# The Classification, Automorphism Group, and Derivation Algebra of the Loop Algebra Related to the Nappi–Witten Lie Algebra

Authors: Xue Chen
Journal: Journal of Mathematics
 (2021)

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## Abstract

SetL≔H4⊗ℂR, R≔ℂt±1, and S≔ℂt±1/mm∈ℤ+. Then, L is called the loop Nappi–Witten Lie algebra. R-isomorphism classes of S/R forms of L are classified. The automorphism group and the derivation algebra of L are also characterized.


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## Body

## 1. Introduction

The conformal field theory (CFT) plays a significant role in many different areas of mathematics and physics. Wess–Zumino–Novikov–Witten (WZNW) models [1] provide interesting examples of CFTs. WZNW models were first studied in the setting of semisimple groups. Later, since the WZNW models based on non-abelian nonsemisimple Lie groups were found to be closely related to the construction of exact string backgrounds, the special types received some attention [2, 3]. Nappi and Witten showed in [3] that the WZNW model (NW model) based on a central extension of the two-dimensional Euclidean group describes the homogeneous four-dimensional space-time corresponding to a gravitational plane wave. The related Lie algebra is called the Nappi–Witten Lie algebra, which is neither abelian nor semisimple. More results on the NW model were given in [4–7].The Nappi–Witten Lie algebraH4 is a four-dimensional Lie algebra over ℂ generated by a,b,c,d with the following Lie brackets:(1)a,b=c,d,a=a,d,b=−b,c,H4=0.There is a nondegenerate invariant symmetric bilinear form, on H4 defined by(2)a,b=1,c,d=1,otherwise,,=0.Set(3)L≔H4⊗Rℂ,R≔ℂt±1,S≔ℂt±1/m,m∈ℤ+.Then,L is a Lie algebra over ℂ under the bracket as follows:(4)x⊗tn1,y⊗tn2=x,y⊗tn1+n2,forx,y∈H4,n1,n2∈ℤ.We callL the loop Nappi–Witten Lie algebra. Recently, the representation theory of Lie algebras related to L was studied in [8, 9].Reference [10] studied the automorphism groups of vertex operator algebras associated with the affine Nappi–Witten algebra H^4. The isomorphism of loop algebras was considered in [11] from the perspective of non-abelian Galois cohomology. Reference [12] gave an explicit description of the algebra of derivations for a large class of algebras defined by étale descent. Motivated by the works mentioned above, we study R-isomorphism classes of S/R forms of L, the automorphism group, and the derivation algebra of L in this paper.To classifyR-isomorphism classes of S/R forms of L in Section 2, we recall the results of automorphisms of H4 in [10]. The following Lemma 1, Theorem 1, Lemma 2, and Lemma 3 will be used in the proof of Theorem 2.Lemma 1.
(see [10]). Let θ∈AutH4. Then,(5)θa,b,c,d=a,b,c,ds00−sβ0s−10−s−1ααβ1−αβ0001,orθa,b,c,d=a,b,c,d0n0nξn−100n−1ρξρ−1ξρ000−1,where s,n∈ℂ∗, α,β,ξ,ρ∈ℂ.Define a linear mapτ of H4 by(6)τa=b,τb=a,τc=−c,τd=−d.It is clear thatτ2=idH4=e and τ∈AutH4. Set G0≔τ,e. Then, G0≅ℤ/2ℤ. Set(7)G1≔ℂ∗×ℂ×ℂ,G≔ℂ∗×ℂ×ℂ×ℤ2ℤ.The multiplication in groupG is defined as follows:(8)s,α,β,e⋅s′,α′,β′,e=ss′,αs′+α′,βs′−1+β′,e,n,ξ,ρ,τ⋅n′,ξ′,ρ′,τ=nn′−1,ρn′−1−ξ′,ξn′−ρ′,e,n,ξ,ρ,τ⋅s,α,β,e=ns−1,ξs−α,ρs−1−β,τ,s,α,β,e⋅n,ξ,ρ,τ=sn,βn−1+ξ,nα+ρ,τ,where s,s′,n,n′∈ℂ∗, α,β,α′,β′,ξ,ρ,ξ′,ρ′∈ℂ. The identity element is 1,0,0,e, and(9)s,α,β,e−1=s−1,−αs−1,−βs,e,n,ξ,ρ,τ−1=n,ρn−1,ξn,τ.Since(10)s,0,0,e⋅1,α,β,e⋅s,0,0,e−1=1,αs−1,βs,e,1,0,0,τ⋅s,α,β,e⋅1,0,0,τ−1=s−1,−β,−α,e,ℂ×ℂ is a normal subgroup of G1 and G1 is a normal subgroup of G. Thus,(11)G1=ℂ∗⋉ℂ×ℂ,G=ℂ∗⋉ℂ×ℂ⋊ℤ2ℤ.Theorem 1.
(see [10]). AutH4≅G=ℂ∗⋉ℂ×ℂ⋊ℤ/2ℤ.By Theorem1, any element of G can be viewed as an automorphism of H4.Lemma 2.
Letϕ=s,α,β,e and ψ=n,ξ,ρ,τ. Then,(12)ϕk=sk,αsk−1+sk−2+…+s+1,βs−k−1+s−k−2+…+s−1+1,e,ψ2k=1,kρn−1−ξ,kξn−ρ,e,ψ2k+1=n,k+1ξ−kρn−1,k+1ρ−knξ,τ,where s,n∈ℂ∗, α,β,ξ,ρ∈ℂ, and k∈ℤ+.Proof.
It follows from (8) immediately.Lemma 3.
.(1)
Ifn,ξ,ρ,τ2k=1,0,0,e for k∈ℤ+, then n,ξ,ρ,τ is conjugate to 1,0,0,τ in AutH4.(2)
Lets,α,β,e≠s,0,0,e. Then, s,α,β,e is conjugate to s,0,0,e in AutH4 if and only if s≠1.(3)
Lets,s′≠1 and s≠s′. Then, s,0,0,e is conjugate to s′,0,0,e in AutH4 if and only if ss′=1.Proof.
.(1)
Ifn,ξ,ρ,τ2k=1,kρn−1−ξ,kξn−ρ,e=1,0,0,e, then ρ=ξn. Choose s,α,β,e such that s2=1/n and nα+β=−ρ. Then,(13)s,α,β,en,ξ,ρ,τs,α,β,e−1=s2n,α+ξ+βn−1s−1,nα+ρ+βs,τ=1,0,0,τ.Thus,n,ξ,ρ,τ is conjugate to 1,0,0,τ in AutH4.(2)
We assume that there existss′,α′,β′,e such that(14)s′,α′,β′,es,α,β,es′,α′,β′,e−1=s,0,0,e.By (8) and (9), we have(15)α′s−1+α=0,β′s−1−1+β=0.Supposes=1. Then, α=β=0 by (15) is a contradiction. Thus, s≠1. Conversely, if s≠1, by (15), α′=α/1−s and β′=β/1−s−1. Then, there exists s′,α′,β′,es′∈ℂ∗ such that (14) holds. So, s,α,β,e is conjugate to s,0,0,e in AutH4.(3)
Since(16)n,ξ,ρ,τs,0,0,en,ξ,ρ,τ−1=s−1,s−1−1ρn−1,s−1ξn,e,s″,α″,β″,es,0,0,es″,α″,β″,e−1=s,s−1α″s″−1,s−1−1β″s″,e,it is clear that s,0,0,e is conjugate to s′,0,0,e if and only if ss′=1.The aim of this paper is to study structures of the loop Nappi–Witten Lie algebraL=H4⊗ℂR. Some ideas we use come from [10–12]. This paper is organized as follows. In Section 2, we classify R-isomorphism classes of S/R forms of L by the first cohomology set. In Section 3, we determine the automorphism group of L. Finally, the derivation algebra of L is also characterized in Section 4.Throughout the paper, the sets of the complex numbers, the nonzero complex numbers, the nonnegative integers, the integers, and the positive integers are denoted byℂ, ℂ∗, ℕ, ℤ, and ℤ+, respectively.


## 2. The First Cohomology SetH1Γ,AutSH4S

We first review some basic notions and assumptions in [11] and then apply them in the class of H4.Letk be an algebraically closed field of characteristic 0. For each m∈ℤ+, we choose a primitive mth root of unity ζm such that these primitive roots of unity are compatible in the sense that ζlml=ζm for all l, m∈ℤ+, and fix the sequence ζmm∈ℤ+ from now on. Suppose that g is a Lie algebra over k and that σ is an automorphism of g of period m (the order of σ might not be equal to m, but of course it is a divisor of m).Let(17)Γ≔ℤmℤ=i¯:i∈ℤ,be the group of integers modulo m, where i¯ denotes the congruence class represented by i. Let(18)R≔kt±1,S≔kt±1/m,be two algebras of Laurent polynomials over k and R⊂S is a k-subalgebra. If we set z=t1/m, then t=zm and(19)R=kz±m,S=kz±1.It is easy to see thatS is an R-algebra.There is an isomorphism ofΓ onto AutRS (the group of automorphisms of S which fixes R) such that(20)i¯↦i¯,andi¯z=ζmiz.LetK∈R,S. Set(21)gK≔g⊗Kk,where gK is a K-module with K-action given by fx⊗h=x⊗fh for x∈g and f,h∈K. Note that gK is also a Lie algebra over K under the bracket as follows:(22)x⊗f,y⊗h=x,y⊗fh,forx,y∈g,f,h∈K.We now recall the definition of the loop algebra ofg relative to σ. Let(23)gi¯=x∈g|σx=ζmix.The loop algebra ofg relative to σ is the subalgebra(24)Lg,σ≔⊕i∈ℤgi¯⊗zki,of the Lie algebra(25)gS=g⊗Sk.Note thatLg,σ is the set of fixed points in gS of the R-algebra automorphism σ⊗1¯−1, and then Lg,σ is an R-subalgebra of gS. Thus, Lg,σ is an R-algebra and hence also a k-algebra. Furthermore, the R-isomorphism class of Lg,σ does not depend on the choice of the period m for σ (see Lemma 2.3 in [11]).To interpret loop algebras asS/R forms of the R-algebra gR, we review the definitions of the S/R form of gR and a certain (non-abelian) first cohomology set and relations between them.AnS/Rform of gR is a Lie algebra ℱ over R such that(26)ℱ⊗SR≅gS,as Lie S-algebras.We now recall some concepts of a (non-abelian) cohomology set.LetG be a group. We say that Γ-acts on G if there is a map i¯,x↦xi¯ of Γ×G⟶G satisfying xj¯i¯=xi¯+j¯, 0¯x=x, and xyi¯=xi¯⋅yi¯ for i¯,j¯∈Γ, x,y∈G.IfΓ acts on a group G, the map u:Γ⟶G, i¯↦ui¯ satisfying(27)ui¯+j¯=ui¯ui¯j¯,is called 1-cocycle, where ui¯j¯ denotes the Γ-action on G. Two 1-cocycles u,v are said to be cohomologous if there exists some f∈G such that(28)vi¯=f−1ui¯fi¯,for all i¯∈Γ. It is an equivalence relation. For every 1-cocycle u, we denote by u the equivalence class containing u. The set of equivalence classes (called cohomology classes) of 1-cocycles from Γ to G is denoted by H1Γ,G, and we call it the (non-abelian) first cohomology set.Define an action ofΓ on AutSgS by(29)τi¯≔id⊗i¯τid⊗i¯−1,for i¯∈Γ and τ∈AutSgS, where AutSgS is the group of automorphisms of gS as a Lie S-algebra.The following result comes from Proposition 3.4 of [11].Proposition 1.
(see [11]). R-isomorphism classes of S/R forms of gR are in one-to-one correspondence with cohomology classes in H1Γ,AutSgS. Explicitly this correspondence is as follows: if u is a 1-cocycle on Γ in AutSgS, then the cohomology class u corresponds to the R-isomorphism classes of the S/R form:(30)gSu≔x∈gS:ui¯id⊗i¯x=xfori¯∈Γ,of gS.Furthermore, Theorem 3.6 in [11] gives our description of loop algebras as S/R forms of gR. In view of Theorem 3.6 in [11], the loop algebra Lg,σ is an S/R-form of gR and is determined by the 1-cocycle:(31)u:Γ⟶AutSgS,i¯⟼σ−i⊗id.So, we can writeLg,σ=gSu, where(32)ui¯=σ−i⊗id,fori¯∈Γ.In the following, setk=ℂ and g=H4. Let S and R be the algebras of Laurent polynomials over ℂ as in (19). We will classify loop algebras of H4 from the view of S/R forms of H4R.Lemma 4.
Letσ,τ∈AutH4 and σm=τm=idH4. We define 1-cocycles u and v on Γ in AutSH4S by(33)ui¯=σ−i⊗id,andvi¯=τ−i⊗id,∀i¯∈Γ.
Ifσ is conjugate to τ in AutH4, then u=v.Proof.
Suppose that there existsh∈AutH4 such that h−1σh=τ. Then, h−1σ−ih=τ−i, ∀i∈ℤ. Set f=h⊗id, then f∈AutSH4S. For any i¯∈Γ and x⊗zn∈H4S, we have(34)f−1ui¯fi¯x⊗zn=h−1⊗idσ−i⊗idid⊗i¯h⊗idid⊗i¯−1x⊗zn=h−1σ−ihx⊗zn=τ−ix⊗zn=vi¯x⊗zn.
Thus,u=v.Theorem 2.
LetΓ=ℤ/mℤ.(i)
Then, there are automorphisms(35)ϕm,r≔ζmr,0,0,e|0≤r≤m2,ψ≔1,0,0,τ,of periodm of H4 in the sense of conjugation. Furthermore, if m∈2ℤ+1, ψ does not exist.(ii)
If we define 1-cocyclesum,r, v on Γ in AutSH4S by(36)um,ri¯=ϕm,r−i⊗id,vi¯=ψ−i⊗id,∀i¯∈Γ,0≤r≤m2,then the corresponding R-isomorphism classes of the S/R form of H4R are as follows:(37)H4Sum,r=Spanℂa⊗zmk+r,b⊗zmk+m−r,c⊗zmk,d⊗zmk|k∈ℤ,0≤r≤m2,H4Sv=Spanℂa−b⊗z2k+1,c⊗z2k+1,d⊗z2k+1,a+b⊗z2k|k∈ℤ.
Furthermore, ifm∈2ℤ+1, neither v nor H4Sv exists.Proof.
.(i)
Letψ=n,ξ,ρ,τ∈AutH4 such that ψm=1,0,0,e. Then, m∈2ℤ+ by Lemma 2, and ψ is conjugate to 1,0,0,τ in AutH4 by Lemma 3(1).Letφ=s,α,β,e∈AutH4 such that(38)φm=sm,αsm−1+sm−2+…+s+1,βs−m−1+s−m−2+…+s−1+1,e=1,0,0,e.Then,(39)φm,r=ζmr,α,β,e,1≤r≤m−1,φm,0=1,0,0,e.For any1≤r≤m−1, φm,r=ζmr,α,β,e is conjugate to ϕm,r=ζmr,0,0,e by Lemma 3(2). Furthermore, ϕm,r=ζmr,0,0,e|0≤r≤m/2 are not conjugate with each other by Lemma 3(3) and clearly not conjugate to 1,0,0,τ.(ii)
It follows from Proposition1 and Lemma 4 by straightforward computation.


## 3. The Automorphism GroupAutℂL of L

In this section, we study the automorphisms ofL, where L=H4⊗ℂR, R=ℂt±1. Let AutRL denote the group of all R-linear automorphisms of L.We first recall the definition of the centroid in [11] that will be used later. Let K be a unital commutative associative algebra over ℂ and T be a Lie algebra over K. Set(40)CtdKT=χ∈EndK−modT|χx,y=x,χyfor allx,y∈T,where CtdKT is called the centroid of T over K. Define λT:K⟶EndK−modT by(41)λTrx=rx,forr∈Kandx∈T.It is clear thatλTK⊆CtdKT. We say that T is central over K if CtdKT=λTK (see [11]).Lemma 5.
Letχ∈CtdℂH4. Then,(42)χx=λ1x,χd=λ1d+ρc,x∈a,b,c,λ1,ρ∈ℂ.Proof.
Letχ∈CtdℂH4. We can assume that(43)χa=λ1a+μ1b,χb=λ2a+μ2b,where λi, μi∈ℂ, i=1,2. Since(44)χa=χd,a=d,χa=χd,a,−χb=χd,b=d,χb=χd,b,χc=χa,b=a,χb=χa,b,we have μ1=λ2=0 and λ1=μ2. Thus,(45)χa=λ1a,χb=λ1b,χc=λ1c,χd=λ1d+ρc,λ1,ρ∈ℂ.Theorem 3.
Ifχ∈CtdℂL, then(46)χx⊗tn=λ0χtx⊗tn,χd⊗tn=λ0χtd⊗tn+αnχtc⊗t0,where x∈a,b,c, λ0χt, αnχt∈R, and n∈ℤ.Proof.
We assume that(47)χa⊗tn=λnχta⊗t0+μnχtb⊗t0,χb⊗tn=δnχta⊗t0+ρnχtb⊗t0,where λnχt,μnχt,δnχt,ρnχt∈R. By(48)d⊗t0,χa⊗tn=χa⊗tn=d⊗tn,χa⊗t0=χd⊗tn,a⊗t0,d⊗t0,χb⊗tn=−χb⊗tn=d⊗tn,χb⊗t0=χd⊗tn,b⊗t0,χa⊗tn,b⊗t0=χc⊗tn=a⊗tn,χb⊗t0,we have(49)μnχt=δnχt=0,λnχt=λ0χttn,ρnχt=ρ0χttn,λ0χt=ρ0χt.
So,(50)χa⊗tn=λ0χta⊗tn,χb⊗tn=λ0χtb⊗tn,χc⊗tn=λ0χtc⊗tn,χd⊗tn=λ0χtd⊗tn+αnχtc⊗t0,λ0χt,αnχt∈R.Corollary 1.
CtdℂL≅R×HomL/L,L,ZL as vector spaces over ℂ, where ZL is the center of L.Proof.
Define the map(51)τ:CtdℂL⟶R×HomLL,L,ZL,χ↦λ0χt,φχ,and φχd⊗tn=αnχtc⊗t0 for any n∈ℤ. It is easy to see that τ is a vector space isomorphism.Corollary 2.
Ifχ∈CtdRL, then(52)χx⊗tn=λ0χtx⊗tn,χd⊗tn=λ0χtd⊗tn+α0χtc⊗tn,where x∈a,b,c, λ0χt, α0χt∈R, and CtdRL is commutative.Proof.
We just need to prove the commutativity ofCtdRL. Let χ1,χ2∈CtdRL. By Theorem 3, it is clear that(53)χ1χ2x⊗tn=χ1λ0χ2tx⊗tn=λ0χ2tλ0χ1tx⊗tn=χ2χ1x⊗tn,for x∈a,b,c and n∈ℤ. Moreover,(54)χ1χ2d⊗tn=χ1λ0χ2td⊗tn+α0χ2tc⊗tn=λ0χ2tλ0χ1td⊗tn+α0χ1tc⊗tn+α0χ2tλ0χ1tc⊗tn=λ0χ1tλ0χ2td⊗tn+α0χ1tλ0χ2t+α0χ2tλ0χ1tc⊗tn=χ2χ1d⊗tn.Corollary 3.
CtdRL=CtdRH4⊗Rℂ≅CtdℂH4⊗Rℂ as algebras over R.Proof.
By Lemma5, we can define φ,ψ∈CtdℂH4 by(55)φx=λx,x∈a,b,c,d,ψy=0,y∈a,b,c,ψd=ρc,λ,ρ∈ℂ.
Define the map(56)η:CtdRL⟶CtdℂH4⊗ℂR,χ↦φ⊗λ0χtλ+ψ⊗α0χtρ,where λ,ρ∈ℂ∗, λ0χt, α0χt∈R. It is easy to check that η is an R-algebra isomorphism.Remark 1.
Corollary3 can also follow from Lemma 5 in [12].Lemma 6.
Ifθ∈AutℂL, then(57)θx⊗tk=λ0λktθx⊗t0,x∈a,b,c,θd⊗tk=λ0λktθd⊗t0+δkt−λ0λktδ0tc⊗t0,where λkt, δkt∈R, k∈ℤ, and λk1+k2t=λ0λk1tλk2t, λ0≔λ0t∈±1.Proof.
Letθ∈AutℂL. We can assume that(58)θa⊗tk=skta⊗t0+mktb⊗t0+αktc⊗t0,θb⊗tk=nkta⊗t0+lktb⊗t0+βktc⊗t0,θd⊗tk=λktd⊗t0+μk1ta⊗t0+μk2tb⊗t0+δktc⊗t0,where skt, mkt, αkt, nkt, lkt, βkt, λkt, μk1t, μk2t, δkt∈R. By(59)θd⊗t0,θa⊗tk=θa⊗tk=θd⊗tk,θa⊗t0,θd⊗t0,θb⊗tk=−θb⊗tk=θd⊗tk,θb⊗t0,we deduce that (60)λ0t=1,mkt=nkt=0,skt=λkts0t,αkt=λktα0t,lkt=λktl0t,βkt=λktβ0t,μk1t=λktμ01t,μk2t=λktμ02t,or(61)λ0t=−1,skt=lkt=0,mkt=−λktm0t,αkt=−λktα0t,nkt=−λktn0t,βkt=−λktβ0t,μk1t=−λktμ01t,μk2t=−λktμ02t.
By (60) and (61) and θc⊗tk=θa⊗tk,θb⊗t0, we have(62)λ0t=1,θa⊗tk=λkts0ta⊗t0+α0tc⊗t0=λktθa⊗t0,θb⊗tk=λktl0tb⊗t0+β0tc⊗t0=λktθb⊗t0,θc⊗tk=λkts0tl0tc⊗t0=λktθc⊗t0,θd⊗tk=λktd⊗t0+λktμ01ta⊗t0+λktμ02tb⊗t0+δktc⊗t0=λktθd⊗t0+δkt−λktδ0tc⊗t0,or(63)λ0t=−1,θa⊗tk=−λktm0tb⊗t0+α0tc⊗t0=−λktθa⊗t0,θb⊗tk=−λktn0ta⊗t0+β0tc⊗t0=−λktθb⊗t0,θc⊗tk=λktm0tn0tc⊗t0=−λktθc⊗t0,θd⊗tk=λktd⊗t0−λktμ01ta⊗t0−λktμ02tb⊗t0+δktc⊗t0=−λktθd⊗t0+δkt+λktδ0tc⊗t0.
Since(64)θd⊗tk1,θa⊗tk2=θa⊗tk1+k2=λ0tλk1+k2tθa⊗t0=λ0tλk1tθd⊗t0+δk1t−λ0tλk1tδ0tc⊗t0,λ0tλk2tθa⊗t0=λ02tλk1tλk2tθa⊗t0,we have λk1+k2t=λ0tλk1tλk2t. The lemma is proved.Define a mapπ:R×HomL/L,L,ZL⟶R by(65)πr,φ=r,for r∈R, φ∈HomL/L,L,ZL. Furthermore, r,0x⊗tk=x⊗rtk for x⊗tk∈L.Lemma 7.
.(1)
Letθ∈AutℂL. Then, θtn,0θ−1∈CtdℂL and πθtn,0θ−1=λ0λnt;(2)
Givenθ∈AutℂL. The linear map ∧:AutℂL⟶AutℂR is defined by the following identity:(66)θctn=λ0λnt⇔πθtn,0θ−1=λ0λnt.
Then,θc∈AutℂR.Proof.
.(1)
For anyX,Y∈L, we have(67)θtn,0θ−1X,Y=θtn,0θ−1X,θ−1Y=θtn,0θ−1X,Y.Then,θtn,0θ−1∈CtdℂL. By Lemma 6, for x∈a,b,c, we have(68)θtn,0θ−1θx⊗tk=θtn,0x⊗tk=θx⊗tn+k=λ0λn+ktθx⊗t0=λ02λntλktθx⊗t0=λ0λntθx⊗tk,θtn,0θ−1θd⊗tk=θd⊗tn+k=λ0λn+ktθd⊗t0+δn+kt−λ0λn+ktδ0tc⊗t0=λ02λntλktθd⊗t0+δn+kt−λ02λntλktδ0tc⊗t0=λ0λntθd⊗tk+δn+kt−λ0λntδtc⊗t0.Then, by (62)-(63) and Theorem 3, we deduce that πθtn,0θ−1=λ0λnt.(2)
By the definition ofθc and Lemma 6, for x∈a,b,c, we have(69)θtn1+tn2,0θ−1θx⊗tk=θtn1+tn2,0x⊗tk=θx⊗tn1+k+tn2+k=λ0λn1t+λn2tθx⊗tk=θctn1+θctn2θx⊗tk,θtn1+tn2,0θ−1θd⊗tk=θd⊗tn1+k+θd⊗tn2+k=λ0λn1tθd⊗tk+δn1+kt−λ0λn1tδktc⊗t0+λ0λn2tθd⊗tk+δn2+kt−λ0λn2tδktc⊗t0=θctn1+θctn2θd⊗tk+∑i=12δni+kt−λ0λnitδktc⊗t0.
Thus,θctn1+tn2=θctn1+θctn2. Similarly, we can get θcμtn=μθctn for any μ∈ℂ. Furthermore, we have(70)θctn1⋅tn2=λ0λn1+n2t=λ02λn1tλn2t=θctn1⋅θctn2.
In the following, we check thatθc−1=θ−1c. By θ∈AutℂL and Lemma 6, we can assume that(71)θ−1c⊗tk=λ0′λk′tθ−1c⊗t0,=λk′ts0′tl0′tc⊗t0=λk′tθ−1c⊗t0,ifλ0′=1,λk′tm0′tn0′tc⊗t0=−λk′tθ−1c⊗t0,ifλ0′=−1,where k∈ℤ, λk′t, s0′t, l0′t, m0′t, n0′t∈R\0, λ0′≔λ0′t∈±1, and λk1+k2′t=λ0′λk1′tλk2′t. Furthermore, it is clear that θ−1tn,0θ∈CtdℂL and πθ−1tn,0θ=λ0′λn′t. Then,(72)θ−1ctn=λ0′λn′t.
Thus, we haveθ−1c⊗tk=θ−1ctkθ−1c⊗t0. For any k∈ℤ, we have(73)c⊗tk=θθ−1c⊗tk=θθ−1ctkθ−1c⊗t0=θc⊗θ−1ctks0′tl0′t,ifλ0′=1,−θc⊗θ−1ctkm0′tn0′t,ifλ0′=−1,=θcθ−1ctks0′tl0′tθc⊗t0,ifλ0′=1,−θcθ−1ctkm0′tn0′tθc⊗t0,ifλ0′=−1,=θcθ−1ctk⋅θcs0′tl0′tθc⊗t0,ifλ0′=1,−θcθ−1ctk⋅θcm0′tn0′tθc⊗t0,ifλ0′=−1,=θcθ−1ctkc⊗t0.
So, we deduce thatθcθ−1ctk=tk. Similarly, we have θ−1cθctk=tk for any k∈ℤ. Therefore, θc−1=θ−1c and θc∈AutℂR.Theorem 4.
For the linear map∧:AutℂL⟶AutℂR with θ↦θc, we have(1)
∧ is a group homomorphism(2)
∧ is surjective(3)
Ker∧≅AutRL⋉ℂ⊗FR, where FR≔τ∈EndℂR|τtk=δkt−tkδ0tfor any k∈ℤ and EndℂR is the set of all linear transformations of RProof.
.(1)
Letθi∈AutℂL and λi0=1, i=1,2. By (62), we can assume that(74)θic⊗tn=λintsi0tli0tc⊗t0=λintθic⊗t0,i=1,2.By Lemma7(2), we have(75)θictn=λint,i=1,2.Then, for anyn∈ℤ,(76)θ1θ2c⊗tn=θ1θ2ctnθ2c⊗t0=θ1c⊗θ2ctns20tl20t=θ1cθ2ctns20tl20tθ1c⊗t0=θ1cθ2ctnθ1cs20tl20tθ1c⊗t0=θ1cθ2ctnθ1θ2c⊗t0.Thus,(77)θ1θ2ctn=θ1cθ2ctn,for anyn∈ℤ.So,θ1θ2c=θ1cθ2c. The assertion can be similarly checked for other cases besides λi0=1,i=1,2. Then, (1) holds.(2)
SinceAutℂR≅ℂ∗⋊ℤ/2ℤ, we have τ∈ℂ∗ or τ∈ℤ/2ℤ for any τ∈AutℂR. If τ∈ℂ∗, we may assume τtn=tn for any n∈ℤ. By Lemma 6, let θ∈AutℂL such that(78)θx⊗tn=tnθx⊗t0,x∈a,b,c,θd⊗tn=tnθd⊗t0+δnt−tnδ0tc⊗t0.Then,θctn=tn=τtn for any n∈ℤ. Thus, θc=τ. If τ is the generator of ℤ/2ℤ, we assume τt=t−1. Let θ∈AutℂL such that(79)θx⊗tn=t−nθx⊗t0,x∈a,b,c,θd⊗tn=t−nθd⊗t0+δnt−t−nδ0tc⊗t0.So,θctn=t−n=τtn for any n∈ℤ and then θc=τ. Therefore, ∧ is surjective.(3)
Letθ∈Ker∧. Then, θc=idR and tk=θctk=λ0λkt for any k∈ℤ. Thus, by Lemma 6, we have(80)θx⊗tk=tkθx⊗t0,x∈a,b,c,θd⊗tk=tkθd⊗t0+δkt−tkδ0tc⊗t0.
Ifδkt−tkδ0t=0 for any k∈ℤ, then θ∈AutRL. If δkt−tkδ0t≠0, define φ⊗τ∈HomH4/H4,H4,ZH4⊗FR by(81)φ⊗τd⊗tk=φd⊗τtk=c⊗δkt−tkδ0t,for anyk∈ℤ,where φd=c, φ⊗τd⊗tk=0 if φd=0, FR≔τ∈EndℂR|τtk=δkt−tkδ0t,for anyk∈ℤ, and EndℂR is the set of all linear transformations of R.
Letθ˜i∈AutRL. By (62)-(63), we can assume that(82)θ˜id⊗t0=λi0d⊗t0+μi01ta⊗t0+μi02tb⊗t0+δi0tc⊗t0,for λi0∈±1,i=1,2. Define the group AutRL⋉HomH4/H4,H4,ZH4⊗FR, whose multiplication is given by(83)θ˜1,φ1⊗τ1θ˜2,φ2⊗τ2=θ˜1θ˜2,λ20φ1⊗τ1+θ˜1φ2⊗τ2.
Furthermore,(84)θ˜,φ⊗τ1,0=θ˜,φ⊗τ,θ˜,φ⊗τ−1=θ˜−1,−λ0θ˜−1φ⊗τ,θ˜,01,φ⊗τθ˜,0−1=1,λ0θ˜φ⊗τ,where θ˜i,φi⊗τi,θ˜,φ⊗τ∈AutRL⋉HomH4/H4,H4,ZH4⊗FR, λi0,λ0∈±1, i=1,2.
Define a mapη:Ker∧⟶AutRL⋉HomH4/H4,H4,ZH4⊗FR by(85)θ↦θ˜,φ⊗τ,for θ˜∈AutRL, φ⊗τ∈HomH4/H4,H4,ZH4⊗FR, and(86)θ˜,φ⊗τx⊗tk=θ˜x⊗tk,θ˜,φ⊗τd⊗tk=θ˜d⊗tk+φ⊗τd⊗tk,for x∈a,b,c and any k∈ℤ. It is easy to see that η is well-defined and bijective. Let θi∈Ker∧ and ηθi=θ˜i,φi⊗τi, i=1,2. By (80)–(86), we have(87)θ1θ2d⊗tk=θ1tkθ˜2d⊗t0+φ2⊗τ2d⊗tk=θ1λ20d⊗tk+μ201ta⊗tk+μ202tb⊗tk+δ20tc⊗tk+θ˜1φ2⊗τ2d⊗tk=tkλ20θ˜1d⊗t0+λ20φ1⊗τ1d⊗tk+tkμ201tθ˜1a⊗t0+tkμ202tθ˜1b⊗t0+tkδ20tθ˜1c⊗t0+θ˜1φ2⊗τ2d⊗tk=tkθ˜1θ˜2d⊗t0+λ20φ1⊗τ1+θ˜1φ2⊗τ2d⊗tk.
Then,ηθ1θ2=θ˜1θ˜2,λ20φ1⊗τ1+θ˜1φ2⊗τ2=ηθ1ηθ2. Thus, η is a group isomorphism. Furthermore, since H4/H4,H4=ℂd and ZH4=ℂc, it is clear that HomH4/H4,H4,ZH4 is isomorphic to ℂ as vector spaces over ℂ. Then, (3) is proved.Corollary 4.
.(88)AutℂL≅AutℂR⋉Ker∧≅AutℂR⋉AutRL⋉ℂ⊗FR.Proof.
It follows from Theorem4.


## 4. The Derivation AlgebraDerℂL of L

In this section, we study the derivations ofL=H4⊗ℂR, where R=ℂt±1. Denote by DerRL the algebra of all R-linear derivations of L.Lemma 8.
Letg∈DerH4. Then,(89)ga=λ1a+δ1c,gb=μ2b+δ2c,gc=λ1+μ2c,gd=−δ2a−δ1b+δ3c,where λ1,μ2,δi∈ℂ, 1≤i≤3.Proof.
Forg∈DerH4, we can assume that(90)ga=λ1a+μ1b+δ1c,gb=λ2a+μ2b+δ2c,gd=αd+λ3a+μ3b+δ3c,where λi,μi,δi,α∈ℂ, 1≤i≤3. Since(91)ga=gd,a+d,ga,−gb=gd,b+d,gb,gc=ga,b+a,gb,we have(92)α=0,μ1=0,λ2=0,μ3=−δ1,λ3=−δ2.
Thus, the lemma is proved.Letdi∈DerH41≤i≤3 be such that(93)d1a=a,d1c=c,d1b=d1d=0,d2b=b,d2c=c,d2a=d2d=0,d3d=c,d3a=d3b=d3c=0.Theorem 5.
DerH4=ℂd1⊕ℂd2⊕ℂd3⊕ℂada⊕ℂadb.Proof.
For anyg∈DerH4, by Lemma 8, we have(94)g=λ1d1+μ2d2+δ3d3+δ2ada−δ1adb.
It is clear thatd1, d2, d3, ada, and adb are linearly independent.Lemma 9.
LetD∈DerℂL. Then,(95)Dx⊗tn=x⊗λ1nt−λ10ttn+tnDx⊗t0,Dd⊗tn=d⊗λ1nt−λ10ttn+tnDd⊗t0+c⊗δ3nt−δ30ttn,where x∈a,b,c, λ1nt, δ3nt∈R, and λ1m+nt=λ1nttm+λ1mttn−λ10ttm+n, m,n∈ℤ.Proof.
ForD∈DerℂL, we may assume that(96)Da⊗tn=λ1nta⊗t0+μ1ntb⊗t0+δ1ntc⊗t0,Db⊗tn=λ2nta⊗t0+μ2ntb⊗t0+δ2ntc⊗t0,Dd⊗tn=αntd⊗t0+λ3nta⊗t0+μ3ntb⊗t0+δ3ntc⊗t0,where λint, μint, δint, αnt∈R, i∈1,2,3. By(97)Da⊗tn=Dd⊗1,a⊗tn+d⊗1,Da⊗tn=Dd⊗tn,a⊗1+d⊗tn,Da⊗1,−Db⊗tn=Dd⊗1,b⊗tn+d⊗1,Db⊗tn=Dd⊗tn,b⊗1+d⊗tn,Db⊗1,Dc⊗tn=Da⊗1,b⊗tn+a⊗1,Db⊗tn=Da⊗tn,b⊗1+a⊗tn,Db⊗1,we have(98)α0t=0,μ1nt=0,λ2nt=0,δ1nt=δ10ttn,δ2nt=δ20ttn,αnt=λ1nt−λ10ttn=μ2nt−μ20ttn,μ3nt=−δ1nt,λ3nt=−δ2nt.
Thus,(99)Da⊗tn=λ1nta⊗t0+δ10ttnc⊗t0=a⊗λ1nt−λ10ttn+tnDa⊗t0,Db⊗tn=μ2ntb⊗t0+δ20ttnc⊗t0=b⊗λ1nt−λ10ttn+tnDb⊗t0,Dc⊗tn=λ1nt+μ20ttnc⊗t0=c⊗λ1nt−λ10ttn+tnDc⊗t0,Dd⊗tn=λ1nt−λ10ttnd⊗t0−δ20ttna⊗t0−δ10ttnb⊗t0+δ3ntc⊗t0=d⊗λ1nt−λ10ttn+tnDd⊗t0+c⊗δ3nt−δ30ttn.
Since(100)Da⊗tn+m=λ1m+nta⊗t0+δ10ttm+nc⊗t0=Dd⊗tn,a⊗tm+d⊗tn,Da⊗tm=a⊗λ1nttm+λ1mttn−λ10ttm+n+c⊗δ10ttm+n,we have λ1m+nt=λ1nttm+λ1mttn−λ10ttm+n. The lemma is proved.Theorem 6.
(i)
LetD∈DerℂL. Then, D,tn,0∈CtdℂL and πD,tn,0=λ1nt−λ10ttn, where π is defined in (65).(ii)
GivenD∈DerℂL. The linear map ρ:DerℂL⟶DerℂR is defined by the following identity: (101)ρDtn=λ1nt−λ10ttn⇔πD,tn,0=λ1nt−λ10ttn.
Then,(1)
ρ is a Lie algebra homomorphism(2)
ρ is surjective(3)
Kerρ≅DerRL+ℂ⊗ER, where ER≔τ∈EndℂR|τtn=δ3nt−δ30ttn,forδ3nt∈Rand anyn∈ℤ, EndℂR is the set of all linear transformations of RProof.
(i)
For anyX,Y∈L, we have(102)D,tn,0X,Y=Dtn,0X,Y−tn,0DX,Y=DtnX,Y−tn,0DX,Y+X,DY=DtnX,Y−tnDX,Y=D,tn,0X,Y.Then,D,tn,0∈CtdℂL. By Lemma 9, for x∈a,b,c and m∈ℤ, we have(103)D,tn,0x⊗tm=Dx⊗tn+m−tn,0Dx⊗tm=x⊗λ1n+mt−λ1mttn=λ1nt−λ10ttnx⊗tm,D,tn,0d⊗tm=Dd⊗tn+m−tn,0Dd⊗tm=d⊗λ1n+mt−λ1mttn+c⊗δ3n+mt−δ3mttn=λ1nt−λ10ttnd⊗tm+δ3n+mt−δ3mttnc⊗t0.Therefore, by Theorem3 and the definition of π in (65), we have πD,tn,0=λ1nt−λ10ttn.(ii)
By (i) and (101), it is easy to see that ρD is a linear transformation of R. For any n,m∈ℤ, we have(104)ρDtn+m=λ1n+mt−λ10ttn+m=λ1nttm+λ1mttn−2λ10ttn+m=ρDtntm+tnρDtm.
Then,ρD∈DerℂR.(1)
ForDi∈DerℂL, by (99), we can assume that(105)Dic⊗t0=c⊗λ10it+μ20it,i=1,2.For anyn∈ℤ, we have(106)D1,D2c⊗tn=c⊗ρD1,D2tn+tnD1,D2c⊗t0=D1c⊗ρD2tn+tnD2c⊗t0−D2c⊗ρD1tn+tnD1c⊗t0=c⊗ρD1ρD2tn+ρD2tnD1c⊗t0+D1tnD2c⊗t0−c⊗ρD2ρD1tn−ρD1tnD2c⊗t0−D2tnD1c⊗t0=c⊗ρD1,ρD2tn+tnc⊗ρD1λ102t+μ202t−tnc⊗ρD2λ101t+μ201t=c⊗ρD1,ρD2tn+tnD1,D2c⊗t0.Thus,ρD1,D2=ρD1,ρD2.(2)
SinceDerℂR=Derℂℂt±1 is a Witt algebra and DerℂR=Spanℂδi=−ti+1d/dt|i∈ℤ, for any δi=−ti+1d/dt∈DerℂR,i∈ℤ, there exists Di=idH4⊗δi∈DerℂL such that ρDi=δi.(3)
For anyD∈Kerρ, we have ρD=0 and by Lemma 9,(107)Dx⊗tn=tnDx⊗t0,x∈a,b,c,Dd⊗tn=tnDd⊗t0+c⊗δ3nt−δ30ttn.
Ifδ3nt−δ30ttn=0 for any n∈ℤ, then D∈DerRL. If δ3nt−δ30ttn≠0, define φ⊗τ∈HomH4/H4,H4,ZH4⊗ER by(108)φ⊗τd⊗tn=φd⊗τtn=c⊗δ3nt−δ30ttn,where φd=c, φ⊗τd⊗tn=0 if φd=0, ER≔τ∈EndℂR|τtn=δ3nt−δ30ttn,forδ3nt∈Rand anyn∈ℤ, EndℂR is the set of all linear transformations of R. So, Kerρ⊆DerRL+HomH4/H4,H4,ZH4⊗ER. Conversely, for any DR∈DerRL, y∈H4, we have DRy⊗tn=tnDRy⊗t0. By Lemma 9 and (101), it follows that ρDR=0 and DerRL⊆Kerρ. For any φ⊗τ∈HomH4/H4,H4,ZH4⊗ER, by the definition of φ⊗τ, it is clear that φ⊗τ∈DerℂL and φ⊗τ∈Kerρ. Therefore, DerRL+HomH4/H4,H4,ZH4⊗ER=Kerρ. Moreover, since H4/H4,H4=ℂd and ZH4=ℂc, it is easy to see that HomH4/H4,H4,ZH4 is isomorphic to ℂ as vector spaces over ℂ. Thus, (3) is proved.Corollary 5.
.(109)DerℂL≅DerℂR+Kerρ≅idH4⊗DerℂR+DerRL+ℂ⊗ER≅CtdℂH4⊗DerℂR⊕DerℂH4⊗R⊕ℂ⊗ERDerℂR.Proof.
The first and second isomorphism relations follow from Theorem6. The third one follows from(110)DerRL=DerRH4⊗ℂR≅DerH4⊗1,H4⊗R=DerH4⊗R.
See also Theorem 7.1 in [13], Theorem 1.1 in [14], and Lemmas 2.7 and 2.8 in [15]. Furthermore, by Lemma 5, we have(111)CtdℂH4⊗DerℂR=idH4⊗DerℂR+ℂ⊗DerℂR.
Then, the corollary is proved.Remark 2.
The last isomorphism relation in Corollary5 overlaps with Theorem 7.1 in [13], Theorem 1.1 in [14], Corollary 2.2 in [16], Theorem 2.9 in [15], and Theorem 4.2 in [12].



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