Bull. Iran. Math. Soc. https://doi.org/10.1007/s41980-018-0035-8 ORIGINAL PAPER

Nonlinear Generalized Lie Triple Higher Derivation on Triangular Algebras Mohammad Ashraf1 · Aisha Jabeen1

Received: 14 March 2018 / Accepted: 18 May 2018 © Iranian Mathematical Society 2018

Abstract Let R be ring with unity. A triangular algebra is an algebra  a commutative  AM of the form A = where A and B are unital algebras over R and M is an 0 B (A, B)-bimodule which is faithful as a left A-module as well as a right B-module. In this paper, we study nonlinear generalized Lie triple higher derivation on A and show that under certain assumptions on A, every nonlinear generalized Lie triple higher derivation on A is of standard form, i.e., each component of a nonlinear generalized Lie triple higher derivation on A can be expressed as the sum of an additive generalized higher derivation and a nonlinear functional vanishing on all Lie triple products on A. Keywords Triangular algebra · Generalized higher derivation · Generalized Lie triple higher derivation Mathematics Subject Classification 16W25 · 15A78

1 Introduction Let R be a commutative ring with unity and A be an algebra over R. For any x, y ∈ A, [x, y](resp. x ◦ y) will denote the commutator x y − yx (resp. anticommutator x y + yx). An R-linear map L : A → A said to be a derivation (resp. Jordan derivation) if L(x y) = L(x)y + x L(y) (resp. L(x ◦ y) = L(x) ◦ y + x ◦ L(y)) holds

B

Aisha Jabeen [email protected] Mohammad Ashraf [email protected]

1

Department of Mathematics, Aligarh Muslim University, Aligarh 202002, India

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for all x, y ∈ A. An R-linear map L : A → A said to be a Lie derivation (resp. Lie triple derivation) if L([x, y]) = [L(x), y] + [x, L(y)]) (resp. L([[x, y], z]) = [[L(x), y], z] + [[x, L(y)], z] + [[x, y], L(z)]) holds for all x, y, z ∈ A. If we drop the linearity assumption in the above definitions, then derivation (resp. Jordan derivation) and Lie derivation (resp. Lie triple derivation) is called a nonlinear derivation (resp. nonlinear Jordan derivation) and nonlinear Lie derivation (resp. nonlinear Lie triple derivation). An R-linear map G : A → A is called a generalized Lie derivation (resp. generalized Lie triple derivation) if there exists a Lie derivation (resp. Lie triple derivation) L such that G([x, y]) = [G(x), y] + [x, L(y)] (resp. G([[x, y], z]) = [[G(x), y], z] + [[x, L(y)], z] + [[x, y], L(z)]) holds for all x, y, z ∈ A. If we drop the linearity of G in the above definitions, then a generalized Lie derivation (resp. generalized Lie triple derivation) is called a nonlinear generalized Lie derivation (resp. nonlinear generalized Lie triple derivation). It can be easily observed that every derivation is a Lie derivation as well as a Jordan derivation. Also, every Lie derivation is a Lie triple derivation while every Lie triple derivation is a generalized Lie triple derivation. However, the converse statements are not true in general. An important formula [[x, y], z] = x ◦ (y ◦ z) − y ◦ (x ◦ z) for all x, y, z ∈ A, shows that every Jordan derivation is also Lie triple derivation. Therefore, studying Lie triple derivations enables us to treat both important classes of Jordan and Lie derivations simultaneously. The study of derivations were further extended to higher derivations (see [8,9,13– 19] and references therein). Let N be the set of nonnegative integers and L = {L n }n∈N be a sequence R-linear mappings L n : A → A such that L 0 = IA . Then L is said L i (x)L j (y) to be a higher derivation (resp. Lie higher derivation) if L n (x y) = i+ j=n  [L i (x), L j (y)]) holds for all x, y ∈ A and L is said to be a (resp. L n ([x, y]) = i+ j=n  [[L i (x), L j (y)], L k (z)] holds Lie triple higher derivation if L n ([[x, y], z]) = i+ j+k=n

for all x, y, z ∈ A. If L is a sequence of nonlinear maps in the above definitions, then L is said to be a nonlinear higher derivation (resp. nonlinear Lie higher derivation) and a nonlinear Lie triple higher derivation respectively. Obviously, every higher derivation is a Lie higher derivation and every Lie higher derivation is a Lie triple higher derivation, but not conversely. A Lie (triple) higher derivation L = {L n }n∈N is said to be standard if each L n = dn + τn , where {dn }n∈N is an additive higher derivation of A and {τn }n∈N is a sequence of R-linear mappings from A into its center Z (A) with each τn vanishes on all (second) commutators of A. It is not difficult to see that L = {L n }n∈N is a Lie (triple) higher derivation of A, but not a higher derivation of A if τn = 0 for some n ∈ N.    am The R-algebra A = T ri(A, M, B) = a ∈ A, m ∈ M, b ∈ B under 0 b the usual matrix operations is called a triangular algebra, where A and B are unital algebras over R and M is a faithful (A, B)-bimodule. The notion of triangular algebra was first introduced by Chase [3] in 1960. Further, in the year 2000, Cheung [4] initiated the study of linear maps on triangular algebras. He described Lie derivations, commuting maps and automorphisms of triangular algebras [5,6]. In the year 2012, Ji et al. [12] studied nonlinear Lie triple derivation on triangular algebras and proved that under

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certain mild assumptions, on triangular algebra A, if a nonlinear map L : A → A satisfies L([[x, y], z]) = [[L(x), y], z] + [[x, L(y)], z] + [[x, y], L(z)] for all x, y, z ∈ A, then L = d + τ where d is an additive derivation of A and τ : A → Z (A) is a map vanishing at Lie triple products [[x, y], z]. Xiao and Wei [18] investigated the nonlinear Lie higher derivations on triangular algebras and obtained that under certain mild assumptions every nonlinear Lie higher derivation L = {L n }n∈N is of standard form, i.e; each component L n (n ≥ 1) can be expressed through an additive higher derivation and nonlinear functionals vanishing on commutators of A. Motivated by these observations, the authors in [1] carried out the study of nonlinear Lie triple higher derivations on triangular algebras and proved that under certain appropriate assumptions nonlinear Lie triple higher derivation L = {L n }n∈N on A is of standard form. Furthermore, in [10] author explored the nonlinear generalized Lie triple derivation on triangular algebra and found that under specific suppositions nonlinear generalized Lie triple derivation has the standard form, that is, it is the sum of additive generalized derivation and a mapping into its center vanishing at second commutator. The above observations inspire us to study nonlinear generalized Lie triple higher derivation on triangular algebras. Let N be the set of nonnegative integers and S = {Gn }n∈N be a sequence of mappings Gn : A → A (not necessarily linear) such that G0 = IA , where IA is the identity element of A. Then for all X, Y, Z ∈ A, S is said to be a (i) nonlinear generalized higher derivation on A if there exists a higher derivation  Gi (X )L j (Y ), L = {L n }n∈N such that L 0 = IA and Gn (X Y ) = i+ j=n

(ii) nonlinear generalized Lie higher derivation on A if there exists a Lie higher  [Gi (X ), L j (Y )], derivation L = {L n }n∈N such that L 0 = IA and Gn ([X, Y ]) = i+ j=n

(iii) nonlinear generalized Lie triple higher derivation on A if there exists a Lie triple higher derivation L = {L n }n∈N such that L 0 = IA and Gn ([[X, Y ], Z ]) =



[[Gi (X ), L j (Y )], L k (Z )].

i+ j+k=n

If S = {Gn }n∈N is the sequence of R-linear maps in the above definitions, then S is said to be a generalized higher derivation (resp. generalized Lie higher derivation) and a generalized Lie triple higher derivation respectively. Moreover, if S = {Gn }n∈N is assumed to be a sequence of additive maps, then S is said to be an additive generalized higher derivation (resp. additive generalized Lie higher derivation) and an additive generalized Lie triple higher derivation respectively. Example 1.1 Let Q be the ring of rational numbers and ST3 (Q) be the algebra of all 3 × 3 strictly ⎛ upper matrices ⎤ over Q. Let us define a map L : ST3 (Q) → ⎡ triangular ⎤⎞ ⎡ 0y 0 0x y ST3 (Q) by L ⎝⎣ 0 0 z ⎦⎠ = ⎣ 0 0 z 2 ⎦ . Then L is a nonlinear Lie triple derivation 000 00 0 n but not a nonlinear Lie derivation on ST3 (Q). Now suppose that L n = Ln! , then L = we define a map {L n }n∈N is a nonlinear Lie triple higher ⎤ ⎛⎡derivation ⎤⎞on ST ⎡ 3 (Q). Further 00 x 0x y G : ST3 (Q) → ST3 (Q) such that G ⎝⎣ 0 0 z ⎦⎠ = ⎣ 0 0 z 2 ⎦ . It can be easily seen 000 00 0

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that G is a nonlinear generalized Lie triple derivation but not a nonlinear generalized n Lie derivation on ST3 (Q). Let us suppose that G n = Gn! . Then G = {G n }n∈N is a nonlinear generalized Lie triple higher derivation on ST3 (Q). In the present paper it is shown that every nonlinear generalized Lie triple higher derivation of A is of standard, i.e., each component Gn can be expressed as the sum of an additive generalized higher derivation and nonlinear functionals vanishing on second commutators of A.

2 Triangular Algebras Throughout this paper, R will always denote a commutative ring with unity. Let A and B be unital algebras over R and let M be an (A, B)-bimodule which is faithful as a left A-module and also as a right B-module. The R-algebra  A = T ri(A, M, B) =

  am a ∈ A, m ∈ M, b ∈ B 0 b

under the usual matrix operations is called a triangular algebra. The center of A is  Z (A) =

  a0 am = mb ∀ m ∈ M . 0b

Define two natural projections πA : A → A and πB : A → B by 

am πA 0 b





=a

and

am πB 0 b

 = b.

Moreover, πA (Z (A)) ⊆ Z (A) and πB (Z (A)) ⊆ Z (B) and there exists a unique algebra isomorphism τ : πA (Z (A)) → πB (Z (A)) such that am = mτ (a) for all a ∈ πA (Z (A)), m ∈ M. Let 1A (resp.1B ) be the identity of the algebra A (resp.B) and let I be the identity of triangular this paper we shall use the following notions:  algebra A. Throughout,  1A 0 0 0 and A11 = PAP, A12 = PAQ, A22 = QAQ. P= , Q = I −P = 0 0 0 1B Thus, A = PAP + PAQ + QAQ = A11 + A12 + A22 where A11 is subalgebra of A isomorphic to A, A22 is subalgebra of A isomorphic to B and A12 is (A11 , A22 )-bimodule isomorphic to M. Also, πA (Z (A)) and πB (Z (A)) are isomorphic to P Z (A)P and Q Z (A)Q respectively. Then there exists an algebra isomorphism τ : P Z (A)P → Q Z (A)Q such that am = mτ (a) for all m ∈ PAQ. From now, we always assume that triangular algebra A = T ri(A, M, B) is 2-torsion free.

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3 Main Result 

 AM Theorem 3.1 Let A = be a triangular algebra and S = {Gn }n∈N be a 0 B nonlinear generalized Lie triple higher derivation on A. If A satisfy: (1) πA (Z (A)) = Z (A) and πB (Z (A)) = Z (B), (2) For x, y ∈ A, if [x, y] ∈ Z (A) then [x, y] = 0. Then S is of standard form. More precisely, there exists an additive generalized higher derivation {χn }n∈N on A and a sequence of nonlinear functionals {h n }n∈N which annihilates all Lie triple products [[X, Y ], Z ] for all X, Y, Z ∈ A such that Gn (T ) = χn (T ) + h n (T ) for all T ∈ A and n ∈ N. In order to prove our main result, we begin with the following: Theorem 3.2 [10, Theorem 2.4.1] Let A = T ri(A, M, B) be a triangular algebra consisting of algebras A, B and a faithful (A, B)-bimodule M. Suppose that A satisfies the following conditions: (1) πA (Z (A)) = Z (A) and πB (Z (A)) = Z (B), (2) For a, b ∈ A, if [a, b] ∈ Z (A) then [a, b] = 0. Then every nonlinear generalized Lie triple derivation G : A → A is of standard form. Remark 3.3 Suppose that L is associated Lie triple higher derivation of S on A. From [1, Theorem 3.1] L = {L n }n∈N is proper, i.e., L n = dn + τn , where {dn }n∈N is an additive higher derivation on A and τn is a mapping from A into its centre which annihilates on all Lie triple products [[X, Y ], Z ]. Now, it can be easily seen that L n (0) = 0, L n (P), L n (Q) ∈ A12 + Z (A); L n (A12 ) ⊆ A12 , L n (A22 ) ⊆ A22 + A12 + Z (A), L n (A11 ) ⊆ A11 + A12 + Z (A); Proof of Theorem 3.1 In order to prove our theorem, we will use the method of induction on n. For n = 1, G1 is nonlinear generalized Lie triple derivation on A. By [10, Theorem 2.4.1] it follows that there exists an additive generalized derivation χ1 and a nonlinear mapping h 1 satisfying h 1 ([[X, Y ], Z ]) = 0 for all X, Y, Z ∈ A such that G1 (T ) = χ1 (T ) + h 1 (T ) for all T ∈ A. Moreover, G1 and χ1 satisfy the following properties:  G1 (0) = 0, G1 (A12 ) ⊆ A12 ; G1 (Aii ) ⊆ Aii + A12 + Z (A), C1 : χ1 (A12 ) ⊆ A12 , χ1 (Aii ) ⊆ Aii + A12 ; We assume that the result holds for all 1 < s < n, n ∈ N. Then there exists i=s and non zero central mapping h an additive generalized higher derivation {χi }i=0 s satisfying h s ([[X, Y ], Z ]) = 0 for all X, Y, Z ∈ A such that Gs (T ) = χs (T ) + h s (T ) for all T ∈ A. Thus Gs and χs satisfy the following properties:  Gs (0) = 0, Gs (A12 ) ⊆ A12 ; Gs (Aii ) ⊆ Aii + A12 + Z (A), Cs : χs (Aii ) ⊆ Aii + A12 . χs (A12 ) ⊆ A12 ;

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Our aim is to show that the above conditions also hold for n. Claim 1 Gn (0) = 0 and Gn (A12 ) ⊆ A12 . Using condition Cs , we find that Gn (0) = Gn ([[0, 0], 0])  [[G p (0), L q (0)], L r (0)] = p+q+r =n

= [[Gn (0), 0], 0] + [[0, L n (0)], 0] + [[0, 0], L n (0)]  [[G p (0), L q (0)], L r (0)] + p+q+r =n p,q,r >0

= 0. As we know that P M12 Q = [[M12 , P], P] for all M12 ∈ A12 , we arrive that Gn (M12 ) = Gn (P M12 Q) = Gn ([[M12 , P], P]) = [[Gn (M12 ), P], P] + [[M12 , L n (P)], P] + [[M12 , P], L n (P)]  [[G p (M12 ), L q (P)], L r (P)]. + p+q+r =n p,q,r >0

Applying condition Cs , it follows that Gn (M12 ) = [[Gn (M12 ), P], P] + [[M12 , L n (P)], P] +[[M12 , P], L n (P)].

(3.1)

Let us assume Gn (M12 ) = R11 + R12 + R22 and L n (P) ∈ A12 + Z (A). Then (3.1) gives us R11 = 0, R22 = 0. Therefore, Gn (A12 ) ⊆ A12 . Claim 2 Gn (Aii ) ⊆ Aii + A12 + Z (A) for i = 1, 2. Since 0 = [[A11 , B22 ], T ] = Gn ([[A11 , B22 ], T ]) = [[Gn (A11 ), B22 ], T ] + [[A11 , L n (B22 )], T ] + [[A11 , B22 ], L n (T )]  [[G p (A11 ), L q (B22 )], L r (T )] + p+q+r =n p,q,r >0

= [[Gn (A11 ), B22 ], T ] + [[A11 , L n (B22 )], T ]  [[G p (A11 ), L q (B22 )], L r (T )]. + p+q+r =n p,q,r >0

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Multiplying both sides by Q, we find that QGn (A11 )Q ∈ Z (B). In the similar manner, on using [[B22 , A11 ], T ] we obtain PGn (B22 )P ∈ Z (A). This implies that τ (PGn (B22 )P) = QGn (A11 )Q. Now we obtain Gn (A11 ) = (PGn (A11 )P − τ −1 (QGn (A11 )Q)) + PGn (A11 )Q + (τ −1 (QGn (A11 )Q) + QGn (A11 )Q) and Gn (B22 ) = (PGn (B22 )P + τ (PGn (B22 )P)) + PGn (B22 )Q +(QGn (B22 )Q − τ (PGn (B22 )P), which gives Gn (Aii ) ⊆ Aii + A12 + Z (A) for i = 1, 2. Remark 3.4 Now we define f n1 (A11 ) = QGn (A11 )Q and f n2 (B22 ) = PGn (B22 )P for any A11 ∈ A11 and B22 ∈ A22 . From Claim 2 it follows that f n1 : A11 → Q Z (A)Q is a mapping such that f n1 ([[X 11 , Y11 ], Z 11 ]) = 0 for all X 11 , Y11 , Z 11 ∈ A11 and f n2 : A22 → P Z (A)P is a nonlinear mapping such that f n2 ([[X 22 , Y22 ], Z 22 ]) = 0 for all X 22 , Y22 , Z 22 ∈ A22 . Now set f n (T ) = f n1 (P T P) + τ −1 ( f n1 (P T P)) + f n2 (QT Q) + τ ( f n2 (QT Q)). Obviously, f n (T ) ∈ Z (A) and f n ([[U, V ], W ]) = 0 for all U, V, W ∈ A. Define a new mapping ωn (T ) = Gn (T ) − f n (T )

for all T ∈ A.

Now from Claims 1 and 2, it is clear that Claim 3 ωn (0) = 0, ωn (A12 ) ⊆ A12 and ωn (Aii ) ⊆ Aii + A12 for i = 1, 2. Claim 4 For any A11 ∈ A11 , M12 ∈ A12 and B22 ∈ A22 ,  (i) ωn (A11 M12 ) = ωn (A11 )M12 + A11 dn (M12 ) +

p+q=n 0< p,q
(ii) ωn (M12 B22 ) = ωn (M12 )B22 + M12 dn (B22 ) +



p+q=n 0< p,q
χ p (A11 )dq (M12 ), χ p (M12 )dq (B22 ).

Since A11 M12 = [[A11 , M12 ], Q], we find that ωn (A11 M12 ) = Gn ([[A11 , M12 ], Q])  ([[G p (A11 ), L q (M12 )], L r (Q)]) = p+q+r =n

= [[ωn (A11 ), M12 ], Q] + [[A11 , dn (M12 )], Q] + [[A11 , M12 ], dn (Q)]  [[χ p (A11 ), dq (M12 )], dr (Q)]. + p+q+r =n p,q,r >0

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As ωn (A11 ) ⊆ A11 + A12 , ωn (M12 ), dn (Q) ∈ A12 and using the condition Cs , the above equation becomes ωn (A11 M12 ) = ωn (A11 )M12 + A11 dn (M12 ) +



χ p (A11 )dq (M12 ).

p+q=n 0< p,q
Also, we know that M12 B22 = [[M12 , B22 ], Q] for M12 ∈ A12 and B22 ∈ A22 . Similarly, we can show that ωn (M12 B22 ) = ωn (M12 )B22 + M12 dn (B22 ) +



χ p (M12 )dq (B22 ).

p+q=n 0< p,q
Claim 5 For any Aii , Bii ∈ Aii (i = 1, 2), ωn (Aii Bii ) = ωn (Aii )Bii + Aii dn (Bii ) +



χ p (Aii )dq (Bii ).

p+q=n 0< p,q
For any A11 , B11 ∈ A11 and M12 ∈ A12 . ωn (A11 B11 M12 ) = ωn ((A11 B11 )M12 ) = ωn (A11 B11 )M12 + A11 B11 dn (M12 )  χ p (A11 B11 )dq (M12 ) + p+q=n 0< p,q
= ωn (A11 B11 )M12 + A11 B11 dn (M12 )  χ p (A11 )dq (B11 )dr (M12 ) + p+q+r =n p,q,r >0

and ωn (A11 B11 M12 ) = ωn (A11 (B11 M12 )) = ωn (A11 )B11 M12 + A11 dn (B11 M12 )  χ p (A11 )dq (B11 M12 ) + p+q=n 0< p,q
= ωn (A11 )B11 M12 + A11 dn (B11 )M12 + A11 B11 dn (M12 )  A11 d p (B11 )dq (M12 ) + p+q=n 0< p,q
+



p+q+r =n p,q,r >0

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χ p (A11 )dq (B11 )dr (M12 )

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= ωn (A11 )B11 M12 + A11 dn (B11 )M12 + A11 B11 dn (M12 )  χ p (A11 )dq (B11 )M12 + p+q=n 0< p,q
+



χ p (A11 )dq (B11 )dr (M12 ).

p+q+r =n p,q,r >0

By the condition Cs , the above expression becomes 

ωn (A11 B11 )M12 = {ωn (A11 )B11 + A11 dn (B11 ) +

χ p (A11 )dq (B11 )}M12 .

p+q=n 0< p,q
Since ωn (A11 ) ⊆ A11 + A12 and PAQ is faithful as a left PAP-module, the above relation implies that ωn (A11 B11 )P = {ωn (A11 )B11 + A11 dn (B11 ) +



χ p (A11 )dq (B11 )}P.

(3.2)

p+q=n 0< p,q
Also, [[A11 , Q], Q] = 0 for all A11 ∈ A11 0 = Gn ([[A11 , Q], Q]) = [[Gn (A11 ), Q], Q] + [[A11 , L n (Q)], Q] + [[A11 , Q], L n (Q)]  [[G p (A11 ), L q (Q)], L r (Q)] + p+q+r =n p,q,r >0

= [[ωn (A11 ), Q], Q] + [[A11 , dn (Q)], Q] + [[A11 , Q], dn (Q)]  [[χ p (A11 ), dq (Q)], dr (Q)]. + p+q+r =n p,q,r >0

Since ωn (A11 ) ⊆ A11 + A12 , dn (Q) ∈ A12 , the above equation implies that 0 = ωn (A11 )Q + A11 dn (Q) +



χ p (A11 )dq (Q) for all A11 ∈ A11 . (3.3)

p+q=n 0< p,q
On substituting A11 by A11 B11 in (3.3), we get 0 = ωn (A11 B11 )Q + A11 B11 dn (Q) +



χ p (A11 B11 )dq (Q).

(3.4)

p+q=n 0< p,q
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Again, note that [[B11 , Q], Q] = 0 for all A11 ∈ A11 0 = L n ([[B11 , Q], Q]) = [[L n (B11 ), Q], Q] + [[B11 , L n (Q)], Q] + [[B11 , Q], L n (Q)]  [[L p (B11 ), L q (Q)], L r (Q)] + p+q+r =n p,q,r >0

= [[dn (B11 ), Q], Q] + [[B11 , dn (Q)], Q] + [[B11 , Q], dn (Q)]  [[d p (B11 ), dq (Q)], dr (Q)]. + p+q+r =n p,q,r >0

This gives us 

0 = dn (B11 )Q + B11 dn (Q) +

d p (B11 )dq (Q)

(3.5)

p+q=n 0< p,q
Now left multiplying A11 in (3.5) and combining it with (3.4) gives 

ωn (A11 B11 )Q +

χ p (A11 )dq (B11 )dr (Q) = A11 dn (B11 )Q.

p+q+r =n 0
This implies that ωn (A11 B11 )Q +

n−1 

χ p (A11 )



dq (B11 )dr (Q) = A11 dn (B11 )Q.

q+r =n− p 0
p=1

Now using the condition Cs , we find that ωn (A11 B11 )Q −

n−1 

χ p (A11 )dn− p (B11 )Q = A11 dn (B11 )Q

p=1

which gives ωn (A11 B11 )Q = A11 dn (B11 )Q +

n−1 

χ p (A11 )dn− p (B11 )Q.

p=1

Hence, ωn (A11 B11 )Q = {ωn (A11 )B11 + A11 dn (B11 ) +

 p+q=n 0< p,q
123

χ p (A11 )dq (B11 )}Q.

(3.6)

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Now adding (3.2) and (3.6), we have 

ωn (A11 B11 ) = ωn (A11 )B11 + A11 dn (B11 ) +

χ p (A11 )dq (B11 ).

p+q=n 0< p,q
Similarly, we can obtain that 

ωn (A22 B22 ) = ωn (A22 )B22 + A22 dn (B22 ) +

χ p (A22 )dq (B22 )

p+q=n 0< p,q
for all A22 , B22 ∈ A22 . Claim 6 For any A11 ∈ A11 , B22 ∈ A22 and M12 , N12 ∈ A12 , (i) ωn (A11 + M12 ) − ωn (A11 ) − ωn (M12 ) ∈ Z (A), (ii) ωn (B22 + N12 ) − ωn (B22 ) − ωn (N12 ) ∈ Z (A). Since A11 X 12 = [[A11 + M12 , X 12 ], Q] for all A11 ∈ A11 and M12 , N12 ∈ A12 , we find that ωn (A11 X 12 ) = Gn ([[A11 + M12 , X 12 ], Q])  [[G p (A11 + M12 ), L q (X 12 )], L r (Q)] = p+q+r =n

= [[ωn (A11 + M12 ), X 12 ], Q] + [[A11 + M12 , dn (X 12 )], Q] + [[A11 + M12 , X 12 ], dn (Q)]  [[χ p (A11 + M12 ), dq (X 12 )], dr (Q)]. + p+q+r =n p,q,r >0

Using the condition Cs and ωn (X 12 ), dn (Q) ∈ A12 , ωn (A11 ) ⊆ A11 + A12 , we get ωn (A11 X 12 ) = [[ωn (A11 + M12 ), X 12 ], Q] + A11 dn (X 12 ). Now using Claim 4(i), we have [[ωn (A11 + M12 ) − ωn (A11 ), X 12 ], Q] = 0.  Assume that ωn (A11 + M12 ) − ωn (A11 ) =

am 0 b

(3.7)

 and by (3.7), we have

ωn (A11 + M12 ) − ωn (A11 ) − P(ωn (A11 + M12 ) − ωn (A11 ))Q ∈ Z (A). (3.8)

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It is clear that P T Q = [[T, P], P] for all T ∈ A. Therefore, P(ωn (A11 + M12 ) − ωn (A11 ))Q = [[ωn (A11 + M12 ) − ωn (A11 ), P], P] = [[Gn (A11 + M12 ), P], P] − [[Gn (A11 ), P], P] = Gn ([[(A11 + M12 ), P], P]) − [[M12 , L n (P)], P]  [[G p (M12 ), L q (P)], L r (P)]. −[[M12 , P], L n (P)] − p+q+r =n p,q,r >0

Applying the condition Cs and using the Claim 1, we have P(ωn (A11 + M12 ) − ωn (A11 ))Q = ωn (M12 ). This implies that ωn (A11 + M12 ) − ωn (A11 ) − ωn (M12 ) ∈ Z (A). Similarly, we can show that ωn (B22 + N12 ) − ωn (B22 ) − ωn (N12 ) ∈ Z (A). Claim 7 ωn is additive on A11 , A12 and A22 respectively. For any M12 , N12 ∈ A12 ωn (M12 + N12 ) = Gn [([P + M12 , Q + N12 ], Q])  [[G p (P + M12 ), L q (Q + N12 )], L r (Q)] = p+q+r =n

= [[ωn (P + M12 ), Q + N12 ], Q] + [[P + M12 , dn (Q + N12 )], Q] +[[P + M12 , Q + N12 ], dn (Q)]  [[χ p (P + M12 ), dq (Q + N12 )], dr (Q)]. + p+q+r =n p,q,r >0

Using the condition Cs , and Claims 3 and 6, we find that ωn (M12 + N12 ) = ωn (M12 ) + ωn (N12 ). Now observe that ωn ((A11 + B11 )M12 ) = ωn (A11 M12 ) + ωn (B11 M12 ) = ωn (A11 )M12 + A11 dn (M12 )  χ p (A11 )dq (M12 ) + p+q=n 0< p,q
+ ωn (B11 )M12 + B11 dn (M12 )  χ p (B11 )dq (M12 ) + p+q=n 0< p,q
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(3.9)

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for all A11 , B11 ∈ A11 and M12 ∈ A12 . On the other hand, ωn ((A11 + B11 )M12 ) = ωn (A11 + B11 )M12 + (A11 + B11 )dn (M12 )  χ p (A11 + B11 )dq (M12 ). (3.10) + p+q=n 0< p,q
Combining (3.9), (3.10) and applying the condition Cs , we have ωn (A11 + B11 )M12 = ωn (A11 )M12 + ωn (B11 )M12 .

(3.11)

Since ωn (A11 ) ⊆ A11 + A12 and PAQ is faithful as a left PAP, (3.11) implies that ωn (A11 + B11 )P = ωn (A11 )P + ωn (B11 )P.

(3.12)

Replace A11 for A11 + B11 in (3.3), to get ωn (A11 + B11 )Q + (A11 + B11 )dn (Q)  χ p (A11 + B11 )dq (Q) = 0 +

(3.13)

p+q=n 0< p,q
for all A11 ∈ A11 . Combining (3.13) with (3.3) and (3.5), we obtain ωn (A11 + B11 )Q = ωn (A11 )Q + ωn (B11 )Q.

(3.14)

Addition of (3.12) and (3.14) implies that ωn (A11 + B11 ) = ωn (A11 ) + ωn (B11 ). Similarly, we can deduce that ωn (A22 + B22 ) = ωn (A22 ) + ωn (B22 ). Claim 8 For any Ai j ∈ Ai j with 1 ≤ i ≤ j ≤ 2, ωn (A11 + A12 + A22 ) − ωn (A11 ) − ωn (A12 ) − ωn (A22 ) ∈ Z (A). For M12 ∈ A12 and A22 ∈ A22 by Claims 3 and 7, we have 0 = ωn (0) = ωn (M12 A22 − M12 A22 ) = ωn (M12 A22 ) + ωn (−M12 A22 ) which implies that ωn (−M12 A22 ) = −ωn (M12 A22 ). By Claim 4 and Claim 7 , we get ωn (A11 M12 − M12 A22 ) = Gn ([[A11 , M12 ], Q]) + Gn ([[A22 , M12 ], Q])  [[G p (A11 ), L q (M12 )], L r (Q)] = p+q+r =n

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+



[[G p (A22 ), L q (M12 )], L r (Q)]

p+q+r =n

= [[ωn (A11 ), M12 ], Q] + [[A11 , dn (M12 )], Q] + [[A11 , M12 ], dn (Q)]  [[χ p (A11 ), dq (M12 )], dr (Q)] + p+q+r =n p,q,r >0

+ [[ωn (A22 ), M12 ], Q] + [[A22 , dn (M12 )], Q]  [[χ p (A22 ), dq (M12 )], dr (Q)]. + [[A22 , M12 ], dn (Q)] +

(3.15)

p+q+r =n p,q,r >0

Also, it is clear that A11 M12 − M12 A22 = [[A11 + A12 + A22 , M12 ], Q] for all Ai j , Mi j ∈ Ai j with i ≤ i ≤ j ≤ 2. Using Claim 3 and by condition Cs , we obtain ωn (A11 M12 − M12 A22 ) = Gn ([[A11 + A12 + A22 , M12 ], Q])  [[G p (A11 + A12 + A22 ), L q (M12 )], L r (Q)] = p+q+r =n

= [[ωn (A11 + A12 + A22 ), M12 ], Q] +[[A11 + A12 + A22 , dn (M12 )], Q] + [[A11 + A12 + A22 , M12 ], dn (Q)]  [[χ p (A11 + A12 + A22 ), dq (M12 )], dr (Q)] + p+q+r =n p,q,r >0

= [[ωn (A11 + A12 + A22 ), M12 ], Q] + A11 dn (M12 ) − dn (M12 )A22  + (χ p (A11 )dq (M12 ) − χ p (M12 )dq (A22 )). p+q=n 0< p,q
(3.16) Combining (3.15) and (3.16) we find that [[ωn (A11 + A12 + A22 ) − ωn (A11 ) − ωn (A22 ), M12 ], Q] = 0.  Let us assume ωn (A11 + A12 + A22 ) − ωn (A11 ) − ωn (A22 ) =

 am . It can be easily 0 b

seen that ωn (A11 + A12 + A22 ) − ωn (A11 ) − ωn (A22 ) −P(ωn (A11 + A12 + A22 ) − ωn (A11 ) − ωn (A22 ))Q ∈ Z (A). To determine the term P(ωn (A11 + A12 + A22 ) − ωn (A11 ) − ωn (A22 ))Q, we use P T Q = [[T, P], P]. Therefore,

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P(ωn (A11 + A12 + A22 ) − ωn (A11 ) − ωn (A22 ))Q = [[ωn (A11 + A12 + A22 ) − ωn (A11 ) − ωn (A22 ), P], P] = [[Gn (A11 + A12 + A22 ), P], P] − [[Gn (A11 ), P], P] −[[Gn (A22 ), P], P] = Gn ([[A11 + A12 + A22 , P], P]) − [[A11 + A12 + A22 , L n (P)], P] −Gn ([[A11 , P], P]) −[[A11 + A12 + A22 , P], L n (P)] + [[A11 , L n (P)], P] + [[A22 , P], L n (P)]  [[G p (A11 + A12 + A22 ), L q (P)], L r (P)] − p+q+r =n p,q,r >0

+



[[G p (A11 ), L q (P)], L r (P)]

p+q+r =n p,q,r >0

+[[A11 , P], L n (P)] + [[A22 , L n (P)], P]  [[G p (A22 ), L q (P)], L r (P)] + p+q+r =n p,q,r >0

= ωn [[A11 + A12 + A22 , P], P] − [[A12 , dn (P)], P] − [[A12 , P], dn (P)]  [[χ p (A11 + A12 + A22 ), dq (P)], dr (P)] − p+q+r =n p,q,r >0

−ωn [[A11 , P], P] − ωn [[A22 , P], P]  + [[χ p (A11 ), dq (P)], dr (P)] p+q+r =n p,q,r >0

+



[[χ p (A22 ), dq (P)], dr (P)]

p+q+r =n p,q,r >0

= ωn (A12 ). Therefore, ωn (A11 + A12 + A22 )−ωn (A11 )−ωn (A12 )−ωn (A22 ) ∈ Z (A) as required. Remark 3.5 Now we establish a mapping gn : A → Z (A) by gn (T ) = ωn (T ) − ωn (P T P) − ωn (P T Q) − ωn (QT Q) and gn ([[U, V ], W ]) = 0 for all U, V, W ∈ A. Then define a mapping χn (T ) = ωn (T )− gn (T ) for all T ∈ A. It is easy to verify the χn satisfies the property χn (A11 + A12 + A22 ) = χn (A11 ) + χn (A12 ) + χn (A22 ). From the definition of χn and gn we find that Gn = ωn + f n = χn + gn + f n = χn + h n where h n = gn + f n . i=n is a higher derivation on triangular algebra A. Claim 9 {χn }i=0

Suppose X, Y ∈ A such that X = X 11 + X 12 + X 22 and Y = Y11 + Y12 + Y22 where X i j , Yi j ∈ Ai j with 1 ≤ i ≤ j ≤ 2. Then

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  χn (X + Y ) = χn (X 11 + X 12 + X 22 ) + (Y11 + Y12 + Y22 )   = χn (X 11 + Y11 ) + (X 12 + Y12 ) + (X 22 + Y22 ) = ωn (X 11 + Y11 ) + ωn (X 12 + Y12 ) + ωn (X 22 + Y22 ) = ωn (X 11 ) + ωn (Y11 ) + ωn (X 12 ) + ωn (Y12 ) + ωn (X 22 ) + ωn (Y22 ) = ωn (X 11 ) + ωn (X 12 ) + ωn (X 22 ) + ωn (Y11 ) + ωn (Y12 ) + ωn (Y22 ) = χn (X 11 + X 12 + X 22 ) + χn (Y11 + Y12 + Y22 ) = χn (X ) + χn (Y ).

(3.17)

By Claims 4 and 5, we have   χn (X Y ) = χn (X 11 + X 12 + X 22 )(Y11 + Y12 + Y22 ) = ωn (X 11 Y11 + X 11 Y12 + X 12 Y22 + X 22 Y22 )  χ p (X 11 )dq (Y11 ) = ωn (X 11 )Y11 + X 11 dn (Y11 ) + p+q=n 0< p
+ωn (X 11 )Y12 + X 11 dn (Y12 )  χ p (X 11 )dq (Y12 ) + ωn (X 12 )Y22 + X 12 dn (Y22 ) + p+q=n 0< p
+



χ p (X 12 )dq (Y22 )

p+q=n 0< p
+ωn (X 22 )Y22 + X 22 dn (Y22 ) +



χ p (X 22 )dq (Y22 ).

(3.18)

p+q=n 0< p
On the other hand, we have χn (X )Y + X dn (Y ) +



χ p (X )dq (Y )

p+q=n 0< p
= (ωn (X 11 ) + ωn (X 12 ) + ωn (X 22 ))Y +



χ p (X 11 )dq (Y11 )

p+q=n 0< p
+X (dn (Y11 ) + dn (Y12 ) + dn (Y22 )) + +



χ p (X 12 )dq (Y11 ) +

p+q=n 0< p
+



p+q=n 0< p
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χ p (X 11 )dq (Y12 ) +

p+q=n 0< p
χ p (X 12 )dq (Y12 ) +

p+q=n 0< p
χ p (X 22 )dq (Y11 ) +



p+q=n 0< p




χ p (X 11 )dq (Y22 )

p+q=n 0< p
χ p (X 12 )dq (Y22 ))

p+q=n 0< p
χ p (X 22 )dq (Y12 ) +



p+q=n 0< p
χ p (X 22 )dq (Y22 ).

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By using the condition Cs , and the Claim 3, we have 

χn (X )Y + X dn (Y ) +

χ p (X )dq (Y )

p+q=n 0< p


= ωn (X 11 )Y11 + X 11 dn (Y11 ) +

χ p (X 11 )dq (Y11 )

p+q=n 0< p
+ ωn (X 11 )Y12 + X 11 dn (Y12 )  χ p (X 11 )dq (Y12 ) + ωn (X 12 )Y22 + X 12 dn (Y22 ) + p+q=n 0< p
+



χ p (X 12 )dq (Y22 )

p+q=n 0< p
+ ωn (X 22 )Y22 + X 22 dn (Y22 ) +



χ p (X 22 )dq (Y22 ).

(3.19)

p+q=n 0< p
Combining (3.18) and (3.19), we get χn (X Y ) = χn (X )Y + X dn (Y ) +



χ p (X )dq (Y ).

p+q=n 0< p
4 Applications In this section, we study the application of Theorem 3.1 to some classical examples of triangular algebra such as upper triangular matrix algebra, nest algebras and block upper triangular matrix algebra. For the definitions of upper triangular matrix algebra, nest algebras and block upper triangular matrix algebra see [4] and references therein. Corollary 4.1 Every nonlinear generalized Lie triple higher derivation has standard form on upper triangular matrix algebra Tn (R). Corollary 4.2 Let X be an infinite dimensional Banach space over the real or complex field F , N be a nest on X which contains a nontrivial element complemented in X and T (N ) be a nest algebra. Then every nonlinear generalized Lie triple higher derivation has standard form on T (N ).

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Corollary 4.3 Let N be a nest of a Hilbert space H dimension greater than 2 and T (N ) be a nontrivial nest algebra. Then every nonlinear generalized Lie triple higher derivation has standard form on T (N ). If Hilbert space H is finite dimensional, then nest algebras are upper block triangular matrices algebras [4]. Corollary 4.4 Every nonlinear generalized Lie triple higher derivation has standard form on block upper triangular matrix algebra Bnk (R).

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