# A Class of Nonlinear Nonglobal Semi-Jordan Triple Derivable Mappings on Triangular Algebras

Authors: Xiuhai Fei; Haifang Zhang
Journal: Journal of Mathematics
 (2021)

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## Abstract

In this paper, we proved that each nonlinear nonglobal semi-Jordan triple derivable mapping on a 2-torsion free triangular algebra is an additive derivation. As its application, we get the similar conclusion on a nest algebra or a 2-torsion free block upper triangular matrix algebra, respectively.


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## Body

## 1. Introduction

LetR be a commutative ring with identity and A a unital algebra over R, Ω=X∈A:X2=0, and Δ be an additive mapping on A. For any X,Y∈A, we denote the Jordan product of X,Y by X°Y=XY+YX. For any X∈A, if 2X=0 implies X=0, then A is said a 2-torsion free algebra. Recall that Δ is called a derivation if ΔXY=ΔXY+XΔY for all X,Y∈A; Δ is called a Jordan derivation if ΔX°Y=ΔX°Y+X°ΔY, for all X,Y∈A; Δ is called a triple derivation if ΔXYZ=ΔXYZ+XΔYZ+XYΔZ, for all X,Y,Z∈A. Δ is called a Jordan triple derivation if ΔX°Y°Z=ΔX°Y°Z+X°ΔY°Z+X°Y°ΔZ, for all X,Y,Z∈A. Furthermore, if Δ is without assumption of additivity in the above definitions, then Δ is said a nonlinear (triple) derivable mapping and a nonlinear Jordan (triple) derivable mapping, respectively. Obviously, every derivation is a Jordan derivation, every derivation is a triple derivation, and every triple derivation is a Jordan triple derivation. However, the inverse statement is not true in general.A natural and very interesting problem that we are dealing with is studying certain conditions on an algebra such that each Jordan (triple) derivation (nonlinear Jordan (triple) derivable mapping) is a derivation.In the past few decades, many mathematicians studied this problem and obtained abundant results. For example, Herstein, in [1], proved that every Jordan derivation on a prime ring not of characteristic 2 is a derivation. This result was extended by Cusack in [2] and Brešar and Vukman in [3] to the case of semiprime, respectively. Zhang, in [4, 5], showed that every Jordan derivation on a nest algebra or a 2-torsion free triangular algebra is an inner derivation or a derivation, respectively. Later, Ghahramani, in [6], extended the result of Zhang and Yu in [5] and proved that, under certain conditions, each Jordan derivation on trivial extension algebras is a sum of a derivation and an antiderivation. For other similar results about Jordan derivations (nonlinear Jordan derivable mappings), we refer the readers to [7–9] and references therein, for more details. With the deepening of research, many research achievements have been obtained about Jordan triple derivations and nonlinear Jordan triple derivable mappings. For example, Bresar, in [10], proved that every Jordan triple derivation on a 2-torsion-free semiprime ring is a derivation. Similar conclusion have been obtained in [11] by Bell and Kappe. Zhao and Li, in [12], proved that every nonlinear ∗-Jordan triple derivation on von Neumann algebras is an additive ∗-derivation. For other similar results about Jordan triple derivations (nonlinear Jordan triple derivable mappings), we refer the readers to [13–15] and references therein, for more details.In 2016, Ashraf and Jabeen, in [15], obtained that if Δ is without the additivity assumption and satisfies(1)ΔXYZ+ZYX=ΔXYZ+XΔYZ+XYΔZ+ΔZYX+ZΔYX+ZYΔX,for all X,Y,Z∈A, then such a Δ is an additive derivation on a 2-torsion-free triangular algebra.In this paper, we call thatΔ is a nonlinear nonglobal semi-Jordan triple derivable mapping on A if Δ is without the additivity assumption and satisfies(2)ΔXYZ+YXZ=ΔXYZ+XΔYZ+XYΔZ+ΔYXZ+YΔXZ+YXΔZ,for all X,Y,Z∈A with XYZ∈Ω.Here, it needs to be pointed out that our above definition is different from Ashraf’s and Jabeen’s in [15]. We will discuss the nonlinear nonglobal semi-Jordan triple derivable mappings on triangular algebras and obtain one main result (see Theorem 1).For convenient reading, we give some basic concepts and properties of triangular algebras as follows.LetA and B be unital algebras over a commutative ring R and M be a unital A,B-bimodule, which is faithful as both are a left A-module and a right B-module. Then, the R-algebra,(3)U=am0b:a∈A,m∈M,b∈B,under the usual matrix operations is called a triangular algebra. We refer the reader to [16] for more details about the triangular algebras. Basic examples of triangular algebras are upper triangular matrix algebras and nest algebras.Let1A and 1B be the identities of the algebras A and B, respectively, and let 1 be the identity of the triangular algebra U. Throughout this paper, we shall use the following notations:(4)P1=1A000,P2=1−P1=0001B.It is clear that the triangular algebraU may be represented as(5)U=P1UP1+P1UP2+P2UP2=A+M+B,where P1UP1 and P2UP2 are subalgebras of U isomorphic to A and B, respectively, and P1UP2 is a P1UP1,P2UP2-bimodule isomorphic to the A,B-bimodule M.


## 2. Nonlinear Nonglobal Semi-Jordan Triple Derivable Mappings on Triangular Algebras

In this section, our main result is Theorem1, and we will show Theorem 1 holds.Theorem 1.
LetU be a 2-torsion-free triangular algebra and Δ be a mapping from U into itself (without assumption of additivity) such that(6)ΔXYZ+YXZ=ΔXYZ+XΔYZ+XYΔZ+ΔYXZ+YΔXZ+YXΔZ,for all X,Y,Z∈U with XYZ∈Ω. Then, Δ is an additive derivation.
In order to prove Theorem1, we introduce Lemmas 1–5 and then prove that Lemmas 1–5 hold. We assume that U be a 2-torsion-free triangular algebra Ω=X∈U:X2=0, and Δ be a nonlinear nonglobal semi-Jordan triple derivable mapping on triangular algebra U.Lemma 1.
Δ
0
=
0 and ΔP1=−ΔP2∈M.Proof.
TakingX=Y=Z=0 in equation (6), we have Δ0=0. Since P1P1P2=0∈Ω, taking X=P1,Y=P1,Z=P2 in equation (6), we obtain(7)0=Δ0=2ΔP1P1P2+2P1ΔP1P2+2P1P1ΔP2=2P1ΔP1P2+2P1ΔP2.
This yields from the property of 2-torsion freeness ofU that(8)P1ΔP2P1=0,P1ΔP1P2+P1ΔP2P2=0.
Similarly, we obtain that(9)P2ΔP1P2=0.
For anyX12∈M, since P2X12P2=0∈Ω, taking X=P2,Y=X12,andZ=P2 in equation (6), we have(10)ΔX12=ΔP2X12P2+X12P2P2=ΔP2X12P2+P2ΔX12P2+P2X12ΔP2+ΔX12P2P2+X12ΔP2P2+X12P2ΔP2=ΔP2X12+P2ΔX12P2+ΔX12P2+X12ΔP2P2+X12P2ΔP2.
Multiplying the above equation from left byP1 and from right by P2, it follows from P1ΔP2P1=0 that 2X12P2ΔP2P2=0. Similarly, we obtain that 2P1ΔP2P1X12=0. Therefore, according to the faithfulness of M and the property of 2-torsion free of U, we have(11)P1ΔP1P1=P2ΔP2P2=0.
So, by equations (8)–(11), we have ΔP1=−ΔP2∈M. The proof is completed.Lemma 2.
For anyX11∈A,X12∈M,X22∈B, then
(i)ΔX11∈A+M and P1ΔX11P2=X11ΔP1
(ii)ΔX22∈M+B and P1ΔX22P2=ΔP2X22
(iii)ΔX12∈MProof.
(i) For anyX11∈A, since X11P2P2=0∈Ω, taking X=X11,Y=Z=P2 in equation (6), then it follows from ΔP2∈M that(12)0=ΔX11P2P2+P2X11P2,=ΔX11P2P2+X11ΔP2P2+X11P2ΔP2+ΔP2X11P2+P2ΔX11P2+P2X11ΔP2,=ΔX11P2+X11ΔP2P2+P2ΔX11P2.
This implies thatP2ΔX11P2=0. Furthermore, multiplying the above equation from left by P1 and from right by P2, it follows from ΔP1=−ΔP2∈M that(13)P1ΔX11P2=−X11ΔP2=X11ΔP1.
Similarly, we can show (ii) holds.
(iii) For anyX12∈M, since X12P1P2=0∈Ω, taking X=X12,Y=P1,andZ=P2 equation (6), we get from ΔP1=−ΔP2∈M that(14)ΔX12=ΔX12P1P2+P1X12P2+ΔX12P1P2+X12ΔP1P2+X12P1ΔP2+ΔP1X12P2+P1ΔX12P2+P1X12ΔP2=P1ΔX12P2.
Therefore,ΔX12∈M. The proof is completed.Lemma 3.
For anyX11,Y11∈A, X12∈M, and X22,Y22∈B, then
(i)ΔX11X22=ΔX11X22+X11ΔX22
(ii)ΔX11X12=ΔX11X12+X11ΔX12
(iii)ΔX12X22=ΔX12X22+X12ΔX22
(iv)ΔX11Y11=ΔX11Y11+X11ΔY11
(v)ΔX22Y22=ΔX22Y22+X22ΔY22Proof.
(i) For anyX11∈A and X22∈B, since X11X22P2=0∈Ω, taking X=X11,Y=X22,andZ=P2 in equation (6), it follows from Lemma 2 that(15)0=ΔX11X22=ΔX11X22P2+X22X11P2=ΔX11X22P2+X11ΔX22P2+X11X22ΔP2+ΔX22X11P2+X22ΔX11P2+X22X11ΔP2=ΔX11X22+X11ΔX22.
(ii) For anyX11∈A and X12∈M, since X12X11P2=0∈Ω, taking X=X12,Y=X11,andZ=P2 in equation (6), we can obtain from ΔP2∈M and Lemma 2 that(16)ΔX11X12=ΔX12X11P2+X11X12P2=ΔX12X11P2+X12ΔX11P2+X12X11ΔP2+ΔX11X12P2+X11ΔX12P2+X11X12ΔP2=ΔX11X12+X11ΔX12.
Similarly, we can show (iii) holds.
(iv) For anyX11,Y11∈AandZ12∈M, by Lemma 2 (ii), on the one hand, we get that(17)ΔX11Y11Z12=ΔX11Y11Z12=ΔX11Y11Z12+X11Y11ΔZ12.
On the other hand,(18)ΔX11Y11Z12=ΔX11Y11Z12=ΔX11Y11Z12+X11ΔY11Z12+X11Y11ΔZ12.
Comparing the above two equations, we obtain(19)ΔX11Y11−ΔX11Y11−X11ΔY11Z12=0.
This yields from the faithfulness ofM and Lemma 2 that(20)P1ΔX11Y11P1=P1ΔX11Y11+X11ΔY11P1=ΔX11Y11+X11ΔY11P1.
Furthermore, by Lemma2 (i), we obtain that(21)P1ΔX11Y11P2=X11Y11ΔP1=X11ΔY11P2.
Therefore, by the above two equations andP2ΔX11Y11P2=0, we get ΔX11Y11=ΔX11Y11+X11ΔY11. Similarly, we can show (v) holds. The proof is completed.Lemma 4.
For anyX11,Y11∈A, X12,Y12∈M, and X22,Y22∈B, then
(i)ΔX11+X12=ΔX11+ΔX12
(ii)ΔX12+X22=ΔX12+ΔX22
(iii)ΔX12+Y12=ΔX12+ΔY12
(iv)ΔX11+Y11=ΔX11+ΔY11
(v)ΔX22+Y22=ΔX22+ΔY22Proof.
(i) For anyX11∈A, X12,Y12∈M, and X22∈B, since P2X11+X12P2=0∈Ω, taking X=P2,Y=X11+X12,andZ=P2 in equation (6), then by ΔP2∈M and P1ΔX11P2=−X11ΔP2, we obtain(22)ΔX12=ΔP2X11+X12P2+X11+X12P2P2=ΔP2X11+X12P2+P2ΔX11+X12P2+P2X11+X12ΔP2+ΔX11+X12P2P2+X11+X12ΔP2P2+X11+X12P2ΔP2=P2ΔX11+X12P2+ΔX11+X12P2+X11ΔP2P2=P2ΔX11+X12P2+ΔX11+X12P2−P1ΔX11P2.
Then, it follows fromΔX12=P1ΔX12P2 and the property of 2-torsion freeness of U that(23)P2ΔX11+X12P2=0,P1ΔX11+X12P2=P1ΔX11P2+ΔX12.
Furthermore, sinceY12X11+X12P2=0∈Ω, taking X=Y12,Y=X11+X12,andZ=P2 in equation (6), then we obtain from Lemma 2, ΔP2∈M, and P2ΔX11+X12P2=0 that(24)ΔX11Y12=ΔY12X11+X12P2+X11+X12Y12P2=ΔY12X11+X12P2+Y12ΔX11+X12P2+Y12X11+X12ΔP2+ΔX11+X12Y12P2+X11+X12ΔY12P2+X11+X12Y12ΔP2=ΔX11+X12Y12+X11ΔY12.
This yields from Lemma3 (ii) that ΔX11+X12−ΔX11Y12=0, and then, by the faithfulness of M, we get that(25)P1ΔX11+X12P1=P1ΔX11P1.
Hence, by equations (23)–(31) and Lemma 2, we get ΔX11+X12=ΔX11+ΔX12. Similarly, we can show (ii) holds.
(iii) For anyX12,Y12∈M, since P1+X12Y12+P2P2=X12+Y12∈Ω, taking X=P1+X12,Y=P2+Y12,andZ=P2 in equation (6), then by Lemmas 2 and 4 (i)-(ii) and ΔP1=−ΔP2∈M, we get that(26)ΔX12+Y12=ΔP1+X12Y12+P2P2+Y12+P2P1+X12P2=ΔP1+X12Y12+P2P2+P1+X12ΔY12+P2P2+P1+X12Y12+P2ΔP2+ΔY12+P2P1+X12P2+Y12+P2ΔP1+X12P2+Y12+P2P1+X12ΔP2=ΔP1+ΔX12+ΔY12+ΔP2=ΔX12+ΔY12.
(iv) For anyX11,Y11∈AandZ12∈M, since Z12X11+Y11P2=0∈Ω, taking X=Z12,Y=X11+Y11,andZ=P2 in equation (6), then it follows from ΔP2∈M that(27)ΔX11Z12+Y11Z12=ΔZ12X11+Y11P2+X11+Y11Z12P2=ΔZ12X11+Y11P2+Z12ΔX11+Y11P2+Z12X11+Y11ΔP2+ΔX11+Y11Z12P2+X11+Y11ΔZ12P2+X11+Y11Z12ΔP2=ΔX11+Y11Z12+X11+Y11ΔZ12.
Therefore, this implies from Lemmas4 (iii) and 3 (ii) that ΔX11+Y11−ΔX11−ΔY11Z12=0, so by the faithfulness of M, we get that(28)P1ΔX11+Y11P1=P1ΔX11P1+P1ΔY11P1.
Furthermore, for anyX11,Y11∈A, we can get from Lemma 2 (i) that(29)P1ΔX11+Y11P2=X11+Y11ΔP1=X11ΔP1+Y11ΔP1=P1ΔX11P2+P1ΔY11P2.
Therefore, we get from above two equations andP2ΔX11+Y11P2=0 that ΔX11+Y11=ΔX11+ΔY11. Similarly, we can show (v) holds. The proof is completed.Lemma 5.
For anyX11∈A,X12∈M, and X22∈B, then ΔX11+X12+X22=ΔX11+ΔX12+ΔX22.Proof.
For anyX11∈A,X12∈M, and X22∈B, since P1X11+X12+X22P2=X12∈Ω, taking X=P1,Y=X11+X12+X22,Y=P1,andZ=P2 in equation (6), then it follows from ΔP1=−ΔP2∈M and Lemma 2 that(30)ΔX12=ΔP1X11+X12+X22P2+X11+X12+X22P1P2=ΔP1X11+X12+X22P2+P1ΔX11+X12+X22P2+P1X11+X12+X22ΔP2+ΔX11+X12+X22P1P2+X11+X12+X22ΔP1P2+X11+X12+X22P1ΔP2=ΔP1X22+P1ΔX11+X12+X22P2+X11ΔP2+X11ΔP1+X11ΔP2=P1ΔX11+X12+X22P2−P1ΔX11P2−P1ΔX22P2.
Hence, we get that(31)P1ΔX11+X12+X22P2=P1ΔX11P2+ΔX12+P1ΔX22P2.
In the following, we will show thatP1ΔX11+X12+X22P1=P1ΔX11P1 and P2ΔX11+X12+X22P2=P2ΔX22P2. Indeed, for any Z12∈M, since P1X11+X12+X22Z12=X11Z12∈Ω, taking X=P1,Y=X11+X12+X22,andZ=Z12 in equation (6), then, by Lemma 2 and ΔP1∈M, we get that(32)Δ2X11Z12=ΔP1X11+X12+X22Z12+X11+X12+X22P1Z12=ΔP1X11+X12+X22Z12+P1ΔX11+X12+X22Z12+P1X11+X12+X22ΔZ12+ΔX11+X12+X22P1Z12+X11+X12+X22ΔP1Z12+X11+X12+X22P1ΔZ12=2P1ΔX11+X12+X22Z12+2X11ΔZ12.
On the contrary, it follows from Lemma4 (iii) and Lemma 3 (ii) that(33)Δ2X11Z12=2ΔX11Z12=2ΔX11Z12+2X11ΔZ12.
Comparing the above two equations, we get that2P1ΔX11+X12+X22−ΔX11P1Z12=0, and then, by the faithfulness of M and the property of 2-torsion free of U, we have(34)P1ΔX11+X12+X22P1=P1ΔX11P1.
Similarly, we can obtain that(35)P2ΔX11+X12+X22P2=P2ΔX22P2.
Therefore, by equations (31)–(35), we get ΔX11+X12+X22=ΔX11+ΔX12+ΔX22. The proof is completed.
In the following, we give the completed proof of Theorem1.Proof of Theorem1.
It follows from Lemmas4 and 5 that Δ is an additive mapping on U. Next, we show that Δ is an derivation on U. Let X=X11+X12+X22 and Y=Y11+Y12+Y22 be arbitrary elements of U, where X11,Y11∈A,X12,Y12∈M, and X22,Y22∈B. Since Δ is an additive mapping on U, then it follows from Lemmas 1–3 that(36)ΔXY=ΔX11Y11+X11Y12+X12Y22+X22Y22=ΔX11Y11+ΔX11Y12+ΔX12Y22+ΔX22Y22=ΔX11Y11+X11ΔY11+ΔX11Y12+X11ΔY12+ΔX12Y22+X12ΔY22+ΔX22Y22+X22ΔY22=ΔX11+X12+X22Y11+Y12+Y22+X11+X12+X22ΔY11+Y12+Y22=ΔXY+XΔY.
Therefore,Δ is an additive derivation on U. The proof is completed.
Next, we give an application of Theorem1 to certain special classes of triangular algebras, such as block upper triangular matrix algebras and nest algebras.
LetR be a commutative ring with identity, and let Mn×kR be the set of all n×k matrices over R. For n≥2 and m≤n, the block upper triangular matrix algebra Tnk¯R is a subalgebra of MnR with the form(37)Mk1RMk1×k2R⋯Mk1×kmR0Mk2R⋯Mk2×kmR⋮⋮⋯vdots00⋯MkmR,where k¯=k1,k2,…,km is an ordered m-vector of positive integer such that k1+k2+⋯+km=n.
A nest of a complex Hilbert spaceH is a chain N of closed subspaces of H containing 0 and H which is closed under arbitrary intersections and closed linear span. The nest algebra associated to N is the algebra:(38)AlgN=T∈BH:TN⊆Nfor allN∈N.
A nestN is called trivial if N=0,H. It is clear that every nontrivial nest algebra is a triangular algebra and every finite dimensional nest algebra is isomorphic to a complex block upper triangular matrix algebra.Corollary 1.
LetTnk¯R be a 2-torsion-free block upper triangular matrix algebra and Δ be a nonlinear nonglobal semi-Jordan triple derivable mapping on Tnk¯R. Then, Δ is an additive derivation.Corollary 2.
LetN be a nontrivial nest of a complex Hilbert space H, AlgN be a nest algebra, and Δ be a nonlinear nonglobal semi-Jordan triple derivable mapping on AlgN. Then, Δ is an additive derivation.



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