# Constructions and Properties for a Finite-Dimensional Modular Lie SuperalgebraKn,m

Authors: Dan Mao; Keli Zheng
Journal: Journal of Mathematics
 (2021)

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## Abstract

In this paper, a finite-dimensional Lie superalgebraKn,m over a field of prime characteristic is constructed. Then, we study some properties of Kn,m. Moreover, we prove that Kn,m is an extension of a simple Lie superalgebra, and if m=n−1, then it is isomorphic to a subalgebra of a restricted Lie superalgebra.


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## Body

## 1. Introduction

In the 1970s, physicists introduced the concept of Lie superalgebra in order to describe supersymmetry (see [1]). Since Lie superalgebra is an important mathematical model of supersymmetry, the research on it has been very active and rich results have been obtained (see [2]). In mathematics, Lie superalgebra is also a natural generalization of Lie algebra. In 1977, Kac completed the classification of finite-dimensional simple Lie superalgebras over a field of characteristic zero (see [3]). The research on Lie superalgebras over a field of characteristic zero has been quite systematic (see [4–6]), but the research on modular Lie superalgebras remains to be perfected (see [7]). Although some mathematicians try to study the classification of modular Lie superalgebras (see [7–13]), the classification problem has still been open. Therefore, it is very important to construct new finite-dimensional modular Lie superalgebras.The finite-dimensional modular Lie superalgebrasWn,m, Hn,m were constructed in [14, 15], respectively. Their natural filtrations are investigated in [16]. The finite-dimensional modular Lie superalgebra Sn,m was given in [17]. Modular Lie superalgebra Kn, which takes the Grassmann algebra as base algebra, was constructed in [18]. Inspired by the above mentioned literatures, this paper constructs a finite-dimensional modular Lie superalgebra of contact type, which is denoted by Kn,m.The remainder of this paper is arranged as follows. A brief summary of the relevant concepts and notations is presented in Section2. In Section 3, we construct the finite-dimensional modular Lie superalgebra Kn,m. In Section 4, we obtain some properties of Kn,m. Moreover, we prove that Kn,m is an extension of Kn,0, and if m=n−1, then it is isomorphic to a subalgebra of KOn−1,n,1¯.


## 2. Preliminaries

Throughout this paper,F denotes an algebraic closed field of characteristic p≥3; n is an integer greater than 3. Apart from the standard notation ℤ, the sets of positive integers and nonnegative integers are denoted by ℕ and ℕ0, respectively. ℤ2=0¯,1¯ denotes the ring of integers modulo 2.LetΛn be the Grassmann algebra over F in n variables x1,x2,…,xn. Set Bk=i1,i2,…,ik|1≤i1<i2<⋯<ik≤n and Bn=∪k=0nBk, where B0=∅. For u=i1,i2,…,ik∈Bk, set u=k, u=i1,i2,…,ik and xu=xi1xi2,…,xik∅=0,x∅=1. Then, xu|u∈Bn is an F-basis of Λn.LetΛn−1 be the Grassmann algebra over F in n−1 variables x1,x2,…,xn−1. Obviously, Λn=Λn−1∧xn.LetD¯k:Λn⟶Wn be the linear map such that for any f∈Λnθ, θ∈ℤ2,(1)D¯kf=∑i=1n−1fiDi+fnxnDn,where fi=−1θxixnDnf+Dif and fn=2f−∑i=1n−1xiDif.LetK¯n=D¯kf|f∈Λn. Then, K¯n is a finite-dimensional Lie superalgebra according to the operations in Wn. Let Kn=K¯n,K¯n. Then, Kn=f|f∈Λn,f≠xu^, where u^=1,…,n−1. In [18], Xin proves that Kn is not a simple Lie superalgebra.LetU=Λn⊗Tm be the tensor product, where Tm is the truncated polynomial algebra satisfying yip=1 for all i=1,2,…,m (see [17]). Then, U is an associative superalgebra with ℤ2-gradation induced by the trivial ℤ2-gradation of Tm and the natural ℤ2-gradation of Λn. Namely, U=U0¯⊕U1¯, where U0¯=Λn0¯⊗Tm and U1¯=Λn1¯⊗Tm.Forf∈Λn and α∈Tm, we abbreviate f⊗α as fα. Then, the elements xuyλ with u∈Bn and λ∈G form an F-basis of U. It is easy to see that U=⊕ni=0Ui is a ℤ-graded superalgebra, where Ui=spanFxuyλ|u∈Bn,u=i,λ∈G. In particular, U0=Tm and Un=spanFxπyλ|λ∈G, where π:=1,2,…,n∈Bn.In this paper, lethgA=A0¯∪A1¯, where A=A0¯⊕A1¯ is a superalgebra. If x is a ℤ2-homogeneous element of A, then degx denotes the ℤ2-degree of x.SetY=1,2,…,n. Given i∈Y, let ∂/∂xi be the partial derivative on Λn with respect to xi. For i∈Y, let Di be the linear transformation on U such that Dixuyλ=∂xu/∂xiyλ for all u∈Bn and λ∈G. Let DerU denote the derivation superalgebra of U (see [11]). Then, Di∈Der1¯U for all i∈Y since ∂/∂xi∈Der1¯Λn (see [7]).Suppose thatu∈Bk⊆Bn and i∈Y. When i∈u, u−i denotes the uniquely determined element of Bk−1 satisfying u−i=u∖i. Then, the number of integers less than i in u is denoted by τu,i. When i∉u, we set τu,i=0 and xu−i=0. Therefore, Dixu=−1τu,ixu−i for all i∈Y and u∈Bn.We definefDg=fDg for f,g∈hgU and D∈hgDerU. Since the multiplication of U is supercommutative, fD is a derivation of U. Let(2)Wn,m=spanFxuyλDi|u∈Bn,λ∈G,i∈Y.Then,Wn,m is a finite-dimensional Lie superalgebra contained in DerU. A direct computation shows that(3)fDi,gDj=fDigDj−−1degfDideggDjgDjfDi,where f,g∈hgU and i,j∈Y.Definition 1 (see [4]).
A Lie superalgebraL is called simple if it does not have any graded ideals which are different from 0 and L and if, moreover, L,L≠0.


## 3. Construction ofKn,m

SetJ=1,…,n−1. Let D˜k:U⟶Wn,m be the linear map such that(4)D˜kf=∑i∈JfiDi+fnxnDn,where f∈hgU, fi=−1degfxixnDnf+Dif,i∈J, and fn=2f−∑i∈JxiDif.LetK¯n,m  = D˜kf|f∈U. Then, K¯n,m is a subspace of Wn,m.Let(5)Gi=Di+xixnDn,∀i∈J,(6)Gn=2xnDn.By direct calculation, we have(7)Gi,Gj=δijGn,Gn,Gj=0,where i,j∈J and δij is the Kronecker delta.Proposition 1.
D˜kf=∑i∈J−1degfGifGi+fGn.Proof.
For anyj∈J and λ∈G, we have(8)∑i∈J−1degfGifGi+fGnxjyλ=∑i∈J−1degfDif+xixnDnfDi+xixnDnxjyλ=−1degfDjf+xjxnDnfyλ=D˜kfxjyλ,∑i∈J−1degfGifGi+fGnxnyλ=∑i∈J−1degfDif+xixnDnfDi+xixnDn+2fxnDnxnyλ=∑i∈J−1degfDif+xixnDnfxixnyλ+2fxnyλ=∑i∈J−1degfDifxixnyλ+2fxnyλ=−∑i∈JxiDif+2fxnyλ=D˜kfxnyλ.
Therefore,(9)D˜kf=∑i∈J−1degfGifGi+fGn.Proposition 2.
Letf∈Uθ and g∈Uμ, where θ,μ∈ℤ2. Then,(10)D˜kf,D˜kg=D˜kf,g,where f,g=D˜kfg−Gnfg.Proof.
LetD˜kf=∑i∈YfiGi,D˜kg=∑j∈YgjGj.
SinceD˜kf=∑i∈J−1degfGifGi+fGn, we get(11)fi=−1θGif,∀i∈J,fn=f,gi=−1μGig,∀i∈J,gn=g.
Therefore,degfi=θ+1,deggi=μ+1, for i∈Y.
Hence,(12)Gifj=Gi−1θGjf=−1θGiGjf.
SinceGi,Gj=δijGn, where i,j∈J, we obtain(13)Gifj=−Gjfi+−1θδijGnf,Gigj=−Gjgi+−1μδijGng.
SetD˜kf,D˜kg=∑j∈YhjGj. For any h∈U, we have(14)D˜kf,D˜kgh=∑i∈YfiGi,∑j∈YgjGjh=∑i,j∈YfiGigjGjh−−1θμgjGjfiGih=∑i,j∈YfiGigjGjh−−1θμ+θ+1gjfiGiGjh−−1θμgjGjfiGih+−1μ+1figjGjGih=∑i,j∈YfiGigjGj−−1θμgjGjfiGih+−1μ+1∑i,j∈YfigjGi,Gjh=∑i,j∈YfiGigjGj−−1θμgjGjfiGih+−1μ+1∑i,j∈YδijfigjGnh.
It follows that(15)D˜kf,D˜kg=∑i∈YfiGi,∑j∈YgjGj=∑i,j∈YfiGigjGj−−1θμgjGjfiGi−−1μ∑i,j∈YδijfigjGn.
For allj∈J, we have(16)hj=∑i∈JfiGigj−−1θμ∑i∈JgiGifj+fGngj−−1θμgGnfj=−∑i∈JfiGjgi+−1θμ∑i∈JgiGjfi+−1μfjGng−−1θμ+θgjGnf+fGngj−−1θμgGnfj=−1θ∑i∈JGjfigi+−1θ+1fiGjgi+−1θ+μGjfGng−−1θμ+θ+μGjgGnf+fGngj−−1θμgGnfj=−1θGj∑i∈Jfigi+−1θ+μGjfGng−−1θ+μ+θGnfGjg+fGn−1μGjg−−1θμgGn−1θGjf=−1θGj∑i∈J−1θGif−1μGig+−1θ+μGjfGng−Gnfg=−1θ+μGj∑i∈J−1θGifGig+fGng−Gnfg=−1θ+μGjD˜kfg−Gnfg=−1θ+μGjf,g.
Then,hj exactly equals to the coefficient of Gj in D˜kf,g.
In addition,(17)hn=∑i∈JfiGig−−1θμ∑i∈JgiGif+fGng−−1θμgGnf−−1μ∑i∈Jfigi=∑i∈J−1θGifGig−−1θμ∑i∈J−1μGigGif+fGng−−1θμgGnf−−1μ∑i∈J−1θGif−1μGig=−−1θμ+μ+θ+1μ+1∑i∈JGifGig+fGng−Gnfg=∑i∈J−1θGifGig+fGng−Gnfg=D˜kfg−Gnfg=f,g.
Then,hn exactly equals to the coefficient of Gn in D˜kf,g.
Therefore,D˜kf,D˜kg=D˜kf,g.
An immediate corollary of this proposition is the following.Corollary 1.
K¯n,m is a subalgebra of Wn,m.
Next, we give another way to expressK¯n,m. We still denote the linear map from U to K¯n,m by D˜k. Namely,(18)D˜k:U⟶K¯n,m.Then, we prove the following proposition.Proposition 3.
K¯n,m≅U.Proof.
Letf∈KerD˜k. We obtain(19)fi=−1degfxixnDnf+Dif=0,∀i∈J,fn=2f−∑i∈JxiDif=0,f=12∑i∈JxiDif=−12∑i∈JxixixnDnf=0.
Therefore,KerD˜k=0. Then, D˜k is injective.
We define an operator, in U. For any f,g∈U, we have(20)f,g=D˜kfg−Gnfg.
By Proposition2, we have(21)D˜kf,g=D˜kf,D˜kg.
Note thatD˜k is injective. Therefore, it can be concluded that U is a Lie superalgebra about the operator , (see [1]). It follows from equation (21) that K¯n,m≅U as Lie superalgebras.
SinceK¯n,m≅U, we use K¯n,m instead of U defined by equation (20). By equations (5) and (6) and Proposition 2, for f,g∈K¯n,m, we have(22)f,g=D˜kfg−Gnfg=∑i∈J−1degfGifGig+fGng−Gnfg=∑i∈J−1degfDifDig+∑i∈J−1degfDifxixnDng+2fxnDng+∑i∈J−1degfxixnDnfDig−2xnDnfg+∑i∈J−1degfxixnDnfxixnDng=2f−∑i∈JxiDifxnDng−−1degfdegg2g−∑i∈JxiDigxnDnf+∑i∈J−1degfDifDig.Definition 2.
Kn,m=K¯n,m,K¯n,m.Proposition 4.
Kn,m=xuyλ|xuyλ∈U,xuyλ≠xu^yλ, where u^=1,…,n−1.Proof.
LetK˜n,m=xuyλ|xuyλ∈U,xuyλ≠xu^yλ. It suffices to prove that Kn,m=K˜n,m. For any f∈K˜n,m, let f=∑t∑λ∈Gctλxutxnδtyλ, where ctλ∈F,ut=t1,…tstj∈J,j=1,…,s. If generator xn is contained in xutxnδtyλ, then δt=1. Otherwise, δt=0.
Firstly, we proveKn,m⊇K˜n,m:(i)
Ifut<n−1, there exist tr∉ut such that(23)xtr,xtrxutxnδtyλ=2xtr−∑i∈JxiDixtrxnDnxtrxutxnδtyλ−−1degxtrdegxtrxutxnδt2xtrxutxnδtyλ−∑i∈JxiDixtrxutxnδtyλxnDnxtr+∑i∈J−1degxtrDixtrDixtrxutxnδtyλ=xtrxtrxutxnyλ−−1τut′,trxutxnδtyλ=−−1τut′,trxutxnδtyλ.ut′=tr,ut,n.Therefore,xutxnδtyλ∈K¯n,m,K¯n,m=Kn,m.(ii)
Ifut=n−1, δt=1, then xutxnδtyλ=xu^xnyλ. Therefore,(24)yλ,xu^xn=−1n−1⋅2xu^xnyλ.
According to (i) and (ii),xutxnδtyλ∈K¯n,m,K¯n,m=Kn,m. Namely, for all f∈K˜n,m, we have f∈Kn,m. Therefore, Kn,m⊇K˜n,m.
Secondly, we proveKn,m⊆K˜n,m:
Note that(25)dimK¯n,m=2npm,dimK˜n,m=2n−1pm,dimK¯n,m−dimK˜n,m=pm.
Therefore, it suffices to prove thatxu^yλ∉K¯n,m,K¯n,m. Without loss of generality, suppose that there exist xu1xnδ1yη,xu2xnδ2yμ∈K¯n,m such that xu1xnδ1yη,xu2xnδ2yμ=xu^yλ, where λ=η+μ.
Ifδ1=δ2=1, then(26)xu1xnδ1yη,xu2xnδ2yμ=2xu1xnδ1yη−∑i∈JxiDixu1xnδ1yηxnDnxu2xnδ2yμ−−1degxu1xnδ1degxu2xnδ22xu2xnδ2yμ−∑i∈JxiDixu2xnδ2yμxnDnxu1xnδ1yη+∑i∈J−1degxu1xnδ1Dixu1xnδ1yηDixu2xnδ2yμ=0.
Ifδ1=1,δ2=0, then xu1xnδ1yη,xu2xnδ2yμ=cuxuxnyλ, where cu∈F,xu∈Αn−1.
Therefore,δ1=δ2=0. Namely,(27)xu1xnδ1yη,xu2xnδ2yμ=xu1yη,xu2yμ=−1degxu1∑i∈J−1τu1,i+τu2,ixu1+u2−2iyλ.
By the definition of the Grassmann algebra, for allu1,u2∈Bn−1, we have(28)∑i∈J−1τu1,i+τu2,ixu1+u2−2iyλ≠xu^yλ.
It follows thatxu^yλ∉K¯n,m,K¯n,m. Then, Kn,m=K¯n,m,K¯n,m⊆K˜n,m.
Therefore,Kn,m=xuyλ|xuyλ∈U,xuyλ≠xu^yλ.


## 4. Some Properties ofKn,m

Proposition 5.
Kn,m does not possess a ℤ-graded structure as Wn,m.Proof.
Suppose thatKn,m has ℤ-gradation:(29)Kn,m=⊕i=−rsKn,mi,where Kn,mi=spanFxuyλ|xu∈Λn,λ∈G,i=iu. Let yλ∈Kn,mt,xnyη∈Kn,mq, where λ,η∈G. Suppose that xiyμ∈Kn,ml, for all i≠n,μ∈G. Since xiyμ,xiyμ=−y2μ, we have t=2l. Since yλ,xnyη=2xnyλ+η, we have t+q=q. Then, t=2l=0. Therefore, xiyμ∈Kn,m0 for all i≠n,μ∈G. Since xiyμ,xnyη=xixnyμ+η and Kn,m0,Kn,mq⊆Kn,mq, we obtain xixnyμ+η∈Kn,mq. For all xj≠xi,j≠n,γ∈G, we have xjyγ∈Kn,m0. Then, xjyγ,xixnyμ+η=xjxixnyγ+μ+η∈Kn,mq, where γ+μ+η∈G. Following the discussion above, we have xuxnyλ∈Kn,mq, where xu∈Λn−1,λ∈G.
On the other hand, letxixjyλ∈Kn,ma, where i≠j≠n,λ∈G. Since xiyμ,xixjyλ=−−1τu,ixjyμ+λ, where u=i,j, we know a=0. Therefore, xixjyλ∈Kn,m0. For all xkyμ∈Kn,m0,k≠n,i,j, we have xkyμ,xkxixjyλ=−−1τu,kxixjyμ+λ, where u=k,i,j. Then, xkxixjyλ∈Kn,m0. Following the discussion above, we have xuyλ∈Kn,m0, where xu∈Λn−1,λ∈G.
Ifxnyη∈Kn,m0, then Kn,m=Kn,m0. If xnyη∈Kn,m1, then Kn,m=Kn,m0⊕Kn,m1.
Therefore,Kn,m does not possess a ℤ-graded structure as Wn,m.Lemma 1 (see [15]).
Let△=∑λ∈Gαλyλ|∑λ∈Gαλ=0,αλ∈F. Then, △ is an ideal of Tm and Tm=△⊕F1.Theorem 1.
LetI0=xua|xu∈Λn,a∈△,xua≠xu^a. Then, I0 is an ideal of Kn,m. Namely, the Lie superalgebra Kn,m is not simple.Proof.
Letxu1a∈I0,xu2b∈Kn,m, where xu1,xu2∈Λn,a∈△,b∈Tm. Then,(30)xu1a,xu2b=2xu1a−∑i∈JxiDixu1axnDnxu2b−−1degxu1degxu22xu2b−∑i∈JxiDixu2bxnDnxu1a+∑i∈J−1degxu1Dixu1aDixu2b=cuxuab,where cu∈F,xu∈Λn. Since △ is an ideal of Tm, we have ab∈△. Then, xu1a,xu2b=cuxuab∈I0. Therefore, I0 is an ideal of Kn,m. Now, we conclude that Kn,m is not a simple Lie superalgebra.
LetKn,m0:=Kn,m/I0. Then,(31)Kn,m0=xu+I0|xu∈Λn,xu≠xu^.
LetKn,0=xu|xu∈Λn,xu≠xu^. Then, we define an operator , in Kn,0. For all f∈Kn,0θ,g∈Kn,0μ, θ,μ∈ℤ2, we define f,g=∑i∈J−1θDifDig. Then, Kn,0 is a simple Lie superalgebra.Theorem 2.
Kn,m0≅Kn,0. Therefore, Kn,m is an extension of the simple Lie superalgebra Kn,0 and I0 is the maximal ideal of Kn,m.Proof.
We define a linear mapσ:Kn,m0⟶Kn,0 such that σxu+I0=xu for all xu+I0∈Kn,m0. Obviously, σ is an isomorphism of linear spaces.
In addition, for allxu1+I0,xu2+I0∈Kn,m0, we have(32)σxu1+I0,xu2+I0=σxu1,xu2+xu1,I0+I0,xu2+I0,I0=σ2xu1−∑i∈JxiDixu1xnDnxu2−−1degxu1degxu22xu2−∑i∈JxiDixu2xnDnxu1+∑i∈J−1degxu1Dixu1Dixu2+I0=σxu1,xu2+I0=xu1,xu2=σxu1+I0,σxu1+I0.
Therefore,σ is a homomorphism of Lie superalgebras. Then, σ is an isomorphism of Lie superalgebras. We consider the following sequence:(33)0⟶iI0⟶ρKn,m⟶σ∘πKn,0⟶00.
In the above sequence,ρ is the embedded map from I0 to Kn,m and π is the natural homomorphism from Kn,m to Kn,m0. Obviously, ρI0=Kerσ∘π. Therefore, the above sequence is an exact sequence. Note that I0 is an ideal of Kn,m and Kn,m=Kn,0⊕I0. Therefore, Kn,m is an extension of the simple Lie superalgebra Kn,0.
LetUn−1,1¯ denote the divided power algebra over F with basis xα|α=…,αi,…∈ℕ0n−1,αi≤p−1. Let Λn−1,n,1¯=Un−1,1¯⊗Λn. For i=1,…,2n−1, let D¯i be the linear transformation of the superalgebra Λn−1,n,1¯ such that(34)D¯ixαxu=xα−εixu,i=1,…,n−1,xα∂xu∂xi,i=n,…,2n−1.
LetWn−1,n,1¯=∑i=12n−1fiD¯i|fi∈Λn−1,n,1¯,i=1,…,2n−1. Then, we define a linear map DKO:Λn−1,n,1¯⟶Wn−1,n,1¯ such that(35)DKOf=∑i=12n−2−1μi′degfD¯i′f+−1degfD¯2n−1fxiD¯i+∑i=12n−2xiDif−2fD¯2n−1,for all f∈Λn−1,n,1¯. Let KOn−1,n,1¯=spanFDKOf∣f∈Λn−1,n,1¯. Then, KOn−1,n,1¯ is a finite-dimensional odd contact Lie superalgebra (see [19]).Proposition 6 (see [15]).
Letm=n−1. Then, Tn−1 is isomorphic to Un−1,1¯.In [15], we can see that the isomorphism ς:Tn−1⟶Un−1,1¯ can be extended to an isomorphism ς˜:U⟶Λn−1,n,1¯.Theorem 3.
Ifm=n−1, then Kn,m is isomorphic to a subalgebra of KOn−1,n,1¯.Proof.
LetKO1n−1,n,1¯=DKOς˜f|f∈U. It is a subalgebra of KOn−1,n,1¯. Then, we define a map φ:Kn,n−1⟶KO1n−1,n,1¯ such that φD˜kf=DKOς˜f for all D˜kf∈Kn,n−1. By virtue of Proposition 6, φ is an isomorphism of Lie superalgebras.



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