Graded Lie algebras with finite polydepth

4 série, t. 36, 2003, p. 793 à 804.

GRADED LIE ALGEBRAS WITH FINITE POLYDEPTH

B

Y

FELIX, S

HALPERIN

J

-C

THOMAS

ABSTRACT. – IfAis a graded connected algebra then we define a new invariant,polydepthA, which is finite ifExt^∗A(M, A)= 0for someA-moduleMof at most polynomial growth. THEOREM1: Iff:X→Y is a continuous map of finite category, and if the orbits ofH_∗(ΩY)acting in the homology of the homotopy fibre grow at most polynomially, thenH_∗(ΩY)has finite polydepth. THEOREM5: IfL is a graded Lie algebra andpolydepthU Lis finite then eitherLis solvable andU Lgrows at most polynomially or else for some integerdand allr,k+d

i=k+1dimLik^r,ksomek(r).

2003 Elsevier SAS

RÉSUMÉ. – Si A est une algèbre graduée connexe nous définissons un nouvel invariant, appelé la profondeur polynomiale noté polydepthA, qui est fini s’il existe unA-module gradué M ayant une croissance au plus polynomiale tel queExt^∗_A(M, A)= 0. THÉORÈME1 : sif:X→Y est une application continue de LS-catégorie finie et si les orbites de l’action de H_∗(ΩY) sur l’homologie de la fibre homotopique de f possèdent une croissance au plus polynomiale alors polydepthH_∗(ΩY) est finie.

THÉORÈME5 : siLest une algèbre de Lie graduée et sipolydepthU Lest fini alors, soitLest résoluble etU Lpossède une croissance au plus polynomiale, soit il existe un entierdtel que pour tout entierron ait k+d

i=k+1dimLik^rpour tous leskplus grands qu’un certaink(r).

2003 Elsevier SAS

We work over a field lk of characteristic different from 2. If V ={Vk} is a graded vector space we denote byV^#={Homlk(Vk, lk)}the dual graded vector space. A graded Lie algebra is a graded vector spaceL, equipped with a bilinear map[,] :Li×Lj→Li+j satisfying

[x, y] + (−1)^ij[y, x] = 0 and

x,[y, z]

=

[x, y], z

+ (−1)^ij

y,[x, z]

forx∈Li,y∈Lj,z∈L. It follows that3[x,[x, x]] = 0forxof odd degree, and so ifchark= 3 we further require that[x,[x, x]] = 0. Finally we consider only graded Lie algebras satisfying L={Li}i1and eachLiis finite dimensional. (Any graded vector spaceV with eachVifinite dimensional is said to have finite type.)

The universal enveloping algebra ofLis denoted byU Land it satisfies the classical Poincaré–

Birkhoff–Witt Theorem (in characteristic3this uses the[x,[x, x]] = 0requirement).

Important examples appear in topology. LetXbe a simply connected topological space with rational homology of finite type. Then the rational homotopy Lie algebraLXofX is defined by

LX=π_∗(ΩX)⊗Q; [,] =Samelson product,

ANNALES SCIENTIFIQUES DE L’ÉCOLE NORMALE SUPÉRIEURE

and the Hurewicz map extends to an isomorphism, [10], U LX

∼=

−→H_∗(ΩX;Q).

Analogously, if X is a finite n-dimensional and r-connected CW complex, then for primes p > n/r,H_∗(ΩX;Fp) =U Efor some graded Lie algebraE[8].

If M is a module over a (graded) algebra A then the grade of M, gradeM, is the least integerq(or∞) such thatExt^q_A(M, A) = 0. And ifV ={Vi}i0is a graded vector space then V has at most polynomial growth if for some constantC, and some non-negative integer,d,

indimViCn^d,n1. In this case the least suchdis called the polynomial bound for the growth ofV and is denoted bypolybd(V). IfV does not have at most polynomial growth we putpolybdV =∞and we say thatV grows faster than any polynomial.

In this paper we combine these two notions in the

DEFINITION. – The polygrade of anA-module,M, is the sum,gradeM + polybdM, and the polydepth ofAis the least integer (or∞) occurring as the polygrade of anA-module.

In the caseA=U Lthe unique augmentationU L→lkmakeslk into aU L-module, and by definition, the grade oflkis the depth ofU L. Sincepolybdlk= 0it follows that:

polydepthU LdepthU L.

(1)

Moreover (cf. Proposition 1.6) ifdimL <∞then equality holds. We shall abuse notation and refer to these invariants respectively aspolydepthLanddepthL.

Note thatExt⁰_{U L}(U L, U L)contains the identity map and so polydepthLpolybdU L.

(2)

Observe as well that for any graded vector spaceM,polybdM = 0if and only ifdimM is finite. ThuspolydepthL= 0if and only ifdepthL= 0, which happens if and only ifLis finite dimensional and concentrated in odd degrees.

Depth has been a useful concept in topology because, on the one hand, Lusternik–

Schnirelmann category satisfies [1]

depthLXcatX

and, on the other hand [3–5], finite depth has important implications for the structure of a graded Lie algebra.

The purpose of this paper is to show that essentially the same implications follow from the weaker hypothesis thatpolydepthLis finite, while simultaneously identifying a larger class of topological spaces and Lie algebras for which the weaker hypothesis holds.

Indeed, we have

THEOREM 1. – IfF→X−→^f Y is a fibration of path-connected spaces, then polydepthH_∗(ΩY)polybdH_∗(F) + catf.

Proof. – The fibration determines an action up to homotopy ofΩY onF, which makesH_∗(F) into anH_∗(ΩY)-module. According to [6],gradeH_∗(F)catf. ✷

Our main structural theorems read:

THEOREM 2. – Let E(L)denote the linear span of elementsx∈Leven such thatadxacts nilpotently on eachy∈L. Then

dimE(L)polydepthL.

THEOREM 3. – The following conditions on a graded Lie algebraLare equivalent (i) Lis the union of solvable ideals andpolydepthLis finite;

(ii) U Lgrows at most polynomially (polybdU Lis finite);

(iii) Levenis finite dimensional, and for some constantC

in

dimLiClog₂n, n1.

In this caseLis solvable.

THEOREM 4. – IfLis a graded Lie algebra of finite polydepth then the union of the solvable ideals ofLis a solvable ideal of finite polydepth.

THEOREM 5. – SupposepolydepthLis finite andLis not solvable. Then there is an integer dsuch that for allr1:

k+d

i=k+1

dimLik^r, k somek(r).

Remark. – In [7] it is shown that ifL=LX whereX is a finite1-connected CW complex, then we may taked= dimXin Theorem 5.

1. Properties of polydepth

LEMMA 1.1. – If M is a module for some graded algebra A of finite type and if Ext^q_A(M, A) = 0thenExt^q_A(A·x, A) = 0for somexin a subquotient module ofM.

Proof. – Recall thatA^#= Homlk(A, lk). ThenExt^q_A(M, A)is the dual ofTor^A_q(M, A^#)and a direct limit argument shows that for somex1, . . . , xn∈M,

Tor^A_q(A·x1+· · ·+A·xn, A^#) = 0.

Now use the exact sequence associated to the inclusion

A·x1+· · ·+A·xn−1∈A·x1+· · ·+A·xn. ✷

COROLLARY 1.2. –PolydepthA is the least m such that polygradeN =m for some monogenicA-moduleN.

Remark. – It follows from the Corollary that we may improve Theorem 1 to the inequality

polydepthH_∗(ΩY)polybd

H_∗(ΩY)·α

+ catf, someα∈H_∗(F).

(3)

PROPOSITION 1.3. – LetLbe a graded Lie algebra.

(i) Each ideal satisfiespolydepthIpolydepthL.

(ii) LetE be a Lie subalgebra ofL. IfLhas finite polydepth and if for eachx∈L/E the orbitU E·xhas at most polynomial growth, thenEhas finite polydepth.

(iii) Fornsufficiently large the sub Lie algebraEgenerated byLnsatisfies polydepthEpolydepthL.

Proof. – (i) This follows from the Hochschild–Serre spectral sequence, converging from Ext^p_{U L/I}(lk,Ext^q_{U I}(M, U L))toExt^p+q_{U L}(M, U L). (Note that sinceU LisU I-free,grade_{U I}(M) is the leastqsuch thatExt^q_{U I}(M, U L) = 0.)

(ii) As in Lemma 1.1,Ext^q_{U L}(M, U L)is dual toTor^{U L}q (M,(U L)^#), and this is the homology of the Cartan–Eilenberg–Chevalley complex∧sL⊗M ⊗(U L)^#. Write L=E⊕V and set Fp=∧sE⊗ ∧^psV ⊗M⊗(U L)^#. This filtration determines a convergent spectral sequence, introduced by Koszul in [9], and which is the Hochschild–Serre spectral sequence whenEis an ideal. The E¹-term of the spectral sequence isTor^{U E}_q (∧^psL/E⊗M,(U L)^#), converging to Tor^{U L}_p+q(M,(U L)^#).

Each element z∈ ∧^psL/E⊗M is contained in a finite sum of U E-modules of the form s(U E·x1)∧ · · · ∧s(U E·xp)⊗M and it follows that

polybd(U E·z)p·polybd(U E·x) + polybd(M)

for somex∈L/E. ChooseM so thatpolydepthL= polygradeM and apply Lemma 1.1 with p+q= gradeM.

(iii) IfExt^p_{U L}(M, U L)is non-zero andpolybd(M)<∞it suffices to chooseE so that the restrictionExt^p_{U L}(M, U L)→Ext^p_{U E}(M, U L)is non-zero ([4], Proposition 3.1). ✷

COROLLARY 1.4 (of the proof of (ii)). – Suppose for somek1thatpolybd(U E·x)k, x∈L/E. ThenpolydepthEkpolydepthL.

COROLLARY 1.5. – LetE be a sub-Lie algebra of a graded Lie algebra L. If Lhas finite polydepth andL/Ehas at most polynomial growth, thenEhas finite polydepth.

Example 1. – Let L(V) be the free Lie algebra on a graded vector spaceV. Then for any graded Lie algebra L,L

L(V)has depth 1. Thus the injection L→L

L(V)shows that each graded Lie algebra is a sub-Lie algebra of a Lie algebra of finite polydepth. The previous corollary gives restriction on a Lie algebraLfor being a sub-Lie algebra of a Lie algebra of finite polydepth,K, when the quotient has at most polynomial growth.

PROPOSITION 1.6. – IfLis a finite dimensional graded Lie algebra then polydepthL= depthL.

Proof. – As observed in the introduction, polydepthLdepthL. On the other hand, by Lemma 1.1, polydepthL= polygradeM for some monogenic module M =U L·x. Now Theorem 3.1 in [3] asserts thatpolygradeM= depthL. ✷

2. Proof of Theorem 2

SupposeI⊂Lis an ideal. IfExt^m_{U L}(M, U L)= 0, thenExt^p_{U L/I}(Tor^{U I}_q (M, k), U L/I)= 0, somep+q=m. (Same proof as in: [3], Lemma 4.3, for the caseM=lk.) By Lemma 1.1 there is a monogenicU L/I-moduleNsuch thatNis a subquotient ofTor^{U I}_q (M, lk), and gradeNp.

Now supposeL/Iis finite dimensional. Then Theorem 3.1 in [3] asserts that gradeN+ polybdN= dim(L/I)even.

On the other hand, write(L/I)even=V ⊕W whereV is the image ofE(L). Letxi∈Leven, yj ∈Lodd and zk ∈E(L) represent respectively bases of W, (L/I)odd and V. Then the elements x^k₁¹· · ·x^k_s^sy^ε₁¹· · ·y_t^εtz₁^m1· · ·z_u^mu, where εi = 0 or 1, represent a basis for U L/I.

Choose thezk to act locally nilpotently in L. Then this basis applied to anyω∈ ∧^qsI, shows thatpolybd(U L/I)·ωdimW. Hence ifu∈ ∧^qsL⊗M represents a generator ofN then

polybdNpolybd(U L/I·u)polybdM+ dimW.

Substitution in the equation above gives

dim(L/I)evengradeN+ polybdM+ dimW gradeM+ polybdM+ dimW.

Choose M so that gradeM + polybd(M) = polydepthL and choose I = L>2k. Then V ∼=E(L)2k and we have

dimE(L)2kpolydepthL.

Since this holds for allkthe theorem is proved.

3. Solvable Lie algebras

LEMMA 3.1. – LetLbe a Lie algebra concentrated in odd degrees. Then ExtU L(−, U L) = HomU L(−, U L).

In particular

polydepthL= polybdU L.

Proof. – Since L =Lodd it is necessarily abelian. Now ExtU L(−, U L) is the dual of Tor^{U L}(−,(U L)^#)and this is the limit ofTor^{U Ln}(−,(U L)^#), which dualizes to

ExtU L_n(−, U L).

Since U Ln is a finite dimensional exterior algebra and U L is U Ln-free it follows that Ext⁺_{U Ln}(−, U L) = 0, and soExt⁺_{U L}(−, U L) = 0.

Finally, sinceExt⁰_{U L}(U L, U L)is non-zero,polydepthLpolybdU L. On the other hand if polydepthL=m <∞, then for someM, we have

Ext^p_{U L}(M, U L) = 0 and polybdM=m−p.

By the above,p= 0 and so there is a non-zeroU L-linear mapf:M→U L. Any f(m)is in someU L<nand iff(m) = 0it follows thatU Ln

∼=

−→U Ln·m. This implies polybdMpolybdU L and polydepthLpolybdU L. ✷

LEMMA 3.2. – LetL be a graded Lie algebra of finite polydepth. IfI is an ideal inL and polybdI <∞thenpolydepthL/I<∞.

Proof. – ChooseM so thatpolygradeM= polydepthL. Ifm= gradeM then it follows (as in [3], proof of Theorem 4.1 for the caseM=lk) that for somep,

Ext^p_{U L/I}

Tor^{U I}_m−p(M, lk), U L/I

= 0.

SinceTor^{U I}m−p(M, lk)is a subquotient of∧^m−psI⊗M it follows that it has polynomial growth at most equal to(m−p) polybdI. ✷

Proof of Theorem 3. –

(i) ⇒ (ii). LetI be the sum of the ideals inL concentrated in odd degrees. Then I is an ideal of this form, necessarily abelian, and L/I has no ideals concentrated in odd degrees.

Moreover polybdU I = polydepthI polydepthL (Lemma 3.1 and Proposition 1.3) and hencepolydepthL/I<∞(Lemma 3.2).

Next we show that every solvable idealJ inL/I is finite dimensional, by induction on the solvlength. Indeed, ifJ is abelian thenJeven=E(J). Since

polydepthJpolydepthL/I (Proposition 1.3),

Theorem 2 asserts thatJevenis finite dimensional. Thus for somer,Jris an ideal concentrated in odd degrees, i.e.Jr= 0.

Now if J has solvlength k then its (k + 1)st derived algebra is abelian and so finite dimensional. Thus for somer,Jrhas solvlengthk−1. By induction,Jis finite dimensional.

By hypothesisL/I is the sum of its solvable ideals. Since these are finite dimensional, each x∈(L/I)evenacts locally nilpotently. Thus(L/I)even=E(L/I), and this is finite dimensional by Theorem 2. ButLeven∼= (L/I)evensinceIis concentrated in odd degrees.

Suppose Leven ⊂ L2n. Since L>2n is an ideal in odd degrees of finite polydepth, polybdU L>n<∞, while triviallypolybdU L/L>2n<∞. HencepolybdU L <∞.

(ii)⇒(iii). ClearlypolybdU LdimLeven, so the latter must be finite. It is trivial from the Poincaré–Birkhoff–Witt theorem that if

indimLi=d(n)then

ind(n)

dim(U L)i2^d(n).

Thus2^d(n)K[nd(n)]^rfor some constantKand some integerr,r1. It follows that d(n)log₂K+rlog₂n+rlog₂d(n)rlog₂n+1

2d(n), nsufficiently large.

(iii)⇒(i). ChooseN so thatI=LN is concentrated in odd degrees. ThenU I is an exterior algebra and so

in

dim(U I)i2

indimLin^C

for some constantC. Thus, sinceL/Iis finite dimensional,polybdU Lis finite. The identity of U Lis inExt⁰_{U L}(U L, U L)and sopolydepthLpolybdU L.

Finally, sinceIis abelian andL/Iis finite dimensional,Litself is solvable. This also proves the last assertion. ✷

Proof of Theorem 4. – This is immediate from Proposition 1.3(i) and Theorem 3. ✷

PROPOSITION 3.3. – SupposeIis a solvable ideal in a Lie algebraLof finite polydepth. Then (i) polydepthL/IpolydepthL;

(ii) dimIevenpolydepthIpolybdU I. Proof. – (i) As noted in the proof of Lemma 3.2,

Ext^p_{U L/I}

Tor^{U I}_m−p(M, lk), U L/I = 0,

where m+ polybdM = polydepthL. Also the Tor is a subquotient of ∧^m−psI⊗M. By Theorem 3

in

dim(∧^m−psI⊗M)i(C1log₂n)^m−p·C2·n^polybdM. Hence

polybd Tor^{U I}m−p(M, lk)polybdM+ 1, and so

polydepthL/IpolydepthM +p+ 1.

Ifp < mthen this givespolydepthL/IpolydepthL. Ifp=mthen Tor^{U I}m−p(M, lk) =M⊗U Ilk.

Hence in this casepolybd Tor^{U I}_m−p(M, lk)polybdM and again polydepthL/IpolydepthL.

(ii) SinceIeven⊂I2n, somen(Theorem 3) we may apply the first assertion to obtain dimIeven= depthI/I>2n (cf. [2])

= polydepthI/I>2n (Proposition 1.6) polydepthI.

The second inequality has already been observed:

polybdU I= polygradeU IpolydepthI. ✷

Example 2. – Consider a Lie algebraLconcentrated in odd degrees with a basis{xi, i1}

satisfying the degree relations

degxi>

j<i

degxj.

Then for eachn,dim(U L)n1. The identity onU Lshows thatpolydepthL= 1.

Example 3. – Consider the graded Lie algebra L = L(a, xn)n2/I, with dega = 2, degxn= 2ⁿ+ 1, and whereIis generated by the relations

(ada)^kxr,(ada)^lxs

= 0, k, l0, r, s2, and adⁿ⁺¹(a)(xn) = 0.

ThenpolybdU L= 2, so thatLhas finite polygrade. On the other hand,Lis solvable but not nilpotent, and is the union of the infinite sequence of the finite dimensional Lie algebrasIN

generated bya, x2, . . . , xN.

PROPOSITION 3.4. – Let L be the direct sum of non-solvable Lie algebras L(i),in. If polydepthL(i)<∞for1in, then

npolydepthL

i

polydepthL(i).

Proof. – We first prove by induction on n that for any U L-module M that has at most polynomial growth, we have

Ext^<nU L(M, U L) = 0.

Consider the Hochschild–Serre spectral sequence Ext^p_{U L(1)}

Tor^U(L(2)_q ^{⊕···⊕L(n))} M,

U

L(2)⊕ · · · ⊕L(n)#

, U L(1)

⇒Ext^p+q_{U L}(M, U L).

(4)

SinceL(1)commutes with the otherL(i)it follows that for each monogenicU L(1)-moduleN that is a subquotient ofTor^U(L(2)_q ^{⊕···⊕L(n))}(M, U(L(2)⊕ · · · ⊕L(n))^#)we have

polybdNpolybdM.

Now since L(1) is not solvable, polybdU L(1) =∞ and the argument in the proof of Lemma 3.1 shows that

Ext⁰_{U L(1)}

N, U L(1)

= 0.

Thus (Lemma 1.1) the left hand in (4) vanishes for p= 0. By induction onn it vanishes for q < n−1and soExt^<n_{U L}(M, U L) = 0. ThuspolydepthLn.

On the other hand, there areU L(i)-modulesM(i)such that polygradeM(i) = polydepthL(i).

Thenn

i=1M(i)is aU L-module that has at most polynomial growth and whose polygrade is the sum of the polygrades of theM(i). ✷

4. Growth of Lie algebras

PROPOSITION 4.1. – LetLbe a non-solvable graded Lie algebra of finite polydepth. Then for each integerr1there is a positive integerd(r)such that

k+d(r)

i=k+1

dimLik^r, ksufficiently large.

Proof. – We distinguish two cases.

Case A:Levencontains an infinite dimensional abelian sub Lie algebraE.

Choosenso thatdimEn(r+ 3) polydepthL. Then there is a finite sequence L=I(0)⊃I(1)⊃ · · · ⊃I(l)

in whichI(j)is an ideal inI(j−1)andI(l)n=En.

By Proposition 1.3,polydepthI(q)polydepthL. Thus without loss of generality we may suppose thatL=I(l), i.e. thatLnis an abelian sub Lie algebra concentrated in even degrees and thatdimLn(r+ 3) polydepthL.

LetM be aU L-module such that

gradeM+ polybdM= polydepthL and putm= gradeM. As observed in the proof of Proposition 1(ii),

Ext^q_{U Ln}(∧^psL>n⊗M, U Ln) = 0, for somep+q=m. It follows that for somez∈ ∧^psL>n⊗M,

polybdU Ln·z+qdimLn

(Theorem 3.1 in [3]). Hence for somex∈L,

p(polybdU Ln·x)dimLn−q−polybdM.

Sincep+q+ polybdM= polydepthLwe conclude that

(2 + polybdU Ln·x)·polydepthLdimLn(r+ 3) polydepthL.

As observed in the introduction, sincedimL=∞,polydepthL >0. It follows that polybd(U Ln·x)r+ 1.

(5)

On the other hand,U Lnis the polynomial algebralk[y1, . . . , ys]on a basisy1, . . . , ysofLn. Because of (5) it is easy to see (induction on s) that this basis can be chosen so that for some w∈U Ln·x,lk[y1, . . . , yr+1]→lk[y1, . . . , yr+1]·wis injective. Putd= ^r+1_i=1degyiand note that

k+degw+d

i=k+degw+1

dimLi

k+d

i=k+1

dimlk[y1, . . . , yr+1]i 1 r!k^r. It follows that for anyr, andksufficiently large,

k+d

i=k+1

dimLik^r−1.

This proves Proposition 5 in case A.

Case B: Every abelian sub Lie algebra ofLevenis finite dimensional.

Let I be the sum of the solvable ideals in L. Then Ieven is finite dimensional and polydepthL/I is finite (Theorem 3 and Proposition 3.3). Thus all abelian sub Lie algebras of L/Iare finite dimensional. Thus it is sufficient to prove case B whenLhas no solvable ideals.

There are now two possibilities: eitherL=Leven, orL has elements of odd degree. In the latter case the sub Lie algebra generated by Loddis an ideal, hence non-solvable and of finite polydepth. Let L(s) denote the sub Lie algebra generated by the firsts linearly independent elements x1, . . . , xs of odd degree. For s sufficiently large, polydepthL(s)polydepthL (Proposition 1) and dimL(s)even >polydepthL (obvious). Thus L(s) cannot be solvable (Theorem 3). In other words, we may assume that either L=Leven or elseLis generated by finitely many elementsx1, . . . , xs of odd degree. In either case set E=Leven, and note that dimEis infinite.

Define a sequence of elementsziand sub Lie algebrasE(i)by settingE(1) =E,ziis a non- zero element inE(i)andE(i+ 1)⊂E(i)is the sub Lie algebra of elements on which adziacts nilpotently.

Since E contains no infinite dimensional abelian Lie algebra some E(N + 1) = 0 and E(1)/E(2)⊕ · · · ⊕E(N)/E(N+ 1)is a graded vector space isomorphic withE.

Put d= degzi and di =d/degzi. Then (adz1)^d1 ⊕ · · · ⊕(adzN)^dN is an injective transformation of E(1)/E(2)⊕ · · · ⊕E(N)/E(N + 1) of degree d. Since this space is isomorphic withEit follows that

t

i=s

dimEi t+d

i=s+d

dimEi and thus

k+d

i=k+1

dimEi d k+d

k+d

i=1

dimEi, k1.

(6)

On the other hand, choosenso that

dimEn(r+ 3)·polydepthL.

(This is possible becauseEis infinite dimensional.) SetI=L>n. LetM be anL-module with polydepthL= polygradeM. As in the proof of Theorem 4.1 in [3],Ext^p_{U L/I}(Tor^{U I}q (M, lk), U L/I) = 0 for p+ q = gradeM. Thus p, q and polybdM are all bounded above by polydepthL. Now Theorem 3.1 of [3] asserts that for someα∈Tor^{U I}_q (M, lk),U L/I·αhas polynomial growth at least equal to dim(L/I)even−p. This means that for some positiveC,

ik

dim(U L/I·α)iCk^{(dim(L/I)even)−p}, ksufficiently large.

SinceTor^{U I}_∗ (M, lk)is the homology of∧^∗sI⊗M, it follows that

ik

dim Tor^{U I}_q (M, lk)i

ik

dimLi

q

ik

dimMi.

But(L/I)even∼=Enand so a quick calculation gives

ik

dimLiKk^r+1, ksufficiently large.

(7)

Finally, recall that eitherL=Levenor elseLis generated by the elements of odd degreexi. In the former caseL=Eand the proposition follows from (6) and (7). In the second case we have

Lodd= [x1, E] +· · ·+ [xs, E] +lkx1+· · ·+lkxs,

and hence (7) yields

ik

Ei K

s+ 1k^r+1+s, ksufficiently large.

Combined with (6) this formula gives the proposition. ✷

Proof of Theorem 5. – SinceLis not solvable we may choosenso that dim(Leven)n>polydepthL (Theorem 3),

and so that the sub Lie algebra generated by Ln satisfies polydepthE polydepthL (Proposition 1.3). ThenEis not solvable (Proposition 3.3).

Letx1, . . . , xsgenerateE(see beginning of case B) and putd= max degxi. LettingU Eact via the adjoint representation onEwe have that

U E[0,q]·E[k+1,k+d]⊃E[k+1,k+q]. For anyr1chooseq=q(r)so that

k+q

i=k+1

dimEik^r+1,

ksufficiently large (Proposition 4.1). Then

k+d

i=k+1

dimLi

k+d

i=k+1

dimEi 1 dimU E[0,q]

k^r+1k^r, ksufficiently large.

Sincedis independent ofr, the theorem is proved. ✷

REFERENCES

[1] FELIXY., HALPERINS., LEMAIREJ.-M., THOMASJ.-C., Modploop space homology, Inventiones math. 95 (1989) 247–262.

[2] FELIXY., HALPERINS., JACOBSONC., LÖFWALLC., THOMASJ.-C., The radical of the homotopy Lie algebra, Amer. J. Math. 110 (1988) 301–322.

[3] FELIXY., HALPERINS., THOMASJ.-C., Lie algebras of polynomial growth, J. London Math. Soc. 43 (1991) 556–566.

[4] FELIXY., HALPERINS., THOMASJ.-C., Hopf algebras of polynomial growth, J. Algebra 125 (1989) 408–417.

[5] FELIXY., HALPERINS., THOMASJ.-C., Engel elements in the homotopy Lie algebra, J. Algebra 144 (1991) 67–78.

[6] FELIXY., HALPERINS., THOMASJ.-C., The category of a map and the grade of a module, Israel J.

Math. 78 (1992) 177–196.

[7] FELIXY., HALPERINS., THOMASJ.-C., Growth and Lie brackets in the homotopy Lie algebra, in:

The Roos Festschrift, vol. 1, Homology Homotopy Appl. 4, no. 2, part 1, 2002, pp. 219–225.

[8] HALPERINS., Universal enveloping algebra and loop space homology, J. Pure Appl. Algebra 83 (1992) 237–282.

[9] KOSZULJ.-L., Homologie et cohomologie des algèbres de Lie, Bull. Soc. Math. France 78 (1950) 65–127.

[10] MILNORJ.W., MOOREJ.C., On the structure of Hopf algebras, Ann. Math. 81 (1965) 211–264.

(Manuscrit reçu le 22 juillet 2002 ; accepté le 24 janvier 2003.)

Yves FELIX Institut de Mathématiques, Université Catholique de Louvain, 1348 Louvain-La-Neuve, Belgique

Stephen HALPERIN College of Computer, Mathematical and Physical Sciences

University of Maryland, College Park, MD 20742-3281, USA

Jean-Claude THOMAS Faculté des Sciences, Université d’Angers,

Bd. Lavoisier, 49045 Angers, France E-mail: [email protected]