Let A be a factor von Neumann algebra acting on a complex Hilbert space H with dim A>1. We prove that if a map δ:A→A satisfies δ([[A,B]∗,C]∗)=[[δ(A),B]∗,C]∗+[[A,δ(B)]∗,C]∗+[[A,B]∗,δ(C)]∗ for any A,B,C∈A with A∗B∗C=0, then δ is an additive ∗-derivation.
Citation: Liang Kong, Chao Li. Non-global nonlinear skew Lie triple derivations on factor von Neumann algebras[J]. AIMS Mathematics, 2022, 7(8): 13963-13976. doi: 10.3934/math.2022771
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Let A be a factor von Neumann algebra acting on a complex Hilbert space H with dim A>1. We prove that if a map δ:A→A satisfies δ([[A,B]∗,C]∗)=[[δ(A),B]∗,C]∗+[[A,δ(B)]∗,C]∗+[[A,B]∗,δ(C)]∗ for any A,B,C∈A with A∗B∗C=0, then δ is an additive ∗-derivation.
Let A be an associative algebra. For A,B∈A, denote by [A,B]=AB−BA the Lie product of A and B. An additive (a linear) map δ:A→A is called a global Lie triple derivation if δ([[A,B],C])=[[δ(A),B],C]+[[A,δ(B)],C]+[[A,B],δ(C)] for all A,B,C∈A. The study of global Lie triple derivations on various algebras has attracted several authors' attention, see for example [2,11,16,17,20]. Next, let δ:A→A be a map (without the additivity (linearity) assumption). δ is called a global nonlinear Lie triple derivation if δ satisfies δ([[A,B],C])=[[δ(A),B],C]+[[A,δ(B)],C]+[[A,B],δ(C)] for all A,B,C∈A. Ji, Liu and Zhao [4] gave the concrete form of global nonlinear Lie triple derivations on triangular algebras. Chen and Xiao [3] investigated global nonlinear Lie triple derivations on parabolic subalgebras of finite-dimensional simple Lie algebras. Very recently, Zhao and Hao [21] paid attention to non-global nonlinear Lie triple derivations. Let F:A×A×A→A be a map and Q be a proper subset of A. δ is called a non-global nonlinear Lie triple derivation if δ satisfies δ([[A,B],C])=[[δ(A),B],C]+[[A,δ(B)],C]+[[A,B],δ(C)] for any A,B,C∈A with F(A,B,C)∈Q. Let M be a finite von Neumann algebra with no central summands of type I1. Zhao and Hao [21] proved that if δ:M→M satisfies δ([[A,B],C])=[[δ(A),B],C]+[[A,δ(B)],C]+[[A,B],δ(C)] for any A,B,C∈M with ABC=0, then δ=d+τ, where d is a derivation from M into itself and τ is a nonlinear map from M into its center such that τ([[A,B],C])=0 with ABC=0.
Let A be an associative ∗-algebra. For A,B∈A, denote by [A,B]∗=AB−BA∗ the skew Lie product of A and B. The skew Lie product arose in representability of quadratic functionals by sesquilinear functionals [12,13]. In recent years, the study related to skew Lie product has attracted some authors' attention, see for example [1,5,6,7,8,9,10,14,15,18,19,22] and references therein. A map δ:A→A (without the additivity (linearity) assumption) is called a global nonlinear skew Lie triple derivation if δ([[A,B]∗,C]∗)=[[δ(A),B]∗,C]∗+[[A,δ(B)]∗,C]∗+[[A,B]∗,δ(C)]∗ for all A,B,C∈A. A map δ:A→A is called an additive ∗-derivation if it is an additive derivation and satisfies δ(A∗)=δ(A)∗ for all A∈A. Li, Zhao and Chen [5] proved that every global nonlinear skew Lie triple derivation on factor von Neumann algebras is an additive ∗-derivation. Taghavi, Nouri and Darvish [15] proved that every global nonlinear skew Lie triple derivation on prime ∗-algebras is additive. Similarly, let F:A×A×A→A be a map and Q be a proper subset of A. If δ satisfies δ([[A,B]∗,C]∗)=[[δ(A),B]∗,C]∗+[[A,δ(B)]∗,C]∗+[[A,B]∗,δ(C)]∗ for any A,B,C∈A with F(A,B,C)∈Q, then δ is called a non-global nonlinear skew Lie triple derivation.
Motivated by the mentioned works, we will concentrate on characterizing a kind of non-global nonlinear skew Lie triple derivations δ on factor von Neumann algebras satisfying δ([[A,B]∗,C]∗)=[[δ(A),B]∗,C]∗+[[A,δ(B)]∗,C]∗+[[A,B]∗,δ(C)]∗ for any A,B,C∈A with A∗B∗C=0.
As usual, C denotes the complex number field. Let H be a complex Hilbert space and B(H) be the algebra of all bounded linear operators on H. Let A⊆B(H) be a factor von Neumann algebra (i.e., the center of A is CI, where I is the identity of A). Recall that A is prime (i.e., for any A,B∈A, AAB={0} implies A=0 or B=0).
The main result is the following theorem.
Theorem 2.1. Let A be a factor von Neumann algebra acting on a complex Hilbert space H with dimA>1. If a map δ:A→A satisfies
δ([[A,B]∗,C]∗)=[[δ(A),B]∗,C]∗+[[A,δ(B)]∗,C]∗+[[A,B]∗,δ(C)]∗ |
for any A,B,C∈A with A∗B∗C=0, then δ is an additive ∗-derivation.
Let P1∈A be a nontrivial projection. Write P2=I−P1, Aij=PiAPj (i,j=1,2). Then A=A11+A12+A21+A22. For any A∈A, A=A11+A12+A21+A22, Aij∈ Aij (i,j=1,2).
Lemma 2.1. (a) δ(Pi)∗=δ(Pi) (i=1,2);
(b) Piδ(Pi)Pj=−Piδ(Pj)Pj (1≤i≠j≤2).
Proof. (a) It is clear that δ(0)=0. For any X21∈A21, it follows from P∗1P∗1X21=0 and [[P1,P1]∗,X21]∗=0 that
0=δ([[P1,P1]∗,X21]∗)=[[δ(P1),P1]∗,X21]∗+[[P1,δ(P1)]∗,X21]∗+[[P1,P1]∗,δ(X21)]∗=−P1δ(P1)∗X21−X21δ(P1)∗+X21δ(P1)P1+P1δ(P1)X21−X21δ(P1)∗P1+X21δ(P1)∗. | (2.1) |
Multiplying (2.1) by P2 from the left and by P1 from the right, we have X21(δ(P1)P1−δ(P1)∗P1)=0. Then by the primeness of A, we get
P1δ(P1)∗P1=P1δ(P1)P1. | (2.2) |
By P∗1P∗2P2=0 and [[P1,P2]∗,P2]∗=0, we have
0=δ([[P1,P2]∗,P2]∗)=[[δ(P1),P2]∗,P2]∗+[[P1,δ(P2)]∗,P2]∗+[[P1,P2]∗,δ(P2)]∗=δ(P1)P2−P2δ(P1)∗P2−P2δ(P1)∗+P2δ(P1)P2+P1δ(P2)P2−P2δ(P2)∗P1. | (2.3) |
Multiplying (2.3) by P2 from both sides, we see that
P2δ(P1)∗P2=P2δ(P1)P2. | (2.4) |
From P∗1P∗1P2=0 and [[P1,P1]∗,P2]∗=0, we have
0=δ([[P1,P1]∗,P2]∗)=[[δ(P1),P1]∗,P2]∗+[[P1,δ(P1)]∗,P2]∗+[[P1,P1]∗,δ(P2)]∗=−P1δ(P1)∗P2+P2δ(P1)P1+P1δ(P1)P2−P2δ(P1)∗P1. | (2.5) |
Multiplying (2.5) by P1 from the left and by P2 from the right, then
P1δ(P1)∗P2=P1δ(P1)P2. | (2.6) |
Multiplying (2.5) by P2 from the left and by P1 from the right, then
P2δ(P1)∗P1=P2δ(P1)P1. | (2.7) |
It follows from (2.2), (2.4), (2.6) and (2.7) that δ(P1)∗=δ(P1). Similarly, we can obtain that δ(P2)∗=δ(P2).
(b) From P∗2P∗1P2=0 and [[P2,P1]∗,P2]∗=0, we have
0=δ([[P2,P1]∗,P2]∗)=[[δ(P2),P1]∗,P2]∗+[[P2,δ(P1)]∗,P2]∗+[[P2,P1]∗,δ(P2)]∗=−P1δ(P2)∗P2+P2δ(P2)P1+P2δ(P1)P2−δ(P1)P2−P2δ(P1)∗P2+P2δ(P1)∗. | (2.8) |
Multiplying (2.8) by P1 from the left and by P2 from the right, we have P1δ(P1)P2=−P1δ(P2)∗P2. Then P1δ(P1)P2=−P1δ(P2)P2 by (a). Similarly, we can obtain that P2δ(P2)P1=−P2δ(P1)P1.
Lemma 2.2. For any Aij∈Aij (1≤i≠j≤2), we have
Pjδ(Aij)Pi=0. |
Proof. Let A12∈A12. For any X12∈A12, since A∗12X∗12P2=0 and [[A12,X12]∗,P2]∗=0, we have
0=δ([[A12,X12]∗,P2]∗)=[[δ(A12),X12]∗,P2]∗+[[A12,δ(X12)]∗,P2]∗+[[A12,X12]∗,δ(P2)]∗=δ(A12)X12−X12δ(A12)∗P2−X∗12δ(A12)∗+P2δ(A12)X∗12+A12δ(X12)P2−P2δ(X12)∗A∗12−X12A∗12δ(P2)+δ(P2)A12X∗12. | (2.9) |
Multiplying (2.9) by P2 from both sides, we have
0=P2δ(A12)X12−X∗12δ(A12)∗P2. | (2.10) |
Replacing X12 with iX12 in (2.10) yields that
0=P2δ(A12)X12+X∗12δ(A12)∗P2. | (2.11) |
Combining (2.10) and (2.11), we see that P2δ(A12)X12=0. Then P2δ(A12)P1=0 by the primeness of A. Similarly, we can obtain that P1δ(A21)P2=0.
Lemma 2.3. For any A12∈A12,B21∈A21, there exist GA12,B21∈A11,KA12,B21∈A22 such that
δ(A12+B21)=δ(A12)+δ(B21)+GA12,B21+KA12,B21. |
Proof. Let T=δ(A12+B21)−δ(A12)−δ(B21). From P∗2(A12+B21)∗P2=P∗2A∗12P2=P∗2B∗21P2=0 and [[P2,B21]∗,P2]∗=0, we have
[[δ(P2),A12+B21]∗,P2]∗+[[P2,δ(A12+B21)]∗,P2]∗+[[P2,A12+B21]∗,δ(P2)]∗=δ([[P2,A12+B21]∗,P2]∗)=δ([[P2,A12]∗,P2]∗)+δ([[P2,B21]∗,P2]∗)=[[δ(P2),A12+B21]∗,P2]∗+[[P2,δ(A12)+δ(B21)]∗,P2]∗+[[P2,A12+B21]∗,δ(P2)]∗, |
which implies
[[P2,T]∗,P2]∗=0. | (2.12) |
Multiplying (2.12) by P1 from the left, we get T12=0. Similarly, T21=0. Let
GA12,B21=T11,KA12,B21=T22. |
Then GA12,B21∈A11,KA12,B21∈A22, and so δ(A12+B21)=δ(A12)+δ(B21)+GA12,B21+KA12,B21.
Lemma 2.4. (a) Pjδ(Pi)Pj=0 (1≤i≠j≤2);
(b) Piδ(Pi)Pi=0 (i=1,2).
Proof. (a) For any X12∈A12, since P∗1X∗12P1=0 and [[P1,X12]∗,P1]∗=0, we have
0=δ([[P1,X12]∗,P1]∗)=[[δ(P1),X12]∗,P1]∗+[[P1,δ(X12)]∗,P1]∗+[[P1,X12]∗,δ(P1)]∗=−X12δ(P1)P1+P1δ(P1)X∗12+P1δ(X12)P1−δ(X12)P1−P1δ(X12)∗P1+P1δ(X12)∗+X12δ(P1)−δ(P1)X∗12. | (2.13) |
Multiplying (2.13) by P1 from the left and by P2 from the right, we have
P1δ(X12)∗P2+X12δ(P1)P2=0. |
It follows from Lemma 2.2 that X12δ(P1)P2=−(P2δ(X12)P1)∗=0. Then P2δ(P1)P2=0. Similarly, P1δ(P2)P1=0.
(b) For any X21∈A21, from (iX21)∗P∗1P1=0, [[iX21,P1]∗,P1]∗=iX∗21+iX21, Lemma 2.1(a) and Lemma 2.3, there exist GiX∗21,iX21∈A11,KiX∗21,iX21∈A22 such that
δ(iX∗21)+δ(iX21)+GiX∗21,iX21+KiX∗21,iX21=δ([[iX21,P1]∗,P1]∗)=[[δ(iX21),P1]∗,P1]∗+[[iX21,δ(P1)]∗,P1]∗+[[iX21,P1]∗,δ(P1)]∗=δ(iX21)P1−P1δ(iX21)∗P1−P1δ(iX21)∗+P1δ(iX21)P1+iX21δ(P1)P1+iP1δ(P1)X∗21+iX21δ(P1)+iX∗21δ(P1)+iδ(P1)X∗21+iδ(P1)X21. | (2.14) |
Multiplying (2.14) by P2 from the left and by P1 from the right, we have
P2δ(iX∗21)P1=2iX21δ(P1)P1+iP2δ(P1)X21. | (2.15) |
By (2.15), Lemma 2.2 and the fact that P2δ(P1)P2=0, we obtain X21δ(P1)P1=0. Then P1δ(P1)P1=0. Similarly, P2δ(P2)P2=0.
Remark 2.1. Let S=P1δ(P1)P2−P2δ(P1)P1. Then S∗=−S by Lemma 2.1. We define a map Δ:A→A by
Δ(X)=δ(X)−[X,S] |
for any X∈A. It is easy to verify that Δ is a map satisfying
Δ([[A,B]∗,C]∗)=[[Δ(A),B]∗,C]∗+[[A,Δ(B)]∗,C]∗+[[A,B]∗,Δ(C)]∗ |
for any A,B,C∈A with A∗B∗C=0. By Lemmas 2.1–2.4, it follows that
(a) Δ(Pi)=0 (i=1,2);
(b) For any Aij∈Aij (1≤i≠j≤2), we have PjΔ(Aij)Pi=0;
(c) For any A12∈A12,B21∈A21, there exist UA12,B21∈A11,VA12,B21∈A22 such that
Δ(A12+B21)=Δ(A12)+Δ(B21)+UA12,B21+VA12,B21. |
Lemma 2.5. Δ(Aii)⊆Aii (i=1,2).
Proof. Let A11∈A11. From A∗11P∗2P2=0, [[A11,P2]∗,P2]∗=0 and Δ(P2)=0, we have
0=Δ([[A11,P2]∗,P2]∗)=[[Δ(A11),P2]∗,P2]∗=Δ(A11)P2−P2Δ(A11)∗P2−P2Δ(A11)∗+P2Δ(A11)P2. | (2.16) |
Multiplying (2.16) by P1 from the left, we get P1Δ(A11)P2=0. Since P∗2A∗11P1=0, [[P2,A11]∗,P1]∗=0 and Δ(P1)=Δ(P2)=0, we have
0=Δ([[P2,A11]∗,P1]∗)=[[P2,Δ(A11)]∗,P1]∗=P2Δ(A11)P1−P1Δ(A11)∗P2. | (2.17) |
Multiplying (2.17) by P2 from the left, we get P2Δ(A11)P1=0. For any X12∈A12, from X∗12A∗11P2=0, [[X12,A11]∗,P2]∗=0 and Δ(P2)=0, we have
0=Δ([[X12,A11]∗,P2]∗)=[[Δ(X12),A11]∗,P2]∗+[[X12,Δ(A11)]∗,P2]∗=−A11Δ(X12)∗P2+P2Δ(X12)A∗11+X12Δ(A11)P2−P2Δ(A11)∗X∗12. | (2.18) |
Multiplying (2.18) by P1 from the left, we get −A11Δ(X12)∗P2+X12Δ(A11)P2=0. It follows from Remark 2.1(b) that X12Δ(A11)P2=A11(P2Δ(X12)P1)∗=0. Then P2Δ(A11)P2=0. Hence Δ(A11)⊆A11. Similarly, Δ(A22)⊆A22.
Lemma 2.6. Δ(Aij)⊆Aij (1≤i≠j≤2).
Proof. Let A12∈A12. Then P2Δ(A12)P1=0 by Remark 2.1(b). For any X12∈A12, from X∗12A∗12P1=0 and Δ(P1)=0, we have
Δ(−A12X∗12+X12A∗12)=Δ([[X12,A12]∗,P1]∗)=[[Δ(X12),A12]∗,P1]∗+[[X12,Δ(A12)]∗,P1]∗=−A12Δ(X12)∗P1+P1Δ(X12)A∗12+X12Δ(A12)P1−Δ(A12)X∗12−P1Δ(A12)∗X∗12+X12Δ(A12)∗. | (2.19) |
Multiplying (2.19) by P2 from the left and by P1 from the right, then by Lemma 2.5, we get P2Δ(A12)X∗12=0. Hence P2Δ(A12)P2=0. Since A∗12X∗12P2=0, [[A12,X12]∗,P2]∗=0 and Δ(P2)=0, we have
0=Δ([[A12,X12]∗,P2]∗)=[[Δ(A12),X12]∗,P2]∗+[[A12,Δ(X12)]∗,P2]∗=Δ(A12)X12−X12Δ(A12)∗P2−X∗12Δ(A12)∗+P2Δ(A12)X∗12+A12Δ(X12)P2−P2Δ(X12)∗A∗12. | (2.20) |
Multiplying (2.20) by P1 from the left and by P2 from the right, then by P2Δ(A12)P2=P2Δ(X12)P2=0, we have P1Δ(A12)X12=0. It follows that P1Δ(A12)P1=0. Therefore Δ(A12)⊆A12. Similarly, Δ(A21)⊆A21.
Lemma 2.7. For any Aii∈Aii,Bij∈Aij,Bji∈Aji (1≤i≠j≤2), we have
(a) Δ(Aii+Bij)=Δ(Aii)+Δ(Bij);
(b) Δ(Aii+Bji)=Δ(Aii)+Δ(Bji).
Proof. (a) Let T=Δ(Aii+Bij)−Δ(Aii)−Δ(Bij). Since (iPj)∗I∗(Aii+Bij)=(iPj)∗I∗Aii=(iPj)∗I∗Bij=0 and [[iPj,I]∗,Aii]∗=0, we have
[[Δ(iPj),I]∗,Aii+Bij]∗+[[iPj,Δ(I)]∗,Aii+Bij]∗+[[iPj,I]∗,Δ(Aii+Bij)]∗=Δ([[iPj,I]∗,Aii+Bij]∗)=Δ([[iPj,I]∗,Aii]∗)+Δ([[iPj,I]∗,Bij]∗)=[[Δ(iPj),I]∗,Aii+Bij]∗+[[iPj,Δ(I)]∗,Aii+Bij]∗+[[iPj,I]∗,Δ(Aii)+Δ(Bij)]∗, |
which implies
[[iPj,I]∗,T]∗=0. | (2.21) |
Multiplying (2.21) by Pi from the left, by Pi from the right, by Pj from both sides, respectively, we get Tij=Tji=Tjj=0. Hence
Δ(Aii+Bij)=Δ(Aii)+Δ(Bij)+Tii. | (2.22) |
For any Xij∈Aij, from (Aii+Bij)∗X∗ijPj=A∗iiX∗ijPj=B∗ijX∗ijPj=0, [[Bij,Xij]∗,Pj]∗=0 and (2.22), we have
[[Δ(Aii)+Δ(Bij)+Tii,Xij]∗,Pj]∗+[[Aii+Bij,Δ(Xij)]∗,Pj]∗+[[Aii+Bij,Xij]∗,Δ(Pj)]∗=[[Δ(Aii+Bij),Xij]∗,Pj]∗+[[Aii+Bij,Δ(Xij)]∗,Pj]∗+[[Aii+Bij,Xij]∗,Δ(Pj)]∗=Δ([[Aii+Bij,Xij]∗,Pj]∗)=Δ([[Aii,Xij]∗,Pj]∗)+Δ([[Bij,Xij]∗,Pj]∗)=[[Δ(Aii)+Δ(Bij),Xij]∗,Pj]∗+[[Aii+Bij,Δ(Xij)]∗,Pj]∗+[[Aii+Bij,Xij]∗,Δ(Pj)]∗. |
This implies
[[Tii,Xij]∗,Pj]∗=0. | (2.23) |
Multiplying (2.23) by Pj from the right, we see that TiiXij=0. Hence Tii=0, and so we obtain (a).
Similarly, we can show that (b) holds.
Lemma 2.8. For any Aij,Bij∈Aij (1≤i≠j≤2), we have
Δ(Aij+Bij)=Δ(Aij)+Δ(Bij). |
Proof. For any A12,B12∈A12, it follows that
[[P1+A12,P2+B12]∗,P2]∗=A12+B12−A∗12−B∗12. | (2.24) |
Then by (2.24) and Remark 2.1(c), there exist UA12+B12,−A∗12−B∗12∈A11, VA12+B12,−A∗12−B∗12 ∈A22 such that
Δ([[P1+A12,P2+B12]∗,P2]∗)=Δ(A12+B12)+Δ(−A∗12−B∗12)+UA12+B12,−A∗12−B∗12+VA12+B12,−A∗12−B∗12. | (2.25) |
From (P1+A12)∗(P2+B12)∗P2=0, Δ(P1)=Δ(P2)=0, (2.25), Lemmas 2.6 and 2.7, we have
Δ(A12+B12)+Δ(−A∗12−B∗12)+UA12+B12,−A∗12−B∗12+VA12+B12,−A∗12−B∗12=Δ([[P1+A12,P2+B12]∗,P2]∗)=[[Δ(A12),P2+B12]∗,P2]∗+[[P1+A12,Δ(B12)]∗,P2]∗=Δ(A12)+Δ(B12)−Δ(A12)∗−Δ(B12)∗. | (2.26) |
Multiplying (2.26) by P1 from the left and by P2 from the right, then by Lemma 2.6 and the fact that UA12+B12,−A∗12−B∗12∈A11,VA12+B12,−A∗12−B∗12∈A22, we see that Δ(A12+B12)=Δ(A12)+Δ(B12). Similarly, we can show that Δ(A21+B21)=Δ(A21)+Δ(B21).
Lemma 2.9. For any Aii,Bii∈Aii (i=1,2), we have
Δ(Aii+Bii)=Δ(Aii)+Δ(Bii). |
Proof. For any A11,B11∈A11,B12∈A12, from A∗11B∗12P2=0, Δ(P2)=0, [[A11,B12]∗,P2]∗=A11B12−B∗12A∗11, Lemmas 2.5, 2.6 and 2.8, we have
Δ(A11B12)+Δ(−B∗12A∗11)=Δ([[A11,B12]∗,P2]∗)=[[Δ(A11),B12]∗,P2]∗+[[A11,Δ(B12)]∗,P2]∗=Δ(A11)B12+A11Δ(B12)−B∗12Δ(A11)∗−Δ(B12)∗A∗11. | (2.27) |
Multiplying (2.27) by P1 from the left and by P2 from the right, we have
Δ(A11B12)=Δ(A11)B12+A11Δ(B12). | (2.28) |
Similarly, we can show that
Δ(A22B21)=Δ(A22)B21+A22Δ(B21). | (2.29) |
For any X12∈A12, it follows from Lemma 2.8 and (2.28) that
Δ(A11+B11)X12+(A11+B11)Δ(X12)=Δ((A11+B11)X12)=Δ(A11X12)+Δ(B11X12)=Δ(A11)X12+A11Δ(X12)+Δ(B11)X12+B11Δ(X12). |
It follows that (Δ(A11+B11)−Δ(A11)−Δ(B11))X12=0. Then Δ(A11+B11)=Δ(A11)+Δ(B11). Similarly, we can show that Δ(A22+B22)=Δ(A22)+Δ(B22).
Lemma 2.10. For any A12∈A12,B21∈A21, we have
Δ(A12+B21)=Δ(A12)+Δ(B21). |
Proof. For any X12∈A12, by X∗12(A12+B21)∗P1=X∗12A∗12P1=X∗12B∗21P1=0,
[[X12,A12+B21]∗,P1]∗=[[X12,A12]∗,P1]∗+[[X12,B21]∗,P1]∗∈A11, |
Remark 2.1(c) and Lemma 2.9, there exist UA12,B21∈A11,VA12,B21∈A22 such that
[[Δ(X12),A12+B21]∗,P1]∗+[[X12,Δ(A12)+Δ(B21)+UA12,B21+VA12,B21]∗,P1]∗+[[X12,A12+B21]∗,Δ(P1)]∗=Δ([[X12,A12+B21]∗,P1]∗)=Δ([[X12,A12]∗,P1]∗)+Δ([[X12,B21]∗,P1]∗)=[[Δ(X12),A12+B21]∗,P1]∗+[[X12,Δ(A12)+Δ(B21)]∗,P1]∗+[[X12,A12+B21]∗,Δ(P1)]∗. |
Then
0=[[X12,UA12,B21+VA12,B21]∗,P1]∗=−VA12,B21X∗12+X12V∗A12,B21. | (2.30) |
Multiplying (2.30) by P1 from the right, we get VA12,B21X∗12=0. Hence VA12,B21=0. Then by Remark 2.1(c), we get
Δ(A12+B21)=Δ(A12)+Δ(B21)+UA12,B21. | (2.31) |
For any X21∈A21, from X∗21(A12+B21)∗P2=X∗21A∗12P2=X∗21B∗21P2=0,
[[X21,A12+B21]∗,P2]∗=[[X21,A12]∗,P2]∗+[[X21,B21]∗,P2]∗∈A22, |
Lemma 2.9 and (2.31), we have
[[Δ(X21),A12+B21]∗,P2]∗+[[X21,Δ(A12)+Δ(B21)+UA12,B21]∗,P2]∗+[[X21,A12+B21]∗,Δ(P2)]∗=Δ([[X21,A12+B21]∗,P2]∗)=Δ([[X21,A12]∗,P2]∗)+Δ([[X12,B21]∗,P2]∗)=[[Δ(X21),A12+B21]∗,P2]∗+[[X21,Δ(A12)+Δ(B21)]∗,P2]∗+[[X21,A12+B21]∗,Δ(P2)]∗, |
which implies
0=[[X21,UA12,B21]∗,P2]∗=−UA12,B21X∗21+X21U∗A12,B21. | (2.32) |
Multiplying (2.32) by P2 from the right, we obtain UA12,B21X∗21=0. Then UA12,B21=0. Hence we obtain the desired result.
Lemma 2.11. For any A11∈A11,B12∈A12,C21∈A21,D22∈A22, we have
(a) Δ(A11+B12+C21)=Δ(A11)+Δ(B12)+Δ(C21);
(b) Δ(B12+C21+D22)=Δ(B12)+Δ(C21)+Δ(D22).
Proof. (a) Let T=Δ(A11+B12+C21)−Δ(A11)−Δ(B12)−Δ(C21). From P∗2(A11+B12+C21)∗P2=P∗2A∗11P2=P∗2B∗12P2=P∗2C∗21P2=0 and [[P2,A11]∗,P2]∗=[[P2,C21]∗,P2]∗=0, we have
[[Δ(P2),A11+B12+C21]∗,P2]∗+[[P2,Δ(A11+B12+C21)]∗,P2]∗+[[P2,A11+B12+C21]∗,Δ(P2)]∗=Δ([[P2,A11+B12+C21]∗,P2]∗)=Δ([[P2,A11]∗,P2]∗)+Δ([[P2,B12]∗,P2]∗)+Δ([[P2,C21]∗,P2]∗)=[[Δ(P2),A11+B12+C21]∗,P2]∗+[[P2,Δ(A11)+Δ(B12)+Δ(C21)]∗,P2]∗+[[P2,A11+B12+C21]∗,Δ(P2)]∗. |
This implies
[[P2,T]∗,P2]∗=0. | (2.33) |
Multiplying (2.33) by P1 from the left, we have T12=0. For any X12∈A12, from P∗1X∗12(A11+B12+C21)=P∗1X∗12A11=P∗1X∗12B12=P∗1X∗12C21=0, [[P1,X12]∗,A11+B12+C21]∗=X12C21−B12X∗12 and Lemma 2.9, we have
[[Δ(P1),X12]∗,A11+B12+C21]∗+[[P1,Δ(X12)]∗,A11+B12+C21]∗+[[P1,X12]∗,Δ(A11+B12+C21)]∗=Δ([[P1,X12]∗,A11+B12+C21]∗)=Δ(X12C21)+Δ(−B12X∗12)=Δ([[P1,X12]∗,A11]∗)+Δ([[P1,X12]∗,B12]∗)+Δ([[P1,X12]∗,C21]∗)=[[Δ(P1),X12]∗,A11+B12+C21]∗+[[P1,Δ(X12)]∗,A11+B12+C21]∗+[[P1,X12]∗,Δ(A11)+Δ(B12)+Δ(C21)]∗, |
which implies
[[P1,X12]∗,T]∗=0. | (2.34) |
Multiplying (2.34) by P1 from both sides, we obtain X12TP1−P1TX∗12=0. Then X12TP1=0 by T12=0. Hence T21=0. Multiplying (2.34) by P2 from the right, we have X12TP2=0 and so T22=0. Let SA11,B12,C21=T11. Then SA11,B12,C21∈A11 and
Δ(A11+B12+C21)=Δ(A11)+Δ(B12)+Δ(C21)+SA11,B12,C21. |
Similarly, there exists a RB12,C21,D22∈A22 such that
Δ(B12+C21+D22)=Δ(B12)+Δ(C21)+Δ(D22)+RB12,C21,D22. | (2.35) |
For any X21∈A21, by [[P2,X21]∗,A11+B12+C21]∗=−A11X∗21+X21A11+X21B12−C21X∗21 and (2.35), there exist a R−A11X∗21,X21A11,X21B12−C21X∗21∈A22 such that
Δ([[P2,X21]∗,A11+B12+C21]∗)=Δ(−A11X∗21)+Δ(X21A11)+Δ(X21B12−C21X∗21)+R−A11X∗21,X21A11,X21B12−C21X∗21. | (2.36) |
From P∗2X∗21(A11+B12+C21)=P∗2X∗21A11=P∗2X∗21B12=P∗2X∗21C21=0, (2.36), Lemmas 2.9 and 2.10, we have
[[Δ(P2),X21]∗,A11+B12+C21]∗+[[P2,Δ(X21)]∗,A11+B12+C21]∗+[[P2,X21]∗,Δ(A11+B12+C21)]∗=Δ([[P2,X21]∗,A11+B12+C21]∗)=Δ(−A11X∗21)+Δ(X21A11)+Δ(X21B12−C21X∗21)+R−A11X∗21,X21A11,X21B12−C21X∗21=Δ(−A11X∗21+X21A11)+Δ(X21B12)+Δ(−C21X∗21)+R−A11X∗21,X21A11,X21B12−C21X∗21=Δ([[P2,X21]∗,A11]∗)+Δ([[P2,X21]∗,B12]∗)+Δ([[P2,X21]∗,C21]∗)+R−A11X∗21,X21A11,X21B12−C21X∗21=[[Δ(P2),X21]∗,A11+B12+C21]∗+[[P2,Δ(X21)]∗,A11+B12+C21]∗+[[P2,X21]∗,Δ(A11)+Δ(B12)+Δ(C21)]∗+R−A11X∗21,X21A11,X21B12−C21X∗21. |
It follows that
[[P2,X21]∗,T]∗=R−A11X∗21,X21A11,X21B12−C21X∗21. | (2.37) |
Multiplying (2.37) by P1 from the right, then by R−A11X∗21,X21A11,X21B12−C21X∗21∈A22, we obtain X21TP1=0. Hence SA11,B12,C21=T11=0, and so Δ(A11+B12+C21)=Δ(A11)+Δ(B12)+Δ(C21).
Similarly, we can show that (b) holds.
Lemma 2.12. For any A11∈A11,B12∈A12,C21∈A21,D22∈A22, we have
Δ(A11+B12+C21+D22)=Δ(A11)+Δ(B12)+Δ(C21)+Δ(D22). |
Proof. Let T=Δ(A11+B12+C21+D22)−Δ(A11)−Δ(B12)−Δ(C21)−Δ(D22). From (A11+B12+C21+D22)∗P∗1P2=A∗11P∗1P2=B∗12P∗1P2=C∗21P∗1P2=D∗22P∗1P2=0 and [[A11+B12+C21+D22,P1]∗,P2]∗=−C∗21+C21, we have
[[Δ(A11+B12+C21+D22),P1]∗,P2]∗+[[A11+B12+C21+D22,Δ(P1)]∗,P2]∗+[[A11+B12+C21+D22,P1]∗,Δ(P2)]∗=Δ([[A11+B12+C21+D22,P1]∗,P2]∗)=Δ([[A11,P1]∗,P2]∗)+Δ([[B12,P1]∗,P2]∗)+Δ([[C21,P1]∗,P2]∗)+Δ([[D22,P1]∗,P2]∗)=[[Δ(A11)+Δ(B12)+Δ(C21)+Δ(D22),P1]∗,P2]∗+[[A11+B12+C21+D22,Δ(P1)]∗,P2]∗+[[A11+B12+C21+D22,P1]∗,Δ(P2)]∗. |
This implies
[[T,P1]∗,P2]∗=0. | (2.38) |
Multiplying (2.38) by P2 from the left, we have T21=0. Similarly, T12=0. For any X12∈A12, from P∗1X∗12(A11+B12+C21+D22)=P∗1X∗12A11=P∗1X∗12B12=P∗1X∗12C21=P∗1X∗12D22=0, [[P1,X12]∗,A11+B12+C21+D22]∗=X12C21−B12X∗12+X12D22−D22X∗12, Lemmas 2.9 and 2.12, we have
[[Δ(P1),X12]∗,A11+B12+C21+D22]∗+[[P1,Δ(X12)]∗,A11+B12+C21+D22]∗+[[P1,X12]∗,Δ(A11+B12+C21+D22)]∗=Δ([[P1,X12]∗,A11+B12+C21+D22]∗)=Δ(X12C21)+Δ(−B12X∗12)+Δ(X12D22−D22X∗12)=Δ([[P1,X12]∗,A11]∗)+Δ([[P1,X12]∗,B12]∗)+Δ([[P1,X12]∗,C21]∗)+Δ([[P1,X12]∗,D22]∗)=[[Δ(P1),X12]∗,A11+B12+C21+D22]∗+[[P1,Δ(X12)]∗,A11+B12+C21+D22]∗+[[P1,X12]∗,Δ(A11)+Δ(B12)+Δ(C21)+Δ(D22)]∗. |
This implies
[[P1,X12]∗,T]∗=0. | (2.39) |
Multiplying (2.39) by P2 from the right, we obtain X12TP2=0. Then T22=0. Similarly, T11=0. Hence we obtain the desired result.
Lemma 2.13. For any Aii,Bii∈Aii,Aij,Bij∈Aij,Bji∈Aji,Bjj∈Ajj (1≤i≠j≤2), we have
(a) Δ(AiiBij)=Δ(Aii)Bij+AiiΔ(Bij);
(b) Δ(AijBjj)=Δ(Aij)Bjj+AijΔ(Bjj);
(c) Δ(AiiBii)=Δ(Aii)Bii+AiiΔ(Bii);
(d) Δ(AijBji)=Δ(Aij)Bji+AijΔ(Bji).
Proof. (a) It follows from (2.28) and (2.29) that (a) holds.
(b) Let A12∈A12,B22∈A22. From A∗12B∗22P2=0, Δ(P2)=0, [[A12,B22]∗,P2]∗=A12B22−B∗22A∗12, Lemmas 2.5, 2.6 and 2.12, we have
Δ(A12B22)+Δ(−B∗22A∗12)=Δ([[A12,B22]∗,P2]∗)=[[Δ(A12),B22]∗,P2]∗)+[[A12,Δ(B22)]∗,P2]∗=Δ(A12)B22+A12Δ(B22)−B∗22Δ(A12)∗−Δ(B22)∗A∗12. | (2.40) |
Multiplying (2.40) by P1 from the left and by P2 from the right, we have Δ(A12B22)=Δ(A12)B22+A12Δ(B22). Similarly, Δ(A21B11)=Δ(A21)B11+A21Δ(B11).
(c) Let A11,B11∈A11,X12∈A12. It follows from (a) that
Δ(A11B11)X12+A11B11Δ(X12)=Δ(A11B11X12)=Δ(A11)B11X12+A11Δ(B11X12)=Δ(A11)B11X12+A11Δ(B11)X12+A11B11Δ(X12). |
It follows that (Δ(A11B11)−Δ(A11)B11−A11Δ(B11))X12=0. Hence Δ(A11B11)=Δ(A11)B11+A11Δ(B11). Similarly, Δ(A22B22)=Δ(A22)B22+A22Δ(B22).
(d) Let A12∈A12,B21∈A21. From B∗21P∗1A12=0, Δ(P1)=0, [[B21,P1]∗,A12]∗=B21A12+A12B21, Lemmas 2.6 and 2.12, we have
Δ(B21A12)+Δ(A12B21)=Δ([[B21,P1]∗,A12]∗)=[[Δ(B21),P1]∗,A12]∗+[[B21,P1]∗,Δ(A12)]∗=Δ(B21)A12+A12Δ(B21)+B21Δ(A12)+Δ(A12)B21. | (2.41) |
Multiplying (2.41) by P1 from both sides, we have Δ(A12B21)=Δ(A12)B21+A12Δ(B21). Similarly, Δ(A21B12)=Δ(A21)B12+A21Δ(B12).
Now, we give the proof of Theorem 2.1 in the following.
Proof of Theorem 2.1. By Lemmas 2.5, 2.6, 2.8, 2.9, 2.12 and 2.13, it is easy to verify that Δ is an additive derivation on A. Let Aij∈Aij (1≤i≠j≤2). By A∗ijP∗jPj=0, Δ(Pj)=0 and Lemma 2.6, we have
Δ(Aij)−Δ(A∗ij)=Δ([[Aij,Pj]∗,Pj]∗)=[[Δ(Aij),Pj]∗,Pj]∗=Δ(Aij)−Δ(Aij)∗. |
It follows that
Δ(A∗ij)=Δ(Aij)∗. | (2.42) |
Let Aii∈Aii, Xji∈Aji (1≤i≠j≤2). Since A∗iiP∗iXji=0, Δ(Pi)=0, [[Aii,Pi]∗,Xji]∗=XjiAii−XjiA∗ii, Lemmas 2.5, 2.6 and 2.13(b), we have
Δ(Xji)Aii+XjiΔ(Aii)−Δ(Xji)A∗ii−XjiΔ(A∗ii)=Δ(XjiAii)−Δ(XjiA∗ii)=Δ([[Aii,Pi]∗,Xji]∗)=[[Δ(Aii),Pi]∗,Xji]+[[Aii,Pi]∗,Δ(Xji)]=Δ(Xji)Aii+XjiΔ(Aii)−Δ(Xji)A∗ii−XjiΔ(Aii)∗. |
It follows that Xji(Δ(A∗ii)−Δ(Aii)∗)=0. Then
Δ(A∗ii)=Δ(Aii)∗. | (2.43) |
For any A∈A, we have A=∑2i,j=1Aij. By (2.42), (2.43) and the additivity of Δ on A, it follows that
Δ(A∗)=2∑i,j=1Δ(A∗ij)=2∑i,j=1Δ(Aij)∗=Δ(A)∗. |
Hence Δ is an additive ∗-derivation. Therefore, δ is an additive ∗-derivation on A by Remark 2.1.
In this paper, we gave the characterization of a kind of non-global nonlinear skew Lie triple derivations on factor von Neumann algebras.
This research is supported by Scientific Research Project of Shangluo University (21SKY104).
The authors declare that there are no conflicts of interest.
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1. | Mohammad Ashraf, Md Shamim Akhter, Mohammad Afajal Ansari, Non-global nonlinear skew Lie n -derivations on *-algebras , 2024, 52, 0092-7872, 3734, 10.1080/00927872.2024.2328802 |