graded lie algebras of maximal class, iii€¦ · ﬁnite quotient: this result was essential to...

Graded Lie Algebras of Maximal Class, III

G. Jurman

Centre for Mathematics and its ApplicationsAustralian National University

Canberra, ACT 0200Australia

E-mail: [email protected]

We describe the isomorphism classes of infinite-dimensionalN-graded Lie algebrasof maximal class generated by their first homogeneous component over fields of charac-teristic two. This complements the analogous work by Caranti and Newman in the oddcharacteristic case.

Key Words:Graded Lie algebras, modular Lie algebras, graded Lie algebras of maximalclass.

1. INTRODUCTION

In their paper [15], A. Shalev and E. Zelmanov started the development of acoclass theory for Lie algebras following the analogous theory for groups. As theypointed out, in late sixties M. Vergne in [16], [17] proved that the algebraic varietyof n-dimensional nilpotent complex Lie algebras has an irreducible component ofdimension exceedingn2 consisting of algebras of maximal class (i.e. of coclassone), which implies that they may not be easily classified. They focussed theirattention on graded Lie algebras generated by their first homogeneous component,and considered the general finite coclass case: they proved that the only just-infinite object in that context has maximal class and it is metabelian.

In the modular case, the situation is far more complicated even for graded alge-bras. In addition to the metabelian algebra (which is the algebra associated with thelower central series of the infinite pro-p-group of maximal class), Shalev [14] builtfor each primep countably many graded Lie algebras of maximal class. These areinsoluble algebras which are obtained by looping some finite-dimensional sim-

The author is a member of INdAM-GNSAGA. He is grateful for his help to A. Caranti, advisor ofthe doctoral dissertation [9] this work is based on. The author is also grateful to M.F. Newman formaking useful suggestions and reading various versions of this paper.

1

2 JURMAN

ple algebras constructed by A.A. Albert and M.S. Frank in [1] by a non-singularderivation. We refer to them as the Albert-Frank-Shalev (AFS, for short) algebras.

In the first paper of this series [4], A. Caranti, S. Mattarei and M.F. Newman tooka further step by building uncountably many algebras of maximal class. They usedtwo techniques: inflation and taking inverse limits. Later, Caranti and Newmanin [5] proved that over every field of odd characteristic all graded Lie algebras ofmaximal class can be obtained by these methods. Actually, every algebra can beobtained either via inflation or taking limits fromAFS-algebras.

As a natural completion of that paper, in the present work we describe a solutionof the isomorphism problem for infinite-dimensionalN-graded Lie algebras ofmaximal class generated by their first homogeneous component when the charac-teristic of the underlying field is two. The main result (Theorem 4.1) stated herediffers essentially from the analogous one in odd characteristic because of theexistence of a further family of infinite-dimensional graded Lie algebras of maxi-mal class in addition to the Albert-Frank-Shalev algebras which are not inflated.These new algebras can be obtained, like theAFS ones, by looping some finite-dimensional simple Lie algebras, called Bi-Zassenhaus algebras, by mean of anon-singular outer derivation. We will denote respectively byB andBl the simpleand the loop structure: their construction and properties are described in [10].C. Carrara proved in [6], [7] thatAFS-algebras can be characterized by a suitablefinite quotient: this result was essential to prove the classification theorem for theodd case; the same property holds forBl-algebras as well. The characterizationtheorem we prove states that by using the two processes of inflation and takinginverse limits, starting fromAFS orBl-algebras, one can construct all infinite-di-mensionalN-graded Lie algebras of maximal class. Note that the soluble limits ofBi-Zassenhaus algebras coincide with the soluble limits for Albert-Frank-Shalevalgebras, so the crucial point is to show that every uninflated algebra is either anAFS or aBl-algebra.

In Section 4 an outline of the proof is summarized: it follows the same steps asin the odd characteristic case; explicit proves are given only for the parts wheredifferences arise: the reader can easily complete the missing proofs by looking atthe corresponding sections in [5].

As in the odd characteristic case, an interesting consequence follows fromour characterization theorem. G. Benkart, A.I. Kostrikin and M.I. Kuznetsovin their papers [2], [12] studied simple modular algebras admitting non-singularderivations and proved that the existence of any non-singular derivation impliesthe existence of a non-singular derivation which preserves a cyclic grading. Acorollary of our main result is that in characteristic two there are only two familiesof finite-dimensional simple algebras that have a non-singular derivation whichpreserves a cycling grading whose components have dimension one, namely theAlbert-Frank and the Bi-Zassenhaus algebras.

As in [5], the proof uses several apparentlyad hocJacobi expansions thatmake it quite computational; even if a more theoretical proof could be found, it

GRADED LIE ALGEBRAS 3

is unlikely that the results can be achieved avoiding the expansions entirely. TheANU p-Quotient Program [8] has as usual played a major role in building smallexamples of finite-dimensional (quotients of) algebras whose structure suggestedthe directions in which to search to develop a general theory and which are theJacobi expansions to employ. However, nowhere does the proof given here relyon machine calculations.

From now on we refer to the paper [4] as I and to [5] as II.

2. PRELIMINARIES

Notations are the same as in II. NowF is a field of characteristicp = 2, whichallows us to ignore signs safely. All commutators are left-normed; moreover wewill widely use without specific mention the exponential shorthand for elementsand sequences

[uvk] = [u v : : : v| {z }k

]

�1; (�2; ��3 )1 = �1; �2; �3 : : : �3| {z }

�

; �2; �3 : : : �3| {z }�

; : : : ;

the generalized Jacobi identity (in the two equivalent expansions)

[v[yu�]] =

�Xi=0

��

i

�[vuiyu��i] =

�Xi=0

��

i

�[vu��iyui] ;

and Lucas’ Theorem (see [13], [11]): ifa =nXi=0

ai � 2i, b =

nXi=0

bi � 2i is the

2-adic expansion of the two integersa andb, then

�a

b

��

nYi=0

�ai

bi

�(mod 2) :

As a direct consequence of Lucas’ Theorem, we have that

�2w � 1

i

�� 1 (mod 2) for 0 � i � 2w � 1 :

Finally, we will use the following identity

2a � 2b =

a�1Xi=b

2i for 0 � a < b : (1)

4 JURMAN

3. BACKGROUND

All the preliminary results we need come from the corresponding section of II,except for some remarks that deserve to be stated explicitly. The first one concernsLemma 3.1 of II: it was originally proved in [3] in the odd case, but its proof isindependent from the characteristicp of the field, so it holds forp = 2, too.

Lemma 3.1. A two-step centralizer other than the first two in order of occur-rence always occurs at the end of a short constituent, and it is followed by anothershort constituent.

Proof. Let q = ph be the parameter of the algebra and letw be the elementat the beginning of the constituent (whose length we denote byl) centralized by atwo-step centralizer different from the first two. Then

0 = [w[yx2q ]] =

�2q

0

�[wyx2q ] = [wxlxx2q�l�2x] :

But l; 2q�l�2� q�1, since they represent constituent lengths: thenl = q�1.

As pointed out in the introduction, we need to introduce a family of infinite-dimensional Lie algebras of maximal class other than the Albert-Frank-Shalevones. We call them the Bi-Zassenhaus loop algebrasBl(g; h). They depend ontwo integer parametersg � 2 andh � 1, and they are described in [10]. They haveonly two distinct two-step centralizers and, in the usual notations, their sequenceof constituent lengths read as

2q; 2q � 1; (2q��1; 2q � 12)1 ;

whereq = 2h and� = 2g � 1.The final remark is about Lemmas 5.3 and 5.7 in I. These lemmas do not hold

when the characteristic is two, as theBl algebras show.

4. LEITFADEN

As anticipated in the introduction, the theorem we will prove can be stated asfollows:

Theorem 4.1. LetL be an infinite-dimensional graded Lie algebra of maxi-mal class over a fieldF of characteristic two, generated by its first homogeneouscomponent.

ThenL is obtained� either via a finite number (possibly zero) of inflation steps from one of the

algebrasAFS(a; b; n; 2)F2F,AFS(a; b;1; 2)F2F orBl(g; h)F2F


� or as the limit of an infinite number of inflation steps.

In the latter case, the algebra one starts with is not relevant.Now the proof of Theorem 4.1 follows as in the odd characteristic case. In

particular, ifL is not inflated then the above steps apply to show thatL itselfis anAFS or aBl-algebra. IfL is inflated, we keep deflatingL and noting thefirst two-step centralizers�i of the deflated algebras: if after a finite numberm

of deflation steps we reach an algebraN that is not inflated, thenN will be anAFS or aBl-algebra, and we can recoverL fromN by inflating with respect to�m; : : : ; �2; �1. If all the algebras we obtain in the process are inflated, thenL

can be obtained as the limit of an infinite number of inflation steps with respect tothe sequence(�i)i�1.

The main part of the proof, which deals with proving that every non-inflatedalgebra of maximal class is anAFS-algebra or aBl-algebra, is split into elevensteps we list below. They resemble those in II.

Lemma 3.1 in I takes care of the caseL metabelian: so we can assumeL notmetabelian. The first three steps are proved in I, too.

The first one is on constituent lengths.

Step 4.1. If L is not metabelian, then the first constituent has length2q,for some powerq = 2h. All other constituents have length2q or 2q � 2�, for0 � � � h.

So letq = 2h be the parameter ofL.The second step characterizes inflated algebras.

Step 4.2. L is inflated if and only if all constituents have even length.

The third step is just a corollary of the previous one.

Step 4.3. If all constituents after the first one are short, thenL is inflated.

From now on, assume that the sequence of constituent lengths ofL includesafter the first long constituent at least another non-short one, that can be eitherlong or intermediate. Then Step 4.4, whose proof is in Section 5 and 6, deals withthe former case.

Step 4.4. SupposeL has two distinct two-step centralizers. If the secondnon-short constituent is long, then all constituents are short or long, and thusL

is inflated.

The next five steps deal with the intermediate case, when the second non-shortconstituent is intermediate. First of all we can describe the initial part of thesequence of constituent lengths ofL by the following step, proved in Section 5.

6 JURMAN

Step 4.5. Suppose the second non-short constituent is an intermediate one,of length2q�2�, for 0 � � < h, and that no two-step centralizer higher than thesecond one occurs before it. Then the initial pattern of constituents is of the form

2q; qr�2; 2q � 2� ;

for some powerr = 2k.

Note that the valuek = 1 takes care of the case when the second constituent isnot short.

Now it is possible to identify the basic elements of the whole sequence ofconstituent lengths by proving (in Section 7) the claim of the next step.

Step 4.6. Suppose there is an initial segment of the sequence of two-stepcentralizers ofL that involves only the first two two-step centralizers, and that thecorresponding segment of the sequence of constituent lengths is

2q; qr�2; 2q � 2� ;

where0 � � < h. Then the algebra has two distinct two-step centralizers andthe sequence of constituent lengths consists of repetitions of the patterns

2q; qr�2 or 2q � 2�; qr�2 :

Now we can identify a first class of algebras.

Step 4.7. SupposeL has two distinct two-step centralizers. If the sequenceof constituent lengths is

2q; qr�2; 2p� 2� ; (qr�2; 2q)1 ;

thenL is obtained from anAFS(a; b;1; 2)F2F via� inflation steps with respectto the first two-step centralizer.

When a further intermediate constituent occurs, it is possible to give somebounds to the occurrences of the patterns shown in Step 4.6.

Step 4.8. Suppose there is an initial segment of the sequence of two-stepcentralizers ofL that involves only the first two two-step centralizers, and that thecorresponding segment of the sequence of constituent lengths is

2q; qr�2; 2q � 2�; (qr�2; 2q)s�1; qr�2; 2q � 2� :

Then eithers = 2l for somel � 0, or r = 2 ands = 2l � 1 for somel � 2.

This is proved in Section 8. Then we can apply the following theorem:


Theorem 4.2. Let L be an infinite-dimensional algebra of maximal classover the fieldF2.

Suppose the sequence of two-step centralizers has an initial segment in whichonly two distinct two-step centralizers occur, and the corresponding initial segmentof the sequence of constituent lengths is

2q; qr�2; 2q � 1; (qr�2; 2q)s�1; qr�2; 2q � 1 ;

with q = 2h, r = 2k and eithers = 2l or r = 2 ands = 2l � 1 for somel � 2.If s = 2l, thenL �= AFS(h; h + k; h + k + l; 2), else ifs = 2l � 1, then

L �= Bl(l; h).In particular, in both casesL has two distinct two-step centralizers.

In the cases = 2l (theAFS case) the theorem has been proved by Carrara inher paper [7]; the case concerning Bi-Zassenhaus algebras is proved in Section 9.

The above theorem and the inflation process allow us to state Step 4.9, whichconcludes the identification of algebras with two distinct two-step centralizers.

Step 4.9. Suppose the sequence of two-step centralizers ofL has an initialsegment involving only the first two two-step centralizers, and that the correspond-ing sequence of constituent lengths begins with

2q; qr�2; 2q � 2�; (qr�2; 2q)s�1; qr�2; 2q � 2� :

ThenL has two distinct two-step centralizers, andL is obtained from anAFS-algebra or aBl-algebra via� inflation steps with respect to the first two-stepcentralizer.

The proofs of the remaining steps, which deal with the case of more than twodistinct two-step centralizers, are unchanged from II, but note that the proof ofStep 4.11 requires Step 4.6 and Step 4.4, and therefore Lemma 3.1.

Step 4.10. Thei-th two-step centralizer in order of occurrence appears forthe first time as

CL1(L2n+1) ;

for somen � i� 1.

Step 4.11. SupposeL has at least three distinct two-step centralizers. Thenall constituents are either short or long, so that by Step4.2the algebra is inflated.

8 JURMAN

5. SEQUENCES OF SHORT CONSTITUENTS

When the second non-short constituent is long, the number of short constituentsbetween the first constituent and the second non-short one can be easily obtainedby repeatingh times the inflation process as in the odd case, obtaining the sameresult.

Lemma 5.1. Suppose the sequence of two-step centralizers ofL has an initialsegment involving only the first two two-step centralizers, and that the correspond-ing sequence of constituent lengths is

2q; qm; 2q :

Thenm = 2r � 3 for some powerr = 2k.

If the second non-short constituent is intermediate, the inflation process aloneis not enough to get an analogous result. First of all we recall an useful formulafrom II.

Calculation Device 5.1. Let 0 6= v 2 Li, for somei > 1, define the ele-mentz = x+y and supposeCi; Ci+1; : : : ; Ci+n�1 2 fFx;Fyg. If [vx1x2 : : : xn]is a non-zero commutator inL, with xi 2 fx; yg, then

[vx1x2 : : : xn] = [vzn] :

The first result tells the parity of the length of the first sequence of consecutiveshort constituents.

Lemma 5.2. Suppose no two-step centralizer other than the first two occursbefore the second non-short constituent which we suppose to be intermediate.Then the number of short constituents between the first constituent and the secondnon-short one is even.

Proof. Write the sequence of two-step centralizers as

2q; qm; 2q � 2� ;

for some0 � � < h. By deflation, we can reduce to the case� = 0. Nowm

cannot be of the form2s + 1 for somes � 0, because of the following Jacobiidentity:

0 = [[yx2q�1(yxq�1)syxq

2�1][yx2q�1(yxq�1)syx

q

2�1]]


= [[yzq(s+2)+q

2�1][yzq(s+2)+

q

2�1]]

= [yz2q(s+2)+q�2y]

+[yz2q(s+2)+q�1] �

sXi=0

�q(s+ 2) + q

2� 1

qi+ q2

�!;

that yields

[yz2q(s+2)+q�2y] = [yx2q�1(yxq�1)myx2q�2y] = 0 ;

sinceq=2 > q=2� 1.

Now we prove a slightly different version of Lemma 5.4 in II; we callminimalintermediatea constituent of length2q � q=2.

Lemma 5.3. SupposeL has an initial segment of the sequence of two-stepcentralizers involving only the first two two-step centralizers, and suppose thecorresponding segment of the sequence of constituent lengths is

2q; qm :

Suppose further that an intermediate constituent in the sequence of two-stepcentralizers ofL is precededby at leastm short ones ending inx.

Then either that intermediate constituent isfollowed by at leastm short onesending inx, or it is minimal intermediate followed by an odd number� < m ofshort constituents ending inx and then by another minimal intermediate one. Thelatter alternative cannot occur if the first non-short constituent after the top longone is intermediate.

Proof. If m = 0 there is nothing to prove, so assumem � 1. Suppose thatthe intermediate constituent (of length2q � 2�) is followed by0 � � < m shortones ending inx. In the odd characteristic case, the condition� � 1 follows fromLemma 5.7 in I; here we must get it from a suitable expansion. By hypothesis2� < q, thus if we callw an element lying2� classes before the beginning of theintermediate constituent, thenw lies inside the last short constituent just beforethe intermediate one. This allows to use the following identity

0 = [w[yx2q�1yxq�1x]]

= [w[yz3q�1x]]

= [w[yz3q�1]x] + [wx[yz3q�1]]

= [wz3qx] �

��3q � 1

2�

�+

�3q � 1

2q

��+[wxz3q�1y]

10 JURMAN

+[wxz3q�1z] �

��3q � 1

2� � 1

�+

�3q � 1

2q � 1

��;

which reduces to

[wz3qx] = [wx2�

yx2q�2��1yxq�1x] = 0 ;

since the last binomial coefficient is the only one which vanishes; thus� � 1.Now follow the inductive proof in II: callv an element at the beginning of the

(m� � � 1)-th constituent before the intermediate one and consider the identity

0 = [v[yzq(m+2)�1x]] = [vzq(m+2)�2�xz2��1x] :

The element[vzq(m+2)�2� ] cannot be centralized by an elementu 6= x; y, sincethe above expansion shows that a constituent of length2� < q would follow.

If it is centralized byx, we have the induction step.If it is centralized byy, the equation above again shows that the length of the

(� + 1)-th constituent isq + 2�, which is an allowed constituent length only if2� = 2h�1 = q=2. This can only happen whenp = 2, so it has no counterpartin II. This means that both the intermediate constituent and the(� + 1)-th oneafter it are minimal, and this is the only case where an intermediate constituentpreceded bym short ones can be followed by a number of short constituents lessthanm (note that whenq = 2 there is only one intermediate constituent, that isclearly minimal). Take an elementw in classq=2 before the first intermediateconstituent and expand the following Jacobi identity:

0 = [w[yx2q�1(yxq�1)�+1x]]

= [w[yzq(�+2)+q�1]x] + [wx[yzq(�+2)+q�1]]

= [wzq(�+3)x] �

�q(� + 2) + q � 1

q2

�+

�+2Xi=2

�q(� + 2) + q � 1

qi

�!(2)

+[wzq(�+3)y]

+[wzq(�+3)z] �

�q(� + 2) + q � 1

q2� 1

�+

�+2Xi=2

�q(� + 2) + q � 1

qi� 1

�!:

Now, �q � 1q2

�and

�q � 1q2� 1

�

are both congruent to one modulo two, so the coefficient of the term[wzq(�+3)x]turns out to be�

� + 2

0

�+

�+2Xi=2

�� + 2

i

�=

�+2Xi=0

�� + 2

i

��

�� + 2

1

�;


and similarly for the coefficient of the term[wzq(�+3)z]:

�� + 2

0

�+

�+2Xi=2

�� + 2

i� 1

�=

�+2Xi=0

�� + 2

i

��

�� + 2

� + 2

�:

Then by using the identity

�+2Xi=0

�� + 2

i

�= (1 + 1)�+2 � 0 (mod 2) ;

we can show that the equation (2) becomes

(� + 2)[wzq(�+3)x] + [wzq(�+3)y] + [wzq(�+3)z] ;

that is

(� + 3)[wzq(�+3)x] = 0 ;

which implies that� is odd since the(� +1)-th constituent after the first interme-diate one is intermediate, too.

Now add the hypothesis that the(m+ 1)-th constituent after the first long oneis intermediate. Take an element� in class one less than the beginning of the(m� �)-th short constituent before the intermediate one: then we get the identity

0 = [�[yx2q�1(yxq�1)m+1y]]

= [�[yzq(m+2)+q�1y]]

= [�[yxq(m+2)+q�1]y] + [�y[yxq(m+2)+q�1]]

= [�zq(m+3)y]

�

0@m��X

j=0

�q(m+ 2) + q � 1

qj

�+

�Xj=0

�q(m+ 2) + q � 1

q(m� � + 1 + j) + q2

�1A+[�yxq(m+2)+q�1y]

+[�yzq(m+3)]

�

0@m��1X

j=0

�q(m+ 2) + q � 1

qj + q � 1

�+

�Xj=0

�q(m+ 2) + q � 1

q(m� � + 1 + j) + q2� 1

�1A :

The coefficient of the first element can be reduced to

m��Xj=0

�m+ 2

j

�+

�Xj=0

�m+ 2

m� � + 1 + j

�

12 JURMAN

and then evaluated by Lucas’ Theorem as

m+1Xj=0

�m+ 2

j

�=

m+2Xj=0

�m+ 2

j

��

�m+ 2

m+ 2

�

= (1 + 1)m+2� 1

� 1 (mod 2) :

Similarly the coefficient of the last element becomes

m��1Xj=0

�m+ 2

j

�+

�Xj=0

�m+ 2

m� � + 1 + j

�;

which is equivalent to

m+2Xj=0

�m+ 2

j

��

�m+ 2

m� �

��

�m+ 2

m+ 2

�= (1 + 1)m+2

�

�m+ 2

m� �

�� 1

� 1 (mod 2) ;

sincem is even (by Lemma 5.2) and� is odd (by expansion (2)). Thus the aboveequation reads

0 = [�zq(m+3)y] + [�yzq(m+2)+q�1y] + [�yzq(m+3)] = [�zq(m+3)+1] ;

and then the last claim of the lemma.

Then we can prove Step 4.5.

Lemma 5.4. Suppose no two-step centralizer other than the first two occursbefore the second non-short constituent. Suppose the sequence of constituentlengths ofL begins as2q; qm; 2q � 2�.

Thenm = 2k � 2 for somek.

Proof. Reduce by inflation to the case� = 0. Let k = [log2(m+ 2)], so thatm + 2 = 2k + �. The number� lies in the range0 � � � 2k � 1 and is evenby Lemma 5.2. Definef = 2k � � � 1 > 0. Suppose nowf � m: then byLemma 5.3 the element

w = [yx2q�1(yxq�1)myx2q�2(yxq�1)fy]

does not vanish. Let

� =

�m+ f + 3

2

�q +

q

2� 1 ;


so thatw has weight2(� + 1) and the element[yz�] lies q=2 classes before theend of the((m+ f + 1)=2)-th short constituent after the first long one. Then wehave:

0 = [[yz�][yz�]]

= [yz2�y]

+[yz2�+1] �

0@

m�f�1

2Xi=0

��

qi+ q2

�+

f�1Xi=0

��

q�m�f+3

2+ i�+ q

2� 1

�1A :

Since � q2� 1q2

�� 0 (mod 2)

and � m+f+32

m�f+32

+ i

�=

�2k

� + i+ 1

�� 0 (mod 2)

since� + i+ 1 � � + 2k � � � 2 + 1 = 2k � 1, the above expansion reduces to

[yz2�y] = w = 0;

a contradiction. Thenf > m, that is� < 1=2: as� is a non-negative integer, it is

zero.

Our version of Lemma 5.5 is slightly different from its analogous in II, since itneeds Lemma 5.3:

Lemma 5.5. Suppose the sequence of two-step centralizers ofL has an initialsegment involving only the first two two-step centralizers, and that the correspond-ing sequence of constituent lengths is

2q; qr�2 ;

wherer is a power of2. Then every long constituent in the sequence of two-stepcentralizers ofL is followed by at leastr � 2 short ones ending inx, until theoccurrence of a pattern: : : ; 2q � q=2; q� ; 2q � q=2 with � < r � 2.

The proof is exactly the same as in II: until the occurrence of an above describedpattern, by Lemma 5.3 and induction we can suppose that a long constituent ispreceded byr� 2 short ones. If it is followed by� < r� 2 short constituents, letv be the element at the end of the(r � 2� �)-th short constituent before the longone. Then the expansion

0 = [v[yzqr�1x]] = [vzqrx]

14 JURMAN

shows that another short one must occur. If the pattern: : : ; 2q� q=2; q� ; 2q� q=2does not occur in the sequence of constituent lengths the thesis applies to all longconstituents: we will show in the next section that this is always the case.

6. ALGEBRAS WITH LONG AND SHORT CONSTITUENTS

By using the results of the previous section, we can now prove Step 4.4. Byhypothesis, the first non-short constituent after the top long one is again long, thusby Lemma 5.1 we know that the sequence of constituent lengths starts as

2q; q2r�3; 2q ;

whereq = 2h andr = 2k, and theh times deflated algebra we get from this onehas parameterr. By Lemma 5.5 and induction, we can assume to have somewherethe pattern

: : : ; qr�2; 2q; q� ; (3)

where� � r � 2. We will show that no intermediate constituent can occur atthe end of the above pattern, for every value of�. As usual, the deflation processallows us to consider only the maximal intermediate case.

Suppose� = r�2 and letv be an element in the middle of the long constituentin the above pattern; sincer � 2, the following expansion from II still holds:

0 = [v[yzq(r+1)�1x]] = [vxq(yxq�1)r�2yx2q�2y] :

The above identity shows that no intermediate constituent is allowed after thepattern (3) for� = r � 2.

Then assume� = r � 1. Whenr > 2 the inequalityr � 2r � 3 is satisfied,so the following expansion holds, wherew is an element in class one less than thebeginning of the long constituent in (3):

0 = [w[yx2q�1(yxq�1)rx]]

= [w[yx2q�1(yxq�1)r ]x]

= [wzq(r+2)x] �

0@�q(r + 1) + q � 1

0

�+

r+1Xj=2

�q(r + 1) + q � 1

qj

�1A :

The coefficient is equivalent modulo two to

�r + 1

0

�+

r+1Xj=2

�r + 1

j

�=

r+1Xj=0

�r + 1

j

��

�r + 1

1

�� 1 ;


sincer is even, hence the expansion reduces to

[wyx2q�1(yxq�1)r�1yxq�1x] = 0 :

Thus in this case there must be another short constituent after the sequence ofr � 1 short ones. Whenr = 2 then� = 1 and the algebra starts as2q; q; 2q; herewe get a weaker result: an intermediate constituent cannot occur after the2q; qin the sequence (3). Suppose that this happens and callv an element in class oneless than the end of the short constituent in (3). Then the expansion

0 = [v[yx2q�1yxq�1x]]

= [v[yz3q�1]x] + [vx[yz3q�1]]

= [vz3qx] �

��3q � 1

1

�+

�3q � 1

2q

��+[vz3q�1y]

+[vz3q�1z] �

��3q � 1

0

�+

�3q � 1

2q � 1

��= [vz3q�1y] + [vz3q�1z]

= [vxyx2q�2yxq�1x]

shows that the intermediate constituent is followed by a short one. But if we nowcallw an element just one class less than the end of the long constituent in (3) weget a contradiction from the following expansion:

0 = [w[yx2q�1yxq�1yxq�1y]]

= [w[yx2q�1yxq�1yxq�1]y]

= [wz4qy] �

��4q � 1

1

�+

�4q � 1

q + 1

�+

�4q � 1

3q

��= [wz4qy]

= [vxyx2q�2yxq�1y]:

Finally, we have to deal with the case� > r � 1. Suppose that after the�short constituents an intermediate one occurs: by deflation, we can suppose thatit is of length2q � 1. By Lemma 5.3, one of the following must occur: eitherthe intermediate constituent is followed by at leastr � 2 short constituents or itis minimal intermediate followed by� < r � 2 short constituents and then byanother minimal intermediate one. In the former case, by deflatingh times thealgebra, we would obtain (as in II) a constituent of length

� + 2 + r � 2 + 1 > 2r ;

16 JURMAN

that is impossible since we saw that the deflated algebra has parameterr (note thatthis holds even forr = 2). The latter case can occur only forq = 2 (since theconstituent of length2q � 1 must be minimal) andr > 2 (since0 � � < r � 2):actually, it cannot occur at all, since we can always get a contradiction to theinitial hypothesis of infinite dimension ofL. So suppose to have the sequence ofconstituent lengths

4; 22r�3; 4 : : : 2r�2; 4; 2�; 3; 2� ; 3 ;

where� < � and consider the expansions below, wherev is the element at thebeginning of the long constituent in the pattern (3).

� � � 2r � 4. Since the(� + 1)-th constituent is short, we can expand

0 = [v[yxxx(yx)�+1x]]

= [v[yz2�+5]x] + [vx[yz2�+5]]

= [vz2�+5x] �

�Xi=0

�2� + 5

2i+ 3

�!

+[vz2�+6y]

+[vz2�+6z] �

�Xi=0

�2� + 5

2i+ 2

�!:

The two coefficient are equivalent, modulo two, respectively to

�+1Xj=1

�2(� + 2) + 1

2j + 1

�and

�+1Xj=1

�2(� + 2) + 1

2j

�;

so both reduce modulo two to

�+1Xj=1

�� + 2

j

�=

�+2Xj=0

�� + 2

j

��

�� + 2

0

��

�� + 2

� + 2

�;

and vanish, yielding

[vxxx(yx)�yxxy] = 0 ;

against the hypothesis of having an intermediate constituent after the� short ones.� � = 2r � 3: First of all we prove that� > 0:

0 = [vxxx(yx)��1y[yxxxyxx]]

= [vxxx(yx)��1y[yz5]x] + [vxxx(yx)��1yx[yz5]]


= [vxxx(yx)��1yz6x] �

��5

1

�+

�5

4

��+[vxxx(yx)��1yz6y]

+[vxxx(yx)��1yz6z] �

��5

0

�+

�5

3

��= [vxxx(yx)��1yz6y] + [vxxx(yx)��1yz6z]

= [vxxx(yx)�yxxyxx]:

Then we show that even� > 0 cannot occur, by using the fact that the(2r�2)-thconstituent is not short:

0 = [vxx[yxxx(yx)2r�2y]]

= [vxx[yz4r�1]y]

= [vxxz4ry] �

2r�3Xi=0

�4r � 1

2i+ 1

�+

�4r � 1

4r � 2

�!

= [vxxx(yx)�yxxyxy] :

� � = 2r � 2: Here we can prove that no intermediate constituent can occurafter the� short ones, by using an already employed relation:

0 = [vxx[yxxx(yx)2r�2y]]

= [vxx[yz4r�1]y]

= [vxxz4ry] �

2r�2Xi=0

�4r � 1

2i+ 1

�!

= [vxxx(yx)�yxxy]:

� � � 2r � 1: A slightly different version of the expansion above shows inthis case that we cannot even have that many short constituents after the long onein (3):

0 = [vx[yxxx(yx)2r�2y]]

= [vx[yz4r�1]y]

= [vxz4ry] �

2r�2Xi=0

�4r � 1

2i+ 2

�!

= [vxxx(yx)2r�1y]:

This proves completely Step 4.4.

18 JURMAN

7. PROOF OF STEP 4.6

The proof of this step is the same as in II, apart from the very last expansion:for completeness’ sake, we briefly summarize the path to get to this point.

The initial part of the sequence of constituent lengths is now

2q; qr�2; 2q � 2� ;

for some0 � � < h, so Lemma 5.3 and Lemma 5.5 state that every non-shortconstituent is followed by at leastr� 2 short ones ending inx. We show here thatthe sequence of constituent lengths consists of repetitions of the patterns

2q; qr�2 or 2q � 2� ; qr�2 :

In particular, forr = 2 this means that there are no short constituents. Moreover,by Lemma 3.1, the above claim implies that no further two-step centralizer canoccur.

By induction, suppose to have a

: : : ; qr�2; 2q; qr�2 (4)

pattern somewhere, and letv be an element at the end of the constituent thatprecedes the long one. Then the constituent after (4) cannot be short, since

0 = [v[yx2q�1(yxq�1)r�1y]] = [vyx2q�1(yxq�1)r�1y] ;

and it cannot be intermediate of length2q � 2 with 6= � since

0 = [v[yx2q�1(yxq�1)r�2yx2q�2 �1y]] = [vyx2q�1(yxq�1)r�2yx2q�2

�1y] :

By using induction again, we can suppose to have a

: : : ; qr�2; 2q � 2�; qr�2 (5)

pattern somewhere, and letv be an element at the end of the last short constituentbefore the intermediate one. To prove that the constituent after (5) cannot be short,first we reduce by induction to the case� = 0, then we expand

0 = [v[yx2q�1(yxq�1)r�1y]] = [vyx2q�2(yxq�1)r�1yx] :

To prove that the constituent after (5) cannot be intermediate of length2q � 2 ,we have to consider separately the cases > � and < �. In the former case,deflate to� = 0 then expand

0 = [v[yx2q�1(yxq�1)r�2yx2q�2 �1y]] = [vyx2q�2(yxq�1)r�2yx2q�2

�1yx] :


The latter case is the one which deserves a deeper investigation. First deflate to = 0, let� = q(r + 1) + q � 2� � 1 and expand from the back

0 = [vyx2��2[yx2q�1(yxq�1)r�2yx2q�2

��1x]]

= [vyx2��2[yz�]x] + [vyx2

��2x[yz�]]

= [vyx2��2z�+1x] �

��

2� � 1

�+

rXi=2

��

qi+ 2� � 2

�!

+[vyx2��2z�+1y]

+[vyx2��2z�+1z] �

��

2�

�+

rXi=2

��

qi+ 2� � 1

�!:

The involved binomial coefficients can be evaluated modulo two by using Lucas’Theorem and the identity (1) as:�

�

2� � 1

��

�r + 1

0

��2h�1 + � � �+ 2� � 1

2� � 1

�� 1 ;�

�

qi+ 2� � 2

��

�r + 1

i

��2h�1 + � � �+ 2� � 1

2� � 2

��

�r + 1

i

�;�

�

qi+ 2� � 1

��

�r + 1

i

��2h�1 + � � �+ 2� � 1

2� � 1

��

�r + 1

i

�;�

�

2�

��

�r + 1

0

��2h�1 + � � �+ 2�+1 + 2��1 + � � �+ 1

2�

�� 0 :

Then, since[vyx2��2z�+1y] = 0, we get

[vyx2q�2��1(yxq�1)r�2yx2q�2yx2

��1x] = 0 ;

because constituents cannot have length2� < q; then the claim follows.

8. PROOF OF STEP 4.8

By Step 4.6, Step 4.7 and deflation, we can suppose to have a sequence ofconstituent lengths which starts as

2q; qr�2; 2q � 1; (qr�2; 2q)��1; qr�2; 2q � 1 : (6)

We have to prove that either� = 2l, for somel � 0, or r = 2 and� = 2l � 1, forsomel � 2. The values1 � � � 4 verifies the thesis, so we can assume� � 5.Moreover, define� = [log2(�)] and� as the exponent of the factor2 in the primedecomposition of the number�.

First of all, we state and prove five lemmas.

20 JURMAN

Lemma 8.1. If � is odd, then another patternqr�2; 2q � 1 occurs at the endof ((6)).

Proof. In view of Step 4.6, either aqr�2; 2q or aqr�2; 2q � 1 can follow (6);then define� = qr(� + 1)=2 + qr=2 + q � 2, so that

[yz�] = [yx2q�1(yxq�1)r�2yx2q�2(yxq�1)r�2(yx2q�1(yxq�1)r�2)��32

yx2q�1(yxq�1)r2�1] ;

and expand

0 = [[yz�][yz�]] = [yz2�y] + [yz2�+1] �� :

If r = 2 the coefficient� is

� =

��12Xj=0

��

2qi

�+

��

2q �+12

� 1

�:

The second term inA vanishes since2q � 1 > 2q � 2, while the sum can beevaluated modulo two as

��12Xj=0

��

2qi

��

��12Xj=0

�2q �+1

2+ 2q � 2

2qi

�

�

��12Xj=0

��+12

i

�

�

�+12Xj=0

��+12

i

��

��+12

�+12

�

� (1 + 1)�+12 � 1

� 1 (mod 2) ;

so that� � 1.Whenr > 2 then

� =

r2�1Xi=0

��

qi

�+

��32Xj=0

r�1Xi=1

��

qrj + qi+ q r2

�

+

r�1Xi=0

��

qr ��12

+ q r2+ qi+ q � 1

�:


The terms in the last sum vanish sinceq � 1 > q � 2; those in the first sum areall zero but the very first one (occurring wheni = 0), since

��

qi

��

�qr �+1

2+ q r

2+ q � 2

qi

��

�r �+1

2

0

��r2

i

��q � 2

0

�(mod 2) ;

and, finally, the remaining sum can be evaluated as follows

��32Xj=0

r�1Xi=1

��

qrj + qi+ q r2

��

��32Xj=0

r�1Xi=1

�qr �+1

2+ q r

2+ q � 2

qrj + qi+ q r2

�

�

��32Xj=0

r�1Xi=1

�r �+1

2+ r

2

rj + i+ r2

��q � 2

0

�

�

��32Xj=0

0@ r

2�1Xi=1

�r �+1

2+ r

2

rj + i+ r2

�

+

r�1Xi= r

2

�r �+1

2+ r

2

rj + i+ r2

�1A

�

��32Xj=0

0@ r

2�1Xi=1

�r �+1

2

rj + (i+ r2)

��r2

0

�

+

r�1Xi= r

2

�r �+1

2

r(j + 1)

��r2

i� r2

�1A (mod 2) ;

now, the first addendum is zero sincer2< r

2+ i < r, while the second one is

non-zero only wheni = r2, so that the sum reduces modulo two to

��32Xj=0

��+12

j + 1

�=

��12Xj=1

��+12

j

�= (1 + 1)

�+12 �

��+12

�+12

��

��+12

0

�� 0 :

Then in both cases the coefficient� is equivalent to one modulo two, so thatthe above expansion reads as

[yz2�x] = [yx2q�1(yxq�1)r�2yx2q�2(yxq�1)r�2(yx2q�1(yxq�1)r�2)��1

yx2q�2(yxq�1)r�2yx2q�2x] = 0 ;

thus eliminating the possibility that the pattern (6) is followed by aqr�2; 2q.

22 JURMAN

Lemma 8.2. If ((6)) is followed by a patternqr�2; 2q � 1, thenr = 2.

Proof. Suppose, by way of contradiction, thatr > 2. By Lemma 5.3 thepatternqr�2; 2q � 1 is followed by at least another short constituent. But thisleads to a contradiction because of the following expansion (a slightly modifiedversion of a similar one occurring in II), wherev is an element in the middle ofthe last long constituent in (6) and� is 2q(r + 1)� 3.

0 = [v[yx2q�1(yxq�1)r�2yx2q�2(yxq�1)r�2yx2q�2y]]

= [v[yz�]y]

= [vz�+1y]

�

r�1Xi=1

��

qi

�+

r�2Xi=0

��

qr + qi+ q � 1

�+

��

2qr + q � 2

�!:

If q > 2 then all the terms but those in the first sum vanish sinceq � 1 andq � 2are bigger thanq � 3, while the first sum reduces to

r�1Xi=1

��

qi

��

r�1Xi=1

�2r + 1

i

��q � 3

0

��

�2r + 1

1

�� 1 (mod 2) :

If q = 2 the integer� is 4r + 1: then, since by Lucas’ Theorem the coefficient

�4r + 1

j

�

is non zero only for the values0; 1; 4r; 4r+1 of j, the only non-vanishing term is

��

2qr + q � 2

�:

Thus in both cases the coefficient of[v[yz�]y] is one, so that the above expansiongives

[v[yz�]y] = [yx2q�1(yxq�1)r�2yx2q�2((yxq�1)r�2yx2q�1)��1

((yxq�1)r�2yx2q�2)2yxq�1y] = 0 ;

as claimed.

Lemma 8.3. If L has

2q; 2q � 1; 2q��1; 2q � 1; 2q � 1; 2q��1; 2q � 1


as its initial sequence of constituent lengths where� is odd, then� > 2� � 2.

Proof. Suppose this is not the case, so that� � 2��2: then��1�(2��) <� � 1; since2� � 2 < � � 1, we can define� as2q � 3 + 2q � 2� andw as theelement one class before the end of the(� � (2� � �))-th long constituent,i.e.

w = [yx2q�1yx2q�2(yx2q�1)��1�(2��)yx2q�2] :

Now expand the following identity:

0 = [w[yx2q�1yx2q�2(yx2q�1)2��2yx2q�2y]]

= [w[yz�y]]

= [w[yz�]y]

= [wz�+1]y] �

2��1��Xi=0

��

2qi+ 1

�+

��

2q(2� � �)

�

+

��1Xi=0

��

2q(2� � � + i) + 2q � 1

�!:

The last sum in the coefficient vanishes since2q � 1 > 2q � 3, while the otherterms can be evaluated modulo two as follows:

2��1��Xi=0

��

2qi+ 1

�+

��

2q(2� � �)

��

2��1��Xi=0

�2�

i

�+

�2�

2� � �

�� 1 ;

since �2�

0

�is the only non-zero term, because of the range where� lies. Then the aboveexpansion reads as

0 = [wz�+1y]

= [yx2q�1yx2q�2(yx2q�1)��1(yx2q�2)2(yx2q�1)��1yx2q�2y] ;

a contradiction.

Lemma 8.4. If L has

2q; qr�2; 2q � 1; (qr�2; 2q)��1; qr�2; 2q � 1; (qr�2; 2q)��1; qr�2; 2q � 1

as its initial sequence of constituent lengths where� is even andr > 2, then� � �.

24 JURMAN

Proof. Again suppose this is not the case. Whenq > 2 it is possible to usethe same expansion as in II; letv be an elementq class before the end of the lastlong constituent in the(qr�2; 2q)��1 pattern and let� = qr(� +1)+2q�3; then[yz�y] is the relation which says that the sequence of constituent lengths is not ofthe form

2q; qr�2; 2q � 1; (qr�2; 2q)��1; qr�2; 2q � 1 ;

and we have

0 = [v[yz�y]]

= [v[yz�]y]

= [yx2q�1(yxq�1)r�2yx2q�2((yxq�1)r�2yx2q�1)��1

(yxq�1)r�2yx2q�2((yxq�1)r�2yx2q�1)��1

(yxq�1)r�2yx2q�2yxq�1y] ;

which proves the claim.Whenq = 2 the expansion we use is still the same, but the coefficient of the

involved term is not zero for a different reason; in fact, in this case, the parameter� is 2�r + 2r + 1 and the coefficient is given by

X�2�

��

�

�;

where� is the set of the indices

� = f2; 2i+ 2; 2r + 1; 2(rt+ r + i) + 1; 2(rt+ 2r) + 1;

2r� + 2i+ 1; 2r� + 2r : i = 1 : : : r � 2; t = 0 : : : � � 2g :

Since �2�r + 2r + 1

2i+ 1

��

��r + r

i

�(mod 2) ;

we have that

X�2�

��

�

��Xj2J

��r + r

j

�(mod 2) ;

where

J = f1 � j � �r + r: j 6= tr � 1; t = 2 : : : �g ;


that is

�r+rXi=1

��r + r

i

��

�Xj=2

��r + r

rj � 1

�� (1 + 1)�r+r �

��r + r

0

�

� 1 (mod 2) ;

since all termsnr � 1 are odd while�r + r is even. Then� � �.

Lemma 8.5. If L has

2q; 2q � 1; 2q��1; 2q � 1; 2q��1; 2q � 1

as its initial sequence of constituent lengths where� is even, then� > 2�, where2� is the highest power of two which divides�.

Proof. Obviously we can assume� � 6. If � is a power of two, then the claimfollows from Theorem 4.2, so suppose� is not a2-power.

When� = 1 the claim follows from the expansion

0 = [[yx2q�1yx2q�2(yx2q�1)�2�1yxq�1][yx2q�1yx2q�2(yx2q�1)

�2�1yxq�1]]

= [[yz2q(�2+1)+q�2][yz2q(

�2+1)+q�2]]

= [yz2q(�+3)�4y]

+[yz2q(�+3)�4z]

�

0@�

2�1Xi=0

�2q(�

2+ 1) + q � 2

2qi+ q

�+

�2q(�

2+ 1) + q � 2

2q �2+ q � 1

�1A :

In fact, all terms in the coefficient of the second element vanish becauseq; q�1 >q � 2, so that the equation reads as

[yz2q(�+3)�4y] = [yx2q�1yx2q�2(yx2q�1)��1(yx2q�2)2y] = 0 :

When2 � � � 2�, we can observe that0 < 2��1 < ��1 and that the integer� � 1 + � � (2� + 1) is not negative and less than� � 1, so we can write thefollowing expansion, by letting� = 2q � (2� + 1) + 2q � 3 andv as an elementat the end of the(� � 1 + � � (2� +1))-th long constituent in the series of� � 1ones:

0 = [v[yx2q�1yx2q�2(yx2q�1)2��1yx2q�2y]]

= [v[yz�y]]

= [v[yz�]y] + [vy[yz�]]

26 JURMAN

= [vz�+1y]

�

0@2�+1��X

i=0

��

2qi

�+

��1Xi=0

��

2q(2� + 1� � + i) + 2q � 1

�1A+[vyz�y]

+[vyz�z]

�

0@2�+1��X

i=0

��

2qi� 1

�+

��1Xi=0

��

2q(2� + 1� � + i) + 2q � 2

�1A :

Since2q � 1; 2q � 2 > 2q � 3 all terms are zero but those in the very first sum,which are congruent modulo two to

�2� + 1

i

�

wherei ranges between0 and2� + 1 � � . Since2 � � � 2�, the upper boundranges between1 and2� � 1, so among the above binomial coefficients the onlynon-zero ones are those corresponding to the values0 and1 of the indexi, andthus the entire sum amounts to zero. Thus the expansion reduces to

0 = [vyz�y] = [yx2q�1yx2q�2(yx2q�1)��1yx2q�2y(yx2q�1)��1yx2q�2y] ;

proving the claim.

Now we can prove the step; we distinguish three cases.

� If � is odd and� + 1 is not a power of two, then in its2-adic expansion

� = 2� +

��1Xi=1

ai2i + 1

not all theai are one, so it makes sense to define

= minf1 � i � �� 1j ai = 0g :

Then� = 2 +1 �A+2 �1 for some positive integerA. Moreover,2 � 2��1 �2� � 2 since� � 5, so that, by Lemma 8.1, Lemma 8.2 and Lemma 8.3 the valueof r is two and the element

[yx2q�1yx2q�2(yx2q�1)��1(yx2q�2)2(yx2q�1)2 �1y]

is not zero.


Now let� = q � 2 + 2q �+1+2

2and expand

0 = [[yx2q�1yx2q�2(yx2q�1)��3+2

2 yxq�1]

[yx2q�1yx2q�2(yx2q�1)��3+2

2 yxq�1]]

= [[yz�][yz�]]

= [yz2�y]

+[yz2�+1]] �

0B@

��1�2

2Xi=0

��

2qi+ q

�+

��

2q �+1�2

2+ q � 1

�

+

2 �2Xi=0

��

2q��+3�2

2+ i�+ q � 2

�1A :

The first sum and the next coefficient in the coefficient are zero becauseq; q�1 >q � 2, while the last sum can be evaluated as follows:

2 �2Xi=0

��

2q��+3�2

2+ i�+ q � 2

��

2 �2Xi=0

� �+1+2

2�+3�2

2+ i

�

�

2 �2Xi=0

� 2 +1�A+2 �1+1+2

2

2 � 1� i

�

�

2 �2Xi=0

�2 � (A+ 1)

2 � 1� i

�(mod 2) :

All terms in the above sum are zero since1 � 2 � 1� i � 2 � 1, so the aboveexpansion reduces to

[yz2�y] = [yx2q�1yx2q�2(yx2q�1)��1(yx2q�2)2(yx2q�1)2 �1y] = 0 ;

proving the first case.� If � is even andr > 2, suppose� is not a power of two: then2�+1+1�� < �

and then, by Lemma 8.4, the element

[yx2q�1(yxq�1)r�2yx2q�2((yxq�1)r�2yx2q�1)��1

(yxq�1)r�2yx2q�2((yxq�1)r�2yx2q�1)2�+1��]

is not zero.

Now let� = q � 2 + q r2+ qr2� and expand

0 = [[yx2q�1(yxq�1)r�2yx2q�2((yxq�1)r�2yx2q�1)2��1(yxq�1)

r2�1]

28 JURMAN

[yx2q�1(yxq�1)r�2yx2q�2((yxq�1)r�2yx2q�1)2��1(yxq�1)

r2�1]]

= [[yz�][yz�]]

= [yz2�y] + [yz2�z] �K :

In evaluatingK, we first encounter coefficients of the form

��

i

�

wherei ranges in the set

nqj: j = 0 : : :

r

2� 1o[n�

qr

2+ q�+ ql + qrj: l = 1 : : : r � 2; j = 0 : : : � � 2� � 1

o;

which are the only ones which gives a non-zero contribute to the coefficient, sinceas soon as we meet the intermediate constituent, the other terms become odd andthen they annihilate the binomial coefficient. Furthermore, among the elementlisted, the shape of� makes the binomial coefficient above to be different fromzero only fori = 0. Then the total coefficient is one and we get

[yz2�x] = [yx2q�1(yxq�1)r�2yx2q�2((yxq�1)r�2yx2q�1)��1

(yxq�1)r�2yx2q�2((yxq�1)r�2yx2q�1)2�+1��] = 0 ;

a contradiction.� Finally, if � is even andr = 2, by Lemma 8.5, the element

[yx2q�1yx2q�2(yx2q�1)��1yx2q�2y(yx2q�1)2��1yx2q�2x]

is not zero. Suppose that� is not a2-power and consider the following expansion,where� = 2q � 2 + 2q

��2+ 2��1

�:

0 = [[yx2q�1yx2q�2(yx2q�1)�2+2��1�1][yx2q�1yx2q�2(yx2q�1)

�2+2��1�1]]

= [[yz�][yz�]]

= [yz2�y]

+[yz2�z] �

0@�

2�2��1Xi=0

��

2qi

�+

2��1Xi=0

��

2q��2� 2��1 + i

�+ 2q � 1

�1A :

The terms in the last sum are zero since2q � 1 > 2q � 2. In the notation ofLemma 8.5, we can write� = 2� � (2n+ 1) for some positive integern. Then the


first sum can be evaluated as follows

�2�2��1Xi=0

�2q � 2 + 2q

��2+ 2��1

�2qi

��

�2�2��1Xi=0

��2+ 2��1

i

�

�

2��nXi=0

�2� � (n+ 1)

i

�

�

nXi=0

�n+ 1

i

�

� (1 + 1)n+1 �

�n+ 1

n+ 1

�(mod 2) ;

which is one; hence the above expansion becomes

[yz2�x] = [yx2q�1yx2q�2(yx2q�1)��1yx2q�2y(yx2q�1)2��1yx2q�2x] = 0 ;

concluding the proof.

9. DETERMINATION OF BI-ZASSENHAUS LOOP ALGEBRAS

In this last section, we prove the part of the Theorem 4.2 concerning Bi-Zassenhaus algebras,i.e. the caser = 2 ands = �.

Proof. By taking� = 0 in Step 4.6, we deduce thatL has only two distincttwo-step centralizersFx andFy and two distinct constituent lengths,2q and2q � 1.

Now we only have to prove thatL andBl(g; h) have the same sequence ofconstituent lengths. This means that, ifw is an element at the beginning of aconstituent, we have to show which one of the elements

[wyx2q�2x] and [wyx2q�2y]

is the generator of the homogeneous component in the corresponding weight.Now we introduce the element

vn = [yx2q�1(yx2q�2(yx2q�1)��1yx2q�2)n] ;

and we suppose that

L2q+1+dn = h[vny]i ;

whered = 2g+h+1 � 2 is the length of the periodic pattern.

30 JURMAN

First we prove that[vny] must be followed by an intermediate constituent. Infact, a graded Lie algebra of maximal classM such that

M2q+1+dn = h[vny]i

where [vny] is followed by a long constituent has finite dimension, since therelations

[yxy] and [v1x] = [yx2q�1yx2q�2(yx2q�1)��1yx2q�2x]

are zero inM , so that we can expand

0 = [vnyx2q�2[yxy]] = [vnyx

2q�1yy]

and

0 = [vn�1y[v1x]]

= [vn�1y[yz2q(�+1)+2q�3x]]

= [vn�1y[yz2q(�+1)+2q�3]x] + [vn�1yx[yz

2q(�+1)+2q�3]]

= [vn�1yz2q(�+1)+2q�2x]

�

0@��1Xi=0

�2q(� + 1) + 2q � 3

2qi+ 2q � 2

�+

�+1Xi=�

�2q(� + 1) + 2q � 3

2qi+ 2q � 3

�1A+[vn�1yxz

2q(�+1)+2q�2]

�

0@��1Xi=0

�2q(� + 1) + 2q � 3

2qi+ 2q � 3

�+

�+1Xi=�

�2q(� + 1) + 2q � 3

2qi+ 2q � 4

�1A= [vn�1yz

2q(�+1)+2q�2x]

= [vnyx2q�1yx] :

Lucas’ Theorem allows us to evaluate the involved binomial coefficients by usingthe following properties, where� is any non-negative integer and� 2 f2; 3; 4g:

�2q(� + 1) + 2q � 3

2q�+ 2q � �

��

�� + 1

�

��2q � 3

2q � �

�(mod 2) ; (7)�

� + 1

�

�=

�2g

�

�� 1 (mod 2) , � = 0; � + 1 :

Then we prove that, after the above cited intermediate constituent, there are atleast�� 1 long ones. Again, we prove that a graded Lie algebra of maximal class


M that coincide withL up to the above intermediate constituent and whose nextintermediate constituent occurs after less than��1 long ones has finite dimension.So let0 � i � � � 2, so that the relation

[yx2q�1yx2q�2(yx2q�1)iyx2q�2y] = 0

holds and suppose that the element

[vnyx2q�2(yx2q�1)iyx2q�2y]

is not zero. Then we have the identities

0 = [vnyx2q�2(yx2q�1)iyx2q�3[yxy]]

= [vnyx2q�2(yx2q�1)iyx2q�2yy]

and

0 = [vn�1yx2q�2(yx2q�1)��1[yx2q�1yx2q�2(yx2q�1)iyx2q�2y]]

= [vn�1yx2q�2(yx2q�1)��1[yz2q(i+2)+2q�3y]]

= [vn�1yx2q�2(yx2q�1)��1y[yz2q(i+2)+2q�3]]

= [vn�1yx2q�2(yx2q�1)��1yz2q(i+2)+2q�2] �

�2q(i+ 2) + 2q � 3

2q � 2

�

+

i+1Xj=1

�2q(i+ 2) + 2q � 3

2qj + 2q � 3

�+

�2q(i+ 2) + 2q � 3

2q(i+ 2) + 2q � 4

�1A= [vn�1yx

2q�2(yx2q�1)��1yz2q(i+2)+2q�2]

= [vnyx2q�2(yx2q�1)iyx2q�2yx] :

In fact, for0 � � � i+ 2 and� 2 f2; 3; 4g, we have�2q(i+ 2) + 2q � 3

2q�+ 2q � �

��

�i+ 2

�

��2q � 3

2q � �

�(mod 2) ;

where the last binomial coefficient vanishes only when� = 2, while

i+1Xj=1

�i+ 2

j

�+

�i+ 2

i+ 2

�=

i+2Xk=1

�i+ 2

k

�

=

i+2Xk=0

�i+ 2

k

��

�i+ 2

0

�

= (1 + 1)i+2 � 1

� 1 (mod 2) :

32 JURMAN

Finally, we have to show that after� � 1 long constituents an intermediate onefollows; this will prove the claim since it will complete the(n + 1)-th period inthe sequence of constituent lengths. Again we prove that the hypothesis that afurther long constituent follows the sequence of� � 1 previous ones cannot leadto an infinite-dimensional algebra. Suppose then that the element

[vn+1xy] = [vnyx2q�2(yx2q�1)�y]

does not vanish; then we can show, by using two already employed relations andthe same considerations of (7), that

0 = [vn+1[yxy]]

= [vnyx2q�2(yx2q�1)��1yx2q�2[yxy]]

= [vyx2q�1(yx2q�1)�yy]

= [vn+1xyy] ;

and

0 = [vn�1yx2q�2(yx2q�1)��1y[v1x]]

= [vn�1yx2q�2(yx2q�1)��1y[yx2q�1yx2q�2(yx2q�1)��1yx2q�2x]]

= [vn�1yx2q�2(yx2q�1)��1y[yz2q(�+1)+2q�3x]]

= [vn�1yx2q�2(yx2q�1)��1y[yz2q(�+1)+2q�3]x]

+[vn�1yx2q�2(yx2q�1)��1yx[yz2q(�+1)+2q�3]]

= [vn�1yx2q�2(yx2q�1)��1yz2q(�+1)+2q�2x]

�

�2q(� + 1) + 2q � 3

2q � 2

�+

�+1Xi=1

�2q(� + 1) + 2q � 3

2qi+ 2q � 3

�!

+[vn�1yx2q�2(yx2q�1)��1yxz2q(�+1)+2q�2]

�

�2q(� + 1) + 2q � 3

2q � 3

�+

�+1Xi=1

�2q(� + 1) + 2q � 3

2q(i+ 1) + 2q � 4

�!

= [vn�1yx2q�2(yx2q�1)��1yz2q(�+1)+2q�2x]

= [vnyx2q�2(yx2q�1)�yx]

= [vn+1xyx] :

The above relations yield

L2q+1+d(n+1) = h[vn+1y]i ;

concluding the induction step and then completing the proof.


REFERENCES1. A. A. Albert and M. S. Frank. Simple Lie algebras of characteristicp. Univ. e Politec. Torino.

Rend. Sem. Mat., 14:117–139, 1954–55.

2. G. Benkart, A. I. Kostrikin, and M. I. Kuznetsov. Finite-dimensional simple Lie algebras with anonsingular derivation.J. Algebra, 171(3):894–916, 1995.

3. A. Caranti and G. Jurman. Quotients of maximal class of thin Lie algebras. The odd characteristiccase.Comm. Algebra, 27(12):5741–5748, 1999.

4. A. Caranti, S. Mattarei, and M. F. Newman. Graded Lie algebras of maximal class.Trans. Amer.Math. Soc., 349(10):4021–4051, 1997.

5. A. Caranti and M. F. Newman. Graded Lie algebras of maximal class, II.J. Algebra, 229:750–784,2000.

6. C. Carrara. (Finite) presentations of loop algebras of Albert-Frank Lie algebras. PhD thesis,Trento, 1998.

7. C. Carrara. (Finite) presentations of the Albert-Frank-Shalev Lie algebras. Boll. Un. Mat. It., toappear, 2000.

8. G. Havas, M. F. Newman, and E. A. O’Brien.ANU p-Quotient Program (version 1.4),written in C, available as a share library withGAP and as part of Magma, or fromhttp://wwwmaths.anu.edu.au/services/ftp.html , School of Mathematical Sci-ences, Canberra, 1997.

9. G. Jurman.On graded Lie algebras in characteristic two. PhD thesis, Trento, 1998.

10. G. Jurman. A family of simple Lie algebras in characteristic two. In preparation, 2000.

11. D. E. Knuth and H. S. Wilf. The power of a prime that divides a generalized binomial coefficient.J. Reine Angew. Math., 396:212–219, 1989.

12. A. I. Kostrikin and M. I. Kuznetsov. Finite-dimensional Lie algebras with a nonsingular derivation.In Algebra and analysis (Kazan, 1994), pages 81–90. de Gruyter, Berlin, 1996.

13. E. Lucas. Sur les congruences des nombres euleriens et des coefficients differentiels des fonctionstrigonometriques, suivant un module premier.Bull. Soc. Math. France, 6:49–54, 1878.

14. A. Shalev. Simple Lie algebras and Lie algebras of maximal class.Arch. Math. (Basel), 63(4):297–301, 1994.

15. A. Shalev and E. I. Zelmanov. Narrow Lie algebras: a coclass theory and a characterization of theWitt algebra.J. Algebra, 189(2):294–331, 1997.

16. M. Vergne. Reductibilite de la variete des algebres de Lie nilpotentes.C. R. Acad. Sci. Paris Ser.A-B, 263:A4–A6, 1966.

17. M. Vergne. Cohomologie des algebres de Lie nilpotentes. Applicationa l’etude de la variete desalgebres de Lie nilpotentes.C. R. Acad. Sci. Paris Ser. A-B, 267:A867–A870, 1968.

graded lie algebras of maximal class, iii€¦ · ﬁnite quotient: this result was essential to...

Documents