# Infinite-Dimensional Modular Lie SuperalgebraΩ

Authors: Xiaoning Xu; Bing Mu
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

All ad-nilpotent elements of the infinite-dimensional Lie superalgebraΩ over a field of positive characteristic are determined. The natural filtration of the Lie superalgebra Ω is proved to be invariant under automorphisms by characterizing ad-nilpotent elements. Then an intrinsic property is obtained by the invariance of the filtration; that is, the integers in the definition of Ω are intrinsic. Therefore, we classify the infinite-dimensional modular Lie superalgebra Ω in the sense of isomorphism.


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## Body

## 1. Introduction

The theory of modular Lie superalgebras has obtained many important results during the last twenty years (e.g., see [1–4]). But the complete classification of the simple modular Lie superalgebras remains an open problem. We know that filtration structures play an important role both in the classification of modular Lie algebras and nonmodular Lie superalgebras (see [5–8]). The natural filtrations of finite-dimensional modular Lie algebras of Cartan type were proved to be invariant in [9, 10]. The similar result was obtained for the infinite-dimensional case [11]. In the case of finite-dimensional modular Lie superalgebras of Cartan type, the invariance of the natural filtration was discussed in [12, 13]. The same conclusion was obtained for some infinite-dimensional modular Lie superalgebras of Cartan type (see [14–17]).In the present paper, we consider the infinite-dimensional modular Lie superalgebraΩ(r,m,q), which was studied in paper [18]. Denote the natural filtration by(Ω(r,m,q)[i])i≥-2. We show that the filtration is invariant under automorphisms by determining ad-nilpotent elements and subalgebras generated by certain ad-nilpotent elements. We are thereby able to obtain an intrinsic characterization of Lie superalgebraΩ(r,m,q).The paper is organized as follows. In Section2, we recall some necessary definitions concerning the modular Lie superalgebraΩ. In Section 3, we establish some technical lemmas which will be used to determine the invariance of the filtration. In Section 4, we prove that the natural filtration(Ω(r,m,q)[i])i≥-2is invariant. Furthermore, we obtain the sufficient and necessary conditions ofΩ(r,m,q)≅Ω(r′,m′,q′); that is, all the Lie superalgebras are classified up to isomorphisms.


## 2. Preliminaries

Throughout the work𝔽denotes an algebraically closed field of characteristicp>3and𝔽is not equal to its prime fieldΠ. Letℤ2={0¯,1¯}be the ring of integers module2. Letℕandℕ0denote the sets of positive integers and nonnegative integers, respectively. Form>0, let𝔼={z1,…,zm}∈𝔽be a subset of𝔽that is linearly independent over the prime fieldΠ, and let H be the additive subgroup generated by𝔼that does not contain1. Ifλ∈H, then we letλ=∑i=1mλiziandyλ=y1λ1⋯ymλm, where0≤λi<p.Givenn∈ℕandr=2n+2, we putM={1,…,r-1}. Letμ1,…,μr-1∈𝔽andμ1=0,μj+μn+j=1,j=2,…,n+1. Ifki∈ℕ0, thenkican be uniquely expressed inp-adic formki=∑v=0∞εv(ki)pv,where0≤εv(ki)<p.We set xiki=∏v=0∞xivεv(ki). We define a truncated polynomial algebra
(1)A=𝔽[x10,x11,…,x20,x21…,xr-1,0,xr-1,1,…,y1,…,ym]such that(2)xijp=0,∀i∈M,j=0,1,…,yip=1,i=1,…,m.Forki,ki′∈ℕ0, it is easy to see that
(3)xikixiki′=xiki+ki′≠0⟺εv(ki)+εv(ki′)<p,v=0,1,….LetQ={(k1,…,kr-1)∣ki∈ℕ0,i∈M}. Ifk=(k1,…,kr-1)∈Q, we setxk=x1k1⋯xr-1kr-1.LetΛ(q)be the Grassmann superalgebra over𝔽inqvariablesξr+1,…,ξr+q, whereq∈ℕandq>1. Denote byΩthe tensor productA⨂𝔽Λ(q). The trivialℤ2-gradation ofAand the naturalℤ2-gradation ofΛ(q)induce aℤ2-gradation ofΩsuch thatΩis an associative superalgebra:
(4)Ω0¯=A⨂𝔽Λ(q)0¯,Ω1¯=A⨂𝔽Λ(q)1¯.Forf∈Aandg∈Λ(q), we abbreviatef⊗gtofg. Let
(5)𝔹k={〈i1,i2,…,ik〉∣r+1≤i1<i2<⋯<ik≤r+q},𝔹(q)=⋃k=0q‍𝔹k,and where𝔹0=⌀. Givenu=〈i1,…,ik〉∈𝔹k, we set|u|=k,{u}={i1,…,ik}andξu=ξi1⋯ξik(|⌀|=0,ξ⌀=1). Then{xkyλξu∣k∈Q,λ∈H,u∈𝔹(q)}is an𝔽-basis ofΩ.If|f|appears in some expression in this paper, we always regardxas aℤ2-homogeneous element and|f|as theℤ2-degree off.Lets=r+q,T={r+1,…,s}andR=M∪T.PutM1={2,…,r-1}.Setei=(δi1,…,δir-1)andxi=xi1=xi0fori∈M. Definei~=0¯, ifi∈M1, andi~=1¯, ifi∈T. Let
(6)i′={i+n,2≤i≤n+1,i-n,n+2≤i≤r-1,i,r+1≤i≤s,[i]={1,2≤i≤n+1,-1,n+2≤i≤r-1,1,r+1≤i≤s.LetD1,D2,…,Dsbe the linear transformations ofΩsuch that
(7)Di(xkyλξu)={ki*xk-eiyλξu,i∈M,xkyλ·∂ξu∂ξi,i∈T,whereki*is the first nonzero number ofε0(ki),ε1(ki),…,εsi(ki). ThenD1,D2,…,Dsare superderivations of the superalgebraΩand|Di|=i~. Set
(8)∂¯=I-∑j∈M1‍μjxj0∂∂xj0-∑j=1m‍zjyj∂∂yj-2-1∑j∈T‍ξj∂∂ξj,whereI is the identity mapping ofΩ. Letf∈Ωbe aℤ2-homogeneous element andg∈Ω; we define a bilinear operation inΩsuch that
(9)[f,g]=D1(f)∂¯(g)-∂¯(f)D1(g)+∑i∈M1∪T[i](-1)i~|f|Di(f)Di′(g).ThenΩbecomes a simple Lie superalgebra. If 2n+4-q≢0(modp), we see thatλ+2-1q-n-2≠0. In the sequel, we always assume that2n+4-q≢0(modp). In some cases, we denoteΩbyΩ(r,m,q)in detail and callΩ(r,m,q)the Lie superalgebra ofΩ-type.Now we give aℤ-gradation ofΩ:Ω=⨁j∈XΩj, where
(10)Ωj=span𝔽{xkyλξu∣∑i∈M1‍ki+2k1+|u|-2=j}.LetΩ[i]=⨁j≥iΩjfor alli≥-2. ThenΩ=Ω[-2]⊃Ω[-1]⊃⋯ are called the natural filtration ofΩ.


## 3. Ad-Nilpotent Elements

LetLbe a Lie superalgebra. Recall that an elementy∈Lis called adL-nilpotent if there exists at∈ℕsuch that(ady)t(L)=0.Ify∈Lis adL-nilpotent, it is also called ad-nilpotent in brief. LetGbe a subset ofL.Putnil(G)={x∈G∣xisadL-nilpotent}, and Nil(G) is the subalgebra ofLgenerated by nil(G).Leta∈ℕ0anda=∑v=0∞εv(a)pvbe thep-adic expression of a, where0≤εv(a)<p.Then
(11)pad(a)=(pad0(a),pad1(a),pad2(a),…)
is called thep-adic sequence ofa, wherepadv(a)=εv(a)for allv∈ℕ0. Fork=(k1,k2,…,kr-1)∈Q, we define thep-adic matrix ofkto be
(12)pad(k)=(pad(k1)pad(k2)⋮pad(kr-1)).
Sincepad(k)is a(r-1)×∞matrix with only finitely many nonzero elements, we can set
(13)ht(k)=max{j∈ℕ∣∃i∈M,padj(ki)≠0}.
Ifz=∑k,λ,u‍αk,λ,uxkyλξu∈Ω is a nonzero element withαk,λ,u∈𝔽, then we may assume
(14)ht(z)=max{ht(k)∣αk,λ,u≠0}.
Forc,d∈ℕ0, we define ∥k∥c,d≔∑i=1r-1∑j=cdpadj(ki)and∥k∥d≔∥k∥0,d.Now for anyt∈ℕandxkyλξu∈Ω, define
(15)𝒮t(xkyλξu)=∥k∥t+2∥k∥1,t+|u|+pad0(k1).Lemma 1.
Letk,k′∈Qandt∈ℕ0. Then the following statements hold. (i)
xkxk′≠0⇔pad(k)+pad(k′)=pad(k+k′).(ii)
Ifki′≠0, then∥k′-ei∥t+2∥k′-ei∥1,t≥∥k′∥t+2∥k′∥1,t-1, wherei∈M1.(iii)
Letxkyλξu∈Ω[1]andt≥ht(xkyλξu). Then𝒮t(xkyλξu)≥3.Proof.
(i) We see that(16)pad(k)+pad(k′)=(pad(k1)pad(k2)⋮pad(kn))+(pad(k1′)pad(k2′)⋮pad(kn′))=(pad(k1)+pad(k1′)pad(k2)+pad(k2′)⋮pad(kn)+pad(kn′)).
Note thatxkxk′≠0⇔εv(ki)+εv(ki′)<p⇔padv(ki)+padv(ki′)<p⇔padv(ki)+padv(ki′)≢0(modp),foralli∈M,v∈ℕ0. By the uniqueness ofp-adic expression, we havexkxk′≠0⇔padv(ki)+padv(ki′)=padv(ki+ki′),foralli∈M,v∈ℕ0.Thus
(17)xkxk′≠0⟺pad(k)+pad(k′)=(pad(k1)+pad(k1′)pad(k2)+pad(k2′)⋮pad(kn)+pad(kn′))=(pad(k1+k1′)pad(k2+k2′)⋮pad(kn+kn′))=pad(k+k′),
as desired.
(ii) Ifpad0(ki′)≠0, thenpad(ki′-1)=(pad0(ki′)-1,pad1(ki′),…). We see that∥k′-ei∥t=∥k′∥t-1,∥k′-ei∥1,t=∥k′∥1,t. So (ii) holds.
Ifpad0(ki′)=0andpadb(ki′)≠0forb≥1, then we can assume that
(18)pad(ki′)=(0,…,0,padb(ki′),padb+1(ki′),…).
Hencepad(ki′-1)=(p-1,…,p-1,padb(ki′),padb+1(ki′),…).
(a) Ift<b, then we get
(19)∥k′-ei∥t+2∥k′-ei∥1,t=(t+1)(p-1)+2t(p-1)>-1,∥k′∥t+2∥k′∥1,t-1=-1.
(b) Ift=bandp>3, then
(20)∥k′-ei∥t+2∥k′-ei∥1,t=t(p-1)+2(b+1)(p-1)+3(3padt(ki′)-1)>3padt(ki′)-1,∥k′∥t+2∥k′∥1,t-1=3padb(ki′)-1.
(c) Ift>b, then
(21)∥k′-ei∥t+2∥k′-ei∥1,t=∥k′-ei∥b+2∥k′-ei∥1,b+3∥k′-ei∥b+1,t≥∥k′∥b+2∥k′∥1,b-1+3∥k′-ei∥b+1,t=∥k′∥t+2∥k′∥1,t-1.
Thus (ii) holds.
(iii) By|k|+2k1+|u|=i+2≥3, we have∑‍(pad0(ki)p0+pad1(ki)p1+⋯)+2pad0(k1)+|u|≥3;that is,∑‍pad0(ki)+2pad0(k1)+|u|≥3.Lemma 2.
Letxkyλξu∈Ω,xk′yηξv∈Ω[1],t≥max{1,ht(xk′yηξv)},i∈M1∪T.Then the following statements hold.(i)
Ifxk′yηξvD1(xkyλξu)≠0, then𝒮t(xk′yηξvD1(xkyλξu))≥𝒮t(xkyλξu)+1.(ii)
IfxkyλξuD1(xk′yηξv)≠0, then𝒮t(xkyλξuD1(xk′yηξv))≥𝒮t(xkyλξu)+1.(iii)
IfDi(xk′yηξv)Di′(xkyλξu)≠0, then𝒮t(Di(xk′yηξv)Di′(xkyλξu))≥𝒮t(xkyλξu)+1.Proof.
(i) Asxk′yηξvk1*xk-e1yλξu=k1*xk′xk-e1yη+λξuξv≠0, xk′xk-e1=xk′+k-e1≠0.Then we have
(22)𝒮t(xk′yηξvD1(xkyλξu))=∥k′+(k-e1)∥t+2∥k′+(k-e1)∥1,t+|u|+|v|+pad0(k1′+(k1-1)).
By the equality above and Lemma 1, we getpad(k′+(k-e1))=pad(k′)+pad(k-e1).Thus
(23)pad0(k1′+(k1-1))=pad0(k1′)+pad0(k1-1),∥k′+(k-e1)∥t=∥k′∥t+∥k-e1∥t,∥k′+(k-e1)∥1,t=∥k′∥1,t+∥k-e1∥1,t.
Hence
(24)𝒮t(xk′yηξvD1(xkyλξu))=∥k′∥t+2∥k′∥1,t+|v|+pad0(k1′)+∥k-e1∥t+2∥k-e1∥1,t+|u|+pad0(k1-1)≥𝒮t(xk′yηξv)+∥k∥t+2∥k∥1,t+|u|+pad0(k1)-1-1≥𝒮t(xkyλξu)+1.
(ii) The proof is completely analogous to (i).
(iii) Fori∈M1, by assumption of this lemma, we havexk′-eixk-ei′≠0, which combined with Lemma 1 yield
(25)𝒮t(Di(xk′yηξv)Di′(xkyλξu))=𝒮t(ki*ki′*x(k′+ei)+(k-ei′)yλ+ηξuξv)=∥(k′-ei)+(k-ei′)∥t+2∥(k′-ei)+(k-ei′)∥1,t+|u|+|v|+pad0(k1+k1′)≥∥k′∥t+2∥k′∥1,t+|v|-1+∥k∥t+2∥k∥1,t+|u|-1+pad0(k1)+pad0(k1′)≥𝒮t(xkyλξu)+1.
Fori∈T, it is easily seen thatεv(k′-ei)+εv(k-ei′)=εv(k′)+εv(k)<p.Also by Lemma 1, we obtain
(26)𝒮t(Di(xk′yηξv)Di′(xkyλξu))=𝒮t(xk′+kyλ+ηξu-〈i〉ξv-〈i′〉)=∥k′+k∥t+2∥k′+k∥1,t+|u-1|+|v-1|+pad0(k1′+k1)≥𝒮t(xkyλξu)+3-2=𝒮t(xkyλξu)+1.
Hence Lemma 2 holds.Lemma 3.
Letxkyλξu∈Ω,xk′yηξv∈Ω[1]andt≥max{1,ht(xk′yηξv)}. Letxk′′yλ′ξu′be a nonzero summand of[xk′yηξv,xkyλξu]. Then𝒮t(xk′′yλ′ξu′)≥𝒮t(xkyλξu)+1.Proof.
By a direct computation, we obtain that(27)[xk′yηξv,xkyλξu]=D1(xk′yηξv)∂(xkyλξu)-D1(xkyλξu)∂(xk′yηξv)+∑i∈M1∪T‍[i](-1)i~|f|Di(xk′yηξv)Di′(xkyλξu),
which satisfies the conditions of Lemma2.Lemma 4.
Ω
[
1
]
⊆
nil
(
Ω
).Proof.
Givent∈ℕ, put
(28)lt=(r-1)(t+1)(p-1)+2(r-1)t(p-1)+(p-1)+n+1.
Clearly, we have𝒮t(xkyλξu)<ltfor all standard basis elementxkyλξuofΩ.
Lett∈ℕsuch thatt≥ht(z).For anyz=Σk′,vck′,vxk′yηξv∈Γ[1]with0≠ck′,v∈𝔽, we have
(29)adz(xkyλξu)=[z,xkyλξu]=D1(z)∂(xkyλξu)-∂(z)D1(xkyλξu)+∑i∈M1∪T‍[i]Di(z)Di′(xkyλξu).
Note thatt≥max{1,ht(xk′yηξv)}.By using Lemma 3 repeatedly we see that(adz)plt(xkyλξu)=0. HenceΩ[1]⊆nil(Ω).Lemma 5.
(i) Iff=∑t=klft∈nil(Ω),whereft∈Ωt, thenft∈nil(Ω).
(ii) Iff=∑t=-2lft∈nil(Ω0¯),thenf-2=0.
(iii) Iff=∑t=-1lft∈nil(Ω0¯), thenf-1=0.
(iv)Nil(Ω[0]∩Ω0¯)=Nil(Ω0∩Ω0¯)+Ω[1]∩Ω0¯.
(v)Nil(Ω0¯)=Nil(Ω[0]∩Ω0¯).Proof.
(i) See Lemma  5 in [14].
(ii) By (i), we see thatf-2is ad-nilpotent. Iff-2≠0, then[yλ,xme1]=-D1(x(me1)∂(yλ))=(λ-1)k1*x(m-1)e1yλfor allm>0andλ≠1. By a direct computation, we obtain(adyλ)m(xme1)=(-1)m(1-λ)mymλ≠0, contradicting the nilpotency off-2. Hencef-2=0, as desired.
(iii) Clearly,f-1is ad-nilpotent by virtue of (i). Iff-1≠0, then we can suppose thatf-1=∑i∈M1γixi≠0, whereγi∈𝔽. Thus there exists someγj≠0.By computation, we have
(30)[Σγixiyλ,xmej]=Σ[i](-1)i~|f|Di(Σγixiyλ)Di′(xmej)=[j]γjkj′*x(m-1)ej′yλ.
Similarly, we get(adf-1)m(xmej′)=[j]m(γj)m(kj′*)mymλ≠0, a contradiction. Consequently,f-1=0.
(iv) Suppose thatf=f0+f[1]is an arbitrary element of nil(Ω[0]∩Ω0¯), wheref0∈Ω0∩Ω0¯,f[1]∈Ω[1]∩Ω0¯. By (1) of this lemma, we see thatf0∈nil(Ω0∩Ω0¯)⊆Nil(Ω0∩Ω0¯). Hence nil(Ω[0]∩Ω0¯)⊆Nil(Ω0∩Ω0¯)+Ω[1]∩Ω0¯and Nil(Ω[0]∩Ω0¯)⊆Nil(Ω0∩Ω0¯)+Ω[1]∩Ω0¯.
Conversely, we haveΩ[1]∩Ω0¯⊆Nil(Ω[0]∩Ω0¯)by means of Lemma 4. Clearly, Nil(Ω0∩Ω0¯)⊆Nil(Ω[0]∩Ω0¯). Thus Nil(Ω0∩Ω0¯)+Ω[1]∩Ω0¯⊆Nil(Ω[0]∩Ω0¯).This shows that Nil(Ω[0]∩Ω0¯)=Nil(Ω0∩Ω0¯)+Ω[1]∩Ω0¯, as desired.
(v) It is obvious that Nil(Ω[0]∩Ω0¯)⊆Nil(Ω0¯).Conversely, we assume thatf=∑t=-2lft∈nil(Ω0¯).
Then by (ii) of this lemma,f-2=0. Hencef=f-1+f[0]∈nil(Ω0¯), wheref-1∈Ω-1∩Ω0¯andf0∈Ω[0]∩Ω0¯. It follows from (iii) thatf-1=0. Nowf=f[0]∈ nil(Ω0¯). Noting thatΩ[1]⊆ nil(Ω), we havef=f[0]⊆nil(Ω[0]∩Ω0¯)⊆Nil(Ω[0]∩Ω0¯). Thus nil(Ω0¯)⊆Nil(Ω[0]∩Ω0¯), and the assertion holds.Lemma 6.
Leti,j∈M1. Suppose thatxkyλξuis an arbitrary standard element ofΩ. Then the following statements hold.(i)
xi2yλ∈nil(Ω0∩Ω0¯).(ii)
If[i]=[j]andi≠j, thenxixjyλ∈nil(Ω0∩Ω0¯).(iii)
If[i]≠[j]andj≠i′, thenxixjyλ∈nil(Ω0∩Ω0¯).(iv)
xixi′yλ∈nil(Ω0∩Ω0¯).Proof.
(i) By a direct computation, we get(31)(adxi2yλ)(xkyλξu)=[xi2yλ,xkyλξu]=-(1-2μi-λ)xi2yλD1(xkyλξu)+2[i]xiyλDi′(xkyλξu)
andxi2yλD1∘xiyλDi′=xiyλDi′∘xi2yλD1. It follows from the binomial theorem that
(32)(adxi2yλ)pn(xkyλξu)=((-(1-2μi-λ)xi2yλD1)pn+(2[i]xiyλDi′)pn)(xkyλξu)=0.
(ii) Also by a direct calculation, we have(33)(adxixjyλ)(xkyλξu)=[xixjyλ,xkyλξu]=-(1-μi-μj-λ)xixjyλD1(xkyλξu)+[i]xjyλDi′(xkyλξu)+[j]xiyλDj′(xkyλξu).
PutA=-(1-μi-μj-λ)xixjyλD1,B=[i]xjyλDi′,andC=[j]xiyλDj′. Obviously,
(34)xixjyλD1(xjyλDi′+xiyλDj′)=(xjyλDi′+xiyλDj′)xixjyλD1,xjyλDi′xiyλDj′=xiyλDj′(xjyλDi′).
Hence(adxixjyλ)pn(xkyλξu)=Apn+Bpn+Cpn=0.
(iii) The proof is completely analogous to (ii).
(iv) According to (i), we see thatxixi′yλ=(1/4)[i,xixiyλ,xi′xi′]∈nil(Ω0∩Ω0¯).Lemma 7.
Suppose thati,j,k∈T are different from each other, andγ,χ∈𝔽. Then
(i)f=γξiξjyλ+χξiξkyλ∈nil(Ω)forγ2+χ2=0.
(ii)ξiξjyλ∈Nil(Ω0∩Ω0¯).Proof.
(i) Setxkyλξube an arbitrary standard element ofΩ. Then
(35)(adf)(xkyλξu)=[f,xkyλξu]=[γξiξjyλ,xkyλξu]+[χξiξkyλ,xkyλξu]=γλξiξjyλD1(xkyλξu)+γξjyλDi(xkyλξu)+γξiyλDj(xkyλξu)+χλξiξjyλD1(xkyλξu)+χξkyλDj(xkyλξu)+χξiyλDk(xkyλξu).
Put
(36)A=γλξiξjyλD1,B=γξjyλDi,C=γξiyλDj,D=χλξiξjyλD1,E=χξkyλDi,F=χξiyλDk.Obviously,
(37)A2=B2=C2=D2=E2=F2=0,AB=BA=AC=CA=AD=DA=AF=FA=BD=DB=BE=EB=CD=DC=CF=FC=DF=FD=0.
Hence(38)(adf)2=AE+EA+BC+CB+BF+FB+CE+EC+DE+ED+EF+FE.
Noting thatγ2+χ2=0, we obtain
(39)(adf)3=BCB+CBC+EFE+FEF.
Similarly,(adf)4=0, and thenf=γξiξjyλ+χξiξkyλ∈nil(Ω).
(ii) Letk∈T∖{i,j}.It follows from (i) that
(40)f1=γξiξjyλ+ξiξkyλ∈Nil(Ω0∩Ω0¯),f2=ξiξj+γξiξk∈Nil(Ω0∩Ω0¯),whereγ2=-1.
Henceξiξjyλ=-(1/2)(γf1-f2)∈Nil(Ω0∩Ω0¯).Lemma 8.
x
i
ξ
j
y
λ
∈
nil
(
Ω
0
∩
Ω
1
¯
)fori∈M1andj∈T.Proof.
By a direct computation, we obtain(41)(adxiξjyλ)(xkyλξu)=[xiξjyλ,xkyλξu]=(-(12-μi-λ)xiξjyλD1+[i]ξjyλDi′-xiyλDj)×(xkyλξu).
PutA=-(1/2-μi-λ)xiξjyλD1,B=[i]ξjyλDi′andC=-xiyλDj. ObservingA2=B2=C2=0andAB=BA=0, we see that
(42)(adxiξjyλ)2=AC+BC+CA+CB,(adxiξjyλ)pn=0,n∈ℕ.
Thusxiξjyλ∈nil(Ω0∩Ω1¯), as required.Lemma 9.
The following statements hold(i)
Nil(Ω0)=span𝔽{xixjyλ,xiξjyλ,ξiξjyλ∣i,j∈M1∪T}.(ii)
Forq≥3,Nil(Ω0∩Ω0¯)=span𝔽{xixjyλ,ξiξjyλ∣i,j∈M1∪T[i],=[j]}.(iii)
Forq=2,Nil(Ω0∩Ω0¯)=span𝔽{xixjyλ∣i,j∈M1}.Proof.
(i) Suppose thatf=γ1x1yλ+∑i,j∈M1∪T(xixjyλ+xiξjyλ+ξiξjyλ)is an arbitrary element of nil(Ω0), whereγ1∈𝔽. Ifγ1≠0, then[γ1x1yλ,yλ]=γ1(1-λ)y2λ. A direct calculation shows that(adf)m(yλ)=γ1m(1-λ)(1-2λ)⋯(1-mλ)ymλ≠0. Thusfis not ad-nilpotent, contradicting the nilpotency off. Henceγ1=0andnil(Ω0)⊆span𝔽{xixjyλ,xiξjyλ,ξiξjyλ∣i,j∈M1∪T}.Obviously,span𝔽{xixjyλ,xiξjyλ,ξiξjyλ∣i,j∈M1∪T}is a subalgebra ofΩ, which yields Nil(Ω0)⊆span𝔽{xixjyλ,xiξjyλ,ξiξjyλ∣i,j∈M1∪T}.
Conversely, we havespan𝔽{xixjyλ,xiξjyλ,ξiξjyλ∣i,j∈M1∪T}⊆Nil(Ω0)by virtue of Lemmas 6, 7, and 8. It follows that Nil(Ω0)=span𝔽{xixjyλ,xiξjyλ,ξiξjyλ∣i,j∈M1∪T}.
(ii) By (i) of the lemma, we have(43)Nil(Ω0∩Ω0¯)⊆Nil(Ω0)∩Ω0¯=span𝔽{xixjyλ,ξiξjyλ∣i,j∈M1∪T,[i]=[j]}.
Conversely, the assertionspan𝔽{xixjyλ,ξiξjyλ∣i,j∈M1∪T,[i]=[j]}⊆Nil(Ω0∩Ω0¯)follows from Lemmas 6 and 7.
(iii) Supposef=γξr+1ξsyλ+∑i,j∈M1γijxixjyλ∈nil(Ω0∩Ω0¯), whereγ,γij∈𝔽. Ifγ≠0, thenadf(ξsyλ)=[f,ξsyλ]=γξr+1y2λand(adf)2(ξsyλ)=γ2ξsy3λ. A direct calculation shows that(adf)2m(ξsyλ)=γ2mξsy2mλ≠0. It follows thatfis not ad-nilpotent. Thusγ=0. Thenf=∑i,j∈M1γijxixjyλ∈nil(Ω0∩Ω0¯)and nil(Ω0∩Ω0¯)⊆span𝔽{xixjyλ∣i,j∈M1}.Note thatspan𝔽{xixjyλ∣i,j∈M1}is a subalgebra ofΩ. Hence Nil(Ω0∩Ω0¯)⊆span𝔽{xixjyλ∣i,j∈M1}and (iii) holds.Letρbe the corresponding representation with respect toΩ0¯moduleΩ-1; that is, ρ(f)=adf|Ω-1for allf∈Ω0. It is easily seen thatρis faithful. Forf∈Ω0, we denote byρ(f)the matrix ofρ(f)relative to the fixed ordered𝔽-basis:
(44){x2yλ,x3yλ,…,xr-1yλ,ξr+1yλ,…,ξsyλ}.Denote by𝔤𝔩(2n,q)the general liner Lie superalgebra of(2n+q)×(2n+q)matrices over𝔽. Leteijdenote the(s-2)×(s-2)matrix whose(i,j)-entry is 1 and 0 elsewhere. LetEndenote the identity matrix of sizen×n. PutG=(0En-En0). Let𝔰𝔭(2n,𝔽)be all the(2n)×(2n)matrices set filled withATG+GA=0. Put
(45)𝒲={(ABCD)∈𝔤𝔩(2n,q)∣A∈𝔰𝔭(2n,𝔽),BTG+C=0,D=-DT(ABCD)}.Set(46)ℒ=𝒲⊕𝔽Es-2.Lemma 10.
(i)ρ(Ω0)=ℒ.
(ii) Iff∈nil(Ω0), thenρ(f)is a nilpotent matrix.Proof.
(i) Leti,j,k∈M1∪T. By computation, we have
(47)[xixj,xkyλ]=[i]δi′kxjyλ+[j]δj′kxiyλ,[x1,xkyλ]=(1-μk-λ)xkyλ.
Henceρ(xixj)=[i]ej,i′+[j]eij′andρ(x1)=(1-μk-λ)Es-2.The other cases are treated similarly. Thus (i) holds.
(ii) Asfis a nilpotent elements, ρ(f)is a nilpotent liner transformation. Then by the definition ofρ, we see thatρ(f)is a nilpotent matrix.Lemma 11.
Iff∈nil(Ω0∩Ω0¯),f≠0, then there exists az∈Ω0∩Ω0¯such that[f,z]∉nil(Ω0∩Ω0¯).Proof.
By Lemma9, we can assume thatf=∑l∈M1γlxl2yλ+∑l,t∈M1,l<tβltxlxtyλ+∑l,t∈T,l<tχltξlξtyλ, whereγl,βlt,χlt∈𝔽.
Supposeγi≠0for somei∈M1andz=xi′2.A direct calculation shows that
(48)[f,z]=[i]4γixixi′yλ+h,
where every item ofhdoes not containxi. Then (ad[f,z])(xi′)=4γiyλxi′. Thus(ad[f,z])n(xi′)=4nγinynλxi′≠0,foralln∈ℕ, which implies that[f,z]is not a nilpotent element.
Ifγi=0for alli∈M1, then we letβij≠0for somei,j∈M1andz=xi′xj′yλ. Also by computation, we have
(49)[f,z]=[j]βijxixi′yλ+h,
where every item ofh does not containxi.Similarly,(ad[f,z])n(xi′)=[j]nβijnynλxi′≠0,foralln∈ℕ, and then[f,z]is not nilpotent.
Hence our assertion follows.
Ifγi=0andβij=0for alli,j∈M1, thenf=∑l,t∈T,l<tχltξlξtyλ∈span𝔽{ξlξtyλ∣l,t∈T,λ∈H}.We see thatρ(f)is a antisymmetric nilpotent matrix. By Lemmas 9(ii) and 10, it is easy to see that there isz∈span𝔽{ξiξjyλ∣i,j∈T}such that[ρ(f),ρ(z)]is not a nilpotent matrix; that is,[f,z]is not an ad-nilpotent element. Hence our assertion holds.Proposition 12.
nil
Ω
=
Ω
[
1
]
⊕
span
𝔽
{
x
i
x
j
y
λ
,
x
i
ξ
j
y
λ
,
ξ
i
ξ
j
y
λ
∣
λ
∈
H
,
i
,
j
∈
M
1
∪
T
}.Proof.
According to Lemma4, we need only to determine all ad-nilpotent elements inΩ-2⊕Ω-1⊕Ω0. For anyn∈ℕ, a direct computation shows that
(50)(adyλ)n=((1-λ)D1)n≠0,(adxiyλ)n=(-(2-1-μi-λ)xiyλD1+[i]yλDi′)n≠0,(adξjyλ)n=(-(2-1-λ)ξjyλD1+yλDj)n≠0,(adx1yλ)n=(yλ∂¯-(1-λ)x1yλD1)n≠0.
Lemmas6, 7, and 8 imply thatxixjyλ,xiξjyλ, and ξiξjyλare ad-nilpotent fori,j∈M1∪T. Hence our assertion holds.


## 4. Filtration and Intrinsic Property

Lemma 13.
Ω
[
0
]
∩
Ω
0
¯
=
Nor
Ω
0
¯
(
Nil
(
Ω
0
¯
)
)andΩ[0]is invariant.Proof.
Firstly, we prove the inclusionΩ[0]∩Ω0¯⊆NorΩ0¯(Nil(Ω0¯)). Lemmas 6–9 show that Nil(Ω0∩Ω0¯)⊲Ω0∩Ω0¯, which combined with (iv) and (v) of Lemma 5, yield
(51)Nil(Ω0¯)=Nil(Ω[0]∩Ω0¯)=Nil(Ω0∩Ω0¯)+Ω[1]∩Ω0¯,[Ω0∩Ω0¯,Nil(Ω[0]∩Ω0¯)]⊆Nil(Ω0∩Ω0¯),[Ω0∩Ω0¯,Ω[1]∩Ω0¯]⊆Ω[1]∩Ω0¯.
Then
(52)[Ω0∩Ω0¯,Nil(Ω0¯)]=[Ω0∩Ω0¯,Nil(Ω0∩Ω0¯)]+[Ω0∩Ω0¯,Ω[1]∩Ω0¯]⊆Nil(Ω0∩Ω0¯)+Ω[1]∩Ω0¯=Nil(Ω0¯),
that is,Ω[0]∩Ω0¯⊆NorΩ0¯(Nil(Ω0¯)).
Let us consider the converse inclusion. Suppose thatf=f-2+f[-1]∈NorΩ0¯(Nil(Ω0¯)), wheref-2∈Ω-2andf[-1]∈Ω[-1]. Iff-2≠0, then[f,x1ξs]=[f-2,x1ξs]+[f[-1],x1ξs]=γf-2ξs+h∉Nil(Ω0¯), whereh∈Ω[0]andγ∈𝔽, a contradiction. Consequently,f-2=0.
Now supposef=f-1+f[0], wheref-1=∑i∈M1γixiyλ∈Ω-1,γi∈𝔽,f[0]∈Ω[0].Ify-1≠0andγj≠0for somej∈M1, then we have[f,xj′2]=2[j]γjxj′yλ+h∉Nil(Ω0¯), whereh∈Ω[0], a contradiction. Thusf-1=0andf=f[0]∈Ω[0]∩Ω0¯.This proves the asserted inclusion.
By the proof above, we know thatΩ[0]is invariant.Lemma 14.
Ω
[
1
]
∩
Ω
0
¯
=
{
f
∈
nil
(
Ω
0
¯
)
∣
[
f
,
Ω
[
0
]
∩
Ω
0
¯
]
⊆
nil
(
Ω
0
¯
)
}andΩ[1]is invariant.Proof.
Letℳ={f∈nil(Ω0¯)∣[f,Ω[0]∩Ω0¯]⊆nil(Ω0¯)}. Supposef∈nil(Ω0¯). By Lemma 5,f-2=0andf-1=0.Then we can assume thatf=f0+f[1]∈ℳ, wheref0∈Ω0∩Ω0¯,f[1]∈Ω[1]∩Ω0¯.Letf0≠0.Clearly,f0∈nil(Ω0).Lemma 5(1) implies thatf0∈nil(Ω0∩Ω0¯). According to Lemma 11, there existsz∈Ω0∩Ω0¯such that[f0,z]∉nil(Ω0∩Ω0¯). Thus[f[1],z]∉nil(Ω0∩Ω0¯)andf[1]is not nilpotent, contradicting the result of Lemma 5. Hencef0=0andf∈Ω[1]∩Ω0¯; that is,ℳ⊆Ω[1]∩Ω0¯.
Conversely,[Ω[1]∩Ω0¯,Ω[0]∩Ω0¯]⊆Ω[1]∩Ω0¯⊆nil(Ω0∩Ω0¯)⊆Nil(Ω0¯)=Nil(Ω[0]∩Ω0¯).It is obvious thatΩ[1]∩Ω0¯⊆ℳand the proof is complete.Lemma 15.
(i)Ω[1]∩Ω1¯={f∈Ω1¯∣[f,Ω1¯]⊆Ω[0]∩Ω0¯}.
(ii)Ω[0]∩Ω1¯=[Ω1¯,Ω[1]∩Ω0¯].
(iii)Ω[-1]={f∈Ω∣[f,Ω[1]]⊆Ω[0]}.Proof.
(i) Put𝒜={f∈Ω1¯∣[f,Ω1¯]⊆Ω[0]∩Ω0¯}.Supposef=f-1+f[0], wheref-1=∑i∈T‍γiξiyλ∈Ω-1∩Ω1¯,f[0]∈Ω[0]∩Ω1¯andγi∈𝔽. Letγj≠0for some onej∈T. Then[f,ξj]=[f-1,ξj]+[f[0],ξj]=γjyλ+[f[0],ξj]∉Ω[-1]∩Ω0¯; that is,[f,ξj]∉Ω[0]∩Ω0¯. This contradictsf∈𝒜. Thusf-1=0.
Letf=f0+f[1]be an arbitrary element of𝒜, wheref0=∑i∈M1,j∈Tγi,jxiξjyλ∈Ω0∩Ω1¯,f[1]∈Ω[1]∩Ω1¯andγi,j∈𝔽. If there is aγit≠0, then[f,ξt]=[f0,ξt]+[f[1],ξt]=γitxiyλ+[f[1],ξt]∉Ω[0]∩Ω0¯, contradictingf∈𝒜. Hencef0=0andf=f[1]∈Ω[1]∩Ω1¯. Consequently,𝒜⊆Ω[1]∩Ω1¯.
Conversely, letf∈Ω[1]∩Ω1¯. Then[f,Ω1¯]⊆[f,Ω[-1]]⊆Ω[0]∩Ω0¯, which shows thatΩ[1]∩Ω1¯⊆𝒜.
(ii) We first prove the inclusionΩ[0]∩Ω1¯⊆[Ω1¯,Ω[1]∩Ω0¯]. Supposexkyλξu∈Ω[0]∩Ω1¯, whereξu=ξiξv. Noting thatx1xkyλξv∈Ω[1]∩Ω0¯andxkyλξu=-2[ξi,x1xkyλξv], we obtainxkyλξu∈[Ω1¯,Ω[1]∩Ω0¯].The converse inclusion is obvious.
(iii) The proof is completely analogous to (ii).Theorem 16.
Ωis transitive.Proof.
Assume the contrary. Suppose that there exists a nonzerof∈Ωlsuch that[f,Ω-1]=0, wherel∈ℕ0. Lettbe the maximal exponent ofx1of all monomial expressions occurring inf. Then we may assume that
(53)f=∑k1=t‍χxkyλξu+∑k1<t‍ωxk′yηξv,
whereχ,ω∈𝔽. For anyj∈M1, we have
(54)0=[f,xj]=∑k1=t‍χ[j′]kj′*xk-ej′yλξu+h,
wherehis the sum of summand that the exponent ofx1is less thant. Since∑k1=t‍χ[j′]kj′*xk-ej′yλξuis linear independence,xk-ej′yλξu=0; that is, kj′=0. For anyj∈T, we get
(55)0=[f,ξj]=∑k1=t‍χxkyλξu-〈j〉+h,
wherehis the sum of summand that the exponent ofx1is less thant. Thus∑k1=t‍χxkyλξu-〈j〉=0; that is,xkyλξudoes not containξjandf=xte1yλ.
Ift=0, thenf∈Ω-2, contradictingl≥0. Lett>0. Then we can suppose
(56)f=αxte1yλ+∑k1=t-1‍χk,λ,uxkyλξu+∑k1<t-1‍ωk′,η,vxk′yηξv,
whereα,χk,λ,u,ωk′,η,v∈𝔽.Forj∈M1, we have
(57)0=[f,xj]=α(1-μj)x(t-1)e1yλxj+∑k1=t-1‍χk,λ,u[j′]kj′*xk-ej′yλξu+h,
wherehis the sum of summand that the exponent ofx1is less thant-1. Thus
(58)α(1-μj)x(t-1)e1yλxj+∑k1=t-1‍χk,λ,u[j′]kj′*xk-ej′yλξu=0.
By αt(1-μj)x(t-1)e1yλxj containsxjand ∑k1=t-1‍χk,λ,u[j′]kj′*xk-ej′yλξu dose not containxj, we havet=0, contradictingt>0. HenceΩis transitive, as required.Theorem 17.
Suppose thatΩandΩ′are the Lie superalgebras ofΩ-type. Ifφis an isomorphism ofΩontoΩ′, thenφ(Ω[i])=Ω[i]′ for alli≥-2.Proof.
As the isomorphismφis an even mapping, we haveφ(Ω0¯)=Ω0¯′andφ(Nil(Ω0¯))=Nil(Ω0¯′). Henceφ(NorΩ0¯(Nil(Ω0¯))=NorΩ′0¯(Nil(Ω0¯′)). By Lemmas 13 and 14, we obtain
(59)φ(Ω[0]∩Ω0¯)=Ω[0]′∩Ω0¯′,φ(Ω[1]∩Ω0¯)=φ({f∈nil(Ω[0])∣[f,Ω[0]∩Ω0¯]⊆nil(Ω[0])})={f∈nil(Ω[0]′)∣[f,Ω[0]′∩Ω0¯′]⊆nil(Ω0¯′)}=Ω[1]′∩Ω0¯′.
Now by virtue of Lemma 15, we get
(60)φ(Ω[0]∩Ω1¯)=φ({f∈Ω[-1]∩Ω1¯∣[f,Ω0¯]⊆Ω[-1]∩Ω1¯})={f∈Ω[-1]′∩Ω1¯′∣[f,Ω0¯′]⊆Ω[-1]′∩Ω1¯′}=Ω[0]′∩Ω1¯′,φ(Ω[-1]∩Ω1¯)=φ([Ω1¯,Ω[1]∩Ω0¯])=[Ω1¯′,Ω[1]∩Ω0¯′]=Ω[-1]′∩Ω1¯′.
It follows thatφ(Ω[0])=Ω[0]′. BecauseΩ0¯⊆Ω[-1], we have
(61)φ(Ω[-1])=φ(Ω[-1]∩Ω0¯+Ω[-1]∩Ω1¯)=φ(Ω0¯+Ω[-1]∩Ω1¯)=Ω0¯′+Ω[-1]′∩Ω1¯′=Ω[-1]′.
SinceΩis transitive by the lemma above, we haveΩ[i+1]={f∈Ω[i]∣[f,Ω[-1]]⊆Ω[i]}for alli≥0. It is easy to show thatφ(Ω[i])=Ω[i]′by induction oni.Theorem 18.
Suppose thatϕis an automorphism ofΩ. Thenϕ(Ω[i])=Ω[i]for all i≥-2, that is; the filtration ofΩis invariant under the automorphism group ofΩ.Proof.
This is a direct consequence of Theorem17.Theorem 19.
LetΩ(r,m,q)andΩ(r′,m′,q′)be the Lie superalgebras ofΩ-type. ThenΩ(r,m,q)≅Ω(r′,m′,q′)if and only ifr=r′,m=m′,q=q′.Proof.
The sufficient condition is obvious. We will prove the necessary condition. Assume thatφ:Ω(r,m,q)→Ω(r′,m′,q′)is an isomorphism of Lie superalgebra. According to Theorem 17, we have
(62)φ(Ω(r,m,q)[-2])=Ω(r′,m′,q′)[-2],φ(Ω(r,m,q)[-1])=Ω(r′,m′,q′)[-1].
Thenφinduces an isomorphism ofℤ2-graded spaces
(63)Ω(r,m,q)[-2]Ω(r,m,q)[-1]⟶Ω(r′,m′,q′)[-2]Ω(r′,m′,q′)[-1].
It is easy to see that
(64)Ω(r,m,q)[-2]Ω(r,m,q)[-1]≅Ω(r,m,q)-2,Ω(r′,m′,q′)[-2]Ω(r′,m′,q′)[-1]≅Ω(r′,m′,q′)-2.
We conclude thatm=m′by the dimension comparison.
Similarly, we also obtain an isomorphism ofℤ2-graded spaces:
(65)Ω(r,m,q)[-1]Ω(r,m,q)[0]≅Ω(r′,m′,q′)[-1]Ω(r′,m′,q′)[0],
and the quotient spaceΩ[-1]/Ω[0]is isomorphic toΩ-1.A comparison of dimensions shows thatr+q=r′+q′. Note thatφ(Ω(r,m,q)α)=Ω(r′,m′,q′)α, whereα∈ℤ2.It follows thatr=r′andq=q′.



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*Source: 923101-2013-09-02.xml*