The groups of automorphisms of the Lie algebras of polynomial vector fields with zero or constant divergence

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polynomial vector fields with zero or constant divergence.

White Rose Research Online URL for this paper:

http://eprints.whiterose.ac.uk/124640/

Version: Accepted Version

Article:

Bavula, V.V. (2017) The groups of automorphisms of the Lie algebras of polynomial vector

fields with zero or constant divergence. Communications in Algebra, 45 (3). pp. 1114-1133.

ISSN 0092-7872

https://doi.org/10.1080/00927872.2016.1175596

[email protected] https://eprints.whiterose.ac.uk/

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The groups of automorphisms of the Lie algebras of

polynomial vector fields with zero or constant divergence

V. V. Bavula (Aut-Div.tex )

Abstract

Let Pn = K[x1, . . . , xn] be a polynomial algebra over a field K of characteristic zero and div0n (respectively, div

c

n) be the Lie algebra of derivations of Pn with zero (respec-tively, constant) divergence. We prove that AutLie(div0n) ≃ AutK₋alg(Pn) (n ≥ 2) and AutLie(div

c

n)≃AutK₋alg(Pn). The Lie algebradiv c

nis a maximal Lie subalgebra of DerK(Pn). Minimal finite sets of generators are found for the Lie algebrasdiv0

nanddiv c n.

Key Words: Group of automorphisms, derivation, the divergence, Lie algebra, automor-phism, locally nilpotent derivation, the Lie algebras of polynomial vector fields with zero or constant divergence.

Mathematics subject classification 2010: 17B40, 17B20, 17B66, 17B65, 17B30.

1

Introduction

In this paper, module means a left module,Kis a field of characteristic zero and K∗ _{is its group}

of units, and the following notation is fixed:

• Pn:=K[x1, . . . , xn] =Lα∈NnKxα is a polynomial algebra overKwhere xα:=x

α1

1 · · ·xαnn

andQn :=K(x1, . . . , xn) is its field of fractions,

• Gn:= AutK−alg(Pn) is the group of automorphisms of the polynomial algebraPn,

• ∂1:= _∂x∂₁, . . . , ∂n :=_∂x∂_n are the partial derivatives (K-linear derivations) of Pn,

• Dn := DerK(Pn) = Lni=1Pn∂i is the Lie algebra of K-derivations of Pn where [∂, δ] :=

∂δ−δ∂,

• Gn:= AutLie(Dn) is the group of automorphisms of the Lie algebraDn,

• δ1 := ad(∂1), . . . , δn := ad(∂n) are the inner derivations of the Lie algebra Dn determined

by∂1, . . . , ∂n (where ad(a)(b) := [a, b]),

• Dn:=Lni=1K∂i,

• Hn:=Ln_i=1KHi whereH1:=x1∂1, . . . , Hn :=xn∂n,

• D′

n:=

Ln

i=1PnHi=Lα∈NnxαHn,

• h:=Ln_i=1Khi whereh1:=∂1x1, . . . , hn:=∂nxn∈EndK(Pn),

• for a derivation∂=Pni=1ai∂i∈Dn, div(∂) :=Pni=1∂a∂xii is thedivergenceof∂,

• div0_n := {∂ ∈ Dn|div(∂) = 0} is the Lie algebra of polynomial vector fields (derivations)

with zero divergence,

• Gn:= AutLie(div0n),

• H′

n:=

Ln−1

• divc_n := {∂ ∈ Dn|div(∂) ∈ K} is the Lie algebra of polynomial vector fields (derivations)

with constant divergence,

• Gc

n:= AutLie(divcn),

• An:=Khx1, . . . , xn, ∂1, . . . , ∂ni=Lα,β∈NnKxα∂β is then’thWeyl algebra,

The groups of automorphisms of the Lie algebras div0_n and divc_n. The aim of the paper is to prove the following two theorems.

16Mar13

Theorem 1.1 Gn=

(

G1/Sh1≃K∗ if n= 1, Gn if n≥2.

Structure of the proof. The case n= 1 is trivial (see Section 2 where the group Sh1 is defined in

(1)). So, letn≥2.

(i)Gn⊆Gn via the group monomorphism (Lemma 2.9.(3))

Gn→Gn, σ7→σ:∂7→σ(∂) :=σ∂σ−1.

(ii) Let σ∈Gn. Then∂1′ :=σ(∂1), . . . , ∂n′ :=σ(∂n) are commuting, locally nilpotent

deriva-tions of the polynomial algebraPn (Lemma 2.14.(1)).

(iii)Tni=1kerPn(∂

′

i) =K(Lemma 2.14.(2)).

(iv) There exists a polynomial automorphismτ∈Gnsuch thatτ σ∈FixG_n(∂1, . . . , ∂n)

(Corol-lary 2.16).

(v) FixG_n(∂₁, . . . , ∂_n) = Sh_n (Proposition 2.13.(3)) where Shndef

Shn:={sλ∈Gn|sλ(x1) =x1+λ1, . . . , sλ(xn) =xn+λn} (1)

is theshift groupof automorphisms of the polynomial algebraPn andλ= (λ1, . . . , λn)∈Kn.

(vi) By (iv) and (v), σ∈Gn, i.e. Gn =Gn.

A16Mar13

Theorem 1.2 Gc_n=Gn.

Structure of the proof. The casen= 1 is trivial (see Section 2). So, letn≥2.

(i)Gn⊆Gcn via the group monomorphism (Lemma 2.9.(4))

Gn→Gcn, σ7→σ:∂7→σ(∂) :=σ∂σ−1.

(ii)div0_n= [divc_n,divc_n] (Lemma 2.10).

(iii) The short exact sequence of group homomorphisms

1→F := FixGc n(div

0

n)→Gcn

res

→Gn→1

is exact (by (i) and Theorem 1.1) where res :σ7→σ|div0

n is the restriction map, see (ii).

(iv) SinceGn=Gn (Theorem 1.1) andGn⊆Gcn (by (i)), the short exact sequence splits G=GF

Gc_n≃Gn⋉F. (2)

Theorem 1.1 was announced in [8] where a sketch of the proof is given based on a study of certain Lie subalgebras ofdiv0_n of finite codimension. Our proof is based on completely different ideas. The groups of automorphisms of infinite dimensional Lie algebras were considered in [2]-[10].

A subalgebra Mof a Lie algebra Gis called amaximalLie subalgebra ifM 6=G andG is the only Lie subalgebra ofG properly containingM.

• (Proposition 2.21) For n≥2, divc_n is a maximal Lie subalgebra of Dn which is also a Gn

-invariant/Gn-invariant Lie subalgebra.

• (Proposition 2.22) For n ≥2, the Gn-module Dn/divcn is simple and infinite dimensional

withEndGn(Dn/div

c

n)≃K.

6Oct13

Theorem 1.3 For n ≥ 2, the set of elements ∂1, x22∂1, x23∂1, . . . , x2n∂1, x21∂2, x21∂3, . . . , x21∂n is a

minimal set of generators for the Lie algebra div0_n.

A6Oct13

Theorem 1.4 For n≥2, the set of elements in Theorem 1.3 together with H1 is a minimal set of generators for the Lie algebra divc_n.

2

Proof of Theorems 1.1 and 1.2

P16AAA In this section, proofs of Theorems 1.1 and 1.2 are given. In the first part of the section some

useful results are proved that are used throughout the paper. The second part of the section can be seen as proofs of Theorem 1.1 and 1.2. The proofs are split into several statements that reflect ‘Structure of the proofs of Theorems 1.1 and and 1.2’ given in the Introduction. As we have seen in the Introduction, Theorem 1.1 is the key point in the proof of Theorem 1.2.

The Lie algebra Dn isZn-graded. The Lie algebra xadbd

Dn =

M

α∈Nn

n

M

i=1

Kxα∂i (3)

is a Zn_{-graded Lie algebra}

Dn =

M

β∈Zn

Dn,β where Dn,β=

M

α−ei=β

Kxα∂i,

i.e. [Dn,α, Dn,β]⊆Dn,α+β for allα, β∈Nn wheree1:= (1,0, . . . ,0), . . . , en:= (0, . . . ,0,1) is the

canonical free basis for the free abelian groupZn_{. This follows from the commutation relations xadbd1}

[xα∂i, xβ∂j] =βixα+β−ei∂j−αjxα+β−ej∂i. (4)

Clearly, for alli, j= 1, . . . , nandα∈Nn, xadbd2

[Hj, xα∂i] =

(

αjxα∂i ifj6=i,

(αi−1)xα∂i ifj=i,

(5)

xadbd3 [∂j, xα∂i] =αjxα−ej∂i. (6)

The support Supp(Dn) := {β ∈ Zn|Dn,β 6= 0} is a submonoid of Zn. Let us find the support

Supp(Dn), the graded componentsDn,β and their dimensions dimKDn,β. For each i= 1, . . . , n,

let Nn,i _{:= {α ∈ N}n_|α

i = 0} and Pn∂i := kerPn(∂i). It follows from the decompositions Pn = P∂i

Dn= n

M

i=1

(P∂i

n ⊕Pnxi)∂i=

n

M

i=1 P∂i

n ∂i⊕

n

M

i=1

PnHi= n

M

i=1 P∂i

n ∂i⊕

M

α∈Nn

xαHn, (7)

Therefore, any derivation∂ =Pn_i=1ai∂i ∈Dn is the unique sum (where ai =bixi+ci, bi ∈Pn

andci∈Pn∂i) dus

∂=

n

X

i=1 biHi+

n

X

i=1

ci∂i. (8)

Hence, Dnb

Supp(Dn) = n

a

i=1

(Nn,i−ei)

a

Nn. (9)

Dnb1

Dn,β=

(

Kxα_∂

i ifβ=α−ei∈Nn,i−ei,

xβ_H

n ifβ∈Nn.

(10)

dimKDn,β=

(

1 ifβ =α−ei∈Nn,i−ei,

n ifβ ∈Nn_.

Let G be an abelian Lie algebra and G∗ _{:= Hom}

K(G, K). AG-module M is called a weight

moduleif

M = M

λ∈G∗

Mλ where Mλ:={m∈M|gm=λ(g)m for all g∈ G}.

The setW(M) :={λ∈ G∗_|M

λ6= 0}is called the set of weightsofM.

The direct sum div0_n =di0_n⊕iv0_n. Recall thatD′

n=

L

α∈NnxαHn. By (7), dus2

div0_n=di0_n⊕iv0_n where di_n0 :=div0_n∩Dn′ and iv0n:= n

M

i=1 P∂i

n ∂i. (11)

We will see that di0_n is a Lie subalgebra of div0_n but iv0_n is not forn ≥ 2. Clearly, di0₁ = 0 and

div0₁=iv0₁=K∂1. There are inclusions

div0₁ ⊂ div0₂⊂ · · · ⊂div0_n⊂ · · ·,

di0₁ ⊂ di0₂⊂ · · · ⊂di0_n ⊂ · · ·,

iv0₁ ⊂ iv0₂⊂ · · · ⊂iv0_n⊂ · · ·.

TheK-linear mapshi =∂1x1, . . . , hn =∂nxn ∈EndK(Pn) are bijections since for all α∈Nn

andi= 1, . . . , n, dus3

hi(xα) = (αi+ 1)xα. (12)

The elements h1, . . . , hn commute, the polynomial algebra Pn is a weighth-module where h :=

⊕n

i=1Khi is an abelian Lie subalgebra of the Lie algebra EndK(Pn) (where [f, g] :=f g−gf) and

the set W(Pn) of weights of the h-module Pn is equal to (1, . . . ,1) +Nn, i.e. W(Pn) = {λ =

(λ1, . . . , λn)|λ∈(1, . . . ,1) +Nn}whereλ(hi) =λifor alli. For each derivation∂=Pn_i=1aiHi ∈

D′

n, dus4

div(∂) =

n

X

i=1

hi(ai). (13)

K-bases fordiv0_n and divc_n. For each pairi6=j, theK-linear map dus5

is a (well-defined) injection: By (13), div(φij(a)) = (hihj−hjhi)(a) = 0, and ifφij(a) = 0 then

hj(a)Hi = hi(a)Hj, and soa= 0 since the maps hi and hj are bijections. For all α∈ Nn and

i6=j, let phiij

θijα :=φij(xα) =xα((αj+ 1)Hi−(αi+ 1)Hj). (15)

In particular,θ0ij =Hi−Hj. Then xaaH

[xα−αiei_x

j∂i, xαii+1∂j] =φji(xα). (16)

It is obvious thatdiv0₁=K∂1 anddivc1=K∂1+KH1.

a17Mar13

Lemma 2.1 Let n≥2. Then

1. di0_n=L_in−₌₁1φi,i+1(Pn).

2. The set of elementsθα

i :=φi,i+1(xα) =xα((αi+1+1)Hi−(αi+1)Hi+1), wherei= 1, . . . , n−1 andα∈Nn_{, is aK-basis fordi}0

n.

3. The set of elements θα

i in statement 2 and xβ∂j, where xβ ∈ Pn∂j and j = 1, . . . , n, is a

K-basis fordiv0_n.

4. The set of elements in statement 3 andHi, wherei is any fixed index in the set{1, . . . , n},

is aK-basis fordivc_n.

Proof. 1. It is obvious thatR:=Pn−i=11φi,i+1(Pn)⊆div0n, see (14). Recall thatdi

0

n=div

0

n∩Dn′

andD′

n =⊕α∈NnxαH_n. By (14) and the fact that the K-linear mapsh₁, . . . , h_n are invertible,

di0_n=R+di0_n∩PnHn.

By (13),di0_n∩PnHn= 0. Therefore,di0n=R.

2. Statement 2 follows from statement 1.

3. Statement 3 follows from statement 2 and (11).

4. Statement 4 follows from statement 3 and the fact thatdivc_n=div0_n⊕KHi,i= 1, . . . , n.

Let θ := x1· · ·xn ∈ Pn. ThenCn := Li∈Nθ

i_H′

n is an abelian Lie subalgebra of div0n that

containsH′

n. We will see thatCn is a Cartan subalgebra of the Lie algebrasdiv0n anddivcn(Lemma

2.3.(3,5)).

By Lemma 2.1.(2,3), forn≥2, divPH

Cn=

n−1

M

i=1

M

m∈N

Kφi,i+1(θm), (17)

divPH1

div0_n=

n

M

i=1

M

α∈Nn,i

Kxα∂i⊕Cn⊕

n−1

M

i=1

M

m∈N

M

α∈Nn

d\{0}

Kφi,i+1(θmxα), (18)

where Nn

d :=∪ni=1Nn,i ={(α1, . . . , αn)∈ Nn|αi = 0 for some i}. We identify the vector space

H′

n ={

Pn

i=1λiHi|Pni=1λi = 0} with its image inKn under theK-linear injection H′n →Kn,

Pn

i=1λiHi 7→(λ1, . . . , λn). So,

H′n ={λ∈Kn|(λ,1) = n

X

i=1

λi= 0}

where 1 := (1,1, . . . ,1) and (λ, µ) :=P_in₌₁λiµi is the standard inner product on Kn. The dual

spaceH′∗

n := HomK(H′n, K) can be identified with the factor space

i.e. [µ](Pn_i=1λiHi) = [µ](λ) := (µ, λ) = Pni=1µiλi. By (18), the H′n-module div0n is a weight

module and the summands in (18) are the (nonzero) weight vectors under the adjoint action of

H′

n ondiv0n,

div0_n= M

[µ]∈W(div0

n)

div0_n,[µ]

wherediv0_n,[µ] :={∂∈div_n0|[H, ∂] = [µ](H)∂ for allH ∈ H′

n} is the weight subspace ofdiv

0

n that

corresponds to the weight [µ] and W =W(div0_n) is the set of weights for div0_n. To simplify the notation we identify the set Nn

d with its isomorphic copy in the factor vector space Kn/K1 via

the mapKn_→Kn_{/K1,λ7→K1. So, WDnb}

W(div0_n) =Ndn. (19)

Dnb1

div0_n,[µ] =

    

Kxα_∂

i⊕Ln−i=11

L

m∈Nφi,i+1(θ

m_xα−ei+1_{) if [µ] = [α−e}

i], α∈Nn,i,

Cn if [µ] = 0,

Ln−1

i=1

L

m∈Nφi,i+1(θ

m_xα_{) otherwise.}

(20)

dimKdiv0n,[µ]=∞ for all [µ]∈W(div0n).

The Lie algebradiv0_n =iv0_n⊕di0_n is the direct sum of its weightH′

n-submodules with Wiv

W(iv0_n) =

n

a

i=1

(Nn,i_−e

i), W(di0n) =Nnd. (21)

ForH =Pn_i=1λiHi∈ Hn andα∈Kn, let

(H, α) = (α, H) =

n

X

i=1 αiλi.

Then, for allα∈Nn_{, div(x}α_{H) = (α+ 1, H). If, in addition,H ∈ H}′

n, that is (H,1) = 0, then,

for all α∈Nn_{, div(x}α_{H) = (α+ 1, H) = (α, H). It follows that di}0

n is the direct sum of vector

spaces Wiv1

di0_n= M

α∈Nn

{KxαH|(H, α+ 1) = 0, H ∈ Hn}. (22)

Let Wiv2

di′_n0:= M

α∈Nn

{KxαH|(H, α) = 0, H ∈ Hn′}=

M

α∈Nn

d

{K[θ]xαH|(H, α) = 0, H ∈ H′n}. (23)

Clearly, forn≥2,di′_n0⊂di_n0and the vector spacedi′_n0is a leftK[θ]-module. We will see shortly that

di′_n0 is a non-Noetherian Lie algebra (Lemma 2.2). Notice thatdi′_n0=C2=⊕i≥0Kθi(H1−H2) = K[θ](H1−H2) whereθ=x1x2.

The commutation relations of the weight vectors in div0_n. By (18) and (19), there are three types of commutation relations of elements from the weight spaces of the Lie algebradiv0_n, see (24), (25) and (26). For allxα_∂

i ∈Pn∂i∂i andxβ∂j∈Pn∂j∂j, xadxbd

[xα∂i, xβ∂j] =

        

φji(xα+β−ei−ej) ifβi6= 0, αj 6= 0

βixα+β−ei∂j ifβi6= 0, αj = 0

−αjxα+β−ej∂i ifβi= 0, αj 6= 0

0 ifβi= 0, αj = 0

(24)

For all elementsxα_{H, x}α′

H′ _∈di0

n, xadxbd1

[xαH, xα′H′] =xα+α′((H, α′)H′−(H′, α)H)∈di0_n (25)

since ((H, α′_)H′_−(H′_{, α)H, α+α}′_{+ 1) = (H, α}′_)(H′_{, α}′_{+ 1)−(H}′_{, α)(H, α+ 1) = 0.}

For allxβ_∈P∂i

n ∂i andxαH ∈di0n, xadxbd2

[xβ∂i, xαH] =αixα+β−eiH−(H, β−ei)xα+β∂i. (26)

If, in additionαi6= 0, then the equality (26) takes the form xadxbd3

[xβ_∂

i, xαH] =xα+β−ei(αiH−(H, β−ei)Hi)∈di0n (27)

since (αiH−(H, β−ei)Hi, α+β−ei+ 1) =−(H, β−ei)(Hi, β−ei+ 1) =−(H, β−ei)·0 = 0.

By (22) and (25),di0_n is aLie subalgebraofdiv0_n which is not an ideal, by(26). By (24),iv0_n is not a Lie algebra forn≥2.

The Lie algebra di′_n0 is not Noetherian, n ≥ 2. Let n ≥ 2. The Lie algebra di′_n0 is a

K[θ]-module,K[θ]di′_n0⊆di′_n0, and the Lie bracket ondi′_n0 is ak[θ]-bilinear: for allp, p′_{∈K[θ] and}

xα_{H, x}α′

H′ _∈di′0

n, ppHa

[pxαH, p′xα′H′] =pp′[xαH, xα′H′]. (28)

So, the Lie algebradi′_n0 is not simple: θi_di′0

n,i∈Nare distinct ideals ofdi′n0.

a3Apr13

Lemma 2.2 Forn≥2, the Lie algebrasdi0_n anddi′_n0 are not Noetherian.

Proof. If n = 2 then di′₂0 = C2 is an infinite dimensional abelian Lie algebra, hence,

non-Noetherian. Let n ≥ 3. Let I be an ideal of the additive monoid Nn _{(I +N}n _{⊆ I). Then} a′(I) :=L_α∈I{xα_{H| (α, H) = 0, H ∈ H}′

n} is an ideal of the Lie algebra di ′0

n, by (25). The map

I7→a′(I) from the set I(Nn_{) of ideals of N}n _{to the setI(di}′0

n) of ideals of the Lie algebradi ′0

n is

an inclusion preserving injection (I1&I2impliesa′(I1)&a′(I2). The setI(Nn) is not Noetherian

(with respect to⊆) hencedi′_n0 is not a Noetherian Lie algebra.

Similarly, by (25), forn≥2, the mapI7→a(I) :=L_α∈I{xα_{H| (α+ 1, H) = 0, H ∈ H}

n}from

the setI(Nn_{) to the setI(di}0

n) of ideals of the Lie algebradi

0

n is an inclusion preserving injection.

Therefore, the Lie algebradi0_n is non-Noetherian.

LetG be a Lie algebra andHbe its Lie subalgebra. ThecentralizerCG(H) :={x∈ G |[x,H] =

0}ofHinGis a Lie subalgebra ofG. In particular,Z(G) :=CG(G) is thecentreof the Lie algebra

G. The normalizerNG(H) := {x∈ G |[x,H] ⊆ H} of H in G is a Lie subalgebra of G, it is the

largest Lie subalgebra ofG that containsHas an ideal.

LetV be a vector space overK. AK-linear mapδ:V →V is called a locally nilpotent map

ifV =S_i≥1ker(δi_{) or, equivalently, for everyv∈V, δ}i_{(v) = 0 for alli≫1. Whenδis a locally}

nilpotent map in V we also say that δ acts locally nilpotently onV. Every nilpotentlinear map

δ, that is δn_{= 0 for some n≥1, is a locally nilpotent map but not vice versa, in general. LetG}

be a Lie algebra. Each elementa∈ Gdetermines the derivation of the Lie algebraG by the rule ad(a) :G → G,b7→[a, b], which is called theinner derivationassociated witha. The set Inn(G) of all the inner derivations of the Lie algebraGis a Lie subalgebra of the Lie algebra (EndK(G),[·,·])

where [f, g] :=f g−gf. There is the short exact sequence of Lie algebras

0→Z(G)→ G→adInn(G)→0,

that is Inn(G)≃ G/Z(G) whereZ(G) is thecentreof the Lie algebraGand ad([a, b]) = [ad(a),ad(b)] for all elements a, b ∈ G. An element a∈ G is called alocally nilpotent element (respectively, a

The Cartan subalgebra Cn of div0n. A nilpotent Lie subalgebraC of a Lie algebra G such

thatC=NG(C) is called aCartan subalgebraofG. We use often the following obvious observation:

An abelian Lie subalgebra that coincides with its centralizer is a maximal abelian Lie subalgebra.

Example. Hn is a Cartan subalgebra of Dn and Hn = CDn(Hn) is a maximal abelian Lie

subalgebra ofDn.

da11Mar13

Lemma 2.3 Let n≥2. Recall thatCn=Li∈Nθ

i_H′

n andθ=x1x2· · ·xn. Then

1. H′

n=Hn∩div0n.

2. Cn=Cdiv0

n(H

′

n)andCn=Cdiv0

n(Cn)is a maximal abelian Lie subalgebra ofdiv 0

n.

3. Cn is a Cartan subalgebra ofdiv0n.

4. Cn+Hn=Cdivc n(H

′

n)andCn=Cdivc

n(Cn)is a maximal abelian Lie subalgebra ofdiv

c n.

5. Ndivc

n(Cn) =Cn+Hn=Ndivcn(Cn+Hn)is a solvable but not nilpotent Lie algebra, and so Cn+Hn andCn are not a Cartan subalgebra ofdivcn.

6. Hn is a Cartan subalgebra ofdivcn andHn=Cdivc

n(Hn)is a maximal abelian Lie subalgebra of divc_n.

Proof. 1. LetH =Pn_i=1λiHi∈ Hn. ThenH ∈ Hn∩div0n iff

Pn

i=1λi = 0 iffH∈ H′n.

2. The fact that Cn=Cdiv0

n(H

′

n) follows from (20). By (25),Cn is an abelian Lie subalgebra

ofdiv0_n. Then the inclusionH′

n⊆Cn=Cdiv0

n(H

′

n) implies the inclusions

Cn=Cdiv0

n(H

′

n)⊇Cdiv0

n(Cn)⊇Cn.

Therefore,Cn=Cdiv0

n(Cn) is a maximal abelian Lie subalgebra of div 0

n.

3. Let N := Ndiv0

n(Cn). By statement 2, Cn ⊆ N. The H

′

n-module div

0

n is weight and

Cn = Cdiv0

n(H

′

n) is the zero weight component of div0n. The inclusion [N,H′n] ⊆ Cn implies

N ⊆Cn, and soN =Cn.

4. Notice thatdivc_n =div0_n⊕KH1,H1∈Cdivc n(H

′

n) andH′n⊆div

0

n. By statement 2,

Cdivc n(H

′

n) =Cdiv0

n(H

′

n)⊕KH1=Cn⊕KH1=Cn+Hn.

Now, the inclusionH′

n⊆Cn implies the inclusions

Cn⊕KH1=Cdivc n(H

′

n)⊇Cdivc

n(Cn)⊇Cn.

Hence Cdivc

n(Cn) =Cn⊕Cdivcn(Cn)∩KH1=Cn and soCn is a maximal abelian Lie subalgebra

ofdivc_n.

5. Let Nc _{:= N}

divc

n(Cn). By statement 4, Cn ⊆N

c_{. The H}′

n-module divcn =div0n⊕KH1

is weight and Cn ⊕KH1 = Cdivc n(H

′

n) is the zero weight component of divcn. The inclusion

[Nc_,H′

n]⊆Cn implies the inclusion [Nc,Hn′]⊆Cn⊕KH1=Cdivc n(H

′

n), and soNc⊆Cn⊕KH1.

Since Cn ⊆ Nc, it follows that Nc = Cn ⊕Nc ∩KH1 = Cn+KH1, i.e. Cn is not a Cartan

subalgebra ofdivc_n.

Clearly, Cn+Hn ⊆ N := Ndivc

n(Cn +Hn). Since H

′

n ⊆ Cn +Hn, [Hn′,N] ⊆ Cn+Hn =

Cdivc n(H

′

n), hence N ⊆Cn+Hn (as Cn+Hn is the zero weight component of the weight Hn′

-moduledivc_n), i.e. N =Cn+Hn. The Lie algebraCn+Hn is solvable but not nilpotent, and so

Cn+Hn is not a Cartan subalgebra ofdivcn.

6. Statement 6 follows from the Example above.

The Lie algebradiv0_nis a weightH′

n-module with respect to the adjoint action. In particular, the

action ofH′

nondiv0nislocally finite dimensional, i.e. for any∂∈div0n, dimK(Pi∈Nad(H

′

n)i(∂))<

∞. We can easily verify that the action of the Cartan subalgebraCn ofdiv0nondiv

0

n is not locally

Pn is a Dn-module. The polynomial algebra Pn is a (left) Dn-module: Dn×Pn → Pn,

(∂, p)7→∂∗p. In more detail, if∂=P_in₌₁ai∂i whereai∈Pn then

∂∗p=

n

X

i=1 ai

∂p ∂xi

.

The fieldK is aDn-submodule ofPn and IkerdiK

PDn

n :=

n

\

i=1

kerPn(∂i) =K. (29)

xa11Mar13

Lemma 2.4 [2] The Dn-module Pn/K is simple with EndDn(Pn/K) = Kid where id is the identity map.

The Gn-module Dn. The Lie algebraDn is aGn-module,

Gn×Dn→Dn, (σ, ∂)7→σ(∂) :=σ∂σ−1.

Every automorphismσ∈Gn is uniquely determined by the elements

x′1:=σ(x1), . . . , x′n:=σ(xn).

LetMn(Pn) be the algebra ofn×nmatrices over Pn. The matrix J(σ) := (J(σ)ij)∈Mn(Pn),

where J(σ)ij = ∂x′

j

∂xi, is called the Jacobian matrix of the automorphism (endomorphism)σ and

its determinantJ(σ) := detJ(σ) is called the Jacobianof σ. So, thej’th column of J(σ) is the

gradient gradx′

j := (

∂x′

j

∂x1, . . . ,

∂x′

j

∂xn)

T _{of the polynomial x}′

j where T stands for the transposition.

Then the derivations

∂1′ :=σ∂1σ−1, . . . , ∂′n:=σ∂nσ−1

are the partial derivatives ofPn with respect to the variablesx′1, . . . , x′n, ddp=dxi

∂1′ = ∂ ∂x′

1

, . . . , ∂n′ =

∂ ∂x′

n

. (30)

Every derivation ∂ ∈ Dn is a unique sum ∂ = Pni=1ai∂i where ai = ∂∗xi ∈ Pn. Let ∂ :=

(∂1, . . . , ∂n)T and∂′:= (∂1′, . . . , ∂′n)T. Then dp=Jnd

∂′ =J(σ)−1∂, i.e. ∂i′= n

X

j=1

(J(σ)−1)ij∂j for i= 1, . . . , n. (31)

In more detail, if ∂′ _{= A∂ where A = (a}

ij) ∈ Mn(Pn), i.e. ∂i′ =

Pn

j=1aij∂j. Then for all i, j= 1, . . . , n,

δij =∂i′∗x′j= n

X

k=1 aik

∂x′ j

∂xk

whereδij is the Kronecker delta function. The equalities above can be written in the matrix form

asAJ(σ) = 1 where 1 is the identity matrix. Therefore,A=J(σ)−1_.

For allσ, τ ∈Gn, Jst=JsJ

J(στ) =J(σ)·σ(J(τ)). (32)

By taking the determinants of both sides of (32), we have a similar equality of the Jacobians:

for allσ, τ ∈Gn. Jst=JsJ1

Properties of the divergence. Recall some of the properties of the divergence map

div :Dn→Pn, ∂= n

X

i=1

ai∂i7→div(∂) = n

X

i=1 ∂ai

∂xi

.

(div-i) div is a K-linear map which is a surjection.

(div-ii) For alla∈Pn and∂∈Dn, div(a∂) =adiv(∂) +∂(a).

(div-iii) For all∂, δ∈Dn, div([∂, δ]) =∂(div(δ))−δ(div(∂)).

(div-iv) Leta1, . . . , an∈Pn;σ:Pn→Pn,xi7→ai,i= 1, . . . , nandJ(a1, . . . , an) :=J(σ) be

the Jacobian ofσ. Then (Proposition 2.3.2, [6])

∂∗ J(a1, . . . , an) =−J(a1, . . . , an)div(∂) + n

X

i=1

J(a1, . . . , ∂∗ai, . . . , an).

(div-v) (Theorem 2.5.5, [6]) If Dn=Ln_i=1Pn∂i′ and∂1′, . . . , ∂n′ commute then

div(∂′1) =· · ·= div(∂n′) = 0.

(div-vi) Let σ : Pn → Pn, xi 7→ x′i, i = 1, . . . , n, be an automorphism of Pn. Then ∂′1 := σ∂1σ−1, . . . , ∂n′ :=σ∂nσ−1 are commuting derivations ofPn such that Dn =Lni=1Pn∂i′ (by (31)

and sinceJ(σ)−1_∈GL

n(Pn)). By (31),∂i=Pni=1(J(σ)−1)ij∂j. By (div-v), n

X

j=1

∂j∗(J(σ)−1)ij = 0 for i= 1, . . . , n.

The divergence commutes with polynomial automorphisms. The following known theorem shows that the divergence commutes with automorphisms, i.e. the divergence map div :

Dn →Pn is a Gn-module homomorphism. We give a short proof.

B16Mar13

Theorem 2.5 For all σ∈Gn and∂∈Dn,

div(σ(∂)) =σ(div(∂)).

Proof. Let ∂ =Pn_i=1ai∂i where ai ∈Pn. Then∂′ =σ∂σ−1 =Pni=1σ(ai)∂i′ where, by (31),

∂′

i=

Pn

j=1(J(σ)−1)ij∂j. Now, by (div-vi),

div(∂′) =

n

X

i,j=1

∂j∗((J(σ)−1)ijσ(ai)) = n

X

i=1

(

n

X

j=1

∂j∗(J(σ)−1)ij)·σ(ai)) + n

X

i=1

n

X

j=1

(J(σ)−1)ij∂j∗σ(ai)

=

n

X

i=1

∂′i∗σ(ai) = n

X

i=1

σ∂iσ−1σ(ai) =σ( n

X

i=1

∂i(ai)) =σ(div(∂)).

11Mar13

Theorem 2.6 _Gn=Gn.

The above theorem was announced in [8] where a sketch of a proof is given, it can also be deduced from [10]. A proof of the above theorem is given in [9] and in ([7], Theorem 3.6), a different approach and a short proof is given in [2]. Some generalizations are given in [4], [3] and [5].

a20Mar13

Proof. The statement follows from Theorem 2.6 and Theorem 2.5.

By (div-iii),div0_n anddivc_n are Lie subalgebras ofDn,div0n is an ideal ofdivcn.

aB16Mar13

Corollary 2.8 The Lie subalgebras div0_n anddivc_n of Dn are alsoGn-submodules of Dn.

Proof. This follows from Theorem 2.5.

By (div-iii), divDnPn

0→div0_n→Dn

div

→Pn→0 (34)

is the short exact sequence of (left)div0_n-modules anddivc_n-modules, i.e. divcom

div([∂, δ]) =∂∗div(δ) (35)

for all∂∈divc_n andδ∈Dn. So, divDnPn1

0→div0_n →divc_ndiv→K→0 (36)

is the short exact sequence of (left)div0_n-modules/divc_n-modules, and so divDnPn2

divc_n=div0_n⊕KHi for i= 1, . . . , n, (37)

since div(Hi) = 1 for alli= 1, . . . , n.

The maximal abelian Lie subalgebra Dn of div0n and divcn.

db11Mar13

Lemma 2.9 1. Cdiv0

n(Dn) =Cdiv c

n(Dn) =Dn and soDn is a maximal abelian Lie subalgebra of div0_n anddivc_n.

2. ([2])FixGn(Dn) = Shn.

3. For n ≥ 2, div0_n is a faithful Gn-module, i.e. the group homomorphism Gn → Gn, σ 7→

σ :∂ 7→ σ∂σ−1_{, is a monomorphism. For n= 1, the groupSh}

1 is the kernel of the group homomorphism G1→G1.

4. divc_n is a faithfulGn-module, i.e. the group homomorphismGn →Gcn,σ7→σ:∂7→σ∂σ−1,

is a monomorphism.

5. FixGn(Dn⊕ H

′

n) ={e} forn≥2.

Proof. 1. SinceCDn(Dn) =Dn, [2], we must haveCdiv0

n(Dn) =Cdiv c

n(Dn) =Dn, and soDn is

a maximal abelian Lie subalgebra ofdiv0_n anddivc_n.

3 and 5. By Corollary 2.8, div0_n is a Gn-submodule ofDn. So, the group homomorphism in

statement 3 is well-defined. The case n= 1 is obvious sinceG1={x7→λx+µ|λ∈K∗, µ∈K}

anddiv0₁=K∂.

So, let n≥2, and σ ∈FixGn(Dn⊕ H

′

n). Then σ∈ FixGn(Dn) = Shn, by statement 2. So, σ(x1) =x1+λ1, . . . , σ(xn) =xn+λn whereλi∈K. Then for alli6=j,

Hi−Hj =σ(Hi−Hj) = (xi+λi)∂i−(xj+λj)∂j =Hi−Hj+λi∂i−λj∂j.

So,λ1=· · ·=λn= 0. This means that σ=e. So, FixGn(Dn⊕ H

′

n) ={e} anddiv0n is a faithful

Gn-module.

4. By Corollary 2.8,divc_nis aGn-submodule ofDn. So, the group homomorphism in statement

4 is well-defined. Now, statement 4 follows from statement 3 forn≥2. Forn= 1, statement 4 is obvious asdivc₁=K∂1+KH1 andG1={x7→λx+µ|λ∈K∗, µ∈K}.

a16Mar13

Proof. The statement is obvious forn= 1 asdivc₁=K∂1+KH1anddiv01=K∂1. So, letn≥2.

By (div-iii), div([divc_n,divc_n]) = 0, and so [divc_n,divc_n]⊆div0_n. Recall thatdiv0_n=iv0_n+di0_n(see (11)),

iv0_n=⊕n

i=1P∂i∂i, [xα∂i, Hi] =xα∂i for allxα∈Pn∂i andi= 1, . . . , n. Henceivn0 ⊆[divcn,divcn].

Finally, for all a∈Pn andi6=j,

[Hk, hj(a)Hi−hi(a)Hj] =hj(Hk∗a)Hi−hi(Hk∗a)Hj.

Hence,{hj(xα)Hi−hi(xα)Hj|06=α∈Nn, i=6 j} ⊆[divcn,div c

n]. Finally,Hi−Hj= [xi∂j, xj∂i]∈

[divc_n,divc_n] for alli6=j. Now, by Lemma 2.1.(3), [divc_n,divc_n] =div0_n. The following lemma is well-known and easy to prove.

c11Mar13

Lemma 2.11 1. Dn is a simple Lie algebra.

2. Z(Dn) ={0}.

3. [Dn, Dn] =Dn.

The next lemma is also known but we give a short elementary proof.

dc11Mar13

Lemma 2.12 1. div0_n is a simple Lie algebra.

2. Z(div0_n) =

(

div0_n if n= 1,

0 if n≥2.

3. [div0_n,div0_n] =

(

0 ifn= 1,

div0_n ifn≥2.

Proof. All three statements are obvious ifn= 1 sincediv0₁=K∂. So, we assume thatn≥2. 1. Let 06=a∈div0_n anda= (a) be the ideal of the Lie algebradiv0_n generated by the element

a. We have to show thata=div0_n. Using the inner derivationsδ1, . . . , δn and ad(Hi−Hj) (where

i6=j) of the Lie algebradiv0_n we see that∂i∈a for somei. ThenDn⊆asince∂j= [∂i, xi∂j] for

all i6=j. Then iv0_n ⊆a sincexα_∂

i = [∂j,(αj+ 1)−1xα+ej∂i] for all i6=j and xα ∈ Pn∂i. Then

H′

n⊆asince [xi∂j, xj∂i] =Hi−Hj for alli6=j. Using the commutation relations Hsthj

[Hs−Ht, hj(xα)Hi−hi(xα)Hj] = (αs−αt)(hj(xα)Hi−hj(xα)Hj) (38)

we see that {hj(xα)Hi −hj(xα)Hj|i 6= j, α ∈ Nn, α 6= (p, p, . . . , p), p = 1,2, . . .} ⊆ a. Let

θ=x1· · ·xn. Finally, by Lemma 2.1.(3),a=div0n since for allp= 1,2, . . .andi6=j, Hsthj1

[xj∂i, hj(xixj−1θp)Hi−hi(xixj−1θp)Hj] = (p+ 2)(hj(θp)Hi−hi(θp)Hj). (39)

In more detail,

LHS = [xj∂i, pxix−j1θpHi−(p+ 2)xixj−1θpHj] =p(p+ 2)θpHi−(p+ 2)(p+ 1)θpHj

+ (p+ 2)θpHi= (p+ 2)(p+ 1)(θpHi−θpHj)

= (p+ 2)(hj(θp)Hi−hi(θp)Hj).

2. Statement 2 follows from statement 1. 3. Statement 3 follows from statement 1.

For anyα= (α1, . . . , αn)∈Nn and anyi6=j, we haveα=αiei+αjej+β whereβ ∈Nnwith

βi =βj= 0. Then taijx

θα

ij = [xαii+1∂j, xα−αiei+ej∂i]. (40)

Indeed, letc be the commutator. Then

c=xβ_[xαi+1

By (40), taijx1

di0_n⊆[iv0_n,iv0_n]. (41)

The inner derivations δ1 = ad(∂1), . . . , dn = ad(∂n) of the Lie algebra Dn are commuting and

locally nilpotent. The Lie algebraDn is a union of vector spaces

Dn=

[

i≥o

Dn,i where Dn,i:={∂∈Dn|δα(∂) = 0 for all α∈Nnsuch that |α|=i+ 1}

where δα ₌ Qn i=1δ

αi

i . Clearly, Dn,i = ⊕nj=1Pn,i∂j where Pn,i := {p ∈ Pn|deg(p) ≤ i} and

[Dn,i, Dn,j]⊆Dn,i+j−1 for alli, j≥0 whereDn,−1:= 0. The inner derivationsδ1, . . . , δn are also

commuting locally nilpotent derivations of the Lie algebras div0_n and divc_n. For each i ∈ N, let

div0_n,i:=div0_n∩Dn,i anddivcn,i:=divcn∩Dn,i. Then, for allα∈Nn andi= 1, . . . , n−1, tiadi

θαi ∈div0n,|α|+1\div0n,|α|. (42)

The automorphisms si,i+1, i = 1, . . . , n−1 (n ≥ 2). The automorphism s = si,i+1 of

the polynomial algebra Pn that swaps the variables xi and xi+1 leaving the rest of the variables

untouched (s(xi) = xi+1 and s(xi+1) = xi) extends uniquely to an automorphism of the Lie

algebraDn. Clearly,sis also an automorphism of the Lie algebrasdiv0n anddivcn (Theorem 2.5).

In particular,

s(∂i) =∂i+1, s(∂i+1) =∂i, s(Hi) =Hi+1ands(Hi+1) =Hi.

Therefore, for allα∈Nn_{, siia}

si,i+1(θαi) =−θ si,i+1(α)

i (43)

where, forα= (α1, . . . , αn),si,i+1(α) = (α1, . . . , αi−1, αi+1, αi, αi+2, . . . , αn). Fori= 1, . . . , n−1;

j∈Nandα∈Nn _withα

i ≥1, ditai

[xji+1∂i, θαi] = (αi+ 1)θα−ei i+jei+1. (44)

Proof. By (27),

LHS = [xj_i+1∂i, xα((αi+1+ 1)Hi−(αi+ 1)Hi+1]

= xα−ei+jei+1_(α

i((αi+1+ 1)Hi−(αi+ 1)Hi+1)−((αi+1+ 1)Hi−(αi+ 1)Hi+1, jei+1−ei)Hi)

= (αi+ 1)xα−ei+jei+1((αi+1+ 1 +j)Hi−αiHi+1)

= (αi+ 1)θiα−ei+jei+1.

Fori= 1, . . . , n−1;j∈Nandα∈Nn _withα

i+1≥1, ditai1

[xji∂i+1, θαi] = (αi+1+ 1)θαi+jei−ei+1. (45)

Proof. By applying the automorphism s =si,i+1 to the equality (44) and using (43) we obtain

(45):

[xj_i∂i+1, θiα] = −s([x j

i+1∂i, θsi(α)] =−s((αi+1+ 1)θis(α)−ei+jei+1

= (αi+1+ 1)θis(s(α)−ei+jei+1)

= (αi+1+ 1)θiα+jei−ei+1.

By (44) and (45), fori= 1, . . . , n−1;j= 1, . . . , nandα∈Nn_{. ditai4}

[∂j, θαi] =

        

(αj+ 1)θiα−ej ifj =i, i+ 1 andαj≥1,

(αi+1+ 1)xα∂i ifj =i, αi= 0,

−(αi+ 1)xα∂i+1 ifj =i+ 1, αi+1= 0, αjθ

α−ej

i otherwise.

dB11Mar13

Proposition 2.13 Let n≥2. Then

1. FixG_n(D_n⊕ H′_n) ={e}.

2. Let σ, τ ∈ Gn. Then σ = τ iff σ(∂i) = τ(∂i) for i = 1, . . . , n and σ(Hj −Hj+1) = τ(Hj−Hj+1)forj= 1, . . . , n−1.

3. FixG_n(∂₁, . . . , ∂_n) = Sh_n.

Proof. 1. Letσ∈F := FixG_n(Dn⊕ Hn′). We have to show thatσ=e. LetF := Fixdiv0

n(σ) := {∂∈div0_n|σ(∂) =∂}. We have to show thatF =div0_n.

(i)Dn+H′n⊆ F: Obvious.

(ii) iv0_n ⊆ F: We have to show that P∂i

n ∂i ⊆ F for all i = 1, . . . , n. Fix i and xα∂i ∈Pn∂i∂i.

If α= 0 then ∂i ∈ Dn ⊆ F, by (i). We use induction on |α| =Pn_i=1αi to prove the statement.

The case |α| = 0 is obvious. So, let |α| ≥1. The element xα_∂

i has weight [α−ei]. The weight

subspace div0_n,[α−ei] is σ-invariant (σ(div0_n,[α−ei]) =div0_n,[α−ei]) and the set {xα_∂

i, θα−ej i+k1|j =

1, . . . , n−1;k = 1,2, . . .} is its K-basis, by (20). The element xα_∂

i belongs to the σ-invariant

subspaceN|α|:=div0n,|α|(=div

0

n∩Dn,|α|). By (42) and (46),

V :=N|α|∩div0n,[α−ei]=Kxα∂i

is aσ-invariant subspace. Therefore,σ(xα_∂

i) =λα,ixα∂i for allxα∂i∈Pn∂i∂i andi= 1, . . . , n for

someλα,i∈K. Clearly,λ0,i= 1 for alli= 1, . . . , nsinceσ(∂i) =∂i. Applying the automorphism

σ to the relations [∂j, xα∂i] = αjxα−ej∂i yields the equalities αj(λα,i−λα−ej,i) = 0. If αj 6= 0

thenλα,i=λα−ej,i= 1 (by induction). Hence,iv 0

n⊆ F.

(iii)di0_n⊆ F: By (ii) and (41).

Therefore,div0_n=di0_n+iv0_n ⊆ F, and sodiv0_n=F, as required. 2. Statement 2 follows from statement 1.

3. Clearly, Shn ⊆ F := FixG_n(∂₁, . . . , ∂_n). Let σ ∈ F and H_i,j′ := σ(H_i,j) where H_i,j :=

Hi−Hj fori6=j. Then

[∂k, Hi,j′ −Hi,j] =σ([∂k, Hi,j])−[∂k, Hi,j] = [∂k, Hi,j]−[∂k, Hi,j] = 0

since [Dn,H′n]⊆ Dn. Then dij :=Hi,j′ −Hi,j ∈ Cdiv0

n(Dn) = Dn, and so dij =

Pn

k=1λkij∂k for

someλk

ij ∈K.

Ifn= 2 thenH′

1,2=H1−H2+λ1∂1−λ2∂2= (x1+λ1)∂1−(x2+λ2)∂2=sλ(H1−H2) where sλ∈Sh2andλ= (λ1, λ2). Hence,s−λ1σ(H1−H2) =H1−H2, i.e. sλ−1σ∈FixG2(D2+H

′

2) ={e},

and soσ=sλ∈Sh2.

Suppose thatn >2. Ifn= 3 then up to action of Sh3, we may assume thatH1′,2=H1,2+λ∂3

andH′

2,3=H2,3+µ∂1+ν∂2 for someλ, µ, ν ∈K. Then 0 = [H1′,2, H2′,3] =−µ∂1+ν∂2+λ∂3and

soλ=µ=ν= 0.

Ifn≥4 then for any four distinct numbersi, j, k, l∈ {1,2, . . . , n} we have the equality

0 = [Hi,j′ , Hk,l′ ] = [Hi,j, dkl]−[Hk,l, dij] =−λikl∂i+λjkl∂j+λkij∂k−λlij∂l.

Therefore,λi

kl= 0 for all distincti,j andk. Then using Shn, we may assume that

H′

1,2=H1,2, H2′,3=H2,3+λ223∂2, H3′,4=H3,4+λ334∂3, . . . , Hn−′ 1,n=Hn−1,n+λn−n−11,n∂n−1.

Then, fori= 2, . . . , n−1, 0 = [H′

i−1,i, Hi,i′ +1] =λii,i+1∂i, i.e. λii,i+1= 0 andHi,i′ +1=Hi,i+1. This

means thatsλσ∈FixG_n(Dn+Hn′) ={e}(statement 2) for somesλ∈Shn, and soσ=s−_λ1∈Shn.

dc13Mar13

Lemma 2.14 Let σ∈Gn,∂1′ :=σ(∂1), . . . , ∂n′ :=σ(∂n); δ1′ := ad(∂1′), . . . , δ′n := ad(∂n′) andD′

1. ∂′

1, . . . , ∂n′ are commuting, locally nilpotent derivations ofPn.

2. Tn_i=1kerPn(∂

′

i) =K.

3. (Embedding trick) Forn≥2, the short sequence of D′_{-modules0→K→P}

n ∆

′

→div0_n(

n

2)

is exact where ∆′(p) :=Qi<j(∂i′(p)∂j′−∂j′(p)∂i′). In particular, for allp∈Pn andα∈Nn,

∆′_∂′α_{(p) =δ}′α_∆′_(p)

where ∂′α_:=Qn

i=1∂

′αi

i andδ′α:=

Qn i=1δ

′αi

i .

4. Let∆′_:P

n→D(

n

2)

n ,p7→Qi<j(∂i′(p)∂j′ −∂j′(p)∂i′). Then ∆′∈DerD′(Pn, D(

n

2)

n ).

Proof. 2. Letλ∈Tn_i=1kerPn(∂

′

i). Then div(λ∂1′) =λdiv(∂′1) +∂1′ ∗λ= 0, i.e. λ∂1′ ∈div 0

n and

λ∂′

1∈Cdiv0

n(∂

′

1, . . . , ∂n′) =σ(Cdiv0

n(∂1, . . . , ∂n)) =σ(Cdiv

0

n(Dn)) =σ(Dn) =σ(

n

M

i=1

K∂i) = n

M

i=1 K∂′

i,

sinceCdiv0

n(Dn) =Dn, Lemma 2.9.(1). Thenλ∈Ksince otherwise the infinite dimensional space

L

i≥0Kλi∂1′ would be a subspace of the finite dimensional spaceσ(Dn).

3. The derivations ∂′

1, . . . , ∂n′ commute since ∂1, . . . , ∂n do. So,D′ is a commutative algebra.

For allp∈Pn and∂∈div0n,

div(a∂) =adiv(∂) +∂(a) =∂(a).

So, the map ∆′ _{is well-defined: div(∂}′

i(p)∂j′ −∂j′(p)∂i′) = (∂j′∂i′ −∂i′∂j′)(a) = 0. Recall that

the vector spaces Pn and div0n are left div

0

n-modules hence they are also left D′-modules since

∂′

1, . . . , ∂n′ ∈div

0

n. The map ∆′ is aD′-homomorphism since, for allp∈Pn,

∆′_∂′

i(p) = [∂i′,∆′(p)], i= 1, . . . , n.

It remains to show that ker(∆′_{) =K. The inclusionK⊆ker(∆}′_{) is obvious.}

(i) ker(∆′_{) ={p∈P}

n|∂i′(p)∂j′ =∂j′(p)∂i′ for alli6=j}: This is obvious. LetP ∂′

i

n := kerPn(∂

′ i).

(ii) ker(∆′_)∩P∂′

i

n = K for i = 1, . . . , n: Given p ∈ Pn. By (i), p ∈ ker(∆′)∩P∂ ′

i

n iff p ∈

ker(∆′_)∩P∂j′

n forj= 1, . . . , n (0 =∂i′(p)∂j′ =∂j′(p)∂i′ ⇒ ∂j′(p) = 0) iff

p∈ker(∆′)∩P∂′1,...,∂′n

n = ker(∆′)∩K=K,

by statement 2. Suppose thatK′ _{:= ker(∆}′_{)\K6=∅, we seek a contradiction.}

(iii) If p ∈ K′ _{then ∂}′

1(p) 6= 0, . . . , ∂n′(p) 6= 0: This follows from (ii) (K′ = ker(∆′)\K =

ker(∆′_)\ker(∆′_)∩P∂′i

n = ker(∆′)\P∂ ′

i

n ).

(iv)P∂1′

n =· · ·=P∂ ′

n

n : Fixp∈K′. By (iii), ∂1′(p)6= 0, . . . , ∂n′(p)6= 0. By (i),

∂i′(p)∂j′ =∂j′(p)∂′i for all i6=j,

and soP∂1′

n =· · ·=P∂ ′

n

n .

(v) P∂1′

n =· · ·=P∂ ′

n

n =K: This statement follows from (iv) and statement 2.

Suppose that ker(∆′_{) 6= K, we seek a contradiction. Fix an element p ∈ ker(∆}′_{)\K. By}

statement 2, ∂′

j(p)6= 0 for some j. For all i 6=j, ∂i′(p)∂j′ =∂j′(p)∂i′ (since p∈ker(∆′)). Hence,

∂′

i(p) 6= 0 for all i. Therefore, for i = 1, . . . , n, ∂i′ = fi∂ for some fi ∈ Pn and a derivation

∂ = Pns=1ps∂s ∈ Dn with gcd(p1, . . . , pn) = 1. For all elements ci ∈ Ci′ := Cdiv0

n(∂

′ i), 0 =

[ci, ∂i′] = ci(fi)∂+fi[ci, ∂], and so [ci, ∂] ∈ Pn∂, by the choice of ∂. By Theorem 1.3, the Lie

algebra div0_n is generated by the set C = Pn_i=1Cdiv0

n(∂i) and also by the set σ(C) =

Hence, [div0_n, ∂]⊆Pn∂. In particular, for alli, [∂i, ∂] =Pjn=1∂i(pj)∂j ∈Pn∂. Hence,∂∈ Dn, by

degree argument. ThenP∂

n 6=K, and soP ∂′

i

n =Pn∂ 6=K. This contradicts to (v).

4. The map ∆′ _{can be seen as the map DpPb}

∆′:Pn →D(

n

2)

n , p7→

Y

i<j

(∂i′(p)∂j′ −∂′j(p)∂i′). (47)

Notice thatPn andD(

n

2)

n are leftPn-modules.

∆′ _{∈ Der}

D′(Pn, D(

n

2)

n ), i.e. the map ∆′ is a D′-derivation from the polynomial algebra Pn to

the leftPn-moduleD(

n

2)

n , i.e. for allp, q∈Pn,

∆′(pq) =q∆′(p) +p∆′(q) :

∆′(pq) = Y

i<j

((q∂i′(p) +p∂i′(q))∂′j−(q∂j′(p) +p∂j′(q))∂i′)

= qY

i<j

(∂i′(p)∂j′−∂j′(p)∂i′) +p

Y

i<j

(∂′i(q)∂j′ −∂j′(q)∂i′)

= q∆′(p) +p∆′(q).

1. The inner derivationsδ1, . . . , δnof the Lie algebradiv0nare commuting and locally nilpotent.

Hence so are the inner derivationsδ1, . . . , δn′, by statements 2 and 3.

A13Mar13

Theorem 2.15 [2] Let ∂′

1, . . . , ∂n′ be commuting, locally nilpotent derivations of the polynomial

algebraPn such thatTni=1kerPn(∂

′

i) =K. Then there exist polynomialsx′1, . . . , x′n ∈Pn such that

con*

∂′

i∗x′j=δij for i, j= 1, . . . , n. (48)

Moreover, the algebra homomorphism

σ:Pn→Pn, x17→x′1, . . . , xn7→x′n

is an automorphism such that∂′

i=σ∂iσ−1= ∂x∂′

i fori= 1, . . . , n.

db13Mar13

Corollary 2.16 Let σ∈Gn. Then τ σ∈FixG_n(∂₁, . . . , ∂_n) for someτ∈G_n.

Proof. By Lemma 2.14, the elements ∂′

1 :=σ(∂1), . . . , ∂n′ :=σ(∂n) satisfy the assumptions of

Theorem 2.15. By Theorem 2.15,∂′

1 :=τ−1(∂1), . . . , ∂n′ :=τ−1(∂n) for someτ ∈Gn. Therefore,

τ σ∈FixG_n(∂₁, . . . , ∂_n).

Proof of Theorem 1.1. Ifn= 1 thendiv0₁=K∂1,G1=T1≃G1/Sh1sinceG1=T1⋉Sh1.

So, let n ≥2. Letσ ∈ Gn. By Corollary 2.16, τ σ ∈ FixG_n(∂1, . . . , ∂n) = Shn (Proposition

2.13.(3)). Therefore,σ∈Gn, i.e. Gn=Gn.

a18Mar13

Lemma 2.17 FixGc n(div

0

n) =

(

Sh1 ifn= 1, {e} ifn≥2.

Proof. If n = 1, σ(∂) = ∂ for some σ ∈ Gc₁ then applying σ to the equality [∂1, H1] = ∂1

yields σ(H1) = H1 +λ∂1 = (x1+λ)∂1 = tλ(H1) for some λ ∈ K where tλ ∈ Sh1. Hence

FixGc

1(div

0

1) = Sh1.

So, let n ≥ 2. Let σ ∈ F := FixGc n(div

0

n), H1′ := σ(H1), . . . , Hn′ := σ(Hn). By (37), it

suffices to show that σ(Hi) = Hi for i = 1, . . . , n. For i 6= j, σ(Hi−Hj) = Hi−Hj, and so

d:=Hi′−Hi=Hj′−Hj. For alli= 1, . . . , n,

So, d ∈ Cdivc

n(Dn) = Dn (Lemma 2.9.(1)) and d =

Pn

i=1λi∂i for some λi ∈ K. The elements

H′

1=H1+d, . . . , Hn′ =Hn+dcommute henced= 0. Therefore,σ=e.

Proof of Theorem 1.2. Ifn= 1 then Gc₁ ≃T1_⋉Sh

1. So, let n≥2. By (2) and Lemma

2.17,Gc_n=Gn.

A17Mar13

Theorem 2.18 For allσ∈Gn,

div(σ(Hi)−Hi) = 0 for i= 1, . . . , n.

Proof. For each i= 1, . . . , n, by (37),σ(Hi) =λi(σ)Hi+di(σ) for some elementsλi(σ)∈K

anddi(σ)∈div0n. By Theorem 2.5,

1 = div(Hi) =σ(div(Hi)) = div(σ(Hi)) = div(λi(σ)Hi+di(σ)) =λi(σ).

The automorphisms of the Lie algebra divc_n preserve divergence.

b17Mar13

Corollary 2.19 1. For allσ∈Gc_n and∂∈divc_n,div(σ(∂)) = div(∂).

2. Every automorphism of the Lie algebra div0_n is extendable to an automorphism of the Lie algebra divc_n, and the extension is unique if n≥2.

Proof. 1. The result follows from Theorem 1.2 and Theorem 2.5. 2. Statement 2 follows from Theorem 1.1 and Theorem 1.2.

b2Apr13

Lemma 2.20 For all n ≥ 2, Pn/K is a simple div0n-module/divcn-module; Enddiv0

n(Pn/K) =

Enddivc

n(Pn/K) = Kid whereid is the identity map. Forn= 1,P1/K is neither a simplediv 0 1 -module nor a simpledivc₁-module.

Proof. If n = 1 then div0₁ = K∂1, divc1 = K∂1+KH1 and Pi≤mKxi1∂1 where m ∈ N are

distinct div0₁/divc₁-submodules of P1. Suppose thatn≥2. It suffices to prove the statement for

div0_n. Let M be a nonzero div0_n-submodule of Pn/K and 0 6= m ∈ M. Using the actions of

∂1, . . . , ∂n∈div0n onmwe obtain an element ofM of the formλxi+Kfor some λ∈K∗. Hence,

xi+K∈M. Thenxj+K=xj∂i∗(xi+K)∈M for allj6=i. So, xα+K∈M for allα∈Nn

with|α|= 1. We use induction on |α| to show that allxα_{+K∈M. Suppose thatm:=|α|>1.}

Ifxα_+k=xm

i +K for somem≥2 andithenxmi +K= (m−1)−1xi(Hi−2Hj)∗(xm−i 1+K).

Then, by applying elements of the typexj∂i wherej =6 ito the elementxmi +K we obtain all the

elementsxα_{+K with|α|=m. Therefore,P}

n/K is a simplediv0n-module/divcn-module.

Letf ∈Enddiv0

n(Pn/K). Then applyingf to the equalities∂i∗(x1+K) =δi1fori= 1, . . . , n,

we obtain the equalities

∂i∗f(x1+K) =δi1 for i= 1, . . . , n.

Hence,f(x1+K)∈Tn_i=2kerPn/K(∂i)∩kerPn/K(∂ 2

1) = (K[x1]/K)∩kerPn/K(∂ 2

1) =K(x1+K).

So,f(x1+K) =λ(x1+K) and sof =λid, by the simplicity of thediv0n-modulePn/K. Therefore,

Enddiv0

n(Pn/K) = Enddivcn(Pn/K) =Kid.

a2Apr13

Proposition 2.21 Forn≥2,divc_n is a maximal Lie subalgebra of Dn. Forn= 1, divc1 is not a maximal Lie subalgebra ofD1. For eachn≥1,divcnis aGn-invariant/Gn-invariant Lie subalgebra

of Dn.

Proof. Forn= 1,divc₁=K∂1+KH1is contained in the Lie subalgebraK∂1+KH1+Kx1H1

ofD1. Suppose thatn≥2. By (34) and (36), divDnPn3

0→divc_n→Dn

div

→Pn/K →0 (49)

is the short exact sequence of divc_n-module. By Lemma 2.20, thedivc_n-module Pn/K is simple.

Then,divc_n is a maximal Lie subalgebra ofDn.

By Theorem 1.2, Theorem 2.6 and Theorem 2.5, divc_n is aGn-invariant/Gn-invariant Lie

c2Apr13

Lemma 2.22 Let n≥2. Then

1. Pn/K is a simpleGn-module with EndGn(Pn/K)≃K.

2. Dn/divcn

div

≃ Pn/K, an isomorphism ofGn-modules.

3. Dn/divcn is a simple Gn-module withEndGn(Pn/K)≃K.

Proof. 1. Let M be a Gn-submodule of Pn properly containing K. We have to show that

M = Pn, i.e. xα ∈ M for all α ∈ Nn. The polynomial algebra Pn = ⊕α∈NnKxα is the direct

sum of 1-dimensional non-isomorphicTn_{-modules. HenceM is a homogeneous submodule of P} n.

Hencexβ_{∈M for someβ6= 0. If β}

i6= 0 then using the automorphismsi :xi7→xi+ 1,xj7→xj,

j 6= i, we see that M ∋ si(xβ)−xβ and degxi(si(x

β_)−xβ_{) =β}

i−1. The module M is closed

under the mapssi−1 fori= 1, . . . , n. Hence, xγ ∈M for all γ≤β whereγ ≤β iffγi≤βi for

all i. In particular, all x1, . . . , xn ∈M. Then applying the automorphismσm : x1 7→xi+xm2, xi 7→xi fori 6= 1, to the element x1, we see thatM ∋(σm−1)(x1) =xm2 for allm ≥1. Then

applying the automorphismx27→Pni=1xi,xi 7→xi fori6= 2, we have (x1+· · ·+xn)m∈M. This

implies that allxα_{∈M, by the homogeneity ofM.}

Letf ∈EndGn(Pn/K). Sincef commutes with the action of the subgroupT

n _ofG

n, we must

have f(xα_{+K) = λ}

α(xα+K) for allα ∈ Nn and some λα ∈ K. In particular, f(x1+K) = λ(x1 +K) for some λ ∈ K. Since f commutes with the action of the symmetric group Sn

(which is obviously a subgroup ofGn),f(xi+K) =λ(xi+K) for alli= 1, . . . , n. Now, we use

induction on|α| show thatf(xα_{+K) =λ(x}α_{+K). The initial case when |α|= 1 has just been}

established. So, let|α|>1. Thenαi >0 for somei, and deg((si−1)xα+K)<|α|. By induction,

f((si−1)xα+K) =λ((si−1)xα+K). Now, it follows from the equality

f(si(xα) +K)−f(xα+K) =f((si−1)xα+K) =λ(si−1)xα+K

thatλα=λ, and sof =λid. Therefore, EndGn(Pn/K) =Kid.

2. Statement 2 follows from (49) and Theorem 2.5. 3. Statement 3 follows from statements 1 and 2.

a14Apr13

Lemma 2.23 The Lie algebradiv0_n is aGn-invariant/Gn-invariant Lie subalgebra of Dn.

Proof. The statement follows from Theorem 1.1, Theorem 2.6 and Theorem 2.5.

Conjecture: Every nonzero homomorphism of the Lie algebradiv0_n is an automorphism.

3

Minimal set of generators for the Lie algebras

div

and

div

TMMSG In this section, the proofs of Theorem 1.3 and Theorem 1.4 are given. _ditai2

[x2i∂i+1, xi+1∂i] =θeii. (50)

In more detail, LHS=x2i∂i−2xixi+1∂i+1=xi(Hi−2Hi+1) =θiei. ditai3

[x2

i+1∂i, xi∂i+1] =−θiei+1. (51)

Similarly, LHS=x2

i+1∂i+1−2xixi+1∂i=−xi+1(2Hi−Hi+1) =−θeii+1.

(i) G = div0_n: To prove that the equality holds we use induction on n. Let n = 2. So,

G=h∂1, x22∂1, x21∂2i. Then∂2, x1∂2, x2∂1∈ G:

x1∂2=1

2[∂1, x

2

1∂2], ∂2= [∂1, x1∂2], x2∂1= 1

2[∂2, x

2 2∂1].

By Lemma 2.1.(3), we have to show that the elementsxi

2∂1,xi1∂2andθ1α=xα((α2+ 1)H2−(α1+

1)H2) belong toG wherei∈Nandα∈N2. By (50) and (51),

θe1

1 = [x21∂2, x2∂1]∈ G, θ1e2 =−[x22∂1, x1∂2]∈ G.

Then using (44) and (45), we see that θα

1 ∈ G for all 06=α∈N2. Forα= 0, θ01 =H1−H2 =

[x1∂2, x2∂1]∈ G. By (26), forj≥1,

G ∋[x2∂1, θ1je2] =−((j+ 1)H1−H2,−e1+e2)xj2+1∂1= (j+ 2)xj2+1∂1.

By symmetry,G ∋(j+ 2)xj₁+1∂2, i.e. G=div0n.

Suppose thatn≥3, and that the equalityG=div0_n′ holds for alln′ such that 2≤n′ < n.

Step 1. {∂i, xνj∂k|i, j, k= 1, . . . , n;j 6=k;ν = 1,2} ⊆ G: Fori= 2, . . . , n,∂i= 1₂[∂1,[∂1, x21∂i]]∈

G andx1∂i = 1₂[∂1, x21∂i] ∈ G. Then xi∂1 = 1₂[∂i, x2i∂1] ∈ G fori = 2, . . . , n. For alli 6=j such

thati6= 1 andj6= 1,xi∂j= [xi∂1, x1∂j]∈ G andx2i∂j= [x2i∂1, x1∂j]∈ G. Fori= 2, . . . , n, fix an

index j such that j6= 1, i. Thenxi∂1= [xi∂j, xj∂1]∈ G and x2i∂1= [xi2∂j, xj∂1]∈ G. The proof

of Step 1 is complete.

Fori= 2, . . . , n, letdiv0_n,ibe the Lie algebradiv0_n−1for the polynomial algebraK[x1, . . . ,xbi, . . . , xn]

(xi is missed).

Step 2. Fori= 1, . . . , n,div0_n,i⊆ G: This follows from Step 1 and induction.

Step 3. di0_n⊆ G: This follows from (25) and Step 2.

Step 4. iv0_n⊆ G: This follows from (26) (whereαi= 0) and Step 3.

(ii)Minimality:

(a) The element∂1cannot be dropped: By Lemma 2.1.(3),div0nis aZ-graded Lie algebra which

is determined by the degree deg in the following way, forxα_∂

i ∈Pn∂i∂i, deg(xα∂i) =|α| −1 and

deg(θα

i) = |α| for all α ∈ Nn and i = 1, . . . , n−1. Clearly, deg(∂1) = −1 (negative) and the

degrees the rest of the generators are equal to 1 (positive). Therefore, the∂1cannot be dropped. (b) The element x2

i∂1 (i= 2, . . . , n) cannot be dropped: Since otherwise the Lie algebra

gen-erated by the remaining elements would belong to ⊕n

j=1K[x1, . . . ,xbi, . . . , xn]∂j (xi is missed), a

contradiction (see (i)).

(c) The element x2

1∂i (i= 2, . . . , n) cannot be dropped: Since otherwise the Lie algebra

gener-ated by the remaining elements would belong to⊕n

j6=iPn∂j, a contradiction (see (i)).

Proof of Theorem 1.4. By Theorem 1.3, the elements in Theorem 1.4 generate the Lie algebradivc_n=div0_n⊕KH1and the elementH1cannot be dropped (by Theorem 1.3).

(a) The element ∂1 cannot be dropped by the same reason as in the proof of Theorem 1.3 as

deg(H1) = 0.

(b) The element x2

i∂1 (i= 2, . . . , n) cannot be droppedby the same reason as in the proof of

Theorem 1.3.

(c) The element x2

1∂i (i= 2, . . . , n) cannot be droppedby the same reason as in the proof of

Theorem 1.3.

Acknowledgements

The work is partly supported by the Royal Society and EPSRC.

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Department of Pure Mathematics University of Sheffield

Hicks Building Sheffield S3 7RH UK