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 $\mathbb{Z}$, the ring of integers, and define a map $L:\mathcal{A}\to \mathcal{A}$ such that $L\left(\begin{array}{ccc}0& x& y\\ 0& 0& z\\ 0& 0& 0\end{array}\right)=\left(\begin{array}{ccc}0& y& 0\\ 0& 0& z\\ 0& 0& 0\end{array}\right).$

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 $\mathbb{N}$ be the set of nonnegative integers and $\Delta ={\left\{{\delta }_n\right\}}_{n\in \mathbb{N}}$ be a sequence of ℛ-linear mappings ${\delta }_n:\mathcal{A}\to \mathcal{A}$ such that ${\delta }_0=i{d}_{\mathcal{A}}$, the identity map on 𝒜. Then Δ is said to be

• (i) a higher derivation on 𝒜 if

  ${\delta }_n(xy)={\displaystyle \sum _{i+j=n}{\delta }_i}(x){\delta }_j(y), for all x,y\in \mathcal{A} \& n\in \mathbb{N};$

• (ii) a Lie higher derivation on 𝒜 if

  ${\delta }_n([x,y])={\displaystyle \sum _{i+j=n}[{\delta }_i}(x),{\delta }_j(y)], for all x,y\in \mathcal{A} \& n\in \mathbb{N};$

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

  ${\delta }_n([[x,y],z])={\displaystyle \sum _{i+j=n}[[{\delta }_i}(x),{\delta }_j(y)],{\delta }_k(z)], for all x,y,z\in \mathcal{A} \& n\in \mathbb{N}.$

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 $\mathcal{L}={\left\{L_n\right\}}_{n\in \mathbb{N}}$ of linear mappings $L_n:\mathcal{A}\to \mathcal{A}$ such that $L_n=\frac{L^n}{n!}$, 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 $\mathfrak{A}=Tri(\mathcal{A},\mathcal{M},\mathcal{B})=\{\left.\left(\begin{array}{cc}a& m\\ 0& b\end{array}\right)\right|a\in \mathcal{A},m\in \mathcal{M},b\in \mathcal{B}\}$ 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 $\delta :\mathcal{A}\to \mathcal{A}$ is derivable at a given point $c\in \mathcal{A}$ if $\delta (x)y+x\delta (y)=\delta (c)$ for every $x,y\in \mathcal{A}$ 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 $\delta :\mathcal{A}\to \mathcal{A}$ is said to be a Lie derivable at a given point $c\in \mathcal{A}$ if $\delta ([x,y])=[\delta (x),y]+[x,\delta (y)]$ for all $x,y\in \mathcal{A}$ with xy=c. Lu and Jing [18] discussed such maps on B(X) where X is a Banach space with $\text{dim}X\ge 3$ 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 $\delta =d+\tau$, where d is a derivation of B(X) and $\tau :B(X)\to ℂI$ 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 $\mathfrak{A}$, if $L:\mathfrak{A}\to \mathfrak{A}$ is an $\mathcal{R}$-linear map satisfying $\delta ([x,y])=[\delta (x),y]+[x,\delta (y)]$ for any $x,y\in \mathfrak{A}$ with xy = 0 (resp. xy = p, where p is a fixed nontrivial idempotent of $\mathfrak{A}$ ), then $\delta =d+\tau ,$ where d is a derivation of $\mathfrak{A}$ and $\tau :\mathfrak{A}\to Z(\mathfrak{A})$ (where $Z(\mathfrak{A})$ is the center of $\mathfrak{A}$ ) 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 $\delta :\mathcal{A}\to \mathcal{A}$ satisfies $\delta ([[x,y],z])=[[\delta (x),y],z]+[[x,\delta (y)],z]+[[x,y],\delta (z)]$ for any $x,y,z\in \mathcal{A}$ 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 $f:\mathcal{A}\to ℂI$ (where $ℂI$ is the center of 𝒜) vanishing at every second commutator [[x, y], z] with xy = 0 (resp. xy = p) such that $\delta (x)=xr-rx+f(x)$ for any $x\in \mathcal{A}.$

An ℛ-linear map δ:𝒜⇒ 𝒜 is Lie triple derivable at a given point c ∈ 𝒜 if $\delta ([[x,y],z])=[[\delta (x),y],z]+[[x,\delta (y)],z]+[[x,y],\delta (z)]$ for all $x,y,z\in \mathcal{A}$ 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 $\mathfrak{A}$ satisfying ${\delta }_n([[x,y],z])={\displaystyle \sum _{i+j+k=n}[[{\delta }_i(x),{\delta }_j(y)],{\delta }_k(z)]}$ for any x,y,z ∈ $\mathfrak{A}$ with xy=0 (resp. xy=p, where p is a fixed nontrivial idempotent).

Throughout the present paper $\mathfrak{A}$ will denote a triangular algebra which is 2-torsion free. Define two natural projections ${\pi }_{\mathcal{A}}:\mathfrak{A}\to \mathcal{A}$ and ${\pi }_{\mathcal{B}}:\mathfrak{A}\to \mathcal{B}$ by ${\pi }_{\mathcal{A}}\left(\begin{array}{cc}a& m\\ 0& b\end{array}\right)=a$ and ${\pi }_{\mathcal{B}}\left(\begin{array}{cc}a& m\\ 0& b\end{array}\right)=b$. The center of $\mathfrak{A}$ coincides with

$$Z(\mathfrak{A})=\{\left.\left(\begin{array}{cc}a& 0\\ 0& b\end{array}\right)\right|a\in \mathcal{A},b\in \mathcal{B},am=mb \text{for all} m\in \mathcal{M}\}.$$

Moreover, ${\pi }_{\mathcal{A}}(Z(\mathfrak{A}))\subseteq Z(\mathcal{A})$ and ${\pi }_{\mathcal{B}}(Z(\mathfrak{A}))\subseteq Z(\mathcal{B})$, and there exists a unique algebra isomorphism $\eta :{\pi }_{\mathcal{B}}(Z(\mathfrak{A}))\to {\pi }_{\mathcal{A}}(Z(\mathfrak{A}))$ such that $\eta (b)m=mb$ for all $m\in \mathcal{M}$.

Let $1_{\mathcal{A}}$ (resp.$1_{\mathcal{B}}$) be the identity of the algebra $\mathcal{A}$ (resp.$\mathcal{B}$) and let $I=\left(\begin{array}{cc}{1}_{\mathcal{A}}& 0\\ 0& 1_{\mathcal{B}}\end{array}\right)$ be the identity of triangular algebra $\mathfrak{A}$. Throughout this paper, we shall use the following notations: $p_1=\left(\begin{array}{cc}{1}_{\mathcal{A}}& 0\\ 0& 0\end{array}\right)$ and $p_2=I-p_1=\left(\begin{array}{cc}0& 0\\ 0& 1_{\mathcal{B}}\end{array}\right)$. Set ${\mathfrak{A}}_{11}=\{\left(\begin{array}{cc}a& 0\\ 0& 0\end{array}\right)a\in \mathcal{A}\}$, ${\mathfrak{A}}_{12}=\{\left(\begin{array}{cc}0& m\\ 0& 0\end{array}\right)m\in \mathcal{M}\}$ and ${\mathfrak{A}}_{22}=\{\left(\begin{array}{cc}0& 0\\ 0& b\end{array}\right)b\in \mathcal{B}\}$. Then we can write $\mathfrak{A}={\mathfrak{A}}_{11}\oplus {\mathfrak{A}}_{12}\oplus {\mathfrak{A}}_{22}$, where ${\mathfrak{A}}_{11}$ is a subalgebra of $\mathfrak{A}$ isomorphic to $\mathcal{A}$, ${\mathfrak{A}}_{22}$ is a subalgebra of $\mathfrak{A}$ isomorphic to $\mathcal{B}$ and ${\mathfrak{A}}_{12}$ is a $({\mathfrak{A}}_{11},{\mathfrak{A}}_{22})$-bimodule isomorphic to the bimodule $\mathcal{M}$. To simplify the notation we will use the following convention: $a_{11}\in \mathcal{A}={\mathfrak{A}}_{11}$, $a_{22}\in \mathcal{B}={\mathfrak{A}}_{22}$ and $a_{12}\in \mathcal{M}={\mathfrak{A}}_{12}$. Then each element $x\in \mathfrak{A}$ can be represented in the form $x=a_{11}+a_{12}+a_{22},$ where $a_{11}\in {\mathfrak{A}}_{11},a_{22}\in {\mathfrak{A}}_{22}$ and $a_{12}\in {\mathfrak{A}}_{12}.$

In what follows, we write $a_{ij}$, it indicates $a_{ij}\in {\mathfrak{A}}_{ij}$ and the corresponding element in $\mathcal{A},\mathcal{B}$ or $\mathcal{M}$. Note that $a_{ij}{a}_{kl}=0$ if $j\ne k$.

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

Lemma 1.1.

Let $\mathfrak{A}$ be a triangular algebra $Tri({\mathfrak{A}}_{11},{\mathfrak{A}}_{12},{\mathfrak{A}}_{22})$. If ${\pi }_{{\mathfrak{A}}_{11}}(Z(\mathfrak{A}))=Z({\mathfrak{A}}_{11})$ and ${\pi }_{{\mathfrak{A}}_{22}}(Z(\mathfrak{A}))=Z({\mathfrak{A}}_{22})$, then there is a unique algebra isomorphism $\eta :Z({\mathfrak{A}}_{22})\to Z({\mathfrak{A}}_{11})$ such that $\eta (b)\oplus b\in Z(\mathfrak{A})$ for any $b\in Z({\mathfrak{A}}_{22})$.

The main result of the present paper states as follows:

Theorem 2.1. Let $\mathfrak{A}=Tri({\mathfrak{A}}_{11},{\mathfrak{A}}_{12},{\mathfrak{A}}_{22})$ be a 2-torsion free triangular algebra consisting of algebras ${\mathfrak{A}}_{11}$ and ${\mathfrak{A}}_{22}$ over a commutative ring $\mathcal{R}$ with unity $1_{{\mathfrak{A}}_{11}}$ and $1_{{\mathfrak{A}}_{22}}$ respectively and ${\mathfrak{A}}_{12}$ be a faithful $({\mathfrak{A}}_{11},{\mathfrak{A}}_{22})$-bimodule, which is faithful as a left ${\mathfrak{A}}_{11}$-module and also as a right ${\mathfrak{A}}_{22}$-module. Suppose that

• (i) ${\pi }_{{\mathfrak{A}}_{11}}(Z(\mathfrak{A}))=Z({\mathfrak{A}}_{11})$ and ${\pi }_{{\mathfrak{A}}_{22}}(Z(\mathfrak{A}))=Z({\mathfrak{A}}_{22})$.

• (ii) For any $a\in {\mathfrak{A}}_{11}$, if $[a,{\mathfrak{A}}_{11}]\in Z({\mathfrak{A}}_{11})$, then $a\in Z({\mathfrak{A}}_{11})$ or for any $b\in {\mathfrak{A}}_{22}$, if $[b,{\mathfrak{A}}_{22}]\in Z({\mathfrak{A}}_{22})$, then $b\in Z({\mathfrak{A}}_{22})$.

If $\Delta ={\left\{{\delta }_n\right\}}_{n\in \mathbb{N}}$ is a sequence of $\mathcal{R}$-linear maps ${\delta }_n:\mathfrak{A}\to \mathfrak{A}$ such that ${\delta }_n[[x,y],z]={\displaystyle \sum _{i+j+k=n}[[{\delta }_i(x),{\delta }_j(y)],{\delta }_k(z)]}$

for all $x,y,z\in \mathfrak{A}$ with xy=0, then for each $n\in \mathbb{N}$,

${\delta }_n=d_n+{\tau }_n$; where $d_n:\mathfrak{A}\to \mathfrak{A}$ is $\mathcal{R}$-linear mapping satisfying $d_n(xy)={\displaystyle \sum _{i+j=n}{d}_i}(x)d_j(y)$ for all $x,y\in \mathfrak{A}$, i.e., $\mathcal{D}={\left\{d_n\right\}}_{n\in \mathbb{N}}$ is a higher derivation of $\mathfrak{A}$ and ${\tau }_n:\mathfrak{A}\to Z(\mathfrak{A})$ is an $\mathcal{R}$-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 ${\delta }_1=\delta$ on $\mathfrak{A}$ there exists a derivation d on $\mathfrak{A}$ and a linear mapping $\tau :\mathfrak{A}\to Z(\mathfrak{A})$ vanishing on second commutators such that $\delta (x)=d(x)+\tau (x)$ for all $x\in \mathfrak{A}$.

Proof of Theorem 2.1. Claim 1. $p_1\delta (p_1)p_1+p_2\delta (p_1)p_2\in Z(\mathfrak{A})$; $\delta (a_{12})=p_1\delta (a_{12})p_2\in {\mathfrak{A}}_{12}$ for any $a_{12}\in {\mathfrak{A}}_{12}$.

Since $a_{12}{p}_1=0$ for any $a_{12}\in {\mathfrak{A}}_{12}$, we have

(2.1)$$\begin{array}{l}\delta (a_{12})=\delta ([[a_{12},p_1],p_1])\\ =[[\delta (a_{12}),p_1],p_1]+[[a_{12},\delta (p_1)],p_1]+[[a_{12},p_1],\delta (p_1)]\\ =\delta (a_{12})p_1-p_1\delta (a_{12})p_1-p_1\delta (a_{12})p_1+p_1\delta (a_{12})+a_{12}\delta (p_1)p_1\\ -a_{12}\delta (p_1)+p_1\delta (p_1)a_{12}-a_{12}\delta (p_1)+\delta (p_1)a_{12}.\end{array}$$

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

$$2(p_1\delta (p_1)a_{12}-a_{12}\delta (p_1)p_2)=0.$$

This implies that $p_1\delta (p_1)a_{12}-a_{12}\delta (p_1)p_2=0,$ that is, $p_1\delta (p_1)p_1{a}_{12}-a_{12}{p}_2\delta (p_1)p_2=0,$ and hence $p_1\delta (p_1)p_1+p_2\delta (p_1)p_2\in Z(\mathfrak{A}).$ By putting $\delta (a_{12})=r_{11}+r_{12}+r_{22}$, $\delta (p_1)=p_{11}+p_{12}+p_{22}$ in (2.1), we get $r_{11}=0=r_{22}$ and so we have $p_i\delta (a_{12})p_i=0$ for $i\in \{1,2\}$, $\delta (a_{12})=p_1\delta (a_{12})p_2.$

Similarly we can get the following result:

Claim 2. $p_1\delta (p_2)p_1+p_2\delta (p_2)p_2\in Z(\mathfrak{A}).$

Claim 3. $\delta (I)=p_1\delta (I)p_1+p_2\delta (I)p_2\in Z(\mathfrak{A}).$

Since $p_1(I-p_1)=0$, we have

$$\begin{array}{l}0=\delta ([[p_1,I-p_1],p_1])\\ =[[\delta (p_1),I-p_1],p_1]+[[p_1,\delta (I-p_1)],p_1]+[[p_1,I-p_1],\delta (p_1)]\\ =[[p_1,\delta (I)],p_1]\\ =-p_1\delta (I)p_2.\end{array}$$

This yields that $\delta (I)=p_1\delta (I)p_1+p_2\delta (I)p_2$. By Claims 1 & 2, we have $\delta (I)=p_1\delta (p_1)p_1+p_2\delta (p_1)p_2+p_1\delta (p_2)p_1+p_2\delta (p_2)p_2\in Z(\mathfrak{A})$.

In the sequel, we define

$$f(x)=\delta (x)+{\delta }_{p_1\delta (p_1)p_2}(x),$$

where ${\delta }_{p_1\delta (p_1)p_2}(x)$ is the inner derivation determined by $p_1\delta (p_1)p_2$, that is,

$${\delta }_{p_1\delta (p_1)p_2}(x)=[p_1\delta (p_1)p_2,x]=p_1\delta (p_1)p_{2}x-x{p}_1\delta (p_1)p_2 \text{for all} x\in \mathfrak{A}.$$

One can verify that

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

for all $x,y,z\in \mathfrak{A}$ with xy=0. Moreover by Claim 1, we have

$$f(p_1)=\delta (p_1)+p_1\delta (p_1)p_2{p}_1-p_1\delta (p_1)p_2=p_1\delta (p_1)p_1+p_2\delta (p_1)p_2\in Z(\mathfrak{A}).$$

By Claim 3, we get $f(I)=\delta (I)\in Z(\mathfrak{A})$. Consequently $f(p_2)=f(I)-f(p_1)\in Z(\mathfrak{A}).$

Clearly for any $a_{12}\in {\mathfrak{A}}_{12}$, ${\delta }_{p_1\delta (p_1)p_2}(a_{12})=[p_1\delta (p_1)p_2,a_{12}]=0$. By Claim 1, we have $f(a_{12})=\delta (a_{12})\in {\mathfrak{A}}_{12}$, and hence

Claim 4. For any $a_{12}\in {\mathfrak{A}}_{12},f(a_{12})\in {\mathfrak{A}}_{12}.$

Claim 5. $f({\mathfrak{A}}_{ii})\subseteq {\mathfrak{A}}_{ii}\oplus {\mathfrak{A}}_{jj}$. There exists an $\mathcal{R}$-linear map ${\tau }_i:{\mathfrak{A}}_{ii}\to Z(\mathfrak{A})$ such that $f(a_{ii})-{\tau }_i(a_{ii})\in {\mathfrak{A}}_{ii}$ for all $a_{ii}\in {\mathfrak{A}}_{ii},where i,j=1,2 and i\ne j.$

First we show for i=1. Since a₁₁p₂=0 for any $a_{11}\in {\mathfrak{A}}_{11}$ and $f(p_2)\in Z(\mathfrak{A})$, it follows that

$$\begin{array}{l}0=f([[a_{11},p_2],p_2])\\ =[[f(a_{11}),p_2],p_2]+[[a_{11},f(p_2)],p_2]+[[a_{11},p_2],f(p_2)]\\ =f(a_{11})p_2-p_{2}f(a_{11})p_2-p_{2}f(a_{11})p_2+p_{2}f(a_{11}).\end{array}$$

Multiplying by p₁ from the left we get $p_{1}f(a_{11})p_2=0$ and hence $f(a_{11})\in {\mathfrak{A}}_{11}\oplus {\mathfrak{A}}_{22}$.

Similarly, we can prove the result for i=2. Thus, $f({\mathfrak{A}}_{ii})\subseteq {\mathfrak{A}}_{ii}\oplus {\mathfrak{A}}_{jj}$.

Now we can write $f(a_{11})=p_{1}f(a_{11})p_1+p_{2}f(a_{11})p_2$. Moreover, since $a_{11}{a}_{22}=0$, for any $a_{22}\in {\mathfrak{A}}_{22}$ and $x\in \mathfrak{A}$, it is easy to check that

$$\begin{array}{l}0=f([[a_{11},a_{22}],x])=[[f(a_{11}),a_{22}],x]+[[a_{11},f(a_{22})],x]\\ =[[f(a_{11}),a_{22}]+[a_{11},f(a_{22})],x].\end{array}$$

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

$$\begin{array}{l}0=p_2[[f(a_{11}),a_{22}]+[a_{11},f(a_{22})],x]p_2\\ =[[p_{2}f(a_{11})p_2,p_2{a}_{22}{p}_2]+[p_2{a}_{11}{p}_2,p_{2}f(a_{22})p_2],p_{2}x{p}_2]\\ =[[p_{2}f(a_{11})p_2,a_{22}],p_{2}x{p}_2].\end{array}$$

This implies that $[p_{2}f(a_{11})p_2,a_{22}]\in Z({\mathfrak{A}}_{22})$. Hence by hypothesis (ii), we find that

$$p_{2}f(a_{11})p_2\in Z({\mathfrak{A}}_{22}).$$

Define ${\tau }_1:{\mathfrak{A}}_{11}\to Z(\mathfrak{A})$ such that ${\tau }_1(a_{11})=\eta (p_{2}f(a_{11})p_2)\oplus p_{2}f(a_{11})p_2$, where η is the map defined in Lemma 1.1. Thus, we get

$\begin{array}{l}f(a_{11})-{\tau }_1(a_{11})=p_{1}f(a_{11})p_1+p_{2}f(a_{11})p_2-\eta (p_{2}f(a_{11})p_2)-p_{2}f(a_{11})p_2\\ =p_{1}f(a_{11})p_1-\eta (p_{2}f(a_{11})p_2)\in {\mathfrak{A}}_{11}.\end{array}$

Since f is $\mathcal{R}$-linear, one can verify that τ₁ is $\mathcal{R}$-linear. Similarly, we can define $\mathcal{R}$-linear map ${\tau }_2:{\mathfrak{A}}_{22}\to Z(\mathfrak{A})$ by ${\tau }_2(a_{22})=p_{1}f(a_{22})p_1\oplus {\eta }^{-1}(p_{1}f(a_{22})p_1)$. Then

$$\begin{array}{l}f(a_{22})-{\tau }_2(a_{22})=p_{1}f(a_{22})p_1+p_{2}f(a_{22})p_2-p_{1}f(a_{22})p_1-{\eta }^{-1}(p_{1}f(a_{22})p_1)\\ =p_{2}f(a_{22})p_2-{\eta }^{-1}(p_{1}f(a_{22})p_1)\in {\mathfrak{A}}_{22}.\end{array}$$

Now, for any$x=a_{11}+a_{12}+a_{22}\in \mathfrak{A}$, we define two $\mathcal{R}$-linear maps $\tau :\mathfrak{A}\to Z(\mathfrak{A})$ and $d:\mathfrak{A}\to \mathfrak{A}$ by

$$\tau (x)={\tau }_1(a_{11})+{\tau }_2(a_{22}) \text{and} d(x)=f(x)-\tau (x) \text{respectively}.$$

Then, $d({\mathfrak{A}}_{ij})\subseteq {\mathfrak{A}}_{ij} for 1\le i\le j\le 2 and d(a_{12})=f(a_{12})$.

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 $x,y\in \mathfrak{A}$. Now, we divide the proof into the following three steps:

Step 1. Since a₁₂a₁₁=0 for any $a_{11}\in {\mathfrak{A}}_{11}$, $a_{12}\in {\mathfrak{A}}_{12}$ and τ(x) is in $Z(\mathfrak{A})$, we have

$$\begin{array}{l}-d(a_{11}{a}_{12})=d([[a_{11},a_{12}],p_1])=f([[a_{11},a_{12}],p_1])\\ =[[f(a_{11}),a_{12}],p_1]+[[a_{11},f(a_{12})],p_1]+[[a_{11},a_{12}],f(p_1)]\\ =[[d(a_{11})+\tau (a_{11}),a_{12}],p_1]+[[a_{11},d(a_{12})],p_1]\\ =-d(a_{11})a_{12}-a_{11}d(a_{12}).\end{array}$$

Hence, $d(a_{11}{a}_{12})=d(a_{11})a_{12}+a_{11}d(a_{12}).$ Similarly, we can get

$$d(a_{12}{a}_{22})=d(a_{12})a_{22}+a_{12}d(a_{22}) \text{for all} a_{12}\in {\mathfrak{A}}_{12},a_{22}\in {\mathfrak{A}}_{22}.$$

Step 2. Let $a_{11},b_{11}\in {\mathfrak{A}}_{11}$. For any $a_{12}\in {\mathfrak{A}}_{12}$, we have

(2.2)$$\begin{array}{l}d(a_{11}{b}_{11}{a}_{12})=d(a_{11})b_{11}{a}_{12}+a_{11}d(b_{11}{a}_{12})\\ =d(a_{11})b_{11}{a}_{12}+a_{11}d(b_{11})a_{12}+a_{11}{b}_{11}d(a_{12}).\end{array}$$

On the other hand,

(2.3)$$d(a_{11}{b}_{11}{a}_{12})=d(a_{11}{b}_{11})a_{12}+a_{11}{b}_{11}d(a_{12}).$$

Comparing (2.2) and (2.3), we have

$$(d(a_{11}{b}_{11})-d(a_{11})b_{11}-a_{11}d(b_{11}))a_{12}=0 \text{for all} a_{12}\in {\mathfrak{A}}_{12}.$$

Since ${\mathfrak{A}}_{12}$ is a faithful left ${\mathfrak{A}}_{11}$-module, we get

$$d(a_{11}{b}_{11})=d(a_{11})b_{11}+a_{11}d(b_{11}) \text{for all} a_{11},b_{11}\in {\mathfrak{A}}_{11}.$$

Similarly, one can arrive at

$$d(a_{22}{b}_{22})=d(a_{22})b_{22}+a_{22}d(b_{22}) \text{for all} a_{22},b_{22}\in {\mathfrak{A}}_{22}.$$

Step 3. Let $x=a_{11}+a_{12}+a_{22}$, $y=b_{11}+b_{12}+b_{22}$ be in $\mathfrak{A}$.

By Steps 1 & 2, we have

$$\begin{array}{l}d(xy)=d((a_{11}+a_{12}+a_{22})(b_{11}+b_{12}+b_{22}))\\ =d(a_{11}{b}_{11})+d(a_{11}{b}_{12})+d(a_{12}{b}_{22})+d(a_{22}{b}_{22})\\ =d(a_{11})b_{11}+a_{11}d(b_{11})+d(a_{11})b_{12}+a_{11}d(b_{12})\\ +d(a_{12})b_{22}+a_{12}d(b_{22})+d(a_{22})b_{22}+a_{22}d(b_{22}).\end{array}$$

On the other hand, $d({\mathfrak{A}}_{ij})\subseteq {\mathfrak{A}}_{ij}$ for $1\le i\le j\le 2$, we have

$$\begin{array}{l}d(x)y+xd(y)=d(a_{11}+a_{12}+a_{22})(b_{11}+b_{12}+b_{22})\\ +(a_{11}+a_{12}+a_{22})d(b_{11}+b_{12}+b_{22})\\ =d(a_{11})b_{11}+a_{11}d(b_{11})+d(a_{11})b_{12}+a_{11}d(b_{12})\\ +d(a_{12})b_{22}+a_{12}d(b_{22})+d(a_{22})b_{22}+a_{22}d(b_{22}).\end{array}$$

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 $x,y,z\in \mathfrak{A}$.

Suppose xy=0. Since $\tau (x)\in Z(\mathfrak{A})$, we have

$$\begin{array}{l}\tau ([[x,y],z])=f([[x,y],z])-d([[x,y],z])\\ =[[f(x),y],z]+[[x,f(y)],z]+[[x,y],f(z)]-d([[x,y],z])\\ =[[d(x)+\tau (x),y],z]+[[x,d(y)+\tau (y)],z]+[[x,y],d(z)+\tau (z)]\\ -d([[x,y],z])\\ =[[d(x),y],z]+[[x,d(y)],z]+[[x,y],d(z)]-d([[x,y],z])\\ =0\end{array}$$

for all $x,y,z\in \mathfrak{A}.$ The proof of our theorem for n=1 is now complete.

Now, suppose that the conclusion holds for all $m<n\in \mathbb{N}$. That is, there exist linear maps $d_m:\mathfrak{A}\to \mathfrak{A}$ and ${\tau }_m:\mathfrak{A}\to Z(\mathfrak{A})$ such that ${\delta }_m(x)=d_m(x)+{\tau }_m(x)$, ${\tau }_m([[x,y],z])=0$ with xy=0 and $d_m(xy)={\displaystyle \sum _{i+j=m}{d}_i}(x)d_j(y)$ for all $x,y,z\in \mathfrak{A}$.

Moreover, ${\delta }_m$ has the following properties:

$$p_1{\delta }_m(p_1)p_1+p_2{\delta }_m(p_1)p_2\in Z(\mathfrak{A});{\delta }_m(a_{12})=p_1{\delta }_m(a_{12})p_2\in {\mathfrak{A}}_{12};$$
$$p_1{\delta }_m(p_2)p_1+p_2{\delta }_m(p_2)p_2\in Z(\mathfrak{A});$$
$${\delta }_m(I)=p_1{\delta }_m(I)p_1+p_2{\delta }_m(I)p_2\in Z(\mathfrak{A}).$$

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

Claim 8. $p_1{\delta }_n(p_1)p_1+p_2{\delta }_n(p_1)p_2\in Z(\mathfrak{A});{\delta }_n(a_{12})=p_1{\delta }_n(a_{12})p_2\in {\mathfrak{A}}_{12}$ for any $a_{12}\in {\mathfrak{A}}_{12}$.

Since $a_{12}{p}_1=0$ for any $a_{12}\in {\mathfrak{A}}_{12}$, by induction hypothesis, we have

(2.4)$$\begin{array}{l}{\delta }_n(a_{12})={\delta }_n([[a_{12},p_1],p_1])\\ =[[{\delta }_n(a_{12}),p_1],p_1]+[[a_{12},{\delta }_n(p_1)],p_1]+[[a_{12},p_1],{\delta }_n(p_1)]\\ +{\displaystyle \sum _{\begin{array}{l}i+j+k=n\\ 0\le i,j,k<n\end{array}}[[{\delta }_i(a_{12}),{\delta }_j(p_1)],{\delta }_k(p_1)]}\\ ={\delta }_n(a_{12})p_1-p_1{\delta }_n(a_{12})p_1-p_1{\delta }_n(a_{12})p_1+p_1{\delta }_n(a_{12})+a_{12}{\delta }_n(p_1)p_1\\ -a_{12}{\delta }_n(p_1)+p_1{\delta }_n(p_1)a_{12}-a_{12}{\delta }_n(p_1)+{\delta }_n(p_1)a_{12}.\end{array}$$

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

$$2(p_1{\delta }_n(p_1)a_{12}-a_{12}{\delta }_n(p_1)p_2)=0.$$

This implies that $p_1{\delta }_n(p_1)a_{12}-a_{12}{\delta }_n(p_1)p_2=0,$ that is, $p_1{\delta }_n(p_1)p_1{a}_{12}-a_{12}{p}_2{\delta }_n(p_1)p_2=0,$ and hence $p_1{\delta }_n(p_1)p_1+p_2{\delta }_n(p_1)p_2\in Z(\mathfrak{A}).$ By putting ${\delta }_n(a_{12})=s_{11}+s_{12}+s_{22}$, ${\delta }_n(p_1)=t_{11}+t_{12}+t_{22}$ in (2.4), we get $s_{11}=0=s_{22}$ and so, we have $p_i{\delta }_n(a_{12})p_i=0$ for $i\in \{1,2\}$. Hence, ${\delta }_n(a_{12})=p_1{\delta }_n(a_{12})p_2.$

Since $p_2{a}_{12}=0$ for any $a_{12}\in {\mathfrak{A}}_{12}$. Similarly, we can get the following result.

Claim 9. $p_1{\delta }_n(p_2)p_1+p_2{\delta }_n(p_2)p_2\in Z(\mathfrak{A}).$

Claim 10. ${\delta }_n(I)=p_1{\delta }_n(I)p_1+p_2{\delta }_n(I)p_2\in Z(\mathfrak{A}).$

Since $p_1(I-p_1)=0$, by using the induction hypothesis, we find that

$$\begin{array}{l}0={\delta }_n([[p_1,I-p_1],p_1])\\ =[[{\delta }_n(p_1),I-p_1],p_1]+[[p_1,{\delta }_n(I-p_1)],p_1]+[[p_1,I-p_1],{\delta }_n(p_1)]\\ +{\displaystyle \sum _{\begin{array}{l}i+j+k=n\\ 0\le i,j,k<n\end{array}}[[{\delta }_i(p_1),{\delta }_j(I-p_1)],{\delta }_k(p_1)]}\\ =p_1{\delta }_n(I)p_1-{\delta }_n(I)p_1-p_1{\delta }_n(I)+p_1{\delta }_n(I)p_1.\end{array}$$

On multiplying by p₁ from the left and by p₂ from the right, we have $p_1{\delta }_n(I)p_2=0,$ and hence we get that ${\delta }_n(I)=p_1{\delta }_n(I)p_1+p_2{\delta }_n(I)p_2$. By Claims 8 & 9, we have ${\delta }_n(I)=p_1{\delta }_n(p_1)p_1+p_2{\delta }_n(p_1)p_2+p_1{\delta }_n(p_2)p_1+p_2{\delta }_n(p_2)p_2\in Z(\mathfrak{A})$.

In the sequel, we define

$$f_n(x)={\delta }_n(x)+{\delta }_{p_1{\delta }_n(p_1)p_2}(x),$$

where ${\delta }_{p_1{\delta }_n(p_1)p_2}(x)$ is the inner derivation determined by $p_1{\delta }_n(p_1)p_2$, that is,

$${\delta }_{p_1{\delta }_n(p_1)p_2}(x)=[p_1{\delta }_n(p_1)p_2,x]=p_1{\delta }_n(p_1)p_{2}x-x{p}_1{\delta }_n(p_1)p_2 \text{for all} x\in \mathfrak{A}.$$

One can verify that

$$\begin{array}{l}{f}_n([[x,y],z])=[[f_n(x),y],z]+[[x,f_n(y)],z]+[[x,y],f_n(z)]\\ +{\displaystyle \sum _{\begin{array}{l}i+j+k=n\\ 0\le i,j,k<n\end{array}}[[f_i(x),f_j(y)],f_k(z)]}\end{array}$$

for all $x,y,z\in \mathfrak{A}$ with xy=0. Moreover by Claim 8, we have

$$f_n(p_1)={\delta }_n(p_1)+p_1{\delta }_n(p_1)p_2{p}_1-p_1{\delta }_n(p_1)p_2=p_1{\delta }_n(p_1)p_1+p_2{\delta }_n(p_1)p_2\in Z(\mathfrak{A}).$$

By Claim 10, we find that $f_n(I)={\delta }_n(I)\in Z(\mathfrak{A})$. Consequently $f_n(p_2)=f_n(I)-f_n(p_1)\in Z(\mathfrak{A}).$ Clearly for any $a_{12}\in {\mathfrak{A}}_{12}$, ${\delta }_{p_1{\delta }_n(p_1)p_2}(a_{12})=[p_1{\delta }_n(p_1)p_2,a_{12}]=0$. By Claim 8, we have $f_n(a_{12})={\delta }_n(a_{12})\in {\mathfrak{A}}_{12}$, and hence

Claim 11. For any $a_{12}\in {\mathfrak{A}}_{12}$, $f_n(a_{12})\in {\mathfrak{A}}_{12}.$

Claim 12. $f_n({\mathfrak{A}}_{ii})\subseteq {\mathfrak{A}}_{ii}\oplus {\mathfrak{A}}_{jj}$. There exists an $\mathcal{R}$-linear map ${\tau }_{ni}:{\mathfrak{A}}_{ii}\to Z(\mathfrak{A})$ such that $f_n(a_{ii})-{\tau }_{ni}(a_{ii})\in {\mathfrak{A}}_{ii}$ for all $a_{ii}\in {\mathfrak{A}}_{ii},where i,j=1,2 and i\ne j.$

First we show the result for i=1. Since a₁₁p₂=0 for any $a_{11}\in {\mathfrak{A}}_{11}$ and $f_n(p_2)\in Z(\mathfrak{A})$, using induction hypothesis, it follows that

$$\begin{array}{l}0=f_n([[a_{11},p_2],p_2])\\ =[[f_n(a_{11}),p_2],p_2]+[[a_{11},f_n(p_2)],p_2]+[[a_{11},p_2],f_n(p_2)]\\ +{\displaystyle \sum _{\begin{array}{l}i+j+k=n\\ 0\le i,j,k<n\end{array}}[[f_i(a_{11}),f_j(p_2)],f_k(p_2)]}\\ =f_n(a_{11})p_2-p_2{f}_n(a_{11})p_2-p_2{f}_n(a_{11})p_2+p_2{f}_n(a_{11}).\end{array}$$

Multiply by p₁ from the left to get $p_1{f}_n(a_{11})p_2=0.$ So, $f_n(a_{11})\in {\mathfrak{A}}_{11}\oplus {\mathfrak{A}}_{22}$.

Similarly we can find the result for i=2. Thus, $f_n({\mathfrak{A}}_{ii})\subseteq {\mathfrak{A}}_{ii}\oplus {\mathfrak{A}}_{jj}$.

Now we can write $f_n(a_{11})=p_1{f}_n(a_{11})p_1+p_2{f}_n(a_{11})p_2$. Moreover, since $a_{11}{a}_{22}=0$ for any $a_{22}\in {\mathfrak{A}}_{22}$ and $x\in \mathfrak{A}$, it is easy to observe that

$$\begin{array}{l}0=f_n([[a_{11},a_{22}],x])=[[f_n(a_{11}),a_{22}],x]+[[a_{11},f_n(a_{22})],x]+[[a_{11},a_{22}],f_n(x)]\\ +{\displaystyle \sum _{\begin{array}{l}i+j+k=n\\ 0\le i,j,k<n\end{array}}[[f_i(a_{11}),f_j(a_{22}),]f_k(x)]}\\ =[[f_n(a_{11}),a_{22}],x]+[[a_{11},f_n(a_{22})],x]\\ =[[f_n(a_{11}),a_{22}]+[a_{11},f_n(a_{22})],x].\end{array}$$

Multiplying by p₂ on both the sides, we get

$$\begin{array}{l}0=p_2[[f_n(a_{11}),a_{22}]+[a_{11},f_n(a_{22})],x]p_2\\ =[[p_2{f}_n(a_{11})p_2,p_2{a}_{22}{p}_2]+[p_2{a}_{11}{p}_2,p_2{f}_n(a_{22})p_2],p_{2}x{p}_2]\\ =[[p_2{f}_n(a_{11})p_2,a_{22}],p_{2}x{p}_2].\end{array}$$

This implies that $[p_2{f}_n(a_{11})p_2,a_{22}]\in Z({\mathfrak{A}}_{22})$. Hence by assumption (ii), we get $p_2{f}_n(a_{11})p_2\in Z({\mathfrak{A}}_{22})$. Define ${\tau }_{n1}:{\mathfrak{A}}_{11}\to Z(\mathfrak{A})$ by ${\tau }_{n1}(a_{11})=\eta (p_2{f}_n(a_{11})p_2)\oplus p_2{f}_n(a_{11})p_2$, where η is the map defined in Lemma 1.1. Thus, we get

$$\begin{array}{l}{f}_n(a_{11})-{\tau }_{n1}(a_{11})=p_1{f}_n(a_{11})p_1+p_2{f}_n(a_{11})p_2-\eta (p_2{f}_n(a_{11})p_2)-p_2{f}_n(a_{11})p_2\\ =p_1{f}_n(a_{11})p_1-\eta (p_2{f}_n(a_{11})p_2)\in {\mathfrak{A}}_{11}.\end{array}$$

Since f_n is ℛ-linear, one can verify that τ_n1 is ℛ-linear. Similarly, we can define ℛ-linear map ${\tau }_{n2}:{\mathfrak{A}}_{22}\to Z(\mathfrak{A})$ by ${\tau }_{n2}(a_{22})=p_1{f}_n(a_{22})p_1\oplus {\eta }^{-1}(p_1{f}_n(a_{22})p_1)$. Then

$$\begin{array}{l}{f}_n(a_{22})-{\tau }_{n2}(a_{22})=p_1{f}_n(a_{22})p_1+p_2{f}_n(a_{22})p_2-p_1{f}_n(a_{22})p_1-{\eta }^{-1}(p_1{f}_n(a_{22})p_1)\\ =p_2{f}_n(a_{22})p_2-{\eta }^{-1}(p_1{f}_n(a_{22})p_1)\in {\mathfrak{A}}_{22}.\end{array}$$

Now, for any $x=a_{11}+a_{12}+a_{22}\in \mathfrak{A}$, we define two ℛ-linear maps ${\tau }_n:\mathfrak{A}\to Z(\mathfrak{A})$ and $d_n:\mathfrak{A}\to \mathfrak{A}$ by

$${\tau }_n(x)={\tau }_{n1}(a_{11})+{\tau }_{n2}(a_{22}) \text{and} d_n(x)=f_n(x)-{\tau }_n(x) \text{respectively}.$$

Then, $d_n({\mathfrak{A}}_{ij})\subseteq {\mathfrak{A}}_{ij} for 1\le i\le j\le 2 and d_n(a_{12})=f_n(a_{12})$.

Claim 13. $d_n(xy)={\displaystyle \sum _{i+j=n}{d}_i}(x)d_j(y)forallx,y\in \mathfrak{A}$

Since $f_n\text{ \& }{\tau }_n$ are ℛ-linear and $d_n(x)=f_n(x)-{\tau }_n(x)$, d_n is an ℛ-linear. It remains to show that $d_n(xy)={\displaystyle \sum _{i+j=n}{d}_i}(x)d_j(y)forallx,y\in \mathfrak{A}$.

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

Step 1. Since $a_{12}{a}_{11}=0$ for any $a_{11}\in {\mathfrak{A}}_{11}$, $a_{12}\in {\mathfrak{A}}_{12}$ and ${\tau }_n(x)$ is in

$Z(\mathfrak{A})$, by induction hypothesis, we have

$$\begin{array}{l}-d_n(a_{11}{a}_{12})=f_n([[a_{11},a_{12}],p_1])\\ =[[f_n(a_{11}),a_{12}],p_1]+[[a_{11},f_n(a_{12})],p_1]+[[a_{11},a_{12}],f_n(p_1)]\\ +{\displaystyle \sum _{\begin{array}{l}i+j+k=n\\ 0\le i,j,k<n\end{array}}[[f_i(a_{11}),f_j(a_{12})],f_k(p_1)]}\\ =[[f_n(a_{11}),a_{12}],p_1]+[[a_{11},f_n(a_{12})],p_1]\\ +{\displaystyle \sum _{\begin{array}{l}i+j+k=n\\ 0\le i,j,k<n\end{array}}[[f_i(a_{11}),f_j(a_{12})],(p_1)]}\\ =[[d_n(a_{11})+{\tau }_n(a_{11}),a_{12}],p_1]+[[a_{11},d_n(a_{12})+{\tau }_n(a_{12})],p_1]\\ +{\displaystyle \sum _{\begin{array}{l}i+j+k=n\\ 0\le i,j,k<n\end{array}}[[d_n(a_{11})+{\tau }_n(a_{11}),d_n(a_{12})+{\tau }_n(a_{12})],p_1]}\\ =[[d_n(a_{11}),a_{12}],p_1]+[[a_{11},d_n(a_{12})],p_1]\\ +{\displaystyle \sum _{\begin{array}{l}i+j+k=n\\ 0\le i,j,k<n\end{array}}[[d_n(a_{11}),d_n(a_{12})],p_1]}\\ =-d_n(a_{11})a_{12}-a_{11}{d}_n(a_{12})-{\displaystyle \sum _{\begin{array}{l}i+j+k=n\\ 0\le i,j,k<n\end{array}}{d}_i(a_{11})d_j(a_{12})}\end{array}$$

Hence, $d_n(a_{11}{a}_{12})={\displaystyle \sum _{i+j=n}{d}_i}(a_{11})d_j(a_{12}).$ Similarly, we can get

$$d_n(a_{12}{a}_{22})={\displaystyle \sum _{i+j=n}{d}_i}(a_{12})d_j(a_{22}) \text{for any} a_{12}\in {\mathfrak{A}}_{12},a_{22}\in {\mathfrak{A}}_{22}.$$

Step 2. Let $a_{11},b_{11}\in {\mathfrak{A}}_{11}$. For any $a_{12}\in {\mathfrak{A}}_{12}$, we have

(2.5)$$\begin{array}{l}{d}_n(a_{11}{b}_{11}{a}_{12})={\displaystyle \sum }_{i+k=n}{d}_i(a_{11}{b}_{11})d_k(a_{12})\\ =d_n(a_{11}{b}_{11})a_{12}+{\displaystyle \sum }_{\begin{array}{c}i+k=n\\ k\ne 0\end{array}}{d}_i(a_{11}{b}_{11})d_k(a_{12})\\ =d_n(a_{11}{b}_{11})a_{12}+{\displaystyle \sum }_{\begin{array}{c}l+t+k=n\\ k\ne 0\end{array}}{d}_l(a_{11})d_t(b_{11})d_k(a_{12}).\end{array}$$

On the other hand,

(2.6)$$\begin{array}{l}{d}_n(a_{11}{b}_{11}{a}_{12})={\displaystyle \sum }_{i+j=n}{d}_i(a_{11})d_j(b_{11}{a}_{12})\\ ={\displaystyle \sum }_{i+j+k=n}{d}_i(a_{11})d_j(b_{11})d_k(a_{12})\\ ={\displaystyle \sum }_{i+j=n}{d}_i(a_{11})d_j(b_{11})a_{12}+{\displaystyle \sum }_{\begin{array}{c}i+j+k=n\\ k\ne 0\end{array}}{d}_i(a_{11})d_j(b_{11})d_k(a_{12}).\end{array}$$

Comparing (2.5) and (2.6), we have

$$(d_n(a_{11}{b}_{11})-{\displaystyle \sum _{i+j=n}{d}_i}(a_{11})d_j(b_{11}))a_{12}=0.$$

Since ${\mathfrak{A}}_{12}$ is a faithful left ${\mathfrak{A}}_{11}$-module, we get

$$d_n(a_{11}{b}_{11})={\displaystyle \sum _{i+j=n}{d}_i}(a_{11})d_j(b_{11}).$$

Similarly, we can calculate

$$d_n(a_{22}{b}_{22})={\displaystyle \sum _{i+j=n}{d}_i}(a_{22})d_j(b_{22}).$$

Step 3. Let $x=a_{11}+a_{12}+a_{22}$, $y=b_{11}+b_{12}+b_{22}$ be in $\mathfrak{A}$. By Steps 1 & 2, we have

$$d_n(xy)={\displaystyle \sum _{i+j=n}{d}_i}(x)d_j(y).$$

Claim 14. ${\tau }_n$ vanishes at second commutator [[x,y],z] with xy=0 for all $x,y,z\in \mathfrak{A}.$

Since xy=0, we find that

$$\begin{array}{l}{\tau }_n([[x,y],z])=f_n([[x,y],z])-d_n([[x,y],z])\\ ={\displaystyle \sum _{i+j+k=n}([[f_i(}x),f_j(y)],f_k(z)])-d_n([[x,y],z])\\ ={\displaystyle \sum _{i+j+k=n}([[d_i(}x)+{\tau }_i(x),d_j(y)+{\tau }_j(y)],d_k(z)+{\tau }_k(z)])\\ -d_n([[x,y],z])\\ ={\displaystyle \sum _{i+j+k=n}([[d_i(}x),d_j(y)],d_k(z)]-d_n([[x,y],z])\\ =0\end{array}$$

for all $x,y,z\in \mathfrak{A}$. The proof is now complete.