Characterizing Derivations on Von Neumann Algebras by Local Actions

Abstract

Let ℳ 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.

1. Introduction

Let 𝒜 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.

Let ℳ 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.

Let ℳ 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 projection 𝑃 with 𝑃=𝐼 on general von Neumann algebras having no central summands of type 𝐼₁.

Theorem 1. Let ℳ 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]). Let ℳ 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]). Let ℳ 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 Theorem 1. By the definitions of core and central carrier, 𝐼−𝑃 is also core free and 𝐼−𝑃=𝐼. For convenience, denote ℳ_𝑖𝑗=𝑃_𝑖ℳ𝑃_𝑗, 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 𝐿(𝐼)𝑃₁=𝑃₁𝐿(𝐼). (1)
For any 𝐴₁₂∈ℳ₁₂, by (𝐼+𝐴₁₂)𝑃₁=𝑃₁, we get 𝐿([𝐼+𝐴₁₂,𝑃₁])=[𝐿(𝐼)+𝐿(𝐴₁₂),𝑃₁]+[𝐼+𝐴₁₂,𝐿(𝑃₁)]; that is, 𝐿𝐴₁₂+𝐿(𝐼)𝑃₁+𝐿𝐴₁₂𝑃₁−𝑃₁𝐿(𝐼)−𝑃₁𝐿𝐴₁₂+𝐴₁₂𝐿𝑃₁−𝐿𝑃₁𝐴₁₂=0. (2) Let 𝐿(𝐴₁₂)=𝑆₁₁+𝑆₁₂+𝑆₂₁+𝑆₂₂. Combining (2) with (1), one gets 0=𝑆₁₁+𝐴₁₂𝐿𝑃₁𝑃₁+𝐴₁₂𝑃₂𝐿𝑃₁𝑃₂−𝑃₁𝐿𝑃₁𝑃₁𝐴₁₂+2𝑆₂₁+𝑆₂₂−𝑃₂𝐿𝑃₁𝑃₁𝐴₁₂. (3) It follows that 𝐴₁₂𝑃₂𝐿(𝑃₁)𝑃₂=𝑃₁𝐿(𝑃₁)𝑃₁𝐴₁₂, which implies 𝑃₁𝐴𝑃₂𝑃₂𝐿𝑃₁𝑃₂+𝑃₁𝐿𝑃₁𝑃₁=𝑃₂𝐿𝑃₁𝑃₂+𝑃₁𝐿𝑃₁𝑃₁𝑃₁𝐴𝑃₂ (4) for all 𝐴∈ℳ.
Similarly, for any 𝐴₂₁∈ℳ₂₁, by using the equation 𝑃₁(𝐼+𝐴₂₁)=𝑃₁, one can show that 𝑃₂𝐴𝑃₁𝑃₂𝐿𝑃₁𝑃₂+𝑃₁𝐿𝑃₁𝑃₁=𝑃₂𝐿𝑃₁𝑃₂+𝑃₁𝐿𝑃₁𝑃₁𝑃₂𝐴𝑃₁ (5) holds for all 𝐴∈ℳ. It follows from (4)-(5) and Lemma 2 that 𝑃₁𝐿(𝑃₁)𝑃₁+𝑃₂𝐿(𝑃₁)𝑃₂∈𝒵(ℳ), as desired.
Let 𝑆=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, 𝐴𝐵=𝑃⟹𝛿([𝐴,𝐵])=[𝛿(𝐴),𝐵]+[𝐴,𝛿(𝐵)]. (6) Moreover, by Claim 1, we have 𝛿𝑃₁=𝐿𝑃₁−𝑃₂𝐿𝑃₁𝑃₁−𝑃₁𝐿𝑃₁𝑃₂=𝑃₁𝐿𝑃₁𝑃₁+𝑃₂𝐿𝑃₁𝑃₂∈𝒵(ℳ). (7)
Claim  2. Consider 𝛿(ℳ_𝑖𝑗)⊆ℳ_𝑖𝑗, 1≤𝑖≠𝑗≤2. Here we only give the proof for ℳ₁₂. For the case ℳ₂₁, the proof is similar.
For any 𝐴₁₂∈ℳ₁₂, let 𝛿(𝐴₁₂)=𝑆₁₁+𝑆₁₂+𝑆₂₁+𝑆₂₂. Since (𝐼+𝐴₁₂)𝑃₁=𝑃₁, we have 𝛿𝐼+𝐴₁₂,𝑃₁=𝛿(𝐼)+𝛿𝐴₁₂,𝑃₁+𝐼+𝐴₁₂,𝛿𝑃₁. (8) 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 𝛿𝑃₁−𝐴₁₂,𝐼+𝐴₁₂=𝛿𝑃₁−𝐴₁₂,𝐼+𝐴₁₂+𝑃₁−𝐴₁₂,𝛿𝐼+𝐴₁₂. (9) By (7) and Claim 2, the above equation yields 𝛿(𝐼)𝐴₁₂=𝐴₁₂𝛿(𝐼); that is, 𝛿(𝐼)𝑃₁𝐴𝑃₂=𝑃₁𝐴𝑃₂𝛿(𝐼)∀𝐴∈ℳ. (10)
Similarly, by the equation (𝐼+𝐴₂₁)(𝑃₁−𝐴₂₁)=𝑃₁, one can show that 𝛿(𝐼)𝑃₂𝐴𝑃₁=𝑃₂𝐴𝑃₁𝛿(𝐼)holds∀𝐴∈ℳ. (11) Equations (10)-(11) and Lemma 2 ensure that 𝛿(𝐼)∈𝒵(ℳ).
Note that, by (7) and Claim 3, one shows 𝛿𝑃₂=𝛿(𝐼)−𝛿𝑃₁∈𝒵(ℳ). (12)
Claim  4. For any 𝐴_𝑖𝑖∈ℳ_𝑖𝑖, there exists some map ℎ_𝑖∶ℳ_𝑖𝑖→𝒵(ℳ) such that 𝛿(𝐴_𝑖𝑖)∈ℳ_𝑖𝑖+ℎ_𝑖(𝐴_𝑖𝑖), 𝑖=1,2.
Firstly, take any invertible 𝐴₁₁∈ℳ₁₁ and let 𝛿(𝐴₁₁)=𝑆₁₁+𝑆₁₂+𝑆₂₁+𝑆₂₂. Since 𝐴⁻¹₁₁𝐴₁₁=𝑃₁, we have 0=𝛿𝐴⁻¹₁₁,𝐴₁₁=𝛿𝐴⁻¹₁₁,𝐴₁₁+𝐴⁻¹₁₁,𝛿𝐴₁₁. (13)
For any 𝐵₂₂∈ℳ₂₂, let 𝛿(𝐵₂₂)=𝑇₁₁+𝑇₁₂+𝑇₂₁+𝑇₂₂. Since 𝑃₁(𝑃₁+𝐵₂₂)=𝑃₁, we have 0=𝛿𝑃₁,𝑃₁+𝐵₂₂=𝛿𝑃₁,𝑃₁+𝐵₂₂+𝑃₁,𝛿𝑃₁+𝛿𝐵₂₂. (14) By (7), the above equation yields 𝑇₁₂=𝑇₂₁=0. (15)
Since (𝐴⁻¹₁₁+𝐵₂₂)𝐴₁₁=𝑃₁, one gets 0=𝛿𝐴⁻¹₁₁+𝐵₂₂,𝐴₁₁=𝛿𝐴⁻¹₁₁+𝛿𝐵₂₂,𝐴₁₁+𝐴⁻¹₁₁+𝐵₂₂,𝛿𝐴₁₁, (16) which and (13) imply 𝛿𝐵₂₂𝐴₁₁−𝐴₁₁𝛿𝐵₂₂+𝐵₂₂𝛿𝐴₁₁−𝛿𝐴₁₁𝐵₂₂=0∀𝐵₂₂∈ℳ₂₂. (17) Let 𝐵₂₂=𝑃₂ in (17), and by (12), we get 𝑃₂𝛿(𝐴₁₁)=𝛿(𝐴₁₁)𝑃₂, and so 𝑆₁₂=𝑆₂₁=0. (18) Thus, combining (15) and (18), (17) reduces to 𝑇₁₁𝐴₁₁−𝐴₁₁𝑇₁₁+𝐵₂₂𝑆₂₂−𝑆₂₂𝐵₂₂=0. It follows that 𝑇₁₁𝐴₁₁=𝐴₁₁𝑇₁₁,𝐵₂₂𝑆₂₂=𝑆₂₂𝐵₂₂ (19) holds for all invertible 𝐴₁₁∈ℳ₁₁ and all 𝐵₂₂∈ℳ₂₂.
Now taking any 𝐴₁₁∈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 𝛿𝐵₂₂=𝑇₁₁+𝑇₂₂=𝑇₂₂+𝑇₁₁−ℎ₂𝐵₂₂+ℎ₂𝐵₂₂∈ℳ₂₂+ℎ₂𝐵₂₂. (20)
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 𝛿𝐴₁₁=𝑃₁𝛿𝐴₁₁𝑃₁+𝑃₂𝛿𝐴₁₁𝑃₂=𝑃₁𝛿𝐴₁₁𝑃₁+𝑃₂𝛿𝐴₁₁𝑃₂−ℎ₁𝐴₁₁+ℎ₁𝐴₁₁∈ℳ₁₁+ℎ₁𝐴₁₁. (21) Claim 4 is true.
Now define two maps ℎ∶ℳ→𝒵(ℳ) and 𝑑∶ℳ→ℳ, respectively, by ℎ(𝐴)=ℎ₁𝐴₁₁+ℎ₂𝐴₂₂,𝑑(𝐴)=𝛿(𝐴)−ℎ(𝐴) (22) for all 𝐴=𝐴₁₁+𝐴₁₂+𝐴₂₁+𝐴₂₂∈ℳ. Then by Claims 2 and 4, we have 𝑑ℳ_𝑖𝑗⊆ℳ_𝑖𝑗,𝑑ℳ_𝑖𝑖⊆ℳ_𝑖𝑖,𝑑ℳ_𝑖𝑗=𝛿ℳ_𝑖𝑗,1≤𝑖≠𝑗≤2. (23)
Claim  5.  𝑑 and ℎ are additive. By the definitions of 𝑑 and ℎ, we only need to verify that 𝑑 is additive on ℳ_𝑖𝑖(𝑖=1,2). In fact, for any 𝐴₁₁,𝐵₁₁∈ℳ₁₁, we have 𝑑𝐴₁₁+𝐵₁₁+ℎ𝐴₁₁+𝐵₁₁=𝛿𝐴₁₁+𝐵₁₁=𝛿𝐴₁₁+𝛿𝐵₁₁=𝑑𝐴₁₁+ℎ𝐴₁₁+𝑑𝐵₁₁+ℎ𝐵₁₁=𝑑𝐴₁₁+𝑑𝐵₁₁+ℎ𝐴₁₁+ℎ𝐵₁₁. (24) That is, 𝑑𝐴₁₁+𝐵₁₁−𝑑𝐴₁₁+𝑑𝐵₁₁=ℎ𝐴₁₁+ℎ𝐵₁₁−ℎ𝐴₁₁+𝐵₁₁∈𝒵(ℳ). (25) 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 that 𝑑 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 𝐴_𝑖𝑖∈ℳ_𝑖𝑖, 𝐴_𝑖𝑗,𝐵_𝑖𝑗∈ℳ_𝑖𝑗, and 𝐵_𝑗𝑗∈ℳ_𝑗𝑗, we have 𝑑(𝐴_𝑖𝑖𝐵_𝑖𝑗)=𝑑(𝐴_𝑖𝑖)𝐵_𝑖𝑗+𝐴_𝑖𝑖𝑑(𝐵_𝑖𝑗) and 𝑑(𝐴_𝑖𝑗𝐵_𝑗𝑗)=𝑑(𝐴_𝑖𝑗)𝐵_𝑗𝑗+𝐴_𝑖𝑗𝑑(𝐵_𝑗𝑗), 1≤𝑖≠𝑗≤2.
In fact, for any invertible 𝐴₁₁∈ℳ₁₁ and any 𝐵₁₂∈ℳ₁₂, since (𝐴⁻¹₁₁+𝐵₁₂)𝐴₁₁=𝑃₁, by (23) and (13), we have −𝑑𝐴₁₁𝐵₁₂=−𝛿𝐴₁₁𝐵₁₂=𝛿𝐴⁻¹₁₁+𝐵₁₂,𝐴₁₁=𝛿𝐴⁻¹₁₁+𝛿𝐵₁₂,𝐴₁₁+𝐴⁻¹₁₁+𝐵₁₂,𝛿𝐴₁₁=𝛿𝐴⁻¹₁₁,𝐴₁₁+𝛿𝐵₁₂,𝐴₁₁+𝐴⁻¹₁₁,𝛿𝐴₁₁+𝐵₁₂,𝛿𝐴₁₁=𝑑𝐵₁₂,𝐴₁₁+𝐵₁₂,𝑑𝐴₁₁=−𝐴₁₁𝑑𝐵₁₂−𝑑𝐴₁₁𝐵₁₂. (26) That is, 𝑑𝐴₁₁𝐵₁₂=𝑑𝐴₁₁𝐵₁₂+𝐴₁₁𝑑𝐵₁₂ (27) for any invertible 𝐴₁₁∈ℳ₁₁ and all 𝐵₁₂∈ℳ₁₂.
Take any 𝐴₁₁∈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 𝑑𝑐𝑃₁𝐵₁₂=𝑑𝑐𝑃₁𝐵₁₂+𝑐𝑃₁𝑑𝐵₁₂𝑑𝑐𝑃₁−𝐴₁₁𝐵₁₂=𝑑𝑐𝑃₁−𝐴₁₁𝐵₁₂+𝑐𝑃₁−𝐴₁₁𝑑𝐵₁₂. (28) Hence, 𝑑(𝐴₁₁𝐵₁₂)=𝑑(𝐴₁₁)𝐵₁₂+𝐴₁₁𝑑(𝐵₁₂) holds for all 𝐴₁₁∈ℳ₁₁ and all 𝐵₁₂∈ℳ₁₂.
Similarly, by 𝐵₁₁(𝐴₂₁+𝐵⁻¹₁₁)=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 𝐴_𝑖𝑖,𝐵_𝑖𝑖∈ℳ_𝑖𝑖, we have 𝑑(𝐴_𝑖𝑖𝐵_𝑖𝑖)=𝑑(𝐴_𝑖𝑖)𝐵_𝑖𝑖+𝐴_𝑖𝑖𝑑(𝐵_𝑖𝑖), 𝑖=1,2.
Let 𝑖≠𝑗. For any 𝐴_𝑖𝑖,𝐵_𝑖𝑖∈ℳ_𝑖𝑖 and any 𝑆_𝑖𝑗∈ℳ_𝑖𝑗, by Step 1, on the one hand, we have 𝑑𝐴_𝑖𝑖𝐵_𝑖𝑖𝑆_𝑖𝑗=𝑑𝐴_𝑖𝑖𝐵_𝑖𝑖𝑆_𝑖𝑗+𝐴_𝑖𝑖𝐵_𝑖𝑖𝑑𝑆_𝑖𝑗; (29) on the other hand, 𝑑𝐴_𝑖𝑖𝐵_𝑖𝑖𝑆_𝑖𝑗=𝑑𝐴_𝑖𝑖𝐵_𝑖𝑖𝑆_𝑖𝑗+𝐴_𝑖𝑖𝑑𝐵_𝑖𝑖𝑆_𝑖𝑗=𝑑𝐴_𝑖𝑖𝐵_𝑖𝑖𝑆_𝑖𝑗+𝐴_𝑖𝑖𝑑𝐵_𝑖𝑗𝑆_𝑖𝑗+𝐴_𝑖𝑖𝐵_𝑖𝑖𝑑𝑆_𝑖𝑗. (30) Comparing the above two equations gets that (𝑑(𝐴_𝑖𝑖𝐵_𝑖𝑖)−𝑑(𝐴_𝑖𝑖)𝐵_𝑖𝑖−𝐴_𝑖𝑖𝑑(𝐵_𝑖𝑖))𝑆_𝑖𝑗=0 holds for all 𝑆_𝑖𝑗∈ℳ_𝑖𝑗; that is, 𝑑𝐴_𝑖𝑖𝐵_𝑖𝑖−𝑑𝐴_𝑖𝑖𝐵_𝑖𝑖−𝐴_𝑖𝑖𝑑𝐵_𝑖𝑖𝑃_𝑖𝑆𝑃_𝑗=0 (31) holds for all 𝑆∈ℳ. Note that 𝑃_𝑗=𝐼. It follows from the definition of the central carrier that span{𝑇𝑃_𝑗(𝑥)∶𝑇∈ℳ,𝑥∈𝐻} is dense in 𝐻. It follows from (23) that 𝑑(𝐴_𝑖𝑖𝐵_𝑖𝑖)=𝑑(𝐴_𝑖𝑖)𝐵_𝑖𝑖+𝐴_𝑖𝑖𝑑(𝐵_𝑖𝑖) holds for all 𝐴_𝑖𝑖,𝐵_𝑖𝑖∈ℳ_𝑖𝑖, as desired.
Note that, by using Step 2 and the fact that 𝑑(ℳ_𝑖𝑖)⊆ℳ_𝑖𝑖 (𝑖=1,2), one can get 𝑑𝑃₁=𝑑𝑃₂=0. (32)
Step  3. Consider 𝑑(𝐴_𝑖𝑗𝐵_𝑗𝑖)=𝑑(𝐴_𝑖𝑗)𝐵_𝑗𝑖+𝐴_𝑖𝑗𝑑(𝐵_𝑗𝑖) for all 𝐴_𝑖𝑗∈ℳ_𝑖𝑗 and 𝐵_𝑗𝑖∈ℳ_𝑗𝑖, 1≤𝑖≠𝑗≤2.
Take any 𝐴₁₂∈ℳ₁₂ and 𝐵₂₁∈ℳ₂₁. Since (𝑃₁+𝐴₁₂−𝐴₁₂𝐵₂₁)(𝑃₁+𝐵₂₁)=𝑃₁, by the definition of 𝑑, Claim 5, (23), and (32), we have 𝑑𝐴₁₂𝐵₂₁−𝐴₁₂−𝐵₂₁−𝐵₂₁𝐴₁₂+𝐵₂₁𝐴₁₂𝐵₂₁+ℎ𝐴₁₂𝐵₂₁−𝐵₂₁𝐴₁₂=𝛿𝐴₁₂𝐵₂₁−𝐴₁₂−𝐵₂₁−𝐵₂₁𝐴₁₂+𝐵₂₁𝐴₁₂𝐵₂₁=𝛿𝑃₁+𝐴₁₂−𝐴₁₂𝐵₂₁,𝑃₁+𝐵₂₁=𝛿𝑃₁+𝐴₁₂−𝐴₁₂𝐵₂₁,𝑃+𝐵₂₁+𝑃₁+𝐴₁₂−𝐴₁₂𝐵₂₁,𝛿𝑃₁+𝐵₂₁=𝑑𝑃₁+𝐴₁₂−𝐴₁₂𝐵₂₁,𝑃₁+𝐵₂₁+𝑃₁+𝐴₁₂−𝐴₁₂𝐵₂₁,𝑑𝑃₁+𝐵₂₁=𝑑𝐴₁₂𝐵₂₁−𝑑𝐴₁₂−𝐵₂₁𝑑𝐴₁₂+𝐵₂₁𝑑𝐴₁₂𝐵₂₁+𝐴₁₂𝑑𝐵₂₁−𝑑𝐵₂₁+𝑑𝐵₂₁𝐴₁₂𝐵₂₁−𝑑𝐵₂₁𝐴₁₂. (33) It follows from Step 1 that 𝑑𝐴₁₂𝐵₂₁−𝐵₂₁𝐴₁₂+ℎ𝐴₁₂𝐵₂₁−𝐵₂₁𝐴₁₂=𝑑𝐴₁₂𝐵₂₁−𝐵₂₁𝑑𝐴₁₂+𝐴₁₂𝑑𝐵₂₁−𝑑𝐵₂₁𝐴₁₂. (34) Multiplying by 𝐴₁₂ from the left side and the right side, respectively, in (34) and applying (23), we get 𝐴₁₂𝑑𝐵₂₁𝐴₁₂−ℎ𝐴₁₂𝐵₂₁−𝐵₂₁𝐴₁₂𝐴₁₂=𝐴₁₂𝐵₂₁𝑑𝐴₁₂+𝐴₁₂𝑑𝐵₂₁𝐴₁₂,𝑑𝐴₁₂𝐵₂₁𝐴₁₂+ℎ𝐴₁₂𝐵₂₁−𝐵₂₁𝐴₁₂𝐴₁₂=𝑑𝐴₁₂𝐵₂₁𝐴₁₂+𝐴₁₂𝑑𝐵₂₁𝐴₁₂. (35) These two equations, together with Step 1, yield −𝑑𝐴₁₂𝐵₂₁𝐴₁₂−ℎ𝐴₁₂𝐵₂₁−𝐵₂₁𝐴₁₂𝐴₁₂=−𝑑𝐴₁₂𝐵₂₁𝐴₁₂+𝐴₁₂𝑑𝐵₂₁𝐴₁₂,−𝐴₁₂𝐵₂₁𝑑𝐴₁₂+ℎ𝐴₁₂𝐵₂₁−𝐵₂₁𝐴₁₂𝐴₁₂=−𝐴₁₂𝑑𝐵₂₁𝐴₁₂+𝐴₁₂𝑑𝐵₂₁𝐴₁₂. (36) Comparing the above two equations, one achieves ℎ𝐴₁₂𝐵₂₁−𝐵₂₁𝐴₁₂𝐴₁₂=0. (37)
Similarly, multiplying by 𝐵₂₁ from the left side and the right side, respectively, in (34), one can verify ℎ𝐴₁₂𝐵₂₁−𝐵₂₁𝐴₁₂𝐵₂₁=0. (38)
Next, we will prove ℎ(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 𝐴₁₂ℎ𝐴₁₂𝐵₂₁−𝐵₂₁𝐴₁₂^∗=𝑉𝐴₁₂ℎ𝐴₁₂𝐵₂₁−𝐵₂₁𝐴₁₂^∗=0. (39) Similarly, one can show 𝐵₂₁ℎ𝐴₁₂𝐵₂₁−𝐵₂₁𝐴₁₂^∗=0. (40) Multiplying by ℎ(𝐴₁₂𝐵₂₁−𝐵₂₁𝐴₁₂)^∗ from the right side in (34), by using (39)-(40), we get 𝑑𝐴₁₂𝐵₂₁−𝐵₂₁𝐴₁₂ℎ𝐴₁₂𝐵₂₁−𝐵₂₁𝐴₁₂^∗+ℎ𝐴₁₂𝐵₂₁−𝐵₂₁𝐴₁₂ℎ𝐴₁₂𝐵₂₁−𝐵₂₁𝐴₁₂^∗=0. (41) Note that, by Step 2, (23), and (39)-(40), we have 𝑑𝐴₁₂𝐵₂₁ℎ𝐴₁₂𝐵₂₁−𝐵₂₁𝐴₁₂^∗=𝑑𝐴₁₂𝐵₂₁𝑃₁ℎ𝐴₁₂𝐵₂₁−𝐵₂₁𝐴₁₂^∗𝑃₁−𝐴₁₂𝐵₂₁𝑑𝑃₁ℎ𝐴₁₂𝐵₂₁−𝐵₂₁𝐴₁₂^∗𝑃₁=−𝐴₁₂𝐵₂₁𝑑𝑃₁ℎ𝐴₁₂𝐵₂₁−𝐵₂₁𝐴₁₂^∗𝑃₁,𝑑𝐵₂₁𝐴₁₂ℎ𝐴₁₂𝐵₂₁−𝐵₂₁𝐴₁₂^∗=𝑑𝐵₂₁𝐴₁₂𝑃₂ℎ𝐴₁₂𝐵₂₁−𝐵₂₁𝐴₁₂^∗𝑃₂−𝐵₂₁𝐴₁₂𝑑𝑃₂ℎ𝐴₁₂𝐵₂₁−𝐵₂₁𝐴₁₂^∗𝑃₂=−𝐵₂₁𝐴₁₂𝑑𝑃₂ℎ𝐴₁₂𝐵₂₁−𝐵₂₁𝐴₁₂^∗𝑃₂. (42)(43) Hence, (41) implies ℎ𝐴₁₂𝐵₂₁−𝐵₂₁𝐴₁₂^∗×ℎ𝐴₁₂𝐵₂₁−𝐵₂₁𝐴₁₂ℎ𝐴₁₂𝐵₂₁−𝐵₂₁𝐴₁₂^∗=0, (44) and so ℎ(𝐴₁₂𝐵₂₁−𝐵₂₁𝐴₁₂)=0. Thus, (34) reduces to 𝑑𝐴₁₂𝐵₂₁−𝐵₂₁𝐴₁₂=𝑑𝐴₁₂𝐵₂₁−𝐵₂₁𝑑𝐴₁₂+𝐴₁₂𝑑𝐵₂₁−𝑑𝐵₂₁𝐴₁₂, (45) 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 that 𝑑 is an additive derivation.
Claim  7. Consider ℎ([𝐴,𝐵])=0 for all 𝐴,𝐵∈ℳ with 𝐴𝐵=𝑃₁.
In fact, for any 𝐴,𝐵∈ℳ with 𝐴𝐵=𝑃₁, we have ℎ=[𝛿(𝐴)]+[(𝐵()=[𝑑(𝐴)(𝐴)]+[(𝐵)(𝐵()=[𝑑(𝐴)]+[(𝐵()([𝐴,𝐵])=𝛿([𝐴,𝐵])−𝑑([𝐴,𝐵]),𝐵𝐴,𝛿)]−𝑑𝐴𝐵−𝐵𝐴−ℎ,𝐵𝐴,𝑑−ℎ)]−𝑑𝐴𝐵−𝐵𝐴,𝐵𝐴,𝑑)]−𝑑𝐴𝐵−𝐵𝐴=0. (46)
Claim  8. The theorem holds.
Indeed, let 𝜑(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. Characterizing 𝜉-Lie Derivations by Local Action

In this section, we will give a complete characterization of additive 𝜉-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. Let ℳ 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 Theorem 1. 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, 𝐿𝑃₁−𝐿𝜉𝑃₁=𝐿𝑃₁𝑃₁−𝜉𝑃₁𝐿𝑃₁+𝑃₁𝐿𝑃₁−𝜉𝐿𝑃₁𝑃₁. (47) Multiplying by 𝑃_𝑖 and 𝑃_𝑗(1≤𝑖≠𝑗≤2) from the left and the right sides, respectively, in (47), one gets 𝑃_𝑖𝐿𝜉𝑃₁𝑃_𝑗=𝜉𝑃_𝑖𝐿𝑃₁𝑃_𝑗,1≤𝑖≠𝑗≤2. (48)
As 𝑃₁𝐼=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, 𝐿𝑃₁−𝐿𝜉𝑃₁=𝐿𝑃₁−𝜉𝐿𝑃₁+𝑃₁𝐿(𝐼)−𝜉𝐿(𝐼)𝑃₁,𝐿𝑃₁−𝐿𝜉𝑃₁=𝐿(𝐼)𝑃₁−𝜉𝑃₁𝐿(𝐼)+𝐿𝑃₁−𝜉𝐿𝑃₁. (49)(50) 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 𝐿𝑃₂𝑃₁−𝜉𝑃₁𝐿𝑃₂+𝑃₂𝐿𝑃₁−𝜉𝐿𝑃₁𝑃₂=0. (51) 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 map 𝛿∶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 𝛿[𝐴,𝐵]_𝜉=[𝛿(𝐴),𝐵]_𝜉+[𝐴,𝛿(𝐵)]_𝜉forany𝐴,𝐵∈ℳwith𝐴𝐵=𝑃=𝑃₁. (52) Moreover, 𝑃₁𝛿(𝐼)𝑃₂=𝑃₂𝛿(𝐼)𝑃₁=𝑃₂𝛿(𝑃₁)𝑃₂=𝑃₁𝛿(𝑃₂)𝑃₁=0 by Claim 1. Thus we obtain 𝛿𝑃₁=𝐿𝑃₁+𝑆𝑃₁−𝑃₁𝑆=𝑃₁𝐿𝑃₁𝑃₁=𝑃₁𝛿𝑃₁𝑃₁−𝑃₁𝑆𝑃₁−𝑃₁𝑆𝑃₁=𝑃₁𝛿𝑃₁𝑃₁∈ℳ₁₁,𝛿𝑃₂=𝐿𝑃₂+𝑆𝑃₂−𝑃₂𝑆=𝑃₂𝐿𝑃₂𝑃₂=𝑃₂𝛿𝑃₂𝑃₂−𝑃₂𝑆𝑃₂−𝑃₂𝑆𝑃₂=𝑃₂𝛿𝑃₂𝑃₂∈ℳ₂₂. (53)(54) Note that (47) is also true for the map 𝛿. So, by (53), one can also get 𝛿𝑃₁+𝛿𝜉𝑃₁−2𝜉𝛿𝑃₁=0. (55)
Claim  2. Consider 𝛿(ℳ_𝑖𝑖)⊆ℳ_𝑖𝑖,𝑖=1,2.  For any 𝐴₂₂∈ℳ₂₂, since 𝑃₁(𝑃₁+𝐴₂₂)=𝑃₁, we have 𝛿𝑃₁,𝑃₁+𝐴_22𝜉=𝛿𝑃₁,𝑃₁+𝐴_22𝜉+𝑃₁,𝛿𝑃₁+𝛿𝐴_22𝜉. (56) By (53) and (55), the above equation reduces to 𝑃₁𝛿𝐴₂₂−𝜉𝛿𝐴₂₂𝑃₁=0. (57) 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 invertible 𝐴₁₁∈ℳ₁₁. By 𝐴₁₁𝐴⁻¹₁₁=𝑃₁ and 𝐴₁₁(𝐴⁻¹₁₁+𝑃₂)=𝑃₁, we get 𝛿𝑃₁−𝛿𝜉𝑃₁=𝛿𝐴₁₁,𝐴⁻¹_11𝜉=𝛿𝐴₁₁,𝐴⁻¹_11𝜉+𝐴₁₁,𝛿𝐴⁻¹_11𝜉,𝛿𝑃₁−𝛿𝜉𝑃₁=𝛿𝐴₁₁,𝐴⁻¹₁₁+𝑃_2𝜉=𝛿𝐴₁₁,𝐴⁻¹₁₁+𝑃_2𝜉+𝐴₁₁,𝛿𝐴⁻¹₁₁+𝛿𝑃_2𝜉. (58) 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, by 𝐴⁻¹₁₁𝐴₁₁=𝑃₁ and (𝐴⁻¹₁₁+𝑃₂)𝐴₁₁=𝑃₁, one can prove 𝑃₂𝛿(𝐴₁₁)𝑃₁=0.
Thus, we have proved that 𝛿(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 𝐴_𝑖𝑗∈ℳ_𝑖𝑗, 1≤𝑖≠𝑗≤2, we have 𝛿(𝑃_𝑖)𝐴_𝑖𝑗=𝐴_𝑖𝑗𝛿(𝑃_𝑗). Moreover, the following statements hold.(1)If 𝜉≠−1, then 𝛿(𝐴_𝑖𝑗)∈ℳ_𝑖𝑗.(2)If 𝜉=−1, then 𝑃₁𝛿(𝐴_𝑖𝑗)𝑃₁=𝑃₂𝛿(𝐴_𝑖𝑗)𝑃₂=0.
First, consider the case 𝐴₁₂∈ℳ₁₂. Since (𝑃₁+𝐴₁₂)𝑃₁=𝑃₁, we have 𝛿𝑃₁+𝐴₁₂,𝑃_1𝜉=𝛿𝑃₁+𝛿𝐴₁₂,𝑃_1𝜉+𝑃₁+𝐴₁₂,𝛿𝑃_1𝜉. (59) By (53) and (55), we get −𝛿𝜉𝐴₁₂=𝛿𝐴₁₂𝑃₁−𝜉𝑃₁𝛿𝐴₁₂−𝜉𝛿𝑃₁𝐴₁₂. (60) 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 𝑃₂𝛿𝐴₁₂𝑃₁=0forthecase𝜉≠−1. (61)
Since (𝑃₁+𝐴₁₂)(𝐼−𝐴₁₂)=𝑃₁, we have 𝛿𝑃₁+𝐴₁₂,𝐼−𝐴_12𝜉=𝛿(𝑃₁)+𝛿(𝐴₁₂),𝐼−𝐴_12𝜉+𝑃₁+𝐴₁₂,𝛿(𝐼)−𝛿𝐴_12𝜉. (62) That is, 𝛿𝑃₁−𝛿𝜉𝑃₁−𝛿𝜉𝐴₁₂=𝛿𝑃₁−𝛿𝑃₁𝐴₁₂+𝛿𝐴₁₂−𝛿𝐴₁₂𝐴₁₂−𝜉𝛿𝑃₁+𝛿𝐴₁₂−𝐴₁₂𝛿𝑃₁−𝐴₁₂𝛿𝐴₁₂+𝑃₁𝛿(𝐼)−𝑃₁𝛿𝐴₁₂+𝐴₁₂𝛿(𝐼)−𝐴₁₂𝛿𝐴₁₂−𝜉𝛿(𝐼)𝑃₁+𝛿(𝐼)𝐴₁₂−𝛿𝐴₁₂𝑃₁−𝛿𝐴₁₂𝐴₁₂. (63)
Since (𝑃₁−𝐴₁₂)(𝐼+𝐴₁₂)=𝑃₁, we have 𝛿𝑃₁−𝐴₁₂,𝐼+𝐴_12𝜉=𝛿(𝑃₁)−𝛿(𝐴₁₂),𝐼+𝐴_12𝜉+𝑃₁−𝐴₁₂,𝛿(𝐼)+𝛿𝐴_12𝜉. (64) That is, 𝛿𝑃₁−𝛿𝜉𝑃₁+𝛿𝜉𝐴₁₂=𝛿𝑃₁+𝛿𝑃₁𝐴₁₂−𝛿𝐴₁₂−𝛿𝐴₁₂𝐴₁₂−𝜉𝛿𝑃₁−𝛿𝐴₁₂+𝐴₁₂𝛿𝑃₁−𝐴₁₂𝛿𝐴₁₂+𝑃₁𝛿(𝐼)+𝑃₁𝛿𝐴₁₂−𝐴₁₂𝛿(𝐼)−𝐴₁₂𝛿𝐴₁₂−𝜉𝛿(𝐼)𝑃₁−𝛿(𝐼)𝐴₁₂+𝛿𝐴₁₂𝑃₁−𝛿𝐴₁₂𝐴₁₂. (65) Comparing (63)-(65) and by (53)-(54), one can obtain −𝛿𝜉𝐴₁₂=−𝛿𝑃₁𝐴₁₂+𝛿𝐴₁₂−𝜉𝛿𝐴₁₂−𝑃₁𝛿𝐴₁₂+𝐴₁₂𝛿𝑃₂−𝜉𝛿𝑃₁𝐴₁₂+𝜉𝛿𝐴₁₂𝑃₁. (66) Then, (60) and (66) yield (1−𝜉)𝛿𝐴₁₂𝑃₁+(1−𝜉)𝑃₁𝛿𝐴₁₂−(1−𝜉)𝛿𝐴₁₂+𝛿𝑃₁𝐴₁₂−𝐴₁₂𝛿𝑃₂=0. (67) Multiplying by 𝑃_𝑖(𝑖=1,2) from both sides in (67), and by (53)-(54), we have 𝑃₁𝛿𝐴₁₂𝑃₁=0,𝑃₂𝛿𝐴₁₂𝑃₂=0; (68) multiplying by 𝑃₁ and 𝑃₂ and from the left and the right sides, respectively, in (67), we have 𝛿𝑃₁𝐴₁₂=𝐴₁₂𝛿𝑃₂. (69) Now combining (61) and (68)-(69), we achieve that the claim holds for any 𝐴₁₂∈ℳ₁₂.
For any 𝐴₂₁∈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 −𝛿𝜉𝐴₂₁=𝑃₁𝛿𝐴₂₁−𝜉𝛿𝐴₂₁𝑃₁−𝜉𝐴₂₁𝛿𝑃₁, (70) 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 𝐴_𝑖𝑗∈ℳ_𝑖𝑗, 1≤𝑖≠𝑗≤2, by Claim 3(1), (60), and (70), we get 𝛿𝜉𝐴_𝑖𝑗=𝜉𝛿𝐴_𝑖𝑗+𝜉𝛿𝑃_𝑖𝐴_𝑖𝑗. (71)
We will complete the proof of Claim 5 by the following several steps.
Step    1. For any 𝐴_𝑖𝑖∈ℳ_𝑖𝑖, 𝐵_𝑗𝑗∈ℳ_𝑗𝑗, 𝐴_𝑖𝑗,𝐵_𝑖𝑗∈ℳ_𝑖𝑗, 1≤𝑖≠𝑗≤2, the following statements hold:(i)𝛿(𝜉𝐴_𝑖𝑖𝐵_𝑖𝑗)=𝜉𝛿(𝐴_𝑖𝑖)𝐵_𝑖𝑗+𝜉𝐴_𝑖𝑖𝛿(𝐵_𝑖𝑗);(ii)𝛿(𝜉𝐴_𝑖𝑗𝐵_𝑗𝑗)=𝜉𝛿(𝐴_𝑖𝑗)𝐵_𝑗𝑗+𝜉𝐴_𝑖𝑗𝛿(𝐵_𝑗𝑗);(iii)𝛿(𝐴_𝑖𝑖𝐵_𝑖𝑗)=𝛿(𝐴_𝑖𝑖)𝐵_𝑖𝑗+𝐴_𝑖𝑖𝛿(𝐵_𝑖𝑗)−𝛿(𝐼)𝐴_𝑖𝑖𝐵_𝑖𝑗;(iv)𝛿(𝐴_𝑖𝑗𝐵_𝑗𝑗)=𝛿(𝐴_𝑖𝑗)𝐵_𝑗𝑗+𝐴_𝑖𝑗𝛿(𝐵_𝑗𝑗)−𝛿(𝐼)𝐴_𝑖𝑗𝐵_𝑗𝑗.
For any 𝐴₁₁∈ℳ₁₁, that is, invertible, and any 𝐵₁₂∈ℳ₁₂, since (𝐴⁻¹₁₁+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 𝛿𝜉𝐴₁₁𝐵₁₂=𝜉𝛿𝐴₁₁𝐵₁₂+𝜉𝐴₁₁𝛿𝐵₁₂holds∀𝐴₁₁∈ℳ₁₁,𝐵₁₂∈ℳ₁₂. (72) Also note that, by (71), 𝛿(𝜉𝐴₁₁𝐵₁₂)=𝜉𝛿(𝐴₁₁𝐵₁₂)+𝜉𝛿(𝑃₁)𝐴₁₁𝐵₁₂. As 𝜉≠0, this and (72) yield 𝛿𝐴₁₁𝐵₁₂=𝛿𝐴₁₁𝐵₁₂+𝐴₁₁𝛿𝐵₁₂−𝛿𝑃₁𝐴₁₁𝐵₁₂. (73)
For any 𝐵₁₁∈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 𝐴_𝑖𝑖∈ℳ_𝑖𝑖, we have 𝛿(𝜉𝐴_𝑖𝑖𝐵_𝑖𝑖)=𝜉𝛿(𝐴_𝑖𝑖)𝐵_𝑖𝑖+𝜉𝐴_𝑖𝑖𝛿(𝐵_𝑖𝑖), 𝑖=1,2.
In fact, let 𝑖≠𝑗. For any 𝐴_𝑖𝑖,𝐵_𝑖𝑖∈ℳ_𝑖𝑖 and any 𝑆_𝑖𝑗∈ℳ_𝑖𝑗, by (71) and Step 1, on the one hand, we have 𝛿𝜉𝐴_𝑖𝑖𝐵_𝑖𝑖𝑆_𝑖𝑗=𝜉𝛿𝐴_𝑖𝑖𝐵_𝑖𝑖𝑆_𝑖𝑗+𝜉𝛿𝑃_𝑖𝐴_𝑖𝑖𝐵_𝑖𝑖𝑆_𝑖𝑗=𝜉𝛿𝐴_𝑖𝑖𝐵_𝑖𝑖𝑆_𝑖𝑗+𝜉𝐴_𝑖𝑖𝐵_𝑖𝑖𝛿𝑆_𝑖𝑗; (74) on the other hand, 𝛿𝜉𝐴_𝑖𝑖𝐵_𝑖𝑖𝑆_𝑖𝑗=𝜉𝛿𝐴_𝑖𝑖𝐵_𝑖𝑖𝑆_𝑖𝑗+𝜉𝛿𝑃_𝑖𝐴_𝑖𝑖𝐵_𝑖𝑖𝑆_𝑖𝑗=𝜉𝛿𝐴_𝑖𝑖𝐵_𝑖𝑖𝑆_𝑖𝑗+𝜉𝐴_𝑖𝑖𝛿𝐵_𝑖𝑖𝑆_𝑖𝑗=𝜉𝛿𝐴_𝑖𝑖𝐵_𝑖𝑖𝑆_𝑖𝑗+𝜉𝐴_𝑖𝑖𝛿𝜉𝜉⁻¹𝐵_𝑖𝑖𝑆_𝑖𝑗=𝜉𝛿𝐴_𝑖𝑖𝐵_𝑖𝑖𝑆_𝑖𝑗+𝜉²𝐴_𝑖𝑖𝛿𝜉⁻¹𝐵_𝑖𝑖𝑆_𝑖𝑗+𝜉𝐴_𝑖𝑖𝐵_𝑖𝑖𝛿𝑆_𝑖𝑗. (75) The above two equations yield that (𝛿(𝐴_𝑖𝑖𝐵_𝑖𝑖)−𝛿(𝐴_𝑖𝑖)𝐵_𝑖𝑖−𝜉𝐴_𝑖𝑖𝛿(𝜉⁻¹𝐵_𝑖𝑖))𝑆_𝑖𝑗=0 holds for all 𝐴_𝑖𝑖,𝐵_𝑖𝑖∈ℳ_𝑖𝑖 and all 𝑆_𝑖𝑗∈ℳ_𝑖𝑗, which implies that (𝛿(𝜉𝐴_𝑖𝑖𝐵_𝑖𝑖)−𝜉𝛿(𝐴_𝑖𝑖)𝐵_𝑖𝑖−𝜉𝐴_𝑖𝑖𝛿(𝐵_𝑖𝑖))𝑆𝑃_𝑗=0 holds for all 𝑆∈ℳ. It follows that 𝛿(𝜉𝐴_𝑖𝑖𝐵_𝑖𝑖)=𝜉𝛿(𝐴_𝑖𝑖)𝐵_𝑖𝑖+𝜉𝐴_𝑖𝑖𝛿(𝐵_𝑖𝑖) as 𝑃_𝑗=𝐼. Step 2 is true.
Step  3. Consider 𝛿(𝐼)=𝐿(𝐼)=0 and 𝛿(𝜉𝐴)=𝜉𝛿(𝐴) for all 𝐴∈ℳ.
Taking 𝐴₁₁=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 𝐴_𝑖𝑖,𝐵_𝑖𝑖∈ℳ_𝑖𝑖, we have 𝛿(𝐴_𝑖𝑖𝐵_𝑖𝑖)=𝛿(𝐴_𝑖𝑖)𝐵_𝑖𝑖+𝐴_𝑖𝑖𝛿(𝐵_𝑖𝑖), 𝑖=1,2.
Let 𝑖≠𝑗. Take any 𝐴_𝑖𝑖,𝐵_𝑖𝑖∈ℳ_𝑖𝑖 and any 𝑆_𝑖𝑗∈ℳ_𝑖𝑗. By Step 1, one gets 𝛿𝐴_𝑖𝑖𝐵_𝑖𝑖𝑆_𝑖𝑗=𝛿𝐴_𝑖𝑖𝐵_𝑖𝑖𝑆_𝑖𝑗+𝐴_𝑖𝑖𝐵_𝑖𝑖𝛿𝑆_𝑖𝑗,𝛿𝐴_𝑖𝑖𝐵_𝑖𝑖𝑆_𝑖𝑗=𝛿𝐴_𝑖𝑖𝐵_𝑖𝑖𝑆_𝑖𝑗+𝐴_𝑖𝑖𝛿𝐵_𝑖𝑖𝑆_𝑖𝑗=𝛿𝐴_𝑖𝑖𝐵_𝑖𝑖𝑆_𝑖𝑗+𝐴_𝑖𝑖𝛿𝐵_𝑖𝑖𝑆_𝑖𝑗+𝐴_𝑖𝑖𝐵_𝑖𝑖𝛿𝑆_𝑖𝑗. (76) Comparing the above two equations, and by Claims 2 and 4, one obtains that 𝛿𝐴_𝑖𝑖𝐵_𝑖𝑖−𝛿𝐴_𝑖𝑖𝐵_𝑖𝑖−𝐴_𝑖𝑖𝛿𝐵_𝑖𝑖𝑆𝑃_𝑗=0 (77) holds for all 𝑆∈ℳ. It follows from the fact 𝑃_𝑗=𝐼 that 𝛿(𝐴_𝑖𝑖𝐵_𝑖𝑖)=𝛿(𝐴_𝑖𝑖)𝐵_𝑖𝑖+𝐴_𝑖𝑖𝛿(𝐵_𝑖𝑖).
Step  5. For any 𝐴_𝑖𝑗∈ℳ_𝑖𝑗 and 𝐵_𝑗𝑖∈ℳ_𝑗𝑖, we have 𝛿(𝐴_𝑖𝑗𝐵_𝑗𝑖)=𝛿(𝐴_𝑖𝑗)𝐵_𝑗𝑖+𝐴_𝑖𝑗𝛿(𝐵_𝑗𝑖), 1≤𝑖≠𝑗≤2.
Take any 𝐴_𝑖𝑗∈ℳ_𝑖𝑗 and 𝐵_𝑗𝑖∈ℳ_𝑗𝑖. Since (𝑃₁+𝐴₁₂−𝐴₁₂𝐵₂₁)(𝑃₁+𝐵₂₁)=𝑃₁, by Claim 3(1), Steps 1 and 3, we have 𝜉𝛿𝐵₂₁𝐴₁₂+𝛿𝐴₁₂𝐵₂₁−𝛿𝐴₁₂𝐵₂₁−𝜉𝐵₂₁𝛿𝐴₁₂+𝐴₁₂𝛿𝐵₂₁−𝜉𝛿𝐵₂₁𝐴₁₂=0. (78) 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 any 𝐴=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 𝛿(𝐴⁻¹₁₁),𝐴₁₁₋₁+𝐴⁻¹₁₁,𝛿(𝐴₁₁)₋₁=0forallinvertible𝐴₁₁∈ℳ₁₁. (79)
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 any 𝐴₁₂∈ℳ₁₂ and 𝐴₂₂∈ℳ₂₂. Since (𝑃₁+𝐴₁₂)(𝑃₁+𝐴₂₂−𝐴₁₂𝐴₂₂)=𝑃₁, by Claim 2 and Claim 3(2), one can verify 𝛿𝐴₁₂=𝛿𝑃₁+𝐴₁₂,𝑃₁+𝐴₂₂−𝐴₁₂𝐴₂₂₋₁=𝛿(𝐴₁₂),𝑃₁+𝐴₂₂−𝐴₁₂𝐴₂₂₋₁+𝑃₁+𝐴₁₂,𝛿(𝐴₂₂)−𝛿(𝐴₁₂𝐴₂₂)₋₁=𝛿𝐴₁₂𝑃₁+𝛿𝐴₁₂𝐴₂₂−𝛿𝐴₁₂𝐴₁₂𝐴₂₂+𝑃₁𝛿𝐴₁₂+𝐴₂₂𝛿𝐴₁₂−𝐴₁₂𝐴₂₂𝛿𝐴₁₂−𝑃₁𝛿𝐴₁₂𝐴₂₂+𝐴₁₂𝛿𝐴₂₂−𝐴₁₂𝛿𝐴₁₂𝐴₂₂−𝛿𝐴₁₂𝐴₂₂𝑃₁−𝛿𝐴₁₂𝐴₂₂𝐴₁₂. (80) Multiplying by 𝑃₁ and 𝑃₂ from the left and the right sides, respectively, in (80), and by Claim 3(2), we have 𝑃₁𝛿𝐴₁₂𝐴₂₂𝑃₂=𝑃₁𝛿𝐴₁₂𝐴₂₂+𝐴₁₂𝛿𝐴₂₂=𝛿𝐴₁₂𝐴₂₂+𝐴₁₂𝛿𝐴₂₂; (81) multiplying by 𝑃₂ and 𝑃₁ from the left and the right sides, respectively, in (80), and by Claim 3(2) again, we have 𝑃₂𝛿𝐴₁₂𝐴₂₂𝑃₁=𝐴₂₂𝛿𝐴₁₂𝑃₁=𝐴₂₂𝛿𝐴₁₂. (82) So 𝛿𝐴₁₂𝐴₂₂=𝑃₁𝛿𝐴₁₂𝐴₂₂𝑃₂+𝑃₂𝛿𝐴₁₂𝐴₂₂𝑃₁=𝛿𝐴₁₂𝐴₂₂+𝐴₁₂𝛿𝐴₂₂+𝐴₂₂𝛿𝐴₁₂. (83)
For any 𝐴₂₁∈ℳ₁₂ and 𝐴₂₂∈ℳ₂₂, by the same argument as above for the equation (𝑃₁+𝐴₂₂−𝐴₂₂𝐴₂₁)(𝑃₁+𝐴₂₁)=𝑃₁, one can check that 𝛿𝐴₂₂𝐴₂₁=𝛿𝐴₂₂𝐴₂₁+𝐴₂₂𝛿𝐴₂₁+𝛿𝐴₂₁𝐴₂₂. (84)
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 invertible 𝐴₁₁∈ℳ₁₁ and any 𝐴₁₂∈ℳ₁₂, since (𝐴⁻¹₁₁+𝐴₁₂)𝐴₁₁=𝑃₁, by Claim 2 and (79), one obtains 𝛿𝐴₁₁𝐴₁₂=𝛿𝐴⁻¹₁₁+𝐴₁₂,𝐴₁₁₋₁=𝛿(𝐴⁻¹₁₁)+𝛿(𝐴₁₂),𝐴₁₁₋₁+𝐴⁻¹₁₁+𝐴₁₂,𝛿(𝐴₁₁)₋₁=𝛿𝐴₁₁𝐴₁₂+𝐴₁₁𝛿𝐴₁₂+𝛿𝐴₁₂𝐴₁₁. (85) Now it is easy to check that 𝛿𝐴₁₁𝐴₁₂=𝛿𝐴₁₁𝐴₁₂+𝐴₁₁𝛿𝐴₁₂+𝛿𝐴₁₂𝐴₁₁holds∀𝐴₁₁∈ℳ₁₁,𝐴₁₂∈ℳ₁₂. (86)
For any invertible 𝐴₁₁∈ℳ₁₁ and any 𝐴₂₁∈ℳ₂₁, as 𝐴₁₁(𝐴₂₁+𝐴⁻¹₁₁)=𝑃₁, by a similar argument to that of the above, one can check that 𝛿𝐴₂₁𝐴₁₁=𝛿𝐴₂₁𝐴₁₁+𝐴₂₁𝛿𝐴₁₁+𝐴₁₁𝛿𝐴₂₁. (87)
Step  3. For any 𝐴_𝑖𝑖∈ℳ_𝑖𝑖, we have 𝛿(𝐴²_𝑖𝑖)=𝛿(𝐴_𝑖𝑖)𝐴_𝑖𝑖+𝐴_𝑖𝑖𝛿(𝐴_𝑖𝑖), 𝑖=1,2.
Let 1≤𝑖≠𝑗≤2. Then, for any 𝐴_𝑖𝑖∈ℳ_𝑖𝑖 and 𝑆_𝑖𝑗∈ℳ_𝑖𝑗, calculating 𝛿(𝐴_𝑖𝑖𝐴_𝑖𝑖𝑆_𝑖𝑗) 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 𝛿𝐴₁₂−𝛿𝐴₁₂𝐴₂₁+𝛿𝐴₂₁+𝛿𝐴₂₁𝐴₁₂=𝛿𝑃₁+𝐴₁₂−𝐴₁₂𝐴₂₁,𝑃₁+𝐴₂₁₋₁=𝛿(𝐴₁₂)−𝛿(𝐴₁₂𝐴₂₁),𝑃₁+𝐴₂₁₋₁+𝑃₁+𝐴₁₂−𝐴₁₂𝐴₂₁,𝛿(𝐴₂₁)₋₁=𝛿𝐴₁₂𝑃₁+𝛿𝐴₁₂𝐴₂₁−𝛿𝐴₁₂𝐴₂₁+𝑃₁𝛿𝐴₁₂−𝛿𝐴₁₂𝐴₂₁+𝐴₂₁𝛿𝐴₁₂+𝑃₁𝛿𝐴₂₁+𝐴₁₂𝛿𝐴₂₁+𝛿𝐴₂₁𝑃₁+𝛿𝐴₂₁𝐴₁₂. (88) Multiplying by 𝑃_𝑖 (𝑖=1,2) from both sides in the above equation, one obtains 𝛿𝐴₁₂𝐴₂₁=𝑃₁𝛿𝐴₁₂𝐴₂₁+𝐴₁₂𝛿𝐴₂₁𝑃₁=𝛿𝐴₁₂𝐴₂₁+𝐴₁₂𝛿𝐴₂₁,𝛿𝐴₂₁𝐴₁₂=𝑃₂𝛿𝐴₂₁𝐴₁₂+𝐴₂₁𝛿𝐴₁₂𝑃₂=𝛿𝐴₂₁𝐴₁₂+𝐴₂₁𝛿𝐴₁₂. (89)(90)
Step  5. For any 𝐴∈ℳ, we have 𝛿(𝐴²)=𝛿(𝐴)𝐴+𝐴𝛿(𝐴). Furthermore, 𝐿 is an additive derivation.
For any 𝐴=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 𝛿𝐴₁₁𝐴⁻¹₁₁+𝐴₁₁𝛿𝐴⁻¹₁₁=0. (91)
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 any 𝐴₂₂,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 implies 𝛿(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 invertible 𝐴₁₁,𝐵₁₁∈ℳ₁₁ and any 𝐴₂₁∈ℳ₂₁, 𝐵₁₂∈ℳ₁₂, since (𝐴₁₁−𝐴₁₁𝐵₁₂)(𝐴⁻¹₁₁+𝐵₁₂+𝑃₂)=𝑃₁ and (𝐵⁻¹₁₁+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 any 𝐴₁₁,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 any 𝐴₁₂∈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 𝛿𝐵₂₁𝐴₁₂𝐶₂₁+𝐵₂₁𝐴₁₂𝛿𝐶₂₁=𝛿𝐵₂₁𝐴₁₂𝐶₂₁=𝛿𝐵₂₁𝐴₁₂𝐶₂₁+𝐵₂₁𝛿𝐴₁₂𝐶₂₁=𝛿𝐵₂₁𝐴₁₂𝐶₂₁+𝐵₂₁𝛿𝐴₁₂𝐶₂₁+𝐵₂₁𝐴₁₂𝛿𝐶₂₁, (92) which implies (𝛿(𝐵₂₁𝐴₁₂)−𝛿(𝐵₂₁)𝐴₁₂−𝐵₂₁𝛿(𝐴₁₂))𝐶𝑃₁=0 for all 𝐶∈ℳ. It follows from the fact 𝑃₁=𝐼 that 𝛿(𝐵₂₁𝐴₁₂)=𝛿(𝐵₂₁)𝐴₁₂+𝐵₂₁𝛿(𝐴₁₂). So Step 5 is true.
Now by Steps 1--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 Theorem 4, we can give different characterizations of additive 𝜉-Lie derivations on von Neumann algebras.

Corollary 5. Let ℳ 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.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

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