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Research article

On L-simulation mappings in partial metric spaces

  • The class of L-contractive mappings was introduced by Cho [12]. In this paper, we provide some fixed point results for such mappings via a control function introduced by Jleli and Samet [14] in the class of partial metric spaces. Some illustrating examples are given.

    Citation: Hassen Aydi, M. A. Barakat, Erdal Karapinar, Zoran D. Mitrović, Tawseef Rashid. On L-simulation mappings in partial metric spaces[J]. AIMS Mathematics, 2019, 4(4): 1034-1045. doi: 10.3934/math.2019.4.1034

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  • The class of L-contractive mappings was introduced by Cho [12]. In this paper, we provide some fixed point results for such mappings via a control function introduced by Jleli and Samet [14] in the class of partial metric spaces. Some illustrating examples are given.


    A Poisson algebra is a triple, (L,,[,]), where (L,) is a commutative associative algebra and (L,[,]) is a Lie algebra that satisfies the following Leibniz rule:

    [x,yz]=[x,y]z+y[x,z],x,y,zL.

    Poisson algebras appear naturally in the study of Hamiltonian mechanics and play a significant role in mathematics and physics, such as in applications of Poisson manifolds, integral systems, algebraic geometry, quantum groups, and quantum field theory (see [7,11,24,25]). Poisson algebras can be viewed as the algebraic counterpart of Poisson manifolds. With the development of Poisson algebras, many other algebraic structures have been found, such as Jacobi algebras [1,9], Poisson bialgebras [20,23], Gerstenhaber algebras, Lie-Rinehart algebras [16,17,26], F-manifold algebras [12], Novikov-Poisson algebras [28], quasi-Poisson algebras [8] and Poisson n-Lie algebras [10].

    As a dual notion of a Poisson algebra, the concept of a transposed Poisson algebra was recently introduced by Bai et al. [2]. A transposed Poisson algebra (L,,[,]) is defined by exchanging the roles of the two binary operations in the Leibniz rule defining the Poisson algebra:

    2z[x,y]=[zx,y]+[x,zy],x,y,zL,

    where (L,) is a commutative associative algebra and (L,[,]) is a Lie algebra.

    It is shown that a transposed Poisson algebra possesses many important identities and properties and can be naturally obtained by taking the commutator in the Novikov-Poisson algebra [2]. There are many results on transposed Poisson algebras, such as those on transposed Hom-Poisson algebras [18], transposed BiHom-Poisson algebras [21], a bialgebra theory for transposed Poisson algebras [19], the relation between 12-derivations of Lie algebras and transposed Poisson algebras [14], the relation between 12-biderivations and transposed Poisson algebras [29], and the transposed Poisson structures with fixed Lie algebras (see [6] for more details).

    The notion of an n-Lie algebra (see Definition 2.1), as introduced by Filippov [15], has found use in many fields in mathematics and physics [4,5,22,27]. The explicit construction of n-Lie algebras has become one of the important problems in this theory. In [3], Bai et al. gave a construction of (n+1)-Lie algebras through the use of n-Lie algebras and some linear functions. In [13], Dzhumadil′daev introduced the notion of a Poisson n-Lie algebra which can be used to construct an (n+1)-Lie algebra under an additional strong condition. In [2], Bai et al. showed that this strong condition for n=2 holds automatically for a transposed Poisson algebra, and they gave a construction of 3-Lie algebras from transposed Poisson algebras with derivations. They also found that this constructed 3-Lie algebra and the commutative associative algebra satisfy the analog of the compatibility condition for transposed Poisson algebras, which is called a transposed Poisson 3-Lie algebra. This motivated them to introduce the concept of a transposed Poisson n-Lie algebra (see Definition 2.2) and propose the following conjecture:

    Conjecture 1.1. [2] Let n2 be an integer and (L,,μn) a transposed Poisson n-Lie algebra. Let D be a derivation of (L,) and (L,μn). Define an (n+1)-ary operation:

    μn+1(x1,,xn+1):=n+1i=1(1)i1D(xi)μn(x1,,ˆxi,,xn+1),x1,,xn+1L,

    where ˆxi means that the i-th entry is omitted. Then, (L,,μn+1) is a transposed Poisson (n+1)-Lie algebra.

    In this paper, based on the identities for transposed Poisson n-Lie algebras given in Section 2, we prove that Conjecture 1.1 holds under a certain strong condition described in Section 3 (see Definition 2.3 and Theorem 3.2).

    Throughout the paper, all vector spaces are taken over a field of characteristic zero. To simplify notations, the commutative associative multiplication () will be omitted unless the emphasis is needed.

    In this section, we first recall some definitions, and then we exhibit a class of identities for transposed Poisson n-Lie algebras.

    Definition 2.1. [15] Let n2 be an integer. An n-Lie algebra is a vector space L, together with a skew-symmetric linear map [,,]:nLL, such that, for any xi,yjL,1in1,1jn, the following identity holds:

    [[y1,,yn],x1,,xn1]=ni=1(1)i1[[yi,x1,,xn1],y1,,ˆyi,,yn]. (2.1)

    Definition 2.2. [2] Let n2 be an integer and L a vector space. The triple (L,,[,,]) is called a transposed Poisson n-Lie algebra if (L,) is a commutative associative algebra and (L,[,,]) is an n-Lie algebra such that, for any h,xiL,1in, the following identity holds:

    nh[x1,,xn]=ni=1[x1,,hxi,,xn]. (2.2)

    Some identities for transposed Poisson algebras in [2] can be extended to the following theorem for transposed Poisson n-Lie algebras.

    Theorem 2.1. Let (L,,[,,]) be a transposed Poisson n-Lie algebra. Then, the following identities hold:

    (1) For any xiL,1in+1, we have

    n+1i=1(1)i1xi[x1,,ˆxi,,xn+1]=0; (2.3)

    (2) For any h,xi,yjL,1in1,1jn, we have

    ni=1(1)i1[h[yi,x1,,xn1],y1,,ˆyi,,yn]=[h[y1,,yn],x1,,xn1]; (2.4)

    (3) For any xi,yjL,1in1,1jn+1, we have

    n+1i=1(1)i1[yi,x1,,xn1][y1,,ˆyi,,yn+1]=0; (2.5)

    (4) For any x1,x2,yiL,1in, we have

    ni=1nj=1,ji[y1,,yix1,,yjx2,,yn]=n(n1)x1x2[y1,y2,,yn]. (2.6)

    Proof. (1) By Eq (2.2), for any 1in+1, we have

    nxi[x1,,xi1,xi+1,,xn+1]=ji[x1,,xi1,xi+1,,xixj,,xn+1].

    Thus, we obtain

    n+1i=1(1)i1nxi[x1,,ˆxi,,xn+1]=n+1i=1n+1j=1,ji(1)i1[x1,,ˆxi,,xixj,,xn+1].

    Note that, for any i>j, we have

    (1)i1[x1,,xj1,xixj,xj+1,,ˆxi,,xn]+(1)j1[x1,,ˆxj,,xi1,xjxi,xi+1,,xn]=(1)i1+(ij1)[x1,,xj1,xj+1,,xi1,xixj,xi+1,,xn]+(1)j1[x1,,ˆxj,,xi1,xjxi,xi+1,,xn]=((1)j2+(1)j1)[x1,,xj1,xj+1,,xi1,xixj,xi+1,,xn]=0,

    which gives n+1i=1n+1j=1,ji(1)i1[x1,,ˆxi,,xixj,,xn+1]=0.

    Hence, we get

    n+1i=1(1)i1nxi[x1,,ˆxi,,xn+1]=0.

    (2) By Eq (2.2), we have

    [h[y1,,yn],x1,,xn1]n1i=1[[y1,,yn],x1,,hxi,,xn1]=nh[[y1,,yn],x1,,xn1],

    and, for any 1jn,

    (1)j1([h[yj,x1,,xn1],y1,,ˆyj,,yn1]+ni=1,ij[[yj,x1,,xn1],y1,,hyi,,ˆyj,,yn1])=(1)j1nh[[yj,x1,,xn1],y1,,ˆyj,,yn1].

    By taking the sum of the above n+1 identities and applying Eq (2.1), we get

    [h[y1,,yn],x1,,xn1]n1i=1[[y1,,yn],x1,,hxi,,xn1]+nj=1(1)j1([h[yj,x1,,xn1],y1,,ˆyj,,yn1]+ni=1,ij[[yj,x1,,xn1],y1,,hyi,,ˆyj,,yn1])=nh[[y1,,yn],x1,,xn1]+nhnj=1(1)j1[[yj,x1,,xn1],y1,,ˆyj,,yn1]=0.

    We denote

    Aj:=ni=1,ij(1)i1[[yi,x1,,xn1],y1,,hyj,,ˆyi,,yn],1jn,Bi:=[[y1,,yn],x1,,hxi,,xn1],1in1.

    Then, the above equation can be rewritten as

    ni=1(1)i1[h[yi,x1,,xn1],y1,,ˆyi,,yn][h[y1,,yn],x1,,xn1]+nj=1Ajn1i=1Bi=0. (2.7)

    By applying Eq (2.1) to Aj,1jn, we have

    Aj=ni=1,ij(1)i1[[yi,x1,,xn1],y1,,hyj,,ˆyi,,yn]=[[y1,,hyj,,yn],x1,,xn1]+(1)j[[hyj,x1,,xn1],y1,,ˆyj,,yn].

    Thus, we get

    nj=1Aj=nj=1[[y1,,hyj,,yn],x1,,xn1]+nj=1(1)j[[hyj,x1,,xn1],y1,,ˆyj,,yn]=n[h[y1,,yn],x1,,xn1]+nj=1(1)j[[hyj,x1,,xn1],y1,,ˆyj,,yn].

    By applying Eq (2.1) to Bi,1in1, we have

    Bi=[[y1,,yn],x1,,hxi,,xn1]=nj=1(1)j1[[yj,x1,,hxi,,xn1],y1,ˆyj,,yn].

    Thus, we get

    n1i=1Bi=n1i=1nj=1(1)j1[[yj,x1,,hxi,,xn1],y1,ˆyj,,yn]=nj=1n1i=1(1)j1[[yj,x1,,hxi,,xn1],y1,ˆyj,,yn].

    Note that, by Eq (2.2), we have

    n1i=1(1)j1[[yj,x1,,hxi,,xn1],y1,ˆyj,,yn]=(1)j1n[h[yj,x1,,xi,,xn1],y1,ˆyj,,yn]+(1)j[[hyj,x1,,xi,,xn1],y1,ˆyj,,yn].

    Thus, we obtain

    n1i=1Bi=nj=1(1)j1n[h[yj,x1,,xi,,xn1],y1,ˆyj,,yn]+nj=1(1)j[[hyj,x1,,xi,,xn1],y1,ˆyj,,yn].

    By substituting these equations into Eq (2.7), we have

    ni=1(1)i1[h[yi,x1,,xn1],y1,,ˆyi,,yn][h[y1,,yn],x1,,xn1]+n[h[y1,,yn],x1,,xn1]+nj=1(1)j[[hyj,x1,,xn1],y1,,ˆyj,,yn]nj=1(1)j1n[h[yj,x1,,xi,,xn1],y1,ˆyj,,yn]nj=1(1)j[[hyj,x1,,xi,,xn1],y1,ˆyj,,yn]=0,

    which implies that

    (n1)(ni=1(1)i[h[yi,x1,,xn1],y1,,ˆyi,,yn]+[h[y1,,yn],x1,,xn1])=0.

    Therefore, the proof of Eq (2.4) is completed.

    (3) By Eq (2.2), for any 1jn+1, we have

    (1)j1n[yj,x1,,xn1][y1,,ˆyj,,yn+1]=n+1i=1,ij(1)j1[y1,,yi[yj,x1,,xn1],,ˆyj,,yn+1].

    By taking the sum of the above n+1 identities, we obtain

    n+1j=1(1)j1n[yj,x1,,xn1][y1,,ˆyj,,yn+1]=n+1j=1n+1i=1,ij(1)j1[y1,,yi[yj,x1,,xn1],,ˆyj,,yn+1].

    Thus, we only need to prove the following equation:

    n+1j=1n+1i=1,ij(1)j1[y1,,yi[yj,x1,,xn1],,ˆyj,,yn+1]=0.

    Note that

    n+1j=1n+1i=1,ij(1)j1[y1,,yi[yj,x1,,xn1],,ˆyj,,yn+1]=n+1i=1n+1j=1,ji(1)j1[y1,,yi[yj,x1,,xn1],,ˆyj,,yn+1]=n+1i=1i1j=1(1)i+j1[yi[yj,x1,,xn1],y1,,ˆyj,,ˆyi,,yn+1]+n+1i=1n+1j=i+1(1)i+j[yi[yj,x1,,xn1],y1,,ˆyi,,ˆyj,,yn+1](2.4)=n+1i=1(1)i[yi[y1,,ˆyi,,yn+1],x1,,xn1](2.3)=0.

    Hence, the conclusion holds.

    (4) By applying Eq (2.2), we have

    n2x1x2[y1,y2,,yn]=nx1nj=1[y1,,yjx2,,yn]=ni=1nj=1,ji[y1,,yix1,,yjx2,,yn]+nj=1[y1,,yjx1x2,,yn]=ni=1nj=1,ji[y1,,yix1,,yjx2,,yn]+nx1x2[y1,,yn],

    which gives

    n(n1)x1x2[y1,y2,,yn]=ni=1nj=1,ji[y1,,yix1,,yjx2,,yn].

    Hence, the proof is completed.

    To prove Conjecture 1.1, we need the following extra condition.

    Definition 2.3. A transposed Poisson n-Lie algebra (L,,[,,]) is called strong if the following identity holds:

    y1[hy2,x1,,xn1]y2[hy1,x1,,xn1]+n1i=1(1)i1hxi[y1,y2,x1,,ˆxi,,xn1]=0 (2.8)

    for any y1,y2,xiL,1in1.

    Remark 2.1. When n=2, the identity is

    y1[hy2,x1]+y2[x1,hy1]+hx1[y1,y2]=0,

    which is exactly Theorem 2.5 (11) in [2]. Thus, in the case of a transposed Poisson algebra, the strong condition always holds. So far, we cannot prove that the strong condition fails to hold for n3.

    Proposition 2.1. Let (L,,[,,]) be a strong transposed Poisson n-Lie algebra. Then,

    y1[hy2,x1,,xn1]hy1[y2,x1,,xn1]=y2[hy1,x1,,xn1]hy2[y1,x1,,xn1] (2.9)

    for any y1,y2,xiL,1in1.

    Proof. By Eq (2.3), we have

    hy1[y2,x1,,xn1]+hy2[y1,x1,,xn1]=n1i=1(1)i1hxi[y1,y2,x1,,ˆxi,,xn1].

    Then, the statement follows from Eq (2.8).

    In this section, we will prove Conjecture 1.1 for strong transposed Poisson n-Lie algebras. First, we recall the notion of derivations of transposed Poisson n-Lie algebras.

    Definition 3.1. Let (L,,[,,]) be a transposed Poisson n-Lie algebra. The linear operation D:LL is called a derivation of (L,,[,,]) if the following holds for any u,v,xiL,1in:

    (1) D is a derivation of (L,), i.e., D(uv)=D(u)v+uD(v);

    (2) D is a derivation of (L,[,,]), i.e.,

    D([x1,,xn])=ni=1[x1,,xi1,D(xi),xi+1,,xn].

    Lemma 3.1. Let (L,,[,,]) be a transposed Poisson n-Lie algebra and D a derivation of (L,,[,,]). For any yiL,1in+1, we have the following:

    (1)

    n+1i=1(1)i1D(yi)D([y1,,ˆyi,,yn+1])=n+1i=1n+1j=1,ji(1)i1D(yi)[y1,,D(yj),,ˆyi,,yn+1]; (3.1)

    (2)

    n+1i=1(1)i1D(yi)D([y1,,ˆyi,,yn+1])=n+1i=1n+1j=1,jin+1k=j+1,ki(1)iyi[y1,,D(yj),,D(yk),,ˆyi,,yn+1], (3.2)

    where, for any i>j, ji denotes the empty sum, which is equal to zero.

    Proof. (1) The statement follows immediately from Definition 3.1.

    (2) By applying Eq (3.1), we need to prove the following equation:

    n+1i=1n+1j=1,ji(1)i1nD(yi)[y1,,D(yj),,ˆyi,,yn+1]=n+1i=1n+1j=1,jin+1k=j+1,ki(1)inyi[y1,,D(yj),,D(yk),,ˆyi,,yn+1].

    For any 1in+1, denote Ai:=nn+1j=1,ji(1)i1D(yi)[y1,,D(yj),,ˆyi,,yn+1]. Then, we have

    n+1i=1n+1j=1,ji(1)i1nD(yi)[y1,,D(yj),,ˆyi,,yn+1]=n+1i=1Ai.

    Note that

    Ai=(1)i1(nD(yi)[D(y1),y2,,ˆyi,,yn+1]+nD(yi)[y1,D(y2),y3,,ˆyi,,yn+1]++nD(yi)[y1,,ˆyi,,yn,D(yn+1)])=(1)i1([D(yi)D(y1),y2,,ˆyi,,yn+1]+n+1k=2,ki[D(y1),y2,,ykD(yi),,ˆyi,,yn+1]+[y1,D(yi)D(y2),y3,,ˆyi,,yn+1]+n+1k=1,k2,i[y1,D(y2),y3,,ykD(yi),,ˆyi,,yn+1]++[y1,,ˆyi,,yn,D(yi)D(yn+1)]+nk=1,ki[y1,,ykD(yi),,ˆyi,,yn,D(yn+1)])=(1)i1n+1j=1,ji[y1,,D(yi)D(yj),,ˆyi,,yn+1]+(1)i1n+1j=1,jin+1k=1,kj,i[y1,,D(yj),,ykD(yi),,ˆyi,,yn+1].

    Thus, we have

    n+1i=1Ai=n+1j=1n+1i=1,ij(1)j1[y1,,D(yj)D(yi),,ˆyj,,yn+1]+n+1i=1n+1j=1,jin+1k=1,ki,j(1)i1[y1,,D(yj),,ykD(yi),,ˆyi,,yn+1]=T1+T2,

    where

    T1:=n+1j=1n+1i=1,ij(1)j1[y1,,D(yj)D(yi),,ˆyj,,yn+1],T2:=n+1i=1n+1j=1,jin+1k=1,ki,j(1)i1[y1,,D(yj),,ykD(yi),,ˆyi,,yn+1].

    Note that

    T1=n+1j,i=1Bji,

    where Bji=(1)j1[y1,,D(yj)D(yi),,ˆyj,,yn+1] for any 1jin+1, and Bii=0 for any 1in+1.

    For any 1i,jn+1, without loss of generality, assume that i<j; then, we have

    Bji+Bij=(1)j1[y1,,D(yj)D(yi),,ˆyj,,yn+1]+(1)i1[y1,,ˆyi,,D(yi)D(yj),,yn+1]=(1)j1[y1,,D(yj)D(yi),,ˆyj,,yn+1]+(1)i1+ji+1[y1,,D(yj)D(yi),,ˆyj,,yn+1]=0,

    which implies that T1=n+1j,i=1Bji=0.

    Thus, we get that n+1i=1Ai=T2.

    We rewrite

    n+1i=1n+1j=1,jin+1k=j+1,ki(1)inyi[y1,,D(yj),,D(yk),,ˆyi,,yn+1]=n+1i=1n+1j=1,jin+1k=j+1,kin+1t=1,tj,k,i(1)i[y1,,D(yj),,D(yk),,ytyi,,ˆyi,,yn+1]+n+1i=1n+1j=1,jin+1k=j+1,ki(1)i[y1,,yiD(yj),,D(yk),,ˆyi,,yn+1]+n+1i=1n+1j=1,jin+1k=j+1,ki(1)i[y1,,D(yj),,yiD(yk),,ˆyi,,yn+1]=M1+M2+M3,

    where

    M1:=n+1i=1n+1j=1,jin+1k=j+1,kin+1t=1,tj,k,i(1)i[y1,,D(yj),,D(yk),,ytyi,,ˆyi,,yn+1],M2:=n+1i=1n+1j=1,jin+1k=j+1,ki(1)i[y1,,yiD(yj),,D(yk),,ˆyi,,yn+1],M3:=n+1i=1n+1j=1,jin+1k=j+1,ki(1)i[y1,,D(yj),,yiD(yk),,ˆyi,,yn+1].

    Note that

    M1=n+1i=1n+1j=1,jin+1k=j+1,kin+1t=1,tj,k,i(1)i[y1,,D(yj),,D(yk),,ytyi,,ˆyi,,yn+1]=n+1i,j,k,t=1Bijkt,

    where

    Bijkt={0,if any two indices are equal or k<j;(1)i[y1,,D(yj),,D(yk),,ytyi,,ˆyi,,yn+1],otherwise.

    For any 1j,kn+1, without loss of generality, assume that t<i; then, we have

    Bijkt+Btjki=(1)i[y1,,D(yj),,D(yk),,ytyi,,ˆyi,,yn+1]+(1)t[y1,,D(yj),,D(yk),,ˆyt,,ytyi,,yn+1]=(1)i[y1,,D(yj),,D(yk),,ytyi,,ˆyi,,yn+1]+(1)t+it1[y1,,D(yj),,D(yk),,ytyi,,ˆyi,,yn+1]=0,

    which implies that M1=0.

    Therefore, we only need to prove the following equation:

    M2+M3=n+1i=1n+1j=1,jin+1k=1,ki,j(1)i1[y1,,D(yj),,ykD(yi),,ˆyi,,yn+1].

    First, we have

    n+1j=1,jin+1k=j+1,ki(1)i[y1,,yiD(yj),,D(yk),,ˆyi,,yn+1]+n+1j=1,jin+1k=j+1,ki(1)i[y1,,D(yj),,yiD(yk),,ˆyi,,yn+1]=n+1k=1,kik1j=1,ji(1)i[y1,,yiD(yj),,D(yk),,ˆyi,,yn+1]+n+1j=1,jin+1k=j+1,ki(1)i[y1,,D(yj),,yiD(yk),,ˆyi,,yn+1]=n+1j=1,jij1k=1,ki(1)i[y1,,yiD(yk),,D(yj),,ˆyi,,yn+1]+n+1j=1,jin+1k=j+1,ki(1)i[y1,,D(yj),,yiD(yk),,ˆyi,,yn+1]=n+1j=1,jin+1k=1,ki,j(1)i[y1,,D(yj),,yiD(yk),,ˆyi,,yn+1].

    Thus,

    M2+M3=n+1i=1n+1j=1,jin+1k=1,ki,j(1)i[y1,,D(yj),,yiD(yk),,ˆyi,,yn+1]=n+1j=1n+1i=1,ijn+1k=1,ki,j(1)i[y1,,D(yj),,yiD(yk),,ˆyi,,yn+1]=n+1j=1n+1i=1,iji1k=1,kj(1)i[y1,,D(yj),,yiD(yk),,ˆyi,,yn+1]+n+1j=1n+1i=1,ijn+1k=i+1,kj(1)i[y1,,D(yj),,ˆyi,,yiD(yk),,yn+1].

    Note that, for any 1jn+1, we have

    n+1i=1,iji1k=1,kj(1)i[y1,,D(yj),,yiD(yk),,ˆyi,,yn+1]=n+1i=1,iji1k=1,kj(1)i[y1,,D(yj),,yk1,yiD(yk),yk+1,,ˆyi,yn+1]=n+1i=1,iji1k=1,kj(1)k1[y1,,D(yj),,ˆyk,,yi1,yiD(yk),yi+1,,yn+1]=n+1i=1,iji1k=1,kj(1)k1[y1,,D(yj),,ˆyk,,yiD(yk),,yn+1].

    Similarly, we have

    n+1i=1,ijn+1k=i+1,kj(1)i[y1,,D(yj),,ˆyi,,yiD(yk),,yn+1]=n+1i=1,ijn+1k=i+1,kj(1)k1[y1,,D(yj),,yiD(yk),,ˆyk,,yn+1].

    Thus,

    M2+M3=n+1j=1n+1i=1,iji1k=1,kj(1)k1[y1,,D(yj),,ˆyk,,yiD(yk),,yn+1]+n+1j=1n+1i=1,ijn+1k=i+1,kj(1)k1[y1,,D(yj),,yiD(yk),,ˆyk,,yn+1]=n+1j=1n+1i=1,ijn+1k=1,ki,j(1)k1[y1,,D(yj),,yiD(yk),,ˆyk,,yn+1]=n+1k=1n+1j=1,jkn+1i=1,ij,k(1)k1[y1,,D(yj),,yiD(yk),,ˆyk,,yn+1]=n+1i=1n+1j=1,jin+1k=1,kj,i(1)i1[y1,,D(yj),,ykD(yi),,ˆyi,,yn+1].

    The proof is completed.

    Theorem 3.1. Let (L,,[,,]) be a strong transposed Poisson n-Lie algebra and D a derivation of (L,,[,,]). Define an (n+1)-ary operation:

    μn+1(x1,,xn+1):=n+1i=1(1)i1D(xi)[x1,,ˆxi,,xn+1] (3.3)

    for any xiL,1in+1. Then, (L,μn+1) is an (n+1)-Lie algebra.

    Proof. For convenience, we denote

    μn+1(x1,,xn+1):=[x1,,xn+1].

    On one hand, we have

    [[y1,,yn+1],x1,,xn](3.3)=n+1i=1(1)i1[D(yi)[y1,,ˆyi,,yn+1],x1,,xn](3.3)=n+1i=1(1)i1D(D(yi)[y1,,ˆyi,,yn+1])[x1,,xn]+n+1i=1nj=1(1)i+j1D(xj)[D(yi)[y1,,ˆyi,,yn+1],x1,,ˆxj,,xn]=n+1i=1(1)i1D2(yi)[y1,,ˆyi,,yn+1][x1,,xn]+n+1i=1(1)i1D(yi)D([y1,,ˆyi,,yn+1])[x1,,xn]+n+1i=1nj=1(1)i+j1D(xj)[D(yi)[y1,,ˆyi,,yn+1],x1,,ˆxj,,xn](3.1)=n+1i=1(1)i1D2(yi)[y1,,ˆyi,,yn+1][x1,,xn]+n+1i=1n+1k=1,ki(1)i1D(yi)[y1,,D(yk),,ˆyi,,yn+1][x1,,xn]+n+1i=1nj=1(1)i+j1D(xj)[D(yi)[y1,,ˆyi,,yn+1],x1,,ˆxj,,xn]=n+1i=1(1)i1D2(yi)[y1,,ˆyi,,yn+1][x1,,xn]+n+1k=1k1i=1(1)k+i1D(yi)[D(yk),y1,,ˆyi,,ˆyk,,yn+1][x1,,xn]+n+1k=1n+1i=k+1(1)i+kD(yi)[D(yk),y1,,ˆyk,,ˆyi,,yn+1][x1,,xn]+n+1i=1nj=1(1)i+j1D(xj)[D(yi)[y1,,ˆyi,,yn+1],x1,,ˆxj,,xn].

    On the other hand, for any 1kn, we have

    (1)k1[[yk,x1,,xn],y1,,ˆyk,,yn+1](3.3)=(1)k1[D(yk)[x1,,xn],y1,,ˆyk,,yn+1]+nj=1(1)j+k1[D(xj)[yk,x1,,ˆxj,,xn],y1,,ˆyk,,yn+1](3.3)=(1)k1D(D(yk)[x1,,xn])[y1,,ˆyk,,yn+1]+k1i=1(1)i+k1D(yi)[D(yk)[x1,,xn],y1,,ˆyi,,ˆyk,,yn+1]+n+1i=k+1(1)i+kD(yi)[D(yk)[x1,,xn],y1,,ˆyk,,ˆyi,,yn+1]+nj=1(1)j+k1D(D(xj)[yk,x1,,ˆxj,,xn])[y1,,ˆyk,,yn+1]+nj=1n+1i=k+1((1)i+jD(yi)[D(xj)[yk,x1,,ˆxj,,xn],y1,,ˆyk,,ˆyi,,yn+1])+nj=1k1i=1((1)i+j1D(yi)[D(xj)[yk,x1,,ˆxj,,xn],y1,,ˆyi,,ˆyk,,yn+1])=(1)k1D2(yk)[x1,,xn][y1,,ˆyk,,yn+1]+k1i=1(1)i+k1D(yi)[D(yk)[x1,,xn],y1,,ˆyi,,ˆyk,,yn+1]+n+1i=k+1(1)i+kD(yi)[D(yk)[x1,,xn],y1,,ˆyk,,ˆyi,,yn+1]+nj=1(1)j+k1D2(xj)[yk,x1,,ˆxj,,xn][y1,,ˆyk,,yn+1]+nj=1n+1i=k+1((1)i+jD(yi)[D(xj)[yk,x1,,ˆxj,,xn],y1,,ˆyk,,ˆyi,,yn+1])+nj=1k1i=1((1)i+j1D(yi)[D(xj)[yk,x1,,ˆxj,,xn],y1,,ˆyi,,ˆyk,,yn+1])+(1)k1D(yk)D([x1,,xn])[y1,,ˆyk,,yn+1]+nj=1(1)j+k1D(xj)D([yk,x1,,ˆxj,,xn])[y1,,ˆyk,,yn+1](3.2)=(1)k1D2(yk)[x1,,xn][y1,,ˆyk,,yn+1]+k1i=1(1)i+k1D(yi)[D(yk)[x1,,xn],y1,,ˆyi,,ˆyk,,yn+1]+n+1i=k+1(1)i+kD(yi)[D(yk)[x1,,xn],y1,,ˆyk,,ˆyi,,yn+1]+nj=1(1)j+k1D2(xj)[yk,x1,,ˆxj,,xn][y1,,ˆyk,,yn+1]+nj=1n+1i=k+1((1)i+jD(yi)[D(xj)[yk,x1,,ˆxj,,xn],y1,,ˆyk,,ˆyi,,yn+1])+nj=1k1i=1((1)i+j1D(yi)[D(xj)[yk,x1,,ˆxj,,xn],y1,,ˆyi,,ˆyk,,yn+1])+nj=1nt=j+1(1)kyk[x1,,D(xj),,D(xt),,xn][y1,,ˆyk,,yn+1]+ni=1nj=1,ji(1)k+ixi[D(yk),x1,,D(xj),,ˆxi,,xn][y1,,ˆyk,,yn+1]+ni=1nj=1,jint=j+1,ti(1)k+ixi[yk,x1,,D(xj),D(xt),,ˆxi,,xn][y1,,ˆyk,,yn+1].

    We denote

    n+1i=1(1)i1[[yi,x1,,xn],y1,,ˆyi,,yn+1]=7i=1Ai,

    where

    A1:=n+1i=1(1)i1D2(yi)[x1,,xn][y1,,ˆyi,,yn+1],A2:=n+1k=1nj=1(1)k+j1D2(xj)[yk,x1,,ˆxj,,xn][y1,,ˆyk,,yn+1],A3:=n+1i=1nj=1nk=j+1(1)iyi[x1,,D(xj),,D(xk),,xn][y1,,ˆyi,,yn+1],A4:=n+1k=1ni=1nj=1,jint=j+1,ti((1)k+ixi[yk,x1,,D(xj),D(xt),,ˆxi,,xn][y1,,ˆyk,,yn+1]),A5:=n+1k=1k1i=1(1)k+i1D(yi)[D(yk)[x1,,xn],y1,,ˆyi,,ˆyk,,yn+1]+n+1k=1n+1i=k+1(1)i+kD(yi)[D(yk)[x1,,xn],y1,,ˆyk,,ˆyi,,yn+1],A6:=n+1k=1ni=1nj=1,ji((1)k+ixi[D(yk),x1,,D(xj),,ˆxi,,xn][y1,,ˆyk,,yn+1]),A7:=n+1k=1nj=1n+1i=k+1((1)k+i+jD(yi)[D(xj)[yk,x1,,ˆxj,,xn],y1,,ˆyk,,ˆyi,,yn+1])+n+1k=1nj=1k1i=1((1)k+i+j1D(yi)[D(xj)[yk,x1,,ˆxj,,xn],y1,,ˆyi,,ˆyk,,yn+1]).

    By Eq (2.5), for fixed j, we have

    n+1k=1(1)k+j1D2(xj)[yk,x1,,ˆxj,,xn][y1,,ˆyk,,yn+1]=0.

    So, we obtain that A2=0.

    By Eq (2.3), for fixed j and k, we have

    n+1i=1(1)iyi[x1,,D(xj),,D(xk),,xn][y1,,ˆyi,,yn+1]=0.

    So, we obtain that A3=0.

    By Eq (2.5), for fixed j and t, we have

    n+1k=1(1)k+ixi[yk,x1,,D(xj),D(xt),,ˆxi,,xn][y1,,ˆyk,,yn+1]=0.

    So, we obtain that A4=0.

    By Eq (2.9), for fixed i and k, we have

    (1)k+i1D(yi)[D(yk)[x1,,xn],y1,,ˆyi,,ˆyk,,yn+1]+(1)i+kD(yk)[D(yi)[x1,,xn],y1,,ˆyi,,ˆyk,,yn+1]=(1)k+i1D(yi)[D(yk),y1,,ˆyi,,ˆyk,,yn+1][x1,,xn]+(1)i+kD(yk)[D(yi),y1,,ˆyi,,ˆyk,,yn+1][x1,,xn].

    Thus, we obtain

    A5=n+1k=1k1i=1(1)k+i1D(yi)[D(yk),y1,,ˆyi,,ˆyk,,yn+1][x1,,xn]+n+1k=1n+1i=k+1(1)i+kD(yi)[D(yk),y1,,ˆyk,,ˆyi,,yn+1][x1,,xn].

    By Eq (2.3), for fixed j and k, we have

    ni=1(1)k+ixi[D(yk),x1,,D(xj),,ˆxi,,xn]=(1)k1D(yk)[x1,,D(xj),,xn]+(1)k+j1D(xj)[D(yk),x1,,xn]=(1)k+jD(yk)[D(xj),x1,,ˆxj,,xn]+(1)k+j1D(xj)[D(yk),x1,,xn].

    Thus, we get

    A6=n+1k=1nj=1(1)k+jD(yk)[D(xj),x1,,ˆxj,,xn][y1,,ˆyk,,yn+1]+n+1k=1nj=1(1)k+j1D(xj)[D(yk),x1,,xn][y1,,ˆyk,,yn+1].

    By Eq (2.4), for fixed j and i, we have

    n+1k=i+1(1)k+i+j1D(yi)[D(xj)[yk,x1,,ˆxj,,xn],y1,,ˆyi,,ˆyk,,yn+1]+i1k=1(1)k+i+jD(yi)[D(xj)[yk,x1,,ˆxj,,xn],y1,,ˆyk,,ˆyi,,yn+1]=(1)j+i1D(yi)[D(xj)[y1,,ˆyi,,yn+1],x1,,ˆxj,,xn].

    So, we obtain

    A7=nj=1n+1i=1(1)j+i1D(yi)[D(xj)[y1,,ˆyi,,yn+1],x1,,ˆxj,,xn].

    By Eq (2.9), we have

    (1)i+jD(yi)[D(xj),x1,,ˆxj,,xn][y1,,ˆyi,,yn+1]+(1)i+j1D(xj)[D(yi),x1,,xn][y1,,ˆyi,,yn+1]+(1)j+i1D(yi)[D(xj)[y1,,ˆyi,,yn+1],x1,,ˆxj,,xn]=(1)j+i1D(xj)[D(yi)[y1,,ˆyi,,yn+1],x1,,ˆxj,,xn].

    So, we get

    A6+A7=n+1i=1nj=1(1)j+i1D(xj)[D(yi)[y1,,ˆyi,,yn+1],x1,,ˆxj,,xn].

    Thus, we have

    7i=1Ai=A1+A5+A6+A7=[[y1,,yn+1],x1,,xn].

    Therefore, (L,μn+1) is an (n+1)-Lie algebra.

    Now, we can prove Conjecture 1.1 for strong transposed Poisson n-Lie algebras.

    Theorem 3.2. With the notations in Theorem 3.1, (L,,μn+1) is a strong transposed Poisson (n+1)-Lie algebra.

    Proof. For convenience, we denote μn+1(x1,,xn+1):=[x1,,xn+1]. According to Theorem 3.1, we only need to prove Eqs (2.2) and (2.8).

    Proof of Eq (2.2). By Eq (3.3), we have

    n+1i=1[x1,,hxi,,xn+1]=D(hx1)[x2,,xn+1]+n+1j=2(1)j1D(xj)[hx1,x2,,ˆxj,,xn+1]D(hx2)[x1,x3,,xn+1]+n+1j=1,j2(1)j1D(xj)[x1,hx2,x3,,ˆxj,,xn+1]++(1)nD(hxn)[x1,,xn]+nj=1(1)j1D(xj)[x1,,ˆxj,,xn,hxn+1]=n+1i=1(1)i1D(hxi)[x1,,ˆxi,,xn]+n+1i=1n+1j=1,ji(1)j1D(xj)[x1,,,hxi,,ˆxj,,xn+1]=n+1i=1(1)i1D(hxi)[x1,,ˆxi,,xn+1]+n+1j=1n+1i=1,ij(1)j1D(xj)[x1,,hxi,,ˆxj,,xn+1]=n+1i=1(1)i1hD(xi)[x1,,ˆxi,,xn+1]+n+1i=1(1)i1xiD(h)[x1,,ˆxi,,xn+1]+n+1j=1n+1i=1,ij(1)j1D(xj)[x1,,hxi,,ˆxj,,xn+1](2.3)=n+1i=1(1)i1hD(xi)[x1,,ˆxi,,xn+1]+n+1j=1n+1i=1,ij(1)j1D(xj)[x1,,hxi,,ˆxj,,xn+1](3.3)=h[x1,,xn+1]+n+1j=1n+1i=1,ij(1)j1D(xj)[x1,,hxi,,ˆxj,,xn+1](2.2)=h[x1,,xn+1]+nhn+1j=1(1)j1D(xj)[x1,,ˆxj,,xn+1](3.3)=h[x1,,xn+1]+nh[x1,,xn+1]=(n+1)h[x1,,xn+1].

    Proof of Eq (2.8). By Eq (3.3), we have

    y1[hy2,x1,,xn]y2[hy1,x1,,xn]+ni=1(1)i1hxi[y1,y2,x1,,ˆxi,,xn]=y1y2D(h)[x1,,xn]+y1hD(y2)[x1,,xn]y1D(x1)[hy2,x2,,xn]+y1D(x2)[hy2,x1,x3,,xn]++(1)ny1D(xn)[hy2,x1,,xn1]y2y1D(h)[x1,,xn]y2hD(y1)[x1,,xn]+y2D(x1)[hy1,x2,,xn]y2D(x2)[hy1,x1,x3,,xn]++(1)n1y2D(xn)[hy1,x1,,xn1]+hx1D(y1)[y2,x2,,xn]hx1D(y2)[y1,x2,,xn]+hx1D(x2)[y1,y2,x3,,xn]++(1)n+1hx1D(xn)[y1,y2,x2,,xn1]hx2D(y1)[y2,x1,x3,,xn]+hx2D(y2)[y1,x1,x3,,xn]hx2D(x1)[y1,y2,x3,,xn]++(1)n+2hx2D(xn)[y1,y2,x1,x3,,xn1]++(1)n1hxnD(y1)[y2,x1,,xn1]+(1)nhxnD(y2)[y1,x1,,xn1]+(1)n+1hxnD(x1)[y1,y2,x2,,xn1]++(1)2n1hxnD(xn1)[y1,y2,x1,,xn2]=y2hD(y1)[x1,,xn]+hx1D(y1)[y2,x2,,xn]+ni=2(1)i1hxiD(y1)[y2,x1,,ˆxi,,xn]+y1hD(y2)[x1,,xn]hx1D(y2)[y1,x2,,xn]+ni=2(1)ihxiD(y2)[y1,x1,,ˆxi,,xn]y1D(x1)[hy2,x2,,xn]+y2D(x1)[hy1,x2,,xn]+ni=2(1)i1hxiD(x1)[y1,y2,x2,,ˆxi,,xn]+y1D(x2)[hy2,x1,x3,,xn]y2D(x2)[hy1,x1,x3,,xn]+hx1D(x2)[y1,y2,x3,,xn]+ni=3(1)ihxiD(x2)[y1,y2,x1,x3,,ˆxi,,xn]+(1)ny1D(xn)[hy2,x1,,xn1]+(1)n1y2D(xn)[hy1,x1,,xn1]+n1j=1(1)n+j1hxjD(xn)[y1,y2,x1,,ˆxj,,xn1]=A1+A2+ni=1Bi,

    where

    A1:=y2hD(y1)[x1,,xn]+ni=1(1)i1hxiD(y1)[y2,x1,,ˆxi,,xn],A2:=y1hD(y2)[x1,,xn]+ni=1(1)ihxiD(y2)[y1,x1,,ˆxi,,xn],

    and, for any 1in,

    Bi:=(1)iy1D(xi)[hy2,x1,,ˆxi,,xn]+(1)i1y2D(xi)[hy1,x1,,ˆxi,,xn]+i1j=1(1)i+j1hxjD(xi)[y1,y2,x1,,ˆxj,,ˆxi,,xn]+nj=i+1(1)i+jhxjD(xi)[y1,y2,x1,,ˆxi,,ˆxj,,xn].

    By Eq (2.3), we have

    A1=hD(y1)(y2[x1,,xn]+ni=1(1)i1xi[y2,x1,,ˆxi,,xn])=0.

    Similarly, we have that A2=0.

    By Eq (2.8), for any 1in, we have

    Bi=(1)iD(xi)(y1[hy2,x1,,ˆxi,,xn]y2[hy1,x1,,ˆxi,,xn]+i1j=1(1)j1hxj[y1,y2,x1,,ˆxj,,ˆxi,,xn]+nj=i+1(1)jhxj[y1,y2,x1,,ˆxi,,ˆxj,,xn])=0.

    Thus, we get

    y1[hy2,x1,,xn]y2[hy1,x1,,xn]+ni=1(1)i1hxi[y1,y2,x1,,ˆxi,,xn]=0.

    The proof is completed.

    Example 3.1. The commutative associative algebra L=k[x1,x2,x3], together with the bracket

    [x,y]:=xD1(y)yD1(x),x,yL.

    gives a transposed Poisson algebra (L,,[,]), where D1=x1 ([2, Proposition 2.2]). Note that the transposed Poisson algebra (L,,[,]) is strong according to Remark 2.5. Now, let D2=x2; one can check that D2 is a derivation of (L,,[,]). Then, there exists a strong transposed Poisson 3-Lie algebra defined by

    [x,y,z]:=D2(x)(yD1(z)zD1(y))+D2(y)(zD1(x)xD1(z))+D2(z)(xD1(y)yD1(x)), x,y,zL.

    We note that [x1,x2,x3]=x3, which is non-zero. The strong condition can be checked as follows:

    For any h,y1,y2,z1,z2L, by a direct calculation, we have

    y1[hy2,z1,z2]=y1z1hD1(z2)D2(y2)y1z2hD1(z1)D2(y2)+y1y2z1D1(z2)D2(h)y1y2z2D1(z1)D2(h)y1y2hD1(z2)D2(z1)+y1z2hD1(y2)D2(z1)+y1y2z2D1(h)D2(z1)+y1y2hD1(z1)D2(z2)y1z1hD1(y2)D2(z2)y1y2z1D1(h)D2(z2),
    y2[hy1,z1,z2]=y2z1hD1(z2)D2(y1)+y2z2hD1(z1)D2(y1)y1y2z1D1(z2)D2(h)+y1y2z2D1(z1)D2(h)+y1y2hD1(z2)D2(z1)y2z2hD1(y1)D2(z1)y1y2z2D1(h)D2(z1)y1y2hD1(z1)D2(z2)+y2z1hD1(y1)D2(z2)+y1y2z1D1(h)D2(z2),
    hz1[y1,y2,z2]=hy2z1D1(z2)D2(y1)hz1z2D1(y2)D2(y1)hy1z1D1(z2)D2(y2)+hz1z2D1(y1)D2(y2)+hy1z1D1(y2)D2(z2)hy2z1D1(y1)D2(z2),
    hz2[y1,y2,z1]=hy2z2D1(z1)D2(y1)+hz1z2D1(y2)D2(y1)+hy1z2D1(z1)D2(y2)hz1z2D1(y1)D2(y2)hy1z2D1(y2)D2(z1)+hy2z2D1(y1)D2(z1).

    Thus, we get

    y1[hy2,z1,z2]y2[hy1,z1,z2]+hz1[y1,y2,z2]hz2[y1,y2,z1]=0.

    We have studied transposed Poisson n-Lie algebras. We first established an important class of identities for transposed Poisson n-Lie algebras, which were subsequently used throughout the paper. We believe that the identities developed here will be useful in investigations of the structure of transposed Poisson n-Lie algebras in the future. Then, we introduced the notion of a strong transposed Poisson n-Lie algebra and derived an (n+1)-Lie algebra from a strong transposed Poisson n-Lie algebra with a derivation. Finally, we proved the conjecture of Bai et al. [2] for strong transposed Poisson n-Lie algebras.

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

    Ming Ding was supported by the Guangdong Basic and Applied Basic Research Foundation (2023A1515011739) and the Basic Research Joint Funding Project of University and Guangzhou City under grant number 202201020103.

    The authors declare that there is no conflict of interest.



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