# Centroids of Lie Supertriple Systems

Authors: Jianrong Peng; Liangyun Chen; Bing Sun
Journal: Advances in Mathematical Physics
 (2015)

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## Abstract

We derive certain structural results concerning
centroids of Lie supertriple systems. Centroids of the
tensor product of a Lie supertriple system and a unital commutative
associative algebra are studied. Furthermore, the centroid of a
tensor product of a simple Lie supertriple system and a polynomial ring
is partly determined.


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## Body

## 1. Introduction

The notion of Lie triple systems already appeared in Cartan’s work on Riemannian geometry, albeit the formal concept was not defined until 1949 by Jacobson in his study of associative algebras that are closed with respect to triple commutators (cf. [1]). The role played by Lie triple systems in the theory of symmetric spaces is parallel to that of Lie algebras in the theory of Lie groups: the tangent space at every point of a symmetric space has the structure of a Lie triple system.As a natural generalization of Lie triple systems, the concept of Lie supertriple systems was introduced in the study of Yang-Baxter equations (cf. [2]). Lie supertriple systems have proved to be of interest in ternary structures appearing in various models in high energy physics, and many important results on Lie supertriple systems have been obtained (cf. [2–7]). In particular, the authors applied Lie supertriple systems to obtain some new solutions of the Yang-Baxter equation (cf. [2]).The centroidΓ(A) of an algebra A is the set of all linear transformations T∈End(A) such that T(xy)=T(x)y=xT(y) for all x,y∈A. Γ(A) is always a unital associative algebra, and if A is “nice” (e.g., unital, prime, or simple), then Γ(A) is a commutative ring of scalars. Centroids are interesting subjects in mathematics, and there has been a great deal of work concerning them. Benkart and Neher studied centroids of Lie algebras in [8] and Melville investigated centroids of nilpotent Lie algebras in [9]. It turns out that results on the centroids of Lie algebras are a key ingredient in the classification of extended affine Lie algebras. The centroids of Lie triple systems were mentioned by Benito et al. in [10]. Results on centroids of Lie triple system, n-Lie algebras, and Lie triple algebras were developed in [11–13].In this paper we present new results concerning the centroids of Lie supertriple systems and give some conclusions of the tensor product of a Lie supertriple system and a unital commutative associative algebra. Furthermore, we completely determine the centroid of the tensor product of a simple Lie supertriple system and a polynomial ring. The paper is organized as follows. In Section2, we list some notions of rings and Lie supertriple systems. In Section 3, we introduce the notion of centroids of Lie supertriple systems. Then we show some properties of centroids of Lie supertriple systems. In Section 4, we investigate the centroid of the tensor product of a Lie supertriple system T and a unital commutative associative algebra A. As an important theorem in this section, we prove that, for a tensor product T~=T⊗R of a simple Lie supertriple system T and a polynomial ring R=F[x1,…,xn], its centroid is Γ(T~)0-=Γ(T)0-⊗R.


## 2. Preliminaries

Throughout this paper, the base fieldF is assumed to be of arbitrary characteristic unless otherwise stated.Definition 1 (see [14]).
An integral domain is a ring with no left or right zero divisors.Definition 2 (see [14]).
An algebraically closed fieldF is a field such that every f(x)∈F[x] is a product of linear polynomials in F[x].Definition 3 (see [15]).
An idempotent of a ring is an elementa such that aa=a.Definition 4 (see [2]).
A Lie supertriple systemT is a Z2-graded vector space with a ternary product [,,] satisfying the following conditions:(1)
d([x,y,z])≡d(x)+d(y)+d(z) (mod 2),(2)
[x,y,z]=-(-1)d(x)d(y)[y,x,z],(3)
(-1)d(x)d(z)[x,y,z]+(-1)d(x)d(y)[y,z,x]+(-1)d(y)d(z)[z,x,y]=0,(4)
[u,v,[x,y,z]]=[[u,v,x],y,z]+(-1)du+dvdx[x,[u,v,y], z]+(-1)du+dvdx+dy[x,y,[u,v,z]],where x, y, z, u, v are homogeneous elements in T. We denote by d(x) the Z2-graded degree of x, ∀x∈Tα, α∈Z2. In the sequel, whenever the notation d(x) appears, it implies that x is a homogeneous element of degree d(x) in T.Definition 5 (see [16]).
LetL=L0-⊕L1- be a superalgebra whose multiplication is denoted by ,. This implies that [Lα,Lβ]⊆Lα+β for all α,β∈Z2. L is a Lie superalgebra if the multiplication satisfies the following identities:(1)
[a,b]=-(-1)αβb,a;(2)
[a,[b,c]]=[[a,b],c]+(-1)αβ[b,[a,c]],for any a∈Lα, b∈Lβ, c∈L,α,β∈Z2.Remark 6.
(1) LetT be a Lie superalgebra. It is clear that if we introduce a triple product [x,y,z] by [x,y,z]=[[x,y],z] for any x,y,z∈T, then T becomes a Lie supertriple system. So in this sense every Lie superalgebra deduces a Lie supertriple system.
(2) Any Lie supertriple systemT can be considered as a subspace of a Lie superalgebra. In fact, set Ls(T) to be the vector space Ls(T)=InnDer(T)⊕T with InnDer(T)={D∣D=ΣL(x,y)}, where L(x,y)z=[x,y,z], for any x,y∈T. The product in Ls(T) is [x1+D1,x2+D2]=(L(x1,x2)+[D1,D2]) + (D1(x2)-(-1)d(D2)d(x1)D2(x1)) for any x1,x2∈T, D1,D2∈InnDer(T). It is easy to see that Ls(T) is a Lie superalgebra.Definition 7 (see [5]).
A derivation of a Lie supertriple systemT is a homogeneous linear map D:T→T such that (1)Dx,y,z=Dx,y,z+-1dDdxx,Dy,z+-1dDdx+dyx,y,Dz,where x,y,z∈T. We denote by d(f) the Z2-graded degree of a homogeneous linear mapping f of T.Definition 8.
An ideal of a Lie supertriple systemT is a Z2-graded subspace I for which [I,T,T]⊆I. Moreover, if [T,I,I]=0, then I is called an abelian ideal of T.Definition 9.
LetI be a nonempty subset of T. One calls CT(I)=x∈T∣L(x,a)=R(a,x)=0, ∀a∈I the centralizer of I in T, where R(a,x)z=(-1)d(z)(d(a)+d(x))[z,a,x]. In particular, CT(T)={x∈T∣[x,T,T]=0} is the center of T, denoted by C(T).


## 3. Centroids of Lie Supertriple Systems

Definition 10.
LetT be a Lie supertriple system over a field F. The centroid of T is the space of F-linear transformations on T given by Γ(T)=ϕ∈EndF(T)∣ϕ([x,y,z])=[ϕ(x),y,z], ∀x,y,z∈T.By the definition of Lie supertriple system we conclude that ifϕ∈Γ(T) and x,y,z∈T, then(2)ϕx,y,z=ϕx,y,z=-1dϕdxx,ϕy,z=-1dϕdx+dyx,y,ϕz.It is clear that the scalar maps will always be in the centroid.Definition 11.
Letϕ∈EndF(T). If ϕ(T)⊆C(T) and ϕ([T,T,T])=0, then ϕ is called a central derivation. The set of all central derivations of T is denoted by ZDer(T).Proposition 12.
Z
D
e
r
(
T
) is an ideal of Γ(T).Proof.
It is easily seen thatZDer(T)⊆Γ(T). For all ϕ∈ZDer(T) and ψ∈Γ(T), we have (3)ϕ,ψx,T,T=ϕψx--1dϕdψψϕx,T,T=ϕψx,T,T--1dϕdψψϕx,T,T=0,ϕ,ψT,T,T=ϕψT,T,T--1dϕdψψϕT,T,T=ϕψT,T,T--1dϕdψψϕT,T,T=0.Hence ZDer(T) is an ideal of Γ(T).Proposition 13.
I
n
n
D
e
r
(
T
) is the subalgebra of EndF(T) and Γ(T) is the centralizer of InnDer(T).Proof.
For anyϕ∈Γ(T) and x,y,z∈T, we have (4)ϕ,Lx,yz=ϕLx,yz--1dϕdLx,yLx,yϕz=ϕx,y,z--1dϕdx+dyLx,yϕz=0,which proves the proposition.IfT is perfect (i.e., T=T(1)=[T,T,T]), then the centroid Γ(T) is supercommutative. Indeed, for all x,y,z∈T,ϕ,ψ∈Γ(T), we have (5)ϕψx,y,z=-1dψ+dxdϕψx,ϕy,z=-1dψ+dxdϕψx,ϕy,z=-1dϕdψψϕx,y,z.Proposition 14.
IfT has no nonzero ideals I,J with [T,I,J]=0 (i.e., T is prime) and C(T)=0, then Γ(T)0- is an integral domain.Proof.
Clearly,id∈Γ(T)0-. If there exist ψ≠0 and ϕ≠0 such that ψφ=0, then there are x,y∈T such that ψ(x)≠0 and ϕ(y)≠0. Then(6)z,ψx,ϕy=-1dψdz-1dz+dxdϕψϕz,x,y=0,∀z∈T.Since C(T)=0, [ϕ(x),T,T]≠0; that is, ϕ[x,T,T]≠0. Therefore, ψ(x) generates a nonzero ideal ψ(Ix)=ψ[[⋯[[x,T,T],T,T]⋯],T,T]. In a similar way, ϕ(y) generates a nonzero ideal ψ(Iy)=ψ[[⋯[[y,T,T],T,T]⋯],T,T] such that [T,ψ(Ix),ϕ(Iy)]=ψϕ[T,Ix,Iy]=0, a contradiction. Hence Γ(T)0- has no zero divisors, and so it is an integral domain.Theorem 15.
IfT is a simple Lie supertriple system over an algebraically closed field F (i.e., T(1)≠0) and T has only two ideals T and 0, then Γ(T)=Fid.Proof.
Letϕ∈Γ(T)⊆EndF(T). Since F is algebraically closed, ϕ has an eigenvalue λ. We denote the corresponding eigenspace by Eλ(ϕ). So Eλ(ϕ)≠0. For any x∈Eλ(ϕ), x=x0+x1, where x0∈T0, x1∈T1, we have ϕ(x)=ϕ(x0)+ϕ(x1)=λx=λx0+λx1, so ϕ(x0)-λx0=ϕ(x1)-λx1∈T0∩T1; that is, ϕ(x0)=λx0, ϕ(x1)=λx1. So we have x0∈T0 and x1∈T1. Therefore Eλ(ϕ) is a Z2-graded vector space. For any v∈Eλ(ϕ),x,y∈T, we have ϕ([v,x,y])=[ϕ(v),x,y]=λ[v,x,y], so [v,x,y]∈Eλ(ϕ). It follows that Eλ(ϕ) is an ideal of T. But T is simple, so Eλ(ϕ)=T; that is, ϕ=λidT. This proves the theorem.WhenΓ(T)=Fid, the Lie supertriple system T is said to be central. Furthermore, if T is simple, T is said to be central simple. Every simple Lie supertriple system is central simple over its centroid.Proposition 16.
LetT be a Lie supertriple system over a field F. Then(1)
T is indecomposable (cannot be written as the direct sum of two nontrivial ideals) if and only if Γ(T) does not contain idempotents except 0 and id;(2)
ifT is perfect, every ϕ∈Γ(T) is supersymmetric with respect to any invariant form on T.Proof.
Consider the following:
(1) Ifϕ∈Γ(T) is an idempotent and satisfies ϕ≠0,id, then ϕ2(x)=ϕ(x), ∀x∈T. We can see that Ker⁡ϕ and Im⁡ϕ are ideals of T. In fact, for any x∈Ker⁡ϕ and y,z∈T, we have x=x0+x1, where x0∈T0 and x1∈T1; then ϕ(x0)∈T0, ϕ(x1)∈T1 and ϕ(x)=ϕ(x0)+ϕ(x1)=0, ϕ(x0)=ϕ(x1)=0, so x0∈Ker⁡ϕ and x1∈Ker⁡ϕ; that is, Ker⁡ϕ is a Z2-graded vector space. ϕ([x,y,z])=[ϕ(x),y,z]=0 implies [x,y,z]∈Ker⁡ϕ. For any x∈Im⁡ϕ, we have x=x0+x1, where x0∈T0 and x1∈T1; there exists a∈T such that x=ϕ(a), a=a0+a1, a0∈T0, a1∈T1, so x=ϕ(a0)+ϕ(a1); that is, x0-ϕ(a0)=x1-ϕ(a1)∈T0∩T1; then x0=ϕ(a0) and x1=ϕ(a1); that is, Im⁡ϕ is a Z2-graded vector space. For any y,z∈Im⁡ϕ, we have [x,y,z]=[ϕ(a),y,z]=ϕ([a,y,z])∈Im⁡ϕ, ∀y,z∈T. Moreover, Ker⁡ϕ∩Im⁡ϕ=0. Indeed, if x∈Ker⁡ϕ∩Im⁡ϕ, then there exists y∈T such that x=ϕ(y) and 0=ϕ(x)=ϕ2(y)=ϕ(y)=x. We have a decomposition x=ϕ(x)+y, ∀x∈T, where ϕ(y)=0. So we have T=Ker⁡ϕ⊕Im⁡ϕ, a contradiction.
On the other hand, supposeT has a decomposition T=T1⊕T2. Then for any x∈T, we have x=x1+x2,xi∈Ti, i=1,2. We choose ϕ∈Γ(T) such that ϕ(x1)=x1 and ϕ(x2)=0. Then ϕ2(x)=ϕ(x1)=x1=ϕ(x). Hence ϕ is an idempotent. By assumption, we know ϕ=0 or ϕ=id. If ϕ=0, then x1=0 implying T=T2. If ϕ=id, then x2=0 implying T=T1.
(2) Letf be an invariant bilinear form on T. Then (7)fx,y,u,v=-1dudv+dudy+dvdyfx,v,u,y,∀x,y,u,v∈T.
SinceT is perfect, for ϕ∈Γ(T), we have (8)fϕx,y,u,v=-1dxdϕfx,ϕy,u,v=-1dxdϕ-1dϕ+dydv+dϕ+dydu+dvdu·fx,v,u,ϕy=-1dxdϕ+dϕ+dydv+dϕ+dydu+dvdu·-1dϕdu+dvfx,ϕv,u,y=-1dxdϕ+dϕ+dydv+dϕ+dydu+dvdu·-1dϕdu+dvfx,ϕv,u,y=-1dxdϕ+dϕ+dydv+dϕ+dydu+dvdu·-1dϕdu+dv·-1dϕ+dvdu+dϕ+dvdy+dydu·fx,y,u,ϕv=-1dϕdx+dy+du·fx,y,u,ϕv,which proves the proposition.Proposition 17.
LetT be a Lie supertriple system over a field F and I a subset of T. Then CT(I) is invariant under Γ(T) and so is any perfect ideal of T.Proof.
For anyϕ∈Γ(T), x∈CT, y∈I, and z∈T, we have(9)Lϕx,yz=ϕx,y,z=ϕx,y,z=0,Ry,ϕxz=-1dzdϕ+dx+dyz,y,ϕx=-1dzdϕ+dx+dy-1dϕdz+dyϕz,y,x=0,which implies that ϕ(x)∈CT(I). So CT(I) is invariant under Γ(T).
LetJ be any perfect ideal of T; then J=[J,J,J]. For any y∈J, there exist a,b,c∈J such that y=[a,b,c] and then we have ϕ(y)=ϕ([a,b,c])=(-1)d(a)d(ϕ)[a,ϕ(b),c]∈[J,T,T]⊆J. Hence J is invariant under Γ(T).Theorem 18.
Letπ:T1→T2 be an epimorphism of Lie supertriple systems. For any f∈EndF(T1;Kerπ):={g∈EndF(T2)∣g(Kerπ)⊆Kerπ} there exists a unique f-∈EndF(T2) satisfying π∘f=f-∘π. Moreover, the following results hold:(1)
The mapπEnd:EndF(T1;Kerπ)→EndF(T2), f↦f- is an algebra homomorphism satisfying the following properties: (10)πEndMultT1=MultT2,πEndΓT1∩EndFT1;Kerπ⊆ΓT2.By restriction, there is an algebra homomorphism(11)πΓ:ΓT1∩EndFT1;Kerπ⟶ΓT2,f⟼f-.IfKerπ=C(T1), then every ϕ∈Γ(T1) leaves Kerπ invariant; hence πΓ is defined on all of Γ(T1).(2)
SupposeT1 is perfect and Kerπ⊆C(T1). Then (12)πΓ:ΓT1∩EndFT1;Kerπ⟶ΓT2,f⟼f-is injective.(3)
IfT1 is perfect, C(T2)=0, and Kerπ⊆C(T1), then πΓ:Γ(T1)→Γ(T2) is an algebra monomorphism.Proof.
Consider the following:
(1) It is easy to see thatπEnd is an algebra homomorphism. Since Ker⁡π is an ideal of T1 and all left and right multiplication operators of T1 leave Ker⁡π invariant, Mult(T1)⊆EndF(T1;Ker⁡π). Furthermore, for the left multiplication operator L(x,y) on T1 we have π∘L(x,y)=L(π(x),π(y))∘π, so πEnd(L(x,y))=L(π(x),π(y)). For the right multiplication we have the analogous formula πEnd(R(x,y))=R(π(x),π(y)). Moreover, π is an epimorphism, so πEnd(Mult(T1))=Mult(T2). Now we show that πEnd(Γ(T1)∩EndF(T1;Ker⁡π))⊆Γ(T2). Let ϕ∈Γ(T1)∩EndF(T1;Ker⁡π). For any x′,y′,z′∈T2 there exist x,y,z∈T1 such that π(x)=x′, π(y)=y′, and π(z)=z′. Then we have (13)ϕ-x′,y′,z′=ϕ-πx,y,z=πϕx,y,z=πx,y,ϕz=πx,πy,ϕ-πz=x′,y′,ϕ-z′,which proves ϕ-∈Γ(T2).
(2) Ifϕ-=0 for ϕ∈Γ(T1)∩EndF(T1;Ker⁡π), then π(ϕ(T1))=ϕ-(π(T1))=0, which means that ϕ(T1)⊆Ker⁡π⊆C(T1). Hence ϕ([x,y,z])=[ϕ(x),y,z]=0, ∀x,y,z∈T1. Furthermore, since T1=T1(1), we can get ϕ=0.
(3) We can see thatπ(C(T1))⊆C(T2)=0 and then C(T1)⊆Ker⁡π. So Ker⁡π⊆C(T1). By (1), we know that πΓ:Γ(T1)→Γ(T2) is a well-defined algebra homomorphism, which is an injection by (2).Proposition 19.
If the characteristic ofF is not 2, then (14)ZDerT=ΓT∩DerT.Proof.
Ifψ∈Γ(T)∩Der(T), then by the definitions of Γ(T) and Der(T), for all x,y,z∈T, we have ψ([x,y,z])=[ψ(x),y,z]+(-1)d(ψ)d(x)[x,ψ(y),z]+(-1)d(ψ)(d(x)+d(y))[x,y,ψ(z)], and ψ([x,y,z])=[ψ(x),y,z]=(-1)d(ψ)d(x)[x,ψ(y),z]=(-1)d(ψ)(d(x)+d(y))[x,y,ψ(z)], so ψ([T,T,T])=0 and ψ(T)⊆C(T). It follows easily that Γ(T)∩Der(T)⊆ZDer(T).
To show the inverse inclusion, letψ∈ZDerT; then 0=ψ([x,y,z])=[ψ(x),y,z]=(-1)d(ψ)d(x)[x,ψ(y),z]=(-1)d(ψ)(d(x)+d(y))[x,y,ψ(z)]. Thus, ψ∈Γ(T)∩Der(T).Lemma 20.
LetI be a nonzero Γ(T)-invariant ideal of T, V(I)={ψ∈Γ(T)∣ψ(I)=0}, and Hom(T/I,CT(I)) the vector space of all linear maps from T/I to CT(I) over F. Define T(I)=f∈Hom(T/I,CT(I))∣f([x-,y-,z-])=[f(x-),y,z]=(-1)d(f)d(x)[x,f(y-),z]=(-1)d(f)(d(x)+d(y))[x,y,f(z-)], for x-,y-,z-∈T/I. Then(1)
T(I) is a subspace of Hom(T/I,CT(I)) and V(I)≅T(I) as vector spaces;(2)
ifΓ(I)=FidI,Γ(T)=FidT⊕V(I) as vector spaces.Proof.
(1) For allf=f0+f1∈T(I), f0∈Hom(T/I,CT(I))0, and f1∈Hom(T/I,CT(I))1, we have (15)fx-,y-,z-=f0+f1x-,y-,z-=fx-,y,z=f0+f1x-,y,z.Hence f0([x-,y-,z-])=[f0(x-),y,z] and f1([x-,y-,z-])=[f1(x-),y,z]; that is, T(I) is a Z2-graded subspace of Hom(T/I,CT(I)). For all ψ∈V(I), ψ=ψ0+ψ1, where ψ0∈Γ(T)0 and ψ1∈Γ(T)1. Also, ψ(x)=ψ0(x)+ψ1(x)=0, ∀x∈I; then ψ0(x)=ψ1(x)=0; that is, V(I) is a Z2-graded vector space. Since ψϕ(I)=ψ(ϕ(I))=0=ϕ(ψ(I))=ϕψ(I), ∀ϕ∈Γ(T), we have ψϕ, ϕψ∈V(I). So V(I) is an ideal of the associative algebra Γ(T). To prove (1) consider the following map α:V(I)→T(I) given by (16)αψy-=ψy,where ψ∈V(I) and y-=y+I∈T/I. The map α is well defined. If y-=y-1, then y-y1∈I, and so ψ(y-y1)=0. It follows easily that α is injective. We now show that α is onto. For every f∈T(I), set ψf:T→T, ψf(x)=f(x-), for all x∈T. It follows from the definition of T(I) that, for all x,y,z∈T, ψf([x,y,z])=f([x-,y-,z-])=[f(x-),y,z]=(-1)d(f)d(x)[x,f(y-),z]=(-1)d(f)(d(x)+d(y))[x,y,f(z-)]. Thus ψf∈Γ(T), and so ψf∈V(I),ψf(I)=0. But α(ψf)=f implies that α is onto. It is easy to see that α preserves operations on vector spaces T/I to CT(I). This proves (1).
We now prove(2). If Γ(I)=FidI, then, for all ψ∈Γ(T),ψ|I=λidI, for some λ∈F. If ψ≠λidT, let φ(x)=λx, for all x∈T; then ψ∈Γ(T) and ψ-φ∈V(I). Clearly, ψ=φ+(ψ-φ). Furthermore, FidT∩V(I)=0, and so (2) is proved.Lemma 21.
LetT be a Lie supertriple system; then φD is a derivation for φ∈Γ(T) and D∈Der(T).Proof.
Ifx,y,z∈T, then(17)φDx,y,z=Dx,y,z+-1dDdx·x,Dy,z+-1dDdx+dyx,y,Dz=φDx,y,z+-1dDdxφx,Dy,z+-1dDdx+dyφx,y,Dz=φDx,y,z+-1dDdx-1dφdxx,φDy,z+-1dDdx+dy-1dφdx+dyx,y,φDz=φDx,y,z+-1dD+dφdxx,φDy,z+-1dD+dφdx+dy-1dφdx+dy·x,y,φDz.Thus φD is a derivation.Theorem 22.
LetT be a Lie supertriple system. Then for all φ∈Γ(T) and D∈Der(T),(1)
Der(g) is contained in the normalizer of Γ(g) in Endg;(2)
Dφ is contained in Γ(T) if and only if φD is a central derivation of T;(3)
φD is a derivation of T if and only if [D,φ] is a central derivation of T.Proof.
(1) For anyφ∈Γ(T), D∈Der(T), and x,y,z∈T, (18)Dφx,y,z=Dφx,y,z=Dφx,y,z+-1dφ+dxdDφx,Dy,z+-1dφ+dx+dydDφx,y,Dz=Dφx,y,z+-1dφ+dxdD-1dDdx·φDx,y,z+-1dφ+dx+dydD·-1dDdx+dyφDx,y,z=Dφx,y,z+-1dφdDφDx,y,z--1dφdDφDx,y,z.So (Dφ-(-1)d(D)d(φ)φD)([x,y,z])=[(Dφ-(-1)d(D)d(φ)φD)(x),y,z]; this is [D,φ]=Dφ-(-1)d(D)d(φ)φD∈Γ(T). This proves (1).
(2) From Lemma21 and (1), Dφ is an element of Γ(T) if and only if φD∈Γ(T)∩Der(T). Thanks to Proposition 19, we get result (2).
(3) It follows from(1), Proposition 19, and Lemma 21 that (3) holds.Now, we study the relationship between the centroid of a decomposable Lie supertriple system and the centroid of its factors.Theorem 23.
Suppose thatT is a Lie supertriple system over F and T=T1⊕T2 with T1,T2 being ideals of T; then(1)
Γ(T)≅Γ(T1)⊕Γ(T2)⊕C(1)⊕C(2), as vector spaces, where (19)C1=φ∈HomT1,T2∣φT1⊆CT2,φT1,T1,T1=0,C2=φ∈HomT2,T1∣φT2⊆CT1,φT2,T2,T2=0;(2)
ZDer(T)≅C(1)⊕C(2)⊕ZDer(T1)⊕ZDer(T2).Proof.
(1) Lettingπi:T→Ti be canonical projections for i=1,2, then π1,π2∈Γ(T) and π1+π2=idT. So we have, for φ∈Γ(T), (20)φ=π1φπ1+π1φπ2+π2φπ1+π2φπ2.Note that πiφπj∈Γ(T) for i,j=1,2. We claim that (21)ΓT=π1ΓTπ1+π1ΓTπ2+π2ΓTπ1+π2ΓTπ2.It suffices to show that π1Γ(T)π1∩π1Γ(T)π2=0 (other cases are similar). For any φ∈π1Γ(T)π1∩π1Γ(T)π2, there exist fi∈Γ(T), i=1,2, such that φ=π1f1π1=π1f2π2. Then φ(x)=π1f2π2(x)=π1f2π2(π2(x))=π1f1(0)=0, for all x∈T, and so φ=0. Let Γ(T)ij=πiΓ(T)πj, i,j=1,2. We now prove (22)ΓT11≅ΓT1,ΓT22≅ΓT2,ΓT12≅C2,ΓT21≅C1.Since φ(T2)=0 for φ∈Γ(T)11, we have φ|T1∈Γ(T1). On the other hand, one can regard Γ(T1) as a subalgebra of Γ(T) by extending any φ0∈Γ(T1) on T2 being equal to zero; that is, φ0(x1)=φ0(x1) and φ0(x2)=0, for all x1∈T1 and x2∈T2. Then φ0∈Γ(T) and φ0∈Γ(T)11. Therefore Γ(T)11≅Γ(T1) with isomorphism σ:Γ(T)11→Γ(T1), σ(φ)=φ|T1 for all φ∈Γ(T)11. Similarly, we have Γ(T)22≅Γ(T2).
Next, we proveΓ(T)12≅C(2). If φ∈Γ(T)12, then there exists φ0 in Γ(T) such that φ=π1φ0π2. For xk=xk1+xk2∈T, where xki∈Ti, i=1,2 and k=1,2,3, we have (23)φx1,x2,x3=π1φ0π2x1,x2,x3=π1φ0x12,x22,x32=π1φ0x12,x22,x32=0and [φ(x1),x2,x3]=φ([x1,x2,x3])=0. Then φ(T)⊆C(T) and φ[T,T,T]=0. It follows that φ|T2(T2)⊆Z(T1),φ|T2([T2,T2,T2])=0 and so φ|T2∈C(2).
Conversely, forφ∈C(2), expanding φ on T (also denoted by φ) by φ[T1,T1,T1]=0, we have π1φπ2=φ and φ∈Γ(T)12. This proves that Γ(T)12 is isomorphic to C(2) with the following isomorphism: (24)τ:ΓT12⟶C2,τφ=φΓT12,for all φ∈Γ(T)12. Similarly, we can prove Γ(T)21≅C(1). Summarizing the above discussion we get (25)ΓT≅ΓT1⊕ΓT2⊕C1⊕C2.This proves (1).
(2) Clearly,C(1)+C(2)+ZDer(T1)+ZDer(T2)⊂ZDer(T). Letting πi:T→Ti be canonical projections for i=1,2, then π1,π2∈Γ(T) and π1+π2=idT. So we have, for φ∈ZDer(T), φ=π1φπ1+π1φπ2+π2φπ1+π2φπ2. Note that πiφπj∈ZDer(T) for i,j=1,2. We claim that ZDer(T)=π1ZDer(T)π1+π1ZDer(T)π2+π2ZDer(T)π1+π2ZDer(T)π2. Let ZDer(T)ij=πiZDer(T)πj, i,j=1,2; we now have ZDer(T)11≅ZDer(T1), ZDer(T)22≅ZDer(T2), ZDer(T)12≅C(2), and ZDer(T)21≅C(1). In the same way as (1), we get (2).A generalized version of the above theorem is stated below without proof.Theorem 24.
Suppose thatT is a Lie supertriple system over F with a decomposition of ideals T=T1⊕T2⊕⋯⊕Tm. Then we have(1)
Γ(T)≅Γ(T1)⊕⋯⊕Γ(Tm)⊕(⊕1≤i≠j≤mCij) as vector spaces, where Cij={φ∈Hom(Ti,Tj)∣φ(Ti)⊆C(Tj),φ[Ti,Ti,Ti]=0, for 1≤i≠j≤m};(2)
ZDer(T)≅ZDer(T1)⊕⋯⊕ZDer(Tm)⊕(⊕1≤i≠j≤mCij).


## 4. Centroid of Tensor Product

Benkart and Neher investigated the centroid of the tensor product of associative algebras and Lie algebras in [8]. In this section we discuss the centroid of the tensor product of a Lie supertriple system and a unital commutative associative algebra. Furthermore, we completely determine the centroid of the tensor product T⊗R of a simple Lie supertriple system T and a polynomial ring R.Definition 25.
LetA be an associative algebra over a field F. The centroid of A is the space of F-linear transformations on A given by Γ(A)={χ∈EndF(A)∣χ(ab)=aχ(b)=χ(a)b,∀a,b∈A}.Proposition 26.
LetT be a Lie supertriple system over F and A a unital commutative associative algebra over F. Then there exists a unique Lie supertriple system structure on T⊗A satisfying [x⊗a1,y⊗a2,z⊗a3]=[x,y,z]⊗a1a2a3, ∀x,y,z∈T, ai∈A, i=1,2,3.Proof.
Foru,v,x,y,z∈T, ai∈A, i=1,2,3,4,5, we have (26)dx⊗a1,y⊗a2,z⊗a3≡dx+dy+dzmod⁡2,x⊗a1,y⊗a2,z⊗a3=x,y,z⊗a1a2a3=--1dxdyy,x,z⊗a2a1a3=--1dxdy·y⊗a2,x⊗a1,z⊗a3=--1dx⊗a1dy⊗a2y⊗a2,x⊗a1,z⊗a3,-1dx⊗a1dz⊗a3x⊗a1,y⊗a2,z⊗a3+-1dx⊗a1dy⊗a2y⊗a2,z⊗a3,x⊗a1+-1dy⊗a2dz⊗a3z⊗a3,x⊗a1,y⊗a2=-1dx⊗a1dz⊗a3x,y,z⊗a1a2a3+-1dx⊗a1dy⊗a2y,z,x⊗a2a3a1+-1dy⊗a2dz⊗a3z,x,y⊗a3a1a2=-1dxdzx,y,z+-1dxdyy,z,x+-1dydzz,x,y⊗a1a2a3=0⊗a1a2a3=0,u⊗a1,v⊗a2,x⊗a3,y⊗a4,z⊗a5=u,v,x,y,z⊗a1a2a3a4a5=u,v,x,y,z+-1du+dvdxx,u,v,y,z+-1du+dvdx+dyx,y,u,v,z⊗a1a2a3a4a5=u,v,x,y,z⊗a1a2a3a4a5+-1du+dvdxx,u,v,y,z⊗a1a2a3a4a5=u⊗a1,v⊗a2,x⊗a3,y⊗a4,z⊗a5+-1du⊗a1+dv⊗a2dx⊗a3x⊗a3,u⊗a1,v⊗a2,y⊗a4,z⊗a5+-1du⊗a1+dv⊗a2dx⊗a3+dy⊗a4x⊗a3,y⊗a4,u⊗a1,v⊗a2,z⊗a5.Thus T⊗A is a Lie supertriple system.It is easy to show that ifT is perfect, then T⊗A is perfect too. Also, for f∈EndF(T) and g∈EndF(A) there exists a unique map f⊗~g∈EndF(T⊗A) such that (27)f⊗~gx⊗a=fx⊗ga,∀x∈T,a∈A.The map should not be confused with the element f⊗g of the tensor product EndF(T)⊗EndF(A). Of course, we have a canonical map (28)ω:EndFT⊗EndFA⟶EndFT⊗A,f⊗g⟼f⊗~g.It is easy to see that if ϕ∈Γ(T) and ψ∈Γ(A), then ϕ⊗~ψ∈Γ(T⊗A). Hence Γ(T)⊗~Γ(A)⊆Γ(T⊗A), where Γ(T)⊗~Γ(A) is the F-span of all endomorphisms ϕ⊗~ψ.Definition 27.
The transformationϕ∈Γ(T⊗A) has finite T-image if, for any a∈A, there exist finitely many a1,…,an∈A such that (29)ϕT⊗Fa⊆T⊗Fa1+⋯+T⊗Fan.It is easy to see thatΓ(T)⊗~Γ(A)⊆{ϕ∈Γ(T⊗A)∣ϕ has finite T-image}. In addition, if T is perfect, ϕ∈Γ(T⊗A) has finite T-image as soon as ϕ(T⊗1)⊆T⊗Fa1+⋯+T⊗Fan for suitable ai∈A.Lemma 28.
LetT be a Lie supertriple system over F and A a unital commutative associative F-algebra. Let {ar}r∈R be a basis of A and ϕ∈Γ(T⊗A). One defines ϕr∈EndF(T) by (30)ϕx⊗1=∑r∈Rϕrx⊗ar.Then all ϕr∈Γ(T).Proof.
Forx1,x2,x3∈T, we have ϕ([x1,x2,x3]⊗1)=∑r∈R‍ϕr([x1,x2,x3])⊗ar and (31)ϕx1,x2,x3⊗1=ϕx1⊗1,x2⊗1,x3⊗1=ϕx1⊗1,x2⊗1,x3⊗1=∑r∈Rϕrx1,x2,x3⊗ar.Hence ϕr([x1,x2,x3])=[ϕr(x1),x2,x3]=[x1,ϕr(x2),x3]=[x1,x2,ϕr(x3)] for all xi∈T, so all ϕr∈Γ(T).Proposition 29.
LetT be a perfect Lie supertriple system over F and A a unital commutative associative F-algebra. Then,(1)
ifT is finite generated as an InnDer(T)-module (or as a Γ(T)-module) or T is central, every ϕ∈Γ(T⊗A) has finite T-image;(2)
if the mapω in Definition 25 is injective, Γ(T)⊗~Γ(A)=ϕ∈Γ(T⊗A)∣ϕhas finite T-image.Proof.
(1) SetI=InnDer(T). Since A is unital, I⊗~id⊆InnDer(T⊗A). Since T is finite generated as an InnDer(T)-module, we can suppose T=Ix1+⋯+Ixn=I1x1+⋯+Inxn for x1,…,xn∈T, I1,…,In∈I. Fix ϕ∈Γ(T⊗A) and a∈A. There exist finite families {xij}⊆T and {aij}⊆A such that ϕ(xi⊗a)=∑j‍xij⊗aij for 1≤i≤n. Hence (32)ϕT⊗a=∑iϕIi⊗~idxi⊗a=-1dϕdIi⊗~id∑iIi⊗~idϕxi⊗a=-1dϕdIi⊗~id∑i,jIi⊗~idxij⊗aij⊆∑i,jT⊗aij.Since T⊗A is perfect, the centroid Γ(T⊗A) is supercommutative. Replacing I by Γ(T) we can use the same argument as above to show that every ϕ∈Γ(T⊗A) has finite T-image if T is a finitely generated Γ(T)-module.
Now, supposeΓ(T)=Fid and T is a torsion-free F-module. Then there exist scalars kr∈F such that ϕr=krid. Hence ϕ(x⊗1)=∑r∈R‍krx⊗ar. Fix x∈T; then almost all krx=0, so almost all kr=0, which in turn implies that ϕ has finite T-image.
(2) In the discussion above, we getΓ(T)⊗~Γ(A)⊆ϕ∈Γ(T⊗A)∣ϕ has finite T-image. So it suffices to prove Γ(T)⊗~Γ(A)⊇ϕ∈Γ(T⊗A)∣ϕ has finite T-image. We suppose that ϕ∈Γ(T⊗A) has finite T-image; then there exists a finite subset J⊆R such that ϕ(x⊗1)=∑r∈J‍ϕr(x)⊗ar. For x1,x2,x3∈T and a∈A, we get (33)ϕx1,x2,x3⊗a=ϕx1⊗1,x2⊗1,x3⊗a=ϕx1⊗1,x2⊗1,x3⊗a=∑r∈Jϕrx1,x2,x3⊗ara=∑r∈Jϕrx1,x2,x3⊗ara.Since T is perfect, we can get ϕ=∑r∈J‍ϕr⊗~λr, where λr is the left multiplication in A by ar. Let {ψs∣s∈S} be an F-basis of Γ(T). Then there exist a finite subset T⊆S and scalars krs∈F(r∈J,s∈T) such that ϕr=∑s∈T‍krsψs. We then get ϕ=∑r∈J‍∑s∈T‍krsψs⊗~λr=∑s∈T‍ψs⊗~ϕs for ϕs=∑r∈J‍krsλr. Now we show that ϕs∈Γ(A). For any x1,x2,x3∈T,a1,a2∈A, we have (34)∑s∈Tψsx1,x2,x3⊗ϕsa1a2=ϕx1,x2,x3⊗a1a2=ϕx1⊗a1,x2⊗a2,x3⊗1=∑s∈Tψsx1⊗ϕsa1,x2⊗a2,x3⊗1=∑s∈Tψsx1,x2,x3⊗ϕsa1a2=∑s∈Tψsx1,x2,x3⊗ϕsa1a2.Because T(1)=T, we can get ∑s∈T‍ψs(x)⊗(ϕs(a1a)-ϕs(a1)a)=0, ∀x∈T, and a1,a2∈A; it follows that ∑s∈T‍ψs⊗~μs=0 where μs∈EndF(A) is defined by μs(a2)=ϕs(a1a2)-ϕs(a1)a2, ∀a2∈A. Since ω is injective, we also have ∑s∈T‍ψs⊗μs=0. So by the linear independence of the ψs, we get that μs=0. Then ψs∈Γ(A). Hence ϕ∈Γ(T)⊗~Γ(A).Next we will determine the centroid of the tensor product of a simple Lie supertriple system and a polynomial ring.LetT be a Lie supertriple system and R=F[x1,…,xn]. Defining the multiplication by [x⊗p,y⊗q,z⊗r]=[x,y,z]⊗pqr, we can make T~=T⊗R into a Lie supertriple system.Now we write out a basis ofΓ(T) in the following manner. First, let E denote the set of central derivations and choose a basis {ek} for E. Second, let {ψj} be a maximal subset of Γ(T) with the restrictions {ψj|[T,T,T]} being linearly independent and Ψ the space spanned by ψj. Then Γ(T)=Ψ+E; {ek} and {ψj} form a basis of Γ(T).Lemma 30.
One hasΓ(T~)⊇Ψ⊗R+E⊗EndF(R).Proof.
Letψ∈Γ(T),m∈R. Then we have (ψ⊗m)([x⊗p,y⊗q,z⊗r])=(ψ⊗m)([x,y,z]⊗pqr)=ψ([x,y,z])⊗mpqr=[ψ(x),y,z]⊗mpqr and [(ψ⊗m)(x⊗p),y⊗q,z⊗r]=[ψ(x)⊗mp,y⊗q,z⊗r]=[ψ(x),y,z]⊗mpqr. Thus ψ⊗m∈Γ(T~).
For anyd∈E and f∈EndF(R),d⊗f∈Γ(T~) since (d⊗f)([x⊗p,y⊗q,z⊗r])=d([x,y,z])⊗f(pqr)=0 and [(d⊗f)(x⊗p),y⊗q,z⊗r]=[d(x)⊗f(p),y⊗q,z⊗r]=[d(x),y,z]⊗f(p)qr=0. Hence Ψ⊗R+E⊗EndF(R)⊆Γ(T~).Theorem 31.
LettingT be a simple Lie supertriple system, R=F[x1,…,xn] and T~=T⊗R. Then Γ(T~)0-=Γ(T)0-⊗R.Proof.
From Lemma30, we get Γ(T)⊗R⊆Γ(T~), so Γ(T)0-⊗R⊆Γ(T~)0-. Now we prove the opposite inclusion.
Let{mi} be a basis for R. Let ψ∈Γ(T~)0-,p∈R,x,y,z∈T and ηi(-,p) be suitable maps in EndF(T) such that we may write (35)ψx⊗p=∑iηix,p⊗mi.Then(36)ψx⊗p,y⊗1,z⊗1=ψx⊗p,y⊗1,z⊗1=∑iηix,p⊗mi,y⊗1,z⊗1=∑iηix,p,y,z⊗mi,while(37)ψx,y,z⊗p=∑iηix,y,z,p⊗mi.So, for each i and p, we have ηi(p,[x,y,z])=[ηi(x,p),y,z], ∀x,y,z∈T. Hence ηi(-,p)∈Γ(T). But Γ(T)=Fid. Therefore, ηi(x,p)=λi(p)x, ∀x∈T, for suitable scalars λi(p).
Thus, now we may writeψ(x⊗p)=∑i‍λi(p)x⊗mi. Since the right hand side is in T⊗R, for each p, we get λi(p)=0 for all except a finite number of i. That is, ψ(x⊗p)=∑in‍x⊗λi(p)mi. Then the map ρ:p↦∑in‍λi(p)mi is well defined. Hence, we have ψ(x⊗p)=x⊗ρ(p). Then we get (38)ψx⊗p,y⊗1,z⊗1=x⊗ρp,y⊗1,z⊗1=x,y,z⊗ρp=-1dψdx⊗px⊗p,ψy⊗1,z⊗1=-1dψdx⊗px⊗p,y⊗ρ1,z⊗1=x,y,z⊗-1dψdx⊗ppρ1=x,y,z⊗pρ1.Choose any triple x,y,z∈T such that [x,y,z]≠0. Then we can conclude that ρ(p)=pρ(1), ∀p∈R. So ρ is determined by its action on 1. Therefore, ρ∈R. Thus, ψ(x⊗p)=(id⊗ρ)(x⊗p), ∀p∈R, x∈T. That is, ψ∈id⊗R⊆Γ(T)0-⊗R. So we conclude that Γ(T~)0-=Γ(T)0-⊗R.



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