Abstract and Applied Analysis

Volume 2013 (2013), Article ID 923101, 10 pages

http://dx.doi.org/10.1155/2013/923101

Research Article

Infinite-Dimensional Modular Lie Superalgebra

Xiaoning Xu and Bing Mu

School of Mathematics, Liaoning University, Shenyang 110036, China

Received 2 July 2013; Accepted 28 July 2013

Academic Editor: Teoman Özer

Copyright © 2013 Xiaoning Xu and Bing Mu. 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.

All ad-nilpotent elements of the infinite-dimensional Lie superalgebra over a field of positive characteristic are determined. The natural filtration of the Lie superalgebra is proved to be invariant under automorphisms by characterizing ad-nilpotent elements. Then an intrinsic property is obtained by the invariance of the filtration; that is, the integers in the definition of are intrinsic. Therefore, we classify the infinite-dimensional modular Lie superalgebra in the sense of isomorphism.

1. Introduction

The theory of modular Lie superalgebras has obtained many important results during the last twenty years (e.g., see [1–4]). But the complete classification of the simple modular Lie superalgebras remains an open problem. We know that filtration structures play an important role both in the classification of modular Lie algebras and nonmodular Lie superalgebras (see [5–8]). The natural filtrations of finite-dimensional modular Lie algebras of Cartan type were proved to be invariant in [9, 10]. The similar result was obtained for the infinite-dimensional case [11]. In the case of finite-dimensional modular Lie superalgebras of Cartan type, the invariance of the natural filtration was discussed in [12, 13]. The same conclusion was obtained for some infinite-dimensional modular Lie superalgebras of Cartan type (see [14–17]).

In the present paper, we consider the infinite-dimensional modular Lie superalgebra, which was studied in paper [18]. Denote the natural filtration by. We show that the filtration is invariant under automorphisms by determining ad-nilpotent elements and subalgebras generated by certain ad-nilpotent elements. We are thereby able to obtain an intrinsic characterization of Lie superalgebra.

The paper is organized as follows. In Section 2, we recall some necessary definitions concerning the modular Lie superalgebra. In Section 3, we establish some technical lemmas which will be used to determine the invariance of the filtration. In Section 4, we prove that the natural filtrationis invariant. Furthermore, we obtain the sufficient and necessary conditions of; that is, all the Lie superalgebras are classified up to isomorphisms.

2. Preliminaries

Throughout the workdenotes an algebraically closed field of characteristicandis not equal to its prime field. Letbe the ring of integers module. Letanddenote the sets of positive integers and nonnegative integers, respectively. For, letbe a subset ofthat is linearly independent over the prime field, and let be the additive subgroup generated bythat does not contain. If, then we letand, where.

Givenand, we put. Letand. If, thencan be uniquely expressed in-adic formwhereWe set We define a truncated polynomial algebra

such that

For, it is easy to see that

Let. If, we set.

Letbe the Grassmann superalgebra overinvariables, whereand. Denote bythe tensor product. The trivial-gradation ofand the natural-gradation ofinduce a-gradation ofsuch thatis an associative superalgebra:

Forand, we abbreviateto. Let

and where. Given, we setand. Thenis an-basis of.

Ifappears in some expression in this paper, we always regardas a-homogeneous element andas the-degree of.

Let,andPutSetandfor. Define, if, and, if. Let

Letbe the linear transformations ofsuch that

whereis the first nonzero number of. Thenare superderivations of the superalgebraand. Set

where is the identity mapping of. Letbe a-homogeneous element and; we define a bilinear operation insuch that

Thenbecomes a simple Lie superalgebra. If , we see that. In the sequel, we always assume that. In some cases, we denotebyin detail and callthe Lie superalgebra of-type.

Now we give a-gradation of:, where

Letfor all. Then are called the natural filtration of.

3. Ad-Nilpotent Elements

Letbe a Lie superalgebra. Recall that an elementis called -nilpotent if there exists asuch thatIfis -nilpotent, it is also called ad-nilpotent in brief. Letbe a subset ofPut, and Nil is the subalgebra ofgenerated by .

Letandbe the-adic expression of a, whereThen

is called the-adic sequence of, wherefor all. For, we define the-adic matrix ofto be

Sinceis amatrix with only finitely many nonzero elements, we can set

If is a nonzero element with, then we may assume

For, we define andNow for anyand, define

Lemma 1. Letand. Then the following statements hold.

(i). (ii) If, then, where.(iii) Letand. Then.

Proof. (i) We see that

Note that. By the uniqueness of-adic expression, we haveThus

as desired.

(ii) If, then. We see that. So (ii) holds.

Ifandfor, then we can assume that

Hence.

(a) If, then we get

(b) Ifand, then

(c) If, then

Thus (ii) holds.

(iii) By, we havethat is,.

Lemma 2. Let Then the following statements hold.

(i) If, then.(ii) If, then.(iii) If, then.

Proof. (i) As , Then we have

By the equality above and Lemma 1, we getThus

Hence

(ii) The proof is completely analogous to (i).

(iii) For, by assumption of this lemma, we have, which combined with Lemma 1 yield

For, it is easily seen thatAlso by Lemma 1, we obtain

Hence Lemma 2 holds.

Lemma 3. Letand. Letbe a nonzero summand of. Then.

Proof. By a direct computation, we obtain that

which satisfies the conditions of Lemma 2.

Lemma 4. .

Proof. Given, put

Clearly, we havefor all standard basis elementof.

Letsuch thatFor anywith, we have

Note thatBy using Lemma 3 repeatedly we see that. Hence.

Lemma 5. (i) Ifwhere, then.

(ii) Ifthen.

(iii) If then.

(iv)   .

(v)   .

Proof. (i) See Lemma  5 in [14].

(ii) By (i), we see thatis ad-nilpotent. If, thenfor alland. By a direct computation, we obtain, contradicting the nilpotency of. Hence, as desired.

(iii) Clearly,is ad-nilpotent by virtue of (i). If, then we can suppose that where. Thus there exists someBy computation, we have

Similarly, we getad, a contradiction. Consequently,.

(iv) Suppose thatis an arbitrary element of nil, where. By (1) of this lemma, we see that. Hence niland Nil.

Conversely, we haveNilby means of Lemma 4. Clearly, Nil. Thus NilThis shows that Nil, as desired.

(v) It is obvious that Nil((Conversely, we assume that

Then by (ii) of this lemma,. Hence, whereand. It follows from (iii) that. Now nil. Noting that nil, we havenil. Thus nil(, and the assertion holds.

Lemma 6. Let. Suppose thatis an arbitrary standard element of. Then the following statements hold.

(i). (ii) Ifand, then.(iii) Ifand, then.(iv).

Proof. (i) By a direct computation, we get

and. It follows from the binomial theorem that

(ii) Also by a direct calculation, we have

Put . Obviously,

Hence.

(iii) The proof is completely analogous to (ii).

(iv) According to (i), we see that.

Lemma 7. Suppose that are different from each other, and. Then

(i)for.

(ii).

Proof. (i) Setbe an arbitrary standard element of. Then

Put

Obviously,

Hence

Noting that, we obtain

Similarly,, and then.

(ii) LetIt follows from (i) that

Hence.

Lemma 8. forand.

Proof. By a direct computation, we obtain

Putand. Observingand, we see that

Thus, as required.

Lemma 9. The following statements hold

(i). (ii)For,.(iii)For,.

Proof. (i) Suppose thatis an arbitrary element of nil, where. If, then. A direct calculation shows that. Thusis not ad-nilpotent, contradicting the nilpotency of. HenceandObviously,is a subalgebra of, which yields Nil.

Conversely, we haveby virtue of Lemmas 6, 7, and 8. It follows that Nil.

(ii) By (i) of the lemma, we have

Conversely, the assertionfollows from Lemmas 6 and 7.

(iii) Suppose where. If, thenand. A direct calculation shows that. It follows thatis not ad-nilpotent. Thus. Thenand nilNote thatis a subalgebra of. Hence Niland (iii) holds.

Letbe the corresponding representation with respect tomodule; that is, for all. It is easily seen thatis faithful. For, we denote bythe matrix ofrelative to the fixed ordered-basis:

Denote bythe general liner Lie superalgebra ofmatrices over. Letdenote thematrix whose-entry is 1 and 0 elsewhere. Letdenote the identity matrix of size. Put. Letbe all thematrices set filled with. Put

Set

Lemma 10. (i)  .

(ii) If, thenis a nilpotent matrix.

Proof. (i) Let. By computation, we have

HenceandThe other cases are treated similarly. Thus (i) holds.

(ii) Asis a nilpotent elements, is a nilpotent liner transformation. Then by the definition of, we see thatis a nilpotent matrix.

Lemma 11. If,, then there exists asuch that.

Proof. By Lemma 9, we can assume that where.

Supposefor someandA direct calculation shows that

where every item ofdoes not contain. Then (ad. Thusad,, which implies thatis not a nilpotent element.

Iffor all, then we letfor someand. Also by computation, we have

where every item of does not containSimilarly,ad,, and thenis not nilpotent.

Hence our assertion follows.

Ifandfor all, thenWe see thatis a antisymmetric nilpotent matrix. By Lemmas 9(ii) and 10, it is easy to see that there issuch thatis not a nilpotent matrix; that is,is not an ad-nilpotent element. Hence our assertion holds.

Proposition 12. .

Proof. According to Lemma 4, we need only to determine all ad-nilpotent elements in. For any, a direct computation shows that

Lemmas 6, 7, and 8 imply that,, and are ad-nilpotent for. Hence our assertion holds.

4. Filtration and Intrinsic Property

Lemma 13. andis invariant.

Proof. Firstly, we prove the inclusion. Lemmas 6–9 show that Nil, which combined with (iv) and (v) of Lemma 5, yield

Then

that is,.

Let us consider the converse inclusion. Suppose that, whereand. If, then, whereand, a contradiction. Consequently,.

Now suppose, whereIfandfor some, then we have, where, a contradiction. ThusandThis proves the asserted inclusion.

By the proof above, we know thatis invariant.

Lemma 14. andis invariant.

Proof. Let. Suppose. By Lemma 5,andThen we can assume that, whereLetClearly,Lemma 5(1) implies that. According to Lemma 11, there existssuch that. Thusandis not nilpotent, contradicting the result of Lemma 5. Henceand; that is,.

Conversely,It is obvious thatand the proof is complete.

Lemma 15. (i)  .

(ii)  .

(iii)  .

Proof. (i) PutSuppose, whereand. Letfor some one. Then; that is,. This contradicts. Thus.

Letbe an arbitrary element of, whereand. If there is a, then, contradicting. Henceand. Consequently,.

Conversely, let. Then, which shows that.

(ii) We first prove the inclusion. Suppose, where. Noting thatand, we obtainThe converse inclusion is obvious.

(iii) The proof is completely analogous to (ii).

Theorem 16. is transitive.

Proof. Assume the contrary. Suppose that there exists a nonzerosuch that, where. Letbe the maximal exponent ofof all monomial expressions occurring in. Then we may assume that

where. For any, we have

whereis the sum of summand that the exponent ofis less than. Sinceis linear independence,; that is, . For any, we get

whereis the sum of summand that the exponent ofis less than. Thus; that is,does not containand.

If, then, contradicting. Let. Then we can suppose

whereFor, we have

whereis the sum of summand that the exponent ofis less than. Thus

By containsand dose not contain, we have, contradicting. Henceis transitive, as required.

Theorem 17. Suppose thatandare the Lie superalgebras of-type. Ifis an isomorphism ofonto, then for all.

Proof. As the isomorphismis an even mapping, we haveand. Hence. By Lemmas 13 and 14, we obtain

Now by virtue of Lemma 15, we get

It follows that. Because, we have

Sinceis transitive by the lemma above, we havefor all. It is easy to show thatby induction on.

Theorem 18. Suppose thatis an automorphism of. Thenfor all , that is; the filtration ofis invariant under the automorphism group of.

Proof. This is a direct consequence of Theorem 17.

Theorem 19. Letandbe the Lie superalgebras of-type. Thenif and only if.

Proof. The sufficient condition is obvious. We will prove the necessary condition. Assume thatis an isomorphism of Lie superalgebra. According to Theorem 17, we have

Theninduces an isomorphism of-graded spaces

It is easy to see that

We conclude thatby the dimension comparison.

Similarly, we also obtain an isomorphism of-graded spaces:

and the quotient spaceis isomorphic toA comparison of dimensions shows that. Note that, whereIt follows thatand.

Acknowledgments

The authors are grateful to the referees for their many valuable comments and suggestions. This work is supported by National Natural Science Foundation of China (no. 11126129) and the PhD Start-up Foundation of Liaoning University of China (no. 2012002).

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