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Research article

Characterizations of generalized Lie n-higher derivations on certain triangular algebras

  • Received: 22 September 2024 Revised: 11 October 2024 Accepted: 11 October 2024 Published: 22 October 2024
  • MSC : 16W25, 15A78

  • The aim of this paper was to provide a characterization of nonlinear generalized Lie n-higher derivations for a certain class of triangular algebras. It was shown that, under some mild conditions, each component Gr of a nonlinear generalized Lie n-higher derivation {Gr}rN of the triangular algebra U could be expressed as the sum of an additive generalized higher derivation and a nonlinear mapping vanishing on all (n1)-th commutators on U.

    Citation: He Yuan, Qian Zhang, Zhendi Gu. Characterizations of generalized Lie n-higher derivations on certain triangular algebras[J]. AIMS Mathematics, 2024, 9(11): 29916-29941. doi: 10.3934/math.20241446

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  • The aim of this paper was to provide a characterization of nonlinear generalized Lie n-higher derivations for a certain class of triangular algebras. It was shown that, under some mild conditions, each component Gr of a nonlinear generalized Lie n-higher derivation {Gr}rN of the triangular algebra U could be expressed as the sum of an additive generalized higher derivation and a nonlinear mapping vanishing on all (n1)-th commutators on U.



    Let R be a commutative ring with unity and A be an algebra over R. Let N be the set of nonnegative integers and L={Lr}rN be a sequence of R-linear mappings Lr:AA such that L0=IdA. If Lr(xy)=i+j=rLi(x)Lj(y) holds for all x,yA, then L is said to be a higher derivation. If Lr([x,y])=i+j=r[Li(x),Lj(y)] holds for all x,yA, then L is said to be a Lie higher derivation. If Lr(Pn(x1,x2,,xn))=i1+i2++in=rPn(Li1(x1),Li2(x2),,Lin(xn)) holds for all x1,x2,,xnA, then L is said to be a Lie n-higher derivation, where pn(x1,x2,,xn) is the (n1)-th commutator.

    In 2003, Cheung [1] studied Lie derivations on triangular algebras and provided sufficient conditions under which every Lie derivation is the sum of a derivation and a linear mapping into its center. Yu and Zhang [2] extended the above result to nonlinear Lie derivations on triangular algebras in 2010. In 2012, Xiao and Wei [3] studied nonlinear Lie higher derivations L={Lr}rN on triangular algebras and obtained, under certain conditions, that each component Lr can be expressed as a sum of an additive higher derivation and a nonlinear mapping that vanishes on all commutators. Qi [4] characterized Lie higher derivations on triangular algebras by acting on zero products and acting on idempotent products. In 2023, Ashraf et al. [5] described Lie-type higher derivations of triangular rings by acting on zero products. In 2011, Benkovič [6] studied generalized Lie derivations on triangular algebras. Subsequently, Lin [7] investigated generalized Lie n-derivations on triangular algebras based on the above results. Benkovič [8] revisited the study of generalized Lie n-derivations by employing commuting and centralizing mappings on triangular algebras. Ashraf and Jabeen [9] considered the nonlinear generalized Lie triple higher derivation L={Lr}rN on triangular algebras and proved that, under certain assumptions, each component Lr can be expressed through an additive generalized higher derivation and a nonlinear mapping that annihilates on all second commutators.

    These observations motivate us to investigate the nonlinear generalized Lie n-higher derivations on triangular algebras.

    Let A and B be unital algebras, and let M be a unital (A,B)-bi-module, which is faithful as a left A-module and also as a right B-module. The set

    U=Tri(A,M,B)={(am0b)|aA,mM,bB}

    is an associative algebra under the usual matrix operations. Let 1A and 1B be identities of the algebras A and B, respectively, then I=(1A001B) is the identity of the triangular algebra U. Throughout this paper, we shall use the following notation:

    P=(1A000),Q=(0001B).

    Accordingly, U can be written as U=PUPPUQQUQ, where PUP is a sub-algebra of U isomorphic to A, QUQ is a sub-algebra of U isomorphic to B, and PUQ is a (PUP,QUQ)-bi-module isomorphic to the bi-module M. In order to simplify, we are going to employ the following symbols U11=PUP, U12=PUQ, U22=QUQ. Define two natural projections πU11:UU11 and πU22:UU22 by πU11(U11+U12+U22)=U11 and πU22(U11+U12+U22)=U22, where U11U11,U12U12,U22U22. It is easy to see that πU11(Z(U)) is a sub-algebra of Z(U11) and that πU22(Z(U)) is a sub-algebra of Z(U22). According to [10, Proposition 3], there exists a unique algebraic isomorphism τ:πU11(Z(U))πU22(Z(U)) such that U11U12=U12τ(U11) for all U11πU11(Z(U)),U12U12. Benkovič [11] proposed the following condition for studying Lie n-derivations on triangular algebras.

    Suppose that A is an associative algebra such that for each aA,

    [[a,A],A]=0[a,A]=0. (2.1)

    The above condition (2.1) is very important for studying mappings on triangular algebras, and at the same time he provided the following two remarks.

    Remark 2.1. [11, Remark 2.1] Let U be a triangular ring. For any UU and for any integer n2, we have

    pn(U,P,,P)=(1)n1PUQandpn(U,Q,,Q)=PUQ.

    Remark 2.2. [11, Remark 2.2] Let UU. If [U,PUQ]=0, then PUP+QUQZ(U).

    The concepts outlined below are necessary for this paper.

    Let A be an associative algebra and Z(A) be the center of A. A mapping F:AA is said to be additive modulo Z(A) if F(x+y)F(x)F(y)Z(A) for all x,yA. If nx=0 implies x=0, for some positive integer nN and arbitrary xA, then A is called n-torsion free.

    We denote the following sequence of polynomials:

    p1(x1)=x1,p2(x1,x2)=[p1(x1),x2]=[x1,x2],p3(x1,x2,x3)=[p2(x1,x2),x3]=[[x1,x2],x3],pn(x1,x2,,xn)=[pn1(x1,x2,,xn1),xn].

    The polynomial pn(x1,x2,,xn) is said to be an (n1)-th commutator (n2).

    Let N be the set of nonnegative integers and G={Gr}rN be a sequence of nonlinear mappings Gr:UU such that G0=IdU. Then, for all x1,x2,,xnU, G is said to be a:

    (ⅰ) Nonlinear generalized higher derivation on U if there exists a higher derivation L={Lr}rN such that L0=IdU and

    Gr(x1x2)=i+j=rGi(x1)Lj(x2).

    (ⅱ) Nonlinear generalized Lie higher derivation on U if there exists a Lie higher derivation L={Lr}rN such that L0=IdU and

    Gr([x1,x2])=i+j=r[Gi(x1),Lj(x2)].

    (ⅲ) Nonlinear generalized Lie n-higher derivation on U if there exists a Lie n-higher derivation L={Lr}rN such that L0=IdU and

    Gr(Pn(x1,x2,,xn))=i1+i2++in=rPn(Gi1(x1),Li2(x2),,Lin(xn)).

    If G is a nonlinear generalized Lie n-higher derivation, then when r=1, G1 is in fact a nonlinear generalized Lie n-derivation, i.e., there exists a Lie n-derivation L1 such that

    G1(Pn(x1,x2,,xn))=Pn(G1(x1),x2,,xn)+ni=2Pn(x1,x2,,L1(xi),,xn)

    for all x1,x2,,xnU. In this case, according to [5, Theorem 3.1], there exist a derivation d1:UU and a mapping τ1:UZ(U) satisfying τ1(Pn(x1,x2,,xn))=0 such that L1(x)=d1(x)+τ1(x) for all xU. Moreover, L1 and d1 satisfy the following properties:

    {L1(P)U12+Z(U),L1(U12)U12,L1(Q)U12+Z(U),L1(I)Z(U),L1(U11)U11+U12+Z(U),L1(U22)U22+U12+Z(U),d1(P),d1(Q)U12,d1(U12)U12,d1(U11)U11+U12,d1(U22)U22+U12.

    Although Lin also used the condition PG(x)Q=0 to describe the nonlinear generalized Lie n-derivations on triangular algebras, we will characterize the nonlinear generalized Lie n-derivations in a new way using the above result. Furthermore, this paper will use the method of induction on r to study the generalized Lie n-higher derivations, and the generalized Lie n-derivations are precisely the case of r=1.

    In Section 2, we present the preliminaries and tools that are necessary. Section 3 investigates the generalized Lie n-derivations in a new manner. Subsequently, the generalized Lie n-higher derivations were studied through induction on r in Section 4.

    This section will prove that a nonlinear generalized Lie n-derivation G1 of the triangular algebras U can be expressed as the sum of an additive generalized derivation and a nonlinear mapping vanishing on all (n1)-th commutators on U. To start, we will provide the following lemma.

    Lemma 3.1. Let U=U11+U12+U22 be a triangular algebra. Suppose that U satisfies the following conditions:

    (i) πU11(Z(U))=Z(U11) and πU22(Z(U))=Z(U22);

    (ii) U11 or U22 satisfies (2.1).

    If G1 is a nonlinear generalized Lie n-derivation on U, then G1=ω1+f1, where ω1:UU is additive modulo Z(U) and f1:UZ(U) satisfies f1(U12)={0}.

    Proof. The definition of G1 implies that

    G1(0)=G1(Pn(0,0,,0))=Pn(G1(0),0,,0)+Pn(0,L1(0),,0)++Pn(0,0,,L1(0))=0.

    Therefore,

    G1(0)=0. (3.1)

    Since Pn(U11,U22,U12,Q,,Q)=0, it follows that

    0=G1(Pn(U11,U22,U12,Q,,Q))=Pn(G1(U11),U22,U12,Q,,Q)+Pn(U11,L1(U22),U12,Q,,Q)++Pn(U11,U22,U12,Q,,L1(Q))=[[G1(U11),U22],U12]+[[U11,L1(U22)],U12].

    It follows from L1(U22)U22+U12+Z(U) that [[G1(U11),U22],U12]=0 for all U11U11,U22U22,U12U12. In view of Remark 2.2, we have QG1(U11)QZ(U22). Likewise, PG1(U22)PZ(U11) can be derived from 0=G1(Pn(U22,U11,U12,Q,,Q)) for all U22U22. In conclusion,

    G1(U11)=PG1(U11)Pτ1(QG1(U11)Q)+PG1(U11)Q+τ1(QG1(U11)Q)+QG1(U11)QU11+U12+Z(U),G1(U22)=PG1(U22)P+τ(PG1(U22)P)+PG1(U22)Q+QG1(U22)Qτ(PG1(U22)P)U22+U12+Z(U).

    It follows that

    G1(Uii)Uii+U12+Z(U),i{1,2}. (3.2)

    Set

    f1(U)=QG1(PUP)Q+τ1(QG1(PUP)Q)+PG1(QUQ)P+τ(PG1(QUQ)P)

    for all UU. Obviously, f1(U)Z(U), f1(U12)={0} and f1(Pn(U1,U2,,Un))=0 for all U1,U2,,UnU. Define a mapping ω1(U)=G1(U)f1(U) for all UU. Next, we will prove that ω1:UU is additive modulo Z(U).

    By U12=Pn(U12,Q,,Q), we obtain

    G1(U12)=G1(Pn(U12,Q,,Q))=Pn(G1(U12),Q,,Q)+Pn(U12,L1(Q),,Q)++Pn(U12,Q,,L1(Q))=PG1(U12)Q

    for all U12U12. Therefore,

    G1(U12)U12. (3.3)

    According to (3.1)–(3.3), we arrive at

    ω1(0)=0,ω1(U12)U12,ω1(Uii)Uii+U12. (3.4)

    For any U11U11, U12U12, on the one hand, it follows from d1(Q)U12 that

    G1(Pn(U11+U12,Q,,Q))=Pn(G1(U11+U12),Q,,Q)+Pn(U11+U12,L1(Q),,Q)++Pn(U11+U12,Q,,L1(Q))=Pn(ω1(U11+U12),Q,,Q)+Pn(U11+U12,d1(Q),,Q)++Pn(U11+U12,Q,,d1(Q))=Pn(ω1(U11+U12),Q,,Q)+Pn(U11+U12,d1(Q),,Q),

    and on the other hand, by (3.1), we get

    G1(Pn(U11+U12,Q,,Q))=G1(Pn(U11,Q,,Q))+G1(Pn(U12,Q,,Q))=Pn(G1(U11),Q,,Q)+Pn(U11,L1(Q),,Q)++Pn(U11,Q,,L1(Q))+Pn(G1(U12),Q,,Q)+Pn(U12,L1(Q),,Q)++Pn(U12,Q,,L1(Q))=Pn(ω1(U11),Q,,Q)+Pn(U11,d1(Q),,Q)++Pn(U11,Q,,d1(Q))+Pn(ω1(U12),Q,,Q)+Pn(U12,d1(Q),,Q)++Pn(U12,Q,,d1(Q))=Pn(ω1(U11),Q,,Q)+Pn(U11,d1(Q),,Q)+Pn(ω1(U12),Q,,Q)+Pn(U12,d1(Q),,Q).

    Comparing the above two relations, we obtain

    0=Pn(ω1(U11+U12)ω1(U11)ω1(U12),Q,,Q)=P(ω1(U11+U12)ω1(U11)ω1(U12))Q.

    Therefore,

    ω1(U11+U12)ω1(U11)ω1(U12)U11+U22

    for all U11U11,U12U12. Based on

    G1(Pn(U11+U12,V12,Q,,Q))=Pn(ω1(U11+U12),V12,Q,,Q)+Pn(U11+U12,d1(V12),Q,,Q)++Pn(U11+U12,V12,,d1(Q))=Pn(ω1(U11+U12),V12,Q,,Q)+Pn(U11+U12,d1(V12),Q,,Q)

    and

    G1(Pn(U11+U12,V12,Q,,Q))=G1(Pn(U11,V12,Q,,Q))+G1(Pn(U12,V12,Q,,Q))=Pn(ω1(U11),V12,Q,,Q)+Pn(U11,d1(V12),Q,,Q)++Pn(U11,V12,Q,,d1(Q))+Pn(ω1(U12),V12,Q,,Q)+Pn(U12,d1(V12),Q,,Q)++Pn(U12,V12,Q,,d1(Q))=Pn(ω1(U11),V12,Q,,Q)+Pn(U11,d1(V12),Q,,Q)+Pn(ω1(U12),V12,Q,,Q)+Pn(U12,d1(V12),Q,,Q),

    and comparing the above two equations, we arrive at

    0=Pn(ω1(U11+U12)ω1(U11)ω1(U12),V12,Q,,Q)=[ω1(U11+U12)ω1(U11)ω1(U12),V12]

    for all U11U11, U12,V12U12. It follows from Remark 2.2 that

    ω1(U11+U12)ω1(U11)ω1(U12)Z(U). (3.5)

    Similarly, we have

    ω1(U22+U12)ω1(U22)ω1(U12)Z(U)

    for all U22U22,U12U12.

    In view of (3.5) and d1(V12)U12, we obtain

    ω1(U12+V12)=G1(Pn(P+U12,Q+V12,Q,,Q))=Pn(ω1(P+U12),Q+V12,Q,,Q)+Pn(P+U12,d1(Q+V12),Q,,Q)++Pn(P+U12,Q+V12,Q,,d1(Q))=Pn(ω1(P)+ω1(U12),Q+V12,Q,,Q)+Pn(P+U12,d1(Q)+d1(V12),Q,,Q)=ω1(P)V12+ω1(P)Q+ω1(U12)+d1(Q)+d1(V12) (3.6)

    for all U12,V12U12. Since

    ω1(V12)=G1(Pn(P,V12,Q,,Q))=Pn(G1(P),V12,Q,,Q)+Pn(P,L1(V12),Q,,Q)++Pn(P,V12,Q,,L1(Q))=ω1(P)V12+d1(V12)

    and

    0=G1(Pn(P,Q,,Q))=Pn(G1(P),Q,,Q)+Pn(P,L1(Q),,Q)++Pn(P,Q,,L1(Q))=ω1(P)Q+d1(Q),

    it follows from (3.6) that

    ω1(U12+V12)=ω1(U12)+ω1(V12) (3.7)

    for all U12,V12U12. On the one hand,

    G1(Pn(U11+V11,U12,Q,,Q))=Pn(G1(U11+V11),U12,Q,,Q)+Pn(U11+V11,L1(U12),Q,,Q)++Pn(U11+V11,U12,Q,,L1(Q))=Pn(ω1(U11+V11),U12,Q,,Q)+Pn(U11+V11,d1(U12),Q,,Q),

    and on the other hand, it follows from (3.7) that

    G1(Pn(U11+V11,U12,Q,,Q))=G1(Pn(U11,U12,Q,,Q))+G1(Pn(V11,U12,Q,,Q))=Pn(G1(U11),U12,Q,,Q)+Pn(U11,L1(U12),Q,,Q)++Pn(U11,U12,Q,,L1(Q))+Pn(G1(V11),U12,Q,,Q)+Pn(V11,L1(U12),Q,,Q)++Pn(V11,U12,Q,,L1(Q))=Pn(ω1(U11),U12,Q,,Q)+Pn(U11,d1(U12),Q,,Q)+Pn(ω1(V11),U12,Q,,Q)+Pn(V11,d1(U12),Q,,Q).

    Comparing the above two relations, we have

    0=Pn(ω1(U11+V11)ω1(U11)ω1(V11),U12,Q,,Q)=(ω1(U11+V11)ω1(U11)ω1(V11))U12.

    Since U12 is faithful as a left U11-module, we get

    ω1(U11+V11)P=ω1(U11)P+ω1(V11)P (3.8)

    for all U11,V11U11. Since Pn(U11,Q,,Q)=0, we arrive at

    0=G1(Pn(U11,Q,,Q))=Pn(G1(U11),Q,,Q)+Pn(U11,L1(Q),,Q)++Pn(U11,Q,,L1(Q))=Pn(ω1(U11),Q,,Q)+Pn(U11,d1(Q),,Q)=ω1(U11)Q+U11d1(Q).

    That is,

    ω1(U11)Q+U11d1(Q)=0 (3.9)

    for all U11U11. Substituting U11+V11 for U11 in (3.9), we obtain

    ω1(U11+V11)Q+(U11+V11)d1(Q)=0. (3.10)

    Using (3.9) and (3.10), we arrive at

    ω1(U11+V11)Qω1(U11)Qω1(V11)Q=0. (3.11)

    According to (3.8) and (3.11), we find

    ω1(U11+V11)=ω1(U11)+ω1(V11) (3.12)

    for all U11,V11U11. Using an analogous manner, we show

    ω1(U22+V22)=ω1(U22)+ω1(V22) (3.13)

    for all U22,V22U22.

    For any U11U11, U12U12, U22U22, on the one hand, we have

    G1(Pn(U11+U12+U22,Q,,Q))=Pn(ω1(U11+U12+U22),Q,,Q)+Pn(U11+U12+U22,d1(Q),,Q)++Pn(U11+U12+U22,Q,,d1(Q))=Pn(ω1(U11+U12+U22),Q,,Q)+Pn(U11+U12+U22,d1(Q),,Q).

    On the other hand, we get

    G1(Pn(U11+U12+U22,Q,,Q))=G1(Pn(U11,Q,,Q))+G1(Pn(U12,Q,,Q))+G1(Pn(U22,Q,,Q))=Pn(ω1(U11),Q,,Q)+Pn(U11,d1(Q),,Q)++Pn(U11,Q,,d1(Q))+Pn(ω1(U12),Q,,Q)+Pn(U12,d1(Q),,Q)++Pn(U12,Q,,d1(Q))+Pn(ω1(U22),Q,,Q)+Pn(U22,d1(Q),,Q)++Pn(U22,Q,,d1(Q))=Pn(ω1(U11),Q,,Q)+Pn(U11,d1(Q),,Q)+Pn(ω1(U12),Q,,Q)+Pn(U12,d1(Q),,Q)+Pn(ω1(U22),Q,,Q)+Pn(U22,d1(Q),,Q).

    Comparing the above two relations, we obtain

    0=Pn(ω1(U11+U12+U22)ω1(U11)ω1(U12)ω1(U22),Q,,Q)=P(ω1(U11+U12+U22)ω1(U11)ω1(U12)ω1(U22))Q.

    Therefore,

    ω1(U11+U12+U22)ω1(U11)ω1(U12)ω1(U22)U11+U22.

    We have

    G1(Pn(U11+U12+U22,V12,,Q))=Pn(ω1(U11+U12+U22),V12,,Q)+Pn(U11+U12+U22,d1(V12),,Q)++Pn(U11+U12+U22,V12,,d1(Q))=Pn(ω1(U11+U12+U22),V12,,Q)+Pn(U11+U12+U22,d1(V12),,Q)

    and

    G1(Pn(U11+U12+U22,V12,,Q))=G1(Pn(U11,V12,,Q))+G1(Pn(U12,V12,,Q))+G1(Pn(U22,V12,,Q))=Pn(ω1(U11),V12,,Q)+Pn(U11,d1(V12),,Q)++Pn(U11,V12,,d1(Q))+Pn(ω1(U12),V12,,Q)+Pn(U12,d1(V12),,Q)++Pn(U12,V12,,d1(Q))+Pn(ω1(U22),V12,,Q)+Pn(U22,d1(V12),,Q)++Pn(U22,V12,,d1(Q))=Pn(ω1(U11),V12,,Q)+Pn(U11,d1(V12),,Q)+Pn(ω1(U12),V12,,Q)+Pn(U12,d1(V12),,Q)+Pn(ω1(U22),V12,,Q)+Pn(U22,d1(V12),,Q).

    Comparing the above two equations, we arrive at

    0=Pn(ω1(U11+U12+U22)ω1(U11)ω1(U12)ω1(U22),V12,,Q)=[ω1(U11+U12+U22)ω1(U11)ω1(U12)ω1(U22),V12]

    for all U11U11, U12,V12U12, U22U22. It follows that

    ω1(U11+U12+U22)ω1(U11)ω1(U12)ω1(U22)Z(U) (3.14)

    for all U11U11, U12U12, U22U22.

    Let U=U11+U12+U22,V=V11+V12+V22 be two elements in U, where Uij,VijUij, 1ij2. Then, it follows from (3.7) and (3.12)–(3.14) that

    ω1(U+V)=ω1((U11+U12+U22)+(V11+V12+V22))=ω1((U11+V11)+(U12+V12)+(U22+V22))=ω1(U11+V11)+ω1(U12+V12)+ω1(U22+V22)+α=ω1(U11)+ω1(V11)+ω1(U12)+ω1(V12)+ω1(U22)+ω1(V22)+α=ω1(U11+U12+U22)+ω1(V11+V12+V22)+β=ω1(U)+ω1(V)+β, (3.15)

    where α,βZ(U), that is, ω1 is additive module Z(U).

    Using Lemma 3.1, we present the main theorem of this section.

    Theorem 3.2. Let U=U11+U12+U22 be a triangular algebra. Suppose that U satisfies the following conditions:

    (i) πU11(Z(U))=Z(U11) and πU22(Z(U))=Z(U22);

    (ii) U11 or U22 satisfies (2.1).

    If G1 is a nonlinear generalized Lie n-derivation on U, then there exists a generalized derivation χ1 and a nonlinear mapping h1 satisfying h1(Pn(U1,U2,,Un))=0 such that G1(U)=χ1(U)+h1(U) for all U,U1,U2,,UnU. Additionally, the mappings G1 and χ1 fulfill the following properties:

    {G1(0)=0,G1(U12)U12,G1(Uii)Uii+U12+Z(U),χ1(U12)U12,χ1(Uii)Uii+U12.

    Proof. According to Lemma 3.1, we define two mappings g1:UZ(U) and χ1:UU by

    g1(U)=ω1(U)ω1(U11)ω1(U12)ω(U22)

    and

    χ1(U)=ω1(U)g1(U)

    for all U=U11+U12+U22U. It follows that

    g1(Pn(U1,U2,,Un))=0,χ1(U11+U12+U22)=ω1(U11)+ω1(U12)+ω1(U22),χ1(U11)=ω1(U11)g1(U11)=ω1(U11)ω1(U11)+ω1(U11)=ω1(U11)

    for all U1,U2,,UnU,U11U11,U12U12,U22U22. Similarly, we have χ1(U12)=ω1(U12) and χ1(U22)=ω1(U22). Therefore,

    χ1(U11+U12+U22)=χ1(U11)+χ1(U12)+χ1(U22) (3.16)

    for all U11U11,U12U12,U22U22. In addition, the following properties can be obtained from (3.1)–(3.4):

    {G1(0)=0,G1(U12)U12,G1(Uii)Uii+U12+Z(U),χ1(U12)U12,χ1(Uii)Uii+U12.

    Let h1(U)=g1(U)+f1(U), then G1(U)=ω1(U)+f1(U)=χ1(U)+h1(U) for all UU. Since g1(Pn(U1,U2,,Un))=f1(Pn(U1,U2,,Un))=0, we have h1(Pn(U1,U2,,Un))=0 for all U1,U2,,UnU. Let U=U11+U12+U22,V=V11+V12+V22 be two elements in U, where Uij,VijUij, 1ij2. Then, it can be inferred from (3.15) and (3.16) that χ1 is additive. Next, we will prove that χ1 is a generalized derivation.

    According to U11U12=Pn(U11,U12,Q,,Q), we obtain

    χ1(U11U12)=G1(Pn(U11,U12,Q,,Q))=Pn(G1(U11),U12,Q,,Q)+Pn(U11,L1(U12),Q,,Q)++Pn(U11,U12,Q,,L1(Q))=Pn(χ1(U11),U12,Q,,Q)+Pn(U11,d1(U12),Q,,Q)++Pn(U11,U12,Q,,d1(Q))=χ1(U11)U12+U11d1(U12) (3.17)

    for all U11U11,U12U12. Similarly, it follows from U12U22=Pn(U12,U22,Q,,Q) that

    χ1(U12U22)=χ1(U12)U22+U12d1(U22) (3.18)

    for all U12U12,U22U22.

    In view of (3.17), we get

    χ1(U11V11U12)=χ1((U11V11)U12)=χ1(U11V11)U12+U11V11d1(U12)

    and

    χ1(U11V11U12)=χ1(U11(V11U12))=χ1(U11)V11U12+U11d1(V11U12)=χ1(U11)V11U12+U11d1(V11)U12+U11V11d1(U12)

    for all U11,V11U11,U12U12. Comparing the above equations, we obtain

    (χ1(U11V11)χ1(U11)V11U11d1(V11))U12=0

    for all U11,V11U11,U12U12. Since U12 is faithful as a left U11-module, we conclude

    χ1(U11V11)P=χ1(U11)V11P+U11d1(V11)P (3.19)

    for all U11,V11U11. Replacing U11 by U11V11 in (3.9), we get

    χ1(U11V11)Q+U11V11d1(Q)=0. (3.20)

    Since L1 is a Lie n-derivation, it follows that

    0=L1(Pn(V11,Q,,Q))=Pn(L1(V11),Q,,Q)+Pn(V11,L1(Q),,Q)++Pn(V11,Q,,L1(Q))=Pn(d1(V11),Q,,Q)+Pn(V11,d1(Q),,Q)++Pn(V11,Q,,d1(Q))=d1(V11)Q+V11d1(Q).

    This implies that

    d1(V11)Q+V11d1(Q)=0 (3.21)

    for all V11U11. Multiplying the left side of (3.21) by U11 and then combining this with (3.20) yields

    χ1(U11V11)Q=χ1(U11)V11Q+U11d1(V11)Q (3.22)

    for all U11,V11U11. Adding (3.19) to (3.22) gives

    χ1(U11V11)=χ1(U11)V11+U11d1(V11) (3.23)

    for all U11,V11U11. In a similar manner, we get

    χ1(U22V22)=χ1(U22)V22+U22d1(V22) (3.24)

    for all U22,V22U22.

    Considering χ1(U11)V22+U11d1(V22)=G1(Pn(U11,V22,Q,,Q))=0 and (3.17), (3.18), (3.23), (3.24), we can conclude that

    χ1(UV)=χ1((U11+U12+U22)(V11+V12+V22))=χ1(U11V11+U11V12+U12V22+U22V22)=χ1(U11V11)+χ1(U11V12)+χ1(U12V22)+χ1(U22V22)=χ1(U11)V11+U11d1(V11)+χ1(U11)V12+U11d1(V12)+χ1(U12)V22+U12d1(V22)+χ1(U22)V22+U22d1(V22)=(χ1(U11)+χ1(U12)+χ1(U22))(V11+V12+V22)+(U11+U12+U22)(d1(V11)+d1(V12)+d1(V22))=χ1(U11+U12+U22)V+Ud1(V11+V12+V22)=χ1(U)V+Ud1(V)

    for all U=U11+U12+U22,V=V11+V12+V22U.

    This section will study the nonlinear generalized Lie n-higher derivations on triangular algebras. First, we present the following result.

    Lemma 4.1. [5, Theorem 4.1] Let U=U11+U12+U22 be an (n1)-torsion free triangular ring such that

    (i) πU11(Z(U))=Z(U11) and πU22(Z(U))=Z(U22);

    (ii) U11 or U22 satisfies (2.1).

    Let {ξr}rN be a family of additive mappings ξr:UU such that for each rN,

    ξr(Pn(U1,U2,,Un))=i1+i2++in=rPn(ξi1(U1),ξi2(U2),,ξin(Un))

    for all U1,U2,,UnU with U1U2Un=0. Then, for each rN,ξr=dr+hr, where {dr}rN is an additive higher derivation on U and {hr}rN is a family of additive mappings hr:UZ(U) vanishing at every (n1)-th commutator Pn(U1,U2,,Un) with U1U2Un=0. Moreover, ξr and dr satisfy the following properties:

    {ξr(P)U12+Z(U),ξr(U12)U12,ξr(Q)U12+Z(U),ξr(I)Z(U),ξr(U11)U11+U12+Z(U),ξr(U22)U22+U12+Z(U),dr(P),dr(Q)U12,dr(U12)U12,dr(U11)U11+U12,dr(U22)U22+U12.

    The main theorem of this section is presented below.

    Theorem 4.2. Let U=U11+U12+U22 be an (n1)-torsion free triangular algebra satisfying

    (i) πU11(Z(U))=Z(U11) and πU22(Z(U))=Z(U22);

    (ii) U11 or U22 satisfies (2.1).

    Let G={Gr}rN be a nonlinear generalized Lie n-higher derivation on U, then Gr=χr+hr, rN, where {χr}rN is an additive generalized higher derivation on U and {hr}rN is a family of nonlinear mappings hr:UZ(U) such that hr(Pn(U1,U2,,Un))=0 for all U1,U2,,UnU.

    To prove the theorem, we will employ an inductive approach with respect to the component index r. If r=1, then the mapping Gr:UU is a nonlinear generalized Lie n-derivation. According to Theorem 3.2, this implies the existence of a generalized derivation χ1 and a nonlinear mapping h1 such that h1(Pn(U1,U2,,Un))=0 and G1(U)=χ1(U)+h1(U) for all U,U1,U2,,UnU. Furthermore, G1 and χ1 satisfy the following properties:

    C1:{G1(0)=0,G1(U12)U12,G1(Uii)Uii+U12+Z(U),χ1(U12)U12,χ1(Uii)Uii+U12.

    We now assume that Theorem 4.2 holds for all 1<t<r, rN. That is, Gt(U)=χt(U)+ht(U) for all UU and 1<t<r, where χt:UU is an additive mapping such that χt(VW)=i+j=tχi(V)dj(W), and ht:UZ(U) is a nonlinear mapping such that ht(Pn(U1,U2,,Un))=0 for all V,W,U1,U2,,UnU, where {dj}jN is an additive higher derivation on U. In addition, Gt and χt satisfy the following properties:

    Ct:{Gt(0)=0,Gt(U12)U12,Gt(Uii)Uii+U12+Z(U),χt(U12)U12,χt(Uii)Uii+U12.

    We will prove that the above properties also hold for r. That is, Gr(U)=χr(U)+hr(U), where χr is an additive mapping satisfying χr(VW)=i+j=rχi(V)dj(W), and hr:UZ(U) is a nonlinear mapping satisfying hr(Pn(U1,U2,,Un))=0 for all U,V,W,U1,U2,,UnU. To begin, we will introduce the following lemma.

    Lemma 4.3. Let U=U11+U12+U22 be a triangular algebra. Suppose that U satisfies the following conditions:

    (i) πU11(Z(U))=Z(U11) and πU22(Z(U))=Z(U22);

    (ii) U11 or U22 satisfies (2.1),

    then the nonlinear mapping Gr mentioned above satisfies Gr=ωr+f1, where ωr:UU is additive modulo Z(U) and f1:UZ(U) satisfies f1(U12)={0}.

    Proof. Applying the property Ct, we get

    Gr(0)=Gr(Pn(0,0,,0))=Pn(Gr(0),0,,0)+Pn(0,Lr(0),,0)++Pn(0,0,,Lr(0))+i1+i2++in=ri1,i2,,in<rPn(Gi1(0),Li2(0),,Lin(0))=0.

    Again, according to the property Ct and Lemma 4.1, we have

    Gr(U12)=Gr(Pn(U12,Q,,Q))=Pn(Gr(U12),Q,,Q)+Pn(U12,Lr(Q),,Q)++Pn(U12,Q,,Lr(Q))+i1+i2++in=ri1,i2,,in<rPn(Gi1(U12),Li2(Q),,Lin(Q))=PGr(U12)Q

    for all U12U12. Therefore, Gr(U12)U12.

    Using the property Ct and Lemma 4.1, we arrive at

    0=Gr(Pn(U11,U22,U12,Q,,Q))=Pn(Gr(U11),U22,U12,Q,,Q)+Pn(U11,Lr(U22),U12,Q,,Q)++Pn(U11,U22,U12,Q,,Lr(Q))+i1+i2++in=ri1,i2,,in<rPn(Gi1(U11),Li2(U22),Li3(U12),,Lin(Q))=[[Gr(U11),U22],U12].

    It follows from Remark 2.2 that QGr(U11)QZ(U22). Similarly, we can obtain PGr(U22)PZ(U11) from 0=Gr(Pn(U22,U11,U12,Q,,Q)). Hence,

    Gr(U11)=PGr(U11)Pτ1(QGr(U11)Q)+PGr(U11)Q+(τ1(QGr(U11)Q)+QGr(U11)Q)U11+U12+Z(U),Gr(U22)=(PGr(U22)P+τ(PGr(U22)P))+PGr(U22)Q+QGr(U22)Qτ(PGr(U22)P)U22+U12+Z(U)

    for all U11U11,U22U22, that is, Gr(Uii)Uii+U12+Z(U) with i{1,2}. Set

    f1(U)=QGr(PUP)Q+τ1(QGr(PUP)Q)+PGr(QUQ)P+τ(PGr(QUQ)P)

    for all UU. Clearly, f1(U12)={0}, f1(U)Z(U), and f1(Pn(U1,U2,,Un))=0 for all U,U1,U2,,UnU. Define a mapping ωr(U)=Gr(U)f1(U) for all UU. It is easy to obtain the following relations:

    ωr(0)=0,ωr(U12)U12,ωr(Uii)Uii+U12.

    Next, we will prove that ωr:UU is additive modulo Z(U).

    On the one hand, since di2(Q),,din(Q))U12, we get

    Gr(Pn(U11+U12,Q,,Q))=Pn(Gr(U11+U12),Q,,Q)+Pn(U11+U12,Lr(Q),,Q)++Pn(U11+U12,Q,,Lr(Q))+i1+i2++in=ri1,i2,,in<rPn(Gi1(U11+U12),Li2(Q),,Lin(Q))=Pn(ωr(U11+U12),Q,,Q)+Pn(U11+U12,dr(Q),,Q)++Pn(U11+U12,Q,,dr(Q))+i1+i2++in=ri1,i2,,in<rPn(χi1(U11+U12),di2(Q),,din(Q))=Pn(ωr(U11+U12),Q,,Q)+Pn(U11+U12,dr(Q),,Q)+i1+i2=r0<i1,i2<r[χi1(U11+U12),di2(Q)]

    for any U11U11, U12U12. On the other hand, according to Gr(0)=0, we arrive at

    Gr(Pn(U11+U12,Q,,Q))=Gr(Pn(U11,Q,,Q))+Gr(Pn(U12,Q,,Q))=Pn(Gr(U11),Q,,Q)+Pn(U11,Lr(Q),,Q)++Pn(U11,Q,,Lr(Q))+Pn(Gr(U12),Q,,Q)+Pn(U12,Lr(Q),,Q)++Pn(U12,Q,,Lr(Q))+i1+i2++in=ri1,i2,,in<rPn(Gi1(U11),Li2(Q),,Lin(Q))+i1+i2++in=ri1,i2,,in<rPn(Gi1(U12),Li2(Q),,Lin(Q))=Pn(ωr(U11),Q,,Q)+Pn(U11,dr(Q),,Q)++Pn(U11,Q,,dr(Q))+Pn(ωr(U12),Q,,Q)+Pn(U12,dr(Q),,Q)++Pn(U12,Q,,dr(Q))+i1+i2++in=ri1,i2,,in<rPn(χi1(U11),di2(Q),,din(Q))+i1+i2++in=ri1,i2,,in<rPn(χi1(U12),di2(Q),,din(Q))=Pn(ωr(U11),Q,,Q)+Pn(U11,dr(Q),,Q)+Pn(ωr(U12),Q,,Q)+Pn(U12,dr(Q),,Q)+i1+i2=r0<i1,i2<r[χi1(U11),di2(Q)]+i1+i2=r0<i1,i2<r[χi1(U12),di2(Q)].

    Comparing the above two relations, we obtain

    0=Pn(ωr(U11+U12)ωr(U11)ωr(U12),Q,,Q)+i1+i2=r0<i1,i2<r[χi1(U11+U12)χi1(U11)χi1(U12),di2(Q)].

    Based on the fact that χi1 is additive, we get

    P(ωr(U11+U12)ωr(U11)ωr(U12))Q=0.

    Therefore,

    ωr(U11+U12)ωr(U11)ωr(U12)U11+U22.

    In the same way, we can obtain

    Gr(Pn(U11+U12,V12,Q,,Q))=Pn(ωr(U11+U12),V12,Q,,Q)+Pn(U11+U12,dr(V12),Q,,Q)++Pn(U11+U12,V12,Q,,dr(Q))+i1+i2++in=ri1,i2,,in<rPn(χi1(U11+U12),di2(V12),,din(Q))=Pn(ωr(U11+U12),V12,Q,,Q)+Pn(U11+U12,dr(V12),Q,,Q)+i1+i2=r0<i1,i2<r[χi1(U11+U12),di2(V12)]

    and

    Gr(Pn(U11+U12,V12,Q,,Q))=Gr(Pn(U11,V12,Q,,Q))+Gr(Pn(U12,V12,Q,,Q))=Pn(ωr(U11),V12,Q,,Q)+Pn(U11,dr(V12),Q,,Q)++Pn(U11,V12,Q,,dr(Q))+Pn(ωr(U12),V12,Q,,Q)+Pn(U12,dr(V12),Q,,Q)++Pn(U12,V12,Q,,dr(Q))+i1+i2++in=ri1,i2,,in<rPn(χi1(U11),di2(V12),,din(Q))+i1+i2++in=ri1,i2,,in<rPn(χi1(U12),di2(V12),,din(Q))=Pn(ωr(U11),V12,Q,,Q)+Pn(U11,dr(V12),Q,,Q)+Pn(ωr(U12),V12,Q,,Q)+Pn(U12,dr(V12),Q,,Q)+i1+i2=r0<i1,i2<r[χi1(U11),di2(V12)]+i1+i2=r0<i1,i2<r[χi1(U12),di2(V12)].

    Comparing the above two equations, we have

    [ωr(U11+U12)ωr(U11)ωr(U12),V12]=0

    for all U11U11, U12,V12U12. Consequently,

    ωr(U11+U12)ωr(U11)ωr(U12)Z(U).

    Analogously, we have

    ωr(U22+U12)ωr(U22)ωr(U12)Z(U).

    Considering

    U12+V12=Pn(P+U12,Q+V12,Q,,Q)

    and

    di2(V12)U12,

    we have

    ωr(U12+V12)=Gr(Pn(P+U12,Q+V12,Q,,Q))=Pn(ωr(P+U12),Q+V12,Q,,Q)+Pn(P+U12,dr(Q+V12),Q,,Q)++Pn(P+U12,Q+V12,Q,,dr(Q))+i1+i2++in=ri1,i2,,in<rPn(χi1(P+U12),di2(Q+V12),,din(Q))=ωr(P)Q+ωr(P)V12+ωr(U12)+dr(Q)+dr(V12)+i1+i2=r0<i1,i2<rχi1(P)di2(Q)+i1+i2=r0<i1,i2<rχi1(P)di2(V12). (4.1)

    Since

    ωr(V12)=Gr(Pn(P,V12,Q,,Q))=Pn(Gr(P),V12,Q,,Q)+Pn(P,Lr(V12),Q,,Q)++Pn(P,V12,Q,,Lr(Q))+i1+i2++in=ri1,i2,,in<rPn(Gi1(P),Li2(V12),,Lin(Q))=ωr(P)V12+dr(V12)+i1+i2=r0<i1,i2<rχi1(P)di2(V12)

    and

    0=Gr(Pn(P,Q,,Q))=Pn(Gr(P),Q,,Q)+Pn(P,Lr(Q),,Q)++Pn(P,Q,,Lr(Q))+i1+i2++in=ri1,i2,,in<rPn(Gi1(P),Li2(Q),,Lin(Q))=ωr(P)Q+dr(Q)+i1+i2=r0<i1,i2<rχi1(P)di2(Q),

    we can derive from (4.1) that

    ωr(U12+V12)=ωr(U12)+ωr(V12) (4.2)

    for all U12,V12U12. On the one hand,

    Gr(Pn(U11+V11,U12,Q,,Q))=Pn(Gr(U11+V11),U12,Q,,Q)+Pn(U11+V11,Lr(U12),Q,,Q)++Pn(U11+V11,U12,Q,,Lr(Q))+i1+i2++in=ri1,i2,,in<rPn(Gi1(U11+V11),Li2(U12),,Lin(Q))=Pn(ωr(U11+V11),U12,Q,,Q)+Pn(U11+V11,dr(U12),Q,,Q)+i1+i2=r0<i1,i2<r[χi1(U11+V11),di2(U12)],

    and on the other hand, in view of (4.2), we have

    Gr(Pn(U11+V11,U12,Q,,Q))=Gr(Pn(U11,U12,Q,,Q))+Gr(Pn(V11,U12,Q,,Q))=Pn(Gr(U11),U12,Q,,Q)+Pn(U11,Lr(U12),Q,,Q)++Pn(U11,U12,Q,,Lr(Q))+Pn(Gr(V11),U12,Q,,Q)+Pn(V11,Lr(U12),Q,,Q)++Pn(V11,U12,Q,,Lr(Q))+i1+i2++in=ri1,i2,,in<rPn(Gi1(U11),Li2(U12),,Lin(Q))+i1+i2++in=ri1,i2,,in<rPn(Gi1(V11),Li2(U12),,Lin(Q))=Pn(ωr(U11),U12,Q,,Q)+Pn(U11,dr(U12),Q,,Q)+Pn(ωr(V11),U12,Q,,Q)+Pn(V11,dr(U12),Q,,Q)+i1+i2=r0<i1,i2<r[χi1(U11),di2(U12)]+i1+i2=r0<i1,i2<r[χi1(V11),di2(U12)].

    As χi1 is additive, comparing the two equations above leads us to obtain

    (ωr(U11+V11)ωr(U11)ωr(V11))U12=0.

    Since U12 is faithful as a left U11-module, we can derive from the above expression that

    ωr(U11+V11)P=ωr(U11)P+ωr(V11)P (4.3)

    for all U11,V11U11. Using Pn(U11,Q,,Q)=0, we deduce

    0=Gr(Pn(U11,Q,,Q))=Pn(Gr(U11),Q,,Q)+Pn(U11,Lr(Q),,Q)++Pn(U11,Q,,Lr(Q))+i1+i2++in=ri1,i2,,in<rPn(Gi1(U11),Li2(Q),,Lin(Q))=Pn(ωr(U11),Q,,Q)+Pn(U11,dr(Q),,Q)++Pn(U11,Q,,dr(Q))+i1+i2++in=ri1,i2,,in<rPn(χi1(U11),di2(Q),,din(Q))=ωr(U11)Q+U11dr(Q)+i1+i2=r0<i1,i2<rχi1(U11)di2(Q). (4.4)

    Replacing U11 with U11+V11 in (4.4), we arrive at

    ωr(U11+V11)Q+(U11+V11)dr(Q)+i1+i2=r0<i1,i2<rχi1(U11+V11)di2(Q)=0. (4.5)

    Again, using (4.4) and (4.5), we obtain

    ωr(U11+V11)Q=ωr(U11)Q+ωr(V11)Q. (4.6)

    Combining (4.3) and (4.6), we find

    ωr(U11+V11)=ωr(U11)+ωr(V11) (4.7)

    for all U11,V11U11. In a similar way, we get

    ωr(U22+V22)=ωr(U22)+ωr(V22) (4.8)

    for all U22,V22U22.

    On the one hand, we get

    Gr(Pn(U11+U12+U22,Q,,Q))=Pn(ωr(U11+U12+U22),Q,,Q)+Pn(U11+U12+U22,dr(Q),,Q)++Pn(U11+U12+U22,Q,,dr(Q))+i1+i2++in=ri1,i2,,in<rPn(χi1(U11+U12+U22),di2(Q),,din(Q))=Pn(ωr(U11+U12+U22),Q,,Q)+Pn(U11+U12+U22,dr(Q),,Q)+i1+i2=r0<i1,i2<r[χi1(U11+U12+U22),di2(Q)],

    and on the other hand, we have

    Gr(Pn(U11+U12+U22,Q,,Q))=Gr(Pn(U11,Q,,Q))+Gr(Pn(U12,Q,,Q))+Gr(Pn(U22,Q,,Q))=Pn(ωr(U11),Q,,Q)+Pn(U11,dr(Q),,Q)++Pn(U11,Q,,dr(Q))+Pn(ωr(U12),Q,,Q)+Pn(U12,dr(Q),,Q)++Pn(U12,Q,,dr(Q))+Pn(ωr(U22),Q,,Q)+Pn(U22,dr(Q),,Q)++Pn(U22,Q,,dr(Q))+i1+i2++in=ri1,i2,,in<rPn(χi1(U11),di2(Q),,din(Q))+i1+i2++in=ri1,i2,,in<rPn(χi1(U12),di2(Q),,din(Q))+i1+i2++in=ri1,i2,,in<rPn(χi1(U22),di2(Q),,din(Q))=Pn(ωr(U11),Q,,Q)+Pn(U11,dr(Q),,Q)+Pn(ωr(U12),Q,,Q)+Pn(U12,dr(Q),,Q)+Pn(ωr(U22),Q,,Q)+Pn(U22,dr(Q),,Q)+i1+i2=r0<i1,i2<r[χi1(U11),di2(Q)]+i1+i2=r0<i1,i2<r[χi1(U12),di2(Q)]+i1+i2=r0<i1,i2<r[χi1(U22),di2(Q)]

    for all U11U11, U12U12, U22U22. Comparing the above two relations, we arrive at

    P(ωr(U11+U12+U22)ωr(U11)ωr(U12)ωr(U22))Q=0.

    Hence,

    ωr(U11+U12+U22)ωr(U11)ωr(U12)ωr(U22)U11+U22.

    We have

    Gr(Pn(U11+U12+U22,V12,,Q))=Pn(ωr(U11+U12+U22),V12,,Q)+Pn(U11+U12+U22,dr(V12),,Q)++Pn(U11+U12+U22,V12,,dr(Q))+i1+i2++in=ri1,i2,,in<rPn(χi1(U11+U12+U22),di2(V12),,din(Q))=Pn(ωr(U11+U12+U22),V12,,Q)+Pn(U11+U12+U22,dr(V12),,Q)+i1+i2=r0<i1,i2<r[χi1(U11+U12+U22),di2(V12)]

    and

    Gr(Pn(U11+U12+U22,V12,,Q))=Gr(Pn(U11,V12,,Q))+Gr(Pn(U12,V12,,Q))+Gr(Pn(U22,V12,,Q))=Pn(ωr(U11),V12,,Q)+Pn(U11,dr(V12),,Q)++Pn(U11,V12,,dr(Q))+Pn(ωr(U12),V12,,Q)+Pn(U12,dr(V12),,Q)++Pn(U12,V12,,dr(Q))+Pn(ωr(U22),V12,,Q)+Pn(U22,dr(V12),,Q)++Pn(U22,V12,,dr(Q))+i1+i2++in=ri1,i2,,in<rPn(χi1(U11),di2(V12),,din(Q))+i1+i2++in=ri1,i2,,in<rPn(χi1(U12),di2(V12),,din(Q))+i1+i2++in=ri1,i2,,in<rPn(χi1(U22),di2(V12),,din(Q))=Pn(ωr(U11),V12,,Q)+Pn(U11,dr(V12),,Q)+Pn(ωr(U12),V12,,Q)+Pn(U12,dr(V12),,Q)+Pn(ωr(U22),V12,,Q)+Pn(U22,dr(V12),,Q)+i1+i2=r0<i1,i2<r[χi1(U11),di2(V12)]+i1+i2=r0<i1,i2<r[χi1(U12),di2(V12)]+i1+i2=r0<i1,i2<r[χi1(U22),di2(V12)].

    Comparing the above two equations, we arrive at

    [ωr(U11+U12+U22)ωr(U11)ωr(U12)ωr(U22),V12]=0

    for all U11U11, U12,V12U12, U22U22. It follows that

    ωr(U11+U12+U22)ωr(U11)ωr(U12)ωr(U22)Z(U)

    for all U11U11, U12U12, U22U22. Similar to (3.15), it can be concluded that ωr is additive modulo Z(U).

    Proof of Theorem 4.2. Similar to the definitions of g1 and χ1, the mappings gr and χr can be defined as

    gr(U)=ωr(U)ωr(U11)ωr(U12)ωr(U22)

    and

    χr(U)=ωr(U)gr(U),

    and it can be obtained that

    χr(U11)=ωr(U11),χr(U12)=ωr(U12),χr(U22)=ωr(U22),χr(U)=χr(U11+U12+U22)=χr(U11)+χr(U12)+χr(U22),

    for all U=U11+U12+U22U. Let hr(U)=gr(U)+f1(U), then Gr(U)=χr(U)+hr(U) and hr(Pn(U1,U2,,Un))=0 for all U,U1,U2,,UnU. Additionally, we can conclude from Lemma 4.3 that χr is additive. Next, we will prove that χr(UV)=p+q=rχp(U)dq(V) for all U,VU, where dq is mentioned in Lemma 4.1.

    Given U11U12=Pn(U11,U12,Q,,Q) and considering the property Ct along with Lemma 4.1, we deduce

    χr(U11U12)=Gr(Pn(U11,U12,Q,,Q))=Pn(Gr(U11),U12,Q,,Q)+Pn(U11,Lr(U12),Q,,Q)++Pn(U11,U12,Q,,Lr(Q))+i1+i2++in=ri1,i2,,in<rPn(Gi1(U11),Li2(U12),Li3(Q),,Lin(Q))=Pn(χr(U11),U12,Q,,Q)+Pn(U11,dr(U12),Q,,Q)++Pn(U11,U12,Q,,dr(Q))+i1+i2++in=ri1,i2,,in<rPn(χi1(U11),di2(U12),di3(Q),,din(Q))=χr(U11)U12+U11dr(U12)+i1+i2=r0<i1,i2<rχi1(U11)di2(U12) (4.9)

    for all U11U11,U12U12. In a similar manner, we get

    χr(U12U22)=χr(U12)U22+U12dr(U22)+i1+i2=r0<i1,i2<rχi1(U12)di2(U22) (4.10)

    for all U12U12,U22U22.

    According to (4.9) and the property Ct, we get

    χr(U11V11U12)=χr((U11V11)U12)=χr(U11V11)U12+U11V11dr(U12)+i1+i2=r0<i1,i2<rχi1(U11V11)di2(U12)=χr(U11V11)U12+U11V11dr(U12)+p+q+s=r0<s<rχp(U11)dq(V11)ds(U12)

    and

    χr(U11V11U12)=χr(U11(V11U12))=χr(U11)V11U12+U11dr(V11U12)+i1+i2=r0<i1,i2<rχi1(U11)di2(V11U12)=χr(U11)V11U12+U11dr(V11)U12+U11V11dr(U12)+q+s=r0<q,s<rU11dq(V11)ds(U12)+p+q+s=r0<p<rχp(U11)dq(V11)ds(U12)=χr(U11)V11U12+U11dr(V11)U12+U11V11dr(U12)+p+q+s=r0<s<rχp(U11)dq(V11)ds(U12)+p+q=r0<p,q<rχp(U11)dq(V11)U12.

    Combining the above two relations, we obtain

    χr(U11V11)U12=χr(U11)V11U12+U11dr(V11)U12+p+q=r0<p,q<rχp(U11)dq(V11)U12.

    Since U12 is faithful as a left U11-module, it follows that

    χr(U11V11)P=χr(U11)V11P+U11dr(V11)P+p+q=r0<p,q<rχp(U11)dq(V11)P (4.11)

    for all U11,V11U11. Using (4.4), we obtain

    χr(U11V11)Q+U11V11dr(Q)+i1+i2=r0<i1,i2<rχi1(U11V11)di2(Q)=0. (4.12)

    Taking into account Lemma 4.1, we conclude

    0=Lr(Pn(V11,Q,,Q))=Pn(Lr(V11),Q,,Q)+Pn(V11,Lr(Q),,Q)++Pn(V11,Q,,Lr(Q))+i1+i2++in=ri1,i2,,in<rPn(Li1(V11),Li2(Q),,Lin(Q))=Pn(dr(V11),Q,,Q)+Pn(V11,dr(Q),,Q)++Pn(V11,Q,,dr(Q))+i1+i2++in=ri1,i2,,in<rPn(di1(V11),di2(Q),,din(Q))=dr(V11)Q+V11dr(Q)+i1+i2=r0<i1,i2<rdi1(V11)di2(Q). (4.13)

    Multiplying the left-hand side of (4.13) by U11 and combining it with (4.12), we get

    χr(U11V11)Q+i1+i2=r0<i1,i2<rχi1(U11V11)di2(Q)=U11dr(V11)Q+i1+i2=r0<i1,i2<rU11di1(V11)di2(Q),

    that is,

    χr(U11V11)Q+q+s=r0<q,s<rU11dq(V11)ds(Q)+p+q+s=r0<p,s<rχp(U11)dq(V11)ds(Q)=U11dr(V11)Q+q+s=r0<q,s<rU11dq(V11)ds(Q).

    Hence,

    χr(U11V11)Q+p+q+s=r0<p,s<rχp(U11)dq(V11)ds(Q)=U11dr(V11)Q (4.14)

    for all U11,V11U11. Based on

    0=dt(V11Q)=q+s=tdq(V11)ds(Q),

    we deduce

    p+q+s=r0<p,s<rχp(U11)dq(V11)ds(Q)=p+q=r0<p,q<rχp(U11)dq(V11)Q. (4.15)

    Combining (4.14) and (4.15), and then using

    χr(U11)V11Q=0,

    we obtain

    χr(U11V11)Q=(χr(U11)V11+U11dr(V11)+p+q=r0<p,q<rχp(U11)dq(V11))Q.

    Applying (4.11) yields that

    χr(U11V11)=χr(U11)V11+U11dr(V11)+p+q=r0<p,q<rχp(U11)dq(V11) (4.16)

    for all U11,V11U11. Analogously,

    χr(U22V22)=χr(U22)V22+U22dr(V22)+p+q=r0<p,q<rχp(U22)dq(V22) (4.17)

    for all U22,V22U22.

    Based on the definition of Gr, it is clear that

    0=Gr(Pn(U11,V22,Q,,Q))=Pn(Gr(U11),V22,Q,,Q)+Pn(U11,Lr(V22),Q,,Q)++Pn(U11,V22,Q,,Lr(Q))+i1+i2++in=ri1,i2,,in<rPn(Gi1(U11),Li2(V22),Li3(Q),,Lin(Q))=Pn(χr(U11),V22,Q,,Q)+Pn(U11,dr(V22),Q,,Q)+p+q=r0<p,q<rχp(U11)dq(V22)=χr(U11)V22+U11dr(V22)+p+q=r0<p,q<rχp(U11)dq(V22).

    Therefore, it follows from the property Ct and (4.9), (4.10), (4.16), (4.17) that

    χr(U)V+Udr(V)+p+q=r0<p,q<rχp(U)dq(V)=(χr(U11)+χr(U12)+χr(U22))(V11+V12+V22)+(U11+U12+U22)dr(V11+V12+V22)+p+q=r0<p,q<rχp(U11)dq(V11)+p+q=r0<p,q<rχp(U11)dq(V12)+p+q=r0<p,q<rχp(U11)dq(V22)+p+q=r0<p,q<rχp(U12)dq(V22)+p+q=r0<p,q<rχp(U22)dq(V22)=χr(U11)V11+χr(U11)V12+χr(U11)V22+χr(U12)V22+χr(U22)V22+U11dr(V11)+U11dr(V12)+U11dr(V22)+U12dr(V22)+U22dr(V22)+p+q=r0<p,q<rχp(U11)dq(V11)+p+q=r0<p,q<rχp(U11)dq(V12)+p+q=r0<p,q<rχp(U11)dq(V22)+p+q=r0<p,q<rχp(U12)dq(V22)+p+q=r0<p,q<rχp(U22)dq(V22)=χr(U11V11)+χr(U11V12)+χr(U12V22)+χr(U22V22)=χr(U11V11+U11V12+U12V22+U22V22)=χr((U11+U12+U22)(V11+V12+V22))=χr(UV)

    for all U=U11+U12+U22,V=V11+V12+V22U.

    The proof of the theorem is completed.

    We used Ashraf et al.'s results on Lie n-derivations to study the nonlinear generalized Lie n-derivations, although Lin also used the condition PG(x)Q=0 to describe the nonlinear generalized Lie n-derivations on triangular algebras. Based on this, we used an inductive method to describe the generalized Lie n-higher derivations. It is shown that, under some mild conditions, each component Gr of a nonlinear generalized Lie n-higher derivation {Gr}rN of the triangular algebra U can be expressed as the sum of an additive generalized higher derivation and a nonlinear mapping vanishing on all (n1)-th commutators on U.

    He Yuan: Conceptualization, Writing-original draft; Qian Zhang: Formal analysis, Writing-original draft; Zhendi Gu: Editing, Writing-original draft. All the contributors have perused and consented to the publishable draft of the manuscript.

    This study was supported by Jilin Science and Technology Department (No. YDZJ202201ZYTS622).

    The authors declare that they have no conflicts of interest.



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