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`Abstract and Applied AnalysisVolume 2013, Article ID 638230, 8 pageshttp://dx.doi.org/10.1155/2013/638230`
Research Article

## The Automorphism Group of the Lie Ring of Real Skew-Symmetric Matrices

1Department of Mathematics, Harbin Institute of Technology, Harbin 150001, China
2Department of Foundation, Harbin Finance University, Harbin 150030, China

Received 11 May 2013; Accepted 17 July 2013

Copyright © 2013 Jinli Xu et al. 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

Denote by the set of all skew-symmetric matrices over the field of real numbers, which forms a Lie ring under the usual matrix addition and the Lie multiplication as , . In this paper, we characterize the automorphism group of the Lie ring .

#### 1. Introduction and Main Result

A Lie ring is defined as a nonassociative ring with multiplication that is anticommutative and satisfies the Jacobi identity. More specifically, we can define a Lie ring to be an abelian group with an operation that has the following properties:(i)biadditive: for all ;(ii)the Jacobi identity: for all ;(iii)for all in ,

It is well known that a Lie algebra can be viewed as a Lie ring. So, the theory of Lie ring can be used in the theory of the Lie algebra. Recall that an automorphism of a Lie ring is a bijective map form onto itself such that and for all . There are a lot of papers that studied the automorphism groups of some fixed Lie rings (or, more for the Lie algebras), see [17].

Note that any associative ring can be made into a Lie ring by defining a bracket operator . Let be the real number field and denote by the rest . Let be the algebra of all matrices over . We denote by the subset of consisting of all skew-symmetric matrices, that is,

It is well known that the set forms a Lie ring under the usual matrix addition and the Lie multiplication as , . In the same way, we know that or (the set of all upper triangular matrices) as well as forms a Lie ring.

Hua [8] gives the form of any automorphism of the Lie ring over a skew field by using the fundamental theorem of geometry of matrices; Dolinar [1] studies the automorphism of the Lie ring of triangular matrices over any field. Jacobson [9] considers the Lie algebra over any algebraically closed field; he gives the form of the automorphism of the Lie algebra for the case if is odd or if is even. Now, let us see a general result on isomorphism of some Lie rings as follows.

Proposition 1 (see [10, 11]). Let and be prime rings with involutions of the first kind and of characteristic not 2. Let and denote, respectively, the skew elements of and . Assume that the dimension of the central closure of over is different from 1, 4, 9, 16, 25, and 64. Then, any Lie isomorphism of onto can be extended uniquely to an associative isomorphism of onto , the associative subrings generated by and , respectively.

Note that the Lie ring is a particular class of the previous setting of skew elements of . So, the previous proposition in fact partially solved the problem to characterize the automorphism group of . However, the problem is still open when takes any positive integer.

The purpose of this paper is to characterize , the automorphism group of the Lie ring , for . Our main result is the following.

Theorem 2. Suppose that is an integer, then if and only if there is a real orthogonal matrix such that
Further, one has , where is the real orthogonal group.

#### 2. Preliminary Results

Now, let us start this section by denoting some notations. Denote by the maximal integer number no more than . Let be the matrix which has 1 in the entry and is elsewhere. Set , and denote by the identity matrix. Note that the notation means that the matrix vanished. Let , .

Suppose that . We call to be commutative if , for all , and call to be maximal commutative if is not only commutative but , for all . Clearly, the maximal commutative subset is a subring of . Suppose that . Set

We denote by and the direct sum and the Kronecker product of and , respectively.

Definition 3. A matrix is called regular if it satisfies the following conditions: (i) when is an even number, there is an orthogonal matrix , and the real numbers with different absolute values, such that
(ii) When is an odd number, there is an orthogonal matrix , and the nonzero real numbers with different absolute values, such that

Now, a subring of is called a regular subring if is maximal commutative and there is a regular matrix in .

For , denote .

Lemma 4 (see [12, 2.5.14]). Suppose that . Then, there are an orthogonal matrix and real numbers such that

Lemma 5 (see [13, 14]). Let be any field, and let denote the space of all alternate matrices over . Then, is an additive surjective mapping of () to itself that preserves rank 2 matrices if and only if is of the following forms:(i), , for all , where , is an invertible matrix, and is a field automorphism of ;(ii)when , is of the form where , , and have the same meaning as before, and is either the identity map or the map:

In the next text, we always assume that is arbitrary.

Lemma 6. Suppose that is a regular subring of . Then, there is an orthogonal matrix and maps , such that

Proof. For every , note that is commutative with every regular matrix in . So, one can obtain the conclusion by Lemma 4.

Corollary 7. Suppose that is a regular subring of , and is a regular matrix. Then,

Lemma 8. Suppose that , are both regular subrings of , and that there is a regular matrix . Then, .

Proof. Note that a regular subring is maximal; the conclusion follows by Corollary 7.

Lemma 9. Both maps and preserve the regular subring. Expressly, for , one has that is a regular matrix if and only if is so.

Proof. Take any regular subring of and a regular matrix . By Lemma 6, we can assume that , where is an orthogonal matrix.
Suppose that satisfying . Then, is a regular matrix in . Since , . This means that . Let be a regular subring containing . By Lemma 8, we only need to prove that . Take any . Then, we see by Lemma 6 that there are such that
So, , and . Hence, . This shows that . Note that is maximal, so we obtain that .
Now, we prove that preserves the regular matrix. Otherwise, suppose that is not a regular matrix, then we will get a contradiction. By the definition, we see that one of the following cases holds.
Case 1. is odd and there is .
Case 2. There is some .
If Case 1 happens, we assume without loss the generality that . We take such that
If Case 2 happens, we assume without loss the generality that . When , we take such that
When , we take such that
On one hand, it is clear that , so we have . On the other hand, ; hence, . Thus, , and so ; this is impossible. Note that is an automorphism; we see that also preserves the regular matrix. The proof is completed.

Lemma 10. Suppose that and . If and , then there is such that

Proof. We can assume without loss the generality by Lemma 4 that , . Hence, we have . If any matrix cannot satisfy the conclusion, then one has . Note that , so we have . This implies that , which contradicts with .

Lemma 11. Let . Then, .

Proof. As , we deduce that . Farther, we have . The desired result follows from the following:

Lemma 12. Suppose that is not a regular matrix. Then, .

Proof. It follows from Lemma 6 that there is an orthogonal matrix such that
Since is not a regular matrix and so is , we see that , for all . If , then we will see that . Otherwise, there is such that , and so . But we know that , this, together with Lemma 11, gives that
This is impossible. Similarly, we can show that if , then , and then we get the conclusion.

Lemma 13. Suppose that and such that . Then, .

Proof. If there is such that , then by Lemma 10 we can choose such that , . Thus, ; . But we see that ; this is impossible. Furthermore,
For any nonzero real number , we replace by in the previous equation. It follows that that is, . Note that is additive. So, is a subspace.
Suppose that where is an orthogonal matrix, , . We first prove the following.
Assertion. If there is an index such that , for all , then .
In fact, for any given , suppose that , then . Now, we assume that the assertion is not true; then there is some index such that . Then, we see by , , and the fact is a space that . This tells us that there is some such that . Thus, , and so we have . But we know that , which contradicts with the conditions of the assertion. This gives that . The assertion is proved.
As is not a regular matrix, one has that is not a regular matrix too. Next, the proof of the lemma is divided into the following cases with respect to .
Case 1. When is odd, note that , so we can assume without loss the generality that . If for some such that , then it follows by Lemma 11 that , which is a contradiction. Now, the lemma follows by using the previous assertion for the index .
Case 2. When is even, assume without loss of the generality that . If , then . In fact, if there is some such that , then one has by Lemma 11 that is a contradiction. If , then . In fact, if there is some such that , then we see by Lemma 11 that ; this is impossible.
When , if there is such that , then . Thus, , which is a contradiction. Now, we get the lemma by using the previous assertion for the index .
When , if there is such that , then since , . This is absurd. As , it is clear that . Hence, we can assume without loss of the generality that . If for some such that , then we have by Lemma 11 that which is a contradiction. The lemma can be shown by using the previous assertion for the index .

Corollary 14. Suppose that and is a subspace with bases which are formed by rank 2 matrices. Then, we have

Proof. Suppose that rank 2 matrices form bases of . Then,
It follows immediately that
If there is such that , then we can choose , not all zero, such that , which is absurd. We see by Lemma 13 that , for all . Thus,
The proof is completed.

Lemma 15. Suppose that and is of rank 2. Then,

Proof. Note that and has bases which are formed by rank 2 matrices. This, together with Corollary 14, proves the conclusion.

Lemma 16. Suppose that are positive real numbers, which are different from one another. Let . Then, we have
In particular, if we let , then and the equation holds if and only if .

Proof. It follows by a direct computation.

Lemma 17. Suppose that preserves the rank 2 matrix subset of . Then, there is a real orthogonal matrix such that

Proof. The proof under the case is obvious. It is not difficult to see that, if , then a surjective map preserving rank 2 matrices still is of the form (i) of Lemma 5. Next, we assume that and assume that has the form (i) of Lemma 5. For distinct , , , it is clear that
Consider the image of ; then, it follows by the form (i) of Lemma 5 that
Hence, by a direct computation and the arbitrariness of , , , it follows that . Clearly, . Note that is the field of real numbers, so we have . Let ; then, the conclusion is obtained.
When and is of the form (ii) of Lemma 5, then we let , , and . Thus, we have by taking the images under for the previous two equations that , which is a contradiction. So, the form (ii) of Lemma 5 does not occur.

#### 3. The Proof of the Main Result

The proof of the main theorem is divided into the following three propositions.

Proposition 18. Suppose that or and . Then, there is an orthogonal matrix such that

Proof. Since is bijective, preserves the rank 2 matrices of or . If , the conclusion is clear. If , then we also can get the conclusion by Lemma 17.

Proposition 19. Suppose that . Then, there is an orthogonal matrix such that

Proof. It is clear that , , and . Note that is regular, so we can assume that , where and is an orthogonal matrix. Without loss of generality, one can assume that
Since the regular subring containing the nonregular matrix is determined by , there are and such that
Therefore,
Suppose that where is a matrix. It follows by that
So, we have , , , and . Note that
Thus,
This, together with , gives that , , . We deduce that
Similarly, we see by and that where . We also have by that
Note that , so we can assume that
Further,
On the other hand, we know by that
By a direct computation with (49) and (50), we have and then . Noting that , we can assume without loss of the generality that (for the case , the proof is similar). We deduce that
If , then , which contradicts Lemma 12. This tells us that .
Again by , we get .
Suppose that
It follows by that
Therefore, we have . Note that . This tells us that . We can assume without loss of the generality that . Then, we get that and . Thus,
For any but fixed , we assert that .
In fact, firstly, by Lemma 12, we can assume that , . So, we have that .
Secondly, noting that and , we deduce that
Furthermore, we see by and that
Thus, . It follows that .
Finally, due to and , one can obtain that
This tells us that , and so we have that , . In other words, , which proves the assertion. Now, we prove that preserves the set of rank 2 matrices on . By applying Lemma 17, we finish the proof.

Proposition 20. Suppose that and . Then, there is an orthogonal matrix such that

Proof. Take any rank 2 matrix . By Lemma 4, we can assume that
Let . Now, we assert that and so that the rank of is 2; that is, we will assert that is a preserver of rank 2 on ; then, we can finish the proof by Lemma 17.
It follows by Lemmas 15 and 16 that
Moreover, we see that . Hence, we have either or . The former means that , as desired. If the latter holds, then it is clear that
In this case, we deduce that and . Hence, the remainder of the proof is the cases (i) , , (ii) , , and (iii) , .
Suppose that . We consider the rank of .
When , it is clear that there is an orthogonal matrix such that and . Without loss of the generality, we can assume that . Note that . If , then one has . Let . As , we can find a matrix such that
If , then . Let . Since , there is a matrix such that
Thanks to , we deduce , which is a contradiction.
When , then for the previous three cases of and , one always has . Note that and are in a common regular subring, and . It follows by Lemma 6 that there is an orthogonal matrix such that and , where , . Due to , we see by Lemma 11 that
Case 1. . We first prove that .
If , then we may as well assume that and . Let , . It is easy to see that . Now, we want to show that , which is a contradiction. Note the following:
So, we know that both and satisfy an equation about the matrix as follows:
That is,
Hence, we get that and . Note that , . After taking the image, we can assume by that
Again by , we see that
We deduce that , . It follows by that . Due to , , one has . This tells us that , which is a contradiction. Similarly, we know that .
Since , it is clear that . When we have that . Note that is not a regular matrix; hence, is not too. Further, one has . This implies that or , which is impossible. Similarly, we deduce that which is also a contradiction.
Case 2. . We first prove that . Otherwise, if , then we have which is a contradiction. In a similar way, we get .
If , we assert that . In fact, if , then by , , we deduce that . Without loss of the generality, we can assume that , , and , . Hence, we see that , which is impossible. Similarly, we deduce that .
Next, we prove when that . Otherwise, by (64) we can assume without loss of the generality that and . Note that , , so we have and . Thus,
But it is clear that , which contradicts with , .
Similarly, we have when that .
Finally, we prove when that . Let . If , then we can find a contradiction similar to the case of . Otherwise, if , then there is an orthogonal matrix such that where , , and . It is easy to see that the three cases , , and cannot simultaneously hold. This means that is impossible.
To sum up the previous arguments, we get that . The proof is completed.

#### Acknowledgments

This paper is dedicated to Professor Chongguang Cao for his good health. The authors wish to thank Professor Xiaomin Tang for introducing and guiding the topic. The authors would also like to thank the referee for his/her careful reading of the paper and valuable comments which greatly improved the readability of the paper. Jinli Xu is supported in part by NSFC(11171294), Natural Science Foundation of Heilongjiang Province of China (Grant no. A201013). Baodong Zheng is supported by the National Natural Science Foundation Grants of China (Grant no. 10871056).

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