Xiaoning XU Xiaojun LI

1 Introduction

During the last fifty years,the theory of Lie superalgebras over fields of characteristic zero has experienced a rather vigorous development in mathematics.For example,Kac[8–9]classified the finite-dimensional simple Lie superalgebras and infinite-dimensional simple linearly compact Lie superalgebras over algebraically closed fields of characteristic zero.The research on modular Lie superalgebras,i.e.,Lie superalgebras over a field of prime characteristic,just began in recent years.The complete classification of the finite-dimensional simple modular Lie superalgebras remains an open problem.However,Many important results on modular Lie superalgebras were obtained(see,e.g.,[1,3–7,10–15,17–22]).

As is well-known,the derivation algebras are very useful subjects in the research of both Lie algebras and Lie superalgebras.In[2,16],the derivation algebras of modular Lie algebras of Cartan type were discussed.Eight families of finite-dimensional simple modular Lie superalgebras of Cartan typeW,S,H,K,HO,KO,SHOandSKOwere constructed and their superderivation algebras were studied in[6,11–13,19,23].

In this paper,we study a class of Lie superalgebras ofO-type over a field of prime characteristic.The article is organized as follows.In Section 2,we give the definition of Lie superalgebras ofO-type and prove that they are simple.In Section 3,the generator sets of these Lie superalgebras are investigated.In Section 4,we first establish some technical lemmas which will be used to determine the homogeneous derivations of Lie superalgebras ofO-type.Then an explicit description of the superderivation algebras is given.

2 Definition and Simplicity

Let F be an algebraically closed field of characteristicp>3 and assume that F is not equal to its prime field Π.Form>0,let E=,···,}∈F be linearly independent over the prime field Π,andHbe the additive subgroup generated by E which dose not contain 1.Ifλ∈H,then letλ=where 0≤

Givenn∈N,letn>2 and:=SetM={1,···,n}.For∈N0,kican be uniquely expressed in thep-adic form=where 0≤

such that=0 for alli∈Mandj=0,1,···,=1 fori=1,···,m.Let

Ifk=(k1,···,kn)∈Q,we letwhere=For 0≤it is easy to see that

Let Λ(n+1)be the Grassmann superalgebras over F inn+1 variablesDenote the tensor product byA:=A(n,n+1,=A(n,⊗Λ(n+1).Obviously,Aare associative superalgebras with a Z2-gradation induced by the trivial Z2-gradation ofA(n,s)and the natural Z2-gradation of Λ(n+1):Forf∈A(n,andg∈Λ(n+1),we abbreviatef⊗gtofg.Fork∈{1,···,n+1},we let

and B(n+1)=where B0:=∅.Givenu=∈Bk,we set{u}={i1,···,ik},|u|=k,

and=Put|∅|=0 and=1.Thenk∈Q,λ∈H,u∈B(n+1)}is an F-basis ofA.

IfLis a superalgebra,thenh(L)denotes the set of all Z2-homogeneous elements ofL,i.e.,h(L)=If|x|occurs in some expression in this paper,then we always regardxas a Z2-homogeneous element and|x|as the Z2-degree ofx.

LetBbe a given Z2-graded vector space over F,andσbe a given homogeneous linear mapping of degree

such thatσ(f)0 for all 0f∈A.It is easy to see thatσ(A)⊆Bis a Z2-graded subspaces.Forσ(f)∈σ(A),one may easily verify thatσ(f)is a Z2-homogeneous element if and only iffis a Z2-homogeneous element ofA,and iff∈Aα,thenσ(f)∈σ(A)whereα∈Z2.

Setei:=fori∈M.PutT={n+1,···,2n+1}andR=M∪T.Define=ifi∈M,andifi∈T.PutT1={n+1,···,2n}.Let

Let(i∈R)be the linear transformations ofσ(A),such that

whereare the linear transformations ofA,such that

whereis the first nonzero number ofThenDiis an even derivation ofAfor anyi∈M,andDiis an odd derivation ofAfor anyi∈T.

Set

whereIis the identity mapping ofA.It is easy to see that

We denoteσ(A)byForσ(f),σ(g)∈h(we de fi ne a bilinear operation in,such that

Theorem 2.1become Lie superalgebras for the operation[,]de fi ned above.

ProofClearlyare superalgebras by(2.2).Letσ(f)∈andσ(h)∈whereα,β,γ∈Z2.Note thatfori∈M∪T1.By(2.2),one may easily verify that[σ(f),σ(g)]=−(−1[σ(g),σ(f)].

Put∂(f)f:=Then we have

We will prove that the operation[,]satisfies the graded Jacobi identity.

According to(2.2),we have

where

Similarly,

where

and

where

Moreover,by a straightforward computation,we can obtain the following equation:

By a careful comparison,we find that the elements on the right-hand side ofa,b,c,a′,b′,c′andcan cancel each other out to be zero.

Therefore

Thusare Lie superalgebras.

Letxi=for alli∈M.Setπ=(π1,···,πn)∈Qandω=+1,···,2n+⟩∈B(n+1).Putω−⟩=+1,···,n+i−1,n+i+1,···,2n+1∈B(n+1).

Lemma 2.1Let f∈A.If Di(f)=0for all i∈R,then f=where aλ∈F.

ProofIf=0 for alli∈R,then we havek=(0,···,0)andu=∅.Henceas desired.

Theorem 2.2Lie superalgebrasare simple.

ProofLetIbe a nonzero ideal of.Assume thatσ(f)is a Z2-homogeneous nonzero element ofI.Suppose thatf==0 forj=0,1.By(2.2),we have

So we can assume=0.Suppose thatf=where=0 fori∈M,j=0,1.Also by(2.2),we get

Thus we can assume thatDi′(f)=0 for alli∈M.Now suppose thatf=where=0 fori∈M,j=0,1,···,t.As

we can assume thatDi(f)=0 for alli∈M.According to Lemma 2.1,f=Iffcontains at least two nonzero terms,we can suppose that

whereaη0,aµ0.Let

Obviously,σ(g)is an element ofIandg∈Awith one term less thanf.Thus we may assume thatσ()∈I.Since 1−λ0,=σ(1)∈I.In particular,−[σ=σ()∈Iand=σ()∈Ifor alli∈M.Then

We will show that∈Ifor allk∈Q,λ∈H,u∈B(n+1).

Case 1is not contained inu.Due to(2.2),we have

Case 2is contained inu.We letwherev∈B(n+1).Suppose that−v=Then

Fori∈M,we have

Case 1 implies that∈I.So∈I.Furthermore,

Again by Case 1,we have∈I. Similarly,by lettingact onwe can obtain∈I.Forj∈Mandji,we get

As∈I,∈I.Similarly,by lettingact onwe can obtain∈I,wherefor alli∈M.ThereforeI=O.The proof is completed.

In the sequel,we denotebyOand write the elementσ(f)asffor simplicity.

Remark 2.1Theorem 2.2 shows thatO=Oare finite-dimensional Lie superalgebras with dimO=We callOthe Lie superalgebras ofO-type.

3 Spanning Sets

Proposition 3.1Let S=|i∈M,0≤∪{yλ|λ∈H}∪j∈T1}.Then Lie superalgebras O are generated by S.

ProofLetYbe the subalgebra generated byS.Firstly,we prove the following:

According to(ii)and the equation above,we have

(iv)∈Yfor 0≤i,j∈M.By virtue of(i),we get

If 1−0(modp),then∈Y.In particular,

If 1−≡0(modp),thenAs 0≤=2.Thus=3.By(i),we have

(v)∈Yfori∈Mandj∈T1.(iii)implies that

It follows that∈Y.Hence

(vi)We will use induction onkto show that∈Y,where∈,1≤t≤k.The conclusion is true for the casek=1 by(iii).Suppose that∈Yforl≤k−1.According to(v),we get

LetIn particular,we have∈Y.

Now we verify that1≤t≤k.The conclusion above and(ii)yield that

In particular,we have∈Y.

(vii)We propose to prove that∈Yby induction onk.

Clearly the assertion is true for the casek=1.Suppose that∈Y.Thus

If 2−1k−10(modp),then∈Y.If−1≡0(modp),then+10(modp).The inductive hypothesis implies that

By virtue of(iv),we see that

In particular,we have∈Y.Since 1−λ0,

(viii)∈Y.By(ii)and(vii),we obtain

(ix∈Y.(v)and(vii)yield that

Hence

Utilizing this procedure continuously,we can obtain∈Y.Thus

Since∈Y,we have

The proof of∈Yis completely similar to the proof above.

Then we prove the result of Proposition 3.1.Letc:=be any basis element ofO.We only need to show thatc∈Y.

Ifu=i.e.,c=then we will provec∈Yby induction on:=Ifdc=0,it follows from(ix)thatc=∈Y.Letdc>0.Then there is ani∈M,such that<.By the induction hypothesis,we have∈Y.Thus

Ifuwe letu=and then

Ifu=ω,i.e.,c=then we still provec∈Yby induction on.If=0,according to(ix),we havec=∈Y.Let>0.Then there is ani∈M,such that<.By the induction hypothesis,we see that∈Y.Hence

Ifu,we letu=The conclusion above and(2.1)yield=0∈YorTherefore,

Clearly,the assertion above is true for allk.By(2.1),we get

Then

Utilizing this procedure continuously,we have∈Y.HenceO⊆Y.ConsequentlyY=O.

4 Superderivations

We know that

Fori∈Z,we let

(2.2)shows thatfor alli,j∈Z.HenceO=are Z-graded Lie superalgebras,whereτ=+n.Clearly,λ∈H}.Iff∈Oi,thenfis called a Z-homogeneous element andiis the Z-degree offwhich is denoted by zd(f).

Let DerαOdenote the linear space of all derivations of degreeαofO,i.e.,

and let DerO:=be the superderivation algebras ofO.Fort∈Z,we let

Then DerO=are Z-graded Lie superalgebras,whereY={−ζ,−ζ+1,···,ζ}andζ=τ+2.Therefore,in order to determine the superderivation algebras DerO,we only need to determineh(DertO)for allt∈Y.

Lemma 4.1Let φ∈h(DerO)and f∈O.Suppose φ()==0for all i∈M and j∈T1.Then φ(f)∈O−2.

ProofLetf==0 for alli∈M,we have

Sinceφ,fαandxiare all Z2-homogeneous elements,∈h(O).Then=0 yieldsφ[]=0 for allα∈,i.e.,=0.Asφ()=0,[φ(),=0 for alli∈M.Similarly,[φ(),=0 for alli∈M.Hence

Leth:=φ()∈O. (2.2)implies that=[φ(fα),1]=0. Moreover,==0 and=Di(h)=0 for alli∈M.ThusDi(h)=0 for alli∈R.By virtue of Lemma 2.1,we geth∈O−2,i.e.,Henceφ(f)∈as desired.

Lemma 4.2Let t∈Zand φ∈h(DertO).If φ(Oj)=0for j=−2,−1,···,s,where s≥ −1and t+s≥ −2,then φ=0.

ProofLetj≥s.We will prove by induction onjthatφ()=0.Letj>sandf∈Oj.It is easy to see that∈Then the assumptionφ()=0 implies thatφ(xi)=φ[f,xi]=φ(==0 for alli∈M.By Lemma 4.1,φ(f)∈O−2.Sincet+j>t+s≥−2,φ(f)∈=0.Soφ()=0,that is,φ(O)=0.Thereforeφ=0.

Proposition 4.1=ad

ProofLetφ∈h(Clearlyφ()=φ()=0.Sinceφ(O0)⊆O−2,we may assume that=withaη∈F.Asη−λ∈H,η−λ1.Letg=andψ=φ−adg.Then considering Z-degree and by(2.2),we obtain

Clearly∈O−2for alli,l,j∈Mandλ∈H.Applyingψto=−we get=0.Similarly,we have

Suppose thatwhere∈F.Note that=0.By applyingψtowe have=0.Thenaθ=0,i.e.,=0.Thusψ()=0.By virtue of Lemma 4.2,ψ=0.Henceφ=adg∈adO−2,as desired.

Ifi∈M,then letτ(i)=.Ifi∈T,then letτ(i)=1.An elementfofOis calledτ(i)-truncated if=0,wherei∈R.

Fori∈R,we define a linear transformationτiofO,such tha(σ(f))=σ((f))and

where we set

By the convention before and the definition above,we still write(σ(f))as(f).Then we have the following lemma directly.

Lemma 4.3(i)If f∈O isµ(i)-truncated,then(f)=f for all i∈R.

(ii)=where i,j∈R with ij.

Lemma 4.4Let∈O,where t1,···,tk∈R.If fiisµ(i)-truncated for i=t1,···,tk,and()=for i,j=t1,···,tk,then there is an f∈L,such that Di(f)=fifor i=,···.

ProofWe will use induction onk.Ifk=1,then letf=By Lemma 4.3(i),we see that(f)=Assume that there isg∈O,such thatfori=,···,Letf=g+(g)).According to Lemma 4.3(ii),we obtain

Asftk−Dtk(g)isµ(tk)-truncated,by virtue of Lemma 4.3(i),we have

The result follows.

Lemma 4.5Assume that φ∈h(DerO).Let=fi=φ(′)+and=for all i∈M.Then the following statements hold:

(a)Di(fj)=for all i,j∈R.

(b)isµ(i)-truncated for all i∈R.

Proof(a)By the assumption,we have

Note that|fi|=|φ|+=|φ|for alli∈M.We will proceed in six steps.

(i)Applyingφto[1,=0 for alli∈M,we obtain

Utilizing(4.1)and(4.3),we get

By(2.2),a direct computation shows that

Soisµ(2n+1)-truncated,i.e.,=0.Thus=for alli∈M.

(ii)Similarly,applyingφto[1,]=0 for alli∈M,we have

From(4.1)–(4.2),we get

Then

Hence=−for all

(iii)Applyingφto=0 for alltogether with(4.3),yields

A direct computation yields

As=for alli∈M,for alli,j∈M.

(iv)Applyingφto=0 for alli,j∈M,and by(4.2),we have

A direct computation shows that

By the claim above,we see that=−

(v)Applyingφto=1 for alli∈M,and by(4.2)–(4.3),we get

A direct computation ensures that

Thus()=for alli∈M.

(vi)Applyingφto=0 for alli∈M,i,and by(4.2)–(4.3),we obtain

By computation,it follows that

ThereforeDj(fi′)=()for alli∈Mandj′∈T1withji.

Now we conclude thatDi(fj)=for alli,j∈R.

(b)By the first part,we obtain=0,that is,isµ(i)-truncated for alli∈T.

Fori∈M,we let=whereedoes not containanddoes not contain.By the assumption of this lemma,we have

As()and()areµ(i)-truncated,(e)=0 for allj∈Rwithji.Noticing thatDi(e)=0,it follows thatDj(e)=0 for allj∈R.Lemma 2.1 yieldse∈O−2.

Applyingφto=1,we get

Putφ=g.Then by(4.1),we obtain

Thus

By the convention before,we see that

Applyingφtofor alli∈M,we have

Utilizing(4.3),we get

A direct computation shows that

Since=0,and

we have

It follows that

Because every term on the right-hand side of the equation above isµ(i)-truncated,(e)=0.Sincee∈O−2,e=0.Thusisµ(i)-truncated for alli∈M.Hence the result holds.

Put ∆ ={θ:H→F|θ(λ+η)=θ(λ)+θ(η),∀λ,η∈H}.Forθ∈∆,we define a linear transformationofO,such that=Clearly∈

Lemma 4.6Let φ∈h(DerO).=0for all i∈M and j∈T1,then there is a θ∈∆,such that φ(yλ)=θ(λ)yλfor all λ∈H.

ProofClearly,φ()==0 for alli∈Mandj∈By Lemma 4.1,we may assumeφ()=with∈F.Applyingφto=(1−λ),we have

It follows thataη=0 for allη∈H{λ}.Thus,whereθ(λ)=Note thatφ(1)==0.Applyingφto=yη,we get=φ().Let=z.By computation,we can conclude that=θ(η)yη.Now applyingφto=(1−we obtain

Furthermore,

As 1−λ0,θ(λ+η)=θ(λ)+θ(η),i.e.,θ∈∆.

Proposition 4.2Let φ∈h(DertO)with t≥−1.Then there exist g∈O and θ∈∆,such that φ=adg+Dθ.

ProofWe first prove that there existg∈Oandθ∈∆,such that(φ−adg−Dθ)(Oj)=0 forj=−2,−1.

In fact,we can suppose thatfiis defined as in Lemma 4.5.Thenandisµ(i)-truncated for alli,j∈R.According to Lemma 4.4,there is anf∈O,such thatDi(f)=for alli∈R.

Letφ1=φ−adf.Note thatDue to Lemma 4.5,we know that

Similarly,φ1()=0 for allj∈T1.Moreover,

By Lemma 4.1,we can supposewithαλ∈F.Putz:=andφ2=φ1−adz.Then

By virtue of Lemma 4.6,there isθ∈∆,such thatLet=φ2−Dθ.Thenφ3()=0 for allλ∈H,that is,=0.Moreover,φ3(xi)==0 for alli∈Mandj∈T1.Since==0 for alli∈Mandj∈T1,Lemma 4.1 yields∈O−2for alli∈M.Similarly,∈∈,∀i∈M,∀j∈T1.Applying=−and by==0,we have

Similarly,=0 for allj∈T1.Thus=0.Lemma 4.2 implies=0.Setg:=f+z.Thenφ=ad(g)+Dθ.

Lemma 4.7Let t>2and φ∈h(Der−tO).Then==0for all j∈,i,l∈M with li.

ProofFori∈M,ifj=i′,letwith∈F.Applyingφto

we have

which combined with=0 and=0 fort>2 yields the following:

Ifε0(t−1)=0,then it is obvious that=0.

Ifε0(t−1)0 and(t−1)∗−10,then=0.When(t−1)∗−1=0,we haveε0(t−1)=1,since(t−1)∗=(t−1)and 0<ε0(t−1)

from the assumption above and=0 fort>2,we get

Hence=0 for allη∈H,that is,=0.

Now letji′andli.By applyingφto

we see that==0.Thus for everyi∈M,we have=0 for alll∈Mandj∈T1withii.

Lemma 4.8Let t>2and φ∈h(DerO).If=0,then=0for all i∈M,j∈T1and0≤k≤

ProofWe proceed in two steps.

(i)We propose to prove that

We first show thatφ()=0 by induction onk.If 0≤ktand=0.Clearly,=0.Applyingφto

we obtain=0 for allν∈Mandj∈T1.Lemma 4.1 ensures that∈=0.It is easy to see that the claim=0 is true for allk≥0.

Then we prove that=0.Ifl=i,then by the argument above,we have

Letli.We use induction onk.Ifkt−1 and=0.By the induction hypothesis andφ()=0,we see that

It follows from Lemma 4.1 that={0}.

Finally,we prove=0 also by induction onk.Ifkt−1 and=0.Similarly,by lettingφact on the equalities below,

we see that the result is zero.Again Lemma 4.1 yields

(ii)Now we return to the proof of this lemma.Ift>3,then∈=0.Putt=3.Then let=∈with∈F.Applyingφto=we have

Since=0 and=0,φ=0.

For 0≤k≤πi,by applyingφto

and by the known results=0,=0 and=0 above,we obtain

Proposition 4.3Let t>2and tpvfor all v∈N.Then h(DeO)={0}.

ProofLetφ∈h(DeO).Considering the Z-degree,we haveSupposewith∈F.By applyingφto

ifε0(t)0,thenIt follows from=0 and=0 that

Ifε0(t)=0,assume thatt=for some 0≤εs(t)

we get=0.Lemma 4.8 implies==0 for alli∈Mandj∈T1.Moreover,=0.According to Proposition 3.1,we see thatφ=0.Henceh(Der−tO)={0}.

Proposition 4.4If t=pvfor some v∈N,thenDeO=

ProofClearly∈Der−tOfor alli∈M.Lett=andφ∈h(Der−tO).It is easy to see thatwith∈F.Ast=,(t)=0.Applyingφto

together with=0,yields

A direct computation implies that

Thus=0 for allη0.Hencewhere=Then

By virtue of Lemma 4.8,we have==0 for alli∈Mandj∈T1.Moreover,ψ()=0.Proposition 3.1 shows thatψ=0.Consequently,φ∈

Ifvi>si,then=0 for alli∈M.By Propositions 4.1–4.4,we obtain the following theorem.

Theorem 4.1DerO=adO⊕{Dθ|θ∈∆}⊕

Theorem 4.2For each algebra in the family,O has no nondegenerate associative form.

ProofAssume thatλis a nondegenerate associative form onO.[23,Proposition 2.3]implies thatλis nonsingular.It follows that0.Sinceλis associative,

Hence(2−=0.As 2−(−1)n0(modp),=0,a contradiction.As a result,Ohas no nonsingular associative form.

Theorem 4.3For each algebra in the family,O is not isomorphic to the simple Lie superalgebras of Cartan type W,S,H,HO,SHO,K,KO,SKO.

ProofRecall that dimO=By means of[13,19],we see that the dimension of modular Lie superalgebrasHOis odd and the dimension of modular Lie superalgebrasHcan not be divided byp.SoOis not isomorphic to modular Lie superalgebrasHandHO,respectively.The outer derivations ofW,S,KandKOare all ad-nilpotent in[6,23],butOpossesses outer derivationsDθwhich are not ad-nilpotent.It follows thatOis not isomorphic to modular Lie superalgebrasW,S,KandKO,respectively.Using Theorem 4.2,we can also prove thatOis not isomorphic to modular Lie superalgebrasSHOandSKO,which possess nondegenerate associative forms on them(see[12]).

AcknowledgementThe authors are grateful to the referees for their many valuable comments and suggestions.

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