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Abstract and Applied Analysis
Volume 2013 (2013), Article ID 923101, 10 pages
Infinite-Dimensional Modular Lie Superalgebra
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.
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.
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 . 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 . 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.
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
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
(ii) If, then. We see that. So (ii) holds.
Ifandfor, then we can assume that
(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
(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.
(iii) If then.
Proof. (i) See Lemma 5 in .
(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,
(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
Proof. (i) Setbe an arbitrary standard element of. Then
Noting that, we obtain
Similarly,, and then.
(ii) LetIt follows from (i) that
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
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
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) .
Proof. (i) PutSuppose, whereand. Letfor some one. Then; that is,. This contradicts. Thus.
Letbe an arbitrary element of