Simple Modules for Modular Lie Superalgebras , , and

Abstract

This paper constructs a series of modules from modular Lie superalgebras , , and over a field of prime characteristic . Cartan subalgebras, maximal vectors of these modular Lie superalgebras, can be solved. With certain properties of the positive root vectors, we obtain that the sufficient conditions of these modules are irreducible -modules, where , , and .

1. Introduction

Giving a broad overview of the present situation, the representation theories of Lie algebras and Lie superalgebras over a field of characteristic 0 have been a remarkable evolution. Kac (see [1]) worked on classification of infinite dimensional simple linearly compact Lie superalgebras. Shchepochkina (see [2]) studied the five exceptional simple Lie superalgebras of vector fields and their fourteen regradings. More detailed description of one of the five simple exceptional Lie superalgebras of vector fields was given (see [3]). The complete proof of the recognition theorem for graded Lie algebras in prime characteristic was given (see [4]). Strade (see [5]) studied simple Lie algebras over fields of positive characteristic and obtained some important results. The classification of finite dimensional modular Lie superalgebras with indecomposable Cartan matrix was given, and the prolongations of the simple finite dimensional Lie algebras and Lie superalgebras with Cartan matrix are studied over algebraically closed fields of characteristic (see [6, 7]).

Su (see [8–12]) got a new class of nongraded simple Lie algebras. It was proved that an irreducible quasifinite module was a module of the intermediate series. He also got some important results about the representation of classical Lie superalgebras. Zhang (see [13, 14]) worked on the graded module of , , over fields of characteristic 0.

There are also a great deal of representative results of Lie algebras over fields of prime characteristic (see [15–18]). For a restricted Cartan-type Lie algebra, restricted simple modules have been determined in the sense that their isomorphism classes have been parametrized. And their dimensions have been computed. Shen (see [19–21]) constructed the graded modules for the Witt, special, and Hamiltonian Lie algebras. Shen determined those simple modules with fundamental dominant weights, except the contact algebra. Holmes (see [15]) solved the remaining problem about the contact algebra. He showed that the simple restricted modules were induced from the restricted universal enveloping algebra for the homogeneous component of degree zero extended trivially to positive components. Hu (see [22]) investigated the graded modules for the graded contact Cartan algebras and . Shu (see [23]) worked on the generalized restricted representations of graded Lie algebras of Cartan type.

However, there are few results about the representations of Lie superalgebras over a field of prime characteristic , called modular Lie superalgebras. Liu (see [24]) established the dimension formula of induced modules and obtained some properties of induced modules. In this paper, we construct modules from Lie superalgebras , , and , induced from homogeneous components of their restricted universal enveloping superalgebras. We intend to show that the generator of these constructed modules belongs to their nonzero submodules.

2. Preliminaries

In this paper, always denotes an algebraically closed field of prime characteristic .

Recall the definition of the restricted Lie superalgebras (see [6]) and the restricted universal enveloping superalgebra (see [25]).

Let be a Lie algebra of characteristic . Then, for every , the operator is a derivation of . If it is an inner derivation for every , that is, if for some element denoted , then the corresponding map is called a -structure on , and the Lie algebra endowed with a -structure is called a restricted Lie algebra. If has no center, then can have not more than one -structure. The Lie algebra possesses a -structure, unique up to the contribution of the center; this -structure is used in the next definition. The notion of a -representation is naturally defined as a linear map such that ; in this case is said to be a -module. Passing to superalgebras, we see that, for any odd , we have for any . So if char , then is always an even derivation for any odd . Now, let be a Lie superalgebra of characteristic . Then, for every , the operator is a derivation of ; that is, -action on is a -representation. For every , the operator is a derivation of . So if, for every , there is such that for any , then we can define for any . We demand that, for any , we have as operators on the whole ; that is, is a restricted -module. Then, the pair of maps is called a -structure or just -structure on , and the Lie superalgebra endowed with a -structure is called a restricted Lie superalgebra.

A pair consisting of an associative -superalgebra with unity and a restricted homomorphism is called a restricted universal enveloping superalgebra if given any associative -superalgebra with unity and any restricted homomorphism ; there exists a unique homomorphism of associative -superalgebra such that . The category of -modules and that of restricted -modules are equivalent. According to the PBW theorem, then the following statement holds: let be a restricted Lie superalgebra. If is a restricted universal enveloping superalgebra and is an ordered basis of over , where , then the elements , , , , and , consist of a basis of over . Sometimes, with no confusion, we will identify with its image in .

Definition 1. Let be a -graded Lie superalgebra over a field of characteristic . Suppose that is a Cartan subalgebra of , where is the set of the th homogenous elements of -graded Lie superalgebra . For and a -module , we set . If , then is called a weight and a nonzero vector in is called a weight vector (of weight ). A nonzero is called a maximal vector (of weight ), provided , where is any positive root vector of .
Let be a -graded Lie superalgebra over . Set , where denotes the homogeneous component of degree in the -graded Lie superalgebra . Then, and . In particular, any -module becomes a -module by letting act trivially. Define , where and denote the restricted universal enveloping superalgebras of and , respectively, and is a simple -module. According to the classical theory, for each weight , there exists a simple -module which is generated by a maximal vector of weight .

In the following, denotes one of three classes of Cartan-type Lie superalgebras , , or . Each of these classes will be described in detail in the following paper.

Remark 2. In this paper, if is a subset of some linear space, then denotes the subspace spanned by the set over .

3. Simple Modules of the Lie Superalgebra

We begin by describing the Lie superalgebra , drawing most of notations and standard results from [26].

Let be an exterior algebra over in variables . Fix , and then becomes -graded if we set , Then, is an associative superalgebra. The multiplication satisfies the rule , in particular, . For , put . Let , where . If , where , then we set . Put . Note that is -graded by , where

For each , let denote the derivation of uniquely determined by the property (=Kronecker delta). Let , be the tuple with th component (=Kronecker delta), and then is a Lie superalgebra, which has a -basis . The bracket product in satisfies where are -homogeneous elements of the Lie superalgebra , and , . Note that . inherits a -gradation from by means of . Consequently, .

Lemma 3. Let be a Cartan subalgebra of . Then, positive root vectors of are .

Proof. Introduce a homomorphism , where is the general linear Lie algebra that sends to (=-matrix with in the -position and zeros elsewhere). Obviously, is an isomorphism of Lie algebras. We know that the Cartan subalgebra of is . Define linear function on the vector space , such that . Then, the positive roots of are . Correspondingly, the positive root vectors are , . By the isomorphism via , we may therefore obtain that the Cartan subalgebra of is . And the positive root vectors are .

According to Definition 1 and Lemma 3, the following statement holds: if is a Cartan subalgebra, is a -module, and , then . Write . A nonzero is a maximal vector (of weight ), provided , .

Theorem 4. If there exist and such that and , then is simple.

Proof. Let be a nonzero submodule of . Now choose . By virtue of , we obtain . Hence, ; namely, . Therefore, we can describe the element form of ; that is, where , . Formula (3) shows that . Then, we obtain the equality in : in particular, In the following, we simply write . Define an order of satisfying that if and only if there exists such that and . We set , where comes from the right side of equality (4). According to the order , we choose the least element . Obviously, . Put . By (5) and (6), we obtain where or . Consequently, .
We will show that is a -module. By virtue of (3), we obtain Thus, we have in . We note that the elements , , are -basis of . Using (8), a straightforward computation shows that Since is a simple -module, we can get that is a simple -module. Since , we can conclude that is nontrivial. must be contained in . Thereby, there exists a maximal vector of weight such that .
Noting that , this yields . According to (3), we have . Then, holds in . By (10), we have where means that is deleted. Since is a -module and is a maximal vector of weight , the first two terms vanish. Formula (11) implies that . Since , we have
If multiplied on the left by the elements , successively, then it yields . By the hypothesis of the theorem, there exists such that , and we thus have Observe that the first term vanishes by the -graded degree of which is 1, furthermore , and the second term is equal to . Since , we have . If we multiply by the elements , consecutively, then we see that . Since -module is generated by , indicates that . Hence, is simple.

4. Simple Modules of the Lie Superalgebra

In the above section, we give the definition of the exterior algebra . We begin by describing , where Putting , we have .

Lemma 5. Let be a Cartan subalgebra of . Then positive root vectors of are .

Proof. Recall the isomorphism via , described above. It induces the isomorphism , where is the special linear Lie algebra.
By calculation, we obtain that the Cartan subalgebra of is equal to . We define linear function on the vector space as before in Lemma 3. Then, the positive roots of are . Correspondingly, the positive root vectors of are . The isomorphism sends to and to , respectively. Therefore, the Cartan subalgebra of is . The positive root vectors are .

In terms of Definition 1 and Lemma 5, the following facts hold: if is a Cartan subalgebra of , is a -module and . Then, . A nonzero is a maximal vector (of weight ), provided , whenever .

Theorem 6. If there exist , such that and , then is simple.

Proof. Let be a nonzero submodule of . Choose contained in . It owns the same form as in ; that is, , where and . The same discussion described above about the Lie superalgebra applies to the Lie superalgebra . We can get a maximal vector such that .
Using (3) and (13), we see that Thus, in . If , we have Since the -graded degree of is 1, it implies that the first term vanishes. By the definition of a maximal vector , it implies that the second and the forth terms vanish. It yields Since , we have .
Multiply by , in turn. By the same calculation as above, it yields , where .
If , then we have It also can be found that the first term vanishes and the second term can conclude by the hypothesis of the Theorem . Multiplying on the left by , then holds. If , then we have Obviously, we can obtain .
Moreover, , for , implies that . And so on, multiply on the left by , consecutively. Finally, it yields .
If , we have ; furthermore, . Imitating the process of calculation, we have . We get the conclusion.

5. Simple Modules of the Lie Superalgebra

Given a linear mapping satisfies where , , and . We can obtain that is an isomorphism of linear spaces. By computation, we know that , where , and We define a bracket product in by means of where , , , and . Pertaining to this bracket product, becomes a Lie superalgebra which is denoted by (see [27]).

Then, is a -graded Lie superalgebra, where , and Put Write .

Lemma 7. is a restricted Lie superalgebra.

Proof. Since is an isomorphism of , we can regard . Obviously, is a subalgebra of . For any , then . Then, the -graded degree of is an even number. If , namely, ,  , or , . By direct calculation, , where or . . If , , we have . In a word, , where . Since is a subalgebra of and is a restricted Lie superalgebra, we can obtain that is a superalgebra.

First we consider the case where is an odd number.

Lemma 8. Let be a Cartan subalgebra of . Then, positive root vectors of are , , where .

Proof. Suppose that , given by is a mapping of vector spaces. It can be verified that is an isomorphism. Let , where . Let . Set . Then, ; namely, , the orthogonal algebra. We can conclude .
By calculation, we obtain that the Cartan subalgebra of is .
We will define linear function the same as before. Then, the positive roots of are and . Correspondingly, the positive root vectors are , respectively. By the isomorphism, we get the positive root vectors of : where So we can obtain that positive root vectors of are for The Cartan subalgebra of is .

In view of Definition 1 and Lemma 8, the following fact holds: suppose is a Cartan subalgebra of , is a -module, and . Then, . A nonzero is a maximal vector (of weight ), provided , whenever .

Theorem 9. If , then is simple.

Proof. Let be a nonzero submodule of . Take and . We note where , , and , or for . Write , in .
By formula (20), we have , for and , . Obviously, , for . And , for . Thus, in , where and . In particular, where . And where .
Put . By (28), we can get , where . Put
Multiplying by and then by (27) and (29), we can obtain , where . And so on, we can conclude that there exists such that , where , . Imitating the discussion as , there exists a maximal vector such that . By (20), we get where does not appear in and . Also we have where occurs in . We will prove in two steps.
First, by (30) and (31), we have Observing these terms, it remains the third nonzero term. With , where does not appear in , then in . Hence, by (30), (31), and (33), then (32) can be adjusted to Secondly, for , then in . Using the induction hypothesis, we get Hence, we get Since , we have . Multiplying on the left by , thus we find that .
Imitate the way above on . First, multiply on the left with . By computation, we can obtain that . Repeating the second step, we can get .
Repeating the process above, we can get . Multiplying on the left by , we then have For , we know that . Following , it implies that . Thus, Moving and expanding these terms of (38), then we have the following identity: Multiplying on the left by , we obtain that . Using the induction hypothesis, finally, we can get the desired formula; namely, . Therefore, is simple.

If is an even number, then we let . Imitating the proof of Lemma 7 and Theorem 9, we can obtain the following theorem.

Theorem 10. If , then is simple.

Conflict of Interests

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

Acknowledgments

This work is supported by the Fundamental Research Funds for the Central Universities of China (Grant no. 14QNJJ003), SRFHLJED (Grant no. 12521157), NSFC (Grant no. 10871057), and NSFJL (Grant no. 20130101068JC).