Volume 2012, Article ID 242736,12pages doi:10.1155/2012/242736
Research Article
Decomposition of Automorphisms of
Certain Solvable Subalgebra of Symplectic Lie Algebra over Commutative Rings
Xing Tao Wang and Lei Zhang
Department of Mathematics, Harbin Institute of Technology, Harbin 150001, China
Correspondence should be addressed to Xing Tao Wang,[email protected] Received 26 March 2012; Accepted 26 April 2012
Academic Editor: Helge Holden
Copyrightq2012 X. T. Wang and L. Zhang. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
LetCl1Rbe the 2l1×2l1matrix symplectic Lie algebra over a commutative ringRwith 2 invertible. ThentCl1R {m01−mm2T
1 | m1is anl1 upper triangular matrix,mT2 m2, over R}is the solvable subalgebra ofCl1R. In this paper, we give an explicit description of the auto- morphism group oftCl1R.
1. Introduction
Classical Lie algebras occupy an important place in matrix algebras. LetRbe a commutative ringR with the identity 1 andR∗ the group of invertible elements inR. LetMnRbe the R-algebra ofnbynmatrices overRthat has a structure of a Lie algebra overRwith bracket operationx, y xy−yxfor anyx, y∈MnR. The symplectic Lie algebra
Cl1R
X|X 0 Il
−Il 0
0 Il
−Il 0
XT 0, X∈M2l2R
1.1
is one of classical Lie algebras, whereT denotes the matrix transpose. It is easy to show that the following subalgebra ofCl1Rsuch that
tCl1R
m1 m2
0 −mT1
|m1 is anl1 upper triangular matrix, mT2 m2 1.2
is solvable.
Lete be the identity matrix inMnRand leteijndenotes the matrix inMnRall of whose entries are 0, except thei, jth entry which is 1. Let
αi,ikeni,ik−enlik1,li1, i1, . . . , l−k1, k0,1, . . . , l, γi,ikeni,lik1eik,li1n , i1, . . . , l−k1, k1, . . . , l,
γiiei,li1n , i1, . . . , l1,
1.3
wheren2l1,l≥1. For discussion latter, we rewritetCl1Ras
tCl1R l
k0 l−k1
i1
Rαi,ikl
k0 l−k1
i1
Rγi,ik. 1.4
Automorphisms of associative algebras have been explored in many articles 1–8.
Encouraged by Dokovi´c9and Cao’s10papers which described the automorphism groups of Lie algebra consisting of all upper triangularn×nmatrices of trace 0 over a connected commutative ring and a commutative ring withninvertible, respectively, in this paper we use similar techniques to those in11to prove that any automorphismψ oftCl1Rcan be uniquely expressed asψ θλD, where θ and λD are inner and diagonal automorphisms, respectively, forl≥1 andRis a commutative ring with 2 invertible. We also give an explicit description of the remaining casel0.
Theorem 1.1. For any automorphismψoftCl1R l≥1there are unique inner and diagonal auto- morphisms,θandλD, respectively, oftCl1Rsuch thatψθλD.
Theorem 1.2. Let I and D be the inner and diagonal automorphism groups, respectively. Then AuttCl1R ID, where AuttCl1Rdenotes the automorphism group oftCl1R.
2. Preliminaries
Let
Pn{αi,ik|i1, . . . , l−k1, k0,1, . . . , l}, Wn
γi,ik|i1, . . . , l−k1, k0,1, . . . , l .
2.1
Then the setPn∪Wnis a basis oftCl1R.
Lemma 2.1. LetHnbe the set generated by the set{αjj, αi,i1, γl1,l1|1≤j≤l1,1≤i≤l}, where n2l1. ThenHntCl1R.
Proof. We only need to show thattCl1R⊆ Hn. It is obvious thatαi,ik ∈Hn, whenk 0,1.
Whenk2, we have
αi,i2 eni,i2−eli3,li1n
eni,i1−eli2,li1n ,
ei1,i2n −eli3,li2n αi,i1, αi1,i2∈Hn.
2.2
Assume thatαi,ik−1∈Hn, thenαi,ik αi,ik−1, αik−1,ik∈Hn, that is,Pn⊆Hn. Sinceγl1,l1∈ Hn, for anyγl−k1,l−k2∈Wn, whenk1,
γl,l1enl,2l2el1,2l1n
enl,l1−en2l2,2l1
, enl1,2l2
αl,l1, γl1,l1
∈Hn.
2.3
Assume that whenkm−1,γl−m2,l−m3 ∈Hn, then whenk m,γl−m1,l−m2 αl−m1,l−m3, γl−m2,l−m3 ∈ Hn, that is,γi,i1 ∈ Hn,i 1, . . . , l. For anyγi,ik ∈ Hn k ≥ 1, whenk 1, γi,i1∈Hn. Whenk≥2,γi,ik αi,ik−1, γik−1,ik∈Hn. Since 2γii αi,i1, γi,i1∈Hn, and 2 is invertible, we haveγii ∈Hnfor 1≤i≤l. ThusWn⊆Hn. BecausePn∪Wnis a basis oftCl1R, we obtaintCl1R⊆Hn.
Now, denotetCl1R bynC0 . Let nC1 nC0 ,nC0 ,nC2 nC1 ,nC1 ,nCj nC1 , nCj−1,j 3, . . . ,2l1. It is not difficult to know
nCj l
kj l−k1
i1
Rαi,ikl
kj
l1
il2−k1/2
Rγ2l−k−i3,i
2l1
kl1
l1−k/2
i1
Rγi,2l−k−i3, 1≤j≤l,
nCj 2l1
kj
l1−k/2
i1
Rγi,2l−k−i3, l1≤j ≤2l1l≥2, nCj 0, 2l2≤j.
2.4
It is easy to check thatnCm ,nCl ⊆nCmlforml≤2l1 ornCm ,nCl 0 forml≥ 2l2. For anyψ ∈ AutnC0 , we haveψnC1 ψnC0 , ψnC0 nC0 ,nC0 nC1 and ψnCj nCj ,j 2, . . . ,2l1. Therefore,ψnCj−1\nCj nCj−1\nCj ,j 1, . . . ,2l1. Note that ifγij ∈Wn, thenγij∈nC2l−i−j3\nC2l−i−j4.
For any maximal idealMofR,R R/Mis a field. The natural homomorphismπ : R → R induces a homomorphism ψM : tCl1R → tCl1R which is surjective. So every
automorphismψ oftCl1Rmay induce an automorphismψ of tCl1R. Using this fact and thatnC2l1 Rγ11 for l ≥ 1, we have thatψγ11 c11γ11, wherec11 ∈ R∗. Otherwise,c11
should be contained in a maximal idealMofR, thenψγ11 0 ontCl1R, whereγ11 is the image ofγ11intCl1R, which is impossible.
Lemma 2.2. Letψbe in AutnC0 . Ifψαjj,ψαj,j1andψγl1,l1are expressed, respectively, as
ψ αjj
l1
i1
ajii αiimodnC1 , j1, . . . , l1, 2.5
ψ αj,j1
l
i1
aji,i1αi,i1cjl1,l1γl1,l1modnC2 , j1, . . . , l, 2.6
ψ γl1,l1
l
i1
al1i,i1αi,i1cl1l1,l1γl1,l1modnC2 , 2.7
then the following matrices are invertible.
iA ajil1×l1, whereajiajii ,j1, . . . , l1,i1, . . . , l1;
iiB bjil1×l1, wherebjiaji,i1,j 1, . . . , l,i1, . . . , l,bj,l1 cjl1,l1,j 1, . . . , l, bl1,ial1i,i1,i1, . . . , landbl1,l1 cl1l1,l1.
Proof. iThatAis invertible follows from the fact thatψ induces an automorphism of the freeR-modulenC0 /nC1 of rankl1 on the basis{αjjnC1 | j 1, . . . , l1}.iiNote that ψinduces an automorphism of the freeR-modulenC1 /nC2 of rankl1 on the basis{αj,j1 nC2 , γl1,l1nC2 | j1, . . . , l}.
Lemma 2.3. Letψ ∈ AutnC0 l ≥ 2. Writeψαjj,ψαj,j1, andψγl1,l1as in2.5–2.7, respectively. Then the following conclusions hold.
iFor 1 ≤ m, k, h≤ l,amh,h1amk,k1 0h /k,amh,h1cml1,l1 0,al1h,h1al1k,k1 0h /k and al1h,h1cl1,l1l1 0.
iiFor 1≤k, h≤l,aihh−aih1,h1aikk−aik1,k1 0 h /k andaihh−aih1,h1ail1,l10 1≤h≤l, herel≥1, where i1, l1.
iiiFor 2 ≤m ≤land 1 ≤ i, k, h ≤l,amii −ami1,i1amhh −amh1,h1amkk −amk1,k1 0 i /h /k /i, herel≥3andamii −ami1,i1amhh −amh1,h1aml1,l101≤i /h≤l.
Proof. iWhenj /m, m1,ψαjj, ψαm,m1 0. So
ami,i1
ajii −aji1,i1
0, cml1,l1ajl1,l10. 2.8
Fromψαmm, ψαm,m1 ψαm,m1andψαm1,m1, ψαm,m1 −ψαm,m1, we have
ami,i1
amii −ami1,i1
ami,i1, 2cml1,l1aml1,l1cl1,l1m ,
ami,i1
am1ii −am1i1,i1
−ami,i1, 2cml1,l1am1l1,l1−cl1,l1m .
2.9
Whenj /l1,ψαjj, ψγl1,l1 0. So
al1i,i1
ajii −aji1,i1
0, cl1l1,l1ajl1,l10. 2.10
Fromψαl1,l1, ψγl1,l1 2ψγl1,l1, we have
al1i,i1
al1ii −al1i1,i1
2al1i,i1, cl1,l1l1 al1l1,l1cl1l1,l1. 2.11
LetC cjil1×l1, wherecjiajii −aji1,i1,j 1, . . . , l1,i1, . . . , l, andcj,l1 2ajl1,l1, j 1, . . . , l1. By Lemma 2.2, detA ∈ R∗, so detC ∈ R∗. Investigating amh,h1amk,k1detC, we may find that hth column and kth column are linearly dependent both are the form 0, . . . ,0,amj,j1,−amj,j1,0, . . . ,0t, j h, kby2.6and2.7, soamh,h1amk,k1detC0. Similarly,
amh,h1cml1,l1detC 0, al1h,h1al1k,k1detC 0h /k andal1h,h1cl1l1,l1detC 0. Then amh,h1
amk,k10h /k,amh,h1cml1,l10,al1h,h1al1k,k10 h /k, andal1h,h1cl1,l1l1 0.
iiWheni1, froma1hh −a1h1,h1a1kk −a1k1,k1detB 0h /k, we havea1hh − a1h1,h1a1kk −a1k1,k1 0 h /k. Similarly, we havea1hh −a1h1,h1a1ll a1l1,l1 01 ≤ h≤l. Whenil1, we get the results similarly.
iiiThe proving process is similar toiandii.
Lemma 2.4. Letψ ∈AutnC0 . Then
iwhenl≥1,ψα12 a112α12modnC2 , wherea112 ∈R∗;
iiif ψα12 a112α12modnC2 , where a112 ∈ R∗, then ψαi,i1 aii,i1αi,i1modnC2 andψγl1,l1 cl1l1,l1γl1,l1modnC2 , whereaii,i1,cl1l1,l1∈R∗.
Proof. iNoting thatαi,i1, γl1,l1∈nC1 \nC2 andγ12∈nC2l \nC2l1, we haveψα12∈nC1 \nC2 and ψγ12 c1212 γ12modnC2l1 ∈ nC2l \nC2l1, where c1212 ∈ R∗. Using 2.7, from ψα11, ψγ12 ψγ12, we havec1212a111 a122 c1212 , that is,a111 a122 1. Writeψα11andψγ11 asψα11
l1
i1a1ii αii modnC1 andψγ11 c11∗ γ11 ∈nC2l1, wherec∗11 ∈R∗. From 2ψγ11
ψα11, ψγ11 2a111c∗11γ11, we havea111 1. Thena122 0. ByLemma 2.3we havea1ii − a1i1,i1 0, i2, . . . , lherel≥2anda1l1,l10. Soa1ii 0, i2, . . . , l1, that is,ψα11
a111 modnC1 . Thenψα12 ψα11, ψα12 a112α12modnC2 anda112 ∈R∗. ByLemma 2.3, iholds.
iiWriteψαj,j1andψγl1,l1as2.6and2.7, respectively. Fromψα13 ψα12, ψα23, we haveψα13 a112a223α13 modnC3 . Sinceα13∈nC2 \nC3 ,ψα13∈nC2 \nC3 . So
a112a223 ∈R∗, that is,a223 ∈R∗. In general, form2, . . . , l, we have
ψα1,m1
m i1
aii,i1α1,m1modnCm1∈nCm \nCm1, m1, . . . , l,
ψ γ1,l1
ψα1,l1, ψ
γl1,l1 l
i1
aii,i1cl1l1,l1γ1,l1modnCl2 ∈nCl1 \nCl2,
2.12
hereaii,i1, cl1l1,l1 should be inR∗,i 1, . . . , l. By Lemma 2.3we have thataji,i1 0i /j,
cjl1,l1 0,j 1, . . . , l, andal1i,i1,i1, . . . , l. Henceψαl,l1andψγl1,l1have the required forms, respectively.
3. The Standard Automorphisms of t
Cl1R
Now let us introduce two types of Lie automorphisms oftCl1R.
(i) Inner Automorphisms
Letr Inaαiji /jorrInaγij. It is easy to check thatryr−1∈nC0 . The mapθr:tCl1R → tCl1Rsuch thatx → rxr−1,x ∈ nC0 , defines an automorphism ofnC0 , which is called an inner automorphismnote thatr is a symplectic matrix defined by
0 Il1
−Il1 0
. We denoteθr
byθaαij,θaγij, respectively. In these cases, we have θaα−1ij θ−aαij,θ−1aγij θ−aγij, respectively, and thatθaαijαii αii−aαij,θaαijαjj αjj aαij,θaαijαkk αkkk /i, j,θaαijαk,k1 αk,k1k /j, i−1,θaαijαj,j1 αj,j1aαi,j1,θaαijαi−1,i αi−1,i−aαi−1,j,θaγiiαii αii−2aγii, θaγi,i1αi,i1 αi,i1−2aγiiandθaαi,i1γi,i1 γi,i12aγii. All inner automorphisms oftCl1R generate a subgroup of AutnC0 , which is denoted byI.
(ii) Diagonal Automorphisms
Letdi ∈ R∗,i 0,1, . . . , l1,d diagd1, . . . , dl1andD diagd, d−1d0. The mapλD: tCl1R → tCl1Rsuch thatx →DxD−1,x∈nC0 , defines an automorphism oftCl1R, which is called a diagonal automorphism. It is clear thatλDλDλDD. So the set of diagonal automor- phisms oftCl1Ris a subgroup of AutnC0 , which is denoted byD.
4. Lemmas for Main Results
Lemma 4.1. Letψ ∈AutnC0 . The following two statements are equivalent:
(i)ψαj,j1 ajj,j1αj,j1modnC2 andψγl1,l1 cl1l1,l1γl1,l1modnC2 , whereajj,j1, cl1l1,l1∈R∗,j 1, . . . , l;
(ii)ψαjj αjj modnC1 , j 1, . . . , l1.
Proof. i⇒ii. Writeψαjjas in2.5. By the process of proving Lemma 2.3, we have a112 a111 −a122 a112,aii,i1ai1ii −ai1i1,i1 −aii,i1,ai1i1,i2ai1i1,i1−ai1i2,i2 ai1i1,i2,i1, . . . , l−1 andall,l1al1ll −al1l1,l1 −all,l1,cl1l1,l1al1l1,l1 cl1,l1l1 . Then we obtain thata111 −a122 1, ai1ii −ai1i1,i1−1,ai1i1,i1−ai1i2,i21,i1, . . . , l−1 andal1l1,l1 1. ByLemma 2.3, we have ajjj 1 1≤j≤l1andajii 0i /j.
ii⇒i. Writeψαj,j1andψγl1,l1, respectively, as in2.6and2.7. Then ψ
αj,j1
ψ αjj
, ψ αj,j1
ajj,j1αj,j1modnC2 , j 1, . . . , l, 2ψ
γl1,l1
ψαl1,l1, ψ
γl1,l1
2cl1,l1l1 γl1,l1modnC2 ,
4.1
that is,ψγl1,l1 cl1l1,l1γl1,l1modnC2 . By the method of modularizing a maximal ideal of Rto a residue field, we know thatajj,j1,cl1l1,l1∈R∗,j1, . . . , l.
Lemma 4.2. Letψbe in AutnC0 . Ifψαjj αjjmodnC1 , then ψα11 α11a112α12modnC2 , ψ
αjj
αjj−aj−1j−1,jαj−1,jajj,j1αj,j1modnC2 , j2, . . . , ll≥2, ψαl1,l1 αl1,l1−all,l1αl,l1cl1l1,l1γl1,l1modnC2 .
4.2
Proof. We expressψαjjas
ψ αjj
αjjl
i1
aji,i1αi,i1cjl1,l1γl1,l1modnC2 , j1, . . . , l1. 4.3
Fromψαjj, ψαkk 0 j /kwe have
ψα11 α11a112α12modnC2 , ψ
αjj
αjjajj−1,jαj−1,jajj,j1αj,j1modnC2 , j 2, . . . , l l≥2,
ψαl1,l1 αl1,l1al1l,l1αl,l1cl1,l1l1 γl1,l1 modnC2 ,
4.4
whereajj,j1aj1j,j1 0, j1, . . . , l.Lemma 4.2is proved.
Lemma 4.3. Letψ be in AutnC0 . If everyψαjjis expressed as the form inLemma 4.2, one may find an inner automorphism
θl
j1
θaj
j,j1αj,j1θ2−1cl1l1,l1γl1,l1 4.5
such that
θψ αjj
αjj modnC2 , j1, . . . , l1. 4.6
Proof. Note thatθ2−1cl1,l1l1 γl1,l1αl1,l1 αl1,l1−cl1l1,l1γl1,l1. Then, byLemma 4.2, it is not difficult to proveLemma 4.3.
Lemma 4.4. Letψbe in AutnC0 . Ifψαjj αjjmodnCk , j1, . . . , l1 1≤k≤l1, then
ψ αjj
αjjajj,jkαj,jk modnCk1 , j 1, . . . ,min{k, l−k1}k≤l, l≥1,
ψ αjj
αjj−aj−kj−k,jαj−k,j ajj,jkαj,jkmodnCk1, jk1, . . . , l−k1
k≤
l1 2
, l≥2
, ψ
αjj
αjj−aj−kj−k,jαj−k,jcj,2l−k−j3j γj,2l−k−j3modnCk1, j l−k2, . . . , l−
k 2
1
k≤ l1
2
, l≥1
, ψ
αjj
αjjcjj,2l−k−j3γj,2l−k−j3modnCk1, j l−k2, . . . , l−
k 2
1
1 l1
2
≤k, l≥2
, ψ
αjj
αjjcjj,2l−k−j3γj,2l−k−j3modnCk1, j l
2 1
k l
21, l≥4, herep even
, ψ
αjj
αjj−aj−kj−k,jαj−k,jcj,2l−k−j3j γj,2l−k−j3modnCk1, j l
22, . . . , l− k
2
1
k l
2 1, l≥4, herepeven
, ψ
αjj
αjjc2l−k−j32l−k−j3,jγ2l−k−j3,j modnCk1, jl−
k 2
2, . . . , k
l2≤k k
2
, l≥3
, ψ
αjj
αjj−aj−kj−k,jαj−k,jc2l−k−j3,j2l−k−j3γ2l−k−j3,j modnCk1, j max
k1, l− k
2
2
, . . . , l1
k≤p, k
2
≥1, l≥2
.
4.7
Proof. We expressψαjj, j1, . . . , l1, as
ψ αjj
αjjl−k1
i1
aji,ikαi,ik l1
il2−k1/2
cj2l−k−i3,iγ2l−k−i3,imodnCk1. 4.8
Whenk1 that is the case inLemma 4.2. The conclusion follows from repeating the process of provingLemma 4.4.
Lemma 4.5. Letψbe in AutnC0 . If everyψαjjbe expressed as the form inLemma 4.4, one may find an inner automorphism
θl−k1
j1
θaj
j,jkαj,jk
l−k/21
jl−k2
θ1δ
jl−1cjj,2l−k−j3γj,2l−k−j3, 4.9
where
δij
1, ij,
0, i /j, hl1−k−1/2k an odd. 4.10
Then
θψ αjj
αjj modnCk1, i1, . . . , l1. 4.11
Proof. Applyθtoψαjjand useLemma 4.4to obtainLemma 4.5.
Lemma 4.6. Letψbe in AutnC0 . Ifψαjj αjjmodnCk , j1, . . . , l1,l1≤k≤2l1l≥1, then
ψ αjj
αjjcjj,2l−k−j3γj,2l−k−j3modnCk1, j 1, . . . , l− k
2
1,
ψ αjj
αjjc2l−k−j32l−k−j3,jγ2l−k−j3,j modnCk1, jl− k
2
2, . . . ,2l−k2, ψ
αjj
αjj modnCk1, j2l−k3, . . . , l1.
4.12
Proof. We expressψαjj, j1, . . . , l1, as
ψ αjj
αjjl−k/21
i1
ci,2l−k−i3j γi,2l−k−i3modnCk1. 4.13
The process of provingLemma 4.6is similar toLemma 4.2.
Lemma 4.7. Letψ be in AutnC0 . If everyψαjjis expressed as the form inLemma 4.6, one may find an inner automorphism
θl−k/21
j1
θ1δ
jh−1cjj,2l−k−j3γj,2l−k−j3, 4.14
wherehl1−k−1/2 (kan odd). Then
θψ αjj
αjj modnCk1. 4.15
Whenk2l1,θψαjj αjj, j 1, . . . , l1.
Proof. It is similar to provingLemma 4.5.
Lemma 4.8. Whenl≥1, letψbe in AutnC0 . Ifψαjj αjj, j 1, . . . , l1, there exists a diagonal automorphismλDsuch thatλDψαj,j1 αj,j1,i1, . . . , l, andλDψγl1,l1 γl1,l1.
Proof. By Lemma 4.1we know that ψαj,j1 ajj,j1αj,j1modnC2 and ψγl1,l1 cl1l1,l1 γl1,l1modnC2 , whereajj,j1,cl1l1,l1 ∈ R∗,j 1, . . . , l. We express ψαj,j1and ψγl1,l1, respectively, as
ψ αj,j1
ajj,j1αj,j1l
k2 l−k1
i1
aji,ikαi,ik
l
k2
l1 il2−k1/2
c2l−k−i3,ij γ2l−k−i3,i
2l1
kl1
l1−k/2
i1
cji,2l−k−i3γi,2l−k−i3, 1≤j≤l,
ψ γl1,l1
cl1l1,l1γl1,l1l
k2 l−k1
i1
al1i,ikαi,ik
l
k2
l1 il2−k1/2
cl12l−k−i3,iγ2l−k−i3,i 2l1
kl1
l1−k/2
i1
cl1i,2l−k−i3γi,2l−k−i3.
4.16
Then
ϕ αj,j1
ϕ
αjj
, ϕ
αj,j1 , ϕ
αj1,j1
ajj,j1αj,j1−cjj,j1γj,j1, j 1, . . . , l.
4.17
In addition,
ϕ αj,j1
ϕ
αjj
,
ajj,j1αj,j1−bjj,j1βj,j1
, ϕ
αj1,j1
ajj,j1αj,j1cjj,j1γj,j1, j1, . . . , l.
4.18
Thuscjj,j10, j1, . . . , l. So ψ
αj,j1
ajj,j1αj,j1, j1, . . . , l. 4.19 Furthermore,
2ψ γl1,l1
ψαl1,l1, ψ
γl1,l1
2cl1l1,l1γl1,l1l−1
i1
al1i,l1αi,l1l−1
i1
cl1i,l1γi,l1. 4.20
Fromψαii, ψγl1,l1 0i /l−1, we haveal1i,l1 0 andcl1i,l1 0,i1,· · ·, l−1, that is, ψ
γl1,l1
cl1,l1l1 γl1,l1. 4.21
Letddiagd1, . . . , dl1andd0cl1l1,l1d2l1, whered1 1, djj
i2aj−i1j−i1,j−i2,j 2, . . . , l1.
ApplyingλDtoψαj,j1,j 1, . . . , l, andψγl1,l1, we get the result.
5. Proofs of Main Results
Proof ofTheorem 1.1. By Lemmas4.3,4.5,4.7, and4.8we haveλDθψαjj αjj, j1, . . . , l1, λDθψαj,j1 αj,j1, j 1, . . . , l, andλDθψγl1,l1 γl1,l1. Since the set{γl1,l1,αl1,l1,αjj, αj,j1 | j 1, . . . , l}generatestCl1R, we know that λDθψ is the identity automorphism of tCl1R. Henceψ θλD−1. The uniqueness of the decomposition follows fromTheorem 1.2.
Proof ofTheorem 1.2. By the first part ofTheorem 1.1we have AuttCl1R ID. For anyx∈ tCl1Randαij∈nC1 we have
λDθaαijλ−1Dx D
Inaαij
D−1xD
Inaαij
−1 D−1
InaλD
αij
x
InaλD
αij
−1 θaλDαijx.
5.1
So λDθaαij θaλDαijλD. Forγij ∈ nC1 , we haveλDθcγij θcλDγijλD. Therefore, I ID.
ObviouslyI ∩ D1. Then,IDID.
6. Discussion for l 0
In this case,tC1 R is generated byα11 and γ11. For any automorphismψ of tC1 R, write ψα11 and ψγ11, respectively, as ψα11 a11α11 c11γ11 and ψγ11 cγ11, where c ∈ R∗. From 2ψγ11 ψα11, ψγ11, we have a11 1. Then θ2−1c11γ11ψα11 α11 and θ2−1c11γ11ψγ11 cγ11. Alsoηcθ2−1c11γ11ψα11 α11 andηcθ2−1c11γ11ψγ11 γ11. Soψ θ−2−1c11γ11
ηc−1.
Acknowledgments
The research was supported partially by National Natural Science Foundation of China Grant nos. 10871056 and 10971150and by Science Research Foundation in Harbin Institute of TechnologyGrant no. HITC200708.
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