Abstract
In this paper, we classify all the finite-dimensional nilpotent Lie superalgebras of multiplier-rank less than or equal to 6 over an algebraically closed field of characteristic zero. We also determine the covers of all the nilpotent Lie superalgebras mentioned above.
1. Introduction
The notion of the multiplier for a finite group arose from Schur’s work on projective representations of groups. There are fruitful results about this topic. However, we only mention that Green proved that for a finite -group of order (see [1]) as well as that Berkovich and Zhou classified all -groups with and , where is defined by (see [2, 3]).
Batten initiated the theory of multipliers and covers for Lie algebras when she studied the second cohomology groups of Lie algebras with coefficients in a 1-dimensional trivial module (see [4]). For a finite-dimensional Lie algebra , she proved that by a free presentation of , where is the underlying field. Moneyhun gave the maximal dimension of (see [5]). Let , which is a nonnegative integer. Batten et al. classified all nilpotent Lie algebras with (see [6]). Hardy and Stitzinger used a different method to get similar results for (see [7, 8]). For filiform Lie algebras, Bosko classified these up to (see [9]).
The notion of the multiplier for Lie algebras may be naturally generalized to the Lie superalgebra case (see [10], for example). Let be a finite-dimensional Lie superalgebra. A Lie superalgebra pair is called a defining pair for provided that and . A defining pair of is said to be maximal if is of maximal super-dimensions. In the case of being a maximal defining pair of , we also call a cover and a (Schur) multiplier of . As in the Lie algebra case, multipliers for Lie superalgebra are unique up to isomorphism, denoted by . Moreover, (see [11]). By Lemma 2.3 of [10], for a Lie superalgebra of super-dimension , we define the super-multiplier-rank of to be and the multiplier-rank of to be
In particular, if and only if is abelian (see Proposition 3.1 of [10]). Hereafter, we write for the super-dimension of a superspace and for a pair of nonnegative integers.
In this paper, we classify all nilpotent Lie superalgebras of multiplier-rank by discussing whether . We also construct the covers of all the nilpotent Lie superalgebras of multiplier-rank .
2. Basics
We assume that all (super)spaces and (super)algebras are over an algebraically closed field of characteristic zero. In , we define a partial order as follows:
We also view as the additive group in the usual way.
To classify all the nilpotent Lie superalgebras of multiplier-rank , we first establish some technical lemmas.
Lemma 1. Let be a nilpotent Lie superalgebra of super-dimension . Then, .
Proof. By Theorem 3.8 of [11], . It follows that
where is the superspace consisting of skew-symmetric bilinear functions of .
Let be a homogeneous basis of and extend it to a homogeneous basis of . Let be the isomorphism from onto the dual space . For , let be a bilinear mapping by for all . Then, . If , then . Then, for all . Hence, is an injection from into . Consequently, By the definition of , we have .☐
Let be the parity functor of superspaces. Note that
Lemma 2. Let be a nilpotent Lie superalgebra. Suppose is a 1-dimensional central ideal of and . The following statements hold. (1)If then (2)If then
Proof. By Lemma 4.9 of [11], we have an exact sequence: Since we have for all and where is a Lie superalgebra monomorphism. Then, and is an injection. Hence, or Furthermore, and (1)If we haveBy the definition of we have (2)If we haveHence, ☐
As in the Lie algebra case (Theorem 1 of [7]), using free presentations of Lie superalgebras, one may prove the following lemma.
Lemma 3. Let and be finite-dimensional Lie superalgebras. Then,
Lemma 4. Let be a finite-dimensional nilpotent Lie superalgebra. Suppose that Then, there exists an ideal of with such that where is an ideal of . (1)If then (2)If then
Proof. Suppose that is a one-dimensional subsuperspace of such that but . Let be a complement to in such that . Then, and .
If then By Lemma 3, we have
Then,
If then Now Therefore, ☐
For convenience, we write for the abelian Lie superalgebra of super-dimension , for the -super-dimensional Heisenberg Lie superalgebra of even center, and for the -super-dimensional Heisenberg Lie superalgebra of odd center (see [12]). In Proposition 4.4 of [10] and section 4 of [13], the authors characterize the multipliers of and :
Hence,
Similarly, let us give a multiplier and cover of .
Case 1. Suppose that Let be a Lie superalgebra with a homogenous basis and multiplications where , and Let be a subsuperspace spanned by Then, is a maximal stem extension of In particular, is a multiplier and is a cover of
Case 2. Suppose that Let be a Lie superalgebra with even basis and odd basis and multiplications where , and and Let be a subsuperspace spanned by Then, is a maximal stem extension of In particular, is a multiplier and is a cover of
Summarizing, we have the following.
Lemma 5. Let be positive integers. Then, is a maximal stem extension of . In particular, is the cover of and the super-dimension of as follows: where
3. Multiplier-Rank Nilpotent Lie Superalgebras
In this section, we will determine all the nilpotent Lie superalgebras of The following theorem is analogous to the one in the Lie algebra case [7], yet it contains more information in our super-case.
Theorem 6. Let be a finite-dimensional, nonabelian, and nilpotent Lie superalgebra of Then, (1)(2) if and only if (3) if and only if (4) if and only if is isomorphic to one of the following Lie superalgebras: (a)(b)(c)
Proof. Let us characterize by discussing whether (1)Suppose that . Let be an ideal contained in with . Take (a)If then one may check that and by Lemma 2. In this case, by (12), we have(b)If , then Lemma 2 yields that . Hence, . Since is not abelian, we have . If then and , an impossibility. Therefore, It remains to consider the case . In this case, we have for some . By (13), we have (2)Suppose that . By Lemma 4, we have where and are ideals of with (a)If , then one may check that there are no algebras satisfying Lemma 2 (2) for , and (b)If , one has to discuss the cases , and (i)Assume that We obtain that and by Lemmas 1 and 4 (1). It follows that by Theorem 5.8 of [10]. Hence,By Lemma 5, (3) is proven. (ii)Assume that We have and by Lemmas 1 and 4 (1). There are no such algebras for by Proposition 5.22 of [10]. Therefore, (1) is proven(iii)Assume that By Lemmas 1 and 4 (1), we have , which yields . It follows that or Hence,Then, (4) holds by Lemma 5. (iv)Assume that We have either or by Lemmas 1 and 4 (1). If we have an impossibility. If then we have This is impossible, because yields by Theorem 5.8 of [10]. Then the, proof is complete☐
4. Multiplier-Rank Nilpotent Lie Superalgebras
In this section, we will determine all the nilpotent Lie superalgebras of We recall that is a Lie algebra with basis and nonzero multiplication and that is a Lie algebra with basis and nonzero multiplication (see [7]).
Theorem 7. Let be a finite-dimensional, nonabelian, and nilpotent Lie superalgebra of Then, (1)(2) if and only if (3) if and only if is isomorphic to one of the following Lie superalgebras: (a)(b)(c)(4) if and only if is isomorphic to one of the following Lie superalgebras: (a)(b)(c)(5) if and only if is isomorphic to one of the following Lie superalgebras: (a)(b)(c)
Proof. Let us characterize by discussing whether (1)Suppose that . We may assume that where and (a)If , one may check that by Lemma 2. Now assume that Then, Since hence or In the first case we obtain that and . Hence, . Then, can be described generally by basis with multiplication given by . To compute the multiplier start withwhere generate the multiplier. A change of variables allows that . Use of the Jacobi identity on all possible triples shows that and . Hence, , contradicting the assumption In the other case , we obtain that . Then, for some . There are no such algebras by (13) for (b)If , then Lemmas 1 and 2 yield and (i)Assume that . As in Lie algebra case (Theorem 2 of [7]), one can determine that(ii)Assume that and . The possible case for is . It follows that and for some . By (12), we haverespectively (2)Suppose that . By Lemma 4, we have where and are ideals of with (a)If , then Lemmas 1 and 4 (2) yield that . Therefore, and Hence,(b)If , then Lemmas 1 and 4 (1) yield that or . As a result of the above, we have proven (1) and (2). To prove (3), (4), and (5), one has to discuss the cases , and (i)Assume that . It is easily checked that there are no such algebras except for and . By Theorem 6, we have andBy Lemma 5, (4) is proven. (ii)Assume that . We may check that and . Then, by Theorem 6. Hence,Then, (3) holds by Lemma 5(iii)Assume that . We have and . By Theorem 6, we have or . Then,The proof is complete.☐
5. Multiplier-Rank Nilpotent Lie Superalgebras
In this section, we will determine all the nilpotent Lie superalgebras of For convenience, let denote a Lie superalgebra with a homogeneous basis and nonzero multiplication
Theorem 8. Let be a finite-dimensional, nonabelian, and nilpotent Lie superalgebra of Then, (1) and (2) if and only if (3) if and only if is isomorphic to one of the following Lie superalgebras: (a)(b)(4) if and only if is isomorphic to one of the following Lie superalgebras: (a)(b)(c)(5) if and only if is isomorphic to one of the following Lie superalgebras: (a)(b)(c)(d)
Proof. Let us characterize by discussing whether (1)Suppose that . We may assume that , where and (a)If , then Lemma 2 readily yields or (i)Assume that Then, from the proof of Theorem 3 of [3], we have(ii)Assume that . This readily yields , , or . In the first case then and for some . By (12), we haveConsider the second case . If and , then there are no such algebras by the previous work. If , then or . If then a computation shows that , contradicting the assumption . If we have by a direct computation, also contradicting the assumption . If , then . By the proof of Theorem 1 of [7], we have and . This contradicts the assumption on If , then , which contradicts the assumption . In the third case we have . By the previous work, we have and . Then, has a homogeneous basis with multiplication given by , where . The Jacobi identity shows that . Since , we have that either or is not zero. Without loss of generality, assume that . Replacing by and by and relabeling, we get a simple multiplication table: To compute the multiplier, we start with where generate the multiplier. By relabeling, we get . Using the Jacobi identity gives . Hence, the multiplier has a homogeneous basis , and then, and . This superalgebra satisfies the requirements. As mentioned above, we have (iii)Assume that . If , then . There are no such algebras by the previous work. Then, It follows from Proposition 4.11 of [10] that(b)If , then Lemma 2 (2) yields that or (i)Assume that Then, . It follows that and If , then Computing the multiplier as before yields and contradicting the assumption If then It is also a contradiction by the previous work(ii)Assume that We have It follows that and Then, or . If then and by computing as above, contradicting the assumption If then has a homogeneous basis with multiplication given by . Computing the multiplier as before yields and also contradicting the assumption (2)Suppose that . By Lemma 4, we have where and are ideals of with (a)If , then Lemma 4(2) yields that and . Then by Theorem 6, we have(b)If , then Lemmas 1 and 4 yield that or . As a result of the above, we have proven (1) and (2). To prove (3), (4), and (5), one has to discuss the cases , and (i)Assume that . Then, . If , then . By Theorem 6, we have . This yields contradicting the assumption . If , then is abelian and an impossibility. If . Then, . By Theorem 7 (4), we have or . Since , we have . Then,By Lemma 5, (3) holds. (ii)Assume that . Then, If computing the multiplier as before yields This contradicts the assumption on Therefore, and Then, is one of the Lie superalgebras listed in Theorem 7 (5). Therefore, is isomorphic to one of the following Lie superalgebras:Then, (4) holds by Lemma 5. (iii)Assume that Then, If , then This is impossible, because the super-dimension of the derived superalgebra of when is not Similarly, if , then a contradiction. Now consider ; we have . By Theorem 7 (3), we haveThe proof is complete.☐
6. Multiplier-Rank Nilpotent Lie Superalgebras
In this section, we will determine all the nilpotent Lie superalgebras of Let us recall that is a Lie algebra with basis and nonzero multiplication For convenience, let denote a Lie superalgebra with a homogeneous basis and nonzero multiplication
Theorem 9. Let be a finite-dimensional, nonabelian, and nilpotent Lie superalgebra of . Then, (1)(2) if and only if (3) if and only if (4) if and only if is isomorphic to one of the following Lie superalgebras: (a)(b)(c)(d)(e)(5) if and only if is isomorphic to one of the following Lie superalgebras: (a)(b)(c)(d)(6) if and only if is isomorphic to one of the following Lie superalgebras: (a)(b)(c)(d)(7) if and only if is isomorphic to one of the following Lie superalgebras: (a)(b)(c)(d)(e)
Proof. Let us characterize by discussing whether (1)Suppose that . We may assume that , where and (a)If , then Lemma 2 readily yields , or In the first case , by Theorem 4 of [7], one can determineIn the remaining cases , and , we obtain that the possible super-dimension of is . It follows that and for some . By (12), we have respectively. (b)If , then Lemma 2 yields that or (i)Assume that Then, . It follows that the only possible super-dimension of is . Hence, we have and . There are no such algebras for by the previous work. If , then or . In the first case then has a homogeneous basis with multiplication . The Jacobi identity shows that . Since and , we have . Take for . If , then we obtain a simple multiplication . Computing the multiplier as before yields and , contradicting the assumption If or is not zero, then without loss of generality, assume that . Replacing by and by and relabeling, we obtain a simple multiplication . Computing the multiplier as before yields and also contradicting the assumption In the other case one may obtain that can be described by a homogeneous basis with multiplication . The Jacobi identity shows that . Take for , relabel and get . Then, and This algebra satisfies the requirements. As mentioned above, we have(ii)Assume that . Then, and by a direct computation. Hence, . It follows that for some . By (12), we have(2)Suppose that . By Lemma 4, we have where and are ideals of with (a)If , then Lemmas 1 and 4 (2) yield that , or . Then, by the previous work (i)Assume that . Then, and Theorem 7 reveals three candidates, only one of which satisfies . Therefore, It yields and contradicts the assumption that (ii)Assume that Then, and , or . Therefore, is isomorphic to one of the following Lie superalgebras:(iii)Assume that Then, and or . Hence,(b)If , then Lemmas 1 and 4 (1) yield that the possible super-multiplier-rank of is , or By the above work, we have proven (1), (2), and (7). To prove (3), (4), (5), and (6), one has to discuss the cases , and (i)Assume that One may obtain that and . By Theorem 8, we have andThen, (3) holds by Lemma 5. (ii)Assume that By Theorem 4 of [7], we have is isomorphic to one of the following Lie superalgebras:By Lemma 5, (4) is proven. (iii)Assume that . Since , it follows that . If , then is abelian and , an impossibility. Therefore, If , then , or by Theorem 8. Therefore is isomorphic to one of the following Lie superalgebras:If , then , or and . These are impossible, because the super-dimension of the derived superalgebras of , and are . Hence, (5) holds by Lemma 5. (iv)Assume that . The only possibility of super-multiplier-rank of is . Then, Theorem 7 reveals four candidates, only three of which satisfy . Therefore, is isomorphic to one of the following Lie superalgebras:The proof is complete.☐
7. Covers
By Theorem 3.3 of [13], we obtain that any two covers of a finite-dimensional Lie superalgebra are isomorphic as ordinary Lie superalgebras. In this section, we will describe the covers of all the nilpotent Lie superalgebras of multiplier-rank .
First, we compute the cover for Let be a homogeneous basis of where and The multiplication is given by the other brackets of basis elements vanishing. Suppose that
is a stem extension of . Then, and . Then, has a homogeneous basis . We may assume that where Without loss of generality, we may suppose that . By the Jacobi identity of Lie superalgebras, we have Then, is spanned by and is spanned by Hence, Now let be a Lie superalgebra with a homogenous basis and multiplication Let be a subsuperspace of spanned by Then,
Since we obtain that is a multiplier and is a cover of
Then, we compute the cover for Let be a Lie superalgebra with a homogenous basis and multiplication Since, is a subalgebra of one may check that is a cover of
Now we compute the cover for Let be a homogeneous basis of where and The multiplication is given by the other brackets of basis elements vanishing. Similarly, one may check that the Lie superalgebra with a homogeneous basis and multiplication is a cover of , which is denoted by
Then, we compute the cover for Since is a subalgebra of one may check that the Lie superalgebra with a homogeneous basis and multiplication is a cover of which is denoted by
When we also obtain a cover for which has a homogenous basis and multiplication
By the previous work, we may obtain a cover for which has a homogenous basis and multiplication When we may obtain a cover for , which has a homogenous basis and multiplication
8. Main Result
Theorem 10. The classification and covers of all the finite-dimensional nilpotent Lie superalgebras of multiplier-rank are listed as follows:
where is a (2, 2)-superdimensional Lie superalgebra with a homogeneous basis and nonzero multiplication is a (2, 2)-superdimensional Lie superalgebra with a homogeneous basis and nonzero multiplication
Proof. By Theorem 5.8 of [10], we obtain all the finite-dimensional nilpotent Lie superalgebras of . In Theorems 6, 7, 8, and 9, all the nilpotent Lie superalgebras of are determined. We only need to describe the covers of all Lie superalgebras of . By Theorem 4.1 of [13], Lemma 5, and Section 7, we may obtain the covers of all the nilpotent Lie superalgebras of . In summary, we obtain Table 1.☐
Data Availability
The data used to support the findings of this study are included within the article.
Conflicts of Interest
The authors declare that they have no conflicts of interest.
Acknowledgments
The second named author (J. Nan) was supported by the NSF of China (11771176). The third named author was supported by the NSF of China (12061029), the NSF of Hainan Province (120RC587), and the NSF of Heilongjiang Province (YQ2020A005).