Characterizations of Nonlinear Lie Derivations of 𝐵(𝑋)

Abstract

Let 𝑋 be an infinite dimensional Banach space, and Φ∶𝐵(𝑋)→𝐵(𝑋) is a nonlinear Lie derivation. Then Φ is the form 𝛿+𝜏 where 𝛿 is an additive derivation of 𝐵(𝑋) and 𝜏 is a map from 𝐵(𝑋) into its center 𝑍_𝐵(𝑋), 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 𝑋 be a Banach space and let 𝐵(𝑋) be the algebra of all bounded linear operators on 𝑋. Some progress has been made towards understanding the Lie derivation recently. In [11], it was shown that letting 𝑋 be a Banach space of dimension greater than 2, if 𝛿∶𝐵(𝑋)→𝐵(𝑋) is a linear map satisfying 𝛿([𝐴,𝐵])=[𝛿(𝐴),𝐵]+[𝐴,𝛿(𝐵)] for any 𝐴,𝐵∈𝐵(𝑋) with 𝐴𝐵=0, then 𝛿=𝑑+𝜏, where 𝑑 is a derivation of 𝐵(𝑋) and 𝜏∶𝐵(𝑋)→𝐶𝐼 is a linear map vanishing at commutators [𝐴,𝐵] with 𝐴𝐵=0. 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 𝒯 be a triangular algebra over a commutative ring 𝑅, if 𝛿∶𝒯→𝒯 is an 𝑅-linear map satisfying 𝛿([𝑥,𝑦])=[𝛿(𝑥),𝑦]+[𝑥,𝛿(𝑦)] for any 𝑥,𝑦∈𝒯 with 𝑥𝑦=0, then 𝛿=𝑑+𝜏, where 𝑑 is a derivation of 𝒯 and 𝜏∶𝒯→𝑍(𝒯) is an 𝑅-linear map vanishing at commutators [𝑥,𝑦] with 𝑥𝑦=0. Bai and Du [14] proved that letting ℳ be a von Neumann algebra with no central summands of type 𝐼₁, if Φ∶ℳ→ℳ is a nonlinear Lie derivation, then Φ is of the form 𝜎+𝜏, where 𝜎 is an additive derivation of ℳ and 𝜏 is a mapping of ℳ into its center 𝑍_ℳ which maps commutators into zero.

Nevertheless, it would seem desirable to investigate nonlinear Lie derivations of 𝐵(𝑋). The present investigation attempts to do this by using the properties of nonlinear Lie derivation and previous research. A map 𝛿∶𝐵(𝑋)→𝐵(𝑋) is called an additive derivation if it is additive and satisfies 𝛿(𝐴𝐵)=𝛿(𝐴)𝐵+𝐴𝛿(𝐵) for any 𝐴,𝐵∈𝐵(𝑋). A linear map 𝛿∶𝐵(𝑋)→𝐵(𝑋) is called a Lie derivation if 𝛿([𝐴𝐵])=[𝛿(𝐴),𝐵]+[𝐴,𝛿(𝐵)] for any 𝐴,𝐵∈𝐵(𝑋), where [𝐴,𝐵]=𝐴𝐵−𝐵𝐴 is the usual Lie product. More generally, a map Φ∶𝐵(𝑋)→𝐵(𝑋) (without the additive assumption) is a nonlinear Lie derivation if Φ([𝐴𝐵])=[Φ(𝐴),𝐵]+[𝐴,Φ(𝐵)] for any 𝐴,𝐵∈𝐵(𝑋).

Recall that 𝐵(𝑋) is called prime algebra if 𝑎𝐵(𝑋)𝑏=0 implies 𝑎=0 or =0. An operator 𝑃∈𝐵(𝑋) is an idempotent provided that 𝑃²=𝑃. In this paper, we suppose that 𝐵(𝑋) is a prime associate algebra. We prove that every nonlinear Lie derivation of 𝐵(𝑋) is the sum of an additive derivation and a map from 𝐵(𝑋) 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 𝑃₁∈𝐵(𝑋) and let 𝑃₂=𝐼−𝑃₁. In what follows, we write 𝒜_𝑖𝑗=𝑃_𝑖𝐵(𝑋)𝑃_𝑗 for, 𝑗=1,2. Then every operator 𝐴∈𝐵(𝑋) can be written as 𝐴=∑²_𝑖,𝑗=1𝐴_𝑖𝑗. Note that 𝐴_𝑖𝑗 notation denotes an arbitrary element of 𝒜_𝑖𝑗.

Definition 1. Letting 𝑆 be a set of 𝐵(𝑋), we call 𝑍_𝑆={𝐴∈𝐵(𝑋)∣[𝐴,𝐵]=0,∀𝐵∈𝑆}(1) the centralizer of 𝑆 about 𝐵(𝑋).
Clearly, 𝑍_𝐵(𝑋) is the center of 𝐵(𝑋).

Lemma 2. Consider that 𝑍_𝒜𝑖𝑗=𝒜_𝑖𝑗+𝑍_𝐵(𝑋),1≤𝑖≠𝑗≤2.

Proof. It is clearly that 𝒜_𝑖𝑗+𝑍_𝐵(𝑋)⊆𝑍_𝒜𝑖𝑗,1≤𝑖≠𝑗≤2.
Conversely, without loss of generality, we proof that 𝑍_𝒜12⊆𝒜₁₂+𝑍_𝐵(𝑋). For any 𝑇=Σ²_𝑖,𝑗=1𝑇_𝑖𝑗∈𝑍_𝒜12, we have 𝑇₁₁𝐴₁₂+𝑇₂₁𝐴₁₂=𝐴₁₂𝑇₂₂+𝐴₁₂𝑇₂₁,∀𝐴₁₂∈𝒜₁₂.(2) Thus 𝑇₂₁𝐴₁₂=𝑇₂₁𝑃₁𝐴𝑃₂=0,∀𝐴∈𝐵(𝑋),(3)𝑇₁₁𝐴₁₂=𝐴₁₂𝑇₂₂,∀𝐴₁₂∈𝒜₁₂.(4) From (3), noting that 𝐵(𝑋) is prime, we obtain 𝑇₂₁𝑃₁=0, that is, 𝑇₂₁=0. By 𝐴₁₁𝐴₁₂∈𝒜₁₂, we have 𝑇₁₁𝐴₁₁𝐴₁₂=𝐴₁₁𝐴₁₂𝑇₂₂=𝐴₁₁𝑇₁₁𝐴₁₂,𝑇₁₁𝐴₁₁−𝐴₁₁𝑇₁₁𝑃₁𝐴𝑃₂=0,∀𝐴∈𝐵(𝑋).(5) Since 𝐵(𝑋) is prime, we have 𝑇₁₁𝐴₁₁=𝐴₁₁𝑇₁₁. We write 𝑇₁₁=𝑍₁𝑃₁, where 𝑍₁∈𝑍_𝐵(𝑋). Since 𝐴₁₂𝐴₂₂∈𝒜₁₂, we have 𝑇₁₁𝐴₁₂𝐴₂₂=𝐴₁₂𝐴₂₂𝑇₂₂=𝐴₁₂𝑇₂₂𝐴₂₂,𝑃₁𝐴𝑃₂𝐴₂₂𝑇₂₂−𝑇₂₂𝐴₂₂=0,∀𝐴∈𝐵(𝑋).(6) Since 𝐵(𝑋) is prime we have 𝑇₂₂𝐴₂₂=𝐴₂₂𝑇₂₂. We write 𝑇₂₂=𝑍₂𝑃₂, where 𝑍₂∈𝑍_𝐵(𝑋). From (4) we have 𝑍₁−𝑍₂𝑃₁𝐴𝑃₂=𝑍₁𝑃₁𝐴𝑃₂−𝑃₁𝐴𝑃₂𝑍₂𝑃₂=𝑇₁₁𝑃₁𝐴𝑃₂−𝑃₁𝐴𝑃₂𝑇₂₂=0.(7) So 𝑍₁−𝑍₂𝑃₁=0.(8) Thus 𝑍₁−𝑍₂𝐴𝑃₁=𝐴𝑍₁−𝑍₂𝑃₁=0,∀𝐴∈𝐵(𝑋).(9) So 𝑍₁=𝑍₂.(10) It follows that 𝑇=𝑇₁₂+𝑍₁𝑃₁+𝑍₂𝑃₂=𝑇₁₂+𝑍₁∈𝒜₁₂+𝑍_𝐵(𝑋).(11)

Lemma 3. Consider that 𝑍_𝐵(𝑋)=𝑍_𝒜12∩𝑍_𝒜21.

Proof. From Lemma 2, we have 𝑍_𝒜12∩𝑍_𝒜21=𝒜₁₂+𝑍_𝐵(𝑋)∩𝒜₂₁+𝑍_𝐵(𝑋)=𝑍_𝐵(𝑋).(12)

Theorem 4. Let 𝑋 be an infinite dimensional Banach space, and Φ∶𝐵(𝑋)→𝐵(𝑋) is a nonlinear Lie derivation. Then Φ is the form 𝛿+𝜏, where 𝛿 is an additive derivation of 𝐵(𝑋) and 𝜏 is a map from 𝐵(𝑋) into its center 𝑍_𝐵(𝑋), which maps commutators into the zero.

Proof. Now we organize the proof in a series of claims.
Claim  1. Consider that Φ(0)=0; moreover, 𝑇_𝑖∈𝐵(𝑋) such that Φ(𝑃_𝑖)−[𝑃_𝑖,𝑇_𝑖]∈𝑍_𝐵(𝑋) and Φ(𝐴_𝑖𝑗)=𝑃_𝑖Φ(𝐴_𝑖𝑗)𝑃_𝑗+[𝐴_𝑖𝑗,𝑇_𝑖], Φ(𝐴_𝑗𝑖)=𝑃_𝑗Φ(𝐴_𝑗𝑖)𝑃_𝑖+[𝐴_𝑗𝑖,𝑇_𝑖] for arbitrarily 𝐴_𝑖𝑗∈𝒜_𝑖𝑗(1≤𝑖≠𝑗≤2).
Obviously Φ(0)=Φ([0,0])=[Φ(0),0]+[0,Φ(0)]=0.
Moreover, for arbitrarily 𝐴_𝑖𝑗, Φ𝐴_𝑖𝑗=Φ𝑃_𝑖,𝐴_𝑖𝑗=Φ𝑃_𝑖,𝐴_𝑖𝑗+𝑃_𝑖,Φ𝐴_𝑖𝑗=Φ𝑃_𝑖𝐴_𝑖𝑗−𝐴_𝑖𝑗Φ𝑃_𝑖+𝑃_𝑖Φ𝐴_𝑖𝑗−Φ𝐴_𝑖𝑗𝑃_𝑖.(13) In (13), the left and right sides are, respectively, multiplied by 𝑃_𝑖 and 𝑃_𝑗, and we have 𝑃_𝑖Φ𝑃_𝑖𝑃_𝑖𝐴_𝑖𝑗=𝐴_𝑖𝑗𝑃_𝑗Φ𝑃_𝑖𝑃_𝑗.(14) In (13), the left and right sides are respectively multiplying by 𝑃_𝑗 and 𝑃_𝑖, and we arrive at 𝑃_𝑗Φ𝐴_𝑖𝑗𝑃_𝑖=0.(15) For arbitrarily 𝐴_𝑗𝑖, Φ𝐴_𝑗𝑖=Φ𝐴_𝑗𝑖,𝑃_𝑖=Φ𝐴_𝑗𝑖,𝑃_𝑖+𝐴_𝑗𝑖,Φ𝑃_𝑖=Φ𝐴_𝑗𝑖𝑃_𝑖−𝑃_𝑖Φ𝐴_𝑗𝑖+𝐴_𝑗𝑖Φ𝑃_𝑖−Φ𝑃_𝑖𝐴_𝑗𝑖.(16) In the previous equation, the left and right sides are, respectively multiplied by 𝑃_𝑗 and 𝑃_𝑖, and we have 𝐴_𝑗𝑖𝑃_𝑖Φ𝑃_𝑖𝑃_𝑖=𝑃_𝑗Φ𝑃_𝑖𝑃_𝑗𝐴_𝑗𝑖.(17) This together with (14) shows that 𝑃_𝑖Φ𝑃_𝑖𝑃_𝑖+𝑃_𝑗Φ𝑃_𝑖𝑃_𝑗,𝐴_𝑖𝑗=𝑃_𝑖Φ𝑃_𝑖𝑃_𝑖+𝑃_𝑗Φ𝑃_𝑖𝑃_𝑗,𝐴_𝑗𝑖=0.(18) From Lemma 3, it is that 𝑃_𝑖Φ𝑃_𝑖𝑃_𝑖+𝑃_𝑗Φ𝑃_𝑖𝑃_𝑗∈𝑍_𝐵(𝑋).(19) Denote 𝑇_𝑖=𝑃_𝑖Φ(𝑃_𝑖)𝑃_𝑗−𝑃_𝑗Φ(𝑃_𝑖)𝑃_𝑖. Therefore Φ𝑃_𝑖−𝑃_𝑖,𝑇_𝑖=𝑃_𝑖Φ𝑃_𝑖𝑃_𝑖+𝑃_𝑗Φ𝑃_𝑖𝑃_𝑗∈𝑍_𝐵(𝑋).(20) At the same time, [𝐴_𝑖𝑗,𝑇_𝑖]=𝑃_𝑗Φ(𝑃_𝑖)𝑃_𝑖𝐴_𝑖𝑗−𝐴_𝑖𝑗𝑃_𝑗Φ(𝑃_𝑖)𝑃_𝑖. From (14), 𝐴_𝑖𝑗,𝑇_𝑖=𝑃_𝑗Φ𝑃_𝑖𝑃_𝑖𝐴_𝑖𝑗−𝐴_𝑖𝑗𝑃_𝑗Φ𝑃_𝑖𝑃_𝑖+𝑃_𝑖Φ𝑃_𝑖𝑃_𝑖𝐴_𝑖𝑗−𝐴_𝑖𝑗𝑃_𝑗Φ𝑃_𝑖𝑃_𝑗=Φ𝑃_𝑖𝐴_𝑖𝑗−𝐴_𝑖𝑗Φ𝑃_𝑖.(21) Because 𝑃_𝑗Φ(𝐴_𝑖𝑗)𝑃_𝑖=0 so by (13) we get Φ(𝐴_𝑖𝑗)=𝑃_𝑖Φ(𝐴_𝑖𝑗)𝑃_𝑗+[𝐴_𝑖𝑗,𝑇_𝑖].
In the same way, we can get Φ(𝐴_𝑗𝑖)=𝑃_𝑗Φ(𝐴_𝑗𝑖)𝑃_𝑖+[𝐴_𝑗𝑖,𝑇_𝑖].
Remark. It is clear that 𝑇→[𝑇,𝑇_𝑖] is an additive derivation. Without loss of generality, we can assume that Φ(𝑃₁)∈𝑍_𝐵(𝑋) and Φ(𝒜_𝑖𝑗)⊆𝒜_𝑖𝑗(1≤𝑖≠𝑗≤2).
Claim  2. Consider that Φ(𝒜_𝑖𝑖)⊆𝒜_𝑖𝑖+𝑍_𝐵(𝑋)(𝑖=1,2).
Let 𝑖=1, for every 𝑇₁₁∈𝒜₁₁, and write Φ(𝑇₁₁)=∑²_𝑖,𝑗=1𝐵_𝑖𝑗. From Φ(𝑃₁)∈𝑍_𝐵(𝑋) and [𝑃₁,𝐴]=[𝐴,𝑃₂], we can get Φ(𝑃₂)∈𝑍_𝐵(𝑋)∶0=Φ𝑇₁₁,𝑃₂=Φ𝑇₁₁,𝑃₂+𝑇₁₁,Φ𝑃₂=Φ𝑇₁₁𝑃₂−𝑃₂Φ𝑇₁₁.(22) From this 𝑃₁Φ(𝑇₁₁)𝑃₂=𝑃₂Φ(𝑇₁₁)𝑃₁=0. So we have 𝐵₁₂=𝐵₂₁=0. For every 𝐴₂₂∈𝒜₂₂, 0=Φ𝑇₁₁,𝐴₂₂=Φ𝑇₁₁,𝐴₂₂+𝑇₁₁,Φ𝐴₂₂=𝐵₂₂𝐴₂₂−𝐴₂₂𝐵₂₂+𝑇₁₁Φ𝐴₂₂−Φ𝐴₂₂𝑇₁₁.(23) Thus, 𝐵₂₂𝐴₂₂−𝐴₂₂𝐵₂₂=0, and so 𝐵₂₂=𝑃₂𝑍 for some ∈𝑍_𝐵(𝑋). Hence Φ𝑇₁₁=𝐵₁₁−𝑃₁𝑍+𝑍∈𝒜₁₁+𝑍_𝐵(𝑋).(24) The other case can be treated similarly.
Claim  3.Φ is almost additive map; that is, for every 𝑇,𝑆∈𝐵(𝑋), Φ(𝑇+𝑆)−Φ(𝑇)−Φ(𝑆)∈𝑍_𝐵(𝑋).
We divided the proof into several steps.
Step  1.   For every 𝑇_𝑖𝑖,𝑇_𝑗𝑖, and 𝑇_𝑖𝑗(1≤𝑖≠𝑗≤2), Φ(𝑇_𝑖𝑖+𝑇_𝑖𝑗)−Φ(𝑇_𝑖𝑖)−Φ(𝑇_𝑖𝑗)∈𝑍_𝐵(𝑋), Φ(𝑇_𝑖𝑖+𝑇_𝑗𝑖)−Φ(𝑇_𝑖𝑖)−Φ(𝑇_𝑗𝑖)∈𝑍_𝐵(𝑋).
Let =1, and let 𝑗=2. For every 𝐴₁₂∈𝒜₁₂,[𝑇₁₁+𝑇₁₂,𝐴₁₂]=[𝑇₁₁,𝐴₁₂]; therefore Φ𝑇₁₁+𝑇₁₂,𝐴₁₂+𝑇₁₁+𝑇₁₂,Φ𝐴₁₂=Φ𝑇₁₁,𝐴₁₂+𝑇₁₁,Φ𝐴₁₂.(25) This yields [Φ(𝑇₁₁+𝑇₁₂)−Φ(𝑇₁₁),𝐴₁₂]=0. From Lemma 2, Φ𝑇₁₁+𝑇₁₂−Φ𝑇₁₁=𝑃₁Φ𝑇₁₁+𝑇₁₂−Φ𝑇₁₁𝑃₂+𝑍(26) for ∈𝑍_𝐵(𝑋).
Since Φ(𝑃₁)∈𝑍_𝐵(𝑋) and [𝑃₁,𝑇]=𝑃₁𝑇𝑃₂ for all 𝑇∈𝒜₁₁⊕𝒜₁₂⊕𝒜₂₂. We have that Φ𝑃₁𝑇𝑃₂=Φ𝑃₁=Φ𝑃,𝑇₁,𝑇+𝑃₁,Φ(𝑇)=𝑃₁Φ(𝑇)𝑃₂.(27)
Then 𝑃₁Φ𝑇₁₁+𝑇₁₂−Φ𝑇₁₁𝑃₂=Φ𝑃₁𝑇₁₁+𝑇₁₂𝑃₂−Φ𝑃₁𝑇₁₁𝑃₂=Φ𝑃₁𝑇₁₁𝑃₂+𝑃₁𝑇₁₂𝑃₂−Φ𝑃₁𝑇₁₁𝑃₂=Φ𝑃₁𝑇₁₂𝑃₂=Φ𝑇₁₂.(28) It shows that Φ(𝑇₁₁+𝑇₁₂)−Φ(𝑇₁₁)=Φ(𝑇₁₂)+𝑍. The rest of the other hand is also the same reason.
Step  2. Φ is additive on 𝒜₁₂ and 𝒜₂₁.
For 𝑇₁₂,𝑆₁₂∈𝒜₁₂. Because 𝑇₁₂+𝑆₁₂=[𝑃₁+𝑇₁₂,𝑃₂+𝑆₁₂], by Step  1, we know that Φ𝑇₁₂+𝑆₁₂=Φ𝑃₁+𝑇₁₂,𝑃₂+𝑆₁₂+𝑃₁+𝑇₁₂,Φ𝑃₂+𝑆₁₂=Φ𝑃₁+Φ𝑇₁₂,𝑃₂+𝑆₁₂+𝑃₁+𝑇₁₂,Φ𝑃₂+Φ𝑆₁₂=Φ𝑇₁₂+Φ𝑆₁₂.(29) The rest is similar.
Step  3. For any 𝑇₁₁∈𝒜₁₁,𝑇₂₂∈𝒜₂₂,Φ(𝑇₁₁+𝑇₂₂)−Φ(𝑇₁₁)−Φ(𝑇₂₂)∈𝑍_𝐵(𝑋).
Let 𝐴₁₂∈𝒜₁₂, and [𝑇₁₁+𝑇₂₂,𝐴₁₂]=𝑇₁₁𝐴₁₂−𝐴₁₂𝑇₂₂. For Step  2, we know that Φ𝑇₁₁+𝑇₂₂,𝐴₁₂+𝑇₁₁+𝑇₂₂,Φ𝐴₁₂=Φ𝑇₁₁𝐴₁₂+Φ−𝐴₁₂𝑇₂₂=Φ𝑇₁₁,𝐴₁₂+Φ𝑇₂₂,𝐴₁₂=Φ𝑇₁₁,𝐴₁₂+𝑇₁₁,Φ𝐴₁₂+Φ𝑇₂₂,𝐴₁₂+𝑇₂₂,Φ𝐴₁₂.(30) So [Φ(𝑇₁₁+𝑇₂₂)−Φ(𝑇₁₁)−Φ(𝑇₂₂),𝐴₁₂]=0. From Lemma 2, Φ𝑇₁₁+𝑇₂₂−Φ𝑇₁₁−Φ𝑇₂₂∈𝒜₁₂+𝑍_𝐵(𝑋).(31) On the other hand, from (27) we get that 𝑃₁(Φ(𝑇₁₁+𝑇₂₂)−Φ(𝑇₁₁)−Φ(𝑇₂₂))𝑃₂=0. So Φ𝑇₁₁+𝑇₂₂−Φ𝑇₁₁−Φ𝑇₂₂∈𝑍_𝐵(𝑋).(32)
Step  4.   For any 𝑇_𝑖𝑖,𝑆_𝑖𝑖∈𝒜_𝑖𝑖(𝑖=1,2),Φ(𝑇_𝑖𝑖+𝑆_𝑖𝑖)−Φ(𝑇_𝑖𝑖)−Φ(𝑆_𝑖𝑖)∈𝑍_𝐵(𝑋).
Let 𝑖=1. For any 𝐴₁₂∈𝒜₁₂,[𝑇₁₁+𝑆₁₁,𝐴₁₂]=𝑇₁₁𝐴₁₂+𝑆₁₁𝐴₁₂. From Step 2, it is that Φ𝑇₁₁+𝑆₁₁,𝐴₁₂+𝑇₁₁+𝑆₁₁,Φ𝐴₁₂=Φ𝑇₁₁𝐴₁₂+Φ𝑆₁₁𝐴₁₂=Φ𝑇₁₁,𝐴₁₂+Φ𝑆₁₁,𝐴₁₂=Φ𝑇₁₁,𝐴₁₂+𝑇₁₁,Φ𝐴₁₂+Φ𝑆₁₁,𝐴₁₂+𝑆₁₁,Φ𝐴₁₂.(33) So [Φ(𝑇₁₁+𝑆₁₁)−Φ(𝑇₁₁)−Φ(𝑆₁₁),𝐴₁₂]=0. From Lemma 2, Φ𝑇₁₁+𝑆₁₁−Φ𝑇₁₁−Φ𝑆₁₁∈𝒜₁₂+𝑍_𝐵(𝑋).(34) On the other hand, from (27) we get that 𝑃₁(Φ(𝑇₁₁+𝑆₁₁)−Φ(𝑇₁₁)−Φ(𝑆₁₁))𝑃₂=0. So Φ𝑇₁₁+𝑆₁₁−Φ𝑇₁₁−Φ𝑆₁₁∈𝑍_𝐵(𝑋).(35) The other case can be treated similarly.
Step  5. Consider that Φ(𝑇_𝑖𝑖+𝑇_𝑗𝑗+𝑇_𝑖𝑗)−Φ(𝑇_𝑖𝑖)−Φ(𝑇_𝑗𝑗)−Φ(𝑇_𝑖𝑗)∈𝑍_𝐵(𝑋).
Let 𝑖=1, and let 𝑗=2. For any 𝐴₁₂∈𝒜₁₂,[𝑇₁₁+𝑇₂₂+𝑇₁₂,𝐴₁₂]=[𝑇₁₁+𝑇₂₂,𝐴₁₂]. By Step 3, it is that Φ𝑇₁₁+𝑇₂₂+𝑇₁₂,𝐴₁₂+𝑇₁₁+𝑇₂₂+𝑇₁₂,Φ𝐴₁₂=Φ𝑇₁₁+Φ𝑇₂₂,𝐴₁₂+𝑇₁₁+𝑇₂₂,Φ𝐴₁₂.(36) So [Φ(𝑇₁₁+𝑇₂₂+𝑇₁₂)−Φ(𝑇₁₁)−Φ(𝑇₂₂),𝐴₁₂]=0. From Lemma 2, it follows that Φ𝑇₁₁+𝑇₂₂+𝑇₁₂−Φ𝑇₁₁−Φ𝑇₂₂=𝑃₁Φ𝑇₁₁+𝑇₂₂+𝑇₁₂−Φ𝑇₁₁−Φ𝑇₂₂𝑃₂+𝑍(37) for some central element 𝑍∈𝑍_𝐵(𝑋). From (27) we get that Φ𝑇₁₂=𝑃₁Φ𝑇₁₁+𝑇₂₂+𝑇₁₂𝑃₂.(38) So Φ𝑇₁₁+𝑇₂₂+𝑇₁₂−Φ𝑇₁₁−Φ𝑇₂₂−Φ𝑇₁₂∈𝑍_𝐵(𝑋).(39) Similarly, Φ(𝑇₂₂+𝑇₁₁+𝑇₂₁)−Φ(𝑇₂₂)−Φ(𝑇₁₁)−Φ(𝑇₂₁)∈𝑍_𝐵(𝑋).
Step  6. For all 𝑇,𝑆∈𝐵(𝑋), Φ(𝑇+𝑆)−Φ(𝑇)−Φ(𝑆)∈𝑍_𝐵(𝑋).
Let 𝑇∈𝐵(𝑋), and write 𝑇=𝑇₁₁+𝑇₁₂+𝑇₂₁+𝑇₂₂. If we know that Φ𝑇₁₁+𝑇₁₂+𝑇₂₁+𝑇₂₂−Φ𝑇₁₁−Φ𝑇₁₂−Φ𝑇₂₁−Φ𝑇₂₂∈𝑍_𝐵(𝑋),(40) then it is right.
For every 𝐴₁₂∈𝒜₁₂,[𝑇₁₁+𝑇₁₂+𝑇₂₁+𝑇₂₂,𝐴₁₂]=[𝑇₁₁+𝑇₂₁+𝑇₂₂,𝐴₁₂]. By Step 5, it shows that Φ𝑇₁₁+𝑇₁₂+𝑇₂₁+𝑇₂₂,𝐴₁₂+𝑇₁₁+𝑇₁₂+𝑇₂₁+𝑇₂₂,Φ𝐴₁₂=Φ𝑇₁₁+Φ𝑇₂₁+Φ𝑇₂₂,𝐴₁₂+𝑇₁₁+𝑇₂₁+𝑇₂₂,Φ𝐴₁₂.(41) So Φ𝑇₁₁+𝑇₁₂+𝑇₂₁+𝑇₂₂−Φ𝑇₁₁−Φ𝑇₂₁−Φ𝑇₂₂,𝐴₁₂=0.(42) Because Φ(𝑇₁₂)∈𝒜₁₂, we clearly have Φ𝑇₁₁+𝑇₁₂+𝑇₂₁+𝑇₂₂−Φ𝑇₁₁−Φ𝑇₁₂−Φ𝑇₂₁−Φ𝑇₂₂,𝐴₁₂=0.(43) Similarly, [𝑇₁₁+𝑇₁₂+𝑇₂₁+𝑇₂₂,𝐴₂₁]=[𝑇₁₁+𝑇₁₂+𝑇₂₂,𝐴₂₁], and we can obtain Φ𝑇₁₁+𝑇₁₂+𝑇₂₁+𝑇₂₂−Φ𝑇₁₁−Φ𝑇₁₂−Φ𝑇₂₁−Φ𝑇₂₂,𝐴₂₁=0.(44) From Lemma 3, Φ𝑇₁₁+𝑇₁₂+𝑇₂₁+𝑇₂₂−Φ𝑇₁₁−Φ𝑇₁₂−Φ𝑇₂₁−Φ𝑇₂₂∈𝑍_𝐵(𝑋).(45) For every 𝐴∈𝐵(𝑋), from Step 6, we define a mapping 𝜏 of 𝐵(𝑋) into 𝑍_𝐵(𝑋) by 𝜏(𝐴)=Φ(𝐴)−Φ𝑃₁𝐴𝑃₁−Φ𝑃₁𝐴𝑃₂−Φ𝑃₂𝐴𝑃₁−Φ𝑃₂𝐴𝑃₂.(46) Then let 𝛿(𝐴)=Φ(𝐴)−𝜏(𝐴).
In the following, we will prove that 𝛿 and 𝜏 are desired maps.
Claim 4. 𝛿 is an additive derivation.
We divide the proof into the following steps.
Step 1. 𝛿 is an additive map.
We only need to show that 𝛿 is an additive on 𝒜_𝑖𝑖. Let 𝐴_𝑖𝑖,𝐴^′_𝑖𝑖∈𝒜_𝑖𝑖, 𝛿𝐴_𝑖𝑖+𝐴^′_𝑖𝑖−𝛿𝐴_𝑖𝑖−𝛿𝐴^′_𝑖𝑖=Φ𝐴_𝑖𝑖+𝐴^′_𝑖𝑖−𝜏𝐴_𝑖𝑖+𝐴^′_𝑖𝑖−Φ𝐴_𝑖𝑖+𝜏𝐴_𝑖𝑖−Φ𝐴^′_𝑖𝑖+𝜏𝐴^′_𝑖𝑖.(47) Thus 𝛿𝐴_𝑖𝑖+𝐴^′_𝑖𝑖−𝛿𝐴_𝑖𝑖−𝛿𝐴^′_𝑖𝑖∈𝑍_𝐵(𝑋)∩𝒜_𝑖𝑖={0},(48) and so 𝛿(𝐴_𝑖𝑖+𝐴^′_𝑖𝑖)=𝛿(𝐴_𝑖𝑖)+𝛿(𝐴^′_𝑖𝑖).
Step  2. For every 𝑇_𝑖𝑖,𝑆_𝑖𝑖∈𝒜_𝑖𝑖, 𝑇_𝑖𝑗,𝑆_𝑖𝑗∈𝒜_𝑖𝑗, 𝑆_𝑗𝑗∈𝒜_𝑗𝑗(1≤𝑖≠𝑗≤2), and 𝛿(𝑇_𝑖𝑖𝑆_𝑖𝑗)=𝛿(𝑇_𝑖𝑖)𝑆_𝑖𝑗+𝑇_𝑖𝑖𝛿(𝑆_𝑖𝑗), 𝛿(𝑇_𝑖𝑗𝑆_𝑗𝑗)=𝛿(𝑇_𝑖𝑗)𝑆_𝑗𝑗+𝑇_𝑖𝑗𝛿(𝑆_𝑗𝑗), 𝛿(𝑇_𝑖𝑖𝑆_𝑖𝑖)=𝛿(𝑇_𝑖𝑖)𝑆_𝑖𝑖+𝑇_𝑖𝑖𝛿(𝑆_𝑖𝑖).
Let 𝑇_𝑖𝑗,𝑆_𝑖𝑗,𝐴_𝑖𝑗∈𝒜_𝑖𝑗(𝑖≠𝑗), 𝛿𝑇_𝑖𝑖𝑆_𝑖𝑗=Φ𝑇_𝑖𝑖𝑆_𝑖𝑗=Φ𝑇_𝑖𝑖,𝑆_𝑖𝑗=Φ𝑇_𝑖𝑖,𝑆_𝑖𝑗+𝑇_𝑖𝑖,Φ𝑆_𝑖𝑗=𝛿𝑇_𝑖𝑖,𝑆_𝑖𝑗+𝑇_𝑖𝑖,𝛿𝑆_𝑖𝑗=𝛿𝑇_𝑖𝑖𝑆_𝑖𝑗+𝑇_𝑖𝑖𝛿𝑆_𝑖𝑗,𝛿𝑇_𝑖𝑗𝑆_𝑗𝑗=Φ𝑇_𝑖𝑗𝑆_𝑗𝑗=Φ𝑇_𝑖𝑗,𝑆_𝑗𝑗=Φ𝑇_𝑖𝑗,𝑆_𝑗𝑗+𝑇_𝑖𝑗,Φ𝑆_𝑗𝑗=𝛿𝑇_𝑖𝑗𝑆_𝑗𝑗+𝑇_𝑖𝑗𝛿𝑆_𝑗𝑗=𝛿𝑇_𝑖𝑗𝑆_𝑗𝑗+𝑇_𝑖𝑗𝛿𝑆_𝑗𝑗.(49) So 𝛿𝑇_𝑖𝑖𝑆_𝑖𝑖𝐴_𝑖𝑗=𝛿𝑇_𝑖𝑖𝑆_𝑖𝑖𝐴_𝑖𝑗+𝑇_𝑖𝑖𝑆_𝑖𝑖𝛿𝐴_𝑖𝑗.(50) On the other hand, 𝛿𝑇_𝑖𝑖𝑆_𝑖𝑖𝐴_𝑖𝑗=𝛿𝑇_𝑖𝑖𝑆_𝑖𝑖𝐴_𝑖𝑗+𝑇_𝑖𝑖𝛿𝑆_𝑖𝑖𝐴_𝑖𝑗=𝛿𝑇_𝑖𝑖𝑆_𝑖𝑖𝐴_𝑖𝑗+𝑇_𝑖𝑖𝛿𝑆_𝑖𝑖𝐴_𝑖𝑗+𝑇_𝑖𝑖𝑆_𝑖𝑖𝛿𝐴_𝑖𝑗.(51) So (𝛿(𝑇_𝑖𝑖𝑆_𝑖𝑖)−𝛿(𝑇_𝑖𝑖)𝑆_𝑖𝑖−𝑇_𝑖𝑖𝛿(𝑆_𝑖𝑖))𝐴_𝑖𝑗=0. Note that 𝐵(𝑋) is prime, so 𝛿𝑇_𝑖𝑖𝑆_𝑖𝑖=𝛿𝑇_𝑖𝑖𝑆_𝑖𝑖+𝑇_𝑖𝑖𝛿𝑆_𝑖𝑖.(52)
Step  3. 𝛿 is an additive derivation on 𝒜₁₁⊕𝒜₁₂⊕𝒜₂₂and𝒜₁₁⊕𝒜₂₁⊕𝒜₂₂.
Let 𝐴=𝐴₁₁+𝐴₁₂+𝐴₂₂ and let 𝐵=𝐵₁₁+𝐵₁₂+𝐵₂₂ be in 𝒜₁₁⊕𝒜₁₂⊕𝒜₂₂. By Steps  1 and 2, we have 𝛿(𝐴𝐵)=𝛿𝐴₁₁+𝐴₁₂+𝐴₂₂𝐵₁₁+𝐵₁₂+𝐵₂₂=𝛿𝐴₁₁𝐵₁₁+𝛿𝐴₁₁𝐵₁₂+𝛿𝐴₁₂𝐵₂₂+𝛿𝐴₂₂𝐵₂₂=𝛿𝐴₁₁𝐵₁₁+𝐴₁₁𝛿𝐵₁₁+𝛿𝐴₁₁𝐵₁₂+𝐴₁₁𝛿𝐵₁₂+𝛿𝐴₁₂𝐵₂₂+𝐴₁₂𝛿𝐵₂₂+𝛿𝐴₂₂𝐵₂₂+𝐴₂₂𝛿𝐵₂₂.(53) From 𝛿(𝒜_𝑖𝑗)⊆𝒜_𝑖𝑗, we have 𝛿(𝐴)(𝐵)𝐴𝐵+𝐴𝛿=𝛿₁₁+𝐴₁₂+𝐴₂₂𝐵₁₁+𝐵₁₂+𝐵₂₂+𝐴₁₁+𝐴₁₂+𝐴₂₂𝛿𝐵₁₁+𝐵₁₂+𝐵₂₂=𝛿𝐴₁₁𝐵₁₁+𝐴₁₁𝛿𝐵₁₁+𝛿𝐴₁₁𝐵₁₂+𝐴₁₁𝛿𝐵₁₂+𝛿𝐴₁₂𝐵₂₂+𝐴₁₂𝛿𝐵₂₂+𝛿𝐴₂₂𝐵₂₂+𝐴₂₂𝛿𝐵₂₂.(54) So 𝛿(𝐴)𝐵+𝐴𝛿(𝐵)=𝛿(𝐴𝐵), and so 𝛿 is an additive derivation on 𝒜₁₁⊕𝒜₁₂⊕𝒜₂₂. So from [15, Theorem 4.1] we can see that 𝛿 is a derivation. Furthermore, due to the main result in [6], there exists an operator 𝑇^′∈𝐵(𝑋) such that 𝛿(𝐴)=𝑇^′𝐴−𝐴𝑇^′(55) for all 𝐴∈𝒜₁₁⊕𝒜₁₂⊕𝒜₂₂.
Similarly, we can know that 𝛿 is an additive derivation on 𝒜₁₁⊕𝒜₂₁⊕𝒜₂₂, so there exists an operator 𝑆^′∈𝐵(𝑋) such that 𝛿(𝐴)=𝑆^′𝐴−𝐴𝑆^′(56) for all 𝐴∈𝒜₁₁⊕𝒜₂₁⊕𝒜₂₂.
Step  4. Consider that 𝛿(𝑇_𝑖𝑗𝑆_𝑗𝑖)=𝛿(𝑇_𝑖𝑗)𝑆_𝑗𝑖+𝑇_𝑖𝑗𝛿(𝑆_𝑗𝑖), for every 𝑇_𝑖𝑗∈𝒜_𝑖𝑗, 𝑆_𝑗𝑖∈𝒜_𝑗𝑖(1≤𝑖≠𝑗≤2).
Now, we write 𝑇^′=∑²_𝑖,𝑗=1𝑇^′_𝑖𝑗 and 𝑆^′=∑²_𝑖,𝑗=1𝑆^′_𝑖𝑗. Then by Step 3, we have that 𝑇^′𝐴₁₁+𝐵₂₂−𝐴₁₁+𝐵₂₂𝑇^′=𝑆^′𝐴₁₁+𝐵₂₂−𝐴₁₁+𝐵₂₂𝑆^′.(57) For all  𝐴₁₁∈𝒜₁₁, for all 𝐵₂₂∈𝒜₂₂, this shows that 𝑇^′−𝑆^′𝐴₁₁=𝐴₁₁𝑇^′−𝑆^′,𝑇^′−𝑆^′𝐵₂₂=𝐵₂₂𝑇^′−𝑆^′.(58) Multiplying the previous two equations by 𝑃₂ and 𝑃₁ from the left respectively, we can know that 𝑇^′₂₁−𝑆^′₂₁𝐴₁₁=0,𝑇^′₁₂−𝑆^′₁₂𝐵₂₂=0.(59) Similarly, 𝑇^′₁₁−𝑆^′₁₁𝐴₁₁=𝐴₁₁𝑇^′₁₁−𝑆^′₁₁,𝑇^′₂₂−𝑆^′₂₂𝐵₂₂=𝐵₂₂𝑇^′₂₂−𝑆^′₂₂.(60) So 𝑇^′₁₂=𝑆^′₁₂,𝑇^′₂₁=𝑆^′₂₁,𝑇^′₁₁−𝑆^′₁₁=𝜆₁𝑃₁,𝑇^′₂₂−𝑆^′₂₂=𝜆₂𝑃₂(61) for some 𝜆₁,𝜆₂∈ℂ.
For every 𝐴₁₂∈𝒜₁₂ and 𝐵₂₁∈𝒜₂₁, using (55), (56), and (61), we get that 𝛿𝐴₁₂𝐵₂₁−𝛿𝐴₁₂𝐵₂₁−𝐴₁₂𝛿𝐵₂₁=𝑇^′𝐴₁₂𝐵₂₁−𝐴₁₂𝐵₂₁𝑇^′−𝑇^′𝐴₁₂−𝐴₁₂𝑇^′𝐵₂₁−𝐴₁₂𝑆^′𝐵₂₁−𝐵₂₁𝑆^′=−𝐴₁₂𝐵₂₁𝑇^′+𝐴₁₂𝑇^′𝐵₂₁−𝐴₁₂𝑆^′𝐵₂₁+𝐴₁₂𝐵₂₁𝑆^′=𝜆₂−𝜆₁𝐴₁₂𝐵₂₁.𝛿𝐵₂₁𝐴₁₂−𝛿𝐵₂₁𝐴₁₂−𝐵₂₁𝛿𝐴₁₂=𝑆^′𝐵₂₁𝐴₁₂−𝐵₂₁𝐴₁₂𝑆^′−𝑆^′𝐵₂₁−𝐵₂₁𝑆^′𝐴₁₂−𝐵₂₁𝑆^′𝐴₁₂−𝐴₁₂𝑆^′=−𝐵₂₁𝐴₁₂𝑆^′+𝐵₂₁𝑆^′𝐴₁₂−𝐵₂₁𝑇^′𝐴₁₂+𝐵₂₁𝐴₁₂𝑇^′=𝜆₂−𝜆₁𝐵₂₁𝐴₁₂.(62) Since at least one of 𝑃₂ or 𝑃₁ is of rank greater than 2, 𝐴^′₁₂∈𝒜₁₂ and 𝐵^′₂₁∈𝒜₂₁ exist such that 𝐴^′₁₂𝐵^′₂₁=0,𝐵^′₂₁𝐴^′₁₂≠0(63) or 𝐵^′₂₁𝐴^′₁₂=0,𝐴^′₁₂𝐵^′₂₁≠0.(64) We suppose that 𝐴^′₁₂𝐵^′₂₁=0 and 𝐵^′₂₁𝐴^′₁₂≠0. From (62), we get 𝜏𝐴^′₁₂,𝐵^′₂₁=Φ𝐴^′₁₂,𝐵^′₂₁−𝛿𝐴^′₁₂,𝐵^′₂₁=Φ𝐴^′₁₂,𝐵^′₂₁+𝐴^′₁₂,Φ𝐵^′₂₁−𝛿𝐴^′₁₂,𝐵^′₂₁=𝛿𝐴^′₁₂,𝐵^′₂₁+𝐴^′₁₂,𝛿𝐵^′₂₁−𝛿𝐴^′₁₂,𝐵^′₂₁=𝛿𝐴^′₁₂𝐵^′₂₁−𝐵^′₂₁𝛿𝐴^′₁₂+𝐴^′₁₂𝛿𝐵^′₂₁−𝛿𝐵^′₂₁𝐴^′₁₂−𝛿𝐴^′₁₂𝐵^′₂₁+𝛿𝐵^′₂₁𝐴^′₁₂=𝜆₁−𝜆₂𝐴^′₁₂𝐵^′₂₁+𝜆₂−𝜆₁𝐵^′₂₁𝐴^′₁₂=𝜆₂−𝜆₁𝐵^′₂₁𝐴^′₁₂.(65) From 𝜏([𝐴^′₁₂,𝐵^′₂₁])∈𝑍_𝐵(𝑋) and 𝐵^′₂₁𝐴^′₁₂≠0, then 𝜆₁=𝜆₂. So 𝛿𝐴₁₂𝐵₂₁=𝛿𝐴₁₂𝐵₂₁+𝐴₁₂𝛿𝐵₂₁,𝛿𝐵₂₁𝐴₁₂=𝛿𝐵₂₁𝐴₁₂+𝐵₂₁𝛿𝐴₁₂(66) for any 𝐴₁₂∈𝒜₁₂ and 𝐵₂₁∈𝒜₂₁. So 𝛿𝑇_𝑖𝑗𝑆_𝑗𝑖=𝛿𝑇_𝑖𝑗𝑆_𝑗𝑖+𝑇_𝑖𝑗𝛿𝑆_𝑗𝑖,(67)𝛿𝑆_𝑗𝑖𝑇_𝑖𝑗=𝛿𝑆_𝑗𝑖𝑇_𝑖𝑗+𝑆_𝑗𝑖𝛿𝑇_𝑖𝑗.(68) Let 𝑇,𝑆∈𝐵(𝑋), and write 𝑇=∑²_𝑖,𝑗=1𝑇_𝑖𝑗 and 𝑆=∑²_𝑖,𝑗=1𝑆_𝑖𝑗, so 𝛿(𝑇𝑆)=𝛿𝑇₁₁𝑆₁₁+𝛿𝑇₁₁𝑆₁₂+𝛿𝑇₁₂𝑆₂₂+𝛿𝑇₁₂𝑆₂₁+𝛿𝑇₂₁𝑆₁₂+𝛿𝑇₂₁𝑆₁₁+𝛿𝑇₂₂𝑆₂₁+𝛿𝑇₂₂𝑆₂₂=𝛿(𝑇)𝑆+𝑇𝛿(𝑆).(69)
Claim  7.  𝜏 sends the commutators into zero: 𝜏=[Φ(𝐴)]+[(𝐵=[𝛿(𝐴)]+[(𝐵([𝐴,𝐵])=Φ([𝐴,𝐵])−𝛿([𝐴,𝐵]),𝐵𝐴,Φ)]−𝛿([𝐴,𝐵]),𝐵𝐴,𝛿)]−𝛿([𝐴,𝐵])=0.(70) The proof is complete.

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).