Derivation algebras and automorphism groups
of a class of deformative super W-algebras Wλs(2,2)
††Supported by NSF grants 11101056, 11271056, 11431010, 11371278 of China.
Hao Wang1), Huanxia Fa2), Junbo Li2)
*1)*Wu Wen-Tsun Key Laboratory of Mathematics and School of Mathematical Sciences,
University of Science and Technology of China, Hefei 230026, China
*2)*School of Mathematics and Statistics, Changshu Institute of Technology, Changshu 215500, China
Abstract.
In this paper, the derivation algebras and automorphism groups of a class of deformative super W-algebras Wλ(2,2) are determined.
Key words:
automorphisms, deformative super W-algebras, derivations.
*MR(2000) Subject Classification: * 17B05, 17B40, 17B65, 17B68.
1 Introduction
It is well known that the W-algebra W(2,2) plays important rolls in many areas of theoretical physics and mathematics which was introduced in [18] for the study of vertex operator algebras generated by weight 2 vectors. It has a basis {Lm,Im∣m∈Z} as a vector space over the complex field C, with the non-trivial Lie brackets [Lm,Ln]=(m−n)Lm+n and [Lm,In]=(m−n)Im+n. Structures and representations of W(2,2) are extensively investigated in the known references, such as [2], [7], [8], [10], [11], [18], [19] and the corresponding references.
Some Lie superalgebras whose even parts are the W(2,2) Lie algebras are constructed in [16]. The deformative super W-algebras Wλs(2,2) (denoted by Lλs just for convenience in the following) investigated in this paper own the same even parts: the W-algebra W(2,2) and are infinite-dimensional Lie superalgebras over the complex field C with the basis {Lm,Im,Gp,Hp∣m∈Z,p∈s+Z}, admitting the following non-vanishing super brackets:
[TABLE]
where s=0 or 21 and λ∈C.
It is known that derivations and automorphisms play important roles in the investigation of structures and representations of the relevant Lie algebras or superalgebras. Many references (i.e., [1],[3],[4],[5],[6],[7],[12],[13],[15]) have focused on derivations and automorphisms of different Lie algebras or superalgebras backgrounds. In this article, the corresponding derivation algebras and automorphism groups are respectively determined in Theorems 2.1 and 3.1.
We briefly recall some notations on Lie superalgebras. Let L=L0⊕L1 be a vector space over C, and all elements below are assumed to be Z2-homogeneous in this subsection, where Z2={0,1}. For x∈L, we always denote [x]∈Z2 to be its parity, i.e., x∈L[x]. An Z2-homogeneous linear map d:L⟶L satisfying d(Li)∈Li+[d] for i∈Z2 and
[TABLE]
is called a homogeneous derivation of parity [d]∈Z2. The derivation d is called even if [d]=0 and odd if [d]=1. Denote by Derp(L) the set of homogeneous derivations of parity p (p∈{0,1}). Denote Der(L)=Der0(L)⊕Der1(L) the set of derivations from L to L.
Denote Inn(L)=Inn0(L)⊕Inn1(L) the set of all inner derivations from L to L, where Innp(L) is the set of homogeneous inner derivations of parity p consisting of \mboxadx (x∈Lp) defined by:
[TABLE]
Denote Zs=Z∪(s+Z), i.e., Zs=Z for s=0 and 21Z for s=21. A Lie superalgebra L is called Zs-graded if L=⊕r∈ZsLr and [Lp,Lq]⊂Lp+q. It is easy to see that the algebras Lλs considered here are Zs-graded with (Lλs)k=CLk⊕CIk⊕CGk⊕CHk (k∈Z) for the case s=0 and (Lλs)k=CLk⊕CIk, (Lλs)21+k=CG21+k⊕CH21+k (k∈Z) for the case s=21.
Denote
[TABLE]
in which r∈Zs. An element d∈Derr(Lλs) is called a homogeneous derivation of degree r, usually denoted as dr. Similarly, one can give the definition of Innr(Lλs) and a homogeneous inner derivation of degree r. Then {\rm Der}({{\mathfrak{L}^{s}_{\lambda}}})\!=\!\mbox{\prod\limits_{r\in{\mathbb{Z}_{s}}}}{\rm Der}_{r}({{\mathfrak{L}^{s}_{\lambda}}}). It should be noticed that for any d=∑r∈Zsdr∈Der(Lλs), the formal sum on the right hand side may be infinite while for any x∈Lλs, d(x)=∑r∈Zsdr(x), the nonzero summands on the right hand side must be finite since Lλs are vector spaces.
A bijective linear map φ: Lλs→Lλs is called an automorphism of Lλs if it satisfies:
[TABLE]
for any x,y∈Lλs,i∈{0,1}.
Denote by Aut(Lλs) the set of all automorphisms of Lλs. It is easy to verify Aut(Lλs) is a group with the multiplication defined by the composition of linear maps.
2 Derivation algebra
The derivation algebra of Lλs can be formulated as the following theorem.
Theorem 2.1**.**
Der(Lλs)=Inn(Lλs)⊕C∂2⊕C∂4⊕δλ,0C∂1⊕δλ,0C∂3,
where ∂1, ∂2 are defined by (2.17) and ∂3, ∂4 are defined by (2.89).
Proof of Theorem 2.1 This theorem follows from Lemmas 2.2 to 2.9. □
Lemma 2.2**.**
Der0(Lλ0)∩Derk(Lλ0)=Inn0(Lλ0)∩Innk(Lλ0), ∀k∈Z∗.
Proof.
For any d0,k∈Der0(Lλ0)∩Derk(Lλ0), we can suppose
[TABLE]
The action of d0,k on both sides of [Ln,Lm]=(n−m)Ln+m gives
[TABLE]
When n=0, we have
[TABLE]
Consider the action of d0,k on both sides of [Ln,Im]=(n−m)In+m, we have
[TABLE]
Taking n=0 and recalling (2.7), we have
[TABLE]
Acting d0,k on both sides of [Ln,Hm]=(2n−m)Hn+m, one has
[TABLE]
When n=0, together with (2.7), one can deduce
[TABLE]
Acting d0,k on both sides of [Ln,Gm]=(2n−m)Gn+m+λ(n+1)Hm+n, we have
[TABLE]
Taking n=0 and recalling (2.7), (2.9), we have
[TABLE]
According to the identities from (2.7) to (2.10), we can deduce
[TABLE]
It is easy to verify that d0,k=\mboxad(kaL0,kLk+kbL0,kIk), which implies
Der0(Lλ0)∩Derk(Lλ0)=Inn0(Lλ0)∩Innk(Lλ0) for any k∈Z∗.
∎
Define ∂1 and ∂2 as follows:
[TABLE]
Then ∂1∈Der0(L00)∩Der0(L00) and ∂2∈Der0(Lλ0)∩Der0(Lλ0) for any λ∈C.
Lemma 2.3**.**
Der0(Lλ0)∩Der0(Lλ0)=Inn0(Lλ0)∩Inn0(Lλ0)⊕Cδλ,0∂1⊕C∂2.
Proof.
For any d0,0∈Der0(Lλ0)∩Der0(Lλ0), we can suppose
[TABLE]
Acting d0,0 on both sides of [Ln,Lm]=(n−m)Ln+m, we obtain
[TABLE]
which implies the following identities
[TABLE]
Then we can deduce
[TABLE]
which together with aL−k,0+aLk,0=aL0,0 gives aL0,0=0. Then
[TABLE]
which together with aL1,0=aL2,0+aL−1,0 gives aL2,0=2aL1,0. Combining (2.26) and (2.27), we know aLk,0=kaL1,0 for k∈Z. Similarly, we can deduce
[TABLE]
Taking the action of d0,0 on both sides of [Ln,Im]=(n−m)In+m, we have
[TABLE]
from which we can deduce
[TABLE]
Using [Ln,Hm]=(2n−m)Hn+m and (1.2), we obtain
[TABLE]
from which we can deduce
[TABLE]
Using [Ln,Gm]=(2n−m)Gn+m+λ(n+1)Hn+m and (1.2), we obtain
[TABLE]
Letting m=0 in the first equation in (2.35), one can obtain
[TABLE]
Letting n=0 in the second equation in (2.35), one can obtain
[TABLE]
Letting m=0 in the second equation in (2.35), one can obtain
[TABLE]
Acting d0,0 on both sides of [In,Gm]=(n−2m)Hn+m, we have
[TABLE]
which imply
[TABLE]
Using [Gn,Gm]=Im+n and (1.2), we obtain
[TABLE]
which gives
[TABLE]
Combining the identities obtained in (2.28)–(2.40), we have deduced the following
[TABLE]
It is easy to see that d0,0=\mboxad(−aL1,0L0−bL1,0I0)+dG0,0∂2+δλ,0cG0,0∂1, which implies
Der0(Lλ0)∩Der0(Lλ0)=Inn0(Lλ0)∩Inn0(Lλ0)⊕Cδλ,0∂1⊕C∂2.
∎
Lemma 2.4**.**
Der1(Lλ0)∩Derk(Lλ0)=Inn1(Lλ0)∩Innk(Lλ0), ∀k∈Z∗.
Proof.
For any d1,k∈Der1(Lλ0)∩Derk(Lλ0), one can suppose
[TABLE]
Using [Ln,Lm]=(n−m)Lm+n and (1.2), we obtain
[TABLE]
Taking n=0, we obtain
[TABLE]
Using [Ln,Im]=(n−m)Im+n and (1.2), we get
[TABLE]
Taking n=0 in (2.49), we obtain
[TABLE]
Acting d1,k on both sides of [Ln,Hm]=(2n−m)Hm+n, we have
[TABLE]
Taking n=0, we obtain
[TABLE]
Using [Ln,Gm]=(2n−m)Gm+n+λ(n+1)Hm+n, we obtain
[TABLE]
Taking n=0 and combining (2.52) with (2.49), we obtian
[TABLE]
Recalling (2.49)–(2.54), one can deduce
[TABLE]
It is easy to verify that
d1,k=\mboxad(kcL0,kGk+(kdL0,k+k2λcL0,k)Hk), which implies Der1(Lλ0)∩Derk(Lλ0)=Inn1(Lλ0)∩Innk(Lλ0) for any k∈Z∗.
∎
Lemma 2.5**.**
Der1(Lλ0)∩Der0(Lλ0)=Inn1(Lλ0)∩Inn0(Lλ0).
Proof.
For any d1,0∈Der1(Lλ0)∩Der0(Lλ0), we can suppose
[TABLE]
Using [Ln,Lm]=(n−m)Ln+m and (1.2), we obtain
[TABLE]
which give cL0,0=0 and
[TABLE]
Then the second equation in (2.65) becomes
[TABLE]
Furthermore, we can deduce
[TABLE]
Using [Ln,Im]=(n−m)In+m and (1.2), we obtain
[TABLE]
from which we can deduce
[TABLE]
Using [Ln,Hm]=(2n−m)Hm+n and (1.2), we have
[TABLE]
which imply
[TABLE]
Using [Ln,Gm]=(2n−m)Gm+n+λ(n+1)Hm+n and (1.2), we obtain
[TABLE]
which imply
[TABLE]
Using [In,Gm]=(n−2m)Hm+n and (1.2), we have
[TABLE]
which imply
[TABLE]
Using [Gn,Gm]=Im+n and (1.2), we have
[TABLE]
Together with (2.73), we can deduce
[TABLE]
which implies
[TABLE]
Combining the identities presented in (2.66)–(2.75), we can deduce
[TABLE]
It is easy to verify that d1,0=\mboxad(−2cL1,0G0+(8λcL1,0−2dL1,0)H0), which imply
Der1(Lλ0)∩Der0(Lλ0)=Inn1(Lλ0)∩Inn0(Lλ0).
∎
Lemma 2.6**.**
Der0(Lλ21)∩Derk(Lλ21)=Inn0(Lλ21)∩Innk(Lλ21)* for any k∈Z∗.*
Proof.
For any d0,k∈Der0(Lλ21)∩Derk(Lλ21), we can suppose
[TABLE]
Acting d0,k on both sides of [Ln,Lm]=(n−m)Ln+m, we obtain
[TABLE]
Taking n=0, one has
[TABLE]
Using [Ln,Im]=(n−m)In+m and (2.5), we obtain
[TABLE]
which give
[TABLE]
Using [Ln,Hm+21]=(2n−1−m)Hn+m+21 and (2.80), we obtain
[TABLE]
which together with (2.81) give
[TABLE]
Using [Ln,Gm+21]=(2n−1−m)Gn+m+21+λ(n+1)Hm+n+21 and (2.80), we obtain
[TABLE]
which together with (2.81) and (2.82) give
[TABLE]
Combining the identities presented in (2.81)–(2.86), we obtain
[TABLE]
It is easy to verify that d0,k=\mboxad(kaL0,kLk+kbL0,kIk), which implies Der0(Lλ21)∩Derk(Lλ21)=Inn0(Lλ21)∩Innk(Lλ21) for any k∈Z∗.
∎
Define ∂3 and ∂4 as follows:
[TABLE]
It is easy to see that ∂3∈Der0(L021)∩Der0(L021) and ∂4∈Der0(Lλ21)∩Der0(Lλ21) for any λ∈C.
Lemma 2.7**.**
Der0(Lλ21)∩Der0(Lλ21)=Inn0(Lλ21)∩Inn0(Lλ21)⊕Cδλ,0∂3⊕C∂4.
Proof.
For any d0,0∈Der0(Lλ21)∩Der0(Lλ21), we can suppose
[TABLE]
Using [Ln,Lm]=(n−m)Ln+m and (1.2), we obtain
[TABLE]
which imply
[TABLE]
Then we can deduce
[TABLE]
which together with aL−k,0+aLk,0=aL0,0 give aL0,0=0 and
[TABLE]
According to aL1,0=aL2,0+aL−1,0 and (2.96), we know aL2,0=2aL1,0. Then aLk,0=kaL1,0 for k∈Z. Similarly, we can deduce
[TABLE]
Using [Ln,Im]=(n−m)In+m and (1.2), we obtain
[TABLE]
which together with (2.97) give
[TABLE]
Using [Ln,Hm+21]=(2n−1−m)Hn+m+21 and (1.2), we have
[TABLE]
which together with (2.97) give
[TABLE]
Using [Ln,Gm+21]=(2n−1−m)Gn+m+21+λ(n+1)Hn+m+21 and (1.2), we obtain
[TABLE]
from which we can deduce
[TABLE]
Acting d0,0 on both sides of [In,Gm+21]=(n−2m−1)Hn+m+21, we obtain
[TABLE]
from which we can deduce
[TABLE]
Using [Gn+21,Gm+21]=Im+n+1 and (1.2), we obtain
[TABLE]
which imply
[TABLE]
Combining the identities presented in (2.97)–(2.113), we can deduce
[TABLE]
It is easy to see that d0,0=\mboxad(−aL1,0L0−bL1,0I0)+dG21,0∂4+21δλ,0bI0,0∂3, which implies
[TABLE]
∎
Lemma 2.8**.**
We have Der1(Lλ21)∩Derk+21(Lλ21)=Inn1(Lλ21)∩Innk+21(Lλ21).
Proof.
For any d1,k∈Der1(Lλ21)∩Derk+21(Lλ21), we can suppose
[TABLE]
Using [Ln,Lm]=(n−m)Ln+m and (1.2), we have
[TABLE]
which imply
[TABLE]
Acting d1,k+21 on both sides of [Ln,Im]=(n−m)In+m, we obtain
[TABLE]
from which we can deduce
[TABLE]
Using [Ln,Hm+21]=(2n−m)Hn+m+21 and (1.2), we obtain
[TABLE]
which imply
[TABLE]
Using [Ln,Gm+21]=(2n−m)Gn+m+21+λ(n+1)Hm+n+21 and (1.2), we obtain
[TABLE]
from which we can deduce
[TABLE]
By (2.126)–(2.131), one has
d1,k+21=\mboxad(k+211cL0,k+21Gk+21+((k+21)2λcL0,k+21+k+21dL0,k+21)Hk+21),
which implies Der1(Lλ21)∩Derk+21(Lλ21)=Inn1(Lλ21)∩Innk+21(Lλ21).
∎
For any d1,0∈Der1(Lλ21)∩Der0(Lλ21), we have d1,0=0, which implies the following lemma.
Lemma 2.9**.**
We have Der1(Lλ21)∩Der0(Lλ21)=Inn1(Lλ21)∩Inn0(Lλ21)={0}.
3 Automorphism groups
Denote by Aut(Lλs) the automorphism group of Lλs and I the subgroup generated by {exp(α\mboxadIk)∣α∈C,k∈Z}. Denote by σ the linear map from Lλs to Lλs satisfying
[TABLE]
where ϵ=1 or −1, k∈Z, μ,α,βϵ,γ∈C, αμ=0 and x2=α2sμ3. Denote by G the set generated (via composition of linear maps) by all such σ.
Theorem 3.1**.**
(1) I is an abelian normal subgroup of Aut(Lλs).
(2) If λ=0, then G is a subgroup of Aut(Lλs), and Aut(Lλs)=I⋊G.
(3) If λ=0, μ=ϵ=1, then G is a subgroup of Aut(Lλs), and Aut(Lλs)=I⋊G.
(4) If λ=0, ϵ=−1, then Aut(Lλs)=I.
Before proving Theorem 3.1, we first give several lemmas.
Lemma 3.2**.**
Denote by I the vector space spanned by {Ik∣k∈Z} over C, we have σ(Ik)∈I.
Proof.
It is trivial according to [Gp,Gq]=Ip+q and (1.5).
∎
Lemma 3.3**.**
Replacing σ with στ for some suitable τ∈I if necessary, and still denote as σ, one can assume
[TABLE]
where α,μ∈C, α=0, μ=0, ϵ=1 or −1, βϵ∈C.
Proof.
The restriction of σ on (Lλs)0 is an element in Aut(W(a,b)) for a=0, b=−1 in [7]. This lemma follows immediately from Theorem 5.2 of [7].
∎
Lemma 3.4**.**
Denote by H the vector space spanned by {Hp∣p∈Zs}. Then σ(Hp)∈H.
Proof.
Considering the actions of σ both sides of [In,Gp]=(n−2p)Hn+p and using Lemma 3.2, we obtain this lemma.
∎
Now we can assume σ(Hp)=∑q∈ZsdHp,HqHq and σ(Gp)=∑q∈Zs(cGp,GqGq+dGp,HqHq). We give the definition of level of elements. Take x=∑p=btapHp∈H,(atab=0) for example: we say t is the top level of x and b is the bottom level of x. By considering the top level and bottom level of [σ(Lm),σ(Hp)]=(2m−p)σ(Hm+p) and [σ(Ln),σ(Gp)]=(2n−p)σ(Gp+n)+λ(n+1)σ(Hn+p), together with Lemmas 3.3 and 3.4, we immediately have the following two lemmas.
Lemma 3.5**.**
σ(Hp)=dHp,HϵpHϵp.
Lemma 3.6**.**
σ(Gp)=cGp,GϵpGϵp+∑qdGp,HqHq.
Lemma 3.7**.**
One has the following identities:
[TABLE]
Proof.
Using [Lm,Hp]=(2m−p)Hm+p, we obtain
[TABLE]
which implies
[TABLE]
Using [Im,Gp]=(m−2p)Hm+p, one has
[TABLE]
from which we can deduce
[TABLE]
According to [Gp,Gq]=Ip+q, one has
[TABLE]
which together with (3.12) gives
[TABLE]
Acting σ on both sides of [L0,Gp]=−pGp+λHp, we obtain
[TABLE]
Compare the top and bottom level of both sides, one can deduce
[TABLE]
Using [Lm,Gp]=(2m−p)Gm+p+λ(m+1)Hm+p, one has
[TABLE]
Taking m=2p in (3.17), one has
[TABLE]
which implies
[TABLE]
According to (3.18), (3.17) becomes
[TABLE]
from which we can deduce dGk+s,Hϵ(k+s)=αkdGs,Hϵs+2kϵαkβϵ(k−2s)cGs,Gϵs.
∎
Proof of Theorem 3.1 (1) For any σ∈Aut(Lλs), we know σ(\mboxadIk)σ−1=\mboxadσ(Ik). According to Lemma 3.2, we know that I is an abelian normal subgroup of Aut(Lλs). And Theorem 3.1 (2)–(4) follows from Lemmas 3.3–3.7 immediately. □
