# Natural Filtrations of Infinite-Dimensional Modular Contact Superalgebras

Authors: Qiang Mu
Journal: Journal of Applied Mathematics
 (2014)

---

## 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.


---

## Body

## 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 ad-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 [10–12], respectively.In this paper, we consider the infinite-dimensional modular contact superalgebraK
(
2
r
+
1
,
n
), which is analogous to the one in the nonmodular situation (see [13]). But since the principal ℤ-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 ad-nilpotent elements and subalgebras generated by certain ad-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 thatr, r
′, n, n
′ are positive integers. Then K
(
2
r
+
1
,
n
)
≅
K
(
2
r
′
+
1
,
n
′
) if and only if (
r
,
n
)
=
(
r
′
,
n
′
).Theorem 3.
Letϕ, ψ be automorphisms of K
(
2
r
+
1
,
n
). Then ϕ
=
ψ if and only if  ϕ
|
K
[
-
1
]
=
ψ
|
K
[
-
1
].The paper is organized as follows. In Section2, we recall the necessary definitions concerning the modular contact superalgebra K
(
2
r
+
1
,
n
). In Section 3, we study the ad-nilpotent elements of K
(
2
r
+
1
,
n
). In Section 4, we complete the proofs of Theorems 1–3.


## 2. Preliminaries

Throughout this paper,𝔽 denotes an algebraic closed field of characteristic p
>
2 and ℤ
2
=
{
0
¯
,
1
¯
} the ring of integers modulo 2. Let ℕ and ℕ
0 denote the sets of positive integers and nonnegative integers, respectively. Given m
∈
ℕ, m
>
1 and α
=
(
α
1
,
α
2
,
…
,
α
m
)
∈
ℕ
0
m, we put |
α
|
=
∑
i
=
1
m
α
i. Let 𝒪
(
m
) denote the divided power algebra over 𝔽 with basis {
x
(
α
)
∣
α
∈
ℕ
0
m
}. For ε
i
=
(
δ
i
1
,
δ
i
2
,
…
,
δ
i
m
), i
=
1,2
,
…
,
m, we abbreviate x
(
ε
i
) to x
i. Let Λ
(
n
) be the exterior superalgebra over 𝔽 in n variables x
m
+
1
,
x
m
+
2
,
…
,
x
s, where s
=
m
+
n. Denote by 𝒪
(
m
,
n
) the tensor product 𝒪
(
m
)
⊗
Λ
(
n
). The trivial ℤ
2-gradation of 𝒪
(
m
) and the natural ℤ
2-gradation of Λ
(
n
) induce a ℤ
2-gradation of 𝒪
(
m
,
n
) such that 𝒪
(
m
,
n
) is an associative superalgebra. For g
∈
𝒪
(
m
) and f
∈
Λ
(
n
), we abbreviate g
⊗
f to g
f. For α
,
β
∈
ℕ
0
m and i
,
j
=
m
+
1
,
m
+
2
,
…
,
s, the following formulas in 𝒪
(
m
,
n
) hold:
(1)
x
(
α
)
x
(
β
)
=
(
α
+
β
α
)
x
(
α
+
β
)
,
x
i
x
j
=
-
x
j
x
i
,
x
(
α
)
x
j
=
x
j
x
(
α
)
,
where
(2)
(
α
+
β
α
)
=
∏
i
=
1
m
(
α
i
+
β
i
α
i
)
.PutY
0
=
{
1,2
,
…
,
m
}, Y
1
=
{
m
+
1
,
m
+
2
,
…
,
s
}, and Y
=
Y
0
∪
Y
1. Let
(3)
𝔹
k
=
{
(
i
1
,
i
2
,
…
,
i
k
)
∣
m
+
1
≤
i
1
<
i
2
<
⋯
<
i
k
≤
s
}
,
𝔹
(
n
)
=
⋃
k
=
0
n
‍
𝔹
k
,
where 𝔹
0
=
⌀. Given u
=
(
i
1
,
i
2
,
…
,
i
k
)
∈
𝔹
k, set |
u
|
=
k, {
u
}
=
{
i
1
,
i
2
,
…
,
i
k
}, and x
u
=
x
i
1
x
i
2
⋯
x
i
k
(
|
⌀
|
=
0
,
x
⌀
=
1
). Then {
x
(
α
)
x
u
∣
α
∈
ℕ
0
m
,
u
∈
𝔹
(
n
)
} is an 𝔽-basis of the infinite-dimensional superalgebra 𝒪
(
m
,
n
).LetD
1
,
D
2
,
…
,
D
s be the linear transformations of 𝒪
(
m
,
n
) such that
(4)
D
i
(
x
(
α
)
x
u
)
=
{
x
(
α
-
ε
i
)
x
u
i
∈
Y
0
x
(
α
)
·
∂
x
u
∂
x
i
i
∈
Y
1
.
Then D
1
,
D
2
,
…
,
D
s are superderivations of the superalgebra 𝒪
(
m
,
n
). Let
(5)
W
(
m
,
n
)
=
{
∑
i
=
1
s
‍
a
i
D
i
∣
a
i
∈
𝒪
(
m
,
n
)
,
i
∈
Y
}
.
Then W
(
m
,
n
) is an infinite-dimensional Lie superalgebra contained in Der
(
𝒪
(
m
,
n
)
) (see [14]). If deg
⁡
(
x
) appears in some expression in this paper, we always regard x as a ℤ
2-homogeneous element and deg
⁡
(
x
) as the ℤ
2-degree of x. Then deg
⁡
(
D
i
)
=
μ
(
i
), where
(6)
μ
(
i
)
=
{
0
¯
i
∈
Y
0
1
¯
i
∈
Y
1
.
The following formula holds in W
(
m
,
n
) (see [14]):
(7)
[
a
D
i
,
b
D
j
]
=
a
D
i
(
b
)
D
j
-
(
-
1
)
deg
⁡
⁡
(
a
D
i
)
deg
⁡
⁡
(
b
D
j
)
b
D
j
(
a
)
D
i
,
where a
,
b
∈
𝒪
(
m
,
n
) and i
,
j
∈
Y.Hereafter letr be a positive integer and let m
=
2
r
+
1. Put J
=
Y
∖
{
m
} and J
0
=
Y
0
∖
{
m
}. For i
∈
J, define
(8)
i
′
=
{
i
+
r
1
≤
i
≤
r
i
-
r
r
<
i
≤
2
r
i
m
<
i
≤
s
,
σ
(
i
)
=
{
1
1
≤
i
≤
r
-
1
r
<
i
≤
2
r
1
i
∈
Y
1
.
Let D
K
:
𝒪
(
m
,
n
)
→
W
(
m
,
n
) be the linear mapping such that
(9)
D
K
(
f
)
=
∑
i
=
1
s
‍
f
i
D
i
,
where
(10)
f
i
=
(
-
1
)
μ
(
i
)
deg
⁡
⁡
(
f
)
(
x
i
D
m
(
f
)
+
σ
(
i
′
)
D
i
′
(
f
)
)
,
∀
i
∈
J
,
f
m
=
2
f
-
∑
i
∈
J
‍
x
i
D
i
(
f
)
.
Then
(11)
[
D
K
(
f
)
,
D
K
(
g
)
]
=
D
K
(
⟦
f
,
g
⟧
)
,
where ⟦
f
,
g
⟧
=
D
K
(
f
)
(
g
)
-
2
D
m
(
f
)
g (see [14]). It follows directly from (11) and the injectivity of D
K that ⟦
,
⟧ defines a Lie multiplication on 𝒪
(
m
,
n
). This Lie superalgebra, denoted by K
(
2
r
+
1
,
n
), is called the infinite-dimensional contact superalgebra. In the sequel, we simply write K for K
(
2
r
+
1
,
n
).The following formula holds inK (see [14]):
(12)
⟦
f
,
g
⟧
=
∑
i
∈
J
‍
σ
(
i
)
(
-
1
)
μ
(
i
)
deg
⁡
⁡
(
f
)
D
i
(
f
)
D
i
′
(
g
)
+
(
2
f
-
∑
i
∈
J
‍
x
i
D
i
(
f
)
)
D
m
(
g
)
-
(
-
1
)
deg
⁡
(
f
)
deg
⁡
(
g
)
(
2
g
-
∑
i
∈
J
‍
x
i
D
i
(
g
)
)
D
m
(
f
)
.
Then K
=
⨁
i
=
-
2
∞
K
[
i
] is a ℤ-graded Lie superalgebra, where
(13)
K
[
i
]
=
span
⁡
𝔽
{
x
(
α
)
x
u
∣
|
α
|
+
α
m
+
|
u
|
=
i
+
2
}
.
Let K
j
=
⨁
i
≥
j
K
[
i
] for j
≥
-
2. Then K
=
K
-
2
⊃
K
-
1
⊃
K
0
⊃
⋯ is referred to as the natural filtration of K.Lemma 4.
K
=
⨁
i
=
-
2
∞
K
[
i
] is transitively graded.Proof.
Assume the contrary, then there existsy
∈
K
[
ℓ
] such that ⟦
y
,
x
j
⟧
=
0 for all j
∈
J, where ℓ
≥
0. Suppose that the largest exponent of x
m among the nonzero summands in the expression of y is equal to t, and write
(14)
y
=
∑
α
,
u
,
α
m
=
t
‍
c
α
,
u
x
(
α
)
x
u
+
∑
β
,
v
,
β
m
<
t
‍
d
β
,
v
x
(
β
)
x
v
,
where c
α
,
u
,
d
β
,
v
∈
𝔽. Hence, for j
∈
J
0,
(15)
0
=
⟦
y
,
x
j
⟧
=
∑
α
,
u
,
α
m
=
t
‍
σ
(
j
′
)
c
α
,
u
x
(
α
-
ε
j
)
x
u
+
h
,
where each exponent of x
m of all nonzero summands in the expression of h is less than t. Then ∑
α
,
u
,
α
m
=
t
σ
(
j
′
)
c
α
,
u
x
(
α
-
ε
j
)
x
u
=
0. Note that all nonzero summands of ∑
α
,
u
,
α
m
=
t
σ
(
j
′
)
c
α
,
u
x
(
α
-
ε
j
)
x
u are 𝔽-linear independent. It follows that each x
(
α
-
ε
j
)
x
u is equal to 0. Hence the exponents of x
j in each x
(
α
)
x
u are equal to 0. Similarly, for j
∈
Y
1, we can prove that x
j does not appear in each x
(
α
)
x
u. Consequently, we see that all x
(
α
)
x
u are of the form x
(
t
ε
m
). If t
=
0, then y
∈
K
[
-
2
], contradicting ℓ
≥
0. Hence t
>
0, and we can write
(16)
y
=
c
t
ε
m
x
(
t
ε
m
)
+
∑
α
,
u
,
α
m
=
t
-
1
‍
c
α
,
u
x
(
α
)
x
u
+
∑
β
,
v
,
β
m
<
t
-
1
‍
d
β
,
v
x
(
β
)
x
v
,
where c
t
ε
m
≠
0. Note that, for j
∈
Y
1,
(17)
0
=
⟦
y
,
x
j
⟧
=
-
c
t
ε
m
x
(
(
t
-
1
)
ε
m
)
x
j
+
∑
α
,
u
,
α
m
=
t
-
1
‍
(
-
1
)
|
u
|
c
α
,
u
x
(
α
)
D
j
(
x
u
)
+
h
,
where each exponent of x
m of the nonzero summands in the expression of h is less than t
-
1. Therefore,
(18)
-
c
t
ε
m
x
(
(
t
-
1
)
ε
m
)
x
j
+
∑
α
,
u
,
α
m
=
t
-
1
‍
(
-
1
)
|
u
|
c
α
,
u
x
(
α
)
D
j
(
x
u
)
=
0
.
Since x
j appears in x
(
(
t
-
1
)
ε
m
)
x
j and does not appear in ∑
α
,
u
,
α
m
=
t
-
1
(
-
1
)
|
u
|
c
α
,
u
x
(
α
)
D
j
(
x
u
), we conclude that t
=
0, a contradiction.


## 3. ad-Nilpotent Elements

Recall thaty
∈
K is called ad-nilpotent if there exists t
∈
ℕ such that (
ad
y
)
t
(
K
)
=
0. For a subset R of K, let nil
(
R
) denote the set of ad-nilpotent elements in R, and let Nil
(
R
) denote the subalgebra of K generated by nil
(
R
).Lemma 5.
Suppose thaty
[
i
]
∈
K
[
i
] for i
≥
-
2. The following statements hold.(1)
Ify
=
∑
i
=
k
t
‍
y
[
i
]
∈
nil
(
K
), then y
[
k
]
∈
nil
(
K
).
(2)
Ify
=
∑
i
=
-
2
t
‍
y
[
i
]
∈
nil
(
K
), then y
[
-
2
]
=
0.
(3)
Ify
=
∑
i
=
-
2
t
‍
y
[
i
]
∈
nil
(
K
0
¯
), then y
[
-
1
]
=
0.
(4)
Ify
=
∑
i
=
-
2
t
‍
y
[
i
]
∈
nil
(
K
0
¯
), then y
[
0
]
∈
nil
(
K
0
¯
).
(5)
Ify
=
∑
i
=
-
2
t
‍
y
[
i
]
∈
nil
(
K
), then y
[
-
1
]
∈
span
⁡
𝔽
{
x
j
∣
j
∈
Y
1
}.Proof.
(
1
) See [15, Lemma 5.1].
(
2
) Suppose that y
[
-
2
]
≠
0. As y is ad-nilpotent, y
[
-
2
] is ad-nilpotent by (
1
). Note that
(19)
(
ad
1
)
k
(
x
(
k
ε
m
)
)
=
(
ad
1
)
k
-
1
(
2
x
(
(
k
-
1
)
ε
m
)
)
=
2
k
≠
0
,
for all k
>
0. This shows that y
[
-
2
] is not ad-nilpotent, a contradiction.
(
3
) By (
1
), we see that y
[
-
1
] is ad-nilpotent. Suppose that y
[
-
1
]
=
∑
i
∈
J
0
a
i
x
i
≠
0, where a
i
∈
𝔽. Then there exists some a
j
≠
0. A direct calculation shows that
(20)
(
ad
y
[
-
1
]
)
k
(
x
(
k
ε
j
′
)
)
=
(
ad
y
[
-
1
]
)
k
-
1
(
σ
(
j
)
a
j
x
(
(
k
-
1
)
ε
j
′
)
)
=
σ
(
j
)
k
a
j
k
≠
0
,
for all k
>
0. It follows that y
[
-
1
] is not ad-nilpotent.
(
4
) is an immediate consequence of (
2
), (
3
), and (
1
). (
5
) follows from (
2
), (
1
), and the proof of (
3
).Leta
∈
ℕ
0 and a
=
∑
l
=
0
∞
a
l
p
l be the p-adic expression of a, where 0
≤
a
l
<
p. Then,
(21)
pad
(
a
)
=
(
pad
0
(
a
)
,
pad
1
(
a
)
,
pad
2
(
a
)
,
…
)
is said to be the p-adic sequence of a, where pad
j
(
a
)
=
a
j for all j
∈
ℕ
0. For α
=
(
α
1
,
α
2
,
…
,
α
m
)
∈
ℕ
0
m, define the p-adic matrix of α to be
(22)
pad
(
α
)
=
(
pad
(
α
1
)
pad
(
α
2
)
⋮
pad
(
α
m
)
)
.
As pad
(
α
) is an m
×
∞ matrix with finitely many nonzero elements,
(23)
ht
(
α
)
=
max
⁡
⁡
{
j
∈
ℕ
0
∣
∃
i
∈
Y
0
:
pad
j
(
α
i
)
≠
0
}
is well defined. Let
(24)
∥
α
∥
b
,
c
=
∑
i
=
1
m
‍
∑
j
=
b
c
‍
pad
j
(
α
i
)
,
for b
∈
ℕ
0 and c
∈
ℕ. We abbreviate ∥
α
∥
0
,
q to ∥
α
∥
q.Suppose thaty
=
∑
α
,
u
c
α
,
u
x
(
α
)
x
u is a nonzero element of K, where c
α
,
u
∈
𝔽. Define
(25)
h
t
(
y
)
=
max
⁡
⁡
{
h
t
(
α
)
∣
c
α
,
u
≠
0
}
.Given thatq
>
0 and x
(
α
)
x
u
∈
K, we define
(26)
𝔉
q
(
x
(
α
)
x
u
)
=
∥
α
∥
q
+
2
∥
α
∥
1
,
q
+
|
u
|
+
pad
0
(
α
m
)
.Lemma 6.
Letα
,
β
∈
ℕ
0
m, i
∈
Y
0, and q
∈
ℕ. Then,(1)
x
(
α
)
x
(
β
)
≠
0 if and only if pad
(
α
)
+
pad
(
β
)
=
pad
(
α
+
β
);
(2)
Ifβ
i
≠
0, then ∥
β
-
ε
i
∥
q
+
2
∥
β
-
ε
i
∥
1
,
q
≥
∥
β
∥
q
+
2
∥
β
∥
1
,
q
-
1;
(3)
Ifx
(
β
)
x
v
∈
K
1 and q
≥
ht
(
x
(
β
)
x
v
), then 𝔉
q
(
x
(
β
)
x
v
)
≥
3.Proof.
(
1
) See [7, Lemma 2.5].
(
2
) First consider the case pad
0
(
β
i
)
≠
0. Then,
(27)
pad
(
β
i
-
1
)
=
(
pad
0
(
β
i
)
-
1
,
pad
1
(
β
i
)
,
…
)
.
It follows that ∥
β
-
ε
i
∥
q
=
∥
β
∥
q
-
1 and ∥
β
-
ε
i
∥
1
,
q
=
∥
β
∥
1
,
q, and thus (
2
) holds.
Next consider the casepad
0
(
β
i
)
=
0. We may assume that
(28)
pad
(
β
i
)
=
(
0
,
…
,
0
,
pad
t
(
β
i
)
,
pad
t
+
1
(
β
i
)
,
…
)
,
where pad
t
(
β
i
)
≠
0 and t
≥
1. Hence,
(29)
pad
(
β
i
-
1
)
=
(
p
-
1
,
…
,
p
-
1
,
pad
t
(
β
i
)
-
1
,
pad
t
+
1
(
β
i
)
,
…
)
.
If q
<
t, then
(30)
∥
β
-
ε
i
∥
q
+
2
∥
β
-
ε
i
∥
1
,
q
=
(
q
+
1
)
(
p
-
1
)
+
2
q
(
p
-
1
)
>
-
1
=
∥
β
∥
q
+
2
∥
β
∥
1
,
q
-
1
.
If q
=
t, noting that p
>
2, then
(31)
∥
β
-
ε
i
∥
q
+
2
∥
β
-
ε
i
∥
1
,
q
=
t
(
p
-
1
)
+
2
(
t
-
1
)
(
p
-
1
)
+
3
(
pad
t
(
β
i
)
-
1
)
≥
2
+
3
pad
t
(
β
i
)
-
3
=
3
pad
t
(
β
i
)
-
1
=
∥
β
∥
q
+
2
∥
β
∥
1
,
q
-
1
.
If q
>
t, then
(32)
∥
β
-
ε
i
∥
q
+
2
∥
β
-
ε
i
∥
1
,
q
=
∥
β
-
ε
i
∥
t
+
2
∥
β
-
ε
i
∥
1
,
t
+
3
∥
β
-
ε
i
∥
t
+
1
,
q
≥
∥
β
∥
t
+
2
∥
β
∥
1
,
t
-
1
+
3
∥
β
-
ε
i
∥
t
+
1
,
q
=
∥
β
∥
q
+
2
∥
β
∥
1
,
q
-
1
.
(
3
) The assumption x
(
β
)
x
v
∈
K
1 implies that |
β
|
+
β
m
+
|
v
|
≥
3. Then it is trivially verified that (
3
) holds.Lemma 7.
Suppose thatx
(
β
)
x
v
∈
K
1, x
(
α
)
x
u
∈
K, q
≥
max
⁡
{
1
,
ht
(
x
(
β
)
x
v
)
}, and i
∈
J. The following statements hold.(1)
Ifx
(
β
)
x
v
D
m
(
x
(
α
)
x
u
)
≠
0, then 𝔉
q
(
x
(
β
)
x
v
D
m
(
x
(
α
)
x
u
)
)
≥
𝔉
q
(
x
(
α
)
x
u
)
+
1.
(2)
Ifx
(
α
)
x
u
D
m
(
x
(
β
)
x
v
)
≠
0, then 𝔉
q
(
x
(
α
)
x
u
D
m
(
x
(
β
)
x
v
)
)
≥
𝔉
q
(
x
(
α
)
x
u
)
+
1.
(3)
IfD
i
(
x
(
β
)
x
v
)
D
i
′
(
x
(
α
)
x
u
)
≠
0, then 𝔉
q
(
D
i
(
x
(
β
)
x
v
)
D
i
′
(
x
(
α
)
x
u
)
)
≥
𝔉
q
(
x
(
α
)
x
u
)
+
1.Proof.
(
1
) The assumption x
(
β
)
x
v
D
m
(
x
(
α
)
x
u
)
≠
0 implies that x
(
β
)
x
(
α
-
ε
m
)
≠
0. By Lemma 6
(
1
), we have
(33)
pad
(
β
+
(
α
-
ε
m
)
)
=
pad
(
β
)
+
pad
(
α
-
ε
m
)
.
Consequently,
(34)
pad
0
(
β
m
+
(
α
m
-
1
)
)
=
pad
0
(
β
m
)
+
pad
0
(
α
m
-
1
)
,
(35)
∥
β
+
(
α
-
ε
m
)
∥
q
=
∥
β
∥
q
+
∥
α
-
ε
m
∥
q
,
(36)
∥
β
+
(
α
-
ε
m
)
∥
1
,
q
=
∥
β
∥
1
,
q
+
∥
α
-
ε
m
∥
1
,
q
.
By (
2
) and (
3
) of Lemma 6, we obtain
(37)
∥
α
-
ε
m
∥
q
+
∥
α
-
ε
m
∥
1
,
q
≥
∥
α
∥
q
+
∥
α
∥
1
,
q
-
1
,
(38)
𝔉
q
(
x
(
β
)
x
v
)
≥
3
.
Combining (34)–(38), we have
(39)
𝔉
q
(
x
(
β
)
x
v
D
m
(
x
(
α
)
x
u
)
)
=
𝔉
q
(
x
(
β
+
(
α
-
ε
m
)
)
x
v
x
u
)
=
∥
β
+
(
α
-
ε
m
)
∥
q
+
2
∥
β
+
(
α
-
ε
m
)
∥
1
,
q
+
|
u
|
+
|
v
|
+
pad
0
(
β
m
+
(
α
m
-
1
)
)
=
∥
β
∥
q
+
∥
α
-
ε
m
∥
q
+
2
∥
β
∥
1
,
q
+
2
∥
α
-
ε
m
∥
1
,
q
+
|
u
|
+
|
v
|
+
pad
0
(
β
m
)
+
pad
0
(
α
m
-
1
)
=
∥
β
∥
q
+
2
∥
β
∥
1
,
q
+
|
v
|
+
pad
0
(
β
m
)
+
∥
α
-
ε
m
∥
q
+
2
∥
α
-
ε
m
∥
1
,
q
+
|
u
|
+
pad
0
(
α
m
-
1
)
≥
𝔉
q
(
x
(
β
)
x
v
)
+
∥
α
∥
q
+
2
∥
α
∥
1
,
q
-
1
+
|
u
|
+
pad
0
(
α
m
)
-
1
≥
𝔉
q
(
x
(
α
)
x
u
)
+
1
.
(
2
) Suppose that x
(
α
)
x
u
D
m
(
x
(
β
)
x
v
)
≠
0. Then,
(40)
𝔉
q
(
x
(
α
)
x
u
D
m
(
x
(
β
)
x
v
)
)
=
𝔉
q
(
x
(
α
+
(
β
-
ε
m
)
)
x
v
x
u
)
=
∥
α
+
(
β
-
ε
m
)
∥
q
+
2
∥
α
+
(
β
-
ε
m
)
∥
1
,
q
+
|
u
|
+
|
v
|
+
pad
0
(
α
m
+
(
β
m
-
1
)
)
=
∥
α
∥
q
+
∥
β
-
ε
m
∥
q
+
2
∥
α
∥
1
,
q
+
2
∥
β
-
ε
m
∥
1
,
q
+
|
u
|
+
|
v
|
+
pad
0
(
α
m
)
+
pad
0
(
β
m
-
1
)
=
∥
α
∥
q
+
2
∥
α
∥
1
,
q
+
|
u
|
+
pad
0
(
α
m
)
+
∥
β
-
ε
m
∥
q
+
2
∥
β
-
ε
m
∥
1
,
q
+
|
v
|
+
pad
0
(
β
m
-
1
)
≥
𝔉
q
(
x
(
α
)
x
u
)
+
∥
β
∥
q
+
2
∥
β
∥
1
,
q
-
1
+
|
v
|
+
pad
0
(
β
m
)
-
1
≥
𝔉
q
(
x
(
α
)
x
u
)
+
1
.
(
3
) Similarly, we have
(41)
𝔉
q
(
D
i
(
x
(
β
)
x
v
)
D
i
′
(
x
(
α
)
x
u
)
)
=
𝔉
q
(
x
(
(
β
-
ε
i
)
+
(
α
-
ε
i
′
)
)
x
v
x
u
)
=
∥
(
β
-
ε
i
)
+
(
α
-
ε
i
′
)
∥
q
+
2
∥
(
β
-
ε
i
)
+
(
α
-
ε
i
′
)
∥
1
,
q
+
|
v
|
+
|
u
|
+
pad
0
(
β
m
+
α
m
)
=
∥
β
-
ε
i
∥
q
+
∥
α
-
ε
i
′
∥
q
+
2
∥
β
-
ε
i
∥
1
,
q
+
2
∥
α
-
ε
i
′
∥
1
,
q
+
|
v
|
+
|
u
|
+
pad
0
(
β
m
)
+
pad
0
(
α
m
)
=
∥
β
-
ε
i
∥
q
+
2
∥
β
-
ε
i
∥
1
,
q
+
∥
α
-
ε
i
′
∥
q
+
2
∥
α
-
ε
i
′
∥
1
,
q
+
|
v
|
+
|
u
|
+
pad
0
(
β
m
)
+
pad
0
(
α
m
)
≥
∥
β
∥
q
+
2
∥
β
∥
1
,
q
-
1
+
∥
α
∥
q
+
2
∥
α
∥
1
,
q
-
1
+
|
v
|
+
|
u
|
+
pad
0
(
β
m
)
+
pad
0
(
α
m
)
=
𝔉
q
(
x
(
β
)
x
v
)
+
𝔉
q
(
x
(
α
)
x
u
)
-
2
≥
𝔉
q
(
x
(
α
)
x
u
)
+
1
.Lemma 8.
Suppose thatx
(
β
)
x
v
∈
K
1, x
(
α
)
x
u
∈
K, and q
≥
max
⁡
{
1
,
ht
(
x
(
β
)
x
v
)
}. Let x
(
α
′
)
x
u
′ be a nonzero summand of  ⟦
x
(
β
)
x
v
,
x
(
α
)
x
u
⟧. Then 𝔉
q
(
x
(
α
′
)
x
u
′
)
≥
𝔉
q
(
x
(
α
)
x
u
)
+
1.Proof.
A direct calculation shows thatx
(
α
′
)
x
u
′ fulfills the conditions of Lemma 7.Given thatq
∈
ℕ, let
(42)
ℓ
q
=
m
(
q
+
1
)
(
p
-
1
)
+
2
m
q
(
p
-
1
)
+
(
p
-
1
)
+
n
+
1
.
Clearly, the inequality 𝔉
q
(
x
(
α
)
x
u
)
<
ℓ
q holds for all x
(
α
)
x
u
∈
K.Lemma 9.
K
1
⊂
nil
(
K
).Proof.
Suppose thaty
=
∑
β
,
u
‍
c
α
,
v
x
(
β
)
x
v is an arbitrary element of K
1, where c
β
,
v
∈
𝔽 and c
β
,
v
≠
0. Let q
∈
ℕ such that q
≥ ht(
y
). Let x
(
α
)
x
u be a standard basis element of K. By using Lemma 8 repeatedly, we see that (
ad
y
)
ℓ
q
(
x
(
α
)
x
u
)
=
0.Lemma 10.
Fori
,
j
∈
J
0, the following statements hold.(1)
x
(
2
ε
i
)
∈
nil
(
K
[
0
]
∩
K
0
¯
).
(2)
Ifσ
(
i
)
=
σ
(
j
) and i
≠
j, then x
i
x
j
∈
nil
(
K
[
0
]
∩
K
0
¯
).
(3)
Ifσ
(
i
)
≠
σ
(
j
) and i
≠
j
′, then x
i
x
j
∈
nil
(
K
[
0
]
∩
K
0
¯
).
(4)
x
i
x
i
′
∈
Nil
(
K
[
0
]
∩
K
0
¯
).Proof.
(
1
) Let x
(
α
)
x
u be a standard basis element of K. A direct calculation shows that
(43)
(
ad
x
(
2
ε
i
)
)
p
(
x
(
α
)
x
u
)
=
(
ad
x
(
2
ε
i
)
)
p
-
1
(
σ
(
i
)
x
i
x
(
α
-
ε
i
′
)
x
u
)
=
σ
(
i
)
p
x
i
p
x
(
α
-
p
ε
i
′
)
x
u
=
0
.
(
2
) Since ad
x
i
x
j
=
σ
(
i
)
(
x
j
D
i
′
+
x
i
D
j
′
) and x
j
D
i
′
∘
x
i
D
j
′
=
x
i
D
j
′
∘
x
j
D
i
′, it follows from the binomial theorem that (
ad
x
i
x
j
)
p
=
σ
(
i
)
p
(
x
j
D
i
′
+
x
i
D
j
′
)
p
=
σ
(
i
)
p
(
(
x
j
D
i
′
)
p
+
(
x
i
D
j
′
)
p
)
=
0.
(
3
) Since ad
x
i
x
j
=
σ
(
i
)
(
x
j
D
i
′
-
x
i
D
j
′
) and x
j
D
i
′
∘
x
i
D
j
′
=
x
i
D
j
′
∘
x
j
D
i
′, we have (
ad
x
i
x
j
)
p
=
σ
(
i
)
p
(
x
j
D
i
′
-
x
i
D
j
′
)
p
=
σ
(
i
)
p
(
(
x
j
D
i
′
)
p
-
(
x
i
D
j
′
)
p
)
=
0.
(
4
) It follows from (
1
) that x
i
x
i
′
=
σ
(
i
)
⟦
x
(
2
ε
i
)
,
x
(
2
ε
i
′
)
⟧
∈
Nil
(
K
[
0
]
∩
K
0
¯
).Lemma 11.
Suppose thatn
≥
3. The following statements hold.
(
1
) Let i
,
j
,
k be distinct elements of Y
1, and leta
,
b
∈
𝔽 be such that a
2
+
b
2
=
0. Then y
=
a
x
i
x
j
+
b
x
i
x
k
∈
nil
(
K
).
(
2
)
x
i
x
j
∈
Nil
(
K
[
0
]
∩
K
0
¯
) holds for all distinct i
,
j
∈
Y
1.Proof.
(
1
) A direct calculation shows that ad
y
=
a
x
j
D
i
-
a
x
i
D
j
+
b
x
k
D
i
-
b
x
i
D
k. For simplicity, we denote a
x
j
D
i
,
-
a
x
i
D
j
,
b
x
k
D
i
,
-
b
x
i
D
k by A
,
B
,
C
,
D, respectively. Clearly,
(44)
A
2
=
B
2
=
C
2
=
D
2
=
0
,
A
C
=
C
A
=
B
D
=
D
B
=
0
.
Then
(45)
(
ad
y
)
2
=
A
B
+
B
A
+
A
D
+
D
A
+
B
C
+
C
B
+
D
C
+
C
D
.
By A
D
A
=
D
A
D
=
B
C
B
=
C
B
C
=
0, we obtain
(46)
(
ad
y
)
3
=
A
B
A
+
A
B
C
+
A
D
C
+
B
A
B
+
B
A
D
+
B
C
D
+
C
B
A
+
C
D
A
+
C
D
C
+
D
A
B
+
D
C
B
+
D
C
D
.
Note that
(47)
A
B
A
=
-
a
2
A
,
B
A
B
=
-
a
2
B
,
C
D
C
=
-
b
2
C
,
D
C
D
=
-
b
2
D
,
a
2
A
=
C
D
A
+
A
D
C
,
a
2
B
=
B
C
D
+
D
C
B
,
b
2
C
=
A
B
C
+
C
B
A
,
b
2
D
=
B
A
D
+
D
A
B
.
It follows that
(48)
A
B
A
+
C
D
A
+
A
D
C
=
0
,
B
A
B
+
B
C
D
+
D
C
B
=
0
,
C
D
C
+
A
B
C
+
C
B
A
=
0
,
D
C
D
+
B
A
C
+
D
A
B
=
0
,
thus proving that (
ad
y
)
3
=
0.
(
2
) Let a
∈
𝔽 such that a
2
=
-
1. Since char
(
𝔽
)
>
2, we have a
2
-
1
=
-
2
≠
0. Let k
∈
Y
1
∖
{
i
,
j
}. Then (
1
) yields
(49)
y
1
=
a
x
i
x
j
+
x
i
x
k
∈
Nil
(
H
[
0
]
∩
H
0
¯
)
,
y
2
=
x
i
x
j
+
a
x
i
x
k
∈
Nil
(
H
[
0
]
∩
H
0
¯
)
.
Hence x
i
x
j
=
-
(
1
/
2
)
(
a
y
1
-
y
2
)
∈
Nil
(
H
[
0
]
∩
H
0
¯
).Lemma 12.
Leti
∈
J
0 and j
∈
Y
1. Then x
i
x
j
∈
nil
(
K
[
0
]
∩
K
1
¯
).Proof.
A direct calculation shows that(50)
ad
x
i
x
j
=
σ
(
i
)
x
j
D
i
′
-
x
i
D
j
,
(
ad
x
i
x
j
)
2
=
-
σ
(
i
)
(
x
i
D
j
∘
x
j
D
i
′
+
x
j
D
i
′
∘
x
i
D
j
)
.
Since
(51)
(
x
i
D
j
∘
x
j
D
i
′
)
∘
(
x
j
D
i
′
∘
x
i
D
j
)
=
0
=
(
x
j
D
i
′
∘
x
i
D
j
)
∘
(
x
i
D
j
∘
x
j
D
i
′
)
,
we have
(52)
(
ad
x
i
x
j
)
2
p
=
-
σ
(
i
)
(
(
x
i
D
j
∘
x
j
D
i
′
)
p
+
(
x
j
D
i
′
∘
x
i
D
j
)
p
)
=
0
.Lemma 13.
(
1
)
Nil
(
K
[
0
]
)
=
span
⁡
𝔽
{
x
i
x
j
∣
i
,
j
∈
J
}.
(
2
)
Nil
(
K
0
)
=
Nil
(
K
[
0
]
)
+
K
1.
(
3
) If n
≥
3, then Nil
(
K
[
0
]
∩
K
0
¯
)
=
span
⁡
𝔽
{
x
i
x
j
∣
i
,
j
∈
J
,
μ
(
i
)
=
μ
(
j
)
}.
(
4
) If n
≤
2, then Nil
(
K
[
0
]
∩
K
0
¯
)
=
span
⁡
𝔽
{
x
i
x
j
∣
i
,
j
∈
J
0
}.
(
5
)
Nil
(
K
0
∩
K
0
¯
)
=
Nil
(
K
[
0
]
∩
K
0
¯
)
+
K
1
∩
K
0
¯.Proof.
(
1
) Let y
=
a
m
x
m
+
∑
i
,
j
∈
J
‍
a
i
j
x
i
x
j be an arbitrary element of nil
(
K
[
0
]
). Suppose that a
m
≠
0. Since (
ad
y
)
t
(
1
)
=
(
-
2
a
m
)
t
≠
0 and ∀
t
∈
ℕ, it follows that y is not ad-nilpotent, a contradiction. Hence a
m
=
0, and therefore nil
(
K
[
0
]
)
⊆
span
⁡
𝔽
{
x
i
x
j
∣
i
,
j
∈
J
}. Noting that span
⁡
𝔽
{
x
i
x
j
∣
i
,
j
∈
J
} is a subalgebra of K, we obtain Nil
(
K
[
0
]
)
⊆
span
⁡
𝔽
{
x
i
x
j
∣
i
,
j
∈
J
}.
Conversely, Lemma10 shows that span
⁡
𝔽
{
x
i
x
j
∣
i
,
j
∈
J
0
}
⊆
Nil
(
K
[
0
]
), and Lemma 12 implies that span
⁡
𝔽
{
x
i
x
j
∣
i
∈
J
0
,
j
∈
Y
1
}
⊆
Nil
(
K
[
0
]
). Moreover, since x
i
x
j
=
⟦
x
1
x
i
,
x
1
′
x
j
⟧ for all i
,
j
∈
Y
1, we have span
⁡
𝔽
{
x
i
x
j
∣
i
,
j
∈
Y
1
}
⊆
Nil
(
K
[
0
]
). Therefore span
⁡
𝔽
{
x
i
x
j
∣
i
,
j
∈
J
}
⊆
Nil
(
K
[
0
]
).
(
2
) It is clear that Nil
(
K
[
0
]
)
⊆
Nil
(
K
0
), which combined with Lemma 9 yields Nil
(
K
[
0
]
)
+
K
1
⊆
Nil
(
K
0
).
On the other hand, suppose thaty
=
y
[
0
]
+
y
1 is an arbitrary element of nil
(
K
0
), where y
[
0
]
∈
K
[
0
] and y
1
∈
K
1. By Lemma 5, we have y
[
0
]
∈
Nil
(
K
[
0
]
), and hence y
=
y
[
0
]
+
y
1
∈
Nil
(
K
[
0
]
)
+
K
1. Since Nil
(
K
[
0
]
)
+
K
1 is a subalgebra of K, it follows that Nil
(
K
0
)
⊆
Nil
(
K
[
0
]
)
+
K
1.
(
3
) By (
1
), we see that Nil
(
K
[
0
]
∩
K
0
¯
)
⊆
Nil
(
K
[
0
]
)
∩
K
0
¯
=
span
⁡
𝔽
{
x
i
x
j
∣
i
,
j
∈
J, μ
(
i
)
=
μ
(
j
)
}. The reverse inclusion follows from Lemmas 10 and 11.
(
4
) Clearly the statement holds when n
=
1. Now we consider the case n
=
2. By (
1
), we can suppose that y
=
a
x
m
+
1
x
s
+
∑
i
,
j
∈
J
0
‍
a
i
j
x
i
x
j is an arbitrary element of nil
(
K
[
0
]
∩
K
0
¯
), where a
,
a
i
j
∈
𝔽. If a
≠
0, a direct calculation shows that (
ad
y
)
2
t
(
x
s
)
=
(
-
1
)
t
a
2
t
x
s
≠
0 for all t
∈
ℕ, contradicting that y is ad-nilpotent. Hence a
=
0 and y
∈
span
⁡
𝔽
{
x
i
x
j
∣
i
,
j
∈
J
0
}, proving nil
(
K
[
0
]
∩
K
0
¯
)
⊆
span
⁡
𝔽
{
x
i
x
j
∣
i
,
j
∈
J
0
}. Since span
⁡
𝔽
{
x
i
x
j
∣
i
,
j
∈
J
0
} is a subalgebra of K, it follows that Nil
(
K
[
0
]
∩
K
0
¯
)
⊆
span
⁡
𝔽
{
x
i
x
j
∣
i
,
j
∈
J
0
}. The reverse inclusion follows from Lemma 10.
(
5
) is completely analogous to the proof of (
2
).Letρ be the corresponding representation with respect to K
[
0
]-module K
[
-
1
]; that is, ρ
(
y
)
=
ad
y
|
K
[
-
1
], ∀
y
∈
K
[
0
]. It is easy to see that ρ is faithful. For y
∈
K
[
0
], we also denote by ρ
(
y
) the matrix of ρ
(
y
) relative to the fixed ordered 𝔽-basis as follows:
(53)
{
x
1
,
x
2
,
…
,
x
m
-
1
,
x
m
+
1
,
…
,
x
s
}
.
Denote by gl
(
2
r
,
n
) the general linear Lie superalgebra of (
2
r
+
n
)
×
(
2
r
+
n
) matrices over 𝔽. Let e
i
j denote the (
s
-
1
)
×
(
s
-
1
) matrix whose (
i
,
j
)-entry is 1 and 0 elsewhere, and G
=
(
0
I
r
-
I
r
0
), where I
r is r
×
r unit matrix. Let sp
(
2
r
,
𝔽
) be the Lie algebra consisting of all 2
r
×
2
r matrices A over 𝔽 satisfying A
T
G
+
G
A
=
0, where A
T is the transpose of A. Set 𝒦
=
ℒ
⊕
𝔽
I
s
-
1; here
(54)
ℒ
=
{
(
A
B
C
D
)
∈
gl
(
2
r
,
n
)
∣
A
∈
sp
(
2
r
,
𝔽
)
,
B
T
G
+
C
=
0
,
D
anti-symmetric
(
A
B
C
D
)
}
.Lemma 14.
(
1
)
ρ
(
K
[
0
]
)
=
𝒦.
(
2
) If y
∈
nil
(
K
[
0
]
), then ρ
(
y
) is a nilpotent matrix.Proof.
(
1
) For i
,
j
∈
J, a direct calculation shows that
(55)
⟦
x
i
x
j
,
x
k
⟧
=
σ
(
i
)
(
-
1
)
μ
(
i
)
(
μ
(
i
)
+
μ
(
j
)
)
δ
i
′
k
x
j
+
σ
(
j
)
(
-
1
)
μ
(
j
)
(
μ
(
i
)
+
μ
(
j
)
)
+
μ
(
i
)
μ
(
j
)
δ
j
′
k
x
i
.
Therefore
(56)
ρ
(
x
i
x
j
)
=
σ
(
i
)
(
-
1
)
μ
(
i
)
+
μ
(
i
)
μ
(
j
)
e
j
^
i
′
^
+
σ
(
j
)
(
-
1
)
μ
(
j
)
e
i
^
j
′
^
,
where
(57)
l
^
=
{
l
if
l
∈
J
0
,
l
-
1
if
l
∈
Y
1
.
Since ⟦
x
m
,
x
k
⟧
=
-
x
k, ∀
k
∈
J, we see that
(58)
ρ
(
x
m
)
=
-
I
s
-
1
.
From (56) and (58), we can easily verify (
1
).
(
2
) follows from the definition of ρ.IfA
=
(
a
i
j
) is an n
×
n antisymmetric matrix over 𝔽, we write Γ
(
A
) to denote ∑
1
≤
i
<
j
≤
n
a
i
j
2. As tr
⁡
(
A
2
)
=
-
2
∑
1
≤
i
<
j
≤
n
a
i
j
2, it is clear that if A is a nilpotent matrix then Γ
(
A
)
=
0.Lemma 15.
Suppose thatn
≥
3. Let A
=
(
a
i
j
) be an antisymmetric matrix over 𝔽 of order n. If A satisfies the following properties:(1)
Γ
(
A
)
=
0;
(2)
Γ
(
[
A
,
B
]
)
=
0 holds for every n
×
n antisymmetric matrix B.
Then,A
=
0.Proof.
LetN
=
{
1,2
,
…
,
n
}. Three cases arise as follows.
Case  1
(
n
>
4
). By the property (
2
) of A, we see that
(59)
Γ
(
[
A
,
E
i
j
-
E
j
i
+
E
k
l
-
E
l
k
]
)
-
Γ
(
[
A
,
E
i
j
-
E
j
i
]
)
-
Γ
(
[
A
,
E
k
l
-
E
l
k
]
)
=
0
holds for all i
,
j
,
k
,
l
∈
N. Therefore, if i
<
j
<
k
<
l, a direct calculation shows that the left-hand side of (59) is equal to 4
a
i
l
a
j
k
-
4
a
i
k
a
j
l. Hence 4
a
i
l
a
j
k
-
4
a
i
k
a
j
l
=
0. Similarly, if i
<
k
<
j
<
l, then -
4
a
i
l
a
k
j
-
4
a
i
k
a
j
l
=
0, and if i
<
k
<
l
<
j, then -
4
a
i
l
a
k
j
+
4
a
i
k
a
l
j
=
0. Thus, for all i
1
<
i
2
<
i
3
<
i
4, we see that a
i
1
i
3
a
i
2
i
4
=
a
i
1
i
4
a
i
2
i
3
=
-
a
i
1
i
2
a
i
3
i
4
=
-
a
i
1
i
3
a
i
2
i
4. Then
(60)
a
i
1
i
2
a
i
3
i
4
=
a
i
1
i
3
a
i
2
i
4
=
a
i
1
i
4
a
i
2
i
3
=
0
.
Denoteη
i
=
∑
t
=
1
n
‍
a
i
t
2 for i
∈
N. If i
≠
j, a direct calculation shows that
(61)
0
=
Γ
(
[
A
,
E
i
j
-
E
j
i
]
)
=
η
i
+
η
j
-
2
a
i
j
2
.
Let k
∈
N
∖
{
i
,
j
}. Then
(62)
η
i
+
η
k
-
2
a
i
k
2
=
0
,
η
j
+
η
k
-
2
a
j
k
2
=
0
.
From the equalities above, we see that η
i
=
a
i
j
2
+
a
i
k
2
-
a
j
k
2 holds for all distinct i
,
j
,
k
∈
N. Pick l
∈
N
∖
{
i
,
j
,
k
}. Then η
i
=
a
i
j
2
+
a
i
l
2
-
a
j
l
2. Hence a
i
l
2
+
a
j
k
2
=
a
i
k
2
+
a
j
l
2 holds for all distinct i
,
j
,
k
,
l
∈
N. It follows that
(63)
a
i
1
i
2
2
+
a
i
3
i
4
2
=
a
i
1
i
3
2
+
a
i
2
i
4
2
=
a
i
2
i
3
2
+
a
i
1
i
4
2
holds for all i
1
<
i
2
<
i
3
<
i
4.
Assume that there exists somea
i
j
≠
0. For distinct k
,
l
∈
N
∖
{
i
,
j
}, we have a
k
l
=
0 by (60). Then we can write the following:
(64)
A
=
∑
t
∈
N
‍
a
i
t
E
i
t
+
∑
t
∈
N
‍
a
j
t
E
j
t
-
∑
t
∈
N
‍
a
i
t
E
t
i
-
∑
t
∈
N
‍
a
j
t
E
t
j
.
A direct calculation shows that
(65)
0
=
Γ
(
[
A
,
E
k
l
-
E
l
k
]
)
=
a
i
l
2
+
a
j
l
2
+
a
i
k
2
+
a
j
k
2
.
Since a
i
l
2
+
a
j
k
2
=
a
j
l
2
+
a
i
k
2 by (63), we obtain a
i
l
2
+
a
j
k
2
=
0. Hence a
i
j
2
=
a
i
j
2
+
a
k
l
2
=
a
i
l
2
+
a
j
k
2
=
0 by (63), contradicting the assumption that a
i
j
≠
0.
Case  2
(
n
=
4
). Note that (60) and (63) hold; that is,
(66)
a
12
a
34
=
a
13
a
24
=
a
14
a
23
=
0
,
a
12
2
+
a
34
2
=
a
13
2
+
a
24
2
=
a
23
2
+
a
14
2
.
Moreover, since 0
=
Γ
(
A
)
-
Γ
(
[
A
,
E
12
-
E
21
]
)
=
a
12
2
+
a
34
2, we see that A
=
0.
Case  3
(
n
=
3
). Since 0
=
Γ
(
A
)
=
a
12
2
+
a
23
2
+
a
13
2 and Γ
(
[
A
,
E
12
-
E
21
]
)
=
a
23
2
+
a
13
2, it follows that a
12
=
0. Similarly a
23
=
a
13
=
0.Lemma 16.
Lety be a nonzero element of nil
(
K
[
0
]
∩
K
0
¯
). Then there exists z
∈
K
[
0
]
∩
K
0
¯ such that ⟦
y
,
z
⟧ is not a
d-nilpotent.Proof.
By Lemma13, we can suppose that
(67)
y
=
∑
l
∈
J
0
‍
a
l
x
(
2
ε
l
)
+
∑
l
,
t
∈
J
0
,
l
<
t
‍
b
l
t
x
l
x
t
+
∑
l
,
t
∈
Y
1
,
l
<
t
‍
c
l
t
x
l
x
t
,
wherea
l
,
b
l
t
,
c
l
t
∈
𝔽. Three cases arise as follows.
Case  1. There exists some a
i
≠
0. Let z
=
x
(
2
ε
i
′
). Then y
,
z
=
σ
(
i
)
a
i
x
i
x
i
′
+
h, where x
i does not appear in the expression of h. Noting that (
ad
⟦
y
,
z
⟧
)
ℓ
(
x
i
′
)
=
a
i
ℓ
x
i
′ for all ℓ
∈
ℕ, we conclude that ⟦
y
,
z
⟧ is not ad-nilpotent.
Case  2. All a
l
=
0, and there exists b
i
j
≠
0. Let z
=
x
i
′
x
j
′. A direct calculation shows that ⟦
y
,
z
⟧
=
σ
(
j
)
a
i
j
x
i
x
i
′
+
h, where x
i does not appear in the expression of h. As (
ad
⟦
y
,
z
⟧
)
ℓ
(
x
i
′
)
=
σ
(
i
)
ℓ
σ
(
j
)
ℓ
a
i
j
ℓ
x
i
′ and ∀
ℓ
∈
ℕ, we see that ⟦
y
,
z
⟧ is not ad-nilpotent.
Case  3. All a
l
=
0 and b
l
t
=
0. So y
∈
span
⁡
𝔽
{
x
i
x
j
∣
i
,
j
∈
Y
1
}. This can happen only if n
≥
3 by Lemma 13
(
4
). Note that ρ
(
y
) is an antisymmetric nilpotent matrix. Then Lemma 15 provides an element z of span
⁡
𝔽
{
x
i
x
j
∣
i
,
j
∈
Y
1
} such that ⟦
ρ
(
y
)
,
ρ
(
z
)
⟧ is not a nilpotent matrix. Hence ⟦
y
,
z
⟧ is not ad-nilpotent.


## 4. Proofs of Theorems

Proof of Theorem1.
We proceed in several steps.
(I)Nil
(
K
0
¯
)
=
Nil
(
K
0
∩
K
0
¯
). Suppose that y is an arbitrary element of nil
(
K
0
¯
). It follows from (
2
) and (
3
) of Lemma 5 that y
∈
K
0, which combined with y
∈
nil
(
K
0
¯
) yields y
∈
nil
(
K
0
∩
K
0
¯
), thus proving nil
(
K
0
¯
)
⊆
nil
(
K
0
∩
K
0
¯
). Hence Nil
(
K
0
¯
)
⊆
Nil
(
K
0
∩
K
0
¯
).
The reverse inclusion is clear.
(II)K
0
∩
K
0
¯
=
Nor
K
0
¯
(
Nil
(
K
0
¯
)
). We first prove K
0
∩
K
0
¯
⊆
Nor
K
0
¯
(
Nil
(
K
0
¯
)
). It follows from (I) that
(68)
Nor
K
0
¯
(
Nil
(
K
0
¯
)
)
=
Nor
K
0
¯
(
Nil
(
K
0
∩
K
0
¯
)
)
.
Therefore
(69)
Nil
(
K
0
∩
K
0
¯
)
⊆
Nor
K
0
¯
(
Nil
(
K
0
∩
K
0
¯
)
)
=
Nor
K
0
¯
(
Nil
(
K
0
¯
)
)
.
By formula (12), we see that ⟦
x
m
,
K
[
0
]
⟧
=
0 and ⟦
x
m
,
K
1
∩
K
0
¯
⟧
⊆
K
1
∩
K
0
¯, proving
(70)
x
m
∈
Nor
K
0
¯
(
Nil
(
K
0
¯
)
)
.
In the case of n
≥
3, by (
3
) and (
5
) of Lemma 13, we obtain K
0
∩
K
0
¯
=
Nil
(
K
0
∩
K
0
¯
)
+
𝔽
x
m. By (69) and (70), we have K
0
∩
K
0
¯
⊆
Nor
K
0
¯
(
Nil
(
K
0
¯
)
). In the case of n
=
2, by (
4
) and (
5
) of Lemma 13, we have K
0
∩
K
0
¯
=
Nil
(
K
0
∩
K
0
¯
)
+
𝔽
x
m
+
𝔽
x
m
+
1
x
s. Note that x
m
+
1
x
s
∈
Nor
K
0
¯
(
Nil
(
K
0
¯
)
), which combined with (69) and (70) yields K
0
∩
K
0
¯
⊆
Nor
K
0
¯
(
Nil
(
K
0
¯
)
). In the case of n
=
1, by (
4
) and (
5
) of Lemma 13, we have K
0
∩
K
0
¯
=
Nil
(
K
0
∩
K
0
¯
)
+
𝔽
x
m. Hence K
0
∩
K
0
¯
⊆
Nor
K
0
¯
(
Nil
(
K
0
¯
)
) by (69) and (70).
Conversely, suppose thaty
=
y
[
-
2
]
+
y
-
1
∈
Nor
K
0
¯
(
Nil
(
K
0
¯
)
), where y
[
-
2
]
∈
K
[
-
2
], y
-
1
∈
K
-
1. If y
[
-
2
]
≠
0, then ⟦
y
,
x
1
x
m
⟧
=
2
y
[
-
2
]
x
1
+
h
∉
Nil
(
K
0
¯
) by Lemma 13
(
5
), contradicting y
∈
Nor
K
0
¯
(
Nil
(
K
0
¯
)
), where h
∈
K
0. Hence y
[
-
2
]
=
0 and we can write y
=
y
[
-
1
]
+
y
0, where y
[
-
1
]
=
∑
i
∈
J
0
‍
a
i
x
i
∈
K
[
-
1
], a
i
∈
𝔽, y
0
∈
K
0. If there is some a
j
≠
0, then ⟦
y
,
x
(
2
ε
j
′
)
⟧
=
σ
(
j
)
a
j
x
j
′
+
h, where h
∈
K
0. Lemma 13
(
5
) shows that ⟦
y
,
x
(
2
ε
j
′
)
⟧
∉
Nil
(
K
0
¯
), contradicting y
∈
Nor
K
0
¯
(
Nil
(
K
0
¯
)
). Hence y
[
-
1
]
=
0 and y
∈
K
0
∩
K
0
¯, proving Nor
K
0
¯
(
Nil
(
K
0
¯
)
)
⊆
K
0
∩
K
0
¯.
(III) LetM
=
{
y
∈
nil
(
K
0
¯
)
∣
⟦
y
,
K
0
∩
K
0
¯
⟧
⊆
nil
(
K
0
¯
)
}. Then M
=
K
1
∩
K
0
¯. Suppose that y
=
y
[
0
]
+
y
1 is an arbitrary element of M, where y
[
0
]
∈
K
[
0
]
∩
K
0
¯ and y
1
∈
K
1
∩
K
0
¯. If y
[
0
]
≠
0, since y
[
0
]
∈
nil
(
K
[
0
]
) by Lemma 5
(
4
), then Lemma 16 provides an element z
∈
K
[
0
]
∩
K
0
¯ such that ⟦
y
[
0
]
,
z
⟧ is not ad-nilpotent. Hence ⟦
y
,
z
⟧ is not ad-nilpotent by Lemma 5
(
4
), contradicting y
∈
M. Therefore y
[
0
]
=
0 and y
∈
K
1
∩
K
0
¯, thus proving M
⊆
K
1
∩
K
0
¯.
On the other hand, since⟦
K
1
∩
K
0
¯
,
K
0
∩
K
0
¯
⟧
⊆
K
1
∩
K
0
¯
⊆
nil
(
K
0
∩
K
0
¯
) by Lemma 9, it follows that K
1
∩
K
0
¯
⊆
M.
(IV) LetQ
=
{
y
∈
K
1
¯
∣
⟦
y
,
K
1
¯
⟧
⊆
K
0
∩
K
0
¯
}. Then Q
=
K
1
∩
K
1
¯. Suppose that y
=
y
[
-
1
]
+
y
0
∈
Q, where y
[
-
1
]
=
∑
i
∈
Y
1
‍
a
i
x
i
∈
K
[
-
1
]
∩
K
1
¯ and y
0
∈
K
0
∩
K
1
¯. If there is some a
j
≠
0, then ⟦
y
,
x
j
⟧
=
-
a
j
+
⟦
y
0
,
x
j
⟧
∉
K
-
1
∩
K
0
¯, and hence ⟦
y
,
x
j
⟧
∉
K
0
∩
K
0
¯, contradicting y
∈
Q. Therefore y
[
-
1
]
=
0, and we can write y
=
y
[
0
]
+
y
1, where y
[
0
]
=
∑
i
∈
J
0
,
j
∈
Y
1
‍
a
i
j
x
i
x
j
∈
K
[
0
]
∩
K
1
¯ and y
1
∈
K
1
∩
K
1
¯. If there exists some a
l
t
≠
0, then ⟦
y
,
x
t
⟧
=
-
∑
i
∈
J
0
‍
a
i
t
x
i
+
⟦
y
1
,
x
t
⟧
∉
K
0
∩
K
0
¯, a contradiction which yields y
[
0
]
=
0 and y
∈
K
1
∩
K
1
¯.
The reverse inclusion follows from the fact thatK
1
¯
⊂
K
-
1.
(V)⟦
K
1
¯
,
K
1
∩
K
0
¯
⟧
=
K
0
∩
K
1
¯. It suffices to show that K
0
∩
K
1
¯
⊆
⟦
K
1
¯
,
K
1
∩
K
0
¯
⟧. Suppose that x
(
α
)
x
u is an arbitrary basis element of K
0
∩
K
1
¯ with x
u
=
x
i
x
v. Note that x
(
α
+
ε
m
)
x
v
∈
K
1
∩
K
0
¯. Since x
(
α
)
x
u
=
⟦
x
i
,
x
(
α
+
ε
m
)
x
v
⟧, it follows that x
(
α
)
x
u
∈
⟦
K
1
¯
,
K
1
∩
K
0
¯
⟧, as desired.
It follows from (II) and (V) thatK
0
=
K
0
∩
K
0
¯
+
K
0
∩
K
1
¯ is invariant under automorphisms of K. By (III) and (IV), we obtain that K
1
=
K
1
∩
K
0
¯
+
K
1
∩
K
1
¯ is invariant. Therefore K
-
1
=
{
x
∈
K
∣
⟦
x
,
K
1
⟧
⊆
K
0
} is invariant. By the transitivity of K, we conclude that
(71)
K
i
+
1
=
{
x
∈
K
i
∣
⟦
x
,
K
-
1
⟧
⊆
K
i
}
,
∀
i
≥
0
.
Hence the natural filtration of K is invariant under automorphisms of K.Proof of Theorem2.
Letφ
:
K
(
2
r
+
1
,
n
)
→
K
(
2
r
′
+
1
,
n
′
) be an isomorphism of Lie superalgebras. Let K and K
′ denote K
(
2
r
+
1
,
n
) and K
(
2
r
′
+
1
,
n
′
), respectively. Since φ
(
K
0
¯
)
=
K
0
¯
′ and φ
(
Nil
(
K
0
¯
)
)
=
Nil
(
K
0
¯
′
), it follows that φ
(
Nor
K
0
¯
(
Nil
(
K
0
¯
)
)
)
=
Nor
K
0
¯
′
(
Nil
(
K
0
¯
′
)
). By (II) in the proof of Theorem 1, we have
(72)
φ
(
K
0
∩
K
0
¯
)
=
φ
(
Nor
K
0
¯
(
Nil
(
K
0
¯
)
)
)
=
Nor
K
0
¯
′
(
Nil
(
K
′
0
¯
)
)
=
K
0
′
∩
K
0
¯
′
.
Consequently
(73)
φ
(
{
y
∈
nil
(
K
0
¯
)
∣
⟦
y
,
K
0
∩
K
0
¯
⟧
⊆
nil
(
K
0
¯
)
}
)
=
{
y
∈
nil
(
K
0
¯
′
)
∣
⟦
y
,
K
0
′
∩
K
0
¯
′
⟧
⊆
nil
(
K
0
¯
′
)
}
.
Applying (III) in the proof of Theorem 1, we see that φ
(
K
1
∩
K
0
¯
)
=
K
1
′
∩
K
0
¯
′, which combined with (V) yields
(74)
φ
(
K
0
∩
K
1
¯
)
=
φ
(
⟦
K
1
¯
,
K
1
∩
K
0
¯
⟧
)
=
⟦
K
1
¯
′
,
K
1
′
∩
K
0
¯
′
⟧
=
K
0
′
∩
K
1
¯
′
.
It follows from (72) and (74) that φ
(
K
0
)
=
K
0
′. Therefore φ induces an isomorphism of ℤ
2-graded spaces φ
~
:
K
/
K
0
→
K
′
/
K
0
′. A comparison of dimensions shows that r
=
r
′ and n
=
n
′. The converse implication is clear.Proof of Theorem3.
It suffices to prove thatϕ
|
K
[
-
1
]
=
ψ
|
K
[
-
1
] implies that ϕ
=
ψ. Since φ
(
1
)
=
φ
(
⟦
x
1
,
x
1
′
⟧
)
=
ψ
(
⟦
x
1
,
x
1
′
⟧
)
=
ψ
(
1
), it follows that φ
|
K
[
-
2
]
=
ψ
|
K
[
-
2
]. We use induction on ℓ to show that
(75)
φ
|
K
[
ℓ
]
=
ψ
|
K
[
ℓ
]
,
∀
ℓ
≥
-
1
.
Assume that ℓ
≥
0 and (75) holds for ℓ
-
1. Suppose that y
∈
K
[
ℓ
] and let z
=
φ
(
y
)
-
ψ
(
y
). We want to prove z
=
0. The induction hypothesis yields that φ
(
x
i
)
=
ψ
(
x
i
) and φ
(
⟦
y
,
x
i
⟧
)
=
ψ
(
⟦
y
,
x
i
⟧
), i
∈
J. Therefore,
(76)
⟦
z
,
φ
(
x
i
)
⟧
=
⟦
φ
(
y
)
-
ψ
(
y
)
,
φ
(
x
i
)
⟧
=
φ
(
⟦
y
,
x
i
⟧
)
-
ψ
(
⟦
y
,
x
i
⟧
)
=
0
.
Since φ
(
K
0
)
=
K
0 and φ
(
K
-
1
)
=
K
-
1 by Theorem 1, φ induces an automorphism φ
~ of the ℤ
2-graded space K
-
1
/
K
0. Consequently there exists a homogeneous basis {
h
1
,
…
,
h
m
-
1
,
h
m
+
1
,
…
,
h
s
} of K
[
-
1
] such that (
x
i
)
≡
h
i
(
mod
⁡
K
0
). Thus there exist g
i
∈
K
0 such that φ
(
x
i
)
=
h
i
-
g
i, i
∈
J. Therefore, (76) shows that ⟦
z
,
h
i
⟧
=
⟦
z
,
g
i
⟧ for all i
∈
J. As z
=
φ
(
y
)
-
ψ
(
y
)
∈
K
0 by Theorem 1, it can be decomposed into z
=
∑
j
=
0
t
z
[
j
], where z
[
j
]
∈
K
[
j
]. Noting that ⟦
z
[
0
]
,
h
i
⟧
∈
K
[
-
1
] and ⟦
z
,
g
i
⟧
∈
K
0, we obtain ⟦
z
[
0
]
,
h
i
⟧
=
0 for all i
∈
J, and hence z
[
0
]
=
0 since K is transitively graded. By induction, we conclude that z
[
j
]
=
0, j
=
0,1
,
…
,
t. Hence, z
=
0, as desired.



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*Source: 601847-2014-04-24.xml*