C OMPOSITIO M ATHEMATICA
I AN S TEWART
The Lie algebra of endomorphisms of an infinite- dimensional vector space
Compositio Mathematica, tome 25, n
o1 (1972), p. 79-86
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79
1. Introduction
The structure of the Lie
algebra
of allendomorphisms
of a finite-dimensional vector space is well known. The purpose of this paper is to
investigate
the infinite-dimensional case, and inparticular
to find thelattice of Lie ideals.
Rosenberg [6]
has carried out theanalogous
pro- gramme for the infinitegeneral linear
group.Notation for Lie
algebras
will follow that of[9, 10].
Let f be any field.Let c be any infinite
cardinal,
with successorc+ .
Let V be a vector space over f of dimension c, and for any infinitecardinal d ~ c+
defineE(c, d)
to be the set of all linear transformations a : V - V such that the
image
of a has dimension d. Then
E(c, d)
is an associativef-algebra.
Undercommutation
[03B1, 03B2]
=03B1/03B2 - 03B203B1 (03B1, 03B2 ~ E(c, d))
it becomes a Liealgebra
which we shall denote L
(c, d).
Inside
L (c, c + )
we let F =L (c, 0),
T = the set ofendomorphisms
oftrace zero
(in
the sense of[9]
p.306),
S = the set of scalarmultiplications
v ~ vk
(v
EV, k
Ef).
We shall prove:THEOREM
(A).
Let L =L (c, c+).
Then the idealsof L are precisely the following:
The lattice
of
ideals has theform
as shown on the next page.Further,
every subidealof L
is anideal,
so that L lies in theclass Z of [9].
An
immediate corollary
of theorem A is thatL(c, c+)
satisfies theminimal condition for
subideals,
Min-si. We shall use this to show that theorem 3.3 of[9]
p. 305 is in a sense bestpossible.
Finally
weapply
our results to prove that any Liealgebra
can beembedded in a
simple
Liealgebra.
80
1 am
grateful
to the referee for manyhelpful
remarks which havesimplified
andimproved
theexposition.
2. The
endomorphism algebra
We attack the
problem through
the associative ideal structure ofE(c, d),
which iseasily
determined.By
Jacobson[5] p.
108 an associativealgebra
A issimple
if andonly
if it issimple
considered as aring.
Thisremark combines with a theorem of Herstein
[3] (see
also Baxter[1])
to
yield:
LEMMA
(1). If
A is asimple
associativef-algebra
and[A, A]
= A thenany proper Lie ideal
of
the Liealgebra
associated with A is contained in the centreof A,
unless A isof
dimension 4 over its centre which isa field of
characteristic 2.
In the
sequel
allalgebras
considered will be infinite-dimensional overtheir centres, so the
exceptional
case never arises.By
aslight
extension ofJacobson
[5] p.
93 theorem 1 we have:LEMMA
(2).
Let c, d beinfinite
cardinals with d ~c+.
Then any non-zeroassociative ideal
of E(c, d)
isof
theform E(c, e)
where0
~ e ~ d.COROLLARY.
If
c ~ d areinfinite
cardinals thenis a
simple
non-commutative associativealgebra.
LEMMA
(3).
Let E =E(c, d)
whereNo
d ~c+.
ThenMore
precisely
we show that there is aunique endomorphism
b of Vsatisfying (1)
such thatand
(hence
b EE).
We
set ai
=alvi and bi
=b|vi.
In view of(3)
the restrictions of(1)
to
X,
toVi-1 (i
>0)
andto Vi (i 0)
arerespectively equivalent
tothe
following équations :
and now the assertion is obvious since
(5)
and(6)
constitute inductive definitions for theb i .
Note that if d =
No
the lemma isfalse,
for then[E, E]
is the set oftrace zero maps which is smaller than E.
For any associative
algebra
A we letZ(A)
denote the centre of A.We then have:
LEMMA
(4). If c ~
d areinfinite cardinals,
thenis trivial
except
when c = d. It then has dimension 1 and consistsof scalar multiplications (modulo E(c, d)).
This follows from :
LEMMA
(5). If
c ~ d areinfinite
cardinals and z EL(c, c+) satisfies
then z E
L(c, d)
+ S.82
The
proof of this
lemma is more intricate than onemight wish,
and willbe
postponed
until later.Putting together
the results so far obtained we have:LEMMA
(6). If c ~
d areinfinite
cardinals then the Liealgebra
is
simple
unless c =d;
when itsonly
nontrivial proper ideal is the centre, which has dimension 1 and consistsof
scalarmultiplications (modulo L(c, d)).
The next result is
implicit
in[9] (p. 310) :
LEMMA
(7).
Let L be a Liealgebra, 03C3
anordinal,
and(G03B1)03B1 ~ 03C3
anascending
series
of
ideals such thatfor
all oc 03C31) G« + l/Gex
issimple non-abelian, 2) CL/G03B1(G03B1+1/G03B1) = G03B1/G03B1.
Then the
only
subidealsof
L are theGex. Consequently
L E Min-si nZ.
PROOF. Let M be a proper subideal of L and let a be the least ordinal such that
G03B1
M. It is easy to see that oc cannot be a limitordinal,
soa =
03B2+1
for some/3.
Thus(M+G03B2)/G03B2
is a subideal ofL/Gp
not con-taining G03B2+1/G03B2.
As the latter is asimple
non-abelian ideal ofL/Gp
wehave
so
by [9] lemma
4.6 p. 309 M centralisesG03B2+1/G03B2, By part (2)
of thehypothesis
M ~G.,
whence M =Gp.
Obviously
L EX,
and L e Min-si since the ordinals are well-ordered.Now we shall show that
L(c, d)
e Min-si nZ.
The presence of tracezero and scalar maps causes
complications,
so westudy
a suitablequotient algebra.
Let L =L(c, d),
letF, S,
T be as in theoremA,
andput
I = F+ S. Then L* =L/I
has anascending
series of idealsfor a suitable ordinal
b;
theLâ being
the ideals(L(c, e) + S)/I arranged
inascending
order.Now I has a series
O ~ T ~ F ~ I of
ideals. But T issimple ([9]
lemma 4.1 p.
306)
andF/T
andIIF
are 1-dimensional. Therefore I e(Min-si)(F) (F) ~ Min-si, by [9]
lemma 2.2 p. 303.By
the samelemma,
in order to prove that L
e Min-si,
it suffices to show that L* e Min-si.This will follow from lemma 7
provided
we can prove thatwhich is
equivalent
to the statement of lemma 5.the set
has
cardinality |B| =
e.PROOF. Let W =
¿ÀeB v03BB. By
definition dim(W) = IBI,
and sinceim(a) ~ W
wehave e ~ |B|.
Ifm
is a basis forim (a),
then eachi03BC
is a linear combination of
finitely
many v03BB(03BB
EB).
ThereforeIBI :9
|Z M| = 0 · e = e
since e is infinite.We now suppose that z is as
above,
and that is a vector space with basis(v03BB)03BB~039B where |039B|
= c.LEMMA
(9).
There exists z’ such thatz’03B103B1
= 0(a
EA), [z’, G] ~ E+ S,
and z-z’ E E+S’.
PROOF. Let M be the set of all
pairs (M, )
where M is a subset of A and is awell-ordering
onM,
such that if a E M thenz. :0
z03B1+1,03B1+1(where
a + 1 is the successor to a in theordering ).
Then vit ispartially
ordered
by
y, where(Ml, 1) (M2, 2)
if andonly
ifMl
is aninitial
segment of M2. Clearly
-4Y is notempty
and satisfies thehypotheses
of Zorn’s lemma. Let
(M, )
be a maximal element of vit.Suppose
for acontradiction that
|M| ~
d. Take an initialsegment
I of M withIII
=d,
and considerwhere
e03B103B2 (a, j8
E039B)
is theelementary
transformationsending v03B1 to v03B2
and all other basis elements to zero.
By hypothesis t E E+ S, yet
(where
termsinvolving
03B1 - 1 for limit ordinals a are deemed to bezero).
Now the coefficient of e03B1, 03B1+1 is z03B103B1-z03B1+1,03B1+1 which is non-zero for d values of a.
By
lemma 8te
E+ S’ which is a contradiction.Thus after
choosing
fewer than d values of a all theremaining
Zaa areequal. Thus 03A3z03B103B1e03B103B1 ~ E+S.
Define z’ =z - E
zaaeaa.84
LEMMA
(10). Suppose
thatz’ e
E+ S. Then there exist subsetsA,
A’of
A and a
bijection 0 :
A - A’ such thatPROOF. Let P be the collection of all
triples (A, A’, 4» satisfying (1)
and
(2). Partially
order Pby
« where(A, A’, 4»« (B, B’, tp)
if andonly if A ~ B, A’ ~ B’,
and03A8|A
=~. By
Zorn’s lemma there is a maxi- mal element(A, A’, 4»
of P. Forbrevity
let~(03B1)
= a’(a
EA).
We claimthat lAI
= d.Suppose
not.Then lAI
= d’ d. LetSince d is infinite we
have |D|
d.By
lemma 8 there must existy’ e (A ~
A’ ~D)
such thatz’,,
=1= 0 for some03B3 ~ y’ (since z’ ~
E+S).
Theny o (A u A’)
sincey’ e
D. Therefore03B3 ~ y’, y e (A u A’ ), y’ ~ (A ~ A’).
Define
Then
(B, B’, 03A8) ~ P
and isgreater
than(A, A’, 4»,
a contradiction.Hence
|A| ~
d as claimed.We may now derive the final contradiction
required
to prove lemma 5.Suppose
for a contradiction thatz’ e
E+ S. Then there exists(A, A’, 4»
as in lemma 10. Define n : V ~ V
by
By
definition 7r e G.By hypothesis
u =[z’, 03C0] ~
E+ S. But for a E A wehave
The coefficients
of v03B1’
isso that
u03B103B1’ ~
0 if a E A. Since1 AI = d
and 03B1 ~ a’ we haveu o E+S,
acontradiction.
Suppose L(c, c+)
propersion. Now L has an
ascending series,
the finite-dimensional factors of which areabelian,
the restsimple.
HenceL/J
issoluble,
so that[L, L]
L, contrary
to lemma 3. Thereforeby
theorem 3.1 of[9]
p. 305 wehave L
~ Z.
We now
proceed
to the:PROOF OF THEOREM
(A).
All the
subalgebras
listed areideals;
theonly
caserequiring
commentbeing (c).
Since L =L(c, c+)
has no ideals of finite codimension(proof
of lemma
12)
the factor(F+S)/T is
central(see [9] p. 305, proof
of theo-rem
3.1).
Therefore anysubspace
X between T and F+S is an ideal.Suppose
now that 7 is an ideal of L. IfI ~
F+S thenby
lemma 11I is in the
given
list. Therefore we may assumeI
F+ S. If I n T ={0}
then
[I, T]
={0}.
But it is easy to see that theonly
elements of L central-ising
everyelementary
transformatione03B103B2 (03B1 ~ fi)
are the elements of S.Hence
I ~
S. Since dim S = 1 we have I ={0}
or S. But T issimple ([9]
lemma 4.1p. 306) so if I ~ T ~ {0}
thenT ~
1. Now I+F+S ~L,
and
by
lemma 11 I+F+S =L(c, d)+S
for some d. If d =No
thenT ~ I F+S,
which is case(c)
of the list. There remains the cased >
bt 0.
Then we have(I+F+S)/(T+S) = (L(c, d)+S)/(T+S)
so that(I+T+S)/(T+S)
is of codimension ~ 1 in(L(c, d)+S)/(T+S) - (L(c, d))/T
which has no proper ideals of finite codimensionby
theargument
of lemma 12. Therefore I+T+S =L(c, d)+S.
Now T ~ Iso we have I+S =
L(c, d)+S.
If I ~L(c, d)+S
and I ~L(c, d)
thenI n
L(c, d)
is of codimension 1 inL(c, d), contradicting
lemma 3. Hence I =L(c, d)
or 1 =L(c, d)+S.
We have
already
remarked(in
lemma12)
that L~ Z; which completes
the
proof
of the theorem.3.
Applications
In
[9]
it isproved
that any Liealgebra satisfying
Min-si andhaving
noideals of finite codimension has an
ascending
series of ideals whose factorsare either infinite-dimensional
simple
or 1-dimensional central. The re-86
sults of theorem A show that the 1-dimensional central factors cannot in
general
bedispensed
with. In[9] this question
was left open. Thealgebras L(c, d)
alsoprovide
newexamples
of Liealgebras
in Min-si nZ.
Following
thegeneral
lines of Scott[7] ]
p. 316 section 11.5.4(for groups)
we can prove:THEOREM
(B). Any
Liealgebra
can be embedded in asimple
Liealgebra.
PROOF. Let K be a Lie
algebra
over a field f.By
Jacobson[4] p.
162 cor.4 K has a faithful
representation by endomorphisms
of a vectorspace V
over f.
By enlarging
Vif necessary we may embed K inL(c+, c+)
for someinfinite cardinal c. If we
split
V into csubspaces
of dimensionc+
and copy the K-action on each of these we may assume that K isrepresented by endomorphisms
whoseimage
hasdimension ~
c. Then thecomposite embedding
maps K into a
simple
Liealgebra.
One
might
ask about Lieanalogies
of otherembedding
theorems for groups. Forexample,
Dark[2] ]
hasproved
that every group can be embedded as a subnormalsubgroup
of aperfect
group.Strangely,
theanalogue
of this is false for Liealgebras -
anexample
may be found in[8],
p. 98REFERENCES W. E. BAXTER
[1] Lie simplicity of a special class of associative rings, Proc. Amer. Math. Soc. 7 (1956)
855-863.
R. S. DARK
[2] On subnormal embedding theorems for groups, J. London Math. Soc. 43 (1968)
387-390.
I. N. HERSTEIN
[3] On the Lie and Jordan rings of a simple associative ring, Amer. J. Math. 77 (1955)
279-285.
N. JACOBSON
[4] Lie algebras, Interscience, New York 1962.
[5] Structure of rings, Amer. Math. Soc. Colloquium Publications XXXVII, Provi- dence, R. I. 1964.
A. ROSENBERG
[6] The infinite general linear group, Ann. of Math. 68 (1958) 278-294.
W. R. SCOTT
[7] Group Theory, Prentice Hall, Englewood Cliffs, New Jersey 1964.
I. N. STEWART
[8] Subideals of Lie algebras, Ph. D. Thesis, University of Warwick 1969.
[9] The minimal condition for subdeals of Lie algebras, Math. Z. 111 (1969) 301-310.
[10] An algebraic treatment of Mal’cev’s theorems concerning nilpotent Lie groups and their Lie algebras, Compositio Math. 22 (1970) 289-312.
(Oblatum 7-1-1971 Mathematics Institute University of Warwick 29-VII-1971 ) Coventry, England