# Characterizing Derivations on Von Neumann Algebras by Local Actions

Authors: Xiaofei Qi; Jia Ji
Journal: Journal of Function Spaces and Applications
 (2013)

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## Abstract

Letℳ be any von Neumann algebra without central summands of type I1 and P a core-free projection with the central carrier I. For any scalar ξ, it is shown that every additive map L on ℳ satisfies L(AB-ξBA)=L(A)B-ξBL(A)+AL(B)-ξL(B)A whenever AB=P if and only if (1) ξ=1, L=φ+h, where φ is an additive derivation and h is a central valued additive map vanishing on AB-BA with AB=P; (2) ξ≠1, L is a derivation with L(ξA)=ξL(A) for each A∈ℳ.


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## Body

## 1. Introduction

Let𝒜 be an algebra over a field 𝔽. For a scalar ξ∈𝔽 and for A,B∈𝒜, we say that A commutes with B up to a factor ξ if AB=ξBA. The notion of commutativity up to a factor for pairs of operators is an important concept and has been studied in the context of operator algebras and quantum groups [1, 2]. Motivated by this, a binary operation [A,B]ξ=AB-ξBA, called ξ-Lie product of A and B, was introduced in [3]. An additive map L:𝒜→𝒜 is called an additive ξ-Lie derivation if L([A,B]ξ)=[L(A),B]ξ+[A,L(B)]ξ holds for all A,B∈𝒜. It is clear that a ξ-Lie derivation is a derivation if ξ=0; is a Lie derivation if ξ=1; is a Jordan derivation if ξ=-1. The structure of ξ-Lie derivations on various operator algebras was discussed by several authors (see [3–5] and the references therein).Recently, the question of under what conditions that a map becomes a (Jordan) derivation attracted much attention of many researchers (see [6–9] and the references therein). For ξ-Lie derivations, an additive (a linear) map L on 𝒜 is said to be ξ-Lie derivable at a point Z∈𝒜 if L([A,B]ξ)=[L(A),B]ξ+[A,L(B)]ξ for any A,B∈𝒜 with [A,B]ξ=Z. Clearly, this definition is only valid for ξ-Lie commutators, that is, the elements of the form Z=[A,B]ξ. For instance, if Z=I and ξ=1, as the unit I may not be a commutator in general, there is no sense to define that L is Lie derivable at I. Since zero is a ξ-Lie commutator for any ξ and any algebra, as a start, Qi and Hou [10] characterized the linear maps Lie derivable at zero between 𝒥-subspace lattice algebras. In [11] Qi et al. considered further the additive maps ξ-Lie derivable at zero on unital prime algebras over a field 𝔽 containing a non-trivial idempotent P. Since factor von Neumann algebras are prime, as a consequence of the result for prime algebras, all additive maps ξ-Lie derivable at zero on factor von Neumann algebras are characterized. Recently, Qi et al. [12] gave a characterization of additive maps ξ-Lie derivable at zero on general von Neumann algebras for all possible ξ.Lu and Jing [13] considered the question of characterizing Lie derivations from another direction. Let X be a Banach space with dimX≥3 and ℬ(X) the algebra of all bounded linear operators acting on X. They [13] showed that, 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 (resp., AB=P, where P is a fixed nontrivial idempotent), then δ=τ+ν, where τ is a derivation of ℬ(X) and ν:ℬ(X)→ℂI is a linear map vanishing at commutators [A,B] with AB=0 (resp., AB=P). Later, this result was generalized to the additive maps on triangular algebras and prime rings in [14, 15], respectively. Let ℳ be a von Neumann algebra without central summands of type I1 and L:ℳ→ℳ an additive map. For ξ-Lie derivations, Qi and Hou [16] showed that L satisfies L([A,B]ξ)=[L(A),B]ξ+[A,L(B)]ξ for any A,B with AB=0 if and only if there exists an additive derivation φ such that (1) ξ=1, L=φ+f, where f:ℳ→𝒵(ℳ) is an additive map vanishing each commutator [A,B] whenever AB=0; (2) ξ=-1, L=φ; (3) ξ=0, L(A)=φ(A)+L(I)A for all A; (4) ξ∉{0,±1}, φ(ξA)=ξL(A), and L(A)=φ(A)+L(I)A for all A.Letℳ be any von Neumann algebra and let A∈ℳ. Recall that the central carrier of A, denoted by A¯, is the intersection of all central projections P such that PA=0. If A is self-adjoint, then the core of A, denoted by A_, is sup{S∈𝒵(ℳ):S=S*,S≤A}. Particularly, if A=P is a projection, it is clear that P_ is the largest central projection ≤P. A projection P is called core free if P_=0. It is easy to see that P_=0 if and only if I-P¯=I. By [17], if ℳ is a von Neumann algebra without central summands of type I1, there exists a nonzero core-free projection P∈ℳ with P¯=I. So, it is easily seen that ℳ is a von Neumann algebra without central summands of type I1 if and only if it has a projection P with P_=0 and P¯=I. Fixed such P. The purpose of the present paper is to give a complete characterization of additive maps L satisfying L([A,B]ξ)=[L(A),B]ξ+[A,L(B)]ξ for any A,B with AB=P on von Neumann algebras without central summands of type I1 for all possible ξ.Letℳ be a von Neumann algebra without central summands of type I1 and P∈ℳ a nonzero core-free projection with P¯=I. Assume that L:ℳ→ℳ is an additive map. In this paper, we prove that L satisfies [L(A),B]ξ+[A,L(B)]ξ=L([A,B]ξ) for any A,B∈ℳ with AB=P if and only if (1) ξ=1, L(A)=φ(A)+h(A) for all A∈ℳ, where φ:ℳ→ℳ is an additive derivation, h:ℳ→𝒵(ℳ) is an additive map vanishing each commutator [A,B] whenever AB=P, and 𝒵(ℳ) denotes the center of ℳ (Theorem 1); (2) ξ≠1, L is an additive derivation satisfying L(ξA)=ξL(A) for all A (Theorem 4).


## 2. Characterizing Lie Derivations by Local Action

In this section, we consider the question of characterizing Lie derivations by local action at a core-free projectionP with P¯=I on general von Neumann algebras having no central summands of type I1.Theorem 1.
Letℳ be a von Neumann algebra without central summands of type I1 and P∈ℳ a nonzero core-free projection with P¯=I. Suppose that L:ℳ→ℳ is an additive map. Then L satisfies L([A,B])=[L(A),B]+[A,L(B)] for any A,B∈ℳ with AB=P if and only if there exists an additive derivation φ:ℳ→ℳ and an additive map h:ℳ→𝒵(ℳ) vanishing each commutator [A,B] whenever AB=P such that L(A)=φ(A)+h(A) for all A∈ℳ, where 𝒵(ℳ) denotes the center of ℳ.We first give two lemmas, which are needed.Lemma 2 (see [17]).
Letℳ be a von Neumann algebra. For projections P,Q∈ℳ, if P¯=Q¯≠0 and P+Q=I, then T∈ℳ commutes with PXQ and QXP for all X∈ℳ implies T∈𝒵(ℳ).Lemma 3 (see [16]).
Letℳ be a von Neumann algebra. Assume that P∈ℳ is a projection with P_=0 and P¯=I. Then PℳP∩𝒵(ℳ)=(I-P)ℳ(I-P)∩𝒵(ℳ)={0}.Proof of Theorem1.
By the definitions of core and central carrier,I-P is also core free and I-P¯=I. For convenience, denote ℳij=PiℳPj, i,j∈{1,2}, where P1=P and P2=I-P. Then ℳ=ℳ11+ℳ12+ℳ21+ℳ22.
The “if” part is obvious. We will prove the “only if” part by checking several claims.
Claim  1. Consider P1L(P1)P1+P2L(P1)P2∈𝒵(ℳ). Since IP1=P1, we have 0=L([I,P1])=[L(I),P1]+[I,L(P1)], and so
(1)L(I)P1=P1L(I).
For anyA12∈ℳ12, by (I+A12)P1=P1, we get L([I+A12,P1])=[L(I)+L(A12),P1]+[I+A12,L(P1)]; that is,
(2)L(A12)+L(I)P1+L(A12)P1-P1L(I)-P1L(A12)+A12L(P1)-L(P1)A12=0.
Let L(A12)=S11+S12+S21+S22. Combining (2) with (1), one gets
(3)0=S11+A12L(P1)P1+A12P2L(P1)P2-P1L(P1)P1A12+2S21+S22-P2L(P1)P1A12.
It follows that A12P2L(P1)P2=P1L(P1)P1A12, which implies
(4)P1AP2(P2L(P1)P2+P1L(P1)P1)=(P2L(P1)P2+P1L(P1)P1)P1AP2
for all A∈ℳ.
Similarly, for anyA21∈ℳ21, by using the equation P1(I+A21)=P1, one can show that
(5)P2AP1(P2L(P1)P2+P1L(P1)P1)=(P2L(P1)P2+P1L(P1)P1)P2AP1
holds for all A∈ℳ. It follows from (4)-(5) and Lemma 2 that P1L(P1)P1+P2L(P1)P2∈𝒵(ℳ), as desired.
LetS=P1L(P1)P2-P2L(P1)P1 and define a map δ:ℳ→ℳ by δ(A)=L(A)+SA-AS for every A∈ℳ. It is easily checked that, for any A,B∈ℳ,
(6)AB=P⟹δ([A,B])=[δ(A),B]+[A,δ(B)].
Moreover, by Claim 1, we have
(7)δ(P1)=L(P1)-P2L(P1)P1-P1L(P1)P2=P1L(P1)P1+P2L(P1)P2∈𝒵(ℳ).
Claim  2. Consider δ(ℳij)⊆ℳij, 1≤i≠j≤2. Here we only give the proof for ℳ12. For the case ℳ21, the proof is similar.
For anyA12∈ℳ12, let δ(A12)=S11+S12+S21+S22. Since (I+A12)P1=P1, we have
(8)δ([I+A12,P1])=[δ(I)+δ(A12),P1]+[I+A12,δ(P1)].
Note that δ(I)=L(I). Then, by (1) and (7), the above equation reduces to S11+2S21+S22=0, which implies S11=S21=S22=0. Hence δ(A12)=S12∈ℳ12.
Claim  3. Consider  δ(I)∈𝒵(ℳ). Take any A12∈ℳ12. Since (P1-A12)(I+A12)=P1, we have
(9)δ([P1-A12,I+A12])=[δ(P1-A12),I+A12]+[P1-A12,δ(I+A12)].
By (7) and Claim 2, the above equation yields δ(I)A12=A12δ(I); that is,
(10)δ(I)P1AP2=P1AP2δ(I)∀A∈ℳ.
Similarly, by the equation(I+A21)(P1-A21)=P1, one can show that
(11)δ(I)P2AP1=P2AP1δ(I)holds∀A∈ℳ.
Equations (10)-(11) and Lemma 2 ensure that δ(I)∈𝒵(ℳ).
Note that, by (7) and Claim 3, one shows
(12)δ(P2)=δ(I)-δ(P1)∈𝒵(ℳ).
Claim  4. For any Aii∈ℳii, there exists some map hi:ℳii→𝒵(ℳ) such that δ(Aii)∈ℳii+hi(Aii), i=1,2.
Firstly, take any invertibleA11∈ℳ11 and let δ(A11)=S11+S12+S21+S22. Since A11-1A11=P1, we have
(13)0=δ([A11-1,A11])=[δ(A11-1),A11]+[A11-1,δ(A11)].
For anyB22∈ℳ22, let δ(B22)=T11+T12+T21+T22. Since P1(P1+B22)=P1, we have
(14)0=δ([P1,P1+B22])=[δ(P1),P1+B22]+[P1,δ(P1)+δ(B22)].
By (7), the above equation yields
(15)T12=T21=0.
Since(A11-1+B22)A11=P1, one gets
(16)0=δ([A11-1+B22,A11])=[δ(A11-1)+δ(B22),A11]+[A11-1+B22,δ(A11)],
which and (13) imply
(17)δ(B22)A11-A11δ(B22)+B22δ(A11)-δ(A11)B22=0hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh∀B22∈ℳ22.
Let B22=P2 in (17), and by (12), we get P2δ(A11)=δ(A11)P2, and so
(18)S12=S21=0.
Thus, combining (15) and (18), (17) reduces to T11A11-A11T11+B22S22-S22B22=0. It follows that
(19)T11A11=A11T11,B22S22=S22B22
holds for all invertible A11∈ℳ11 and all B22∈ℳ22.
Now taking anyA11∈ℳ11, there is a scalar c such that cP1-A11 is invertible. Then (19) implies T11A11=A11T11 for all A11∈ℳ11, and so T11∈𝒵(ℳ11). Note that, for T11∈𝒵(ℳ11), there must be some h2(B22)∈𝒵(ℳ) such that h2(B22)-T11∈𝒵(ℳ22). So
(20)δ(B22)=T11+T22=T22+(T11-h2(B22))+h2(B22)∈ℳ22+h2(B22).
On the other hand, by (19) and the arbitrariness of B22∈ℳ22, one can show that S22∈𝒵(ℳ22), and so δ(A11)=S11+S22∈ℳ11+𝒵(ℳ22) for all invertible A11∈ℳ11. Now consider any A11∈ℳ11. There is a scalar c such that cP1-A11 is invertible. Also note that cP1 is invertible in ℳ11. Hence δ(A11)=P1δ(A11)P1+P2δ(A11)P2∈ℳ11+𝒵(ℳ22) for all A11∈ℳ11. Again, for P2δ(A11)P2∈𝒵(ℳ22), there is some h1(A11)∈𝒵(ℳ) such that h1(A11)-P2δ(A11)P2∈𝒵(ℳ11). It follows that
(21)δ(A11)=P1δ(A11)P1+P2δ(A11)P2=P1δ(A11)P1+(P2δ(A11)P2-h1(A11))+h1(A11)∈ℳ11+h1(A11).
Claim 4 is true.
Now define two mapsh:ℳ→𝒵(ℳ) and d:ℳ→ℳ, respectively, by
(22)h(A)=h1(A11)+h2(A22),d(A)=δ(A)-h(A)
for all A=A11+A12+A21+A22∈ℳ. Then by Claims 2 and 4, we have
(23)d(ℳij)⊆ℳij,d(ℳii)⊆ℳii,d(ℳij)=δ(ℳij),1≤i≠j≤2.
Claim  5.  d and h are additive. By the definitions of d and h, we only need to verify that d is additive on ℳii(i=1,2). In fact, for any A11,B11∈ℳ11, we have
(24)d(A11+B11)+h(A11+B11)=δ(A11+B11)=δ(A11)+δ(B11)=d(A11)+h(A11)+d(B11)+h(B11)=d(A11)+d(B11)+(h(A11)+h(B11)).
That is,
(25)d(A11+B11)-(d(A11)+d(B11))=h(A11)+h(B11)-h(A11+B11)∈𝒵(ℳ).
Since ℳ11∩𝒵(ℳ)={0} by Lemma 3, one sees d(A11+B11)=d(A11)+d(B11), and consequently, h(A11)+h(B11)=h(A11+B11).
Similarly, one can prove thatd is additive on ℳ22.
Claim  6.  d is a derivation; that is, d(AB)=d(A)B+Ad(B) for all A,B∈ℳ.
We will complete the proof of the claim by three steps.
Step  1. For any Aii∈ℳii, Aij,Bij∈ℳij, and Bjj∈ℳjj, we have d(AiiBij)=d(Aii)Bij+Aiid(Bij) and d(AijBjj)=d(Aij)Bjj+Aijd(Bjj), 1≤i≠j≤2.
In fact, for any invertibleA11∈ℳ11 and any B12∈ℳ12, since (A11-1+B12)A11=P1, by (23) and (13), we have
(26)-d(A11B12)=-δ(A11B12)=δ([A11-1+B12,A11])=[δ(A11-1)+δ(B12),A11]+[A11-1+B12,δ(A11)]=[δ(A11-1),A11]+[δ(B12),A11]+[A11-1,δ(A11)]+[B12,δ(A11)]=[d(B12),A11]+[B12,d(A11)]=-A11d(B12)-d(A11)B12.
That is,
(27)d(A11B12)=d(A11)B12+A11d(B12)
for any invertible A11∈ℳ11 and all B12∈ℳ12.
Take anyA11∈ℳ11. There is a scalar c such that cP1-A11 is invertible. Note that cP1 is also invertible in ℳ11. So by (27) one gets
(28)d(cP1B12)=d(cP1)B12+cP1d(B12)d((cP1-A11)B12)=d(cP1-A11)B12+(cP1-A11)d(B12).
Hence, d(A11B12)=d(A11)B12+A11d(B12) holds for all A11∈ℳ11 and all B12∈ℳ12.
Similarly, byB11(A21+B11-1)=P1, one can show that d(A21B11)=d(A21)B11+A21d(B11) holds for all A21∈ℳ21 and B11∈ℳ11; by (P1+A22-A22B21)(P1+B21)=P1, one can show that d(A22B21)=d(A22)B21+A22d(B21) holds for all A22∈ℳ22 and B21∈ℳ21; and by using the relation (P1+A12)(P1+B22-A12B22)=P1, one can obtain that d(A12B22)=d(A12)B22+A12d(B22) holds for all A12∈ℳ12 and B22∈ℳ22.
Step  2. For any Aii,Bii∈ℳii, we have d(AiiBii)=d(Aii)Bii+Aiid(Bii), i=1,2.
Leti≠j. For any Aii,Bii∈ℳii and any Sij∈ℳij, by Step 1, on the one hand, we have
(29)d(AiiBiiSij)=d(AiiBii)Sij+AiiBiid(Sij);
on the other hand,
(30)d(AiiBiiSij)=d(Aii)BiiSij+Aiid(BiiSij)=d(Aii)BiiSij+Aiid(Bij)Sij+AiiBiid(Sij).
Comparing the above two equations gets that (d(AiiBii)-d(Aii)Bii-Aiid(Bii))Sij=0 holds for all Sij∈ℳij; that is,
(31)(d(AiiBii)-d(Aii)Bii-Aiid(Bii))PiSPj=0
holds for all S∈ℳ. Note that Pj¯=I. It follows from the definition of the central carrier that span{TPj(x):T∈ℳ,x∈H} is dense in H. It follows from (23) that d(AiiBii)=d(Aii)Bii+Aiid(Bii) holds for all Aii,Bii∈ℳii, as desired.
Note that, by using Step 2 and the fact thatd(ℳii)⊆ℳii (i=1,2), one can get
(32)d(P1)=d(P2)=0.
Step  3. Consider d(AijBji)=d(Aij)Bji+Aijd(Bji) for all Aij∈ℳij and Bji∈ℳji, 1≤i≠j≤2.
Take anyA12∈ℳ12 and B21∈ℳ21. Since (P1+A12-A12B21)(P1+B21)=P1, by the definition of d, Claim 5, (23), and (32), we have
(33)d(A12B21-A12-B21-B21A12+B21A12B21)+h(A12B21-B21A12)=δ(A12B21-A12-B21-B21A12+B21A12B21)=δ([P1+A12-A12B21,P1+B21])=[δ(P1+A12-A12B21),P+B21]+[P1+A12-A12B21,δ(P1+B21)]=[d(P1+A12-A12B21),P1+B21]+[P1+A12-A12B21,d(P1+B21)]=d(A12)B21-d(A12)-B21d(A12)+B21d(A12B21)+A12d(B21)-d(B21)+d(B21)A12B21-d(B21)A12.
It follows from Step 1 that
(34)d(A12B21-B21A12)+h(A12B21-B21A12)=d(A12)B21-B21d(A12)+A12d(B21)-d(B21)A12.
Multiplying by A12 from the left side and the right side, respectively, in (34) and applying (23), we get
(35)A12d(B21A12)-h(A12B21-B21A12)A12=A12B21d(A12)+A12d(B21)A12,d(A12B21)A12+h(A12B21-B21A12)A12=d(A12)B21A12+A12d(B21)A12.
These two equations, together with Step 1, yield
(36)-d(A12)B21A12-h(A12B21-B21A12)A12=-d(A12B21)A12+A12d(B21)A12,-A12B21d(A12)+h(A12B21-B21A12)A12=-A12d(B21A12)+A12d(B21)A12.
Comparing the above two equations, one achieves
(37)h(A12B21-B21A12)A12=0.
Similarly, multiplying byB21 from the left side and the right side, respectively, in (34), one can verify
(38)h(A12B21-B21A12)B21=0.
Next, we will proveh(A12B21-B21A12)=0. To do this, for any A12, let A12=V|A12| be its polar decomposition. Then (37) implies h(A12B21-B21A12)|A12|=0, and so |A12|h(A12B21-B21A12)*=0. It follows that
(39)A12h(A12B21-B21A12)*=V|A12|h(A12B21-B21A12)*=0.
Similarly, one can show
(40)B21h(A12B21-B21A12)*=0.
Multiplying by h(A12B21-B21A12)* from the right side in (34), by using (39)-(40), we get
(41)d(A12B21-B21A12)h(A12B21-B21A12)*+h(A12B21-B21A12)h(A12B21-B21A12)*=0.
Note that, by Step 2, (23), and (39)-(40), we have
(42)d(A12B21)h(A12B21-B21A12)*=d(A12B21P1h(A12B21-B21A12)*P1)-A12B21d(P1h(A12B21-B21A12)*P1)=-A12B21d(P1h(A12B21-B21A12)*P1),(43)d(B21A12)h(A12B21-B21A12)*=d(B21A12P2h(A12B21-B21A12)*P2)-B21A12d(P2h(A12B21-B21A12)*P2)=-B21A12d(P2h(A12B21-B21A12)*P2).
Hence, (41) implies
(44)h(A12B21-B21A12)*×h(A12B21-B21A12)h(A12B21-B21A12)*=0,
and so h(A12B21-B21A12)=0. Thus, (34) reduces to
(45)d(A12B21-B21A12)=d(A12)B21-B21d(A12)+A12d(B21)-d(B21)A12,
which implies that d(A12B21)=d(A12)B21+A12d(B21) and d(B21A12)=B21d(A12)+d(B21)A12 hold for all A12∈ℳ12 and B21∈ℳ21, as desired.
Now combining Steps  1–3, it is easily checked thatd is an additive derivation.
Claim  7. Consider h([A,B])=0 for all A,B∈ℳ with AB=P1.
In fact, for anyA,B∈ℳ with AB=P1, we have
(46)h([A,B])=δ([A,B])-d([A,B])=[δ(A),B]+[A,δ(B)]-d(AB-BA)=[d(A)-h(A),B]+[A,d(B)-h(B)]-d(AB-BA)=[d(A),B]+[A,d(B)]-d(AB-BA)=0.
Claim  8. The theorem holds.
Indeed, letφ(A)=d(A)-(SA-AS) for all A∈ℳ; then, by the definitions of δ and d, we have L(A)=φ(A)+h(A) for all A∈ℳ. It is easy to check that φ is an additive derivation on ℳ.
The proof is finished.


## 3. Characterizingξ-Lie Derivations by Local Action

In this section, we will give a complete characterization of additiveξ-Lie derivations for ξ≠1 by local action at some core-free projection P with P¯=I on general von Neumann algebras having no central summands of type I1.The following is the main result of this section.Theorem 4.
Letℳ be a von Neumann algebra without central summands of type I1 and P∈ℳ a nonzero core-free projection with P¯=I. Suppose that L:ℳ→ℳ is an additive map and ξ is a scalar with ξ≠1. Then L satisfies L([A,B]ξ)=[L(A),B]ξ+[A,L(B)]ξ for any A,B∈ℳ with AB=P if and only if L is an additive derivation and L(ξA)=ξL(A) for all A∈ℳ.Proof.
Still, the “if” part is clear. For the “only if” part, we use the same symbols as that in the proof of Theorem1. In the sequel, we always assume that ξ≠1 and L:ℳ→ℳ is an additive map satisfying L([A,B]ξ)=[L(A),B]ξ+[A,L(B)]ξ for A,B∈ℳ with AB=P=P1. We will prove the “only if” part by several claims.
Claim  1.  Consider P1L(I)P2=P2L(I)P1=0 and P2L(P1)P2=P1L(P2)P1=0. Since P1P1=P1, we have L([P1,P1]ξ)=[L(P1),P1]ξ+[P1,L(P1)]ξ; that is,
(47)L(P1)-L(ξP1)=L(P1)P1-ξP1L(P1)+P1L(P1)-ξL(P1)P1.
Multiplying by Pi and Pj(1≤i≠j≤2) from the left and the right sides, respectively, in (47), one gets
(48)PiL(ξP1)Pj=ξPiL(P1)Pj,1≤i≠j≤2.
AsP1I=IP1=P1, L([P1,I]ξ)=[L(P1),I]ξ+[P1,L(I)]ξ, and L([I,P1]ξ)=[L(I),P1]ξ+[I,L(P1)]ξ, that is,
(49)L(P1)-L(ξP1)=L(P1)-ξL(P1)+P1L(I)-ξL(I)P1,(50)L(P1)-L(ξP1)=L(I)P1-ξP1L(I)+L(P1)-ξL(P1).
Multiplying by P1 and P2 from the left and the right sides, respectively, in (49) and combining with (48), one gets P1L(I)P2=0; multiplying by P2 and P1 from the left and the right sides, respectively, in (50) and combining with (48), one gets P2L(I)P1=0.
On the other hand, since(P1-P2)P1=P1, we have L([P1-P2,P1]ξ)=[L(P1)-L(P2),P1]ξ+[P1-P2,L(P1)]ξ, which and (47) yield
(51)L(P2)P1-ξP1L(P2)+P2L(P1)-ξL(P1)P2=0.
This implies (1-ξ)P1L(P2)P1=0 and (1-ξ)P2L(P1)P2=0. It follows from the fact that ξ≠1 that P1L(P2)P1=0 and P2L(P1)P2=0. Claim 1 is true.
Define a mapδ:ℳ→ℳ by δ(A)=L(A)+SA-AS for each A∈ℳ, where S=P1L(P1)P2-P2L(P1)P1. It is easily verified that δ is also an additive map satisfying
(52)δ([A,B]ξ)=[δ(A),B]ξ+[A,δ(B)]ξforanyA,B∈ℳwithAB=P=P1.
Moreover, P1δ(I)P2=P2δ(I)P1=P2δ(P1)P2=P1δ(P2)P1=0 by Claim 1. Thus we obtain
(53)δ(P1)=L(P1)+SP1-P1S=P1L(P1)P1=P1δ(P1)P1-P1(SP1-P1S)P1=P1δ(P1)P1∈ℳ11,(54)δ(P2)=L(P2)+SP2-P2S=P2L(P2)P2=P2δ(P2)P2-P2(SP2-P2S)P2=P2δ(P2)P2∈ℳ22.
Note that (47) is also true for the map δ. So, by (53), one can also get
(55)δ(P1)+δ(ξP1)-2ξδ(P1)=0.
Claim  2. Consider δ(ℳii)⊆ℳii,i=1,2.  For any A22∈ℳ22, since P1(P1+A22)=P1, we have
(56)δ([P1,P1+A22]ξ)=[δ(P1),P1+A22]ξ+[P1,δ(P1)+δ(A22)]ξ.
By (53) and (55), the above equation reduces to
(57)P1δ(A22)-ξδ(A22)P1=0.
Multiplying by P2 from the right side in (57), one gets P1δ(A22)P2=0; multiplying by P1 from both sides in (57), one gets P1δ(A22)P1=0.
Similarly, by(P1+A22)P1=P1, one can show P2δ(A22)P1=0. Thus, we have proved that δ(A22)∈ℳ22 for all A22∈ℳ22.
Next, take any invertibleA11∈ℳ11. By A11A11-1=P1 and A11(A11-1+P2)=P1, we get
(58)δ(P1)-δ(ξP1)=δ([A11,A11-1]ξ)=[δ(A11),A11-1]ξ+[A11,δ(A11-1)]ξ,δ(P1)-δ(ξP1)=δ([A11,A11-1+P2]ξ)=[δ(A11),A11-1+P2]ξ+[A11,δ(A11-1)+δ(P2)]ξ.
Comparing the two above equations, and by (54), one has δ(A11)P2-ξP2δ(A11)=0, which implies P1δ(A11)P2=P2δ(A11)P2=0 as ξ≠1.
Similarly, byA11-1A11=P1 and (A11-1+P2)A11=P1, one can prove P2δ(A11)P1=0.
Thus, we have proved thatδ(A11)∈ℳ11 if A11∈ℳ11 is invertible. Now for any A11∈ℳ11, we can find a scalar c such that cP1-A11 is invertible in ℳ11. It follows from the preceding case that δ(cP1-A11), δ(cP1)∈ℳ11. Therefore, δ(A11)=δ(cP1)-δ(cP1-A11)∈ℳ11 for each A11∈ℳ11. Claim 2 holds.
Claim  3. For any Aij∈ℳij, 1≤i≠j≤2, we have δ(Pi)Aij=Aijδ(Pj). Moreover, the following statements hold.(1)
Ifξ≠-1, then δ(Aij)∈ℳij.(2)
Ifξ=-1, then P1δ(Aij)P1=P2δ(Aij)P2=0.
First, consider the caseA12∈ℳ12. Since (P1+A12)P1=P1, we have
(59)δ([P1+A12,P1]ξ)=[δ(P1)+δ(A12),P1]ξ+[P1+A12,δ(P1)]ξ.
By (53) and (55), we get
(60)-δ(ξA12)=δ(A12)P1-ξP1δ(A12)-ξδ(P1)A12.
Multiplying by P2 and P1 from the left and the right sides, respectively, in (60), we get P2δ((1+ξ)A12)P1=0. Since A12 is arbitrary, replacing A12 by (1+ξ)-1A12 if ξ≠-1, one achieves
(61)P2δ(A12)P1=0forthecaseξ≠-1.
Since(P1+A12)(I-A12)=P1, we have
(62)δ([P1+A12,I-A12]ξ)=[δ(P1)+δ(A12),I-A12]ξ+[P1+A12,δ(I)-δ(A12)]ξ.
That is,
(63)δ(P1)-δ(ξP1)-δ(ξA12)=δ(P1)-δ(P1)A12+δ(A12)-δ(A12)A12-ξ(δ(P1)+δ(A12)-A12δ(P1)-A12δ(A12))+P1δ(I)-P1δ(A12)+A12δ(I)-A12δ(A12)-ξ(δ(I)P1+δ(I)A12-δ(A12)P1-δ(A12)A12).
Since(P1-A12)(I+A12)=P1, we have
(64)δ([P1-A12,I+A12]ξ)=[δ(P1)-δ(A12),I+A12]ξ+[P1-A12,δ(I)+δ(A12)]ξ.
That is,
(65)δ(P1)-δ(ξP1)+δ(ξA12)=δ(P1)+δ(P1)A12-δ(A12)-δ(A12)A12-ξ(δ(P1)-δ(A12)+A12δ(P1)-A12δ(A12))+P1δ(I)+P1δ(A12)-A12δ(I)-A12δ(A12)-ξ(δ(I)P1-δ(I)A12+δ(A12)P1-δ(A12)A12).
Comparing (63)-(65) and by (53)-(54), one can obtain
(66)-δ(ξA12)=-δ(P1)A12+δ(A12)-ξδ(A12)-P1δ(A12)+A12δ(P2)-ξδ(P1)A12+ξδ(A12)P1.
Then, (60) and (66) yield
(67)(1-ξ)δ(A12)P1+(1-ξ)P1δ(A12)-(1-ξ)δ(A12)+δ(P1)A12-A12δ(P2)=0.
Multiplying by Pi(i=1,2) from both sides in (67), and by (53)-(54), we have
(68)P1δ(A12)P1=0,P2δ(A12)P2=0;
multiplying by P1 and P2 and from the left and the right sides, respectively, in (67), we have
(69)δ(P1)A12=A12δ(P2).
Now combining (61) and (68)-(69), we achieve that the claim holds for any A12∈ℳ12.
For anyA21∈ℳ21, by the relations P1(P1+A21)=P1, (I-A21)(P1+A21)=P1, and (I+A21)(P1-A21)=P1, and by using a similar argument to that of the above, one can show that
(70)-δ(ξA21)=P1δ(A21)-ξδ(A21)P1-ξA21δ(P1),
and the claim also holds for A21.
Claim  4.  L is an additive derivation with L(ξA)=ξL(A) for each A∈ℳ.
We will prove the claim by considering three cases.
Case  1. Consider ξ≠0,-1. In this case, for any Aij∈ℳij, 1≤i≠j≤2, by Claim 3(1), (60), and (70), we get
(71)δ(ξAij)=ξδ(Aij)+ξδ(Pi)Aij.
We will complete the proof of Claim 5 by the following several steps.
Step    1. For any Aii∈ℳii, Bjj∈ℳjj, Aij,Bij∈ℳij, 1≤i≠j≤2, the following statements hold:(i)
δ(ξAiiBij)=ξδ(Aii)Bij+ξAiiδ(Bij);(ii)
δ(ξAijBjj)=ξδ(Aij)Bjj+ξAijδ(Bjj);(iii)
δ(AiiBij)=δ(Aii)Bij+Aiiδ(Bij)-δ(I)AiiBij;(iv)
δ(AijBjj)=δ(Aij)Bjj+Aijδ(Bjj)-δ(I)AijBjj.
For anyA11∈ℳ11, that is, invertible, and any B12∈ℳ12, since (A11-1+B12)A11=P1, by Claim 2, Claim 3(1), and (58), we get δ(ξA11B12)=ξδ(A11)B12+ξA11δ(B12). Now for any A11∈ℳ11, note that there exists some c such that cP1-A11 is invertible in ℳ11. It is easy to check that
(72)δ(ξA11B12)=ξδ(A11)B12+ξA11δ(B12)holds∀A11∈ℳ11,B12∈ℳ12.
Also note that, by (71), δ(ξA11B12)=ξδ(A11B12)+ξδ(P1)A11B12. As ξ≠0, this and (72) yield
(73)δ(A11B12)=δ(A11)B12+A11δ(B12)-δ(P1)A11B12.
For anyB11∈ℳ11 and A21∈ℳ21, one can similarly check that δ(ξA21B11)=ξδ(A21)B11+ξA21δ(B11) and δ(A21B11)=δ(A21)B11+A21δ(B11)-A21B11δ(I); for any A22,B22∈ℳ22, A12∈ℳ12, and B21∈ℳ21, since (P1+A12)(P1+B22-A12B22)=P1 and (P1+A22-A22B21)(P1+A21)=P1, by the definition of δ, Claim 3(1), (55), and (71), one can check that Step 1 holds.
Step  2. For any Aii∈ℳii, we have δ(ξAiiBii)=ξδ(Aii)Bii+ξAiiδ(Bii), i=1,2.
In fact, leti≠j. For any Aii,Bii∈ℳii and any Sij∈ℳij, by (71) and Step 1, on the one hand, we have
(74)δ(ξAiiBiiSij)=ξδ(AiiBiiSij)+ξδ(Pi)AiiBiiSij=ξδ(AiiBii)Sij+ξAiiBiiδ(Sij);
on the other hand,
(75)δ(ξAiiBiiSij)=ξδ(AiiBiiSij)+ξδ(Pi)AiiBiiSij=ξδ(Aii)BiiSij+ξAiiδ(BiiSij)=ξδ(Aii)BiiSij+ξAiiδ(ξξ-1BiiSij)=ξδ(Aii)BiiSij+ξ2Aiiδ(ξ-1Bii)Sij+ξAiiBiiδ(Sij).
The above two equations yield that (δ(AiiBii)-δ(Aii)Bii-ξAiiδ(ξ-1Bii))Sij=0 holds for all Aii,Bii∈ℳii and all Sij∈ℳij, which implies that (δ(ξAiiBii)-ξδ(Aii)Bii-ξAiiδ(Bii))SPj=0 holds for all S∈ℳ. It follows that δ(ξAiiBii)=ξδ(Aii)Bii+ξAiiδ(Bii) as Pj¯=I. Step 2 is true.
Step  3. Consider δ(I)=L(I)=0 and δ(ξA)=ξδ(A) for all A∈ℳ.
TakingA11=B11=P1 in Step 2, one obtains δ(ξP1)=2ξδ(P1), which and (55) yield δ(P1)=0. Note that δ(P2)A21=A21δ(P1) for each A21 (Claim 3). It follows that δ(P2)A21=δ(P2)P2AP1=0, which implies δ(P2)=0. So δ(I)=δ(P1)+δ(P2)=0. Now it is obvious that δ(ξA)=ξδ(A) for all A∈ℳ by (71) and Step 2.
Step  4. For any Aii,Bii∈ℳii, we have δ(AiiBii)=δ(Aii)Bii+Aiiδ(Bii), i=1,2.
Leti≠j. Take any Aii,Bii∈ℳii and any Sij∈ℳij. By Step 1, one gets
(76)δ(AiiBiiSij)=δ(AiiBii)Sij+AiiBiiδ(Sij),δ(AiiBiiSij)=δ(Aii)BiiSij+Aiiδ(BiiSij)=δ(Aii)BiiSij+Aiiδ(Bii)Sij+AiiBiiδ(Sij).
Comparing the above two equations, and by Claims 2 and 4, one obtains that
(77)(δ(AiiBii)-δ(Aii)Bii-Aiiδ(Bii))SPj=0
holds for all S∈ℳ. It follows from the fact P¯j=I that δ(AiiBii)=δ(Aii)Bii+Aiiδ(Bii).
Step  5. For any Aij∈ℳij and Bji∈ℳji, we have δ(AijBji)=δ(Aij)Bji+Aijδ(Bji), 1≤i≠j≤2.
Take anyAij∈ℳij and Bji∈ℳji. Since (P1+A12-A12B21)(P1+B21)=P1, by Claim 3(1), Steps 1 and 3, we have
(78)ξδ(B21A12)+δ(A12)B21-δ(A12B21)-ξB21δ(A12)+A12δ(B21)-ξδ(B21)A12=0.
It follows that δ(A12B21)=δ(A12)B21+A12δ(B21) and δ(B21A12)=δ(B21)A12+B21δ(A12) as ξ≠0.
Step  6. For any A,B∈ℳ, we have δ(AB)=δ(A)B+Aδ(B). Furthermore, L is an additive derivation.
For anyA=A11+A12+A21+A22, B=B11+B12+B21+B22∈ℳ, by the additivity of δ, and Steps 1–5, one can easily check that δ(AB)=δ(A)B+Aδ(B). Note that L(A)=δ(A)-SA+AS for all A∈ℳ. So L is an additive derivation.
Case  2. Consider ξ=-1. In this case, by (55), we get δ(P1)=0. Then (58) implies
(79)[δ(A11-1),A11]-1+[A11-1,δ(A11)]-1=0hhhhhhhhhforallinvertibleA11∈ℳ11.
Step  1. For any A12∈ℳ12, A21∈ℳ12 and A22∈ℳ22, we have δ(A12A22)=δ(A12)A22+A12δ(A22)+A22δ(A12) and δ(A22A21)=δ(A22)A21+A22δ(A21)+δ(A21)A22.
Take anyA12∈ℳ12 and A22∈ℳ22. Since (P1+A12)(P1+A22-A12A22)=P1, by Claim 2 and Claim 3(2), one can verify
(80)δ(A12)=δ([P1+A12,P1+A22-A12A22]-1)=[δ(A12),P1+A22-A12A22]-1+[P1+A12,δ(A22)-δ(A12A22)]-1=δ(A12)P1+δ(A12)A22-δ(A12)A12A22+P1δ(A12)+A22δ(A12)-A12A22δ(A12)-P1δ(A12A22)+A12δ(A22)-A12δ(A12A22)-δ(A12A22)P1-δ(A12A22)A12.
Multiplying by P1 and P2 from the left and the right sides, respectively, in (80), and by Claim 3(2), we have
(81)P1δ(A12A22)P2=P1δ(A12)A22+A12δ(A22)=δ(A12)A22+A12δ(A22);
multiplying by P2 and P1 from the left and the right sides, respectively, in (80), and by Claim 3(2) again, we have
(82)P2δ(A12A22)P1=A22δ(A12)P1=A22δ(A12).
So
(83)δ(A12A22)=P1δ(A12A22)P2+P2δ(A12A22)P1=δ(A12)A22+A12δ(A22)+A22δ(A12).
For anyA21∈ℳ12 and A22∈ℳ22, by the same argument as above for the equation (P1+A22-A22A21)(P1+A21)=P1, one can check that
(84)δ(A22A21)=δ(A22)A21+A22δ(A21)+δ(A21)A22.
Step  2. For any A21∈ℳ21, A12∈ℳ12, and A11∈ℳ11, we have δ(A11A12)=δ(A11)A12+A11δ(A12)+δ(A12)A11 and δ(A21A11)=δ(A21)A11+A21δ(A11)+A11δ(A21).
For any invertibleA11∈ℳ11 and any A12∈ℳ12, since (A11-1+A12)A11=P1, by Claim 2 and (79), one obtains
(85)δ(A11A12)=δ([A11-1+A12,A11]-1)=[δ(A11-1)+δ(A12),A11]-1+[A11-1+A12,δ(A11)]-1=δ(A11)A12+A11δ(A12)+δ(A12)A11.
Now it is easy to check that
(86)δ(A11A12)=δ(A11)A12+A11δ(A12)+δ(A12)A11holds∀A11∈ℳ11,A12∈ℳ12.
For any invertibleA11∈ℳ11 and any A21∈ℳ21, as A11(A21+A11-1)=P1, by a similar argument to that of the above, one can check that
(87)δ(A21A11)=δ(A21)A11+A21δ(A11)+A11δ(A21).
Step  3. For any Aii∈ℳii, we have δ(Aii2)=δ(Aii)Aii+Aiiδ(Aii), i=1,2.
Let1≤i≠j≤2. Then, for any Aii∈ℳii and Sij∈ℳij, calculating δ(AiiAiiSij) by two different ways, one can easily check that Step 3 is true.
Step  4. For any A12∈ℳ12 and A21∈ℳ21, we have δ(A12A21)=δ(A12)A21+A12δ(A21) and δ(A21A12)=δ(A21)A12+A21δ(A12).
In fact, since(P1+A12-A12A21)(P1+A21)=P1, by Claim 2, Claim 3(2), and Step 2, one can verify
(88)δ(A12)-δ(A12A21)+δ(A21)+δ(A21A12)=δ([P1+A12-A12A21,P1+A21]-1)=[δ(A12)-δ(A12A21),P1+A21]-1+[P1+A12-A12A21,δ(A21)]-1=δ(A12)P1+δ(A12)A21-δ(A12A21)+P1δ(A12)-δ(A12A21)+A21δ(A12)+P1δ(A21)+A12δ(A21)+δ(A21)P1+δ(A21)A12.
Multiplying by Pi (i=1,2) from both sides in the above equation, one obtains
(89)δ(A12A21)=P1δ(A12)A21+A12δ(A21)P1=δ(A12)A21+A12δ(A21),(90)δ(A21A12)=P2δ(A21)A12+A21δ(A12)P2=δ(A21)A12+A21δ(A12).
Step  5. For any A∈ℳ, we have δ(A2)=δ(A)A+Aδ(A). Furthermore, L is an additive derivation.
For anyA=A11+A12+A21+A22∈ℳ, by the additivity of δ and Steps 1–4, it is easy to verify δ(A2)=δ(A)A+Aδ(A). Note that L(A)=δ(A)+AS-SA for each A. So L is an additive Jordan derivation. Now by [18], L is an additive derivation because von Neumann algebras are semiprime and 2-torsion free.
Case  3. Consider ξ=0. In this case, δ in fact satisfies AB=P⇒δ(A)B+Aδ(B)=δ(P). So, by (55), δ(P1)=0. Then, (58) yields
(91)δ(A11)A11-1+A11δ(A11-1)=0.
Step  1. For any A12∈ℳ12, B21∈ℳ21, and A22,B22∈ℳ22, we have δ(A12B22)=δ(A12)B22+A12δ(B22) and δ(A22B21)=δ(A22)B21+A22δ(B21).
Note that(P1+A12)(P1+B22+A12B22)=P1 and (P1+A22-A22B21)(P1+B21)=P1. By Claim 3(1), it is easy to check that Step 1 is true.
Step  2. For any A22,B22∈ℳ22, we have δ(A22B22)=δ(A22)B22+A22δ(B22).
Take anyA22,B22∈ℳ22 and any S21. By using Step 1 and calculating δ(A22B22S21) by two different ways, Step 2 is easily checked.
Note that Step 2 impliesδ(P2)=0.
Step  3. For any A11,B11∈ℳ11, A21∈ℳ21, and B12∈ℳ12, we have δ(A11B12)=δ(A11)B12+A11δ(B12) and δ(A21B11)=δ(A21)B11+A21δ(B11).
First, for any invertibleA11,B11∈ℳ11 and any A21∈ℳ21, B12∈ℳ12, since (A11-A11B12)(A11-1+B12+P2)=P1 and (B11-1+A21+P2)(B11-A21B11)=P1, by Claim 2, Claim 3(1), the fact δ(P2)=0, and (91), one can obtain that Step 3 holds. Note that, for any A11∈ℳ11, there exists a scalar c such that cP1-A11 is invertible in ℳ11. It is easy to check that Step 3 also holds for all A11 and B11∈ℳ11.
Step  4. For any A11,B11∈ℳ11, we have δ(A11B11)=δ(A11)B11+A11δ(B11).
Take anyA11,B11∈ℳ11 and any S12∈ℳ12. By Step 3, calculating δ(A11B11S12) by two different ways, Step 4 is true.
Step  5. For any A12∈ℳ12 and B21∈ℳ21, we have δ(A12B21)=δ(A12)B21+A12δ(B21) and δ(B21A12)=B21δ(A12)+δ(B21)A12.
For anyA12∈ℳ12 and B21∈ℳ21, since (P1+A12-A12B21)(P1+B21)=P1, by Claim 2, Claim 3(1), and the fact δ(P1)=0, we get δ(A12B21)=δ(A12)B21+A12δ(B21). Thus, by Steps 1 and 3, for any C12∈ℳ, we have
(92)δ(B21A12)C21+B21A12δ(C21)=δ(B21A12C21)=δ(B21)A12C21+B21δ(A12C21)=δ(B21)A12C21+B21δ(A12)C21+B21A12δ(C21),
which implies (δ(B21A12)-δ(B21)A12-B21δ(A12))CP1=0 for all C∈ℳ. It follows from the fact P¯1=I that δ(B21A12)=δ(B21)A12+B21δ(A12). So Step 5 is true.
Now by Steps1–5 and similar arguments as that in the proof of Theorem 1, one can check that δ is a derivation. As L(A)=δ(A)-SA+AS for each A, L is also a derivation.
The proof of the theorem is finished.By Theorem4, we can give different characterizations of additive ξ-Lie derivations on von Neumann algebras.Corollary 5.
Letℳ be a von Neumann algebra without central summands of type I1. Suppose that L:ℳ→ℳ is an additive map and ξ is a scalar with ξ≠1. Then, the following statements are equivalent.(1)
L is a ξ-Lie derivation.(2)
L is a derivation and L(ξA)=ξL(A) for each A∈ℳ.(3)
L satisfies L([A,B]ξ)=[L(A),B]ξ+[A,L(B)]ξ for any A,B∈ℳ with AB=P, where P is some fixed projection with P_=0 and P¯=I.



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*Source: 407427-2013-12-29.xml*