# Classification of Nilpotent Lie Superalgebras of Multiplier-Rank Less than or Equal to 6

Authors: Shuang Lang; Jizhu Nan; Wende Liu
Journal: Advances in Mathematical Physics
 (2021)

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## Abstract

In this paper, we classify all the finite-dimensional nilpotent Lie superalgebras of multiplier-rank less than or equal to 6 over an algebraically closed field of characteristic zero. We also determine the covers of all the nilpotent Lie superalgebras mentioned above.


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## Body

## 1. Introduction

The notion of the multiplierMG for a finite group G arose from Schur’s work on projective representations of groups. There are fruitful results about this topic. However, we only mention that Green proved that MG≤p1/2nn−1 for a finite p-group G of order pn (see [1]) as well as that Berkovich and Zhou classified all p-groups with tG=0,1 and 2, where tG is defined by MG=p1/2nn−1−tG (see [2, 3]).Batten initiated the theory of multipliers and covers for Lie algebras when she studied the second cohomology groups of Lie algebras with coefficients in a 1-dimensional trivial module (see [4]). For a finite-dimensional Lie algebra L, she proved that ML≅H2L,F by a free presentation of L, where F is the underlying field. Moneyhun gave the maximal dimension of ML (see [5]). Let tL=1/2nn−1−dimML, which is a nonnegative integer. Batten et al. classified all nilpotent Lie algebras with tL=0,1,2 (see [6]). Hardy and Stitzinger used a different method to get similar results for tL≤8 (see [7, 8]). For filiform Lie algebras, Bosko classified these up to tL=16 (see [9]).The notion of the multiplier for Lie algebras may be naturally generalized to the Lie superalgebra case (see [10], for example). Let L be a finite-dimensional Lie superalgebra. A Lie superalgebra pair K,M is called a defining pair for L provided that L≅K/M and M⊂ZK∩K2. A defining pair K,M of L is said to be maximal if K is of maximal super-dimensions. In the case of K,M being a maximal defining pair of L, we also call K a cover and M a (Schur) multiplier of L. As in the Lie algebra case, multipliers for Lie superalgebra L are unique up to isomorphism, denoted by ML. Moreover, ML≅H2L,F (see [11]). By Lemma 2.3 of [10], for a Lie superalgebra L of super-dimension m,n, we define the super-multiplier-rank of L to be
(1)smrL=12mm−1+12nn+1,mn−sdimML,and the multiplier-rank of L to be
(2)mrL=smrL.In particular,smrL=0,0 if and only if L is abelian (see Proposition 3.1 of [10]). Hereafter, we write sdimV for the super-dimension of a superspace V and a,b=a+b for a pair a,b of nonnegative integers.In this paper, we classify all nilpotent Lie superalgebras of multiplier-rank≤6 by discussing whether ZL⊆L2. We also construct the covers of all the nilpotent Lie superalgebras of multiplier-rank ≤6.


## 2. Basics

We assume that all (super)spaces and (super)algebras are over an algebraically closed fieldF of characteristic zero. In ℤ×ℤ, we define a partial order as follows:
(3)m,n≤k,l⇔m≤k,n≤l.We also viewℤ×ℤ as the additive group in the usual way.To classify all the nilpotent Lie superalgebras of multiplier-rank≤6, we first establish some technical lemmas.Lemma 1.
LetL be a nilpotent Lie superalgebra of super-dimension m,n. Then, sdimL2≤smrL.Proof.
By Theorem 3.8 of [11], ML≅H2L,F. It follows that
(4)sdimZ2L,F≤sdimC2L,F≤12mm−1+12nn+1,mn,where C2L,F is the superspace consisting of skew-symmetric bilinear functions of L.
Letx1,⋯,xs∣y1,⋯,yt be a homogeneous basis of L2 and extend it to a homogeneous basis of L:x1,⋯,xs,xs+1,⋯,xm∣y1,⋯,yt,yt+1,⋯,yn. Let x↦x∗ be the isomorphism from L onto the dual space L∗. For x∗∈L∗, let x^:L×L⟶F be a bilinear mapping by x^y,z=x∗y,z for all y,z∈L. Then, x^∈B2L,F. If x^=0, then x∗L,L=0. Then, x=0 for all x∈L2. Hence, x↦x^ is an injection from L2 into B2L,F. Consequently, sdimML≤1/2mm−1+1/2nn+1,mn−sdimL2. By the definition of smrL, we have sdimL2≤smrL.☐LetΠ be the parity functor of superspaces. Note that
(5)sdimV+sdimΠV=dimV,dimV.Lemma 2.
LetL be a nilpotent Lie superalgebra. Suppose K is a 1-dimensional central ideal of L and K⊆L2. The following statements hold.
(1)
IfsdimK=1,0, then smrL/K+sdimL2≤smrL(2)
IfsdimK=0,1, then smrL/K+sdimΠL2+0,1≤smrLProof.
By Lemma 4.9 of [11], we have an exact sequence:
(6)0⟶HomLK,F⟶IHomL,F⟶RHomK,F⟶TMLK⟶IML⟶δLL2⊗K.
SinceK⊆L2, we have Rfk=f∘ik∈fL2=0 for all k∈K and f∈HomL,F, where i:K⟶L is a Lie superalgebra monomorphism. Then, R=0 and T is an injection. Hence, sdimimT=sdimK=1,0 or 0,1. Furthermore, sdimML=sdimimI+sdimimδ and sdimML/K=sdimimI+sdimimT.(1)
IfsdimimT=sdimK=1,0, we have(7)sdimML+1,0=sdimML+sdimimT=sdimimI+sdimimδ+sdimimT=sdimMLK+sdimimδ≤sdimL/L2⊗K+sdimMLK=sdimLL2+sdimMLK.
By the definition ofsmrL, we have smrL/K+sdimL2≤smrL.(2)
IfsdimimT=sdimK=0,1, we have(8)sdimML+0,1≤sdimLL2⊗K+sdimMLK=sdimΠLL2+sdimMLK.
Hence,smrL/K+sdimΠL2+0,1≤smrL.☐As in the Lie algebra case (Theorem 1 of [7]), using free presentations of Lie superalgebras, one may prove the following lemma.Lemma 3.
LetA and B be finite-dimensional Lie superalgebras. Then,
(9)sdimMA⊕B=sdimMA+sdimMB+sdimAA2⊗BB2.Lemma 4.
LetL be a finite-dimensional nilpotent Lie superalgebra. Suppose that ZLÚL2. Then, there exists an ideal A of L with dimA=1 such that L=M⊕A, where M is an ideal of L.
(1)
IfsdimA=1,0, then smrL=smrM+sdimM2(2)
IfsdimA=0,1, then smrL=smrM+sdimΠM2+1,0Proof.
Suppose thatA is a one-dimensional subsuperspace of L such that A⊆ZL but AÚL2. Let M be a complement to A in L such that L2⊆M. Then, L=M⊕A and MA=0,0.
IfsdimA=1,0, then sdimM/M2⊗A=sdimM−sdimM2. By Lemma 3, we have
(10)sdimML=sdimMA⊕M=sdimMA+sdimMM+sdimAA2⊗MM2.
Then,smrL=smrM+sdimM2.
IfsdimA=0,1, then sdimM/M2⊗A=sdimΠM−sdimΠM2. Now 1/2mm−1+1/2nn+1,mn−smrL=1/2mm−1+1/2n−1n,mn−1−smrM+n−1,m−sdimΠM2. Therefore, smrL=smrM+sdimΠM2+1,0.☐For convenience, we writeAbm,n for the abelian Lie superalgebra of super-dimension m,n, Hp,q for the 2p+1,q-super-dimensional Heisenberg Lie superalgebra of even center, and Hk for the k,k+1-super-dimensional Heisenberg Lie superalgebra of odd center (see [12]). In Proposition 4.4 of [10] and section 4 of [13], the authors characterize the multipliers of Hp,q and Hk:
(11)sdimMHp,q=2p2−p+12q2+12q−1,2pq,p+q≥2,0,0,p=0,q=1,2,0,p=1,q=0,sdimMHk=k2,k2−1,k≥2,1,1,k=1.Hence,(12)smrHp,q=2p+1,q=sdimHp,q,p+q≥2,(13)smrHk=k+1,k+1,k≥2,smrH1=1,2.Similarly, let us give a multiplier and cover ofHp,q⊕Abs,t.Case 1.
Suppose thatp,q=1,0. Let H1,0⊕Abs,t^ be a Lie superalgebra with a homogenous basis
(14)x,y,z,ak,γk,δ,ηk,θ,λc,n,σd,o∣bm,εm,ϑm,μk,m,and multiplications
(15)x,y=z,x,ak=γk,x,z=δ,x,bm=εm,y,ak=ηk,y,z=θ,y,bm=ϑm,ac,an=λc,n,ak,bm=μk,m,bd,bo=σd,o,where 1≤c,k,n≤s,1≤d,m,o≤t, and c<n,d≤o. Let MH1,0⊕Abs,t^ be a subsuperspace spanned by γk,δ,ηk,θ,λc,n,σd,o∣εm,ϑm,μk,m. Then,
(16)0⟶MH1,0⊕Abs,t^⟶H1,0⊕Abs,t^⟶H1,0⊕Abs,t⟶0,is a maximal stem extension of H1,0⊕Abs,t. In particular, MH1,0⊕Abs,t^ is a multiplier and H1,0⊕Abs,t^ is a cover of H1,0⊕Abs,t.Case 2.
Suppose thatp,q≠1,0. Let Hp,q⊕Abs,t^ be a Lie superalgebra with even basis
(17)xi,yi,z,ak,αe,j,αi,βf,h,γi,k,ζe,j,ηi,k,λc,n,σd,o,ςm,l,ωu,r,ρl,and odd basis
(18)wl,bm,εi,m,εi,l,ϑi,m,ιi,l,μk,m,νk,l,and multiplications
(19)2xe,xj=αe,j,xi,yi=z+αi,xf,yh=βf,h,xi,ak=γi,k,xi,bm=εi,m,xi,wl=εi,l,ye,yj=ζe,j,yi,ak=ηi,k,yi,bm=ϑi,m,yi,wl=ιi,l,ac,an=λc,n,ak,bm=μk,m,ak,wl=νk,l,bd,bo=σd,o,bm,wl=ςm,l,wu,wr=ωu,r,wl,wl=z+ρl,where 1≤e,f,h,i,j≤p,1≤c,k,n≤s,1≤d,m,o≤t,1≤l,r,u≤q, and e<j,f≠h,c<n,d≤o, and u<r. Let MHp,q⊕Abs,t^ be a subsuperspace spanned by
(20)αe,j,αi,βf,h,γi,k,ζe,j,ηi,k,λc,n,σd,o,ςm,l,ωu,r,ρl∣εi,m,εi,l,ϑi,m,ιi,l,μk,m,νk,l.
Then,(21)0⟶MHp,q⊕Abs,t^⟶Hp,q⊕Abs,t^⟶Hp,q⊕Abs,t⟶0,is a maximal stem extension of Hp,q⊕Abs,t. In particular, MHp,q⊕Abs,t^ is a multiplier and Hp,q⊕Abs,t^ is a cover of Hp,q⊕Abs,t.Summarizing, we have the following.Lemma 5.
Letp,q,s,t be positive integers. Then,
(22)0⟶MHp,q⊕Abs,t^⟶Hp,q⊕Abs,t^⟶Hp,q⊕Abs,t⟶0,is a maximal stem extension of Hp,q⊕Abs,t. In particular, Hp,q⊕Abs,t^ is the cover of Hp,q⊕Abs,t and the super-dimension of MHp,q⊕Abs,t as follows:
(23)12s2+12t2+32s+12t+2,st+2t,p,q=1,0,p~+q~+s~+t~−1,2pq+2pt+st+sq,p,q≠1,0,where p~=2p2−p+2ps,q~=1/2q2+1/2q,s~=1/2s2−1/2s,t~=1/2t2+1/2t+tq.


## 3. Multiplier-Rank3 Nilpotent Lie Superalgebras

In this section, we will determine all the nilpotent Lie superalgebras ofmrL=3. The following theorem is analogous to the one in the Lie algebra case [7], yet it contains more information in our super-case.Theorem 6.
LetL be a finite-dimensional, nonabelian, and nilpotent Lie superalgebra of mrL=3. Then,
(1)
smrL≠0,3(2)
smrL=1,2 if and only if L≅H0,2(3)
smrL=3,0 if and only if L≅H1,0⊕Ab2,0(4)
smrL=2,1 if and only if L is isomorphic to one of the following Lie superalgebras:
(a)
H1,0⊕Ab1,1(b)
H0,1⊕Ab1,0(c)
H1Proof.
Let us characterizeL by discussing whether ZL⊆L2.(1)
Suppose thatZL⊆L2. Let K be an ideal contained in ZL with dimK=1. Take H=L/K(a)
IfsdimK=1,0, then one may check that smrL=1,2 and sdimL2=1,0 by Lemma 2. In this case, by (12), we have(24)L≅H0,2.(b)
IfsdimK=0,1, then Lemma 2 yields that smrL=2,1. Hence, smrH+sdimΠL2≤2,0. Since L is not abelian, we have sdimΠL2≠0,0. If sdimΠL2=2,0, then sdimH2=0,1 and smrH=0,0, an impossibility. Therefore, sdimΠL2≠2,0. It remains to consider the case sdimΠL2=1,0. In this case, we have L≅Hk for some k. By (13), we have L≅H1(2)
Suppose thatZLÚL2. By Lemma 4, we have L=M⊕A, where A and M are ideals of L with dimA=1(a)
IfsdimA=0,1, then one may check that there are no algebras satisfying Lemma 2 (2) for smrL=3,0,2,1,1,2, and 0,3(b)
IfsdimA=1,0, one has to discuss the cases smrL=3,0,0,3,2,1, and 1,2(i)
Assume thatsmrL=3,0. We obtain that smrM=2,0 and sdimM2=1,0 by Lemmas 1 and 4 (1). It follows that M≅H1,0⊕Ab1,0 by Theorem 5.8 of [10]. Hence,(25)L≅H1,0⊕Ab2,0.
By Lemma5, (3) is proven.
(ii)
Assume thatsmrL=0,3. We have smrM=0,2 and sdimM2=0,1 by Lemmas 1 and 4 (1). There are no such algebras for smrM=0,2, by Proposition 5.22 of [10]. Therefore, (1) is proven(iii)
Assume thatsmrL=2,1. By Lemmas 1 and 4 (1), we have smrM=1,1, which yields sdimM2=1,0. It follows that M≅H1,0⊕Ab0,1 or H0,1. Hence,(26)L≅H1,0⊕Ab1,1orH0,1⊕Ab1,0.
Then, (4) holds by Lemma5.
(iv)
Assume thatsmrL=1,2. We have either smrM=1,2 or smrM=1,1 by Lemmas 1 and 4 (1). If smrM=1,2, we have sdimM2=0,0, an impossibility. If smrM=1,1, then we have sdimM2=0,1. This is impossible, because smrM=1,1 yields sdimM2=1,0 by Theorem 5.8 of [10]. Then the, proof is complete☐


## 4. Multiplier-Rank4 Nilpotent Lie Superalgebras

In this section, we will determine all the nilpotent Lie superalgebras ofmrL=4. We recall that L3,4,1,4 is a Lie algebra with basis x,y,z,r and nonzero multiplication x,y=z,x,z=r and that L4,5,2,4 is a Lie algebra with basis x,y,z,r,c and nonzero multiplication x,y=z,x,c=r (see [7]).Theorem 7.
LetL be a finite-dimensional, nonabelian, and nilpotent Lie superalgebra of mrL=4. Then,
(1)
smrL≠0,4(2)
smrL=1,3 if and only if L≅H0,3(3)
smrL=2,2 if and only if L is isomorphic to one of the following Lie superalgebras:
(a)
H1,0⊕Ab0,2(b)
H0,1⊕Ab0,1(c)
H1⊕Ab1,0(4)
smrL=4,0 if and only if L is isomorphic to one of the following Lie superalgebras:
(a)
H1,0⊕Ab3,0(b)
L3,4,1,4(c)
L4,5,2,4(5)
smrL=3,1 if and only if L is isomorphic to one of the following Lie superalgebras:
(a)
H1,0⊕Ab2,1(b)
H0,1⊕Ab2,0(c)
H1,1Proof.
Let us characterizeL by discussing whether ZL⊆L2.(1)
Suppose thatZL⊆L2. We may assume that K⊆ZL, where dimK=1 and H≅L/K(a)
IfsdimK=0,1, one may check that smrL=2,2 by Lemma 2. Now assume that smrL=2,2. Then, sdimΠL2+smrH≤2,1. Since sdimH2≤smrH, hence sdimL2=1,1 or 0,1. In the first case sdimL2=1,1, we obtain that sdimH2=1,0 and smrH=1,0. Hence, H≅H1,0. Then, L can be described generally by basis x,y,z∣k with multiplication given by x,y=z. To compute the multiplier start with(27)x,y=z+s1,x,z=s2,x,k=s3,y,z=s4,y,k=s5,z,k=s6,k,k=s7,where s1,s2,⋯,s7 generate the multiplier. A change of variables allows that s1=0. Use of the Jacobi identity on all possible triples shows that s6=0 and ML=s2,s4,s7∣s3,s5. Hence, smrL=1,1, contradicting the assumption smrL=2,2. In the other case sdimL2=0,1, we obtain that L2=ZL. Then, L≅Hk for some k. There are no such algebras by (13) for smrL=2,2(b)
IfsdimK=1,0, then Lemmas 1 and 2 yield smrL=4,0,3,1 and 1,3(i)
Assume thatsmrL=4,0. As in Lie algebra case (Theorem 2 of [7]), one can determine that(28)L≅L3,4,1,4orL4,5,2,4.(ii)
Assume thatsmrL=3,1 and 1,3. The possible case for sdimL2 is 1,0. It follows that L2=ZL and L≅Hp,q for some p,q. By (12), we have(29)L≅H1,1,H0,3,
respectively(2)
Suppose thatZLÚL2. By Lemma 4, we have L=A⊕M, where A and M are ideals of L with dimA=1(a)
IfsdimA=0,1, then Lemmas 1 and 4 (2) yield that smrL=2,2. Therefore, smrM=1,1 and sdimM2=1,0. Hence,(30)L≅H1,0⊕Ab0,2orH0,1⊕Ab0,1.(b)
IfsdimA=1,0, then Lemmas 1 and 4 (1) yield that smrL=4,0,2,2 or 3,1. As a result of the above, we have proven (1) and (2). To prove (3), (4), and (5), one has to discuss the cases smrL=4,0,2,2, and 3,1(i)
Assume thatsmrL=4,0. It is easily checked that there are no such algebras except for sdimM2=1,0 and smrM=3,0. By Theorem 6, we have M≅H1,0⊕Ab2,0 and(31)L≅H1,0⊕Ab3,0.
By Lemma5, (4) is proven.
(ii)
Assume thatsmrL=2,2. We may check that smrM=2,1 and sdimM2=0,1. Then, M≅H1 by Theorem 6. Hence,(32)L≅H1⊕Ab1,0.
Then, (3) holds by Lemma5(iii)
Assume thatsmrL=3,1. We have smrM=2,1 and sdimM2=1,0. By Theorem 6, we have M≅H1,0⊕Ab1,1 or H0,1⊕Ab1,0. Then,(33)L≅H1,0⊕Ab2,1orH0,1⊕Ab2,0.
The proof is complete.☐


## 5. Multiplier-Rank5 Nilpotent Lie Superalgebras

In this section, we will determine all the nilpotent Lie superalgebras ofmrL=5. For convenience, let L1 denote a Lie superalgebra with a homogeneous basis x,k∣z,y and nonzero multiplication x,y=z,z,y=k.Theorem 8.
LetL be a finite-dimensional, nonabelian, and nilpotent Lie superalgebra of mrL=5. Then,
(1)
smrL≠0,5 and 2,3(2)
smrL=1,4 if and only if L≅H0,4(3)
smrL=5,0 if and only if L is isomorphic to one of the following Lie superalgebras:
(a)
H1,0⊕Ab4,0(b)
H2,0(4)
smrL=4,1 if and only if L is isomorphic to one of the following Lie superalgebras:
(a)
H1,0⊕Ab3,1(b)
H0,1⊕Ab3,0(c)
H1,1⊕Ab1,0(5)
smrL=3,2 if and only if L is isomorphic to one of the following Lie superalgebras:
(a)
H1,0⊕Ab1,2(b)
H0,1⊕Ab1,1(c)
H1,2(d)
L1Proof.
Let us characterizeL by discussing whether ZL⊆L2.(1)
Suppose thatZL⊆L2. We may assume that K⊆ZL, where dimK=1 and H≅L/K(a)
IfsdimK=1,0, then Lemma 2 readily yields smrL=5,0,3,2 or 1,4(i)
Assume thatsmrL=5,0. Then, from the proof of Theorem 3 of [3], we have(34)L≅H2,0.(ii)
Assume thatsmrL=3,2. This readily yields sdimL2=1,0, 2,0, or 1,1. In the first case sdimL2=1,0, then L2=ZL and L≅Hp,q for some p,q. By (12), we have(35)L≅H1,2.
Consider the second casesdimL2=2,0. If smrH=0,2 and 0,1, then there are no such algebras by the previous work. If smrH=1,1, then H≅H1,0⊕Ab0,1 or H0,1. If H≅H1,0⊕Ab0,1, then a computation shows that smrL=5,2, contradicting the assumption smrL=3,2. If H≅H0,1, we have smrL=2,1 by a direct computation, also contradicting the assumption smrL=3,2. If smrH=1,0, then H≅H1,0. By the proof of Theorem 1 of [7], we have L≅L3,4,1,4 and smrL=4,0. This contradicts the assumption on smrL. If smrH=1,2, then H≅H0,2, which contradicts the assumption smrL=3,2. In the third case sdimL2=1,1, we have sdimH2=0,1. By the previous work, we have smrH=2,1 and H≅H1. Then, L has a homogeneous basis x,k∣z,y with multiplication given by x,y=z,z,z=a1k,y,y=a2k,z,y=a3k, where a1,a2,a3∈F,k∈ZL. The Jacobi identity shows that a1=0. Since sdimL2=1,1, we have that either a2 or a3 is not zero. Without loss of generality, assume that a3≠0. Replacing z by −1/2a3z and y by −2y+a2/a3z and relabeling, we get a simple multiplication table:
(36)x,y=z,z,y=k.
To compute the multiplier, we start with(37)x,k=s1,x,z=s2,x,y=z+s3,k,z=s4,k,y=s5,z,z=s6,z,y=k+s7,y,y=s8,where s1,s2,⋯,s8 generate the multiplier. By relabeling, we get s3=s7=0. Using the Jacobi identity gives s1=s4=s6=0. Hence, the multiplier has a homogeneous basis s8∣s2,s5, and then, sdimML=1,2 and smrL=3,2. This superalgebra satisfies the requirements. As mentioned above, we have
(38)L≅L1.(iii)
Assume thatsmrL=1,4. If sdimL2≥1,0, then smrH≤0,3. There are no such algebras by the previous work. Then, sdimL2=1,0. It follows from Proposition 4.11 of [10] that(39)L≅H0,4.(b)
IfsdimK=0,1, then Lemma 2 (2) yields that smrL=3,2 or 2,3(i)
Assume thatsmrL=3,2. Then, sdimΠL2+smrH≤3,1. It follows that sdimL2=1,1 and smrH≤2,0. If smrH=2,0, then H≅H1,0⊕Ab1,0. Computing the multiplier as before yields sdimML=5,3 and smrL=2,1, contradicting the assumption smrL=3,2. If smrH=1,0, then H≅H1,0. It is also a contradiction by the previous work(ii)
Assume thatsmrL=2,3. We have sdimΠL2+smrH≤2,2. It follows that sdimL2=1,1 and smrH=1,1. Then, H≅H1,0⊕Ab0,1 or H0,1. If H≅H1,0⊕Ab0,1, then sdimML=4,2 and smrL=2,4 by computing ML as above, contradicting the assumption smrL=2,3. If H≅H0,1, then L has a homogeneous basis z∣x,k with multiplication given by x,x=z. Computing the multiplier as before yields sdimML=1,1 and smrL=2,1, also contradicting the assumption smrL=2,3(2)
Suppose thatZLÚL2. By Lemma 4, we have L=A⊕M, where A and M are ideals of L with dimA=1(a)
IfsdimA=0,1, then Lemma 4(2) yields that smrL=3,2 and smrM=2,1. Then by Theorem 6, we have(40)L≅H1,0⊕Ab1,2orH0,1⊕Ab1,1.(b)
IfsdimA=1,0, then Lemmas 1 and 4 yield that smrL=5,0,4,1 or 3,2. As a result of the above, we have proven (1) and (2). To prove (3), (4), and (5), one has to discuss the cases smrL=5,0,4,1, and 3,2(i)
Assume thatsmrL=5,0. Then, sdimM2≤2,0. If sdimM2=2,0, then smrM=3,0. By Theorem 6, we have M≅H1,0⊕Ab2,0. This yields sdimM2=1,0, contradicting the assumption sdimM2=2,0. If sdimM2=0,0, then M is abelian and smrM=5,0, an impossibility. If sdimM2=1,0. Then, smrM=4,0. By Theorem 7 (4), we have M≅H1,0⊕Ab3,0,L3,4,1,4 or L4,5,2,4. Since sdimM2=1,0, we have M≅H1,0⊕Ab3,0. Then,(41)L≅H1,0⊕Ab4,0.
By Lemma5, (3) holds.
(ii)
Assume thatsmrL=4,1. Then, 0,0<sdimM2≤2,0. If sdimM2=2,0, computing the multiplier as before yields smrL=2,1. This contradicts the assumption on smrL. Therefore, sdimM2=1,0 and smrM=3,1. Then, M is one of the Lie superalgebras listed in Theorem 7 (5). Therefore, L is isomorphic to one of the following Lie superalgebras:(42)H1,0⊕Ab3,1,H0,1⊕Ab3,0,H1,1⊕Ab1,0.
Then, (4) holds by Lemma5.
(iii)
Assume thatsmrL=3,2. Then, 0,0<sdimM2≤1,1. If sdimM2=0,1, then smrM=3,1. This is impossible, because the super-dimension of the derived superalgebra of M when smrM=3,1 is not 0,1. Similarly, if sdimM2=1,1, then smrM=2,1, a contradiction. Now consider sdimM2=1,0; we have smrM=2,2. By Theorem 7 (3), we have(43)L≅H1,0⊕Ab1,2orH0,1⊕Ab1,1.
The proof is complete.☐


## 6. Multiplier-Rank6 Nilpotent Lie Superalgebras

In this section, we will determine all the nilpotent Lie superalgebras ofmrL=6. Let us recall that L4,5,1,6 is a Lie algebra with basis x,y,z,r,c and nonzero multiplication x,y=z,x,z=r,y,c=r. For convenience, let L2 denote a Lie superalgebra with a homogeneous basis a,z∣x,k and nonzero multiplication x,x=z,a,x=k.Theorem 9.
LetL be a finite-dimensional, nonabelian, and nilpotent Lie superalgebra of mrL=6. Then,
(1)
smrL≠0,6(2)
smrL=1,5 if and only if L≅H0,5(3)
smrL=2,4 if and only if L≅H0,4⊕Ab1,0(4)
smrL=6,0 if and only if L is isomorphic to one of the following Lie superalgebras:
(a)
H1,0⊕Ab5,0(b)
H2,0⊕Ab1,0(c)
L3,4,1,4⊕Ab1,0(d)
L4,5,2,4⊕Ab1,0(e)
L4,5,1,6(5)
smrL=5,1 if and only if L is isomorphic to one of the following Lie superalgebras:
(a)
H1,0⊕Ab4,1(b)
H0,1⊕Ab4,0(c)
H1,1⊕Ab2,0(d)
H2,1(6)
smrL=4,2 if and only if L is isomorphic to one of the following Lie superalgebras:
(a)
H1,0⊕Ab2,2(b)
H0,1⊕Ab2,1(c)
H1,2⊕Ab1,0(d)
H1,1⊕Ab0,1(7)
smrL=3,3 if and only if L is isomorphic to one of the following Lie superalgebras:
(a)
H1,0⊕Ab0,3(b)
H0,1⊕Ab0,2(c)
H1,3(d)
L2(e)
H2Proof.
Let us characterizeL by discussing whether ZL⊆L2.(1)
Suppose thatZL⊆L2. We may assume that K⊆ZL, where dimK=1 and H≅L/K(a)
IfsdimK=1,0, then Lemma 2 readily yields smrL=6,0,5,1,3,3, or 1,5. In the first case smrL=6,0, by Theorem 4 of [7], one can determine(44)L≅L4,5,1,6.
In the remaining casessmrL=5,1,3,3, and 1,5, we obtain that the possible super-dimension of L2 is 1,0. It follows that L2=ZL and L≅Hp,q for some p,q. By (12), we have
(45)L≅H2,1,H1,3,H0,5,respectively.
(b)
IfsdimK=0,1, then Lemma 2 yields that smrL=3,3 or 2,4(i)
Assume thatsmrL=3,3. Then, sdimΠL2+smrH≤3,2. It follows that the only possible super-dimension of L2 is 1,1. Hence, we have sdimH2=1,0 and smrH≤2,1. There are no such algebras for smrH<2,1 by the previous work. If smrH=2,1, then H≅H1,0⊕Ab1,1 or H0,1⊕Ab1,0. In the first case H≅H1,0⊕Ab1,1, then L has a homogeneous basis x,y,z,a∣b,k with multiplication x,y=z,x,b=a2k,y,b=a3k,z,b=a4k,a,b=a5k. The Jacobi identity shows that a4=0. Since a∉L2 and ZL⊆L2, we have a5≠0. Take 1/a5a for a. If a2=a3=0, then we obtain a simple multiplication x,y=z,a,b=k. Computing the multiplier as before yields sdimML=5,3 and smrL=4,5, contradicting the assumption smrL=3,3. If a2 or a3 is not zero, then without loss of generality, assume that a2≠0. Replacing x by 1/a2x and y by −a3x+a2y and relabeling, we obtain a simple multiplication x,y=z,x,b=k,a,b=k. Computing the multiplier as before yields sdimML=6,1 and smrL=3,7, also contradicting the assumption smrL=3,3. In the other case H≅H0,1⊕Ab1,0, one may obtain that L can be described by a homogeneous basis a,z∣x,k with multiplication x,x=z,a,x=a2k,z,x=a3k. The Jacobi identity shows that a3=0. Take 1/a2a for a, relabel and get x,x=z,a,x=k. Then, sdimML=1,1 and smrL=3,3. This algebra satisfies the requirements. As mentioned above, we have(46)L≅L2.(ii)
Assume thatsmrL=2,4. Then, sdimΠL2+smrH≤2,3 and sdimL2=0,1 by a direct computation. Hence, L2=ZL. It follows that L≅Hk for some k. By (12), we have(47)L≅H2.(2)
Suppose thatZLÚL2. By Lemma 4, we have L=A⊕M, where A and M are ideals of L with dimA=1(a)
IfsdimA=0,1, then Lemmas 1 and 4 (2) yield that smrL=5,1,4,2, or 3,3. Then, sdimΠM2=0,1 by the previous work
(i)
Assume thatsmrL=5,1. Then, smrM=4,0 and Theorem 7 reveals three candidates, only one of which satisfies sdimM2=1,0. Therefore, L≅H1,0⊕Ab3,1. It yields smrL=4,1 and contradicts the assumption that smrL=5,1(ii)
Assume thatsmrL=4,2. Then, smrM=3,1 and M≅H1,0⊕Ab2,1,H0,1⊕Ab2,0, or H1,1. Therefore, L is isomorphic to one of the following Lie superalgebras:(48)H1,0⊕Ab2,2,H0,1⊕Ab2,1,H1,1⊕Ab0,1.(iii)
Assume thatsmrL=3,3. Then, smrM=2,2 and M≅H1,0⊕Ab0,2 or H0,1⊕Ab0,1. Hence,(49)L≅H1,0⊕Ab0,3orH0,1⊕Ab0,2.(b)
IfsdimA=1,0, then Lemmas 1 and 4 (1) yield that the possible super-multiplier-rank of L is 2,4,6,0,5,1, or 4,2. By the above work, we have proven (1), (2), and (7). To prove (3), (4), (5), and (6), one has to discuss the cases smrL=2,4,6,0,5,1, and 4,2(i)
Assume thatsmrL=2,4. One may obtain that smrM=1,4 and sdimM2=1,0. By Theorem 8, we have M≅H0,4 and(50)L≅H0,4⊕Ab1,0.
Then, (3) holds by Lemma5.
(ii)
Assume thatsmrL=6,0. By Theorem 4 of [7], we have L is isomorphic to one of the following Lie superalgebras:(51)L3,4,1,4⊕Ab1,0,H1,0⊕Ab5,0,L4,5,2,4⊕Ab1,0,H2,0⊕Ab1,0.
By Lemma5, (4) is proven.
(iii)
Assume thatsmrL=5,1. Since sdimM2≤smrM, it follows that smrM≥3,1. If smrM=5,1, then M is abelian and sdimM2=0,0, an impossibility. Therefore, smrM≠5,1. If smrM=4,1, then M≅H1,0⊕Ab3,1,H0,1⊕Ab3,0, or H1,1⊕Ab1,0 by Theorem 8. Therefore L is isomorphic to one of the following Lie superalgebras:(52)H1,0⊕Ab4,1,H0,1⊕Ab4,0,H1,1⊕Ab2,0.
IfsmrM=3,1, then M≅H1,0⊕Ab2,1,H0,1⊕Ab2,0, or H1,1 and sdimM2=2,0. These are impossible, because the super-dimension of the derived superalgebras of H1,0⊕Ab2,1,H0,1⊕Ab2,0, and H1,1 are 1,0. Hence, (5) holds by Lemma 5.
(iv)
Assume thatsmrL=4,2. The only possibility of super-multiplier-rank of M is 3,2. Then, Theorem 7 reveals four candidates, only three of which satisfy sdimM2=1,0. Therefore, L is isomorphic to one of the following Lie superalgebras:(53)H1,0⊕Ab2,2,H0,1⊕Ab2,1,H1,2⊕Ab1,0.
The proof is complete.☐


## 7. Covers

By Theorem 3.3 of [13], we obtain that any two covers of a finite-dimensional Lie superalgebra are isomorphic as ordinary Lie superalgebras. In this section, we will describe the covers of all the nilpotent Lie superalgebras of multiplier-rank ≤6.First, we compute the cover forL3,4,1,4⊕Ab1,0. Let x1,x2,x3,x4,u∣xi=u=0¯,i=1,2,3,4 be a homogeneous basis of L3,4,1,4⊕Ab1,0, where x1,x2,x3,x4∈L3,4,1,4 and u∈Ab1,0. The multiplication is given by x1,x2=x3,x1,x3=x4, the other brackets of basis elements vanishing. Suppose that
(54)0⟶M⟶K⟶L3,4,1,4⊕Ab1,0⟶0,is a stem extension ofL3,4,1,4⊕Ab1,0. Then, M⊂K2∩ZK and K/M≅L3,4,1,4⊕Ab1,0. Then, K/M has a homogeneous basis x1+M,x2+M,x3+M,x4+M,u+M. We may assume that
(55)x1,x2=x3+a1,x1,x3=x4+a2,x1,x4=a3,x2,x3=a4,x2,x4=a5,x3,x4=a6,x1,u=a7,x2,u=a8,x3,u=a9,x4,u=a10,where a1,⋯,a10∈M1. Without loss of generality, we may suppose that a1=a2=0. By the Jacobi identity of Lie superalgebras, we have a5=a6=a9=a10=0. Then, M is spanned by a3,a4,a7,a8 and K is spanned by x1,x2,x3,x4,u,a3,a4,a7,a8. Hence, sdimM≤4,0. Now let L3,4,1,4⊕Ab1,0^ be a Lie superalgebra with a homogenous basis x,y,z,r,u,a,b,c,d and multiplication x,y=z,x,z=r,x,r=a,y,z=b,x,u=c,y,u=d. Let ML3,4,1,4⊕Ab1,0^ be a subsuperspace of L3,4,1,4⊕Ab1,0^ spanned by a,b,c,d. Then,
(56)ML3,4,1,4⊕Ab1,0^⊆ZL3,4,1,4⊕Ab1,0^∩L3,4,1,4⊕Ab1,02^,L3,4,1,4⊕Ab1,0ML3,4,1,4⊕Ab1,0≅L3,4,1,4⊕Ab1,0.SincesdimML3,4,1,4⊕Ab1,0^=4,0, we obtain that ML3,4,1,4⊕Ab1,0^ is a multiplier and L3,4,1,4⊕Ab1,0^ is a cover of L3,4,1,4⊕Ab1,0.Then, we compute the cover forL3,4,1,4. Let L3,4,1,4^ be a Lie superalgebra with a homogenous basis x,y,z,r,a,b and multiplication x,y=z,x,z=r,x,r=a,y,z=b. Since, L3,4,1,4 is a subalgebra of L3,4,1,4⊕Ab1,0, one may check that L3,4,1,4^ is a cover of L3,4,1,4.Now we compute the cover forL4,5,2,4⊕Ab1,0. Let x,y,z,c,r,v∣x=y=z=c=r=v=0¯ be a homogeneous basis of L4,5,2,4⊕Ab1,0, where x,y,z,c,r∈L4,5,2,4 and v∈Ab1,0. The multiplication is given by x,y=z,x,c=r, the other brackets of basis elements vanishing. Similarly, one may check that the Lie superalgebra with a homogeneous basis x,y,z,c,r,v,a1,a2,a3,a4,a5,a6,a7,a8 and multiplication
(57)x,y=z,x,z=a1,x,c=rx,r=a2,y,z=a3,y,c=a4,c,r=a5,x,v=a6,y,v=a7,c,v=a8is a cover of L4,5,2,4⊕Ab1,0, which is denoted by L4,5,2,4⊕Ab1,0^.Then, we compute the cover forL4,5,2,4. Since L4,5,2,4 is a subalgebra of L4,5,2,4⊕Ab1,0, one may check that the Lie superalgebra with a homogeneous basis x,y,z,c,r,a1,a2,a3,a4,a5 and multiplication x,y=z,x,z=a1,x,c=r,x,r=a2,y,z=a3,y,c=a4,c,r=a5 is a cover of L4,5,2,4, which is denoted by L4,5,2,4^.WhensmrL=6,0, we also obtain a cover L4,5,1,6^ for L4,5,1,6, which has a homogenous basis x,y,z,r,c,a1,a2,a3,a4 and multiplication
(58)x,y=z,x,z=r,x,r=a1,x,c=a2,y,z=a3,y,c=r+a4.By the previous work, we may obtain a coverL1^ for L1, which has a homogenous basis x,k,s8∣z,y,s2,s5 and multiplication x,y=z,z,y=k,x,z=s2,k,y=s5,y,y=s8. When smrL=3,3, we may obtain a cover L2^ for L2, which has a homogenous basis a,z,b∣c,x,k and multiplication a,x=k,a,k=c,x,x=z,x,k=b.


## 8. Main Result

Theorem 10.
The classification and covers of all the finite-dimensional nilpotent Lie superalgebrasL of multiplier-rank ≤6 are listed as follows:
whereL1 is a (2, 2)-superdimensional Lie superalgebra with a homogeneous basis x,k∣z,y and nonzero multiplication x,y=z,z,y=k,L2 is a (2, 2)-superdimensional Lie superalgebra with a homogeneous basis a,z∣x,k and nonzero multiplication x,x=z,a,x=k.Proof.
By Theorem 5.8 of [10], we obtain all the finite-dimensional nilpotent Lie superalgebras L of mrL≤2. In Theorems 6, 7, 8, and 9, all the nilpotent Lie superalgebras of mrL=3,4,5,6 are determined. We only need to describe the covers of all Lie superalgebras of mrL≤6. By Theorem 4.1 of [13], Lemma 5, and Section 7, we may obtain the covers of all the nilpotent Lie superalgebras of mrL≤6. In summary, we obtain Table 1.☐Table 1
The classification and covers.
smrLLCover0,0Any abelian Lie superalgebraAbs,t,s,t∈ℕAb^s,t1,0H1,0H^1,02,0H1,0⊕Ab1,0H1,0⊕Ab1,0^(1, 1)H1,0⊕Ab0,1H1,0⊕Ab1,0^H0,1H^0,11,2H0,2H^0,23,0H1,0⊕Ab2,0H1,0⊕Ab2,0^(2,1)H1,0⊕Ab1,1H1,0⊕Ab1,1^H0,1⊕Ab1,0H0,1⊕Ab1,0^H1H^11,3H0,3H^0,3(2,2)H1,0⊕Ab0,2H1,0⊕Ab0,2^H0,1⊕Ab0,1H0,1⊕Ab0,1^(4,0)H1,0⊕Ab3,0H1,0⊕Ab3,0^L3,4,1,4L3,4,1,4^L4,5,2,4L4,5,2,4^(3,1)H1,0⊕Ab2,1H1,0⊕Ab2,1^H0,1⊕Ab2,0H0,1⊕Ab2,0^H1,1H^1,11,4H0,4H^0,4(5,0)H1,0⊕Ab4,0H1,0⊕Ab4,0^H2,0H^2,0(4,1)H1,0⊕Ab3,1H1,0⊕Ab3,1^H0,1⊕Ab3,0H0,1⊕Ab3,0^H1,1⊕Ab1,0H1,1⊕Ab1,0^(3,2)H1,0⊕Ab1,2H1,0⊕Ab1,2^H0,1⊕Ab1,1H0,1⊕Ab1,1^H1,2H^1,2L1L1^2,4H0,4⊕Ab1,0H0,4⊕Ab1,0^(6,0)H1,0⊕Ab5,0H1,0⊕Ab5,0^H2,0⊕Ab1,0H2,0⊕Ab1,0^L3,4,1,4⊕Ab1,0L3,4,1,4⊕Ab1,0^L4,5,2,4⊕Ab1,0L4,5,2,4⊕Ab1,0^L4,5,1,6L4,5,1,6^(5,1)H1,0⊕Ab4,1H1,0⊕Ab5,0^H0,1⊕Ab4,0H0,1⊕Ab4,0^H1,1⊕Ab2,0H1,1⊕Ab2,0^H2,1H^2,1(4,2)H1,0⊕Ab2,2H1,0⊕Ab2,2^H0,1⊕Ab2,1H0,1⊕Ab2,1^H1,2⊕Ab1,0H1,2⊕Ab1,0^H1,1⊕Ab0,1H1,1⊕Ab0,1^(3,3)H1,0⊕Ab0,3H1,0⊕Ab0,3^H0,1⊕Ab0,2H0,1⊕Ab0,2^H1,3H^1,3L2L2^H2H^21,5H0,5H^0,5



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