Generalized Derivations of Hom–Lie Triple Systems

Abstract

In this paper, we give some properties of the generalized derivation algebra $\mathrm{GDer}(T)$ of a Hom–Lie triple systems T. In particular, we prove that $\mathrm{GDer}(T) = \mathrm{QDer}(T) + \mathrm{QC}(T)$, the sum of the quasiderivation algebra and the quasicentroid. We also prove that $\mathrm{QDer}(T)$ can be embedded as derivations in a larger Hom–Lie triple system. General results on centroids of Hom–Lie triple systems are also developed in this paper.

1 Introduction

Lie triple systems arose initially in Cartan’s study of Riemannian geometry, but whose concept was introduced by Jacobson in 1949 to study subspaces of associative algebras closed under triple commutators [[u, v], w]. 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.

Hom-type algebras have been investigated by many authors (cf. [2, 3, 5, 6, 9, 11, 13–16]). In particular, the notion of Hom–Lie triple systems was introduced by Yau (cf. [17]). Hom–Lie triple systems are a generalization of Lie triple systems whose classical ternary Jacobi identity is twisted by two linear maps. When these twisted maps are both equal to the identity map, Hom–Lie triple systems degenerate to Lie triple systems.

As is well known, derivation and generalized derivation algebras are very important subjects both in the research of Lie algebras and Lie triple systems. The most important and systematic research on the generalized derivation algebra of a Lie algebra was due to Leger and Luks (cf. [7]). In [7], some nice properties of the generalized derivation algebra and their subalgebras, for example, of the quasiderivation algebra and of the centroid have been obtained. In particular, they investigated the structure of the generalized derivation algebra and characterized the Lie algebras satisfying certain conditions. Meanwhile, they also pointed that there exist some connections between quasiderivations and cohomology of Lie algebras. Melville dealt particularly with the centroids of nilpotent Lie algebras (cf. [12]). In [8], some new results concerning the centroids of Lie triple systems were proved, and some conclusions of the tensor product of a Lie triple system and a unital commutative associative algebra are given.

The purpose of this paper is to generalize some beautiful results in [7, 8] to the generalized derivation algebra of a Hom–Lie triple system. In this paper, we mainly study the derivation algebra $\mathrm{Der}(T)$, the center derivation algebra ZDer(T), the quasiderivation algebra QDer(T), the centroid C(T), and the generalized derivation algebra GDer(T) of a Hom–Lie triple system T.

We proceed as follows. In Sect. 2, we recall some basic definitions and propositions which will be used in what follows. In Sect. 3, we study generalized derivation algebra and their Hom-subalgebras, showing that the quasicentroid of a Hom–Lie triple system is a Hom–Lie algebra if only and if it is a Hom-associative algebra. In Sect. 4, we prove that the quasiderivations of T can be embedded as derivations in a larger Hom–Lie triple system and obtain a direct sum decomposition of $\mathrm{Der}(\tilde{T})$ when the annihilator of T is equal to zero. In Sect. 5, we introduce the notion of centroids of Hom–Lie triple systems and we show some properties of centroids of Lie triple systems. In Sect. 6, we investigate the centroid of the tensor product of a Hom–Lie triple system T and a unital commutative associative algebra A.

2 Preliminaries

Definition 2.1

[15] A vector space T together with a trilinear map $(x, y, z)\rightarrow [x, y, z]$ is called a Lie triple system if

• (1) $[x, x, z]=0,$

• (2) $[x, y, z]+ [y, z, x]+[z, x, y]=0,$

• (3) $[x, y, [z, u, v]]=[[x, y, z], u, v]+[z, [x, y, u], v]+[z, u, [x, y, v]],$

for all $x, y, z, u, v \in T$.

Definition 2.2

[10] A Hom–Lie triple system $(T, [\cdot , \cdot , \cdot ], \alpha =(\alpha _{1}, \alpha _{2}))$ consists of a vector space T over a field $\mathbb {F}$, a trilinear map $[\cdot , \cdot , \cdot ]:T\times T\times T\rightarrow T$, and two linear maps $\alpha _{i}: T\rightarrow T$ for $i=1, 2$, called twisted maps, such that for all $x, y, z, u, v \in T$, $(1)\quad [x, x, z]=0,$

• (2) $[x, y, z]+ [y, z, x]+[z, x, y]=0,$

• (3) $[\alpha _{1}(x), \alpha _{2}(y), [z, u, v]]=[[x, y, z], \alpha _{1}(u), \alpha _{2}(v)]\,+\,[\alpha _{1}(z), [x, y, u], \alpha _{2}(v)]\quad +[\alpha _{1}(z), \alpha _{2}(u), [x, y, v]].$

In particular, if $\alpha =\alpha _{1}=\alpha _{2}$ preserves the bracket, (i.e., $\alpha [x, y, z]=[\alpha (x), \alpha (y), \alpha (z)]$, $\forall x, y, z\in T$), then we call $(T, [\cdot , \cdot , \cdot ], \alpha )$ a multiplicative Hom–Lie triple system.

A morphism $f:(T, [\cdot , \cdot , \cdot ], \alpha =(\alpha _{1}, \alpha _{2}))\rightarrow (T', [\cdot , \cdot , \cdot ], \alpha '=(\alpha '_{1}, \alpha '_{2}))$ of Hom–Lie triple systems is a linear map satisfying $f([x, y, z])=[f(x), f(y), f(z)]$ and $f\circ \alpha =\alpha ' \circ f$ for $i = 1, 2$. An isomorphism is a bijective morphism.

When the twisted maps $\alpha _{i}$ are both equal to the identity map, a Hom–Lie triple system is a Lie triple system, so Lie triple systems are special examples of Hom–Lie triple systems [10].

Definition 2.3

Let $(T, [\cdot , \cdot , \cdot ], \alpha )$ be a multiplicative Hom–Lie triple system and define the following subvector space $\mho$ of $\mathrm{End}(T)$ consisting of linear maps D on T as

$$\begin{aligned} \mho =\{D\in \mathrm{End}(L) |~D\alpha =\alpha D \} \end{aligned}$$

and

$$\begin{aligned} \tilde{\alpha }:\mho \rightarrow \mho ;~\tilde{\alpha }(D)=\alpha D. \end{aligned}$$

Then $(\mho , [\cdot , \cdot ], \tilde{\alpha })$ is a Hom–Lie algebra over $\mathbb {F}$ with the bracket

$$\begin{aligned}{}[D_{1}, D_{2}]=D_{1}D_{2}-D_{2}D_{1}, \end{aligned}$$

for all $D_{1}, D_{2}\in \mho .$

Definition 2.4

Let $(T, [\cdot , \cdot , \cdot ], \alpha )$ be a multiplicative Hom–Lie triple system. A linear map $D:T\rightarrow T$ is said to be an $\alpha ^{k}$-derivation of T(where $k\in {\mathbb {N}}, ~k\ge 0$), if it satisfies for all $x, y, z\in T,$

$$\begin{aligned} D\alpha= & {} \alpha D,\\&[D(x), \alpha ^{k}(y), \alpha ^{k}(z)]+[\alpha ^{k}(x), D(y), \alpha ^{k}(z)]\\&+\,[\alpha ^{k}(x), \alpha ^{k}(y), D(z)]=D([x, y, z]). \end{aligned}$$

We denote the set of all $\alpha ^{k}$-derivations by $\mathrm{Der}_{\alpha ^{k}}(T)$, then $\mathrm{Der}(T)=\bigoplus _{k\ge 0}\mathrm{Der}_{\alpha ^{k}}(T)$ provided with the commutator and the following map

$$\begin{aligned} \tilde{\alpha }:\mathrm{Der}(T)\rightarrow \mathrm{Der}(T);~~\tilde{\alpha }(D)=D\alpha \end{aligned}$$

is a Hom-subalgebra of $\mho$ and is called the derivation algebra of T.

Definition 2.5

An endomorphism $D\in \mathrm{End}(T)$ is said to be a generalized $\alpha ^{k}$-derivation of T, if there exist two endomorphisms $D', D''\in \mathrm{End}(T)$ such that

$$\begin{aligned} D\alpha= & {} \alpha D, ~D^{'}\alpha =\alpha D^{'}, ~D^{''}\alpha =\alpha D^{''}, ~D^{'''}\alpha =\alpha D^{'''},\nonumber \\&[D(x), \alpha ^{k}(y), \alpha ^{k}(z)]+[\alpha ^{k}(x), D^{'}(y), \alpha ^{k}(z)]\nonumber \\&+\,[\alpha ^{k}(x), \alpha ^{k}(y), D^{''}(z)]=D^{'''}([x, y, z]), \end{aligned}$$

(2.1)

for all $x, y, z\in T.$

Definition 2.6

An endomorphism $D\in \mathrm{End}(T)$ is said to be an $\alpha ^{k}$-quasiderivation, if there exists an endomorphism $D'\in \mathrm{End}(T)$ such that

$$\begin{aligned} D\alpha= & {} \alpha D, ~D^{'}\alpha =\alpha D^{'}\nonumber \\&[D(x), \alpha ^{k}(y), \alpha ^{k}(z)]+[\alpha ^{k}(x), D(y), \alpha ^{k}(z)]\nonumber \\&+\,[\alpha ^{k}(x), \alpha ^{k}(y), D(z)]=D^{'}([x, y, z]), \end{aligned}$$

(2.2)

for all $x, y, z\in T.$

Let $\mathrm{GDer}_{\alpha ^{k}}(T)$ and $\mathrm{QDer}_{\alpha ^{k}}(T)$ be the sets of generalized $\alpha ^{k}$-derivations and of $\alpha ^{k}$-quasiderivations, respectively. That is

$$\begin{aligned} \mathrm{GDer}(T)=\bigoplus _{k\ge 0}\mathrm{GDer}_{\alpha ^{k}}(T), \quad \mathrm{QDer}(T)=\bigoplus _{k\ge 0}\mathrm{QDer}_{\alpha ^{k}}(T). \end{aligned}$$

It is easy to verify that both $\mathrm{GDer}(T)$ and $\mathrm{QDer}(T)$ are Hom-subalgebras of $\mho$ (See Proposition 3.1).

Definition 2.7

If $\mathrm{C}(T)=\bigoplus _{k\ge 0}\mathrm{C}_{\alpha ^{k}}(T),$ with $\mathrm{C}_{\alpha ^{k}}(T)$ consisting of $D\in \mathrm{End}(T)$ satisfying

$$\begin{aligned} D\alpha&= \alpha D, \\ [D(x), \alpha ^{k}(y), \alpha ^{k}(z)]&=[\alpha ^{k}(x), D(y), \alpha ^{k}(z)]=[\alpha ^{k}(x), \quad \alpha ^{k}(y), D(z)]=D([x, y, z]), \end{aligned}$$

for all $x, y, z\in T,$ then $\mathrm{C}(T)$ is called an $\alpha ^{k}$-centroid of T.

Definition 2.8

If $\mathrm{QC}(T)=\bigoplus _{k\ge 0}\mathrm{QC}_{\alpha ^{k}}(T)$ with $\mathrm{QC}_{\alpha ^{k}}(T)$ consisting of $D\in \mathrm{End}(T)$ such that

$$\begin{aligned}&D\alpha =\alpha D,\\&[D(x), \alpha ^{k}(y), \alpha ^{k}(z)]=[\alpha ^{k}(x), D(y), \alpha ^{k}(z)]=[\alpha ^{k}(x), \alpha ^{k}(y), D(z)], \end{aligned}$$

for all $x, y, z\in T,$ then $\mathrm{QC}(T)$ is called an $\alpha ^{k}$-quasicentroid of T.

Define $\mathrm{ZDer}(T):=\bigoplus _{k\ge 0}\mathrm{Der}_{\alpha ^{k}}(T)$, where $\mathrm{Der}_{\alpha ^{k}}(T)$ consists of $D\in \mathrm{End}(T)$ such that

$$\begin{aligned} D\alpha= & {} \alpha D, \\&[D(x), \alpha ^{k}(y), \alpha ^{k}(z)]=D([x, y, z])=0, \end{aligned}$$

for all $x, y, z\in T,$ then $\mathrm{ZDer}(T)$ is called an $\alpha ^{k}$-central derivation of T.

It is easy to verify that

$$\begin{aligned} \mathrm{ZDer}(T)\subseteq \mathrm{Der}(T)\subseteq \mathrm{QDer}(T)\subseteq \mathrm{GDer}(T)\subseteq \mathrm{End}(T). \end{aligned}$$

Definition 2.9

Let $(T, [\cdot , \cdot , \cdot ], \alpha )$ be a multiplicative Hom–Lie triple system. If $\mathrm{Z}(T)=\{x\in T| [x, y, z]=0, \forall y, z\in T\},$ then $\mathrm{Z}(T)$ is called the center of T.

3 Generalized derivation algebra and their Hom-subalgebras

First, we give some basic properties of the center derivation algebra, the quasiderivation algebra and the generalized derivation algebra of a Hom–Lie triple system.

Proposition 3.1

If $(T, [\cdot , \cdot , \cdot ], \alpha )$ is a multiplicative Hom–Lie triple system, then the following statements hold:

• (1) $\mathrm{GDer}(T), \mathrm{QDer}(T)$ and $\mathrm{C}(T)$ are Hom-subalgebras of $\mho$.

• (2) $\mathrm{ZDer}(T)$ is a Hom-ideal of $\mathrm{Der}(T)$.

Proof

(1) Assume that $D_{1}\in \mathrm{GDer}_{\alpha ^{k}}(T), ~D_{2}\in \mathrm{GDer}_{\alpha ^{s}}(T). ~\forall x, y, z\in T,$ we have

$$\begin{aligned}&\quad [(\tilde{\alpha }(D_{1}))(x), \alpha ^{k+1}(y), \alpha ^{k+1}(z)]\\&=[(D_{1}\alpha )(x), \alpha ^{k+1}(y), \alpha ^{k+1}(z)]=\alpha [D_{1}(x), \alpha ^{k}(y), \alpha ^{k}(z)]\\&=\alpha (D'''_{1}([x, y, z])-[\alpha ^{k}(x), D'_{1}(y), \alpha ^{k}(z)]-[\alpha ^{k}(x), \alpha ^{k}(y), D''_{1}(z)])\\&=\tilde{\alpha }(D_{1}^{'''})([x, y, z])-[\alpha ^{k+1}(x), \tilde{\alpha }(D_{1}^{'})(y), \alpha ^{k+1}(z)]\\&\quad -[\alpha ^{k+1}(x), \alpha ^{k+1}(y), \tilde{\alpha }(D_{1}^{''})(z)]. \end{aligned}$$

Since both $\tilde{\alpha }(D_{1}^{'''})$, $\tilde{\alpha }(D_{1}^{''})$and $\tilde{\alpha }(D_{1}^{'})$ are in $\mathrm{End}(T), ~\tilde{\alpha }(D_{1})\in \mathrm{GDer}_{\alpha ^{k+1}}(T).$

We also have

$$\begin{aligned}&\quad [D_{1}D_{2}(x), \alpha ^{k+s}(y), \alpha ^{k+s}(z)]\\&=D^{'''}_{1}[D_{2}(x), \alpha ^{s}(y), \alpha ^{s}(z)]-[\alpha ^{k}(D_{2}(x)), D^{'}_{1}(\alpha ^{s}(y)), \alpha ^{k+s}(z)]\\&\quad -\,[\alpha ^{k}(D_{2}(x)), \alpha ^{k+s}(y), D^{''}_{1}(\alpha ^{s}(z))]\\&=D'''_{1}(D'''_{2}([x, y, z])-[\alpha ^{s}(x), D^{'}_{2}(y), \alpha ^{s}(z)]-[\alpha ^{s}(x), \alpha ^{s}(y), D^{''}_{2}(z)])\\&\quad -\,[\alpha ^{k}(D_{2}(x)), D^{'}_{1}(\alpha ^{s}(y)), \alpha ^{k+s}(z)]-[\alpha ^{k}(D_{2}(x)), \alpha ^{k+s}(y), D^{''}_{1}(\alpha ^{s}(z))]\\&=D'''_{1}D'''_{2}([x, y, z])-D'''_{1}([\alpha ^{s}(x), D^{'}_{2}(y), \alpha ^{s}(z)])-D'''_{1}([\alpha ^{s}(x), \alpha ^{s}(y), D^{''}_{2}(z)])\\&\quad -\,[\alpha ^{k}(D_{2}(x)), D^{'}_{1}(\alpha ^{s}(y)), \alpha ^{k+s}(z)]-[\alpha ^{k}(D_{2}(x)), \alpha ^{k+s}(y), D^{''}_{1}(\alpha ^{s}(z))]\\&=D'''_{1}D'''_{2}([x, y, z])-[D_{1}(\alpha ^{s}(x)), \alpha ^{k}(D^{'}_{2}(y)), \alpha ^{k+s}(z)]\\&\quad -\,[\alpha ^{k+s}(x), D^{'}_{1}D^{'}_{2}(y), \alpha ^{k+s}(z)]\\&\quad -\,[\alpha ^{k+s}(x), \alpha ^{k}(D^{'}_{2}(y)), D^{''}_{1}(\alpha ^{s}(z))]-[D_{1}(\alpha ^{s}(x)), \alpha ^{k+s}(y), \alpha ^{k}(D^{''}_{2}(z))]\\&\quad -\,[\alpha ^{k+s}(x), D^{'}_{1}(\alpha ^{s}(y)), \alpha ^{k}(D^{''}_{2}(z))]-[\alpha ^{k+s}(x), \alpha ^{k+s}(y), D''_{1}D''_{2}(z)]\\&\quad -\,[\alpha ^{k}(D_{2}(x)), D^{'}_{1}(\alpha ^{s}(y)), \alpha ^{k+s}(z)]-[\alpha ^{k}(D_{2}(x)), \alpha ^{k+s}(y), D^{''}_{1}(\alpha ^{s}(z))] \end{aligned}$$

and

$$\begin{aligned}&[D_{2}D_{1}(x), \alpha ^{k+s}(y), \alpha ^{k+s}(z)]\\&=D^{'''}_{2}[D_{1}(x), \alpha ^{k}(y), \alpha ^{k}(z)]-[\alpha ^{s}(D_{1}(x)), D^{'}_{2}(\alpha ^{k}(y)), \alpha ^{k+s}(z)]\\&\quad -\,[\alpha ^{s}(D_{1}(x)), \alpha ^{k+s}(y), D^{''}_{2}(\alpha ^{k}(z))]\\&=D'''_{2}(D'''_{1}([x, y, z])-[\alpha ^{k}(x), D^{'}_{1}(y), \alpha ^{k}(z)]-[\alpha ^{k}(x), \alpha ^{k}(y), D^{''}_{1}(z)])\\&\quad -\,[\alpha ^{s}(D_{1}(x)), D^{'}_{2}(\alpha ^{k}(y)), \alpha ^{k+s}(z)]-[\alpha ^{s}(D_{1}(x)), \alpha ^{k+s}(y), D^{''}_{2}(\alpha ^{k}(z))]\\&=D'''_{2}D'''_{1}([x, y, z])-D'''_{2}([\alpha ^{k}(x), D^{'}_{1}(y), \alpha ^{k}(z)])-D'''_{2}([\alpha ^{k}(x), \alpha ^{k}(y), D^{''}_{1}(z)])\\&\quad -\,[\alpha ^{s}(D_{1}(x)), D^{'}_{2}(\alpha ^{k}(y)), \alpha ^{k+s}(z)]-[\alpha ^{s}(D_{1}(x)), \alpha ^{k+s}(y), D^{''}_{2}(\alpha ^{k}(z))]\\&=D'''_{2}D'''_{1}([x, y, z])-[D_{2}(\alpha ^{k}(x)), \alpha ^{s}(D^{'}_{1}(y)), \alpha ^{k+s}(z)]\\&\quad -\,[\alpha ^{k+s}(x), D^{'}_{2}D^{'}_{1}(y), \alpha ^{k+s}(z)]\\&\quad -\,[\alpha ^{k+s}(x), \alpha ^{s}(D^{'}_{1}(y)), D^{''}_{2}(\alpha ^{k}(z))]-[D_{2}(\alpha ^{k}(x)), \alpha ^{k+s}(y), \alpha ^{s}(D^{''}_{1}(z))]\\&\quad -\,[\alpha ^{k+s}(x), D^{'}_{2}(\alpha ^{k}(y)), \alpha ^{s}(D^{''}_{1}(z))]-[\alpha ^{k+s}(x), \alpha ^{k+s}(y), D''_{2}D''_{1}(z)]\\&\quad -\,[\alpha ^{s}(D_{1}(x)), D^{'}_{2}(\alpha ^{k}(y)), \alpha ^{k+s}(z)]-[\alpha ^{s}(D_{1}(x)), \alpha ^{k+s}(y), D^{''}_{2}(\alpha ^{k}(z))]. \end{aligned}$$

Thus for all $x, y, z\in L$, it follows that

$$\begin{aligned} {[}[D_{1}, D_{2}](x), \alpha ^{k+s}(y), \alpha ^{k+s}(z){]}= & {} [D^{'''}_{1}, D^{'''}_{2}]([x, y, z])\\&-\,[\alpha ^{k+s}(x), [D^{'}_{1}, D^{'}_{2}](y), \alpha ^{k+s}(z)]\\&-\,[\alpha ^{k+s}(x), \alpha ^{k+s}(y), [D^{''}_{1}, D^{''}_{2}](z)]. \end{aligned}$$

Obviously, $[D'_{1}, D'_{2}]$, $[D''_{1}, D''_{2}]$, and $[D'''_{1}, D'''_{2}]$ are contained in $\mathrm{End}(T)$, so $[D_{1}, D_{2}]\in \mathrm{GDer}_{\alpha ^{k+s}}(T)$ and $\mathrm{GDer}(T)$ is a Hom-subalgebra of $\mho$.

Similarly, $\mathrm{QDer}(T)$ is a Hom-subalgebra of $\mho$.

Assume that $D_{1}\in \mathrm{C}_{\alpha ^{k}}(T), D_{2}\in \mathrm{C}_{\alpha ^{s}}(T)$. For all $x, y, z\in T,$ we have

$$\begin{aligned} \tilde{\alpha }(D_{1})([x, y, z])= & {} D_{1}(\alpha ([x, y, z]))=D_{1}([\alpha (x), \alpha (y), \alpha (z)])\\= & {} [\alpha ^{k+1}(x), D_{1}(\alpha (y)), \alpha ^{k+1}(z)]=[\alpha ^{k+1}(x), \tilde{\alpha }(D_{1})(y), \alpha ^{k+1}(z)], \end{aligned}$$

and so $\tilde{\alpha }(D_{1})\in \mathrm{C}_{\alpha ^{k+1}}(T).$ Note that

$$\begin{aligned}{}[D_{1}, D_{2}]([x, y, z])= & {} D_{1}D_{2}([]x, y, z)-D_{2}D_{1}([x, y, z])\\= & {} D_{1}([D_{2}(x), \alpha ^{s}(y), \alpha ^{s}(z)])-D_{2}([D_{1}(x), \alpha ^{k}(y), \alpha ^{k}(z)])\\= & {} [[D_{1}, D_{2}](x), \alpha ^{k+s}(y), \alpha ^{k+s}(z)]. \end{aligned}$$

Similarly,

$$\begin{aligned}{}[D_{1}, D_{2}]([x, y, z])= & {} [\alpha ^{k+s}(x), [D_{1}, D_{2}](y), \alpha ^{k+s}(z)], \end{aligned}$$

and

$$\begin{aligned}{}[D_{1}, D_{2}]([x, y, z])= & {} [\alpha ^{k+s}(x), \alpha ^{k+s}(y), [D_{1}, D_{2}](z)]. \end{aligned}$$

Then $[D_{1}, D_{2}]\in \mathrm{C}_{\alpha ^{k+s}}(T)$. Thus $\mathrm{C}(T)$ is a Hom-subalgebra of $\mho$.

(2) Assume that $D_{1}\in \mathrm{ZDer}_{\alpha ^{k}}(T), D_{2}\in \mathrm{Der}_{\alpha ^{s}}(T)$. $\forall x, y, z\in T,$ we have

$$\begin{aligned} \tilde{\alpha }(D_{1})([x, y, z])= & {} D_{1}(\alpha ([x, y, z]))=D_{1}([\alpha (x), \alpha (y), \alpha (z)])=0\\= & {} [D_{1}(\alpha (x)), \alpha ^{k+1}(y), \alpha ^{k+1}(z)]=[\tilde{\alpha }(D_{1})(x), \alpha ^{k+1}(y), \alpha ^{k+1}(z)], \end{aligned}$$

Hence $\tilde{\alpha }(D_{1})\in \mathrm{ZDer}_{\alpha ^{k+1}}(T).$ Note that

$$\begin{aligned}{}[D_{1}, D_{2}]([x, y, z])&=D_{1}D_{2}([x, y, z])-D_{2}D_{1}([x, y, z])\\&=D_{1}([D_{2}(x), \alpha ^{s}(y), \alpha ^{s}(z)])+D_{1}([\alpha ^{s}(x), D_{2}(y), \alpha ^{s}(z)])\\&\quad +\,D_{1}([\alpha ^{s}(x), \alpha ^{s}(y), D_{2}(z)])-D_{2}([D_{1}(x), \alpha ^{k}(y), \alpha ^{k}(z)])=0, \end{aligned}$$

and

$$\begin{aligned}&[[D_{1}, D_{2}](x), \alpha ^{s+k}(y), \alpha ^{s+k}(z)]\\&=[D_{1}D_{2}(x), \alpha ^{s+k}(y), \alpha ^{s+k}(z)]-[D_{2}D_{1}(x), \alpha ^{s+k}(y), \alpha ^{s+k}(z)]\\&=D_{1}([D_{2}(x), \alpha ^{s}(y), \alpha ^{s}(z)])-D_{2}([D_{1}(x), \alpha ^{k}(y), \alpha ^{k}(z)])\\&\quad +\,[\alpha ^{s}(D_{1}(x)), D_{2}(\alpha ^{k}(y)), \alpha ^{s+k}(z)]+[\alpha ^{s}(D_{1}(x)), \alpha ^{s+k}(y), D_{2}(\alpha ^{k}(z))]=0, \end{aligned}$$

then $[D_{1}, D_{2}]\in \mathrm{ZDer}_{\alpha ^{k+s}}(T)$. Thus $\mathrm{ZDer}(T)$ is a Hom-ideal of $\mathrm{Der}(T)$. $\square$

Lemma 3.2

If $(T, [\cdot , \cdot , \cdot ], \alpha )$ is a multiplicative Hom–Lie triple system, then

• (1) $[\mathrm{Der}(T), \mathrm{C}(T)]\subseteq \mathrm{C}(T).$

• (2) $[\mathrm{QDer}(T), \mathrm{QC}(T)]\subseteq \mathrm{QC}(T).$

• (3) $[\mathrm{QC}(T), \mathrm{QC}(T)]\subseteq \mathrm{QDer}(T).$

• (4) $\mathrm{C}(T)\subseteq \mathrm{QDer}(T).$

Proof

(1)   Assume that $D_{1}\in \mathrm{GDer}_{\alpha ^{k}}(T), D_{2}\in \mathrm{C}_{\alpha ^{s}}(T)$. For all $x, y, z\in T,$ we have

$$\begin{aligned}&[D_{1}, D_{2}]([x, y, z])\\&\quad =D_{1}D_{2}([x, y, z])-D_{2}D_{1}([x, y, z])\\&\quad =D_{1}([D_{2}(x), \alpha ^{s}(y), \alpha ^{s}(z)])-D_{2}([D_{1}(x), \alpha ^{k}(y), \alpha ^{k}(z)])\\&\qquad -\,D_{2}([\alpha ^{k}(x), D_{1}(y), \alpha ^{k}(z)])-D_{2}([\alpha ^{k}(x), \alpha ^{k}(y), D_{1}(z)])\\&\quad =[D_{1}D_{2}(x), \alpha ^{k+s}(y), \alpha ^{k+s}(z)]+[\alpha ^{k}(D_{2}(x)), \alpha ^{s}(D_{1}(y)), \alpha ^{k+s}(z)]\\&\qquad +\,[\alpha ^{s}(D_{2}(x)), \alpha ^{k+s}(y), \alpha ^{s}(D_{1}(z))]-[D_{1}D_{2}(x), \alpha ^{k+s}(y), \alpha ^{k+s}(z)]\\&\qquad -\,[\alpha ^{k}(D_{2}(x)), \alpha ^{s}(D_{1}(y)), \alpha ^{k+s}(z)]-[D_{2}(\alpha ^{k}(x)), \alpha ^{k+s}(y), \alpha ^{s}(D_{1}(z))]\\&\quad =[[D_{1}, D_{2}](x), \alpha ^{k+s}(y), \alpha ^{k+s}(z)]. \end{aligned}$$

Similarly,

$$\begin{aligned}{}[D_{1}, D_{2}]([x, y, z])=[\alpha ^{k+s}(x), [D_{1}, D_{2}](y), \alpha ^{k+s}(z)], \end{aligned}$$

and

$$\begin{aligned}{}[D_{1}, D_{2}]([x, y, z])=[\alpha ^{k+s}(x), \alpha ^{k+s}(y)[D_{1}, D_{2}](z)]. \end{aligned}$$

Then $[D_{1}, D_{2}]\in \mathrm{C}_{\alpha ^{k+s}}(T)$, so $[\mathrm{Der}(T), \mathrm{C}(T)]\subseteq \mathrm{C}(T).$

• (2) Similar to the proof of (1).

• (3) Assume that $D_{1}\in \mathrm{QC}_{\alpha ^{k}}(T), D_{2}\in \mathrm{C}_{\alpha ^{s}}(T).$ For all $x, y, z\in T,$ we have

  $$\begin{aligned}&[[D_{1}, D_{2}](x), \alpha ^{k+s}(y), \alpha ^{k+s}(z)]+[\alpha ^{k+s}(x), [D_{1}, D_{2}](y), \alpha ^{k+s}(z)]\\&\quad +\,[\alpha ^{k+s}(x), \alpha ^{k+s}(y), [D_{1}, D_{2}](z)]\\&=[D_{1}D_{2}(x), \alpha ^{k+s}(y), \alpha ^{k+s}(z)]+[\alpha ^{k+s}(x), D_{1}D_{2}(y), \alpha ^{k+s}(z)]\\&\quad +\,[\alpha ^{k+s}(x), \alpha ^{k+s}(y), D_{1}D_{2}(z)]-[D_{2}D_{1}(x), \alpha ^{k+s}(y), \alpha ^{k+s}(z)]\\&\quad -\,[\alpha ^{k+s}(x), D_{2}D_{1}(y), \alpha ^{k+s}(z)]-[\alpha ^{k+s}(x), \alpha ^{k+s}(y), D_{2}D_{1}(z)]. \end{aligned}$$

It is easy to verify

$$\begin{aligned}{}[D_{1}D_{2}(x), \alpha ^{k+s}(y), \alpha ^{k+s}(z)]= & {} [\alpha ^{k}(D_{2}(x)), D_{1}(\alpha ^{s}(y)), \alpha ^{k+s}(z)]\\= & {} [\alpha ^{k+s}(x), D_{2}D_{1}(y), \alpha ^{k+s}(z)], \end{aligned}$$

and

$$\begin{aligned}{}[\alpha ^{k+s}(x), \alpha ^{k+s}(y), D_{1}D_{2}(z)]= & {} [\alpha ^{s}(D_{1}(x)), \alpha ^{k+s}(y), D_{2}(\alpha ^{k}(z))]\\= & {} [D_{2}D_{1}(x), \alpha ^{k+s}(y), \alpha ^{k+s}(z)]. \end{aligned}$$

Hence

$$\begin{aligned}&[[D_{1}, D_{2}](x), \alpha ^{k+s}(y), \alpha ^{k+s}(z)]+[\alpha ^{k+s}(x), [D_{1}, D_{2}](y), \alpha ^{k+s}(z)]\\&\quad +\,[\alpha ^{k+s}(x), \alpha ^{k+s}(y), [D_{1}, D_{2}](z)]=0, \end{aligned}$$

and $[D_{1}, D_{2}]\in \mathrm{QDer}_{\alpha ^{k+s}}(T)$.

(4)   Assume that $D\in \mathrm{QC}_{\alpha ^{k}}(T).$ Then for all $x, y, z\in T,$ we have

$$\begin{aligned} D([x, y, z])= & {} [D(x), \alpha ^{k}(y), \alpha ^{k}(z)]=[\alpha ^{k}(x), D(y), \alpha ^{k}(z)]\\= & {} [\alpha ^{k}(x), \alpha ^{k}(y), D(z)]. \end{aligned}$$

So

$$\begin{aligned} 3D([x, y, z])= & {} [D(x), \alpha ^{k}(y), \alpha ^{k}(z)]+[\alpha ^{k}(x), D(y), \alpha ^{k}(z)]\\&+\,[\alpha ^{k}(x), \alpha ^{k}(y), D(z)], \end{aligned}$$

which means $D\in \mathrm{QDer}_{\alpha ^{k}}(T)$. $\square$

Theorem 3.3

Let $(T, [\cdot , \cdot , \cdot ], \alpha )$ be a multiplicative Hom–Lie triple system, $\alpha$ a surjection and $\mathrm{Z}(T)$ the center of T. Then $[\mathrm{C}(T), \mathrm{QC}(T)]\subseteq \mathrm{End}(T, \mathrm{Z}(T)).$ Moreover, if $\mathrm{Z}(T)=\{0\},$ then $[\mathrm{C}(T), \mathrm{QC}(T)]=\{0\}.$

Proof

Assume that $D_{1}\in \mathrm{C}_{\alpha ^{k}}(T), D_{2}\in \mathrm{QC}_{\alpha ^{s}}(T).$ For all $x\in T$,  since $\alpha$ is surjection, $\forall y, z\in T, ~\exists y', z' \in T,$ such that $y=\alpha ^{k+s}(y'), z=\alpha ^{k+s}(z'),$ then

$$\begin{aligned}&[[D_{1}, D_{2}](x), y, z]=[[D_{1}, D_{2}](x), \alpha ^{k+s}(y'), \alpha ^{k+s}(z')]\\&=[D_{1}D_{2}(x), \alpha ^{k+s}(y'), \alpha ^{k+s}(z')]-[D_{2}D_{1}(x), \alpha ^{k+s}(y'), \alpha ^{k+s}(z')]\\&=D_{1}([D_{2}(x), \alpha ^{s}(y'), \alpha ^{s}(z')])-[\alpha ^{s}(D_{1}(x)), D_{2}(\alpha ^{k}(y')), \alpha ^{s+k}(z')]\\&=D_{1}([D_{2}(x), \alpha ^{s}(y'), \alpha ^{s}(z')])-D_{1}([\alpha ^{s}(x), D_{2}(y'), \alpha ^{s}(z')])\\&=D_{1}([D_{2}(x), \alpha ^{s}(y'), \alpha ^{s}(z')]-[\alpha ^{s}(x), D_{2}(y'), \alpha ^{s}(z')])=0. \end{aligned}$$

Hence $[D_{1}, D_{2}](x)\in \mathrm{Z}(T)$, and $[D_{1}, D_{2}]\in \mathrm{End}(T, \mathrm{Z}(T))$ as desired. Furthermore, if $\mathrm{Z}(T)=\{0\},$ it is clear that $[\mathrm{C}(T), \mathrm{QC}(T)]=\{0\}.$ $\square$

Theorem 3.4

If $(T, [\cdot , \cdot , \cdot ], \alpha )$ is a multiplicative Hom–Lie triple system over a field ${\mathbb {F}}$ of characteristic $\not =2$, then $\mathrm{ZDer}(T)=\mathrm{C}(T)\cap \mathrm{Der}(T).$

Proof

Assume that $D\in \mathrm{C}_{\alpha ^{k}}(T)\cap \mathrm{Der}_{\alpha ^{k}}(T)$. For all $x, y, z\in T,$ we have

$$\begin{aligned} D([x, y, z])= & {} [D(x), \alpha ^{k}(y), \alpha ^{k}(z)]+[\alpha ^{k}(x), D(y), \alpha ^{k}(z)]\\&+\,[\alpha ^{k}(x), \alpha ^{k}(y), D(z)], \end{aligned}$$

and

$$\begin{aligned} D([x, y, z])= & {} [D(x), \alpha ^{k}(y), \alpha ^{k}(z)]=[\alpha ^{k}(x), D(y), \alpha ^{k}(z)]\\= & {} [\alpha ^{k}(x), \alpha ^{k}(y), D(z)]. \end{aligned}$$

Then $2D([x, y, z])=0,$ so $D([x, y, z])=0$ since char${\mathbb {F}}\not =2$. Hence $D\in \mathrm{ZDer}_{\alpha ^{k}}(T)$ and $\mathrm{C}(T)\cap \mathrm{Der}(T)\subseteq \mathrm{ZDer}(T).$

On the other hand, assume that $D\in \mathrm{ZDer}_{\alpha ^{k}}(T)$, for all $x, y, z\in T,$ we have $D([x, y, z])=[D(x), \alpha ^{k}(y), \alpha ^{k}(z)]=0.$ It is easy to verify $D\in \mathrm{C}_{\alpha ^{k}}(T)\cap \mathrm{Der}_{\alpha ^{k}}(T)$ and $\mathrm{ZDer}(T)\subseteq \mathrm{C}(T)\cap \mathrm{Der}(T).$ $\square$

By Theorem 2.3 in [18], if $(T, [\cdot , \cdot ])$ is a Lie superalgebra with $~\mathrm{Z}(T)=\{0\},$ where $\mathrm{Z}(T)$ is the center of T, then $\mathrm{C}(T)=\mathrm{QDer}(T)\cap \mathrm{QC}(T).$ But it is not true in case that $(T, [\cdot , \cdot ], \alpha )$ is a multiplicative Hom–Lie superalgebra (See [19, Example2.5] ). Either is it not true in case that $(T, [\cdot , \cdot , \cdot ], \alpha )$ is a multiplicative Hom–Lie triple system.

Example 3.5

Let $\{x_{1}, x_{2}, x_{3}\}$ be a basis of a 3-dimensional linear space T over ${\mathbb {F}}$. The following bracket and linear map $\alpha$ on T define a Hom–Lie triple system over ${\mathbb {F}}$:

with $[x_{i}, x_{j}, x_{k}]=-[x_{j}, x_{i}, x_{k}]$, for $i, j, k\in \{1, 2, 3\}$ and others equal to zero.

Define $D:T\rightarrow T$ by

$$\begin{aligned} D(x_{1})=x_{1}, ~D(x_{2})=2^{k}x_{2}, ~D(x_{3})=2^{k}x_{3}. ~~(k\in {\mathbb {Z}}_+) \end{aligned}$$

It is obvious that $\mathrm{Z}(T)=0.$ $\forall y\in T,$ suppose $y=ax_{1}+bx_{2}+cx_{3}.$ Define $D^{'}\in \mathrm{End}(T)$ by

$$\begin{aligned} D^{'}(x_{1})=2^{k+1}x_{1}, ~D^{'}(x_{2})=2^{k+1 }x_{2}, ~D^{'}(x_{3})=2^{k+1}2^{k}x_{3}. \end{aligned}$$

It is obvious that for $i, j, l=1, 2, 3,$

$$\begin{aligned} D^{'}([x_{i}, x_{j}, x_{l}])= & {} [D(x_{i}), \alpha ^{k}(x_{j}), \alpha ^{k}(x_{l})]+[\alpha ^{k}(x_{i}), D(x_{j}), \alpha ^{k}(x_{l})]\\&+\,[\alpha ^{k}(x_{i}), \alpha ^{k}(x_{j}), D(x_{l})], \end{aligned}$$

and

$$\begin{aligned}{}[D(x_{i}), \alpha ^{k}(x_{j}), \alpha ^{k}(x_{l})]=[\alpha ^{k}(x_{i}), D(x_{j}), \alpha ^{k}(x_{l})]=[\alpha ^{k}(x_{i}), \alpha ^{k}(x_{j}), D(x_{l})]. \end{aligned}$$

So for all $k\in {\mathbb {Z}}_+$,

$$\begin{aligned} D\in \mathrm{QDer}(T)\cap \mathrm{QC}(T). \end{aligned}$$

While

$$\begin{aligned}&D([x_{1}, x_{2}, x_{3}])=D(x_{2})=2^{k}x_{2}.\\&\quad [D(x_{1}), \alpha ^{k}(x_{2}), \alpha ^{k}(x_{3})]=[x_{1}, 2^{k}x_{2}, 2^{k}x_{3}]=2^{k+1}x_{2}. \end{aligned}$$

So $D([x_{1}, x_{2}, x_{3}])\not = [D(x_{1}), \alpha ^{k}(x_{2}), \alpha ^{k}(x_{3})],$ that means $D\not \in \mathrm{C}(T).$

Definition 3.6

[1] Let $(L, \mu , \alpha )$ be a Hom-algebra.

• (1) The Hom-associator of L is the trilinear map $as_{\alpha }:L\times L\times L\rightarrow L$ defined as

  $$\begin{aligned} as_{\alpha }=\mu \circ (\mu \otimes \alpha -\alpha \otimes \mu ). \end{aligned}$$

  In terms of elements, the map $as_{\alpha }$ is given by

  $$\begin{aligned} as_{\alpha }(x, y, z)=\mu (\mu (x, y), \alpha (z))-\mu (\alpha (x), \mu (y, z)) \end{aligned}$$

  for all $x, y, z\in L.$

• (2) Let T be a Hom-algebra over a field $\mathbf K$ of characteristic $\not =2$ with an even bilinear multiplication $\circ .$ If $\alpha :L\rightarrow L$ is a linear map, then $(L, \circ , \alpha )$ is a Hom–Jordan algebra if the following identities

  $$\begin{aligned}&x\circ y=y\circ x, \\&as_{\alpha }(x\circ y, \alpha (z), \alpha (w))+as_{\alpha }(y\circ w, \alpha (z), \alpha (x))\\&as_{\alpha }(w\circ x, \alpha (z), \alpha (y))=0, \end{aligned}$$

  hold for all $x, y, z\in L.$

Proposition 3.7

[1] Let $(T, [\cdot , \cdot , \cdot ], \alpha )$ be a multiplicative Hom–Lie triple system, with the operation $D_{1}\bullet D_{2}=D_{1}D_{2}+D_{2}D_{1},$ for all elements $D_{1}, D_{2}\in \mho ,$ the triple $(\mho , \bullet , \alpha )$ is a Hom–Jordan algebra.

Corollary 3.8

Let $(T, [\cdot , \cdot , \cdot ], \alpha )$ be a multiplicative Hom–Lie triple system, with the operation $D_{1}\bullet D_{2}=D_{1}D_{2}+D_{2}D_{1},$ for all elements $D_{1}, D_{2}\in \mathrm{QC}(T)$. Then the triple $(\mathrm{QC}(T), \bullet , \alpha )$ is a Hom–Jordan algebra.

Proof

We need only to show that $D_{1}\bullet D_{2}\in \mathrm{QC}(T)$, for all $D_{1}\in \mathrm{QC_{\alpha ^{k}}}(T), D_{2}\in \mathrm{QC_{\alpha ^{s}}}(T)$ and $x, y, z\in T,$ we have

$$\begin{aligned}&[D_{1}\bullet D_{2}(x), \alpha ^{s+k}(y), \alpha ^{s+k}(z)]\\&\quad =[D_{1}D_{2}(x), \alpha ^{s+k}(y), \alpha ^{s+k}(z)]+[D_{2}D_{1}(x), \alpha ^{s+k}(y), \alpha ^{s+k}(z)]\\&\quad =[\alpha ^{k}(D_{2}(x)), D_{1}(\alpha ^{s}(y), \alpha ^{s+k}(z)]+[\alpha ^{s}(D_{1}(x)), D_{2}(\alpha ^{k}(y), \alpha ^{s+k}(z)]\\&\quad =[D_{2}(\alpha ^{k}(x)), \alpha ^{s}(D_{1}(y), \alpha ^{s+k}(z)]+[D_{1}(\alpha ^{s}(x)), \alpha ^{k}(D_{2}(y), \alpha ^{s+k}(z)]\\&\quad =[\alpha ^{s+k}(x), D_{2}D_{1}(y), \alpha ^{s+k}(z)]+[\alpha ^{s+k}(x), D_{1}D_{2}(y), \alpha ^{s+k}(z)]\\&\quad =[\alpha ^{s+k}(x), D_{1}\bullet D_{2}(y), \alpha ^{s+k}(z)]. \end{aligned}$$

Similarly, $[D_{1}\bullet D_{2}(x), \alpha ^{s+k}(y), \alpha ^{s+k}(z)]=[\alpha ^{s+k}(x), \alpha ^{s+k}(y), D_{1}\bullet D_{2}(z)]$. Then $D_{1}\bullet D_{2}\in \mathrm{QC_{\alpha ^{s+k}}}(T)$ and $\mathrm{QC}(T)$ is a Jordan algebra. $\square$

Theorem 3.9

If $(T, [\cdot , \cdot ], \alpha )$ is a multiplicative Hom–Lie triple system over ${\mathbb {F}}$, then we have

• (1) If char ${\mathbb {F}}\not =2$, then $\mathrm{QC}(T)$ is a Hom–Lie algebra with $[D_{1}, D_{2}]= D_{1}D_{2}-D_{2}D_{1}$ if and only if $\mathrm{QC}(T)$ is also a Hom-associative algebra with respect to composition.

• (2) If char ${\mathbb {F}}\not =3$, $\alpha$ is a surjection and $\mathrm{Z}(T)=\{0\}$, then $\mathrm{QC}(T)$ is a Hom–Lie algebra if and only if $~[\mathrm{QC}(T), \mathrm{QC}(T)]=0.$

Proof

(1) $(\Leftarrow )$ For all $D_{1}, D_{2}\in \mathrm{QC}(T),$ we have $D_{1}D_{2}\in \mathrm{QC}(T)$ and $D_{2}D_{1}\in \mathrm{QC}(T)$, so $[D_{1}, D_{2}]=D_{1}D_{2}-D_{2}D_{1}\in \mathrm{QC}(T).$ Hence, $\mathrm{QC}(T)$ is a Hom–Lie algebra.

$(\Rightarrow )$ Note that $D_{1}D_{2}=D_{1}\bullet D_{2}+\frac{[D_{1}, D_{2}]}{2}$ and by Proposition 2.8, we have $D_{1}\bullet D_{2}\in \mathrm{QC}(T), ~ [D_{1}, D_{2}]\in \mathrm{QC}(T)$. It follows that $D_{1}D_{2}\in \mathrm{QC}(T)$ as desired.

(2) $(\Rightarrow )$   Assume that $D_{1}\in \mathrm{QC_{\alpha ^{k}}}(T), D_{2}\in \mathrm{QC_{\alpha ^{s}}}(T),$ for all $x, y, z\in T$, there are $y', z'\in T,$ such that $y=\alpha ^{s+k}(y'), z=\alpha ^{s+k}(z')$ since $\alpha$ is a surjection. $\mathrm{QC}(T)$ is a Hom–Lie algebra, so $[D_{1}, D_{2}]\in \mathrm{QC_{\alpha ^{k+s}}}(T)$, then

$$\begin{aligned}{}[[D_{1}, D_{2}](x), y, z]= & {} [[D_{1}, D_{2}](x), \alpha ^{s+k}(y'), \alpha ^{s+k}(z')]\\= & {} [\alpha ^{s+k}(x), [D_{1}, D_{2}](y'), \alpha ^{s+k}(z')]\\= & {} [\alpha ^{s+k}(x), \alpha ^{s+k}(y'), [D_{1}, D_{2}](z')]. \end{aligned}$$

From the proof of Lemma 3.2 (5), we have

$$\begin{aligned}{}[[D_{1}, D_{2}](x), y, z]= & {} [[D_{1}, D_{2}](x), \alpha ^{s+k}(y'), \alpha ^{s+k}(z')]\\= & {} -[\alpha ^{s+k}(x), [D_{1}, D_{2}](y'), \alpha ^{s+k}(z')]\\&-\,[\alpha ^{s+k}(x), \alpha ^{s+k}(y'), [D_{1}, D_{2}](z')]. \end{aligned}$$

Hence $3[[D_{1}, D_{2}](x), y, z]=0.$ We have $[[D_{1}, D_{2}](x), y, z]=0,$ i.e., $[D_{1}, D_{2}]=0$ since char ${\mathbb {F}}\not =3$.

$(\Leftarrow )$ It is clear. $\square$

4 The quasiderivations of Hom–Lie triple systems

In this section, we will prove that the quasiderivations of T can be embedded as derivations in a larger Hom–Lie triple system and obtain a direct sum decomposition of Der(T) when the annihilator of T is equal to zero.

Proposition 4.1

Let $(T, [\cdot , \cdot , \cdot ], \alpha )$ be a Hom–Lie triple system over ${{\mathbb {F}}}$ and t an indeterminate. We define $\breve{T}:= \{\Sigma (x\otimes t+y\otimes t^{3})| x, y\in T\}, ~\breve{\alpha }(\breve{T}):=\{\Sigma (\alpha (x)\otimes t+\alpha (y)\otimes t^{2}):x, y\in T\}$. Then $\breve{T}$ is a Hom–Lie triple system with the operation $[x\otimes t^{i}, y\otimes t^{j}, z\otimes t^{k}]=[x, y, z]\otimes t^{i+j+k},$ for all $x, y, z\in T, i, j\in \{1, 3\}$.

Proof

For all $x, y, z, u, v\in T$ and $i, j, k, m, n \in \{1, 3\},$ we have

$$\begin{aligned} \begin{array}{ll} [x\otimes t^{i}, y\otimes t^{j}, z\otimes t^{k}]&{}=[x, y, z]\otimes t^{i+j+k}\\ &{}=-[y, x, z]\otimes t^{i+j+k}\\ &{}=-[y\otimes t^{j}, x\otimes t^{i}, z\otimes t^{k}], \end{array} \end{aligned}$$

$$\begin{aligned}&\ \ [x\otimes t^{i}, y\otimes t^{j}, z\otimes t^{k}]+[y\otimes t^{j}, z\otimes t^{k}, x\otimes t^{i}]+[z\otimes t^{k}, x\otimes t^{i}, y\otimes t^{j}]\\&=[x, y, z]\otimes t^{i+j+k}+[y, z, x]\otimes t^{i+j+k}+[z, x, y]\otimes t^{i+j+k}\\&=([x, y, z]+[y, z, x]+[z, x, y])\otimes t^{i+j+k}=0, \end{aligned}$$

and

$$\begin{aligned}&[\breve{\alpha }(x\otimes t^{i}), \breve{\alpha }(y\otimes t^{j}), [z\otimes t^{k}, u\otimes t^{m}, v\otimes t^{n}]]\\&=[\alpha (x), \alpha (y), [z, u, v]]\otimes t^{i+j+k+m+n}\\&=([[x, y, z], \alpha (u), \alpha (v)]+[\alpha (z), [x, y, u], \alpha (v)]\\&\quad +\,[\alpha (z), \alpha (u), [x, y, v]])\otimes t^{i+j+k+m+n}\\&=[[x\otimes t^{i}, y\otimes t^{j}, z\otimes t^{k}], \breve{\alpha }(u\otimes t^{m}), \breve{\alpha }(v\otimes t^{n})]\\&\quad +\,[\breve{\alpha }(z\otimes t^{k}), [x\otimes t^{i}, y\otimes t^{j}, u\otimes t^{m}], \breve{\alpha }(v\otimes t^{n})]\\&\quad +\,[\breve{\alpha }(z\otimes t^{k}), \breve{\alpha }(u\otimes t^{m}), [x\otimes t^{i}, y\otimes t^{j}, v\otimes t^{n}]]. \end{aligned}$$

Hence $\breve{T}$ is a Hom–Lie triple system. $\square$

For convenience, we write $xt(xt^{3})$ in place of $x\otimes t(x\otimes t^{3}).$

If U is a subspace of T such that $T=U\oplus [T, T, T],$ then

$$\begin{aligned} \breve{T}=Tt+Tt^{3}=Tt+Ut^{3}+[T, T, T]t^{3}, \end{aligned}$$

Now we define a map $\varphi :\mathrm{QDer}(T)\rightarrow \mathrm{End}(\breve{T})$ satisfying

$$\begin{aligned} \varphi (D)(at+ut^{2}+bt^{2})=D(a)t+D'(b)t^{3}, \end{aligned}$$

where $D\in \mathrm{QDer}_{\alpha ^{k}}($T),  and $D'$ is in Eq. (2.2), $a\in T, u\in U, b\in [T, T, T]$.

Proposition 4.2

$T, \breve{T}, \varphi$ are defined as above. Then

• (1) $\varphi$ is injective and $\varphi (D)$ does not depend on the choice of $D'$.

• (2) $\varphi (\mathrm{QDer}(T))\subseteq \mathrm{Der}(\breve{T}).$

Proof

(1) If $\varphi (D_{1})=\varphi (D_{2}),$ then for all $a\in T, b\in [T, T, T]$, and $u\in U,$ we have

$$\begin{aligned} \varphi (D_{1})(at+ut^{3}+bt^{3})=\varphi (D_{2})(at+ut^{3}+bt^{3}), \end{aligned}$$

that is

$$\begin{aligned} D_{1}(a)t+D'_{2}(b)t^{3}= D_{2}(a)t+D'_{2}(b)t^{3}, \end{aligned}$$

so $D_{1}(a)=D_{2}(a).$ Hence $D_{1}=D_{2},$ and $\varphi$ is injective.

Suppose that there exists $D''$ such that

$$\begin{aligned} \varphi (D)(at+ut^{3}+bt^{3})=D(a)t+D''(b)t^{3}, \end{aligned}$$

and

$$\begin{aligned}&[D(x), \alpha ^{k}(y), \alpha ^{k}(z)]+[\alpha ^{k}(x), D(y), \alpha ^{k}(z)]\\&\quad +\,[\alpha ^{k}(x), \alpha ^{k}(y), D(z)]=D''([x, y, z]), \end{aligned}$$

then we have

$$\begin{aligned} D'([x, y, z])=D''([x, y, z]), \end{aligned}$$

thus $D'(b)=D''(b).$ Hence

$$\begin{aligned} \varphi (D)(at+ut^{3}+bt^{3})=D(a)t+D'(b)t^{3}=D(a)t+D''(b)t^{3}, \end{aligned}$$

which implies $\varphi (D)$ is determined by D.

(2) We have $[xt^{i}, yt^{j}, zt^{k}]=[x, y, z]t^{i+j+k}=0,$ for all $i+j+k\ge 4$. Thus, to show $\varphi (D)\in \mathrm{Der}(\breve{T}),$ we need only to check the validness of the following equation

$$\begin{aligned} \begin{array}{ll} \varphi (D)([xt, yt, zt])&{}=[\varphi (D)(xt), \breve{\alpha }^{k}(yt), \breve{\alpha }^{k}(zt)]+[\breve{\alpha }^{k}(xt), \varphi (D)(yt), \breve{\alpha }^{k}(zt)]\\ &{}\quad +\,[\breve{\alpha }^{k}(xt), \breve{\alpha }^{k}(yt), \varphi (D)(zt)]. \end{array} \end{aligned}$$

For all $x, y, z\in T,$ we have

$$\begin{aligned}&\varphi (D)([xt, yt, zt])=\varphi (D)([x, y, z]t^3)=D'([x, y, z])t^3\\&=([D(x), \alpha ^{k}(y), \alpha ^{k}(z)]+[\alpha ^{k}(x), D(y), \alpha ^{k}(z)]+[\alpha ^{k}(x), \alpha ^{k}(y), D(z)])t^3\\&=[D(x)t, \alpha ^{k}(y)t, \alpha ^{k}(z)t]+[\alpha ^{k}(x)t, D(y)t, \alpha ^{k}(z)t]+[\alpha ^{k}(x)t, \alpha ^{k}(y)t, D(z)t]\\&=[\varphi (D)(xt), \breve{\alpha }(yt), \breve{\alpha }(zt)]+[\breve{\alpha }(xt), \varphi (D)(yt), \breve{\alpha }(zt)]+[\breve{\alpha }(xt), \breve{\alpha }(yt), \varphi (D)(zt)]. \end{aligned}$$

Therefore, for all $D\in \mathrm{QDer}_{\alpha ^{k}}(T)$, we have $\varphi (D)\in \mathrm{Der}_{\alpha ^{k}}(\breve{T})$. $\square$

Proposition 4.3

Let T be a Hom–Lie triple system. $\mathrm{Z}(T)=\{0\}$ and $\breve{T}, ~\varphi$ are defined as above. Then $\mathrm{Der}(\breve{T})=\varphi (\mathrm{QDer}(T))\dotplus \mathrm{ZDer}(\breve{T}).$

Proof

Since $\mathrm{Z}(T)=\{0\}$, we have $\mathrm{Z}(\breve{T})=Tt^3.$ For all $g\in \mathrm{Der}(\breve{T}),$ we have $g(\mathrm{Z}(\breve{T}))\subseteq \mathrm{Z}(\breve{T}),$ hence $g(Ut^3)\subseteq g(\mathrm{Z}(\breve{T}))\subseteq \mathrm{Z}(\breve{T})=Tt^3.$ Now we define a map $f:Tt+Ut^3+[T, T, T]t^3\rightarrow Tt^3$ by

$$\begin{aligned} \ f(x)=\left\{ \begin{array}{ll}g(x)\cap Tt^3, &{} x\in Tt ;\\ g(x), &{} x\in Ut^3 ;\\ 0, &{} x\in [T, T]t^3. \end{array}\right. \end{aligned}$$

It is clear that f is linear. Note that

$$\begin{aligned}&f([\breve{T}, \breve{T}, \breve{T}])=f([T, T, T]t^3)=0,\\&\quad [f(\breve{T}), \breve{\alpha }^{k}(\breve{T}), \breve{\alpha }^{k}(\breve{T})]\subseteq [Tt^3, \alpha ^{k}(T)t+\alpha ^{k}(T)t^3, \alpha ^{k}(T)t+\alpha ^{k}(T)t^3]=0, \end{aligned}$$

hence $f\in \mathrm{ZDer}_{\alpha ^{k}}(\breve{T}).$ Since

$$\begin{aligned} (g-f)(Tt)=g(Tt)-g(Tt)\cap Tt^{3}=g(Tt)-Tt^{3}\subseteq Tt, ~ (g-f)(Ut^3)=0, \end{aligned}$$

and

$$\begin{aligned} (g-f)([T, T, T]t^3)=g([\breve{T}, \breve{T}, \breve{T}])\subseteq [\breve{T}, \breve{T}, \breve{T}]=[T, T, T]t^3, \end{aligned}$$

there exist $D, D'$ in Eq. (2.2) such that for all $a\in T, ~b\in [T, T, T]$,

$$\begin{aligned} (g-f)(at)=D(a)t, ~ (g-f)(bt^3)=D'(b)t^3. \end{aligned}$$

Since $(g-f)\in \mathrm{Der}(\breve{T})$ and by the definition of $\mathrm{Der}(\breve{T})$, we have

$$\begin{aligned}&[(g-f)(a_1t), \breve{\alpha }^{k}(a_2t), \breve{\alpha }^{k}(a_3t)]+[\breve{\alpha }^{k}(a_1t), (g-f)(a_2t), \breve{\alpha }^{k}(a_3t)]\\&+\,[\breve{\alpha }^{k}(a_1t), \breve{\alpha }^{k}(a_2t), (g-f)(a_3t)]=(g-f)([a_1t, a_2t, a_3t]). \end{aligned}$$

for all $a_1, a_2, a_3\in T.$ Hence

$$\begin{aligned}&[D(a_1), \alpha ^{k}(a_2)t, \alpha ^{k}(a_3)t]+[\alpha ^{k}(a_1)t, D(a_2), \alpha ^{k}(a_3)t]\\&+\,[\alpha ^{k}(a_1)t, \alpha ^{k}(a_2)t, D(a_3)]=D'([a_1, a_2, , a_3])t^{3}. \end{aligned}$$

Thus $D\in \mathrm{QDer}_{\alpha ^{k}}(T).$ Therefore, $g-f=\varphi (D)\in \varphi (\mathrm{QDer}(T))$, so $\mathrm{Der}(\breve{T})\subseteq \varphi (\mathrm{QDer}(T))+\mathrm{ZDer}(\breve{T}).$ By Proposition 4.2 (2) we have $\mathrm{Der}(\breve{T})=\varphi (\mathrm{QDer}(T))+\mathrm{ZDer}(\breve{T}).$

For all $f\in \varphi (\mathrm{QDer}(T))\cap \mathrm{ZDer}(\breve{T})$, there exists an element $D\in \mathrm{QDer}(T)$ such that $f=\varphi (D).$ Then

$$\begin{aligned} f(at+ut^3+bt^3)=\varphi (D)(at+ut^3+bt^3)=D(a)t+D'(b)t^3, \end{aligned}$$

for all $a\in T, b\in [T, T, T].$

On the other hand, since $f\in \mathrm{ZDer}(\breve{T}),$ we have

$$\begin{aligned} f(at+bt^3+ut^3)\in \mathrm{Z}(\breve{T})=Tt^3. \end{aligned}$$

That is to say, $D(a)=0,$ for all $a\in T$ and so $D=0.$ Hence $f=0.$

Therefore $\mathrm{Der}(\breve{T})=\varphi (\mathrm{QDer}(T))\dotplus \mathrm{ZDer}(\breve{T})$ as desired. $\square$

5 The Centroid of Hom–Lie Triple System

Proposition 5.1

Let $(T, [\cdot , \cdot , \cdot ], \alpha )$ be a multiplicative Hom–Lie triple system with a surjection $\alpha$. If T has no nonzero ideals I, J with $[T, I, J]=0$, i.e., T is prime, then $\mathrm{C}(T)$ is an integral domain.

Proof

First $\mathrm{id} \in \mathrm{C}(T)$. If there exist $0\ne \psi \in \mathrm{C}_{\alpha ^{k}}(T), ~ 0\ne \varphi \in \mathrm{C}_{\alpha ^{s}}(T)$ such that $\psi \varphi =0$, then there exists $x, y, x^{'}, y^{'}\in T$ such that $\psi (x)=\psi (\alpha ^{s}(x^{'}))\ne 0$ and $\varphi (y)=\varphi (\alpha ^{k}(y^{'}))\ne 0$.

Then $\psi \varphi ([T, x^{'}, y^{'}])=\psi [\alpha ^{s}(T), \alpha ^{s}(x^{'}), \varphi (y^{'})]=[\alpha ^{s+k}(T), \psi (\alpha ^{s}(x^{'})), \varphi (\alpha ^{k}(y^{'}))]=[T, \psi (x), \varphi (y)]=0$. Let $I_{x}, ~I_{y}$ be ideals spanned by x,  y, respectively, then $\psi (I_{x})$ and $\varphi (I_{y})$ are also ideals of T. T is prime means $\mathrm{Z}(T)=\{0\},$ so $[\psi (x), T, T]\ne \{0\}$ and $[\varphi (y), T, T]\ne \{0\}.$ Therefore $\psi (I_{x})$ and $\varphi (I_{y})$ are two nonzero ideals of T such that $[T, \psi (I_{x}), \varphi (I_{y})]=0$, respectively, a contradiction. Hence $\mathrm{C}(T)$ have no zero divisor, it is an integral domain. $\square$

Proposition 5.2

If $(T, [\cdot , \cdot , \cdot ], \alpha )$ is a simple Hom–Lie triple system over an algebraically closed field $\mathbf {F}$, i.e., $T^{(1)}\ne 0$, and T has only two ideals T and 0, then $\mathrm{C}(T)=\mathbb {F}$id if and only if $\alpha =\mathrm{\pm id}$.

Proof

$(\Leftarrow )$ $\alpha =\mathrm{\pm id}$ means T is a Lie triple system, so one gets $\mathrm{C}(T)=\mathbb {F}$id (See [3,  Theorem 1]).

$(\Rightarrow )$ For all $k\in {\mathbb {N}}^{+},$ $\forall ~0\ne \psi \in \mathrm{C}_{\alpha ^{k}}(T),$ we have $\psi =\mu \mathrm{id}, ~\mu \in {\mathbb {F}}, ~\mu \ne 0.$ So $\forall x, y, z\in T,$ we have $\mu [x, y, z]=\psi ([x, y, z])=[\psi (x), \alpha ^{k}(y), \alpha ^{k}(z)]=\mu [x, \alpha ^{k}(y), \alpha ^{k}(z)].$ Hence $[x, y, z]=[x, \alpha ^{k}(y), \alpha ^{k}(z)]$ for all $k\in {\mathbb {N}}^{+}.$

Since $\mathbb {F}$ is algebraically closed, $\alpha$ has an eigenvalue $\lambda$. We denote the corresponding eigenspace by $E_{\lambda }(\alpha )$. So $E_{\lambda }(\alpha )\ne 0$. Let $k=1,$ for any $x\in E_{\lambda }(\alpha ), y, z\in T$, we have $\alpha ([x, y, z])=[\alpha (x), \alpha (y), \alpha (z)]=\lambda [x, \alpha (y), \alpha (z)]=\lambda [x, y, z]$, so $[x, y, z]\in E_{\lambda }(\alpha )$. It follows that $E_{\lambda }(\alpha )$ is an ideal of T. But T is simple, so $E_{\lambda }(\alpha )=T$, i.e., $\alpha =\lambda$id. Then we have for all $x, y, z\in T, ~k=1,$ $[x, y, z]=[x, \alpha (y), \alpha (z)]=[x, \lambda y, \lambda z]=\lambda ^{2}[x, y, z],$ so $\lambda ^{2}=1$ and $\alpha =\pm$id. This proves the theorem. $\square$

When $\mathrm{C}(T)=\mathbb {F}$id, the Lie triple system T is said to be central. Furthermore, if T is simple, T is said to be central simple. From Theorem 4.2, one gets every central simple Hom–Lie triple system over an algebraically closed field is a Lie triple system.

Proposition 5.3

Let $(T, [\cdot , \cdot , \cdot ], \alpha )$ be a multiplicative Hom–Lie triple system over a field $\mathbb {F}$. Then

• (1) If $\alpha$ is a surjection, then T is indecomposable (cannot be written as the direct sum of two nontrivial ideals) if and only if $\mathrm{C}(T)$ does not contain idempotents except 0 and id.

• (2) If T is perfect, then every $\psi \in \mathrm{C}_{\alpha ^{k}}(T)~(k\ge 0)$ is ${\alpha ^{k}}$-symmetric with respect to any invariant form on T.

Proof: (1) $(\Rightarrow )$ If there exists $\psi \in \mathrm{C}_{\alpha ^{k}}(T)$ is an idempotent and satisfies $\psi \ne 0, \mathrm{id}$, then $\psi ^{2}(x)=\psi (x)$, $\forall x\in T$. We can see that $\mathrm{ker}\psi$ and $\mathrm{Im}\psi$ are ideals of T. In fact, for any $x\in \mathrm{Ker}\psi$ and $y, z \in T$, we have $\psi ([x, y, z])=[\psi (x), \alpha ^{k}(y), \alpha ^{k}(z)]=0$, which implies $[x, y, z]\in \mathrm{Ker}\psi$. For any $x\in \mathrm{Im}\psi$, there exists $a\in T$ such that $x=\psi (a)$. for all $y, z\in T$, there are $y^{'}, z^{'}$ such that $y=\alpha ^{k}(y^{'}), ~z=\alpha ^{k}(z^{'}).$ Then $[x, y, z]=[\psi (a), \alpha ^{k}(y^{'}), \alpha ^{k}(z^{'})]=\psi ([a, y^{'}, z^{'}])\in \mathrm{Im}\psi$. Moreover, $\mathrm{Ker}\psi \cap \mathrm{Im}\psi =0$. Indeed, if $x\in \mathrm{Ker}\psi \cap \mathrm{Im}\psi$, then there exists $y\in T$ such that $x=\psi (y)$ and $0=\psi (x)=\psi ^{2}(y)=\psi (y)=x$. We have a decomposition $x=\psi (x)+y$, $\forall x\in T$, where $\psi (y)=0$. So we have $T=\mathrm{Ker}\psi \dotplus \mathrm{Im}\psi$, a contradiction.

$(\Leftarrow )$ Similar with [8, Proposition1].

(2) Let f be an invariant $\mathbb {F}$-bilinear form on T. Then $f([a, b, c], d)=f(a, [d, c, b])$ $\forall a, b, c, d\in T$. Since T is perfect, let $\psi \in \mathrm{C}_{\alpha ^{k}}(T)~(k\ge 0)$, then we have

$$\begin{aligned} f(\psi ([a, b, c]), \alpha ^{k}(d))= & {} f([\alpha ^{k}(a), \psi (b), \alpha ^{k}(c)], \alpha ^{k}(d))\\= & {} f(\alpha ^{k}(a), [\alpha ^{k}(d), \alpha ^{k}(c), \psi (b)])\\= & {} f([\alpha ^{k}(a), \psi ([d, c, b])])\\= & {} f(\alpha ^{k}(a), [\psi (d), \alpha ^{k}(c), \alpha ^{k}(b)])\\= & {} f(\alpha ^{k}([a, b, c]), \psi (d)). \end{aligned}$$

Proposition 5.4

Let $(T, [\cdot , \cdot , \cdot ], \alpha )$ be a Hom–Lie triple system over a field $\mathbb {F}$ and I be an $\alpha$-invariant subspace of T. Then $\mathrm{Z}_{T}(I)$ is invariant under $\mathrm{C}(T)$, so is any perfect ideal of T.

Proof

For any $\psi \in \mathrm{C}_{\alpha ^{k}}(T)$ and $x\in \mathrm{Z}_{T}(I)$, $\forall y\in I, z\in T$, there are $y^{'}, z^{'}$ such that $y=\alpha ^{k}(y^{'}), ~z=\alpha ^{k}(z^{'}).$ Then

$$\begin{aligned} L(\psi (x), y)(z)=[\psi (x), y, z]=[\psi (x), \alpha ^{k}(y^{'}), \alpha ^{k}(z^{'})]=\psi ([x, y^{'}, z^{'}])=0, \end{aligned}$$

and

$$\begin{aligned} R(y, \psi (x))(z)=[z, y, \psi (x)]=[\alpha ^{k}(z^{'}), \alpha ^{k}(y^{'}), \psi (x)]=\varphi ([z^{'}, y^{'}, x])=0, \end{aligned}$$

which implies that $\psi (x)\in \mathrm{Z}_{T}(I)$. So $\mathrm{Z}_{T}(I)$ is invariant under $\mathrm{C}(T)$.

Let J be any perfect ideal of T, then $J=[J, J, J]$. For any $y\in J$, there exists $a, b, c\in J$ such that $y=[a, b, c]$, then we have $\psi (y)=\psi ([a, b, c])=[\alpha ^{k}(a), \psi (b), \alpha ^{k}(c)] \in [J, T, T]\subseteq J$. Hence J is invariant under $\mathrm{C}(T)$. $\square$

Theorem 5.5

Let $(T_{1}, [\cdot , \cdot , \cdot ], \alpha _{1})$ and $(T_{2}, [\cdot , \cdot , \cdot ], \alpha _{2})$ be two Hom–Lie triple systems over field $\mathbb {F}$ with $\alpha _{1}$ a surjection. Let $\pi : T_{1} \rightarrow T_{2}$ be an epimorphism of Hom–Lie triple systems. For any $f\in \mathrm{End_{\mathbf {F}}}(T_{1};\mathrm{Ker}\pi ):=\{g\in \mathrm{End_{\mathbf {F}}}(T_{1})|g(\mathrm{Ker}\pi )\subseteq \mathrm{Ker}\pi \}$ there exists a unique $\bar{f}\in \mathrm{End_{\mathbf {F}}}(T_{2})$ satisfying $\pi \circ f=\bar{f}\circ \pi$. Moreover, the following results hold:

(1)   The map $\pi _\mathrm{End}: \mathrm{End_{\mathbb {F}}}(T_{1};\mathrm{Ker}\pi )\rightarrow \mathrm{End_{\mathbb {F}}}(T_{2})$, $f\mapsto \bar{f}$ is a Hom-algebra homomorphism with the following properties:

$$\begin{aligned} \pi _\mathrm{End}(\mathrm{Mult}(T_{1}))=\mathrm{Mult}(T_{2}), ~\pi _\mathrm{End}(\mathrm{C}(T_{1})\cap \mathrm{End_{\mathbb {F}}}(T_{1}; \mathrm{Ker}\pi ))\subseteq \mathrm{C}(T_{2}). \end{aligned}$$

By restriction, there is a Hom-algebra homomorphism

$$\begin{aligned} \pi _{\mathrm{C}}: \mathrm{C}(T_{1})\cap \mathrm{End_{\mathbb {F}}}(T_{1}; \mathrm{Ker}\pi )\rightarrow \mathrm{C}(T_{2}), f\mapsto \bar{f}. \end{aligned}$$

If $\mathrm{Ker}\pi =Z(T_{1})$, then every $\phi \in \mathrm{C}(T_{1})$ leaves $\mathrm{Ker}\pi$ invariant, hence $\pi _{\mathrm{C}}$ is defined on all of $\mathrm{C}(T_{1})$.

(2)   Suppose $T_{1}$ is perfect and $\mathrm{Ker}\pi \subseteq Z(T_{1})$. Then $\pi _{\mathrm{C}}: \mathrm{C}(T_{1})\cap \mathrm{End_{\mathbb {F}}}(T_{1}; \mathrm{Ker}\pi ) \rightarrow \mathrm{C}(T_{2}), f\mapsto \bar{f}$ is injective.

(3)   If $T_{1}$ is perfect, $Z(T_{2})=0$ and $\mathrm{Ker}\pi \subseteq Z(T_{1})$, then $\pi _{\mathrm{C}}: \mathrm{C}(T_{1})\rightarrow \mathrm{C}(T_{2})$ is a Hom-algebra monomorphism.

Proof

(1) It is easy to see that $\pi _{\mathrm{End}}$ is a Hom-algebra homomorphism. In fact, for all $f, g\in \mathrm{End_{\mathbb {F}}}(T_{1};\mathrm{Ker}\pi )$, we have $\pi (fg)=(\bar{f}\pi )g=\bar{f}\bar{g}\pi ,$ so $\pi _\mathrm{End}(fg)=\pi _\mathrm{End}(f)\pi _\mathrm{End}(g).$ Meanwhile, $\pi (\alpha _{1} f)=(\alpha _{2}\pi )f=(\alpha _{2}\bar{f})\pi$ i.e., $\pi _{\mathrm{End}}\alpha _{1}(f)=\alpha _{2}\pi _{\mathrm{End}}(f)$, so $\pi _{\mathrm{End}}\alpha _{1}=\alpha _{2}\pi _{\mathrm{End}}.$

$\mathrm{Ker}\pi$ is an ideal of $T_{1}$ and all left and right multiplication operators of $T_{1}$ leave $\mathrm{Ker}\pi$ invariant. In fact, for all $x, y\in T, ~z\in \mathrm{Ker}\pi ,$ $\pi (L(x, y)(z))=\pi ([x, y, z])=[\pi (x), \pi (y), \pi (z)]=0.$ Similarly, $\pi (R(x, y)(z))=0.$ So $\mathrm{Mult}(T_{1})\subseteq \mathrm{End_{\mathbb {F}}}(T_{1}; \mathrm{Ker}\pi )$.

Furthermore, for the left multiplication operator L(x, y) on $T_{1}$ we have $\pi \cdot L(x, y)=L(\pi (x), \pi (y))\cdot \pi$, so $\pi _\mathrm{End}(L(x, y))=L(\pi (x), \pi (y))$. For the right multiplication we have the analogous formula $\pi _\mathrm{End}(R(x, y))=R(\pi (x), \pi (y))$. Moreover, $\pi$ is an epimorphism, so $\pi _\mathrm{End}(\mathrm{Mult}(T_{1}))=\mathrm{Mult}(T_{2})$.

Now we show that $\pi _\mathrm{End}(\mathrm{C}(T_{1})\cap \mathrm{End_{\mathbb {F}}}(T_{1}; \mathrm{Ker}\pi ))\subseteq \mathrm{C}(T_{2})$. Let $\phi \in \mathrm{C}_{\alpha ^{k}}(T_{1})\cap \mathrm{End_{\mathbb {F}}}(T_{1}; \mathrm{Ker}\pi )$. For any $x', y', z'\in T_{2}$ there exist $x, y, z\in T_{1}$ such that $\pi (x)=x', \pi (y)=y', \pi (z)=z'$. Then we have ${\bar{\phi }}([x', y', z'])=\bar{\phi }(\pi ([x, y, z]))=\pi (\phi ([x, y, z]))= \pi ([\alpha ^{k}(x), \alpha ^{k}(y), \phi (z)])=[\alpha ^{k}\pi (x), \alpha ^{k}\pi (y), \bar{\phi }(\pi (z))] =[\alpha ^{k}(x'), \alpha ^{k}(y'), \bar{\phi }(z')],$ which proves $\bar{\phi }\in \mathrm{C}_{\alpha ^{k}}(T_{2})$. Meanwhile, $\bar{\alpha }_{i}$ denotes an endomorphism on $\mathrm{C}(T_{i}), ~i=\{1, 2\}$, we show that $\pi _{\mathrm{C}}\cdot \bar{\alpha }_{1}=\bar{\alpha }_{2}\pi _{\Gamma }.$ In fact, for all $\phi \in \pi _\mathrm{End}(\mathrm{C}(T_{1})\cap \mathrm{End_{\mathbb {F}}}(T_{1}),$ we have $\pi \phi \alpha _{1}=\bar{\phi }\pi \alpha _{1}=\alpha _{2}\bar{\phi }\pi =\alpha _{2}(\pi _{\mathrm{C}}(\phi ))\pi .$ i.e., $\pi (\bar{\alpha }_{1}\phi )=\bar{\alpha }_{2}\pi _\mathrm{C}(\phi )\pi$. If $\mathrm{Ker}\pi =\mathrm{C}(T_{1})$, it is obvious that $\phi \in \mathrm{C}(T_{1})$ leaves $\mathrm{Ker}\pi$ invariant.

(2) If $\bar{\phi }=0$ for $\phi \in \mathrm{C}(T_{1})\cap \mathrm{End_{\mathbf {F}}}(T_{1}; \mathrm{Ker}\pi )$, then $\pi (\phi (T_{1}))=\bar{\phi }(\pi (T_{1}))=0$, which means that $\phi (T_{1})\subseteq \mathrm{Ker}\pi \subseteq C(T_{1})$. Hence $\phi ([x, y, z])=[\phi (x), \alpha ^{k}(y), \alpha ^{k}(z)]=0$, $\forall x, y, z\in T_{1}$. Furthermore, since $T_{1}=T_{1}^{(1)}$, we can get $\phi =0$.

(3) We can see that $\pi (\mathrm{Z}(T_{1}))\subseteq \pi (\mathrm{Z}(T_{2}))=0$. In fact, for all $y, z\in T_{2}$, there are $y^{'}, z^{'}\in T_{2}$ such that $y=\alpha ^{k}(y^{'}), ~z=\alpha ^{k}(z^{'}).$ And for all $x\in \mathrm{Z}(T_{1})$, $[\pi (x), y, z]=[\pi (x), \pi (y^{'}), \pi (z^{'})]=\pi ([x, y^{'}, z^{'}])=0.$ Hence $\mathrm{Z}(T_{1})\subseteq \mathrm{Ker}\pi$. So $\mathrm{Ker}\pi \subseteq \mathrm{Z}(T_{1})$. By (1), we know that $\pi _\mathrm{C}: \mathrm{C}(T_{1})\rightarrow \mathrm{C}(T_{2})$ is a well-defined Hom-algebra homomorphism, which is a injection by (2). $\square$

6 Centroid of Tensor Product

Benkart and Neher investigated the centroid of associative algebras and Lie algebras in [4]. In this section, we discuss centroid of the tensor product of a Lie triple system and a unital commutative associative algebra. Furthermore, we completely determine the centroid of the tensor product $T\otimes R$ of a simple Lie triple system T and a polynomial ring R.

Definition 6.1

Let A be an associative algebra over a field $\mathbb {F}$. The centroid of A is the space of $\mathbb {F}$-linear transforms on A given by $\mathrm{C}(A)=\{\psi \in \mathrm{End}(A)|\psi (ab)=a\psi (b)=\psi (a)b, \ \forall a, b\in A\}.$

Let $(T, [\cdot , \cdot , \cdot ], \alpha )$ be a Hom–Lie triple system over $\mathbb {F}$ and A be a unital commutative associative algebra over $\mathbb {F}$. There exists a unique Hom–Lie triple system structure on $T\otimes A$ satisfying $[x\otimes a, y\otimes b, z\otimes c]=[x, y, z]\otimes abc$ and $\breve{\alpha }(x\otimes a)=\alpha (x)\otimes a$ for $x, y, z\in T, a, b, c\in A$. We can show that if T is perfect, then $T\otimes A$ is perfect too. Also, for $D\in \mathrm{End}(T)$ and $\psi \in \mathrm{End}(A)$ there exists a unique map $D\tilde{\otimes }\psi \in \mathrm{End}(T\otimes A)$ such that

$$\begin{aligned} (D\tilde{\otimes }\psi )(x\otimes a)=D(x)\otimes \psi (a), \forall x\in T, a\in A. \end{aligned}$$

The map should not be confused with the element $D\otimes \psi$ of the tensor product $\mathrm{End}(T)\otimes \mathrm{End}(A)$. Of course, we have a canonical map $\omega : \mathrm{End}(T)\otimes \mathrm{End}(A)\rightarrow \mathrm{End}(T\otimes A): D\otimes \psi \mapsto D\tilde{\otimes }\psi .$ It is easy to see that if $D\in \mathrm{C}(T)$ and $\psi \in \mathrm{C}(A)$, then $D\tilde{\otimes }\psi \in \mathrm{C}(T\otimes A)$. Hence $\mathrm{C}(T)\tilde{\otimes } \mathrm{C}(A)\subseteq \mathrm{C}(T\otimes A)$, where $\mathrm{C}(T)\tilde{\otimes } \mathrm{C}(A)$ is the $\mathbb {F}$-span of all endomorphism $D\tilde{\otimes }\psi$.

Next we will determine the centroid of the tensor product of a central simple Hom–Lie triple system and a polynomial ring.

Let $(T, [\cdot , \cdot , \cdot ], \alpha )$ be a central simple Hom–Lie triple system over an algebraically closed field $\mathbb {F}$ and $R=\mathbb {F}[x_{1}, \cdots , x_{n}]$. By defining the multiplication as $[x\otimes p, y\otimes q, z\otimes r]=[x, y, z]\otimes pqr$, we can make $\tilde{T}=T\otimes R$ into a Hom–Lie triple system.

Theorem 6.1

If $(T, [\cdot , \cdot , \cdot ], \alpha )$ is a central simple Hom–Lie triple system over an algebraically closed field $\mathbb {F}$, $R=\mathbb {F}[x_{1}, \cdots , x_{n}]$ and $\tilde{T}=T\otimes R$. Then $\mathrm{C}(\tilde{T})=\mathrm{C}(T)\tilde{\otimes }R$.

Proof

From Theorem 4.2, it is easy to verify that a central simple Hom–Lie triple system over an algebraically closed field $\mathbb {F}$ is a Lie triple system. So discussion of the tensor product of a central simple Hom–Lie triple system and a polynomial ring turns to be the same as tensor product of a Lie triple system and a polynomial ring. By [8], the theorem follows. $\square$