# Characterizing Jordan Derivable Maps on Triangular Rings by Local Actions

Authors: Hoger Ghahramani; Mohammad Nader Ghosseiri; Tahereh Rezaei
Journal: Journal of Mathematics
 (2022)

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## Abstract

Suppose thatT=TriA,ℳ,ℬ is a 2-torsion free triangular ring, and S=A,B|AB=0,A,B∈T∪A,X|A∈T,X∈P,Q, where P is the standard idempotent of T and Q=I−P. Let δ:T⟶T be a mapping (not necessarily additive) satisfying, A,B∈S⇒δA∘B=A∘δB+δA∘B, where A∘B=AB+BA is the Jordan product of T. We obtain various equivalent conditions for δ, specifically, we show that δ is an additive derivation. Our result generalizes various results in these directions for triangular rings. As an application, δ on nest algebras are determined.


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## Body

## 1. Introduction

Letℛ be a ring and δ:ℛ⟶ℛ be a mapping (not necessarily additive). δ is called a derivable map if δab=aδb+δab for all a,b∈ℛ. Moreover, δ is called a Jordan derivable map if δa∘b=a∘δb+δa∘b for all a,b∈ℛ, where a∘b=ab+ba is the Jordan product of ℛ. An additive derivable mapping δ is called additive derivation. If δ is an additive Jordan derivable mapping, then it is called an additive Jordan derivation. These defined maps are important classes of maps on the rings and there has been many studies on them from different directions, and here we mention some of these study routes which is interesting for us.One of the interesting issues is the study of relationship between the additive and multiplicative structure of maps on the rings. In this line of investigation, first Martinadle [1] considered some conditions on a ring ℛ, proved that any multiplicative bijection map of ℛ is additive. Then the question, what maps on a ring ℛ are automatically additive was considered and different results were obtained in this line, we refer the reader to [2, 3] and references therein for more details. Especially, it has been proved on special rings that every derivable map or Jordan derivable map is additive, for instance, see [4–6].Obviously, any additive derivation is an additive Jordan derivation, but the converse may not hold in general (see [7]). Another interesting study routes on derivations and Jordan derivations is: on what rings (algebras) is any (linear) additive Jordan derivation is (linear) additive derivation? The first result in this way was obtained by Herstein [8], which proved that on 2-torsion-free prime rings, any additive Jordan derivation is an additive derivation. Later in [9], this result was generalized for 2-torsion free semiprime rings, and after that, this result was proved for other various rings (algebras) or the structure of additive (linear) Jordan derivations on some rings characterized in terms of additive (linear) derivations (see [7, 10–13] and references therein).Another way to study derivable maps (additive derivations) and Jordan derivable maps (additive Jordan derivations) is to study them according to local conditions. One of these local conditions is studying the maps which operate on special pairs of elements of a ringℛ like special maps. More precisely, assume that S⊆ℛ×ℛ, in this line of investigation, considering those maps that for all a,b∈S, operate like derivable maps (additive derivations) or Jordan derivable maps (additive Jordan derivations). First Brešar in [14] proved that if δ is an additive map on a unital ring ℛ, that contains a nontrivial idempotent and aδb+δab=0 for all a,b∈ℛ with ab=0, then δa=τa+ca, where τ is an additive derivation on ℛ and c belongs to the center of ℛ. Following this line of investigation, derivations and Jordan derivations (additive or nonadditive) at zero products or another special pairs of several rings or algebras has been studied and considerable results has been achieved. For instance, see [12, 15–21] and the references therein.In the research lines mentioned above, some considerable results on prime rings or semiprime rings have been achieved. Of course this study routes have been established on some non-semiprime rings or operator algebras (especially nest algebras), of which we can mention triangular rings as one of the most important ones. In the following we introduce this ring and hint at some results on it. LetA and ℬ be unital rings and ℳ be a unital A,ℬ-bimodule, which is faithful as a left A-module and also as a right ℬ-module. The triangular ring Tri A,ℳ,ℬ is as follows:(1)TriA,ℳ,ℬ=am0b:a∈A,m∈ℳ,b∈ℬ,under the usual matrix operations. Tri A,ℳ,ℬ is a unital ring with identity I=1A001ℬ, where 1A, 1ℬ are identities of A and ℬ, respectively. This ring contains important class of rings like upper triangular block matrices over a unital ring ℛ, especially the ring of upper triangular matrices over a unital ring ℛ, some nest algebras on Banach spaces, especially nest algebras on Hilbert spaces. Note that, if A and ℬ be unital algebras over a commutative ring C and ℳ be a unital faithful A,ℬ-bimodule, then Tri A,ℳ,ℬ is an algebra over C. Zhang in [22] proved that any linear Jordan derivation on a 2-torsion free triangular algebra Tri A,ℳ,ℬ is a linear derivation and in [18] this result is obtained for additive Jordan derivations on 2-torsion free triangular rings. In [23] has been shown that any Jordan drivable map on a 2-torsion free triangular algebra is an additive derivation, which is a generalization of result of [22]. In [24] it has been proved that a linear map δ on Tnℂ (all n×n upper triangular matrices over the complex field ℂ), satisfying the following equation:(2)A,B∈Tnℂ,AB=0⇒δA∘B=δA∘B+A∘δB,is a linear derivation. In [21] it has been shown that a mapping δ (not necessarily additive) on TnF (F is a field and n≥3) satisfying the following equation:(3)A,B∈TnF,AB=0⇒δAB=δAB+AδB,is an additive derivation. Let T=TriA,ℳ,ℬ be a triangular ring and consider the subset S of T×T as follows:(4)S=A,B|AB=0,A,B∈T∪A,X|A∈T,X∈P,Q,whereP=1A000 is the standard idempotent in T and Q=I−P. In this paper we consider a mapping δ (not necessarily additive) on T which satisfies the following condition:(5)A,B∈S⇒δA∘B=A∘δB+δA∘B,and prove that if T is 2-torsion free, then δ is an additive derivation. Note that if the mapping δ on T is derivable, Jordan derivable, additive Jordan derivation or δ is an additive map on T satisfying(6)A,B∈T,AB=0⇒δA∘B=A∘δB+δA∘B,then δ satisfies (49) (see proof of Theorem 1). So our result generalizes various results in these directions for triangular rings, especially each of the results of [21, 23], Theorem 4.2 (for G=0), [22]. Next theorem is the main result of our paper.Theorem 1.
Suppose thatT=TriA,ℳ,ℬ is a 2-torsion free triangular ring and δ:T⟶T is a mapping (not necessarily additive). Let S⊆T×T be as follows:(7)S=A,B|AB=0,A,B∈T∪A,X|A∈T,X∈P,Q,where P∈T is the standard idempotent and Q=I−P. Then the following are equivalent:(i)
A,B∈S⇒δA∘B=A∘δB+δA∘B;(ii)
A,B∈T⇒δA∘B=A∘δB+δA∘B;(iii)
δ is additive and A,B∈T,AB=0⇒δA°B=A∘δB+δA∘B;(iv)
δ is an additive Jordan derivation;(v)
δ is a derivable map;(vi)
δ is an additive derivation.The proof of this theorem will be given in Section3. In the above theorem, we have considered the 2-torsion free condition. The necessity of this condition can be a question of interest.Letx be a fixed element of the ring ℛ. The mapping Ix:ℛ⟶ℛ defined by Ixa=ax−xaa∈ℛ is an additive derivation which is called inner derivation. On nest algebras, derivations can be characterized in terms of inner derivations. According to this point, Theorem 1 can be obtained in more specific on nest algebras. We present this result in Section 2 as an application of Theorem 1 on nest algebras.


## 2. Application to Nest Algebras

LetX be a (real or complex) Banach space, let BX be the Banach algebra of all bounded linear operators on X. A nest N on X is a chain of closed (under norm topology) subspaces of X with 0 and X in N such that for every family Nα of elements of N, both ∩Nα and ∨Nα (closed linear span of Nα) belong to N. The nest algebra associated to the nest N, denoted by alg N is as follows:(8)algN=T∈ℬX:TN⊆Nfor anyN∈N.We say thatN is nontrivial whenever N≠0,X. If N is trivial, then alg N=ℬX.Remark 1.
LetN is a nontrivial nest on X and N∈N with N≠0 and N≠X, is complemented. Then there exists an idempotent P∈algN such that ranP=N, and the nest algebra alg N has a representation as the following triangular algebra.(9)algN=TriPalgNP,PalgNI−P,I−PalgNI−P,where I is the identity operator. PalgNP and I−PalgNI−P are unital algebras with unities P and I−P, respectively, and PalgNI−P is a faithful unital PalgNP,I−PalgNI−P-bimodule.Since any closed (under norm topology) subspace of a Hilbert space is complemented,it follows that for any nontrivial nestN on a Hilbert space ℋ, each N∈N with N≠0 and N≠ℋ, is complemented. Thus, every nontrivial nest algebra on a Hilbert space satisfies the conclusion in Remark 1.We have the following result on nest algebras.Theorem 2.
LetN be a nest on a Banach space X, and there exists a nontrivial element N in N which is complemented in X. Suppose that δ:algN⟶algN is a mapping (not necessarily additive), and S⊆algN×algN is as follows:(10)S=A,B|AB=0,A,B∈algN∪A,X|A∈algN,X∈P,Q,where P∈algN is the idempotent with ranP=N and Q=I−P. Then the following are equivalent:(i)
A,B∈S⇒δA∘B=A∘δB+δA∘B;(ii)
A,B∈S⇒δA∘B=A∘δB+δA∘B;(iii)
δ is additive and A,B∈algN,AB=0⇒δA∘B=A∘δB+δA∘B;(iv)
δ is an additive Jordan derivation;(v)
δ is a derivable map;(vi)
δ is an additive derivation.Suppose, further, thatX is an infinite-dimensional Banach space. Then the above conditions are also equivalent to:(vii)
δ is an inner derivation.Proof.
From Remark 1, algN is a triangular algebra, and all the assumptions of Theorem 1 hold. So all cases (i) to (vi) are equal. If X is an infinite-dimensional Banach space, then by [25] every additive derivation of alg N is linear. From [26], Theorem 2 any linear derivation of a nest algebra on a Banach space is continuous and by [27] all continuous linear derivations of a nest algebra on a Banach space are inner derivations (see also [28], Theorem 2.3). Given this, it is proved that if X is an infinite-dimensional Banach space, condition (vii) is equivalent to condition (vi). The proof is complete.By Theorem2, we have the following corollary.Corollary 1.
LetN be a nontrivial nest on a Hilbert space ℋ. Suppose that δ:algN⟶algN is a mapping (not necessarily additive), and S⊆algN×algN is as follows(11)S=A,B|AB=0,A,B∈algN∪A,X|A∈algN,X∈P,Q,where P∈algN is the is the orthogonal projection on a nontrivial element N∈N, and Q=I−P. Then the following are equivalent:(i)
A,B∈S⇒δA∘B=A∘δB+δA∘B;(ii)
A,B∈algN⇒δA∘B=A∘δB+δA∘B;(iii)
δ is additive and A,B∈algN,AB=0⇒δA∘B=A∘δB+δA∘B;(iv)
δ is an additive Jordan derivation;(v)
δ is a derivable map;(vi)
δ is an additive derivation.Suppose, further, thatℋ is an infinite-dimensional Banach space. Then the above conditions are also equivalent to:(vii)
δ is an inner derivation.Note that ifℋ is a Hilbert space, and dimℋ<∞, then there exist additive derivations of the nest algebra which are not inner (see [29]).


## 3. Proof of Theorem1

In this section weassume thatT=TriA,ℳ,ℬ is a 2-torsion free triangular ring, and P=1A000 is the standard idempotent of T and Q=I−P=0001B which is also an idempotent. Also we put(12)T11=a000|a∈A,T12=0m00|m∈ℳ,T22=000b|b∈ℬ.It is obvious thatT=T11⊕T12⊕T22 and for any A∈T we have(13)A=PAP+PAQ+QAQ,QAP=0.Proof of Theorem1:The following statements are clear:vi⇒i, vi⇒ii, vi⇒iii, vi⇒iv, vi⇒v, ii⇒i and iv⇒i. We just prove the next items and so that the proof is complete.(iii)⇒(i): It is enough to prove that for all A∈T we have(14)δA∘P=A∘δP+δA∘P,δA∘Q=A∘δQ+δA∘Q.Define the additive mappingτ:T⟶T by τA=δA−IδPA. Given that IδP is an additive derivation, it is easily checked that(15)A,B∈T,AB=0⇒τA∘B=A∘τB+τA∘B.Also,PτPQ=0.SincePQ=QP=0, thus(16)0=τP∘Q=τP∘Q+P∘τQ.Multiplying both sides of (16) by Q, we arrive at 2QτPQ=0, and hence QτPQ=0. Since PAQP=0 for all A∈T, it follows that(17)τPAQ=τPAQ∘P=τPAQ∘P+PAQ∘τP.By (17) and the fact that QτPQ=0 we get(18)τPAQ=τPPAQ+P∘τPAQ.Multiplying both sides of (18) by P, we conclude that PτPAQP=0. Now multiplying (18) from the left by P, it yields PτPPAQ=0 for all A∈T. Since M is faithful as a left A-module, we conclude that PτPP=0. By using the results obtained, τP=0. In view of (16) and the fact that τP=0, we get P∘τQ=0, thus τQ=QτQQ. Given that QPAQ=0 and with the same argument above, we can prove that QτQQ=0 and hence τQ=0.For allA∈T we have PAPQ=QPAP=0, and hence(19)0=τPAP∘Q=τPAP∘Q+PAP∘τQ.It follows fromτQ=0 that τPAP∘Q=0 for all A∈T. Thus(20)τPAP∘P+PAP∘τP=τPAP∘P=τPAP∘P+τPAP∘Q=2τPAP=τPAP∘P,for all A∈T. Now according to the results we have(21)τA∘P=τPAP∘P+τPAQ∘P+τQAQ∘P=τPAP∘P+PAP∘τP+τPAQ∘P+PAQ∘τP+τQAQ∘P+QAQ∘τP=τA∘P+A∘τP,for all A∈T. By a similar argument as given above, we can prove that(22)τA∘Q=τA∘Q+A∘τQ,for all A∈T. Since IδP is an additive derivation, the condition i is obtained for δ.(v)⇒(i): Since δ is derivable, so(23)δ0=δ00+0δ0,and therefore for all A,B∈T with AB=0, we have(24)0=δAB=δAB+AδB,δBA=δBA+BδA.Thus(25)δA∘B=δBA=δAB+δBA=δA∘B+A∘δB.Defineτ:T⟶T by τA=δA−IδPA. So that τ is a derivable map and PτPQ=0. It follows from τP=τP2=τPP+PτP that PτPP=0 and QτPQ=0. Hence τP=0. Since τ is derivable, then τ0=0 and τI=0. So τQ=0.For allA∈T, we have(26)τPAQ=τPPAQ+PτPAQ=PτPAQ,so QτPAQQ=0. Also(27)τPAQ=τPAQQ+PAQτQ=τPAQQ,thus PτPAQP=0. Therefore(28)τPAQ=PτPAQQ,for all A∈T. Further(29)τPAQ=τP+PAQQ=τP+PAQQ+P+PAQτQ=τP+PAQQ,0=τP=τP+PAQP=τP+PAQP+P+PAQτP=τP+PAQP.By adding two recent statements, we arrive at the following equation:(30)τP+PAQ=τPAQ,for all A∈T.So for allA,B∈T, we have(31)τPAQ+PBQ=τP+PAQPBQ+Q=τP+PAQPBQ+Q+P+PAQτPBQ+Q=τPAQPBQ+Q+P+PAQτPBQ.From the above equality and (28) we find that(32)τPAQ+PBQ=τPAQ+τPBQ,for all A,B∈T.We have(33)0=τPAPQ=τPAPQ+PAPτQ=τPAPQ,so that(34)τPAP=PτPAPP,for all A∈T.It follows from (32) that(35)τPAP+PBPPCQ=τPAPCQ+PBPCQ=τPAPCQ+τPBPCQ=τPAPPCQ+PAPτPCQ+τPBPPCQ+PBPτPCQ.On the other hand(36)τPAP+PBPPCQ=τPAP+PBPPCQ+PAP+PBPτPCQ.Comparing recent two statements we get(37)τPAP+PBP−τPAP−τPBPPCQ=0,for all A,B,C∈T. From the above statement, (34) and faithfulness of ℳ as a left A-module, we conclude that(38)τPAP+PBP=τPAP+τPBP,for all A,B∈T.By (38), we have(39)τA∘PP−τAPP−τPAP=τA∘PP−A∘PτP−τAP+APτP−τPAP+PAτP=τ2PAP−τPAP−τPAP=0,for all A∈T. Beside(40)τA∘PQ−τAPQ−τPAQ=τA∘PQ−A∘PτQ+APτQ−τPAQ+PAτQ=0.By adding two recent statements, we obtain(41)τA∘P=τAP+τPA,for all A∈T. Now by the above identity and the fact that τ is a derivable map, we arrive at(42)τA∘P=τA∘P+A∘τP,for all A∈T. According to this statement and that IδP is a derivation, we conclude that(43)δA∘P=δA∘P+A∘δP,for all A∈T. By the simillar argument as above, we can prove that(44)δA∘Q=δA∘Q+A∘δQ,for all A∈T. Therefore δ satisfies (i).(i)⇒(vi): Define the mapping τ:T⟶T by τA=δA−IδPA and easily can be seen that τ satisfies(45)A,B∈S⇒τA∘B=τA∘B+A∘τB.AlsoPτPQ=0.We prove thatτ is an additive derivation through the following steps.Step 1.
τ0=0.Proof.
Since0∘0=0 then(46)τ0=τ0∘0+τ0∘0=0.Step 2.
For allA∈T(47)τPAQ=PτPAQQ.Proof.
For anyA∈T we have PAQ,P∈S. So that(48)τPAQ=τPAQ∘P=τPAQ∘P+PAQ∘τP=τPAQP+PτPAQ+PAQτP+τPPAQ.
If we multiplying both sides of (48) by P, we arrive at PτPAQP=0 and if we multiplying both sides of (48) by Q, we obtain QτPAQQ=0. Therefore τPAQ=PτPAQQ for all A∈T.Step 3.
τP=0 and τQ=0.Proof.
SinceQ,P∈S, it follows that(49)0=τQ∘P=τQ∘P+Q∘τP.
So(50)τQP+PτQ+QτP+τPQ=0.
By multiplying both sides of (50) by Q, we have QτPQ=0. Now multiplying (50) from the left by P and from the right by Q, we see that PτPPAQ=0 for all A∈T. Faithfulness of ℳ implies that PτPP=0. By the results obtained and that PτPQ=0, we conclude that τP=0. From this result and (50), we have τQP+PτQ=0 and so PτQP=0 and PτQQ=0. Since for all A∈T, Q,PAQ∈S, so we have(51)τPAQ=τQ∘PAQ=τQ∘PAQ+Q∘τPAQ.
Multiplying both sides of the above identity, we arrive atPAQτQQ=0, for all A∈T. By faithfulness of ℳ, we have QτQQ=0, so τQ=0.Step 4.
For allA∈T(52)τPAP=PτPAPP,τQAQ=QτQAQQ.Proof.
SincePAP,Q∈S for all A∈T, from Step 3, we have(53)0=τPAP∘Q=τPAP∘Q+PAP∘τQ=τPAPQ+QτPAP.
HenceQτPAPQ=0 and PτPAPQ=0. So(54)τPAP=PτPAPP,for all A∈T. Given that QAQ,P∈S for all A∈T, using Step 3 and same argument as above, we can prove that(55)τQAQ=QτQAQQ,for all A∈T.Step 5.
For allA,B∈T(56)QτPAP+PBQQ=0,PτPAQ+QBQP=0.Proof.
SinceQ,PAP+PBQ∈S and PAQ+QBQ,P∈S, for all A,B∈T, from Step 3 we obtain(57)τPBQ=τQ∘PAP+PBQ=QτPAP+PBQ+τPAP+PBQQ,τPAQ=τPAQ+QBQ∘P=τPAQ+QBQP+PτPAQ+QBQ.
In the above, multiplying both sides of first statement byQ and multiplying both sides of second statement by P and according to Step 2, the desired result is obtained.Step 6.
For allA,B∈T(58)τPAPBQ=τPAPPBQ+PAPτPBQ,τPAQBQ=τPAQQBQ+PAQτQBQ.Proof.
SincePBQ,PAP∈S for all A,B∈T, so that(59)τPAPBQ=τPBQ∘PAP=τPBQPAP+PAPτPBQ+PBQτPAP+τPAPPBQ.
From Steps2 and 4, we find that τPBQP=0 and QτPAP=0, hence(60)τPAPBQ=τPAPPBQ+PAPτPBQ,for all A,B∈T.
SinceQBQ,PAQ∈S for all A,B∈T, from Steps 2, 4 and similar argument as above, we conclude that(61)τPAQBQ=τPAQQBQ+PAQτQBQ,for all A,B∈T.Step 7.
For allA,B∈T(62)τPAPBP=τPAPPBP+PAPτPBP,τQAQBQ=τQAQQBQ+QAQτQBQ.Proof.
According to Step6, for all A,B∈T we have(63)τPAPBPCQ=τPAPBPPCQ+PAPBPτPCQ.
On the other hand(64)τPAPBPCQ=PAPτPBPCQ+τPAPPBPCQ=PAPτPBPPCQ+PAPBPτPCQ+τPAPPBPCQ.
Comparing above statements and according to Step4, we get(65)PτPAPBP−τPAPPBP−PAPτPBPPCQ=0.
Now by the faithfulness ofℳ, we have(66)PτPAPBP−τPAPPBP−PAPτPBPP=0.
So(67)τPAPBP−τPAPPBP−PAPτPBPP=0,for all A,B∈T. It follows from Step 4 that(68)τPAPBP−τPAPPBP−PAPτPBPQ=0,for all A,B∈T. By adding two recent statements we arrive at(69)τPAPBP=τPAPPBP+PAPτPBP,for all A,B∈T. Using Steps 4, 6 and similar arguments as above, we can show that(70)τQAQBQ=τQAQQBQ+QAQτQBQ,for all A,B∈T.Step 8.
For allA,B∈T(71)τPAP+PBQ=τPAP+τPBQ,τPAQ+QBQ=τPAQ+τQBQ.Proof.
For allA,B∈T we have PCQ,PAP+PBQ∈S. So(72)τPCQ∘PAP+PBQ=τPCQ∘PAP+PBQ+PCQ∘τPAP+PBQ.
From Step2, we get(73)τPCQ∘PAP+PBQ=PAPτPCQ+PCQτPAP+PBQ+τPAP+PBQPCQ.
Beside by Step6, we have(74)τPCQ∘PAP+PBQ=τPAPCQ=τPAPPCQ+PAPτPCQ.
Comparing two recent statements and multiplying outcome relation from the left byP and from the right by Q, and using Step 5, we arrive at(75)PτPAP+PBQPCQ=τPAPPCQ,for all A,B∈T. By faithfulness of ℳ, we conclude that(76)PτPAP+PBQP=PτPAPP.
From the above statement, Steps4 and 5, we obtain(77)τPAP+PBQ−τPAP=PτPAP+PBQ−τPAPQ.
So that(78)τPAP+PBQ−τPAP=τPAP+PBQ−τPAP°Q=τPAP+PBQ°Q,for all A,B∈T. Since Q,PAP+PBQ∈S for all A,B∈T, we have(79)τPBQ=τQ°PAP+PBQ=τQ°PAP+PBQ+Q°τPAP+PBQ=Q°τPAP+PBQ.
Comparing the above statements, we get(80)τPAP+PBQ=τPAP+τPBQ,for all A,B∈T.
SincePAQ+QBQ,PCQ∈S and PAQ+QBQ,P∈S for all A,B∈T, by the same computation as above, we arrive at(81)τPAQ+QBQ=τPAQ+τQBQ,for all A,B∈T.Step 9.
τ on T11, T12, T22 is additive.Proof.
For allA,B∈T, we have PBQ+Q,PAQ+P∈S. So that(82)τPAQ+PBQ=τPBQ+Q°PAQ+P=τPBQ+Q°PAQ+P+PBQ+Q°τPAQ+P.
Now from Steps2, 3 and 8, we get(83)τPAQ+PBQ=τPAQ+τPBQ,for all A,B∈T. Therefore τ is additive on T12.
From Step6, we have(84)τPAP+PBPPCQ=τPAP+PBPPCQ+PAP+PBPτPCQ.
On the other hand, according to additivity ofτ on T12 and from Step 6, we have(85)τPAP+PBPPCQ=τPAPCQ+PBPCQ=τPAPCQ+τPBPCQ=τPAPPCQ+PAPτPCQ+τPBPPCQ+PBPτPCQ.
By comparing the above statements and faithfulness ofℳ, we conclude that(86)PτPAP+PBPP=PτPAPP+PτPBPP,for all A,B∈T. Now Step 4 implies that τ is additive on T11.
By the similar argument as above we conclude thatτ is also additive on T22.Step 10.
For allA∈T.(87)τA=τPAP+τPAQ+τQAQ.Proof.
For allA∈T we have A,P∈S. So that(88)τA°P=τA°P+A°τP=τA°P.
On the other hand(89)τA°P=τPAP°P+PAQ°P+QAQ°P=τPAP°P+PAQ°P.
SincePAP,P∈S, PAQ,P∈S and from Steps 3, 4 and 8, we conclude that(90)τA°P=τPAP°P+τPAQ°P=τPAP°P+τPAQ°P=τPAP°P+τPAQ°P+τQAQ°P.
Thus(91)τA−τPAP−τPAQ−τQAQ°P=0.
Now for allA∈T, since A,Q∈S and by the similar argument as above we prove that(92)τA−τPAP−τPAQ−τQAQ°Q=0.
Therefore we get the required result.Step 11.
δ is an additive derivation.Proof.
From Steps9, 10, for all A,B∈T we have(93)τA+B=τPAP+PBP+PAQ+PBQ+QAQ+QBQ=τPAP+PBP+τPAQ+PBQ+τQAQ+QBQ=τPAP+τPBP+τPAQ+τPBQ+τQAQ+τQBQ=τA+τB.
Thereforeτ is additive on T.
It follows from Steps2, 4, 6 and 7 that(94)τAB=τAB+AτB,for all A,B∈T. Hence τ is an additive derivation. Since IδP is also an additive derivation, so δ=τ+IδP is an additive derivation and the proof is complete.


### 3.1. Declarations

(i)
The authors declare that no funds, grants, or other support were received during the preparation of this manuscript.(ii)
The authors have no relevant financial or non-financial interests to disclose.


## 3.1. Declarations

(i)
The authors declare that no funds, grants, or other support were received during the preparation of this manuscript.(ii)
The authors have no relevant financial or non-financial interests to disclose.



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*Source: 9941760-2022-12-20.xml*