Lie (Jordan) Centralizer on Graded Rings

Abstract

In this paper, we provide the necessary and sufficient conditions for a Lie centralizer to be proper and a Jordan centralizer to be a centralizer on graded rings. Since every Lie centralizer naturally induces an anti-symmetric mapping and every Jordan centralizer naturally induces a symmetric mapping, our results provide the underlying graded structures reflected in these mappings. As applications, we recover known results for trivial extension algebras and triangular algebras, and additionally characterize Lie (Jordan) centralizers on exterior algebras, whose operations are inherently anti-symmetric.

1. Introduction and Preliminaries

Centralizers and Lie (Jordan) centralizers are classes of mappings widely studied in the analysis of algebraic structures. Recall that an additive mapping $\varphi$ on a ring $\mathcal{R}$ is called a centralizer if for all $a,b\in \mathcal{R}$:

$$\varphi \left(ab\right)=\varphi \left(a\right)b=a\varphi \left(b\right).$$

(1)

An additive mapping J on a ring $\mathcal{R}$ is called a Jordan centralizer if for all $a,b\in \mathcal{R}$:

$$J(a\circ b)=J\left(a\right)\circ b=a\circ J\left(b\right),$$

(2)

where $a\circ b=ab+ba$. An additive mapping L on a ring $\mathcal{R}$ is called a Lie centralizer if for all $a,b\in \mathcal{R}$:

$$L\left(\right[a,b\left]\right)=\left[L\right(a),b]=[a,L(b\left)\right],$$

(3)

where $[a,b]=ab-ba$ denotes the Lie product. A Lie centralizer L is called a proper Lie centralizer if it can be expressed as $L=\varphi +\theta$, where $\varphi$ is a centralizer and $\theta$ is an additive map from the ring to its center. Clearly, all centralizers are Jordan centralizers and Lie centralizers, with Jordan centralizers being symmetric mappings and Lie centralizers being anti-symmetric mappings. However, Lie (Jordan) centralizers are not necessarily centralizers. It is natural and interesting to find some conditions under which a Lie (Jordan) centralizer map is a centralizer map. In [1], Jabeen studies Lie (Jordan) centralizers on generalized matrix algebras, establishing the necessary and sufficient conditions for a Lie centralizer map to be proper and proving that every Jordan centralizer is a centralizer. In [2], the authors obtain some conditions under which a Jordan centralizer on a trivial extension algebra is a centralizer. In [3], Ajda and Wu present characterizations of Lie centralizers on triangular rings without units. Additionally, ref. [4] provided a description of Lie centralizers at an arbitrary but fixed point on triangular algebras. Further studies on various mappings over generalized matrix algebras are presented in [5,6] and references therein.

In fact, trivial extension algebras, triangular algebras and some generalized matrix algebras are all graded rings. In this paper, we focus on the conditions under which a Lie centralizer is proper and a Jordan centralizer coincides with a centralizer on graded rings.

Throughout this paper, unless explicitly stated otherwise, all rings are assumed to be associative and to possess an identity element (denoted by 1). Additionally, all indices $i,j,k$ represent non-negative integers. A ring $\mathcal{R}$ is called graded (by the natural numbers) if one can write $\mathcal{R}$ as a direct sum (as an abelian group),

$$\mathcal{R}=\underset{i}{⨁}{\mathcal{R}}_i={\mathcal{R}}_0\oplus {\mathcal{R}}_1\oplus {\mathcal{R}}_2\oplus \dots$$

such that for all non-negative integers m and n we have

$${\mathcal{R}}_m{\mathcal{R}}_n\subseteq {\mathcal{R}}_{m+n}.$$

The nonzero elements of ${\displaystyle \bigcup _{i⩾0}}{\mathcal{R}}_i$ are called homogeneous elements of $\mathcal{R}$, a nonzero $r\in {\mathcal{R}}_i$ is said to be homogeneous of degree i, and we write $\mathrm{deg}\left(r\right)=i$. Any nonzero $r\in \mathcal{R}$ has a unique expression as a sum of homogeneous elements $r={\displaystyle \sum _i}{r}_i$ where $r_i\in {\mathcal{R}}_i$, and the sum is finite. For more details on graded rings, one can refer to reference [7]. For convenience, we introduce two maps:

$${\pi }_i:\mathcal{R}\to {\mathcal{R}}_i\mathrm{via}r\mapsto r_i$$

and

$$\inf \left(r\right)=\left\{\begin{array}{cc}\inf \{i\mid r_i\ne 0\}\hfill & r\ne 0,\hfill \\ +\infty \hfill & r=0,\hfill \end{array}\right.$$

(4)

which means the smallest degree of homogeneous elements.

2. Centralizer and Jordan Centralizer on Graded Rings

In this section, we describe the structures of centralizers and Jordan centralizers on graded rings, and provide the necessary and sufficient conditions for a Jordan centralizer to be a centralizer.

To facilitate the proofs in the subsequent sections, we first introduce several preliminary propositions and lemmas.

Proposition 1.

Let $\mathcal{R}=\underset{i}{⨁}{\mathcal{R}}_i$ be a graded ring. Let $r\in \mathcal{R}$ and $b_j\in {\mathcal{R}}_j$. Then we have the following properties:

(1) 

${\pi }_i$ is additive.

(2) 

${\pi }_{k+j}\left(r{b}_j\right)={\pi }_k\left(r\right)b_j$ and ${\pi }_{k+j}\left(b_{j}r\right)=b_j{\pi }_k\left(r\right)$.

(3) 

${\pi }_{k+j}\left([r,b_j]\right)=[{\pi }_k\left(r\right),b_j]$.

(4) 

${\pi }_{k+j}(r\circ b_j)={\pi }_k\left(r\right)\circ b_j$.

Proof. 

(1) is clear and (4) is similar to (3). It suffices to prove (2) and (3).

For (2),

$${\pi }_{k+j}\left(r{b}_j\right)={\pi }_{k+j}\left(\sum _i{r}_i{b}_j\right)=\sum _i{\pi }_{k+j}\left(r_i{b}_j\right).$$

Since each term $r_i{b}_j\in {\mathcal{R}}_{i+j}$,

$$\sum _i{\pi }_{k+j}\left(r_i{b}_j\right)=r_k{b}_j={\pi }_k\left(r\right)b_j.$$

The proof of ${\pi }_{k+j}\left(b_{j}r\right)=b_j{\pi }_k\left(r\right)$ is similar.

For (3), by (1) and (2)

$$\begin{array}{cc}\hfill {\pi }_{k+j}\left([r,b_j]\right)& ={\pi }_{k+j}(r{b}_j-b_{j}r)\hfill \\ \hfill & ={\pi }_{k+j}\left(r{b}_j\right)-{\pi }_{k+j}\left(b_{j}r\right)\hfill \\ \hfill & ={\pi }_k\left(r\right)b_j-b_j{\pi }_k\left(r\right)\hfill \\ \hfill & =[{\pi }_k\left(r\right),b_j].\hfill \end{array}$$

The proof is completed. □

Proposition 2.

Let $\mathcal{R}={\displaystyle \underset{i}{⨁}}{\mathcal{R}}_i$ be a graded ring. When $r\in \mathcal{R}$ and $b_j\in {\mathcal{R}}_j$, we have the following properties:

(1) 

$\inf (a+b)⩾\min \left\{\inf \right(a),\inf (b\left)\right\}$ if $a,b\in \mathcal{R}$.

(2) 

$\inf \left(r{b}_j\right)⩾\inf \left(r\right)+j$ and $\inf \left(b_{j}r\right)⩾\inf \left(r\right)+j$.

(3) 

$\inf \left([r,b_j]\right)⩾\inf \left(r\right)+j$.

(4) 

$\inf (r\circ b_j)⩾\inf \left(r\right)+j$.

Proof. 

(1) The proof is clear.

(2) By repeatedly applying (1), we can obtain:

$$\inf \left(r{b}_j\right)=\inf \left(\sum _i{r}_i{b}_j\right)⩾\min \{\inf \left(r_0{b}_j\right),\inf \left(r_1{b}_j\right),\inf \left(r_2{b}_j\right),\dots \}.$$

By (4), $\inf \left(r_i{b}_j\right)=\left\{\begin{array}{cc}i+j\hfill & \mathrm{if}{r}_i{b}_j\ne 0\hfill \\ +\infty \hfill & \mathrm{if}{r}_i{b}_j=0\hfill \end{array}\right.$, which implies $\inf \left(r{b}_j\right)⩾\inf \left(r\right)+j$.

(3) Combining (1) and (2), we deduce:

$$\inf \left([r,b_j]\right)=\inf (r{b}_j-b_{j}r)⩾\min \{\inf \left(r{b}_j\right),\inf (-b_{j}r)\}=\inf \left(r{b}_j\right)⩾\inf \left(r\right)+j.$$

(4) The proof is similar to (3). □

The structure of Lie (Jordan) centralizers is closely related to that of the center in graded rings. We will now present a detailed analysis of the center’s structure in graded rings. For any ring $\mathcal{R}$, the center of a subset $\mathcal{S}\subseteq \mathcal{R}$ is denoted by:

$$Z\left(\mathcal{S}\right):=\left\{r\in \mathcal{R}\mid rs=sr\mathrm{for}\mathrm{all}s\in \mathcal{S}\right\}.$$

Lemma 1.

Let $\mathcal{R}={\displaystyle \underset{i}{⨁}}{\mathcal{R}}_i$ be a graded ring. Then we have:

$$Z\left(\mathcal{R}\right)=\left\{r\in \mathcal{R}\mid {\pi }_k\left(r\right)\in Z\left({\mathcal{R}}_j\right)\mathit{for}\mathit{all}j,k\right\}.$$

Proof. 

Let $r\in Z\left(\mathcal{R}\right)$ and $b_j\in {\mathcal{R}}_j$, by Proposition 1 and hypothesis,

$${\pi }_k\left(r\right)b_j={\pi }_{k+j}\left(r{b}_j\right)={\pi }_{k+j}\left(b_{j}r\right)=b_j{\pi }_k\left(r\right).$$

This implies ${\pi }_k\left(r\right)\in Z\left({\mathcal{R}}_j\right)$. If $r\in \mathcal{R}$ and ${\pi }_k\left(r\right)\in Z\left({\mathcal{R}}_j\right)$ for all $j,k$. Then, for any $s\in \mathcal{R}$,

$$rs=\sum _k\sum _j{\pi }_k\left(r\right){\pi }_j\left(s\right)=\sum _k\sum _j{\pi }_j\left(s\right){\pi }_k\left(r\right)=sr.$$

Thus $r\in Z\left(\mathcal{R}\right)$. □

Definition 1.

An additive mapping ϕ on a ring $\mathcal{R}$ is called a centralizer if for all $a,b\in \mathcal{R}$:

$$\varphi \left(ab\right)=\varphi \left(a\right)b=a\varphi \left(b\right).$$

Lemma 2.

Let ϕ be an additive map of a graded ring $\mathcal{R}={\displaystyle \underset{i}{⨁}}{\mathcal{R}}_i$, then the following conditions are equivalent:

(1) 

ϕ is a centralizer;

(2) 

$\varphi \left(a_i\right)=\varphi \left(1\right)a_i=a_i\varphi \left(1\right)$ for all $a_i\in {\mathcal{R}}_i$;

(3) 

${\pi }_{k+i}\left(\varphi \left(a_i\right)\right)={\pi }_k\left(\varphi \left(1\right)\right)a_i=a_i{\pi }_k\left(\varphi \left(1\right)\right)$ and ${\pi }_j\left(\varphi \left(a_i\right)\right)=0$ when $j<i$ for all $a_i\in {\mathcal{R}}_i$.

Proof. 

$\left(1\right)\Rightarrow \left(2\right)$: By (1), let $a=a_i\in {\mathcal{R}}_i$, $b=1\in {\mathcal{R}}_0$; we obtain:

$$\varphi \left(a_i\right)=\varphi \left(1\right)a_i=a_i\varphi \left(1\right).$$

$\left(2\right)\Rightarrow \left(1\right)$: Let $a={\displaystyle \sum _i}{a}_i$, $b={\displaystyle \sum _j}{b}_j\in \mathcal{R}$, where $a_i\in {\mathcal{R}}_i,b_j\in {\mathcal{R}}_j$. Then we have:

$$ab=\sum _i\sum _j{a}_i{b}_j.$$

By additive of $\varphi$ and assumption, we obtain:

$$\varphi \left(ab\right)=\varphi \left(\sum _i\sum _j{a}_i{b}_j\right)=\sum _i\sum _j\varphi \left(a_i{b}_j\right)=\sum _i\sum _j\varphi \left(1\right)a_i{b}_j.$$

Similarly,

$$\varphi \left(a\right)b=\varphi \left(\sum _i{a}_i\right)\sum _j{b}_j=\sum _i\sum _j\varphi \left(a_i\right)b_j=\sum _i\sum _j\varphi \left(1\right)a_i{b}_j$$

and

$$a\varphi \left(b\right)=\sum _i{a}_i\varphi \left(\sum _j{b}_j\right)=\sum _i\sum _j{a}_i\varphi \left(b_j\right)=\sum _i\sum _j\varphi \left(1\right)a_i{b}_j.$$

It is clear that

$$\varphi \left(ab\right)=\varphi \left(a\right)b=a\varphi \left(b\right).$$

$\left(2\right)\Rightarrow \left(3\right)$: Let $a_i\in {\mathcal{R}}_i$. By assumption and Proposition 1, we have:

$${\pi }_{k+i}\left(\varphi \left(a_i\right)\right)={\pi }_{k+i}\left(\varphi \left(1\right)a_i\right)={\pi }_k\left(\varphi \left(1\right)\right)a_i.$$

Similarly, ${\pi }_{k+i}\left(\varphi \left(a_i\right)\right)=a_i{\pi }_k\left(\varphi \left(1\right)\right)$. In fact,

$$\inf \left(\varphi \left(a_i\right)\right)=\inf \left(\varphi \left(1\right)a_i\right)⩾\inf \left(\varphi \left(1\right)\right)+i⩾i,$$

we know that if $j<i$, then

$${\pi }_j\left(\varphi \left(a_i\right)\right)=0.$$

$\left(3\right)\Rightarrow \left(2\right)$: Let $a_i\in {\mathcal{R}}_i$. It is easy to determine $\inf \left(\varphi \left(a_i\right)\right)⩾i$. By assumption and Proposition 1,

$${\pi }_{k+i}\left(\varphi \left(a_i\right)\right)={\pi }_k\left(\varphi \left(1\right)\right)a_i={\pi }_{k+i}\left(\varphi \left(1\right)a_i\right).$$

Thus, $\varphi \left(a_i\right)=\varphi \left(1\right)a_i$. □

Definition 2.

An additive mapping J on a ring $\mathcal{R}$ is called a Jordan centralizer if for all $a,b\in \mathcal{R}$:

$$J(a\circ b)=J\left(a\right)\circ b=a\circ J\left(b\right),$$

where $a\circ b=ab+ba$.

Lemma 3.

Let J be an additive map of a graded ring $\mathcal{R}={\displaystyle \underset{i}{⨁}}{\mathcal{R}}_i$, then the following conditions are equivalent:

(1) 

J is a Jordan centralizer;

(2) 

$J(a_i\circ b_j)=J\left(a_i\right)\circ b_j$ for all $a_i\in {\mathcal{R}}_i$, $b_j\in {\mathcal{R}}_j$;

(3) 

${\pi }_{k+j}\left(J(a_i\circ b_j)\right)={\pi }_k\left(J\left(a_i\right)\right)\circ b_j$ and ${\pi }_k\left(J(a_i\circ b_j)\right)=0$ when $k<\max \{i,j\}$ for all $a_i\in {\mathcal{R}}_i$, $b_j\in {\mathcal{R}}_j$.

Proof. 

$\left(1\right)\Rightarrow \left(2\right)$: The proof is clear by (2).

$\left(2\right)\Rightarrow \left(1\right)$: Let $a={\displaystyle \sum _i}{a}_i$, $b={\displaystyle \sum _j}{b}_j\in \mathcal{R}$. Then we have:

$$a\circ b=\sum _i\sum _j(a_i\circ b_j).$$

Because J is additive, then we obtain:

$$J(a\circ b)=J\left(\sum _i\sum _j(a_i\circ b_j)\right)=\sum _i\sum _{j}J(a_i\circ b_j).$$

Similarly,

$$J\left(a\right)\circ b=J\left(\sum _i{a}_i\right)\circ \sum _j{b}_j=\sum _i\sum _j(J\left(a_i\right)\circ bj).$$

Since $a\circ b=b\circ a$, it is clear that

$$J(a\circ b)=J\left(a\right)\circ b=a\circ J\left(b\right).$$

$\left(2\right)\Rightarrow \left(3\right)$: Let $a_i\in {\mathcal{R}}_i$ and $b_j\in {\mathcal{R}}_j$. With $J(a_i\circ b_j)=J\left(a_i\right)\circ b_j$ and Proposition 1, it follows that

$${\pi }_{k+j}\left(J(a_i\circ b_j)\right)={\pi }_{k+j}(J\left(a_i\right)\circ b_j)={\pi }_k\left(J\left(a_i\right)\right)\circ b_j.$$

By Proposition 2, we know that $\inf \left(J(a_i\circ b_j)\right)=\inf (J\left(a_i\right)\circ b_j)⩾j$. Thus

$${\pi }_k\left(J(a_i\circ b_j)\right)=0,$$

(5)

for $k<j$ and if $k<i$, then

$${\pi }_k\left(J(b_j\circ a_i)\right)={\pi }_k\left(J(a_i\circ b_j)\right)=0.$$

(6)

Combining Equations (5) and (6), we conclude that

$${\pi }_k\left(J(a_i\circ b_j)\right)=0,$$

for $k<\max \{i,j\}$.

$\left(3\right)\Rightarrow \left(2\right)$: Let $a_i\in {\mathcal{R}}_i$, $b_j\in {\mathcal{R}}_j$, then

$$\inf \left(J(a_i\circ b_j)\right)⩾\max \{i,j\}.$$

Then ${\pi }_{k+j}J(a_i\circ b_j)={\pi }_k\left(J\left(a_i\right)\right)\circ b_j$ implies that

$$J(a_i\circ b_j)=J\left(a_i\right)\circ b_j.$$

The proof is completed. □

Theorem 1.

Let $\mathcal{R}={\displaystyle \underset{i}{⨁}}{\mathcal{R}}_i$ be a 2-torsion free graded ring, and J be a Jordan centralizer on $\mathcal{R}$, then J is a centralizer if and only if ${\pi }_{k}J\left(1\right)\in Z\left({\mathcal{R}}_j\right)$ for all $i,k$.

Proof. 

By Lemmas 1 and 3.(2), we have:

$$2J\left(a_i\right)=J(1\circ a_i)=J\left(1\right)\circ a_i=J\left(1\right)a_i+a_{i}J\left(1\right)=2J\left(1\right)a_i$$

for all $a_i\in {\mathcal{R}}_i$. Which means that $J\left(a_i\right)=J\left(1\right)a_i$ for all $a_i\in {\mathcal{R}}_i$. By Lemma 2, J is a centralizer.

The converse is clear. □

3. Lie Centralizer on Graded Rings

In this section, we describe the structure of Lie centralizers on graded rings and provide the necessary and sufficient conditions for a Lie centralizer to be proper.

Definition 3.

An additive mapping L on a ring $\mathcal{R}$ is called a Lie centralizer if for all $a,b\in \mathcal{R}$:

$$L\left(\right[a,b\left]\right)=\left[L\right(a),b]=[a,L(b\left)\right],$$

where $[a,b]=ab-ba$ denotes the Lie product.

Lemma 4.

Let L be an additive map of a graded ring $\mathcal{R}={\displaystyle \underset{i}{⨁}}{\mathcal{R}}_i$, then the following conditions are equivalent:

(1) 

L is a Lie centralizer;

(2) 

$L[a_i,b_j]=[L\left(a_i\right),b_j]$ for all $a_i\in {\mathcal{R}}_i$, $b_j\in {\mathcal{R}}_j$;

(3) 

${\pi }_{k+j}\left(L[a_i,b_j]\right)=[{\pi }_k\left(L\left(a_i\right)\right),b_j]$ and ${\pi }_k\left(L[a_i,b_j]\right)=0$ for all $a_i\in {\mathcal{R}}_i$, $b_j\in {\mathcal{R}}_j$ if $k<\max \{i,j\}$.

Proof. 

$\left(1\right)\Rightarrow \left(2\right)$: The proof is clear.

$\left(2\right)\Rightarrow \left(1\right)$: Let $a={\displaystyle \sum _i}{a}_i,b={\displaystyle \sum _j}{b}_j\in \mathcal{R}$, where $a_i\in {\mathcal{R}}_i,b_j\in {\mathcal{R}}_j$. Then we have:

$$[a,b]=\sum _i\sum _j[a_i,b_j].$$

Because L is additive, then we obtain:

$$L[a,b]=L\left(\sum _i\sum _j[a_i,b_j]\right)=\sum _i\sum _{j}L[a_i,b_j].$$

Similarly,

$$[L\left(a\right),b]=\left[L\left(\sum _i{a}_i\right),\sum _j{b}_j\right]=\sum _i\sum _j[L\left(a_i\right),b_j].$$

By Theorem 2.1 in [8], it is easy to determine that

$$L[a,b]=\left[L\right(a),b]=[a,L(b\left)\right].$$

$\left(2\right)\Rightarrow \left(3\right)$: Let $a_i\in {\mathcal{R}}_i$ and $b_j\in {\mathcal{R}}_j$. As per $L[a_i,b_j]=[L\left(a_i\right),b_j]$ and Proposition 1, it follows that

$${\pi }_{k+j}\left(L[a_i,b_j]\right)={\pi }_{k+j}\left([L\left(a_i\right),b_j]\right)=[{\pi }_k\left(L\left(a_i\right)\right),b_j].$$

By Proposition 2, we know that $\inf \left(L[a_i,b_j]\right)=\inf \left([L\left(a_i\right),b_j]\right)⩾j$. Thus if $k<j$, then

$${\pi }_k\left(L[a_i,b_j]\right)=0$$

(7)

and if $k<i$, then

$${\pi }_k\left(L[b_j,a_i]\right)=-{\pi }_k\left(L[a_i,b_j]\right)=0.$$

(8)

Combining Equations (7) and (8), we conclude that if $k<\max \{i,j\}$, then

$${\pi }_k\left(L[a_i,b_j]\right)=0.$$

$\left(3\right)\Rightarrow \left(2\right)$: Let $a_i\in {\mathcal{R}}_i$ and $b_j\in {\mathcal{R}}_j$, then

$$\inf \left(L[a_i,b_j]\right)⩾\max \{i,j\}.$$

Then, ${\pi }_{k+j}\left(L[a_i,b_j]\right)=[{\pi }_k\left(L\left(a_i\right)\right),b_j]$ implies that

$$L[a_i,b_j]=[L\left(a_i\right),b_j].$$

The proof is completed. □

Lemma 5.

Let T be an additive map of a graded ring $\mathcal{R}={\displaystyle \underset{i}{⨁}}{\mathcal{R}}_i$, then the following conditions are equivalent:

(1) 

$T[a,b]=0$ for all $a,b\in \mathcal{R}$;

(2) 

$T[a_i,b_j]=0$ for all $a_i\in {\mathcal{R}}_i$ and $b_j\in {\mathcal{R}}_j$.

Proof. 

It suffices to prove $\left(2\right)\Rightarrow \left(1\right)$: Let $a={\displaystyle {\displaystyle \sum _i}}{a}_i,b={\displaystyle {\displaystyle \sum _j}}{b}_j\in \mathcal{R}$, then

$$T[a,b]=\sum _i\sum _{j}T[a_i,b_j]=0.$$

The proof is completed. □

The main result of this article is as follows:

Theorem 2.

Let $\mathcal{R}={\displaystyle \underset{i}{⨁}}{\mathcal{R}}_i$ be a graded ring and L be a Lie centralizer on $\mathcal{R}$. Then the following conditions are equivalent:

(1) 

L is proper; that is, there exists a centralizer $\varphi :\mathcal{R}\to \mathcal{R}$ such that $L=\varphi +\theta$, where θ vanishes at commutators;

(2) 

For all k, there exist finitely many nonzero $r_k\in Z\left(\mathcal{R}\right)\bigcap {\mathcal{R}}_k$ such that ${\pi }_{k+i}\left(L\left(a_i\right)\right)-r_k{a}_i\in Z\left(\mathcal{R}\right)$ for all $a_i\in {\mathcal{R}}_i$, and ${\pi }_j\left(L\left(a_i\right)\right)\in Z\left(\mathcal{R}\right)$ for all $a_i\in \mathcal{R}$ if $j<i$.

Proof. 

$\left(1\right)\Rightarrow \left(2\right)$: Let $r_k={\pi }_k\left(\varphi \left(1\right)\right)\in {\mathcal{R}}_k$. By Lemma 2, it is easy to check $r_k\in Z\left(\mathcal{R}\right)$ and ${\pi }_{k+i}\left(\varphi \left(a_i\right)\right)=r_k{a}_i$. By Lemma 1, we have:

$${\pi }_{k+i}\left(\theta \left(a_i\right)\right)={\pi }_{k+i}\left((L-\varphi )\left(a_i\right)\right)={\pi }_{k+i}\left(L\left(a_i\right)\right)-{\pi }_{k+i}\left(\varphi \left(a_i\right)\right)={\pi }_{k+i}\left(L\left(a_i\right)\right)-r_k{a}_i\in Z\left(\mathcal{R}\right).$$

If $j<i$, by Lemma 2, ${\pi }_j\left(\varphi \left(a_i\right)\right)=0$, and hence

$${\pi }_j\left(L\left(a_i\right)\right)={\pi }_j\left(\varphi \left(a_i\right)\right)+{\pi }_j\left(\theta \left(a_i\right)\right)={\pi }_j\left(\theta \left(a_i\right)\right)\in Z\left(\mathcal{R}\right).$$

$\left(2\right)\Rightarrow \left(1\right)$: Define $\varphi :\mathcal{R}\to \mathcal{R}$ by ${\pi }_j\left(\varphi \left(a\right)\right)={\displaystyle \sum _{k+i=j}}{r}_k{a}_i$. It is easy to determine $\varphi$ is well-defined. For any $a,b\in \mathcal{R}$, we have

$$\begin{array}{cc}\hfill \varphi (a+b)& =\sum _j{\pi }_j\left(\varphi (a+b)\right)\hfill \\ \hfill & =\sum _j\sum _{k+i=j}\left(r_k(a_i+b_i)\right)\hfill \\ \hfill & =\sum _j\sum _{k+i=j}(r_k{a}_i+r_k{b}_i)\hfill \\ \hfill & =\sum _j\sum _{k+i=j}{r}_k{a}_i+\sum _j\sum _{k+i=j}{r}_k{b}_i\hfill \\ \hfill & =\sum _j{\pi }_j\left(\varphi \left(a\right)\right)+\sum _j{\pi }_j\left(\varphi \left(b\right)\right)\hfill \\ \hfill & =\varphi \left(a\right)+\varphi \left(b\right),\hfill \end{array}$$

which means that $\varphi$ is additive. Clearly,

$${\pi }_k\left(\varphi \left(1\right)\right)=r_k.$$

Note that for any $a_i\in {\mathcal{R}}_i$

$${\pi }_{k+i}\left(\varphi \left(a_i\right)\right)=r_k{a}_i={\pi }_k\left(\varphi \left(1\right)\right)a_i$$

and

$$\inf \left(\varphi \left(a_i\right)\right)=\inf \left(\sum _k{r}_k{a}_i\right)⩾i.$$

And hence $\varphi$ is a centralizer on $\mathcal{R}$ by Lemma 2. Furthermore, let $\theta =L-\varphi$, then $\theta$ is an additive map since L and $\varphi$ are both additive. For all $a_i\in {\mathcal{R}}_i$ and $b_j\in {\mathcal{R}}_j$,

$$\theta [a_i,b_j]=L[a_i,b_j]-\varphi [a_i,b_j]=[(L-\varphi )\left(a_i\right),b_j]=[\theta \left(a_i\right),b_j]=0,$$

by Lemmas 1 and 4. Thus $\theta$ vanishes at commutators by Lemma 5. □

4. Applications to Specific Algebras

In this section, we use the theorems established in this paper to provide the necessary and sufficient conditions for a Lie centralizer to be proper and for a Jordan centralizer to be a centralizer in specific triangular algebras, trivial extension algebras, and exterior algebras, and to construct nontrivial proper Lie centralizers.

At first, we can construct the nontrivial proper Lie centralizer on graded rings by Theorem 2.

Example 1.

Let $\mathcal{A}=\mathcal{R}\left[x\right]{/}_{\langle x^n\rangle }$, where R is the algebra consisting of $n\times n$ matrices over the real number field $\mathbb{R}$. It is well known that

$$\mathcal{A}=\mathcal{R}\oplus \mathcal{R}x\oplus \mathcal{R}{x}^2\oplus \dots \oplus \mathcal{R}{x}^{n-1}.$$

Let I be the identity matrix, then we define that

$$r_k={\pi }_k\left(\varphi \left(I\right)\right)={\lambda }_{k}I{x}^k$$

where ${\lambda }_k\in \mathbb{R}$ and for all $A\in \mathcal{R}$

$${\pi }_j\left(\theta \left(A{x}^i\right)\right)=tr\left(A\right)I{x}^j$$

where $tr\left(A\right)$ is a trace of A. Define

$$\varphi \left(A{x}^i\right)=\sum _{k=0}^{n-1-i}{r}_{k}A{x}^i,\mathrm{for}\mathrm{any}A{x}^i\in \mathcal{R}{x}^i$$

and

$$\theta \left(A{x}^i\right)=\sum _{j=0}^{n-1}{\pi }_j\left(\theta \left(A{x}^i\right)\right),\mathrm{for}\mathrm{any}A{x}^i\in \mathcal{R}{x}^i.$$

Then $\varphi$ is a centralizer on $\mathcal{A}$ by Lemma 2 and $\theta$ vanishes at commutators. Then $L=\varphi +\theta$ is a proper Lie centralizer.

Corollary 1.

Let $\mathcal{R}=\mathcal{A}\oplus \mathcal{M}$ be a trivial extension algebra. If there exists a nontrivial idempotent p such that $pmq=m$ for all $m\in \mathcal{M}$ with $q=1-p$:

(1) 

If $\mathcal{R}$ is a 2-torsion free algebra and $\mathcal{M}$ is a faithful bimodule, then every Jordan centralizer is a centralizer.

(2) 

Lie centralizer L is proper if and only if there exists $r_0\in \mathcal{A}\cap Z\left(\mathcal{R}\right)$ such that ${\pi }_{\mathcal{A}}L\left(a\right)-r_{0}a\in Z\left(\mathcal{R}\right)$ for all $a\in \mathcal{A}$.

Proof. 

(1) Clearly, $\mathcal{R}={\mathcal{R}}_0\oplus {\mathcal{R}}_1$ becomes a graded ring with ${\mathcal{R}}_0=\mathcal{A}$, ${\mathcal{R}}_1=\mathcal{M}$ and ${\mathcal{R}}_i=0$ for $i>1$. Let J be a Jordan centralizer, then for all $a\in \mathcal{A}$ and $m\in \mathcal{M}$,

$$\begin{array}{cc}\hfill 2{\pi }_{1}J\left(am\right)& ={\pi }_{1}J((ap+pa)m+m(ap+pa))\hfill \\ \hfill & ={\pi }_{1}J((a\circ p)\circ m)\hfill \\ \hfill & ={\pi }_{0}J(a\circ p)\circ m\hfill \\ \hfill & ={\pi }_{0}J\left(a\right)\circ p\circ m\hfill \\ \hfill & =2{\pi }_{0}J\left(a\right)m.\hfill \end{array}$$

Thus ${\pi }_{1}J\left(am\right)={\pi }_{0}J\left(a\right)m$, which means that ${\pi }_{1}J\left(m\right)={\pi }_{0}J\left(1\right)m$. Similarly, ${\pi }_{1}J\left(m\right)=m{\pi }_{0}J\left(1\right)$. In fact,

$$\begin{array}{cc}\hfill 0& ={\pi }_{1}J\left(am\right)-{\pi }_{0}J\left(1\right)am\hfill \\ \hfill & ={\pi }_{0}J\left(a\right)m-{\pi }_{0}J\left(1\right)am\hfill \\ \hfill & =({\pi }_{0}J\left(a\right)-{\pi }_{0}J\left(1\right)a)m.\hfill \end{array}$$

Since $\mathcal{M}$ is a faithful bimodule, ${\pi }_{0}J\left(a\right)-{\pi }_{0}J\left(1\right)a=0$. Similarly, ${\pi }_{0}J\left(a\right)-a{\pi }_{0}J\left(1\right)=0$. By Lemma 2.5 in [2], ${\pi }_{1}J\left(1\right)\in Z\left(\mathcal{R}\right)$. From the above, by Theorem 1, J is a centralizer.

(2) By Theorem 2, it is sufficient to prove the ’only if’ part. Let L be a Lie centralizer. By assumption,

$$L\left(m\right)=L[p,m]=\left[L\right(p),m]=L\left(p\right)m-mL\left(p\right).$$

In fact,

$$0=L[p,p]=[L\left(p\right),p]=L\left(p\right)p-pL\left(p\right)={\pi }_{0}L\left(p\right)p-p{\pi }_{0}L\left(p\right)-{\pi }_{1}L\left(p\right).$$

which implies that

$${\pi }_{1}L\left(p\right)=0.$$

Then

$$L\left(m\right)={\pi }_{0}L\left(p\right)m-m{\pi }_{0}L\left(p\right)=r_{0}pm-m{r}_{0}p=r_{0}m.$$

It is easy to know that

$${\pi }_{1}L\left(a\right)={\pi }_{1}L(paq+pap+qap+qaq).$$

By Lemma 5,

$${\pi }_{1}L\left(paq\right)={\pi }_{1}L[p,paq]={\pi }_{1}L\left(p\right)paq-paq{\pi }_{1}L\left(p\right)=0$$

and

$$0={\pi }_{1}L[pap,p]={\pi }_{1}L\left(pap\right)p-p{\pi }_1\left(pap\right)=-{\pi }_{1}L\left(pap\right).$$

Similarly, ${\pi }_{1}L\left(qap\right)=0$ and ${\pi }_{1}L\left(qaq\right)=0$. Then we have ${\pi }_{1}L\left(a\right)=0$. Let $r_1=0$; by Theorem 2, L is a proper Lie centralizer. □

Remark 1.

The conditions in Corollary 1(1) differ from those in Theorem 2.11 in [2].

The following results can also be obtained from Corollary 3.3 in [3] and Theorem 4.3 in [1]. We prove the results by Theorems 1 and 2.

Corollary 2.

Let $\mathcal{R}=\left(\begin{array}{cc}\mathcal{A}& \mathcal{M}\\ 0& \mathcal{B}\end{array}\right)$ be a triangular algebra. Suppose that $\mathcal{M}$ is a faithful ($\mathcal{A}$,$\mathcal{B}$)-bimodule.

(1) 

If $\mathcal{R}$ is a 2-torsion free triangular algebra, then every Jordan centralizer is a centralizer.

(2) 

If ${\pi }_{\mathcal{A}}\left(Z\left(\mathcal{R}\right)\right)=Z\left(\mathcal{A}\right)$ and ${\pi }_{\mathcal{B}}\left(Z\left(\mathcal{R}\right)\right)=Z\left(\mathcal{B}\right)$, then every Lie centralizer is a proper Lie centralizer.

Proof. 

(1) Clearly, $J\left(1\right)\in Z\left(\mathcal{R}\right)$. By Theorem 1, J is a centralizer.

(2) Considering

$${\mathcal{R}}_0=\left(\begin{array}{cc}\mathcal{A}& 0\\ 0& \mathcal{B}\end{array}\right),{\mathcal{R}}_1=\left(\begin{array}{cc}0& \mathcal{M}\\ 0& 0\end{array}\right),{\mathcal{R}}_i=0\mathrm{for}i>1,$$

it is easy to check that $\mathcal{R}={\displaystyle \underset{i}{⨁}}{\mathcal{R}}_i={\mathcal{R}}_0\oplus {\mathcal{R}}_1$ becomes a graded ring. For any Lie centralizer L on $\mathcal{R}$, Lemma 5 and Lemma 2.3 in [3] imply that

$$L\left(\begin{array}{cc}a& 0\\ 0& b\end{array}\right)=\left(\begin{array}{cc}{L}_{\mathcal{A}}^{\mathcal{A}}\left(a\right)+L_{\mathcal{A}}^{\mathcal{B}}\left(b\right)& 0\\ 0& L_{\mathcal{B}}^{\mathcal{A}}\left(a\right)+L_{\mathcal{B}}^{\mathcal{B}}\left(b\right)\end{array}\right),$$

where $L_{\mathcal{A}}^{\mathcal{A}}:\mathcal{A}\to \mathcal{A}$, $L_{\mathcal{B}}^{\mathcal{A}}:\mathcal{A}\to Z\left(\mathcal{B}\right)$, $L_{\mathcal{A}}^{\mathcal{B}}:\mathcal{B}\to Z\left(\mathcal{A}\right)$ and $L_{\mathcal{B}}^{\mathcal{B}}:\mathcal{B}\to \mathcal{B}$.

By Proposition 1.1 in [9], there exists a unique ring isomorphism $\omega :Z\left(\mathcal{A}\right)\to Z\left(\mathcal{B}\right)$ such that $am=m\omega \left(a\right)$ where $a\in \mathcal{A}$ for all $m\in \mathcal{M}$. Then, let

$$r_0=\left(\begin{array}{cc}{L}_{\mathcal{A}}^{\mathcal{A}}\left(1_{\mathcal{A}}\right)-{\omega }^{-1}\left(L_{\mathcal{B}}^{\mathcal{A}}\left(1_{\mathcal{A}}\right)\right)& 0\\ 0& L_{\mathcal{B}}^{\mathcal{B}}\left(1_{\mathcal{B}}\right)-\omega \left(L_{\mathcal{A}}^{\mathcal{B}}\left(1_{\mathcal{B}}\right)\right)\end{array}\right)\mathrm{and}{r}_1=0.$$

Let $\left(\begin{array}{cc}a& 0\\ 0& b\end{array}\right)\in {\mathcal{R}}_0$ and $\left(\begin{array}{cc}0& m\\ 0& 0\end{array}\right)\in {\mathcal{R}}_1$, we have

$$\begin{array}{cc}\hfill & {\pi }_{0}L\left(\begin{array}{cc}a& 0\\ 0& b\end{array}\right)-r_0\left(\begin{array}{cc}a& 0\\ 0& b\end{array}\right)\hfill \\ \hfill & =\left(\begin{array}{cc}\begin{array}{c}{L}_{\mathcal{A}}^{\mathcal{A}}\left(a\right)+L_{\mathcal{A}}^{\mathcal{B}}\left(b\right)\\ -(L_{\mathcal{A}}^{\mathcal{A}}\left(1_{\mathcal{A}}\right)-{\omega }^{-1}\left(L_{\mathcal{B}}^{\mathcal{A}}\left(1_{\mathcal{A}}\right)\right))a\end{array}& 0\\ 0& \begin{array}{c}{L}_{\mathcal{B}}^{\mathcal{A}}\left(a\right)+L_{\mathcal{B}}^{\mathcal{B}}\left(b\right)\\ -(L_{\mathcal{B}}^{\mathcal{B}}\left(1_{\mathcal{B}}\right)-\omega \left(L_{\mathcal{A}}^{\mathcal{B}}\left(1_{\mathcal{B}}\right)\right))b\end{array}\end{array}\right).\hfill \end{array}$$

In fact, by

$$L\left(\begin{array}{cc}0& m\\ 0& 0\end{array}\right)=L\left[\left(\begin{array}{cc}{1}_{\mathcal{A}}& 0\\ 0& 0\end{array}\right),\left(\begin{array}{cc}0& m\\ 0& 0\end{array}\right)\right]=\left(\begin{array}{cc}0& L_{\mathcal{A}}^{\mathcal{A}}\left(1_{\mathcal{A}}\right)m-m{L}_{\mathcal{B}}^{\mathcal{A}}\left(1_{\mathcal{A}}\right)\\ 0& 0\end{array}\right)$$

and

$$L\left(\begin{array}{cc}0& m\\ 0& 0\end{array}\right)=L\left[\left(\begin{array}{cc}0& 0\\ 0& 1_{\mathcal{B}}\end{array}\right),\left(\begin{array}{cc}0& m\\ 0& 0\end{array}\right)\right]=\left(\begin{array}{cc}0& L_{\mathcal{A}}^{\mathcal{B}}\left(1_{\mathcal{B}}\right)m-m{L}_{\mathcal{B}}^{\mathcal{B}}\left(1_{\mathcal{B}}\right)\\ 0& 0\end{array}\right),$$

we have:

$$\begin{array}{cc}\hfill & (L_{\mathcal{A}}^{\mathcal{A}}\left(a\right)+L_{\mathcal{A}}^{\mathcal{B}}\left(b\right)-(L_{\mathcal{A}}^{\mathcal{A}}\left(1_{\mathcal{A}}\right)-{\omega }^{-1}\left(L_{\mathcal{B}}^{\mathcal{A}}\left(1_{\mathcal{A}}\right)\right))a)m\hfill \\ \hfill & -m(L_{\mathcal{B}}^{\mathcal{A}}\left(a\right)+L_{\mathcal{B}}^{\mathcal{B}}\left(b\right)-(L_{\mathcal{B}}^{\mathcal{B}}\left(1_{\mathcal{B}}\right)-\omega \left(L_{\mathcal{A}}^{\mathcal{B}}\left(1_{\mathcal{B}}\right)\right))b)\hfill \\ \hfill & =(L_{\mathcal{A}}^{\mathcal{A}}\left(a\right)m-m{L}_{\mathcal{B}}^{\mathcal{A}}\left(a\right))+(L_{\mathcal{A}}^{\mathcal{B}}\left(b\right)m-m{L}_{\mathcal{B}}^{\mathcal{B}}\left(b\right))\hfill \\ \hfill & -(L_{\mathcal{A}}^{\mathcal{A}}\left(1_{\mathcal{A}}\right)am-am{L}_{\mathcal{B}}^{\mathcal{A}}\left(1_{\mathcal{A}}\right))-(L_{\mathcal{A}}^{\mathcal{B}}\left(1_{\mathcal{B}}\right)mb-mb{L}_{\mathcal{B}}^{\mathcal{B}}\left(1_{\mathcal{B}}\right))\hfill \\ \hfill & =\left({\pi }_1\left[L\left(\begin{array}{cc}a& 0\\ 0& 0\end{array}\right),\left(\begin{array}{cc}0& m\\ 0& 0\end{array}\right)\right]+{\pi }_1\left[L\left(\begin{array}{cc}0& 0\\ 0& b\end{array}\right),\left(\begin{array}{cc}0& m\\ 0& 0\end{array}\right)\right]\right)\hfill \\ \hfill & -{\pi }_{1}L\left(\begin{array}{cc}0& am\\ 0& 0\end{array}\right)-{\pi }_{1}L\left(\begin{array}{cc}0& mb\\ 0& 0\end{array}\right)\hfill \\ \hfill & =\left({\pi }_{1}L\left(\begin{array}{cc}0& am\\ 0& 0\end{array}\right)+{\pi }_{1}L\left(\begin{array}{cc}0& mb\\ 0& 0\end{array}\right)\right)-{\pi }_{1}L\left(\begin{array}{cc}0& am\\ 0& 0\end{array}\right)-{\pi }_{1}L\left(\begin{array}{cc}0& mb\\ 0& 0\end{array}\right)\hfill \\ \hfill & =0.\hfill \end{array}$$

Thus, ${\pi }_{0}L\left(\begin{array}{cc}a& 0\\ 0& b\end{array}\right)-r_0\left(\begin{array}{cc}a& 0\\ 0& b\end{array}\right)\in Z\left(\mathcal{R}\right)$. Similarly, it is easy to check

$${\pi }_{1}L\left(\begin{array}{cc}0& m\\ 0& 0\end{array}\right)-r_0\left(\begin{array}{cc}0& m\\ 0& 0\end{array}\right)\in Z\left(\mathcal{R}\right),{\pi }_{0}L\left(\begin{array}{cc}0& m\\ 0& 0\end{array}\right)=0\mathrm{and}{\pi }_{1}L\left(\begin{array}{cc}a& 0\\ 0& b\end{array}\right)-r_1\left(\begin{array}{cc}a& 0\\ 0& b\end{array}\right)=0.$$

By Theorem 2, L is a proper Lie centralizer. □

An exterior algebra is a classical graded ring with extensive applications in various fields. For more details on exterior algebras, readers are referred to [10]. Let V be an n-dimensional vector space over a field $\mathbb{F}$, and let $\{e_1,e_2,\dots ,e_n\}$ be its basis. The exterior algebra $\bigwedge \left(V\right)$ of V is defined as the quotient algebra of the tensor algebra $T\left(V\right)$, where

$$T\left(V\right)=\underset{k}{⨁}{T}^{k}V=\mathbb{F}\oplus V\oplus (V\otimes V)\oplus (V\otimes V\otimes V)\oplus \dots ,$$

by the two-sided ideal I generated by all elements of the form $x\otimes x$ such that $x\in V$. Symbolically,

$$\bigwedge \left(V\right):=T\left(V\right)/I.$$

The k-th exterior power of V, denoted ${\bigwedge }^k\left(V\right)$, is a vector space, whose elements are referred to as k-vectors

$$x_1\wedge x_2\wedge \dots \wedge x_k,x_i\in V,i=1,2,\dots ,k.$$

The set

$$\{e_{i_1}\wedge e_{i_2}\wedge \dots \wedge e_{i_k}\mid 1⩽i_1<i_2<\dots <i_k⩽n\}$$

is a basis of ${\bigwedge }^k\left(V\right)$. Any element of the exterior algebra can be written as a sum of k-vectors. Hence, as a vector space, the exterior algebra is a direct sum

$$\bigwedge \left(V\right)={\bigwedge }^0\left(V\right)\oplus {\bigwedge }^1\left(V\right)\oplus {\bigwedge }^2\left(V\right)\oplus \dots \oplus {\bigwedge }^n\left(V\right),$$

where ${\bigwedge }^0\left(V\right)=\mathbb{F}$ and ${\bigwedge }^1\left(V\right)=V$. Thus $\bigwedge \left(V\right)$ is a graded ring with ${\bigwedge }^i\left(V\right)=0$ when $i>n$. Finally, the exterior product of $x\in {\bigwedge }^i\left(V\right)$ and $y\in {\bigwedge }^j\left(V\right)$ satisfies the following three properties:

1.

$x\wedge y$ is distributive over addition and scalar multiplication;

2.

Associativity: $(x\wedge y)\wedge z=x\wedge (y\wedge z)$ for all $z\in {\bigwedge }^k\left(V\right)$;

3.

$x\wedge y={(-1)}^{ij}y\wedge x$, so two vectors of odd degrees anticommute; otherwise the vectors commute.

It is easy to check that $Z(\bigwedge \left(V\right))=\left({\displaystyle \underset{i\mathrm{is}\mathrm{even}}{⨁}}{\bigwedge }^i\left(V\right)\right)\oplus {\bigwedge }^n\left(V\right)$. By Theorems 1 and 2, we obtain the following corollary.

Corollary 3.

Let $\bigwedge \left(V\right)$ be an exterior algebra where V is an n-dimensional vector space over a field $\mathbb{F}$, L be a Lie centralizer, and J be a Jordan centralizer.

1. 

J is a centralizer if and only if ${\pi }_k\left(J\left(1\right)\right)=0$ if $k<n$ with k is odd.

2. 

The following conditions are equivalent:

(1) 

L is a proper Lie centralizer;

(2) 

For every even integer k, there exists an element $r_k\in {\bigwedge }^k\left(V\right)$ such that ${\pi }_{k+i}\left(L\left(a_i\right)\right)=r_k\wedge a_i$ whenever $k+i<n$, which holds for all odd integers i and $a_i\in {\bigwedge }^i\left(V\right)$. And ${\pi }_j\left(L\left(a_i\right)\right)=0$ if $j<i$ with j is odd.