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##  eISSN 0454-8124
pISSN 1225-6951

### Article

Kyungpook Mathematical Journal 2020; 60(4): 683-710

Published online December 31, 2020

### Characterizations of Lie Triple Higher Derivations of Triangular Algebras by Local Actions

Mohammad Ashraf, Mohd Shuaib Akhtar* and Aisha Jabeen

Department of Mathematics, Aligarh Muslim University, Aligarh, India
e-mail : mashraf80@hotmail.com, mshuaibakhtar@gmail.com and ajabeen329@gmail.com

Received: April 24, 2019; Accepted: February 25, 2020

Let $N$ be the set of nonnegative integers and $A$ be a 2-torsion free triangular algebra over a commutative ring $R$ . In the present paper, under some lenient assumptions on $A$ , it is proved that if $Δ = { δ n } n ∈ ℕ$ is a sequence of $R$ -linear mappings $δ n : A → A$ satisfying $δ n ( [ [ x , y ] , z ] ) = ∑ i + j + k = n [ [ δ i ( x ) , δ j ( y ) ] , δ k ( z ) ]$ for all $x , y , z ∈ A$ with $x y = 0$ (resp. $x y = p$ , where p is a nontrivial idempotent of $A$ ), then for each $n ∈ ℕ$ , $δ n = d n + τ n$ ; where $d n : A → A$ is $R$ -linear mapping satisfying $d n ( x y ) = ∑ i + j = n d i ( x ) d j ( y )$ for all $x , y ∈ A$ , i.e. $D = { d n } n ∈ ℕ$ is a higher derivation on $A$ and $τ n : A → Z ( A )$ (where $Z ( A )$ is the center of $A$ ) is an $R$ -linear map vanishing at every second commutator $[ [ x , y ] , z ]$ with xy=0 (resp. $x y = p$ ).

Keywords: triangular algebra, Lie higher derivation, Lie triple higher derivation

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

• (ii) a Lie higher derivation on 𝒜 if

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

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  in 1960. Further, in the year 2000, Cheung  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  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  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  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 δn 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

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 $b∈Z(A22)$.

### 2. Characterizations of Lie Triple Higher Derivations by Action on Zero Product

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

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

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,

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

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

Then, .

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

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

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

On the other hand,

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

Comparing (2.2) and (2.3), we have

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

Similarly, one can arrive at

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. 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 δn 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

$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

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

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,

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

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

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

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

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 fn is ℛ-linear, one can verify that τn1 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

Then, .

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

Since are ℛ-linear and $dn(x)=fn(x)−τn(x)$, dn 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

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

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

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

On the other hand,

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

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.

### 3. Characterizations of Lie Triple Higher Derivations by Action on Idem-potent Product

The proof of the following theorem shares the same outline as that of Theorem 2.1 but requires different technique.

### Theorem 3.1.

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. 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 $b∈Z(A22)$.

• (iii) For every $a∈A11$, there exists an integer t such that $t1A11−a$ is invertible.

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

For the proof of Theorem 3.1, we proceed by induction on n. We provide the proof, for n=1, through several claims. Indeed, we show that for every Lie triple derivation $δ1=δ$ on $A$ there exist 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 3.1. Claim 1. $p1δ(p1)p1+p2δ(p1)p2∈Z(A)$; $δ(a12)=p1δ(a12)p2∈A12$ for any $a12∈A12$.

Since $(p1+a12)p1=p1$ for any $a12∈A12$, we have

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

On multiplying the above equality by p1 and p2 from the left and the right respectively, we get $2(p1δ(p1)a12−a12δ(p1)p2)=0.$ This gives that $p1δ(p1)a12−a12δ(p1)p2=0,$ that is, $p1δ(p1)p1a12−a12p2δ(p1)p2=0.$ It follows that $p1δ(p1)p1+p2δ(p1)p2∈Z(A)$. By putting $δ(a12)=r11+r12+r22$, $δ(p1)=p11+p12+p22$ in (3.1), we get $r11=0=r22$ and so we have $piδ(a12)pi=0$ for $i∈{1,2}$. Hence, $δ(a12)=p1δ(a12)p2.$ Now, define

$f(x)=δ(x)+δp1δ(p1)p2(x),$

where $δp1δ(p1)p2$ is the inner derivation determined by $p1δ(p1)p2$. Then, we have

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

and $f([[x,y],z])=[[f(x),y],z]+[[x,f(y)],z]+[[x,y],f(z)]$ for all $x,y,z∈A$ with xy=p. Moreover, for any $a12∈A12$, by Claim 1, we have

$f(a12)=δ(a12)+δp1δ(p1)p2(a12)=δ(a12)∈A12.$

Claim 2. $f(I)=p1f(I)p1+p2f(I)p2∈Z(A)$ and $f(p2)∈Z(A).$

Since Ip1=p1, we have

$0=f([[I,p1],p1])=[[f(I),p1],p1]+[[I,f(p1)],p1]+[[I,p1],f(p1)]=f(I)p1−p1f(I)p1−p1f(I)p1+p1f(I).$

This yields that $p1(f(I)p1−p1f(I)p1−p1f(I)p1+p1f(I))p2=0$ and hence we find that $p1f(I)p2=0.$

So, we get $f(I)=p1f(I)p1+p2f(I)p2.$ For any $a12∈A12$, since $(p1−a12)(I+a12)=p1$, we have

$−f(a12)=f([[p1−a12,I+a12],p1]) =[[f(p1)−f(a12),I+a12],p1]+[[p1−a12,f(I)+f(a12)],p1] +[[p1−a12,I+a12],f(p1)] =−f(a12)p1−a12f(p1)p1−p1f(p1)a12+p1f(a12)a12+a12f(p1) +p1f(I)p1+p1f(a12)p1−a12f(I)p1−f(I)p1−p1f(I) −p1f(a12)+a12f(I)+p1f(I)p1−p1f(I)a12+p1f(a12)p1) −p1f(a12)a12+a12f(p1)−f(p1)a12.$

On multiplying above equality by p1 and p2 from the left and the right respectively, we get $a12f(I)p2−p1f(I)a12=0$ and $a12p2f(I)p2−p1f(I)p1a12=0.$

his implies that $f(I)=p1f(I)p1+p2f(I)p2∈Z(A)$. Consequently, $f(p2)=f(I)−f(p1)∈Z(A)$.

Claim 3. $f(Aii)⊆Aii⊕Ajj$. There exists an $R$-linear map $τi:Aii→Z(A)$ such that $f(aii)−τi(aii)∈Aii$ for all .

First we show for i=1. Suppose that a11 is invertible in $A11$, that is, there exists an element $a11−1∈A11$ such that $a11a11−1=a11−1a11=p1$. From $a11a11−1=p1$ and $(a11−1+p2)a11=p1$, we have

$0=f([[a11−1,a11],p1])=[[f(a11−1),a11],p1]+[[a11−1,f(a11)],p1]+[[a11−1,a11],f(p1)]$

and hence by Claim 2,

$0=f([[a11−1+p2,a11],p1])=[[f(a11−1)+f(p2),a11],p1]+[[a11−1+p2,f(a11)],p1]+[[a11−1+p2,a11],f(p1)]=[[f(a11−1),a11],p1]+[[a11−1,f(a11)],p1]+[[a11−1,a11],f(p1)]+[[p2,f(a11)],p1]=p2f(a11)p1+p1f(a11)p2.$

On multiplying by p1 from the left and by p2 from the right, we get $p1f(a11)p2=0$, and hence we find that $f(a11)⊆A11⊕A22.$

If a11 is not invertible in $A11$, by the hypothesis (iii), there exists an integer t such that $tp1−a11$ is invertible in $A11$. It follows from the preceding case that $f(tp1−a11)∈A11⊕A22$. Therefore, we have

$f(a11)=tf(p1)−f(tp1−a11)∈A11⊕A22.$

Similarly, we can prove the result for i=2. Now we can write $f(a11)=p1f(a11)p1+p2f(a11)p2$. First suppose that a11 is invertible in $A11$ with inverse element $a11−1$. Note that $a11a11−1=p1$ and $(a11−1+a22)a11=p1$, we get

$0=f([[a11−1,a11],x])=[[f(a11−1),a11],x]+[[a11−1,f(a11)],x]+[[a11−1,a11],f(x)],$

and hence

$0=f([[a11−1+a22,a11],x])=[[f(a11−1)+f(a22),a11],x]+[[a11−1+a22,f(a11)],x]+[[a11−1+a22,a11],f(x)]=[[f(a22),a11],x]+[[a22,f(a11)],x]=[[f(a11),a22]+[a11,f(a22)],x].$

Multiplying by p2 on both the sides, 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 get $p2f(a11)p2∈Z(A22)$.

If a11 is not invertible in $A11$, by the hypothesis (iii), there exists an integer t such that $(tp1−a11)$ is invertible in $A11$. It follows from the preceding case that

$0=f([[a22,tp1−a11],x])=[[f(a22),tp1−a11],x]+[[a22,tf(p1)−f(a11)],x]+[[a22,tp1−a11],f(x)]=−[[f(a22),a11],x]−[[a22,f(a11)],x].$

Multiplying by p2 on both the sides, 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 get $p2f(a11)p2∈Z(A22)$.

Define $τ1:A11→Z(A)$ by $τ1(a11)=η(p2f(a11)p2)⊕p2f(a11)p2$, 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 an 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 any $x=a11+a12+a22∈A$, we define two $R$-linear mappings $τ:A→Z(A)$ and $d:A→A$ by $τ(x)=τ1(a11)+τ2(a22)$ and $d(x)=f(x)−τ(x)$ respectively. Then, .

Claim 4. d is a derivation.

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

Step 1. If a11 is invertible in $A11$ with inverse element $a11−1$, then $(a11−1+a11−1a12)a11=p1$ for any $a11∈A11$, $a12∈A12$, we have

$−d(a12)=d([[a11,a11−1+a11−1a12],p1])=f([[a11,a11−1+a11−1a12],p1]) =[[f(a11),a11−1+a11−1a12],p1]+[[a11,f(a11−1)+f(a11−1a12)],p1] +[[a11,a11−1+a11−1a12],f(p1)].$

Since $[[f(a11),a11−1],p1]+[[a11,f(a11−1)],p1]+[[a11,a11−1],f(p1)]=0$, we have

$−d(a12)=[[f(a11),a11−1a12],p1]+[[a11,f(a11−1a12)],p1]=[[d(a11),a11−1a12],p1]+[[a11,d(a11−1a12)],p1]=−d(a11)a11−1a12−a11d(a11−1a12).$

Hence, $d(a12)=d(a11)a11−1a12+a11d(a11−1a12).$ Replacing a12 by $a11a12$, we arrive at

$d(a11a12)=d(a11)a12+a11d(a12).$

For any $a11∈A11$, let $tp1−a11$ be invertible in $A11$. Then

$d((tp1−a11)a12)=d(tp1−a11)a12+(tp1−a11)d(a12).$

Since $d(p1a12)=d(p1)a12+p1d(a12)$, we have

$d(a11a12)=d(a11)a12+a11d(a12).$

Step 2. Let $a12∈A12$ and $a22∈A22$. Observe that $(p1+a12)(p1+a22−a12a22)=p1$ and $(p1+a22−a12a22)(p1+a12)=p1+a12$. Since $f(p1)∈Z(A)$, we have

$−d(a12)=d([[p1+a22−a12a22,p1+a12],p1]) =f([[p1+a22−a12a22,p1+a12],p1]) =[[f(p1)+f(a22)−f(a12a22),p1+a12],p1] +[[p1+a22−a12a22,f(p1)+f(a12)],p1] +[[p1+a22−a12a22,p1+a12],f(p1)] =[[d(a22)−d(a12a22),p1+a12],p1]+[[p1+a22−a12a22,d(a12)],p1] =a12d(a22)−d(a12a22)−d(a12)+d(a12)a22.$

Thus, $d(a12a22)=d(a12)a22+a12d(a22)$ for any $a12∈A12,a22∈A22$.

With the same approach as used in the proof of Claim 6 of Theorem 2.1, we can get:

Step 3. For any $a11,b11∈A11$ and $a22,b22∈A22,$

• (i) $d(a11b11)=d(a11)b11+a11d(b11)$,

• (ii) $d(a22b22)=d(a22)b22+a22d(b22)$.

Step 4. $d(xy)=d(x)y+xd(y)$ for all $x,y∈A.$

Claim 5. τ vanishes at second commutator [[x,y],z] with xy=p for all $x,y,z∈A$.

Since xy=p, we find that

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

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

Now, suppose that the conclusion holds for all $m. 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=p and $dm(xy)=∑ i+j=mdi(x)dj(y)$ for all $x,y,z∈A$.

Moreover, $δm$ has the following properties:

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

We shall show that $δn$ also satisfies the similar properties. We prove this through the following claims:

Claim 6. $p1δn(p1)p1+p2δn(p1)p2∈Z(A)$; $δn(a12)=p1δn(a12)p2∈A12$ for any $a12∈A12$.

Since $(p1+a12)p1=p1$ for any $a12∈A12$, by induction hypothesis, we have

$δn(a12)=δn([[p1+a12,p1],p1]) =[[δn(p1)+δn(a12),p1],p1])+[[p1+a12,δn(p1)],p1] +[[p1+a12,p1],δn(p1)]+∑ i+j+k=n 0≤i,j,k

On multiplying by p1 from the left and by p2 from the right in the above equation, we get $2(p1δn(p1)a12−a12δn(p1)p2)=0$. This gives that $p1δn(p1)a12−a12δn(p1)p2=0,$ that is, $p1δn(p1)p1a12−a12p2δn(p1)p2=0.$ It follows that $p1δn(p1)p1+p2δn(p1)p2∈Z(A)$. By putting $δn(a12)=s11+s12+s22$, $δn(p1)=t11+t12+t22$ in (3.2), we get $s11=0=s22$ and so, we have $piδn(a12)pi=0$ for $i∈{1,2}$. Hence, $δn(a12)=p1δn(a12)p2.$

Now, define

$fn(x)=δn(x)+δp1δn(p1)p2(x),$

where $δp1δn(p1)p2$ is the inner derivation determined by $p1δn(p1)p2$. Then, we have

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

and

$fn([[x,y],z])=[[fn(x),y],z]+[[x,fn(y)],z]+[[x,y],fn(z)] +∑ i+j+k=n 0≤i,j,k

for all $x,y,z∈A$ with xy=p. Moreover, for any $a12∈A12$, by Claim 6, we have

$fn(a12)=δn(a12)+δp1δn(p1)p2(a12)=δn(a12)∈A12.$

Claim 7. $fn(I)=p1fn(I)p1+p2fn(I)p2∈Z(A)$ and $fn(p2)∈Z(A)$.

Since $Ip1=p1$, using induction hypothesis, we have

$0=fn([[I,p1],p1]) =[[fn(I),p1],p1]+[[I,fn(p1)],p1]+[[I,p1],fn(p1)] +∑ i+j+k=n 0≤i,j,k

On multiplying by p1 and by p2 from the left and the right respectively, we get $p1fn(I)p2=0.$ Hence, we find that $fn(I)=p1fn(I)p1+p2fn(I)p2.$ For any $a12∈A12$, since $(p1−a12)(I+a12)=p1$, using induction hypothesis, we have

$−fn(a12)=fn([[p1−a12,I+a12],p1]) =[[fn(p1)−fn(a12),I+a12],p1]+[[p1−a12,fn(I)+fn(a12)],p1] +[[p1−a12,I+a12],fn(p1)] +∑ i+j+k=n 0≤i,j,k

Further, multiply by p1 from the left and by p2 from the right, we find that

$a12fn(I)p2−p1fn(I)a12=0a12p2fn(I)p2−p1fn(I)p1a12=0.$

This implies that $fn(I)=p1fn(I)p1+p2fn(I)p2∈Z(A)$. Consequently, $fn(p2)=fn(I)−fn(p1)∈Z(A)$.

Claim 8. $fn(Aii)⊆Aii⊕Ajj$. There exists an ℛ-linear map $τni:Aii→Z(A)$ such that $fn(aii)−τni(aii)∈Aii$ for all .

First we show the result for i=1. Suppose that a11 is invertible in $A11$, that is, there exists an element $a11−1∈A11$ such that $a11a11−1=a11−1a11=p1$. From $a11a11−1=p1$ and $(a11−1+p2)a11=p1$, by induction hypothesis, we have

$0=fn([[a11−1,a11],p1])=[[fn(a11−1),a11],p1]+[[a11−1,fn(a11)],p1]+[[a11−1,a11],fn(p1)]+∑ i+j+k=n 0≤i,j,k

and hence by Claim 7,

$0=fn([[a11−1+p2,a11],p1]) =[[fn(a11−1)+fn(p2),a11],p1]+[[a11−1+p2,fn(a11)],p1] +[[a11−1+p2,a11],fn(p1)]+∑ i+j+k=n 0≤i,j,k

This yields that $p1(p2fn(a11)p1+p1fn(a11)p2)p2=0$ and hence we find that $p1fn(a11)p2=0$. From this we get $fn(a11)⊆A11⊕A22$.

On the other hand if a11 is not invertible in $A11$, by the hypothesis (iii), there exists an integer t such that tp1-a11 is invertible in $A11$. It follows from the preceding case that $fn(tp1−a11)∈A11⊕A22$. Therefore, we have $fn(a11)=tfn(p1)−fn(tp1−a11)∈A11⊕A22.$ Similarly, we can prove that for i=2.

Now we can write $fn(a11)=p1fn(a11)p1+p2fn(a11)p2$. First, suppose that a11 is invertible in $A11$ with inverse element $a11−1$. Note that $a11a11−1=p1$ and $(a11−1+a22)a11=p1$, using induction hypothesis, we get

$0=fn([[a11−1,a11],x]) =[[fn(a11−1),a11],x]+[[a11−1,fn(a11)],x]+[[a11−1,a11],fn(x)] +∑ i+j+k=n 0≤i,j,k

and hence

$0=fn([[a11−1+a22,a11],x]) =[[fn(a11−1)+fn(a22),a11],x]+[[a11−1+a22,fn(a11)],x] +[[a11−1+a22,a11],fn(x)]+∑ i+j+k=n 0≤i,j,k

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 hypothesis (ii), we get $p2fn(a11)p2∈Z(A22)$.

If a11 is not invertible in $A11$, by the hypothesis (iii), there exists an integer t such that $(tp1−a11)$ is invertible in $A11$. It follows from the preceding case that

$0=fn([[a22,tp1−a11],x]) =[[fn(a22),tp1−a11],x]+[[a22,tfn(p1)−fn(a11)],x]+[[a22,tp1−a11],fn(x)] +∑ i+j+k=n 0≤i,j,k

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 hypothesis (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 fn is ℛ-linear, one can verify that $τn1$ 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

Then, .

Claim 9. $dn(xy)=∑ i+j=ndi(x)dj(y)$ for all $x,y∈A$.

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

We divide the proof into the following three Steps:

Step 1. If a11 is invertible in $A11$ with inverse element $a11−1$, then $(a11−1+a11−1a12)a11=p1$ for any $a11∈A11$, $a12∈A12$, we have

$−dn(a12)=dn([[a11,a11−1+a11−1a12],p1])=fn([[a11,a11−1+a11−1a12],p1]) =[[fn(a11),a11−1+a11−1a12],p1]+[[a11,fn(a11−1)+fn(a11−1a12)],p1] +[[a11,a11−1+a11−1a12],fn(p1)] +∑ i+j+k=n 0≤i,j,k

Since,

$0=[[fn(a11),a11−1],p1]+[[a11,fn(a11−1)],p1]+[[a11,a11−1],fn(p1)] +∑ i+j+k+=n 0≤i,j,k

we find that

Hence, $dn(a12)=∑ i+j=ndi(a11)dj(a11−1a12).$ Replacing a12 by $a11a12$, we arrive at

$dn(a11a12)=∑ i+j=ndi(a11)dj(a12).$

For any $a11∈A11$, let $tp1−a11$ be invertible in $\mathfrak{A}_{11}$. Then

$dn((tp1−a11)a12)=∑ i+j=ndi(tp1−a11)dj(a12).$

Since $dn(p1a12)=∑ i+j=ndi(p1)dj(a12)$, we have $dn(a11a12)=∑ i+j=ndi(a11)dj(a12).$

Step 2. Let $a12∈A12$ and $a22∈A22$. Observe that $(p1+a12)(p1+a22−a12a22)=p1$ and $(p1+a22−a12a22)(p1+a12)=p1+a12$. Since $fn(p1)∈Z(A)$, we have

$−dn(a12)=dn([[p1+a22−a12a22,p1+a12],p1]) =fn([[p1+a22−a12a22,p1+a12],p1]) =[[fn(p1)+fn(a22)−fn(a12a22),p1+a12],p1] +[[p1+a22−a12a22,fn(p1)+fn(a12)],p1] +[[p1+a22−a12a22,p1+a12],fn(p1)] +∑ i+j+k=n 0≤i,j,k

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

Using the same approach as used in the proof of Claim 13 of Theorem 2.1, we find that

Step 3. For any $a11,b11∈A11$ and $a22,b22∈A22,$

• (i) $dn(a11b11)=∑ i+j=ndi(a11)dj(b11),$

• (ii) $dn(a22b22)=∑ i+j=ndi(a22)dj(b22).$

Step 4. $dn(xy)=∑ i+j=ndi(x)dj(y)$ for all $x,y∈A.$

Claim 10. τn vanishes at second commutator [[x,y],z] with xy=p for all $x,y,z∈A.$

Since xy=p, 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.

As an application of Theorems 2.1 & 3.1, we consider the nest algebra case. We know that every nontrivial nest algebra is a triangular algebra (see ), which satisfies the conditions of Theorems 2.1 & 3.1 and hence we have the following results.

### Theorem 4.1.

Let 𝒩 be an arbitrary nontrivial nest on a Hilbert space T of dimension greater than 2, Alg𝒩 be the associated nest algebra. If $Δ={δn}n∈ℕ$ is a sequence of ℛ-linear maps $δn:AlgN→AlgN$ satisfying $δn[[x,y],z]=∑ i+j+k=n[[δi(x),δj(y)],δk(z)]$ for all $x,y,z∈AlgN$ with xy=0. Then for each $n∈ℕ$, where $H={hn}n∈ℕ$ is an inner higher derivation on $AlgN$ and $τn:AlgN→FI$ (where $FI$ is the center of $AlgN$) is an $R$-linear map vanishing at the second commutator [[x,y],z] with xy=0.

Proof. Since $N$ is nontrivial nest, the associated nest algebra is a triangular algebra which satisfies the conditions of Theorem 2.1. Then there exists a higher derivation $D={dn}n∈ℕ$ of $AlgN$ and a linear map $τn:AlgN→FI$ vanishing at the second commutator [[x,y],z] with xy=0 such that for each $n∈ℕ$, Since every higher derivation on $AlgN$ is inner (see [10, 21]), there is an inner higher derivation $H={hn}n∈ℕ$ on $AlgN$. This implies that for each $n∈ℕ$

### Theorem 4.2.

Let 𝒩 be a nontrivial nest on a Hilbert space T of dimension greater than 2, Alg𝒩 be the associated nest algebra and p be a nontrivial projection in 𝒩. If $Δ={δn}n∈ℕ$ is a sequence of ℛ-linear maps $δn:AlgN→AlgN$ satisfying $δn[[x,y],z]=∑ i+j+k=n[[δi(x),δj(y)],δk(z)]$ for all $x,y,z∈AlgN$ with xy=p. Then for each $n∈ℕ$, where $H={hn}n∈ℕ$ is an inner higher derivation on $AlgN$ and $τn:AlgN→FI$ (where $FI$ is the center of $AlgN$) is an $R$-linear map vanishing at the second commutator [[x,y],z] with xy=p.

Proof. Let $A11=pAlgNp$, $A22=(I−p)AlgN(I−p)$ and $A12=pAlgN(I−p)$. Then $A11$ and $A22$ are unital algebras with unit element p and I-p respectively and $AlgN=Tri(A11,A12,A22)$ is a triangular algebra. Also $AlgN$ satisfies the conditions of Theorem 3.1, then there exists a higher derivation $D={dn}n∈ℕ$ of $AlgN$ and a linear map $τn:AlgN→FI$ vanishing at the second commutator [[x,y],z] with xy=p such that for each $n∈ℕ$, Since every higher derivation on $AlgN$ is inner (see [10, 21]) there exists an inner higher derivation $H={hn}n∈ℕ$ on $AlgN$ such that for each $n∈ℕ$,

If Hilbert space T is finite dimensional, then nest algebras are upper block triangular matrices algebras .

### Theorem 4.3.

Let $Bn(R)$ be a proper block upper triangular matrix algebra over a commutative ring ℛ. If $Δ={δn}n∈ℕ$ is a sequence of ℛ-linear maps $δn:Bn(R)→Bn(R)$ satisfying $δn([[x,y],z])=∑ i+j+k=n[[δi(x),δj(y)],δk(z)]$ for all $x,y,z∈Bn(R)$ with xy=0 (resp. xy=p, p be a nontrivial projection in $Bn(R))$. Then for each $n∈ℕ$, where $H={hn}n∈ℕ$ is an inner higher derivation on $Bn(R)$ and $τn:Bn(R)→FI$ (where $FI$ is the center of $Bn(R)$) is an ℛ-linear map vanishing at the second commutator [[x,y],z] with xy=0 (resp. xy=p).

Proof. It can be easily seen that conditions of Theorems 2.1 & 3.1 hold for block upper triangular matrix algebra and from [21, Proposition 2.6] all higher derivations of $Bn(R)$ are inner. Hence δn is the sum of an inner higher derivation $hn:Bn(R)→Bn(R)$ and a functional $τn:Bn(R)→FI$ that vanishes on all second commutators of $Bn(R).$

The authors are indebted to the referee for his/her helpful comments and suggestions.

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