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Abstract and Applied Analysis
Volume 2013 (2013), Article ID 404219, 9 pages
The Centroid of a Lie Triple Algebra
School of Mathematics and Statistics, Northeast Normal University, Changchun 130024, China
Received 21 June 2013; Accepted 11 July 2013
Academic Editor: Emrullah Yaşar
Copyright © 2013 Xiaohong Liu and Liangyun Chen. 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.
General results on the centroids of Lie triple algebras are developed. Centroids of the tensor product of a Lie triple algebra and a unitary commutative associative algebra are studied. Furthermore, the centroid of the tensor product of a simple Lie triple algebra and a polynomial ring is completely determined.
In recent years, Lie triple algebras (i.e., Lie-Yamaguti algebras or general Lie triple systems) have attracted much attention in Lie theories. They contain Lie algebras and Lie triple systems as special cases ([1–6]). So, it is of vital importnce to study some properties of Lie triple algebras. The concept of Lie triple algebra has been introduced, originally, by Yamaguti as general Lie triple system by himself, Sagle, and others. Since Lie triple algebras are generalization of Lie algebras and Lie triple system, it is natural for us to imagine whether or not some results of Lie algebras and Lie triple system hold in Lie triple algebras. Now, as a generalization of Lie triple algebra, Hom-Lie-Yamaguti was introduced by Lister in .
Benkart and Neher studied centroid of Lie algebras in , and Melville investigated centroid of nilpotent Lie algebras in . It turns out that result on the centroid of Lie algebras is a key ingredient in the classification of extended affine Lie algebras. The centroids of Lie triple systems were mentioned by Benito et al. . Now, some results on centroids of Lie triple system and -Lie algebras were developed in [11, 12].
In this paper we present new results concerning the centroids of Lie triple algebras and give some conclusion of the tensor product of a Lie triple algebra and a unitary commutative associative algebra. Furthermore, we completely determine the centroid of the tensor product of a simple Lie triple algebra and a polynomial ring. The organization of the rest of this paper is as follows. Section 1 is for basic notions and facts on Lie triple algebras. Section 2 is devoted to the structures and properties of the centroids of Lie triple algebras. Section 3 describes the structures of the centroids of tensor product of Lie triple algebras.
Definition 1 (see ). A Lie triple algebra (also called a Lie-Yamaguti algebra or a general Lie triple system) is a vector space over an arbitrary field with a bilinear map denoted by of into and a ternary map denoted by of into satisfying the following axioms:(a), (b), (c), (d), (e), (f), where .
Remark 2. Any Lie algebra is a Lie triple algebra relative to and , for . If , for all , the axioms stated earlier reduce to that of Lie algebra, and if , for all , the axioms stated earlier reduce to that of Lie triple system. In this sense, the Lie triple algebra is a more general concept than that of the Lie algebra and Lie triple system.
Definition 3 (see [1, 2]). A derivation of a Lie triple algebra is a linear transformation of into satisfying the following conditions:(1),
(2), for all .
Let be the set of all derivation of ; then is regarded as a subalgebra of the general Lie algebra and is called the derivation algebra of .
Definition 4. A Lie triple subalgebra of is called an ideal if and .
Definition 5. A Lie triple algebra is perfect if .
Definition 6. Let be a nonempty subset of ; we call , for all the centralizer of in . In particular, is the center of .
Definition 7. Suppose that is an ideal of the Lie triple algebra , on the quotient vector space Define the operator: , where . Then is also a Lie triple algebra, and it is called a quotient algebra of by .
3. The Centroids of Lie Triple Algebras
Definition 8. Let be a Lie triple algebra over a field . The centroid of is the transform on given by .
By (a)–(f), we can conclude that if , then we have , for all .
From the definition, it is clear that the scalars will always be in the centroid.
Proposition 9. If is perfect, then the centroid is commutative.
Proof. For all , we have , .
Proposition 10. Let be a Lie triple algebra over a field and a subset of . Then;(1) is invariant under , (2) every perfect ideal of is invariant under .
Proof. For any , , , , we have , , and , . Therefore, , which implies that is invariant under .
Let be any perfect ideal of ; then . For any , there exist , such that , and then we have . Hence is invariant under .
Definition 11. Let and ; then is called a central derivation.
The set of all central derivation of is denoted by . Clearly, and is an ideal of . A more precise relationship is summarized as follows.
Next, we will develop some general results on centroids of Lie triple algebra.
Proposition 12. If has no nonzero ideals and with , then is an integral domain.
Proof. Clearly, . If there exist , such that , then there exist such that and . Then, . Therefore, and can span two nonzero ideals of such that , which is a contradiction. Hence, has no zero divisor; it is an integral domain.
Theorem 13. If is a simple Lie triple algebra over an algebraically closed field , then .
Proof. Let . Since is algebraically closed, has an eigenvalue . We denote the corresponding eigenspace to be , so , for any , and we have , so . It follows that is an ideal of . But, is simple, so ; that is, . This proves the theorem.
When , the Lie triple algebra is said to be central. Furthermore, if is simple, is said to be central simple. Every simple Lie triple algebra is central simple over its centroid.
Proposition 14. Let be a Lie triple algebra over a field ; then(1) is indecomposable if and only if does not contain idempotents except 0 and ; (2) if is perfect, then every is symmetric with respect to any invariant form on .
Proof. If there exists which is an idempotent and satisfies , then , for all . We assert that and are ideals of . In fact, for any and , we have , which implies . For any , there exists such that . Then, we have
This proves our assertion. Moreover, . Indeed, if , then there exists such that and . We have a decomposition , for all , where . So, we have , which is 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 have or . If , then , implying . If , then , implying .
Let be an invariant -bilinear form on . Then , , for all . Since is perfect, for , we have
The result follows.
Remark 15. Set ; then we call , and the left multiplication operator of ; and right multiplication operator of . Denote by the subalgebra of generated by the left and right multiplication operators of . Then, is the centralizer of .
Theorem 16. Let be an epimorphism of Lie triple algebra; for any , there exists a unique satisfying . Moreover, the following results hold.(1) The map , is a homomorphism with the following properties: . There is a homomorphism .
If , then every leaves invariant; that is, is defined on all of .(2) Suppose is perfect and ; then is injective.(3) If is perfect, , and , then is a monomorphism.
Proof. It is easy to see that is a 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 . Then, we have
which proves .
If for, , then , which means that . Hence, , . Furthermore, since is a perfect ideal, we can get .
We can see that , which follows that . So, . From (1), we know that is a well-defined homomorphism, which is an injection by (2).
Proposition 17. If the characteristic of is not , then
Proof. If , then by the definition of and , for all , we have , and , so and . It follows easily that .
To show the inverse inclusion, let ; then . Thus, .
This implies that .
Lemma 18. Let be a nonzero -invariant ideal of , , and let be the vector space of all linear maps from to over . Define , where . Then one has the following.(1) is a subspace of and as vector spaces.(2) If , then as vector spaces.
Proof. It is easily seen that is an ideal of the associative algebra . To prove (1) 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 fairly easy to see that preserves operations on vector spaces from to . This proves (1).
We now prove (2). If , then for all , for some . If , let , for all ; then and . Clearly, . Furthermore, , and so (2) is proved.
Lemma 19. Let be a Lie triple algebra; then is a derivation for .
Proof. If , then Thus, is a derivation.
Theorem 20. Let be a Lie triple algebra, and for any , then one has the following.(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. For any , and ,
Then, we get . On the other hand, we have So, ; this is . This proves (1). From Lemma 19 and (1), is an element of if and only if . Thanks to Proposition 17, we get the result (2). It follows from (1), Proposition 17, and Lemma 19 that the result holds.
Now, we study the relationship between the centroid of a decomposable Lie triple algebra and the centroid of its factors.
Theorem 21. Suppose that is a Lie triple algebra over and with and being ideals of ; then as vector spaces, where , .
Proof. Let 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 that
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, , for all . Then and . Therefore, with isomorphism , for all . Similarly, we have .
Next, we prove that . If there exists in such that . For , where and , we have Then, and . It follows that , and so .
Conversely, for , expending on (also denoted by ) or by , we have and . This proves that is isomorphic to with the following isomorphism: for all . Similarly, we can prove that . Summarizing the aforesaid discussion, we have
The proof is completed.
A generalized version of Theorem 21 is stated next without proof.
Theorem 22. Suppose that is a Lie triple algebra over and with a decomposition of ideals . Then, one has as vector spaces, where , for .
4. The Centroids of Tensor Product of Lie Triple Algebras
Definition 23. Let be an associative algebra over a field , where the centroid of is the space of -linear transforms on given by ; then is an associative algebra of . If is be a finite-dimensional Lie triple algebra over , let be a tensor product of the underlying vector spaces and . Then, over a field with respect to the following 2-ary and 3-ary multilinear operation: where . This Lie triple algebra is called the tensor product of and . 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 as the following: It is easy to see that if and , then . Hence, , where is the -span of all endomorphisms .
Definition 24. The transformation is said to have finite -image, if for any , there exist finitely many such that It is easy to see that -. In addition, has finite -image, if for suitable .
Lemma 25. Let be a Lie triple algebra over and a unitary commutative associative over . Let be a basis of , and let . Define by then all .
Proof. For , we have Hence, one has for all . So all .
Proposition 26. Let be a perfect Lie triple algebra over , and let be a unitary commutative associative over which is free as a -module. Then, one has the following.(1)If is perfect, then is perfect too.(2)If is finitely generated as a -module (or as a -module), or if is a central and a torsion-free -module, then every has finite -image.(3)If is a free -module and the map of Definition 23 is injective, then has finite -.
Proof. The definition of perfect shows that (1) holds. Let . It follows that from is unital. Since is finitely generated as a -module, we can suppose for . Fix and . There exist finite families and such that
for . Hence,
Since is perfect, the centroid is commutative. By replacing with we can use the same argument aforementione to show that every has finite -image, if is a finitely generated -module.
Now, suppose that and that is a torsion-free -module. Then there exist scalars such that . Hence, Fix ; then almost all and almost all , which in turn implies that has finite -image.
From the aforementione discussion above, we get So, it suffices to prove that . We suppose that has finite -image, and then there exists a finite subset such that equation in Lemma 25 becomes as follows: For and , we get Since is perfect, we can get where is the left multiplication in by . Let be a basis of . Then, there exist a finite subset and scalars such that We then get Now we show that . For any , we have Since , we have and . It follows that where is defined by , for all . 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 triple algebra and a polynomial ring. Here after we study the centroid of Lie triple algebra over a filed . Let be a basis of the central derivation and a maximal subset of such that and are linear independent. Then, we have the following result.
Theorem 27. Let denote the subspace of spanned by . Then is a basis of and as vector spaces.
Proof. Since and are linear independent, is linear independent in . By definition of , the is independent in . For since is a basis of vector spaces , and is a basis of vector spaces , there exist , ( is a finite set of positive integers) such that We then have If , then It follows that is a central derivation. So there exist , ( is a finite set of positive integers) such that Therefore, The proof is completed.
Lemma 28. Let be a simple Lie triple algebra, , and ; then one has .
Proof. Let . Then, we have Thus . For any and , we have Hence .
Theorem 29. Let be a simple Lie triple algebra, , and . Then .
Proof. From Lemma 28, we get . Now we prove the opposite inclusion. Let be a basis for , and let be suitable maps in such that