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Advances in Mathematical Physics
Volume 2015, Article ID 354341, 11 pages
http://dx.doi.org/10.1155/2015/354341
Research Article

Structures and Low Dimensional Classifications of Hom-Poisson Superalgebras

1School of Mathematics and Statistics, Northeast Normal University, Changchun 130024, China
2School of Applied Sciences, Jilin Engineering Normal University, Changchun 130052, China

Received 4 May 2015; Accepted 4 August 2015

Academic Editor: Hoshang Heydari

Copyright © 2015 Qingcheng Zhang 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

Hom-Poisson superalgebras can be considered as a deformation of Poisson superalgebras. We prove that Hom-Poisson superalgebras are closed under tensor products. Moreover, we show that Hom-Poisson superalgebras can be described using only the twisting map and one binary operation. Finally, all algebra endomorphisms on 2-dimensional complex Poisson superalgebras are computed, and their associated Hom-Poisson superalgebras are described explicitly.

1. Introduction

Poisson algebras are used in many fields in mathematics and physics. In mathematics, Poisson algebras play a fundamental role in Poisson geometry [1], quantum groups [2], and deformation of commutative associative algebras. In physics, Poisson algebras are a major part of deformation quantization, Hamiltonian mechanics [3], and topological field theories [4]. Poisson-like structures are also used in the study of vertex operator algebras [5]. Poisson superalgebras can be seen as a direct generalization of Poisson algebras. Remm show that a Poisson superalgebra can be described using only one binary operation in [6, 7].

Recently, a twisted generalization of noncommutative Poisson algebras, called Hom-noncommutative Poisson algebras, are studied in [8]. In a noncommutative Hom-Poisson algebra, there exists a twisted map, a Hom-Lie bracket and a Hom-associative product. The associativity, the Jacobi identity, and the Leibniz identity are considered as their Hom-type analogues in a noncommutative Hom-Poisson algebra. The purpose of this paper is to introduce and study a twisted generalization of Poisson superalgebras, called Hom-Poisson superalgebras.

This paper is organized as follows. In Section 2, we recall the construction of the Hom-Lie superalgebras. In Section 3, we give the definition of Hom-Poisson superalgebras. We show that starting with a Poisson superalgebra and an even Poisson superalgebra endomorphism, a Hom-Poisson superalgebra can be constructed. Moreover, we prove that Hom-Poisson superalgebras are closed under tensor products. In Section 4, we show that a Hom-Poisson superalgebra can be described using only the twisting map and one binary operation. Section 5 is devoted to classifying all the algebra endomorphisms on all the 2-dimensional complex Poisson superalgebras and 2-dimensional Hom-Poisson superalgebras.

Throughout the paper is the field of complex numbers. All algebras and vector spaces are considered over .

2. Hom-Lie and Hom-Associative Superalgebras

In this section, we first recall the notion of Hom-Lie superalgebras and then give some construction of the Hom-Lie superalgebras.

Definition 1 (see [9]). A Hom-associative superalgebra is a triple consisting of a -graded vector space , an even bilinear map , and an even homomorphism of algebras satisfying for all homogeneous elements .

Definition 2 (see [9]). A Hom-Lie superalgebra is a triple consisting of a -graded vector space , an even bilinear map and an even homomorphism of algebras satisfyingfor all homogeneous elements

Let and be two Hom-Lie superalgebras. An even homomorphism is said to be a morphism of Hom-Lie superalgebras if

Morphisms of Hom-associative superalgebras are defined similarly.

The following theorem provides a method to construct a Hom-Lie superalgebra by a Lie superalgebra and an even homomorphism of Lie superalgebras.

Proposition 3 (see [9]). Let be a Lie superalgebra and let be an even endomorphism of Lie superalgebras. Then is a Hom-Lie superalgebra, where .
Moreover, suppose that is another Lie superalgebra and is an even endomorphism of Lie superalgebras. If is a morphism of Lie superalgebras satisfying , then is a morphism of Hom-Lie superalgebras.

Example 4 (see [9]). From the orthosymplectic Lie superalgebra , where is spanned by and is spanned by The defining nonzero relations are Let define a linear map by then is an even Lie superalgebra automorphism; by Proposition 3, we obtain a family of Hom-Lie superalgebras . These Hom-Lie superalgebras are not Lie superalgebras for .

3. Hom-Poisson Superalgebras

Definition 5. A Hom-Poisson superalgebra is a tuple consisting of a -graded vector space , two even bilinear maps , and an even homomorphism of algebras , where , , satisfying the following axioms.(1) is a supercommutative Hom-associative superalgebra.(2) is a Hom-Lie superalgebra.(3)The Hom-Leibniz superidentity holds for all homogeneous elements .

Definition 6. Let and be two Hom-Poisson superalgebras. An even homomorphism is said to be a morphism of Hom-Poisson superalgebras if

Proposition 7. Let be a Hom-associative superalgebra and be a binary operation on defined bythen is a Hom-Poisson superalgebra with the same twisting map

Proof. It is straightforward.

From Proposition 7, there is the following construction of Hom-Poisson superalgebras by Poisson superalgebras and homomorphisms.

Theorem 8. Let be a Poisson superalgebra and let be an even endomorphism of Poisson superalgebras. Then is a Hom-Poisson superalgebra, where and .
Moreover, suppose that is another Poisson superalgebra and is an even endomorphism of Poisson superalgebras. If is a morphism of Poisson superalgebras satisfying , then is a morphism of Hom-Poisson superalgebras.

Proof. By Proposition 3, is a Hom-Lie superalgebra.
We will show that satisfies the axioms (1) and (3) of Definition 5. In fact Hence is a Hom-Poisson superalgebra. Then second assertion follows from

This theorem provides a method to construct a Hom-Poisson superalgebra by a Poisson superalgebra and an even homomorphism of Poisson superalgebras.

Example 9. Let be a -dimensional -graded vector space, where is generated by and is generated by and nonzero products are given by then is a Poisson superalgebra. For any , we consider the homomorphism defined by By Theorem 8, for any , there is the corresponding Hom-Poisson superalgebra with the nonzero products It is not a Poisson superalgebra when

Example 10. Let be a -dimensional -graded vector space, where is generated by and is generated by and the nonzero products are given by then is a Poisson superalgebra. For any , we consider the homomorphism defined by By Theorem 8, for any , there is the corresponding Hom-Poisson superalgebra with the nonzero products It is not a Poisson superalgebra when

We know Hom-Poisson algebras are closed under tensor products ([8] Theorem ). Here we aim to consider it in detail in the superalgebra case.

Theorem 11. Let be Hom-Poisson superalgebras for , and let . Define the operations and by for Then is a Hom-Poisson superalgebra.

Proof. The is a supercommutative Hom-associative superalgebra following from the supercommutativity and Hom-associativity of both . Also, the supercommutativity of the and the antisupersymmetry of the imply the antisupersymmetry of . It remains to prove the Hom-Jacobi superidentity and the Hom-Leibniz superidentity in .
To simplify the typography, we abbreviate , and using juxtaposition and drop the subscripts in and . Pick Then where ,  + Consider where Considerwhere Using the Hom-Jacobi superidentity in and the supercommutativity and Hom-associativity in , we haveLikewise, using the supercommutativity and Hom-associativity in and the Hom-Jacobi superidentity in , we obtainUsing the Hom-Leibniz superidentity in , then we have This shows that satisfies the Hom-Jacobi superidentity: Finally, we check the Hom-Leibniz superidentity in . Using the Hom-associativity and the Hom-Leibniz superidentity in the , we have Therefore, we have

Setting in Theorem 11, we obtain the result about Poisson superalgebras.

Corollary 12. Let be Poisson superalgebras for , and let . Define the operations by for Then is a Poisson superalgebra.

4. Admissible Hom-Poisson Superalgebras

A Poisson algebra has two binary operations, the Lie bracket and the commutative associative product. It is shown in [10] that Poisson algebras can be described using only one binary operation via the polarization-depolarization process. Moreover, the result of Poisson algebras is extended to Hom-Poisson algebras in [8]. In other words, the paper shows that a Hom-Poisson algebra can be described using only the twisting map and one binary operation. The purpose of this section is to extend this alternative description of Poisson algebras to Hom-Poisson superalgebras.

Definition 13. An admissible Hom-Poisson superalgebra is a Hom-superalgebra satisfyingwhere , for any homogeneous elements , the identity (32) is called the Hom-Remm identity.

Remark 14. In particular, taking , we find the notion of admissible Poisson superalgebra presented in [8].

Theorem 15. Let be a double Hom-superalgebra. Then is a Hom-Poisson superalgebra if and only if there exists on a nonassociative product such that is an admissible Hom-Poisson superalgebra.

Proof. Assume that is a Hom-Poisson superalgebra. Consider the multiplication We deduce that Thus the associativity condition can be denoted by where Likewise, the Hom-Poisson bracket can be denoted by and the Hom-super Jacobi condition The Hom-Leibniz superidentity can be denoted by Let us consider the vector Then We deduce that the product satisfies for any homogeneous elements
Conversely, assume that the produce of the nonassociative product satisfies for any homogeneous elements Let , , be the elements of defined in the first part, respectively, in relation to the Hom-associativity, Hom-super Jacobi, and Hom-super Leibniz relations. We have

Taking in Theorem 15, we obtain the following result, which is Theorem in [7].

Corollary 16. Let be a double superalgebra. Then is a Poisson superalgebra if and only if there exists on a nonassociative product satisfying where , for any homogeneous elements

Definition 17. A Hom-nonassociative superalgebra is called Hom-superflexive if the multiplication satisfies for any homogeneous elements , where is called a Hom-associator of the multiplication.

Proposition 18. Let be a Hom-Poisson superalgebra. Then the Hom-Remm product defining the Hom-Poisson superalgebra structure is Hom-superflexive.

Proof. Let We have

Taking in Proposition 18, we obtain the following result, which is Proposition in [7].

Corollary 19. Let be a Poisson superalgebra. Then the Remm product defining Poisson superalgebra structure is superflexive.

Remark 20. The deformation cohomology of Hom-Poisson superalgebras can be computed with the Hom-Remm identity, which is similar to the method in [11]. This content is not primary in the paper, we do not have a detailed discussion here.

5. A Classification of 2-Dimensional Hom-Poisson Superalgebras

In this section, we only consider that is nontrivial. denotes one of the 2-dimensional admissible Poisson superalgebra types. denotes one of the homomorphism types corresponding to . denotes one of the 2-dimensional admissible Hom-Poisson superalgebra types corresponding to . In the following, the products equal to zero are omitted.

Lemma 21. Let be an admissible Poisson superalgebra and let be an even Poisson superalgebra endomorphism. Then is an admissible Hom-Poisson superalgebra, where .

Proof. It is straightforward by Definition 13.

Lemma 22 (see [7]). Let be a 2-dimensional admissible Poisson superalgebra with a basis , where . Then is one of the following types:, , , , .., , , ., , .

Proof. LetBy Corollary 16, we haveNow we consider the cases as follows.
Case  1. If , then ; hence we have .
Case  2. If , then .
Subcase  2.1. If , then .
If , then we have .
If , then ; hence we have .
Subcase  2.2. If , then , ; hence we have .

Remark 23. Lemma in [6] has been obtained; however, there is one minor inaccuracy on the product operation in that proof.

Lemma 24. Let be a 2-dimensional admissible Poisson superalgebra with dim and dim . Then an even homomorphism of type is as follows:

Proof. Let From is an even homomorphism, we obtain By Lemma 22 and (50) and (51), we obtain Case  1. If , then we have .
Case  2. If , then we consider two cases as follows.
If , then ; hence we have .
If , then , ; hence we have .

Corollary 25. Let be a 2-dimensional admissible Hom-Poisson superalgebra. Then with respect to is as follows:, .., , , , , .

Proof. Apply Lemmas 21, 22, and 24.

Lemma 26. Let be a 2-dimensional admissible Poisson superalgebra with and . Then an even homomorphism of type is as follows:

Proof. By Lemma 22 and (20) and (21), we obtain .
Case  1. Suppose that , we have .
Case  2. Suppose that , we consider two cases as follows.
If , then we have
If , then ; hence we have .

Corollary 27. Let be a 2-dimensional admissible Hom-Poisson superalgebra. Then with respect to is as follows:  .  .  , .

Proof. Apply Lemmas 21, 22, and 26.

Lemma 28. Let be a 2-dimensional admissible Poisson superalgebra with dim and dim . Then an even homomorphism of type is as follows:

Proof. By Lemma 22 and (20) and (21), we obtain Case  1. If , then ; hence we have .
Case  2. If , then ; hence we have .
Case  3. If , then we consider two cases as follows.
If , we have .
If , then ; hence we have .
Case  4. If , then ; hence we have .

Corollary 29. Let be a 2-dimensional admissible Hom-Poisson superalgebra. Then with respect to is as follows:  .  , , , .  .  , .  , , , , .

Proof. Apply Lemmas 21, 22, and 28.

Lemma 30. Let be a 2-dimensional admissible Poisson superalgebra with and . Then an even homomorphism of type is as follows:

Proof. By Lemma 22 and (20) and (21), we obtain .
Case  1. If , then we have .
Case  2. If , then ; hence we have .

Corollary 31. Let be a 2-dimensional admissible Hom-Poisson superalgebra. Then with respect to is as follows:  .  , , , .

Proof. Apply Lemmas 21, 22, and 30.

Theorem 32. Let be a 2-dimensional admissible Poisson superalgebra with and . Then an even homomorphism of is as follows:

Proof. Apply Lemmas 2430.

Corollary 33. Let be a 2-dimensional admissible Hom-Poisson superalgebra. Then with respect to is as follows:  , .  .  , , , , , .  .  .  , .  .  , , , .  .  , .  , ,