# Nonlinear Jordan Derivable Mappings of Generalized Matrix Algebras by Lie Product Square-Zero Elements

Authors: Xiuhai Fei; Haifang Zhang
Journal: Journal of Mathematics
 (2021)

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## Abstract

The aim of the paper is to give a description of nonlinear Jordan derivable mappings of a certain class of generalized matrix algebras by Lie product square-zero elements. We prove that under certain conditions, a nonlinear Jordan derivable mappingΔ of a generalized matrix algebra by Lie product square-zero elements is a sum of an additive derivation δ and an additive antiderivation f. Moreover, δ and f are uniquely determined.


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## Body

## 1. Introduction

Throughout the paper, by an algebra we shall mean an algebra over a fixed unital commutative ringℛ. Let A be a unital algebra, P be a fixed element of A, Ω=A∈A:A2=0, and Δ:A⟼A be an additive (resp., without assumption of additivity) mapping. For any A,B∈A, denote the Jordan product (resp., Lie product) of AandB by A∘B=AB+BA (resp., A,B=AB−BA). For any A∈A, if 2A=0, implying A=0, then A is said to be a 2-torsion free algebra. Recall that Δ is called an additive derivation (resp., nonlinear derivable mapping) if ΔAB=ΔAB+AΔB for all A,B∈A. It is called an additive antiderivation (resp., nonlinear antiderivable mapping) if ΔAB=ΔBA+BΔA for all A,B∈A. It is called an additive Jordan derivation (resp., nonlinear Jordan derivable mapping) if ΔA∘B=ΔA∘B+A∘ΔB for all A,B∈A. It is called a P point derivable mapping (resp., P point nonlinear derivable mapping) if ΔAB=ΔAB+AΔB for all A,B∈A with AB=P. Similarly, we have the definitions of P point Jordan derivable mapping and P point nonlinear Jordan derivable mapping. In this paper, if Δ is without assumption of additivity and satisfies(1)ΔA∘B=ΔA∘B+A∘ΔB,for all A,B∈A with A,B∈Ω, then we say Δ is a nonlinear Jordan derivable mapping of A by Lie product square-zero elements. Obviously, every additive derivation or additive antiderivation is an additive Jordan derivation. However, the inverse statement is not true in general (see [1]).A natural and very interesting problem that we are dealing with is studying those conditions on a ring or an algebra such that every additive Jordan derivation or every nonlinear Jordan derivable map is an additive derivation.In the past few decades, the research on this problem has attracted the attention of many mathematicians. For examples, Herstein in [2] proved that every additive Jordan derivation on a prime ring not of characteristic 2 is an additive derivation. Later, this result was extended by Cusack in [3] and Brešar and Vukman in [4] to the case of semiprime ring, respectively. Zhang and Yu in [5, 6] showed that every additive Jordan derivation on a nest algebra and a 2-torsion free triangular algebra is an inner derivation and an additive derivation, respectively. For some conclusions about P point Jordan derivable mapping, we refer the readers to [7–9] and references therein for more details. Lu in [10] showed that every nonlinear Jordan derivable mapping on a 2-torsion free semiprime ring is an additive derivation. Ashraf and Jabeen in [11] showed that every nonlinear Jordan triple derivable mapping on a 2-torsion free triangular algebra is an additive derivation. In particular, Benkovič in [1] proved that every additive Jordan derivation from an upper triangular matrix algebra to its bimodule is a sum of an additive derivation and an additive antiderivation. Li et al. in [12] proved that under certain conditions every additive Jordan derivation on a generalized matrix algebra is a sum of an additive derivation and an additive antiderivation. For other similar results, we refer the readers to [13, 14] and references therein for more details.Inspired by the above references, in this paper, we study the nonlinear Jordan derivable mapping of generalized matrix algebras by Lie product square-zero elements, we get that under certain conditions, a nonlinear Jordan derivable mapping of a generalized matrix algebra by Lie product square-zero elements is a sum of an additive derivation and an additive antiderivation.


## 2. Generalized Matrix Algebras

LetA and ℬ be two unital algebras with unit elements 1A and 1ℬ, respectively, ℳ be a faithful A,ℬ-bimodule, and N be a ℬ,A-bimodule. Then,(2)G=AℳNℬ=AMNB:A∈A,M∈ℳ,N∈N,B∈ℬis a ℛ-algebra under matrix-like addition and multiplication, satisfying the following commutative diagrams:(3) (4)where ΦMN:ℳ⊗ℬN⟼A and ΨNM:N⊗Aℳ⟼ℬ are two bimodule homomorphisms. Such a ℛ-algebra G is called a generalized matrix algebra. For convenience in reading, we give the multiplication of generalized matrix algebra as follows:(5)AMNBA′M′N′B′=AA′+ΦMNM⊗N′AM′+MB′NA′+BN′BB′+ΨNMN⊗M′,for all A,A′∈A,M,M′∈ℳ,N,N′∈N, and B,B′∈ℬ. Furthermore, when N=0, then G is a triangular algebra. The most common examples of generalized matrix algebras are triangular algebras and full matrix algebras (see [15, 16] for details).Considering a generalized matrix algebraG, let 1 be the unit element of G. Set(6)P1=1A000,P2=1−P1=0001ℬ,and Gij=PiGPj1≤i,j≤2. It is clear that if ΦMNM⊗N′=0=ΨNMN⊗M′, then G can be represented as(7)G=A+ℳ+N+ℬ=G11+G12+G21+G22.Hence, for anyA∈G, A can be represented as A=A11+A12+A21+A22, where Aij∈Gij1≤i,j≤2, and we can easily check that AijBji=0,∀Aij∈Gij,andBji∈Gji1≤i≠j≤2 (i.e., MN=NM=0).


## 3. Main Results

In this paper, our main result is the following theorem.Theorem 1.
LetG be a 2-torsion free generalized matrix algebra and Δ be a nonlinear Jordan derivable mapping of G by Lie product square-zero elements. If G satisfies MN=NM=0, then there exist an additive derivation δ and an additive antiderivation f of G, such that ΔA=δA+fA for all A∈G. Moreover, δ and f are uniquely determined.In order to prove Theorem1, we introduce some lemmas. Next, we assume that G is a 2-torsion generalized matrix algebra, ℳ is a faithful A,ℬ-bimodule and N is a ℬ,A-bimodule (i.e., G12 is a faithful G11,G22-bimodule and G21 is a G22,G11-bimodule), MN=NM=0 (i.e., AijBji=0,∀Aij∈Gij,andBji∈Gji1≤i≠j≤2), Ω=A∈G:A2=0, and Δ is a nonlinear Jordan derivable mapping of G by Lie product square-zero elements.Lemma 1.
For anyAii∈Gii,Aij,Bij∈Gij,Cji∈Gji1≤i≠j≤2, then(i)
Δ0=0(ii)
ΔPi=PiΔPiPj+PjΔPiPi, ΔP1=−ΔP2, and Aij∘ΔP1=Aij∘ΔP2=0(iii)
ΔAij=PiΔAijPj+PjΔAijPi and Aij∘ΔBij=Aij∘ΔCji=0(iv)
PjΔAiiPj=0,PiΔAiiPj=AiiΔPi and PjΔAiiPi=ΔPiAiiProof.
(i)
SinceΔ is a nonlinear Jordan derivable mapping of G by Lie product square-zero elements, then for any A,B∈G with A,B∈Ω, we have(8)ΔA∘B=ΔA∘B+A∘ΔB.Therefore, takingA=B=0 in equation (8), we have Δ0=Δ0∘0+0∘Δ0=0.(ii)
SincePi,Pj=0∈Ω1≤i≠j≤2 and taking A=PiandB=Pj in equation (8), we have(9)0=ΔPi∘Pj=ΔPiPj+PjΔPi+PiΔPj+ΔPjPi.Multiplying equation (9) from both the sides by Pj and then by the property of 2-torsion freeness of G, we have(10)PjΔPiPj=0.Multiplying equation (9) from the left by Pi and from the right by Pj, we get(11)PiΔPiPj=−PiΔPjPj.Next, we show thatPiΔPiPi=0 (1≤i≤2). Indeed, for any X12∈G12, since P1,X12=X12∈Ω and taking A=P1andB=X12 in equation (8), we obtain(12)ΔX12=ΔP1∘X12=ΔP1∘X12+P1∘ΔX12=ΔP1X12+X12ΔP1+P1ΔX12+ΔX12P1.Multiplying equation (12) from the left by P1 and from the right by P2 and then by equation (10), we have P1ΔP1X12=P1ΔP1P1X12=0; similarly, we can get that X12ΔP2P2=X12P2ΔP2P2=0; therefore, we obtain from the faithfulness of G12 that(13)P1ΔP1P1=0,P2ΔP2P2=0.Thus, we obtain from equations (10)–(12) that ΔPi=PiΔPiPj+PjΔPiPi (1≤i≠j≤2) and ΔP1=−ΔP2, and then, it follows from MN=NM=0 that Aij∘ΔP1=Aij∘ΔP2=0 for all Aij∈Gij1≤i≠j≤2.(iii)
For anyAij∈GijandCji∈Gji(1≤i≠j≤2), since Pi,Aij=Aij∈Ω and taking A=PiandB=Aij in equation (8), it follows from Lemma 1 (ii) that(14)ΔAij=ΔPi∘Aij=ΔPi∘Aij+Pi∘ΔAij=PiΔAij+ΔAijPi.Multiplying (14) by Pi and Pj from both sides, respectively, we get that(15)PjΔAijPj=PiΔAijPi=0.This implies thatΔAij=PiΔAijPj+PjΔAijPi (1≤i≠j≤2), and then by MN=NM=0, we get Aij∘ΔBij=Aij∘ΔCji=0.(iv)
For anyAii∈Gii (1≤i≤2), since Pj,Aii=0∈Ω (1≤i≠j≤2) and taking A=PjandB=Aii in equation (8), we get(16)0=ΔPj∘Aii=ΔPjAii+AiiΔPj+PjΔAii+ΔAiiPj.
Multiplying equation (16) by Pj from both sides and then by the property of 2-torsion freeness of G, we have PjΔAiiPj=0.
Multiplying equation (16) from the left by Pi and from the right by Pj, it follows from Lemma 1 (ii) that(17)PiΔAiiPj=−AiiΔPjPj=AiiΔPiPj=AiiΔPi.
Similarly, we can show thatPjΔAiiPi=ΔPiAii holds. The proof is completed.Lemma 2.
For anyAii,Bii∈Gii,Aij,Bij∈Gij,Aji∈Gji,andAjj∈Gjj1≤i≠j≤2, then(i)
ΔAii+Aij=ΔAii+ΔAij(ii)
ΔAij+Ajj=ΔAij+ΔAjj(iii)
ΔAij+Bij=ΔAij+ΔBij(iv)
ΔAij+Aji=ΔAij+ΔAji(v)
ΔAii+Bii=ΔAii+ΔBii(vi)
ΔAii+Aij+Aji=ΔAii+ΔAij+ΔAji(vii)
ΔAii+Aij+Aji+Ajj=ΔAii+ΔAij+ΔAji+ΔAjjProof.
(i)
For anyAii∈GiiandAij∈Gij1≤i≠j≤2, since Aii+Aij,Pj=Aij∈Ω, taking A=Aii+AijandB=Pj in equation (8), and so by Lemma 1 (ii)–(iv), we have(18)ΔAij=ΔAii+Aij∘Pj=ΔAii+Aij∘Pj+Aii+Aij∘ΔPj=ΔAii+AijPj+PjΔAii+Aij+AiiΔPj+ΔPjAii=ΔAii+AijPj+PjΔAii+Aij−PiΔAiiPj−PjΔAiiPi.This yields that(19)ΔAij+PiΔAiiPj+PjΔAiiPi=ΔAii+AijPj+PjΔAii+Aij.Multiplying equation (19) by Pj from both sides and then by the property of 2-torsion freeness of G and Lemma 1 (iii), we have(20)PjΔAii+AijPj=0.Multiplying equation (19) from the left by Pi and from the right by Pj, we get(21)PiΔAii+AijPj=PiΔAiiPj+PiΔAijPj.Multiplying equation (19) from the left by Pj and from the right by Pi, we get(22)PjΔAii+AijPi=PjΔAiiPi+PjΔAijPi.Next, we show thatPiΔAii+AijPi=PiΔAiiPi. Indeed, for any A11∈G11,A12,X12∈G12, since A11+A12,X12=A11,X12=A11X12∈Ω, on the one hand, taking A=A11+A12,B=X12 in equation (8), and then by Lemma 1 (iii), we get(23)ΔA11X12=ΔA11+A12∘X12=ΔA11+A12∘X12+A11+A12∘ΔX12=ΔA11+A12X12+X12ΔA11+A12+A11ΔX12+ΔX12A11.On the other hand, takingA=A11andB=X12 in equation (8), we get that(24)ΔA11X12=ΔA11∘X12=ΔA11∘X12+A11∘ΔX12=ΔA11X12+X12ΔA11+A11ΔX12+ΔX12A11.Comparing equations (23) and (24), we get(25)ΔA11+A12−ΔA11X12+X12ΔA11+A12−ΔA11=0.Multiplying equation (25) from the left by P1 and from the right by P2, then by equation (20) and Lemma 1 (iv), we get(26)P1ΔA11+A12−ΔA11P1X12=0.Similarly, for anyX12∈G12,A21∈G21,andA22∈G22, we can get that(27)X12P2ΔA22+A21−ΔA22P2=0.This yields from the faithfulness ofG12 that(28)P1ΔA11+A12P1=P1ΔA11P1andP2ΔA22+A21P2=P2ΔA22P2.Therefore, it follows from equations (20)–(28) and Lemma 1 (iii)-(iv) that ΔAii+Aij=ΔAii+ΔAij.(ii)
Similar to (i), we can show that (ii) holds.(iii)
For anyAij,Bij∈Gij1≤i≠j≤2, since Pi+Aij,Bij+Pj=Aij+Bij∈Ω, taking A=Pi+Aij,B=Bij+Pj in equation (8), and so by Lemma 2 (i)-(ii) and Lemma 1 (ii)–(iv), we have(29)ΔAij+Bij=ΔPi+Aij∘Bij+Pj=ΔPi+Aij∘Bij+Pj+Pi+Aij∘ΔBij+Pj=ΔPi+ΔAij∘Bij+Pj+Pi+Aij∘ΔBij+ΔPj=ΔPi∘Pj+ΔAij∘Pj+Pi∘ΔBij+Pi∘ΔPj=ΔPi+ΔAij+ΔBij+ΔPj=ΔAij+ΔBij.(iv)
For anyAij∈GijandAji∈Gji1≤i≠j≤2, it follows from MN=NM=0 that Pj+Aij,Aji+Pi=−Aij+AijAji−AjiAij+Aji=−Aij+Aij∈Ω; taking A=Pj+AijandB=Aji+Pi in equation (8) and we get from MN=NM=0, Lemma 2 (i)-(ii), and Lemma 1 (ii)–(iv) that(30)ΔAij+Aji=ΔPj+Aij∘Aji+Pi=ΔPj+Aij∘Aji+Pi+Pj+Aij∘ΔAji+Pi=ΔPj+ΔAij∘Aji+Pi+Pj+Aij∘ΔAji+ΔPi=ΔPj∘Pi+ΔAij∘Pi+Pj∘ΔAji+Pj∘ΔPi=ΔPj+ΔAij+ΔAji+ΔPi=ΔAij+ΔAji.(v)
For anyA11,B11∈G11, it follows from Lemma 1 (iv) that(31)ΔA11=P1ΔA11P1+A11ΔP1+ΔP1A11,(32)ΔB11=P1ΔB11P1+B11ΔP1+ΔP1B11,(33)ΔA11+B11=P1ΔA11+B11P1+A11+B11ΔP1+ΔP1A11+B11.In the following, we show thatP1ΔA11+B11P1=P1ΔA11P1+P1ΔB11P1. Indeed, for any A11,B11∈G11,X12∈G12, on the one hand, since A11+B11,X12=A11+B11X12∈Ω and taking A=A11+B11andB=X12 in equation (8), then we get(34)ΔA11+B11X12=ΔA11+B11∘X12+A11+B11∘ΔX12.On the other hand, sinceA11,X12=A11X12∈Ω and B11,X12=B11X12∈Ω, we get, respectively,(35)ΔA11X12=ΔA11∘X12=ΔA11∘X12+A11∘ΔX12,(36)ΔB11X12=ΔB11∘X12=ΔB11∘X12+B11∘ΔX12.Therefore, it follows from Lemma2 (iii) and equations (34)–(36) that(37)ΔA11+B11−ΔA11−ΔB11X12+X12ΔA11+B11−ΔA11−ΔB11=0.Multiplying equation (37) from the left by P1 and from the right by P2 and then by Lemma 1 (iv), we get(38)P1ΔA11+B11−ΔA11−ΔB11P1X12=0.Therefore, by the faithfulness ofG12, we get that(39)P1ΔA11+B11P1=P1ΔA11P1+P1ΔB11P1.Hence, we get from equations (31)–(33) and (39) and Lemma 1 (iv) that ΔA11+B11=ΔA11+ΔB11. Similarly, we can show that ΔA22+B22=ΔA22+ΔB22 holds for all A22,B22∈G22.(vi)
For anyAii∈Gii,Aij∈Gij,andAji∈Gji1≤i≠j≤2, it follows from MN=NM=0 that Aii+Aij+Aji,Pj=Aij−Aji∈Ω; taking A=Aii+Aij+Aji,B=Pj in equation (8) and then by Lemma 1 (ii)–(iv) and Lemma 2 (iv), we obtain(40)ΔAij+ΔAji=ΔAij+Aji=ΔAii+Aij+Aji∘Pj=ΔAii+Aij+Aji∘Pj+Aii+Aij+Aji∘ΔPj=ΔAii+Aij+AjiPj+PjΔAii+Aij+Aji+AiiΔPj+ΔPjAii=ΔAii+Aij+AjiPj+PjΔAii+Aij+Aji−PiΔAiiPj−PjΔAiiPi.This implies that(41)PiΔAii+Aij+AjiPj=PiΔAiiPj+PiΔAijPj+PiΔAjiPj,(42)PjΔAii+Aij+AjiPi=PjΔAiiPi+PjΔAijPi+PjΔAjiPi,(43)PjΔAii+Aij+AjiPj=0.In the following, we show thatPiΔAii+Aij+AjiPi=PiΔAiiPi. Indeed, for any A11∈G11,A12,X12∈G12,A21∈G21, on the one hand, it follows from MN=NM=0 that A11+A12+A21,X12=A11X12−X12A21+A21X12=A11X12∈Ω; taking A=A11+A12+A21,B=X12 in equation (8) and so by MN=NM=0 and Lemma 1 (iii), we have(44)ΔA11X12=ΔA11+A12+A21∘X12=ΔA11+A12+A21∘X12+A11+A12+A21∘X12=ΔA11+A12+A21∘X12+A11∘X12.On the other hand, sinceA11,X12=A11X12∈Ω, we get(45)ΔA11X12=ΔA11∘X12=ΔA11∘X12+A11∘X12.Comparing equations (44) and (45), we get(46)ΔA11+A12+A21−ΔA11X12+X12ΔA11+A12+A21−ΔA11=0.Multiplying equation (46) from the left by P1 and from the right by P2 and then by Lemma 1 (iv), equation (43), and the faithfulness of G12, we get(47)P1ΔA11+A12+A21P1=P1ΔA11P1.Similarly, for anyA22∈G22,A12∈G12,andA21∈G21, we can get that(48)P2ΔA22+A12+A21P2=P2ΔA22P2.Hence, for anyAii∈Gii,Aij∈Gij,andAji∈Gji1≤i≠j≤2, we get that(49)PiΔAii+Aij+AijPi=PiΔAiiPi.Therefore, it follows from equations (41)–(48) and Lemma 1 (iii)-(iv) that ΔAii+Aij+Aji=ΔAii+ΔAij+ΔAji.(vii)
For anyAii∈Gii,Aij∈Gij,Aji∈Gji,andAjj∈Gjj1≤i≠j≤2, it follows from MN=NM=0 that Aii+Aij+Aji+Ajj,Pj=Aij−Aji∈Ω; taking A=Aii+Aij+Aji+Ajj,B=Pj in equation (8) and so by Lemma 2 (vi) and Lemma 1 (iv), we have(50)ΔAij+ΔAji+2ΔAjj=ΔAij+Aji+2Ajj=ΔAii+Aij+Aji+Ajj∘Pj=ΔAii+Aij+Aji+Ajj∘Pj+Aii+Aij+Aji+Ajj∘ΔPj=ΔAii+Aij+Aji+Ajj∘Pj+Aii+Ajj∘ΔPj.
Similarly, we obtain(51)2ΔAii+ΔAij+ΔAji=ΔAii+Aij+Aji+Ajj∘Pi+Aii+Ajj∘ΔPi.
Therefore, it follows from (50) and (51), Lemma 1 (ii), and the property of 2-torsion freeness of G that ΔAii+Aij+Aji+Ajj=ΔAii+ΔAij+ΔAji+ΔAjj. The proof is completed.Lemma 3.
Δ is an additive mapping.Proof.
For anyA,B∈G, let A=A11+A12+A21+A22 and B=B11+B12+B21+B22, where Aij,Bij∈Gij1≤i,j≤2, and by Lemma 2, we have(52)ΔA+B=ΔA11+A12+A21+A22+B11+B12+B21+B22=ΔA11+B11+A12+B12+A21+B21+A22+B22=ΔA11+B11+ΔA12+B12+ΔA21+B21+ΔA22+B22=ΔA11+ΔB11+ΔA12+ΔB12+ΔA21+ΔB21+ΔA22+ΔB22=ΔA11+A12+A21+A22+ΔB11+B12+B21+B22=ΔA+ΔB.
Therefore,Δ is an additive mapping on G.The proof is completed.Remark 1.
For anyA∈G, we define a mapping f:G⟼G as(53)fA=P2ΔP1AP2P1+P1ΔP2AP1P2.
Then, by the definition off, we can easily obtain that(54)fAii=0andfAij=PjΔAijPi,for all Aii∈Gii,Aij∈Gij(1≤i≠j≤2).
Next, we will show thatf is an additive antiderivation on G. First, we introduce Lemma 4 and get that f is an additive mapping, and then, we introduce Lemmas 5 and 6 and show that f is an additive antiderivation on G.Lemma 4.
Letf:G⟼G be as in Remark 1. Then, f is an additive mapping.Proof.
For anyA,B∈G, since we have shown that Δ is an additive mapping in Lemma 3 and then by the definition of f, we obtain that(55)fA+B=P2ΔP1A+BP2P1+P1ΔP2A+BP1P2=P2ΔP1AP2+P1BP2P1+P1ΔP2AP1+P2BP1P2=P2ΔP1AP2P1+P2ΔP1BP2P1+P1ΔP2AP1P2+P1ΔP2BP1P2=P2ΔP1AP2P1+P1ΔP2AP1P2+P2ΔP1BP2P1+P1ΔP2BP1P2=fA+fB.
Therefore,f is an additive mapping on G.The proof is completed.Lemma 5.
Letf:G⟼G be as in Remark 1. Then, for any Aii,Bii∈Gii,Aij,Bij∈Gij,Bji∈Gji,andBjj∈Gjj1≤i≠j≤2,(i)
fAiiBii=fBiiAii+BiifAii(ii)
fAiiBjj=fBjjAii+BjjfAii(iii)
fAijBji=fBjiAij+BjifAij(iv)
fAijBij=fBijAij+BijfAij(v)
fAiiBji=fBjiAii+BjifAii(vi)
fAjiBjj=fBjjAji+BjjfAji(vii)
fAiiBij=fBijAii+BijfAii(viii)
fAjiBii=fBiiAji+BiifAjiProof.
(i)
For anyAii,Bii∈Gii1≤i≤2, it follows from fAiiBii=0=fAii=fBii that fAiiBii=fBiiAii+BiifAii.(ii)
Similar to (i), we can showholds.(iii)
For anyAij∈GijandBji∈Gji1≤i≠j≤2, on the one hand, it follows from MN=NM=0 that fAijBji=f0=0. On the other hand, by the definition of f, we have fBji=PiΔBjiPj and fAij=PjΔAijPi and so fBjiAij=0=BjifAij; therefore, we have fAijBji=fBjiAij+BjifAij.(iv)
For anyAij,Bij∈Gij1≤i≠j≤2, on the one hand, we have fAijBij=f0=0. On the other hand, by the definition of f, we have fAij=PjΔAijPi and fBij=PjΔBijPi, and then, it follows from MN=NM=0 that fBijAij=PjΔBijPiAij=0 and BijfAij=BijPjΔAijPi=0 and so fAijBij=fBijAij+BijfAij.(v)
Similar to (iv), we can show that (v) holds.(vi)
Similar to (iv), we can show that (vi) holds.(vii)
For anyAii∈GiiandBij∈Gij1≤i≠j≤2, since Aii,Bij=AiiBij∈Ω, we have(56)ΔAiiBij=ΔAii∘Bij=ΔAii∘Bij+Aii∘ΔBij=ΔAiiBij+BijΔAii+AiiΔBij+ΔBijAii.
Hence, by the definition off, we have(57)fAiiBij=PjΔAiiBijPi=PjΔBijAii=fBijAii.
Therefore, we obtain fromfAii=0 that fAiiBij=fBijAii+BijfAii. Similarly, we can show viii holds. The proof is completed.Lemma 6.
Letf:G⟼G be as in Remark 3.1. Then, f is an additive antiderivation.Proof.
For anyA,B∈G, let A=A11+A12+A21+A22 and B=B11+B12+B21+B22, where Aij,Bij∈Gij1≤i,j≤2, and we obtain from Lemmas 4 and 5 that(58)fAB=fA11+A12+A21+A22B11+B12+B21+B22=fA11B11+fA11B12+fA11B21+fA11B22+fA12B11+fA12B12+fA12B21+fA12B22+fA21B11+fA21B12+fA21B21+fA21B22+fA22B11+fA22B12+fA22B21+fA22B22=fB11A11+A12+A21+A22+fB12A11+A12+A21+A22+fB21A11+A12+A21+A22+fB22A11+A12+A21+A22+B11+B12+B21+B22fA11+B11+B12+B21+B22fA12+B11+B12+B21+B22fA21+B11+B12+B21+B22fA22=fBA+BfA.
Therefore,f is an additive antiderivation on G. The proof is completed.Remark 2.
For anyA∈G, we define a mapping δ:G⟼G as(59)δA=ΔA−fA.
Then, we obtain from Lemmas3 and 4 that δ is an additive mapping on G.
In the following, we will introduce Lemmas7–9 and show that δ is an additive derivation on G.Lemma 7.
Letδ:G⟼G be as in Remark 2. Then, for any Aii∈GiiandAij∈Gij1≤i≠j≤2,(i)
δP1=−δP2 and δPi=PiδPiPj+PjδPiPi(ii)
δAij=PiδAijPj(iii)
PjδAiiPj=0,PiδAiiPj=AiiδPi and PjδAiiPi=δPiAiiProof.
By Lemma1 and Remarks 1 and 2, we can easily check that Lemma 7 holds. The proof is completed.Lemma 8.
Letδ:G⟼G be as in Remark 2. Then, for any Aii,Bii∈Gii,Aij,Bij∈Gij,Aji,Bji∈Gji,andBjj∈Gjj1≤i≠j≤2,(i)
δAiiBij=δAiiBij+AiiδBij(ii)
δAijBjj=δAijBjj+AijδBjj(iii)
δAiiBii=δAiiBii+AiiδBii(iv)
δAiiBjj=δAiiBjj+AiiδBjj(v)
δAijBji=δAijBji+AijδBji(vi)
δAijBij=δAijBij+AijδBij(vii)
δAiiBji=δAiiBji+AiiδBji(viii)
δAjiBjj=δAjiBjj+AjiδBjjProof.
(i)
For anyAii,Bij∈Gij1≤i≠j≤2, on the one hand, since Aii,Bij=AiiBij∈Ω, we get that(60)ΔAiiBij=ΔAiiBij+BijΔAii+AiiΔBij+ΔBijAii.And then, it follows fromMN=NM=0, Lemma 1, and Remark 1 that(61)ΔAiiBij=ΔAiiBij+AiiΔBij+fBijAii.On the other hand, we obtain from Lemma6 and fAii=0 that(62)fAiiBij=fBijAii.Therefore, by Remark2, equations (61) and (62), AiifBij=AiiPjΔBijPi=0, and fAii=0, we get(63)δAiiBij=ΔAiiBij−fAiiBij=ΔAiiBij+AiiΔBij=ΔAii−fAiiBij+AiiΔBij−fBij=δAiiBij+AiiδBij.(ii)
Similarly, we can show that (ii) holds.(iii)
For anyA11,B11∈G11andY12∈G12, by Lemma 8 (i), on the one hand, we get(64)δA11B11Y12=δA11B11Y12=δA11B11Y12+A11B11δY12.On the other hand, we have(65)δA11B11Y12=δA11B11Y12=δA11B11Y12+A11δB11Y12+A11B11δY12.Comparing equations (64) and (65), we get(66)δA11B11−δA11B11−A11δB11Y12=0.This yields from the faithfulness ofG12 that(67)P1δA11B11P1=P1δA11B11+A11δB11P1.Furthermore, by Lemma7 (i) and (iii), we have(68)P1δA11B11P2+P2δA11B11P1=A11B11δP1+δP1A11B11=A11δB11P2+P2δA11B11.Therefore, we obtain from Lemma7 (iii) and equations (67) and (68) that(69)δA11B11=P1δA11B11P1+P1δA11B11P2+P2δA11B11P1=P1δA11B11+A11δB11P1+A11δB11P2+P2δA11B11=deltaA11B11+A11δB11.Similarly, we can showδA22B22=δA22B22+A22δB22 holds for all A22,B22∈G22.(iv)
For anyAii∈GiiandBjj∈Gjj1≤i≠j≤2, by Lemma 1 (iv), we have(70)ΔAiiBjj+AiiΔBjj∈GijandBjjΔAii+ΔBjjAii∈Gji.Furthermore, sinceAii,Bjj=0∈Ω, we get(71)0=ΔAii∘Bjj=ΔAii∘Bjj+Aii∘ΔBjj=ΔAiiBjj+BjjΔAii+AiiΔBjj+ΔBjjAii=ΔAiiBjj+AiiΔBjj+BjjΔAii+ΔBjjAii.And so, this yields from equation (70) that(72)ΔAiiBjj+AiiΔBjj=0.Therefore, it follows from equation (72) and δAiiBjj=δ0=0=fAii=fBjj that(73)δAiiBjj=ΔAii−fAiiBjj+AiiΔBjj−fBjj=δAiiBjj+AiiδBjj.(v)
For anyAij∈GijandBji∈Gji1≤i≠j≤2, it follows from MN=NM=0 and Lemma 7 (ii) that(74)δAijBji=δ0=0.And,(75)δAijBji=PiδAijPjBji=0andAijδBji=AijPjδBjiPi=0.Thus, we obtain from (74) and (75) that(76)δAijBji=δAijBji+AijδBji.(vi) Similar to (v), we can show that (vi) holds.(vii)
For anyAii∈Gii,Aij∈Gij,andBji∈Gji1≤i≠j≤2, on the one hand, we get(77)δAiiBji=δ0=0.On the other hand, it follows fromMN=NM=0 and Lemma 7 (ii)-(iii) that(78)δAiiBji=PiδAiiPjBji=0andAiiδBji=AiiPjδBjiPi=0.Comparing equations (77) and (78), we get(79)δAiiBji=δAiiBji+AiiδBji.(viii)
Similar to (vii), we can show that (viii) holds. The proof is completed.Lemma 9.
Letδ:G⟼G be as in Remark 2. Then, δ is an additive derivation.Proof.
For anyA,B∈G, let A=A11+A12+A21+A22 and B=B11+B12+B21+B22, where Aij,Bij∈Gij1≤i,j≤2; since δ is an additive mapping and then by Lemma 8, we have(80)δAB=δA11+A12+A21+A22B11+B12+B21+B22=δA11B11+δA11B12+δA11B21+δA11B22+δA12B11+δA12B12+δA12B21+δA12B22+δA21B11+δA21B12+δA21B21+δA21B22+δA22B11+δA22B12+δA22B21+δA22B22=δA11B11+B12+B21+B22+δA12B11+B12+B21+B22+δA21B11+B12+B21+B22+δA22B11+B12+B21+B22+A11+A12+A21+A22δB11+A11+A12+A21+A22δB12+A11+A12+A21+A22δB21+A11+A12+A21+A22δB22=δAB+AδB.
Therefore,δ is an additive derivation on G. The proof is completed.Proof.
of Theorem 1. For anyA∈G, let A=A11+A12+A21+A22, where Aij∈Gij (1≤i,j≤2); by Remarks 1 and 2 and Lemmas 6 and 9, we obtain that(81)ΔA=δA+fA,where δ is an additive derivation and f is an additive antiderivation, respectively. Furthermore, fAii=0 for all Aii∈Gii(1≤i≤2).
In the following, we check thatδ and f are unique. Let d:G⟼G be an additive derivation and h:G⟼G be an additive antiderivation such that hAii=0 for all Aii∈Gii (1≤i≤2). Suppose that(82)ΔA=dA+hA.
Then, for anyA∈G, by Theorem 1, we get that δA+fA=ΔA=dA+hA and so δA−dA=hA−fA for all A∈G; this yields from δ and d two additive derivations that h−f is an additive derivation; moreover, by Remark 1, we get h−fAii=0 for all Aii∈Gii (1≤i≤2).
Next, we show thath−fAij=0 for all Aij∈Gij (1≤i≠j≤2). Indeed, for any Aij∈Gij (1≤i≠j≤2), on the one hand, since h−f is an additive derivation and hPi=fPi=0(1≤i≤2) and so we get that(83)h−fAij=h−fPiAij=h−fPiAij+Pih−fAij=Pih−fAij.
On the other hand, sinceh and f are two additive antiderivations and hPj=fPj=0(1≤j≤2), we get that(84)h−fAij=h−fAijPj=hAijPj−fAijPj=hPjAij+PjhAij−fPjAij−PjfAij=PjhAij−PjfAij=Pjh−fAij.
Thus, this yields from equations (83) and (84) that h−fAij=0 for all Aij∈Gij (1≤i≠j≤2). Therefore, h−fA=h−fA11+h−fA12+h−fA21+h−fA22=0 and so h=f and δ=d. The proof is completed.Remark 3.
By the above lemmas, Remarks1 and 2, and the proof of Theorem 1, we can easily obtain that if Δ is a nonlinear Jordan derivable mapping of G by Lie product square-zero elements, then the following statements are equivalent:(i)
Δ is an additive derivation(ii)
PjΔAijPi=0,∀Aij∈Gij1≤i≠j≤2(iii)
ΔAij∈Gij, ∀Aij∈Gij1≤i≠j≤2
Next, we give an application of Remark3 to triangular algebras and we obtain that every nonlinear Jordan derivable mapping of a triangular algebra by Lie product square-zero elements is an additive derivation.Corollary 1.
LetA and ℬ be unital algebras, ℳ be a unital A,ℬ-bimodule, which is faithful as both a left A-module and a right ℬ-module, U=TriA,ℳ,ℬ=am0b:a∈A,m∈ℳ,b∈ℬ be a 2-torsion free triangular algebra, and Δ:U⟼U be a nonlinear Jordan derivable mapping of U by Lie product square-zero elements; then, Δ is an additive derivation.Proof.
of Corollary1. Let 1A and 1ℬ be the identities of the algebras A and ℬ, respectively, and let 1 be the identity of the triangular algebra U. We denote(85)P1=1A000by the standard idempotent ofU,P2=1−P1=0001ℬ,(86)Uij=PiUPjfor1≤i,j≤2.
It is clear that the triangular algebraU may be represented as(87)U=P1UP1+P1UP2+P2UP2=A+ℳ+ℬ.
For anyA∈U, let A=A11+A12+A22, where Aij∈Uij (1≤i,j≤2); then for any A12∈U12, since U21=0 and by Lemma 1 (iii), we get(88)ΔA12=P1ΔA12P2+P2ΔA12P1=P1ΔA12P2.
Therefore, we obtain from Remark3 (iii) that Δ is an additive derivation. The proof is completed.



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