# The Centroid of a Lie Triple Algebra

Authors: Xiaohong Liu; Liangyun Chen
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

General results on the centroids of Lie triple algebras are developed. Centroids of the tensor product of a Lie triple algebra and a unitary commutative associative algebra are studied. Furthermore, the centroid of the tensor product of a simple Lie triple algebra and a polynomial ring is completely determined.


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## Body

## 1. Introduction

In recent years, Lie triple algebras (i.e., Lie-Yamaguti algebras or general Lie triple systems) have attracted much attention in Lie theories. They contain Lie algebras and Lie triple systems as special cases ([1–6]). So, it is of vital importnce to study some properties of Lie triple algebras. The concept of Lie triple algebra has been introduced, originally, by Yamaguti as general Lie triple system by himself, Sagle, and others. Since Lie triple algebras are generalization of Lie algebras and Lie triple system, it is natural for us to imagine whether or not some results of Lie algebras and Lie triple system hold in Lie triple algebras. Now, as a generalization of Lie triple algebra, Hom-Lie-Yamaguti was introduced by Lister in [7].Benkart and Neher studied centroid of Lie algebras in [8], and Melville investigated centroid of nilpotent Lie algebras in [9]. It turns out that result on the centroid of Lie algebras is a key ingredient in the classification of extended affine Lie algebras. The centroids of Lie triple systems were mentioned by Benito et al. [10]. Now, some results on centroids of Lie triple system and n-Lie algebras were developed in [11, 12].In this paper we present new results concerning the centroids of Lie triple algebras and give some conclusion of the tensor product of a Lie triple algebra and a unitary commutative associative algebra. Furthermore, we completely determine the centroid of the tensor product of a simple Lie triple algebra and a polynomial ring. The organization of the rest of this paper is as follows. Section1 is for basic notions and facts on Lie triple algebras. Section 2 is devoted to the structures and properties of the centroids of Lie triple algebras. Section 3 describes the structures of the centroids of tensor product of Lie triple algebras.


## 2. Preliminaries

Definition 1 (see [1]).
A Lie triple algebra (also called a Lie-Yamaguti algebra or a general Lie triple system)𝔤 is a vector space over an arbitrary field F with a bilinear map denoted by xy of 𝔤×𝔤 into 𝔤 and a ternary map denoted by [x,y,z] of 𝔤×𝔤×𝔤 into 𝔤 satisfying the following axioms:(a)
xx=0,(b)
[x,x,y]=0,(c)
(xy)z+(yz)x+(zx)y+[x,y,z]+[y,z,x]+[z,x,y]=0,(d)
[xy,z,w]+[yz,x,w]+[zx,y,w]=0,(e)
[x,y,zw]=([x,y,z])w+z([x,y,w]),(f)
([x,y,v])([z,w,v])=[[x,y,z],w,v]+[z[x,y,w],v],where x,y,z,w,v∈𝔤.Remark 2.
Any Lie algebra is a Lie triple algebra relative toxy=[x,y] and D(x,y)z=[[x,y],z], for x,y,z∈𝔤. If D(x,y)z=0, for all x,y,z∈𝔤, the axioms stated earlier reduce to that of Lie algebra, and if xy=0, for all x,y∈𝔤, the axioms stated earlier reduce to that of Lie triple system. In this sense, the Lie triple algebra is a more general concept than that of the Lie algebra and Lie triple system.Definition 3 (see [1, 2]).
A derivation of a Lie triple algebra𝔤 is a linear transformation D of 𝔤 into 𝔤 satisfying the following conditions:(1)
D(xy)=(Dx)y+x(Dy),(2)
D([x,y,z])=[Dx,y,z]+[x,Dy,z]+[x,y,Dz], for all x,y,z∈𝔤.
LetDer(𝔤) be the set of all derivation of 𝔤; then Der(𝔤) is regarded as a subalgebra of the general Lie algebra gl(𝔤) and is called the derivation algebra of 𝔤.Definition 4.
A Lie triple subalgebraI of 𝔤 is called an ideal if 𝔤I⊆I and [𝔤,I,𝔤]⊆I.Definition 5.
A Lie triple algebra𝔤 is perfect if 𝔤=𝔤𝔤=[𝔤,𝔤,𝔤].Definition 6.
LetI be a nonempty subset of 𝔤; we call Z𝔤(I)={x∈𝔤∣xa=[x,a,y]=0,  for all  a∈I,y∈𝔤} the centralizer of I in 𝔤. In particular, Z(𝔤)={x∈𝔤∣x𝔤=[x,𝔤,𝔤]=0} is the center of 𝔤.Definition 7.
Suppose thatI is an ideal of the Lie triple algebra 𝔤, on the quotient vector space
(1)gI={x=x+I∣∀x∈g}.
Define the operator: D(x¯,y¯)(z¯)=D(x,y)z¯,x¯y¯=xy¯, where x,y,z∈g. Then g/I is also a Lie triple algebra, and it is called a quotient algebra of 𝔤 by I.


## 3. The Centroids of Lie Triple Algebras

Definition 8.
Let𝔤 be a Lie triple algebra over a field F. The centroid of 𝔤 is the transform on 𝔤 given by Γ(𝔤)={ψ∈EndF(𝔤)∣ψ([x,y,z])=[ψ(x),y,z],ψ(xy)=(ψ(x))y,for allx,y,z,∈𝔤}.
By (a)–(f), we can conclude that ifψ∈Γ(𝔤), then we have ψ([x,y,z])=[ψ(x),y,z]=[x,ψ(y),z]=[x,y,ψ(z)],ψ(xy)=(ψ(x))y=x(ψ(y)), for all x,y,z,∈𝔤.
From the definition, it is clear that the scalars will always be in the centroid.Proposition 9.
If𝔤 is perfect, then the centroid Γ(𝔤) is commutative.Proof.
For allx,y,z∈𝔤,ϕ,ψ∈Γ(𝔤), we have ϕψ([x,y,z])=[ψ(x),ϕ(y),z]=ψϕ([x,y,z]),  ϕψ(xy)=ψ(x)ϕ(y)=ψϕ(xy).Proposition 10.
Let𝔤 be a Lie triple algebra over a field F and B a subset of 𝔤. Then;(1)
Z𝔤(B)is invariant underΓ(𝔤),(2)
every perfect ideal of𝔤is invariant underΓ(𝔤).Proof.
(
1
) For any ψ∈Γ(𝔤),  x∈Z𝔤(B),  y∈B,  z∈𝔤, we have [ψ(x),y,z]=[ψ(x),y,z]=ψ[x,y,z]=0,  (ψ(x))z=ψ(xz)=0, and [z,y,ψ(x)]=ψ[z,y,x]=0,  z(ψ(x))=ψ(zx)=0. Therefore, ψ(x)∈Γ(𝔤), which implies that Z𝔤(B) is invariant under Γ(𝔤).
(
2
) Let J be any perfect ideal of 𝔤; then J=JJ=[J,J,J]. For any y∈J, there exist a,b,c,d,e∈J, such that y=ab=[c,d,e], and then we have ψ(y)=ψ(ab)=(ψ(a))b∈J𝔤⊆J,ψ(y)=ψ([c,d,e])=[c,ψ(d),e]∈[J,𝔤,𝔤]⊆J. Hence J is invariant under Γ(𝔤).Definition 11.
Letφ(𝔤)⊆Z(𝔤) and φ(𝔤𝔤)=φ([𝔤,𝔤,𝔤])=0; then φ is called a central derivation.The set of all central derivation of𝔤 is denoted by C(𝔤). Clearly, C(𝔤)⊆Γ(𝔤) and C(𝔤) is an ideal of Γ(𝔤). A more precise relationship is summarized as follows.Next, we will develop some general results on centroids of Lie triple algebra.Proposition 12.
If𝔤 has no nonzero ideals I and J with [𝔤,I,J]=0,IJ=0, then Γ(𝔤) is an integral domain.Proof.
Clearly,id∈Γ(𝔤). If there exist ψ,φ∈Γ(𝔤), ψ≠0,ϕ≠0 such that ψφ=0, then there exist x,y∈𝔤 such that ψ(x)≠0 and ϕ(y)≠0. Then, [𝔤,ψ(x),ϕ(y)]=ψϕ([𝔤,x,y])=0,(ψ(x))(ϕ(y))=ψϕ(xy)=0. Therefore, ψ(x) and ϕ(y) can span two nonzero ideals I,J of 𝔤 such that [𝔤,I,J]=IJ=0, which is a contradiction. Hence, Γ(𝔤) has no zero divisor; it is an integral domain.Theorem 13.
If𝔤 is a simple Lie triple algebra over an algebraically closed field F, then Γ(𝔤)=Fid.Proof.
Letϕ∈Γ(𝔤)⊆EndF(𝔤). Since F is algebraically closed, ϕ has an eigenvalue λ. We denote the corresponding eigenspace to be Eλ(ϕ)={v∈𝔤∣ϕ(v)=λ(v)}, so Eλ(ϕ)≠0, for any v∈Eλ(ϕ),x,y∈𝔤, and we have ϕ([v,x,y])=[ϕ(v),x,y]=λ[v,x,y],ϕ((vx))=ϕ(v)x=(λv)x=λ(vx), so [v,x,y]∈Eλ(ϕ),vx∈Eλ(ϕ). It follows that Eλ(ϕ) is an ideal of 𝔤. But, 𝔤 is simple, so Eλ(ϕ)=𝔤; that is, Eλ(ϕ)=λid𝔤. This proves the theorem.WhenΓ(𝔤)=Fid, the Lie triple algebra 𝔤 is said to be central. Furthermore, if 𝔤 is simple, 𝔤 is said to be central simple. Every simple Lie triple algebra is central simple over its centroid.Proposition 14.
Let𝔤 be a Lie triple algebra over a field F; then(1)
𝔤is indecomposable if and only ifΓ(𝔤)does not contain idempotents except 0 andid;(2)
if𝔤is perfect, then everyϕ∈Γ(𝔤)is symmetric(f([a,b,c],d)=f(a[d,c,b]),f(ab,c)=f(a,bc))with respect to any invariant form on𝔤.Proof.
(
1
)  If there exists ϕ∈Γ(𝔤) which is an idempotent and satisfies ϕ≠0,id, then ϕ2(x)=ϕ(x), for all x∈𝔤. We assert that Kerϕ and Imϕ are ideals of 𝔤. In fact, for any x∈Kerϕ and y,z∈𝔤, we have ϕ([x,y,z])=[ϕ(x),y,z]=0,ϕ(xy)=(ϕ(x))y=0y=0, which implies [x,y,z]∈Kerϕ,xy∈Kerϕ. For any x∈Imϕ, there exists a∈𝔤 such that x=ϕ(a). Then, we have
(2)[x,y,z]=[ϕ(a),y,z]=ϕ([a,y,z])∈Imϕ,∀y,z∈𝔤,xy=ϕ(a)y=ϕ(ay)∈Imϕ,∀y∈𝔤.
This proves our assertion. Moreover, Kerϕ∩Imϕ=0. Indeed, if x∈Kerϕ∩Imϕ, then there exists y∈𝔤 such that x=ϕ(y) and 0=ϕ(x)=ϕ2(y)=ϕ(y)=x. We have a decomposition x=ϕ(x)+y, for all x∈𝔤, where ϕ(y)=0. So, we have 𝔤=Kerϕ⊕Imϕ, which is a contradiction.
On the other hand, suppose𝔤 has a decomposition 𝔤=𝔤1⊕𝔤2. Then for any x∈𝔤, we have x=x1+x2, xi∈𝔤i,  i=1,2. We choose ϕ∈Γ(𝔤) 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 have ϕ=0 or ϕ=id. If ϕ=0, then x1=0, implying 𝔤=𝔤2. If ϕ=id, then x2=0, implying 𝔤=𝔤1.
(
2
)  Let f be an invariant F-bilinear form on 𝔤. Then f([a,b,c],d)=f(a[d,c,b]),  f(ab,c)=f(a,bc),  for all  a,b,c,d∈𝔤. Since 𝔤 is perfect, for ϕ∈Γ(𝔤), we have
(3)f(ϕ([a,b,c]),d)=f([a,ϕ(b),c],d)=f(a,[d,c,ϕ(b)])=f([a,ϕ([d,c,b])])=f(a,[ϕ(d),c,b])=f([a,b,c],ϕ(d)),f(ϕ(ab),c)=f(a(ϕ(b)),c)=f(a,ϕ(b)c)=f(a,bϕ(c))=f(ab,ϕ(c)).
The result follows.Remark 15.
SetL(x)(y)=xy,L(x,y)(z)=[x,y,z],R(x)(y)=yx,R(x,y)(z)=[z,x,y],forallx,y,z,∈𝔤; then we call L(x), and L(x,y) the left multiplication operator of 𝔤;  R(x) and R(x,x) right multiplication operator of 𝔤. Denote by Mult(𝔤)=MultF(𝔤) the subalgebra of EndF(𝔤) generated by the left and right multiplication operators of 𝔤. Then, Γ(𝔤) is the centralizer of Mult(𝔤).Theorem 16.
Letπ:𝔤1→𝔤2 be an epimorphism of Lie triple algebra; for any f∈EndF(𝔤1;Kerπ):={g∈EndF(𝔤2)∣g(Kerπ)⊆Kerπ}, there exists a unique f¯∈EndF(𝔤2) satisfying π∘f=f¯∘π. Moreover, the following results hold.(1)
The mapπEnd:EndF(𝔤1;Kerπ)→EndF(𝔤2), f↦f¯ is a homomorphism with the following properties:πEnd(Mult(𝔤1))=Mult(𝔤2),πEnd(Γ(𝔤1)∩EndF(𝔤1;Kerπ))⊆Γ(𝔤2). There is a homomorphism πΓ:Γ(𝔤1)∩EndF(𝔤1;Kerπ)→Γ(𝔤2),f↦f¯.
IfKerπ=Z(𝔤1), then every ϕ∈Γ(𝔤1) leaves Kerπ invariant; that is, πΓ is defined on all of Γ(𝔤1).(2)
Suppose𝔤1 is perfect and Kerπ⊆Z(𝔤1); then πΓ:Γ(𝔤1)∩EndF(𝔤1;Kerπ)→Γ(𝔤2),f↦f¯ is injective.(3)
If𝔤1 is perfect, Z(𝔤2)=0, and Kerπ⊆Z(𝔤1), then πΓ:Γ(𝔤1)→Γ(𝔤2) is a monomorphism.Proof.
(
1
) It is easy to see that πEnd is a homomorphism. Since Kerπ is an ideal of 𝔤1 and all left and right multiplication operators of 𝔤1 leave Kerπ invariant, Mult(𝔤1)⊆EndF(𝔤1;Kerπ). Furthermore, for the left multiplication operator L(x,y),L(x) on 𝔤1, we have π∘L(x,y)=L(π(x),π(y))∘π,π∘L(x)=L(π(x))∘π, so πEnd(L(x,y))=L(π(x),π(y)),πEnd(L(x))=L(π(x)). For the right multiplication, we have the analogous formula πEnd(R(x,y))=R(π(x),π(y)),πEnd(R(x))=R(π(x)). Moreover, π is an epimorphism, so πEnd(Mult(𝔤1))=Mult(𝔤2). Now, we show that πEnd(Γ(𝔤1)∩EndF(𝔤1;Kerπ))⊆Γ(𝔤2). Let ϕ∈Γ(𝔤1)∩EndF(𝔤1;Kerπ). For any x′,y′,z′∈𝔤2, there exist x,y,z∈𝔤1 such that π(x)=x′,π(y)=y′,π(z)=z′. Then, we have
(4)ϕ¯([x′,y′,z′])=ϕ¯(π([x,y,z]))=π(ϕ([x,y,z]))=π([x,ϕ(y),z])=[π(x),ϕ¯(π(y)),π(z)]=[x′,ϕ¯(y′),z′]=π([ϕ(x),y,z])=[ϕ¯(π(x)),π(y),π(z)]=[ϕ¯(x′),y′,z′]=π([x,y,ϕ(z)])=[π(x),π(y),ϕ¯(π(z))]=[x′,y′,ϕ¯(z′)],ϕ¯(x′y′)=ϕ¯(π(x)π(y))=ϕ¯(π(xy))=πϕ(xy)=(ϕ¯πx)(π(y))=(ϕ¯(x′))y′=π(x(ϕ(y)))=π(x)(πϕ(y))=x′(ϕ¯π(y))=x′(ϕ¯(y′)),
which proves ϕ¯∈Γ(𝔤2).
(
2
) If ϕ¯=0 for, ϕ∈Γ(𝔤1)∩EndF(𝔤1;Kerπ), then π(ϕ(𝔤1))=ϕ¯(π(𝔤1))=0, which means that ϕ(𝔤1)⊆Kerπ⊆Z(𝔤1). Hence, ϕ([x,y,z])=[ϕ(x),y,z]=0,ϕ(xy)=(ϕ(x)y)=0, forallx,y,z∈𝔤1. Furthermore, since 𝔤1 is a perfect ideal, we can get ϕ=0.
(
3
) We can see that π(Z(𝔤1))⊆Z(𝔤2)=0, which follows that Z(𝔤1)⊆Kerπ. So, Kerπ⊆Z(𝔤1). From (1), we know that πΓ:Γ(𝔤1)→Γ(𝔤2) is a well-defined homomorphism, which is an injection by (2).Proposition 17.
If the characteristic ofF is not 2, then
(5)C(𝔤)=Γ(𝔤)∩Der(𝔤).Proof.
Ifψ∈Γ(𝔤)∩Der(𝔤), then by the definition of Γ(𝔤) and Der(𝔤), for all x,y,z∈𝔤, we have ψ(xy)=(ψ(x))y+x(ψ(y)),ψ([x,y,z])=[ψ(x),y,z]+[x,ψ(y),z]+[x,y,ψ(z)], and ψ(xy)=(ψ(x))y=x(ψ(y)),ψ([x,y,z])=[ψ(x),y,z]=[x,ψ(y),z]=[x,y,ψ(z)], so ψ(𝔤𝔤)=ψ([𝔤,𝔤,𝔤])=0 and ψ(𝔤)⊆C(𝔤). It follows easily that Γ(𝔤)∩Der(𝔤)⊆C(𝔤).
To show the inverse inclusion, letψ∈C(𝔤); then 0=ψ([x,y,z])=[ψ(x),y,z]=[x,ψ(y),z]=[x,y,ψ(z)],0=ψ(xy)=(ψ(x))y=x(ψ(y)). Thus, ψ∈Γ(𝔤)∩Der(𝔤).
This implies thatC(𝔤)=Γ(𝔤)∩Der(𝔤).Lemma 18.
LetI be a nonzero Γ(𝔤)-invariant ideal of 𝔤, V(I)={ψ∈Γ(𝔤)ψ(I)=0}, and let Hom(𝔤/I,Z𝔤(I)) be the vector space of all linear maps from 𝔤/I to Z𝔤(I) over F. Define T(I)={f∈Hom(𝔤/I,Z𝔤(I))∣f([x¯,y¯,z¯])=[f(x¯),y,z]=[x,f(y¯),z]=[x,y,f(z¯)],f(x¯y¯)=(f(x¯))y=x(f(y¯))}, where x¯,y¯,z¯∈𝔤/I. Then one has the following.(1)
T(I) is a subspace of Hom(𝔤/I,Z𝔤(I)) and V(I)≅T(I) as vector spaces.(2)
IfΓ(I)=FidI, then Γ(𝔤)=Fid𝔤⊕V(I) as vector spaces.Proof.
(
1
) It is easily seen that V(I) is an ideal of the associative algebra Γ(𝔤). To prove (1) consider the following map α:V(I)→T(I) given by
(6)α(ψ)(y¯)⟼ψ(y),
where ψ∈V(I) and y¯=y+I∈𝔤/I. The map α is well defined. If y¯=y1¯, 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:𝔤→𝔤,ψf(x)=f(x¯), for all x∈𝔤. It follows from the definition of T(I) that, for all x,y,z∈𝔤, ψf(xy)=f(x¯y¯)=(f(x¯))y=x(fy¯),ψf([x,y,z])=f([x¯,y¯,z¯])=[f(x¯),y,z]=[x,f(y¯),z]=[x,y,f(z¯)]. Thus, ψf∈Γ(𝔤), and so ψf∈V(I),ψf(I)=0. But, α(ψf)=f implies that α is onto. It is fairly easy to see that α preserves operations on vector spaces from 𝔤/I to Z𝔤(I). This proves (1).
We now prove (2). IfΓ(I)=FidI, then for all ψ∈Γ(𝔤),ψ∣I=λidI, for some λ∈F. If ψ≠λid𝔤, let φ(x)=λx, for all x∈𝔤; then ψ∈Γ(𝔤) and ψ-φ∈V(I). Clearly, ψ=φ+(ψ-φ). Furthermore, Fid𝔤∩V(I)=0, and so (2) is proved.Lemma 19.
Let𝔤 be a Lie triple algebra; then φD is a derivation for φ∈Γ(𝔤),D∈Der(𝔤).Proof.
Ifx,y,z∈𝔤, then
(7)φD(xy)=φ((D(x))y+x(D(y)))=(φD(x))y+x(φD(y)),φD([x,y,z])=φ([D(x),y,z]mmm+[x,D(y),z]+[x,y,D(z)])=[φD(x),y,z]+[x,φD(y),z]+[x,y,φD(z)].
Thus, φD is a derivation.Theorem 20.
Let𝔤 be a Lie triple algebra, and for any φ∈Γ(𝔤),D∈Der(𝔤), then one has the following.(1)
Der(𝔤) is contained in the normalizer of Γ(𝔤) in End(𝔤).(2)
Dφ is contained in Γ(𝔤) if and only if φD is a central derivation of 𝔤.(3)
φD is a derivation of 𝔤 if and only if [D,φ] is a central derivation of 𝔤.Proof.
For anyφ∈Γ(𝔤),D∈Der(𝔤), and x,y,z∈𝔤,
(8)Dφ(xy)=D(φ(x)y)=(Dφ(x))y+(φ(x))(Dy)=(Dφ(x))y+φD(xy)-(φD(x))y.
Then, we get(Dφ-φD)(xy)=((Dφ-φD)(x))y. On the other hand, we have
(9)Dφ([x,y,z])=D([φ(x),y,z])=[Dφ(x),y,z]+[φ(x),D(y),z]+[φ(x),y,D(z)]=[Dφ(x),y,z]+φD([x,y,z])-[φD(x),y,z].
So, (Dφ-φD)([x,y,z])=[(Dφ-φD)(x),y,z]; this is [D,φ]=Dφ-φD∈Γ(𝔤). This proves (1). From Lemma 19 and (1), Dφ is an element of Γ(𝔤) if and only if φD∈Γ(𝔤)∩Der(𝔤). Thanks to Proposition 17, we get the result (2). It follows from (1), Proposition 17, and Lemma 19 that the result holds.Now, we study the relationship between the centroid of a decomposable Lie triple algebra and the centroid of its factors.Theorem 21.
Suppose that𝔤 is a Lie triple algebra over F and 𝔤=𝔤1⊕𝔤2 with 𝔤1 and 𝔤2 being ideals of 𝔤; then
(10)Γ(𝔤)≅Γ(𝔤1)⊕Γ(𝔤2)⊕C(1)⊕C(2),
as vector spaces, where C(1)={φ∈Hom(𝔤1,𝔤1)∣φ(𝔤1)⊆Z(𝔤2),φ(𝔤2𝔤2)=φ[𝔤2,𝔤2,𝔤2]=0}, C(2)={φ∈Hom(𝔤2,𝔤2)∣φ(𝔤2)⊆Z(𝔤1),φ(𝔤1𝔤1)=φ[𝔤1,𝔤1,𝔤1]=0}.Proof.
Letπi:𝔤→𝔤i be canonical projections for i=1,2; then π1,π2∈Γ(𝔤) and π1+π2=id𝔤. So we have for φ∈Γ(𝔤),
(11)φ=π1φπ1+π1φπ2+π2φπ1+π2φπ2.
Note that πiφπj∈Γ(𝔤) for i,j=1,2. We claim that
(12)Γ(𝔤)=π1Γ(𝔤)π1+π1Γ(𝔤)π2+π2Γ(𝔤)π1+π2Γ(𝔤)π2.
It suffices to show that π1Γ(𝔤)π1∩π1Γ(𝔤)π2 (other cases are similar). For any φ∈π1Γ(𝔤)π1∩π1Γ(𝔤)π2, there exist fi∈Γ(𝔤),i=1,2, such that φ=π1f1π1=π1f2π2. Then, φ(x)=π1f2π2(x)=π1f2π2(π2(x))=π1f1(0)=0, for all x∈𝔤, and so φ=0. Let Γ(𝔤)ij=πiΓ(𝔤)πj,i,j=1,2. We now prove that
(13)Γ(𝔤)11≅Γ(𝔤1),Γ(𝔤)22≅Γ(𝔤2),Γ(𝔤)12≅C2,Γ(𝔤)21≅C1.
Since φ(𝔤2)=0 for φ∈Γ(𝔤)11, we have φ∣𝔤1∈Γ(𝔤1). On the other hand, one can regard Γ(𝔤1) as a subalgebra of Γ(𝔤) by extending any φ0∈Γ(𝔤1) on 𝔤2 being equal to zero; that is, φ0(x1)=φ0(x1),φ0(x2)=0, for all x1∈𝔤1,x2∈𝔤2. Then φ0∈Γ(𝔤) and φ0∈Γ(𝔤)11. Therefore, Γ(𝔤)11≅Γ(𝔤1) with isomorphism σ:Γ(𝔤)11→Γ(𝔤1),σ(φ)=φ∣𝔤1, for all φ∈Γ(𝔤)11. Similarly, we have Γ(𝔤)22≅Γ(𝔤2).
Next, we prove thatΓ(𝔤)12≅C2. If φ∈Γ(𝔤)12 there exists φ0 in Γ(𝔤) such that φ=π1φ0π2. For xk=xk1+xk2∈𝔤, where xki∈𝔤i,i=1,2 and k=1,2,3, we have
(14)φ(x1x2)=π1φ0π2(x1x2)=π1φ0(x12x22)=π1((φ0x12)x22)=0,φ([x1,x2,x3])=π1φ0π2([x1,x2,x3])=π1φ0([x12,x22,x32])=π1([φ0x12,x22,x32])=0,(φ(x1))x2=φ(x1x2)=0,[φ(x1),x2,x3]=φ([x1,x2,x3])=0.
Then, φ(𝔤)⊆Z(𝔤) and φ(𝔤𝔤)=φ[𝔤,𝔤,𝔤]=0. It follows that φ∣𝔤2(𝔤2)⊆Z(𝔤1),φ∣𝔤2(𝔤2𝔤2)=φ∣𝔤2([𝔤2,𝔤2,𝔤2])=0, and so φ∣𝔤2∈C2.
Conversely, forφ∈C2, expending φ on 𝔤 (also denoted by φ) or by φ(𝔤1𝔤1)=φ[𝔤1,𝔤1,𝔤1]=0, we have π1φπ2=φ and φ∈Γ(𝔤)12. This proves that Γ(𝔤)12 is isomorphic to C2 with the following isomorphism:
(15)τ:Γ(𝔤)12⟶C2,τ(φ)=φ∣Γ(𝔤)12,
for all φ∈Γ(𝔤)12. Similarly, we can prove that Γ(𝔤)21≅C1. Summarizing the aforesaid discussion, we have
(16)Γ(𝔤)≅Γ(𝔤1)⊕Γ(𝔤2)⊕C(1)⊕C(2).
The proof is completed.A generalized version of Theorem21 is stated next without proof.Theorem 22.
Suppose that𝔤 is a Lie triple algebra over F and with a decomposition of ideals 𝔤=𝔤1⊕𝔤2⊕⋯⊕𝔤m. Then, one has
(17)Γ(𝔤)≅Γ(𝔤1)⊕⋯⊕Γ(𝔤m)⊕(⊕1≤i≠j≤mCij),
as vector spaces, where Cij={φ∈Hom(𝔤i,𝔤j)∣φ(𝔤i)⊆Z(𝔤j),φ(𝔤i𝔤i)=φ[𝔤i,𝔤i,𝔤i]=0, for 1≤i≠j≤m}.


## 4. The Centroids of Tensor Product of Lie Triple Algebras

Definition 23.
LetA be an associative algebra over a field F, where the centroid of A is the space of F-linear transforms on A given by Γ(A)={φ∈EndF(A)φ(ab)=aφ(b)=φ(a)b,foralla,b∈A}; then Γ(A) is an associative algebra of EndF(A). If 𝔤 is be a finite-dimensional Lie triple algebra over F, let 𝔤⊗A be a tensor product of the underlying vector spaces A and 𝔤. Then, 𝔤⊗A over a field F with respect to the following 2-ary and 3-ary multilinear operation:
(18)[x1⊗a1,x2⊗a2,x3⊗a3]=[x1,x2,x3]⊗a1a2a3,(x1⊗a1)(x2⊗a2)=(x1x2)⊗a1a2,
where xi∈𝔤,ai∈A,i=1,2,3. This Lie triple algebra 𝔤⊗A is called the tensor product of A and 𝔤. For f∈EndF(𝔤) and φ∈EndF(A), there exists a unique map f⊗~φ∈EndF(𝔤⊗A) such that
(19)(f⊗~φ)(x⊗a)=f(x)⊗φ(a),∀x∈𝔤,a∈A.
The map should not be confused with the element f⊗φ of the tensor product EndF(𝔤)⊗EndF(A). Of course, we have a canonical map as the following:
(20)ω:EndF(𝔤)⊗EndF(A)⟶EndF(𝔤⊗A):f⊗φ⟼f⊗~g.
It is easy to see that if ϕ∈Γ(𝔤) and ψ∈Γ(A), then ϕ⊗~ψ∈Γ(𝔤⊗A). Hence, Γ(𝔤)⊗~Γ(A)⊆Γ(𝔤⊗A), where Γ(𝔤)⊗~Γ(A) is the F-span of all endomorphisms ϕ⊗~ψ.Definition 24.
The transformationϕ∈Γ(𝔤⊗A) is said to have finite 𝔤-image, if for any a∈A, there exist finitely many a1,…,an∈A such that
(21)ϕ(𝔤⊗Fa)⊆𝔤⊗Fa1+⋯+𝔤⊗Fan.
It is easy to see that Γ(𝔤)⊗~Γ(A)⊆{ϕ∈Γ(𝔤⊗A)∣ϕhasfinite𝔤-image}. In addition, ϕ∈Γ(𝔤⊗A) has finite 𝔤-image, if ϕ(𝔤⊗1)⊆𝔤⊗Fa1+⋯+𝔤⊗Fan for suitable ai∈A.Lemma 25.
Let𝔤 be a Lie triple algebra over F and A a unitary commutative associative over F. Let {ar}r∈ℜ be a basis of A, and let ϕ∈Γ(𝔤⊗A). Define ϕr∈EndF(𝔤) by
(22)ϕ(x⊗1)=∑r∈ℜϕr(x)⊗ar,
then all ϕr∈Γ(𝔤).Proof.
Forx1,x2,x3∈𝔤, we have
(23)ϕ([x1,x2,x3]⊗1)=∑r∈ℜ‍ϕr([x1,x2,x3])⊗ar,ϕ(x1x2)=∑r∈ℜϕr(x1x2)⊗ar,ϕ([x1,x2,x3]⊗1)=ϕ([x1⊗1,x2⊗1,x3⊗1])=[ϕ(x1⊗1),x2⊗1,x3⊗1]=∑r∈ℜ‍[ϕr(x1),x2,x3]⊗ar=[x1⊗1,ϕ(x2⊗1),x3⊗1]=∑r∈ℜ‍[x1,ϕr(x2),x3]⊗ar=[x1⊗1,x2⊗1,ϕ(x3⊗1)]=∑r∈ℜ‍[x1,x2,ϕr(x3)]⊗ar,(ϕ(x1x2)⊗1)=ϕ((x1⊗1)(x2⊗1))=∑r∈ℜ‍ϕr(x1x2)⊗ar=(ϕ(x1⊗1))(x2⊗1)=∑r∈ℜ‍(ϕr(x1)x2)⊗ar=(x1⊗1)(ϕ(x2⊗1))=∑r∈ℜ‍(x1(ϕr(x1)))⊗ar.
Hence, one has
(24)ϕr([x1,x2,x3])=[ϕr(x1),x2,x3]=[x1,ϕr(x2),x3]=[x1,x2,ϕr(x3)],ϕr(x1x2)=(ϕr(x1))x2=x1(ϕr(x2)),
for all xi∈𝔤. So all ϕr∈Γ(𝔤).Proposition 26.
Let𝔤 be a perfect Lie triple algebra over F, and let A be a unitary commutative associative over F which is free as a F-module. Then, one has the following.(1)
If𝔤is perfect, then𝔤⊗Ais perfect too.(2)
If𝔤is finitely generated as aMult(𝔤)-module (or as aΓ(𝔤)-module), or if𝔤is a central and a torsion-freeF-module, then everyϕ∈Γ(𝔤⊗A)has finite𝔤-image.(3)
IfΓ(𝔤)is a freeF-module and the mapωof Definition23 is injective, thenΓ(𝔤)⊗~Γ(A)={ϕ∈Γ(𝔤⊗A)∣ϕhas finite𝔤-image}.Proof.
The definition of perfect shows that (1) holds. LetM=Mult(𝔤). It follows that M⊗~id⊆Mult(𝔤⊗A) from A  is unital. Since 𝔤 is finitely generated as a Mult(𝔤)-module, we can suppose 𝔤=Mx1+⋯+Mxn for x1,…,xn∈𝔤. Fix ϕ∈Γ(𝔤⊗A) and a∈A. There exist finite families {xij}⊆𝔤 and {aij}⊆A such that
(25)ϕ(xi⊗a)=∑j‍xij⊗aij,
for 1≤i≤n. Hence,
(26)ϕ(𝔤⊗a)=∑i‍ϕ((M⊗~id)(xi⊗a))=∑i,j‍(M⊗~id)(xij⊗aij)⊆∑i,j‍𝔤⊗aij.
Since 𝔤⊗A is perfect, the centroid Γ(𝔤⊗A) is commutative. By replacing M with Γ(𝔤) we can use the same argument aforementione to show that every ϕ∈Γ(T⊗A) has finite 𝔤-image, if 𝔤 is a finitely generated Γ(𝔤)-module.
Now, suppose thatΓ(𝔤)=Fid and that 𝔤 is a torsion-free F-module. Then there exist scalars kr∈F such that ϕr=krid. Hence,
(27)ϕ(x⊗1)=∑r∈ℜ‍krx⊗ar.
Fix x∈𝔤; then almost all krx=0 and almost all kr=0, which in turn implies that ϕ has finite 𝔤-image.
From the aforementione discussion above, we get(28)Γ(𝔤)⊗~Γ(A)⊆{ϕ∈Γ(𝔤⊗A)∣ϕhasfinite𝔤-image}.
So, it suffices to prove that Γ(𝔤)⊗~Γ(A)⊇{ϕ∈Γ(𝔤⊗A)∣ϕhasfinite𝔤-image}. We suppose that ϕ∈Γ(𝔤⊗A) has finite 𝔤-image, and then there exists a finite subset 𝔍⊆ℜ such that equation in Lemma 25 becomes as follows:
(29)ϕ(x⊗1)=∑r∈𝔍‍ϕr(x)⊗ar.
For x1,x2,x3∈𝔤 and a∈A, we get
(30)ϕ([x1,x2,x3]⊗a)=ϕ([x1⊗1,x2⊗1,x3⊗a])=[ϕ(x1⊗1),x2⊗1,x3⊗a]=∑r∈𝔍‍[ϕr(x1),x2,x3]⊗ara=∑r∈𝔍‍ϕr([x1,x2,x3])⊗ara,ϕ((x1x2)⊗a)=ϕ((x1⊗1)(x2⊗a))=(ϕ(x1⊗1))(x2⊗a)=∑r∈𝔍‍(ϕr(x1)x2)⊗ara=∑r∈𝔍‍ϕr(x1x2)⊗ara.
Since 𝔤 is perfect, we can get
(31)ϕ=∑r∈𝔍ϕr⊗~λr,
where λr is the left multiplication in A by ar. Let {ψs∣s∈𝔖} be a basis of Γ(𝔤). Then, there exist a finite subset 𝔅⊆𝔖 and scalars krs∈F(r∈𝔍,s∈𝔅) such that
(32)ϕr=∑s∈𝔅‍krsψs.
We then get
(33)ϕ=∑r∈𝔍‍∑s∈𝔅‍krsψs⊗~λr=∑s∈𝔅ψs⊗~ϕs,forϕs=∑r∈𝔍krsλr.
Now we show that ϕs∈Γ(A). For any x1,x2,x3∈𝔤,a1,a2∈A, we have
(34)∑s∈𝔅‍ψs([x1,x2,x3])⊗ϕs(a1a2)=ϕ([x1,x2,x3]⊗a1a2)=[ϕ(x1⊗a1),x2⊗a2,x3⊗1]=[∑s∈𝔅‍ψs(x1)⊗ϕs(a1),x2⊗a2,x3⊗1]=∑s∈𝔅‍[ψs(x1),x2,x3]⊗ϕs(a1)a2=∑s∈𝔅‍ψs([x1,x2,x3])⊗ϕs(a1)a2,∑s∈𝔅‍ψs(x1x2)⊗ϕs(a1)=ϕ((x1x2)⊗a1)=(ϕ(x1⊗a1)(x2⊗1))=∑s∈𝔅‍(ψs(x1))(x2)⊗ϕs(a1)1=∑s∈𝔅‍ψs(x1x2)⊗ϕs(a1)1.
Since [𝔤𝔤,𝔤]=𝔤𝔤=𝔤, we have
(35)∑s∈𝔅‍ψs(x)⊗(ϕs(a1a)-ϕs(a1)a)=0,forallx∈𝔤 and a1,a2∈A. It follows that
(36)∑s∈𝔅‍ψs⊗~μs=0,
where μs∈EndF(A) is defined by μs(a2)=ϕs(a1a2)-ϕs(a1)a2, for all a2∈A. Since ω is injective, we also have
(37)∑s∈𝔅‍ψs⊗μs=0.
So by the linear independence of the ψs, we get that μs=0. Then ψs∈Γ(A). Hence ϕ∈Γ(𝔤)⊗~Γ(A).Next, we will determine the centroid of the tensor product of a simple Lie triple algebra and a polynomial ring. Here after we study the centroid of Lie triple algebra over a filedF. Let {ek} be a basis of the central derivation C(𝔤) and {φj} a maximal subset of Γ(𝔤) such that {φj∣𝔤𝔤} and {φj∣[𝔤,𝔤,𝔤]} are linear independent. Then, we have the following result.Theorem 27.
LetΨ denote the subspace of Γ(𝔤) spanned by {φj}. Then {ek,φj} is a basis of Γ(𝔤) and Γ(𝔤)=Ψ⊕C(𝔤) as vector spaces.Proof.
Since{φj∣𝔤𝔤} and {φj∣[𝔤,𝔤,𝔤]} are linear independent, {φj} is linear independent in Γ(𝔤). By definition of {ek,φj}, the {ek,φj} is independent in Γ(𝔤). For φ∈Γ(𝔤) since {φj∣𝔤𝔤} is a basis of vector spaces {φ∣𝔤𝔤∣φ∈Γ(𝔤)}, and {φj∣[𝔤,𝔤,𝔤]} is a basis of vector spaces {φ∣[𝔤,𝔤,𝔤]∣φ∈Γ(𝔤)}, there exist gs∈F, s∈J (J is a finite set of positive integers) such that
(38)φ∣[𝔤,𝔤,𝔤]=∑s∈J‍gsφs∣[𝔤,𝔤,𝔤],φ∣𝔤𝔤=∑s∈J‍gsφs∣𝔤𝔤.
We then have
(39)(φ-∑s∈J‍gsφs)∣[𝔤,𝔤,𝔤]=0,(φ-∑s∈J‍gsφs)∣𝔤𝔤=0.
If y1,y2,y3∈𝔤, then
(40)0=(φ-∑s∈J‍gsφs)([y1,y2,y3])=[(φ-∑s∈J‍gsφs)(y1),y2,y3],0=(φ-∑s∈J‍gsφs)(y1y2)=((φ-∑s∈J‍gsφs)(y1))y2.
It follows that
(41)(φ-∑s∈J‍gsφs)(𝔤)⊆Z(𝔤),φ-∑s∈J‍gsφs,
is a central derivation. So there exist ri∈F, i∈I (I is a finite set of positive integers) such that
(42)φ-∑s∈J‍gsφs=∑i∈I‍rici.
Therefore,
(43)φ=∑s∈J‍gsφs+∑i∈I‍rici.
The proof is completed.Lemma 28.
Let𝔤 be a simple Lie triple algebra, R=F[x1,…,xn], and 𝔤~=𝔤⊗R; then one has Γ(𝔤~)⊇Ψ⊗R+C(g)⊗EndF(R).Proof.
Letψ∈Γ(𝔤),m∈R. Then, we have
(44)(ψ⊗m)([x⊗p,y⊗q,z⊗r])=(ψ⊗m)([x,y,z]⊗pqr)=ψ([x,y,z])⊗mpqr=[ψ(x),y,z]⊗mpqr=[x,ψ(y),z]⊗mpqr=[x,y,ψ(z)]⊗mpqr,[(ψ⊗m)(x⊗p),y⊗q,z⊗r]=[ψ(x),y,z]⊗mpqr,[x⊗p,(ψ⊗m)(y⊗q),z⊗r]=[x,ψ(y),z]⊗mpqr,[x⊗p,y⊗q,(ψ⊗m)(z⊗r)]=[x,y,ψ(z)]⊗mpqr,(ψ⊗m)((x⊗p)(y⊗q))=(ψ⊗m)((xy)⊗pq)=ψ(xy)⊗mpq=(ψ(x)y)⊗mpq,(45)((ψ⊗m)(x⊗p))(y⊗q)=(ψ(x)⊗mp)(y⊗q)=(ψ(x))y⊗mpq.
Thus ψ⊗m∈Γ(𝔤~). For any d∈C(g) and f∈EndF(R),d⊗f∈Γ(𝔤~), we have
(46)(d⊗f)([x⊗p,y⊗q,z⊗r])=d([x,y,z])⊗f(pqr)=0,[(d⊗f)(x⊗p),y⊗q,z⊗r]=[d(x)⊗f(p),y⊗q,z⊗r]=[d(x),y,z]⊗f(p)qr=0,[(x⊗p)(d⊗f)(y⊗q),z⊗r]=[x⊗p,d(y)⊗f(q),z⊗r]=[x,d(y),z]⊗pf(q)r=0,[x⊗p,y⊗q,(d⊗f)(z⊗r)]=[(x⊗p)y⊗q,d(z)⊗f(r)]=[x,y,d(z)]⊗pqf(r)=0,(d⊗f)((x⊗p)(y⊗q))=d(xy)⊗f(pq)=0,((d⊗f)(x⊗p))(y⊗q)=(d(x)⊗f(p))(y⊗q)=(d(x)y)⊗f(p)q=0,(x⊗p)((d⊗f)(y⊗q))=(x⊗p)(d(y)⊗f(q))=(xd(y))⊗pf(q)=0.
Hence Ψ⊗R+C(g)⊗EndF(R)⊆Γ(𝔤~).Theorem 29.
Let𝔤 be a simple Lie triple algebra, R=F[x1,…,xn], and 𝔤~=𝔤⊗R. Then Γ(𝔤~)=Γ(𝔤)⊗R.Proof.
From Lemma28, we get Γ(𝔤)⊗R⊆Γ(𝔤~). Now we prove the opposite inclusion. Let {mi} be a basis for R,ψ∈Γ(𝔤~),p∈R,x,y,z∈𝔤, and let ηi(-,p) be suitable maps in EndF(𝔤) such that
(47)ψ(x⊗p)=∑i‍ηi(x,p)⊗mi.
Then,
(48)ψ([x⊗p,y⊗1,z⊗1])=[ψ(x⊗p),y⊗1,z⊗1]=[∑i‍ηi(x,p)⊗mi,y⊗1,z⊗1]=∑i‍[ηi(x,p),y,z]⊗mi=ψ([x⊗1,y⊗p,z⊗1])=[x⊗1,ψ(y⊗p),z⊗1]=[x⊗1,∑i‍ηi(y,p)⊗mi,z⊗1]=∑i‍[x,ηi(y,p),z]⊗mi=ψ([x⊗1,y⊗1,z⊗p])=[x⊗1,y⊗1,ψ(z⊗p)]=[x⊗1,y⊗1,∑i‍ηi(z,p)⊗mi]=∑i‍[x,y,ηi(z,p)]⊗mi,ψ((x⊗p)(y⊗1))=∑i‍(ηi(x,p)⊗mi)(y⊗1)=∑‍(ηi(x,p)y)⊗mi=(x⊗p)(ψ(y⊗1))=∑i‍(x,ηi(y,p))⊗mi,
while
(49)ψ([x,y,z]⊗p)=∑i‍ηi([x,y,z],p)⊗mi,ψ((xy)⊗p)=∑i‍ηi((xy),p)⊗mi.
So, for each i and p, we have ηi(p[x,y,z])=[ηi(x,p),y,z]=[x,ηi(y,p),z]=[x,y,ηi(z,p)], ηi(p,(xy))=(ηi(x,p),y)=(x,ηi(y,p)),forallx,y,z∈𝔤. Hence, ηi(-,p)∈Γ(𝔤). But Γ(𝔤)=Fid. Therefore, ηi(x,p)=λi(p)x, for all x∈𝔤, for suitable scalars λi(p).
Thus, we may write(50)ψ(x⊗p)=∑i‍λi(p)x⊗mi.
Since the right-hand side is in 𝔤⊗R, for each p, we get λi(p)=0 for all except for a finite number of i. That is,
(51)ψ(x⊗p)=∑i=1n‍x⊗λi(p)mi.
Then, the map
(52)ρ:p⟼∑i=1n‍λi(p)mi
is well defined. Hence, we have ψ(x⊗p)=x⊗ρ(p). Thus,
(53)[ψ(x⊗p),y⊗1,z⊗1]=[x⊗ρ(p),y⊗1,z⊗1]=[x,y,z]⊗ρ(p),[x⊗p,ψ(y⊗1),z⊗1]=[x⊗p,y⊗ρ(1),z⊗1]=[x,y,z]⊗pρ(1),(ψ(x⊗p)(y⊗1))=(x⊗ρ(p))(y⊗1)=(xy)⊗ρ(p)=(x⊗p)(ψ(y⊗1))=(x⊗p)(y⊗ρ(1))=(xy)⊗pρ(1).
Choose x,y,z∈𝔤 such that xy≠0,[x,y,z]≠0. Then we can conclude that ρ(p)=pρ(1), forallp∈R. So ρ is determined by its action on 1. Therefore, ρ∈R. Thus,
(54)ψ(x⊗p)=(id⊗ρ)(x⊗p),∀p∈R,x∈𝔤.
That is, ψ∈id⊗R≅Γ(𝔤)⊗R. So, we have Γ(𝔤~)=Γ(𝔤)⊗R.



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