Abstract

The natural filtration of the infinite-dimensional contact superalgebra over an algebraic closed field of positive characteristic is proved to be invariant under automorphisms by characterizing ad-nilpotent elements and the subalgebras generated by certain ad-nilpotent elements. Moreover, we obtain an intrinsic characterization of contact superalgebras and a property of automorphisms of these Lie superalgebras.

1. Introduction

Filtration structures play an important role in the classification of modular Lie algebras (see [1, 2]) and nonmodular Lie superalgebras (see [3, 4]), respectively. We know that the Lie algebras and Lie superalgebras of Cartan type possess a natural filtration structure. The natural filtrations of finite-dimensional modular Lie algebras of Cartan type were proved to be invariant in [5, 6]. In the infinite-dimensional case, the same conclusion was proved in [7], by determining -nilpotent elements. In the case of Lie superalgebras of Cartan type, the invariance of the natural filtrations of some Lie superalgebras was proved in [8, 9]. Similar results for Lie superalgebras of generalized Cartan type were obtained in [1012], respectively.

In this paper, we consider the infinite-dimensional modular contact superalgebra , which is analogous to the one in the nonmodular situation (see [13]). But since the principal -gradations of Lie superalgebras of Cartan type are different (see [13]), most results and proofs for other Lie superalgebras cannot be applied to contact superalgebras. Therefore the corresponding results and proofs for contact superalgebras have to be established separately. By determining the -nilpotent elements and subalgebras generated by certain -nilpotent elements, we prove the main result of this paper.

Theorem 1. The natural filtration of the infinite-dimensional contact Lie superalgebra is invariant under automorphisms.

Thereby, one obtains the following theorems.

Theorem 2. Suppose that , , , are positive integers. Then if and only if .

Theorem 3. Let , be automorphisms of . Then if and only if  .

The paper is organized as follows. In Section 2, we recall the necessary definitions concerning the modular contact superalgebra . In Section 3, we study the ad-nilpotent elements of . In Section 4, we complete the proofs of Theorems 13.

2. Preliminaries

Throughout this paper, denotes an algebraic closed field of characteristic and the ring of integers modulo 2. Let and denote the sets of positive integers and nonnegative integers, respectively. Given , and , we put . Let denote the divided power algebra over with basis . For , , we abbreviate to . Let be the exterior superalgebra over in variables , where . Denote by the tensor product . The trivial -gradation of and the natural -gradation of induce a -gradation of such that is an associative superalgebra. For and , we abbreviate to . For and , the following formulas in hold: where

Put , , and . Let where . Given , set , , and . Then is an -basis of the infinite-dimensional superalgebra .

Let be the linear transformations of such that Then are superderivations of the superalgebra . Let Then is an infinite-dimensional Lie superalgebra contained in (see [14]). If appears in some expression in this paper, we always regard as a -homogeneous element and as the -degree of . Then , where The following formula holds in (see [14]): where and .

Hereafter let be a positive integer and let . Put and . For , define Let be the linear mapping such that where Then where (see [14]). It follows directly from (11) and the injectivity of that defines a Lie multiplication on . This Lie superalgebra, denoted by , is called the infinite-dimensional contact superalgebra. In the sequel, we simply write for .

The following formula holds in (see [14]): Then is a -graded Lie superalgebra, where Let for . Then is referred to as the natural filtration of .

Lemma 4. is transitively graded.

Proof. Assume the contrary, then there exists such that for all , where . Suppose that the largest exponent of among the nonzero summands in the expression of is equal to , and write where . Hence, for , where each exponent of of all nonzero summands in the expression of is less than . Then . Note that all nonzero summands of are -linear independent. It follows that each is equal to . Hence the exponents of in each are equal to . Similarly, for , we can prove that does not appear in each . Consequently, we see that all are of the form . If , then , contradicting . Hence , and we can write where . Note that, for , where each exponent of of the nonzero summands in the expression of is less than . Therefore, Since appears in and does not appear in , we conclude that , a contradiction.

3. ad-Nilpotent Elements

Recall that is called -nilpotent if there exists such that . For a subset of , let denote the set of -nilpotent elements in , and let denote the subalgebra of generated by .

Lemma 5. Suppose that for . The following statements hold.(1)If , then .(2)If , then .(3)If , then .(4)If , then .(5)If , then .

Proof. See [15, Lemma 5.1].
Suppose that . As is -nilpotent, is -nilpotent by . Note that for all . This shows that is not -nilpotent, a contradiction.
By , we see that is -nilpotent. Suppose that , where . Then there exists some . A direct calculation shows that for all . It follows that is not -nilpotent.
is an immediate consequence of , , and . follows from , , and the proof of .

Let and be the -adic expression of , where . Then, is said to be the -adic sequence of , where for all . For , define the -adic matrix of to be As is an matrix with finitely many nonzero elements, is well defined. Let for and . We abbreviate to .

Suppose that is a nonzero element of , where . Define

Given that and , we define

Lemma 6. Let , , and . Then,(1) if and only if ;(2)If , then ;(3)If and , then .

Proof. See [7, Lemma 2.5].
First consider the case . Then, It follows that and , and thus holds.
Next consider the case . We may assume that where and . Hence, If , then If , noting that , then If , then
The assumption implies that . Then it is trivially verified that holds.

Lemma 7. Suppose that , , , and . The following statements hold.(1)If , then .(2)If , then .(3)If , then .

Proof. The assumption implies that . By Lemma 6, we have Consequently, By and of Lemma 6, we obtain Combining (34)–(38), we have
Suppose that . Then,
Similarly, we have

Lemma 8. Suppose that , , and . Let be a nonzero summand of  . Then .

Proof. A direct calculation shows that fulfills the conditions of Lemma 7.

Given that , let Clearly, the inequality holds for all .

Lemma 9. .

Proof. Suppose that is an arbitrary element of , where and . Let such that ht. Let be a standard basis element of . By using Lemma 8 repeatedly, we see that .

Lemma 10. For , the following statements hold.(1).(2)If   and , then .(3)If   and , then .(4).

Proof. Let be a standard basis element of . A direct calculation shows that
Since and , it follows from the binomial theorem that .
Since and , we have .
It follows from that .

Lemma 11. Suppose that . The following statements hold.
Let be distinct elements of , and let be such that . Then .
holds for all distinct .

Proof. A direct calculation shows that . For simplicity, we denote by , respectively. Clearly, Then By , we obtain Note that It follows that thus proving that .
Let such that . Since , we have . Let . Then yields Hence .

Lemma 12. Let and . Then .

Proof. A direct calculation shows that Since we have

Lemma 13. .
.
If , then .
If , then .
.

Proof. Let be an arbitrary element of . Suppose that . Since and , it follows that is not -nilpotent, a contradiction. Hence , and therefore . Noting that is a subalgebra of , we obtain .
Conversely, Lemma 10 shows that , and Lemma 12 implies that . Moreover, since for all , we have . Therefore .
It is clear that , which combined with Lemma 9 yields .
On the other hand, suppose that is an arbitrary element of , where and . By Lemma 5, we have , and hence . Since is a subalgebra of , it follows that .
By , we see that , . The reverse inclusion follows from Lemmas 10 and 11.
Clearly the statement holds when . Now we consider the case . By , we can suppose that is an arbitrary element of , where . If , a direct calculation shows that for all , contradicting that is -nilpotent. Hence and , proving . Since is a subalgebra of , it follows that . The reverse inclusion follows from Lemma 10.
is completely analogous to the proof of .

Let be the corresponding representation with respect to -module ; that is, , . It is easy to see that is faithful. For , we also denote by the matrix of relative to the fixed ordered -basis as follows: Denote by the general linear Lie superalgebra of matrices over . Let denote the matrix whose -entry is and elsewhere, and , where is unit matrix. Let be the Lie algebra consisting of all matrices over satisfying , where is the transpose of . Set ; here

Lemma 14. .
If , then is a nilpotent matrix.

Proof. For , a direct calculation shows that Therefore where Since , , we see that From (56) and (58), we can easily verify .
follows from the definition of .

If is an antisymmetric matrix over , we write to denote . As , it is clear that if is a nilpotent matrix then .

Lemma 15. Suppose that . Let be an antisymmetric matrix over of order . If satisfies the following properties:(1);(2) holds for every antisymmetric matrix .
Then, .

Proof. Let . Three cases arise as follows.
Case  1 . By the property of , we see that holds for all . Therefore, if , a direct calculation shows that the left-hand side of (59) is equal to . Hence . Similarly, if , then , and if , then . Thus, for all , we see that . Then
Denote for . If , a direct calculation shows that Let . Then From the equalities above, we see that holds for all distinct . Pick . Then . Hence holds for all distinct . It follows that holds for all .
Assume that there exists some . For distinct , we have by (60). Then we can write the following: A direct calculation shows that Since by (63), we obtain . Hence by (63), contradicting the assumption that .
Case  2 . Note that (60) and (63) hold; that is, Moreover, since , we see that .
Case  3 . Since and , it follows that . Similarly .

Lemma 16. Let be a nonzero element of . Then there exists such that is not -nilpotent.

Proof. By Lemma 13, we can suppose that
where . Three cases arise as follows.
Case  1. There exists some . Let . Then , where does not appear in the expression of . Noting that for all , we conclude that is not -nilpotent.
Case  2. All , and there exists . Let . A direct calculation shows that , where does not appear in the expression of . As and , we see that is not -nilpotent.
Case  3. All and . So . This can happen only if by Lemma 13. Note that is an antisymmetric nilpotent matrix. Then Lemma 15 provides an element of such that is not a nilpotent matrix. Hence is not -nilpotent.

4. Proofs of Theorems

Proof of Theorem 1. We proceed in several steps.
(I) . Suppose that is an arbitrary element of . It follows from and of Lemma 5 that , which combined with yields , thus proving . Hence .
The reverse inclusion is clear.
(II) . We first prove . It follows from (I) that Therefore By formula (12), we see that and , proving In the case of , by and of Lemma 13, we obtain . By (69) and (70), we have . In the case of , by and of Lemma 13, we have . Note that , which combined with (69) and (70) yields . In the case of , by and of Lemma 13, we have . Hence by (69) and (70).
Conversely, suppose that , where , . If , then by Lemma 13, contradicting , where . Hence and we can write , where , , . If there is some , then , where . Lemma 13 shows that , contradicting . Hence and , proving .
(III) Let . Then . Suppose that is an arbitrary element of , where and . If , since by Lemma 5, then Lemma 16 provides an element such that is not -nilpotent. Hence is not -nilpotent by Lemma 5, contradicting . Therefore and , thus proving .
On the other hand, since by Lemma 9, it follows that .
(IV) Let . Then . Suppose that , where and . If there is some , then , and hence , contradicting . Therefore , and we can write , where and . If there exists some , then , a contradiction which yields and .
The reverse inclusion follows from the fact that .
(V) . It suffices to show that . Suppose that is an arbitrary basis element of with . Note that . Since , it follows that , as desired.
It follows from (II) and (V) that is invariant under automorphisms of . By (III) and (IV), we obtain that is invariant. Therefore is invariant. By the transitivity of , we conclude that Hence the natural filtration of is invariant under automorphisms of .

Proof of Theorem 2. Let be an isomorphism of Lie superalgebras. Let and denote and , respectively. Since and , it follows that . By (II) in the proof of Theorem 1, we have Consequently Applying (III) in the proof of Theorem 1, we see that , which combined with (V) yields It follows from (72) and (74) that . Therefore induces an isomorphism of -graded spaces . A comparison of dimensions shows that and . The converse implication is clear.

Proof of Theorem 3. It suffices to prove that implies that . Since , it follows that . We use induction on to show that Assume that and (75) holds for . Suppose that and let . We want to prove . The induction hypothesis yields that and , . Therefore, Since and by Theorem 1, induces an automorphism of the -graded space . Consequently there exists a homogeneous basis of such that . Thus there exist such that , . Therefore, (76) shows that for all . As by Theorem 1, it can be decomposed into , where . Noting that and , we obtain for all , and hence since is transitively graded. By induction, we conclude that , . Hence, , as desired.

Conflict of Interests

The author declares that there is no conflict of interests regarding the publication of this paper.