional Cartan series were investigated by J. X. Yuan, L. P. Sun and W. D. Liu [19]. In the present paper, we characterize the Hom-Lie superalgebra structures on the five infinite-dimensional exceptional simple Lie superalgebras of vector fields. Taking advantage of the ℤ-grading structures and the transitivity, we analyse the Hom-Lie superalgebra structures through computing the Hom-Lie superalgebra structures on the 0-th and (−1)-th ℤ-components. The main result of this study shows that the Hom-Lie superalgebra structures on the five infinite-dimensional exceptional simple Lie superalgebras of vector fields must be trivial.

2 Preliminaries

Throughout the paper, we will use the following notations. 𝔽 denotes an algebraically closed field of characteristic zero. ℤ2 := {0, 1} is the additive group of two-elements. The symbols |x| and zd(x) denote the ℤ2-degree of a ℤ2-homogeneous element x and the ℤ-degree of a ℤ-homogeneous element x, respectively. Notation 〈v1, …, vk〉 is used to represent the linear space spanned by v1, …, vk over 𝔽. Now, let us review the five infinite dimensional exceptional simple Lie superalgebras and their ℤ-grading structures adopted in this paper. More detailed descriptions can be found in [18, 20].

We construct a Weisfeiler filtration of L by using open subspaces:

L=L−h⊃L−h+1⊃…⊃L0⊃L1⊃…

where L is a simple linearly compact Lie superalgebra. Let 𝔤i = Lj/Li+1, GrL = ⊕i=−h∞gi . In [18], all possible choices for nonpositive part 𝔤≤ 0 = ⊕i=−h∞gi of the associated graded Lie superalgebra of L were derived. The five exceptional simple Lie superalgebras are as follows (the subalgebra 𝔤≤ 0 is written below as the h+1- tuple(𝔤−h, 𝔤−h+1, … 𝔤−1,𝔤0)):

E(5,10):(F5∗,∧2(F5),sl(5))E(3,6):(F3∗,F3⊗F2,gl(3)⊕sl(2))E(3,8):(F2,F3∗,F3⊗F2,gl(3)⊕sl(2))E(1,6):(F,Fn,cso(n),n≥1,n≠2)E(4,4):(F4|4,P^(4))

As we know, the five exceptional simple Lie superalgebras are transitive and irreducible. In particular, 𝔤−i = g−1i for i ≥ 1 and the transitivity will be used frequently in this paper:

transitivity: if x∈gi,i≥0, then [x,g−1]=0 implies x=0. (1)

To study the Hom-Lie superalgebras structure, we recall the following definitions [16].

A Hom-Lie superalgbra is a triple (𝔤, [⋅, ⋅], σ) consisting of a ℤ2-graded vector space 𝔤, an even bilinear mapping [⋅, ⋅] : 𝔤 × 𝔤 → 𝔤 and an even linear mapping: σ : 𝔤 → 𝔤 satisfying:

σ[x,y]=[σ(x),σ(y)], (2)

[x,y]=−(−1)|x||y|[y,x], (3)

(−1)|x||z|[σ(x),[y,z]]+(−1)|y||x|[σ(y),[z,x]]+(−1)|z||y|[σ(z),[x,y]]=0. (4)

for all homogeneous elements x, y, z ∈ 𝔤.

An even linear mapping σ on a Lie superalgebra 𝔤 is called a Hom-Lie superalgebra structure on 𝔤 if (𝔤, [⋅, ⋅], σ) is a Hom-Lie superalgebra. In particular, σ is called trivial if σ = id|g.

It is obvious that σ is graded and if σ is a Hom-Lie superalgebra structure on a simple Lie superalgebra 𝔤, then σ must be a monomorphism.

Lemma 2.1

Let 𝔤 = ⊕i≥−h𝔤i be a transitive Lie superalgebra. If σ is a Hom-Lie superalgebra structure on 𝔤 and σ|𝔤−1 = id|𝔤−1, then σ(x)−x ∈ 𝔤−k, where x ∈ 𝔤0, k ≥ 1.

Proof

For any y ∈ 𝔤−1, from σ|𝔤−1 = id|𝔤−1 and (2), one can deduce

[σ(x),y]=[σ(x),σ(y)]=σ[x,y]=[x,y].

Then [σ(x)−x, y] = 0. So, [σ(x)−x, 𝔤−1] = 0. The transitivity(1) of 𝔤 and the gradation of σ imply that σ(x)−x ∈ 𝔤−k, k ≥ 1. □

Lemma 2.2

Let 𝔤 = ⊕i≥−h𝔤i be a transitive and irreducible Lie superalgebra. If σ is a Hom-Lie superalgebra structure on 𝔤, and satisfies

σ|go⊕g−1=id|go⊕g−1,

then

σ=id|g.

Proof

Let i ≥ 1. Equation (2) and 𝔤−i = g−1i imply

σ|g−i=id|g−i. (5)

Suppose x∈gi,y,z∈9≤0=⊕i=−h0gi, by (4) and (5), one can deduce

[σ(x),[y,z]]=[x,[y,z]].

Since 𝔤 is transitive and irreducible, then 𝔤−1 = [𝔤0, 𝔤−1]. Together with Equation (4), it implies

[σ(x)−x,g≤0]=0.

Then σ(x)−x = 0, σ|𝔤i = id|𝔤i. Thus, we have σ = id|𝔤, that is, σ is trivial. □

According to Lemma 2.2, to study the Hom-Lie superalgebra structures on five exceptional simple Lie superalgebras, we can begin with their 0-th and (−1)-th components. In the remaining of this paper, we will directly use Equation (4) without further remarks.

3 Hom-Lie superalgebra structures on five exceptional simple Lie superalgebras

3.1 E(4, 4)

From [20], there is a unique irreducible gradation over E(4, 4), such that E(4, 4) is the only inconsistent gradated algebra (its (−1)-th component is not purely odd) of the five exceptional simple Lie superalgebras. The 0-th component E(4, 4)0 is isomorphic to P^ (4), which is the unique nontrivial central extension of P(4) . As E(4, 4)0- module, E(4, 4)−1 is isomorphic to the natural module 𝔽4|4. Throughout the rest of the paper, we will use the notations which are introduced in [21] for E(4, 4),.

Let C = (cij) ∈ 𝔤𝔩4(𝔽) be a skew-symmetric matrix. C~=(c~ij)=(εijklckl) stands for the Hodge dual of C, where εijkl is the symbol of permutation (1234)↦(ijkl) . For short, write i′ : = i+4, i = 1, 2, 3, 4. Fix a basis of P^ (4) : 𝓐 ∪ 𝓑 ∪ 𝓒 ∪ 𝓘, where

A:={Eij−Ej′i′,1≤i≠j≤4},B:={Eij′+Eji′,1≤i≤j≤4},C:={Ei′j−Ej′i−(E~i′j−E~j′i),1≤i<j≤4},I:={I,Ejj−Ei+1,i+1−(Ei′i′−Ei′+1,i′+1),1≤i≤3},

I is the unit matrix.

Suppose E(4, 4)−1 = 〈vi, vi′|i = 1, …, 4〉, where |vi| = 0, |vi′| = 1.

Proposition 3.1

If σ is a Hom-Lie superalgebra structure on E(4, 4), then

σ|E(4,4)−1=id|E(4,4)−1.

Proof

Let i = 1, …, 8, k, j = 1, …, 4 and i ≠ k′, j ≠ i. By using Equation (4), one can deduce

[σ(vi),vk]=[σ(vi),[Ekj−Ej′k′,vj]]=0. (6)

Similarly, one has

[σ(vi),vk′]=−[σ(vi),[Ejk−Ek′j′,vj′]]=0 (7)

and

[σ(vk),vk′]=−[σ(vk),[Ek′j−Ej′k−(E~k′j−E~j′k),vj]]=−[σ(vj),vj′].

The arbitrariness of k, j in the last equation shows

[σ(vi),vi′]=0,i=1,...,4. (8)

It follows from Equations (6)-(8) that

[σ(E(4,4)−1),E(4,4)−1]=0.

Using the transitivity (1), one gets σ(E(4, 4)−1) ⊆ E(4, 4)−1. Recall that |σ| = 0, one may suppose

σ(vi)=∑m=18λmvm,λm,∈F.

Now, for distinct i, j, k, l = 1, 2, 3, 4, suppose (1234)↦(ijkl) is an even permutation and

c=Ek′j−Ej′k−(Eli′−Eil′)∈C,a=Ejl−El′j′∈A.

Obviously, [a, vi] = 0, [vi, c] = 0. Then

0=[σ(vi),[c,a]]=[σ(vi),Ek′l−El′k−(Eij′−Eji=−λlvk′+λkvl′.

Hence, λk = λl = 0, and then σ(vi) = λjvi, i = 1, 2, 3, 4.

The equation

0=[σ(vi′),[Ekj−Ej′k′,Ejl′+Elj=[σ(vi′),Ekl′+Elk′]=λl′vk+λk′vl

implies λk′ = λl′ = 0. One has σ(vi′) = λi′vi′.

At last, let us prove σ(vi) = vi and σ(vi′) = vi′. For distinct i, j, k, put

h=Ejj−Ej+1,j+1−(Ekk−Ek+1,k+1)∈I,a=Eji−Ej′i′∈A.

Clearly,

[σ(h),vj]=[σ(h),[a,vi]]=−[σ(vi),a]=λivj. (9)

Recall that σ is monomorphic. Let σ−1 denote a left inverse of σ. Then

σ−1[σ(h),vj]=[h,σ−1(vj)]=λj−1[h,vj]=λj−1vj.

It implies that

[σ(h),vj]=σ(λj−1vj)=vj. (10)

Comparing (9) with (10), one gets λj = 1. Analogously, one gets σ(vi′) = vi′.

Summing up the above, we have proved σ|E(4, 4)−1 = id|E(4, 4)−1. □

Proposition 3.2

If σ is a Hom-Lie superalgebra structure on E(4, 4), then

σ|E(4,4)0=id|E(4,4)0.

Proof

Put x ∈ E(4, 4)0, v ∈ E(4, 4)−1. Using Proposition 3.1 and Lemma 2.1, one can deduce σ(x)-x ∈ E(4, 4)−1. Note that 𝓐 and 𝓘 are generated (Lie product) by 𝓑 and 𝓒, it is sufficient to prove the cases for x ∈ 𝓑 and x ∈ 𝓒.

Case 1: Let x = Eij′+Eji′ ∈ 𝓑. Noting that |σ| = 0, one may suppose

σ(x)=x+∑m=14λm′vm′,λm′∈F.

Put

c=Ek′l−El′k−(E~k′l−E~l′k)∈C,b=Ekk′∈B,

where k, l ≠ i, j. Then

0=[σ(x),[b,c]]=[Eij′+Eji′+∑m=14λm′vm′,Ekl−El′k′]=λk′vl′.

It implies that λk′ = 0, k ≠ i, j. Then,

σ(x)=x+λi′vi′+λj′vj′. (11)

Now, put

c=Ei′k−Ek′i−(E~i′k−E~k′i′)∈C,b=Eii′∈B,k≠i,j.

Using Equation (11), one may suppose σ(b) = b+μi′vi′, μi′ ∈ 𝔽, then

0=[σ(x),a]=[Eij′+Eji′+λi′vi′+λj′vj′,Eik−Ek′i′]=λi′vk′.

So λi′ = 0. That is, σ(x) = x+λj′vj′. Similarly, put

c=Ej′k−Ek′j−(E~j′k−E~k′j′)∈C,b=Ejj′∈B.

One obtains λj′ = 0. Thus, we have σ(x) = x for any x ∈ 𝓑.

Case 2: Let x = Ei′j−Ej′i−(Ekl′−Elk′) ∈ 𝓒, where εijkl = 1. Noting that |σ| = 0, one may assume σ(x)=x+∑m=14μm′vm′,μm′∈F.

Firstly, put

c=Ek′l−El′k−(Eji′−Eij′)∈C,b=Ekk′∈B,

where εijkl = 1. Equation (4) and the result σ(b) = b obtained in Case 1 imply

0=[σ(x),[c,b]]=[x+∑m=14μm′vm′,Ekl−El′k′]=c0+μk′vl′,c0∈C.

The equation above shows μk′ = 0. By the arbitrariness of k ≠ i, j, one has

σ(x)=x+μi′vi′+μj′vj′. (12)

Secondly, put

c=El′j−Ej′l−(Eik′−Ek′i)∈C,b=Ejj′.

Using Equation (12), one may suppose

σ(c)=c+γl′vl′+γj′vj′,γl′,γj′∈F.

On one hand,

[σ(x),[b,c]]=−[σ(b),[c,x]]−[σ(c),[x,b]]=El′i−Ei′l−(Ekj′−Ejk′)−γj′vi′.

On the other hand,

[σ(x),[b,c]]=[Ei′j−Ej′i−(Ekl′−Elk′)+μi′vi′+μj′vj′,El′j′−Ejl]=El′i−Ei′l−(Ekj′−Ejk′)−μj′vl′.

Since i ≠ l, one has γj′ = μj′ = 0. Then σ(x) = x+λi′vi′.

At last, put

c=Ei′k−Ek′i−(Elj′−Ejl′)∈C,b=Eii′∈B.

We can obtain λi′ = 0 in the same way. Thus, σ(x) = x is proved.

Summing up, we have proved σ(x) = x for any x ∈ E(4, 4)0, that is, σ|E(4, 4)0 = id|E(4, 4)0. □

3.2 E(3, 6), E(5, 10) and E(3, 8)

Before studying the Hom-Lie superalgebra structures on E(3, 6), E(5, 10) and E(3, 8), we would like to review their algebraic structures briefly [18, 20, 22, 23]. From [18] we know that the even part E(5, 10)0 of E(5, 10) is isomorphic to the Lie algebra S5, which consists of divergence 0 polynomial vector fields on 𝔽5, i.e., polynomial vector fields annihilating the volume form dx1∧ dx2∧…∧ dx5. As S5-module, the odd part E(5, 10)1 of E(5, 10) is isomorphic to dΩ1(5), the space of closed polynomial differential 2-forms on 𝔽5. In the following, we keep the notations from [22]:

dij:=dxi∧dxj,∂i:=∂/∂xi.

We write each element D of E(5, 10)0 as

D=∑i=15fi∂i, where fi∈F[[x1,x2,...,x5]],∑i=15∂i(fi)=0,

and denote E ∈ E(5, 10)1 through

E=∑i,j=15fijdij, where fij∈F[[x1,x2,...,x5]],dE=0. (13)

The bracket in E(5, 10)1 is defined by

[fdij,gdkl]=εtijklfg∂t,

where εtijkl is the sign of permutation (tijkl) when {tijkl} = {12345} and zero otherwise. The bracket of E(5, 10)0 with E(5, 10)1 is defined by the usual action of vector fields on differential forms. The irreducible consistent ℤ-gradation over E(5, 10) is defined by letting (see [22])

zd(xi)=2,zd(d)=−52,zd(dxi)=−12.

Then E(5, 10) = ⊕i≥−2𝔤i, where

go≃sl(5)=〈xi∂j,xk∂k−xk+1∂k+1|i,j=1,…,5,i≠j,k=1,…,4〉,g−1≃∧2F5=〈dij|1≤i<j≤5〉,9−2≃F5∗=〈∂j|i=1,...,5〉.

The exceptional simple Lie superalgebra E(3, 6) is a subalgebra of E(5, 10) . The irreducible consistent ℤ-gradation over E(3, 6) is induced by the ℤ-gradation of E(5, 10) above. Let h1 = x1∂1−x2∂2, h2 = x2∂2−x3∂3, h3 = x4∂4−x5∂5, h4 = −x2∂2−x3∂3+2x5∂5. Then E(3, 6) = ⊕i≥−2𝔤i, where

go≃gl(3)⊕sl(2)=〈xi∂j,xk∂l,hm|i,j=1,2,3,i≠j,k,l=4,5,k≠l,m=1,...,4〉,g−1≃F3⊗F2=〈dij|i=1,2,3,j=4,5〉,g−2≃F3∗=〈∂j|i=1,2,3〉.

The exceptional simple Lie superalgebra E(3, 8), which is strikingly similar to E(3, 6), carries a unique irreducible consistent ℤ-gradation [23] defined by

zd(xi)=−zd(∂i)=zd(dxi)=2,i=1,2,3;zd(x4)=zd(x5)=−3.

Then E(3, 8) = ⊕i≥−3𝔤i, where

gj=E(3,6)j,j=0,−1,−2;g−3≃F2≃〈dx4,dx5〉. (14)

Hereafter, 𝔤 denotes E(5, 10), E(3, 6) or E(3, 8) unless otherwise noted. We establish a technical lemma, which can be verified directly.

Lemma 3.3

If σ is a Hom-Lie superalgebra structure on 𝔤, then

  1. [σ(dij),dil]=0,if dij,dil∈E(5,10),−[σ(dil),dij],if dij,dil∈E(3,6)orE(3,8);

  2. [σ(dij), dkj] = 0, in particular, [σ(dij), dij] = 0;

  3. [σ(dij), dkl] = [dij, σ(dkl)] for distinct i, j, k, l.

Proposition 3.4

Let 𝔤 be Lie superalgebra E(5, 10), E(3, 6) or E(3, 8). If σ is a Hom-Lie superalgebra structure on 𝔤, then

σ|g−1=id|g−1.

Proof

First, let us prove σ|𝔤−1 = λid|𝔤−1, where λ ∈ 𝔽.

Case 1 : 𝔤 = E(5, 10). Noting that the gradation over E(5, 10) is consistent and |σ| = 0, |dij| = 1, one may suppose

σ(dij)=∑1≤m<n≤5fmndmn, where fmn∈F[[x1,…,x5]].

By Lemma 3.3 (1), one has

[σ(dij),dil]=∑1≤m<n≤5fmndmn,dil=∑m,n≠i,lfmndmn,dil=∑m,n≠i,l[fmndmn,dil]=∑m,n≠i,lεqmnilfmn∂q=0.

The arbitrariness of l shows

fmn=0, where m,n≠i.

Similarly, by Lemma 3.3 (2), one gets

fmn=0, where m,n≠j.

Thus,

σ(dij)=fijdij.

In view of σ(dij) ∈ E(5, 10)1 ≃ dΩ1(5) and (13), one has

0=d(σ(dij))=d(fijdij)=∑m=15∂m(fij)dxm∧dxi∧dxj.

Therefore, ∂m(fij) = 0 for m ≠ i, j, that is fij ∈ 𝔽[[xi, xj So one may suppose σ(dkl) = fkldkl, where fkl ∈ 𝔽[[xk, xl Then

[σ(dij),dkl]=[fijdij,dkl]=εqijklfij∂q,[σ(dkl),dij]=[fkldkl,dij]=εqklijfkl∂q.

For distinct i, j, k, l, Lemma 3.3 (3) implies fij = fkl∈ F. Noting that there exists t such that t ≠ i, j, k, l, one may obtain fkt = fij in the same way. Then fij = λ ∈ 𝔽 for any i, j = 1, 2, 3, 4, 5 and i ≠ j. Thus, we proved σ(dij) = λ dij.

Case 2 : 𝔤 = E(3, 6) or E(3, 8). Noting that the gradation over 𝔤 is consistent and |σ| = 0, |dij| = 1, one may suppose

σ(dij)=∑m,nfmndmn, where m=1,2,3,n=4,5,fmn∈N[[x1,…,x5]].

By Lemma 3.3 (2), one can deduce

[σ(dij),dkj]=[∑m,nfmndmn,dkj]=∑m≠k,n≠jfmndmn,dkj=0.

It implies fmt = 0, m ≠ k, t ≠ j. By the arbitrariness of k, one may suppose

σ(dij)=∑m=13fmjdmj,σ(dil)=∑m=13fmldml,fmj,fml∈F[[x1,…,x5].

Noting that [σ(dil), dkj] = [σ(dil), [xk∂j, dij]] = −[σ(dij), dkl], the simple computations show fmj = fml for m = 1, 2, 3. In view of σ(dij) ∈ 𝔤1 ⊆ dΩ1(5) and (13), one could further deduce

fmj=fml∈F[[x1,x2,x3]]. (15)

For distinct i, k, q = 1, 2, 3,

0=[σ(dij),[xi∂q,xq∂k]]=[σ(dij),xj∂k]=∑m=13fmjdmj,xj∂k.=−∑m=13(xj∂k(fmj)dmj+fmjxj∂k(dmj))=−(xj∂k(fij)+fkj)dij−xj∂k(fkj)dkj−xj∂k(fqj)dqj.

Therefore,

fkj=−xj∂k(fij). (16)

∂k(fkj)=∂k(fqj)=0, (17)

From (15)-(17) and the arbitrariness of k, one has fmj = 0, m ≠ i. Thus, one may suppose

σ(dij)=fidij,σ(dkl)=fkdkl,where fi∈F[[xi]],fk∈F[[xk]].

Then, for distinct i, j, k, l, q,

[σ(dij),dkl]=[fidij,dkl]=εqijklfi∂q,[dij,σ(dkl)]=[dij,fkdkl,]=εqijklfk∂q.

By Lemma 3.3 (3), fi = fk ∈ 𝔽. Thus, we have proved σ(dij) = λ dij for some λ ∈ 𝔽.

Now, let us prove λ = 1 for σ|𝔤−1 = λid|𝔤−1. Suppose x = xi∂l, y = xl∂q, z = dqj. Then

[σ(x),dlj]=[σ(x),[y,z]]=−[σ(y),[z,x]]−[σ(z),[x,y]]=λdij.

Noting that σ is a monomorphism, one may suppose σ−1 is a left linear inverse of σ. Then the equation above implies

σ−1[σ(x),dlj]=[x,λ−1dlj]=λ−1dij=dij.

Thus, λ = 1, that is, σ(dij) = dij. The proof is complete.□

Proposition 3.5

Let 𝔤 be Lie superalgebra E(5, 10), E(3, 6) or E(3, 8). If σ is a Hom-Lie superalgebra structure on 𝔤, then

σ|g0=id|g0.

Proof

Case 1 : 𝔤 = E(5, 10). Note that E(5, 10)0 is generated (Lie product) by {xi ∂j|i, j = 1, …, 5, i ≠ j}, it is sufficient to prove σ(x) = x only for x = xi ∂j. By Proposition 3.4, Lemma 2.1 and |σ| = 0, we have σ(xj ∂j) − xj∂j ∈ 𝔤−2. Then one may suppose

σ(xi∂j)=xi∂j+∑m=15λm∂m,λm∈F.

For distinct i, j, q, k, l,

0=[σ(xi∂j),[xq∂k,xk∂l]]=x+∑m=15λm∂m,xq∂l=λq∂l.

Hence, λq = 0. The arbitrariness of q ≠ i, j shows

σ(xi∂j)=xi∂j+λi∂i+λj∂j.

Similarly,

0=[σ(xi∂j),[xi∂l,xl∂k]]=[x+λi∂i+λj∂j,xi∂k]=λi∂k

implies λj = 0. Therefore,

σ(xi∂j)=xi∂j+λj∂j. (18)

On one hand,

[σ(xi∂j),[xj∂i,xi∂k]]=[xi∂j+λj∂j,xj∂k]=xi∂k+λj∂k. (19)

On the other hand, using (4) and (18), one can derive

[σ(xi∂j),[xj∂i,xi∂k]]=−[σ(xi∂k),xi∂i−xj∂j]=xi∂k. (20)

Comparing (19) with (20), one gets λj = 0. By (18), we have σ(xj∂j) = xj∂j.

Case 2 : 𝔤 = E(3, 6) or E(3, 8). Firstly, let us show σ(xi∂j) = xi∂j for any xi∂j∈ 𝔤0. As before, one may suppose

σ(xi∂j)=xi∂j+∑m=13λm∂m,λm∈F.

Case 2.1 : Let i, j = 1, 2, 3. For distinct i, j, k, by (4),

0=[σ(xi∂j),[xi∂k,xk∂j]]=[σ(xi∂j),xi∂j]=xi∂j+∑m=13λm∂m,xi∂j=λi∂j.

Consequently, λi = 0. One may suppose σ(xk ∂j) = xk∂j + ∑m≠k3μm∂m, On one hand,

[σ(xi∂j),xk∂i]=xi∂j+∑m≠iλm∂m,xk∂i=−xk∂j+λk∂i.

On the other hand, from (4) one deduces

[σ(xi∂i),xk∂i]=[σ(xi∂j),[xk∂j,xj∂i]]=−[σ(xk∂j),[xj∂i,xi∂j]]−[σ(xj∂i),[xi∂j,xk∂j]]=−xk∂j−μj∂j+μi∂i.

Comparing the two equations above, one gets μj = 0 and λk = μi. In view of the arbitrariness of i, j, k, one may assume

σ(xi∂j)=xi∂j+λ∂k,σ(xk∂j)=xk∂j+λ∂i,i,j,k are distinct.

Then

σ(xi∂j)=σ[xi∂k,xk∂j]=[σ(xi∂k),σ(xk∂j)]=xi∂j−λ∂k.

The two equations above imply λ = 0, and then σ(xj ∂j) = xj∂j, i, j = 1, 2, 3.

Case 2.2 : Let i, j = 4, 5. For distinct k, l, q = 1, 2, 3,

0=[σ(xi∂j),[xk∂q,xq∂l]]=[σ(xi∂j),xk∂l]=[xi∂j+∑m=13λm∂m,xk∂l]=λk∂l.

Therefore, λk = 0. The arbitrariness of k implies σ(xi ∂j) = xi∂j.

Next, we will prove σ(hm) = hm for m = 1, 2, 3, 4. Similarly, suppose σ(hm) = hm+∑k=13λk∂k. Noting that hm is an element of the basis of 𝔤0’s Cartan subalgebra, one may assume [hm, xi∂j] = γ xi∂j, γ∈ 𝔽, i, j = 1, 2, 3. By the result obtained in Case 2.1, one can deduce

σ[hm,xi∂j]=γσ(xi∂j)=γxi∂j.

On the other hand,

σ[hm,xi∂j]=[σ(hm),σ(xi∂j)]=hm+∑k=13λk∂k,xi∂j=γxi∂j+λi∂j.

The two equations above imply that λi = 0 for i = 1, 2, 3. Thus, we have σ(hm) = hm for m = 1, 2, 3, 4.

Summing up, we have proved σ(x) = x for any x ∈ 𝔤0, that is, σ|𝔤0 = id|𝔤0.□

3.3 E(1,6)

The exceptional simple Lie superalgebra E(1, 6) is a subalgebra of K(1, 6). The principal gradation over K(1, 6) (see [20]) induces an irreducible consistent ℤ-gradation over E(1, 6). Moreover, E(1, 6) has the same non-positive ℤ-graded components as K(1, 6) : E(1, 6)j = K(1, 6)j, j ≤ 0. Let x be even and ξi, i = 1, …, 6, be odd indeterminates. Then

E(1,6)0≃cspo(6)=〈x,ξiξj|i,j=1,…,6,i≠j〉,E(1,6)−1≃F6=〈ξi|i=1,…,6〉,E(1,6)−2≃F.

For f, g ∈ E(1, 6), the bracket product is defined as

[f,g]=2f−∑i=16ξi∂i(f)∂x(g)−(−1)|f||g|2g−∑i=16ξi∂i(g)∂x(f)+(−1)|f|∑i=16∂i(f)∂j(g).

Proposition 3.6

If σ is a Hom-Lie superalgebra structure on E(1, 6), then

σ|E(1,6)−1=id|E(1,6)−1.

Proof

For i, j, k = 1, …, 6 and j ≠ i, k, by (4),

[σ(ξi),ξk]=[σ(ξi),[ξjξk,ξj]]=δk,i[σ(ξj),ξj].

It implies that [σ(ξi), ξk] = 0 for any k, i = 1, …, 6. From |σ| = 0 and the transitivity of E(1, 6), one can deduce σ(E(1, 6)−1) ⊆ E(1, 6)−1. So one may assume σ(ξi) = ∑k=16λkξk,λk ∈ 𝔽. By using Equation (4), for distinct i, j, s, t = 1, …, 6, one can deduce

0=[σ(ξi),[ξjξt,ξsξt]]=[σ(ξi),ξjξs]=λjξs−λsξj.

Then λk = 0 for any k ≠ i. That is σ(ξi) = λiξi. By using Equation (4) again, one can deduce

−λj=[σ(ξi),[ξjξi,ξj]]=[σ(ξi),ξi]=[λiξi,ξi]=−λi.

Moreover, one knows σ(ξi) = λξi for any i = 1, …, 6. The remaining work is to prove λ = 1. For distinct i, j, t,

[σ(ξtξj),ξt]=[σ(ξtξj),[ξiξt,ξi]]=−[σ(ξi),[ξtξj,ξiξt]]=[λξi,ξjξi]=λξj. (21)

Suppose σ−1 is a left inverse of the monomorphism σ. Then

σ−1([σ(ξtξj),ξt])=[ξtξj,σ−1(ξt)]=λ−1ξj.

Hence

[σ(ξtξj),ξt]=σ(λ−1ξj)=ξj. (22)

Equations (21) and (22) imply λ = 1. The proof is completed.□

Proposition 3.7

If σ is a Hom-Lie superalgebra structure on E(1, 6), then

σ|E(1,6)0=id|E(1,6)0.

Proof

For any y ∈ E(1, 6)0, by Proposition 3.6 and Lemma 2.1, σ(y) − y ∈ E(1, 6)−2 ≃ 𝔽. So, one may suppose σ(y) = y + λ, λ ∈ 𝔽. Then

[σ(y),x]=[y+λ,x]=[λ,x]=2λ. (23)

If y = x, noting that x is an element of a basis of E(1, 6)0’s Cartan subalgebra, one may assume

[σ(x),x]=μσ(x)=μ(x+λ),μ∈F. (24)

Comparing (23) and (24), one has μ = 0, and then λ = 0, that is σ(x) = x.

If y = ξiξj,

[σ(ξiξj),x]=[σ(ξiξj),σ(x)]=σ[ξiξj,x]=0. (25)

From (23) and (25), one gets λ = 0 again. The proof is complete.□

3.4 The main result

Propositions 3.1, 3.2, 3.4-3.7 show that all the Hom-Lie superalgebra structures on the 0-th and (−1)-th ℤ-components of each infinite-dimensional exceptional simple Lie superalgebra are trivial. Then combining this with Lemma 2.2, we immediately have:

Theorem 3.8

There is only the trivial Hom-Lie superalgebra structure on each exceptional simple Lie superalgebra.

Acknowledgement

This work is supported by Natural Science Foundation of Heilongjiang province (No.A2015017) and National Natural Science Foundation of China (No.11471090). The authors would like to thank deeply the referees for their detailed and helpful suggestions and comments.

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Received: 2016-9-13
Accepted: 2017-9-14
Published Online: 2017-11-13

© 2017 Sun and Liu

This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 License.

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https://doi.org/10.1515/math-2017-0112
Keywords for this article
Exceptional Lie superalgebra; Hom-Lie superalgebra structure; Gradation; Transitivity
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  279. Topical Issue on Cyber-security Mathematics
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  282. Feedback equivalence of convolutional codes over finite rings
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