Research Article | Open Access

# Characterizations of Nonlinear Lie Derivations of

**Academic Editor:**Chunrui Zhang

#### Abstract

Let be an infinite dimensional Banach space, and is a nonlinear Lie derivation. Then Φ is the form where is an additive derivation of and is a map from into its center , which maps commutators into the zero.

#### 1. Introduction

The local actions of derivations of operator algebras are still not completely understood. One is the well-known local derivation problem. The other is to study conditions under derivations of operator algebras that can be completely determined by the action on some sets of operators. In recent years there has been a great interest in the study of Lie derivations of operator algebras (e.g., see [1–6]). In particular, the structure of additive or linear Lie derivation on ring or algebra has been studied by many authors (e.g., see [7–10]).

Let be a Banach space and let be the algebra of all bounded linear operators on . Some progress has been made towards understanding the Lie derivation recently. In [11], it was shown that letting be a Banach space of dimension greater than 2, if is a linear map satisfying for any with , then , where is a derivation of and is a linear map vanishing at commutators with . In [12], it was shown that every nonlinear Lie derivation of triangular algebras is the sum of an additive derivation and a map into its center sending commutators to zero. Ji and Qi [13] proved that letting be a triangular algebra over a commutative ring , if is an -linear map satisfying for any with , then , where is a derivation of and is an -linear map vanishing at commutators with . Bai and Du [14] proved that letting be a von Neumann algebra with no central summands of type , if is a nonlinear Lie derivation, then is of the form , where is an additive derivation of and is a mapping of into its center which maps commutators into zero.

Nevertheless, it would seem desirable to investigate nonlinear Lie derivations of . The present investigation attempts to do this by using the properties of nonlinear Lie derivation and previous research. A map is called an additive derivation if it is additive and satisfies for any . A linear map is called a Lie derivation if for any , where is the usual Lie product. More generally, a map (without the additive assumption) is a nonlinear Lie derivation if for any .

Recall that is called prime algebra if implies or . An operator is an idempotent provided that . In this paper, we suppose that is a prime associate algebra. We prove that every nonlinear Lie derivation of is the sum of an additive derivation and a map from into its center sending commutators to the zero. We want to mention here that there is no additive assumed.

#### 2. Result and Proof

Fix a nontrivial idempotent and let . In what follows, we write for, . Then every operator can be written as . Note that notation denotes an arbitrary element of .

*Definition 1. *Letting be a set of , we call
the centralizer of about .

Clearly, is the center of .

Lemma 2. *Consider that .*

*Proof. *It is clearly that .

Conversely, without loss of generality, we proof that . For any , we have
Thus
From (3), noting that is prime, we obtain , that is, . By , we have
Since is prime, we have . We write , where . Since , we have
Since is prime we have . We write , where . From (4) we have
So
Thus
So
It follows that

Lemma 3. *Consider that .*

*Proof. *From Lemma 2, we have

Theorem 4. *Let be an infinite dimensional Banach space, and is a nonlinear Lie derivation. Then is the form , where is an additive derivation of and is a map from into its center , which maps commutators into the zero.*

*Proof. *Now we organize the proof in a series of claims. *Claim* * 1. * Consider that ; moreover, such that and , for arbitrarily .

Obviously .

Moreover, for arbitrarily , In (13), the left and right sides are, respectively, multiplied by and , and we have In (13), the left and right sides are respectively multiplying by and , and we arrive at For arbitrarily , In the previous equation, the left and right sides are, respectively multiplied by and , and we have This together with (14) shows that From Lemma 3, it is that Denote . Therefore At the same time, . From (14), Because so by (13) we get .

In the same way, we can get .

*Remark.*It is clear that is an additive derivation. Without loss of generality, we can assume that and .

*Claim*

*Consider that .*

*2.*Let , for every , and write . From and , we can get From this . So we have . For every , Thus, , and so for some . Hence The other case can be treated similarly.

*Claim*

*3.*is almost additive map; that is, for every , .

We divided the proof into several steps.

*Step*

*For every , and , , .*

*1.*Let , and let . For every ; therefore This yields . From Lemma 2, for .

Since and for all . We have that

Then It shows that . The rest of the other hand is also the same reason.

*Step*

*is additive on and .*

*2.*For . Because , by Step 1, we know that The rest is similar.

*Step*

*For any .*

*3.*Let , and . For Step 2, we know that So . From Lemma 2, On the other hand, from (27) we get that . So

*Step*

*For any .*

*4.*Let . For any . From Step 2, it is that So . From Lemma 2, On the other hand, from (27) we get that . So The other case can be treated similarly.

*Step*

*Consider that .*

*5.*Let , and let . For any . By Step 3, it is that So . From Lemma 2, it follows that for some central element . From (27) we get that So Similarly, .

*Step*

*For all , .*

*6.*Let , and write . If we know that then it is right.

For every . By Step 5, it shows that So Because , we clearly have Similarly, , and we can obtain From Lemma 3, For every , from Step 6, we define a mapping of into by Then let .

In the following, we will prove that and are desired maps.

*Claim*is an additive derivation.

*4.*We divide the proof into the following steps.

*Step*is an additive map.

*1.*We only need to show that is an additive on . Let , Thus and so .

*Step*

*For every , , , and , , .*

*2*.Let , So On the other hand, So . Note that is prime, so

*Step*

*is an additive derivation on .*

*3.*Let and let be in . By Steps 1 and 2, we have From , we have So , and so is an additive derivation on . So from [15, Theorem 4.1] we can see that is a derivation. Furthermore, due to the main result in [6], there exists an operator such that for all .

Similarly, we can know that is an additive derivation on , so there exists an operator such that for all .

*Step*

*Consider that , for every , .*

*4.*Now, we write and . Then by Step 3, we have that For all , for all , this shows that Multiplying the previous two equations by and from the left respectively, we can know that Similarly, So for some .

For every and , using (55), (56), and (61), we get that Since at least one of or is of rank greater than 2, and exist such that or We suppose that and . From (62), we get From and , then . So for any and . So Let , and write and , so

*Claim*

*sends the commutators into zero: The proof is complete.*

*7*.#### Acknowledgments

The authors would like to thank his tutor Professor Baodong Zheng for his valuable suggestions and encouragement. This work was supported by Mudanjiang city science and technology plan Projects (G2013n1446) and Mudanjiang Normal Colllege Youth academic backbone Projects (G201209).

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#### Copyright

Copyright © 2013 Donghua Huo et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.