Review Recurring Topics

From Hearing Sounds to Recognizing Phonemes: Primary Auditory Cortex is A Truly Perceptual Language Area

  • The aim of this article is to present a systematic review about the anatomy, function, connectivity, and functional activation of the primary auditory cortex (PAC) (Brodmann areas 41/42) when involved in language paradigms. PAC activates with a plethora of diverse basic stimuli including but not limited to tones, chords, natural sounds, consonants, and speech. Nonetheless, the PAC shows specific sensitivity to speech. Damage in the PAC is associated with so-called “pure word-deafness” (“auditory verbal agnosia”). BA41, and to a lesser extent BA42, are involved in early stages of phonological processing (phoneme recognition). Phonological processing may take place in either the right or left side, but customarily the left exerts an inhibitory tone over the right, gaining dominance in function. BA41/42 are primary auditory cortices harboring complex phoneme perception functions with asymmetrical expression, making it possible to include them as core language processing areas (Wernicke’s area).

    Citation: Byron Bernal, Alfredo Ardila. From Hearing Sounds to Recognizing Phonemes: Primary Auditory Cortex is A Truly Perceptual Language Area[J]. AIMS Neuroscience, 2016, 3(4): 454-473. doi: 10.3934/Neuroscience.2016.4.454

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  • The aim of this article is to present a systematic review about the anatomy, function, connectivity, and functional activation of the primary auditory cortex (PAC) (Brodmann areas 41/42) when involved in language paradigms. PAC activates with a plethora of diverse basic stimuli including but not limited to tones, chords, natural sounds, consonants, and speech. Nonetheless, the PAC shows specific sensitivity to speech. Damage in the PAC is associated with so-called “pure word-deafness” (“auditory verbal agnosia”). BA41, and to a lesser extent BA42, are involved in early stages of phonological processing (phoneme recognition). Phonological processing may take place in either the right or left side, but customarily the left exerts an inhibitory tone over the right, gaining dominance in function. BA41/42 are primary auditory cortices harboring complex phoneme perception functions with asymmetrical expression, making it possible to include them as core language processing areas (Wernicke’s area).


    Let R be a commutative ring with identity, and A a unital algebra over R. For any X,YA, denote the Jordan product of X,Y by XY=XY+YX. An additive mapping Δ from A into itself is called a derivation (resp., anti-derivation) if Δ(XY)=Δ(X)Y+XΔ(Y) (resp., Δ(XY)=Δ(Y)X+YΔ(X)) for all X,YA. It is called a Jordan derivation if Δ(XY)=Δ(X)Y+XΔ(Y) for all X,YA. It is called a Jordan triple derivation if Δ(XYZ)=Δ(X)YZ+XΔ(Y)Z+XYΔ(Z) for all X,Y,ZA. Obviously, every derivation or anti-derivation is a Jordan derivation. However, the inverse statement is not true in general (see [1]). If a Jordan derivation or Jordan triple derivation is not a derivation, then it is said to be proper. Otherwise, it is said to be improper.

    In the past few decades, the problem of characterizing the structure of Jordan derivations and Jordan triple derivations has attracted the attention of many mathematical workers and has achieved some important research results. For example, Herstein in [2] proved that every Jordan derivation on a prime ring not of characteristic 2 is a derivation. This result was extended by Cusack in [3] and Brešar in [4] to the case of semiprime. Zhang in [5,6] showed that every Jordan derivation on a nest algebra or a 2-torsion free triangular algebra is a inner derivation or a derivation, respectively. Later, Hoger in [7] extended the result of Zhang in [6] and proved that, under certain conditions, each Jordan derivation on trivial extension algebras is a sum of a derivation and an anti-derivation. In addition, there have been many research results on Jordan triple derivations, as shown in references [8,9,10,11].

    Definition 1.1. Let R be a commutative ring with identity, A a unital algebra over R, N0 be the set of all nonnegative integers, and D={dn}nN0 be a family of additive maps on A such that d0=idA (the identity map on A). D is said to be:

    (i) a higher derivation if for each nN0,

    dn(XY)=p+q=ndp(X)dq(Y)

    for all X,YA;

    (ii) a higher anti-derivation if for each nN0,

    dn(XY)=p+q=ndp(Y)dq(X)

    for all X,YA;

    (iii) a higher Jordan derivation if for each nN0,

    dn(XY)=p+q=ndp(X)dq(Y)

    for all X,YA;

    (iv) a higher Jordan triple derivation if for each nN0,

    dn(XYZ)=p+q+r=ndp(X)dq(Y)dr(Z)

    for all X,Y,ZA.

    If a higher Jordan derivation or a higher Jordan triple derivation is not a higher derivation, then it is said to be proper. Otherwise, it is said to be improper. With the deepening of research on this topic, many research achievements have been obtained about higher Jordan derivations and higher Jordan triple derivations. For example, Xiao and Wei in [12] proved that every higher Jordan derivation on triangular algebras is a higher derivation; Fu, Xiao, and Du in [13] extended this conclusion, and proved that every nonlinear higher Jordan derivation on triangular algebras is a higher derivation. Later, Vishki, Mirzavaziri, and Moafian in [14] proved that, under certain conditions, every higher Jordan derivation on trivial extension algebras is a higher derivation, and this conclusion further extended the works of the authors of references [12,13]. Salih and Haetinger in [15] proved that, under certain conditions, every higher Jordan triple derivation on prime rings is a higher derivation. Ashraf and Jabeenin [16] proved that every nonlinear higher Jordan triple derivable mapping on triangular algebras is a higher derivation.

    In this paper, we are interested in describing the form of higher Jordan triple derivation on trivial extension algebras. As a main result, we give conditions under which each higher Jordan triple derivation on trivial extension algebras is a sum of a higher derivation and a higher anti-derivation. This result extends the study of Jordan derivation on trivial extension algebras [7], Jordan triple derivations on -type trivial extension algebras [17], and Jordan higher derivations on trivial extension algebras [14].

    Let R be a commutative ring with identity, A a unital algebra over R and M be an A-bimodule. Then the direct product AM together with the pairwise addition, scalar product, and algebra multiplication defined by

    (a,m)(b,n)=(ab,an+mb)(a,bA,m,nM)

    is an R-algebra with a unity (1,0) denoted by

    T=AM={(a,m):aA,mM}

    and T is called a trivial extension algebra.

    An important example of trivial extension algebra is the triangular algebra which was introduced by Cheung in [18]. Let A and B be unital algebras over a commutative ring R, and M be a unital (A,B)-bimodule, which is faithful as both a left A-module and a right B-module. Then, the R-algebra

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

    under the usual matrix operations is called a triangular algebra. Basic examples of triangular algebras are upper triangular matrix algebras and nest algebras.

    It is well-known that every triangular algebra can be viewed as a trivial extension algebra. Indeed, denote by AB the direct product as an R-algebra, and then M is viewed as an AB-bimodule with the module action given by (a,b)m=am and m(a,b)=mb for all (a,b)AB and mM. Then triangular algebra U is isomorphic to trivial extensions algebra T=(AB)M. However, a trivial extension algebra is not necessarily a triangular algebra. For more details about trivial extension algebras, we refer the readers to [19,20,21].

    The following notations will be used in our paper: Let R be a commutative ring with identity, A a unital algebra over R, M an A-bimodule, T=AM be a 2-torsion free trivial extension algebra (i.e., for any XT, 2X={0} implies X=0), and denote by 1 and 0 are the unity and zero of T=AM, respectively.

    We say T=AM is a -type trivial extension algebra if A has a non-trivial idempotent element e and f=1e such that

    (i) eMf=M;

    (ii) exeM={0} implies exe=0,xA;

    (iii) Mfxf={0} implies fxf=0,xA;

    (iv) exfye=0=fxeyf=0,x,yA.

    For convenience, in the following we let P1=(e,0), P2=(f,0), and

    Tij=PiTPj( 1ij2).

    It is not hard to see that the trivial extension algebra T may be represented as

    T=P1TP1+P1TP2+P2TP1+P2TP2=T11+T12+T21+T22.

    Then every element AT may be represented as A=A11+A12+A21+A22, where AijTij(1ij2). In the following, we give a property of -type trivial extension algebras (see Lemma 1.1).

    Lemma 1.1. [17] Let T be a -type trivial extension algebra and 1ij2. Then,

    (i) for any A11T11, if A11T12=0, then A11=0;

    (ii) for any A22T22, if T12A22=0, then A22=0;

    (iii) AijBji=AiiBji=AijBii=0, Aii,BiiTii,AijTij,BjiTji.

    For ease of reading, we provide the main conclusions of reference [17] as follows:

    Theorem 1.1. [17] Let T=AM be a 2-torsion free -type trivial extension algebra and Δ be a Jordan triple derivation on T. Then, there exists a derivation D and an anti-derivation φ on T, respectively, such that

    Δ(A)=D(A)+φ(A)

    for all AT.

    The main result of this paper is the following theorem:

    Theorem 2.1. Let T=AM be a 2-torsion free -type trivial extension algebra, and D={dn}nN0 be a higher Jordan triple derivation on T. Then, there exists a higher derivation G={gn}nN0 and a higher anti-derivation F={fn}nN0 on T, respectively, such that

    dn(X)=gn(X)+fn(X)

    for any n1 and XT.

    In order to prove Theorem 2.1, we shall establish Theorems 2.2 and 2.3 in the following. We assume that T is a -type trivial extension algebra, N0 is the set of all nonnegative integers, and D={dn}nN0 is a higher Jordan triple derivation on T.

    In [17], it is proved that if d1 is a Jordan triple derivation on T, then for all AijTij (1i,j2), d1 satisfies the following properties (L):

    (i) d1(P1)=d1(P2);

    (ii) d1(P1)=P1d1(P1)P2+P2d1(P1)P1 and d1(P2)=P1d1(P2)P2+P2d1(P2)P1;

    (iii) P2d1(A11)P2=0, P1d1(A11)P2=A11d1(P1) and P2d1(A11)P1=d1(P1)A11;

    (iv) P1d1(A22)P1=0,P1d1(A22)P2=d1(P2)A22 and P2d1(A22)P1=A22d1(P2);

    (v) d1(A12)=P1d1(A12)P2+P2d1(A12)P1 and d1(A21)=P1d1(A21)P2+P2d1(A21)P1;

    (vi) d1(P1)d1(P2)=d1(P1)d1(A12)=d1(P1)d1(A21)=d1(P2)d1(A12)=d1(P2)d1(A21)=0;

    (vii) d1(A12)d1(A12)=d1(A21)d1(A21)=d1(A12)d1(A21)=0.

    Now, for all AijTij (1i,j2), we assume that dk (1k<n) satisfy the properties L. In the following, we show that dn satisfies the properties L.

    Lemma 2.1. Let D={dn}nN0 be a higher Jordan triple derivation on T. Then, for each n1, and for any A11T11,A12T12,A21T21, A22T22,

    (i) dn(P1)=P1dn(P1)P2+P2dn(P1)P1 and dn(P2)=P1dn(P2)P2+P2dn(P2)P1;

    (ii) dn(P1)=dn(P2);

    (iii) P2dn(A11)P2=0, P1dn(A11)P2=A11dn(P1) and P2dn(A11)P1=dn(P1)A11;

    (iv) P1dn(A22)P1=0,P1dn(A22)P2=dn(P2)A22 and P2dn(A22)P1=A22dn(P2);

    (v) dn(A12)=P1dn(A12)P2+P2dn(A12)P1 and dn(A21)=P1dn(A21)P2+P2dn(A21)P1;

    (vi) dn(P1)dn(P2)=dn(P1)dn(A12)=dn(P1)dn(A21)=dn(P2)dn(A12)=dn(P2)dn(A21)=0;

    (vii) dn(A12)dn(A12)=dn(A21)dn(A21)=dn(A12)dn(A21)=0.

    Proof. (i) For each n1 and for any X,Y,ZT, by the definition of D={dn}nN0, we get

    dn(XYZ)=p+q+r=ndp(X)dq(Y)dr(Z). (2.1)

    Taking X=Y=Z=P1 in Eq (2.1), we assume that dk (1k<n) satisfy the properties L, and then it follows from Lemma 1.1 (iii) that

    4dn(P1)=p+q+r=ndp(P1)dq(P1)dr(P1)=p+q+r=n,1p,q,rdp(P1)dq(P1)dr(P1)+q+r=n,1q,rP1dq(P1)dr(P1)+p+r=n,1p,rdp(P1)P1dr(P1)+p+q=n,1p,qdp(P1)dq(P1)P1+dn(P1)P1P1+P1dn(P1)P1+P1P1dn(P1)=dn(P1)P1P1+P1dn(P1)P1+P1P1dn(P1)=4P1dn(P1)P1+4P1dn(P1)+4dn(P1)P1.

    This yields from the 2-torsion freeness of T that

    P1dn(P1)P1=P2dn(P1)P2=0.

    Similarly, we get that

    P1dn(P2)P1=P2dn(P2)P2=0.

    Therefore, dn(P1)=P1dn(P1)P2+P2dn(P1)P1 and dn(P2)=P1dn(P2)P2+P2dn(P2)P1.

    (ii) For each n1, taking X=P1,Y=P2,Z=P1 in Eq (2.1), we assume that dk (1k<n) satisfy the properties L, then by Lemma 1.1 (iii) and Lemma 2.1 (i), we get that

    0=p+q+r=ndp(P1)dq(P2)dr(P1)=p+q+r=n,1p,q,rdp(P1)dq(P2)dr(P1)+q+r=n,1q,rP1dq(P2)dr(P1)+p+r=n,1p,rdp(P1)P2dr(P1)+p+q=n,1p,qdp(P1)dq(P2)P1+dn(P1)P2P1+P1dn(P2)P1+P1P2dn(P2)=dn(P1)P2P1+P1dn(P2)P1+P1P2dn(P2)={dn(P1)P2+P2dn(P1)}P1+{P1dn(P2)+dn(P2)P1}P1=P1dn(P1)P2+P2dn(P1)P1+P1dn(P2)+dn(P2)P1+2P1dn(P2)P1=P1dn(P1)P2+P2dn(P1)P1+P1dn(P2)P2+P2dn(P2)P1=dn(P1)+dn(P2).

    (iii)(iv) For each n1 and for any A11T11, taking X=A11,Y=Z=P2 in Eq (2.1), we assume that dk (1k<n) satisfy the properties L, and then by Lemma 1.1 (iii) and Lemma 2.1 (i,ii), we get that

    0=p+q+r=ndp(A11)dq(P2)dr(P2)=p+q+r=n,1p,q,rdp(A11)dq(P2)dr(P2)+q+r=n,1q,rA11dq(P2)dr(P2)+p+r=n,1p,rdp(A11)P2dr(P2)+p+q=n,1p,qdp(A11)dq(P2)P2+dn(A11)P2P2+A11dn(P2)P2+A11P2dn(P2)=dn(A11)P2P2+A11dn(P2)P2+A11P2dn(P2)={dn(A11)P2+P2dn(A11)}P2+{A11dn(P2)+dn(P2)A11}P2={dn(A11)P2+P2dn(A11)+2P2dn(A11)P2}+{A11dn(P2)+P2dn(P2)A11}=dn(A11)P2+P2dn(A11)+A11dn(P2)+dn(P2)A11.

    This implies that P2dn(A11)P2=0. and

    P1dn(A11)P2=A11dn(P1) and P2dn(A11)P1=dn(P1)A11.

    Similarly, for each n1 and for any A22T22, we get that P1dn(A22)P1=0, P1dn(A22)P2=dn(P2)A22 and P2dn(A22)P1=A22dn(P2).

    (v) For each n1 and for any A12T12, taking X=P1,Y=A12,Z=P2 in Eq (2.1), we assume that dk (1k<n) satisfy the properties L, and then by Lemma 1.1 (iii) and Lemma 2.1 (i,ii), we get that

    dn(A12)=p+q+r=ndp(P1)dq(A12)dr(P2)=p+q+r=n,1p,q,rdp(P1)dq(A12)dr(P2)+q+r=n,1q,rP1dq(A12)dr(P2)+p+r=n,1p,rdp(P1)A12dr(P2)+p+q=n,1p,qdp(P1)dq(A12)P2+dn(P1)A12P2+P1dn(A12)P2+P1A12dn(P2)=P1dn(A12)P2=P1dn(A12)P2+P2dn(A12)P1.

    Similarly, for each n1 and for any A21T21, we get that dn(A21)=P1dn(A21)P2+P2dn(A21)P1.

    (vi) For each n1 and for any A12T12, A21T21, by Lemma 1.1 (iii) and Lemma 2.1 (i,ii,v), we can easily check that (vi) holds. Similarly, we show (vii) holds. The proof is complete.

    Theorem 2.2. Let F={fn}nN0 be a sequence of mappings on T (with f0=ifT). For each n1 and XT, define

    fn(X)=P2dn(P1XP2)P1+P1dn(P2XP1)P2.

    Then, F is a higher anti-derivation on T.

    It is clear that fn(Aii)=0 and fn(Aij)=Pjfn(Aij)Pi for each n1, and for any AiiTii,AijTij (1ij2).

    In the following, we show that F={fn}nN0 is a higher anti-derivation, i.e., for each n1 and for any X,YT, fn satisfies fn(XY)=p+q=nfp(Y)fq(X). For this, we introduce Lemmas 2.2 and 2.3, and prove Lemmas 2.2 and 2.3.

    Lemma 2.2. Let fn:TT be defined as in Theorem 2.2. Then, for each n1 and for any Aii,BiiTii,Aij,BijTij,BjiTji,BjjTjj (1ij2),

    (i) fn(AiiBii)=p+q=nfp(Bii)fq(Aii);

    (ii) fn(AiiBjj)=p+q=nfp(Bjj)fq(Aii);

    (iii) fn(AiiBji)=p+q=nfp(Bji)fq(Aii);

    (iv) fn(AijBii)=p+q=nfp(Bii)fq(Aij);

    (v) fn(AijBij)=p+q=nfp(Bij)fq(Aij);

    (vi) fn(AijBji)=p+q=nfp(Bji)fq(Aij).

    Proof. (i) For any n1 and Aii,BiiTii (1i2), we get from fn(AiiBii)=fn(Aii)=fn(Bii)=0 that

    fn(AiiBii)=p+q=nfp(Bii)fq(Aii).

    Similarly, we can show (ii) holds.

    (iii) For each n1 and for any AiiTii,BjiTji (1ij2), on the one hand, we have fn(AiiBji)=fn(0)=0. On the other hand, it follows from fn(Aii)=0 and fn(Bji)=Pifn(Bji)Pj that

    p+q=nfp(Bji)fq(Aii)=p+q=n,1p,qfp(Bji)fq(Aii)+fn(Bji)Aii+Bjifn(Aii)=fn(Bji)Aii=(Pifn(Bji)Pj)Aii=0.

    Therefore, fn(AiiBji)=p+q=nfp(Bji)fq(Aii). Similarly, we get (iv).

    (v) For each n1 and for any Aij,BijTij (1ij2), on the one hand, we have fn(AijBij)=fn(0)=0. On the other hand, we get from fn(Bij)fn(Aij)={Pjfn(Bij)Pi}{Pjfn(Aij)Pi}=0 and Lemma 1.1 (iii) that

    p+q=nfp(Bij)fq(Aij)=p+q=n,1p,qfp(Bij)fq(Aij)+fn(Bij)Aij+Bijfn(Aij)=fn(Bij)Aij+Bijfn(Aij)={Pjfn(Bij)Pi}Aij+Bij{Pjfn(Aij)Pi}=0.

    Therefore, fn(AijBij)=p+q=nfp(Bij)fq(Aij). Similarly, we get (vi). The proof is complete.

    Lemma 2.3. Let fn:TT be defined as in Theorem 2.2. Then, for each n1 and for any AiiTii,BijTij,BjjTjj (1ij2),

    (i) fn(AiiBij)=p+q=nfp(Bij)fq(Aii);

    (ii) fn(AijBjj)=p+q=nfp(Bjj)fq(Aij).

    Proof. (i) For each n1 and for any AiiTii,BijTij (1ij2), it follows from AiiBij=AiiBijPj and Lemma 2.1 that

    fn(AiiBij)=Pjdn(AiiBij)Pi=Pjdn(AiiBijPj)Pi=Pj{p+q+r=ndp(Aii)dq(Bij)dr(Pj)}Pi=Pj{p+q+r=n,1p,q,rdp(Aii)dq(Bij)dr(Pj)}Pi+Pj{q+r=n,1q,rAiidq(Bij)dr(Pj)}Pi+Pj{p+r=n,1r,tdp(Aii)Bijdr(Pj)}Pi+Pj{p+q=n,1r,s,tdp(Aii)dq(Bij)Pj}Pi+Pj{dn(Aii)BijPj}Pi+Pj{Aiidn(Bij)Pj}Pi+Pj{AiiBijdn(Pj)}Pi=Pj{dn(Aii)BijPj}Pi+Pj{Aiidn(Bij)Pj}Pi+Pj{AiiBijdn(Pj)}Pi=Pj{dn(Aii)BijPj+Pjdn(Aii)Bij+Bijdn(Aii)Pj}Pi+Pj{Aiidn(Bij)Pj+Pjdn(Bij)Aii}Pi+Pj{AiiBijdn(Pj)}Pi=Pjdn(Bij)AiiPi=fn(Bij)Aii. (2.2)

    On the other hand, it is follows from fn(Aii)=0 (n1) that fnk(Bij)dk(Aii)=0, so we get

    fn(AiiBij)=fn(Bij)Aii+fn1(Bij)f1(Aii)+fn2(Bij)f2(Aii)++Bijfn(Aii)=p+q=nfp(Bij)fq(Aii).

    Similarly, we get (ii). The proof is complete.

    In the following, we give the completed proof of Theorem 2.2:

    Proof of Theorem 2.2. For each n1, let A=A11+A12+A21+A22 and B=B11+B12+B21+B22 be arbitrary elements of T, where Aij,BijTij (1i,j2). Then, it follows from Lemmas 2.2 and 2.3 that

    fn(AB)=fn((A11+A12+A21+A22)(B11+B12+B21+B22))=fn(A11B11)+fn(A11B12)+fn(A11B21)+fn(A11B22)+fn(A12B11)+fn(A12B12)+fn(A12B21)+fn(A12B22)+fn(A21B11)+fn(A21B12)+fn(A21B21)+fn(A21B22)+fn(A22B11)+fn(A22B12)+fn(A22B21)+fn(A22B22)=p+q=nfp(B11)fq(A11)+p+q=nfp(B11)fq(A12)+p+q=nfp(B11)fq(A21)+p+q=nfp(B11)fq(A22)+p+q=nfp(B12)fq(A11)+p+q=nfp(B12)fq(A12)+p+q=nfp(B12)fq(A21)+p+q=nfp(B12)fq(A22)+p+q=nfp(B21)fq(A11)+p+q=nfp(B21)fq(A12)+p+q=nfp(B21)fq(A21)+p+q=nfp(B21)fq(A22)+p+q=nfp(B22)fq(A11)+p+q=nfp(B22)fq(A12)+p+q=nfp(B22)fq(A21)+p+q=nfp(B22)fq(A22)=p+q=nfp(B)fq(A).

    Therefore, F={fn}nN0 is a higher anti-derivation on T. The proof is complete.

    Theorem 2.3. Let G={gn}nN0 be a sequence of mappings on T (with g0=igT). For each n1 and for any XT, define

    gn(X)=dn(X)fn(X).

    Then, G is a higher derivation on T.

    Next, we show that G={gn}nN0 is a higher derivation on T. In order to prove G is a higher derivation, we introduce Lemmas 2.4–2.6, and then, using the mathematical induction, we prove Lemmas 2.4–2.6.

    In [12] Theorem 1.3, we have proved that if g1=d1f1, then g1 is a derivation on T, i.e., for any X,YT, g1 satisfies

    g1(XY)=g1(X)Y+Xg1(Y)=p+q=1gp(X)gq(Y).

    Therefore, in the following, we assume that

    gk(XY)=p+q=kgp(X)gq(Y) (2.3)

    for each 1k<n and X,YT. Next, we prove that Lemmas 2.4–2.6 hold.

    By the definitions of F={fn}nN0 and G={gn}nN0, and by Lemma 2.1, we can easily check that the following Lemma holds:

    Lemma 2.4. Let gn:TT be defined as in Theorem 2.3. Then, for each n1 and for any AiiTii,AijTij (1ij2),

    (i) gn(Pi)=gn(Pj) and gn(Pi)=Pign(Pi)Pj+Pjgn(Pi)Pi;

    (ii) Pjgn(Aii)Pj=0,Pign(Aii)Pj=Aiign(Pi) and Pjgn(Aii)Pi=gn(Pi)Aii;

    (iii) gn(Aij)=Pign(Aij)Pj.

    Lemma 2.5. Let gn:TT be defined as in Theorem 2.3. Then, for each n1, and for any Aii,BiiTii, BjjTjj,Aij,BijTij (1ij2),

    (i) gn(AiiBij)=p+q=ngp(Aii)gq(Bij);

    (ii) gn(AijBjj)=p+q=ngp(Aij)gq(Bjj);

    (iii) gn(AiiBii)=p+q=ngp(Aii)gq(Bii);

    (iv) gn(AiiBjj)=p+q=ngp(Aii)gqBjj).

    Proof. (i) For each n1 and for any AiiTii,BijTij (1ij2), taking X=Aii,Y=Bij,Z=Pj in Eq (2.1), and by Lemma 1.1 (iii) and Lemma 2.1, we get

    dn(AiiBij)=dn(AiiBijPj)=p+q+r=ndp(Aii)dq(Bij)dr(Pj)=p+q+r=n,1p,q,rdp(Aii)dq(Bij)dr(Pj)+q+r=n,1q,rAiidq(Bij)dr(Pj)+p+r=n,1p,rdp(Aii)Bijdr(Pj)+p+q=n,1p,qdp(Aii)dq(Bij)Pj+dn(Aii)BijPj+Aiidn(Bij)Pj+AiiBijdn(Pj)=p+q=n,1p,qdp(Aii)dq(Bij)Pj+dn(Aii)BijPj+Aiidn(Bij)Pj=p+q=n,1p,qdp(Aii)dq(Bij)+dn(Aii)Bij+Aiidn(Bij)+dn(Bij)Aii=p+q=ndp(Aii)dq(Bij)+dn(Bij)Aii.

    Therefore, it follows from Eq (2.2), with fn(Aii)=0 and fn(Aij)=Pjfn(Aij)Pi (n1), that

    gn(AiiBij)=dn(AiiBij)fn(AiiBij)=p+q=ndp(Aii)dq(Bij)+dn(Bij)Aiidn(Bij)Aii=p+q=n,1p,qdp(Aii)dq(Bij)+dn(Aii)Bij+Aiidn(Bij)=p+q=n,1p,q{dp(Aii)fp(Aii)}dq(Bij)+{dn(Aii)fn(Aii)}Bij+Aii{dn(Bij)fn(Bij)}=p+q=n,1p,qgp(Aii)dq(Bij)+gn(Aii)Bij+Aiign(Bij)=p+q=n,1p,qgp(Aii){dq(Bij)fq(Bij)}+gn(Aii)Bij+Aiign(Bij)=p+q=n,1p,qgp(Aii)gq(Bij)+gn(Aii)Bij+Aiign(Bij)=p+q=ngp(Aii)gq(Bij).

    Similarly, we get that (ii) holds.

    (iii) For each n1 and for any Aii,BiiTii,XijTij (1ij2), by Lemma 2.5 (i) and Eq (2.3), on the one hand, we get

    gn(AiiBiiXij)=gn((AiiBii)Xij)=p+q=n,1qgp(AiiBii)gq(Xij)+gn(AiiBii)Xij=p+q=n,1q{r+s=pgr(Aii)gs(Bii)}gq(Xij)+gn(AiiBii)Xij=r+s+q=n,1qgr(Aii)gs(Bii)gq(Xij)+gn(AiiBii)Xij.

    On the other hand, we have

    gn(AiiBiiXij)=gn(Aii(BiiXij))=p+q=ngp(Aii)gq(BiiXij)=p+q=ngp(Aii)r+s=qgr(Bii)gs(Xij)=p+r+s=ngp(Aii)gr(Bii)gs(Xij)=p+r+s=n,1sgp(Aii)gr(Bii)gs(Xij)+p+r=ngp(Aii)gr(Bii)Xij.

    Comparing the above two equations, we get

    {gn(AiiBii)p+r=ngp(Aii)gr(Bii)}Xij=0,XijTij(1ij2).

    This yields from Lemma 1.1 (i) that

    Pign(AiiBii)Pi=Pi{p+r=ngp(Aii)gr(Bii)}Pi. (2.4)

    Next, we show that

    Pign(AiiBii)Pj=Pi{p+r=ngp(Aii)gr(Bii)}Pj and Pjgn(AiiBii)Pi=Pj{p+r=ngp(Aii)gr(Bii)}Pi.

    Indeed, for each n1 and for any Aii,BiiTii (1ij2), taking X=Aii,Y=Z=Pj in Eq (2.1), by Lemma 2.1, we get

    0=dn(AiiPjPj)=p+q+r=ndp(Aii)dq(Pj)dr(Pj)=p+q+r=n,1p,q,rdp(Aii)dq(Pj)dr(Pj)+q+r=n,1q,rAiidq(Pj)dr(Pj)+p+r=n,1p,rdp(Aii)Pjdr(Pj)+p+q=n,1p,qdp(Aii)dq(Pj)Pj+dn(Aii)PjPj+Aiidn(Pj)Pj+AiiPjdn(Pj)=p+q=n,1p,qdp(Aii)dq(Pj)Pj+dn(Aii)PjPj+Aiidn(Pj)Pj=p+q=n,1p,q{dp(Aii)dq(Pj)+dq(Pj)dp(Aii)}Pj+dn(Aii)Pj+Pjdn(Aii)+Aiidn(Pj)+dn(Pj)Aii=p+q=ndp(Aii)dq(Pj)+p+q=ndq(Pj)dp(Aii).

    Therefore, we get from p+q=ndp(Aii)dq(Pj)Tij and p+q=ndq(Pj)dp(Aii)Tji that

    p+q=ndp(Aii)dq(Pj)=0 and  p+q=ndq(Pj)dp(Aii)=0.

    So we get from fk(Aii)=0 and fk(Pj)=0 (k1) that

    0=p+q=ndp(Aii)dq(Pj)=p+q=n,1p,qdp(Aii)dq(Pj)+dn(Aii)Pj+Aiidn(Pj)=p+q=n,1p,q(dp(Aii)fp(Aii))(dq(Pj)fq(Pj))+(dn(Aii)fn(Aii))Pj+Aii(dn(Pj)fn(Pj))=p+q=n,1p,qgp(Aii)gq(Pj)+gn(Aii)Pj+Aiign(Pj)=p+q=ngp(Aii)gq(Pj)=p+q=n,1qgp(Aii)gq(Pj)+gn(Aii)Pj=p+q=n,1qgp(Aii)gq(Pi)+gn(Aii)Pj=p+q=ngp(Aii)gq(Pi)+gn(Aii)Pi+gn(Aii)Pj=p+q=ngp(Aii)gq(Pi)+gn(Aii).

    Therefore,

    gn(Aii)=p+q=ngp(Aii)gq(Pi). (2.5)

    For each n1 and for any Aii,BiiTii, by Eq (2.5), we get

    gn(AiiBii)=p+q=ngp(AiiBii)gq(Pi)=p+q=n,1qgp(AiiBii)gq(Pi)+gn(AiiBii)Pi=p+q=n,1q{r+s=pgr(Aii)gs(Bii)}gq(Pi)+gn(AiiBii)Pi=r+s+q=n,1qgr(Aii)gs(Bii)gq(Pi)+gn(AiiBii)Pi=nr=0gr(Aii){s+q=nr,1qgs(Bii)gq(Pi)}+gn(AiiBii)Pi=nr=0gr(Aii){s+q=nrgs(Bii)gq(Pi)gnr(Bii)Pi}+gn(AiiBii)Pi=nr=0gr(Aii){gnr(Bii)gnr(Bii)Pi}+gn(AiiBii)Pi=nr=0gr(Aii)gnr(Bii)nr=0gr(Aii)gnr(Bii)Pi+gn(AiiBii)Pi=p+q=ngp(Aii)gq(Bii)+p+q=ngp(Aii)gq(Bii)Pi+gn(AiiBii)Pi.

    This implies that

    Pign(AiiBii)Pj=Pi{p+q=ngp(Aii)gq(Bii)}Pj. (2.6)

    Similarly, we get

    Pjgn(AiiBii)Pi=Pj{p+q=ngp(Aii)gq(Bii)}Pi. (2.7)

    Therefore, by Eqs (2.4), (2.6), (2.7) and Lemma 2.4 (ii), we get that

    gn(AiiBii)=Pi{p+q=ngp(Aii)gq(Bii)}Pi+Pi{p+q=ngp(Aii)gq(Bii)}Pj+Pj{p+q=ngp(Aii)gq(Bii)}Pi=p+q=ngp(Aii)gq(Bii).

    (iv) For each n1 and for any AiiTii,BjjTjj (1ij2), taking X=Aii,Y=Bjj,Z=Pj (1ij2) in Eq (2.1), we get from Lemma 2.1 that

    0=dn(AiiBjjPj)=p+q+r=ndp(Aii)dq(Bjj)dr(Pj)=p+q+r=n,1p,q,rdp(Aii)dq(Bjj)dr(Pj)+q+r=n,1q,rAiidq(Bjj)dr(Pj)+p+r=n,1p,rdp(Aii)Bjjdr(Pj)+p+q=n,1p,qdp(Aii)dq(Bjj)Pj+dn(Aii)BjjPj+Aiidn(Bjj)Pj+AiiBjjdn(Pj)=p+q=n,1p,qdp(Aii)dq(Bjj)Pj+dn(Aii)BjjPj+Aiidn(Bjj)Pj=p+q=n,1p,qdp(Aii)dq(Bjj)Pj+p+q=n,1p,qdp(Bjj)dq(Aii)Pj+dn(Aii)Bjj+Bjjdn(Aii)+Aiidn(Bjj)+dn(Bjj)Aii=p+q=n,1p,qdp(Aii)dq(Bjj)+p+q=n,1p,qdp(Bjj)dq(Aii)+dn(Aii)Bjj+Bjjdn(Aii)+Aiidn(Bjj)+dn(Bjj)Aii=p+q=ndp(Aii)dq(Bjj)+p+q=ndp(Bjj)dq(Aii).

    Hence, we get from p+q=ndp(Aii)dq(Bjj)Tij and p+q=ndp(Bjj)dq(Aii)Tji (1ij2) that

    p+q=ndp(Aii)dq(Bjj)=p+q=ndp(Bjj)dq(Aii)=0.

    Therefore, it follows from gk(Aii)=dk(Aii) and gk(Bjj)=dk(Bjj) (k1) that

    gn(AiiBjj)=0=p+q=ndp(Aii)dq(Bjj)=p+q=ngp(Aii)gq(Bjj).

    The proof is complete.

    Lemma 2.6. Let gn:TT be defined as in Theorem 2.3. Then for each n1 and for any AiiTii,BjjTjj,Aij,BijTij,Aji,BjiTji (1ij2),

    (i) gn(AijBji)=p+qgp(Aij)gq(Bji);

    (ii) gn(AijBij)=p+q=ngp(Aij)gq(Bij);

    (iii) gn(AiiBji)=p+q=ngp(Aii)gq(Bji);

    (iv) gn(AjiBjj)=p+q=ngp(Aji)gq(Bjj).

    Proof. (i) For each n1 and for any AijTij,BjiTji (1ij2), it follows from Lemma 1 (iii) that AijBji=0, and therefore we get

    gn(AijBji)=gn(0)=0.

    On the other hand, by Lemma 1 (iii) and Lemma 2.4 (iii), we have gp(Aij)gq(Bji)=0, therefore we get

    gn(AijBji)=0=p+q=ngp(Aij)gq(Bji).

    Similarly, we get that (ii) holds.

    (iii) For each n1 and for any AiiTii,BjiTji (1ij2), by Lemma 1 (iii) and Lemma 2.4 (ii,iii), we get gp(Aii)gq(Bji)=0, and therefore we get

    gn(AiiBji)=0=p+q=ngp(Aii)gq(Bji).

    Similarly, we get (iv) holds. The proof is complete.

    In the following, we complete the proof of Theorem 2.3.

    Proof of Theorem 2.3. For any n1, let A=A11+A12+A21+A22 and B=B11+B12+B21+B22 be arbitrary elements of T, where Aij,BijTij (1i,j2). It follows from Lemmas 2.4–2.6 that

    gn(AB)=gn((A11+A12+A21+A22)(B11+B12+B21+B22))=gn(A11B11)+gn(A11B12)+gn(A11B21)+gn(A11B22)+gn(A12B11)+gn(A12B12)+gn(A12B21)+gn(A12B22)+gn(A21B11)+gn(A21B12)+gn(A21B21)+gn(A21B22)+gn(A22B11)+gn(A22B12)+gn(A22B21)+gn(A22B22)=p+q=ngp(A11)gq(B11)+p+q=ngp(A11)gq(B12)+p+q=ngp(A11)gq(B21)+p+q=ngp(A11)gq(B22)+p+q=ngp(A12)gq(B11)+p+q=ngp(A12)gq(B12)+p+q=ngp(A12)gq(B21)+p+q=ngp(A12)gq(B22)+p+q=ngp(A21)gq(B11)+p+q=ngp(A21)gq(B12)+p+q=ngp(A21)gq(B21)+p+q=ngp(A21)gq(B22)+p+q=ngp(A22)gq(B11)+p+q=ngp(A22)gq(B21)+p+q=ngp(A22)gq(B21)+p+q=ngp(A22)gq(B22)=p+q=ngp(A11+A12+A21+A22)gq(B11+B12+B21+B22)=p+q=ngp(A)gq(B).

    Therefore, G={gn}nN0 is a higher derivation on T. The proof is complete.

    Next, we show that Theorem 2.1 holds.

    Proof of Theorem 2.1. For each n1 and for any A,BT, by Theorems 2.2 and 2.3, we obtain that

    dn(A)=gn(A)+fn(A),

    where G={gn}nN0 is a higher derivation and F={fn}nN0 is a higher anti-derivation from T into itself such that fn(Aii)=0 for all AiiTii (1i2). The proof is complete.

    Remark 2.1. Let D={dn}nN0 be a higher Jordan triple derivation from T into itself. Then, by Theorems 2.1 and 2.2, we obtain that the following statements are equivalent.

    (i) D={dn}nN0 is a higher derivation;

    (ii) Pjdn(Aij)Pi=0 for each n1 and for any AijTij (1ij2);

    (iii) dn(Aij)Tij for each n1 and for any AijTij (1ij2).

    In the following, we show that every higher Jordan triple derivation on triangular algebras is a higher derivation.

    Corollary 2.1. Let A and B be unital algebras over a commutative ring R and M be a unital (A,B)-bimodule, which is faithful as both a left A-module and a right B-module, and U be the 2-torsion free triangular algebra, and D={dn}nN0 be a higher Jordan triple derivation on U. Then D={dn}nN0 is a higher derivation.

    Proof of Corollary 2.1. Let 1A and 1B be the identities of the algebras A and B, respectively, and let 1 be the identity of the triangular algebra U. We denote

    P1=(1A000) by the standard idempotent of U,  P2=1P1=(0001B)

    and

    Uij=PiUPj for 1ij2.

    It is clear that the triangular algebra U may be represented as

    U=P1UP1+P1UP2+P2UP2=A+M+B.

    Here 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.

    By the definition of triangular algebra U, we can easily check that U is a -type trivial extension algebra, and so if U is a 2-torsion free triangular algebra, then for any n1, A=A11+A12+A22U, where AijUij (1i,j2), we get from Theorem 2.1 that

    dn(A)=gn(A)+fn(A).

    Where G={gn}nN0 is a higher derivation and F={fn}nN0 is a higher anti-derivation from U into itself such that fn(Aii)=0 for all AiiUii (1i2). Next, we show that fn(A12)=0 for each n1 and for any A12U12.

    Indeed, for any A12U12, it follows from Lemma 2.1 (v) and U21={0} that

    dn(A12)=P1dn(A12)P2+P2dn(A12)P1=P1dn(A12)P2.

    And then we obtain from the definition of fn in Theorem 2.2 that fn(A12)=P2dn(A12)P1=0. Therefore, for any AU, fn(A)=0, so D={gn}nN0 is a higher derivation. The proof is complete.

    Next, we give an application of Corollary 2.1 to certain special classes of triangular algebras, such as block upper triangular matrix algebras and nest algebras.

    Let R be a commutative ring with identity and let Mn×k(R) be the set of all n×k matrices over R. For n2 and mn, the block upper triangular matrix algebra Tˉkn(R) is a subalgebra of Mn(R) with the form

    (Mk1(R)Mk1×k2(R)Mk1×km(R)0Mk2(R)Mk2×km(R)00Mkm(R)),

    where ˉk=(k1,k2,,km) is an ordered m-vector of positive integers such that k1+k2++km=n.

    A nest of a complex Hilbert space H is a chain N of closed subspaces of H containing {0} and H, which is closed under arbitrary intersections and closed linear span, and B(H) is the algebra of all bounded linear operators on H. The nest algebra associated with N is the algebra

    AlgN={TB(H):TNN, for all NN}.

    A nest N 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 2.2. Let Tˉkn(R) be a 2-torsion free block upper triangular matrix algebra, and D={dn}nN0 be a higher Jordan triple derivation on Tˉkn(R). Then, D={dn}nN0 is a higher derivation.

    Corollary 2.3. Let N be a nontrivial nest of a complex Hilbert space H, AlgN a nest algebra, and D={dn}nN0 a higher Jordan triple derivation on AlgN. Then, D={dn}nN0 is a higher derivation.

    The authors declare that they have not used artificial intelligence tools in the creation of this article.

    This research is supported by the National Natural Science Foundation of China (No.11901451), Talent Project Foundation of Yunnan Provincial Science and Technology Department (No.202105AC160089), Natural Science Foundation of Yunnan Province (No.202101BA070001198), and Basic Research Foundation of Yunnan Education Department (No.2021J0915).

    The authors declare that there are no conflicts of interest regarding the publication of this paper.

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