Infinite-Dimensional Modular Lie Superalgebra Ω

Abstract

All ad-nilpotent elements of the infinite-dimensional Lie superalgebra Ω 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.

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 Section 2, 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,…,𝑧_𝑚}∈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=∑^𝑚_𝑖=1𝜆_𝑖𝑧_𝑖and𝑦^𝜆=𝑦^𝜆1₁⋯𝑦^𝜆𝑚_𝑚, where0≤𝜆_𝑖<𝑝.

Given𝑛∈ℕand𝑟=2𝑛+2, we put𝑀={1,…,𝑟−1}. Let𝜇₁,…,𝜇_𝑟−1∈𝔽and𝜇₁=0,𝜇_𝑗+𝜇_𝑛+𝑗=1,𝑗=2,…,𝑛+1. If𝑘_𝑖∈ℕ₀, then𝑘_𝑖can be uniquely expressed in𝑝-adic form𝑘_𝑖=∑^∞_𝑣=0𝜀_𝑣(𝑘_𝑖)𝑝^𝑣,where0≤𝜀_𝑣(𝑘_𝑖)<𝑝.We set 𝑥^𝑘𝑖_𝑖=∏^∞_𝑣=0𝑥^{𝜀𝑣(𝑘𝑖)}_𝑖𝑣. We define a truncated polynomial algebra 𝐴=𝔽𝑥₁₀,𝑥₁₁,…,𝑥₂₀,𝑥₂₁…,𝑥_𝑟−1,0,𝑥_𝑟−1,1,…,𝑦₁,…,𝑦_𝑚 (1)

such that 𝑥^𝑝_𝑖𝑗𝑦=0,∀𝑖∈𝑀,𝑗=0,1,…,_𝑝𝑖=1,𝑖=1,…,𝑚. (2)

For𝑘_𝑖,𝑘^′_𝑖∈ℕ₀, it is easy to see that 𝑥^𝑘𝑖_𝑖𝑥^𝑘′𝑖_𝑖=𝑥^{𝑘𝑖+𝑘′𝑖}_𝑖≠0⟺𝜀_𝑣𝑘_𝑖+𝜀_𝑣𝑘^′_𝑖<𝑝,𝑣=0,1,…. (3)

Let𝑄={(𝑘₁,…,𝑘_𝑟−1)∣𝑘_𝑖∈ℕ₀,𝑖∈𝑀}. If𝑘=(𝑘₁,…,𝑘_𝑟−1)∈𝑄, we set𝑥^𝑘=𝑥^𝑘1₁⋯𝑥^𝑘𝑟−1_𝑟−1.

LetΛ(𝑞)be the Grassmann superalgebra over𝔽in𝑞variables𝜉_𝑟+1,…,𝜉_𝑟+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: Ω₀=𝐴_𝔽Λ𝑞₀,Ω₁=𝐴_𝔽Λ𝑞₁. (4)

For𝑓∈𝐴and𝑔∈Λ(𝑞), we abbreviate𝑓⊗𝑔to𝑓𝑔. Let 𝔹_𝑘=𝑖₁,𝑖₂,…,𝑖_𝑘∣𝑟+1≤𝑖₁<𝑖₂<⋯<𝑖_𝑘,𝔹𝑞=≤𝑟+𝑞_𝑞𝑘=0𝔹_𝑘, (5)

and where𝔹₀=Ø. Given𝑢=⟨𝑖₁,…,𝑖_𝑘⟩∈𝔹_𝑘, we set|𝑢|=𝑘,{𝑢}={𝑖₁,…,𝑖_𝑘}and𝜉^𝑢=𝜉_𝑖1⋯𝜉_𝑖𝑘(|Ø|=0,𝜉^Ø=1). Then{𝑥^𝑘𝑦^𝜆𝜉^𝑢∣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}.Set𝑒_𝑖=(𝛿_𝑖1,…,𝛿_𝑖𝑟−1)and𝑥_𝑖=𝑥¹_𝑖=𝑥_𝑖0for𝑖∈𝑀. Define𝑖̃=0, if𝑖∈𝑀₁, and𝑖̃=1, if𝑖∈𝑇. Let 𝑖^′=[𝑖]=𝑖+𝑛,2≤𝑖≤𝑛+1,𝑖−𝑛,𝑛+2≤𝑖≤𝑟−1,𝑖,𝑟+1≤𝑖≤𝑠,1,2≤𝑖≤𝑛+1,−1,𝑛+2≤𝑖≤𝑟−1,1,𝑟+1≤𝑖≤𝑠. (6)

Let𝐷₁,𝐷₂,…,𝐷_𝑠be the linear transformations ofΩsuch that 𝐷_𝑖𝑥^𝑘𝑦^𝜆𝜉^𝑢=𝑘^∗_𝑖𝑥^{𝑘−𝑒𝑖}𝑦^𝜆𝜉^𝑢𝑥,𝑖∈𝑀,_𝑘𝑦^𝜆⋅𝜕𝜉^𝑢𝜕𝜉_𝑖,𝑖∈𝑇, (7)

where𝑘^∗_𝑖is the first nonzero number of𝜀₀(𝑘_𝑖),𝜀₁(𝑘_𝑖),…,𝜀_𝑠𝑖(𝑘_𝑖). Then𝐷₁,𝐷₂,…,𝐷_𝑠are superderivations of the superalgebraΩand|𝐷_𝑖|=𝑖̃. Set 𝜕=𝐼−_𝑗∈𝑀1𝜇_𝑗𝑥_𝑗0𝜕𝜕𝑥_𝑗0−^𝑚_𝑗=1𝑧_𝑗𝑦_𝑗𝜕𝜕𝑦_𝑗−2⁻¹_𝑗∈𝑇𝜉_𝑗𝜕𝜕𝜉_𝑗, (8)

where𝐼 is the identity mapping ofΩ. Let𝑓∈Ωbe aℤ₂-homogeneous element and𝑔∈Ω; we define a bilinear operation inΩsuch that 𝑓,𝑔=𝐷₁𝑓𝜕𝑔−𝜕𝑓𝐷₁𝑔+_{𝑖∈𝑀1∪𝑇}[𝑖)𝑖̃](−1^|𝑓|𝐷_𝑖𝑓𝐷_𝑖′𝑔. (9)

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∈𝑋Ω_𝑗, where Ω_𝑗=span_𝔽𝑥^𝑘𝑦^𝜆𝜉^𝑢∣_𝑖∈𝑀1𝑘_𝑖+2𝑘₁+|𝑢|−2=𝑗. (10)

LetΩ_[𝑖]=⨁_𝑗≥𝑖Ω_𝑗for 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 ad_𝐿-nilpotent if there exists a𝑡∈ℕsuch that(ad𝑦)^𝑡(𝐿)=0.If𝑦∈𝐿is ad_𝐿-nilpotent, it is also called ad-nilpotent in brief. Let𝐺be a subset of𝐿.Putnil(𝐺)={𝑥∈𝐺∣𝑥isad_𝐿-nilpotent}, and Nil(𝐺) is the subalgebra of𝐿generated by nil(𝐺).

Let𝑎∈ℕ₀and𝑎=∑^∞_𝑣=0𝜀_𝑣(𝑎)𝑝^𝑣be the𝑝-adic expression of a, where0≤𝜀_𝑣(𝑎)<𝑝.Then pad(𝑎)=pad₀(𝑎),pad₁(𝑎),pad₂(𝑎),… (11) is called the𝑝-adic sequence of𝑎, wherepad_𝑣(𝑎)=𝜀_𝑣(𝑎)for all𝑣∈ℕ₀. For𝑘=(𝑘₁,𝑘₂,…,𝑘_𝑟−1)∈𝑄, we define the𝑝-adic matrix of𝑘to be pad(𝑘)=pad𝑘₁pad𝑘₂⋮pad𝑘_𝑟−1. (12) Sincepad(𝑘)is a(𝑟−1)×∞matrix with only finitely many nonzero elements, we can set ℎ𝑡(𝑘)=max𝑗∈ℕ∣∃𝑖∈𝑀,pad_𝑗𝑘_𝑖≠0. (13) If𝑧=∑_{𝑘,𝜆,𝑢}𝛼_{𝑘,𝜆,𝑢}𝑥^𝑘𝑦^𝜆𝜉^𝑢∈Ω is a nonzero element with𝛼_{𝑘,𝜆,𝑢}∈𝔽, then we may assume ℎ𝑡(𝑧)=maxℎ𝑡(𝑘)∣𝛼_{𝑘,𝜆,𝑢}≠0. (14) For𝑐,𝑑∈ℕ₀, we define ‖𝑘‖_𝑐,𝑑∶=∑^𝑟−1_𝑖=1∑^𝑑_𝑗=𝑐pad_𝑗(𝑘_𝑖)and‖𝑘‖_𝑑∶=‖𝑘‖_0,𝑑.Now for any𝑡∈ℕand𝑥^𝑘𝑦^𝜆𝜉^𝑢∈Ω, define 𝒮_𝑡𝑥^𝑘𝑦^𝜆𝜉^𝑢=‖𝑘‖_𝑡+2‖𝑘‖_1,𝑡+|𝑢|+pad₀𝑘₁. (15)

Lemma 1. Let𝑘,𝑘^′∈𝑄and𝑡∈ℕ₀. Then the following statements hold. (i)𝑥^𝑘𝑥^𝑘′≠0⇔pad(𝑘)+pad(𝑘^′)=pad(𝑘+𝑘^′). (ii) If𝑘^′_𝑖≠0, then‖𝑘^′−𝑒_𝑖‖_𝑡+2‖𝑘^′−𝑒_𝑖‖_1,𝑡≥‖𝑘^′‖_𝑡+2‖𝑘^′‖_1,𝑡−1, where𝑖∈𝑀₁.(iii) Let𝑥^𝑘𝑦^𝜆𝜉^𝑢∈Ω_[1]and𝑡≥ℎ𝑡(𝑥^𝑘𝑦^𝜆𝜉^𝑢). Then𝒮_𝑡(𝑥^𝑘𝑦^𝜆𝜉^𝑢)≥3.

Proof. (i) We see that pad(𝑘)+pad𝑘^′=pad𝑘₁pad𝑘₂⋮pad𝑘_𝑛+pad𝑘^′₁pad𝑘^′₂⋮pad𝑘^′_𝑛=pad𝑘₁+pad𝑘^′₁pad𝑘₂+pad𝑘^′₂⋮pad𝑘_𝑛+pad𝑘^′_𝑛. (16)
Note that𝑥^𝑘𝑥^𝑘′≠0⇔𝜀_𝑣(𝑘_𝑖)+𝜀_𝑣(𝑘^′_𝑖)<𝑝⇔pad_𝑣(𝑘_𝑖)+pad_𝑣(𝑘^′_𝑖)<𝑝⇔pad_𝑣(𝑘_𝑖)+pad_𝑣(𝑘^′_𝑖)≢0(mod𝑝),forall𝑖∈𝑀,𝑣∈ℕ₀. By the uniqueness of𝑝-adic expression, we have𝑥^𝑘𝑥^𝑘′≠0⇔pad_𝑣(𝑘_𝑖)+pad_𝑣(𝑘^′_𝑖)=pad_𝑣(𝑘_𝑖+𝑘^′_𝑖),forall𝑖∈𝑀,𝑣∈ℕ₀.Thus 𝑥^𝑘𝑥^𝑘′≠0⟺pad(𝑘)+pad𝑘^′=pad𝑘₁+pad𝑘^′₁pad𝑘₂+pad𝑘^′₂⋮pad𝑘_𝑛+pad𝑘^′_𝑛=pad𝑘₁+𝑘^′₁pad𝑘₂+𝑘^′₂⋮pad𝑘_𝑛+𝑘^′_𝑛=pad𝑘+𝑘^′, (17)
as desired.
(ii) Ifpad₀(𝑘^′_𝑖)≠0, thenpad(𝑘^′_𝑖−1)=(pad₀(𝑘^′_𝑖)−1,pad₁(𝑘^′_𝑖),…). We see that‖𝑘^′−𝑒_𝑖‖_𝑡=‖𝑘^′‖_𝑡−1,‖𝑘^′−𝑒_𝑖‖_1,𝑡=‖𝑘^′‖_1,𝑡. So (ii) holds.
Ifpad₀(𝑘^′_𝑖)=0andpad_𝑏(𝑘^′_𝑖)≠0for𝑏≥1, then we can assume that pad𝑘^′_𝑖=0,…,0,pad_𝑏𝑘^′_𝑖,pad_𝑏+1𝑘^′_𝑖,…. (18)
Hencepad(𝑘^′_𝑖−1)=(𝑝−1,…,𝑝−1,pad_𝑏(𝑘^′_𝑖),pad_𝑏+1(𝑘^′_𝑖),…).
(a) If𝑡<𝑏, then we get 𝑘^′−𝑒_𝑖𝑡+2𝑘^′−𝑒_𝑖1,𝑡=()𝑘𝑡+1𝑝−1+2𝑡𝑝−1>−1,_′𝑡+2𝑘^′_1,𝑡−1=−1. (19)
(b) If𝑡=𝑏and𝑝>3, then 𝑘^′−𝑒_𝑖𝑡+2𝑘^′−𝑒_𝑖1,𝑡()3=𝑡𝑝−1+2𝑏+1𝑝−1+3pad_𝑡𝑘^′_𝑖−1>3pad_𝑡𝑘^′_𝑖𝑘−1,_′𝑡+2𝑘^′_1,𝑡−1=3pad_𝑏𝑘^′_𝑖−1. (20)
(c) If𝑡>𝑏, then 𝑘^′−𝑒_𝑖𝑡+2𝑘^′−𝑒_𝑖1,𝑡=𝑘^′−𝑒_𝑖𝑏+2𝑘^′−𝑒_𝑖1,𝑏+3𝑘^′−𝑒_{𝑖𝑏+1,𝑡}≥𝑘^′_𝑏+2𝑘^′_1,𝑏−1+3𝑘^′−𝑒_{𝑖𝑏+1,𝑡}=𝑘^′_𝑡+2𝑘^′_1,𝑡−1. (21)
Thus (ii) holds.
(iii) By|𝑘|+2𝑘₁+|𝑢|=𝑖+2≥3, we have∑(pad₀(𝑘_𝑖)𝑝⁰+pad₁(𝑘_𝑖)𝑝¹+⋯)+2pad₀(𝑘₁)+|𝑢|≥3;that is,∑pad₀(𝑘_𝑖)+2pad₀(𝑘₁)+|𝑢|≥3.

Lemma 2. Let 𝑥^𝑘𝑦^𝜆𝜉^𝑢∈Ω,𝑥^𝑘′𝑦^𝜂𝜉^𝑣∈Ω_[1],𝑡≥max{1,ℎ𝑡(𝑥^𝑘′𝑦^𝜂𝜉^𝑣)},𝑖∈𝑀₁∪𝑇.Then the following statements hold.(i) If𝑥^𝑘′𝑦^𝜂𝜉^𝑣𝐷₁(𝑥^𝑘𝑦^𝜆𝜉^𝑢)≠0, then𝒮_𝑡(𝑥^𝑘′𝑦^𝜂𝜉^𝑣𝐷₁(𝑥^𝑘𝑦^𝜆𝜉^𝑢))≥𝒮_𝑡(𝑥^𝑘𝑦^𝜆𝜉^𝑢)+1.(ii) If𝑥^𝑘𝑦^𝜆𝜉^𝑢𝐷₁(𝑥^𝑘′𝑦^𝜂𝜉^𝑣)≠0, then𝒮_𝑡(𝑥^𝑘𝑦^𝜆𝜉^𝑢𝐷₁(𝑥^𝑘′𝑦^𝜂𝜉^𝑣))≥𝒮_𝑡(𝑥^𝑘𝑦^𝜆𝜉^𝑢)+1.(iii) If𝐷_𝑖(𝑥^𝑘′𝑦^𝜂𝜉^𝑣)𝐷_𝑖′(𝑥^𝑘𝑦^𝜆𝜉^𝑢)≠0, then𝒮_𝑡(𝐷_𝑖(𝑥^𝑘′𝑦^𝜂𝜉^𝑣)𝐷_𝑖′(𝑥^𝑘𝑦^𝜆𝜉^𝑢))≥𝒮_𝑡(𝑥^𝑘𝑦^𝜆𝜉^𝑢)+1.

Proof. (i) As 𝑥^𝑘′𝑦^𝜂𝜉^𝑣𝑘^∗₁𝑥^𝑘−𝑒1𝑦^𝜆𝜉𝑢=𝑘^∗₁𝑥^𝑘′𝑥^𝑘−𝑒1𝑦^𝜂+𝜆𝜉^𝑢𝜉^𝑣≠0, 𝑥^𝑘′𝑥^𝑘−𝑒1=𝑥^{𝑘′+𝑘−𝑒1}≠0.Then we have 𝒮_𝑡𝑥^𝑘′𝑦^𝜂𝜉^𝑣𝐷₁𝑥^𝑘𝑦^𝜆𝜉^𝑢=𝑘^′+𝑘−𝑒_1𝑡+2𝑘^′+𝑘−𝑒_11,𝑡+|𝑢|+|𝑣|+pad₀𝑘^′₁+𝑘₁−1. (22) By the equality above and Lemma 1, we getpad(𝑘^′+(𝑘−𝑒₁))=pad(𝑘^′)+pad(𝑘−𝑒₁).Thus pad₀𝑘^′₁+𝑘₁−1=pad₀𝑘^′₁+pad₀𝑘₁,𝑘−1_′+𝑘−𝑒_1𝑡=𝑘^′_𝑡+𝑘−𝑒_1𝑡,𝑘^′+𝑘−𝑒_11,𝑡=𝑘^′_1,𝑡+𝑘−𝑒_11,𝑡. (23) Hence 𝒮_𝑡𝑥^𝑘′𝑦^𝜂𝜉^𝑣𝐷₁𝑥^𝑘𝑦^𝜆𝜉^𝑢=𝑘^′_𝑡+2𝑘^′_1,𝑡+|𝑣|+pad₀𝑘^′₁+𝑘−𝑒_1𝑡+2𝑘−𝑒_11,𝑡+|𝑢|+pad₀𝑘₁−1≥𝒮_𝑡𝑥^𝑘′𝑦^𝜂𝜉^𝑣+‖𝑘‖_𝑡+2‖𝑘‖_1,𝑡+|𝑢|+pad₀𝑘₁−1−1≥𝒮_𝑡𝑥^𝑘𝑦^𝜆𝜉^𝑢+1. (24)
(ii) The proof is completely analogous to (i).
(iii) For𝑖∈𝑀₁, by assumption of this lemma, we have𝑥^{𝑘′−𝑒𝑖}𝑥^{𝑘−𝑒𝑖′}≠0, which combined with Lemma 1 yield 𝒮_𝑡𝐷_𝑖𝑥^𝑘′𝑦^𝜂𝜉^𝑣𝐷_𝑖′𝑥^𝑘𝑦^𝜆𝜉^𝑢=𝒮_𝑡𝑘^∗_𝑖𝑘^∗_𝑖′𝑥^{(𝑘′+𝑒𝑖)+(𝑘−𝑒𝑖′)}𝑦^𝜆+𝜂𝜉^𝑢𝜉^𝑣=𝑘^′−𝑒_𝑖+𝑘−𝑒_𝑖′𝑡+2𝑘^′−𝑒_𝑖+𝑘−𝑒_{𝑖′1,𝑡}+|𝑢|+|𝑣|+pad₀𝑘₁+𝑘^′₁≥𝑘^′_𝑡+2𝑘^′_1,𝑡+|𝑣|−1+‖𝑘‖_𝑡+2‖𝑘‖_1,𝑡+|𝑢|−1+pad₀𝑘₁+pad₀𝑘^′₁≥𝒮_𝑡𝑥^𝑘𝑦^𝜆𝜉^𝑢+1. (25)
For𝑖∈𝑇, it is easily seen that𝜀_𝑣(𝑘^′−𝑒_𝑖)+𝜀_𝑣(𝑘−𝑒_𝑖′)=𝜀_𝑣(𝑘^′)+𝜀_𝑣(𝑘)<𝑝.Also by Lemma 1, we obtain 𝒮_𝑡𝐷_𝑖𝑥^𝑘′𝑦^𝜂𝜉^𝑣𝐷_𝑖′𝑥^𝑘𝑦^𝜆𝜉^𝑢=𝒮_𝑡𝑥^𝑘′+𝑘𝑦^𝜆+𝜂𝜉^{𝑢−⟨𝑖⟩}𝜉^{𝑣−⟨𝑖′⟩}=𝑘^′+𝑘_𝑡+2𝑘^′+𝑘_1,𝑡+||+||+𝑢−1𝑣−1pad₀𝑘^′₁+𝑘₁≥𝒮_𝑡𝑥^𝑘𝑦^𝜆𝜉^𝑢+3−2=𝒮_𝑡𝑥^𝑘𝑦^𝜆𝜉^𝑢+1. (26) Hence Lemma 2 holds.

Lemma 3. Let𝑥^𝑘𝑦^𝜆𝜉^𝑢∈Ω,𝑥^𝑘′𝑦^𝜂𝜉^𝑣∈Ω_[1]and𝑡≥max{1,ℎ𝑡(𝑥^𝑘′𝑦^𝜂𝜉^𝑣)}. Let𝑥^𝑘′′𝑦^𝜆′𝜉^𝑢′be a nonzero summand of[𝑥^𝑘′𝑦^𝜂𝜉^𝑣,𝑥^𝑘𝑦^𝜆𝜉^𝑢]. Then𝒮_𝑡(𝑥^𝑘′′𝑦^𝜆′𝜉^𝑢′)≥𝒮_𝑡(𝑥^𝑘𝑦^𝜆𝜉^𝑢)+1.

Proof. By a direct computation, we obtain that 𝑥^𝑘′𝑦^𝜂𝜉^𝑣,𝑥^𝑘𝑦^𝜆𝜉^𝑢=𝐷₁𝑥^𝑘′𝑦^𝜂𝜉^𝑣𝜕𝑥^𝑘𝑦^𝜆𝜉^𝑢−𝐷₁𝑥^𝑘𝑦^𝜆𝜉^𝑢𝜕𝑥^𝑘′𝑦^𝜂𝜉^𝑣+_{𝑖∈𝑀1∪𝑇}[𝑖)𝑖̃](−1^|𝑓|𝐷_𝑖𝑥^𝑘′𝑦^𝜂𝜉^𝑣𝐷_𝑖′𝑥^𝑘𝑦^𝜆𝜉^𝑢, (27)
which satisfies the conditions of Lemma 2.

Lemma 4. Ω_[1]⊆nil(Ω).

Proof. Given𝑡∈ℕ, put 𝑙_𝑡=()()𝑡+𝑟−1)(𝑡+1𝑝−1+2𝑟−1𝑝−1𝑝−1+𝑛+1. (28)
Clearly, we have𝒮_𝑡(𝑥^𝑘𝑦^𝜆𝜉^𝑢)<𝑙_𝑡for all standard basis element𝑥^𝑘𝑦^𝜆𝜉^𝑢ofΩ.
Let𝑡∈ℕsuch that𝑡≥ℎ𝑡(𝑧).For any𝑧=Σ_𝑘′,𝑣𝑐_𝑘′,𝑣𝑥^𝑘′𝑦^𝜂𝜉^𝑣∈Γ_[1]with0≠𝑐_𝑘′,𝑣∈𝔽, we have ad𝑧𝑥^𝑘𝑦^𝜆𝜉^𝑢=𝑧,𝑥^𝑘𝑦^𝜆𝜉^𝑢=𝐷₁(𝑧)𝜕𝑥^𝑘𝑦^𝜆𝜉^𝑢−𝜕(𝑧)𝐷₁𝑥^𝑘𝑦^𝜆𝜉^𝑢+_{𝑖∈𝑀1∪𝑇}[𝑖]𝐷_𝑖(𝑧)𝐷_𝑖′𝑥^𝑘𝑦^𝜆𝜉^𝑢. (29)
Note that𝑡≥max{1,ℎ𝑡(𝑥^𝑘′𝑦^𝜂𝜉^𝑣)}.By using Lemma 3 repeatedly we see that(ad𝑧)^𝑝𝑙𝑡(𝑥^𝑘𝑦^𝜆𝜉^𝑢)=0. HenceΩ_[1]⊆nil(Ω).

Lemma 5. (i) If𝑓=∑^𝑙_𝑡=𝑘𝑓_𝑡∈nil(Ω),where𝑓_𝑡∈Ω_𝑡, then𝑓_𝑡∈nil(Ω).
(ii) If𝑓=∑^𝑙_𝑡=−2𝑓_𝑡∈nil(Ω₀),then𝑓₋₂=0.
(iii) If𝑓=∑^𝑙_𝑡=−1𝑓_𝑡∈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[𝑦^𝜆,𝑥^𝑚𝑒1]=−𝐷₁(𝑥^(𝑚𝑒1)𝜕(𝑦^𝜆))=(𝜆−1)𝑘^∗₁𝑥^{(𝑚−1)𝑒1}𝑦^𝜆for all𝑚>0and𝜆≠1. By a direct computation, we obtain(ad𝑦^𝜆)^𝑚(𝑥^𝑚𝑒1)=(−1)^𝑚(1−𝜆)^𝑚𝑦^𝑚𝜆≠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∈𝑀1𝛾_𝑖𝑥_𝑖≠0, where𝛾_𝑖∈𝔽. Thus there exists some𝛾_𝑗≠0.By computation, we have Σ𝛾_𝑖𝑥_𝑖𝑦^𝜆,𝑥^𝑚𝑒𝑗[𝑖)𝑖̃=Σ](−1^|𝑓|𝐷_𝑖Σ𝛾_𝑖𝑥_𝑖𝑦^𝜆𝐷_𝑖′𝑥^𝑚𝑒𝑗=𝑗𝛾_𝑗𝑘^∗_𝑗′𝑥^{(𝑚−1)𝑒𝑗′}𝑦^𝜆. (30)
Similarly, we get(ad𝑓₋₁)^𝑚(𝑥^{𝑚𝑒𝑗′})=[𝑗]^𝑚(𝛾_𝑗)^𝑚(𝑘^∗_𝑗′)^𝑚𝑦^𝑚𝜆≠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]∩Ω₀and 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=∑^𝑙_𝑡=−2𝑓_𝑡∈nil(Ω₀).
Then by (ii) of this lemma,𝑓₋₂=0. Hence𝑓=𝑓₋₁+𝑓_[0]∈nil(Ω₀), where𝑓₋₁∈Ω₋₁∩Ω₀and𝑓₀∈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 that𝑥^𝑘𝑦^𝜆𝜉^𝑢is an arbitrary standard element ofΩ. Then the following statements hold.(i)𝑥²_𝑖𝑦^𝜆∈nil(Ω₀∩Ω₀). (ii) If[𝑖]=[𝑗]and𝑖≠𝑗, then𝑥_𝑖𝑥_𝑗𝑦^𝜆∈nil(Ω₀∩Ω₀).(iii) If[𝑖]≠[𝑗]and𝑗≠𝑖^′, then𝑥_𝑖𝑥_𝑗𝑦^𝜆∈nil(Ω₀∩Ω₀).(iv)𝑥_𝑖𝑥_𝑖′𝑦^𝜆∈nil(Ω₀∩Ω₀).

Proof. (i) By a direct computation, we get ad𝑥²_𝑖𝑦^𝜆𝑥^𝑘𝑦^𝜆𝜉^𝑢=𝑥²_𝑖𝑦^𝜆,𝑥^𝑘𝑦^𝜆𝜉^𝑢=−1−2𝜇_𝑖−𝜆𝑥²_𝑖𝑦^𝜆𝐷₁𝑥^𝑘𝑦^𝜆𝜉^𝑢+2[𝑖]𝑥_𝑖𝑦^𝜆𝐷_𝑖′𝑥^𝑘𝑦^𝜆𝜉^𝑢 (31)
and𝑥²_𝑖𝑦^𝜆𝐷₁∘𝑥_𝑖𝑦^𝜆𝐷_𝑖′=𝑥_𝑖𝑦^𝜆𝐷_𝑖′∘𝑥²_𝑖𝑦^𝜆𝐷₁. It follows from the binomial theorem that ad𝑥²_𝑖𝑦^𝜆𝑝𝑛𝑥^𝑘𝑦^𝜆𝜉^𝑢=−1−2𝜇_𝑖−𝜆𝑥²_𝑖𝑦^𝜆𝐷₁^𝑝𝑛+2[𝑖]𝑥_𝑖𝑦^𝜆𝐷_𝑖′^𝑝𝑛𝑥^𝑘𝑦^𝜆𝜉^𝑢=0. (32)
(ii) Also by a direct calculation, we have ad𝑥_𝑖𝑥_𝑗𝑦^𝜆𝑥^𝑘𝑦^𝜆𝜉^𝑢=𝑥_𝑖𝑥_𝑗𝑦^𝜆,𝑥^𝑘𝑦^𝜆𝜉^𝑢=−1−𝜇_𝑖−𝜇_𝑗−𝜆𝑥_𝑖𝑥_𝑗𝑦^𝜆𝐷₁𝑥^𝑘𝑦^𝜆𝜉^𝑢+[𝑖]𝑥_𝑗𝑦^𝜆𝐷_𝑖′𝑥^𝑘𝑦^𝜆𝜉^𝑢+𝑗𝑥_𝑖𝑦^𝜆𝐷_𝑗′𝑥^𝑘𝑦^𝜆𝜉^𝑢. (33)
Put 𝐴=−(1−𝜇_𝑖−𝜇_𝑗−𝜆)𝑥_𝑖𝑥_𝑗𝑦^𝜆𝐷₁,𝐵=[𝑖]𝑥_𝑗𝑦^𝜆𝐷_𝑖′,and𝐶=[𝑗]𝑥_𝑖𝑦^𝜆𝐷_𝑗′. Obviously, 𝑥_𝑖𝑥_𝑗𝑦^𝜆𝐷₁𝑥_𝑗𝑦^𝜆𝐷_𝑖′+𝑥_𝑖𝑦^𝜆𝐷_𝑗′=𝑥_𝑗𝑦^𝜆𝐷_𝑖′+𝑥_𝑖𝑦^𝜆𝐷_𝑗′𝑥_𝑖𝑥_𝑗𝑦^𝜆𝐷₁,𝑥_𝑗𝑦^𝜆𝐷_𝑖′𝑥_𝑖𝑦^𝜆𝐷_𝑗′=𝑥_𝑖𝑦^𝜆𝐷_𝑗′𝑥_𝑗𝑦^𝜆𝐷_𝑖′. (34)
Hence(ad𝑥_𝑖𝑥_𝑗𝑦^𝜆)^𝑝𝑛(𝑥^𝑘𝑦^𝜆𝜉^𝑢)=𝐴^𝑝𝑛+𝐵^𝑝𝑛+𝐶^𝑝𝑛=0.
(iii) The proof is completely analogous to (ii).
(iv) According to (i), we see that𝑥_𝑖𝑥_𝑖′𝑦^𝜆=(1/4)[𝑖,𝑥_𝑖𝑥_𝑖𝑦^𝜆,𝑥_𝑖′𝑥_𝑖′]∈nil(Ω₀∩Ω₀).

Lemma 7. Suppose that 𝑖,𝑗,𝑘∈𝑇 are different from each other, and𝛾,𝜒∈𝔽. Then
(i)𝑓=𝛾𝜉_𝑖𝜉_𝑗𝑦^𝜆+𝜒𝜉_𝑖𝜉_𝑘𝑦^𝜆∈nil(Ω)for𝛾²+𝜒²=0.
(ii)𝜉_𝑖𝜉_𝑗𝑦^𝜆∈Nil(Ω₀∩Ω₀).

Proof. (i) Set𝑥^𝑘𝑦^𝜆𝜉^𝑢be an arbitrary standard element ofΩ. Then ad𝑓𝑥^𝑘𝑦^𝜆𝜉^𝑢=𝑓,𝑥^𝑘𝑦^𝜆𝜉^𝑢=𝛾𝜉_𝑖𝜉_𝑗𝑦^𝜆,𝑥^𝑘𝑦^𝜆𝜉^𝑢+𝜒𝜉_𝑖𝜉_𝑘𝑦^𝜆,𝑥^𝑘𝑦^𝜆𝜉^𝑢=𝛾𝜆𝜉_𝑖𝜉_𝑗𝑦^𝜆𝐷₁𝑥^𝑘𝑦^𝜆𝜉^𝑢+𝛾𝜉_𝑗𝑦^𝜆𝐷_𝑖𝑥^𝑘𝑦^𝜆𝜉^𝑢+𝛾𝜉_𝑖𝑦^𝜆𝐷_𝑗𝑥^𝑘𝑦^𝜆𝜉^𝑢+𝜒𝜆𝜉_𝑖𝜉_𝑗𝑦^𝜆𝐷₁𝑥^𝑘𝑦^𝜆𝜉^𝑢+𝜒𝜉_𝑘𝑦^𝜆𝐷_𝑗𝑥^𝑘𝑦^𝜆𝜉^𝑢+𝜒𝜉_𝑖𝑦^𝜆𝐷_𝑘𝑥^𝑘𝑦^𝜆𝜉^𝑢. (35) Put 𝐴=𝛾𝜆𝜉_𝑖𝜉_𝑗𝑦^𝜆𝐷₁,𝐵=𝛾𝜉_𝑗𝑦^𝜆𝐷_𝑖,𝐶=𝛾𝜉_𝑖𝑦^𝜆𝐷_𝑗,𝐷=𝜒𝜆𝜉_𝑖𝜉_𝑗𝑦^𝜆𝐷₁,𝐸=𝜒𝜉_𝑘𝑦^𝜆𝐷_𝑖,𝐹=𝜒𝜉_𝑖𝑦^𝜆𝐷_𝑘. (36)Obviously, 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. (37)
Hence ad𝑓²=𝐴𝐸+𝐸𝐴+𝐵𝐶+𝐶𝐵+𝐵𝐹+𝐹𝐵+𝐶𝐸+𝐸𝐶+𝐷𝐸+𝐸𝐷+𝐸𝐹+𝐹𝐸. (38)
Noting that𝛾²+𝜒²=0, we obtain ad𝑓³=𝐵𝐶𝐵+𝐶𝐵𝐶+𝐸𝐹𝐸+𝐹𝐸𝐹. (39)
Similarly,(ad𝑓)⁴=0, and then𝑓=𝛾𝜉_𝑖𝜉_𝑗𝑦^𝜆+𝜒𝜉_𝑖𝜉_𝑘𝑦^𝜆∈nil(Ω).
(ii) Let𝑘∈𝑇⧵{𝑖,𝑗}.It follows from (i) that 𝑓₁=𝛾𝜉_𝑖𝜉_𝑗𝑦^𝜆+𝜉_𝑖𝜉_𝑘𝑦^𝜆∈NilΩ₀∩Ω₀,𝑓₂=𝜉_𝑖𝜉_𝑗+𝛾𝜉_𝑖𝜉_𝑘∈NilΩ₀∩Ω₀,where𝛾²=−1. (40)
Hence𝜉_𝑖𝜉_𝑗𝑦^𝜆=−(1/2)(𝛾𝑓₁−𝑓₂)∈Nil(Ω₀∩Ω₀).

Lemma 8. 𝑥_𝑖𝜉_𝑗𝑦^𝜆∈nil(Ω₀∩Ω₁)for𝑖∈𝑀₁and𝑗∈𝑇.

Proof. By a direct computation, we obtain ad𝑥_𝑖𝜉_𝑗𝑦^𝜆𝑥^𝑘𝑦^𝜆𝜉^𝑢=𝑥_𝑖𝜉_𝑗𝑦^𝜆,𝑥^𝑘𝑦^𝜆𝜉^𝑢=−12−𝜇_𝑖−𝜆𝑥_𝑖𝜉_𝑗𝑦^𝜆𝐷₁+[𝑖]𝜉_𝑗𝑦^𝜆𝐷_𝑖′−𝑥_𝑖𝑦^𝜆𝐷_𝑗×𝑥^𝑘𝑦^𝜆𝜉^𝑢. (41) Put𝐴=−(1/2−𝜇_𝑖−𝜆)𝑥_𝑖𝜉_𝑗𝑦^𝜆𝐷₁,𝐵=[𝑖]𝜉_𝑗𝑦^𝜆𝐷_𝑖′and𝐶=−𝑥_𝑖𝑦^𝜆𝐷_𝑗. Observing𝐴²=𝐵²=𝐶²=0and𝐴𝐵=𝐵𝐴=0, we see that ad𝑥_𝑖𝜉_𝑗𝑦^𝜆2=𝐴𝐶+𝐵𝐶+𝐶𝐴+𝐶𝐵,ad𝑥_𝑖𝜉_𝑗𝑦^𝜆𝑝𝑛=0,𝑛∈ℕ. (42)
Thus𝑥_𝑖𝜉_𝑗𝑦^𝜆∈nil(Ω₀∩Ω₁), as required.

Lemma 9. The following statements hold(i)Nil(Ω₀)=span_𝔽{𝑥_𝑖𝑥_𝑗𝑦^𝜆,𝑥_𝑖𝜉_𝑗𝑦^𝜆,𝜉_𝑖𝜉_𝑗𝑦^𝜆∣𝑖,𝑗∈𝑀₁∪𝑇}. (ii)For𝑞≥3,Nil(Ω₀∩Ω₀)=span_𝔽{𝑥_𝑖𝑥_𝑗𝑦^𝜆,𝜉_𝑖𝜉_𝑗𝑦^𝜆∣𝑖,𝑗∈𝑀₁∪𝑇[𝑖],=[𝑗]}.(iii)For𝑞=2,Nil(Ω₀∩Ω₀)=span_𝔽{𝑥_𝑖𝑥_𝑗𝑦^𝜆∣𝑖,𝑗∈𝑀₁}.

Proof. (i) Suppose that𝑓=𝛾₁𝑥₁𝑦^𝜆+∑_{𝑖,𝑗∈𝑀1∪𝑇}(𝑥_𝑖𝑥_𝑗𝑦^𝜆+𝑥_𝑖𝜉_𝑗𝑦^𝜆+𝜉_𝑖𝜉_𝑗𝑦^𝜆)is an arbitrary element of nil(Ω₀), where𝛾₁∈𝔽. If𝛾₁≠0, then[𝛾₁𝑥₁𝑦^𝜆,𝑦^𝜆]=𝛾₁(1−𝜆)𝑦^2𝜆. A direct calculation shows that(ad𝑓)^𝑚(𝑦^𝜆)=𝛾^𝑚₁(1−𝜆)(1−2𝜆)⋯(1−𝑚𝜆)𝑦^𝑚𝜆≠0. Thus𝑓is not ad-nilpotent, contradicting the nilpotency of𝑓. Hence𝛾₁=0andnil(Ω₀)⊆span_𝔽{𝑥_𝑖𝑥_𝑗𝑦^𝜆,𝑥_𝑖𝜉_𝑗𝑦^𝜆,𝜉_𝑖𝜉_𝑗𝑦^𝜆∣𝑖,𝑗∈𝑀₁∪𝑇}.Obviously,span_𝔽{𝑥_𝑖𝑥_𝑗𝑦^𝜆,𝑥_𝑖𝜉_𝑗𝑦^𝜆,𝜉_𝑖𝜉_𝑗𝑦^𝜆∣𝑖,𝑗∈𝑀₁∪𝑇}is a subalgebra ofΩ, which yields Nil(Ω₀)⊆span_𝔽{𝑥_𝑖𝑥_𝑗𝑦^𝜆,𝑥_𝑖𝜉_𝑗𝑦^𝜆,𝜉_𝑖𝜉_𝑗𝑦^𝜆∣𝑖,𝑗∈𝑀₁∪𝑇}.
Conversely, we havespan_𝔽{𝑥_𝑖𝑥_𝑗𝑦^𝜆,𝑥_𝑖𝜉_𝑗𝑦^𝜆,𝜉_𝑖𝜉_𝑗𝑦^𝜆∣𝑖,𝑗∈𝑀₁∪𝑇}⊆Nil(Ω₀)by virtue of Lemmas 6, 7, and 8. It follows that Nil(Ω₀)=span_𝔽{𝑥_𝑖𝑥_𝑗𝑦^𝜆,𝑥_𝑖𝜉_𝑗𝑦^𝜆,𝜉_𝑖𝜉_𝑗𝑦^𝜆∣𝑖,𝑗∈𝑀₁∪𝑇}.
(ii) By (i) of the lemma, we have NilΩ₀∩Ω₀⊆NilΩ₀∩Ω₀=span_𝔽𝑥_𝑖𝑥_𝑗𝑦^𝜆,𝜉_𝑖𝜉_𝑗𝑦^𝜆∣𝑖,𝑗∈𝑀₁∪𝑇,[𝑖]=𝑗. (43)
Conversely, the assertionspan_𝔽{𝑥_𝑖𝑥_𝑗𝑦^𝜆,𝜉_𝑖𝜉_𝑗𝑦^𝜆∣𝑖,𝑗∈𝑀₁∪𝑇,[𝑖]=[𝑗]}⊆Nil(Ω₀∩Ω₀)follows from Lemmas 6 and 7.
(iii) Suppose𝑓=𝛾𝜉_𝑟+1𝜉_𝑠𝑦^𝜆+∑_{𝑖,𝑗∈𝑀1}𝛾_𝑖𝑗𝑥_𝑖𝑥_𝑗𝑦^𝜆∈nil(Ω₀∩Ω₀), where𝛾,𝛾_𝑖𝑗∈𝔽. If𝛾≠0, thenad𝑓(𝜉_𝑠𝑦^𝜆)=[𝑓,𝜉_𝑠𝑦^𝜆]=𝛾𝜉_𝑟+1𝑦^2𝜆and(ad𝑓)²(𝜉_𝑠𝑦^𝜆)=𝛾²𝜉_𝑠𝑦^3𝜆. A direct calculation shows that(ad𝑓)^2𝑚(𝜉_𝑠𝑦^𝜆)=𝛾^2𝑚𝜉_𝑠𝑦^2𝑚𝜆≠0. It follows that𝑓is not ad-nilpotent. Thus𝛾=0. Then𝑓=∑_{𝑖,𝑗∈𝑀1}𝛾_𝑖𝑗𝑥_𝑖𝑥_𝑗𝑦^𝜆∈nil(Ω₀∩Ω₀)and nil(Ω₀∩Ω₀)⊆span_𝔽{𝑥_𝑖𝑥_𝑗𝑦^𝜆∣𝑖,𝑗∈𝑀₁}.Note thatspan_𝔽{𝑥_𝑖𝑥_𝑗𝑦^𝜆∣𝑖,𝑗∈𝑀₁}is a subalgebra ofΩ. Hence Nil(Ω₀∩Ω₀)⊆span_𝔽{𝑥_𝑖𝑥_𝑗𝑦^𝜆∣𝑖,𝑗∈𝑀₁}and (iii) holds.

Let𝜌be the corresponding representation with respect toΩ₀moduleΩ₋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: 𝑥₂𝑦^𝜆,𝑥₃𝑦^𝜆,…,𝑥_𝑟−1𝑦^𝜆,𝜉_𝑟+1𝑦^𝜆,…,𝜉_𝑠𝑦^𝜆. (44)

Denote by𝔤𝔩(2𝑛,𝑞)the general liner Lie superalgebra of(2𝑛+𝑞)×(2𝑛+𝑞)matrices over𝔽. Let𝑒_𝑖𝑗denote the(𝑠−2)×(𝑠−2)matrix whose(𝑖,𝑗)-entry is 1 and 0 elsewhere. Let𝐸_𝑛denote the identity matrix of size𝑛×𝑛. Put𝐺=^0𝐸𝑛_−𝐸𝑛0. Let𝔰𝔭(2𝑛,𝔽)be all the(2𝑛)×(2𝑛)matrices set filled with𝐴^𝑇𝐺+𝐺𝐴=0. Put (),𝐵𝒲=𝐴𝐵𝐶𝐷∈𝔤𝔩2𝑛,𝑞∣𝐴∈𝔰𝔭2𝑛,𝔽_𝑇𝐺+𝐶=0,𝐷=−𝐷^𝑇. (45)

Set ℒ=𝒲⊕𝔽𝐸_𝑠−2. (46)

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 𝑥_𝑖𝑥_𝑗,𝑥_𝑘𝑦^𝜆=[𝑖]𝛿_𝑖′𝑘𝑥_𝑗𝑦^𝜆+𝑗𝛿_𝑗′𝑘𝑥_𝑖𝑦^𝜆,𝑥₁,𝑥_𝑘𝑦^𝜆=1−𝜇_𝑘−𝜆𝑥_𝑘𝑦^𝜆. (47)
Hence𝜌(𝑥_𝑖𝑥_𝑗)=[𝑖]𝑒_𝑗,𝑖′+[𝑗]𝑒_𝑖𝑗′and𝜌(𝑥₁)=(1−𝜇_𝑘−𝜆)𝐸_𝑠−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∩Ω₀such that[𝑓,𝑧]∉nil(Ω₀∩Ω₀).

Proof. By Lemma 9, we can assume that𝑓=∑_𝑙∈𝑀1𝛾_𝑙𝑥²_𝑙𝑦^𝜆+∑_{𝑙,𝑡∈𝑀1,𝑙<𝑡}𝛽_𝑙𝑡𝑥_𝑙𝑥_𝑡𝑦^𝜆+∑_{𝑙,𝑡∈𝑇,𝑙<𝑡}𝜒_𝑙𝑡𝜉_𝑙𝜉_𝑡𝑦^𝜆, where𝛾_𝑙,𝛽_𝑙𝑡,𝜒_𝑙𝑡∈𝔽.
Suppose𝛾_𝑖≠0for some𝑖∈𝑀₁and𝑧=𝑥²_𝑖′.A direct calculation shows that 𝑓,𝑧=[𝑖]4𝛾_𝑖𝑥_𝑖𝑥_𝑖′𝑦^𝜆+ℎ, (48)
where every item ofℎdoes not contain𝑥_𝑖. Then (ad[𝑓,𝑧])(𝑥_𝑖′)=4𝛾_𝑖𝑦^𝜆𝑥_𝑖′. Thus(ad[𝑓,𝑧])^𝑛(𝑥_𝑖′)=4^𝑛𝛾^𝑛_𝑖𝑦^𝑛𝜆𝑥_𝑖′≠0,forall𝑛∈ℕ, which implies that[𝑓,𝑧]is not a nilpotent element.
If𝛾_𝑖=0for all𝑖∈𝑀₁, then we let𝛽_𝑖𝑗≠0for some𝑖,𝑗∈𝑀₁and𝑧=𝑥_𝑖′𝑥_𝑗′𝑦^𝜆. Also by computation, we have 𝑓,𝑧=𝑗𝛽_𝑖𝑗𝑥_𝑖𝑥_𝑖′𝑦^𝜆+ℎ, (49)
where every item of ℎ does not contain𝑥_𝑖.Similarly,(ad[𝑓,𝑧])^𝑛(𝑥_𝑖′)=[𝑗]^𝑛𝛽^𝑛_𝑖𝑗𝑦^𝑛𝜆𝑥_𝑖′≠0,forall𝑛∈ℕ, and then[𝑓,𝑧]is not nilpotent.
Hence our assertion follows.
If𝛾_𝑖=0and𝛽_𝑖𝑗=0for all𝑖,𝑗∈𝑀₁, then𝑓=∑_{𝑙,𝑡∈𝑇,𝑙<𝑡}𝜒_𝑙𝑡𝜉_𝑙𝜉_𝑡𝑦^𝜆∈span_𝔽{𝜉_𝑙𝜉_𝑡𝑦^𝜆∣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∈span_𝔽{𝜉_𝑖𝜉_𝑗𝑦^𝜆∣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. nilΩ=Ω_[1]⊕span_𝔽{𝑥_𝑖𝑥_𝑗𝑦^𝜆,𝑥_𝑖𝜉_𝑗𝑦^𝜆,𝜉_𝑖𝜉_𝑗𝑦^𝜆∣𝜆∈𝐻,𝑖,𝑗∈𝑀₁∪𝑇}.

Proof. According to Lemma 4, we need only to determine all ad-nilpotent elements inΩ₋₂⊕Ω₋₁⊕Ω₀. For any𝑛∈ℕ, a direct computation shows that ad𝑦^𝜆𝑛=(1−𝜆)𝐷₁^𝑛≠0,ad𝑥_𝑖𝑦^𝜆𝑛=−2⁻¹−𝜇_𝑖−𝜆𝑥_𝑖𝑦^𝜆𝐷₁+[𝑖]𝑦^𝜆𝐷_𝑖′^𝑛≠0,ad𝜉_𝑗𝑦^𝜆𝑛=−2⁻¹−𝜆𝜉_𝑗𝑦^𝜆𝐷₁+𝑦^𝜆𝐷_𝑗^𝑛≠0,ad𝑥₁𝑦^𝜆𝑛=𝑦^𝜆𝜕−(1−𝜆)𝑥₁𝑦^𝜆𝐷₁^𝑛≠0. (50)
Lemmas 6, 7, and 8 imply that𝑥_𝑖𝑥_𝑗𝑦^𝜆,𝑥_𝑖𝜉_𝑗𝑦^𝜆, and 𝜉_𝑖𝜉_𝑗𝑦^𝜆are ad-nilpotent for𝑖,𝑗∈𝑀₁∪𝑇. Hence our assertion holds.

4. Filtration and Intrinsic Property

Lemma 13. Ω_[0]∩Ω₀=Nor_Ω0(Nil(Ω₀))andΩ_[0]is invariant.

Proof. Firstly, we prove the inclusionΩ_[0]∩Ω₀⊆Nor_Ω0(Nil(Ω₀)). Lemmas 6–9 show that Nil(Ω₀∩Ω₀)⊲Ω₀∩Ω₀, which combined with (iv) and (v) of Lemma 5, yield NilΩ₀=NilΩ_[0]∩Ω₀=NilΩ₀∩Ω₀+Ω_[1]∩Ω₀,Ω₀∩Ω₀,NilΩ_[0]∩Ω₀⊆NilΩ₀∩Ω₀,Ω₀∩Ω₀,Ω_[1]∩Ω₀⊆Ω_[1]∩Ω₀. (51) Then Ω₀∩Ω₀,NilΩ₀=Ω₀∩Ω₀,NilΩ₀∩Ω₀+Ω₀∩Ω₀,Ω_[1]∩Ω₀⊆NilΩ₀∩Ω₀+Ω_[1]∩Ω₀=NilΩ₀, (52)
that is,Ω_[0]∩Ω₀⊆Nor_Ω0(Nil(Ω₀)).
Let us consider the converse inclusion. Suppose that𝑓=𝑓₋₂+𝑓_[−1]∈Nor_Ω0(Nil(Ω₀)), where𝑓₋₂∈Ω₋₂and𝑓_[−1]∈Ω_[−1]. If𝑓₋₂≠0, then[𝑓,𝑥₁𝜉_𝑠]=[𝑓₋₂,𝑥₁𝜉_𝑠]+[𝑓_[−1],𝑥₁𝜉_𝑠]=𝛾𝑓₋₂𝜉_𝑠+ℎ∉Nil(Ω₀), whereℎ∈Ω_[0]and𝛾∈𝔽, a contradiction. Consequently,𝑓₋₂=0.
Now suppose𝑓=𝑓₋₁+𝑓_[0], where𝑓₋₁=∑_𝑖∈𝑀1𝛾_𝑖𝑥_𝑖𝑦^𝜆∈Ω₋₁,𝛾_𝑖∈𝔽,𝑓_[0]∈Ω_[0].If𝑦₋₁≠0and𝛾_𝑗≠0for some𝑗∈𝑀₁, then we have[𝑓,𝑥²_𝑗′]=2[𝑗]𝛾_𝑗𝑥_𝑗′𝑦^𝜆+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]∩Ω₀={𝑓∈nil(Ω₀)∣[𝑓,Ω_[0]∩Ω₀]⊆nil(Ω₀)}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∩Ω₀such 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∈𝑇𝛾_𝑖𝜉_𝑖𝑦^𝜆∈Ω₋₁∩Ω₁,𝑓_[0]∈Ω_[0]∩Ω₁and𝛾_𝑖∈𝔽. Let𝛾_𝑗≠0for some one𝑗∈𝑇. Then[𝑓,𝜉_𝑗]=[𝑓₋₁,𝜉_𝑗]+[𝑓_[0],𝜉_𝑗]=𝛾_𝑗𝑦^𝜆+[𝑓_[0],𝜉_𝑗]∉Ω_[−1]∩Ω₀; that is,[𝑓,𝜉_𝑗]∉Ω_[0]∩Ω₀. This contradicts𝑓∈𝒜. Thus𝑓₋₁=0.
Let𝑓=𝑓₀+𝑓_[1]be an arbitrary element of𝒜, where𝑓₀=∑_{𝑖∈𝑀1,𝑗∈𝑇}𝛾_𝑖,𝑗𝑥_𝑖𝜉_𝑗𝑦^𝜆∈Ω₀∩Ω₁,𝑓_[1]∈Ω_[1]∩Ω₁and𝛾_𝑖,𝑗∈𝔽. If there is a𝛾_𝑖𝑡≠0, then[𝑓,𝜉_𝑡]=[𝑓₀,𝜉_𝑡]+[𝑓_[1],𝜉_𝑡]=𝛾_𝑖𝑡𝑥_𝑖𝑦^𝜆+[𝑓_[1],𝜉_𝑡]∉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]. Suppose𝑥^𝑘𝑦^𝜆𝜉^𝑢∈Ω_[0]∩Ω₁, where𝜉^𝑢=𝜉_𝑖𝜉^𝑣. Noting that𝑥₁𝑥^𝑘𝑦^𝜆𝜉^𝑣∈Ω_[1]∩Ω₀and𝑥^𝑘𝑦^𝜆𝜉^𝑢=−2[𝜉_𝑖,𝑥₁𝑥^𝑘𝑦^𝜆𝜉^𝑣], we obtain𝑥^𝑘𝑦^𝜆𝜉^𝑢∈[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∈Ω_𝑙such that[𝑓,Ω₋₁]=0, where𝑙∈ℕ₀. Let𝑡be the maximal exponent of𝑥₁of all monomial expressions occurring in𝑓. Then we may assume that 𝑓=_𝑘1=𝑡𝜒𝑥^𝑘𝑦^𝜆𝜉^𝑢+_𝑘1<𝑡𝜔𝑥^𝑘′𝑦^𝜂𝜉^𝑣, (53) where𝜒,𝜔∈𝔽. For any𝑗∈𝑀₁, we have 0=𝑓,𝑥_𝑗=_𝑘1=𝑡𝜒𝑗^′𝑘^∗_𝑗′𝑥^{𝑘−𝑒𝑗′}𝑦^𝜆𝜉^𝑢+ℎ, (54) whereℎis the sum of summand that the exponent of𝑥₁is less than𝑡. Since∑_𝑘1=𝑡𝜒[𝑗^′]𝑘^∗_𝑗′𝑥^{𝑘−𝑒𝑗′}𝑦^𝜆𝜉^𝑢is linear independence,𝑥^{𝑘−𝑒𝑗′}𝑦^𝜆𝜉^𝑢=0; that is, 𝑘_𝑗′=0. For any𝑗∈𝑇, we get 0=𝑓,𝜉_𝑗=_𝑘1=𝑡𝜒𝑥^𝑘𝑦^𝜆𝜉^{𝑢−⟨𝑗⟩}+ℎ, (55)
whereℎis the sum of summand that the exponent of𝑥₁is less than𝑡. Thus∑_𝑘1=𝑡𝜒𝑥^𝑘𝑦^𝜆𝜉^{𝑢−⟨𝑗⟩}=0; that is,𝑥^𝑘𝑦^𝜆𝜉^𝑢does not contain𝜉_𝑗and𝑓=𝑥^𝑡𝑒1𝑦^𝜆.
If𝑡=0, then𝑓∈Ω₋₂, contradicting𝑙≥0. Let𝑡>0. Then we can suppose 𝑓=𝛼𝑥^𝑡𝑒1𝑦^𝜆+_{𝑘1=𝑡−1}𝜒_{𝑘,𝜆,𝑢}𝑥^𝑘𝑦^𝜆𝜉^𝑢+_{𝑘1<𝑡−1}𝜔_{𝑘′,𝜂,𝑣}𝑥^𝑘′𝑦^𝜂𝜉^𝑣, (56) where𝛼,𝜒_{𝑘,𝜆,𝑢},𝜔_{𝑘′,𝜂,𝑣}∈𝔽.For𝑗∈𝑀₁, we have 0=𝑓,𝑥_𝑗=𝛼1−𝜇_𝑗𝑥^{(𝑡−1)𝑒1}𝑦^𝜆𝑥_𝑗+_{𝑘1=𝑡−1}𝜒_{𝑘,𝜆,𝑢}𝑗^′𝑘^∗_𝑗′𝑥^{𝑘−𝑒𝑗′}𝑦^𝜆𝜉^𝑢+ℎ, (57) whereℎis the sum of summand that the exponent of𝑥₁is less than𝑡−1. Thus 𝛼1−𝜇_𝑗𝑥^{(𝑡−1)𝑒1}𝑦^𝜆𝑥_𝑗+_{𝑘1=𝑡−1}𝜒_{𝑘,𝜆,𝑢}𝑗^′𝑘^∗_𝑗′𝑥^{𝑘−𝑒𝑗′}𝑦^𝜆𝜉^𝑢=0. (58) By 𝛼_𝑡(1−𝜇_𝑗)𝑥^{(𝑡−1)𝑒1}𝑦^𝜆𝑥_𝑗 contains𝑥_𝑗and ∑_{𝑘1=𝑡−1}𝜒_{𝑘,𝜆,𝑢}[𝑗^′]𝑘^∗_𝑗′𝑥^{𝑘−𝑒𝑗′}𝑦^𝜆𝜉^𝑢 dose not contain𝑥_𝑗, 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)=Ω^′₀and𝜑(Nil(Ω₀))=Nil(Ω^′₀). Hence𝜑(Nor_Ω0(Nil(Ω₀))=Nor_Ω′0(Nil(Ω^′₀)). By Lemmas 13 and 14, we obtain 𝜑Ω_[0]∩Ω₀=Ω^′_[0]∩Ω^′₀,𝜑Ω_[1]∩Ω₀=𝜑𝑓∈nilΩ_[0]∣𝑓,Ω_[0]∩Ω₀⊆nilΩ_[0]=𝑓∈nilΩ^′_[0]∣𝑓,Ω^′_[0]∩Ω^′₀⊆nilΩ^′₀=Ω^′_[1]∩Ω^′₀. (59) Now by virtue of Lemma 15, we get 𝜑Ω_[0]∩Ω₁=𝜑𝑓∈Ω_[−1]∩Ω₁∣𝑓,Ω₀⊆Ω_[−1]∩Ω₁=𝑓∈Ω^′_[]−1∩Ω^′₁∣𝑓,Ω^′₀⊆Ω^′_[]−1∩Ω^′₁=Ω^′_[0]∩Ω^′₁,𝜑Ω_[−1]∩Ω₁=𝜑Ω₁,Ω_[1]∩Ω₀=Ω^′₁,Ω_[1]∩Ω^′₀=Ω^′_[]−1∩Ω^′₁. (60) It follows that𝜑(Ω_[0])=Ω^′_[0]. BecauseΩ₀⊆Ω_[−1], we have 𝜑Ω_[−1]=𝜑Ω_[−1]∩Ω₀+Ω_[−1]∩Ω₁=𝜑Ω₀+Ω_[−1]∩Ω₁=Ω^′₀+Ω^′_[]−1∩Ω^′₁=Ω^′_[]−1. (61) 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 Theorem 17.

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 𝜑Ω𝑟,𝑚,𝑞_[−2]=Ω𝑟^′,𝑚^′,𝑞^′_[−2],𝜑Ω𝑟,𝑚,𝑞_[−1]=Ω𝑟^′,𝑚^′,𝑞^′_[−1]. (62) Then𝜑induces an isomorphism ofℤ₂-graded spaces Ω𝑟,𝑚,𝑞_[−2]Ω𝑟,𝑚,𝑞_[−1]⟶Ω𝑟^′,𝑚^′,𝑞^′_[−2]Ω𝑟^′,𝑚^′,𝑞^′_[−1]. (63) It is easy to see that Ω𝑟,𝑚,𝑞_[−2]Ω𝑟,𝑚,𝑞_[−1]≅Ω𝑟,𝑚,𝑞₋₂,Ω𝑟^′,𝑚^′,𝑞^′_[−2]Ω𝑟^′,𝑚^′,𝑞^′_[−1]≅Ω𝑟^′,𝑚^′,𝑞^′₋₂. (64)
We conclude that𝑚=𝑚^′by the dimension comparison.
Similarly, we also obtain an isomorphism ofℤ₂-graded spaces: Ω𝑟,𝑚,𝑞_[−1]Ω𝑟,𝑚,𝑞_[0]≅Ω𝑟^′,𝑚^′,𝑞^′_[−1]Ω𝑟^′,𝑚^′,𝑞^′_[0], (65)
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^′.

Acknowledgments

The authors are grateful to the referees for their many valuable comments and suggestions. This work is supported by National Natural Science Foundation of China (no. 11126129) and the PhD Start-up Foundation of Liaoning University of China (no. 2012002).