# Conformal Super-Biderivations on Lie Conformal Superalgebras

Authors: Zhongxian Huang
Journal: Journal of Mathematics
 (2021)

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## Abstract

In this paper, the conformal super-biderivations of two classes of Lie conformal superalgebras are studied. By proving some general results on conformal super-biderivations, we determine the conformal super-biderivations of the loop super-Virasoro Lie conformal superalgebra and Neveu–Schwarz Lie conformal superalgebra. Especially, any conformal super-biderivation of the Neveu–Schwarz Lie conformal superalgebra is inner.


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## Body

## 1. Introduction

Lie conformal superalgebras, introduced by Kac in [1], encode the singular part of the operator product expansion of chiral fields in conformal field theory. The conformal super-algebras play important roles in quantum field theory, vertex algebras, integrable systems, and so on and have drawn much attention in the branches of physics and mathematics. Finite simple Lie conformal superalgebras were classified by Fattori and Kac in [2], and their representation theories were developed in [3–5]. Moreover, some infinite Lie conformal superalgebras were also studied, such as loop super-Virasoro Lie conformal superalgebra [6] and Lie conformal superalgebras of Block type [7]. Other results on Lie conformal superalgebras can be seen in [8, 9].In recent years, biderivations have been extensively studied for various algebra structures [10–14]. The authors in [15, 16] generalized biderivations of Lie algebras to the concept of super-biderivations of superalgebras independently. The authors in [17] studied super-biderivations on the super Galilean conform algebra. The conformal biderivations of the loop Wa,b Lie conformal algebra and loop Virasoro Lie conformal algebra are determined in [18].As a generalization of conformal biderivations of Lie conformal algebras and a parallel concept of super-biderivations of Lie superalgebras, we introduce the concept of conformal super-biderivations on Lie conformal superalgebras. We hope that biderivations would contribute to the development of structure theories of Lie conformal superalgebras. This is our motivation to present this paper.In this paper, we concentrate on the loop super-Virasoro Lie conformal superalgebracls (see [6]), which is defined as a C∂-module cls=⊕i∈ℤℂ∂Li⊕⊕i∈ℤC∂Gi with a C∂-basis Li,Gi|i∈ℤ and λ-brackets given by(1)LiλLj=∂+2λLi+j,LiλGj=∂+32λGi+j,GiλLj=12∂+32λGi+j,GiλGj=2Li+j,for any i,j∈ℤ. In section 2, we recall definition of Lie conformal superalgebras. Some general results about conformal super-biderivations are obtained in section 3. In section 4, we determine the conformal super-biderivations of cls and Neveu–Schwarz Lie conformal superalgebra NS, and all conformal super-biderivations of the NS are inner.Throughout this paper, all vector spaces are over the complex fieldℂ.


## 2. Preliminaries

In this section, we recall definition about Lie conformal superalgebra in [1, 4, 7, 8].A vector spaceV is called Z2-graded algebra if V=V0¯⊕V1¯, and x∈Vi¯ is called Z2-homogenous and we write x=i¯.Definition 1.
A Lie conformal superalgebraR=R0¯⊕R1¯ is a Z2-graded C∂-module with a C-linear map R⊗R⟶ℂλ⊗R,a⊗b⟶aλb, called the λ-bracket satisfying RθλRτ⊆Rθ+τλ and the following axioms:(1)
Conformal sesquilinearity:∂aλb=−λaλb, aλ∂b=∂+λaλb.(2)
Skew-supersymmetry:aλb=−−1abb−∂−λa.(3)
Jacobi identity:aλbμc=aλbλ+μc+−1abbμaλc, for all a,b,c∈R,θ,τ∈Z2, where λ is an indeterminate and ∂ is a derivation of the λ-bracket.
From conformal sesquilinearity, we can conclude that(2)f∂aλb=f−λaλb,aλf∂b=f∂+λaλb,∀f∂∈ℂ∂.
By Definition1, the loop super-Virasoro Lie conformal superalgebra cls=cls0¯⊕cls1¯ with cls0¯=⊕i∈ℤℂ∂Li and cls1¯=⊕i∈ℤℂ∂Gi, and(3)clsi¯λclsj¯⊆clsi¯+j¯λ,for i¯,j¯∈ℤ2.Example 1.
(see [9]). Neveu–Schwarz Lie conformal superalgebra NS is a free Z2-graded C∂-module NS=ℂ∂L⊕ℂ∂G with the following conditions:(4)LλL=∂+2λL,LλG=∂+32λG,GλL=12∂+32λG,GλG=L,where NS0¯=ℂ∂L and NS1¯=ℂ∂G.


## 3. Conformal Super-Biderivations

Definition 2.
LetR be a Lie conformal superalgebra. We call a conformal bilinear map φλ:R×R⟶Rλ a conformal super-biderivation of R if it satisfies the following equations:(5)φλx,y=−−1xyφ−∂−λy,x,(6)φλx,yμz=φλx,yλ+μz+−1xyyμφλx,z,(7)φλ+μxμy,z=xμφλy,z−−1xyyλφμx,z,for all homogeneous x,y,z∈R.Remark 1.
Equation (6) is equivalent to equation (7).Proof.
We suppose equation (7) satisfies. On the one hand, using equation (5), we have(8)φλ+μyμz,x=−−1yx+zxφ−∂−λ−μx,yμz.
On the other hand, using conformal sesquilinearity, we obtain(9)φλ+μyμz,x=yμφλz,x−−1yzzλφμy,x,=−−1xzyμφ−∂−λx,z+−1yz+yxzλφ−∂−μx,y.
Then(10)−1yx+zxφ−∂−λ−μx,yμz=−1xzyμφ−∂−λx,z−−1yz+yxzλφ−∂−μx,y.
Replaceλ,μ by −∂−λ−μ,μ, respectively, and by conformal sesquilinearity, we have(11)−1yxφλx,yμz=yμφλx,z+−1yz+yx+zyφλx,yλ+μz.
This implies that equation (6) is satisfied. The reverse conclusion follows similarly.
We call the conformal super-biderivationφλa the inner conformal super-biderivation if there exists a fixed complex number a such that φλax,y=axλy.
To avoid lengthy notations, we let(12)Φλ,μ,γx,y,w,v=−1wyφμx,yμ+γwλv−xμyμ+γφλw,v.Lemma 1.
Letϕλ be a conformal super-biderivation of Lie conformal superalgebra R. Then,(13)Φλ,μ,γx,y,w,v=−1xwΦμ,λ,γw,y,x,v,for any homogeneous x,y,w,v∈R.Proof.
Firstly, from the definition of conformal super-biderivation, we have(14)ϕλ+μxμw,yγv=xμϕλw,yγv−−1xwwλϕμx,yγv=xμϕλw,yλ+γv+−1wyxμyγϕλw,v−−1xwwλφμx,yμ+γv−−1xw+xywλyγφμx,v,ϕλ+μxμw,yγv=ϕλ+μxμw,yλ+μ+γv+−1yx+ywyγϕλ+μxμw,v=xμϕλw,yλ+μ+γv−−1xwwλϕμx,yλ+μ+γv+−1yx+ywyγxμφλw,v−−1yx+yw+xwyγwλφμx,v.
Using the Jacobi identity of Lie conformal superalgebras, we obtain(15)xμφλw,yλ+γv=xμφλw,yλ+μ+γv+−1xw+xyφλw,yλ+γxμv,xμyγϕλw,v=xμyμ+γϕλw,v+−1xyyγxμϕλw,v,wλφμx,yμ+γv=wλφμx,yλ+μ+γv+−1xw+wyφλx,yμ+γwλv,wλyγϕμx,v=wλyλ+γϕμx,v+−1wyyγwλϕμx,v.
Therefore, we get(16)−1xw+xyϕλw,yλ+γxμv+−1wyxμyμ+γϕλw,v=−1wyφμx,yμ+γwλv+−1xw+xywλyλ+γφμx,v,which implies(17)−1wyϕμx,yμ+γwλv−xμyμ+γϕλw,v=−1xw+xyφλw,yλ+γxμv−wλyλ+γφμx,v.
Thus,(18)Φλ,μ,γx,y,w,v=−1xwΦμ,λ,γw,y,x,v,for any homogeneous x,y,w,v∈R.Lemma 2.
Letϕλ be a conformal super-biderivation of Lie conformal superalgebra R. Then,(i)
ϕμx,yμ+γwλv=xμyμ+γϕλw,v, for all homogeneous x,y,w,v∈R(ii)
Ifxλy=0, then φλx,y∈ZRλR, where ZRλR is the center of RλRProof.
By skew-supersymmetry and conformal sesquilinearity, we have(19)Φλ,μ,γx,y,w,v=−1wyϕμx,yμ+γwλv−xμyμ+γϕλw,v=−−1wy+xyϕ−∂−μy,xμ+γwλv−y−∂−μxμ+γϕλw,v=−−1wy+xyϕγy,xμ+γwλv−yγxμ+γϕλw,v=−−1xyΦλ,γ,μy,x,w,v.
Note that(20)Φλ,μ,γx,y,w,v=−1xwΦμ,λ,γw,y,x,v=−−1xw+wyΦμ,γ,λy,w,x,v=−−1xw+wy+yxΦγ,μ,λx,w,y,v.
On the other hand, we have(21)Φλ,μ,γx,y,w,v=−−1xyΦλ,γ,μy,x,w,v=−−1xy+ywΦγ,λ,μw,x,y,v=−1xy+yw+xwΦγ,μ,λx,w,y,v.
Thus,Φλ,μ,γx,y,w,v=0; this implies that (22)φμx,yμ+γwλv=xμyμ+γφλw,v,
for any homogeneousx,y,w,v∈R. This implies (ii) directly follows from (i).


## 4. Conformal Super-Biderivations ofcls

Theorem 1.
Every conformal super-biderivationϕλ on the cls has the following forms:(23)φλLi,Lj=∂+2λ∑t∈ℤat−i−jLt,φλLi,Gj=∂+32λ∑t∈ℤat−i−jGt,φλGi,Lj=12∂+32λ∑t∈ℤat−i−jGt,φλGi,Gj=2∑t∈ℤat−i−jLt,for all i,j∈ℤ, where ak for any k∈ℤ are complex numbers.Proof.
Letϕλ be a conformal super-biderivation on the cls. We shall complete the proof by verifying the following four claims.Claim 1.
There exist some complex numbersak for any k∈ℤ such that(24)φλLi,Lj=∂+2λ∑t∈ℤat−i−jLt,for all i,j∈ℤ.
For anyi,j∈ℤ, we may assume that(25)φλLi,Lj=∑k∈ℤfijk∂,λLk+∑k∈ℤgijk∂,λGk,where fijk∂,λ,gijk∂,λ∈ℂ∂,λ. By Lemma 2(i), we have(26)φμLi,Ljμ+γLiλLj=LiμLjμ+γφλLi,Lj.
Furthermore, we note that(27)∑k∈ℤfijk∂,μLk+∑k∈ℤgijk∂,μGkμ+γ∂+2λLi+j=∂+2μLi+jμ+γ∑k∈ℤfijk∂,λLk+∑k∈ℤgijk∂,λGk.
That is,(28)∑k∈ℤfijk−μ−γ,μLkμ+γ∂+2λLi+j+∑k∈ℤgijk−μ−γ,μGkμ+γ∂+2λLi+j=∑k∈ℤfijk∂+μ+γ,λ∂+2μLi+jμ+γLk+∑k∈ℤgijk∂+μ+γ,λ∂+2μLi+jμ+γGk.
Therefore, we have(29)∂+μ+γ+2λ∑k∈ℤfijk−μ−γ,μLkμ+γLi+j+∂+μ+γ+2λ∑k∈ℤgijk−μ−γ,μGkμ+γLi+j=μ−γ∑k∈ℤfijk∂+μ+γ,λLi+jμ+γLk+μ−γ∑k∈ℤgijk∂+μ+γ,λLi+jμ+γGk.
It follows that(30)∂+μ+γ+2λ∂+2μ+2γ∑k∈ℤfijk−μ−γ,μLi+j+k+∂+μ+γ+2λ12∂+32μ+32γ∑k∈ℤgijk−μ−γ,μGi+j+k=μ−γ∂+2μ+2γ∑k∈ℤfijk∂+μ+γ,λLi+j+k+μ−γ∂+32μ+32γ∑k∈ℤgijk∂+μ+γ,λGi+j+k,which implies (31)∂+μ+γ+2λ∑k∈ℤfijk−μ−γ,μLi+j+k=μ−γ∑k∈ℤfijk∂+μ+γ,λLi+j+k,(32)∂+μ+γ+2λ12∂+32μ+32γ∑k∈ℤgijk−μ−γ,μGi+j+k=μ−γ∂+32μ+32γ∑k∈ℤgijk∂+μ+γ,λGi+j+k.
From (32), we have(33)∂+μ+γ+2λ12∂+32μ+32γgijk−μ−γ,μ=μ−γ∂+32μ+32γgijk∂+μ+γ,λ.
By comparing the degrees of∂ on both sides of (33), we can suppose that(34)gijk∂,λ=c0λ+c1λ∂,for some c0λ,c1λ∈ℂλ. Substituting this formula to (33), we can obtain(35)∂+μ+γ+2λ12∂+32μ+32γc0μ+c1μ−μ−γ=μ−γ∂+32μ+32γc0λ+c1λ∂+μ+γ.
Comparing the coefficients ofγ3, one can deduce c1μ−c1λ=0. Thus, c1λ∈ℂ which we denote by c1. Therefore, (35) can be written as(36)∂+μ+γ+2λ12∂+32μ+32γc0μ+c1−μ−γ=μ−γ∂+32μ+32γc0λ+c1∂+μ+γ.
We getc0μ=2c1μ. Considering the coefficients of ∂, we have(37)12∂2c1μ−γ=c1μ−γ∂2,which implies c1=0 and gijk∂,λ=0.
By (31), we obtain(38)∂+μ+γ+2λfijk−μ−γ,μ=μ−γfijk∂+μ+γ,λ.
We suppose that(39)fijk∂,λ=a0λ+a1λ∂,for some a0λ,a1λ∈ℂλ. Thus,(40)∂+μ+γ+2λa0μ+a1μ−μ−γ=μ−γa0λ+a1λ∂+μ+γ.
By comparing the coefficients of∂ on both sides of (40), one can deduce(41)a0μ+a1μ−μ−γ=a1λμ−γ.
Hence,a1μ∈ℂ, and we denote it by a1. We also get a0λ=2λa1. Therefore, fijk∂,λ=a1∂+2λ. We denote a1 by aijk. Thus,(42)φλLi,Lj=∂+2λ∑k∈ℤaijkLk,where aijk∈ℂ.
Furthermore, by Lemma2(i), we have(43)φμLi,Ljμ+γLkλLl=LiμLjμ+γφλLk,Ll.
Therefore,(44)μ−γ∑t∈ℤaijtLtμ+γLkλLl=∂+μ+γ+2λ∑s∈ℤaklsLiμLjμ+γLs.
That is,(45)μ−γ∑t∈ℤaijtLtμ+γ∂+2λLk+l=∂+μ+γ+2λ∑s∈ℤakls∂+2μLi+jμ+γLs.
We get(46)∑t∈ℤaijtLt+k+l=∑s∈ℤaklsLi+j+s,which shows that aijt=aklt+k+l−i−j. Hence, a00t=aklt+k+l, denoted by at. Thus,(47)φλL0,L0=∂+2λ∑t∈ℤatLt.
Then, we conclude that(48)φλLi,Lj=∂+2λ∑t∈ℤat−i−jLt,for all i,j∈ℤ, where ak for any k∈ℤ are complex numbers.Claim 2.
φλLi,Gj=∂+3/2λ∑k∈ℤak−i−jGk for all i,j∈ℤ.
For anyi,j∈ℤ, we suppose that(49)φλLi,Gj=∑k∈ℤdijk∂,λLk+∑k∈ℤhijk∂,λGk,where dijk∂,λ,hijk∂,λ∈ℂ∂,λ. By Lemma 2(i), we have(50)φμLi,Ljμ+γLiλGj=LiμLjμ+γφλLi,Gj.
Therefore,(51)∂+2μ∑k∈ℤak−i−jLkμ+γLiλGj=LiμLjμ+γ∑k∈ℤdijk∂,λLk+∑k∈ℤhijk∂,λGk.
That is,(52)μ−γ∑k∈ℤak−i−jLkμ+γLiλGj=∑k∈ℤdijk∂+μ+γ,λLiμLjμ+γLk+∑k∈ℤhijk∂+μ+γ,λLiμLjμ+γGk.
We get(53)μ−γ∑k∈ℤak−i−jLkμ+γ∂+32λGi+j=∑k∈ℤdijk∂+μ+γ,λ∂+2μLi+jμ+γLk+∑k∈ℤhijk∂+μ+γ,λ∂+2μLi+jμ+γGk.
Then(54)∂+μ+γ+32λ∑k∈ℤak−i−jLkμ+γGi+j=∑k∈ℤdijk∂+μ+γ,λLi+jμ+γLk+∑k∈ℤhijk∂+μ+γ,λLi+jμ+γGk.
Hence,(55)∂+μ+γ+32λ∑k∈ℤak−i−j∂+32μ+32γGi+j+k=∂+2μ+2γ∑k∈ℤdijk∂+μ+γ,λLi+j+k+∂+32μ+32γ∑k∈ℤhijk∂+μ+γ,λGi+j+k,which implies dijk∂,λ=0 and(56)∂+μ+γ+32λ∑k∈ℤak−i−jGi+j+k=∑k∈ℤhijk∂+μ+γ,λGi+j+k.
We get(57)∂+μ+γ+32λak−i−j=hijk∂+μ+γ,λ.
Then,(58)hijk∂,λ=∂+32λak−i−j.
We obtain(59)φλLi,Gj=∂+32λ∑k∈ℤak−i−jGk,for all i,j∈ℤ, where ak for any k∈ℤ are complex numbers.Claim 3.
φλGi,Lj=1/2∂+3/2λ∑k∈ℤak−i−jGk for all i,j∈ℤ.
For anyi,j∈ℤ, we suppose that(60)φλGi,Lj=∑k∈ℤpijk∂,λLk+∑k∈ℤqijk∂,λGk,where pijk∂,λ,qijk∂,λ∈ℂ∂,λ. By Lemma 2(i), we have(61)φμLi,Ljμ+γGiλLj=LiμLjμ+γφλGi,Lj.
Thus,(62)∂+2μ∑k∈ℤak−i−jLkμ+γGiλLj=LiμLjμ+γ∑k∈ℤpijk∂,λLk+∑k∈ℤqijk∂,λGk.
That is,(63)μ−γ∑k∈ℤak−i−jLkμ+γGiλLj=∑k∈ℤpijk∂+μ+γ,λLiμLjμ+γLk+∑k∈ℤqijk∂+μ+γ,λLiμLjμ+γGk.
Hence,(64)μ−γ∑k∈ℤak−i−jLkμ+γ12∂+32λGi+j=∑k∈ℤpijk∂+μ+γ,λ∂+2μLi+jμ+γLk+∑k∈ℤqijk∂+μ+γ,λ∂+2μLi+jμ+γGk.
We get(65)12∂+12μ+12γ+32λ∑k∈ℤak−i−jLkμ+γGi+j=∑k∈ℤpijk∂+μ+γ,λLi+jμ+γLk+∑k∈ℤqijk∂+μ+γ,λLi+jμ+γGk.
Then,(66)12∂+12μ+12γ+32λ∑k∈ℤak−i−j∂+32μ+32γGi+j+k=∂+2μ+2γ∑k∈ℤpijk∂+μ+γ,λLi+j+k+∂+32μ+32γ∑k∈ℤqijk∂+μ+γ,λGi+j+k,which implies pijk∂,λ=0 and(67)12∂+12μ+12γ+32λ∑k∈ℤak−i−jGi+j+k=∑k∈ℤqijk∂+μ+γ,λGi+j+k.
It follows(68)12∂+12μ+12γ+32λak−i−j=qijk∂+μ+γ,λ.
This shows that(69)qijk∂,λ=12∂+32λak−i−j
Hence,(70)φλGi,Lj=12∂+32λ∑k∈ℤak−i−jGk,for all i,j∈ℤ, where ak for any k∈ℤ are complex numbers.Claim 4.
φλGi,Gj=2∑k∈ℤak−i−jLk, for all i,j∈ℤ.
For anyi,j∈ℤ, we suppose that(71)φλGi,Gj=∑k∈ℤuijk∂,λLk+∑k∈ℤvijk∂,λGk,where uijk∂,λ,vijk∂,λ∈ℂ∂,λ. By Lemma 2(i), we have(72)φμLi,Ljμ+γGiλGj=LiμLjμ+γφλGi,Gj.
Therefore,(73)∂+2μ∑k∈ℤak−i−jLkμ+γGiλGj=LiμLjμ+γ∑k∈ℤuijk∂,λLk+∑k∈ℤvijk∂,λGk.
That is,(74)μ−γ∑k∈ℤak−i−jLkμ+γGiλGj=∑k∈ℤuijk∂+μ+γ,λLiμLjμ+γLk+∑k∈ℤvijk∂+μ+γ,λLiμLjμ+γGk.
Hence,(75)μ−γ∑k∈ℤak−i−jLkμ+γ2Li+j=∑k∈ℤuijk∂+μ+γ,λ∂+2μLi+jμ+γLk+∑k∈ℤvijk∂+μ+γ,λ∂+2μLi+jμ+γGk.
Thus, we have(76)2∑k∈ℤak−i−jLkμ+γLi+j=∑k∈ℤuijk∂+μ+γ,λLi+jμ+γLk+∑k∈ℤvijk∂+μ+γ,λLi+jμ+γGk.
It follows(77)2∂+2μ+2γ∑k∈ℤak−i−jLi+j+k=∂+2μ+2γ∑k∈ℤuijk∂+μ+γ,λLi+j+k+∂+32μ+32γ∑k∈ℤvijk∂+μ+γ,λGi+j+k,which implies vijk∂,λ=0 and(78)2∑k∈ℤak−i−jLi+j+k=∑k∈ℤuijk∂+μ+γ,λLi+j+k,hence(79)2ak−i−j=uijk∂+μ+γ,λ.
We can getuijk∂,λ=2ak−i−j and φλGi,Gj=2∑k∈ℤak−i−jLk, for all i,j∈ℤ, where ak for any k∈ℤ are complex numbers.Theorem 2.
Every conformal super-biderivationϕλ on NS has the following forms:(80)φλL,L=a∂+2λL,φλL,G=a∂+32λG,φλG,L=a12∂+32λG,φλG,G=aL,where a are complex numbers. Therefore, any conformal super-biderivation on NS is inner.The proof of this theorem is similar to that of Theorem1.



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