# Enveloping Lie Superalgebras and Killing-Ricci Forms of Bol Superalgebras

Authors: Sylvain Attan
Journal: Journal of Mathematics
 (2021)

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## Abstract

In this paper, enveloping Lie superalgebras of Bol superalgebras is introduced. The notion of Killing-Ricci forms and invariant forms of these superalgebras is investigated as generalization of the one of Bol algebras.


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## Body

## 1. Introduction

Bol algebras were introduced [1] in the context of a study of smooth Bol loops. The algebras play the same role with respect to Bol loops as Lie algebras do with respect to Lie groups or Malcev algebras to Moufang loops [2].A vector spaceV equipped with a trilinear operation [,,] is called a Lie triple system if a,a,b=0, a,b,c+b,c,a+c,a,b=0,x,y,a,b,c=x,y,a,b,c+a,x,y,b,c+a,b,x,y,c for all x,y,a,b,c∈V. A (left) Bol algebra V,⋅,,, is a Lie triple system V,,, with an additional bilinear skew-symmetric operation ⋅ satisfying a,b,c⋅d=a,b,c⋅d+c⋅a,b,d+c,d,a⋅b+a⋅b⋅c⋅d. A related notion is that of a Lie triple algebra, introduced under the name generalized Lie triple systems, by Yamaguti [3], and called later as Lie Yamaguti algebras [4].From the standard enveloping Lie algebra of a given Bol algebra, the notion of Killing-Ricci form and invariant form for a Bol algebra are introduced and studied in [5].Aℤ2-graded generalization of Lie algebras, called Lie superalgebras, is considered in [6, 7], while a ℤ2-graded generalization of Lie Yamaguti algebras called Lie Yamaguti superalgebras was first considered in [8] and generalizes Lie supertriple systems [9]. The reader may refer to [10] for applications of Lie supertriple systems in physics. As Lie Yamaguti superalgebras, Bol superalgebras first introduced in [11] may also be viewed as a generalization of Lie supertriple systems. For relations between Malcev superalgebras and Bol superalgebras, one may refer to [12].As a part of the general theory of superalgebras, the notion of Killing form of Lie algebras is extended to one of Lie triple systems [13], Lie superalgebras [7], Lie supertriple systems [14, 15], and next Lie Yamaguti superalgebras [16].In this paper, we define enveloping Lie superalgebras and the Killing-Ricci form of Bol superalgebras and study this Killing-Ricci form, which could be seen as a generalization of the one of Bol algebras [5] and the Killing form of Lie supertriple systems [14, 15]. Unlike Bol algebras, in Bol superalgebras, there is an odd subspace, which is not a Bol subalgebra. This complicates the work more than Bol algebras case.The rest of this paper is organized as follows. In section2, we first recall some basics on Lie and Malcev superalgebras as well as Lie supertriple systems and Bol superalgebras. In Section 3, we define the notion of pseudo-superderivations (Definition 8), study their properties (Lemma 1), and introduce the notion of enveloping Lie superalgebras of Bol superalgebras (Definition 12). In Section 4, the Killing-Ricci form of Bol superalgebras is defined (Definition 13) and some of its properties are investigated (Theorem 1, Proposition 3, and Lemma 3). In the next section, the invariant form of Bol superalgebras is defined (Definition 14) and some results are obtained (Lemma 4 and Theorem 2).Throughout this paper, all vector superspaces and superalgebras are finite dimensional over a fixed ground fieldK of characteristic 0.


## 2. Some Basics on Superalgebras

We recall here some useful definitions and examples of Lie supertriple systems as well as the one of Bol superalgebras. These examples are obtained from the relation between Malcev superalgebras and Bol superalgebras, which could be found in [12].Now, letM be a linear superspace over K, that is, a ℤ2-graded linear space with a direct sum M=M0⊕M1. The elements of Mj, j∈ℤ2, are said to be homogeneous of parity j. The parity of a homogeneous element x is denoted by x¯. For all i,j∈ℤ2, i+j will always means that this sum is calculated modulo 2. If N=N0⊕N1 is another superspace, a linear map f:M⟶N is said to be of degree r∈ℤ2 if fMi⊆Ni+r for all i∈ℤ2. If f is of degree r=0, that is, fMi⊆Ni for all i∈ℤ2, then f is said to be an even linear map. An algebra A,, is called a superalgebra if the underlying vector space is ℤ2-graded, i.e., A=A0⊕A1 and if furthermore Ai,Aj⊂Ai+j. For any binary operation, we will sometimes use juxtaposition in order to reduce the number of braces; i.e., for “⋅”, xy⋅z means x⋅y⋅z.Definition 1.
A Lie superalgebra is the superalgebra (A=A0⊕A1,,) satisfying the superskew-symmetry and the super-Jacobi identities that is(1)x,y=−−1x¯y¯y,x,(2)x,y,z+−1x¯y¯+z¯y,z,x+−1z¯x¯+y¯z,x,y=0,for all x,y,z∈ℋA. In terms of the super-Jacobian,(3)SJx,y,z=x,y,z+−1x¯y¯+z¯y,z,x+−1z¯x¯+y¯z,x,y.
The super-Jacobi identity is written asSJx,y,z=0 for all x,y,z∈ℋA.
Another class of superalgebras that is of interest in this paper is the one of Malcev superalgebras.Definition 2.
(see [17, 18]). A superalgebraM=M0⊕M1,,is called a Malcev superalgebra if it satisfies the following superidentities:(4)x,y=−−1x¯y¯y,xsuperskew−symmetry,x,y,z,t−x,y,z,t−−1y¯z¯+t¯x,z,t,y−−1t¯y¯+z¯x,t,y,z=−1y¯z¯x,z,y,tsuper−Malcevidentity,for x,y,z,t∈ℋM.Definition 3.
(1)
A supertriple system is a coupleS=S0⊕S1,,,consisting of aℤ2-gradedK-vector spaceS=S0⊕S1and aK-trilinear map [„], satisfyingSi,Sj,Sk⊂Si+j+kfor alli,j,kinℤ2such that for allx,y,z,∈ℋS, the following equations hold:(5)x,y,z=−−1x¯y¯y,x,zleftsuperskew−symmetry,x,y,z+−1x¯y¯+z¯y,z,x+−1z¯x¯+y¯z,x,y=0,superternary−Jacobiidentity.(2)
A Lie supertriple system [9, 19] is a supertriple system S=S0⊕S1,,, such that the superternary Nambu identity(6)x,y,u,v,w=x,y,u,v,w+−1u¯x¯+y¯u,x,y,v,w+−1x¯+y¯u¯+v¯u,v,x,y,w,holds for all x,y,u,v,w∈ℋS.Example 1.
(see [9, 12, 19]). LetL=L0⊕L1,,be a Lie superalgebra, thenL,,,is a Lie supertriple system where for allx,y,z∈ℋL,x,y,z≔x,y,z.Definition 4.
(see [11, 12]). A Bol superalgebra is a tripleB=B0⊕B1,⋅,,,consisting of a superspaceB=B0⊕B1, a linear map⋅:B⊗2⟶BsatisfyingBi⋅Bj⊆Bi+j, and a trilinear map,,:B⊗3⟶BsatisfyingBi,Bj,Bk⊆Bi+j+k,i,j,k∈ℤ2, such that the following holds:BS1x⋅y=−−1x¯y¯y⋅x.BS2x,y,z=−−1x¯y¯y,x,z.BS3x,y,z+−1x¯y¯+z¯y,z,x+−1z¯x¯+y¯z,x,y=0.BS4x,y,u,v,w=x,y,u,v,w+−1u¯x¯+y¯u,x,y,v,w+−1x¯+y¯u¯+v¯u,v,x,y,w.BS5x,y,u⋅v=−1u¯x¯+y¯u⋅x,y,v+x,y,u⋅v+−1x¯+y¯u¯+v¯u,v,x⋅y+x⋅y⋅u⋅vholds for allx,y,u,v,w∈ℋS.Remark 1.
A Bol superalgebra is a Lie triple supersystemB=B0⊕B1,,,with a superskew-symmetric binary map.such thatBi.Bj⊆Bi+j(7)x,y,u⋅v=−1u¯x¯+y¯u⋅x,y,v+x,y,u⋅v+−1x¯+y¯u¯+v¯u,v,x⋅y+x⋅y⋅u⋅v,holds for all x,y,u,v∈ℋB.In [12], we proved that any Malcev superalgebra (M=M0⊕M1) equipped with a trilinear operation [,,] where(8)x,y,z≔132x⋅y⋅z−−1x¯y¯+z¯y⋅z⋅x−−1z¯x¯+y¯z⋅x⋅y,for all x,y,z∈ℋM becomes a Bol superalgebra M=M0⊕M1,⋅,,,. This allows us to get the following example of Bol superalgebra from an example of a Malcev superalgebra [20].Example 2.
LetL22,2,⋅be a non-Lie Malcev superalgebra [20] defined with respect to a basise1,e2,e3,e4, whereL22,20¯=spane1,e2andL22,21¯=spane3,e4with the nonzero products given bye1⋅e2=e2,e1⋅e3=e3,e1⋅e4=−e4,ande2⋅e3=−e4. Then,L23,1,⋅,,,is a four-dimensional Bol superalgebra withL22,20¯=spane1,e2andL22,21¯=spane3,e4, where the nonzero products are given bye1⋅e2=e2,e1⋅e3=e3,e1⋅e4=−e4,e2⋅e3=−e4ande1,e2,e1=−e2,e1,e3,e1=−e3,e1,e4,e1=−e4.From [12], we also get the following example.Example 3.
(see [12]). There is a four-dimensional Bol superalgebraL23,1,⋅,,,withL23,10¯=spane1,e2,e3andL23,11¯=spane4, where the nonzero products are given bye1⋅e3=e1,e2⋅e3=e1+e2,e3⋅e4=e4,e4⋅e4=e1ande1,e3,e3=e1,e2,e3,e3=2e1+e2,e3,e4,e3=−e4.Definition 5.
LetB=B0⊕B1be a Bol superalgebra. A graded subspaceH=H0⊕H1ofBis a subsuperalgebra ofBifHi⋅Hj⊆Hi+jandHi,Hj,Hk⊆Hi+j+k, for alli,j,k∈ℤ2.Definition 6.
A subsuperalgebraH=H0⊕H1of a Bol superalgebraBis an invariant subsuperalgebra (resp., an ideal) ofBifB,B,H⊆H(resp.,BH⊆HandB,B,H⊆H).IfH is an ideal of B, it is an invariant subsuperalgebra of B. Obviously, the center of a Bol superalgebra B defined by ZB=x∈B,xy=0andx,y,z=0,∀y,z∈B is an ideal of B.Definition 7.
LetA=A0⊕A1andB=B0⊕B1be two Bol superalgebras. An even linear mapf:A⟶Bis called a morphism of Bol superalgebras iffxy=fxfyandfx,y,z=fx,fy,fzfor allx,y,z∈B0∪B1.Recall [6] that if V=V0⊕V1 is a vector superspace, then the set of the linear mappings of V into itself which are homogeneous of degree r is denoted by EndrV=f∈EndV,fVi⊆Vr+i, and we obtain an associative superalgebra EndV=End0V⊕End1V. The bracket f,g=fg−−1f¯g¯gf makes EndV into a Lie superalgebra which we denote by lV or lm,n where m=dimV0 and n=dimV1. Let e1,…,em,em+1,…,em+n be a basis of V. In this basis, the matrix of f∈lm,n is expressed as αβγδ, α being an m×m−, δ an n×n−, β an m×n−, and γ an n×m− matrix. The matrices of even elements have the form α00δ and those of odd ones 0βγ0. For f=αβγδ, the supertrace of M is defined by strM=trα−trδ and does not depend on the choice of a homogeneous basis. We have strf,g=0; that is, strfg=−1f¯g¯strgf and strhfh−1=strf.


## 3. Enveloping Lie Superalgebras of a Bol Superalgebra

As derivations for algebras, superderivations of different superalgebras are an important subject of study in superalgebras and diverse area. They appear in many fields of mathematics and physics. In particular, they allow the construction of new superalgebras structures. In the case of Bol superalgebras, instead of superderivations, we have the notion of pseudo-superderivations. They generalize pseudoderivations for Bol algebras [2] and superderivations for Lie supertriple systems [19] and allow the construction of enveloping Lie superalgebras of Bol superalgebras.Definition 8.
LetB=B0⊕B1be a Bol superalgebra. A linear mapP∈EndrBis called a pseudo-superderivation of companiona∈Brr∈ℤ2if, for anyx,y,z∈B0∪B1:(9)Px,y,z=Px,y,z+−1rx¯x,Py,z+−1rx¯+y¯x,y,Pz,(10)Pxy=−1rx¯xPy+Pxy+−1rx¯+y¯x,y,a+a⋅xy.Remark 2.
Note that a pseudo-superderivationPcan have more than one companion. Let denote the set of all companions of a pseudo-superderivationPbyComP.LetpSrB be the set of all pseudo-superderivations of degree r and pSB=pS0B⊕pS1B. Furthermore, let PSrB=P,a,P∈pSrB,a∈ComP and PSB=PS0B⊕PS1B.For any pseudo-superderivationsP,Q∈pS0B∪pS1B of a Bol superalgebra B, consider the supercommutator given by P,Q≔PQ−−1P¯Q¯QP. Then, we have the following.Lemma 1.
LetP,a∈PSrB,Q,b∈PSrB,R,c∈PSsB, andλ∈K. Then, the following holds:(1)
P,a+λQ,b≔P+λQ,a+λb∈PSrB.(2)
P,a,R,c≔P,R,Pc−−1rsRa−ac∈PSr+sB.Proof.
LetP,a∈PSrB, Q,b∈PSrB, R,c∈PSsB, and λ∈K. The first statement is a straightforward computation. For the second statement, pick x,y,z,u,v∈B0∪B1. Then, using repeatedly 9 for P and R, we prove 9 for P,R as follows:(11)P,Rx,y,w=PRx,y,w−−1rsRPx,y,w=PRx,y,z+−1sx¯x,Ry,z+−1sx¯+y¯x,y,Rz−−1rsRPx,y,z+−1rx¯x,Py,z+−1rx¯+y¯x,y,Pz=PRx,y,z+−1rx¯+sRx,Py,z+−1rs+x¯+y¯Rx,y,Pz+−1sx¯Px,Ry,z+−1r+sx¯x,PRy,z+−1sx¯+rx¯+y¯+sx,Ry,Pz+−1sx¯+y¯Px,y,Rz+−1sx¯+y¯+rx¯x,Py,Rz+−1r+sx¯+y¯x,y,PRz−−1rsRPx,y,z+−1sx¯+rPx,Ry,z+−1sr+x¯+y¯Px,y,Rz+−1rx¯Rx,Py,z+−1r+sx¯x,RPy,z+−1rx¯+sx¯+y¯+rx,Py,Rz+−1rx¯+y¯Rx,y,Pz+−1rx¯+y¯+sx¯x,Ry,Pz+−1r+sx¯+y¯x,y,RPz=P,Rx,y,z+−1r+sx¯x,P,Ry,z+−1r+sx¯+y¯x,y,P,Rz.
Thus, we get (9) for P,R. To prove, (10) for P,R, note that if we use repeatedly (10) for P, we get(12)Pc⋅xy=−1rsc⋅Pxy+Pc⋅xy+−1rx¯+y¯+sc,xy,a+a⋅c⋅xy=+−1rx¯+sc⋅xPy+−1rsc⋅Pxy+−1rx¯+y¯+sc⋅x,y,a+−1rsc⋅a⋅xy+Pc⋅xy+−1rx¯+y¯+sc,xy,a+a⋅c⋅xy,and therefore, by (9) and (12) for P and R, we get(13)P,Rxy=PRxy−−1rsRPxy=P−1sx¯xRy+Rxy+−1sx¯+y¯x,y,c+c⋅xy−−1rsR−1rx¯xPy+Pxy+−1rx¯+y¯x,y,a+a⋅xy=−1r+sx¯xPRy+−1sx¯PxRy+−1rx¯+y¯+s+sx¯x,Ry,a+−1sx¯a⋅xRy+−1rx¯+sRxPy+PRxPy+−1rx¯+y¯+sRx,y,aa⋅Rxy+−1sx¯+y¯Px,y,c+−1sx¯+y¯+rx¯x,Py,c+−1r+sx¯+y¯x,y,Pc+−1rx¯+sc⋅xPy+−1rsc⋅Pxy+−1rx¯+y¯+sc⋅x,y,a+−1rsc⋅a⋅xy+Pc⋅xy+−1rx¯+y¯+sc,xy,a+a⋅c⋅xy−−1rs−1r+sx¯xRPy+−1rx¯RxPy+−1sx¯+y¯+r+rx¯x,Py,c+−1rx¯c⋅xPy+−1sx¯+rPxRy+RPxy+−1sx¯+y¯+rPx,y,c+c⋅Pxy+−1rx¯+y¯Rx,y,a+−1rx¯+y¯+sx¯x,Ry,a+−1r+sx¯+y¯x,y,Ra+−1sx¯+ra⋅xRy+−1rsa⋅Rxy+−1sx¯+y¯+ra⋅x,y,c+−1rsa⋅c⋅xy+Ra⋅xy−1sx¯+y¯+ra,xy,c+c⋅a⋅xy=−1r+sx¯x⋅P,Ry+P,Rx⋅y+−1r+sx¯+y¯x,y,Pc−−1rsRa+Pc−−1rsRa⋅xy+−1rx¯+y¯+sc,xy,a−−1sx¯+y¯a,xy,c+−1rx¯+y¯+scx,y,a−−1sx¯+y¯ax,y,c=−1r+sx¯x⋅P,Ry+P,Rx⋅y+−1r+sx¯+y¯x,y,Pc−−1rsRa+Pc−−1rsRa⋅xy−−1r+sx¯+y¯+rsxy,c,a−−1sx¯+y¯a,xy,c−−1r+sx¯+y¯x,y,ac−−1sx¯+y¯ax,y,cbyBS1andBS2=−1r+sx¯x⋅P,Ry+P,Rx⋅y+−1r+sx¯+y¯x,y,Pc−−1rsRa−ac+Pc−−1rsRa−ac⋅xybyBS1−BS3andBS5,and then, we get (10). This ends the proof.
Now, we can prove the following result.Proposition 1.
LetB=B0⊕B1be a Bol superalgebra. Then,PSB,,is a Lie superalgebra.Proof.
The proof follows by Lemma1 and the fact that pSB is a Lie superalgebra since it is straightforward to check that pSB is a subsuperalgebra of the Lie superalgebra EndB.Definition 9.
PSBis called the enlarged Lie superalgebra of pseudo-superderivations of a Bol superalgebraB.LetB=B0⊕B1 be a Bol superalgebra. For any x,y∈B0∪B1, denote by Dx,y, the endomorphism of B defined by Dx,yz≔x,y,z for all z∈B. We have, for any x,y∈B0∪B1, r∈ℤ2, Dx,yBr∈Br+x¯+y¯; that is, Dx,y is a linear map of degree x¯+y¯. Moreover, it comes from BS4 and BS5 that(14)Dx,yu,v,w=Dx,yu,v,w+−1u¯x¯+y¯u,Dx,yv,w+−1x¯+y¯u¯+v¯u,v,Dx,ywDx,yu⋅v=−1u¯x¯+y¯u⋅Dx,yv+Dx,yu⋅v+−1x¯+y¯u¯+v¯u,v,x⋅y+x⋅y⋅u⋅v,for any x,y,u,v in B0∪B1. It follows that Dx,y is pseudo-superderivation of degree x¯+y¯ and companion xy, called inner pseudo-superderivation of B.Now, one can reformulate the definition of a Bol superalgebra in the following manner.Definition 10.
A vector superspaceB=B0⊕B1equipped with a bilinear operationx,y↦x⋅ysatisfyingBi⋅Bj⊆Bi+jand a trilinear operationx,y,z↦x,y,zsatisfyingBi,Bj,Bk⊆Bi+j+k;i,j,k∈ℤ2, is called a Bol superalgebra if the following holds:BS1x⋅y=−−1x¯y¯y⋅x.BS2x,y,z=−−1x¯y¯y,x,z.BS3x,y,z+−1x¯y¯+z¯y,z,x+−1z¯x¯+y¯z,x,y=0, and the endomorphismDx,y:z↦x,y,zis its pseudo-superderivation with a companionxyfor allx,y∈B0∪B1andz∈B0∪B1.LetipSrB,B be the vector space spanned by all inner pseudo-superderivations Dx,yx,y∈B0∪B1andx¯+y¯=r∈ℤ2.We can define naturally aℤ2−gradation by setting ipSB,B=ipS0B,B⊕ipS1B,B. Evidently, ipSB,B is a subsuperalgebra of the Lie superalgebra pSB. Accordingly, IPSB,B can be introduced as the set of all pairs P,c, where P∈ipSB,B and c∈ComP. Evidently, IPSB,B is a subsuperalgebra of the Lie superalgebra PSB.Definition 11.
IPSB,Bis called the enlarged Lie superalgebra of inner pseudo-superderivations.LetB=B0⊕B1 be a Bol superalgebra and H=H0⊕H1 be a subsuperalgebra of PSB such that IPSB,B⊆H. For i∈ℤ2, let ℒiB=Bi⊕Hi and define a new superbracket operation in ℒHB=ℒ0B⊕ℒ1B=B⊕H as follows: for any x,y∈B0∪B1, P,a,Q,b∈H0∪H1:(15)x,y≔Dx,y,xy,(16)P,a,x≔−−1a¯x¯x,P,a≔Px,(17)P,a,Q,b≔P,Q,Pb−−1a¯b¯Qa−ab.Then, we have the following.Proposition 2.
ℒHB,,is a Lie superalgebra.Proof.
It is clear that the operation [,] is supersymmetric. For the Jacobi superidentity, there are many cases to distinguish. First, for allx,y,u in B0∪B1, P,a,Q,b in H0∪H1, we get(18)x,y,P,a+−1x¯y¯+a¯y,P,a,x+−1a¯x¯+y¯P,a,x,yu=x,y,Pu−−1a¯x¯+y¯Px,y,u−−1x¯a¯+y¯+a¯y¯Py,x,u+−1a¯x¯+y¯Px,y,u=−−1a¯x¯+y¯−−1a¯x¯+y¯x,y,Pu+Px,y,u−−1a¯x¯x,Py,u−Px,y,u=0by9,i.e., x,y,P,a+−1x¯y¯+a¯y,P,a,x+−1a¯x¯+y¯P,a,x,y=0.(19)P,a,x,Q,b+−1a¯x¯+b¯x,Q,b,P,a+−1b¯a¯+x¯Q,b,P,a,x=PQx−−1a¯b¯QPx+−1a¯x¯+b¯Qx,P,a−−1x¯b¯Px,Q,b=PQx−−1a¯b¯QPx−PQx+−1a¯b¯QPx=0.
Next, the other cases when the three elements are inB0∪B1 or H0∪H1 follow from HBS3 and the fact that H is a Lie superalgebra.Definition 12.
An enveloping superalgebra of a Bol superalgebraBis a Lie superalgebraℒHBdefined above. TakingH=PSB, we obtain the maximal enveloping superalgebra, and takingH=IPSB,B, we obtain the minimal (standard) enveloping superalgebra.The following result will be used in the last section.Lemma 2.
LetKbe an ideal of a Bol superalgebraB. Then,K=K⊕IPSB,Kis an ideal of the standard enveloping superalgebraL=B⊕IPSB,B.Proof.
It suffices to prove thatK,L⊆K, which is a straightforward computation.


## 4. Killing-Ricci Forms of Bol Superalgebras

The definition of the Killing-Ricci form given here for Bol superalgebras stems from [5], where the Killing-Ricci form for Bol algebras is defined following [21] as the restriction of the Killing form of the standard enveloping Lie algebra of the given Bol algebra to this later. Let B=B0⊕B1 be an n-dimensional Bol superalgebra and LB=B0⊕IPS0B,B⊕B1⊕IPS1B,B≔B⊕IPSB,B its standard enveloping Lie superalgebra. Let α be the Killing form of LB and β, the restriction of α to B×B. From [5], we introduce the following definition.Definition 13.
The formβis called the Killing-Ricci form of the Bol superalgebraB.For anyx,y∈B0∪B1, define the endomorphism Rx,y of the vector superspace B by Rx,yz=−1z¯x¯+y¯z,x,y=−1z¯x¯+y¯Dz,xy for all z∈B0∪B1. It is clear that Rx,y is of degree x¯+y¯.The next result gives an explicit expression ofβ.Theorem 1.
For everyx,y∈B0∪B1,(20)βx,y=strRx,y+−1x¯y¯Ry,x.Proof.
Letei, fi, ui, and vi be bases for B0,B1, IPS0B,B, and IPS1B,B, respectively. It suffices to prove (20) for all elements x,y in the basis, i.e., βei,ej=αei,ej, βei,fj=αei,fj, and βfi,fj=αfi,fj. For these bases, we express the operations of B and ISPB,B, using the tensor notation (i.e., repeated indices imply summation), as follows:(21)Dei,ej=Rijmum,Dei,fj=Sijmvm,Dfi,fj=Tijmum,um,ei=umei=Cmijej,vm,ei=vmei=Hmijfj,um,fi=umfi=Kmijfj,vm,fi=vmfi=Lmijej.
SinceRei,fjB0⊆B1 and Rei,fjB1⊆B0, we have strRei,fj+Rfj,ei=0 and then βei,fj=0=αei,fj (consistency property of α). Hence, it remains to show that βei,ej=αei,ej and βfi,fj=αfi,fj. The identities (15) and (16) imply the following:(22)LeiLejek=ei,ej,ek=ei,Dej,ek=Dek,ejei=RkjmCmitet,LeiLejfk=ei,ej,fk=ei,Dej,fk=−Dej,fkei=−SjkmHmitft,LeiLejum=ei,ej,um=−ei,umej=−Cmjkei,ek=−CmjkRiktut,LeiLejvm=ei,ej,vm=−ei,vmej=−Hmjkei,ek=−HmjkSiktvt.
Thus, from the relationβei,ej=αei,ej=strLeiLej, we get(23)βei,ej=RkjmCmik+SjkmHmik−CmjkRikm+HmjkSikm.
From the other hand, we have(24)Rei,ejek=Dek,eiej=−Dei,ekej=−Rikmumej=−RikmCmjtet,Rei,ejfk=Dfk,eiej=−Dei,fkej=−Sikmvmej=−SikmHmjtft.
By interchangingi and j, we have(25)Rej,eiek=−RjkmCmitet=RkjmCmitet,andRej,eifk=−RjkmHmitft.
Then, we get(26)strRei,ej+−1ei¯ej¯Rej,ei=−RikmCmjk+SikmHmjk+RkjmCmik+RjkmHmik.
From (23) and (26), we obtain(27)βei,ej=strRei,ej+−1ei¯ej¯Rej,ei.
Again, the identities (15) and (16) imply the following:(28)LfiLfjek=fi,fj,ek=−fi,Dek,fj=−Dek,fjfi=−Skjmvmfi=−SkjmLmitet,LfiLfjfk=fi,fj,fk=fi,Dfj,fk=−Dfj,fkfi=−Tjkmumfi=−TjkmKmitft,LfiLfjuk=fi,fj,uk=−fi,ukfj=−Kkjmfi,fm=−KkjmDfi,fm=−KkjmTimsus,LfiLfjvk=fi,fj,vk=fi,vkfj=Lkjmfi,em=−LkjmDem,fi=−LkjmSmisvs.
Again, from the relationβfi,fj=αfi,fj=strLfiLfj, we get(29)βfi,fj=−SkjmLmik+TjkmKmik−KkjmTimk+LkjmSmik.
From the other hand, we get(30)Rfi,fjek=Dek,fifj=Skimvmfj=SkimLmjses,and,Rfi,fjfk=Dfk,fifj=Tkimumfj=TkimKmjsfs.
By interchangingi and j, we get(31)Rfj,fiek=SkjmLmises,andRfj,fifk=TkjmKmisfs=TjkmKmisfs.
Next, we get(32)strRfi,fj+−1fi¯fj¯Rfj,fi=strRfi,fj−Rfj,fi,=SkimLmjk−TkimKmjk−SkjmLmik+TjkmKmik=SmikLkjm−TmikKkjm−SkjmLmik+TjkmKmik,and therefore from (29) and (32), we obtain(33)βfi,fj=strRfi,fj+−1fi¯fj¯Rfj,fi,where by relation (20) is proved.Remark 3.
Recall that ifB,⋅,,,is a (left) Bol algebra, then the Killing-Ricci form onBis defined asβx,y=trRx,y+Ry,xwhich we deduce from the one for (right) Bol algebra [5] whereRx,yz=z,x,y. So, if a Bol superalgebraBis reduced to a Bol algebra,βas in Theorem 1is the Killing-Ricci form of the Bol algebraB.Proposition 3.
LetB=B0⊕B1be a Bol superalgebra with a Killing-Ricci formβ. Then, the following holds:(1)
βx,y=−1x¯y¯βy,x for all x,y∈B0∪B1 (supersymmetry).(2)
βB0,B1=0 (consistence).(3)
βAx,Ay=βx,y, A∈AutB.(4)
βx,y,z,u=−−1z¯x¯+y¯βz,x,y,u, for all x,y,z,u∈B0∪B1.Proof.
We know [6] that if α is a Killing form of any Lie superalgebra, then α satisfies the relations 1−3 of the proposition above. Then, the relations 1−3 follow from the fact that the Killing-Ricci form β is the restriction to B of the Killing form of the standard enveloping superalgebra LB of B. For the relation (4), pick x,y,z,u∈B0∪B1 and denote by α the Killing form of LB. Since LB is a Lie superalgebra, then α satisfies(34)αx,y,z=−−1x¯y¯αy,x,z.
Then,(35)βx,y,z,u=αx,y,z,u=−−1z¯x¯+y¯αz,x,y,u=−−1z¯x¯+y¯βz,x,y,u.
Hence, the proposition is proved.Remark 4.
If we consider the Bol superalgebraBas a Lie supertriple system with the ternary operation [″], that we denote byBS, then the relation(4)in Proposition 3 says that the Killing-Ricci form ofBis an invariance form ofBS.The following result will be used below.Lemma 3.
LetKbe a Killing-Ricci form of a Bol superalgebraB. Then, the following conditions are equivalent:(36)Kx,y,z,u=−−1z¯x¯+y¯Kz,x,y,u,Kx,y,z,u=−−1y¯z¯+u¯Kx,z,u,y,Kx,y,z,u=−1x¯y¯+z¯u¯Ky,x,u,z,for all x,y,z,u∈B0∪B1.Proof.
We know that, in any Lie supertriple system with an invariant form, the relations (39)–(41) are inequivalent [14]. The proof then follows from the fact that the Killing-Ricci form of B is an invariant form of BS.


## 5. Invariant Forms of Bol Superalgebras

In this section, we introduce the concept of invariant forms of Bol superalgebras as generalization of those of Bol algebras and Lie supertriple systems.Definition 14.
An invariant formbof a Bol superalgebraB=B0⊕B1is a supersymmetric bilinear form onBsatisfying the identities(37)bxy,z=−−1x¯y¯by,xz,(38)bx,y,z,u=−−1y¯z¯+u¯bx,z,u,y,for allx,y,z,u∈B0∪B1.Remark 5.
IfBis reduced to a Lie supertriple system (resp., a (left) Bol algebra), thenbis reduced to an invariant form to a Lie supertriple system [14, 15], (resp., a (left) Bol algebra) which can be deduced from one of the (right) Bol algebras [5].Letb be an invariant form of a Bol superalgebra B=B0⊕B1. Then, b is an invariant form of BS, and by Lemma 3, b satisfies the following equivalent conditions:(39)bx,y,z,u=−−1z¯x¯+y¯bz,x,y,u,(40)bx,y,z,u=−−1y¯z¯+u¯bx,z,u,y,(41)bx,y,z,u=−1x¯y¯+z¯u¯by,x,u,z,for all x,y,z,u∈B0∪B1.Definition 15.
Letbbe an invariant form of a Bol superalgebraBandVbe a subset ofB. The orthogonalV⊥ofVwith respect tobis defined byV⊥=x∈B,bx,y=0,∀y∈V. The invariant formbis nondegenerate ifB⊥=0.Lemma 4.
Letbbe an invariant form of a Bol superalgebraB.Then, the following holds:(1)
B+B,B,B⊥=ZB if b is nondegenerate.(2)
IfI is an ideal of B then I⊥ is an ideal of B. In particular, B⊥ is an ideal of B.Proof.
Pick a homogeneous elementu in B+B,B,B⊥. Then, for any x,y,z in B0∪B1, we get bu,xy=0 and bu,x,y,z=0. This implies by (37) and (38) that −1u¯x¯bxu,y=0 and −1z¯x¯+y¯bu,z,x,y=0. As b is nondegenerate, we obtain xu=0 and u,z,x=0 for anyx,z in B0∪B1, and this gives u∈ZB.
Conversely, ifu∈ZB, we have for all x,y,z,v,w in B, bu,xy+z,v,w=bu,xy+bu,z,v,w=0 and u∈B+B,B,B⊥ whence B+B,B,B⊥=ZB.
Now, suppose thatI is an ideal of B that is BI⊆I, B,I,B⊆I; then for any homogeneous elements x,y∈B, u∈I⊥, and v∈I, we get bxu,v=−−1u¯x¯bu,xv=0 and bx,u,y,v=−−1x¯u¯bu,x,y,v=−1x¯u¯+y¯+v¯bu,y,v,x=0 by (21). Then, BI⊥⊆I⊥ and B,I⊥,B⊆I⊥, i.e., I⊥ is an ideal of B.
We can now prove the following result.Theorem 2.
LetB=B0⊕B1be a Bol superalgebra such thatstrDx,yLz=0for allx,y,z∈B. Then, the Killing-Ricci formβofBis nondegenerate if and only if the standard enveloping superalgebraLB=B⊕IPSB,Bis a semisimple Lie superalgebra.Proof.
Letα be the Killing form of the Lie superalgebra LB. Then, for all x,y,z∈B, αDx,y,z=strLDx,yLz=strDx,yLz, that is, strDx,yLz=0 if and only if(42)αDx,y,z=0.
Then, by (42) and the invariance of α, we have for x,y,u,v∈B0∪B1,(43)αDx,y,Du,v=αx,y,u,v,=−−1x¯y¯αy,x,u,v=−1x¯u¯+v¯+y¯αy,u,v,x=−1x¯u¯+v¯+y¯βy,u,v,x.
Thus, ifβ is nondegenerate, the restriction of α to IPSB,B×IPSB,B is nondegenerate, and therefore, α is nondegenerate.
Now by contradiction, suppose thatβ is degenerate. Then, by Lemma 4, B⊥ is a nonzero ideal of B, and therefore, B⊥⊕IPSB,B⊥ is a nonzero ideal of LB by Lemma 2.
By the identities (42) and (43), we obtain(44)αB⊥⊕IPSB,B⊥,B⊕IPSB,B=αB⊥,B+αB⊥,IPSB,B+αIPSB,B⊥,B+αIPSB,B⊥,IPSB,B=αB⊥,B+αB,B⊥,B,B=0.
It follows thatα is nondegenerate, and therefore, LB is nondegenerate, which ends the theorem.



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