# Simple Modules for Modular Lie SuperalgebrasW

Authors: Zhu Wei; Qingcheng Zhang; Yongzheng Zhang; Chunyue Wang
Journal: Advances in Mathematical Physics
 (2015)

---

## Abstract

This paper constructs a series of modules from modular Lie superalgebrasW
(
0
∣
n
), S
(
0
∣
n
), and K
(
n
) over a field of prime characteristic p
≠
2. 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 L-modules, where L
=
W
(
0
∣
n
), S
(
0
∣
n
), and K
(
n
).


---

## Body

## 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 p
>
2 (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 W, S, H 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 K
(
m
,
n
) and K
(
n
). 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 characteristicp
≠
2, 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 W
(
0
∣
n
), S
(
0
∣
n
), and K
(
n
), induced from homogeneous components of their restricted universal enveloping superalgebras. We intend to show that the generator 1
⊗
m of these constructed modules belongs to their nonzero submodules.


## 2. Preliminaries

In this paper,F always denotes an algebraically closed field of prime characteristic p
≠
2.Recall the definition of the restricted Lie superalgebras (see [6]) and the restricted universal enveloping superalgebra (see [25]).LetL be a Lie algebra of characteristic p
≠
2. Then, for every x
∈
L, the operator a
d
x
p is a derivation of L. If it is an inner derivation for every x
∈
L, that is, if a
d
x
p
=
a
d
x
[
p
] for some element denoted x
[
p
], then the corresponding map [
p
]
:
x
→
x
[
p
] is called a p-structure on L, and the Lie algebra L endowed with a p-structure is called a restricted Lie algebra. If L has no center, then L can have not more than one p-structure. The Lie algebra g
l
(
n
) possesses a p-structure, unique up to the contribution of the center; this p-structure is used in the next definition. The notion of a p-representation is naturally defined as a linear map ρ
:
L
→
g
l
(
V
) such that ρ
(
x
[
p
]
)
=
ρ
(
x
)
[
p
]; in this case V is said to be a p-module. Passing to superalgebras, we see that, for any odd D
∈
D
e
r
(
A
), we have (1)
D
2
n
a
,
b
=
∑
n
l
D
2
l
a
,
D
2
n
-
2
l
bfor any a
,
b
∈
A. So if char p
≠
2, then D
2
p is always an even derivation for any odd D
∈
D
e
r
(
A
). Now, let L be a Lie superalgebra of characteristic p
≠
2. Then, for every x
∈
L
0
¯, the operator a
d
x
p is a derivation of L; that is, L
0
¯-action on L
1
¯ is a p-representation. For every x
∈
L
1
¯, the operator a
d
x
2
p
=
a
d
x
2
p is a derivation of L. So if, for every x
∈
L
0
¯, there is x
[
p
]
∈
L
0
¯ such that a
d
x
p
=
a
d
x
[
p
] for any x
∈
L
0
¯, then we can define x
[
2
p
]
≔
(
x
2
)
[
p
] for any x
∈
L
1
¯. We demand that, for any x
∈
L
0
¯, we have a
d
x
p
=
a
d
x
[
p
] as operators on the whole L; that is, L
1
¯ is a restricted L
0
¯-module. Then, the pair of maps (2)
p
:
L
0
¯
⟶
L
0
¯
x
⟼
x
p
,
2
p
:
L
1
¯
⟶
L
0
¯
x
⟼
x
2
pis called a p
∣
2
p-structure or just p-structure on L, and the Lie superalgebra L endowed with a p-structure is called a restricted Lie superalgebra.A pair(
u
(
L
)
,
i
) consisting of an associative F-superalgebra with unity and a restricted homomorphism i
:
L
→
u
(
L
)
- is called a restricted universal enveloping superalgebra if given any associative F-superalgebra A with unity and any restricted homomorphism f
:
L
→
A
-; there exists a unique homomorphism f
¯
:
u
(
L
)
→
A of associative F-superalgebra such that f
¯
∘
i
=
f. The category of u
(
L
)-modules and that of restricted L-modules are equivalent. According to the PBW theorem, then the following statement holds: let (
L
,
[
p
]
) be a restricted Lie superalgebra. If (
u
(
L
)
,
i
) is a restricted universal enveloping superalgebra and (
l
j
)
j
∈
J
0
∪
(
f
j
)
j
∈
J
1 is an ordered basis of L over F, where l
j
∈
L
0
¯
,
f
j
∈
L
1
¯, then the elements i
(
l
j
1
)
s
1
i
(
l
j
2
)
s
2
⋯
i
(
l
j
n
)
s
n
i
(
f
i
1
)
i
(
f
i
2
)
⋯
i
(
f
i
m
), j
1
<
⋯
<
j
n, 0
≤
s
k
≤
p
-
1, 1
≤
k
≤
n, and i
1
<
⋯
<
i
m, consist of a basis of u
(
L
) over F. Sometimes, with no confusion, we will identify L with its image i
(
L
) in u
(
L
).Definition 1.
LetL be a Z-graded Lie superalgebra over a field of characteristic p. Suppose that H is a Cartan subalgebra of L
0, where L
0 is the set of the 0th homogenous elements of Z-graded Lie superalgebra L. For λ
∈
H
∗ and a u
(
L
0
)-module V, we set V
λ
≔
{
v
∈
V
∣
h
·
v
=
λ
(
h
)
v
,
∀
h
∈
H
}. If V
λ
≠
0, then λ is called a weight and a nonzero vector v in V
λ is called a weight vector (of weight λ). A nonzero v
∈
V
λ is called a maximal vector (of weight λ), provided x
·
v
=
0, where x is any positive root vector of L
0.
LetL
=
Σ
i
∈
Z
L
i be a Z-graded Lie superalgebra over F. Set N
+
≔
∑
i
>
0
L
i, where L
i denotes the homogeneous component of degree i in the Z-graded Lie superalgebra L. Then, N
+
⊲
N
+
+
L
0
≔
L
+ and L
+
/
N
+
≅
L
0. In particular, any L
0-module becomes a L
+-module by letting N
+ act trivially. Define M
L
(
S
)
≔
u
(
L
)
⊗
u
(
L
+
)
S, where u
(
L
) and u
(
L
+
) denote the restricted universal enveloping superalgebras of L and L
+, respectively, and S is a simple u
(
L
0
)-module. According to the classical theory, for each weight λ, there exists a simple u
(
L
0
)-module S
(
λ
) which is generated by a maximal vector of weight λ.In the following,L denotes one of three classes of Cartan-type Lie superalgebras W
(
0
∣
n
), S
(
0
∣
n
), or K
(
n
). Each of these classes will be described in detail in the following paper.Remark 2.
In this paper, ifA is a subset of some linear space, then 〈
A
〉 denotes the subspace spanned by the set A over F.


## 3. Simple Modules of the Lie SuperalgebraW
(
0
∣
n
)

We begin by describing the Lie superalgebraW
(
0
∣
n
), drawing most of notations and standard results from [26].LetΛ
(
n
) be an exterior algebra over F in n variables x
1
,
…
,
x
n. Fix n
∈
N, and then Λ
(
n
) becomes Z
2-graded if we set deg
⁡
x
i
=
1
¯, i
=
1
,
…
,
n
. Then, Λ
(
n
) is an associative superalgebra. The multiplication satisfies the rule x
i
x
j
=
-
x
j
x
i, in particular, x
i
x
i
=
0. For k
=
1
,
…
,
n, put B
k
≔
{
(
i
1
,
…
,
i
k
)
∣
1
≤
i
1
<
i
2
<
⋯
<
i
k
≤
n
}. Let B
(
n
)
=
⋃
i
=
0
n
B
k, where B
0
=
∅. If u
=
(
i
1
,
i
2
,
…
,
i
r
)
∈
B
r, where 1
≤
i
1
<
i
2
<
⋯
<
i
r
≤
n, then we set x
u
=
x
i
1
x
i
2
⋯
x
i
r. Put x
∅
=
1. Note that Λ
(
n
) is Z-graded by Λ
(
n
)
r
:
=
〈
{
x
u
∣
u
∈
B
r
}
〉, where r
=
0,1
,
…
,
n
.For each1
≤
i
≤
n, let D
i denote the derivation of Λ
(
n
) uniquely determined by the property D
i
(
x
j
)
=
(
∂
/
∂
x
i
)
(
x
j
)
=
δ
i
j (=Kronecker delta). Let ε
k
,
1
≤
k
≤
n, be the n tuple with jth component δ
j
k (=Kronecker delta), and then W
=
W
(
0
∣
n
)
:
=
{
∑
i
=
1
n
a
i
D
i
∣
a
i
∈
Λ
(
n
)
} is a Lie superalgebra, which has a F-basis {
x
u
D
i
∣
u
∈
B
(
n
)
}. The bracket product in W satisfies (3)
x
u
D
i
,
x
v
D
j
=
x
u
D
i
x
v
D
j
-
-
1
d
e
g
x
u
D
i
d
e
g
x
v
D
j
x
v
D
j
x
u
D
i
,where x
u
D
i
,
x
v
D
j are Z
2-homogeneous elements of the Lie superalgebra W, and d
e
g
(
x
u
D
i
)
=
d
e
g
(
x
u
)
+
1, d
e
g
(
x
v
D
j
)
=
d
e
g
(
x
v
)
+
1. Note that d
e
g
(
D
i
)
=
1
¯
∈
Z
2. W inherits a Z-gradation from Λ
(
n
) by means of W
k
=
Σ
j
Λ
(
n
)
k
+
1
D
j. Consequently, W
=
∑
i
=
-
1
n
-
1
W
i.Lemma 3.
LetH
=
∑
i
=
1
n
F
x
i
D
i be a Cartan subalgebra of W
0. Then, positive root vectors of W
0 are {
x
i
D
j
∣
1
≤
i
<
j
≤
n
}.Proof.
Introduce a homomorphismφ
:
W
0
→
g
l
n
(
F
), where g
l
n
(
F
) is the general linear Lie algebra that sends x
i
D
j to E
i
j (=n
×
n-matrix with 1 in the (
i
,
j
)-position and zeros elsewhere). Obviously, φ is an isomorphism of Lie algebras. We know that the Cartan subalgebra H
1 of g
l
n
(
F
) is 〈
{
E
i
i
∣
i
=
1
,
…
,
n
}
〉. Define linear function Λ
j on the vector space 〈
{
E
11
,
…
,
E
n
n
}
〉, such that Λ
j
(
E
i
i
)
=
δ
j
i. Then, the positive roots of g
l
n
(
F
) are {
Λ
i
-
Λ
j
}
i
<
j. Correspondingly, the positive root vectors are E
i
j, 1
≤
i
<
j
≤
n. By the isomorphism W
0
≅
g
l
n
(
F
) via x
i
D
j
↦
E
i
j, we may therefore obtain that the Cartan subalgebra H of W
0 is 〈
{
x
i
D
i
∣
i
=
1
,
…
,
n
}
〉. And the positive root vectors are {
x
i
D
j
∣
1
≤
i
<
j
≤
n
}.According to Definition1 and Lemma 3, the following statement holds: if H is a Cartan subalgebra, V is a u
(
W
0
)-module, and λ
∈
H
∗, then V
λ
=
{
v
∈
V
∣
x
i
D
i
·
v
=
λ
(
x
i
D
i
)
v
,
1
≤
i
≤
n
}. Write λ
i
:
=
λ
(
x
i
D
i
). A nonzero v
∈
V
λ is a maximal vector (of weight λ), provided x
i
D
j
·
v
=
0, ∀
1
≤
i
<
j
≤
n.Theorem 4.
If there existi
,
j
∈
{
1
,
…
,
n
} and j
<
i such that λ
i
≠
1 and λ
j
≠
0, then M
W
(
S
(
λ
)
) is simple.Proof.
LetM
′ be a nonzero submodule of M
W
(
S
(
λ
)
)
=
u
(
W
)
⊗
u
(
W
+
)
S
(
λ
). Now choose 0
≠
v
∈
M
′. By virtue of u
(
W
+
)
·
S
(
λ
)
=
u
(
W
0
)
·
S
(
λ
), we obtain u
(
W
)
·
S
(
λ
)
=
(
u
(
W
0
)
+
u
(
W
-
1
)
)
·
S
(
λ
)
⊆
u
(
W
-
1
)
·
S
(
λ
). Hence, M
W
(
S
(
λ
)
)
=
u
(
W
)
⊗
u
(
W
+
)
S
(
λ
)
=
(
u
(
W
-
1
)
+
u
(
W
0
)
)
⊗
u
(
W
+
)
S
(
λ
)
⊆
u
(
W
-
1
)
⊗
u
(
W
+
)
S
(
λ
); namely, M
W
(
S
(
λ
)
)
=
u
(
W
-
1
)
⊗
u
(
W
+
)
S
(
λ
). Therefore, we can describe the element form of v; that is, (4)
v
=
Σ
β
∈
A
c
β
i
D
1
β
1
⋯
i
D
n
β
n
⊗
s
β
,where A
≔
{
β
=
(
β
1
,
…
,
β
n
)
∣
β
i
=
0
o
r
1
,
i
=
1
,
…
,
n
}
⊂
Z
n, c
(
β
)
∈
F. Formula (3) shows that [
D
k
,
D
j
]
=
0. Then, we obtain the equality in u
(
W
): (5)
i
D
k
i
D
j
=
-
i
D
j
i
D
k
;in particular,(6)
i
D
j
2
=
0
.In the following, we simply write i
(
D
j
)
=
D
j. Define an order of A satisfying that β
=
(
β
1
,
…
,
β
n
)
<
β
′
=
(
β
1
′
,
…
,
β
n
′
) if and only if there exists k
∈
{
1,2
,
…
,
n
} such that β
i
=
β
i
′
∀
i
>
k and β
k
<
β
k
′. We set B
≔
{
β
∈
A
∣
c
(
β
)
≠
0
}, where c
(
β
) comes from the right side of equality (4). According to the order A, we choose the least element η
=
(
η
1
,
…
,
η
n
)
∈
B. Obviously, c
(
η
)
≠
0. Put y
≔
∏
j
=
1
n
D
j
1
-
η
j. By (5) and (6), we obtain (7)
y
v
=
∏
j
=
1
n
D
j
1
-
η
j
Σ
β
∈
A
c
β
D
1
β
1
⋯
D
n
β
n
⊗
s
β
=
α
c
η
∏
j
=
1
n
D
j
⊗
s
η
∈
M
′
,where α
=
1 or -
1. Consequently, ∏
j
=
1
n
D
j
⊗
s
η
∈
M
′.
We will show that∏
j
=
1
n
D
j
⊗
S
(
λ
) is a u
(
W
0
)-module. By virtue of (3), we obtain [
x
k
D
l
,
D
i
]
=
-
δ
i
k
D
l
. Thus, we have (8)
x
k
D
l
D
i
=
D
i
x
k
D
l
-
δ
i
k
D
l
,in u
(
W
). We note that the elements x
k
D
l, ∀
1
≤
k
,
l
≤
n, are F-basis of W
0. Using (8), a straightforward computation shows that (9)
x
k
D
l
·
∏
j
=
1
n
D
j
⊗
S
λ
=
∏
j
=
1
k
-
1
D
j
-
D
l
+
D
k
x
k
D
l
D
k
+
1
⋯
D
n
⊗
S
λ
=
-
∏
j
=
1
k
-
1
D
j
D
l
∏
j
=
k
+
1
n
D
j
+
∏
j
=
1
n
D
j
x
k
D
l
⊗
S
λ
⊆
∏
j
=
1
n
D
j
⊗
S
λ
.
SinceS
(
λ
) is a simple u
(
W
0
)-module, we can get that ∏
j
=
1
n
D
j
⊗
S
(
λ
) is a simple u
(
W
0
)-module. Since ∏
j
=
1
n
D
j
⊗
s
η
∈
M
′
∩
∏
j
=
1
n
D
j
⊗
S
(
λ
), we can conclude that M
′
∩
∏
j
=
1
n
D
j
⊗
S
(
λ
) is nontrivial. ∏
j
=
1
n
D
j
⊗
S
(
λ
) must be contained in M
′. Thereby, there exists a maximal vector m of weight λ such that ∏
j
=
1
n
D
j
⊗
m
∈
M
′.
Noting thati
≠
1, this yields (
x
1
x
i
D
i
)
∏
j
=
1
n
D
j
⊗
m
∈
M
′. According to (3), we have [
x
1
x
i
D
i
,
D
j
]
=
δ
1
j
x
i
D
i
-
δ
i
j
x
1
D
i. Then, (10)
x
1
x
i
D
i
D
j
=
-
D
j
x
1
x
i
D
i
+
δ
1
j
x
i
D
i
-
δ
i
j
x
1
D
iholds in u
(
W
). By (10), we have (11)
x
1
x
i
D
i
∏
j
=
1
n
D
j
⊗
m
=
-
D
1
x
1
x
i
D
i
+
x
i
D
i
D
2
⋯
D
n
⊗
m
=
-
1
i
-
1
D
1
⋯
D
i
-
1
x
1
x
i
D
i
D
i
⋯
D
n
⊗
m
+
D
2
⋯
D
i
-
1
x
i
D
i
D
i
⋯
D
n
⊗
m
=
-
1
n
D
1
⋯
D
n
x
1
x
i
D
i
⊗
m
+
-
1
i
D
1
⋯
D
i
^
⋯
D
n
x
1
D
i
⊗
m
+
D
2
⋯
D
n
x
i
D
i
⊗
m
-
D
2
⋯
D
n
⊗
m
,where D
i
^ means that D
i is deleted. Since S
(
λ
) is a u
(
W
0
)-module and m is a maximal vector of weight λ, the first two terms vanish. Formula (11) implies that (
x
1
x
i
D
i
)
∏
j
=
1
n
D
j
⊗
m
=
(
λ
i
-
1
)
D
2
⋯
D
n
⊗
m. Since λ
i
≠
1, we have D
2
⋯
D
n
⊗
m
∈
M
′
.
IfD
2
⋯
D
n
⊗
m multiplied on the left by the elements x
2
x
i
D
i
,
x
3
x
i
D
i
,
…
,
x
i
-
1
x
i
D
i, successively, then it yields D
i
⋯
D
n
⊗
m
∈
M
′. By the hypothesis of the theorem, there exists 1
≤
j
<
i
≤
n such that λ
j
≠
0, and we thus have (12)
x
i
x
j
D
j
D
i
⋯
D
n
⊗
m
=
-
D
i
x
i
x
j
D
j
+
x
j
D
j
D
i
+
1
⋯
D
n
⊗
m
.Observe that the first term vanishes by the Z-graded degree of x
i
x
j
D
j which is 1, furthermore x
i
x
j
D
j
·
m
=
0, and the second term is equal to λ
j
D
i
+
1
⋯
D
n
⊗
m. Since λ
j
≠
0, we have D
i
+
1
⋯
D
n
⊗
m
∈
M
′. If we multiply D
i
+
1
⋯
D
n
⊗
m by the elements x
i
+
1
x
j
D
j
,
…
,
x
n
x
j
D
j, consecutively, then we see that 1
⊗
m
∈
M
′. Since u
(
W
0
)-module M
W
(
S
(
λ
)
) is generated by 1
⊗
m, 1
⊗
m
∈
M
′ indicates that M
W
(
S
(
λ
)
)
=
M
′. Hence, M
W
(
S
(
λ
)
) is simple.


## 4. Simple Modules of the Lie SuperalgebraS
(
0
∣
n
)

In the above section, we give the definition of the exterior algebraΛ
(
n
). We begin by describing S
=
S
(
n
)
≔
〈
{
D
i
j
(
a
)
∣
a
∈
Λ
(
n
)
,
i
,
j
=
1
,
…
,
n
}
〉, where (13)
D
i
j
a
=
D
i
a
D
j
+
D
j
a
D
i
.Putting S
i
=
W
i
∩
S, we have W
-
1
=
S
-
1.Lemma 5.
LetH
=
∑
i
=
1
n
-
1
F
D
i
+
1
,
i
(
x
i
+
1
x
i
) be a Cartan subalgebra of S
0. Then positive root vectors of S
0 are {
x
i
D
j
∣
1
≤
i
<
j
≤
n
}.Proof.
Recall the isomorphismW
0
≅
g
l
n
(
F
) via x
i
D
j
↦
E
i
j, described above. It induces the isomorphism ψ
:
S
0
≅
s
l
(
n
,
F
)
,
D
i
j
(
x
k
x
l
)
↦
δ
i
k
E
l
j
-
δ
i
l
E
k
j
+
δ
j
k
E
l
i
-
δ
j
l
E
k
i, where s
l
(
n
,
F
) is the special linear Lie algebra.
By calculation, we obtain that the Cartan subalgebraH
1 of s
l
(
n
,
F
) is equal to 〈
{
E
i
i
-
E
i
+
1
,
i
+
1
∣
i
=
1
,
…
,
n
-
1
}
〉. We define linear function Λ
j on the vector space 〈
{
E
11
,
…
,
E
n
n
}
〉 as before in Lemma 3. Then, the positive roots of s
l
(
n
,
F
) are {
Λ
i
-
Λ
j
}
i
<
j. Correspondingly, the positive root vectors of s
l
(
n
,
F
) are E
i
j
,
1
≤
i
<
j
≤
n. The isomorphism ψ sends -
(
1
/
2
)
D
j
j
(
x
i
x
j
)
=
x
i
D
j to E
i
j and D
i
+
1
,
i
(
x
i
+
1
x
i
)
=
x
i
D
i
-
x
i
+
1
D
i
+
1 to E
i
,
i
-
E
i
+
1
,
i
+
1, respectively. Therefore, the Cartan subalgebra of S
0 is 〈
{
x
i
D
i
-
x
i
+
1
D
i
+
1
∣
i
=
1
,
…
,
n
-
1
}
〉. The positive root vectors are {
x
i
D
j
∣
1
≤
i
<
j
≤
n
}.In terms of Definition1 and Lemma 5, the following facts hold: if H is a Cartan subalgebra of S
0, V is a u
(
S
0
)-module and λ
∈
H
∗. Then, V
λ
=
{
v
∈
V
∣
D
i
+
1
,
i
(
x
i
+
1
x
i
)
·
v
=
λ
i
v
,
1
≤
i
≤
n
-
1
}. A nonzero v
∈
V
λ is a maximal vector (of weight λ), provided x
i
D
j
·
v
=
0, whenever 1
≤
i
<
j
≤
n.Theorem 6.
If there existi
,
j, such that j
-
i
>
1 and λ
i
≠
0
,
λ
j
≠
0, then M
S
(
S
(
λ
)
) is simple.Proof.
LetM
′ be a nonzero submodule of M
S
(
S
(
λ
)
). Choose 0
≠
v contained in M
′. It owns the same form as in W; that is, v
=
Σ
β
∈
A
c
(
β
)
i
(
D
1
)
β
1
⋯
i
(
D
n
)
β
n
⊗
s
β, where β
=
(
β
1
,
…
,
β
n
) and c
(
β
)
∈
F. The same discussion described above about the Lie superalgebra W applies to the Lie superalgebra S. We can get a maximal vector m such that ∏
j
=
1
n
D
j
⊗
m
∈
M
′.
Using (3) and (13), we see that (14)
D
k
,
D
i
j
a
=
-
D
i
j
D
k
a
.Thus, D
i
j
(
a
)
D
k
=
(
-
1
)
d
(
a
)
[
D
k
D
i
j
(
a
)
+
D
i
j
(
D
k
(
a
)
)
] in u
(
S
). If i
≠
1, we have (15)
D
i
,
i
+
1
x
1
x
i
x
i
+
1
·
∏
j
=
1
n
D
j
⊗
m
=
-
D
1
·
D
i
,
i
+
1
x
1
x
i
x
i
+
1
·
D
2
⋯
D
n
⊗
m
-
D
i
,
i
+
1
x
i
x
i
+
1
·
D
2
⋯
D
n
⊗
m
=
-
1
i
-
1
D
1
⋯
D
i
-
1
D
i
,
i
+
1
x
1
x
i
x
i
+
1
D
i
⋯
D
n
⊗
m
-
D
2
⋯
D
i
-
1
D
i
,
i
+
1
x
i
x
i
+
1
D
i
⋯
D
n
⊗
m
=
-
1
i
D
1
⋯
D
i
D
i
,
i
+
1
x
1
x
i
x
i
+
1
D
i
+
1
⋯
D
n
⊗
m
+
-
1
i
+
1
D
1
⋯
D
i
-
1
D
i
,
i
+
1
x
1
x
i
+
1
D
i
+
1
⋯
D
n
⊗
m
-
D
2
⋯
D
i
D
i
,
i
+
1
x
i
x
i
+
1
D
i
+
1
⋯
D
n
⊗
m
-
D
2
⋯
D
i
-
1
D
i
,
i
+
1
x
i
+
1
D
i
+
1
⋯
D
n
⊗
m
=
-
1
i
+
1
D
1
⋯
D
i
+
1
D
i
,
i
+
1
x
1
x
i
x
i
+
1
D
i
+
2
⋯
D
n
⊗
m
+
-
1
i
+
1
D
1
⋯
D
i
D
i
,
i
+
1
x
1
x
i
D
i
+
2
⋯
D
n
⊗
m
+
-
1
i
+
2
D
1
⋯
D
i
-
1
D
i
,
i
+
1
x
1
D
i
+
2
⋯
D
n
⊗
m
-
D
2
⋯
D
i
+
1
D
i
,
i
+
1
x
i
x
i
+
1
D
i
+
2
⋯
D
n
⊗
m
+
-
1
i
+
1
D
1
⋯
D
i
^
D
i
+
1
D
i
,
i
+
1
x
1
x
i
+
1
D
i
+
2
⋯
D
n
⊗
m
+
D
2
⋯
D
i
D
i
,
i
+
1
x
i
D
i
+
2
⋯
D
n
⊗
m
-
D
2
⋯
D
n
⊗
m
.Since the Z-graded degree of x
1
x
i
x
i
+
1is 1, it implies that the first term vanishes. By the definition of a maximal vector m, it implies that the second and the forth terms vanish. It yields (16)
D
i
,
i
+
1
x
1
x
i
x
i
+
1
·
∏
i
=
1
n
D
i
⊗
m
=
-
D
2
⋯
D
n
D
i
,
i
+
1
x
i
x
i
+
1
⊗
m
=
λ
i
D
2
⋯
D
n
⊗
m
.Since λ
i
≠
0, we have D
2
⋯
D
n
⊗
m
∈
M
′.
MultiplyD
2
⋯
D
n
⊗
m by D
i
,
i
+
1
x
2
x
i
x
i
+
1
,
…
,
D
i
,
i
+
1
(
x
i
-
1
x
i
x
i
+
1
), in turn. By the same calculation as above, it yields D
i
⋯
D
n
⊗
m
∈
M
′, where 1
<
i
<
n.
Ifj
<
i
-
1, then we have (17)
D
j
,
j
+
1
x
i
x
j
x
j
+
1
D
i
⋯
D
n
⊗
m
=
-
1
n
-
i
+
1
D
i
D
i
+
1
⋯
D
n
D
j
,
j
+
1
x
i
x
j
x
j
+
1
⊗
m
-
D
i
+
1
⋯
D
n
D
j
,
j
+
1
x
j
x
j
+
1
⊗
m
.It also can be found that the first term vanishes and the second term can conclude D
i
+
1
⋯
D
n
⊗
m
∈
M
′ by the hypothesis of the Theorem λ
j
≠
0. Multiplying D
i
+
1
⋯
D
n
⊗
m on the left by D
j
,
j
+
1
x
i
+
1
x
j
x
j
+
1
,
…
,
D
j
,
j
+
1
(
x
n
x
j
x
j
+
1
), then 1
⊗
m
∈
M
′ holds. If j
>
i
+
1, then we have (18)
D
j
,
j
+
1
x
i
x
j
x
j
+
1
D
i
⋯
D
n
⊗
m
=
-
1
j
-
i
·
D
i
⋯
D
j
-
1
D
j
,
j
+
1
x
i
x
j
x
j
+
1
D
j
⋯
D
n
⊗
m
-
D
j
,
j
+
1
x
j
x
j
+
1
D
i
+
1
⋯
D
n
⊗
m
=
-
1
j
-
i
+
1
·
D
i
⋯
D
j
D
j
,
j
+
1
x
i
x
j
x
j
+
1
D
j
+
1
⋯
D
n
⊗
m
+
-
1
j
-
i
·
D
i
⋯
D
j
-
1
D
j
,
j
+
1
x
i
x
j
+
1
D
j
+
1
⋯
D
n
⊗
m
-
D
i
+
1
⋯
D
j
-
1
D
j
D
j
,
j
+
1
x
j
x
j
+
1
D
j
+
1
⋯
D
n
⊗
m
-
D
i
+
1
⋯
D
j
-
1
D
j
,
j
+
1
x
j
+
1
D
j
+
1
⋯
D
n
⊗
m
=
-
1
n
-
i
+
1
D
i
⋯
D
n
D
j
,
j
+
1
x
i
x
j
x
j
+
1
⊗
m
-
1
j
-
i
+
2
D
i
⋯
D
j
+
1
^
⋯
D
n
D
j
,
j
+
1
x
i
x
j
·
-
1
j
-
i
+
1
D
i
⋯
D
j
^
⋯
D
n
x
i
D
j
⊗
m
-
D
i
+
1
⋯
D
n
D
j
,
j
+
1
x
j
x
j
+
1
⊗
m
=
λ
j
D
i
+
1
⋯
D
n
⊗
m
.Obviously, we can obtain D
i
+
1
⋯
D
n
⊗
m
∈
M
′.
Moreover,D
j
,
j
+
1
(
x
i
+
1
x
j
x
j
+
1
)
D
i
+
1
⋯
D
n
⊗
m, for j
≠
i
+
1, implies that D
i
+
2
⋯
D
n
⊗
m
∈
M
′. And so on, multiply D
i
+
2
⋯
D
n
⊗
m
∈
M
′ on the left by D
i
,
i
+
1
x
i
+
2
x
i
x
i
+
1
,
…
,
D
i
,
i
+
1
(
x
n
x
i
x
i
+
1
), consecutively. Finally, it yields 1
⊗
m
∈
M
′.
Ifi
=
1, we have D
j
,
j
+
1
(
x
1
x
j
x
j
+
1
)
·
∏
i
=
1
n
D
i
⊗
m
=
λ
j
D
2
⋯
D
n
⊗
m
∈
M
′; furthermore, D
2
⋯
D
n
⊗
m
∈
M
′. Imitating the process of calculation, we have 1
⊗
m
∈
M
′. We get the conclusion.


## 5. Simple Modules of the Lie SuperalgebraK
(
n
)

Given a linear mappingD
k
~
:
Λ
(
n
)
→
W
(
n
) satisfies (19)
D
k
~
a
=
∑
i
=
1
n
-
1
a
i
D
i
+
a
n
x
n
D
n
,where a
∈
Λ
(
n
)
α, a
i
=
(
-
1
)
α
(
x
i
x
n
D
n
(
a
)
+
D
i
(
a
)
), and a
n
=
2
a
-
∑
i
=
1
n
-
1
x
i
D
i
(
a
). We can obtain that D
k
~
:
Λ
(
n
)
→
D
k
~
(
Λ
(
n
)
) is an isomorphism of linear spaces. By computation, we know that [
D
k
~
(
a
)
,
D
k
~
(
b
)
]
=
D
k
~
(
a
,
b
), where a
,
b
=
D
k
~
(
a
)
(
b
)
-
(
-
1
)
α
β
2
b
x
n
D
n
(
a
)
,
a
∈
Λ
(
n
)
α
,
b
∈
Λ
(
n
)
β
,
α, and β
∈
Z
2
. We define a bracket product in Λ
(
n
) by means of (20)
a
,
b
=
2
a
-
∑
i
=
1
n
-
1
x
i
D
i
a
x
n
D
n
b
-
-
1
α
β
2
b
-
∑
i
=
1
n
-
1
x
i
D
i
b
x
n
D
n
a
+
∑
i
=
1
n
-
1
-
1
α
D
i
a
D
i
b
,where a
∈
Λ
(
n
)
α, b
∈
Λ
(
n
)
β, α, and β
∈
Z
2. Pertaining to this bracket product, Λ
(
n
) becomes a Lie superalgebra which is denoted by K
(
n
) (see [27]).Then,K
(
n
)
=
∑
j
≥
-
2
K
j is a Z-graded Lie superalgebra, where K
j
≔
〈
{
x
u
∣
j
=
u
+
δ
(
u
,
n
)
-
2
}
〉, and (21)
δ
u
,
n
≔
0
,
n
∉
u
,
1
,
n
∈
u
.Put(22)
j
v
=
j
+
q
,
1
≤
j
≤
q
,
j
-
q
,
q
+
1
≤
j
≤
2
q
.Write K
≔
K
(
n
).Lemma 7.
K
(
n
) is a restricted Lie superalgebra.Proof.
SinceD
k
~ is an isomorphism of Λ
(
n
)
→
D
k
~
(
Λ
(
n
)
), we can regard K
(
n
)
=
{
D
k
~
(
a
)
∣
a
∈
Λ
(
n
)
}. Obviously, K
(
n
) is a subalgebra of W
(
0
∣
n
). For any a
∈
Λ
(
n
)
0
¯, then D
k
~
(
a
)
∈
K
0
¯
⊆
W
0
¯. Then, the Z-graded degree of a is an even number. If x
u
=
2, namely, x
u
=
x
k
x
l,  k
≠
l
≠
n, or x
u
=
x
k
x
n, k
≠
n. By direct calculation, (
D
k
~
(
x
k
x
l
)
)
p
=
c
D
k
~
(
x
k
x
l
), where c
=
1 or -
1. (
D
k
~
(
x
k
x
n
)
)
p
=
0. If x
u
=
2
t, t
∈
{
2,3
,
…
}, we have (
D
k
~
(
x
u
)
)
p
=
0. In a word, (
D
k
~
(
a
)
)
p
∈
K
0
¯, where a
∈
Λ
(
n
)
0
¯. Since K
(
n
) is a subalgebra of W
(
n
) and W
(
n
) is a restricted Lie superalgebra, we can obtain that K
(
n
) is a superalgebra.First we consider the case wheren
=
2
q
+
1 is an odd number.Lemma 8.
LetH
=
〈
{
μ
x
j
x
j
v
,
x
n
∣
μ
∈
F
,
j
=
1
,
…
,
q
}
〉 be a Cartan subalgebra of K
0. Then, positive root vectors of K
0 are {
(
1
/
2
)
x
j
x
i
+
(
μ
/
2
)
x
j
v
x
i
+
(
μ
/
2
)
x
i
v
x
j
+
(
1
/
2
)
x
j
v
x
i
v, x
j
x
i
+
μ
x
i
v
x
j
+
x
i
x
j
v
+
x
i
v
x
j
v
∣
1
≤
i
<
j
≤
q
}, where v
μ
2
=
-
1.Proof.
Suppose thatφ
:
K
0
→
L
=
{
A
∈
M
n
-
1
(
F
)
∣
A
t
+
A
=
0
}
⊕
F, given by (23)
x
i
x
j
⟼
E
j
i
-
E
i
j
,
i
≠
j
,
i
,
j
≠
n
,
x
n
⟼
1
∈
F
,is a mapping of vector spaces. It can be verified that φ is an isomorphism. Let P
=
I
q
(
1
/
2
)
I
q
-
μ
I
q
(
μ
/
2
)
I
q, where μ
2
=
-
1. Let M
≔
P
t
P
=
0
I
q
I
q
0. Set L
(
P
)
=
{
P
-
1
A
P
∣
A
∈
L
}. Then, A
∈
L
(
P
)
⇔
M
A
+
A
t
M
=
0; namely, A
∈
o
(
q
), the orthogonal algebra. We can conclude K
0
≅
L
⊕
F
≅
o
(
q
)
⊕
F.
By calculation, we obtain that the Cartan subalgebraH
1 of o
(
q
) is H
1
=
〈
{
E
j
j
-
E
j
v
j
v
∣
j
=
1
,
…
,
q
}
〉.
We will define linear functionΛ
j the same as before. Then, the positive roots of o
(
q
) are {
Λ
i
-
Λ
j
}
1
≤
i
<
j
≤
q and {
Λ
i
+
Λ
j
}
1
≤
i
<
j
≤
q. Correspondingly, the positive root vectors are E
i
j
-
E
j
v
i
v
,
E
i
j
v
-
E
j
i
v
,
1
≤
i
<
j
≤
q, respectively. By the isomorphism, we get the positive root vectors of L:(24)
P
E
i
j
-
E
j
v
i
v
P
-
1
=
1
2
E
i
j
-
E
j
i
+
μ
2
E
i
j
v
-
E
j
v
i
+
μ
2
E
j
i
v
-
E
i
v
j
+
1
2
E
i
v
j
v
-
E
j
v
i
v
,
P
E
i
j
v
-
E
j
i
v
P
-
1
=
E
i
j
-
E
j
i
+
μ
E
j
i
v
-
E
i
v
j
+
μ
E
j
v
i
-
E
i
j
v
+
E
j
v
i
v
-
E
i
v
j
v
,where 1
≤
i
<
j
≤
n
. So we can obtain that positive root vectors of K
0 are(25)
1
2
x
j
x
i
+
μ
2
x
j
v
x
i
+
μ
2
x
i
v
x
j
+
1
2
x
j
v
x
i
v
,
x
j
x
i
+
μ
x
i
v
x
j
+
x
i
x
j
v
+
x
i
v
x
j
v
,for 1
≤
i
<
j
≤
q
. The Cartan subalgebra of K
0 is 〈
{
μ
x
i
v
x
i
,
x
n
∣
1
≤
i
≤
q
}
〉.In view of Definition1 and Lemma 8, the following fact holds: suppose H is a Cartan subalgebra of K
0, V is a u
(
K
0
)-module, and λ
∈
H
∗. Then, V
λ
=
{
w
∈
V
∣
(
x
j
x
j
v
)
·
w
=
λ
j
w
,
x
n
·
w
=
λ
n
w
,
j
=
1
,
…
,
q
}. A nonzero w
∈
V
λ is a maximal vector (of weight λ), provided (
x
i
-
μ
x
i
v
)
x
j
·
w
=
0, whenever 1
≤
i
<
j
≤
q.Theorem 9.
Ifλ
n
≠
0, then M
K
(
S
(
λ
)
) is simple.Proof.
LetM
′ be a nonzero submodule of M
K
(
S
(
λ
)
). Take w
∈
M
′ and w
≠
0. We note (26)
w
=
Σ
β
∈
A
c
β
i
x
1
β
1
⋯
i
x
n
-
1
β
n
-
1
i
1
β
0
⊗
s
β
,where β
=
(
β
1
,
…
,
β
n
-
1
,
β
0
), c
(
β
)
∈
F, and A
≔
{
a
=
Σ
k
a
k
ε
k
∣
a
k
=
0, or 1 for 0
≤
k
≤
n
-
1,0
≤
a
n
≤
p
-
1
}
⊂
Z
n. Write i
(
x
j
)
=
x
j, i
(
1
)
=
x
0 in u
(
K
).
By formula (20), we have [
x
i
,
x
j
]
=
0, for i
≠
j and i, j
≠
n. Obviously, [
x
i
,
x
i
]
=
-
1, for i
≠
n. And [
1
,
x
i
]
=
0, for i
≠
n. Thus, (27)
x
i
x
j
=
-
x
j
x
i
,in u
(
K
), where i
≠
j and i
,
j
≠
n. In particular, (28)
2
x
i
2
=
-
x
0
,where i
≠
n. And (29)
x
0
x
i
=
x
i
x
0
,where i
≠
n.
Putα
0
=
min
⁡
{
t
0
∣
w
=
Σ
β
∈
A
c
(
β
)
x
1
t
1
⋯
x
n
-
1
t
n
-
1
x
0
t
0
⊗
s
β
,
c
(
β
)
≠
0
}. By (28), we can get x
0
p
-
1
-
α
0
·
w
=
Σ
β
′
∈
A
c
β
′
x
1
t
1
⋯
x
n
-
1
t
n
-
1
x
0
p
-
1
⊗
s
β
′
∈
M
′, where β
′
=
(
t
1
,
…
,
t
n
-
1
,
α
0
). Put α
1
=
min
⁡
{
t
1
∣
Σ
β
′
∈
A
c
(
β
′
)
x
1
t
1
⋯
x
n
-
1
t
n
-
1
x
0
p
-
1
⊗
s
β
′
,
c
(
β
′
)
≠
0
}
.
MultiplyingΣ
β
′
∈
A
c
(
β
′
)
x
1
t
1
⋯
x
n
-
1
t
n
-
1
x
0
p
-
1
⊗
s
β
′ by x
1
1
-
α
1 and then by (27) and (29), we can obtain Σ
β
′
′
∈
A
c
(
β
′
′
)
x
1
⋯
x
n
-
1
t
n
-
1
x
0
p
-
1
⊗
s
β
′
′
∈
M
′, where β
′
′
=
(
α
1
,
t
2
,
…
,
t
n
-
1
,
α
0
). And so on, we can conclude that there exists η such that c
(
η
)
x
1
x
2
⋯
x
n
-
1
x
0
p
-
1
⊗
s
η
∈
M
′, where η
=
(
α
1
,
α
2
,
…
,
α
n
-
1
,
α
0
), c
(
η
)
≠
0. Imitating the discussion as W, there exists a maximal vector m such that x
1
x
2
⋯
x
n
-
1
x
0
p
-
1
⊗
m
∈
M
′. By (20), we get (30)
x
i
,
x
u
x
n
=
x
i
x
u
x
n
,where x
i does not appear in x
u and x
i
≠
x
n. Also we have (31)
x
i
,
x
u
x
n
=
x
u
-
ε
i
x
n
,where x
i occurs in x
u. We will prove x
2
x
3
⋯
x
n
-
1
x
0
p
-
1
⊗
m
∈
M
′ in two steps.
First, by (30) and (31), we have (32)
x
1
x
n
x
1
x
2
⋯
x
n
-
1
x
0
p
-
1
⊗
m
=
x
1
x
1
x
n
-
x
n
x
2
⋯
x
n
-
1
x
0
p
-
1
⊗
m
=
x
1
x
2
x
1
x
n
x
3
⋯
x
n
-
1
x
0
p
-
1
⊗
m
+
x
1
x
1
x
2
x
n
x
3
⋯
x
n
-
1
x
0
p
-
1
⊗
m
+
x
2
x
n
x
3
⋯
x
n
-
1
x
0
p
-
1
⊗
m
-
x
2
x
n
x
3
⋯
x
n
-
1
x
0
p
-
1
⊗
m
.
Observing these terms, it remains the third nonzero term. With[
1
,
x
u
x
n
]
=
2
x
u
x
n, where x
n does not appear in x
u, then (33)
x
0
x
u
x
n
-
x
u
x
n
x
0
=
2
x
u
x
n
,in u
(
K
). Hence, by (30), (31), and (33), then (32) can be adjusted to (34)
x
2
x
3
⋯
x
n
-
1
x
n
x
0
p
-
1
⊗
m
∈
M
′
.
Secondly, for[
1
,
x
n
]
=
2
x
n, then x
n
x
0
=
(
x
0
-
2
)
x
n in u
(
K
). Using the induction hypothesis, we get (35)
x
n
x
0
p
-
1
=
x
0
-
2
p
-
1
x
n
.Hence, we get (36)
x
2
x
3
⋯
x
n
-
1
x
n
x
0
p
-
1
⊗
m
=
x
2
x
3
⋯
x
n
-
1
x
0
-
2
p
-
1
x
n
⊗
m
=
x
2
x
3
⋯
x
n
-
1
x
0
-
2
p
-
1
⊗
λ
n
m
∈
M
′
.
Sinceλ
n
≠
0, we have x
2
x
3
⋯
x
n
-
1
(
x
0
-
2
)
p
-
1
⊗
m
∈
M
′. Multiplying x
2
x
3
⋯
x
n
-
1
(
x
0
-
2
)
p
-
1
⊗
m on the left by x
0
p
-
1, thus we find that x
2
⋯
x
n
-
1
x
0
p
-
1
⊗
m
∈
M
′.
Imitate the way above onx
2
⋯
x
n
-
1
x
0
p
-
1
⊗
m. First, multiply x
2
⋯
x
n
-
1
x
0
p
-
1
⊗
m on the left with x
2
x
n. By computation, we can obtain that x
3
⋯
x
n
x
0
p
-
1
⊗
m
∈
M
′. Repeating the second step, we can get x
3
⋯
x
n
-
1
x
0
p
-
1
⊗
m
∈
M
′.
Repeating the process above, we can getx
n
-
1
x
0
p
-
1
⊗
m
∈
M
′. Multiplying x
n
-
1
x
0
p
-
1
⊗
m on the left by x
n
-
1
x
n, we then have (37)
x
n
-
1
x
n
x
n
-
1
x
0
p
-
1
⊗
m
=
x
n
-
1
x
n
-
1
x
n
x
0
p
-
1
+
x
n
x
0
p
-
1
⊗
m
=
x
n
x
0
p
-
1
⊗
m
=
x
0
-
2
p
-
1
x
n
⊗
m
∈
M
′
.For λ
n
≠
0, we know that (
x
0
-
2
)
p
-
1
⊗
m
∈
M
′. Following x
0
p
-
1
(
x
0
-
2
)
p
-
1
⊗
m
∈
M
′, it implies that x
0
p
-
1
⊗
m
∈
M
′. Thus, (38)
x
n
x
0
p
-
1
⊗
m
=
x
0
-
2
p
-
1
x
n
⊗
m
∈
M
′
.Moving and expanding these terms of (38), then we have the following identity:(39)
-
2
p
-
1
λ
n
⊗
m
=
x
n
x
0
p
-
1
⊗
m
-
x
0
p
-
1
+
⋯
+
-
p
x
0
-
2
p
-
2
⊗
m
.Multiplying (40)
-
2
p
-
1
λ
n
⊗
m
=
x
n
x
0
p
-
1
⊗
m
-
x
0
p
-
1
+
⋯
+
-
p
x
0
-
2
p
-
2
⊗
mon the left by x
0
p
-
2, we obtain that x
0
p
-
2
⊗
m
∈
M
′. Using the induction hypothesis, finally, we can get the desired formula; namely, 1
⊗
m
∈
M
′. Therefore, M
K
(
S
(
λ
)
) is simple.Ifn
=
2
q
+
2 is an even number, then we let P
1
=
P
1. Imitating the proof of Lemma 7 and Theorem 9, we can obtain the following theorem.Theorem 10.
Ifλ
n
≠
0, then M
K
(
S
(
λ
)
) is simple.



---

*Source: 250570-2015-07-09.xml*