# Characterizations of Nonlinear Lie Derivations ofB(X)

Authors: Donghua Huo; Baodong Zheng; Hongyu Liu
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

LetX 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 ZB(X), which maps commutators into the zero.


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## Body

## 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]).LetX 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 Zℳ which maps commutators into zero.Nevertheless, it would seem desirable to investigate nonlinear Lie derivations ofB(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 thatB(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 idempotentP1∈B(X) and let P2=I-P1. In what follows, we write 𝒜ij=PiB(X)Pj for, j=1,2. Then every operator A∈B(X) can be written as A=∑i,j=12Aij. Note that Aij notation denotes an arbitrary element of 𝒜ij.Definition 1.
LettingS be a set of B(X), we call
(1)ZS={A∈B(X)∣[A,B]=0,∀B∈S}
the centralizer of S about B(X).
Clearly,ZB(X) is the center of B(X).Lemma 2.
Consider thatZ𝒜ij=𝒜ij+ZB(X),1≤i≠j≤2.Proof.
It is clearly that𝒜ij+ZB(X)⊆Z𝒜ij,1≤i≠j≤2.
Conversely, without loss of generality, we proof thatZ𝒜12⊆𝒜12+ZB(X). For any T=Σi,j=12Tij∈Z𝒜12, we have
(2)T11A12+T21A12=A12T22+A12T21,∀A12∈𝒜12.
Thus
(3)T21A12=T21P1AP2=0,∀A∈B(X),(4)T11A12=A12T22,∀A12∈𝒜12.
From (3), noting that B(X) is prime, we obtain T21P1=0, that is, T21=0. By A11A12∈𝒜12, we have
(5)T11A11A12=A11A12T22=A11T11A12,(T11A11-A11T11)P1AP2=0,∀A∈B(X).
Since B(X) is prime, we have T11A11=A11T11. We write T11=Z1P1, where Z1∈ZB(X). Since A12A22∈𝒜12, we have
(6)T11A12A22=A12A22T22=A12T22A22,P1AP2(A22T22-T22A22)=0,∀A∈B(X).
Since B(X) is prime we have T22A22=A22T22. We write T22=Z2P2, where Z2∈ZB(X). From (4) we have
(7)(Z1-Z2)P1AP2=Z1P1AP2-P1AP2Z2P2=T11P1AP2-P1AP2T22=0.
So
(8)(Z1-Z2)P1=0.
Thus
(9)(Z1-Z2)AP1=A(Z1-Z2)P1=0,∀A∈B(X).
So
(10)Z1=Z2.
It follows that
(11)T=T12+Z1P1+Z2P2=T12+Z1∈𝒜12+ZB(X).Lemma 3.
Consider thatZB(X)=Z𝒜12∩Z𝒜21.Proof.
From Lemma2, we have
(12)Z𝒜12∩Z𝒜21=(𝒜12+ZB(X))∩(𝒜21+ZB(X))=ZB(X).Theorem 4.
LetX 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 ZB(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, Ti∈B(X) such that Φ(Pi)-[Pi,Ti]∈ZB(X) and Φ(Aij)=PiΦ(Aij)Pj+[Aij,Ti], Φ(Aji)=PjΦ(Aji)Pi+[Aji,Ti] for arbitrarily Aij∈𝒜ij(1≤i≠j≤2).
ObviouslyΦ(0)=Φ([0,0])=[Φ(0),0]+[0,Φ(0)]=0.
Moreover, for arbitrarilyAij,
(13)Φ(Aij)=Φ([Pi,Aij])=[Φ(Pi),Aij]+[Pi,Φ(Aij)]=Φ(Pi)Aij-AijΦ(Pi)+PiΦ(Aij)-Φ(Aij)Pi.
In (13), the left and right sides are, respectively, multiplied by Pi and Pj, and we have
(14)PiΦ(Pi)PiAij=AijPjΦ(Pi)Pj.
In (13), the left and right sides are respectively multiplying by Pj and Pi, and we arrive at
(15)PjΦ(Aij)Pi=0.
For arbitrarily Aji,
(16)Φ(Aji)=Φ([Aji,Pi])=[Φ(Aji),Pi]+[Aji,Φ(Pi)]=Φ(Aji)Pi-PiΦ(Aji)+AjiΦ(Pi)-Φ(Pi)Aji.
In the previous equation, the left and right sides are, respectively multiplied by Pj and Pi, and we have
(17)AjiPiΦ(Pi)Pi=PjΦ(Pi)PjAji.
This together with (14) shows that
(18)[PiΦ(Pi)Pi+PjΦ(Pi)Pj,Aij]=[PiΦ(Pi)Pi+PjΦ(Pi)Pj,Aji]=0.
From Lemma 3, it is that
(19)PiΦ(Pi)Pi+PjΦ(Pi)Pj∈ZB(X).
Denote Ti=PiΦ(Pi)Pj-PjΦ(Pi)Pi. Therefore
(20)Φ(Pi)-[Pi,Ti]=PiΦ(Pi)Pi+PjΦ(Pi)Pj∈ZB(X).
At the same time, [Aij,Ti]=PjΦ(Pi)PiAij-AijPjΦ(Pi)Pi. From (14),
(21)[Aij,Ti]=PjΦ(Pi)PiAij-AijPjΦ(Pi)Pi+PiΦ(Pi)PiAij-AijPjΦ(Pi)Pj=Φ(Pi)Aij-AijΦ(Pi).
Because PjΦ(Aij)Pi=0 so by (13) we get Φ(Aij)=PiΦ(Aij)Pj+[Aij,Ti].
In the same way, we can getΦ(Aji)=PjΦ(Aji)Pi+[Aji,Ti].
Remark. It is clear that T→[T,Ti] is an additive derivation. Without loss of generality, we can assume that Φ(P1)∈ZB(X) and Φ(𝒜ij)⊆𝒜ij(1≤i≠j≤2).
Claim  2. Consider that Φ(𝒜ii)⊆𝒜ii+ZB(X)(i=1,2).
Leti=1, for every T11∈𝒜11, and write Φ(T11)=∑i,j=12Bij. From Φ(P1)∈ZB(X) and [P1,A]=[A,P2], we can get Φ(P2)∈ZB(X):(22)0=Φ([T11,P2])=[Φ(T11),P2]+[T11,Φ(P2)]=Φ(T11)P2-P2Φ(T11).
From this P1Φ(T11)P2=P2Φ(T11)P1=0. So we have B12=B21=0. For every A22∈𝒜22,
(23)0=Φ([T11,A22])=[Φ(T11),A22]+[T11,Φ(A22)]=B22A22-A22B22+T11Φ(A22)-Φ(A22)T11.
Thus, B22A22-A22B22=0, and so B22=P2Z for some ∈ZB(X). Hence
(24)Φ(T11)=B11-P1Z+Z∈𝒜11+ZB(X).
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)∈ZB(X).
We divided the proof into several steps.
Step  1.   For every Tii,Tji, and Tij(1≤i≠j≤2), Φ(Tii+Tij)-Φ(Tii)-Φ(Tij)∈ZB(X), Φ(Tii+Tji)-Φ(Tii)-Φ(Tji)∈ZB(X).
Let=1, and let j=2. For every A12∈𝒜12,[T11+T12,A12]=[T11,A12]; therefore
(25)[Φ(T11+T12),A12]+[T11+T12,Φ(A12)]=[Φ(T11),A12]+[T11,Φ(A12)].
This yields [Φ(T11+T12)-Φ(T11),A12]=0. From Lemma 2,
(26)Φ(T11+T12)-Φ(T11)=P1(Φ(T11+T12)-Φ(T11))P2+Z
for ∈ZB(X).
SinceΦ(P1)∈ZB(X) and [P1,T]=P1TP2 for all T∈𝒜11⊕𝒜12⊕𝒜22. We have that
(27)Φ(P1TP2)=Φ([P1,T])=[Φ(P1),T]+[P1,Φ(T)]=P1Φ(T)P2.
Then(28)P1(Φ(T11+T12)-Φ(T11))P2=Φ(P1(T11+T12)P2)-Φ(P1(T11)P2)=Φ(P1T11P2+P1T12P2)-Φ(P1T11P2)=Φ(P1T12P2)=Φ(T12).
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.
ForT12,S12∈𝒜12. Because T12+S12=[P1+T12,P2+S12], by Step  1, we know that
(29)Φ(T12+S12)=[Φ(P1+T12),P2+S12]+[P1+T12,Φ(P2+S12)]=[Φ(P1)+Φ(T12),P2+S12]+[P1+T12,Φ(P2)+Φ(S12)]=Φ(T12)+Φ(S12).
The rest is similar.
Step  3. For any T11∈𝒜11,T22∈𝒜22,Φ(T11+T22)-Φ(T11)-Φ(T22)∈ZB(X).
LetA12∈𝒜12, and [T11+T22,A12]=T11A12-A12T22. For Step  2, we know that
(30)[Φ(T11+T22),A12]+[T11+T22,Φ(A12)]=Φ(T11A12)+Φ(-A12T22)=Φ([T11,A12])+Φ([T22,A12])=[Φ(T11),A12]+[T11,Φ(A12)]+[Φ(T22),A12]+[T22,Φ(A12)].
So [Φ(T11+T22)-Φ(T11)-Φ(T22),A12]=0. From Lemma 2,
(31)Φ(T11+T22)-Φ(T11)-Φ(T22)∈𝒜12+ZB(X).
On the other hand, from (27) we get that P1(Φ(T11+T22)-Φ(T11)-Φ(T22))P2=0. So (32)Φ(T11+T22)-Φ(T11)-Φ(T22)∈ZB(X).
Step  4.  For any Tii,Sii∈𝒜ii(i=1,2),Φ(Tii+Sii)-Φ(Tii)-Φ(Sii)∈ZB(X).
Leti=1. For any A12∈𝒜12,[T11+S11,A12]=T11A12+S11A12. From Step 2, it is that
(33)[Φ(T11+S11),A12]+[T11+S11,Φ(A12)]=Φ(T11A12)+Φ(S11A12)=Φ([T11,A12])+Φ([S11,A12])=[Φ(T11),A12]+[T11,Φ(A12)]+[Φ(S11),A12]+[S11,Φ(A12)].
So [Φ(T11+S11)-Φ(T11)-Φ(S11),A12]=0. From Lemma 2,
(34)Φ(T11+S11)-Φ(T11)-Φ(S11)∈𝒜12+ZB(X).
On the other hand, from (27) we get that P1(Φ(T11+S11)-Φ(T11)-Φ(S11))P2=0. So
(35)Φ(T11+S11)-Φ(T11)-Φ(S11)∈ZB(X).
The other case can be treated similarly.
Step  5. Consider that Φ(Tii+Tjj+Tij)-Φ(Tii)-Φ(Tjj)-Φ(Tij)∈ZB(X).
Leti=1, and let j=2. For any A12∈𝒜12,[T11+T22+T12,A12]=[T11+T22,A12]. By Step 3, it is that
(36)[Φ(T11+T22+T12),A12]+[T11+T22+T12,Φ(A12)]=[Φ(T11)+Φ(T22),A12]+[T11+T22,Φ(A12)].
So [Φ(T11+T22+T12)-Φ(T11)-Φ(T22),A12]=0. From Lemma 2, it follows that
(37)Φ(T11+T22+T12)-Φ(T11)-Φ(T22)=P1(Φ(T11+T22+T12)-Φ(T11)-Φ(T22))P2+Z
for some central element Z∈ZB(X). From (27) we get that
(38)Φ(T12)=P1Φ(T11+T22+T12)P2.
So
(39)Φ(T11+T22+T12)-Φ(T11)-Φ(T22)-Φ(T12)∈ZB(X).
Similarly, Φ(T22+T11+T21)-Φ(T22)-Φ(T11)-Φ(T21)∈ZB(X).
Step  6. For all T,S∈B(X), Φ(T+S)-Φ(T)-Φ(S)∈ZB(X).
LetT∈B(X), and write T=T11+T12+T21+T22. If we know that
(40)Φ(T11+T12+T21+T22)-Φ(T11)-Φ(T12)-Φ(T21)-Φ(T22)∈ZB(X),
then it is right.
For everyA12∈𝒜12,[T11+T12+T21+T22,A12]=[T11+T21+T22,A12]. By Step 5, it shows that
(41)[Φ(T11+T12+T21+T22),A12]+[T11+T12+T21+T22,Φ(A12)]=[Φ(T11)+Φ(T21)+Φ(T22),A12]+[T11+T21+T22,Φ(A12)].
So
(42)[Φ(T11+T12+T21+T22)-Φ(T11)-Φ(T21)-Φ(T22),A12]=0.
Because Φ(T12)∈𝒜12, we clearly have
(43)[Φ(T11+T12+T21+T22)-Φ(T11)-Φ(T12)-Φ(T21)-Φ(T22),A12]=0.
Similarly, [T11+T12+T21+T22,A21]=[T11+T12+T22,A21], and we can obtain
(44)[Φ(T11+T12+T21+T22)-Φ(T11)-Φ(T12)-Φ(T21)-Φ(T22),A21]=0.
From Lemma 3,
(45)Φ(T11+T12+T21+T22)-Φ(T11)-Φ(T12)-Φ(T21)-Φ(T22)∈ZB(X).
For every A∈B(X), from Step 6, we define a mapping τ of B(X) into ZB(X) by
(46)τ(A)=Φ(A)-Φ(P1AP1)-Φ(P1AP2)-Φ(P2AP1)-Φ(P2AP2).
Then let δ(A)=Φ(A)-τ(A).
In the following, we will prove thatδ and τ are desired maps.
Claim4.
δ is an additive derivation.
We divide the proof into the following steps.
Step1.
δ is an additive map.
We only need to show thatδ is an additive on 𝒜ii. Let Aii,Aii′∈𝒜ii,
(47)δ(Aii+Aii′)-δ(Aii)-δ(Aii′)=Φ(Aii+Aii′)-τ(Aii+Aii′)-Φ(Aii)+τ(Aii)-Φ(Aii′)+τ(Aii′).
Thus
(48)δ(Aii+Aii′)-δ(Aii)-δ(Aii′)∈ZB(X)∩𝒜ii={0},
and so δ(Aii+Aii′)=δ(Aii)+δ(Aii′).
Step  2. For every Tii,Sii∈𝒜ii, Tij,Sij∈𝒜ij, Sjj∈𝒜jj(1≤i≠j≤2), and δ(TiiSij)=δ(Tii)Sij+Tiiδ(Sij), δ(TijSjj)=δ(Tij)Sjj+Tijδ(Sjj), δ(TiiSii)=δ(Tii)Sii+Tiiδ(Sii).
LetTij,Sij,Aij∈𝒜ij(i≠j),
(49)δ(TiiSij)=Φ(TiiSij)=Φ([Tii,Sij])=[Φ(Tii),Sij]+[Tii,Φ(Sij)]=[δ(Tii),Sij]+[Tii,δ(Sij)]=δ(Tii)Sij+Tiiδ(Sij),δ(TijSjj)=Φ(TijSjj)=Φ([Tij,Sjj])=[Φ(Tij),Sjj]+[Tij,Φ(Sjj)]=[δ(Tij)Sjj]+[Tijδ(Sjj)]=δ(Tij)Sjj+Tijδ(Sjj).
So
(50)δ(TiiSiiAij)=δ(TiiSii)Aij+TiiSiiδ(Aij).
On the other hand,
(51)δ(TiiSiiAij)=δ(Tii)SiiAij+Tiiδ(SiiAij)=δ(Tii)SiiAij+Tiiδ(Sii)Aij+TiiSiiδ(Aij).
So (δ(TiiSii)-δ(Tii)Sii-Tiiδ(Sii))Aij=0. Note that B(X) is prime, so (52)δ(TiiSii)=δ(Tii)Sii+Tiiδ(Sii).
Step  3.δ is an additive derivation on 𝒜11⊕𝒜12⊕𝒜22and𝒜11⊕𝒜21⊕𝒜22.
LetA=A11+A12+A22 and let B=B11+B12+B22 be in 𝒜11⊕𝒜12⊕𝒜22. By Steps  1 and 2, we have
(53)δ(AB)=δ[(A11+A12+A22)(B11+B12+B22)]=δ(A11B11)+δ(A11B12)+δ(A12B22)+δ(A22B22)=δ(A11)B11+A11δ(B11)+δ(A11)B12+A11δ(B12)+δ(A12)B22+A12δ(B22)+δ(A22)B22+A22δ(B22).
From δ(𝒜ij)⊆𝒜ij, we have
(54)δ(A)B+Aδ(B)=δ(A11+A12+A22)(B11+B12+B22)+(A11+A12+A22)δ(B11+B12+B22)=δ(A11)B11+A11δ(B11)+δ(A11)B12+A11δ(B12)+δ(A12)B22+A12δ(B22)+δ(A22)B22+A22δ(B22).
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
(55)δ(A)=T′A-AT′
for all A∈𝒜11⊕𝒜12⊕𝒜22.
Similarly, we can know thatδ is an additive derivation on 𝒜11⊕𝒜21⊕𝒜22, so there exists an operator S′∈B(X) such that
(56)δ(A)=S′A-AS′
for all A∈𝒜11⊕𝒜21⊕𝒜22.
Step  4. Consider that δ(TijSji)=δ(Tij)Sji+Tijδ(Sji), for every Tij∈𝒜ij, Sji∈𝒜ji(1≤i≠j≤2).
Now, we writeT′=∑i,j=12Tij′ and S′=∑i,j=12Sij′. Then by Step 3, we have that
(57)T′(A11+B22)-(A11+B22)T′=S′(A11+B22)-(A11+B22)S′.
For all  A11∈𝒜11, for all B22∈𝒜22, this shows that
(58)(T′-S′)A11=A11(T′-S′),(T′-S′)B22=B22(T′-S′).
Multiplying the previous two equations by P2 and P1 from the left respectively, we can know that
(59)(T21′-S21′)A11=0,(T12′-S12′)B22=0.
Similarly,
(60)(T11′-S11′)A11=A11(T11′-S11′),(T22′-S22′)B22=B22(T22′-S22′).
So
(61)T12′=S12′,T21′=S21′,T11′-S11′=λ1P1,T22′-S22′=λ2P2
for some λ1,λ2∈ℂ.
For everyA12∈𝒜12 and B21∈𝒜21, using (55), (56), and (61), we get that
(62)δ(A12B21)-δ(A12)B21-A12δ(B21)=T′A12B21-A12B21T′-(T′A12-A12T′)B21-A12(S′B21-B21S′)=-A12B21T′+A12T′B21-A12S′B21+A12B21S′=(λ2-λ1)A12B21.δ(B21A12)-δ(B21)A12-B21δ(A12)=S′B21A12-B21A12S′-(S′B21-B21S′)A12-B21(S′A12-A12S′)=-B21A12S′+B21S′A12-B21T′A12+B21A12T′=(λ2-λ1)B21A12.
Since at least one of P2 or P1 is of rank greater than 2, A12′∈𝒜12 and B21′∈𝒜21 exist such that
(63)A12′B21′=0,B21′A12′≠0
or
(64)B21′A12′=0,A12′B21′≠0.
We suppose that A12′B21′=0 and B21′A12′≠0. From (62), we get
(65)τ([A12′,B21′])=Φ([A12′,B21′])-δ([A12′,B21′])=[Φ(A12′),B21′]+[A12′,Φ(B21′)]-δ([A12′,B21′])=[δ(A12′),B21′]+[A12′,δ(B21′)]-δ([A12′,B21′])=δ(A12′)B21′-B21′δ(A12′)+A12′δ(B21′)-δ(B21′)A12′-δ(A12′B21′)+δ(B21′A12′)=(λ1-λ2)A12′B21′+(λ2-λ1)B21′A12′=(λ2-λ1)B21′A12′.
From τ([A12′,B21′])∈ZB(X) and B21′A12′≠0, then λ1=λ2. So
(66)δ(A12B21)=δ(A12)B21+A12δ(B21),δ(B21A12)=δ(B21)A12+B21δ(A12)
for any A12∈𝒜12 and B21∈𝒜21. So
(67)δ(TijSji)=δ(Tij)Sji+Tijδ(Sji),(68)δ(SjiTij)=δ(Sji)Tij+Sjiδ(Tij).
Let T,S∈B(X), and write T=∑i,j=12Tij and S=∑i,j=12Sij, so (69)δ(TS)=δ(T11S11)+δ(T11S12)+δ(T12S22)+δ(T12S21)+δ(T21S12)+δ(T21S11)+δ(T22S21)+δ(T22S22)=δ(T)S+Tδ(S).
Claim  7.  τ sends the commutators into zero:
(70)τ([A,B])=Φ([A,B])-δ([A,B])=[Φ(A),B]+[A,Φ(B)]-δ([A,B])=[δ(A),B]+[A,δ(B)]-δ([A,B])=0.
The proof is complete.



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*Source: 245452-2013-03-26.xml*