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Thesis for the Degree of Ph. D.
MICHKGAN STATE UNEVERSITY
HYO CHflL MYUNG
1970

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FLEXIBLE LIE-ADMISSIBLE ALGEBRAS

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ABSTRACT
FLEXIBLE LIE-ADMIS S IBLE ALGEBRAS
BY

Hyo Chul Myung

The purpose of this thesis is a study of noncommutah
tive, flexible nilalgebras. For an algebra A, the alge-
bra A- is defined as the same vector space as A, but
with a multiplication given by [x,y] = xy - yx for all
x and y in A. An algebra A is said to be Lie;

admissible if A“ is a Lie algebra.

 

Assume that A is a finite dimensional, flexible,
Lie-admissible algebra over a field F of characteristic
# 2. Let Aa denote the root space of A_ for a split
Cartan subalgebra H of A_ corresponding to the root

a. The main results center around the following theorems:

Theorem 1: Suppose that A- has a split, abelian
Cartan subalgebra H which is a nil A-subalgebra. If
Ad is one-dimensional for every d #'O and the center of

A- is 0, then A is a Lie algebra isomorphic to A—.

Theorem 2: Let F be of characteristic 0 and the

 

solvable radical N of A- be nilpotent. Then a Levi;
factor S of A- is an ideal of A if and only if S
has a split Cartan subalgebra H being a nil A-subalgebra

and [N,H] = O.

Hyo Chul Myung

Theorem 3: Suppose that A is power-associative

 

and that A- is reductive such that there exists a split
Cartan subalgebra H of A- with h3 = O for all h in

H. Then the center of A- is an ideal of A.

Theorem 4: If A is simple and x3 = O for all

 

x in A, then A is a Lie algebra.

Theorem 5: Let F be of characteristic 0. If A

 

is a central simple nilalgebra of dimension 5 4, then A
is either commutative or a Lie algebra, and in the latter

case A is the 3-dimensional simple Lie algebra.

Theorem 6: If A is a noncommutative nilalgebra of

 

dimension 3 4 such that A— is nilpotent, then A is
nilpotent such that all products of any 4 elements in A

are 0.

FLEXIBLE LIE-ADMISSIBLE ALGEBRAS

BY

Hyo Chul Myung

A THESIS

Submitted to
Michigan State University
in partial fulfillment of the requirements
for the degree of

DOCTOR OF PHILOSOPHY

Department of MathematiCs

1970

ACKNOWLEDGMENT

The author wishes to express his sincere gratitude

to Professor M. L. Tomber for suggesting the problem and

for his stimulating guidance during the research.

ii

To Myung-Mi

iii

TABLE OF CONTENTS

Page
ACKNOWLEDGMENTS . . . . . . . . . . . . . . . . . . . ii
INTRODUCTION . . . . . . . . . . . . . . . . . . . . 1
CHAPTER
I. GENERAL RESULTS IN CASE Rad A- # O . . . 4
II. SOME SIMPLE NILALGEBRAS . . . . . . . . . . . 25

III. NILALGEBRAS OF DIMENSION LOWER THAN 5 . . . . 37

BIBLIOGRAPHY . . . . . . . . . . . . . . . . . . . . 45

iv

INTRODUCTI ON

In 1948, A. A. Albert [1] first introduced flexible
Lie-admissible algebras. He also defined Jordan-admissible
algebras and proved that any finite dimensional, flexible
algebra A such that A+ is a simple Jordan algebra is
either quasi—associative or a Jordan algebra. Albert pro-
posed to investigate an analogous problem for flexible Lie-
admissible algebras. In 1962, Laufer and Tomber [9] par-
tially solved this problem by showing that if A is a
finite dimensional, flexible, power-associative algebra
over an algebraically closed field of characteristic 0
such that A- is a simple Lie algebra, then A is a Lie
algebra isomorphic to A-. In Chapter I, we extend this
result to the case of Rad A_ #’O as well as generalize
the result of Laufer and Tomber.

Although a Levi-factor S of A- mayhbe an ideal
of A-, S is not an ideal of A even in case A is a
nilalgebra. In terms of a Cartan subalgebra of S, we
give a necessary and sufficient condition that the Levi-
factor S of A- be an ideal of A when$N§ad A- is
nilpotent. This result provides some interesting applica—

tions to the case that A' is reductive.

The class of flexible algebras includes all commuta—
tive and anticommutative algebras, in particular, Jordan
and Lie algebras, and alternative algebras, as well as the
generalized standard algebras of R. Schafer [13]. The gen-
eralized standard algebras generalize both alternative al-
gebras and standard algebras of Albert [1]. Any standard
algebra is flexible Lie—admissible. Schafer [13] proved
that any generalized standard nilalgebra is nilpotent and
hence there is no simple generalized standard nilalgebra.
Simple Lie algebras give examples that being nil does not
imply nilpotency for flexible Lie-admissible algebras. If
a simple, flexible algebra A is not nil, then various
structures are known, while very few results are known if
A is nil. If A- is a simple Lie algebra, then it is
shown in Oehmke [10] that A is a simple nilalgebra, and
in this case. A is a Lie algebra- At the present time,
the only known simple, flexible, Lie—admissible nilalgebras
are Lie algebras. Since any commutative algebra is trivi—
ally Lie-admissible, these remarks suggest the conjecture
that any simple, flexible, Lie-admissible nilalgebra is
either a Lie algebra or commutative. In Chapter II, we
treat in part this conjecture and deal with certain nil—
algebras whose minus-algebras are nilpotent.

In Chapter III, by using the known structures of Lie
algebras of dimension 4, we determine all 4-dimensional,
flexible, Lie—admissible nilalgebras whose minus -algebras

are not non—nilpotent, solvable Lie algebras. This enables

3

us to show that any noncommutative, flexible nilalgebra
A of dimension 5 4 such that A- is a nilpotent Lie

algebra is nilpotent.

CHAPTER I
GENERAL RESULTS IN CASE Rad A_ 7’ 0

Flexible Lie-admissible algebras were first intro—
duced by A. A. Albert in [1]. As an analog to the result
that Albert has proved for a flexible algebra A such
that A+ is a simple Jordan algebra, in [9] Laufer and
Tomber proved that if A is a finite dimensional, flexible,
power-associative algebra over an algebraically closed field
of characteristic 0 such that A- is a simple Lie alge-
bra, then A is a Lie algebra isomorphic to A-. In this
case it is shown that A is a simple nilalgebra. The’
proof mostly depends on the structure of the simple Lie
algebra A-. In this chapter we generalize the result of
Laufer and Tomber in point of view of the structure of Lie
algebra. A- and Obtain some general results in case that
A- has non-zero.radical.

We briefly introduce the standard terminologies, def-
initions,.and notations in nonassociative algebras.

For an algebra A, the algebra A- is defined as
the same vector space as A, but with a multiplication
given by [x,y] = xy - yx where juxtaposition denotes mul-
tiplication in A. An algebra A is said to beagle:

admissible if A_ is a Lie algebra, that is,
4

 

(1 1) [x.[y.ZJ] + [2:[XoY]] + [y.[z.X]] = 0
and

(x.Y] + [YIX] = 0
for all x, y, z in A. It is shown in [1] that an algebra
[x.YJ = [Rx - Lx'Ry ' Ly]
where Ra and La are right and left multiplications by

A is Lie—admissible if and only if R

a in the algebra A.

An algebra A is called flexible if it satisfies the

flexible law

(1.2) (xy)x = x(yx),

or equivalently

(1.3) LR =RL
x x x x

for all x, y in A. The linearizations of (1.2) and

(1.3) implies

(1.4) (Xy)z + (zy)x = X(y2) + 2(YX)
and
(1.5) L -LL=R -RR

xy y x yx y X'
respectively. In [1] Albert has shown that an algebra is

- L ].
X Y Y
If we denote the linear transformation RX - Lx by DX = D(x),

a flexible Liesadmissible algebra if and only if
1.6 R = R , R
( ) [x.y] [

then (1.6) may be written as RxD(y) = [Rx' Dy] or In

a flexible.Lie-admissible algebra- Dy‘ is a derivation.
An algebra .A, is said to be power—associative.if the
subalgebra generated by x is associative for every x in IL

It is also shown in [1] that a flexible algebra A of

characteristic # 2,3,5 is power-associative if and only
if x2x2 = x3x for all x in A. An element x in a
power-associative algebra A is called nilpotent in case
there is a positive integer n such that xn = 0. A power—
associative algebra is called nil_if every element is nil-
potent. An algebra A is called simple in case A has
no proper ideals and A2 #’O.

In this chapter we will make extensive use of the
standard terminologies and the known results of Lie algebras

as found, for example, in Chevalley [5], Jacobson [8], and

Seligman [14].

We first prove a sequence of lemmas which will be used
throughout this thesis. In this chapter we assume that A
denotes a finite dimensional, flexible, Lie-admissible al—

gebra over a field F.

Lemma 1.1: Let H be a split Cartan subalgebra of

 

A- and Am the root space of A- for H corresponding
to the root d. Then

0 if a + B is not a root;
AA:

a B A

a + B If a + B 15 a root.

In particular, H is a subalgebra of A.

Proof: Let D be a derivation of A such that the
characteristic roots of D belong to the base field F.
Let A be a characteristic root of D and AA = {x E A‘

x(D - AI)k = O for some k]. Then it is known that

0 if A + u is not a Characteristic root:

AAA 5
“ AX + if A + u is a characteristic root.
Since Dh is a derivation of A for every h in H and
= A , A C C =
Au hQH mm 0% AmmABm) Awh) + am) A<a+e>cm

for all h in H and hence AaAB C hEL A(a + B)(h) = Aa +£3-
Since H. is the root space corresponding to the root 0,

this in particular implies HH C H. This proves the lemma.

In the following lemmas we assume that the character-
istic of F is not 2. AG denotes the root space of A-
for a split Cartan subalgebra H corresponding to the root

a and F[x] the subalgebra of A generated by x in A.

Lemma 1.2: Let H be a split Cartan subalgebra of

 

A_ and h an element of H such that Dy is a scalar on
Aa. a #’O, for all y in F[h]. If h3 = 0, then

[Aa, h2] = O for all a #0.

Proof: Let a be any non-zero root of A- for H.

Let Dh = Rh - Lh = 1E and th = uE on Ad for some
A, u in F, where E is the identity map on Aa' Since
h4 = O, (1.5) implies that L 2 = R 2 and hence
2 2
h h
(1.7) u(2R 2 - uE) = O.
h

. 3 . . _
Since h - O, (1.5) implies Lthh — haRh, hence

(ha - LIE) (Rh - XE) = haRh, that 13,

(1.8) - Aha -Lfifil-+ AuE = O.

Now suppose that u y’o. From (1.7) we have 2R 2 - uE = O
h

and combine this with (1.8) to obtain u(AE - 2Rh) = 0.
Since u #'0, this implies
(1.9) 2Rh = RE.
2 _ 2
On the other hand, from L 2 - Lh — R 2 - Rh or (R 2 - uE)
h h h
- (Rh - AB)2 = R - ha , we have SE = 1(2Rh - AB).

2
h
Hence by (1.9), u = O and this is a contradiction. There-

fore we have u = O or D 2 = O on Ad for any non-zero

h
root a. This completes the proof.

Lemma 1.3: Let H be the same as in Lemma 1.2.

 

Suppose that H is power-associative and that h is an

element of H such that both Dy and Ry are scalars on

Aa' a #’O, for all y in F[h]. If h4 = 0, then
[Aa' h3] = O for all a #'0.
Proof: Let a be any non—zero root of A— for H.
Let D = R - = AE, D = R - L = HE and R = VE
h h Lh h2 h2 h2 h2
on A0 for some 1, u, v in F. Since h4 = O, (1.5)
implies L 22 = R 22 and so “(2R 2 - uE) = O, that is,
h h h
u(2v — u) = 0. Suppose u #'0, then u = 2v. Since
h5 = 0, equation (1.5) implies L 3L 2 = R 3R 2. Hence,
h h h h
it follows that L 3(R 2 — uE) = R 3R 2. Thus (v - u)L 3
h h h h h
= OR 3, and since u = 2v, this implies L 3 = - R 3.
h h h
Hence L - L. = R — R gives 2R - u( - AE)
h3 Lh h2 h3 Rh h2 h3 Rh
- Ath = 0, Since L 2 = (v — u)E and Lh = Rh - XE.

h

Therefore

(1.10) 4Rh3 - u.(2Rh - AB) = O.
. 4
USIng h = O and (1.5), L = R and hence
h3Lh hBR‘h

2

R 3(2Rh - AB) = 0. Thus by (1.10), this implies 4(R 3) ==O,
h h
and since R is a scalar on A , this gives R = O
h3 a h3
and L 3 = 0. Hence D 3 = O on A . Now suppose u = O.
h h a 2
Then R 2 = L 2 = vE, and hence (1.5) applied to h and
h h
h3 implies VD 3 = 0. Now (1.5) applied to h and h2
h
gives D 3 = th. Therefore, since D 3 is a scalar on
h h
A , D = O on A . This completes the proof.
a h3 a

Lemma 1.4: Let H be a split Cartan subalgebra of

 

A- such that Dh is a scalar on Aa' a y’o, for all h
in H. If H is a zero algebra, that is, hh' = O for

all h, h' in H, then Rh + Lh = O for all h in H.

Proof: Let Dh = Rh - Lh = XE on Aa' a #'0.
Since h2 = O, Lh2 = ha and hence
(1.11) 1(2Rh - AB) = 0.
If A = O, Rh = Lh' Since a is a non-zero root of A-

for H, we may then choose h' 6 H such that Dh' = uE #'O

on Aa' By (1.5), Lth, = Rth, , and so RhD , = O.

0 and in parti-

But since Dh' = HE #'0, this implies Rh
cular 2Rh = AB = 0. If A #'O, 2Rh = 1E by (1.11). Hence
we Obtain 2Rh = AB on Aa' and therefore Rh + Lh = O

on A. This proves the lemma.

10

Theorem 1.5: Suppose that A— has a split, abelian

 

Cartan subalgebra H which is a nil subalgebra of A
and F is of Characteristic #’2. If AG is one-
dimensional for any a #’O and the center of A- is 0,

then A is a Lie algebra isomorphic to A—.

Proof: Since H is finite dimensional, there is a
t > 1 such that ht = O for all h e H. We first show
that H is a zero algebra. Suppose that t 2,3 and let
n be the least integer such that 3n.z t. For any element

h of H, let g = h“

and then g3 = 0. Since Ad for
a #’O is one—dimensional, it follows from Lemmas 1.1 and
1.2 that g2 belongs to the center of A_, and hence
g2 = O or g2 = h2n = 0. If 2n > 4, let m be the

least positive integer with 3m 2.2n and then m < n. The

above argument implies h2m = 0. Hence by repeated appli-
cations of this, it follows that either h4 = O or h2 = 0.
If h4 = O, h3 = O by Lemma 1.3 and so h2 = 0. Since

H is abelian, it follows that H is a zero algebra. There-
fore if Aa = Fx and [x,h] = d(h)x, then xh = - hx

= %a(h)x by Lemma 1.4. Let a, B be any non—zero roots.

If a + B is not a root, then AaAB = ABAa = O by Lemma
1.1. Now suppose that a + B is a root. If a + B = 0,
Choose h E H with a = a(h) #’O and let xh = WX, so

hx = (W - a)x. Equation (1.4) applied to h, x, y implies

that (hx)y - h(xy) + (yx)h - y(xh) = O, and since xy

and ya: are in H, it follows that yx = --g[x,y]. Therefore

11

if a + B is a root, we may set
(1.12) xy - yx = 12, xy = uz, yx = (u — A)z,
where AB = Fy and z is a non—zero element in Au + B.
Choose h 6 H with B(h) #'O and let a = a(h), B = B(h).
Then

1 l
(1.13) Xh = - hx = 30x, zh = - hz = 31a + B)z.
By (1.4), (hx)y — h(xy) + (yx)h — y(xh) = O, and by (1.12)
and (1.13) this implies that %{-au + (a + B)u + (u-—X)(a4—B)
- OL(u - 1)]2 = 0. Since 2 7! 0, this implies B(Zu- 1) =0
and so A = 2n. Therefore xy = - yx = %{x,y] and this

holds for all x, y in A. This completes the proof.

In [9] Laufer and Tomber have proved that if A-
is semi-simple over an algebraically closed field of char—
acteristic 0, then A is a direct sum of simple ideals
Ai of A such that Ai- are simple Lie algebras. In
this case it follows from Oehmke [10] that if A is power-
associative, then it is a nilalgebra, and therefore A—
satisfies the conditions in Theorem 1.5- Hence Theorem 1.5
generalizes the result of Laufer and Tomber. In Theorem
1.5 the conditions that H is abelian, dim Aa = 1 for
a #'O, and center A- = 0 do not seem to be strong enough
to imply that H is a nil subalgebra of A. The condition
that the center of A_ is O is essential in the theorem.

Examples for these two facts are given in the following:

12

Example 1.1: Let F be of characteristic #’2.

 

Let A(d) be an algebra over F with basis (x, y, h}
such that the multiplications are given by

xh = x, yh = %(a + 1)Y.

by = %(l — 00y. h2 = h.
and all other products are 0, where a #'O, l in F.

Then A(a) is a flexible algebra. In fact, if u==Ax4—uy+-Vh

is any element of A(a),

v O O O O O

R = O l<1+a>v O L = O —1—(1-a)v O
u 2 ' u 2

O %(1 - am v A %(l + 00p. v

relative to (x, y, h}. A direct computation shows that
LuRu = RuLu and hence A(a) is flexible. Then A(a)_ is
given by
[X: Y] = O. [X: h] = X: [Y1h] = GY.

and hence it follows that A(d)- is a solvable Lie algebra.
Therefore A(a) is flexible Lie-admissible. Since a #'O,]q
A(a)— = Fh + Fx + Fy is the Cartan decomposition of A(a)_
for Cartan subalgebra H = Fh, and Fx and Fy are the
root spaces corresponding to roots 1 and a. The center
of A(a)- is 0, but H is not a nilalgebra. We notice
that A(a) is isomorphic to A(B) if and only if a = B,
since A(a)- is isomorphic to A(B)- if and only if a = B.

The above example tells us that the algebra A— in

Theorem 1.5 need not be semi-simple.

Example 1.2: Let F be of characteristic #'2. Let

A be an algebra over F with basis (x, y, h, 2] such

13

that the multiplications are given by

xy = z + %h, yx = z - %h, xh = - hx = %x,
yh = — hy = - %y, h2 = - z,

and all other products are 0. In Chapter III, Theorem 3.1,
we will show that this algebra is flexible Lie-admissible.
Then H = Fh + F2 is an abelian Cartan subalgebra of A-,
and A_ = H + Fx + Fy is the Cartan decomposition of A-
for H where Fx and Fy are the root spaces corres-
ponding to roots 1 and - 1, respectively. The multipli-
cation table shows that H is a nilalgebra such that

u3 = O for all u in H. We note that the center of A-

is Fz. But A is not a Lie algebra, since h2 = - z #'0.

A finite dimensional Lie algebra L over a field
F of characteristic #'2, 3 is called classical if:
(1) the center of L is O; (2) L = [L, L]: (3) L has
an abelian Cartan subalgebra H (called a classical Cartan
subalgebra), relative to which: a) L = ZlLa where ad h
is a scalar on La for any h in H aid a: b) if a #'O

is a root, [La' L is one-dimensional: c) if a and

-a]
B are roots and B #’O, then not all a + kB are roots.
The family of classical Lie algebras includes the family

of finite dimensional, split semi-simple Lie algebras over

a field of characteristic 0. For any classical Lie algebra
L, it is shown in Seligman [14, p. 30] that La for

a #’O is one-dimensional. Therefore from Theorem 1.5

we Obtain:

14

Corollary 1.6: Suppose that A- is a classical
Lie algebra such that A- has a classical Cartan subalgebra
which is a nil subalgebra of A. Then A is a Lie algebra
isomorphic to A_.

Let F be of characteristic 0 and S a Levi-
factor of A—. Then S is not in general an ideal of A
although it is an ideal of A-. In fact, let B be the

algebra of 2 x 2 matrices over F and let

[Z DH: ‘31] «[3 2]-

Then S = Fx + Fy + Fh is a Levi-factor of B_ which is

X
II

an ideal of B_ but S is not a subalgebra of B. Here
we notice that a Cartan subalgebra Fh of S is not a

subalgebra of B. Next, let A be the algebra given in
Example 1.2. Then S = Fx + Fy + Fh is a Levi-factor of
A— which is an ideal of A—, and Fh is a Cartan sub-
algebra of S. But S is not a subalgebra of A, since

h2 = — z ¢ S. In this case, we notice that Fh is not a

subalgebra of A. But h is nilpotent, that is, h3 = O,
and A is not associative, since (yh)h #'Yh2 = 0.
Now, we wish to give a necessary and sufficient con-

dition that a Levi—factor of A- is an ideal of A when

the solvable radical of A_ is nilpotent.

15

Theorem 1.7: Suppose that A is a finite dimensional,

 

flexible, Lie-admissible algebra over an algebraically closed
field F of characteristic 0 such that the solvable rad-
ical N of A_ is nilpotent. Then a Levi-factor S of

A- is an ideal of A if and only if S has a Cartan sub-
algebra H that is a nil subalgebra of A and such that

[H, N] = O. In this case, N is a subalgebra of A, and
furthermore if A is simple, then either A- is nilpotent

or A is a Lie algebra.

Proof: If the Levi—factor S is an ideal of A, then
S has a Cartan'subalgebra H satisfying the conditions
since [8, N] = 0. NOW suppose that the Levi-factor S
of A- has a Cartan subalgebra H satisfying the conditions.
We show that [S, N] = O and hence S is an ideal of A-.
Let Sac be the root space of S for H corresponding
to the root do # O and then so = Fx. If h is an ele-
ment of H such that ao(h) # O,0 by (1.1), ao(h)[N, x]
= EN. ao(h)x] c [N, [x, h]] c [[N, h], x] + [[x, N], h] = O.
Hence it follows that [N, x] = O, and so [N, s] = 0. It
follows from Dixmier [6, p. 20] that there exists a Cartan
subalgebra H' of A- such that H' = H + H' n N and
AG = Ad n S + Aa n N where Ad is the root space of A-
for H' corresponding to the root G. Since [N, H] =.O
and H and N are nilpotent, it follows that H + N

is nilpotent in A—. Since H' is a maximal, nilpotent

subalgebra of A", this implies that H' = H + N. Therefore,

16

for any non-zero root a, Ad 9 S and hence S D H + Z: A ,

a#b
that is, S = H + Z: A . Since a(z) = O for all z in N
ayb
and a, this implies that S = H + Z) A is the Cartan

a#b
decomposition of S for H, and hence AG is one-

dimensional for all a #'0. Since H is a commutative,

nil subalgebra of A and the center of S is 0, from
Lemmas 1.2 and 1.3 it follows that H is a zero algebra.
Hence by Lemma 1.4, Lh + Rh = O on S for all h in

H, and so

(1.14) ~ xh = - hx = %[x, h] = %a(h)x

for x in Ad. We first prove that S is a subalgebra

of A. If a and B are non-zero roots, then AdAB C S

if a + B is a non-zero root. Let x and y be non-zero
elements of Ad and A-a' respectively. Then by Lemma

1.1, xy 6 H', and let xy = h + z for h in H and

z in N. Since a is a non-zero root, by (1.14) we may
choose h' in H such that yh' = - h'y = 1y with A #'O
in F. By (1.4), (xy)h' + (h'y)x = x(yh') + h'(yx) and
hence (h + z)h' — Ayx = Axy + h'(h"*+ 2) since yx = h"+-z
for some h" in H. Since H is a zero algebra and

[N, H] = 0, this implies that 1(xy + yx) = O and xy+yx=O
since A #'0. Hence h + h" + 22 = O and so h + h" = z = 0,
because H + N is a vector space direct sum. Therefore

xy = - yx belongs to H, and hence this proves that S

is‘a subalgebra of A. By Theorem 1.5, S is a Lie sub-
algebra of A. Now let a be any non-zero root and [x,yn'h]

a basis of the 3-dimensional simple subalgebra of S

17

corresponding to a such that

(1.15) xh = x, yh = — y, xy = h.

Let 2 be any element of N and then xz = zx = 1x,

yz = zy = uy for A, u in F, since [N, S] = 0. Let

hz = zh = Vh + u where H' = Fh @iB (vector space direct
sum) and u E B. Then by (1.4), (xy)z + (zy)x = x(yz) + z(yx)
and by (1.15) this implies 2(Vh + u) = 2uh and so v = u,
u = 0. Similarly (yx)z + (zx)y = y(xz) + z(xy) and (1.15)
imply v = 1. Therefore S is an ideal of A and moreover
we have

(1.16) xz = zx = Ax, yz = zy = 1y, hz = zh = Ah

for z in N and non—zero root a, where A in F
depends on 2 and a.

Since [S, N] = O and the center of S is 0, it
follows that N = {x E A ] [x, S] = 0}. Hence by (1.6),
[xy, s] = x[y, s] + [x, s]y = O for all x, y in N and
s in S. Therefore N is a subalgebra of A. If A-
is not nilpotent, then S #'0. Hence if A is simple and
A- is not nilpotent, then A = S and so A is.a Lie

algebra. This completes the proof of Theorem 1.7.

Corollary 1.8: Let A, N, S, and H be the same
as in Theorem 1.7. Then for any non-zero root a, there
exists a subalgebra Ba of A such that:

)

(l) the Lie algebra Ba- = N + S(a is a Levi—
decomposition of Ba- where 8(a) is the

3-dimensional simple Lie algebra, and

18

(a)

(2) the multiplications between N and S are
given by 52 = 28 = fa(z)s for all z in N
and s in 3(a), where fa is a linear function

on N.

a) are subalgebras of A

(a)

Proof: Since N and S(

(a) (a) (a)

and NS = S N c S , Ba = N + S is a subalgebra

of A. Since N is also the radical of Ba- and S(a)

is a simple Lie algebra, it follows that the Lie algebra

(a)

N + S is a Levi-decomposition of Ba-' It follows

from (1.16) that fa is a linear function on N.

Corollary 1.9: Let A and N be the same as in

 

Theorem 1.7. If A is power—associative and N is a nil
subalgebra of A, then the linear function fa is O for

any a # O, and hence N is an ideal of A.
For the proof we first prove the following lemma:

Lemma 1.10: Suppose that A is a finite dimensional,

 

flexible, power-associative algebra over a field of charac-
teristic 0. If x is a nilpotent element of A such

that DX is nilpotent, then both R.X and LX are nilpotent.

Proof: Let A+ be the algebra defined as the same'
vector space as A but with a multiplication given by
x-y = %(xy + yx). Then A+ is a commutative and power—

associative algebra. If x is nilpotent in A, so is

19

x in A+. Hence it follows from Gerstenhaber [7] that

l . . .
TX —‘-2-(RX + Lx) is nilpotent. The fleXible law LXRX—RXLX

implies that DX commutes with Tx' Hence if DX is nil—

potent, so are T +-l D = R and T --l D = L . This
x 2 x x X 2 x x

proves the lemma.

Proof of Corollary 1.9: Let 2 be an element of

 

N and a a non-zero root. Since D2 is nilpotent, so
is Rz by Lemma 1.10. By Corollary 1.8, fa(z) is a
characteristic root of Rz, and fa(z) = 0. Hence N is

an ideal of A.

A finite dimensional Lie algebra L over a field of

characteristic 0 is called reductive if ad L is com-

 

pletely reducible. It is shown in Chevalley [5, p. 255]

that the definition is equivalent to the fact that the quo-
tient algebra L/Z of L by the center Z of L is
semi-simple. Then the radical R of L coincides with

Z. In fact, (R + Z)/Z is a solvable ideal of a semi-
simple algebra L/Z and hence (R + Z)/Z = 0 or R = Z.

Let S be a Levi-factor of L and then [L, L] = [S, S] = S,
since S is semi-simple. Hence if L is reductive, then

L has unique Levi-factor [L, L]. Therefore, as an imme—

diate consequence of Theorem 1.7, we obtain:

Corollary 1.11: Suppose that A— is a reductive
Lie algebra over a field of characteristic 0 such that

[A, A] has a split Cartan subalgebra being a-nil subalgebra

20

of A. Then [A, A] is an ideal of A and so a Lie alge-
bra. Moreover if A is simple then A is either commuta-

tive or a Lie algebra.

Theorem 1.12: Suppose that A is power—associative

 

and A- is reductive over a field of characteristic. 0

such that there exists a split Cartan subalgebra H of

A_ with h3 = 0 for all h in H. Then the center Z

of A_ is an ideal of A. Moreover if A is simple,

then it is a Lie algebra.

Proof: Since Aa for a #’0 is one—dimensional

and. H is abelian, it follows from Lemma 1.2 that h2

belongs to Z for every h in H. Let hl and h2 be
2 2
1 + 2hlh2 + h2

is in Z, and hence hlh2 6 Z, i.e., H2 C Z. Since every

element of Z is nilpotent, it follows from Lemma 1.10

any elements of H. Then (hl + h2)2 = h

that ZAa = AaZ = 0 for a #'0. Since Z G H, this implies
that Z is an ideal of (A. If A is simple, then either

Z = 0 or Z = A. If Z = A, then A is a commutative

algebra with x3 = 0 for all x in A. Since A is

commutative, the linearization of x2x = 0 implies
(xy)z + (yz)x-+ (zx)y = O, or equivalently

R +RR +RR :0.
xy XY yx

2

Then R = - 2Rx and hence R commutes with Rx'

2 2
x x

that is, (xy)x2 = x(yxz) for all x and y in A. A
is thus a Jordan algebra. However, since there is no simple

Jordan nilalgebra of finite dimension, we‘must have Z = 0.

21

Therefore it follows from Theorem 1.5 that A is a.Lie

algebra. This completes the proof.

An element x of A is called an absolute zero

divisor of A if xA = Ax = 0.

Corollary 1.13: Suppose that A is power—associative

 

and the radical R of A- is one-dimensional. If there
exists a split Cartan subalgebra H of A- with h3 = 0
for all h in H, then R is the set of absolute zero

divisors of A.

Proof: We first show that A- is reductive. Let
S be a Levi—factor of A- and R = Fz. Let H' be a
split Cartan subalgebra of S. If [H', R] #'0, it follows
that H' is also a Cartan subalgebra of A—. Let SQ

be the root space of S for H' corresponding to the

root 0 and A- = R + ZLSa. Since [R, H'] #'0, 2 belongs

a .
to a non—zero root O of A- for H'. Since R is an
ideal of A , [2, Sa] 15 contained in R 0 Sa+O for

a #0, and hence [2, 50‘] = O for a 7’0. Let a 5/0

and h = [x, y] #'0 for x in so and y in S-a’ By
(1.1), [z,[x, y]] = [2, h] = O(h)z = [y,[x, z]] + [x,[z, y]]=0
for all a #'0, and hence O = 0. This is a contradiction

and therefore [H', R] = 0. Hence A— is reductive. It
follows from Theorem 1.12 that RH G R and RAa = AaR = 0

for a #'0. Since H is a commutative nilalgebra containing

R and R is one-dimensional, by Lemma 1.10 this implies

RH = HR = 0. This completes the proof.

22

We notice that the algebra A in Theorem 1.12 and
Corollary 1.13 need not be a Lie algebra. The algebra
given in Example 1.2 in fact satisfies the conditions above

but is not a Lie algebra.

We now wish to give examples of algebras satisfying

the conditions in Theorem 1.7, which are not Lie algebras.

Example 1.3: Let F be of characteristic #'2. Let

 

A be the algebra given by
(1.17) $2 = 25 = as for all s in S and 22 = Bz, a,
B 6 F, where S is the 3-dimensional simple Lie algebra
and z is a basic element of A. Let (x, y, h] be a
basis of S such that

xh = x, yh = -y, xy = h.

Let u = Ax + uy + vh + 02 be any element of A. The

multiplication table gives

 

 

,
[v + a0 0 u 0-
O — v + as - A O
Ru:
- A u. an 0
L a1 au av Ba]
and
I
F-v + a0 0 -u 0
0 v+-aO A 0
Lu =
A r-u ad 0
ax au av BO]

 

 

23

relative to the basis (x, y, h, 2]. Direct computations
show RuLu = LuRu and hence A is flexible. AI is also

a Lie algebra and A— = Fz +.S_ where F2 is the center

of A‘. Hence A is flexible Lie-admissible. If a==B==O,
then A is a Lie algebra isomorphic to A-. Conversely

if A is a Lie algebra, then B = 0 from 22 = Bz.

(xy)z + (zx)y + (yz)x = 0 implies a = 0. Therefore A

is a Lie algebra if and only if a = B = 0. Let the algebra
given in (1.17) denote by A(a, B). If B # 0, then Zkkn‘o)
is not isomorphic to A(a, B) for every a, since A(a, B)
has a non-zero idempotent B—lz but A(a, 0) does not.

We note that A(a, 0) is power—associative if and only if
a = 0. If B # 0, then A(aB-l, l) is isomorphic to

A(a, B). Let A(a) = A(a, l) and then 2 is a non-zero
idempotent of A(a). If (s + xz)2 = s + xz for A # 0

in F and s # 0 in S, then a = l and x = l. 7Con—

'5
%, then s + z is a non-zero idempotent
of A(%) for every 5 in S- Hence 2 is only non—zero

versely if a =

idempotent of A(a) if and only if a #.%. Suppose that
A(a) is isomorphic to A(B). If either a =-% or B = %3
then a = B = %- from the above remark. Let. a #~% and

B #-%. Then by the remark z is mapped into itself under
the isomorphism. l and B are only characteristic roots
of R2 in A(B), but a is also a characteristic root of
R2 in A(B) since A(a) a A(B). Similarly B is a char—

acteristic root of Rz in A(a). Hence a = B and so we

24

proved that A(a) is isomorphic to A(B) if and only if

a = B. Therefore we obtain a family of non-Lie algebras

satisfying the conditions in Theorem 1.7. A(a) is not in
3 u2u2

general power-associative. In fact, u u = holds for

every u in A(a) if and only if 2a3 - 3a2 + a = 0, that

is, A(a) is power-associative if and only if a = O, a=r%,
or a = 1
Correction

 

In the proof of Theorem 2 of the paper of Laufer and
Tomber [9], the step (i), which proves the Cartan subalgebra
H of A- is a nil subalgebra of A, is not correct. As—
suming that H is not a nilalgebra, there exists an idem-
potent e # 0 in H. Setting ex - xe = ax, ex = Bx and
ey - ye = -ay, ey = B'y where a # 0 in F and x # 0
and y # 0 belong to the root a(H) and -a(H), respec-
tively, they conclude B + B' = 0 by claiming that xy or
yx belongs to the root B + B' relative to e. But since
Re and Le are not in general derivations of A, B + B'
may not be a root. In fact, (1.5) applied to e implies
R - R: = Le - L: and this applied to x, y in turn implies
B =-%(l + a) and B' = %(1 - a). Thus B + B' = l. HOwever,
if A is not nilalgebra, it follows from Oehmke [10] that

A has an identity element. But since A— is simple, this

is impossible. Therefore A is nil and so is H.

CHAPTER II
SOME SIMPLE NILALGEBRAS

The class of flexible algebras includes all commuta-
tive and anticommutative algebras, in particular, Jordan
and Lie algebras, and alternative algebras, as well as the
noncommutative Jordan algebras introduced by Schafer [11].
All algebras we consider here are finite dimensional over
a field of characteristic #’2. In [13] Schafer defined
the generalized standard algebras which include all alterna-
tive algebras and-the standard algebras of Albert [1].
Any commutative Jordan algebra is included in the class of
generalized standard algebras. It is shown in Schafer [13]
that any generalized standard algebra is a noncommutative
Jordan algebra A, i.e., a flexible algebra with (xzy)x==x20pd
for all x, y in A. Hence the class of noncommutative
Jordan algebras includes most of the well known nonassocia-
tive algebras. Any generalized standard algebra is power-
associative. Any standard algebra is flexible Lie-admissible.
However, there is no implication between the class of gener-
alized standard algebras and the class of flexible Lie-
admissible algebras. In fact, flexible Lie-admissibility
does not imply power-associativity as shown by the algebras
given in Example 1.3. Any nonassociative alternative algebra

25

26

of characteristic ¥'2, 3 is not flexible Lie-admissible,
since it is shown in Albert [1] that an alternative algebra
of characteristic #'2, 3 is Lie-admissible if and only
if it is associative.

It is also shown in [13] that any generalized stand-
ard nilalgebra is nilpotent, where an algebra A is called

nilpotent in case there exists an integer t such that

 

every product of t elements of A, no matter how asso-
ciated, is 0. Hence there are no simple generalized
standard nilalgebras. Simple Lie algebras give examples

that the result is not valid for noncommutative Jordan .
algebras and flexible Lie-admissible algebras. The nil—
potency of flexible nilalgebras is not known even in case

of commutative, power-associative nilalgebras. The exis—
tence of simple, commutative nilalgebras has not been settled,
but the relatively simple case of dimension 5 3 was settled

by Gerstenhaber [7] in the following:

Theorem: Let A be a commutative power-associative
nilalgebra over a field of characteristic 0. If A is

of dimension S 3, then it is nilpotent.

Hence there is no simple, commutative, power—associative
nilalgebra of dimension 5 3.

Let A now be a noncommutative, flexible, simple nil—
algebra. All simple, anticommutative algebras are such

algebras. R. Block [2] has recently proved the theorem:

27

Theorem: Let A be a finite dimensional flexible
algebra over a field of characteristic # 2. If A is
Simple and not anticommutative, then either A+ is simple

or A has an identity element 1.

Here the algebra A+ is defined as the same vector
space as A but with a multiplication given by
x-y = %(xy + yx), and hence A+ is a commutative algebra.
If A is a power—associative nilalgebra, then so is Af.
Therefore, if A is a simple, flexible nilalgebra and not
anticommutative, then A+ is a commutative, simple nil-
algebra by this theorem. Thus one might conjecture that
a noncommutative, flexible, simple nilalgebra is anti-
commutative. At the present time, no proof is known.

A Jordan algebra is a commutative algebra A satis-
fying the Jordan identity (xy)x2 = x(yxz) for all x, y

in A. An algebra A is said to be Jordan-admissible in

 

case A+ is a Jordan algebra. It is shown in [1] that
any noncommutative Jordan algebra is Jordan-admissible.
The above remarks lead one to seek a sufficient
condition that a noncommutative, flexible, simple nilalgebra
A be anticommutative. If, in particular, A is a simple
noncommutative Jordan nilalgebra, then it follows from the
above theorem that it is anticommutative, since A+ is
a Jordan nilalgebra and there is no simple Jordan nilalgebra.
Laufer and Tomber have concluded that if A is power-associative

and A- is a simple Lie algebra over an algebraically closed

28

field of Characteristic 0, then A is anticommutative.
In this case the simplicity of A- implies that A is a
simple nilalgebra. Since any commutative algebra is triv-
ially Lie—admissible, this result suggests the conjecture
that any simple, flexible, Lie-admissible nilalgebra is
either commutative or anticommutative and in the latter case
A is a Lie algebra. In Chapter I we have already discussed
such simple algebras A under certain restrictions on the
Lie algebras A—. In most of the cases we assumed only that
a Cartan subalgebra H of A- is a nil subalgebra of A.
The Cartan decomposition Of A- by H then worked as an
analog to the idempotent decomposition in flexible, power-
associative, non-nil algebras.

The principal purpose of this chapter is to answer
in part the above questions. We will also deal with certain
nilalgebras whose minus-algebras are nilpotent Lie algebras.
The proofs are mainly based on the two theorems above, the
structures of A-, and the results in Chapter I.

We will assume throughout this chapter that A is a
finite dimensional algebra over a field F and all nilalge-

bras are power—associative.

Theorem 2.1: Let A be a simple, flexible algebra over

a field of characteristic # 2 such that x3 = 0 for all x

 

in A. Then A is anticommutative. Furthermore, if A is

Lie-admissible, then it is a Lie algebra.

29

Proof: We first show that A is Jordan—admissible.

The linearization of xzx = 0 implies (x-y)-z + (z-x)-y

+ (y-z)~x = 0, or equ1valently Tx-y + TXTy + TyTx = 0
1 __1_
where xoy — 2(xy + yx) and TX — 2(Rx + Lx)' Then
T 2 = - 2TX2 and hence T 2 commutes with TX, that is,
x x

(x-y)-x2 = x-(y.x2) for all x, y in A. A+ is thus a

Jordan algebra. Suppose that A is not anticommutative.
Then since A is simple, it follows, from the result of
Block quoted above, that A+ is simple. Since A+ is a
nilalgebra and there is no simple Jordan nilalgebra, this
is a contradiction. This completes the proof.

A simple, flexible nilalgebra may be both Jordan-
admissible and Lie-admissible as shown by simple Lie algebras.
Since it follows from the theorem of Block that any simple,
flexible, Jordan—admissible nilalgebra is anticommutative,

we obtain:

Theorem 2.2: A simple, flexible nilalgebra of charac—

 

teristic # 2 is both JOrdan-admissible and Lie-admissible

if and only if it is a Lie algebra.

In general, the fact that un = 0, n > 3, for all u

in A does not imply that A is Jordan—admissible, as

shown by the followings:

Example 2.1: Let A be a 4-dimensional algebra of

characteristic # 2 given by

3O

xz = zx = zh = hz = 22 = hx = y,

and all other products are 0, where (x, y, 2, h} is a

basis of A. Then A is a flexible nilalgebra such that

u4 = 0 for all u in A, but x3 = y # 0. A is also
Lie-admissible, since [A, [A, A]] = 0. Let TX=r%(R -rL L

Relative to the basis (x, y, 2, h},

 

 

 

 

[O O 1 O1 [O 1 O 0
O O O O O O O O
TX: ’ T2:
O 1 O O x O 1 O O
1
LO 3 O 0] L0 1 O 0]

Then TXTX2 # TXZTX’ Since the (l, 2)-entry of TXTX2 is

l but that of T 2TX is 0. Hence A is not a Jordan-
x

admissible algebra.

Theorem 2.3: Let A be an n—dimensional, flexible,

 

Lie-admissible nilalgebra over a field F of characteristic
0 such that A- is nilpotent. If A- has an abelian
ideal of dimension n - 1, then the center Z of A— is
an ideal of A.

For the proof we first prove:

Lemma 2.4: Let A be an n-dimensional, flexible,

 

Lie-admissible nilalgebra over the field F. If I\ is not
commutative and B is an abelian ideal of A- of dimension

n - 1, then B is an ideal of A.

31

Proof: Writing A_ e B @ Fh as a vector space,
[A, A] = [h, B] # 0 since A is not commutative. Let g
be a non-zero element of [h, B] and then [h, z] = g for
some 2 in B. Let x be any element of B and
xh = u + ah for u in B and a in F. Since B is
abelian, [xh, z] = x[h, z] + [x, z]h implies xg = gx = ag.
Since DX2 = 0 and g # 0, a = 0 by Lemma 1.10, and

hence xh is in B. Likewise hx is in B and this

proves the lemma.

Proof of Theorem 2.3: If A is commutative, then
it is trivial. Let A be noncommutative. Since A- is
nilpotent, n 2 3. Let B be an abelian ideal of A—
of dimension n - 1. Writing A— = B 0 Fh, [B, h] # 0
and Dh induces a nilpotent linear transformation on B,
since A— is a nilpotent Lie algebra. Then B can be

expressed as a direct sum of subspaces B = M1 6 M2 0 ...(3N&

such that each“ M1 is a cyclic subspace of B relative to

2 0.. I . l . I
Dh, n1 n2 2 2 nr where n1 is the dimension of M1,

and n1 is the nil index of Dh in B. Let

x x ] be a basis of Mi *with [xi

[Xi,1' i,2' ... , h]

= x. and x.
i,k [_i,ni

0 and [B, h] # 0, it follows that the center Z of A—

i'ni 'k-l'

,'h] = O, k = 2, ... , ni. Since [B, B]

is the subspace generated by {Xl,nl' X2,n2' ... , xr,nr}°
By Lemma 2.4, hxi,ni = Xi,nih belongs to B. Then
[hxi,n.' h] = hEXi,n , h] + [h, h1xi,n. = 0 implies

l i i

(2.1) hxi.ni = Xi,nih E Z, i = 1, 2, ... , r.
Since [B, h] # 0, n1 2 2, and let p be such that
n1 2 n2 2 ... 2 np 2 2 and np+l = ... = nr = 1. Let x

be any element of B. Since xh and hx belong to B

by Lemma 2.4, if i s p then 0 = [xi n —1’ xh] =
' i

x[xi,n.-l’ h] = XXi,n. and Similarly xi,n x = 0. Hence

1 i 1
(2.2) Bxi,ni = Xi,niB = 0, 1.: l, ... , p.
If 3 > p then by (2.1), 0 = [Xi,k' xj,n.hJ = xj'nj[xi'k,ln]
= xj,njxi,k+1 where l = l, 2, ... , p, 1 s k s ni - 1.
Hence

x. x. = x. x. = 0,

(2.3) 3,1 i,k lql< 3,1

1 s i s p, j > p, 2 s k s ni.

Let i s p and then [Xj,1xi,l' h] = xj,1[xi,l' h] —
Xj,lxi,2 = 0 by (2.3) if j > p. The equations
[xj,lxi,1' h] = 0 imply that Xj,lxi,1 = xi,lxj,l belong
to Z for j > p and 1 s i s p. Therefore, since Z is
a subalgebra of A, by (2.1), (2.2), and (2.3) this implies
that Z is an ideal of A. This proves Theorem 2.3.

Theorem 2.3 and Lemma 2.4 are not valid for non-nil

algebras as shown by the following:

Example 2.2: Let A be a 4-dimensional algebra with

 

multiplication given by
x2 = 2x, xy = y + z,
yx = y, xz = zx = y = z,

uh = hu = u for all u in A,

and all other products are 0, where (X, y, z, h} is a

33

basis of A. Then A is flexible and A- is a nilpotent
Lie algebra since [A, [A, A]] = 0. Fy + F2 + Fh is an
abelian ideal of A_ and F2 + Fh is the center of A-,
but none of them is an ideal of A. In this case A is not

associative since x2y # x(xy).

If A is a nilalgebra of dimension n over a field
F then xn + l = 0 for all x in A. If K is an exten-
sion field of F, let AK be the scalar extension of A
to an algebra over K. If F has an infinitely many ele—
ments, it follows from Braun and Koecher [4, p. 23] that
AK is also a nilalgebra over K. If A is a flexible Lie—
admissible algebra over F of characteristic # 2, 3, then
so is AK for any extension K of F since (1.6) is a
bilinear identity. An algebra A over F is called central

simple in case AK is simple over K for every extension

K of F.

Lemma 2.5: Let A be a 4—dimensional, flexible nil-

 

algebra over an algebraically closed field F of Character-
istic 0 such that A- is a solvable Lie algebra. If A

is not commutative, then it is not simple.

Proof: If A- is nilpotent, it follows from Bourbaki
[3, p. 120] that A- has an abelian ideal of dimension 3.
Hence by Theorem 2.3, A is not simple. Now suppose that
A- is not nilpotent and let H be a Cartan subalgebra of
A-. Let Aa denote the root space of A_ for H corres-

ponding to root a # 0. Then the dimension of 11 is 1,2L or 3.

34

If the dimension Of H is 3, then it follows from Lemma 1.1
that Aa for a # O is an ideal Of A. The proof continues

by examining the two other possibilities for H.

Case I: dim H = l. The possible Cartan decompositions
of A_ for H are A’ = H + Ad, A’ = H + Ag + AB, and

A— = H + Aa + A + Ay' In the first case Aa is an ideal

B
Of A. In the last case it follows from Theorem 1.5 that

A is a solvable Lie algebra and not simple. In the remaining

case we want to show that Aa + A is an ideal of A. If

B

a + B # 0, then by Lemma 1.1, Aa + AB is an ideal of A.

Now suppose a + B = 0 and let H = Fh. We may then assume

Aa = Fx + Fy and AB = Fz. Setting zh = yz, then
hz = (y + a)z since zh — hz = - az. Since h2 = 0, (1.4)
applied to z, h, h implies y2 = (y + a)2 and a = - 2y.

It follows that

(2.4) zh = - hz = — %az.

Since xz and zx belong to H and h2 = 0, (1.4) applied
to x, 2, h and (2.4) imply xz + zx = 0, and similarly

yz + zy = 0. If [x, z] ¢ 0, the derived algebra [A, A]

Of A- contains h and z, and hence [A, A] is not a
nilpotent Lie algebra. Since the derived algebra of a solva-
ble Lie algebra is nilpotent, this is a contradiction.

Therefore we obtain [x, z] = 0 and similarly [y, z] = 0.

Hence A A = A A = 0 and A + A is an ideal of A.
a B B a a B
Case II: dim H = 2. H is a commutative nil subalgebra

of A and u3 = 0 for all u in H. The possible Cartan

35

decompositions of A- for H are A = H + Aa and

A_ = H + Aa + A In the first case Aa is an ideal of

5°

A. In the second case both Aa and A are one—dimensional.

B
Let Z in this case be the center of A-. If Z = 0, then
it follows from Theorem 1.5 that A is a Lie algebra and
not simple. If Z # 0, then it follows from Lemmas 1.2

and 1.10 that HH : Z and ZAa = ZAB = 0. Since Z is

contained in H, this implies that Z is an ideal of A.

Therefore A is not simple and this proves the lemma.

Theorem 2.6: Let A be a central simple, flexible,

 

Lie—admissible nilalgebra over a field F of Characteristic
0. If the dimension of A is not greater than 4, then A
is either commutative or a Lie algebra, and in the latter

case A is the 3-dimensional simple Lie algebra.

Proof: Let K be the algebraic closure of F. Then
by the remarks above AK is a simple algebra satisfying the
conditions in the theorem. Hence without loss of generality
we may assume that the field F is algebraically closed.

If the dimension Of A is not greater than 3, then by the
result of Gerstenhaber the algebra A+ is nilpotent. Hence
it follows from the theorem of Block that A is anticommu-
tative. Therefore A is in this case the 3-dimensional
simple Lie algebra. Now suppose that A is a noncommutative
algebra of dimension 4. If A- is a solvable Lie algebra,

from Lemma 2.5 it follows that A is not simple. If A-

is not solvable, then A- is a reductive Lie algebra with

36

one-dimensional center Z. In this case A— has a Cartan
subalgebra Of dimension 2. Hence it follows from Theorem 1.12
that Z is an ideal of A. Consequently A must be
commutative if the dimension of A is 4. This completes

the proof.

Remark on Theorem 2.6: (l) The condition that A

 

is central may be replaced by the fact that if A— is a
non-nilpotent Lie algebra of dimension 4, then it has a
split Cartan subalgebra. (2) Although there is no simple,
flexible nilalgebra of dimension 2, we may prove that any
noncommutative, flexible, Lie-admissible nilalgebra of
dimension 2 is a Lie algebra. Indeed, in this case A has

a basis [x, y} with xy — yx = x. The equation [x, x2]

= [y, y2] = 0 imply x2 = y2 = 0. From [x, xy] = x[x, y]
= x2 = 0, we obtain xy = Ax for some 1 in F. Since
y2 = 0, (1.4) applied to x, y, y implies (xy)y = y(yx)
and this gives 12 = (A - l)2 or A =-l. Hence A is a

-2

Lie algebra.

CHAPTER III
NILALGEBRAS OF DIMENSION LOWER THAN 5

Let L be a 4-dimensional Lie algebra over an alge-
braically closed field of characteristic 0. Then L is
completely known unless it is non-nilpotent solvable. In
this Chapter, by using the known structures of Lie algebras
of dimension 4, we determine all 4-dimensional, flexible,
Lie-admissible nilalgebras whose minus—algebras are not non;
nilpotent, solvable Lie algebras. We also prove that such
algebras whose minus-algebras are non-abelian nilpotent are
nilpotent.

We assume throughout this chapter that A denotes a
finite dimensional, flexible, Lie-admissible algebra over
a field F of characteristic 0 and that all nilalgebras

are power-associative.

Theorem 3.1: Let F be algebraically closed. Then
A is a 4—dimensional nilalgebra over F such that A‘ is
not solvable if and only if it is either a non—solvable Lie

algebra or given by

XY = z + 1h. yx = Z - 1h. Xh = - hX = AX:
2 2 2
Yh = "' by = " '12; : h2 = " Z,

and all.other products are 0, where [x, y, h, z] is a basis

of A. In this case A is a nilalgebra with u3 =0 for all u inA.
37

38

Proof: Suppose that A is a 4-dimensional nilalgebra
such that A- is not solvable. Then A- = Fz + S is a
Levi-decomposition Of A- with the 3-dimensional simple Lie
algebra S and the center Fz of A-. Let (x, y, h] be
a basis of S with [x, h] = x, [y, h] = - y, [x, y] = h.
Then H = Fz + Fh is a Cartan subalgebra of A- and thus
a subalgebra of A. Hence v3 = 0 for all v in H and
it follows from Corollary 1.13 that z is an absolute zen)

2

divisor of A. By Lemma 1.2, h = az for a in F.

Hence O = [x, h2] = h[x, h] + [x, th = hx + xh and this

with [x, h] = x implies xh = - hx = %X, and similarly
hy = - yh = %y. Since x and y belong to roots 1 and

— l of A_ for H, respectively, we obtain xy = Bz + yh
and yx = Bz + (y - 1)h. Then (xy)h + (hy)x = x(yh) + h(yx)
implies B = - a and y =-%. If a = 0, then A is a Lie
algebra isomorphic to A-. If a # 0, then we replace - az
by z to obtain the algebra given in the theorem. Conver-
sely suppose that A is given by the multiplication table
stated in the theorem. Let u = 1x + By + uh + Oz be any

element of A. The matrix representations of Ru and Lu

relative to the basis [x, y, h, z} are

l. 1 j
[- 2v 0 3L1 11
l l
-—v -—A A
R = 2 2
u 11X 1 O _
2 2)): V
L O 0.1

 

 

39

and
P .1 ,1 W
2V 2“ L1
0 'lv 'lx A
_ 2 2
L _
u -l->. -1 _
2 2*1 V
_ O O O O _]

 

 

Then both R L and L R are equal to
u u u u

[ Ml—VZ “L12 wv O ‘1
1 -12 Au-vz -1v 0
Z 1v Liv 2Au 0

L O O O O]

 

 

and thus A is flexible. We obtain u2 = (ZXH - v2)z and
hence U3 = 0. From the multiplication table, it follows
that A_ is isomorphic to the Lie algebra Fz + S and
thus A is Lie-admissible. This completes the proof.

The algebra in Example 1.2 is obtained from the alge-

bra above.

Theorem 3.2: A is a 4—dimensional nilalgebra suCh

 

that A- is a nilpotent Lie algebra with one-dimensional

center if and only if it is given by

x2 = az, xh = -%y + Bz,
hx = ly + Bz, yh = - hy = -lz,
2 2
h2 = yz, a, B, y in F,

and all other products are 0, where [x, y, z, h] is a basis

4

of A. In this case A is a nilalgebra with u 0 for

all u in A and a Lie algebra if and only if a = B = y = O.

40

Proof: Suppose that A is a nilalgebra such that
A- is a nilpotent Lie algebra with one-dimensional center.
Then it follows from Bourbaki [3, p. 120] that AI has a
basis (x, y, 2, h] such that [h, x] = y, [h, y] = z and
all other products under [ , ] are 0. Let B = Fx + Fy + F2
and Z = Fz. Then B is an abelian ideal of A_ and Z
is the center of A-. Hence it follows from Theorem 2.3
and Lemma 2.4 that B and Z are ideals of A. By Lemma 1.10,
z is an absolute zero divisor of A. Since xh is in .B,
0 = [xh, x] = x[h, x] = xy and thus xy = yx = 0. [h, yh] =
[h, y]h = zh = 0 implies yh = 12 for x in F. From
[xh, h] = [x, h]h = — yh, we obtain xh = xy + Hz for u

in F. We use [h, x2] = xy + yx = 0 to obtain x2 = az

for a in F. The equations [h, h2] = 0 and [h2, y] =
hz + zh = 0 give h2 = az. Therefore, since by - yh = z
and yh = xz, 0 = [h2, x] = hy + yh = (21 + 1) z and thus
x = --%. Hence A has the multiplication table given in

the theorem. Conversely, suppose that A is the algebra
given in the theorem. If Tu = xx + uy + vz + Oh is any

element of A, then

 

 

[.0 -%o (ix‘+ BO 0‘1
1
O O -20 O
R =
u 0 O O O
1 1
BO ‘2-)\ BA+YG+§U OJ

and

 

 

[O é-o ax + Bv O1
O O '%C O
L =
‘1 O O O O
1 1
L0 -'2—)\ El + YO-é—U OJ
relative to [x, y, 2, h]. A direct calculation shows

RuLu = LuRu and thus A is flexible. It follows from the
multiplication table that A— is a nilpotent Lie algebra

with one-dimensional center Fz. Moreover we obtain

R 3 = L 3
u u

0 and A2 ; Fy + Fz and so AZA2 = 0. There-

fore u3u = u2u2 = 0 for all u in A and thus completes

the proof.

Theorem 3.3: A is a 4-dimensional nilalgebra such

 

that A— is a nilpotent Lie algebra with two-dimensional

center if and only if it is given by

x2 = ay + Bz, xz = zx = yy,
xh=5y+lz. hx= (6+1)y+lz.
z2 = uy, zh = hz = vy, h2 = oy + T2
. . 2 2 2
for a, B, y, 6, A: Lit v, 0, T In F With L18 = LIA = “T = O,

and all other products are 0, where [x, y, z, h} is a basis

of A. In this case A is a nilalgebra with u4 = 0 for
all u in A and a Lie algebra if and only if 5 = -v%

and all other parameters are 0.

Proof: Suppose that A is a 4—dimensional nilalgebra
such that A- is nilpotent and has two-dimensional center.

Then it also follows from Bourbaki [3] that A- has a basis

42

[x, y, z, h] such that [h, x] = y and all products under
[ , ] are 0. Let B = Fx + Fy + F2 and Z = Fy + Fz.
Then B is an abelian ideal of A- and Z is the center
of A_. Thus B and Z are ideals of A. Hence

0 = [h, xy] = [h, x]y = y2 and similarly yz = zy = 0.
Since xy and x2 are in Z and B, respectively,

[h, x2] = 2xy implies that xy and x2 are in Fy and
Fy + Fz, respectively. Thus xy = yx = 0 by Lemma 1.10

and x2 = ay + Bz for a, B in F. It follows, from

[h2, x] h[h, x] + [h, x]h = 2yh, that yh = hy belongs

to Fy. This implies that yh = hy = 0 and h2==Oy-+Tz

for O, T in F. Hence [h, xh] [h, x]h = yh = 0 and

this gives xh = by + 12 and hx = (6 + 1)y + AZ for 5, A

I

in F. Setting zx = xz = yy + y'z for y, y in F, then
(xz)x = y’xz and thus y’ = 0 by Lemma 1.10. Similarly

hz = zh = vy. Since Z is a nilalgebra of dimension 2,

3

z 0 and this implies z2 = uy. Since x belongs to B,

xzx2 = (ay + B2)2 = B222 = Bzuz = 0. Since h belongs to
the subalgebra Fy + Fz + Fh, hzh2 = 0 implies pt 2 = 0.
Similarly, we obtain “X2 = 0 from (xh)2(xh)2 = 0. There-

fore A is the algebra given in the theorem. A similar

computation as in Theorem 3.2 shows that A is flexible

and Ru3 = Lu3 = 0 for all u in A. From the multipli-
cation table we obtain that A2 : Fy + Fz and A2A2 :F- uy.
If u # 0, then 'B = A = T = 0 and thus A2A2 = Fy2 = 0.
Hence u3u = u2u2 = 0 for all u in A. If 6 = --%

and all other parameters are 0, then A is a Lie algebra

 

43

isomorphic to A—. If A is Lie, then 6 = --% from
hx + xh = 0 and all other parameters are 0 and thus

completes the proof.

Theorems 3.1, 3.2, and 3.3 determine all 4-dimensional
nilalgebras whose minus-algebras are not non-nilpotent
solvable. None of these algebras is associative, since
(hh)x # h(hx) in Theorems 3.1 and 3.2 and in Theorem 3.3

with u = v = 0 and y = T = 1.

Theorem 3.4: If A is a noncommutative nilalgebra

 

of dimension S 4 such that A- is nilpotent, then A is
nilpotent such that all products of any four elements in

A are 0.

Proof: If A is a nilpotent Lie algebra of dimen-
sion S 2, A- is abelian. Hence the dimension of A is
in this case 3 or 4. If the dimension of A is 4, then

A is either the algebra in Theorem 3.2 or in Theorem 3.3.

2 2

In Theorem 3.2 we see that A c Fy + Fz and that AA and

2 . . . .
A A are contained in Fz. Since 2 is an absolute zero

2A) = (AZA)A = A(AA2)=

A2A2 = 0. In Theorem 3.3 we also notice that A2 : Fy + Fz

divisor of A, it follows that A(A

and that AA2 and A2A are contained in Fy. Since y

is in this case an absolute zero divisor of A, we Obtain
the same result. Now suppose that the dimension of A is
3. Then A- has a basis [x, y, z) such that [x, y] = z,

[x, z] = [y, z] = 0, and F2 is the center of A-. By

44

Theorem 2.3, F2 is an ideal of A and hence z is an
absolute zero divisor of A. Hence [xy, x] = x[y, x] =

— xz = 0 and [xy, y] = [x, y]y = zy = 0, and these imply
that xy and yx belong to Fz. From [x2, x] = [x2, y] =
2x2 = 0, it follows that x2 belongs to Fz, and similarly
y2 is in Fz. Therefore A2 9: F2 and thus A2A=AA2 = 0.

This completes the proof.

Theorem 3.4 is not valid in case A- is not nilpotent,
as shown by the algebra in Theorem 3.1 and non-nilpotent,
solvable Lie algebras.

At the present time, Theorem 3.4 for arbitrary dimen-

sion is not settled.

[l]

[2]

[3]
[4]

[5]

[6]

[7]

[8]
[9]

[10]

[ll]

[12]

[13]

[14]

BIBLIOGRAPHY

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

A. A. Albert, Power-associative rings, Trans. Amer. Math.
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R. E. Block, Determination pg 51 for the simple flexible
algebras, Proc. Nat. Acad. Sci. U.S.A. 61(1968),394-397.

N. Bourbaki, Algébres g3 Lie, Hermann, Paris, 1960.

H. Braun and M. Koecher, Jordan-algebren, Springer-

Verlag, Berlin, 1966.

C. Chevalley, Théorie des groupes.dg Lie, Hermann,
Paris, 1968.

J. Dixmier, Sous-algebres g; Cartan pp decompositions
gg_Levi dans les algébres dp Lie, Trans. Roy. Soc.
Canada, Sec. III, (3) 50(1956), 17-21.

M. Gerstenhaber, Qp_nilalgebras and linear varieties g;
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N. Jacobson, Lie Algebras, Interscience, New York, 1962.

P. J. Laufer and M. L. Tomber, Some Lie admissible
algebras, Canad. J. Math. 14(1962), 287-292.

R. H. Oehmke, 9p flexible algebras, Ann. of Math. 68(1958),
221-230. .

R. D. Schafer, Noncommutative JOrdan algebras pg charac-
teristic 9, Proc. Amer. Math. Soc. 6(1955), 472-475.

, Ag introduction pp nonassociative algebras.
Academic Press, New York, 1966.
, Generalized standard algebras, J. of
Algebra, 12(1969), 386-417.
G. B. Seligman, MOdular Lie algebras, Springer-Verlag,

 

 

Berlin, 1967.

45

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