Abstract

Let be a ring having unit 1. Denote by the center of . Assume that the characteristic of is not 2 and there is an idempotent element such that . It is shown that, under some mild conditions, a map is a multiplicative Lie triple derivation if and only if for all , where is an additive derivation and is a map satisfying for all . As applications, all Lie (triple) derivations on prime rings and von Neumann algebras are characterized, which generalize some known results.

1. Introduction

Let be an associative ring with the center . For any element , we set . Recall that a map is a multiplicative derivation or nonlinear derivation if , for all , is a multiplicative Lie derivation, if , for all , and is a multiplicative Lie triple derivation, if , for all . Particularly, if is additive (linear), then above maps are, respectively, additive (linear) derivations, additive (linear) Lie derivations, and additive (linear) Lie triple derivations. We often omit “linear” for “linear derivations.”

The structure of additive (linear) derivations and additive (linear) Lie (triple) derivations on rings or algebras has been studied by many authors. Bre ar in [1] proved that every additive Lie derivation on a prime ring with characteristic not 2 can be decomposed as , where is an additive derivation from into its central closure and is an additive map of into the extended centroid sending commutators to zero. Mathieu and Villena [2] showed that every linear Lie derivation on a -algebra is standard, that is, can be decomposed as the form , where is a derivation and is a central valued linear map vanishing at each commutator. In [3] Qi and Hou proved that the same is true for additive Lie derivations of nest algebras on Banach spaces. Miers [4] showed that every linear Lie triple derivation on , a von Neumann algebra with no central summands of type , is of the form , where is a derivation and is a central valued linear map vanishing at every Lie triple products . Recently, Wang and Lu [5] described the structure of linear Lie triple derivations on -subspace lattice algebras. For other results, see [610] and the references therein.

For the study of multiplicative derivations and multiplicative Lie (triple) derivations, Daif [11] initially proved that each multiplicative derivation on a 2-torsion free prime ring containing a nontrivial idempotent is additive. Yu and Zhang [12] showed that every multiplicative Lie derivation on triangular algebras is the sum of an additive derivation and a map into its center sending commutators to zero, and, later, Ji et al. [13] generalized this result to the case of multiplicative Lie triple derivations. Assume that is a nontrivial nest on a Banach space over the complex field which contains a nontrivial complemented element, and Alg is the associated nest algebra. Li and Fang in [14] obtained the same result as the above for multiplicative Lie triple derivations on Alg .

The purpose of the present paper is to consider the problem of characterizing nonlinear Lie triple derivations on general rings.

Let be a ring having unit and an idempotent element , and let denote the center of . Assume that the characteristic of is not 2 and satisfies that and . Let be a multiplicative Lie triple derivation. We show that if and do not contain nonzero central ideals, then , for all , where is a central element depending on and (Theorem 1); furthermore, if also satisfies that, for , , and , then , for all , where is an additive derivation and is a map satisfying , for all (Theorem 2). As applications, some characterizations of multiplicative (additive) Lie (triple) derivations on prime rings and von Neumann algebras are obtained, respectively (Corollaries 37).

2. Main Results and Corollaries

The following are our main results in this paper.

Theorem 1. Let be a ring having unit 1 and an idempotent element . Assume that the characteristic of is not 2 and satisfies the following two conditions:(i) and ;(ii) and do not contain nonzero central ideals.
Assume that is a multiplicative Lie triple derivation. Then ,  for all , where is a central element depending on and .

Moreover, if the ring in Theorem 1 also satisfies that, for , , and , then has more concrete form.

Theorem 2. Let be a ring having unit and an idempotent element . Assume that the characteristic of is not 2 and satisfies the following two conditions:(i) and ;(ii) and do not contain nonzero central ideals;(iii)for , , and .
Then a map is a multiplicative Lie triple derivation if and only if , for all , where is an additive derivation and is a map satisfying , for all .

Recall that a ring is prime if, for any , implies or .

Applying Theorem 2 to prime rings, we have the following result.

Corollary 3. Let be a prime ring having unit and a nontrivial idempotent, and let be a map. If the characteristic of is not 2 and , are noncommutative, then the following two statements are equivalent.(1) is a multiplicative Lie triple derivation.(2)There exist an additive derivation and a map satisfying for all such that for all .

Proof. Let be a nontrivial idempotent. It is obvious that satisfies the condition (i) in Theorem 2.
Claim. If is a central ideal of a noncommutative prime ring , then .
Take any . Since is central, we have for all . Thus, for any , one gets and so Since is noncommutative, there exist two elements such that . It follows from the primeness of that . The claim holds.
Since is prime, both and are prime. Thus, by the above claim, the condition (ii) in Theorem 2 is satisfied.
Now, for any fixed , define two maps and , respectively, by for all . It is clear that both and are derivations. Posner in [9, Theorem 2] proved that, if is a derivation of a noncommutative prime such that, for all , is in the center of , then is the zero derivation. Thus, by [9, Theorem 2], for , if is noncommutative and , then ; that is, . Similarly, if is noncommutative and , then . Hence the condition (iii) in Theorem 2 is also satisfied.
Now, by Theorem 2, the corollary is true.

Let be the algebra of all bounded linear operators acting on a complex Hilbert space . Recall that a von Neumann algebra is a subalgebra of some satisfying , where ([15]). For , the central carrier of , denoted by , is the intersection of all central projections such that . If is self-adjoint, then the core of , denoted by , is sup . Particularly, if is a projection, it is clear that is the largest central projection . A projection is core-free if . It is easy to see that if and only if ([16]).

Applying Theorem 2 to von Neumann algebras, we have the following corollary.

Corollary 4. Let be a von Neumann algebra without central summands of type and a map. Then the following statements are equivalent.(1) is a multiplicative Lie triple derivation.(2)There exist an additive derivation and a map vanishing at each Lie triple product such that , for all .

Proof. Assume that is a von Neumann algebra without central summands of type . Then, by [16], there exists a nonzero core-free projection with . Fix such and note that . It follows from the definition of the central carrier that both span and span are dense in . So and .
Bre ar and Miers [17] proved that if such that , then . This implies that has no nonzero central ideals. Note that and are also von Neumann algebras without central summands of type . So both and have no nonzero central ideals.
Finally, for , if and , by the Kleinecke-Shirokov theorem ([18]), both and are central quasinilpotent for all . Hence for all ; that is, and .
Thus, if has no central summands of type , by what the above stated, satisfies the corresponding assumptions (i)–(iii) in Theorem 2. By Theorem 2, the corollary is true.

Note that a multiplicative Lie derivation must be a multiplicative Lie triple derivation. So the following corollary is immediate.

Corollary 5. Let be a von Neumann algebra without central summands of type and a map. Then the following statements are equivalent.(1) is a multiplicative Lie derivation.(2)There exist an additive derivation and a map vanishing at each commutator such that , for all .

If in Corollaries 4 and 5 is additive, more can be said. In fact, a complete characterization of additive Lie (triple) derivations on any von Neumann algebras can be obtained, which is a slight generalization of the corresponding result in [4].

Corollary 6. Let be a von Neumann algebra and an additive map. Then the following statements are equivalent.(1) is a Lie triple derivation.(2)There exist an additive derivation and an additive map vanishing at each Lie triple product such that , for all .

Proof. Clearly, one only needs to check (1) (2). In the following assume that is an additive Lie triple derivation.
Take the central projection , so that, with respect to the space decomposition , , where is of type and has no central summands of type . Note that we may have . Then can be decomposed as
Claim. .
For any and any , we have , and so for all . Hence is central quasinilpotent, which implies for all . It follows that . The claim holds.
Since is a central projection and is of type , we have By the above claim, , for all , and so
Define a map by Then, for any , we have It is easy to prove that is an additive Lie triple derivation on . So, by Corollary 4, there exist an additive derivation and an additive map vanishing at each Lie triple product such that Let and . By (4)–(9), one gets that is an additive derivation on and is an additive map satisfying for all such that holds for all . Hence (2) is true and the proof is finished.

Particularly, for additive Lie derivations, we have the following corollary.

Corollary 7. Let be a von Neumann algebra and an additive map. Then the following statements are equivalent.(1) is a Lie derivation.(2)There exist an additive derivation and an additive map vanishing at each commutator such that for all .

3. The Proof of Main Results

In this section, we will give proofs of our main results, Theorems 1 and 2.

In the sequel, assume that is a unital ring and containing an idempotent satisfying and . It is clear that . Write and . Then can be written as , where ( ).

We first give several lemmas, which are needed to prove the main results.

Lemma 8 (see [3, Lemma 3.1]). The center of is

By Lemma 8, it is easily seen that if , then and .

Lemma 9. Let . If and hold for all and all , then .

Proof. Write . Since for all , we have Multiplying and from the left and right in the equation, one gets . Multiplying from both sides in the equation, one has ; that is, for all . This implies .
Similarly, from the relation for all , one can prove and . It follows from Lemma 8 that .

Lemma 10. Let . If   holds for all ( ), then .

Proof. For , since , we have for all . It follows from Lemma 8 that .

Lemma 11. For , if ( ), then .

Proof. For , assume that . So for all , where . It follows from the assumption on that .

Lemma 12 (see [19, Lemma 4]). Let be a ring having unit 1 and an idempotent element . Assume that satisfies that , which implies that . For , if for all , then there exists an element such that .

Applying Lemma 12 to our ring , we get that, for any , if holds for all , then there exists such that , .

Now, we are in the position to give the proofs of Theorems 1 and 2.

Proof of Theorem 1. We will prove the theorem by a series of claims.
Claim 1. .
By the definition of , we have
Claim 2. ,   .
Let . For any and , we have Multiplying by and from the left and the right in the above equation, respectively, one gets for all , which implies since the characteristic of is not 2.
Similarly, for any , by the relation , one can check that Applying Lemma 8 to (14) and (15), one obtains for . The claim holds.
Now let for all . Then is also a multiplicative Lie triple derivation and satisfies as Claim 2. Thus, without loss of generality, we may assume
Claim 3. .
By Claim 2, . So we only need to check . In fact, by (16), we have which implies . Hence .
Claim 4. , .
Here, we only give the proof for . The proof for the other inclusion is similar.
For any , by (16), (13) can be reduced to Taking any and any , one obtains which implies . Note that So, by (18), one achieves It follows from char   that for all . By Lemma 8, one can get for all . It is easily checked that is an ideal of . So, by the assumption (ii), , and so by the condition (i). This and (18) imply that , as desired.
Claim 5. For any , we have , .
By (16) and Claims 3 and 4, it is clear that
Claim 6. For any , we have , .
Let and . For any , by (16) and Claims 3 and 4, one has that is, . Note that by Claim 4. So Similarly, for any , by the relations and , one can check Combining (24) and (25) and by Lemma 10, we achieve Thus, to complete the proof of the claim, we still need to check . In fact, by (16), we have It follows that , which implies .
Claim 7. is additive on , .
Take any ( ). By (16) and Claims 3–6, one obtains
Claim 8. For , we have , , and .
Let . For any and , by (16) and Claim 3, one has A simple calculation yields , , and . So the claim is true.
Claim 9. For any and , we have .
Take any and write . Note that, by Claim 8, one has . So, to prove the claim, one only needs to check .
To do this, taking any , by Claims and 7, we have Combining the above two equations yields
Similar to the above discussion, one can also prove that Now, by Lemma 10, it follows that , and the claim holds.
Claim 10. For any , any , and any , we have(i) ;(ii) .
For any and any , since , by Claim 4, one can easily get ; that is, Write . Then (33) implies that Letting in (33), a simple calculation yields So (34) is reduced to , which and Lemma 8 imply that holds for all . Note that is an ideal of . It follows from the assumption (ii) that , and so On the other hand, by using the equation , one can also show Combining (35)–(37) and Lemma 8, we get that (i) holds.
The proof of (ii) is similar and we omit it here.
Claim 11. For any ( ), we have(i) and ;(ii) and .
Here, we only give the proof of (i). The proof of (ii) is similar.
For any ( ), by Claim 3, one gets which, together with Claim 4, implies and , as desired.
Claim 12. For any ( ), we have(i) ;(ii) .
Still, we only give the proof of (i).
Take any , , and . Firstly, by Claims 8, 9, and 11, there exists some such that
Next, for any , by Claims 3 and 4, one has Comparing the above equations, we get
Finally, for any , a similar argument to the above achieves
Now, combining (39)–(42) and Lemma 10, the claim holds.
Claim 13. For any , we have .
Take any and write For any and any , by Claims 4, 10, and 12, one has Now the above two equations and Lemma 9 imply .
Claim 14. For any , we have ,   .
Let and be arbitrary. For any , by (16) and Claims 3 and 4, we have It follows that
Similarly, one can check
Now, together with Lemma 10 and Claim 8, (46)-(47) imply that the claim holds.
Claim 15. for all . Therefore, Theorem 1 holds.
In fact, by Claim 7 and Claims 13 and 14, it is easily seen that the claim is true.
The proof of the theorem is complete.

Proof of Theorem 2. The “if” part is clear. We will prove the “only if” part by several claims.
Claim 1. For any , there exists a map such that , .
We only give the proof for here. The proof for any is similar.
For any , by Claim 8 in the proof of Theorem 1, we have Now, taking any and any , one has which implies . It follows that . By the assumption (iii) in the theorem, one obtains for all ; that is, for all . It follows from Lemma 12 that for some . Hence The claim holds.
Now define two maps and , respectively, by for all . Then, by Claim 4 in the proof of Theorem 1 and Claim 1, we have that Moreover, for .
Claim 2. is additive on .
By Claim 7 in the proof of Theorem 1, is additive on and .
Let . For any , by Claim 14 in the proof of Theorem 1, (52), and the definition of , we have It follows from Lemma 11 that ; that is, is additive on , .
Now, for any , by what the above proved, one gets Hence is additive on .
Claim 3. is a derivation.
We will divide the proof of the claim into five steps.
Step 1. For any and any , we have , .
For any and ( ), by the definition of and (33), one obtains By a similar argument to that of Step 1, one can show the following.
Step 2. For any and any , we have , .
Step 3. For any , we have , .
Take any and any ( ). By Step 1, one has The above two equations yield ; that is, for all . Note that . It follows that .
Step 4. For any and any , we have , .
For any and any ( ), noting that , by (52), Claim 2, and the definitions of and , we have Multiplying from both sides in the above equation, and noting that (52), one obtains Define a set It is easily seen that . Also note that, for any , Thus, is a central ideal of . It follows from the assumption (ii) that . So for all and , .
Step 5. For any , we have ; that is, is a derivation.
By Claim 2 and Steps 1–4, it is easily checked that the step is true.
Claim 4. satisfies for all .
In fact, for any , by the definition of and Claim 3, we have The proof of the theorem is complete.

Conflict of Interests

The author declares that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

The author wishes to express her thanks to the referees for their helpful comments and suggestions. This work is partially supported by National Natural Science Foundation of China (11101250) and Youth Foundation of Shanxi Province (2012021004).