# The Natural Filtration of Finite Dimensional Modular Lie Superalgebras of Special Type

Authors: Keli Zheng; Yongzheng Zhang
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

This paper is concerned with the natural filtration of Lie superalgebraS(n,m) of special type over a field of prime characteristic. We first construct the modular Lie superalgebra S(n,m). Then we prove that the natural filtration of S(n,m) is invariant under its automorphisms.


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## Body

## 1. Introduction

Although many structural features of nonmodular Lie superalgebras (see [1–3]) are well understood, there seem to be very few general results on modular Lie superalgebras. The treatment of modular Lie superalgebras necessitates different techniques which are set forth in [4, 5]. In [6], four series of modular graded Lie superalgebras of Cartan type were constructed, which are analogous to the finite dimensional modular Lie algebras of Cartan type [7] or the four series of infinite dimensional Lie superalgebras of Cartan type defined by even differential forms over a field of characteristic zero [8]. Recent works on the modular Lie superalgebras of Cartan type can also be found in [9–13] and references therein.It is well known that filtration techniques are of great importance in the structure and the classification theories of Lie (super)algebras (see [1, 2, 14, 15]). For some classes of modular Lie (super)algebras, the filtrations have been well investigated, for example, the natural filtrations of finite dimensional modular Lie algebras of Cartan type [16, 17] and of finite dimensional simple modular Lie superalgebras W, S, and H of Cartan type [18, 19].The original motivation for this paper comes from the researches of structures for the finite dimensional modular Lie superalgebrasW(n,m) and H(n,m), which were first introduced in [20, 21], respectively. The starting point of our studies is to construct a class of finite dimensional modular Lie superalgebras of special type, which is denoted by S(n,m). A brief summary of the relevant concepts and notations in the finite dimensional modular Lie superalgebras S(n,m) is presented in Section 2. In Section 3, by using the ad-nilpotent elements of S(n,m), we show that the natural filtration of S(n,m) is invariant under its automorphisms.


## 2. Preliminaries

Throughout this paper,𝔽 denotes an algebraic closed field of characteristic p>2, and  n is an integer greater than 3. In addition to the standard notation ℤ, we write ℕ and ℕ0 to denote the sets of positive integers and nonnegative integers, respectively.LetΛ(n) be the Grassmann algebra over 𝔽 in n variables x1,x2,…,xn. Set 𝔹k={〈i1,i2,…,ik〉∣1≤i1<i2<⋯<ik≤n} and 𝔹(n)=⋃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∈𝔹(n)} is an 𝔽-basis of Λ(n).LetΠ denote the prime field of  𝔽; that is, Π={0,1,…,p-1}. Suppose that the set {z1,z2,…,zm} is a Π-linearly independent finite subset of 𝔽. Let G={∑i=1m‍λizi∣λi∈Π}. Then G is an additive subgroup of 𝔽. Let 𝔽[y1,y2,…,ym] be the truncated polynomial algebra satisfying yip=1 for all i=1,2,…,m. For every element λ=∑i=1m‍λizi∈G, define yλ=y1λ1y2λ2⋯ymλm. Then yλyη=yλ+η for all λ,η∈G. Let 𝕋(m) denote 𝔽[y1,y2,…,ym]. Then 𝕋(m)={∑λ∈G‍aλyλ∣aλ∈𝔽}. Let 𝒰=Λ(n)⊗𝕋(m). Then 𝒰 is an associative superalgebra with ℤ2-gradation induced by the trivial ℤ2-gradation of 𝕋(m) and the natural ℤ2-gradation of Λ(n); that is, 𝒰=𝒰0-⊕𝒰1-, where 𝒰0-=Λ(n)0-⊗𝕋(m) and 𝒰1-=Λ(n)1-⊗𝕋(m).Forf∈Λ(n) and α∈𝕋(m), we abbreviate f⊗α as fα. Then the elements xuyλ with u∈𝔹(n) and λ∈G form an 𝔽-basis of 𝒰. It is easy to see that 𝒰=⊕i=0n𝒰i is a ℤ-graded superalgebra, where 𝒰i=span𝔽{xuyλ∣u∈𝔹(n),|u|=i,λ∈G}. In particular, 𝒰0=𝕋(m) and 𝒰n=span𝔽{xπyλ∣λ∈G}, where π:=〈1,2,…,n〉∈𝔹(n).In this paper, ifA=A0-⊕A1- is a superalgebra (or ℤ2-graded linear space), let DerA be the derivation superalgebra of A (see [1] or [2] for the definition) and hg(A)=A0-∪A1-; that is, hg(A) is the set of all ℤ2-homogeneous elements of A. If degx occurs in some expression, we regard x as a ℤ2-homogeneous element and degx as the ℤ2-degree of x. Let A=⊕i=-rnAi be a ℤ-graded superalgebra. If x∈Ai, then we call x a ℤ-homogeneous element and i the ℤ-degree of x and set zd(x)=i.SetY={1,2,…,n}. Given that i∈Y, let ∂/∂xi be the partial derivative on Λ(n) with respect to xi. For i∈Y, let Di be the linear transformation on 𝒰 such that Di(xuyλ)=(∂xu/∂xi)yλ for all u∈𝔹(n) and λ∈G. Then Di∈Der1-𝒰 for all i∈Y since ∂/∂xi∈Der1-(Λ(n)).Suppose thatu∈𝔹k⊆𝔹(n) and i∈Y. When i∈{u}, we denote the uniquely determined element of  𝔹k-1 satisfying {u-〈i〉}={u}∖{i} by u-〈i〉 and denote the number of integers less than i in {u} by τ(u,i). When i∉{u}, we set τ(u,i)=0 and xu-〈i〉=0. Therefore, Di(xu)=(-1)τ(u,i)xu-〈i〉 for any i∈Y and u∈𝔹(n).We define(fD)(g)=fD(g) for f, g∈hg(𝒰) and D∈hg(Der𝒰). Since the multiplication of 𝒰 is supercommutative, it follows that fD is a derivation of 𝒰. Let
(1)W(n,m)=span𝔽{xuyλDi∣u∈𝔹(n),λ∈G,i∈Y}.
Then W(n,m) is a finite dimensional Lie superalgebra contained in Der𝒰. A direct computation shows that
(2)[fDi,gDj]=fDi(g)Dj-(-1)degfDideggDjgDj(f)Di,
where f, g∈hg(𝒰) and i, j∈Y.LetDr1r2:𝒰→W(n,m) be the linear map such that for every f∈hg(𝒰) and r1,r2∈Y,
(3)Dr1r2(f)=∑i=12‍friDri,
where fr1=-Dr2(f) and fr2=-Dr1(f). It is easy to see that Dr1r2 is an even linear map. Let S(n,m)={Dij(f)∣f∈𝒰,i,j∈Y}. Then S(n,m) is a finite dimensional Lie superalgebra with a ℤ-gradation S(n,m)=⊕r=-1n-2Sr(n,m), where Sr(n,m)={Dij(xuyλ)∣u∈𝔹(n),|u|=r+2,λ∈G,i,j∈Y}. In this paper, S(n,m) is called the Lie superalgebra of special type.By the definition of linear mapDr1r2, the following equalities are easy to verify:
(4)Dii(f)=-2Di(f)Di,Dij(f)=Dji(f),[Dk,Dij(f)]=-Dij(Dk(f)),(5)[Ds1s2(f),Dr1r2(g)]=∑i,j=12‍(-1)degfDsirj(fsigrj),
where f, g∈hg(𝒰); i, j, k∈Y; and fsi, grj and as in (3). The equality (5) shows that S(n,m) is a subalgebra of W(n,m). Hereafter, S(n,m) and Si(n,m) will be simply denoted by S and Si, respectively.PutA={Dij(xπyλ)∣i,j∈Y,λ∈G} and B={Dij(xkyη)∣i,j,k∈Y,η∈G}.Proposition 1.
The Lie superalgebraS is generated by A∪B.Proof.
Suppose thatA∪B generate the subalgebra Q of S. Since A and B are subsets of S, it follows that Q⊆S.
Next we will consider the reverse inclusion.
It is easy to see thatDki(xkyλ)=-yλDi for all distinct elements i, k of Y and λ∈G. Therefore, zd(Dki(xkyλ))=-1 and S-1⊆Q.
A direct calculation shows that(6)[Dij(xπyλ),Dkl(xkyη)]=[-Di(xπyλ)Dj-Di(xπyλ)Dj,-yηDl]=(-1)n(DiDl(xπyλ+η)Dj+DjDl(xπyλ+η)Di)=-(-1)nDij(Dl(xπyλ+η))∈S,
for all distinct elements i, j, k, l of Y and λ, η∈G. It follows from zd(Dij(Dl(xπyλ+η)))=n-3 that Sn-3⊆Q.
For distinct elementsi, j, k, l, g of Y and λ, η, ζ∈G, we have
(7)[Dij(Dl(xπyλ+η)),Dkg(xkyζ)]=(-1)n+1Dij(DgDl(xπyλ+η+ζ))
and zd(Dij(DgDl(xπyλ+η+ζ)))=n-4. Thus Sn-4⊆Q.
By the same methods above, we may obtainDij(xuyλ)∈S for u∈𝔹(n); that is, Si⊆Q for 1≤i≤n-5.
According toDii(xixjxkyλ)=-2xjxkyλDi∈S1 and xkyλ+ηDi∈S0, we have
(8)xkyλ+ηDi=[xjxkyλDi,yηDj]∈Q.
Hence S0⊆Q.
In conclusion,S⊆Q. Therefore, the desired result follows immediately.


## 3. The Natural Filtration ofS(n,m)

Adopting the notion of  [22], the element x of Lie superalgebra S is called ad-nilpotent if adx is a nilpotent linear transformation. The set of all ad-nilpotent elements of S is denoted by nil(S). Let S(j)=⊕i≥jSi. Then
(9)S=S(-1)⊇S(0)⊇S(1)⊇⋯⊇S(n-2)⊇S(n-1)=0
is a descending filtration of S, which is called the natural filtration of S. We also call {S(k)∣k∈ℤ} a filtration of S for short, where S(k)=S if k≤-1 and S(k)=0 if k≥n-2. Since S is ℤ-graded and finite dimensional, we may easily obtain S-1⊆nil(S) and S(1)⊆nil(S).LetMn(𝔽) denote the set of all n×n matrices over 𝔽. Notice that dim𝕋(m)=pm. Without loss of generality, we may suppose that {y1,…,ypm} is a standard 𝔽-basis of 𝕋(m). If z=∑i,j=1n∑q=1pmaijqxiyqDj∈S0, where aijq∈𝔽, then let ρ(z)=(A1⋱Apm)npm×npm, where Aq=(aijq)n×n∈Mn(𝔽).Lemma 2.
Suppose thatz=∑i,j=1n‍∑q=1pm‍aijqxiyqDj∈S0. If z is ad-nilpotent, then ρ(z) is a nilpotent matrix.Proof.
LetΓ be the representation of S0 with values in S-1. Then Γ(z)=adz and the matrix of Γ(z) over the basis {y1D1,…,y1Dn,…,ypmD1,…,ypmDn} of S-1 is A=(-(A1)t⋱-(Apm)t)npm×npm, where Aq=(aijq)n×n∈Mn(𝔽). Since z is ad-nilpotent, the representation Γ(z) is a nilpotent linear transformation. It implies that A is nilpotent. Therefore, ρ(z)=-At is a nilpotent matrix.Lemma 3.
Letz=∑i=kn-1zi, where zi∈Si and k≤n-1. If z∈nil(S) and k≥0, then zk∈nil(S).Proof.
Suppose thatz=zk+z′, where zk∈Sk and z′∈⊕i=k+1n-1Si⊆S(k+1). Since z∈nil(S), we may assume that (adz)t=0. Let x be a ℤ-homogeneous element of S with ℤ-degree i. Then (adz)t(x)=0. On the other hand,
(10)(adz)t(x)=(ad(zk+z′))t(x)=(adzk)t(x)+h,
which implies (adzk)t(x)+h=0. It is easy to see that (adzk)t(x)∈S(kt+i) and h∈S(kt+i+1)=⊕j≥kt+i+1Sj. Thus (adzk)t(x)=0. Since x is an arbitrary ℤ-homogeneous element of S, we have (adzk)t(S)=0. Then (adzk)t=0; that is, zk∈nil(S).Suppose thatEij denotes the n×n matrix whose (i,j) element is 1 and otherwise is zero. Obviously,
(11)EijEkl=δjkEil,
where δjk is the Kronecker delta.Ifz=∑i,j=1n∑q=1pmaijqxiyqDj∈S0, where aijq∈𝔽, then
(12)ρ(z)=∑i,j=1n‍aij1Eij+∑i,j=n+12n‍aij2Eij+⋯+∑i,j=n(pm-1)+1npm‍aijpmEij.LetΔ={z∈nil(S)∣adz(S)⊆nil(S)}.Lemma 4.
Suppose thatz=∑i=-1n-2zi, where zi∈Si. If z∈Δ, then z-1=0.Proof.
Suppose that0≠z-1=∑i=1n∑q=1pmaiqyqDi, where aiq∈𝔽. Let ajq≠0 and j, k, l∈Y such that i, j, k are distinct. We may assume that d=[z-1,Dkl(xkxlxj)]. A direct calculation shows that
(13)d=[∑i=1n‍∑q=1pm‍aiqyqDi,-xlxjDl+xkxjDk]=-∑q=1pm‍(alqxjyqDl-ajqxlyqDl-akqxjyqDk+ajqxkyqDk).
By equalities (11) and (12), we have
(14)(ρ(d))t=(-1)t((-1)t(aj1)tEll+(aj1)tEkk+(-1)t-1al1(aj1)t-1Ejl-ak1(aj1)t-1Ejk+(-1)t(a(j+n)2)tE(l+n)(l+n)+(a(j+n)2)tE(k+n)(k+n)+(-1)t-1a(l+n)2(a(j+n)1)t-1E(j+n)(l+n)-a(k+n)2(a(j+n)2)t-1E(j+n)(k+n)+⋯+(-1)t(a(j+pm-n)pm)tE(l+pm-n)(l+pm-n)+(a(j+pm-n)pm)tE(k+pm-n)(k+pm-n)+(-1)t-1a(l+pm-n)pm(a(j+pm-n)pm)t-1×E(j+pm-n)(l+pm-n)-a(k+pm-n)pm(a(j+pm-n)pm)t-1E(j+pm-n)(k+pm-n)).
Since (aj1)t≠0, we have (ρ(d))t≠0. So ρ(d) is not a nilpotent matrix. By Lemma 2, it follows that d∉nil(S). By Lemma 3, we have [z,Dkl(xkxlxj)]∉nil(S). Then z∉Δ. It contradicts z∈Δ. This proves our assertion.Lemma 5.
Letz=∑i=-1n-2‍zi, where zi∈Si. If z∈Δ, then z0=0.Proof.
Assume thatz0≠0. Let z0=∑i,j=1n‍∑q=1pm‍aijqxiyqDj, aijq∈𝔽, and
(15)l=min{i∣aijλ≠0,i,j∈Y},t=min{j∣aijλ≠0,i,j∈Y}.
(i) Suppose thatl≤t. Let
(16)k=max{j∣aljλ≠0,j∈Y}.
Then alkq≠0. It is easy to see that t≤k. Since l≤t, we have l≤k. Therefore,
(17)z0=∑j=tk‍∑q=1pm‍aljqxlyqDj+∑i=l+1n‍∑j=tn‍∑q=1pm‍aijqxiyqDj.
Assume that l=k. It follows from t≤k that t≤l. Then we have t=l which implies that
(18)z0=∑q=1pm‍allqxlyqDl+∑i=l+1n‍∑j=tn‍∑q=1pm‍aijqxiyqDj.
Therefore,
(19)ρ(z0)=all1Ell+∑i=l+1n‍∑j=tn‍aij1Eij+a(l+n)(l+n)2E(l+n)(l+n)+∑i=l+1+n2n‍∑j=t+n2n‍aij2Eij+⋯+a(l+n(pm-1))(l+n(pm-1))pmE(l+n)(l+n)+∑i=l+1+n(pm-1)npm‍∑j=t+n(pm-1)npm‍aijpmEij=(A1B1C1⋱ApmBpmCpm)npm×npm,
where Ak=a(l+(k-1)n)(l+(k-1)n)qE(l+(k-1)n)(l+(k-1)n) is an (l+(k-1)n)×(l+(k-1)n) matrix and q∈{1,…,pm}. Since all1≠0, we have A1 not being a nilpotent matrix. Then ρ(z0) is not a nilpotent matrix and z0∉nil(S). Lemma 3 shows that z∉nil(S). It is a contradiction of to z∈Δ; that is, l<k.
Suppose thath∈Y and h≠l, k. Let d=[z0,xkDl]. By equality (2), we obtain
(20)d=∑q=1pm‍(alkqxlyqDl+∑i=l+1n‍aikqxiyqDl-∑j=tk‍aljqxkyqDj).
Since l<k, ρ(d) also has the matrix form as ρ(z0), it follows from alk1≠0 that A1 is not a nilpotent matrix. Then ρ(d) is not nilpotent. So z∉nil(S) and [z,xkDl]∉nil(S). It is a contradiction of z∈Δ.
(ii) Suppose thatt<l. Let k=max{i∣aitλ≠0} and d′=[z,xtDk]. Imitating (i), we may prove that ρ(d′) is also not nilpotent. Then the desired result follows.Lemma 6.
(i)
If z∈S0∩nil(S) and h∈S(1), then z+h∈nil(S).
(ii)
Suppose that i, j are distinct elements of Y; then xiyλDj∈nil(S) for all λ∈G.
(iii)
Suppose that i, j, k are distinct elements of Y; then axjyλDk+bxiyηDk∈nil(S), where a, b∈𝔽 and λ, η are arbitrary elements of G.Proof.
(i) A direct verification shows that{adz}∪{adS(1)} is a weakly closed subset of nilpotent elements of pl(S), where pl(S) is the general linear Lie superalgebra of S. It was shown in [23, Theorem 1 of Chapter II] that each element of span𝔽({adz}∪{adS(1)}) is a nilpotent linear transformation of S. Then adz+adh is nilpotent. So z+h is ad-nilpotent.
(ii) To prove(adxiyλDj)p=0, we may assume without loss of generality that i<j. Set η to be an arbitrary element of G. If k≠i, then
(21)(adxiyλDj)2(xuyηDk)=[xiyλDj,[xiyλDj,xuyηDk]]=(-1)τ(u,j)[xiyλDj,xixu-〈j〉yλ+ηDk]=0.
In the case of k=i, we have
(22)(adxiyλDj)3(xuyηDk)=[xiyλDj,[xiyλDj,[xiyλDj,xuyηDi]]]=[xiyλDj,[xiyλDj,(-1)τ(u,j)xixu-〈j〉yλDi-xuyλ+ηDj]]=(-1)τ(u,j)[xiyλDj,-xixu-〈j〉yλDj-xixu-〈j〉y2λ+ηDj]=0.
Forp>2 we obtain (adxiyλDj)p(xuyηDk)=0. Therefore (adxiyλDj)p(S)=0. This yields (adxiyλDj)p=0. Thus xiyλDj∈nil(S).
(iii) According to(ii) and [xjyλDk,xiyηDk]=0, {adxjyλDk,adxiyηDk} is a weakly closed subset of nilpotent elements of pl(S). So axjyλDk+bxiyηDk∈nil(S), where a, b∈𝔽.Lemma 7.
Ifi, j, k are distinct elements of   Y, then xixjyλDk∈Δ for all λ∈G.Proof.
Suppose thatl∈Y∖{i,j,k}. Then xixjyλDk∈S(1)⊆nil(S). Let z=∑i=-1n-2zi, where zi∈Si. Assume that [xixjyλDk,z]=f0+f1, where f0=[xixjyλDk,z-1]∈S0 and f1∈S(1). Let z-1=∑l=1n∑η∈G‍alηyηDl. Then
(23)f0=[xixjyλDk,∑l=1n‍∑η∈G‍alηyηDl]=∑η∈G‍(aiηxjyλ+ηDk-ajηxiyλ+ηDk).
By (iii) of Lemma 6, we have f0∈S0∩nil(S). By (i) of Lemma 6, it follows that f0+f1∈nil(S). We finally obtain xixjyλDk∈Δ for all λ∈G.LetQ={z∈nil(S)∣adz(Δ)⊆Δ}.Lemma 8.
Q
=
S
(
1
).Proof.
By the definition ofΔ, we have S(2)⊆Δ. Lemmas 4 and 5 show that Δ⊆S(1). Then [S(1),Δ]⊆[S(1),S(1)]⊆S(2)⊆Δ. Thus S(1)⊆Q.
Next we will proveQ⊆S(1). Let z∈Q and z=∑i=-1n-2zi, where zi∈Si. Assume that z-1=∑l=1n∑λ∈G‍alλyλDl≠0, alλ∈𝔽. Without loss of generality, we may suppose that ai≠0. Let d=xixjyηDk, where i, j, k are distinct elements of Y and η is an arbitrary element of G. By Lemma 7, we have d∈Δ. Let [z,d]=h0+h1, where h0=[z-1,d]∈S0 and h1∈S(1). Since ai≠0, we have h0=∑λ∈G(aiλxjyλ+ηDk-ajλxiyλ+ηDk)≠0. Lemma 5 implies that h0+h1∉Δ. It is a contradiction of z∈Q. Hence z-1=0.
Assume that0≠z0=∑i,j=1n∑q=1pm‍aijλxiyqDj, aijq∈𝔽, and suppose that l and t are as the definitions in (15). We may suppose that l≤t (the proof is similar to the case t<l) and let k be as the definition in (16). In a similar way to the first part of the proof in Lemma 5, we have l<k. Suppose that h∈Y∖{l,k,t} and d1=xkxhDl. Lemma 7 shows that d1∈Δ. Let [z,d1]=g1+g2, where g1=[z0,d1]∈S1 and g2∈S(2). Using equality (2), we have
(24)g1=∑q=1pm‍(alkqxlxhyqDl-∑i=l+1n‍aihqxixkyqDl-∑j=tk‍aljqxkxhyqDj).
If h<t, then aihq=0 in the above equality, where i∈Y∖{1,…,l-1}. Thus
(25)[Dh,g1]=-∑q=1pm‍(alkqxlyqDl+∑i=l+1n‍aihqxiyqDl+ahhqxkyqDl-aljqxkyqDj∑i=l+1n‍).
By equality (12), the matrix ρ([Dh,g1]) has the matrix form as in Lemma 5. Since alkq≠0, A1 is not a nilpotent matrix. It implies that ρ([Dh,g1]) is not nilpotent. Hence [Dh,g1]∉nil(S). Lemma 3 shows that [Dh,g1+g2]∉nil(S); that is, [Dh,g1+g2]∉Δ. It contradicts z∈Q. Thus z0=0. Therefore, z∈S(1) and Q⊆S(1).According to the fact thatΔ and Q are invariant subspaces under the automorphisms of S and Lemma 8, S(1) is also invariant under the automorphisms of S. Since
(26)S(0)={x∈S∣[x,S(1)]⊆S(1)},S(i)={x∈Si-1∣[x,S]⊆S(i-1)},i≥1,
we may easily obtain the following theorem.Theorem 9.
The natural filtration ofS is invariant under the automorphisms of S.Let𝔖i=S(i)/S(i+1) for -1≤i≤n-2. Then 𝔖i is a ℤ-graded space. Suppose that 𝔖:=⊕i=-1n-2𝔖i; then 𝔖 is also a ℤ-graded space. Let x+S(i+1)∈𝔖i and y+S(j+1)∈𝔖j. Define
(27)[x+S(i+1),y+S(j+1)]:=[x,y]+S(i+j+1).
It is easy to see that the definition above is reasonable. There exists a linear expansion such that 𝔖 has an operator [,]. A direct verification shows that 𝔖 is a Lie superalgebra with respect to the operator [,]. The Lie superalgebras 𝔖 is called a Lie superalgebra induced by the natural filtration of S.Lemma 10.
𝔖
≅
S.Proof.
Letϕ:S→𝔖 be a linear map such that ϕ(x)=x+S(i+1), where x∈S(i)∖S(i+1). A direct verification shows that ϕ is a homomorphism of Lie superalgebras. Suppose that y∈kerϕ. If y≠0, then there exists i≥-1 such that y∈S(i)∖S(i+1). Since ϕ(y)=0, we have y+S(i+1)=0. Hence y∈S(i+1). That shows that y=0. Thus, kerϕ=0. Therefore, ϕ is a monomorphism. It follows from the fact S is finite dimensional that ϕ is an isomorphism.The definition ofϕ shows that
(28)ϕ(Si)={x+S(i+1)∣x∈Si}={x+S(i+1)∣x∈S(i)}=S(i)/S(i+1)=𝔖i,i≥-1.Suppose thatm, n, m′, n′ are elements of ℕ0 and n, n′ are greater than 3. In a similar way to S, the Lie superalgebra S(n′,m′) will be simply denoted by S′. According to the definitions of Δ, Q, and 𝔖 in S, the Δ′, Q′, and 𝔖′ in S′ are also defined by the same method, respectively.Proposition 11.
Suppose thatS≅S′ and σ is an isomorphism from S to S′; then σ(S(i))=S(i)′ for all i≥-1.Proof.
It is clear thatσ(S(-1))=S(-1)′ and σ(nil(S))=nil(S′). A direct verification shows that σ(Δ)=Δ′. Hence σ(Q)=Q′. By virtue of Lemma 8, we have Q=S(1) and Q′=S(1)′. Thus σ(S(1))=S(1)′. By equalities (26), the desired result σ(S(i))=S(i)′ for all i≥-1 is obtained.Lemma 12.
Suppose thatS≅S′ and σ is an isomorphism from S to S′; then σ induces an isomorphism σ~ from 𝔖 to 𝔖′ such that σ~(𝔖i)=𝔖i′ for all i≥-1.Proof.
Define a linear mapσ~:𝔖→𝔖′ such that
(29)σ~(x+S(i+1))=σ(x)+S(i+1)′,
where x+S(i+1)∈𝔖i. Using Proposition 11, the definition of σ~ is reasonable and
(30)σ~([x+S(i+1),y+S(j+1)])=σ([x,y])+S(i+j+1)′=[σ(x)+S(i+1)′,σ(y)+S(j+1)′]=[σ~(x+S(i+1)′),σ~(y+S(j+1)′)].
Thus σ~ is a homomorphism from 𝔖 to 𝔖′. Clearly, σ~(𝔖i)=𝔖i′ for all i≥-1. It follows that σ~ is an epimorphism.
Suppose thaty∈kerσ~; then y∈𝔖. So we may suppose that y=∑i=-1n-2yi and yi∈𝔖i. Since 𝔖i=S(i)/S(i+1), let yi=zi+S(i+1), where zi∈S(i). Hence σ~(yi)=σ(zi)+S(i+1)′. It follows from σ~(y)=0 that ∑i=-1n-2σ~(yi)=0. Thus σ~(yi)=0; that is, σ(zi)+S(i+1)′=0. It follows that σ(zi)∈S(i+1)′. By Proposition 11, we have zi∈σ-1(S(i+1)′)=S(i+1). Then yi=zi+S(i+1)=0 for -1≤i≤n-2. Therefore, y=0 and kerσ~=0. Consequently, σ~ is an isomorphism induced by σ such that σ~(𝔖i)=𝔖i′ for all i≥-1.Theorem 13.
S
≅
S
′ if and only if m=m′ and n=n′.Proof.
Because the sufficiency is obvious, it suffices to prove the necessity. Suppose thatϕ:S→𝔖 is the isomorphism given in the proof of Lemma 10. Similarly, there also exists the ϕ′:S′→𝔖′. According to the equality (28) and Lemma 12, we have
(31)ϕ(Si)=𝔖i,ϕ′(Si′)=𝔖i′,σ~(𝔖i)=𝔖i′
for -1≤i≤n-2. Let ψ=(ϕ′)-1σ~ϕ. Then
(32)ψ(Si)=(ϕ′)-1σ~ϕ(Si)=(ϕ′)-1σ~(𝔖i)=(ϕ′)-1(𝔖i′)=Si′.
In particular, ψ(S-1)=S-1′. It follows from dimS-1=dimS-1′ that npm=n′pm′. By virtue of the definition of Si, we have
(33)S0=span𝔽{Dij(xkxlyλ)∈S∣i,j,k,l∈Y,λ∈G}.
Thus dimS0=(n2-1)pm. Similarly, dimS0′=(n′2-1)pm′. According to dimS0=dimS0′ and npm=n′pm′, we have n=n′. In conclusion, the proof is completed.



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*Source: 891241-2013-08-01.xml*