#### 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 [18]. 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