ideally constrained lie algebras

Journal of Algebra 253 (2002) 31–49

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Ideally constrained Lie algebras

Norberto Gaviolia,∗ and Valerio Montib

a Dipartimento di Matematica Pura ed Applicata, Università degli Studi dell’Aquila,Via Vetoio, I-67010 Coppito (L’Aquila) AQ, Italy

b Dipartimento di Metodi e Modelli Matematici per le Scienze Applicate, Università degli Studi diRoma ‘La Sapienza’, via Scarpa, 16, I-00161 Roma, Italy

Received 2 October 2000

Communicated by George Glauberman

Abstract

In this paper we deal with graded Lie algebrasL such that there exists a positive integerr

such that for every positive integeri and for every homogeneous idealI � Li the inclusionI ⊇ Li+r−1 holds. The solvable case and ther = 1 case receive a special attention. 2002Elsevier Science (USA). All rights reserved.

Keywords:Graded Lie algebras; Thin Lie algebras; Ideally constrained Lie algebras

1. Introduction

Graded Lie algebras whose lattice of homogeneous ideals satisfies some‘narrowness’ conditions have been studied in recent years. Some of theseconditions were first imposed as obvious counterparts of analogous conditionsabout (pro-)p-groups and their lattice of (open) normal subgroups. The theory of‘narrow’ Lie algebras became later a topic of independent interest rather than amere translation of the (pro-)p-groups case. Indeed the group case and the Liealgebra case can be rather different: as an example, Caranti et al. [1] showed thatthere are uncountably many non-solvable Lie algebras of maximal class, whileevery (pro-)p-group of finite coclass is solvable (see [2,3]).

* Corresponding author.E-mail addresses:[email protected] (N. Gavioli), [email protected] (V. Monti).

0021-8693/02/$ – see front matter 2002 Elsevier Science (USA). All rights reserved.PII: S0021-8693(02)00057-1

32 N. Gavioli, V. Monti / Journal of Algebra 253 (2002) 31–49

Here we studyideally r-constrainedLie algebras, i.e. graded Lie algebrasL generated by their first homogeneous component, in whichI � Li impliesI ⊇ Li+r−1 for every homogeneous idealI and positive integeri. These algebrasarise as a natural generalization of thin Lie algebras (namely 2-generated ideally1-constrained Lie algebras), see [4]. Bonmassar and Scoppola studied in [5]normally constrainedpro-p-groups, that are the group-theoretic analogue ofideally 1-constrained Lie algebras. In that paper they showed that, forp odd,a pro-p-groupG of class at least 3 is normally constrained if and only if theLie algebra associated to the lower central series ofG is ideally 1-constrained.There is no evidence of the existence of a similar relationship between the ideallyr-constrained Lie algebras and their group-theoretic analogue.

After some general remarks about ideallyr-constrained Lie algebras, inSection 3 we focus our attention on the solvable case. In Theorem 17 we showthat in order to verify that a solvable Lie algebraL is ideally r-constrainedit is sufficient to verify that the same is true for a suitable nilpotent quotientof L. Moreover, the Lie algebra structure ofL is almost entirely determinedby this quotient. This allows us to prove that a solvable ideallyr-constrainedLie algebra whose nilpotency class is large enough is a quotient of a uniquelydetermined universal solvable non-nilpotent ideallyr-constrained Lie algebra(see Theorem 22). Moreover, we prove that a finitely generated non-metabeliansolvable ideallyr-constrained Lie algebra whose nilpotency class is large enoughhas a basis field of finite characteristic and this characteristic satisfies an additionalcondition (see Proposition 19 for details). This suggests that ‘most’ finitelygenerated solvable ideallyr-constrained Lie algebras are metabelian. We thenclassify in Section 4 metabelian 2-generated non-nilpotent ideallyr-constrainedLie algebra.

Finally, Section 5 is devoted to a closer investigation of ideally 1-constrainedLie algebras: in particular, we show that a finitely generated ideally 1-constrainedLie algebra of nilpotency class at least 4 over a finite field can be regarded as thecentral quotient of a 2-generated ideally 1-constrained Lie algebra over a finiteextension of the base field (see Corollary 33). This result is mainly a rewritingof a similar statement about the Lie algebra associated to a normally constrainedpro-p-group, proved by Bonmassar and Scoppola [5].

2. Notation and elementary results

Throughout this paper aLie algebrais a graded Lie algebraL= ⊕i�1Li over

a fieldF , generated byL1. Accordingly a homomorphism of algebras is a gradedhomomorphism of Lie algebras, an ideal is a homogeneous ideal and a quotientof a Lie algebraL is a quotient ofL over a homogeneous idealI : asL/I isgenerated by its elements of degree 1 we get thatL/I is a Lie algebra in the senseused above. IfI is an ideal ofL, we writeIi for I ∩Li .

N. Gavioli, V. Monti / Journal of Algebra 253 (2002) 31–49 33

Obviously, a Lie algebraL is finitely generated if and only if dimL1 is finite.We use the symbolLk to denote thekth term of the lower central series ofL, andthe symbolL(k) to denote thekth term of the derived series ofL. A Lie algebraLis nilpotent ifLk = {0} for some positive integerk: in this case we say thatL has(nilpotency) classc if Lc = {0} andLc+1 = {0}. If a Lie algebraL is not nilpotentwe say thatL has infinite (nilpotency) class.

For a subsetX of a Lie algebraL define recursively[X,rL1] to be the subspace[X,L1] if r = 1 and[X,rL1] = [[X, (r − 1)L1],L1] if r > 1.

The finite field withq elements is denoted by GF(q).For a setS letFF (S) (or simplyF(S)) be the free Lie algebra over the fieldF

on the setS, with the grading induced by deg(x)= 1 for x ∈ S.For the convenience of the reader we collect here some elementary results

which we shall often use without further reference.

Proposition 1. LetL be a Lie algebra. Then:

(1) Li+1 = [Li,L1] for every positive integeri;(2) if Ik is a subspace ofLk andI is the ideal generated byIk , thenI = ⊕

i�k Ii ,whereIi = [Ii−1,L1] for i > k;

(3) thekth termLk of the lower central series ofL is⊕

i�k Li;(4) L has finite nilpotency classc if and only ifLc = {0} andLc+1 = {0}: in this

caseLi = {0} for all i > c.

We need the following easy proposition.

Proposition 2. LetL be a Lie algebra and letr be a positive integer. The followingconditions are equivalent:

(1) for every nonzero idealI of L there exists a positive integeri such thatLi ⊇ I ⊇ Li+r ;

(2) for every idealI of L and for every positive integeri one has thatI ⊆ Li orI ⊇ Li+r−1;

(3) [x, rL1] = Ldegx+r for every nonzero homogeneous elementx of L.

Proof. Conditions (1) and (2) are trivially equivalent. If condition (2) holds thenthe idealI generated by a nonzero homogeneous elementx of degreei is notcontained inLi+1 and then it containsLi+r . As Ii+r = [x, rL1] condition (3) issatisfied. Finally, assume that condition (3) holds. IfI is a nonzero ideal, letibe the maximum integer such thatI ⊆ Li and letx be a nonzero homogeneouselement inI of degreei: thereforeI ⊇ [x, rL1] = Li+r so thatI ⊇ Li+r andcondition (1) holds. ✷


Definition 3. A Lie algebra is said to beideally r-constrainedif it satisfies oneof the equivalent conditions of Proposition 2. We say that a Lie algebra isideallyfinitely constrainedif it is ideally r-constrained for some positive integerr.

The first elementary properties of ideally finitely constrained Lie algebras arecollected in the following immediate remark.

Remark 4. LetL be an ideallyr-constrained Lie algebra. Then:

(1) every quotient ofL is ideallyr-constrained;(2) every proper quotient ofL has finite class;(3) if L has infinite class, then the intersection of two nonzero ideals ofL is

nonzero;(4) if L has finite nilpotency classc thenLmax(1,c−r+1) ⊇ Z(L) ⊇ Lc , if L has

infinite class then it is centerless;(5) for everyi we have dimLi � (dimL1)

r .

We remind the reader that a Lie algebraL is said to bejust infinite ifdimL = ∞ and every proper quotient ofL has finite dimension. Thereforea finitely generated ideallyr-constrained Lie algebra is just infinite or finitedimensional.

Leedham-Green, Plesken and Klaas have introduced the concept ofobliquityof a pro-p-group (see Definition 1.5 in [6]). This suggests a similar definition forthe obliquity of a Lie algebra.

Definition 5. Let L be a Lie algebra and leti > 0. Defineµi(L) to be theintersection ofLi+1 with the intersection of all ideals ofL not contained inLi+1.Theobliquityo(L) of L is max{dim(Li+1/µi(L)) | i > 0} if the maximum exists;otherwiseo(L)= ∞.

We now point out that for a finitely generated Lie algebra the property of‘having finite obliquity’ is equivalent to that of ‘being ideally finitely constrained.’

Proposition 6. If L is a Lie algebra of finite obliquityr thenL is ideally(r + 1)-constrained. IfL is a d-generated ideallyr-constrained Lie algebra thenL hasobliquity at mostrdr .

Proof. Assume thatL has obliquity r. Let I be a nonzero ideal ofL andlet i be the maximum integer such thatI ⊆ Li . ThereforeI ⊇ µi(L) so thatdim(Li+1/(Li+1 ∩ I)) � r. This implies thatLj = Ij for some integerj , suchthati + 1 � j � i + r + 1, so thatI ⊇ Lj ⊇ Li+r+1. Conversely assume thatL is


an ideallyr-constrainedd-generated Lie algebra: in this case every ideal which isnot contained inLi+1 containsLi+r . This means thatµi(L)⊇ Li+r ; therefore

dim(Li+1/µi(L)

)�

i+r∑i+1

dimLj � rdr

by Remark 4.4. ✷

3. Solvable ideally finitely constrained Lie algebras

We give some further notation and definitions.

Notation 7. We denote by the symbolLj the class of Lie algebrasL such thatLj

is abelian.

Definition 8. The degreedeg(I) of a nonzero idealI of a Lie algebraL is theminimum integern such thatIn = {0}.

Definition 9. The abelian radicalof a Lie algebraL, denoted by RadAb(L), isthe sum of all the abelian ideals ofL. The j th abelian radicalof L, denotedby RadjAb(L), is the sum of all the abelian ideals ofL containingLj . If

RadAb(L) = {0} (respectively RadjAb(L) = {0}) we define theabelian degree

degAb(L) of L (respectively thej th abelian degreedegjAb(L) of L) to be the

degree of RadAb(L) (respectively of RadjAb(L)).

We note that RadjAb(L) = {0} if and only if L ∈ Lj . We can now give somecharacterizations of solvable ideally finitely constrained Lie algebras.

Proposition 10. If L is an ideally finitely constrained Lie algebra, then thefollowing conditions are equivalent:

(1) L is solvable;(2) L contains a nonzero abelian ideal;(3) RadAb(L) = {0};(4) L ∈ Lj for some positive integerj .

Proof. It is obvious that (1) implies (2), and that (2) and (3) are equivalent,while (2) implies (4) by Proposition 2. AsL(k) ⊆ L2k−1

we easily see that (4)implies (1). ✷

By combining this result with Remark 4.4 we get the following corollary.


Corollary 11. If L is a non-solvable ideally finitely constrained Lie algebra thenCL(I)= {0} for every nonzero idealI .

An easy application of Proposition 2 gives following remark.

Remark 12. Let L ∈ Lj+1 − Lj be a solvable ideallyr-constrained Lie algebra.

Thenj − r + 1 � degAb(L) � degj+1Ab (L) � j + 1.

Under suitable hypothesis the abelian radical of an ideally finitely constrainedLie algebra is abelian, so that there exists a unique maximal abelian ideal.

Proposition 13. Let L ∈ Lj+1 be an ideallyr-constrained Lie algebra of class

at least 2j + r. ThenRadj+1Ab (L) is abelian andRadj+1

Ab (L) = CL(Lj+1). If,

moreover,L has class at least2j + 2r − 1 thenRadj+1Ab (L)= RadAb(L).

Proof. We prove the first statement by showing thatCL(Lj+1) is abelian. By way

of contradiction we assume that this is false. Two non-commuting homogeneouselementsw and y of CL(L

j+1) necessarily have degree at mostj , so that[[w,y], rL1] = {0}. If we choosew andy such that degw + degy is as largeas possible, then, for every homogeneousx ∈L, we obtain

[w,y, x] = −[x,w,y] − [y, x,w] = 0,

and this implies[[w,y], rL1] = {0}, a contradiction.To prove the second statement we consider an abelian idealI of L and we

assume, by way of contradiction, thatI � CL(Lj+1). In particular,I � Lj+1,

so thatI ⊇ Lj+r . If u is a homogeneous element inI − CL(Lj+1) and v is

a homogeneous element inLj+1 − CL(u), thenu /∈ Lj+1 and v /∈ I , so thatdegu � j and degv � j + r − 1. This implies that[[u,v], rL1] = {0}. If wechooseu andv of maximal degree we obtain[u,v, x] = −[x,u, v]− [v, x,u] = 0for every homogeneousx ∈L, a contradiction. ✷

The following lemma can be proved by a straightforward application of Jacobiidentity.

Lemma 14. Let I be an abelian ideal of a Lie algebraL. If y is a homogeneouselement ofL such that[y,L] ⊆ I thenCI (y) and[I, y] are ideals.

We need two more lemmas.

Lemma 15. If L is a Lie algebra andh is a positive integer such thatL/Lh+1

is ideallyr-constrained, then for every integeri such thati + r − 1 � h and forevery idealI , one has thatI ⊆ Li or I ⊇ Li+r−1.


Proof. Assuming thatI � Li , by passing to the quotientL/Lh+1, we deduce thatI + Lh+1 containsLi+r−1. ThenI ⊇ Li+r−1, and by Proposition 1.1 it followsthatI ⊇ Li+r−1. ✷Lemma 16. Let L andL be Lie algebras belonging toLj+1 and letρ : L → L

be a homomorphism. Ifkerρ ⊆ L2j+2 andI is an abelian ideal ofL containingLj+1 then Iρ−1 is an abelian ideal ofL. In particular, if ρ is surjective thenRadj+1

Ab (L) = Radj+1Ab (L)ρ−1.

Proof. Let J be maximal in the family of the abelian ideals ofL that containLj+1 and are contained inIρ−1. By way of contradiction assumeJ = Iρ−1: lety be a homogeneous element inIρ−1 − J of maximum degree. As degy � j

and[y,J ] ⊆ kerρ ⊆ L2j+2 it follows that, for i � j + 1, we have[y,Ji] = {0},i.e.CJ (y)∩ Li = Ji . In particular,CJ (y) containsLj+1: by Lemma 14CJ (y) isan ideal so thatCJ (y) ⊇ Lj+1. ThereforeCJ (y) = J andJ + Fy would be anabelian ideal contradicting our choice ofJ . ✷

We are now ready to prove the following theorem.

Theorem 17. Let L ∈ Lj+1 − Lj be a Lie algebra of(j + 1)th abelian degreed and classc at least j + d + 2r − 1, where r is a positive integer. PutR := Radj+1

Ab (L). If L/Lj+d+2r is ideallyr-constrained then:

(1) L is ideallyr-constrained;(2) R is abelian andLj � R;(3) if y ∈ Lj −R thenady induces an injective linear mapping fromRk in Rk+j

for d � k � c − j − r;(4) if y ∈Lj −R thenady induces a surjective linear mapping fromRk in Rk+j

for k � d + r;(5) dimLk = dimLk+j for d + r � k � c− j − r.

Proof. By Lemma 15 an abelian ideal of degreed containsLd+r which thenturns out to be abelian. AsLj is not abelian this impliesd + r � j + 1. ApplyingProposition 13 toL/Lj+d+2r and Lemma 16 to the canonical projection ofL overL/Lj+d+2r we get (2).

If y ∈ Lj − R, thenR + Fy is a non-abelian ideal ofL, that is[R,y] = {0}.Moreover, Lemma 14 shows thatCR(y) and[R,y] are ideals ofL. Assume, byway of contradiction, that[R,y] � Lj+d+r : then[R,y] ⊆ Lj+d+1 (Lemma 15).This implies that[Rd,y] = {0}, i.e. CR(y) ⊇ Rd , thereforeCR(y) ⊇ Ld+r

(Lemma 15). In particular we have[R,y]j+d+2r−1 = [Rd+2r−1, y] = {0}, so that[R,y] ⊆ Lj+d+r (Lemma 15). Combining this with the inclusionCR(y)⊇ Ld+r


we would get[R,y] = {0}, a contradiction. Therefore[R,y] ⊇ Lj+d+r andstatement (4) is proved.

We now prove by induction oni that an idealI of degreei containsLi+r .This is true fori � j + d + r − 1 in force of Lemma 15 and the hypothesis.If i � j + d + r we can easily check that the elementsx of R such that[x, y] ∈ I form an idealI∗ sinceR is abelian. By definition[I∗, y] ⊆ I sothat statement (4) shows that[I∗, y] = I and that degI∗ � i − j . By inductivehypothesisI∗ ⊇ Li+r−j , so thatI contains[Li+r−j , y]. As [Li+r−j , y] = Li+r

by statement (4), we get statement (1).To prove statement (3) we may assume, without loss of generality, thatc is

finite. Statement (4) implies thatLc−j � CR(y): then CR(y) ⊆ Lc−j−r+1 asclaimed.

Finally statement (5) follows readily from statements (3) and (4) (note that asd � j + 1− r thenR ⊇ Ld+r ). ✷

Statement (1) of Theorem 17 implies the following corollary.

Corollary 18. Let L ∈ Lj+1 − Lj and letd := degj+1Ab (L). ThenL is ideally r-

constrained if and only ifL/Lj+d+2r is ideallyr-constrained.

We also have the next proposition.

Proposition 19. Let L ∈ Lj+1 − Lj (j � 2) be a finitely generated ideallyr-constrained Lie algebra of abelian degreed and classc at least2j + d + 2r − 1.Then charF is a prime number dividingdimF Ld+r/Lj+d+r (in particular,charF � j (dimL1)

r by Remark4.5).

Proof. By Proposition 13 we know that degj+1Ab (L) = d . We use statement (2)

of Theorem 17 and we choosey ∈ Lj − Radj+1Ab (L). Let A be the polynomial

ring F [X]/(X2). The quotientM := Ld+r/L2j+d+r can be regarded as a rightA-module if we identify the multiplication forX with the map induced by ady.By Proposition 13 Radj+1

Ab (L) is abelian and it containsLd+r by Remark 12:the Jacobi identity then shows that, for everyx ∈ L, the maps ady and adxcommute onLd+r , thus ad induces a homomorphismof (non-graded) Lie algebrasfrom L into EndAM. Fix anF -basisB for

⊕j+d+r−1d+r Li : by statements (3) and

(4) of Theorem 17 the basisB induces a basisB for M as anA-module. Therepresentative matrix of the endomorphism ofM induced by ady with respect tothe basisB is X · Id. As y ∈ L2 this matrix has trace 0 and we get our claim.✷

The following theorem shows that in a solvable ideally finitely constrained Liealgebra of infinite class, the homogeneous components of large degree can be


regarded as vector spaces over an extension of the base field: we obtain in thisway a lower bound for their dimension over the base field.

Theorem 20. Let L ∈ Lj+1 − Lj be an ideallyr-constrained Lie algebra ofinfinite class, letd := degAb(L), and assume thatk � d + r. ThenLk can beregarded as aK-vector space whereK is a field extension ofF . TheK-vectorspace and theF -vector space structures onLk are compatible,adx is aK-linearmapping ofLd+r in itself for all x ∈ L anddimF K � dimF (Lj /RadAb(L)j ).

Proof. Let A be the ring of the degree-preservingF -linear mappings ofLd+r initself commuting with adx for all x ∈L. If ϕ is a nonzero element ofA then Imϕ

and kerϕ are ideals ofL.For everyk � d + r we denote byϕk the endomorphism ofLk induced byϕ.

If kerϕ = {0} then, by Remark 4.3,Lk ⊆ Imϕ ∩ kerϕ for sufficiently largek, thatis ϕk is surjective andϕk = 0, a contradiction. Thereforeϕ is injective. Choosey ∈ Lj − RadAb(L): Theorem 17 shows that, fork � d + r, ady induces anisomorphism betweenLk andLk+j commuting withϕ, so thatϕk is surjectiveif and only if ϕk+j is surjective. AsLk ⊆ Imϕ for sufficiently largek, thismeans thatϕ is surjective. Soϕ is bijective and, obviously,ϕ−1 ∈ A. ThenAis a skew field. PutK := Z(A): to complete the proof it remains to show thatdimF K � dimF (Lj /RadAb(L)j ). In order to do this, denote byπ the ho-momorphism fromLd+r in Lj+d+r induced by ady: this is an isomorphismby Theorem 17. Consider the linear mappingS from Lj into A definedby w �→ (adw)π−1: a direct computation shows thatS is well defined andImS ⊆Z(A), while kerS = CL(L

d+r )j = Radd+rAb (L)j = RadAb(L)j by Propo-

sition 13. ✷Theorem 17 shows that the homogeneous components of large degree of

a solvable ideallyr-constrained Lie algebraL are periodic, i.e., with the samenotation as in the cited theorem, one has thatLk

∼= Lk+j for large values ofk.This suggests the possibility to extend these algebras by periodically ‘adding’homogeneous components. This idea is carried on in the following two theorems.

Theorem 21. Let L be a solvable ideallyr-constrained Lie algebra belongingto Lj+1. Suppose further thatL has class at least

m := max(j + d + 3r − 1,2j + r + 1) whered := degAb(L).

Then there exists an ideallyr-constrained Lie algebraUj+1r (L) ∈ Lj+1 and

a surjective homomorphism of Lie algebrasπ :Uj+1r (L) → L such that for

every ideally r-constrained Lie algebraL ∈ Lj+1 and for every surjectivehomomorphism of Lie algebrasρ : L → L there exists a unique homomorphismσ :Uj+1

r (L) → L such thatπ = σρ. Moreover Uj+1r (L) is unique (up to

isomorphism), it is not a proper quotient of any ideallyr-constrained Lie algebraof Lj+1 anddegAb(U

j+1r (L))= degAb(L).


Proof. Let j ′ be the minimal integer such thatL ∈ Lj ′+1: clearly j ′ � j .Remark 12 yieldsd + r � j ′ +1. By hypothesisL has then class at least 2j ′ +2r,

so that Proposition 13 shows thatR := Radj′+1

Ab (L) is the unique maximal abelianideal ofL.

Choose a basisS for L1 and consider the free Lie algebraF on the setS. Letϕ be the homomorphism fromF ontoL induced by the identity ofS. Let I bethe ideal ofF generated by[F j+1,F j+1] and by the homogeneous elements ofkerϕ of degree at mostm − r. SetUj+1

r (L) := F/I (we shall write simplyUfor Uj+1

r (L)) and letπ :U → L be the surjective homomorphism induced byϕ.As kerπ ⊆ Um−r+1 ⊆ U2j+2, Lemma 16 shows thatRπ−1 is an abelian idealof U . In particular,U ∈ Lj ′+1 and degAb(U) = degj

′+1Ab (U) = d . By construction

U/Uj ′+d+2r � L/Lj ′+d+2r so that statement (1) of Theorem 17 implies thatU isideally r-constrained.

If L ∈ Lj+1 is an ideallyr-constrained Lie algebra andρ : L → L is a surjec-tive homomorphism of Lie algebras, then there exists a homomorphismν :F → L

such thatνρ = ϕ. As kerρ � Lm we have that kerρ ⊆ Lm−r+1: this implies thatI ⊆ kerν and that there exists a homomorphismσ :U → L such thatπ = σρ.This homomorphism is unique sinceπ induces an isomorphism betweenL1 andL1 andρ induces an isomorphism betweenL1 andL1. The other statements areobvious. ✷Theorem 22. Let L ∈ Lj+1 − Lj be an ideallyr-constrained Lie algebra ofabelian degreed and class at leastmax(3j + d + 2r − 1, j + d + 3r − 1). ThenUj+1r (L) has infinite class.

Proof. By way of contradiction we assume thatU := Uj+1r (L) has finite classc,

and we claim thatU is a proper quotient of an ideallyr-constrained Lie algebraU ∈ Lj+1, against Theorem 21. We note thatU ∈ Lj+1 − Lj , so that weuse statement (2) of Theorem 17 and we fix an elementy ∈ Uj − R, whereR := RadAb(U). ClearlyCR(y) ⊇ Uc−j+1, while statement (3) of Theorem 17yields CR(y) ⊆ Uc−j−r+1 so that if we choose a homogeneous elementx inCR(y) of minimum degreet we have

c − j − r + 1 � t � c− j + 1. (1)

We now give a construction forU : the idea is to ‘add’ toU a nontrivial elementand to identify this element with[x, y]. As a vector space we setU := ⊕

Ui ,where Ui := Ui for i = j + t and Uj+t := Uj+t ⊕ E, with E a vector spaceof dimension 1. Note thatU becomes a subspace ofU in an obvious way. Wedenote byπ : U → U the projection on the first componentU = U ⊕ E. Wefix now a complementHt of Fx in Ut , and we setH := Ht ⊕ ⊕

i =t Ui so that

U = Fx⊕H : let τ : U → Fx be the projection on the first component. We denoteby ψ : U → E the composition ofτ with an isomorphismυ from Fx ontoE.


Finally, by Theorem 17, we know that ady induces an isomorphism between⊕t−1d+r Ui and

⊕j+t−1j+d+r Ui : we extend the inverse of this isomorphism to a linear

mappingθ of U in R by lettingθ = 0 overUi for i < j + d + r or i > j + t − 1.We are now ready to define a Lie product inU . If z andw are elements ofU weset:

[z,w] := [zπ,wπ] + [zθ,wπ]ψ + [zπ,wθ ]ψ. (2)

This product is bilinear, respects the grading ofU and satisfies the identity[z, z] = 0 for everyz ∈ U . To prove thatU is a Lie algebra it remains only to verifythat the Jacobi identity holds. Before doing this, we collect some elementaryproperties of the product just defined:

(1) [z,w]π = [zπ,wπ] for everyz, w ∈ U ;(2) [z,w] = 0 if z andw belong toR ⊕E;(3) [E, U] = {0};(4) [z,w,v] = [z,w,v]π + [[z,w]πθ, vπ]ψ + [[z,w]π,vθ ]ψ .

If we can prove that the product satisfies the Jacobi identity, then property (1)yields thatU is a (proper) quotient ofU , while property (2) yieldsU ∈ Lj+1

and degj+1Ab (U) = d . As a consequence Theorem 17 implies thatU is ideally

r-constrained becauseU/Uj+d+2r � U/Uj+d+2r . To complete the proof of thetheorem we are only left to show that the Jacobi identity holds.

It is enough to consider three homogeneous elementsz,w andv of U . If two (orthree) of them have degree at leastj + 1 then, by property (2), the Jacobi identitytrivially holds. The same is true, by property (1), if degz+ degw+ degv = j + t .We only have to consider the case degz + degw + degv = j + t , degw � j ,degv � j .

By property (4) our claim is equivalent to:[[zπ,wπ]θ, vπ]ψ + [wπ,vπ, zθ ]ψ + [[vπ, zπ]θ,wπ]

ψ = 0. (3)

The hypothesis onz, w andv imply thatt − j � degz � j + t −2: combining thehypothesis onc with inequality (1) we may deduce that degz � j + d + r. Thenthe definition ofθ yields zπ = [zθ, y], so that an easy application of the Jacobiidentity (inU of course) gives

[zπ,wπ] = [zθ, y,wπ] = [zθ,wπ,y]. (4)

As before we may observe thatd + r � deg[zθ,wπ] � t − 1, so that we obtain[zθ,wπ] = [zθ,wπ,y]θ . By combining this relation with relation (4) we get[zπ,wπ]θ = [zθ,wπ]. Similarly [vπ, zπ]θ = [vπ, zθ ]. By using these equalitiesrelation (3) reduces to:

[zθ,wπ,vπ]ψ + [wπ,vπ, zθ ]ψ + [vπ, zθ,wπ]ψ = 0

which is an immediate consequence of the Jacobi identity inU . ✷


4. 2-generated metabelian ideally finitely constrained Lie algebras ofinfinite class

In this section we shall classify the 2-generated metabelian ideally finitelyconstrained Lie algebras of infinite class. We begin by considering the freemetabelian Lie algebraM in 2 generatorsx andy with the grading induced byM1 := 〈x, y〉. It is well known that a basis forMi , with i � 2, is given by thei − 1 Lie commutators

[y, x, . . . , x︸︷︷︸j+1 times

, y, . . . , y︸︷︷︸i−2−j times

] = [y, x](adx)j (ady)i−2−j

for 0 � j � i − 2. If c ∈ M2 then [c, y, x] = [c, x, y]: this allows us to definean action of the polynomial ring in 2 indeterminatesF [X,Y ] on M2 by settingcX := [c, x] andcY := [c, y]. By comparing the dimensions of the homogeneouscomponents we see thatM2 is a freeF [X,Y ]-module of rank 1 with basis[y, x]. In particular, every homogeneous elementc of M2 can be written uniquelyas [y, x]f (X,Y ) wheref (X,Y ) is a homogeneous polynomial and the idealgenerated byc is the set of the elements[y, x]f (X,Y )g(X,Y ) asg(X,Y ) variesin F [X,Y ]. We now introduce the following:

Notation 23. If f (X,Y ) is a nonzero homogeneous polynomial ofF [X,Y ] ofdegree at least 1 we denote byM(f (X,Y )) the quotient Lie algebra ofM overthe ideal generated by[y, x]f (X,Y ).

We are now ready to prove the following theorem.

Theorem 24.

(1) If f (X,Y ) has degreed thendimM(f (X,Y ))i = i − 1 for 2 � i � d andM(f (X,Y ))i = d for i � d + 1;

(2) M(f (X,Y ))�M(g(X,Y )) if and only if there exists a matrixA ∈ GL(2,F )and an elementk ∈ F ∗ such thatf ((X,Y )A)= kg(X,Y );

(3) M(f (X,Y )) is ideally finitely constrained if and only iff (X,Y ) isirreducible. More precisely:(a) if f (X,Y ) has degreed > 1 then M(f (X,Y )) is ideally (d − 1)-

constrained,(b) if f (X,Y ) has degree1 then M(f (X,Y )) is the unique infinite-

dimensional metabelian Lie algebra of maximal class;(4) if L is a 2-generated metabelian ideally finitely constrained Lie algebra

of infinite class thenL � M(f (X,Y )) for some irreducible polynomialf (X,Y ).


Proof. If f (X,Y ) is a homogeneous polynomial of degreed andI is the idealgenerated by[y, x]f (X,Y ) then the above remarks show that dimIi = i − d − 1for i � d + 2 and dimIi = 0 otherwise. Statement (1) follows readily.

Statement (2) easily follows from the fact thatM(f (X,Y ))�M(g(X,Y )) ifand only if there exists a basisx ′, y ′ of M1 and a nonzero elementh of F suchthat[y, x]f (X,Y )= h[y ′, x ′]g(X′, Y ′) whereX′ = adx ′, Y ′ = ady ′.

Consider a reducible homogeneous polynomialf (X,Y ) and letg(X,Y ) bea divisor of f (X,Y ) satisfying 0< degg(X,Y ) < degf (X,Y ) and necessar-ily homogeneous. HenceM(f (X,Y )) has a proper quotient isomorphic toM(g(X,Y )), which is of infinite class, so thatM(f (X,Y )) is not ideally finitelyconstrained by Remark 4.

Conversely, consider an irreducible homogeneous polynomialf (X,Y ) ofdegreed : if d = 1, then, by statement (1),M(f (X,Y )) is a Lie algebra ofmaximal class which is well known to be ideally 1-constrained. Assume thenthatd � 2 and letz ∈ M(f (X,Y ))i − {0} for a positive integeri. If i = 1 then[z,M(f (X,Y ))1] = M(f (X,Y ))2, so that[z, (d − 1)L1] = M(f (X,Y ))d . Ifi > 1 thenz = [y, x]g(X,Y )+ ([y, x]f(X,Y )) whereg(X,Y ) is a homogeneouspolynomial of degreei − 2 which is not a multiple off (X,Y ). Let J be theideal ofM generated by[y, x]g(X,Y ): thenJi+d−1 is the set of the elements[y, x]g(X,Y )h(X,Y ) whereh(X,Y ) ranges over the set of the homogeneouspolynomials of degreed−1. As a direct consequence we get that dimJi+d−1 = d

and thatJi+d−1 does not contain any element of([y, x]f (X,Y )). Then in thequotientM(f (X,Y )) we get that dim[z, (d − 1)L1] = d : by statement (1) thismeans that[z, (d − 1)L1] =M(f (X,Y ))i+d−1. This proves statement (3).

Finally let L be as in statement (4): asM is not ideally finitely constrained(for it has proper quotients of infinite class), it follows thatL is a proper quotientof M. SoL = M/I for a nonzero idealI : let [y, x]f (X,Y ) be a homogeneouselement ofI of minimum degreed + 2. We note that dimLd+1 = d . Let J bethe ideal generated by[y, x]f (X,Y ): if I � J then there existsz ∈ Ii − J forsome integeri � d + 2. In this case dimLi < dimM(f (X,Y ))i . By statement(1) dimM(f (X,Y ))i = d , so that dimLi < dimLd+1 contradicting statement(3) of Theorem 17. ThenI = J andL �M(f (X,Y )). ✷

5. Ideally 1-constrained Lie algebras

In this section we shall study in detail ideally 1-constrained Lie algebras.

Proposition 25. An ideally1-constrained nilpotent Lie algebraL has no non-homogeneous ideals.

Proof. The proof is by induction on the classc of L, the casec = 1 being trivial.Assume thatc > 1 and letI be an ideal ofL. We apply the inductive hypothesis


to L/Lc and deduce that there existsi such thatLi ⊇ I + Lc � Li+1. Thenthere existxi ∈ Li − {0} andxc ∈ Lc such thatxi + xc ∈ I . ThereforeI contains[xi +xc,L1] = [xi,L1] = Li+1. Since the ideal generated byLi+1 isLi+1 we getLi ⊇ I ⊇ Li+1. This obviously implies thatI is homogeneous.✷Proposition 26. LetL be a finitely generated ideally1-constrained Lie algebra ofclass at least2. ThendimL1 is even and, if the fieldF is finite,dimL1 � 2 dimL2.

Proof. Let x1, x2, . . . , xn be a basis ofL1 andw1,w2, . . . ,wr be a basis ofL2.Let [xj , xk] = ∑r

h=1αjhkwh, with αjhk ∈ F . For everyΓ = (γ1, γ2, . . . , γn) ∈ Fn

and every∆ = (δ1, δ2, . . . , δr) ∈ Fr we define the matricesD(Γ ) ∈ M(r,n,F )

andC(∆) ∈M(n,F ) by settingD(Γ )hk = ∑j γjαjhk andC(∆)jk = ∑

h δhαjhk .An elementx belongs toLi − {0} if and only if x = ∑

j γjxj for someΓ = (γ1, γ2, . . . , γn) ∈ Fn − {(0,0, . . . ,0)}. The subspace[x,L1] is generatedby {[x, xk]}nk=1, that is by {∑j,h γjαjhkwh}k. These elements generateL2 ifand only if rkD(Γ ) = r. The matrixD(Γ ) has rankr for everyΓ = 0 if andonly if ∆D(Γ ) = 0 for every∆ = 0 and everyΓ = 0. As ∆D(Γ ) = Γ C(∆)

this happens if and only ifC(∆) is a nonsingular matrix for every∆ = 0.By constructionC(∆) is an alternating matrix,1 so thatn is even. Moreover,detC(∆) = (PfC(∆))2. So PfC(∆) is a homogeneous polynomial of degreen/2, in ther unknownsδ1, δ2, . . . , δr , with no nontrivial root. The theorem ofChevalley–Warning implies thatr � n/2 if F is finite. ✷

The following example shows that in the second statement of Proposition 26the hypothesis thatF is finite cannot be dropped.

Example 27. LetK be any field: consider the fieldK(x,y) of the quotients of thering K[x, y] of the polynomials in the unknownsx andy with coefficients inK.We use same the notation and arguments as in Proposition 26, and we consider analgebraL over the fieldK(x,y) of class 2, with dimL1 = 4, dimL2 = 3, wherewe define the constantsαjhk so that the matrixC(∆) is the following:

C(∆)=

0 δ3 δ1 δ2x

−δ3 0 −δ2 δ1−δ1 δ2 0 −δ3y

−δ2x −δ1 δ3y 0

.

It can be easily seen that actuallyL is a Lie algebra. The Pfaffian of the matrixC(∆) is PfC(∆) = δ2

1 + δ22x + δ2

3y. By a standard argument follows that PfC(∆)

vanishes if and only ifδ1 = δ2 = δ3 = 0. ThereforeL is ideally 1-constrained.

1 We recall that a matrixC is said alternating ifC +Ct = 0, and the elements of the main diagonalof C are 0. The latter condition is redundant in characteristic other than 2. The Pfaffian PfC of analternating matrixC is a polynomial in its entries such that detC = (PfC)2.


In the particular casesK = Q or K = R, we can substitute 1 tox and y

and get a matrix whose Pfaffian isδ21 + δ2

2 + δ23, that vanishes if and only if

δ1 = δ2 = δ3 = 0. Then the second statement of Proposition 26 does not holdeven if the base field isQ or R.

To proceed in our analysis we need the following lemma.

Lemma 28. LetL be an ideally1-constrained Lie algebra.

(1) If x ∈Li − {0}, thendimCL1(x)+ dimLi+1 = dimL1.(2) If L is finitely generated anddimLi+1 = 1 then

dimCL1(Li)= dimL1 − dimLi.

(3) If L is finitely generated andLi+1 = {0} then

dimCL1(Li)� dimL1 − dimLi.

Proof. The first statement clearly follows from the equality[x,L1] = Li+1.In order to prove the second statement take a basisx1, x2, . . . , xr for Li : asCL1(Li) = ⋂r

j=1CL1(xj ) and dimCL1(xj ) = dimL1 − 1 for everyj , we findthat dimCL1(Li) � dimL1 − dimLi . On the other hand, the adjoint mappinginduces an injective linear mapping fromLi into hom(L1/CL1(Li),Li+1) sothat dimLi � dimL1 − dimCL1(Li). Finally, the third statement follows easilyfrom the second one by taking a quotient ofL over an ideal with codimension 1in Li+1. ✷

We now give a lower bound for the dimension of the homogeneous componentsof a finitely generated ideally 1-constrained Lie algebra.

Theorem 29. Let L be a finitely generated ideally1-constrained Lie algebra ofclassc such thatdimL1 = 2n. Thenn � dimLi � 2n for everyi, i < c.

Proof. The second inequality is just Remark 4.4. In order to prove the first one,we may assume thatc is finite and proceed by induction onc, the casesc = 1,2being trivial. So letc > 2: inductive hypothesis implies that dimLi � n fori � c−2. By way of contradiction letr := dimLc−1 < n. By passing to a suitablequotient ofL if necessary, we may assume that dimLc = 1. SetC := CL1(Lc−1)

and choosex ∈Lc−2 −{0}: Lemma 28 then yields dimC = dimCL1(x)= 2n− r.ThereforeC ∩ CL1(x) = {0}: asCL(Lc−1) ∩ CL(x) is an ideal (it is the cen-tralizer of the idealLc−1 + Fx) this means thatCL(Lc−1) ∩ CL(x) ⊇ L2.This is true for everyx ∈ Lc−2 so that [L2,Lc−2] = {0}. We know that[Lc−2,L1,C] = [Lc−1,C] = {0}, and[L1,C,Lc−2] = [L2,Lc−2] = {0}, so that


[C,Lc−2,L1] = {0}. This implies that [C,Lc−2] = {0} since, otherwise,[C,Lc−2,L1] = Lc. Therefore

dimCL1(Lc−2)� dimC = 2n− r > dimL1 − dimLc−2

contradicting Lemma 28. ✷Corollary 30. If L is a finitely generated ideally1-constrained Lie algebra ofclass at least3, and the fieldF is finite, thendimL1 = 2 dimL2.

Proof. This is an easy consequence of Theorem 29 and Proposition 26.✷Lemma 31. LetL be a finitely generated ideally1-constrained Lie algebra suchthatdimL1 = 2n, dimL2 = n, dimL3 � n (by Theorem29and Corollary30 thisis always the case ifL has class at least4 and the fieldF is finite). If x ∈L1 −{0}thenCL1(x) is an abelian subalgebra ofL.

Proof. If y, z ∈ CL1(x) then [y, z] ∈ CL2(x), so that we may prove thelemma assuming thatCL2(x) = {0}: in this case, as dimL2 � dimL3, we get[x,L2] = L3. We denote byL the quotient ofL over an ideal containing[x,L2]and of codimension 1 inL3 and we use bar notation consistently. Asx ∈CL(L

2),which is abelian by Proposition 13, we have thatCL1

(x) ⊇ CL1(L2). Lemma 28

gives dimCL1(L2) = dimCL1

(x) = n so thatCL1(L2) = CL1

(x). This implies

that[CL1(x),CL1

(x)] = {0}: asL/L3 � L/L3 we get the thesis. ✷The construction of the field given in the following theorem is the Lie algebra

counterpart of a result of Verardi (see [7]).

Theorem 32. Let L be a finitely generated ideally1-constrained Lie algebra ofclass2 such thatdimL1 = 2n anddimL2 = n. Suppose thatCL1(x) is an abeliansubalgebra ofL for everyx ∈ L1 − {0}. Then there exists a field extensionKof F of degreen and a gradedF -isomorphism of algebras betweenL and the2-generated free Lie algebra of class2 overK.

Proof. LetR be theF -algebra whose elements are the degree preservingF -linearmappingsϕ of L in itself commuting with adl for all l ∈ L. Chooseϕ ∈ R:the definition ofR implies that kerϕ and Imϕ are ideals ofL. In particular,kerϕ ⊇ L2 or kerϕ ⊆ L2. In the former case Imϕ ∩L2 = {0}, so that Imϕ = {0},that isϕ = 0. In the latter caseϕ induces a bijective endomorphism onL1, so thatImϕ ⊇ L1, which implies that Imϕ = L andϕ is an isomorphism; moreover itcan be easily checked thatϕ−1 belongs toR. ThenR is a division algebra andK := Z(R) is an extension field ofF . The algebraL can be regarded as a LieK-algebra, and this structure extends the LieF -algebra structure. Moreover,

n = dimF L2 = dimK L2 · dimF K;


if dimF K � n, then dimF K = n, dimK L2 = 1, dimK L1 = 2, andL is the2-generated free LieK-algebra of class 2. To complete the proof of the theoremwe are only left to show that dimF K � n.

We fix a, b ∈ L1 such that[a, b] = 0, and we putA := CL1(a), B := CL1(b).By hypothesisA and B centralize every elementc ∈ A ∩ B. By Lemma 28,dimA = dimB = n, and dimCL1(c) = n if c = 0. This means thatA = B ifA ∩ B = {0}, but this contradicts[a, b] = 0. SoA ∩ B = {0} andL1 = A ⊕ B.Moreover, ada induces an isomorphismα of F -vector spaces betweenB andL2,and adb induces an isomorphismβ of F -vector spaces betweenA andL2.

For everyx ∈ A andy ∈ B we have[[y, x]α−1 + x, a + y] = [[y, x]α−1, a

] + [x, y] = 0,

then[y, x]α−1 + x ∈ CL1(a + y). By hypothesisCL1(a + y) is abelian, so thatfor everyx, x ′ ∈ A andy ∈B it results[[y, x]α−1 + x, [y, x ′]α−1 + x ′] = 0,

that is[[y, x ′]α−1, x] = [[y, x]α−1, x ′]. (5)

Similarly:[[x, y ′]β−1, y] = [[x, y]β−1, y ′] (6)

for everyx ∈ A, y, y ′ ∈ B.Consider the decompositionL =A⊕B ⊕L2 and the projectionsπ1 :L →A,

π2 :L →B andπ3 :L→L2 so thatl = lπ1+ lπ2+ lπ3 for everyl ∈ L. For everyt ∈ A we define a linear mappingϕt of L in itself as follows:

lϕt := [[t, b]α−1, lπ1]β−1 + [lπ2, t]α−1 + [

lπ3α−1, t

].

Note thatLiϕt ⊆ Li for i = 1,2. If l′ ∈ L, by a straightforward computation weget

lϕt adl′ = [[[t, b]α−1, lπ1]β−1, l′π2

] + [[lπ2, t]α−1, l′π1]

and

l adl′ϕt = [[lπ1, l′π2]α−1, t

] + [[lπ2, l′π1]α−1, t

].

We apply relation (5) settingx = t , x ′ = l′π1 andy = lπ2 and get[[lπ2, t]α−1, l′π1] = [[lπ2, l

′π1]α−1, t].

We now consider the following chain of equalities:[[[t, b]α−1, lπ1]β−1, l′π2

] = [[l′π2, lπ1]β−1, [t, b]α−1]= [

t,[[l′π2, lπ1]β−1, b

]α−1]

= [t, [l′π2, lπ1]α−1],


where the first equality follows by relation (6) (settingx = lπ1, y = [t, b]α−1,y ′ = l′π2), and the second one follows by relation (5) (settingx = [l′π2, lπ1]β−1,x ′ = t , y = b). We may then conclude thatlϕt adl′ = l adl′ϕt for everyl, l′ ∈ L,i.e. ϕt ∈ R. Finally, it can be easily shown thatϕt ∈ Z(R) = K and that themapping sendingt to ϕt is a linear injective mapping fromA into R. This yieldsdimF K � n, as required. ✷Corollary 33. LetL be a finitely generated ideally1-constrained Lie algebra ofclassc � 4 over a finite fieldF = GF(q) such thatdimF L1 = 2n. Then thereexists a2-generated Lie algebraL over the fieldK = GF(qn) and a surjectivehomomorphismf of F -algebras fromL ontoL such thatL has the same classof L andf induces an isomorphism betweenLi andLi for i < c. Moreover, ifL ∈ Lj for somej we may assume that alsoL ∈ Lj .

Proof. Let F :=FK(X,Y ). By Lemma 31 and Theorem 32 it follows thatL/L3

is F -isomorphic toF/F3. Proposition C in [8] implies thatL is a quotient ofFover a homogeneousF -idealI .

As L is ideally 1-constrained, a homogeneous elementa of F of degreei < c

does not belong toI if and only if [a,F1]+ Ii+1 =Fi+1. So, leta ∈ Ii with i < c

andk ∈ K: then

[ka,F1] + Ii+1 = [a, kF1] + Ii+1 ⊆ [a,F1] + Ii+1 =Fi+1

so thatka ∈ Ii . ThenIi is aK-vector space fori < c. Consider now theK-idealJ of F generated by

⊕i<c Ii and byFc+1 if L has finite class, only by

⊕i<c Ii

otherwise and letL := F/J . It is now straightforward to see thatL satisfies ourrequests. IfLj is abelian we note that[Lj ,Lj ] ⊆ kerf : we can replaceJ byJ + [F j ,F j ] in the previous argument.✷

Acknowledgments

The authors thank Carlo M. Scoppola (University of L’Aquila) for a numberof comments and suggestions and for a very careful reading of the final versionof the manuscript.

References

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[4] A. Caranti, S. Mattarei, M.F. Newman, C.M. Scoppola, Thin groups of prime-power order andthin Lie algebras, Quart. J. Math. Oxford (2) 47 (1996) 279–296.

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