Decomposition of Automorphisms of Certain Solvable Subalgebra of Symplectic Lie Algebra over Commutative Rings

Abstract

Let 𝐶_𝑙+1(𝑅) be the 2(𝑙+1)×2(𝑙+1) matrix symplectic Lie algebra over a commutative ring 𝑅 with 2 invertible. Then 𝐭^(𝐶)_𝑙+1(𝑅)  =  {(^𝑚1𝑚2_{0−𝑚𝑇1}) ∣ 𝑚₁ is an 𝑙+1 upper triangular matrix, 𝑚^𝑇₂=𝑚₂,over𝑅} is the solvable subalgebra of 𝐶_𝑙+1(𝑅). In this paper, we give an explicit description of the automorphism group of 𝐭^(𝐶)_𝑙+1(𝑅).

1. Introduction

Classical Lie algebras occupy an important place in matrix algebras. Let 𝑅 be a commutative ring 𝑅 with the identity 1 and 𝑅^∗ the group of invertible elements in 𝑅. Let 𝑀_𝑛(𝑅) be the 𝑅-algebra of 𝑛 by 𝑛 matrices over 𝑅 that has a structure of a Lie algebra over 𝑅 with bracket operation [𝑥,𝑦]=𝑥𝑦−𝑦𝑥 for any 𝑥,𝑦∈𝑀_𝑛(𝑅). The symplectic Lie algebra 𝐶_𝑙+1(𝑅)=𝑋∣𝑋0𝐼_𝑙−𝐼_𝑙0+0𝐼_𝑙−𝐼_𝑙0𝑋^𝑇=0,𝑋∈𝑀_2𝑙+2(𝑅) (1.1) is one of classical Lie algebras, where 𝑇 denotes the matrix transpose. It is easy to show that the following subalgebra of 𝐶_𝑙+1(𝑅) such that 𝐭^(𝐶)_𝑙+1(𝑅)=𝑚₁𝑚₂0−𝑚^𝑇₁∣𝑚₁isan𝑙+1uppertriangularmatrix,𝑚^𝑇₂=𝑚₂ (1.2) is solvable.

Let 𝑒 be the identity matrix in 𝑀_𝑛(𝑅) and let 𝑒^(𝑛)_𝑖𝑗 denotes the matrix in 𝑀_𝑛(𝑅) all of whose entries are 0, except the (𝑖,𝑗)th entry which is 1. Let 𝛼_{𝑖,𝑖+𝑘}=𝑒^(𝑛)_{𝑖,𝑖+𝑘}−𝑒^(𝑛)_{𝑙+𝑖+𝑘+1,𝑙+𝑖+1}𝛾,𝑖=1,…,𝑙−𝑘+1,𝑘=0,1,…,𝑙,_{𝑖,𝑖+𝑘}=𝑒^(𝑛)_{𝑖,𝑙+𝑖+𝑘+1}+𝑒^(𝑛)_{𝑖+𝑘,𝑙+𝑖+1}𝛾,𝑖=1,…,𝑙−𝑘+1,𝑘=1,…,𝑙,_𝑖𝑖=𝑒^(𝑛)_{𝑖,𝑙+𝑖+1},𝑖=1,…,𝑙+1, (1.3) where 𝑛=2(𝑙+1), 𝑙≥1. For discussion latter, we rewrite 𝐭^(𝐶)_𝑙+1(𝑅) as 𝐭^(𝐶)_𝑙+1(𝑅)=^𝑙_𝑘=0^{𝑙−𝑘+1}_𝑖=1𝑅𝛼_{𝑖,𝑖+𝑘}+^𝑙_𝑘=0^{𝑙−𝑘+1}_𝑖=1𝑅𝛾_{𝑖,𝑖+𝑘}. (1.4)

Automorphisms of associative algebras have been explored in many articles [1–8]. Encouraged by Doković [9] and Cao's [10] papers which described the automorphism groups of Lie algebra consisting of all upper triangular 𝑛×𝑛 matrices of trace 0 over a connected commutative ring and a commutative ring with 𝑛 invertible, respectively, in this paper we use similar techniques to those in [11] to prove that any automorphism 𝜓 of 𝐭^(𝐶)_𝑙+1(𝑅) can be uniquely expressed as 𝜓=𝜃𝜆_𝐷, where 𝜃 and 𝜆_𝐷 are inner and diagonal automorphisms, respectively, for 𝑙≥1 and 𝑅 is a commutative ring with 2 invertible. We also give an explicit description of the remaining case 𝑙=0.

Theorem 1.1. For any automorphism 𝜓 of 𝐭^(𝐶)_𝑙+1(𝑅)(𝑙≥1) there are unique inner and diagonal automorphisms, 𝜃 and 𝜆_𝐷, respectively, of 𝐭^(𝐶)_𝑙+1(𝑅) such that 𝜓=𝜃𝜆_𝐷.

Theorem 1.2. Let ℐ and 𝒟 be the inner and diagonal automorphism groups, respectively. Then Aut(𝐭^(𝐶)_𝑙+1(𝑅))=ℐ⋉𝒟, where   Aut(𝐭^(𝐶)_𝑙+1(𝑅)) denotes the automorphism group of 𝐭^(𝐶)_𝑙+1(𝑅).

2. Preliminaries

Let 𝑃_𝑛=𝛼_{𝑖,𝑖+𝑘},𝑊∣𝑖=1,…,𝑙−𝑘+1,𝑘=0,1,…,𝑙_𝑛=𝛾_{𝑖,𝑖+𝑘}∣𝑖=1,…,𝑙−𝑘+1,𝑘=0,1,…,𝑙. (2.1) Then the set 𝑃_𝑛∪𝑊_𝑛 is a basis of 𝐭^(𝐶)_𝑙+1(𝑅).

Lemma 2.1. Let 𝐻_𝑛 be the set generated by the set {𝛼_𝑗𝑗,𝛼_𝑖,𝑖+1,𝛾_{𝑙+1,𝑙+1}∣1≤𝑗≤𝑙+1,1≤𝑖≤𝑙}, where 𝑛=2(𝑙+1). Then 𝐻_𝑛=𝐭^(𝐶)_𝑙+1(𝑅).

Proof. We only need to show that 𝐭^(𝐶)_𝑙+1(𝑅)⊆𝐻_𝑛. It is obvious that 𝛼_{𝑖,𝑖+𝑘}∈𝐻_𝑛, when 𝑘=0,1. When 𝑘=2, we have 𝛼_𝑖,𝑖+2=𝑒^(𝑛)_𝑖,𝑖+2−𝑒^(𝑛)_{𝑙+𝑖+3,𝑙+𝑖+1}=𝑒^(𝑛)_𝑖,𝑖+1−𝑒^(𝑛)_{𝑙+𝑖+2,𝑙+𝑖+1},𝑒^(𝑛)_{𝑖+1,𝑖+2}−𝑒^(𝑛)_{𝑙+𝑖+3,𝑙+𝑖+2}=𝛼_𝑖,𝑖+1,𝛼_{𝑖+1,𝑖+2}∈𝐻_𝑛. (2.2) Assume that 𝛼_{𝑖,𝑖+𝑘−1}∈𝐻_𝑛, then 𝛼_{𝑖,𝑖+𝑘}=[𝛼_{𝑖,𝑖+𝑘−1},𝛼_{𝑖+𝑘−1,𝑖+𝑘}]∈𝐻_𝑛, that is, 𝑃_𝑛⊆𝐻_𝑛. Since 𝛾_{𝑙+1,𝑙+1}∈𝐻_𝑛, for any 𝛾_{𝑙−𝑘+1,𝑙−𝑘+2}∈𝑊_𝑛, when 𝑘=1, 𝛾_𝑙,𝑙+1=𝑒^(𝑛)_𝑙,2𝑙+2+𝑒^(𝑛)_{𝑙+1,2𝑙+1}=𝑒^(𝑛)_𝑙,𝑙+1−𝑒^(𝑛)_{2𝑙+2,2𝑙+1},𝑒^(𝑛)_{𝑙+1,2𝑙+2}=𝛼_𝑙,𝑙+1,𝛾_{𝑙+1,𝑙+1}∈𝐻_𝑛. (2.3) Assume that when 𝑘=𝑚−1, 𝛾_{𝑙−𝑚+2,𝑙−𝑚+3}∈𝐻_𝑛, then when 𝑘=𝑚, 𝛾_{𝑙−𝑚+1,𝑙−𝑚+2}=[𝛼_{𝑙−𝑚+1,𝑙−𝑚+3},𝛾_{𝑙−𝑚+2,𝑙−𝑚+3}]∈𝐻_𝑛, that is, 𝛾_𝑖,𝑖+1∈𝐻_𝑛, 𝑖=1,…,𝑙. For any 𝛾_{𝑖,𝑖+𝑘}∈𝐻_𝑛(𝑘≥1), when 𝑘=1, 𝛾_𝑖,𝑖+1∈𝐻_𝑛. When 𝑘≥2, 𝛾_{𝑖,𝑖+𝑘}=[𝛼_{𝑖,𝑖+𝑘−1},𝛾_{𝑖+𝑘−1,𝑖+𝑘}]∈𝐻_𝑛. Since 2𝛾_𝑖𝑖=[𝛼_𝑖,𝑖+1,𝛾_𝑖,𝑖+1]∈𝐻_𝑛, and 2 is invertible, we have 𝛾_𝑖𝑖∈𝐻_𝑛 for 1≤𝑖≤𝑙. Thus 𝑊_𝑛⊆𝐻_𝑛. Because 𝑃_𝑛∪𝑊_𝑛 is a basis of 𝐭^(𝐶)_𝑙+1(𝑅), we obtain 𝐭^(𝐶)_𝑙+1(𝑅)⊆𝐻_𝑛.

Now, denote 𝐭^(𝐶)_𝑙+1(𝑅) by 𝐧₀^(𝐶). Let 𝐧₁^(𝐶)=[𝐧₀^(𝐶),𝐧₀^(𝐶)], 𝐧₂^(𝐶)=[𝐧₁^(𝐶),𝐧₁^(𝐶)], 𝐧_𝑗^(𝐶)=[𝐧₁^(𝐶),𝐧^(𝐶)_𝑗−1], 𝑗=3,…,2𝑙+1. It is not difficult to know 𝐧_𝑗^(𝐶)=^𝑙_𝑘=𝑗^{𝑙−𝑘+1}_𝑖=1𝑅𝛼_{𝑖,𝑖+𝑘}+^𝑙_𝑘=𝑗^𝑙+1_{[]𝑖=𝑙+2−(𝑘+1)/2}𝑅𝛾_{2𝑙−𝑘−𝑖+3,𝑖}+^2𝑙+1_𝑘=𝑙+1^{𝑙+1−[𝑘/2]}_𝑖=1𝑅𝛾_{𝑖,2𝑙−𝑘−𝑖+3}𝐧,1≤𝑗≤𝑙,_𝑗(𝐶)=^2𝑙+1_𝑘=𝑗^{𝑙+1−[𝑘/2]}_𝑖=1𝑅𝛾_{𝑖,2𝑙−𝑘−𝑖+3}(),𝐧,𝑙+1≤𝑗≤2𝑙+1𝑙≥2_𝑗(𝐶)=0,2𝑙+2≤𝑗. (2.4)

It is easy to check that [𝐧_𝑚^(𝐶),𝐧_𝑙^(𝐶)]⊆𝐧^(𝐶)_𝑚+𝑙 for 𝑚+𝑙≤2𝑙+1 or [𝐧_𝑚^(𝐶),𝐧_𝑙^(𝐶)]=0 for 𝑚+𝑙≥2𝑙+2. For any 𝜓∈  Aut(𝐧₀^(𝐶)), we have 𝜓(𝐧₁^(𝐶))=[𝜓(𝐧₀^(𝐶)),𝜓(𝐧₀^(𝐶))]=[𝐧₀^(𝐶),𝐧₀^(𝐶)]=𝐧₁^(𝐶) and 𝜓(𝐧_𝑗^(𝐶))=𝐧_𝑗^(𝐶), 𝑗=2,…,2𝑙+1. Therefore, 𝜓(𝐧^(𝐶)_𝑗−1⧵𝐧_𝑗^(𝐶))=𝐧^(𝐶)_𝑗−1⧵𝐧_𝑗^(𝐶), 𝑗=1,…,2𝑙+1. Note that if 𝛾_𝑖𝑗∈𝑊_𝑛, then 𝛾_𝑖𝑗∈𝐧^(𝐶)_{2𝑙−𝑖−𝑗+3}⧵𝐧^(𝐶)_{2𝑙−𝑖−𝑗+4}.

For any maximal ideal 𝑀 of 𝑅, 𝑅=𝑅/𝑀 is a field. The natural homomorphism 𝜋∶𝑅→𝑅 induces a homomorphism 𝜓_𝑀∶𝐭^(𝐶)_𝑙+1(𝑅)→𝐭^(𝐶)_𝑙+1(𝑅) which is surjective. So every automorphism 𝜓 of 𝐭^(𝐶)_𝑙+1(𝑅) may induce an automorphism 𝜓 of 𝐭^(𝐶)_𝑙+1(𝑅). Using this fact and that 𝐧^(𝐶)_2𝑙+1=𝑅𝛾₁₁ (for 𝑙≥1), we have that 𝜓(𝛾₁₁)=𝑐₁₁𝛾₁₁, where 𝑐₁₁∈𝑅^∗. Otherwise, 𝑐₁₁ should be contained in a maximal ideal 𝑀 of 𝑅, then 𝜓(𝛾₁₁)=0 on 𝐭^(𝐶)_𝑙+1(𝑅), where 𝛾₁₁ is the image of 𝛾₁₁ in 𝐭^(𝐶)_𝑙+1(𝑅), which is impossible.

Lemma 2.2. Let 𝜓 be in Aut(𝐧₀^(𝐶)). If 𝜓(𝛼_𝑗𝑗), 𝜓(𝛼_𝑗,𝑗+1) and 𝜓(𝛾_{𝑙+1,𝑙+1}) are expressed, respectively, as 𝜓𝛼_𝑗𝑗=^𝑙+1_𝑖=1𝑎^(𝑗)_𝑖𝑖𝛼_𝑖𝑖mod𝐧₁^(𝐶)𝜓𝛼,𝑗=1,…,𝑙+1,_𝑗,𝑗+1=^𝑙_𝑖=1𝑎^(𝑗)_𝑖,𝑖+1𝛼_𝑖,𝑖+1+𝑐̃^(𝑗)_{𝑙+1,𝑙+1}𝛾_{𝑙+1,𝑙+1}mod𝐧₂^(𝐶)𝜓𝛾,𝑗=1,…,𝑙,_{𝑙+1,𝑙+1}=^𝑙_𝑖=1𝑎^(𝑙+1)_𝑖,𝑖+1𝛼_𝑖,𝑖+1+𝑐̂^(𝑙+1)_{𝑙+1,𝑙+1}𝛾_{𝑙+1,𝑙+1}mod𝐧₂^(𝐶), (2.5)(2.6)(2.7) then the following matrices are invertible.(i)𝐴=(𝑎_𝑗𝑖)_{(𝑙+1)×(𝑙+1)}, where 𝑎_𝑗𝑖=𝑎^(𝑗)_𝑖𝑖, 𝑗=1,…,𝑙+1, 𝑖=1,…,𝑙+1;(ii)𝐵=(𝑏_𝑗𝑖)_{(𝑙+1)×(𝑙+1)}, where 𝑏_𝑗𝑖=𝑎^(𝑗)_𝑖,𝑖+1, 𝑗=1,…,𝑙, 𝑖=1,…,𝑙, 𝑏_𝑗,𝑙+1=𝑐̃^(𝑗)_{𝑙+1,𝑙+1}, 𝑗=1,…,𝑙, 𝑏_𝑙+1,𝑖=𝑎^(𝑙+1)_𝑖,𝑖+1, 𝑖=1,…,𝑙 and 𝑏_{𝑙+1,𝑙+1}=𝑐̂^(𝑙+1)_{𝑙+1,𝑙+1}.

Proof. (i) That 𝐴 is invertible follows from the fact that 𝜓 induces an automorphism of the free 𝑅-module 𝐧₀^(𝐶)/𝐧₁^(𝐶) of rank 𝑙+1 on the basis {𝛼_𝑗𝑗+𝐧₁^(𝐶)∣𝑗=1,…,𝑙+1}. (ii) Note that 𝜓 induces an automorphism of the free 𝑅-module 𝐧₁^(𝐶)/𝐧₂^(𝐶) of rank 𝑙+1 on the basis {𝛼_𝑗,𝑗+1+𝐧₂^(𝐶),𝛾_{𝑙+1,𝑙+1}+𝐧₂^(𝐶)∣𝑗=1,…,𝑙}.

Lemma 2.3. Let 𝜓∈  Aut(𝐧₀^(𝐶))(𝑙≥2). Write 𝜓(𝛼_𝑗𝑗), 𝜓(𝛼_𝑗,𝑗+1), and 𝜓(𝛾_{𝑙+1,𝑙+1}) as in (2.5)–(2.7), respectively. Then the following conclusions hold.(i)For 1≤𝑚,𝑘,ℎ≤𝑙, 𝑎^(𝑚)_ℎ,ℎ+1𝑎^(𝑚)_𝑘,𝑘+1=0(ℎ≠𝑘), 𝑎^(𝑚)_ℎ,ℎ+1𝑐̃^(𝑚)_{𝑙+1,𝑙+1}=0, 𝑎^(𝑙+1)_ℎ,ℎ+1𝑎^(𝑙+1)_𝑘,𝑘+1=0(ℎ≠𝑘)  and 𝑎^(𝑙+1)_ℎ,ℎ+1𝑐̂^(𝑙+1)_{𝑙+1,𝑙+1}=0.(ii)For 1≤𝑘,ℎ≤𝑙, (𝑎^(𝑖)_ℎℎ−𝑎^(𝑖)_ℎ+1,ℎ+1)(𝑎^(𝑖)_𝑘𝑘−𝑎^(𝑖)_{𝑘+1,𝑘+1})=0(ℎ≠𝑘)   and (𝑎^(𝑖)_ℎℎ−𝑎^(𝑖)_ℎ+1,ℎ+1)𝑎^(𝑖)_{𝑙+1,𝑙+1}=0(1≤ℎ≤𝑙,here𝑙≥1), where  𝑖=1,𝑙+1.(iii)For 2≤𝑚≤𝑙 and 1≤𝑖,𝑘,ℎ≤𝑙, (𝑎^(𝑚)_𝑖𝑖−𝑎^(𝑚)_{𝑖+1,𝑖+1})(𝑎^(𝑚)_ℎℎ−𝑎^(𝑚)_ℎ+1,ℎ+1)(𝑎^(𝑚)_𝑘𝑘−𝑎^(𝑚)_{𝑘+1,𝑘+1})=0(𝑖≠ℎ≠𝑘≠𝑖,here𝑙≥3) and (𝑎^(𝑚)_𝑖𝑖−𝑎^(𝑚)_{𝑖+1,𝑖+1})(𝑎^(𝑚)_ℎℎ−𝑎^(𝑚)_ℎ+1,ℎ+1)𝑎^(𝑚)_{𝑙+1,𝑙+1}=0(1≤𝑖≠ℎ≤𝑙).

Proof. (i) When 𝑗≠𝑚,𝑚+1, [𝜓(𝛼_𝑗𝑗),𝜓(𝛼_𝑚,𝑚+1)]=0. So 𝑎^(𝑚)_𝑖,𝑖+1𝑎^(𝑗)_𝑖𝑖−𝑎^(𝑗)_{𝑖+1,𝑖+1}=0,𝑐̃^(𝑚)_{𝑙+1,𝑙+1}𝑎^(𝑗)_{𝑙+1,𝑙+1}=0. (2.8) From [𝜓(𝛼_𝑚𝑚),𝜓(𝛼_𝑚,𝑚+1)]=𝜓(𝛼_𝑚,𝑚+1) and [𝜓(𝛼_{𝑚+1,𝑚+1}),𝜓(𝛼_𝑚,𝑚+1)]=−𝜓(𝛼_𝑚,𝑚+1), we have 𝑎^(𝑚)_𝑖,𝑖+1𝑎^(𝑚)_𝑖𝑖−𝑎^(𝑚)_{𝑖+1,𝑖+1}=𝑎^(𝑚)_𝑖,𝑖+1,2𝑐̃^(𝑚)_{𝑙+1,𝑙+1}𝑎^(𝑚)_{𝑙+1,𝑙+1}=𝑐̃^(𝑚)_{𝑙+1,𝑙+1},𝑎^(𝑚)_𝑖,𝑖+1𝑎^(𝑚+1)_𝑖𝑖−𝑎^(𝑚+1)_{𝑖+1,𝑖+1}=−𝑎^(𝑚)_𝑖,𝑖+1,2𝑐̃^(𝑚)_{𝑙+1,𝑙+1}𝑎^(𝑚+1)_{𝑙+1,𝑙+1}=−𝑐̃^(𝑚)_{𝑙+1,𝑙+1}. (2.9) When 𝑗≠𝑙+1, [𝜓(𝛼_𝑗𝑗),𝜓(𝛾_{𝑙+1,𝑙+1})]=0. So 𝑎^(𝑙+1)_𝑖,𝑖+1𝑎^(𝑗)_𝑖𝑖−𝑎^(𝑗)_{𝑖+1,𝑖+1}=0,𝑐̂^(𝑙+1)_{𝑙+1,𝑙+1}𝑎^(𝑗)_{𝑙+1,𝑙+1}=0. (2.10) From [𝜓(𝛼_{𝑙+1,𝑙+1}),𝜓(𝛾_{𝑙+1,𝑙+1})]=2𝜓(𝛾_{𝑙+1,𝑙+1}), we have 𝑎^(𝑙+1)_𝑖,𝑖+1𝑎^(𝑙+1)_𝑖𝑖−𝑎^(𝑙+1)_{𝑖+1,𝑖+1}=2𝑎^(𝑙+1)_𝑖,𝑖+1,𝑐̂^(𝑙+1)_{𝑙+1,𝑙+1}𝑎^(𝑙+1)_{𝑙+1,𝑙+1}=𝑐̂^(𝑙+1)_{𝑙+1,𝑙+1}. (2.11) Let 𝐶=(𝑐_𝑗𝑖)_{(𝑙+1)×(𝑙+1)}, where 𝑐_𝑗𝑖=𝑎^(𝑗)_𝑖𝑖−𝑎^(𝑗)_{𝑖+1,𝑖+1}, 𝑗=1,…,𝑙+1, 𝑖=1,…,𝑙, and 𝑐_𝑗,𝑙+1=2𝑎^(𝑗)_{𝑙+1,𝑙+1},  𝑗=1,…,𝑙+1. By Lemma 2.2, det𝐴∈𝑅^∗, so det𝐶∈𝑅^∗. Investigating 𝑎^(𝑚)_ℎ,ℎ+1𝑎^(𝑚)_𝑘,𝑘+1det𝐶, we may find that ℎth column and 𝑘th column are linearly dependent (both are the form (0,…,0,𝑎^(𝑚)_𝑗,𝑗+1,−𝑎^(𝑚)_𝑗,𝑗+1,0,…,0)^𝑡,𝑗=ℎ,𝑘) by (2.6) and (2.7), so 𝑎^(𝑚)_ℎ,ℎ+1𝑎^(𝑚)_𝑘,𝑘+1det𝐶=0. Similarly, 𝑎^(𝑚)_ℎ,ℎ+1𝑐̃^(𝑚)_{𝑙+1,𝑙+1}det𝐶=0, 𝑎^(𝑙+1)_ℎ,ℎ+1𝑎^(𝑙+1)_𝑘,𝑘+1det𝐶=0(ℎ≠𝑘) and 𝑎^(𝑙+1)_ℎ,ℎ+1𝑐̂^(𝑙+1)_{𝑙+1,𝑙+1}det𝐶=0. Then 𝑎^(𝑚)_ℎ,ℎ+1𝑎^(𝑚)_𝑘,𝑘+1=0(ℎ≠𝑘), 𝑎^(𝑚)_ℎ,ℎ+1𝑐̃^(𝑚)_{𝑙+1,𝑙+1}=0, 𝑎^(𝑙+1)_ℎ,ℎ+1𝑎^(𝑙+1)_𝑘,𝑘+1=0(ℎ≠𝑘), and 𝑎^(𝑙+1)_ℎ,ℎ+1𝑐̂^(𝑙+1)_{𝑙+1,𝑙+1}=0.
(ii) When 𝑖=1, from (𝑎⁽¹⁾_ℎℎ−𝑎⁽¹⁾_ℎ+1,ℎ+1)(𝑎⁽¹⁾_𝑘𝑘−𝑎⁽¹⁾_{𝑘+1,𝑘+1})det𝐵=0(ℎ≠𝑘), we have (𝑎⁽¹⁾_ℎℎ−𝑎⁽¹⁾_ℎ+1,ℎ+1)(𝑎⁽¹⁾_𝑘𝑘−𝑎⁽¹⁾_{𝑘+1,𝑘+1})=0(ℎ≠𝑘). Similarly, we have (𝑎⁽¹⁾_ℎℎ−𝑎⁽¹⁾_ℎ+1,ℎ+1)(𝑎⁽¹⁾_𝑙𝑙+𝑎⁽¹⁾_{𝑙+1,𝑙+1})=0(1≤ℎ≤𝑙). When 𝑖=𝑙+1, we get the results similarly.
(iii) The proving process is similar to (i) and (ii).

Lemma 2.4. Let 𝜓∈Aut(𝐧₀^(𝐶)). Then(i)when 𝑙≥1, 𝜓(𝛼₁₂)=𝑎⁽¹⁾₁₂𝛼₁₂mod𝐧₂^(𝐶), where 𝑎⁽¹⁾₁₂∈𝑅^∗;(ii)if 𝜓(𝛼₁₂)=𝑎⁽¹⁾₁₂𝛼₁₂mod𝐧₂^(𝐶),  where 𝑎⁽¹⁾₁₂∈𝑅^∗, then 𝜓(𝛼_𝑖,𝑖+1)=𝑎^(𝑖)_𝑖,𝑖+1𝛼_𝑖,𝑖+1mod𝐧₂^(𝐶) and 𝜓(𝛾_{𝑙+1,𝑙+1})=𝑐̂^(𝑙+1)_{𝑙+1,𝑙+1}𝛾_{𝑙+1,𝑙+1}mod𝐧₂^(𝐶),  where 𝑎^(𝑖)_𝑖,𝑖+1,𝑐̂^(𝑙+1)_{𝑙+1,𝑙+1}∈𝑅^∗.

Proof. (i) Noting that 𝛼_𝑖,𝑖+1,𝛾_{𝑙+1,𝑙+1}∈𝐧₁^(𝐶)⧵𝐧₂^(𝐶) and 𝛾₁₂∈𝐧^(𝐶)_2𝑙⧵𝐧^(𝐶)_2𝑙+1, we have 𝜓(𝛼₁₂)∈𝐧₁^(𝐶)⧵𝐧₂^(𝐶) and 𝜓(𝛾₁₂)=𝑐̂⁽¹²⁾₁₂𝛾₁₂mod𝐧^(𝐶)_2𝑙+1∈𝐧^(𝐶)_2𝑙⧵𝐧^(𝐶)_2𝑙+1, where 𝑐̂⁽¹²⁾₁₂∈𝑅^∗. Using (2.7), from [𝜓(𝛼₁₁),𝜓(𝛾₁₂)]=𝜓(𝛾₁₂), we have 𝑐̂⁽¹²⁾₁₂(𝑎⁽¹⁾₁₁+𝑎⁽¹⁾₂₂)=𝑐̂⁽¹²⁾₁₂, that is, 𝑎⁽¹⁾₁₁+𝑎⁽¹⁾₂₂=1. Write 𝜓(𝛼₁₁) and 𝜓(𝛾₁₁) as 𝜓(𝛼₁₁)=∑^𝑙+1_𝑖=1𝑎⁽¹⁾_𝑖𝑖𝛼_𝑖𝑖mod𝐧₁^(𝐶) and 𝜓(𝛾₁₁)=𝑐^∗₁₁𝛾₁₁∈𝐧^(𝐶)_2𝑙+1, where 𝑐^∗₁₁∈𝑅^∗. From 2𝜓(𝛾₁₁)=[𝜓(𝛼₁₁),𝜓(𝛾₁₁)]=2𝑎⁽¹⁾₁₁𝑐^∗₁₁𝛾₁₁, we have 𝑎⁽¹⁾₁₁=1. Then 𝑎⁽¹⁾₂₂=0. By Lemma 2.3 we have 𝑎⁽¹⁾_𝑖𝑖−𝑎⁽¹⁾_{𝑖+1,𝑖+1}=0,𝑖=2,…,𝑙  (here  𝑙≥2) and 𝑎⁽¹⁾_{𝑙+1,𝑙+1}=0. So 𝑎⁽¹⁾_𝑖𝑖=0,𝑖=2,…,𝑙+1, that is, 𝜓(𝛼₁₁)=𝑎⁽¹⁾₁₁mod𝐧₁^(𝐶). Then 𝜓(𝛼₁₂)=[𝜓(𝛼₁₁),𝜓(𝛼₁₂)]=𝑎⁽¹⁾₁₂𝛼₁₂mod𝐧₂^(𝐶) and 𝑎⁽¹⁾₁₂∈𝑅^∗. By Lemma 2.3, (i) holds.
(ii) Write 𝜓(𝛼_𝑗,𝑗+1) and 𝜓(𝛾_{𝑙+1,𝑙+1}) as (2.6) and (2.7), respectively. From 𝜓(𝛼₁₃)=[𝜓(𝛼₁₂),𝜓(𝛼₂₃)], we have 𝜓(𝛼₁₃)=𝑎⁽¹⁾₁₂𝑎⁽²⁾₂₃𝛼₁₃mod𝐧₃^(𝐶). Since 𝛼₁₃∈𝐧₂^(𝐶)⧵𝐧₃^(𝐶), 𝜓(𝛼₁₃)∈𝐧₂^(𝐶)⧵𝐧₃^(𝐶). So 𝑎⁽¹⁾₁₂𝑎⁽²⁾₂₃∈𝑅^∗, that is, 𝑎⁽²⁾₂₃∈𝑅^∗. In general, for 𝑚=2,…,𝑙, we have 𝜓𝛼_1,𝑚+1=^𝑚_𝑖=1𝑎^(𝑖)_𝑖,𝑖+1𝛼_1,𝑚+1mod𝐧^(𝐶)_𝑚+1∈𝐧_𝑚^(𝐶)⧵𝐧^(𝐶)_𝑚+1𝜓𝛾,𝑚=1,…,𝑙,_1,𝑙+1=𝜓𝛼_1,𝑙+1,𝜓𝛾_{𝑙+1,𝑙+1}=^𝑙_𝑖=1𝑎^(𝑖)_𝑖,𝑖+1𝑐̂^(𝑙+1)_{𝑙+1,𝑙+1}𝛾_1,𝑙+1mod𝐧^(𝐶)_𝑙+2∈𝐧^(𝐶)_𝑙+1⧵𝐧^(𝐶)_𝑙+2, (2.12) here 𝑎^(𝑖)_𝑖,𝑖+1,𝑐̂^(𝑙+1)_{𝑙+1,𝑙+1} should be in 𝑅^∗, 𝑖=1,…,𝑙. By Lemma 2.3 we have that 𝑎^(𝑗)_𝑖,𝑖+1=0(𝑖≠𝑗), 𝑐̃^(𝑗)_{𝑙+1,𝑙+1}=0, 𝑗=1,…,𝑙, and 𝑎^(𝑙+1)_𝑖,𝑖+1, 𝑖=1,…,𝑙. Hence 𝜓(𝛼_𝑙,𝑙+1) and 𝜓(𝛾_{𝑙+1,𝑙+1}) have the required forms, respectively.

3. The Standard Automorphisms of t^(𝐶)_𝑙+1(𝑅)

Now let us introduce two types of Lie automorphisms of 𝐭^(𝐶)_𝑙+1(𝑅).

(i) Inner Automorphisms
Let 𝑟=𝐼_𝑛+𝑎𝛼_𝑖𝑗(𝑖≠𝑗) or 𝑟=𝐼_𝑛+𝑎𝛾_𝑖𝑗. It is easy to check that 𝑟𝑦𝑟⁻¹∈𝐧₀^(𝐶). The map 𝜃_𝑟: 𝐭^(𝐶)_𝑙+1(𝑅)→𝐭^(𝐶)_𝑙+1(𝑅) such that 𝑥↦𝑟𝑥𝑟⁻¹, 𝑥∈𝐧₀^(𝐶), defines an automorphism of 𝐧₀^(𝐶), which is called an inner automorphism  (note that 𝑟 is a symplectic matrix defined by 0𝐼_𝑙+1−𝐼_𝑙+10). We denote 𝜃_𝑟 by 𝜃_{𝑎𝛼𝑖𝑗}, 𝜃_{𝑎𝛾𝑖𝑗}, respectively. In these cases, we have 𝜃⁻¹_{𝑎𝛼𝑖𝑗}=𝜃_{−𝑎𝛼𝑖𝑗}, 𝜃⁻¹_{𝑎𝛾𝑖𝑗}=𝜃_{−𝑎𝛾𝑖𝑗}, respectively, and that 𝜃_{𝑎𝛼𝑖𝑗}(𝛼_𝑖𝑖)=𝛼_𝑖𝑖−𝑎𝛼_𝑖𝑗, 𝜃_{𝑎𝛼𝑖𝑗}(𝛼_𝑗𝑗)=𝛼_𝑗𝑗+𝑎𝛼_𝑖𝑗, 𝜃_{𝑎𝛼𝑖𝑗}(𝛼_𝑘𝑘)=𝛼_𝑘𝑘(𝑘≠𝑖,𝑗), 𝜃_{𝑎𝛼𝑖𝑗}(𝛼_𝑘,𝑘+1)=𝛼_𝑘,𝑘+1(𝑘≠𝑗,𝑖−1), 𝜃_{𝑎𝛼𝑖𝑗}(𝛼_𝑗,𝑗+1)=𝛼_𝑗,𝑗+1+𝑎𝛼_𝑖,𝑗+1, 𝜃_{𝑎𝛼𝑖𝑗}(𝛼_{𝑖−1,𝑖})=𝛼_{𝑖−1,𝑖}−𝑎𝛼_{𝑖−1,𝑗}, 𝜃_{𝑎𝛾𝑖𝑖}(𝛼_𝑖𝑖)=𝛼_𝑖𝑖−2𝑎𝛾_𝑖𝑖, 𝜃_{𝑎𝛾𝑖,𝑖+1}(𝛼_𝑖,𝑖+1)=𝛼_𝑖,𝑖+1−2𝑎𝛾_𝑖𝑖 and 𝜃_{𝑎𝛼𝑖,𝑖+1}(𝛾_𝑖,𝑖+1)=𝛾_𝑖,𝑖+1+2𝑎𝛾_𝑖𝑖. All inner automorphisms of 𝐭^(𝐶)_𝑙+1(𝑅) generate a subgroup of Aut(𝐧₀^(𝐶)), which is denoted by ℐ.

(ii) Diagonal Automorphisms
Let 𝑑_𝑖∈𝑅^∗, 𝑖=0,1,…,𝑙+1, 𝑑=diag(𝑑₁,…,𝑑_𝑙+1) and 𝐷=diag(𝑑,𝑑⁻¹𝑑₀). The map 𝜆_𝐷: 𝐭^(𝐶)_𝑙+1(𝑅)→𝐭^(𝐶)_𝑙+1(𝑅) such that 𝑥↦𝐷𝑥𝐷⁻¹, 𝑥∈𝐧₀^(𝐶), defines an automorphism of 𝐭^(𝐶)_𝑙+1(𝑅), which is called a diagonal automorphism. It is clear that 𝜆_𝐷𝜆_𝐷=𝜆_𝐷𝐷. So the set of diagonal automorphisms of 𝐭^(𝐶)_𝑙+1(𝑅) is a subgroup of Aut(𝐧₀^(𝐶)), which is denoted by 𝒟.

4. Lemmas for Main Results

Lemma 4.1. Let 𝜓∈Aut(𝐧₀^(𝐶)). The following two statements are equivalent:
(i) 𝜓(𝛼_𝑗,𝑗+1)=𝑎^(𝑗)_𝑗,𝑗+1𝛼_𝑗,𝑗+1mod𝐧₂^(𝐶) and 𝜓(𝛾_{𝑙+1,𝑙+1})=𝑐̂^(𝑙+1)_{𝑙+1,𝑙+1}𝛾_{𝑙+1,𝑙+1}mod𝐧₂^(𝐶), where 𝑎^(𝑗)_𝑗,𝑗+1,𝑐̂^(𝑙+1)_{𝑙+1,𝑙+1}∈𝑅^∗, 𝑗=1,…,𝑙;
(ii) 𝜓(𝛼_𝑗𝑗)=𝛼_𝑗𝑗mod𝐧₁^(𝐶),𝑗=1,…,𝑙+1.

Proof. (i)⇒(ii). Write 𝜓(𝛼_𝑗𝑗) as in (2.5). By the process of proving Lemma 2.3, we have 𝑎⁽¹⁾₁₂(𝑎⁽¹⁾₁₁−𝑎⁽¹⁾₂₂)=𝑎⁽¹⁾₁₂, 𝑎^(𝑖)_𝑖,𝑖+1(𝑎^(𝑖+1)_𝑖𝑖−𝑎^(𝑖+1)_{𝑖+1,𝑖+1})=−𝑎^(𝑖)_𝑖,𝑖+1, 𝑎^(𝑖+1)_{𝑖+1,𝑖+2}(𝑎^(𝑖+1)_{𝑖+1,𝑖+1}−𝑎^(𝑖+1)_{𝑖+2,𝑖+2})=𝑎^(𝑖+1)_{𝑖+1,𝑖+2}, 𝑖=1,…,𝑙−1 and 𝑎^(𝑙)_𝑙,𝑙+1(𝑎^(𝑙+1)_𝑙𝑙−𝑎^(𝑙+1)_{𝑙+1,𝑙+1})=−𝑎^(𝑙)_𝑙,𝑙+1, 𝑐̂^(𝑙+1)_{𝑙+1,𝑙+1}𝑎^(𝑙+1)_{𝑙+1,𝑙+1}=𝑐̂^(𝑙+1)_{𝑙+1,𝑙+1}. Then we obtain that 𝑎⁽¹⁾₁₁−𝑎⁽¹⁾₂₂=1, 𝑎^(𝑖+1)_𝑖𝑖−𝑎^(𝑖+1)_{𝑖+1,𝑖+1}=−1, 𝑎^(𝑖+1)_{𝑖+1,𝑖+1}−𝑎^(𝑖+1)_{𝑖+2,𝑖+2}=1, 𝑖=1,…,𝑙−1 and 𝑎^(𝑙+1)_{𝑙+1,𝑙+1}=1. By Lemma 2.3, we have 𝑎^(𝑗)_𝑗𝑗=1(1≤𝑗≤𝑙+1) and 𝑎^(𝑗)_𝑖𝑖=0(𝑖≠𝑗).
(ii)⇒(i). Write 𝜓(𝛼_𝑗,𝑗+1) and 𝜓(𝛾_{𝑙+1,𝑙+1}), respectively, as in (2.6) and (2.7). Then 𝜓𝛼_𝑗,𝑗+1=𝜓𝛼_𝑗𝑗,𝜓𝛼_𝑗,𝑗+1=𝑎^(𝑗)_𝑗,𝑗+1𝛼_𝑗,𝑗+1mod𝐧₂^(𝐶)𝛾,𝑗=1,…,𝑙,2𝜓_{𝑙+1,𝑙+1}=𝜓𝛼_{𝑙+1,𝑙+1},𝜓𝛾_{𝑙+1,𝑙+1}=2𝑐̂^(𝑙+1)_{𝑙+1,𝑙+1}𝛾_{𝑙+1,𝑙+1}mod𝐧₂^(𝐶), (4.1) that is, 𝜓(𝛾_{𝑙+1,𝑙+1})=𝑐̂^(𝑙+1)_{𝑙+1,𝑙+1}𝛾_{𝑙+1,𝑙+1}mod𝐧₂^(𝐶). By the method of modularizing a maximal ideal of 𝑅 to a residue field, we know that 𝑎^(𝑗)_𝑗,𝑗+1,𝑐̂^(𝑙+1)_{𝑙+1,𝑙+1}∈𝑅^∗, 𝑗=1,…,𝑙.

Lemma 4.2. Let 𝜓 be in Aut(𝐧₀^(𝐶)). If 𝜓(𝛼_𝑗𝑗)=𝛼_𝑗𝑗mod𝐧₁^(𝐶), then 𝜓𝛼₁₁=𝛼₁₁+𝑎⁽¹⁾₁₂𝛼₁₂mod𝐧₂^(𝐶),𝜓𝛼_𝑗𝑗=𝛼_𝑗𝑗−𝑎^(𝑗−1)_{𝑗−1,𝑗}𝛼_{𝑗−1,𝑗}+𝑎^(𝑗)_𝑗,𝑗+1𝛼_𝑗,𝑗+1mod𝐧₂^(𝐶)(),𝜓𝛼,𝑗=2,…,𝑙𝑙≥2_{𝑙+1,𝑙+1}=𝛼_{𝑙+1,𝑙+1}−𝑎^(𝑙)_𝑙,𝑙+1𝛼_𝑙,𝑙+1+𝑐^(𝑙+1)_{𝑙+1,𝑙+1}𝛾_{𝑙+1,𝑙+1}mod𝐧₂^(𝐶). (4.2)

Proof. We express 𝜓(𝛼_𝑗𝑗) as 𝜓𝛼_𝑗𝑗=𝛼_𝑗𝑗+^𝑙_𝑖=1𝑎^(𝑗)_𝑖,𝑖+1𝛼_𝑖,𝑖+1+𝑐^(𝑗)_{𝑙+1,𝑙+1}𝛾_{𝑙+1,𝑙+1}mod𝐧₂^(𝐶),𝑗=1,…,𝑙+1. (4.3) From [𝜓(𝛼_𝑗𝑗),𝜓(𝛼_𝑘𝑘)]=0(𝑗≠𝑘) we have 𝜓𝛼₁₁=𝛼₁₁+𝑎⁽¹⁾₁₂𝛼₁₂mod𝐧₂^(𝐶),𝜓𝛼_𝑗𝑗=𝛼_𝑗𝑗+𝑎^(𝑗)_{𝑗−1,𝑗}𝛼_{𝑗−1,𝑗}+𝑎^(𝑗)_𝑗,𝑗+1𝛼_𝑗,𝑗+1mod𝐧₂^(𝐶)(),𝜓𝛼,𝑗=2,…,𝑙𝑙≥2_{𝑙+1,𝑙+1}=𝛼_{𝑙+1,𝑙+1}+𝑎^(𝑙+1)_𝑙,𝑙+1𝛼_𝑙,𝑙+1+𝑐^(𝑙+1)_{𝑙+1,𝑙+1}𝛾_{𝑙+1,𝑙+1}mod𝐧₂^(𝐶), (4.4) where 𝑎^(𝑗)_𝑗,𝑗+1+𝑎^(𝑗+1)_𝑗,𝑗+1=0,𝑗=1,…,𝑙. Lemma 4.2 is proved.

Lemma 4.3. Let 𝜓 be in Aut(𝐧₀^(𝐶)). If every 𝜓(𝛼_𝑗𝑗) is expressed as the form in Lemma 4.2, one may find an inner automorphism 𝜃=^𝑙_𝑗=1𝜃_{𝑎(𝑗)𝑗,𝑗+1𝛼𝑗,𝑗+1}𝜃_{2−1𝑐(𝑙+1)𝑙+1,𝑙+1𝛾𝑙+1,𝑙+1} (4.5) such that 𝜃𝜓𝛼_𝑗𝑗=𝛼_𝑗𝑗mod𝐧₂^(𝐶),𝑗=1,…,𝑙+1. (4.6)

Proof. Note that 𝜃_{2−1𝑐(𝑙+1)𝑙+1,𝑙+1𝛾𝑙+1,𝑙+1}(𝛼_{𝑙+1,𝑙+1})=𝛼_{𝑙+1,𝑙+1}−𝑐^(𝑙+1)_{𝑙+1,𝑙+1}𝛾_{𝑙+1,𝑙+1}. Then, by Lemma 4.2, it is not difficult to prove Lemma 4.3.

Lemma 4.4. Let 𝜓 be in Aut(𝐧₀^(𝐶)). If 𝜓(𝛼_𝑗𝑗)=𝛼_𝑗𝑗mod𝐧_𝑘^(𝐶),𝑗=1,…,𝑙+1(1≤𝑘≤𝑙+1), then 𝜓𝛼_𝑗𝑗=𝛼_𝑗𝑗+𝑎^(𝑗)_{𝑗,𝑗+𝑘}𝛼_{𝑗,𝑗+𝑘}mod𝐧^(𝐶)_𝑘+1{),𝜓𝛼,𝑗=1,…,min𝑘,𝑙−𝑘+1}(𝑘≤𝑙,𝑙≥1_𝑗𝑗=𝛼_𝑗𝑗−𝑎^{(𝑗−𝑘)}_{𝑗−𝑘,𝑗}𝛼_{𝑗−𝑘,𝑗}+𝑎^(𝑗)_{𝑗,𝑗+𝑘}𝛼_{𝑗,𝑗+𝑘}mod𝐧^(𝐶)_𝑘+1,𝑗=𝑘+1,…,𝑙−𝑘+1𝑘≤𝑙+12,𝜓𝛼,𝑙≥2_𝑗𝑗=𝛼_𝑗𝑗−𝑎^{(𝑗−𝑘)}_{𝑗−𝑘,𝑗}𝛼_{𝑗−𝑘,𝑗}+𝑐^(𝑗)_{𝑗,2𝑙−𝑘−𝑗+3}𝛾_{𝑗,2𝑙−𝑘−𝑗+3}mod𝐧^(𝐶)_𝑘+1,𝑘𝑗=𝑙−𝑘+2,…,𝑙−2+1𝑘≤𝑙+12,𝜓𝛼,𝑙≥1_𝑗𝑗=𝛼_𝑗𝑗+𝑐^(𝑗)_{𝑗,2𝑙−𝑘−𝑗+3}𝛾_{𝑗,2𝑙−𝑘−𝑗+3}mod𝐧^(𝐶)_𝑘+1,𝑘𝑗=𝑙−𝑘+2,…,𝑙−2+11+𝑙+12,𝜓𝛼≤𝑘,𝑙≥2_𝑗𝑗=𝛼_𝑗𝑗+𝑐^(𝑗)_{𝑗,2𝑙−𝑘−𝑗+3}𝛾_{𝑗,2𝑙−𝑘−𝑗+3}mod𝐧^(𝐶)_𝑘+1,𝑙𝑗=2𝑙+1𝑘=2,𝜓𝛼+1,𝑙≥4,here𝑝even_𝑗𝑗=𝛼_𝑗𝑗−𝑎^{(𝑗−𝑘)}_{𝑗−𝑘,𝑗}𝛼_{𝑗−𝑘,𝑗}+𝑐^(𝑗)_{𝑗,2𝑙−𝑘−𝑗+3}𝛾_{𝑗,2𝑙−𝑘−𝑗+3}mod𝐧^(𝐶)_𝑘+1,𝑙𝑗=2𝑘+2,…,𝑙−2𝑙+1𝑘=2,𝜓𝛼+1,𝑙≥4,here𝑝even_𝑗𝑗=𝛼_𝑗𝑗+𝑐^{(2𝑙−𝑘−𝑗+3)}_{2𝑙−𝑘−𝑗+3,𝑗}𝛾_{2𝑙−𝑘−𝑗+3,𝑗}mod𝐧^(𝐶)_𝑘+1,𝑘𝑗=𝑙−2𝑘+2,…,𝑘𝑙+2≤𝑘+2,𝜓𝛼,𝑙≥3_𝑗𝑗=𝛼_𝑗𝑗−𝑎^{(𝑗−𝑘)}_{𝑗−𝑘,𝑗}𝛼_{𝑗−𝑘,𝑗}+𝑐^{(2𝑙−𝑘−𝑗+3)}_{2𝑙−𝑘−𝑗+3,𝑗}𝛾_{2𝑙−𝑘−𝑗+3,𝑗}mod𝐧^(𝐶)_𝑘+1,𝑘𝑗=max𝑘+1,𝑙−2𝑘+2,…,𝑙+1𝑘≤𝑝,2≥1,𝑙≥2. (4.7)

Proof. We express 𝜓(𝛼_𝑗𝑗),𝑗=1,…,𝑙+1, as 𝜓𝛼_𝑗𝑗=𝛼_𝑗𝑗+^{𝑙−𝑘+1}_𝑖=1𝑎^(𝑗)_{𝑖,𝑖+𝑘}𝛼_{𝑖,𝑖+𝑘}+^𝑙+1_{)]𝑖=𝑙+2−[(𝑘+1/2}𝑐^(𝑗)_{2𝑙−𝑘−𝑖+3,𝑖}𝛾_{2𝑙−𝑘−𝑖+3,𝑖}mod𝐧^(𝐶)_𝑘+1. (4.8) When 𝑘=1 that is the case in Lemma 4.2. The conclusion follows from repeating the process of proving Lemma 4.4.

Lemma 4.5. Let 𝜓 be in Aut(𝐧₀^(𝐶)). If every 𝜓(𝛼_𝑗𝑗) be expressed as the form in Lemma 4.4, one may find an inner automorphism 𝜃=^{𝑙−𝑘+1}_𝑗=1𝜃_{𝑎(𝑗)𝑗,𝑗+𝑘𝛼𝑗,𝑗+𝑘}^{𝑙−[𝑘/2]+1}_{𝑗=𝑙−𝑘+2}𝜃_{(1+𝛿𝑗𝑙)−1𝑐(𝑗)𝑗,2𝑙−𝑘−𝑗+3𝛾𝑗,2𝑙−𝑘−𝑗+3}, (4.9) where 𝛿_𝑖𝑗=()()1,𝑖=𝑗,0,𝑖≠𝑗,ℎ=𝑙+1−𝑘−1/2𝑘𝑎𝑛𝑜𝑑𝑑. (4.10) Then 𝜃𝜓𝛼_𝑗𝑗=𝛼_𝑗𝑗mod𝐧^(𝐶)_𝑘+1,𝑖=1,…,𝑙+1. (4.11)

Proof. Apply 𝜃 to 𝜓(𝛼_𝑗𝑗) and use Lemma 4.4 to obtain Lemma 4.5.

Lemma 4.6. Let 𝜓 be in Aut(𝐧₀^(𝐶)). If 𝜓(𝛼_𝑗𝑗)=𝛼_𝑗𝑗mod𝐧_𝑘^(𝐶),𝑗=1,…,𝑙+1, 𝑙+1≤𝑘≤2𝑙+1(𝑙≥1), then 𝜓𝛼_𝑗𝑗=𝛼_𝑗𝑗+𝑐^(𝑗)_{𝑗,2𝑙−𝑘−𝑗+3}𝛾_{𝑗,2𝑙−𝑘−𝑗+3}mod𝐧^(𝐶)_𝑘+1𝑘,𝑗=1,…,𝑙−2𝜓𝛼+1,_𝑗𝑗=𝛼_𝑗𝑗+𝑐^{(2𝑙−𝑘−𝑗+3)}_{2𝑙−𝑘−𝑗+3,𝑗}𝛾_{2𝑙−𝑘−𝑗+3,𝑗}mod𝐧^(𝐶)_𝑘+1𝑘,𝑗=𝑙−2𝜓𝛼+2,…,2𝑙−𝑘+2,_𝑗𝑗=𝛼_𝑗𝑗mod𝐧^(𝐶)_𝑘+1,𝑗=2𝑙−𝑘+3,…,𝑙+1. (4.12)

Proof. We express 𝜓(𝛼_𝑗𝑗),𝑗=1,…,𝑙+1, as 𝜓𝛼_𝑗𝑗=𝛼_𝑗𝑗+^{𝑙−[𝑘/2]+1}_𝑖=1𝑐^(𝑗)_{𝑖,2𝑙−𝑘−𝑖+3}𝛾_{𝑖,2𝑙−𝑘−𝑖+3}mod𝐧^(𝐶)_𝑘+1. (4.13) The process of proving Lemma 4.6 is similar to Lemma 4.2.

Lemma 4.7. Let 𝜓 be in Aut(𝐧₀^(𝐶)). If every 𝜓(𝛼_𝑗𝑗) is expressed as the form in Lemma 4.6, one may find an inner automorphism 𝜃=^{𝑙−[𝑘/2]+1}_𝑗=1𝜃_{(1+𝛿𝑗ℎ)−1𝑐(𝑗)𝑗,2𝑙−𝑘−𝑗+3𝛾𝑗,2𝑙−𝑘−𝑗+3}, (4.14) where ℎ=𝑙+1−(𝑘−1)/2 (𝑘 an odd). Then 𝜃𝜓𝛼_𝑗𝑗=𝛼_𝑗𝑗mod𝐧^(𝐶)_𝑘+1. (4.15) When 𝑘=2𝑙+1, 𝜃𝜓(𝛼_𝑗𝑗)=𝛼_𝑗𝑗,𝑗=1,…,𝑙+1.

Proof. It is similar to proving Lemma 4.5.

Lemma 4.8. When 𝑙≥1, let 𝜓 be in Aut(𝐧₀^(𝐶)). If 𝜓(𝛼_𝑗𝑗)=𝛼_𝑗𝑗,𝑗=1,…,𝑙+1, there exists a diagonal automorphism 𝜆_𝐷 such that 𝜆_𝐷𝜓(𝛼_𝑗,𝑗+1)=𝛼_𝑗,𝑗+1, 𝑖=1,…,𝑙, and 𝜆_𝐷𝜓(𝛾_{𝑙+1,𝑙+1})=𝛾_{𝑙+1,𝑙+1}.

Proof. By Lemma 4.1 we know that 𝜓(𝛼_𝑗,𝑗+1)=𝑎^(𝑗)_𝑗,𝑗+1𝛼_𝑗,𝑗+1mod𝐧₂^(𝐶) and 𝜓(𝛾_{𝑙+1,𝑙+1})=𝑐̂^(𝑙+1)_{𝑙+1,𝑙+1}𝛾_{𝑙+1,𝑙+1}mod𝐧₂^(𝐶), where 𝑎^(𝑗)_𝑗,𝑗+1,𝑐̂^(𝑙+1)_{𝑙+1,𝑙+1}∈𝑅^∗, 𝑗=1,…,𝑙. We express 𝜓(𝛼_𝑗,𝑗+1) and 𝜓(𝛾_{𝑙+1,𝑙+1}), respectively, as 𝜓𝛼_𝑗,𝑗+1=𝑎^(𝑗)_𝑗,𝑗+1𝛼_𝑗,𝑗+1+^𝑙_𝑘=2^{𝑙−𝑘+1}_𝑖=1𝑎^(𝑗)_{𝑖,𝑖+𝑘}𝛼_{𝑖,𝑖+𝑘}+^𝑙_𝑘=2^𝑙+1_{[]𝑖=𝑙+2−(𝑘+1)/2}𝑐̃^(𝑗)_{2𝑙−𝑘−𝑖+3,𝑖}𝛾_{2𝑙−𝑘−𝑖+3,𝑖}+^2𝑙+1_𝑘=𝑙+1^{𝑙+1−[𝑘/2]}_𝑖=1𝑐̃^(𝑗)_{𝑖,2𝑙−𝑘−𝑖+3}𝛾_{𝑖,2𝑙−𝑘−𝑖+3}𝜓𝛾,1≤𝑗≤𝑙,_{𝑙+1,𝑙+1}=𝑐̂^(𝑙+1)_{𝑙+1,𝑙+1}𝛾_{𝑙+1,𝑙+1}+^𝑙_𝑘=2^{𝑙−𝑘+1}_𝑖=1𝑎^(𝑙+1)_{𝑖,𝑖+𝑘}𝛼_{𝑖,𝑖+𝑘}+^𝑙_𝑘=2^𝑙+1_{[]𝑖=𝑙+2−(𝑘+1)/2}𝑐̂^(𝑙+1)_{2𝑙−𝑘−𝑖+3,𝑖}𝛾_{2𝑙−𝑘−𝑖+3,𝑖}+^2𝑙+1_𝑘=𝑙+1^{𝑙+1−[𝑘/2]}_𝑖=1𝑐̂^(𝑙+1)_{𝑖,2𝑙−𝑘−𝑖+3}𝛾_{𝑖,2𝑙−𝑘−𝑖+3}. (4.16) Then 𝜑𝛼_𝑗,𝑗+1=𝜑𝛼_𝑗𝑗,𝜑𝛼_𝑗,𝑗+1,𝜑𝛼_{𝑗+1,𝑗+1}=𝑎^(𝑗)_𝑗,𝑗+1𝛼_𝑗,𝑗+1−𝑐̃^(𝑗)_𝑗,𝑗+1𝛾_𝑗,𝑗+1,𝑗=1,…,𝑙. (4.17) In addition, 𝜑𝛼_𝑗,𝑗+1=𝜑𝛼_𝑗𝑗,𝑎^(𝑗)_𝑗,𝑗+1𝛼_𝑗,𝑗+1−𝑏̃^(𝑗)_𝑗,𝑗+1𝛽_𝑗,𝑗+1,𝜑𝛼_{𝑗+1,𝑗+1}=𝑎^(𝑗)_𝑗,𝑗+1𝛼_𝑗,𝑗+1+𝑐̃^(𝑗)_𝑗,𝑗+1𝛾_𝑗,𝑗+1,𝑗=1,…,𝑙. (4.18) Thus 𝑐̃^(𝑗)_𝑗,𝑗+1=0,𝑗=1,…,𝑙. So 𝜓𝛼_𝑗,𝑗+1=𝑎^(𝑗)_𝑗,𝑗+1𝛼_𝑗,𝑗+1,𝑗=1,…,𝑙. (4.19) Furthermore, 2𝜓𝛾_{𝑙+1,𝑙+1}=𝜓𝛼_{𝑙+1,𝑙+1},𝜓𝛾_{𝑙+1,𝑙+1}=2𝑐̂^(𝑙+1)_{𝑙+1,𝑙+1}𝛾_{𝑙+1,𝑙+1}+^𝑙−1_𝑖=1𝑎^(𝑙+1)_𝑖,𝑙+1𝛼_𝑖,𝑙+1+^𝑙−1_𝑖=1𝑐̂^(𝑙+1)_𝑖,𝑙+1𝛾_𝑖,𝑙+1. (4.20) From [𝜓(𝛼_𝑖𝑖),𝜓(𝛾_{𝑙+1,𝑙+1})]=0(𝑖≠𝑙−1), we have 𝑎^(𝑙+1)_𝑖,𝑙+1=0 and 𝑐̂^(𝑙+1)_𝑖,𝑙+1=0, 𝑖=1,⋯,𝑙−1, that is, 𝜓𝛾_{𝑙+1,𝑙+1}=𝑐̂^(𝑙+1)_{𝑙+1,𝑙+1}𝛾_{𝑙+1,𝑙+1}. (4.21) Let 𝑑=diag(𝑑₁,…,𝑑_𝑙+1) and 𝑑₀=𝑐̂^(𝑙+1)_{𝑙+1,𝑙+1}𝑑²_𝑙+1, where 𝑑₁=1,𝑑_𝑗=∏^𝑗_𝑖=2𝑎^{(𝑗−𝑖+1)}_{𝑗−𝑖+1,𝑗−𝑖+2}, 𝑗=2,…,𝑙+1. Applying 𝜆_𝐷 to 𝜓(𝛼_𝑗,𝑗+1), 𝑗=1,…,𝑙, and 𝜓(𝛾_{𝑙+1,𝑙+1}), we get the result.

5. Proofs of Main Results

Proof of Theorem 1.1. By Lemmas 4.3, 4.5, 4.7, and 4.8 we have 𝜆_𝐷𝜃𝜓(𝛼_𝑗𝑗)=𝛼_𝑗𝑗,𝑗=1,…,𝑙+1, 𝜆_𝐷𝜃𝜓(𝛼_𝑗,𝑗+1)=𝛼_𝑗,𝑗+1,𝑗=1,…,𝑙, and 𝜆_𝐷𝜃𝜓(𝛾_{𝑙+1,𝑙+1})=𝛾_{𝑙+1,𝑙+1}. Since the set {𝛾_{𝑙+1,𝑙+1}, 𝛼_{𝑙+1,𝑙+1}, 𝛼_𝑗𝑗, 𝛼_𝑗,𝑗+1∣𝑗=1,…,𝑙} generates 𝐭^(𝐶)_𝑙+1(𝑅), we know that 𝜆_𝐷𝜃𝜓 is the identity automorphism of 𝐭^(𝐶)_𝑙+1(𝑅). Hence 𝜓=𝜃′𝜆_𝐷−1. The uniqueness of the decomposition follows from Theorem 1.2.

Proof of Theorem 1.2. By the first part of Theorem 1.1 we have Aut(𝐭^(𝐶)_𝑙+1(𝑅))=ℐ𝒟. For any 𝑥∈𝐭^(𝐶)_𝑙+1(𝑅) and 𝛼_𝑖𝑗∈𝐧₁^(𝐶) we have 𝜆_𝐷𝜃_{𝑎𝛼𝑖𝑗}𝜆_𝐷⁻¹(𝑥)=𝐷𝐼_𝑛+𝑎𝛼_𝑖𝑗𝐷⁻¹𝑥𝐷𝐼_𝑛+𝑎𝛼_𝑖𝑗⁻¹𝐷⁻¹=𝐼_𝑛+𝑎𝜆_𝐷𝛼_𝑖𝑗𝑥𝐼_𝑛+𝑎𝜆_𝐷𝛼_𝑖𝑗⁻¹=𝜃_{𝑎𝜆𝐷(𝛼𝑖𝑗)}(𝑥). (5.1) So 𝜆_𝐷𝜃_{𝑎𝛼𝑖𝑗}=𝜃_{𝑎𝜆𝐷(𝛼𝑖𝑗)}𝜆_𝐷. For 𝛾_𝑖𝑗∈𝐧₁^(𝐶), we have 𝜆_𝐷𝜃_{𝑐𝛾𝑖𝑗}=𝜃_{𝑐𝜆𝐷(𝛾𝑖𝑗)}𝜆_𝐷. Therefore, ℐ⊲ℐ𝒟. Obviously ℐ∩𝒟=1. Then, ℐ𝒟=ℐ⋉𝒟.

6. Discussion for 𝑙=0

In this case, 𝐭₁^(𝐶)(𝑅) is generated by 𝛼₁₁ and 𝛾₁₁. For any automorphism 𝜓 of 𝐭₁^(𝐶)(𝑅), write 𝜓(𝛼₁₁) and 𝜓(𝛾₁₁), respectively, as 𝜓(𝛼₁₁)=𝑎₁₁𝛼₁₁+𝑐₁₁𝛾₁₁ and 𝜓(𝛾₁₁)=𝑐𝛾₁₁, where 𝑐∈𝑅^∗. From 2𝜓(𝛾₁₁)=[𝜓(𝛼₁₁),𝜓(𝛾₁₁)], we have 𝑎₁₁=1. Then 𝜃_{2−1𝑐11𝛾11}𝜓(𝛼₁₁)=𝛼₁₁ and 𝜃_{2−1𝑐11𝛾11}𝜓(𝛾₁₁)=𝑐𝛾₁₁. Also 𝜂_𝑐𝜃_{2−1𝑐11𝛾11}𝜓(𝛼₁₁)=𝛼₁₁ and 𝜂_𝑐𝜃_{2−1𝑐11𝛾11}𝜓(𝛾₁₁)=𝛾₁₁. So 𝜓=𝜃_{−2−1𝑐11𝛾11}𝜂_𝑐−1.

Acknowledgments

The research was supported partially by National Natural Science Foundation of China (Grant nos. 10871056 and 10971150) and by Science Research Foundation in Harbin Institute of Technology (Grant no. HITC200708).