International Journal of Mathematics and Mathematical Sciences

Volume 2012 (2012), Article ID 242736, 12 pages

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

Received 26 March 2012; Accepted 26 April 2012

Academic Editor: Helge Holden

Copyright © 2012 Xing Tao Wang and Lei 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.

Abstract.

Let be the matrix symplectic Lie algebra over a commutative ring with 2 invertible. Then   =  { ∣  is an upper triangular matrix, is the solvable subalgebra of . In this paper, we give an explicit description of the automorphism group of .

1. Introduction

Classical Lie algebras occupy an important place in matrix algebras. Let be a commutative ring with the identity 1 and the group of invertible elements in . Let be the -algebra of by matrices over that has a structure of a Lie algebra over with bracket operation for any . The symplectic Lie algebra

is one of classical Lie algebras, where denotes the matrix transpose. It is easy to show that the following subalgebra of such that

is solvable.

Let be the identity matrix in and let denotes the matrix in all of whose entries are 0, except the th entry which is 1. Let

where , . For discussion latter, we rewrite as

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 matrices of trace 0 over a connected commutative ring and a commutative ring with invertible, respectively, in this paper we use similar techniques to those in [11] to prove that any automorphism of can be uniquely expressed as , where and are inner and diagonal automorphisms, respectively, for and is a commutative ring with 2 invertible. We also give an explicit description of the remaining case .

Theorem 1.1. For any automorphism of there are unique inner and diagonal automorphisms, and , respectively, of such that .

Theorem 1.2. Let and be the inner and diagonal automorphism groups, respectively. Then Aut, where    denotes the automorphism group of .

2. Preliminaries

Let

Then the set is a basis of .

Lemma 2.1. Let be the set generated by the set , where . Then .

Proof. We only need to show that . It is obvious that , when . When , we have

Assume that , then , that is, . Since , for any , when ,

Assume that when , , then when , , that is, , . For any , when , . When , . Since , and 2 is invertible, we have for . Thus . Because is a basis of , we obtain .

Now, denote by . Let , , , . It is not difficult to know

It is easy to check that for or for . For any  , we have and , . Therefore, , . Note that if , then .

For any maximal ideal of , is a field. The natural homomorphism induces a homomorphism which is surjective. So every automorphism of may induce an automorphism of . Using this fact and that (for ), we have that , where . Otherwise, should be contained in a maximal ideal of , then on , where is the image of in , which is impossible.

Lemma 2.2. Let be in . If , and are expressed, respectively, as

then the following matrices are invertible.

(i), where , , ;(ii), where , , , , , , and .

Proof. (i) That is invertible follows from the fact that induces an automorphism of the free -module of rank on the basis . (ii) Note that induces an automorphism of the free -module of rank on the basis .

Lemma 2.3. Let  . Write , , and as in (2.5)–(2.7), respectively. Then the following conclusions hold.

(i)For , , ,  and .(ii)For ,    and , where  .(iii)For and , and .

Proof. (i) When , . So

From and , we have

When , . So

From , we have

Let , where , , , and ,  . By Lemma 2.2, det, so det. Investigating , we may find that th column and th column are linearly dependent (both are the form by (2.6) and (2.7), so . Similarly, , and . Then , , , and .

(ii) When , from , we have . Similarly, we have . When , we get the results similarly.

(iii) The proving process is similar to (i) and (ii).

Lemma 2.4. Let . Then

(i)when , , where ;(ii)if ,  where , then and ,  where .

Proof. (i) Noting that and , we have and , where . Using (2.7), from , we have , that is, . Write and as and , where . From , we have . Then . By Lemma 2.3 we have   (here  ) and . So , that is, . Then and . By Lemma 2.3, (i) holds.

(ii) Write and as (2.6) and (2.7), respectively. From , we have . Since , . So , that is, . In general, for , we have

here should be in , . By Lemma 2.3 we have that , , , and , . Hence and have the required forms, respectively.

3. The Standard Automorphisms of

Now let us introduce two types of Lie automorphisms of .

(i) Inner Automorphisms

Let or . It is easy to check that . The map : such that , , defines an automorphism of , which is called an inner automorphism  (note that is a symplectic matrix defined by ). We denote by , , respectively. In these cases, we have , , respectively, and that , , , , , , , and . All inner automorphisms of generate a subgroup of Aut, which is denoted by .

(ii) Diagonal Automorphisms

Let , , and . The map : such that , , defines an automorphism of , which is called a diagonal automorphism. It is clear that . So the set of diagonal automorphisms of is a subgroup of Aut, which is denoted by .

4. Lemmas for Main Results

Lemma 4.1. Let . The following two statements are equivalent:

(i) and , where , ;

(ii) .

Proof. (i)⇒(ii). Write as in (2.5). By the process of proving Lemma 2.3, we have , , , and , . Then we obtain that , , , and . By Lemma 2.3, we have and .

(ii)⇒(i). Write and , respectively, as in (2.6) and (2.7). Then

that is, . By the method of modularizing a maximal ideal of to a residue field, we know that , .

Lemma 4.2. Let be in . If , then

Proof. We express as

From we have

where . Lemma 4.2 is proved.

Lemma 4.3. Let be in . If every is expressed as the form in Lemma 4.2, one may find an inner automorphism

such that

Proof. Note that . Then, by Lemma 4.2, it is not difficult to prove Lemma 4.3.

Lemma 4.4. Let be in . If , then

Proof. We express , as

When that is the case in Lemma 4.2. The conclusion follows from repeating the process of proving Lemma 4.4.

Lemma 4.5. Let be in . If every be expressed as the form in Lemma 4.4, one may find an inner automorphism

where

Then

Proof. Apply to and use Lemma 4.4 to obtain Lemma 4.5.

Lemma 4.6. Let be in . If , , then

Proof. We express , as

The process of proving Lemma 4.6 is similar to Lemma 4.2.

Lemma 4.7. Let be in . If every is expressed as the form in Lemma 4.6, one may find an inner automorphism

where ( an odd). Then

When , .

Proof. It is similar to proving Lemma 4.5.

Lemma 4.8. When , let be in . If , there exists a diagonal automorphism such that , , and .

Proof. By Lemma 4.1 we know that and , where , . We express and , respectively, as

Then

In addition,

Thus . So

Furthermore,

From , we have and , , that is,

Let and , where , . Applying to , , and , 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 , , and . Since the set , , , generates , we know that is the identity automorphism of . Hence . 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 . For any and we have

So . For , we have . Therefore, . Obviously . Then, .

6. Discussion for

In this case, is generated by and . For any automorphism of , write and , respectively, as and , where . From , we have . Then and . Also and . So .

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).

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