The finite-dimensional modular Lie superalgebra Ω

Journal of Algebra 321 (2009) 3601–3619

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Journal of Algebra www.elsevier.com/locate/jalgebra

The finite-dimensional modular Lie superalgebra Ω ✩ Yongzheng Zhang ∗ , Qingcheng Zhang Department of Mathematics, Northeast Normal University, Changchun 130024, China

a r t i c l e

i n f o

Article history: Received 6 February 2007 Available online 2 April 2009 Communicated by Vera Serganova Keywords: Modular Lie superalgebra Grassmann superalgebra Derivation superalgebra

a b s t r a c t The finite-dimensional modular Lie superalgebra Ω is constructed. The simplicity of Ω is proved. Its derivation superalgebra is determined. Then it is obtained that Ω is not isomorphic to any known Z -graded modular Lie superalgebra of Cartan type. © 2009 Elsevier Inc. All rights reserved.

1. Introduction As is well known, the research pertaining to Lie superalgebras over a field of characteristic zero has been a remarkable progress (see [1–4]). But the research on modular Lie superalgebras, i.e., Lie superalgebras over a field of prime characteristic, just began in recent years. In [5] modular Lie superalgebras are investigated in order to describe the simple Lie algebras over a field of characteristic two. In [10] the restricted enveloping algebras of modular Lie superalgebras are discussed. In [6,7,13–16] modular Lie superalgebras W , S , H , HO and K of Cartan-type are studied. We know that the derivation algebras of modular Lie algebras are important subjects in the research of modular Lie algebras. The derivation algebras of modular Lie algebras of Cartan-type are investigated in [1,10]. In [11] the derivation algebras of Lie algebras of generalized H -type are determined. Similarly, the derivation superalgebras of modular Lie superalgebras are also very interesting subjects. Recently, in [6,9,13,14] the derivation superalgebras of modular Lie superalgebras W , S , H , HO and K of Cartan-type are investigated. This paper is arranged as follows. In Section 2 we construct the finite-dimensional modular Lie superalgebra Ω and prove its simplicity. In Section 3, first we describe the Z -gradation and Z 2 gradation of the derivation superalgebra of Ω . Then we prove some indispensable lemmas and give the desired generator set of Ω which shall be applied to prove the main results. Finally we give ✩

*

This work was supported by the National Natural Science Foundation of China (10871057, 10701019). Corresponding author. E-mail addresses: [email protected] (Y. Zhang), [email protected] (Q. Zhang).

0021-8693/$ – see front matter © 2009 Elsevier Inc. All rights reserved. doi:10.1016/j.jalgebra.2009.01.038

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Y. Zhang, Q. Zhang / Journal of Algebra 321 (2009) 3601–3619

the derivation superalgebra of Ω and point out that Ω is not isomorphic to any known modular Lie superalgebras of Cartan-type. 2. Construction and simplicity In this paper F always denotes a field of characteristic p > 3 and F is not equal to its prime field Π . It is easily seen that there is a finite subset E in F such that E is linearly independent over Π and the additive subgroup H generated by E does not contain 1. Let N be the set of positive integers, and N 0 the set of non-negative integers. Given n ∈ N, r = 2n + 2. Suppose that μ1 , . . . , μr ∈ F such that μ1 = 0, μr = 1, μ j + μn+ j = 1, j = 2, . . . , n + 1. Let G be the additive subgroup of F generated by {μ j | j = 1, . . . , r } and {h0 = 1, h1 , . . . , hd } be a basis of G over Π . Set M = {1, . . . , r − 1}. Suppose that | E | = m and si ∈ N 0 , i = 1, . . . , r. We define a truncated polynomial algebra

¯ = F [x10 , x11 , . . . , x1s1 , . . . , xr0 , xr1 , . . . , xrsr , y 1 , . . . , ym ] A such that p

sr = d ;

xr j = 1,

j = 0, 1, . . . , d;

p

xi j = 0,

d

p

∀i ∈ M , j = 0, 1, . . . , si ; d

y i = 1,

i = 1, . . . , m.

Let kr = i =0 kri h i ∈ G, where 0  kri < p. Put xr r = i =0 xriri . For i ∈ M, we let πi = p si +1 − 1. If ki is an integer such that 0  ki  πi , then ki can be uniquely expressed in p-adic form: k

ki =

si 

k

ε v (ki ) p v , where 0  ε v (ki ) < p .

v =0 k

We set xi i =

si

ε v (ki )

v =0 xi v k

k

s . For 0  ki , ki  πi and ki = vi=0 ε v (ki ) p v , it is easy to see that ki +ki

xi i xi i = xi

= 0

 

ε v (ki ) + ε v ki < p , v = 0, 1, . . . , si .

⇔

(1) ¯

Let Q¯ = {(k1 , . . . , kr ) | 0  ki  πi , i ∈ M ; kr ∈ G }. If k¯ = (k1 , . . . , kr ) ∈ Q¯ , then let xk = x11 · · · xr r .

m

k

λ1

k

λ

Since | E | = m, we can suppose E = { z1 , . . . , zm }. If λ = i =1 λi zi ∈ H , then let y λ = y 1 · · · ymm . Given q ∈ N \ {1}. Let Λ(q) be the Grassmann superalgebra over F in q variables ξr +1 , . . . , ξr +q . ¯ := A¯ ⊗ Λ(q). Let Z 2 := {0¯ , 1¯ } denote the ring of integers modulo 2. Denote the tensor product by Ω Obviously, Ω¯ is an associative superalgebra with a Z 2 -gradation induced by the trivial Z 2 -gradation ¯ and the natural Z 2 -gradation of Λ(q): of A

Ω¯ 0¯ = A¯ ⊗ F Λ(q)0¯ ,

Ω¯ 1¯ = A¯ ⊗ F Λ(q)1¯ .

¯ g ∈ Λ(q), then we simply write f ⊗ g as f g. For k ∈ {1, . . . , q}, we set If f ∈ A, 



B k = i 1 , i 2 , . . . , i k  r + 1  i 1 < i 2 < · · · < i k  r + q and B (q) =

q k=0



B k , where B 0 = φ . If u = i 1 , . . . , ik ∈ B k , we let |u | = k, {u } = {i 1 , . . . , ik } and ξ u = ¯

ξi 1 · · · ξik . Put |φ| = 0 and ξ φ = 1. Then {ξ u | u ∈ B (q)} is an F -basis of Λ(q). So {xk y λ ξ u | k¯ ∈ Q¯ , λ ∈ H , ¯. u ∈ B (q)} is an F -basis of Ω ¯ such that D r (xk¯ y λ ξ u ) = kr xk¯ −¯er y λ ξ u . If kr ∈ G , Put e¯ i = (δi1 , . . . , δir ), i = 1, . . . , r . Let D r ∈ End(Ω) kr k l k l k l e¯r 0 φ ¯ clearly xr = x y ξ ∈ Ω . For kr , lr ∈ G , it is easy to see that D r (xr r xrr ) = D r (xr r )xrr + xr r D r (xrr ). So D r is an even derivation of Ω¯ . ¯ such that D i (xk¯ y λ ξ u ) = k∗ xk¯ −¯ei y λ ξ u , where k∗ is the first nonIf i ∈ M , then we let D i ∈ End(Ω) i i zero number of ε0 (ki ), ε1 (ki ), . . . , εsi (ki ).

Y. Zhang, Q. Zhang / Journal of Algebra 321 (2009) 3601–3619

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Lemma 2.1. Let 0  k, l  πi , i ∈ M. Then



k

l





k



l

k

 l

D i xi i xii = D i xi i xii + xi i D i xii .

(2)

Proof. Set t = min{ v | ε v (ki ) = 0}, s = min{ v | ε v (li ) = 0}. (i) If there is a u ∈ {0, 1, . . . , si } such that εu (ki ) + εu (li ) > p, then εu (ki − 1) + εu (li )  p, εu (ki ) + εu (li − 1)  p . From (1) we know that both sides of (2) are zero. (ii) If there is a u ∈ {0, 1, . . . , si } such that εu (ki ) + εu (li ) = p, the left side of (2) is zero by (1) and u  max{t , s}. If u > max{t , s}, then

εu (ki − 1) = εu (ki ),

εu (li − 1) = εu (li ).

So

εu (ki − 1) + εu (li ) = p = εu (ki ) + εu (li − 1). Thus the right side of (2) is zero. If u = t, u > s, then t > s. Therefore εs (ki − 1) = p − 1. So

εs (ki ) = 0. It follows that

εs (ki − 1) + εs (li )  ( p − 1) + εs (li )  p . Clearly

εt (ki ) + εt (li − 1) = εu (ki ) + εu (li ) = p . k −1 l

k

l −1

By virtue of (1), xi i xii = 0 = xi i xii . Then the right side of (2) is zero. Similarly, we can treat the case for u = s, u > t. If u = t = s, then ki ∗ + li ∗ = εu (ki ) + εu (li ) = p . Thus k −1 li xi

k i ∗ xi i

l −1

+ li ∗ xi i xii k

  k +l −1 = ki ∗ + li ∗ xi i i = 0.

Hence (2) holds. (iii) Suppose that ε v (ki )+ ε v (li ) < p, where v = 0, 1, . . . , si . If t < s, then (ki + li )∗ = ki ∗ and εt (li − 1)  p . So xki i xlii −1 = 0. It follows that k +li −1

(ki + li )∗ xi i

k +li −1

= k i ∗ xi i

k −1 li xi

= k i ∗ xi i

l −1

+ li ∗ xi i xii k

εt (ki )+

.

Then (2) holds. If t = s, it is easy to see that k +li −1

(ki + li )∗ xi i Then (2) also holds.

k −1 li xi

= k i ∗ xi i

l −1

+ li ∗ xi i xii k

.

2

If L is a superalgebra, then h( L ) denotes the set of all Z 2 -homogeneous elements of L, i.e., h( L ) = L 0¯ ∪ L 1¯ . If |x| occurs in some expression in this paper, we always regard x as a Z 2 -homogeneous element and |x| as the Z 2 -degree of x.

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Y. Zhang, Q. Zhang / Journal of Algebra 321 (2009) 3601–3619

¯ for any i ∈ M. Set s = r + q By virtue of Lemma 2.1, we know that D i is an even derivation of Ω ∂ ¯ , i.e., D i is a derivation , i ∈ T , then D i is an odd derivation of Ω and T = {r + 1, . . . , s}. Let D i = ∂ξ i

¯ Define ˜i = 0, ¯ if 1  i  r, and ˜i = 1, ¯ if i ∈ T . of Ω¯ whose Z 2 -degree is 1. Let ⎧ ⎨ i + n if 2  i  n + 1, i  = i − n if n + 2  i  r , ⎩ i if r + 1  i  s, ⎧ if 2  i  n + 1, ⎨1 [i ] = −1 if n + 2  i  r , ⎩ 1 if r + 1  i  s.

Set 1 = r, r  = 1, [1] = 1 and [r ] = −1. Put S = {1, 2, . . . , s}. Clearly S = {i  | i ∈ S }, ˜i = i˜ and ˜ [i  ](−1)i = −[i ], ∀i ∈ S . Then we can define a bilinear operation [ , ] in Ω¯ such that

[ f , g] =

s 

˜

¯ g ∈ Ω. ¯ f ∈ h(Ω),

[i ](−1)i | f | D i ( f ) D i  ( g ),

(3)

i =1

¯ is a Lie superalgebra. Theorem 2.2. For the operation [ , ] defined in (3), Ω ¯ β and Proof. We only prove that the operation [ , ] satisfies graded Jacobi identity. Let f ∈ Ω¯ α , g ∈ Ω ¯ γ , where α , β, γ ∈ Z 2 . Then h∈Ω   (−1)αγ f , [ g , h] = (−1)αγ

 f,

s 

 ˜i β

[i ](−1) D i ( g ) D i  (h)

i =1

= (−1)αγ

s 

  ˜ ˜ [i ][ j ](−1) j α +i β D j ( f ) D j  D i ( g ) D i  (h)

i , j =1

=

s 

˜

˜

[i ][ j ](−1)αγ + j α +i β D j ( f ) D j  D i ( g ) D i  (h)

i , j =1

+

s 

˜

˜

˜ ˜

[i ][ j ](−1)αγ + j α +i β+ j (i +β) D j ( f ) D i ( g ) D j  D i  (h)

i , j =1

=

s 

˜

˜

[i  ][ j ](−1)αγ + j α +i β D j ( f ) D j  D i  ( g ) D i (h)

i , j =1

+

s 

˜

˜

˜ ˜

[ j ][i ](−1)αγ +i α + j β+i ( j +β) D i ( f ) D j ( g ) D i  D j  (h)

i , j =1

=

s 

˜

˜ ˜

˜

˜ ˜

[i  ][ j ](−1)γ β+ j α +i α +i +i γ + j i D i (h) D j ( f ) D i  D j  ( g )

i , j =1

+

s  i , j =1

˜

˜

˜ ˜j +i˜ β

[i ][ j ](−1)αγ +i α + j β+i

D i ( f ) D j ( g ) D i  D j  (h).

Y. Zhang, Q. Zhang / Journal of Algebra 321 (2009) 3601–3619

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Similarly, s    ˜ ˜ ˜ ˜ ˜ ˜ (−1)β α g , [h, f ] = [i  ][ j ](−1)αγ + j β+i β+i +i α + j i D i ( f ) D j ( g ) D i  D j  (h) i , j =1 s 

+

˜

˜ j˜ +i˜ γ

˜

[i ][ j ](−1)α β+i β+ j γ +i

D i ( g ) D j (h) D i  D j  ( f ),

i , j =1 s    ˜ ˜ ˜ ˜ ˜ ˜ (−1)β γ h, [ f , g ] = [i  ][ j ](−1)β α + j γ +i γ +i +i β+ j i D i ( g ) D j (h) D i  D j  ( f ) i , j =1 s 

+

˜

˜ ˜j +i˜ α

˜

[i ][ j ](−1)β γ +i γ + j α +i

D i (h) D j ( f ) D i  D j  ( g ).

i , j =1

˜ Utilizing above three equalities and ˜i = i˜ , [i  ](−1)i = −[i ], i = 1, 2, . . . , s, we obtain that

      (−1)αγ f , [ g , h] + (−1)β α g , [h, f ] + (−1)β γ h, [ f , g ] = 0.

2

Set M 1 = M \ {1}. By virtue of the formula (3) we have



¯



¯

xk y λ ξ u , xl y η ξ v =

s   ¯   ¯  ˜ u [i ](−1)i |ξ | D i xk y λ ξ u D i  xl y η ξ v i =1

¯

¯

¯

¯

= k∗1l1 xk−¯e1 xl−¯e1 y λ+η ξ u ξ v − k1 l∗1 xk−¯e1 xl−¯e1 y λ+η ξ u ξ v  ¯ ¯ + [i ]k∗i l∗i  xk−¯ei xl−¯ei y λ+η ξ u ξ v i∈M1

     u ¯ ¯ + (−1)|ξ | xk xl y λ+η D i ξ u D i ξ v .

(4)

i∈T

¯ such that Let D be the linear transformation of Ω 





¯

D xk y λ ξ u = k1 +





¯

k j μ j + λ + 2−1 |u | − 1 xk y λ ξ u .

(5)

j∈M1

By the definition of D and Eq. (4), noticing that



¯

¯



 



¯



¯



μ j + μ j = 1, j ∈ M, we get that 

¯

¯

D xk y λ ξ u , xl y η ξ v = D xk y λ ξ u , xl y η ξ v + xk y λ ξ u , D xl y η ξ v



= k 1 + l 1 +





   ¯   ¯ (k j + l j )μ j + λ + η + 2−1 |u | + | v | − 2 xk y λ ξ u , xl y η ξ v .

j∈M1

¯ . Hence V := { f ∈ Ω¯ | D ( f ) = 0} is a subalgebra of Ω¯ . If Then D is an even derivation of Ω 



¯

ak¯ λu xk y λ ξ u ∈ V , where 0 = ak¯ λu ∈ F , then D ( ¯

¯

¯

ak¯ λu xk y λ ξ u ) = 0. So



¯

¯

ak¯ λu D (xk y λ ξ u ) = 0. Since

ak¯ λu = 0, D (xk y λ ξ u ) = 0. It follows that V = xk y λ ξ u | D (xk y λ ξ u ) = 0 . Thus



¯

 

V = xk y λ ξ u  k1 = 1 −

 j∈M1



k j μ j − λ − 2−1 |u | .

(6)

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Y. Zhang, Q. Zhang / Journal of Algebra 321 (2009) 3601–3619

˜ be the F -algebra obtained from A¯ by omitting variables xr0 , xr1 , . . . , xrd . Then Ω˜ := A˜ ⊗ ∧(q) is Let A ¯. a subalgebra of associative superalgebra Ω Put Q = {k = (k1 , . . . , kr −1 ) | 0  ki  πi , ∀i ∈ M }. Obviously, {xk y λ ξ u | k ∈ Q , λ ∈ G , u ∈ B (q)} ˜ . If k¯ = (k1 , . . . , kr ) ∈ Q¯ , then we set that δ(k¯ ) := (k1 , . . . , kr −1 ). Let σ : V → Ω˜ is an F -basis of Ω ¯ xδ(l) y η ξ v . Hence

shows that such that

¯

¯

¯

¯

¯

σ (xk y λ ξ u ) = xδ(k) y λ ξ u . If σ (xk y λ ξ u ) = σ (xl y η ξ v ), then xδ(k) y λ ξ u = δ(k¯ ) = δ(¯l), λ = η and u = v . Eq. (6) implies that k1 = l1 . Therefore k¯ = ¯l. This

be a linear mapping such that

σ is an isomorphism mapping of linear spaces. We define a bilinear operation [ , ] in Ω˜ 

¯



¯



¯



¯

xδ(k) y λ ξ u , xδ(l) y η ξ v := σ xk y λ ξ u , xl y η ξ v .

(7)

˜ is a Lie superalgebra which is isomorphic to V . If k¯ ∈ Q¯ , then we simply write δ(k¯ ) as k. Set Then Ω e i = (δi1 , . . . , δir −1 ), ∀i ∈ M 1 . By virtue of (7) and (4), it follows that 





¯

¯

xk y λ ξ u , xl y η ξ v = σ xk y λ ξ u , xl y η ξ v



   l j μ j − η − 2−1 | v | xk−e1 xl y λ+η ξ u ξ v = k∗1 1 − j∈M1



+







− l∗1 1 −

k j μ j − λ − 2−1 |u | xk xl−e1 y λ+η ξ u ξ v

j∈M1

[i ]k∗i l∗i  xk−ei xl−ei y λ+η ξ u ξ v

i∈M1

     u + (−1)|ξ | xk yl y λ+η D i ξ u D i ξ v .

(8)

i∈T

Let I be the identity mapping. Set

∂¯ = I −



μ j x j0

j∈M1

∂ ∂ x j0

−

m 

zj yj

j =1

 ∂ ∂ − 2−1 ξj . ∂yj ∂ξ j j∈T

Eq. (8) yields that







 









xk y λ ξ u , xl y η ξ v = D 1 xk y λ ξ u ∂¯ xl y η ξ v − ∂¯ xk y λ ξ u D 1 xl y η ξ v

+





    ˜ u [i ](−1)i |ξ | D i xk y λ ξ u D i  xl y η ξ v .

i∈M1 ∪T

˜ , g ∈ Ω˜ , we have Consequently, for f ∈ h(Ω) ¯ g ) − ∂( ¯ f )D1(g) + [ f , g ] = D 1 ( f )∂(



˜

[i ](−1)i | f | D i ( f ) D i  ( g ).

i∈M1 ∪T

Let xi = x1i = xi0 , ∀i ∈ M. Set

π = (π1 , . . . , πr −1 ), ω = r + 1, . . . , s ∈ B (q).

Lemma 2.3. If λ ∈ H , 2n + 4 − q ≡ 0 (mod p ), then λ + 2−1 q − n − 2 = 0.

(9)

Y. Zhang, Q. Zhang / Journal of Algebra 321 (2009) 3601–3619

3607

Proof. Assume that λ + 2−1 q − n − 2 = 0. Then 2λ = 2n + 4 − q. Since 2n + 4 − q ≡ 0 (mod p ), 2n + 4 − q is a non-zero element in Π . So 2λ is a non-zero element in Π . Thus there is t ∈ {1, . . . , p − 1} such that t (2λ) = 1 ∈ Π ⊆ F . Because H is an additive group and λ ∈ H , t (2λ) = (2t )λ ∈ H . So 1 ∈ H . It / H. 2 contradicts to 1 ∈ If 2n + 4 − q ≡ 0 (mod p ), then we write Ω˜ as Ω . In the following, without particular statement we always assume that 2n + 4 − q ≡ 0 (mod p ). Let R = M ∪ T .



Lemma 2.4. Let f ∈ Ω . If D i ( f ) = 0, ∀i ∈ R, then f =

λ λ∈ H aλ y , where aλ

∈ F.

Proof. If D i (xk y λ ξ u ) = 0, ∀i ∈ R, then k = (0, . . . , 0), u = φ . Thus xk y λ ξ u = y λ . Hence the lemma holds. 2 Theorem 2.5. Ω is a finite-dimensional simple Lie superalgebra. Proof. Let I be a non-zero ideal of Ω , 0 = f ∈ Ω . Let f = xt1 f 0 + xt1−1 f 1 + · · · + f t , where f 0 =  0, D 1 ( f j ) = 0, j = 0, 1, . . . , t. Utilizing the formula (9), we have

  (ad 1)t ( f ) = (ad 1)t −1 [1, f ] = (ad 1)t −1 − D 1 ( f ) = · · · = (−1)t D t1 ( f )  t  t  j   j t t = (−1) D 1 x1 f 0 = (−1)t D t1 x1 f 0 . j =1

j =1

j

If j < t, it is easy to check that D t1 (x1 ) = 0 and D t1

 t

x1 =

D t1−1



 ∗ t −1

t x1

∗

∗

= t (t − 1)



D 1 xt1−2



 t

= · · · = (−1)

t 

 j

∗

.

j =1

t

Hence (ad 1)t ( f ) = (−1)t ( j =1 j ∗ ) f 0 . So f 0 ∈ I . Let f 0 = xli g 0 + xli−1 g 1 + · · · + gl , where i ∈ M 1 , g 0 = 0 and D i ( g j ) = 0, j = 0, 1, . . . , l. Because

 l

l

(ad xi  ) ( f 0 ) = [i ]

l 

 j

∗

g0 ,

j =1

we have g 0 ∈ I . Thus we can suppose that g 0 ∈ I and D i ( g 0 ) = 0, ∀i ∈ M. If D i ( g 0 ) = 0, where i ∈ T , then we can suppose that g 0 = ξi h0 + h1 , where i ∈ T , h0 = 0, D i (h0 ) = D i (h1 ) = 0. Then h0 = −[ξi , g 0 ] ∈ I . Thus we can suppose that D i (h0 ) = 0, ∀i ∈ R. By virtue of Lemma 2.4, h0 = λ∈ H aλ y λ . If h0 contains at least two non-zero terms, then we can suppose that



h 0 = aη y η + aμ y μ +

aλ y λ ,

λ∈ H \{η,μ}

where aη = 0, aμ = 0. Let h0 : = [x1 , h0 ] − (1 − η)h0

= (η − μ)aμ y μ +



(η − λ)aλ y λ .

λ∈ H \{η,μ}

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Y. Zhang, Q. Zhang / Journal of Algebra 321 (2009) 3601–3619 

Obviously h0 is an element of I with one term less than h0 . Utilizing this procedure continu/ H , 1 − λ = 0. Then 1 = ously, we can suppose that aλ y λ ∈ I , aλ = 0. Then y λ ∈ I . Since 1 ∈ (1 − λ)−1 [x1 y −λ , y λ ] ∈ I . If k1 < π1 , then xk y λ ξ u = (k1 + 1)∗ [xk+e1 y λ ξ u , 1] ∈ I . In particular xi ∈ I and ξ j ∈ I , ∀i ∈ M, ∀ j ∈ T . If there is i ∈ M 1 such that ki < πi , then

  [i ](ki + 1)∗ xk y λ ξ u = xk+ei y λ ξ u , xi  − (1 − μi  )k∗1 xk+ei −1 y λ ξ u ∈ I . Therefore xk y λ ξ u ∈ I . If there is i ∈ T \ {u }, then xk y λ ξ u = −[xk y λ ξ u ξi , ξi ] ∈ Y . By virtue of Lemma 2.3, λ+ 2−1 q − n − 2 = 0. Hence xπ y λ ξ ω = (λ+ 2−1 q − n − 2)−1 [xπ y λ ξ ω , x1 ] ∈ I . Consequently I = Ω . The proof is completed. 2 3. Derivation superalgebra of Ω First we recall some useful definitions which shall appear frequently in this section. Let L = L 0¯ ⊕ L 1¯ be a Lie superalgebra and let V be a subspace of L. If V = ( V ∩ L 0¯ ) ⊕ ( V ∩ L 1¯ ), then V is called a Z 2 -graded subspace of L. If L is a Lie superalgebra and L = i ∈ Z L i , where L i is a Z 2 -graded subspace superalgebra. of L and [ L i , L j ] ⊆ L i + j , for any i , j ⊆ L , then L is called a Z -graded Lie  In Section 2 we constructed the Lie superalgebra Ω . Recall that Ω = α ∈ Z 2 Ωα ,

  Ωα = span F xk y λ ξ u  k ∈ Q , λ ∈ H , u ∈ B (q), |u | = α .

(10)

      Ωi := span F xk y λ ξ u  ki + 2k1 + |u | − 2 = i .

(11)

For i ∈ Z , we let

i∈M1

Eqs. (10) and (11) show that Ωi = (Ω i ∩ Ω0¯ ) ⊕ (Ωi ∩ Ω1¯ ), ∀i ∈ Z . From Eq. (8) we obtain that [Ωi , Ω j ] ⊆ Ωi + j , ∀i , j ∈ Z . Hence Ω = i ∈ Z Ωi is a Z -graded Lie superalgebra. Set Ω(i ,α ) := Ωi ∩ Ωα , where i ∈ Z , α ∈ Z 2 . Then Ω = (i ,α )∈ Z × Z 2 Ω(i ,α ) is also Z × Z 2 -graded. Let f ∈ Ω. If f ∈ Ωi , then f is called Z -homogeneous  element and i is the Z -degree of f . Frequently the Z -degree of f is denoted by zd( f ). Put τ = i ∈ M 1 πi + 2π1 + q − 2. If i < −2 or i > τ , it is easy to see that Ωi = 0. Let X = {−2, −1, . . . , τ }. Then Ω = λ ∈ H }. The following lemma comes from Ref. [12].



Lemma 3.1. Let G be an abelian group, V = g ∈G V g and W = linear spaces. The Hom F ( V , W ) inherits a G-gradation by Hom F ( V , W ) g :=







g ∈G

i∈ X

Ωi . Clearly Ω−2 = span F { y λ |

W g the finite-dimensional G-graded





ϕ ∈ Hom F ( V , W )  ϕ ( V h ) ⊆ V g +h , ∀h ∈ G .

Let pl Ω denote the general linear Lie superalgebra which consists of all linear transformations of Ω . Note that Ω is both Z 2 -graded and Z -graded. From Lemma 3.1 we can get the following lemma immediately. Lemma 3.2. (i) pl Ω =



α ∈ Z 2 plα Ω, where

pl Ω := α







ϕ ∈ pl Ω  ϕ (Ωβ ) ⊆ Ωα +β , ∀β ∈ Z 2 .

Y. Zhang, Q. Zhang / Journal of Algebra 321 (2009) 3601–3619

(ii) pl Ω =



t∈ Z

plt Ω , where pl Ω :=



t

Let

3609





ϕ ∈ pl Ω  ϕ (Ωi ) ⊆ Ωt +i , ∀i ∈ Z .

ϕ ∈ plα Ω , where α ∈ Z 2 . If 







ϕ [x, y ] = ϕ (x), y + (−1)α |x| x, ϕ ( y ) , ∀x ∈ h(Ω), ∀ y ∈ Ω , then ϕ is called a derivation of degree of all derivations of degree α of Ω , where α ∈ Z 2 . Set

α of Ω . Let Derα Ω denote the linear space

Der Ω := Der Ω ⊕ Der Ω. 0¯

1¯

We call Der Ω the derivation superalgebra of Ω. For t ∈ Z , we let



Der Ω := {ϕ ∈ Der Ω  ϕ (Ωi ) ⊆ Ωt +i , ∀i ∈ Z }. t

Obviously Dert Ω ⊆ plt Ω. Put δ = τ + 2 and Y = {−δ, −δ + 1, . . . , δ}. / X for If t ∈ Z \ Y , then t + i ∈ any i ∈ X . Hence plt Ω = Dert Ω = 0, ∀t ∈ Z \ Y . In particular, pl Ω = t ∈Y plt Ω. Lemma 3.3. Der Ω =



t ∈Y

Dert Ω.





Proof. Clearly From (ii) t ∈Y Dert Ω ⊆ Der Ω. We shall  prove that Derα Ω ⊆ t ∈Y Dert Ω, ∀α ∈ Z 2 .  of Lemma 3.2, we have Derα Ω ⊆ plα Ω ⊆ t ∈Y plt Ω. For any ϕ ∈ Der α Ω, we have ϕ = t ∈Y ϕt , where ϕt ∈ plt Ω. By virtue of (i) of Lemma 3.2, we can suppose that ϕt = ϕt 0¯ + ϕt 1¯ , where ϕt 0¯ ∈ pl0¯ Ω , ϕt 1¯ ∈ pl1¯ Ω. Then

ϕ=

   (ϕt 0¯ + ϕt 1¯ ) = ϕt 0¯ + ϕt 1¯ . t ∈Y

t ∈Y

(12)

t ∈Y







By virtue of Eq. (12) and ϕ ∈ plα Ω, we have ϕ = t ∈Y ϕt α . Then t ∈Y ϕt = t ∈Y ϕt α . Therefore ϕt = ϕt α , ∀t ∈ Y . Particularly |ϕt | = α , ∀t ∈ Y . Let f i ∈ Ω(i,β) and g j ∈ Ω j , where i , j ∈ X , β ∈ Z 2 . Then

 t ∈Y

ϕt [ f i , g j ] =

 



ϕt [ f i , g j ] = ϕ [ f i , g j ]

t ∈Y

   = ϕ ( f i ), g j + (−1)α β f i , ϕ ( g j )      = ϕt ( f i ), g j + (−1)α β f i , ϕt ( g j ) t ∈Y

=



t ∈Y

  ϕt ( f i ), g j + (−1)α β f i , ϕt ( g j ) . 

(13)

t ∈Y

For any f ∈ Ωβ =



i∈ X

Ω(i ,β) , g ∈ Ω =



f =

j∈ X

 i∈ X

Ω j , we can suppose that

fi,

g=

 j∈ X

g j,

(14)

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Y. Zhang, Q. Zhang / Journal of Algebra 321 (2009) 3601–3619

where f i ∈ Ω(i ,β) , g j ∈ Ω j . Utilizing Eqs. (13) and (14), it is easy to check that









ϕt [ f , g ] = ϕt ( f ), g + (−1)α β f , ϕt ( g ) . Hence ϕt ∈ (Der Ω) ∩ (plt Ω) =  Dert Ω , ∀t ∈ Y . Then ∀α ∈ Z 2 . Consequently Der Ω ⊆ t ∈Y Dert Ω. 2

ϕ∈



t ∈Y

Dert Ω. So Derα (Ω) ⊆



t ∈Y

Dert Ω ,

Lemma 3.4. Dert Ω is a Z 2 -graded subspace of Der Ω , ∀t ∈ Y . Proof. Let ϕ ∈ Dert Ω. Then ϕ ∈ Der Ω = Der0¯ Ω ⊕ Der1¯ Ω. So ϕ = ϕ0¯ + ϕ1¯ , where ϕ 0¯ ∈ Der 0¯ Ω , ϕ1¯ ∈ Der1¯ Ω. Suppose that f ∈ Ω(i ,α ) , where i ∈ X , α ∈ Z 2 . Obviously ϕ0¯ ( f ) ∈ Ωα = j ∈ X Ω( j ,α ) . Set   ϕ0¯ ( f ) = j∈ X f j , where f j ∈ Ω( j,α ) . Similarly we can suppose that ϕ1¯ ( f ) = j∈ X g j , where g j ∈ Ω( j,α +1¯ ) . So

  ( f j + g j) ∈ Ω j. j∈ X

(15)

j∈ X

On the other hand



( f j + g j) =

j∈ X



fj +

j∈ X



g j = ϕ0¯ ( f ) + ϕ1¯ ( f ) = ϕ ( f ) ∈ Ωt +i .

(16)

j∈ X

Considering Eqs. (15) and (16) we can obtain that f j + g j = 0,

∀ j ∈ X \ t + i.

(17)

Since f j ∈ Ωα , g j ∈ Ωα +1¯ , Eq. (17) shows that f j = g j = 0, ∀ j ∈ X \ t + i . Thus ϕ0¯ ( f ) = f t +i ∈ Ωt +i . So ϕ0¯ (Ω(i ,α ) ) ⊆ Ωt +i , ∀α ∈ Z 2 . It follows that ϕ0¯ (Ωi ) ⊆ Ωt +i , ∀i ∈ X . Then ϕ0¯ ∈ plt Ω. Thus ϕ0¯ ∈ (plt Ω) ∩ (Der0¯ Ω). Similarly, ϕ1¯ ∈ (plt Ω) ∩ (Der1¯ Ω). Therefore Der Ω = (pl Ω ∩ Der Ω) ⊕ (pl Ω ∩ Der Ω). t

t

0¯

1¯

t

(18)

It is easy to see that

(pl Ω) ∩ (Der Ω) = (pl Ω) ∩ (Der Ω ∩ Der Ω) t

α

α

t

= (pl Ω ∩ Der Ω) ∩ Der Ω α

t

= Der Ω ∩ Der Ω, t

α

∀α ∈ Z 2 .

(19)

By virtue of Eqs. (18) and (19), we have Der Ω ⊆ (Der Ω ∩ Der Ω) ⊕ (Der Ω ∩ Der Ω). t

t

0¯

It follows that Dert Ω is a Z 2 -graded subspace of Der Ω.

1¯

t

2

In the following we shall determine the derivation superalgebra Der Ω of Ω . By virtue of Lemmas 3.3 and 3.4, we only need to determine the h(Dert Ω), ∀t ∈ Y .

Y. Zhang, Q. Zhang / Journal of Algebra 321 (2009) 3601–3619

Lemma 3.5. Let ϕ ( f ) ∈ Ω−2 .

3611

ϕ ∈ h(Der Ω), f ∈ Ω. If ϕ (xi ) = ϕ [ f , xi ] = ϕ (ξ j ) = ϕ [ f , ξ j ] = 0, ∀i ∈ M 1 , ∀ j ∈ T , then 

Proof. Let f = α ∈ Z 2 f α , where f α  ∈ Ωα . We shall provethat ϕ ( f α ) ∈ Ω−2 , ∀α ∈ Z 2 . By virtue Since ϕ , f α and xi of ϕ [ f , xi ] = 0, ∀i ∈ M 1 , we have ϕ [ α ∈ Z 2 f α , xi ] = 0. So α ∈ Z 2 ϕ [ f α , xi ] = 0.  are all Z 2 -homogeneous elements, ϕ [ f α , xi ] is Z 2 -homogeneous. Then equation α ∈ Z 2 ϕ [ f α , xi ] = 0 shows that ϕ [ f α , xi ] = 0, ∀α ∈ Z 2 . Hence









ϕ ( f α ), xi + (−1)|ϕ |α f α , ϕ (xi ) = 0.

Noticing that

ϕ (xi ) = 0, we have [ϕ ( f α ), xi ] = 0, ∀i ∈ M 1 . Thus 





ϕ ( f α ), 1 = [i ] ϕ ( f α ), [xi , xi ]



      = [i ] ϕ ( f α ), xi , xi  + [i ] xi , ϕ ( f α ), xi  = 0.

It follows that Di









ϕ ( f α ) = [i ] ϕ ( f α ), xi = 0, ∀i ∈ M 1 .

Similarly, we have [ϕ ( f α ), ξ j ] = 0 and D j (ϕ ( f α )) = 0, ∀ j ∈ T . By virtue of Lemma 2.4, Hence ϕ ( f ) ∈ Ω−2 . 2

ϕ ( f α ) ∈ Ω−2 .

Lemma 3.6. Let t ∈ Z and ϕ ∈ h(Dert Ω). If ϕ (Ω j ) = 0, j = −2, −1, . . . , k, where k  −1 and k + t  −2, then ϕ = 0. Proof. Let j  k. We shall prove ϕ (Ω j ) = 0 by induction on j . Let j > k and f ∈ Ω j . Assume that ϕ (Ω j−1 ) = 0. Clearly [ f , xi ], [ f , ξ j ] ∈ Ω j−1 . So

ϕ [ f , xi ] = ϕ [ f , ξ j ] = 0, ∀i ∈ M 1 , ∀ j ∈ T . Since ϕ (Ω−1 ) = 0, ϕ (xi ) = ϕ (ξ j ) = 0. By virtue of Lemma 3.5, ϕ ( f ) ∈ Ω−2 . Then ϕ ( f ) ∈ Ω−2 ∩ Ωt + j . Since t + j > t + k  −2, Ω−2 ∩ Ωt + j = 0. So ϕ ( f ) = 0. Thus ϕ (Ω j ) = 0. Induction is completed. Consequently ϕ = 0. 2 If i ∈ M , then let τ (i ) = πi . If i ∈ T , then let τ (i ) = 1. An element f of Ω is called τ (i )-truncated τ (i ) if D i ( f ) = 0, where i ∈ M ∪ T . For i ∈ M , we define a linear mapping ρi : Ω → Ω such that







ρi xk y λ ξ u = (ki + 1)∗

−1

xk+e i y λ ξ u ,

where we set xk+e i = 0 for k + e i ∈ / Q . For i ∈ T , we define a linear mapping ρi : Ω → Ω such that ρi (xk y λ ξ u ) = xk y λ ξi ξ u . From the definitions we can obtain the following lemma directly. Lemma 3.7. (i) If f ∈ Ω is τ (i )-truncated, then D i ρi ( f ) = f , ∀i ∈ R . ˜˜

(ii) D i ρ j = (−1)i j ρ j D i , where i , j ∈ R, i = j .

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Y. Zhang, Q. Zhang / Journal of Algebra 321 (2009) 3601–3619

Lemma 3.8. Let f t1 , . . . , f tk ∈ Ω, where t 1 , . . . , tk ∈ R . If f i is τ (i )-truncated, i = t 1 , . . . , tk , and D i ( f j ) = ˜˜

(−1)i j D j ( f i ), i , j = t 1 , . . . , tk , then there is f ∈ Ω such that D i ( f ) = f i , i = t 1 , . . . , tk . Proof. Induction on k. If k = 1, then let f = ρt1 ( f t1 ). By virtue of (i) of Lemma 3.7, D t1 ( f ) = D t1 ρt1 ( f ) = f t1 . Assume that there is g ∈ Ω such that D i ( g ) = f i , i = t 1 , . . . , tk−1 . Let f = g + ρtk ( f tk − D tk ( g )). For i = t 1 , . . . , tk−1 , utilizing (ii) of Lemma 3.7 and the hypothesis of this lemma, we have





D i ( f ) = f i + D i ρtk f tk − D tk ( g )

  ˜ ˜ = f i + (−1)tk i ρtk D i ( f tk ) − D i D tk ( g )   ˜˜ ˜˜ ˜ ˜ = f i + (−1)tk i ρtk (−1)itk D tk ( f i ) − (−1)itk D tk D i ( g ) = fi. Since f tk − D tk ( g ) is

τ (tk )-truncated, by virtue of Lemma 3.7 we have 



D tk ( f ) = D tk ( g ) + D tk ρtk f tk − D tk ( g )

= D tk ( g ) + f tk − D tk ( g ) 2

= f tk . ˜˜

Lemma 3.9. Let f i ∈ Ω , ∀i ∈ R . If D i ( f j ) = (−1)i j D j ( f i ), ∀i , j ∈ R , then following statements hold: (i) f i is τ (i )-truncated, ∀i ∈ T . π π (ii) exi i , e ∈ Ω−2 , is the only term containing xi i that is possible to arise in f i . Proof. By virtue of the hypothesis of this lemma, we have 2D i ( f i ) = 0, ∀i ∈ T . Then (i) holds. We shall π π prove (ii). Let f i ∈ Ω. we can suppose that f i = exi i + h, where i ∈ M, e ∈ Ω and D i (e ) = 0 = D i i (h). Since i ∈ M, ˜i = 0¯ . For any j ∈ R , by the assumption of this lemma we have π

D i ( f j ) = D j ( f i ) = D j (e )xi i + D j (h). Because D i ( f j ) and D j (h) are both e ∈ Ω−2 . 2 Lemma 3.10. Suppose that

(−1)|ϕ | ϕ (ξ

τ (i )-truncated, D j (e ) = 0, ∀ j ∈ R . By virtue of Lemma 2.4,

ϕ ∈ h(Der Ω). Let f 1 = ϕ (1), f i = [i ]ϕ (xi ) − [i ]μi f 1 xi , ∀i ∈ M 1 ; f i =

|ϕ | −1 f ξ , 1 i i ) − (−1) 2

˜˜

∀i ∈ T . Then D i ( f j ) = (−1)i j D j ( f i ), ∀i , j ∈ R.

Proof. By the assumption we have

Applying

ϕ ( xi  ) = [ i ] f i + μi f 1 xi  , ∀i ∈ M 1 ,

(20)

ϕ (ξi ) = (−1)|ϕ | f i + 2−1 f 1 ξi , ∀i ∈ T .

(21)

ϕ to the equality [1, ξi ] = 0, we obtain that [ϕ (1), ξi ] + [1, ϕ (ξi )] = 0. Then   [ f 1 , ξi ] + 1, (−1)|ϕ | f i + 2−1 f 1 ξi = 0.

Y. Zhang, Q. Zhang / Journal of Algebra 321 (2009) 3601–3619

3613

Utilizing Eq. (8), a direct operation shows that 2−1 D 1 ( f 1 )ξi + (−1)| f 1 | D i ( f 1 ) − (−1)|ϕ | D 1 ( f i ) − 2−1 D 1 ( f 1 ξi ) = 0. Noting that | f 1 | = |ϕ (1)| = |ϕ |, we get that D 1 ( f i ) = D i ( f 1 ), ∀i ∈ T . Applying have





ϕ to [ξi , ξ j ] = −δi j 1 we





ϕ (ξi ), ξ j + (−1)|ϕ | ξi , ϕ (ξ j ) = −δi j f 1 , ∀i , j ∈ T .

Utilizing Eq. (21), we obtain that

(−1)|ϕ | 2−1 D 1 ( f i )ξ j + (−1)|ϕ | (−1)| f i | D j ( f i ) + 4−1 D 1 ( f 1 ξi )ξ j + (−1)| f 1 ξi | 2−1 D j ( f 1 ξi ) − 2−1 ξi D 1 ( f j ) + (−1)|ξi | D i ( f j ) − (−1)|ϕ | 4−1 ξi D 1 ( f 1 ξ j ) + (−1)|ϕ | (−1)|ξi | 2−1 D i ( f 1 ξ j ) = −δi j f 1 .

(22)

Note that D 1 ( f i ) = D i ( f 1 ), ∀i ∈ T , |ϕ | = | f 1 | and |ϕ | + 1¯ = | f i |. By virtue of Eq. (22), we obtain that D i ( f j ) = − D j ( f i ), ∀i , j ∈ T . Similarly, applying ϕ to the equalities [1, xi  ] = 0 and [xi  , x j  ] = 

δ ji  [i ]1 respectively and utilizing Eq. (20) we get that D i ( f 1 ) = D 1 ( f i ) and D i ( f j ) = D j ( f i ), ∀i , j ∈ M 1 . Applying ϕ to [xi  , ξ j ] = 0 we obtain D i ( f j ) = D j ( f i ), ∀i ∈ M 1 , ∀ j ∈ T . Hence the lemma holds. 2 Lemma 3.11. Let ϕ ∈ h(Der Ω) and f i be defined according to Lemma 3.10, ∀i ∈ R . Then f i is τ (i )-truncated, ∀i ∈ R. Proof. If i ∈ T , by Lemmas 3.10 and 3.9 we know that f i is we obtain that [ϕ (x1 ), 1] + [x1 , f 1 ] = f 1 . Hence f1 = D1

τ (i )-truncated. Applying ϕ to [x1 , 1] = 1,





¯ f 1 ) − x1 D 1 ( f 1 ). ϕ (x1 ) + ∂( π

By virtue of Lemma 3.9 we can set f 1 = ex1 1 + g 1 , where e ∈ Ω−2 , g 1 is π and x1 D 1 to the equality f 1 = ex1 1 + g 1 respectively, we obtain that

¯ e )xπ1 − ∂( ¯ g 1 ), ¯ f 1 ) = ∂( ∂( 1

τ (1)-truncated. Applying ∂¯

π

x1 D 1 ( f 1 ) = −ex1 1 + x1 D 1 ( g 1 ).

Then f1 = D1









¯ e )xπ1 − ∂( ¯ g ) − x1 D 1 ( f 1 ), ϕ (x1 ) + ∂( 1

π

ex1 1 = x1 D 1 ( g 1 ) − x1 D 1 ( f 1 ). Above two equalities show that π

g 1 = f 1 − ex1 1 = D 1





¯ e )xπ1 + ∂( ¯ g 1 ) − x1 D 1 ( g 1 ). ϕ (x1 ) + ∂( 1

¯ g 1 ) and x1 D 1 ( g 1 ) are all τ (1)-truncated, ∂( ¯ e ) = 0. Thus e = 0. Consequently Since g 1 , D 1 (ϕ (x1 )), ∂( f 1 = g 1 is τ (1)-truncated. Similarly, applying ϕ to [x1 , xi  ] = μi xi  , ∀i ∈ M 1 , we can obtain that f i is τ (i )-truncated, ∀i ∈ M 1 . 2

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Y. Zhang, Q. Zhang / Journal of Algebra 321 (2009) 3601–3619

Let  = {θ : H → F | θ(λ + η) = θ(λ) + θ(η), ∀λ, η ∈ H }. For θ ∈ , we define a linear transformation D θ of Ω by means of D θ (xk y λ ξ u ) = θ(λ)xk y λ ξ u . It is easy to see that D θ ∈ Der0¯ Ω. Lemma 3.12. Let ϕ ∈ h(Der Ω). If ϕ (x1 ) = ϕ (xi ) = ϕ (ξ j ) = 0, ∀i ∈ M 1 , ∀ j ∈ T , then there is θ ∈  such that ϕ ( y λ ) = θ(λ) y λ , ∀λ ∈ H . Proof. By virtue of Lemma 3.5 we can let

(1 − λ) y λ we obtain that aη

ϕ ( yλ ) =



η η∈ H aη y . Let λ ∈ H . Applying

ϕ to [x1 , y λ ] = θ(λ) = aλ . Applying ϕ

= 0, ∀η ∈ H \ {λ}. = to [x1 y η , 1] = y η we have D 1 (ϕ (x1 y η )) = θ(η) y η . Then ϕ (x1 y η ) = θ(η)x1 y η + g , where g ∈ Ω and D 1 ( g ) = 0. Again applying ϕ to [x1 y η , y λ ] = (1 − λ) y η+λ we get that Hence ϕ ( y λ )

θ(λ) y λ , where

    θ(η)x1 y η + g , y λ + x1 y η , θ(λ) y λ = (1 − λ)θ(η + λ) y η+λ . Then

(1 − λ)θ(η) y η+λ + (1 − λ)θ(λ) y η+λ = (1 − λ)θ(η + λ) y η+λ . Since 1 − λ = 0, θ(η + λ) = θ(η) + θ(λ). Thus θ ∈ . Lemma 3.13. If j = −2, −1.

2

ϕ ∈ h(Der Ω), then there are g ∈ Ω and θ ∈  such that (ϕ − ad g − D θ )(Ω j ) = 0,

Proof. Let f i , i ∈ R , be the elements of Ω which are defined according to Lemma 3.10. By Lemmas 3.10, 3.11 and 3.8, there is f ∈ Ω such that D i ( f ) = f i , ∀i ∈ R . Let ϕ1 = ϕ − ad f . By virtue of Eq. (20)

ϕ1 ( x i ) = ϕ ( x i ) − [ f , x i ]

  = ϕ (xi ) − D 1 ( f )μi  xi + [i  ] D i  ( f )   = ϕ ( xi ) − μi  f 1 xi + [i  ] f i  = 0,

 ∀i ∈ M 1 . Similarly, ϕ1 (ξ j ) = 0, ∀ j ∈ T . By virtue of Lemma 3.5, ϕ1 (x1 ) = λ∈ H aλ y λ . Let z =  −1 λ and ϕ = ϕ − ad z. Then ϕ (x ) = 0 and ϕ (x ) = ϕ (ξ ) = 0, ∀i ∈ M , 2 1 2 1 2 i 2 j 1 λ∈ H (λ − 1) aλ y ∀ j ∈ T . By Lemma 3.12, ϕ2 ( y λ ) = θ(λ) y λ , where θ ∈ . Let ϕ3 = ϕ2 − D θ . Then ϕ3 ( y λ ) = 0, ∀λ ∈ H . Hence ϕ3 (Ω−2 ) = 0 and ϕ3 (x1 ) = ϕ3 (xi ) = ϕ3 (ξ j ) = 0, ∀i ∈ M 1 , ∀ j ∈ T . By Lemma 3.5, ϕ3 (xi y λ ), ϕ3 (xi xi ) ∈ Ω−2 . Note that [xi y λ , Ω−2 ] = {0} = [Ω−2 , xi xi ]. Applying ϕ to xi y λ = [i ][xi y λ , xi xi  ], we obtain ϕ3 (xi y λ ) = 0, ∀i ∈ M 1 . Similarly, ϕ3 (ξ j y λ ) = 0, ∀ j ∈ T . Consequently ϕ3 (Ω−1 ) = 0. Clearly ϕ3 = ϕ − ad( f + z) − D θ . 2 Lemma 3.14. Let ϕ ∈ h(Dert Ω), where t  −1. Then there are f ∈ Ω and θ ∈  such that ϕ = ad f + D θ . Proof. This is a direct consequence of Lemmas 3.13 and 3.6.

2

Lemma 3.15. Der−2 Ω = ad Ω−2 .



ϕ ∈ h(Der−2 Ω). Then ϕ (Ω0 ) ⊆ Ω−2 . Suppose that ϕ (x1 y λ ) = η∈ H aη y η . Since η − λ ∈ H ,  η − λ = 1. Let g = η∈ H (η − λ − 1)−1 aη y η and ϕ1 = ϕ − ad g . We have ϕ1 (x1 y λ ) = 0. Obviously ϕ1 (xi xi ), ϕ1 (xi x j y λ ) ∈ Ω−2 . Applying ϕ1 to xi x j y λ = [i  ][xi xi , xi x j y λ ] we obtain that ϕ1 (xi x j y λ ) = 0,

Proof. Let

where j = i  . Similarly, it follows that

Y. Zhang, Q. Zhang / Journal of Algebra 321 (2009) 3601–3619











3615



ϕ1 xi ξl y λ = ϕ1 ξl ξ j y λ = ϕ1 xi xi y λ = 0, ∀i ∈ M 1 , ∀l, j ∈ T , ∀λ ∈ H . Thus

ϕ1 (Ω0 ) = 0. By Lemma 3.6 we have ϕ1 = 0. Hence ϕ = ad g ∈ ad Ω−2 . 2

Lemma 3.16. Let ϕ ∈ h(Der−t Ω), where t > 2. Suppose that i ∈ M 1 . If ϕ (xti ) = 0, then ϕ (xli ) = 0, 0  l  πi . Proof. If l < t , then that l > t and ϕ Ωl−2−t = 0. 2

ϕ (xli ) ∈ Ωl−2−t = 0. Let l  t . We shall prove ϕ (xli ) = 0 by induction on l. Assume = 0. By virtue of Lemma 3.5 we obtain that ϕ (xli ) ∈ Ω−2 . Hence ϕ (xli ) ∈ Ω−2 ∩

(xli−1 )

Lemma 3.17. Let S l = {x1s , 0  s  l; ξ j , j ∈ T ; y λ , λ ∈ H }, where 2  l 

π1 . Suppose that L is the

subalgebra of Ω generated by S l . Then xl1 ξ j ∈ L, ∀ j ∈ T .

Proof. Clearly x1 ξ j = [x21 , ξ j ] ∈ L . If l∗ = 2, then xl1 ξ j = 2(l∗ − 2)−1 [xl1 , x1 ξ j ] ∈ L . Let l∗ = 2. Let λ ∈ H \ {0}. By virtue of 1 ∈ / H , it follows that 1 − λ, 1 − 2λ and 3 + 2λ are all non-zero elements. Then





x1 ξ j y λ = (1 − λ)−1 (1 − 2λ)−1 x21 , x1 ξ j , y λ



∈ L.

Moreover



 

xl1 ξ j = −2−1 λ−1 (1 + λ)−1 (3 + 2λ)−1 xl1 , x1 ξ j y λ , x21 , y −λ



∈ L.

2

Lemma 3.18. Let t > 2 and ϕ ∈ h(Der−t Ω). (i) If ϕ (xk1 ) = 0, where k = [ 2t ] is the largest integer equal or less than (ii) If t is an old number, then ϕ

(xl1 )

= 0, 0  l  π1 .

t , 2

then ϕ (xl1 ) = 0, 0  l  π1 .

Proof. (i) If l < k, then 2l < t and zd(ϕ (xl1 )) = −t + 2l − 2  −3. Hence

ϕ (xl1 ) = 0. Let l  k. We shall

ϕ = 0 by induction on l. Assume that ϕ = 0, s = 0, 1, . . . , l. By Lemma 3.17 we have ϕ (xl1 ξ j ) = 0, ∀ j ∈ T . Since zd(ϕ (xi ξ j ))  −3, ϕ (xi ξ j ) = 0. Then

prove

(xl1 )

(x1s )









ϕ xl1 xi = −ϕ xl1 ξ j , xi ξ j = 0, ∀xi ∈ M 1 . Utilizing Lemma 3.5 we have ϕ (xl1+1 ) ∈ Ω−2 . Because zd(xl1+1 ) + zd(ϕ ) = 2l − t, serve that 2l − t  2k − t  −1. Then



ϕ (xl1+1 ) ∈ Ω2l−t . Ob-



ϕ xl1+1 ∈ Ω−2 ∩ Ω2l−t = {0}. Induction is completed. (ii) If t is an old number, then zd(ϕ (xk1 )) = 2k − 2 − t = −3, where k = [ 2t ]. Hence virtue of (i),

ϕ

(xl1 )

= 0, where 0  l  π1 . 2

ϕ (xk1 ) = 0. By

k

Theorem 3.19. Let S = {xi i | i ∈ M , 0  ki  πi } ∪ { y λ | λ ∈ H } ∪ {ξ j | j ∈ T }. Then Ω is generated by S. Proof. Let Y be the subalgebra generated by S. First we prove the following statements: k 1) By Lemma 3.17 we know that x11 ξ j ∈ Y , k1 < π1 , j ∈ T . 2) xri xis = ((r + 1)∗ )−1 ((s + 1)∗ )−1 [xri +1 , xis+1 ] ∈ Y , r < πi , s < πi  , i ∈ M 1 .

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Y. Zhang, Q. Zhang / Journal of Algebra 321 (2009) 3601–3619

π π

3) xi i xi  i ∈ Y , i ∈ M 1 . In fact, if

μi = −1, by virtue of 1), 

πi 

xi i ξ j = (1 + μi )−1 xi ξ j , xi π

If

∈ Y.

μi = −1, by virtue of 1) and 2), 

 πi −1

xi i ξ j = −2[i  ] x2i xi  , xi π

π

π π

, x1 ξ j



∈ Y.

π

π

Similarly xi  i ξ j ∈ Y . Hence xi i xi  i = −[xi i ξ j , xi  i ξ j ] ∈ Y . 4) ξ ω ∈ Y . In fact, by induction on k we can obtain ξ j 1 ξ j 2 · · · ξ jk ∈ Y , where j 1 , . . . , jk ∈ T . Hence ω ξ ∈ Y. k k 5) By virtue of 1), x11 y λ = −(1 − λ)−1 [x11 ξ j , [x1 ξ j , y λ ]] ∈ Y , k1  π1 , λ ∈ H . k

6) x11 y λ ξ ω ∈ Y , k1 < π1 , λ ∈ H . If q = 2, by virtue of 5)



x11 y λ ξ ω = (k1 + 1)∗ k

−1 

1 − 2−1 q

−1 



k +1 λ

y ,ξω ∈ Y .

x11

If q = 2, then q = ξr +1 ξr +2 . For k1 < π1 − 1,



x11 y λ ξ ω = 4 (k + 2)∗ k

−1  −1  k1 +2 λ   x1 y , ξr +1 , ξr +2 ∈ Y . (k + 1)∗

Then

  π −1 λ ω π −2 y ξ = (3 − 2λ)−1 x21 , x1 1 y λ ξ ω ∈ Y .

x1 1

7) xxi xi  ∈ Y , i ∈ M 1 . Let λ ∈ H \ {0}. If

μi = 0, by virtue of 2) and 5), 



1 xi xi  y λ = 2−1 [i ]μ− xi x2i  , x1 y λ , xi i



∈ Y.

Moreover





x1 xi xi  = 2−1 λ−1 (1 + 2λ)−1 x1 y λ , x21 , xi xi  y λ If



∈ Y.

μi = 0, then μi = 0. Similarly we can obtain that x1 xi xi ∈ Y . 8) x1 xi ξ ω ∈ Y , ∀i ∈ M 1 . If μi  = 0, from 1) we have 



1 2 ω x1 xi ξ ω = 2−1 μ− x1 ξ , xi ∈ Y . i

If

μi = 0, then μi = 0. Similarly x1 xi ξ ω ∈ Y By 2) and 6) 







x1 xi ξ ω = 2−1 [i  ] x1 xi  ξ ω , x2i − 2−1 x1 ξ ω , x2i xi  ∈ Y . 9) x1 1 y λ ξ ω ∈ Y , ∀λ ∈ H . Let a = 2 − 2−1 q − λ. If a = 0, from 5) and 6), π

 π



x1 1 y λ ξ ω = a−1 x1 1 y λ , x1 ξ ω ∈ Y . π

Suppose that a = 0. Since 1 ∈ / G, λ = 0. By 6) and 7), π −1

x1 1

 π

xi xi  ξ ω = x1 1

 ξ , x1 xi xi  ∈ Y .

−1 ω

Y. Zhang, Q. Zhang / Journal of Algebra 321 (2009) 3601–3619

Let

3617

μi = 0 (otherwise μi = 0, the proof is similar). By virtue of 8),  π



1 x1 1 xi ξ ω = μ− x1 1 , x1 xi ξ ω ∈ Y . i

π

Then

 π

π



π −1

x1 1 ξ ω = [i ] x1 1 xi ξ ω , xi  − [i ]μi x1 1

xi xi  ξ ω ∈ Y .

10) Let



ρ1 (k1 , λ) = xk11 y λ ξ ω ,

ρi (k1 , λ) = xk11

i  π j π j

x j x j

 yλ ξ ω ,

j =2

where i  2. Then

ρi (k1 , λ) ∈ Y , k1  π1 , ∀λ ∈ H . We shall prove 10) by induction on i. By virtue of 9), assumption of induction and 3),



ρi (k1 , λ) = 2−1 (k1 + 1)∗

−1 

ρ1 (k1 , λ) ∈ Y . Let i > 1. If k1 < π1 , by the π 

ρi−1 (k1 + 1, λ), xπi i xi i ∈ Y .

Suppose that k1 = π1 . Let b = i + 1 − 2−1 q − λ If b = 0, then





ρi (π1 , λ) = 2−1 b−1 x21 , ρi (π1 − 1, λ) ∈ Y . If b = 0, then i = n + 1 by Lemma 2.3. Hence i < n + 1. We can suppose that we have

μi+1 = o. Utilizing b = 0





1 ρi (π1 , λ)xi+1 = −2−1 μ− x2 x , ρi (π1 − 1, λ) ∈ Y . i +1 1 i +1

By virtue of 7),







ρi (π1 − 1, λ)xi+1 x(i+1) = x1 xi+1 x(i+1) , xπ1 1 , ρi (0, λ) ∈ Y , 



ρi (π1 , λ) = [i + 1] ρi (π1 , λ)xi+1 , x(i+1) − [i + 1]μi+1 ρi (π1 − 1, λ) ∈ Y . k λ ω Proof of Theorem 3.19. Let z := xk y λ ξ u  be any basis  element. If u = ω , i.e., z = x y ξ , then we shall prove z ∈ Y by induction on l z := i ∈ M 1 πi − i ∈ M 1 ki . If l z = 0, then z = ρn+1 (k1 , λ) ∈ Y by virtue of 10). Let l z > 0. Then there is i ∈ M 1 such that ki < πi . By the assumption of induction, z := xk+e i y λ ξ ω ∈ Y and g := xk−e1 +e i xi  y λ ξ ω ∈ Y . Since [ z , xi  ] = k∗1 μi g + [i ](k + 1)∗ z, z ∈ Y . The induction is completed. If u = ω , then we can let {ω} \ {u } = { j 1 , . . . , jt }. By virtue of 10),









xk y λ ξ u = c ξ j 1 , . . . , ξ jt , xk y λ ξ ω · · · ∈ Y , where c = 1 or −1. Then Ω ⊆ Y . Consequently Y = Ω .

2

Corollary 3.20. Let t > 2 and ϕ ∈ h(Der−t Ω). Suppose that ϕ (xti ) = 0, ∀i ∈ M 1 . If t is an odd number, then

ϕ = 0. If t is an even number and ϕ (xk11 ) = 0, where k1 = 2t , then ϕ = 0.

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Y. Zhang, Q. Zhang / Journal of Algebra 321 (2009) 3601–3619

Proof. This is a direct consequence of Lemmas 3.16, 3.18 and Theorem 3.19.

2

Lemma 3.21. Let t > 2. If there is not any v ∈ N such that t = p v or 2p v , then h(Der−t Ω) = 0. Proof. Let ϕ ∈ h(Der−t Ω). First we shall ϕ (xti ) = 0, ∀i ∈ M 1 . If t > πi , then xti = 0. Let t  πi . If ε0 (t ) = 0, then ε0 (t − 1) = p − 1. Hence ε0 (1) + ε0 (t − 1) < p. Clearly ε v (1) + ε v (t − 1) = ε v (t ) < p for any v > 0. The expression (1) implies that xi xti −1 = xti . Hence

 Considering Z -degree we have



xi xi  , xti = [i  ]t ∗ xi xti −1 = [i  ]t ∗ xti .

(23)

ϕ (xti ) ∈ Ω−2 . Applying ϕ to (23), we obtain that ϕ (xti ) = 0. If ε0 (t ) = 0,

then we let l = max{ v | ε v (t ) = 0}. The assumption of this theorem ensures that t = pl . Considering t − pl +1

pl

pl

t − pl +1

Z -degree we get ϕ (xi xi  ) = ϕ (xi ) = 0. Applying ϕ to [xi xi  , xi ] = [i  ]xti , we obtain that t ϕ (xi ) = 0. If t is an odd number, Corollary 3.20 implies that ϕ = 0. Let t = 2k1 be an even number. Clearly

ϕ (xk11 ) ∈ Ω−2 . Let ϕ (xk11 ) = 

k





λ λ∈ H aλ y . If

ε0 (k1 ) = 0, from (1) we have x1 xk11 −1 = xk11 . Then k −1

k

x1 , x11 = x11 − ε0 (k1 )x1 x11

  = 1 − ε0 (k1 ) xk11 .

(24)

Applying ϕ to the equality (24) we have (1 − λ)aλ = (1 − ε0 (k1 ))aλ , ∀λ ∈ H . Since 1 ∈ / H , aλ = 0, ∀λ ∈ H . Then ϕ (xk11 ) = 0. If ε0 (k1 ) = 0, we let s = max{ v | ε v (k1 ) = 0}. Then s > 0. The assumption s s t = 2p s implies that k = p s . Since ε0 ( p s − 1) + ε0 (k1 − p s + 1) = p, from (1) x p −1 xk1 − p +1 = 0. It follows that

 Applying

k − p s +1

x11

ps 

, x1

k −ps ps x1

= x11

p s −1 k1 − p s +1 x1

+ x1

= xk11 .

(25)

ϕ to (25) we obtain that ϕ (xk11 ) = 0. Corollary 3.20 yields that ϕ = 0. 2 pv

Lemma 3.22. If t = p v , v > 0, then Der−t Ω = D i

pv

| i ∈ M 1 . If t = 2p v , v > 0, then Der−t Ω = D 1 .

pv

Proof. It is easy to see that D i ∈ Der−t Ω , ∀i ∈ M. Let t = p v and ϕ ∈ Der−t Ω . Then ϕ (xti ) =  h i ( y ) ∈ Ω−2 , ∀i ∈ M 1 . Set h i ( y ) = λ∈ H ai λ y λ . Applying ϕ to [x1 , xti ] = xti we obtain that h i ( y ) = ai0 1, t ai0 ∈ F . Let ψ = ϕ − i ∈ M 1 ai0 D i . Then ϕ (xti ) = 0, ∀i ∈ M 1 . Corollary 3.20 implies that ψ = 0. Hence v

ϕ ∈ D ip | i ∈ M 1 .

Let t = 2p v and ϕ ∈ Der−t Ω . Set s = 2t = p v . Then ϕ (x1s ) = h( y ) ∈ Ω−2 . Applying ϕ to [x1 , x1s ] = x1s we have ϕ (x1s ) = c1, where c ∈ F . Then (ϕ − c D 1s )(x1s ) = 0. By Corollary 3.20 we obtain that ϕ = c D 1s ∈ D 1s . 2 pvi

Let W 1 = { D θ | θ ∈ }. Then W 1 is m-dimensional linear space. If v i > si , then D i Let W 2 = theorem.

p Di

vi

= 0, ∀i ∈ M .

| ∀i ∈ M , 0 < v i  si . By Lemmas 3.14, 3.15, 3.21 and 3.22 we obtain the following

Theorem 3.23. Der Ω = ad Ω ⊕ W 1 ⊕ W 2 . Let 2n + 4 − q ≡ 0 (mod p ). Ω ∗ denotes the commutator subalgebra of Ω˜ . It can be proved that Ω ∗ = xk y λ ξ u | (k, λ, u ) = (π , 0, ω) and Ω ∗ is a simple Lie superalgebra. Completely imitating the proof of the Theorem 3.19 we can obtain the following theorem. Theorem 3.24. Der Ω ∗ = ad Ω ∗ ⊕ ad xπ ξ ω ⊕ W 1 ⊕ W 2 .

Y. Zhang, Q. Zhang / Journal of Algebra 321 (2009) 3601–3619

3619

Theorem 3.25. Ω is not isomorphic to the known modular Lie superalgebras of Cartan type.



Proof. It is easy to see that dim Ω = 2q pl , where l = i ∈ M (si + 1) + m. From Refs. [6,13,14] we know that the dimensions of modular Lie superalgebras S , H and HO are not divided by p. From Refs. [8,14] we know that every outer derivation of modular Lie superalgebras W and K is ad-nilpotent. But Ω possesses the outer derivation D θ which is not ad-nilpotent. Consequently Ω is not isomorphic to known modular Lie superalgebras of Cartan type. 2 Acknowledgment The authors are grateful to the referee for his suggestions of correction. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16]

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