The Centroid of a Lie Triple Algebra

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.

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. Section 1 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 to 𝑥𝑦=[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 algebra 𝔤 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.
Let Der(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 subalgebra 𝐼 of 𝔤 is called an ideal if 𝔤𝐼⊆𝐼 and [𝔤,𝐼,𝔤]⊆𝐼.

Definition 5. A Lie triple algebra 𝔤 is perfect if 𝔤=𝔤𝔤=[𝔤,𝔤,𝔤].

Definition 6. Let 𝐼 be a nonempty subset of 𝔤; we call 𝑍_𝔤(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 that 𝐼 is an ideal of the Lie triple algebra 𝔤, on the quotient vector space 𝑔𝐼=𝑥=𝑥+𝐼∣∀𝑥∈𝑔. (1) Define the operator: 𝐷(𝑥,𝑦)(𝑧)=𝐷(𝑥,𝑦)𝑧,𝑥𝑦=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. Let 𝔤 be a Lie triple algebra over a field 𝐅. The centroid of 𝔤 is the transform on 𝔤 given by Γ(𝔤)={𝜓∈End_𝐅(𝔤)∣𝜓([𝑥,𝑦,𝑧])=[𝜓(𝑥),𝑦,𝑧],𝜓(𝑥𝑦)=(𝜓(𝑥))𝑦,forall𝑥,𝑦,𝑧,∈𝔤}.
By (a)–(f), we can conclude that if 𝜓∈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. If 𝔤 is perfect, then the centroid Γ(𝔤) is commutative.

Proof. For all 𝑥,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. Let 𝔤 be a Lie triple algebra over a field 𝐅 and 𝐵 a subset of 𝔤. Then;(1)𝑍_𝔤(𝐵) is invariant under Γ(𝔤), (2) every perfect ideal of 𝔤 is invariant under Γ(𝔤).

Proof. (1) For any 𝜓∈Γ(𝔤),  𝑥∈𝑍_𝔤(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 𝑍_𝔤(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. Let 𝜑(𝔤)⊆𝑍(𝔤) and 𝜑(𝔤𝔤)=𝜑([𝔤,𝔤,𝔤])=0; then 𝜑 is called a central derivation.

The set of all central derivation of 𝔤 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. If 𝔤 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. If 𝔤 is a simple Lie triple algebra over an algebraically closed field 𝐅, then Γ(𝔤)=𝐅id.

Proof. Let 𝜙∈Γ(𝔤)⊆End_𝐅(𝔤). Since 𝐅 is algebraically closed, 𝜙 has an eigenvalue 𝜆. We denote the corresponding eigenspace to be 𝐸_𝜆(𝜙)={𝑣∈𝔤∣𝜙(𝑣)=𝜆(𝑣)}, so 𝐸_𝜆(𝜙)≠0, for any 𝑣∈𝐸_𝜆(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]∈𝐸_𝜆(𝜙),𝑣𝑥∈𝐸_𝜆(𝜙). It follows that 𝐸_𝜆(𝜙) is an ideal of 𝔤. But, 𝔤 is simple, so 𝐸_𝜆(𝜙)=𝔤; that is, 𝐸_𝜆(𝜙)=𝜆id_𝔤. This proves the theorem.

When Γ(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. Let 𝔤 be a Lie triple algebra over a field 𝐅; then(1)𝔤 is indecomposable if and only if  Γ(𝔤) does not contain idempotents except 0 and id; (2) if 𝔤 is perfect, then every 𝜙∈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 on 𝔤.

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 =𝜙(𝑎)(𝑎)𝑥,𝑦,𝑧,𝑦,𝑧=𝜙𝑎,𝑦,𝑧∈Im𝜙,∀𝑦,𝑧∈𝔤,𝑥𝑦=𝜙𝑦=𝜙𝑎𝑦∈Im𝜙,∀𝑦∈𝔤. (2) 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, suppose 𝔤 has a decomposition 𝔤=𝔤₁⊕𝔤₂. Then for any 𝑥∈𝔤, we have 𝑥=𝑥₁+𝑥₂, 𝑥_𝑖∈𝔤_𝑖,  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 𝑓𝜙(𝑏)(𝑏𝜙(𝑑)[](𝑑),𝑓𝜙()𝑎𝜙(𝑏)(𝑏)𝑐(𝑐)(𝑐).([𝑎,𝑏,𝑐]),𝑑=𝑓𝑎,𝜙,𝑐,𝑑=𝑓𝑎,𝑑,𝑐,𝜙)=𝑓𝑎,𝜙([𝑑,𝑐,𝑏])=𝑓𝑎,,𝑐,𝑏=𝑓𝑎,𝑏,𝑐,𝜙𝑎𝑏,𝑐=𝑓,𝑐=𝑓𝑎,𝜙=𝑓𝑎,𝑏𝜙=𝑓𝑎𝑏,𝜙 (3)
The result follows.

Remark 15. Set 𝐿(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)=Mult_𝐅(𝔤) the subalgebra of End_𝐅(𝔤) generated by the left and right multiplication operators of 𝔤. Then, Γ(𝔤) is the centralizer of Mult(𝔤).

Theorem 16. Let 𝜋∶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 map 𝜋_End∶End_𝐅(𝔤₁;Ker𝜋)→End_𝐅(𝔤₂), 𝑓↦𝑓 is a homomorphism with the following properties: 𝜋_End(Mult(𝔤₁))=Mult(𝔤₂),𝜋_End(Γ(𝔤₁)∩End_𝐅(𝔤₁;Ker𝜋))⊆Γ(𝔤₂). There is a homomorphism 𝜋_Γ∶Γ(𝔤₁)∩End_𝐅(𝔤₁;Ker𝜋)→Γ(𝔤₂),𝑓↦𝑓.
If Ker𝜋=𝑍(𝔤₁), then every 𝜙∈Γ(𝔤₁) leaves Ker𝜋 invariant; that is, 𝜋_Γ is defined on all of Γ(𝔤₁).(2) Suppose 𝔤₁ is perfect and Ker𝜋⊆𝑍(𝔤₁); then 𝜋_Γ∶Γ(𝔤₁)∩End_𝐅(𝔤₁;Ker𝜋)→Γ(𝔤₂),𝑓↦𝑓 is injective.(3) If 𝔤₁ is perfect, 𝑍(𝔤₂)=0, and Ker𝜋⊆𝑍(𝔤₁), then 𝜋_Γ∶Γ(𝔤₁)→Γ(𝔤₂) is a monomorphism.

Proof. (1) It is easy to see that 𝜋_End is a homomorphism. Since Ker𝜋 is an ideal of 𝔤₁ and all left and right multiplication operators of 𝔤₁ leave Ker𝜋 invariant, Mult(𝔤₁)⊆End_𝐅(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 𝜋_End(𝐿(𝑥,𝑦))=𝐿(𝜋(𝑥),𝜋(𝑦)),𝜋_End(𝐿(𝑥))=𝐿(𝜋(𝑥)). For the right multiplication, we have the analogous formula 𝜋_End(𝑅(𝑥,𝑦))=𝑅(𝜋(𝑥),𝜋(𝑦)),𝜋_End(𝑅(𝑥))=𝑅(𝜋(𝑥)). Moreover, 𝜋 is an epimorphism, so 𝜋_End(Mult(𝔤₁))=Mult(𝔤₂). Now, we show that 𝜋_End(Γ(𝔤₁)∩End_𝐅(𝔤₁;Ker𝜋))⊆Γ(𝔤₂). Let 𝜙∈Γ(𝔤₁)∩End_𝐅(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 𝜙𝑥^′,𝑦^′,𝑧^′=𝜙𝜋𝜙𝑦=𝜋(𝑥),𝑥,𝑦,𝑧=𝜋𝑥,𝑦,𝑧=𝜋𝑥,𝜙,𝑧𝜙𝜋𝑦(𝑧=𝑥,𝜋)_′,𝜙𝑦^′,𝑧^′𝜙(𝑥)==𝜋,𝑦,𝑧𝜙(𝜋(𝑥)),𝜋𝑦,𝜋(𝑧)=𝜙𝑥^′,𝑦^′,𝑧^′=𝜋𝑥,𝑦,𝜙(𝑧)=𝜋(𝑥),𝜋𝑦,𝜙(𝜋(𝑧=𝑥))_′,𝑦^′,𝜙𝑧^′,𝜙𝑥^′𝑦^′=𝜙𝜋(𝑥)𝜋𝑦=𝜙𝜋=𝑥𝑦=𝜋𝜙𝑥𝑦𝜙𝜋𝑥𝜋𝑦=𝜙𝑥^′𝑦^′𝑥𝜙𝑦(𝑥)𝑦=𝜋=𝜋𝜋𝜙=𝑥^′𝜙𝜋𝑦=𝑥^′𝜙𝑦^′, (4) which proves 𝜙∈Γ(𝔤₂).
(2) If 𝜙=0 for, 𝜙∈Γ(𝔤₁)∩End_𝐅(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 𝜋_Γ∶Γ(𝔤₁)→Γ(𝔤₂) is a well-defined homomorphism, which is an injection by (2).

Proposition 17. If the characteristic of  𝐅 is not 2, then 𝐶(𝔤)=Γ(𝔤)∩Der(𝔤). (5)

Proof. If 𝜓∈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, let 𝜓∈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 that 𝐶(𝔤)=Γ(𝔤)∩Der(𝔤).

Lemma 18. Let 𝐼 be a nonzero Γ(𝔤)-invariant ideal of 𝔤, 𝑉(𝐼)={𝜓∈Γ(𝔤)𝜓(𝐼)=0}, and let Hom(𝔤/𝐼,𝑍_𝔤(𝐼)) be the vector space of all linear maps from 𝔤/𝐼 to 𝑍_𝔤(𝐼) over 𝐅. Define 𝑇(𝐼)={𝑓∈Hom(𝔤/𝐼,𝑍_𝔤(𝐼))∣𝑓([𝑥,𝑦,𝑧])=[𝑓(𝑥),𝑦,𝑧]=[𝑥,𝑓(𝑦),𝑧]=[𝑥,𝑦,𝑓(𝑧)],𝑓(𝑥𝑦)=(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,𝑍_𝔤(𝐼)) and 𝑉(𝐼)≅𝑇(𝐼) as vector spaces.(2) If  Γ(𝐼)=𝐅id_𝐼, then Γ(𝔤)=𝐅id_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 𝛼𝜓𝑦⟼𝜓𝑦, (6) 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 𝜓_𝑓∶𝔤→𝔤,𝜓_𝑓(𝑥)=𝑓(𝑥), for all 𝑥∈𝔤. It follows from the definition of 𝑇(𝐼) that, for all 𝑥,𝑦,𝑧∈𝔤, 𝜓_𝑓(𝑥𝑦)=𝑓(𝑥𝑦)=(𝑓(𝑥))𝑦=𝑥(𝑓𝑦),𝜓_𝑓([𝑥,𝑦,𝑧])=𝑓([𝑥,𝑦,𝑧])=[𝑓(𝑥),𝑦,𝑧]=[𝑥,𝑓(𝑦),𝑧]=[𝑥,𝑦,𝑓(𝑧)]. Thus, 𝜓_𝑓∈Γ(𝔤), and so 𝜓_𝑓∈𝑉(𝐼),𝜓_𝑓(𝐼)=0. But, 𝛼(𝜓_𝑓)=𝑓 implies that 𝛼 is onto. It is fairly easy to see that 𝛼 preserves operations on vector spaces from 𝔤/𝐼 to 𝑍_𝔤(𝐼). This proves (1).
We now prove (2). If Γ(𝐼)=𝐅id_𝐼, then for all 𝜓∈Γ(𝔤),𝜓∣_𝐼=𝜆id_𝐼, for some 𝜆∈𝐅. If 𝜓≠𝜆id_𝔤, let 𝜑(𝑥)=𝜆𝑥, for all 𝑥∈𝔤; then 𝜓∈Γ(𝔤) and 𝜓−𝜑∈𝑉(𝐼). Clearly, 𝜓=𝜑+(𝜓−𝜑). Furthermore, 𝐅id_𝔤∩𝑉(𝐼)=0, and so (2) is proved.

Lemma 19. Let 𝔤 be a Lie triple algebra; then 𝜑𝐷 is a derivation for 𝜑∈Γ(𝔤),𝐷∈Der(𝔤).

Proof. If 𝑥,𝑦,𝑧∈𝔤, then (𝐷(𝑥𝐷𝑦=(𝑥)𝑦𝑦,𝐷(𝑥)+𝑦+(𝑧=(𝑥)+𝑦+(𝑧.𝜑𝐷𝑥𝑦=𝜑))𝑦+𝑥𝜑𝐷+𝑥𝜑𝐷𝜑𝐷𝑥,𝑦,𝑧=𝜑,𝑦,𝑧𝑥,𝐷,𝑧𝑥,𝑦,𝐷)𝜑𝐷,𝑦,𝑧𝑥,𝜑𝐷,𝑧𝑥,𝑦,𝜑𝐷) (7) Thus, 𝜑𝐷 is a derivation.

Theorem 20. Let 𝔤 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 any 𝜑∈Γ(𝔤),𝐷∈Der(𝔤), and 𝑥,𝑦,𝑧∈𝔤, 𝜑(𝑥)𝑦=(𝑥)𝜑(𝑥)=(𝑥)−(𝑥)𝐷𝜑𝑥𝑦=𝐷𝐷𝜑𝑦+𝐷𝑦𝐷𝜑𝑦+𝜑𝐷𝑥𝑦𝜑𝐷𝑦. (8)
Then, we get (𝐷𝜑−𝜑𝐷)(𝑥𝑦)=((𝐷𝜑−𝜑𝐷)(𝑥))𝑦. On the other hand, we have 𝜑(𝑥)=(𝑥)+𝜑(𝑥)𝑦+𝜑(𝑥)(𝑧=(𝑥)−(𝑥).𝐷𝜑𝑥,𝑦,𝑧=𝐷,𝑦,𝑧𝐷𝜑,𝑦,𝑧,𝐷,𝑧,𝑦,𝐷)𝐷𝜑,𝑦,𝑧+𝜑𝐷𝑥,𝑦,𝑧𝜑𝐷,𝑦,𝑧 (9) 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 that 𝔤 is a Lie triple algebra over 𝐅 and 𝔤=𝔤₁⊕𝔤₂ with 𝔤₁ and 𝔤₂ being ideals of 𝔤; then Γ(𝔤)≅Γ𝔤₁⊕Γ𝔤₂⊕𝐶(1)⊕𝐶(2), (10) 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. Let 𝜋_𝑖∶𝔤→𝔤_𝑖 be canonical projections for 𝑖=1,2; then 𝜋₁,𝜋₂∈Γ(𝔤) and 𝜋₁+𝜋₂=id_𝔤. So we have for 𝜑∈Γ(𝔤), 𝜑=𝜋₁𝜑𝜋₁+𝜋₁𝜑𝜋₂+𝜋₂𝜑𝜋₁+𝜋₂𝜑𝜋₂. (11) Note that 𝜋_𝑖𝜑𝜋_𝑗∈Γ(𝔤) for 𝑖,𝑗=1,2. We claim that Γ(𝔤)=𝜋₁Γ(𝔤)𝜋₁+𝜋₁Γ(𝔤)𝜋₂+𝜋₂Γ(𝔤)𝜋₁+𝜋₂Γ(𝔤)𝜋₂. (12) 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 𝑓_𝑖∈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=𝜋_𝑖Γ(𝔤)𝜋_𝑗,𝑖,𝑗=1,2. We now prove that Γ(𝔤)₁₁≅Γ𝔤₁,Γ(𝔤)₂₂≅Γ𝔤₂,Γ(𝔤)₁₂≅𝐶₂,Γ(𝔤)₂₁≅𝐶₁. (13) 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 that Γ(𝔤)₁₂≅𝐶₂. If 𝜑∈Γ(𝔤)₁₂ there exists 𝜑₀ in Γ(𝔤) such that 𝜑=𝜋₁𝜑₀𝜋₂. For 𝑥_𝑘=𝑥¹_𝑘+𝑥²_𝑘∈𝔤, where 𝑥^𝑖_𝑘∈𝔤_𝑖,𝑖=1,2 and 𝑘=1,2,3, we have 𝜑𝑥₁𝑥₂=𝜋₁𝜑₀𝜋₂𝑥₁𝑥₂=𝜋₁𝜑₀𝑥²₁𝑥²₂=𝜋₁𝜑₀𝑥²₁𝑥²₂𝜑𝑥=0,₁,𝑥₂,𝑥₃=𝜋₁𝜑₀𝜋₂𝑥₁,𝑥₂,𝑥₃=𝜋₁𝜑₀𝑥²₁,𝑥²₂,𝑥²₃=𝜋₁𝜑₀𝑥²₁,𝑥²₂,𝑥²₃𝜑𝑥=0,₁𝑥₂=𝜑𝑥₁𝑥₂𝜑𝑥=0,₁,𝑥₂,𝑥₃=𝜑𝑥₁,𝑥₂,𝑥₃=0. (14) 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, for 𝜑∈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: 𝜏∶Γ(𝔤)₁₂⟶𝐶₂,𝜏𝜑=𝜑∣Γ(𝔤)₁₂, (15) for all 𝜑∈Γ(𝔤)₁₂. Similarly, we can prove that Γ(𝔤)₂₁≅𝐶₁. Summarizing the aforesaid discussion, we have Γ(𝔤)≅Γ𝔤₁⊕Γ𝔤₂⊕𝐶(1)⊕𝐶(2). (16)
The proof is completed.

A generalized version of Theorem 21 is stated next without proof.

Theorem 22. Suppose that 𝔤 is a Lie triple algebra over 𝐅 and with a decomposition of ideals 𝔤=𝔤₁⊕𝔤₂⊕⋯⊕𝔤_𝑚. Then, one has Γ(𝔤)≅Γ𝔤₁⊕⋯⊕Γ𝔤_𝑚⊕⊕_{1≤𝑖≠𝑗≤𝑚}𝐶_𝑖𝑗, (17) as vector spaces, where 𝐶_𝑖𝑗={𝜑∈Hom(𝔤_𝑖,𝔤_𝑗)∣𝜑(𝔤_𝑖)⊆𝑍(𝔤_𝑗),𝜑(𝔤_𝑖𝔤_𝑖)=𝜑[𝔤_𝑖,𝔤_𝑖,𝔤_𝑖]=0, for 1≤𝑖≠𝑗≤𝑚}.

4. The Centroids of Tensor Product of Lie Triple Algebras

Definition 23. Let 𝐴 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∈End_𝐅(𝐴)𝜑(𝑎𝑏)=𝑎𝜑(𝑏)=𝜑(𝑎)𝑏,forall𝑎,𝑏∈𝐴}; then Γ(𝐴) is an associative algebra of End_𝐅(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: 𝑥₁⊗𝑎₁,𝑥₂⊗𝑎₂,𝑥₃⊗𝑎₃=𝑥₁,𝑥₂,𝑥₃⊗𝑎₁𝑎₂𝑎₃,𝑥₁⊗𝑎₁𝑥₂⊗𝑎₂=𝑥₁𝑥₂⊗𝑎₁𝑎₂, (18) where 𝑥_𝑖∈𝔤,𝑎_𝑖∈𝐴,𝑖=1,2,3. This Lie triple algebra 𝔤⊗𝐴 is called the tensor product of 𝐴 and 𝔤. For 𝑓∈End_𝐅(𝔤) and 𝜑∈End_𝐅(𝐴), there exists a unique map 𝑓⊗𝜑∈End_𝐅(𝔤⊗𝐴) such that 𝑓⊗𝜑(𝑥⊗𝑎)=𝑓(𝑥)⊗𝜑(𝑎),∀𝑥∈𝔤,𝑎∈𝐴. (19) The map should not be confused with the element 𝑓⊗𝜑 of the tensor product End_𝐅(𝔤)⊗End_𝐅(𝐴). Of course, we have a canonical map as the following: 𝜔∶End_𝐅(𝔤)⊗End_𝐅(𝐴)⟶End_𝐅(𝔤⊗𝐴)∶𝑓⊗𝜑⟼𝑓⊗𝑔. (20) 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 transformation 𝜙∈Γ(𝔤⊗𝐴) is said to have finite 𝔤-image, if for any 𝑎∈𝐴, there exist finitely many 𝑎₁,…,𝑎_𝑛∈𝐴 such that 𝜙(𝔤⊗𝐅𝑎)⊆𝔤⊗𝐅𝑎₁+⋯+𝔤⊗𝐅𝑎_𝑛. (21) 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⊗𝐅𝑎_𝑛 for suitable 𝑎_𝑖∈𝐴.

Lemma 25. Let 𝔤 be a Lie triple algebra over 𝐅 and 𝐴 a unitary commutative associative over 𝐅. Let {𝑎_𝑟}_𝑟∈ℜ be a basis of 𝐴, and let 𝜙∈Γ(𝔤⊗𝐴). Define 𝜙_𝑟∈End_𝐅(𝔤) by 𝜙(𝑥⊗1)=_𝑟∈ℜ𝜙_𝑟(𝑥)⊗𝑎_𝑟, (22) then all 𝜙_𝑟∈Γ(𝔤).

Proof. For 𝑥₁,𝑥₂,𝑥₃∈𝔤, we have 𝜙𝑥₁,𝑥₂,𝑥₃⊗1=_𝑟∈ℜ𝜙_𝑟𝑥₁,𝑥₂,𝑥₃⊗𝑎_𝑟,𝜙𝑥₁𝑥₂=_𝑟∈ℜ𝜙_𝑟𝑥₁𝑥₂⊗𝑎_𝑟,𝜙𝑥₁,𝑥₂,𝑥₃⊗1=𝜙𝑥₁⊗1,𝑥₂⊗1,𝑥₃=𝜙𝑥⊗1₁⊗1,𝑥₂⊗1,𝑥₃⊗1=_𝑟∈ℜ𝜙_𝑟𝑥₁,𝑥₂,𝑥₃⊗𝑎_𝑟=𝑥₁⊗1,𝜙𝑥₂⊗1,𝑥₃⊗1=_𝑟∈ℜ𝑥₁,𝜙_𝑟𝑥₂,𝑥₃⊗𝑎_𝑟=𝑥₁⊗1,𝑥₂⊗1,𝜙𝑥₃⊗1=_𝑟∈ℜ𝑥₁,𝑥₂,𝜙_𝑟𝑥₃⊗𝑎_𝑟,𝜙𝑥₁𝑥₂⊗1=𝜙𝑥₁⊗1𝑥₂⊗1=_𝑟∈ℜ𝜙_𝑟𝑥₁𝑥₂⊗𝑎_𝑟=𝜙𝑥₁⊗1𝑥₂⊗1=_𝑟∈ℜ𝜙_𝑟𝑥₁𝑥₂⊗𝑎_𝑟=𝑥₁⊗1𝜙𝑥₂⊗1=_𝑟∈ℜ𝑥₁𝜙_𝑟𝑥₁⊗𝑎_𝑟. (23) Hence, one has 𝜙_𝑟𝑥₁,𝑥₂,𝑥₃=𝜙_𝑟𝑥₁,𝑥₂,𝑥₃=𝑥₁,𝜙_𝑟𝑥₂,𝑥₃=𝑥₁,𝑥₂,𝜙_𝑟𝑥₃,𝜙_𝑟𝑥₁𝑥₂=𝜙_𝑟𝑥₁𝑥₂=𝑥₁𝜙_𝑟𝑥₂, (24) for all 𝑥_𝑖∈𝔤. So all 𝜙_𝑟∈Γ(𝔤).

Proposition 26. Let 𝔤 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)If 𝔤 is perfect, then 𝔤⊗𝐴 is perfect too.(2)If 𝔤 is finitely generated as a Mult(𝔤)-module (or as a Γ(𝔤)-module), or if 𝔤 is a central and a torsion-free 𝐅-module, then every 𝜙∈Γ(𝔤⊗𝐴) has finite 𝔤-image.(3)If Γ(𝔤) is a free 𝐅-module and the map 𝜔 of Definition 23 is injective, then Γ(𝔤)⊗Γ(𝐴)={𝜙∈Γ(𝔤⊗𝐴)∣𝜙  has finite  𝔤-𝑖𝑚𝑎𝑔𝑒}.

Proof. The definition of perfect shows that (1) holds. Let 𝑀=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+⋯+𝑀𝑥_𝑛 for 𝑥₁,…,𝑥_𝑛∈𝔤. Fix 𝜙∈Γ(𝔤⊗𝐴) and 𝑎∈𝐴. There exist finite families {𝑥_𝑖𝑗}⊆𝔤 and {𝑎_𝑖𝑗}⊆𝐴 such that 𝜙𝑥_𝑖⊗𝑎=_𝑗𝑥_𝑖𝑗⊗𝑎_𝑖𝑗, (25) for 1≤𝑖≤𝑛. Hence, 𝜙(𝔤⊗𝑎)=_𝑖𝜙(𝑀⊗id)𝑥_𝑖=⊗𝑎_𝑖,𝑗(𝑀⊗id)𝑥_𝑖𝑗⊗𝑎_𝑖𝑗⊆_𝑖,𝑗𝔤⊗𝑎_𝑖𝑗. (26) 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 that Γ(𝔤)=𝐅id and that 𝔤 is a torsion-free 𝐅-module. Then there exist scalars 𝑘_𝑟∈𝐅 such that 𝜙_𝑟=𝑘_𝑟id. Hence, 𝜙(𝑥⊗1)=_𝑟∈ℜ𝑘_𝑟𝑥⊗𝑎_𝑟. (27) Fix 𝑥∈𝔤; then almost all 𝑘_𝑟𝑥=0 and almost all 𝑘_𝑟=0, which in turn implies that 𝜙 has finite 𝔤-image.
From the aforementione discussion above, we get Γ(𝔤)⊗Γ(𝐴)⊆𝜙∈Γ(𝔤⊗𝐴)∣𝜙hasfinite𝔤-image. (28) 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: 𝜙(𝑥⊗1)=_𝑟∈𝔍𝜙_𝑟(𝑥)⊗𝑎_𝑟. (29) For 𝑥₁,𝑥₂,𝑥₃∈𝔤 and 𝑎∈𝐴, we get 𝜙𝑥₁,𝑥₂,𝑥₃⊗𝑎=𝜙𝑥₁⊗1,𝑥₂⊗1,𝑥₃=𝜙𝑥⊗𝑎₁⊗1,𝑥₂⊗1,𝑥₃⊗𝑎=_𝑟∈𝔍𝜙_𝑟𝑥₁,𝑥₂,𝑥₃⊗𝑎_𝑟𝑎=_𝑟∈𝔍𝜙_𝑟𝑥₁,𝑥₂,𝑥₃⊗𝑎_𝑟𝜙𝑥𝑎,₁𝑥₂⊗𝑎=𝜙𝑥₁⊗1𝑥₂=𝜙𝑥⊗𝑎₁⊗1𝑥₂⊗𝑎=_𝑟∈𝔍𝜙_𝑟𝑥₁𝑥₂⊗𝑎_𝑟𝑎=_𝑟∈𝔍𝜙_𝑟𝑥₁𝑥₂⊗𝑎_𝑟𝑎. (30) Since 𝔤 is perfect, we can get 𝜙=_𝑟∈𝔍𝜙_𝑟⊗𝜆_𝑟, (31) where 𝜆_𝑟 is the left multiplication in 𝐴 by 𝑎_𝑟. Let {𝜓_𝑠∣𝑠∈𝔖} be a basis of Γ(𝔤). Then, there exist a finite subset 𝔅⊆𝔖 and scalars 𝑘_𝑟𝑠∈𝐅(𝑟∈𝔍,𝑠∈𝔅) such that 𝜙_𝑟=_𝑠∈𝔅𝑘_𝑟𝑠𝜓_𝑠. (32) We then get 𝜙=_{𝑟∈𝔍𝑠∈𝔅}𝑘_𝑟𝑠𝜓_𝑠⊗𝜆_𝑟=_𝑠∈𝔅𝜓_𝑠⊗𝜙_𝑠,for𝜙_𝑠=_𝑟∈𝔍𝑘_𝑟𝑠𝜆_𝑟. (33) Now we show that 𝜙_𝑠∈Γ(𝐴). For any 𝑥₁,𝑥₂,𝑥₃∈𝔤,𝑎₁,𝑎₂∈𝐴, we have _𝑠∈𝔅𝜓_𝑠𝑥₁,𝑥₂,𝑥₃⊗𝜙_𝑠𝑎₁𝑎₂=𝜙𝑥₁,𝑥₂,𝑥₃⊗𝑎₁𝑎₂=𝜙𝑥₁⊗𝑎₁,𝑥₂⊗𝑎₂,𝑥₃=⊗1_𝑠∈𝔅𝜓_𝑠𝑥₁⊗𝜙_𝑠𝑎₁,𝑥₂⊗𝑎₂,𝑥₃⊗1=_𝑠∈𝔅𝜓_𝑠𝑥₁,𝑥₂,𝑥₃⊗𝜙_𝑠𝑎₁𝑎₂=_𝑠∈𝔅𝜓_𝑠𝑥₁,𝑥₂,𝑥₃⊗𝜙_𝑠𝑎₁𝑎₂,_𝑠∈𝔅𝜓_𝑠𝑥₁𝑥₂⊗𝜙_𝑠𝑎₁=𝜙𝑥₁𝑥₂⊗𝑎₁=𝜙𝑥₁⊗𝑎₁𝑥₂⊗1=_𝑠∈𝔅𝜓_𝑠𝑥₁𝑥₂⊗𝜙_𝑠𝑎₁1=_𝑠∈𝔅𝜓_𝑠𝑥₁𝑥₂⊗𝜙_𝑠𝑎₁1. (34) Since [𝔤𝔤,𝔤]=𝔤𝔤=𝔤, we have _𝑠∈𝔅𝜓_𝑠(𝑥)⊗𝜙_𝑠𝑎₁𝑎−𝜙_𝑠𝑎₁𝑎=0, (35)forall𝑥∈𝔤 and 𝑎₁,𝑎₂∈𝐴. It follows that _𝑠∈𝔅𝜓_𝑠⊗𝜇_𝑠=0, (36) where 𝜇_𝑠∈End_𝐅(𝐴) is defined by 𝜇_𝑠(𝑎₂)=𝜙_𝑠(𝑎₁𝑎₂)−𝜙_𝑠(𝑎₁)𝑎₂, for all 𝑎₂∈𝐴. Since 𝜔 is injective, we also have _𝑠∈𝔅𝜓_𝑠⊗𝜇_𝑠=0. (37) So by the linear independence of the 𝜓_𝑠, we get that 𝜇_𝑠=0. Then 𝜓_𝑠∈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 filed 𝐅. Let {𝑒_𝑘} be a basis of the central derivation 𝐶(𝔤) and {𝜑_𝑗} a maximal subset of Γ(𝔤) such that {𝜑_𝑗∣_𝔤𝔤} and {𝜑_𝑗∣_{[𝔤,𝔤,𝔤]}} are linear independent. Then, we have the following result.

Theorem 27. Let Ψ denote the subspace of Γ(𝔤) spanned by {𝜑_𝑗}. Then {𝑒_𝑘,𝜑_𝑗} is a basis of Γ(𝔤) and Γ(𝔤)=Ψ⊕𝐶(𝔤) as vector spaces.

Proof. Since {𝜑_𝑗∣_𝔤𝔤} and {𝜑_𝑗∣_{[𝔤,𝔤,𝔤]}} are linear independent, {𝜑_𝑗} is linear independent in Γ(𝔤). By definition of {𝑒_𝑘,𝜑_𝑗}, the {𝑒_𝑘,𝜑_𝑗} is independent in Γ(𝔤). For 𝜑∈Γ(𝔤) since {𝜑_𝑗∣_𝔤𝔤} is a basis of vector spaces {𝜑∣_𝔤𝔤∣𝜑∈Γ(𝔤)}, and {𝜑_𝑗∣_{[𝔤,𝔤,𝔤]}} is a basis of vector spaces {𝜑∣_{[𝔤,𝔤,𝔤]}∣𝜑∈Γ(𝔤)}, there exist 𝑔_𝑠∈𝐅, 𝑠∈𝐽 (𝐽 is a finite set of positive integers) such that 𝜑∣_{[𝔤,𝔤,𝔤]}=_𝑠∈𝐽𝑔_𝑠𝜑_𝑠∣_{[𝔤,𝔤,𝔤]},𝜑∣_𝔤𝔤=_𝑠∈𝐽𝑔_𝑠𝜑_𝑠∣_𝔤𝔤. (38) We then have 𝜑−_𝑠∈𝐽𝑔_𝑠𝜑_𝑠∣_{[𝔤,𝔤,𝔤]}=0,𝜑−_𝑠∈𝐽𝑔_𝑠𝜑_𝑠∣_𝔤𝔤=0. (39) If 𝑦₁,𝑦₂,𝑦₃∈𝔤, then 0=𝜑−_𝑠∈𝐽𝑔_𝑠𝜑_𝑠𝑦₁,𝑦₂,𝑦₃=𝜑−_𝑠∈𝐽𝑔_𝑠𝜑_𝑠𝑦₁,𝑦₂,𝑦₃,0=𝜑−_𝑠∈𝐽𝑔_𝑠𝜑_𝑠𝑦₁𝑦₂=𝜑−_𝑠∈𝐽𝑔_𝑠𝜑_𝑠𝑦₁𝑦₂. (40) It follows that 𝜑−_𝑠∈𝐽𝑔_𝑠𝜑_𝑠(𝔤)⊆𝑍(𝔤),𝜑−_𝑠∈𝐽𝑔_𝑠𝜑_𝑠, (41) is a central derivation. So there exist 𝑟_𝑖∈𝐅, 𝑖∈𝐼 (𝐼 is a finite set of positive integers) such that 𝜑−_𝑠∈𝐽𝑔_𝑠𝜑_𝑠=_𝑖∈𝐼𝑟_𝑖𝑐_𝑖. (42) Therefore, 𝜑=_𝑠∈𝐽𝑔_𝑠𝜑_𝑠+_𝑖∈𝐼𝑟_𝑖𝑐_𝑖. (43) The proof is completed.

Lemma 28. Let 𝔤 be a simple Lie triple algebra, 𝑅=𝐅[𝑥₁,…,𝑥_𝑛], and 𝔤=𝔤⊗𝑅; then one has Γ(𝔤)⊇Ψ⊗𝑅+𝐶(𝑔)⊗End_𝐅(𝑅).

Proof. Let 𝜓∈Γ(𝔤),𝑚∈𝑅. Then, we have ==𝜓(𝑥)=𝑦=(𝑧=𝜓(𝑥)=𝑦(=(𝑧==𝜓(𝑥)𝑦=𝜓(𝑥)=𝜓(𝑥𝜓⊗𝑚𝑥⊗𝑝,𝑦⊗𝑞,𝑧⊗𝑟𝜓⊗𝑚𝑥,𝑦,𝑧⊗𝑝𝑞𝑟=𝜓𝑥,𝑦,𝑧⊗𝑚𝑝𝑞𝑟,𝑦,𝑧⊗𝑚𝑝𝑞𝑟𝑥,𝜓,𝑧⊗𝑚𝑝𝑞𝑟𝑥,𝑦,𝜓)⊗𝑚𝑝𝑞𝑟,𝜓⊗𝑚𝑥⊗𝑝,𝑦⊗𝑞,𝑧⊗𝑟,𝑦,𝑧⊗𝑚𝑝𝑞𝑟,𝑥⊗𝑝,𝜓⊗𝑚𝑦⊗𝑞,𝑧⊗𝑟𝑥,𝜓,𝑧⊗𝑚𝑝𝑞𝑟,𝑥⊗𝑝,𝑦⊗𝑞,𝜓⊗𝑚𝑧⊗𝑟)𝑥,𝑦,𝜓)⊗𝑚𝑝𝑞𝑟,𝜓⊗𝑚𝑥⊗𝑝𝑦⊗𝑞𝜓⊗𝑚𝑥𝑦⊗𝑝𝑞=𝜓𝑥𝑦⊗𝑚𝑝𝑞⊗𝑚𝑝𝑞,𝜓⊗𝑚𝑥⊗𝑝𝑦⊗𝑞⊗𝑚𝑝𝑦⊗𝑞)𝑦⊗𝑚𝑝𝑞. (44)(45) Thus 𝜓⊗𝑚∈Γ(𝔤). For any 𝑑∈𝐶(𝑔) and 𝑓∈End_𝐅(𝑅),𝑑⊗𝑓∈Γ(𝔤), we have =𝑑(𝑥)𝑝=𝑑(𝑥)𝑝=𝑦𝑞=𝑦𝑞(=(𝑧)(𝑟=(𝑧(𝑟)=𝑑(𝑥)𝑝=𝑑(𝑥)𝑦𝑝=𝑑𝑦𝑞=𝑦𝑞𝑑⊗𝑓𝑥⊗𝑝,𝑦⊗𝑞,𝑧⊗𝑟=𝑑𝑥,𝑦,𝑧⊗𝑓𝑝𝑞𝑟=0,𝑑⊗𝑓𝑥⊗𝑝,𝑦⊗𝑞,𝑧⊗𝑟⊗𝑓,𝑦⊗𝑞,𝑧⊗𝑟,𝑦,𝑧⊗𝑓𝑞𝑟=0,𝑥⊗𝑝𝑑⊗𝑓𝑦⊗𝑞,𝑧⊗𝑟𝑥⊗𝑝,𝑑⊗𝑓,𝑧⊗𝑟𝑥,𝑑,𝑧⊗𝑝𝑓𝑟=0,𝑥⊗𝑝,𝑦⊗𝑞,𝑑⊗𝑓𝑧⊗𝑟)𝑥⊗𝑝𝑦⊗𝑞,𝑑⊗𝑓)𝑥,𝑦,𝑑)⊗𝑝𝑞𝑓=0,𝑑⊗𝑓𝑥⊗𝑝𝑦⊗𝑞=𝑑𝑥𝑦⊗𝑓𝑝𝑞=0,𝑑⊗𝑓𝑥⊗𝑝𝑦⊗𝑞⊗𝑓𝑦⊗𝑞⊗𝑓𝑞=0,𝑥⊗𝑝𝑑⊗𝑓𝑦⊗𝑞𝑥⊗𝑝⊗𝑓𝑥𝑑⊗𝑝𝑓=0. (46) Hence Ψ⊗𝑅+𝐶(𝑔)⊗End_𝐅(𝑅)⊆Γ(𝔤).

Theorem 29. Let 𝔤 be a simple Lie triple algebra, 𝑅=𝐅[𝑥₁,…,𝑥_𝑛], and 𝔤=𝔤⊗𝑅. Then Γ(𝔤)=Γ(𝔤)⊗𝑅.

Proof. From Lemma 28, we get Γ(𝔤)⊗𝑅⊆Γ(𝔤). Now we prove the opposite inclusion. Let {𝑚_𝑖} be a basis for 𝑅,𝜓∈Γ(𝔤),𝑝∈𝑅,𝑥,𝑦,𝑧∈𝔤, and let 𝜂_𝑖(−,𝑝) be suitable maps in End_𝐅(𝔤) such that 𝜓𝑥⊗𝑝=_𝑖𝜂_𝑖𝑥,𝑝⊗𝑚_𝑖. (47) Then, 𝜓=𝜓=𝑥⊗𝑝,𝑦⊗1,𝑧⊗1𝑥⊗𝑝,𝑦⊗1,𝑧⊗1_𝑖𝜂_𝑖𝑥,𝑝⊗𝑚_𝑖,𝑦⊗1,𝑧⊗1=_𝑖𝜂_𝑖𝑥,𝑝,𝑦,𝑧⊗𝑚_𝑖===𝜓𝑥⊗1,𝑦⊗𝑝,𝑧⊗1𝑥⊗1,𝜓𝑦⊗𝑝,𝑧⊗1𝑥⊗1,_𝑖𝜂_𝑖𝑦,𝑝⊗𝑚_𝑖=,𝑧⊗1_𝑖𝑥,𝜂_𝑖𝑦,𝑝,𝑧⊗𝑚_𝑖===𝜓𝑥⊗1,𝑦⊗1,𝑧⊗𝑝𝑥⊗1,𝑦⊗1,𝜓𝑧⊗𝑝𝑥⊗1,𝑦⊗1,_𝑖𝜂_𝑖𝑧,𝑝⊗𝑚_𝑖=_𝑖𝑥,𝑦,𝜂_𝑖𝑧,𝑝⊗𝑚_𝑖,𝜓𝑥⊗𝑝𝑦⊗1=_𝑖𝜂_𝑖𝑥,𝑝⊗𝑚_𝑖𝜂𝑦⊗1=_𝑖𝑥,𝑝𝑦⊗𝑚_𝑖=𝑥⊗𝑝𝜓𝑦⊗1=_𝑖𝑥,𝜂_𝑖𝑦,𝑝⊗𝑚_𝑖, (48) while 𝜓𝑥,𝑦,𝑧⊗𝑝=_𝑖𝜂_𝑖𝑥,𝑦,𝑧,𝑝⊗𝑚_𝑖,𝜓𝑥𝑦⊗𝑝=_𝑖𝜂_𝑖𝑥𝑦,𝑝⊗𝑚_𝑖. (49) So, for each 𝑖 and 𝑝, we have 𝜂_𝑖(𝑝[𝑥,𝑦,𝑧])=[𝜂_𝑖(𝑥,𝑝),𝑦,𝑧]=[𝑥,𝜂_𝑖(𝑦,𝑝),𝑧]=[𝑥,𝑦,𝜂_𝑖(𝑧,𝑝)], 𝜂_𝑖(𝑝,(𝑥𝑦))=(𝜂_𝑖(𝑥,𝑝),𝑦)=(𝑥,𝜂_𝑖(𝑦,𝑝)),forall𝑥,𝑦,𝑧∈𝔤. Hence, 𝜂_𝑖(−,𝑝)∈Γ(𝔤). But Γ(𝔤)=𝐅id. Therefore, 𝜂_𝑖(𝑥,𝑝)=𝜆_𝑖(𝑝)𝑥, for all 𝑥∈𝔤, for suitable scalars 𝜆_𝑖(𝑝).
Thus, we may write 𝜓𝑥⊗𝑝=_𝑖𝜆_𝑖𝑝𝑥⊗𝑚_𝑖. (50) Since the right-hand side is in 𝔤⊗𝑅, for each 𝑝, we get 𝜆_𝑖(𝑝)=0 for all except for a finite number of 𝑖. That is, 𝜓𝑥⊗𝑝=^𝑛_𝑖=1𝑥⊗𝜆_𝑖𝑝𝑚_𝑖. (51) Then, the map 𝜌∶𝑝⟼^𝑛_𝑖=1𝜆_𝑖𝑝𝑚_𝑖 (52) is well defined. Hence, we have 𝜓(𝑥⊗𝑝)=𝑥⊗𝜌(𝑝). Thus, 𝜓=𝑝=𝑝,=(1)=(1),𝜓=𝑝=𝑝=𝜓=(1)=(1).𝑥⊗𝑝,𝑦⊗1,𝑧⊗1𝑥⊗𝜌,𝑦⊗1,𝑧⊗1𝑥,𝑦,𝑧⊗𝜌𝑥⊗𝑝,𝜓𝑦⊗1,𝑧⊗1𝑥⊗𝑝,𝑦⊗𝜌,𝑧⊗1𝑥,𝑦,𝑧⊗𝑝𝜌𝑥⊗𝑝𝑦⊗1𝑥⊗𝜌𝑦⊗1𝑥𝑦⊗𝜌𝑥⊗𝑝𝑦⊗1𝑥⊗𝑝𝑦⊗𝜌𝑥𝑦⊗𝑝𝜌 (53) 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, 𝜓𝑥⊗𝑝=id⊗𝜌𝑥⊗𝑝,∀𝑝∈𝑅,𝑥∈𝔤. (54) That is, 𝜓∈id⊗𝑅≅Γ(𝔤)⊗𝑅. So, we have Γ(𝔤)=Γ(𝔤)⊗𝑅.

Acknowledgments

The authors would like to thank the referee for valuable comments and suggestions on this paper. The works was supported by NNSF of China (no. 11171057), Natural Science Foundation of Jilin province (no. 201115006), the Fundamental Research Funds for the Central Universities (no. 12SSXT139), and Scientific Research Foundation for Returned Scholars, Ministry of Education of China.