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Review

A systematic review on the incidence and risk factors of surgical site infections following hepatopancreatobiliary (HPB) surgery

  • Received: 13 January 2022 Revised: 31 March 2022 Accepted: 06 April 2022 Published: 20 April 2022
  • Background 

    Surgical site infections (SSI) are one of the most common hospital acquired infections and result in increased morbidity, mortality and financial burden on health services. The incidence of SSIs are not clearly defined and infection rates as varied as 20%–40% have been reported. The aim of this study was to systematically review the incidence and risk factors of SSI following HPB surgery.

    Methods 

    The database of Medline (via PubMed) was systematically searched from 2013–2022. Articles were screened using the PRISMA statement and those that met the inclusion criteria were included in the study.

    Results 

    Sixteen studies were eligible for inclusion in this systematic review. The average incidence of SSI was 29.8%. Key risk factors identified included male gender, open surgery, preoperative biliary stenting and obesity.

    Conclusions 

    The incidence of SSI following HPB surgery varied, but it is generally high. A variety of pre-disposing patient factors can affect infection rates following HPB surgery. The results from this study suggest that perhaps laparoscopic surgery should be used where possible, and that there should be an awareness that gender, obesity and the use of stents may increase the incidence of SSIs following these operations.

    Citation: Lucy E. Chambers, Aali J. Sheen, Kathryn A. Whitehead. A systematic review on the incidence and risk factors of surgical site infections following hepatopancreatobiliary (HPB) surgery[J]. AIMS Bioengineering, 2022, 9(2): 123-144. doi: 10.3934/bioeng.2022010

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  • Background 

    Surgical site infections (SSI) are one of the most common hospital acquired infections and result in increased morbidity, mortality and financial burden on health services. The incidence of SSIs are not clearly defined and infection rates as varied as 20%–40% have been reported. The aim of this study was to systematically review the incidence and risk factors of SSI following HPB surgery.

    Methods 

    The database of Medline (via PubMed) was systematically searched from 2013–2022. Articles were screened using the PRISMA statement and those that met the inclusion criteria were included in the study.

    Results 

    Sixteen studies were eligible for inclusion in this systematic review. The average incidence of SSI was 29.8%. Key risk factors identified included male gender, open surgery, preoperative biliary stenting and obesity.

    Conclusions 

    The incidence of SSI following HPB surgery varied, but it is generally high. A variety of pre-disposing patient factors can affect infection rates following HPB surgery. The results from this study suggest that perhaps laparoscopic surgery should be used where possible, and that there should be an awareness that gender, obesity and the use of stents may increase the incidence of SSIs following these operations.



    The Schrödinger-Virasoro algebra is an infinite-dimensional Lie algebra that was introduced (see, e.g., [10]) in the context of non-equilibrium statistical physics. In [21], the author give a representation of the Schrödinger-Virasoro algebra by using vertex algebras, and introduced an extension of the Schrödinger-Virasoro algebra. To be precise, for ε{0,12}, the Schrödinger-Virasoro algebra SV(ε) is a Lie algebra with the C basis

    {Li,Hj,Ii|iZ,jε+Z}

    and Lie brackets

    [Lm,Ln]=(mn)Lm+n,[Lm,Hn]=(12mn)Hm+n,[Lm,In]=nIm+n,[Hm,Hn]=(mn)Im+n,[Hm,In]=[Im,In]=0.

    The Lie algebra SV(12) is called the original Schrödinger-Virasoro algebra, and SV(0) is called the twisted Schrödinger-Virasoro algebra. Recently, the theory of the structure and representations of both original and twisted Schrödinger-Virasoro algebra has been investigated in a series of studies. For instance, the Lie bialgebra structures, (bi)derivations, automorphisms, 2-cocycles, vertex algebra representations and Whittaker modules were investigated in [9,11,14,15,21].

    Post-Lie algebras were introduced around 2007 by B. Vallette [22], who found the structure in a purely operadic manner as the Koszul dual of a commutative trialgebra. Post-Lie algebras have arose the interest of a great many authors, see [4,5,12,13]. One of the most important problems in the study of post-Lie algebras is to find the post-Lie algebra structures on the (given) Lie algebras. In [13,18,20], the authors determined all post-Lie algebra structures on sl(2,C) of special linear Lie algebra of order 2, the Witt algebra and the W-algebra W(2,2) respectively.

    In this paper, we shall study the graded post-Lie algebra structures on the Schrödinger-Virasoro algebra. We only study the twisted Schrödinger-Virasoro algebra SV(0), the case for the original Schrödinger-Virasoro algebra SV(12) is similar. For convenience we denote S=SV(0). It should be noted that the commutative post-Lie algebra structures on S already are given by [11], we will consider the general case.

    Throughout this paper, we denote by Z the set of all integers. For a subset S of Z and a fixed integer k, denote S=S{0}, S>k={tS|t>k}, S<k={tS|t<k}, Sk={tS|tk} and Sk={tS|tk}. We assume that the field in this paper always is the complex number field C.

    The paper is organized as follows. In Section 2, we give general results on post-Lie algebras and some lemmas which will be used to our proof. In Section 3, we completely characterize the graded post-Lie algebra structures on Schrödinger-Virasoro algebra S. In Section 4, by using the post-Lie algebra structures we characterize the forms of the homogeneous Rota-Baxter operator on S.

    We will give the essential definitions and results as follows.

    Definition 2.1. A post-Lie algebra (V,,[,]) is a vector space V over a field k equipped with two k-bilinear products xy and [x,y] satisfying that (V,[,]) is a Lie algebra and

    [x,y]z=x(yz)y(xz)x,yz, (1)
    x[y,z]=[xy,z]+[y,xz] (2)

    for all x,yV, where x,y=xyyx. We also say that (V,,[,]) is a post-Lie algebra structure on the Lie algebra (V,[,]). If a post-Lie algebra (V,,[,]) satisfies xy=yx for all x,yV, then it is called a commutative post-Lie algebra.

    Suppose that (L,[,]) is a Lie algebra. Two post-Lie algebras (L,[,],1) and (L,[,],2) on the Lie algebra L are called to the isomorphic if there is an automorphism τ of the Lie algebra (L,[,]) satisfies

    τ(x1y)=τ(x)2τ(y),x,yL.

    Remark 1. The left multiplications of the post-Lie algebra (V,[,],) are denoted by L, i.e., we have L(x)(y)=xy for all x,yV. By (2), we see that all operator L(x) are Lie algebra derivations of the Lie algebra (V, [, ]).

    Lemma 2.2. [15] Denote by Der(S) and by Inn(S) the space of derivations and the space of inner derivations of S respectively. Then

    Der(S)=Inn(S)CD1CD2CD3

    where D1,D2,D3 are outer derivations defined by

    D1(Ln)=0,D1(Hn)=Hn,D1(In)=2In,D2(Ln)=nIn,D2(Hn)=0,D2(In)=0,D3(Ln)=In,D3(Hn)=0,D3(In)=0.

    Since the Schrödinger-Virasoro algebra S is graded, we suppose that the post-Lie algebra structure on the Schrödinger-Virasoro algebra S to be graded. Namely, we mainly consider the post-Lie algebra structure on Schrödinger-Virasoro algebra S which satisfies

    LmLn=ϕ(m,n)Lm+n, (3)
    LmHn=φ(m,n)Hm+n, (4)
    LmIn=χ(m,n)Im+n, (5)
    HmLn=ψ(m,n)Hm+n, (6)
    HmHn=ξ(m,n)Im+n, (7)
    ImLn=θ(m,n)Im+n, (8)
    HmIn=ImHn=ImIn=0, (9)

    for all m,nZ, where ϕ, φ, χ, ψ, ξ, θ are complex-valued functions on Z×Z.

    We start with the crucial lemma.

    Lemma 3.1. There exists a graded post-Lie algebra structure on S satisfying (3)-(9) if and only if there are complex-valued functions f,g,h on Z and complex numbers a,μ such that

    ϕ(m,n)=(mn)f(m), (10)
    φ(m,n)=(m2n)f(m)+δm,0μ, (11)
    χ(m,n)=nf(m)+2δm,0μ, (12)
    ψ(m,n)=(n2m)h(m), (13)
    ξ(m,n)=(mn)h(m), (14)
    θ(m,n)=mg(m)+δm,0na, (15)
    (mn)(f(m+n)(1+f(m)+f(n))f(n)f(m))=0, (16)
    (mn)δm+n,0μ(1+f(m)+f(n))=0, (17)
    (m2n)(h(m+n)(1+f(m)+h(n))f(m)h(n))=0, (18)
    nδm+n,0a(1+f(m)+g(n))=0, (19)
    n(m+n)(g(m+n)(1+f(m)+g(n))f(m)g(n))    =δn,0m2a(f(m)g(m)), (20)
    (mn)δm+n,0a(1+h(m)+h(n))=0, (21)
    (mn)(g(m+n)(1+h(m)+h(n))h(m)h(n))=0. (22)

    Proof. Suppose that there exists a graded post-Lie algebra structure satisfying (3)-(9) on S. By Remark 1, L(x) is a derivation of S. It follows by Lemma 2.2 that there are a linear map ψ from S into itself and linear functions α,β,γ on S such that

    xy=(adψ(x)+α(x)D1+β(x)D2+γ(x)D3)(y)=[ψ(x),y]+α(x)D1(y)+β(x)D2(y)+γ(x)D3(y)

    where Di,i=1,2,3 are given by Lemma 2.2. This, together with (3)-(9), gives that

    LmLn=[ψ(Lm),Ln]+β(Lm)nIn+γ(Lm)In=ϕ(m,n)Lm+n, (23)
    LmHn=[ψ(Lm),Hn]+α(Lm)Hn=φ(m,n)Hm+n, (24)
    LmIn=[ψ(Lm),In]+α(Lm)2In=χ(m,n)Im+n, (25)
    HmLn=[ψ(Hm),Ln]+β(Hm)nIn+γ(Hm)In=ψ(m,n)Hm+n, (26)
    HmHn=[ψ(Hm),Hn]+α(Hm)Hn=ξ(m,n)Im+n, (27)
    HmIn=[ψ(Hm),In]+α(Hm)2In=0, (28)
    ImLn=[ψ(Im),Ln]+β(Im)nIn+γ(Im)In=θ(m,n)Im+n, (29)
    ImHn=[ψ(Im),Hn]+α(Im)Hn=0, (30)
    ImIn=[ψ(Im),In]+α(Im)2In=0. (31)

    Let

    ψ(Lm)=iZa(m)iLi+iZb(m)iHi+iZc(m)iIi,ψ(Hm)=iZd(m)iLi+iZe(m)iHi+iZf(m)iIi,ψ(Im)=iZg(m)iLi+iZh(m)iHi+iZx(m)iIi

    where a(m)i,b(m)i,c(m)i,d(m)i,e(m)i,f(m)i,g(m)i,h(m)i,x(m)iC for all iZ. Then by (23)-(31), similar to the proof of [18], we obtain that (10)-(22) hold.

    The "if'' part is a direct checking. The proof is completed.

    Lemma 3.2. Let f,g,h be complex-valued functions on Z and μ,aC satisfying (18) and (20). Then we have

    g(n),h(n){0,1}for everyn0. (32)

    Proof. By letting m=0 in (18) and (20), respectively, we have nh(n)(1+h(n))=0 and n2g(n)(1+g(n))=0. This implies (32).

    Lemma 3.3. Let f,g,h be complex-valued functions on Z and μ,a be complex numbers satisfying (17)-(22). If f(Z)=0, then we have μ=a=0 and

    g(Z)=h(Z)=0org(Z)=h(Z)=1.

    Proof. Since f(Z)=0, we take m=n=1 in (17) and (19) we have μ=0 and

    a(1+g(1))=0. (33)

    By letting n=0 and m=1 in (20) we deduce that ag(1)=0. This, together with (33), implies a=0. As μ=a=0, so Equations (18), (20) and (22) become to

    (m2n)(h(m+n)(1+h(n))=0, (34)
    n(m+n)(g(m+n)(1+g(n))=0, (35)
    (mn)(g(m+n)h(m)h(n)+h(m)g(m+n)+h(n)g(m+n))=0. (36)

    We now prove the following four claims:

    Claim 1. If h(1)=0, then h(Z)=0.

    By (34) with n=1 we see that h(m+1)=0 for all m2. It follows that h(Z{3})=0. Since h(2)=0, by taking n=2,m=1 in (34) we have 32h(3)=0, which implies h(3)=0. We obtain h(Z)=0.

    Claim 2. If h(1)=1, then h(Z)=1.

    By (34) with m+n=1 we see that h(n)=1 for all nZ with 13n20. This means that h(Z)=1.

    Claim 3. If g(1)=0, then g(Z)=0.

    By (35) with n=1 we see that g(m+1)=0 for all m1. It follows that g(Z)=0.

    Claim 4. If g(1)=1, then g(Z)=1.

    By (35) with m+n=1 we see that g(n)=1 for all n0. This means that g(Z)=1.

    Now we consider the values of h(1) and g(1) according to (32).

    Case i. If h(1)=g(1)=0, then by Claims 1 and 3 we have h(Z)=0 and g(Z)=0. According to (36) with n=1 and m=1 we know g(0)=0. This means that g(Z)=0.

    Case ii. If h(1)=g(1)=1, then by Claims 2 and 4 we have h(Z)=1 and g(Z)=1. According to (36) with n=1 and m=1 we see that 1+g(0)=0 and so that g(0)=1. This implies g(Z)=1.

    Case iii. If h(1)=0, g(1)=1, then we will get a contradiction. In fact, by Claims 1 and 4, we have h(Z)=0 and g(Z)=1. From (36) with m=2,n=1 we see that g(1)=0 which contradicts g(1)=1.

    Case iv. If h(1)=1, g(1)=0, then we will also get a contradiction. In fact, by Claims 2 and 3, we have h(Z)=1 and g(Z)=0. From (36) with with m=2, n=1 we see that g(1)=1 which contradicts g(1)=0. The proof is completed.

    Lemma 3.4. Let f,g,h be complex-valued functions on Z and μ,a be complex numbers satisfying (17)-(22). If f(Z2)=1,f(Z1)=0, then μ=a=0 and g, h must satisfy one of the following forms:

    (i) g(Z)=h(Z)=0;

    (ii) g(Z)=h(Z)=1;

    (iii) h(Z0)=0, h(Z1)=1 and

    g(Z1)=0, g(Z1)=1, g(0)=ˆλ for some ˆλC.

    Proof. By f(Z2)=1,f(Z1)=0, similar to the proof of Lemma 3.3, we know μ=a=0. From this, we have by (18), (20) and (22) that

    h(m+n)(h(n)+1)=0 if m1,m2n0, (37)
    g(m+n)(g(n)+1)=0 if m1,n0,m+n0, (38)
    g(m+n)(1+h(m)+h(n))=h(m)h(n) if mn. (39)

    We first prove the following six claims:

    Claim 1. If h(1)=0, then h(Z)=0.

    By (37) with n=1 we see that h(m+1)=0 for all m210 with m1. Hence, we deduce that h(Z2)=0. Note that h(2)=0, by (37) with n=2 we see that h(m+2)=0 for all m220 with m1. We now have h(Z3)=0. If we repeat this process, we see that h(Zk)=0 for all k=1,2,3,. Note that k1(Zk)=Z, so one has h(Z)=0.

    Claim 2. If h(1)=1, then h(Z)=1.

    By (37) with m+n=1 we see that h(n)=h(1m)=1 for all 3m2+10 with m1. This deduces that h(Z2)=1. Note that h(2)=1, by (37) with m+n=2 we see that h(m2)=1 for all 3m2+20 with m1. Thus, h(Z3)=1. If we repeat this process, we see that h(Zk)=1 for all k=1,2,3,. Note that k1(Zk)=Z, so one has h(Z)=1.

    Claim 3. If h(1)=1, then h(Z1)=1.

    By (37) with m+n=1 we see that h(n)=h(1m)=1 for all 3m210 with m1. This implies h(Z1)=1.

    Claim 4. If h(1)=0, then h(Z0)=0.

    By (37) with n=1 we see that h(m1)=0 for all m2 with m1. It follows that h(Z0{3})=0. Let m=1,n=2 in (37), from m2n we have h(3)=0. Therefore, we get h(Z0)=0.

    Next, similar to Claims 1 and 3, we from (38) obtain the following claims.

    Claim 5. If g(1)=0, then g(Z)=0.

    Claim 6. If g(1)=1, then g(Z1)=1.

    Now we discuss the values of h(1) and h(1). By (32), h(1),h(1){1,0}.

    Case i. When h(1)=0.

    By Claim 1 we have h(Z)=0. According to (39), one has g(m+n)=0 for any m,nZ with mn. This implies g(Z)=0.

    Case ii. When h(1)=1.

    By Claim 2 we have h(Z)=1. According to (39), one has g(m+n)=1 for any m,nZ with mn. This implies g(Z)=1.

    Case iii. When h(1)=1 and h(1)=0.

    By Claims 3 and 4 we have h(Z0)=0 and h(Z1)=1. This, together with (39), yields g(m+n)=0 for any m,nZ with m,n0 and mn, and g(m+n)=1 for any m,nZ with m,n1 and mn. Consequently, we obtain g(Z1)=0 and g(Z3)=1. By (32), g(1){1,0}. If g(1)=0, then Claim 5 tells us that g(Z)=0 which contracts g(Z3)=1. Therefore, we have g(1)=1. From this with Claim 6 we have g(Z1)=1. Let g(0)=ˆλ for some ˆλC.

    It is easy to check that the values of g given in Cases i-iii above are consistent with (38). They give the conclusions (i), (ii) and (iii) respectively. The proof is completed.

    Lemma 3.5. Let f,g,h be complex-valued functions on Z and μ,a be complex numbers satisfying (17)-(22). If f(Z>0)=1,f(Z<0)=0 and f(0)=c for some cC, then there are λ,ˆτC such that μ,a, g, h must be one of the following forms:

    (i) a=0, μC and g(Z)=h(Z)=0;

    (ii) a=0, μC and g(Z)=h(Z)=1;

    (iii) μC, h(Z>0)=1, h(Z<0)=0, h(0)=λ and g(Zk)=1, g(Zk1)=0

    for some k{2,1,1,2,3}, g(0)=ˆτ and a=0 when k1;

    (iv) a=0, μC and h(Zt)=1, h(Zt1)=0 for some tZ{0,1} and

    g(Zs)=1, g(Zs1)=0 for some s{2t2,2t1,2t,2t+1,2t+2}.

    Proof. Take m=n0 in (18)-(22), one has

    h(0)(1+f(n)+h(n))=f(n)h(n), for all n0, (40)
    a(1+f(n)+g(n))=0, for all n0, (41)
    a(1+h(n)+h(n))=0, for all n0, (42)
    g(0)(1+h(n)+h(n))=h(n)h(n), for all n0. (43)

    Note that f(Z>0)=1,f(Z<0)=0 and f(0)=c for some cC. It is follows by (18), (20) and (22) that

    h(n)(h(m+n)+1)=0 for all m>0,m2n0; (44)
    h(m+n)(h(n)+1)=0 for all m<0,m2n0; (45)
    g(n)(g(m+n)+1)=0 for all m>0,n0,m+n0; (46)
    g(m+n)(g(n)+1)=0 for all m<0,n0,m+n0; (47)
    g(m+n)(1+h(m)+h(n))=h(m)h(n) for all mn. (48)

    For any tZ, we first prove some claims as follows.

    Claim 1. If h(t)=0, then h(Zt)=0.

    In fact, by (44) with n=tm we deduce h(tm)=0 for all m>0 with m23t. This implies h(Zt{13t})=0. On the other hand, by (45) with n=t we see that h(m+t)=0 for all m<0 with m2t. This gives that h(Zt{3t})=0. Clearly, 3t13t since t0. Thereby, we obtain h(Zt)=0.

    Claim 2. If h(t)=1, then h(Zt)=1.

    This proof is similar to Claim 1 by using (44) and (45). Also, similar to Claims 1 and 2, by (46) and (47) we can obtain the following two claims:

    Claim 3. If g(t)=0, then g(Zt)=0.

    Claim 4. If g(t)=1, then g(Zt)=1.

    According to (32), by Claims 1 and 2, h must be one of the following forms:

    (1) h(Z)=0;

    (2) h(Z)=1;

    (3) h(Z>0)=1, h(Z<0)=0 and h(0)=λ for some λC;

    (4) h(Zt)=1, h(Zt1)=0 for some tZ{0,1}.

    In view of the above result, the next proof will be divided into the following cases.

    Case i. When h(Z)=0.

    By taking n=1 in (40), one has h(0)=0. Hence we see that h(Z)=0. This together with (48) yields g(Z)=0. In addition, we have by (43) that a=0.

    Case ii. When h(Z)=1.

    By taking n=1 in (40), one has h(0)=1. Hence we see that h(Z)=1. This together with (48) yields g(Z)=1. In addition, by (43) we get a=0.

    Case iii. When h(Z>0)=1, h(Z<0)=0 and h(0)=λ for some λC.

    By (48) we see that g(m+n)=1 for any m,nZ with m,n>0 and mn, and g(m+n)=0 for any m,nZ with m,n<0 and mn. Consequently, we obtain g(Z3)=0 and g(Z3)=1. By (32), g(i){1,0} for i{2,1,1,2}. In view of Claims 3 and 4, we can assume that g(k)=1 and g(k1)=0 for some k{2,1,1,2,3}. In all, by Claims 3 and 4 we get g(Zk)=1 and g(Zk1)=0. Next, if k{1,2} then by taking n=k in (41) we have a=0; and if k{2,3} then by taking n=k1 in (41) we also have a=0. But a can be arbitrary if k=1.

    Case iv. When h(Zt)=1, h(Zt1)=0 for some tZ{0,1}.

    Note that t2 or t1, then by taking n=1 in (42) we have a=0. Next, by(48) we see that g(m+n)=1 for any m,nZ with m,nt and mn, and g(m+n)=0 for any m,nZ with m,nt1 and mn. Consequently, we obtain g(Z2t3)=0 and g(Z2t+1)=1. By (32), g(i){1,0} for i{2t2,2t1,2t,2t+1}. In view of Claims 3 and 4, we can assume that g(s)=1 and g(s1)=0 for some s{2t2,2t1,2t,2t+1,2t+2}. Note that 0{2t2,2t1,2t,2t+1} since t0,1, by Claims 3 and 4 we get g(Zs)=1 and g(Zs1)=0. The proof is completed.

    Lemma 3.6. Let f,g,h be complex-valued functions on Z and μ,a be complex numbers. Then (17)-(22) hold if and only if f,g,h,a,μ meet one of the situations listed in Table 2.

    Table 2.  Values of f,g,h satisfying (16)-(22), where a,μC, k{2,1,1,2,3}, tZ{0,1} and s{2t2,2t1,2t,2t+1,2t+2}.
    Cases f(n) from Table 1 a, μ h(n),g(n)
    WP11 P1 a=μ=0 h(Z)=g(Z)=0
    WP12 P1 a=μ=0 h(Z)=g(Z)=1
    WP21 P2 a=μ=0 h(Z)=g(Z)=0
    WP22 P2 a=μ=0 h(Z)=g(Z)=1
    WPc31,μ Pc3 a=0 and μ h(Z)=g(Z)=0
    WPc32,μ Pc3 a=0 and μ h(Z)=g(Z)=1
    WPc3,k3,μ Pc3 a=0 and μ h(Z>0)=1, h(Z<0)=0 and
    g(Zk)=1, g(Zk1)=0
    WPc3,k=14,a,μ Pc3 a and μ h(Z>0)=1, h(Z<0)=0 and
    g(Z>0)=1, g(Z<0)=0
    WPc3,s,t5,μ Pc3 a=0 and μ h(Zt)=1, h(Zt1)=0 and
    g(Zs)=1, g(Zs1)=0
    WPc41,μ Pc4 a=0 and μ h(Z)=g(Z)=0
    WPc42,μ Pc4 a=0 and μ h(Z)=g(Z)=1
    WPc4,k3,μ Pc4 a=0 and μ h(Z>0)=0, h(Z<0)=1 and
    g(Zk)=0, g(Zk1)=1
    WPc4,k=14,a,μ Pc4 a and μ h(Z>0)=0, h(Z<0)=1 and
    g(Z>0)=0, g(Z<0)=1
    WPc4,s,t5,μ Pc4 a=0 and μ h(Zt)=0, h(Zt1)=1 and
    g(Zs)=0, g(Zs1)=1
    WP51 P5 a=μ=0 h(Z)=g(Z)=0
    WP52 P5 a=μ=0 h(Z)=g(Z)=1
    WP53 P5 a=μ=0 h(Z0)=0, h(Z1)=1 and
    g(Z1)=0, g(Z1)=1
    WP61 P6 a=μ=0 h(Z)=g(Z)=0
    WP62 P6 a=μ=0 h(Z)=g(Z)=1
    WP63 P6 a=μ=0 h(Z0)=1, h(Z1)=0 and
    g(Z1)=1, g(Z1)=0
    WP71 P7 a=μ=0 h(Z)=g(Z)=0
    WP72 P7 a=μ=0 h(Z)=g(Z)=1
    WP73 P7 a=μ=0 h(Z0)=1, h(Z1)=0 and
    g(Z1)=1, g(Z1)=0
    WP81 P8 a=μ=0 h(Z)=g(Z)=0
    WP82 P8 a=μ=0 h(Z)=g(Z)=1
    WP83 P8 a=μ=0 h(Z0)=0, h(Z1)=1 and
    g(Z1)=0, g(Z1)=1

     | Show Table
    DownLoad: CSV

    Proof. The proof of the "if" direction can be directly verified. We now prove the "only if" direction. In view of f satisfying (16), by Theorem 2.4 of [10] we know that f is determined by Table 1.

    Table 1.  Values of f satisfying (16), where cC.
    Cases f(n)
    P1 f(Z)=0
    P2 f(Z)=1
    Pc3 f(Z>0)=1,f(Z<0)=0andf(0)=c
    Pc4 f(Z>0)=0,f(Z<0)=1andf(0)=c
    P5 f(Z2)=1andf(Z1)=0
    P6 f(Z2)=0andf(Z1)=1
    P7 f(Z1)=0andf(Z2)=1
    P8 f(Z1)=1andf(Z2)=0

     | Show Table
    DownLoad: CSV

    When f takes the form of Case P1 in Table 1, by the results of Lemma 3.3, we see that μ,a,g,h must satisfy the condition of Cases WP11 and WP12 in Table 2. From Lemma 3.3, Cases WP11,i=1,2 is easy to say. In the same way, when f takes the form of Case P2 in Table 1, then we obtain the forms of Cases WP21 and WP22 in Table 2.

    When f takes the form of Case Pc3 in Table 1, by the results of Lemma 3.5, we see that μ,a,g,h must satisfy the one condition of Cases WPc3i,μ,i=1,2, WPc3,k3,μ, WPc3,k=14,a,μ and WPc3,s,t5,μ in Table 2. From Lemma 3.5, the results of Cases WPc3i,μ,i=1,2 are easily obtained; and Case WPc3,k3,μ satisfies μC, h(Z>0)=1, h(Z<0)=0, h(0)=λ and g(Zk)=1, g(Zk1)=0, for some k{2,1,1,2,3}, g(0)=ˆτ with a=0 when k1 and a is arbitrary if k=1; Case WPc3,k=14,a,μ satisfies μC, h(Z>0)=1, h(Z<0)=0, h(0)=λ and g(Z>0)=1, g(Z<0)=0 for some k=1, g(0)=ˆτ; Case WPc3,s,t5,μ satisfies a=0, μC and h(Zt)=1, h(Zt1)=0 for some tZ{0,1} and g(Zs)=1, g(Zs1)=0 for some s{2t2,2t1,2t,2t+1,2t+2}. In the same way, when f takes the form of Case Pc4 in Table 1, then we obtain the results of Cases WPc4i,μ,i=1,2, WPc4,k3,μ, WPc4,k=14,a,μ and WPc4,s,t5,μ in Table 2, respectively.

    When f takes the form of Case P5 in Table 1, by the results of Lemma 3.4, we see that μ,a,g,h must satisfy the condition of Cases WP5i,i=1,2,3 in Table 2. From Lemma 3.4, the results of Cases WP5i,i=1,2, are easily obtained; and for Case WP53, we get h(Z0)=0, h(Z1)=1 and g(Z1)=0, g(Z1)=1, g(0)=ˆλ for some ˆλC. Similarly, when f takes the form of Case Pk,k=6,7,8 in Table 1, then we obtain the forms of Cases WPki, i=1,2,3, k=6,7,8 in Table 2. The proof is completed.

    Lemma 3.7. Let (P(ϕi,φi,χi,ψi,ξi,θi),i), i=1,2 be two algebras with the same linear space as S and equipped with C-bilinear products xiy such that

    LmiLn=ϕi(m,n)Lm+n,LmiHn=φi(m,n)Hm+n,LmiIn=χi(m,n)Im+n,HmiLn=ψi(m,n)Hm+n,HmiHn=ξi(m,n)Im+n,ImiLn=θi(m,n)Im+n,HmiIn=ImiHn=ImiIn=0

    for all m,nZ, where ϕi,φi,χi,ψi,ξi,θi, i=1,2 are complex-valued functions on Z×Z. Furthermore, let τ:P(ϕ1,φ1,χ1,ψ1,ξ1,θ1)P(ϕ2,φ2,χ2,ψ2,ξ2,θ2) be a linear map determined by

    τ(Lm)=Lm,τ(Hm)=Hm,τ(Im)=Im

    for all mZ. In addition, suppose that (P(ϕ1,φ1,χ1,ψ1,ξ1,θ1),[,],1) is a post-Lie algebra. Then (P(ϕ2,φ2,χ2,ψ2,ξ2,θ2),[,],,2) is a post-Lie algebra and τ is an isomorphism on post-Lie algebras if and only if

    {ϕ2(m,n)=ϕ1(m,n);φ2(m,n)=φ1(m,n);χ2(m,n)=χ1(m,n);ψ2(m,n)=ψ1(m,n);ξ2(m,n)=ξ1(m,n);θ2(m,n)=θ1(m,n). (49)

    Proof. Clearly, τ is a Lie automorphism of S. Suppose (P(ϕ2,φ2,χ2,ψ2,ξ2,θ2),[,],2) is a post-Lie algebra and τ:P(ϕ1,φ1,χ1,ψ1,ξ1,θ1)P(ϕ2,φ2,χ2,ψ2,ξ2,θ2) is a post-Lie isomorphism. Then we have

    τ(LmiLn)=ϕi(m,n)L(m+n),τ(LmiHn)=φi(m,n)H(m+n),τ(LmiIn)=χi(m,n)I(m+n),τ(HmiLn)=ψi(m,n)H(m+n),τ(HmiHn)=ξi(m,n)I(m+n),τ(ImiLn)=θi(m,n)I(m+n)

    for i=1,2. This tell us that that (49) holds. Conversely, we first suppose that (49) hold. Then, by using Lemma 3.1 and (ϕ1,φ1,χ1,ψ1,ξ1,θ1,[,],1) is a post-Lie algebra, we know that there are complex-valued functions f1,g1,h1 on Z and complex numbers a1,μ1 satisfying (10)-(22) with replacing (ϕ,φ,χ,ψ,ξ,θ,f,g,h,μ,a) by (ϕ1,φ1,χ1,ψ1,ξ1,θ1,f1,g1,h1,μ1,a1). Next, let f2(m)=f1(m), g2(m)=g1(m), h2(m)=h1(m), μ2=μ1 and a2=a1, then we see that (10)-(22) hold with replacing (ϕ,φ,χ,ψ,ξ,θ,f,g,h,μ,a) by (ϕ2,φ2,χ2,ψ2,ξ2,θ2,f1,g1,h1,μ1,a1). By Lemma 3.1, P(ϕ2,φ2,χ2,ψ2,ξ2,θ2) is a post-Lie algebra.

    The remainder is to prove that τ is an isomorphism between post-Lie algebra. But one has

    τ(Lm1Ln)=ϕ1(m,n)L(m+n)=ϕ2(m,n)L(m+n)=τ(Lm)2τ(Ln),τ(Lm1Hn)=φ1(m,n)H(m+n)=φ2(m,n)H(m+n)=τ(Lm)2τ(Hn),τ(Lm1In)=χ1(m,n)I(m+n)=χ2(m,n)I(m+n)=τ(Lm)2τ(In),τ(Hm1Ln)=ψ1(m,n)H(m+n)=ψ2(m,n)H(m+n)=τ(Hm)2τ(Ln),τ(Hm1Hn)=φ1(m,n)I(m+n)=φ2(m,n)I(m+n)=τ(Hm)2τ(Hn),
    τ(Im1Ln)=θ1(m,n)I(m+n)=ϕ2(m,n)I(m+n)=τ(Im)2τ(Ln)

    and τ(Hm1In)=τ(Hm)2τ(In)=0, τ(Im1Hn)=τ(Im)2τ(Hn) = 0, τ(Im1In)=τ(Im)2τ(In) = 0. The proof is completed.

    Theorem 3.8. A graded post-Lie algebra structure on S satisfying (3)-(9) must be one of the following types, for all m,nZ (in every case ImHn=HmIn=ImIn=0),

    (WP11): LmP11Ln=0, LmP11Hn=0, LmP11In=0, HmP11Ln=0, HmP11Hn=0, ImP11Ln=0;

    (WP12): LmP12Ln=0, LmP12Hn=0, LmP12In=0, HmP12Ln=(n2m)Hm+n, HmP12Hn=(nm)Im+n, ImP12Ln=mIm+n;

    (WP21): LmP21Ln=(nm)Lm+n, LmP21Hn=(nm2)Hm+n, LmP21In=nIm+n, HmP21Ln=0, HmP21Hn=0, ImP21Ln=0;

    (WP22): LmP22Ln=(nm)Lm+n, LmP22Hn=(nm2)Hm+n, LmP22In=nIm+n, HmP22Ln=(n2m)Hm+n, HmP22Hn=(nm)Im+n, ImP22Ln=mIm+n;

    (WPc3,s,k,ti,a,μ,λ): i=1,2,3,4,5

    LmPc3iLn={(nm)Lm+n,m>0,ncLn,m=0,0,m<0;

    LmPc3iHn={(nm2)Hm+n,m>0,(nc+μ)Hn,m=0,0,m<0;

    LmPc3iIn={nIm+n,m>0,(nc+2μ)In,m=0,0,m<0;

      HmPc3iLn=δi,2(n2m)Hm+n

         +(δi,3+δi,4){(n2m)Hm+n,m>0,n2λHn,m=0,0,m<0;

         +δi,5{(n2m)Hm+n,mt,0,mt1;

      HmPc3iHn=δi,2(nm)Im+n

         +(δi,3+δi,4){(nm)Im+n,m>0,nλIn,m=0,0,m<0;

         +δi,5{(nm)Im+n,mt,0,mt1;

      ImPc3iLn=δi,2(m)Im+n

         +δi,3{mIm+n,mk,0,mk1;

         +δi,4{mIm+n,m>0,naIn,m=0,0,m<0;

         +δi,5{mIm+n,ms,0,ms1;

    (WPc4,s,k,ti,a,μ,λ): i=1,2,3,4,5

           LmPc4iLn={(nm)Lm+n,m<0,ncLn,m=0,0,m>0;

           LmPc4iHn={(nm2)Hm+n,m<0,(nc+μ)Hn,m=0,0,m>0;

          LmPc4iIn={nIm+n,m<0,(nc+2μ)In,m=0,0,m>0;

    HmPc4iLn=δi,2(n2m)Hn+m

         +(δi,3+δi,4){0,m>0,n2λHn,m=0,(n2m)Hm+n,m<0;

         +δi,5{0,mt,(n2m)Hm+n,mt1;

    HmPc4iHn=δi,2(nm)In+m

         +(δi,3+δi,4){0,m>0,nλIn,m=0,(nm)Im+n,m<0;

         +δi,5{0,mt,(nm)Im+n,mt1;

    ImPc4iLn=δi,2(m)In+m

         +δi,3{0,mk,mIm+n,mk1;

         +δi,4{0,m>0,naIn,m=0,mIm+n,m<0;

         +δi,5{0,ms,mIm+n,ms1;

    (WP5i): i=1,2,3,

           LmP5iLn={(nm)Lm+n,m2,0,m1;

           LmP5iHn={(nm2)Lm+n,m2,0,m1;

          LmP5iIn={nIm+n,m2,0,m1;

    HmP5iLn=δi,2(n2m)Hm+n

         +δi,3{0,m0,(n2m)Hm+n,m1;

    HmP5iHn=δi,2(nm)Im+n

         +δi,3{0,m0,(nm)Im+n,m1;

    ImP5iLn=δi,2(m)Im+n

         +δi,3{0,m0,mIm+n,m1;

    (WP6i): i=1,2,3,

           LmP6iLn={(nm)Lm+n,m1,0,m2;

           LmP6iHn={(nm2)Hm+n,m1,0,m2;

          LmP6iIn={nIm+n,m1,0,m2;

    HmP6iLn=δi,2(n2m)Hm+n

         +δi,3{(n2m)Hm+n,m0,0,m1;

    HmP6iHn=δi,2(nm)Im+n

         +δi,3{(nm)Im+n,m0,0,m1;

    ImP6iLn=δi,2(m)Im+n

         +δi,3{mIm+n,m1,0,m0;

    (WP7i): i=1,2,3,

           LmP7iLn={(nm)Lm+n,m2,0,m1;

           LmP7iHn={(nm2)Hm+n,m2,0,m1;

          LmP7iIn={nIm+n,m2,0,m1;

    HmP7iLn=δi,2(n2m)Hm+n

         +δi,3{(n2m)Hm+n,m0,0,m1;

    HmP7iHn=δi,2(nm)Im+n

         +δi,3{(nm)Im+n,m0,0,m1;

    ImP7iLn=δi,2(m)Im+n

         +δi,3{mIm+n,m1,0,m0;

    (WP8i): i=1,2,3,

           LmP8iLn={(nm)Lm+n,m1,0,m2;

           LmP8iHn={(nm2)Hm+n,m1,0,m2;

           LmP8iIn={nIm+n,m1,0,m2;

    HmP8iLn=δi,2(n2m)Hm+n

         +δi,3{0,m0,(n2m)Hm+n,m1

    HmP8iHn=δi,2(nm)Im+n

         +δi,3{0,m0,(nm)Im+n,m1

    ImP8iLn=δi,2(m)Im+n

         +δi,3{0,m0,mIm+n,m1

    where c,a,μ,λC, k{2,1,1,2,3}, tZ{0,1} and s{2t2,2t1,2t,2t+1,2t+2}. Conversely, the above types are all the graded post-Lie algebra structures satisfying (3)-(9) on S. Furthermore, the post-Lie algebras WPc3,s,k,ti,a,μ,λ, WP5j and WP6j are isomorphic to the post-Lie algebras WPc4,s,k,ti,a,μ,λ, WP7j and WP8j, i=1,2,3,4,5 and j=1,2,3 respectively, and other post-Lie algebras are not mutually isomorphic.

    Proof. Suppose that (S,[,],) is a class of post-Lie algebra structures satisfying (3)-(9) on the Schrödinger-Virasoro algebra S. By Lemma 3.3-3.5, there are complex-valued functions f, g, h on Z and complex numbers μ,a such that one of 26 cases in Table 2 holds. From this with Lemma 3.1, we obtain 26 classes of graded post-Lie algebra structures on S. We claim that h(0)=λ and g(0)=ˆτ in WPcj,s,k,ti,a,μ,λ,j=3,4 and i=1,2,3,4,5 and g(0)=ˆλ in WPji, j=5,6,7,8 and i=1,2,3. We claim that g(0)=ˆλ and g(0)=ˆτ will not appear in every structures, when m=0, for example, in Case WP5i, i=1,2,3, then ImP53Ln=0ˆλI0+n=0, one has ImP53Ln=0 for m0, and in Case WPc3,s,k,ti,a,μ,λ, i=1,2,3,4,5, then HmP33,λLn=(n20)λH0+n=0, one has HmP33,λLn=n2λHn for m=0. Hence we can obtain 26 classes of graded post-Lie algebra structures on S listed in the theorem.

    Conversely, every type of the 26 cases means that there are complex-valued functions f and g, h on Z and complex numbers a,μ such that (10)-(15) hold and, the Equations (16)-(22) are easily verified. Thus, by Lemma 3.1 we see that they are the all graded post-Lie algebra structures satisfying (3)-(9) on the Schrödinger-Virasoro algebra S.

    Finally, by Lemma 3.7 with maps LmLm, HmHm, ImIm we know that the post-Lie algebras WPc3,s,k,ti,a,μ,λ, WP5j and WP6j are isomorphic to the post-Lie algebras WPc4,s,k,ti,a,μ,λ, WP7j and WP8j, i=1,2,3,4,5 and j=1,2,3 respectively. Clearly, the other post-Lie algebras are not mutually isomorphic. The proof is completed.

    The Rota-Baxter algebra was introduced by the mathematician Glen E. Baxter [2] in 1960 in his probability study, and was popularized mainly by the work of Rota [G. Rota1, G. Rota2] and his school. Recently, the Rota-Baxter algebra relation were introduced to solve certain analytic and combinatorial problem and then applied to many fields in mathematics and mathematical physics (see [6,7,19,23] and the references therein). Now let us recall the definition of Rota-Baxter operator.

    Definition 4.1. Let L be a complex Lie algebra. A Rota-Baxter operator of weight λC is a liner map R:LL satisfying

    [R(x),R(y)]=R([R(x),y]+[x,R(y)])+λR([x,y]),x,yL. (50)

    Note that if R is a Rota-Baxter operator of weight λ0, then λ1R is a Rota-Baxter operator of weight 1. Therefore, one only needs to consider Rota-Baxter operators of weight 0 and 1.

    In this section, we mainly consider the homogeneous Rota-Baxter operator R of weight 1 on the Schrödinger-Virasoro S given by

    R(Lm)=f(m)Lm,   R(Hm)=h(m)Hm,   R(Im)=g(m)Im (51)

    for all mZ, where f,g,h are complex-valued functions on Z.

    Lemma 4.2. (see [1]) Let (L,[,]) be a Lie algebra and R:LL a Rota-Baxter operator of weight 1. Define a new operation xy=[R(x),y] on L. Then (L,[,],) is a post-Lie algebra.

    Theorem 4.3. A homogeneous Rote-Baxrer operator R of weight 1 satisfying (51) on the Schrödinger-Virasoro S must be one of the following types

    (RP11): R(Lm)=0,R(Hn)=0,R(In)=0;

    (RP12): R(Lm)=0,R(Hn)=Hn,R(In)=In;

    (RP21): R(Lm)=Lm,R(Hn)=0,R(In)=0;

    (RP22): R(Lm)=Lm,R(Hn)=Hn,R(In)=In;

    (RPc31): R(Lm)={Lm,  m>0,cL0,  m=0,0,m<0; R(Hn)=0, R(In)=0;

    (RPc32): R(Lm)={Lm,  m>0,cL0,  m=0,0,m<0; R(Hn)=Hn, R(In)=In;

    (RPc3,k3,ˆτ,λ): R(Lm)={Lm,  m>0,cL0,  m=0,0,m<0; R(Hn)={Hn,  n>0,λH0,  n=0,0,n<0;

    R(In)={In,  nk,ˆτI0,  n=0,0,nk1;

    (RPc3,s,t5): R(Lm)={Lm,  m>0,cL0,  m=0,0,m<0; R(Hn)={Hn,  nt,0,  nt1;

    R(In)={In,  ns,0,  ns1;

    (RPc41): R(Lm)={Lm,  m<0,cL0,  m=0,0,m>0; R(Hn)=0, R(In)=0;

    (RPc42): R(Lm)={Lm,  m<0,cL0,  m=0,0,m>0; R(Hn)=Hn, R(In)=In;

    (RPc4,k3,ˆτ,λ): R(Lm)={Lm,  m<0,cL0,  m=0,0,m>0; R(Hn)={0,  n>0,λH0,  n=0,Hn,  n<0;

    R(In)={0,  nk,ˆτI0,  n=0,In,  nk1;

    (RPc4,s,t5): R(Lm)={Lm,  m>0,cL0,  m=0,0,m<0; R(Hn)={0,  nt,Hn,  nt1;

    R(In)={0,  ns,In,  ns1;

    (RP51): R(Lm)={Lm,  m2,0,  m1; R(Hn)=0, R(In)=0;

    (RP52): R(Lm)={Lm,  m2,0,  m1; R(Hn)=Hn, R(In)=In;

    (RP53,ˆλ): R(Lm)={Lm,  m2,0,  m1; R(Hn)={0,  n0,Hn,  n1;

    R(In)={0,  n1,ˆλI0,  n=0,In,  n1;

    (RP61): R(Lm)={Lm,m1,0,  m2; R(Hn)=0, R(In)=0;

    (RP62): R(Lm)={Lm,m1,0,  m2; R(Hn)=Hn, R(In)=In;

    (RP63,ˆλ): R(Lm)={Lm,m1,0,  m2; R(Hn)={Hn,n0,0,  n1;

    R(In)={In,n1,ˆλI0,  n=0,0,  n1;

    (RP71): R(Lm)={Lm,m2,0,  m1; R(Hn)=0, R(In)=0;

    (RP72): R(Lm)={Lm,m2,0,  m1; R(Hn)=Hn, R(In)=In;

    (RP73,ˆλ): R(Lm)={Lm,m2,0,  m1; R(Hn)={0,  n1,Hn,  n0;

    R(In)={0,n1,ˆλI0,  n=0,In,  n1;

    (RP81): R(Lm)={Lm,m1,0,  m2; R(Hn)=0, R(In)=0;

    (RP82): R(Lm)={Lm,m1,0,  m2, R(Hn)=Hn, R(In)=In;

    (RP83,ˆλ): R(Lm)={Lm,m1,0,  m2, R(Hn)={Hn,n1,0,  n0;

    R(In)={In,n1,ˆλI0,  n=0,0,  n1

    for all m,nZ, where c,λ,ˆλ,ˆτC, k{2,1,1,2,3} with k1, tZ{0,1} and s{2t2,2t1,2t,2t+1,2t+2}.

    Proof. In view of Lemma 4.2, if we define a new operation xy=[R(x),y] on S, then (S,[,],) is a post-Lie algebra. By (51), we have

    LmLn=[R(Lm),Ln]=(mn)f(m)Lm+n, (52)
    LmHn=[R(Lm),Hn]=(m2n)f(m)Hm+n, (53)
    LmIn=[R(Lm),In]=nf(m)Im+n, (54)
    HmLn=[R(Hm),Ln]=(n2m)h(m)Hm+n, (55)
    HmHn=[R(Hm),Hn]=(mn)h(m)Im+n, (56)
    ImLn=[R(Im),Ln]=mg(m)Im+n (57)

    and ImHn=[R(Im),Hn]=HmIn=[R(Hm),In]=ImIn=[R(Im),In]=0 for all m,nZ. This means that (S,[,],) is a graded post-Lie algebras structure satisfying (3)-(9) with ϕ(m,n)=(mn)f(m), φ(m,n)=(m2n)f(m), χ(m,n)=nf(m), ψ(m,n)=(n2m)h(m), ξ(m,n)=(mn)h(m) and θ(m,n)=mg(m).

    A similar discussion to Lemma 3.1 gives

    (mn)(f(m+n)f(n)f(m)+f(m)f(m+n)+f(n)f(m+n))=0,(m2n)(h(m+n)f(m)h(n)+f(m)h(m+n)+h(n)h(m+n))=0,n(m+n)(g(m+n)(1+f(m)+g(n))f(m)g(n))=0,(mn)(g(m+n)h(m)h(n)+h(m)g(m+n)+h(n)g(m+n))=0.

    From this we conclude that Equations (10)-(22) hold with a=μ=0. In the same way of Lemma 3.6, we see that f,g,h must satisfy Table 2 with a=μ=0. This excludes Cases WPc3,k=14,a,μ and WPc4,k=14,a,μ. Thus, f, g, h must be of the 24 cases listed in Table 2 with a=μ=0, which can yield the 24 forms of R one by one. It is easy to verify that every form of R listed in the above is a Rota-Baxter operator of weight 1 satisfying (51). The proof is completed.

    The natural question is: how we can characterize the Rota-Baxter operators of weight zero on the Schrödinger-Virasoro S? This is related to the so called pre-Lie algebra which is a class of Lie-admissible algebras whose commutators are Lie algebras. Pre-Lie algebras appeared in many fields in mathematics and physics under different names like left-symmetric algebras, Vinberg algebras and quasi-associative algebras (see the survey article [3] and the references therein). Now we recall the definition of pre-Lie algebra as follows.

    Definition 4.4. A pre-Lie algebra A is a vector space A with a bilinear product satisfying

    (xy)zx(yz)=(yx)zy(xz),x,y,zA. (58)

    As a parallel result of Lemma 4.2, one has the following conclusion.

    Proposition 1. (see [8]) Let (L,[,]) be a Lie algebra with a Rota-Baxter operator R of weight 0 on it. Define a new operation xy=[R(x),y] for any x,yL. Then (L,) is a pre-Lie algebra.

    Using a similar method on classification of Rota-Baxter operators of weight 1 in the above subsection, by Proposition 1 we can get the forms of Rota-Baxter operators of weight zero when the corresponding structure of pre-Lie algebra are known. For example, consider the homogeneous Rota-Baxter operator R of weight zero on the Schrödinger-Virasoro algebra S satisfying (51). According to Proposition 1, if we define a new operation xy=[R(x),y] on S, then (S,) is a pre-Lie algebra. By (51), we have Equations (52)-(57) hold. At this point we can apply the relevant results on pre-Lie algebra satisfying (52)-(57). But the classification of graded pre-Lie algebra structures on S is also an unsolved problem, as far as we know. In fact, we can direct characterize the Rota-Baxter operators of weight zero on the Schrödinger-Virasoro S satisfying (51) following the approach of [6]. Due to limited space, it will not be discussed here.

    We would like to express our sincere thanks to the anonymous referees for their careful reading and valuable comments towards the improvement of this article.



    Author contributions



    Lucy E. Chambers was responsible for the acquisition, analysis and interpretation of data for the work. Aali J. Sheen and Kathryn A. Whitehead were responsible for the conception and design of the work. All authors were responsible for the drafting and revision of the data and final approval of the version to be published.

    Conflict of interest



    The author declares no conflicts of interest in this paper.

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