Let ℛ be a commutative ring with unity, 𝒜 be an algebra over ℛ and Z(𝒜) be the center of 𝒜. Recall that an ℛ-linear map d:𝒜⇒𝒜 is called a derivation on 𝒜 if d(xy)=d(x)y+xd(y) holds for all x,y∈𝒜. An ℛ-linear map δ:𝒜⇒𝒜 is called a Lie derivation on 𝒜 if δ([x,y])=[δ(x),y]+[x,δ(y)] holds for all x,y∈𝒜, where [x,y]=xy-yx is the usual Lie product. An ℛ-linear map δ:𝒜⇒𝒜 is called a Lie triple derivation on 𝒜 if δ([[x,y],z])=[[δ(x),y],z]+[[x,δ(y)],z]+[[x,y],δ(z)] holds for all x,y,z∈𝒜. It is easy to check that every derivation on 𝒜 is a Lie derivation on 𝒜 and that every Lie derivation on 𝒜 is a Lie triple derivation on 𝒜. However, the converse need not be true in general. For example if we consider the algebra 𝒜 of all 3 × 3 strictly upper triangular matrices over Z, the ring of integers, and define a map L:A→A such that L0xy00z000=0y000z000.

Then it can easily be seen that L is a Lie triple derivation on 𝒜 which is neither a derivation nor a Lie derivation on 𝒜. Now, let N be the set of nonnegative integers and Δ={δn}n∈ℕ be a sequence of ℛ-linear mappings δn:A→A such that δ0=idA, the identity map on 𝒜. Then Δ is said to be

• (i) a higher derivation on 𝒜 if

  δn(xy)=∑ i+j=nδi(x)δj(y),  for all  x,y∈A &  n∈ℕ;

• (ii) a Lie higher derivation on 𝒜 if

  δn([x,y])=∑ i+j=n[δi(x),δj(y)],  for all  x,y∈A &  n∈ℕ;

• (iii) a Lie triple higher derivation on 𝒜 if

  δn([[x,y],z])=∑ i+j=n[[δi(x),δj(y)],δk(z)],  for all  x,y,z∈A &  n∈ℕ.

It is also easy to observe that there exists Lie triple higher derivation on an algebra 𝒜 which is not a Lie higher derivation on 𝒜. For example consider the algebra 𝒜 of all 3 × 3 strictly upper triangular matrices over the field 𝒬 of rational numbers, and consider the sequence L={Ln}n∈ℕ of linear mappings Ln:A→A such that Ln=Lnn!, where L is Lie triple derivation on 𝒜 which is not a Lie derivation on 𝒜. Then by using induction on n, it can be easily verified that ℒ is a Lie triple higher derivation on 𝒜 but not a Lie higher derivation on 𝒜.

The ℛ-algebra A=Tri(A,M,B)={am0ba∈A,m∈M,b∈B} under the usual matrix operations is called a triangular algebra, where 𝒜 and 𝓑 are unital algebras over ℛ and ℳ is an (𝒜,𝓑)-bimodule. Recall that a left (resp. right) 𝒜-module ℳ is faithful if aℳ=0 (resp. ℳa=0) implies that a=0 for every a∈ 𝒜. The notion of triangular ring was first introduced by Chase [5] in 1960. Further, in the year 2000, Cheung [7] initiated the study of linear maps on triangular algebras. He described Lie derivations, commuting maps and automorphisms of triangular algebras (see for reference [8, 9]).

In the recent years, derivation and Lie derivation have been studied by several authors (see [1, 2, 3, 4, 9, 10, 12, 14, 16, 19, 20]) in various directions. One direction of investigation is to study the conditions under which derivations and Lie derivations can be completely determined by the action on some subsets of 𝒜. We say that an ℛ-linear map δ:A→A is derivable at a given point c∈A if δ(x)y+xδ(y)=δ(c) for every x,y∈A with xy=c and such c is called a derivable point of 𝒜. This kind of maps were discussed by several authors (see [6, 15, 22]). Similarly, an ℛ-linear map δ:A→A is said to be a Lie derivable at a given point c∈A if δ([x,y])=[δ(x),y]+[x,δ(y)] for all x,y∈A with xy=c. Lu and Jing [18] discussed such maps on B(X) where X is a Banach space with dimX≥3 and B(X) is the algebra of all bounded linear operators acting on X and proved that if δ is Lie derivable at c=0 (resp. c=p, where p is a fixed nontrivial idempotent of B(X)), then δ=d+τ, where d is a derivation of B(X) and τ:B(X)→ℂI is a linear map vanishing at every commutator [x,y] with xy=0 (resp. xy=p). Ji and Qi [13] investigated this problem on triangular algebras and obtained that under some mild conditions on A, if L:A→A is an R-linear map satisfying δ([x,y])=[δ(x),y]+[x,δ(y)] for any x,y∈A with xy = 0 (resp. xy = p, where p is a fixed nontrivial idempotent of A ), then δ=d+τ, where d is a derivation of A and τ:A→Z(A) (where Z(A) is the center of A ) is an ℛ-linear map vanishing at commutators [x, y] with xy = 0 (resp. xy = p). Furthermore, in [17] Liu analysed Lie triple derivation on factor von Neumann algebra 𝒜 of dimension greater than one and found that if a linear map δ:A→A satisfies δ([[x,y],z])=[[δ(x),y],z]+[[x,δ(y)],z]+[[x,y],δ(z)] for any x,y,z∈A with xy = 0 (resp. xy = p, where p is a fixed nontrivial projection of 𝒜), then there exist an operator r ∈ 𝒜 and a linear map f:A→ℂI (where ℂI is the center of 𝒜) vanishing at every second commutator [[x, y], z] with xy = 0 (resp. xy = p) such that δ(x)=xr−rx+f(x) for any x∈A.

An ℛ-linear map δ:𝒜⇒ 𝒜 is Lie triple derivable at a given point c ∈ 𝒜 if δ([[x,y],z])=[[δ(x),y],z]+[[x,δ(y)],z]+[[x,y],δ(z)] for all x,y,z∈A with xy=c. It is obvious that the condition of being a Lie triple derivable map at some point is much weaker than the condition of being a Lie triple derivation. So far, there has been no result on the study of the local actions of Lie triple derivations on triangular algebras. Motivated by these observations, the purpose of the present paper is to characterize the additive mapping δ_n on triangular algebra A satisfying δn([[x,y],z])=∑ i+j+k=n[[δi(x),δj(y)],δk(z)] for any x,y,z ∈ A with xy=0 (resp. xy=p, where p is a fixed nontrivial idempotent).

Throughout the present paper A will denote a triangular algebra which is 2-torsion free. Define two natural projections πA:A→A and πB:A→B by πAam0b=a and πBam0b=b. The center of A coincides with

Z(A)={a00ba∈A,b∈B,am=mb for all m∈M}.

Moreover, πA(Z(A))⊆Z(A) and πB(Z(A))⊆Z(B), and there exists a unique algebra isomorphism η:πB(Z(A))→πA(Z(A)) such that η(b)m=mb for all m∈M.

Let 1A (resp.1B) be the identity of the algebra A (resp.B) and let I=1A001B be the identity of triangular algebra A. Throughout this paper, we shall use the following notations: p1=1A000 and p2=I−p1=0001B . Set A11={a000a∈A}, A12={0m00m∈M} and A22={000bb∈B}. Then we can write A=A11⊕A12⊕A22, where A11 is a subalgebra of A isomorphic to A, A22 is a subalgebra of A isomorphic to B and A12 is a (A11,A22)-bimodule isomorphic to the bimodule M. To simplify the notation we will use the following convention: a11∈A=A11, a22∈B=A22 and a12∈M=A12. Then each element x∈A can be represented in the form x=a11+a12+a22, where a11∈A11,a22∈A22 and a12∈A12.

In what follows, we write aij, it indicates aij∈Aij and the corresponding element in A,B or M. Note that aijakl=0 if j≠k.

The proof of the following lemma can be seen in [8, Propositon 3].

Lemma 1.1.

Let A be a triangular algebra Tri(A11,A12,A22). If πA11(Z(A))=Z(A11) and πA22(Z(A))=Z(A22), then there is a unique algebra isomorphism η:Z(A22)→Z(A11) such that η(b)⊕b∈Z(A) for any b∈Z(A22).

The main result of the present paper states as follows:

Theorem 2.1. Let A=Tri(A11,A12,A22) be a 2-torsion free triangular algebra consisting of algebras A11 and A22 over a commutative ring R with unity 1A11 and 1A22 respectively and A12 be a faithful (A11,A22)-bimodule, which is faithful as a left A11-module and also as a right A22-module. Suppose that

• (i) πA11(Z(A))=Z(A11) and πA22(Z(A))=Z(A22).

• (ii) For any a∈A11, if [a,A11]∈Z(A11), then a∈Z(A11) or for any b∈A22, if [b,A22]∈Z(A22), then b∈Z(A22).

If Δ={δn}n∈ℕ is a sequence of R-linear maps δn:A→A such that δn[[x,y],z]=∑ i+j+k=n[[δi(x),δj(y)],δk(z)]

for all x,y,z∈A with xy=0, then for each n∈ℕ,

δn=dn+τn; where dn:A→A is R-linear mapping satisfying dn(xy)=∑ i+j=ndi(x)dj(y) for all x,y∈A, i.e., D={dn}n∈ℕ is a higher derivation of A and τn:A→Z(A) is an R-linear map vanishing at every second commutator [[x,y],z] with xy=0.

The proof of Theorem 2.1 is based on the induction on n. We provide the proof, for n=1, through several claims. Indeed, we show that under the given assumptions of our theorem every Lie triple derivation δ1=δ on A there exists a derivation d on A and a linear mapping τ:A→Z(A) vanishing on second commutators such that δ(x)=d(x)+τ(x) for all x∈A.

Proof of Theorem 2.1. Claim 1. p1δ(p1)p1+p2δ(p1)p2∈Z(A); δ(a12)=p1δ(a12)p2∈A12 for any a12∈A12.

Since a12p1=0 for any a12∈A12, we have

(2.1)δ(a12)=δ([[a12,p1],p1])   =[[δ(a12),p1],p1]+[[a12,δ(p1)],p1]+[[a12,p1],δ(p1)]   =δ(a12)p1−p1δ(a12)p1−p1δ(a12)p1+p1δ(a12)+a12δ(p1)p1    −a12δ(p1)+p1δ(p1)a12−a12δ(p1)+δ(p1)a12.

On multiplying the above equality from left by p₁ and right by p₂, we get

2(p1δ(p1)a12−a12δ(p1)p2)=0.

This implies that p1δ(p1)a12−a12δ(p1)p2=0, that is, p1δ(p1)p1a12−a12p2δ(p1)p2=0, and hence p1δ(p1)p1+p2δ(p1)p2∈Z(A). By putting δ(a12)=r11+r12+r22, δ(p1)=p11+p12+p22 in (2.1), we get r11=0=r22 and so we have piδ(a12)pi=0 for i∈{1,2}, δ(a12)=p1δ(a12)p2.

Similarly we can get the following result:

Claim 2. p1δ(p2)p1+p2δ(p2)p2∈Z(A).

Claim 3. δ(I)=p1δ(I)p1+p2δ(I)p2∈Z(A).

Since p1(I−p1)=0, we have

0=δ([[p1,I−p1],p1])=[[δ(p1),I−p1],p1]+[[p1,δ(I−p1)],p1]+[[p1,I−p1],δ(p1)]=[[p1,δ(I)],p1]=−p1δ(I)p2.

This yields that δ(I)=p1δ(I)p1+p2δ(I)p2. By Claims 1 & 2, we have δ(I)=p1δ(p1)p1+p2δ(p1)p2+p1δ(p2)p1+p2δ(p2)p2∈Z(A).

In the sequel, we define

f(x)=δ(x)+δp1δ(p1)p2(x),

where δp1δ(p1)p2(x) is the inner derivation determined by p1δ(p1)p2, that is,

δp1δ(p1)p2(x)=[p1δ(p1)p2,x]=p1δ(p1)p2x−xp1δ(p1)p2  for all  x∈A.

One can verify that

f([[x,y],z])=[[f(x),y],z]+[[x,f(y)],z]+[[x,y],f(z)]

for all x,y,z∈A with xy=0. Moreover by Claim 1, we have

f(p1)=δ(p1)+p1δ(p1)p2p1−p1δ(p1)p2=p1δ(p1)p1+p2δ(p1)p2∈Z(A).

By Claim 3, we get f(I)=δ(I)∈Z(A). Consequently f(p2)=f(I)−f(p1)∈Z(A).

Clearly for any a12∈A12, δp1δ(p1)p2(a12)=[p1δ(p1)p2,a12]=0. By Claim 1, we have f(a12)=δ(a12)∈A12, and hence

Claim 4. For any a12∈A12,f(a12)∈A12.

Claim 5. f(Aii)⊆Aii⊕Ajj. There exists an R-linear map τi:Aii→Z(A) such that f(aii)−τi(aii)∈Aii for all aii∈Aii,where i,j=1,2 and i≠j.

First we show for i=1. Since a₁₁p₂=0 for any a11∈A11 and f(p2)∈Z(A), it follows that

0=f([[a11,p2],p2])=[[f(a11),p2],p2]+[[a11,f(p2)],p2]+[[a11,p2],f(p2)]=f(a11)p2−p2f(a11)p2−p2f(a11)p2+p2f(a11).

Multiplying by p₁ from the left we get p1f(a11)p2=0 and hence f(a11)∈A11⊕A22.

Similarly, we can prove the result for i=2. Thus, f(Aii)⊆Aii⊕Ajj.

Now we can write f(a11)=p1f(a11)p1+p2f(a11)p2. Moreover, since a11a22=0, for any a22∈A22 and x∈A, it is easy to check that

0=f([[a11,a22],x])=[[f(a11),a22],x]+[[a11,f(a22)],x]=[[f(a11),a22]+[a11,f(a22)],x].

Multiplying by p₂ on both the sides of the above equation, we get

0=p2[[f(a11),a22]+[a11,f(a22)],x]p2=[[p2f(a11)p2,p2a22p2]+[p2a11p2,p2f(a22)p2],p2xp2]=[[p2f(a11)p2,a22],p2xp2].

This implies that [p2f(a11)p2,a22]∈Z(A22). Hence by hypothesis (ii), we find that

p2f(a11)p2∈Z(A22).

Define τ1:A11→Z(A) such that τ1(a11)=η(p2f(a11)p2)⊕p2f(a11)p2, where η is the map defined in Lemma 1.1. Thus, we get

f(a11)−τ1(a11)=p1f(a11)p1+p2f(a11)p2−η(p2f(a11)p2)−p2f(a11)p2      =p1f(a11)p1−η(p2f(a11)p2)∈A11.

Since f is R-linear, one can verify that τ₁ is R-linear. Similarly, we can define R-linear map τ2:A22→Z(A) by τ2(a22)=p1f(a22)p1⊕η−1(p1f(a22)p1). Then

f(a22)−τ2(a22)=p1f(a22)p1+p2f(a22)p2−p1f(a22)p1−η−1(p1f(a22)p1)      =p2f(a22)p2−η−1(p1f(a22)p1)∈A22.

Now, for anyx=a11+a12+a22∈A, we define two R-linear maps τ:A→Z(A) and d:A→A by

τ(x)=τ1(a11)+τ2(a22)     and     d(x)=f(x)−τ(x)  respectively.

Then, d(Aij)⊆Aij   for  1≤i≤j≤2  and  d(a12)=f(a12).

Claim 6. d is a derivation.

Since f & τ are ℛ-linear and d(x)=f(x)-τ(x), d is ℛ-linear. It remains to show that d(xy)=d(x)y+xd(y), for all x,y∈A. Now, we divide the proof into the following three steps:

Step 1. Since a₁₂a₁₁=0 for any a11∈A11, a12∈A12 and τ(x) is in Z(A), we have

−d(a11a12)=d([[a11,a12],p1])=f([[a11,a12],p1])=[[f(a11),a12],p1]+[[a11,f(a12)],p1]+[[a11,a12],f(p1)]=[[d(a11)+τ(a11),a12],p1]+[[a11,d(a12)],p1]=−d(a11)a12−a11d(a12).

Hence, d(a11a12)=d(a11)a12+a11d(a12). Similarly, we can get

d(a12a22)=d(a12)a22+a12d(a22)  for all  a12∈A12,a22∈A22.

Step 2. Let a11,b11∈A11. For any a12∈A12, we have

(2.2)d(a11b11a12)=d(a11)b11a12+a11d(b11a12)=d(a11)b11a12+a11d(b11)a12+a11b11d(a12).

On the other hand,

(2.3)d(a11b11a12)=d(a11b11)a12+a11b11d(a12).

Comparing (2.2) and (2.3), we have

(d(a11b11)−d(a11)b11−a11d(b11))a12=0  for all  a12∈A12.

Since A12 is a faithful left A11-module, we get

d(a11b11)=d(a11)b11+a11d(b11)  for all  a11,b11∈A11.

Similarly, one can arrive at

d(a22b22)=d(a22)b22+a22d(b22)  for all  a22,b22∈A22.

Step 3. Let x=a11+a12+a22, y=b11+b12+b22 be in A.

By Steps 1 & 2, we have

d(xy)=d((a11+a12+a22)(b11+b12+b22))  =d(a11b11)+d(a11b12)+d(a12b22)+d(a22b22)  =d(a11)b11+a11d(b11)+d(a11)b12+a11d(b12)   +d(a12)b22+a12d(b22)+d(a22)b22+a22d(b22).

On the other hand, d(Aij)⊆Aij for 1≤i≤j≤2, we have

d(x)y+xd(y)=d(a11+a12+a22)(b11+b12+b22)      +(a11+a12+a22)d(b11+b12+b22)     =d(a11)b11+a11d(b11)+d(a11)b12+a11d(b12)      +d(a12)b22+a12d(b22)+d(a22)b22+a22d(b22).

Hence, we find that d(xy)=d(x)y+xd(y), i.e., d is a derivation.

Claim 7. τ vanishes at second commutator [[x,y],z] with xy=0 for all x,y,z∈A.

Suppose xy=0. Since τ(x)∈Z(A), we have

τ([[x,y],z])=f([[x,y],z])−d([[x,y],z])     =[[f(x),y],z]+[[x,f(y)],z]+[[x,y],f(z)]−d([[x,y],z])     =[[d(x)+τ(x),y],z]+[[x,d(y)+τ(y)],z]+[[x,y],d(z)+τ(z)]      −d([[x,y],z])     =[[d(x),y],z]+[[x,d(y)],z]+[[x,y],d(z)]−d([[x,y],z])     =0

for all x,y,z∈A. The proof of our theorem for n=1 is now complete.

Now, suppose that the conclusion holds for all m<n∈ℕ. That is, there exist linear maps dm:A→A and τm:A→Z(A) such that δm(x)=dm(x)+τm(x), τm([[x,y],z])=0 with xy=0 and dm(xy)=∑ i+j=mdi(x)dj(y) for all x,y,z∈A.

Moreover, δm has the following properties:

p1δm(p1)p1+p2δm(p1)p2∈Z(A);δm(a12)=p1δm(a12)p2∈A12;
p1δm(p2)p1+p2δm(p2)p2∈Z(A);
δm(I)=p1δm(I)p1+p2δm(I)p2∈Z(A).

We will show that δ_n also satisfies the similar properties. We prove this through the following claims:

Claim 8. p1δn(p1)p1+p2δn(p1)p2∈Z(A);δn(a12)=p1δn(a12)p2∈A12 for any a12∈A12.

Since a12p1=0 for any a12∈A12, by induction hypothesis, we have

(2.4)δn(a12)=δn([[a12,p1],p1]) =[[δn(a12),p1],p1]+[[a12,δn(p1)],p1]+[[a12,p1],δn(p1)]  +∑ i+j+k=n 0≤i,j,k<n [[δi(a12),δj(p1)],δk(p1)] =δn(a12)p1−p1δn(a12)p1−p1δn(a12)p1+p1δn(a12)+a12δn(p1)p1  −a12δn(p1)+p1δn(p1)a12−a12δn(p1)+δn(p1)a12.

On multiplying the above equality by p₁ and p₂ from left and right respectively, we get

2(p1δn(p1)a12−a12δn(p1)p2)=0.

This implies that p1δn(p1)a12−a12δn(p1)p2=0, that is, p1δn(p1)p1a12−a12p2δn(p1)p2=0, and hence p1δn(p1)p1+p2δn(p1)p2∈Z(A). By putting δn(a12)=s11+s12+s22, δn(p1)=t11+t12+t22 in (2.4), we get s11=0=s22 and so, we have piδn(a12)pi=0 for i∈{1,2}. Hence, δn(a12)=p1δn(a12)p2.

Since p2a12=0 for any a12∈A12. Similarly, we can get the following result.

Claim 9. p1δn(p2)p1+p2δn(p2)p2∈Z(A).

Claim 10. δn(I)=p1δn(I)p1+p2δn(I)p2∈Z(A).

Since p1(I−p1)=0, by using the induction hypothesis, we find that

0=δn([[p1,I−p1],p1]) =[[δn(p1),I−p1],p1]+[[p1,δn(I−p1)],p1]+[[p1,I−p1],δn(p1)]  +∑ i+j+k=n 0≤i,j,k<n [[δi(p1),δj(I−p1)],δk(p1)] =p1δn(I)p1−δn(I)p1−p1δn(I)+p1δn(I)p1.

On multiplying by p₁ from the left and by p₂ from the right, we have p1δn(I)p2=0, and hence we get that δn(I)=p1δn(I)p1+p2δn(I)p2. By Claims 8 & 9, we have δn(I)=p1δn(p1)p1+p2δn(p1)p2+p1δn(p2)p1+p2δn(p2)p2∈Z(A).

In the sequel, we define

fn(x)=δn(x)+δp1δn(p1)p2(x),

where δp1δn(p1)p2(x) is the inner derivation determined by p1δn(p1)p2, that is,

δp1δn(p1)p2(x)=[p1δn(p1)p2,x]=p1δn(p1)p2x−xp1δn(p1)p2  for all  x∈A.

One can verify that

fn([[x,y],z])=[[fn(x),y],z]+[[x,fn(y)],z]+[[x,y],fn(z)]+∑ i+j+k=n 0≤i,j,k<n [[fi(x),fj(y)],fk(z)]

for all x,y,z∈A with xy=0. Moreover by Claim 8, we have

fn(p1)=δn(p1)+p1δn(p1)p2p1−p1δn(p1)p2=p1δn(p1)p1+p2δn(p1)p2∈Z(A).

By Claim 10, we find that fn(I)=δn(I)∈Z(A). Consequently fn(p2)=fn(I)−fn(p1)∈Z(A). Clearly for any a12∈A12, δp1δn(p1)p2(a12)=[p1δn(p1)p2,a12]=0. By Claim 8, we have fn(a12)=δn(a12)∈A12, and hence

Claim 11. For any a12∈A12, fn(a12)∈A12.

Claim 12. fn(Aii)⊆Aii⊕Ajj. There exists an R-linear map τni:Aii→Z(A) such that fn(aii)−τni(aii)∈Aii for all aii∈Aii,where i,j=1,2 and i≠j.

First we show the result for i=1. Since a₁₁p₂=0 for any a11∈A11 and fn(p2)∈Z(A), using induction hypothesis, it follows that

0=fn([[a11,p2],p2]) =[[fn(a11),p2],p2]+[[a11,fn(p2)],p2]+[[a11,p2],fn(p2)] +∑ i+j+k=n 0≤i,j,k<n [[fi(a11),fj(p2)],fk(p2)] =fn(a11)p2−p2fn(a11)p2−p2fn(a11)p2+p2fn(a11).

Multiply by p₁ from the left to get p1fn(a11)p2=0. So, fn(a11)∈A11⊕A22.

Similarly we can find the result for i=2. Thus, fn(Aii)⊆Aii⊕Ajj.

Now we can write fn(a11)=p1fn(a11)p1+p2fn(a11)p2. Moreover, since a11a22=0 for any a22∈A22 and x∈A, it is easy to observe that

0=fn([[a11,a22],x])=[[fn(a11),a22],x]+[[a11,fn(a22)],x]+[[a11,a22],fn(x)]  +∑ i+j+k=n 0≤i,j,k<n [[fi(a11),fj(a22),]fk(x)] =[[fn(a11),a22],x]+[[a11,fn(a22)],x] =[[fn(a11),a22]+[a11,fn(a22)],x].

Multiplying by p₂ on both the sides, we get

0=p2[[fn(a11),a22]+[a11,fn(a22)],x]p2=[[p2fn(a11)p2,p2a22p2]+[p2a11p2,p2fn(a22)p2],p2xp2]=[[p2fn(a11)p2,a22],p2xp2].

This implies that [p2fn(a11)p2,a22]∈Z(A22). Hence by assumption (ii), we get p2fn(a11)p2∈Z(A22). Define τn1:A11→Z(A) by τn1(a11)=η(p2fn(a11)p2)⊕p2fn(a11)p2, where η is the map defined in Lemma 1.1. Thus, we get

fn(a11)−τn1(a11)=p1fn(a11)p1+p2fn(a11)p2−η(p2fn(a11)p2)−p2fn(a11)p2=p1fn(a11)p1−η(p2fn(a11)p2)∈A11.

Since f_n is ℛ-linear, one can verify that τ_n1 is ℛ-linear. Similarly, we can define ℛ-linear map τn2:A22→Z(A) by τn2(a22)=p1fn(a22)p1⊕η−1(p1fn(a22)p1). Then

fn(a22)−τn2(a22)=p1fn(a22)p1+p2fn(a22)p2−p1fn(a22)p1−η−1(p1fn(a22)p1)=p2fn(a22)p2−η−1(p1fn(a22)p1)∈A22.

Now, for any x=a11+a12+a22∈A, we define two ℛ-linear maps τn:A→Z(A) and dn:A→A by

τn(x)=τn1(a11)+τn2(a22)     and     dn(x)=fn(x)−τn(x)  respectively.

Then, dn(Aij)⊆Aij   for  1≤i≤j≤2  and  dn(a12)=fn(a12).

Claim 13. dn(xy)=∑ i+j=ndi(x)dj(y)forallx,y∈A

Since fn & τn are ℛ-linear and dn(x)=fn(x)−τn(x), d_n is an ℛ-linear. It remains to show that dn(xy)=∑ i+j=ndi(x)dj(y)forallx,y∈A.

Now, we divide the proof into the following three steps:

Step 1. Since a12a11=0 for any a11∈A11, a12∈A12 and τn(x) is in

Z(A), by induction hypothesis, we have

−dn(a11a12)=fn([[a11,a12],p1])    =[[fn(a11),a12],p1]+[[a11,fn(a12)],p1]+[[a11,a12],fn(p1)]     +∑ i+j+k=n 0≤i,j,k<n [[fi(a11),fj(a12)],fk(p1)]    =[[fn(a11),a12],p1]+[[a11,fn(a12)],p1]     +∑ i+j+k=n 0≤i,j,k<n [[fi(a11),fj(a12)],(p1)]    =[[dn(a11)+τn(a11),a12],p1]+[[a11,dn(a12)+τn(a12)],p1]     +∑ i+j+k=n 0≤i,j,k<n [[dn(a11)+τn(a11),dn(a12)+τn(a12)],p1]    =[[dn(a11),a12],p1]+[[a11,dn(a12)],p1]     +∑ i+j+k=n 0≤i,j,k<n [[dn(a11),dn(a12)],p1]    =−dn(a11)a12−a11dn(a12)−∑ i+j+k=n 0≤i,j,k<n di(a11)dj(a12)

Hence, dn(a11a12)=∑ i+j=ndi(a11)dj(a12). Similarly, we can get

dn(a12a22)=∑ i+j=ndi(a12)dj(a22)  for any  a12∈A12,a22∈A22.

Step 2. Let a11,b11∈A11. For any a12∈A12, we have

(2.5)dn(a11b11a12)=∑i+k=ndi(a11b11)dk(a12)=dn(a11b11)a12+∑ i+k=n k≠0 di(a11b11)dk(a12)=dn(a11b11)a12+∑ l+t+k=n k≠0 dl(a11)dt(b11)dk(a12).

On the other hand,

(2.6)dn(a11b11a12)=∑i+j=ndi(a11)dj(b11a12)=∑i+j+k=ndi(a11)dj(b11)dk(a12)=∑i+j=ndi(a11)dj(b11)a12+∑ i+j+k=n k≠0 di(a11)dj(b11)dk(a12).

Comparing (2.5) and (2.6), we have

(dn(a11b11)−∑ i+j=ndi(a11)dj(b11))a12=0.

Since A12 is a faithful left A11-module, we get

dn(a11b11)=∑ i+j=ndi(a11)dj(b11).

Similarly, we can calculate

dn(a22b22)=∑ i+j=ndi(a22)dj(b22).

Step 3. Let x=a11+a12+a22, y=b11+b12+b22 be in A. By Steps 1 & 2, we have

dn(xy)=∑ i+j=ndi(x)dj(y).

Claim 14. τn vanishes at second commutator [[x,y],z] with xy=0 for all x,y,z∈A.

Since xy=0, we find that

τn([[x,y],z])=fn([[x,y],z])−dn([[x,y],z])     =∑i+j+k=n([[fi(x),fj(y)],fk(z)])−dn([[x,y],z])     =∑i+j+k=n([[di(x)+τi(x),dj(y)+τj(y)],dk(z)+τk(z)])      −dn([[x,y],z])     =∑i+j+k=n([[di(x),dj(y)],dk(z)]−dn([[x,y],z])     =0

for all x,y,z∈A. The proof is now complete.