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

Abstract

This paper is concerned with the natural filtration of Lie superalgebra 𝑆(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.

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 superalgebras 𝑊(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.

Let Λ(𝑛) be the Grassmann algebra over 𝔽 in 𝑛 variables 𝑥₁,𝑥₂,…,𝑥_𝑛. Set 𝔹_𝑘={⟨𝑖₁,𝑖₂,…,𝑖_𝑘⟩∣1≤𝑖₁<𝑖₂<⋯<𝑖_𝑘≤𝑛} and 𝔹(𝑛)=⋃^𝑛_𝑘=0𝔹_𝑘, where 𝔹₀=∅. For 𝑢=⟨𝑖₁,𝑖₂,…,𝑖_𝑘⟩∈𝔹_𝑘, set |𝑢|=𝑘, {𝑢}={𝑖₁,𝑖₂,…,𝑖_𝑘} and 𝑥^𝑢=𝑥_𝑖1𝑥_𝑖2⋯𝑥_𝑖𝑘(|∅|=0,𝑥^∅=1). Then {𝑥^𝑢∣𝑢∈𝔹(𝑛)} is an 𝔽-basis of Λ(𝑛).

Let Π denote the prime field of  𝔽; that is, Π={0,1,…,𝑝−1}. Suppose that the set {𝑧₁,𝑧₂,…,𝑧_𝑚} is a Π-linearly independent finite subset of 𝔽. Let 𝐺={∑^𝑚_𝑖=1𝜆_𝑖𝑧_𝑖∣𝜆_𝑖∈Π}. Then 𝐺 is an additive subgroup of 𝔽. Let 𝔽[𝑦₁,𝑦₂,…,𝑦_𝑚] be the truncated polynomial algebra satisfying 𝑦^𝑝_𝑖=1 for all 𝑖=1,2,…,𝑚. For every element 𝜆=∑^𝑚_𝑖=1𝜆_𝑖𝑧_𝑖∈𝐺, define 𝑦^𝜆=𝑦^𝜆1₁𝑦^𝜆2₂⋯𝑦^𝜆𝑚_𝑚. Then 𝑦^𝜆𝑦^𝜂=𝑦^𝜆+𝜂 for all 𝜆,𝜂∈𝐺. Let 𝕋(𝑚) denote 𝔽[𝑦₁,𝑦₂,…,𝑦_𝑚]. Then 𝕋(𝑚)={∑_𝜆∈𝐺𝑎_𝜆𝑦^𝜆∣𝑎_𝜆∈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).

For 𝑓∈Λ(𝑛) and 𝛼∈𝕋(𝑚), we abbreviate 𝑓⊗𝛼 as 𝑓𝛼. Then the elements 𝑥^𝑢𝑦^𝜆 with 𝑢∈𝔹(𝑛) and 𝜆∈𝐺 form an 𝔽-basis of 𝒰. It is easy to see that 𝒰=⊕^𝑛_𝑖=0𝒰_𝑖 is a ℤ-graded superalgebra, where 𝒰_𝑖=span_𝔽{𝑥^𝑢𝑦^𝜆∣𝑢∈𝔹(𝑛),|𝑢|=𝑖,𝜆∈𝐺}. In particular, 𝒰₀=𝕋(𝑚) and 𝒰_𝑛=span_𝔽{𝑥^𝜋𝑦^𝜆∣𝜆∈𝐺}, where 𝜋∶=⟨1,2,…,𝑛⟩∈𝔹(𝑛).

In this paper, if 𝐴=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=⊕^𝑛_𝑖=−𝑟𝐴_𝑖 be a ℤ-graded superalgebra. If 𝑥∈𝐴_𝑖, then we call 𝑥 a ℤ-homogeneous element and 𝑖 the ℤ-degree of 𝑥 and set 𝑧𝑑(𝑥)=𝑖.

Set 𝑌={1,2,…,𝑛}. Given that 𝑖∈𝑌, let 𝜕/𝜕𝑥_𝑖 be the partial derivative on Λ(𝑛) with respect to 𝑥_𝑖. For 𝑖∈𝑌, let 𝐷_𝑖 be the linear transformation on 𝒰 such that 𝐷_𝑖(𝑥^𝑢𝑦^𝜆)=(𝜕𝑥^𝑢/𝜕𝑥_𝑖)𝑦^𝜆 for all 𝑢∈𝔹(𝑛) and 𝜆∈𝐺. Then 𝐷_𝑖∈Der₁𝒰 for all 𝑖∈𝑌 since 𝜕/𝜕𝑥_𝑖∈Der₁(Λ(𝑛)).

Suppose that 𝑢∈𝔹_𝑘⊆𝔹(𝑛) and 𝑖∈𝑌. When 𝑖∈{𝑢}, we denote the uniquely determined element of  𝔹_𝑘−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 𝑥^{𝑢−⟨𝑖⟩}=0. Therefore, 𝐷_𝑖(𝑥^𝑢)=(−1)^{𝜏(𝑢,𝑖)}𝑥^{𝑢−⟨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 𝑊(𝑛,𝑚)=span_𝔽𝑥^𝑢𝑦^𝜆𝐷_𝑖∣𝑢∈𝔹(𝑛),𝜆∈𝐺,𝑖∈𝑌. (1) Then 𝑊(𝑛,𝑚) is a finite dimensional Lie superalgebra contained in Der𝒰. A direct computation shows that 𝑓𝐷_𝑖,𝑔𝐷_𝑗=𝑓𝐷_𝑖𝑔𝐷_𝑗−(−1)^{deg𝑓𝐷𝑖deg𝑔𝐷𝑗}𝑔𝐷_𝑗𝑓𝐷_𝑖, (2) where 𝑓, 𝑔∈ℎ𝑔(𝒰) and 𝑖, 𝑗∈𝑌.

Let 𝐷_𝑟1𝑟2∶𝒰→𝑊(𝑛,𝑚) be the linear map such that for every 𝑓∈ℎ𝑔(𝒰) and 𝑟₁,𝑟₂∈𝑌, 𝐷_𝑟1𝑟2𝑓=²_𝑖=1𝑓_𝑟𝑖𝐷_𝑟𝑖, (3) where 𝑓_𝑟1=−𝐷_𝑟2(𝑓) and 𝑓_𝑟2=−𝐷_𝑟1(𝑓). It is easy to see that 𝐷_𝑟1𝑟2 is an even linear map. Let 𝑆(𝑛,𝑚)={𝐷_𝑖𝑗(𝑓)∣𝑓∈𝒰,𝑖,𝑗∈𝑌}. Then 𝑆(𝑛,𝑚) is a finite dimensional Lie superalgebra with a ℤ-gradation 𝑆(𝑛,𝑚)=⊕^𝑛−2_𝑟=−1𝑆_𝑟(𝑛,𝑚), where 𝑆_𝑟(𝑛,𝑚)={𝐷_𝑖𝑗(𝑥^𝑢𝑦^𝜆)∣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 map 𝐷_𝑟1𝑟2, the following equalities are easy to verify: 𝐷_𝑖𝑖𝑓=−2𝐷_𝑖𝑓𝐷_𝑖,𝐷_𝑖𝑗𝑓=𝐷_𝑗𝑖𝑓,𝐷_𝑘,𝐷_𝑖𝑗𝑓=−𝐷_𝑖𝑗𝐷_𝑘𝑓,𝐷_𝑠1𝑠2𝑓,𝐷_𝑟1𝑟2𝑔=²_𝑖,𝑗=1(−1)^deg𝑓𝐷_{𝑠𝑖𝑟𝑗}𝑓_𝑠𝑖𝑔_𝑟𝑗, (4)(5) where 𝑓, 𝑔∈ℎ𝑔(𝒰); 𝑖, 𝑗, 𝑘∈𝑌; and 𝑓_𝑠𝑖, 𝑔_𝑟𝑗 and as in (3). The equality (5) shows that 𝑆(𝑛,𝑚) is a subalgebra of 𝑊(𝑛,𝑚). Hereafter, 𝑆(𝑛,𝑚) and 𝑆_𝑖(𝑛,𝑚) will be simply denoted by 𝑆 and 𝑆_𝑖, respectively.

Put 𝐴={𝐷_𝑖𝑗(𝑥^𝜋𝑦^𝜆)∣𝑖,𝑗∈𝑌,𝜆∈𝐺} and 𝐵={𝐷_𝑖𝑗(𝑥_𝑘𝑦^𝜂)∣𝑖,𝑗,𝑘∈𝑌,𝜂∈𝐺}.

Proposition 1. The Lie superalgebra 𝑆 is generated by 𝐴∪𝐵.

Proof. Suppose that 𝐴∪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 that 𝐷_𝑘𝑖(𝑥_𝑘𝑦^𝜆)=−𝑦^𝜆𝐷_𝑖 for all distinct elements 𝑖, 𝑘 of 𝑌 and 𝜆∈𝐺. Therefore, 𝑧𝑑(𝐷_𝑘𝑖(𝑥_𝑘𝑦^𝜆))=−1 and 𝑆₋₁⊆𝑄.
A direct calculation shows that 𝐷_𝑖𝑗𝑥^𝜋𝑦^𝜆,𝐷_𝑘𝑙𝑥_𝑘𝑦^𝜂=−𝐷_𝑖𝑥^𝜋𝑦^𝜆𝐷_𝑗−𝐷_𝑖𝑥^𝜋𝑦^𝜆𝐷_𝑗,−𝑦^𝜂𝐷_𝑙=(−1)^𝑛𝐷_𝑖𝐷_𝑙𝑥^𝜋𝑦^𝜆+𝜂𝐷_𝑗+𝐷_𝑗𝐷_𝑙𝑥^𝜋𝑦^𝜆+𝜂𝐷_𝑖=−(−1)^𝑛𝐷_𝑖𝑗𝐷_𝑙𝑥^𝜋𝑦^𝜆+𝜂∈𝑆, (6) for all distinct elements 𝑖, 𝑗, 𝑘, 𝑙 of 𝑌 and 𝜆, 𝜂∈𝐺. It follows from 𝑧𝑑(𝐷_𝑖𝑗(𝐷_𝑙(𝑥^𝜋𝑦^𝜆+𝜂)))=𝑛−3 that 𝑆_𝑛−3⊆𝑄.
For distinct elements 𝑖, 𝑗, 𝑘, 𝑙, 𝑔 of 𝑌 and 𝜆, 𝜂, 𝜁∈𝐺, we have 𝐷_𝑖𝑗𝐷_𝑙𝑥^𝜋𝑦^𝜆+𝜂,𝐷_𝑘𝑔𝑥_𝑘𝑦^𝜁=(−1)^𝑛+1𝐷_𝑖𝑗𝐷_𝑔𝐷_𝑙𝑥^𝜋𝑦^{𝜆+𝜂+𝜁} (7) and 𝑧𝑑(𝐷_𝑖𝑗(𝐷_𝑔𝐷_𝑙(𝑥^𝜋𝑦^{𝜆+𝜂+𝜁})))=𝑛−4. Thus 𝑆_𝑛−4⊆𝑄.
By the same methods above, we may obtain 𝐷_𝑖𝑗(𝑥^𝑢𝑦^𝜆)∈𝑆 for 𝑢∈𝔹(𝑛); that is, 𝑆_𝑖⊆𝑄 for 1≤𝑖≤𝑛−5.
According to 𝐷_𝑖𝑖(𝑥_𝑖𝑥_𝑗𝑥_𝑘𝑦^𝜆)=−2𝑥_𝑗𝑥_𝑘𝑦^𝜆𝐷_𝑖∈𝑆₁ and 𝑥_𝑘𝑦^𝜆+𝜂𝐷_𝑖∈𝑆₀, we have 𝑥_𝑘𝑦^𝜆+𝜂𝐷_𝑖=𝑥_𝑗𝑥_𝑘𝑦^𝜆𝐷_𝑖,𝑦^𝜂𝐷_𝑗∈𝑄. (8) Hence 𝑆₀⊆𝑄.
In conclusion, 𝑆⊆𝑄. Therefore, the desired result follows immediately.

3. The Natural Filtration of 𝑆(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≥𝑗𝑆_𝑖. Then 𝑆=𝑆₍₋₁₎⊇𝑆₍₀₎⊇𝑆₍₁₎⊇⋯⊇𝑆_(𝑛−2)⊇𝑆_(𝑛−1)=0 (9) 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).

Let 𝑀_𝑛(𝔽) denote the set of all 𝑛×𝑛 matrices over 𝔽. Notice that dim𝕋(𝑚)=𝑝^𝑚. Without loss of generality, we may suppose that {𝑦₁,…,𝑦_𝑝𝑚} is a standard 𝔽-basis of 𝕋(𝑚). If 𝑧=∑^𝑛_𝑖,𝑗=1∑^𝑝𝑚_𝑞=1𝑎_𝑖𝑗𝑞𝑥_𝑖𝑦_𝑞𝐷_𝑗∈𝑆₀, where 𝑎_𝑖𝑗𝑞∈𝔽, then let 𝜌(𝑧)=^𝐴1⋱_{𝐴𝑝𝑚𝑛𝑝𝑚×𝑛𝑝𝑚}, where 𝐴_𝑞=(𝑎_𝑖𝑗𝑞)_𝑛×𝑛∈𝑀_𝑛(𝔽).

Lemma 2. Suppose that 𝑧=∑^𝑛_𝑖,𝑗=1∑^𝑝𝑚_𝑞=1𝑎_𝑖𝑗𝑞𝑥_𝑖𝑦_𝑞𝐷_𝑗∈𝑆₀. If 𝑧 is ad-nilpotent, then 𝜌(𝑧) is a nilpotent matrix.

Proof. Let Γ be the representation of 𝑆₀ with values in 𝑆₋₁. Then Γ(𝑧)=ad𝑧 and the matrix of Γ(𝑧) over the basis {𝑦₁𝐷₁,…,𝑦₁𝐷_𝑛,…,𝑦_𝑝𝑚𝐷₁,…,𝑦_𝑝𝑚𝐷_𝑛} of 𝑆₋₁ is 𝐴=^{−(𝐴1)𝑡⋱}_{−(𝐴𝑝𝑚)𝑡𝑛𝑝𝑚×𝑛𝑝𝑚}, where 𝐴_𝑞=(𝑎_𝑖𝑗𝑞)_𝑛×𝑛∈𝑀_𝑛(𝔽). Since 𝑧 is ad-nilpotent, the representation Γ(𝑧) is a nilpotent linear transformation. It implies that 𝐴 is nilpotent. Therefore, 𝜌(𝑧)=−𝐴^𝑡 is a nilpotent matrix.

Lemma 3. Let 𝑧=∑^𝑛−1_𝑖=𝑘𝑧_𝑖, where 𝑧_𝑖∈𝑆_𝑖 and 𝑘≤𝑛−1. If 𝑧∈nil(𝑆) and 𝑘≥0, then 𝑧_𝑘∈nil(𝑆).

Proof. Suppose that 𝑧=𝑧_𝑘+𝑧^′, where 𝑧_𝑘∈𝑆_𝑘 and 𝑧^′∈⊕^𝑛−1_𝑖=𝑘+1𝑆_𝑖⊆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, (ad𝑧)^𝑡(𝑥)=ad𝑧_𝑘+𝑧^′𝑡(𝑥)=ad𝑧_𝑘^𝑡(𝑥)+ℎ, (10) which implies (ad𝑧_𝑘)^𝑡(𝑥)+ℎ=0. It is easy to see that (ad𝑧_𝑘)^𝑡(𝑥)∈𝑆_{(𝑘𝑡+𝑖)} and ℎ∈𝑆_{(𝑘𝑡+𝑖+1)}=⊕_{𝑗≥𝑘𝑡+𝑖+1}𝑆_𝑗. Thus (ad𝑧_𝑘)^𝑡(𝑥)=0. Since 𝑥 is an arbitrary ℤ-homogeneous element of 𝑆, we have (ad𝑧_𝑘)^𝑡(𝑆)=0. Then (ad𝑧_𝑘)^𝑡=0; that is, 𝑧_𝑘∈nil(𝑆).

Suppose that 𝐸_𝑖𝑗 denotes the 𝑛×𝑛 matrix whose (𝑖,𝑗) element is 1 and otherwise is zero. Obviously, 𝐸_𝑖𝑗𝐸_𝑘𝑙=𝛿_𝑗𝑘𝐸_𝑖𝑙, (11) where 𝛿_𝑗𝑘 is the Kronecker delta.

If 𝑧=∑^𝑛_𝑖,𝑗=1∑^𝑝𝑚_𝑞=1𝑎_𝑖𝑗𝑞𝑥_𝑖𝑦_𝑞𝐷_𝑗∈𝑆₀, where 𝑎_𝑖𝑗𝑞∈𝔽, then 𝜌(𝑧)=^𝑛_𝑖,𝑗=1𝑎_𝑖𝑗1𝐸_𝑖𝑗+^2𝑛_{𝑖,𝑗=𝑛+1}𝑎_𝑖𝑗2𝐸_𝑖𝑗+⋯+^𝑛𝑝𝑚_{𝑝𝑖,𝑗=𝑛𝑚−1+1}𝑎_{𝑖𝑗𝑝𝑚}𝐸_𝑖𝑗. (12)

Let Δ={𝑧∈nil(𝑆)∣ad𝑧(𝑆)⊆nil(𝑆)}.

Lemma 4. Suppose that 𝑧=∑^𝑛−2_𝑖=−1𝑧_𝑖, where 𝑧_𝑖∈𝑆_𝑖. If 𝑧∈Δ, then 𝑧₋₁=0.

Proof. Suppose that 0≠𝑧₋₁=∑^𝑛_𝑖=1∑^𝑝𝑚_𝑞=1𝑎_𝑖𝑞𝑦_𝑞𝐷_𝑖, where 𝑎_𝑖𝑞∈𝔽. Let 𝑎_𝑗𝑞≠0 and 𝑗, 𝑘, 𝑙∈𝑌 such that 𝑖, 𝑗, 𝑘 are distinct. We may assume that 𝑑=[𝑧₋₁,𝐷_𝑘𝑙(𝑥_𝑘𝑥_𝑙𝑥_𝑗)]. A direct calculation shows that 𝑑=^𝑛𝑝_𝑖=1^𝑚_𝑞=1𝑎_𝑖𝑞𝑦_𝑞𝐷_𝑖,−𝑥_𝑙𝑥_𝑗𝐷_𝑙+𝑥_𝑘𝑥_𝑗𝐷_𝑘=−^𝑝𝑚_𝑞=1𝑎_𝑙𝑞𝑥_𝑗𝑦_𝑞𝐷_𝑙−𝑎_𝑗𝑞𝑥_𝑙𝑦_𝑞𝐷_𝑙−𝑎_𝑘𝑞𝑥_𝑗𝑦_𝑞𝐷_𝑘+𝑎_𝑗𝑞𝑥_𝑘𝑦_𝑞𝐷_𝑘. (13)
By equalities (11) and (12), we have 𝜌(𝑑)^𝑡=(−1)^𝑡(−1)^𝑡𝑎_𝑗1^𝑡𝐸_𝑙𝑙+𝑎_𝑗1^𝑡𝐸_𝑘𝑘+(−1)^𝑡−1𝑎_𝑙1𝑎_𝑗1^𝑡−1𝐸_𝑗𝑙−𝑎_𝑘1𝑎_𝑗1^𝑡−1𝐸_𝑗𝑘+(−1)^𝑡𝑎_(𝑗+𝑛)2^𝑡𝐸_{(𝑙+𝑛)(𝑙+𝑛)}+𝑎_(𝑗+𝑛)2^𝑡𝐸_{(𝑘+𝑛)(𝑘+𝑛)}+(−1)^𝑡−1𝑎_(𝑙+𝑛)2𝑎_(𝑗+𝑛)1^𝑡−1𝐸_{(𝑗+𝑛)(𝑙+𝑛)}−𝑎_(𝑘+𝑛)2𝑎_(𝑗+𝑛)2^𝑡−1𝐸_{(𝑗+𝑛)(𝑘+𝑛)}+()+⋯−1^𝑡𝑎_{(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)}. (14) Since (𝑎_𝑗1)^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,𝐷_𝑘𝑙(𝑥_𝑘𝑥_𝑙𝑥_𝑗)]∉nil(𝑆). Then 𝑧∉Δ. It contradicts 𝑧∈Δ. This proves our assertion.

Lemma 5. Let 𝑧=∑^𝑛−2_𝑖=−1𝑧_𝑖, where 𝑧_𝑖∈𝑆_𝑖. If 𝑧∈Δ, then 𝑧₀=0.

Proof. Assume that 𝑧₀≠0. Let 𝑧₀=∑^𝑛_𝑖,𝑗=1∑^𝑝𝑚_𝑞=1𝑎_𝑖𝑗𝑞𝑥_𝑖𝑦_𝑞𝐷_𝑗, 𝑎_𝑖𝑗𝑞∈𝔽, and 𝑙=min𝑖∣𝑎_𝑖𝑗𝜆≠0,𝑖,𝑗∈𝑌,𝑡=min𝑗∣𝑎_𝑖𝑗𝜆≠0,𝑖,𝑗∈𝑌. (15)
(i) Suppose that 𝑙≤𝑡. Let 𝑘=max𝑗∣𝑎_𝑙𝑗𝜆≠0,𝑗∈𝑌. (16) Then 𝑎_𝑙𝑘𝑞≠0. It is easy to see that 𝑡≤𝑘. Since 𝑙≤𝑡, we have 𝑙≤𝑘. Therefore, 𝑧₀=^𝑘𝑝_𝑗=𝑡^𝑚_𝑞=1𝑎_𝑙𝑗𝑞𝑥_𝑙𝑦_𝑞𝐷_𝑗+^𝑛𝑛_𝑖=𝑙+1^𝑝_𝑗=𝑡^𝑚_𝑞=1𝑎_𝑖𝑗𝑞𝑥_𝑖𝑦_𝑞𝐷_𝑗. (17) Assume that 𝑙=𝑘. It follows from 𝑡≤𝑘 that 𝑡≤𝑙. Then we have 𝑡=𝑙 which implies that 𝑧₀=^𝑝𝑚_𝑞=1𝑎_𝑙𝑙𝑞𝑥_𝑙𝑦_𝑞𝐷_𝑙+^𝑛𝑛_𝑖=𝑙+1^𝑝_𝑗=𝑡^𝑚_𝑞=1𝑎_𝑖𝑗𝑞𝑥_𝑖𝑦_𝑞𝐷_𝑗. (18) Therefore, 𝜌𝑧₀=𝑎_𝑙𝑙1𝐸_𝑙𝑙+^𝑛𝑛_{𝑖=𝑙+1𝑗=𝑡}𝑎_𝑖𝑗1𝐸_𝑖𝑗+𝑎_{(𝑙+𝑛)(𝑙+𝑛)2}𝐸_{(𝑙+𝑛)(𝑙+𝑛)}+^2𝑛_{𝑖=𝑙+1+𝑛}^2𝑛_{𝑗=𝑡+𝑛}𝑎_𝑖𝑗2𝐸_𝑖𝑗+⋯+𝑎_{(𝑙+𝑛(𝑝𝑚−1))(𝑙+𝑛(𝑝𝑚−1))𝑝𝑚}𝐸_{(𝑙+𝑛)(𝑙+𝑛)}+^𝑛𝑝𝑚_{𝑝𝑖=𝑙+1+𝑛𝑚−1}^𝑛𝑝𝑚_{𝑝𝑗=𝑡+𝑛𝑚−1}𝑎_{𝑖𝑗𝑝𝑚}𝐸_𝑖𝑗=𝐴₁𝐵₁𝐶₁⋱𝐴_𝑝𝑚𝐵_𝑝𝑚𝐶_{𝑝𝑚𝑛𝑝𝑚×𝑛𝑝𝑚}, (19) where 𝐴_𝑘=𝑎_{(𝑙+(𝑘−1)𝑛)(𝑙+(𝑘−1)𝑛)𝑞}𝐸_{(𝑙+(𝑘−1)𝑛)(𝑙+(𝑘−1)𝑛)} is an (𝑙+(𝑘−1)𝑛)×(𝑙+(𝑘−1)𝑛) matrix and 𝑞∈{1,…,𝑝^𝑚}. Since 𝑎_𝑙𝑙1≠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 that ℎ∈𝑌 and ℎ≠𝑙, 𝑘. Let 𝑑=[𝑧₀,𝑥_𝑘𝐷_𝑙]. By equality (2), we obtain 𝑑=^𝑝𝑚_𝑞=1𝑎_𝑙𝑘𝑞𝑥_𝑙𝑦_𝑞𝐷_𝑙+^𝑛_𝑖=𝑙+1𝑎_𝑖𝑘𝑞𝑥_𝑖𝑦_𝑞𝐷_𝑙−^𝑘_𝑗=𝑡𝑎_𝑙𝑗𝑞𝑥_𝑘𝑦_𝑞𝐷_𝑗. (20) Since 𝑙<𝑘, 𝜌(𝑑) also has the matrix form as 𝜌(𝑧₀), it follows from 𝑎_𝑙𝑘1≠0 that 𝐴₁ is not a nilpotent matrix. Then 𝜌(𝑑) is not nilpotent. So 𝑧∉nil(𝑆) and [𝑧,𝑥_𝑘𝐷_𝑙]∉nil(𝑆). It is a contradiction of 𝑧∈Δ.
(ii) Suppose that 𝑡<𝑙. Let 𝑘=max{𝑖∣𝑎_𝑖𝑡𝜆≠0} and 𝑑^′=[𝑧,𝑥_𝑡𝐷_𝑘]. 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 𝑥_𝑖𝑦^𝜆𝐷_𝑗∈nil(𝑆) for all 𝜆∈𝐺.
(iii) Suppose that 𝑖, 𝑗, 𝑘 are distinct elements of 𝑌; then 𝑎𝑥_𝑗𝑦^𝜆𝐷_𝑘+𝑏𝑥_𝑖𝑦^𝜂𝐷_𝑘∈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 span_𝔽({ad𝑧}∪{ad𝑆₍₁₎}) is a nilpotent linear transformation of 𝑆. Then ad𝑧+adℎ is nilpotent. So 𝑧+ℎ is ad-nilpotent.
(ii) To prove (ad𝑥_𝑖𝑦^𝜆𝐷_𝑗)^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 ad𝑥_𝑖𝑦^𝜆𝐷_𝑗²𝑥^𝑢𝑦^𝜂𝐷_𝑘=𝑥_𝑖𝑦^𝜆𝐷_𝑗,𝑥_𝑖𝑦^𝜆𝐷_𝑗,𝑥^𝑢𝑦^𝜂𝐷_𝑘=(−1)^{𝜏(𝑢,𝑗)}𝑥_𝑖𝑦^𝜆𝐷_𝑗,𝑥_𝑖𝑥^{𝑢−⟨𝑗⟩}𝑦^𝜆+𝜂𝐷_𝑘=0. (21) In the case of 𝑘=𝑖, we have ad𝑥_𝑖𝑦^𝜆𝐷_𝑗³𝑥^𝑢𝑦^𝜂𝐷_𝑘=𝑥_𝑖𝑦^𝜆𝐷_𝑗,𝑥_𝑖𝑦^𝜆𝐷_𝑗,𝑥_𝑖𝑦^𝜆𝐷_𝑗,𝑥^𝑢𝑦^𝜂𝐷_𝑖=𝑥_𝑖𝑦^𝜆𝐷_𝑗,𝑥_𝑖𝑦^𝜆𝐷_𝑗,(−1)^{𝜏(𝑢,𝑗)}𝑥_𝑖𝑥^{𝑢−⟨𝑗⟩}𝑦^𝜆𝐷_𝑖−𝑥^𝑢𝑦^𝜆+𝜂𝐷_𝑗=(−1)^{𝜏(𝑢,𝑗)}𝑥_𝑖𝑦^𝜆𝐷_𝑗,−𝑥_𝑖𝑥^{𝑢−⟨𝑗⟩}𝑦^𝜆𝐷_𝑗−𝑥_𝑖𝑥^{𝑢−⟨𝑗⟩}𝑦^2𝜆+𝜂𝐷_𝑗=0. (22)
For 𝑝>2 we obtain (ad𝑥_𝑖𝑦^𝜆𝐷_𝑗)^𝑝(𝑥^𝑢𝑦^𝜂𝐷_𝑘)=0. Therefore (ad𝑥_𝑖𝑦^𝜆𝐷_𝑗)^𝑝(𝑆)=0. This yields (ad𝑥_𝑖𝑦^𝜆𝐷_𝑗)^𝑝=0. Thus 𝑥_𝑖𝑦^𝜆𝐷_𝑗∈nil(𝑆).
(iii) According to (ii) and [𝑥_𝑗𝑦^𝜆𝐷_𝑘,𝑥_𝑖𝑦^𝜂𝐷_𝑘]=0, {ad𝑥_𝑗𝑦^𝜆𝐷_𝑘,ad𝑥_𝑖𝑦^𝜂𝐷_𝑘} is a weakly closed subset of nilpotent elements of 𝑝𝑙(𝑆). So 𝑎𝑥_𝑗𝑦^𝜆𝐷_𝑘+𝑏𝑥_𝑖𝑦^𝜂𝐷_𝑘∈nil(𝑆), where 𝑎, 𝑏∈𝔽.

Lemma 7. If  𝑖, 𝑗, 𝑘 are distinct elements of  𝑌, then 𝑥_𝑖𝑥_𝑗𝑦^𝜆𝐷_𝑘∈Δ for all 𝜆∈𝐺.

Proof. Suppose that 𝑙∈𝑌⧵{𝑖,𝑗,𝑘}. Then 𝑥_𝑖𝑥_𝑗𝑦^𝜆𝐷_𝑘∈𝑆₍₁₎⊆nil(𝑆). Let 𝑧=∑^𝑛−2_𝑖=−1𝑧_𝑖, where 𝑧_𝑖∈𝑆_𝑖. Assume that [𝑥_𝑖𝑥_𝑗𝑦^𝜆𝐷_𝑘,𝑧]=𝑓₀+𝑓₁, where 𝑓₀=[𝑥_𝑖𝑥_𝑗𝑦^𝜆𝐷_𝑘,𝑧₋₁]∈𝑆₀ and 𝑓₁∈𝑆₍₁₎. Let 𝑧₋₁=∑^𝑛_𝑙=1∑_𝜂∈𝐺𝑎_𝑙𝜂𝑦^𝜂𝐷_𝑙. Then 𝑓₀=𝑥_𝑖𝑥_𝑗𝑦^𝜆𝐷_𝑘,^𝑛_{𝑙=1𝜂∈𝐺}𝑎_𝑙𝜂𝑦^𝜂𝐷_𝑙=_𝜂∈𝐺𝑎_𝑖𝜂𝑥_𝑗𝑦^𝜆+𝜂𝐷_𝑘−𝑎_𝑗𝜂𝑥_𝑖𝑦^𝜆+𝜂𝐷_𝑘. (23) 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 𝑥_𝑖𝑥_𝑗𝑦^𝜆𝐷_𝑘∈Δ for all 𝜆∈𝐺.

Let 𝑄={𝑧∈nil(𝑆)∣ad𝑧(Δ)⊆Δ}.

Lemma 8. 𝑄=𝑆₍₁₎.

Proof. By the definition of Δ, 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 prove 𝑄⊆𝑆₍₁₎. Let 𝑧∈𝑄 and 𝑧=∑^𝑛−2_𝑖=−1𝑧_𝑖, where 𝑧_𝑖∈𝑆_𝑖. Assume that 𝑧₋₁=∑^𝑛_𝑙=1∑_𝜆∈𝐺𝑎_𝑙𝜆𝑦^𝜆𝐷_𝑙≠0, 𝑎_𝑙𝜆∈𝔽. Without loss of generality, we may suppose that 𝑎_𝑖≠0. Let 𝑑=𝑥_𝑖𝑥_𝑗𝑦^𝜂𝐷_𝑘, 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 𝑎_𝑖≠0, we have ℎ₀=∑_𝜆∈𝐺(𝑎_𝑖𝜆𝑥_𝑗𝑦^𝜆+𝜂𝐷_𝑘−𝑎_𝑗𝜆𝑥_𝑖𝑦^𝜆+𝜂𝐷_𝑘)≠0. Lemma 5 implies that ℎ₀+ℎ₁∉Δ. It is a contradiction of 𝑧∈𝑄. Hence 𝑧₋₁=0.
Assume that 0≠𝑧₀=∑^𝑛_𝑖,𝑗=1∑^𝑝𝑚_𝑞=1𝑎_𝑖𝑗𝜆𝑥_𝑖𝑦_𝑞𝐷_𝑗, 𝑎_𝑖𝑗𝑞∈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=𝑥_𝑘𝑥_ℎ𝐷_𝑙. 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 𝑔₁=^𝑝𝑚_𝑞=1𝑎_𝑙𝑘𝑞𝑥_𝑙𝑥_ℎ𝑦_𝑞𝐷_𝑙−^𝑛_𝑖=𝑙+1𝑎_𝑖ℎ𝑞𝑥_𝑖𝑥_𝑘𝑦_𝑞𝐷_𝑙−^𝑘_𝑗=𝑡𝑎_𝑙𝑗𝑞𝑥_𝑘𝑥_ℎ𝑦_𝑞𝐷_𝑗. (24) If ℎ<𝑡, then 𝑎_𝑖ℎ𝑞=0 in the above equality, where 𝑖∈𝑌⧵{1,…,𝑙−1}. Thus 𝐷_ℎ,𝑔₁=−^𝑝𝑚_𝑞=1𝑎_𝑙𝑘𝑞𝑥_𝑙𝑦_𝑞𝐷_𝑙+^𝑛_𝑖=𝑙+1𝑎_𝑖ℎ𝑞𝑥_𝑖𝑦_𝑞𝐷_𝑙+𝑎_ℎℎ𝑞𝑥_𝑘𝑦_𝑞𝐷_𝑙−𝑎_𝑙𝑗𝑞𝑥_𝑘𝑦_𝑞𝐷_𝑗. (25) By equality (12), the matrix 𝜌([𝐷_ℎ,𝑔₁]) has the matrix form as in Lemma 5. Since 𝑎_𝑙𝑘𝑞≠0, 𝐴₁ is not a nilpotent matrix. It implies that 𝜌([𝐷_ℎ,𝑔₁]) is not nilpotent. Hence [𝐷_ℎ,𝑔₁]∉nil(𝑆). Lemma 3 shows that [𝐷_ℎ,𝑔₁+𝑔₂]∉nil(𝑆); that is, [𝐷_ℎ,𝑔₁+𝑔₂]∉Δ. It contradicts 𝑧∈𝑄. Thus 𝑧₀=0. Therefore, 𝑧∈𝑆₍₁₎ and 𝑄⊆𝑆₍₁₎.

According to the fact that Δ and 𝑄 are invariant subspaces under the automorphisms of 𝑆 and Lemma 8, 𝑆₍₁₎ is also invariant under the automorphisms of 𝑆. Since 𝑆₍₀₎=𝑥∈𝑆∣𝑥,𝑆₍₁₎⊆𝑆₍₁₎,𝑆_(𝑖)=𝑥∈𝑆_𝑖−1∣[𝑥,𝑆]⊆𝑆_(𝑖−1),𝑖≥1, (26) we may easily obtain the following theorem.

Theorem 9. The natural filtration of 𝑆 is invariant under the automorphisms of 𝑆.

Let 𝔖_𝑖=𝑆_(𝑖)/𝑆_(𝑖+1) for −1≤𝑖≤𝑛−2. Then 𝔖_𝑖 is a ℤ-graded space. Suppose that 𝔖∶=⊕^𝑛−2_𝑖=−1𝔖_𝑖; then 𝔖 is also a ℤ-graded space. Let 𝑥+𝑆_(𝑖+1)∈𝔖_𝑖 and 𝑦+𝑆_(𝑗+1)∈𝔖_𝑗. Define 𝑥+𝑆_(𝑖+1),𝑦+𝑆_(𝑗+1)∶=𝑥,𝑦+𝑆_{(𝑖+𝑗+1)}. (27) 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. Let 𝜙∶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 of 𝜙 shows that 𝜙𝑆_𝑖=𝑥+𝑆_(𝑖+1)∣𝑥∈𝑆_𝑖=𝑥+𝑆_(𝑖+1)∣𝑥∈𝑆_(𝑖)=𝑆_(𝑖)/𝑆_(𝑖+1)=𝔖_𝑖,𝑖≥−1. (28)

Suppose that 𝑚, 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 that 𝑆≅𝑆^′ and 𝜎 is an isomorphism from 𝑆 to 𝑆^′; then 𝜎(𝑆_(𝑖))=𝑆^′_(𝑖) for all 𝑖≥−1.

Proof. It is clear that 𝜎(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 that 𝑆≅𝑆^′ and 𝜎 is an isomorphism from 𝑆 to 𝑆^′; then 𝜎 induces an isomorphism 𝜎 from 𝔖 to 𝔖^′ such that 𝜎(𝔖_𝑖)=𝔖^′_𝑖 for all 𝑖≥−1.

Proof. Define a linear map 𝜎∶𝔖→𝔖^′ such that 𝜎𝑥+𝑆_(𝑖+1)=𝜎(𝑥)+𝑆^′_()𝑖+1, (29) where 𝑥+𝑆_(𝑖+1)∈𝔖_𝑖. Using Proposition 11, the definition of 𝜎 is reasonable and 𝜎𝑥+𝑆_(𝑖+1),𝑦+𝑆_(𝑗+1)=𝜎𝑥,𝑦+𝑆^′_𝑖+𝑗+1=𝜎(𝑥)+𝑆^′_()𝑖+1,𝜎𝑦+𝑆^′_𝑗+1=𝜎𝑥+𝑆^′_()𝑖+1,𝜎𝑦+𝑆^′_𝑗+1. (30) Thus 𝜎 is a homomorphism from 𝔖 to 𝔖^′. Clearly, 𝜎(𝔖_𝑖)=𝔖^′_𝑖 for all 𝑖≥−1. It follows that 𝜎 is an epimorphism.
Suppose that 𝑦∈ker𝜎; then 𝑦∈𝔖. So we may suppose that 𝑦=∑^𝑛−2_𝑖=−1𝑦_𝑖 and 𝑦_𝑖∈𝔖_𝑖. Since 𝔖_𝑖=𝑆_(𝑖)/𝑆_(𝑖+1), let 𝑦_𝑖=𝑧_𝑖+𝑆_(𝑖+1), where 𝑧_𝑖∈𝑆_(𝑖). Hence 𝜎(𝑦_𝑖)=𝜎(𝑧_𝑖)+𝑆^′_(𝑖+1). It follows from 𝜎(𝑦)=0 that ∑^𝑛−2_𝑖=−1𝜎(𝑦_𝑖)=0. Thus 𝜎(𝑦_𝑖)=0; that is, 𝜎(𝑧_𝑖)+𝑆^′_(𝑖+1)=0. It follows that 𝜎(𝑧_𝑖)∈𝑆^′_(𝑖+1). By Proposition 11, we have 𝑧_𝑖∈𝜎⁻¹(𝑆^′_(𝑖+1))=𝑆_(𝑖+1). Then 𝑦_𝑖=𝑧_𝑖+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(𝔖_𝑖)=𝔖^′_𝑖 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 that 𝜙∶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 𝜙𝑆_𝑖=𝔖_𝑖,𝜙^′𝑆^′_𝑖=𝔖^′_𝑖,𝜎𝔖_𝑖=𝔖^′_𝑖 (31) for −1≤𝑖≤𝑛−2. Let 𝜓=(𝜙^′)⁻¹𝜎𝜙. Then 𝜓𝑆_𝑖=𝜙^′−1𝜎𝜙𝑆_𝑖=𝜙^′−1𝜎𝔖_𝑖=𝜙^′−1𝔖^′_𝑖=𝑆^′_𝑖. (32) In particular, 𝜓(𝑆₋₁)=𝑆^′₋₁. It follows from dim𝑆₋₁=dim𝑆^′₋₁ that 𝑛𝑝^𝑚=𝑛^′𝑝^𝑚′. By virtue of the definition of 𝑆_𝑖, we have 𝑆₀=span_𝔽𝐷_𝑖𝑗𝑥_𝑘𝑥_𝑙𝑦^𝜆∈𝑆∣𝑖,𝑗,𝑘,𝑙∈𝑌,𝜆∈𝐺. (33) Thus dim𝑆₀=(𝑛²−1)𝑝^𝑚. Similarly, dim𝑆^′₀=(𝑛^′2−1)𝑝^𝑚′. According to dim𝑆₀=dim𝑆^′₀ and 𝑛𝑝^𝑚=𝑛^′𝑝^𝑚′, we have 𝑛=𝑛^′. In conclusion, the proof is completed.

Acknowledgments

The authors thank Yang Jiang for the helpful comments and suggestions. They also give their special thanks to the referees for many helpful suggestions. This work was supported by the National Natural Science Foundation of China (Grant no. 11171055) and the Fundamental Research Funds for the Central Universities (no. 12SSXT139).