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

Authors: Xing Tao Wang; Lei Zhang
Journal: International Journal of Mathematics and Mathematical Sciences
 (2012)

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## Abstract

LetCl+1(R) be the 2(l+1)×2(l+1) matrix symplectic Lie algebra over a commutative ring R with 2 invertible. Then tl+1CR  =  {m-1m-20-m-1T ∣ m̅1 is an l+1 upper triangular matrix, m̅2T=m̅2,overR} is the solvable subalgebra of Cl+1(R). In this paper, we give an explicit description of the automorphism group of tl+1(C)(R).


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## Body

## 1. Introduction

Classical Lie algebras occupy an important place in matrix algebras. LetR be a commutative ring R with the identity 1 and R* the group of invertible elements in R. Let Mn(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∈Mn(R). The symplectic Lie algebra
(1.1)Cl+1(R)={X∣X(0Il-Il0)+(0Il-Il0)XT=0,X∈M2l+2(R)}
is one of classical Lie algebras, where T denotes the matrix transpose. It is easy to show that the following subalgebra of Cl+1(R) such that
(1.2)tl+1(C)(R)={(m-1m-20-m-1T)∣m-1isanl+1uppertriangularmatrix,m-2T=m-2}
is solvable.Lete be the identity matrix in Mn(R) and let eij(n) denotes the matrix in Mn(R) all of whose entries are 0, except the (i,j)th entry which is 1. Let
(1.3)αi,i+k=ei,i+k(n)-el+i+k+1,l+i+1(n),i=1,…,l-k+1,k=0,1,…,l,γi,i+k=ei,l+i+k+1(n)+ei+k,l+i+1(n),i=1,…,l-k+1,k=1,…,l,γii=ei,l+i+1(n),i=1,…,l+1,
where n=2(l+1), l≥1. For discussion latter, we rewrite tl+1(C)(R) as
(1.4)tl+1(C)(R)=∑k=0l∑i=1l-k+1Rαi,i+k+∑k=0l∑i=1l-k+1Rγi,i+k.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 tl+1(C)(R) can be uniquely expressed as ψ=θλD, where θ and λD 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 automorphismψ of tl+1(C)(R)(l≥1) there are unique inner and diagonal automorphisms, θ and λD, respectively, of tl+1(C)(R) such that ψ=θλD.Theorem 1.2.
Letℐ and 𝒟 be the inner and diagonal automorphism groups, respectively. Then   Aut(tl+1(C)(R))=ℐ⋉𝒟, where   Aut(tl+1(C)(R)) denotes the automorphism group of tl+1(C)(R).


## 2. Preliminaries

Let(2.1)Pn={αi,i+k∣i=1,…,l-k+1,k=0,1,…,l},Wn={γi,i+k∣i=1,…,l-k+1,k=0,1,…,l}.
Then the set Pn∪Wn is a basis of tl+1(C)(R).Lemma 2.1.
LetHn be the set generated by the set {αjj,αi,i+1,γl+1,l+1∣1≤j≤l+1,1≤i≤l}, where n=2(l+1). Then Hn=tl+1(C)(R).Proof.
We only need to show thattl+1(C)(R)⊆Hn. It is obvious that αi,i+k∈Hn, when k=0,1. When k=2, we have
(2.2)αi,i+2=ei,i+2(n)-el+i+3,l+i+1(n)=[(ei,i+1(n)-el+i+2,l+i+1(n)),(ei+1,i+2(n)-el+i+3,l+i+2(n))]=[αi,i+1,αi+1,i+2]∈Hn.
Assume that αi,i+k-1∈Hn, then αi,i+k=[αi,i+k-1,αi+k-1,i+k]∈Hn, that is, Pn⊆Hn. Since γl+1,l+1∈Hn, for any γl-k+1,l-k+2∈Wn, when k=1,
(2.3)γl,l+1=el,2l+2(n)+el+1,2l+1(n)=[(el,l+1(n)-e2l+2,2l+1(n)),el+1,2l+2(n)]=[αl,l+1,γl+1,l+1]∈Hn.
Assume that when k=m-1, γl-m+2,l-m+3∈Hn, then when k=m, γl-m+1,l-m+2=[αl-m+1,l-m+3,γl-m+2,l-m+3]∈Hn, that is, γi,i+1∈Hn, i=1,…,l. For any γi,i+k∈Hn(k≥1), when k=1, γi,i+1∈Hn. When k≥2, γi,i+k=[αi,i+k-1,γi+k-1,i+k]∈Hn. Since 2γii=[αi,i+1,γi,i+1]∈Hn, and 2 is invertible, we have γii∈Hn for 1≤i≤l. Thus Wn⊆Hn. Because Pn∪Wn is a basis of tl+1(C)(R), we obtain tl+1(C)(R)⊆Hn.Now, denotetl+1(C)(R) by n0(C). Let n1(C)=[n0(C),n0(C)], n2(C)=[n1(C),n1(C)], nj(C)=[n1(C),nj-1(C)], j=3,…,2l+1. It is not difficult to know
(2.4)nj(C)=∑k=jl∑i=1l-k+1Rαi,i+k+∑k=jl∑i=l+2-[(k+1)/2]l+1Rγ2l-k-i+3,i+∑k=l+12l+1∑i=1l+1-[k/2]Rγi,2l-k-i+3,1≤j≤l,nj(C)=∑k=j2l+1∑i=1l+1-[k/2]Rγi,2l-k-i+3,l+1≤j≤2l+1(l≥2),nj(C)=0,2l+2≤j.It is easy to check that[nm(C),nl(C)]⊆nm+l(C) for m+l≤2l+1 or [nm(C),nl(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(nj(C))=nj(C), j=2,…,2l+1. Therefore, ψ(nj-1(C)∖nj(C))=nj-1(C)∖nj(C), j=1,…,2l+1. Note that if γij∈Wn, then γij∈n2l-i-j+3(C)∖n2l-i-j+4(C).For any maximal idealM of R, R-=R/M is a field. The natural homomorphism π:R→R- induces a homomorphism ψM:tl+1(C)(R)→tl+1(C)(R-) which is surjective. So every automorphism ψ of tl+1(C)(R) may induce an automorphism ψ- of tl+1(C)(R-). Using this fact and that n2l+1(C)=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-(γ-11)=0 on tl+1(C)(R-), where γ-11 is the image of γ11 in tl+1(C)(R-), which is impossible.Lemma 2.2.
Letψ be in Aut(n0(C)). If ψ(αjj), ψ(αj,j+1) and ψ(γl+1,l+1) are expressed, respectively, as
(2.5)ψ(αjj)=∑i=1l+1aii(j)αiimod⁡n1(C),j=1,…,l+1,(2.6)ψ(αj,j+1)=∑i=1la~i,i+1(j)αi,i+1+c~l+1,l+1(j)γl+1,l+1mod⁡n2(C),j=1,…,l,(2.7)ψ(γl+1,l+1)=∑i=1la^i,i+1(l+1)αi,i+1+c^l+1,l+1(l+1)γl+1,l+1mod⁡n2(C),
then the following matrices are invertible.(i)
A=(aji)(l+1)×(l+1), where aji=aii(j), j=1,…,l+1, i=1,…,l+1;(ii)
B=(bji)(l+1)×(l+1), where bji=a~i,i+1(j), j=1,…,l, i=1,…,l, bj,l+1=c~l+1,l+1(j), j=1,…,l, bl+1,i=a^i,i+1(l+1), i=1,…,l  and bl+1,l+1=c^l+1,l+1(l+1).Proof.
(i) ThatA 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 {αjj+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 {αj,j+1+n2(C),γl+1,l+1+n2(C)∣j=1,…,l}.Lemma 2.3.
Letψ∈  Aut(n0(C))(l≥2). Write ψ(αjj), ψ(αj,j+1), and ψ(γl+1,l+1) as in (2.5)–(2.7),  respectively. Then the following conclusions hold.(i)
For1≤m,k,h≤l, a~h,h+1(m)a~k,k+1(m)=0(h≠k), a~h,h+1(m)c~l+1,l+1(m)=0, a^h,h+1(l+1)a^k,k+1(l+1)=0(h≠k)  and a^h,h+1(l+1)c^l+1,l+1(l+1)=0.(ii)
For1≤k,h≤l, (ahh(i)-ah+1,h+1(i))(akk(i)-ak+1,k+1(i))=0(h≠k)   and (ahh(i)-ah+1,h+1(i))al+1,l+1(i)=0(1≤h≤l,herel≥1), where  i=1,l+1.(iii)
For2≤m≤l and 1≤i,k,h≤l, (aii(m)-ai+1,i+1(m))(ahh(m)-ah+1,h+1(m))(akk(m)-ak+1,k+1(m))=0(i≠h≠k≠i,herel≥3) and (aii(m)-ai+1,i+1(m))(ahh(m)-ah+1,h+1(m))al+1,l+1(m)=0(1≤i≠h≤l).Proof.
(i) Whenj≠m,m+1, [ψ(αjj),ψ(αm,m+1)]=0. So
(2.8)a~i,i+1(m)(aii(j)-ai+1,i+1(j))=0,c~l+1,l+1(m)al+1,l+1(j)=0.
From [ψ(αmm),ψ(αm,m+1)]=ψ(αm,m+1) and [ψ(αm+1,m+1),ψ(αm,m+1)]=-ψ(αm,m+1), we have
(2.9)a~i,i+1(m)(aii(m)-ai+1,i+1(m))=a~i,i+1(m),2c~l+1,l+1(m)al+1,l+1(m)=c~l+1,l+1(m),a~i,i+1(m)(aii(m+1)-ai+1,i+1(m+1))=-a~i,i+1(m),2c~l+1,l+1(m)al+1,l+1(m+1)=-c~l+1,l+1(m).
When j≠l+1, [ψ(αjj),ψ(γl+1,l+1)]=0. So
(2.10)a^i,i+1(l+1)(aii(j)-ai+1,i+1(j))=0,c^l+1,l+1(l+1)al+1,l+1(j)=0.
From [ψ(αl+1,l+1),ψ(γl+1,l+1)]=2ψ(γl+1,l+1), we have
(2.11)a^i,i+1(l+1)(aii(l+1)-ai+1,i+1(l+1))=2a^i,i+1(l+1),c^l+1,l+1(l+1)al+1,l+1(l+1)=c^l+1,l+1(l+1).
Let C=(cji)(l+1)×(l+1), where cji=aii(j)-ai+1,i+1(j), j=1,…,l+1, i=1,…,l, and cj,l+1=2al+1,l+1(j),  j=1,…,l+1. By Lemma 2.2, detA∈R*, so detC∈R*. Investigating a~h,h+1(m)a~k,k+1(m)detC, we may find that hth column and kth column are linearly dependent  (both are the form (0,…,0,a~j,j+1(m),-a~j,j+1(m),0,…,0)t,j=h,k) by (2.6) and (2.7), so a~h,h+1(m)a~k,k+1(m)detC=0. Similarly, a~h,h+1(m)c~l+1,l+1(m)detC=0, a^h,h+1(l+1)a^k,k+1(l+1)detC=0(h≠k) and a^h,h+1(l+1)c^l+1,l+1(l+1)detC=0. Then a~h,h+1(m)a~k,k+1(m)=0(h≠k), a~h,h+1(m)c~l+1,l+1(m)=0, a^h,h+1(l+1)a^k,k+1(l+1)=0(h≠k), and a^h,h+1(l+1)c^l+1,l+1(l+1)=0.
(ii) Wheni=1, from (ahh(1)-ah+1,h+1(1))(akk(1)-ak+1,k+1(1))detB=0(h≠k), we have (ahh(1)-ah+1,h+1(1))(akk(1)-ak+1,k+1(1))=0(h≠k). Similarly, we have (ahh(1)-ah+1,h+1(1))(all(1)+al+1,l+1(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.
Letψ∈Aut(n0(C)). Then(i)
whenl≥1, ψ(α12)=a~12(1)α12modn2(C), where a~12(1)∈R*;(ii)
ifψ(α12)=a~12(1)α12modn2(C),  where a~12(1)∈R*, then ψ(αi,i+1)=a~i,i+1(i)αi,i+1modn2(C) and ψ(γl+1,l+1)=c^l+1,l+1(l+1)γl+1,l+1modn2(C),  where a~i,i+1(i),c^l+1,l+1(l+1)∈R*.Proof.
(i) Noting thatαi,i+1,γl+1,l+1∈n1(C)∖n2(C) and γ12∈n2l(C)∖n2l+1(C), we have ψ(α12)∈n1(C)∖n2(C) and ψ(γ12)=c^12(12)γ12modn2l+1(C)∈n2l(C)∖n2l+1(C), where c^12(12)∈R*. Using (2.7), from [ψ(α11),ψ(γ12)]=ψ(γ12), we have c^12(12)(a11(1)+a22(1))=c^12(12), that is, a11(1)+a22(1)=1. Write ψ(α11) and ψ(γ11) as ψ(α11)=∑i=1l+1‍aii(1)αiimodn1(C) and ψ(γ11)=c11*γ11∈n2l+1(C), where c11*∈R*. From 2ψ(γ11)=[ψ(α11),ψ(γ11)]=2a11(1)c11*γ11, we have a11(1)=1. Then a22(1)=0. By Lemma 2.3 we have aii(1)-ai+1,i+1(1)=0,i=2,…,l  (here  l≥2) and al+1,l+1(1)=0. So aii(1)=0,i=2,…,l+1, that is, ψ(α11)=a11(1)modn1(C). Then ψ(α12)=[ψ(α11),ψ(α12)]=a~12(1)α12modn2(C) and a~12(1)∈R*. By Lemma 2.3, (i) holds.
(ii) Writeψ(αj,j+1) and ψ(γl+1,l+1) as (2.6) and (2.7), respectively. From ψ(α13)=[ψ(α12),ψ(α23)], we have ψ(α13)=a~12(1)a~23(2)α13modn3(C). Since α13∈n2(C)∖n3(C), ψ(α13)∈n2(C)∖n3(C). So a~12(1)a~23(2)∈R*, that is, a~23(2)∈R*. In general, for m=2,…,l, we have
(2.12)ψ(α1,m+1)=∏i=1ma~i,i+1(i)α1,m+1mod⁡nm+1(C)∈nm(C)∖nm+1(C),m=1,…,l,ψ(γ1,l+1)=[ψ(α1,l+1),ψ(γl+1,l+1)]=∏i=1la~i,i+1(i)c^l+1,l+1(l+1)γ1,l+1mod⁡nl+2(C)∈nl+1(C)∖nl+2(C),
here a~i,i+1(i),c^l+1,l+1(l+1) should be in R*, i=1,…,l. By Lemma 2.3 we have that a~i,i+1(j)=0(i≠j), c~l+1,l+1(j)=0, j=1,…,l, and a^i,i+1(l+1), i=1,…,l. Hence ψ(αl,l+1) and ψ(γl+1,l+1) have the required forms, respectively.


## 3. The Standard Automorphisms oftl+1(C)(R)

Now let us introduce two types of Lie automorphisms oftl+1(C)(R).(i) Inner Automorphisms
Letr=In+aαij(i≠j) or r=In+aγij. It is easy to check that ryr-1∈n0(C). The map θr: tl+1(C)(R)→tl+1(C)(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 (0Il+1-Il+10)). We denote θr by θaαij, θaγij, respectively. In these cases, we have θaαij-1=θ-aαij, θaγij-1=θ-aγij, respectively, and that θaαij(αii)=αii-aαij, θaαij(αjj)=αjj+aαij, θaαij(αkk)=αkk(k≠i,j), θaαij(αk,k+1)=αk,k+1(k≠j,i-1), θaαij(αj,j+1)=αj,j+1+aαi,j+1, θaαij(αi-1,i)=αi-1,i-aαi-1,j, θaγii(αii)=αii-2aγii, θaγi,i+1(αi,i+1)=αi,i+1-2aγii and θaαi,i+1(γi,i+1)=γi,i+1+2aγii. All inner automorphisms of tl+1(C)(R) generate a subgroup of Aut(n0(C)), which is denoted by ℐ.(ii) Diagonal Automorphisms
Letdi∈R*, i=0,1,…,l+1, d=diag(d1,…,dl+1) and D=diag(d,d-1d0). The map λD: tl+1(C)(R)→tl+1(C)(R) such that x↦DxD-1, x∈n0(C), defines an automorphism of tl+1(C)(R), which is called a diagonal automorphism. It is clear that λDλD-=λDD-. So the set of diagonal automorphisms of tl+1(C)(R) is a subgroup of Aut(n0(C)), which is denoted by 𝒟.


## 4. Lemmas for Main Results

Lemma 4.1.
Letψ∈Aut(n0(C)). The following two statements are equivalent:
(i)ψ(αj,j+1)=a~j,j+1(j)αj,j+1modn2(C) and  ψ(γl+1,l+1)=c^l+1,l+1(l+1)γl+1,l+1modn2(C), where  a~j,j+1(j),c^l+1,l+1(l+1)∈R*, j=1,…,l;
(ii)ψ(αjj)=αjjmodn1(C),j=1,…,l+1.Proof.
(i)⇒(ii). Write ψ(αjj) as in (2.5). By the process of proving Lemma 2.3, we have a~12(1)(a11(1)-a22(1))=a~12(1), a~i,i+1(i)(aii(i+1)-ai+1,i+1(i+1))=-a~i,i+1(i), a~i+1,i+2(i+1)(ai+1,i+1(i+1)-ai+2,i+2(i+1))=a~i+1,i+2(i+1), i=1,…,l-1 and a~l,l+1(l)(all(l+1)-al+1,l+1(l+1))=-a~l,l+1(l), c^l+1,l+1(l+1)al+1,l+1(l+1)=c^l+1,l+1(l+1). Then we obtain that a11(1)-a22(1)=1, aii(i+1)-ai+1,i+1(i+1)=-1, ai+1,i+1(i+1)-ai+2,i+2(i+1)=1, i=1,…,l-1 and al+1,l+1(l+1)=1. By Lemma 2.3, we have ajj(j)=1(1≤j≤l+1) and aii(j)=0(i≠j).
(ii)⇒(i). Write ψ(αj,j+1) and ψ(γl+1,l+1), respectively, as in (2.6) and (2.7). Then
(4.1)ψ(αj,j+1)=[ψ(αjj),ψ(αj,j+1)]=a~j,j+1(j)αj,j+1mod⁡n2(C),j=1,…,l,2ψ(γl+1,l+1)=[ψ(αl+1,l+1),ψ(γl+1,l+1)]=2c^l+1,l+1(l+1)γl+1,l+1mod⁡n2(C),
that is, ψ(γl+1,l+1)=c^l+1,l+1(l+1)γl+1,l+1modn2(C). By the method of modularizing a maximal ideal of R to a residue field, we know that a~j,j+1(j),c^l+1,l+1(l+1)∈R*, j=1,…,l.Lemma 4.2.
Letψ be in Aut(n0(C)). If ψ(αjj)=αjjmodn1(C), then
(4.2)ψ(α11)=α11+a12(1)α12mod⁡n2(C),ψ(αjj)=αjj-aj-1,j(j-1)αj-1,j+aj,j+1(j)αj,j+1mod⁡n2(C),j=2,…,l(l≥2),ψ(αl+1,l+1)=αl+1,l+1-al,l+1(l)αl,l+1+cl+1,l+1(l+1)γl+1,l+1mod⁡n2(C).Proof.
We expressψ(αjj) as
(4.3)ψ(αjj)=αjj+∑i=1lai,i+1(j)αi,i+1+cl+1,l+1(j)γl+1,l+1mod⁡n2(C),j=1,…,l+1.
From [ψ(αjj),ψ(αkk)]=0(j≠k) we have
(4.4)ψ(α11)=α11+a12(1)α12mod⁡n2(C),ψ(αjj)=αjj+aj-1,j(j)αj-1,j+aj,j+1(j)αj,j+1mod⁡n2(C),j=2,…,l(l≥2),ψ(αl+1,l+1)=αl+1,l+1+al,l+1(l+1)αl,l+1+cl+1,l+1(l+1)γl+1,l+1mod⁡n2(C),
where aj,j+1(j)+aj,j+1(j+1)=0,j=1,…,l. Lemma 4.2 is proved.Lemma 4.3.
Letψ be in Aut(n0(C)). If every ψ(αjj) is expressed as the form in Lemma 4.2, one may find an inner automorphism
(4.5)θ=∏j=1lθaj,j+1(j)αj,j+1θ2-1cl+1,l+1(l+1)γl+1,l+1
such that
(4.6)θψ(αjj)=αjjmod⁡n2(C),j=1,…,l+1.Proof.
Note thatθ2-1cl+1,l+1(l+1)γl+1,l+1(αl+1,l+1)=αl+1,l+1-cl+1,l+1(l+1)γl+1,l+1. Then, by Lemma 4.2, it is not difficult to prove Lemma 4.3.Lemma 4.4.
Letψ be in  Aut(n0(C)). If  ψ(αjj)=αjjmodnk(C),j=1,…,l+1(1≤k≤l+1), then
(4.7)ψ(αjj)=αjj+aj,j+k(j)αj,j+kmod⁡nk+1(C),j=1,…,min⁡{k,l-k+1}(k≤l,l≥1),ψ(αjj)=αjj-aj-k,j(j-k)αj-k,j+aj,j+k(j)αj,j+kmod⁡nk+1(C),j=k+1,…,l-k+1(k≤[l+12],l≥2),ψ(αjj)=αjj-aj-k,j(j-k)αj-k,j+cj,2l-k-j+3(j)γj,2l-k-j+3mod⁡nk+1(C),j=l-k+2,…,l-[k2]+1(k≤[l+12],l≥1),ψ(αjj)=αjj+cj,2l-k-j+3(j)γj,2l-k-j+3mod⁡nk+1(C),j=l-k+2,…,l-[k2]+1(1+[l+12]≤k,l≥2),ψ(αjj)=αjj+cj,2l-k-j+3(j)γj,2l-k-j+3mod⁡nk+1(C),j=l2+1(k=l2+1,l≥4,herepeven),ψ(αjj)=αjj-aj-k,j(j-k)αj-k,j+cj,2l-k-j+3(j)γj,2l-k-j+3mod⁡nk+1(C),j=l2+2,…,l-[k2]+1(k=l2+1,l≥4,herepeven),ψ(αjj)=αjj+c2l-k-j+3,j(2l-k-j+3)γ2l-k-j+3,jmod⁡nk+1(C),j=l-[k2]+2,…,k(l+2≤k+[k2],l≥3),ψ(αjj)=αjj-aj-k,j(j-k)αj-k,j+c2l-k-j+3,j(2l-k-j+3)γ2l-k-j+3,jmod⁡nk+1(C),j=max⁡{k+1,l-[k2]+2},…,l+1(k≤p,[k2]≥1,l≥2).Proof.
We expressψ(αjj),j=1,…,l+1, as
(4.8)ψ(αjj)=αjj+∑i=1l-k+1ai,i+k(j)αi,i+k+∑i=l+2-[(k+1)/2]l+1c2l-k-i+3,i(j)γ2l-k-i+3,imod⁡nk+1(C).
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.
Letψ be in Aut(n0(C)). If every ψ(αjj)  be expressed as the form in Lemma 4.4, one may find an inner automorphism
(4.9)θ=∏j=1l-k+1θaj,j+k(j)αj,j+k∏j=l-k+2l-[k/2]+1θ(1+δjl)-1cj,2l-k-j+3(j)γj,2l-k-j+3,
where
(4.10)δij={1,i=j,0,i≠j,h=l+1-(k-1)/2(kanodd).
Then
(4.11)θψ(αjj)=αjjmod⁡nk+1(C),i=1,…,l+1.Proof.
Applyθ to ψ(αjj) and use Lemma 4.4 to obtain Lemma 4.5.Lemma 4.6.
Letψ be in  Aut(n0(C)). If ψ(αjj)=αjjmodnk(C),j=1,…,l+1, l+1≤k≤2l+1(l≥1), then
(4.12)ψ(αjj)=αjj+cj,2l-k-j+3(j)γj,2l-k-j+3mod⁡nk+1(C),j=1,…,l-[k2]+1,ψ(αjj)=αjj+c2l-k-j+3,j(2l-k-j+3)γ2l-k-j+3,jmod⁡nk+1(C),j=l-[k2]+2,…,2l-k+2,ψ(αjj)=αjjmod⁡nk+1(C),j=2l-k+3,…,l+1.Proof.
We expressψ(αjj),j=1,…,l+1, as
(4.13)ψ(αjj)=αjj+∑i=1l-[k/2]+1ci,2l-k-i+3(j)γi,2l-k-i+3mod⁡nk+1(C).
The process of proving Lemma 4.6 is similar to Lemma 4.2.Lemma 4.7.
Letψ be in Aut(n0(C)). If every ψ(αjj) is expressed as the form in Lemma 4.6, one may find an inner automorphism
(4.14)θ=∏j=1l-[k/2]+1θ(1+δjh)-1cj,2l-k-j+3(j)γj,2l-k-j+3,
where h=l+1-(k-1)/2 (k an odd).  Then
(4.15)θψ(αjj)=αjjmod⁡nk+1(C).
When  k=2l+1, θψ(αjj)=αjj,j=1,…,l+1.Proof.
It is similar to proving Lemma4.5.Lemma 4.8.
Whenl≥1, let ψ be in Aut(n0(C)). If ψ(αjj)=αjj,j=1,…,l+1, there exists a diagonal automorphism λD such that λDψ(αj,j+1)=αj,j+1, i=1,…,l, and λDψ(γl+1,l+1)=γl+1,l+1.Proof.
By Lemma4.1 we know that ψ(αj,j+1)=a~j,j+1(j)αj,j+1modn2(C) and ψ(γl+1,l+1)=c^l+1,l+1(l+1)γl+1,l+1modn2(C), where a~j,j+1(j),c^l+1,l+1(l+1)∈R*, j=1,…,l. We express ψ(αj,j+1) and ψ(γl+1,l+1), respectively, as
(4.16)ψ(αj,j+1)=a~j,j+1(j)αj,j+1+∑k=2l∑i=1l-k+1a~i,i+k(j)αi,i+k+∑k=2l∑i=l+2-[(k+1)/2]l+1c~2l-k-i+3,i(j)γ2l-k-i+3,i+∑k=l+12l+1∑i=1l+1-[k/2]c~i,2l-k-i+3(j)γi,2l-k-i+3,1≤j≤l,ψ(γl+1,l+1)=c^l+1,l+1(l+1)γl+1,l+1+∑k=2l‍∑i=1l-k+1‍a^i,i+k(l+1)αi,i+k+∑k=2l∑i=l+2-[(k+1)/2]l+1c^2l-k-i+3,i(l+1)γ2l-k-i+3,i+∑k=l+12l+1∑i=1l+1-[k/2]c^i,2l-k-i+3(l+1)γi,2l-k-i+3.
Then
(4.17)φ(αj,j+1)=[φ(αjj),[φ(αj,j+1),φ(αj+1,j+1)]]=a~j,j+1(j)αj,j+1-c~j,j+1(j)γj,j+1,j=1,…,l.
In addition,
(4.18)φ(αj,j+1)=[φ(αjj),[(a~j,j+1(j)αj,j+1-b~j,j+1(j)βj,j+1),φ(αj+1,j+1)]]=a~j,j+1(j)αj,j+1+c~j,j+1(j)γj,j+1,j=1,…,l.
Thus c~j,j+1(j)=0,j=1,…,l. So
(4.19)ψ(αj,j+1)=a~j,j+1(j)αj,j+1,j=1,…,l.
Furthermore,
(4.20)2ψ(γl+1,l+1)=[ψ(αl+1,l+1),ψ(γl+1,l+1)]=2c^l+1,l+1(l+1)γl+1,l+1+∑i=1l-1a^i,l+1(l+1)αi,l+1+∑i=1l-1c^i,l+1(l+1)γi,l+1.
From [ψ(αii),ψ(γl+1,l+1)]=0(i≠l-1), we have a^i,l+1(l+1)=0 and c^i,l+1(l+1)=0, i=1,⋯,l-1, that is,
(4.21)ψ(γl+1,l+1)=c^l+1,l+1(l+1)γl+1,l+1.
Let d=diag(d1,…,dl+1) and d0=c^l+1,l+1(l+1)dl+12, where d1=1,dj=∏i=2j‍a~j-i+1,j-i+2(j-i+1), j=2,…,l+1. Applying λD to ψ(αj,j+1), j=1,…,l, and ψ(γl+1,l+1), we get the result.


## 5. Proofs of Main Results

Proof of Theorem1.1.
By Lemmas4.3, 4.5, 4.7, and 4.8 we have λDθψ(αjj)=αjj,j=1,…,l+1, λDθψ(αj,j+1)=αj,j+1,j=1,…,l, and λDθψ(γl+1,l+1)=γl+1,l+1. Since the set {γl+1,l+1, αl+1,l+1, αjj, αj,j+1∣j=1,…,l} generates tl+1(C)(R), we know that λDθψ is the identity automorphism of tl+1(C)(R). Hence ψ=θ′λD-1. The uniqueness of the decomposition follows from Theorem 1.2.Proof of Theorem1.2.
By the first part of Theorem1.1 we have Aut(tl+1(C)(R))=ℐ𝒟. For any x∈tl+1(C)(R) and αij∈n1(C) we have
(5.1)λDθaαijλD-1(x)=D(In+aαij)D-1xD(In+aαij)-1D-1=(In+aλD(αij))x(In+aλD(αij))-1=θaλD(αij)(x).
So λDθaαij=θaλD(αij)λD. For γij∈n1(C), we have λDθcγij=θcλD(γij)λD. Therefore,  ℐ⊲ℐ𝒟. Obviously ℐ∩𝒟=1. Then, ℐ𝒟=ℐ⋉𝒟.


## 6. Discussion forl=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 θ2-1c11γ11ψ(α11)=α11 and θ2-1c11γ11ψ(γ11)=cγ11. Also ηcθ2-1c11γ11ψ(α11)=α11 and ηcθ2-1c11γ11ψ(γ11)=γ11. So ψ=θ-2-1c11γ11ηc-1.



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*Source: 242736-2012-07-30.xml*