Centroids of Lie Supertriple Systems

Abstract

We derive certain structural results concerning centroids of Lie supertriple systems. Centroids of the tensor product of a Lie supertriple system and a unital commutative associative algebra are studied. Furthermore, the centroid of a tensor product of a simple Lie supertriple system and a polynomial ring is partly determined.

1. Introduction

The notion of Lie triple systems already appeared in Cartan’s work on Riemannian geometry, albeit the formal concept was not defined until 1949 by Jacobson in his study of associative algebras that are closed with respect to triple commutators (cf. [1]). The role played by Lie triple systems in the theory of symmetric spaces is parallel to that of Lie algebras in the theory of Lie groups: the tangent space at every point of a symmetric space has the structure of a Lie triple system.

As a natural generalization of Lie triple systems, the concept of Lie supertriple systems was introduced in the study of Yang-Baxter equations (cf. [2]). Lie supertriple systems have proved to be of interest in ternary structures appearing in various models in high energy physics, and many important results on Lie supertriple systems have been obtained (cf. [2–7]). In particular, the authors applied Lie supertriple systems to obtain some new solutions of the Yang-Baxter equation (cf. [2]).

The centroid of an algebra is the set of all linear transformations such that for all . is always a unital associative algebra, and if is “nice” (e.g., unital, prime, or simple), then is a commutative ring of scalars. Centroids are interesting subjects in mathematics, and there has been a great deal of work concerning them. Benkart and Neher studied centroids of Lie algebras in [8] and Melville investigated centroids of nilpotent Lie algebras in [9]. It turns out that results on the centroids of Lie algebras are a key ingredient in the classification of extended affine Lie algebras. The centroids of Lie triple systems were mentioned by Benito et al. in [10]. Results on centroids of Lie triple system, -Lie algebras, and Lie triple algebras were developed in [11–13].

In this paper we present new results concerning the centroids of Lie supertriple systems and give some conclusions of the tensor product of a Lie supertriple system and a unital commutative associative algebra. Furthermore, we completely determine the centroid of the tensor product of a simple Lie supertriple system and a polynomial ring. The paper is organized as follows. In Section 2, we list some notions of rings and Lie supertriple systems. In Section 3, we introduce the notion of centroids of Lie supertriple systems. Then we show some properties of centroids of Lie supertriple systems. In Section 4, we investigate the centroid of the tensor product of a Lie supertriple system and a unital commutative associative algebra . As an important theorem in this section, we prove that, for a tensor product of a simple Lie supertriple system and a polynomial ring , its centroid is .

2. Preliminaries

Throughout this paper, the base field is assumed to be of arbitrary characteristic unless otherwise stated.

Definition 1 (see [14]). An integral domain is a ring with no left or right zero divisors.

Definition 2 (see [14]). An algebraically closed field is a field such that every is a product of linear polynomials in .

Definition 3 (see [15]). An idempotent of a ring is an element such that .

Definition 4 (see [2]). A Lie supertriple system is a -graded vector space with a ternary product satisfying the following conditions:(1) (mod 2),(2),(3),(4), ,where , , , , are homogeneous elements in . We denote by the -graded degree of , , . In the sequel, whenever the notation appears, it implies that is a homogeneous element of degree in .

Definition 5 (see [16]). Let be a superalgebra whose multiplication is denoted by . This implies that for all . is a Lie superalgebra if the multiplication satisfies the following identities:(1);(2),for any , , .

Remark 6. (1) Let be a Lie superalgebra. It is clear that if we introduce a triple product by for any , then becomes a Lie supertriple system. So in this sense every Lie superalgebra deduces a Lie supertriple system.
(2) Any Lie supertriple system can be considered as a subspace of a Lie superalgebra. In fact, set to be the vector space with , where , for any . The product in is + for any , . It is easy to see that is a Lie superalgebra.

Definition 7 (see [5]). A derivation of a Lie supertriple system is a homogeneous linear map such that where . We denote by the -graded degree of a homogeneous linear mapping of .

Definition 8. An ideal of a Lie supertriple system is a -graded subspace for which . Moreover, if , then is called an abelian ideal of .

Definition 9. Let be a nonempty subset of . One calls , the centralizer of in , where . In particular, is the center of , denoted by .

3. Centroids of Lie Supertriple Systems

Definition 10. Let be a Lie supertriple system over a field . The centroid of is the space of -linear transformations on given by , .

By the definition of Lie supertriple system we conclude that if and , then

It is clear that the scalar maps will always be in the centroid.

Definition 11. Let . If and , then is called a central derivation. The set of all central derivations of is denoted by

Proposition 12. is an ideal of .

Proof. It is easily seen that . For all and , we have Hence is an ideal of .

Proposition 13. is the subalgebra of and is the centralizer of .

Proof. For any and , we have which proves the proposition.

If is perfect (i.e., ), then the centroid is supercommutative. Indeed, for all , we have

Proposition 14. If has no nonzero ideals with (i.e., is prime) and , then is an integral domain.

Proof. Clearly, . If there exist and such that , then there are such that and . Then Since , ; that is, . Therefore, generates a nonzero ideal . In a similar way, generates a nonzero ideal such that , a contradiction. Hence has no zero divisors, and so it is an integral domain.

Theorem 15. If is a simple Lie supertriple system over an algebraically closed field (i.e., ) and has only two ideals and , then .

Proof. Let . Since is algebraically closed, has an eigenvalue . We denote the corresponding eigenspace by . So . For any , , where , , we have , so ; that is, , . So we have and . Therefore is a -graded vector space. For any , we have , so . It follows that is an ideal of . But is simple, so ; that is, . This proves the theorem.

When , the Lie supertriple system is said to be central. Furthermore, if is simple, is said to be central simple. Every simple Lie supertriple system is central simple over its centroid.

Proposition 16. Let be a Lie supertriple system over a field . Then(1) is indecomposable (cannot be written as the direct sum of two nontrivial ideals) if and only if does not contain idempotents except and ;(2)if is perfect, every is supersymmetric with respect to any invariant form on .

Proof. Consider the following:
(1) If is an idempotent and satisfies , then , . We can see that and are ideals of . In fact, for any and , we have , where and ; then , and , , so and ; that is, is a -graded vector space. implies . For any , we have , where and ; there exists such that , , , , so ; that is, ; then and ; that is, is a -graded vector space. For any , we have , . Moreover, . Indeed, if , then there exists such that and . We have a decomposition , , where . So we have , a contradiction.
On the other hand, suppose has a decomposition . Then for any , we have ,, . We choose such that and . Then . Hence is an idempotent. By assumption, we know or id. If , then implying . If , then implying .
(2) Let be an invariant bilinear form on . Then Since is perfect, for , we have which proves the proposition.

Proposition 17. Let be a Lie supertriple system over a field and a subset of . Then is invariant under and so is any perfect ideal of .

Proof. For any , , , and , we have which implies that . So is invariant under .
Let be any perfect ideal of ; then . For any , there exist such that and then we have . Hence is invariant under .

Theorem 18. Let be an epimorphism of Lie supertriple systems. For any there exists a unique satisfying . Moreover, the following results hold:(1)The map , is an algebra homomorphism satisfying the following properties:  By restriction, there is an algebra homomorphism  If , then every leaves invariant; hence is defined on all of .(2)Suppose is perfect and . Then  is injective.(3)If is perfect, , and , then is an algebra monomorphism.

Proof. Consider the following:
(1) It is easy to see that is an algebra homomorphism. Since is an ideal of and all left and right multiplication operators of leave invariant, . Furthermore, for the left multiplication operator on we have , so . For the right multiplication we have the analogous formula . Moreover, is an epimorphism, so . Now we show that . Let . For any there exist such that , , and . Then we have which proves .
(2) If for , then , which means that . Hence , . Furthermore, since , we can get .
(3) We can see that and then . So . By (1), we know that is a well-defined algebra homomorphism, which is an injection by (2).

Proposition 19. If the characteristic of is not , then

Proof. If , then by the definitions of and , for all , we have , and , so and It follows easily that
To show the inverse inclusion, let ; then Thus, .

Lemma 20. Let be a nonzero -invariant ideal of , , and the vector space of all linear maps from to over . Define , for . Then(1) is a subspace of and as vector spaces;(2)if , as vector spaces.

Proof. (1) For all , , and , we have Hence and ; that is, is a -graded subspace of . For all , , where and . Also, , ; then ; that is, is a -graded vector space. Since , , we have , . So is an ideal of the associative algebra . To prove consider the following map given by where and . The map is well defined. If , then , and so . It follows easily that is injective. We now show that is onto. For every , set , , for all . It follows from the definition of that, for all , . Thus , and so . But implies that is onto. It is easy to see that preserves operations on vector spaces to . This proves .
We now prove . If , then, for all , for some . If , let , for all ; then and . Clearly, . Furthermore, , and so is proved.

Lemma 21. Let be a Lie supertriple system; then is a derivation for and .

Proof. If , then Thus is a derivation.

Theorem 22. Let be a Lie supertriple system. Then for all and ,(1) is contained in the normalizer of in ;(2) is contained in if and only if is a central derivation of ;(3) is a derivation of if and only if is a central derivation of .

Proof. (1) For any , , and , So ; this is . This proves .
(2) From Lemma 21 and , is an element of if and only if . Thanks to Proposition 19, we get result .
(3) It follows from , Proposition 19, and Lemma 21 that holds.

Now, we study the relationship between the centroid of a decomposable Lie supertriple system and the centroid of its factors.

Theorem 23. Suppose that is a Lie supertriple system over and with being ideals of ; then(1), as vector spaces, where (2).

Proof. (1) Letting be canonical projections for , then and . So we have, for , Note that for . We claim that It suffices to show that (other cases are similar). For any , there exist , , such that . Then , for all , and so . Let , . We now prove Since for , we have . On the other hand, one can regard as a subalgebra of by extending any on being equal to zero; that is, and , for all and . Then and . Therefore with isomorphism , for all . Similarly, we have .
Next, we prove . If , then there exists in such that . For , where , and , we have and Then and . It follows that and so .
Conversely, for , expanding on (also denoted by ) by , we have and . This proves that is isomorphic to with the following isomorphism: for all . Similarly, we can prove . Summarizing the above discussion we get This proves .
(2) Clearly, Letting be canonical projections for , then and . So we have, for , . Note that for . We claim that Let , ; we now have , , , and . In the same way as (1), we get (2).

A generalized version of the above theorem is stated below without proof.

Theorem 24. Suppose that is a Lie supertriple system over with a decomposition of ideals Then we have(1) as vector spaces, where , for ;(2)

4. Centroid of Tensor Product

Benkart and Neher investigated the centroid of the tensor product of associative algebras and Lie algebras in [8]. In this section we discuss the centroid of the tensor product of a Lie supertriple system and a unital commutative associative algebra. Furthermore, we completely determine the centroid of the tensor product of a simple Lie supertriple system and a polynomial ring .

Definition 25. Let be an associative algebra over a field . The centroid of is the space of -linear transformations on given by

Proposition 26. Let be a Lie supertriple system over and a unital commutative associative algebra over . Then there exists a unique Lie supertriple system structure on satisfying , , , .

Proof. For , , , we have Thus is a Lie supertriple system.

It is easy to show that if is perfect, then is perfect too. Also, for and there exists a unique map such that The map should not be confused with the element of the tensor product . Of course, we have a canonical map It is easy to see that if and , then . Hence , where is the -span of all endomorphisms .

Definition 27. The transformation has finite -image if, for any , there exist finitely many such that

It is easy to see that has finite -image In addition, if is perfect, has finite -image as soon as for suitable .

Lemma 28. Let be a Lie supertriple system over and a unital commutative associative -algebra. Let be a basis of and . One defines by Then all .

Proof. For , we have and Hence for all , so all .

Proposition 29. Let be a perfect Lie supertriple system over and a unital commutative associative -algebra. Then,(1)if is finite generated as an -module (or as a -module) or is central, every has finite -image;(2)if the map in Definition 25 is injective, has finite -image.

Proof. (1) Set . Since is unital, . Since is finite generated as an -module, we can suppose for , . Fix and . There exist finite families and such that for . Hence Since is perfect, the centroid is supercommutative. Replacing by we can use the same argument as above to show that every has finite -image if is a finitely generated -module.
Now, suppose and is a torsion-free -module. Then there exist scalars such that . Hence . Fix ; then almost all , so almost all , which in turn implies that has finite -image.
(2) In the discussion above, we get has finite -image So it suffices to prove has finite -image. We suppose that has finite -image; then there exists a finite subset such that For and , we get Since is perfect, we can get , where is the left multiplication in by . Let be an -basis of . Then there exist a finite subset and scalars such that . We then get for . Now we show that . For any , we have Because , we can get , , and ; it follows that where is defined by , . Since is injective, we also have . So by the linear independence of the , we get that . Then . Hence .

Next we will determine the centroid of the tensor product of a simple Lie supertriple system and a polynomial ring.

Let be a Lie supertriple system and . Defining the multiplication by , we can make into a Lie supertriple system.

Now we write out a basis of in the following manner. First, let denote the set of central derivations and choose a basis for . Second, let be a maximal subset of with the restrictions being linearly independent and the space spanned by . Then ; and form a basis of .

Lemma 30. One has .

Proof. Let . Then we have and Thus .
For any and since and Hence .

Theorem 31. Letting be a simple Lie supertriple system, and . Then .

Proof. From Lemma 30, we get , so . Now we prove the opposite inclusion.
Let be a basis for . Let and be suitable maps in such that we may write Then while So, for each and , we have , . Hence . But . Therefore, , , for suitable scalars .
Thus, now we may write Since the right hand side is in , for each , we get for all except a finite number of . That is, . Then the map is well defined. Hence, we have Then we get Choose any triple such that . Then we can conclude that , . So is determined by its action on 1. Therefore, . Thus, , , . That is, . So we conclude that

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

The authors would like to thank the referee for valuable comments and suggestions on this paper. This research was supported by NNSF of China (nos. 11171055 and 11471090) and the Fundamental Research Funds for the Central University (no. 14ZZ2221).